# Help solving taylor series.

• Dec 14th 2010, 09:30 AM
goblinf
Help solving taylor series.
I've just started working with Taylor series and can't figure out how to find the series of the following function:

g(x)=$\displaystyle \int_0^x [2e^v+3e^{-v}] dv$

Could anyone please show me the solution?
• Dec 14th 2010, 09:39 AM
tonio
Quote:

Originally Posted by goblinf
I've just started working with Taylor series and can't figure out how to find the series of the following function:

g(x)=$\displaystyle \int_0^x [2e^v+3e^{-v}] dv$

Could anyone please show me the solution?

$\displaystyle \int\limits_0^x [2e^v+3e^{-v}] dv=2e^x-2-3e^{-x}+3=2e^x-3e^{-x}+1$ . Now just remember that

$\displaystyle \displaystyle{e^x=\sum\limits^\infty_{k=0}\frac{x^ k}{k!}\,,\,\,\forall x\in\mathbb{R}}$

Tonio
• Dec 14th 2010, 09:59 AM
goblinf
And for which values of x does the series converge?
• Dec 14th 2010, 11:32 AM
TheCoffeeMachine
Quote:

Originally Posted by goblinf
And for which values of x does the series converge?

Tonio stated that after the series, surely? It holds for all $\displaystyle x$.
• Dec 15th 2010, 01:29 AM
goblinf
Quote:

Originally Posted by TheCoffeeMachine
Tonio stated that after the series, surely? It holds for all $\displaystyle x$.

Could you please describe this a bit more. How do you see that it holds for all $\displaystyle x$?
• Dec 15th 2010, 01:31 AM
Prove It
It's well known that the Taylor series for $\displaystyle \displaystyle e^x$ is convergent for all $\displaystyle \displaystyle x$.

But this is also relatively easy to show using the ratio test.