I have a function $f$ that is defined as $f(x)=g(x)/h(x)$. I am trying to find the Maclaurin series of the $f$ function, knowing that I perfectly know the Maclaurin series of $g$ and $h$. I also know that the radius of convergency of my two series are infinite. What are my possibilities ?
I have to say that both of the Maclaurin functions are not made with integer polynomials, for example :
$$g(x) = \sum_{n=0}^{+\infty} \frac{4^n}{n!}x^{-\frac{1+2n}{2}}$$
Particularly, the second term, $h$ gives me a strong dificulty: two sums are inside of it:
$$h(x) = \sum_{n=0}^{+\infty} \left( \frac{2^{2n+1}}{n!}x^{\frac{-(2n+1)}{2}}\left( 1-\sum_{k=0}^{+\infty}\begin{pmatrix}2n+1 \\2k\end{pmatrix}x^{2k}\right) \right)$$
I tried to invert one of the Maclaurin series in order to use the Cauchy's product formula:
$$\sum_{n=0}^{+\infty} g_n \sum_{n=0}^{+\infty} h_n = \sum_{n=0}^{+\infty} \sum_{k=0}^{n} g_k \, h_{n-k}$$
but the result that I have obtained does not converge to my function. How should I do this?
Best Answer
First, $$ \begin{align} g(x) &=\sum_{n=0}^\infty\frac{4^n}{n!}x^{-n-\frac12}\tag{1a}\\ &=\frac1{\sqrt{x}}e^{4/x}\tag{1b} \end{align} $$ This has an essential singularity at $x=0$. Therefore, it does not have a Maclaurin series.
Next, note that $$ \sum\limits_{k=0}^\infty\binom{2n+1}{2k}x^{2k}=\frac{(1+x)^{2n+1}+(1-x)^{2n+1}}2\tag2 $$ Therefore, $$ \begin{align} h(x) &=\sum_{n=0}^\infty\frac{4^n}{n!}x^{-n-\frac12}\left(2-(1+x)^{2n+1}-(1-x)^{2n+1}\right)\tag{3a}\\ &=\frac1{\sqrt{x}}\left(2e^{4/x}-(1+x)e^{\frac4x(1+x)^2}-(1-x)e^{\frac4x(1-x)^2}\right)\tag{3b} \end{align} $$ Again, this function has an essential singularity at $x=0$. Therefore, it also does not have a Maclaurin series.
However, $$ \begin{align} \frac{h(x)}{g(x)} &=2-(1+x)e^{4(x+2)}-(1-x)e^{4(x-2)}\tag{4a}\\ &=2-e^{4x}\!\left((1+x)e^8+(1-x)e^{-8}\right)\tag{4b}\\[3pt] &=2-2e^{4x}(\cosh(8)+x\sinh(8))\tag{4c} \end{align} $$ Therefore, $$ \frac{g(x)}{h(x)}=\frac1{2-2e^{4x}(\cosh(8)+x\sinh(8))}\tag5 $$ which does have a Maclaurin series, but a pretty complicated one. Here are the first few terms: $$ \small-\frac1{2\cosh(8)-2}+\frac{4\cosh(8)+\sinh(8)}{2(\cosh(8)-1)^2}x-\frac{\cosh^2(4)(9\cosh(8)+4\sinh(8)-1)}{8\sinh^6(4)}x^2+\dots\tag6 $$