need help

[marvellover]

how can i solve the gaussian function in one dimension using laplace transform and fourier transform i don't understand pls help me

 

[HallsofIvy]

What do you mean by "solve a function"? Typically we "solve an equation" or "solve a problem" but the gaussian function isn't either of those.

 

[marvellover]

I meant any equation that is considered gaussian in 1 D

 

[Deleted member 4993]

I meant any equation that is considered gaussian in 1 D

Please give us an example of:

Gaussian equation in 1 D that you need to solve, using laplace transform and fourier transform.

 

[marvellover]

I want to know how to solve the Gaussian integral in 1D and 2D by the Laplace transform and the Fourier transform

1D gaussian equation -

2D gaussian equation -

where x and y can be any random variable

 

[topsquark]

I think you mean
$$G(x) = \dfrac{1}{ \pi } e^{-(x- x_m)^2/ ( \Delta x )^2}$$
$$G(x, y) = \dfrac{1}{ \pi ^2 } e^{-(x- x_m)^2/ ( \Delta x )^2} ~ e^{-(y- y_m)^2/ ( \Delta y )^2}$$
This is a bit different from how the Gaussian function is usually written but you can fix that if you need to. Any Calculus methods will work the same way.

Take G(x, y) and modify it a bit, just to show the method.
$$G(x, y) = A e^{-x^2} e^{-y^2}$$
$$\int_{- \infty }^{ \infty } \int_{- \infty }^{ \infty } A e^{-x^2 - y^2 } ~ dx ~ dy$$
Convert to polar coordinates:
$$= \int_{0}^{ \infty } \int_{0}^{2 \pi } A e^{-r^2} ~ r ~ dr ~ d \theta$$
You can now use integration by parts.

The G(x) integration is done in exactly the same way. $$G(x) G(y) = G(x, y)$$, so $$\int G(x) ~ dx = \sqrt{ \int G(x, y) dx dy}$$
-Dan

 

[marvellover]

THANK YOU SO MUCH SIR

 

[yoscar04]

In addition, the Fourier transform of a Gaussian is also a Gaussian (in the k space), you may write down the definition of the Fourier transform and reduce your problem to the integral of a Gaussian that can be solved using a method similar to the one exposed by one of the forum members to compute the Laplace transform.