问:
Suppose B is diagonalizable, then $e^{B}$ is diagonalizable and $e^{\lambda}$ is an eigenvalue of $e^{B}$ if and only if $\lambda$ is an eigenval of B
As stated above, Suppose B is diagonalizable, then $e^{B}$ is diagonalizable and $e^{\lambda}$ is an eigenvalue of $e^{B}$ if and only if $\lambda$ is an eigenvalue of B. I'm struggling to determine if this is true or not, is there proof or theorem to justify this one way or the other?

答: I'll assume $B$ is a $n\times n$ matrix with complex entries. If $B$ is diagonalizable, then we might as well diagonalize it, so we will work in a basis where $B$ is a diagonal matrix. As you probably know, the exponential of $B$ may be defined by the power series $$e^B=\sum_{k=0}^{\infty}\frac{1}{k!}B^k=I + B + \frac{B^2}{2!}+(...)  ,$$ where $I$ is appropriate Identity matrix. The identity is of course diagonal and, since the product of diagonal matrices is also diagonal, so is every other term in the series ($B^2$ is diagonal. If you assume $B^k$ is diagonal, then $B^{k+1}$ will be as well). Thus the exponential matrix is a sum of diagonal matrices, and hence diagonal in the same basis. We can say a bit more. There's a theorem in linear algebra which states that an operator is diagonalizable if and only if it has $n$ linearly independent eigenvectors. So we may write $B=\rm{diag}(\lambda_1,...\lambda_n)$, where the diagonal entries are eigenvalues of $B$ (not necessarily distinct or real). Let $v$ be an eigenvector of $B$. We have $Bv=\lambda v$ for some $\lambda$. By the definition presented above, we have $$e^{B}v=\sum_{k=0}^{\infty}\frac{1}{k!}B^k v= \left(1 + \lambda+ \frac{1}{2}\lambda^2+... \right)v=e^{\lambda}v,$$ where I used the fact that $B^k v=\lambda B^{k-1}v=...=\lambda^k v$. So $e^{\lambda}$ in indeed an eigenvalue of $e^B$. The converse, however, is not true. In fact, the exponential is periodic, so $e^{\lambda}=e^{(\lambda + 2\pi iN)}$, for any integer $N$. Thus $e^{(\lambda + 2\pi iN)}$ is an eigenvalue of $e^B$ , but of course that doesn't imply $\lambda + 2\pi iN$ is an eigenvalue of $B$.


问:
Quadratic equation and inequality of solution
I have a quadratic equation, $bx^2 + 2ax + b =0$, with, $a>b>0$.
I can solve this as followings,
$x_{+} = \frac{- a +\sqrt{a^2 - b^2}}{b},x_{-} = \frac{- a -\sqrt{a^2 - b^2}}{b},$

In my text book, an inequality equation, $x_{+}^{2} < 1 < x_{-}^2$, is written with the solution, but I cannot derive the equation. If you can, please teach me the derivation.
Best regards,

答:
From $bx^2+2ax+b=0$, we can compute the discriminant, of which $4(a^2-b^2)>0$, and we conclude that there are two distinct roots. Furthermore since $a>0$, $b>0$, the roots must be negative.

Hence $x_- < x_+<0$

Also,

$$x^2+\frac{2a}{b}+1=0$$

From the constant term,we can conclude that $x_+x_-=1$,  hence $x_- < -1 < x_+<0$.

Hence, we must have $x_+^2 <1 < x_-^2$.

ContentLicense: CC BY-SA 4.0


问:
Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis?
Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.

Basis step: The statement is true for $n=1$ since: $y^{1} + \frac{1}{y^{1}} = y^{n} + \frac{1}{y^{n}} \in \mathbb{Z}$.

Induction hypothesis: Assume that $y^{k} + \frac{1}{y^{k}}$ and $y^{k-1} + \frac{1}{y^{k-1}}$ are integers for all $n \geq 1$.

Induction step: We show that $y^{k+1} + \frac{1}{y^{k+1}}$ is true.

$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (\frac{y^{k}}{y} + \frac{y}{y^{k}})$

$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (y^{k-1} + \frac{1}{y^{k-1}})$

Since it is given that $y + \frac{1}{y} \in \mathbb{Z}$ and by the induction hypothesis, $y^{k} + \frac{1}{y^{k}}$, $y^{k-1} + \frac{1}{y^{k-1}} \in \mathbb{Z}$:

$\Rightarrow (\mathbb{Z})(\mathbb{Z})-\mathbb{Z}=\mathbb{Z}$ (Closure property for integers)

Therefore, $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.

This is the only way I was able to do the proof. However, I want to know if I am allowed to assume that $P(k-1)$ is also true in the induction hypothesis since as far as I know we can only assume that $P(k)$ is true. Would also appreciate feedback and alternative solutions for my proof!

答:
You don't actually require strong induction,
and strong induction on the statement
"$y^{n} + \frac{1}{y^{n}}$ is an integer"
requires a little more than you were able to show.

The problem lies in establishing $P(2)$. If you only have $P(1),$ and your inductive step requires two previous cases, you are stuck.
Strong induction is not a get-out-of-jail-free card you can play in this situation;
if it were, a lot of nonsensical "fake proofs" would actually be provable.

What I think you actually need here is just a "bigger statement".
Instead of saying $P(n)$ is "$y^{n} + \frac{1}{y^{n}}$ is an integer",
say that $P(n)$ is "$y^{n} + \frac{1}{y^{n}}$ is an integer and
$y^{n+1} + \frac{1}{y^{n+1}}$ is an integer."
Then in order to do the inductive step, that is, to show that $P(n)$ implies $P(n+1)$, write $P(n+1)$. It is

$$ y^{n+1} + \frac{1}{y^{n+1}} \text{ is an integer and }
y^{n+2} + \frac{1}{y^{n+2}} \text{ is an integer.} $$

The first half ($y^{n+1} + \frac{1}{y^{n+1}}$ is an integer)
is easy, because $A \land B \implies B$ and in this proof
$P(n)$ is of the form $A \land B$ where $B$ is
"$y^{n+1} + \frac{1}{y^{n+1}}$ is an integer."

And the inductive step you already found proves the second half
($y^{n+2} + \frac{1}{y^{n+2}}$ is an integer).

Now to get started, you just need to fill in a base case. But you need two parts to this base case; $y + \frac1y$ is only half of a suitable base case.
To complete the base case, you could prove that
$y^2 + \frac{1}{y^2}$ is an integer
(where you're allowed to use the fact that $y + \frac1y$ is an integer;
just remember this is part of proving the base case, not an inductive step).

Another way to get a complete base case would be to prove that
$y^0 + \frac{1}{y^0}$ is an integer. That is, your base case would be $P(0).$
I think this should be easy enough to prove; in fact you can even show
which integer $y^0 + \frac{1}{y^0}$ is.

A way to do it that actually does use strong induction and just the
statement for $y + \frac1y$ as a base case is to separate your inductive step into two cases. One case applies when $k + 1 = 2$; in this case you must prove $P(2)$ using only $P(1).$ The other case applies when $k + 1 > 2$;
then, since $k - 1 \geq 1,$ you know $P(k-1)$ is one of the statements
$P(1),P(2),\ldots,P(k),$ so by strong induction you get to assume it for the inductive step.
To make your proof use this method you would have to say you are invoking strong induction, and you would have to work out the case $k+1=2.$


问:
Solving IVPs with one-sided initial condition using Laplace Transform
Consider an initial value problem of first order linear ODE.
$$\frac{\mathrm{d}}{\mathrm{d}t}y(t)+2y(t)=e^{-t}H(t),\lim_{t \to 0^-}y(t)=2$$
where $H(t)$ is the Heaviside function. Let $L[y: t \to s]=Y(s)$ and applying Laplace Transform to both side of the equation.
$$sY(s)-y(0)+2Y(s)=\frac{1}{s+1}.$$
Then I find $y(t)$ to be
$$Y(s)=\frac{1}{(s+1)(s+2)}+\frac{1}{s+2}y(0) \Rightarrow y(t)=y(0)e^{-2t}+(e^{-t}-e^{-2t}).$$
My main question is whether it is appropriate to use $y(0^-)$ in this case as the value of $y(0)$, which in this case only the value of left sided limit is given.

答:
Yes because in the definition of Laplace transform we can state:

$$L\{f(t) \}=\int_0^{\infty}e^{-st}f(t)dt=\int_{0^-}^{\infty}e^{-st}f(t)dt$$

Since with Impulse response and Heaviside response, our functions are not differentiable at $t=0$, so we take Laplace Transform from $t=0^-$, without loss of generality.

This not a rigorous explanation by any means, but it should show you why we chose $0^-$ instead of $0$


问:
Understanding unfamiliar notation related to algebraic geometry and combinatorics
I'm reading a paper$^\dagger$ and having a bit of trouble understanding some of their notation. The notation is introduced on the bottom of the first page.

Let $N \cong \mathbf{Z}^n$ be a lattice, and let $P \subseteq N_{\mathbf{Q}} := N \otimes _{\mathbf{Z}} {\mathbf{Q}}$ be an $n$-dimensional lattice polytope

Does anybody know what this notation means, specifically what it means to have the integers or rationals as a subscript?

$^\dagger$ Alexander M. Kasprzyk, Benjamin Nill, Reflexive polytopes of higher index and the number $12$.

答:
Are you familiar with tensor products in general? Because the subscript for lattices is defined right there: $N_{\mathbb Q}$ is the $\mathbb Q$-vector space obtained by tensoring $N$ with the rationals, a process often referred to as extending scalars. Basically, think of $\mathbb Z^n$ as sitting in $\mathbb Q^n$, and now consider a set of generators for $N$; under this inclusion you can think of them as vectors and consider their $\mathbb Q$-span. In particular, if $N$ has full rank then $N_{\mathbb Q}$ is the whole $\mathbb Q^n$.

The symbol $\otimes_{\mathbb Z}$ just means that we are taking the tensor product over $\mathbb Z$, which is a necessary part of the data of a tensor product; roughly it means that we are "building the new $\mathbb Q$-structure over the skeleton of the existing $\mathbb Z$-structure on $N$, as opposed to any more specialized or fundamental structure (of which there are not really examples in this case, but if you consider e.g. two complex vector spaces and take their tensor product over $\mathbb Q$, that would be less specialized than tensoring over $\mathbb C$ but more specialized than tensoring over $\mathbb Z$).


问:
Proving that, for integers $n$, $m$, $p$, if $n| 72m–p$ and $n|8m$, then $n|p$
I'm trying to write a proof but I have been stuck for the past few hours trying to make progress.

I have to prove that

For all integers $n$, $m$, and $p$, if $n| (72m–p)$ and $n| (8m)$, then $n|p$.

I tried substituting $72m-p$ for $x$, and $8m$ for $y$, but I'm not really sure where to go from there.

答:
$${\displaystyle 72m-p\equiv 0{\pmod {n}}}$$
$${\displaystyle 9(8m)\equiv p{\pmod {n}}}$$
$${\displaystyle 8m\equiv 0{\pmod {n}}}$$
$${\displaystyle \text{Therefore: }9(0)\equiv p{\pmod {n}}}$$
$${\displaystyle p\equiv 0{\pmod {n}}}$$
$$\rightarrow n|p$$


问:
Number of ways to apply a function taking any number of inputs and producing one output to $n$ items
I have a function, fn that can take any number of inputs and produces one output. The order of the inputs doesn't matter.

However, if just one input is passed, the function just returns it as is, making this redundant. So to get something interesting, two or more inputs must be passed.

Now, I have $n$ objects on which to apply this function. I could either pass the first two as inputs to fn, get the result then pass the remaining $n-2$ as inputs to get the second result and finally combine the two to get the final answer. Or, I could pass all $n$ of them in one go and get the final answer. How many ways to get the final answer are there?

When we have three objects for example, there seem to be 4 ways to do this, forming four trees (see below). Also, is there a name for these kinds of trees?

答:
You're looking at irreducible rooted trees with $n$ labeled leaves.... but more commonly, these are just evolutionary trees or phylogenetic trees with unlabelled interior nodes and $n$ labelled tips.

Enumerations and recurrence relations are given by J. Felsenstein ("The number of evolutionary trees", Systematic Zoology, 27 (1978), 27-33.)  A scanned copy is available in the notes to OEIS A005373 (which is not, however, this sequence).

If I'm reading Table 1 correctly in the scanned paper, the number of such trees with $n$ nodes is $1,1,4,26,236,2752,39208,\ldots$ ... which leads us to the actual sequence of counts, at OEIS A000311.


问:
A tensor question with a modulo group

Im failing to see how the statement could be true, as, if we consider a=15, b=17 then we should have 15 ⊗ 5 = 17 ⊗ 3. I cant seem to get those two to be equal.

答: Writting $n \textrm{ mod } 12$ as $\bar{n}$, note that $\bar{n} = n\bar1$. Then using the bilinearity of $-\otimes-$ we get that $a \otimes \bar b$ and $b \otimes \bar a$ both are equal to $ab(1 \otimes \bar1)$.

Title: A tensor question with a modulo group


问:
Self-Paced Graduate Math Courses for Independent Study
Does anyone know of any graduate math courses that are self-paced, for independent study?

I am a high school math teacher at a charter school in Texas. While I am quite happy with where I am right now, but my goal is to earn at least 18 graduate credits on math subjects so that I can teach higher-level math, and be certified as dual-credit math teacher. (HS class where students earn high school as well as college credits at the same time.) I am aware of many online graduate classes offered by some respectable universities, but all of them are semester-based, which may not be very feasible since my full-time teaching is extremely demanding, not to mention that math is anything but a casual subject.

I welcome any suggestions even for programs from outside of US, as long as they are accredited and conducted in English. (For example, I was told that the college system in the Philippine is an exact "copy cat" of the US.) For your information, I am quite comfortable studying independently, in fact, I took lots of prerequisite math classes successfully under this study mode. By the way, last year I took GRE for this purpose, my verbal + quantitative score is a decent 1200 under old scoring scale.

Thank you very much for your time and help.

答: For a decent selection of grad courses and to whet your appetite, ocw.mit.edu : MIT open course ware. </p>

ContentLicense: CC BY-SA 3.0


问:
Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$
I am trying to show that  $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:

\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.

Body: Hint:$$\frac{x^{2x}}{(x+1)^{x+1}}=x^{2x-(x+1)}\frac{1}{\left(1+ \frac{1}{x}  \right)^{x+1}} \sim\frac{x^{x-1}}{e} $$


问:
Methods of Calculating Great Circle Distances
I wish to calculate the shortest distance between two points on the Earth. What are the methods of doing this?

So far, the ones I know are:

Haversine formula, which assumes a spherical Earth - great circle.

Vincenty's formulae, which assumes an oblate spheroid and is therefore more accurate than the Haversine

I was wondering how does calculus of variations and geodesics come in. Specifically, I read this paper. Is the method of calculus of variations any more accurate than Vincenty's or the Haversine? Or is this a totally different method altogether?

I would appreciate it if someone could explain so that a 15 year old would understand, thanks.

答:
"Geodesic" is just the name we give to the shortest path on a surface between two points.

For planes or 3-dimensional Euclidean space, geodesics are just straight lines.

On a sphere, the geodesics turn out to be "great circles". That is, a circle resulting from intersecting the sphere with a plane that passes through the sphere's center.

For an oblate spheroid, the geodesics are tougher to identify. Any plane passing through both poles of the spheroid will produce a geodesic. The equator is also a geodesic. But if you consider two points that not directly north and south of each other, and do not both lie on the equator, the geodesic between them is not easily described.

The "calculus of variations" is a technique for finding geodesics, but it only works indirectly. (Actually the calculus of variations is useful for finding many things, but here we are only concerned with geodesics.) The basic idea behind the calculus of variations is that since a geodesic is the shortest path between two points, if we modify it anywhere, no matter how small the change, the length of the path should increase. If we use a parameter to control the size of the change, then the length of the path is a function of the parameter which has its minimum value when size of the modification is $0$.

If you don't know anything about calculus, then how this works requires a considerable amount of new concepts, far more than could be reasonably included here. Consider it voodoo magic which produces a weird equation that tells us what geodesics should look like.

If you know any calculus, then you may recall that at its minimum, a well-behaved function has a derivative of $0$. By setting the derivative of the length with respect to the variation parameter to $0$, we get a "differential equation" that the geodesic must solve. How to solve differential equations is a major concern (they pop up everywhere when math is applied), so we have developed a lot of tools for that purpose.

So the calculus of variations gives us the differential equation, but it is solving the differential equation, given the two endpoints, that provides the geodesic between them. And it is the geodesic that allows us to determine the distance along the surface between the endpoints.

For planes, this is all simple to work out, and just leads back to the Euclidean distance formula $$s = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2+ (z_1 - z_2)^2}$$

For spheres, the math requires trigonometric functions, but still results in the relatively simple haversine formula.

For oblate spheroids, there is no nice general formula. Instead we use numerical techniques for estimating the distance to any desired precision. This is what Vincenty's formulae are doing. This is actually less of a problem than people suppose. We cannot calculate square roots or trig functions to perfect accuracy either and use similar numerical methods when we calculate with them.

The more complex the surface is, the harder it is to calculate distance along it.

Which brings us to the real issue in your question. The problem with using the haversine formula to calculate distances on Earth is not that the formula is inaccurate. The haversine formula is perfectly accurate for calculating distances on spheres. The inaccuracy is in treating the surface of the Earth as a perfect sphere. It is closer in shape to an oblate spheroid. So using Vincenty's formulae for oblate spheroids to calculate distance provides a better accuracy.

But Earth is also not exactly an oblate spheroid. To improve on Vincenty's formulae, we need a more accurate model of the Earth. That is not a mathematical problem. It is a Geoscience problem. You have to measure the Earth and develop of model of its surface that meets the standard of accuracy you desire. You are not going to find a simple description for it such as "oblate spheroid".

It takes only 3 numbers to completely describe a plane: 2 angles to indicate the direction perpendicular to the plane, and the distance from the plane to the origin.

It takes 4 numbers to completely describe a sphere: the 3 coordinates of the its center, and its radius.

It takes 7 numbers to completely describe an oblate spheroid: 3 coordinates for its center, 2 angles to determine the direction of its axis, and the major and minor radii.

Depending on how closely you want to model the Earth, it will likely take many thousands of numbers to describe the surface. Most likely you would want a triangulation of the surface using spherical triangles of varying radii for faces. Accurately calculating distances on such a surface would be a major undertaking, and best done by starting with the great circle route and tweaking it on each face.

And this begs the question "What exactly is the surface of the Earth anyway?" Do we include every mountain? Every hill? Every blade of grass? The more detailed your surface is, the greater the distance between any two points on it will be. If you were to look at the "blades of grass" level, you would discover that the "circumference of the Earth" is many times greater than 40,000 kilometers/25,000 miles, and constantly changing.  How you answer that question depends very much on what your distance calculations will be used for. A hiker, a car traveller, and an airplane passenger have very different ideas about the distance from New York to Los Angeles.


问:
Gallery of unlabelled trees with n vertices
Can anyone point me to a gallery (printed or online) of unlabelled trees, sorted according to their order (i.e., number of vertices)?  That is, for each order n in oeis.org/A000055 (up to maybe n=11 or so) I'd like to see a visualization of all the trees.

I found Gary Fredericks' online database Small Simple Graphs but there are a few problems which make it unsuitable for my purposes:  1) It's for graphs in general as opposed to trees.  (Though maybe there is some way of using the filters in this tool to suppress the graphs with cycles?)  2) It works only for n up to 9.  3) It shows only five graphs at a time, though I'd like to see all the trees for a given n on the same web page (or printed page).

Alternatively can someone point me to some code which can automatically generate images of such trees, preferably in a vector format?  Then I could arrange them myself into a single web page or LaTeX document.

答:
geng which comes with nauty can generate these trees (along with other classes of graphs) very quickly; they can be viewed with showg.  It will give a list of adjacencies and it's straightforward to write one's own script to convert it to one's desired format.  The command is e.g.

```<code>geng 7 6:6 -c
</code>```

for $7$-node trees.

Here's the 6 to 8 vertex trees below (it could easily extend this table to 15-node graphs, and beyond that with a bit of effort):

Body:
geng which comes with nauty can generate these trees (along with other classes of graphs) very quickly; they can be viewed with showg.  It will give a list of adjacencies and it's straightforward to write one's own script to convert it to one's desired format.  The command is e.g.

geng 7 6:6 -c

for $7$-node trees.

Here's the 6 to 8 vertex trees below (it could easily extend this table to 15-node graphs, and beyond that with a bit of effort):


答:
Taking the $\log$ of both sides for big enough $x$
$$\begin{align*}\log \frac{x^{2x}}{(x+1)^{x+1}}&=2x\log x-(x+1)\log(x+1)\\&=x\log \frac{x^2}{x+1}-\log(x+1)\\&\geq x\log \frac{x^2}{2x}-\log(2x)\\&=\log \frac{x^{x-1}}{2^{x+1}}.\end{align*}$$
Now take $e$ to both sides to get, for large enough $x$
$$\frac{x^{x}}{(x+1)^{x+1}}\geq \frac{x^{x-1}}{2^{x+1}}\geq\frac{2^{2x-2}}{2^{x+1}}\to\infty .$$

Title: Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$

Body:
I am trying to show that  $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:

\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.


问:
Bounding a somewhat complicated integral (exponential of a polynomial)
I am interested in bounding the following integral, where $a>0$ is a constant:

$$\int\limits_0^a \exp\left(\left(x^2 - \frac{2a^4+3}{4a^2}\right)^2\right) dx$$

I first conjectured that

$$\int\limits_0^a \exp\left(\left(x^2 - \frac{ba^4+c}{a^2}\right)^2 \right) dx \le a^{-1}\exp \left( \left( \frac{ba^4+c}{a^2}\right)^2 \right)$$

when $c \ge 0$ and $b > 0$. However, this is false (for example, take $c = 0$, $b = 9/20$, and plug in $a = 2$). However, this inequality does appear to hold when $b \ge \frac{1}{2}$, from the numerical tests I have done. I am not sure how to prove it, however. The expressions that I have seen for exponentials of quartics are quite complicated and do not easily admit an upper bound (see for instance Exponential of a Quartic). Does anyone have any ideas?

[Note: In fact, what I'm really interested in is the regime $a \rightarrow \infty$, but so far the upper bound I have guessed appears to hold up for all $a > 0$.]

Updated wording: I am not just looking for any bound on this integral, but I indeed would like to prove the bound that I have proposed above. This was not made clear in my original post; I apologize for that.

答:
Another approach I can think of that could provide more insight would be computing a Taylor Remainder for the expansion around $a=0$ of:

$$
F(a) = \int_0^a f(a,x)dx = \int_0^a \exp(g(a,x))dx
$$

$$
F(a) = F(0) + F'(a) + F''(\xi)/2
$$
Naturally $F(0) = 0$. So:

For that, we can use Leibniz Integral Rule:
$$
\frac{dF}{da} |_{x=a} = f(a,a) \times 1 + \int_0^a \frac{\partial f(a,x)}{\partial a}dx  = W(a)
$$

Computing $f(a,a)$:
$$
f(a,a) = \exp \left(\frac{(1-b)a^4 -c}{a^2} \right)^2
$$
Now the partial derivative:
$$
\frac{\partial f(a,x)}{\partial a} = f(a,x) \times \frac{\partial g(a,x)}{\partial a} = f(a,x) \times 2 \left( \left( x^2 - \frac{ba^4+c}{a^2}\right) \left( -2ab +2\frac{c}{a^3}\right) \right)
$$

Well... I'll just use Mathematica:
$$
W(a) = \int_0^a \frac{\left(a^2-3\right) e^{\left(\frac{3}{4 a^4}-\frac{1}{2 a^2}+x^2\right)^2}
   \left(4 a^4 x^2-2 a^2+3\right)}{2 a^9} \, dx+e^{\frac{\left(4 a^6-2 a^2+3\right)^2}{16
   a^8}}
$$

The Remainder term will look like this:
$$
\frac{e^{\left(\frac{3}{4 \xi ^4}+\xi ^2-\frac{1}{2 \xi ^2}\right)^2} \left(\xi ^6+\xi
   ^2-3\right) \left(4 \xi ^6-2 \xi ^2+3\right)}{\xi ^9}+\int_0^{\xi }
   \frac{e^{\left(\frac{3-2 \xi ^2}{4 \xi ^4}+x^2\right)^2} \left(-162 \xi ^2-24 \xi ^{14}
   x^2-36 \xi ^6 \left(8 x^2+1\right)+9 \xi ^4 \left(24 x^2+13\right)+4 \xi ^{12} \left(4
   x^4+30 x^2+5\right)-2 \xi ^{10} \left(48 x^4+8 x^2+63\right)+2 \xi ^8 \left(72 x^4+60
   x^2+83\right)+81\right)}{4 \xi ^{18}} \, dx
$$

So, I call it quits on this attempt, but I do admit it is starting to look plausible, but for a very limited range on $a$, such that $a^{-1}$ is actually a big number and thus $a^{-1} K $ for big enough $K$ can be an actual bound... But thinking like this...

Yeah... Nope, I don't think this would work...


问:
Volume integral $\int_{0}^{2\pi}\int_{0}^{1}\sqrt{r^2-2r\cos(t)+1}\>r dr dt$
I need to calculate the volume of the solid
$$E=\{(x,y,z)\in\mathbb{R}^{3}\mid x^2+y^2-2y\leq0,0\leq z\leq\sqrt{(x-1)^2+(y-1)^2}\}$$
I know
$$x^2+y^2-2y=0$$
can be rewritten in polar coordinates
$$x=r\cos(t),\quad y-1=r\sin(t)$$
with $0\leq r\leq1,0\leq t\leq2\pi$. Then the volume integral for the solid is
$$\int_{0}^{2\pi}\int_{0}^{1}\sqrt{r^2-2r\cos(t)+1}\>r dr \space dt$$
But I have no idea how to evaluate this integral.

答:
Observe that the resulting integral for the volume

$$V= \int_{0}^{2\pi}\int_{0}^{1}\sqrt{r^2-2r\cos t+1}\>r \space dr \space dt$$

is on the integrand $\sqrt{r^2-2r\cos t+1}$, which is the distance between the fixed point $(1,0)$ on the edge of the unit disk and any point $(r, \theta)$ over the unit disk. Then, with the polar origin chosen at the fixed point, the integral can be equivalently expressed as

$$V= \int_{-\frac\pi2}^{\frac\pi2}\int_{0}^{2\cos\theta}r\>rdr \space dt= \frac{16}3\int_{0}^{\frac\pi2}\cos^3tdt=\frac{32}9
$$


问:
Show $x \mapsto \frac{x}{\Vert x\Vert^2}$ is orientation reversing.
I wanted to show that the map $\mathbb{R}^n- \{0\} → \mathbb{R}^n- \{0\},\ x\mapsto \frac{x}{\Vert x\Vert^2}$ is orientation reversing. But i don't know how to tackle it appropriately.

The solution proposed the following:

At $p = (1,0,...,0)$ the differential of the given map is the identity
on $0 \oplus \mathbb{R}^{n-1} \subset T_p\mathbb{R}^n$ while it is
multiplication by $-1$ on $\mathbb{R}  \oplus 0 \subset
 T_p\mathbb{R}^{n}$.

Then they justify it by connectedness of $\mathbb{R}^{n}$ for $n$ greater than $1$.

I am not sure whether I fully understand the solution. For simplicity let's assume $n = 2$, then at $(1,0)$ the differential of the map is $(-1,1)$, similarly $(-1,1,1...,1)$ for arbitrary $n$. How is it the identity on $0 \oplus \mathbb{R}^{n-1} \subset T_p\mathbb{R}^n$?

Could someone elaborate? I tried to look up similar problems in my books but couldn't find any. I would really like to understand this.

Thanks for any help!

答:
Calculate the matrix representing the differential in standard coordinates in $\mathbb{R}^n$, that is, the Jacobian of the given map: The derivative of the $j$-th component of $\frac{x}{ \| x \|^2}$ with respect to $x^i$, is $\frac{ \delta_i^j \| x \|^2 - 2 x_i x^j }{ \| x \|^4}$.  Evaluated at the point $(1, 0, \dots, 0)$, this gives zero if $i \ne j$, gives $-1$ if $i=j=1$, and $1$ if $i = j \ne 1$, so the Jacobian is block diagonal with the upper left $1\times 1$ block being $-1$ and the lower right $(n-1)\times (n-1)$ block being the identity, which is what the solution you quote is saying in a nicer way.

Makes sense geometrically too, as the map is an inversion in the unit sphere in $\mathbb{R}^n$, and at any point $p$, any tangent vectors perpendicular to $p$ are simply rescaled by $1/\|p\|^2$, whereas the ones parallel to $p$ switch direction.


问:
Derivative map of the diagonal inclusion map on manifolds
I was trying to work through a problem(#10 of $\S$1.2) in Guillemin and Pollack's book $\textit{differential topology. }$ The problem is given as follows.

Let $f: X\longrightarrow  X\times X$ be the mapping $f(x)=(x,x)$.
  Check that $df_x(v)=(v,v)$. Here $X\subset \mathbf R^m$ is a manifold.

My attempt so far has been:

First we parametrise an open neighbourhood of $x\in X$ and $(x,x) \in X\times X$ locally by $\phi$ and $\phi \times \phi$ into open subsets $U\subset\mathbf R^m$ and $U\times U$ (we use $\phi(0)=x$ for simplicity). This gives the commuting diagram as follows:

$$
\begin{array}[c]{ccc}
X\;\;&\stackrel{f}{\longrightarrow}&X\times X\\
\downarrow\scriptstyle{\phi}&&\downarrow\scriptstyle{\phi \times \phi}\\
U\;\;&\stackrel{h}{\longrightarrow}&U\times U
\end{array}
$$

$$
\begin{array}[c]{ccc}
T_x(X)&\stackrel{df_x}{\longrightarrow}&T_{(x,x)}(X\times X)\\
\downarrow\scriptstyle{d\phi_0}&&\downarrow\scriptstyle{d\phi_0 \times d\phi_0}\\
\mathbf R^m\;\;&\stackrel{dh_0}{\longrightarrow}&\mathbf R^m \times \mathbf R^m
\end{array}
$$

According to the definition (or the commuting diagram above), $df_x=(d\phi_0 \times d\phi_0) \circ dh_0 \circ d\phi_0$.

However I have no idea how to proceed after that. If I want to calculate $df_x$, I have to know what $d\phi_0$ is first... But since we let $\phi(0)=x$, what would the derivative map of that be (since $\phi(0)=x$ just means we send a specific point $0$ to a specific point $x$, it doesn't tell us anything about the expression of this parameterisation)?

Thanks everyone for the help!

答:
Hoping it helps you, I am expanding A.Bellmunt's comment.

Let be $x$ a point of $X$, a submanifold of $\mathbb R^n$, and $\phi:X\to\mathbb R^m$ a local coordinate chart centered at $x$ (i.e. $\phi(x)=0$).

Therefore $\phi\times\phi:X\times X\to\mathbb R^m\times\mathbb R^m$ is a local coordinates chart centered at $(x,x)$ (i.e. $(\phi\times\phi)(x,x)=(0,0)$).

Now we get the local expression $f=(\phi\times\phi)^{-1}\circ \widetilde{f}\circ\phi$, where $\widetilde{f}$ is the linear map $$\widetilde{f}:u\in\mathbb R^m\to(u,u)\in\mathbb R^m\times\mathbb R^m.$$
Therefore:

$\widetilde f$ is linear, so it coincides with $d_0\widetilde f$, and

if $v\in T_xM \overset{d_x\phi}{\longrightarrow}\tilde v\in\mathbb R^m$, then $(v,v)\in T_{(x,x)}X\times X\overset{d_{(0,0)}(\phi\times\phi)}{\longrightarrow}(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m$,

and by 1. and 2. we get immediately the searched expression of $d_xf$, i.e.:
$$\begin{array}{ccc}
v\in T_xM&\overset{d_xf}{\longrightarrow}&(v,v)\in T_{(x,x)}X\times X\\
\downarrow d_x\phi&&\downarrow d_{(x,x)}(\phi\times\phi)\\
\tilde v\in\mathbb R^m&\overset{d_0\widetilde f}{\longrightarrow}&(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m\end{array}$$

Dear Evariste, answering the supplementary question in your comment:

given any curve $\gamma=(\gamma_1,\gamma_2)$ in $M\times M$, if we take the time-derivative of $(\phi\times\phi)\circ\gamma=(\phi\circ\gamma_1)\times(\phi\circ\gamma_2)$, then, by the chain rule, we get $$\begin{aligned}(d\phi\times d\phi)\circ\gamma'&=(d\phi\circ\gamma_1')\times(d\phi\times\gamma_2')=\dfrac{d}{dt}((\phi\circ\gamma_1)\times(\phi\circ\gamma_2))\\&=\dfrac{d}{dt}((\phi\times\phi)\circ\gamma)=d(\phi\times\phi)\circ\gamma'.\end{aligned}$$ By the arbitrariness of $\gamma$, we derive the desired identity.


问:
Does $\partial_{\varepsilon_k} f(x^k)$ approximate $ \partial f (\bar{x})$ for $ x^k \to \bar{x} $ and $\varepsilon_{k} \to 0 $?
Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is convex and globally Lipschitz continuous.

Let $(x^k) \subset \mathbb{R}^n$ be a sequence such that $ x^k \to \bar{x} $ for some $\bar{x} \in \mathbb{R}^n$.

Assume $(\varepsilon_k) \subset \mathbb{R}_{+}$ a sequence of positive scalars such that $\varepsilon_{k} \to 0 \, .$

Does the following hold:

$$ \operatorname{dist} \left( \, \partial_{\varepsilon_k} f(x^k) \, , \, \partial f (\bar{x}) \, \right):= \sup_{g \in \partial_{\varepsilon_k} f(x^k)} \, \, \inf_{h \in  \partial f (\bar{x})} \, \| g - h \|_2  \to 0 $$

where

$\partial_{\varepsilon_k} f(\cdot)$ denotes the $\varepsilon_k$-subdifferential of $f$

$\partial f(\cdot)$ denotes the regular convex subdifferential of $f$

$\| \cdot \|$ denotes the Euclidean distance.

The following is all from Convex Analysis and Minimization Algorithms II and might be of use:

Definition 1.1.1 Given $x \in \mathbb{R}^n$ the vector $s \in \mathbb{R}^{n}$ is called an $\varepsilon$ -subgradient of $f$ at $x$ when the following property holds:
$$
f(y) \geqslant f(x)+\langle s, y-x\rangle-\varepsilon \quad \text { for all } y \in \mathbb{R}^{n}
$$

It follows immediately from the definition that

$\partial_{\varepsilon} f(x) \subset \partial_{\varepsilon^{\prime}} f(x)$ whenever $\varepsilon \leqslant \varepsilon^{\prime}$

$\partial f(x)=\partial_{0} f(x)=\cap\left\{\partial_{\varepsilon} f(x): \varepsilon>0\right\}\left[=\lim _{\varepsilon \downarrow 0} \partial_{\varepsilon} f(x)\right] $

Theorem 1.1.4 For $\varepsilon \geqslant 0, \partial_{\varepsilon} f(x)$ is a closed convex set, which is nonempty and bounded.

Proposition 4.1.1 Let $\left\{\left(\varepsilon_{k}, x_{k}, s_{k}\right)\right\}$ be a sequence converging to $(\varepsilon, x, s),$ with $s_{k} \in$ $\partial_{\varepsilon_{k}} f\left(x_{k}\right)$ for all $k .$ Then $s \in \partial_{\varepsilon} f(x)$.

Proposition 4.1.2 Let $\delta>0$ and $L$ be such that $f$ is Lipschitzian with constant $L$ on some ball $B(x, \delta),$ where $x \in \mathbb{R}^n$. Then, for all $\delta^{\prime}<\delta$
$$
\|s\| \leqslant L+\frac{\varepsilon}{\delta-\delta^{\prime}}
$$
whenever $s \in \partial_{\varepsilon} f(y),$ with $y \in B\left(x, \delta^{\prime}\right)$.

As a result, the multifunction $\partial_{\varepsilon} f$ is outer semi-continuous, just as is the exact subdifferential.

Theorem 4.1.3 Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a convex Lipschitzian function on $\mathbb{R}^{n} .$ Then there exists $K>0$ such that, for all $x, x^{\prime}$ in $\mathbb{R}^{n}$ and $\varepsilon, \varepsilon^{\prime}$ positive:
$$
\Delta_{H}\left(\partial_{\varepsilon} f(x), \partial_{\varepsilon^{\prime}} f\left(x^{\prime}\right)\right) \leqslant \frac{K}{\min \left\{\varepsilon, \varepsilon^{\prime}\right\}}\left(\left\|x-x^{\prime}\right\|+\left|\varepsilon-\varepsilon^{\prime}\right|\right)
$$

This result  implies the inner semi-continuity of $(x, \varepsilon) \longmapsto \partial_{\varepsilon} f(x)$ for a Lipschitz-continuous $f$.

In particular, for fixed $\varepsilon>0$

$$\partial_{\varepsilon} f(y) \subset \partial_{\varepsilon} f(x)+\|y-x\| B(0, \frac{K}{\varepsilon}) \quad \text{for all } x \text{ and } y \, .$$

Corollary 4.1.5 Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be convex. For any $\delta \geqslant 0,$ there is $K_{\delta}>0$ such that
$$\Delta_{H}\left(\partial_{\varepsilon} f(x), \partial_{\varepsilon} f\left(x^{\prime}\right)\right) \leqslant \frac{K_{\delta}}{\varepsilon}\left\|x-x^{\prime}\right\| \quad \text{ for all } x \text{ and } x^{\prime} \in B(0, \delta) \, .$$

Theorem 4.2.1  Let be given $f: \mathbb{R}^n \to \mathbb{R}$, $x \in \mathbb{R}^n$ and $\varepsilon \geqslant 0 .$ For any $\eta>0$ and $s \in \partial_{\varepsilon} f(x),$ there exist $x_{\eta} \in B(x, \eta)$ and
$s_{\eta} \in \partial f\left(x_{\eta}\right)$ such that $\left\|s_{\eta}-s\right\| \leqslant \varepsilon / \eta$.

This result can be written in a set formulation:
$$
\partial_{\varepsilon} f(x) \subset \bigcap_{\eta>0} \bigcup_{\|y-x\| \leqslant \eta}\left\{\partial f(y)+B(0, \frac{\varepsilon}{\eta})\right\} .
$$

The following one is not from the above mentioned source but easy to prove:

Theorem Assume $f: \mathbb{R}^n \to \mathbb{R}$ is convex and let $x \in \mathbb{R}^n$. Then for every $\varepsilon > 0$ there is a $\delta > 0$ such that

$$
\bigcup_{y \in B_{\delta}(x)} \partial f(y) \subset \partial_{\varepsilon} f(x) \,.
$$

答:
Your Theorem 4.2.1 and the upper semicontinuity of Clarke differential is enough to answer the question.

We want to prove that $$\lim \sup_{g \in \partial_{\varepsilon_k} f(x^k)} \, \, \inf_{h \in  \partial f (\bar{x})} \, \| g - h \|_2 = 0. \tag{*}$$ The main idea is to note that, since $$\partial_{\epsilon} f (x) \subset \cup_{\|y-x\| \leq \sqrt{\epsilon}}\  \partial f (y) + B\left(0,\sqrt{\epsilon}\right),$$ each $g \in \partial_{\varepsilon_k} f(x^k)$ is an approximation of some Clarke differential close to $x$.

Aiming to prove (*), let $\epsilon>0$ be given, there exists $\delta>0$ such that for all $x' \in B(\bar{x},\delta)$ implies that
$$\partial f(x') \subset \partial f(\bar{x})+B\left(0,\dfrac{\epsilon}{2}\right)  \tag{1} $$ due to upper semicontinuity of Clarke differential. Choosing $k_0$ such that for each $k \geq k_0$ it's true that $$\sqrt{\epsilon_k}<\dfrac{\min\{\delta,\epsilon\}}{2} \tag{2}$$ and
$$\|x^{k} - \bar{x}\| < \dfrac{\delta}{2}, \tag{3}$$ given $g \in \partial_{\epsilon_k} f(x^{k}) $, by the Theorem 4.2.1, there exist $y^{k}$ such that $$\|y^{k} - x^{k}\| \leq \sqrt{\epsilon_k}$$ and $$g \in \partial f (y^{k}) + B\left(0,\dfrac{\epsilon_k}{\sqrt{\epsilon_k}}\right) = \partial f (y^{k})+ B\left(0,\sqrt{\epsilon_k}\right).\tag{4}$$ Since $\|y^{k} - x^{k}\| < \dfrac{\delta}{2}$ and $\|x^{k} - \bar{x}\| < \dfrac{\delta}{2}$ by (2) and (3), we have $$\|y^{k} - \bar{x}\| \leq \|y^{k} - x^{k}\|+\|x^{k} - \bar{x}\|<\delta$$ which means, by (1), $$
\partial_{\epsilon_k} f(y^{k}) \subset \partial f(\bar{x})+B\left(0,\dfrac{\epsilon}{2}\right).$$ Consequently, by (4),
\begin{align}
g \in \partial f (y^{k})+ B\left(0,\sqrt{\epsilon_k}\right) \subset & \partial f(\bar{x})+B\left(0,\dfrac{\epsilon}{2}\right) + B\left(0,\sqrt{\epsilon_k}\right) \\
\subset & \partial f(\bar{x})+B\left(0,\dfrac{\epsilon}{2}\right) + B\left(0,\dfrac{\epsilon}{2}\right) \\
\subset & \partial f(\bar{x})+B\left(0,\epsilon \right).\end{align} Finally,
$$\inf_{h \in \partial f(\bar{x})} \|g - h\|<\epsilon $$ for all $g\in \partial_{\epsilon_k} f (x^{k})$ which means $$\sup_{g\in \partial_{\epsilon_k} f(x^{k})} \inf_{h \in \partial f(\bar{x})} \|g - h\| \leq \epsilon.$$


问:
Isomorphism of schemes restricts down to local isomorphism
I just want to make sure I am understanding the definitions I'm learning correctly.

Suppose we have an isomorphism of schemes $\pi: X \to Y$, and suppose we have some affine cover of $X$, $\{U_i\}$. Is it true that $\pi|_{U_i}$ is an isomorphism for each $i$?

答:
Yes. The restriction of a homeomorphism is a homeomorphism so that $\pi|_{U_i}:U_i\to \pi(U_i)$ is a homeomorphism. One just needs to check that it is an isomorphism on the structure sheaves, but an isomorphism of sheaves is determined at the stalk level.

Denote $\mathcal{O}_{U_i}=\mathcal{O}_X|_{U_i}$ and $\mathcal{O}_{\pi(U_i)}$ analogously for $\mathcal{O}_{Y}|_{\pi(U_i)}$. Given $x\in U_i$ and $\pi(x)\in \pi(U_i)$, then the stalk map $\mathcal{O}_{\pi(U_i),y}\to \mathcal{O}_{U_i,x}$ of $\pi|_{U_i}$ agrees with the stalk map $\mathcal{O}_{\pi(U_i),y}\to \mathcal{O}_{U,x}$ for $\pi$. So, the map $\pi|_{U_i}$ is also an isomorphism on stalks. Hence, we are done.


问:
If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).
I tried Googling for the keywords "the theory of odd perfect numbers" and one of the search results that came up was this document, titled ON THE DIVISORS OF THE SUM OF A GEOMETRICAL SERIES WHOSE FIRST TERM IS UNITY AND COMMON RATIO ANY POSITIVE OR NEGATIVE INTEGER.  (The author turns out to be J. J. Sylvester.)

Let me copy the first two paragraphs from that document:

A REDUCED Fermatian, $\frac{r^p - 1}{r - 1}$, is obviously only another name for the sum of a geometrical series whose first term is unity and common ratio an integer, $r$.

If $p$ is a prime number, it is easily seen that the above reduced Fermatian will not be divisible by $p$, unless $r - 1$ is so, in which case (unless $p$ is $2$) it will be divisible by $p$, but not by $p^2$.

Now, let me translate the second paragraph into modern language, fitted to the case of the divisor-sum function of the special prime power component of an odd perfect number $N = q^k n^2$, where $q$ is the special prime and satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  That is, we will be considering
$$\sigma(q^k) = \frac{q^{k+1} - 1}{q - 1},$$
where $\sigma(x)=\sigma_1(x)$ is the classical divisor sum function of $x$.

(Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.)

Let us now see what we have got.  Translating to modern language (in the context of $\sigma(q^k)$):

PROPOSITION: If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).

Here is my:

QUESTION: What happens to the PROPOSITION when $k=1$?

MY ATTEMPT

Suppose that the Descartes-Frenicle-Sorli Conjecture holds.  Then $k=1$.

Then $k+1=2$ is prime.  Also, since $q \equiv 1 \pmod 4$, then $q - 1 \equiv 0 \pmod 4$, so that $2 = (k + 1) \mid (q - 1)$.  Then $\sigma(q^k)$ is divisible by $(k+1)=2$ but not by $(k+1)^2=4$, which is true.  So what's with the UNLESS?

Here is a proof for $\sigma(q^k) \equiv 2 \pmod 4$:

First, since $q \equiv 1 \pmod 4$, then
$$\sigma(q^k) = 1 + q + \ldots + q^k \equiv k + 1 \pmod 4.$$
Next, since $k \equiv 1 \pmod 4$, then
$$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$
QED

答:
This is deduced due to a famous lemma often called as "LTE lemma" (Lifting the exponents). The lemma is as follows:

If $p\not=2$ is a prime number and $a,b$ are positive integers such that $p|a-b$, then for every positive integer $n$ we have : $V_p(a^n-b^n)=V_p(a-b)+V_p(n)$

The lemma also holds for $p=2$ but it's slightly different in that case. This theorem can be easily proved using induction.

Anyways using LTE lemma on your problem, since $p=k+1|q-1$, then $$V_p(\frac{q^{k+1}-1}{q-1})=V_p(q^{k+1}-1)-V_p(q-1)=V_p(q-1)+V_p(k+1)-V_p(q-1)=V_p(k+1)$$
But we know that $V_p(k+1)=V_p(p)=1$ so finally $V_p(\frac{q^{k+1}-1}{q-1})=1$ hence the sum of divisors is divisible by $p$ but not by $p^2$.

P.S. Here is almost everything on the lemma


问:
Mean squared error and Chebyshev's inequality
How do I prove the following using Chebyshev's inequality?

Let $X$ and $Y$ be two random variables satisfying that $\mathbb{E}\left[(Y-X)^2\right]=0$, then $P(X\neq Y)=0$.

答: I believe that in this case Markov's inequality is more appropriate. Let $\varepsilon > 0.$ Then $\mathbb{P}(\vert X - Y \vert \geq \varepsilon) = \mathbb{P}(\vert X - Y \vert^2 \geq \varepsilon^2) \leq \frac{\mathbb{E}[(X - Y)^2]}{\varepsilon^2} = 0.$ Thus the sets $\vert X - Y \vert^{-1}([\varepsilon, \infty))$ have probability $0$ and are nested. It follows that $\mathbb{P}(\vert X - Y \vert \neq 0) = \lim_{\varepsilon \to 0} \mathbb{P}(\vert X - Y \vert \geq \varepsilon) = 0,$ i.e. $\mathbb{P}(X \neq Y) = 0.$ I hope this helps. :)


问:
Isomorphic quotient groups
Suppose that $G$ is a finite group, and that $H$ and $K$ are normal subgroups of $G$ with trivial intersection, and suppose that $H$ and $K$ are isomorphic. Is it true that the quotient groups $G/H$ and $G/K$ are isomorphic? </p>

答:
As Mariano has shown, the answer is a clear no.

The best repair I can think of is the following: suppose that $H$ and $K$ are normal subgroups of a group $G$ such that there exists an automorphism $\varphi: G \rightarrow G$ with $\varphi(H) = K$.  Then $G/H \cong G/K$: indeed, the isomorphism is induced by $\varphi$.  In particular, the above condition holds if $H$ and $K$ are conjugate subgroups of $G$, which is useful enough to be worth remembering.

Note finally that the condition that $H \cap K = \{e\}$ seems to have nothing to do with anything: it neither helps nor hurts the desired conclusion, so far as I can see.

Title: Isomorphic quotient groups


问:
Let the pair $(X, Y)$ have the bivariate normal distribution. Show that $aX + bY$ has a univariate normal distribution
Let the pair $(X, Y)$ have the bivariate normal distribution of (6.76), and let $a, b \in \mathbb{R}$.
Show that $aX + bY$ has a univariate normal distribution, possibly with zero variance.

$$g(x,y)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}e^{-\frac{1}{2}Q(x,y)} \text{ for } x,y \in \mathbb{R}, \tag{6.76} $$

where

$$Q(x,y)=\frac{1}{1-\rho^2}\left[\left(\frac{x-\mu_1}{\sigma_1} \right)^2-2\rho \left(\frac{x-\mu_1}{\sigma_1}\right) \left(\frac{y-\mu_2}{\sigma_2}\right) + \left(\frac{y-\mu_2}{\sigma_2}\right)^2 \right]$$ for $\mu_1,\mu_2 \in \mathbb{R}, \sigma_1,\sigma_2 >0, -1<\rho<1$.

I have no idea how to do this problem. I was thinking maybe trying to split $g(x,y)$ into separate marginal density functions but that math seems way too complicated considering what $g$ is.

答:
We do the following reversible transformation on the random vector
$$
\left[\begin{array}{l}
U \\
Y
\end{array}\right]=\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{l}
X \\
Y
\end{array}\right]
$$
where require
$$
\left|\begin{array}{cc}
a & b \\
-b & a
\end{array}\right|=a^{2}+b^{2} \neq 0
$$
This means $a$ and $b$ are not zero at the same time. Now what we know is the PDF of  random vector $\left[\begin{array}{l}
X \\
Y
\end{array}\right] $
,that is $g(x,y)$ as mentioned above.
Then we can get the PDF of  random vector $\left[\begin{array}{l}
U \\
V
\end{array}\right] $ , denote it by $q(u,v)$.
Note the Jacobi Determinant $|J| = \frac{1}{a^2+b^2}$. To simplify the result, without the generality, we can assume $a = \cos\alpha, b=\sin\alpha$ where $0\le\alpha\le2\pi$ , after compulation, we can get the following fomular
$$
\begin{aligned}
q(u, v) &=g(u \cos \alpha-v \sin \alpha, u \sin \alpha+v \cos \alpha) \\
&=\frac{1}{2 \pi \sigma_{1} \sigma_{2} \sqrt{1-\rho^{2}}} \exp \left\{-\frac{1}{2\left(1-\rho^{2}\right)}\left(A u^{2}-2 B u v+C v^{2}\right)\right\}
\end{aligned}
$$
where
$$
\begin{array}{l}
A=\frac{\cos ^{2} \alpha}{\sigma_{1}^{2}}-2 \rho \frac{\cos \alpha \sin \alpha}{\sigma_{1} \sigma_{2}}+\frac{\sin ^{2} \alpha}{\sigma_{2}^{2}} \\
B=\frac{\cos \alpha \sin \alpha}{\sigma_{1}^{2}}-\rho \frac{\sin ^{2} \alpha-\cos ^{2} \alpha}{\sigma_{1} \sigma_{2}}-\frac{\cos \alpha \sin \alpha}{\sigma_{2}^{2}} \\
C=\frac{\sin ^{2} \alpha}{\sigma_{1}^{2}}+2 \rho \frac{\cos \alpha \sin \alpha}{\sigma_{1} \sigma_{2}}+\frac{\cos ^{2} \alpha}{\sigma_{2}^{2}}
\end{array}
$$
Now we get the following conclusion:

1. The random vector obtained by the coordinate rotation transformation of the two-dimensional normal vector still obeys the normal distribution.

Now what we want to know is the PDF of random variable $U = aX+bY$ which is just Marginal probability density function of $g(x,y)$ , the Join Probability Density Fuction

There is another conclusion about Normal Distribution as following.

2. The marginal distribution of the bivariate normal distribution is still the normal distribution

That what we just want to get!!!

Hope my answer can help you!!!


问:
Arranging numbers in an array (Swedish Math Olympiad 1986)
Consider an $m\times n$ array of real numbers.  Let $d>0$.  Suppose that the difference between the maximum number and the minimum number in each row is at most $d$.  We then sort the numbers in each column so that the maximum element is in the first row and the minimum element is in the  last row (so each column is arranged in decreasing order).  Show that the difference between the maximum number and the minimum number in each row (after the change) is still $\le d$.

Source : Swedish Math Olympiad 1986, also Principles and Techniques in Combinatorics

Here is my attempt.  Let $M_k$ and $m_k$ be the original largest and smallest elements of the $k$th row.  Let $M'_k$ and $m'_k$ be the new largest and smallest elements of the $k$th row.  We have $M'_1\ge M'_2 \ge ... \ge M'_m$ and $m_1'\ge m_2' \ge ... \ge m'_m$.  Let's try induction on $k$.

For $k=1$, suppose that $M'_1$ was originally in row $i$.  Then $M'_1\le M_i$.  We know that $m'_1$ is in the same column as some element originally from row $i$.  Let that element be $a_i$.  Therefore $m'_1\ge a_i\ge m_i$ so $M'_1-m'_1\le M_i-m_i\le d$.

Let $1<k\le m$.  Suppose that $M'_k$ was originally in row $i$ so $M'_k\le M_i$.  We know that $m'_k$ is in the same column as some element $a_i$ originally from row $i$.  Suppose that $a_i$ is in row $j$ after the change.  If $j\ge k$, then $m'_k\ge a_i\ge m_i$ so $M'_k-m'_k\le M_i-m_i\le d$.  We now assume $j<k$.  What to do here?

BTW in the book "Principles and Techniques in Combinatorics" this problem is in chapter 3 which is about pigeonhole principle.  So if you have a solution with pigeonhole that'd be great.

答:
Here is another approach, though it again doesn't use PP.

We first prove the 2 column case. Consider the bipartite graph, with vertices represented by these values in each column, and we draw an edge between 2 values in different columns if their difference is at most $d$.

Observe that each $a_i$ is connected to (at least the) corresponding $b_i$.

For a subset $C$ consisting of vertices in a column, let $Im(C)$ be the vertices that they are connected with. Then, the previous observation establishes $ |C| \leq |Im(C)|$.

Now, when we re-arrange the columns, suppose that $a_i'$ is not connected to $b_i'$.

$a_i'$ is connected to some $b_k'$. Let the minimal index be $K$, and $ i \neq K$.

WLOG $i < K$. (The case of $i>K$ can be dealt with similarly.)

Then, with $C = \{b_1', b_2', \ldots b_{K-1}'\}$, $Im(C) \subset \{a_1', a_2', \ldots a_{i-1}'\}$.

This gives us $|C| > |Im(C)|$, which is a contradiction.

Hence, $a_i'$ is connected to $b_i'$, so their difference is at most $d$.

Now, for the general $n-$column case, the above establishes that for any 2 columns, the rearranged columns have entries with difference at most $d$ in each row. Hence, this is true for the entire row.


答:
Define $\vec{f}(x)=\frac{\vec{x}}{x\cdot x}$, which coordinate-wise is
$$
f_j(x)=\frac{x_j}{x\cdot x}\tag1
$$
The Jacobian of $f$ is
$$
\partial_if_j(x)=\frac{\delta_{ij}x\cdot x-2x_ix_j}{(x\cdot x)^2}\tag2
$$
Written as matrices, this is
$$
\frac{I-2\frac{xx^T}{x^Tx}}{x^Tx}\tag3
$$
The matrix $I-2\frac{xx^T}{x^Tx}$ is a reflection in the direction of $x$; that is, locally, the map flips in the radial direction and scales by the inverse of the square of the distance from the origin.

The reflection reverses the orientation.

Title: Show $x \mapsto \frac{x}{\Vert x\Vert^2}$ is orientation reversing.

Body:
I wanted to show that the map $\mathbb{R}^n- \{0\} → \mathbb{R}^n- \{0\},\ x\mapsto \frac{x}{\Vert x\Vert^2}$ is orientation reversing. But i don't know how to tackle it appropriately.

The solution proposed the following:

At $p = (1,0,...,0)$ the differential of the given map is the identity
on $0 \oplus \mathbb{R}^{n-1} \subset T_p\mathbb{R}^n$ while it is
multiplication by $-1$ on $\mathbb{R}  \oplus 0 \subset
 T_p\mathbb{R}^{n}$.

Then they justify it by connectedness of $\mathbb{R}^{n}$ for $n$ greater than $1$.

I am not sure whether I fully understand the solution. For simplicity let's assume $n = 2$, then at $(1,0)$ the differential of the map is $(-1,1)$, similarly $(-1,1,1...,1)$ for arbitrary $n$. How is it the identity on $0 \oplus \mathbb{R}^{n-1} \subset T_p\mathbb{R}^n$?

Could someone elaborate? I tried to look up similar problems in my books but couldn't find any. I would really like to understand this.

Thanks for any help!


问:
Term for number only divisible by 1, itself, and its square root
Do numbers only divisible by 1, themselves, and their square roots have a specific term?

Seems like they're all squares of primes: 1, 9, 25, 49, 121, 169...
I feel like I'm missing something really obvious, I apologize if that's the case. Searching using the obvious keywords didn't bring anything directly referencing this class of numbers to light.

答: They are the numbers such that $k = p^2$, where $p$ is a prime, and also $1$.


问:
Suppose B is diagonalizable, then $e^{B}$ is diagonalizable and $e^{\lambda}$ is an eigenvalue of $e^{B}$ if and only if $\lambda$ is an eigenval of B
As stated above, Suppose B is diagonalizable, then $e^{B}$ is diagonalizable and $e^{\lambda}$ is an eigenvalue of $e^{B}$ if and only if $\lambda$ is an eigenvalue of B. I'm struggling to determine if this is true or not, is there proof or theorem to justify this one way or the other?

答:
Let $B$ be diagonalised as $B=UDU^*$, $U$ is a unitary matrix and $D=diag(\lambda_1, \lambda_2, \cdots, \lambda_n)$. Then Note that $e^B=e^{UDU^*}=Ue^DU^*$ and $e^D=diag(e^{\lambda_1}, e^{\lambda_2}, \cdots, e^{\lambda_n})$ via the expansion $e^A=\sum_{i=0}^\infty \frac{A^i}{i!}$. Now since $e^B=U \begin{pmatrix}
e^{\lambda_1} & 0 & \cdots  & 0\\
0  & e^{\lambda_2} & \cdots & 0 \\
  & \cdots &   \\
0 & 0 & \cdots & e^{\lambda_n}
\end{pmatrix} U^*=U \begin{pmatrix}
e^{\lambda_1+2m_1 \pi i} & 0 & \cdots  & 0\\
0  & e^{\lambda_2+2m_2 \pi i} & \cdots & 0 \\
  & \cdots &   \\
0 & 0 & \cdots & e^{\lambda_n+2m_n\pi i}
\end{pmatrix} U^*$   for positive integers $m_1, m_2, \cdots, m_n$, you can see that only "$\Rightarrow$" is true.


问:
Why isn't the surface area of a shell in a sphere the difference between two spherical caps?
In geometry, a spherical cap or spherical dome is a portion of a sphere or of a ball cut off by a plane. It is also a spherical segment of one base, i.e., bounded by a single plane. [(Wikipedia)][1]

[In a proof of the Shell Theorem, it is stated that the area of a thin ring of shell of a sphere is 2πR2sin(θ)dθ][2]

The thing is, I can actually understand the derivation of this formula but even so, I'm stuck on why it can not be the difference of caps, as:

$2πRh_1 - 2πRh_2$

Is this something similar to the volume of a cone not being $πr²h$?
[1]: https://en.wikipedia.org/wiki/Spherical_cap
[2]: https://i.stack.imgur.com/h9LMs.png

答:
Let us use $z$ instead of $h$. Assume $0\le z_1< z_2 \le R$, then notice that
\begin{align}
\text{area of the ring}=-2\pi R(z_2-z_1) = -2\pi R\Delta z.
\end{align}
Infinitesimally, we have that
\begin{align}
\text{area of the ''thin" ring}=-2\pi R\ dz.
\end{align}
Since $z = R\cos\phi$, then we have that
\begin{align}
dz = \frac{dz}{dR} dR+\frac{dz}{d\phi} d\phi = -R\sin\phi d\phi
\end{align}
which means
\begin{align}
\text{area of the ''thin" ring}=2\pi R^2\sin\phi d\phi.
\end{align}


问:
Solving a Linear Congruence
I've been trying to solve the following linear congruence with not much success:

$19\equiv 21x\pmod{26}$

If anyone could point me to the solution I'd be grateful, thanks in advance

答:
Hint: $\,{\rm mod}\ 26\!:\ \dfrac{19}{21} \equiv \dfrac{-7\cdot 1\ \ \,}{7\cdot 3}\equiv -\dfrac{27}{3}\equiv -9.\ \ \ $  Alternatively

note: $\,{\rm mod}\ 26\!:\ \dfrac{19}{21} \equiv \dfrac{-7}{-5} \equiv -7\left(\dfrac{-1}5\right) \equiv -7\left(\dfrac{25}5\right) \equiv -35 \equiv -9$

While such ad-hoc techniques often work quickly for small numbers, for larger numbers one should resort to algorithms such as the extended Euclidean algorithm.  Also, keep in mind that modular fractions uniquely exist only when the denominator is coprime to the modulus.

Title: Solving a Linear Congruence


问:
Find an angle to accelerate at to most quickly go from one movement vector to another
Okay, so this is with respect to game design, so that’s where I’m coming from (please try to use smol words, I am no mathematician)

I have a 2D space ship. Its velocity is defined by vector A, let’s say (1i,0j), and the ship’s current position is (0,0). I also have a point P, let’s define it as (0,3). I’m trying to figure out how to select an angle at which the ship can be constantly accelerated at a rate of 1 unit/sec^2 in order to go from traveling along vector A to having a velocity that is directed exactly at point P. Ideally, this angle would result in the ship reaching the correct vector before it crosses point P in order for it to have time to turn around and negatively accelerate.
I’m assuming I could find a way to plot a parabola using the calculus and stuff but it’s a little beyond me at the moment.
Any help is appreciated :)

答:
To answer this problem we will generalize the equation for acceleration in one direction. From kinematics, we know that:
$$x_f = x_0 + v_{x0}\,t + \frac{1}{2}\,a_x\,t^2$$
Solving for acceleration, we find that:
$$a_x = \frac{2\,(x_f - x_0 - v_{x0}\,t)}{t^2}$$
We will say that the initial position is always zero relative to the position of the ship, meaning that $x_0 = 0$. This simplifies our equation to:
$$a_x = \frac{2\,(x_f - v_{x0}\,t)}{t^2} = 2\,x_f\,t^{-2} - 2\,v_{x0}\,t^{-1}$$
In order to find the optimal acceleration, we must find the critical points for $t$ by taking the time derivative:
$$\frac{da_x}{dt} = -4\,x_f\,t^{-3} + 2\,v_{x0}\,t^{-2}$$
Looking at a graph, we can see that this is equal to zero only when $t = 2$ meaning:
$$a_x = \frac{2\,[x_f - v_{x0}\,(2)]}{(2)^2} = \frac{x_f - 2\,v_{x0}}{2}$$
This equation is the optimal acceleration required in a single direction to get to a desired location where $x_f$ is the distance from the ship to the destination and $v_{x0}$ is the initial velocity in the same direction. To find the magnitude and direction of the acceleration vector in 2D space:
$$a = \sqrt{a_x^2 + a_y^2}$$
$$\theta = \arctan{\left(\frac{a_y}{a_x}\right)}$$
You should always check that the angle makes sense with the components of your acceleration vector. If they do not match up, you can usually add $180^{\circ}$ to your angle value.


问:
Solving $A = 94\left(1 - e^{-0.43 B}\right)^{0.52}$ for $B$
I am not a regular math user. I need to solve an equation to apply on timeseries variables.

I have value of $A$ but don't have value of $B$ variable. Here, time series variables (mentioned as variable $A$) with known values are canopy cover values (range $40$ to $100$). Whereas variable $B$ (unknown values) are the leaf area index (LAI). The output LAI values (variable $B$) might be in $0.5$ to $10$.

The equation:
$$ A = 94\left(1 - e^{-0.43 B}\right)^{0.52}$$
(In MS Excel: A = 94*POWER(1-EXP(-0.43*B),0.52)})

What would be the equation that can be applied to get value of "$B = \text{expression}$"?

This might be quite simple, but for me it's not. Please help.

答:
The equation is easily solved for $B$ in terms of $A$ by "reversing" each of the operations that appears on the right-hand side:

$$ A = 94(1 - e^{-0.43 B})^{0.52} $$

$$ A/94 = (1 - e^{-0.43 B})^{0.52} $$

$$ (A/94)^{1/0.52} = 1 - e^{-0.43 B} $$

$$ e^{-0.43 B} = 1 - (A/94)^{1/0.52} $$

$$ -0.43 B = \ln (1 - (A/94)^{1/0.52}) $$

Therefore:

$$ B = \frac{\ln (1 - (A/94)^{1/0.52})}{-0.43} $$

This derivation of $B$ and especially the last line make it clear that we want $A$ to be nonnegative and less than $94$ in order for there to be a corresponding real value of $B$.  Indeed the last line gives $B$ as the ratio of two negative numbers, so that $B\gt 0$ when $0 \lt A \lt 94$.  This is mostly consistent with what the Question says about "canopy cover values" except those in the range for $A$ at or above $94$.


问:
number of valuation ring of $\Bbb Q_p((T))$
I would like to know the number of valuation rings of  $\Bbb Q_p((T))$.
I know $\Bbb Q_p$ has $2$ valuation rings, that is,$\Bbb Q_p$ and $\Bbb Z_p$.
Every algebraic extension of $\Bbb Q_p$ has more than $2$ valuation rings because of extension theorem on valuation.But $\Bbb Q_p((T))$ is not algebraic over $\Bbb Q_p$,I am at a loss.

How many valuation rings of  $\Bbb Q_p((T))$ are there?

Thank you in advance.

答:
A valuation ring as you define it in a comment can also be described via a valuation, i.e. a surjective homomorphism $v: K^\times \twoheadrightarrow \Gamma$ onto a totally ordered abelian group $\Gamma$.

You should first of all note (also w.r.t. your previous question) that if we stay in this generality, almost all fields you can reasonably think of have infinitely many valuation rings, although you might not be able to see them:

Namely, let $K$ be any (!) field that embeds into $\mathbb C$ (and unless you have a good point against the axiom of choice, this includes all $\mathbb Q_p$ and also the example $\mathbb Q_p((T))$ in this question). Now there are isomorphisms (unless you have a good point against the axiom of choice) $\mathbb C \simeq \mathbb C_\ell$ where $\mathbb C_\ell$ is the completion of the algebraic closure of the $\ell$-adic numbers $\mathbb Q_\ell$ for your second favourite prime $\ell$. Said field $\mathbb C_\ell$ comes naturally equipped with its $\ell$-adic valuation $v_\ell$ with value group (written multiplicatively) $\Gamma = \ell^\mathbb Q$. The restriction of this valuation along the isomorphisms and to the field $K$ is a valuation on $K$, which when restricted to $\mathbb Q$ gives the $\ell$-adic valuation. In particular, all these valuations are distinct and thus we have (modulo choice) infinitely many distinct valuation rings in $K$.

So that kind of dooms your hope to find a field with three (or any finite number of) valuation rings, at least in characteristic $0$. (Update: As reuns comments and I spelled out in an answer to your previous question, actually every field that is not algebraic over some $\mathbb F_p$ has infinitely many valuation rings -- and those that are algebraic over some $\mathbb F_p$ of course have only one, namely themselves.)

Good news is, if you restrict your search to valuation rings whose value group $\Gamma$ (from now on written additively) is discrete in the sense that it has a smallest element $>0_\Gamma$ (which unfortunately is not the same as what we call a DVR; the difference is that here I allow "discrete" value groups of higher rank, i.e. basically $\mathbb Z^n$), then I think you've found a match here.

To see that, adapt reuns' beautiful answer: The $n$-divisibility of that big subgroup $(1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]], \cdot) \subset K^\times$ still shows (because of our "dsicreteness" assumption) that all its elements must have value $0$, and of course so do all roots of unity; if $v(p)=0$ continue as in his proof and either get the trivial valuation $w_0(K^\times)=0$ i.e.

$$R_0=K$$

or the discrete (rank 1) valuation $w_1(\sum a_i T^i)=\min(i: a_i \neq 0)$ i.e.

$$R_1= \mathbb Q_p[[T]].$$

But now we could also have the option that $v(p) > 0$ as long as we make sure (cf. reuns' first bullet) that $v(p^{-r}T)$ is always $\ge 0$, meaning that the value of $T$ has to be "infinitely bigger" than that of $p$. That forces us basically to have $\mathbb Z \oplus \mathbb Z$ as value group, with the generator $v(T) := (1,0)$ "infinitely bigger" than the other generator $v(p)=(0,1)$, i.e. the lexicographic order. Call this rank 2 valuation $w_2 : K^\times \rightarrow \mathbb Z \oplus \mathbb Z $, it is explicitly given on $x= \sum a_i T^i$ as

$$w_2(x) := \left(w_1(x), v_p(a_{w_1(x)})\right).$$

Its valuation ring is

$$R_2 = \{x \in K: w_2(x) \ge (0,0)\} = \{\sum_{i \ge 0} a_i X^i \in \mathbb Q_p[[X]]: a_0 \in \mathbb Z_p\}$$

and there you got your three possible valuation rings with "discrete" value groups (of ranks $0,1,2$) $R_0, R_1, R_2$.

Compare the other answers to Concrete examples of valuation rings of rank two. for another example.


问:
Derivative of a conditional density
I would like to find the derivative of $f(y|u)$ with respect to c where $f(y|u)$ follows $N(x,a^2)$ and $x=g(c)$. So far I got
$f'(y|u)$ = $-0.5a^2N(x,a^2)d/dc(y|u-x)$. But how do I take the derivative of $y|u$ if I don't know that. Only I know is that $y|u$ follows normal and $y=x+e$ where $e$ is random noise.

答: The density of $N(g(c), a^2)$ evaluated at some number $y$ is $\frac{1}{\sqrt{2\pi a^2}} e^{-(y-g(c))^2/(2a^2)}$. Treat $y$ and $a$ as fixed, and take the derivative of this expression with respect to $c$. The result should involve both $g(c)$ and $g'(c)$.

Title: Derivative of a conditional density


问:
Finding ML estimate with constant?
I had a (truncated at zero) Poisson distribution with pmf $f(x;\theta)=\frac{e^{-\theta}\theta^{x}}{x!(1-e^{-\theta})}$ and $x=1, 2, ...$ and $\theta>0$ where $\theta$ is the parameter and $X$ is the r.v.

My lecturer asked how we'd find the ML estimate of $f(k;\theta)$ with fixed integer $k>0$, and I'm really not sure how you'd do that.

The likelihood function generally involves a product of pmfs for each random variable in a sample, whereas this has a random sample of $1$. So does that mean the likelihood function would look like $L(\theta)=f(k;\theta)=\frac{e^{-\theta}\theta^{k}}{k!(1-e^{-\theta})}$? And then you could find the derivative as $\frac{\partial}{\partial \theta}L(\theta)=\frac{1}{k!}\frac{\partial}{\partial \theta}(\frac{\theta^k e^{-\theta}}{1-e^{-\theta}})$. But, I can see that (after using a calculator) $\frac{1}{k!}\frac{\partial}{\partial \theta}(\frac{\theta^k e^{-\theta}}{1-e^{-\theta}})=0$ has no solution for $\theta$, so I must have gone wrong somewhere.

Any tips would be appreciated.

答:
Most of the times it is easier to work with the log-likelihood function, so in this case that would be $l(k, \theta) = - \theta + k \log \theta - \log k! - \log(1 - e^{-\theta}).$ We differentiate $l$ with respect to $\theta$ and equate it to $0$ to obtain $$\frac{\partial l}{\partial \theta} (k, \theta) = -1 + \frac{k}{\theta} - \frac{e^{-\theta}}{1 - e^{- \theta}} = 0,$$ so $\frac{k}{\theta} = \frac{1}{1 - e^{-\theta}}$ or $\frac{\theta}{1 - e^{- \theta}} = k.$ A little analysis helps us conclude that for all $k$ there is a unique solution to this equation, call it $\theta_k.$ Moreover, the graph of the function $g(\theta) = \frac{\theta}{1 - e^{-\theta}}$ has the asymptote $y = x$ at $\infty,$ so $\theta_k$ is approximately $k$ if $k$ is large enough. I hope this helps. :)

Edit: Note that $\theta = k - k e^{-\theta}$ means that $\theta e^\theta = k e^\theta - k,$ so $e^{\theta}(\theta - k) = -k,$ from which we get $(\theta - k) e^{\theta - k} = - k e^{-k}.$ Then $\theta - k = W(-k e^{-k}),$ so $\theta_k = k + W(-k e^{-k}),$ where $W$ denotes the Lambert W-function.

Title: Finding ML estimate with constant?


问:
Solving the recurrence relation $T(n)=3T\Big(\frac{n}{2}\Big)+n$

Use the expand, guess and verify approach to solve the following recurrence relation
$$\cases{T(1) = 1\\T(n) = 3T\Big(\frac{n}{2}\Big)+n & for $n > 1$}$$

I think it's defined for $n=2^m$, $m>1$, so here's my attempt: $$T(n)=3T\Big(\frac{n}{2}\Big)+n=T(n)=3^2T\Big(\frac{n}{2^2}\Big)+\frac{3}{2}+n=3^3T\Big(\frac{n}{2^3}\Big)+\frac{3^2}{2^2}n+\frac{3}{2}n+n$$

I guess that after $k$ steps, we get $$T(n)=3^kT\Big(\frac{n}{2^k}\Big)+n\sum_{i=0}^{k-1} \Big(\frac{3}{2}\Big)^i$$

Setting $k=\log n$, $$T(n)=3^{\log n}+n\sum_{i=0}^{\log (n)-1} \Big(\frac{3}{2}\Big)^i$$

I'm having a hard time proving the previous formula by induction. I tried this:

Base case: $$T(1)=3^{0}+1\sum_{i=0}^{0-1} \Big(\frac{3}{2}\Big)^i =(?) 1$$

Is that summation equal to zero? Those indices got me wondering. Also, assuming everything is correct so far, I got stuck when trying to write down the induction step.

Inductive step: If $T(k)$ is true, then $T(2k)$ must be true too. So, $$T(2k)=3T(k)+2k$$ $$T(2k)=3^{\log 2k}+3k\sum_{i=0}^{\log (k)-1} \Big(\frac{3}{2}\Big)^i + 2k$$ The last step doesn't seem to be correct. Any help, please? Thanks in advance.

答:
So indeed,
$$
\begin{split}
T(n)
 &= 3T\left(\frac{n}{2}\right) + n \\
 &= 3 \left[3T\left(\frac{n}{4}\right) + \frac{n}{2}\right] + n\\
 &= 3^2 T\left(\frac{n}{2^2}\right) + \frac{3n}{2} + n\\
 &= 3^3 T\left(\frac{n}{2^3}\right) + \frac{3^2}{2^2} n + \frac{3n}{2} + n\\
 &= \ldots \\
 &= 3^k T\left(\frac{n}{2^k}\right) + n \sum_{i=0}^{k-1} \left(\frac32\right)^i\\
 &= 3^k T\left(\frac{n}{2^k}\right) + n \frac{(3/2)^k - 1}{3/2-1}\\
 &= 3^k T\left(\frac{n}{2^k}\right) + 2n \left[\left(\frac32\right)^k - 1\right]\\
 &= 3^k \left[T\left(\frac{n}{2^k}\right) + \frac{2n}{2^k} \right] - 2n\\
\end{split}
$$
Finally, substituting $k= \log_2 n = \lg n \iff n = 2^k$, we get
$$
T(n)
 = 3^{\lg n} \left[T(1)+2\right]-2n
 = 3 \cdot 3^{\lg n}-2n
 = 3^{\lg (2n)}-2n.
$$

For the inductive step, note $\lg(2n) = \lg 2 + \lg n = 1 + \lg n$, so you assume $T(n) = 3^{\lg (2n)}-2n$ and prove that $T(2n) = 3^{\lg (4n)} - (4n)$.


问:
Hartogs' Theorem

Show that the well-ordering theorem is equivalent to:

  For any two sets $X$ and $Y$ there is an injection of $X$ to $Y$ or $Y$ to $X$.

I have only some ideas for the "$\Leftarrow$" direction:

If we have $f:X\to Y$ is injective, then follows ((5.2) in Jech) that $|X|\geq |Y|$.

(By choosing $f^{-1}(\{y\})$ for each $y\in Y$.)

Since we can choose $X$ and $Y$ arbitrarily, I thought we can then compare all elements in $X$ and also find a least element. But I don't really know how to get there formally.

For the other direction I have no idea yet.

I'd be very happy for any hint how to approach this proof!

答:
Hint: For a set $X$ show that there exists a least ordinal $\kappa$ such that $\kappa\nleq|X|$. In order to show its existence you will need to show it is the range of a function whose domain is $\mathcal{P(P(}X))$ (and use replacement).

Now we have that $\kappa\nleq|X|$ so $|X|<\kappa$. Conclude Zermelo's theorem.

Title: Hartogs' Theorem


答:
Hint:

$$\begin{align}T(n) &= 3T\left(\dfrac n2\right) + n\tag{1} \\ T(n) &= 3^{\log n} + n\sum_{i=0}^{\log (n)-1} \left(\frac{3}{2}\right)^i\tag{2}\end{align}$$

If you set $n = 1$ in as your base case, then you end up with $T\left(\dfrac 12\right)$ in $(1)$ which does not mean anything. Instead, set $n = 2$ as your base case. Then, $(1)$ becomes $$T(2) = 3\times T(1) + 2 = 3\times 1 + 2 = 5$$ and $(2)$ becomes $$T(2) = 3^{\log 2} + 2\times\left(\dfrac 32\right)^0 = 3^1+2\times1 = 5$$ and you can see that $(1)\equiv(2)$. The base case, therefore, holds.

Title: Solving the recurrence relation $T(n)=3T\Big(\frac{n}{2}\Big)+n$

Body:
Use the expand, guess and verify approach to solve the following recurrence relation
$$\cases{T(1) = 1\\T(n) = 3T\Big(\frac{n}{2}\Big)+n & for $n > 1$}$$

I think it's defined for $n=2^m$, $m>1$, so here's my attempt: $$T(n)=3T\Big(\frac{n}{2}\Big)+n=T(n)=3^2T\Big(\frac{n}{2^2}\Big)+\frac{3}{2}+n=3^3T\Big(\frac{n}{2^3}\Big)+\frac{3^2}{2^2}n+\frac{3}{2}n+n$$

I guess that after $k$ steps, we get $$T(n)=3^kT\Big(\frac{n}{2^k}\Big)+n\sum_{i=0}^{k-1} \Big(\frac{3}{2}\Big)^i$$

Setting $k=\log n$, $$T(n)=3^{\log n}+n\sum_{i=0}^{\log (n)-1} \Big(\frac{3}{2}\Big)^i$$

I'm having a hard time proving the previous formula by induction. I tried this:

Base case: $$T(1)=3^{0}+1\sum_{i=0}^{0-1} \Big(\frac{3}{2}\Big)^i =(?) 1$$

Is that summation equal to zero? Those indices got me wondering. Also, assuming everything is correct so far, I got stuck when trying to write down the induction step.

Inductive step: If $T(k)$ is true, then $T(2k)$ must be true too. So, $$T(2k)=3T(k)+2k$$ $$T(2k)=3^{\log 2k}+3k\sum_{i=0}^{\log (k)-1} \Big(\frac{3}{2}\Big)^i + 2k$$ The last step doesn't seem to be correct. Any help, please? Thanks in advance.


问:
How to determine the Galois group knowing the following
Let's suppose I have the polynomial $p$ of degree $n$:

$p(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$

Let's also suppose I have another polynomial $f$ of degree $m$, which their roots $r$ satisfy a linear relation, for instance, the following one:

$r_f = r_{p_n} + r_{p_{n-1}}$

Is there any relation between their groups? Any references on this particular matter?

EDIT

I have read this https://www.researchgate.net/publication/348814816_Matrices_and_Galois_Theory and I am attempted to conclude the following. Given a field $P$ an its extension (generated by the roots of $p(x)$) $P(p_1, p_2, \ldots, p_n)$ contains the another extension $P(p_1+p_2, p_1 +p_3, \ldots, p_{n-1}+p_n)$, which turns out to be the field of $f(x)$. Therefore,

$ H = P(p_1+p_2, p_1 +p_3, \ldots, p_{n-1}+p_n) \subseteq P(p_1, p_2, \ldots, p_n) = P$

Is this reasoning correct? Thanks

答: If the linear relation has rational coefficients (of course, any two sets of complex numbers satisfy relations if the coefficients can be arbitrary), then yes, the extension given by $f$ is contained in the extension given by $p$ implying a relationship between Galois group by the fundamental theorem of Galois theory. If the relation matrix is invertible, then the extensions are the same, as are the Galois group.


问:
Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?)
I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral

$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$

I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.

In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) - \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) - \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) - \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.

Any other approaches or suggestions to this challenging integral would be welcome.

答:
Approximating the integral using rational approximants

Using integration by parts, we easily get
$$I = \int_0^{\pi/2} \ln(1 - x\cot x) \mathrm{d} x
= - \frac{\pi^3}{24} + \frac{\pi}{2}\ln 2 - \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
$I \approx -3.353337262889472...$

We focus on
$$I_1 = \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
We may approximate $I_1$ by rational approximants to $\tan x$ at $0$.

Pade $(5, 4)$ approximant $\tan x \approx \frac{x(x^4 - 105x^2 + 945)}{15x^4 - 420x^2 + 945}
\triangleq g(x)$

We can bound the error by
$$0 \le \frac{x^3}{g(x) - x} - \frac{x^3}{\tan x - x} \le g(\pi/2) \approx 0.000045.$$
We have
$$I_1 \approx \int_0^{\pi/2} \frac{x^3}{g(x) - x} \mathrm{d} x =
- \frac{5}{112}\pi^3 + \frac{165}{56}\pi - \frac{243}{56}\sqrt{10}
\operatorname{arctanh} \frac{\pi\sqrt{10}}{30}.$$
Then we have $I \approx -3.353344689$ with relative error $< 0.000003$.


问:
Tensor product used in this notation
A connection form is a Lie-algebra-valued one form $\omega\in\Omega^1(P,\mathfrak{g})$ on a principal bundle P. I have also seen this written this as $\Omega^1(P)\otimes\mathfrak{g}$, but I do not see the reasoning behind this notation.

答:
If $M$ is a manifold and $V$ a vector space, you can think of a "$V$-valued one-form" on $M$ as element of $T^*M \otimes V$ in the following way. If $(x_1,\dots,x_n)$ are local coordinates on $M$ and $e_1,\dots,e_m$ is a basis of $V$, then the one-form which sends the basis vector $\frac{\partial}{\partial x_i}$ (of $TM$) to the basis vector $e_j$ (of V) can be thought of as the element $dx_i \otimes e_j$.

In other words, if $\omega \in T^*M$ is an ordinary one-form, and $v \in V$, and $X \in TM$ a tangent vector, then $\omega \otimes v$ is the $V$-valued one-form which acts on $X$ via

$$ (\omega \otimes v)(X) = \omega(X) \, v $$

Title: Tensor product used in this notation


问:
About a metric's symmetric properties...
Hello fellow mathematicians!

I was wondering why most metrics I have seen in Relativity are symmetric, also a property I have started to notice in my differential geometry book.

I guess the definition through the "dot-product" may be the reason why...

$$ g_{\alpha \beta}=\vec{x}_\alpha\cdot \vec{x}_\beta  $$

But:

Are there non-symmetric metrics?

What can be said about them?

Also, how do we call those metrics which only have non-zero components in the main diagonal?

And perhaps the question that has been bothering me a lot: What can be said about a metric having non-zero components out off the main diagonal?

I have been having thoughts on physical implications but since I'm not that far into Relativity nor Diff. Geometry, I can't really come up with something yet.

My main book has been Kreysizg's Differential Geometry, which may not be suited for some questions I have been thinking off, yet it has been an awesome book to start with! Any further recommendations?

答: You usually want a metric to "measure" distance in some way.  One property we want to be preserved is that the distance from A to B is the same as the distance from B to A.  This property is exactly what symmetry in the form means.  So if a metric is not symmetric then it doesn't really do anything useful.</p>


问:
Possible real extension of tetration, or ones with similar growth rate; what makes it difficult?
This question arose when I read a (very introductory) googology book. Tetration is essentially repeated exponentiation (right-associative), just like how multiplication is repeated addition, defined for integers $a>0,n\geq0$: $$a\uparrow\uparrow n=\begin{cases}1, n=0\\ a^{a\uparrow\uparrow(n-1)}, n>0\end{cases}$$

My question is whether there are nice extensions of this operation to the reals, like how the gamma function extends the factorial. Personally I don't expect it to be extended nicely to the complex numbers (given how $i^i$ is not even nicely defined), but maybe for reals we could use monotonicity (though very loose) to bound it somehow. I have seen an answer here but it seems to be a uniqueness theorem and an algorithm, and I wonder if there are more "explicit" constructions (using elementary functions, integrals, elliptic functions, etc). I may have overlooked something though.

I also considered a weaker version: can we find a simpler function that mimics the growth rate of tetration (and is also defined on the reals)? This should probably be an easier task, so in exchange I would like to see more "elementary" constructions. (By growth rate, I think big-O is too strict and maybe FGH is better, but I'm open to more reasonable options.)

At last, if we don't currently have an answer as satisfactory for tetrations as for factorial, what's the difficulty here? Is it because the recursive relation is more complicated? Or do we need stronger arithmetic to even prove that tetration is a function? (I don't know much about the latter though, and only listed it here as a possibility)

答:
Extensions of tetration

Counter-intuitively it appears that extending tetration to the complex numbers is easier than extending tetration to the real numbers. Likely because the complex numbers have more degrees of freedom.

Special functions for extending tetration - Faà Di Bruno's formula and the Bell polynomials are invaluable in extending tetration. They provide exact solutions to tetration of whole numbers. Most extensions of tetration to the real and complex numbers are only approximations.

As a member of the Ackermann function, tetration denotes a unique growth rate.

The difficulty with finding special functions for constructing tetration lies with the powers being so successful in constructing functions. But the umbral calculus emphasizes other fundamental processes besides the powers, like composition.

Consider iterated functions as iterated composition.

Faà Di Bruno's formula is
$$D^mf(g(z)) = \sum_{\pi(m)} \frac{m!}{k_1! \cdots k_m!}
(D^kf)(g(z))
\left(\frac{Dg(z)}{1!}\right)^{k_1} \cdots
\left(\frac{D^ng(z)}{m!}\right)^{k_m}$$

where $\pi(m)$ denotes a partition of $m$, usually denoted by $1^{k_1}2^{k_2}\cdots n^{k_m}$, with $k_1+2k_2+ \cdots mk_m=k$; where $k_i$ is the number of parts of size $i$. The partition function $p(m)$ is a decategorized version of $\pi(m)$, the function $\pi(m)$ enumerates the integer partitions of $m$, while $p(m)$ is the cardinality of the enumeration of $\pi(m)$.

Mathematica code for Faà Di Bruno's formula
$$\small D[f[g[z]], \{z, m\}] = \
\underset{K[1]=0}{\overset{n}{\sum }}f^{(K[1])}[g[z]]\ \text{BellY}\left[m,K[1],\text{Table}\left[g^{(K[2])}[z],\{K[2],1,-K[1]+m+1\}\right]\right]$$

$$D^m \ a\uparrow \uparrow z = D^m \ a^{(a\uparrow \uparrow {(z-1)})} = \sum_{\pi(m)} \frac{m!}{k_1! \cdots k_m!}
(D^k \ {a^z})(a^{(z-1)})
\left(\frac{D \ ^{z-1}a}{1!}\right)^{k_1} \cdots
\left(\frac{D^n \ ^{z-1}a}{m!}\right)^{k_m}$$

$$a\uparrow \uparrow b=\sum_{m=1}^{\infty} \frac{1}{m!} \ D^m \ (a\uparrow \uparrow b) \ (b-a\uparrow \uparrow \infty)^m$$

Fixed Points and Infinite Solutions

When the Lyapunov multipliers are a root of unity, a special solution is called for. Since there are infinite roots of unity, there are infinite solutions.

$f(0)=f_0=0$ - Fixed Point

$q \in \mathbb{Q}$, $n \in \mathbb{N^+}$, $\lambda \in \mathbb{S1}$

$\lambda = e^{2 q \pi i}$ - root of unity (symmetry)

$f'(0)=1$ - Abel's Functional Equation produces series of polynomials

$$f^t(z)=z+\frac{1}{2}t f_2 (z-f_0)^2 +\frac{1}{12}(3(t^2-t)f_2^2+2tf_3)(z-f_0)^3+\ldots$$

$f'(0)\ne \lambda^n$ - Schröder's Functional Equation produces series of exponentials

\begin{eqnarray}
        f^t(z)=f_0 &+& \lambda ^t (z-f_0)+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) f_2}{2 (-1+\lambda )} (z-f_0)^2 \\
        & + & \frac{1}{6}  \left(\frac{3 \lambda ^{-2+t} \left(-1+\lambda ^t\right) \left(-\lambda +\lambda ^t\right) f_2^2}{(-1+\lambda )^2 (1+\lambda )}+\frac{\lambda ^{-1+t} \left(-1+\lambda ^{2 t}\right) f_3}{-1+\lambda ^2}\right) (z-f_0)^3+\ldots
        \end{eqnarray}

$f'(0)= \lambda^n$ - Neutral Fixed Points

Neutral Fixed Points in Tetration

$e^{e^{2 \pi i x-{e^{2 \pi i x}}}}$

Tetration Mandelbrot Fractal

\begin{eqnarray}
        {}^t a = a_o & + & \lambda ^t\left(1-a_o\right)+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) \text{Log}\left(a_o\right){}^2}{2 (-1+\lambda )}\left(1-a_o\right){}^2 \\
        & + &\frac{1}{6}\text{  }\left(\frac{3 \lambda ^{-2+t} \left(-1+\lambda ^t\right) \left(-\lambda +\lambda ^t\right)\text{  }\text{Log}\left(a_o\right){}^4}{(-1+\lambda )^2 (1+\lambda )}\\
+\frac{\lambda ^{-1+t} \left(-1+\lambda ^t\right) \left(1+\lambda ^t\right)\text{  }\text{Log}\left(a_o\right){}^3}{(-1+\lambda ) (1+\lambda )}\right)\left(1-a_o\right){}^3+\ldots
        \end{eqnarray}

Convergence

The convergence of each type of fixed point needs to be considered in order to prove tetration as a whole always converges.

Convergence can be proven by functional equation.

Points described by the Schröder's Functional Equation are dense in the complex plane. It's fractal spirals are exponential spirals, excluding negative powers like $z^{-1}$. But repeated exponentiation is convergent.

On the other hand are the roots of unity which manifest as the symmetry of the exponential function. For each root there are an infinite number of "larger" associated numbers from the Schröder's Functional Equation that are convergent.


问:
What is the following limit?
How can I calculate the following limit?
What do I miss here? What am I doing wrong?

$$\begin{align*}
\lim_{x\rightarrow0}\frac{6x\cdot\sin x-6\cdot\sin\left(x^{2}\right)+x^{4}}{x^{5}\left(e^{x}-1\right)}= & \lim_{x\rightarrow0}\frac{6x\cdot\sin x}{x^{5}\left(e^{x}-1\right)}-\lim_{x\rightarrow0}\frac{6\cdot\sin\left(x^{2}\right)}{x^{5}\left(e^{x}-1\right)}+\lim_{x\rightarrow0}\frac{x^{4}}{x^{5}\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin x}{x}}_{\rightarrow1}-\lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin\left(x^{2}\right)}{x^{2}}}_{\rightarrow1}+\lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\infty\neq\frac{21}{20}.
\end{align*}$$

According to WolframAlpha $21/20$ is the solution. What am I doing wrong?

答:
Using the limit $\lim_{x\to 0}(e^x-1)/x=1$ the denominator of the expression under limit can be safely replaced by $x^6$. Next we add and subtract $6x^2$ in numerator and express it as $$6(x^2-\sin x^2)+6x\sin x-6x^2+x^4$$ This allows us to split the desired limit as a sum of two limits the first one of which is $$6\lim_{x\to 0}\frac{x^2-\sin x^2}{x^6}=6\lim_{t\to 0}\frac{t-\sin t} {t^3}=1$$ via L'Hospital's Rule (just applying once) or Taylor series.

The other limit we need to evaluate is $$\lim_{x\to 0}\frac{6\sin x - 6x+x^3}{x^5}$$ We can apply L'Hospital's Rule once to get the expression $$\frac{3x^2-6(1-\cos x)} {5x^4}$$ and putting $x=2t$ this transforms into $$\frac {3}{20}\cdot\frac{t^2-\sin^2t}{t^4}=\frac{3}{20}\cdot\frac{t+\sin t} {t} \cdot\frac{t-\sin t} {t^3}$$ and this tends to $(3/20)(2)(1/6)=1/20$.

The desired limit is thus $1+(1/20)=21/20$. You should see that the limit has been evaluated using just two applications of L'Hospital's Rule.

Also observe that the the term $6x^2$ was added and subtracted in numerator by observing the term $-6\sin x^2$ and knowing that this could lead to a split as $6(x^2-\sin x^2)$. Whenever you see that an expression can be split into multiple terms always try to ensure that at least one of the terms after split has a finite limit. Then that particular term can be handled without knowing anything about limiting behavior of other terms.

Title: What is the following limit?


问:
differentiating a Summation for the mean energy - Statistical Physics
I am currently taking a Physics course Statistical Physics, and this pop up regularly:

this is from Fundamental of statistical and thermal physics

My question is: I understand Z is the partition theorem equal to the summation from zero to infinity of exponential to the - beta times the energy.
How is the derivative related to the summation?
why can I go from the left side of the equation meaning the side where both summations are present to the right side where theres a partial derivative of z with respect to beta.

what is the math process to go from summations to derivatives?

答:
In physics, $Z$ is the partition function given by
$$Z = \sum_{i}e^{-\beta E_i}$$ where i = 1,2,3.. $\infty$ is the index of the microstates of the system.

Now if you differentiate w.r.t. $\beta$, you get
$$\frac{\delta Z}{\delta \beta} = -E_i Z$$
This is true because the derivative of the sum is equal to sum of derivatives. And the derivative of each term in the summation is equal to $E_i e^{-\beta E_i}$, for a given microstate $i$.

$$ E_i = -\frac{1}{Z}\frac{\delta Z}{\delta \beta}$$

$\frac{1}{Z}$ can also be written as $\frac{\delta}{\delta \beta} ln Z$. Hence the relation. Hope this helps.


问:
The set of Involutary Matrices is a smooth regular submanifold?

Let $M_n(\mathbb R)=M_{n\times n}(\mathbb R)$ be the vector space of $n\times n$ matrices with entries in $\mathbb R$.
Let $S=\big\{ A\in M_2(\mathbb R)\,\big|\, A^2=I,\, A\ne \pm I\big\}$.

Show that $S$ is a smooth regular submanifold of $M_2(\mathbb R)$.

I am using the Regular Value Theorem stated in this question: Regular Value Theorem Using Implicit Function Theorem in Calculus.

I tried to construct a map
$f: M_2(\mathbb{R}) \to M_2(\mathbb{R})$
$$f : A \mapsto A^2 - I$$

The above mapping can be expressed as:
$$f(a, b, c, d) = (a^2+b c - 1, a b+b d, a c+c d, b c+d^2 - 1)$$
I can calculate the derivative to be
$$Df(a, b, c, d) = \left(
    \begin{array}{cccc}
    2 a & c & b & 0 \\
    b & a+d & 0 & b \\
    c & 0 & a+d & c \\
    0 & c & b & 2 d \\
    \end{array}
    \right)$$

However, there are some element in $S$ that is not a regular point. For example, $B = \begin{pmatrix} 0 &1 \\ 1 & 0 \end{pmatrix}$ with $f(B) = 0$ where the $Df(B) = \left(
\begin{array}{cccc}
 0 & 1 & 1 & 0 \\
 1 & 0 & 0 & 1 \\
 1 & 0 & 0 & 1 \\
 0 & 1 & 1 & 0 \\
\end{array}
\right)$ where the rank is 2.

It seems that the Regular Value Theorem cannot be used to conclude that $S$ is a smooth regular submanifold. Is there a another way to do it?

Any thoughts are appreciated!

答:
Let me set
$$ S_{+} = \{ A \in M_n(\mathbb{R}) \, | \, A^2 = I \}, \,\,\, S = S_{+} \setminus \{ \pm I \}. $$
I'll discuss the case $n = 2$ first. You have defined a (natural) smooth function $f \colon M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$  such that $S_{+} = f^{-1} \left( 0_{2 \times 2} \right)$ (and not $S$!) is a pre-image of a point. If $0_{2 \times 2}$ would have been a regular value of $f$ this would imply that $S_{+}$ is a zero-dimensional submanifold. Unfortunately $S_{+} = S \cup \{ \pm I \}$ is not a submanifold but in fact the union of two points $\{ \pm I \}$ and a two-dimensional submanifold $S$. So in order to show $S$ is a submanifold using the regular value theorem we need to find a function whose range is two-dimensional and not four dimensional.

Let's try to describe $S$ more explicitly. Write
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \,\,\,
A^2 = \begin{pmatrix} a^2 + bc & b(a+d) \\ c(a+d) & bc + d^2 \end{pmatrix}. $$
Then if $A^2 = I$ there are two options:

If $a + d \neq 0$ then $b = c = 0$ and so $A$ is digonal with $a,b = \pm 1$.

If $a + d = 0$ then $A$ has the form
$$ \begin{pmatrix} a & b \\ c & -a \end{pmatrix} $$
with $a^2 + bc = 1$.

This shows that
$$ S = \left \{ \left. \begin{pmatrix} a & b \\ c & d \end{pmatrix}  \right| a + d = 0, \, bc - ad = 1 \right \} $$
which coincides with Jason Devito's description of $S$ as the set of matrices which satisfy $\operatorname{tr}(A) = 0$ and $\det(A) = -1$. Hence it is natural to try and define $f \colon M_2(\mathbb{R}) \rightarrow \mathbb{R}^2$ by $f(A) = (a + d, bc - ad)$ and verify that $(0,1)$ is a regular value of $f$. Identifying $M_2(\mathbb{R})$ with $\mathbb{R}^4$ in the natural order corresponding to $(a,b,c,d)$ we get that

$$ Df|_{(a,b,c,d)} = \begin{pmatrix} 1 & 0 & 0 & 1 \\ -d & c & b & -a \end{pmatrix}. $$

Since $a + d = 0$ on $S$ the rows of $Df$ are indeed linearly independent on $S$ so $(0,1)$ is a regular value. Identifying $S$ with the manifold $ \{ (a,b,c) \in \mathbb{R}^3 \, | \, a^2 + bc = 1 \} $ and using the classification of the level sets of quadratic forms, one can see that $S$ is a one-sheeted hyperboloid (and so diffeomorphic to $\mathbb{R} \times S^1$), again agreeing with Jason Devito's guess.

This handles the case of $2 \times 2$ matrices. What about the general case? The equation $A^2 = I$ implies that the minimal polynomial of $A$ must divide $x^2 - 1 = (x-1)(x+1)$ and in particular $A$ is diagonalizable with $\pm 1$'s on the diagonal. Given $0 \leq k \leq n$, set
$$D_k = \operatorname{diag}(\underbrace{1,\dots,1}_{k \textrm{ times}}, -1, \dots, -1) \in M_n(\mathbb{R}). $$
A matrix $A$ satisfies $A^2 = I$ iff it is similar to $D_k$ for some $0 \leq k \leq n$. Hence, in general, $S_{+}$ is the disjoint union of $n + 1$ conjugacy classes and each conjugacy class is regular submanifold (see here) of codimension $k^2 + (n-k)^2$. When $n = 2$, $S$ is precisely the conjugacy class of $\operatorname{diag}(-1,1)$ and it can be described as the set of matrices whose characteristic polynomial is $(x-1)(x+1) = x^2 - 1$. Since the characteristic polynomial of a $2 \times 2$ matrix is $x^2 - \operatorname{tr}(A)x + \det(A)$ we again arrive to the description that $S_{+}$ is the set of matrices which satisfy $\operatorname{tr}(A) = 0$ and $\det(A) = -1$. This allows us to think of $S_{+}$ as the inverse image of $x^2 - 1$ under the map which sends a matrix to its characteristic polynomial and gives us another method of proving that $S_{+}$ is an embedded submanifold (using this answer).


问:
Let $a_n = \frac{n^p}{e^n}$ Show that $a_n$ converges for all $p>0$
In my work on this problem I've set up a limit:
$$\lim _{x\to \infty }\left(\frac{x^p}{e^x}\right)$$

I can see that for any number of p, successive applications of L'Hopital's Rule will eventually bring the numerator to a finite number, if p is an integer. If p is a fraction successive applications of L'Hopital's rule will eventually bring the $x$ term from the numerator to the denominator. Therefore in each case the limit will equal zero because the denominator's $e^x$ term will still be the same.

How do I express this logic in math form?

答: If you want you can take $f(y)=y^\frac{1}{p}$ and apply it to your quantity.  You can then use continuity of f and its inverse to use l'hopital's rule once.  i.e.  use L'hopitals rule on $\frac{x}{e^{x/p}}$ then raise it to the power of p and use the fact that $y^p$ is continuous so that $\lim\limits_{x \to \infty}(\frac{x}{e^{x/p}})^p=\left(\lim\limits_{x \to \infty}(\frac{x}{e^{x/p}})\right)^p$


问:
Hilbert's Nullstellensatz for analytic functions
In Demailly's book complex analytic and differential geometry, p97, he states that $F_{V(F), 0} = \sqrt{F}$, where $F$ is an ideal of the ring of holomorphic functions in an open subset of $\mathbb C^{n}$ containing $0$, $V(F)$ are the common zeros of the elements in $F$, and $\sqrt{F}$ is the radical of $F$. What confuses me is that the statement seems to suggest $0 \in V(F)$ which does not seem to be correct. One can simply take the ideal generated by a holomorphic function which does not vanish at $0$.

答: The germ of a holomorphic function that does not vanish at $0$ is invertible because as a power series, it has an inverse in the ring of formal power series, which can then be shown to be convergent as well. So the ideal generated by that would be the whole ring. Its vanishing set $V(F)$ would then be empty.


Title: Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?)

Body:
I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral

$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$

I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.

In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) - \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) - \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) - \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.

Any other approaches or suggestions to this challenging integral would be welcome.

Body:
I prefer to add another set of comments.

Suppose that we use
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot (x)) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot (x))^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n}$$ where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot (x))^n \, dx$$ It seems possible to compute the $a_n$'s for any value of $n$. I have not been able to find, even emprically, any relation between them.

Now, let us consider
$$b_k=\sum_{n = 1}^k \frac{a_n}{n}$$  The problem is that the convergence is extremely slow
$$\left(
\begin{array}{ccc}
k & b_k &  {b_k}/{b_{k-1}} \\
 1 & 1.088793044 &  \\
 2 & 1.531621992 & 1.406715445 \\
 3 & 1.787024517 & 1.166752976 \\
 4 & 1.958303531 & 1.095845923 \\
 5 & 2.083382818 & 1.063871246 \\
 6 & 2.179888546 & 1.046321649 \\
 7 & 2.257271880 & 1.035498757 \\
 8 & 2.321115965 & 1.028283738 \\
 9 & 2.374959149 & 1.023197111 \\
 10 & 2.421167052 & 1.019456294 \\
 20 & 2.680973995 & 1.006279984 \\
 30 & 2.800600958 & 1.003289782 \\
 40 & 2.873001074 & 1.002089208 \\
 50 & 2.922816842 & 1.001472276 \\
 60 & 2.959780104 & 1.001107498 \\
 70 & 2.988610619 & 1.000871248 \\
 80 & 3.011910979 & 1.000708125 \\
 90 & 3.031249532 & 1.000590014 \\
 100 & 3.047635130 & 1.000501298
\end{array}
\right)$$


问:
$H$ and $K$ are supposed to be closed in $X$, Why?
Let $X$ be a topological space and $A$ a closed subspace of $X$. Let $H$ and $K$ be closed in $A$. Now $H$ and $K$ are supposed to be closed in $X$, Why?</p>

答:
The complement of $H$ in $A$ is open in $A$, so is the intersection of some open set $U$ in $X$ with $A$. The complement of $H$ in $X$ is then the union of $U$ and the complement of $A$ in $X$, both of which are open.

Exactly the same argument works for $K$ - there seems to be no reason to state the problem with two subsets, so I assume this comes from some larger context.


问:
$I_n=\int_0^1 \frac{x^n}{\sqrt{1+x^2}}$, find $nI_n$
Let $$I_n=\int_0^1 \frac{x^n}{\sqrt{1+x^2}} dx$$ for $n\in \mathbb{N}$
Find $$\lim_{n\to \infty} nI_n$$

My approach:
Let me explain my observations briefly.
Note that $I_{n-1}-I_n=\int_0^1 \frac{x^{n-1}(1-x)}{\sqrt{1+x^2}} dx \ge 0$

Hence $I_n$ is decreasing.Also $I_n\ge 0$.So, $I_n$ is convergent. Let $\lim_{n\to \infty} I_n =l$

Now $I_n+I_{n-2} =\frac{\sqrt{2}}{n-1} -\frac{2}{n-1} I_n$
(By parts +little bit calculation)

Applying limit to the above equation we get,
$l=0$.

Also from the above equation we have,
$(n-1)I_n+(n-1)I_{n-2}=\sqrt{2}-2I_n$.
If $J_n := nI_n$, then
$\frac{n-1}{n} J_n +\frac{n-1}{n-2} J_{n-2}=\sqrt{2}-2I_n$

Yes I am only now one step away from getting $\lim_{n\to \infty} J_n=\frac{1}{\sqrt{2}}$

But I am unable to prove that $J_n$ is convergent

Any idea??

答: [Write $\frac {nx^{n}} {\sqrt {1+x^{2}}}$ as $ ({nx^{n-1}}) ( \frac x{\sqrt {1+x^{2}}})$]. Integration by parts gives $\int_0^{1}\frac {nx^{n}} {\sqrt {1+x^{2}}} dx=x^{n}\frac x {\sqrt {1+x^{2}}}|_0^{1}-\int_0^{1} x^{n}\frac d {dx} \frac x {\sqrt {1+x^{2}}}dx$. DCT shows that the second term tends to $0$. Hence the limit is $\frac 1 {\sqrt 2}$.