# Thread: Finding New Taylor Series from Old Ones

1. ## Finding New Taylor Series from Old Ones

Use a know taylor series to find the taylor series about c=0 for the given function and find its radius of convergence.

$\displaystyle f(x)=x sin(x)$

well, the taylor series for sin(x) is

$\displaystyle sin(x)= \sum\frac{-1^k}{2k+1!}x^{2k+1}$

So I know I plug in the x's, but I don't know which ones since there's one before and one inside of the sin function.

2. Originally Posted by kl.twilleger
Use a know taylor series to find the taylor series about c=0 for the given function and find its radius of convergence.

$\displaystyle f(x)=x sin(x)$

well, the taylor series for sin(x) is

$\displaystyle sin(x)= \sum\frac{-1^k}{2k+1!}x^{2k+1}$

So I know I plug in the x's, but I don't know which ones since there's one before and one inside of the sin function.
You have:

$\displaystyle \sin(x)= \sum_{k=0}^{\infty}\frac{-1^k}{(2k+1)!}x^{2k+1}$

so:

$\displaystyle f(x)=x \sin(x)= \sum_{k=0}^{\infty}\frac{-1^k}{(2k+1)!}x^{2k+2}$

Now adjust the indices to taste.

This has the same radius of convergence as the series for $\displaystyle \sin(x)$

CB