问:
Is there a way to find the value of the largest entry of a hidden vector of integers?
Consider the following set-up:

A person has a vector of with integer entries, which is not known to
you. The person refuses to reveal the vector to you; however, you may
supply another vector with integer entries, and they will tell you the
result of taking the dot product with the hidden vector.

A straightforward way to extract the value at the $i$-th index is to supply the vector $(0, 0, \ldots, 1, \ldots, 0)$ - where the number $1$ occurs at the $i$-th position. Doing this for all values of $i$ lets you recover the hidden vector in number of steps equal to the length of the vector.

Is there a shorter way to just extract the absolute value of the largest entry (or an upper bound for the same) in this set-up? For example, if it is known that all the entries happen to be positive integers, one can supply the vector $(1, 1, \ldots, 1)$ and obtain an upper bound for the largest entry in a single step.

Also, if it happens that it is not possible to do this in fewer than $n$ steps (where $n$ is the length of the vector), could you please refer me to a proof for why this is the case?

Another variant of the question - is it possible to determine the positions of the negative numbers in the vector in fewer than $n$ steps?

Edit: I came across a variant of this question a while ago. The set-up was almost identical, except the vector had positive integer entries instead of just integer entries. The solution involved finding an upper bound $M$ for the largest term (exactly as I have outlined above, and then taking the dot product with the vector $(1, M, M^2, \ldots, M^{n-1})$ to get a number whose base-$M$ expansion was the entries of the vector.

I found this really interesting, and I was wondering if this could be generalized to either the full set of integers or the rationals. Since it seems unlikely that one will be able to determine the full vector in fewer than $n$ steps, I am interested in knowing whether it will be possible to determine a simpler property - such as the positions of the negative entries - in fewer than $n$ steps.

答:
Without any assumption on the vector, you cannot do this in fewer than $n$ attempts.  Let the true secret vector be $\mathbf{a}$. It's enough to show that for any $\mathbf{u}_1, \ldots, \mathbf{u}_{n-1} \in \mathbb{Z}^n$ there exists another vector $\mathbf{b}$ in $\mathbb{Z}^n$ with different maximum entry and such that $\mathbf{a} \cdot \mathbf{u}_i = \mathbf{b}\cdot \mathbf{u}_i$ for all $i$.  That way, no matter what your $n-1$ "questions" were, you can never tell if the answer was $\mathbf{a}$ or $\mathbf{b}$.

The span of $\mathbf{u}_1, \ldots, \mathbf{u}_{n-1}$ in $\mathbb{Q}^n$ has dimension at most $n-1$, so its orthogonal complement has dimension at least 1. The orthogonal complement has a nonzero integer vector $\mathbf{v}$, because if you pick any nonzero element you can multiply it by the lcm of the denominators of its entries to get an integer vector. Add a large enough multiple of $\mathbf{v}$ to $\mathbf{a}$ so that you get a vector with a different maximum entry - call this $\mathbf{b}$.  Now $\mathbf{a} \cdot \mathbf{u}_i$ and $\mathbf{b}\cdot \mathbf{u}_i$ are equal for all $i$, because they differ by something orthogonal to the $u_i$, but $\mathbf{a}$ and $\mathbf{b}$ have different maximum entries.

The same argument applies to locating the negative entries: by adding or subtracting a large enough multiple of an integer vector in the orthogonal complement to $\mathbf{a}$ you can obtain a vector with negative entries in different places with the same dot products with the $\mathbf{u}_i$s.

This is closely related to the "guess the polynomial" game which has appeared on this site before: your enemy has a secret polynomial $p(x) \in \mathbb{Z}[x]$ which you have to guess and will only answer questions of the form "What is $p(r)$?" for $r \in \mathbb{Q}$. If the coefficients are nonnegative then you can do it in two steps: ask for $p(1)$ to get a bound on the coefficients then for $p(M)$ where $M$ is sufficiently large that the trick in the OP works.  The only difference is that you are not given the degree of the polynomial whereas in this question you know $n$.


问:
Weyl's theorem over non-algebraically closed fields
Weyl's theorem states that every finite-dimensional representation of a semisimple Lie algebra is completely reducible, meaning it is a direct sum of irreducible submodules. In Humphreys' book Introduction to Lie Algebras and Representation Theory (section 6.3), this is proved in the case that the base field is algebraically closed of characteristic 0. Similarly, in Hall's book Lie Groups, Lie Algebras, and Representations (section 10.3), the base field is assumed to be $\mathbb{C}$.

Does Weyl's theorem hold over fields that are not algebraically closed? In this answer from 6 years ago, Dietrich Burde says "one can apply an argument, that the statement is already true for $K$ if it was true for the algebraic closure of $K$." Is there such an argument, and if so, what is it? I know that in general, an irreducible representation may become reducible over a field extension, so it's not clear to me why complete reducibility over $\overline{K}$ would imply complete reducibility over $K$.

答:
Here are two references for this version of Weyl's theorem (characteristic 0 but not algebraically closed):

Jacobson's Lie Algebras does not assume the base field is algebraically closed. Weyl's theorem is Theorem 8 in Chapter III.7, and the proof is via Whitehead's lemma.

It is also Theorem 7.8.11 in Weibel's An Introduction to Homological Algebra. The proof uses cohomology.

ContentLicense: CC BY-SA 4.0


问:
Quick Volumes with Lines Question
So I was given the following prompt with the graph:

"Let $R$ be the region bounded above by the graph of $y=1-x^2$ and below by the graph of $y=x^2-1$, for $-1\le x\le1$, as shaded in the figure above. What is the volume of the solid generated when the region $R$ is revolved about the horizontal line $y=3$?"

I understand that the bounds of the integral that would solve for the volume would be from $-1$ to $1$, but I'm a bit confused about what the equation for the volume might look like here. I was thinking it might look something like the following: $\int((1-x^2+3)^2-(x^2-1+3)^2)$, but I'm not too sure. Any help would be appreciated!

答:
Instead of rotating around the line $y=3$, we translate everything $3$ units down and solve
$$\pi\int_{-1}^1 \left[\left(x^2-4\right)^2-\left(-x^2-2\right)^2\right]\,dx=16\pi$$

alternative Pappus' theorem

Compute di area $A$ between the "blue" parabolas, $A=\frac83$  and multiply by the circumference $C=6\pi$ described by the centroid $O$ of the region $A$

$$V=(6\pi)\cdot \frac83=16\pi$$

$$...$$

Title: Quick Volumes with Lines Question


问:
How do I find a general formula for $\int_0^\infty e^{-(ax^2+bx+c)} dx$
Setting our integral to $I$ and squaring it:
$$
I = \int_0^\infty e^{-(ax^2+bx+c)} dx
$$
$$
I^2 = e^{-2c}\iint_{\mathbb{R}^2} e^{-(a(x^2+y^2) + b(x+y))}dxdy
$$
I want to convert to polar co-ordinates but I get an ugly expression and depending on the values of $a$ and $b$ we will not obtain circular symmetry:
$$
I^2 =e^{-2c} \int_0^{2\pi}\int_0^\infty e^{-ar^2 - br\sqrt{1+2\sin(\theta)}}rdrd\theta
$$
Are my bounds incorrect and if so how can I change them?

答: Hint. First let $x=y-b/2a$ in the original integral $I$ before squaring.


问:
Does the function have a positive lower bound?
Let $\zeta$ be an irrational number and $p, q, n$ be natural numbers. The function $g$ is defined as:

$$g_n(\zeta,p,q)=p^n|p\zeta-q|.$$

Does there exist a positive $r$ such that:

$$\forall p\in\mathbb{N}-\{0\},\ \forall q\in\mathbb{N},\ g_n(\zeta,p,q)\geq r?$$

Is the result relevant to $\zeta$ and $n$?

答:
COMMENT.-The inequality $\left|\pi -\dfrac qp\right|>\dfrac{1}{p^{42}}$, valid for all rational $\dfrac qp$ with $p≥2$), is not recent, it is due to Mahler (1903-1988) and was published in 1953 twice in The Philosophical Transactions of the Royal Society, 245 (England) and in Indagationes mathematicae, of the Royal Dutch Mathematical Society, 15 (Holland). This inequality, of an unusual character until the date of its discovery,("striking inequality" comments Alan Baker) indicates, with the limitations of its time, in what way rationals cannot approach $\pi$: every rational "close" to π determines its "distance from π", in the sense of never at a distance less than or equal to the power $42$ of its denominator.

Definition.- The measure of irrationality of a real α is the smallest of all reals $\mu$ for which there is a positive constant $A$ such that for all rational $\dfrac qp\ne \alpha$  with
$p\gt 0$ we have $$|\alpha- \dfrac qp|\gt\dfrac{A}{p^{\space\mu}}$$
Theorem (Liouville).-For every algebraic number $\alpha$, real of degree $n≥2$ there exists a positive number $c(α)$ such that $\left|\alpha-\dfrac qp\right|>\dfrac{c(α)}{p^n}$ is true for all rational  $\dfrac qp$  ($\space p>0$).

In other words, the measure of irrationality of the algebraic irrational $\alpha$ cannot be greater than its degree (it is quite less, but you had to wait until 1955 to find out). It was this theorem that allowed the Liouville transcendental numbers to be explicitly defined.

I give a succinct account of the progress in the value of $n$.

►Joseph Liouville (1809-1882), French: approximation $n+ϵ$, in 1844.

►Axel Thue (1863-1922), Norwegian: approximation $\dfrac n2 + 1 + ϵ$, in 1909.

►Carl Ludwig Siegel (1896-1981), German: approximation $2\sqrt n + ϵ$, in 1929. Siegel conjectured the best bound later found by Roth.

Freeman John Dyson (1923-), British-American: approximation $\sqrt{2n} + ϵ$, in 1947.

Klaus Friedrich Roth (1925-), German-British: $2 + ϵ$ approximation, in 1955. The best level, a profound result that earned Roth the Fields Medal. "The achievement is one that speacks for itself: it closes a chapter , and a new chapter is now opened. Roth’s theorem settles a question which is both of a fundamental nature and of extreme difficulty. It will stand as a landmark in mathematics for as long as mathematics is cultivated" (Harold Davenport, from his presentation by Roth for the Fields Medal at the Edinburgh International Congress, 1958).


问:
Derive a formula for the nth term of a sequence (most likely a Beatty sequence)
In this MIT Putnam seminar problem set, the following question can be found (it's problem 8):

Define a sequence $a_1 < a_2 < · · ·$ of positive integers as follows. Pick $a_1 = 1$. Once $a_1, . . . , a_n$ have been chosen, let $a_{n+1}$ be the least positive integer not already chosen and not of the form $a_i +i$ for $1 \leq i \leq n$. Thus $a_1 + 1 = 2$ is not allowed, so $a_2 = 3$. The first few terms of the sequence are $1,3,4,6,8,9,11,12,14,16,17,19,....$. Find a simple formula for $a_n$. Your formula should enable you, for instance, to compute $a_{1,000,000}$.

I tried to look for pattern but I could not find any. After I looked at OEIS, it seems to be a Beatty sequence $a_n = \lfloor n\phi \rfloor$ where $\phi = \frac{\sqrt{5}+1}{2}$.

I am wondering how one could "derive" the formula. It is quite odd that the derivation gives greatest integer function of a number as an answer!

答:
Comments only:

Since you have already known the answer, the following is just hindsight.  First of all we know that $\{ a_n \}$ and $\{ b_n = a_n+n \}$ is a partition of $\mathbb N$. On the other hand if a formula of $\{ a_n \}$ ensures that $\{a_n\}$ and $\{a_n+n\}$ partition $\mathbb N$ then we are done (Edit: not quite, we still need $a_{n-1}+n-1>a_n$, but the formula we derive indeed satisfies this condition.)

Next, it's probably natural (hindsight of course) to conjecture that the proportion of $\{ a_n \}$ and $\{ b_n \}$ are asymptotically constant, i.e.,

$$\lim_{n\to\infty} \frac{n}{a_n} = \alpha, \lim_{n\to\infty} \frac{n}{b_n}=1 -\alpha \\
\implies 1-\alpha = \lim_{n\to\infty} \frac{\frac{n}{a_n}}{1+\frac{n}{a_n}}=\frac{\alpha}{1+\alpha}\implies \alpha = \frac{1+\sqrt 5}{2}$$

By examining a number of $a_i$'s and $b_i$'s we further conjecture
$$a_n = \left\lfloor \alpha n\right\rfloor$$

Finally, we can apply either one of the proofs of the Rayleigh_theorem to conclude $\{a_n\}$ and $\{ b_n \}$ indeed partition $\mathbb N$.


问:
Maximum number of non-congruent triangles with side lengths integers and less than $9$

Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.

I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ of them but after that I gave up. Is there any systematic approach to question?

Body:
We do some systematic counting.

First the isosceles (including equilateral) ones. If the base is $1$, other two sides can be any from $1$ to $8$. If base is $2$, other two can be any from $2$ to $8$. If base is $3$ - $2$ to $8$. Base $4$ - $3$ to $8$. Base $5$ - $3$ to $8$. Base $6$ - $4$ to $8$. Base $7$ - $4$ to $8$. Base $8$ - $5$ to $8$. Total $$8+7+7+6+6+5+5+4=48$$

For scalene, we write the triplet of side lengths in decreasing order respecting triangle inequality.

$[8,7,(6 \,\text{to}\, 2)]. [8,6,(5\,\text{to}\, 3)]. [8, 5, 4]. [8,4,✗].$

$[7,6,(5 \,\text{to}\, 2)]. [7,5,(4 \,\text{to}\, 3)]. [7, 4, ✗].$

$[6,5,(4 \,\text{to}\, 2)]. [6,4,3]. [6, 3, ✗].$

$[5,4,(3 \,\text{to}\, 2)]. [5,3,✗].$

$[4,3,2]. [4,2,✗].$

$[3,2,✗].$

Sum : $5+3+1+4+2+3+1+2+1=22$

Grand total : $\boxed{48+22=70}$


答:
The number of non congruent non degenerate triangles
with integer sides with largest side equal to $\,n+1\,$
is the OEIS sequence A002623.
For $\,n=7\,$ the answer is $70.$ The OEIS entry has a
comment which justifies the interpretation as count of
triangles

Also number of triples (x,y,z) with 0 < x <= y <= z <= n + 1, x + y > z. - Ralf Steiner, Feb 06 2020

The formula

A002620(n+3) = a(n+1) - a(n). - Michael Somos, Sep 04 1999

means that the first differences of the triangle counts
is the quarter-squares sequence A002620.

Alternatively, the
OEIS sequence A173196
has two extra zeros at the beginning of the sequence.
Thus it is the number of non congruent non degenerate
triangles with integer sides with largest side $\,<n.\,$
The value for $\,n=9\,$ is again $70$. The
triangle counts arranged as a number triangle
OEIS sequence A003983
with $\,n\,$ the largest side and $\,m\,$ the
smallest side makes the pattern clear.

```<code> n\m 1  2  3  4  5  6  7  8  9 10   row sum accum sum
 -+------------------------------   ------- ---------
 1|  1                                  1        1
 2|  1  1                               2        3
 3|  1  2  1                            4        7
 4|  1  2  2  1                         6       13
 5|  1  2  3  2  1                      9       22
 6|  1  2  3  3  2  1                  12       34
 7|  1  2  3  4  3  2  1               16       50
 8|  1  2  3  4  4  3  2  1            20       70
 9|  1  2  3  4  5  4  3  2  1         25       95
10|  1  2  3  4  5  5  4  3  2  1      30      125
</code>```

Title: Maximum number of non-congruent triangles with side lengths integers and less than $9$

Body:
Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.

I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ of them but after that I gave up. Is there any systematic approach to question?

Body:
The number of non congruent non degenerate triangles
with integer sides with largest side equal to $\,n+1\,$
is the OEIS sequence A002623.
For $\,n=7\,$ the answer is $70.$ The OEIS entry has a
comment which justifies the interpretation as count of
triangles

Also number of triples (x,y,z) with 0 < x <= y <= z <= n + 1, x + y > z. - Ralf Steiner, Feb 06 2020

The formula

A002620(n+3) = a(n+1) - a(n). - Michael Somos, Sep 04 1999

means that the first differences of the triangle counts
is the quarter-squares sequence A002620.

Alternatively, the
OEIS sequence A173196
has two extra zeros at the beginning of the sequence.
Thus it is the number of non congruent non degenerate
triangles with integer sides with largest side $\,<n.\,$
The value for $\,n=9\,$ is again $70$. The
triangle counts arranged as a number triangle
OEIS sequence A003983
with $\,n\,$ the largest side and $\,m\,$ the
smallest side makes the pattern clear.

 n\m 1  2  3  4  5  6  7  8  9 10   row sum accum sum
 -+------------------------------   ------- ---------
 1|  1                                  1        1
 2|  1  1                               2        3
 3|  1  2  1                            4        7
 4|  1  2  2  1                         6       13
 5|  1  2  3  2  1                      9       22
 6|  1  2  3  3  2  1                  12       34
 7|  1  2  3  4  3  2  1               16       50
 8|  1  2  3  4  4  3  2  1            20       70
 9|  1  2  3  4  5  4  3  2  1         25       95
10|  1  2  3  4  5  5  4  3  2  1      30      125


答:
When we saw this problem years ago, the idea was to involve Beatty sequences. IIRC, We couldn't solve it without using Beatty sequences (or its ideas in some form).

Here's how the wishful thinking went:

Let the sequence be $a_n$.

Define $ b_n = a_n + n$, which is the complement.

IF (and that's a big wishful if) $a_n = \lfloor \alpha \times n \rfloor$, then $b_n = \lfloor (\alpha+1)n\rfloor$

Hence, invoking Beatty sequences, we want $\frac{1}{\alpha} + \frac{1}{ \alpha + 1 } = 1$ for positive irrational $ \alpha$.

This gives $ \alpha = \frac{ \sqrt{5} + 1 } { 2}$ .

It remains to verify that $ a_n, b_n$ are indeed these sequences, which you use Beatty sequence proof again to demonstrate.

Title: Derive a formula for the nth term of a sequence (most likely a Beatty sequence)

Body:
In this MIT Putnam seminar problem set, the following question can be found (it's problem 8):

Define a sequence $a_1 < a_2 < · · ·$ of positive integers as follows. Pick $a_1 = 1$. Once $a_1, . . . , a_n$ have been chosen, let $a_{n+1}$ be the least positive integer not already chosen and not of the form $a_i +i$ for $1 \leq i \leq n$. Thus $a_1 + 1 = 2$ is not allowed, so $a_2 = 3$. The first few terms of the sequence are $1,3,4,6,8,9,11,12,14,16,17,19,....$. Find a simple formula for $a_n$. Your formula should enable you, for instance, to compute $a_{1,000,000}$.

I tried to look for pattern but I could not find any. After I looked at OEIS, it seems to be a Beatty sequence $a_n = \lfloor n\phi \rfloor$ where $\phi = \frac{\sqrt{5}+1}{2}$.

I am wondering how one could "derive" the formula. It is quite odd that the derivation gives greatest integer function of a number as an answer!


问:
Eigenvalues of a 2x2 block matrix with invertible diagonal blocks
There is something I do not understand for eigenvalues of block matrices.
If we consider a matrix \begin{equation} M = \begin{pmatrix} A & B \\ C & D\end{pmatrix}. \end{equation} Here each block is a $N\times N$ matrix. We know that, provided $A$ is not singular i.e. $\det(A) \neq 0$, the determinant of $M$ is obtained using the Schur's complement of block $A$ of the matrix $M$ via
\begin{equation} \det(M) = \det(A)\cdot \det(D - CA^{-1}B) \, . \end{equation}
Thus from this equation we know that the eigenvalues of block $A$ are also eigenvalues of $M$ because they cancel the caracteristic polynomial of $M$ as well, right ?

By symmetry, if $D$ is not singular, i.e. $\det(D) \neq 0$, the determinant of $M$ is also
\begin{equation} \det(M) = \det(D)\cdot\det(A - BD^{-1}C) \, .\end{equation}
Thus we also learn from this that eigenvalues of $D$ are also eigenvalues of $M$ for the same reason explained previsouly. Therefore the spectrum of $M$ should be composed of the $N$ eigenvalues of $A$ and the $N$ eigenvalues of $D$. In practice, this is not correct. Could someone explain me what I am missing here please ? It should be something obvious and trivial I suppose, but I am totally missing it.

Anyway, can we extract some interesting information about the eigenvalues of $M$ in the case of 4 squared blocks of the same size, which are all invertible but different ?

Thank you !

答:
No, it is not necessarily the case that the eigenvalues of $A$ (or $D$) are eigenvalues of $M$.

Note that $\lambda$ is an eigenvalue of $A$ iff $A - \lambda I$ fails to be invertible.
If $\lambda$ is not an eigenvalue of $A$, then the characteristic polynomial of $M$ can be written in the form
$$
\det(M - \lambda I) = \det(A - \lambda I) \cdot \det([D - C(A - \lambda I)^{-1}B] - \lambda I).
$$
Note that the expression on the right is only defined if $(A - \lambda I)$ is invertible.

That said, the expression on the right certainly gives us some idea about the characteristic polynomial of $M$: because the expression on the right is equal to the characteristic polynomial except at the finitely many values of $\lambda$ where it is undefined and because polynomials are continuous, we can say the following. If $\mu$ is an eigenvalue of $A$, then
$$
\det(M - \mu I) = \lim_{\lambda \to \mu} \det(M - \lambda I) = \lim_{\lambda \to \mu}\det(A - \lambda I) \det(D - C(A - \lambda I)^{-1}B).
$$
However, note that the fact that $\det(A - \mu I) = 0$ does not necessarily imply that the limit on the right is equal to zero. In particular, if $\lim_{\lambda \to \mu}\det(D - C(A - \lambda I)^{-1}B)$ is infinite, then "plugging in $\mu$" gives us the indeterminate expression $0 \cdot \infty$.

An illustrative example: suppose that $N = 1$ and $A = B = C = D = 1$. Note that the determinant of a $1 \times 1$ matrix is simply the entry of this matrix. We find that
$$
M = \pmatrix{1&1\\1&1}, \quad \det(M - \lambda I) = \lambda^2 - 2\lambda = \lambda(\lambda
- 2).
$$
On the other hand, we compute
$$
\det(A - \lambda I) = 1 - \lambda, \quad \det(D - C(A - \lambda I)^{-1}C) = 1 - \frac{1}{1 - \lambda} = \frac{-\lambda}{1-\lambda}.
$$
We find that
$$
\lim_{\lambda \to 1} \det(A - \lambda I) \det(D - C(A - \lambda I)^{-1}C) =
\lim_{\lambda \to 1} \frac{-\lambda(1 - \lambda)}{1-\lambda}.
$$
Although it is indeed the case that $\det(A - 1 \cdot I) = 1 - 1 = 0$, it is not the case that the above limit is equal to zero. The limit of the above expression is $-1$, which coincides with the result of $\det(M - 1 \cdot I)$.


问:
Solving a system of Diophantine equations
For a problem that I'm working on, I need to solve the following system of Diophantine equations:-
$a^3+40033=d$, $b^3+39312=d$, $ c^3+4104 = d$ where $a,b,c>0$ are all DISTINCT positive integers, and $a,b,c$ ∉ {$ 2, 9, 15, 16, 33, 34$}

How does one go about solving this? Is brute-force the only possible way? Or could there be a case that no integer solutions exist for this equation?

Also, are there any online computing engines, that allow me to set constraints, and solve Diophantine equations of this sort?

Any and all help is appreciated! Thanks!

答:
The only solutions are $(a,b,c,d)=(2,9,33,40041),\ (15,16,34,43408)$ but you have in your post explicitly ruled these out. If you follow the hint by Goos, you get to
$$b^3-a^3=721, \\ (b-a)(b^2+ba+a^2)=7\cdot 103,$$
so that either $b-a=1$ or $b-a=7$ which lead to the two solutions above when the other factor is set to $721$ or $103$ respectively.

Maybe you had found the solutions, but your question is about why there are not any others. If so I hope this answer helps see that.

ContentLicense: CC BY-SA 3.0


问:
If $X_n$ converges to X in probability then $X_n$ converges to X a.s.
We know that if $X_n$ converges to X a.s. then $X_n$ converges to X in probability in the space ( Ω, A, P) . But the converse is not true in general . We can construct  such examples on any probability space (Ω, A, P) which has a “non-atomic part”, meaning that, there is an $Ω_0$ ∈ A with P($Ω_0$)>0, such that, for any A ∈ A, A ⊂ $Ω_0$, with P(A)>0 and any 0< δ <1, one has B ∈A, B ⊂A with P(B)= δP(A).
But when the probability space is “purely atomic”, not only do we have no counter-example as above, but indeed, convergence in probability implies a.s. convergence. My question is how can I show that if (Ω, A, P) be a discrete probability space, that is, Ω is a countable set and A consists of all subsets of Ω. Then for random variables $X_n$, n ≥ 0 and X on (Ω, A, P), $X_n$ converges to X in probability then $X_n$ converges to X a.s.

答:
We can do this by contrapositive. Assume that $X_n\not\to X$ a.s., then there exists a set $B$ with $\mathbb{P}(B)>0$ where $B$ is the set of all elements $i$ where $X_n(i)\not\to X(i)$. Since $\Omega$ is assumed to be countable, then $B$ is a countable set. Now the $\sigma-$algebra $\mathcal{A}=\mathcal{P}(\Omega)$ by the problem, so we can write $B$ in the following manner.
$$B= \bigcup_{i\in B}\{i\} $$
These singleton sets are disjoint and we have countable number of disjoint sets, thus
$$\mathbb{P}(B)=\sum_{i\in B}\mathbb{P}(\{i\}) $$
Since $\mathbb{P}(B)>0$ and the measure is positive, then there must exist some $j\in B$ such that $\mathbb{P}(\{j\})>0$. Otherwise you would have that $\mathbb{P}(B)=0$.

Since $j\in B$, then $X_n(j)\not\to X(j)$. This implies that for any $\epsilon > 0$, for any $N\in\mathbb{N}$ we can find $n\geq N$ such that $|X_n(j)-X(j)|>\epsilon$. So by monotonicity of $\mathbb{P}$ we have that for any $\epsilon>0$ that
$$\lim_{n\to\infty}\mathbb{P}(\{i: |X_n(i)-X(i)|>\epsilon\}) \geq \mathbb{P}(\{j\})>0$$
Hence $X_n$ does not converge to $X$ in probability.


问:
In how many ways can the 10 vacant positions be filled?
There are 10 vacant positions in the central cabinet of ministers. There is a total of 350 MPs in the ruling party. A cabinet post can be given to any of these MPs and an MP can take more than one cabinet position. In how many ways can the 10 vacant positions be filled?</p>

答:
If each MP can take on several positions, argue that the first position can be filled in $350$ ways, the second position can be filled in $350$ ways (this includes the MP taking the first post), and so on. In all, $350^{10}$.

If each MP can occupy just one position, the first one can be filled in $350$ ways, which leaves $350 - 1$ for the second, up to $350 - 9$ for the last. In all, $350 \cdot 349 \cdot \dotsm \cdot 341 = 350^{\underline{10}}$. This last product is called a falling power, the notation I use is the one advocated by Knuth. There are others. You could also write it as $350! / 340!$.


答:
Let $(\Omega, A, \mathbf{P})$ be a discrete probability space, $\{\omega_n: n \in \mathbb{N}\} \subseteq \Omega$ with $\sum_{n=1}^\infty \mathbf{P}(\{\omega_n\}) = 1$. Assume $(X_n)_n, X$ are random variables on $\Omega$ such that $X_n$ does not converge almost sure to $X$. We show that $X_n$ cannot converge to $X$ in probability neiter.

As $X_n \not \rightarrow X$ a.s. there is some $\omega := \omega_n \in \Omega$ such that $X_n(\omega) \not \rightarrow X(\omega)$ for $(n \rightarrow \infty)$ and $\mathbb{P}(\omega) =: \delta > 0$. Note that here we use that $\Omega$ is discrete. In particular it follows that for every $\epsilon > 0$ there are infintely many $n \in \mathbb{N}$ with $$|X_n(\omega) - X(\omega)| > \epsilon. $$ Hence we obtain $$\mathbf{P}(|X_n - X| > \epsilon) \geq \mathbb{P}(\{\omega\}) = \delta > 0 $$ for arbitrarily large $n \in \mathbb{N}$. Thus $X_n$ cannot converge to $X$ in probability.

Title: If $X_n$ converges to X in probability then $X_n$ converges to X a.s.

Body:
We know that if $X_n$ converges to X a.s. then $X_n$ converges to X in probability in the space ( Ω, A, P) . But the converse is not true in general . We can construct  such examples on any probability space (Ω, A, P) which has a “non-atomic part”, meaning that, there is an $Ω_0$ ∈ A with P($Ω_0$)>0, such that, for any A ∈ A, A ⊂ $Ω_0$, with P(A)>0 and any 0< δ <1, one has B ∈A, B ⊂A with P(B)= δP(A).
But when the probability space is “purely atomic”, not only do we have no counter-example as above, but indeed, convergence in probability implies a.s. convergence. My question is how can I show that if (Ω, A, P) be a discrete probability space, that is, Ω is a countable set and A consists of all subsets of Ω. Then for random variables $X_n$, n ≥ 0 and X on (Ω, A, P), $X_n$ converges to X in probability then $X_n$ converges to X a.s.


问:
Find the number of unequivalent number of configuration on a square lattice
Say I have a $4 \times 4$ lattice on which I want to pick say 5 sites. Let's say for instance that I want to put 5 one's inside of the following zero matrix :

$$ \left[ \array{ 0 \, 0 \, | \, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 \\ \hline 0  \, 0 \,|\, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 } \right]$$

On the matrix, the horizontal and vertical lines in the middle are supposed to be "mirror symmetry axes" of the system. That is the following configurations are all considered to be equivalent :

$$ \left[ \array{ 0 \, 0 \, | \, 1 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 \\ \hline 0  \, 0 \,|\, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 } \right]$$

$$ \left[ \array{ 0 \, 1 \, | \, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 \\ \hline 0  \, 0 \,|\, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 } \right]$$

$$ \left[ \array{ 0 \, 0 \, | \, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 \\ \hline 0  \, 0 \,|\, 0 \, 0 \\ 0 \, 0 \, | \, 1 \, 0 } \right]$$

$$ \left[ \array{ 0 \, 0 \, | \, 0 \, 0 \\ 0 \, 0 \, | \, 0 \, 0 \\ \hline 0  \, 0 \,|\, 0 \, 0 \\ 0 \, 1 \, | \, 0 \, 0 } \right]$$

By mirror symmetry axes, I mean that I don't impose the configuration to be symmetric (especially with 5 sites !) but simply that they belong to the same equivalence class of configurations.

How many distinct configurations (i.e. equivalence classes) can I expect to find ?

Is there a way to generalize the solution to any $ n \times n$ lattice and $m$ sites ?

答:
Alright.  Your group of symmetries is of size 4.  Which means that any equivalence class (=orbit) of a configuration has to have size which is a factor of 4 -- 1, 2, or 4.
So let's count equivalence classes of each size separately.
For your case -- $n = 4$, $m = 5$, all orbits are of size 4 (not too difficult to see), so there are $\frac{1}{4} \binom{4^2}{5} = 1092$ distinct configurations.

Likewise, if $n$ is even, and $m$ is odd, all orbits are of size 4 and we have $\frac{1}{4} \binom{n^2}{m}$ distinct configurations.

Now let's break out the other cases.

$n$ even, $m$ even, not divisible by 4.  There are 3 distinct subgroups of symmetries of size 2 (mirror reflection about each line, and 180-degree rotation).  Any configuration with $m$ not divisible by 4 can be symmetric under at most one of these.  Each symmetry subgroup will contribute $\frac{1}{2}\binom{n^2/2}{m/2}$ distinct configurations (orbit size is 2, which gives us the $\frac{1}{2}$). How many non-symmetric configurations are there?  Well, there are $\binom{n^2}{m}$ configurations in total, so $\binom{n^2}{m} - 3\binom{n^2/2}{m/2}$ are non-symmetric and lie in orbits of size 4.

Total distinct configurations: $\frac{1}{4} \binom{n^2}{m} + \frac{3}{4}\binom{n^2/2}{m/2}$.  (Check this against n=2, m=2 -- two distinct "domino" configurations, one "diagonal" configuration, $3 = 1/4 * 6 + 3/4 * 2$.)

n even, m divisible by 4.  Now a configuration could be symmetric under one of the 3 subgroups, under the full symmetry group, or not symmetric. Fully symmetric configurations: $\binom{n^2/4}{m/4}$, in orbits of size 1. For each subgroup, we now have $\binom{n^2/2}{m/2} - \binom{n^2/4}{m/4}$ configurations that are not symmetric under the full group, in orbits of size 2. And for non-symmetric configurations, we have $\binom{n^2}{m} - 3\binom{n^2/2}{m/2} + 2\binom{n^2/4}{m/4}$, in orbits of size 4 (check that the total number of configurations is correct).

Total distinct configurations: $A + 3\frac{1}{2}B + \frac{1}{4}C = \frac{1}{4} \binom{n^2}{m} + \frac{3}{4}\binom{n^2/2}{m/2}$ again. (Check $n=2, m=4$: only one configuration, the full board, and $1 = 1/4 \binom{4}{4} + 3/4 \binom{2}{2}$.)

Now we move on to $n$ odd. Here the sites in the lattice are not all equivalent -- the central site has an orbit of size 1.
If $m$ is $3\pmod{4}$ or $2\pmod{4}$, we have (like the $2\pmod{4}$ case above) configurations that are symmetric under subgroups but none under the full group -- odd $m$ must include the central site. Each subgroup therefore contributes $\frac{1}{2}\binom{(n^2-1)/2}{\lfloor m/2\rfloor}$ distinct configurations and the total number of distinct configurations is $\frac{1}{4}\binom{n^2}{m} + \frac{3}{4}\binom{(n^2-1)/2}{\lfloor m/2 \rfloor}$.

If $m$ is $0\pmod{4}$ or $1\pmod{4}$, we are in a situation similar to the "m divisible by 4" case above.  Again, odd $m$ means that symmetric configurations must include the center site. But we have the same number as before: $\frac{1}{4}\binom{n^2}{m} + \frac{3}{4}\binom{(n^2-1)/2}{\lfloor m/2 \rfloor}$ distinct configurations.


问:
Partial sum of Geometric sequence
Can someone explain to me how the following summation goes from the left to the right:

$\sum_{i=1}^{r}n(1-p)^{i-1} = \frac{n(1-(1-p)^r)}{p}$.

I have used the formula for a Geometric series. My common ratio is: $(1-p)$. I get the following:

$\sum_{i=1}^{r}n(1-p)^{i-1} = n\cdot\frac{1-(1-p^r)}{1-(1-p)}$.

From here, I simply don't know how to get to $\frac{n(1-(1-p)^r)}{p}$

答:
In your question, $n$ is a constant, which has nothing to do with $i$. The first implication of what you have done is incorrect.

Note that $n(1-p)^{i-1}$ means $n\cdot ((1-p)^{i-1})$, and

$$
\sum_{i=1}^{r}n(1-p)^{i-1}=n\sum_{i=1}^{r}(1-p)^{i-1}
$$

all you need to work out is to find a formula for
$$
\sum_{i=1}^{r}x^{i-1}
$$
and then set $x=1-p$.

Title: Partial sum of Geometric sequence


问:
Do we get a section on a ringed space $\mathcal{A}$ by mapping each element $x$ to the multiplicative unit element $1_x$ of $\mathcal{A}_x$?
Suppose $(X, \mathcal{A})$ is a ringed space (where $\mathcal{A}$ is assumed to be a sheaf of unital, commutative rings over $X$) and consider the map
\begin{eqnarray}
\psi \colon X \to \mathcal{A} \qquad X \ni x \mapsto 1_x \in \mathcal{A}_x
\end{eqnarray}

Is it true that this map must always be continuous, i.e. that $\psi$ is a section?

答: The element $1_x\in\mathcal{A}_x$ is just the image of the unit $1\in\mathcal{A}(X)$ under the restriction map $\mathcal{A}(X)\to\mathcal{A}_x$.  So, your map is continuous, since it corresponds to the section $1\in\mathcal{A}(X)$.


答:
Put $L$ to be the longest side, an $x,y$ the other two.

We can limit ourselves in considering $x \le y$.

Then we shall have
$$
\left\{ \matrix{
  0 \le x \le y \le L \hfill \cr
  L \le x + y \hfill \cr}  \right.
$$
which is the area within the triangle $VCB$
in this sketch

and which is equivalent to triangle $OVA$, which
in turn is half of the triangle $OAC$.

Now, the number of integer points inside the right triangle $OAC$ with sides $[0,L]$, including the borders i.e. the equalities
is
$$
N_{\,OAC}  = 1 + 2 +  \cdots  + \left( {L + 1} \right)
 = \left( \matrix{  L + 2 \cr   2 \cr}  \right)
$$
while those on the segment $OV$ are
$$
N_{\,OV}  = \left\lfloor {L/2} \right\rfloor  + 1
$$

So the number we are looking for is
$$ \bbox[lightyellow] {
\eqalign{
  & N = \left( {{{N_{\,OAC}  - N_{\,OV} } \over 2}} \right) + N_{\,OV}
  = {1 \over 2}\left( {N_{\,OAC}  + N_{\,OV} } \right) =   \cr
  &  = {1 \over 2}\left( {\left( \matrix{  L + 2 \cr   2 \cr}  \right)
 + \left\lfloor {L/2} \right\rfloor  + 1} \right) \cr}
}$$

which for $L= 1, \ldots,10$ gives
$$
2, 4, 6, 9, 12, 16, 20, 25, 30, 36
$$

If instead you are not going to consider the degenerated cases $x=0, \; x+y=L$ then it is easy
to deduct from the above the corresponding side lengths.


问:
Find an equivalent condition to $\operatorname{tr}(A^{\ast}B)=0$ for two complex hermitian matrices $A, B$ with rank of $B$ 1.
I have two matrices $A,B$ given that both are positive semi-definite, complex. Another given is that $B$ has rank $1$ and is hermitian.
Additionally $B=ww^{\ast}$ where $w\in \mathbb{C}^{m}$ with $w_k=e^{-2\pi i u k}$.

Now I want to find a condition that implies (or better still is equivalent to) $\operatorname{tr}(A^{\ast}B)=0$. Here $\ast$ stands for the complex cunjugate transpose. This condition should if somehow possible be only dependent on $A$.
Originally I thought maybe $\det(A)=0$ would suffice but have not been able to prove this.

答:
Note that
$$
\operatorname{tr}(A^*B) = \operatorname{tr}(Aww^*) = \operatorname{tr}(w^*Aw) = w^*Aw.
$$
So of course, this trace is zero iff $w^*Aw = 0$. Moreover, because $A$ is positive semidefinite, it can be shown that $w^*Aw = 0 \iff Aw = 0$ (I give some proofs of this result in my earlier post here).

That is, we can say that $\operatorname{tr}(A^*B) = 0$ if and only if $w \in \ker(A)$.  That is, this trace is zero iff we have
$$
\sum_{k=1}^n a_{jk}e^{- 2 \pi i uk} = 0 \quad j = 1, \dots, n,
$$
where $a_{jk}$ denotes the $j,k$ entry of $A$.


问:
Prove that $d$ is a gcd of $a_1, ..., a_n$ iff $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$.
To anyone encountering this problem: As the comments and answer correctly point out, if $R$ is a commutative ring with identity, and $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$, then $d$ is a gcd of $a_1, ..., a_n$. If, additionally, $R$ is Noetherian (for instance if $R$ is a PID), then this becomes an equivalence. However, if $R$ is not Noetherian this may fail (as in the example given by Exodd).

Furthermore, in any commutative ring $R$ with identity, the following stronger statement is true: $d\in R$ is a gcd of $a_1, ..., a_n$ and is equal to $r_1 a_1 + \cdots + r_n a_n$ for some $r_i\in R$ iff $\langle d\rangle = \langle a_1 \rangle + \cdots + \langle a_n\rangle$.

Let $R$ be a commutative ring with identity, and let $a_1, ..., a_n\in R$. I would like to prove that $d\in R$ is a gcd of $a_1, ..., a_n$ iff $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$.

If $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$, then $\langle a_i\rangle \subseteq\langle d\rangle$, so $d\mid a_i$ for all $i$. Furthermore, if $c\mid a_i$ for all $i$, then $\langle a_i\rangle\subseteq\langle c\rangle$, so $\langle a_1 \rangle + \cdots + \langle a_n\rangle\subseteq\langle c\rangle$, or $\langle d\rangle\subseteq\langle c\rangle$, so $c\mid d$, implying $d$ is a gcd of $a_1, ..., a_n$.

I'm finding the other direction trickier. I thought I had an argument but I can't quite make it work. First, if $d \mid a_i$ then $\langle a_i\rangle \subseteq\langle d\rangle$ for all $i$, so $\langle a_1\rangle + \cdots +\langle a_n\rangle \subseteq \langle d\rangle$.

My idea was this: let $J = \{I\subseteq \langle a_1\rangle + \cdots + \langle a_n\rangle\ |\ I\ \text{principal ideal}\}$, ordered by inclusion. $\color{red}{\text{Then $J$ has a maximal element by Zorn's lemma.}}$ This is not actually true, as only finite chains have upper bounds in $J$, unless $R$ is a PID. That is, if we have an arbitrary collection of principal ideals $\{\langle a\rangle\}_{a\in A}$, totally ordered by inclusion, then we cannot guarantee that $\bigcup_{a\in A}\langle a\rangle$ is also a principal ideal (it is an ideal however).

If we roll with it, letting $\langle d\rangle$ be a maximal element of $J$, clearly as $\langle a_i\rangle\subseteq \langle a_1\rangle + \cdots + \langle a_n\rangle$ we must have $\langle a_i\rangle\subseteq\langle d\rangle$, so $d \mid a_i$. Furthermore, if $c \mid a_i$, then $\langle a_1\rangle + \cdots + \langle a_n\rangle \subseteq\langle c\rangle$, so $\langle d\rangle\subseteq\langle c\rangle$, and thus $c \mid d$. So $d$ is a gcd of $a_1, ..., a_n$.

Can I modify my argument to make this work? Is there a simpler route I've missed?

答:
Let $a_1 = x_1$ and $a_2 = x_2$ in $\mathbb K[x_1,x_2]$, for any field $\mathbb K$.
The only gcd of $a_1,a_2$ are the elements of $\mathbb K^*$, but
$$
<1> = \mathbb K[x_1,x_2] \ne <x_1,x_2> = <x_1> + <x_2>
$$


问:
Smooth mapping on $ \mathbb R$
I am going through differentiable manifolds and come across a problem:

How to construct a smooth mapping $f: \mathbb R \rightarrow  \mathbb R$ such that

$f^{-1}(0)=0$

$f^{'}(0) \neq 0$

$\forall \epsilon >0,f^{-1}(-\epsilon ,\epsilon )$ is not homeomorphic to $(-\epsilon ,\epsilon )$

Any help?

答:
SUGGESTION (not a full answer): Something that oscillates a lot would be a good choice. You might want to start looking at

$$
f(x) = x^2 \sin \frac1x
$$
(where $f(0)$ is defined to be 0, but I didn't want to write out cases).
Now $f'(0) = 0$, so it doesnt quite work. But if
$$
g(x) = f(x) + x
$$
you might have something to work with...

Title: Smooth mapping on $ \mathbb R$


问:
approximating a power law into a polynomial
We are trying to find the coefficients A,B and C such that  x^q=A(q-1)x^2+B(q-2)x+C(q-1)(q-2), were 1<=q<=2.
I have tried the matlab code below but not sure to implement constrain  1<=q<=2</p>
```<code>clc
for l = [1 2 inf]
    alpha=1.7;
    fprintf('For norm %d\n', l)
    fprintf('Coefficients    c1    c2    c3\n')
    for n = [5 10 100]
        i = 1:n ;
        x = 0 + i/n ;
        c = [1 1 1] ;

        %Difference function that we want to minimize
        g = @(c) (alpha-1).*c(1).*x.^2 +(alpha-2).*c(2).*x ...
                +(alpha-1).*(alpha-2).*c(3) - x.^alpha;
        f_norm = @(c) norm(g(c), l) ;
        C = fminsearch(f_norm, c);
        fprintf('n = %d      ', n)
        fprintf('%f     %f\n', C(1),   C(2),    C(3))
        % Compare plot of e^x and p(x).
        p = @(x) (alpha-1).*C(1).*x.^2 +(alpha-2).*C(2).*x ...
                +(alpha-1).*(alpha-2).*C(3);
        xx = linspace(0,2,1e5);
        figure;
        plot(xx, p(xx), '--r', xx,  xx.^alpha);
        str = sprintf('Plot with n = %d, for norm %d', n,l);
        title(str,'FontSize',24)
        xlabel('x','FontSize',20)
        ylabel('y','FontSize',20)
        legend('p2 approximation','power model');
    end
end
<span class="math-container">```</span>
</code>```

答:
Approximate in what sense? Agrees at some selected points, minimizes e.g. the square of the difference over the ranges, ...? As stated, the problem is woefully underdetermined.

What will be the use of the approximation?

Why is the (simple!) function $x^q$ not good enough?


问:
How many ways to colour a tetrahedron with monochromatic triangles.
I'm trying to find how many different ways there are to colour the edges of a regular tetrahedron with n colours such that there are no monochromatic triangles.

Certainly for one triangle there must be n choose 3 ways but I'm not quite sure how to generalise this to a tetrahedron.

Any help would be much appreciated!

答:
For one triangle there are actually $n^3-n$ edge colourings, including reflections and rotations.

There are $6$ edges in a tetrahedron, $n^6$ ways to colour the edges, $4$ triangles, $n^4$ colourings with any given triangle monocromatic, $6$ pairs of triangles, $n^2$ colourings with any given pair of triangles both monochromatic, 4 ways to pick $3$ triangles, $n$ colourings with $3$ triangles, hence all triangles, monochromatic, $1$ way to pick $4$ triangles and $n$ colourings with $4$ triangles monochromatic.

The number of $n$-colourings without monochromatic triangles is then
$n^6-4n^4+6n^2-3n$.


问:
Quick Volume with Vertical Lines Question
So I was given the following prompt with the graph:

"Let $M$ be the region in the first quadrant bounded by the graphs of $y=\arctan(x)$ and $y=\frac{\pi}{4}x$, as shown in the figure below. What is the volume of the solid generated when region $M$ is revolved around the vertical line $x=2$?"

So I guess I'm just a bit confused about what my volume equation might look like here. I understand that the equations here would both have to be expressed in terms of $y$, but I'm a bit lost about where to go after that. Any help would be appreciated!

答:
Hint:

The volume of the solid generated when region $M$ is revolved around the vertical line $x=2$ is the same of the volume you get rotating $M'$ around the orizontal line $y=2$ where $M'$ is defined as:
$$M' := \{(x,y) \in \mathbb{R}^2 : 0\le x \le \frac{\pi}{4}, \ \tan x \le y \le \frac{4}{\pi} x\}$$

Or better, it is the same of the volume you get rotating $M''$ around the orizontal line $y=0$ where $M''$ is defined as:
$$M'' := \{(x,y) \in \mathbb{R}^2 : 0\le x \le \frac{\pi}{4}, \ 2- \frac{4}{\pi} x  \le y \le 2-\tan x \} $$

The Volume you are searching for is then

$$ V = \pi \int_0^{\pi/4} {\left( {{{\left[ 2-\tan x \right]}^2} – {{\left[ 2- \frac{4}{\pi} x \right]}^2}} \right)dx} = \pi +\frac{\pi^2}{6}-2\pi\log 2 \sim 0.43$$


问:
Help w/ Proof That's Using Contour Integration
I am reading a proof in Feedback Systems by Astrom, for the Bode Sensitivity Integral, pg 339.  I am stuck on a specific part of the proof.

He is evaluating an integral along a contour which makes up the imaginary axis.  He has the following:

\begin{align}
 -i\int_{-iR}^{iR} \log(S(iw))\,dw
&=-i\int_{-iR}^{iR}\log\left | S(iw) \right |\,dw -i\int_{-iR}^{iR}\angle S(iw)\,dw\\
&=-2i\int_{0}^{iR}\log\left | S(iw) \right |\,dw
\end{align}

Taking a look at the second, or middle expression above, he makes the claim the second term, or imaginary part of the logarithm, is odd, while the first term is even.

Given that the angle on the upper part of the y-ordinate is $\pi/2$ and the lower part is $-pi/2$, the fact that the second term is odd seems obvious, so this integral vanishes.

I would like to show that the first term is even for any complex function.

This is easy to show for a simple example like $$ f(z) = Re^{i \theta} $$

I am struggling to show it for any function $$ f(z)=f(Re^{i \theta}) $$

I wrote a MATLAB script that allows me to put in any $f(z)$ I can think of and its always true to the claim.  I would just like to "prove" it.

EDIT: By the way, I found a link to the proof, page 339 in the following PDF:
https://www.cds.caltech.edu/~murray/books/AM05/pdf/am08-complete_22Feb09.pdf

答:
With the nudge I got from J.G above, I came up with the following.  I still don't feel good about this.  Not even sure it's correct.  If anyone has improvements, I am all ears!

We want to show that the integrand term is even in the following integral, for any function S(jw)
$$ j\int_{-jR}^{jR}ln\left|S\left(jw\right)\right|dw $$
This would imply the following about the integrand.
$$ ln\left|S\left(jw\right)\right|=ln\left|S\left(-jw\right)\right| $$
Given that S is a complex function of z, the following should also be true.
$$ S\left(jw\right)=S\left(-jw\right)\rightarrow S\left(Re^{j\theta}\right)=S\left(Re^{-j\theta}\right)=S\left(Re^{j\theta}\right)^\ast  $$

We also have the following general properties when considering the magnitude and complex conjugate of complex numbers.
$$ \left|z\right|=\sqrt{zz^*}=\sqrt{z^*z}=\left|z^*\right| $$
This generalizes to any complex function of z, as follows.
$$\left|f\left(z\right)\right|=\sqrt{f\left(z\right)f\left(z\right)^\ast}=\sqrt{f\left(z\right)^\ast f\left(z\right)}=\left|f\left(z\right)^\ast\right| $$

We took advantage of the fact that complex numbers are commutative in the last two equations. Circling back around, the second equation above is the condition that must be satisfied for the integrand to be an even function. The properties of S(jw) shown in the third equation are immediately implied.  We also know from the general properties of complex numbers that the magnitudes of a complex function and its complex conjugate are indeed equal.  Therefore, we come back around to the first equation showing that the integrand is indeed an even function.  This allows us to write the integrand as the following.

$$ 2j\int_{0}^{jR}ln\left|S\left(jw\right)\right|dw  $$


问:
Show that R has uncountable number of dense subsets
Following comes Topology Without Tears by Morris

Let S and T be non-empty subsets of a topological space (X, τ) with S ⊆ T.

(iii)  if S is dense in X, then T is dense in X.

What I want to prove is:

Show that R has uncountable number of dense
subsets using(iii).

Comments:
Since S is uncountable P(S) is uncountable
so no bijection exists.I can’t see how a
bijection exists.

Try

Let S={$a+b\sqrt{2}$,a,b$\in N$,
$a+b\pi$,a,b$\in N$}

Let the set T={Q$\cup$$ Q^{c}$ ,(Q$\cup I)^{c}$ $\cup $ (Q $\cup $ I)}

So each set in S is distinct uncountable and
dense in R

So each set in T is distinct uncountable and
dense in R

Since S$\subset T$ and by (iii) this shows
there are $2^{\mathfrak{c}}$ dense sets

If I have walked off the “deep end”, I would
like to know how to do this stuff correctly.

答: Union $\mathbb{Q}$ with a given irrational number to get a set T, certainly T is dense in $\mathbb{R}$ (this works since $\mathbb{Q}$ is countable implies that the irrationals form an uncountable set).


答:
It's quite simple.

$ \mathbb{Q}$ is dense in $\mathbb{R}.$

The function $f: \mathbb{R}$ \ $\mathbb{Q} \to \mathbb{R}$ defined by $f(i) = \mathbb{Q} \cup \{i\}$ is injective and the domain is uncountable, therefore the image is also uncountable. Every element in the image of $f$ is a superset of $\mathbb{Q}$, which we know is dense in $\mathbb{R}$.

So by $(iii)$, the image of $f$ is a collection of uncountable dense subsets of $\mathbb{R}.$

Body:
Following comes Topology Without Tears by Morris

Let S and T be non-empty subsets of a topological space (X, τ) with S ⊆ T.

(iii)  if S is dense in X, then T is dense in X.

What I want to prove is:

Show that R has uncountable number of dense
subsets using(iii).

Comments:
Since S is uncountable P(S) is uncountable
so no bijection exists.I can’t see how a
bijection exists.

Try

Let S={$a+b\sqrt{2}$,a,b$\in N$,
$a+b\pi$,a,b$\in N$}

Let the set T={Q$\cup$$ Q^{c}$ ,(Q$\cup I)^{c}$ $\cup $ (Q $\cup $ I)}

So each set in S is distinct uncountable and
dense in R

So each set in T is distinct uncountable and
dense in R

Since S$\subset T$ and by (iii) this shows
there are $2^{\mathfrak{c}}$ dense sets

If I have walked off the “deep end”, I would
like to know how to do this stuff correctly.


问:
About Methods to Prove That Two Sets Are Equal
When it comes to proving that two sets are equal, say $A = B$, we're usually told that we have to prove that $A \subset B$ and $B \subset A$. However, I'm under the impression that this strategy isn't unique. Two sets are equal if they have the same elements, and this commonly used strategy is just one way to prove that.

A better way to prove that two sets are equal, at least in my opinion, is to use set notation. However, because I never see this being used, I'm unsure if this method is correct. Example:

Prove that $ (A \cup B) \times C = (A \times C) \cup (B \times C) $.

Proof. We have,
\begin{align}
(A \cup B) \times C &= \{x; x \in A \operatorname{or} B \} \times \{y; y \in C \} \\
&= \{(x, y); x \in A \operatorname{or} B, y \in C \} \\
&= \{(x, y); x \in A, y \in C \} \cup \{(x, y); x \in B, y \in C \} \\
&= (A \times C) \cup (B \times C).
\end{align}

So the question is, is this method correct? Thanks in advance.

答: Both methods can work (even though they are sort of the same), and there are many other methods for proving that two sets are equal. Often times, especially in algebra, $S\subset R$, where R is a set equipped with binary structures. and S satisfies some certain algebraic property that forces it to be equal to R (e.g. if R is a unital commutative ring and I is an ideal, $I = R$ iff I contains a unit). Moral of the story, it depends on the context which method is "best" or "better."


答:
This series of equalities is correct, and is kind of hiding the argument that one is a subset of the other. It covers both directions of containment. For example, to show that $(A \cup B) \times C \subseteq (A \times C) \cup (B \times C)$, replace all $=$ with $\subseteq$, and then for showing that $(A \cup B) \times C \supseteq (A \times C) \cup (B \times C)$, replace all $=$ with $\supseteq$. These replacements work because equality implies containment on either side. So the given method is correct and sort of basically hides the usual method.

When it comes to equality of sets, at least at the very basic level, you will be using this and only this method. Sometimes, you can switch things up : for example, if you want to prove that $A=B$ you can try proving instead that $A^c = B^c$. Or you could go for a mixture : if $A \subset B$ and $A^c \subset B^c$ then $A=B$ as well. You could also prove that $A \cap B^c = \emptyset$ and $A^c \cap B = \emptyset$.

The advantage of using complement/union/intersection modifications is that sometimes a set $A$ is better understood by its complement (e.g. $A = $ "non-multiples of $6$" then the complement of $A$ is "multiples of $6$" which are easier to list), and sometimes two sets are difficult to understand but it is easy to understand their intersection. But it all boils down to that kind of reasoning in the end.

Title: About Methods to Prove That Two Sets Are Equal

Body:
When it comes to proving that two sets are equal, say $A = B$, we're usually told that we have to prove that $A \subset B$ and $B \subset A$. However, I'm under the impression that this strategy isn't unique. Two sets are equal if they have the same elements, and this commonly used strategy is just one way to prove that.

A better way to prove that two sets are equal, at least in my opinion, is to use set notation. However, because I never see this being used, I'm unsure if this method is correct. Example:

Prove that $ (A \cup B) \times C = (A \times C) \cup (B \times C) $.

Proof. We have,
\begin{align}
(A \cup B) \times C &= \{x; x \in A \operatorname{or} B \} \times \{y; y \in C \} \\
&= \{(x, y); x \in A \operatorname{or} B, y \in C \} \\
&= \{(x, y); x \in A, y \in C \} \cup \{(x, y); x \in B, y \in C \} \\
&= (A \times C) \cup (B \times C).
\end{align}

So the question is, is this method correct? Thanks in advance.


问:
Two Definitions of the Weyl Tensor
I'm reading "Textbook in Tensor Calculus and Differential Geometry" by Prasun Kumar Nayak and came across the Weyl tensor/projective curvature tensor $C_{kijl}$. The book states that
$$C_{kijl}=R_{kijl}+\frac{1}{1-N}(g_{kj}R_{il}-g_{kl}R_{ij}) \tag{1}$$

However I found on Wikipedia that
$$C_{iklm}=R_{iklm}+\frac{1}{N-2}(R_{im}g_{kl}-R_{il}g_{km}+R_{kl}g_{im}-R_{km}g_{il})+\frac{1}{(N-1)(N-2)}R(g_{il}g_{km}-g_{im}g_{kl})  \tag{2}$$

It was not immediately obvious to me that both of these are equivalent to each other?

答:
The first equation $C_{kijl}=R_{kijl}+\frac{1}{1-N}(g_{kj}R_{il}-g_{kl}R_{ij})$ is not the usual Weyl tensor. It is Weyl projective curvature tensor and it is different from Weyl tensor. It usually denoted by $W_2$. See definition 2.1 here.

In the NAYAK's book you should read "projective curvature tensor or Weyl tensor" as "projective curvature tensor or projective Weyl tensor".

Title: Two Definitions of the Weyl Tensor


问:
Chain rule Differentiation help
Can someone help me differentiate the function $$\frac{\ln(x)+\sin(x)}{x^2}$$ With respect to $x$? I tryed applying the chain rule but I keep geting lost.</p>

答:
First of all, there's no need to use the chain rule. The chain rule is for when you have a function of a function, for example $\operatorname{e}^{\sin x}$ or $(\sin x - \cos x)^2$. In this case, the quotient rule will suffice. Notice

Recall that the quotient rule says that

$$\left( \frac{\operatorname{f}}{\operatorname{g}} \right)' = \frac{\operatorname{f}'\!\operatorname{g}-\operatorname{f}\!\operatorname{g}'}{\operatorname{g}^2}$$

In your question $\operatorname{f}(x) = \ln x + \sin x$ and $\operatorname{g}(x) = x^2$. Recall that $(\ln x)' = 1/x$, $(\sin x)' = \cos x$ and $(x^2)' = 2x$. Putting all of this together:

\begin{array}{ccc}
\left( \frac{\ln x + \sin x}{x^2}\right)' &=& \frac{(1/x+\cos x)x^2-(\ln x + \sin x)(2x)}{(x^2)^2} \\ \\
&=& \frac{x+x^2\cos x - 2x\ln x - 2x\sin x}{x^4} \\ \\
&=& \frac{1}{x^3} + \frac{\cos x}{x^2}-2\frac{\ln x}{x^3} - 2\frac{\sin x}{x^3}
\end{array}

Title: Chain rule Differentiation help


问:
Example of a continuous function with discontinuous quadratic variation
Let $f: [0,\infty)\to \mathbb{R}$. The quadratic variation of $f$, if it exists, is defined as the function $\langle f\rangle: [0,\infty) \to \mathbb{R}$ with
$$  \langle f\rangle_t := \lim_{n\to \infty} \sum_{t_i \in \pi_n(t)} \left( f(t_{i+1}) - f(t_i) \right)^2  $$
for $t \in [0,\infty)$ where $\{\pi_n(t): n\in \mathbb{N}\}$ is a sequence of refining partitions of $[0,t]$.

I am looking for an example of a function $f$ such that $f$ is continuous and its quadratic variation $\langle f\rangle$ exists, but $\langle f \rangle$ is not continuous.

Motivation: In probability lecture notes, one sometimes reads (e.g. in the context of the pathwise Ito formula) that a path $X(\omega)$ is assumed to be continuous with continuous quadratic variation. Therefore, I would like to understand why it is necessary to explicitly demand the quadratic variation to be continuous if this is desired.

答:
An example is given in Appendix 5.2 of Coquet, Jakubowski, Mémin, and Słominski, Natural decomposition of processes and weak Dirichlet processes, pp. 81–116, Springer, Berlin, Heidelberg, 2006.

For completeness, I replicate the example below: we restrict ourselves to the time interval $[0,1]$. Let $f \in C[0,1]$ be defined by $f(t) = 0$ when $t = 1 - 2^{1-2p}$ and  $f(t) = \frac{1}{p}$ when $t = 1 - 2^{-2p}$, where $p \in \mathbb{Z}^+$. We complete the construction of $f$ by linearly interpolating between these points. The graph of $f$ looks like a sequence of shrinking scalene triangles as you move forward in time.

By construction, is is clear that $f$ is of bounded variation on all intervals $[0,t]$ with $t < 1$; hence its quadratic variation on those intervals vanish.

By considering the quadratic variation on $[0,1]$ along the sample points $T = \{ 1 - 2^{-2k} \, : k \geq 1 \}$, you get that this quantity is infinite.

Finally, you may construct a sequence of refining partitions of $[0,1]$, say $(\pi_n)_{n=1}^\infty$, given by:

\begin{align*}
\pi_n = \bigcup_{j \leq 2^{2n} - 1} \{j 2^{-2n}\} \cup \bigcup_{k \geq n} \{ 1 - 2^{-2k} \}
\end{align*}

You can show that along this sequence, the pathwise quadratic variation takes up its entire mass at $t=1$.


问:
Does a convex function sandwiched between two quadratic functions have Lipschitz derivatives?
Assume we have a convex function $f:\mathbb{R}^n\to \mathbb{R}$ and that it holds,
$$c_1 ||x||^2 \leq f(x) \leq c_2 ||x||^2.$$
It looks like $f$ should behave quadratically around $0$, but can we proof that it has Lipschitz derivatives in a neighbourhood $U$ of $0$? Or some other bound on the derivatives near $0$?

I know Lipschitz gradients would be a consequence of L-smoothness in the sense that $\forall x,y\in U$,
$$f(y)\leq f(x) + \nabla f(x)^\top (y-x)+\dfrac{L}{2}||x-y||^2 $$
but this seems to be a stronger assumption as $f$ has to be upper bounded by a tangential quadratic function at every point in $U$ and not just in $0$.

答:
The best you can do is a Lipschitz estimate at zero itself, that is, there is an $A > 0$ such that
\begin{equation*}
\sup \left\{ \|p\| \, \mid \, p \in \partial f(x), \, \, x \in B(0,r) \right\} \leq A r.
\end{equation*}
(Notice $f$ is differentiable at $0$ and $Df(0) = 0$.)  In fact, there is also a lower bound:
\begin{equation*}
\|p\| \geq c_{1}\|x\| \quad \text{if} \, \, p \in \partial f(x).
\end{equation*}

The lower bound is much simpler to obtain.  If $x \in \mathbb{R}^{n} \setminus \{0\}$ and $p \in \partial f(x)$, then
\begin{equation*}
0 = f(0) \geq f(x) - \langle p, x \rangle \geq c_{1} \|x\|^{2} - \langle p, x \rangle.
\end{equation*}
Hence $\langle p, \frac{x}{\|x\|} \rangle \geq c_{1}\|x\|$, implying $\|p\| \geq c_{1} \|x\|$.

To obtain the upper bound, fix $r > 0$ and assume that $\|x\| \leq r$.  If $p \in \partial f(x)$ and $\xi \in S^{n-1}$, then
\begin{equation*}
c_{2} \|x + r \xi\|^{2} \geq f(x + r \xi) \geq f(x) + r \langle p, \xi \rangle \geq c_{1} \|x\|^{2} + r \langle p, \xi \rangle.
\end{equation*}
Thus,
\begin{equation*}
\langle p, \xi \rangle \leq \frac{1}{r} \left( c_{2} \|x\|^{2} + c_{2} r^{2} + 2 c_{2} r \langle x, \xi \rangle - c_{1} \|x\|^{2} \right) \leq (4c_{2} - c_{1}) r.
\end{equation*}
This gives us the desired upper bound with $A = 4c_{2} - c_{1}$.

The reason this is the best we could hope for (i.e. we can't get $Df$ Lipschitz in a neighborhood of $0$) is $c_{1} < c_{2}$.  Therefore, as soon as you get away from $0$, $f$ has a positive amount of space to wiggle around --- if $f$ can wiggle around a bunch, then certainly $Df$ can't be constrained, there's no reason for it to be continuous.

To make this more precise, let's go to $n = 1$.  Let $h : [0,\infty] \to [c_{1},c_{2}]$ be a non-decreasing function with a sequence of discontinuities converging at zero. Define $g : \mathbb{R} \to \mathbb{R}$ with $g(x) = -g(-x)$ by
\begin{equation*}
g(x) = 2 h(x) x.
\end{equation*}
$g$ is non-decreasing.  Therefore, the function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = \int_{0}^{x} g(s) \, ds$ is convex.  Furthermore, if $x > 0$, then
\begin{equation*}
c_{1}|x|^{2} \leq f(x) \leq c_{2} |x|^{2}.
\end{equation*}
Since $f(x) = f(-x)$, $f$ satisfies the desired inequalities, but $f'$ is not continuous in any neighborhood of zero.

We can generalize the previous example to $n > 1$ simpy by setting $F(x) = f(\|x\|)$.


问:
Conformal maps and möbius transformations
Is it possible to find a möbius transformation mapping $\mathbb{C}$ to the unit disc $D(0,1)$

my attempt:

I think that since $\mathbb{C}$ isn't bounded by any circlines but the unit disc is bounded by a circle centre $0$ radius $1$ then we can't find such a composition of mobius transformations to do so.

Is this correct, any guidance would be helpful.

答:
No holomorphic function can map the complex plane to a bounded open set. See
Liouville's theorem.</p>


答: Yes, Mobius transformations send generalized circles to generalized circles. A generalized circle is a cirlce or a line. If there was a Mobius transformation sending $\mathbb{C}$ to the disk, then its inverse would also be a Mobius transformation and would send the unit circle to some other generalized circle, exactly as you say. Since Mobius transformations are continuous the unit circle will be mapped on one of the sides of the generalized circle.

Title: Conformal maps and möbius transformations

Body:
Is it possible to find a möbius transformation mapping $\mathbb{C}$ to the unit disc $D(0,1)$

my attempt:

I think that since $\mathbb{C}$ isn't bounded by any circlines but the unit disc is bounded by a circle centre $0$ radius $1$ then we can't find such a composition of mobius transformations to do so.

Is this correct, any guidance would be helpful.


问:
when does a functor map products into products?
Motivation: wikipedia claims, that in algebraic topology, there holds: $\pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y)$ and $\pi_1(X\vee Y)\cong\pi_1(X)\ast\pi_1(Y)$. A similar statement holds for arbitrary products and one-point unions, making the (covariant) fundamental group functor $\pi_1:\mathrm{TOP}^0 / h-\mathrm{TOP}^0\rightarrow GRPS$ preserve products and coproducts.

I'm guessing the same holds for the functors $\pi_k$ (homotopy groups), $H_k$ (homology groups)?

Definitions:

Examples:
In the category of sets, groups, rings, $R$-modules, vector spaces, topological spaces,etc, the product is the cartesian product. In the category of sets and topological spaces, the coproduct is the disjoint union/topological sum. In the category of groups, the coproduct is the free product $\ast$. In the abelian group / $R$-modules / vector spaces category, it is the direct sum $\oplus$. In the topological pointed spaces category, it is the one-point union $\vee$.

Question: I would very much like to prove this in a general way, so I'd like to know the following:
Theorem???: Suppose $F:\underline{A}\rightarrow\underline{B}$ is a covariant / contravariant functor. What are some (reasonably general) sufficient conditions on $F,\underline{A},\underline{B}$, that make $F$ send (products to products and coproducts to coproducts) / (products to coproducts and coproducts to products ), i.e.
$$F \text{ covariant }\Rightarrow F(\prod_{i\in I}A_i)=\prod_{i\in I}F(A_i)\text{ and }F(\coprod_{i\in I}A_i)=\coprod_{i\in I}F(A_i);$$
$$F \text{ contravariant }\Rightarrow F(\prod_{i\in I}A_i)=\coprod_{i\in I}F(A_i)\text{ and }F(\coprod_{i\in I}A_i)=\prod_{i\in I}F(A_i)?$$

The sufficient conditions that I'm looking for are primarily intended for important standard functors, such as $\pi_k$, $H_k$, $H^k$, tangent and cotangent bundle functor, ... so that I can prove the results in one sweep.

Counterexample: In the category of division rings / fields, we have $\mathbb{Z}_2\times\mathbb{Z}_2 = \mathbb{Z}_2$, thus the forgetful functor to the category of sets doesn't preserve products.

答: The preservation of (co)products is related to the existence of a left (or right) adjoint; see e.g. wikipedia on the Adjoint functor theorem.  If you know the existence of an adjoint (and some functors are essentially defined as adjoints, such as free groups, and tensor products), this provides a convenient way to deduce that the functor preserves products (or coproducts, as the case may be).


问:
Frobenius norm of an orthogonal projector
So, I've tried multiplying this out to see if there's a pattern. I originally thought that the orthogonal projector was made up of orthonormal basis because the equation for it is sigma qiqi*, but that doesn't work.

I've also tried using root(trace(PP*)) = root (trace(P)) , but I can't see a pattern there that helps me either.

I have a feeling that this should be one (some intuition). How should I go about doing this?

Thanks

答:
Hint 1: Orthogonal projectors are self-adjoint, meaning $P^\ast =P$.

Compute $tr(PP^\ast)=tr(P^2)=\_\_\_\_$

(This is probably closer to the question you were asking, and I just didn't see it before.)

Hint 2: If you listed all the components of $P$ out in a row vector $q$ with entries $q_i$, then $tr(PP^\ast)$ is indeed $qq^\ast$. To see this, you just need to write down what the diagonal entries of $PP^\ast$ are, and then sum them together.


问:
How to show that $\mathbb{R}$ in the language of rings does not admit quantifier elimination?
I am studying for an exam in an introductory mathematical logic course. Suppose we are working in the language of rings $\mathcal{L}=(+,-,\cdot,0,1)$. Then $\mathbb{R}$ does not admit quantifier elimination.

First of all I am a bit unsure of what it means for a structure to admit quantifier elimination, but I assume that $\mathbb{R}$ is shorthand for $\operatorname{Th}(\mathbb{R})=\{\phi \text{ an } \mathcal{L}\text{-sentence}\mid \mathbb{R}\models \phi\}$, since in this case it would make sense (quantifier elimination is defined for theories). Please correct me if this is a false assumption.

So now I want to show that $\mathbb{R}$ does not admit quantifier elimination. It seems difficult to choose a specific formula which has no equivalent quantifier-free formula and then showing this is the case. I do however have the following theorem at my disposal:

Let $T$ be an $\mathcal{L}$-theory, $n\geq 1$ a natural number and $\phi(x_1,\ldots,x_n)$ an $\mathcal{L}$-formula. Then the following are equivalent:

There is a quantifier-free $\mathcal{L}$ formula $\psi(x_1,\ldots,x_n)$ such that $\phi$ and $\psi$ are equivalent in $T$.

Let $\mathcal{M},\mathcal{N}$ be models of $T$ and let $\mathcal{A}$ be a common substructure. Then for any $\overline{a}\in A^n$ one has $\mathcal{M}\models \phi[\overline a]\iff \mathcal{N}\models \phi[\overline a]$.

If we use this, it suffices to find some elementary equivalent sub- or superstructure in which there is a non-true formula which becomes true or vice versa.

For example, one can take $\mathbb{R}_\text{alg}= \{r\in\mathbb{R}\mid \exists 0\neq p(X)\in \mathbb{Q}[X] \text{ such that } p(r)=0\}$. I tried looking at the formula $\exists x x^2+y=z$ but I believe this to be equivalent in both models.

So can anyone help me find a formula which is not equivalent in both models (or in other cases where we do not choose $\mathbb{R}_\text{alg}$). Or is there perhaps a different method which is better suited for this problem? I would prefer a method which is applicable in similar situations as I believe the one I describe is, but any proof is welcome.

答: Your idea of using a formula defining $\leq$ is a good one, but it will be easier to work with a similar formula that has only one free variable; I suggest $\exists x\,(x^2=y)$.  I also think it's easier (at least in this case) to ignore the model-theoretic criterion and prove directly that $\exists x\,(x^2=y)$ isn't equivalent, in $\mathbb R$, to any quantifier-free formula with only the variable $y$. The key ingredient of that proof will be that (1) any such quantifier-free formula is just a propositional combination of polynomial equations about $y$ (with integer coefficients, but that won't be needed), (2)  polynomial equation about $y$ is either satisfied by all values of $y$ or by only finitely many (here's where it's pleasant to have just one variable), and (3) therefore a quantifier-free, one-variable formula defines a finite or cofinite subset of $\mathbb R$.


问:
Product of Levi-Civita symbol is determinant?
I am confused with how can one write product of Levi-Civita symbol as determinant? I want to prove 'epsilon-delta' identity and found this questions answers it. But I am stuck at product of Levi-Civita symbol

Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory

$$
\varepsilon_{ijk}\varepsilon_{lmn} = \begin{vmatrix}
    \delta_{il} & \delta_{im} & \delta_{in} \\
    \delta_{jl} & \delta_{jm} & \delta_{jn} \\
    \delta_{kl} & \delta_{km} & \delta_{kn} \\
  \end{vmatrix}
$$

This Wikipedia article gives the above relation. I am confused how Product of Levi-Civita symbol is a determinant. Can someone explain?

答:
It is important to understand that usually the determinant is a scalar
but in this case it is not. Recall the cross product $A \times B$,
can be computed as a determinant and it is a vector. We will interpret
as determinant to the regular expansion of determinants but this time
instead of a scalar entries we have 2-rank tensor entries.

We expand the determinant as follows:
\begin{eqnarray*}
  \left |
  \begin{array}{ccc}
    \delta_m^r & \delta_n^r & \delta_p^r \\
    \delta_m^s & \delta_n^s & \delta_p^s \\
    \delta_m^t & \delta_n^t & \delta_p^t
  \end{array}
  \right |
  &=& \delta_m^r \delta_n^s \delta_p^t + \delta_n^r \delta_p^s \delta_m^t +
  \delta_p^r \delta_m^s \delta_n^t
  - \delta_p^r \delta_n^s \delta_m^t - \delta_n^r \delta_m^s \delta_p^t - \delta_m^r \delta_p^s
  \delta_n^t .
\end{eqnarray*}
We define
\begin{eqnarray*}
  a_{mnp}^{rst} =
  &=&
  \delta_m^r \delta_n^s \delta_p^t  +
\delta_m^r \delta_n^s \delta_p^t  + \delta_m^r \delta_n^s \delta_p^t  -
\delta_m^r \delta_n^s \delta_p^t  - \delta_m^r \delta_n^s \delta_p^t
-\delta_m^r \delta_n^s \delta_p^t
\end{eqnarray*}
where we commuted factors so $m,n,p$ are in that order in all subsscripts.
We recognize that the sequence of delta operations $a_{mnp}^{rst}$ is a completely skew-symmetri
system in $r,s,t$. That is, for example, let us reverse the order or $s$ and $t$ and find
\begin{eqnarray*}
  a^{rts}_{mnp} &=&
  \delta_m^r \delta_n^t \delta_p^s  +
  \delta_m^s \delta_n^r \delta_p^t  + \delta_m^t \delta_n^s \delta_p^r  -
  \delta_m^s \delta_n^t \delta_p^r  - \delta_m^t \delta_n^r \delta_p^s  -\delta_m^r \delta_n^s
  \delta_p^t \\
  &=& -\delta_m^r \delta_n^s \delta_p^t  -\delta_m^t \delta_n^r \delta_p^s -\delta_m^s \delta_n^t
  \delta_p^r + \delta_m^t \delta_n^s \delta_p^r + \delta_m^s \delta_n^r \delta_p^t + \delta_m^r
  \delta_n^t \delta_p^s \\
  &=& -a^{rst}_{mnp}
\end{eqnarray*}
where we fixed the order by using the commutative law of the sum (with sign) operation .
Likewise we can interchange the order of $r,s$ and $r,t$ an get
\begin{eqnarray*}
  a_{mnp}^{srt} = -a_{mnp}^{rst} \quad  \text{and} \quad
  a_{mnp}^{tsr} = -a_{mnp}^{rst}.
\end{eqnarray*}
Hence $a_{mnp}^{rst}$ ($m,n,p$ fixed) is completely skew-symmetric in $r,s,t$ and  so
\begin{eqnarray*}
  a_{mnp}^{rst} = \epsilon^{rst} a_{mnp}^{123}
\end{eqnarray*}
with
\begin{eqnarray*}
  a_{mnp}^{123}  =
  \delta_m^1 \delta_n^2 \delta_p^3  +
  \delta_m^3 \delta_n^1 \delta_p^2  + \delta_m^2 \delta_n^3 \delta_p^1  -
  \delta_m^3 \delta_n^2 \delta_p^1  - \delta_m^2 \delta_n^1 \delta_p^3  -\delta_m^1 \delta_n^3
  \delta_p^2 .
\end{eqnarray*}
This is equal to $1$ if $mpn$ is an even permutation of $(1 \; 2 \; 3)$ or
$-1$ if $mpn$ is an odd permutation of $(1 \; 2 \; 3)$. That is,
\begin{eqnarray*}
  a_{mnp}^{rst} = \epsilon^{rst} \epsilon_{mnp}
\end{eqnarray*}
if $m=n$ or $m=p$ we would get 0. Then

\begin{eqnarray*}
  \left |
  \begin{array}{ccc}
    \delta_m^r & \delta_n^r & \delta_p^r \\
    \delta_m^s & \delta_n^s & \delta_p^s \\
    \delta_m^t & \delta_n^t & \delta_p^t
  \end{array}
  \right |
  &=& \epsilon^{rst} \epsilon_{mnp} .
\end{eqnarray*}


答:
$$f(x)=x\left(\sin^2(x)+\frac1{1+x^2}\right)$$

There is only one value of $x$ such that $f(x)=0$

$f'(0)=1\ne0$

$f$ oscillates between values arbitrarily close to $0$ and arbitrarily large values.

The last property guarantees that $f^{-1}((-\varepsilon,\varepsilon))$ is made of a union of infinitely many disjoint open intervals, hence it's not homeomorphic to $(-\varepsilon,\varepsilon)$, for any $\varepsilon>0$.

With the example in my comment above, $f(x)=\dfrac{x}{1+x^2}$, the constraints where only partially fulfilled: for a large enough $\varepsilon>0$, $f^{-1}((-\varepsilon,\varepsilon))=\Bbb R$, because that $f$ is bounded. Here it doesn't happen.

Title: Smooth mapping on $ \mathbb R$

Body:
I am going through differentiable manifolds and come across a problem:

How to construct a smooth mapping $f: \mathbb R \rightarrow  \mathbb R$ such that

$f^{-1}(0)=0$

$f^{'}(0) \neq 0$

$\forall \epsilon >0,f^{-1}(-\epsilon ,\epsilon )$ is not homeomorphic to $(-\epsilon ,\epsilon )$

Any help?


问:
How to find the center of a spiral given only the physical object?
I have a spiral cam from a belt sander tension assembly that needs to have a hole bored in the exact center and I am unsure of how to find it mathematically. I have the object and can measure the diameter at various points but I am stuck at how to get it into a mathematical equation

答:
Since it's not a circle, the center is not well-defined; there may be different points you could reasonably call the 'center' depending on what properties you like.

One reasonable choice is the centroid, i.e. center of mass. You could find the centroid by moving a line/straightedge until the shape balances perfectly on its end, and then do this again with another line at an angle to it. The intersection of the two lines is the centroid.  You can check that the shape should balance perfectly on this point. See plumb-line method at wikipedia.

If you want a purely mathematical way to find the center of mass, you will need to provide precise equations for the shape.  Then you can use calculus methods.


问:
20 Gifts and four stacks, How many Gifts in each pile?
There are twenty gifts stacked up into $4$ piles.
The first pile has $3$ less than the second pile.
The second pile has $2$ more than the third pile.
The fourth pile has twice as many as the second pile.
How many gifts are in each pile?

This is for my $8$ year old son... Embarassing that I can't figure it out.

答:
Besides André’s approach, I can also imagine a bright child starting with an empty first pile, three gifts in the second pile, and one gift in the third, easily fulfilling the first three requirements, and then working out that there must be six gifts in the fourth pile. That’s the smallest arrangement that meets the requirements on the relative sizes of the piles; unfortunately, it uses only $10$ of the $20$ gifts.

The first three piles change in step with one another: add one gift to one of them, and you clearly must add one gift to each of them. And every time you add one gift to the second pile, you must add two gifts to the fourth pile. Thus, adding one gift to any of the first three piles forces you to add a total of five gifts. Aha! Do this twice, and you’ve used all $20$ gifts. You’ve now added two gifts to each of the first three piles and four gifts to the fourth pile, so the piles now contain $2$, $5$, $3$, and $10$ gifts, respectively.


问:
Integrate the following functions with respect to $x$
Integrate the following functions with respect to $x$:

$5x^4+3x+10$

$x^3-2x^2+9x$

For context I have recently completed my university degree and decided to go over some first year mathematics only to find i'm really struggling to remember integration, these are two questions that I have written down but I am struggling to remember the method behind it. I post this question hoping someone would be able to do some form of step-by-step calculation.

答:
Do you know the power rule?
$$
\frac{d}{dx}(ax^n)=anx^{n-1}
$$
I remember this rule using the mantra 'multiply by the power, subtract one from the power'. Since integration is differentiation in reverse, we have to 'add one to the power, divide by the power', so
$$
\int ax^n \, dx= \frac{ax^{n+1}}{n+1} + C\, .
$$
Use the power rule, along with the sum rule,
$$
\int f(x)+g(x) \, dx = \int f(x) \, dx + \int g(x) \, dx \, ,
$$
to solve your problem. You can write constants such as $10$ as $10x^{0}$.


问:
If $X_n$ converges to X in probability then $X_n$ converges to X a.s.
We know that if $X_n$ converges to X a.s. then $X_n$ converges to X in probability in the space ( Ω, A, P) . But the converse is not true in general . We can construct  such examples on any probability space (Ω, A, P) which has a “non-atomic part”, meaning that, there is an $Ω_0$ ∈ A with P($Ω_0$)>0, such that, for any A ∈ A, A ⊂ $Ω_0$, with P(A)>0 and any 0< δ <1, one has B ∈A, B ⊂A with P(B)= δP(A).
But when the probability space is “purely atomic”, not only do we have no counter-example as above, but indeed, convergence in probability implies a.s. convergence. My question is how can I show that if (Ω, A, P) be a discrete probability space, that is, Ω is a countable set and A consists of all subsets of Ω. Then for random variables $X_n$, n ≥ 0 and X on (Ω, A, P), $X_n$ converges to X in probability then $X_n$ converges to X a.s.

答:
Let's review the definitions.  To say that $X_n$ converges to $X$ in probability means that for any $\epsilon > 0$,

$$\lim\limits_{n \to \infty} P(|X_n - X| \geq \epsilon) = 0.$$

To say that $X_n$ converges to $X$ almost surely means that

$$P\Big( \{ s \in \Omega : \lim\limits_{n \to \infty} X_n(s) = X(s)\}\Big) = 1.$$

Proposition: For a discrete probability space $(\Omega, A, P)$, and random variables $X, X_n : n\in \mathbb N$, the following are equivalent:

(i): $X_n$ converges to $X$ in probability.

(ii): $X_n$ converges to $X$ almost surely.

(iii): $X_n$ converges to $X$ pointwise at all points of nonzero measure, which is to say that for every $s \in \Omega$ such that $P(s) > 0$, we have $\lim\limits_{n \to \infty} X_n(s) = X(s)$.

(i) $\Rightarrow$ (iii): Suppose that there exists an $s_0 \in \Omega$ with $\eta := P(s_0) > 0$ for which we do not have $\lim\limits_{n \to \infty} X_n(s_0) = X(s_0)$.  Then there exists an $\epsilon > 0$ such that $|X_n(s_0) - X(s_0)| \geq \epsilon$ for infinitely many $n$.  If we set

$$E_n = \{ s \in \Omega : |X_n(s) - X(s)| \geq \epsilon\}$$

then we have $s_0 \in E_n$ for infinitely many $n$.  In particular, $P(E_n) \geq P(s_0) = \eta$ for infinitely many $n$.  Therefore,

$$\lim\limits_{n \to \infty} P(E_n) = \lim\limits_{n \to \infty} P(|X_n - X| \geq \epsilon)$$
cannot be zero.  This shows that $X_n$ does not converge to $X$ in probability.

(iii) $\Rightarrow$ (ii): Suppose that $X_n$ does not converge to $X$ almost surely.  Then

$$P \Big( \{s \in \Omega : \lim\limits_{n \to \infty} X_n(s) = X(s)\} \Big) < 1.$$

Since $\Omega$ is a discrete probability space, the only way this can happen is if there exists an $s_0 \in \Omega$, with $P(s_0) > 0$, for which $\lim\limits_{n \to \infty} X_n(s) \neq X(s)$.

(ii) $\Rightarrow$ (i): This is well known.


问:
Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$
It is also given that $abc = 1$.

I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $

How to go further

答:
We will prove the homogeneous inequality (where we used $abc=1$) $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[6]{abc}$$ Since equality is attainable for $a=b=c$, this would show that the desired $k$ is $3$. Observe that, in virtue of AM-GM $$\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{b}}+\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{c}}}6\geqslant \sqrt[6]{abc}\iff\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{b}}+\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{c}}}2\geqslant 3\cdot \sqrt[6]{abc} $$
Also, AM-GM yields $$\frac{2\cdot\frac{a}{\sqrt{b}}+2\cdot \frac{a}{\sqrt{c}}+\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}}{6}\geqslant \sqrt[6]{a^3}=\sqrt{a}$$ Adding the first inequality and the cyclic variations of the latter one, we arrive at the desired inequality.


问:
Expected area covered by circles inside a square
$n$ circles of radius $r$ are to be placed randomly inside a square of side length $l$. What is the expected area the circles will enclose?

Edit: As @Mirko suggested, we define random placement of the circles by picking points from a square of side length $l-2r$ concentric to the initial one from a uniform distribution and using each one as a center for a specific circle.

The problem I'm facing is that I do not know how to statistically account for overlaps with the circles. I have tried to divide the square into little square "cells" of side $2r$ such that each circle can exist in one of the cells and obtained an expression for the probability of $k$ circles to exist with no overlap. However, this approach drastically understates the actual probability as it doesn't account for circles existing in between cells.

Any alternative approach would be greatly appreciated.

答:
So we're assuming $l\ge2r$. Let's take the square in Cartesian coordinates to be $S:=[-l/2,\,l/2]^2$. The point $(x,\,y)$ therein falls within a circle centred at $(u,\,v)\in S^\prime:=[-(l/2-r),\,l/2-r]$ iff$$(x-u)^2+(y-v)^2\le r^2.$$Let $f(x,\,y)$ denote the probability of this when we take $U,\,V$ to be Uniform IIDs on their support; the same treatment of $X,\,Y$ makes the probability a random point lies in at least one of the circles$$1-\int_S(1-f)^ndxdy,$$so multiplying this by $\pi l^2$ gives the circles' union's mean area. Note the $n\to\infty$ limit is less than $\pi l^2$ because some points have $f=0\implies\lim_{n\to\infty}(1-f)^n=1$. In particular, a point in $S$ may be as far as $r\sqrt{2}$ from the nearest point in $S^\prime$.

So now we have two tasks: compute $f$, then compute $\int_S(1-f)^ndxdy$. Note $f$ is the proportion of the area of $S^\prime$ within a distance $r$ of $(x,\,y)$. While obtaining $f$ as a piecewise function is feasible, I suspect the integrals are numerical for large $n$.


问:
Random graphs as random variable

The $G (n, p)$ model, due to Erdös and Rényi, has two parameters, $n$
and $p$. Here $n$ is the number of vertices of the graph and $p$ is
the edge probability. For each pair of distinct vertices, $v$ and $w,
> p$ is the probability that the edge $(v,w)$ is present. The presence
of each edge is statistically independent of all other edges. The
graph-valued random variable with these parameters is denoted by $G(n,p)$. When we refer to “the graph $G(n,p)$”, we mean one realization
of the random variable

Random graphs are random variables?

I thought that $G(n,p)$ is a random variable with $\Omega\rightarrow \mathbb{R}$, where $\Omega$ is $\mathcal{P}\{1,...,n\}^2,\mathcal{P}\{\mathcal{P}\{1,...,n\}^2\},1/|\mathcal{P}\{1,...,n\}^2|$ and $G(n,p)(\omega)= |\omega|^p|\{1,...,n\}^2|-|\omega|^{1-p}$. But if that would be the case I don t know how I can formally verify that the event that an edge is present is stochastically independed that an oher edge is present. I have chosen the model to look at all possible sets of edges ie all graphs possible, the author of this paper (J.Spencer) mentioned a graph-valued random variable hence I thought I have to find a probabiliypace where an elemen represents a graph.

If someone could tell me how I can verify the stochastic independence of the existence of two edges and how the random graphs are modeled as random variables I would really appreciate it

答:
Random graphs are a random variable, but not a real-valued random variable: a graph-valued one. $G(n,p)$ is a function from the sample space $\Omega$ to the set of all $2^{\binom n2}$ graphs on a fixed vertex set $\{v_1, v_2, \dots, v_n\}$.

So what's the sample space, then? We could be lazy and just take $\Omega$ to be the same set of graphs, so that $G(n,p)$ is the identity function. Generally, we try not to be too specific about what $\Omega$ is, though, because we might want other sources of randomness involved. We are happy as long as the distribution of $G(n,p)$ is correct: as long as $$\Pr[G(n,p) = G] = p^{|E(G)|} (1-p)^{\binom n2 - |E(G)|}.$$

We might also want to think of the graph evolution process, in which case $\Omega = [0,1]^{\binom n2}$. For each outcome $\omega \in \Omega$, $G(n,p)(\omega)$ includes the $i^{\text{th}}$ edge (order the edges however you like) if $\omega_i < p$. Here, we can define multiple graph valued random variables $G(n,p)$ and $G(n,p')$ in the same probability space.

Title: Random graphs as random variable


问:
Prove series form of fractional harmonic numbers
Let $H_\alpha$ be the $\alpha$th fractional harmonic number so that
$$
H_\alpha = \int_0^1 \frac{1-x^\alpha}{1-x}\,\text dx.
$$

I want to directly show
$$
H_\alpha = \sum_{k=1}^\infty \frac{\alpha}{k(k+\alpha)}.
$$

I know this must be true because $\psi(1  + \alpha) = -\gamma + H_\alpha$ and $\psi(1 + \alpha) = -\gamma + \sum_{k=1}^\infty \frac{\alpha}{n(n+\alpha)}$ where $\psi$ is the digamma function but I haven't managed to prove it.

I've proven this for $\alpha \in \mathbb N$ because in this case $(1-x^\alpha)/(1-x) = \sum_{i=0}^{\alpha-1}x^i$ and I can split up $\frac{\alpha}{k(k+\alpha)} = \frac{1}{k} - \frac{1}{k+\alpha}$ to get telescoping since eventually $1/k' = 1/(k+\alpha)$. But if $\alpha \notin \mathbb N$ then I get neither the factorization nor the telescoping so it seems those tricks only help in $\mathbb N$. Even if $\alpha = p/q \in \mathbb Q$ the telescoping fails and I don't see any way to generalize my approach in $\mathbb N$, so it seems an entirely different approach may be needed. So how can I prove this directly? I've also tried a few different series for $\frac{1-x^\alpha}{1-x}$ but no luck so far.

I'm sure there are many ways to show this using fancy properties of $\psi$ and other special functions, but I'm trying to prove it directly. I'm only bringing up $\psi$ for context. Thanks a lot for any help.

答:
Working the problem in the opposite direction has geometrical meaning in calculating the Euler-Mascheroni constant $\gamma$.  The areas $\frac{1}{\lfloor x \rfloor}-\frac{1}{x}$ on each interval $x \in \left[n,n+1\right]$ can be collected together on $\left[0,1\right]$ to provide more uniform behavior.  Your $H_\alpha$ function gives the total sum of these differences at offset $\alpha$ from all integers $n$:

$$
H_\alpha = \sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k+\alpha}
$$

Going further and integrating this on $\left[0,1\right]$ would give us $\gamma$ but clearing the disjointed summation to give a more continuous, unified view is a natural next step:

$$
\sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k+\alpha} = \sum_{k=1}^\infty \int_0^1 t^k - t^{k+\alpha} dt \\
= \int_0^1 \left( \sum_{k=1}^\infty t^k - \sum_{k=1}^\infty t^{k+\alpha} \right) dt \\
= \int_0^1 \frac {1}{1-t} - \frac {t^\alpha}{1-t} dt \\
=\int_0^1 \frac {1-t^\alpha}{1-t} dt
$$


问:
Diffeomorphic smooth structures on a topological manifold doesn‘t necessarily imply that the identity is a diffeomorphism
Let $M$ be a topological manifold. I can equip  $M$ with a smooth atlas $\mathscr{A}$ and a smooth atlas $\mathscr{B}$. If there exists a diffeomorphism $f$ between the two resulting smooth manifolds, it doesn‘t necessarily follow that the identity is a diffeomorphism. (This would imply that $\mathscr{A}\cup\mathscr{B}$ is a smooth atlas and I know counterexamples.)

But the composition of two diffeomorphisms is again a diffeomorphism and we have $\;\text{id}_M=f^{-1}\circ f$. Where am I thinking wrong?

答: If $f : (M, \mathscr{A}) \to (M, \mathscr{B})$ is a diffeomorphism, then $f^{-1} : (M, \mathscr{B}) \to (M, \mathscr{A})$ is a diffeomorphism and hence $\operatorname{id}_M = f^{-1}\circ f : (M, \mathscr{A}) \to (M, \mathscr{A})$ is a diffeomorphism. However, that tells us nothing about whether $\operatorname{id}_M : (M, \mathscr{A}) \to (M, \mathscr{B})$ is a diffeomorphism or not.


问:
Prove that the integral of a rapidly oscillating function is $0$
Claim: If $n$ is a nonzero integer, then $$\int_{-\pi/2}^{\pi/2} \exp[{2in(x+\tan x)}] \ \mathrm{d}x = 0$$

Using Euler's identity $e^{ix}= \cos x + i \sin x$, the fact that $\sin$ is an odd function so integrated over a symmetric integral gives $0$, and the parity of $\cos$, one gets

$$\int_{0}^{\pi/2} \! \! \cos(2n(x+\tan x)) \ \mathrm{d}x$$

I tried using substitution, but it didn't get me anywhere. Is there maybe a general way to approach these rapidly oscillating integrals to show that the positive and negative areas cancel out?

答:
Integrate $f(z) = \exp[2in(z+\tan z)]$ around the rectangular contour with vertices $-\pi/2,\pi/2,-\pi/2+ iR, \pi/2+iR$. For the moment, ignore the singularity at $z=\pm \pi/2$.

Since $n$ is an integer, $f(z+\pi) = f(z)$, hence integrals on two vertical line cancel each other, hence
$$\int_{-\pi/2}^{\pi/2} f(z) dz = \int_{-\pi/2}^{\pi/2} f(z+iR) dz$$ for any $R>0$. But $$\int_{-\pi/2}^{\pi/2} f(z+iR) dz = e^{-2nR}\int_{-\pi/2}^{\pi/2} \exp[2in(x+\tan(x+iR))]dx$$
last integral is bounded as $R\to \infty$ because $\tan$ is bounded at large imaginary argument, letting $R\to \infty$ gives $\int_{-\pi/2}^{\pi/2} f(z) dz = 0$ when $n> 0$.

Now we take account of the singularity at $z=\pm \pi/2$. Let $r$ be small positive, let $C_1$ be the circle $\pi/2 + re^{i\theta}$ with $\pi/2 \leq \theta\leq \pi$. Then I claim $\int_{C_1} f(z) dz \to 0$ as $r\to 0$. Indeed, $$\tan(\pi/2 + re^{i\theta}) = -\cot(re^{i\theta}) = -r^{-1}e^{-i\theta} + O(r)$$
so $$\int_{C_1} f(z) dz = ir\int_{\pi/2}^{\pi} \left( \exp[-2inr^{-1}e^{-i\theta}] + O(1)\right) d\theta$$
If $n>0$, then $\Re(-2inr^{-1}e^{-i\theta})\leq 0$ for $\theta$ in range, the integral tends to $0$ as $r\to 0$. This take cares the singularity at $z=\pi/2$. Similar procedure handles the singularity at $z=-\pi/2$.

So we rigorously proved, when $n$ is positive integer, $$\lim_{\varepsilon\to 0^+} \int_{-\pi/2 + \varepsilon}^{\pi/2 - \varepsilon} f(x) dx = 0$$
The same is true for $n$ negative, by looking at real and imaginary part.

A very analogous integral $$\int_{-\pi/2}^{\pi/2} \exp[2i(x-\tan x)] dx \approx 1.70067$$ is nonzero. Curious readers can identify which of step above argument fails.


问:
A question on ramification groups in Milne
https://www.jmilne.org/math/CourseNotes/ANT301.pdf
Look at page 123 of these notes.

Consider this
$$\sigma \in G_0 \Leftrightarrow |\sigma(\alpha)-\alpha| < \pi ^0, \forall \alpha \in B \Leftrightarrow v_{\pi}(\sigma(\alpha)-\alpha) \geq 1, \forall \alpha \in B \Leftrightarrow \sigma(\alpha) \cong \alpha \mod \mathfrak{p}, \forall \alpha \in B,$$
so $G_0$ is the usual inertia group that I already knew.

Is it wise to do the same for $G_{-1}$? Consider this
$$\sigma \in G_{-1} \Leftrightarrow |\sigma(\alpha)-\alpha| < \pi ^{-1}, \forall \alpha \in B \Leftrightarrow v_{\pi}(\sigma(\alpha)-\alpha) \geq 0, \forall \alpha \in B \Leftrightarrow \sigma(\mathfrak{p})= \mathfrak{p}, \forall \alpha \in B,$$
I can not do the last implication. If I can do it, then $G_{-1}$ is the usual decomposition group that I already knew.

$$v_{\pi}(\sigma(\alpha)-\alpha) \geq 0, \forall \alpha \in B \Leftrightarrow \sigma(\mathfrak{p})= \mathfrak{p}, \forall \alpha \in B$$
Why we have this implication?

I can just show this
$$v_{\pi}(\sigma(\alpha)-\alpha) \geq 0, \forall \alpha \in B \Rightarrow \sigma(\mathfrak{p}) \subseteq \mathfrak{p}, \forall \alpha \in B.$$
If the valuation is preserved under the action of $\sigma$, $v(\sigma(\pi))=v(\pi)$, then I can show this
$$v_{\pi}(\sigma(\alpha)-\alpha) \geq 0, \forall \alpha \in B \Rightarrow \sigma(\mathfrak{p}) = \mathfrak{p}, \forall \alpha \in B.$$

答: I found the answer myself after an hour. For the direct implication observe that $\sigma(\mathfrak{p})$ is another prime, but in local fields there is only one unique prime. The reverse implication is quite easy, both of $\alpha$ and $\sigma(\alpha)$ are integral, so their valuation is nonnegative. In local fields the decomposition group is the whole Galois group.


问:
Question about Lagrange Error Bounds Derivation
My calculus class recently went over Lagrange error bounds, but I have a couple questions about the derivation of the equation. Here's the derivation I've found online (I know it's not too rigorous) alongside with any questions I have about any of the steps.

Let $E(x)=f(x)-P_{n,a}(x)$ for all $x\in[a,b]$ where $P_{n,a}(x)$ denotes the nth degree polynomial approximation centered at a and b is the value of x we're approximating $f(x)$ at.

By the properties of the taylor expaonsion, $E^{(k)}(a)=f^{(k)}(a)-P^{(k)}_{n,a}(a)=0$ for all $k \in \mathbb{Z}, k \leq n$.

Does this relationship come from differentiating the definition of the error bound k times and then using  x=a since the derivatives at the point must be equal? I think that's where it comes from, but I'm not entirely sure.

Since the $(n+1)$th derivative of a polynomial of degree n is $0$,
$E^{(n+1)}(x)=f^{(n+1)}(x)-P^{(n+1)}_{n,a}=f^{(n+1)}(x)$. From this it follows that $|E^{(n+1)}(x)|=|f^{(n+1)}(x)|$. Assuming that $f^{(n+1)}(x)$ is continuous on $[a,b]$, there must exist some M such that $|E^{(n+1)}(x)|=|f^{(n+1)}(x)|\leq M$.

I know the fact that M has to exist comes from the Boundedness theorem, and intutively $f$ can't "go off to infinity" if it's continuous, but is there a more rigorous way to state this? Should I wait until I learn some analysis before trying to prove it? Most of the proofs I've seen use theorems from real analysis which I haven't learned yet.

The next step is to antidifferentiate both sides of the inequality to get
$\int_{ }^{ }\left|E^{\left(n+1\right)}\left(x\right)\right|dx\le\int_{ }^{ }Mdx$ and because $\int_{ }^{ }\left|f\left(x\right)\right|dx\ge\left|\int_{ }^{ }f\left(x\right)dx\right|$, this means $\left|\int_{ }^{ }E^{\left(n\right)}\left(x\right)dx\right|\le Mx+C$ for some constant C.

I have 2 questions about this step. The first is, how do we know that antidifferentiating on both sides of the inequality preserve the inequality? My first thought was that if $f(x)\leq g(x)$ then $g(x)-f(x)$ is strictly positive, but I don't think that this means the antiderivative of $g(x)-f(x)$ is strictly positive (the example I thought about was $x^2$ and $\frac{1}{3}x^3+C$). So how would we show that antidifferentiating preserves the inequality?

The second question I had on this step was about the absolute value inequality. Intuitively $\left|\int_{ }^{ }f\left(x\right)dx\right|\le\int_{ }^{ }\left|f\left(x\right)dx\right|$ makes sense if thinking about the "area under the curve" idea, but is there a more formal way to say this? Is extending the area under the curve idea from definite integrals to indefinite ones even valid?

Finally, the last step in most of the derivations is using $|E^{n}(a)|\leq0$ and then saying $0=Mx+C \Rightarrow c=-Ma$. The rest of the steps are just repeating this process until you get to $|E(x)|\leq \frac{(M)(x-a)^{n+1}}{(n+1)!}$.

I have one question about this, any value of C greater than $-Ma$ would make the inequality $|E^{n}(x)|\leq Mx+c$, so how do we choose C to be equal to $-Ma$ specifically? Some sites say that we need to find the minimal value of C, which is -Ma, but how can we be sure that this minimal value will satisfy the inequality for other values of x besides a?

Those are all the questions I had about Lagrange Error Bounds. I hope the answers to these questions aren't too obvious/trivial (I'm not the greatest at mathematics, so they might be dumb questions I'm not sure).
Any clarification would really help.
Thanks for reading

答:
You are just differentiating the definition of $E$ $k$ times. The important thing is that $P_{n,a}(a),P_{n,a}'(a),\dots,P_{n,a}^{(n)}(a)$ are the same as $f(a),f'(a),\dots,f^{(n)}(a)$. Achieving this property was why we constructed $P_{n,a}$ the way we did in the first place.

The fact that a continuous function on a closed interval is bounded is a relatively basic theorem in real analysis, called the extreme value theorem in most books I have seen. If you don't know it then you can just take it for granted.

This step of doing an indefinite integral is, in my opinion, very bad style. It should really be done with definite integrals all the way through. The point is that you know $E(a)=0,E'(a)=0,\dots,E^{(n)}(a)=0$, so you can repeatedly apply the fundamental theorem of calculus to get $E(x)=\int_a^x E'(y_1) dy_1 = \int_a^x \int_a^{y_1} E''(y_2) dy_2 dy_1 = \dots$, continuing the pattern until you've gotten down to $E^{(n+1)}(y_{n+1})$ as the integrand. Then you take absolute values on the outside, bring absolute values inside (as you mentioned), apply $|E^{(n+1)}| \leq M$, and then you can just perform all $n+1$ integrals explicitly. (Using integrals is not the only way to do it, by the way; you can find this derivation written up using the mean value theorem, for instance.) Just as a small example to show the pattern, say $n=2$, then

$$E(x)=\int_a^x E'(y_1) dy = \int_a^x \int_a^{y_1} E''(y_2) dy_2 dy_1. \\
|E(x)| \leq \int_a^x \int_a^{y_1} |E''(y_2)| dy_2 dy_1 \leq \int_a^x \int_a^{y_1} M dy_2 dy_1 \\
= \int_a^x M(y_1-a) dy_1 = \frac{1}{2} M (x-a)^2.$$

I think your other questions are all technicalities that arise from trying to force indefinite integrals where they do not belong. Using definite integrals, you don't have to think about these things.