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Taylor and Maclaurin Series:

$\left(\begin{array}{l}{k} \\ {n}\end{array}\right) =\frac{k(k-1)(k-2)(k-3) \dots}{n !}$       binomial series

If $k$ is any real number, $|x|<1 \Rightarrow$ Then

$(1+x)^{k} =\sum_{n=0}^{\infty} \left(\begin{array}{l}{k} \\ {n}\end{array}\right) x^{n}= 1+k_{n}+\frac{k(k-1)}{2 !} x^{2}+\frac{k(k-1)(k-2)}{3!} x^{3} \cdots$

we call $\left(\begin{array}{l}{k} \\ {n}\end{array}\right)$ binomial Coefficients

Find the Maclaurin series for the function $f(x)=\frac{1}{\sqrt{4-x}}$ and its radius of convergence.

binomial series $\Rightarrow(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{array}{l}{k} \\ {n}\end{array}\right) x^{n}$

$f(x)=\frac{1}{\sqrt{4-x}}=\frac{1}{\sqrt{4\left(1-\frac{x}{4}\right)}}=\frac{1}{2 \sqrt{1-\frac{x}{4}}}=\frac{1}{2} \frac{1}{\left(1-\frac{x}{4}\right)^\frac{1}{2}}=\frac{1}{2} \cdot\left(1-\frac{x}{4}\right)^{-\frac{1}{2}}$

$k=-\frac{1}{2}$ , replace $x$ by $-\frac{x}{4}$

$\frac{1}{\sqrt{4-x}}= \frac{1}{2}\left(1-\frac{x}{4}\right)^{-\frac{1}{2}}= \frac{1}{2} \sum_{n=0}^{\infty}\left(\begin{array}{c}{-\frac{1}{2}} \\ {n}\end{array}\right)\left(-\frac{x}{4}\right)^{n}$

$=\frac{1}{2} \left[1+\left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right)+\right. \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2 !}\left(-\frac{x}{4}\right)^{2}+ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3 !}\left(-\frac{x}{4}\right)^{3} +\ldots\ldots1]$

The series is converges when $\left|-\frac{x}{4}\right|<1 \Rightarrow|x|<4$

Radius of convergence $=4$

Find the Taylor series for $f(x)=(x-1)^{4} \mathrm{e}^{x}$ about $x=1.$

$f(x)=(x-1)^{4} e^{x}$ let $g(x)=e^{x}$ at $x=1$    $c=1$

$\begin{array}{l|l}{g(x)=e^{x}} & {g(1)=e} \\ {g^{\prime}(x)=e^{x}} & {g^{\prime}(1)=e} \\ {g^{\prime \prime}(x)=e^{x}} & {g^{\prime \prime}(1)=e}\end{array}$

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

$g(x)=e+\frac{e}{1!}(x-1)+\frac{e}{2 !}(x-1)^{2}-\cdots$

$g(x)=e \sum_{n=0}^{\infty} \frac{(x-1)^{n}}{n !}$

$f(x)=(x-1)^{4} e \sum_{n=0}^{\infty} \frac{(x-1)^{n}}{n !}$

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