I've just started working with Taylor series and can't figure out how to find the series of the following function: g(x)=$\displaystyle \int_0^x [2e^v+3e^{-v}] dv$ about x=0 Could anyone please show me the solution?
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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)=$\displaystyle \int_0^x [2e^v+3e^{-v}] dv$ about x=0 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
And for which values of x does the series converge?
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 $\displaystyle x$.
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 $\displaystyle x$?
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.
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