TITLE: Confidence Intervals: When to use what formula. I'm completely confused
QUESTION [1 upvotes]: I'm working on a few confidence interval questions, each with slight differences that make it very hard to determine what to use.
One question asks:

An electric scale gives a reading equal to the true weight plus a random
  error that is normally distributed with mean 0 mg and standard deviation  = 0.1 mg. Suppose
  that the results of five successive weighings (in mg) of the same object are as follows:
Heading  3.142, 3.163, 3.155, 3.150, 3.141 .
a) Compute a 95 percent confidence interval estimate of the true weight.
b) Compute a 99 percent confidence interval estimate of the true weight.

So here, the true mean is known. The standard deviation is known and all stats associated with the sample is known. What formula is used in this situation? The student-t? I tried but it's long and I think there's a faster way.
The other question asks:

Each of 16 science students independently measured the melting point of
  lead. The sample mean of these measurements was 330.2 degrees centigrade.
a) If the standard deviation of such measurements is known to be 14, find a 99 percent two sided confidence interval estimate of the true melting point of lead.
b) Suppose that the population variance is not known in advance. If the sample standard
  deviation is 15.4 degrees centigrade, compute a 99 percent two-sided confidence interval
  of the true melting point of lead.

In one situation we have the Standard Deviation, in the other we only have the sample standard deviation. What is used here?

REPLY [1 votes]: First question: Imagine doing the weighing $5$ times. Let random variables $X_1,\dots, X_5$ be the results. Now look at the random variable $Y=(X_1+\cdots+X_5)/5$.
This random variable is normal, with mean the true mean, and standard deviation equal to $\dfrac{1}{\sqrt{5}}$ times the standard deviation of any $X_i$, in this case $0.1$.
Now compute the sample mean, that is, add up the observations, divide by $5$. Suppose the result is $a$. Then our symmetric $95\%$ confidence interval is the interval that goes from $a-1.96\dfrac{0.1}{\sqrt{5}}$ to $a+1.96\dfrac{0.1}{\sqrt{5}}$. The $99\%$ confidence interval is obtained similarly, with $2.57$ replacing $1.96$. I assume you know where these numbers come from.
For the situation where the population standard deviation is not known, one uses Student's $t$-distribution with $n-1$ degrees of freedom instead of the normal. Information can be found here. You will need to locate appropriate tables. (For largish $n$, there are no tables, but in that case one can safely assume that the sample standard deviation is the true standard deviation.)


TITLE: Is gaussian elimination "used"?
QUESTION [2 upvotes]: According to the book Linear algebra and its applications by Strang, "(The) good method is Gaussian Elimination. This is the algorithm that is constantly used to solve large systems of equations".
Is it really constantly used to solve large systems of equations, or are there better algorithms for it? If yes, which algorithms are usually used to actually solve them?

REPLY [0 votes]: It depends on what you mean by large. On most regular computers you can solve dense systems up to 5k x 5k fairly easily. If the matrix is ill-conditioned you can use other algorithms and if it pretty large it is probably better to use iterative methods.


TITLE: Does TIR obeys laws of reflection?
QUESTION [0 upvotes]: After reaching the critical angle, do Snell's laws and laws of reflection hold good?

REPLY [1 votes]: Yes, Snell's law of reflection holds. Also the Fresnel expressions hold, even if there is extinction so that the refractive index is complex valued.

REPLY [0 votes]: Well Snell's law of refraction gives the value of the sine of the angle of refraction>1. The sine of any angle cannot be greater than 1.
So , even if one say it can hold, it will not give any value that can be further worked upon. So instead light gets reflected, and yes the basic law of reflection hold.


TITLE: How to solve $3\sin^3x-5\sin x\cos x+2\cos^2 x=0$?
QUESTION [0 upvotes]: Solve $3\sin^3x-5\sin x\cos x+2\cos^2 x=0$.
It should use simple identities, but no identity I used helped me. There has to be a trick but I don't seem to find it. I could really use any kind of help.

REPLY [3 votes]: here is way to solve the equation $$3\sin^3t-5\sin t\cos t+2\cos^2 t=0 $$ we will introduce $$x = \cos t, y = \sin t, x^2 + y^2 = 1\tag 1$$ now we have
$$0=3y^3 - 5xy + 2x^2 = 3y(1-x^2)-5xy+2x^2=-y(3x^2+5x-3)+2x^2 .$$ soving for $y$ gives $$y= \frac{2x^2}{3x^2 + 5x-3}\tag 2 $$
the graphs of $(1)$ and $(2)$ intersect at four points in the first, third and fourth quadrants. the coordinates are  $$x = 0.912, 0.563, 0.937, 0.418.$$ the corresponding angles in radians are $$0.422, 0.9727,3.498, 5.144 $$


TITLE: the discriminant of a function field extension
QUESTION [3 upvotes]: Let $f:X\to \mathbf{P}^1$ be a branched cover of smooth projective connected curves over $\overline{\mathbf{Q}}$, where $\mathbf{P}^1$ is the projective line over $\overline{\mathbf{Q}}$.
What relations are there between the discriminant of the function field extension associated to $f:X\to \mathbf{P}^1$ and other more common "invariants" of the cover like the ramification divisor, etc. ?

REPLY [2 votes]: The ramification divisor on $X$ is $\sum_x (e_x-1)[x]$ where $e_x$ is the ramification index of $O_{P^1,f(x)}\to O_{X,x}$. The different of the extension $O_{P^1,f(x)}\to O_{X,x}$ is $e_x-1$ (see e.g. Serre, Local fields). The discriminant at $y\in P^1$ is $\sum_{x\in f^{-1}(y)} (e_x-1)$ (Serre, op. cit.).
In summary, if the ramification divisor is $R:=\sum_x (e_x-1)[x]$ on $X$, the  discriminant divisor is $\sum_y(\sum_{x\in f^{-1}(y)} (e_x-1))[y]=f_*R$.
It is very important here that the base field has characteristic 0.


TITLE: A question on roots of a Quadratic.
QUESTION [4 upvotes]: If roots of the equation ${ax}^2+bx+c=0$ are of the form $$\frac{\alpha}{\alpha-1},\frac{\alpha+1}{\alpha}$$
Then the value of $$({a+b+c})^2$$
I have no clue how to approach this one, any help is appreciated!

REPLY [5 votes]: From the form:
$\frac{1}{x} + x = 2$ => $x^2 + x -2 = 0$, and first equation is $kx^2 + kx - 2k = 0$.
So, $(a+b+c)^2=0$


TITLE: Is there a simple series representation for following sequence?
QUESTION [0 upvotes]: I'm given an odd number defined by the variable $N$. I'm trying to identify a function relating the following sequences;
Given $N=7$
$A = \{1,2,3,4,5,6\}$
NOTE: A = 1:N-1 always
$B = \{6,6,4,4,2,2\}$
$F(N) = B\dots$ what is $F(N)$ ?
No, this is not a homework question. It's related to central difference formulations. $N$ is the "stencil" size, $A$ is the derivative, and $B$ is the order of accuracy. I would like to present this pattern as a simple function and I can't seem to find an explicit relation. Also, if there is one that is more generic to include forward/backward differences, that would be even better. Basically, I want to relate the derivatives, stencil sizes, and orders of accuracy. I'm not looking for an algorithm...just a general formula that I could present on a slide or in a paper instead of writing a full table.

EDIT: Just in case there is something more generic, for forward/backward $N$ need not be odd and the sequences would follow
$A = \{1,2,3,4,5,6\}$
NOTE again: A = 1:N-1 always, but B is different
$B = \{6,5,4,3,2,1\}$ or simply $B = \{N-1:-1:1\}$
EDIT 2: The following additional examples are given to help expose the pattern for B
Given $N=11$
$B = \{10,10,8,8,6,6,4,4,2,2\}$
Given $N=5$
$B = \{4,4,2,2\}$

REPLY [0 votes]: For central differences, let's consider the cases when $m \in A$ is odd and even separately.
For odd $m$, we have stencils of the form
$$
\frac{\partial^m f}{\partial x^m} \Big|_{x_0} = \frac{1}{\Delta x^m} \sum_{k=1}^{(N-1)/2} c_k^{(m)} (f_k - f_{-k}) + \mathcal{O}\left( \frac{\partial^N f}{\partial x^N} \Big|_{x_0} \frac{\Delta x^N}{\Delta x^m} \right)
$$
and thus the order of accuracy is $N-m$.
For even $m$, we have stencils of the form
$$
\frac{\partial^m f}{\partial x^m} \Big|_{x_0} = \frac{1}{\Delta x^m} \left\{ c^{(m)}_0 f_0 + \sum_{k=1}^{(N-1)/2} c_k^{(m)} (f_k + f_{-k})  \right\} + \mathcal{O}\left( \frac{\partial^{N+1} f}{\partial x^{N+1}} \Big|_{x_0} \frac{\Delta x^{N+1}}{\Delta x^m} \right)
$$
and thus the order of accuracy is $N+1-m$.
The relation you are looking for is thus
$$
F(N,m) =
\begin{cases}
N-m & \text{for odd } m \\
N+1-m & \text{for even } m.
\end{cases}
$$
In the general case, just write down the truncation error of the stencil to find the relation among the derivative, the stencil size and the order of accuracy.


TITLE: How many unique rows of pebbles are there?
QUESTION [1 upvotes]: A valid row of pebbles has $n$ white, and $n$ black pebbles. Thus, its
  length is $2n$.
Two rows of pebbles are considered equivalent if one can be produced
  from another by any combination of inversions (replacing all whites with
  blacks and vice versa) and flips (reverse the order of pebbles).
How many unique rows of pebbles are there for some $n$?
If we want our rows of pebbles to look mixed, we can say
  that a mixed row of pebbles can't have more than $2$ consecutive
  pebbles of the same color.
(mixed case $m=1$ is trivial with only one solution per $n$)
Then, how many rows of unique mixed pebbles are there for some
  $n$? (mixed case $m=2$)

We can represent pebbles with $0$'s and $1$'s.
I've tried generating examples for some $n$'s by iterating binary numbers;
We have solutions for first few $n$'s, for unique rows:
$$  1, 3, 7, 23, 71, 252, 890, 3299, 12283, 46508, \dots$$
Edit: Corrected thanks to Jeremy Dover from comments: Now,  this sequence is in OEIS.

If we eliminate ones with $3$ or more same consecutive pebbles from previous examples;
Then we have the number of unique mixed ones for $n=1,\dots,10,\dots$ below ($m=2$):
$$ 1 ,3 ,5 ,12 ,25 ,61 ,140 ,347 ,841 ,2108 , \dots$$
Edit: Also corrected. This sequence still does not appear in OEIS.

Q: How can we find these values for unique mixed examples algebraically?

P.S. Is there any significantly faster way to generate all solutions for some $n$? The outputs with all examples for the second sequence so far: pastebin.com ; Where second column is the decimal value and third is the difference between consecutive examples.

Also, would it be interesting to generalize this and find solutions for rows of size $k\times n$ where $k$ is the number of distinct
  colors of pebbles, and also consider $m$ consecutive pebbles allowed?
   How does one calculate/compute a solution for some $(k,n,m)$?

A follow up; If we were to have $2n$ rows of pebbles together to form a "pebble garden", how many unique mixed pebble gardens are there?
That is, now we also need to check the columns, as well as rows, for the same number of pebbles and for no three consecutive pebbles of same color. We also need to consider both horizontal and vertical flips and the 90° degree rotations for uniqueness, then this follow up is equivalent to the following question:
How many distinct Unruly boards are there?

P.S. We could also stack the gardens into "pebble cubes", and continue to "pebble tesseracts", and continue to... $N$ dimensional unique pebble structures? But given the follow up has zero progress posed there as of this moment, this is perhaps still an overreach.

REPLY [2 votes]: For  starters   we  verify  the   formula  that  is  given   at  OEIS
A045723  which treats  the problem  without
adjacency  constraints. We  solve the  case  of $n$  instances of  $k$
colors for both $n\times k$ even and  odd. What we have here is a case
of Power  Group Enumeration with the  slot permutations consisting
of the identity  and the reflection. We represent the  color scheme by
by a  sequence of $k$ blocks  of $n$ identical colors,  $n\times k$ in
total. The object permutation group acting on these is obtained by the
simultaneous action  of the symmetric  group $S_k$ on the  $k$ blocks,
which encodes the fact that the  colors are swappable, and a vector of
$k$ permutations $\gamma_1,\ldots, \gamma_k$  from $S_n$ acting on the
constituents  of  the blocks,  which  encode  the  fact that  the  $n$
instances  of   the  $k$  colors   are  the  same.   Denote   this  by
$Z(F_{n,k}).$
For Power Group Enumeration we require the cycle index of the slot
permutation group which is
$$Z(Q_{n,k}) = \frac{1}{2} (a_1^{nk} + a_2^{nk/2})$$
when $n\times k$ is even and
$$Z(Q_{n,k}) = \frac{1}{2} (a_1^{nk} + a_1 a_2^{(nk-1)/2})$$
when $n\times  k$ is odd.   We must cover  these two with  cycles from
permutations from  $Z(F_{n,k})$ where we  are using a set  cover, i.e.
all colors  must be present.  Now  for the first one  there clearly is
only one possibility,  coverings using fixed points  from the identity
permutation.   Since   there  are  $k!\times  n!^k$   permutations  in
$Z(F_{n,k})$ we get a contribution of
$$\frac{(nk)!}{k! \times n!^k}.$$
The power of  $a_2$ requires more work and must  be simplified to make
for a  feasible computation.  Note  once more that  in a set  cover we
must use all cycles from the  permutation being sourced for the cover.
This clearly  requires for  the two permutations  to have  exactly the
same  cycle  structure.   Therefore we  are  extracting  $[a_2^{nk/2}]
Z(F_{n,k})$  and  $[a_1  a_2^{(nk-1)/2}] Z(F_{n,k}).$  Supposing  that
$\beta$  is the  permutation from  $S_k$  we see  that it  must be  an
involution, consisting of fixed points and two-cycles only, since this
is  the structure  of the  coefficient being  extracted and  we cannot
return to the  starting point of a cycle without  having made at least
one turn. Now  if $\beta$ fixes a certain block  then the contribution
from  that block  (permutation  $\gamma$) to  the  cycle index  enters
identically and hence it must be an involution as well.  One the other
hand if two blocks are on  a two-cycle the cycles where their elements
reside must have  even length wich implies they  are two-cycles.  This
means that  the corresponding  permutations $\gamma_a$  and $\gamma_b$
are inverses  of each other and  the joint action with  $\beta$ splits
everything into two-cycles. Using the exponential formula we introduce
$$Z(S_{q,\mid m}) =
[w^q] \exp\left(\sum_{d|m} a_d \frac{w^d}{d}\right)$$
we thus obtain
$$[a_2^{nk/2}] Z(F_{n,k}) =
[a_2^{nk/2}] Z(S_{k,\mid 2})
\left(Z(S_{n,\mid 2}), \frac{1}{n!} a_2^n\right)$$
and similar  for $n\times k$  odd. Here  a cycle index  evaluated with
arguments signifies  substitution into $a_1,  a_2$ etc.  It remains to
observe that we have two choices  for each cycle being covered when we
cover the  two-cycles and  any permutation of  the source  cycles from
$Z(F_{n,k})$ is  valid.  We  finally get for  the desired  closed form
when $n\times k$ is even
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2} \left(\frac{(nk)!}{k! \times n!^k}
+ (nk/2)! \times 2^{nk/2} \times
[a_2^{nk/2}] Z(S_{k,\mid 2})
\left(Z(S_{n,\mid 2}), \frac{a_2^n}{n!}\right) \right)}$$
and
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2} \left(\frac{(nk)!}{k! \times n!^k}
+ ((nk-1)/2)! \times 2^{(nk-1)/2} \times
[a_1 a_2^{(nk-1)/2}] Z(S_{k,\mid 2})
\left(Z(S_{n,\mid 2}), \frac{a_2^n}{n!}\right) \right)}$$
otherwise. We can compute these values with the following Maple code.

with(numtheory);
with(combinat);

pet_cycleind_symm :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

pet_cycleind_symm_nk_div2 :=
proc(n, k)
option remember;
local idx_col, idx_slot, res;

    if n=1 and k=1 then return a[1] fi;

    if n=1 or k=1 then
        return pet_cycleind_symm(n*k);
    fi;

    idx_col := coeftayl(exp(a[1]*z+a[2]/2*z^2), z=0, k);
    idx_slot := coeftayl(exp(a[1]*z+a[2]/2*z^2), z=0, n);

    res :=
    subs({a[1]=idx_slot, a[2]=a[2]^n/n!},
         idx_col);

    expand(res);
end;

EXA :=
proc(n, k)
option remember;
local idx;

    if n=1 or k=1 then return 1 fi;

    idx := pet_cycleind_symm_nk_div2(n, k);

    if type(n*k, even) then
        1/2*((n*k)!*coeff(idx, a[1], n*k) +
             ((n*k)/2)!*2^(n*k/2)*
             coeff(idx, a[2], n*k/2));
    else
        1/2*((n*k)!*coeff(idx, a[1], n*k) +
             ((n*k-1)/2)!*2^((n*k-1)/2)*
             coeff(coeff(idx, a[2], (n*k-1)/2), a[1], 1));
    fi;
end;

EXB :=
proc(n, k)
option remember;
local idx;

    if n=1 or k=1 then return 1 fi;

    idx := pet_cycleind_symm_nk_div2(n, k);

    if type(n*k, even) then
        1/2*((n*k)!/k!/n!^k +
             ((n*k)/2)!*2^(n*k/2)*
             coeff(idx, a[2], n*k/2));
    else
        1/2*((n*k)!/k!/n!^k +
             ((n*k-1)/2)!*2^((n*k-1)/2)*
             coeff(coeff(idx, a[2], (n*k-1)/2), a[1], 1));
    fi;
end;

We get for two swappable colors the sequence
$$1, 3, 7, 23, 71, 252, 890, 3299, 12283, 46508,
\\ 176870, 677294, 2602198, \ldots $$
which matches  the data from  the OP.  With three swappable  colors we
find
$$1, 11, 148, 2955, 63231, 1430912, 33259920,
\\ 788827215, 18989544145, 462583897776, \ldots $$
and with four the data are
$$1, 65, 7780, 1315825, 244448316, 48099214856,
\\ 9844135755168, 2074189508907945, \ldots $$
The  initial segments  of these  may  be verified  with the  following
enumeration routine.

#! /usr/bin/perl -w
#

sub recurse {
    my ($n, $k, $kf, $rest, $sofar, $orbits) = @_;

    if(scalar(@$sofar) == $n*$k){
        my $sofar_rev = [ reverse(@$sofar) ];

        my @orbit;

        for(my $idx = 0; $idx < $kf; $idx++){
            my @permcols = ( 0 .. ($k-1) );

            for(my ($d, $ind) = ($k, $idx);
                $d > 1; $d--){
                my $pos = $ind % $d;

                if($pos != $d-1){
                    my $tmp = $permcols[$pos];
                    $permcols[$pos] = $permcols[$d-1];
                    $permcols[$d-1] = $tmp;
                }

                $ind = ($ind - $pos) / $d;
            }

            my $conf = join '',  map {
                chr(ord('A') + $permcols[$_])
            } @$sofar;

            push @orbit, $conf;

            $conf = join '',  map {
                chr(ord('A') + $permcols[$_])
            } @$sofar_rev;

            push @orbit, $conf;
        }

        my @sorted = sort(@orbit);
        $orbits->{$sorted[0]} = 1;
        return;
    }

    for(my $col = 0; $col < $k; $col++){
        if($rest->[$col] > 0){
            $rest->[$col]--;
            push @$sofar, $col;

            recurse($n, $k, $kf, $rest, $sofar, $orbits);

            pop @$sofar;
            $rest->[$col]++;
        }
    }
}

MAIN : {
    my $mx = shift || 5;
    my $k = shift || 2;

    my $kf = 1;
    for(my $f=2; $f <= $k; $f++){
        $kf *= $f;
    }

    $| = 1;

    for(my $n=1; $n <= $mx; $n++){
        my $orbits = {};

        recurse($n, $k, $kf, [($n) x $k],
                [], $orbits);

        print " " if $n > 1;
        print scalar(keys(%$orbits));
    }

    print "\n";
}

There  is   another  example   of  PGE   at  the   following  MSE
link.


TITLE: What does it mean, geometrically speaking to find the limit of a matrix raised to power?
QUESTION [1 upvotes]: I know from 3B1B's videos that a matrix essentially refers to rotating and skewing graphs, with the matrix giving us the new coordinates of the unit vectors. So, if
$$A=
\begin{bmatrix}
a& 0  \\
0 & b \\
\end{bmatrix}
$$,
with $a>1$ and $b>1$
then we have that ${\lim \limits_{n \to \infty}(A^n)}^{-1}=O$.
Where 'O' is the null matrix
Therefore, the entire system collapses onto the origin in the limiting case.
However, I'm unable to visualize all intermediate points of this transformation. Is there any program that will allow me to do this and is there any way to do this manually?
In short, what would the equivalent of the graphical representation of such a transformation look like?

REPLY [1 votes]: I won’t share many programs for visualizing this, mostly because that depends on what exactly about the limit you want to visualize. Mathematica is useful for sketching diagrams. I’ve never used SageMath, but that is a free alternative.
Here’s a way to think about the situation manually.
A matrix $A: \mathbb{R}^2 \to \mathbb{R}^2$ is determined entirely by where it sends $e_1 = \binom{1}{0}$ and $e_2 = \binom{0}{1}$:
$$
  A\left(x e_1 + y e_2\right) = \begin{pmatrix}
    a & b\\
    c & d
  \end{pmatrix}\begin{pmatrix}
    x \\ y
  \end{pmatrix} = (ax + by) e_1 + (cx + dy)e_2.
$$
In the case of your diagonal matrix $ A = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$, we see that:
$$
  A(x e_1 + y e_2) = (ax)e_1 + (by) e_2.
$$
$A$ acts by multiplying the coordinates independently. Therefore $A^n (x e_1 + y e_2) = (a^n x) e_1 + (b^n y) e_2$. The inverse of $A^n$ is the matrix which sends $(a^n x) e_1 + (b^n y) e_2$ to the starting input: $x e_1 + y e_2$. We see that we need to divide the first coordinate by $a^n$ and the second by $b^n$:
$$
  \left(A^n\right)^{-1} = \begin{pmatrix}
    a^{-n} & 0 \\
    0 & b^{-n}
  \end{pmatrix}
$$
In the comments you mention that $|a| > 1$ and $|b| > 1$, therefore $\lim_{n\to\infty} a^{-n} = \lim_{n\to\infty} b^{-n} = 0$. Thus, as we let $n \to \infty$, the entries of $(A^n)^{-1}$ go to zero. This determines what the limit matrix does to $e_1$ and $e_2$: since all the entries go to zero:
$$
  \lim_{n\to\infty} \left( A^n \right)^{-1} = 0.
$$
Alternatively, since we know what the entries of $(A^n)^{-1}$ are, we can see what the intermediate points are:
$$
  (A^n)^{-1} (x e_1 + y e_2) = \left(\frac{x}{a^n} e_1 + \frac{y}{b^n} e_2\right).
$$
Visually, $(A^n)^{-1}$ shrinks the horizontal axis by $a^n$ and the vertical axis by $b^n$. In the limit, both axes are shrunk by a factor of $\infty$. That is to say, the limit matrix sends every point to the origin.


TITLE: Physical interpretation of the retarded vs. Feynman propagators?
QUESTION [16 upvotes]: We calculate the real-space propagator $\Delta(x)$ for a free real scalar field $\varphi(x)$ with mass $m$ by performing the Fourier transform (using sign convention +---)
$$\Delta(x) = \int \frac{d^3k\, d\omega}{(2 \pi)^4} \frac{e^{-i k \cdot x}}{k^2 - m^2}$$
Depending on whether we deform the contour of integration for the $\omega$ integral to the the same or opposite sides of the two poles at $\omega = \pm \sqrt{{\bf  k}^2 + m^2}$, we get either the retarded or the Feynman propagator (or the advanced or the anti-time-ordered propagator, but let's ignore those two options).
From the Wikipedia article on propagator, the retarded propagator is
$$G_\text{ret}(x) = i\, \langle [ \varphi(x), \varphi(0) ] \rangle\, \Theta(x_0)= \frac{1}{2 \pi} \delta(\tau^2) - \frac{m\, J_1(m \tau)}{4 \pi \tau}$$
if $x$ is inside the origin's future light cone, and zero otherwise, where $\Theta$ is the step function, $\delta$ the Dirac delta function, $\tau$ the proper time $\sqrt{x \cdot x}$, and $J_1$ the Bessel function.  The Feynman propagator is
$$G_F(x) = -i\, \langle \mathcal{T} \varphi(x) \varphi(0) \rangle = \begin{cases} -\frac{1}{4 \pi} \delta(s) + \frac{m}{8 \pi \sqrt{s}} H^{(2)}_1(m \sqrt{s}) \qquad &\text{if } s \geq 0 \\ -\frac{i m }{4 \pi^2 \sqrt{-s}} K_1(m \sqrt{-s}) \qquad &\text{if } s < 0 \end{cases},$$
where $\mathcal{T}$ is the time-ordering symbol, $s$ the spacetime interval $x \cdot x$, $H^{(2)}$ the Hankel function, and $K_1$ the modified Bessel function.
Looking at the operator expectation values, it's clear that the Feynman propagator is the right one to use for calculating probabilities of past-to-future propagation.  But looking at the actual functional expressions, $G_\text{ret}$ naively "looks" more causally correct, because it vanishes outside the light cone, as we'd naively expect for a particle propagator.  (I know, I know, the spacelike correlations decay exponentially and don't actually violate causality because they can't transmit causal influence, etc. etc.)
Does the retarded propagator $G_\text{ret}$ have any physical significance?  I understand why locality requires $G_\text{ret}(x) \equiv 0$ for $x$ outside the lightcone, but it seems a bit strange that once we enforce this, we then completely ignore its value inside the light cone.  But I don't know how to intepret the $\langle \varphi(0) \varphi(x) \rangle$ part of $G_\text{ret}$ if $x^0$ is positive.

REPLY [9 votes]: The convolution $G_{ret}*f$ of the retarded propagator $G_{ret}$ with a source term $f$ which vanishes sufficiently far in the past is the unique solution of the inhomogeneous Klein-Gordon equation with source term $f$ which vanishes in the far past. It is necessarily a nontrivial superposition of positive and negative energy solutions at all times when it is not identically zero, precisely because it vanishes outside the future light cone of the support of $f$.
The convolution $G_F*f$ of the Feynman propagator $G_F$ with a source term $f$ which vanishes outside a bounded space-time region, on its turn, is a solution of the inhomogeneous Klein-Gordon equation which is of positive energy outside the past light cone of the support of $f$ (i.e. in the far future) and of negative energy outside the future light cone of the support of $f$ (i.e. in the far past). Because of this, it cannot vanish in any nonvoid open region of space-time, unlike $G_{ret}*f$.
The relation between the position-space support and the momentum-space support of $G_{ret}$ and $G_F$ can be seen as a consequence of the uncertainty principle, which restricts the regions where a distribution can vanish in position and momentum space (the latter after a Fourier transform).


TITLE: Taylor Series of $\ln(1+x^3)/[\tan(x)-x]$
QUESTION [0 upvotes]: Hi i want to find the Taylor series of the function $$\frac{\ln(1+x^3)}{[\tan(x)-x]}$$ at $0$ until n=3
i have $\ln(1+x^3)=x^3-\frac{x^6}{2}+\frac{x^9}{3}+o(x^9)$ and $\tan(x)-x=\frac13 x^3+o(x^3)$
by devision I found that $$\frac{\ln(1+x^3)}{\tan(x)-x}=3-\frac{3}{2}x^3+o(x^3)$$
but I don't find the same result as here.

REPLY [0 votes]: In fact the issue is that you simply divided by $x^3$ anf forgot the $o()$ stuff.
This should be $\frac1{x^3}\times \frac 1{1+o(1)}=\frac 1{x^3}\times(1-o(1))$
All resulting terms smaller than $3\times o(1)=o(1)$ should be ignored in calculation below:
$\begin{align}
\dfrac{x^3-\frac 12x^6+\frac 13x^3+o(x^3)}{\frac 13 x^3+o(x^3)} &=\dfrac{3-\frac 32x^3+x^6+o(x^6)}{1+o(1)}=\left(3-\frac 32x^3+x^6+o(x^6)\right)\times(1-o(1))\\\\
&\require{cancel}=3-o(1)-\cancel{\frac 32x^3}+\cancel{o(x^3)}+\cancel{x^6}-\cancel{o(x^6)}+\cancel{o(x^6)}-\cancel{o(x^6)}\\\\&=3+o(1)\end{align}$
So to keep a development up to $o(x^3)$ you need to keep at least this degree on the denominator after division by $x^3$, this means up to $o(x^6)$ while developing $(\tan x-x)$.


TITLE: Differentiability vs Having a Derivative
QUESTION [7 upvotes]: In some calculus texts, one comes across the following statement - "A function is differentiable, if it has a derivative and a function has a derivative, if it is differentiable.". Two such texts are  - A treatise on Advanced Calculus by Philip Franklin(this book can be found on archive.org - check section 69 - Page 109) and Advanced Calculus - An Introduction to Linear Analysis by Leonard F. Richardson - Chapter 4 - Page 101.
Analytical proofs are also given to justify these statements.
This raises the question - What is the physical difference between being differentiable and having a derivative?
The book by Leonard Richardson gives some analytical explanation about the existence of a linear function, which approximates the function at points very close to x. If such a function exists, then the function is said to be differentiable(and then he goes on to prove analytically that the function has a derivative and vice versa). The problem is that the whole proof is anlaytic, and is not illustrated using a practical example. When we say that a function is differentiable, we generally check for the existence of its derivative (i.e the limit of its differential coefficient). But this text 'implies' that its possible to check the differentiability of a function by checking for the existence of such a linear function(whose slope as you've guessed becomes the derivative in the limit). So what essentially is the difference, between this approach and the standard approach? It seems purely algebraic to me....Or is there more to it?
A related question is - what's the difference between Caratheodory's definition of the derivative and the standard definition of the derivative?. The difference seems purely algebraic to me.I would like to know if there's more to it.(I think the answer to this question is closely related to the first, but since Caratheodory's definition is rarely covered in most standard calculus texts, I've decided to gain as much information as possible, on this subject.)
I am not a mathematics major, so there are limits to my theoretical thinking! I would greatly appreciate if practical examples are provided to illustrate these concepts.

REPLY [1 votes]: For the second question, there is no difference between the two definitions - this is a simple exercise. For the first, there is also no difference. You've defined differentiability through the exist of a linear mapping which approximates the function near $x.$ The derivative is simply the matrix which represents this function; the proof given in Franklin's book that the two are equivalent is because he essentially defines differentiable by multiplying both sides of the limit definition of a derivative by $\Delta x,$ which goes to $0,$ so it may not be entirely clear you can do this (if you're not used to this sort of thing). Anyway, there's no difference.
For your comment above - if you define a fruit with certain characteristics as an apple, then you have a fruit which has characteristics that are easily seen to be equivalent, you need to actually prove that the characteristics are equivalent before you can conclude it's an apple. This is what they're doing - it's a trivial statement, but for the sake of a "rigorous" textbook, they've done it.
As an edit: If you have a LINEAR function that approximates the function $f$ you are looking at arbitrarily close to a point, this is the tangent line in $2$-dimensions - now, the function has a slope; this is the derivative. In higher dimensions, you will have a hyperplane that approximates this point - hyperplanes are translates of a linear mapping; linear mappings admit a matrix representation. This is your derivative.


TITLE: Integral with limit
QUESTION [0 upvotes]: I am trying to proof that
$$
f : [a,b] \rightarrow \mathbb{C} \ \mbox{is continuous} \
\longrightarrow
\underset{n \rightarrow \infty}{\lim} \int_a^b f(x) e^{-inx} \ dx = 0
$$
where $ [a,b] $ is a compact interval.
Integrating by parts
$$
\int_a^b f(x) e^{-inx} \ dx =
\left[\frac{f(x) e^{-inx}}{-in} \right]_a^b - \frac{1}{in} \int_a^b f'(x) e^{-inx} \ dx
$$
and we have $ \underset{n \rightarrow \infty}{\lim} [\frac{f(x) e^{-inx}}{-in}]_a^b = 0 $ because $ e^{-inx} $ is bounded.
But I dont know what to do with the expression $ \int_a^b f'(x) e^{-inx} \ dx $, since we have not any hypothesis about continity or boundary of $ f' $.

REPLY [1 votes]: This is the famous Riemann-Lebesgue lemma - or at least, it's a case of that lemma, with the function $f$ more restricted than it has to be. The general strategy of all the proofs I know: First, prove it for sufficiently nice $f$. Second, show that nice enough $f$ are dense in the space of functions $f$ lives in. Finally, split an arbitrary $f$ in that space into the sum of something nice and something small. Both parts go to zero, so their sum does, and we win.
This integration by parts is one of those proofs I know, and it's the one I prefer. You've got step 1 there; the theorem is proved for $f\in C^1[a,b]$. The next step is to show that $C^1$ functions are dense in $C[a,b]$ under the appropriate norm. Here, we can approximate anything continuous on $[a,b]$ uniformly by $C^1$ functions. There are lots of ways to do this - Stone-Weierstrass, fitting splines, smoothing with a convolution - the fact that continuous functions on the closed interval are uniformly continuous gives us lots of flexibility. This is all very standard, of course, and can easily be looked up.
Then, for that final step, we split $f=g+h$ for an arbitrary continuous $f$, where $g$ is $C^1$ and $h$ is small. What, exactly, do we mean by "small"? Integrate
$$\int_a^b e^{-inx}(g(x)+h(x))\,dx = \int_a^b e^{-inx}g(x)\,dx + \int_a^b e^{-inx}h(x)\,dx$$
The first term there will go to zero as $n\to\infty$, so we need a bound for the second term. For that, we use that $|e^{inx}|=1$, so
$$\left|\int_a^b e^{-inx}h(x)\,dx\right|\le \int_a^b\left|e^{-inx}h(x)\right|\,dx \le \int_a^b |h(x)|\,dx$$
Oh. That's the 1-norm. What we really needed here was for the $C^1$ functions to be dense in the $L^1$ sense. Fortunately, it's a finite interval, and a uniformly convergent sequence also converges in $L^1$, or indeed any $L^p$. So then, choose $g\in C^1$ so that $\sup_x |f(x)-g(x)|\le \frac{\epsilon}{2(b-a)}$, and $\int_a^b |f-g|\le \frac{\epsilon}{2}$. Then choose $N$ large enough that $\left|\int_a^b e^{-inx}g(x)\,dx\right| <\frac{\epsilon}{2}$ for all $n>N$, and we get that $\int_a^b e^{-inx}f(x)\,dx < \epsilon$ for all $n>N$. That $\epsilon$ was arbitrary, and we have the desired convergence to zero.


text:
\section{Preliminaries} \label{sec:prelims}

\paragraph{Notation} We use bold faces to denote vectors and matrices. By $\vec{x}_{-j}$ we denote the vector $\vec{x}$ that involves all coordinates but $j$. We use $\textrm{vec}(\vec{A})$ to denote the standard vectorization of matrix $\vec{A}$. Moreover, we use $\norm{\textrm{vec}(\vec{A})}_{1,\textrm{off}}$ to denote the $\ell_1$ norm of $\textrm{vec}(\vec{A})$ by excluding the diagonal entries of matrix $A$ and $\textrm{nz}(\vec{A})$ for the number of non-zero entries of matrix $\vec{A}$. We denote by $\mathcal{S}_{d\times d}$ the set of symmetric matrices.

\paragraph{Norms} For a $d \times d$ matrix $\vec{A}$,
\begin{align*}
\norm{\vec{A}}_2 = \max_{\norm{\vec{x}}_2 = 1} \norm{\vec{Ax}}_2, \: \norm{\vec{A}}_{\infty} = \max_{j \in [n]} \sum_{i=1}^n |A_{ij}|, \: \norm{\vec{A}}_F = \sqrt{\sum_{i=1}^n \sum_{j=1}^n A_{ij}^2}.
\end{align*}
When $\vec{A}$ is a symmetric matrix we have that $\norm{\vec{A}}_2 \le \norm{\vec{A}}_{\infty} \le \norm{\vec{A}}_F \le \sqrt{n}\norm{\vec{A}}_2 \le \sqrt{n}\norm{\vec{A}}_\infty$. For a vector $\vec{x} \in \mathbb{R}^d$ we also have,
\begin{align*}
\norm{\vec{x}}_2 = \sqrt{\sum_{i=1}^d {x}_i^2}, \: \norm{\vec{x}}_{\infty} = \max_{j \in [d]} |{x}_j| , \: \norm{\vec{x}}_1 = \sum_{i=1}^d |{x}_i|.
\end{align*}
It holds that $\ell_{1}$ is the dual of $\ell_{\infty}$ and for $\vec{x},\vec{y} \in \mathbb{R}^d$ one can have $\vec{x}^T \vec{y} \leq \norm{\vec{x}}_1 \norm{\vec{y}}_{\infty}$ (H\"older's inequality).

\paragraph{Truncated Gaussian Distribution}
For a measurable set $S$ with parameters $\vec{\mu}, \vec{\Sigma}$, the density function for the truncated Gaussian distribution $\mathcal{N}(\vec{\mu},\vec{\Sigma}, S)$ with mean $\vec{\mu}$ and covariance $\vec{\Sigma}$ is defined as follows:
\begin{align}
\mathcal{N}(\vec{\mu},\vec{\Sigma},S;\vec{x})\defeq
{
	\begin{cases}
	\dfrac{\mathcal{N}(\vec{\mu},\vec{\Sigma};\vec{x})}{\int_{S}\mathcal{N}(\vec{\mu},\vec{\Sigma};\vec{x}) d\vec{x}} \;,\; \vec{x} \in S\\
	0                                            \;,\; \vec{x} \notin S
	\end{cases}
}
\end{align}
where
\[
\mathcal{N}(\vec{\mu},\vec{\Sigma};\vec{x}) \defeq \frac{1}{\sqrt{2\pi \textrm{det}(\vec{\Sigma})}}\exp\left(-\frac{1}{2} (\vec{x}-\vec{\mu})^T \vec{\Sigma}^{-1} (\vec{x}-\vec{\mu})\right).
\]
Throughout this paper, we will assume that the set $S$ can be accessed through a {\em membership oracle}.
\begin{definition}[Membership oracle] Let $S \subset \mathbb{R}^d$ be a measurable set. A membership oracle of $S$ is a function that given an arbitrary $\vec{x} \in \mathbb{R}^d$, it returns yes if it belongs to the set, otherwise no (i.e., it implements the indicator function of $S$). We assume oracle access to the indicator of $S$.
\end{definition}

\paragraph{Precision matrix and sparsity}
Let $G=(V,E)$ be an undirected graph with $V= [d]$. A random vector $\vec{x} \in \mathbb{R}^d$ is said to be distributed according to the (undirected) Gaussian Graphical model with graph $G$ if $\vec{X}$ has a multivariate Gaussian distribution $\mathcal{N}(\vec{\mu}, \vec{\Sigma})$ with
\begin{align}
\left(  \vec{\Sigma^{-1}} \right)_{ij} = 0 \; \; \forall \: (i,j) \notin E,
\end{align}
$\vec{\Sigma}^{-1}$ which we denote by $\vec{\Theta}$ is known as the precision matrix. In our results, the sample complexity depends on the number of non-zero entries of $\vec{\Sigma}^{-1}$, i.e., $\textrm{nz}(\vec{\Sigma}^{-1})$.

\paragraph{Strong Convexity}
\begin{lemma}[folklore, see p.309 of \cite{Wbook}]\label{lem:wainwright99} Suppose that a differentiable function $f : \mathbb{R}^d \to \mathbb{R}$ is $\kappa$-strongly convex in the sense that
\begin{equation}\label{eq:9.100a}
f(\vec{y}) \geq f(\vec{x}) + (\vec{y}-\vec{x})^T\nabla f(\vec{x}) +\frac{\kappa}{2} \norm{\vec{y}-\vec{x}}_2^2 \textrm{ for all }\vec{x},\vec{y} \in \mathbb{R}^d.
\end{equation}
It holds that
\begin{equation}
(\vec{y}-\vec{x})^T\left(\nabla f(\vec{y})-\nabla f(\vec{x})\right) \geq \kappa \norm{\vec{y}-\vec{x}}_2^2 \textrm{ for all }\vec{x},\vec{y} \in \mathbb{R}^d.
\end{equation}
\end{lemma}

\begin{lemma}\label{lem:wainwright910} Suppose that $f : \mathbb{R}^d \to \mathbb{R}$ is a twice differentiable, convex function that is locally $\kappa$-strongly convex around $\vec{x}$, in the sense that the lower bound \ref{eq:9.100a}  holds for all vectors $\vec{z}$ in the ball $\mathbb{B}_2 = \{\vec{z}: \norm{\vec{z}-\vec{x}}_2 \leq \rho\}$. It holds that
\begin{equation}
(\vec{y}-\vec{x})^T\left(\nabla f(\vec{y})-\nabla f(\vec{x})\right) \geq \rho\kappa \norm{\vec{y}-\vec{x}}_2 \textrm{ for all }\vec{y} \in \mathbb{R}^d \backslash \mathbb{B}_{2}.
\end{equation}
\end{lemma}
\begin{proof}
Let $\vec{y} \in \mathbb{R}^d \backslash \mathbb{B}_2$ and $\vec{x}_t = t(\vec{y}-\vec{x}) + \vec{x}$ and $g(t) = \nabla f(\vec{x}_t)$. The derivative is given by $g'(t) = \nabla ^2 f(\vec{x}_t) (\vec{y}-\vec{x})$. Let $0 \leq b \leq 1$ be such that $\norm{\vec{x}_b - \vec{x}}_2 = \rho$ and observe that $\rho = b\norm{\vec{y}-\vec{x}}_2$.
From fundamental theorem of calculus we get
\begin{align*}
(\vec{y} - \vec{x})^T(\nabla f(\vec{y}) - \nabla f(\vec{x})) &= (\vec{y} - \vec{x})^T(g(1) - g(0))\\
&=\int_0^{1} (\vec{y}-\vec{x})^T \nabla^2 f(\vec{x}_t) (\vec{y}-\vec{x}) dt\\&\geq
\int_0^{b} (\vec{y}-\vec{x})^T \nabla^2 f(\vec{x}_t) (\vec{y}-\vec{x}) dt \textrm{ since $f$ is convex}\\&\geq
\int_0^{b} \kappa\norm{\vec{y}-\vec{x}}_2^2 dt \textrm{ since $f$ is }\kappa-\textrm{strongly convex}\\&=
b\kappa\norm{\vec{y}-\vec{x}}_2^2 = \kappa \rho \norm{\vec{y}-\vec{x}}_2,
\end{align*}
and the claim follows.
\end{proof}

\begin{comment}
One important well-known fact (can be easily derived from Schur complement) is that the random variable $\vec{X}_i$ conditioned on the rest of the variables (the neighbors) follows also a Gaussian distribution. More precisely
\begin{equation}\label{eqn:untruncated_cond}
\vec{X}_i | \vec{X}_{-i} \sim \vec{\mu}_i - \frac{1}{\vec{\Theta}_{ii}} \vec{\Theta}_{-i} (\vec{X}_{-i}-\vec{\mu}_{-i}) + \mathcal{N}\left(0, \frac{1}{\vec{\Theta}_{ii}}\right),
\end{equation}
where $\vec{\Theta}_{-i}$ denotes the $i$-th row of $\vec{\Theta}$ by removing entry $\vec{\Theta}_{ii}$.
A similar result can be shown for truncated Graphical models (see Lemma \ref{lem:conditionaltruncated}) and this is a very crucial fact for our structural learning result (Theorem \ref{thm:main2}).

\paragraph{Stochastic Gradient Descent} Stochastic Gradient Descent (SGD) is arguably one of the most commonly used first order methods for the task of optimizing a convex function. It can be summarized in Algorithm 1.
\footnote{\textcolor{red}{Formatting of the SGD algorithm}}
\begin{algorithm}[h!]
	\KwData{Function $f$, Parameter $M$}
	\KwResult{Solve $\min _{\vec{w} \in \mathcal{W}}f(\vec{w})$ }
     $\vec{w}_1 = 0$
    \For{$t=1;t\leq M;t++$}
    {
        Choose a random vector $\vec{v}_t$ s.t $\mathbb{E}[\vec{v}|\vec{w}_t] \in \partial f(\vec{w}_t)$\;
        Set $\eta_t = \frac{1}{\lambda t}$\;
        Set $\vec{w}_{t+1/2} = \vec{w}_t - \eta_t \vec{v}_t$\;
        Set $\vec{w}_{t+1} = \arg\min _{\vec{w} \in \mathcal{W}} \norm{\vec{w} - \vec{w}_{t+1/2}}_2^2$\;
    }
    \Return{$\frac{1}{M} \sum_{t=1}^M \vec{w}_t$}
	\caption{Projected Stochastic Gradient Descent}
\label{alg:SGD}
\end{algorithm}

The following theorem can be shown for SGD.
\begin{theorem}[Theorem 14.11 in \cite{understanding}] Assume that f is $\lambda$-strongly convex and that $\vec{v}_t$ is a random vector such that $\mathbb{E}[\vec{v}_t|\vec{w}_{t-1}] \in \partial f(\vec{w}_{t-1})$, $\mathbb{E}\left[\norm{\vec{v}_t}_2^2\right] \leq \rho^2$. Let $\vec{w}^* \in \arg\min_{w \in \mathcal{W}} f(w)$ (for some convex set $\mathcal{W}$). It holds that
\begin{equation}
\mathbb{E}\left[\frac{1}{M} \sum_{i=1}^M \vec{w}_i\right] - f(\vec{w}^*) \leq \frac{\rho^2}{2\lambda M}(1+\log M).
\end{equation}
\end{theorem}
\end{comment}

config: arxiv


text:
\begin{document}

\begin{abstract}
We prove an Allard-type regularity theorem for free-boundary minimal surfaces in Lipschitz domains locally modelled on convex polyhedra.  We show that if such a minimal surface is sufficiently close to an appropriate free-boundary plane, then the surface is $C^{1,\alpha}$ graphical over this plane.  We apply our theorem to prove partial regularity results for free-boundary minimizing hypersurfaces, and relative isoperimetric regions.
\end{abstract}
\maketitle

\tableofcontents

\input{prelim.tex}

\section{Main theorem}\label{sec:main}

Our main Theorem \ref{thm:main} is the following Allard-type regularity result, which says loosely that whenever an integral varifold $M$ has free-boundary in $\Omega = \Phi(\OmegaRef = \OmegaRefN \times \R)$, and: $\Phi$ is sufficiently close to the identity, $M$ has sufficiently small mean-curvature, and $M$ is sufficiently varifold close to the ``horizontal'' plane $\R^n \times \{0\}$, then $\spt M$ is a $C^{1,\alpha}$ perturbation of $\OmegaRefN$.

In general $\Omega$ is curved, and so $\spt M$ will not be graphical over a particularly ``nice'' subdomain of $\R^n$.  Instead, it is more convenient and precise to look at $\Phi^{-1}(\spt M)$, which will be a graph over $\R^n \cap \OmegaRef \cap B_{1/32}$.  We do not lose anything in our estimates by doing this, as even before this transformation we must use $|D^2\Phi|_{C^0}$ to control the tilting of tangent planes of $\spt M$.

For various reasons we in fact want to allow not only the diffeomorphism $\Phi$ to change, but also the reference domain $\OmegaRef$ (mainly because different points in $\Phi(\OmegaRef)$ will be modeled on different polyhedral cone domains, even when staying in the same stratum; see Section \ref{sec:changing-type}).  For this reason we in fact consider domains of the form $\Phi(\Psi(\OmegaRef))$, where $\Phi$ is a diffeomorphism close to $Id$, and $\Psi$ is a linear map close to $Id$.  Our constants $\delta, c, \alpha$ will be uniform in $\Phi, \Psi$, but in each particular case our ``reference'' polyhedral domain will be $\Psi(\OmegaRef)$.  It may be easier to parse Theorem \ref{thm:main} by considering the case when $\Psi = Id$, $p = 0$, $V = \R^n$, in which case $q = 0$, $W = \R^n$, $T_0\Omega = \OmegaRef$.

\begin{theorem}[Allard-type regularity]\label{thm:main}
Let $\OmegaRef$ be a polyhedral cone domain.  There are constants $\delta(\OmegaRef)$, $c(\OmegaRef)$, $\alpha(\OmegaRef)$ so that the following holds.  Let $\Omega = \Phi(\Psi(\OmegaRef)) \cap B_1 \in \cD_\delta(\OmegaRef)$, and take $M \in \cIH(\Omega, B_1)$.  Assume there is a $V \in \cP_{T_0\Omega}$, $p \in V^\perp$, so that
\begin{equation}\label{eqn:main-hyp1}
E := \int_{B_1} d_{p + V}^2 d\mu_M + ||H||_{L^\infty(B_1; \mu_M)}^2 + |D^2\Phi|_{C^0(B_2)}^2 \leq \delta^2
\end{equation}
and
\begin{equation}\label{eqn:main-hyp2}
\theta_M(0, 1) \leq (3/2) \Theta_{T_0\Omega}, \quad \spt M \cap B_{1/512} \neq \emptyset.
\end{equation}
Then if we set $q = \Phi^{-1}(p)$, $W = D\Phi^{-1}|_p V$, we can find a function $f : (q + W) \cap\Psi(\OmegaRef) \cap B_{1/128}(q) \to W^\perp$ satisfying
\begin{equation}\label{eqn:main-concl1}
|f|_{C^{1,\alpha}} \leq c E^{1/2},
\end{equation}
so that
\begin{equation}\label{eqn:main-concl2}
\Phi^{-1}(\spt M) \cap B_{1/256} \subset \Phi^{-1}(\spt M) \cap B_{1/128}(q) = \graph_{q + W}(f) \cap B_{1/128}(q).
\end{equation}
\end{theorem}

Some comments are in order.
\begin{remark}
Even though $\OmegaRef$ is convex, $\Omega$ need not be.
\end{remark}

\begin{remark}
$\alpha$ can in fact be chosen to be anything in some interval $(0, e - 1)$ (where $e = e(\OmegaRef) > 1$ as in Proposition \ref{prop:expansion} is determined by the Neumann eigenvalue expansion of $\Omega_0^l \cap \del B_1$), provided $\delta$ and $c$ are taken to depend on $\alpha$.  For example, when $\OmegaRef$ is $\R^{n+1}$ or a half-plane, then any $\alpha \in (0, 1)$ is admissible.  When $\OmegaRef = W^2 \times \R^{n-1}$ is a wedge with angle $\gamma < \pi$, then we can take $\alpha \in (0, \min\{1, \pi/\gamma - 1\})$.  See Remark \ref{rem:expansion} for more details.
\end{remark}

\begin{remark}
We state and prove Theorem \ref{thm:main} in codimension-one Euclidean space.  However when $l = 1$ our proof carries over verbatim to higher codimension and ambient manifolds, giving an alternate proof of \cite{GrJo}.  When $l \geq 2$, the proof carries over except for two estimates in Section \ref{sec:first-var}, which continue to hold if one knows a priori that $\spt M$ is contained in some $(n+1)$-dimensional submanifold.  See Section \ref{sec:codim} for details.
\end{remark}

\begin{remark}
If one assumes $\theta_M(0, 1) \leq (1+\delta) \Theta_{T_0\Omega}$ then Theorem \ref{thm:main} holds for varifolds which are only rectifiable, but have a lower density bound $\theta_M \geq 1$ $\mu_M$-a.e.  This requires only minor modifications of the proof. (Specifically, in the contradiction arguments of Proposition \ref{prop:graph} and Lemma \ref{lem:sharp-mass-bound}, the choice of constant $\delta_1$ in Corollary \ref{cor:decay}, and the choice of constants in the induction argument of of Theorem \ref{thm:total-decay}.).  In a similar vein, Theorem \ref{thm:main} also holds if we assume $H_M \in L^p(\mu_M)$, for $p > n$ instead of $p = \infty$.  In this case our constants would depend on $p$ also.
\end{remark}

Our regularity Theorem \ref{thm:main} requires $V$ to be ``horizontal,'' in the sense that if $\Psi(\OmegaRef) = \OmegaRefN \times \R$, then $V \supset \OmegaRefN$.  When $V$ is instead ``vertical,'' in the sense that $V \supset \{0\} \times \R$, then regularity as in Theorem \ref{thm:main} can fail.  Below is an example illustrating this.
\begin{example}\label{ex:vertical-plane}
Let $W \subset \R^2$ be the wedge $\{ r e^{i\theta} : r \in [0, \infty), -\pi/6 \leq \theta \leq \pi/6 \}$, and let $\bY \subset \R^2$ be the cone over $\{ 1, e^{i 2\pi/3}, e^{-i 2\pi/3}\}$, consisting of three rays meeting at the origin at $120^0$.  Then for every $\eps > 0$, the integral varifold $M_\eps$ given by integrating over $(\eps + \bY ) \cap W$ is stationary with free-boundary in $W$.  As $\eps \to 0$, then $M_\eps$ converges as varifolds to the ``vertical'' plane $P = \{ x \geq 0, y = 0 \}$, but none of the $M_\eps$ are $C^1$ perturbations of $P$.

One can construct a similar example by restricted the tetrahedal cone to a $0$-symmetric domain in $\R^3$ consisting of the intersection of $4$ half-spces.
\end{example}
Example \ref{ex:vertical-plane} is a little contrived, but we expect one should be able to construct smooth counterexamples.  However, we would not expect these examples to be minimizing (in the sense of integral currents).  Relatedly, when $\OmegaRef$ is a wedge, or has dihedral angles $\leq \pi/2$, then the vertical planes are not minimizing (Lemma \ref{lem:minz-planes}).  For more general convex $\OmegaRef$, or when $\OmegaRef$ is a non-convex wedge, this may fail (see Examples \ref{ex:non-convex}, \ref{ex:possible-minz-plane}).

When $\Omega$ is non-convex, Theorem \ref{thm:main} can fail also, even when $M$ is in some sense minimizing.
\begin{example}\label{ex:non-convex}
\cite[Theorem 1]{HildebrandtSauvignyminimalIII} implies the following: Suppose $\Omega = W^2 \times \R$, where $W^2$ is a wedge with angle $> \pi$, and $\Gamma$ is any smooth curve in $\sint\Omega \cap \{ x^2 + y^2 = 1\}$, such that $\Gamma$ meets $\del\Omega$ only at its endpoints $\{p_1, p_2\}$, and the height function $x_3|_\Gamma$ has no maxima away from the endpoints.  Let $B_+ = \{ (x, y) \in \R^2 : x^2 + y^2 < 1, y > 0 \}$.  Then there is a map $F \in C^0(\overline{B_+}, \Omega) \cap W^{1,2}(\overline{B_+}, \Omega)$ such that
\begin{enumerate}
	\item $F$ minimizes the Dirichlet energy, and $F(B_+)$ is a smooth minimal surface;
	\item $F$ maps the semi-circle $\del B_+ \cap \{y > 0\}$ monotonically to $\Gamma$, $\pi_{\R^2 \times \{0\}} \circ F$ maps the interval $[-1, 1]$ monotonically into to $\del W$, and $\pi_{\R^2} \circ F$ maps $B_+$ diffeomorphically to $\sint\Omega \cap B_1$;
	\item writing $F(\overline{B_+}) \cap \{0\}\times \R = \{q\}$, then $F^{-1}(q) = [a_1, a_2]$ is an interval of positive length, and $F$ extends smoothly to $\overline{B_+} \setminus \{ -1, 1, a_1, a_2\}$;
	\item on $(-1, 1)\setminus [a_1, a_2]$, $F$ meets $\del\Omega$ orthogonally;
	\item on $(a_1, a_2)$, the unit normal of $F$ is horizontal.
\end{enumerate}

These items imply that $M = [F(\overline{B_+})] \in \cIVT_2(\Omega, \R^3\setminus \Gamma)$, and has zero mean curvature.  By choosing $\Gamma$ to be contained in a very thin slab $\R^2\times [-\eps, \eps]$, the maximum principle implies $F(\overline{B_+})$ is contained in this slab also.  Therefore we can arrange so that $M \llcorner B_{1/2}$ is arbitrarily varifold close to the multiplicity-one horiztonal plane $[\R^2] \llcorner B_{1/2}$, but $\spt M$ will never be graphical over $\R^2$ at $0$.
\end{example}

\subsection{Outline of proof}

Our strategy to prove Theorem \ref{thm:main} is to show the following excess decay estimate: for all $x \in \spt M \cap B_{1/16}$ there is a plane $V_x \in \cP_{T_x\Omega}$ so that
\begin{equation}\label{eqn:outline-1}
\sup_{z \in B_r(x) \cap \spt M} r^{-2} d(z, x + V_x)^2 \leq c(\OmegaRef) r^{2\alpha} E \quad \forall 0 < r < 1/4,
\end{equation}
where $\alpha(\OmegaRef) \in (0, 1)$.  From \eqref{eqn:outline-1} it follows easily that $\spt M \cap B_{1/32}$, and hence $\Phi^{-1}(\spt M) \cap B_{1/64}$, is a $C^{1,\alpha}$ graph with norm controlled by $c E^{1/2}$.

We prove \eqref{eqn:outline-1} in two steps.  In step one (Section \ref{sec:decay}), given any fixed model cone $\OmegaRef$, we prove a decay like \eqref{eqn:outline-1} with $x = 0$.  Loosely speaking we show that $\spt M \cap B_1$ resembles a $W^{1,2}$ harmonic function in $\OmegaRef$ with Neumann boundary conditions.  By understanding these linear solutions (by a Fourier expansion and an eigenvalue estimate), we can prove $C^{1,\alpha}$ decay.  This basic idea goes back to DeGiorgi, who proved interior regularity by a similar ``excess decay'' strategy, and is implemented in a fashion closer to our style in \cite{All}, \cite{Simon1}, \cite{EdSp}.

In step two (Section \ref{sec:reg}), we exploit the polyhedral structure of $\OmegaRef$ to show that for every $x \in B_{1/4}$, and $0 < r < 1/4$, we can find radii $r = r_0^+ \geq r_0^- \geq r_1^+ \geq r_1^- \ldots \geq r_{n+1}^+ \geq r_{n+1}^- = 0$, such that when $s \in [r_i^-, r_i^+]$ then $\Omega \cap B_s(x)$ is modelled on some $T_{z_i}\OmegaRef$, and $r_i^- / r_{i-1}^+ \geq 1/c(\OmegaRef)$.  In other words, outside of finitely many scales (controlled only by $\OmegaRef)$, $\Omega \cap B_s(x)$ is modelled on some polyhedral cone of the form $T_z\OmegaRef$.  Since there are only finitely many tangents $T_z\OmegaRef$, we can therefore inductively prove decay by our first step in each interval $[r_i^-, r_i^+]$, and extend decay from one interval to the next by enlarging our constant $c$ in \eqref{eqn:outline-1} by a controlled amount.

The key technique hurdle in proving both steps is to show that $M$ has controlled first variation $\delta M$.  Our hypotheses imply $\delta M$ is only bounded in directions tangential to $\Omega$, but we need to establish both that $\delta M$ is bounded in all directions, and an a priori trace-like estimate for $||\delta M||$ in terms of $||M||$ (Theorem\ref{thm:first-var}).  This is the main point where we use convexity of $\OmegaRef$.

A priori control on $\delta M$ gives good compactness for sequences of such $M$, and allows us to prove a sharp $L^\infty-L^2$ estimate, and a uniform lower density bound.  The $L^\infty$ estimate is crucial in Step 1, to show that $M$ doesn't ``concentrate'' near $\del\Omega$ at the scale of excess (and therefore can be well approximated by the interior ``graphical'' region).  The lower density bound means that in various (blow-up) limits the varifolds do not disappear, as they might in the non-convex case.

We elaborate on these steps below.  For simplicity, in our outline we will assume $\Omega = \OmegaRef$, and $H = 0$.

\textbf{Step 1: Decay towards a single cone model.}  We wish to prove an excess decay of the following type: if $M$ is sufficiently varifold close in $B_1$ to a horizontal plane $(p + V) \cap \Omega$, then there is a new horizontal plane $p' + V'$ and a $\theta(\Omega)$ so that
\begin{equation}\label{eqn:outline-2}
E(M, p' + V', 0, \theta) \leq \frac{1}{2} E(M, p + V, 0, 1).
\end{equation}
Here $E(M, p + V, x, r) = r^{-n-2} \int_{B_r(x)} d_{p + V}(z)^2 d\mu_M(z)$.  (Proposition \ref{prop:decay} is phrased in terms of the $L^2$ excess, which is a little more convenient, but in this setting the $L^\infty$ and $L^2$ excesses are the same; see \ref{cor:height-w12-bound}).  By iterating \eqref{eqn:outline-2}, we obtain an estimate like
\[
E(M, p'' + V'', 0, r) \leq c(\Omega) r^{2\alpha} E^\infty(M, p + V, 0, 1) \quad \forall d(\spt M, 0) \leq r \leq 1.
\]
This is our main decay estimate in Step 1.

We prove \eqref{eqn:outline-2} by contradiction.  We assume there is a sequence of $M_i$ such that $M_i \to [(p + V) \cap \Omega]$ in $B_1$ as varifolds, such that
\[
\inf_{p' + V'} E(M_i, p' + V', 0, \theta) \geq \frac{1}{2} E(M_i, p + V, 0, 1/2) =: E_i \quad ( \to 0 \text{ as } i \to \infty).
\]
On larger and larger sets $U_i \subset\subset \sint\Omega \cap (p + V)$ (for $U_i \to \sint\Omega \cap (p + V)$) we can write $\spt M_i \cap U_i = \graph_{p + V}(u_i)$, where $u_i$ satisfy the minimal surface equation.  By setting $v_i = E_i^{-1/2} u_i$, then the $v_i$ are locally bounded in $L^\infty$, and after passing to a subsequence we get convergence
\[
v_i \to v,
\]
where the \emph{Jacobi field} $v$ satisfies
\[
\Delta v  = 0, \quad \del_n v = 0, \quad \int_{B_1(0) \cap (p + V) \cap \Omega} v^2 \leq 1.
\]

We now prove two key technical facts: First, by the sharp $L^\infty$ bound \eqref{eqn:sharp-height}, we have strong convergence
\[
E_i^{-1} E(M_i, p + V, 0, r) \to r^{-n-2}\int_{B_r(0)\cap (p + V) \cap \Omega} v^2 \quad \forall 0 < r < 1/4.
\]
Second, the $W^{1,2}$ estimate of \eqref{eqn:W12-bound}, a sharp eigenvalue estimate for convex domains in the sphere (Theorem \ref{thm:evalue}), and a standard Fourier-type decomposition, we can expand $v$ as above like
\[
v = a + b \cdot x + O(r^{1+\alpha}),
\]
where $b$ lies in some direction of translational symmetry of $\Omega$.  In other words, the eigenvalue bound of Theorem \ref{thm:evalue} implies that any free-boundary plane is integrable through rotations.

We can now repeat the blow-up argument with $p_i + V_i = \graph_{p + V}( E_i a + E_i b\cdot x)$ in place of $p + V$, and obtain the Jacobi field
\[
v' = v - a - b\cdot x = O(r^{1+\alpha}).
\]
Hence for $i >> 1$ we have
\[
E_i^{-1} E(M_i, p_i + V_i, 0, r) \leq 2 r^{-n-2} \int_{B_r(0) \cap (p + V) \cap\Omega} v^2 \leq c(\Omega) r^{2\alpha},
\]
which is a contradiction for $r(\Omega)$ small.

This general strategy is very robust, and has been implemented in many other contexts.  However in any given situation there are typically two key technical issues to address: strong convergence in norm of the non-linear problem to the linear problem; and decay of the linear problem.  We handle these in our situation by our sharp $L^\infty$ bound and our sharp eigenvalue estimate.

\textbf{Step 2: Decay across cone models.} We prove the general decay \eqref{eqn:outline-1} by using Step 1 and an inductive argument on the strata.  We first observe (Lemma \ref{lem:B}) that every point in $\Omega$ is locally modelled on some (other) polyhedral cone.  Precisely, there is a $B(\Omega)$ so that if $x \in \del_i\Omega$, and $r \leq d(x,\del_{i-1} \Omega)$, then
\[
B_{B r}(x) \cap \Omega = B_{B r}(x) \cap (x + T_x\Omega).
\]
(When $\Omega \in \cD_\delta(\OmegaRef)$ is only a perturbation of a polyhedral cone a similar statement holds.)

Now given $x \in \del_i\Omega$, we choose points $\tilde x_j \in \del_{i_j} \Omega$ and radii
\[
1 = r^-_{J+1} \geq r^+_J \geq r^-_J \geq \ldots \geq r_0^+ \geq r_0^- = 0,
\]
so that $\Omega \cap B_r(\tilde x_j)$ is modelled on some fixed polyhedral cone $T_{z_j}\OmegaRef$, when $r \in [r_j^-, r_j^+]$, and $r_j^-/r_{j-1}^+ \geq 1/c(\OmegaRef)$.  The degree of symmetry $i_j$ is strictly decreasing as $j$ increases, so $J \leq n$.  We will apply Step 1 to get decay $r_j^+ \to r_j^-$, and then give up a controlled number of scales to ``decay'' $r_j^- \to r_{j-1}^+$.

Some care must be taken in construting these points/radii, since for each $j$ we need that:
\begin{enumerate}
\item $r_j^+ \leq B d(\tilde x_j, \del_{i_j - 1}\Omega)$, so that $\Omega \cap B_{r_j^+}(\tilde x_j)$ looks close to a polyhedral cone;
\item $|x - \tilde x_j| \leq r_j^-$, so that we can apply Step 1 in $B_{r_j^+}(x_j)$ to get decay down to $r_j^-$;
\item $B_{r_j^+}(\tilde x_j) \subset B_{r_{j+1}^{-}/2}(\tilde x_{j+1})$, but $r_j^+ \geq c r_{j+1}^-$, so that we can control the mass and excess in $B_{r_j^+}(\tilde x_j)$ in terms of the mass/excess in $B_{r_{j+1}^{-}}(\tilde x_{j+1})$.
\end{enumerate}

Once we construct these $\tilde x_j$, $r_j^\pm$, we inductively prove a statement of the form: for each $j$, there is a $V_j \in \cP_{T_{\tilde x_j}\Omega}$ so that
\[
E(M, p_j + V_j, \tilde x_j, r) \leq \Lambda_j E r^{2\alpha} \quad \forall r_j^- \leq r \leq 1/4,
\]
and
\[
\theta_M(\tilde x_j, r) \leq (7/4) \Theta_{T_{\tilde x_j}\Omega} \quad \forall r_j^- \leq r \leq r_j^+.
\]
The excess control ensures $M \llcorner B_{r}(\tilde x_j)$ looks like a plane with some multiplicity, the mass control ensures this multiplicity is $\leq 1$, and the fact that $x \in B_{r_j^-/2}(\tilde x_j)$ ensures this multiplicity is $\geq 1$.  The proof of this statement follows in a fairly straightforward way from Lemma \ref{lem:B} and Step 1.  An extra argument by contradiction (Lemma \ref{lem:sharp-mass-bound}) is required to ensure the mass control at scale $B_{r_j^-}(\tilde x_j)$ carries over to scale $B_{r_{j-1}^+}(\tilde x_{j-1})$.

\textbf{Minimizers} Here we apply our regularity theorem to minimizing currents/isoperimetric sets of finite perimeter.  We classify low-dimensional minimizing cones, and prove a compactness theorem, which together with our regularity result implies a partial regularity theorem by standard dimension reducing techniques.

The compactness is fairly straightforward -- we just need to adapt an argument of Gruter \cite{Gr} to ensure mass cannot accumulate near the boundary.  The main step is classifying low-dimensional minimizing cones, which we prove by induction on the number of symmetries of $\Omega$.  The idea is as follows: Assume that any minimizing $T$ in $\Omega_0 \times \R^{m-1}$ is a horizontal plane.  Then if $T$ is minimizing in $\Omega_0\times \R^m$, by induction and our regularity theorem $T$ is a $C^{1,\alpha}$ surface away from $0$.  Under certain circumstances, then we can boost this to $C^{2,\alpha}$ regularity, and thereby adapt Simons' classical argument to prove $T$ must be planar.  Then, again in certain circumstances, a cut-and-paste argument implies this plane must be horizontal.

Unfortunately, even in low dimensions we start running into issues.  The barrier to adapting Simons' argument when $\Omega_0$ is more than $3$ dimensional is the lack of $C^{2,\alpha}$ regularity in $3$-dimensional cones.  We suspect this holds if the dihedral angles are at most $\pi/2$, but this requires a Neumann eigenvalue estimate for spherical domains which is not known.  On the flip side, for general dihedral angles it is not clear that every minimizing plane need be horizontal -- when $\Omega_0$ is only an intersection of $3$ half-spaces in $\R^3$, it is plausible there are non-horizontal minimizers.  Taken together, we get a codimension $2$ bound for the singular set in general domains, a codimension $3$ bound in domains with dihedral angles $\leq \pi/2$, and the (sharp) codimension $7$ bound in certain special cases (e.g when all dihedral angles are $=\pi/2$).

\input{first-var.tex}

\input{evals.tex}

\input{decay.tex}

\input{reg3.tex}

\input{codim.tex}

\input{minz2.tex}

\input{app-first-var.tex}

\bibliographystyle{plain}
\bibliography{refs}

\end{document}


TITLE: Does $\pi_1$ have a right adjoint?
QUESTION [15 upvotes]: Eilenberg and Mac Lane showed that given a group $G$ there exists a pointed topological space $X_G$ such that $\pi(X_G,\bullet)\cong G$. It is obviously a way to "invert direction" to the functor $\pi_1\colon \mathbf{Top}^\bullet\to \mathbf{Grp}$ to a functor $\mathcal K\colon \mathbf{Grp}\to \mathbf{Top}^\bullet$ such that $\pi_1(\mathcal K(G),\bullet)\cong G$ (almost by definition). This is equivalent to say that there exists a natural transformation (equivalence, in this case) between $\pi_1\circ\mathcal K$ and $\mathbf{1}_{\mathbf{Grp}}$, which turns out to resemble some sort of counity.
It would be wonderful if I could define an adjunction between the two categories in exam, given by the two functors. Everytime I try to think about some sort of unity to this hypotetical adjunction I poorly fail: considering the vast literature in the field of algebraic topology, I believe in only two possible cases. The first, nothing interesting arises from this adjunction. The second, there is no sort of adjunction.
The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among unity and counity, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi_1\circ\mathcal K\to \mathbf{1}_{\mathbf{Grp}}$ is an equivalence: can I conclude that it is universal?

REPLY [2 votes]: This question was answered a long time ago, but let me mention a couple of relevant papers which invest on Fosco's intuition and partially confirms it.

Paul C. Kainen, Weak adjoint functors.  P.C. Math Z (1971) 122: 1.

The main point of the paper is the observation that $\mathsf{K}( -, n)$ is a weak  adjoint for $\pi_n$. The paper insists many subtleties, like (non-)preservation of limits and (non-)functoriality of the weak adjoints.  Very soon Kainen changed research topic, but it is worthy to mention also the paper below.

Paul C. Kainen, Universal coefficient theorems for generalized homology and stable cohomotopy, Pacific J. Math. 37 (1971) 397– 407.


TITLE: If a Parseval-frame is orthogonal is it actually a base?
QUESTION [1 upvotes]: Let $A$ be a unital C*-algebra and $\mathcal{E}$ a Hilbert A-module. If $\{m_j\}$ is a finite Parseval frame for $\mathcal{E}$ and $\{m_j\}$ is orthogonal, is it actually a base? I suspect that's not the case because $\langle m_i,m_i \rangle b=0$ doesn't necessarily imply that $b=0$.
Note: $\{m_j\}$ is a Parseval frame if for all $m\in \mathcal{E}$ we have that
$$m=\sum_j m_j\langle m_j ,m \rangle$$

REPLY [0 votes]: I suppose, by "base" you mean a linearly independent set which spans $\mathcal E$ as a complex vector space.
You are right. Let $A$ be any unital C*-algebra. Consider $A$ as the standard right Hilbert $A$-module. Then $\{1_A\}$ is a frame but clearly not a base, if $A$ has complex dimension greater than one.


TITLE: Why is the effect of Coulomb's force the same on more than two charges?
QUESTION [1 upvotes]: As we have developed the idea of electric field lines from Coulomb's law. If two equal and opposite charges are in plane, all the effect of electric field vectors adds to line joining the two points, which means all the electric field lines of forces contribute to Coulomb's force.
When more than two charges are in plane, the electric field lines distribute between charges, which means all electric field lines do not contribute to apply a force between the charges. It should be reduced by this analogy. But we use Coulomb's force the same, whether two charges or more charges. Why? It should be reduced if more than two charges are in plane by distribution of electric field lines.
Simple

How can charge $q_1$ apply the same amount of force, be it one other charge in space or multiple ($n$ charges)?
How is it possible?

REPLY [0 votes]: If I'm reading your question and your comments correctly, you seem to be thinking that if there are two charges $q_1$ and $q_2$, then the force on $q_2$ due to $q_1$ is because of all the field lines coming out of $q_1$. That's not true.
To find the force on $q_2$ due to $q_1$ from the field lines, you first need to find the electric field from the field lines. The direction of the electric field is just tangent to the field line at the point where $q_2$ is located. To find the magnitude of the electric field at the point where $q_2$ is located, you need to follow a procedure. If you want to know why this procedure works, I suggest you read up on Gauss's law. If you have a convention of $n$ lines per coulomb where $n$ is a very big number, then take a very small area element $dA$ centered about the point and count the number of field lines passing through $dA$. The electric field at that point is
$$E = \frac{\text{# of lines passing through } dA}{n \epsilon_0 dA}$$
Then the force on $q_2$ is just $q_2E$. The larger you take $n$ to be, the smaller you are allowed to take $dA$ to be. So really you can take $dA$ to be as small as you like. The takeaway from this is that the electric field strength and hence the force on $q_2$ due to $q_1$ is only dependent on the field lines of $q_1$ very close to $q_2$. All the field lines of $q_1$ are not responsible for the force on $q_2$.


TITLE: Finding UMVUE from Lehmann-Sheffe Directly
QUESTION [0 upvotes]: I am having some trouble with an example from the book I am following.
Let $X_1,X_2,...,X_n$ for $n>2$ be an iid set of $N(\mu,\sigma^2)$ random variables with, $\mu\in\mathbb{R}$ and $\sigma^2\in\mathbb{R}^+$, both unknown. Under this circumtance $T=(\overline{X}_n,S_n^2)=(Y,W)$, the bi-variate statistic containing the sample mean and sample variance, is sufficient and complete for $(\mu,\sigma^2)$, and $\overline{X}_n$ and $\frac{nS_n^2}{\sigma^2}$ are independent, $N(\mu,\sigma^2/n)$ and $\chi^2_{n-1}$ random variables, respectively.
Because of the Lehmann-Scheffe lemma, an unbiased function $h(Y,W)$ for $\mu$ is the almost surely unique UMVUE of $\mu$. Therefore, such a function $h$ must satisfy
\begin{equation}
\int_0^\infty f(w)\left[\int_{-\infty}^\infty h(y,w)f(y)dy\right]dw=\mu\nonumber
\end{equation}
where $f(y)$ and $f(w)$ are the pdf of $Y$ and $W$ respectively.
I have tried several of the techniques like differentiating several times both sides by $\mu$, and using completeness in order to find the condition that leads to
\begin{equation}
h(Y,W)=Y\text{ } a.s. P_{\mu,\sigma^2}, \forall (\mu,\sigma^2)\in\mathbb{R}\times\mathbb{R}^+
\end{equation}
I can not seem to find the conditions required for the last equation to hold.
Any suggestions? is there any general rule that should be applied?
Best regards,
JMJulio

REPLY [1 votes]: It was quite simple. Thanks to all who took the trouble to look at it.
Since the joint cdf of $Y$ and $Z$ is of bounded variation, the derivative wrt $\mu$ of the left hand side of
\begin{equation}
E_{\mu,\sigma^2}\left[h(Y,Z)\right]=\mu
\end{equation}
interchanges with both integrals, leading to
\begin{equation}
E_{\mu,\sigma^2}\left[h(Y,Z)\frac{n}{\sigma^2}(y-\mu)\right]=1
\end{equation}
Therefore,
\begin{equation}
E_{\mu,\sigma^2}\left[h(Y,Z)Y\right]=E_{\mu,\sigma^2}\left[Y^2\right]
\end{equation}
and applying completeness to
\begin{equation}
E_{\mu,\sigma^2}\left[h(Y,Z)Y-Y^2\right]=0
\end{equation}
this leads to $h(Y,Z)Y=Y^2$ a.s. $P_{\mu,\sigma^2}$ $\forall\mu,\sigma^2\in\mathbb{R}\times\mathbb{R}^+$, from where
\begin{equation}
h(Y,Z)=Y \text{ a.s. } P_{\mu,\sigma^2} \forall\mu,\sigma^2\in\mathbb{R}\times\mathbb{R}^+
\end{equation}


text:
resembles residual blocks in ResNets \cite{chen2018neural} establishes the simple, yet profound connection between the worlds of deep learning and differential equations \cite{haber2017stable}. The evolution of an initial condition $\mathbf{x}_0 \in {\mathbb{R}^n}$ from $t_0$ to $t$ is given by the integral expression,
$$
\mathbf{x}_t(\theta) = \mathbf{x}_0 + \int_{t_0}^t f(\mathbf{x}(s), s, \theta) ds . \label{eq:IVP_ODE}
$$ The corresponding flow operator defined by, $$\phi_t(\mathbf{x}_0; \theta) = \mathbf{x}_t(\theta),$$ is a parametric map ${\mathbb{R}^d} \mapsto {\mathbb{R}^d}$. As such, it provides a hypothesis space for function estimation in machine learning, and may be viewed as the continuous limit of ResNet-like architectures \cite{he2016deep}.

Reversible deep neural architecture schemes enable activations to be re-derived from those of a subsequent layer, removing the need to store them in memory \cite{gomez2017reversible}. For a Neural ODE, by construction a reversible map, loss function gradients can be computed via the adjoint sensitivity method with constant memory cost independent of depth. This decoupling of depth and memory has major implications for applications involving large inputs such as video, graphs, or 3D data \cite{davis2020time}.

\section{Stable Nonautonomous Neural ODE systems}

Consider an ordinary differential equation (ODE) defined as:
\begin{equation}
\label{ode}
\frac{d\mathbf{x}(t)}{dt} = f_{\theta}(\mathbf{x}_{t},t),
\end{equation}
with initial condition $\mathbf{x}(0) = \mathbf{x}_{0} \in \mathbb{R}^{d}$ and where $f_{\theta}:\mathbb{R}^{d} \times \mathbb{R} \rightarrow \mathbb{R}^{d}$ for $\theta \in \mathbb{R}^{l}$. Flows that are solutions to ODEs from Equation \ref{ode} can naturally be regarded as continuous generalizations of discrete \mathrm{ResNet} architectures of the form:
\begin{equation}
\mathbf{x}_{i+1} = \mathbf{x}_{i} + \sigma(\mathbf{W}_{i}\mathbf{x}_{i} + \mathbf{b}_{i}),
\end{equation}
for matrices $\mathbf{W}_{1},...,\mathbf{W}_{L} \in \mathbb{R}^{d \times d}$, bias vectors $\mathbf{b}_{1},...,\mathbf{b}_{L} \in \mathbb{R}^{d}$ and function $\sigma: \mathbb{R} \rightarrow \mathbb{R}$.

Considering \mathrb{ResNet}s through the lens of ODEs reveals several advantages, including: \textbf{(1)} it provides means to define deep $\mathrm{ResNet}$ architectures encoded by a compact sets of parameters (independent of the number of hidden layers $L$); \textbf{(2)} it gives rise to several new variants of computational schemes corresponding to different integrators for flows that are solutions of Equation \ref{ode}.

Standard neural ODEs, where $f_{\theta}$ is a neural network parameterized only by $\theta$, and not taking as input time $t$, correspond to autonomous systems with very limited expressiveness. Appending time $t$ as an additional input dimension helps, but the capacity of the system is still limited.
Another approach is to consider computational systems defined by ODEs of the form :
\begin{equation}
\label{ode_sigma}
\frac{d\mathbf{x}(t)}{dt} = \sigma(\mathbf{W}^{\theta}(t)\mathbf{x}_{t} + \mathbf{b}(t)),
\end{equation}
where $\mathbf{W}^{\theta}(t):\mathbb{R} \rightarrow \mathbb{R}^{d \times d}$ is a matrix-valued function parameterized by $\theta \in \mathbb{R}^{l}$. Several classes of functions $\mathbf{W}^{\theta}(t)$ have been proposed such as those based on  trigonometric polynomials:
\begin{equation}
\mathtbf{W}^{\theta}(t)[i,j] = \sum_{k=0}^{d} a^{i,j}_{k}\sin^{k}(t) +
\sum_{k=0}^{d} b^{i,j}_{k}\cos^{k}(t),
\end{equation}
where $\theta=[a^{i,j}_{k},b^{i,j}_{k}]^{i,j \in \{0,...,d-1\}}_{k=0,1,...,d}$ and $d$ is a hyperparameter controlling the capacity of the model.

Neural ODEs are deep computational systems and thus can suffer from the gradient vanishing/explosion problem. In particular, all previously described neural ODEs are characterized by vanishing/exploding gradients and in turn, slow training.

In this work we propose nonautonomous neural ODE systems with \textbf{provably} stable gradients.
To do that, we encode functions $W^{\theta}(t)$ via matrix-flows on the orthogonal group $\mathcal{O}(d)$, where $\theta \in \mathbb{R}^{l}$ are the parameters of the flow.

From the computational point of view, inference in such systems involves  computing simultaneously two flows: matrix-flow $\mathcal{F}_{\mathrm{matrix}}$ on $\mathcal{O}(d)$ and the main flow $\mathcal{F}_{\mathrm{main}}$ on $\mathbb{R}^{d}$ defined by $\mathcal{F}_{\mathrm{matrix}}$.

\subsection{Designing Stable Neural ODEs}

To design Neural ODEs with stable gradients, we first discretize Equation \ref{ode_sigma} using the standard Euler's formula to get:
\begin{equation}
\mathbf{x}_{\frac{i+1}{N}} = \mathbf{x}_{\frac{i}{N}} + \frac{1}{N} \sigma \left(\mathbf{W}^{\theta} \left(\frac{i}{N} \right)\mathbf{x}_{\frac{i}{N}}+\mathbf{b}_{\frac{i}{N}} \right),
\end{equation}

where $i=0,1,.,,,N-1$ and $N \in \mathbb{N}_{+}$ is fixed. We assume that integration is conducted on the interval $[0,1]$.
Denote: $\mathbf{a}^{N}_{i}=\mathbf{x}_{\frac{i}{N}}$, $\mathbf{c}^{N}_{i}=\mathbf{b}_{\frac{i}{N}}$, $\mathbf{V}^{N}_{i}=\mathbf{W}^{\theta}(\frac{i}{N})$ and
$\mathbf{z}^{N}_{i+1} = \mathbf{V}^{N,\theta}_{i+1}\mathbf{a}^{N}_{i}+\mathbf{c}^{N}_{i+1}$.
We obtain the following discrete dynamical system:
\begin{equation}
\label{recurs}
\mathbf{a}^{N}_{i+1} = \mathbf{a}^{N}_{i} + \frac{1}{N}\sigma \left(\mathbf{z}^{N}_{i+1} \right) ,
\end{equation}
for $i=0,1,...,N-1$.

Let $\mathcal{L} = \mathcal{L}(\mathbf{x}_{1})=\mathcal{L}(\mathbf{a}^{N}_{N})$ by the loss function.
Our first goal is to compute $\frac{\partial L}{\partial \mathbf{a}^{N}_{i}}$ for $i=0,...,N-1$.

Using Equation \ref{recurs}, we get:
\begin{align}
\begin{split}
\frac{\partial L}{\partial \mathbf{a}^{N}_{i}} =
\frac{\partial L}{\partial \mathbf{a}^{N}_{i+1}}
\frac{\partial \mathbf{a}^{N}_{i+1}}{\partial \mathbf{a}^{N}_{i}}=
\frac{\partial L}{\partial \mathbf{a}^{N}_{i+1}}(\mathbf{I}_{d}+\frac{1}{N}\mathrm{diag}(\sigma^{\prime}(\mathbf{z}^{N}_{i+1}))\frac{\partial \mathbf{z}^{N}_{i+1}}{\partial \mathbf{a}^{N}_{i}}) = \\
\frac{\partial L}{\partial \mathbf{a}^{N}_{i+1}}(\mathbf{I}_{d}+\frac{1}{N}\mathrm{diag}(\sigma^{\prime}(\mathbf{z}^{N}_{i+1})) \mathbf{V}^{N,\theta}_{i+1})
\end{split}
\end{align}

Therefore we conclude that:
\begin{equation}
\frac{\partial L}{\partial \mathbf{a}^{N}_{i}} = \frac{\partial L}{\partial \mathbf{a}^{N}_{N}}
\prod_{l=i+1}^{N}
(\mathbf{I}_{d}+\frac{1}{N}\mathrm{diag}(\sigma^{\prime}(\mathbf{z}^{N}_{l})) \mathbf{V}^{N,\theta}_{l})
\end{equation}

{\color{blue} Valerii: "l" is already used as a dimension of $\theta$}

{\color{blue} Valerii: Either the order of multiplication should be reversed (if $\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}$ is "horizontal"), or $\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}$ should be on the right (if $\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}}$ is "vertical").}

Note that if $\sigma(x) = |x|$ then
$\mathrm{diag}(\sigma^{\prime}(\mathbf{z}^{N}_{l}))$ is a diagonal matrix with nonzero entries taken from $\{-1, +1\}$, in particular $D \in \mathcal{O}(d)$, where $\mathcal{O}(d)$ stands for the \textit{orthogonal group}.

Define: $\mathbf{G}^{N}_{l} = \mathrm{diag}(\sigma^{\prime}(\mathbf{z}^{N}_{l}))\mathbf{V}^{N,\theta}_{l}$.
Thus we have:
\begin{equation}
\frac{\partial L}{\partial \mathbf{a}^{N}_{i}} = \frac{\partial L}{\partial \mathbf{a}^{N}_{N}}
\prod_{l=i+1}^{N}
(\mathbf{I}_{d}+\frac{1}{N}\mathbf{G}^{N}_{l})
\end{equation}

\begin{lemma}[Stable Gradients Lemma]
If $\mathbf{G}^{N}_{l} \in \mathcal{O}(d)$ for every $N,l$ then for any $\epsilon>0$
the following holds:
\begin{equation}
(\frac{1}{e} - \epsilon)\|\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}\|_{2} \leq \|\frac{\partial L}{\partial \mathbf{a}^{N}_{i}}\|_{2} \leq e
\|\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}\|_{2}
\end{equation}
for $i=0,1,2,...,N-1$, where the first inequality is true for $N$ large enough ($N \geq N(\epsilon)$).
\end{lemma}

\begin{proof}
We have:
\begin{align}
\begin{split}
\prod_{l=i+1}^{N}
(\mathbf{I}_{d}+\frac{1}{N}\mathbf{G}^{N}_{l}) =
\sum_{\{l_{1},...,l_{k}\} \subseteq \{i+1,...,N\}} \frac{1}{N^{k}}  \mathbf{G}^{N}_{l_{1}} \cdot ... \cdot \mathbf{G}^{N}_{l_{k}}
\end{split}
\end{align}

Therefore we can conclude that:
\begin{align}
\begin{split}
\|\prod_{l=i+1}^{N}
(\mathbf{I}_{d}+\frac{1}{N}\mathbf{G}^{N}_{l})\|_{2} \leq
\sum_{\{l_{1},...,l_{k}\} \subseteq \{i+1,...,N\}} \frac{1}{N^{k}}  \|\mathbf{G}^{N}_{l_{1}} \cdot ... \cdot \mathbf{G}^{N}_{l_{k}}\|_{2} \\  = \sum_{k=0}^{N-i} \frac{1}{N^{k}}{N-i \choose k} = \sum_{k=0}^{N-i}\frac{(N-i-k+1)\cdot...\cdot (N-i)}{N^{k}k!} \leq \sum_{k=0}^{N-i} \frac{1}{k!} \leq e,
\end{split}
\end{align}
where we used the fact that $\mathcal{O}(d)$ is a group under matrix-multiplication and $\|\mathbf{G}\|_{2} = 1$ for every $\mathbf{G} \in \mathcal{O}(d)$.
That proves inequality: $\|\frac{\partial L}{\partial \mathbf{a}^{N}_{i}}\|_{2} \leq e
\|\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}\|_{2}$.

To prove that: $\|\frac{\partial L}{\partial \mathbf{a}^{N}_{i}}\|_{2} \geq (\frac{1}{e}-\epsilon)
\|\frac{\partial L}{\partial \mathbf{a}^{N}_{N}}\|_{2}$, for large enough $N \geq N(\epsilon)$, it suffices to observe that if $\mathbf{G} \in \mathcal{O}(d)$, then for any $\mathbf{v} \in \mathbb{R}^{d}$ we have (by triangle inequality):
\begin{equation}
\|(\mathbf{I}_{d} + \frac{1}{N}\mathbf{G})\mathbf{v}\|_{2} \geq
\|\mathbf{v}\|_{2} - \|\frac{1}{N}\mathbf{Gv}\|_{2}=(1-\frac{1}{N})\|\mathbf{v}\|_{2}.
\end{equation}

The inequality then follows immediately from the fact that sequence: $\{a_{N}:N=1,2,3,...\}$ defined as: $a_{N}=(1-\frac{1}{N})^{N}$ has limit $e^{-1}$.

\end{proof}


TITLE: Nth derivative proof
QUESTION [1 upvotes]: I am given the function f(x) = $\sqrt{3x+5}$
I have calculated the expression of the nth derivative to be
$f^{(n)}(x)=3^n\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2n-3}{2}\right)(3x+5)^{-(2n-1)/2}$
How would I prove this expression to be true by induction?

REPLY [0 votes]: The base case is $n = 1$, which is true by
$f'(x) = \frac 32 (3x+5)^{-1/2} = 3^1\frac 12(3x+5)^{-(2-1)/2}$
Now differentiate $f^{(n)}(x)$, and if the resulting expression is $f^{(n+1)}(x)=3^{(n+1)}\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2(n+1)-3}{2}\right)(3x+5)^{-(2(n+1)-1)/2}$
Then you've proven that the expression is true, by induction.


TITLE: Cubic equation with two complex roots and one real root?
QUESTION [2 upvotes]: Find the real root of the equation $z^3 + z + 10 = 0$ given that one complex root is $1 – 2i$.
I've realized that the roots are $(1-2i), (1+2i)$, and a real number we'll call $a$.
So using the theorem got me $(z-1-2i)(z-1+2i)(z-x)$.
No idea on where to go next.

REPLY [5 votes]: The polynomial is monic (lead coefficient $1$). The coefficient of $z^2$ is therefore the negative of the sum of the roots. This coefficient is $0$.
The two known roots have sum $2$, so the missing root must be $-2$.


TITLE: To prove that $ \bigcup_{i \in I} \wp(A_i) \subseteq \wp (\bigcup_{i \in I} A_i ) $
QUESTION [0 upvotes]: To prove that $ \bigcup_{i \in I} \wp(A_i) \subseteq \wp (\bigcup_{i \in I} A_i ) $
My Attempt   :
Let $ A \in  \bigcup_{i \in I} \wp(A_i) $.
So $\exists i \in I (A \in \wp (A_i))$
i.e $\exists i \in I (A \subseteq (A_i))$
Let $ y \in A$ be arbitrary element of A
$\exists i \in I (y \in A_i)$
$ y \in \bigcup_{i \in I} A_i$
So $ A \subseteq \bigcup_{i \in I} A_i $
i.e $ A \in \wp (\bigcup_{i \in I} A_i) $
Is this correct ? Thank you very much
PS $ \wp $ denotes power set

REPLY [2 votes]: It is correct, but the way it is written is a bit clumsy. It is better to not use too much quantifiers. Here is how I would write the proof:
Let $A \in \bigcup_{i \in I} P(A_i)$. Then there is $k \in I$ with $A \in P(A_k)$, which means $A \subseteq A_k$. But since $A_k \subseteq \bigcup_{i \in I} A_i$, we have $A \subseteq  \bigcup_{i \in I} A_i$ and thus $A \in P(\bigcup_{i \in I} A_i)$.


TITLE: Prove the functions are unique in a volume, vector calculus problem
QUESTION [5 upvotes]: I am working through the following problem, but finding it hard to know where to go.
Using the Divergence theorem and the following identities
$\nabla \cdot (A \times B) = B\cdot(\nabla \times A) - A\cdot(\nabla \times B)$
$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla ^2 A$
In a volume V, enclosed by a surface S, the vector fields X and Y satisfy the coupled equations
$\nabla \times \nabla \times X = X+Y$
$\nabla \times \nabla \times Y = Y-X$
If the values of $\nabla \times X$ and $\nabla \times Y$ are given on S, show that X and Y are unique in V.
I am assuming that I need to show that $\nabla ^2 X$ and $\nabla ^2 Y$ are equal to zero and that X and Y are zero on S to satisfy the uniqueness theorem for Poisson's equation. But am unsure of a good way to get there, so before I write my scribbles if someone could point me in the write direction it would be great.
Any help, pointing in the right direction would be very helpful.
EDIT: Fixed the second expression, original didn't make sense.

REPLY [0 votes]: With this I am not even sure what the question is
but here is the answer I give to prove $X \neq Y$:
Sum the two equations
$$2Y = \nabla \times \nabla \times X + \nabla \times \nabla \times Y$$
Subtract the two equations
$$2X = \nabla \times \nabla \times X - \nabla \times \nabla \times Y$$
Using $\nabla \times \nabla \times A = \nabla(\nabla \cdot A) - \nabla^2 A$
\begin{align}
2Y &= \nabla(\nabla \cdot X) − \nabla ^2 X + \nabla(\nabla \cdot Y) − \nabla ^2 Y\\
2Y &= \nabla\big(\nabla \cdot(X+Y)\big) − \nabla ^2 (X+Y)
\end{align}
and then resub $\nabla \times \nabla \times X = X+Y$
$$
2Y = \nabla\big(\nabla \cdot (\nabla \times \nabla \times X)\big)− \nabla ^2 (\nabla \times \nabla \times X)
$$
similarly
$$
-2X = \nabla\big(\nabla \cdot (\nabla \times \nabla \times Y)\big) − \nabla ^2 (\nabla \times \nabla \times Y)
$$
Therefore if $X=Y$
$$\nabla \times X = -\nabla \times Y$$
which implys $Y=0$ and $X=0$.
Therefore $X = Y$ only in trivial case.


TITLE: How to extend rational parametrization of the circle to three dimensions?
QUESTION [1 upvotes]: I recently became aware of the rational parametrization of the circle in two dimensions:
$$\left(\frac{1-m^2}{1+m^2}, \frac{2m}{1+m^2}\right)$$
for a unit circle centered on the origin.
I'm interested in extending this to an arbitrary circle in three dimensions, in particular given the center, normal and radius.
I'd like the result to still be a rational parametrization.

REPLY [2 votes]: Pick two unit vectors $\mathbf u$ and $\mathbf v$ so that together with the specified normal they form an orthonormal frame. The parametrization you want is $$\mathbf c + \frac{1-m^2}{1+m^2}r\mathbf u+\frac{2m}{1+m^2}r\mathbf v,$$
where $\mathbf c$ and $r$ are the center and radius of the circle. What we're doing is simply taking the $xy$ plane that the original circle lies in, and mapping the origin to $\mathbf c$ and the $x$ and $y$ axes to $r\mathbf u$ and $r\mathbf v$.


TITLE: Finding the vertical and horizontal asymptotes of the function $f(x) = \frac{e^x(x + 1)}{e^{2x}(x^2 - 1)}$
QUESTION [1 upvotes]: I am to find the vertical and horizontal asymptotes of this given function:
$$f(x) = \frac{e^x(x + 1)}{e^{2x}(x^2 - 1)}$$
To find the vertical asymptote,I think I equate the bottom line to zero and whatever my $x$ gives is the vertical asymptote?
But I don't know how to solve $e^{2x}(x^2-1) = 0$. Is it going to be $x = +1,-1$?  Hence, the vertical asymptotes are $1$ and $-1$.
To find horizontal asymptotes, I think I should find the limit as $x \to \infty$ and $x \to -\infty$.
I don't know how to go about this.
Can someone please help?

REPLY [1 votes]: You can just calculate the value of $y$ at $x\to-\infty \;\text{and} \; x
\to\infty$
$$\lim_{x\to-\infty}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=-\infty$$
$$\lim_{x\to\infty}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=0$$
As limit exists for $x\to\infty$,
$$y=0$$
is the horizontal asymptote.
For vertical asymptote, you must find values of $x$ for which $y\to\infty$.
$$y=f(x)=\frac{e^x(x+1)}{e^{2x}(x^2-1)}$$
As denominator goes to $0$ for $x=\pm1$, we will check limit of $y$ at these values.
$$\lim_{x\to-1}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=-\frac{e}{2}$$
$$\lim_{x\to1}\frac{e^x(x+1)}{e^{2x}(x^2-1)}=\infty$$
Since $y
\to\infty$ for $x=1$, $$x=1$$ is the vertical asymptote.
$x=-1$, is not the vertical asymptote as limit of $y$ is defined at this value.


TITLE: Is there a geometric intepretation of the trace of tensor on a Riemannian manifold?
QUESTION [2 upvotes]: For a long time I thought the trace for matrices were just an elementary function with nice properties. But it is much more than that and really should be think of as a geometrical object (see Geometric interpretation of trace). For example considering the trace of a square matrix $M$ associated to a application  $\phi_M \colon E\rightarrow F$ with $\dim E= \dim F = n$ doesn't make sense if $E\neq F$ and one should use the Trace only for endomorphism.
Now we consider tensors on a Riemannian manifold. There is the notion of "partial trace" : for a tensor $ T^{j_1\cdots j_n}_{k_1,\cdots k_m}$ we write
$$"\text{Trace}(T)"=T^{j_1\cdots j_{n-1}}_{k_1,k_{m-1}} :=\sum_i T^{j_1\cdots j_{n-1},i}_{k_1,k_{m-1},i}$$ when the trace has been made on the $n$ and $m$ coordinates. But now, as before, I understand the definition and I can use it to do computations but I just have no insight why this is a interesting object and what is the geometry behind it. How do I know I am not writting something that just doesn't make sense ?
An example of particular interest (and also fundamental in physics for general relativity) is the Riemann curvature tensor / Ricci tensor / scalar curvature.
$$ R^i_{jkl}\rightarrow R_{jl}\rightarrow R$$
My geometric intepretation of the Rieman curvature tensor is with parallele transport on small loops. I also know about the scalar curvature and the volume of the sphere. But I really can't see why one should be the trace of the other or in physics why the massive particles (which have no idea about tensors) should use the trace for the law of gravitation...

REPLY [2 votes]: Quite generally, you implement the partial trace operation if you wish to discard local information in a coordinate-invariant way. On a manifold, the Ricci tensor gives you the average value of the sectional curvature along a particular direction, and its trace discards the directional information by averaging the curvature over the different directions.
A different context, which you may find instructive, appears in quantum physics, where the state of a system is described by a positive-definite matrix called the density matrix $\rho$. If the system is composed of two subsystems $A$ and $B$ and you wish to discard information about subsystem $B$ (for example, because it is far removed and you do not have access to it), then the partial trace $\rho_A=\sum_{i\in B}\rho_{i,i}$ describes the state of the system $A$ by itself.


TITLE: Why can't we use Gauss' law on a rod of length $2L$?
QUESTION [1 upvotes]: Suppose we have a rod of length $2L$, and we want to compute the electric field in at P, which is on the perpendicular bisector of the rod, as shown below:

Why isn't Gauss' law applicable here? My professor stated that this is because there is electric field at the rod's ends, and also brought up a an example of a infinite length rod, with a cylinder placed around it and said that Gauss' law was applicable there because we didn't have any electric field parallel to our surface. However none of these examples helped me. Can anyone elaborate on them, or give me another example?

REPLY [1 votes]: in the first look, I thought there may be a chance to use Gauss's theorem for there is a region near the point P, the electric field is in the $z$ direction. What if we make Gaussian surface, the tin-can so shallow, as shown in the following figure:

Applying Gauss's theorem to this surface
$$ \tag{1}
   2\pi r \Delta x E(r) = \frac{\lambda \Delta x}{\epsilon_o};\\
E(r) = \frac{\lambda}{2\pi \epsilon_o r}
$$
Where $\lambda$ is the linear charge density.
Now we will compute the electric field by direct integration to check the result.
$$
E(r) = \frac{\lambda}{4\pi \epsilon_o}
\int_{-L}^L \frac{r dx}{\left( r^2 + x^2 \right)^{3/2} }.
$$
Change variable to $\tan \theta = \frac{x}{r}$, we can carry out the integration:
$$ \tag{2}
E(r) = \frac{\lambda}{2\pi \epsilon_o r} \frac{L}{\sqrt{L^2 + r^2}}\approx \frac{\lambda}{2\pi \epsilon_o r} \left( 1 - \frac{r^2}{2 L^2}\right).
$$
Comparing Eq.(1) and Eq.(2), we conclude that a vicinity of point P having electric field in the outward $r$ direction is not good enough to apply Gauss's theorem in the finite length of line charge. Its validity requires additional condition, either $L\to\infty$, or the point P is very close to the line charge, $r \ll L$.


TITLE: Hilbert space of two particles of spins $j_1$ and $j_2$, noninteracting versus interacting
QUESTION [2 upvotes]: For a system of two noninteracting particles of spins $j_1$ and $j_2$, the joint Hilbert space $\mathcal{V}$ is the tensor product of the individual Hilbert spaces $\mathcal{V}_1$ and $\mathcal{V}_2$. Notationally, $\mathcal{V}=\mathcal{V}_1\otimes\mathcal{V}_2$ which is spanned by the $(2j_1+1)(2j_2+1)$ product states or their linear combinations.

What is the Hilbert space when there is an interaction between the particles? Do the product states still serve as a basis which spans the space?

REPLY [0 votes]: There are two “naturally” useful bases for the same Hilbert space.  The space ${\cal V}_1\otimes {\cal V}_2$ is reducible as a direct sum of $SU(2)$-invariant subspaces.
The first basis is the uncoupled basis with $\vert j_1m_1\rangle \vert j_2, m_2\rangle$ as basis states.
The second is the coupled basis $\vert JM\rangle$, with $j_1+j_2\ge J\ge \vert j_1-j_2\vert$, where states are expressed as linear combinations
$$
\vert JM\rangle=\sum_{m_1m_2} C^{JM}_{j_1 m_1;j_2 m_2}
\vert j_1m_1\rangle \vert j_2m_2\rangle\, .
$$
Since states $\vert JM\rangle$ have “good angular momentum” quantum numbers $J$, one would expect that this basis is useful whenever you need to refer to the total angular momentum of the system.  In this basis, the actual momentum operator act within each $SU(2)$-invariant subspace.
Contrariwise the uncoupled basis would be useful whenever single-particle properties are best to described the system.


TITLE: If I know Double negation law, De Morgan laws and implication negation is it enough for negating any compound propositions or there are more?
QUESTION [1 upvotes]: I know:

Double negation law:
$ \lnot(\lnot p)\equiv p $

De Morgan laws:
$ \lnot (p \wedge q) \equiv \lnot p\vee \lnot q $
$ \lnot (p \vee q) \equiv \lnot p \wedge \lnot q $

Implication negation:
$\lnot (p \Rightarrow q) \equiv p \wedge \lnot q$

Question: Is it enough for negating any compound propositions or there are more law/rules?

REPLY [2 votes]: If by "negate" you mean "put in negation normal form" - the question is trivial otherwise, since "$\neg\varphi$" is always the negation of $\varphi$ - then the answer is yes.
For example, to simplify $\neg(a\wedge (b\vee c))$ we would proceed as follows:

$\neg(a\wedge P)\equiv\neg a\vee\neg P$.
Taking $P$ to be $b\vee c$ above, we get $\neg P\equiv \neg b\wedge \neg c$.
Putting these together we get $\neg(a\wedge(b\vee c))\equiv \neg a\vee (\neg b\wedge\neg c)$.

Of course, this example doesn't constitute a proof; to actually prove the result you need to argue via induction on formula complexity.


TITLE: Solution $x^{73}+x^{48}+x^{26} \equiv 0$ mod $13$
QUESTION [1 upvotes]: I am trying to solve $x^{73}+x^{48}+x^{26} \equiv 0$ mod $13$. Can I say that By Fermat’s Little Theorem I have $x^{12} \equiv 1$ mod $13$? If I am correct remaining part will be simple but otherwise I have no idea how to deal with it.

REPLY [4 votes]: Using Euler's theorem (assuming $x$ is a positive natural number less than $13$ :-)(as the comments below point out $x\equiv 0\mod13\;$ is a solution)
$$
x^{a}\equiv x^{a\mod \phi(13)}\equiv x^{a\mod 12}\mod 13
$$
Therefore our equation turns into
$$
x+1+x^2\equiv 0\mod 13
$$
A little experimentation leads to the solution $x=3$ and this leads to the factorization
$$x^2+x+1=(x-9)(x-3)\mod13$$


TITLE: Let $R$ be a ring. Define a circle composition ◦ in R by $a ◦ b =a+b-ab$, $a, b ∈ R$.
QUESTION [1 upvotes]: For (b) the neutral element is $0$ since $a◦0 = a + 0-a0 = 0$ and $0◦a = 0 + a-0a = 0$, then element is invertible for all a in S must be $a◦ 1 = a + 1-a1 = 1$ and $1◦a = 1 + a-1a = 1$. I have not been able to prove the closure law of ◦, could someone help me?
For Part (C), let $\phi :U(R)\rightarrow S$. $\phi$ is given by $a \mapsto a+x-ax$ ?

REPLY [10 votes]: Hint $\ $ The calculations become simpler and more intuitive if you employ the following observation. Any set bijection $\,h\,:\,R'\to R\,$ serves to transport the ring structure of $\,(R,+,*,0,1),$ to $\,(R',\oplus,\otimes,0',1') \,$ by defining the operations in $\,R'\,$ to make $\,h\,$ be a ring isomorphism $$\begin{align} &a \oplus b\, =\ h^{-1}(h(a) + h(b)),\quad 0' = h^{-1}(0)\\
 &a \otimes b\, =\, h^{-1}(h(a)\, * h(b)),\quad\, 1' = h^{-1}(1)\end{align}\qquad \qquad$$
Yours is special case $\ h(x) = 1\!-\!x\ $ so $\,\ a\otimes b = 1-(1\!-\!a)(1\!-\!b) = a+b-ab$