How do I calculate the Jacobian for a convolution?

  • Context: MHB 
  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Calculation Jacobian
Click For Summary

Discussion Overview

The discussion revolves around calculating the Jacobian for a transformation related to convolution, specifically the transformation defined by \( t = \tau + p \) and \( \tau = \tau \). Participants explore the correct formulation of the Jacobian and its implications for the convolution process.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a Jacobian calculation using the determinant of a matrix derived from partial derivatives, initially arriving at a value of -2.
  • Another participant clarifies the transformation and suggests that the Jacobian should be calculated as \( \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \), resulting in -1.
  • There is a mention of the independence of \( \tau \) from \( p \) in the context of the transformation.
  • Participants express willingness to assist further if there are additional questions or misunderstandings.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct value of the Jacobian, as one participant's calculation leads to -2 while another's leads to -1. The discussion reflects differing interpretations of the transformation.

Contextual Notes

The discussion highlights potential confusion regarding the independence of variables and the formulation of the Jacobian, which may depend on specific assumptions about the transformation.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hello! :o

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?
 
Physics news on Phys.org
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

$$
\frac{\partial (t, \tau)}{\partial (\tau, p)}
=
\begin{vmatrix}
\frac{\partial t}{\partial \tau} & \frac{\partial t}{\partial p} \\
\frac{\partial \tau}{\partial \tau} & \frac{\partial \tau}{\partial p}
\end{vmatrix}
=
\begin{vmatrix}
1 & 1 \\
1 & 0
\end{vmatrix}
= -1.
$$
 
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)
 
I'm glad to help. If you encounter more difficulties or didn't understand what I meant, please ask! :)
 
Fantini said:
I'm glad to help. If you encounter more difficulties or didn't understand what I meant, please ask! :)

Ok! :D
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K