泰勒級數

泰勒級數（又叫泰勒展開式，英文係Taylor Series）係將光滑函數展開為多項式嘅方法。其中一個好處係，容易做微積分。

定義

${\displaystyle f(x)}$ 喺參考點 ${\displaystyle x=a}$ 嘅泰勒展開式：

${\displaystyle f(x)={\sum _{n=0}^{\infty }}{f^{(n)}(a) \over n!}(x-a)^{n}=f(a)+f'(a)(x-a)+{f''(a) \over 2!}(x-a)^{2}+{f^{(3)}(a) \over 3!}(x-a)^{3}+...}$

當 ${\displaystyle a=0}$ 嗰陣，又叫麥克勞林級數。

例子

• ${\displaystyle e^{x}={\sum _{n=0}^{\infty }}{x^{n} \over n!}=1+x+{x^{2} \over 2!}+{x^{3} \over 3!}+{x^{4} \over 4!}+...}$
• ${\displaystyle \cos(x)={\sum _{n=0}^{\infty }}(-1)^{n}{x^{2n} \over (2n)!}=1-{x^{2} \over 2!}+{x^{4} \over 4!}-{x^{6} \over 6!}+...}$
• ${\displaystyle \sin(x)={\sum _{n=0}^{\infty }}(-1)^{n}{x^{2n+1} \over (2n+1)!}=x-{x^{3} \over 3!}+{x^{5} \over 5!}-{x^{7} \over 7!}+...}$

證明

假設一個函數可以寫做無限項嘅多項式：

${\displaystyle f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+...+a_{n}x^{n}+...}$

${\displaystyle f(0)=0!\ a_{0}}$

將函數微分1階：

${\displaystyle f'(x)=1\ a_{1}+2\ a_{2}\ x+3\ a_{3}\ x^{2}+4\ a_{4}\ x^{3}+...+n\ a_{n}\ x^{n-1}+...}$

${\displaystyle f'(0)=1!\ a_{1}}$

2階微分：

${\displaystyle f''(x)=(2\times 1)\ a_{2}+(3\times 2)\ a_{3}\ x+(4\times 3)\ a_{4}\ x^{2}+(5\times 4)\ a_{5}\ x^{3}+...+[n\times (n-1)]\ a_{n}\ x^{n-2}+...}$

${\displaystyle f''(0)=2!\ a_{2}}$

3階微分：

${\displaystyle f^{(3)}(x)=P_{3}^{3}\ a_{3}\ +P_{3}^{4}\ a_{4}\ x+P_{3}^{5}\ a_{5}\ x^{2}+P_{3}^{6}\ a_{6}\ x^{3}+...+P_{3}^{n}\ a_{n}\ x^{n-3}+...}$

${\displaystyle f^{(3)}(0)=3!\ a_{3}}$

k階微分：

${\displaystyle f^{(k)}(x)=P_{k}^{k}\ a_{k}\ +P_{k}^{k+1}\ a_{k+1}\ x+P_{k}^{k+2}\ a_{k+2}\ x^{2}+P_{k}^{k+3}\ a_{k+3}\ x^{3}+...+P_{k}^{n}\ a_{n}\ x^{n-k}+...}$

${\displaystyle f^{(k)}(0)=k!\ a_{k}}$

n階微分：

${\displaystyle f^{(n)}(x)=P_{n}^{n}\ a_{n}+P_{n}^{n+1}\ a_{n+1}\ x+...}$

${\displaystyle f^{(n)}(0)=n!\ a_{n}}$

所以原函數 ${\displaystyle f(x)}$ 每項嘅係數可以寫做：

${\displaystyle a_{k}={f^{(k)}(0) \over k!}\ ,\ (k=0,1,2,...,n,n+1,...)}$

再配返每項嘅變數可以寫做：

${\displaystyle a_{k}\ x^{k}={f^{(k)}(0) \over k!}x^{k}\ ,\ (k=0,1,2,...,n,n+1,...)}$

所以原函數可以寫做：

${\displaystyle f(x)={\sum _{k=0}^{\infty }}a_{k}\ x^{k}={\sum _{k=0}^{\infty }}{f^{(k)}(0) \over k!}x^{k}}$ (麥克勞林級數)

另外，令 ${\displaystyle g(x)}$ 係 ${\displaystyle f(x)}$ 向右平移 ${\displaystyle a}$ 個單位：

${\displaystyle g(x)=f(x-a)={\sum _{k=0}^{\infty }}a_{k}\ (x-a)^{k}={\sum _{k=0}^{\infty }}{f^{(k)}(a) \over k!}(x-a)^{k}}$ (泰勒級數)