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Find the Maclaurin series of $e^{x^{2}}$

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

Replace $x$ by $x^{2} \Rightarrow$

$e^{x^{2}}=1+\frac{x^{2}}{1!}+\frac{\left(x^{2}\right)^{2}}{2 !}+\frac{\left(x^{2}\right)^{3}}{3 !}+\cdots$

$e^{x^{2}}=\sum_{n=0}^{\infty} \frac{x^{2 n}}{n !}$

Find a power series representation for $f(x)=\frac{1}{l-x}$

use long diviston: $\frac{1}{1-x}=1+x+x^{2}+x^{3}$

$f(x)=\frac{1}{1-x}=1+x +x^{2}+x^{3} \cdots=\sum_{n=0}^{\infty} x^{n}\quad (G \cdot s)$

It is convergent only if $|r|=|x|<1 \quad ∴ x \in(-1,1)$

$∴ \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}$ , Interval of Convergence $(1 . c)=(-1,1)$

Radius of Convergence $=1$

Solution 2: using Maclaurin Series:

Maclaurin Series$\frac{1}{1-x}=f(0)+\frac{f'{(0)}}{1 !}+\frac{f''{(0)}}{{2!}} x^{2}$

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

$=1+x+x^{2}+x^{3}\ldots$

$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$ using Ratio test $, u_{n}=x^{n}$

$u_{n}+1=x^{n+1} \ , \lim _{n \rightarrow \infty} |\frac{u_ n+1}{u_n}|= \lim _{n \rightarrow \infty} |\frac{x^{n+1}}{x^{n}}|= \lim _{n\rightarrow \infty}|x|=|x|<1$

$\sum_{n=0}^{\infty} x^{n}= \frac{1}{1-x}$ is $(A . c)$ at ${x} \in(-1, 1)$ ,  Raduis $=1$

Find the Taylor series for $e^{x}$ at $x=2$

$f^{n}{(2)}=e^{2}$

$c=2$

Taylor Series definition:

$e^{x}=\sum_{n=0}^{\infty} \frac{f^{n}{(2)}}{n !}(x-2)^{n}=\sum_{n=0}^{\infty} \frac{e^{x}}{n !}(x-2)^{n}$

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

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

$=0<1$

$A \cdot C$ for all $x \in R$

$I \cdot C=(-\infty, \infty)$

$R \cdot C=\infty$