MHB How do I calculate the Jacobian for a convolution?

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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
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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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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