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Laplace Transform and Gamma function


Johnny5

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Can someone express for me the relationship between the two. I want to do some review. Also, I don't know what latex symbol can be used to represent the Laplace transform, so I'd like to see that. And also the symbol for a Fourier transform as well.

 

Thanks

 

PS: I used to use the Gamma function all the time, but I forget it now. I could google it, but I'd rather discuss it here.

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The only one I vaguely remember seeing is here. That is:

 

[math]\text{LT}(t^p) = \frac{\Gamma(p+1)}{s^{p+1}}[/math].

 

To be perfectly honest with you, I've not used it much myself. I was planning to do an essay on the topic, but I couldn't find enough interesting material at the right kind of level to include.

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In regards to Laplace/Fourier transforms, I tend to just use FT(.) or LT(.). You can do these nicely by using \text{FT}.

 

I was hoping for that script L, that's on the site you gave the link to.

I want to be consistent in symbolism.

 

 

[math] \mathcal{F} [/math]

 

[math] \mathcal{L} [/math]

 

Ok so, here then is the definition of the Gamma function:

 

[math] \Gamma (n+1) \equiv \int_{x=0}^{x= \infty} x^n e^{-x} dx [/math]

 

And here is the definition of the Laplace transform of f(t):

 

[math] \mathcal{L} [ f(t) ] \equiv \int_{t=0}^{t= \infty} f(t) e^{-St} dt[/math]

 

Let x become St in the definition of the Gamma function. Hence

[math] \Gamma (n+1) = \int_{St=0}^{St= \infty} (St)^n e^{-St} d(St) [/math]

 

Let S be finite, and nonzero, therefore we can divide both sides of the upper and lower limits of integration by S to obtain:

 

[math] \Gamma (n+1) = \int_{t=0}^{t= \infty} (St)^n e^{-St} d(St) [/math]

 

Let S be constant in time, or abstractly let d(St) = S dt, hence:

 

[math] \Gamma (n+1) = \int_{t=0}^{t= \infty} (St)^n e^{-St} Sdt [/math]

 

Therefore:

 

[math] \Gamma (n+1) = \int_{t=0}^{t= \infty} S^{n+1} t^n e^{-St} dt [/math]

 

Since S is constant in time, and n is constant in time, we can pull Sn+1 outside of the integral to obtain:

 

[math] \Gamma (n+1) = S^{n+1} \int_{t=0}^{t= \infty} t^n e^{-St} dt [/math]

 

Since S is nonzero, we can divide both sides of the equation above by Sn+1 to obtain:

 

[math] \frac{\Gamma (n+1)}{S^{n+1} } = \int_{t=0}^{t= \infty} t^n e^{-St} dt [/math]

 

We can now recognize the RHS as the Laplace transform of tn, therefore:

 

[math] \frac{\Gamma (n+1)}{S^{n+1} } = \mathcal{L} [ t^n ]

[/math]

 

This is exactly the relation given by Dave, only with n instead of p. :)

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To be perfectly honest with you' date=' I've not used it much myself. I was planning to do an essay on the topic, but I couldn't find enough interesting material at the right kind of level to include.[/quote']

 

I think you should write an essay, I wouldn't mind seeing some of it.

 

Regards

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Unfortunately, I changed my essay title after doing a bit of research. The new title is "Chaotic Systems and Strange Attractors" - only about 4,000 words, but it's enough :P

 

I found the Gamma function to be very interesting, but it was very much labour intensive trying to find enough material that was "different" and easy enough that I could quickly digest it and get it into the essay. I did have quite a lot of fun scouring the journals though.

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Does it exist any explicit formula for the inverse of the gamma function (within some suitable region)?

 

I have (without succes) tried to derive an inverse with Laplace transformation, the inverse of Laplace transformation, other integral transformations, with functional equations, etc etc.

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Does it exist any explicit formula for the inverse of the gamma function (within some suitable region)?

 

I have (without succes) tried to derive an inverse with Laplace transformation' date=' the inverse of Laplace transformation, other integral transformations, with functional equations, etc etc.[/quote']

 

There is an inverse for the laplace transform, so yes. I used to know that stuff backwards and fowards. Ah someone else will have to show you.

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There is an inverse for the laplace transform, so yes. I used to know that stuff backwards and fowards. Ah someone else will have to show you.

 

Here's the inverse Laplace transform formula.

 

[math]f(t) = \frac{1}{2\pi j}\int_{\sigma - j\infty}^{\sigma + j\infty}F(s)e^{st}ds[/math], where [math]s = \sigma + j\omega[/math], and [math]F(s)[/math] is the Laplace transform of [math]f(t)[/math].

 

To evaluate this integral as it stands, one needs to use techniques from complex analysis, specifically, contour integration.

 

It's worth noting that most people never have any use for this formula, since one can basically use lookup tables to compute Laplace transforms, and use the same tables, combined with partial fractions to compute inverse Laplace transforms.

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Sorry! I was somewhat unclear.

 

I mean that I have tried to use Laplace transforms and inverse Laplace transforms, to derive an inverse to the gamma function.

 

Like this:

 

[math]

\ Gamma (n+1) = n! = \int_{x=0}^{x= \infty} x^n e^{-x} dx

[/math]

 

with inverse:

 

[math]

\ Gammainverse (n!) = n+1.

[/math]

 

But what is the explicit formula for Gammainverse (an integral, etc)?

Could it be of the form:

 

[math]

\ Gammainverse(s) = \int_{t=0}^{t= \infty} f(s,t) dt

[/math]

 

with f some suitable function? If so, how do I prove this?

With any integral transform?

 

That is what I have failed to derive.

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Here's the inverse Laplace transform formula.

 

[math]f(t) = \frac{1}{2\pi j}\int_{\sigma - j\infty}^{\sigma + j\infty}F(s)e^{st}ds[/math]' date=' where [math']s = \sigma + j\omega[/math], and [math]F(s)[/math] is the Laplace transform of [math]f(t)[/math].

 

Dapthar, do you know how to derive the formula for the inverse Laplace transform? It certainly doesn't just pop into one's head.

 

Thank you

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Dapthar' date=' do you know how to derive the formula for the inverse Laplace transform? It certainly doesn't just pop into one's head.

 

Thank you[/quote']This PDF has a fairly good derivation of the inverse Laplace transform.

 

Link: http://wwwteach.phy.bris.ac.uk/Level3/phys30600/CourseMaterials/Laplace.pdf (There isn't a typo in the link, as there isn't supposed to be a dot after the 'www'.)

 

Just skip down to page 5, under the Bromwich Inversion Formula heading.

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This PDF has a fairly good derivation of the inverse Laplace transform.

 

Link: http://wwwteach.phy.bris.ac.uk/Level3/phys30600/CourseMaterials/Laplace.pdf (There isn't a typo in the link' date=' as there isn't supposed to be a dot after the 'www'.[/i'])

 

Just skip down to page 5, under the Bromwich Inversion Formula heading.

 

In this post, I'm going to attempt to understand the derivation in that link you gave, thanks. :)

 

OK I am starting with the:

 

Bromwich Inversion Formula

 

The derivation begins by stating the definition of the Laplace transform of a function of t:

 

Definition (Laplace Transform):

 

[math] \mathcal{L} [ f(t) ] \equiv \int_{t=0}^{t= \infty} f(t) e^{-St} dt[/math]

 

Now, in the attached derivation, the author uses the letter p, instead of the letter S, but because I used the letter S earlier in this thread, I am going to stick with S.

 

In the attachment, the author states that p is a real variable. A real variable, is a symbol which when instantiated, must be instantiated with a real number. In other words, a real variable is a symbol which denotes a real number, but that symbol is to be considered a variable quantity, and not a constant.

 

Apparently, the crucial step in deriving a formula for the inverse of the Laplace transform, is to allow S (or p in the attachment), to range over the complex numbers.

 

Recall that the real numbers are considered to be a subset of the complex numbers. We can write this as follows:

 

[math] \mathbb{R} \subset \mathbb{C} [/math]

 

So the assumption which this entire derivation appears to be founded on, is the following:

 

[math] S \in \mathbb{C} [/math]

 

That is read, "S is an element of the complex numbers"

 

Recall that a complex number is usually expressed in the following form:

 

[math] a + bi [/math]

 

Where a,b are elements of the real number system, and i denotes the square root of negative one. The chief difference between the complex number system, and the real number system, is that the real number system is ordered, and the complex number system is not. In other words suppose that x,y are elements of the real numbers. The the following statement must be true: (x<y XOR x=y XOR x>y). But now suppose that x,y are arbitrary complex numbers... it is not necessarily true that (x<y XOR x=y XOR x>y). It must be true if the two numbers chosen happened to be real numbers, but it probably isn't true if the two numbers both have nonzero imaginary components, and are not equivalent. You wont be able to say the one number is less than the other. This fact stems from a contradiction that arises when one assumes that i is a number, but I won't go into it right now. Let me continue trying to follow this man's derivation of the formula for the inverse of the Laplace transform...

 

He continues by saying this...

 

"By assuming that p (in our case S) is complex, we can use the known formula for the inverse of the Fourier transform."

 

My question would be, "known to whom?" Apparently, he has pushed the derivation of the one, onto the derivation of the other.

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