31. 设 $f(x)$ 是 $[0,1]$ 上的二次连续可微函数, 满足 $f(0)=f(1)=0$, $f'(0)=1$, $f'(1)=0$. 证明 $\int_0^1 (f''(x))^2 dx\geqslant 4$.
Posted by haifeng on 2017-06-23 18:08:42 last update 2022-04-19 09:41:03 | Answers (1) | 收藏
Posted by haifeng on 2017-06-23 18:08:42 last update 2022-04-19 09:41:03 | Answers (1) | 收藏
Posted by haifeng on 2017-03-14 09:08:52 last update 2017-03-14 09:09:54 | Answers (0) | 收藏
设 $f\in C^2[a,b]$, 且 $f(a)=f'(a)=f'(b)=0$, 记 $M=\max_{x\in[a,b]}|f(x)|$. 则有
\[
M^2\leqslant\frac{(b-a)^3}{\pi}\int_a^b |f''(x)|^2 dx.
\]
[hint]
利用 Wirtinger 不等式 以及问题1912的结论.
Posted by haifeng on 2017-03-14 08:52:58 last update 2017-03-14 09:56:48 | Answers (1) | 收藏
设 $f\in C^1[a,b]$, 且 $f(a)=0$, 证明
\[
\sup_{x\in[a,b]}|f(x)|\leqslant\sqrt{b-a}\cdot\sqrt{\int_a^b (f'(x))^2 dx}.
\]
特别的, 如果 $a=0, b=1$, 则有 $\sup_{x\in[0,1]}|f(x)|\leqslant\sqrt{\int_0^1 (f'(x))^2 dx}.$ (见问题1863)
References:
Paulo Ney de Souza, Jorge-Nuno Silva, Berkeley Problems in Mathematics,Third Edition.
Posted by haifeng on 2017-03-14 08:38:32 last update 2022-04-20 10:42:51 | Answers (0) | 收藏
假设 $g\in C^1[a,b]$, 且 $g(a)=g(b)=0$. 则有
\[
\int_a^b g^2(t)dt\leqslant\Bigl(\frac{b-a}{\pi}\Bigr)^2\int_a^b|g'(t)|^2 dt.
\]
Remark:
高维的 Wirtinger 不等式又称为球面上的 Poincaré 不等式.
References:
https://arxiv.org/abs/1407.6871
https://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions
Wirtinger's inequality is seen as the one-dimensional version of Friedrichs' inequality.
Posted by haifeng on 2017-03-13 18:02:41 last update 2017-03-14 10:40:03 | Answers (1) | 收藏
求 \[\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx.\]
[相关的积分]
证明:
\[
\int_0^1 \frac{1}{\sqrt{1-x^4}}dx\cdot\int_0^1 \frac{x^2}{\sqrt{1-x^4}}dx=\frac{\pi}{4}.
\]
如果我们记 $f(x)=\frac{1}{\sqrt{1-x^4}}$, $g(x)=\frac{x^2}{\sqrt{1-x^4}}$, 则
\[
\begin{aligned}
f(x)+g(x)=\frac{1+x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1+x^2}{1-x^2}},\\
f(x)-g(x)=\frac{1-x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1-x^2}{1+x^2}},\\
\end{aligned}
\]
于是 $(f(x)+g(x))(f(x)-g(x))=1$, 因此原积分
\[
\begin{split}
\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx&=\int_0^1\sqrt{\frac{1-x^2}{1+x^2}}\\
&=\int_0^1(f(x)-g(x))dx\\
&=\int_0^1 \frac{1}{\sqrt{1-x^4}}dx-\int_0^1 \frac{x^2}{\sqrt{1-x^4}}dx
\end{split}
\]
References:
吉米多维奇, 《数学分析习题集题解》(五) 3872.
[分析]
设 $x^2=\sin\theta$, $\theta\in[0,\frac{\pi}{2}]$, 则 $x=\sqrt{\sin\theta}$, $dx=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$.
\[
\begin{split}
\text{原式}&=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\theta}}{1+\sin\theta}\cdot\frac{\cos\theta}{2\sqrt{\sin\theta}}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2\theta}{2\sqrt{\sin\theta}(1+\sin\theta)}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\sin^2\theta}{2\sqrt{\sin\theta}(1+\sin\theta)}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\sin\theta}{2\sqrt{\sin\theta}}d\theta\\
&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta}d\theta,
\end{split}
\]
其中积分
\[
\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta}d\theta\stackrel{t=\sqrt{\sin\theta}}{=}2\int_0^1\frac{t^2}{\sqrt{1-t^4}}dt
\]
是一个椭圆积分, 可以参考问题998 .
Posted by haifeng on 2016-12-27 07:07:47 last update 2016-12-27 07:07:47 | Answers (1) | 收藏
设 $I(x)=\int_{\pi}^{2\pi}\frac{y\sin(xy)}{y-\sin y}dy$, 求 $\int_{0}^{1}I(x)dx$.
Posted by haifeng on 2016-08-23 14:54:56 last update 2016-08-23 14:54:56 | Answers (1) | 收藏
计算定积分
\[
\int_0^1 \frac{x^2\arcsin x}{\sqrt{1-x^2}}dx
\]
Posted by haifeng on 2016-04-29 04:51:05 last update 2016-04-29 04:51:05 | Answers (1) | 收藏
\[
\int_1^2 \frac{1}{x(1+x^n)}dx
\]
Posted by haifeng on 2016-04-29 04:43:34 last update 2016-04-29 04:43:34 | Answers (1) | 收藏
\[
\int_1^3 \sqrt{(3-x)(x-1)}dx
\]
Posted by haifeng on 2016-04-29 04:39:31 last update 2020-11-01 14:20:27 | Answers (1) | 收藏
\[
\int_0^1 (1-x^2)^8 dx
\]
一般的, 证明积分
\[
\int_0^1 (1-x^2)^n dx=\frac{(2n)!!}{(2n+1)!!}
\]
除了这里利用递推关系求出, 还可以应用积分余项. 请参考梅加强著《数学分析》例5.7.1