MHB How do I calculate the Jacobian for a convolution?

Click For Summary
To calculate the Jacobian for the transformation t=τ+p, the correct approach involves recognizing that τ is independent of p. The Jacobian is given by the determinant of the matrix formed by the partial derivatives, resulting in a value of -1. This contrasts with the initial assumption that the Jacobian would equal -2, which was incorrect. The discussion highlights the importance of correctly identifying independent variables in transformations. Understanding these concepts is crucial for accurate calculations in convolution.
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?
 
Mathematics 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
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 6 ·
Replies
6
Views
730
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K