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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 \(=2\)

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