How do I calculate the Jacobian for a convolution?

In summary, the conversation discusses the calculation of the Jacobian for a transformation t=τ+p and its relation to cylindrical coordinates. It is explained that since τ is independent of p, the Jacobian is equal to -1. The conversation ends with the offer for further clarification if needed.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hello! :eek:

Having the transformation t=τ+p, I want to calculate the jacobian $\frac{J(t,τ)}{J(τ,p)}$.
Isn't it $$ \frac{J(t,τ)}{J(τ,p)}=\begin{vmatrix}
\frac{ \vartheta t}{\vartheta τ}& \frac{\vartheta t}{\vartheta p} \\
\frac{\vartheta τ}{\vartheta τ} & \frac{\varthetaτ }{\vartheta p}
\end{vmatrix}=\begin{vmatrix}
1& 1 \\
1 & -1
\end{vmatrix}=-1-1=-2$$? But the absolute value of the Jacobian for the convolution is $1$..What did I wrong?
 
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  • #2
The transformation you actually have is this:

$$
\begin{cases}
t = \tau + p, \\
\tau = \tau.
\end{cases}
$$

Seems tautological? Yes, but it's the same way you do cylindrical coordinates:

$$
\begin{cases}
x = r \cos \theta, \\
y = r \sin \theta, \\
z = z.
\end{cases}
$$

We see that $\tau$ is independent of $p$. Therefore your Jacobian is

$$
\frac{\partial (t, \tau)}{\partial (\tau, p)}
=
\begin{vmatrix}
\frac{\partial t}{\partial \tau} & \frac{\partial t}{\partial p} \\
\frac{\partial \tau}{\partial \tau} & \frac{\partial \tau}{\partial p}
\end{vmatrix}
=
\begin{vmatrix}
1 & 1 \\
1 & 0
\end{vmatrix}
= -1.
$$
 
  • #3
Fantini said:
The transformation you actually have is this:

$$
\begin{cases}
t = \tau + p, \\
\tau = \tau.
\end{cases}
$$

Seems tautological? Yes, but it's the same way you do cylindrical coordinates:

$$
\begin{cases}
x = r \cos \theta, \\
y = r \sin \theta, \\
z = z.
\end{cases}
$$

We see that $\tau$ is independent of $p$. Therefore your Jacobian is

$$
\frac{\partial (t, \tau)}{\partial (\tau, p)}
=
\begin{vmatrix}
\frac{\partial t}{\partial \tau} & \frac{\partial t}{\partial p} \\
\frac{\partial \tau}{\partial \tau} & \frac{\partial \tau}{\partial p}
\end{vmatrix}
=
\begin{vmatrix}
1 & 1 \\
1 & 0
\end{vmatrix}
= -1.
$$

Ok! Thanks for your help! (Yes)
 
  • #4
I'm glad to help. If you encounter more difficulties or didn't understand what I meant, please ask! :)
 
  • #5
Fantini said:
I'm glad to help. If you encounter more difficulties or didn't understand what I meant, please ask! :)

Ok! :D
 

FAQ: How do I calculate the Jacobian for a convolution?

What is a Jacobian in mathematics?

The Jacobian is a mathematical concept that is used to describe the rate of change in a system of multiple variables. It is commonly used in calculus, differential equations, and other branches of mathematics.

How is the Jacobian matrix calculated?

The Jacobian matrix is calculated by taking the partial derivatives of a set of equations with respect to a set of variables. Each element of the matrix represents the rate of change of one variable with respect to another.

What is the importance of the Jacobian in scientific research?

The Jacobian is important in scientific research because it allows scientists to understand and describe the behavior of complex systems. It is used in a wide range of fields, including physics, engineering, and economics.

Can the Jacobian be negative?

Yes, the Jacobian can be negative. This indicates that the variables in the system are changing in opposite directions. For example, if the Jacobian of a system is -1, it means that when one variable increases, the other decreases by the same amount.

How is the Jacobian used in optimization problems?

The Jacobian is used in optimization problems to find the minimum or maximum value of a function. By setting the Jacobian equal to zero, the critical points of the function can be found, which can then be used to determine the optimal solution.

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