# Series expansion

1. Nov 17, 2009

### matematikawan

I'm trying to work something on inverse Laplace transform. I need to express a transfer function F(s) to the form
$$F(s)=\frac{s^{-1} (a_0 + a_1s^{-1} + a_2s^{-2}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }$$

I can easily do it for rational function e.g.
$$\frac{s^3+2s^2+3s+1}{s+4}= \frac{s^{-1} (1+2s^{-1}+3s^{-2}+s^{-3})}{s^{-3}+4s^{-4}}$$

for some indicial e.g
$$\frac{1}{\sqrt{s^2+s}}=\frac{1}{s}(1+s^{-1})^{-1/2}$$
Expand using the binomial theorem.

My problem is how to express irrational functions such as s-3/2 or
$$\frac{e^{-\sqrt{s}}}{s}$$
to the form of F(s).

2. Nov 18, 2009

### g_edgar

$F(s)$ is regular at infinity (in the sense of complex variables) but neither $s^{-3/2}$ nor $e^{-\sqrt{s}}/s$ is.

3. Nov 19, 2009

### matematikawan

That is an interesting observation that you make. If you could please explain more what regular at infinity means. I'm do not know much about complex variables theory. That's why I want to avoid dealing with Bromwich integral.

My idea is to repeatedly use
$$L^{-1} { \frac{F(s)}{s} }=\int_0^t f(t) dt$$
for evaluating inverse transform.

From Laplace transform table I know that $s^{-3/2}$ and $e^{-2\sqrt{s}}/s$ can be inverted. That's why I expect the functions could be express as F(s)

4. Nov 19, 2009

### g_edgar

OK, the function

$$F(s)=\frac{s^{-1} (a_0 + a_1s^{-1} + a_2s^{-2}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }$$

has the property that $s^kF(s)$ converges to a finite nonzero limit as $s \to \infty$ for some integer $k$ . Neither of the two functions in question has that property.

5. Nov 19, 2009

### gato_

There is no standart way for those transforms. You have to proceed from the integral form:
$$\frac{1}{2\pi}PV \int_{-\infty}^{\infty}{exp(st)s^{-3/2}ds}=\frac{1}{2\pi}2\sqrt{t}\int_{0}^{\infty}{exp(z)z^{-1/2-1}dz}=2\sqrt{t}\frac{1}{2\pi}\Gamma[-1/2]=-2\sqrt{\frac{t}{\pi}}$$
(for t>0)

Last edited: Nov 19, 2009
6. Nov 20, 2009

### matematikawan

Actually I'm trying to solve pde not inverse Laplace transform per se. You know after we transform the pde wrt time and solve the transformed ode in s-domain, we usually obtain quite complicated F(s). It is this F(s) that I want to invert numerically.

But from what have you have shown, may be I can learnt something from it. Just need further clarification. Why is your inverse transform for s-3/2 has a negative sign. The entry http://www.vibrationdata.com/Laplace.htm" [Broken] has no such sign. And why is your integral form different from entry 1.2 of the table. Where has the integration end limit $c+i\infty$ gone?

Last edited by a moderator: May 4, 2017
7. Nov 20, 2009

### matematikawan

So the converse statement is not true for those cases. Thanks g_edgar. I have to sort out this first. May be I will change the form to

$$F(s)=\frac{s^{-1} (a_0 + a_1s^{-0.5} + a_2s^{-1}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }$$

I will try to figure out how to handle this case and come back again if I have a problem.

8. Nov 26, 2009

### gato_

you are right abount the factor in the limits of the integral (missed another i in the denominator ). The difference in sign is probably because i chose the wrong contour

9. Nov 26, 2009

### gato_

Apparently it is easier to prove the converse, I am not sure about the inversion formula. Now,
$$\int_{0}^{\infty}t^{\nu}e^{-st}dt=\frac{1}{\nu+1}\int_{0}^{\infty}z^{\nu}e^{-z}dz=\frac{\Gamma(\nu+1)}{s^{\nu+1}}$$
The correct form of the integral I posted previously is then an alternative definition for $$\Gamma$$:
$$\frac{1}{\Gamma(\nu+1)}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{s}}{s^{\nu+1}}$$
Which I didn't kow