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• Notes

Taylor and Maclaurin Series:

i.e $f(x)=\sum_{n=0}^{\infty} a{n}(x-c)^{n}\quad ,\quad |x<-c|<r$Region of Convergence

then $a n=\frac{f^{n}{(c)}}{n !} \quad ,\quad f(x)=\sum_{n=0}^{\infty} \frac{f^{n}{(c)}}{n !}(x-a)^{n}$

$f(x)=f(c)+\frac{f^{\prime}(c)}{1 !}(x-c)+\frac{f^{\prime \prime}(c)}{2 !}(x-c)^{2}+\cdots$

Taylor series of the function $F$ at $C$

If $c=0$

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

$f(x)=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots$

Maclawrin Series of $F$

$f(0)^{(0)}=f(0)$

② $0 !=1$

$\sum_{n=2}^{\infty} f(n)=\sum_{n=1}^{\infty} f(n+1)$                                $n=2 \quad n+1=2 \rightarrow n=1$

④ $\sum_{n=1}^{\infty} f(n)=\sum_{n=2}^{\infty} f(n-1)$                               $n=1 \quad n-1=1 \quad \rightarrow n=2$

Find the Maclaurin Series of the function $f(x)=e^{x}$ and it's radius of convergence.

Maclaurin series  $f(x)=\sum_{n=0}^{\infty} \frac{f^{n}(0)}{n !} x^{n}=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots$

Maclaurin Series $(M \cdot s)$ of $e^{x}$ is $\sum_{n=0}^{\infty} \frac{f^{n}{(0)}}{n !} x^{n}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$

$=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots \cdots$

To check convergence, use Ratio test:

$\lim _{n \rightarrow \infty} |\frac{u_{n+1}}{u_{n}} |, \sum_{n=0}^{\infty} \frac{x^{n}}{n !} u_{n}=\frac{x^{n}}{n !} \quad, \quad u_{n+1}=\frac{x^{n+1}}{(n+1) !}$

$\lim _{n \rightarrow \infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow \infty}\left|\frac{x^{n+1}}{(n+1) !} \cdot \frac{n !}{x^{n}}\right|=\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !}\left|\frac{x^{n+1}}{x^{n}}\right|$

$\lim _{n \rightarrow \infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow \infty} \frac{n !}{(n+1) n !}|x| =|x| \lim _{n \rightarrow \infty} \frac{1}{n+1}=|x| \lim \frac{1}{\infty}$

$\lim _{n \rightarrow \infty}\left|\frac{u_ n+1}{u_ n}\right| =0<1 \rightarrow \sum_{n=0}^{\infty} \frac{x^{4}}{n !}$ is $A \cdot c$ for all $x$

$e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}$ , region of convergence $=(-\infty, \infty)$

Radius$=\infty$

$e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}=1+\frac{x}{1!}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots$

① If $x=1 \Rightarrow e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots =\sum_{n=0}^{\infty} \frac{1}{n !}$

② If we replace $(x)$ by $(-x)$

$e^{-x}=1+\frac{-x}{1!}+\frac{x^{2}}{2 !}-\frac{x^{3}}{31}+\dots=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{n !}$

③ $\frac{e^{x}+e^{-x}}{2} :$ $e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3 !}+\cdots$

$e^{-x}=1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots$

$\frac{e^{x}+e^{-x}}{2}= 1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\cdots= \sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !}=\cos h (x)$

but $\cos h (x)=\frac{e^{x}+e^{-x}}{2}$

$\lim _{n \rightarrow \infty} | \frac{u_ n+1}{u_ n} |= |\frac{x^{2(n+1)}}{(2(n+1)) !} \cdot \frac{(2 n) !}{x^{2 n}}| =\frac{(2 n) !}{(2 n+2) !} .\left|\frac{{x^{2 n+2}}}{x^{2 n}}\right|$

$=\frac{2 n !}{(2 n+2)(2 n+1)(2 n) !} .|\frac{x^{2}}{1}| =x^{2} \lim _{n \rightarrow \infty}|\frac{1}{(2 n+2)(2 n+1)}|=0$

$\sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !}$is Converngent for all $x \in R$

$\cos h (x)=\sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !}$ , Interval of Converagence $=(-\infty, \infty)$

Radius of Convergence $=\infty$

$\sin h (x)= \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !}$ ,  Interval of Converagence $=(-\infty, \infty)$

Radius of Convergence $=\infty$