Power Representation of A Series

Question

I am asked to find the power representation of the series $ f(x)=xln(2+3x)+1/(2-x)^2 $ about a = 0. I took the power series of $ xln(2+3x) $ and $ 1/(2-x)^2 $ separately and added them together to give me: $ xln(2)+\sum_1^\infty((-1)^{n-1}3^nx^{n+1}) /n2^n + 1/4\sum_1^\infty(n-1)(x/2)^{n-2} $ But im unsure how to make it all under one sum

Answer 1

\begin{align}&x\ln(2)+\sum_{n=1}^\infty((-1)^{n-1}3^nx^{n+1}) /n2^n + 1/4\sum_{n=1}^\infty(n-1)(x/2)^{n-2} \\&=x\ln(2)+\sum_{n=1}^\infty((-1)^{n-1}3^nx^{n+1}) /n2^n + 1/4\sum_{m=1}^\infty(m-1)(x/2)^{m-2} \\&=x\ln(2)+\sum_{p=2}^\infty((-1)^{p-2}3^nx^{p}) /(p-1)2^{p-1} + 1/4\sum_{q=-1}^\infty(q+1)(x/2)^{q} \end{align}

where I let $p=n+1$ and $q=m-2$ in the last equality.

To combine then handle the case when $q=-1, 0,1$ separately, and you can combine both terms now.

Answer 2

Hint: The power series expansion of $\frac{1}{(2-x)^2}$ at $x=0$ is \begin{align*} \frac{1}{(2-x)^2}&=\frac{1}{4}\cdot\frac{1}{\left(1-\frac{x}{2}\right)^2}\\ &=\frac{1}{4}\sum_{n=0}^\infty\binom{-2}{n}\left(-\frac{x}{2}\right)^n =\frac{1}{4}\sum_{n=0}^\infty\binom{n+1}{n}\left(\frac{x}{2}\right)^n\\ &=\frac{1}{4}\sum_{n=0}^\infty\frac{n+1}{2^n}x^n \end{align*}

Here we use the binomial series expansion and the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

Note: Valid representations of a power series expansion at $x=0$ are for instance \begin{align*} \sum_{n=0}^\infty a_n x^n\qquad\text{or}\qquad a_0+a_1x+\sum_{n=2}^\infty a_nx^n\tag{1} \end{align*}

A representation putting everything in one sum is not always preferable. Sometimes it might be rather cumbersome to put everything in one sum and a representation as it is indicated in the right-hand side of (1) is more convenient. Nevertheless we usually collect all terms with equal power of $x$ into one term.

We obtain this way \begin{align*} \color{blue}{x}&\color{blue}{\ln(2+3x)+\frac{1}{(2-x)^2}}\\ &=(\ln 2)x+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{3}{2}\right)^nx^{n+1} +\frac{1}{4}\sum_{n=0}^\infty \frac{n+1}{2^{n}}x^n\\ &=(\ln 2)x+\sum_{n=2}^\infty \frac{(-1)^{n}}{n-1}\left(\frac{3}{2}\right)^{n-1}x^{n} +\sum_{n=0}^\infty \frac{n+1}{2^{n+2}}x^n\tag{2}\\ &\,\,\color{blue}{=\frac{1}{4}+\left(\frac{1}{4}+\ln 2\right)x +\sum_{n=2}^\infty\left( \frac{(-1)^{n}}{n-1}\left(\frac{3}{2}\right)^{n-1}+\frac{n+1}{2^{n+2}}\right)x^n}\tag{3} \end{align*}

Comment:

• In (2) we shift the index of the left series by one to obtain a repesentation with $x^n$ and we multiply the denominator of the coefficient of the right-hand series with $4$.

• In (3) we collect terms with equal power of $x$ keeping the constant and linear term out of the series for convenience only.