Marks of Maths

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 2021/01/04  Share

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证明Jensen不等式

一致连续与导数

2021/1/5 Daily Problem:

Problem 1:

$\int\sqrt{1+\frac{1}{4x}}\mathrm{d}x\\$

Solution:

$\int\sqrt{1+\frac{1}{4x}}\mathrm{d}x\\ =\frac{1}{2}\int\sqrt{\frac{4x+1}{x}}\mathrm{d}x\\ =\frac{1}{2}\int\frac{4x+1}{\sqrt{x(4x+1)}}\mathrm{d}x\\ =\frac{1}{2}\int\frac{4x+\frac{1}{2}}{\sqrt{x(4x+1)}}\mathrm{d}x +\frac{1}{4}\cdot\frac{\mathrm{d}x}{\sqrt{4x^2+x}}\\ =\frac{1}{2}\sqrt{x(4x+1)}+C_1+\frac{1}{8}\cdot\frac{\mathrm{d}(8x+1)} {\sqrt{(8x+1)^2-1}}\\ = \frac{1}{2}\cdot\sqrt{x(4x+1)}+\frac{1}{8}\ln \left|{8x+1}+4\sqrt{4x^2+x}\right|+C$

Problem 2:

$\int{\frac{x\cdot\cos{x}}{\sin^3{x}}}\mathrm{d}x$

Solution:

$令：I=\int{\frac{x\cdot\cos{x}}{\sin^3{x}}}\mathrm{d}x =\int{x\cdot\cot{x}\cdot\csc^2{x}}\mathrm{d}x\\ =-\int{x\cdot{\cot{x}}}\mathrm{d}\cot{x}\\ =-\left[x\cdot{\cot^2{x}}-\int{\left[\cot{x}\cdot\left({\cot{x}-x\cdot\csc^2{x}}\right)\right]\mathrm{d}x}\right]\\ =-x\cdot{\cot^2{x}}+\int{\cot^2{x}\mathrm{d}x}-I\\ =>2\cdot{I}=-x\cdot{\cot^2{x}}+\int{\cot^2{x}\mathrm{d}x} \\=-x\cdot{\cot^2{x}}+\int{(\csc^2{x}-1)\mathrm{d}x} \\=-x\cdot{\cot^2{x}}-\cot{x}-x+C\\ 故：I=-\frac{1}{2}\left(x\cdot{\cot^2{x}}+\cot{x}+x\right)+C\\$

参数方程求导

$有： \left\{ \begin{array}{ll} y=u(t)\hbox{,}\\ x=v(t)\hbox{.} \end{array} \right.\\ 则：\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}y}{\mathrm{d}t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x};\\ \frac{\mathrm{d^2}y}{\mathrm{d}x^2} =\frac{\mathrm{d}}{\mathrm{d}x}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)} {\mathrm{d}t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x} =\frac{\mathrm{d}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)} {\mathrm{d}t}/\frac{\mathrm{d}x}{\mathrm{d}t}\\ (须知x, y对t的导数是易得的,在题中x,y对t求导)$

2020/1/6 Daily Problem:

Problem 1:

$f(x)在[a,b]连续,(a,b)可导，\\证明：\exists\xi\in(a,b),s.t.f'(\xi)=\frac{f(\xi)-f(a)}{b-\xi}.$

Solution:

$分析:不妨先令x=\xi，对要证明的式子变形:\\ f'(x)(b-x)=f(x)-f(a);\\ 即:[f'(x)-f(a)]'(b-x)+(b-x)'[f(x)-f(a)];\\ 不妨构造辅助函数:\\F(x)=[f(x)-f(a)](b-x);\\ 再用Rolle定理,以下证明就十分容易了.\\ 类似的还有，证明:f'(\xi)\pm{g'(\xi)f(\xi)};\\ 构造辅助函数f'(x)e^{\pm{g(x)}};\\ 有时要结合零点存在性定理.$

Problem 2:

$设f(x)在[a,b]上二阶可导，f'(a)=f'(b)=0.\\证明：\exists\xi\in(a,b),s.t. \frac{4}{(b-a)^2}\left|f(b)-f(a)\right|\le\left|f''(\xi)\right|.$

Solution:

$f(x)在x=a和x=b处的带Lagrange余项的Taylor公式分别为:\\ f(x)=f(a)+\frac{f''(\xi_1)}{2}(x-a)^2;\\ f(x)=f(b)+\frac{f''(\xi_2)}{2}(x-b)^2;\\ 故f(\frac{a+b}{2})=f(a)+\frac{f''(\xi_1)}{8}(b-a)^2\\ =f(b)+\frac{f''(\xi_2)}{8}(b-a)^2;\\ => f(b)-f(a)=\frac{f''(\xi_1)-f''(\xi_2)}{8}(b-a)^2;\\ \therefore \left|f(b)-f(a)\right|=\left|\frac{f''(\xi_1)-f''(\xi_2)}{8}(b-a)^2\right| \le\left|\frac{f''(\xi)}{4}(b-a)^2\right|;\\ 其中：f''(\xi)=\max\{|f''(\xi_1)|,|f''(\xi_2)|\}.$

Problem 3:

$\int\frac{1+\sin{x}}{1+\cos{x}}\cdot{e^x}\mathrm{d}x$

Solution:

$I= \int\frac{1+2\sin{\frac{x}{2}}\cdot\cos{\frac{x}{2}}}{2\cos^2{\frac{x}{2}}}\mathrm{d}x\\ =\frac{1}{2} \int{ \sec^2{\frac{x}{2}}e^x\mathrm{d}x+\int{\tan{\frac{x}{2}}}\mathrm{d}x }\\ =\int{e^x}\mathrm{d}\tan{\frac{x}{2}}+\int{\tan{\frac{x}{2}}}\cdot{e^x}\mathrm{d}x\\ =\tan{\frac{x}{2}}\cdot{e^x}+C$

Cantor定理

闭区间上的连续函数一致连续。

Lipschitz条件

$<=>一致连续\\ 若定义在I上的有一阶导的函数f(x)\\ 满足|f'(x)|\le{L},(L为一正常数).\\ 则称该函数Lipschitz连续.\\ 通俗地说，导函数有界，则原函数一致连续.$

连续函数的性质

$(1)有界性: |f(x)|\le{M};\\ (2)最值性:f(\xi)\le{f(x)\le{f(\eta)}};\\ (3)零点存在定理;\\ (4)介值性:R_f\in[m,M]\\ (f(x)可以取遍最大值到最小值之间的每一个值);$

Author: Endsieg77

Link: https://endsieg77.github.io/2021/01/04/Marks-of-Maths/

Post on: January 4th 2021, 10:14:15 pm

Update on: January 13th 2021, 2:43:55 am

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CATALOG

1. 2021/1/5 Daily Problem:
1. 1.1. Problem 1:
2. 1.2. Problem 2:
2. 参数方程求导
3. 2020/1/6 Daily Problem:
4. Cantor定理
5. Lipschitz条件
6. 连续函数的性质

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