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• Notes

Representation of functions as Power Series :

$\sin (x)=\frac{x}{1 !}-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !} \cdots=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n+1}}{(2 n+1) !}$

$\sum_{n=0}^{\infty} x^{n}=1+x+x^{2}+x^{3} =\frac{1}{1-x} \quad,|x|<1$

$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}, \quad|x|<1$

Express $\frac{1}{1+x^{2}}$ as the sum of a power series and find the interval of convergence.

We know that $\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$

replacing $x$ by $-x^{2}$ we get $\rightarrow$

$\frac{1}{1+x^{2}}=\frac{1}{1-\left(-x^{2}\right)}= =\sum_{n=0}^{\infty}\left(-x^{2}\right)^{n}=\sum_{n=0}^{\infty}(-1)^{n} x^{2 n}=1-x^{2}+x^{4}-x^{6}+x^{8}\cdots$

Because it is a geometric series, it Converges when

$\left|-x^{2}\right|<1 \quad \Rightarrow x^{2}<1 \Rightarrow|x|<1$

Interval of Convergence $=(-1,1)$

Radius of Convergene $=1$

Find a power series representation for $\frac{1}{x+2}$

We know that $\frac{1}{1-x}=\sum x^{n} \quad,|x|<1$

but we have $\frac{1}{x+2}=\frac{1}{2+x}=\frac{1}{2-(-x)}=\frac{1}{2\left[1-\left(-\frac{x}{2}\right)\right]}$

$\frac{1}{x+2}=\frac{1}{2} \sum_{n=0}^{\infty}\left(\frac{-x}{2}\right)^{n}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n+1}} x^{n}$                (Geometric series)

Converges when $\left|\frac{-x}{2}\right|<1 \quad \Rightarrow|x|<2$

Interval of Convergenee is $(-2,2)$

Radius of Convergence $=2$