# Math Help - Laplace transform from Taylor Series

1. ## Laplace transform from Taylor Series

Problem: Using the Taylor series for $\sin t$, and assuming that the Laplace transform of this series can be computed term by term, verify that $\mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1}.$

Attempt at solution:
So I start with the the taylor series of $\sin t = \displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n t^{2n + 1}}{(2n + 1)!}$,

Then using the fact that $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}},$ where $n \in \mathbb{Z}^+$, I am able to get to this step:

$\mathcal{L}\{\sin t\} = F(s) = \displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n}{s^{2(n+1)}}$

However here I'm stumped, I'm guessing there's some well known summation result I should be using to get to the final step, but the ones I've looked up I haven't been able to successfully use them to get the result I want.

Any suggestions would greatly be appreciated. Thanks in advance.

2. Alright, sorry about this, I was able to solve it. I wish I could delete the thread, however here's the solution just in case anyone is wondering I guess.

$F(s) = \displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n}{s^{2(n+1)}} = \displaystyle\sum^{\infty}_{n=0} \frac{1}{s^2}\left(\frac{-1}{s^2}\right)^n
$

We know $\displaystyle\sum^{\infty}_{n=0} ar^n = \frac{a}{1 - r}$. Here $a = \frac{1}{s^2}$ and $r = \frac{-1}{s^2}$. Thus $\displaystyle\sum^{\infty}_{n=0} \frac{1}{s^2}\left(\frac{-1}{s^2}\right)^n = \frac{\frac{1}{s^2}}{1 + \frac{1}{s^2}} = \frac{1}{s^2 + 1}$ as desired.

3. Nice signature!