人工智能教程 - 數學基礎課程1.1 - 數學分析（一）34-35 泰勒級數

泰勒級數

$C_{N+1}=\frac{NC_N+1(C_N+1)}{N+1}-\frac{(N+1)C_N+1}{N+1}$

Center of mass of N+1 blocks

X-coordinate

$C_{N+1}=C_N+\frac{1}{N+1}$

$C_1=1$
$C_2=1+\frac{1}{2}$
$C_3=C_2+\frac{1}{3}=1+\frac{1}{2}+\frac{1}{3}$
$C_N=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{N}$

$C_N=S_N$
$lnN<S_N<(lnN)+1$
$as \ \ N\rightarrow \infty,lnN\rightarrow \infty \ \ and \ \ S_N \rightarrow \infty$

Power Series

$1+x+x^2+x^3+...=\frac{1}{1-x}$

|x|<1 (geometric series)

(converge)

$(1+x+x^2) =S$
$x+x^2+x^3+... =Sx$
$1=S-Sx =S(1-x)$

Resoning incomplete because it requires S exists

General Power Series

$a_0+a_1x+a_2x^2+a_3x^3+... =\sum_{n=0}^{\infty}a_nx^n$

|x|<R(radius of convergence)

(-R<x<R)

where series convergence $\sum a_nx^n$ diverge

|x|>R very delicate borderline(很微妙的邊界)—not used by us

$|a_nx^n|\rightarrow 0$ exponentially fast |x|<R

$|a_nx^n|\neq \rightarrow 0$ for |x|>R

Rules for conergant power series are just like polynomials

f(x)+g(x),f(x).g(x),f(g(x)),f(x)/g(x)

$\frac{d}{dx}f(x),\int f(x)dx$

$\frac{d}{dx}(a_0+a_1x+a_2x^2+a_3x^3+...)=a_1+2a_2x+3a_3x^2+...$
$\int(a_0+a_1x+a_2x^2+...)dx=c+a_0+a_1x^2/2+a_2x^3/3+...$

Taylor’s formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$

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級數

$\LARGE f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...$

Ex: （Euler歐拉）

	x=0
$f(x)=e^x$	1
$f'(x)=e^x$	1
$f''(x)=e^x$	1

$\LARGE e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

Ex2:

$\LARGE \frac{1}{1+x}=1-x+x^2-x^3+...$

Ex3:

$\LARGE sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$

New Power Series From Old

1). Multiply

$\LARGE xsin(x)=x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-...$

2). Differentiate求導

$\LARGE cos(x)=sin'(x)=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-...\LARGE =1-\frac{x^2}{2}+\frac{x^4}{4!}-...$

3). Integrate:

$\LARGE ln(1+x)=\int_{0}^{x}\frac{dt}{1+t}$

$\LARGE (x>-1)=\int_0^x(1-t+t^2+t^3+...)dt$

$\LARGE =[t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+...]|_0^x$

$\LARGE ln(1+x)=[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...](R=1)$

4).Substitue:

$\LARGE e^{-t^2}$$\LARGE (x=-t^2;in \ \ e^x)$

$\LARGE =1-t^2+\frac{t^4}{2!}+\frac{t^6}{3!}-...$

$\LARGE Er f(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dx\LARGE =\frac{2}{\sqrt{\pi}}(x-\frac{x^3}{3}+\frac{x^5}{5.2!}-\frac{7^3}{7.3!}+...)$