Midterm Revision

The way to prove a series is uniform convergence
1. Theorem (Cauchy Criterium for uniform convergence of series)
2. Theorem (Weierstrass M-test)
3. Definition
4. If we need to check whether it is uniform convergent in a interval, we may use
  1. Assume that it converges absolutely at x = x_0Then it converges uniformly on the closed interval [-c, c], where c = |x_0|.
    
    Then it converges uniformly on the closed interval [-c, c], where c = |x_0|. 2. Theorem (Abel's theorem)
The way to find the radius of convergence of a series
1. D'Alembert Criterion (test)
2. Root test (Cauchy)
3. Remember to check the endpoints of intervals (Use comparison test)
Taylor Series

\[ T_{N,x_0}(x)=f\left(x_{0}\right)+\frac{f^{\prime}\left(x_{0}\right)}{1!}\left(x- x_{0}\right)+\frac{f''\left(x_{0}\right)}{2!}\left(x-x_{0}\right)^{2}+\cdots+\frac{f^{\left(N\right)}\left(x_{0}\right)}{N!} \left(x-x_{0}\right)^{N} \]

$f(x) - T_{N,a}(x) = \frac{f^{(N+1)}(c)}{(N+1)!} (x-a)^{N+1} \quad \text{for some } c \text{ between } a \text{ and } x$ (Bounded error)

$\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$

$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$

$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

For $\ln(x)$, we know $\ln(x)^{\prime}=\frac{1}{x}=\sum_{n=0}^{\infty}\left(1-x\right)^{n}=\sum_{n=0}^{\infty} \left(-1\right)^{n}\left(x-1\right)^{n}$

Then $\ln(x)=\int\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(x-1\right)^{n}=\sum_{n=0} ^{\infty}\left(-1\right)^{n}\int\left(x-1\right)^{n}=\sum_{n=0}^{\infty}\left(-1\right )^{n}\frac{\left(x-1\right)^{n+1}}{n+1}$ (centered at $x=1$)

$\ln(x+1)=\sum_{n=0}^{\infty}\left(-1\right)^{n}\frac{\left(x\right)^{n+1}}{n+1}$
The way to prove $f$ is integrable
1. Definition: $U(f)=L(f)$
2. Show $U(f, P_{\epsilon}) - L(f, P_{\epsilon}) < \epsilon$
3. Show $\lim\limits_{n \to +\infty}L(f, P_{n}) = \lim\limits_{n \to +\infty}U(f, P_{n}) $
4. Show $f$ is continuous on a closed interval
5. Theorem (Integrable Limit theorem)
Theorem (Fundamental Theorem of Calculus)

Theorem
Theorem (Decomposition of real rational functions)

Theorem (Integration by parts)

Theorem (Substitution)
The length of its graph in $[a, b]$ is $\displaystyle \int_a^b \sqrt{1 + (f'(x))^2} \, dx$
$\displaystyle\int_{0}^{1} \frac{1}{\sqrt{x}} dx$ where $I = (0, 1]$

Then consider $\displaystyle\lim_{c \to 0^+} \int_{c}^{1} \frac{1}{\sqrt{x}} dx$, we have $\displaystyle\int_{c}^{1} \frac{1}{\sqrt{x}} dx = \displaystyle\int_{c}^{1} x^{-1/2} dx = \frac{x^{1/2}}{1/2}\Big|_c^1 = 2\sqrt{x}\Big|_c^1 = 2 - 2\sqrt{c}$

Then $\displaystyle\lim_{c \to 0^+} 2 - 2\sqrt{c} = 2$, thus $\displaystyle\int_{0}^{1}\frac{1}{\sqrt{x}}dx$ converges to $2$
$\int \sec(x) dx$ $=$ $\ln(|\sec(x)+\tan(x)|)+c$

Remember $\sec(x) = \frac{1}{\cos(x)}$ and $\arctan(x)'=\frac{1}{1+x^{2}}$

We want to use integration by parts. How do we choose $u$ and $v'$?
We pick $u$ in the following order:
(1) inverse of trigonometric functions $\arcsin, \arccos,...$
(2) logarithmic function $\ln(x)$
(3) algebraic function $x^n, \sqrt{x}, ...$
(4) trigonometric function $\sin(x), \cos(x), ...$
(5) exponential functions $e^x, 2^x, ...$

$\int \sec^3(x) dx = \int \sec(x) \sec^2(x) dx$

Let $u = \sec(x) \implies du = \tan(x)\sec(x) dx$
$v' = \sec^2(x) \implies v = \tan(x)$

Now, by parts

$\int \sec^3(x) dx = \sec(x)\tan(x) - \int \tan(x) (\tan(x)\sec(x)) dx$
$= \sec(x)\tan(x) - \int \tan^2(x) \sec(x) dx$
$= \sec(x)\tan(x) - \int (\sec^3(x) - \sec(x)) dx$
$= \sec(x)\tan(x) - \int \sec^3(x) dx + \int \sec(x) dx$

So $\int \sec^{3}(x) dx = \frac{1}{2}(\sec(x)\tan(x) + \ln|\sec(x) + \tan(x)|) + C$

Because $\tan^2(x)\sec(x) = \frac{\sin^2(x)}{\cos^2(x)} \sec(x) = \frac{(1-\cos^2(x))}{\cos^2(x)} \sec(x) = (\sec^2(x) - 1) \sec(x)$
(20 points)
Prove that the function $f(x) = \begin{cases} \frac{1}{q}, & \text{if } x = \frac{p}{q} \text{ for } p \in \mathbb{Z}, q \in \mathbb{N}, \text{ and } p, q \text{ are mutually prime} \\ 0, & \text{if } x \text{ is irrational}. \end{cases}$ is integrable on $[0,1]$.

Consider $P_{n}=\{0,\frac{1}{n},\frac{2}{n},\ldots,1\}$, then $m_{i}=\inf\{f(x):x\in[x_{i-1},x_{i}]\}=0$ since irrational number is dense in any compact set.

Then $M_{i}=\sup\{f(x):x\in[x_{i-1},x_{i}]\}$ and $U(f,P_{n})-L\left(f,P_{n}\right)=\sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i} -x_{i-1}\right)=\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)$

Let's $B_{N}=\{1,\frac{1}{2},\frac{1}{3},\frac{2}{3},...,\frac{N-1}{N}\}$ with $|B_N|=m$, then consider $P_{n}=\{x_0=0,x_1,...,x_{n-1},x_n=1\}$

We know $B_{N}\subset [0,1]$, then $\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)=\sum_{\left\lbrack x_{i-1},x_{i}\right\rbrack\cap B_{N}\neq\emptyset}M_{i}\left(x_{i}-x_{i-1}\right)+\sum_{\left\lbrack x_{i-1},x_{i}\right\rbrack\cap B_{N}=\emptyset}M_{i}^{\prime}\left(x_{i}-x_{i-1}\right)$

Since $M_{i}=\sup\{f(x):x\in[x_{i-1},x_{i}]\cap B_{N}\}=\frac{1}{i-1}$ and $M_{i}^{\prime}=\sup\{f(x):x\in[x_{i-1},x_{i}]\setminus B_{N}\}\leq\frac{1}{N+1}$

And we know there are at most $2m$ subintervals such that $\left\lbrack x_{i-1},x_{i}\right\rbrack\cap B_{N}\neq\emptyset$ because one point in $B_N$ can be in two intervals

Then $\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)\leq2m\cdot1\cdot\max\left(x_{i}-x_{i-1} \right)+\frac{1}{N+1}\sum\left(x_{i}-x_{i-1}\right)\leq2m\cdot1\cdot\max\left(x_{i} -x_{i-1}\right)+\frac{1}{N+1}\left(1-0\right)\leq2m\cdot1\cdot\max\left(x_{i}-x_{i-1} \right)+\frac{1}{N+1}$

Then let $\max\left(x_{i}-x_{i-1}\right)<\frac{\varepsilon}{4m}$ and $\frac{1}{N+1}<\frac{\varepsilon}{2}$, then we have $\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)<\varepsilon$

Thus $U(f,P_{n})-L\left(f,P_{n}\right)<\varepsilon$, then $f$ is integrable

泰勒展开精度问题不四舍五入