Taylor/Mclaurin Series of a Product

We can find the Taylor or Mclaurin series of a product of functions  
\[f(x), \; g(x)\]
  about a point by collecting powers of  
\[(x-x_0)\]
  or  
\[x\]
  in the product of the Taylor or 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  
\[g(x)\]
  us  
\[\sum^{\infty}_{m=0} b_m(x-x_0)^m\]
  then the Taylor series for  
\[f(x)g(x)\]
  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  
\[e^x sinx\]
  up to  
\[x^3\]
.
\[e^x=\sum^{\infty}_{n=0} \frac{x^n}{n!}=1 + x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+...=1+x+ \frac{x^2}{2}+ \frac{x^3}{6}+...\]

\[sinx=\sum^{\infty}_{m=0} \frac{(-1)^mx^{2m+1}}{(2m+1)!}=x - \frac{x^3}{3!}+...=x- \frac{x^3}{6}+...\]

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

Add comment

Security code
Refresh