MHB Problem of the Week #64 - June 17th, 2013

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Let
\[f(t)=\begin{cases}\dfrac{\sin t}{t}, & t\neq 0\\ 1, & t=0.\end{cases}\]

(a) Find the Taylor series for $f$ about $0$.
(b) Assuming that the Laplace transform can be computed term by term, verify that $\mathcal{L}\{f(t)\}=\arctan(1/s)$ for $s>1$.

-----

 

[Chris L T521]

This week's question was correctly answered by Sudharaka. You can find his solution below.

Spoiler

(a)

\begin{eqnarray}

f(t)&=&\begin{cases}\frac{\sin t}{t}, & t\neq 0\\ 1, & t=0\end{cases}\\

&=&\begin{cases}\frac{1}{t}\sum_{n=0}^{\infty} \frac{(-1)^n t^{2n+1}}{(2n+1)!}, & t\neq 0\\ 1, & t=0\end{cases}\\

&=&\begin{cases}\sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{(2n+1)!}, & t\neq 0\\ 1, & t=0\end{cases}\\

\therefore f(t) &=&\sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{(2n+1)!}

\end{eqnarray}

(b)

\begin{eqnarray}

\mathcal{L}\{ f(t) \}&=&\sum_{n=0}^{\infty} \frac{(-1)^n \mathcal{L}\{t^{2n}\}}{(2n+1)!}\\

&=&\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{(2n)!}{s^{2n+1}}\right) \mbox{ where }s>0\\

&=&\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}\left(\frac{1}{s}\right)^{2n+1} \mbox{ where }s>0\\

\therefore \mathcal{L}\{ f(t) \}&=& \arctan\left(\frac{1}{s}\right) \mbox{ where }\frac{1}{s}\leq 1

\end{eqnarray}

That is,

\[\mathcal{L}\{ f(t) \}= \arctan\left(\frac{1}{s}\right) \mbox{ where }s\geq 1\]

Note: Here we have assumed \(s\) to be a real number. Although the Laplace transform is defined for complex \(s\) as well.