3
$\begingroup$

Find the inverse Laplace transform

$F(t)=\mathcal{L}^{-1}(s^{-\frac{1}{2}}e^{-\frac{1}{s}})$

using each of the following techniques:

  1. Expand the exponential in a Taylor series about s=∞, and take inverse Laplace transforms term by term (this is allowable since the series is uniformly convergent.).

  2. Sum the resultant series in terms of elementary functions.

$\endgroup$
  • $\begingroup$ That looks doable. What part have you problems with? $\endgroup$ – Daniel Fischer Oct 20 '13 at 18:17
  • $\begingroup$ Expanding the exponential in a Taylor series about s=∞. $\endgroup$ – Ravi Oct 20 '13 at 18:19
  • $\begingroup$ Ah. You know $e^z = \sum\limits_{n=0}^\infty \frac{1}{n!}z^n$, I suppose. Just insert $z = -\frac1s$ and you have it. $\endgroup$ – Daniel Fischer Oct 20 '13 at 18:21
  • $\begingroup$ I am just stuck here: after expanding I am getting the series as :1/s^.5 -1/s^1.5 + 1/s^2.5 ..so on.Now, I think I just need to manipulate it, so that I can use this formula:1/s^(n+1)=(t^n)/n!..But somehow could not able to translate into that form.Any help please? $\endgroup$ – Ravi Oct 20 '13 at 19:36
  • $\begingroup$ What is the inverse Laplace transform of $s^{-\alpha}$ generally? $\endgroup$ – Daniel Fischer Oct 20 '13 at 19:39
0
$\begingroup$

you should use the gamma function to find the inverse of each of the 1/sqrt(s) instead of the basic integer factorial. then take out a common factor of sqrt(t) and find a general series sum from n=0 to infinity. you also seem to be missing a 1/n! in the series.

then use the given gamma formula from the standard formulae to sub for the Gamma(n+1/2) function. you can then use wolfram to get a standard function that the series will converge to.

Then suggest what I can do for part b :)

$\endgroup$
  • $\begingroup$ I have done this:We know, Taylor Series:〖 e〗^z=∑_(n=0)^∞▒1/n! z^n Here, we take z=-1/s So, L^(-1) {s^(-1/2) e^(-1/s)} =L^(-1) {〖(-z)〗^(-1/2) 1/n! z^n} =- L^(-1) {1/n! 1/√z z^n} =-[L^(-1) { z^(-1/2) z^1} - L^(-1) {1/2! z^(-1/2) z^2} + … ] =-[L^(-1) {1/1! z^(1/2) } - L^(-1) {1/2! z^(3/2) } + L^(-1) {1/2! z^(5/2) }-…] $\endgroup$ – Ravi Oct 21 '13 at 11:21
  • $\begingroup$ Now what?I am not sure about Gama Function?Is there anyway I can do inverse Laplace transform of s−α generally? $\endgroup$ – Ravi Oct 21 '13 at 11:22
  • $\begingroup$ You should have $s^{-1/2}*exp(-1/s) = 1/s^{1/2} - 1/s^{3/2} + 1/(2!s^{5/2})+...$ etc.. then $L^{-1} {1/s^n} = t^{n-1}/Gamma(n)$ from Spiegel. Apply the inverse to the series and solve one by one. then take out a common factor of $1/t^{0.5}$ $\endgroup$ – Boris Oct 21 '13 at 13:31
  • $\begingroup$ The series sum should be $\sum_{n=0}^\infty (-t)^n/(Gamma(n+1/2)*n!)$ Enter the given formula for gamma(n+1/2), simplify a little and enter that series into wolfram. you should get a fairly standard function $\endgroup$ – Boris Oct 21 '13 at 13:39
  • $\begingroup$ How you are deducing that the series sum should be ∑∞n=0(−t)n/(Gamma(n+1/2)∗n!)?Also Gamma (n+1/2),I could not find any formula for that.But as in Wolfram I got:(2 (n+0.5)!)/(2 n+1) $\endgroup$ – Ravi Oct 21 '13 at 16:43
0
$\begingroup$

$\frac{\cos 2(\sqrt{t})}{\sqrt{\pi t}}$

$\endgroup$
  • $\begingroup$ I know the answer can be determined directly from the Wolfram, but that is not what I am suppose to do.:) $\endgroup$ – Ravi Oct 22 '13 at 8:54
  • $\begingroup$ It is also in the tables from Spiegel $\endgroup$ – anon Oct 22 '13 at 22:49
  • $\begingroup$ I know,but I have to 'proof' it! $\endgroup$ – Ravi Oct 23 '13 at 18:45
0
$\begingroup$

We know that $$ \displaystyle \mathcal{L} \{t^{n-1}\} = \frac{\Gamma(n)}{s^{n}}, n>0 ,s>0 $$ $$ \frac{1}{s^n} = \frac{\displaystyle \mathcal{L} \{t^{n-1}\}}{\Gamma(n)} $$ $$ \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^n}\} = \frac{t^{n-1}}{\Gamma(n)} $$ Therefore $$ \displaystyle \mathcal{L^{-1}} \{ s^{-1/2} e^{-1/s} \} = \displaystyle \mathcal{L^{-1}} \{ \frac{1}{s^{1/2}} - \frac{1}{s^{3/2}} + \frac{1}{(2!s^{5/2})} + ... + (-1)^{n}\frac{1}{n! s^{n+\frac{1}{2}}} \} $$ $$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^{n+\frac{1}{2}}}\} $$ $$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{t^{n-\frac{1}{2}}}{\Gamma(n+\frac{1}{2})} $$ But after this I don't know how to simplify

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.