Finding the coefficients of the power series representation given a function

functionspower seriessequences-and-seriestaylor expansion

I'm trying to solve this problem:

Find the coefficients $a_k$ and the interval of convergence of the series: $f(x)=\frac{arctan(x)}{x+1}=\sum_{k=0}^\infty a_k(2x+1)^k$

So far what I've tried is to get the series representation of $arctan(x)$ centered in $(x-\frac{1}{2})$ like this:
$\frac{1}{1+x^2}=\frac{1}{\frac{5}{4}+(x^2-\frac{1}{4})}=\frac{1}{\frac{5}{4}}\frac{1}{1-(-\frac{4}{5}(x^2-\frac{1}{4}))}=\frac{4}{5}\sum_{k=0}^\infty (\frac{4}{5}(-(x^2-\frac{1}{4})))^k=\frac{4}{5}\sum_{k=0}^\infty(-1)^k\frac{4^k}{5^k}(x+\frac{1}{2})^k(x-\frac{1}{2})^k$

However I'm not sure how to proceed from here. Oh also I can easily get a series representation of $\frac{1}{1+x}$ of the form $\sum_{k=0}^\infty a_k(2x+1)^k$, but in that case I don't know how to incorporate the $arctan(x)$.

Best Answer

It is convenient to make a partial fraction decomposition since the linear terms in the denominator are then easier to expand.

We expand $\frac{d}{dx}\arctan(x)$ at $x=-\frac{1}{2}$ and obtain \begin{align*} \color{blue}{\frac{d}{dx} \arctan (x)}&=\frac{1}{1+x^2}=\frac{1}{(1+ix)(1-ix)}\\ &=\frac{1}{2}\,\frac{1}{1-ix}+\frac{1}{2}\,\frac{1}{1+ix}\\ &=\frac{1}{2}\,\frac{1}{1+\frac{i}{2}-i\left(x+\frac{1}{2}\right)} +\frac{1}{2}\,\frac{1}{1-\frac{i}{2}+i\left(x+\frac{1}{2}\right)}\\ &=\frac{1}{2}\,\frac{1}{\frac{2+i}{2}\left(1-\frac{2i}{2+i}\left(x+\frac{1}{2}\right)\right)} +\frac{1}{2}\,\frac{1}{\frac{2-i}{2}\left(1+\frac{2i}{2-i}\left(x+\frac{1}{2}\right)\right)}\\ &=\frac{1}{2+i}\sum_{k=0}^\infty\left(\frac{i}{2+i}\right)^k(2x+1)^k\\ &\qquad+\frac{1}{2-i}\sum_{k=0}^\infty(-1)^k\left(\frac{i}{2-i}\right)^k(2x+1)^k\tag{1}\\ &=\sum_{k=0}^\infty\frac{1}{5^{k+1}}\left(i^k(2-i)^{k+1}+(-i)^k(2+i)^{k+1}\right)(2x+1)^k\tag{2}\\ &=\sum_{k=0}^\infty\frac{1}{5^{k+1}}\left((1+2i)^{k}(2-i)+(1-2i)^{k}(2+i)\right)(2x+1)^k\\ &=\sum_{k=0}^\infty\frac{1}{5^{k+1}} \sum_{l=0}^k\binom{k}{l}\left((2i)^l(2-i)+(-2i)^l(2+i)\right)(2x+1)^k\tag{3}\\ &=\sum_{k=0}^\infty\frac{1}{5^{k+1}} \left(2\sum_{l=0}^k\binom{k}{l}(2i)^l\left(1+(-1)^l\right)\right.\\ &\qquad\qquad\qquad\quad \left.+i\sum_{l=0}^k\binom{k}{l}(2i)^l\left(1-(-1)^l\right)\right)(2x+1)^k\tag{4}\\ &=\sum_{k=0}^\infty\frac{1}{5^{k+1}} \left(\sum_{l=0}^k\binom{k}{2l}2^{2l+2}(-1)^l\right.\\ &\qquad\qquad\qquad\quad \left.+\sum_{l=0}^k\binom{k}{2l+1}2^{2l+2}(-1)^l\right)(2x+1)^k\tag{5}\\ &\,\,\color{blue}{=\sum_{k=0}^\infty\frac{1}{5^{k+1}}\sum_{l=0}^k\binom{k+1}{2l+1}2^{2l+2}(-1)^l(2x+1)^k}\tag{6} \end{align*}

Comment:

  • In (1) we can make the geometric series expansion and get a wanted representation in the form $\sum_{i=0}^\infty c_k(2x+1)^k$. We make some further transformations/simplifications to get rid of the imaginary unit.

  • In (2) we expand the terms of the series with the complex conjugate to obtain $\left(\frac{1}{2+i}\right)^{k+1}=\left(\frac{2-i}{(2+i)(2-i)}\right)^{k+1}=\frac{1}{5^{k+1}}(2-i)^{k+1}$ and similarly with the other part.

  • In (3) we make a binomial expansion and collect the terms conveniently in the next step.

  • In (4) we take even index $2l$ in the left-hand sum and odd index $2l+1$ in the right hand sum, since other indices do not contribute.

  • In (5) we use the binomial identity $\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}$.

In order to derive the series expansion of $\arctan (x)$ at $x=-\frac{1}{2}$ we can integrate (6) termwise and set the constant term accordingly. We obtain

\begin{align*} \arctan(x)&=-\arctan\left(\frac{1}{2}\right) +\sum_{k=0}^\infty\frac{1}{5^{k+1}}\sum_{l=0}^k\binom{k+1}{2l+1}\frac{2^{2l+2}(-1)^l}{k+1}(2x+1)^{k+1}\\ &=-\arctan\left(\frac{1}{2}\right) +\sum_{k=0}^\infty\frac{1}{5^{k+1}}\sum_{l=0}^k\binom{k}{2l}\frac{2^{2l+2}(-1)^l}{2l+1}(2x+1)^{k+1}\\ &=-\arctan\left(\frac{1}{2}\right) +\sum_{k=1}^\infty\frac{1}{5^{k}}\sum_{l=0}^{k-1}\binom{k-1}{2l}\frac{2^{2l+2}(-1)^l}{2l+1}(2x+1)^{k}\tag{7}\\ \end{align*}

Expansion of $\frac{1}{x+1}$ at $x=-\frac{1}{2}$ is not that hard. We have \begin{align*} \frac{1}{x+1}=\frac{1}{\frac{1}{2}\left(1+(2x+1)\right)}=2\sum_{k=0}^\infty(-1)^k(2x+1)^k\tag{8} \end{align*}

Finally multiplying (7) with (8) we get the desired expansion: \begin{align*} \frac{\arctan(x)}{x+1}&=\sum_{k=0}^\infty \left(\color{blue}{-2\arctan\left(\frac{1}{2}\right)(-1)^k}\right.\\ &\qquad\qquad\left.\color{blue}{+\sum_{m=1}^k\frac{1}{5^m}\sum_{l=0}^{m-1}\binom{m-1}{2l}\frac{2^{2l+3}(-1)^l}{2l+1}(-1)^{k-m}}\right) (2x+1)^k \end{align*} with the wanted $a_k$ marked in blue.

The radius of convergence is $\frac{1}{2}$ since the nearest singularity at the center $x=-\frac{1}{2}$ of the series expansion is $-1$.

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