How Do You Compute the Inverse Laplace Transform of a Power Series?

[nacho-man]

Please refer to the attachment.

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!

 

[chisigma]

Please refer to the attachment.

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-man]

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

 

[chisigma]

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

$\chi$ $\sigma$

 

[blue_raver22]

I understand that the inverse Laplace transform is a mathematical operation that is used to convert a function from the Laplace domain to the time domain. It is denoted as L^-1 and is the inverse of the Laplace transform, denoted as L. The Laplace transform is a powerful tool in engineering, physics, and other scientific fields for solving differential equations and analyzing systems.

In this particular question, we are given the function e^x and asked to find its inverse Laplace transform. To do this, we can use the definition of the Laplace transform, which states that:

L{f(t)} = F(s) = ∫f(t)e^-st dt

Where s is the complex variable and t is the time variable. In the case of e^x, we have:

L{e^x} = ∫e^x e^-st dt

To solve this integral, we can use the power series expansion of e^x given by:

e^x = 1 + x + x^2/2! + x^3/3! + ...

Substituting this into our integral, we get:

L{e^x} = ∫(1 + x + x^2/2! + x^3/3! + ...)e^-st dt

= ∫e^-st + xe^-st + x^2/2!e^-st + x^3/3!e^-st + ... dt

= ∫e^-st dt + ∫xe^-st dt + ∫x^2/2!e^-st dt + ∫x^3/3!e^-st dt + ...

Using the definition of the Laplace transform, we can rewrite this as:

L{e^x} = 1/s + 1/s^2 + 2!/s^3 + 3!/s^4 + ...

= ∑n=0∞n!/s^(n+1)

This is the inverse Laplace transform of e^x. However, the question also asks us to take the inverse transform term by term. This means that we need to use the inverse Laplace transform on each term separately. This will give us:

L^-1{1/s} = 1

L^-1{1/s^2} = t

L^-1{2!/s^3} = 2!/2!