Taylor/Mclaurin Series of a Quotient
\[\frac{f(x)}{g(x)}\]
of functions \[f(x), \; g(x)\]
about a point by collecting powers of \[(x-x_0)\]
or \[x\]
after using the binomial expansion to express \[\frac{1}{g(x)} \]
in the form \[\sum^{\infty}_{n=0}a_n(x-x_0)^n\]
or \[\sum^{\infty}_{n=0}a_nx^n\]
for a Mclaurin series.If the Taylor series for
\[f(x)\]
is \[\sum^{\infty}_{n=0} a_n(x-x_0)^n\]
and the Taylor series for \[\frac{1}{g(x)}\]
us \[\sum^{\infty}_{m=0} b_m(x-x_0)^m\]
then the Taylor series for \[\frac{f(x)}{g(x)}\]
is \[\sum^{\infty}_{k=0} c_k (x-x_0)^k c\]
where \[c_k= \sum_{m+n=k} a_nb_m\]
and similarly for the Mclaurin series.Example: Find the Mclaurin series for
\[\frac{e^x}{cosx}\]
up to \[x^3\]
.\[e^x=\sum^{\infty}_{n=0} \frac{x^n}{n!}=1 + x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+...=1+x+ \frac{x^2}{2}+ \frac{x^3}{6}+...\]
\[\frac{1}{cpsx}=\frac{1}{\sum^{\infty}_{m=0} \frac{(-1)^{m}x^{2m}}{(2m)!}}=(1- \frac{x^2}{2!}+...)^{-1}=1+ \frac{x^2}{2}+...\]
\[\begin{equation} \begin{aligned}\frac{e^x}{cosx} &= (1+x+ \frac{x^2}{2}+ \frac{x^3}{6}+...)(1+ \frac{x^2}{2}+...) \\ &= 1+(1)x+( \frac{1}{2} + \frac{1}{2} )x^2+(\frac{1}{6} + \frac{1}{2})x^3+... \\ &=1 +x+x^2 + \frac{2x^3}{3}+.... \end{aligned} \end{equation} \]
