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We can find the Taylor or 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  
\[x^3\]
.
\[e^x=\sum^{\infty}_{n=0} \frac{x^n}{n!}=1 + x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+...=1+x+ \frac{x^2}{2}+ \frac{x^3}{6}+...\]

\[\frac{1}{cpsx}=\frac{1}{\sum^{\infty}_{m=0} \frac{(-1)^{m}x^{2m}}{(2m)!}}=(1- \frac{x^2}{2!}+...)^{-1}=1+ \frac{x^2}{2}+...\]

\[\begin{equation} \begin{aligned}\frac{e^x}{cosx} &= (1+x+ \frac{x^2}{2}+ \frac{x^3}{6}+...)(1+ \frac{x^2}{2}+...) \\ &= 1+(1)x+( \frac{1}{2} + \frac{1}{2} )x^2+(\frac{1}{6} + \frac{1}{2})x^3+... \\ &=1 +x+x^2 + \frac{2x^3}{3}+.... \end{aligned} \end{equation} \]