From 19428cb7dcf72fb16f09f86766ff815fdfa4a83e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Fri, 16 Jan 2026 12:59:26 +0800 Subject: [PATCH 01/15] vault backup: 2026-01-16 12:59:26 --- 素材/未命名.md | 1 - 1 file changed, 1 deletion(-) diff --git a/素材/未命名.md b/素材/未命名.md index 084f72d..a8ccd73 100644 --- a/素材/未命名.md +++ b/素材/未命名.md @@ -7,7 +7,6 @@ 则在 $(a,b)$ 内至少存在一点 $\xi$，使得 $f'(\xi)=0$。 罗尔定理的几何意义为：满足条件的函数曲线在区间内至少有一条水平切线。 它是拉格朗日中值定理（$f(b)-f(a)=f'(\xi)(b-a)$）当 $f(a)=f(b)$ 时的特例。 -fdfduhfidhf ### **适用条件** 罗尔定理的核心适用题型是证明导函数方程 $f'(\xi)=0$ 在区间 $(a,b)$ 内有根以及衍生的相关证明题。 具体可分为以下几类： From 387d2f2a4482c35c4dd7bc150a52cca635e99255 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Fri, 16 Jan 2026 12:59:39 +0800 Subject: [PATCH 02/15] vault backup: 2026-01-16 12:59:39 --- 素材/未命名.md | 107 -------------------------------------------- 1 file changed, 107 deletions(-) delete mode 100644 素材/未命名.md diff --git a/素材/未命名.md b/素材/未命名.md deleted file mode 100644 index a8ccd73..0000000 --- a/素材/未命名.md +++ /dev/null @@ -1,107 +0,0 @@ -## **罗尔定理** -### **原理** -若函数 f(x) 满足以下三个条件： -在闭区间 $[a,b]$ 上连续； -在开区间 $(a,b)$ 内可导； -区间端点函数值相等，即 $f(a)=f(b)$； -则在 $(a,b)$ 内至少存在一点 $\xi$，使得 $f'(\xi)=0$。 -罗尔定理的几何意义为：满足条件的函数曲线在区间内至少有一条水平切线。 -它是拉格朗日中值定理（$f(b)-f(a)=f'(\xi)(b-a)$）当 $f(a)=f(b)$ 时的特例。 -### **适用条件** -罗尔定理的核心适用题型是证明导函数方程 $f'(\xi)=0$ 在区间 $(a,b)$ 内有根以及衍生的相关证明题。 -具体可分为以下几类： -1.直接证明 $f'(\xi)$=0 存在根 -题目给出函数 f(x) 在 $[a,b]$ 上的连续性、$(a,b)$ 内的可导性，且满足 $f(a)=f(b)$，直接应用罗尔定理证明存在 $\xi\in(a,b)$ 使得 $f'(\xi)=0$。 -2.构造辅助函数证明导函数相关方程有根 -对于形如 $f'(\xi)+g(\xi)f(\xi)=0$、$f''(\xi)=0$ 等方程，需构造满足罗尔定理条件的辅助函数 $F(x)$，通过 $F(a)=F(b)$ 推导 $F'(\xi)=0$，进而等价转化为目标方程。 -3.结合多次罗尔定理证明高阶导数零点存在 -若函数 f(x) 有 n+1 个点的函数值相等，可多次应用罗尔定理，证明其 n 阶导数 $f^{(n)}(\xi)=0$ 在对应区间内有根。 -4.证明函数恒为常数（反证法结合罗尔定理） -若 $f'(x)\equiv0$ 在区间内成立，可通过反证法假设存在两点函数值不等，结合罗尔定理推出矛盾，进而证明函数为常数。 - -### **例题** ->[!example] 例1 -设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 - -**解析**： -设辅助函数 $\varphi(x) = x^n f(x)$，则 $\varphi(0)=0$，$\varphi(1)=0$。由罗尔定理，存在 $\xi \in (0,1)$，使得 $\varphi'(\xi)=0$即 -$$ -n\xi^{n-1} f(\xi) + \xi^n f'(\xi) = 0 -$$ -两边除以 $\xi^{n-1}$ ($\xi>0$)，得 $nf(\xi) + \xi f'(\xi) = 0$。 - - - ->[!example] 例2 -设函数 $f(x)$ 在 $[a,b]$ 上可导，且 -$$f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0，$$ -试证明 $f'(x) = 0$ 在 $(a,b)$ 内至少有两个根。 - - -**解析**： -由导数极限定理及 $f'_+(a)f'_-(b) > 0$，知在 $a$ 右侧和 $b$ 左侧，$f(x)$ 的符号相同，不妨设 $f'_+(a)>0$，$f'_-(b)>0$。则在 $a$ 右侧附近 $f(x)>0$，在 $b$ 左侧附近 $f(x)>0$。由于 $f(a)=f(b)=0$，由极值点的费马定理，$f(x)$ 在 $(a,b)$ 内至少有一个极大值点，该点处导数为零。又因为 $f(x)$ 在 $[a,b]$ 上连续，在 $(a,b)$ 内可导，且 $f(a)=f(b)$，由罗尔定理至少存在一点 $c \in (a,b)$ 使 $f'(c)=0$。结合极大值点处的导数零点，可知至少有两个导数为零的点。 - - - ->[!example] 例3 -设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 -$$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： -(1)至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； -(2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 - -**解析**： -(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理及罗尔定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 -(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 - - - - -## **拉格朗日中值定理** -### **原理** -若函数 f(x) 满足两个条件: -在闭区间 $[a,b]$ 上连续； -在开区间 $(a,b)$ 内可导； -则在 $(a,b)$ 内至少存在一点 $\xi$，使得 -$f(b)-f(a)=f'(\xi)(b-a)$ -也可写成等价形式 $f'(\xi)=\dfrac{f(b)-f(a)}{b-a}$。 -是罗尔定理的推广，同时也是柯西中值定理的特例。其几何意义为：满足条件的函数曲线在区间 (a,b) 内，至少存在一点的切线与连接端点 (a,f(a)) 和 (b,f(b)) 的弦平行。 - -### **适用条件** -拉格朗日中值定理的核心适用题型是建立函数增量与导数的关联，进行不等式的证明，这是最常见的题型。通过对目标函数在指定区间上应用拉格朗日中值定理，得到 $f(b)-f(a)=f'(\xi)(b-a)$，再利用导数 $f'(\xi)$ 的取值范围（有界性、正负性）放大或缩小式子，推导不等式。 - -### **例题** ->[!example] 例1 -设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 -$$f(0) = 0, \quad |f'(x)| \leq 1,$$证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 - -**解析**： -对任意 $x \in (-1,1)$，由拉格朗日中值定理，存在 $\xi$ 介于 $0$ 与 $x$ 之间，使得 -$$ -f(x) - f(0) = f'(\xi)(x-0) -$$ -即 $f(x) = f'(\xi) x$。由于 $|f'(\xi)| \leq 1$，$|x| < 1$，故 $|f(x)| = |f'(\xi)| \cdot |x| < 1$。 - - - - ->[!example] 例2 -设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 -$$f(x_1 + x_2) < f(x_1) + f(x_2)$$ - -**解析**： -不妨设 $0 < x_1 < x_2$。由拉格朗日中值定理： -$$ -f(x_1+x_2)-f(x_2) = f'(\xi_1)x_1, \quad \xi_1 \in (x_2, x_1+x_2) -$$ -$$ -f(x_1)-f(0) = f'(\xi_2)x_1, \quad \xi_2 \in (0, x_1) -$$ -于是 -$$ -f(x_1+x_2)-f(x_2)-f(x_1) = [f'(\xi_1)-f'(\xi_2)]x_1 -$$ -对 $f'(x)$ 在 $[\xi_2,\xi_1]$ 上应用拉格朗日中值定理，存在 $\xi \in (\xi_2,\xi_1)$，使 -$$ -f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 -$$ -故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 \ No newline at end of file From 68d83d62405f94138815a2503dde8468ed5caa87 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Fri, 16 Jan 2026 13:07:41 +0800 Subject: [PATCH 03/15] vault backup: 2026-01-16 13:07:41 --- 素材/罗尔定理和拉格朗日定理.md | 108 -------------------- 1 file changed, 108 deletions(-) delete mode 100644 素材/罗尔定理和拉格朗日定理.md diff --git a/素材/罗尔定理和拉格朗日定理.md b/素材/罗尔定理和拉格朗日定理.md deleted file mode 100644 index 20c295a..0000000 --- a/素材/罗尔定理和拉格朗日定理.md +++ /dev/null @@ -1,108 +0,0 @@ -## **罗尔定理** -### **原理** -若函数 f(x) 满足以下三个条件： -在闭区间 $[a,b]$ 上连续； -在开区间 $(a,b)$ 内可导； -区间端点函数值相等，即 $f(a)=f(b)$； -则在 $(a,b)$ 内至少存在一点 $\xi$，使得 $f'(\xi)=0$。 -罗尔定理的几何意义为：满足条件的函数曲线在区间内至少有一条水平切线。 -它是拉格朗日中值定理（$f(b)-f(a)=f'(\xi)(b-a)$）当 $f(a)=f(b)$ 时的特例。 - -### **适用条件** -罗尔定理的核心适用题型是证明导函数方程 $f'(\xi)=0$ 在区间 $(a,b)$ 内有根以及衍生的相关证明题。 -具体可分为以下几类： -1.直接证明 $f'(\xi)$=0 存在根 -题目给出函数 f(x) 在 $[a,b]$ 上的连续性、$(a,b)$ 内的可导性，且满足 $f(a)=f(b)$，直接应用罗尔定理证明存在 $\xi\in(a,b)$ 使得 $f'(\xi)=0$。 -2.构造辅助函数证明导函数相关方程有根 -对于形如 $f'(\xi)+g(\xi)f(\xi)=0$、$f''(\xi)=0$ 等方程，需构造满足罗尔定理条件的辅助函数 $F(x)$，通过 $F(a)=F(b)$ 推导 $F'(\xi)=0$，进而等价转化为目标方程。 -3.结合多次罗尔定理证明高阶导数零点存在 -若函数 f(x) 有 n+1 个点的函数值相等，可多次应用罗尔定理，证明其 n 阶导数 $f^{(n)}(\xi)=0$ 在对应区间内有根。 -4.证明函数恒为常数（反证法结合罗尔定理） -若 $f'(x)\equiv0$ 在区间内成立，可通过反证法假设存在两点函数值不等，结合罗尔定理推出矛盾，进而证明函数为常数。 - -### **例题** ->[!example] 例1 -设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 - -**解析**： -设辅助函数 $\varphi(x) = x^n f(x)$，则 $\varphi(0)=0$，$\varphi(1)=0$。由罗尔定理，存在 $\xi \in (0,1)$，使得 $\varphi'(\xi)=0$即 -$$ -n\xi^{n-1} f(\xi) + \xi^n f'(\xi) = 0 -$$ -两边除以 $\xi^{n-1}$ ($\xi>0$)，得 $nf(\xi) + \xi f'(\xi) = 0$。 - - - ->[!example] 例2 -设函数 $f(x)$ 在 $[a,b]$ 上可导，且 -$$f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0，$$ -试证明 $f'(x) = 0$ 在 $(a,b)$ 内至少有两个根。 - - -**解析**： -由导数极限定理及 $f'_+(a)f'_-(b) > 0$，知在 $a$ 右侧和 $b$ 左侧，$f(x)$ 的符号相同，不妨设 $f'_+(a)>0$，$f'_-(b)>0$。则在 $a$ 右侧附近 $f(x)>0$，在 $b$ 左侧附近 $f(x)>0$。由于 $f(a)=f(b)=0$，由极值点的费马定理，$f(x)$ 在 $(a,b)$ 内至少有一个极大值点，该点处导数为零。又因为 $f(x)$ 在 $[a,b]$ 上连续，在 $(a,b)$ 内可导，且 $f(a)=f(b)$，由罗尔定理至少存在一点 $c \in (a,b)$ 使 $f'(c)=0$。结合极大值点处的导数零点，可知至少有两个导数为零的点。 - - - ->[!example] 例3 -设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 -$$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： -(1)至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； -(2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 - -**解析**： -(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理及罗尔定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 -(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 - - - - -## **拉格朗日中值定理** -### **原理** -若函数 f(x) 满足两个条件: -在闭区间 $[a,b]$ 上连续； -在开区间 $(a,b)$ 内可导； -则在 $(a,b)$ 内至少存在一点 $\xi$，使得 -$f(b)-f(a)=f'(\xi)(b-a)$ -也可写成等价形式 $f'(\xi)=\dfrac{f(b)-f(a)}{b-a}$。 -是罗尔定理的推广，同时也是柯西中值定理的特例。其几何意义为：满足条件的函数曲线在区间 (a,b) 内，至少存在一点的切线与连接端点 (a,f(a)) 和 (b,f(b)) 的弦平行。 - -### **适用条件** -拉格朗日中值定理的核心适用题型是建立函数增量与导数的关联，进行不等式的证明，这是最常见的题型。通过对目标函数在指定区间上应用拉格朗日中值定理，得到 $f(b)-f(a)=f'(\xi)(b-a)$，再利用导数 $f'(\xi)$ 的取值范围（有界性、正负性）放大或缩小式子，推导不等式。 - -### **例题** ->[!example] 例1 -设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 -$$f(0) = 0, \quad |f'(x)| \leq 1,$$证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 - -**解析**： -对任意 $x \in (-1,1)$，由拉格朗日中值定理，存在 $\xi$ 介于 $0$ 与 $x$ 之间，使得 -$$ -f(x) - f(0) = f'(\xi)(x-0) -$$ -即 $f(x) = f'(\xi) x$。由于 $|f'(\xi)| \leq 1$，$|x| < 1$，故 $|f(x)| = |f'(\xi)| \cdot |x| < 1$。 - - - - ->[!example] 例2 -设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 -$$f(x_1 + x_2) < f(x_1) + f(x_2)$$ - -**解析**： -不妨设 $0 < x_1 < x_2$。由拉格朗日中值定理： -$$ -f(x_1+x_2)-f(x_2) = f'(\xi_1)x_1, \quad \xi_1 \in (x_2, x_1+x_2) -$$ -$$ -f(x_1)-f(0) = f'(\xi_2)x_1, \quad \xi_2 \in (0, x_1) -$$ -于是 -$$ -f(x_1+x_2)-f(x_2)-f(x_1) = [f'(\xi_1)-f'(\xi_2)]x_1 -$$ -对 $f'(x)$ 在 $[\xi_2,\xi_1]$ 上应用拉格朗日中值定理，存在 $\xi \in (\xi_2,\xi_1)$，使 -$$ -f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 -$$ -故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 \ No newline at end of file From 24f211ea13d3d36e74cc93cdc44da72a7256b478 Mon Sep 17 00:00:00 2001 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z{^olwbQp#I8LMHBB&2iW0B?2zuUcm%p4>pgP2MLX=E|&y``}WvG#}0 zQ++FW1gbyov{xpcda@c2d(EFTo;YW4UB7}ncmR(7CaeoNPkeGOnpk5T?e|+3BY1gT z*UrD50p<=!%gI$D1&mN3+0;$c_J3`ZrEbNJX|g^l`gd5o`_FIs>H6RDDt-(!@OC=y z7@y;nJPXV=bd!}rsk(7F`c~z4k}J6o8v~P8On*H2*Rx~1>VF2u|IoX_inR3YAT+mZ zP0kUzoOpjeV^CzQO+4Xs)1SQ-UW}BjQQr~hf0DEg-;Fh!|7+iCR(;|F56=91L27rr zV-sm7%J-^g^!_>YTtly|k&@y@mma*4DiC~z=y(RQh!>|1R2YqJDm=}!kVFR$Vce<4zKV&T4h`vB-Zw=DD1k^+s3Q$CvHIa05irquks z5dG=fEMF!jCSth7(KJcN)&*AOlNJG;Q|q~zTlArdSgAHn{* zzBC)@$CCkJLYJ>TT)EbIo#y7lgmQ|Au9eAR{3m_J=u&`2)A_I1*y7$F_T})ODVh2{ m^#009CZhyjF8K4-XIA%y-u@nUge#N*KkCZ6m2N3moccfD1;vv9 diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index d666c1b..50d0ebf 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -136,7 +136,7 @@ $$ --- >[!example] 例3 -设 $f(x)$ 在 $[0, 1]$ 上可导，且$f(1) = 2\int_0^{1/2} e^{\frac{x^2}{2}-x} f(x) dx$ +设 $f(x)$ 在 $[0, 1]$ 上可导，且$f(1) = 2\int_0^{1/2} \text{e}^{\frac{x^2}{2}-x} f(x) dx$ 证明：存在 $\xi \in (0, 1)$ 使得：$f'(\xi) = (1-\xi) f(\xi)$ **解析**： @@ -411,7 +411,7 @@ $$ >[!example] 例2 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内二阶可导，又若 $f(x)$ 的图形与联结 $A(a, f(a))$，$B(b, f(b))$ 两点的弦交于点 $C(c, f(c))$ ($a \leq c \leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 -**分析：![[微分中值定理图.png]]** +**分析：![[多次运用 微分中值定理.png]] 二阶导的零点就是图像的拐点，从图中能直观地看出来，函数图像的凹凸性确实发生了改变。现在的问题就是如何证明。 首先可以很直观地看到，函数图像应当有两条与直线$AB$平行的切线，由拉格朗日中值定理也可以证明这一点。这样，$f'(x)$就在不同地方取到了相同的函数值，这就想到用罗尔定理，从而可以证明题中结论。 From 42e962cdfcb2e5760d5e9009a54ef48bd0e8ab56 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 17 Jan 2026 10:08:22 +0800 Subject: [PATCH 05/15] vault backup: 2026-01-17 10:08:22 --- 编写小组/讲义/微分中值定理（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index d666c1b..ce442ce 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -360,7 +360,7 @@ $$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$ >[!example] 例3 设 $0 < a < b$，证明存在 $\xi \in (a, b)$，使得： -$$f(b)-f(a) = \frac{3\xi^2}{a^2+ab+b^2} f'(\xi)(b-a)$$ +$$f(b) - f(a) = \frac{f'(\xi)}{3\xi^2} \cdot (b - a)(a^2 + ab + b^2)$$ **解析**： 将等式变形为： From 52b211df3630cf2aed5fa35b1edfa7621a53bd49 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Sat, 17 Jan 2026 10:12:45 +0800 Subject: [PATCH 06/15] vault backup: 2026-01-17 10:12:44 --- 编写小组/讲义/微分中值定理（解析版）.md | 6 ++---- 1 file changed, 2 insertions(+), 4 deletions(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 50d0ebf..c5dab63 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -118,7 +118,7 @@ $$ --- >[!example] 例2 -设 $f(x)$ 在 $[a, b]$ 上三阶可导，且 $f(a) = f'(a) = f(b) = 0$。 +设 $f(x)$ 在 $[a, b]$ 上三阶可导，且 $f(a) = f'(a) = f(b) =f''(b)= 0$。 证明：存在 $\xi \in (a, b)$ 使得：$f'''(\xi) + k f''(\xi) = 0$ **解析**： @@ -129,9 +129,7 @@ $$ 因此构造辅助函数： $$ H(x) = \text{e}^{kx} f''(x) -$$ -由条件可推知存在 $\eta_1, \eta_2 \in (a, b)$ 使 $f''(\eta_1) = f''(\eta_2) = 0$，从而 $H(\eta_1)=H(\eta_2)=0$。 -对 $H(x)$ 应用罗尔定理即得证。 +$$由$f(a)=f(b)=0$及罗尔定理知，存在$c\in(a,b),f'(c)=0$；又$f'(a)=0$，则存在$d\in(a,c),f''(d)=0$；又$f''(b)=0$，知$H(d)=H(b)=0$，得存在$\xi\in(d,b)\subset(a,b),H'(\xi)=0\Rightarrow f'''(\xi)+kf''(\xi)=0$。 --- From 6301558a944056dc444da5e8e057f0843424322c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Sat, 17 Jan 2026 10:14:47 +0800 Subject: [PATCH 07/15] vault backup: 2026-01-17 10:14:47 --- 编写小组/讲义/微分中值定理.md | 2 +- 编写小组/讲义/微分中值定理（解析版）.md | 8 ++------ 2 files changed, 3 insertions(+), 7 deletions(-) diff --git a/编写小组/讲义/微分中值定理.md b/编写小组/讲义/微分中值定理.md index 3493a06..dfcdf8c 100644 --- a/编写小组/讲义/微分中值定理.md +++ b/编写小组/讲义/微分中值定理.md @@ -116,7 +116,7 @@ $$ 需进一步寻找另一个点 $c$ 使 $F(c)=0$，才可应用罗尔定理。通常需结合题目其他条件（如积分中值定理、零点定理等）找出该点。 >[!example] 例2 -设 $f(x)$ 在 $[a, b]$ 上三阶可导，且 $f(a) = f'(a) = f(b) = 0$。 +设 $f(x)$ 在 $[a, b]$ 上三阶可导，且 $f(a) = f'(a) = f(b) =f''(b)= 0$。 （有改动） 证明：存在 $\xi \in (a, b)$ 使得：$f'''(\xi) + k f''(\xi) = 0$ ```text diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index c5dab63..1ec1fb1 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -123,13 +123,9 @@ $$ **解析**： 结论可写为： -$$ -\bigl[ \text{e}^{kx} f''(x) \bigr]' \big|_{x=\xi} = 0 -$$ +$$\bigl[ \text{e}^{kx} f''(x) \bigr]' \big|_{x=\xi} = 0$$ 因此构造辅助函数： -$$ -H(x) = \text{e}^{kx} f''(x) -$$由$f(a)=f(b)=0$及罗尔定理知，存在$c\in(a,b),f'(c)=0$；又$f'(a)=0$，则存在$d\in(a,c),f''(d)=0$；又$f''(b)=0$，知$H(d)=H(b)=0$，得存在$\xi\in(d,b)\subset(a,b),H'(\xi)=0\Rightarrow f'''(\xi)+kf''(\xi)=0$。 +$$H(x) = \text{e}^{kx} f''(x)$$由$f(a)=f(b)=0$及罗尔定理知，存在$c\in(a,b),f'(c)=0$；又$f'(a)=0$，则存在$d\in(a,c),f''(d)=0$；又$f''(b)=0$，知$H(d)=H(b)=0$，得存在$\xi\in(d,b)\subset(a,b),H'(\xi)=0\Rightarrow f'''(\xi)+kf''(\xi)=0$。 --- From 50bdb2244e9d01abb7cfa527734c47ecec8703a2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 17 Jan 2026 10:17:18 +0800 Subject: [PATCH 08/15] vault backup: 2026-01-17 10:17:18 --- 编写小组/讲义/微分中值定理.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/微分中值定理.md b/编写小组/讲义/微分中值定理.md index 3493a06..8bba791 100644 --- a/编写小组/讲义/微分中值定理.md +++ b/编写小组/讲义/微分中值定理.md @@ -421,7 +421,7 @@ $$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$ >[!example] 例3 设 $0 < a < b$，证明存在 $\xi \in (a, b)$，使得： -$$f(b)-f(a) = \frac{3\xi^2}{a^2+ab+b^2} f'(\xi)(b-a)$$ +$$f(b) - f(a) = \frac{f'(\xi)}{3\xi^2} \cdot (b - a)(a^2 + ab + b^2)$$ ```text From a4235948952743a3b78b70a822af2235e8741e6a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= Date: Sat, 17 Jan 2026 10:22:22 +0800 Subject: [PATCH 09/15] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/柯西中值定理与常见辅助函数.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/柯西中值定理与常见辅助函数.md b/素材/柯西中值定理与常见辅助函数.md index 76332e8..24886bf 100644 --- a/素材/柯西中值定理与常见辅助函数.md +++ b/素材/柯西中值定理与常见辅助函数.md @@ -29,7 +29,7 @@ $$ >[!example] 例1 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $a>0$。证明存在 $\xi \in (a, b)$，使得： -$$\frac{f(b)-f(a)}{b-a} = \xi f'(\xi) \cdot \frac{\ln(b/a)}{b-a}$$ +$$f(b)-f(a) = \xi f'(\xi) \cdot \ln(b/a)$$ **解析**： 将等式变形为： From 164d814a357359dd1da7818158d108fed10e95da Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 17 Jan 2026 10:22:35 +0800 Subject: [PATCH 10/15] vault backup: 2026-01-17 10:22:35 --- 编写小组/讲义/微分中值定理.md | 23 ----------------------- 1 file changed, 23 deletions(-) diff --git a/编写小组/讲义/微分中值定理.md b/编写小组/讲义/微分中值定理.md index 15e5a92..32c3aa6 100644 --- a/编写小组/讲义/微分中值定理.md +++ b/编写小组/讲义/微分中值定理.md @@ -419,29 +419,6 @@ $$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$ ``` ->[!example] 例3 -设 $0 < a < b$，证明存在 $\xi \in (a, b)$，使得： -$$f(b) - f(a) = \frac{f'(\xi)}{3\xi^2} \cdot (b - a)(a^2 + ab + b^2)$$ - -```text - - - - - - - - - - - - - - - - - -``` ## 多次运用中值定理 From 14c9a96e83a8eb94f634f5e041b50f3eea59273f Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 17 Jan 2026 10:22:58 +0800 Subject: [PATCH 11/15] vault backup: 2026-01-17 10:22:58 --- .../微分中值定理（解析版）.md | 18 ------------------ 1 file changed, 18 deletions(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 88e2c99..9d7cea2 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -350,24 +350,6 @@ $$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$ f'(\xi) = \frac{a+b}{2\eta} f'(\eta) $$ ---- - ->[!example] 例3 -设 $0 < a < b$，证明存在 $\xi \in (a, b)$，使得： -$$f(b) - f(a) = \frac{f'(\xi)}{3\xi^2} \cdot (b - a)(a^2 + ab + b^2)$$ - -**解析**： -将等式变形为： -$$ -\frac{f(b)-f(a)}{b^3 - a^3} = \frac{f'(\xi)}{3\xi^2} -$$ -取 $g(x) = x^3$，则 $g'(x) = 3x^2 \neq 0$ 在 $(a, b)$ 内成立。 -由柯西中值定理，存在 $\xi \in (a, b)$ 使得： -$$ -\frac{f(b)-f(a)}{b^3 - a^3} = \frac{f'(\xi)}{3\xi^2} -$$ -整理后即得所求。 - ## 多次运用中值定理 多次运用中值定理一般有如下特征： From 8150b6e4323eaf83a174106d23a8ddf11e15b03d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= Date: Sat, 17 Jan 2026 10:25:06 +0800 Subject: [PATCH 12/15] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/讲义/微分中值定理.md | 2 +- 编写小组/讲义/微分中值定理（解析版）.md | 2 +- 2 files changed, 2 insertions(+), 2 deletions(-) diff --git a/编写小组/讲义/微分中值定理.md b/编写小组/讲义/微分中值定理.md index 15e5a92..a9f8f41 100644 --- a/编写小组/讲义/微分中值定理.md +++ b/编写小组/讲义/微分中值定理.md @@ -371,7 +371,7 @@ $$ >[!example] 例1 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $a>0$。证明存在 $\xi \in (a, b)$，使得： -$$\frac{f(b)-f(a)}{b-a} = \xi f'(\xi) \cdot \frac{\ln(b/a)}{b-a}$$ +$$f(b)-f(a) = \xi f'(\xi) \cdot ln(b/a)$$ ```text diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 88e2c99..85af572 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -306,7 +306,7 @@ $$ >[!example] 例1 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $a>0$。证明存在 $\xi \in (a, b)$，使得： -$$\frac{f(b)-f(a)}{b-a} = \xi f'(\xi) \cdot \frac{\ln(b/a)}{b-a}$$ +$$f(b)-f(a) = \xi f'(\xi) \cdot ln(b/a)$$ **解析**： 将等式变形为： From 3aa992882096e5303044960b860cd4c95b66d365 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Sat, 17 Jan 2026 10:31:08 +0800 Subject: [PATCH 13/15] vault backup: 2026-01-17 10:31:08 --- 编写小组/讲义/微分中值定理（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 9d7cea2..f6099ae 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -387,7 +387,7 @@ $$ >[!example] 例2 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内二阶可导，又若 $f(x)$ 的图形与联结 $A(a, f(a))$，$B(b, f(b))$ 两点的弦交于点 $C(c, f(c))$ ($a \leq c \leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 -**分析：![[多次运用 微分中值定理.png]] +**分析：**![[多次运用 微分中值定理.png]] 二阶导的零点就是图像的拐点，从图中能直观地看出来，函数图像的凹凸性确实发生了改变。现在的问题就是如何证明。 首先可以很直观地看到，函数图像应当有两条与直线$AB$平行的切线，由拉格朗日中值定理也可以证明这一点。这样，$f'(x)$就在不同地方取到了相同的函数值，这就想到用罗尔定理，从而可以证明题中结论。 From 272fd0c704e92b5b61e0b563664c29b843c75bb4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Sat, 17 Jan 2026 10:32:12 +0800 Subject: [PATCH 14/15] vault backup: 2026-01-17 10:32:12 --- 笔记分享/LaTeX（KaTeX）特殊输入.md | 6 +++++- 编写小组/讲义/微分中值定理（解析版）.md | 2 +- 2 files changed, 6 insertions(+), 2 deletions(-) diff --git a/笔记分享/LaTeX（KaTeX）特殊输入.md b/笔记分享/LaTeX（KaTeX）特殊输入.md index 877d957..c05450f 100644 --- a/笔记分享/LaTeX（KaTeX）特殊输入.md +++ b/笔记分享/LaTeX（KaTeX）特殊输入.md @@ -19,6 +19,10 @@ $\sum\limits_{i=2}^{n}a_i, \prod\limits_{i=1}^{n}a_i$ $\int_1^5 x\mathrm{d}x$ +如果像让积分符号更加好看，就在前面加上\displaystyle，如 + +$\displaystyle\int_1^5f(x)\text{d}x$ + 积分符号的上下记号，用limits套： -$\int\limits_{L}(x+y)\mathrm{d}s$ +$\displaystyle\int\limits_{L}(x+y)\mathrm{d}s$ diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index f2ef579..9201e1d 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -9,7 +9,7 @@ tags: ### **原理** -在证明与导数相关的等式或不等式时，常通过构造辅助函数，将原问题转化为对某个函数应用中值定理（如罗尔定理、拉格朗日定理等）。构造辅助函数的核心思想是：**将待证等式视为某个函数求导后的结果**。 +在证明与导数相关的等式或不等式时，常通过构造辅助函数，将原问题转化为对某个函数应用中值定理（如罗尔定理、拉格朗日定理等）。构造辅助函数的核心思想是：**将待证等式视为某个函数求导（或多次求导）后的结果**。 ### **常见构造类型** From 36477298a9bcf8e197a0819106d541dc356390c5 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Sat, 17 Jan 2026 10:33:21 +0800 Subject: [PATCH 15/15] vault backup: 2026-01-17 10:33:21 --- 编写小组/讲义/微分中值定理.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/微分中值定理.md b/编写小组/讲义/微分中值定理.md index a174474..426bef9 100644 --- a/编写小组/讲义/微分中值定理.md +++ b/编写小组/讲义/微分中值定理.md @@ -9,7 +9,7 @@ tags: ### **原理** -在证明与导数相关的等式或不等式时，常通过构造辅助函数，将原问题转化为对某个函数应用中值定理（如罗尔定理、拉格朗日定理等）。构造辅助函数的核心思想是：**将待证等式视为某个函数求导后的结果**。 +在证明与导数相关的等式或不等式时，常通过构造辅助函数，将原问题转化为对某个函数应用中值定理（如罗尔定理、拉格朗日定理等）。构造辅助函数的核心思想是：**将待证等式视为某个函数求导（或多次求导）后的结果**。 ### **常见构造类型**