Jacobian of the transformation

In summary, the conversation discusses the process of performing an integral of the form \int_0^a {\int_0^b {f\left( {x - y} \right)dxdy}} and suggests using variable substitutions to simplify the integration. However, it is pointed out that the final form of the integral does not make sense and may require dividing the area into two separate integrals. Additionally, the conversation also mentions a particular book that presents a different approach to solving the integral, but the reasoning behind the second equality sign is unclear.
  • #1
Niles
1,866
0

Homework Statement


Hi

I wish to perform an integral of the form

[tex]
\int_0^a {\int_0^b {f\left( {x - y} \right)dxdy} }
[/tex]

What I do first is to define s := x-y, and ds = dx. Then we get

[tex]
\int_0^a {\int_{-y}^{b-y} {f\left( {s} \right)dsdy} }
[/tex]

Then I can define t := x+y, so dt = dy. Then I get

[tex]
\int_{x}^{x+a} {\int_{-y}^{b-y} {f\left( {s} \right)dsdt} }
[/tex]

I also have to multiply by 2, since it is the Jacobian of the transformation. But look at the limits: It doesn't seem to make things easier. Where am I going wrong?
 
Last edited:
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  • #2


Well, what, exactly do you want to do with that? You obviously can't "do" the integral without knowing what f is. However, your limits of integration in the final integral makes no sense. From the first form, since the limits of integration are constants, the integral does not depend on x or y but your last form for the integral clearly does.

The region of integration in the x,y plane is a rectangle with sides parallel to the x and y axes. In the u,v plane it is still a rectangle but with sides at 45 degrees to the u and v axes. Depending on what a and b are, you would probably have to divide the area into two separate integrations. In any case, I don't see that change of variable helping.
 
  • #3


The reason why I am asking is because in my book they write the following

[tex]
I = \int_0^a {dt_1 \int_0^a {dt_2 \,F(x,t_1 - t_2 )} } F(y,t_1 - t_2 ) = a\int_0^a {dt\,\,F(x,t)F(y, - t)}
[/tex]

I cannot quite see how the second equality-sign comes about. I thought it came from substitution, as above.
 

What is the Jacobian of a transformation?

The Jacobian of a transformation is a matrix of partial derivatives that describes the rate of change of a set of variables with respect to another set of variables. It is commonly used in multivariate calculus and differential geometry to study transformations between coordinate systems.

How is the Jacobian of a transformation calculated?

The Jacobian is calculated by taking the partial derivatives of each output variable with respect to each input variable and arranging them in a matrix. It is also possible to calculate the Jacobian using the chain rule for multivariable functions.

What is the significance of the Jacobian of a transformation?

The Jacobian provides important information about the local behavior of a transformation, such as its orientation and scale. It is also used to calculate the change of variables in integrals and to determine whether a transformation is invertible.

Can the Jacobian of a transformation be negative?

Yes, the Jacobian can be negative if the transformation involves an orientation-reversing change. This means that the transformation flips the orientation of the coordinate system, resulting in a negative Jacobian determinant.

How is the Jacobian of a transformation used in machine learning?

The Jacobian is an important tool in machine learning for determining the gradient of a function, which is used in optimization algorithms. It is also used in neural networks for backpropagation, where the Jacobian matrix is used to update the weights of the network.

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