TITLE: Convolution $f \ast f$, where $f \in L^{2}(\mathbb{R}^d)$ but $f \not \in L^{1}(\mathbb{R}^d)$
QUESTION [2 upvotes]: Suppose that I have a real-valued function $f$ that is in $L^{2}(\mathbb{R}^d)$ but not in $L^{1}(\mathbb{R}^d)$. It is known that $\mathcal{F}^{-1}(f) \in L^{p}(\mathbb{R}^d)$ for all $p \in \mathbb{N}$. I want to know if this is sufficient enough to deduce that the expression
$$\displaystyle \| f \ast f \|_{2}$$
is finite. Since $f$ is in $L^2$, then by Plancherel's theorem we have
$$\|f \ast f\|_{2} = \| \mathcal{F}(f \ast f) \|_2,$$
where $\mathcal{F}$ denotes the Fourier transform. I'd like to use the convolution theorem at this point, but as far as I understand, we would need $f$ to be in $L^1$. This suggests that $f$ does not have a Fourier transform in the traditional sense. Where can we go from here?
It is also known that $f$ is the Fourier transform of a characteristic function. In general, it is not true that $f \ast f \in L^{2}(\mathbb{R}^d)$. But for which functions does it hold?

REPLY [0 votes]: One of the ideas would be (napkin math incoming!) to take $f\in L^{4/3}$, then $\hat f \in L^4$, then $|\hat f|^2 \in L^2$, then by the inverse FT we can hope that $f\ast f \in L^2$.


TITLE: Does wind speed things up or accelerate things?
QUESTION [3 upvotes]: This question may seem odd, but I can't think of anything better. So I'll go straight to the point.
Let's say there's a projectile in air going east, shot at a certain angle, with a certain speed. Then there's the wind blowing, for let's say 5 meters per second in the same direction as the projectile.
So how will the wind affect the projectile? Will it accelerate for 5 meters per second every second it lasts in the air, or will the speed just increase by 5 meters per second all through out? And what if the wind goes exactly the opposite direction? Will it have a negative effect? Should I also take into consideration the drag the object makes, the mass, the terminal velocity, or if there are other things I've missed?
I don't mind if you give formulas. It would also help. But I just wanted to know, in simple terms because I'm not really that inclined with terms used in physics, the effects of a blowing wind to a projectile motion.
Thanks in advance and sorry if I made some errors or misunderstanding with this post or if I used the wrong tags. Feel free to edit it.
Thanks again,
Vincent :)

REPLY [3 votes]: It does the same thing as if there were no wind but the gun (which shoots East) is traveling at 5 m/s West when it shoots.


TITLE: Use Lucas' test with $a=7$ and prove $71$ is prime
QUESTION [0 upvotes]: My working so far:
$71-1=70$ and Prime factors of $70$ are $2 \times 5 \times 7$
Check $a=7$:
$7^{(\frac{70}{2})} \equiv 7^{35} \equiv x (mod 71)$
How do I find $x$? Usually I would use Fermat's little theorem and firstly find $\phi(71)$ except 71 is prime (giving the value as $71-1=70$ but this is a fact that we're trying to prove in the first place!!
I could of course use my calculator to calculate it, but this assumes the numbers aren't too extremely horrible.
How else do you calculate this nicely?

REPLY [0 votes]: Why don't we start with the minimum positive integer that is $>1$ and reach $7^{35}$ as the multiplication by $2$ is much easier.
$2^5=32,2^7=128\equiv -14{\pmod {71}}$
Now $2^8\equiv -28{\pmod {71}},2^{16}\equiv 784\equiv 3,2^{32}\equiv 9$
So, $2^{35}=2^{32}2^3\equiv 9\cdot 8 {\pmod {71}} \equiv 1$
If $m\equiv 2^n{\pmod {71}},$ where $n$ is any positive integer,
$m^{35}=(2^n)^{35}=(2^{35})^n\equiv (1)^n{\pmod {71}}=1$
$\implies (-m)^{35}=m^{35}(-1)^{35}\equiv-1{\pmod {71}}$
$\implies (-m)^5≢1{\pmod {71}}$  and $(-m)^7≢1$
We just need to show that, $(-m)^2 ≢1{\pmod {71}}$
If $n=6,m=64\implies  -m=-64\equiv 7{\pmod {71}}$, which is $a$, according to the given problem.
Now $a^2=7^2=49 ≢1{\pmod {71}}$,
so by Lucas test, $71$ is prime.
This method can be employed to any prime of the form $2(2r+1)+1\equiv 3 \pmod 4$


TITLE: Defining acceleration in gravity-free space
QUESTION [0 upvotes]: Without information from outside a closed spaceship, an astronaut cannot distinguish A from B.
A) In gravity-free space, the floor accelerates upwards at $a=g$ and hits a dropped watch.
B) On earth's surface, a dropped watch accelerates downwards at $a=g$ and hits the floor.
My question is, what is the acceleration relative to in A?  Since acceleration is the second derivative of displacement with respect to time, what is the displacement relative to?  If no other object is used to define the displacement then how can we know the spaceship is accelerating?  If another object is used to define displacement, then wouldn't there be a gravitational force affecting the spaceship, if only weakly?  The concept of "inertial reference frame" seems like circular logic in this case.

REPLY [2 votes]: My question is, what is the acceleration relative to in A?

Acceleration isn't relative, it's absolute. You can detect it with an accelerometer.

If another object is used to define displacement, then wouldn't there be a gravitational force affecting the spaceship, if only weakly?

One way to build an accelerometer is to use a free-falling test mass as your other object. The gravitational effect can be made as small as desired by making the test mass small enough.


TITLE: Linear independence of $X=\{\begin{bmatrix}x_{1}\\ y_{1} \end{bmatrix},...,\begin{bmatrix} x_{k}\\ y_{k} \end{bmatrix}\} \subset \mathbb R^{n+m}$
QUESTION [0 upvotes]: Show that if the set $T=\{x_1,...,x_k\} \subset \mathbb R^n$ is linearly independent, then $$X=\{\begin{bmatrix}
           x_{1} \\
           y_{1}
         \end{bmatrix},...,\begin{bmatrix}
           x_{k} \\
           y_{k}
         \end{bmatrix}\} \subset \mathbb R^{n+m}\tag{$m \ge 1$}
$$
is also linearly independent.

I assumed that $T=\{x_1,...,x_k\} \subset \mathbb R^n$ is linearly independent, then for some scalars $\lambda_1,...,\lambda_k$ let $$\sum_{j=1}^{k}\lambda_j\begin{bmatrix}
           x_{j} \\
           y_{j}
         \end{bmatrix}=0$$
Which is equivalent to $$\sum_{j=1}^{k}\lambda_j\begin{bmatrix}
           x_{j} \\
           0
         \end{bmatrix}+\sum_{j=1}^{k}\lambda_j\begin{bmatrix}
           0 \\
           y_{j}
         \end{bmatrix}=0$$
Now how to use the assumption to show that all such scalars are zero?

REPLY [0 votes]: By definition, $$\lambda \left [\begin{array}{c}x\\y\end{array} \right ] = \left [\begin{array}{c}\lambda x \\ \lambda y\end{array} \right ] \quad\mbox{and}\quad \left [\begin{array}{c}x \\y\end{array} \right ] + \left [\begin{array}{c}u\\v\end{array} \right ] = \left [\begin{array}{c}x+u \\ y+v\end{array} \right ]$$
where $x,y,u,v\in\mathbb R^n$ and $\lambda\in\mathbb R$. Effectively, these are matrix arithmetic rules. Conclude that
$$\left [\begin{array}{c}0\\0\end{array} \right ]= \sum \lambda _k \left [\begin{array}{c}x_k\\y_k\end{array} \right ] = \left [\begin{array}{c}\sum \lambda _kx_k \\ \sum \lambda _ky_k\end{array} \right ], $$
where $0\in \mathbb R^n$ is the zero. In particular, $\sum \lambda _kx_k = 0$.


TITLE: Factoring inequalities on Double Summation (Donald Knuth's Concrete Mathematics)
QUESTION [3 upvotes]: If you have the Concrete mathematics book please refer to page 40 and 41.
So how come this given sum
$$
\sum_{1 \le j < k + j \le n} \frac{1}{k}
$$
becomes
$$
\sum_{1\le k \le n}\sum_{1\le j \le n-k} \frac{1}{k}
$$
?
I do not understand how this is proven since to my understanding i am following from the book's method of factoring inequalities,
$$
[1\le j < k+j \le n] = [1 \le j \le n][j < k+j\le n] \text{ or } [1\le k+j \le n][1 \le j < k+j]
$$
And it doesn't look like the double summation above. I Tried simplifying the two above and it leads me to nowhere near the answer.

REPLY [2 votes]: If $1\le j<k+j\le n$, then $1\le j$ and $k+j\le n$; the latter inequality is equivalent to $j\le n-k$, so $1\le j<k+j\le n$ implies that $1\le j\le n-k$, the inequality governing the inner summation. Thus, we want to find some expression such that
$$[1\le j<k+j\le n]=[\text{expression}][1\le j\le n-k]\;.$$
What part(s) of $1\le j<k+j\le n$ can we not infer from $1\le j\le n-k$? The inequality $1\le j\le n-k$ says everything about $j$ that can be inferred from $1\le j<k+j\le n$, and it also implies that $k\le n-1$, but it imposes no lower bound on $k$; the missing factor will have to take care of that.
The original inequality $1\le j<k+j\le n$ implies that $1\le k$, since $j<k+j$, and that $k<n$, since $k+1\le k+j\le n$; thus, it implies that $1\le k\le n-1$. The inequality $1\le j\le n-k$ gives us half of that, but it doesn’t say that $1\le k$. Thus, we should have
$$[1\le j<k+j\le n]=[1\le k][1\le j\le n-k]\;,$$
and you can fairly easily check that this is indeed the case. If we use this decomposition, we get
$$\sum_{1\le j<k+j\le n}\frac1k=\sum_{1\le k}\,\sum_{1\le j\le n-k}\frac1k\;.$$
This is technically correct, but it’s harder than necessary to work with. The inner summation is non-zero only for $k\le n-1$, so we might as well add this condition to the outer sum as well:
$$[1\le j<k+j\le n]=[1\le k\le n-1][1\le j\le n-k]\;,$$
and
$$\sum_{1\le j<k+j\le n}\frac1k=\sum_{1\le k\le n-1}\,\sum_{1\le j\le n-k}\frac1k\;.$$
This is a perfectly reasonable way to rewrite the original summation as a double summation, but as the marginal note in Concrete Mathematics points out, it’s a little messier than necessary. There is no harm in simplifying the outer condition to $1\le k\le n$ to get
$$[1\le j<k+j\le n]=[1\le k\le n][1\le j\le n-k]$$
and
$$\sum_{1\le j<k+j\le n}\frac1k=\sum_{1\le k\le n}\,\sum_{1\le j\le n-k}\frac1k\;:$$
when $k=n$, the inner summation is $0$ anyway.


TITLE: All solutions to Quadratic matrix polynomials
QUESTION [1 upvotes]: I am after two things:
1- algorithms for finding all solutions of possibly large quadratic matrix equations of the form $AX^2+BX+C=0$
2- (if possible) software implementing the algorithms
Everything except Grobner basis based methods.
Thanks,
Pat

REPLY [1 votes]: Normally, $X$ is a vector. So, $X^2$ does not make sense much.
I hack your question a bit.
$$A^T X + X A - X B R^{-1} B^T X + Q = 0$$
It is called Algebraic Riccati Equation.
What software for solving it?
C++ or MATLAB


TITLE: Components of the vector product
QUESTION [1 upvotes]: consider a parallelogram $ABCD$ with $A(3,-2,-1)$, $B(2,1,3)$ and $C(0,4,1)$
find the coordinates of $D$ and calculate the area of this parallelogram

REPLY [1 votes]: a parallelogram can be divided into two triangles of equal area by a diagonal,that is if you find area of triangle ABC and then multiply it by two, you will get area of parallelogram.
area of triangle is half of product of two sides and angle between them, in your case area of triangle is 1/2{(AB)(BC)sin*} where * is the angle between A and B, and you know that (AB)(BC)sin* = ABxBC(cross product), twice of area of this triangle is area of your parallelogram,hence area of parallelogram becomes(ABxBC).solve ABxBC and get the answer.


TITLE: A functional equation defined on the real axis.
QUESTION [3 upvotes]: $$
f: \mathbb{R} \to \mathbb{R}\qquad \frac{f(x+y)}{x+y} = \frac{f(x)-f(y)}{x-y}, \qquad \forall x,y\in \mathbb{R}, \left|x\right| \neq \left|y\right|
$$
Can I prove anything interesting about this function? I need to find it.

REPLY [3 votes]: Since $|x|\not = |y|$, I'll let $x\notin \{-\frac{1}{2}, 0, \frac{1}{2}\}$ for the following substitutions
Observe first that, for $x\to -x$ and $y=x+1$ $$\frac{f(1)}{1}=-\frac{f(-x)-f(x+1)}{2x+1}\iff f(-x)-f(x+1)=-(2x+1)·f(1)\tag{I}$$
Letting $x\to x+1\text{ and }y=x-1$  we obtain $$\frac{f(2x)}{2x}=\frac{f(x+1)-f(x-1)}{2}\iff f(x+1)-f(x-1)=\frac{f(2x)}{x}\tag{2}$$
Finally $x\to x-1$ and $y\to-x$: $$\frac{f(-1)}{-1}=\frac{f(x-1)-f(-x)}{2x-1}\iff f(x-1)-f(-x)=-f(-1)·(2x-1)\tag{3}$$
The trick now is to add $(1), (2)$ and $(3)$ since the LHS turns $0$

$$\begin{array}a &0&=-(2x+1)·f(1)+\frac{f(2x)}{x}-f(-1)·(2x-1)\\ \iff&0&=-2x^2·f(1)-x·f(1)+f(2x)-2x^2·f(-1)+x·f(-1)\\ \iff &-f(2x)&=-2x^2·\big(f(1)+f(-1)\big)-x·\big(f(1)-f(-1)\big)\\ \iff& f(x)& =\frac{1}{2}·x^2·\big(f(1)+f(-1)\big)+\frac{1}{2}·x·\big(f(1)-f(-1)\big) \end{array}$$

Which is basically a quadratic equation in $x$. Can you end it now?


TITLE: what is the radius of convergence of $\sum_{n=1}^\infty \cos^n(n^2+\ln(n)) x^n$?
QUESTION [1 upvotes]: I want to calculate the radius of convergence of the power series  $$\sum_{n=1}^\infty \cos^n(n^2+\ln(n)) x^n$$ so I tried to use Hadamard's formula, but the problem is how to calculate $\limsup_{n \to \infty} \cos(n^2+\ln(n))$.
Can someone help me?

REPLY [2 votes]: $|\cos(x)| \le 1$, so the radius of convergence is at least $1$.  Showing it is exactly $1$ will be somewhat harder.  I think you can use van der Corput's difference theorem to show that
$n^2 + \ln(n)$ is uniformly distributed mod $2\pi$.


text:
\section{Additional numerical results} \label{appendix:extra_discussion_and_num_results}

\subsection{Structured random example with i.i.d.\ observation error} \label{appendix:num_results_iid_error}

Consider the linear inverse problem as described in  \Cref{subsec:num_results_random}, but with  i.i.d.\  observation error. In \Cref{fig:spectrum_iid_obs_error,fig:batch_effect_subplot_iid_obs_error,fig:compare_algorithms_4subplot_iid_obs_error,fig:random_sampling_error_4x3subplot_iid_obs_error,fig:diff_MI_random_sampling_error_4x2subplot_iid_obs_error}
we present numerical investigations analogous to those in \Cref{subsec:num_results_random}.

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/spectrum_iid_obs_error}
  	 \caption{Spectrum of the relevant operators of the inverse problem with i.i.d.\  observation error. The solid line is the median across 1000 random instances of the forward model. The whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum eigen/singular value. The prior and observation error covariances are fixed and not random.}
  	\label{fig:spectrum_iid_obs_error}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/batch_effect_subplot_iid_obs_error}
	\caption{Performance of each greedy heuristic for different batch sizes for the case of inverse problem with i.i.d.\  observation error. The batch size ranges from single index selection to one shot approach. The solid line is the median across 1000 random instances for the forward model.}
  	\label{fig:batch_effect_subplot_iid_obs_error}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/compare_algorithms_4subplot_iid_obs_error}
	\caption{A comparative study of the greedy heuristics for four different batch sizes for the case of inverse problem with i.i.d.\  observation error. The solid line is the median across 1000 random instances for the forward model.}
  	\label{fig:compare_algorithms_4subplot_iid_obs_error}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/random_sampling_error_4x3subplot_iid_obs_error}
	\caption{Performance of the greedy heuristics for different batch sizes across all 1000 random instances of the forward model for the case of inverse problem with i.i.d.\  observation error. The solid line is the median; the whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum mutual information captured.}
  	\label{fig:random_sampling_error_4x3subplot_iid_obs_error}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/diff_MI_random_sampling_error_4x2subplot_iid_obs_error}
  	 \caption{MM based batch greedy approaches compared against the standard batch greedy heuristic for different batch sizes across all 1000 random instances of the forward model for the case of inverse problem with i.i.d.\  observation error. The solid line is the median; the whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum mutual information captured.}
  	\label{fig:diff_MI_random_sampling_error_4x2subplot_iid_obs_error}
\end{figure}

\subsection{Structured random example with exponential prior covariance} \label{appendix:num_results_exp_prior}

Consider the linear inverse problem as described in  \Cref{subsec:num_results_random}, but with exponential kernel used to define the prior on the inference parameters. In \Cref{fig:spectrum_corr_obs_error_exp_prior,fig:batch_effect_subplot_corr_obs_error_exp_prior,fig:compare_algorithms_4subplot_corr_obs_error_exp_prior,fig:random_sampling_error_4x3subplot_corr_obs_error_exp_prior,fig:diff_MI_random_sampling_error_4x2subplot_corr_obs_error_exp_prior}
we present numerical investigations analogous to those in \Cref{subsec:num_results_random}.

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/spectrum_corr_obs_error_exp_prior}
  	 \caption{Spectrum of the relevant operators of the inverse problem with exponential prior covariance. The solid line is the median across 1000 random instances of the forward model. The whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum eigen/singular value. The prior and observation error covariances are fixed and not random.}
  	\label{fig:spectrum_corr_obs_error_exp_prior}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/batch_effect_subplot_corr_obs_error_exp_prior}
	\caption{Performance of each greedy heuristic for different batch sizes for the case of inverse problem with exponential prior covariance. The batch size ranges from single index selection to one shot approach. The solid line is the median across 1000 random instances for the forward model.}
  	\label{fig:batch_effect_subplot_corr_obs_error_exp_prior}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/compare_algorithms_4subplot_corr_obs_error_exp_prior}
	\caption{A comparative study of the greedy heuristics for four different batch sizes for the case of inverse problem with exponential prior covariance. The solid line is the median across 1000 random instances for the forward model.}
  	\label{fig:compare_algorithms_4subplot_corr_obs_error_exp_prior}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/random_sampling_error_4x3subplot_corr_obs_error_exp_prior}
	\caption{Performance of the greedy heuristics for different batch sizes across all 1000 random instances of the forward model for the case of inverse problem with exponential prior covariance. The solid line is the median; the whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum mutual information captured.}
  	\label{fig:random_sampling_error_4x3subplot_corr_obs_error_exp_prior}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/diff_MI_random_sampling_error_4x2subplot_corr_obs_error_exp_prior}
  	 \caption{MM based batch greedy approaches compared against the standard batch greedy heuristic for different batch sizes across all 1000 random instances of the forward model for the case of inverse problem with exponential prior covariance. The solid line is the median; the whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum mutual information captured.}
  	\label{fig:diff_MI_random_sampling_error_4x2subplot_corr_obs_error_exp_prior}
\end{figure}

\subsection{Structured random example with dimension of inference parameters greater than observations} \label{appendix:num_results_param_greater_than_obs}

We  consider a linear inverse problem with random linear forward model similar to the one described in \Cref{subsec:num_results_random}, but now the dimension of parameters is set to $100$, $X \in \mathbb{R}^{100}$, while cardinality of the candidate set of observations is fixed at $50$, $Y \in \mathbb{R}^{50}$.
We fix the prior, $\Gamma_X$ and observation error, $\Gamma_{Y|X}$, covariances using squared exponential kernel with correlation length $0.021$ and $0.042$ respectively. In \Cref{fig:spectrum_param_number_greater_than_obs,fig:batch_effect_subplot_param_number_greater_than_obs,fig:compare_algorithms_4subplot_param_number_greater_than_obs,fig:random_sampling_error_4x3subplot_param_number_greater_than_obs,fig:diff_MI_random_sampling_error_4x2subplot_param_number_greater_than_obs}
we present numerical investigations analogous to those in \Cref{subsec:num_results_random}.

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/spectrum_param_number_greater_than_obs}
  	 \caption{Spectrum of the relevant operators of the inverse problem with number of parameters greater than observations. The solid line is the median across 1000 random instances of the forward model. The whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum eigen/singular value. The prior and observation error covariances are fixed and not random.}
  	\label{fig:spectrum_param_number_greater_than_obs}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/batch_effect_subplot_param_number_greater_than_obs}
	\caption{Performance of each greedy heuristic for different batch sizes for the case of inverse problem with number of parameters greater than observations. The batch size ranges from single index selection to one shot approach. The solid line is the median across 1000 random instances for the forward model.}
  	\label{fig:batch_effect_subplot_param_number_greater_than_obs}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/compare_algorithms_4subplot_param_number_greater_than_obs}
	\caption{A comparative study of the greedy heuristics for four different batch sizes for the case of inverse problem with number of parameters greater than observations. The solid line is the median across 1000 random instances for the forward model.}
  	\label{fig:compare_algorithms_4subplot_param_number_greater_than_obs}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/random_sampling_error_4x3subplot_param_number_greater_than_obs}
	\caption{Performance of the greedy heuristics for different batch sizes across all 1000 random instances of the forward model for the case of inverse problem with number of parameters greater than observations. The solid line is the median; the whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum mutual information captured.}
  	\label{fig:random_sampling_error_4x3subplot_param_number_greater_than_obs}
\end{figure}

\begin{figure}[h!]
  	\centering \includegraphics[width=\textwidth,angle=0]{figures/random_prob/diff_MI_random_sampling_error_4x2subplot_param_number_greater_than_obs}
  	 \caption{MM based batch greedy approaches compared against the standard batch greedy heuristic for different batch sizes across all 1000 random instances of the forward model for the case of inverse problem with number of parameters greater than observations. The solid line is the median; the whiskers capture the interquantile range (10\% to 90\%),  and the $\star$ marks the maximum and minimum mutual information captured.}
  	\label{fig:diff_MI_random_sampling_error_4x2subplot_param_number_greater_than_obs}
\end{figure}

config: arxiv


TITLE: Capacitance of two cocentric spheres, contradicting results
QUESTION [3 upvotes]: Suppose we are given two conducting, cocentric spheres of radius $a_1$ and $a_2$ respectively. The inner sphere with charge $q$, the outer sphere with charge $-q$.
I can calculate the capacitance of this system by calculating the potential difference $U$ between the plates and then use the definition $C = q / U$. This is easy enough and indeed gives me the right result
$$C = \dfrac{4\pi \epsilon_0}{a_1^{-1} - a_2^{-1}}$$
But now I tried to caluclate $C$ by finding out the energy stored in the system in two different ways:
First we have the expression
$$W_{el} = \frac12 \frac {q^2}C$$
for the energy stored in any capacitor. Now I wanted to compare this to the energy stored in the electric field (since that's where the energy is, right?!) $$E(r) = \begin{cases}\frac1{4\pi \epsilon_0} \frac q{r^2} & a_1 < r < a_2 \\ 0 & \text{otherwise}\end{cases}$$
to derive the above formula for $C$ again. The energy density is $w_{el} = \frac {\epsilon_0} 2 E^2$, therefore the energy stored in the electric field is
\begin{eqnarray*}
W_{el} &=& \int w_{el} \;\mathrm{d}V \\
&=& 4\pi \int_{a_1}^{a_2} \frac{\epsilon_0}2 \left(\frac1{4\pi \epsilon_0} \frac q{r^2}\right)^2 r \; \mathrm dr \\
&=& \frac1{4\pi \epsilon_0}\frac{q^2}{2} \int_{a_1}^{a_2} \frac{\mathrm d r}{r^3} \\
&=& \frac1{4\pi \epsilon_0}\frac{q^2}{2} \frac14\left(\frac 1{a_1^4} - \frac1{a_2^4} \right)
\end{eqnarray*}
Comparing the two expressions gives
$$C = \dfrac{16\pi \epsilon_0}{a_1^{-4} - a_2^{-4}}$$
So my question is: Why is this wrong? Where is the energy in this capacitor stored, if not in the electric field (as it doesn't seem to be - unless I have made a mistake in deriving the result somewhere...)?
Edit: I also noticed, that the second result can be rewritten as
$$C = \dfrac{4\pi \epsilon_0}{a_1^{-1} - a_2^{-1}}\frac 4{a_1^{-3} + a_1^{-2}a_2^{-1} + a_1^{-1}a_2^{-2} + a_2^{-3}}$$
but I don't know whether this has any significance.
Thanks for reading and any help will be greatly appreciated! :)

REPLY [3 votes]: Two things stand out to my eye:
1) In your volume integral, when you switched to spherical coordinates, you put in a $4 \pi r \ dr$ where there should be a $4 \pi r^2 dr$.
2) Your evaluation of the 1-D integral is not quite right.
I think these two are enough to explain the discrepancy.


TITLE: How would I go about finding the "new" depth of the water, so that I could subtract the cone's volume from it?
QUESTION [0 upvotes]: A cylinder of radius 2 contains water to a depth ℎ. A cone with base radius  and perpendicular height  is lowered into the water. When the bottom of the cone is resting on the bottom of the beaker the water just covers the top of the cone. Find an expression for the initial depth of the water ℎ in terms of .
How could I find the new depth, so that I can define $h$ as $new depth - volume of cone$?

REPLY [2 votes]: The volume of water is $V_w=\pi(2a)^2 h$. The volume of the cone is $V_c=\frac13\pi (a)^2 a$.
To find the new height of the water $h^*$ use the equation:
$$
\pi(2a)^2 h^*=V_w+V_c\implies h^*=\frac{V_w+V_c}{\pi(2a)^2}=h+\frac1{12}a.
$$
Can you take it from here?


TITLE: Show Circle Group $ \mathbb{T} $ isomorphic to $\mathbb { C } ^ { * } / \mathbb { R } ^ { * } $
QUESTION [0 upvotes]: I'm trying to think of an extension to this question, which asks to show whether $\mathbb { C } ^ { * } / \mathbb { R } ^ { + } \simeq \mathbb { T } $. They do it using the First Isomorphism Theorem.
I think my postulate should be true because $ \mathbb { R } ^ { + } \simeq  { R } ^ { * }  $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism
$$ \phi : \mathbb { C } ^ { * } \mapsto \mathbb{T} $$
Any suggestions?

REPLY [0 votes]: Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)


TITLE: Is continuous spectrum closed?
QUESTION [1 upvotes]: I’m learning spectrum theory. I know the definition of point or continuous spectrum, knowing that spectrum of a bounded (or unbounded but self-adjoint) operator consisting only these two type. I have seen several operator with a closed continuous spectrum, but I don’t think it has to be closed when the point spectrum isn’t empty. Can someone show me a counter-example? And, does a continuous spectral point have to be a limit point of continuous spectral points?
Thank you so much !

REPLY [3 votes]: No, the continuous spectrum need not be closed. Let $f$ be defined on $[-1,1]$ by
$$
  f(x) = \left\{\matrix{
  0, & \text{if } -1\leq x\leq0,\cr
  x, & \text{if } 0\leq x\leq1,\hfill
  }\right.
  $$
and let $T$ be the corresponding multiplication operator on $L^2[-1,1]$, namely
$$
  T(\xi)|_x = f(x)\xi(x).
  $$
Then the spectrum of $T$ is $[0,1]$, the point spectrum is $\{0\}$, and
the continuous spectrum is $(0,1]$.
The answer to the second question is also no.  If $T$ is the diagonal operator on $\ell^2$, with diagonal entries $\{1/n: n\geq1\}$, then the point spectrum is $\{1/n: n\geq1\}$ and the continuous spectrum is $\{0\}$, and clearly 0 is not the limit of continuous spectral points (different from itself).


TITLE: Prove that $\sup\{f(x) : x < s\} ≤ \inf\{f(y) : y > s\}$ and both values are real numbers given an increasing function
QUESTION [0 upvotes]: Suppose there's an increasing function $f : R → R,$ indicating that $f(x) ≤ f(y)$ whenever $x ≤ y.$
My question is: given $s ∈ R,$ how would I show that $\sup\{f(x) : x < s\} ≤ \inf\{f(y) : y > s\}$ and also that both values are real numbers ?

REPLY [1 votes]: You have the inequality $f(x) \leq f(y)$ for all $x \leq y$, so take $x_n$ strictly increasing to $s$ and $y_n$ strictly decreasing to $s$. Necessarily $x_n < s < y_n$ for each $n$, so $f(x_n) \leq f(y_n)$, so $\sup\{f(x): x < s\} = \lim_n f(x_n) \leq \lim_n f(y_n) = \inf\{f(y): y > s\}$.


TITLE: Prove the relation between relative velocity ,momentum and energy
QUESTION [1 upvotes]: In Chapter 8 of F.Mandl's book Quantum field theory, during the derivation of the differential cross section, the following relation is used:
$$E_1E_2v_{rel}=\sqrt{(p_1p_2)^2-m_1^2m_2^2} \,\, ,$$
where $v_{rel}$ is the relative velocity between two colliding particles, $m_1$ and $m_2$ are the rest masses.
How is this equation derived?

REPLY [1 votes]: Consider the following results:

From the definition of scalar product of four vectors,
$$ \tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$
The usual dispersion relations:
$$ \tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$
The velocity $\textbf{v}_i$ in terms of momentum and energy:
$$ \tag{3} \textbf{v}_i = \textbf{p}_i / E_i .$$
From the collinearity of the particles:
$$ \tag{4} \textbf{p}_1 \cdot \textbf{p}_2 = |\textbf{p}_1| \, | \textbf{p}_2|. $$
The relative velocity $v_{rel} \equiv |\textbf{v}_1-\textbf{v}_2|$ can be written as:
$$ \tag{5} v_{rel} E_1 E_2 = \left| \textbf{p}_1 E_2 - \textbf{p}_2 E_1 \right|.$$
To see this, we use (3):
$$ v_{rel} \equiv |\textbf{v}_1-\textbf{v}_2|
= \left| \frac{\textbf{p}_1}{E_1} - \frac{\textbf{p}_2}{E_2} \right|
= \frac{1}{E_1 E_2} \left| \textbf{p}_1 E_2 - \textbf{p}_2 E_1 \right|
$$

Now, starting from (1), expanding the energy squared with (2) and using (4), we have:
$$ \tag{6} (p_1 p_2)^2 - m_1^2 m_2^2 = 2 (\textbf{p}_1 \cdot \textbf{p}_2)^2 + m_1^2 | \textbf{p}_2 |^2 + m_2^2 | \textbf{p}_1|^2 - 2 E_1 E_2 \textbf{p}_1 \cdot \textbf{p}_2.$$
Now you can just square (5) again using (2) and (4) and check that what you get is equal to the RHS of (6).


TITLE: Relations & Cartesian product explanation
QUESTION [3 upvotes]: I'm confused about this notation:
R c A x B

(R is a subset of the Cartesian product of A and B).
Cartesian product of AxB looks like this, right?
    1       2        3
x (x,1)   (x,2)    (x,3)
y (y,1)   (y,2)    (y,3)
z (z,1)   (z,2)    (z,3)

Where 123 index is B, and xyz index is A.
And R in the notation means 'Relation' (I guess).
But how can a relation be a subset of AxB?
Or is it just a way of writing that R is defined in the context of AxB?
For example: R is defined like this:
R = {(x,y) e Z x Z | y = x^2 + 2x - 3}

And that this definition of R occurs for the Cartesian product only?
I'm lost here, can someone explain this to me?
I'm also confused about the following notations:
If R c AxB is a relation, and (a,b) is an element of that relation, we note this: aRb instead of this: (a,b) e R.
And what is the notation

UcA

?
And, what exactly is a relation?
Thanks a lot!

REPLY [2 votes]: First of all, I wish to point out that your example definition of a relation:
$$R = \{(x,y)\in\mathbb Z\times\mathbb Z\mid y=x^2+2x-3\}$$
IS a subset of $\mathbb Z\times\mathbb Z$.
All of its elements are members of $\mathbb Z\times\mathbb Z$ according to your own definition.
Note that each element of $\mathbb Z$ is a scalar whereas each element of $\mathbb Z\times\mathbb Z$ is a tuple.

If the above is confusing in any way, I suggest consulting a good dictionary and thoroughly clearing up the terms "tuple," "scalar," "relation."

However, a brief explanation, not intended to be rigorous:
A scalar is something that's considered as a whole; it's not considered as having parts.
(Note: Anything can always be considered as having parts, but the parts don't need to be considered for every application.  Outside the realm of math, consider a car dealer's inventory: each car is a single entry.  A car parts dealer considers car parts, but even then each part is ultimately comprised of molecules and atoms, but this fact is irrelevant.  In much the same way, a scalar is considered to be indivisible at the level of abstraction we're dealing with, whether it is or isn't indivisible in other abstractions.)
You know what a set is, by your question.  (I hope.)  A tuple is an ordered set.  Usually we are only concerned with tuples that all have the same number of elements.  "Ordered pair" is another word for "tuple."  Bigger tuples might be called "3-tuples," "4-tuples," etc.

Since you are probably used to considering relations as either "true" or "false"—such as the relation "is greater than"—consider set-theoretic relations the same way.
Let's say $R$ is the "is greater than" relation.  Then we have, in your notation:
$$R = \{(x,y)\in\mathbb Z\times\mathbb Z\mid x>y\}$$
Note that this consists of a set of ordered pairs taken from the Cartesian product $\mathbb Z\times\mathbb Z$ according to a certain rule.  In other words, if the "greater than" relation evaluates to "true," we include the tuple in our set $R$; if it evaluates to "false," we exclude the tuple.
So $R$ already IS a subset of the Cartesian product.  Set theory is just a way to formalize this.
And thus you're saying:
$$x>y \iff (x,y)\in R$$
Which is to say:
$$xRy\iff (x,y)\in R$$
And in both of these definitions, $x\in\mathbb Z\land y\in\mathbb Z$, which is to say $(x,y)\in\mathbb Z\times\mathbb Z$, so $R\subseteq\mathbb Z$.

In other words, your question is just about notation.  I hope that helps.

I will add that a Cartesian product has no indices in set theory.  Sets are unordered.  A Cartesian product of two sets is also a set, so it's unordered.  The elements of the set (the elements of the Cartesian product) are themselves ordered pairs, but there is no order between pairs.  In other words, the Cartesian product is a set of tuples; it's NOT a matrix.
Matrix multiplication is a different thing.  Matrices ARE ordered; sets (including Cartesian products) are not ordered.


TITLE: Combinatorial proof of a binomial coefficient summation: $\sum_{k=1}^n \binom nk \binom n{k-1} = \frac12\binom{2n+2}{n+1} - \binom{2n}n$
QUESTION [17 upvotes]: Let $n$ and $k$ be integers with $1 \leq k \leq n$. Show that:
$$\sum_{k=1}^n {n \choose k}{n \choose k-1} = \frac12{2n+2 \choose n+1} - {2n \choose n}$$
I was told this is supposed to use a combinatorial proof and while I'm not that comfortable with that, many similar proofs use a mathematical proof that shows equality. Any guidance would be much appreciated.

REPLY [0 votes]: Both sides count the number of ways of choosing $n+1$ items from the set $\{1,2,3,\ldots,2n\}$.
The $k^\text{th}$ term in the sum on the left counts selections where $k$ items are chosen from the first half of the list, i.e. from $\{1,2,\ldots,n\}$ and $n+1-k$ items are chosen from the second half of the list, i.e. from $\{n+1,n+2,\ldots,2n\}$ (which means that $k-1$ items from the second half are not chosen). By the pigeonhole principle, at least one item must come from the first half, which explains why the sum starts with $k=1$.
We are choosing approximately half of the items in the set. We could make it exactly half by expanding the set we are selecting from to include $0$ and $2n+1$. To obtain the desired count we would then exclude those selections in which either $0$ or $2n+1$ (or both) were among the chosen. Expanding the set is nice since, with $n+1$ items chosen, there are also $n+1$ items unchosen, and we may partition the set of selections into pairs, each consisting of a selection and its complement. (The complement of a selection is the selection where the designations "chosen" and "unchosen" are interchanged). In each pair, one selection will include $0$ and the other will not. So we may eliminate the selections containing $0$ simply by taking the selection from each pair that does not contain $0$ and discarding the one that does. This explains the factor of $\frac{1}{2}$ in the first term. From those selections that remain, we now subtract those that contain $2n+1$ (and hence $n$ items from $\{1,2,\ldots,2n\}$.


TITLE: Showing the circular part of the Bromwich contour vanishes
QUESTION [5 upvotes]: I've been using this example $$\mathcal{L}^{-1} \{\frac{3}{(s^2+9)^2}\}(t)$$
Actually calculating the inverse is fine, but I got stuck trying to prove that the circular part of the Bromwich contour goes to $0$. If the angle to the top of the circular path from the x-axis is $\alpha$, then the whole path rotates $2(\pi-\alpha)$ :
$$\int_{0}^{2(\pi-\alpha)}{\frac{3e^{st}}{(s^2+9)^2}\,ds}$$
$s=Re^{i\theta},ds=iRe^{i\theta}d\theta$
$$3iR\int_{0}^{2(\pi-\alpha)}{\frac{e^{Rte^{i\theta}}}{(R^2e^{i2\theta}+9)^2}e^{i\theta}\, d\theta}$$
Using $|\int_{a}^{b}{f(x)\, dx}|\leq\int_{a}^{b}{|f(x)|\, dx}$ and the reverse triangle inequality on the denominator :
$$\begin{align}\left|3iR\int_{0}^{2(\pi-\alpha)}{\frac{e^{Rte^{i\theta}}}{(R^2e^{i2\theta}+9)^2}e^{i\theta}\,d\theta}\right|&\leq 3iR\int_{0}^{2(\pi-\alpha)}{\left|e^{Rte^{i\theta}}\right|\frac{1}{\left|R^2e^{i2\theta}+9\right|^2}\left|e^{i\theta}\right|\, d\theta}\\&\leq3iR\int_{0}^{2(\pi-\alpha)}{\left|e^{Rt\cos(\theta)}\right|\left|e^{iRt\sin(\theta)}\right|\frac{1}{|R^2-9|^2}\, d\theta}\\&\leq\frac{3iR}{|R^2-9|}\int_{0}^{2(\pi-\alpha)}{\left|e^{Rt\cos(\theta)}\right|\, d\theta}
\end{align}$$
How can I continue this to show that the expression goes to $0$ as $R\to\infty$? I'm worried that the use of the reverse triangle inequality and the bounds of integration are incorrect.
Edit : The contour looks like the image on below, which is from here

REPLY [5 votes]: The Bromwich contour lies on the straight-line path $s=c+iy$, $c>0$ and $y\in (-\infty,\infty)$. So, the semicircular contour $C_R$ is defined by $s=Re^{i\theta}$ where $\theta\in [\pi/2-\arcsin(c/R) ,3\pi/2+\arcsin(c/R)]$. Thus, we have for $R>3$
$$\left|\int_{\pi/2-\arcsin(c/R)}^{3\pi/2+\arcsin(c/R)}\frac{3e^{Rte^{i\theta}}}{((R^2e^{i2\theta})^2+9)^2}\,iRe^{i\theta}\,d\theta\right|\le \frac{3R}{(R^2-9)}\int_{\pi/2-\arcsin(c/R)}^{3\pi/2+\arcsin(c/R)}e^{Rt\cos(\theta)}\,d\theta\tag1$$
The integral on the right-hand side of $(1)$ can be estimated as follows.
$$\begin{align}
\int_{\pi/2-\arcsin(c/R)}^{3\pi/2+\arcsin(c/R)}e^{Rt\cos(\theta)}\,d\theta&=2\int_{\pi/2-\arcsin(c/R)}^{\pi}e^{R\cos(\theta)}\,d\theta\\\\
&=2\int_{-\arcsin(c/R)}^{\pi/2}e^{-Rt\sin(\theta)}\,d\theta\\\\
&=2\int_0^{\arcsin(c/R)}e^{Rt\sin(\theta)}\,d\theta+2\int_0^{\pi/2}e^{-Rt\sin(\theta)}\,d\theta\\\\
&\le 2\arcsin(c/R)e^{ct}+2\int_0^{\pi/2}e^{-Rt(2\theta/\pi)}\,d\theta\\\\
&=2\arcsin(c/R)e^{ct}+\frac{\pi(1-e^{-Rt})}{Rt}
\end{align}$$
which clearly goes to $0$ as $R$ tends to infinity.


TITLE: Prove that $\lim_{n\to \infty}S_n = \infty$
QUESTION [4 upvotes]: Let $(x_n),(y_n)$ be two sequences of positive real numbers, with $x_n \to \infty$ and $$S_n=\frac{x_n}{x_n+y_1}+\frac{x_n}{2x_n+y_2}+\dots+\frac{x_n}{nx_n+y_n}$$
  Prove that $\lim_{n\to \infty}S_n = \infty$

I tried to write the given sum as $S_n=x_n \left(\frac{1}{x_n+y_1}+\frac{1}{2\left(x_n+\frac{y_2}{2}\right)}+\dots+\frac{1}{n\left(x_n+\frac{y_n}{n}\right)} \right)$ and managed to prove the claim when $\left(\frac{y_n}{n}\right)$ is bounded. For the case when it is unbounded, I tried to use the sequence $a_n=\max \{\frac{y_1}{1}, \dots , \frac{y_n}{n} \}$, for which $a_n \to \infty$, in order to get $S_n \geq \frac{x_n}{x_n+a_n}\left(1+\frac{1}{2}+\dots+\frac{1}{n} \right)$, but I couldn't finish.
Also, I tried writing the sum as $S_n=\frac{1}{1+\frac{y_1}{x_n}}+\frac{1}{2+\frac{y_2}{x_n}}+\dots+\frac{1}{n+\frac{y_n}{x_n}}$, but nothing came out of it...

REPLY [4 votes]: Let $i=10^{100000}$, choose $N$ such that for each $n\geq N$, we have $$\frac{y_1}{x_n}< 1,\quad \frac{y_2}{x_n}< 1, \quad \cdots \quad \frac{y_i}{x_n}<1$$
This is possible because $x_n\to\infty$. Then for $n\geq N$, we have
$$\begin{aligned}S_n &> \frac{x_n}{x_n + y_1} + \frac{x_n}{2x_n + y_2}+\cdots+\frac{x_n}{ix_n + y_i} \\ &> \frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{i+1} = \frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{10^{100000}+1} \end{aligned}$$
Let $i$ be still larger number to conlcude $S_n\to \infty$.


TITLE: If four boxes are arranged left to right and there are two possible states for each of two events in each box, how many configurations are possible?
QUESTION [0 upvotes]: So... I have 4 boxes. The order of these boxes are important. So I define them as 1st box, 2nd box, 3rd box, 4th box, from left to right.
There are two unique events that can happen to these boxes.
They can be in state A1 or A2.
At the same time, they can be in state B1 or B2.
What are all possible configurations ever for this?
So one configuration could be: Box 1 --> A1, B1 | Box 2 --> A1, B1 | Box 3 --> A1, B1 | Box 4 --> A1, B1
Another one could be: Box 1 --> A1, B1 | Box 2 --> A2, B2 | Box 3 --> A1, B1 | Box 4 --> A2, B1

REPLY [1 votes]: You basically have a binary number: $bbbbaaaa$. Where both $b$ and $a$ can take either $0$ or $1$. The 1st $a$ and first $b$ relate to the state of the 1st box. Your 2 examples would be represented by numbers: $00000000$ and $01000101$. This is an 8 bit number, and therefore there are $2^8$ combinations.
As far as the code goes, since it's just a number and you need $n$th digit to find the state of each box, you can simply iterate over numbers from $0$ to $2^8$ and look at the bits of the number. Here's an example in Java:
class Permutations {
    public static void main(String[] args) {
        for(int i = 0; i < 256; i++) {
            String binary = padLeftWithZeros(Integer.toBinaryString(i));
            System.out.println(binary + ", Box #2, B-state: " + isBitSet(i, 6));
        }
    }

    private static boolean isBitSet(int mask, int bitIndex) {
        int bitFilled = 1 << bitIndex;// fill nth bit with 1
        return (mask & bitFilled) > 0;// check mask has nth bit 1 or 0
    }
    private static String padLeftWithZeros(String s) {
        return String.format("%8s", s).replaceAll(" ", "0");
    }
}


TITLE: What is the adjoint of the operator?
QUESTION [1 upvotes]: What is the adjoint of the operator $\frac{d}{dx}$in quantum mechanics?
I think the answer should be $\frac{d}{dx}$ because it has no imaginary term.

REPLY [2 votes]: In terms of wave functions, we have (for $p=-i\hbar\frac{d}{dx}$)
\begin{align}
\int_{-\infty}^\infty dx \phi(x)^* \left(\hat p \psi(x)\right)&=
\int_{-\infty}^\infty dx \phi(x)^* \left(-i\hbar\frac{d}{dx}\psi(x)\right)\, ,\\
&= -i\hbar\phi^*(x)\psi(x)\Bigl\vert_{-\infty}^\infty-(-i\hbar)
\int_{-\infty}^\infty dx \left(\frac{d}{dx}\phi(x)\right)^* \psi(x)
\end{align}
using integration by parts.  It is an assumption that $\psi(\pm\infty)\to 0$ so this removes the boundary term and we're left with
$$
\int_{-\infty}^\infty dx \phi(x)^* \left(\hat p \psi(x)\right)=
\int_{-\infty}^\infty dx \left(-i\hbar\frac{d}{dx}\phi(x)\right)^* \psi(x)
=\int_{-\infty}^\infty dx \left(\hat p\phi(x)\right)^* \psi(x) \, .
$$
This shows that $\hat p=-i\hbar\frac{d}{dx}$, with the condition $\psi(\pm\infty)\to 0$, is hermitian (or its own adjoint).  There are subtleties tied technical issues but with "reasonable" functions this will work.
You can work out by yourself what happens for $\frac{d}{dx}$ given the discussion on $\hat p$.  Moreover, it is very good to reflect on the assumption $\psi(\pm\infty)\to 0$ and other technicalities in the context of for instance this discussion on the quantum infinite well.


TITLE: Differentiability of an action of the group of invertible elements of a $C^{*}$-algebra $\mathcal{A}$ on the dual of $\mathcal{A}$
QUESTION [3 upvotes]: I am studying the actions of Banach-Lie groups on Banach manifolds, and I am not able to concretely evaluate the differentiability properties of a specific action.
Let $\mathcal{A}$ be a unital $C^{*}$-algebra, and $\mathcal{A}^{*}$ its topological dual, which is a complex Banach space with norm:
$$
\left|\left|\omega\right|\right|:=\sup\left\{|\omega(\mathbf{A})|\,;\mathbf{A}\in\mathcal{A}\colon||\mathbf{A}||=1\right\}
$$
Let us denote with $\mathcal{G}(\mathcal{A})$ the set of all invertible elements in $\mathcal{A}$.
This set is a Banach-Lie group with respect to the multiplication operation inherited from $\mathcal{A}$.
This group naturally acts on $\mathcal{A}$ as follows:
$$
\mathbf{A}\mapsto c_{\mathbf{G}}(\mathbf{A}):=\mathbf{G}\,\mathbf{A}\,\mathbf{G}^{\dagger}\,.
$$
By means of this action, an action on $\mathcal{A}^{*}$ is obtained:
$$
\omega\mapsto \alpha_{\mathbf{G}}(\omega)
$$
$$
\left(\alpha_{\mathbf{G}}(\omega)\right)(\mathbf{A}):=\omega\left(c_{\mathbf{G}}(\mathbf{A})\right)=\omega\left(\mathbf{G}\,\mathbf{A}\,\mathbf{G}^{\dagger}\right)\,.
$$
It is easy to see that this action is linear and preserves hermiticity.
I am interested in the differentiability properties of this action.
By definition, the action is differentiable if the map:
$$
\mathcal{G}(\mathcal{A})\times\mathcal{A}^{*}\rightarrow\mathcal{A}^{*}
$$
$$
(\mathbf{G}\,;\omega)\mapsto \alpha_{G}\omega
$$
is differentiable.
I know the abstract definition of differentiability (Frechet derivative) of a map between Banach manifolds, however, I do not know how to concretely proceed to investigate the subject.
Specifically, a map $f:V\rightarrow W$ between Banach space is differentiable at $x_{0}\in V$ if, for all $\epsilon>0$ there exist $\delta_{\epsilon}>0$ and a bounded linear map $Df_{x_{0}}:V\rightarrow W$ such that:
$$
||f(x-x_{0}) - f(x_{0}) - Df_{x_{0}}(x-x_{0})||<\epsilon
$$
whenever $||x-x_{0}||<\delta_{\epsilon}$.
Since $\mathcal{G}(\mathcal{A})$ is only a Banach manifold, I need a coordinates system in order to use the definition of Frechet derivative, and I do not know what kind of coordinates system I should use.
Furthermore, even if I knew it, I would need an "educated guess" of the explicit face of the derivative $Df_{x_{0}}$ to actually calculate the limit in the definition of the Frechet derivative.
Are there other, more smart ways to proceed?
Thank You in advance.
EDIT
With a little effort, I was able to prove that the linear map $\alpha_{\mathbf{G}}:\mathcal{A}^{*}\rightarrow\mathcal{A}^{*}$ is bounded, and thus continuous, for all $\mathbf{G}\in\mathcal{G}(\mathcal{A})$.
This means that $\alpha_{\mathbf{G}}$ is analytic for all $\mathbf{G}$ (it is a continuous 1-homogeneous polynomial), but I do not know if it is of some relevance.
EDIT 2
According to Andreas Cap's answer, I tried to work out the details without using the $t$ parameter.
Let $U_{\mathbf{G}}\subset\mathcal{G}(\mathcal{A})$ be a neighbourhood of $\mathbf{G}$ such that $\mathbf{G}+\mathbf{H}\in U$ for all $\mathbf{H}\in U$.
Then we can write:
$$
\alpha(\mathbf{G}+\mathbf{H}\,;\omega + \tau)(\mathbf{A}) - \alpha(\mathbf{G}\,;\omega)(\mathbf{A})=\omega(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger}) + \omega(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger}) + \tau(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger}) + \tau(\mathbf{G}\mathbf{A}\mathbf{G}^{\dagger})
$$
The linear functional introduced by Andreas Cap is:
$$
D\alpha_{\mathbf{G}\,;\omega}(\mathbf{H}\,;\tau)(\mathbf{A})=\omega(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger}) + \tau(\mathbf{G}\mathbf{A}\mathbf{G}^{\dagger})
$$
and thus:
$$
\alpha(\mathbf{G}+\mathbf{H}\,;\omega + \tau)(\mathbf{A}) - \alpha(\mathbf{G}\,;\omega)(\mathbf{A}) - D\alpha_{\mathbf{G}\,;\omega}(\mathbf{H}\,;\tau)(\mathbf{A})= \tau(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger})+ (\omega+\tau)(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger})
$$
Let us call this functional $\gamma$.
The norm of $\gamma$ is:
$$
||\gamma||=\sup_{\mathbf{A}\neq\mathbf{0}}\frac{|\gamma(\mathbf{A})|}{||\mathbf{A}||}
$$
Then:
$$
|\gamma(\mathbf{A})|\leq|\tau(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger})| + |(\omega+\tau)(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger})|\leq
$$
$$
\leq2||\tau||\,||\mathbf{G}||\,||\mathbf{A}||\,||\mathbf{H}|| + ||\omega||\,||\mathbf{H}||^{2}\,||\mathbf{A}|| + ||\tau||\,||\mathbf{H}||^{2} \,||\mathbf{A}||
$$
which means:
$$
||\gamma||=2||\tau||\,||\mathbf{G}||\,||\mathbf{H}|| + ||\omega||\,||\mathbf{H}||^{2} + ||\tau||\,||\mathbf{H}||^{2}
$$
The action $\alpha$ is differentiable at $(\mathbf{G}\,;\omega)$ if, for all $\epsilon>0$ there exists $\delta_{\epsilon}>0$ such that:
$$
\frac{||\gamma||}{||(\mathbf{H}\,;\tau)||}<\epsilon
$$
whenever $||(\mathbf{H}\,;\tau)||<\delta_{\epsilon}$.
However, we have:
$$
\frac{||\gamma||}{||(\mathbf{H}\,;\tau)||}=\frac{2||\tau||\,||\mathbf{G}||\,||\mathbf{H}|| + ||\omega||\,||\mathbf{H}||^{2} + ||\tau||\,||\mathbf{H}||^{2}}{||\mathbf{H}|| +||\tau||}
$$
and I am not sure that this is less than $\epsilon$.

REPLY [1 votes]: I think this is easier than it initially looks. First of all, $\mathcal G(\mathcal A)$ is an open subset in the Banach space $\mathcal A$, so there is no need to worry about coordinates. On Banach spaces, differentiation basically works as you are used to in open subsets of $\mathbb R^n$. Moreover, the maps you consider all are induces by linear (or conjugate linear) and bilinear maps.
Since $\mathcal G(\mathcal A)$ is open, any of the curves $G+tB$ stays in $\mathcal G(\mathcal A)$. Now for $t\in\mathbb R$, bilinearity of the multiplication gives $(G+tB)A(G+tB)^\dagger=GAG^\dagger+t(BAG^\dagger+GAB^\dagger)+t^2BAB^\dagger$ and hence
$$
(\omega+t\tau)((G+tB)A(G+tB)^\dagger=\omega(GAG^\dagger)+t\big(\omega(BAG^\dagger+GAB^\dagger)+\tau(GAG^\dagger)\big)+O(t^2)
$$
which shows that $Df(G,\omega)(B,\tau)$ is the linear functional mapping $A$ to $\omega(BAG^\dagger+GAB^\dagger)+\tau(GAG^\dagger)$.


TITLE: The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
QUESTION [13 upvotes]: I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.$$ A natural question to ask is whether such a formula exists for $(R^{-1})_{js}^{ir}=\langle R^{-1}, u^i_j \otimes u^r_s \rangle$. An obvious guess would be to take the inverse of the matrix $[R_{js}^{ir}]_{i,r,j,s}$.
That is, to guess that $$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$
is equal to $$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$ This guess is confirmed by the fact that
$$
\delta_{ij}\delta_{rs} = \langle R R^{-1},u^i_j \otimes u^r_s\rangle = \sum_{k,l} \langle R,u^i_k \otimes u^r_l \rangle \langle R^{-1},u^k_j \otimes u^l_s \rangle= \sum_{k,l} R_{kl}^{ir} (R^{-1})^{kl}_{js}.
$$
The matrix inverse is easy to calculate and gives us the formula
$$\langle R, u^i_j \otimes u^r_s \rangle  = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.
$$
Now let's try and test this result: As is very well known $(S \otimes $id)$R = R^{-1}$. Thus, $$\langle R^{-1}, u^i_j \otimes u^r_s \rangle = \langle R, S(u^i_j) \otimes u^r_s \rangle.$$ In the case of $N=2$, $i=j=r=s=1$, we have $S(u^1_1) = u^2_2$, and so,
$$
(R^{-1})^{11}_{11}=\langle R^{-1},u^1_1 \otimes u^1_1 \rangle = \langle R,S(u^1_1) \otimes u^1_1 \rangle = \langle R,u^2_2 \otimes u^1_1 \rangle  = R^{21}_{21} = 1.
$$
But the formula above gives us that $(R^{-1})^{11}_{11}$ is equal to
$$ q^{-\delta_{11}}\delta_{11}\delta_{11} - (q-q^{-1})\theta (1-1)\delta_{11}\delta_{11} = q^{-1},$$
Moreover, performing the analogous calculations for the other possible values of $i,j,r,s$ we get two different matrices: Using the general formula we get
$$
\left(
\begin{array} {cccc}
                q^{-1} & 0              & 0 & 0     \\\
                0      & 1              & 0 & 0     \\\
  0      & -\lambda             & 1 & 0  \\\
  0      & 0              & 0 & q^{-1} \\\
\end{array}
\right);
$$
and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get
$$
\left(
\begin{array} {cccc}
                1  &  0  & 0 & -\lambda     \\\
                0  &  q  & 0 & 0           \\\
  0  &  0  & q & 0           \\\
  0  &  0  & 0 & 1           \\\
\end{array}
\right),
$$
where $\lambda = (q-q^{-1})$.
I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text

REPLY [9 votes]: I think the only issue here is a harmless error in your calculation and that there is a normalization of the $R$-matrix for $U_q(sl_N)$ by a factor of $q^{1/2}$ which you have omitted (See 8.4.2 of Klymik Schmudgen).
First, I get $(R^{-1})^{21}_{12} = -q^{-1} R^{21}_{12}$,
because $\langle(S\otimes id)(R),a^2_1\otimes a^1_2\rangle = \langle R,S(a^2_1)\otimes a^1_2 \rangle = -q^{-1} \langle R,a^2_1\otimes a^1_2\rangle$, using that $S(b)=-q^{-1} b$ from Proposition 4.1.2.3 of Klymik Schmudgen.  So the actual matrix you should get should just be $q^{-1}$ times what you had expected to get.
Now the factor of q^{-1} here is because you had multiplied the actual universal R matrix by $q^{1/2}$ and so $(\lambda A)^{-1} = \lambda^{-1} A^{-1}$, so there's a factor of $\lambda^2$ as a discrepancy.
I hope this helps!


TITLE: How to find the 3-direction Lorentz boost transformation on Dirac spinor?
QUESTION [1 upvotes]: I am struggling to work out correct Lorentz transformation for a boost in the 3-direction on a Dirac spinor, $u(p)$. According to Peskin & Schroeder pg. 46, I need to use the equations:
$$S^{0i} = -\frac{i}{2}\pmatrix{\sigma ^i &0 \\ 0 & -\sigma ^i} \hspace{10mm} (3.26)$$
$$\Lambda _{\frac{1}{2}} = exp\left(-\frac{i}{2}\omega _{\mu \nu} S^{\mu \nu} \right) \hspace{6mm} (3.30) $$
Where (3.26) gives me the generator for the boost and (3.30) gives the transformation.
My Attempt
Using the above,
$$S^{03} = -\frac{i}{2}\pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3}$$
$$ \Lambda _{\frac{1}{2}} = exp \left (-\frac{1}{4}\omega_{03}\pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3} \right)  $$
Then the transformation would be:
$$ u(p) =  exp \left [-\frac{1}{4}\omega_{03}\pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3} \right] u(p) $$
The Issue
The book states, on the first line of (3.49), that the transformation is instead:
$$ u(p) = exp \left[-\frac{1}{2}\eta \pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3} \right] u(p)$$
Where $\eta$ is the rapidity (whatever that is). I can't see where the $\omega_{\mu \nu}$ has gone, do they group it in with the $\eta$? Thanks!

REPLY [1 votes]: Let us evaluate the sum in the exponential:
\begin{equation}
-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}=-\frac{i}{2}(\omega_{03}S^{03}+\omega_{30}S^{30}).
\end{equation}
Due to $S^{\mu\nu}$ and $\omega_{\mu\nu}$ being antisymmetric, we get:
\begin{equation}
-\frac{i}{2}(\omega_{03}S^{03}+\omega_{30}S^{30})=-\frac{i}{2}(\omega_{03}S^{03}+(-\omega_{03})(-S^{03})) = -i\,\omega_{03}S^{03}.
\end{equation}
Now set $\omega_{03}=\eta$ (on page 40 in Peskin & Schroeder, they do the same for a boost in the $x$-direction, with $\omega_{01}=\eta$ instead). Thus:
\begin{equation}
\exp\left(-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}\right) = \exp\left(-i\eta\left(-\frac{i}{2}\begin{pmatrix}\sigma^i&0\\0&-\sigma^i\end{pmatrix}\right)\right)=\exp\left(-\frac{1}{2}\eta\begin{pmatrix}\sigma^i&0\\0&-\sigma^i\end{pmatrix}\right).
\end{equation}


TITLE: Sums of two perfect squares
QUESTION [2 upvotes]: Show that if $q$ is a number that can be expressed as the sum of two perfect squares, then $2q$ and $5q$ can also be expressed as the sum of two perfect squares.

EDIT: I've recently revisited this problem and I found an elementary answer which I posted as an answer below.

REPLY [0 votes]: I found an elementary method that is a little more straightforward:
$$q=a^2 + (a+s)^2$$
$$q = 2a^2 + 2as + s^2$$
$$2q = 4a^2 + 4as + 2s^2$$
$$2q = 4a^2 + 4as + s^2 + s^2$$
$$2q = (2a+s)^2 + s^2$$

$$q=a^2 + (a+s)^2$$
$$q = 2a^2 + 2as + s^2$$
$$5q = 10a^2 + 10as + 5s^2$$
$$5q = 9a^2 + 6as + s^2 + a^2 + 4as + 4s^2$$
$$5q = (3a+s)^2 + (a+2s)^2$$


TITLE: d'Alembert's Solution: Should anything more be added to it?
QUESTION [3 upvotes]: I have this initial value problem:
$$ u_{tt}- c^2u_{xx} = 0; \,0 < x < 1; \,t > 0; $$
$$ u(x, 0) = 0; \,u_t(x, 0) = 1; 0  <= x  <= 1; $$
$$ u(0, t) = u(1, t) = 0; t >= 0: $$
Solving it using d'Alembert's Solution, answer I get is,
$$ u(x,y) = t $$
But that does not satisfy last condition. Putting x 0 or 1 should make u 0. But that is not the case here. Am I missing something? Do we have to add something after d'Alembert's Solution?

REPLY [1 votes]: D'Alambert's formula is for problems on the whole line (infinite string). Your problem is on a bounded interval (finite string) with boundary conditions.


TITLE: Problem on Combinations allowing repetition with no missing element.
QUESTION [0 upvotes]: Let's consider a problem :
Find all five-digit combinations from the set $\{1,2,3\}$ with no missing digits.
I don't understand how the solution for this problem can be mapped to
${k-1 \choose n-1}$
Where $k$ is length of combination and $n$ is number of elements.
Is there any other intuitive method to solve this problem ?

REPLY [1 votes]: This problem can be mapped to the stars and bars problem. You have the $k$ stars and $n-1$ bars. The total combinations is ${ k + n - 1} \choose {n-1}$ if you allow empty urns. In your case, you don't want to allow empty urns, and this is equivalent to saying you want a star in each urn. Thus, you can just place a star in each urn and then use the regular stars and bars method to partition the remaining stars, allowing empty urns.
Therefore, you now only have $k - n$ stars to partition, and ${k  - 1} \choose {n-1}$ total combinations.