Maclaurin series of a quotient of functions

Question

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?

Answer 1

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: $$ \small-\frac1{2\cosh(8)-2}+\frac{4\cosh(8)+\sinh(8)}{2(\cosh(8)-1)^2}x-\frac{\cosh^2(4)(9\cosh(8)+4\sinh(8)-1)}{8\sinh^6(4)}x^2+\dots\tag6 $$

Answer 2

This is done by a process similar to polynomial long division. If denominator is divisible by $x$, then the answer is not a power series, but a Laurent series in fact.

I'll show a concrete example of how it is calculated, namely if we know

\begin{align*} \sin x&=x-{x^3\over 6}+{x^5\over 120}\mp \dots \\ \cos x&=1-{x^2\over 2}+{x^4\over 24}\mp \dots \end{align*}

then the quotient ${\sin x \over {\cos x}} = \tan x$. First, take the lowest terms of the numerator and denominator, in this case $x$ and $1$, and divide them to get the first term of the quotient

\begin{align*} \tan x &= x+\dots \end{align*}

Now subtract $x$ times the denominator from the numerator

\begin{align*} \sin x - x\cos x&=\left(x-{x^3\over 6}\pm \dots\right) -x\left(1-{x^2\over 2}\pm \dots\right)\\ &= {x^3\over 3}-{x^5\over 30}\pm \dots \end{align*}

Now we repeat: take the lowest term in the top and bottom, $x^3\over 3$ and $1$, and divide them for the next term of $\tan x$.

\begin{align*} \tan x &= x+{x^3\over 3}\dots \end{align*}

Now we subtract $x^3\over 3$ times the denominator from the numerator again

\begin{align*} \left({x^3\over 3}-{x^5\over 30}\pm \dots\right) -{x^3\over 3}\left(1-{x^2\over 2}+...\right)&={2x^5\over 15}\pm \dots \end{align*}

Your series is a Laurent series in the variable $x^{1\over 2}$, so you can apply the same process to it.