Inverse Laplace Transform

nacho

Active member

For part a)
so far I have:
$e^x = 1 + \frac{x}{1!} + ...+ \frac{x^n}{n!}$
So
$S^\frac{-1}{2}e^\frac{-1}{S} = S^\frac{-1}{2}(1 -\frac{1}{S} + \frac{1}{2!S^2} - \frac{1}{3!S^3} + \frac{1}{4!S^4} + ... - ...)$

I don't think my $S^\frac{-1}{2}$ on the outside is correct though.
I don't know why it says 'take the inverse transform term by term', since this is a never ending series.

Little lost for part 2. They haven't covered it in lectures yet, I briefly recall this being related to the cauchy integral formula... or something similar we did earlier.

Any help is appreciated,

Thanks!

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chisigma

Well-known member

For part a)
so far I have:
$e^x = 1 + \frac{x}{1!} + ...+ \frac{x^n}{n!}$
So
$S^\frac{-1}{2}e^\frac{-1}{S} = S^\frac{-1}{2}(1 -\frac{1}{S} + \frac{1}{2!S^2} - \frac{1}{3!S^3} + \frac{1}{4!S^4} + ... - ...)$

I don't think my $S^\frac{-1}{2}$ on the outside is correct though.
I don't know why it says 'take the inverse transform term by term', since this is a never ending series.

Little lost for part 2. They haven't covered it in lectures yet, I briefly recall this being related to the cauchy integral formula... or something similar we did earlier.

Any help is appreciated,

Thanks!
The expansion of F(s) is...

$\displaystyle \frac{e^{- \frac{1}{s}}}{\sqrt{s}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{s^{n + \frac{1}{2}}\ n!}\ (1)$

... and because is...

$\displaystyle \mathcal{L}^{-1} \{\frac{1}{s^{n + \frac{1}{2}}}\} = \frac{t^{n - \frac{1}{2}}}{\Gamma(n + \frac{1}{2})} = \frac{(2\ t)^{n}}{\sqrt{\pi\ t}\ (2\ n - 1)!!}\ (2)$

... is also...

$\displaystyle \mathcal{L}^{-1} \{\frac{e^{- \frac{1}{s}}}{\sqrt{s}}\} = \frac{1}{\sqrt{\pi\ t}}\ \sum_{n=0}^{\infty} \frac{(-1)^{n}\ (2\ t)^{n}}{n!\ (2 n - 1)!!} = \frac{\cos 2\ \sqrt{t}}{\sqrt{\pi\ t}}\ (3)$

Kind regards

$\chi$ $\sigma$

nacho

Active member
Ok, what's with the double factorial? I have never seen that before and this worries me

chisigma

Well-known member
Ok, what's with the double factorial? I have never seen that before and this worries me
$\displaystyle (2 n-1) \cdot (2 n-3) \cdot ... \cdot 3 \cdot 1 = (2 n-1)!!$

$\displaystyle 2 n \cdot (2 n - 2) \cdot ...\cdot 2 = (2\ n)!!$

Kind regards

$\chi$ $\sigma$