TITLE: Distributional Limits of Fourier Transforms
QUESTION [0 upvotes]: This is a problem for a practice applied analysis qualifying exam:
Let $f(x,y)$ be the indicator function of the rectangle $R = [-A,A]\times [-B,B]$ in $\mathbb{R}^2$. Compute $\hat f$, the Fourier transform of $f$. Evaluate the limit of $\hat f$ in the sense of distributions as $A \to \infty$ for fixed $B$.
After some fun manipulation, I calculated  $\hat f (x_1, x_2) = \frac{2(\sin A x_1)(\sin B x_2)}{\pi x_1 x_2}$. I'm having trouble with the limit of $\hat f$ in the sense of distributions. It seems counterproductive in this case to use $\hat f \phi = f \hat \phi$ for $\phi \in \mathcal{D}(\mathbb{R}^2)$, as we just did all the work to find $\hat f$. However, when I've tried to calculate the quantity directly, I haven't made any meaningful progress.
$$\lim_{A \to \infty} \hat f \phi = \int_{\mathbb{R}^2} \hat f(x) \phi(x) dx = \int_{\mathbb{R}^2} \frac{2(\sin A x_1)(\sin B x_2)}{\pi x_1 x_2} \phi(x_1, x_2) dx_1dx_2$$

REPLY [1 votes]: Not a full solution, but hopefully a help.
Let's do a variable change:
$$\begin{align}
\left\langle \frac{\sin Ax}{x}, \phi(x) \right\rangle
& = \int \frac{\sin Ax}{x} \phi(x) dx \\
& = \{ \text{ set $\xi=Ax$ } \} \\
& = \int \frac{\sin\xi}{\xi/A} \phi\left(\frac{\xi}{A}\right) \frac{d\xi}{A} \\
& = \int \frac{\sin\xi}{\xi} \phi\left(\frac{\xi}{A}\right) d\xi
\end{align}$$
Now, $\phi(\xi/A)\to\phi(0)$ as $A\to\infty$, so if swapping integral and limit can be justified,
$$
\left\langle \frac{\sin Ax}{x}, \phi(x) \right\rangle
\to \int \frac{\sin\xi}{\xi} \phi(0) d\xi
= \to \int \frac{\sin\xi}{\xi} d\xi \, \phi(0) = C \phi(0) = \langle C\delta(x), \phi(x) \rangle
$$
where $C = \int \frac{\sin\xi}{\xi} d\xi$.
Thus,
$$ \frac{\sin Ax}{x} \to C\delta $$
So, if you can justify swapping integral and limit, then you only have to determine $C$ and take this to your two-dimensional case.


TITLE: prove homomorphic image has order 4
QUESTION [1 upvotes]: Let $G$ be a group of order $20$ show that $G$ has homomorphic image of order $4$.
From Cauchy theorem we have elements $a,b \in G$ of orders $2,5$ respectively then
from first isomorphic theorem homomorphic image is isomorphic with $G/kerf$ for some homomorphism $f:G \to im(f)$
so if $kerf=<b>$ then will be ok but to show that I need $<b>$ is normal subgroup of $G$ and I have some problems with that

REPLY [1 votes]: $G/\langle b\rangle$ is such a homomorphic image. The subgroup generated by $b$ is normal by the Sylow's theorems. Indeed the number $n$ of subgroups of order $5$ is congruent to $1\mod 5$, and a divisor of $20/5=4$. Hence $n=1$, which proves the subgroup is normal.


TITLE: How to show that $\delta^3(\vec{x}-\vec{a})=\lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}$
QUESTION [1 upvotes]: I hope to show that:
$$\delta^3(\vec{x}-\vec{a})=\lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}$$
I want to show by:
$$\int^{+\infty}_{-\infty} f(\vec{x}) \lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}\, d\vec{x}=f(\vec{a})$$
$$\lim_{\alpha\to0}\int^{+\infty}_{-\infty} f(\vec{x})  \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}\, d\vec{x}=f(\vec{a})$$
But I do not know how to proceed. Thank you very much.

REPLY [0 votes]: In THIS ANSWER, I provided a rigorous development to show that the Dirac Delta can be regularized as
$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$
where
$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} $$
Here, we will use a formal approach that can easily be made rigorous as in the aforementioned post.

The interpretation of the regularization of the Dirac Delta given by
$$\delta(\vec r-\vec a)\sim \lim_{\alpha \to 0}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}$$
is that for any smooth test function $f(\vec r)$, we have
$$\lim_{\alpha \to 0}\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV=f(\vec a)$$
We split the integral into the sum of two integrals
$$\begin{align}
\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV&=\int_{\mathscr{R}^3-B_{\vec a}(\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV\\\\
&+\int_{B_{\vec a}(\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV
\end{align}$$
where $B_{\vec a}$ is a sphere of radius $\delta$ centered at $\vec a$.  Note that as $\alpha \to 0$, the first integral vanishes.  Heuristically, for small  $\delta$, the second integral can be approximated by exploiting the continuity of $f$.  Then, we can write
$$\begin{align}
\lim_{\alpha \to 0}\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV&=f(\vec a)\lim_{\alpha \to 0}\int_{B_{\vec a}(\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,dV\\\\
&=f(\vec a)\lim_{\alpha \to 0}\frac{\alpha}{\pi^2}\int_0^{2\pi}\int_0^\pi \int_0^{\delta}\frac{1}{(\alpha^2+r^2)^2}r^2\sin(\theta)\,dr\,d\theta\,d\phi\\\\
&=f(\vec a)\lim_{\alpha \to 0}\left(\frac{\alpha}{\pi^2}\,(2\pi)\,(2)\,\frac{\pi}{4\alpha}\right)\\\\
&=f(\vec a)
\end{align}$$
as was to be shown!
Therefore, we have for any smooth test function $f(\vec r)$
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\alpha \to 0}\int_{\mathscr{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV=f(\vec a)}$$
and it is in this sense that
$$\bbox[5px,border:2px solid #C0A000]{\delta(\vec r-\vec a)\sim \lim_{\alpha \to 0}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}}$$


TITLE: Force on charged particle when moving along magnetic field lines
QUESTION [1 upvotes]: I was thinking, what happens to the force on a charged particle when it is moving along the magnetic field lines?
I am familiar with the right-hand-rule and it seems to me that RHR does not apply in this situation. Since the RHR, the field lines, direction of particle moving and the force on the particle are all perpendicular to each other.
But the question I am asking is, what happens to the force when the direction of a particle moving is parallel to the field lines or along the field lines?
Here is a picture of what I mean:

The electron, moving down or along the field lines with a velocity of v.

REPLY [2 votes]: Because $\mathbf F=q\mathbf v\times\mathbf B$, if $\mathbf v$ is parallel to $\mathbf B$ then $\mathbf F=0$ and the particle experiences no magnetic force.
This (which uses a RHR to get the directions right) is always valid.


TITLE: Doubt regarding conditional statement in mathematical logic
QUESTION [0 upvotes]: Conditional statement is represented as $p\to q$. Its truth table is given as:
$$
\begin {array}{|c|c|c|}
\hline
p & q & p\to q\\
\hline
T&T&T\\\hline
T&F&F\\\hline
F&T&T\\\hline
F&F&T\\\hline
\hline
\end{array}$$
As an explanation, my textbook gave the following example:

If $x$ is an odd integer, then $4$ divides $x^2-1$.
If $x$ is an odd integer, then $4$ does not divide $x^2-1$.
If $x$ is not an odd integer, then $4$ divides $x^2-1$.
If $x$ is not an odd integer, then $4$ does not divide $x^2-1$.

It says that (2) is false, while others are true, because in all the three statements, the conclusion is a fact.
I don't see how (3) is true. $4$ does not divide $x^2-1$ when $x$ is even. What am I missing?

REPLY [2 votes]: To understand $p\to q$, I always think of it as stating the $p\to q$ statement and then cycling through the 4 possibilities of $p$ and $q$.
So let's say $p$ is the statement, "It is raining today."
And $q$ is the statement, "The store is selling umbrellas."
I tell you $p\to q$.  This means if $p$ is true, then $q$ must be true.  Now, let's go through the possibilities and see if any of them contradict this statement.

It is raining today and the store is selling umbrellas.
It is raining today and the store is not selling umbrellas.
It is not raining today and the store is selling umbrellas.
It is not raining today and the store is not selling umbrellas.

Can you see that #2 is the only one that contradicts $p\to q$?

REPLY [0 votes]: Hint: Since $p \rightarrow q$ is false for the if-then statement (2), then (3) is the negation of a false statement (2). Once this point is taken, then one applies the table to the negation of a false if-then statement and the solution will be taken from there.

REPLY [0 votes]: The four claims don't really have a truth value before someone decides on a value for $x$.
If, for example, we set $x=7$, then (1), (3) and (4) are indeed true and (2) false -- but not for the reason you quote. The reason why (4) is true here is that the antecedent is false -- namely, "7 is not an odd integer" is false.
On the other hand, if we set $x=6$, then (1) and (2) are automatically true, (4) is true too, but (3) is now false.

Note that the meaning of $\to$ is "designed" to be useful under a $\forall$ quantifier. So we can say that $\forall x\in\mathbb Z:(1)$ and $\forall x\in\mathbb Z:(4)$ are both true, because (1) and (4) are true both when $x$ is odd and when $x$ is even.
But $\forall x\in\mathbb Z:(2)$ and $\forall x\in\mathbb Z:(3)$ are false -- it is only some but not all possible $x$ that make e.g. (2) true.


TITLE: Specifying complex domain in Wolfram Alpha
QUESTION [1 upvotes]: Hopefully this kind of question is ok, have seen a couple of other WA-queries that hasn't been downvoted. Apologies if not.
I'm trying to find the max/min of a complex function over certain domains, the unit disk or circle for example. It doesn't seem as though Wolfram Alpha understands what I'm trying to say though.
For example if I try $4z^2 +1, |z|=1$, WA just ignores the absolute value restriction. I've tried mod(z), abs(z), $\{|z|=1\}$ and some other stuff. Documentation only seems to contain examples over the reals. Any guesses?

REPLY [2 votes]: Optimization problems generally assume real input an output.  Accordingly, WolframAlpha assumes this type of input.  You can get around this issue by recasting your problem in terms of real real variables.  For example:
minimize |4 (x + I*y)^2 + 1| for x^2+y^2==1

or
minimize |4*exp(I*t) + 1|


TITLE: Confused with this SVD problem: Does it matter which singular vectors you choose?
QUESTION [4 upvotes]: I am trying to decompose the following matrix using the Singular Value Decomposition (SVD): $$A = \begin{bmatrix}
    4 & 4\\
    -3 & 3\\
    \end{bmatrix} = U\Sigma V^T$$
Here is my work (I know this is far from the most efficient way to do SVD, but please follow along my way):

Finding $\Sigma$ and $V$:
$$A^T A = \begin{bmatrix}25 & 7\\7 & 25\\\end{bmatrix} \quad\text{with } \lambda_1 = 32, v_1 = \begin{bmatrix}1/\sqrt2\\1/\sqrt2\\\end{bmatrix} \quad\text{and } \lambda_2 = 18, v_2 = \begin{bmatrix}1/\sqrt2\\-1/\sqrt2\\\end{bmatrix}$$
So, $$V = \begin{bmatrix}1/\sqrt2 & 1/\sqrt2\\1/\sqrt2 & -1/\sqrt2\\\end{bmatrix} \quad \text{and} \quad \Sigma = \begin{bmatrix} \sqrt{32} & 0 \\ 0 & \sqrt{18}\\\end{bmatrix} $$
Finding $U$:
$$A A^T = \begin{bmatrix}32 & 0\\0 & 18\\\end{bmatrix} \quad\text{with } \lambda_1 = 32, u_1 = \begin{bmatrix}1\\0\\\end{bmatrix} \quad\text{and } \lambda_2 = 18, u_2 = \begin{bmatrix}0\\1\\\end{bmatrix}$$
So, $$U = \begin{bmatrix}32 & 0\\0 & 18\\\end{bmatrix} $$

However, $$U \Sigma V^T = \begin{bmatrix}4 & 4\\3 & -3\\\end{bmatrix} \neq A$$
Did I do something wrong?

Next attempt:

This time, I used $u_2 = \begin{bmatrix}0\\-1\\\end{bmatrix}$ instead of $\begin{bmatrix}0\\1\\\end{bmatrix}$. So, $U = \begin{bmatrix}1 & 0\\0 & -1\\\end{bmatrix}$.
Now, it seems to work: $$U \Sigma V^T = \begin{bmatrix}4 & 4\\-3 & 3\\\end{bmatrix} = A.$$

So my question is: Does it matter which singular vectors you choose for $U$ and $V$? In other words, if you find a singular vector $x$ with unit length, how do you know to choose $x$ or $-x$? I know that in  Eigenvalue decomposition, it didn't matter because you can change the diagonal matrix $\Lambda$ accordingly. What about in SVD?

REPLY [1 votes]: Short Answer: The convention of choosing non-negative singular values locks our answer for $U$.

Explanation: The square of the singular values ($\sigma_i$) of a matrix $A$ is equal to the eigenvalues of $A^T A$ ($\lambda_i$), which are all non-negative (since $A^T A$ is positive semidefinite). There are two solutions to $\sigma_i^2 = \lambda_i$: the positive root ($\sigma_i = \sqrt\lambda_i$), and the negative root ($\sigma_i = -\sqrt\lambda_i$). If we were free to choose either one, then we could also choose either $u_i$ or $-u_i$ as the $i^{th}$ left singular vector, since we can adjust the sign of $\sigma_i$ in the diagonal matrix $\Sigma$ to match the final calculation for $A = U \Sigma V^T$. However, since it is a convention to use the positive root in $\Sigma$, our choice for $u_i$ is locked by $A$ and $v_i$: $$u_i = \frac{A v_i}{\sigma_i}.$$
In the example above, $U = \begin{bmatrix}
    1 & 0\\
    0 & 1\\
    \end{bmatrix}$ works fine if we choose $\sigma_2 = -\sqrt{18}$. Then:
$$U \Sigma V^T = \begin{bmatrix}
    1 & 0\\
    0 & 1\\
    \end{bmatrix} \begin{bmatrix}
    \sqrt{32} & 0\\
    0 & -\sqrt{18}\\
    \end{bmatrix} \begin{bmatrix}
    1/\sqrt2 & 1/\sqrt2\\
    1/\sqrt2 & -1/\sqrt2\\
    \end{bmatrix} = \begin{bmatrix}
    4 & 4\\
    -3 & 3\\
    \end{bmatrix} = A$$
However, since we choose the positive root for $\sigma_2$ by convention, we would have to change the sign of either $u_2$ or $v_2$ (the latter would flip the sign of the second row of $V^T$):
$$U \Sigma V^T = \begin{bmatrix}
    1 & 0\\
    0 & -1\\
    \end{bmatrix} \begin{bmatrix}
    \sqrt{32} & 0\\
    0 & \sqrt{18}\\
    \end{bmatrix} \begin{bmatrix}
    1/\sqrt2 & 1/\sqrt2\\
    1/\sqrt2 & -1/\sqrt2\\
    \end{bmatrix} = \begin{bmatrix}
    4 & 4\\
    -3 & 3\\
    \end{bmatrix} = A$$
$$\text{or}$$
$$U \Sigma V^T = \begin{bmatrix}
    1 & 0\\
    0 & 1\\
    \end{bmatrix} \begin{bmatrix}
    \sqrt{32} & 0\\
    0 & \sqrt{18}\\
    \end{bmatrix} \begin{bmatrix}
    1/\sqrt2 & 1/\sqrt2\\
    -1/\sqrt2 & 1/\sqrt2\\
    \end{bmatrix} = \begin{bmatrix}
    4 & 4\\
    -3 & 3\\
    \end{bmatrix} = A$$


TITLE: How to compute the modular multiplicative inverse on WolframAlpha?
QUESTION [2 upvotes]: $e = 17$
$\varphi(3233) = (61 - 1)(53 - 1) = 3120$
Compute d, the modular multiplicative inverse of e (mod φ(n)) yielding
    d = 2753.
http://en.wikipedia.org/wiki/RSA_%28algorithm%29#A_working_example
I tried this:
"modular inverse of  (17 (mod φ(3233)))"
http://www.wolframalpha.com/input/?i=modular+inverse+of++%2817+%28mod+%CF%86%283233%29%29%29
It resulted into 1/17 and not 2753. Where's the mistake?

REPLY [2 votes]: Wolfram Alpha can use Mathematica commands, so when in doubt look into the Mathematica documentation. In this case, you can use

PowerMod[17,-1,EulerPhi[3233]]

to get the correct result.


TITLE: Proving two consecutive odd primes have at least 3 prime divisors.
QUESTION [0 upvotes]: Prove that the sum of two consecutive odd primes has at least three prime divisors (not necessarily different).

REPLY [7 votes]: Well, if $P$ and $Q$ are consecutive odd primes, then $P+Q$ is even, so $2$ divides $P+Q$. Hence, $P+Q=2R$, for some $R$. If $R$ were prime, then $R$ were a prime between $Q$ and $P$, a contradiction.


TITLE: What is the smallest positive integer $n > 1$ such that $3^n$ ends with $003$?
QUESTION [3 upvotes]: What is the smallest positive integer $n > 1$  such that $3^n$ ends
with $003$?

Hello! I hope you are doing great. I was doing some number theory and solving the above question but I could not. Any help would be appreciated.

Here's what I've done so far:
Since $3^n$ ends with $003$, hence, $3^{n-1}$ should end with $001$. Since the units digit of the power is $1$, $n-1$ is a multiple of $4$.
Also note that $125 | 3^{n} - 003$. Not sure how this would help.

That's it. I have not made any more progress.
Thank You

REPLY [2 votes]: As we need $3^m\equiv001\pmod{1000}\equiv1\pmod{10},$
$4\mid m$
$$3^{4n}=(10-1)^{2n}=(1-10)^{2n}\equiv1-\binom{2n}110+\binom{2n}210^2\pmod{1000}$$
We need $$-20n+100n(2n-1)\equiv0\pmod{1000}$$
$25$ must divide $n(5n-3)$
As $5\nmid(5n-3),25$  must divide $n$