How do I calculate the Jacobian for a convolution?

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The discussion focuses on calculating the Jacobian for a convolution transformation defined by the equations t=τ+p and τ=τ. The correct Jacobian determinant is derived as -1, which arises from the matrix formed by the partial derivatives of t and τ with respect to τ and p. The transformation illustrates the independence of τ from p, similar to cylindrical coordinates. The absolute value of the Jacobian for the convolution is confirmed to be 1, indicating a proper understanding of the transformation's properties.

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mathmari
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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?
 
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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
 

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