From b04df6f4f651fc081740b216fc6b4ce94b108978 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:44:49 +0800 Subject: [PATCH 1/6] vault backup: 2026-01-14 17:44:49 --- ...线性方程组的解与秩的不等式.md | 26 ++-------- ...的解与秩的不等式（解析版）.md | 52 +------------------ 2 files changed, 6 insertions(+), 72 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index bfed081..673e8d9 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -407,9 +407,9 @@ $$ #### 2. 再证$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ - 考虑$C$的行向量： - 设$A = \begin{bmatrix} a_1^T \\ a_2^T \\ \vdots \\ a_m^T \end{bmatrix}$，则 + 设$A = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix}$，则 $$ - C = \begin{bmatrix} a_1^T B \\ a_2^T B \\ \vdots \\ a_m^T B \end{bmatrix}. + C = \begin{bmatrix} a_1 B \\ a_2 B \\ \vdots \\ a_m B \end{bmatrix}. $$ 因此$C$的每一行都是$B$的行向量的线性组合。 - 所以$C$的行空间是$B$的行空间的子空间，故 @@ -490,15 +490,6 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -更严格地，我们可以直接写： -$$ -\operatorname{rank}(M) \ge \operatorname{rank}\begin{bmatrix} -A \\ -I_n -\end{bmatrix} + \operatorname{rank}\begin{bmatrix} -I_n & B -\end{bmatrix} - n -$$ 但更简单的常用方法是利用： $$ \operatorname{rank}\begin{bmatrix} @@ -511,16 +502,7 @@ $$ --- #### 4. 从变换后的矩阵得到下界 -观察变换后的矩阵： -$$ -\operatorname{rank}\begin{bmatrix} -A & -AB \\ -I_n & O -\end{bmatrix} -\ge \operatorname{rank}\begin{bmatrix} -I_n & O -\end{bmatrix} + \operatorname{rank}([-AB]) -$$ + 实际上更直接的方法是注意到： $$ \operatorname{rank}\begin{bmatrix} @@ -530,7 +512,7 @@ I_n & O = \operatorname{rank}\begin{bmatrix} O & -AB \\ I_n & O -\end{bmatrix} \quad (\text{列变换}) +\end{bmatrix} \quad (\text{行变换}) $$ 即： $$ diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 6bccef1..f413cce 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -519,15 +519,6 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -更严格地，我们可以直接写： -$$ -\operatorname{rank}(M) \ge \operatorname{rank}\begin{bmatrix} -A \\ -I_n -\end{bmatrix} + \operatorname{rank}\begin{bmatrix} -I_n & B -\end{bmatrix} - n -$$ 但更简单的常用方法是利用： $$ \operatorname{rank}\begin{bmatrix} @@ -540,16 +531,7 @@ $$ --- #### 4. 从变换后的矩阵得到下界 -观察变换后的矩阵： -$$ -\operatorname{rank}\begin{bmatrix} -A & -AB \\ -I_n & O -\end{bmatrix} -\ge \operatorname{rank}\begin{bmatrix} -I_n & O -\end{bmatrix} + \operatorname{rank}([-AB]) -$$ + 实际上更直接的方法是注意到： $$ \operatorname{rank}\begin{bmatrix} @@ -559,7 +541,7 @@ I_n & O = \operatorname{rank}\begin{bmatrix} O & -AB \\ I_n & O -\end{bmatrix} \quad (\text{列变换}) +\end{bmatrix} \quad (\text{行变换}) $$ 即： $$ @@ -645,33 +627,3 @@ D = 0 & 1 & 0 & 0 \end{pmatrix}$$ **答案**： (B) 12 - ->[!example] 例3 - 设 $f(x)$ 在 $(-\infty, +\infty)$ 内二阶可导，且 $f''(x) \neq 0$。 -（1）证明：对于任何非零实数 $x$，存在唯一的 $\theta(x)$ ($0<\theta(x)<1$)，使得 - $$f(x) = f(0) + x f'(x\theta(x));$$ - （2）求 - $$\lim_{x \to 0} \theta(x).$$ - -解： -1. 证： 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 - -$$ -f(x)=f(0)+x f'(x\theta(x)). -$$ - -如果这样的 $\theta(x)$ 不唯一，则存在 $\theta_{1}(x)$ 与 $\theta_{2}(x)$ $(\theta_{1}(x)<\theta_{2}(x))$，使得 $f'(x\theta_{1}(x))=f'(x\theta_{2}(x))$，由罗尔定理，存在一点 $\xi$，使得 $f''(\xi)=0$，这与 $f''(x)\neq 0$ 矛盾。所以 $\theta(x)$ 是唯一的。 - -2. 解 注意到 $f''(0)=\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x\theta(x)}$，又知 - -$$ -\begin{aligned} -\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x} -&= \lim_{x\rightarrow 0} \frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \\ -&= \lim_{x\rightarrow 0} \frac{f(x)-f(0)-x f'(0)}{x^{2}} \\ -&= \lim_{x\rightarrow 0} \frac{f'(x)-f'(0)}{2x} \\ -&= \frac{f''(0)}{2}, -\end{aligned} -$$ - -所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 \ No newline at end of file From 850c10194729d9a417ddbaa0d2ccaa3eb68f8fb2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Wed, 14 Jan 2026 17:45:46 +0800 Subject: [PATCH 2/6] vault backup: 2026-01-14 17:45:46 --- .../微分中值定理（解析版）.md | 74 +++++++++++++++++++ 1 file changed, 74 insertions(+) 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 b52eaf9..e9dd243 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 @@ -5,3 +5,77 @@ tags: **内部资料，禁止传播** **编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 +### 多次运用中值定理 +多次运用中值定理一般有如下特征： +1. 有多个中值（如$\xi,\eta$两个中值）； +2. 有二阶导出现。 +这种题目一般会比较难，而且通常要结合其他方法，比如反证法、构造函数等。 + +多次用中值定理又分为两种： +1. 在同一区间（或一个区间包含另一个区间）上用不同的中值定理； +2. 在相邻区间上对原函数和导函数用同一个（也可能是不同的）中值定理。 + +>[!example] 例1 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $0 < a < b$，试证存在 $\xi, \eta \in (a, b)$，使得 $$f'(\xi) = \frac{a + b}{2\eta} f'(\eta).$$ + +**分析：** + 首先注意到题目要求我们证明的式子中有两个中值（即在区间内的某个点及其函数值），故想到可能会用多次中值定理。右边有$\frac{f'(\eta)}{\eta}$的形式，一般会想到用拉格朗日或者柯西，又题中没有哪个值让$f$等于$0$，所以大概率就是柯西中值定理。 + 令$g(x)=x^2$，则有$$\exists\eta\in(a,b),\frac{f'(\eta)}{g'(\eta)}=\frac{f'(\eta)}{2\eta}=\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f(b)-f(a)}{b^2-a^2}$$ + 再对比要求的式子，知道再用一次拉格朗日中值定理就行。 + +**解**： +对 $f(x)$ 和 $g(x)=x^2$ 在 $[a,b]$ 上应用柯西中值定理，得存在 $\eta \in (a,b)$，使得 +$$ +\frac{f(b)-f(a)}{b^2-a^2} = \frac{f'(\eta)}{2\eta} +$$ +整理得 +$$ +\frac{f(b)-f(a)}{b-a} = \frac{a+b}{2\eta} f'(\eta) +$$ +再对 $f(x)$ 在 $[a,b]$ 上应用拉格朗日中值定理，存在 $\xi \in (a,b)$，使得 +$$ +\frac{f(b)-f(a)}{b-a} = f'(\xi) +$$ +比较两式即得结论。 + +>[!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：** +![[Pasted image 20260114164542.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。 + +**解**： +弦 $AB$ 的方程为 +$$ +y = f(a) + \frac{f(b)-f(a)}{b-a}(x-a) +$$ +由条件，$f(c) = f(a) + \frac{f(b)-f(a)}{b-a}(c-a)$。分别对 $f(x)$ 在 $[a,c]$ 和 $[c,b]$ 上应用拉格朗日中值定理，存在 $\xi_1 \in (a,c)$，$\xi_2 \in (c,b)$，使得 +$$ +f'(\xi_1) = \frac{f(c)-f(a)}{c-a} = \frac{f(b)-f(a)}{b-a} +$$ +$$ +f'(\xi_2) = \frac{f(b)-f(c)}{b-c} = \frac{f(b)-f(a)}{b-a} +$$ +故 $f'(\xi_1)=f'(\xi_2)$。再对 $f'(x)$ 在 $[\xi_1,\xi_2]$ 上应用罗尔定理，存在 $\xi \in (\xi_1,\xi_2) \subset (a,b)$，使 $f''(\xi)=0$。 + +>[!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 $f(x) > x^2$。 + +**分析：** + 初看没有什么思路，涉及到二阶导一般会用泰勒展开或者对导数用中值定理。我们对一阶导和二阶导的性质一概不清楚，所以应该考虑对导数用中值定理。 + 直接对$f(x)$用中值定理吗？不是，这样子我们得不出任何的结论。或许我们应该构造一个新的函数，让我们更容易研究一些。观察要证的式子，由于我们完全不知道导数的性质，所以考虑函数$g(x)=f(x)-x^2$，有$g'(x)=f'(x)-2x,g''(x)=f''(x)-2$，且$g(0)=0,g(1)=0,g(\frac{1}{2})>0$.看到这些$0$就舒服很多了。注意这里应该是多次运用中值定理中的第2种情况，因为题目很明显给了我们分割区间的一个点：$\large{\frac{1}{2}}$. + +**解**： +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。在$[0,\frac{1}{2}],[\frac{1}{2},1]$上分别用拉格朗日中值定理得$\exists\eta_1\in(0,\frac{1}{2}),\eta_2\in(\frac{1}{2},1)$，使得$$g'(\eta_1)=\frac{g(1/2)-g(0)}{1/2}>0,g'(\eta_2)=\frac{g(1)-g(1/2)}{1/2}<0$$对$g'(x)$在区间$[\eta_1,\eta_2]$上用拉格朗日中值定理得，$\exists\xi\in(\eta_1,\eta_2)$，使得$$g''(\xi)=\frac{g'(\eta_2)-g'(\eta_1)}{\eta_2-\eta_1}<0,$$即$$f''(\xi)<2.$$ +(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，故$g(x_0)\le0$。由$g(0)=0,g(1)=0,g(1/2)>0,\exists\alpha\in(0,1),g(\alpha)=0$.故由罗尔中值定理，$$\exists\beta_1\in(0,\alpha),\beta_2\in(\alpha,1),g'(\beta_1)=g'(\beta_2)=0,$$从而$$\exists\beta\in(\beta_1,\beta_2),g''(\beta)=0$$这与$f''(x)\neq2$，即$g''(x)\neq0$矛盾，故结论成立。 + +**题后总结：** + 1. 注意观察要证等式的形式，如果需要用到中值定理，就看看应该是哪个中值定理，比如等于某个确定的值一般就是罗尔，有多个函数（有些比较荫蔽，尤其是幂函数）一般是柯西，其他一些奇奇怪怪的情况基本上是拉格朗日 + 2. 注意分析和综合相结合的方法，从结论向前推一推，推不动了再从条件往后推一推，这是证明题的很重要的思路 + 3. 数形结合可以给我们很大的信心并提供思路，但是也要小心用 + From 91148e072c8e4c90e7919fe09657e1b45a83567c Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:45:59 +0800 Subject: [PATCH 3/6] vault backup: 2026-01-14 17:45:59 --- ...罗尔定理与拉格朗日中值定理.md | 33 ++++++++++++++++++- 1 file changed, 32 insertions(+), 1 deletion(-) diff --git a/素材/罗尔定理与拉格朗日中值定理.md b/素材/罗尔定理与拉格朗日中值定理.md index 3b19eb2..b657011 100644 --- a/素材/罗尔定理与拉格朗日中值定理.md +++ b/素材/罗尔定理与拉格朗日中值定理.md @@ -103,4 +103,35 @@ $$ $$ f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 $$ -故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 \ No newline at end of file +故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 + + +>[!example] 例3 + 设 $f(x)$ 在 $(-\infty, +\infty)$ 内二阶可导，且 $f''(x) \neq 0$。 +（1）证明：对于任何非零实数 $x$，存在唯一的 $\theta(x)$ ($0<\theta(x)<1$)，使得 + $$f(x) = f(0) + x f'(x\theta(x));$$ + （2）求 + $$\lim_{x \to 0} \theta(x).$$ + +解： +1. 证： 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 + +$$ +f(x)=f(0)+x f'(x\theta(x)). +$$ + +如果这样的 $\theta(x)$ 不唯一，则存在 $\theta_{1}(x)$ 与 $\theta_{2}(x)$ $(\theta_{1}(x)<\theta_{2}(x))$，使得 $f'(x\theta_{1}(x))=f'(x\theta_{2}(x))$，由罗尔定理，存在一点 $\xi$，使得 $f''(\xi)=0$，这与 $f''(x)\neq 0$ 矛盾。所以 $\theta(x)$ 是唯一的。 + +2. 解 注意到 $f''(0)=\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x\theta(x)}$，又知 + +$$ +\begin{aligned} +\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x} +&= \lim_{x\rightarrow 0} \frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \\ +&= \lim_{x\rightarrow 0} \frac{f(x)-f(0)-x f'(0)}{x^{2}} \\ +&= \lim_{x\rightarrow 0} \frac{f'(x)-f'(0)}{2x} \\ +&= \frac{f''(0)}{2}, +\end{aligned} +$$ + +所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 \ No newline at end of file From 1c0748ad13bd8ba782b1daa000c9bfbb342648de Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:47:42 +0800 Subject: [PATCH 4/6] vault backup: 2026-01-14 17:47:42 --- .../线性方程组的解与秩的不等式（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index f413cce..6ab1770 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -519,7 +519,7 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -但更简单的常用方法是利用： +利用： $$ \operatorname{rank}\begin{bmatrix} A & O \\ From d997c228c4a5741f5f5d3b82aa0ff1a255c9f267 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:48:06 +0800 Subject: [PATCH 5/6] vault backup: 2026-01-14 17:48:06 --- 编写小组/讲义/线性方程组的解与秩的不等式.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 673e8d9..6d5f365 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -490,7 +490,7 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -但更简单的常用方法是利用： +利用： $$ \operatorname{rank}\begin{bmatrix} A & O \\ From 4e5fa271417f2d2e3a6995229b2cdf15a7d3de72 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Wed, 14 Jan 2026 17:50:53 +0800 Subject: [PATCH 6/6] vault backup: 2026-01-14 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\leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 -**分析：** -![[Pasted image 20260114164542.png]] +**分析：![[微分中值定理图.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。