2024-2025 微积分 ( 甲 )I 期末考试部分思路 ¶

14. 已知数列 \(\{a_n\},\{b_n\}\) 满足 \(e^{a_n}=a_n + e^{b_n}\)，其中 \(0 < a_n < \dfrac{1}{n^2}\)。

证明 :

(1) \(0 < b_n < \dfrac{3a_n^2}{4}\)

(2) \(\lim\limits_{n \to \infty}(\dfrac{b_1}{a_1}+\dfrac{b_2}{a_2}+\cdots+\dfrac{b_n}{a_n})\) 存在。

(1)\(\iff 1<e^{b_n} = e^{a_n}-a_n < e^{\frac{3a_n^2}{4}}\)

左边显然成立 (\(e^x\ge x+1\))，下面考虑右边的不等式

由题：\(0 < a_n < \dfrac{1}{n^2}\le 1\)

\(e^{a_n}-a_n < e^{\frac{3a_n^2}{4}}\)

\(\Leftarrow e^{a_n}-a_n < \dfrac{3a_n^2}{4} +1 \le e^{\frac{3a_n^2}{4}}\)

\(\Leftarrow e^x-\dfrac{3}{4}x^2-x-1 <0, \forall x\in(0,1)\)

设 \(f(x) = e^x-\dfrac{3}{4}x^2-x-1,f'(x) = e^x-\dfrac{3}{2}x-1,f''(x) = e^x-\dfrac{3}{2}\)

\(f'(x)\) 在 \((0,1)\) 上最大值在端点处取到，\(f'(0) = 0,f'(1) = e-2.5<0\)

\(\therefore f(x)\) 在 \((0,1)\) 上递减，\(f(x)\le f(0) = 0\)

(2) \(\dfrac{b_i}{a_i}< \dfrac{3}{4}a_i<\dfrac{3}{4i^2}, i\in \mathbb{N^+}\)

用柯西收敛定理容易证明

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15. 已知函数 \(f(x)\) 在区间 \([0,1]\) 上有 2 阶导数，且 \(f''(x)<0\)，\(f(0)=f(1)=0\)。证明：

(1) \(f'(x)\) 在区间 \((0,1)\) 内存在唯一零点 \(x_0\)，且当 \(x\in(0,1)\) 时 \(f(x)>0\)

(2) \(\exists x_1\in(0,x_0)\)，\(x_2\in(x_0,1)\)，使得 \(f(x_1)=f(x_2)=\dfrac{f(x_0)}{2}\)，且 \(\displaystyle\int_{0}^{1}f(x)dx<f(x_0)(x_2 - x_1)\)

(1) \(f(1) = f(0) = 0 \Rightarrow \exists x_0\in(0,1), f'(x_0) = 0\)(Rolle)

\(\because f''(x)<0\) 所以一阶导数递减，\(0<x<x_0, f'(x)>0;x_0<x<1,f'(x)<0\)

\(\therefore f(x)\) 在 \((0,x_0)\) 上递增，在 \((x_0,1)\) 上递减

\(f(0) = f(1) = 0\Rightarrow f(x)>0, \forall x\in(0,1)\)

(2) 设 \(g(x) = f(x)-\dfrac{f(x_0)}{2}\)

\(g(x)\) 在 \((0,x_0)\) 上递增，在 \((x_0,1)\) 上递减

\(g(0)<0,g(x_0)>0,g(1)<0\Rightarrow\exists x_1\in(0,x_0),x_2\in(x_0,1)\)s.t. \(g(x_1) = g(x_2) = 0\)

至于第二问的不等式，本质上是比较面积大小

\(\displaystyle\int_{0}^{1}f(x)\mathrm{d}x = \left(\displaystyle\int_{0}^{x_1}+\displaystyle\int_{x_1}^{x_2}+\displaystyle\int_{x_2}^{1}\right)f(x)\mathrm{d}x\)

\(f(x_0)(x_2-x_1 ) = \displaystyle\int_{x_1}^{x_2}f(x_0)\mathrm{d}x = \displaystyle\int_{x_1}^{x_2}f(x_0)-f(x)\mathrm{d}x + \displaystyle\int_{x_1}^{x_2}f(x)\mathrm{d}x\)

\(\therefore\displaystyle\int_{0}^{1}f(x)dx<f(x_0)(x_2 - x_1)\)

\(\iff\displaystyle\int_{0}^{x_1}f(x)\mathrm{d}x+\displaystyle\int_{x_2}^{1}f(x)\mathrm{d}x < \displaystyle\int_{x_1}^{x_2}f(x_0)-f(x)\mathrm{d}x\)

\(\Leftarrow \displaystyle\int_{0}^{x_1}f(x)\mathrm{d}x < \displaystyle\int_{x_1}^{x_0}f(x_0)-f(x)\mathrm{d}x 且 \displaystyle\int_{x_2}^{1}f(x)\mathrm{d}x < \displaystyle\int_{x_0}^{x_2}f(x_0)-f(x)\mathrm{d}x\)

先证明 \(2x_1<x_0\iff x_1<\dfrac{x_0}{2} \iff f(x_1)= \dfrac{f(x_0)}{2} \lt f(\dfrac{x_0}{2})\)

这就是琴升不等式 \(\dfrac{f(x_0)+f(0)}{2} \lt f\left(\dfrac{x_0+0}{2}\right)\), 证略

构造 \(h_1(x) = f(x_0)-f(2x_1-x),x\le x_1\)

显然 \(h_1(2x_1-x_0) = 0,\displaystyle\int_{2x_1-x_0}^{x_1}h_1(x)\mathrm{d}x = \displaystyle\int_{x_1}^{x_0}f(x_0)-f(x)\mathrm{d}x\)

下面证明，\(0<x\le x_1\) 时，\(h_1(x)\ge f(x)\)

\(h(x) = h_1(x)-f(x) = f(x_0)-f(2x_1-x) -f(x),h(x_1) = 0\)

\(h'(x) = f'(2x_1-x)-f'(x),h'(x_1) = 0\)

\(h''(x) = -f''(2x_1-x)-f''(x) >0\)

\(\therefore h'(x)<0 ,h(x)\) 递减，\(h(x)\ge h(x_1) = 0\)

\(\therefore h_1(x)\ge f(x)\Rightarrow \displaystyle\int_{0}^{x_1}h_1(x)\mathrm{d}x > \displaystyle\int_0^{x_1}f(x)\mathrm{d}x\)

又 \(2x_1-x_0<x_0\) 时 \(h_1(x)>0\)

\(\therefore \displaystyle\int_{2x_1-x_0}^{x_1}h_1(x)\mathrm{d}x = \displaystyle\int_{x_1}^{x_0}f(x_0)-f(x)\mathrm{d}x > \displaystyle\int_0^{x_1}f(x)\mathrm{d}x\)

第一部分证明完毕，第二部分同理

所以最后的结论成立