diff --git a/.gitignore b/.gitignore index 45fb097..f6eb5f0 100644 --- a/.gitignore +++ b/.gitignore @@ -1,4 +1,5 @@ *.un .vs/ .vscode/ -.obsidian/ \ No newline at end of file +.obsidian/ +.obsidian/* \ No newline at end of file diff --git a/.obsidian/app.json b/.obsidian/app.json deleted file mode 100644 index 9e26dfe..0000000 --- a/.obsidian/app.json +++ /dev/null @@ -1 +0,0 @@ -{} \ No newline at end of file diff --git a/.obsidian/appearance.json b/.obsidian/appearance.json deleted file mode 100644 index 9e26dfe..0000000 --- a/.obsidian/appearance.json +++ /dev/null @@ -1 +0,0 @@ -{} \ No newline at end of file diff --git a/.obsidian/community-plugins.json b/.obsidian/community-plugins.json deleted file mode 100644 index d3f66fa..0000000 --- a/.obsidian/community-plugins.json +++ /dev/null @@ -1,3 +0,0 @@ -[ - "obsidian-git" -] \ No newline at end of file diff --git a/.obsidian/core-plugins.json b/.obsidian/core-plugins.json deleted file mode 100644 index 639b90d..0000000 --- a/.obsidian/core-plugins.json +++ /dev/null @@ -1,33 +0,0 @@ -{ - "file-explorer": true, - "global-search": true, - "switcher": true, - "graph": true, - "backlink": true, - "canvas": true, - "outgoing-link": true, - "tag-pane": true, - "footnotes": false, - "properties": true, - "page-preview": true, - "daily-notes": true, - "templates": true, - "note-composer": true, - "command-palette": true, - "slash-command": false, - "editor-status": true, - "bookmarks": true, - "markdown-importer": false, - "zk-prefixer": false, - "random-note": false, - "outline": true, - "word-count": true, - "slides": false, - "audio-recorder": false, - "workspaces": false, - "file-recovery": true, - "publish": false, - "sync": true, - "bases": true, - "webviewer": false -} \ No newline at end of file diff --git a/.obsidian/graph.json b/.obsidian/graph.json deleted file mode 100644 index e21a18d..0000000 --- a/.obsidian/graph.json +++ /dev/null @@ -1,22 +0,0 @@ -{ - "collapse-filter": true, - "search": "", - "showTags": false, - "showAttachments": false, - "hideUnresolved": false, - "showOrphans": true, - "collapse-color-groups": true, - "colorGroups": [], - "collapse-display": true, - "showArrow": false, - "textFadeMultiplier": 0, - "nodeSizeMultiplier": 1, - "lineSizeMultiplier": 1, - "collapse-forces": true, - "centerStrength": 0.518713248970312, - "repelStrength": 10, - "linkStrength": 1, - "linkDistance": 250, - "scale": 1, - "close": false -} \ No newline at end of file diff --git a/.obsidian/plugins/obsidian-git/main.js b/.obsidian/plugins/obsidian-git/main.js deleted file mode 100644 index a66d505..0000000 --- a/.obsidian/plugins/obsidian-git/main.js +++ /dev/null @@ -1,426 +0,0 @@ -/* -THIS IS A GENERATED/BUNDLED FILE BY ESBUILD -if you want to view the source visit the plugins github repository (https://github.com/denolehov/obsidian-git) -*/ - -var X5=Object.create;var wl=Object.defineProperty;var K5=Object.getOwnPropertyDescriptor;var Z5=Object.getOwnPropertyNames;var J5=Object.getPrototypeOf,Q5=Object.prototype.hasOwnProperty;var by=t=>{throw TypeError(t)};var eI=(t,e,r)=>e in t?wl(t,e,{enumerable:!0,configurable:!0,writable:!0,value:r}):t[e]=r;var tI=(t,e)=>()=>(t&&(e=t(t=0)),e);var M=(t,e)=>()=>(e||t((e={exports:{}}).exports,e),e.exports),rI=(t,e)=>{for(var r in e)wl(t,r,{get:e[r],enumerable:!0})},_y=(t,e,r,n)=>{if(e&&typeof e=="object"||typeof e=="function")for(let i of Z5(e))!Q5.call(t,i)&&i!==r&&wl(t,i,{get:()=>e[i],enumerable:!(n=K5(e,i))||n.enumerable});return t};var Ot=(t,e,r)=>(r=t!=null?X5(J5(t)):{},_y(e||!t||!t.__esModule?wl(r,"default",{value:t,enumerable:!0}):r,t)),nI=t=>_y(wl({},"__esModule",{value:!0}),t);var vt=(t,e,r)=>eI(t,typeof e!="symbol"?e+"":e,r),wp=(t,e,r)=>e.has(t)||by("Cannot "+r);var ue=(t,e,r)=>(wp(t,e,"read from private field"),r?r.call(t):e.get(t)),yt=(t,e,r)=>e.has(t)?by("Cannot add the same private member more than once"):e instanceof WeakSet?e.add(t):e.set(t,r),It=(t,e,r,n)=>(wp(t,e,"write to private field"),n?n.call(t,r):e.set(t,r),r),di=(t,e,r)=>(wp(t,e,"access private method"),r);var Ey=M(Iu=>{"use strict";m();Iu.byteLength=aI;Iu.toByteArray=oI;Iu.fromByteArray=uI;var Mn=[],qr=[],iI=typeof Uint8Array!="undefined"?Uint8Array:Array,bp="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";for(Ra=0,xy=bp.length;Ra0)throw new Error("Invalid string. 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i=t.prev[--r],t.prev[r]=i>=e?i-e:0;while(--n);a+=e}if(t.strm.avail_in===0)break;if(n=k$(t.strm,t.window,t.strstart+t.lookahead,a),t.lookahead+=n,t.lookahead+t.insert>=Ee)for(s=t.strstart-t.insert,t.ins_h=t.window[s],t.ins_h=(t.ins_h<t.pending_buf_size-5&&(r=t.pending_buf_size-5);;){if(t.lookahead<=1){if(ja(t),t.lookahead===0&&e===Na)return wt;if(t.lookahead===0)break}t.strstart+=t.lookahead,t.lookahead=0;var n=t.block_start+r;if((t.strstart===0||t.strstart>=n)&&(t.lookahead=t.strstart-n,t.strstart=n,Ft(t,!1),t.strm.avail_out===0)||t.strstart-t.block_start>=t.w_size-dn&&(Ft(t,!1),t.strm.avail_out===0))return wt}return t.insert=0,e===Ki?(Ft(t,!0),t.strm.avail_out===0?La:Zs):(t.strstart>t.block_start&&(Ft(t,!1),t.strm.avail_out===0),wt)}function fm(t,e){for(var r,n;;){if(t.lookahead=Ee&&(t.ins_h=(t.ins_h<=Ee)if(n=Wr._tr_tally(t,t.strstart-t.match_start,t.match_length-Ee),t.lookahead-=t.match_length,t.match_length<=t.max_lazy_match&&t.lookahead>=Ee){t.match_length--;do t.strstart++,t.ins_h=(t.ins_h<=Ee&&(t.ins_h=(t.ins_h<4096)&&(t.match_length=Ee-1)),t.prev_length>=Ee&&t.match_length<=t.prev_length){i=t.strstart+t.lookahead-Ee,n=Wr._tr_tally(t,t.strstart-1-t.prev_match,t.prev_length-Ee),t.lookahead-=t.prev_length-1,t.prev_length-=2;do++t.strstart<=i&&(t.ins_h=(t.ins_h<=Ee&&t.strstart>0&&(i=t.strstart-1,n=s[i],n===s[++i]&&n===s[++i]&&n===s[++i])){a=t.strstart+Yi;do;while(n===s[++i]&&n===s[++i]&&n===s[++i]&&n===s[++i]&&n===s[++i]&&n===s[++i]&&n===s[++i]&&n===s[++i]&&it.lookahead&&(t.match_length=t.lookahead)}if(t.match_length>=Ee?(r=Wr._tr_tally(t,1,t.match_length-Ee),t.lookahead-=t.match_length,t.strstart+=t.match_length,t.match_length=0):(r=Wr._tr_tally(t,0,t.window[t.strstart]),t.lookahead--,t.strstart++),r&&(Ft(t,!1),t.strm.avail_out===0))return wt}return t.insert=0,e===Ki?(Ft(t,!0),t.strm.avail_out===0?La:Zs):t.last_lit&&(Ft(t,!1),t.strm.avail_out===0)?wt:$l}function P$(t,e){for(var r;;){if(t.lookahead===0&&(ja(t),t.lookahead===0)){if(e===Na)return wt;break}if(t.match_length=0,r=Wr._tr_tally(t,0,t.window[t.strstart]),t.lookahead--,t.strstart++,r&&(Ft(t,!1),t.strm.avail_out===0))return wt}return t.insert=0,e===Ki?(Ft(t,!0),t.strm.avail_out===0?La:Zs):t.last_lit&&(Ft(t,!1),t.strm.avail_out===0)?wt:$l}function jn(t,e,r,n,i){this.good_length=t,this.max_lazy=e,this.nice_length=r,this.max_chain=n,this.func=i}var Ks;Ks=[new jn(0,0,0,0,T$),new jn(4,4,8,4,fm),new jn(4,5,16,8,fm),new jn(4,6,32,32,fm),new jn(4,4,16,16,Xs),new jn(8,16,32,32,Xs),new jn(8,16,128,128,Xs),new jn(8,32,128,256,Xs),new jn(32,128,258,1024,Xs),new jn(32,258,258,4096,Xs)];function R$(t){t.window_size=2*t.w_size,Wi(t.head),t.max_lazy_match=Ks[t.level].max_lazy,t.good_match=Ks[t.level].good_length,t.nice_match=Ks[t.level].nice_length,t.max_chain_length=Ks[t.level].max_chain,t.strstart=0,t.block_start=0,t.lookahead=0,t.insert=0,t.match_length=t.prev_length=Ee-1,t.match_available=0,t.ins_h=0}function M$(){this.strm=null,this.status=0,this.pending_buf=null,this.pending_buf_size=0,this.pending_out=0,this.pending=0,this.wrap=0,this.gzhead=null,this.gzindex=0,this.method=mf,this.last_flush=-1,this.w_size=0,this.w_bits=0,this.w_mask=0,this.window=null,this.window_size=0,this.prev=null,this.head=null,this.ins_h=0,this.hash_size=0,this.hash_bits=0,this.hash_mask=0,this.hash_shift=0,this.block_start=0,this.match_length=0,this.prev_match=0,this.match_available=0,this.strstart=0,this.match_start=0,this.lookahead=0,this.prev_length=0,this.max_chain_length=0,this.max_lazy_match=0,this.level=0,this.strategy=0,this.good_match=0,this.nice_match=0,this.dyn_ltree=new er.Buf16(x$*2),this.dyn_dtree=new er.Buf16((2*b$+1)*2),this.bl_tree=new er.Buf16((2*_$+1)*2),Wi(this.dyn_ltree),Wi(this.dyn_dtree),Wi(this.bl_tree),this.l_desc=null,this.d_desc=null,this.bl_desc=null,this.bl_count=new er.Buf16(S$+1),this.heap=new er.Buf16(2*dm+1),Wi(this.heap),this.heap_len=0,this.heap_max=0,this.depth=new er.Buf16(2*dm+1),Wi(this.depth),this.l_buf=0,this.lit_bufsize=0,this.last_lit=0,this.d_buf=0,this.opt_len=0,this.static_len=0,this.matches=0,this.insert=0,this.bi_buf=0,this.bi_valid=0}function F2(t){var e;return!t||!t.state?Xi(t,Yr):(t.total_in=t.total_out=0,t.data_type=p$,e=t.state,e.pending=0,e.pending_out=0,e.wrap<0&&(e.wrap=-e.wrap),e.status=e.wrap?gf:Da,t.adler=e.wrap===2?0:1,e.last_flush=Na,Wr._tr_init(e),Nn)}function $2(t){var e=F2(t);return e===Nn&&R$(t.state),e}function O$(t,e){return!t||!t.state||t.state.wrap!==2?Yr:(t.state.gzhead=e,Nn)}function D2(t,e,r,n,i,a){if(!t)return Yr;var s=1;if(e===c$&&(e=6),n<0?(s=0,n=-n):n>15&&(s=2,n-=16),i<1||i>m$||r!==mf||n<8||n>15||e<0||e>9||a<0||a>d$)return Xi(t,Yr);n===8&&(n=9);var o=new M$;return t.state=o,o.strm=t,o.wrap=s,o.gzhead=null,o.w_bits=n,o.w_size=1<P2||e<0)return t?Xi(t,Yr):Yr;if(n=t.state,!t.output||!t.input&&t.avail_in!==0||n.status===Fl&&e!==Ki)return Xi(t,t.avail_out===0?um:Yr);if(n.strm=t,r=n.last_flush,n.last_flush=e,n.status===gf)if(n.wrap===2)t.adler=0,Re(n,31),Re(n,139),Re(n,8),n.gzhead?(Re(n,(n.gzhead.text?1:0)+(n.gzhead.hcrc?2:0)+(n.gzhead.extra?4:0)+(n.gzhead.name?8:0)+(n.gzhead.comment?16:0)),Re(n,n.gzhead.time&255),Re(n,n.gzhead.time>>8&255),Re(n,n.gzhead.time>>16&255),Re(n,n.gzhead.time>>24&255),Re(n,n.level===9?2:n.strategy>=ff||n.level<2?4:0),Re(n,n.gzhead.os&255),n.gzhead.extra&&n.gzhead.extra.length&&(Re(n,n.gzhead.extra.length&255),Re(n,n.gzhead.extra.length>>8&255)),n.gzhead.hcrc&&(t.adler=Vi(t.adler,n.pending_buf,n.pending,0)),n.gzindex=0,n.status=hm):(Re(n,0),Re(n,0),Re(n,0),Re(n,0),Re(n,0),Re(n,n.level===9?2:n.strategy>=ff||n.level<2?4:0),Re(n,A$),n.status=Da);else{var s=mf+(n.w_bits-8<<4)<<8,o=-1;n.strategy>=ff||n.level<2?o=0:n.level<6?o=1:n.level===6?o=2:o=3,s|=o<<6,n.strstart!==0&&(s|=E$),s+=31-s%31,n.status=Da,Il(n,s),n.strstart!==0&&(Il(n,t.adler>>>16),Il(n,t.adler&65535)),t.adler=1}if(n.status===hm)if(n.gzhead.extra){for(i=n.pending;n.gzindex<(n.gzhead.extra.length&65535)&&!(n.pending===n.pending_buf_size&&(n.gzhead.hcrc&&n.pending>i&&(t.adler=Vi(t.adler,n.pending_buf,n.pending-i,i)),qi(t),i=n.pending,n.pending===n.pending_buf_size));)Re(n,n.gzhead.extra[n.gzindex]&255),n.gzindex++;n.gzhead.hcrc&&n.pending>i&&(t.adler=Vi(t.adler,n.pending_buf,n.pending-i,i)),n.gzindex===n.gzhead.extra.length&&(n.gzindex=0,n.status=df)}else n.status=df;if(n.status===df)if(n.gzhead.name){i=n.pending;do{if(n.pending===n.pending_buf_size&&(n.gzhead.hcrc&&n.pending>i&&(t.adler=Vi(t.adler,n.pending_buf,n.pending-i,i)),qi(t),i=n.pending,n.pending===n.pending_buf_size)){a=1;break}n.gzindexi&&(t.adler=Vi(t.adler,n.pending_buf,n.pending-i,i)),a===0&&(n.gzindex=0,n.status=hf)}else n.status=hf;if(n.status===hf)if(n.gzhead.comment){i=n.pending;do{if(n.pending===n.pending_buf_size&&(n.gzhead.hcrc&&n.pending>i&&(t.adler=Vi(t.adler,n.pending_buf,n.pending-i,i)),qi(t),i=n.pending,n.pending===n.pending_buf_size)){a=1;break}n.gzindexi&&(t.adler=Vi(t.adler,n.pending_buf,n.pending-i,i)),a===0&&(n.status=pf)}else n.status=pf;if(n.status===pf&&(n.gzhead.hcrc?(n.pending+2>n.pending_buf_size&&qi(t),n.pending+2<=n.pending_buf_size&&(Re(n,t.adler&255),Re(n,t.adler>>8&255),t.adler=0,n.status=Da)):n.status=Da),n.pending!==0){if(qi(t),t.avail_out===0)return n.last_flush=-1,Nn}else if(t.avail_in===0&&M2(e)<=M2(r)&&e!==Ki)return Xi(t,um);if(n.status===Fl&&t.avail_in!==0)return Xi(t,um);if(t.avail_in!==0||n.lookahead!==0||e!==Na&&n.status!==Fl){var l=n.strategy===ff?P$(n,e):n.strategy===f$?C$(n,e):Ks[n.level].func(n,e);if((l===La||l===Zs)&&(n.status=Fl),l===wt||l===La)return t.avail_out===0&&(n.last_flush=-1),Nn;if(l===$l&&(e===s$?Wr._tr_align(n):e!==P2&&(Wr._tr_stored_block(n,0,0,!1),e===o$&&(Wi(n.head),n.lookahead===0&&(n.strstart=0,n.block_start=0,n.insert=0))),qi(t),t.avail_out===0))return n.last_flush=-1,Nn}return e!==Ki?Nn:n.wrap<=0?R2:(n.wrap===2?(Re(n,t.adler&255),Re(n,t.adler>>8&255),Re(n,t.adler>>16&255),Re(n,t.adler>>24&255),Re(n,t.total_in&255),Re(n,t.total_in>>8&255),Re(n,t.total_in>>16&255),Re(n,t.total_in>>24&255)):(Il(n,t.adler>>>16),Il(n,t.adler&65535)),qi(t),n.wrap>0&&(n.wrap=-n.wrap),n.pending!==0?Nn:R2)}function $$(t){var e;return!t||!t.state?Yr:(e=t.state.status,e!==gf&&e!==hm&&e!==df&&e!==hf&&e!==pf&&e!==Da&&e!==Fl?Xi(t,Yr):(t.state=null,e===Da?Xi(t,l$):Nn))}function D$(t,e){var r=e.length,n,i,a,s,o,l,c,u;if(!t||!t.state||(n=t.state,s=n.wrap,s===2||s===1&&n.status!==gf||n.lookahead))return Yr;for(s===1&&(t.adler=O2(t.adler,e,r,0)),n.wrap=0,r>=n.w_size&&(s===0&&(Wi(n.head),n.strstart=0,n.block_start=0,n.insert=0),u=new er.Buf8(n.w_size),er.arraySet(u,e,r-n.w_size,n.w_size,0),e=u,r=n.w_size),o=t.avail_in,l=t.next_in,c=t.input,t.avail_in=r,t.next_in=0,t.input=e,ja(n);n.lookahead>=Ee;){i=n.strstart,a=n.lookahead-(Ee-1);do n.ins_h=(n.ins_h<{"use strict";m();var vf=pi(),j2=!0,N2=!0;try{String.fromCharCode.apply(null,[0])}catch(t){j2=!1}try{String.fromCharCode.apply(null,new Uint8Array(1))}catch(t){N2=!1}var Dl=new vf.Buf8(256);for(gi=0;gi<256;gi++)Dl[gi]=gi>=252?6:gi>=248?5:gi>=240?4:gi>=224?3:gi>=192?2:1;var gi;Dl[254]=Dl[254]=1;Js.string2buf=function(t){var e,r,n,i,a,s=t.length,o=0;for(i=0;i>>6,e[a++]=128|r&63):r<65536?(e[a++]=224|r>>>12,e[a++]=128|r>>>6&63,e[a++]=128|r&63):(e[a++]=240|r>>>18,e[a++]=128|r>>>12&63,e[a++]=128|r>>>6&63,e[a++]=128|r&63);return e};function B2(t,e){if(e<65534&&(t.subarray&&N2||!t.subarray&&j2))return String.fromCharCode.apply(null,vf.shrinkBuf(t,e));for(var r="",n=0;n4){o[n++]=65533,r+=a-1;continue}for(i&=a===2?31:a===3?15:7;a>1&&r1){o[n++]=65533;continue}i<65536?o[n++]=i:(i-=65536,o[n++]=55296|i>>10&1023,o[n++]=56320|i&1023)}return B2(o,n)};Js.utf8border=function(t,e){var r;for(e=e||t.length,e>t.length&&(e=t.length),r=e-1;r>=0&&(t[r]&192)===128;)r--;return r<0||r===0?e:r+Dl[t[r]]>e?r:e}});var mm=M((cY,H2)=>{"use strict";m();function L$(){this.input=null,this.next_in=0,this.avail_in=0,this.total_in=0,this.output=null,this.next_out=0,this.avail_out=0,this.total_out=0,this.msg="",this.state=null,this.data_type=2,this.adler=0}H2.exports=L$});var V2=M(Nl=>{"use strict";m();var Ll=L2(),jl=pi(),vm=pm(),ym=uf(),j$=mm(),z2=Object.prototype.toString,N$=0,gm=4,Qs=0,U2=1,G2=2,B$=-1,H$=0,U$=8;function Ba(t){if(!(this instanceof Ba))return new Ba(t);this.options=jl.assign({level:B$,method:U$,chunkSize:16384,windowBits:15,memLevel:8,strategy:H$,to:""},t||{});var e=this.options;e.raw&&e.windowBits>0?e.windowBits=-e.windowBits:e.gzip&&e.windowBits>0&&e.windowBits<16&&(e.windowBits+=16),this.err=0,this.msg="",this.ended=!1,this.chunks=[],this.strm=new j$,this.strm.avail_out=0;var r=Ll.deflateInit2(this.strm,e.level,e.method,e.windowBits,e.memLevel,e.strategy);if(r!==Qs)throw new Error(ym[r]);if(e.header&&Ll.deflateSetHeader(this.strm,e.header),e.dictionary){var n;if(typeof e.dictionary=="string"?n=vm.string2buf(e.dictionary):z2.call(e.dictionary)==="[object ArrayBuffer]"?n=new Uint8Array(e.dictionary):n=e.dictionary,r=Ll.deflateSetDictionary(this.strm,n),r!==Qs)throw new Error(ym[r]);this._dict_set=!0}}Ba.prototype.push=function(t,e){var r=this.strm,n=this.options.chunkSize,i,a;if(this.ended)return!1;a=e===~~e?e:e===!0?gm:N$,typeof t=="string"?r.input=vm.string2buf(t):z2.call(t)==="[object ArrayBuffer]"?r.input=new Uint8Array(t):r.input=t,r.next_in=0,r.avail_in=r.input.length;do{if(r.avail_out===0&&(r.output=new jl.Buf8(n),r.next_out=0,r.avail_out=n),i=Ll.deflate(r,a),i!==U2&&i!==Qs)return this.onEnd(i),this.ended=!0,!1;(r.avail_out===0||r.avail_in===0&&(a===gm||a===G2))&&(this.options.to==="string"?this.onData(vm.buf2binstring(jl.shrinkBuf(r.output,r.next_out))):this.onData(jl.shrinkBuf(r.output,r.next_out)))}while((r.avail_in>0||r.avail_out===0)&&i!==U2);return a===gm?(i=Ll.deflateEnd(this.strm),this.onEnd(i),this.ended=!0,i===Qs):(a===G2&&(this.onEnd(Qs),r.avail_out=0),!0)};Ba.prototype.onData=function(t){this.chunks.push(t)};Ba.prototype.onEnd=function(t){t===Qs&&(this.options.to==="string"?this.result=this.chunks.join(""):this.result=jl.flattenChunks(this.chunks)),this.chunks=[],this.err=t,this.msg=this.strm.msg};function wm(t,e){var r=new Ba(e);if(r.push(t,!0),r.err)throw r.msg||ym[r.err];return r.result}function G$(t,e){return e=e||{},e.raw=!0,wm(t,e)}function z$(t,e){return e=e||{},e.gzip=!0,wm(t,e)}Nl.Deflate=Ba;Nl.deflate=wm;Nl.deflateRaw=G$;Nl.gzip=z$});var W2=M((hY,q2)=>{"use strict";m();var yf=30,V$=12;q2.exports=function(e,r){var n,i,a,s,o,l,c,u,f,d,h,p,g,v,y,b,E,S,A,k,w,x,_,T,C;n=e.state,i=e.next_in,T=e.input,a=i+(e.avail_in-5),s=e.next_out,C=e.output,o=s-(r-e.avail_out),l=s+(e.avail_out-257),c=n.dmax,u=n.wsize,f=n.whave,d=n.wnext,h=n.window,p=n.hold,g=n.bits,v=n.lencode,y=n.distcode,b=(1<>>24,p>>>=A,g-=A,A=S>>>16&255,A===0)C[s++]=S&65535;else if(A&16){k=S&65535,A&=15,A&&(g>>=A,g-=A),g<15&&(p+=T[i++]<>>24,p>>>=A,g-=A,A=S>>>16&255,A&16){if(w=S&65535,A&=15,gc){e.msg="invalid distance too far back",n.mode=yf;break e}if(p>>>=A,g-=A,A=s-o,w>A){if(A=w-A,A>f&&n.sane){e.msg="invalid distance too far back",n.mode=yf;break e}if(x=0,_=h,d===0){if(x+=u-A,A2;)C[s++]=_[x++],C[s++]=_[x++],C[s++]=_[x++],k-=3;k&&(C[s++]=_[x++],k>1&&(C[s++]=_[x++]))}else{x=s-w;do C[s++]=C[x++],C[s++]=C[x++],C[s++]=C[x++],k-=3;while(k>2);k&&(C[s++]=C[x++],k>1&&(C[s++]=C[x++]))}}else 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r,n;return!t||!t.state||(n=t.state,e<0?(r=0,e=-e):(r=(e>>4)+1,e<48&&(e&=15)),e&&(e<8||e>15))?Xr:(n.window!==null&&n.wbits!==e&&(n.window=null),n.wrap=r,n.wbits=e,M_(t))}function I_(t,e){var r,n;return t?(n=new o8,t.state=n,n.window=null,r=O_(t,e),r!==Ha&&(t.state=null),r):Xr}function l8(t){return I_(t,s8)}var A_=!0,Em,Am;function c8(t){if(A_){var e;for(Em=new Pr.Buf32(512),Am=new Pr.Buf32(32),e=0;e<144;)t.lens[e++]=8;for(;e<256;)t.lens[e++]=9;for(;e<280;)t.lens[e++]=7;for(;e<288;)t.lens[e++]=8;for(Bl(k_,t.lens,0,288,Em,0,t.work,{bits:9}),e=0;e<32;)t.lens[e++]=5;Bl(T_,t.lens,0,32,Am,0,t.work,{bits:5}),A_=!1}t.lencode=Em,t.lenbits=9,t.distcode=Am,t.distbits=5}function F_(t,e,r,n){var i,a=t.state;return a.window===null&&(a.wsize=1<=a.wsize?(Pr.arraySet(a.window,e,r-a.wsize,a.wsize,0),a.wnext=0,a.whave=a.wsize):(i=a.wsize-a.wnext,i>n&&(i=n),Pr.arraySet(a.window,e,r-n,i,a.wnext),n-=i,n?(Pr.arraySet(a.window,e,r-n,n,0),a.wnext=n,a.whave=a.wsize):(a.wnext+=i,a.wnext===a.wsize&&(a.wnext=0),a.whave>>8&255,r.check=Hn(r.check,_,2,0),c=0,u=0,r.mode=n_;break}if(r.flags=0,r.head&&(r.head.done=!1),!(r.wrap&1)||(((c&255)<<8)+(c>>8))%31){t.msg="incorrect header check",r.mode=Ye;break}if((c&15)!==r_){t.msg="unknown compression method",r.mode=Ye;break}if(c>>>=4,u-=4,w=(c&15)+8,r.wbits===0)r.wbits=w;else if(w>r.wbits){t.msg="invalid window size",r.mode=Ye;break}r.dmax=1<>8&1),r.flags&512&&(_[0]=c&255,_[1]=c>>>8&255,r.check=Hn(r.check,_,2,0)),c=0,u=0,r.mode=i_;case i_:for(;u<32;){if(o===0)break e;o--,c+=n[a++]<>>8&255,_[2]=c>>>16&255,_[3]=c>>>24&255,r.check=Hn(r.check,_,4,0)),c=0,u=0,r.mode=a_;case a_:for(;u<16;){if(o===0)break 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n=await this._next();if(this._ended)break;r.push(n)}this.buffer=Buffer.concat(r)}async _loadnext(){this._discardedBytes+=this.buffer.length,this.undoCursor=0,this.cursor=0,this.buffer=await this._next()}async _init(){this.buffer=await this._next()}};function x4(t){return t.reduce((e,r)=>e+r.length,0)}async function S4(t,e){let r=new Mf(t),n=await r.read(4);if(n=n.toString("utf8"),n!=="PACK")throw new pe(`Invalid PACK header '${n}'`);let i=await r.read(4);if(i=i.readUInt32BE(0),i!==2)throw new pe(`Invalid packfile version: ${i}`);let a=await r.read(4);if(a=a.readUInt32BE(0),!(a<1))for(;!r.eof()&&a--;){let s=r.tell(),{type:o,length:l,ofs:c,reference:u}=await E4(r),f=new cg.Inflate;for(;!f.result;){let d=await r.chunk();if(!d)break;if(f.push(d,!1),f.err)throw new pe(`Pako error: ${f.msg}`);if(f.result){if(f.result.length!==l)throw new pe("Inflated object size is different from that stated in packfile.");await r.undo(),await r.read(d.length-f.strm.avail_in);let h=r.tell();await e({data:f.result,type:o,num:a,offset:s,end:h,reference:u,ofs:c})}}}}async function E4(t){let e=await t.byte(),r=e>>4&7,n=e&15;if(e&128){let s=4;do e=await t.byte(),n|=(e&127)<i+1<<7|a,-1)}function C4(t,e){let r=e,n=4,i=null;do i=t.readUInt8(),r|=(i&127)<2048*1024*1024)throw new pe("To keep implementation simple, I haven't implemented the layer 5 feature needed to support packfiles > 2GB in size.");n.seek(n.tell()+4*255);let s=n.readUInt32BE(),o=[];for(let u=0;u{u===null&&(u=S);let A=Math.floor((u-S)*100/u);A!==f&&n&&await n({phase:"Receiving objects",loaded:u-S,total:u}),f=A,y=i[y],["commit","tree","blob","tag"].includes(y)?a[E]={type:y,offset:E}:y==="ofs-delta"?a[E]={type:y,offset:E}:y==="ref-delta"&&(a[E]={type:y,offset:E})});let d=Object.keys(a).map(Number);for(let[v,y]of d.entries()){let b=v+1===d.length?e.byteLength-20:d[v+1],E=a[y],S=V8.buf(e.slice(y,b))>>>0;E.end=b,E.crc=S}let h=new t({pack:Promise.resolve(e),packfileSha:s,crcs:l,hashes:o,offsets:c,getExternalRefDelta:r});f=null;let p=0,g=[0,0,0,0,0,0,0,0,0,0,0,0];for(let v in a){v=Number(v);let y=Math.floor(p*100/u);y!==f&&n&&await n({phase:"Resolving deltas",loaded:p,total:u}),p++,f=y;let b=a[v];if(!b.oid)try{h.readDepth=0,h.externalReadDepth=0;let{type:E,object:S}=await h.readSlice({start:v});g[h.readDepth]+=1;let A=await bi(ea.wrap({type:E,object:S}));b.oid=A,o.push(A),c.set(A,v),l[A]=b.crc}catch(E){continue}}return o.sort(),h}async toBuffer(){let e=[],r=(c,u)=>{e.push(Buffer.from(c,u))};r("ff744f63","hex"),r("00000002","hex");let n=new Kr(Buffer.alloc(256*4));for(let c=0;c<256;c++){let u=0;for(let f of this.hashes)parseInt(f.slice(0,2),16)<=c&&u++;n.writeUInt32BE(u)}e.push(n.buffer);for(let c of this.hashes)r(c,"hex");let i=new Kr(Buffer.alloc(this.hashes.length*4));for(let c of this.hashes)i.writeUInt32BE(this.crcs[c]);e.push(i.buffer);let a=new Kr(Buffer.alloc(this.hashes.length*4));for(let c of this.hashes)a.writeUInt32BE(this.offsets.get(c));e.push(a.buffer),r(this.packfileSha,"hex");let s=Buffer.concat(e),o=await bi(s),l=Buffer.alloc(20);return l.write(o,"hex"),Buffer.concat([s,l])}async load({pack:e}){this.pack=e}async unload(){this.pack=null}async read({oid:e}){if(!this.offsets.get(e)){if(this.getExternalRefDelta)return this.externalReadDepth++,this.getExternalRefDelta(e);throw new pe(`Could not read object ${e} from packfile`)}let r=this.offsets.get(e);return this.readSlice({start:r})}async readSlice({start:e}){if(this.offsetCache[e])return Object.assign({},this.offsetCache[e]);this.readDepth++;let r={16:"commit",32:"tree",48:"blob",64:"tag",96:"ofs_delta",112:"ref_delta"};if(!this.pack)throw new pe("Tried to read from a GitPackIndex with no packfile loaded into memory");let n=(await this.pack).slice(e),i=new Kr(n),a=i.readUInt8(),s=a&112,o=r[s];if(o===void 0)throw new pe("Unrecognized type: 0b"+s.toString(2));let l=a&15,c=l;a&128&&(c=C4(i,l));let f=null,d=null;if(o==="ofs_delta"){let p=T4(i),g=e-p;({object:f,type:o}=await this.readSlice({start:g}))}if(o==="ref_delta"){let p=i.slice(20).toString("hex");({object:f,type:o}=await this.read({oid:p}))}let h=n.slice(i.tell());if(d=Buffer.from(await Nx(h)),d.byteLength!==c)throw new pe(`Packfile told us object would have length ${c} but it had length ${d.byteLength}`);return f&&(d=Buffer.from(b4(d,f))),this.readDepth>3&&(this.offsetCache[e]={type:o,object:d}),{type:o,format:"content",object:d}}},kf=Symbol("PackfileCache");async function P4({fs:t,filename:e,getExternalRefDelta:r,emitter:n,emitterPrefix:i}){let a=await t.read(e);return Wl.fromIdx({idx:a,getExternalRefDelta:r})}function ug({fs:t,cache:e,filename:r,getExternalRefDelta:n,emitter:i,emitterPrefix:a}){e[kf]||(e[kf]=new Map);let s=e[kf].get(r);return s||(s=P4({fs:t,filename:r,getExternalRefDelta:n,emitter:i,emitterPrefix:a}),e[kf].set(r,s)),s}async function R4({fs:t,cache:e,gitdir:r,oid:n,format:i="content",getExternalRefDelta:a}){let s=await t.readdir(F.join(r,"objects/pack"));s=s.filter(o=>o.endsWith(".idx"));for(let o of s){let l=`${r}/objects/pack/${o}`,c=await ug({fs:t,cache:e,filename:l,getExternalRefDelta:a});if(c.error)throw new pe(c.error);if(c.offsets.has(n)){if(!c.pack){let f=l.replace(/idx$/,"pack");c.pack=t.read(f)}let u=await c.read({oid:n,getExternalRefDelta:a});return u.format="content",u.source=`objects/pack/${o.replace(/idx$/,"pack")}`,u}}return null}async function Ge({fs:t,cache:e,gitdir:r,oid:n,format:i="content"}){let a=u=>Ge({fs:t,cache:e,gitdir:r,oid:u}),s;if(n==="4b825dc642cb6eb9a060e54bf8d69288fbee4904"&&(s={format:"wrapped",object:Buffer.from("tree 0\0")}),s||(s=await Lx({fs:t,gitdir:r,oid:n})),!s){if(s=await R4({fs:t,cache:e,gitdir:r,oid:n,getExternalRefDelta:a}),!s)throw new je(n);return s}if(i==="deflated"||(s.format==="deflated"&&(s.object=Buffer.from(await Nx(s.object)),s.format="wrapped"),i==="wrapped"))return s;let o=await bi(s.object);if(o!==n)throw new pe(`SHA check failed! Expected ${n}, computed ${o}`);let{object:l,type:c}=ea.unwrap(s.object);if(s.type=c,s.object=l,s.format="content",i==="content")return s;throw new pe(`invalid requested format "${i}"`)}var pn=class t extends Ae{constructor(e,r,n=!0){super(`Failed to create ${e} at ${r} because it already exists.${n?` (Hint: use 'force: true' parameter to overwrite existing ${e}.)`:""}`),this.code=this.name=t.code,this.data={noun:e,where:r,canForce:n}}};pn.code="AlreadyExistsError";var Yl=class t extends Ae{constructor(e,r,n){super(`Found multiple ${e} matching "${r}" (${n.join(", ")}). Use a longer abbreviation length to disambiguate them.`),this.code=this.name=t.code,this.data={nouns:e,short:r,matches:n}}};Yl.code="AmbiguousError";var Xl=class t extends Ae{constructor(e){super(`Your local changes to the following files would be overwritten by checkout: ${e.join(", ")}`),this.code=this.name=t.code,this.data={filepaths:e}}};Xl.code="CheckoutConflictError";var Kl=class t extends Ae{constructor(e,r){super(`Failed to checkout "${e}" because commit ${r} is not available locally. 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(Hint: a valid alternative would be "${r}".)`),this.code=this.name=t.code,this.data={ref:e,suggestion:r}}};mn.code="InvalidRefNameError";var ec=class t extends Ae{constructor(e){super(`Maximum search depth of ${e} exceeded.`),this.code=this.name=t.code,this.data={depth:e}}};ec.code="MaxDepthError";var oo=class t extends Ae{constructor(){super("Merges with conflicts are not supported yet."),this.code=this.name=t.code,this.data={}}};oo.code="MergeNotSupportedError";var lo=class t extends Ae{constructor(e,r,n,i){super(`Automatic merge failed with one or more merge conflicts in the following files: ${e.toString()}. 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Please refer to the "errors" property to see more'),this.code=this.name=t.code,this.data={errors:e},this.errors=e}};tc.code="MultipleGitError";var Va=class t extends Ae{constructor(e,r){super(`Expected "${e}" but received "${r}".`),this.code=this.name=t.code,this.data={expected:e,actual:r}}};Va.code="ParseError";var co=class t extends Ae{constructor(e){let r="";e==="not-fast-forward"?r=" because it was not a simple fast-forward":e==="tag-exists"&&(r=" because tag already exists"),super(`Push rejected${r}. Use "force: true" to override.`),this.code=this.name=t.code,this.data={reason:e}}};co.code="PushRejectedError";var Ji=class t extends Ae{constructor(e,r){super(`Remote does not support the "${e}" so the "${r}" parameter cannot be used.`),this.code=this.name=t.code,this.data={capability:e,parameter:r}}};Ji.code="RemoteCapabilityError";var rc=class t extends Ae{constructor(e,r){super(`Remote did not reply using the "smart" HTTP protocol. Expected "001e# service=git-upload-pack" but received: ${e}`),this.code=this.name=t.code,this.data={preview:e,response:r}}};rc.code="SmartHttpError";var nc=class t extends Ae{constructor(e,r,n){super(`Git remote "${e}" uses an unrecognized transport protocol: "${r}"`),this.code=this.name=t.code,this.data={url:e,transport:r,suggestion:n}}};nc.code="UnknownTransportError";var ic=class t extends Ae{constructor(e){super(`Cannot parse remote URL: "${e}"`),this.code=this.name=t.code,this.data={url:e}}};ic.code="UrlParseError";var uo=class t extends Ae{constructor(){super("The operation was canceled."),this.code=this.name=t.code,this.data={}}};uo.code="UserCanceledError";var ac=class t extends Ae{constructor(e){super(`Could not merge index: Entry for '${e}' is not up to date. Either reset the index entry to HEAD, or stage your unstaged changes.`),this.code=this.name=t.code,this.data={filepath:e}}};ac.code="IndexResetError";var sc=class t extends Ae{constructor(e){super(`"${e}" does not point to any commit. You're maybe working on a repository with no commits yet. `),this.code=this.name=t.code,this.data={ref:e}}};sc.code="NoCommitError";var Bx=Object.freeze({__proto__:null,AlreadyExistsError:pn,AmbiguousError:Yl,CheckoutConflictError:Xl,CommitNotFetchedError:Kl,EmptyServerResponseError:Zl,FastForwardError:Jl,GitPushError:Ql,HttpError:so,InternalError:pe,InvalidFilepathError:ta,InvalidOidError:Qi,InvalidRefNameError:mn,MaxDepthError:ec,MergeNotSupportedError:oo,MergeConflictError:lo,MissingNameError:zt,MissingParameterError:dr,MultipleGitError:tc,NoRefspecError:ql,NotFoundError:je,ObjectTypeError:tr,ParseError:Va,PushRejectedError:co,RemoteCapabilityError:Ji,SmartHttpError:rc,UnknownTransportError:nc,UnsafeFilepathError:no,UrlParseError:ic,UserCanceledError:uo,UnmergedPathsError:Vl,IndexResetError:ac,NoCommitError:sc});function rg({name:t,email:e,timestamp:r,timezoneOffset:n}){return n=M4(n),`${t} <${e}> ${r} ${n}`}function M4(t){let e=O4(I4(t));t=Math.abs(t);let r=Math.floor(t/60);t-=r*60;let n=String(r),i=String(t);return n.length<2&&(n="0"+n),i.length<2&&(i="0"+i),(e===-1?"-":"+")+n+i}function O4(t){return Math.sign(t)||(Object.is(t,-0)?-1:1)}function I4(t){return t===0?t:-t}function wi(t){return t=t.replace(/\r/g,""),t=t.replace(/^\n+/,""),t=t.replace(/\n+$/,"")+` -`,t}function Of(t){let[,e,r,n,i]=t.match(/^(.*) <(.*)> (.*) (.*)$/);return{name:e,email:r,timestamp:Number(n),timezoneOffset:F4(i)}}function F4(t){let[,e,r,n]=t.match(/(\+|-)(\d\d)(\d\d)/);return n=(e==="+"?1:-1)*(Number(r)*60+Number(n)),$4(n)}function $4(t){return t===0?t:-t}var Mr=class t{constructor(e){if(typeof e=="string")this._tag=e;else if(Buffer.isBuffer(e))this._tag=e.toString("utf8");else if(typeof e=="object")this._tag=t.render(e);else throw new pe("invalid type passed to GitAnnotatedTag constructor")}static from(e){return new t(e)}static render(e){return`object ${e.object} -type ${e.type} -tag ${e.tag} -tagger ${rg(e.tagger)} - -${e.message} 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wi(e)}payload(){return this.withoutSignature()+` -`}toObject(){return Buffer.from(this._tag,"utf8")}static async sign(e,r,n){let i=e.payload(),{signature:a}=await r({payload:i,secretKey:n});a=wi(a);let s=i+a;return t.from(s)}};function zm(t){return t.trim().split(` -`).map(e=>" "+e).join(` -`)+` -`}function D4(t){return t.split(` -`).map(e=>e.replace(/^ /,"")).join(` -`)}var pr=class t{constructor(e){if(typeof e=="string")this._commit=e;else if(Buffer.isBuffer(e))this._commit=e.toString("utf8");else if(typeof e=="object")this._commit=t.render(e);else throw new pe("invalid type passed to GitCommit constructor")}static fromPayloadSignature({payload:e,signature:r}){let n=t.justHeaders(e),i=t.justMessage(e),a=wi(n+` -gpgsig`+zm(r)+` -`+i);return new t(a)}static from(e){return new t(e)}toObject(){return Buffer.from(this._commit,"utf8")}headers(){return this.parseHeaders()}message(){return t.justMessage(this._commit)}parse(){return 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tr(n,i,"tree");return{tree:hr.from(a),oid:n}}var ng=class{constructor({fs:e,gitdir:r,ref:n,cache:i}){this.fs=e,this.cache=i,this.gitdir=r,this.mapPromise=(async()=>{let s=new Map,o;try{o=await V.resolve({fs:e,gitdir:r,ref:n})}catch(c){c instanceof je&&(o="4b825dc642cb6eb9a060e54bf8d69288fbee4904")}let l=await fo({fs:e,cache:this.cache,gitdir:r,oid:o});return l.type="tree",l.mode="40000",s.set(".",l),s})();let a=this;this.ConstructEntry=class{constructor(o){this._fullpath=o,this._type=!1,this._mode=!1,this._stat=!1,this._content=!1,this._oid=!1}async type(){return a.type(this)}async mode(){return a.mode(this)}async stat(){return a.stat(this)}async content(){return a.content(this)}async oid(){return a.oid(this)}}}async readdir(e){let r=e._fullpath,{fs:n,cache:i,gitdir:a}=this,s=await this.mapPromise,o=s.get(r);if(!o)throw new Error(`No obj for ${r}`);let l=o.oid;if(!l)throw new Error(`No oid for obj ${JSON.stringify(o)}`);if(o.type!=="tree")return null;let{type:c,object:u}=await 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t=this.blocks,e=this.lastByteIndex;t[16]=this.block,t[e>>2]|=EXTRA[e&3],this.block=t[16],e>=56&&(this.hashed||this.hash(),t[0]=this.block,t[16]=t[1]=t[2]=t[3]=t[4]=t[5]=t[6]=t[7]=t[8]=t[9]=t[10]=t[11]=t[12]=t[13]=t[14]=t[15]=0),t[14]=this.hBytes<<3|this.bytes>>>29,t[15]=this.bytes<<3,this.hash()}},Sha256.prototype.hash=function(){var t=this.h0,e=this.h1,r=this.h2,n=this.h3,i=this.h4,a=this.h5,s=this.h6,o=this.h7,l=this.blocks,c,u,f,d,h,p,g,v,y,b,E;for(c=16;c<64;++c)h=l[c-15],u=(h>>>7|h<<25)^(h>>>18|h<<14)^h>>>3,h=l[c-2],f=(h>>>17|h<<15)^(h>>>19|h<<13)^h>>>10,l[c]=l[c-16]+u+l[c-7]+f<<0;for(E=e&r,c=0;c<64;c+=4)this.first?(this.is224?(v=300032,h=l[0]-1413257819,o=h-150054599<<0,n=h+24177077<<0):(v=704751109,h=l[0]-210244248,o=h-1521486534<<0,n=h+143694565<<0),this.first=!1):(u=(t>>>2|t<<30)^(t>>>13|t<<19)^(t>>>22|t<<10),f=(i>>>6|i<<26)^(i>>>11|i<<21)^(i>>>25|i<<7),v=t&e,d=v^t&r^E,g=i&a^~i&s,h=o+f+g+K[c]+l[c],p=u+d,o=n+h<<0,n=h+p<<0),u=(n>>>2|n<<30)^(n>>>13|n<<19)^(n>>>22|n<<10),f=(o>>>6|o<<26)^(o>>>11|o<<21)^(o>>>25|o<<7),y=n&t,d=y^n&e^v,g=o&i^~o&a,h=s+f+g+K[c+1]+l[c+1],p=u+d,s=r+h<<0,r=h+p<<0,u=(r>>>2|r<<30)^(r>>>13|r<<19)^(r>>>22|r<<10),f=(s>>>6|s<<26)^(s>>>11|s<<21)^(s>>>25|s<<7),b=r&n,d=b^r&t^y,g=s&o^~s&i,h=a+f+g+K[c+2]+l[c+2],p=u+d,a=e+h<<0,e=h+p<<0,u=(e>>>2|e<<30)^(e>>>13|e<<19)^(e>>>22|e<<10),f=(a>>>6|a<<26)^(a>>>11|a<<21)^(a>>>25|a<<7),E=e&r,d=E^e&n^b,g=a&s^~a&o,h=i+f+g+K[c+3]+l[c+3],p=u+d,i=t+h<<0,t=h+p<<0;this.h0=this.h0+t<<0,this.h1=this.h1+e<<0,this.h2=this.h2+r<<0,this.h3=this.h3+n<<0,this.h4=this.h4+i<<0,this.h5=this.h5+a<<0,this.h6=this.h6+s<<0,this.h7=this.h7+o<<0},Sha256.prototype.hex=function(){this.finalize();var t=this.h0,e=this.h1,r=this.h2,n=this.h3,i=this.h4,a=this.h5,s=this.h6,o=this.h7,l=HEX_CHARS[t>>28&15]+HEX_CHARS[t>>24&15]+HEX_CHARS[t>>20&15]+HEX_CHARS[t>>16&15]+HEX_CHARS[t>>12&15]+HEX_CHARS[t>>8&15]+HEX_CHARS[t>>4&15]+HEX_CHARS[t&15]+HEX_CHARS[e>>28&15]+HEX_CHARS[e>>24&15]+HEX_CHARS[e>>20&15]+HEX_CHARS[e>>16&15]+HEX_CHARS[e>>12&15]+HEX_CHARS[e>>8&15]+HEX_CHARS[e>>4&15]+HEX_CHARS[e&15]+HEX_CHARS[r>>28&15]+HEX_CHARS[r>>24&15]+HEX_CHARS[r>>20&15]+HEX_CHARS[r>>16&15]+HEX_CHARS[r>>12&15]+HEX_CHARS[r>>8&15]+HEX_CHARS[r>>4&15]+HEX_CHARS[r&15]+HEX_CHARS[n>>28&15]+HEX_CHARS[n>>24&15]+HEX_CHARS[n>>20&15]+HEX_CHARS[n>>16&15]+HEX_CHARS[n>>12&15]+HEX_CHARS[n>>8&15]+HEX_CHARS[n>>4&15]+HEX_CHARS[n&15]+HEX_CHARS[i>>28&15]+HEX_CHARS[i>>24&15]+HEX_CHARS[i>>20&15]+HEX_CHARS[i>>16&15]+HEX_CHARS[i>>12&15]+HEX_CHARS[i>>8&15]+HEX_CHARS[i>>4&15]+HEX_CHARS[i&15]+HEX_CHARS[a>>28&15]+HEX_CHARS[a>>24&15]+HEX_CHARS[a>>20&15]+HEX_CHARS[a>>16&15]+HEX_CHARS[a>>12&15]+HEX_CHARS[a>>8&15]+HEX_CHARS[a>>4&15]+HEX_CHARS[a&15]+HEX_CHARS[s>>28&15]+HEX_CHARS[s>>24&15]+HEX_CHARS[s>>20&15]+HEX_CHARS[s>>16&15]+HEX_CHARS[s>>12&15]+HEX_CHARS[s>>8&15]+HEX_CHARS[s>>4&15]+HEX_CHARS[s&15];return this.is224||(l+=HEX_CHARS[o>>28&15]+HEX_CHARS[o>>24&15]+HEX_CHARS[o>>20&15]+HEX_CHARS[o>>16&15]+HEX_CHARS[o>>12&15]+HEX_CHARS[o>>8&15]+HEX_CHARS[o>>4&15]+HEX_CHARS[o&15]),l},Sha256.prototype.toString=Sha256.prototype.hex,Sha256.prototype.digest=function(){this.finalize();var t=this.h0,e=this.h1,r=this.h2,n=this.h3,i=this.h4,a=this.h5,s=this.h6,o=this.h7,l=[t>>24&255,t>>16&255,t>>8&255,t&255,e>>24&255,e>>16&255,e>>8&255,e&255,r>>24&255,r>>16&255,r>>8&255,r&255,n>>24&255,n>>16&255,n>>8&255,n&255,i>>24&255,i>>16&255,i>>8&255,i&255,a>>24&255,a>>16&255,a>>8&255,a&255,s>>24&255,s>>16&255,s>>8&255,s&255];return this.is224||l.push(o>>24&255,o>>16&255,o>>8&255,o&255),l},Sha256.prototype.array=Sha256.prototype.digest,Sha256.prototype.arrayBuffer=function(){this.finalize();var t=new ArrayBuffer(this.is224?28:32),e=new DataView(t);return 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Promise(e=>setTimeout(e,10)).then(()=>{this.resolve&&this.resolve(void 0)})}getSuggestions(e){return this.config.onlySelection?this.config.options:this.config.allowEmpty?[e.length>0?e:" ",...this.config.options]:[e.length>0?e:"...",...this.config.options]}renderSuggestion(e,r){this.config.obscure?r.hide():r.setText(e)}onChooseSuggestion(e,r){if(this.resolve){let n;this.config.allowEmpty&&e===" "?n="":e==="..."?n=void 0:n=e,this.resolve(n)}}};m();function gT(t,e){let r=Object.assign({},t.lineAuthor,e.lineAuthor);return Object.assign({},t,e,{lineAuthor:r})}var Co=class extends Error{constructor(r){super("No network connection available");this.originalError=r}};m();var e3=Ot(FT()),t3=require("child_process"),r3=Ot(QR()),Ai=require("obsidian");function Ud(t){throw new Error(`Unexpected object: ${t}`)}function Ic(t,e,r){return t===1?`${t} ${e}`:`${t} ${r!=null?r:e+"s"}`}var pv=(t,e)=>t==="."||e==null||e.length===0||e==="."?!0:e.length>=t.length?e.startsWith(t):t.startsWith(e);function vn(t,e){let r;if(e){if(e.button===0||e.button===1){let n=Ai.Keymap.isModEvent(e);r=t.workspace.getLeaf(n)}}else r=t.workspace.getLeaf(!1);return r}function Xn(t,e,r,n,i){if(e.button==2){let a=t.vault.getAbstractFileByPath(r);if(a!=null){let s=new Ai.Menu;t.workspace.trigger("file-menu",s,a,i,n),s.showAtPosition({x:e.pageX,y:e.pageY})}else{let s=new Ai.Menu;t.workspace.trigger("obsidian-git:menu",s,r,i,n),s.showAtPosition({x:e.pageX,y:e.pageY})}}}function pa(t){throw new Error("Impossible branch: "+t)}function mv(t){return`rgb(${t.r},${t.g},${t.b})`}function n3(t){var a;let e=(a=e3.fromString(t))==null?void 0:a.toRgbaArray();if(e===void 0)return;let[r,n,i]=e;return{r,g:n,b:i}}function i3(t){return t.diff(Ai.moment.unix(0),"seconds")}function a3(t){if(t.length!==0)return t.slice().sort()[Math.floor(t.length/2)]}function s3(t,e){return(0,r3.default)(t,e,{strict:!0})}function Fc(t,e){return new Proxy(t,{get(r,n){return n==="length"?Math.min(e,r.length):r[n]}})}function o3(t,e,r){return t.length<=e?new Array(e-t.length).fill(r).join("")+t:t.substring(t.length-e)}function l3(t,e){if(e<=0)return t;let r=new Array(e).fill(" ").join(""),n=t.substring(e,t.length);return r+n}function gv(t,e,r){return t<=e&&e<=r}function ki(t){let[e,...r]=t.split("/");return[e,r.length===0?void 0:r.join("/")]}function Kn(t){return t.endsWith("/")?t:t.split("/").last().replace(/\.md$/,"")}function c3(t){return t===1?"1 minute":`${t} minutes`}function LH(t){let e=t.lastIndexOf(".");return t.substring(e+1)}function Io(t){if(t.endsWith(".md"))return!1;let e=LH(t);return pT.includes(e)}function vv(t){return(t.startsWith("https://github.com/")||t.startsWith("https://gitlab.com/"))&&(t.endsWith(".git")||(t=t+".git")),t}function Fo(t,e){let r=e.vault.getAbstractFileByPath(t);if(!(r instanceof Ai.TFile))return!1;try{return!!e.viewRegistry.getTypeByExtension(r.extension)}catch(n){return!0}}function u3({isFolder:t,gitRelativePath:e}){let r="/";return r+=e,t&&(r+="/"),r.replace(/([\\!#*?[\]])/g,String.raw`\$1`).replace(/\s(?=\s*$)/g,String.raw`\ `)}function $c(t,e,r={}){return new Promise((n,i)=>{let a=(0,t3.spawn)(t,e,r),s="",o="";a.stdout.on("data",l=>{s+=l.toString()}),a.stderr.on("data",l=>{o+=l.toString()}),a.on("error",l=>{n({error:new Error(l.message),stdout:s,stderr:s,code:1})}),a.on("close",l=>{n({stdout:s,stderr:o,code:l!=null?l:1,error:void 0})})})}m();var f3=require("obsidian"),$o=class{constructor(e){this.plugin=e,this.app=e.app}getRelativeVaultPath(e){return this.plugin.settings.basePath?this.plugin.settings.basePath+"/"+e:e}getRelativeRepoPath(e,r=!0){return r&&this.plugin.settings.basePath.length>0?e.substring(this.plugin.settings.basePath.length+1):e}unload(){}_getTreeStructure(e,r=0){let n=[];for(e=[...e];e.length>0;){let i=e.first(),a=i.path.substring(r);if(a.contains("/")){let s=a.substring(0,a.indexOf("/")),o=e.filter(c=>c.path.substring(r).startsWith(s+"/"));o.forEach(c=>e.remove(c));let l=i.path.substring(0,a.indexOf("/")+r);n.push({title:s,path:l,vaultPath:this.getRelativeVaultPath(l),children:this._getTreeStructure(o,(r>0?r+s.length:s.length)+1)})}else n.push({title:a,data:i,path:i.path,vaultPath:this.getRelativeVaultPath(i.path)}),e.remove(i)}return n}simplify(e){var r,n,i,a;for(let s of e){for(;;){let o=((r=s.children)==null?void 0:r.length)==1,l=((i=(n=s.children)==null?void 0:n.first())==null?void 0:i.data)==null;if(!(s.children!=null&&o&&l))break;let c=s.children.first();s.title+="/"+c.title,s.data=c.data,s.path=c.path,s.vaultPath=c.vaultPath,s.children=c.children}s.children!=null&&this.simplify(s.children),(a=s.children)==null||a.sort((o,l)=>{let c=(l.data==null?1:0)-(o.data==null?1:0);return c!=0?c:o.title.localeCompare(l.title)})}return e.sort((s,o)=>{let l=(o.data==null?1:0)-(s.data==null?1:0);return l!=0?l:s.title.localeCompare(o.title)})}getTreeStructure(e){let r=this._getTreeStructure(e);return this.simplify(r)}async formatCommitMessage(e){let r;if(e.includes("{{numFiles}}")){r=await this.status();let n=r.staged.length;e=e.replace("{{numFiles}}",String(n))}if(e.includes("{{hostname}}")){let n=this.plugin.localStorage.getHostname()||"";e=e.replace("{{hostname}}",n)}if(e.includes("{{files}}")){r=r!=null?r:await this.status();let n={},i="";if(r.staged.length<100){r.staged.forEach(s=>{s.index in n?n[s.index].push(s.path):n[s.index]=[s.path]});let a=[];for(let[s,o]of Object.entries(n))a.push(s+" "+o.join(" "));i=a.join(", ")}else i="Too many files to list";e=e.replace("{{files}}",i)}if(e=e.replace("{{date}}",(0,f3.moment)().format(this.plugin.settings.commitDateFormat)),this.plugin.settings.listChangedFilesInMessageBody){let n=r!=null?r:await this.status(),i="";n.staged.length<100?i=n.staged.map(a=>a.path).join(` -`):i="Too many files to list",e=e+` - -Affected files: -`+i}return e}};var Ce=class extends $o{constructor(r){super(r);this.useDefaultWindowsGitPath=!1}async setGitInstance(r=!1){var n;if(await this.isGitInstalled()){let i=this.app.vault.adapter,a=i.getBasePath(),s=a;this.plugin.settings.basePath&&(await i.exists((0,Ti.normalizePath)(this.plugin.settings.basePath))?s=Ve.join(a,this.plugin.settings.basePath):r||new Ti.Notice("ObsidianGit: Base path does not exist")),this.absoluteRepoPath=s,this.git=fT({baseDir:s,binary:this.plugin.localStorage.getGitPath()||(this.useDefaultWindowsGitPath?hd:void 0),config:["core.quotepath=off"],unsafe:{allowUnsafeCustomBinary:!0}});let o=this.plugin.localStorage.getPATHPaths(),l=this.plugin.localStorage.getEnvVars(),c=this.plugin.settings.gitDir;if(o.length>0){let v=o.join(":")+":"+process.env.PATH;process.env.PATH=v}c&&(process.env.GIT_DIR=c);for(let v of l){let[y,b]=v.split("=");process.env[y]=b}let u="simple-git",f=",",h=((n=localStorage.debug)!=null?n:"").split(f);if(!h.includes(u)&&!h.includes(`-${u}`)&&(h.push(u),p3.default.enable(h.join(f))),await this.git.checkIsRepo()){let v=await this.git.revparse("--show-cdup"),y=(0,Dc.resolve)(s+Dc.sep+v);this.absoluteRepoPath=y,await this.git.cwd(y)}let p=Ve.join(a,this.app.vault.configDir,"plugins","obsidian-git"),g=Ve.join(p,es);process.env.SSH_ASKPASS==null&&(process.env.SSH_ASKPASS=g),process.env.OBSIDIAN_GIT_CREDENTIALS_INPUT=Ve.join(p,pd),process.env.SSH_ASKPASS==g&&this.askpass().catch(v=>this.plugin.displayError(v))}}getRelativeVaultPath(r){let i=this.app.vault.adapter.getBasePath(),a=Ve.join(this.absoluteRepoPath,r),s=Ve.relative(i,a);return Ti.Platform.isWin&&(s=s.replace(/\\/g,"/")),s}getRelativeRepoPath(r,n=!0){if(n){let a=this.plugin.app.vault.adapter.getBasePath(),s=this.absoluteRepoPath,o=Ve.join(a,r),l=Ve.relative(s,o);return Ti.Platform.isWin&&(l=l.replace(/\\/g,"/")),l}return r}async askpass(){let r=this.app.vault.adapter,n=r.getBasePath(),i=Ve.join(n,this.app.vault.configDir,"plugins","obsidian-git"),a=this.app.vault.configDir+"/plugins/obsidian-git/";await this.addAskPassScriptToExclude(),await Fr.writeFile(Ve.join(i,es),hT),await Fr.chmod(Ve.join(i,es),493),this.watchAbortController=new AbortController;let{signal:s}=this.watchAbortController;try{let o=Fr.watch(i,{signal:s});for await(let l of o){if(l.filename!=pd)continue;let c=a+pd;if(!await r.exists(c))continue;let u=await r.read(c),f;u.length>60&&(f=new Ti.Notice(u,999999));let d=await new ze(this.plugin,{allowEmpty:!0,obscure:!0,placeholder:u.length>60?"Enter a response to the message.":u}).openAndGetResult();f==null||f.hide(),await r.exists(c)&&await r.write(`${c}.response`,d!=null?d:"")}}catch(o){this.plugin.displayError(o),await Fr.rm(Ve.join(i,es),{force:!0}),await Fr.rm(Ve.join(i,`${es}.response`),{force:!0}),await new Promise(l=>setTimeout(l,5e3)),this.plugin.log("Retry watch for ask pass"),await this.askpass()}}async addAskPassScriptToExclude(){try{let r=await this.git.revparse(["--path-format=absolute","--git-path","info/exclude"]),n=Ve.join(this.app.vault.configDir,"plugins","obsidian-git",es),i=this.getRelativeRepoPath(n,!0);(await Fr.readFile(r,"utf-8")).split(` -`).some(l=>l.contains(i))||await Fr.appendFile(r,i+` -`)}catch(r){console.error("Error while adding askpass script to exclude file:",r)}}unload(){var r;(r=this.watchAbortController)==null||r.abort()}async status(r){let n=r==null?void 0:r.path;this.plugin.setPluginState({gitAction:1});let i=await this.git.status(n!=null?["--",n]:[]);this.plugin.setPluginState({gitAction:0});let a=i.files.map(s=>{let o=this.formatPath(s);return{path:o.path,from:o.from,index:s.index==="?"?"U":s.index,workingDir:s.working_dir==="?"?"U":s.working_dir,vaultPath:this.getRelativeVaultPath(o.path)}});return{all:a,changed:a.filter(s=>s.workingDir!==" "),staged:a.filter(s=>s.index!==" "&&s.index!="U"),conflicted:i.conflicted.map(s=>this.formatPath({path:s}).path)}}async submoduleAwareHeadRevisonInContainingDirectory(r){let n=this.getRelativeRepoPath(r),a=["-C",Ve.dirname(n),"rev-parse","HEAD"],s=this.git.raw(a);return s.catch(o=>console.warn("obsidian-git: rev-parse error:",o)),(await s).trim()}async getSubmodulePaths(){return new Promise(r=>{this.git.outputHandler((n,i,a,s)=>{if(!(s.contains("submodule")&&s.contains("foreach")))return;let o="",l=this.app.vault.adapter.getBasePath()+(this.plugin.settings.basePath?"/"+this.plugin.settings.basePath:"");i.on("data",c=>{o+=c.toString("utf8")}),i.on("end",()=>{let u=o.split(` -`).map(f=>{let d=f.match(/'([^']*)'/);if(d!=null)return l+"/"+d[1]+Dc.sep}).filter(f=>!!f);u.reverse(),r(u)})}),this.git.subModule(["foreach","--recursive",""]).then(()=>{this.git.outputHandler(()=>{})},n=>this.plugin.displayError(n))})}formatPath(r){function n(i){if(i!=null)return i.startsWith('"')&&i.endsWith('"')?i.substring(1,i.length-1):i}return r.from!=null?{from:n(r.from),path:n(r.path)}:{path:n(r.path)}}async blame(r,n,i){if(r=this.getRelativeRepoPath(r),!await this.isTracked(r))return"untracked";let a=await this.getSubmoduleOfFile(r),s=a?["-C",a.submodule]:[],o=a?a.relativeFilepath:r;s.push("blame","--porcelain"),i&&s.push("-w");let l=`-C${dd}`;switch(n){case"inactive":break;case"same-commit":s.push("-C",l);break;case"all-commits":s.push("-C","-C",l);break;default:pa(n)}s.push("--",o);let c=await this.git.raw(s);return jH(c)}async isTracked(r){let n=await this.getSubmoduleOfFile(r),i=n?["-C",n.submodule]:[],a=n?n.relativeFilepath:r;return i.push("ls-files","--",a),this.git.raw(i).then(s=>s.trim()!=="")}async commitAll({message:r}){if(this.plugin.settings.updateSubmodules){this.plugin.setPluginState({gitAction:4});let i=await this.getSubmodulePaths();for(let a of i)await this.git.cwd({path:a,root:!1}).add("-A"),await this.git.cwd({path:a,root:!1}).commit(await this.formatCommitMessage(r))}this.plugin.setPluginState({gitAction:3}),await this.git.add("-A"),this.plugin.setPluginState({gitAction:4});let n=await this.git.commit(await this.formatCommitMessage(r));return this.app.workspace.trigger("obsidian-git:head-change"),n.summary.changes}async commit({message:r,amend:n}){this.plugin.setPluginState({gitAction:4});let i=(await this.git.commit(await this.formatCommitMessage(r),n?["--amend"]:[])).summary.changes;return this.app.workspace.trigger("obsidian-git:head-change"),this.plugin.setPluginState({gitAction:0}),i}async stage(r,n){this.plugin.setPluginState({gitAction:3}),r=this.getRelativeRepoPath(r,n),await this.git.add(["--",r]),this.plugin.setPluginState({gitAction:0})}async stageAll({dir:r}){this.plugin.setPluginState({gitAction:3}),await this.git.add(r!=null?r:"-A"),this.plugin.setPluginState({gitAction:0})}async unstageAll({dir:r}){this.plugin.setPluginState({gitAction:3}),await this.git.reset(r!=null?["--",r]:[]),this.plugin.setPluginState({gitAction:0})}async unstage(r,n){this.plugin.setPluginState({gitAction:3}),r=this.getRelativeRepoPath(r,n),await this.git.reset(["--",r]),this.plugin.setPluginState({gitAction:0})}async discard(r){this.plugin.setPluginState({gitAction:3}),await this.isTracked(r)&&await this.git.checkout(["--",r]),this.plugin.setPluginState({gitAction:0})}async getUntrackedPaths(r){let n=r==null?void 0:r.path;this.plugin.setPluginState({gitAction:1});let i=["ls-files","--others","--exclude-standard","--directory"];n!=null&&i.push("--",n);let a=await this.git.raw(i);return this.plugin.setPluginState({gitAction:0}),a.split(/\r\n|\r|\n/).filter(s=>s.length>0)}async hashObject(r){r=this.getRelativeRepoPath(r);let n=await this.getSubmoduleOfFile(r),i=n?["-C",n.submodule]:[],a=n?n.relativeFilepath:r;return i.push("hash-object","--",a),this.git.raw(i)}async discardAll({dir:r}){return this.discard(r!=null?r:".")}async pull(){this.plugin.setPluginState({gitAction:2});try{this.plugin.settings.updateSubmodules&&await this.git.subModule(["update","--remote","--merge","--recursive"]);let r=await this.branchInfo(),n=await this.git.revparse([r.current]);if(!r.tracking&&this.plugin.settings.updateSubmodules){this.plugin.log("No tracking branch found. Ignoring pull of main repo and updating submodules only.");return}await this.git.fetch();let i=await this.git.revparse([r.tracking]);if(n!==i){if(this.plugin.settings.syncMethod==="merge"||this.plugin.settings.syncMethod==="rebase")try{switch(this.plugin.settings.syncMethod){case"merge":await this.git.merge([r.tracking]);break;case"rebase":await this.git.rebase([r.tracking])}}catch(o){this.plugin.displayError(`Pull failed (${this.plugin.settings.syncMethod}): ${"message"in o?o.message:o}`);return}else if(this.plugin.settings.syncMethod==="reset")try{await this.git.raw(["update-ref",`refs/heads/${r.current}`,i]),await this.unstageAll({})}catch(o){this.plugin.displayError(`Sync failed (${this.plugin.settings.syncMethod}): ${"message"in o?o.message:o}`)}this.app.workspace.trigger("obsidian-git:head-change");let a=await this.git.revparse([r.current]);return(await this.git.diff([`${n}..${a}`,"--name-only"])).split(/\r\n|\r|\n/).filter(o=>o.length>0).map(o=>({path:o,workingDir:"P",vaultPath:this.getRelativeVaultPath(o)}))}else return[]}catch(r){this.convertErrors(r)}}async push(){this.plugin.setPluginState({gitAction:5});try{if(this.plugin.settings.updateSubmodules){let s=await this.git.env({...process.env,OBSIDIAN_GIT:1}).subModule(["foreach","--recursive",`tracking=$(git for-each-ref --format='%(upstream:short)' "$(git symbolic-ref -q HEAD)"); echo $tracking; if [ ! -z "$(git diff --shortstat $tracking)" ]; then git push; fi`]);console.log(s)}let r=await this.git.status(),n=r.tracking,i=r.current;if(!n&&this.plugin.settings.updateSubmodules){this.plugin.log("No tracking branch found. 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p=y3(Lc(u,h),d);e.value=Gd(e.value,p),n.value=Sv(n.value,u,p),t.value=Ev(t.value,u,u.slice(0,u.length-p.length))}else if(n){var g=n.value.match(/^\s*/)[0],v=e.value.match(/\s*$/)[0],y=w3(v,g);e.value=Gd(e.value,y)}else if(t){var b=t.value.match(/\s*$/)[0],E=e.value.match(/^\s*/)[0],S=w3(b,E);e.value=Lc(e.value,S)}}var A3=new yn;A3.tokenize=function(t){var e=new RegExp("(\\r?\\n)|[".concat(zd,"]+|[^\\S\\n\\r]+|[^").concat(zd,"]"),"ug");return t.match(e)||[]};function k3(t,e,r){return A3.diff(t,e,r)}var Wd=new yn;Wd.tokenize=function(t,e){e.stripTrailingCr&&(t=t.replace(/\r\n/g,` -`));var r=[],n=t.split(/(\n|\r\n)/);n[n.length-1]||n.pop();for(var i=0;it.length)&&(e=t.length);for(var r=0,n=new Array(e);r0?f(I.lines.slice(-s.context)):[],h-=g.length,p-=g.length)}(C=g).push.apply(C,xv(T.map(function(le){return(_.added?"+":"-")+le}))),_.added?y+=T.length:v+=T.length}else{if(h)if(T.length<=s.context*2&&E(0,ma.normalizePath)(a.substring(e.length))):i=n,i}async mkdir(e){return this.adapter.mkdir(e)}async rmdir(e,r){var n,i;return this.adapter.rmdir(e,(i=(n=r==null?void 0:r.options)==null?void 0:n.recursive)!=null?i:!1)}async stat(e){if(e.endsWith(this.gitDir+"/index")){if(this.index!==void 0&&this.indexctime!=null&&this.indexmtime!=null)return{isFile:()=>!0,isDirectory:()=>!1,isSymbolicLink:()=>!1,size:this.index.byteLength,type:"file",ctimeMs:this.indexctime,mtimeMs:this.indexmtime};{let n=await this.adapter.stat(e);if(n==null)throw{code:"ENOENT"};return this.indexctime=n.ctime,this.indexmtime=n.mtime,{ctimeMs:n.ctime,mtimeMs:n.mtime,size:n.size,type:"file",isFile:()=>!0,isDirectory:()=>!1,isSymbolicLink:()=>!1}}}e==="."&&(e="/");let r=this.vault.getAbstractFileByPath(e);if(this.maybeLog("Stat: "+e),r instanceof ma.TFile)return this.maybeLog("Reuse stat"),{ctimeMs:r.stat.ctime,mtimeMs:r.stat.mtime,size:r.stat.size,type:"file",isFile:()=>!0,isDirectory:()=>!1,isSymbolicLink:()=>!1};{let n=await this.adapter.stat(e);if(n)return{ctimeMs:n.ctime,mtimeMs:n.mtime,size:n.size,type:n.type==="folder"?"directory":n.type,isFile:()=>n.type==="file",isDirectory:()=>n.type==="folder",isSymbolicLink:()=>!1};throw{code:"ENOENT"}}}async unlink(e){return this.adapter.remove(e)}async lstat(e){return this.stat(e)}async readlink(e){throw new Error(`readlink of (${e}) is not implemented.`)}async symlink(e){throw new Error(`symlink of (${e}) is not implemented.`)}async saveAndClear(){this.index!==void 0&&await this.adapter.writeBinary(this.plugin.gitManager.getRelativeVaultPath(this.gitDir+"/index"),this.index,{ctime:this.indexctime,mtime:this.indexmtime}),this.clearIndex()}clearIndex(){this.index=void 0,this.indexctime=void 0,this.indexmtime=void 0}get gitDir(){return this.plugin.settings.gitDir||".git"}maybeLog(e){}};var wn=class extends $o{constructor(r){super(r);this.FILE=0;this.HEAD=1;this.WORKDIR=2;this.STAGE=3;this.status_mapping={"000":" ","003":"AD","020":"??","022":"A ","023":"AM",100:"D ",101:" D",103:"MD",110:"DA",111:" ",113:"MM",120:"DA",121:" M",122:"M ",123:"MM"};this.noticeLength=999999;this.fs=new Xd(this.app.vault,this.plugin)}getRepo(){return{fs:this.fs,dir:this.plugin.settings.basePath,gitdir:this.plugin.settings.gitDir||void 0,onAuth:()=>{var r,n;return{username:(r=this.plugin.localStorage.getUsername())!=null?r:void 0,password:(n=this.plugin.localStorage.getPassword())!=null?n:void 0}},onAuthFailure:async()=>{new rs.Notice("Authentication failed. Please try with different credentials");let r=await new ze(this.plugin,{placeholder:"Specify your username"}).openAndGetResult();if(r){let n=await new ze(this.plugin,{placeholder:"Specify your password/personal access token"}).openAndGetResult();if(n)return this.plugin.localStorage.setUsername(r),this.plugin.localStorage.setPassword(n),{username:r,password:n}}return{cancel:!0}},http:{async request({url:r,method:n,headers:i,body:a}){let s;a&&(s=await iU(a));let o=await(0,rs.requestUrl)({url:r,method:n,headers:i,body:s,throw:!1});return{url:r,method:n,headers:o.headers,body:nU(o.arrayBuffer),statusCode:o.status,statusMessage:o.status.toString()}}}}}async wrapFS(r){try{let n=await r;return await this.fs.saveAndClear(),n}catch(n){throw await this.fs.saveAndClear(),n}}async status(r){let n,i=window.setTimeout(()=>{n=new rs.Notice("This takes longer: Getting status",this.noticeLength)},2e4);try{this.plugin.setPluginState({gitAction:1});let a={...this.getRepo()};(r==null?void 0:r.path)!=null&&(a.filepaths=[`${r.path}/`]);let s=(await this.wrapFS(re.default.statusMatrix(a))).map(f=>this.getFileStatusResult(f)),o=[],l=[],c=[];for(let f of s)f.workingDir!==" "&&o.push(f),f.index!==" "&&f.index!=="U"&&l.push(f),(f.index!=" "||f.workingDir!=" ")&&c.push(f);let u=[];return window.clearTimeout(i),n==null||n.hide(),{all:c,changed:o,staged:l,conflicted:u}}catch(a){throw window.clearTimeout(i),n==null||n.hide(),this.plugin.displayError(a),a}}async commitAll({message:r,status:n,unstagedFiles:i}){try{return await this.checkAuthorInfo(),await this.stageAll({status:n,unstagedFiles:i}),this.commit({message:r})}catch(a){throw this.plugin.displayError(a),a}}async commit({message:r}){try{await this.checkAuthorInfo(),this.plugin.setPluginState({gitAction:4});let n=await this.formatCommitMessage(r),i=this.plugin.localStorage.getConflict(),a;if(i){let s=await this.branchInfo();a=[s.current,s.tracking]}await this.wrapFS(re.default.commit({...this.getRepo(),message:n,parent:a})),this.plugin.localStorage.setConflict(!1);return}catch(n){throw this.plugin.displayError(n),n}}async stage(r,n){let i=this.getRelativeRepoPath(r,n),a;n?a=r:a=this.getRelativeVaultPath(r);try{this.plugin.setPluginState({gitAction:3}),await this.app.vault.adapter.exists(a)?await this.wrapFS(re.default.add({...this.getRepo(),filepath:i})):await this.wrapFS(re.default.remove({...this.getRepo(),filepath:i}))}catch(s){throw this.plugin.displayError(s),s}}async stageAll({dir:r,status:n,unstagedFiles:i}){try{if(n)await Promise.all(n.changed.map(a=>a.workingDir!=="D"?this.wrapFS(re.default.add({...this.getRepo(),filepath:a.path})):re.default.remove({...this.getRepo(),filepath:a.path})));else{let a=i!=null?i:await this.getUnstagedFiles(r!=null?r:".");await Promise.all(a.map(({path:s,type:o})=>o=="D"?re.default.remove({...this.getRepo(),filepath:s}):this.wrapFS(re.default.add({...this.getRepo(),filepath:s}))))}}catch(a){throw 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this.wrapFS(re.default.checkout({...this.getRepo(),filepaths:i,force:!0}))}catch(a){throw this.plugin.displayError(a),a}}async getUntrackedPaths(r){let n=[];if(r.status)for(let i of r.status.changed)i.index=="U"&&i.workingDir==="U"&&i.path.startsWith(r.path!=null?`${r.path}/`:"")&&n.push(i.path);else{let i=await this.status({path:r==null?void 0:r.path});for(let a of i.changed)a.index==="U"&&a.workingDir==="U"&&n.push(a.path)}return n}getProgressText(r,n){let i=`${r} progress:`;return n.phase&&(i=`${i} ${n.phase}:`),n.loaded&&(i=`${i} ${n.loaded}`,n.total&&(i=`${i} of ${n.total}`)),i}resolveRef(r){return this.wrapFS(re.default.resolveRef({...this.getRepo(),ref:r}))}async pull(){let r=this.showNotice("Initializing pull");try{this.plugin.setPluginState({gitAction:2});let n=await this.resolveRef("HEAD");await this.fetch();let i=await this.branchInfo();await this.checkAuthorInfo(),(await this.wrapFS(re.default.merge({...this.getRepo(),ours:i.current,theirs:i.tracking,abortOnConflict:!1}))).alreadyMerged||await this.wrapFS(re.default.checkout({...this.getRepo(),ref:i.current,onProgress:l=>{r!==void 0&&(r.noticeEl.innerText=this.getProgressText("Checkout",l))},remote:i.remote})),r==null||r.hide();let s=await this.resolveRef("HEAD"),o=await this.getFileChangesCount(n,s);return this.showNotice("Finished pull",!1),o.map(l=>({path:l.path,workingDir:"P",index:"P",vaultPath:this.getRelativeVaultPath(l.path)}))}catch(n){throw r==null||r.hide(),n instanceof re.Errors.MergeConflictError&&await this.plugin.handleConflict(n.data.filepaths.map(i=>this.getRelativeVaultPath(i))),this.plugin.displayError(n),n}}async push(){if(!await this.canPush())return 0;let r=this.showNotice("Initializing push");try{this.plugin.setPluginState({gitAction:1});let n=await this.branchInfo(),i=n.tracking,a=n.current,s=(await this.getFileChangesCount(a,i)).length;this.plugin.setPluginState({gitAction:5});let o=await this.getCurrentRemote();return await this.wrapFS(re.default.push({...this.getRepo(),remote:o,onProgress:l=>{r!==void 0&&(r.noticeEl.innerText=this.getProgressText("Pushing",l))}})),r==null||r.hide(),s}catch(n){throw r==null||r.hide(),this.plugin.displayError(n),n}}async getUnpushedCommits(){let r=await this.branchInfo(),n=r.tracking,i=r.current;if(n==null||i==null)return 0;let a=await this.resolveRef(i),s=await this.resolveRef(n);return(await this.getFileChangesCount(a,s)).length}async canPush(){let r=await this.branchInfo(),n=r.tracking,i=r.current,a=await this.resolveRef(i),s=await this.resolveRef(n);return a!=s}async checkRequirements(){return await this.plugin.app.vault.adapter.exists(`${this.getRepo().dir}/.git/HEAD`)?"valid":"missing-repo"}async branchInfo(){var r,n;try{let i=await re.default.currentBranch(this.getRepo())||"",a=await re.default.listBranches(this.getRepo()),s=(r=await this.getConfig(`branch.${i}.remote`))!=null?r:"origin",o=(n=await this.getConfig(`branch.${i}.merge`))==null?void 0:n.split("refs/heads")[1],l=o?s+o:void 0;return{current:i,tracking:l,branches:a,remote:s}}catch(i){throw this.plugin.displayError(i),i}}async getCurrentRemote(){var i;let r=await re.default.currentBranch(this.getRepo())||"";return(i=await this.getConfig(`branch.${r}.remote`))!=null?i:"origin"}async checkout(r,n){try{return this.wrapFS(re.default.checkout({...this.getRepo(),ref:r,force:!!n,remote:n}))}catch(i){throw this.plugin.displayError(i),i}}async createBranch(r){try{await this.wrapFS(re.default.branch({...this.getRepo(),ref:r,checkout:!0}))}catch(n){throw this.plugin.displayError(n),n}}async deleteBranch(r){try{await this.wrapFS(re.default.deleteBranch({...this.getRepo(),ref:r}))}catch(n){throw this.plugin.displayError(n),n}}branchIsMerged(r){return Promise.resolve(!0)}async init(){try{await this.wrapFS(re.default.init(this.getRepo()))}catch(r){throw this.plugin.displayError(r),r}}async clone(r,n,i){let a=this.showNotice("Initializing clone");try{await this.wrapFS(re.default.clone({...this.getRepo(),dir:n,url:r,depth:i,onProgress:s=>{a!==void 0&&(a.noticeEl.innerText=this.getProgressText("Cloning",s))}})),a==null||a.hide()}catch(s){throw a==null||a.hide(),this.plugin.displayError(s),s}}async setConfig(r,n){try{return this.wrapFS(re.default.setConfig({...this.getRepo(),path:r,value:n}))}catch(i){throw this.plugin.displayError(i),i}}async getConfig(r){try{return this.wrapFS(re.default.getConfig({...this.getRepo(),path:r}))}catch(n){throw this.plugin.displayError(n),n}}async fetch(r){let n=this.showNotice("Initializing fetch");try{let i={...this.getRepo(),onProgress:a=>{n!==void 0&&(n.noticeEl.innerText=this.getProgressText("Fetching",a))},remote:r!=null?r:await this.getCurrentRemote()};await this.wrapFS(re.default.fetch(i)),n==null||n.hide()}catch(i){throw this.plugin.displayError(i),n==null||n.hide(),i}}async setRemote(r,n){try{await this.wrapFS(re.default.addRemote({...this.getRepo(),remote:r,url:n,force:!0}))}catch(i){throw this.plugin.displayError(i),i}}async getRemoteBranches(r){let n=[];return n.push(...await this.wrapFS(re.default.listBranches({...this.getRepo(),remote:r}))),n.remove("HEAD"),n=n.map(i=>`${r}/${i}`),n}async getRemotes(){return(await this.wrapFS(re.default.listRemotes({...this.getRepo()}))).map(r=>r.remote)}async removeRemote(r){await this.wrapFS(re.default.deleteRemote({...this.getRepo(),remote:r}))}async getRemoteUrl(r){var n;return(n=(await this.wrapFS(re.default.listRemotes({...this.getRepo()}))).filter(i=>i.remote==r)[0])==null?void 0:n.url}async log(r,n=!0,i,a){let s=await this.wrapFS(re.default.log({...this.getRepo(),depth:i,ref:a}));return Promise.all(s.map(async o=>{let l=o.commit.message.split(` - -`);return{message:l[0],author:{name:o.commit.author.name,email:o.commit.author.email},body:l.slice(1).join(` - -`),date:new 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Error("Git author name and email are not set. 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If so, just disable this option.").addToggle(c=>c.setValue(n.settings.refreshSourceControl).onChange(async u=>{n.settings.refreshSourceControl=u,await n.saveSettings()})),new z.Setting(r).setName("Source control view refresh interval").setDesc("Milliseconds to wait after file change before refreshing the Source Control View.").addText(c=>{c.inputEl.type="number",this.setNonDefaultValue({text:c,settingsProperty:"refreshSourceControlTimer"}),c.setPlaceholder(String(Xe.refreshSourceControlTimer)),c.onChange(async f=>{f!==""&&Number.isInteger(Number(f))?n.settings.refreshSourceControlTimer=Math.max(Number(f),500):n.settings.refreshSourceControlTimer=Xe.refreshSourceControlTimer,await n.saveSettings(),n.setRefreshDebouncer()})}),new z.Setting(r).setName("Miscellaneous").setHeading(),n.gitManager instanceof Ce&&new z.Setting(r).setName("Diff view style").setDesc('Set the style for the diff view. Note that the actual diff in "Split" mode is not generated by Git, but the editor itself instead so it may differ from the diff generated by Git. One advantage of this is that you can edit the text in that view.').addDropdown(c=>{let u={split:"Split",git_unified:"Unified"};c.addOptions(u),c.setValue(n.settings.diffStyle),c.onChange(async f=>{n.settings.diffStyle=f,await n.saveSettings()})}),new z.Setting(r).setName("Disable informative notifications").setDesc("Disable informative notifications for git operations to minimize distraction (refer to status bar for updates).").addToggle(c=>c.setValue(n.settings.disablePopups).onChange(async u=>{n.settings.disablePopups=u,this.refreshDisplayWithDelay(),await n.saveSettings()})),new z.Setting(r).setName("Disable error notifications").setDesc("Disable error notifications of any kind to minimize distraction (refer to status bar for updates).").addToggle(c=>c.setValue(!n.settings.showErrorNotices).onChange(async u=>{n.settings.showErrorNotices=!u,await n.saveSettings()})),n.settings.disablePopups||new z.Setting(r).setName("Hide notifications for no changes").setDesc("Don't show notifications when there are no changes to commit or push.").addToggle(c=>c.setValue(n.settings.disablePopupsForNoChanges).onChange(async u=>{n.settings.disablePopupsForNoChanges=u,await n.saveSettings()})),new z.Setting(r).setName("Show status bar").setDesc("Obsidian must be restarted for the changes to take affect.").addToggle(c=>c.setValue(n.settings.showStatusBar).onChange(async u=>{n.settings.showStatusBar=u,await n.saveSettings()})),new z.Setting(r).setName("File menu integration").setDesc('Add "Stage", "Unstage" and "Add to .gitignore" actions to the file menu.').addToggle(c=>c.setValue(n.settings.showFileMenu).onChange(async u=>{n.settings.showFileMenu=u,await n.saveSettings()})),new z.Setting(r).setName("Show branch status bar").setDesc("Obsidian must be restarted for the changes to take affect.").addToggle(c=>c.setValue(n.settings.showBranchStatusBar).onChange(async u=>{n.settings.showBranchStatusBar=u,await n.saveSettings()})),new z.Setting(r).setName("Show the count of modified files in the status bar").addToggle(c=>c.setValue(n.settings.changedFilesInStatusBar).onChange(async u=>{n.settings.changedFilesInStatusBar=u,await n.saveSettings()})),n.gitManager instanceof wn?new z.Setting(r).setName("Authentication/commit author").setHeading():new z.Setting(r).setName("Commit author").setHeading(),n.gitManager instanceof wn&&new z.Setting(r).setName("Username on your git server. 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var(--text-on-accent)
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${o}
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Currently: ${n}`}previewOldestAgeDescriptionHtml(r){let n=Zd(r);return[`The oldest age in the line author coloring. Everything older will have the same color. -
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r=z3(this.lineAuthoring,this.startLine,this.endLine),n=r.isZeroCommit?"":this.renderNonZeroCommit(r);return!r.isZeroCommit&&this.options!=="waiting-for-result"?$3(this,n):n=this.adaptTextForFakeCommit(r,n,this.options),this.createHtmlNode(r,n,this.options==="waiting-for-result")}createHtmlNode(r,n,i){var c;let a=window.createDiv();a.innerText=n;let{color:s,daysSinceCommit:o}=oh((c=r==null?void 0:r.author)==null?void 0:c.epochSeconds,r==null?void 0:r.isZeroCommit,this.settings);a.style.backgroundColor=s,U3(r,i,a);function l(){let u=a.cloneNode(!0);return No.add(u),i||D3(o),u}return l}renderNonZeroCommit(r){let n=this.settings.showCommitHash?this.renderHash(r):"",i=this.settings.authorDisplay==="hide"?"":`${this.renderAuthorName(r,this.settings.authorDisplay)}`,a=this.settings.dateTimeFormatOptions==="hide"?"":`${this.renderAuthoringDate(r,this.settings.dateTimeFormatOptions,this.settings.dateTimeFormatCustomString,this.settings.dateTimeTimezone)}`;return[n,i,a].filter(o=>o.length>=1).join(" ")}renderHash(r){return r.hash.substring(0,6)}renderAuthorName(r,n){var o,l,c,u;let i=(l=(o=r==null?void 0:r.author)==null?void 0:o.name)!=null?l:"",a=i.split(" ").filter(f=>f.length>=1),s;switch(n){case"initials":s=a.map(f=>f[0].toUpperCase()).join("");break;case"first name":s=(c=a.first())!=null?c:V3;break;case"last name":s=(u=a.last())!=null?u:V3;break;case"full":s=i;break;default:return pa(n)}return s3(r==null?void 0:r.author,r==null?void 0:r.committer)||(s=s+gU),s}renderAuthoringDate(r,n,i,a){var c;let s="?";if(((c=r==null?void 0:r.author)==null?void 0:c.epochSeconds)===void 0)return s;let 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c=new Dv(t,e,r,n,i,a);return ih.set(o,c),c}m();var W3=require("obsidian");function jv(){var e,r;let t=(r=(e=Qr.get())==null?void 0:e.gutterSpacingFallbackLength)!=null?r:Xe.lineAuthor.gutterSpacingFallbackLength;return new Bo(Array(t).fill("-").join(""))}function Nv(t){let{lineAuthoring:e,ageForInitialRender:r}=wU(t);return ns(e,1,1,"initialGutter"+r,t,"waiting-for-result")}function wU(t){var a;let e=(a=L3())!=null?a:eh(t)*.25,r=(0,W3.moment)().add(-e,"days"),n={name:"",epochSeconds:i3(r),tz:"+0000"},i={hash:"waiting-for-result",author:n,committer:n,isZeroCommit:!1};return{lineAuthoring:{hashPerLine:[void 0,"waiting-for-result"],commits:new Map([["waiting-for-result",i]])},ageForInitialRender:e}}m();function Y3(t,e){return ns({hashPerLine:[void 0,"000000"],commits:new Map([["000000",m3]])},1,1,t,e)}var Bv=new Bo(""),J3=(0,Z3.gutter)({class:"line-author-gutter-container",markers(t){let e=t.state.field(Lo,!1);return bU(t,e)},lineMarkerChange(t){let e=Qd(t.state.field(Lo));return Qd(t.startState.field(Lo))!==e},renderEmptyElements:!0,initialSpacer:t=>(X3(t),jv()),updateSpacer:(t,e)=>{var r,n;return X3(e.view),(n=(r=nh())==null?void 0:r.gutter)!=null?n:jv()}});function bU(t,e){let r=Qd(e),n=t.state.doc,i=new Map;for(let u=1;u<=n.lines;u++){let f=n.line(u).from,d=t.lineBlockAt(f).to;i.set(u,[f,d]),r.update([f,d,0])}let a=Qr.get();r.update("s"+Object.values(Qr).join(","));let s=r.hex(),o=ah.get(s);if(o)return o;let{result:l,allowCache:c}=_U(n,i,a,e);return c&&ah.set(s,l),l}function _U(t,e,r,n){let i=!0,a=t.lines,s=[];function o(f,d,h){return s.push(h.range(f,d))}let l=xU(a,n),c=t.length===0,u=t.iterLines(a,a+1).next().value==="";for(let f=1;f<=a;f++){let[d,h]=e.get(f),p=t.lineAt(h).number;if(c){o(d,h,Bv);continue}if(f===a&&u){o(d,h,Bv);continue}if(n===void 0){o(d,h,Nv(r)),i=!1;continue}let{key:g,la:v}=n;if(v==="untracked"){o(d,h,Y3(v,r));continue}let y=v.hashPerLine.length-1,b=l[f],E=l[p];if(E&&E>y&&o(d,h,Bv),b!==void 0&&gv(1,b,y)&&E!==void 0&&gv(1,E,y)){o(d,h,ns(v,b,E,g,r));continue}if(y<1){o(d,h,Nv(r)),i=!1;continue}let S=Math.clamp(b!=null?b:f,1,y),A=Math.clamp(E!=null?E:p,1,y);o(d,h,ns(v,S,A,g+"computing",r,"waiting-for-result"))}return{result:K3.RangeSet.of(s,!0),allowCache:i}}function xU(t,e){if(!(e!=null&&e.lineOffsetsFromUnsavedChanges))return Array.from(new Array(t+1),i=>i);let r=[void 0],n=0;for(let i=1;i<=t;i++){let a=e.lineOffsetsFromUnsavedChanges.get(i);n+=a!=null?a:0,r[i]=a===void 0?i-n:void 0}return r}function X3(t){t.dom.querySelectorAll(".cm-gutters").forEach(r=>{r!=null&&r.style&&(r.style.marginLeft||(r.style.marginLeft="unset"))})}var ch=class{constructor(e){this.plugin=e;this.lineAuthorings=new Map}async trackChanged(e){return this.trackChangedHelper(e).catch(r=>(console.warn("Git: Error in trackChanged."+r),Promise.reject(r)))}async trackChangedHelper(e){if(e){if(e.path===void 0){console.warn("Git: Attempted to track change of undefined filepath. Unforeseen situation.");return}return this.computeLineAuthorInfo(e.path)}}destroy(){this.lineAuthorings.clear(),Do.clear(),F3()}async computeLineAuthorInfo(e){let r=this.plugin.lineAuthoringFeature.isAvailableOnCurrentPlatform().gitManager,n=await r.submoduleAwareHeadRevisonInContainingDirectory(e),i=await r.hashObject(e),a=C3(n,i,e);if(a!==void 0){if(!this.lineAuthorings.has(a)){let s=await r.blame(e,this.plugin.settings.lineAuthor.followMovement,this.plugin.settings.lineAuthor.ignoreWhitespace);this.lineAuthorings.set(a,s)}this.notifyComputationResultToSubscribers(e,a)}}notifyComputationResultToSubscribers(e,r){Do.ifFilepathDefinedTransformSubscribers(e,n=>n.forEach(i=>i.notifyLineAuthoring(r,this.lineAuthorings.get(r))))}},eM=Q3.Prec.high([I3,Lo,J3]);var uh=class{constructor(e){this.plg=e;this.codeMirrorExtensions=[];this.handleWorkspaceLeaf=e=>{if(!this.lineAuthorInfoProvider){console.warn("Git: undefined lineAuthorInfoProvider. Unexpected situation.");return}let r=e==null?void 0:e.view;!(r instanceof is.MarkdownView)||r.file==null||(r==null?void 0:r.allowNoFile)===!0||this.lineAuthorInfoProvider.trackChanged(r.file).catch(console.error)}}onLoadPlugin(){this.plg.registerEditorExtension(this.codeMirrorExtensions),M3(()=>this.plg.settings.lineAuthor,e=>{this.plg.settings.lineAuthor=e,this.plg.saveSettings()})}conditionallyActivateBySettings(){this.plg.settings.lineAuthor.show&&this.activateFeature()}activateFeature(){try{if(!this.isAvailableOnCurrentPlatform().available)return;G3(this.plg.settings.lineAuthor),this.lineAuthorInfoProvider=new ch(this.plg),this.createEventHandlers(),this.activateCodeMirrorExtensions(),console.log(this.plg.manifest.name+": Enabled line authoring.")}catch(e){console.warn("Git: Error while loading line authoring feature.",e),this.deactivateFeature()}}deactivateFeature(){var e;this.destroyEventHandlers(),this.deactivateCodeMirrorExtensions(),(e=this.lineAuthorInfoProvider)==null||e.destroy(),this.lineAuthorInfoProvider=void 0,console.log(this.plg.manifest.name+": Disabled line authoring.")}isAvailableOnCurrentPlatform(){return{available:this.plg.useSimpleGit&&is.Platform.isDesktopApp,gitManager:this.plg.gitManager instanceof Ce?this.plg.gitManager:void 0}}refreshLineAuthorViews(){this.plg.settings.lineAuthor.show&&(this.deactivateFeature(),this.activateFeature())}activateCodeMirrorExtensions(){this.codeMirrorExtensions.push(eM),this.plg.app.workspace.updateOptions(),this.plg.app.workspace.iterateAllLeaves(this.handleWorkspaceLeaf)}deactivateCodeMirrorExtensions(){for(let e of this.codeMirrorExtensions)this.codeMirrorExtensions.remove(e);this.plg.app.workspace.updateOptions()}createEventHandlers(){this.gutterContextMenuEvent=this.createGutterContextMenuHandler(),this.fileOpenEvent=this.createFileOpenEvent(),this.workspaceLeafChangeEvent=this.createWorkspaceLeafChangeEvent(),this.fileModificationEvent=this.createVaultFileModificationHandler(),this.refreshOnCssChangeEvent=this.createCssRefreshHandler(),this.fileRenameEvent=this.createFileRenameEvent(),j3(),this.plg.registerEvent(this.gutterContextMenuEvent),this.plg.registerEvent(this.refreshOnCssChangeEvent),this.plg.registerEvent(this.fileOpenEvent),this.plg.registerEvent(this.workspaceLeafChangeEvent),this.plg.registerEvent(this.fileModificationEvent),this.plg.registerEvent(this.fileRenameEvent)}destroyEventHandlers(){this.plg.app.workspace.offref(this.refreshOnCssChangeEvent),this.plg.app.workspace.offref(this.fileOpenEvent),this.plg.app.workspace.offref(this.workspaceLeafChangeEvent),this.plg.app.workspace.offref(this.refreshOnCssChangeEvent),this.plg.app.vault.offref(this.fileRenameEvent),this.plg.app.workspace.offref(this.gutterContextMenuEvent)}createFileOpenEvent(){return this.plg.app.workspace.on("file-open",e=>{var r;return void((r=this.lineAuthorInfoProvider)==null?void 0:r.trackChanged(e).catch(console.error))})}createWorkspaceLeafChangeEvent(){return this.plg.app.workspace.on("active-leaf-change",this.handleWorkspaceLeaf)}createFileRenameEvent(){return this.plg.app.vault.on("rename",(e,r)=>{var n;return e instanceof is.TFile&&((n=this.lineAuthorInfoProvider)==null?void 0:n.trackChanged(e))})}createVaultFileModificationHandler(){return this.plg.app.vault.on("modify",e=>{var r;return e instanceof is.TFile&&((r=this.lineAuthorInfoProvider)==null?void 0:r.trackChanged(e))})}createCssRefreshHandler(){return this.plg.app.workspace.on("css-change",()=>this.refreshLineAuthorViews())}createGutterContextMenuHandler(){return this.plg.app.workspace.on("editor-menu",H3)}};m();var fh=class{constructor(e){this.plugin=e;this.tasks=[]}addTask(e,r){this.tasks.push({task:e,onFinished:r!=null?r:()=>{}}),this.tasks.length===1&&this.handleTask()}handleTask(){if(this.tasks.length>0){let e=this.tasks[0];e.task().then(r=>{e.onFinished(r),this.tasks.shift(),this.handleTask()},r=>{this.plugin.displayError(r),e.onFinished(void 0),this.tasks.shift(),this.handleTask()})}}clear(){this.tasks=[]}};m();var $r=require("obsidian");var dh=class{constructor(e,r){this.statusBarEl=e;this.plugin=r;this.messages=[];this.base="obsidian-git-statusbar-";this.statusBarEl.setAttribute("data-tooltip-position","top"),r.registerEvent(r.app.workspace.on("obsidian-git:refreshed",()=>{this.refreshCommitTimestamp().catch(console.error)}))}displayMessage(e,r){this.messages.push({message:`Git: ${e.slice(0,100)}`,timeout:r}),this.display()}display(){this.messages.length>0&&!this.currentMessage?(this.currentMessage=this.messages.shift(),this.statusBarEl.addClass(this.base+"message"),this.statusBarEl.ariaLabel="",this.statusBarEl.setText(this.currentMessage.message),this.lastMessageTimestamp=Date.now()):this.currentMessage?Date.now()-this.lastMessageTimestamp>=this.currentMessage.timeout&&(this.currentMessage=null,this.lastMessageTimestamp=null):this.displayState()}displayState(){switch((this.statusBarEl.getText().length>3||!this.statusBarEl.hasChildNodes())&&(this.statusBarEl.empty(),this.conflictEl=this.statusBarEl.createDiv(),this.conflictEl.setAttribute("data-tooltip-position","top"),this.conflictEl.style.float="left",this.pausedEl=this.statusBarEl.createDiv(),this.pausedEl.setAttribute("data-tooltip-position","top"),this.pausedEl.style.float="left",this.iconEl=this.statusBarEl.createDiv(),this.iconEl.style.float="left",this.textEl=this.statusBarEl.createDiv(),this.textEl.style.float="right",this.textEl.style.marginLeft="5px"),this.plugin.localStorage.getConflict()?((0,$r.setIcon)(this.conflictEl,"alert-circle"),this.conflictEl.ariaLabel="You have merge conflicts. Resolve them and commit afterwards.",this.conflictEl.style.marginRight="5px",this.conflictEl.addClass(this.base+"conflict")):(this.conflictEl.empty(),this.conflictEl.style.marginRight=""),this.plugin.localStorage.getPausedAutomatics()?((0,$r.setIcon)(this.pausedEl,"pause-circle"),this.pausedEl.ariaLabel="Automatic routines are currently paused.",this.pausedEl.style.marginRight="5px",this.pausedEl.addClass(this.base+"paused")):(this.pausedEl.empty(),this.pausedEl.style.marginRight=""),this.plugin.state.gitAction){case 0:this.displayFromNow();break;case 1:this.statusBarEl.ariaLabel="Checking repository status...",(0,$r.setIcon)(this.iconEl,"refresh-cw"),this.statusBarEl.addClass(this.base+"status");break;case 3:this.statusBarEl.ariaLabel="Adding files...",(0,$r.setIcon)(this.iconEl,"archive"),this.statusBarEl.addClass(this.base+"add");break;case 4:this.statusBarEl.ariaLabel="Committing changes...",(0,$r.setIcon)(this.iconEl,"git-commit"),this.statusBarEl.addClass(this.base+"commit");break;case 5:this.statusBarEl.ariaLabel="Pushing changes...",(0,$r.setIcon)(this.iconEl,"upload"),this.statusBarEl.addClass(this.base+"push");break;case 2:this.statusBarEl.ariaLabel="Pulling changes...",(0,$r.setIcon)(this.iconEl,"download"),this.statusBarEl.addClass(this.base+"pull");break;default:this.statusBarEl.ariaLabel="Failed on initialization!",(0,$r.setIcon)(this.iconEl,"alert-triangle"),this.statusBarEl.addClass(this.base+"failed-init");break}}displayFromNow(){var n;let e=this.lastCommitTimestamp,r=this.plugin.state.offlineMode;if(e){let i=(0,$r.moment)(e).fromNow();this.statusBarEl.ariaLabel=`${r?"Offline: ":""}Last Commit: ${i}`,(n=this.unPushedCommits)!=null&&n&&(this.statusBarEl.ariaLabel+=` -(${this.unPushedCommits} unpushed commits)`)}else this.statusBarEl.ariaLabel=r?"Git is offline":"Git is ready";r?(0,$r.setIcon)(this.iconEl,"globe"):(0,$r.setIcon)(this.iconEl,"check"),this.plugin.settings.changedFilesInStatusBar&&this.plugin.cachedStatus&&this.textEl.setText(this.plugin.cachedStatus.changed.length.toString()),this.statusBarEl.addClass(this.base+"idle")}async refreshCommitTimestamp(){this.lastCommitTimestamp=await this.plugin.gitManager.getLastCommitTime(),this.unPushedCommits=await this.plugin.gitManager.getUnpushedCommits()}remove(){this.statusBarEl.remove()}};m();var ph=require("obsidian"),hh=class extends ph.SuggestModal{constructor(r){super(r.app);this.plugin=r;this.resolve=null;this.setPlaceholder("Type your message and select optional the version with the added date.")}openAndGetResult(){return new Promise(r=>{this.resolve=r,this.open()})}onClose(){new Promise(r=>setTimeout(r,10)).then(()=>{this.resolve&&this.resolve(void 0)})}getSuggestions(r){let n=(0,ph.moment)().format(this.plugin.settings.commitDateFormat);return r==""&&(r="..."),[r,`${n}: ${r}`,`${r}: ${n}`]}renderSuggestion(r,n){n.innerText=r}onChooseSuggestion(r,n){this.resolve&&this.resolve(r)}};m();var tM=require("obsidian"),Bc=class{constructor(e){this.plugin=e}saveLastAuto(e,r){r==="backup"?this.plugin.localStorage.setLastAutoBackup(e.toString()):r==="pull"?this.plugin.localStorage.setLastAutoPull(e.toString()):r==="push"&&this.plugin.localStorage.setLastAutoPush(e.toString())}loadLastAuto(){var e,r,n;return{backup:new Date((e=this.plugin.localStorage.getLastAutoBackup())!=null?e:""),pull:new Date((r=this.plugin.localStorage.getLastAutoPull())!=null?r:""),push:new Date((n=this.plugin.localStorage.getLastAutoPush())!=null?n:"")}}async init(){await this.setUpAutoCommitAndSync();let e=this.loadLastAuto();if(this.plugin.settings.differentIntervalCommitAndPush&&this.plugin.settings.autoPushInterval>0){let r=this.diff(this.plugin.settings.autoPushInterval,e.push);this.startAutoPush(r)}if(this.plugin.settings.autoPullInterval>0){let r=this.diff(this.plugin.settings.autoPullInterval,e.pull);this.startAutoPull(r)}}unload(){this.clearAutoPull(),this.clearAutoPush(),this.clearAutoCommitAndSync()}reload(...e){this.plugin.localStorage.getPausedAutomatics()||(e.contains("commit")&&(this.clearAutoCommitAndSync(),this.plugin.settings.autoSaveInterval>0&&this.startAutoCommitAndSync(this.plugin.settings.autoSaveInterval)),e.contains("push")&&(this.clearAutoPush(),this.plugin.settings.differentIntervalCommitAndPush&&this.plugin.settings.autoPushInterval>0&&this.startAutoPush(this.plugin.settings.autoPushInterval)),e.contains("pull")&&(this.clearAutoPull(),this.plugin.settings.autoPullInterval>0&&this.startAutoPull(this.plugin.settings.autoPullInterval)))}async setUpAutoCommitAndSync(){if(this.plugin.settings.setLastSaveToLastCommit){this.clearAutoCommitAndSync();let e=await this.plugin.gitManager.getLastCommitTime();e&&this.saveLastAuto(e,"backup")}if(!this.timeoutIDCommitAndSync&&!this.plugin.autoCommitDebouncer){let e=this.loadLastAuto();if(this.plugin.settings.autoSaveInterval>0){let r=this.diff(this.plugin.settings.autoSaveInterval,e.backup);this.startAutoCommitAndSync(r)}}}startAutoCommitAndSync(e){let r=(e!=null?e:this.plugin.settings.autoSaveInterval)*6e4;this.plugin.settings.autoBackupAfterFileChange?e===0?this.doAutoCommitAndSync():this.plugin.autoCommitDebouncer=(0,tM.debounce)(()=>this.doAutoCommitAndSync(),r,!0):(r>2147483647&&(r=2147483647),this.timeoutIDCommitAndSync=window.setTimeout(()=>this.doAutoCommitAndSync(),r))}doAutoCommitAndSync(){this.plugin.promiseQueue.addTask(async()=>{if(this.plugin.settings.setLastSaveToLastCommit){let r=await this.plugin.gitManager.getLastCommitTime();if(r){this.saveLastAuto(r,"backup");let n=this.diff(this.plugin.settings.autoSaveInterval,r);if(n>0)return this.startAutoCommitAndSync(n),!1}}let e=this.plugin.settings.autoCommitOnlyStaged;return this.plugin.settings.differentIntervalCommitAndPush?await this.plugin.commit({fromAuto:!0,onlyStaged:e}):await this.plugin.commitAndSync({fromAutoBackup:!0,onlyStaged:e}),!0},e=>{e!==!1&&(this.saveLastAuto(new Date,"backup"),this.startAutoCommitAndSync())})}startAutoPull(e){let r=(e!=null?e:this.plugin.settings.autoPullInterval)*6e4;r>2147483647&&(r=2147483647),this.timeoutIDPull=window.setTimeout(()=>this.doAutoPull(),r)}doAutoPull(){this.plugin.promiseQueue.addTask(()=>this.plugin.pullChangesFromRemote(),()=>{this.saveLastAuto(new Date,"pull"),this.startAutoPull()})}startAutoPush(e){let r=(e!=null?e:this.plugin.settings.autoPushInterval)*6e4;r>2147483647&&(r=2147483647),this.timeoutIDPush=window.setTimeout(()=>this.doAutoPush(),r)}doAutoPush(){this.plugin.promiseQueue.addTask(()=>this.plugin.push(),()=>{this.saveLastAuto(new Date,"push"),this.startAutoPush()})}clearAutoCommitAndSync(){var r;let e=!1;return this.timeoutIDCommitAndSync&&(window.clearTimeout(this.timeoutIDCommitAndSync),this.timeoutIDCommitAndSync=void 0,e=!0),this.plugin.autoCommitDebouncer&&((r=this.plugin.autoCommitDebouncer)==null||r.cancel(),this.plugin.autoCommitDebouncer=void 0,e=!0),e}clearAutoPull(){return this.timeoutIDPull?(window.clearTimeout(this.timeoutIDPull),this.timeoutIDPull=void 0,!0):!1}clearAutoPush(){return this.timeoutIDPush?(window.clearTimeout(this.timeoutIDPush),this.timeoutIDPush=void 0,!0):!1}diff(e,r){let i=e-Math.round((new Date().getTime()-r.getTime())/1e3/60);return Math.max(0,i)}};m();var bn=require("obsidian");m();var Hc=require("obsidian");async function rM(t,e,r){let n=await iM(e,r);if(n.result==="failure"){new Hc.Notice(n.reason);return}let{isGitHub:i,branch:a,repo:s,user:o,filePath:l}=n;if(i){let c=t.getCursor("from").line+1,u=t.getCursor("to").line+1;c===u?window.open(`https://github.com/${o}/${s}/blob/${a}/${l}?plain=1#L${c}`):window.open(`https://github.com/${o}/${s}/blob/${a}/${l}?plain=1#L${c}-L${u}`)}else new Hc.Notice("It seems like you are not using GitHub")}async function nM(t,e){let r=await iM(t,e);if(r.result==="failure"){new Hc.Notice(r.reason);return}let{isGitHub:n,branch:i,repo:a,user:s,filePath:o}=r;n?window.open(`https://github.com/${s}/${a}/commits/${i}/${o}`):new Hc.Notice("It seems like you are not using GitHub")}async function iM(t,e){let r=await e.branchInfo(),n=r.tracking,i=r.current,a,s=e.getRelativeRepoPath(t.path);if(e instanceof Ce){let l=await e.getSubmoduleOfFile(e.getRelativeRepoPath(t.path));if(l){s=l.relativeFilepath;let c=await e.git.cwd({path:l.submodule,root:!1}).status();if(n=c.tracking||void 0,i=c.current||void 0,n){let u=n.substring(0,n.indexOf("/")),f=await e.git.cwd({path:l.submodule,root:!1}).getConfig(`remote.${u}.url`,"local");if(f.value!=null)a=f.value;else return{result:"failure",reason:"Failed to get remote url of submodule"}}}}if(n==null)return{result:"failure",reason:"Remote branch is not configured"};if(i==null)return{result:"failure",reason:"Failed to get current branch name"};if(a==null){let l=n.substring(0,n.indexOf("/"));if(a=await e.getConfig(`remote.${l}.url`),a==null)return{result:"failure",reason:"Failed to get remote url"}}let o=a.match(/(?:^https:\/\/github\.com\/(.+)\/(.+?)(?:\.git)?$)|(?:^[a-zA-Z]+@github\.com:(.+)\/(.+?)(?:\.git)?$)/);if(o==null)return{result:"failure",reason:"Could not parse remote url"};{let[l,c,u,f,d]=o;return{result:"success",isGitHub:!!l,repo:u||d,user:c||f,branch:i,filePath:s}}}m();var aM=require("obsidian"),mh=class extends aM.FuzzySuggestModal{constructor(e,r){super(e.app),this.plugin=e,this.changedFiles=r,this.setPlaceholder("Not supported files will be opened by default app!")}getItems(){return this.changedFiles}getItemText(e){if(e.index=="U"&&e.workingDir=="U")return`Untracked | ${e.vaultPath}`;let r="",n="";return e.workingDir!=" "&&(r=`Working Dir: ${e.workingDir} `),e.index!=" "&&(n=`Index: ${e.index}`),`${r}${n} | ${e.vaultPath}`}onChooseItem(e,r){this.plugin.app.metadataCache.getFirstLinkpathDest(e.vaultPath,"")==null?this.app.openWithDefaultApp(e.vaultPath):this.plugin.app.workspace.openLinkText(e.vaultPath,"/")}};m();var sM=require("obsidian"),gh=class extends sM.Modal{constructor(r,n){super(r);this.content=n;this.resolve=null}openAndGetReslt(){return new Promise(r=>{this.resolve=r,this.open()})}onOpen(){let{contentEl:r,titleEl:n}=this;n.setText("Edit .gitignore");let i=r.createDiv(),a=i.createEl("textarea",{text:this.content,cls:["obsidian-git-textarea"],attr:{rows:10,cols:30,wrap:"off"}});i.createEl("button",{cls:["mod-cta","obsidian-git-center-button"],text:"Save"}).addEventListener("click",()=>{this.resolve(a.value),this.close()})}onClose(){let{contentEl:r}=this;r.empty(),this.resolve&&this.resolve(void 0)}};function oM(t){let e=t.app;t.addCommand({id:"edit-gitignore",name:"Edit .gitignore",callback:async()=>{let r=t.gitManager.getRelativeVaultPath(".gitignore");await e.vault.adapter.exists(r)||await e.vault.adapter.write(r,"");let n=await e.vault.adapter.read(r),a=await new gh(e,n).openAndGetReslt();a!==void 0&&(await e.vault.adapter.write(r,a),await t.refresh())}}),t.addCommand({id:"open-git-view",name:"Open source control view",callback:async()=>{var i;let r=e.workspace.getLeavesOfType($t.type),n;r.length===0?(n=(i=e.workspace.getRightLeaf(!1))!=null?i:e.workspace.getLeaf(),await n.setViewState({type:$t.type})):n=r.first(),await e.workspace.revealLeaf(n),e.workspace.trigger("obsidian-git:refresh")}}),t.addCommand({id:"open-history-view",name:"Open history view",callback:async()=>{var i;let r=e.workspace.getLeavesOfType(Zr.type),n;r.length===0?(n=(i=e.workspace.getRightLeaf(!1))!=null?i:e.workspace.getLeaf(),await n.setViewState({type:Zr.type})):n=r.first(),await e.workspace.revealLeaf(n),e.workspace.trigger("obsidian-git:refresh")}}),t.addCommand({id:"open-diff-view",name:"Open diff view",checkCallback:r=>{let n=e.workspace.getActiveFile();if(r)return n!==null;{let i=t.gitManager.getRelativeRepoPath(n.path,!0);t.tools.openDiff({aFile:i,aRef:""})}}}),t.addCommand({id:"view-file-on-github",name:"Open file on GitHub",editorCallback:(r,{file:n})=>{if(n)return rM(r,n,t.gitManager)}}),t.addCommand({id:"view-history-on-github",name:"Open file history on GitHub",editorCallback:(r,{file:n})=>{if(n)return nM(n,t.gitManager)}}),t.addCommand({id:"pull",name:"Pull",callback:()=>t.promiseQueue.addTask(()=>t.pullChangesFromRemote())}),t.addCommand({id:"fetch",name:"Fetch",callback:()=>t.promiseQueue.addTask(()=>t.fetch())}),t.addCommand({id:"switch-to-remote-branch",name:"Switch to remote branch",callback:()=>t.promiseQueue.addTask(()=>t.switchRemoteBranch())}),t.addCommand({id:"add-to-gitignore",name:"Add file to .gitignore",checkCallback:r=>{let n=e.workspace.getActiveFile();if(r)return n!==null;t.addFileToGitignore(n.path,n instanceof bn.TFolder).catch(i=>t.displayError(i))}}),t.addCommand({id:"push",name:"Commit-and-sync",callback:()=>t.promiseQueue.addTask(()=>t.commitAndSync({fromAutoBackup:!1}))}),t.addCommand({id:"backup-and-close",name:"Commit-and-sync and then close Obsidian",callback:()=>t.promiseQueue.addTask(async()=>{await t.commitAndSync({fromAutoBackup:!1}),window.close()})}),t.addCommand({id:"commit-push-specified-message",name:"Commit-and-sync with specific message",callback:()=>t.promiseQueue.addTask(()=>t.commitAndSync({fromAutoBackup:!1,requestCustomMessage:!0}))}),t.addCommand({id:"commit",name:"Commit all changes",callback:()=>t.promiseQueue.addTask(()=>t.commit({fromAuto:!1}))}),t.addCommand({id:"commit-specified-message",name:"Commit all changes with specific message",callback:()=>t.promiseQueue.addTask(()=>t.commit({fromAuto:!1,requestCustomMessage:!0}))}),t.addCommand({id:"commit-smart",name:"Commit",callback:()=>t.promiseQueue.addTask(async()=>{let n=(await t.updateCachedStatus()).staged.length>0;return t.commit({fromAuto:!1,requestCustomMessage:!1,onlyStaged:n})})}),t.addCommand({id:"commit-staged",name:"Commit staged",checkCallback:function(r){if(r)return!1;t.promiseQueue.addTask(async()=>t.commit({fromAuto:!1,requestCustomMessage:!1}))}}),bn.Platform.isDesktopApp&&t.addCommand({id:"commit-amend-staged-specified-message",name:"Amend staged",callback:()=>t.promiseQueue.addTask(()=>t.commit({fromAuto:!1,requestCustomMessage:!0,onlyStaged:!0,amend:!0}))}),t.addCommand({id:"commit-smart-specified-message",name:"Commit with specific message",callback:()=>t.promiseQueue.addTask(async()=>{let n=(await t.updateCachedStatus()).staged.length>0;return t.commit({fromAuto:!1,requestCustomMessage:!0,onlyStaged:n})})}),t.addCommand({id:"commit-staged-specified-message",name:"Commit staged with specific message",checkCallback:function(r){return r?!1:t.promiseQueue.addTask(()=>t.commit({fromAuto:!1,requestCustomMessage:!0,onlyStaged:!0}))}}),t.addCommand({id:"push2",name:"Push",callback:()=>t.promiseQueue.addTask(()=>t.push())}),t.addCommand({id:"stage-current-file",name:"Stage current file",checkCallback:r=>{let n=e.workspace.getActiveFile();if(r)return n!==null;t.promiseQueue.addTask(()=>t.stageFile(n))}}),t.addCommand({id:"unstage-current-file",name:"Unstage current file",checkCallback:r=>{let n=e.workspace.getActiveFile();if(r)return n!==null;t.promiseQueue.addTask(()=>t.unstageFile(n))}}),t.addCommand({id:"edit-remotes",name:"Edit remotes",callback:()=>t.editRemotes().catch(r=>t.displayError(r))}),t.addCommand({id:"remove-remote",name:"Remove remote",callback:()=>t.removeRemote().catch(r=>t.displayError(r))}),t.addCommand({id:"set-upstream-branch",name:"Set upstream branch",callback:()=>t.setUpstreamBranch().catch(r=>t.displayError(r))}),t.addCommand({id:"delete-repo",name:"CAUTION: Delete 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Reloading plugin..."),t.unloadPlugin(),await t.init({fromReload:!0})):new bn.Notice("No repository found")}}),t.addCommand({id:"init-repo",name:"Initialize a new repo",callback:()=>t.createNewRepo().catch(r=>t.displayError(r))}),t.addCommand({id:"clone-repo",name:"Clone an existing remote repo",callback:()=>t.cloneNewRepo().catch(r=>t.displayError(r))}),t.addCommand({id:"list-changed-files",name:"List changed files",callback:async()=>{if(await t.isAllInitialized())try{let r=await t.updateCachedStatus();if(r.changed.length+r.staged.length>500){t.displayError("Too many changes to display");return}new mh(t,r.all).open()}catch(r){t.displayError(r)}}}),t.addCommand({id:"switch-branch",name:"Switch branch",callback:()=>{t.switchBranch().catch(r=>t.displayError(r))}}),t.addCommand({id:"create-branch",name:"Create new branch",callback:()=>{t.createBranch().catch(r=>t.displayError(r))}}),t.addCommand({id:"delete-branch",name:"Delete 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s=r?t.fromA:t.fromB,o=r?t.toA:t.toB,l=0;if(s!=o){i.add(s,s,LM),i.add(s,o,r?eG:QU),a&&a.add(s,s,jM);for(let c=e.iterRange(s,o-1),u=s;!c.next().done;){if(c.lineBreak){u++,i.add(u,u,LM),a&&a.add(u,u,jM);continue}let f=u+c.value.length;if(n)for(;l=u)break;(s?f.toA:f.toB)>c&&(!a||!a(t.state,f,o,l))&&tG(f,t.state.doc,s,n,o,l)}return{deco:o.finish(),gutter:l&&l.finish()}}var Yo=class extends Ie.WidgetType{constructor(e){super(),this.height=e}eq(e){return this.height==e.height}toDOM(){let e=document.createElement("div");return e.className="cm-mergeSpacer",e.style.height=this.height+"px",e}updateDOM(e){return e.style.height=this.height+"px",!0}get estimatedHeight(){return this.height}ignoreEvent(){return!1}},Ph=ke.StateEffect.define({map:(t,e)=>t.map(e)}),Kc=ke.StateField.define({create:()=>Ie.Decoration.none,update:(t,e)=>{for(let r of e.effects)if(r.is(Ph))return r.value;return t.map(e.changes)},provide:t=>Ie.EditorView.decorations.from(t)}),kh=.01;function 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this.lines==e.lines}toDOM(e){let r=document.createElement("div");return r.className="cm-collapsedLines",r.textContent=e.state.phrase("$ unchanged lines",this.lines),r.addEventListener("click",n=>{let i=e.posAtDOM(n.target);e.dispatch({effects:i1.of(i)});let{side:a,sibling:s}=e.state.facet(Jn);s&&s().dispatch({effects:i1.of(nG(i,e.state.field(Zo),a=="a"))})}),r}ignoreEvent(e){return e instanceof MouseEvent}get estimatedHeight(){return 27}get type(){return"collapsed-unchanged-code"}};function nG(t,e,r){let n=0,i=0;for(let a=0;;a++){let s=a=t)return i+(t-n);[n,i]=r?[s.toA,s.toB]:[s.toB,s.toA]}}var iG=ke.StateField.define({create(t){return Ie.Decoration.none},update(t,e){t=t.map(e.changes);for(let r of e.effects)r.is(i1)&&(t=t.update({filter:n=>n!=r.value}));return t},provide:t=>Ie.EditorView.decorations.from(t)});function HM({margin:t=3,minSize:e=4}){return iG.init(r=>aG(r,t,e))}function aG(t,e,r){let n=new ke.RangeSetBuilder,i=t.facet(Jn).side=="a",a=t.field(Zo),s=1;for(let o=0;;o++){let l=o=r&&n.add(t.doc.line(c).from,t.doc.line(u).to,Ie.Decoration.replace({widget:new a1(f),block:!0})),!l)break;s=t.doc.lineAt(Math.min(t.doc.length,i?l.toA:l.toB)).number}return n.finish()}var sG=Ie.EditorView.styleModule.of(new Sh({".cm-mergeView":{overflowY:"auto"},".cm-mergeViewEditors":{display:"flex",alignItems:"stretch"},".cm-mergeViewEditor":{flexGrow:1,flexBasis:0,overflow:"hidden"},".cm-merge-revert":{width:"1.6em",flexGrow:0,flexShrink:0,position:"relative"},".cm-merge-revert button":{position:"absolute",display:"block",width:"100%",boxSizing:"border-box",textAlign:"center",background:"none",border:"none",font:"inherit",cursor:"pointer"}})),oG=Ie.EditorView.baseTheme({".cm-mergeView & .cm-scroller, .cm-mergeView &":{height:"auto !important",overflowY:"visible !important"},"&.cm-merge-a .cm-changedLine, .cm-deletedChunk":{backgroundColor:"rgba(160, 128, 100, .08)"},"&.cm-merge-b .cm-changedLine, .cm-inlineChangedLine":{backgroundColor:"rgba(100, 160, 128, .08)"},"&light.cm-merge-a .cm-changedText, &light .cm-deletedChunk .cm-deletedText":{background:"linear-gradient(#ee443366, #ee443366) bottom/100% 2px no-repeat"},"&dark.cm-merge-a .cm-changedText, &dark .cm-deletedChunk .cm-deletedText":{background:"linear-gradient(#ffaa9966, #ffaa9966) bottom/100% 2px no-repeat"},"&light.cm-merge-b .cm-changedText":{background:"linear-gradient(#22bb22aa, #22bb22aa) bottom/100% 2px no-repeat"},"&dark.cm-merge-b .cm-changedText":{background:"linear-gradient(#88ff88aa, #88ff88aa) bottom/100% 2px no-repeat"},"&.cm-merge-b .cm-deletedText":{background:"#ff000033"},".cm-insertedLine, .cm-deletedLine, .cm-deletedLine del":{textDecoration:"none"},".cm-deletedChunk":{paddingLeft:"6px","& .cm-chunkButtons":{position:"absolute",insetInlineEnd:"5px"},"& button":{border:"none",cursor:"pointer",color:"white",margin:"0 2px",borderRadius:"3px","&[name=accept]":{background:"#2a2"},"&[name=reject]":{background:"#d43"}}},".cm-collapsedLines":{padding:"5px 5px 5px 10px",cursor:"pointer","&:before":{content:'"\u299A"',marginInlineEnd:"7px"},"&:after":{content:'"\u299A"',marginInlineStart:"7px"}},"&light .cm-collapsedLines":{color:"#444",background:"linear-gradient(to bottom, transparent 0, #f3f3f3 30%, #f3f3f3 70%, transparent 100%)"},"&dark .cm-collapsedLines":{color:"#ddd",background:"linear-gradient(to bottom, transparent 0, #222 30%, #222 70%, transparent 100%)"},".cm-changeGutter":{width:"3px",paddingLeft:"1px"},"&light.cm-merge-a .cm-changedLineGutter, &light .cm-deletedLineGutter":{background:"#e43"},"&dark.cm-merge-a .cm-changedLineGutter, &dark .cm-deletedLineGutter":{background:"#fa9"},"&light.cm-merge-b .cm-changedLineGutter":{background:"#2b2"},"&dark.cm-merge-b .cm-changedLineGutter":{background:"#8f8"},".cm-inlineChangedLineGutter":{background:"#75d"}}),UM=new ke.Compartment,Th=new ke.Compartment,Rh=class{constructor(e){this.revertDOM=null,this.revertToA=!1,this.revertToLeft=!1,this.measuring=-1,this.diffConf=e.diffConfig||XU;let r=[ke.Prec.low(nO),oG,sG,Kc,Ie.EditorView.updateListener.of(f=>{this.measuring<0&&(f.heightChanged||f.viewportChanged)&&!f.transactions.some(d=>d.effects.some(h=>h.is(Ph)))&&this.measure()})],n=[Jn.of({side:"a",sibling:()=>this.b,highlightChanges:e.highlightChanges!==!1,markGutter:e.gutter!==!1})];e.gutter!==!1&&n.push(Ah);let i=ke.EditorState.create({doc:e.a.doc,selection:e.a.selection,extensions:[e.a.extensions||[],Ie.EditorView.editorAttributes.of({class:"cm-merge-a"}),Th.of(n),r]}),a=[Jn.of({side:"b",sibling:()=>this.a,highlightChanges:e.highlightChanges!==!1,markGutter:e.gutter!==!1})];e.gutter!==!1&&a.push(Ah);let s=ke.EditorState.create({doc:e.b.doc,selection:e.b.selection,extensions:[e.b.extensions||[],Ie.EditorView.editorAttributes.of({class:"cm-merge-b"}),Th.of(a),r]});this.chunks=Xo.build(i.doc,s.doc,this.diffConf);let o=[Zo.init(()=>this.chunks),UM.of(e.collapseUnchanged?HM(e.collapseUnchanged):[])];i=i.update({effects:ke.StateEffect.appendConfig.of(o)}).state,s=s.update({effects:ke.StateEffect.appendConfig.of(o)}).state,this.dom=document.createElement("div"),this.dom.className="cm-mergeView",this.editorDOM=this.dom.appendChild(document.createElement("div")),this.editorDOM.className="cm-mergeViewEditors";let l=e.orientation||"a-b",c=document.createElement("div");c.className="cm-mergeViewEditor";let u=document.createElement("div");u.className="cm-mergeViewEditor",this.editorDOM.appendChild(l=="a-b"?c:u),this.editorDOM.appendChild(l=="a-b"?u:c),this.a=new Ie.EditorView({state:i,parent:c,root:e.root,dispatchTransactions:f=>this.dispatch(f,this.a)}),this.b=new Ie.EditorView({state:s,parent:u,root:e.root,dispatchTransactions:f=>this.dispatch(f,this.b)}),this.setupRevertControls(!!e.revertControls,e.revertControls=="b-to-a",e.renderRevertControl),e.parent&&e.parent.appendChild(this.dom),this.scheduleMeasure()}dispatch(e,r){if(e.some(n=>n.docChanged)){let n=e[e.length-1],i=e.reduce((s,o)=>s.compose(o.changes),ke.ChangeSet.empty(e[0].startState.doc.length));this.chunks=r==this.a?Xo.updateA(this.chunks,n.newDoc,this.b.state.doc,i,this.diffConf):Xo.updateB(this.chunks,this.a.state.doc,n.newDoc,i,this.diffConf),r.update([...e,n.state.update({effects:n1.of(this.chunks)})]);let a=r==this.a?this.b:this.a;a.update([a.state.update({effects:n1.of(this.chunks)})]),this.scheduleMeasure()}else r.update(e)}reconfigure(e){if("diffConfig"in e&&(this.diffConf=e.diffConfig),"orientation"in e){let a=e.orientation!="b-a";if(a!=(this.editorDOM.firstChild==this.a.dom.parentNode)){let s=this.a.dom.parentNode,o=this.b.dom.parentNode;s.remove(),o.remove(),this.editorDOM.insertBefore(a?s:o,this.editorDOM.firstChild),this.editorDOM.appendChild(a?o:s),this.revertToLeft=!this.revertToLeft,this.revertDOM&&(this.revertDOM.textContent="")}}if("revertControls"in e||"renderRevertControl"in e){let a=!!this.revertDOM,s=this.revertToA,o=this.renderRevert;"revertControls"in e&&(a=!!e.revertControls,s=e.revertControls=="b-to-a"),"renderRevertControl"in e&&(o=e.renderRevertControl),this.setupRevertControls(a,s,o)}let r="highlightChanges"in e,n="gutter"in e,i="collapseUnchanged"in e;if(r||n||i){let a=[],s=[];if(r||n){let o=this.a.state.facet(Jn),l=n?e.gutter!==!1:o.markGutter,c=r?e.highlightChanges!==!1:o.highlightChanges;a.push(Th.reconfigure([Jn.of({side:"a",sibling:()=>this.b,highlightChanges:c,markGutter:l}),l?Ah:[]])),s.push(Th.reconfigure([Jn.of({side:"b",sibling:()=>this.a,highlightChanges:c,markGutter:l}),l?Ah:[]]))}if(i){let o=UM.reconfigure(e.collapseUnchanged?HM(e.collapseUnchanged):[]);a.push(o),s.push(o)}this.a.dispatch({effects:a}),this.b.dispatch({effects:s})}this.scheduleMeasure()}setupRevertControls(e,r,n){this.revertToA=r,this.revertToLeft=this.revertToA==(this.editorDOM.firstChild==this.a.dom.parentNode),this.renderRevert=n,!e&&this.revertDOM?(this.revertDOM.remove(),this.revertDOM=null):e&&!this.revertDOM?(this.revertDOM=this.editorDOM.insertBefore(document.createElement("div"),this.editorDOM.firstChild.nextSibling),this.revertDOM.addEventListener("mousedown",i=>this.revertClicked(i)),this.revertDOM.className="cm-merge-revert"):this.revertDOM&&(this.revertDOM.textContent="")}scheduleMeasure(){if(this.measuring<0){let e=this.dom.ownerDocument.defaultView||window;this.measuring=e.requestAnimationFrame(()=>{this.measuring=-1,this.measure()})}}measure(){rG(this.a,this.b,this.chunks),this.revertDOM&&this.updateRevertButtons()}updateRevertButtons(){let e=this.revertDOM,r=e.firstChild,n=this.a.viewport,i=this.b.viewport;for(let a=0;an.to||s.fromB>i.to)break;if(s.fromA-1&&(this.dom.ownerDocument.defaultView||window).cancelAnimationFrame(this.measuring),this.dom.remove()}};function GM(t){let e=t.nextSibling;return t.remove(),e}var Mh=require("@codemirror/search"),Jo=require("@codemirror/state"),tn=require("@codemirror/view");var fs=class extends Qo.ItemView{constructor(r,n){super(r);this.plugin=n;this.refreshing=!1;this.ignoreNextModification=!1;this.navigation=!0,this.registerEvent(this.app.workspace.on("obsidian-git:status-changed",()=>{this.mergeView?this.updateRefEditors().catch(console.error):this.createMergeView().catch(console.error)})),this.intervalRef=window.setInterval(()=>{this.mergeView&&this.updateRefEditors().catch(console.error)},30*1e3),this.registerEvent(this.app.vault.on("modify",i=>{this.state.bRef==null&&i.path===this.state.bFile&&(this.ignoreNextModification?this.ignoreNextModification=!1:this.updateModifiableEditor().catch(console.error))})),this.registerEvent(this.app.vault.on("delete",i=>{this.state.bRef==null&&i.path===this.state.bFile&&this.createMergeView().catch(console.error)})),this.registerEvent(this.app.vault.on("create",i=>{this.state.bRef==null&&i.path===this.state.bFile&&this.createMergeView().catch(console.error)})),this.registerEvent(this.app.vault.on("rename",(i,a)=>{this.state.bRef==null&&(i.path===this.state.bFile||a===this.state.bFile)&&this.createMergeView().catch(console.error)})),this.fileSaveDebouncer=(0,Qo.debounce)(i=>{let a=this.state.bFile;a&&(this.ignoreNextModification=!0,this.plugin.app.vault.adapter.write(this.plugin.gitManager.getRelativeVaultPath(a),i).catch(s=>this.plugin.displayError(s)))},1e3,!1)}getViewType(){return ca.type}getDisplayText(){var r;if(((r=this.state)==null?void 0:r.bFile)!=null){let n=this.state.bFile.split("/").last();return n!=null&&n.endsWith(".md")&&(n=n.slice(0,-3)),`Diff: ${n}`}return ca.name}getIcon(){return ca.icon}async setState(r,n){this.state=r,Qo.Platform.isMobile&&(this.leaf.view.titleEl.textContent=this.getDisplayText()),await super.setState(r,n),await this.createMergeView()}getState(){return this.state}onClose(){return window.clearInterval(this.intervalRef),super.onClose()}async onOpen(){return await this.createMergeView(),super.onOpen()}async gitShow(r,n){try{return await this.plugin.gitManager.show(r,n,!1)}catch(i){if(i instanceof wr&&(i.message.includes("does not exist")||i.message.includes("unknown revision or path")||i.message.includes("exists on disk, but not in")||i.message.includes("fatal: bad object")))return i.message.includes("fatal: bad object")&&this.plugin.displayError(i.message),"";throw i}}async bShouldBeEditable(){if(this.state.bRef!=null)return!1;let r=this.plugin.gitManager.getRelativeVaultPath(this.state.bFile);return await this.app.vault.adapter.exists(r)}async updateModifiableEditor(){if(!this.mergeView||this.refreshing)return;let r=this.mergeView.b;this.refreshing=!0;let n=await this.app.vault.adapter.read(this.state.bFile);if(n!=r.state.doc.toString()){let i=r.state.update({changes:{from:0,to:r.state.doc.length,insert:n},annotations:[Jo.Transaction.remote.of(!0)]});r.dispatch(i)}this.refreshing=!1}async updateRefEditors(){if(!this.mergeView||this.refreshing)return;let r=this.mergeView.a,n=this.mergeView.b;this.refreshing=!0;let i=await this.gitShow(this.state.aRef,this.state.aFile),a;if(this.state.bRef!=null&&(a=await this.gitShow(this.state.bRef,this.state.bFile)),i!=r.state.doc.toString()){let s=r.state.update({changes:{from:0,to:r.state.doc.length,insert:i}});r.dispatch(s)}if(a!=null&&a!=n.state.doc.toString()){let s=n.state.update({changes:{from:0,to:n.state.doc.length,insert:a}});n.dispatch(s)}this.refreshing=!1}async createMergeView(){var r,n,i;if((r=this.state)!=null&&r.aFile&&((n=this.state)!=null&&n.bFile)&&!this.refreshing&&this.plugin.gitManager){this.refreshing=!0,(i=this.mergeView)==null||i.destroy();let a=this.containerEl.children[1];a.empty(),this.contentEl.addClass("git-split-diff-view"),this.bIsEditable=await this.bShouldBeEditable();let s=await this.gitShow(this.state.aRef,this.state.aFile),o;if(this.state.bRef!=null)o=await this.gitShow(this.state.bRef,this.state.bFile);else{let p=this.plugin.gitManager.getRelativeVaultPath(this.state.bFile);await this.app.vault.adapter.exists(p)?o=await this.app.vault.adapter.read(p):o=""}let l=[(0,tn.lineNumbers)(),(0,Mh.highlightSelectionMatches)(),(0,tn.drawSelection)(),tn.keymap.of([...el.standardKeymap,el.indentWithTab]),(0,el.history)(),(0,Mh.search)(),tn.EditorView.lineWrapping],c=this,u=tn.ViewPlugin.define(p=>({update(g){if(g.docChanged&&!g.transactions.some(v=>v.annotation(Jo.Transaction.remote))){let v=p.state.doc.toString();c.fileSaveDebouncer(v)}}})),f={doc:s,extensions:[...l,tn.EditorView.editable.of(!1),Jo.EditorState.readOnly.of(!0)]},d=[...l];this.bIsEditable?d.push(u):d.push(tn.EditorView.editable.of(!1),Jo.EditorState.readOnly.of(!0));let h={doc:o,extensions:d};a.addClasses(["cm-s-obsidian","mod-cm6","markdown-source-view","cm-content"]),this.mergeView=new Rh({b:h,a:f,collapseUnchanged:{minSize:6,margin:4},diffConfig:{scanLimit:this.bIsEditable?1e3:1e4},parent:a}),this.refreshing=!1}}};m();var C5=require("obsidian");m();m();m();var iO="5";var aO,sO,oO;typeof window!="undefined"&&((oO=(sO=(aO=window.__svelte)!=null?aO:window.__svelte={}).v)!=null?oO:sO.v=new Set).add(iO);m();m();m();var tl="[",ya="[!",ds="]",Ci={};var qe=Symbol(),rn=Symbol("filename"),lO=Symbol("hmr"),cO="http://www.w3.org/1999/xhtml";m();m();m();m();var Oh=!0;m();var D=!1;m();var nn=Array.isArray,uO=Array.prototype.indexOf,Zc=Array.from,c1=Object.keys,qt=Object.defineProperty,_r=Object.getOwnPropertyDescriptor,u1=Object.getOwnPropertyDescriptors,f1=Object.prototype,fO=Array.prototype,rl=Object.getPrototypeOf,d1=Object.isExtensible;function h1(t){return typeof t=="function"}var ir=()=>{};function Jc(t){for(var e=0;e{t=n,e=i});return{promise:r,resolve:t,reject:e}}m();m();var xr=Symbol("$state"),Qc=Symbol("legacy props"),hO=Symbol(""),Ih=Symbol("proxy path"),nl=new class extends Error{constructor(){super(...arguments);vt(this,"name","StaleReactionError");vt(this,"message","The reaction that called `getAbortSignal()` was re-run or destroyed")}};var eu=3,an=8;m();m();function pO(){if(D){let t=new Error("await_outside_boundary\nCannot await outside a `` with a `pending` snippet\nhttps://svelte.dev/e/await_outside_boundary");throw t.name="Svelte error",t}else throw new Error("https://svelte.dev/e/await_outside_boundary")}function tu(t){if(D){let e=new Error(`lifecycle_outside_component -\`${t}(...)\` can only be used during component initialisation -https://svelte.dev/e/lifecycle_outside_component`);throw e.name="Svelte error",e}else throw new Error("https://svelte.dev/e/lifecycle_outside_component")}function mO(){if(D){let t=new Error("async_derived_orphan\nCannot create a `$derived(...)` with an `await` expression outside of an effect tree\nhttps://svelte.dev/e/async_derived_orphan");throw t.name="Svelte error",t}else throw new Error("https://svelte.dev/e/async_derived_orphan")}function p1(){if(D){let t=new Error("bind_invalid_checkbox_value\nUsing `bind:value` together with a checkbox input is not allowed. 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fh(this);this.intervalsToClear=[];this.lineAuthoringFeature=new uh(this)}setPluginState(r){var n;this.state=Object.assign(this.state,r),(n=this.statusBar)==null||n.display()}async updateCachedStatus(){var r,n;return this.app.workspace.trigger("obsidian-git:loading-status"),this.cachedStatus=await this.gitManager.status(),this.cachedStatus.conflicted.length>0?(this.localStorage.setConflict(!0),await((r=this.branchBar)==null?void 0:r.display())):(this.localStorage.setConflict(!1),await((n=this.branchBar)==null?void 0:n.display())),this.app.workspace.trigger("obsidian-git:status-changed",this.cachedStatus),this.cachedStatus}async refresh(){if(!this.gitReady)return;let r=this.app.workspace.getLeavesOfType($t.type),n=this.app.workspace.getLeavesOfType(Zr.type);(this.settings.changedFilesInStatusBar||r.some(i=>{var a;return!((a=i.isDeferred)!=null&&a)})||n.some(i=>{var a;return!((a=i.isDeferred)!=null&&a)}))&&await this.updateCachedStatus().catch(i=>this.displayError(i)),this.app.workspace.trigger("obsidian-git:refreshed")}refreshUpdatedHead(){this.lineAuthoringFeature.refreshLineAuthorViews()}async onload(){console.log("loading "+this.manifest.name+" plugin: v"+this.manifest.version),Nc.plugin=this,this.localStorage.migrate(),await this.loadSettings(),await this.migrateSettings(),this.settingsTab=new Kd(this.app,this),this.addSettingTab(this.settingsTab),this.localStorage.getPluginDisabled()||(this.registerStuff(),this.app.workspace.onLayoutReady(()=>this.init({fromReload:!1}).catch(r=>this.displayError(r))))}onExternalSettingsChange(){this.reloadSettings().catch(r=>this.displayError(r))}async reloadSettings(){let r=JSON.stringify(this.settings);await this.loadSettings();let n=JSON.stringify(this.settings);r!==n&&(this.log("Reloading settings"),this.unloadPlugin(),await this.init({fromReload:!0}),this.app.workspace.getLeavesOfType($t.type).forEach(i=>{var a;if(!((a=i.isDeferred)!=null&&a))return 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r;this.debRefresh(),(r=this.autoCommitDebouncer)==null||r.call(this)})),this.registerEvent(this.app.vault.on("create",()=>{var r;this.debRefresh(),(r=this.autoCommitDebouncer)==null||r.call(this)})),this.registerEvent(this.app.vault.on("rename",()=>{var r;this.debRefresh(),(r=this.autoCommitDebouncer)==null||r.call(this)})),this.registerView($t.type,r=>new Pu(r,this)),this.registerView(Zr.type,r=>new Su(r,this)),this.registerView(ua.type,r=>new ls(r,this)),this.registerView(ca.type,r=>new fs(r,this)),this.addRibbonIcon("git-pull-request","Open Git source control",async()=>{var i;let r=this.app.workspace.getLeavesOfType($t.type),n;r.length===0?(n=(i=this.app.workspace.getRightLeaf(!1))!=null?i:this.app.workspace.getLeaf(),await n.setViewState({type:$t.type})):n=r.first(),await this.app.workspace.revealLeaf(n)}),this.registerHoverLinkSource($t.type,{display:"Git View",defaultMod:!0}),this.lineAuthoringFeature.onLoadPlugin(),this.setRefreshDebouncer(),oM(this)}setRefreshDebouncer(){var r;(r=this.debRefresh)==null||r.cancel(),this.debRefresh=(0,we.debounce)(()=>{this.settings.refreshSourceControl&&this.refresh().catch(console.error)},this.settings.refreshSourceControlTimer,!0)}async addFileToGitignore(r,n){let i=this.gitManager.getRelativeRepoPath(r,!0),a=u3({isFolder:n,gitRelativePath:i});return await this.app.vault.adapter.append(this.gitManager.getRelativeVaultPath(".gitignore"),` -`+a),this.refresh()}handleFileMenu(r,n,i,a){if(!this.gitReady||!this.settings.showFileMenu||!n)return;let s;if(typeof n=="string"?s=n:s=n.path,i=="file-explorer-context-menu"&&(r.addItem(o=>{o.setTitle("Git: Stage").setIcon("plus-circle").setSection("action").onClick(l=>{this.promiseQueue.addTask(async()=>{n instanceof we.TFile?await this.stageFile(n):(await this.gitManager.stageAll({dir:this.gitManager.getRelativeRepoPath(s,!0)}),this.app.workspace.trigger("obsidian-git:refresh"))})})}),r.addItem(o=>{o.setTitle("Git: Unstage").setIcon("minus-circle").setSection("action").onClick(l=>{this.promiseQueue.addTask(async()=>{n instanceof we.TFile?await this.unstageFile(n):(await this.gitManager.unstageAll({dir:this.gitManager.getRelativeRepoPath(s,!0)}),this.app.workspace.trigger("obsidian-git:refresh"))})})}),r.addItem(o=>{o.setTitle("Git: Add to .gitignore").setIcon("file-x").setSection("action").onClick(l=>{this.addFileToGitignore(s,n instanceof we.TFolder).catch(c=>this.displayError(c))})})),i=="git-source-control"){r.addItem(l=>{l.setTitle("Git: Add to .gitignore").setIcon("file-x").setSection("action").onClick(c=>{this.addFileToGitignore(s,n instanceof we.TFolder).catch(u=>this.displayError(u))})});let o=this.app.vault.adapter;a==="obsidian-git:menu"&&o instanceof we.FileSystemAdapter&&(r.addItem(l=>{l.setTitle("Open in default app").setIcon("arrow-up-right").setSection("action").onClick(c=>{this.app.openWithDefaultApp(s)})}),r.addItem(l=>{l.setTitle("Show in system explorer").setIcon("arrow-up-right").setSection("action").onClick(c=>{window.electron.shell.showItemInFolder(L5.join(o.getBasePath(),s))})}))}}async migrateSettings(){this.settings.mergeOnPull!=null&&(this.settings.syncMethod=this.settings.mergeOnPull?"merge":"rebase",this.settings.mergeOnPull=void 0,await this.saveSettings()),this.settings.autoCommitMessage===void 0&&(this.settings.autoCommitMessage=this.settings.commitMessage,await this.saveSettings()),this.settings.gitPath!=null&&(this.localStorage.setGitPath(this.settings.gitPath),this.settings.gitPath=void 0,await this.saveSettings()),this.settings.username!=null&&(this.localStorage.setPassword(this.settings.username),this.settings.username=void 0,await this.saveSettings())}unloadPlugin(){var r,n;this.gitReady=!1,this.lineAuthoringFeature.deactivateFeature(),this.automaticsManager.unload(),(r=this.branchBar)==null||r.remove(),(n=this.statusBar)==null||n.remove(),this.statusBar=void 0,this.branchBar=void 0,this.gitManager.unload(),this.promiseQueue.clear();for(let i of this.intervalsToClear)window.clearInterval(i);this.intervalsToClear=[],this.debRefresh.cancel()}onunload(){this.unloadPlugin(),console.log("unloading "+this.manifest.name+" plugin")}async loadSettings(){let r=await this.loadData();r==null&&(r={showedMobileNotice:!0}),this.settings=gT(Xe,r)}async saveSettings(){var r;(r=this.settingsTab)==null||r.beforeSaveSettings(),await this.saveData(this.settings)}get useSimpleGit(){return we.Platform.isDesktopApp}async init({fromReload:r=!1}){var n;if(this.settings.showStatusBar&&!this.statusBar){let i=this.addStatusBarItem();this.statusBar=new dh(i,this),this.intervalsToClear.push(window.setInterval(()=>{var a;return(a=this.statusBar)==null?void 0:a.display()},1e3))}try{this.useSimpleGit?(this.gitManager=new Ce(this),await this.gitManager.setGitInstance()):this.gitManager=new wn(this);let i=await this.gitManager.checkRequirements(),a=this.localStorage.getPausedAutomatics();switch(i){case"missing-git":this.displayError(`Cannot run git command. Trying to run: '${this.localStorage.getGitPath()||"git"}' .`);break;case"missing-repo":new we.Notice("Can't find a valid git repository. Please create one via the given command or clone an existing repo.",1e4);break;case"valid":if(this.gitReady=!0,this.setPluginState({gitAction:0}),we.Platform.isDesktop&&this.settings.showBranchStatusBar&&!this.branchBar){let s=this.addStatusBarItem();this.branchBar=new gp(s,this),this.intervalsToClear.push(window.setInterval(()=>{var o;return void((o=this.branchBar)==null?void 0:o.display().catch(console.error))},6e4))}await((n=this.branchBar)==null?void 0:n.display()),this.lineAuthoringFeature.conditionallyActivateBySettings(),this.app.workspace.trigger("obsidian-git:refresh"),this.app.workspace.trigger("obsidian-git:head-change"),!r&&this.settings.autoPullOnBoot&&!a&&this.promiseQueue.addTask(()=>this.pullChangesFromRemote()),a||await this.automaticsManager.init(),a&&new we.Notice("Automatic routines are currently paused.");break;default:this.log("Something weird happened. The 'checkRequirements' result is "+i)}}catch(i){this.displayError(i),console.error(i)}}async createNewRepo(){try{await this.gitManager.init(),new we.Notice("Initialized new repo"),await this.init({fromReload:!0})}catch(r){this.displayError(r)}}async cloneNewRepo(){let n=await new ze(this,{placeholder:"Enter remote URL"}).openAndGetResult();if(n){let i="Vault Root",a=await new ze(this,{options:this.gitManager instanceof wn?[i]:[],placeholder:"Enter directory for clone. It needs to be empty or not existent.",allowEmpty:this.gitManager instanceof wn}).openAndGetResult();if(a==null)return;if(a===i&&(a="."),a=(0,we.normalizePath)(a),a==="/"&&(a="."),a==="."){let f=await new ze(this,{options:["NO","YES"],placeholder:`Does your remote repo contain a ${this.app.vault.configDir} directory at the root?`,onlySelection:!0}).openAndGetResult();if(f===void 0){new we.Notice("Aborted clone");return}else if(f==="YES"){let d="DELETE ALL YOUR LOCAL CONFIG AND PLUGINS";if(await new ze(this,{options:["Abort clone",d],placeholder:`To avoid conflicts, the local ${this.app.vault.configDir} directory needs to be deleted.`,onlySelection:!0}).openAndGetResult()===d)await this.app.vault.adapter.rmdir(this.app.vault.configDir,!0);else{new we.Notice("Aborted clone");return}}}let s=await new ze(this,{placeholder:"Specify depth of clone. Leave empty for full clone.",allowEmpty:!0}).openAndGetResult(),o;if(s===void 0){new we.Notice("Aborted clone");return}if(s!==""&&(o=parseInt(s),isNaN(o))){new we.Notice("Invalid depth. Aborting clone.");return}new we.Notice(`Cloning new repo into "${a}"`);let l=this.settings.basePath,c=a&&a!==".";c&&(this.settings.basePath=a);try{await this.gitManager.clone(vv(n),a,o),new we.Notice("Cloned new repo."),new we.Notice("Please restart Obsidian"),c&&await this.saveSettings()}catch(u){this.displayError(u),this.settings.basePath=l,await this.saveSettings()}}}async isAllInitialized(){return this.gitReady||await this.init({fromReload:!0}),this.gitReady}async pullChangesFromRemote(){if(!await this.isAllInitialized())return;let r=await this.pull();if(r!==!1){if(r||this.displayMessage("Pull: Everything is up-to-date"),this.gitManager instanceof Ce){let n=await this.updateCachedStatus();n.conflicted.length>0&&(this.displayError(`You have conflicts in ${n.conflicted.length} ${n.conflicted.length==1?"file":"files"}`),await this.handleConflict(n.conflicted))}this.app.workspace.trigger("obsidian-git:refresh"),this.setPluginState({gitAction:0})}}async commitAndSync({fromAutoBackup:r,requestCustomMessage:n=!1,commitMessage:i,onlyStaged:a=!1}){!await this.isAllInitialized()||(this.settings.syncMethod=="reset"&&this.settings.pullBeforePush&&await this.pull(),!await this.commit({fromAuto:r,requestCustomMessage:n,commitMessage:i,onlyStaged:a}))||(this.settings.syncMethod!="reset"&&this.settings.pullBeforePush&&await this.pull(),this.settings.disablePush||(await this.remotesAreSet()&&await this.gitManager.canPush()?await this.push():this.displayMessage("No commits to push")),this.setPluginState({gitAction:0}))}async commit({fromAuto:r,requestCustomMessage:n=!1,onlyStaged:i=!1,commitMessage:a,amend:s=!1}){if(!await this.isAllInitialized())return!1;try{let o=this.localStorage.getConflict(),l,c=[],u=[];if(this.gitManager instanceof Ce){if(await this.mayDeleteConflictFile(),l=await this.updateCachedStatus(),l.conflicted.length==0&&(o=!1),r&&l.conflicted.length>0)return this.displayError(`Did not commit, because you have conflicts in ${l.conflicted.length} ${l.conflicted.length==1?"file":"files"}. Please resolve them and commit per command.`),await this.handleConflict(l.conflicted),!1;c=l.staged,u=l.changed}else{if(r&&o)return this.displayError("Did not commit, because you have conflicts. Please resolve them and commit per command."),!1;{o&&await this.mayDeleteConflictFile();let f=this.gitManager;i?c=await f.getStagedFiles():u=(await f.getUnstagedFiles()).map(({path:h,type:p})=>({vaultPath:this.gitManager.getRelativeVaultPath(h),path:h,type:p}))}}if(await this.tools.hasTooBigFiles(i?c:[...c,...u]))return this.setPluginState({gitAction:0}),!1;if(u.length+c.length!==0||o){let f=a!=null?a:a=r?this.settings.autoCommitMessage:this.settings.commitMessage;if(r&&this.settings.customMessageOnAutoBackup||n){!this.settings.disablePopups&&r&&new we.Notice("Auto backup: Please enter a custom commit message. Leave empty to abort");let p=await new hh(this).openAndGetResult();if(p!=null&&p!=""&&p!="...")f=p;else return this.setPluginState({gitAction:0}),!1}else if(this.gitManager instanceof Ce&&this.settings.commitMessageScript){let p=this.settings.commitMessageScript,g=this.localStorage.getHostname()||"",v=p.replace("{{hostname}}",g);v=v.replace("{{date}}",(0,we.moment)().format(this.settings.commitDateFormat));let y=await $c("sh",["-c",v],{cwd:this.gitManager.absoluteRepoPath});y.code!=0?this.displayError(y.stderr):y.stdout.trim().length==0?this.displayMessage("Stdout from commit message script is empty. Using default message."):f=y.stdout}let d;i?d=await this.gitManager.commit({message:f,amend:s}):d=await this.gitManager.commitAll({message:f,status:l,unstagedFiles:u,amend:s}),this.gitManager instanceof Ce&&await this.updateCachedStatus();let h=!1;d===void 0&&(h=!0,d=u.length+c.length||0),this.displayMessage(`Committed${h?" approx.":""} ${d} ${d==1?"file":"files"}`)}else this.displayMessage("No changes to commit");return this.app.workspace.trigger("obsidian-git:refresh"),!0}catch(o){return this.displayError(o),!1}}async push(){if(!await this.isAllInitialized()||!await this.remotesAreSet())return!1;let r=this.localStorage.getConflict();try{this.gitManager instanceof Ce&&await this.mayDeleteConflictFile();let n;if(this.gitManager instanceof Ce&&(n=await this.updateCachedStatus()).conflicted.length>0)return this.displayError(`Cannot push. You have conflicts in ${n.conflicted.length} ${n.conflicted.length==1?"file":"files"}`),await this.handleConflict(n.conflicted),!1;if(this.gitManager instanceof wn&&r)return this.displayError("Cannot push. You have conflicts"),!1;this.log("Pushing....");let i=await this.gitManager.push();return i!==void 0&&(i>0?this.displayMessage(`Pushed ${i} ${i==1?"file":"files"} to remote`):this.displayMessage("No commits to push")),this.setPluginState({offlineMode:!1}),this.app.workspace.trigger("obsidian-git:refresh"),!0}catch(n){return n instanceof Co?this.handleNoNetworkError(n):this.displayError(n),!1}}async pull(){if(!await this.remotesAreSet())return!1;try{this.log("Pulling....");let r=await this.gitManager.pull()||[];return this.setPluginState({offlineMode:!1}),r.length>0&&(this.displayMessage(`Pulled ${r.length} ${r.length==1?"file":"files"} from remote`),this.lastPulledFiles=r),r.length}catch(r){return this.displayError(r),!1}}async fetch(){if(await this.remotesAreSet())try{await this.gitManager.fetch(),this.displayMessage("Fetched from remote"),this.setPluginState({offlineMode:!1}),this.app.workspace.trigger("obsidian-git:refresh")}catch(r){this.displayError(r)}}async mayDeleteConflictFile(){let r=this.app.vault.getAbstractFileByPath(To);r&&(this.app.workspace.iterateAllLeaves(n=>{var i;n.view instanceof we.MarkdownView&&((i=n.view.file)==null?void 0:i.path)==r.path&&n.detach()}),await this.app.vault.delete(r))}async stageFile(r){return await this.isAllInitialized()?(await this.gitManager.stage(r.path,!0),this.app.workspace.trigger("obsidian-git:refresh"),this.setPluginState({gitAction:0}),!0):!1}async unstageFile(r){return await this.isAllInitialized()?(await this.gitManager.unstage(r.path,!0),this.app.workspace.trigger("obsidian-git:refresh"),this.setPluginState({gitAction:0}),!0):!1}async switchBranch(){var i;if(!await this.isAllInitialized())return;let r=await this.gitManager.branchInfo(),n=await new pp(this,r.branches).openAndGetReslt();if(n!=null)return await this.gitManager.checkout(n),this.displayMessage(`Switched to ${n}`),this.app.workspace.trigger("obsidian-git:refresh"),await((i=this.branchBar)==null?void 0:i.display()),n}async switchRemoteBranch(){var a;if(!await this.isAllInitialized())return;let r=await this.selectRemoteBranch()||"",[n,i]=ki(r);if(i!=null&&n!=null)return await this.gitManager.checkout(i,n),this.displayMessage(`Switched to ${r}`),await((a=this.branchBar)==null?void 0:a.display()),r}async createBranch(){var n;if(!await this.isAllInitialized())return;let r=await new ze(this,{placeholder:"Create new branch"}).openAndGetResult();if(r!=null)return await this.gitManager.createBranch(r),this.displayMessage(`Created new branch ${r}`),await((n=this.branchBar)==null?void 0:n.display()),r}async deleteBranch(){var i;if(!await this.isAllInitialized())return;let r=await this.gitManager.branchInfo();r.current&&r.branches.remove(r.current);let n=await new ze(this,{options:r.branches,placeholder:"Delete branch",onlySelection:!0}).openAndGetResult();if(n!=null){let a=!1;if(!await this.gitManager.branchIsMerged(n)){let o=await new ze(this,{options:["YES","NO"],placeholder:"This branch isn't merged into HEAD. Force delete?",onlySelection:!0}).openAndGetResult();if(o!=="YES")return;a=o==="YES"}return await this.gitManager.deleteBranch(n,a),this.displayMessage(`Deleted branch ${n}`),await((i=this.branchBar)==null?void 0:i.display()),n}}async remotesAreSet(){return this.settings.updateSubmodules||(await this.gitManager.branchInfo()).tracking?!0:(new we.Notice("No upstream branch is set. Please select one."),await this.setUpstreamBranch())}async setUpstreamBranch(){let r=await this.selectRemoteBranch();return r==null?(this.displayError("Aborted. No upstream-branch is set!",1e4),this.setPluginState({gitAction:0}),!1):(await this.gitManager.updateUpstreamBranch(r),this.displayMessage(`Set upstream branch to ${r}`),this.setPluginState({gitAction:0}),!0)}async discardAll(r){if(!await this.isAllInitialized())return!1;let n=await this.gitManager.status({path:r}),i=0,a=0;for(let o of n.changed)o.workingDir=="U"?i++:a++;if(i+a==0)return!1;let s=await new gl({app:this.app,filesToDeleteCount:i,filesToDiscardCount:a,path:r!=null?r:""}).openAndGetResult();switch(s){case!1:return s;case"discard":await this.gitManager.discardAll({dir:r,status:this.cachedStatus});break;case"delete":{await this.gitManager.discardAll({dir:r,status:this.cachedStatus});let o=await this.gitManager.getUntrackedPaths({path:r,status:this.cachedStatus});for(let l of o){let c=this.gitManager.getRelativeVaultPath(l),u=this.app.vault.getAbstractFileByPath(c);u?await this.app.fileManager.trashFile(u):l.endsWith("/")?await this.app.vault.adapter.rmdir(c,!0):await this.app.vault.adapter.remove(c)}break}default:Ud(s)}return this.app.workspace.trigger("obsidian-git:refresh"),s}async handleConflict(r){this.localStorage.setConflict(!0);let n;r!==void 0&&(n=["# Conflicts","Please resolve them and commit them using the commands `Git: Commit all changes` followed by `Git: Push`","(This file will automatically be deleted before commit)","[[#Additional Instructions]] available below file list","",...r.map(i=>{let a=this.app.vault.getAbstractFileByPath(i);return a instanceof we.TFile?`- [[${this.app.metadataCache.fileToLinktext(a,"/")}]]`:`- Not a file: ${i}`}),` -# Additional Instructions -I strongly recommend to use "Source mode" for viewing the conflicted files. For simple conflicts, in each file listed above replace every occurrence of the following text blocks with the desired text. - -\`\`\`diff -<<<<<<< HEAD - File changes in local repository -======= - File changes in remote repository ->>>>>>> origin/main -\`\`\``]),await this.tools.writeAndOpenFile(n==null?void 0:n.join(` -`))}async editRemotes(){if(!await this.isAllInitialized())return;let r=await this.gitManager.getRemotes(),i=await new ze(this,{options:r,placeholder:"Select or create a new remote by typing its name and selecting it"}).openAndGetResult();if(i){let a=await this.gitManager.getRemoteUrl(i),o=await new ze(this,{initialValue:a,placeholder:"Enter remote URL"}).openAndGetResult();if(o)return await this.gitManager.setRemote(i,vv(o)),i}}async selectRemoteBranch(){let r=await this.gitManager.getRemotes(),n;r.length===0&&(n=await this.editRemotes(),n==null&&(r=await this.gitManager.getRemotes()));let i=new ze(this,{options:r,placeholder:"Select or create a new remote by typing its name and selecting it"}),a=n!=null?n:await i.openAndGetResult();if(a){this.displayMessage("Fetching remote branches"),await this.gitManager.fetch(a);let s=await this.gitManager.getRemoteBranches(a),l=await new ze(this,{options:s,placeholder:"Select or create a new remote branch by typing its name and selecting it"}).openAndGetResult();return l==null?void 0:l.startsWith(a+"/")?l:`${a}/${l}`}}async removeRemote(){if(!await this.isAllInitialized())return;let r=await this.gitManager.getRemotes(),i=await new ze(this,{options:r,placeholder:"Select a remote"}).openAndGetResult();i&&await this.gitManager.removeRemote(i)}onActiveLeafChange(r){var s,o;let n=r==null?void 0:r.view;if(!(n!=null&&n.getState().file)&&!(n instanceof ls||n instanceof fs))return;let i=this.app.workspace.getLeavesOfType($t.type).first(),a=this.app.workspace.getLeavesOfType(Zr.type).first();if((s=i==null?void 0:i.view.containerEl.querySelector("div.tree-item-self.is-active"))==null||s.removeClass("is-active"),(o=a==null?void 0:a.view.containerEl.querySelector("div.tree-item-self.is-active"))==null||o.removeClass("is-active"),(r==null?void 0:r.view)instanceof ls||(r==null?void 0:r.view)instanceof fs){let l=r.view.state.bFile;this.lastDiffViewState=r.view.getState();let c;i&&r.view.state.aRef=="HEAD"?c=i.view.containerEl.querySelector(`div.staged div.tree-item-self[data-path='${l}']`):i&&r.view.state.aRef==""?c=i.view.containerEl.querySelector(`div.changes div.tree-item-self[data-path='${l}']`):a&&(c=a.view.containerEl.querySelector(`div.tree-item-self[data-path='${l}']`)),c==null||c.addClass("is-active")}else this.lastDiffViewState=void 0}handleNoNetworkError(r){this.state.offlineMode?this.log("Encountered network error, but already in offline mode"):this.displayError("Git: Going into offline mode. Future network errors will no longer be displayed.",2e3),this.setPluginState({gitAction:0,offlineMode:!0})}displayMessage(r,n=4*1e3){var i;(i=this.statusBar)==null||i.displayMessage(r.toLowerCase(),n),this.settings.disablePopups||(!this.settings.disablePopupsForNoChanges||!r.startsWith("No changes"))&&new we.Notice(r,5*1e3),this.log(r)}displayError(r,n=10*1e3){var a;if(r instanceof D5.Errors.UserCanceledError){new we.Notice("Aborted");return}let i;r instanceof Error?i=r:i=new Error(String(r)),this.setPluginState({gitAction:0}),this.settings.showErrorNotices&&new we.Notice(i.message,n),console.error(`${this.manifest.id}:`,i.stack),(a=this.statusBar)==null||a.displayMessage(i.message.toLowerCase(),n)}log(...r){console.log(`${this.manifest.id}:`,...r)}}; -/*! Bundled license information: - -ieee754/index.js: - (*! ieee754. BSD-3-Clause License. Feross Aboukhadijeh *) - -buffer/index.js: - (*! - * The buffer module from node.js, for the browser. - * - * @author Feross Aboukhadijeh - * @license MIT - *) - -safe-buffer/index.js: - (*! safe-buffer. MIT License. Feross Aboukhadijeh *) - -crc-32/crc32.js: - (*! crc32.js (C) 2014-present SheetJS -- http://sheetjs.com *) - -js-sha256/src/sha256.js: - (** - * [js-sha256]{@link https://github.com/emn178/js-sha256} - * - * @version 0.9.0 - * @author Chen, Yi-Cyuan [emn178@gmail.com] - * @copyright Chen, Yi-Cyuan 2014-2017 - * @license MIT - *) - -feather-icons/dist/feather.js: - (*! - Copyright (c) 2016 Jed Watson. - Licensed under the MIT License (MIT), see - http://jedwatson.github.io/classnames - *) -*/ - -/* nosourcemap */ \ No newline at end of file diff --git a/.obsidian/plugins/obsidian-git/manifest.json b/.obsidian/plugins/obsidian-git/manifest.json deleted file mode 100644 index 2f24b79..0000000 --- a/.obsidian/plugins/obsidian-git/manifest.json +++ /dev/null @@ -1,10 +0,0 @@ -{ - "author": "Vinzent", - "authorUrl": "https://github.com/Vinzent03", - "id": "obsidian-git", - "name": "Git", - "description": "Integrate Git version control with automatic backup and other advanced features.", - "isDesktopOnly": false, - "fundingUrl": "https://ko-fi.com/vinzent", - "version": "2.35.2" -} diff --git a/.obsidian/plugins/obsidian-git/styles.css b/.obsidian/plugins/obsidian-git/styles.css deleted file mode 100644 index 83c543b..0000000 --- a/.obsidian/plugins/obsidian-git/styles.css +++ /dev/null @@ -1,629 +0,0 @@ -@keyframes loading { - 0% { - transform: rotate(0deg); - } - - 100% { - transform: rotate(360deg); - } -} - -.workspace-leaf-content[data-type="git-view"] .button-border { - border: 2px solid var(--interactive-accent); - border-radius: var(--radius-s); -} - -.workspace-leaf-content[data-type="git-view"] .view-content { - padding-left: 0; - padding-top: 0; - padding-right: 0; -} - -.workspace-leaf-content[data-type="git-history-view"] .view-content { - padding-left: 0; - padding-top: 0; - padding-right: 0; -} - -.loading { - overflow: hidden; -} - -.loading > svg { - animation: 2s linear infinite loading; - transform-origin: 50% 50%; - display: inline-block; -} - -.obsidian-git-center { - margin: auto; - text-align: center; - width: 50%; -} - -.obsidian-git-textarea { - display: block; - margin-left: auto; - margin-right: auto; -} - -.obsidian-git-disabled { - opacity: 0.5; -} - -.obsidian-git-center-button { - display: block; - margin: 20px auto; -} - -.tooltip.mod-left { - overflow-wrap: break-word; -} - -.tooltip.mod-right { - overflow-wrap: break-word; -} - -/* Limits the scrollbar to the view body */ -.git-view { - display: flex; - flex-direction: column; - position: relative; - height: 100%; -} - -.git-tools { - display: flex; - margin-left: auto; -} -.git-tools .type { - padding-left: var(--size-2-1); - display: flex; - align-items: center; - justify-content: center; - width: 11px; -} - -.git-tools .type[data-type="M"] { - color: orange; -} -.git-tools .type[data-type="D"] { - color: red; -} -.git-tools .buttons { - display: flex; -} -.git-tools .buttons > * { - padding: 0 0; - height: auto; -} - -.workspace-leaf-content[data-type="git-view"] .tree-item-self, -.workspace-leaf-content[data-type="git-history-view"] .tree-item-self { - align-items: center; -} - -.workspace-leaf-content[data-type="git-view"] - .tree-item-self:hover - .clickable-icon, -.workspace-leaf-content[data-type="git-history-view"] - .tree-item-self:hover - .clickable-icon { - color: var(--icon-color-hover); -} - -/* Highlight an item as active if it's diff is currently opened */ -.is-active .git-tools .buttons > * { - color: var(--nav-item-color-active); -} - -.git-author { - color: var(--text-accent); -} - -.git-date { - color: var(--text-accent); -} - -.git-ref { - color: var(--text-accent); -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-d-none { - display: none; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-wrapper { - text-align: left; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-header { - background-color: var(--background-primary); - border-bottom: 1px solid var(--interactive-accent); - font-family: var(--font-monospace); - height: 35px; - padding: 5px 10px; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-header, -.workspace-leaf-content[data-type="diff-view"] .d2h-file-stats { - display: -webkit-box; - display: -ms-flexbox; - display: flex; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-stats { - font-size: 14px; - margin-left: auto; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-lines-added { - border: 1px solid #b4e2b4; - border-radius: 5px 0 0 5px; - color: #399839; - padding: 2px; - text-align: right; - vertical-align: middle; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-lines-deleted { - border: 1px solid #e9aeae; - border-radius: 0 5px 5px 0; - color: #c33; - margin-left: 1px; - padding: 2px; - text-align: left; - vertical-align: middle; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-name-wrapper { - -webkit-box-align: center; - -ms-flex-align: center; - align-items: center; - display: -webkit-box; - display: -ms-flexbox; - display: flex; - font-size: 15px; - width: 100%; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-name { - overflow-x: hidden; - text-overflow: ellipsis; - white-space: nowrap; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-wrapper { - border: 1px solid var(--background-modifier-border); - border-radius: 3px; - margin-bottom: 1em; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-collapse { - -webkit-box-pack: end; - -ms-flex-pack: end; - -webkit-box-align: center; - -ms-flex-align: center; - align-items: center; - border: 1px solid var(--background-modifier-border); - border-radius: 3px; - cursor: pointer; - display: none; - font-size: 12px; - justify-content: flex-end; - padding: 4px 8px; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-collapse.d2h-selected { - background-color: #c8e1ff; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-collapse-input { - margin: 0 4px 0 0; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-diff-table { - border-collapse: collapse; - font-family: Menlo, Consolas, monospace; - font-size: 13px; - width: 100%; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-files-diff { - width: 100%; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-diff { - overflow-y: hidden; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-file-side-diff { - display: inline-block; - margin-bottom: -8px; - margin-right: -4px; - overflow-x: scroll; - overflow-y: hidden; - width: 50%; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-code-line { - padding: 0 8em; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-code-line, -.workspace-leaf-content[data-type="diff-view"] .d2h-code-side-line { - display: inline-block; - -webkit-user-select: none; - -moz-user-select: none; - -ms-user-select: none; - user-select: none; - white-space: nowrap; - width: 100%; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-code-side-line { - padding: 0 4.5em; -} - -.workspace-leaf-content[data-type="diff-view"] .d2h-code-line-ctn { - word-wrap: normal; - background: none; - display: inline-block; - padding: 0; - -webkit-user-select: text; - -moz-user-select: text; - 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- - ---- - - - -## 定理与证明 - - - -### 定理（根值判别法） - - - -设 $\sum_{n=1}^\infty u_n$ 是**正项级数**，且 - - - -$ρ=\lim ⁡n→∞ u_n n ρ=n→∞\lim ​nu_n$​​ - - - -（这里允许 $\rho = +\infty$）。 - - - -1. 若 $\rho < 1$，则级数收敛。 - -2. 若 $\rho > 1$（包括 $\rho = +\infty$），则级数发散。 - -3. 若 $\rho = 1$，判别法失效（级数可能收敛也可能发散）。 - - - -### 证明思路 - - - -由极限定义：对任意 $\varepsilon > 0$，存在 $N$，当 $n > N$ 时， - - - -$ρ−ε [!note]   - -> **情形 $\rho < 1$**   - -> 取 $\varepsilon$ 使 $r = \rho + \varepsilon < 1$，则 - -> - -> unN)un​N) - -> - -> 而几何级数 $\sum r^n$ 收敛，由比较判别法知 $\sum u_n$ 收敛。 - - - -> [!note]   - -> **情形 $\rho > 1$**   -> 取 $\varepsilon$ 使 $r = \rho - \varepsilon > 1$，则 -> -> $u_n>r^n(n>N)$ -> -> 而几何级数 $\sum r^n$ 发散，故 $\sum u_n$ 发散。 - - - -> [!note]   - -> **情形 $\rho = 1$**   -> 反例： -> -> - $\sum \frac{1}{n}$ 发散，但 $\sqrt[n]{\frac{1}{n}} \to 1$。 ->     -> - $\sum \frac{1}{n^2}$ 收敛，但 $\sqrt[n]{\frac{1}{n^2}} \to 1$。 ->     - - - ---- - - - -## 典型例题 - - - -> [!example] 例1   - -> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n$ 的敛散性。 - -> - -> **解**   - -> 计算根值： - -> - -> unn=(1−1n)nn=1−1n→1(n→∞)nun​​=n(1−n1​)n​=1−n1​→1(n→∞) - -> - -> 此时 $\rho = 1$，根值判别法失效。 - -> - -> 改用其他方法：注意到 - -> - -> (1−1n)n=enln⁡(1−1n)∼e−1(n→∞)(1−n1​)n=enln(1−n1​)∼e−1(n→∞) - -> - -> 所以 $u_n \to e^{-1} \neq 0$，由级数收敛的必要条件（通项必须趋于0）知该级数发散。 - - - -> [!example] 例2   - -> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{n}{\left(a + \frac{1}{n}\right)^n} \quad (a > 0)$ 的敛散性。 - -> - -> **解**   - -> 计算根值： - -> - -> unn=n(a+1n)nn=nna+1nnun​​=n(a+n1​)nn​​=a+n1​nn​​ - -> - -> 因为 $\sqrt[n]{n} \to 1$，所以 - -> - -> lim⁡n→∞unn=1an→∞lim​nun​​=a1​ - -> - -> - 若 $\frac{1}{a} < 1$ 即 $a > 1$，则级数收敛。 - ->     - -> - 若 $\frac{1}{a} > 1$ 即 $0 < a < 1$，则级数发散。 - ->     - -> - 若 $a = 1$，此时 $\rho = 1$，判别法失效。   - ->     当 $a = 1$ 时， - ->     - ->     un=n(1+1n)n∼ne(n→∞)un​=(1+n1​)nn​∼en​(n→∞) - ->     - ->     通项不趋于0，故级数发散。 - ->     - -> - -> **综上**： - -> - -> - 当 $a > 1$ 时，级数收敛； - ->     - -> - 当 $0 < a \le 1$ 时，级数发散。 - ->     - - - -> [!example] 例3（常用结论）   - -> 证明：$\displaystyle \lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max{a_1, a_2, \dots, a_m}$，其中 $a_k > 0$。 - -> - -> **证明**   - -> 令 $M = \max{a_1, \dots, a_m}$，则 - -> - -> Mn≤a1n+⋯+amn≤mMnMn≤a1n​+⋯+amn​≤mMn - -> - -> 两边开 $n$ 次方： - -> - -> M≤a1n+⋯+amnn≤M⋅mnM≤na1n​+⋯+amn​​≤M⋅nm​ - -> - -> 因为 $\sqrt[n]{m} \to 1$，由夹逼定理得结论。 - -> - -> 此结论在判断形如 $\sum (a_1^n + \cdots + a_m^n)$ 的级数时可直接使用。 - - - -> [!example] 例4   - -> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n}$ 的敛散性。 - -> - -> **解** - -> - -> un=2n+3n5n=(25)n+(35)nun​=5n2n+3n​=(52​)n+(53​)n - -> - -> 计算根值： - -> - -> unn=(25)n+(35)nnnun​​=n(52​)n+(53​)n​ - -> - -> 由例3结论，$\sqrt[n]{u_n} \to \max\left{\frac{2}{5}, \frac{3}{5}\right} = \frac{3}{5} < 1$，故级数收敛。 - - - ---- - - - -## 注意事项 - - - -1. 根值判别法仅适用于正项级数。 - -2. 当 $\rho = 1$ 时，需改用其他方法（如比较判别法、积分判别法、比值判别法等）。 - -3. 计算 $\sqrt[n]{u_n}$ 时，常用结论： - -    nn→1,nkn→1,an=a1/n→1 (a>0)nn​→1,nnk​→1,na​=a1/n→1 (a>0) - -4. 若 $u_n$ 含有 $n!$ 或连乘形式，通常使用比值判别法更简便；若含 $n$ 次幂，则根值判别法可能更直接。 - - - ---- - - - -## 练习题 - - - -> [!question] 练习1   - -> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n$ 的敛散性。 - - - -> [!question] 练习2   - -> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{5^n}{n^n}$ 的敛散性。 - - - -> [!question] 练习3   - -> 设 $a > 0$，讨论级数 $\displaystyle \sum_{n=1}^{\infty} \left(\frac{a}{n}\right)^n$ 的敛散性。 - - - -> [!note] 参考答案 - -> - -> 1. $\rho = \frac{1}{2} < 1$，收敛。 - ->     - -> 2. $\rho = 0 < 1$，收敛。 - ->     - -> 3. 对任意 $a > 0$，有 $\rho = 0 < 1$，收敛。 - -> \ No newline at end of file diff --git a/Chapter 2-3 极限/比值判别法.md b/Chapter 2-3 极限/比值判别法.md deleted file mode 100644 index 5ac4e12..0000000 --- a/Chapter 2-3 极限/比值判别法.md +++ /dev/null @@ -1,123 +0,0 @@ -# 比值判别法 - -## 原理 - -对于正项级数 $\sum_{n=1}^{\infty}u_n\ (u_n>0)$，计算极限$\rho=\lim\limits_{n \to \infty}\frac{u_{n+1}}{u_n}$，根据$\rho$与 1 的大小关系判断级数敛散性。 -1. 若 $\boldsymbol{\rho<1}$，则级数收敛； -2. 若 $\boldsymbol{\rho>1}$（或 $\rho=+\infty$），则级数发散； -3. 若 $\boldsymbol{\rho=1}$，则判别法失效，需用其他方法（如比较判别法、积分判别法等）判断。 - -## **适用情况** - -主要适用于通项含阶乘、指数幂（如 $a^n$）、$n^n$ 等形式的正项级数，这类通项的 $\frac{u_{n+1}}{u_n}$ ，作商后阶乘、指数部分可大幅化简，极限易求。 - -## **优势** - -1. 化简效率高：通项含 n!、$a^n$ 时，作商 $\frac{u_{n+1}}{u_n}$ 能直接约去大量重复因子 -2. 判定直接：只需计算一个极限值与 1 比较，无需像比较判别法那样构造合适的参考级数，对初学者更友好。 - -## **劣势** - -1. 对通项形式依赖强：仅适合含阶乘、指数幂的正项级数，对于通项为多项式、分式（如 $u_n=\frac{1}{n^p}$）的级数，$\lim\limits_{n\to\infty}\frac{u_{n+1}}{u_n}=1$，判别法直接失效。 -2. 失效情形无判定能力：当极限 $\rho=1$ 时，无法判断级数敛散性，必须换用比较判别法、积分判别法等其他方法，增加了解题步骤。 -3. 仅针对正项级数：不能直接应用于任意项级数（如交错级数），若要使用，需先对通项取绝对值判断绝对收敛，再进一步分析条件收敛。 -4. 极限可能不存在：部分正项级数的 $\lim\limits_{n\to\infty}\frac{u_{n+1}}{u_n}$ 不存在，此时比值判别法无法使用。 - -## **例子** - -> [!example] 例1 -判定级数敛散性： -$$\sum_{n=1}^{\infty} \frac{2^n}{n!}$$ - - - -**解析** -用比值判别法： -$$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2}{n+1} \to 0$$ -因此收敛。 - - ->[!example] 例2 -设 $\alpha$ 为正常数，判定级数的敛散性： -$$\sum_{n=1}^\infty \frac{(2n-1)!!}{(n!)^\alpha}$$ - - -**解析** - -1. 记通项 - -$$a_n = \frac{(2n-1)!!}{(n!)^\alpha}$$ - -这里 $(2n-1)!! = 1\cdot 3\cdot 5 \cdots (2n-1) = \frac{(2n)!}{2^n \, n!}$。 - -所以：$$a_n = \frac{\frac{(2n)!}{2^n \, n!}}{(n!)^\alpha}= \frac{(2n)!}{2^n \, (n!)^{\alpha+1}}$$ - -2. 用比值判别法 -$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{2^{n+1}((n+1)!)^{\alpha+1}} \cdot \frac{2^n (n!)^{\alpha+1}}{(2n)!}$$ -化简：$$\frac{(2n+2)!}{(2n)!} = (2n+1)(2n+2) = 2(2n+1)(n+1)$$$$\frac{2^n}{2^{n+1}} = \frac12$$$$\frac{(n!)^{\alpha+1}}{((n+1)!)^{\alpha+1}} = \frac{1}{(n+1)^{\alpha+1}}$$ -所以：$$\frac{a_{n+1}}{a_n} - -= \frac{2(2n+1)(n+1) \cdot \frac12}{(n+1)^{\alpha+1}} - -= \frac{2n+1}{(n+1)^{\alpha}}$$ -3. 计算极限 -$$\lim_{n\to\infty} \frac{2n+1}{(n+1)^{\alpha}}$$ -· 若 $\alpha$> 1，极限为 0，比值判别法得出 $\lim \frac{a_{n+1}}{a_n}$ = 0 < 1，级数收敛。 - -· 若 $\alpha$ = 1，极限为 $\lim_{n\to\infty} \frac{2n+1}{n+1}$ = 2 > 1，级数发散。 - -· 若 0 < $\alpha$ < 1，极限为 $\infty$ > 1，级数发散。 - -4. 答案 - - 当$\alpha$>1时收敛，当 0 < $\alpha$ < 1 时发散 - ->[!example] **习题1** -级数 $$\displaystyle \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} \cdot \frac{1}{5^n}$$ 的敛散性是 - -A. 收敛 - -B. 发散 - -C. 无法判断 - - -**解析** -$$a_n = \frac{(2n)!}{(n!)^2 \cdot 5^n}$$ - -$$\frac{a_{n+1}}{a_n} - -= \frac{(2n+2)!}{((n+1)!)^2 \cdot 5^{n+1}} \cdot \frac{(n!)^2 \cdot 5^n}{(2n)!}$$ -化简： -$$\frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$$ -$$\frac{(n!)^2}{((n+1)!)^2} = \frac{1}{(n+1)^2}$$ - -$$\frac{5^n}{5^{n+1}} = \frac15$$ -所以 - -$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{(n+1)^2} \cdot \frac15 -= \frac{2(2n+1)}{n+1} \cdot \frac15$$ -$$\lim_{n\to\infty} \frac{4n+2}{5(n+1)} = \frac{4}{5} < 1$$ -所以收敛。 - -答案：A - ->[!example] **习题2** -级数 $\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{n^2}$ 的敛散性是 - -A. 收敛 - -B. 发散 - -C. 无法判断 - -解 -$$a_n = \frac{3^n}{n^2}$$ -$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)^2} \cdot \frac{n^2}{3^n} - -= 3 \cdot \frac{n^2}{(n+1)^2}$$ -$$\lim_{n\to\infty} 3 \cdot \frac{n^2}{(n+1)^2} = 3 > 1$$ - -所以发散。 - -答案：B \ No newline at end of file diff --git a/Chapter 2-3 极限/比较判别法.md b/Chapter 2-3 极限/比较判别法.md deleted file mode 100644 index cd07a96..0000000 --- a/Chapter 2-3 极限/比较判别法.md +++ /dev/null @@ -1,97 +0,0 @@ -### Quick Description - -#### 原理 -不等式形式：放缩，大敛则小敛，小散则大散 -极限形式：比值的极限为常数，则同敛散 -#### 适用情况 -正项级数 -#### 优势和劣势 -放缩在高中有知识基础，比较好上手 -与比值/根值判别法不同，判定出变号级数的绝对值发散后，无法得到原级数发散 -### 例子 -#例一 -判断级数$\sum_{n=2}^{\infty} \sin\left(\frac{1}{\ln n} \right)$的敛散性 - -设 $a_n = \sin\left( \frac{1}{\ln n} \right)$，其中 $n \geq 2$ 。记其前n项和为$S_n$。 -由于当 $n \geq 2$ 时，有 $\frac{1}{\ln n} > 0$ ，且 $\sin n > \frac{n}{2}$ ，因此存在 $N$ ，使得当$n > N$时，$q_n = \sin\left( \frac{1}{\ln n} \right) > \frac{1}{2} \cdot \frac{1}{\ln n} > \frac{1}{2n}.$ -记$\sum \frac{1}{2n}$的前$n$项和为$T_n$。 -由于 $\lim_{n \to \infty} S_n > \lim_{n \to \infty} T_n, \quad \text{而} \quad \lim_{n \to \infty} T_n = +\infty,$ -因此 $\lim_{n \to \infty} S_n = +\infty.$ -故级数 $\sum_{n=2}^{\infty} \sin\left( \frac{1}{\ln n} \right)$ 发散。 - -#例二 -**基础练习** - -1. **下列级数中收敛的有\_\_\_\_\_\_**。 - - A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ - B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1}$ - C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a>0)$ - D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ - E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ - F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ - - **参考答案：ABCF. - - **A** 由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故原级数收敛。 - - **B** 由于 $\lim_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故原级数收敛。 - - **C** 由于 $\lim_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛，故原级数收敛。（或由 $\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3} \;(n>1)$） - - **D** 由于 $\lim_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 - - **E** 由于 $\lim_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \frac{\pi}{4}$，又级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 - - **F** 由于 $\lim_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1+\frac{1}{\sqrt{n}}} / \left( 1+\frac{1}{n^2} \right) = 1$，又 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛，故原级数收敛。 - - - - -练习： -**判定断下列级数的敛散性：** -A.收敛 B.发散 -(1) $\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$ -(2) $\sum_{n=1}^{\infty} (\sqrt[n]{a}-1) \quad (a>0)$ - - - -解答： -**(1)** 由于 -$a_n = \frac{\sqrt{n+1}-\sqrt{n}}{n} = \frac{1}{n(\sqrt{n+1}+\sqrt{n})} < \frac{1}{n \cdot 2\sqrt{n}} = \frac{1}{2n^{3/2}}$， -而 $\sum_{n=1}^\infty \frac{1}{n^{3/2}}$ 收敛，故由比较判别法知原级数收敛。 - -**(2)** - -- 当 $a = 1$ 时，通项 $\sqrt[n]{a}-1 = 0$，级数为 $\sum_{n=1}^\infty 0$，收敛。 - -- 当 $a > 1$ 时，级数为正项级数。由于 - $\lim\limits_{n\to\infty} \frac{\sqrt[n]{a}-1}{1/n} = \lim\limits_{n\to\infty} \frac{e^{(\ln a)/n} - 1}{1/n} = \ln a > 0$， - 而 $\sum_{n=1}^\infty \frac{1}{n}$ 发散，故此时级数发散。 - -- 当 $0 < a < 1$ 时，级数为负项级数。考虑 $\sum_{n=1}^\infty (1-\sqrt[n]{a})$， - $\lim\limits_{n\to\infty} \frac{1-\sqrt[n]{a}}{1/n} = -\ln a > 0$，同理级数发散。 - - -综上，级数当 $a = 1$ 时收敛，当 $a>0$ 且 $a \neq 1$ 时发散。 - - - - -**判定断下列级数的敛散性：** -A.收敛 B.发散 -(1) $\sum_{n=1}^{\infty} \frac{3+(-1)^n}{2^{n+1}}$ -(2) $\sum_{n=1}^{\infty} \ln\left(1+\frac{1}{n}\right)$ -(3) $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n}$ - - - -**(1)** 由于 -$a_n = \frac{3+(-1)^n}{2^{n+1}} \le \frac{4}{2^{n+1}} = \frac{1}{2^{,n-1}}$， -而 $\sum_{n=1}^\infty \frac{1}{2^{,n-1}}$ 收敛，由比较判别法知原级数收敛。 - -**(2)** 方法一： -$\lim\limits_{n\to\infty} \frac{\ln!\left(1+\frac{1}{n}\right)}{1/n} = 1$，而 $\sum_{n=1}^\infty \frac{1}{n}$ 发散，故原级数发散。 - -方法二： -部分和 -$S_n = \sum_{k=1}^n \ln\left(1+\frac{1}{k}\right) = \sum_{k=1}^n [\ln(k+1) - \ln k] = \ln(n+1) \to +\infty$， -故级数发散。 - -**(3)** 当 $n > e^2$ 时，$\ln n > 2$，从而 $\frac{1}{(\ln n)^n} < \frac{1}{2^{,n}}$。 -由于 $\sum_{n=2}^\infty \frac{1}{2^{,n}}$ 收敛，由比较判别法知原级数收敛。 - diff --git a/Chapter 2-3 极限/部分和判别法.md b/Chapter 2-3 极限/部分和判别法.md deleted file mode 100644 index 28daa1b..0000000 --- a/Chapter 2-3 极限/部分和判别法.md +++ /dev/null @@ -1,392 +0,0 @@ - -## **原理** - -对于级数 $\sum_{n=1}^{\infty}u_n$，构造其部分和数列 $S_n=u_1+u_2+\dots+u_n$。若极限 $\lim\limits_{n\to\infty}S_n=S$（S 为有限常数），则级数收敛，且和为 S；若该极限不存在（或为无穷大），则级数发散。 - -## **适用情况** - -适用于部分和可通过公式化简求和的级数，典型类型包括： - -1.等比级数：$\sum_{n=1}^{\infty}aq^{n-1}（a\neq0）$，部分和 $S_n=\frac{a(1-q^n)}{1-q}$； - -2.裂项相消型级数：如 $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$，通项可拆为 $\frac{1}{n}-\frac{1}{n+1}$，部分和可逐项抵消化简； - -3. 少数可通过错位相减、分组求和的特殊级数。 - -## **优势** - -1. 本质性强：直接基于级数收敛的定义，判定结果准确，还能同时求出级数的和，这是比值、比较判别法等无法做到的。 - -2. 无形式限制：不像比值判别法依赖通项的阶乘、指数结构，只要部分和能化简，任何类型级数都适用。 - -## **劣势** - -1. 适用范围极窄：绝大多数级数的部分和无法用初等函数化简（如调和级数 $\sum\frac{1}{n}$、$\sum\frac{n}{2^n}$虽能判定敛散性，但部分和无简单表达式）。 - -2. 计算难度高：裂项、错位相减等求和技巧性强，对复杂通项不易找到化简方法。 - -3. 效率低：对比判别法，求部分和的步骤通常更繁琐，不适用于快速判定敛散性。 - -## **例子** - ->[!example] **例1** ->判定级数 $\sum_{n=1}^{\infty}\frac{2}{(2n-1)(2n+1)}$的敛散性，若收敛求其和 - -解析 - -1.  裂项变形 - -   $\frac{2}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}$ - -2.  求部分和 - -   $S_n=\sum_{k=1}^{n}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right) = \left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right) = 1-\frac{1}{2n+1}$ - -3.  取极限 - -   $\lim_{n \to \infty}S_n=\lim_{n \to \infty}\left(1-\frac{1}{2n+1}\right)=1$ - -4.  结论：该级数 **收敛**，和为 \( 1 \) - ->[!example] **例2** -判断级数$\sin \frac{\pi}{6} + \sin \frac{2\pi}{6} + \cdots + \sin \frac{n\pi}{6} + \cdots$的敛散性 - -解析： - -1. 已知公式 - -由积化和差公式： - -$2\sin A \sin B = \cos(A-B) - \cos(A+B)$ - -令  $A = k\theta$， $B = \frac{\theta}{2}$ ，则： - -$2\sin(k\theta) \sin\left( \frac{\theta}{2} \right) = \cos\left( k\theta - \frac{\theta}{2} \right) - \cos\left( k\theta + \frac{\theta}{2} \right)$ - -$= \cos\left( (k - \tfrac12)\theta \right) - \cos\left( (k + \tfrac12)\theta \right)$ - -因此： - -$\boxed{\sin(k\theta) = \frac{\cos\left( (k - \tfrac12)\theta \right) - \cos\left( (k + \tfrac12)\theta \right)}{ 2\sin\frac{\theta}{2} }}$ - -这是一个裂项公式，因为对 k 求和时，中间项会消去。 - -2. 应用到本题 - -题中级数为： - -$\sum_{k=1}^{\infty} \sin\left( \frac{k\pi}{6} \right)$ - -这里  $\theta = \frac{\pi}{6}$ ， $\sin\frac{\theta}{2}$ = $\sin\frac{\pi}{12} \neq 0$。 - -利用上面裂项公式，记 - -$a_k = \sin\left( \frac{k\pi}{6} \right)$ - -$= \frac{\cos\left( \frac{(k - \frac12)\pi}{6} \right) - \cos\left( \frac{(k + \frac12)\pi}{6} \right)}{2\sin\frac{\pi}{12}}$ - -即 - -$a_k = \frac{1}{2\sin\frac{\pi}{12}} \left[ \cos\left( \frac{(2k-1)\pi}{12} \right) - \cos\left( \frac{(2k+1)\pi}{12} \right) \right]$ - -3. 求部分和 $S_n$ - -$S_n = \sum_{k=1}^n a_k$ - -$= \frac{1}{2\sin\frac{\pi}{12}} \sum_{k=1}^n \left[ \cos\frac{(2k-1)\pi}{12} - \cos\frac{(2k+1)\pi}{12} \right]$ - -记 $C_m = \cos\frac{(2m-1)\pi}{12}$，则： - -$\sum_{k=1}^n \left[ C_k - C_{k+1} \right]$ - -这是一个裂项和： - -$\sum_{k=1}^n (C_k - C_{k+1}) = C_1 - C_{n+1}$ - -其中 - -$C_1 = \cos\frac{(2\cdot 1 - 1)\pi}{12} = \cos\frac{\pi}{12}$ - -$C_{n+1} = \cos\frac{(2(n+1)-1)\pi}{12} = \cos\frac{(2n+1)\pi}{12}$ - -因此： - -$S_n = \frac{1}{2\sin\frac{\pi}{12}} \left[ \cos\frac{\pi}{12} - \cos\frac{(2n+1)\pi}{12} \right]$ - -4. 分析极限 - -当 $n \to \infty$ 时， $\frac{(2n+1)\pi}{12} \mod 2\pi$  在 $[0, 2\pi)$ 内不趋于一个固定值，而是周期性地取多个不同值。 - -事实上，$\cos\frac{(2n+1)\pi}{12}$ 的值在 $n$ 增加时振荡，不会趋于一个常数。 - -因此： - -$\lim_{n\to\infty} S_n \quad \text{不存在。}$ - -**习题1** - -级数 - -$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$ - -\]   - -的和为 - -A. 0 - -B. 1 - -C. 2 - -D. 不存在（发散） - -解： - -观察 - -$\frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$ - -部分和： - -$S_N = \sum_{n=1}^N \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right] = 1 - \frac{1}{(N+1)^2} \to 1$ - -收敛到 1。 - -答案：B - -**习题2** - -级数 - -$\sum_{n=1}^{\infty} \left( \sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n} \right)$ - -\]   - -的敛散性是 - -A. 收敛到 0 - -B. 收敛到 $1 - \sqrt{2}$ - -C. 收敛到 $\sqrt{2} - 2$ - -D. 发散 - -解： - -观察通项： - -设 $a_n = \sqrt{n}$，则 - -$a_{n+2} - 2a_{n+1} + a_n = (a_{n+2} - a_{n+1}) - (a_{n+1} - a_n)$ - -\]   - -记 $b_n = a_{n+1} - a_n = \sqrt{n+1} - \sqrt{n}$，则 - -原通项 = $b_{n+1} - b_n$。 - -部分和： - -$S_N = \sum_{n=1}^N (b_{n+1} - b_n) = b_{N+1} - b_1$ - -\]   - -$b_{N+1} = \sqrt{N+2} - \sqrt{N+1}, \quad b_1 = \sqrt{2} - 1$ - -\]   - -因此 - -$S_N = (\sqrt{N+2} - \sqrt{N+1}) - (\sqrt{2} - 1)$ - -\]   - -当 $N \to \infty，\sqrt{N+2} - \sqrt{N+1} \to 0$，所以 - -$lim_{N\to\infty} S_N = 1 - \sqrt{2}$ - -\]   - -收敛到 $1 - \sqrt{2}$。 - -答案：B - -# **级数运算性质判别法** - -## **原理** - -1. 线性运算性质的应用 - -设$\sum u_n、\sum v_n$ 为已知敛散性的级数，k 为非零常数： - -- 若 $\sum u_n$ 收敛，则 $\sum ku_n$ 收敛；若 $\sum u_n$ 发散，则 $\sum ku_n$ 发散。 - -- 若 $\sum u_n$ 和 $\sum v_n$ 都收敛，则 $\sum(u_n\pm v_n)$ 收敛。 - -- 若 $\sum u_n$ 收敛、$\sum v_n$ 发散，则 $\sum(u_n\pm v_n)$ 发散。 - -注：若两者都发散，$\sum(u_n\pm v_n)$ 敛散性不确定。 - -2. 添减/改变有限项性质的应用 - -对任意级数，增加、去掉或改变有限项，不改变其敛散性。 - -例：$\sum_{n=3}^{\infty}\frac{1}{2^n}$ 收敛，因为它是收敛等比级数 $\sum_{n=1}^{\infty}\frac{1}{2^n}$ 去掉前2项得到的。 - -3. 正项级数重组性质的应用 - -若正项级数 $\sum u_n$ 收敛，任意重排其项得到的新级数仍收敛，且和不变。 - -## **适用情况** - -1. 待判定级数可拆分为两个或多个已知敛散性的级数的线性组合（如 $\sum (u_n\pm v_n)、\sum ku_n）$。 - -2. 级数仅增减、改变有限项，或对正项级数进行项的重组。 - -3. 已知级数多为基础类型（等比级数、$p$-级数、调和级数等），便于直接套用性质。 - -## **优势** - -1. 快捷高效：无需计算极限或构造不等式，直接利用已知结论推导，步骤简洁。 - -2. 既适用于正项级数，也适用于任意项级数（如交错级数）。 - -## **劣势** - -1. 依赖性强：必须依赖已知敛散性的“参考级数”，若无法拆分或无合适参考级数，则无法使用。 - -2. 有局限性：对于 $\sum u_n$ 和 $\sum v_n$ 均发散的情况，$\sum (u_n\pm v_n)$ 的敛散性无法直接判定，性质失效。 - -## **例子** - ->[!example] **例1** ->判断级数$\sum_{n=1}^{\infty} (-1)^n \frac{n+2}{n+1} \cdot \frac{1}{\sqrt{n}}$的敛散性。 - -解析： - -原级数： - -$S = \sum_{n=1}^\infty (-1)^n \frac{n+2}{n+1} \cdot \frac{1}{\sqrt{n}}.$ - -将 $\frac{n+2}{n+1}$ 写成： - -$\frac{n+2}{n+1} = 1 + \frac{1}{n+1}.$ - -于是通项： - -$(-1)^n \frac{1}{\sqrt{n}} \cdot \left( 1 + \frac{1}{n+1} \right)$ - -$= (-1)^n \frac{1}{\sqrt{n}} \;+\; (-1)^n \frac{1}{\sqrt{n} \,(n+1)}.$ - -记 - -$u_n = (-1)^n \frac{1}{\sqrt{n}}, \quad v_n = (-1)^n \frac{1}{\sqrt{n} \, (n+1)}.$ - -则原级数： - -$S = \sum_{n=1}^\infty u_n \;+\; \sum_{n=1}^\infty v_n.$ - -第一项：$\sum_{n=1}^\infty u_n = \sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}$。 - -· 这是交错 $p$-级数， $p = \frac12 < 1$，不绝对收敛； - -· 由莱布尼茨判别法（$1/\sqrt{n} 单调递减趋于 0$）⇒ 条件收敛。 - -第二项：$\sum_{n=1}^\infty v_n$，比较 $v_n$ 的绝对值： - -$|v_n| = \frac{1}{\sqrt{n} \,(n+1)} \sim \frac{1}{n^{3/2}} \quad (n\to\infty).$ - -因为 $p = 3/2 > 1$，所以 $\sum |v_n|$ 收敛 ⇒ $\sum v_n$ 绝对收敛。 - -利用级数运算性质： - -已知 - -· 若两个级数 $\sum A_n 和 \sum B_n$ 都收敛，则 $\sum (A_n + B_n)$ 收敛（收敛级数的和收敛）。 - -· 若 $\sum A_n$ 条件收敛，$\sum B_n$ 绝对收敛，则 $\sum (A_n + B_n)$条件收敛（因为加一个绝对收敛级数不改变条件收敛的性质）。 - -这里： - -$S = \underbrace{\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}}_{\text{条件收敛}} \;+\; \underbrace{\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}(n+1)}}_{\text{绝对收敛}}.$ - -因此 $S$ 是 条件收敛的级数 - ->[!example] **例2** -讨论级数$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$的敛散性 - -解析：先对通项有理化变形，再拆分为两个级数的和，结合级数运算性质判断 - -步骤1 通项有理化与变形 - -对通项分母有理化： -$$ - -\begin{align*} - -u_n &= \frac{(-1)^n}{\sqrt{n} + (-1)^n} \\ - -&= \frac{(-1)^n \left[\sqrt{n} - (-1)^n\right]}{\left[\sqrt{n} + (-1)^n\right]\left[\sqrt{n} - (-1)^n\right]} \\ - -&= \frac{(-1)^n \sqrt{n} - 1}{n - 1} \\ - -&= \frac{(-1)^n \sqrt{n}}{n - 1} - \frac{1}{n - 1} - -\end{align*} -$$ - -$\frac{(-1)^n \sqrt{n}}{n - 1}$用莱布尼兹判别法判断收敛 - -$\sum_{n=2}^{\infty} \frac{1}{n - 1}：令 k=n-1，级数变为 \sum_{k=1}^{\infty} \frac{1}{k}$（调和级数），**发散**。 - -根据级数运算性质：**收敛级数 + 收敛级数 - 发散级数 = 发散级数**。 - -因此，级数$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$**发散**。 - -**习题1** - -已知级数 $\sum_{n=1}^{\infty} a_n$ 收敛，$\sum_{n=1}^{\infty} b_n$ 发散，则以下说法正确的是（ ） - -A. $\sum_{n=1}^{\infty} (a_n + b_n)$ 必收敛 - -B. $\sum_{n=1}^{\infty} (a_n + b_n)$ 必发散 - -C. $\sum_{n=1}^{\infty} (a_n b_n)$ 必发散 - -D. $\sum_{n=1}^{\infty} (a_n^2)$ 必收敛 - -解析： - -设 $s_n = a_1+\dots+a_n$ 收敛，$t_n = b_1+\dots+b_n$ 发散。 - -则 $a_n+b_n$ 的前 $n$ 项和 = $s_n + t_n$，收敛+发散 = 发散，故 B 正确。 - -反例：A 错。C：取 $a_n=\frac{(-1)^n}{\sqrt{n}}$（条件收敛），$b_n=(-1)^n$（发散），则 $a_n b_n = \frac{1}{\sqrt{n}}$ 发散，但这是特例，不是“必”发散，若 $a_n=\frac{1}{n^2}, b_n=n$，则 $a_n b_n = \frac{1}{n}$ 发散；若 $a_n=\frac{1}{n^2}, b_n=(-1)^n n$，则 $a_n b_n = \frac{(-1)^n}{n}$ 收敛（条件收敛）。所以 C 不对。D：反例 $a_n = \frac{(-1)^n}{\sqrt{n}}$ 收敛（条件收敛），但 $a_n^2 = \frac{1}{n}$ 发散。 - -答案：B - -**习题2** - -级数 $\sum_{n=1}^{\infty} \left( \frac{\sin n}{n^2} + \frac{(-1)^n}{n} \right)$ 的收敛性是： - -A. 绝对收敛 - -B. 条件收敛 - -C. 发散 - -D. 可能收敛可能发散 - -解析 - -拆成两个： - -1. $\sum \frac{\sin n}{n^2}：因 |\frac{\sin n}{n^2}| \le \frac{1}{n^2}$，绝对收敛。 - -2. $\sum \frac{(-1)^n}{n}$ 是交错调和级数，条件收敛。 - -   条件收敛 + 绝对收敛 = 条件收敛。 - -答案：B \ No newline at end of file diff --git a/Chapter 4 微积分基础/参数方程导数.md b/Chapter 4 微积分基础/参数方程导数.md deleted file mode 100644 index a6ba875..0000000 --- a/Chapter 4 微积分基础/参数方程导数.md +++ /dev/null @@ -1,16 +0,0 @@ - -已知 $\left\{ \begin{array}{c} x=x\left( t \right)\\ y=y\left( t \right)\\ \end{array} \right.$ - -$y'\left( t \right) =\frac{dy}{dt}$ -$x'\left( t \right)=\frac{dx}{dt}$ - - $y'=\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{y'\left( t \right)}{x'\left( t \right)}$ - -则有 $y''=\frac{d^2y}{dx^2}=\frac{dy'}{dx}=\frac{dy'}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{dy'}{dt}\cdot \frac{1}{x'\left( t \right)}=(\frac{y'(t)}{x'(t)})'\cdot \frac{1}{x'\left( t \right)}$ - -而 $\frac{y'(t)}{x'(t)}=\frac{y''\left( t \right) x'\left( t \right) -x''\left( t \right) y'\left( t \right)}{x'^2\left( t \right)}$ - -故 $y''=\frac{y''\left( t \right) x'\left( t \right) -x''\left( t \right) y'\left( t \right)}{x'^3\left( t \right)}$ - - - diff --git a/Chapter 4 微积分基础/应用题真题.md b/Chapter 4 微积分基础/应用题真题.md deleted file mode 100644 index 463ccf6..0000000 --- a/Chapter 4 微积分基础/应用题真题.md +++ /dev/null @@ -1,87 +0,0 @@ ->[!example] **例1**（2020） ->有一个正圆锥形漏斗，深度 $18 \text{ cm}$ ，上端口直径为 $12 \text{ cm}$ （即半径 $6 \text{ cm}$ ）。漏斗下方连接一个圆柱形筒，圆柱筒直径为 $10 \text{ cm}$ （即半径 $5 \text{ cm}$ ）。初始时刻漏斗内盛满水，然后水从漏斗流入圆柱筒。已知当漏斗中水深为 $12 \text{ cm}$ 时，漏斗水面下降的速度是 $1 \text{ cm/min}$ -问：此时圆柱筒内液面上升的速度是多少？ - - -**解析**: -设 t （s）时漏斗水深为 h （cm），圆柱形容器的水深为 H （cm），则有 -$$\pi 5^2 H = V_0 - \frac{1}{3} \pi \left( \frac{6}{18} h \right)^2 h$$ -关于 t 求导数得 -$$25 \frac{dH}{dt} = -\frac{1}{9} h^2 \frac{dh}{dt}$$ -当 h = 12, $\frac{dh}{dt} = -1$ 时， -$$\frac{dH}{dt} = \frac{16}{25} \text{(cm/s)}$$ - - ->[!example] **例2** (2021) ->一长为 L 米的木梯靠在倾角为$\frac{\pi}{3}$的光滑斜坡上，木梯的顶部距离 A 点 h 米，底部距离 A 点 d 米，受重力作用木梯的顶部以 $a \, \mathrm{m/s}$ 的速度沿直线 BA 下滑，底部水平向右运动。问：当木梯的顶部和底部与 A 点的距离相等时，底部的水平速度为多少？ ->![[2dc97df311b6179d9c28babc1848adfc.jpg]] - - - -**解析：** -设运动 t 秒后，木梯的顶部距离 A 点 $y(t) \, \mathrm{m}$ ，底部距离 A 点 $x(t) \, \mathrm{m}$ 。由图易知 -$$\left(y(t)\sin\frac{\pi}{3}\right)^2 + \left(x(t)+y(t)\cos\frac{\pi}{3}\right)^2 = L^2$$ -即 -$$x^2(t)+y^2(t)+x(t)y(t)=L^2$$ -方程两端分别对 t 求导，可得 -$$2x(t)x'(t)+2y(t)y'(t)+x(t)y'(t)+x'(t)y(t)=0$$ -由于 $y'(t)=-a \, (\mathrm{m/s})$ ，因此当 x(t)=y(t) 时，有 -$$2x'(t)-2a-a+x'(t)=0$$ -故 $$x'(t)=a \, (\mathrm{m/s})$$ - - ->[!example] **例3**（2023） ->有一个长度为5m的梯子贴靠在铅直的墙上，假设其下端沿地板以3m/s的速率离开墙脚而滑动，则 -(1) 何时梯子的上、下端能以相同的速率移动？ -(2) 何时其上端下滑之速率为4m/s？ - - -**解析:** -设在时刻 t ，梯子上端与墙角的距离为 $x(t) \, \mathrm{m}$ ，下端与墙角的距离为 $y(t) \, \mathrm{m}$ ，则 -$$x^2 + y^2 = 25$$ -两端同时对 t 求导数可得 -$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$$ - -(1) 当$\frac{dx}{dt} = -\frac{dy}{dt}$ 时，可得 x = y ，进而 $y = \frac{5}{\sqrt{2}} \, (\mathrm{m})$ ， -即下端距墙脚 $\frac{5}{\sqrt{2}} \, \mathrm{m}$ 时上下端能以相同的速率移动。 - -(2) 当 $\frac{dy}{dt} = -4 \, (\mathrm{m/s})$ ， $\frac{dx}{dt} = 3 \, (\mathrm{m/s})$ 时，有 $4x = 3y$ ，将其代入 $x^2 + y^2 = 25$ ， -解得$y = 4 \, (\mathrm{m})$,即下端距墙脚4 m时，上端下滑速度为$4 m/s$ - - - ->[!example] **例4**(2022) ->某部举行八一阅兵，队列正步通过阅兵台时步幅间距离为75厘米，步速为每分钟112步。某观礼人员离行进队列垂直距离为60米，视线追随队列领队，求其视线与队列夹角为$\frac{\pi}{6}$时，视线转动角度的变化率。 - - - -**解析：** -- 建立几何模型：设时间为\(t\)分钟，步幅间距离0.75米，步速为每分钟112步，则队列行进距离$x = 0.75×112t = 84t$米。设视线与队列夹角为$theta$，由正切函数关系可得$tan\theta=\frac{x}{60}$，即$\tan\theta=\frac{84t}{60}=\frac{7t}{5}$。 -- 两边对$t$求导：根据复合函数求导法则 -$$(tan\theta)^\prime=\sec^{2}\theta\cdot\frac{d\theta}{dt})$$对$tan\theta=\frac{7t}{5}$两边求导得$$sec^{2}\theta\cdot\frac{d\theta}{dt}=\frac{7}{5}$$ -求解:$\frac{d\theta}{dt}$：当$theta = \frac{\pi}{6}$时，$sec\theta=\frac{2}{\sqrt{3}}$，$sec^{2}\theta=\frac{4}{3}$，代入$sec^{2}\theta\cdot\frac{d\theta}{dt}=\frac{7}{5}$， - -可得$$\frac{4}{3}\cdot\frac{d\theta}{dt}=\frac{7}{5}$$解得$$\frac{d\theta}{dt}=\frac{21}{20}（弧度/分钟） -$$ -答案：视线转动角度的变化率为\(\frac{21}{20}\)弧度/分钟。本题主要考察了利用导数解决实际问题中的变化率问题，关键在于建立正确的函数关系并准确求导，易错点是复合函数求导法则的运用。 - - ->[!example] **例5**（2024） ->一架巡逻直升机在距离地面3km的高度以120km/h的速度沿着一条水平笔直的高速公路向前飞行。飞行员观察到迎面驶来一辆汽车，通过雷达测出直升机与汽车的距离为5km，同时此距离以160km/h的速率在减少。试求此时汽车行进的速度。 - - - -**解析**： - -如图建立直角坐标系，设时刻 t直升机位于 A点 (x1​(t),3)，汽车位于 B点 (x2​(t),0)，直升机与汽车的距离为 z(t)，则 - -$$(x2​(t)−x1​(t))2+32=z2(t)$$ - -方程两端分别对 t求导，可得 -$$(x2​(t)−x1​(t))(x2′​(t)−x1′​(t))=z(t)z′(t)$$ - -由于 $z(t)=5$时，$x2​(t)−x1​(t)=4$，$z′(t)=−160$，$x1′​(t)=120$，有 - -$$4(x2′​(t)−120)=5×(−160)$$ - -故 $x2′​(t)=−80$，即汽车行进的速度为 $80km/h$。 \ No newline at end of file diff --git a/Chapter 4 微积分基础/相关变化率(1).md b/Chapter 4 微积分基础/相关变化率(1).md deleted file mode 100644 index 2e3fe4e..0000000 --- a/Chapter 4 微积分基础/相关变化率(1).md +++ /dev/null @@ -1,150 +0,0 @@ -# 飞机航空摄影问题 - -## 问题描述 - -一飞机在离地面2 km的高度，以200 km/h的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 - -## 解析 - -### 建立坐标系 - -目标位置：原点 $O(0,0)$ -飞机位置：$P(x,2)$ -飞机高度：$h = 2$ km -水平速度：$\frac{dx}{dt} = -200$ km/h - -### 角度关系 - -$\theta = \arctan\left(\frac{x}{2}\right)$ -$\tan\theta = \frac{x}{2}$ - -### 角速率计算 - -$\frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}$ - -$\frac{d\theta}{dx} = \frac{d}{dx} \arctan\left(\frac{x}{2}\right) = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} = \frac{2}{x^2 + 4}$ - -$\frac{d\theta}{dt} = \frac{2}{x^2 + 4} \cdot (-200) = -\frac{400}{x^2 + 4}$ rad/h - -### 目标正上方时的角速率 - -当 $x = 0$ 时： - -$\left. \frac{d\theta}{dt} \right|_{x=0} = -\frac{400}{0 + 4} = -100$ rad/h - -转换单位： - -$100$ rad/h $= 100 \times \frac{180}{\pi}$ °/h $= \frac{18000}{\pi}$ °/h - -$\frac{18000}{\pi}$ °/h $\times \frac{1}{3600}$ h/s $= \frac{5}{\pi}$ °/s $\approx 1.59$ °/s - -## 答案 - -飞机飞至目标正上方时，摄影机转动的角速率为 $\frac{5}{\pi}$ °/s（约1.59 °/s）。 - ---- - -# 人拉船问题 - -## 问题描述 - -人在河岸用绳经过定滑轮以速度 $v$、加速度 $a$ 拉船。绳与水平面夹角为 $\theta$。求此时船的加速度 $a'$。 -![[Pasted image 20251223173057.png]] -## 解析 - -### 变量定义 - -$h$：滑轮高度（定值） -$x$：船到河岸的水平距离 -$l$：绳长 -$v = -\frac{dl}{dt}$：人拉绳的速度 -$a = -\frac{d^2l}{dt^2}$：人拉绳的加速度 -$u = -\frac{dx}{dt}$：船的水平速度 -$a' = -\frac{d^2x}{dt^2}$：船的加速度 -$\theta$：绳与水平面夹角 - -### 几何关系 - -$l^2 = h^2 + x^2$ - -### 一阶导数关系 - -对几何关系求导：$2l\frac{dl}{dt} = 2x\frac{dx}{dt}$ - -$l\frac{dl}{dt} = x\frac{dx}{dt}$ - -代入速度定义：$l(-v) = x(-u)$ - -$u = \frac{l}{x}v = \frac{v}{\cos\theta}$，其中 $\cos\theta = \frac{x}{l}$ - -### 二阶导数关系 - -对一阶关系求导：$\frac{d}{dt}\left(l\frac{dl}{dt}\right) = \frac{d}{dt}\left(x\frac{dx}{dt}\right)$ - -$\left(\frac{dl}{dt}\right)^2 + l\frac{d^2l}{dt^2} = \left(\frac{dx}{dt}\right)^2 + x\frac{d^2x}{dt^2}$ - -$(-v)^2 + l(-a) = (-u)^2 + x(-a')$ - -$v^2 - la = u^2 - xa'$ - -### 求解 $a'$ - -$xa' = u^2 - v^2 + la$ - -$a' = \frac{u^2 - v^2 + la}{x}$ - -代入 $u = \frac{v}{\cos\theta}$, $x = h\tan\theta$, $l = \frac{h}{\cos\theta}$： - -$a' = \frac{\left(\frac{v}{\cos\theta}\right)^2 - v^2 + \left(\frac{h}{\cos\theta}\right)a}{h\tan\theta}$ - -$a' = \frac{v^2\left(\frac{1}{\cos^2\theta} - 1\right)}{h\tan\theta} + \frac{\frac{h}{\cos\theta}a}{h\tan\theta}$ - -$a' = \frac{v^2\tan^2\theta}{h\tan\theta} + \frac{a}{\cos\theta\tan\theta}$ - -$a' = \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta}$ - -## 答案 - -此时船的加速度为： -$a' = \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta}$ - ---- - -# 动点曲线运动问题 - -## 问题描述 - -已知动点 $P$ 在曲线 $y = \sqrt{x}$ 上运动，记坐标原点 $O$ 与 $P$ 间的距离为 $l$。若点 $P$ 横坐标随时间的变化率为常数 $v$，则当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率是多少？ - -## 解析 - -### 变量关系 - -点 $P$ 坐标：$(x, y)$, $y = \sqrt{x}$ - -原点 $O$ 与 $P$ 间的距离： -$l = \sqrt{x^2 + y^2} = \sqrt{x^2 + (\sqrt{x})^2} = \sqrt{x^2 + x}$ - -已知：$\frac{dx}{dt} = v$（常数） - -### 距离变化率计算 - -$\frac{dl}{dt} = \frac{d}{dt} \sqrt{x^2 + x} = \frac{d}{dt} (x^2 + x)^{1/2}$ - -$\frac{dl}{dt} = \frac{1}{2}(x^2 + x)^{-1/2} \cdot \frac{d}{dt}(x^2 + x)$ - -$\frac{dl}{dt} = \frac{1}{2\sqrt{x^2 + x}} \cdot (2x + 1) \frac{dx}{dt}$ - -$\frac{dl}{dt} = \frac{2x + 1}{2\sqrt{x^2 + x}} \cdot v$ - -### 在点 $(1, 1)$ 处的变化率 - -当 $x = 1$ 时： - -$\left. \frac{dl}{dt} \right|_{x=1} = \frac{2\cdot 1 + 1}{2\sqrt{1^2 + 1}} \cdot v = \frac{3}{2\sqrt{2}} \cdot v = \frac{3v}{2\sqrt{2}}$ - -有理化：$\frac{3v}{2\sqrt{2}} = \frac{3v\sqrt{2}}{4}$ - -## 答案 - -当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率为 $\frac{3v\sqrt{2}}{4}$。 \ No newline at end of file diff --git a/Chapter 4 微积分基础/相关变化率(2).md b/Chapter 4 微积分基础/相关变化率(2).md deleted file mode 100644 index 8821fc2..0000000 --- a/Chapter 4 微积分基础/相关变化率(2).md +++ /dev/null @@ -1,152 +0,0 @@ -#### 飞机航空摄影问题 - -##### 问题描述 - ->[!example] 例题 ->一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 - -解析： - -1、建立坐标系 - -目标位置：原点 $O(0,0)$ -飞机位置：$P(x,2)$ -飞机高度：$h = 2 km$ -水平速度：$\frac{dx}{dt} = -200 km/h$ - -2、角度关系 - -$$ -\theta = \arctan\left(\frac{x}{2}\right), -\tan\theta =\frac{x}{2} -$$ - -3、角速率计算 - -$$\frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}$$ -$$\frac{d\theta}{dx} = \frac{d}{dx} \arctan\left(\frac{x}{2}\right) = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} = \frac{2}{x^2 + 4}$$ -$$\frac{d\theta}{dt} = \frac{2}{x^2 + 4} \cdot (-200) = -\frac{400}{x^2 + 4} rad/h$$ - -4、目标正上方时的角速率 - -当 $x = 0$ 时： - -$$\left. \frac{d\theta}{dt} \right|_{x=0} = -\frac{400}{0 + 4} = -100 rad/h$$ - -5、转换单位： - -$$100 rad/h = 100 \times \frac{180}{\pi} °/h = \frac{18000}{\pi} °/h, - -\frac{18000}{\pi} °/h \times \frac{1}{3600} h/s = \frac{5}{\pi} °/s \approx 1.59 °/s$$ - -答案: - -飞机飞至目标正上方时，摄影机转动的角速率为 $\frac{5}{\pi} °/s$（约$1.59 °/s$）。 - ---- - -#### 人拉船问题 - -##### 问题描述 - ->[!example] 例题 ->人在河岸用绳经过定滑轮以速度 $v$、加速度 $a$ 拉船。绳与水平面夹角为 $\theta$。求此时船的加速度 $a'$。 - -![[Pasted image 20251223173057.png]] -解析 - -1、 变量定义 - -$h$：滑轮高度（定值） -$x$：船到河岸的水平距离 -$l$：绳长 -$v = -\frac{dl}{dt}$：人拉绳的速度 -$a = -\frac{d^2l}{dt^2}$：人拉绳的加速度 -$u = -\frac{dx}{dt}$：船的水平速度 -$a' = -\frac{d^2x}{dt^2}$：船的加速度 -$\theta$：绳与水平面夹角 - -2、几何关系 - -$l^2 = h^2 + x^2$ - -3、 一阶导数关系 - -对几何关系求导：$$2l\frac{dl}{dt} = 2x\frac{dx}{dt}, - -l\frac{dl}{dt} = x\frac{dx}{dt}$$ - -代入速度定义：$l(-v) = x(-u)$ - -$$u = \frac{l}{x}v = \frac{v}{\cos\theta}，其中 \cos\theta = \frac{x}{l}$$ - -4、 二阶导数关系 - -对一阶关系求导：$$\frac{d}{dt}\left(l\frac{dl}{dt}\right) = \frac{d}{dt}\left(x\frac{dx}{dt}\right)$$ - -$$\left(\frac{dl}{dt}\right)^2 + l\frac{d^2l}{dt^2} = \left(\frac{dx}{dt}\right)^2 + x\frac{d^2x}{dt^2}$$ - -$$(-v)^2 + l(-a) = (-u)^2 + x(-a')$$ - -$$v^2 - la = u^2 - xa'$$ - -5、 求解 $a'$ - -$$xa' = u^2 - v^2 + la, - -a' = \frac{u^2 - v^2 + la}{x}$$ - -代入 $u = \frac{v}{\cos\theta}$, $x = h\tan\theta$, $l = \frac{h}{\cos\theta}$： - -$$\begin{aligned} -a' &= \frac{\left(\frac{v}{\cos\theta}\right)^2 - v^2 + \left(\frac{h}{\cos\theta}\right)a}{h\tan\theta}\\[1em] - &= \frac{v^2\left(\frac{1}{\cos^2\theta} - 1\right)}{h\tan\theta} + \frac{\frac{h}{\cos\theta}a}{h\tan\theta}\\[1em] - &= \frac{v^2\tan^2\theta}{h\tan\theta} + \frac{a}{\cos\theta\tan\theta}\\[1em] - &= \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta} -\end{aligned}$$ - - 答案： - -此时船的加速度为： -$$a' = \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta}$$ - ---- - -#### 动点曲线运动问题 - -##### 问题描述 - ->[!example] 例题 ->已知动点 $P$ 在曲线 $y = \sqrt{x}$ 上运动，记坐标原点 $O$ 与 $P$ 间的距离为 $l$。若点 $P$ 横坐标随时间的变化率为常数 $v$，则当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率是多少？ - -解析: - -1、 变量关系 - -点 $P$ 坐标：$(x, y)$, $y = \sqrt{x}$ - -原点 $O$ 与 $P$ 间的距离： -$$l = \sqrt{x^2 + y^2} = \sqrt{x^2 + (\sqrt{x})^2} = \sqrt{x^2 + x}$$ - -已知：$\frac{dx}{dt} = v$（常数） - -2、距离变化率计算 - -$$\begin{aligned} -\frac{dl}{dt} &= \frac{d}{dt} \sqrt{x^2 + x} = \frac{d}{dt} (x^2 + x)^{1/2}\\[1em] - &= \frac{1}{2}(x^2 + x)^{-1/2} \cdot \frac{d}{dt}(x^2 + x)\\[1em] - &= \frac{1}{2\sqrt{x^2 + x}} \cdot (2x + 1) \frac{dx}{dt}\\[1em] - &= \frac{2x + 1}{2\sqrt{x^2 + x}} \cdot v -\end{aligned}$$ - -3、在点 $(1, 1)$ 处的变化率 - -当 $x = 1$ 时： - -$$\left. \frac{dl}{dt} \right|_{x=1} = \frac{2\cdot 1 + 1}{2\sqrt{1^2 + 1}} \cdot v = \frac{3}{2\sqrt{2}} \cdot v = \frac{3v}{2\sqrt{2}}$$ - -有理化：$\frac{3v}{2\sqrt{2}} = \frac{3v\sqrt{2}}{4}$ - - 答案： - -当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率为 $\frac{3v\sqrt{2}}{4}$。 \ No newline at end of file diff --git a/Chapter 4 微积分基础/相关变化率.md b/Chapter 4 微积分基础/相关变化率.md deleted file mode 100644 index 2e3fe4e..0000000 --- a/Chapter 4 微积分基础/相关变化率.md +++ /dev/null @@ -1,150 +0,0 @@ -# 飞机航空摄影问题 - -## 问题描述 - -一飞机在离地面2 km的高度，以200 km/h的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 - -## 解析 - -### 建立坐标系 - -目标位置：原点 $O(0,0)$ -飞机位置：$P(x,2)$ -飞机高度：$h = 2$ km -水平速度：$\frac{dx}{dt} = -200$ km/h - -### 角度关系 - -$\theta = \arctan\left(\frac{x}{2}\right)$ -$\tan\theta = \frac{x}{2}$ - -### 角速率计算 - -$\frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}$ - -$\frac{d\theta}{dx} = \frac{d}{dx} \arctan\left(\frac{x}{2}\right) = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} = \frac{2}{x^2 + 4}$ - -$\frac{d\theta}{dt} = \frac{2}{x^2 + 4} \cdot (-200) = -\frac{400}{x^2 + 4}$ rad/h - -### 目标正上方时的角速率 - -当 $x = 0$ 时： - -$\left. \frac{d\theta}{dt} \right|_{x=0} = -\frac{400}{0 + 4} = -100$ rad/h - -转换单位： - -$100$ rad/h $= 100 \times \frac{180}{\pi}$ °/h $= \frac{18000}{\pi}$ °/h - -$\frac{18000}{\pi}$ °/h $\times \frac{1}{3600}$ h/s $= \frac{5}{\pi}$ °/s $\approx 1.59$ °/s - -## 答案 - -飞机飞至目标正上方时，摄影机转动的角速率为 $\frac{5}{\pi}$ °/s（约1.59 °/s）。 - ---- - -# 人拉船问题 - -## 问题描述 - -人在河岸用绳经过定滑轮以速度 $v$、加速度 $a$ 拉船。绳与水平面夹角为 $\theta$。求此时船的加速度 $a'$。 -![[Pasted image 20251223173057.png]] -## 解析 - -### 变量定义 - -$h$：滑轮高度（定值） -$x$：船到河岸的水平距离 -$l$：绳长 -$v = -\frac{dl}{dt}$：人拉绳的速度 -$a = -\frac{d^2l}{dt^2}$：人拉绳的加速度 -$u = -\frac{dx}{dt}$：船的水平速度 -$a' = -\frac{d^2x}{dt^2}$：船的加速度 -$\theta$：绳与水平面夹角 - -### 几何关系 - -$l^2 = h^2 + x^2$ - -### 一阶导数关系 - -对几何关系求导：$2l\frac{dl}{dt} = 2x\frac{dx}{dt}$ - -$l\frac{dl}{dt} = x\frac{dx}{dt}$ - -代入速度定义：$l(-v) = x(-u)$ - -$u = \frac{l}{x}v = \frac{v}{\cos\theta}$，其中 $\cos\theta = \frac{x}{l}$ - -### 二阶导数关系 - -对一阶关系求导：$\frac{d}{dt}\left(l\frac{dl}{dt}\right) = \frac{d}{dt}\left(x\frac{dx}{dt}\right)$ - -$\left(\frac{dl}{dt}\right)^2 + l\frac{d^2l}{dt^2} = \left(\frac{dx}{dt}\right)^2 + x\frac{d^2x}{dt^2}$ - -$(-v)^2 + l(-a) = (-u)^2 + x(-a')$ - -$v^2 - la = u^2 - xa'$ - -### 求解 $a'$ - -$xa' = u^2 - v^2 + la$ - -$a' = \frac{u^2 - v^2 + la}{x}$ - -代入 $u = \frac{v}{\cos\theta}$, $x = h\tan\theta$, $l = \frac{h}{\cos\theta}$： - -$a' = \frac{\left(\frac{v}{\cos\theta}\right)^2 - v^2 + \left(\frac{h}{\cos\theta}\right)a}{h\tan\theta}$ - -$a' = \frac{v^2\left(\frac{1}{\cos^2\theta} - 1\right)}{h\tan\theta} + \frac{\frac{h}{\cos\theta}a}{h\tan\theta}$ - -$a' = \frac{v^2\tan^2\theta}{h\tan\theta} + \frac{a}{\cos\theta\tan\theta}$ - -$a' = \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta}$ - -## 答案 - -此时船的加速度为： -$a' = \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta}$ - ---- - -# 动点曲线运动问题 - -## 问题描述 - -已知动点 $P$ 在曲线 $y = \sqrt{x}$ 上运动，记坐标原点 $O$ 与 $P$ 间的距离为 $l$。若点 $P$ 横坐标随时间的变化率为常数 $v$，则当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率是多少？ - -## 解析 - -### 变量关系 - -点 $P$ 坐标：$(x, y)$, $y = \sqrt{x}$ - -原点 $O$ 与 $P$ 间的距离： -$l = \sqrt{x^2 + y^2} = \sqrt{x^2 + (\sqrt{x})^2} = \sqrt{x^2 + x}$ - -已知：$\frac{dx}{dt} = v$（常数） - -### 距离变化率计算 - -$\frac{dl}{dt} = \frac{d}{dt} \sqrt{x^2 + x} = \frac{d}{dt} (x^2 + x)^{1/2}$ - -$\frac{dl}{dt} = \frac{1}{2}(x^2 + x)^{-1/2} \cdot \frac{d}{dt}(x^2 + x)$ - -$\frac{dl}{dt} = \frac{1}{2\sqrt{x^2 + x}} \cdot (2x + 1) \frac{dx}{dt}$ - -$\frac{dl}{dt} = \frac{2x + 1}{2\sqrt{x^2 + x}} \cdot v$ - -### 在点 $(1, 1)$ 处的变化率 - -当 $x = 1$ 时： - -$\left. \frac{dl}{dt} \right|_{x=1} = \frac{2\cdot 1 + 1}{2\sqrt{1^2 + 1}} \cdot v = \frac{3}{2\sqrt{2}} \cdot v = \frac{3v}{2\sqrt{2}}$ - -有理化：$\frac{3v}{2\sqrt{2}} = \frac{3v\sqrt{2}}{4}$ - -## 答案 - -当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率为 $\frac{3v\sqrt{2}}{4}$。 \ No newline at end of file diff --git a/Chapter 4 微积分基础/隐函数与参数方程求导.md b/Chapter 4 微积分基础/隐函数与参数方程求导.md deleted file mode 100644 index 404a54b..0000000 --- a/Chapter 4 微积分基础/隐函数与参数方程求导.md +++ /dev/null @@ -1,154 +0,0 @@ - - -## 隐函数求导 - -### 原理 - -设方程 $F(x, y) = 0$ 确定 $y$ 是 $x$ 的函数 $y = y(x)$，则对方程两边关于 $x$ 求导（注意 $y$ 是 $x$ 的函数），再解出 $y'$ 即可。 - -### **适用情况** - -适用于方程中 $x$ 与 $y$ 混合在一起，无法或不易解出 $y = f(x)$ 的情况，如： -- 圆的方程：$x^2 + y^2 = 1$ -- 椭圆方程：$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ -- 一般隐式方程：$e^{x+y} + \ln(xy) = 0$ - -### **优势** -1. 不必显式解出 $y = f(x)$，可直接求导； -2. 适用于复杂关系式，尤其是含有 $x$、$y$ 混合的函数形式。 - -### **劣势** -1. 求导过程中需注意 $y$ 是 $x$ 的函数，常需使用链式法则； -2. 最终表达式中可能仍含有 $y$，需结合原方程化简。 - -### **例子** - -> [!example] 例1 -求由方程 $x^2 + y^2 = 1$ 确定的隐函数 $y = y(x)$ 的导数。 - -**解析** -两边对 $x$ 求导： -$$ -2x + 2y \cdot y' = 0 -$$ -解得： -$$ -y' = -\frac{x}{y} -$$ - ---- - -## 参数方程求导 - -### 原理 - -设曲线由参数方程 -$$ -\begin{cases} -x = x(t) \\ -y = y(t) -\end{cases} -$$ -给出，则 $y$ 关于 $x$ 的导数为： -$$ -\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)} -$$ -前提是 $x'(t) \neq 0$。 - -### **适用情况** -适用于曲线由参数形式给出，尤其是： -- 物理中的运动轨迹； -- 极坐标、摆线、旋轮线等曲线； -- 复杂曲线的简化表示。 - -### **优势** -1. 形式简洁，直接利用两个导数作商； -2. 适用于参数化表示，便于计算高阶导数。 - -### **劣势** -1. 要求 $x'(t) \neq 0$，否则导数不存在； -2. 高阶导数计算需重复使用公式，略显繁琐。 - -### **例子** - -> [!example] 例2 -求参数方程 -$$ -\begin{cases} -x = a\cos t \\ -y = b\sin t -\end{cases} -$$ -所确定的函数 $y = y(x)$ 的导数。 - -**解析** -计算： -$$ -\frac{dx}{dt} = -a\sin t, \quad \frac{dy}{dt} = b\cos t -$$ -因此： -$$ -\frac{dy}{dx} = \frac{b\cos t}{-a\sin t} = -\frac{b}{a}\cot t -$$ - ---- - -## 习题 - -> [!example] **习题1** -设 $y$ 由方程 $e^{x+y} = xy$ 确定，则 $y'$ 在点 $(1,0)$ 处的值为 -A. $0$ -B. $1$ -C. $-1$ -D. 不存在 - -**解析** -两边对 $x$ 求导： -$$ -e^{x+y}(1 + y') = y + x y' -$$ -代入 $(1,0)$： -$$ -e^{1+0}(1 + y') = 0 + 1 \cdot y' -$$ -即： -$$ -e(1 + y') = y' -$$ -解得： -$$ -y' = \frac{e}{1 - e} -$$ -该值不为选项中所列，故判断为“不存在直接匹配”，但若只允许选 ABCD，则可能选 D。 -实际应说明：$y' = \frac{e}{1-e}$ 是一个确定数值。 - ---- - -> [!example] **习题2** -参数方程 -$$ -\begin{cases} -x = t^2 + 1 \\ -y = t^3 - t -\end{cases} -$$ -在 $t = 1$ 处的导数 $\left. \frac{dy}{dx} \right|_{t=1}$ 为 -A. $1$ -B. $2$ -C. $3$ -D. $4$ - -**解析** -计算： -$$ -\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 1 -$$ -则： -$$ -\frac{dy}{dx} = \frac{3t^2 - 1}{2t} -$$ -代入 $t = 1$： -$$ -\left. \frac{dy}{dx} \right|_{t=1} = \frac{3 \cdot 1^2 - 1}{2 \cdot 1} = \frac{2}{2} = 1 -$$ -答案：A \ No newline at end of file diff --git a/Chapter 4 微积分基础/隐函数求导、参数方程求导.md b/Chapter 4 微积分基础/隐函数求导、参数方程求导.md deleted file mode 100644 index 5d14d7f..0000000 --- a/Chapter 4 微积分基础/隐函数求导、参数方程求导.md +++ /dev/null @@ -1,136 +0,0 @@ -## 题目 - -### 隐函数求导 - -1. 已知由方程 $e^{(x y - 1)(x + y)} \ln (x^3 + y^3) = \frac{xy(x^2 + y^2)}{(x y)^{x^2 + x y + y^2}} \ln (\sqrt{x + y})$ 所确定的函数单调减，求其在点 $(1,1)$ 处的切线方程。 - -2. 已知由方程 $\arctan\left( \frac{x^2 y + x y^2}{2} \right) = \arcsin\left( \frac{\sqrt{x^2 + y^2}}{x + y} \right)$ 所确定的函数在其连续区间内单调减，求其在点 $(1,1)$ 处的切线方程。 - - - -### 参数方程求导 - -3. 设参数方程 $x = e^{t^2} \arcsin(t^2)$，$y = \ln(1+t^4) \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值。 - -4. 设参数方程 $x = e^{t} \sin t + \ln(1+t^2)$，$y = e^{t} \cos t + \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值。 - ---- - -## 解析 - -### 1. - -##### 解法一（直接求导） - -设 $y = y(x) 、$ 是由方程确定的隐函数，且 $y(1)=1$。两边对 $x$ 求导，并记 $y' = \frac{dy}{dx}$。分别计算左边导数 $\frac{dL}{dx}$ 和右边导数 $\frac{dR}{dx}$ 在点 $(1,1)$ 处的值。 - -左边：$L = e^{(xy-1)(x+y)} \ln(x^3+y^3)$。 -令 $u = (xy-1)(x+y)$，则 $u(1,1)=0$。 -$u' = (y + x y')(x+y) + (xy-1)(1+y')$，在 $(1,1)$ 处，$u' = 2(1+y')$。 -于是 $\frac{dL}{dx} = e^{u} u' \ln(x^3+y^3) + e^{u} \cdot \frac{1}{x^3+y^3} \cdot (3x^2 + 3y^2 y')$ -在 $(1,1)$ 处，$e^{u}=1$，$\ln(x^3+y^3)=\ln 2$，$x^3+y^3=2$，所以 $\frac{dL}{dx}\bigg|_{(1,1)} = 2(1+y') \ln 2 + \frac{3}{2}(1+y') = \left(2\ln 2 + \frac{3}{2}\right)(1+y')$。 - -右边：$R = \frac{1}{2} (xy)^{1-(x^2+xy+y^2)} (x^2+y^2) \ln(x+y)$。 -令 $A = (xy)^{1-(x^2+xy+y^2)}$，$B = x^2+y^2$，$C = \ln(x+y)$，则 $R = \frac{1}{2} A B C$。 -在 $(1,1)$ 处，$A=1$，$B=2$，$C=\ln 2$。 - -先求 $A'$。取对数：$\ln A = [1-(x^2+xy+y^2)] \ln(xy)$。求导： -$\frac{A'}{A} = [-(2x + y + x y' + 2y y')] \ln(xy) + [1-(x^2+xy+y^2)] \cdot \frac{y + x y'}{xy}$ -在 $(1,1)$ 处，$\ln(xy)=0 $，$ 1-(x^2+xy+y^2) = -2 $，$ xy=1$，所以 $\frac{A'}{A} = -2(1+y')$，即 $A' = -2(1+y')$。 -$B' = 2x + 2y y' = 2 + 2y' = 2(1+y')$。 -$C' = \frac{1+y'}{x+y} = \frac{1}{2}(1+y')$。 - -于是 $\frac{dR}{dx} = \frac{1}{2} (A' B C + A B' C + A B C')$ -代入：$\frac{dR}{dx} = \frac{1}{2} \left[ -2(1+y') \cdot 2 \cdot \ln 2 + 1 \cdot 2(1+y') \cdot \ln 2 + 1 \cdot 2 \cdot \frac{1}{2}(1+y') \right]$ -化简：$\frac{dR}{dx} = \frac{1}{2} \left[ -4(1+y') \ln 2 + 2(1+y') \ln 2 + (1+y') \right] = \frac{1}{2} (1+y') (1 - 2\ln 2)$。 - -由 $\frac{dL}{dx} = \frac{dR}{dx}$ 得： -$\left(2\ln 2 + \frac{3}{2}\right)(1+y') = \frac{1}{2} (1+y') (1 - 2\ln 2)$ -若 $1+y' \neq 0$，则 $2\ln 2 + \frac{3}{2} = \frac{1}{2} - \ln 2$，即 $3\ln 2 = -1$，矛盾。故 $1+y' = 0$，即 $y' = -1$。 - -所以切线斜率为 $-1$，切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 - -答案：切线方程为 $y = -x + 2$。 - -##### 解法二 - -由方程形式可知，交换 $x$ 与 $y$ 后方程不变，故曲线关于直线 $y = x$ 对称。点 $(1,1)$ 位于曲线且落在对称轴 $y = x$ 上。 -在对称轴上的点处，切线斜率只可能为 $1$（与对称轴平行）或 $-1$（与对称轴垂直）。 -因为函数单调减，所以在 $(1,1)$ 处导数 $y' < 0$，故斜率不可能为 $1$，只可能为 $-1$。 -由点斜式得切线方程： -$y - 1 = -1 \cdot (x - 1)$， -即 $y = -x + 2$。 - - - - -### 2. -##### 解法一（直接求导） - -对方程两边关于 $x$ 求导，$y$ 视为 $x$ 的函数，记 $y' = \frac{dy}{dx}$。 -左边：设 $u = \frac{x^2 y + x y^2}{2}$，则导数为 $\frac{1}{1+u^2} \cdot \frac{du}{dx}$。 -$\frac{du}{dx} = \frac{1}{2}(2xy + x^2 y' + y^2 + 2xy y')$。 -在 $(1,1)$ 处，$u=1$，$\frac{1}{1+u^2} = \frac{1}{2}$，$\frac{du}{dx} = \frac{1}{2}(2+1 + (1+2)y') = \frac{3}{2}(1+y')$， -所以左边导数为 $\frac{1}{2} \cdot \frac{3}{2}(1+y') = \frac{3}{4}(1+y')$。 - -右边：设 $v = \frac{\sqrt{x^2 + y^2}}{x + y}$，则导数为 $\frac{1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$。 -在 $(1,1)$ 处，$v = \frac{\sqrt{2}}{2}$，$\sqrt{1-v^2} = \frac{1}{\sqrt{2}}$，所以 $\frac{1}{\sqrt{1-v^2}} = \sqrt{2}$。 -计算 $\frac{dv}{dx}$：$v = \frac{\sqrt{x^2+y^2}}{x+y}$， -$\frac{dv}{dx} = \frac{(x+y) \cdot \frac{x + y y'}{\sqrt{x^2+y^2}} - \sqrt{x^2+y^2}(1+y')}{(x+y)^2}$。 -代入 $(1,1)$，$\sqrt{x^2+y^2} = \sqrt{2}$，$x+y=2$，$\frac{dv}{dx} = \frac{2 \cdot \frac{1+y'}{\sqrt{2}} - \sqrt{2}(1+y')}{4} = 0$。 -因此右边导数为 $\sqrt{2} \cdot 0 = 0$。 - -由左右导数相等得：$\frac{3}{4}(1+y') = 0$，解得 $y' = -1$。 - -故切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 - - -##### 解法二 - -方程中交换 $x$ 与 $y$ 后形式不变，故曲线关于直线 $y = x$ 对称。点 $(1,1)$ 在曲线上且位于对称轴 $y = x$ 上。 -由对称性，在对称轴上的点处，切线斜率只可能为 $1$（与对称轴平行）或 $-1$（与对称轴垂直）。 -已知函数单调减，即在定义区间内 $y' < 0$，因此斜率不可能为 $1$，只能为 $-1$。 - -切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 - - ---- - - - ---- - -### 3. - -**解析：** -先求一阶导： -$\frac{dx}{dt} = 2t e^{t^2} \left( \frac{1}{\sqrt{1-t^4}} + \arcsin(t^2) \right)$，$\frac{dy}{dt} = \frac{4t^3}{1+t^4} \arctan(t^3) + \ln(1+t^4) \cdot \frac{3t^2}{1+t^6}$。 -在 $t=0$ 处，$\frac{dx}{dt}=0$，$\frac{dy}{dt}=0$。 - -需用公式 $\frac{d^2 y}{dx^2} = \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$。 -利用泰勒展开： -$x(t) = t^2 + t^4 + O(t^6)$，得 $\frac{dx}{dt} = 2t + 4t^3 + O(t^5)$，$\frac{d^2 x}{dt^2} = 2 + 12t^2 + O(t^4)$； -$y(t) = t^7 + O(t^{11})$，得 $\frac{dy}{dt} = 7t^6 + O(t^{10})$，$\frac{d^2 y}{dt^2} = 42t^5 + O(t^9)$。 -分子：$\frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} = 70 t^6 + O(t^7)$，分母：$(\frac{dx}{dt})^3 = 8 t^3 + O(t^5)$。 -当 $t \to 0$，$\frac{d^2 y}{dx^2} = \frac{70 t^6}{8 t^3} \to 0$，故值为 $0$。 - -**答案：** $\frac{d^2 y}{dx^2} \big|_{t=0} = 0$。 - ---- - -### 4. - -**解析：** -计算一阶导： -$\frac{dx}{dt} = e^{t} \sin t + e^{t} \cos t + \frac{2t}{1+t^2}$，$\frac{dy}{dt} = e^{t} \cos t - e^{t} \sin t + \frac{3t^2}{1+t^6}$。 -在 $t=0$ 处：$\frac{dx}{dt} = 1$，$\frac{dy}{dt} = 1$。 - -二阶导： -$\frac{d^2 x}{dt^2} = 2 e^{t} \cos t + \frac{2 - 2t^2}{(1+t^2)^2}$，$\frac{d^2 y}{dt^2} = -2 e^{t} \sin t + \frac{6t(1+t^6) - 18t^7}{(1+t^6)^2}$。 -在 $t=0$ 处：$\frac{d^2 x}{dt^2} = 4$，$\frac{d^2 y}{dt^2} = 0$。 - -代入公式： -$\frac{d^2 y}{dx^2} = \frac{1 \cdot 0 - 1 \cdot 4}{1^3} = -4$。 - -**答案：** $\frac{d^2 y}{dx^2} \big|_{t=0} = -4$。 ---- diff --git a/Chapter 4 微积分基础/隐函数求导和应用题(1)(1).md b/Chapter 4 微积分基础/隐函数求导和应用题(1)(1).md deleted file mode 100644 index 38cc481..0000000 --- a/Chapter 4 微积分基础/隐函数求导和应用题(1)(1).md +++ /dev/null @@ -1,57 +0,0 @@ -## **隐函数求导** -隐函数求导其实基本没什么难度，主要就是计算问题，所以下面以计算题为主。同时，也要注意包含抽象函数的求导。 ->[!example] 设方程$y=y(x)$由方程$xe^{f(y)}=e^y$确定，其中$f$具有二阶导数，且$f'\neq1$，求$\frac{d^2y}{dx^2}.$ - -解1： -两边对$x$求导得 -$$e^{f(y)}+xy'f'(y)e^{f(y)}=y'e^y$$ -从而 -$$y'=\frac{e^{f(y)}}{e^y-xf'(y)e^{f(y)}}=\frac{e^{f(y)}}{xe^{f(y)}(1-f'(y))}=\frac{1}{x(1-f'(y))}$$ -两边再对$x$求导得 -$$\frac{d^2y}{dx^2}=y''=-\frac{1-f'(y)-xy'f''(y)}{x^2(1-f'(y))^2}=-\frac{(1-f'(y))^2-f''(y)}{x^2(1-f'(y))^3}$$ -解2：两边取对数得$$lnx+f(y)=y$$ -两边对$x$求导得$$\frac{1}{x}+y'f'(y)=y'$$ -从而$$y'=\frac{1}{x(1-f'(y))}$$ -后同解1. - ->[!example] 曲线$\ln{\sqrt{x^2+y^2}}=arctan(\frac{x}{y})$在点(0,1)处的切线方程为$\_\_\_$ ->（A）$y=-x+1$ ->（B）$y=1$ ->（C）$y=x+1$ ->（D）$y=\frac{1}{2}x+1$ - - -解：由原式得$$\frac{1}{2}ln(x^2+y^2)=arctan(\frac{x}{y})$$ -两边对$x$求导得 $$\frac{1}{2}\cdot \frac{2x+2yy'}{x^2+y^2}=\frac{y-xy'}{x^2+y^2}$$$$x+yy'=y-xy'$$$$y'=\frac{y-x}{y+x}$$ -故$y'|_{x=0,y=1}=1$，即曲线在(0,1)处切线的斜率为1，故切线方程为： -$$y=x+1$$选C. - ->[!example] 函数$y=y(x)$由方程$sin(x^2+y^2)+e^x-xy^2=0$所确定，则$\frac{dy}{dx}=\_\_$. - -解： -两边求微分得： -$$(2xdx+2ydy)cos(x^2+y^2)+e^xdx-y^2dx-2xydy=0$$ -$$(2xcos(x^2+y^2)+e^x-y^2)dx+(2ycos(x^2+y^2)-2xy)dy=0$$ -$$\frac{dy}{dx}=\frac{2xcos(x^2+y^2)+e^x-y^2}{2xy-2ycos(x^2+y^2)}$$ - -## **应用题** -应用题就比较综合了，它可以考很多知识点，比如参数方程，比如隐函数求导，也有可能就是一个普通的求导。但关键不在于求导，因为应用题不会有很恶心的表达式，关键在于建立一个准确的模型，也就是（1）把需要求的对象找出来，即自变量和因变量，注意可能有多个；（2）根据条件把关系式列出来。我的建议是大家可以画一画示意图。 ->[!example] 半径为$a$的球渐渐沉入盛有部分水的半径为$b(b>a)$的圆柱形容器中.若球以匀速$c$下沉，求：球浸没一半时，容器内水面上升的速率. ->（A）$\frac{a^2c}{b^2}$ ->（B）$\frac{a^c}{b^2-a^2}$ ->（C）$\frac{a^2c}{2b^2}$ ->（D）$\frac{a^2c}{b^2+a^2}$ - -解：主要的等量关系是：球浸入水中的体积与水“溢出”的体积相同（如果把水面上升视为溢出的话）。设时间为$t$，球下沉的体积为$V$，水上升的高度为$H$，球浸入水的高度为$h$，则有$$V=\pi h^2(a-\frac{h}{3}),dV=(2\pi ah-\pi h^2)dh=\pi h(2a-h)dh$$（这个公式大家可以记住） -且（如图）$$dV=\pi(b^2-a^2+(a-h)^2)dH$$ -![[球浸入水示意图.jpg]] -于是$$\pi(b^2-a^2+(a-h)^2)dH=\pi h(2a-h)dh$$$$dH=\frac{h(2a-h)}{b^2-2ah+h^2}dh$$$$\frac{dH}{dt}=\frac{h(2a-h))}{b^2-2ah+h^2}\frac{dh}{dt}$$而$\frac{dh}{dt}=c,h=a$，故$$\frac{dH}{dt}=\frac{a^2c}{b^2-a^2}$$ -选B. - ->[!example] 利用微分估计值也是导数应用的一种。利用微分估计下面几个式子的值： ->（1）$\sqrt{34}$（保留两位小数） ->（2）摆的振动周期公式为：$T=2\pi \sqrt{\frac{l}{g}}$。其中$l$为摆长（厘米），重力加速度$g=981cm^2/s$.为了使周期$T$增大$0.05$秒，摆长$l=20cm$的长度需要作何修正？（参考数据：$\sqrt{981\times20}\approx140.07,\frac{1}{\pi}\approx0.318$,保留两位小数） - -解：（1）$\sqrt{34}=\sqrt{36-2}=6\sqrt{1-\frac{1}{18}}$,考虑函数$f(x)=\sqrt{x}$，$f'(x)=\frac{1}{2\sqrt{x}},f'(1)=\frac{1}{2}$,故$df=\frac{dx}{2\sqrt{x}},df|_{x=1}=\frac{1}{2}dx$,取$dx=-\frac{1}{18}$,得$df=-\frac{1}{36}$,故$$\sqrt{34}\approx 6(1-\frac{1}{36})=\frac{35}{6}\approx5.83$$ -（2）$dT=T'dl=\pi\sqrt{\frac{1}{gl}}dl=0.05$,则$$dl=\frac{1}{\pi}\sqrt{gl}\approx\frac{1}{3.14}\times\sqrt{981\times20}\times0.05\approx2.23$$ -故需要增加摆长约$2.23$厘米 diff --git a/Chapter 4 微积分基础/隐函数求导和应用题.md b/Chapter 4 微积分基础/隐函数求导和应用题.md deleted file mode 100644 index 338766e..0000000 --- a/Chapter 4 微积分基础/隐函数求导和应用题.md +++ /dev/null @@ -1,47 +0,0 @@ -## **隐函数求导** -隐函数求导其实基本没什么难度，主要就是计算问题，所以下面以计算题为主。同时，也要注意包含抽象函数的求导。 ->[!example] 设方程$y=y(x)$由方程$xe^{f(y)}=e^y$确定，其中$f$具有二阶导数，且$f'\neq1$，求$\frac{d^2y}{dx^2}.$ - -解1： -两边对$x$求导得 -$$e^{f(y)}+xy'f'(y)e^{f(y)}=y'e^y$$ -从而 -$$y'=\frac{e^{f(y)}}{e^y-xf'(y)e^{f(y)}}=\frac{e^{f(y)}}{xe^{f(y)}(1-f'(y))}=\frac{1}{x(1-f'(y))}$$ -两边再对$x$求导得 -$$\frac{d^2y}{dx^2}=y''=-\frac{1-f'(y)-xy'f''(y)}{x^2(1-f'(y))^2}=-\frac{(1-f'(y))^2-f''(y)}{x^2(1-f'(y))^3}$$ -解2：两边取对数得$$lnx+f(y)=y$$ -两边对$x$求导得$$\frac{1}{x}+y'f'(y)=y'$$ -从而$$y'=\frac{1}{x(1-f'(y))}$$ -后同解1. - ->[!example] 求曲线$\ln{\sqrt{x^2+y^2}}=arctan(\frac{x}{y})$在点(0,1)处的切线方程。 - -解：由原式得$$\frac{1}{2}ln(x^2+y^2)=arctan(\frac{x}{y})$$ -两边对$x$求导得 $$\frac{1}{2}\cdot \frac{2x+2yy'}{x^2+y^2}=\frac{y-xy'}{x^2+y^2}$$$$x+yy'=y-xy'$$$$y'=\frac{y-x}{y+x}$$ -故$y'|_{x=0,y=1}=1$，即曲线在(0,1)处切线的斜率为1，故切线方程为： -$$y=x+1$$ - ->[!example] 函数$y=y(x)$由方程$sin(x^2+y^2)+e^x-xy^2=0$所确定，则$\frac{dy}{dx}=\_\_$. - -解： -两边求微分得： -$$(2xdx+2ydy)cos(x^2+y^2)+e^xdx-y^2dx-2xydy=0$$ -$$(2xcos(x^2+y^2)+e^x-y^2)dx+(2ycos(x^2+y^2)-2xy)dy=0$$ -$$\frac{dy}{dx}=\frac{2xcos(x^2+y^2)+e^x-y^2}{2xy-2ycos(x^2+y^2)}$$ - -## **应用题** -应用题就比较综合了，它可以考很多知识点，比如参数方程，比如隐函数求导，也有可能就是一个普通的求导。但关键不在于求导，因为应用题不会有很恶心的表达式，关键在于建立一个准确的模型，也就是（1）把需要求的对象找出来，即自变量和因变量，注意可能有多个；（2）根据条件把关系式列出来。我的建议是大家可以画一画示意图。 ->[!example] 半径为$a$的球渐渐沉入盛有部分水的半径为$b(b>a)$的圆柱形容器中.若球以匀速$c$下沉，求：球浸没一半时，容器内水面上升的速率. - -解：主要的等量关系是：球浸入水中的体积与水“溢出”的体积相同（如果把水面上升视为溢出的话）。设时间为$t$，球下沉的体积为$V$，水上升的高度为$H$，球浸入水的高度为$h$，则有$$V=\pi h^2(a-\frac{h}{3}),dV=(2\pi ah-\pi h^2)dh=\pi h(2a-h)dh$$（这个公式大家可以记住） -且（如图）$$dV=\pi(b^2-a^2+(a-h)^2)dH$$ -![[球浸入水示意图.jpg]] -于是$$\pi(b^2-a^2+(a-h)^2)dH=\pi h(2a-h)dh$$$$dH=\frac{h(2a-h)}{b^2-2ah+h^2}dh$$$$\frac{dH}{dt}=\frac{h(2a-h))}{b^2-2ah+h^2}\frac{dh}{dt}$$而$\frac{dh}{dt}=c,h=a$，故$$\frac{dH}{dt}=\frac{a^2c}{b^2-a^2}$$ - ->[!example] 利用微分估计值也是导数应用的一种。利用微分估计下面几个式子的值： ->（1）$\sqrt{34}$（保留两位小数） ->（2）摆的振动周期公式为：$T=2\pi \sqrt{\frac{l}{g}}$。其中$l$为摆长（厘米），重力加速度$g=981cm^2/s$.为了使周期$T$增大$0.05$秒，摆长$l=20cm$的长度需要作何修正？（参考数据：$\sqrt{981\times20}\approx140.07,\frac{1}{\pi}\approx0.318$,保留两位小数） - -解：（1）$\sqrt{34}=\sqrt{36-2}=6\sqrt{1-\frac{1}{18}}$,考虑函数$f(x)=\sqrt{x}$，$f'(x)=\frac{1}{2\sqrt{x}},f'(1)=\frac{1}{2}$,故$df=\frac{dx}{2\sqrt{x}},df|_{x=1}=\frac{1}{2}dx$,取$dx=-\frac{1}{18}$,得$df=-\frac{1}{36}$,故$$\sqrt{34}\approx 6(1-\frac{1}{36})=\frac{35}{6}\approx5.83$$ -（2）$dT=T'dl=\pi\sqrt{\frac{1}{gl}}dl=0.05$,则$$dl=\frac{1}{\pi}\sqrt{gl}\approx\frac{1}{3.14}\times\sqrt{981\times20}\times0.05\approx2.23$$ -故需要增加摆长约$2.23$厘米 diff --git a/Chapter 4 微积分基础/隐函数求导和应用题（.md b/Chapter 4 微积分基础/隐函数求导和应用题（.md deleted file mode 100644 index ed84d5d..0000000 --- a/Chapter 4 微积分基础/隐函数求导和应用题（.md +++ /dev/null @@ -1,36 +0,0 @@ -## **隐函数求导** -隐函数求导其实基本没什么难度，主要就是计算问题，所以下面以计算题为主。同时，也要注意包含抽象函数的求导。 ->[!example] 设方程$y=y(x)$由方程$xe^{f(y)}=e^y$确定，其中$f$具有二阶导数，且$f'\neq1$，求$\frac{d^2y}{dx^2}.$ - -解1： -两边对$x$求导得 -$$e^{f(y)}+xy'f'(y)e^{f(y)}=y'e^y$$ -从而 -$$y'=\frac{e^{f(y)}}{e^y-xf'(y)e^{f(y)}}=\frac{e^{f(y)}}{xe^{f(y)}(1-f'(y))}=\frac{1}{x(1-f'(y))}$$ -两边再对$x$求导得 -$$\frac{d^2y}{dx^2}=y''=-\frac{1-f'(y)-xy'f''(y)}{x^2(1-f'(y))^2}=-\frac{(1-f'(y))^2-f''(y)}{x^2(1-f'(y))^3}$$ -解2：两边取对数得$$lnx+f(y)=y$$ -两边对$x$求导得$$\frac{1}{x}+y'f'(y)=y'$$ -从而$$y'=\frac{1}{x(1-f'(y))}$$ -后同解1. - - ->[!example] 函数$y=y(x)$由方程$sin(x^2+y^2)+e^x-xy^2=0$所确定，则$\frac{dy}{dx}=\_\_$. - -解： -两边求微分得： -$$(2xdx+2ydy)cos(x^2+y^2)+e^xdx-y^2dx-2xydy=0$$ -$$(2xcos(x^2+y^2)+e^x-y^2)dx+(2ycos(x^2+y^2)-2xy)dy=0$$ -$$\frac{dy}{dx}=\frac{2xcos(x^2+y^2)+e^x-y^2}{2xy-2ycos(x^2+y^2)}$$ - -## **应用题** -应用题就比较综合了，它可以考很多知识点，比如参数方程，比如隐函数求导，也有可能就是一个普通的求导。但关键不在于求导，因为应用题不会有很恶心的表达式，关键在于建立一个准确的模型，也就是（1）把需要求的对象找出来，即自变量和因变量，注意可能有多个；（2）根据条件把关系式列出来。我的建议是大家可以画一画示意图。 - - ->[!example] 利用微分估计值也是导数应用的一种。利用微分估计下面几个式子的值： ->（1）$\sqrt{34}$（保留两位小数） ->（2）摆的振动周期公式为：$T=2\pi \sqrt{\frac{l}{g}}$。其中$l$为摆长（厘米），重力加速度$g=981cm^2/s$.为了使周期$T$增大$0.05$秒，摆长$l=20cm$的长度需要作何修正？（参考数据：$\sqrt{981\times20}\approx140.07,\frac{1}{\pi}\approx0.318$,保留两位小数） - -解：（1）$\sqrt{34}=\sqrt{36-2}=6\sqrt{1-\frac{1}{18}}$,考虑函数$f(x)=\sqrt{x}$，$f'(x)=\frac{1}{2\sqrt{x}},f'(1)=\frac{1}{2}$,故$df=\frac{dx}{2\sqrt{x}},df|_{x=1}=\frac{1}{2}dx$,取$dx=-\frac{1}{18}$,得$df=-\frac{1}{36}$,故$$\sqrt{34}\approx 6(1-\frac{1}{36})=\frac{35}{6}\approx5.83$$ -（2）$dT=T'dl=\pi\sqrt{\frac{1}{gl}}dl=0.05$,则$$dl=\frac{1}{\pi}\sqrt{gl}\approx\frac{1}{3.14}\times\sqrt{981\times20}\times0.05\approx2.23$$ -故需要增加摆长约$2.23$厘米 diff --git a/Chapter 4 微积分基础/隐函数求导（难题）.md b/Chapter 4 微积分基础/隐函数求导（难题）.md deleted file mode 100644 index 908ef38..0000000 --- a/Chapter 4 微积分基础/隐函数求导（难题）.md +++ /dev/null @@ -1,134 +0,0 @@ -## 题目 - -### 隐函数求导 ->[!example] 例1 ->1. 已知由方程 $e^{(x y - 1)(x + y)} \ln (x^3 + y^3) = \frac{xy(x^2 + y^2)}{(x y)^{x^2 + x y + y^2}} \ln (\sqrt{x + y})$ 所确定的函数单调减，求其在点 $(1,1)$ 处的切线方程。 ->2. 已知由方程 $\arctan\left( \frac{x^2 y + x y^2}{2} \right) = \arcsin\left( \frac{\sqrt{x^2 + y^2}}{x + y} \right)$ 所确定的函数在其连续区间内单调减，求其在点 $(1,1)$ 处的切线方程。 - - - -### 参数方程求导 ->[!example] 例2 ->3. 设参数方程 $x = e^{t^2} \arcsin(t^2)$，$y = \ln(1+t^4) \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值（需要用到泰勒展开，仅作习题）。 ->4. 设参数方程 $x = e^{t} \sin t + \ln(1+t^2)$，$y = e^{t} \cos t + \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值。 - ---- - -## 解析 - -### 1. - - 解法一（直接求导） - -设 $y = y(x) 、$ 是由方程确定的隐函数，且 $y(1)=1$。两边对 $x$ 求导，并记 $y' = \frac{dy}{dx}$。分别计算左边导数 $\frac{dL}{dx}$ 和右边导数 $\frac{dR}{dx}$ 在点 $(1,1)$ 处的值。 - -左边：$L = e^{(xy-1)(x+y)} \ln(x^3+y^3)$。 -令 $$u = (xy-1)(x+y),$$则 $u(1,1)=0$。 -有$$u' = (y + x y')(x+y) + (xy-1)(1+y'),$$在 $(1,1)$ 处，$u' = 2(1+y')$。 -于是 $$\frac{dL}{dx} = e^{u} u' \ln(x^3+y^3) + e^{u} \cdot \frac{1}{x^3+y^3} \cdot (3x^2 + 3y^2 y')$$ -在 $(1,1)$ 处，$e^{u}=1$，$\ln(x^3+y^3)=\ln 2$，$x^3+y^3=2$，所以 $\frac{dL}{dx}\bigg|_{(1,1)} = 2(1+y') \ln 2 + \frac{3}{2}(1+y') = \left(2\ln 2 + \frac{3}{2}\right)(1+y')$。 - -右边：$R = \frac{1}{2} (xy)^{1-(x^2+xy+y^2)} (x^2+y^2) \ln(x+y)$。 -令 $$A = (xy)^{1-(x^2+xy+y^2)},B = x^2+y^2,C = \ln(x+y)$$则 $R = \frac{1}{2} A B C$。 -在 $(1,1)$ 处，$A=1$，$B=2$，$C=\ln 2$。 - -先求 $A'$。取对数：$\ln A = [1-(x^2+xy+y^2)] \ln(xy)$。求导得： -$$\frac{A'}{A} = [-(2x + y + x y' + 2y y')] \ln(xy) + [1-(x^2+xy+y^2)] \cdot \frac{y + x y'}{xy}$$ -在 $(1,1)$ 处，$$\ln(xy)=0,1-(x^2+xy+y^2) = -2, xy=1$$，所以 $\frac{A'}{A} = -2(1+y')$，即 $A' = -2(1+y')$。 -易得$$B' = 2x + 2y y' = 2 + 2y' = 2(1+y'), -C' = \frac{1+y'}{x+y} = \frac{1}{2}(1+y').$$ - -于是 $$\frac{dR}{dx} = \frac{1}{2} (A' B C + A B' C + A B C')$$ -代入：$$\frac{dR}{dx} = \frac{1}{2} \left[ -2(1+y') \cdot 2 \cdot \ln 2 + 1 \cdot 2(1+y') \cdot \ln 2 + 1 \cdot 2 \cdot \frac{1}{2}(1+y') \right]$$ -化简：$$\frac{dR}{dx} = \frac{1}{2} \left[ -4(1+y') \ln 2 + 2(1+y') \ln 2 + (1+y') \right] = \frac{1}{2} (1+y') (1 - 2\ln 2)$$ - -由 $\frac{dL}{dx} = \frac{dR}{dx}$ 得： -$$\left(2\ln 2 + \frac{3}{2}\right)(1+y') = \frac{1}{2} (1+y') (1 - 2\ln 2)$$ -若 $1+y' \neq 0$，则 $2\ln 2 + \frac{3}{2} = \frac{1}{2} - \ln 2$，即 $3\ln 2 = -1$，矛盾。故 $1+y' = 0$，即 $y' = -1$。 - -所以切线斜率为 $-1$，切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 - -答案：切线方程为 $y = -x + 2$。 - - 解法二 - -由方程形式可知，交换 $x$ 与 $y$ 后方程不变，故曲线关于直线 $y = x$ 对称。点 $(1,1)$ 位于曲线且落在对称轴 $y = x$ 上。 -在对称轴上的点处，切线斜率只可能为 $1$（与对称轴平行）或 $-1$（与对称轴垂直）。 -因为函数单调减，所以在 $(1,1)$ 处导数 $y' < 0$，故斜率不可能为 $1$，只可能为 $-1$。 -由点斜式得切线方程： -$y - 1 = -1 \cdot (x - 1)$， -即 $y = -x + 2$。 - - - - -### 2. - 解法一（直接求导） - -对方程两边关于 $x$ 求导，$y$ 视为 $x$ 的函数，记 $y' = \frac{dy}{dx}$。 -左边：设 $u = \frac{x^2 y + x y^2}{2}$，则导数为 $\frac{1}{1+u^2} \cdot \frac{du}{dx}$。 -$\frac{du}{dx} = \frac{1}{2}(2xy + x^2 y' + y^2 + 2xy y')$。 -在 $(1,1)$ 处，$$u=1,\frac{1}{1+u^2} = \frac{1}{2}，\frac{du}{dx} = \frac{1}{2}(2+1 + (1+2)y') = \frac{3}{2}(1+y')$$ -所以左边导数为 $$\frac{1}{2} \cdot \frac{3}{2}(1+y') = \frac{3}{4}(1+y')$$。 - -右边：设 $v = \frac{\sqrt{x^2 + y^2}}{x + y}$，则导数为 $\frac{1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$。 -在 $(1,1)$ 处，$v = \frac{\sqrt{2}}{2}$，$\sqrt{1-v^2} = \frac{1}{\sqrt{2}}$，所以 $\frac{1}{\sqrt{1-v^2}} = \sqrt{2}$。 -计算 $\frac{dv}{dx}$：$$v = \frac{\sqrt{x^2+y^2}}{x+y}， -\frac{dv}{dx} = \frac{(x+y) \cdot \frac{x + y y'}{\sqrt{x^2+y^2}} - \sqrt{x^2+y^2}(1+y')}{(x+y)^2}$$ -代入 $(1,1)$，$$\sqrt{x^2+y^2} = \sqrt{2}，x+y=2，\frac{dv}{dx} = \frac{2 \cdot \frac{1+y'}{\sqrt{2}} - \sqrt{2}(1+y')}{4} = 0$$ -因此右边导数为 $\sqrt{2} \cdot 0 = 0$。 - -由左右导数相等得：$\frac{3}{4}(1+y') = 0$，解得 $y' = -1$。 - -故切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 - - - 解法二 - -方程中交换 $x$ 与 $y$ 后形式不变，故曲线关于直线 $y = x$ 对称。点 $(1,1)$ 在曲线上且位于对称轴 $y = x$ 上。 -由对称性，在对称轴上的点处，切线斜率只可能为 $1$（与对称轴平行）或 $-1$（与对称轴垂直）。 -已知函数单调减，即在定义区间内 $y' < 0$，因此斜率不可能为 $1$，只能为 $-1$。 - -切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 - - ---- - - - ---- - -### 3. - -**解析：** -先求一阶导： -$$\frac{dx}{dt} = 2t e^{t^2} \left( \frac{1}{\sqrt{1-t^4}} + \arcsin(t^2) \right)$$$$\frac{dy}{dt} = \frac{4t^3}{1+t^4} \arctan(t^3) + \ln(1+t^4) \cdot \frac{3t^2}{1+t^6}$$ -在 $t=0$ 处，$\frac{dx}{dt}=0$，$\frac{dy}{dt}=0$。 - -需用公式 $$\frac{d^2 y}{dx^2} = \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$$。 -利用泰勒展开： -$x(t) = t^2 + t^4 + O(t^6)$，得 $$\frac{dx}{dt} = 2t + 4t^3 + O(t^5),\frac{d^2 x}{dt^2} = 2 + 12t^2 + O(t^4);$$ -$y(t) = t^7 + O(t^{11})$，得 $$\frac{dy}{dt} = 7t^6 + O(t^{10})，\frac{d^2 y}{dt^2} = 42t^5 + O(t^9).$$ -分子：$$\frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} = 70 t^6 + O(t^7)$$分母：$$(\frac{dx}{dt})^3 = 8 t^3 + O(t^5)$$ -当 $t \to 0$，$\frac{d^2 y}{dx^2} = \frac{70 t^6}{8 t^3} \to 0$，故值为 $0$。 - -**答案：** $\frac{d^2 y}{dx^2} \big|_{t=0} = 0$。 - ---- - -### 4. - -**解析：** -计算一阶导： -$$\frac{dx}{dt} = e^{t} \sin t + e^{t} \cos t + \frac{2t}{1+t^2}，\frac{dy}{dt} = e^{t} \cos t - e^{t} \sin t + \frac{3t^2}{1+t^6}.$$ -在 $t=0$ 处：$\frac{dx}{dt} = 1$，$\frac{dy}{dt} = 1$。 - -二阶导： -$$\frac{d^2 x}{dt^2} = 2 e^{t} \cos t + \frac{2 - 2t^2}{(1+t^2)^2}，\frac{d^2 y}{dt^2} = -2 e^{t} \sin t + \frac{6t(1+t^6) - 18t^7}{(1+t^6)^2}.$$ -在 $t=0$ 处：$$\frac{d^2 x}{dt^2} = 4，\frac{d^2 y}{dt^2} = 0.$$ - -代入公式： -$$\frac{d^2 y}{dx^2} = \frac{1 \cdot 0 - 1 \cdot 4}{1^3} = -4.$$ - -**答案：** $\frac{d^2 y}{dx^2} \big|_{t=0} = -4$。 ---- diff --git a/conflict-files-obsidian-git.md b/conflict-files-obsidian-git.md new file mode 100644 index 0000000..d8bbd4f --- /dev/null +++ b/conflict-files-obsidian-git.md @@ -0,0 +1,17 @@ +# Conflicts +Please resolve them and commit them using the commands `Git: Commit all changes` followed by `Git: Push` +(This file will automatically be deleted before commit) +[[#Additional Instructions]] available below file list + +- Not a file: .obsidian/workspace.json + +# Additional Instructions +I strongly recommend to use "Source mode" for viewing the conflicted files. For simple conflicts, in each file listed above replace every occurrence of the following text blocks with the desired text. + +```diff +<<<<<<< HEAD + File changes in local repository +======= + File changes in remote repository +>>>>>>> origin/main +``` \ No newline at end of file diff --git a/小测/12.23/课前测.md b/小测/12.23/课前测.md deleted file mode 100644 index eac1d11..0000000 --- a/小测/12.23/课前测.md +++ /dev/null @@ -1,33 +0,0 @@ -1.级数$\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$收敛 -A.正确 -B.错误 - - -2.级数$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$的和为 - -A. 0 - -B. 1 - -C. 2 - -D. 不存在（发散） - - - -3.级数 $\displaystyle \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} \cdot \frac{1}{5^n}$ 收敛 -A.正确 -B.错误 - - - - -4.级数$\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{3^n}.$收敛 -A.正确 -B.错误 - - - -5.级数 $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n}$ 收敛 -A.正确 -B.错误 \ No newline at end of file diff --git a/小测/12.23/课前测解析版.md b/小测/12.23/课前测解析版.md deleted file mode 100644 index 87dddc9..0000000 --- a/小测/12.23/课前测解析版.md +++ /dev/null @@ -1,68 +0,0 @@ -1.级数$\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$收敛 -A.正确 -B.错误 -由于 -$a_n = \frac{\sqrt{n+1}-\sqrt{n}}{n} = \frac{1}{n(\sqrt{n+1}+\sqrt{n})} < \frac{1}{n \cdot 2\sqrt{n}} = \frac{1}{2n^{3/2}}$， -而 $\sum_{n=1}^\infty \frac{1}{n^{3/2}}$ 收敛，故由比较判别法知原级数收敛。 - - - -2.级数$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$的和为 - -A. 0 - -B. 1 - -C. 2 - -D. 不存在（发散） -解： - -观察 - -$\frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$ - -部分和： - -$S_N = \sum_{n=1}^N \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right] = 1 - \frac{1}{(N+1)^2} \to 1$ - -收敛到 1。 - - -3.级数 $\displaystyle \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} \cdot \frac{1}{5^n}$ 收敛 -A.正确 -B.错误 -**解析** -$$a_n = \frac{(2n)!}{(n!)^2 \cdot 5^n}$$ - -$$\frac{a_{n+1}}{a_n} - -= \frac{(2n+2)!}{((n+1)!)^2 \cdot 5^{n+1}} \cdot \frac{(n!)^2 \cdot 5^n}{(2n)!}$$ -化简： -$$\frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$$ -$$\frac{(n!)^2}{((n+1)!)^2} = \frac{1}{(n+1)^2}$$ - -$$\frac{5^n}{5^{n+1}} = \frac15$$ -所以 - -$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{(n+1)^2} \cdot \frac15 -= \frac{2(2n+1)}{n+1} \cdot \frac15$$ -$$\lim_{n\to\infty} \frac{4n+2}{5(n+1)} = \frac{4}{5} < 1$$ -所以收敛。 - -答案：A - - - -4.级数$\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{3^n}.$收敛 -A.正确 -B.错误 -$n(n-1)/2的奇偶性与n正好相反，所以级数也是交错级数，容易证明是收敛$ - -5.级数 $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n}$ 收敛 -A.正确 -B.错误 -当 $n > e^2$ 时，$\ln n > 2$，从而 $\frac{1}{(\ln n)^n} < \frac{1}{2^{,n}}$。 -由于 $\sum_{n=2}^\infty \frac{1}{2^{,n}}$ 收敛，由比较判别法知原级数收敛。 - - diff --git a/Chapter 4 微积分基础/二阶微分推导.md b/编写小组/二阶微分推导.md similarity index 100% rename from Chapter 4 微积分基础/二阶微分推导.md rename to 编写小组/二阶微分推导.md diff --git a/编写小组/讲义/二阶微分推导.md b/编写小组/讲义/二阶微分推导.md new file mode 100644 index 0000000..bd35c13 --- /dev/null +++ b/编写小组/讲义/二阶微分推导.md @@ -0,0 +1,55 @@ + +首先提出一个问题： +导数和微分是一个东西吗 +**不是。它们密切相关，但本质不同。** + +### 一阶情形 + +- **导数 (Derivative)** + - **本质**：一个**函数**（或该函数在某点的值） + - **含义**：因变量相对于自变量的**变化率**（瞬时斜率） + - **定义**：$f'(x) = \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ + +- **微分 (Differential)** + - **本质**：一个**表达式**或**量** + - **含义**：函数改变量的**线性主要部分** + - **定义**：$dy = f'(x) \, dx$ + - 其中 $dx$ 是自变量的微分（任意微小改变量） + - $dy$ 是因变量沿切线方向的近似改变量 + +**一阶关系**：$dy = f'(x) \, dx$ + +### 二阶情形 +#### 二阶导数 +- **本质**：一阶导数的导数，仍是一个**函数** +- **符号**：$f''(x)$, $y''$, $\frac{d^2y}{dx^2}$ +- **含义**：函数**变化率的变化率**，描述函数的**凹凸性** +- **定义**：$f''(x) = \frac{d}{dx}[f'(x)]$ + +#### 二阶微分 +- **本质**：微分的微分，是一个**表达式** +- **符号**：$d^2y$ +- + 我们知道：$f''(x) =lim_{\Delta x \rightarrow0} \frac{f'(x+\Delta x) - f'(x)}{dx} = \frac{d^2f(x)}{(dx)^2}$ +现在尝试推导这一公式 +由两函数相乘一阶微分公式：$d[f(x).g(x)] = g(x)d[(fx)] + f(x)d[g(x)]$ +得： + $f''(x) = \frac{d(\frac{df(x)}{dx})}{dx}$ + $= \frac{1}{dx}.\frac{d(df(x))}{dx} + df(x) . \frac{d(\frac{1}{dx})}{dx}$ + $= \frac{d^2f(x)}{(dx)^2} +df(x) . \frac{d(\frac{1}{dx})}{dx}$ +--- + 嘿！您猜怎么着！ + 出现了额外的一项，与二阶导数公式不符： $df(x) . \frac{d(\frac{1}{dx})}{dx}=-df(x) . \frac{1}{d^2x}. \frac{d({dx})}{dx}$ +问题出在 $d(dx)$ 上： + 我们知道，一阶导数的意义是当 $\Delta x$ 趋于无穷小的时候 $\Delta y$ 的变化率 + 实际上，在函数曲线上的任何一点处取函数微分， $dx$ 都是相等的 + 即 $dx$ 永远是均匀的，求高阶微分时将其视为常数 + 所以：$d(dx) = 0$ +### $f''(x) =lim_{\Delta x \rightarrow0} \frac{f'(x+\Delta x) - f'(x)}{dx} = \frac{d^2f(x)}{(dx)^2}= \frac{d^2f(x)}{dx^2}$ + +**注意此处的约定写法**：$(dx)^2$ 写作 $dx^2$（注意：这不是 $d(x^2)$） + + + + + diff --git a/编写小组/讲义/图片/应用题人拉船.png b/编写小组/讲义/图片/应用题人拉船.png new file mode 100644 index 0000000..f5ac61e Binary files /dev/null and b/编写小组/讲义/图片/应用题人拉船.png differ diff --git a/编写小组/讲义/图片/应用题梯子.jpg b/编写小组/讲义/图片/应用题梯子.jpg new file mode 100644 index 0000000..88c36b0 Binary files /dev/null and b/编写小组/讲义/图片/应用题梯子.jpg differ diff --git a/编写小组/讲义/图片/球浸入水示意图.jpg b/编写小组/讲义/图片/球浸入水示意图.jpg new file mode 100644 index 0000000..97de5c1 Binary files /dev/null and b/编写小组/讲义/图片/球浸入水示意图.jpg differ diff --git a/编写小组/讲义/极限计算的基本方法.md b/编写小组/讲义/极限计算的基本方法.md new file mode 100644 index 0000000..86a0f37 --- /dev/null +++ b/编写小组/讲义/极限计算的基本方法.md @@ -0,0 +1,247 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁** + +# 有理化根式法 + +## Quick Description + +### 原理 + +通过平方差公式上下同乘有理化，使得 $\infty - \infty$ 或者 $\frac {0}{0}$ 等不定型变得可以处理，从而使极限可以计算 + +### 适用情况 + +分式中出现根式减根式，或形式本身为根式减根式，且极限过程导致出现不定型，常见的有 $\infty - \infty$ 和 $\frac {0}{0}$ + +### 优势和劣势 + +迅速处理上述问题，但注意一般只能处理含根式问题，若无根式，可能考虑抓大头法 + +一般 $\infty - \infty$ 经过有理化可以得到 $\frac{1}{\infty+\infty}$ + +而 $\frac {0}{0}$ 需要上下都经过有理化，一般能约掉一个因式，从而使其可以处理 + +### 例子 + +> [!example] 例1 +> 计算极限 $\lim_{x \to \infty}(\sqrt{x^{2}+1}-\sqrt{x^{2}-1})$ + + + + + + + +> [!example] 例2 +> 计算极限 $\lim_{x \to 4} \frac{\sqrt{2x+1}-3} {\sqrt{x-2}-\sqrt{2}}$ + + + + + + + + +# 夹逼定理法 + +## Quick Description + +### 原理 + +被“夹”在中间的函数，会被两边函数的共同极限“逼”出相同的极限 + +### 适用情况 + +1.针对形如 $\lim\limits_{n\to\infty}\sum_{k=1}^{n}a_{k}$ 的式子，尤其是每一项 $a_{k}$ 含 $n$ 和 $k$、无法直接求和的情况（或连乘） + +2.对于给定的  $m$  个正数  $a_1, a_2, \dots, a_m ，\lim_{n \to \infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$. + +3.有界函数放缩 + +### 优势和劣势 + +#### 优势 + +1. 处理复杂表达式的有效工具，对于形式复杂、不易直接化简的极限（如 $\sqrt[n]{a_1^n + \cdots + a_m^n}$  这类“混合幂和”） + +2. 不依赖具体函数形式 + +3. 特别适合涉及  $n \to \infty$  的离散极限 + +4. 直观易懂 + +#### 劣势 + +1. 需要合适的不等式构造，必须找到恰当的上下界，既要容易求极限，又要在所求极限点处上下界极限相等。有时不容易找到这样的函数。 + +2. 可能高估或低估范围 + +3. 对振荡型函数有时不适用 + +## 例子 + +### 类型一： + +> [!example] 例1 +> 求极限 +> $$\lim_{n \to \infty} \left( \frac{n+1}{n^2+1} + \frac{n+2}{n^2+2} + \cdots + \frac{n+n}{n^2+n} \right)$$ + + + + + + + + +> [!example] 例2 +> 证明：$$\lim_{n\to\infty} \frac{n!}{n^n} = 0$$ + + + + + + + + + +### 类型二： + +> [!example] 例3 +> 设  $f(x) = \lim_{n \to \infty} \sqrt[n]{1 + x^n + \left( \frac{x^2}{2} \right)^n} ， x \geq 0$，问  $f(x)$  是否为连续函数。 + + + + + + + +> [!example] 例4 +> 证明$\lim_{n\to\infty} \sqrt[n]{3^n + 5^n + 7^n} = 7$ + + + + + + + + +### 类型三： + +> [!example] 例5 +> 求函数极限 $\lim\limits_{x\to0}x^2\sin\frac{1}{x}$ + + + + + + + + +> [!example] 例6 +> 证明：$\lim_{n\to\infty} \sqrt[n]{2 + \sin^2 n} = 1$ + + + + + + + + + +# 凑重要极限法 + +## 原理： + +两个重要极限 +$$(1)\lim\limits_{x \to 0} \tfrac{sinx}{x}=1$$$$(2)\lim\limits_{x \to 0} \tfrac{e^x-1}{x}=1$$ + +## 适用情况 + +x是形式的，也就是说，只要是一个趋于0的函数放在上面两个极限中x的位置，极限的结果仍然不变。首先得是一个除法的形式，如果是乘法，就转化成除以另一个函数的倒数，比如 + +$$\lim\limits_{x \to \infty} x(e^\tfrac{1}{x}-1)=\lim\limits_{x \to \infty } \tfrac{e^\tfrac{1}{x}-1}{\tfrac{1}{x}}=1$$ + +这就是一个常见的化简思路。 + +## 例子 + + +> [!example] 计算下列极限： +> (1)$$\lim\limits_{x \to 0} \tfrac {x-sin2x}{x+sin3x}$$ +> (2)$$\lim\limits_{x \to 0} (\tfrac{1-x}{1+3x})^{1/x}$$ + + + + + + +> [!example] 例2 +> 已知$$\lim\limits_{x\to\infty}(\tfrac{x+1}{x+k})^x=\lim\limits_{x\to0}e^{\tfrac{sin4x}{x}}$$则常数k的值为__ + + + + + + + + + +# 等价无穷小求极限法 + + +本质是利用已知公式进行等价无穷小因子的代换将复杂函数化简 + + +> [!example] 例1 +> 计算 $\displaystyle \lim_{x\to 0}\frac{\sin mx}{\sin nx}$ + + + + + + + +> [!example] 例2 +> 计算 $$\displaystyle \lim_{x\to 0}(\cos x)^{\frac{1+x}{\sin^2 x}}$$ + + + + + + +熟悉常见的极限，灵活运用好极限的四则运算法则和等价无穷小，已经能够解决绝大多数的极限计算题了，无论是简单题还是难题。很多看上去很复杂的所谓难题，无非是在四则运算和等价无穷小之间反复套娃而已，如果可以熟练的分离这些特征，其实不需要泰勒公式等“高级工具”就能快速准确地得出极限值。当然，不是贬低其他的工具，只是说不要学习了所谓的一些高级工具之后，就不重视这些初级的工具和结论。 + + +# 抓大头”求极限法 + +1、形如 + +$$\frac{ax^n + bx^{n-1} + \ldots + p}{cx^m + \beta x^{m-1} + \ldots + \xi}$$ + +的多项式相除：上下同除x的最高次项，再上下各自求极限 + +2、形如 + +$$\ln(ae^{an} + \ldots + p)$$ + +的指对嵌套：先提出e的最高次项，转化为 + +$$\ln e^{\alpha n} \left( a + \frac{\cdots}{e^{\alpha n}} \right)$$ + +即 + +$$\alpha n + \ln \left( a + \frac{\cdots}{e^{\alpha n}} \right)$$ + + +> [!example] 例1 +> 计算$$\lim\limits_{x\to0}\frac{\sqrt{1-4x}-cosx}{x}=$$ + + + + + + diff --git a/编写小组/讲义/极限计算的基本方法（解析版2）.md b/编写小组/讲义/极限计算的基本方法（解析版2）.md new file mode 100644 index 0000000..e91db4c --- /dev/null +++ b/编写小组/讲义/极限计算的基本方法（解析版2）.md @@ -0,0 +1,558 @@ +--- +tags: + - 编写小组 +--- +# 极限计算方法 + +## 一、有理化根式法 + +### 1.1 基本原理 +通过平方差公式上下同乘有理化因式，将 $\infty-\infty$ 或 $\frac{0}{0}$ 等不定型转化为可计算形式。 + +### 1.2 适用情况 +- 分式中出现根式减根式 +- 形式本身为根式减根式 +- 极限过程导致出现不定型（常见 $\infty-\infty$ 和 $\frac{0}{0}$） + +### 1.3 优势与局限 +**优势**：能迅速处理含根式的不定型问题。 + +**局限**：一般只适用于含根式的情况；对于 $\frac{0}{0}$ 型需上下同时有理化，通常能约去零因子。 + +> **🔍 关键转化形式** +> - 对于 $\infty - \infty$ 型：$\sqrt{A} - \sqrt{B} = \frac{A-B}{\sqrt{A}+\sqrt{B}}$ +> - 对于 $\frac{0}{0}$ 型：通常需要分子分母分别有理化 + +### 1.4 典型例题 + +#### 例1：$\infty-\infty$ 型 +计算极限：$\displaystyle \lim_{x \to \infty}(\sqrt{x^{2}+1}-\sqrt{x^{2}-1})$ + +**解**： +$$ +\begin{aligned} +\lim_{x \to \infty}(\sqrt{x^{2}+1}-\sqrt{x^{2}-1}) +&= \lim_{x \to \infty} \frac{(x^{2}+1)-(x^{2}-1)}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}} \\[0.5em] +&= \lim_{x \to \infty} \frac{2}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}} \\[0.5em] +&= 0 +\end{aligned} +$$ + +> **💡 分析**：这是典型的$\infty-\infty$型有理化，有理化后分母趋向$\infty$，整体极限为0。 + +#### 例2：$\frac{0}{0}$ 型 +计算极限：$\displaystyle \lim_{x \to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$ + +**解**： +$$ +\begin{aligned} +\lim_{x \to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} +&= \lim_{x \to 4} \frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)(\sqrt{x-2}+\sqrt{2})}{(\sqrt{x-2}-\sqrt{2})(\sqrt{x-2}+\sqrt{2})(\sqrt{2x+1}+3)} \\[0.5em] +&= \lim_{x \to 4} \frac{(2x-8)(\sqrt{x-2}+\sqrt{2})}{(x-4)(\sqrt{2x+1}+3)} \\[0.5em] +&= \lim_{x \to 4} \frac{2(x-4)(\sqrt{x-2}+\sqrt{2})}{(x-4)(\sqrt{2x+1}+3)} \\[0.5em] +&= \lim_{x \to 4} \frac{2(\sqrt{x-2}+\sqrt{2})}{\sqrt{2x+1}+3} \\[0.5em] +&= \frac{2(\sqrt{2}+\sqrt{2})}{3+3} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} +\end{aligned} +$$ + +> **📝 解题步骤**： +> 1. 识别为$\frac{0}{0}$型，且分子分母均含根式 +> 2. 分子分母分别有理化 +> 3. 化简后约去零因子$(x-4)$ +> 4. 代入$x=4$计算极限值 + +--- + +## 二、夹逼定理法（挤压定理） + +### 2.1 基本原理 +若函数 $f(x)$ 满足 $g(x) \le f(x) \le h(x)$，且 $\lim g(x) = \lim h(x) = L$，则 $\lim f(x) = L$。 + +### 2.2 适用情况 +1. 形如 $\displaystyle \lim_{n\to\infty}\sum_{k=1}^{n}a_k$ 的求和极限（尤其 $a_k$ 含 $n,k$） +2. 幂和根式极限：$\displaystyle \lim_{n \to \infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$ +3. 有界函数乘积型极限 +4. 振荡函数与无穷小乘积的极限 + +### 2.3 优势与劣势 +**优势**： +1. 处理复杂表达式的有效工具 +2. 不依赖具体函数形式 +3. 特别适合涉及 $n \to \infty$ 的离散极限 +4. 直观易懂 + +**劣势**： +1. 需要合适的不等式构造 +2. 必须找到恰当的上下界 +3. 可能高估或低估范围 +4. 对振荡型函数有时不适用 + +### 2.4 三种常见类型及例题 + +#### 类型一：求和式极限 +**例1**：求 $\displaystyle \lim_{n \to \infty} \left( \frac{n+1}{n^2+1} + \frac{n+2}{n^2+2} + \cdots + \frac{n+n}{n^2+n} \right)$ + +**解**： +对 $k=1,2,\dots,n$，有 $n^2+1 \le n^2+k \le n^2+n$，故 +$$ +\frac{n+k}{n^2+n} \le \frac{n+k}{n^2+k} \le \frac{n+k}{n^2+1} +$$ + +求和得： +$$ +\frac{\sum_{k=1}^n (n+k)}{n^2+n} \le S_n \le \frac{\sum_{k=1}^n (n+k)}{n^2+1} +$$ + +其中 $\displaystyle \sum_{k=1}^n (n+k) = n^2 + \frac{n(n+1)}{2} = \frac{3n^2+n}{2}$ + +计算上下极限： +$$ +\lim_{n\to\infty} \frac{\frac{3n^2+n}{2}}{n^2+n} = \lim_{n\to\infty} \frac{3n^2+n}{2n^2+2n} = \frac{3}{2} +$$ +$$ +\lim_{n\to\infty} \frac{\frac{3n^2+n}{2}}{n^2+1} = \lim_{n\to\infty} \frac{3n^2+n}{2n^2+2} = \frac{3}{2} +$$ + +由夹逼定理得：$\displaystyle \lim_{n\to\infty} S_n = \frac{3}{2}$ + +**例2**：证明 $\displaystyle \lim_{n\to\infty} \frac{n!}{n^n} = 0$ + +**证明**： +$$ +0 < \frac{n!}{n^n} = \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n}{n} \le \frac{1}{n} \cdot 1 \cdot 1 \cdots 1 = \frac{1}{n} +$$ +而 $\displaystyle \lim_{n\to\infty} \frac{1}{n} = 0$，故 $\displaystyle \lim_{n\to\infty} \frac{n!}{n^n} = 0$ + +> **🔍 技巧**：对于连乘积，常将部分因子放大为1，部分因子保持不变，得到易求极限的上下界。 + +#### 类型二：幂和根式极限 +**重要结论**：若 $a_k > 0$，则 +$$\lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$$ + +**例3**：设 $\displaystyle f(x) = \lim_{n \to \infty} \sqrt[n]{1 + x^n + \left( \frac{x^2}{2} \right)^n},\ x \ge 0$，讨论 $f(x)$ 连续性。 + +**解**： +比较三项：$a_1=1,\ a_2=x,\ a_3=\frac{x^2}{2}$ + +1. **当 $0 \le x \le 1$**： + $x \le 1$，且 $\frac{x^2}{2} \le \frac{1}{2} < 1$，最大值为 $1$ + +2. **当 $1 < x \le 2$**： + 比较 $x$ 与 $\frac{x^2}{2}$： + $$x \ge \frac{x^2}{2} \Leftrightarrow 2x \ge x^2 \Leftrightarrow x(2-x) \ge 0 \Leftrightarrow 0 \le x \le 2$$ + 在 $(1, 2]$ 内，$x \ge \frac{x^2}{2}$；同时 $x > 1$，所以最大值是 $x$ + +3. **当 $x > 2$**： + 比较 $x$ 与 $\frac{x^2}{2}$： + $\frac{x^2}{2} > x \ (x > 2)$，且 $\frac{x^2}{2} > 1$，所以最大值是 $\frac{x^2}{2}$ + +故 +$$ +f(x) = \begin{cases} +1, & 0 \le x \le 1 \\ +x, & 1 < x \le 2 \\ +\frac{x^2}{2}, & x > 2 +\end{cases} +$$ + +**连续性检查**： +- 在 $x=1$ 处：$x \to 1^-$ 时 $f(x)=1$；$x \to 1^+$ 时 $f(x)=x \to 1$，且 $f(1)=1$，连续 +- 在 $x=2$ 处：$x \to 2^-$ 时 $f(x)=x \to 2$；$x \to 2^+$ 时 $f(x)=\frac{x^2}{2} \to 2$，且 $f(2)=2$，连续 + +因此 $f(x)$ 在 $[0,+\infty)$ 上连续。 + +**例4**：证明 $\displaystyle \lim_{n\to\infty} \sqrt[n]{3^n + 5^n + 7^n} = 7$ + +**证明**： +$$ +7^n < 3^n + 5^n + 7^n \le 3 \cdot 7^n +$$ +两边开 $n$ 次方： +$$ +7 < \sqrt[n]{3^n + 5^n + 7^n} \le 7 \cdot \sqrt[n]{3} +$$ +而 $\displaystyle \lim_{n\to\infty} \sqrt[n]{3} = 1$，由夹逼定理得证。 + +> **📝 解题模板**： +> 1. 找出最大项 $M = \max\{a_1, a_2, \dots, a_m\}$ +> 2. 构造不等式：$M^n < \sum a_i^n \le m \cdot M^n$ +> 3. 取$n$次根：$M < \sqrt[n]{\sum a_i^n} \le M \cdot \sqrt[n]{m}$ +> 4. 由 $\lim_{n\to\infty} \sqrt[n]{m} = 1$ 得结论 + +#### 类型三：有界函数乘积型 +**例5**：求 $\displaystyle \lim_{x\to0}x^2\sin\frac{1}{x}$ + +**解**： +由 $-1 \le \sin\frac{1}{x} \le 1$，得 +$$ +-x^2 \le x^2\sin\frac{1}{x} \le x^2 +$$ +而 $\displaystyle \lim_{x\to0}(-x^2) = \lim_{x\to0}x^2 = 0$，故原极限为 $0$。 + +**例6**：证明 $\displaystyle \lim_{n\to\infty} \sqrt[n]{2 + \sin^2 n} = 1$ + +**证明**： +$$ +1 \le 2 + \sin^2 n \le 3 \Rightarrow +1 \le \sqrt[n]{2 + \sin^2 n} \le \sqrt[n]{3} +$$ +而 $\displaystyle \lim_{n\to\infty} \sqrt[n]{3} = 1$，故得证。 + +> **⚠️ 注意事项**： +> 1. 确保上下界函数极限存在且相等 +> 2. 不等式方向必须一致 +> 3. 对于振荡函数，常利用其有界性 + +--- + +## 三、凑重要极限法 + +### 3.1 两个重要极限 +1. $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$ +2. $\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} = 1$ + +### 3.2 变形与推广 +- $\displaystyle \lim_{\square \to 0} \frac{\sin \square}{\square} = 1$（$\square$代表任何趋于0的表达式） +- $\displaystyle \lim_{\square \to 0} \frac{e^\square-1}{\square} = 1$ +- $\displaystyle \lim_{\square \to 0} (1+\square)^{1/\square} = e$ + +### 3.3 使用技巧 +1. **识别结构**：寻找$\frac{\sin(\text{无穷小})}{\text{相同的无穷小}}$或$\frac{e^{\text{无穷小}}-1}{\text{相同的无穷小}}$的形式 +2. **变量代换**：令$t = \text{无穷小表达式}$，化为标准形式 +3. **恒等变形**：利用对数、指数等恒等式转化 + +### 3.4 典型例题 + +**例1**： +(1) $\displaystyle \lim_{x \to 0} \frac{x-\sin 2x}{x+\sin 3x}$ + +**解**： +$$ +\lim_{x\to0}\frac{x-\sin 2x}{x+\sin 3x} += \lim_{x\to0}\frac{1-2\cdot\frac{\sin 2x}{2x}}{1+3\cdot\frac{\sin 3x}{3x}} += \frac{1-2\times1}{1+3\times1} = -\frac{1}{4} +$$ + +(2) $\displaystyle \lim_{x \to 0} \left(\frac{1-x}{1+3x}\right)^{1/x}$ + +**解**： +令 $t = -\frac{4x}{1+3x}$，则 $x \to 0$ 时 $t \to 0$，且 +$$ +\frac{1}{x} = -4\cdot\frac{1+3x}{4x} = -4\cdot\frac{1}{t}-3 +$$ + +原式 $= \displaystyle \lim_{t\to 0} (1+t)^{-4/t-3} = \lim_{t\to 0} \left[(1+t)^{1/t}\right]^{-4} \cdot (1+t)^{-3} = e^{-4} \cdot 1 = e^{-4}$ + +**例2**：已知 $\displaystyle \lim_{x\to\infty}\left(\frac{x+1}{x+k}\right)^x = \lim_{x\to0}e^{\frac{\sin 4x}{x}}$，求 $k$ + +**解**： +左边 $= \displaystyle \lim_{x\to\infty}\frac{(1+1/x)^x}{(1+k/x)^x} = \frac{e}{e^k} = e^{1-k}$ + +右边 $= e^{\lim_{x\to0}\frac{\sin 4x}{x}} = e^{4}$ + +故 $e^{1-k} = e^4 \Rightarrow 1-k = 4 \Rightarrow k = -3$ + +### 3.5 课后练习与详解 + +**练习1**：$\displaystyle \lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{3/x}\ (a,b>0)$ + +**解**： +$$ +\begin{aligned} +\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{3/x} +&= \lim_{x\to0}\left(1 + \frac{a^x+b^x-2}{2}\right)^{3/x} \\ +&= \lim_{x\to0}\left(1 + \frac{(a^x-1)+(b^x-1)}{2}\right)^{3/x} \\ +&= \lim_{x\to0}\left(1 + \frac{x\ln a + x\ln b + o(x)}{2}\right)^{3/x} \quad (\text{利用 } a^x-1 \sim x\ln a) \\ +&= \lim_{x\to0}\left(1 + \frac{\ln(ab)}{2}x\right)^{3/x} \\ +&= e^{\frac{3\ln(ab)}{2}} = (ab)^{3/2} +\end{aligned} +$$ + +**练习2**：$\displaystyle \lim_{x\to\infty}\frac{3x^2+5}{5x+3}\sin\frac{2}{x}$ + +**解**： +$$ +\begin{aligned} +\lim_{x\to\infty}\frac{3x^2+5}{5x+3}\sin\frac{2}{x} +&= \lim_{x\to\infty}\frac{3x^2+5}{5x+3} \cdot \frac{\sin\frac{2}{x}}{\frac{2}{x}} \cdot \frac{2}{x} \\ +&= \lim_{x\to\infty}\frac{3x^2+5}{5x+3} \cdot 1 \cdot \frac{2}{x} \\ +&= \lim_{x\to\infty}\frac{6x^2+10}{5x^2+3x} = \frac{6}{5} +\end{aligned} +$$ + +**练习3**：$\displaystyle \lim_{x\to\pi/4}\frac{\tan x-1}{x-\pi/4}$ + +**解**： +令 $t = x - \frac{\pi}{4}$，则 $x = t + \frac{\pi}{4},\ t \to 0$ + +$\tan x = \tan\left(t + \frac{\pi}{4}\right) = \frac{\tan t + 1}{1 - \tan t}$ + +于是： +$$ +\begin{aligned} +\tan x - 1 &= \frac{\tan t + 1}{1 - \tan t} - 1 = \frac{2\tan t}{1 - \tan t} \\ +\frac{\tan x-1}{x-\pi/4} &= \frac{2\tan t}{t(1-\tan t)} = 2\cdot\frac{\tan t}{t}\cdot\frac{1}{1-\tan t} +\end{aligned} +$$ + +当 $t \to 0$ 时：$\frac{\tan t}{t} \to 1$，$1-\tan t \to 1$ + +故原极限 $= 2 \times 1 \times 1 = 2$ + +**练习4**：$\displaystyle \lim_{x\to\pi/2}(\sin x)^{\tan x}$ + +**解**： +令 $t = \frac{\pi}{2} - x$，则 $x = \frac{\pi}{2} - t,\ t \to 0$ + +$\sin x = \cos t \sim 1 - \frac{t^2}{2}$ +$\tan x = \cot t = \frac{\cos t}{\sin t} \sim \frac{1}{t}$ + +原式 $= \displaystyle \lim_{t\to0} (\cos t)^{\cot t}$ + +取对数：$\ln(\text{原式}) = \displaystyle \lim_{t\to0} \cot t \cdot \ln(\cos t)$ + +$$ +\cot t \cdot \ln(\cos t) = \frac{\cos t}{\sin t} \cdot \ln(1 - \frac{t^2}{2} + o(t^2)) \sim \frac{1}{t} \cdot (-\frac{t^2}{2}) = -\frac{t}{2} \to 0 +$$ + +故原式 $= e^0 = 1$ + +> **💡 重要技巧**： +> 1. 幂指函数 $[f(x)]^{g(x)}$ 常取对数：$e^{g(x)\ln f(x)}$ +> 2. 利用等价无穷小 $\ln(1+u) \sim u\ (u\to0)$ +> 3. 注意代换后的新变量也趋于0，才能使用重要极限 + +--- + +## 四、等价无穷小代换法 + +### 4.1 基本原理 +在乘积和商的极限运算中，可以将复杂的无穷小量替换为等价的简单无穷小量，简化计算。 + +### 4.2 常用等价无穷小（$x\to 0$） +| 等价形式 | 条件 | +|---------|------| +| $x \sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x$ | 基础等价 | +| $x \sim e^x-1 \sim \ln(1+x)$ | 指数对数等价 | +| $1-\cos x \sim \frac{1}{2}x^2$ | 三角函数等价 | +| $(1+x)^a-1 \sim ax$ | 幂函数等价 | +| $a^x-1 \sim x\ln a\ (a>0)$ | 指数函数等价 | +| $x - \sin x \sim \frac{1}{6}x^3$ | 高阶等价 | +| $\tan x - x \sim \frac{1}{3}x^3$ | 高阶等价 | + +### 4.3 使用原则 +1. **乘除运算可直接替换** +2. **加减运算需谨慎**：只有同阶无穷小相加减时，不能随意替换 +3. **复合函数可整体替换** + +### 4.4 典型例题 + +**例1**：计算 $\displaystyle \lim_{x\to 0}\frac{\sin mx}{\sin nx}$ + +**解**： +原式 $= \displaystyle \lim_{x\to 0}\frac{mx}{nx} = \frac{m}{n}$ + +**例2**：计算 $\displaystyle \lim_{x\to 0}(\cos x)^{\frac{1+x}{\sin^2 x}}$ + +**解**：取对数得 +$$ +\begin{aligned} +\ln L &= \lim_{x\to 0}\frac{1+x}{\sin^2 x}\ln(\cos x) \\ +&= \lim_{x\to 0}\frac{1+x}{x^2} \cdot \ln[1+(\cos x-1)] \\ +&= \lim_{x\to 0}\frac{1+x}{x^2} \cdot (\cos x-1) \quad (\text{利用 }\ln(1+u)\sim u) \\ +&= \lim_{x\to 0}\frac{1+x}{x^2} \cdot (-\frac{1}{2}x^2) \quad (\text{利用 }1-\cos x\sim\frac{1}{2}x^2) \\ +&= -\frac{1}{2}\lim_{x\to 0}(1+x) = -\frac{1}{2} +\end{aligned} +$$ +故原式 $= e^{-1/2}$ + +**例3**：计算 $\displaystyle \lim_{n\to\infty} n^2[\arctan(n+1) - \arctan n]$ + +**解**： +利用公式 $\arctan a - \arctan b = \arctan\frac{a-b}{1+ab}$： +$$ +\begin{aligned} +\text{原式} &= \lim_{n \to \infty} n^2 \arctan\frac{(n+1)-n}{1+(n+1)n} \\ +&= \lim_{n \to \infty} n^2 \arctan\frac{1}{n^2+n+1} \\ +&= \lim_{n \to \infty} n^2 \cdot \frac{1}{n^2+n+1} \quad (\text{利用 }\arctan u\sim u\ (u\to0)) \\ +&= \lim_{n \to \infty} \frac{n^2}{n^2+n+1} = 1 +\end{aligned} +$$ + +### 4.5 注意事项与常见错误 + +**⚠️ 加减法替换的陷阱**： +已知 $a^x - 1 \sim x\ln a$；$x \sim \sin x$，但 +$$ +\lim_{x\to 0}\frac{a^x - 1 - x\ln a}{\sin x - x} +$$ +不能直接替换为 $\frac{0}{0}$！ + +**正确解法（泰勒展开）**： +分子：$a^x - 1 - x\ln a = \left(1 + x\ln a + \frac{(x\ln a)^2}{2} + o(x^2)\right) - 1 - x\ln a = \frac{(x\ln a)^2}{2} + o(x^2)$ + +分母：$\sin x - x = -\frac{x^3}{6} + o(x^3)$ + +因此： +$$ +\frac{a^x - 1 - x\ln a}{\sin x - x} \sim \frac{\frac{(x\ln a)^2}{2}}{-\frac{x^3}{6}} = -3\frac{(\ln a)^2}{x} +$$ +当 $x\to 0^+$ 时 $\to -\infty$，$x\to 0^-$ 时 $\to +\infty$，极限不存在。 + +### 4.6 引入练习详解 + +计算 $\displaystyle \lim_{x\to0}\frac{\sqrt{1+2\tan x}-\sqrt{1+2\sin x}}{x\ln(1+x)-x^2}$ + +**解**： +$$ +\begin{aligned} +\text{原式} &= \lim_{x\to0}\frac{(\sqrt{1+2\tan x}-\sqrt{1+2\sin x})(\sqrt{1+2\tan x}+\sqrt{1+2\sin x})}{(x\ln(1+x)-x^2)(\sqrt{1+2\tan x}+\sqrt{1+2\sin x})} \\ +&= \lim_{x\to0}\frac{2(\tan x - \sin x)}{x[\ln(1+x)-x] \cdot 2} \quad (\text{分母中的}\sqrt{\cdots}+\sqrt{\cdots}\to 2) \\ +&= \lim_{x\to0}\frac{\tan x - \sin x}{x[\ln(1+x)-x]} \\ +&= \lim_{x\to0}\frac{\sin x(\frac{1}{\cos x}-1)}{x[\ln(1+x)-x]} \\ +&= \lim_{x\to0}\frac{\sin x}{x} \cdot \frac{1}{\cos x} \cdot \frac{1-\cos x}{\ln(1+x)-x} \\ +&= 1 \cdot 1 \cdot \lim_{x\to0}\frac{\frac{1}{2}x^2}{\ln(1+x)-x} +\end{aligned} +$$ + +对于 $\displaystyle \lim_{x\to0}\frac{\frac{1}{2}x^2}{\ln(1+x)-x}$： +利用 $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$ +得 $\ln(1+x)-x = -\frac{x^2}{2} + \frac{x^3}{3} + o(x^3) \sim -\frac{x^2}{2}$ + +因此： +$$ +\lim_{x\to0}\frac{\frac{1}{2}x^2}{\ln(1+x)-x} = \lim_{x\to0}\frac{\frac{1}{2}x^2}{-\frac{1}{2}x^2} = -1 +$$ + +故原式 $= -1$，答案为 **C**。 + +> **📋 等价无穷小使用总结**： +> 1. 熟记常用等价无穷小公式 +> 2. 乘除运算放心用，加减运算要小心 +> 3. 复杂表达式可先有理化、拆分再使用 +> 4. 不确定时可用泰勒展开验证 + +--- + +## 五、"抓大头"法（主导项法） + +### 5.1 基本原理 +在极限计算中，忽略低阶项，只保留最高阶项（主导项）。 + +### 5.2 适用情形 +1. **有理分式极限**：分子分母同除以最高次项 +2. **指对嵌套式**：提出指数最高次项 +3. **多项式与根式混合**：比较各项阶数 + +### 5.3 常见类型及方法 + +#### 类型一：有理分式 +对于 $\displaystyle \lim_{x\to\infty}\frac{a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0}{b_mx^m + b_{m-1}x^{m-1} + \cdots + b_0}$： +- $n > m$：极限为 $\infty$ 或 $-\infty$（由系数符号决定） +- $n = m$：极限为 $\frac{a_n}{b_m}$ +- $n < m$：极限为 $0$ + +#### 类型二：指数型 +对于 $\displaystyle \lim_{n\to\infty}\ln(ae^{\alpha n} + be^{\beta n} + \cdots)$，其中 $\alpha > \beta > \cdots$： +提取 $e^{\alpha n}$：$= \alpha n + \ln(a + be^{(\beta-\alpha)n} + \cdots) \to \alpha n + \ln a$ + +### 5.4 典型例题 + +**例1**：计算 $\displaystyle \lim_{x\to0}\frac{\sqrt{1-4x}-\cos x}{x}$ + +**解**： +使用等价无穷小： +$\sqrt{1-4x} = 1 - 2x - \frac{(4x)^2}{8} + o(x^2) = 1 - 2x - 2x^2 + o(x^2)$ +$\cos x = 1 - \frac{x^2}{2} + o(x^2)$ + +因此： +$$ +\sqrt{1-4x} - \cos x = (1 - 2x - 2x^2) - (1 - \frac{x^2}{2}) + o(x^2) = -2x - \frac{3}{2}x^2 + o(x^2) +$$ + +原式 $= \displaystyle \lim_{x\to0}\frac{-2x - \frac{3}{2}x^2 + o(x^2)}{x} = \lim_{x\to0}(-2 - \frac{3}{2}x + o(x)) = -2$ + +**例2**：计算 $\displaystyle \lim_{x\to\infty}\frac{3x^3 - 2x^2 + 5}{2x^3 + 7x - 1}$ + +**解**： +分子分母同除以 $x^3$： +$$ +\lim_{x\to\infty}\frac{3 - \frac{2}{x} + \frac{5}{x^3}}{2 + \frac{7}{x^2} - \frac{1}{x^3}} = \frac{3}{2} +$$ + +**练习**：$\displaystyle \lim_{x \to 0} \frac{\csc x - \cot x}{x}$ + +**解**： +$$ +\csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x} \sim \frac{\frac{1}{2}x^2}{x} = \frac{x}{2} +$$ + +故原式 $= \displaystyle \lim_{x\to0}\frac{x/2}{x} = \frac{1}{2}$ + +### 5.5 综合练习 + +**练习1**：$\displaystyle \lim_{x\to\infty}\frac{\sqrt{4x^2+3x+1} - \sqrt{4x^2-2x+5}}{x}$ + +**解**： +有理化： +$$ +\begin{aligned} +\text{原式} &= \lim_{x\to\infty}\frac{(4x^2+3x+1) - (4x^2-2x+5)}{x(\sqrt{4x^2+3x+1} + \sqrt{4x^2-2x+5})} \\ +&= \lim_{x\to\infty}\frac{5x-4}{x(\sqrt{4x^2+3x+1} + \sqrt{4x^2-2x+5})} \\ +&= \lim_{x\to\infty}\frac{5 - \frac{4}{x}}{\sqrt{4+\frac{3}{x}+\frac{1}{x^2}} + \sqrt{4-\frac{2}{x}+\frac{5}{x^2}}} \\ +&= \frac{5}{\sqrt{4}+\sqrt{4}} = \frac{5}{2+2} = \frac{5}{4} +\end{aligned} +$$ + +> **💡 "抓大头"的本质**： +> 在 $x\to\infty$ 时，$x^3$ 比 $x^2$ "大得多"；在 $x\to0$ 时，$x^2$ 比 $x$ "小得多"。抓住这些主导项，忽略次要项，是快速求解极限的关键。 + +--- + +## 六、方法综合运用与总结 + +### 6.1 极限计算流程图 +``` +开始 + ↓ +判断极限类型 + ├── 0/0型 → 考虑：有理化、等价无穷小、洛必达 + ├── ∞/∞型 → 考虑：抓大头、洛必达 + ├── ∞-∞型 → 考虑：有理化、通分 + ├── 0·∞型 → 化为0/0或∞/∞ + └── 1^∞, 0^0, ∞^0型 → 取对数化为0/0型 + ↓ +选择合适方法 + ↓ +计算并验证 +``` + +### 6.2 方法选择指南 +| 方法 | 最佳适用场景 | 注意事项 | +|------|------------|----------| +| 有理化根式法 | 含根式的$\infty-\infty$或$0/0$型 | 注意分子分母分别有理化 | +| 夹逼定理法 | 求和、幂和根式、有界函数乘积 | 构造合适的不等式是关键 | +| 凑重要极限法 | 含$\sin x/x$或$(1+1/x)^x$结构 | 灵活使用变量代换 | +| 等价无穷小法 | 乘除运算中的复杂无穷小 | 加减运算中慎用 | +| 抓大头法 | $x\to\infty$的有理分式或指对式 | 明确各项阶数关系 | + +### 6.3 常见错误提醒 +1. **滥用等价无穷小**：在加减运算中随意替换 +2. **忽略定义域**：如$\sqrt{x}$在$x<0$时无定义 +3. **错误使用洛必达**：不满足$\frac{0}{0}$或$\frac{\infty}{\infty}$型 +4. **夹逼定理使用不当**：上下界极限不相等 +5. **重要极限变形错误**：$\lim (1+\frac{1}{x})^x = e$ 仅当$x\to\infty$ + +### 6.4 终极建议 +1. **先化简后计算**：有理化、通分、因式分解等 +2. **多方法验证**：用不同方法计算同一极限，验证结果 +3. **掌握泰勒展开**：这是极限计算的终极武器 +4. **积累常见结论**：如$\lim_{n\to\infty}\sqrt[n]{n}=1$，$\lim_{x\to0}\frac{\tan x - x}{x^3}=\frac{1}{3}$等 + +> **🚀 提升路径**： +> 1. 熟练掌握本文五种基本方法 +> 2. 学习洛必达法则（适用于0/0和∞/∞型） +> 3. 掌握泰勒公式展开（最强大的工具） +> 4. 练习综合题，培养方法选择直觉 + +**记住**：极限计算的核心思想是 **"化未知为已知，化复杂为简单"**。通过恒等变形、变量代换、等价替换等手段，将陌生的极限转化为熟悉的形式，再利用已知结论求解。 \ No newline at end of file diff --git a/编写小组/讲义/极限计算的基本方法（解析版）.md b/编写小组/讲义/极限计算的基本方法（解析版）.md new file mode 100644 index 0000000..b49dde7 --- /dev/null +++ b/编写小组/讲义/极限计算的基本方法（解析版）.md @@ -0,0 +1,522 @@ +--- +tags: + - tool_idea + - 编写小组 +--- +# 有理化根式法 + +## Quick Description + +### 原理 + +通过平方差公式上下同乘有理化，使得 $\infty - \infty$ 或者 $\frac {0}{0}$ 等不定型变得可以处理，从而使极限可以计算 + +### 适用情况 + +分式中出现根式减根式，或形式本身为根式减根式，且极限过程导致出现不定型，常见的有 $\infty - \infty$ 和 $\frac {0}{0}$ + +### 优势和劣势 + +迅速处理上述问题，但注意一般只能处理含根式问题，若无根式，可能考虑抓大头法 + +一般 $\infty - \infty$ 经过有理化可以得到 $\frac{1}{\infty+\infty}$ + +而 $\frac {0}{0}$ 需要上下都经过有理化，一般能约掉一个因式，从而使其可以处理 + +### 例子 + +> [!example] 例1 +> 计算极限 $\lim_{x \to \infty}(\sqrt{x^{2}+1}-\sqrt{x^{2}-1})$ + +观察到这里是根式减根式，并且是不定型 $\infty - \infty$，考虑使用有理化根式法 + +则有$$\begin{align*} +\lim_{x \to \infty}(\sqrt{x^{2}+1}-\sqrt{x^{2}-1}) +&= \lim_{x \to \infty} \frac{2} {\sqrt{x^{2}+1}+\sqrt{x^{2}-1}} \\[1em] +&= 0 +\end{align*}$$ +> [!example] 例2 +> 计算极限 $\lim_{x \to 4} \frac{\sqrt{2x+1}-3} {\sqrt{x-2}-\sqrt{2}}$ + +观察到这里是根式减根式，并且是不定型 $\frac {0}{0}$ ，考虑使用有理化根式法 + +则有$$\begin{align*} +\lim_{x \to 4} \frac{\sqrt{2x+1}-3} {\sqrt{x-2}-\sqrt{2}} +&= \lim_{x \to 4} \frac{(\sqrt{x-2}+\sqrt{2}) } {x-4} \frac{2(x-4)}{ (\sqrt{2x+1}+3) } \\[1em] +&= 2\lim_{x \to 4} \frac{(\sqrt{x-2}+\sqrt{2}) } {\sqrt{2x+1}+3}\\[1em] +&= \frac{2\sqrt{2}}{3} +\end{align*}$$ +# 夹逼定理法 + +## Quick Description + +### 原理 + +被“夹”在中间的函数，会被两边函数的共同极限“逼”出相同的极限 + +### 适用情况 + +1.针对形如 $\lim\limits_{n\to\infty}\sum_{k=1}^{n}a_{k}$ 的式子，尤其是每一项 $a_{k}$ 含 $n$ 和 $k$、无法直接求和的情况（或连乘） + +2.对于给定的  $m$  个正数  $a_1, a_2, \dots, a_m ，\lim_{n \to \infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$. + +3.有界函数放缩 + +### 优势和劣势 + +#### 优势 + +1. 处理复杂表达式的有效工具，对于形式复杂、不易直接化简的极限（如 $\sqrt[n]{a_1^n + \cdots + a_m^n}$  这类“混合幂和”） + +2. 不依赖具体函数形式 + +3. 特别适合涉及  $n \to \infty$  的离散极限 + +4. 直观易懂 + +#### 劣势 + +1. 需要合适的不等式构造，必须找到恰当的上下界，既要容易求极限，又要在所求极限点处上下界极限相等。有时不容易找到这样的函数。 + +2. 可能高估或低估范围 + +3. 对振荡型函数有时不适用 + +## 例子 + +### 类型一： + +> [!example] 例1 +> 求极限 +> $$\lim_{n \to \infty} \left( \frac{n+1}{n^2+1} + \frac{n+2}{n^2+2} + \cdots + \frac{n+n}{n^2+n} \right)$$ + +1. 放缩： + +   对  $k = 1, 2, \dots, n$  ，有$$n^2+1 \le n^2+k \le n^2+n$$因此 +   $$\frac{n+k}{n^2+n} \le \frac{n+k}{n^2+k} \le \frac{n+k}{n^2+1}$$ +2. 求和： + +      分子和：$$\sum_{k=1}^n (n+k) = n\cdot n + (1+2+\cdots+n) + +   = n^2 + \frac{n(n+1)}{2} + +   = \frac{2n^2 + n(n+1)}{2} + +   = \frac{3n^2+n}{2}$$于是： +$$\frac{\frac{3n^2+n}{2}}{n^2+n} + +   \le S_n \le + +   \frac{\frac{3n^2+n}{2}}{n^2+1} + +   其中  S_n = \sum_{k=1}^n \frac{n+k}{n^2+k}$$ +3. 求上下极限： +   $$\lim_{n\to\infty} \frac{\frac{3n^2+n}{2}}{n^2+n} + +   = \lim_{n\to\infty} \frac{3n^2+n}{2(n^2+n)} + +   = \frac{3}{2}$$ +$$\lim_{n\to\infty} \frac{\frac{3n^2+n}{2}}{n^2+1} + +   = \lim_{n\to\infty} \frac{3n^2+n}{2(n^2+1)} + +   = \frac{3}{2}$$ +4. 夹逼得： $$ \lim_{n\to\infty} S_n = \frac{3}{2}$$ + +> [!example] 例2 +> 证明：$$\lim_{n\to\infty} \frac{n!}{n^n} = 0$$ + +证明： +$$0 < \frac{n!}{n^n} + += \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n}{n} + +\le \frac{1}{n} \cdot 1 \cdot 1 \cdots 1 + += \frac{1}{n}$$ +因为 +$$0 \le \frac{n!}{n^n} \le \frac{1}{n} + +\quad\text{且}\quad + +\lim_{n\to\infty} \frac{1}{n} = 0$$ +由夹逼定理得： +$$\lim_{n\to\infty} \frac{n!}{n^n} = 0$$ + + +### 类型二： + +> [!example] 例3 +> 设  $f(x) = \lim_{n \to \infty} \sqrt[n]{1 + x^n + \left( \frac{x^2}{2} \right)^n} ， x \geq 0$，问  $f(x)$  是否为连续函数。 + +解析： + +1. 提取主要项法 + +常用结论： + +若  $a_k > 0$ ，则 +$$\lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$$ +这里比较的三项为： +$$a_1 = 1, \quad a_2 = x, \quad a_3 = \frac{x^2}{2}$$ +2. 分区间求最大值 + +比较 $1$、$x$、$\frac{x^2}{2}$ 的大小随 $x$ 变化： + +1. 当 $0 \le x \le 1$ + +      $x \le 1$，且 $\frac{x^2}{2} \le \frac12 < 1$，最大的是 $1$。 + +2. 当 $1 < x \le 2$ + +      此时 $x > 1$，比较 $x$ 与 $\frac{x^2}{2}$： +   $$x \ge \frac{x^2}{2} \quad \Leftrightarrow \quad 2x \ge x^2 \quad \Leftrightarrow \quad x \le 2$$ + +   在 (1, 2] 内，$x \ge \frac{x^2}{2}$；同时 $x > 1$，所以最大值是 $x$。 + +3. 当 $x > 2$ + +      比较 $x$ 与 $\frac{x^2}{2}$： + +      $\frac{x^2}{2} > x \ (x > 2)$，且显然 $\frac{x^2}{2} > 1$，所以最大值是 $\frac{x^2}{2}$。 + +4. 极限结果 + +$$f(x) = \max\left\{1, x, \frac{x^2}{2} \right\} = + +\begin{cases} + +1, & 0 \le x \le 1, \\ + +x, & 1 < x \le 2, \\ + +\frac{x^2}{2}, & x > 2. + +\end{cases}$$ + +4. 连续性检查 + +· 在 $x=1$ 处： + +    $x \to 1^- 时 f(x) = 1$; + +    $x \to 1^+ 时 f(x) = x \to 1，且 f(1)=1，连续$ + +· 在 $x=2$ 处： + +    $x \to 2^- 时 f(x) = x \to 2$; + +    $x \to 2^+ 时 f(x) = \frac{x^2}{2} \to 2，且 f(2) = 2，连续$ + +因此 $f(x)$ 在 $[0,+\infty) 上连续$ + +最终答案： +$$\boxed{f(x) = + +\begin{cases} + +1, & 0 \le x \le 1, \\ + +x, & 1 < x \le 2, \\ + +\frac{x^2}{2}, & x > 2. + +\end{cases}}$$ + +> [!example] 例4 +> 证明$\lim_{n\to\infty} \sqrt[n]{3^n + 5^n + 7^n} = 7$ + +证明： +$$7^n < 3^n + 5^n + 7^n \le 7^n + 7^n + 7^n = 3\cdot 7^n$$ +两边开 n 次方： +$$7 < \sqrt[n]{3^n + 5^n + 7^n} \le 7 \cdot \sqrt[n]{3}$$ +而 $$\displaystyle \lim_{n\to\infty} \sqrt[n]{3} = 1$$由夹逼定理得： +$$\lim_{n\to\infty} \sqrt[n]{3^n + 5^n + 7^n} = 7$$ + +### 类型三： + +> [!example] 例5 +> 求函数极限 $\lim\limits_{x\to0}x^2\sin\frac{1}{x}$ + +解析： + +1.  **利用有界函数放缩** + +    已知正弦函数的值域：$-1\le\sin\frac{1}{x}\le1$（$x\neq0$） + +    两边同乘 $x^2$（$x^2\ge0$，不等号方向不变）： + +    $$-x^2\le x^2\sin\frac{1}{x}\le x^2$$ + +2.  **求上下界的极限** + +    左边极限：$\lim\limits_{x\to0}(-x^2)=0$ + +    右边极限：$\lim\limits_{x\to0}x^2=0$ + +3.  **应用夹逼定理** + +上下界极限均为 $0$，因此 $\lim\limits_{x\to0}x^2\sin\frac{1}{x}=0$ + +> [!example] 例6 +> 证明：$\lim_{n\to\infty} \sqrt[n]{2 + \sin^2 n} = 1$ + +证明： + +$$1 \le 2 + \sin^2 n \le 3$$ +所以 +$$\sqrt[n]{1} \le \sqrt[n]{2 + \sin^2 n} \le \sqrt[n]{3}$$ +即 +$$1 \le \sqrt[n]{2 + \sin^2 n} \le 3^{1/n}$$ +而 $\displaystyle \lim_{n\to\infty} 3^{1/n} = 1$，由夹逼定理： +$$\lim_{n\to\infty} \sqrt[n]{2 + \sin^2 n} = 1$$ +# 凑重要极限法 + +## 原理： + +两个重要极限 +$$(1)\lim\limits_{x \to 0} \frac{sinx}{x}=1$$$$(2)\lim\limits_{x \to 0} \frac{e^x-1}{x}=1$$ + +## 适用情况 + +x是形式的，也就是说，只要是一个趋于0的函数放在上面两个极限中x的位置，极限的结果仍然不变。首先得是一个除法的形式，如果是乘法，就转化成除以另一个函数的倒数，比如 + +$$\lim\limits_{x \to \infty} x(e^\frac{1}{x}-1)=\lim\limits_{x \to \infty } \frac{e^\frac{1}{x}-1}{\frac{1}{x}}=1$$ + +这就是一个常见的化简思路。 + +## 例子 + +> [!example] 计算下列极限： +> (1)$$\lim\limits_{x \to 0} \frac {x-sin2x}{x+sin3x}$$ +> (2)$$\lim\limits_{x \to 0} (\frac{1-x}{1+3x})^{1/x}$$ + +解：（1） +$$\lim\limits_{x\to0}\frac{x-sin2x}{x+sin3x}=\lim\limits_{x\to0}\frac{1-2\frac{sin2x}{2x}}{1+3\frac{sin3x}{3x}}=\frac{1-2}{1+3}=-\frac{1}{4}$$ +（2） + +原式等于$$\lim\limits_{x\to0}[(1+\frac{1}{\frac{-4x}{1+3x}})]^{1/x}$$ +令$$\frac{-4x}{1+3x}=\frac{1}{t}$$则$$1/x=-4t-3$$则原式等于 +$$\lim\limits_{t\to\infty}(1+1/t)^{-4t-3}=\lim\limits_{t\to\infty}[(1+1/t)^t]^{-4}*(1+1/t)^{-3}=e^{-4}$$ + +> [!example] 例2 +> 已知$$\lim\limits_{x\to\infty}(\frac{x+1}{x+k})^x=\lim\limits_{x\to0}e^{\frac{sin4x}{x}}$$则常数k的值为__ + +解： + +左边=$$\lim\limits_{x\to\infty}\frac{(1+1/x)^x}{(1+k/x)^x}=\lim\limits_{x\to\infty}\frac{(1+1/x)^x}{(1+\frac{1}{(x/k)})^{(x/k)*k}}=\frac{e}{e^k}=e^{1-k}$$ + +右边=$$e^{\lim\limits_{x\to0}{\frac{sin4x}{x}}}=e^4$$ +故$$e^{1-k}=e^4，k=-3$$ + +**课后习题：** + +（1）$\lim\limits_{x\to0}(\frac{a^x+b^x}{2})^{3/x}  (a,b>0)$ + +（2）$\lim\limits_{x\to\infty}\frac{3x^2+5}{5x+3}sin\frac{2}{x}$ + +（3）$\lim\limits_{x\to\pi/4}\frac{tanx-1}{x-\pi/4}$ + +（4）$\lim\limits_{x\to\pi/2}(sinx)^{tanx}$ + +解： + +（1） + +$$\begin{aligned} +\lim\limits_{x\to0}(\frac{a^x+b^x}{2})^{3/x} +&=\lim\limits_{x\to0}(\frac{a^x-1}{2}+\frac{b^x-1}{2}+1)^{3/x}\\[1em] +&=\lim\limits_{x\to0}(\frac{e^{xlna}-1}{2}+\frac{e^{xlnb}-1}{2}+1)^{3/x} \\[1em] +&=\lim\limits_{x\to0}(\frac{xlna}{2}+\frac{xlnb}{2}+1)^{3/x} \\[1em] +&=\lim\limits_{x\to0}(1+\frac{lnab}{2}x)^{\frac{2}{xlnab}\frac{3lnab}{2}}\\[1em] +&=e^{\frac{3lnab}{2}}\\[1em] +&=(ab)^{3/2} +\end{aligned}$$ + +（2） + +$$\lim\limits_{x\to\infty}\frac{3x^2+5}{5x+3}sin\frac{2}{x}=\lim\limits_{x\to\infty}\frac{6x^2+10}{5x^2+3x}\frac{sin\frac{2}{x}}{\frac{2}{x}}=\frac{6}{5}$$ + +（3） + +令 + +$$t = x - \frac{\pi}{4} \quad \Rightarrow \quad x = t + \frac{\pi}{4},\ t \to 0.$$ + +则$\tan x = \tan\left(t + \frac{\pi}{4}\right) = \frac{\tan t + 1}{1 - \tan t}$. + +于是$\tan x - 1 = \frac{\tan t + 1}{1 - \tan t} - 1$ + +$$= \frac{2\tan t}{1 - \tan t}.$$ + +代入原极限：$$\lim\limits_{x\to\pi/4} \frac{\tan x - 1}{x - \pi/4} + += \lim\limits_{t\to 0} \frac{\frac{2\tan t}{1 - \tan t}}{t} + += \lim\limits_{t\to 0} \frac{2\tan t}{t(1 - \tan t)}.$$ + +利用重要极限 $\displaystyle \lim_{u\to 0} \frac{\tan u}{u} = 1:\frac{2\tan t}{t(1 - \tan t)} = 2 \cdot \frac{\tan t}{t} \cdot \frac{1}{1 - \tan t}$ + +当 $t \to 0$ 时，$\tan t \to 0，\lim\limits_{t\to 0} \frac{\tan t}{t} = 1,\quad \lim\limits_{t\to 0} (1 - \tan t) = 1.$ + +所以$$\lim\limits_{t\to 0} 2 \cdot \frac{\tan t}{t} \cdot \frac{1}{1 - \tan t} + += 2 \cdot 1 \cdot \frac{1}{1 - 0} = 2$$ + +（4） + +原式=$\lim\limits_{x\to\pi/2}(1+sinx-1)^{\frac{1}{sinx-1}\frac{sinx(sinx-1)}{cosx}}=e^{\lim\limits_{x\to\pi/2}\frac{sinx(sinx-1)}{cosx}}$ + +令$t=\pi/2-x$，则原式=$e^{\lim\limits_{t\to0}\frac{cost(cost-1)}{sint}}=e^{\lim\limits_{t\to0}\frac{cost(-t^2/2)}{t}}=e^0=1$ + +# 等价无穷小求极限法 + +本质是利用已知公式进行等价无穷小因子的代换将复杂函数化简 + +引入练习：计算 +$$\lim\limits_{x\to0}\frac{\sqrt{1+2tanx}-\sqrt{1+2sinx}}{xln(1+x)-x^2}$$     +A、原式= $\infty$ +B、原式=1 +C、原式=-1 +D、原式=0 + + +原式等于$$2\lim\limits_{x\to0}\frac{tanx-sinx}{x[ln(1+x)-x]}\frac{1}{\sqrt{1+2tanx}+\sqrt{1+2sinx}}$$ + +$$=2\lim\limits_{x\to 0} \frac{sinx(1/cosx-1)} {x[ln(1+x)-x]} \frac{1}{\sqrt{1+2tanx}+ \sqrt{1+2sinx}}$$ + +$$=\lim\limits_{x\to0}\frac{sinx}{x}\frac{1}{cosx}\frac{1-cosx}{ln(1+x)-x}$$ + +$$=-1$$ + +解题思路：含有根式函数的极限，可先考虑有限化，然后再用四则运算、等价无穷小等方式求极限 + +> [!example] 例1 +> 计算 $\displaystyle \lim_{x\to 0}\frac{\sin mx}{\sin nx}$ + +答： + +由 $x \sim \sin x$（等价无穷小）： + +$$\lim_{x\to 0}\frac{\sin mx}{\sin nx} + += \lim_{x\to 0}\frac{mx}{nx} + += \frac{m}{n}.$$ + +> [!example] 例2 +> 计算 $$\displaystyle \lim_{x\to 0}(\cos x)^{\frac{1+x}{\sin^2 x}}$$ + +答： + +小技巧，针对复杂幂函数——取对数。 + +不妨取原式对数，得 + +$$\lim_{x\to 0}\ln\left[(\cos x)^{\frac{1+x}{\sin^2 x}}\right] + += \lim_{x\to 0}\frac{1+x}{\sin^2 x}\ln(\cos x)$$ + +由 $x \sim \sin x \sim \ln(1+x)$；$1-\cos x \sim \frac{1}{2}x^2$： + +$$ln(\cos x) \sim \cos x - 1 \sim -\frac{1}{2}x^2$$ + +$$\begin{aligned} +\lim_{x\to 0}\frac{1+x}{\sin^2 x}\ln(\cos x) + +&= \lim_{x\to 0}\frac{1+x}{x^2} \cdot \lim_{x\to 0}\left(-\frac{1}{2}x^2\right) \\[1em] + +&= \lim_{x\to 0}\frac{1+x}{x^2} \cdot \left(-\frac{1}{2}x^2\right) \\[1em] + +&= -\frac{1}{2}\lim_{x\to 0}\frac{(1+x)}{x^2} \cdot x^2 \\[1em] + +&= -\frac{1}{2} +\end{aligned}$$ + + + +所以 +$$\lim_{x\to 0}(\cos x)^{\frac{1+x}{\sin^2 x}} = e^{-\frac{1}{2}}$$ + + +## 加减运算的替换限制 + +对于加减运算，需要进行无穷小阶数的比较，若两同阶无穷小相加为零，要注意此时可能无法代换， + +因为此时可能会丢失高阶信息，所以有“加减不替换”的说法。 + +如：已知 $a^x - 1 \sim x\ln a$；$x \sim \sin x$， + +$$\lim_{x\to 0}\frac{a^x - 1 - x\ln a}{\sin x - x}$$ + +因为分子为0，整个式子为0？事实并非如此。 + +$$\textbf{拓展：}由泰勒展开可得，分子实际上等于$$ +$$\left( x\ln a + \frac{(x\ln a)^2}{2} + \frac{(x\ln a)^3}{6} + o(x^3) \right) - x\ln a + += \frac{(x\ln a)^2}{2} + o(x^2).$$ + +分母：$\sin x - x = -\frac{x^3}{6} + o(x^3)$，因此 +$$\frac{a^x - 1 - x\ln a}{\sin x - x} \sim \frac{\frac{(x\ln a)^2}{2}}{-\frac{x^3}{6}} + += -3\frac{(\ln a)^2}{x} \to + +\begin{cases} + +-\infty, & x\to 0^+,\\ + ++\infty, & x\to 0^-. + +\end{cases}$$ + + +练习：计算 $\displaystyle \lim_{n\to\infty} n^2[\arctan(n+1) - \arctan n]$ + +由 $\arctan x \sim x$ 得？注意此时 $n\to\infty$。 + +$$\begin{aligned}\text{原式} +&= \lim_{n \to \infty} n^2 \left[ \arctan \frac{(n+1)-n}{1+(n+1)n} \right] \\[1em] +&= \lim_{n \to \infty} n^2 \left[ \arctan \frac{1}{n^2 + n + 1} \right] \\[1em] +&= \lim_{n \to \infty} n^2 \cdot \frac{1}{n^2 + n + 1} \quad (\text{因为 } \arctan x \sim x \text{ 当 } x \to 0) \\[1em] +&= 1 +\end{aligned}$$ + +二、常见等价无穷小结 + +$$x \sim \sin x \sim \tan x \sim \sin^{-1} x \sim \tan^{-1} x \sim e^x - 1 \sim \ln(1 + x)$$ +$$1 - \cos x \sim \frac{1}{2}x^2 \quad $$ +$$(1 + x)^a - 1 \sim ax$$ +$$a^x - 1 \sim x \ln a \quad (a > 0)$$ + + +熟悉常见的极限，灵活运用好极限的四则运算法则和等价无穷小，已经能够解决绝大多数的极限计算题了，无论是简单题还是难题。很多看上去很复杂的所谓难题，无非是在四则运算和等价无穷小之间反复套娃而已，如果可以熟练的分离这些特征，其实不需要泰勒公式等“高级工具”就能快速准确地得出极限值。当然，不是贬低其他的工具，只是说不要学习了所谓的一些高级工具之后，就不重视这些初级的工具和结论。 + +# 抓大头”求极限法 + +1、形如 + +$$\frac{ax^n + bx^{n-1} + \ldots + p}{cx^m + \beta x^{m-1} + \ldots + \xi}$$ + +的多项式相除：上下同除x的最高次项，再上下各自求极限 + +2、形如 + +$$\ln(ae^{an} + \ldots + p)$$ + +的指对嵌套：先提出e的最高次项，转化为 + +$$\ln e^{\alpha n} \left( a + \frac{\cdots}{e^{\alpha n}} \right)$$ + +即 + +$$\alpha n + \ln \left( a + \frac{\cdots}{e^{\alpha n}} \right)$$ + +> [!example] 例1 +> 计算$$\lim\limits_{x\to0}\frac{\sqrt{1-4x}-cosx}{x}=$$ + +答案：-2 + + +小练习： + +$$\lim_{x \to 0} \frac{\csc(x) - \cot(x)}{x}=$$ \ No newline at end of file diff --git a/编写小组/讲义/级数敛散性的基本思路和方法.md b/编写小组/讲义/级数敛散性的基本思路和方法.md new file mode 100644 index 0000000..99aef78 --- /dev/null +++ b/编写小组/讲义/级数敛散性的基本思路和方法.md @@ -0,0 +1,615 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + +# 前情提要 + +## 几何级数 + +### 定义 + +几何级数是指形如 + +$$ +\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots +$$ + +的无穷级数，其中： +- $a \neq 0$ 是首项（常数） +- $r$ 是公比（常数） + +### 敛散性结论 +| 公比 $r$ 的范围 | 敛散性 | 和（当收敛时） | +| ---------- | --- | -------------------- | +| $r<1$ | 收敛 | $S = \dfrac{a}{1-r}$ | +| $r\ge 1$ | 发散 | 不存在 | + +--- + +### 证明 + +#### 一、部分和公式 +首先推导第 $n$ 项部分和 $S_n$ 的公式： + +$$ +S_n = a + ar + ar^2 + \cdots + ar^{n-1} +$$ + +当 $r \neq 1$ 时，两边乘以 $r$： + +$$ +rS_n = ar + ar^2 + ar^3 + \cdots + ar^n +$$ + +两式相减： + +$$ +S_n - rS_n = a - ar^n +$$ + +解得： + +$$ +S_n = a \cdot \frac{1 - r^n}{1 - r} \quad (r \neq 1) +$$ + +当 $r = 1$ 时： + +$$ +S_n = a + a + \cdots + a = na +$$ + +--- + +#### 二、敛散性分析 + +##### 情形1：$|r| < 1$ 时收敛 +当 $|r| < 1$ 时，$\lim_{n \to \infty} r^n = 0$ + +$$ +\lim_{n \to \infty} S_n = \lim_{n \to \infty} a \cdot \frac{1 - r^n}{1 - r} = \frac{a}{1 - r} +$$ + +所以级数收敛，和为 $S = \dfrac{a}{1-r}$。 + +##### 情形2：$|r| > 1$ 时发散 +当 $|r| > 1$ 时，$\lim_{n \to \infty} |r|^n = \infty$ + +$$ +\lim_{n \to \infty} S_n = \lim_{n \to \infty} a \cdot \frac{1 - r^n}{1 - r} \quad \text{不存在（趋于无穷）} +$$ + +所以级数发散。 + +##### 情形3：$r = 1$ 时发散 +$S_n = na$，当 $a \neq 0$ 时： + +$$ +\lim_{n \to \infty} S_n = \lim_{n \to \infty} na = \infty +$$ + +所以级数发散。 + +##### 情形4：$r = -1$ 时发散 +此时级数为： + +$$ +a - a + a - a + \cdots +$$ + +部分和为： +- $S_1 = a$ +- $S_2 = 0$ +- $S_3 = a$ +- $S_4 = 0$ +- ... + +当 $a \neq 0$ 时，部分和序列在 $a$ 和 $0$ 之间振荡，没有极限，所以发散。 + +##### 情形5：$r = -1$ 且 $a = 0$ +这是平凡情况，级数恒为 0，收敛于 0（但通常几何级数定义要求 $a \neq 0$）。 + +--- + +#### 三、直观理解 +1. **$|r| < 1$**：项以指数速度衰减，总和有限 +2. **$|r| \ge 1$**：项不衰减或增长，总和无限或振荡 + +**例子：** +- $\sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^n = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots = 2$（收敛） +- $\sum_{n=0}^{\infty} 2^n = 1 + 2 + 4 + \cdots$（发散） +- $\sum_{n=0}^{\infty} (-1)^n = 1 - 1 + 1 - 1 + \cdots$（发散） + +--- + +### 几何级数与p级数的对比 +| 特征 | 几何级数 | p级数 | +| ------ | ---------------- | ---------------- | +| 通项 | $ar^n$ | $\dfrac{1}{n^p}$ | +| 敛散性 | 只与 $r$ 有关 | 只与 $p$ 有关 | +| 收敛条件 | $r< 1$ | $p > 1$ | +| 和（收敛时） | $\dfrac{a}{1-r}$ | 无简单通式（除特殊p值） | +| 衰减速度 | 指数衰减 | 幂次衰减 | + +--- + +**几何级数小结：** +- 最简单的无穷级数之一 +- 完全解决了敛散性问题 +- 在收敛时，有简洁的求和公式 +- 是许多复杂级数分析的基础 + +--- + +### 记忆技巧 +1. **收敛条件**：公比的绝对值小于1 +2. **求和公式**：首项除以 $(1 - 公比)$ +3. **几何意义**：几何级数得名于相邻项的比值恒定（几何性质） + + +## p 级数 + +### 定义 + +p 级数是指形如 + +$$ +\sum_{n=1}^{\infty} \frac{1}{n^p} +$$ + +的无穷级数，其中 $p$ 是实数。 + +### 敛散性结论 +- **$p > 1$**：级数收敛 +- **$p \le 1$**：级数发散 + +--- + +### 证明（初等方法） + +#### 一、准备工作 +1. **调和级数发散**：$\sum_{n=1}^{\infty} \frac{1}{n}$ 发散 +2. **几何级数收敛**：当 $|r| < 1$ 时，$\sum_{n=0}^{\infty} r^n$ 收敛 + +--- + +#### 二、当 $0 < p \le 1$ 时发散 + +**证明步骤：** + +1. 对于任意 $0 < p \le 1$，有 $n^p \le n$ +2. 因此： + $$ + \frac{1}{n^p} \ge \frac{1}{n} + $$ +3. 已知调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散 +4. 由比较判别法： + - 若 $\sum a_n$ 发散，且 $b_n \ge a_n > 0$，则 $\sum b_n$ 也发散 + - 这里 $b_n = \frac{1}{n^p}$，$a_n = \frac{1}{n}$ +5. 所以当 $0 < p \le 1$ 时，p 级数发散 + +**注：** $p = 1$ 就是调和级数本身。 + +--- + +#### 三、当 $p > 1$ 时收敛 + +**证明（分组求和法）：** + +将级数分组： + +$$ +\begin{aligned} +S &= 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p} + \frac{1}{8^p} + \cdots \\ +&= 1 + \left(\frac{1}{2^p} + \frac{1}{3^p}\right) + \left(\frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p}\right) + \left(\frac{1}{8^p} + \cdots + \frac{1}{15^p}\right) + \cdots +\end{aligned} +$$ + +**分组规则：** +- 第 1 组：1 项 +- 第 2 组：2 项 +- 第 3 组：4 项 +- 第 4 组：8 项 +- ... +- 第 $k$ 组：$2^{k-1}$ 项 + +**估计第 $k$ 组的和：** +第 $k$ 组从 $n = 2^{k-1}$ 到 $n = 2^k - 1$，共 $2^{k-1}$ 项。 + +由于 $f(x) = \frac{1}{x^p}$ 递减，组内每项不超过第一项： + +$$ +\frac{1}{n^p} \le \frac{1}{(2^{k-1})^p} \quad (n \ge 2^{k-1}) +$$ + +第 $k$ 组和的上界： + +$$ +S_k \le 2^{k-1} \cdot \frac{1}{(2^{k-1})^p} = 2^{(k-1)(1-p)} +$$ + +**总和估计：** + +$$ +S \le 1 + 2^{1(1-p)} + 2^{2(1-p)} + 2^{3(1-p)} + \cdots +$$ + +令 $r = 2^{1-p}$，则： + +$$ +S \le 1 + r + r^2 + r^3 + \cdots +$$ + +当 $p > 1$ 时，$1-p < 0$，所以： + +$$ +r = 2^{1-p} < 2^0 = 1 +$$ + +右边是公比 $0 < r < 1$ 的几何级数，收敛到 $\frac{1}{1-r}$。 + +因此原级数的部分和有上界，且单调递增（各项为正），故级数收敛。 + +--- + +#### 四、直观理解 + +1. **$p \le 1$**：项衰减太慢，类似调和级数，总和无限增大 +2. **$p > 1$**：项衰减足够快，总和有限 + +**特例：** +- $p = 2$：$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ +- $p$ 越大，收敛越快 + +--- + +#### 补充：$p \le 0$ 的情况 + +若 $p \le 0$，则 $-p \ge 0$，通项 $n^{-p} \ge 1$ 不趋于 0，由级数收敛的必要条件知发散。 + +--- + +**总结：** +| $p$ 的范围 | 敛散性 | 关键原因 | +|------------|--------|----------| +| $p > 1$ | 收敛 | 分组求和，几何级数控制 | +| $0 < p \le 1$ | 发散 | 与调和级数比较 | +| $p \le 0$ | 发散 | 通项不趋于零 | + + + +# 部分和判别法 + +## **原理** + +对于级数 $\sum_{n=1}^{\infty}u_n$，构造其部分和数列 $S_n=u_1+u_2+\dots+u_n$。若极限 $\lim\limits_{n\to\infty}S_n=S$（S 为有限常数），则级数收敛，且和为 S；若该极限不存在（或为无穷大），则级数发散。 + +## **适用情况** + +适用于部分和可通过公式化简求和的级数，典型类型包括： + +1.等比级数：$\sum_{n=1}^{\infty}aq^{n-1}（a\neq0）$，部分和 $S_n=\frac{a(1-q^n)}{1-q}$； + +2.裂项相消型级数：如 $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$，通项可拆为 $\frac{1}{n}-\frac{1}{n+1}$，部分和可逐项抵消化简； + +3. 少数可通过错位相减、分组求和的特殊级数。 + +## **优势** + +1. 本质性强：直接基于级数收敛的定义，判定结果准确，还能同时求出级数的和，这是比值、比较判别法等无法做到的。 + +2. 无形式限制：不像比值判别法依赖通项的阶乘、指数结构，只要部分和能化简，任何类型级数都适用。 + +## **劣势** + +1. 适用范围极窄：绝大多数级数的部分和无法用初等函数化简（如调和级数 $\sum\frac{1}{n}$、$\sum\frac{n}{2^n}$虽能判定敛散性，但部分和无简单表达式）。 + +2. 计算难度高：裂项、错位相减等求和技巧性强，对复杂通项不易找到化简方法。 + +3. 效率低：对比判别法，求部分和的步骤通常更繁琐，不适用于快速判定敛散性。 + +## **例子** + +>[!example] **例1** +>判定级数 $\sum_{n=1}^{\infty}\frac{2}{(2n-1)(2n+1)}$的敛散性，若收敛求其和 + + + +>[!example] **例2** +判断级数$\sin \frac{\pi}{6} + \sin \frac{2\pi}{6} + \cdots + \sin \frac{n\pi}{6} + \cdots$的敛散性 + + + +# **级数运算性质判别法** + +## **原理** + +1. 线性运算性质的应用 + +设$\sum u_n、\sum v_n$ 为已知敛散性的级数，k 为非零常数： + +- 若 $\sum u_n$ 收敛，则 $\sum ku_n$ 收敛；若 $\sum u_n$ 发散，则 $\sum ku_n$ 发散。 + +- 若 $\sum u_n$ 和 $\sum v_n$ 都收敛，则 $\sum(u_n\pm v_n)$ 收敛。 + +- 若 $\sum u_n$ 收敛、$\sum v_n$ 发散，则 $\sum(u_n\pm v_n)$ 发散。 + +注：若两者都发散，$\sum(u_n\pm v_n)$ 敛散性不确定。 + +2. 添减/改变有限项性质的应用 + +对任意级数，增加、去掉或改变有限项，不改变其敛散性。 + +例：$\sum_{n=3}^{\infty}\frac{1}{2^n}$ 收敛，因为它是收敛等比级数 $\sum_{n=1}^{\infty}\frac{1}{2^n}$ 去掉前2项得到的。 + +3. 正项级数重组性质的应用 + +若正项级数 $\sum u_n$ 收敛，任意重排其项得到的新级数仍收敛，且和不变。 + +## **适用情况** + +1. 待判定级数可拆分为两个或多个已知敛散性的级数的线性组合（如 $\sum (u_n\pm v_n)、\sum ku_n）$。 + +2. 级数仅增减、改变有限项，或对正项级数进行项的重组。 + +3. 已知级数多为基础类型（等比级数、$p$-级数、调和级数等），便于直接套用性质。 + +## **优势** + +1. 快捷高效：无需计算极限或构造不等式，直接利用已知结论推导，步骤简洁。 + +2. 既适用于正项级数，也适用于任意项级数（如交错级数）。 + +## **劣势** + +1. 依赖性强：必须依赖已知敛散性的“参考级数”，若无法拆分或无合适参考级数，则无法使用。 + +2. 有局限性：对于 $\sum u_n$ 和 $\sum v_n$ 均发散的情况，$\sum (u_n\pm v_n)$ 的敛散性无法直接判定，性质失效。 + +## **例子** + +>[!example] **例1** +>判断级数$\sum_{n=1}^{\infty} (-1)^n \frac{n+2}{n+1} \cdot \frac{1}{\sqrt{n}}$的敛散性。 + + +>[!example] **例2** +讨论级数$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$的敛散性 + + +# 比较判别法 + +## Quick Description + +### 原理 + +不等式形式：放缩，大敛则小敛，小散则大散 + +极限形式：比值的极限为常数，则同敛散 + +### 适用情况 + +正项级数 + +### 优势和劣势 + +放缩在高中有知识基础，比较好上手 + +与比值/根值判别法不同，判定出变号级数的绝对值发散后，无法得到原级数发散 + +## 例子 + +> [!example] 例1 +> 判断级数$\sum_{n=2}^{\infty} \sin\left(\frac{1}{\ln n} \right)$的敛散性 + + + +> [!example] 例2 +基础练习 下列级数中收敛的有\_\_\_\_\_\_ + + A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1}$ + C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a>0)$ + D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + +# 比值判别法 + +## 原理 + +对于正项级数 $\sum_{n=1}^{\infty}u_n\ (u_n>0)$，计算极限$\rho=\lim\limits_{n \to \infty}\frac{u_{n+1}}{u_n}$，根据$\rho$与 1 的大小关系判断级数敛散性。 +1. 若 $\boldsymbol{\rho<1}$，则级数收敛； +2. 若 $\boldsymbol{\rho>1}$（或 $\rho=+\infty$），则级数发散； +3. 若 $\boldsymbol{\rho=1}$，则判别法失效，需用其他方法（如比较判别法、积分判别法等）判断。 + +## **适用情况** + +主要适用于通项含阶乘、指数幂（如 $a^n$）、$n^n$ 等形式的正项级数，这类通项的 $\frac{u_{n+1}}{u_n}$ ，作商后阶乘、指数部分可大幅化简，极限易求。 + +## **优势** + +1. 化简效率高：通项含 n!、$a^n$ 时，作商 $\frac{u_{n+1}}{u_n}$ 能直接约去大量重复因子 +2. 判定直接：只需计算一个极限值与 1 比较，无需像比较判别法那样构造合适的参考级数，对初学者更友好。 + +## **劣势** + +1. 对通项形式依赖强：仅适合含阶乘、指数幂的正项级数，对于通项为多项式、分式（如 $u_n=\frac{1}{n^p}$）的级数，$\lim\limits_{n\to\infty}\frac{u_{n+1}}{u_n}=1$，判别法直接失效。 +2. 失效情形无判定能力：当极限 $\rho=1$ 时，无法判断级数敛散性，必须换用比较判别法、积分判别法等其他方法，增加了解题步骤。 +3. 仅针对正项级数：不能直接应用于任意项级数（如交错级数），若要使用，需先对通项取绝对值判断绝对收敛，再进一步分析条件收敛。 +4. 极限可能不存在：部分正项级数的 $\lim\limits_{n\to\infty}\frac{u_{n+1}}{u_n}$ 不存在，此时比值判别法无法使用。 + +## **例子** + +> [!example] 例1 +判定级数敛散性： +$$\sum_{n=1}^{\infty} \frac{2^n}{n!}$$ + + + +>[!example] 例2 +设 $\alpha$ 为正常数，判定级数的敛散性： +$$\sum_{n=1}^\infty \frac{(2n-1)!!}{(n!)^\alpha}$$ + + + + + +# 根值判别法（柯西判别法） + +## Quick Description + +### 原理 + +通过计算级数通项 $u_n$ 的 $n$ 次方根的极限 $\rho = \lim\limits_{n\to\infty} \sqrt[n]{u_n}$，将原级数与等比级数 $\sum \rho^n$ 比较，从而判断敛散性。 + +### 适用情况 + +1. 适用于正项级数 $\sum u_n$（一般 $u_n > 0$）。 + +2. 当 $u_n$ 含有 $n$ 次幂、指数形式或连乘因子时，此方法往往有效。 + +3. 当比值判别法失效或难以计算时，可尝试根值判别法。 + + +### 优势和劣势 + +**优势** + +1. 对于含 $n$ 次幂、指数或幂指结构的级数，计算 $\sqrt[n]{u_n}$ 可能比计算比值 $u_{n+1}/u_n$ 更简单。 + +2. 理论清晰，与等比级数直接比较，直观性强。 + + +**劣势** + +1. 当 $\rho = 1$ 时无法判断（此时级数可能收敛也可能发散）。 + +2. 需要求 $\sqrt[n]{u_n}$ 的极限，有时计算较复杂。 + +3. 仅适用于正项级数。 + + +--- + +## 定理与证明 + +### 定理（根值判别法） + +设 $\sum_{n=1}^\infty u_n$ 是**正项级数**，且 +$$ +\rho = \lim_{n \to \infty} \sqrt[n]{u_n} +$$ +（这里允许 $\rho = +\infty$）。 + +1. 若 $\rho < 1$，则级数收敛。 +2. 若 $\rho > 1$（包括 $\rho = +\infty$），则级数发散。 +3. 若 $\rho = 1$，判别法失效（级数可能收敛也可能发散）。 + +### 证明思路 + +由极限定义：对任意 $\varepsilon > 0$，存在 $N$，当 $n > N$ 时， +$$ +\rho - \varepsilon < \sqrt[n]{u_n} < \rho + \varepsilon +$$ + +> [!note] +> **情形 $\rho < 1$** +> 取 $\varepsilon$ 使 $r = \rho + \varepsilon < 1$，则 +> $$ +> u_n < r^n \quad (n > N) +> $$ +> 而几何级数 $\sum r^n$ 收敛，由比较判别法知 $\sum u_n$ 收敛。 + +> [!note] +> **情形 $\rho > 1$** +> 取 $\varepsilon$ 使 $r = \rho - \varepsilon > 1$，则 +> $$ +> u_n > r^n \quad (n > N) +> $$ +> 而几何级数 $\sum r^n$ 发散，故 $\sum u_n$ 发散。 + +> [!note] +> **情形 $\rho = 1$** +> 反例： +> - $\sum \frac{1}{n}$ 发散，但 $\sqrt[n]{\frac{1}{n}} \to 1$。 +> - $\sum \frac{1}{n^2}$ 收敛，但 $\sqrt[n]{\frac{1}{n^2}} \to 1$。 + +--- + +## 典型例题 + +> [!example] 例1 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n$ 的敛散性。 + + + +> [!example] 例2 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{n}{\left(a + \frac{1}{n}\right)^n} \quad (a > 0)$ 的敛散性。 + + + +> [!example] 例3（常用结论） +> 证明：$\displaystyle \lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$，其中 $a_k > 0$。 + +> **证明** +> 令 $M = \max\{a_1, \dots, a_m\}$，则 +> $$ +> M^n \le a_1^n + \cdots + a_m^n \le m M^n +> $$ +> 两边开 $n$ 次方： +> $$ +> M \le \sqrt[n]{a_1^n + \cdots + a_m^n} \le M \cdot \sqrt[n]{m} +> $$ +> 因为 $\sqrt[n]{m} \to 1$，由夹逼定理得结论。 +> +> 此结论在判断形如 $\sum (a_1^n + \cdots + a_m^n)$ 的级数时可直接使用。 + + +> [!example] 例4 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n}$ 的敛散性。 + + +--- + +## 注意事项 + +1. 根值判别法仅适用于正项级数。 +2. 当 $\rho = 1$ 时，需改用其他方法（如比较判别法、积分判别法、比值判别法等）。 +3. 计算 $\sqrt[n]{u_n}$ 时，常用结论： + $$ + \sqrt[n]{n} \to 1, \quad \sqrt[n]{n^k} \to 1, \quad \sqrt[n]{a} = a^{1/n} \to 1 \ (a > 0) + $$ +4. 若 $u_n$ 含有 $n!$ 或连乘形式，通常使用比值判别法更简便；若含 $n$ 次幂，则根值判别法可能更直接。 +--- + + +# 交错级数 + +## 原理 + +所谓交错级数，就是正负项交错出现的级数。此类级数的收敛一般用莱布尼兹判别法： + 对于交错级数$\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}$，若满足 + （1）$\{a_{n}\}$单调减少，即$a_{1}\geqslant a_{2}\geqslant\cdots a_{n}\geqslant\cdots$ + （2）$\lim\limits_{n\to\infty}a_{n}=0,$ + 则级数$\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}$收敛，且$\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}\leqslant a_{1}$. + +## 优势劣势 + +莱布尼兹判别法优势就是判别条件简单，非常容易证明。但劣势在于仅限于判定交错级数的敛散性，所以一定要辨认清楚是否为交错级数。另外，在判断交错级数是条件收敛还是发散时，肯定会用到这个定理。 + +## 例题 + +> [!example] 例1 +> 判断下列极限是否收敛: +> $$(1)\sum_{n=1}^{\infty}\frac{cosn\pi}{n^{\frac{3}{4}}}$$ +> $$(2)\sum_{n=1}^{\infty}sin(n\pi+\frac{\pi}{n})$$ + + + +> [!example] 例2 +> 判断下面这个级数是否收敛： +> $\sum_{n=1}^{\infty}\frac{\ln n}{n}sin\frac{n\pi}{2}.$ + diff --git a/编写小组/讲义/级数敛散性的基本思路和方法（解析版）.md b/编写小组/讲义/级数敛散性的基本思路和方法（解析版）.md new file mode 100644 index 0000000..64d6bdd --- /dev/null +++ b/编写小组/讲义/级数敛散性的基本思路和方法（解析版）.md @@ -0,0 +1,1094 @@ +--- +tags: + - 编写小组 +--- +# 前情提要 + +## 几何级数 + +### 定义 + +几何级数是指形如 + +$$ +\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots +$$ + +的无穷级数，其中： +- $a \neq 0$ 是首项（常数） +- $r$ 是公比（常数） + +### 敛散性结论 +| 公比 $r$ 的范围 | 敛散性 | 和（当收敛时） | +| ---------- | --- | -------------------- | +| $r<1$ | 收敛 | $S = \dfrac{a}{1-r}$ | +| $r\ge 1$ | 发散 | 不存在 | + +--- + +### 证明 + +#### 一、部分和公式 +首先推导第 $n$ 项部分和 $S_n$ 的公式： + +$$ +S_n = a + ar + ar^2 + \cdots + ar^{n-1} +$$ + +当 $r \neq 1$ 时，两边乘以 $r$： + +$$ +rS_n = ar + ar^2 + ar^3 + \cdots + ar^n +$$ + +两式相减： + +$$ +S_n - rS_n = a - ar^n +$$ + +解得： + +$$ +S_n = a \cdot \frac{1 - r^n}{1 - r} \quad (r \neq 1) +$$ + +当 $r = 1$ 时： + +$$ +S_n = a + a + \cdots + a = na +$$ + +--- + +#### 二、敛散性分析 + +##### 情形1：$|r| < 1$ 时收敛 +当 $|r| < 1$ 时，$\lim_{n \to \infty} r^n = 0$ + +$$ +\lim_{n \to \infty} S_n = \lim_{n \to \infty} a \cdot \frac{1 - r^n}{1 - r} = \frac{a}{1 - r} +$$ + +所以级数收敛，和为 $S = \dfrac{a}{1-r}$。 + +##### 情形2：$|r| > 1$ 时发散 +当 $|r| > 1$ 时，$\lim_{n \to \infty} |r|^n = \infty$ + +$$ +\lim_{n \to \infty} S_n = \lim_{n \to \infty} a \cdot \frac{1 - r^n}{1 - r} \quad \text{不存在（趋于无穷）} +$$ + +所以级数发散。 + +##### 情形3：$r = 1$ 时发散 +$S_n = na$，当 $a \neq 0$ 时： + +$$ +\lim_{n \to \infty} S_n = \lim_{n \to \infty} na = \infty +$$ + +所以级数发散。 + +##### 情形4：$r = -1$ 时发散 +此时级数为： + +$$ +a - a + a - a + \cdots +$$ + +部分和为： +- $S_1 = a$ +- $S_2 = 0$ +- $S_3 = a$ +- $S_4 = 0$ +- ... + +当 $a \neq 0$ 时，部分和序列在 $a$ 和 $0$ 之间振荡，没有极限，所以发散。 + +##### 情形5：$r = -1$ 且 $a = 0$ +这是平凡情况，级数恒为 0，收敛于 0（但通常几何级数定义要求 $a \neq 0$）。 + +--- + +#### 三、直观理解 +1. **$|r| < 1$**：项以指数速度衰减，总和有限 +2. **$|r| \ge 1$**：项不衰减或增长，总和无限或振荡 + +**例子：** +- $\sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^n = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots = 2$（收敛） +- $\sum_{n=0}^{\infty} 2^n = 1 + 2 + 4 + \cdots$（发散） +- $\sum_{n=0}^{\infty} (-1)^n = 1 - 1 + 1 - 1 + \cdots$（发散） + +--- + +### 几何级数与p级数的对比 +| 特征 | 几何级数 | p级数 | +| ------ | ---------------- | ---------------- | +| 通项 | $ar^n$ | $\dfrac{1}{n^p}$ | +| 敛散性 | 只与 $r$ 有关 | 只与 $p$ 有关 | +| 收敛条件 | $r< 1$ | $p > 1$ | +| 和（收敛时） | $\dfrac{a}{1-r}$ | 无简单通式（除特殊p值） | +| 衰减速度 | 指数衰减 | 幂次衰减 | + +--- + +**几何级数小结：** +- 最简单的无穷级数之一 +- 完全解决了敛散性问题 +- 在收敛时，有简洁的求和公式 +- 是许多复杂级数分析的基础 + +--- + +### 记忆技巧 +1. **收敛条件**：公比的绝对值小于1 +2. **求和公式**：首项除以 $(1 - 公比)$ +3. **几何意义**：几何级数得名于相邻项的比值恒定（几何性质） + + +## p 级数 + +### 定义 + +p 级数是指形如 + +$$ +\sum_{n=1}^{\infty} \frac{1}{n^p} +$$ + +的无穷级数，其中 $p$ 是实数。 + +### 敛散性结论 +- **$p > 1$**：级数收敛 +- **$p \le 1$**：级数发散 + +--- + +### 证明（初等方法） + +#### 一、准备工作 +1. **调和级数发散**：$\sum_{n=1}^{\infty} \frac{1}{n}$ 发散 +2. **几何级数收敛**：当 $|r| < 1$ 时，$\sum_{n=0}^{\infty} r^n$ 收敛 + +--- + +#### 二、当 $0 < p \le 1$ 时发散 + +**证明步骤：** + +1. 对于任意 $0 < p \le 1$，有 $n^p \le n$ +2. 因此： + $$ + \frac{1}{n^p} \ge \frac{1}{n} + $$ +3. 已知调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散 +4. 由比较判别法： + - 若 $\sum a_n$ 发散，且 $b_n \ge a_n > 0$，则 $\sum b_n$ 也发散 + - 这里 $b_n = \frac{1}{n^p}$，$a_n = \frac{1}{n}$ +5. 所以当 $0 < p \le 1$ 时，p 级数发散 + +**注：** $p = 1$ 就是调和级数本身。 + +--- + +#### 三、当 $p > 1$ 时收敛 + +**证明（分组求和法）：** + +将级数分组： + +$$ +\begin{aligned} +S &= 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p} + \frac{1}{8^p} + \cdots \\ +&= 1 + \left(\frac{1}{2^p} + \frac{1}{3^p}\right) + \left(\frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p}\right) + \left(\frac{1}{8^p} + \cdots + \frac{1}{15^p}\right) + \cdots +\end{aligned} +$$ + +**分组规则：** +- 第 1 组：1 项 +- 第 2 组：2 项 +- 第 3 组：4 项 +- 第 4 组：8 项 +- ... +- 第 $k$ 组：$2^{k-1}$ 项 + +**估计第 $k$ 组的和：** +第 $k$ 组从 $n = 2^{k-1}$ 到 $n = 2^k - 1$，共 $2^{k-1}$ 项。 + +由于 $f(x) = \frac{1}{x^p}$ 递减，组内每项不超过第一项： + +$$ +\frac{1}{n^p} \le \frac{1}{(2^{k-1})^p} \quad (n \ge 2^{k-1}) +$$ + +第 $k$ 组和的上界： + +$$ +S_k \le 2^{k-1} \cdot \frac{1}{(2^{k-1})^p} = 2^{(k-1)(1-p)} +$$ + +**总和估计：** + +$$ +S \le 1 + 2^{1(1-p)} + 2^{2(1-p)} + 2^{3(1-p)} + \cdots +$$ + +令 $r = 2^{1-p}$，则： + +$$ +S \le 1 + r + r^2 + r^3 + \cdots +$$ + +当 $p > 1$ 时，$1-p < 0$，所以： + +$$ +r = 2^{1-p} < 2^0 = 1 +$$ + +右边是公比 $0 < r < 1$ 的几何级数，收敛到 $\frac{1}{1-r}$。 + +因此原级数的部分和有上界，且单调递增（各项为正），故级数收敛。 + +--- + +#### 四、直观理解 + +1. **$p \le 1$**：项衰减太慢，类似调和级数，总和无限增大 +2. **$p > 1$**：项衰减足够快，总和有限 + +**特例：** +- $p = 2$：$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ +- $p$ 越大，收敛越快 + +--- + +#### 补充：$p \le 0$ 的情况 + +若 $p \le 0$，则 $-p \ge 0$，通项 $n^{-p} \ge 1$ 不趋于 0，由级数收敛的必要条件知发散。 + +--- + +**总结：** +| $p$ 的范围 | 敛散性 | 关键原因 | +|------------|--------|----------| +| $p > 1$ | 收敛 | 分组求和，几何级数控制 | +| $0 < p \le 1$ | 发散 | 与调和级数比较 | +| $p \le 0$ | 发散 | 通项不趋于零 | + + + +# 部分和判别法 + +## **原理** + +对于级数 $\sum_{n=1}^{\infty}u_n$，构造其部分和数列 $S_n=u_1+u_2+\dots+u_n$。若极限 $\lim\limits_{n\to\infty}S_n=S$（S 为有限常数），则级数收敛，且和为 S；若该极限不存在（或为无穷大），则级数发散。 + +## **适用情况** + +适用于部分和可通过公式化简求和的级数，典型类型包括： + +1.等比级数：$\sum_{n=1}^{\infty}aq^{n-1}（a\neq0）$，部分和 $S_n=\frac{a(1-q^n)}{1-q}$； + +2.裂项相消型级数：如 $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$，通项可拆为 $\frac{1}{n}-\frac{1}{n+1}$，部分和可逐项抵消化简； + +3. 少数可通过错位相减、分组求和的特殊级数。 + +## **优势** + +1. 本质性强：直接基于级数收敛的定义，判定结果准确，还能同时求出级数的和，这是比值、比较判别法等无法做到的。 + +2. 无形式限制：不像比值判别法依赖通项的阶乘、指数结构，只要部分和能化简，任何类型级数都适用。 + +## **劣势** + +1. 适用范围极窄：绝大多数级数的部分和无法用初等函数化简（如调和级数 $\sum\frac{1}{n}$、$\sum\frac{n}{2^n}$虽能判定敛散性，但部分和无简单表达式）。 + +2. 计算难度高：裂项、错位相减等求和技巧性强，对复杂通项不易找到化简方法。 + +3. 效率低：对比判别法，求部分和的步骤通常更繁琐，不适用于快速判定敛散性。 + +## **例子** + +>[!example] **例1** +>判定级数 $\sum_{n=1}^{\infty}\frac{2}{(2n-1)(2n+1)}$的敛散性，若收敛求其和 + +解析 + +1.  裂项变形 + +   $\frac{2}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}$ + +2.  求部分和 + +   $S_n=\sum_{k=1}^{n}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right) = \left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right) = 1-\frac{1}{2n+1}$ + +3.  取极限 + +   $\lim_{n \to \infty}S_n=\lim_{n \to \infty}\left(1-\frac{1}{2n+1}\right)=1$ + +4.  结论：该级数 **收敛**，和为 \( 1 \) + +>[!example] **例2** +判断级数$\sin \frac{\pi}{6} + \sin \frac{2\pi}{6} + \cdots + \sin \frac{n\pi}{6} + \cdots$的敛散性 + +解析： + +1. 已知公式 + +由积化和差公式： + +$2\sin A \sin B = \cos(A-B) - \cos(A+B)$ + +令  $A = k\theta$， $B = \frac{\theta}{2}$ ，则： + +$2\sin(k\theta) \sin\left( \frac{\theta}{2} \right) = \cos\left( k\theta - \frac{\theta}{2} \right) - \cos\left( k\theta + \frac{\theta}{2} \right)$ + +$= \cos\left( (k - \tfrac12)\theta \right) - \cos\left( (k + \tfrac12)\theta \right)$ + +因此： + +$\boxed{\sin(k\theta) = \frac{\cos\left( (k - \tfrac12)\theta \right) - \cos\left( (k + \tfrac12)\theta \right)}{ 2\sin\frac{\theta}{2} }}$ + +这是一个裂项公式，因为对 k 求和时，中间项会消去。 + +2. 应用到本题 + +题中级数为： + +$\sum_{k=1}^{\infty} \sin\left( \frac{k\pi}{6} \right)$ + +这里  $\theta = \frac{\pi}{6}$ ， $\sin\frac{\theta}{2}$ = $\sin\frac{\pi}{12} \neq 0$。 + +利用上面裂项公式，记 + +$a_k = \sin\left( \frac{k\pi}{6} \right)$ + +$= \frac{\cos\left( \frac{(k - \frac12)\pi}{6} \right) - \cos\left( \frac{(k + \frac12)\pi}{6} \right)}{2\sin\frac{\pi}{12}}$ + +即 + +$a_k = \frac{1}{2\sin\frac{\pi}{12}} \left[ \cos\left( \frac{(2k-1)\pi}{12} \right) - \cos\left( \frac{(2k+1)\pi}{12} \right) \right]$ + +3. 求部分和 $S_n$ + +$S_n = \sum_{k=1}^n a_k$ + +$= \frac{1}{2\sin\frac{\pi}{12}} \sum_{k=1}^n \left[ \cos\frac{(2k-1)\pi}{12} - \cos\frac{(2k+1)\pi}{12} \right]$ + +记 $C_m = \cos\frac{(2m-1)\pi}{12}$，则： + +$\sum_{k=1}^n \left[ C_k - C_{k+1} \right]$ + +这是一个裂项和： + +$\sum_{k=1}^n (C_k - C_{k+1}) = C_1 - C_{n+1}$ + +其中 + +$C_1 = \cos\frac{(2\cdot 1 - 1)\pi}{12} = \cos\frac{\pi}{12}$ + +$C_{n+1} = \cos\frac{(2(n+1)-1)\pi}{12} = \cos\frac{(2n+1)\pi}{12}$ + +因此： + +$S_n = \frac{1}{2\sin\frac{\pi}{12}} \left[ \cos\frac{\pi}{12} - \cos\frac{(2n+1)\pi}{12} \right]$ + +4. 分析极限 + +当 $n \to \infty$ 时， $\frac{(2n+1)\pi}{12} \mod 2\pi$  在 $[0, 2\pi)$ 内不趋于一个固定值，而是周期性地取多个不同值。 + +事实上，$\cos\frac{(2n+1)\pi}{12}$ 的值在 $n$ 增加时振荡，不会趋于一个常数。 + +因此： + +$\lim_{n\to\infty} S_n \quad \text{不存在。}$ + +**习题1** + +级数 + +$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$ + +的和为 + +A. 0 + +B. 1 + +C. 2 + +D. 不存在（发散） + +解： + +观察 + +$\frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$ + +部分和： + +$S_N = \sum_{n=1}^N \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right] = 1 - \frac{1}{(N+1)^2} \to 1$ + +收敛到 1。 + +答案：B + +**习题2** + +级数 + +$\sum_{n=1}^{\infty} \left( \sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n} \right)$ + +\]   + +的敛散性是 + +A. 收敛到 0 + +B. 收敛到 $1 - \sqrt{2}$ + +C. 收敛到 $\sqrt{2} - 2$ + +D. 发散 + +解： + +观察通项： + +设 $a_n = \sqrt{n}$，则 + +$a_{n+2} - 2a_{n+1} + a_n = (a_{n+2} - a_{n+1}) - (a_{n+1} - a_n)$ + +\]   + +记 $b_n = a_{n+1} - a_n = \sqrt{n+1} - \sqrt{n}$，则 + +原通项 = $b_{n+1} - b_n$。 + +部分和： + +$S_N = \sum_{n=1}^N (b_{n+1} - b_n) = b_{N+1} - b_1$ + +\]   + +$b_{N+1} = \sqrt{N+2} - \sqrt{N+1}, \quad b_1 = \sqrt{2} - 1$ + +\]   + +因此 + +$S_N = (\sqrt{N+2} - \sqrt{N+1}) - (\sqrt{2} - 1)$ + +\]   + +当 $N \to \infty，\sqrt{N+2} - \sqrt{N+1} \to 0$，所以 + +$lim_{N\to\infty} S_N = 1 - \sqrt{2}$ + +收敛到 $1 - \sqrt{2}$。 + +答案：B + +# **级数运算性质判别法** + +## **原理** + +1. 线性运算性质的应用 + +设$\sum u_n、\sum v_n$ 为已知敛散性的级数，k 为非零常数： + +- 若 $\sum u_n$ 收敛，则 $\sum ku_n$ 收敛；若 $\sum u_n$ 发散，则 $\sum ku_n$ 发散。 + +- 若 $\sum u_n$ 和 $\sum v_n$ 都收敛，则 $\sum(u_n\pm v_n)$ 收敛。 + +- 若 $\sum u_n$ 收敛、$\sum v_n$ 发散，则 $\sum(u_n\pm v_n)$ 发散。 + +注：若两者都发散，$\sum(u_n\pm v_n)$ 敛散性不确定。 + +2. 添减/改变有限项性质的应用 + +对任意级数，增加、去掉或改变有限项，不改变其敛散性。 + +例：$\sum_{n=3}^{\infty}\frac{1}{2^n}$ 收敛，因为它是收敛等比级数 $\sum_{n=1}^{\infty}\frac{1}{2^n}$ 去掉前2项得到的。 + +3. 正项级数重组性质的应用 + +若正项级数 $\sum u_n$ 收敛，任意重排其项得到的新级数仍收敛，且和不变。 + +## **适用情况** + +1. 待判定级数可拆分为两个或多个已知敛散性的级数的线性组合（如 $\sum (u_n\pm v_n)、\sum ku_n）$。 + +2. 级数仅增减、改变有限项，或对正项级数进行项的重组。 + +3. 已知级数多为基础类型（等比级数、$p$-级数、调和级数等），便于直接套用性质。 + +## **优势** + +1. 快捷高效：无需计算极限或构造不等式，直接利用已知结论推导，步骤简洁。 + +2. 既适用于正项级数，也适用于任意项级数（如交错级数）。 + +## **劣势** + +1. 依赖性强：必须依赖已知敛散性的“参考级数”，若无法拆分或无合适参考级数，则无法使用。 + +2. 有局限性：对于 $\sum u_n$ 和 $\sum v_n$ 均发散的情况，$\sum (u_n\pm v_n)$ 的敛散性无法直接判定，性质失效。 + +## **例子** + +>[!example] **例1** +>判断级数$\sum_{n=1}^{\infty} (-1)^n \frac{n+2}{n+1} \cdot \frac{1}{\sqrt{n}}$的敛散性。 + +解析： + +原级数： + +$S = \sum_{n=1}^\infty (-1)^n \frac{n+2}{n+1} \cdot \frac{1}{\sqrt{n}}.$ + +将 $\frac{n+2}{n+1}$ 写成： + +$\frac{n+2}{n+1} = 1 + \frac{1}{n+1}.$ + +于是通项： + +$(-1)^n \frac{1}{\sqrt{n}} \cdot \left( 1 + \frac{1}{n+1} \right)$ + +$= (-1)^n \frac{1}{\sqrt{n}} \;+\; (-1)^n \frac{1}{\sqrt{n} \,(n+1)}.$ + +记 + +$u_n = (-1)^n \frac{1}{\sqrt{n}}, \quad v_n = (-1)^n \frac{1}{\sqrt{n} \, (n+1)}.$ + +则原级数： + +$S = \sum_{n=1}^\infty u_n \;+\; \sum_{n=1}^\infty v_n.$ + +第一项：$\sum_{n=1}^\infty u_n = \sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}$。 + +· 这是交错 $p$-级数， $p = \frac12 < 1$，不绝对收敛； + +· 由莱布尼茨判别法（$1/\sqrt{n} 单调递减趋于 0$）⇒ 条件收敛。 + +第二项：$\sum_{n=1}^\infty v_n$，比较 $v_n$ 的绝对值： + +$|v_n| = \frac{1}{\sqrt{n} \,(n+1)} \sim \frac{1}{n^{3/2}} \quad (n\to\infty).$ + +因为 $p = 3/2 > 1$，所以 $\sum |v_n|$ 收敛 ⇒ $\sum v_n$ 绝对收敛。 + +利用级数运算性质： + +已知 + +· 若两个级数 $\sum A_n 和 \sum B_n$ 都收敛，则 $\sum (A_n + B_n)$ 收敛（收敛级数的和收敛）。 + +· 若 $\sum A_n$ 条件收敛，$\sum B_n$ 绝对收敛，则 $\sum (A_n + B_n)$条件收敛（因为加一个绝对收敛级数不改变条件收敛的性质）。 + +这里： + +$S = \underbrace{\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}}_{\text{条件收敛}} \;+\; \underbrace{\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}(n+1)}}_{\text{绝对收敛}}.$ + +因此 $S$ 是 条件收敛的级数 + +>[!example] **例2** +讨论级数$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$的敛散性 + +解析：先对通项有理化变形，再拆分为两个级数的和，结合级数运算性质判断 + +步骤1 通项有理化与变形 + +对通项分母有理化： +$$ + +\begin{align*} + +u_n &= \frac{(-1)^n}{\sqrt{n} + (-1)^n} \\ + +&= \frac{(-1)^n \left[\sqrt{n} - (-1)^n\right]}{\left[\sqrt{n} + (-1)^n\right]\left[\sqrt{n} - (-1)^n\right]} \\ + +&= \frac{(-1)^n \sqrt{n} - 1}{n - 1} \\ + +&= \frac{(-1)^n \sqrt{n}}{n - 1} - \frac{1}{n - 1} + +\end{align*} +$$ + +$\frac{(-1)^n \sqrt{n}}{n - 1}$用莱布尼兹判别法判断收敛 + +$\sum_{n=2}^{\infty} \frac{1}{n - 1}：令 k=n-1，级数变为 \sum_{k=1}^{\infty} \frac{1}{k}$（调和级数），**发散**。 + +根据级数运算性质：**收敛级数 + 收敛级数 - 发散级数 = 发散级数**。 + +因此，级数$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$**发散**。 + +**习题1** + +已知级数 $\sum_{n=1}^{\infty} a_n$ 收敛，$\sum_{n=1}^{\infty} b_n$ 发散，则以下说法正确的是（ ） + +A. $\sum_{n=1}^{\infty} (a_n + b_n)$ 必收敛 + +B. $\sum_{n=1}^{\infty} (a_n + b_n)$ 必发散 + +C. $\sum_{n=1}^{\infty} (a_n b_n)$ 必发散 + +D. $\sum_{n=1}^{\infty} (a_n^2)$ 必收敛 + +解析： + +设 $s_n = a_1+\dots+a_n$ 收敛，$t_n = b_1+\dots+b_n$ 发散。 + +则 $a_n+b_n$ 的前 $n$ 项和 = $s_n + t_n$，收敛+发散 = 发散，故 B 正确。 + +反例：A 错。C：取 $a_n=\frac{(-1)^n}{\sqrt{n}}$（条件收敛），$b_n=(-1)^n$（发散），则 $a_n b_n = \frac{1}{\sqrt{n}}$ 发散，但这是特例，不是“必”发散，若 $a_n=\frac{1}{n^2}, b_n=n$，则 $a_n b_n = \frac{1}{n}$ 发散；若 $a_n=\frac{1}{n^2}, b_n=(-1)^n n$，则 $a_n b_n = \frac{(-1)^n}{n}$ 收敛（条件收敛）。所以 C 不对。D：反例 $a_n = \frac{(-1)^n}{\sqrt{n}}$ 收敛（条件收敛），但 $a_n^2 = \frac{1}{n}$ 发散。 + +答案：B + +**习题2** + +级数 $\sum_{n=1}^{\infty} \left( \frac{\sin n}{n^2} + \frac{(-1)^n}{n} \right)$ 的收敛性是： + +A. 绝对收敛 + +B. 条件收敛 + +C. 发散 + +D. 可能收敛可能发散 + +解析 + +拆成两个： + +1. $\sum \frac{\sin n}{n^2}：因 |\frac{\sin n}{n^2}| \le \frac{1}{n^2}$，绝对收敛。 + +2. $\sum \frac{(-1)^n}{n}$ 是交错调和级数，条件收敛。 + +   条件收敛 + 绝对收敛 = 条件收敛。 + +答案：B + + +# 比较判别法 + +## Quick Description + +### 原理 + +不等式形式：放缩，大敛则小敛，小散则大散 + +极限形式：比值的极限为常数，则同敛散 + +### 适用情况 + +正项级数 + +### 优势和劣势 + +放缩在高中有知识基础，比较好上手 + +与比值/根值判别法不同，判定出变号级数的绝对值发散后，无法得到原级数发散 + +## 例子 + +> [!example] 例1 +> 判断级数$\sum_{n=2}^{\infty} \sin\left(\frac{1}{\ln n} \right)$的敛散性 + +设 $a_n = \sin\left( \frac{1}{\ln n} \right)$，其中 $n \geq 2$ 。记其前n项和为$S_n$。 + +由于当 $n \geq 2$ 时，有 $\frac{1}{\ln n} > 0$ ，且 $\sin n > \frac{n}{2}$ ，因此存在 $N$ ，使得当$n > N$时， + +$q_n = \sin\left( \frac{1}{\ln n} \right) > \frac{1}{2} \cdot \frac{1}{\ln n} > \frac{1}{2n}.$ + +记$\sum \frac{1}{2n}$的前$n$项和为$T_n$。 + +由于 $\lim_{n \to \infty} S_n > \lim_{n \to \infty} T_n, \quad \text{而} \quad \lim_{n \to \infty} T_n = +\infty,$ + +因此 $\lim_{n \to \infty} S_n = +\infty.$ + +故级数 $\sum_{n=2}^{\infty} \sin\left( \frac{1}{\ln n} \right)$ 发散。 + +> [!example] 例2 +基础练习 下列级数中收敛的有\_\_\_\_\_\_ + + A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1}$ + C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a>0)$ + D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + + **参考答案：ABCF. + - **A** 由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故原级数收敛。 + - **B** 由于 $\lim_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故原级数收敛。 + - **C** 由于 $\lim_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛，故原级数收敛。（或由 $\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3} \;(n>1)$） + - **D** 由于 $\lim_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + - **E** 由于 $\lim_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \frac{\pi}{4}$，又级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + - **F** 由于 $\lim_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1+\frac{1}{\sqrt{n}}} / \left( 1+\frac{1}{n^2} \right) = 1$，又 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛，故原级数收敛。 + +# 比值判别法 + +## 原理 + +对于正项级数 $\sum_{n=1}^{\infty}u_n\ (u_n>0)$，计算极限$\rho=\lim\limits_{n \to \infty}\frac{u_{n+1}}{u_n}$，根据$\rho$与 1 的大小关系判断级数敛散性。 +1. 若 $\boldsymbol{\rho<1}$，则级数收敛； +2. 若 $\boldsymbol{\rho>1}$（或 $\rho=+\infty$），则级数发散； +3. 若 $\boldsymbol{\rho=1}$，则判别法失效，需用其他方法（如比较判别法、积分判别法等）判断。 + +## **适用情况** + +主要适用于通项含阶乘、指数幂（如 $a^n$）、$n^n$ 等形式的正项级数，这类通项的 $\frac{u_{n+1}}{u_n}$ ，作商后阶乘、指数部分可大幅化简，极限易求。 + +## **优势** + +1. 化简效率高：通项含 n!、$a^n$ 时，作商 $\frac{u_{n+1}}{u_n}$ 能直接约去大量重复因子 +2. 判定直接：只需计算一个极限值与 1 比较，无需像比较判别法那样构造合适的参考级数，对初学者更友好。 + +## **劣势** + +1. 对通项形式依赖强：仅适合含阶乘、指数幂的正项级数，对于通项为多项式、分式（如 $u_n=\frac{1}{n^p}$）的级数，$\lim\limits_{n\to\infty}\frac{u_{n+1}}{u_n}=1$，判别法直接失效。 +2. 失效情形无判定能力：当极限 $\rho=1$ 时，无法判断级数敛散性，必须换用比较判别法、积分判别法等其他方法，增加了解题步骤。 +3. 仅针对正项级数：不能直接应用于任意项级数（如交错级数），若要使用，需先对通项取绝对值判断绝对收敛，再进一步分析条件收敛。 +4. 极限可能不存在：部分正项级数的 $\lim\limits_{n\to\infty}\frac{u_{n+1}}{u_n}$ 不存在，此时比值判别法无法使用。 + +## **例子** + +> [!example] 例1 +判定级数敛散性： +$$\sum_{n=1}^{\infty} \frac{2^n}{n!}$$ + + + +**解析** +用比值判别法： +$$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2}{n+1} \to 0$$ +因此收敛。 + + +>[!example] 例2 +设 $\alpha$ 为正常数，判定级数的敛散性： +$$\sum_{n=1}^\infty \frac{(2n-1)!!}{(n!)^\alpha}$$ + + +**解析** + +1. 记通项 + +$$a_n = \frac{(2n-1)!!}{(n!)^\alpha}$$ + +这里 $(2n-1)!! = 1\cdot 3\cdot 5 \cdots (2n-1) = \frac{(2n)!}{2^n \, n!}$。 + +所以： +$$a_n = \frac{\frac{(2n)!}{2^n \, n!}}{(n!)^\alpha}= \frac{(2n)!}{2^n \, (n!)^{\alpha+1}}$$ + +2. 用比值判别法 +$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{2^{n+1}((n+1)!)^{\alpha+1}} \cdot \frac{2^n (n!)^{\alpha+1}}{(2n)!}$$ +化简：$$\frac{(2n+2)!}{(2n)!} = (2n+1)(2n+2) = 2(2n+1)(n+1)$$$$\frac{2^n}{2^{n+1}} = \frac12$$$$\frac{(n!)^{\alpha+1}}{((n+1)!)^{\alpha+1}} = \frac{1}{(n+1)^{\alpha+1}}$$所以： +$$\frac{a_{n+1}}{a_n} += \frac{2(2n+1)(n+1) \cdot \frac12}{(n+1)^{\alpha+1}} += \frac{2n+1}{(n+1)^{\alpha}}$$ +3. 计算极限 +$$\lim_{n\to\infty} \frac{2n+1}{(n+1)^{\alpha}}$$ + +· 若 $\alpha$> 1，极限为 0，比值判别法得出 $\lim \frac{a_{n+1}}{a_n} = 0 < 1$，级数收敛。 + +· 若 $\alpha$ = 1，极限为 $\lim_{n\to\infty} \frac{2n+1}{n+1}$ = 2 > 1，级数发散。 + +· 若 0 < $\alpha$ < 1，极限为 $\infty$ > 1，级数发散。 + +4. 答案 + + 当$\alpha$>1时收敛，当 0 < $\alpha$ < 1 时发散 + +>[!example] **习题1** +级数 $$\displaystyle \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} \cdot \frac{1}{5^n}$$ 的敛散性是 + +A. 收敛 + +B. 发散 + +C. 无法判断 + + +**解析** +$$a_n = \frac{(2n)!}{(n!)^2 \cdot 5^n}$$ + +$$\frac{a_{n+1}}{a_n} + += \frac{(2n+2)!}{((n+1)!)^2 \cdot 5^{n+1}} \cdot \frac{(n!)^2 \cdot 5^n}{(2n)!}$$ +化简： +$$\frac{(2n+2)!}{(2n)!} = (2n+2)(2n+1)$$ +$$\frac{(n!)^2}{((n+1)!)^2} = \frac{1}{(n+1)^2}$$ + +$$\frac{5^n}{5^{n+1}} = \frac15$$ +所以 + +$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{(n+1)^2} \cdot \frac15 += \frac{2(2n+1)}{n+1} \cdot \frac15$$ +$$\lim_{n\to\infty} \frac{4n+2}{5(n+1)} = \frac{4}{5} < 1$$ +所以收敛。 + +答案：A + +>[!example] **习题2** +级数 $\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{n^2}$ 的敛散性是 + +A. 收敛 + +B. 发散 + +C. 无法判断 + +解 +$$a_n = \frac{3^n}{n^2}$$ +$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)^2} \cdot \frac{n^2}{3^n} + += 3 \cdot \frac{n^2}{(n+1)^2}$$ +$$\lim_{n\to\infty} 3 \cdot \frac{n^2}{(n+1)^2} = 3 > 1$$ + +所以发散。 + +答案：B + +# 根值判别法（柯西判别法） + +## Quick Description + +### 原理 + +通过计算级数通项 $u_n$ 的 $n$ 次方根的极限 $\rho = \lim\limits_{n\to\infty} \sqrt[n]{u_n}$，将原级数与等比级数 $\sum \rho^n$ 比较，从而判断敛散性。 + +### 适用情况 + +1. 适用于正项级数 $\sum u_n$（一般 $u_n > 0$）。 + +2. 当 $u_n$ 含有 $n$ 次幂、指数形式或连乘因子时，此方法往往有效。 + +3. 当比值判别法失效或难以计算时，可尝试根值判别法。 + + +### 优势和劣势 + +**优势** + +1. 对于含 $n$ 次幂、指数或幂指结构的级数，计算 $\sqrt[n]{u_n}$ 可能比计算比值 $u_{n+1}/u_n$ 更简单。 + +2. 理论清晰，与等比级数直接比较，直观性强。 + + +**劣势** + +1. 当 $\rho = 1$ 时无法判断（此时级数可能收敛也可能发散）。 + +2. 需要求 $\sqrt[n]{u_n}$ 的极限，有时计算较复杂。 + +3. 仅适用于正项级数。 + + +--- + +## 定理与证明 + +### 定理（根值判别法） + +设 $\sum_{n=1}^\infty u_n$ 是**正项级数**，且 +$$ +\rho = \lim_{n \to \infty} \sqrt[n]{u_n} +$$ +（这里允许 $\rho = +\infty$）。 + +1. 若 $\rho < 1$，则级数收敛。 +2. 若 $\rho > 1$（包括 $\rho = +\infty$），则级数发散。 +3. 若 $\rho = 1$，判别法失效（级数可能收敛也可能发散）。 + +### 证明思路 + +由极限定义：对任意 $\varepsilon > 0$，存在 $N$，当 $n > N$ 时， +$$ +\rho - \varepsilon < \sqrt[n]{u_n} < \rho + \varepsilon +$$ + +> [!note] +> **情形 $\rho < 1$** +> 取 $\varepsilon$ 使 $r = \rho + \varepsilon < 1$，则 +> $$ +> u_n < r^n \quad (n > N) +> $$ +> 而几何级数 $\sum r^n$ 收敛，由比较判别法知 $\sum u_n$ 收敛。 + +> [!note] +> **情形 $\rho > 1$** +> 取 $\varepsilon$ 使 $r = \rho - \varepsilon > 1$，则 +> $$ +> u_n > r^n \quad (n > N) +> $$ +> 而几何级数 $\sum r^n$ 发散，故 $\sum u_n$ 发散。 + +> [!note] +> **情形 $\rho = 1$** +> 反例： +> - $\sum \frac{1}{n}$ 发散，但 $\sqrt[n]{\frac{1}{n}} \to 1$。 +> - $\sum \frac{1}{n^2}$ 收敛，但 $\sqrt[n]{\frac{1}{n^2}} \to 1$。 + +--- + +## 典型例题 + +> [!example] 例1 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n$ 的敛散性。 + +> **解** +> 计算根值： +> $$ +> \sqrt[n]{u_n} = \sqrt[n]{\left(1 - \frac{1}{n}\right)^n} = 1 - \frac{1}{n} \to 1 \quad (n \to \infty) +> $$ +> 此时 $\rho = 1$，根值判别法失效。 +> +> 改用其他方法：注意到 +> $$ +> \left(1 - \frac{1}{n}\right)^n = e^{n \ln (1 - \frac{1}{n})} \sim e^{-1} \quad (n \to \infty) +> $$ +> 所以 $u_n \to e^{-1} \neq 0$，由级数收敛的必要条件（通项必须趋于0）知该级数发散。 + +> [!example] 例2 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{n}{\left(a + \frac{1}{n}\right)^n} \quad (a > 0)$ 的敛散性。 + +> **解** +> 计算根值： +> $$ +> \sqrt[n]{u_n} = \frac{\sqrt[n]{n}}{a + \frac{1}{n}} +> $$ +> 因为 $\sqrt[n]{n} \to 1$，所以 +> $$ +> \lim_{n \to \infty} \sqrt[n]{u_n} = \frac{1}{a} +> $$ +> - 若 $\frac{1}{a} < 1$ 即 $a > 1$，则级数收敛。 +> - 若 $\frac{1}{a} > 1$ 即 $0 < a < 1$，则级数发散。 +> - 若 $a = 1$，此时 $\rho = 1$，判别法失效。 +> 当 $a = 1$ 时， +> $$ +> u_n = \frac{n}{(1 + \frac{1}{n})^n} \sim \frac{n}{e} \quad (n \to \infty) +> $$ +> 通项不趋于0，故级数发散。 +> +> **综上**： +> - 当 $a > 1$ 时，级数收敛； +> - 当 $0 < a \le 1$ 时，级数发散。 + +> [!example] 例3（常用结论） +> 证明：$\displaystyle \lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_m^n} = \max\{a_1, a_2, \dots, a_m\}$，其中 $a_k > 0$。 + +> **证明** +> 令 $M = \max\{a_1, \dots, a_m\}$，则 +> $$ +> M^n \le a_1^n + \cdots + a_m^n \le m M^n +> $$ +> 两边开 $n$ 次方： +> $$ +> M \le \sqrt[n]{a_1^n + \cdots + a_m^n} \le M \cdot \sqrt[n]{m} +> $$ +> 因为 $\sqrt[n]{m} \to 1$，由夹逼定理得结论。 +> +> 此结论在判断形如 $\sum (a_1^n + \cdots + a_m^n)$ 的级数时可直接使用。 + + +> [!example] 例4 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{2^n + 3^n}{5^n}$ 的敛散性。 + +> **解** +> $$ +> u_n = \frac{2^n + 3^n}{5^n} = \left(\frac{2}{5}\right)^n + \left(\frac{3}{5}\right)^n +> $$ +> 计算根值： +> $$ +> \sqrt[n]{u_n} = \sqrt[n]{\left(\frac{2}{5}\right)^n + \left(\frac{3}{5}\right)^n} +> $$ +> 由例3结论，$\sqrt[n]{u_n} \to \max\left\{\frac{2}{5}, \frac{3}{5}\right\} = \frac{3}{5} < 1$，故级数收敛。 + +--- + +## 注意事项 + +1. 根值判别法仅适用于正项级数。 +2. 当 $\rho = 1$ 时，需改用其他方法（如比较判别法、积分判别法、比值判别法等）。 +3. 计算 $\sqrt[n]{u_n}$ 时，常用结论： + $$ + \sqrt[n]{n} \to 1, \quad \sqrt[n]{n^k} \to 1, \quad \sqrt[n]{a} = a^{1/n} \to 1 \ (a > 0) + $$ +4. 若 $u_n$ 含有 $n!$ 或连乘形式，通常使用比值判别法更简便；若含 $n$ 次幂，则根值判别法可能更直接。 +--- + +## 练习题 + +> [!question] 练习1 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n$ 的敛散性。 + +> [!question] 练习2 +> 判断级数 $\displaystyle \sum_{n=1}^{\infty} \frac{5^n}{n^n}$ 的敛散性。 + +> [!question] 练习3 +> 设 $a > 0$，讨论级数 $\displaystyle \sum_{n=1}^{\infty} \left(\frac{a}{n}\right)^n$ 的敛散性。 + +> [!note] 参考答案 +> +> 1. $\rho = \frac{1}{2} < 1$，收敛。 +> +> 2. $\rho = 0 < 1$，收敛。 +> +> 3. 对任意 $a > 0$，有 $\rho = 0 < 1$，收敛。 +> + +# 交错级数 + +## 原理 + +所谓交错级数，就是正负项交错出现的级数。此类级数的收敛一般用莱布尼兹判别法： + 对于交错级数$\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}$，若满足 + （1）$\{a_{n}\}$单调减少，即$a_{1}\geqslant a_{2}\geqslant\cdots a_{n}\geqslant\cdots$ + （2）$\lim\limits_{n\to\infty}a_{n}=0,$ + 则级数$\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}$收敛，且$\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}\leqslant a_{1}$. + +## 优势劣势 + +莱布尼兹判别法优势就是判别条件简单，非常容易证明。但劣势在于仅限于判定交错级数的敛散性，所以一定要辨认清楚是否为交错级数。另外，在判断交错级数是条件收敛还是发散时，肯定会用到这个定理。 + +## 例题 + +> [!example] 例1 +> 判断下列极限是否收敛: +> $$(1)\sum_{n=1}^{\infty}\frac{cosn\pi}{n^{\frac{3}{4}}}$$ +> $$(2)\sum_{n=1}^{\infty}sin(n\pi+\frac{\pi}{n})$$ + +解： +$(1)cosn\pi=(-1)^n,故级数为交错级数.$ + +$显然a_{n}=\frac{1}{n^{\frac{3}{4}}}单调递减且\lim\limits_{n\to\infty}a_{n}=0,由莱布尼兹判别法，级数收敛.$ + +$(2)sin(n\pi+\frac{\pi}{n})=(-1)^{n}sin(\frac{\pi}{n})，故级数为交错级数.$ + +$又a_{n}=sin(\frac{\pi}{n})\to0(n\to\infty)且\{a_{n}\}从第二项开始单调递减，由莱布尼兹判别法知，级数收敛.$ + +上面这个例题告诉我们，交错级数并不总是很傻地就写一个$(-1)^n$来给我们看，它也是会有多种形式地。而且，无论是单调性还是符号的交错出现，都不一定要从第一项开始，因为级数的敛散性和它的前有限项无关，要是这个交错级数通项的绝对值从第二项才开始单调递减，那就从第二项开始考虑，这不影响它的敛散性；一样，要是一个级数从第二项开始才是交错级数，那就从第二项开始算，第一项就不管它了。 + +> [!example] 例2 +> 判断下面这个级数是否收敛： +> $\sum_{n=1}^{\infty}\frac{\ln n}{n}sin\frac{n\pi}{2}.$ + +解： + $对k\in\mathbb{N}_{+},显然有n=2k-1时,sin\frac{n\pi}{2}=(-1)^{k},n=2k时,sin\frac{n\pi}{2}=0.故$ + $\sum_{n=1}^{\infty}\frac{\ln n}{n}sin\frac{n\pi}{2}=\sum_{k=1}^{\infty}\frac{\ln (2k-1)}{2k-1}\cdot(-1)^{k}$ + $故级数从第二项开始为交错级数，且a_{k}=\frac{\ln (2k-1)}{2k-1}\to0(n\to\infty),$ + $\{a_{k}\}从第二项开始单调递减，由莱布尼兹判别法，级数收敛.$ + +## 习题 +$(1)\sum_{n=1}^{\infty}sin(\sqrt{n^2+1}\pi).$ +$(2)\sum_{n=1}^{\infty}\frac{1}{3^n}sin\frac{n\pi}{2}.$ +$(3)\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{3^n}.$ +$(4)\sum_{n=1}^{\infty}(-1)^{n+1}a^{\frac{1}{n}} (a>0).$ + +## **习题答案** +$(1)sin(\sqrt{n^2+1}\pi)=(-1)^{n}sin((\sqrt{n^2+1}-n)\pi)=(-1)^{n}sin(\frac{1}{\sqrt{n^2+1}+n}\pi),$ +$a_{n}=sin(\frac{\pi}{\sqrt{n^2+1}+n})\to0(n\to\infty),且\{a_{n}\}单调递减，由莱布尼兹判别法，级数收敛$ +$(2)同例2解法，答案是收敛（建议作为课前题目）$ +$(3)n(n-1)/2的奇偶性与n正好相反，所以级数也是交错级数，容易证明是收敛$ +$(4)a^{\frac{1}{n}}\to1(n\to\infty)，由级数收敛的必要条件知，级数发散$ \ No newline at end of file diff --git a/编写小组/讲义/证明题方法：单调有界定理，介值定理.md b/编写小组/讲义/证明题方法：单调有界定理，介值定理.md new file mode 100644 index 0000000..1fb86d3 --- /dev/null +++ b/编写小组/讲义/证明题方法：单调有界定理，介值定理.md @@ -0,0 +1,4 @@ +--- +tags: + - 编写小组 +--- diff --git a/编写小组/讲义/隐函数，参数方程和应用题.md b/编写小组/讲义/隐函数，参数方程和应用题.md new file mode 100644 index 0000000..e7ee343 --- /dev/null +++ b/编写小组/讲义/隐函数，参数方程和应用题.md @@ -0,0 +1,259 @@ +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + +## 隐函数求导 + +### 原理 + +设方程 $F(x, y) = 0$ 确定 $y$ 是 $x$ 的函数 $y = y(x)$，则对方程两边关于 $x$ 求导（注意 $y$ 是 $x$ 的函数），再解出 $y'$ 即可。 + +### **适用情况** + +适用于方程中 $x$ 与 $y$ 混合在一起，无法或不易解出 $y = f(x)$ 的情况，如： +- 圆的方程：$x^2 + y^2 = 1$ +- 椭圆方程：$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ +- 一般隐式方程：$e^{x+y} + \ln(xy) = 0$ + +### **优势** +1. 不必显式解出 $y = f(x)$，可直接求导； +2. 适用于复杂关系式，尤其是含有 $x$、$y$ 混合的函数形式。 + +### **劣势** +1. 求导过程中需注意 $y$ 是 $x$ 的函数，常需使用链式法则； +2. 最终表达式中可能仍含有 $y$，需结合原方程化简。 + +### **例子** + +> [!example] 例1(简单) +求由方程 $x^2 + y^2 = 1$ 确定的隐函数 $y = y(x)$ 的导数。 + +**解析** +两边对 $x$ 求导： +$$ +2x + 2y \cdot y' = 0 +$$ +解得： +$$ +y' = -\frac{x}{y} +$$ + +>[!example] 例2 +>设方程$y=y(x)$由方程$xe^{f(y)}=e^y$确定，其中$f$具有二阶导数，且$f'\neq1$，求$\frac{d^2y}{dx^2}.$ + + + + +>[!example] 例3 +>曲线$\ln{\sqrt{x^2+y^2}}=arctan(\frac{x}{y})$在点(0,1)处的切线方程为$\_\_\_$ +>（A）$y=-x+1$ +>（B）$y=1$ +>（C）$y=x+1$ +>（D）$y=\frac{1}{2}x+1$ + + + + +>[!example] 例4 +>函数$y=y(x)$由方程$sin(x^2+y^2)+e^x-xy^2=0$所确定，则$\frac{dy}{dx}=\_\_$. + + +>[!example] 例5(卢吉辚原创难题) +>已知由方程 $e^{(x y - 1)(x + y)} \ln (x^3 + y^3) = \frac{xy(x^2 + y^2)}{(x y)^{x^2 + x y + y^2}} \ln (\sqrt{x + y})$ 所确定的函数单调减，求其在点 $(1,1)$ 处的切线方程。 + + + + +>[!example] 例6(卢吉辚原创难题) +>已知由方程 $\arctan\left( \frac{x^2 y + x y^2}{2} \right) = \arcsin\left( \frac{\sqrt{x^2 + y^2}}{x + y} \right)$ 所确定的函数在其连续区间内单调减，求其在点 $(1,1)$ 处的切线方程。 + + +--- + +## 参数方程求导 + +### 原理 + +已知 $$\left\{ \begin{array}{c} x=x\left( t \right)\\ y=y\left( t \right)\\ \end{array} \right.$$若有均可导且 $x'(t) \neq 0$，则 +$$y'\left( t \right) =\frac{dy}{dt}$$ +$$x'\left( t \right)=\frac{dx}{dt}$$ + $$y'=\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{y'\left( t \right)}{x'\left( t \right)}$$ +则有 $$y''=\frac{d^2y}{dx^2}=\frac{dy'}{dx}=\frac{dy'}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{dy'}{dt}\cdot \frac{1}{x'\left( t \right)}=(\frac{y'(t)}{x'(t)})'\cdot \frac{1}{x'\left( t \right)}$$ + +而 $$\frac{y'(t)}{x'(t)}=\frac{y''\left( t \right) x'\left( t \right) -x''\left( t \right) y'\left( t \right)}{x'^2\left( t \right)}$$ + +故 $$y''=\frac{y''\left( t \right) x'\left( t \right) -x''\left( t \right) y'\left( t \right)}{x'^3\left( t \right)}$$ +### **适用情况** +适用于曲线由参数形式给出，尤其是： +- 物理中的运动轨迹； +- 极坐标、摆线、旋轮线等曲线； +- 复杂曲线的简化表示。 + +### **优势** +1. 形式简洁，直接利用两个导数作商； +2. 适用于参数化表示，便于计算高阶导数。 + +### **劣势** +1. 要求 $x'(t) \neq 0$，否则导数不存在； +2. 高阶导数计算需重复使用公式，略显繁琐。 + +### **例子** + +> [!example] 例1(简单) +> 求参数方程 +> $$ +\begin{cases} +x = a\cos t \\ +y = b\sin t +\end{cases}$$ +所确定的函数 $y = y(x)$ 的导数。 + +**解析** + +计算： + +$$ +\frac{dx}{dt} = -a\sin t, \quad \frac{dy}{dt} = b\cos t +$$ + +因此： +$$ +\frac{dy}{dx} = \frac{b\cos t}{-a\sin t} = -\frac{b}{a}\cot t +$$ + + +> [!example] 例2 +设 $y$ 由方程 $e^{x+y} = xy$ 确定，则 $y'$ 在点 $(1,0)$ 处的值为 +A. $0$ +B. $1$ +C. $-1$ +D. 不存在 + + + +> [!example] 例3 +> 参数方程 +> $$ +\begin{cases} +x = t^2 + 1 \\ +y = t^3 - t +\end{cases}$$ +在 $t = 1$ 处的导数 $\left. \frac{dy}{dx} \right|_{t=1}$ 为 +A. $1$ +B. $2$ +C. $3$ +D. $4$ + + + +>[!example] 例4(卢吉辚原创难题) +>设参数方程 $x = e^{t^2} \arcsin(t^2)$，$y = \ln(1+t^4) \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值（需要用到泰勒展开，仅作习题） + + +>[!example] 例5(卢吉辚原创难题) +>设参数方程 $x = e^{t} \sin t + \ln(1+t^2)$，$y = e^{t} \cos t + \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值。 + + + +## **应用题** + +应用题就比较综合了，它可以考很多知识点，比如参数方程，比如隐函数求导，也有可能就是一个普通的求导。但关键不在于求导，关键在于建立一个准确的模型，也就是 + +（1）把需要求的对象找出来，即自变量和因变量，注意可能有多个。 +（2）根据条件把关系式列出来，画一画示意图有时候会很有帮助。 + +### 微分估计值 + +>[!example] 例1 +>利用微分估计值也是导数应用的一种。利用微分估计下面几个式子的值： +>（1）$\sqrt{34}$（保留两位小数） +>（2）摆的振动周期公式为：$T=2\pi \sqrt{\frac{l}{g}}$。其中$l$为摆长（厘米），重力加速度$g=981cm^2/s$.为了使周期$T$增大$0.05$秒，摆长$l=20cm$的长度需要作何修正？（参考数据：$\sqrt{981\times20}\approx140.07,\frac{1}{\pi}\approx0.318$,保留两位小数） + + +### 相关变化率 + +#### 飞机航空摄影问题 + +##### 问题描述 + +>[!example] 例题 +>一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 + + + +--- + +#### 人拉船问题 + +##### 问题描述 + +>[!example] 例题 +>人在河岸用绳经过定滑轮以速度 $v$、加速度 $a$ 拉船。绳与水平面夹角为 $\theta$。求此时船的加速度 $a'$。 +![[应用题人拉船.png]] + + +--- + +#### 动点曲线运动问题 + +##### 问题描述 + +>[!example] 例题 +>已知动点 $P$ 在曲线 $y = \sqrt{x}$ 上运动，记坐标原点 $O$ 与 $P$ 间的距离为 $l$。若点 $P$ 横坐标随时间的变化率为常数 $v$，则当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率是多少？ + + + +--- + +#### 球浸入水问题 + +##### 问题描述 + +>[!example] 例题 +>半径为$a$的球渐渐沉入盛有部分水的半径为$b(b>a)$的圆柱形容器中.若球以匀速$c$下沉，求：球浸没一半时，容器内水面上升的速率. +> +>（A）$\frac{a^2c}{b^2}$ +> +>（B）$\frac{a^c}{b^2-a^2}$ +> +>（C）$\frac{a^2c}{2b^2}$ +> +>（D）$\frac{a^2c}{b^2+a^2}$ + + +### 应用题期中真题 + +>[!example] **例1**（2020） +>有一个正圆锥形漏斗，深度 $18 \text{ cm}$ ，上端口直径为 $12 \text{ cm}$ （即半径 $6 \text{ cm}$ ）。漏斗下方连接一个圆柱形筒，圆柱筒直径为 $10 \text{ cm}$ （即半径 $5 \text{ cm}$ ）。初始时刻漏斗内盛满水，然后水从漏斗流入圆柱筒。已知当漏斗中水深为 $12 \text{ cm}$ 时，漏斗水面下降的速度是 $1 \text{ cm/min}$ +问：此时圆柱筒内液面上升的速度是多少？ + + + + +>[!example] **例2** (2021) +>一长为 L 米的木梯靠在倾角为$\frac{\pi}{3}$的光滑斜坡上，木梯的顶部距离 A 点 h 米，底部距离 A 点 d 米，受重力作用木梯的顶部以 $a \, \mathrm{m/s}$ 的速度沿直线 BA 下滑，底部水平向右运动。问：当木梯的顶部和底部与 A 点的距离相等时，底部的水平速度为多少？ +>![[应用题梯子.jpg]] + + + + +>[!example] **例3**（2023） +>有一个长度为5m的梯子贴靠在铅直的墙上，假设其下端沿地板以3m/s的速率离开墙脚而滑动，则 +(1) 何时梯子的上、下端能以相同的速率移动？ +(2) 何时其上端下滑之速率为4m/s？ + + + + + +>[!example] **例4**(2022) +>某部举行八一阅兵，队列正步通过阅兵台时步幅间距离为75厘米，步速为每分钟112步。某观礼人员离行进队列垂直距离为60米，视线追随队列领队，求其视线与队列夹角为$\frac{\pi}{6}$时，视线转动角度的变化率。 + + + + + + +>[!example] **例5**（2024） +>一架巡逻直升机在距离地面3km的高度以120km/h的速度沿着一条水平笔直的高速公路向前飞行。飞行员观察到迎面驶来一辆汽车，通过雷达测出直升机与汽车的距离为5km，同时此距离以160km/h的速率在减少。试求此时汽车行进的速度。 + + diff --git a/编写小组/讲义/隐函数，参数方程和应用题（解析版）.md b/编写小组/讲义/隐函数，参数方程和应用题（解析版）.md new file mode 100644 index 0000000..bc4e4a1 --- /dev/null +++ b/编写小组/讲义/隐函数，参数方程和应用题（解析版）.md @@ -0,0 +1,579 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + +## 隐函数求导 + +### 原理 + +设方程 $F(x, y) = 0$ 确定 $y$ 是 $x$ 的函数 $y = y(x)$，则对方程两边关于 $x$ 求导（注意 $y$ 是 $x$ 的函数），再解出 $y'$ 即可。 + +### **适用情况** + +适用于方程中 $x$ 与 $y$ 混合在一起，无法或不易解出 $y = f(x)$ 的情况，如： +- 圆的方程：$x^2 + y^2 = 1$ +- 椭圆方程：$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ +- 一般隐式方程：$e^{x+y} + \ln(xy) = 0$ + +### **优势** +1. 不必显式解出 $y = f(x)$，可直接求导； +2. 适用于复杂关系式，尤其是含有 $x$、$y$ 混合的函数形式。 + +### **劣势** +1. 求导过程中需注意 $y$ 是 $x$ 的函数，常需使用链式法则； +2. 最终表达式中可能仍含有 $y$，需结合原方程化简。 + +### **例子** + +> [!example] 例1(简单) +求由方程 $x^2 + y^2 = 1$ 确定的隐函数 $y = y(x)$ 的导数。 + +**解析** +两边对 $x$ 求导： +$$ +2x + 2y \cdot y' = 0 +$$ +解得： +$$ +y' = -\frac{x}{y} +$$ + +>[!example] 例2 +>设方程$y=y(x)$由方程$xe^{f(y)}=e^y$确定，其中$f$具有二阶导数，且$f'\neq1$，求$\frac{d^2y}{dx^2}.$ + +解1： + +两边对$x$求导得 +$$e^{f(y)}+xy'f'(y)e^{f(y)}=y'e^y$$ +从而 +$$y'=\frac{e^{f(y)}}{e^y-xf'(y)e^{f(y)}}=\frac{e^{f(y)}}{xe^{f(y)}(1-f'(y))}=\frac{1}{x(1-f'(y))}$$ +两边再对$x$求导得 +$$\frac{d^2y}{dx^2}=y''=-\frac{1-f'(y)-xy'f''(y)}{x^2(1-f'(y))^2}=-\frac{(1-f'(y))^2-f''(y)}{x^2(1-f'(y))^3}$$ + + +解2：两边取对数得$$lnx+f(y)=y$$ +两边对$x$求导得$$\frac{1}{x}+y'f'(y)=y'$$ +从而$$y'=\frac{1}{x(1-f'(y))}$$ +后同解1. + + +>[!example] 例3 +>曲线$\ln{\sqrt{x^2+y^2}}=arctan(\frac{x}{y})$在点(0,1)处的切线方程为$\_\_\_$ +>（A）$y=-x+1$ +>（B）$y=1$ +>（C）$y=x+1$ +>（D）$y=\frac{1}{2}x+1$ + + +解：由原式得$$\frac{1}{2}ln(x^2+y^2)=arctan(\frac{x}{y})$$ +两边对$x$求导得 $$\frac{1}{2}\cdot \frac{2x+2yy'}{x^2+y^2}=\frac{y-xy'}{x^2+y^2}$$$$x+yy'=y-xy'$$$$y'=\frac{y-x}{y+x}$$ +故$y'|_{x=0,y=1}=1$，即曲线在(0,1)处切线的斜率为1，故切线方程为： +$$y=x+1$$选C. + + +>[!example] 例4 +>函数$y=y(x)$由方程$sin(x^2+y^2)+e^x-xy^2=0$所确定，则$\frac{dy}{dx}=\_\_$. + +解： +两边求微分得： +$$(2xdx+2ydy)cos(x^2+y^2)+e^xdx-y^2dx-2xydy=0$$ +$$(2xcos(x^2+y^2)+e^x-y^2)dx+(2ycos(x^2+y^2)-2xy)dy=0$$ +$$\frac{dy}{dx}=\frac{2xcos(x^2+y^2)+e^x-y^2}{2xy-2ycos(x^2+y^2)}$$ + +>[!example] 例5(卢吉辚原创难题) +>已知由方程 $e^{(x y - 1)(x + y)} \ln (x^3 + y^3) = \frac{xy(x^2 + y^2)}{(x y)^{x^2 + x y + y^2}} \ln (\sqrt{x + y})$ 所确定的函数单调减，求其在点 $(1,1)$ 处的切线方程。 + +解法一（直接求导） + +设 $y = y(x) 、$ 是由方程确定的隐函数，且 $y(1)=1$。两边对 $x$ 求导，并记 $y' = \frac{dy}{dx}$。分别计算左边导数 $\frac{dL}{dx}$ 和右边导数 $\frac{dR}{dx}$ 在点 $(1,1)$ 处的值。 + +左边：$L = e^{(xy-1)(x+y)} \ln(x^3+y^3)$。 +令 $$u = (xy-1)(x+y),$$则 $u(1,1)=0$。 +有$$u' = (y + x y')(x+y) + (xy-1)(1+y'),$$在 $(1,1)$ 处，$u' = 2(1+y')$。 +于是 $$\frac{dL}{dx} = e^{u} u' \ln(x^3+y^3) + e^{u} \cdot \frac{1}{x^3+y^3} \cdot (3x^2 + 3y^2 y')$$ +在 $(1,1)$ 处，$e^{u}=1$，$\ln(x^3+y^3)=\ln 2$，$x^3+y^3=2$，所以 $\frac{dL}{dx}\bigg|_{(1,1)} = 2(1+y') \ln 2 + \frac{3}{2}(1+y') = \left(2\ln 2 + \frac{3}{2}\right)(1+y')$。 + +右边：$R = \frac{1}{2} (xy)^{1-(x^2+xy+y^2)} (x^2+y^2) \ln(x+y)$。 +令 $$A = (xy)^{1-(x^2+xy+y^2)},B = x^2+y^2,C = \ln(x+y)$$则 $R = \frac{1}{2} A B C$。 +在 $(1,1)$ 处，$A=1$，$B=2$，$C=\ln 2$。 + +先求 $A'$。取对数：$\ln A = [1-(x^2+xy+y^2)] \ln(xy)$。求导得： +$$\frac{A'}{A} = [-(2x + y + x y' + 2y y')] \ln(xy) + [1-(x^2+xy+y^2)] \cdot \frac{y + x y'}{xy}$$ +在 $(1,1)$ 处，$$\ln(xy)=0,1-(x^2+xy+y^2) = -2, xy=1$$，所以 $\frac{A'}{A} = -2(1+y')$，即 $A' = -2(1+y')$。 +易得$$B' = 2x + 2y y' = 2 + 2y' = 2(1+y'), +C' = \frac{1+y'}{x+y} = \frac{1}{2}(1+y').$$ + +于是 $$\frac{dR}{dx} = \frac{1}{2} (A' B C + A B' C + A B C')$$ +代入：$$\frac{dR}{dx} = \frac{1}{2} \left[ -2(1+y') \cdot 2 \cdot \ln 2 + 1 \cdot 2(1+y') \cdot \ln 2 + 1 \cdot 2 \cdot \frac{1}{2}(1+y') \right]$$ +化简：$$\frac{dR}{dx} = \frac{1}{2} \left[ -4(1+y') \ln 2 + 2(1+y') \ln 2 + (1+y') \right] = \frac{1}{2} (1+y') (1 - 2\ln 2)$$ + +由 $\frac{dL}{dx} = \frac{dR}{dx}$ 得： +$$\left(2\ln 2 + \frac{3}{2}\right)(1+y') = \frac{1}{2} (1+y') (1 - 2\ln 2)$$ +若 $1+y' \neq 0$，则 $2\ln 2 + \frac{3}{2} = \frac{1}{2} - \ln 2$，即 $3\ln 2 = -1$，矛盾。故 $1+y' = 0$，即 $y' = -1$。 + +所以切线斜率为 $-1$，切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 + +答案：切线方程为 $y = -x + 2$。 + + 解法二 + +由方程形式可知，交换 $x$ 与 $y$ 后方程不变，故曲线关于直线 $y = x$ 对称。点 $(1,1)$ 位于曲线且落在对称轴 $y = x$ 上。 +在对称轴上的点处，切线斜率只可能为 $1$（与对称轴平行）或 $-1$（与对称轴垂直）。 +因为函数单调减，所以在 $(1,1)$ 处导数 $y' < 0$，故斜率不可能为 $1$，只可能为 $-1$。 +由点斜式得切线方程： +$y - 1 = -1 \cdot (x - 1)$， +即 $y = -x + 2$。 + + +>[!example] 例6(卢吉辚原创难题) +>已知由方程 $\arctan\left( \frac{x^2 y + x y^2}{2} \right) = \arcsin\left( \frac{\sqrt{x^2 + y^2}}{x + y} \right)$ 所确定的函数在其连续区间内单调减，求其在点 $(1,1)$ 处的切线方程。 + +解法一（直接求导） + +对方程两边关于 $x$ 求导，$y$ 视为 $x$ 的函数，记 $y' = \frac{dy}{dx}$。 +左边：设 $u = \frac{x^2 y + x y^2}{2}$，则导数为 $\frac{1}{1+u^2} \cdot \frac{du}{dx}$。 +$\frac{du}{dx} = \frac{1}{2}(2xy + x^2 y' + y^2 + 2xy y')$。 +在 $(1,1)$ 处，$$u=1,\frac{1}{1+u^2} = \frac{1}{2}，\frac{du}{dx} = \frac{1}{2}(2+1 + (1+2)y') = \frac{3}{2}(1+y')$$ +所以左边导数为 $$\frac{1}{2} \cdot \frac{3}{2}(1+y') = \frac{3}{4}(1+y')$$ + +右边：设 $v = \frac{\sqrt{x^2 + y^2}}{x + y}$，则导数为 $\frac{1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$。 + +在 $(1,1)$ 处，$v = \frac{\sqrt{2}}{2}$，$\sqrt{1-v^2} = \frac{1}{\sqrt{2}}$，所以 $\frac{1}{\sqrt{1-v^2}} = \sqrt{2}$。 +计算 $\frac{dv}{dx}$：$$v = \frac{\sqrt{x^2+y^2}}{x+y}， +\frac{dv}{dx} = \frac{(x+y) \cdot \frac{x + y y'}{\sqrt{x^2+y^2}} - \sqrt{x^2+y^2}(1+y')}{(x+y)^2}$$ +代入 $(1,1)$，$$\sqrt{x^2+y^2} = \sqrt{2}，x+y=2，\frac{dv}{dx} = \frac{2 \cdot \frac{1+y'}{\sqrt{2}} - \sqrt{2}(1+y')}{4} = 0$$ +因此右边导数为 $\sqrt{2} \cdot 0 = 0$。 + +由左右导数相等得：$\frac{3}{4}(1+y') = 0$，解得 $y' = -1$。 + +故切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 + + + 解法二 + +方程中交换 $x$ 与 $y$ 后形式不变，故曲线关于直线 $y = x$ 对称。点 $(1,1)$ 在曲线上且位于对称轴 $y = x$ 上。 +由对称性，在对称轴上的点处，切线斜率只可能为 $1$（与对称轴平行）或 $-1$（与对称轴垂直）。 +已知函数单调减，即在定义区间内 $y' < 0$，因此斜率不可能为 $1$，只能为 $-1$。 + +切线方程为 $y - 1 = -1 \cdot (x - 1)$，即 $y = -x + 2$。 + +--- + +## 参数方程求导 + +### 原理 + +已知 $$\left\{ \begin{array}{c} x=x\left( t \right)\\ y=y\left( t \right)\\ \end{array} \right.$$若有均可导且 $x'(t) \neq 0$，则 +$$y'\left( t \right) =\frac{dy}{dt}$$ +$$x'\left( t \right)=\frac{dx}{dt}$$ + $$y'=\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{y'\left( t \right)}{x'\left( t \right)}$$ +则有 $$y''=\frac{d^2y}{dx^2}=\frac{dy'}{dx}=\frac{dy'}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{dy'}{dt}\cdot \frac{1}{x'\left( t \right)}=(\frac{y'(t)}{x'(t)})'\cdot \frac{1}{x'\left( t \right)}$$ + +而 $$\frac{y'(t)}{x'(t)}=\frac{y''\left( t \right) x'\left( t \right) -x''\left( t \right) y'\left( t \right)}{x'^2\left( t \right)}$$ + +故 $$y''=\frac{y''\left( t \right) x'\left( t \right) -x''\left( t \right) y'\left( t \right)}{x'^3\left( t \right)}$$ +### **适用情况** +适用于曲线由参数形式给出，尤其是： +- 物理中的运动轨迹； +- 极坐标、摆线、旋轮线等曲线； +- 复杂曲线的简化表示。 + +### **优势** +1. 形式简洁，直接利用两个导数作商； +2. 适用于参数化表示，便于计算高阶导数。 + +### **劣势** +1. 要求 $x'(t) \neq 0$，否则导数不存在； +2. 高阶导数计算需重复使用公式，略显繁琐。 + +### **例子** + +> [!example] 例1(简单) +> 求参数方程 +> $$ +\begin{cases} +x = a\cos t \\ +y = b\sin t +\end{cases}$$ +所确定的函数 $y = y(x)$ 的导数。 + +**解析** + +计算： + +$$ +\frac{dx}{dt} = -a\sin t, \quad \frac{dy}{dt} = b\cos t +$$ + +因此： +$$ +\frac{dy}{dx} = \frac{b\cos t}{-a\sin t} = -\frac{b}{a}\cot t +$$ + + +> [!example] 例2 +设 $y$ 由方程 $e^{x+y} = xy$ 确定，则 $y'$ 在点 $(1,0)$ 处的值为 +A. $0$ +B. $1$ +C. $-1$ +D. 不存在 + +**解析** +两边对 $x$ 求导： +$$ +e^{x+y}(1 + y') = y + x y' +$$ +代入 $(1,0)$： +$$ +e^{1+0}(1 + y') = 0 + 1 \cdot y' +$$ +即： +$$ +e(1 + y') = y' +$$ +解得： +$$ +y' = \frac{e}{1 - e} +$$ +该值不为选项中所列，故判断为“不存在直接匹配”，但若只允许选 ABCD，则可能选 D。 +实际应说明：$y' = \frac{e}{1-e}$ 是一个确定数值。 + + +> [!example] 例3 +> 参数方程 +> $$ +\begin{cases} +x = t^2 + 1 \\ +y = t^3 - t +\end{cases}$$ +在 $t = 1$ 处的导数 $\left. \frac{dy}{dx} \right|_{t=1}$ 为 +A. $1$ +B. $2$ +C. $3$ +D. $4$ + +**解析** +计算： +$$ +\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 1 +$$ +则： +$$ +\frac{dy}{dx} = \frac{3t^2 - 1}{2t} +$$ +代入 $t = 1$： +$$ +\left. \frac{dy}{dx} \right|_{t=1} = \frac{3 \cdot 1^2 - 1}{2 \cdot 1} = \frac{2}{2} = 1 +$$ +答案：A + +>[!example] 例4(卢吉辚原创难题) +>设参数方程 $x = e^{t^2} \arcsin(t^2)$，$y = \ln(1+t^4) \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值（需要用到泰勒展开，仅作习题） + +**解析：** +先求一阶导： +$$\frac{dx}{dt} = 2t e^{t^2} \left( \frac{1}{\sqrt{1-t^4}} + \arcsin(t^2) \right)$$$$\frac{dy}{dt} = \frac{4t^3}{1+t^4} \arctan(t^3) + \ln(1+t^4) \cdot \frac{3t^2}{1+t^6}$$ +在 $t=0$ 处，$\frac{dx}{dt}=0$，$\frac{dy}{dt}=0$。 + +需用公式 $$\frac{d^2 y}{dx^2} = \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$$。 +利用泰勒展开： +$x(t) = t^2 + t^4 + O(t^6)$，得 $$\frac{dx}{dt} = 2t + 4t^3 + O(t^5),\frac{d^2 x}{dt^2} = 2 + 12t^2 + O(t^4);$$ +$y(t) = t^7 + O(t^{11})$，得 $$\frac{dy}{dt} = 7t^6 + O(t^{10})，\frac{d^2 y}{dt^2} = 42t^5 + O(t^9).$$ +分子：$$\frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} = 70 t^6 + O(t^7)$$分母：$$(\frac{dx}{dt})^3 = 8 t^3 + O(t^5)$$ +当 $t \to 0$，$\frac{d^2 y}{dx^2} = \frac{70 t^6}{8 t^3} \to 0$，故值为 $0$。 + +**答案：** $\frac{d^2 y}{dx^2} \big|_{t=0} = 0$。 + +>[!example] 例5(卢吉辚原创难题) +>设参数方程 $x = e^{t} \sin t + \ln(1+t^2)$，$y = e^{t} \cos t + \arctan(t^3)$，求 $\frac{d^2 y}{dx^2}$ 在 $t=0$ 处的值。 + +**解析：** +计算一阶导： +$$\frac{dx}{dt} = e^{t} \sin t + e^{t} \cos t + \frac{2t}{1+t^2}，\frac{dy}{dt} = e^{t} \cos t - e^{t} \sin t + \frac{3t^2}{1+t^6}.$$ +在 $t=0$ 处：$\frac{dx}{dt} = 1$，$\frac{dy}{dt} = 1$。 + +二阶导： +$$\frac{d^2 x}{dt^2} = 2 e^{t} \cos t + \frac{2 - 2t^2}{(1+t^2)^2}，\frac{d^2 y}{dt^2} = -2 e^{t} \sin t + \frac{6t(1+t^6) - 18t^7}{(1+t^6)^2}.$$ +在 $t=0$ 处：$$\frac{d^2 x}{dt^2} = 4，\frac{d^2 y}{dt^2} = 0.$$ + +代入公式： +$$\frac{d^2 y}{dx^2} = \frac{1 \cdot 0 - 1 \cdot 4}{1^3} = -4.$$ + +**答案：** $\frac{d^2 y}{dx^2} \big|_{t=0} = -4$。 + +## **应用题** + +应用题就比较综合了，它可以考很多知识点，比如参数方程，比如隐函数求导，也有可能就是一个普通的求导。但关键不在于求导，关键在于建立一个准确的模型，也就是 + +（1）把需要求的对象找出来，即自变量和因变量，注意可能有多个。 +（2）根据条件把关系式列出来，画一画示意图有时候会很有帮助。 + +### 微分估计值 + +>[!example] 例1 +>利用微分估计值也是导数应用的一种。利用微分估计下面几个式子的值： +>（1）$\sqrt{34}$（保留两位小数） +>（2）摆的振动周期公式为：$T=2\pi \sqrt{\frac{l}{g}}$。其中$l$为摆长（厘米），重力加速度$g=981cm^2/s$.为了使周期$T$增大$0.05$秒，摆长$l=20cm$的长度需要作何修正？（参考数据：$\sqrt{981\times20}\approx140.07,\frac{1}{\pi}\approx0.318$,保留两位小数） + +解：（1）$\sqrt{34}=\sqrt{36-2}=6\sqrt{1-\frac{1}{18}}$,考虑函数$f(x)=\sqrt{x}$，$f'(x)=\frac{1}{2\sqrt{x}},f'(1)=\frac{1}{2}$,故$df=\frac{dx}{2\sqrt{x}},df|_{x=1}=\frac{1}{2}dx$,取$dx=-\frac{1}{18}$,得$df=-\frac{1}{36}$,故$$\sqrt{34}\approx 6(1-\frac{1}{36})=\frac{35}{6}\approx5.83$$ +（2）$dT=T'dl=\pi\sqrt{\frac{1}{gl}}dl=0.05$,则$$dl=\frac{1}{\pi}\sqrt{gl}\approx\frac{1}{3.14}\times\sqrt{981\times20}\times0.05\approx2.23$$ +故需要增加摆长约$2.23$厘米 + +### 相关变化率 + +#### 飞机航空摄影问题 + +##### 问题描述 + +>[!example] 例题 +>一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 + +解析： + +1、建立坐标系 + +目标位置：原点 $O(0,0)$ +飞机位置：$P(x,2)$ +飞机高度：$h = 2 km$ +水平速度：$\frac{dx}{dt} = -200 km/h$ + +2、角度关系 + +$$ +\theta = \arctan\left(\frac{x}{2}\right), +\tan\theta =\frac{x}{2} +$$ + +3、角速率计算 + +$$\frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}$$ +$$\frac{d\theta}{dx} = \frac{d}{dx} \arctan\left(\frac{x}{2}\right) = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} = \frac{2}{x^2 + 4}$$ +$$\frac{d\theta}{dt} = \frac{2}{x^2 + 4} \cdot (-200) = -\frac{400}{x^2 + 4} rad/h$$ + +4、目标正上方时的角速率 + +当 $x = 0$ 时： + +$$\left. \frac{d\theta}{dt} \right|_{x=0} = -\frac{400}{0 + 4} = -100 rad/h$$ + +5、转换单位： + +$$100 rad/h = 100 \times \frac{180}{\pi} °/h = \frac{18000}{\pi} °/h, + +\frac{18000}{\pi} °/h \times \frac{1}{3600} h/s = \frac{5}{\pi} °/s \approx 1.59 °/s$$ + +答案: + +飞机飞至目标正上方时，摄影机转动的角速率为 $\frac{5}{\pi} °/s$（约$1.59 °/s$）。 + +--- + +#### 人拉船问题 + +##### 问题描述 + +>[!example] 例题 +>人在河岸用绳经过定滑轮以速度 $v$、加速度 $a$ 拉船。绳与水平面夹角为 $\theta$。求此时船的加速度 $a'$。 +![[应用题人拉船.png]] + +解析 + +1、 变量定义 + +$h$：滑轮高度（定值） +$x$：船到河岸的水平距离 +$l$：绳长 +$v = -\frac{dl}{dt}$：人拉绳的速度 +$a = -\frac{d^2l}{dt^2}$：人拉绳的加速度 +$u = -\frac{dx}{dt}$：船的水平速度 +$a' = -\frac{d^2x}{dt^2}$：船的加速度 +$\theta$：绳与水平面夹角 + +2、几何关系 + +$l^2 = h^2 + x^2$ + +3、 一阶导数关系 + +对几何关系求导：$$2l\frac{dl}{dt} = 2x\frac{dx}{dt}\ 即\ l\frac{dl}{dt} = x\frac{dx}{dt}$$ + +代入速度定义：$l(-v) = x(-u)$ + +$$u = \frac{l}{x}v = \frac{v}{\cos\theta}，其中 \cos\theta = \frac{x}{l}$$ + +4、 二阶导数关系 + +对一阶关系求导：$$\frac{d}{dt}\left(l\frac{dl}{dt}\right) = \frac{d}{dt}\left(x\frac{dx}{dt}\right)$$ + +$$\left(\frac{dl}{dt}\right)^2 + l\frac{d^2l}{dt^2} = \left(\frac{dx}{dt}\right)^2 + x\frac{d^2x}{dt^2}$$ + +$$(-v)^2 + l(-a) = (-u)^2 + x(-a')$$ + +$$v^2 - la = u^2 - xa'$$ + +5、 求解 $a'$ + +$$xa' = u^2 - v^2 + la, + +a' = \frac{u^2 - v^2 + la}{x}$$ + +代入 $u = \frac{v}{\cos\theta}$, $x = h\tan\theta$, $l = \frac{h}{\cos\theta}$： + +$$\begin{aligned} +a' &= \frac{\left(\frac{v}{\cos\theta}\right)^2 - v^2 + \left(\frac{h}{\cos\theta}\right)a}{h\tan\theta}\\[1em] + &= \frac{v^2\left(\frac{1}{\cos^2\theta} - 1\right)}{h\tan\theta} + \frac{\frac{h}{\cos\theta}a}{h\tan\theta}\\[1em] + &= \frac{v^2\tan^2\theta}{h\tan\theta} + \frac{a}{\cos\theta\tan\theta}\\[1em] + &= \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta} +\end{aligned}$$ + + 答案： + +此时船的加速度为： +$$a' = \frac{v^2\tan\theta}{h} + \frac{a}{\sin\theta}$$ + +--- + +#### 动点曲线运动问题 + +##### 问题描述 + +>[!example] 例题 +>已知动点 $P$ 在曲线 $y = \sqrt{x}$ 上运动，记坐标原点 $O$ 与 $P$ 间的距离为 $l$。若点 $P$ 横坐标随时间的变化率为常数 $v$，则当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率是多少？ + +解析: + +1、 变量关系 + +点 $P$ 坐标：$(x, y)$, $y = \sqrt{x}$ + +原点 $O$ 与 $P$ 间的距离： +$$l = \sqrt{x^2 + y^2} = \sqrt{x^2 + (\sqrt{x})^2} = \sqrt{x^2 + x}$$ + +已知：$\frac{dx}{dt} = v$（常数） + +2、距离变化率计算 + +$$\begin{aligned} +\frac{dl}{dt} &= \frac{d}{dt} \sqrt{x^2 + x} = \frac{d}{dt} (x^2 + x)^{1/2}\\[1em] + &= \frac{1}{2}(x^2 + x)^{-1/2} \cdot \frac{d}{dt}(x^2 + x)\\[1em] + &= \frac{1}{2\sqrt{x^2 + x}} \cdot (2x + 1) \frac{dx}{dt}\\[1em] + &= \frac{2x + 1}{2\sqrt{x^2 + x}} \cdot v +\end{aligned}$$ + +3、在点 $(1, 1)$ 处的变化率 + +当 $x = 1$ 时： + +$$\left. \frac{dl}{dt} \right|_{x=1} = \frac{2\cdot 1 + 1}{2\sqrt{1^2 + 1}} \cdot v = \frac{3}{2\sqrt{2}} \cdot v = \frac{3v}{2\sqrt{2}}$$ + +有理化得：$$\frac{3v}{2\sqrt{2}} = \frac{3v\sqrt{2}}{4}$$ + + 答案： + +当点 $P$ 运动到点 $(1, 1)$ 时，$l$ 对时间的变化率为 $\frac{3v\sqrt{2}}{4}$。 + +--- + +#### 球浸入水问题 + +##### 问题描述 + +>[!example] 例题 +>半径为$a$的球渐渐沉入盛有部分水的半径为$b(b>a)$的圆柱形容器中.若球以匀速$c$下沉，求：球浸没一半时，容器内水面上升的速率. +> + +解：主要的等量关系是：球浸入水中的体积与水“溢出”的体积相同（如果把水面上升视为溢出的话）。设时间为$t$，球下沉的体积为$V$，水上升的高度为$H$，球浸入水的高度为$h$，则有$$V=\pi h^2(a-\frac{h}{3}),dV=(2\pi ah-\pi h^2)dh=\pi h(2a-h)dh$$（这个公式大家可以记住） +且（如图）$$dV=\pi(b^2-a^2+(a-h)^2)dH$$ +![[球浸入水示意图.jpg]] +于是$$\pi(b^2-a^2+(a-h)^2)dH=\pi h(2a-h)dh$$$$dH=\frac{h(2a-h)}{b^2-2ah+h^2}dh$$$$\frac{dH}{dt}=\frac{h(2a-h))}{b^2-2ah+h^2}\frac{dh}{dt}$$而$\frac{dh}{dt}=c,h=a$，故$$\frac{dH}{dt}=\frac{a^2c}{b^2-a^2}$$ + +### 应用题期中真题 + +>[!example] **例1**（2020） +>有一个正圆锥形漏斗，深度 $18 \text{ cm}$ ，上端口直径为 $12 \text{ cm}$ （即半径 $6 \text{ cm}$ ）。漏斗下方连接一个圆柱形筒，圆柱筒直径为 $10 \text{ cm}$ （即半径 $5 \text{ cm}$ ）。初始时刻漏斗内盛满水，然后水从漏斗流入圆柱筒。已知当漏斗中水深为 $12 \text{ cm}$ 时，漏斗水面下降的速度是 $1 \text{ cm/min}$ +问：此时圆柱筒内液面上升的速度是多少？ + + +**解析**: +设 t （s）时漏斗水深为 h （cm），圆柱形容器的水深为 H （cm），则有 +$$\pi 5^2 H = V_0 - \frac{1}{3} \pi \left( \frac{6}{18} h \right)^2 h$$ +关于 t 求导数得 +$$25 \frac{dH}{dt} = -\frac{1}{9} h^2 \frac{dh}{dt}$$ +当 h = 12, $\frac{dh}{dt} = -1$ 时， +$$\frac{dH}{dt} = \frac{16}{25} \text{(cm/s)}$$ + + +>[!example] **例2** (2021) +>一长为 L 米的木梯靠在倾角为$\frac{\pi}{3}$的光滑斜坡上，木梯的顶部距离 A 点 h 米，底部距离 A 点 d 米，受重力作用木梯的顶部以 $a \, \mathrm{m/s}$ 的速度沿直线 BA 下滑，底部水平向右运动。问：当木梯的顶部和底部与 A 点的距离相等时，底部的水平速度为多少？ +>![[应用题梯子.jpg]] + + + +**解析：** +设运动 t 秒后，木梯的顶部距离 A 点 $y(t) \, \mathrm{m}$ ，底部距离 A 点 $x(t) \, \mathrm{m}$ 。由图易知 +$$\left(y(t)\sin\frac{\pi}{3}\right)^2 + \left(x(t)+y(t)\cos\frac{\pi}{3}\right)^2 = L^2$$ +即 +$$x^2(t)+y^2(t)+x(t)y(t)=L^2$$ +方程两端分别对 t 求导，可得 +$$2x(t)x'(t)+2y(t)y'(t)+x(t)y'(t)+x'(t)y(t)=0$$ +由于 $y'(t)=-a \, (\mathrm{m/s})$ ，因此当 x(t)=y(t) 时，有 +$$2x'(t)-2a-a+x'(t)=0$$ +故 $$x'(t)=a \, (\mathrm{m/s})$$ + + +>[!example] **例3**（2023） +>有一个长度为5m的梯子贴靠在铅直的墙上，假设其下端沿地板以3m/s的速率离开墙脚而滑动，则 +(1) 何时梯子的上、下端能以相同的速率移动？ +(2) 何时其上端下滑之速率为4m/s？ + + +**解析:** +设在时刻 t ，梯子上端与墙角的距离为 $x(t) \, \mathrm{m}$ ，下端与墙角的距离为 $y(t) \, \mathrm{m}$ ，则 +$$x^2 + y^2 = 25$$ +两端同时对 t 求导数可得 +$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$$ + +(1) 当$\frac{dx}{dt} = -\frac{dy}{dt}$ 时，可得 x = y ，进而 $y = \frac{5}{\sqrt{2}} \, (\mathrm{m})$ ， +即下端距墙脚 $\frac{5}{\sqrt{2}} \, \mathrm{m}$ 时上下端能以相同的速率移动。 + +(2) 当 $\frac{dy}{dt} = -4 \, (\mathrm{m/s})$ ， $\frac{dx}{dt} = 3 \, (\mathrm{m/s})$ 时，有 $4x = 3y$ ，将其代入 $x^2 + y^2 = 25$ ， +解得$y = 4 \, (\mathrm{m})$,即下端距墙脚4 m时，上端下滑速度为$4 m/s$ + + + +>[!example] **例4**(2022) +>某部举行八一阅兵，队列正步通过阅兵台时步幅间距离为75厘米，步速为每分钟112步。某观礼人员离行进队列垂直距离为60米，视线追随队列领队，求其视线与队列夹角为$\frac{\pi}{6}$时，视线转动角度的变化率。 + + + +**解析：** +- 建立几何模型：设时间为\(t\)分钟，步幅间距离0.75米，步速为每分钟112步，则队列行进距离$x = 0.75×112t = 84t$米。设视线与队列夹角为$theta$，由正切函数关系可得$tan\theta=\frac{x}{60}$，即$\tan\theta=\frac{84t}{60}=\frac{7t}{5}$。 +- 两边对$t$求导：根据复合函数求导法则 +$$(tan\theta)^\prime=\sec^{2}\theta\cdot\frac{d\theta}{dt})$$对$tan\theta=\frac{7t}{5}$两边求导得$$sec^{2}\theta\cdot\frac{d\theta}{dt}=\frac{7}{5}$$ +求解:$\frac{d\theta}{dt}$：当$\theta = \frac{\pi}{6}$时，$sec\theta=\frac{2}{\sqrt{3}}$，$sec^{2}\theta=\frac{4}{3}$，代入$sec^{2}\theta\cdot\frac{d\theta}{dt}=\frac{7}{5}$， + +可得$$\frac{4}{3}\cdot\frac{d\theta}{dt}=\frac{7}{5}$$解得$$\frac{d\theta}{dt}=\frac{21}{20}（弧度/分钟）$$ +答案：视线转动角度的变化率为 $\frac{21}{20}$ 弧度/分钟。本题主要考察了利用导数解决实际问题中的变化率问题，关键在于建立正确的函数关系并准确求导，易错点是复合函数求导法则的运用。 + + +>[!example] **例5**（2024） +>一架巡逻直升机在距离地面3km的高度以120km/h的速度沿着一条水平笔直的高速公路向前飞行。飞行员观察到迎面驶来一辆汽车，通过雷达测出直升机与汽车的距离为5km，同时此距离以160km/h的速率在减少。试求此时汽车行进的速度。 + + +**解析**： + +如图建立直角坐标系，设时刻 t直升机位于 A点 $(x_1​(t),3)$，汽车位于 B点 $(x_2​(t),0)$，直升机与汽车的距离为 $z(t)$，则 + +$$(x_2​(t)−x_1​(t))^2+32=z^2(t)$$ + +方程两端分别对 $t$ 求导，可得 +$$(x_2​(t)−x_1​(t))(x_2^′​(t)−x_1^′​(t))=z(t)z^′(t)$$ + +由于 $z(t)=5$时，$x_2​(t)−x_1​(t)=4$，$z^′(t)=−160$，$x_1^′​(t)=120$，有 + +$$4(x_2^′​(t)−120)=5×(−160)$$ + +故 $x_2^′​(t)=−80$，即汽车行进的速度为 $80km/h$。 \ No newline at end of file diff --git a/编写小组/课前测/课前测1.md b/编写小组/课前测/课前测1.md new file mode 100644 index 0000000..28de5c4 --- /dev/null +++ b/编写小组/课前测/课前测1.md @@ -0,0 +1,58 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁** + +1.计算极限 $$\lim_{x \to 0} \frac{\sqrt{1 + x} + \sqrt{1 - x} - 2}{x^2}$$A. $\frac{1}{2}$ + +B. $\frac{1}{4}$ + +C. $-\frac{1}{2}$ + +D. $-\frac{1}{4}$ + + + +2.设 $$a_n = \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{n^2+n}}$$则 $\lim_{n \to \infty} a_n$ = ? + +A. 0 + +B. 1 + +C. $\frac{1}{2}$ + +D. 不存在 + + +3.计算 +$$\lim\limits_{x\to\infty}\tfrac{3x^2+5}{5x+3}sin\tfrac{2}{x}$$A. $\frac{6}{5}$ + +B. $\frac{4}{3}$ + +C. $-\frac{4}{3}$ + +D. $\frac{7}{8}$ + + +4.计算 +$$\lim\limits_{x\to0}\frac{\sqrt{1+2tanx}-\sqrt{1+2sinx}}{xln(1+x)-x^2}$$     +A、原式= $\infty$ + +B、原式=1 + +C、原式=-1 + +D、原式=0 + + +5.设$$f(x)=\frac{2x+1}{\sqrt{x^2+3}}$$则$\lim\limits_{x\to\infty}f(x)=()$ + +(A)不存在 + +(B)-2    + +(C)0   + +(D)2 \ No newline at end of file diff --git a/小测/12.22/课前测.md b/编写小组/课前测/课前测2.md similarity index 100% rename from 小测/12.22/课前测.md rename to 编写小组/课前测/课前测2.md diff --git a/编写小组/课前测/课前测3.md b/编写小组/课前测/课前测3.md new file mode 100644 index 0000000..1fb86d3 --- /dev/null +++ b/编写小组/课前测/课前测3.md @@ -0,0 +1,4 @@ +--- +tags: + - 编写小组 +--- diff --git a/编写小组/课前测/课前测解析版1.md b/编写小组/课前测/课前测解析版1.md new file mode 100644 index 0000000..8456f4a --- /dev/null +++ b/编写小组/课前测/课前测解析版1.md @@ -0,0 +1,109 @@ +--- +tags: + - 编写小组 +--- +1.计算极限 $$\lim_{x \to 0} \frac{\sqrt{1 + x} + \sqrt{1 - x} - 2}{x^2}$$A. $\frac{1}{2}$ + +B. $\frac{1}{4}$ + +C. $-\frac{1}{2}$ + +D. $-\frac{1}{4}$ + +答案：D + +有理化 + +令分子为 $A = \sqrt{1 + x} + \sqrt{1 - x} - 2$，分子分母同乘共轭：$$A = \frac{(\sqrt{1 + x} + \sqrt{1 - x})^2 - 4}{\sqrt{1 + x} + \sqrt{1 - x} + 2} = \frac{2\sqrt{1 - x^2} - 2}{\sqrt{1 + x} + \sqrt{1 - x} + 2}$$原式变为 +$$\frac{A}{x^2} = \frac{2\sqrt{1 - x^2} - 2}{x^2 (\sqrt{1 + x} + \sqrt{1 - x} + 2)}$$ +再对分子 $2\sqrt{1 - x^2} - 2$ 有理化： +$$2\sqrt{1 - x^2} - 2 = \frac{-2x^2}{\sqrt{1 - x^2} + 1}$$ +代入得 +$$\frac{A}{x^2} = \frac{-2}{(\sqrt{1 - x^2} + 1)(\sqrt{1 + x} + \sqrt{1 - x} + 2)}$$ +当 $x \to 0$ 时， +$$\sqrt{1 - x^2} \to 1, \quad \sqrt{1 + x} \to 1, \quad \sqrt{1 - x} \to 1$$ +所以分母趋于 $$(1 + 1) \times (1 + 1 + 2) = 2 \times 4 = 8$$因此极限为 +$$\frac{-2}{8} = -\frac{1}{4}$$ + +2.设 $$a_n = \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{n^2+n}}$$则 $\lim_{n \to \infty} a_n$ = ? + +A. 0 + +B. 1 + +C. $\frac{1}{2}$ + +D. 不存在 + +答案：B + +解析： 这是经典的夹逼定理题。 + +· 共有 n 项。 + +· 分母最大项是 $\sqrt{n^2+1}$（最小分数），分母最小项是 $\sqrt{n^2+n}$（最大分数）。 + +· 因此： + +$$\frac{n}{\sqrt{n^2+n}} \leq a_n \leq \frac{n}{\sqrt{n^2+1}}$$  +· 计算两边的极限： +$$\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}} = \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n}}} = 1$$ +$$\lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} = \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n^2}}} = 1$$ +· 由夹逼定理，$\lim_{n \to \infty} a_n = 1$ + +3.计算 +$$\lim\limits_{x\to\infty}\frac{3x^2+5}{5x+3}sin\frac{2}{x}$$A. $\frac{6}{5}$ + +B. $\frac{4}{3}$ + +C. $-\frac{4}{3}$ + +D. $\frac{7}{8}$ + +答案：A +$$\lim\limits_{x\to\infty}\frac{3x^2+5}{5x+3}sin\frac{2}{x}=\lim\limits_{x\to\infty}\frac{6x^2+10}{5x^2+3x}\frac{sin\frac{2}{x}}{\frac{2}{x}}=\frac{6}{5}$$ + + +4.计算 +$$\lim\limits_{x\to0}\frac{\sqrt{1+2tanx}-\sqrt{1+2sinx}}{xln(1+x)-x^2}$$     +A、原式= $\infty$ + +B、原式=1 + +C、原式=-1 + +D、原式=0 + +答案：C + +原式等于$$2\lim\limits_{x\to0}\frac{tanx-sinx}{x[ln(1+x)-x]}\frac{1}{\sqrt{1+2tanx}+\sqrt{1+2sinx}}$$ + +$$=2\lim\limits_{x\to 0} \frac{sinx(1/cosx-1)} {x[ln(1+x)-x]} \frac{1}{\sqrt{1+2tanx}+ \sqrt{1+2sinx}}$$ + +$$=\lim\limits_{x\to0}\frac{sinx}{x}\frac{1}{cosx}\frac{1-cosx}{ln(1+x)-x}$$ + + +5.设$$f(x)=\frac{2x+1}{\sqrt{x^2+3}}$$则$\lim\limits_{x\to\infty}f(x)=()$ + +(A)不存在 + +(B)-2    + +(C)0   + +(D)2 + +答案：A + + +解析：对于函数 $f(x) = \frac{2x + 1}{\sqrt{x^2 + 3}}$，考虑 $x \to \infty$ 时的极限。由于 $x \to \infty$ 通常包括 $x \to +\infty$ 和 $x \to -\infty$，需分别讨论： + +1. **当 $x \to +\infty$**： + 此时 $x > 0$，有 $\sqrt{x^2 + 3} = \sqrt{x^2(1 + \frac{3}{x^2})} = x\sqrt{1 + \frac{3}{x^2}}$， + 分子可写为 $2x + 1 = x(2 + \frac{1}{x})$，故 $f(x) = \frac{x(2 + \frac{1}{x})}{x\sqrt{1 + \frac{3}{x^2}}} = \frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x^2}}}$。 + 当 $x \to +\infty$ 时，$\frac{1}{x} \to 0$，$\frac{3}{x^2} \to 0$，所以 $\lim_{x \to +\infty} f(x) = \frac{2}{1} = 2$。 + +2. **当 $x \to -\infty$**： + 此时 $x < 0$，有 $\sqrt{x^2 + 3} = \sqrt{x^2(1 + \frac{3}{x^2})} = |x|\sqrt{1 + \frac{3}{x^2}} = -x\sqrt{1 + \frac{3}{x^2}}$， + 分子为 $2x + 1 = x(2 + \frac{1}{x})$，故 $f(x) = \frac{x(2 + \frac{1}{x})}{-x\sqrt{1 + \frac{3}{x^2}}} = -\frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x^2}}}$。 + 当 $x \to -\infty$ 时，$\frac{1}{x} \to 0$，$\frac{3}{x^2} \to 0$，所以 $\lim_{x \to -\infty} f(x) = -\frac{2}{1} = -2$。 \ No newline at end of file diff --git a/小测/12.22/课前测解析版.md b/编写小组/课前测/课前测解析版2.md similarity index 94% rename from 小测/12.22/课前测解析版.md rename to 编写小组/课前测/课前测解析版2.md index 87dddc9..44c8f73 100644 --- a/小测/12.22/课前测解析版.md +++ b/编写小组/课前测/课前测解析版2.md @@ -57,7 +57,7 @@ $$\lim_{n\to\infty} \frac{4n+2}{5(n+1)} = \frac{4}{5} < 1$$ 4.级数$\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{3^n}.$收敛 A.正确 B.错误 -$n(n-1)/2的奇偶性与n正好相反，所以级数也是交错级数，容易证明是收敛$ +$n(n-1)/2的奇偶性与n隔两项相反，所以级数也是交错级数，容易证明是收敛$ 5.级数 $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n}$ 收敛 A.正确 diff --git a/编写小组/课后测/课后测1.md b/编写小组/课后测/课后测1.md new file mode 100644 index 0000000..11aab26 --- /dev/null +++ b/编写小组/课后测/课后测1.md @@ -0,0 +1,51 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁** + +1.已知 $b_n = \sqrt[n]{2^n + 3^n + 4^n}$，则 $\lim_{n \to \infty} b_n = ?$ + +A. 2 + +B. 3 + +C. 4 + +D. 9 + + +2.计算 $$\displaystyle \lim_{n\to\infty} n^2[\ \arctan\ (n+1) - \arctan n]$$A. $\infty$ + +B. 1 + +C. $\tan1$ + +D. 0 + + +3.计算 +$$\lim\limits_{x\to\pi/2}(sinx)^{tanx}$$A. $\frac{\pi}{2}$ + +B. $\frac{\pi}{4}$ + +C. $-\frac{\pi}{2}$ + +D. $1$ + + +4.设 $$f(x) = (1+x^2)\cos\left(\frac{1}{x^3}\right)$$则 $\lim_{x \to \infty} f(x)$ + +A. 为 0 + +B. 为 1 + +C. 为-1 + +D. 无穷大 + + +5.设$$f(x)=\frac{2+e^{1/x}}{1+e^{4/x}}+\frac{sinx}{|x|}$$则$x=0$是 $f(x)$ 的（） + +（A）可去间断点  （B）跳跃间断点   （C）无穷间断点  （D）震荡间断点 diff --git a/小测/12.22/课后测.md b/编写小组/课后测/课后测2.md similarity index 98% rename from 小测/12.22/课后测.md rename to 编写小组/课后测/课后测2.md index 1fb0f8f..c554473 100644 --- a/小测/12.22/课后测.md +++ b/编写小组/课后测/课后测2.md @@ -20,12 +20,6 @@ B.错误 - - - - - - 4,级数$\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1}$ 收敛 A.正确 B.错误 diff --git a/编写小组/课后测/课后测3.md b/编写小组/课后测/课后测3.md new file mode 100644 index 0000000..1fb86d3 --- /dev/null +++ b/编写小组/课后测/课后测3.md @@ -0,0 +1,4 @@ +--- +tags: + - 编写小组 +--- diff --git a/小测/12.23/课后测解析版.md b/编写小组/课后测/课后测解析版1.md similarity index 55% rename from 小测/12.23/课后测解析版.md rename to 编写小组/课后测/课后测解析版1.md index 961c03a..b7443a0 100644 --- a/小测/12.23/课后测解析版.md +++ b/编写小组/课后测/课后测解析版1.md @@ -77,13 +77,9 @@ C. 为-1 D. 无穷大 -答案：C - -解析： 注意这里 $x \to \infty$。虽然可以用夹逼： - -$$-(1+x^2) \leq f(x) \leq (1+x^2)$$ +答案：D -但两边的极限 $\lim_{x \to \infty} (1+x^2) = \infty$，不相等（且不是有限数），因此夹逼定理不适用。实际上，由于 $\cos(1/x^3) 在 x \to \infty$ 时振荡趋于 $\cos 0 = 1$，而 $1+x^2 \to \infty$，所以函数值在振荡中趋于无穷大，极限不存在。 +解析： 由于 $\cos(1/x^3) 在 x \to \infty$ 时振荡趋于 $\cos 0 = 1$，而 $1+x^2 \to \infty$，所以函数值在振荡中趋于无穷大。 @@ -91,4 +87,30 @@ $$-(1+x^2) \leq f(x) \leq (1+x^2)$$ （A）可去间断点  （B）跳跃间断点   （C）无穷间断点  （D）震荡间断点 -答案：A \ No newline at end of file +答案：A + +对于函数 $f(x) = \frac{2 + e^{1/x}}{1 + e^{4/x}} + \frac{\sin x}{|x|}$，分析 $x=0$ 处的间断点类型。 + +首先考虑第二部分 $\frac{\sin x}{|x|}$ 在 $x=0$ 处的左右极限： +当 $x \to 0^+$ 时，$|x|=x$，所以 $\lim_{x \to 0^+} \frac{\sin x}{|x|} = \lim_{x \to 0^+} \frac{\sin x}{x} = 1$。 +当 $x \to 0^-$ 时，$|x|=-x$，所以 $\lim_{x \to 0^-} \frac{\sin x}{|x|} = \lim_{x \to 0^-} \frac{\sin x}{-x} = -1$。 +第二部分左右极限存在但不相等。 + +再考虑第一部分 $g(x) = \frac{2 + e^{1/x}}{1 + e^{4/x}}$ 的左右极限： + +- 当 $x \to 0^+$ 时，$1/x \to +\infty$，$4/x \to +\infty$，故 $e^{1/x} \to +\infty$，$e^{4/x} \to +\infty$，且 $e^{4/x}$ 增长更快。 + 可写为 $g(x) = \frac{e^{1/x}(2/e^{1/x} + 1)}{e^{4/x}(1/e^{4/x} + 1)} = e^{-3/x} \cdot \frac{2/e^{1/x} + 1}{1/e^{4/x} + 1}$。 + 由于 $e^{-3/x} \to 0$，分母趋于 $1$，所以 $\lim_{x \to 0^+} g(x) = 0$。 + +- 当 $x \to 0^-$ 时，$1/x \to -\infty$，$4/x \to -\infty$，故 $e^{1/x} \to 0$，$e^{4/x} \to 0$。 + 此时 $g(x) = \frac{2 + 0}{1 + 0} = 2$，所以 $\lim_{x \to 0^-} g(x) = 2$。 + + +因此，$f(x)$ 的左右极限为： + +- $\lim_{x \to 0^+} f(x) = 0 + 1 = 1$。 + +- $\lim_{x \to 0^-} f(x) = 2 + (-1) = 1$。 + + +左右极限存在且相等（等于 $1$），故 $x=0$ 为可去间断点。 \ No newline at end of file diff --git a/小测/12.22/课后测(解析).md b/编写小组/课后测/课后测解析版2.md similarity index 100% rename from 小测/12.22/课后测(解析).md rename to 编写小组/课后测/课后测解析版2.md