Help solving taylor series.

**goblinf** · December 14th 2010, 09:30 AM

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

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

about x=0

Could anyone please show me the solution?

**tonio** · December 14th 2010, 09:39 AM

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

about x=0

Could anyone please show me the solution?

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

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

Tonio

**goblinf** · December 14th 2010, 09:59 AM

And for which values of x does the series converge?

**TheCoffeeMachine** · December 14th 2010, 11:32 AM

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 $x$ .

**goblinf** · December 15th 2010, 01:29 AM

Originally Posted by TheCoffeeMachine

Tonio stated that after the series, surely? It holds for all $x$ .

Could you please describe this a bit more. How do you see that it holds for all $x$ ?

**Prove It** · December 15th 2010, 01:31 AM

It's well known that the Taylor series for $\displaystyle e^x$ is convergent for all $\displaystyle x$ .

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

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