中值定理 10 ¶
常数 K 值法 ¶
简要介绍 ¶
涉及中值定理的题，往往需要构造辅助函数，然而有一些辅助函数的构造又令人费解。常数 K 值法是适用于一类中值定理题目的构造方法，通过将中值变元之外的常数部分设为 k，然后再利用主元法或对称式和罗尔定理就可以完成证明。
例题 ¶
例 1 ¶
\(设f(x)为[a,b]上的可导函数，求证：\exists\varepsilon\in (a,b)满足：\)
\(证明1: 设k = \dfrac{f(b)-f(a)}{b-a} .........①，\iff 证明f'(\varepsilon) = k\) \(由①得：f(b) - f(a) - k(b-a) =0\)
\(设F(x) = f(x) - f(a) - k(x - a)......(\textbf{主元法})\) \(则F(b) = F(a) = 0\)
\(由罗尔定理: \exists\varepsilon\in(a,b)，F'(\varepsilon) = 0\iff f'(\varepsilon) = k\)
Tip
使用主元法，往往观察到变量的对称性，而且有一个天然条件 \(F(a) = 0\)
\(证明2: 设k = \dfrac{f(b)-f(a)}{b-a} .........①，\iff 证明f'(\varepsilon) = k\) \(由①得：f(b) - kb = f(a) - ka......(\textbf{对称式})\)
\(设F(x) = f(x) - kx \Rightarrow F(a) = F(b)\)
\(由罗尔定理: \exists\varepsilon\in(a,b)，F'(\varepsilon) = 0\iff f'(\varepsilon) = k\)
Tip
构造对称式，有点像高中学的 " 同构 "
例 2 ¶
\(b>a>0，设f(x)为[a,b]上的可导函数，求证：\exists\varepsilon\in (a,b)满足：\)
\(设\dfrac{bf(a)-af(b)}{b-a} = k\Rightarrow bf(a) - af(b) = k(b-a)\)
\(同时除以ab\Rightarrow \dfrac{f(a)}{a} - \dfrac{k}{a} = \dfrac{f(b)}{b} - \dfrac{k}{b}\)
\(设F(x) = \dfrac{f(x)}{x} - \dfrac{k}{x} \Rightarrow F(a) = F(b)\) \(\exists\varepsilon\in (a,b), F'(\varepsilon) = 0\)
\(F'(x) = \dfrac{xf'(x) - f(x)+k}{x^2} \Rightarrow k = f(\varepsilon) - \varepsilon f(\varepsilon)\)
练习 ¶
请用主元法完成例 2
Answer
\(\dfrac{bf(a) - af(b)}{b - a} = k\)
\(bf(a) - af(b) = k(b - a)\)
\((1) \text{ 若令 } F(x) = [xf(a) - af(x)] - k(x - a)\)
\(F(a) = 0 \quad F(b) = 0\)
\(F'(x) = f(a) - af'(x) - k\)
\(f(a) - af'(\varepsilon) = k\)
\(\text{应该要证明 } k = f(\varepsilon) - \varepsilon f'(\varepsilon)\) \(\text{所以证明失败}\)
\(xf(a) - af(x) = k(x - a)\)
\(\text{求导以后要清去} f(a) \text{，所以要进行变形}\)
\(\dfrac{f(a)}{a} - \dfrac{f(x)}{x} = \dfrac{k(x - a)}{ax}\)
\(\text{令 } F(x) = \dfrac{f(a)}{a} - \dfrac{f(x)}{x} - k\left(\dfrac{1}{a} - \dfrac{1}{x}\right)\)
\(F(a) = 0 \quad F(b) = 0\)
\(f'(\varepsilon) = -\dfrac{\varepsilon f(\varepsilon) + f'(\varepsilon)}{\varepsilon^2} - \dfrac{k^2}{\varepsilon^2} = 0\)
练习 ¶
\(b>a>0，设f(x)为[a,b]上的可导函数，求证\exists\varepsilon\in (a,b)\ s.t.\)
Answer
\(设k = \dfrac{ab[f(b) - f(a)]}{b-a}\)
\(则f(b) - f(a) = \dfrac{k(b-a)}{ab} = \dfrac{k}{a} - \dfrac{k}{b}\Rightarrow f(b)+\dfrac{k}{b} = f(a)+\dfrac{k}{a}\)
\(设F(x) = f(x)+\dfrac{k}{x}\Rightarrow F(a) = F(b)\Rightarrow\) \(\exists\varepsilon\ s.t.\ F'(\varepsilon) = 0 \Rightarrow f'(\varepsilon) - \dfrac{k}{\varepsilon^2} = 0\Rightarrow k = \varepsilon^2f'(\varepsilon)\)
例 3 ¶
\(已知a<b<c, f(x)二阶可导，证明\exists\xi\in(a,b)，使\)
\(设k = \dfrac{f(a)}{(a-c)(a-b)}+\dfrac{f(b)}{(b-c)(b-a)}+\dfrac{f(c)}{(c-b)(c-a)}\)
\(同乘(a-b)(b-c)(c-a)\Rightarrow f(a)(c-b)+f(b)(a-c)+f(c)(b-a) = k(a-b)(b-c)(c-a)\)
\(\textbf{以a为主元}\) \(设F(x) = f(x)(c-b)+f(b)(x-c)+f(c)(b-x) - k(x-b)(b-c)(c-x)\) \(F(a) = F(b) = F(c) = 0\)
\(由罗尔定理\exists m_1\in(a,b),m_2\in(b,c), F'(m_1) = F'(m_2) = 0\)
\(由罗尔定理\exists\xi\in(m_1,m_2),F''(\xi) = 0\)
\(F''(x) = (c-b)f''(x) - (c-b)2k \Rightarrow f''(\xi) = 2k\)
例 4 ¶
\(设f在[a,b]上二阶可导，f(a) = f(b) = 0 求证：对于某个x\in(a,b)，存在\xi\in(a,b)，使得：\)
\(设k = \dfrac{f(x)}{(x-a)(x-b)}(把x当作某个固定值c)\) \(令F(t) = f(t) - k(t-a)(t-b)\)
\(F(a) = F(x) = F(b) = 0\Rightarrow \exists m_1\in(a,x),m_2\in(x,b)\ s.t.F'(m_1) = F'(m_2) = 0\)
\(\therefore\exists\xi\in(a,b) \Rightarrow F''(\xi) = 0即f''(\xi) = 2k\)
练习 ¶
设 \(a_1 < a_2 < \cdots < a_n\) 为 \(n\) 个不同的实数，\(f(x)\) 在 \([a_1, a_n]\) 上有 \(n\) 阶导数，且 \(f(a_1) = f(a_2) = \cdots = f(a_n) = 0\)。求证：对\(\forall c \in [a_1, a_n]\)，\(\exists \xi \in (a_1, a_n)\)使得
Answer
(1) 若 \(c = a_k (k = 1,2,\cdots,n)\)，则 \(f(c) = 0\)；从而，对 \(\forall \xi \in (a_1, a_n)\) 结论均成立。
(2) 若 \(c \neq a_k\) ( 对 \(\forall k = 1,2,\cdots,n\))，记 \(\dfrac{f(c)}{(c - a_1)(c - a_2)\cdots(c - a_n)} = k\)。
令 \(F(x) = f(x) - k(x - a_1)(x - a_2)\cdots(x - a_n)\)，则：\(F(x)\) 在 \([a_1, a_n]\) 上有 \(n\) 阶导数
且 \(F(a_1) = F(a_2) = \cdots = F(a_n) = 0\)，\(F(c) = 0\)，即 \(F(x)\) 在 \([a_1, a_n]\) 有 \((n + 1)\) 个零点。
由 Rolle 定理，\(\exists \xi \in (a_1, a_n)\) 使得 \(F^{(n)}(\xi) = 0\)；而 \(F^{(n)}(x) = f^{(n)}(x) - n!k\)。
因此，\(\exists \xi \in (a_1, a_n)\) 使得 \(f(c) = \dfrac{(c - a_1)(c - a_2)\cdots(c - a_n)}{n!}f^{(n)}(\xi)\)。
Summary
常数 k 值法通过把一大堆常数设成 k，不仅简化了题面，还对构造函数有着辅助作用。构造函数需要观察函数的对称性质。