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• Notes

Representation of functions as Power series:

If $f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$

$∴ f^{\prime}(x)=\sum_{n=1}^{\infty} n \ {a_n} x^{n-1}$

If $f(x)=\sum_{n=0}^{\infty} a_{n} (x-c )^{n}$

$∴ f^{\prime}(x)=\sum_{n=0}^{\infty} n \ {a_n}(x-c)^{n-1}$

$∴ \int f(x) d x=\sum_{n=0}^{\infty} \frac{a_{n} x^{n+1}}{n+1}+c$          $∴ \int f(x) d x=\sum_{n=0}^{\infty} \frac{a_{n}(x-c)^{n+1}}{n+1}+c$

$∴ \frac{d}{d x}\left[\sum_{n=0}^{\infty} a_{n}(x-c)^{n}\right]=\sum_{n=0}^{\infty} \frac{d}{d x}[{a_n} (x-c)^{n} ]$

$∴ \int\left[\sum_{n=0}^{\infty} a_{n}(x-c)^{n}\right] d x=\sum_{n=0}^{\infty} \int a_{n}(x-c)^{n} d x$

Express $\frac{1}{(1-x)^{2}}$ as a power series.

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

$\frac{1}{(1-x)^{2}}=\sum_{n=0}^{\infty} n\ x^{n-1}$

Radius of convergence is still the same $r=1$

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

we know that $f^{\prime}(x)=\frac{d}{d x}\left[\tan ^{-1}(x)\right]=\frac{1}{1+x^{2}}$

$\frac{1}{1+x^{2}}=\frac{1}{1-\left(-x^{2}\right)}=\sum\left(-x^{2}\right)^{n}$

$\frac{1}{1+x^{2}}=\sum\left(-x^{2}\right)^{n}=1-x^{2}+x^{4}-x^{6}+x^{8}+\ldots$

$\tan ^{-1}(x)=\int \frac{1}{1+x^{2}}=\int\left(1-x^{2}+x^{4}-x^{6}+x^{8} \cdots \right) d x$

$\tan ^{-1}(x)=\int \frac{1}{1+x^{2}}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\frac{x^{9}}{9}+\cdots+c$

To Find $C$ $x=0 \quad \Rightarrow c=\tan ^{-1}(0)=0$

$\tan ^{-1}(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\frac{x^{9}}{9}+\cdots =\int\left[\sum\left(-x^{2}\right)^{n}\right] d x$

$\tan ^{-1}(x)=\sum \int\left(-{x^{2}}\right)^{n} d x=\sum \int(-1)^{n}\left(x^{2}\right)^{n} d x=\sum \int(-1)^{n}\left(x^{2 n}\right) d x$                $\left(x^{a}\right)^{b}=x^{a b}$

$\tan ^{-1}(x)=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2 n+1}}{2 n+1}$

radius of Convergence $=1$