TITLE: Proof systems and their hierarchy
QUESTION [4 upvotes]: Why ZFC is placed in top of the proof system hierarchy? How it can p-simulate other systems?

REPLY [6 votes]: It is an open problem if there is an optimal propositional proof system. Therefore we don't know if ZFC as a propositional proof system is optimal either.
ZFC as propositional proof system can p-simulate any propositional proof system whose soundness (if there is proof for a formula then the formula is true) as a propositional proof system is provable in ZFC. The trick (which I think is due to Steve Cook) is based on the fact that proofs are concrete finite objects, if there is a proof for a formula in a proof system, then ZFC can prove its existence (i.e. ZFC is $\Sigma_1$-complete), combining this with the provability of the soundness we derive the truth of the encoded formula in ZFC. The rest of the argument is translating this first order proof in ZFC to a propositional proof in ZFC as propositional proof system and proving the equivalence of a propositional formula with the formula itself in the proof system. This can be done in $\mathsf{TC^0}$-Frege and any system that contains it. See Logical Foundations of Proof Complexity, 2010 by Cook and Nguyen for the details.


TITLE: Taylor series expansion of the cotangent function and denominators of the Riemann zeta function at positive even integers
QUESTION [4 upvotes]: Consider the Taylor series expansion of the cotangent function:
$$
\frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} - \frac{2 x^5}{945} - \frac{x^7}{4725} - \frac{2 x^9}{93555} - \frac{1382 x^{11}}{638512875} - \frac{4 x^{13}}{18243225} + O(x^{15}).
$$
Now let $\zeta(s)$ represent the Riemann zeta function defined for positive integer $s>1$ as usual by:
$$
\zeta(s) = \sum_{n=1}^{ \infty} \frac{1}{n^s}.
$$
At positive even integers $2s$ we have the identity:
$$
{\displaystyle A_{n}\zeta (2n)=B_{n}\pi ^{2n}\,\!},
$$
where the sequence $A_n$ ($A002432$ in the OEIS) represents the denominators in $\zeta(2s)$. Starting from $s=3$ these begin with:

$\zeta (6)=1+{\frac  {1}{2^{6}}}+{\frac  {1}{3^{6}}}+\cdots ={\frac  {\pi ^{6}}{945}}$
$\zeta (8)=1+{\frac  {1}{2^{8}}}+{\frac  {1}{3^{8}}}+\cdots ={\frac  {\pi ^{8}}{9450}}$
$\zeta (10)=1+{\frac  {1}{2^{{10}}}}+{\frac  {1}{3^{{10}}}}+\cdots ={\frac  {\pi ^{{10}}}{93555}}$
$\zeta (12)=1+{\frac  {1}{2^{{12}}}}+{\frac  {1}{3^{{12}}}}+\cdots ={\frac  {691\pi ^{{12}}}{638512875}}$
$\zeta (14)=1+{\frac  {1}{2^{{14}}}}+{\frac  {1}{3^{{14}}}}+\cdots ={\frac  {2\pi ^{{14}}}{18243225}}$

Note that the denominators of the Taylor series expansion above almost coincide (except for the first several terms) with the values of $A_n$. Below is the sequence of $A_n$, in which the numbers that appear as denominators in the Taylor expansion of $\cot x$ are underlined:
$$
6, 90, \underline{945}, 9450, \underline{93555}, \underline{638512875}, \underline{18243225}, 325641566250, \underline{38979295480125}, \underline{1531329465290625}, \underline{13447856940643125}, \underline{201919571963756521875}, \underline{11094481976030578125}, \underline{564653660170076273671875}, \underline{5660878804669082674070015625}, \ldots
$$
What explains this? Is this just a conincidence?

REPLY [1 votes]: The Taylor series for the cotangent function is
$$\cot(x)=\sum_{n=0}^\infty \frac{(-1)^n 2^{2n}B_{2n}}{(2n)!}\,x^{2n-1}$$
whereas we have for $n\ge 1$
$$\zeta(2n)=\frac{(-1)^{n+1}2^{2n}B_{2n}}{2(2n)!}\,\pi^{2n}$$
So, we see that the coefficients $a_n$ of the Taylor series for the cotangent function can be written as
$$a_n=-2\frac{\zeta(2n)}{\pi^{2n}}\tag1$$
for $n\ge1$.

Note that we can recover the partial fraction expansion representation of the cotangent function from the Taylor series with coefficients $a_n$ as given in $(1)$.  Proceeding, we have
$$\begin{align}
\cot(x)&=\frac1x+\sum_{n=1}^\infty a_nx^{2n-1}\\\\
&=\frac1x-2\sum_{n=1}^\infty \frac{\zeta(2n)}{\pi^{2n}}\,x^{2n-1}\\\\
&=\frac1x-\frac2x\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^{2n}}\left(\frac x\pi\right)^{2n}\\\\
&=\frac1x-\frac2x \sum_{k=1}^\infty \sum_{n=1}^\infty  \left(\frac {x}{k\pi}\right)^{2n}\\\\
&=\frac1x-\frac2x \sum_{k=1}^\infty \frac{(x/k\pi)^2}{1-(x/k\pi)^2}\\\\
&=\frac1x+2x\sum_{k=1}^\infty \frac{1}{x^2-(k\pi)^2}
\end{align}$$
as was to be shown!


TITLE: A problem on the equation $\bar{\partial} g=f$ in complex analysis
QUESTION [2 upvotes]: I'm reading Voisin's famous book Hodge theory and Complex algebraic geometry, page 30. And in this section Voisin proved the following fact: given a smooth function $f$, we can solve the equation $\bar{\partial} g=f$ locally. To be more precise, we can suppose that $f$ is of compact support and write down the explicit formula for $u$ as:
$$
u(z)=\frac{1}{2 i \pi} \int_{\mathbb{C}} \frac{f(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta}.
$$
Of course this looks very reasonable. But I also read Hormander's famous book Introduction to complex analysis in several variables. In page 30, theorem 2.3.1 (1990 edition), he wrote a remark, which said $\bar{\partial} g=f$ needn't have a solution even when $f$ is of compact support! He said [take an arbitrary $f$ with nonzero Lebesgue integral on $\mathbb{C}$].
I am very much confused for the conclusions on two masters' books look like quite different. Can anyone explain why Hormander said we can [take an arbitrary $f$ with nonzero Lebesgue integral on $\mathbb{C}$] as a counterexample? Or have I misunderstood anything? Thanks in advance!

REPLY [2 votes]: A comment that got too long but answering the three questions in the last two comments of the OP:
For the first question (here since $ n \ge 2$ we have a form of type $(0,1)$ with compact support $f=(f_1,..f_k)$ which satisfies the condition $\bar \partial f=0$ and we construct $u$ with compact support st $\bar \partial u=f$ - in one dimension there is no condition of course since automatically $\bar \partial (fd\bar z)=0$ as $d\bar z \wedge d\bar z=0$))
we construct $u$ like in the OP in the first variable only: $u(z_1,z_2,..z_n)=\frac{1}{2 i \pi} \int_{\mathbb{C}} \frac{f_1(\zeta, z_2,..z_n)}{\zeta-z_1} d \zeta \wedge d \bar{\zeta}$; then obviously $u=0$ when $|z_2+..z_n| \to \infty$ as the integrand is zero (and then by analytic continuation since $u$ is analytic outside the compact support of $f$ we have that $u=0$ there, hence it has indeed compact support).
It is easy to see that $u$ works and gives $\bar \partial_j u=f_j$ as by defintion we have this for $j=1$ and then we have $\bar \partial f=0$ which is equivalent to $\bar \partial_j f_k=\bar \partial_k f_j$, so we differentiate under the integral sign and switch $\bar \partial_j f_1$ with $\bar \partial_1 f_k$ and again apply the one dimensional case.
In general, we can add to $u$ any analytic function $g$ and we still have $\bar \partial (u+g)=f$, but the point is that we can construct one such $u$ with compact support, not that every solution has compact support (which is of course not true)
For the second question, we need $f$ with compact support so the integral exists, we can differentiate at will under the integral sign, we can switch series with the integral etc (obviously weaker conditions work and part of the theory is to understand that).
For the third question - notice that if $z \to \infty$, $zu(z) \to \frac{1}{\pi}\int_{\mathbb C} fdA$ which is then non-zero, so $u$ cannot be zero for large $|z|$ when the Lebesgue integral of $f$ is not zero, hence $u$ doesn't have compact support in general. The difference between $n=1$ and $n \ge 2$ is then clear and the two statements in the OP are not contradictory as they refer to two different situations.
($u(z)=\frac{1}{2 i \pi} \int_{\mathbb{C}} \frac{f(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta}=-\frac{1}{\pi}\int_{\mathbb C}\frac{f(\zeta)}{\zeta-z}dA=\frac{1}{\pi}\sum_{k \ge 0}\int_{\mathbb C}f(\zeta)\zeta^k/z^{k+1}dA$ hence $zu(z) \to \frac{1}{\pi}\int_{\mathbb C} fdA$ as noted)


TITLE: Do plasmas really not experience a $\nabla \vec B$ drift even though the individual particles do?
QUESTION [1 upvotes]: I'm currently working through Chen's Introduction to Plasma Physics and Controlled Fusion, and I just got to the chapter on the fluid theory of plasmas. Chen claims that the $\nabla\vec B$ drift does not exist for fluids. I have trouble understanding this as each particle would experience this drift. Is this a limitation of the fluid theory of plasmas or do plasmas as a whole really not experience it, even though the particles individually do? If it is not a limitation of the theory, how does this make sense?

REPLY [0 votes]: I finally read the part in Section 3.4 of Chen's book (Introduction to Plasma Physics and Controlled Fusion, Volume 1: Plasma Physics) to which you refer.  I think he is implying the following though I must admit I do not like the hand-waviness of the argument.  For reference, I have the 2nd Edition.
If you take a continuous, isotropic Maxwellian and evolve it with the Vlasov equation in the absence of electric fields and the magnetic field gradients are gradual (i.e., the gradient scale length is larger than the gyroradius of the particles), then the particle distribution will remain an isotropic Maxwellian.  A $\nabla B$-drift will generate anisotropies in the distribution function (i.e., oblateness orthogonal the quasi-static magnetic field), thus changing the distribution function away from an isotropic Maxwellian.
Note that you must also neglect sources and losses in the evolution of the distribution function described above.  It's a lengthy way of saying, rather obtusely, that stationary, static magnetic fields cannot do work on charged particles.  Another part of this that is not shown for some reason is that in describing the dynamics, Chen does not explicitly delineate between kinetic and fluid.  That is, he mentions the bulk fluid velocity (see the following for discussion of velocity moments https://physics.stackexchange.com/a/218643/59023) but does not really explain in detail why the fluid velocity would not show a $\nabla B$-drift.  He provides a highly idealized, single-particle picture to explain the reason why the $\nabla B$-drift does not arise, but it seems like a highly selective choice of a physical region to perform the ensemble average over which to calculate the fluid moments.
The problem is that a flow of plasma incident on a magnetic field gradient will induce a $\nabla B$-drift, which illustrates one of the many major weaknesses in fluid models.  Again, I think Chen is trying to state that if you calculate the fluid moments prior to examining any dynamics, then the $\nabla B$-drift will not arise.
One way to handle this if you are going to use a fluid approximation is to do something called gyroaveraging – ensemble time average over a gyroperiod for each species.  This will give you guiding center motions and is a starting point for something called gyrokinetics.  This method will elicit $\nabla B$-drifts.
In short, you are right to be confused by this discussion (assuming it has not improved in the 3rd Edition).
As an aside, I took a class that used this book once as well and found myself continually confused and frustrated by the numerous errors, typos, and opaque descriptions of processes.  I would recommend the following books to supplement and provide, hopefully, better explanations and details:

Plasma Confinement by Hazeltine and Meiss
Introduction to Plasma Physics by Gurnett and Bhattacharjee
Fundamentals of Plasma Physics by Bellan


TITLE: If the absolute value of each function in a sequence has a Lebesgue integral that is bounded by a fixed number, will the limit function be integrable?
QUESTION [1 upvotes]: We have sequence of measurable functions $f_1, f_2, \dots$ such that $f_n \rightarrow f$ a.e. and $f$ measurable. If we know that $\int |f_n|d\mu < B$ for all $n$ ($B$ is fixed and finite). Also assume that $\mu[\Omega]$ is finite. Is this enough to imply that all the $f_n$'s and $f$ are integrable? How or counterexample?

REPLY [2 votes]: Edited. Note that for Lebesgue integration, a function $g$ is integrable (the integral $\int g\,d\mu$ exists and is finite) if and only if $g\in\mathcal{L}^1(\mu)$, if and only if $\int |g|\,d\mu$ is finite; so it makes no sense to both assume that $|f_n|$ are integrable (which you do when you write $\int |f_n|\,d\mu\lt B$) and to ask whether $f_n$ are integrable; they have to be for your hypothesis to make sense.
Detour Removed
Since $|f_n|\to |f|$ almost everywhere, by Fatou's Lemma $|f|$ is integrable and
$$\int |f|\,d\mu \leq \liminf_{n\to\infty}\int |f_n|\,d\mu.$$
So $|f|$ (hence $f$) is in $\mathcal{L}^1(\mu)$, as desired.
Finite measure of $\Omega$ is not required.


TITLE: Remainder term for Maclaurin's $\sin x$ expansion
QUESTION [1 upvotes]: We know that for the Maclaurin's series $$\sum_{k=0}^{n}\frac{ f^{k}(0) }{(n+1)!}x^{k}$$ the remainder term is given by the following formula:
$$R_{n} = \frac{\left | f^{(n+1)}(z) \right |x^{n+1} }{(n+1)!}$$
I want to calculate $\sin x$ using Maclaurin's expansion for $$\sin x = \sum^w_{k=0} \frac{(-1)^k}{(2k+1)!} x^{2k+1} $$
and then i want to calculate the remainder term using $w$. Now considering that $n=2w +1$, I think that the remainder term should look like this:
$$R_{w} = \frac{\left | f^{(2w+1+1)}(z) \right |x^{2w+1+1} }{(2w+1+1)!}$$ where $f(x) = \sin x$.
Would this formula for the remainder term be correct?

REPLY [0 votes]: Another way without Taylor remainder term:
The sine and cosine series are alternating. By the Leibniz test and its error estimate, the error is always smaller than the last term (if the last term is smaller than the next-to-last term), i.e., for $x>0$ one has the inclusion
$$
x-\frac{x^3}6\pm …- \frac{x^{4m-1}}{(4m-1)!}\le \sin x\le x-\frac{x^3}6\pm …+ \frac{x^{4n+1}}{(4n+1)!}
$$
for arbitrary $m,n$ and especially for $m=n$ and $m=n+1$ under the condition $x<4n$.


TITLE: Which version of Hahn-Banach is this proof using?
QUESTION [1 upvotes]: I am referring to this: https://en.wikipedia.org/wiki/Goldstine_theorem
Which version of Hahn-Banach Theorem is the proof using?
As I am aware, there are different versions of Hahn-Banach (https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem), e.g. geometric Hahn Banach, etc, however I am not sure which version this proof is using.
Thanks for any help.

REPLY [1 votes]: Define $\phi \colon U + \def\span{\operatorname{span}}\span\{x\} \to \mathbf R$ by
$$ \phi(u + \lambda x) = \lambda \def\d{\operatorname{dist}}\d(x,U) $$
Then $\phi$ is bounded, as
$$ \def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}
  \abs{\phi(u+\lambda x)} = \abs{\lambda} \abs{\d(x,U)} \le
\abs{\lambda} \norm{x - (-\lambda^{-1}u)} = \norm{u+\lambda x} $$
Hence $\norm{\phi} \le 1$. By Hahn-Banach (the extension version) there is an $\Phi \in X^*$ such that $\norm{\Phi} = \norm{\phi} \le 1$ and $\Phi|_{U+\span \{x\}} = \phi$, that is $\Phi|_U = 0$ and $\Phi(x) = \d(x,U) \ge 1+\delta$.


TITLE: In an A.P , first term = 2 & sum of first 5 terms is 1/4th of sum of next five terms. Write the equation & find d.
QUESTION [0 upvotes]: Q: In an A.P ,  first term = 2 & sum of first 5 terms is 1/4th of sum of next five terms. Write the equation & find d.
My solution:
$S_5$ =$\frac{1}{4}$*{$S_{10}$- $S_5$}. Q says that it is equal to sum of $\frac{1}{4}$ th of next 5 terms I.e summation of 6th , 7th , 8th , 9th , 10th term. It doesn’t not say it equals to $S_{10}$.
According to my textbook , Answer is :
$S_5$ = $\frac{1}{4}$ * $S_{10}$ .
How is this right ?

REPLY [0 votes]: Let $a_n$ be the nth term and d be the common difference. Then according to your question
\begin{align*}
S_5= 5a+10d = 1/4(5a+35d) \\
20a+40d = 5a +35d \\
a=2 \implies 40+40d = 10+35d \\
d= -6
\end{align*}
What has been written in your textbook maybe a typo and please check your calculations. It should be $S_{10}= 5S_5$.


TITLE: Integers $x,y$ so that $2y^2+3\mid x^2-2$
QUESTION [0 upvotes]: The question is

Do there exist integers $x,y$ so that $2y^2+3$ divides $x^2-2$?

Apparently, this problem can be solved with quadratic reciprocity. Is there any simpler way?

REPLY [0 votes]: Note that $2 y^2 + 3 \equiv 3 \mod 8$ if $y$ is even, $\equiv 5 \mod 8$ if $y$ is odd.  Therefore $2 y^2 + 3$ must have some prime factor $\equiv 3$ or $5 \mod 8$.
But $2$ is not a quadratic residue for such a prime: see e.g. here.


TITLE: How to prove Linear Independence
QUESTION [0 upvotes]: How to prove the set $S=\{x,|x|\}$ is linearly independent. Where S is a subset of set of real valued functions on $\mathbb{R}$.
Thank You.

REPLY [0 votes]: If $c_1 x + c_2 |x| = 0$, for all $x \in \Bbb{R}$, then $c_1 (1) + c_2|1| = 0 \implies c_1 = -c_2$, and substituting in $x=-1$ we have $c_1 = c_2$.  The only simultaneous solution to
$$
c_1 + c_2 = 0, \\
-c_1 + c_2 = 0
$$
is $c_1 = c_2 = 0$, so yes they're linearly independent.


TITLE: Finding the basis transition matrix $A$ from $U$ into $V$ satisfying $V=AU$.
QUESTION [0 upvotes]: Consider the vector space $\mathbb{R^3}$ and the bases $U=\{e_1,e_2,e_3\}$ and $V=\{e_1, e_1+e_2, e_1+e_2+e_3\}$. I want to find the basis transition matrix $A$ from $U$ into $V$ satisfying $V=AU$. Also i want to find the coordinate vectors of the given vector $(1,2,3) \in \mathbb{R^3}$ with respect to the bases $U$ and $V$, respectively, and relate these vectors with the matrix $A$.
What I have done: I define a map $T:U \longrightarrow V$ such that $T(e_1)=e_1$, $T(e_2)=e_1 + e_2$ and $T(e_3)=e_1 + e_2 + e_3$. I've denoted $e_1=v_1$, $e_1+e_2=v_2$ and $e_1 + e_2 + e_3 =v_3$. Then, I've obtained that $e_1=v_1$, $e_2=v_2-v_1$ and $e_3=v_3-v_2$. I guess that the transition matrix A will be that
$
   A=
  \left[ {\begin{array}{cc}
   1 & 1 & 1\\
   0 & 1 & 1\\
   0 & 0 & 1\\
  \end{array} } \right]
$. Here I am confused. I suppose, something is wrong. Is there any algorithm to solve this type questions?

REPLY [0 votes]: The way I always found convenient to find the matrix of basis transition is to consider it as the identity map
$$\text{Id}: \mathbb{R}^3 \to \mathbb{R}^3,$$
where the first space has basis $U$ and the second space has basis $V$. If you then form the matrix as you usually do, you should find the matrix
$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}$$
as you did. How did we find this? Well, if you form the matrix corresponding to some linear transformation, you consider the image of the basisvectors in $U$ under that transformation and look for the coordinates of these images with respect to the basis $V$. So the coordinates of the image of the first basisvector of $U$ form the first column, the coordinates of the image of the second basisvector of $U$ form the second column, and so on.
I am not quite sure what you did with the part where you considered $T$, nor how you could conclude from it how $A$ looks.

REPLY [0 votes]: Your result is correct. The matrix that you search simply has as columns the components of the vectors of the new basis expressed in the old basis.


TITLE: Area between lines and parabolas (in description)
QUESTION [0 upvotes]: Find the area of the region enclosed by the parabola $y=x^2-3x$, the line $y=2x$, and the line $y=x$.
This time, I could not break up the area into more parts, so I did not know how to calculate the area. Does anyone know how to do this?

REPLY [4 votes]: Hint:

\begin{align*}
\text{Area}&=\int_0^4\left(2x-x\right)dx+\int_4^5\left[2x-\left(x^2-3x\right)\right]dx
\end{align*}


TITLE: Path independence of the line integral of the electrostatic field created by an arbitrary charge distribution
QUESTION [0 upvotes]: Suppose we have the following situation:

A positive arbitrarily shaped charge distribution.
 The line integral of the total electric field, generated by the whole distribution, from $A$ to $B$ is independent of the path joining them .
Why is it so?
The line integral of $E_{1}\,$, the $E$-field generated by $dq_{1}$, depends solely on the radial distance $\Delta r_{1}$, here measured from $dq_{1}$, that the path covers. The same goes for the line integral of $E_{2}$.
$E_{1}+E_{2}$ is not radial, so why is its line integral path independent?

REPLY [1 votes]: Mathematically, you can use Stokes theorem to show that because the curl of the electric field is zero (just one of Maxwell's equations), that the integral you indicate is also independent of the path between your two end points (: apologies for the lack of equations


TITLE: Extending a polynomial function on an interval to be infinitely differentiable on all of R
QUESTION [0 upvotes]: If $ f:(a,b) \to \mathbb{R}$ is a polynomial function, can it be extended to $ g:\mathbb{R} \to \mathbb{R}$ such that g is infinitely many times differentiable and it is NOT the same polynomial?
What if we also assume that g has to be a polynomial, but not the same as f?
EDIT: I wasn't clear enough in the last question. Obviously g cannot be a different polynomial on all of the real line. But could it be f on (a,b) and then a different polynomial on the rest of $ \mathbb{R} $?

REPLY [1 votes]: Yes; there are well-known $C^\infty$ functions with compact support.  That is, a function which is nonzero on a bounded interval, infinitely differentiable, but which are zero outside that interval.
So, to get a new $g$, just take the original polynomial $f$, then add a $C^\infty$ function to it whose "support" is outside $(a,b)$.
Note that $g$ cannot be made to be a polynomial, since then $f-g$ is a polynomial with infinitely many zeroes (namely, all of $(a,b)$) and is therefore the zero polynomial.


TITLE: Show there is a set $H$ which is the intersection of countable many open sets containing $A$ with $\mu(H \setminus A)=0$
QUESTION [1 upvotes]: Let $A \subset [0,1]$ be a Lebesgue measurable set and let $\mu$ denote the Lebesgue measure.
I am trying to show that there is a set $H$ which is the intersection of countable many open sets containing $A$ with $\mu(H \setminus A)=0$
This is the proof.

For each $n=1,2,\ldots$ there is an open set $G_n$ containing $A$ such that
  $\mu(G_n \setminus A) < \frac{1}{n}$
This follows from the fact that the Lebesgue outer measure $\mu(A)$ is
  the infimum of $\mu(G)$ over open sets $G$ containing $A$, and from
  measurability of $A$, $\mu(G_n \setminus A) = \mu(G_n)-\mu(A)$.

Why is $\mu(G_n \setminus A) < \frac{1}{n}$? Where does $\frac{1}{n}$
come from? And what is the explanation above trying to say?

Let $H=\bigcap^\infty_{n=1} G_n$. $H$ need not be open but it is the
  intersection of countably many open sets containing $A$ in particular it
  is Borel and Lebesgue measurable.

Why does $H=\bigcap^\infty_{n=1} G_n$? And how do we know it is Borel and Lebesgue measurable?

$\mu(H \setminus A) <\frac{1}{n}$ for every $n$ so $\mu(H \setminus A)=0$

In general what is this proof trying to say?

REPLY [1 votes]: It looks as if you're trying to read a proof without knowing the basic definitions and without knowing some basic conventions.
As for "Why does $H=\bigcap^\infty_{n=1} G_n$?", the answer is that they just said "Let $H=\bigcap^\infty_{n=1} G_n$".  That just means they're giving a name to this intersection of sets; they're calling it $H$.
You quoted a definition: Lebesgue outer measure $\mu(A)$ is the infimum of $\mu(G)$ over open sets $G$ containing $A$.  That means $\mu(A)$ is the largest number that does not exceed the measure of any open set $G$ "containing" $A$. To say $G$ "contains" $A$ means that $A\subseteq G$, i.e. every member of $A$ is a member of $G$.  To say that $\mu(A)$ is the largest number not exceeding the measure of any open set $G$ containing $A$ implies that every number larger than $\mu(A)$ does exceed the measure of some open set $G$ containing $A$.  Thus for every number larger than $\mu(A)$ there is an open set $G$ containing $A$ whose measure is less than that number larger than $\mu(A)$.  In particular, if $n\in\{1,2,3,\ldots\}$, then there is an open set $G_n$ containing $A$ whose measure is less than $\mu(A)+\frac 1 n$.  The reason for concerning ourselves with $1/n$ for $n\in\{1,2,3,\ldots\}$ rather than with arbitrary positive numbers is that we want a set of positive numbers that is countably infinite rather than uncountable.  Any for reasons that appear later in the proof, we want a set of such numbers that is bounded below by $0$ but is not bounded below by any positive number.  The set $\{1/1, 1/2, 1/3,\ldots\}$ qualifies.
To understand how we know $H$ is Borel measurable, you need to recall what "Borel measurable" means.  A Borel-measurable set, or "Borel set", is any set that can be made by starting with open sets and applying complementation, countable unions, and countable intersections.  $H$ was defined to be a countable intersection of open sets.
Say the measure of $A$ is $0.8$.  Then the outer measure of $A$ and the inner measure of $A$ are both $0.8$.  That outer measure is $0.8$ means that for every number $0.8+\varepsilon$ bigger than $0.8$, there is some open set containing $A$ whose measure is less than $0.8+\varepsilon$, but there is no open set containing $A$ whose measure is less than $0.8$.  That means in particular

There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 2$; and
There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 3$; and
There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 4$; and
There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 5$;
and so on.

The intersection of all of these must have measure at least $0.8$ since it contains $A$.  But the measure of the intersection must be less than $0.8 + \frac 1 2$, and less than $0.8+\frac13$ and less than $0.8+\frac14$, and so on.  The only number not less than $0.8$ that is less than all of those, is $0.8$.  Hence the measure of the intersection must be $0.8$.  And the intersection is the intersection of countably many open sets.


TITLE: Proof that eigenvalues of the Coulomb hamiltonian are not nonnegative
QUESTION [3 upvotes]: Let's consider the Coulomb Hamiltonian
$$
-\Delta - \frac{1}{|x|}$$
in $\mathbb{R}^3$.
It is known that eigenvalues of the Coulomb Hamiltonian are negative.
I know it has negative eigenvalues, but I can't prove it hasn't nonnegative eigenvalues.
That is, I want to prove the following.
If smooth function $u$ satisfies the following
$$\int_{\mathbb{R}^3
} |u(x)|^2 dx + \int_{\mathbb{R}^3
} |\nabla u(x)|^2 dx ＜ \infty$$ and
$$-\Delta u - \frac{u}{|x|} = \lambda u$$ for some $\lambda \geq 0$, then $u=0$.
Any advice would be appreciated.

REPLY [1 votes]: As I recall, there's a theorem that bound states with potentials that go to zero at infinity necessarily have negative total energy, otherwise the wave function is not normalizable. This implies eigenvalues of the energy will be non-positive. The remaining question is whether the ground state has 0 energy. Waves in a bounded region of space give rise to interference which tends to limit solutions to the associated wave equation, so bound states imply discrete Eigenvalues.
Now a Coulomb potential has scattering states, i.e. unbounded: Coulomb Scattering. These have a continuous spectrum of positive eigenvalues.


TITLE: Example of a maximum likelihood estimator that is not a sufficient statistic
QUESTION [3 upvotes]: I am currently researching on providing some bounds on estimation using some information theoretic tools (I won't expend on that here for now, I may make a post about it later) and turns out that given a phenomenon $X$, an observation $Y$, then $\hat{x}(Y)$, the maximum likelihood estimator of $X$ based on $Y$, may apparently not be a sufficient statistic and this is a something I would like to study, the answer to this post states that such an example exists when $Y$ consists of samples that are not i.i.d but fails to provide such an example and I haven't found anything about it. Have anyone seen something of the sort ?

REPLY [4 votes]: If $T$ is a sufficient statistic for $\theta$ and a unique MLE of $\theta$ exists, then the MLE must be a function of $T$.

So if you can find a situation where there can be several maximum likelihood estimators, there remains a possibility that you can choose one MLE that might not be a function of a sufficient statistic alone.
A simple example to consider is the $U(\theta,\theta+1)$ distribution.
Consider i.i.d random variables $X_1,X_2,\ldots,X_n$ having the above distribution.
Then the likelihood function given the sample $(x_1,\ldots,x_n)$ is
$$L(\theta)=\prod_{i=1}^n \mathbf1_{\theta<x_i<\theta+1}=\mathbf1_{\theta<x_{(1)},x_{(n)}<\theta+1}\quad,\,\theta\in\mathbb R$$
A sufficient statistic for $\theta$ is $$T(X_1,\ldots,X_n)=(X_{(1)},X_{(n)})$$
And an MLE of $\theta$ is any $\hat\theta$ satisfying $$\hat\theta<X_{(1)},X_{(n)}<\hat\theta+1$$
or equivalently, $$X_{(n)}-1<\hat\theta<X_{(1)} \tag{1}$$
Choose $$\hat\theta'= (\sin^2 X_1)(X_{(n)}-1) + (\cos^2 X_1)(X_{(1)})$$
Then $\hat\theta'$ satisfies $(1)$ but it does not depend on the sample only through $T$.


TITLE: Does $c^n(n!+c^n)\lt (n+c^2)^n$ hold for all positive integers $n$ and $c\gt 0$?
QUESTION [2 upvotes]: I am  not sure whether the following inequality is true? Some small $n$ indicates it is true.
Let $n$ be a positive integer and $c\gt0$, then
$$c^n(n!+c^n)\lt(n+c^2)^n.$$

REPLY [1 votes]: For $n=1$, you have $c+c^2 < 1+c^2$, which is equivalent to $c < 1$, hence always false because you supposed $c > 0$.
For $n=2$, you have $2c^2 + c^4 < c^4 +4c^2 + 4$, which is equivalent to $2c^2 > -4$ (always true).
Aryabhatta's example ensures us that we cannot conjecture that the inequality will always be true for $n > 2$.


TITLE: Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2?
QUESTION [0 upvotes]: Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2?

I thought that if there exists a limit on both sides then it would be differentiable?

REPLY [1 votes]: If a function is differentiable, then it implies that it is continuous. However, the opposite is not true.
For example:  |x| is not differentiable at x=0, although it's continuous.
Also for this function $\sqrt|x-2|$, the right hand limit at x=2 and left hand limit at x=2 does not match, so limit doesn't exist at x=2, hence not differentiable.
$$\\
\textbf{LHL: } \lim_{h\rightarrow 2^{-}} \frac{ f(2+h)-f(2)}{h} = \lim_{h\rightarrow 2^{-}} \frac{ \sqrt (2+h-2)-\sqrt(2-2)}{h}\\ =\lim_{h\rightarrow 2^{-}} \frac{ \sqrt (h)-\sqrt(0)}{h}\\ =\lim_{h\rightarrow 2^{-}} \frac{ \sqrt (h)}{h} = \frac{1}{\sqrt 2}
\\
 \textbf{RHL: } \lim_{h\rightarrow 2^{+}} \frac{ f(2+h)-f(2)}{h} = \lim_{h\rightarrow 2^{+}} \frac{ \sqrt (2-h-2)-\sqrt(2-2)}{h}\\ =\lim_{h\rightarrow 2^{+}} \frac{ \sqrt (-h)-\sqrt(0)}{h}\\ =\lim_{h\rightarrow 2^{+}} \frac{ \sqrt (-h)}{h} = \text{not defined}\\ \\\\ \because \textbf{RHL} \neq \textbf{LHL}
$$
This is not a differentiable at x=2
Another simple way to check for differentiability is that there should not be a sharp corner in the curve.
In the plot of the $\sqrt|x-2|$, this corner occurs at $ x=2 $ (as shown in the figure)
plot here


TITLE: What is the moduli functor $\mathbb{P}Ext^1(L,M)$ represents
QUESTION [1 upvotes]: Let $X$ be an algebraic space and $L,M$ are vector bundles with rank $n,m$.
Then, It is known that $\mathbb{P}Ext^1(L,M)$ is a parameter space for isomorphism classes of vector bundle of rank $n+m$, which is obtained by extension of $L$ by $M$.
My Question is, What is moduli functor $\mathbb{P}Ext^1(L,M)$ represents?.
In the book of Huybrechts & Lehn, there is a familly, which they call "universal familly" $\mathcal{U}$ on X $\times$ $\mathbb{P}Ext^1(L,M)$, that is
$0 \to (\pi_1)^*M\otimes(\pi_2)^*\mathcal{O}(1) \to \mathcal{U} \to (\pi_1)^*L \to 0$.
So, my second question is that is this really a universal family for moduli functor that $\mathbb{P}Ext^1(L,M)$?.
Thank you for read my questions.

REPLY [1 votes]: You answered your own question.  Let $F:\text{Alg. Sp.} \to \text{Sets}$ be the functor that associates to every algebraic space $T$ the set of equivalence classes of triples $(\mathcal{L},\mathcal{V},q,p)$ of an invertible sheaf $\mathcal{L}$ on $T$, a locally free sheaf $\mathcal{V}$ on $X\times T$, a homomorphism of coherent sheaves on $X\times T$,
$$ q:\text{pr}_X^*M\otimes_{\mathcal{O}_{X\times T}} \text{pr}_T^*\mathcal{L} \to \mathcal{V},$$
and a homomorphism of coherent sheaves on $X\times T$,
$$ p:\mathcal{V}\to \text{pr}_X^* L,$$
such that the following sequence is a short exact sequence,
$$
0 \to \text{pr}_X^*M\otimes_{\mathcal{O}_{X\times T}}\text{pr}_T^*\mathcal{L} \to \mathcal{V} \to \text{pr}_X^*L \to 0.
$$
An equivalence from the triple $(\mathcal{L},\mathcal{V},q,p)$ to a second equivalence $(\mathcal{L}',\mathcal{V}',q',p')$ is a pair $(\alpha,\beta)$ of an isomorphism $\alpha:\mathcal{L}\to \mathcal{L}'$ of invertible sheaves on $T$ and an isomorphism $\beta:\mathcal{V}\to \mathcal{V}'$ of locally free sheaves on $X\times T$ such that both $p'\circ \beta$ equals $p$ and such that $\beta\circ q$ equals $q'\circ \text{Id}_M\otimes \alpha$.
The functor of equivalence classes of triples is represented by the universal object $\mathcal{U}$.
Edit. I forgot to exclude the trivial extension.  We need to add the hypothesis that for every geometric point of $T$, the corresponding extension of $L$ by $M$ is nontrivial.


TITLE: Give a natural deduction proof of $\varphi\vdash\top$, where $\varphi$ is any formula
QUESTION [3 upvotes]: As in the title, the question is

Give a natural deduction proof of $\varphi\vdash\top$, where $\varphi$ is any formula.

Could I do this proof by deriving  $\varphi \rightarrow \top$ with $ \rightarrow$-introduction rule which says that if you can derive a formula $\psi$ from a formula $\varphi$, then $\varphi \rightarrow \psi$ is true. So, with this derivation I will use $\varphi$  as an assumption and derive from $\varphi$, or am I thinking wrong? I prefer hints before solution!

REPLY [1 votes]: Here is one way that might be proven using explosion (X), negation introduction (¬I), contradiction introduction (⊥I) and conditional introduction (→I).
To implement this in this proof checker I let $\top$ be the same as $\neg \bot$ and  $P$ is any formula corresponding to $\psi$ in the question.

A shorter proof also works using negation introduction (¬I) and conditional introduction (⊥I):

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


TITLE: Different definitions for $\limsup$
QUESTION [1 upvotes]: I am wondering if the following various definitions of $\limsup$ are equivalent:
$$\displaystyle{\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \sup_{m \geq n} \{ x_m \}}$$
$$\displaystyle{\limsup_{n \to \infty} x_n = \lim_{n \to \infty} (\sup_{m \geq n} x_m)}$$
$$\displaystyle{\limsup_{n \to \infty} x_n = \inf_{n \geq 0} \sup_{m \geq n} x_m}$$
I pulled these from various sources and mostly I am interested in the different usages of the parentheses and how does one "read" these definitions step by step.

REPLY [3 votes]: The short answer is yes, they are all equivalent (assuming as a commenter mentioned that you replace $x\longrightarrow \infty$ with $n\longrightarrow \infty$).
Intuitively, limsup is a limit of a sequence, where the sequence's n-th term is given by sup$_{m\geq n}x_m$ (presuming we started out with some given initial sequence $(x_0, x_1, x_2, x_3, …)$ that is bounded). For instance, we get the 4-th term of the sequence by looking at the terms of the original sequence from $x_4$ onwards, and taking the supremum of those values. For the 5-th term, take the supremum of all terms of the original sequence from $x_5$ onwards, etc. Since successive terms of this new sequence are being formed by taking the supremum of progressively smaller subsets of the original sequence, we can conclude that this new sequence is (monotonically) decreasing. Thus it is also bounded above (the first term is the largest), and we know it is bounded below since the original sequence is bounded below. A sequence being bounded and decreasing is enough to infer that it converges, and so we can take its limit, and we call this value the limsup$(x_n)$ (since it is the limit of a sequence formed by taking the supremums of subsets of some original sequence $(x_n)$).
This fundamental intuition should show you that all of the definitions you gave are equivalent. The first two are trivially equivalent - the only difference is notational, in terms of where you prefer to place parentheses. But the last one is equivalent to the first two by the above. We noted that the supremum sequence was decreasing and bounded, and so it has a well-defined limit and infimum, and these values are the same. Therefore, $\text{lim}_{n\longrightarrow\infty}\text{sup}_{m\geq n}x_m = \text{inf}_{n\geq 0}\text{sup}_{m\geq n}x_m$.
In a nutshell, first form the sup-sequence, then take its infimum and its limit, and these values are the same, and the value is called the limsup.
Edit: On the left hand side, you wrote limsup$_{n\longrightarrow\infty}(x_n)$, but that doesn't really make sense (based on what I just wrote above). It is just limsup$(x_n)$. But still on the right-hand-side you do want to replace $x\longrightarrow\infty$ with $n\longrightarrow\infty$ for that value to make sense.


TITLE: Find the points on the parabola $\,y=x^2$ that are closest to the point $\,(0,8)$
QUESTION [0 upvotes]: Find the points on the parabola $y=x^2$ that are closest to the point $(0,8)$.

Having trouble figuring out where to start.

REPLY [3 votes]: One way of doing this problem without calculus is to place the center of a circle at (0,8), and gradually expand the circle. Eventually, the circle will touch the parabola at exactly 2 points (because the circle is centered on the parabola's axis of symmetry). These points are (-c,d), and (c,d). Notice that vertical component (aka $y$) only has a single solution (d).

To find the solution, we have to solve the following system.
$$x^2+(y-8)^2=r^2$$
$$y=x^2$$
substiuting we have
$$y + (y-8)^2=r^2$$
$$y+(y^2-16y+64)-r^2=0$$
$$y^2-15y+(64-r^2)=0$$
$$y=\frac{15\pm\sqrt{15^2-4(64-r^2)}}{2}$$
Recall that we said that $y$ has a single solution, which means that the discriminant ($15^2-4(64-r^2)$) is zero. Thus we have
$$y=\frac{15\pm\sqrt{0}}{2}=\frac{15}{2}$$
and
$$x=\pm\sqrt{\frac{15}{2}}$$
so the points are $(-\sqrt{15/2},15/2)$ and $(\sqrt{15/2},15/2)$