TITLE: Convolution Integral and dummy variable
QUESTION [0 upvotes]: I have no idea how to find a dummy variable or start with this problem. I'm given the function:
$$
{\rm g}\left(\, x\,\right)
=\int^{x}_{0}\left(\, x - t\,\right){\rm f}\left(\,t\,\right)\,{\rm d}t
$$

Compute ${\rm g}\left(\,x\,\right)$ assuming ${\rm f}\left(\, x\,\right) = x$.

Second, completely independent of $a$ find the dummy variable.

REPLY [0 votes]: I agree with RandomGuy. In convolution the dummy variable is the one that is always integrated out. In what i've learned from taking a signals course the convolution shifts one of the signals using a dummy variable and then uses integration to "detect" where the two functions overlap, because in essence integration is addition.
Imagine the two signals on the same graph. Move one signal and every time it overlaps(area) with the second signal that means whatever you have as an integrand exists. Then add up all of these integrands for every overlap that occurs.


TITLE: Differentiation of Scalar with respect to Vector, Second Order, involving Hadamard Products
QUESTION [1 upvotes]: I need to calculate the Gradient of the scalar $y^Ty$, with respect to vector $x$:
$$y= (((R((G^T((Exl^T)\circ P))\circ W))\circ S)\circ H)l$$
$$\frac{\partial y^Ty}{\partial x}=\frac{2}{u^Tu}jj^Tx?$$
where $y$, $x$ and $l$ are vectors and the rest of the terms are matrices:
*Dimensions: R:(g * g), G:(e* g), E(e*k), x(1*k), l(p*1), P(e*p), W,S,H(g *p)
Following this elegant solution involving traces, it can be shown that:
$$u^Ty=j:x \implies u^Ty=j^Tx$$ where: $$j=E^T((G((R^T((ul^T)\circ H\circ S))\circ W))\circ P)l$$
I am using a little trick to calculate the Gradient of $y^Ty$ with respect to $x$:
$$y^Tyu^Tu=yy^T:u^Tu = u^Tyy^T:u^T=u^Ty:u^Ty \implies$$
$$y^Tyu^Tu=(J^Tx)^Tj^Tx=x^Tjj^Tx \implies$$
$$y^Ty=\frac{1}{u^Tu}x^Tjj^Tx\implies$$
$$\frac{\partial y^Ty}{\partial x}=\frac{2}{u^Tu}jj^Tx$$
I am getting to the same solution, applying product rule:
$$\partial (y^Ty)=\partial (y^T)y+(y^T(\partial (y))^T)^T$$
$$\frac{\partial y^Tu}{\partial x}=\frac{\partial u^Ty}{\partial x}=j$$
$$\partial (y^Tu)=\partial (y^T)u+(y^T(\partial (u))^T)^T \implies \frac{\partial(y^T)}{\partial x}=ju^T(uu^T)^+$$
$$\partial (u^Ty)=\partial (u^T)y+(u^T(\partial (y))^T)^T \implies \frac{\partial(y)}{\partial x}=ju^T(uu^T)^+$$
I believe we can write $y$ as: $$y=(uu^T)^+uj^Tx$$
From the above equations, I get again:
$$\frac{\partial y^Ty}{\partial x}=\frac{2}{u^Tu}jj^Tx$$
I think that I am doing sth wrong, because combining this derivation with a couple of others, I end up getting many solutions, despite having more equations than unknowns. My hunch is that where I have $jj^T$, I should be having a full rank $(k)$ matrix instead. Any ideas are welcome! Thanks.
(I had a typo in the dimensions of $E$; Apologies if this caused trouble to anyone before fixed)

REPLY [1 votes]: Vectorization followed by diagonalization will prove to be a useful operation, so let's create a notation for it.
$$\eqalign{
{\cal P} &= {\rm Diag}\Big({\rm vec}(P)\Big) \\
{\cal W} &= {\rm Diag}\Big({\rm vec}(W)\Big) \\
}$$
Define the matrices
$$\eqalign{
A &= Exl^T\odot P \\
B &= W\odot G^TA \\
C &= S\odot H \\
}$$
Then consider their factorizations/vectorizations.
$$\eqalign{
C &= \sum_k \sigma_ku_kv_k^T, \quad U_k = {\rm Diag}(\sigma_ku_k),\;
V_k = {\rm Diag}(v_k) \\
C\odot Z &= \sum_kU_kZV_k \\
a &= {\rm vec}(A) \\
  &= {\rm vec}(P)\odot{\rm vec}(Exl^T) \\
  &= {\rm vec}(P)\odot(l\otimes E)x \\
  &= {\cal P} (l\otimes E)x \\
b &= {\rm vec}(B) \\
  &= {\rm vec}(W)\odot{\rm vec}(G^TA) \\
  &= {\cal W}(I\otimes G^T)a \\
  &= {\cal W}(I\otimes G^T){\cal P} (l\otimes E)x \\
  &= Qx \\
}$$
Use all of this to write $y$ in a nicer form.
$$\eqalign{
y
 &= (S\odot H\odot R(W\odot(G^T(Exl^T\odot P))))l \\
 &= (S\odot H\odot R(W\odot(G^TA)))l \\
 &= (C\odot RB)l \\
 &= \sum_k {\rm vec}\Big(U_k RB V_k l\Big)
  = \sum_k (l^TV_k\otimes U_kR)\,b \\
 &= Jb \\&= JQx \\
}$$
With this nice expression for $y$, finding the gradient of the objective function is simple.
$$\eqalign{
\phi &= y^Ty \\ &= y:y \\
d\phi &= 2y:dy \\
 &= 2y:JQ\,dx \\
 &= 2\,Q^TJ^Ty:dx \\
\frac{\partial \phi}{\partial x}
 &= 2\,Q^TJ^Ty \\
 &= 2\,(l^T\otimes E^T)\;{\cal P}\;(I\otimes G)\;{\cal W}\;
 \sum_k\Big(V_kl\otimes R^TU_k\Big)\;y \\
}$$
To relate this to my previous comments, note that $\;M=JQ$.
Update
The variables $(U_k,V_k)$ can be removed from the final expression since
$$\eqalign{
\sum_k\Big(V_kl\otimes R^TU_k\Big)\;y
 &= {\rm vec}\bigg(\sum_k R^TU_kyl^TV_k\bigg) \\
 &= {\rm vec}\Big(R^T\big(yl^T\odot C\big)\Big)
  = {\rm vec}\Big(R^TYCL\Big) \\
&{where}\quad Y = {\rm Diag}(y),\; L = {\rm Diag}(l) \\
}$$
This means that you don't need to perform an SVD decompostion to utilize the result.


TITLE: Michelson-Morley Experiment: A.P. French's book on Special Relativity
QUESTION [1 upvotes]: I am doing a self-study of A.P. French's book on Special relativity and I am stuck at the Michelson-Morley experiment. I am having a few difficulties in the derivation in the book,

On page 54, it is mentioned that the ether wind and light add up using vector addition and that the resultant velocity relative to the interferometer is given by $\sqrt{c^2 - v^2}$. I do not understand this since if the ether wind is parallel to the horizontal arm, shouldn't the velocity along the vertical arm remain independent of the ether wind speed? Am I missing something here?
On page 64, there is a remark that the MM experiment shows the null result for any random orientation of the arm relative to the horizontal. However, when I try to follow the steps for the usual MM experiment, my equations does not solve the equation $$t_1-t_2 = \frac{2(l_{10} - l_{20})/c}{\sqrt{1 - \frac{v^2}{c^2}}}$$ derived earlier in the literature. Did anybody solve this? If requested, I will post my solution here for better clarity.

REPLY [0 votes]: The derivation for the equation you mentioned above, hope this helps.


TITLE: Homotopy type of the complement to a subvariety of $\mathbb C^n$
QUESTION [7 upvotes]: Let $V^k\subset \mathbb C^n$ be a sub variety, such that all its irreducible components have dimension $\ge k$. Is it true that $\mathbb C^n\setminus V^k$ has homotopy type of a CW complex of dimension $\le 2n-k-1$?
Comments. 1)This is true for $k=n-1$, since in this case $\mathbb C^n\setminus V^{n-1}$ is affine. Case $k=0$ is trivial.
2) This question would help to answer:
An analogue of Lefschetz hyperplane theorem for complements to subvarieties in $\mathbb C^n$ ?

REPLY [8 votes]: Take $n=4$ and let $V = \{ z_1=z_2=0 \} \cup \{ z_3=z_4=0 \}$. I claim that $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$, which has nontrivial homology in degree $6$, contrary to your supposed bound, which is in degree $5$.
Note that $\mathbb{C}^4 \setminus V = \left( \mathbb{C}^2 \setminus \{ (0,0) \} \right)^2$. Taking the quotient by $\mathbb{R}_{+}$, we see that $\mathbb{C}^2 \setminus \{ (0,0) \}$ is
homotopic to $S^3$, so $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$.

I can prove the required cohomology vanishing if you require that $V$ be Cohen-Macaulay.
Write $U$ for $\mathbb{C}^4 \setminus V$. We have the Hodge-de Rham spectral sequence: $H^q(U, \Omega^p) \implies H^{p+q}(U, \mathbb{C})$. Singe $U$ is an open subset of $\mathbb{C}^n$, we have $\Omega^p \cong \mathcal{O}^{\oplus \binom{n}{p}}$ so $H^q(U, \Omega^p) \cong H^q(U, \mathcal{O})^{\bigoplus \binom{n}{p}}$.
We can identify $H^q(U, \mathcal{O})$ with a local cohomology module of $V$, which the Cohen-Macaulay condition should force to be $0$ for $q > n-k-1$. So $H^q(\Omega^p)$'s will  be zero for $q>n-k-1$. Then the spectral sequence immediately forces cohomology to vanish for $p+q > n+(n-k-1)$, as you desired.
I have no idea of how to get a statement in homotopy out of the Cohen-Macaulay condition.


TITLE: Prove: If $A \subset B$ then $P(A) \le P(B)$ and $P(B-A) = P(B)-P(A)$
QUESTION [2 upvotes]: I'm trying to prove the following theorem using the axioms quoted below.

Theorem: If $A \subset B$ then $P(A) \le P(B)$ and $P(B-A) = P(B)-P(A)$

Axiom 1: For every event $A$ in the class $C$,
$$P(A)  \ge 0$$
Axiom 2: For the sure or certain event $S$ in the class $C$,
$$P(S) = 1$$
Axiom 3: For any number of mutually exclusive events $A_1,
A_2,\cdots$, in the class C,
$$P(A_1 \cup A_2 \cup \cdots) = P(A_1)+P(A_2) +\cdots $$

For the first part, because $B = (B\setminus A) \cup (B \cap A)$ we have  $P(B) = P((B\setminus A) \cup (B \cap A))$ $= P(B\setminus A)+P(B \cap A) \ge  0+P(B \cap A) \ge P(A) $ therefore $P(B) - P(A) \ge 0$. Hence $P(B) \ge P(A)$.
However, I can't get anywhere for the second part. I know that $P(B-A) = P(B \cap A')$ but I don't know how to turn the intersection into union so that I can use axiom 3. How do I prove this?

REPLY [1 votes]: Notice that $B-A$ and $A\cap B=A$ are mutually exclusive and their union is $B$. One of your axioms handles that.

REPLY [1 votes]: Hint: Since $A\subseteq B,$ then $A\cap B=A.$