问:
Finding Common Eigenvalues and Eigenvectors via Commutative Algebra?
I am interested in the following question:

Question: Let $H(\lambda)$ be a Hermitian matrix whose entries are polynomials of a parameter $\lambda\in\mathbb C$. Consider all pairs of $(\lambda, \lambda')\in \mathbb C^2$ that satisfies a polynomial relation $F(\lambda, \lambda')=0$. For what $(\lambda, \lambda')$ among these pairs, the matrices $H(\lambda)$ and $H(\lambda')$ has a common eigenvalue and a corresponding common eigenvectors (up to constant multiples)?

I want to know an algorithm to answer this question.

Motivation: I want to solve a certain type of PDE on a region $\Omega$, which is a disjoint union of $\Omega_1, \Omega_2$. The most general solution of PDE in each subregion is known. To construct the solution on the whole $\Omega$, I have to impose the continuity on the boundary of $\Omega_1$ and $\Omega_2$. This matter reduces to the above problem!

The most brute force approach to this problem would be listing all eigenvalues and eigenvectors, and finding the common one. However, since the matrix size is large, the characteristic polynomial is of large degree. Then, one cannot explicitly find the roots of the polynomial!

I suspect that there is a more algebraic approach (using commutative algebra, Grobner basis, etc) to this problem. Although I have bare knowledge of commutative algebra, I know that the existence of common roots of two polynomials can be easily determined by computing the resultants. Hence, if one only cares about the eigenvalues, then the resultant would be an answer. But what about eigenvectors also?

Remark: I have a Mathematica code for the explicit form of $H(\lambda)$ and $F(\lambda, \lambda')$.

答:
One idea is as follows: let $n$ denote the size of $H(\lambda),H(\lambda')$. Introduce a variable $x$, which will stand for the common eigenvalue of $H(\lambda)$ and $H(\lambda')$. The matrices $H(\lambda), H(\lambda')$ have common eigenvalue $x$ with some associated common eigenvector $v$ if and only if

$$
\operatorname{rank} \pmatrix{H(\lambda) - xI\\ H(\lambda') - xI} \leq n-1.
$$

Note that a matrix $M$ (with at least $n$ rows and columns) will have rank at most $n-1$ if and only if all of its submatrices of size $n \times n$ have determinant zero. Thus, the condition above can be expressed by setting all $\binom{2n}{n}$ such determinants equal to zero.

If we introduce the additional condition $F(\lambda,\lambda') = 0$, then we can characterize the desired set of values $(x,\lambda,\lambda')$ as the solution to a system of polynomial equations. From there, one could apply some standard approach.


问:
Showing $\left(\frac{z+i}{z-i}\right)^n = -1$ implies $z$ is real
I have shown the following identity: $$ \frac{1+e^{i \theta}}{1-e^{i \theta}} = \frac{1}{2}\cot\left(\frac{\theta}{2}\right) $$ And I now need to use this to show that the the equation: $$ \left(\frac{z+i}{z-i}\right)^n = -1$$ has $\Im(z) = 0$.

I'm not sure where to get started on this, I have found a way to do it by conjugating the 2nd equation on both sides and showing that z is equal to its own conjugate, however this does not use the original identity, and I'd really like to see how it follows from that.

Any help would be much appreciated.

答:
Something is wrong with your identity  as  $$\frac{1+e^{i2t}}{1-e^{2it}}=\frac{1+\cos2t+i\sin2t}{1-\cos2t-i\sin2t}=\frac{2\cos^2t+i2\sin t\cos t}{2\sin^2t-i2\sin t\cos t}$$

$$=\frac{2\cos t(\cos t+i\sin t)}{-2i\sin t(i\sin t+\cos t)}=i\cot t$$

Now using De Moivre's formula , $$\frac{z+i}{z-i}=(-1)^{\frac1n}=e^{\frac{(2r+1)i\pi}n}$$ where $0\le r<n$

Applying componendo and dividendo, $$\frac z{-i}=\frac{e^{\frac{(2r+1)i\pi}n}+1}{e^{\frac{(2r+1)i\pi}n}-1}=i \cot \frac{(2r+1)\pi}{2n}$$

$$\implies z= \cot \frac{(2r+1)\pi}{2n}$$

ContentLicense: CC BY-SA 3.0


问:
$f(x)=\log|2\sin(\frac{x}{2})| \in L^2(-\pi,\pi)$
I need to prove that $f(x)=\log|2\sin(\frac{x}{2})| \in L^2(-\pi,\pi)$. I tried to integrate $f^2(x)$ but it seems too difficult...

Thank you for any help!

答:
Hint:

This may help you deal with a simpler integral. Notice that
$$\log|2\sin(x/2)|=\log\Big|\frac{\sin(x/2)}{x/2}x\Big|=\log\Big|\frac{\sin(x/2)}{x/2}\Big| + \log|x|$$

Now, the function $f_1:x\mapsto\frac{\sin(x/2)}{x/2}$ is away from $0$ in the $[-\pi,\pi]$ so the $\log\circ f_1$ is nice and square integrable.

The problem is now reduced to see whether $\log|x|$ is square integrable in $[0,\pi]$ (this function is even), for your initial integral is the sum of two functions and so, the square integrability of both implies the square integrability of the sum.

For $\int^\pi_0\log^2|x|\,dx$, you may try using integration by parts.

ContentLicense: CC BY-SA 4.0


问:
If $\cos A = \sin B$ , does $A+B = 90$ degrees?

If $\cos A = \sin B$ , does $A+B = 90$ degrees?

My school teacher says that this works but I was unable to understand how. I have been unable to find a proof for this. How does this work?

Edit I should add that this is for an acute case

答:
But this isn't the only solution.
$$\cos A=\sin(A+pi/2)=\sin B \implies B=A+\pi/2 +2k\pi$$
for $k=0$, $B=A+\pi/2  \iff B-A= \pi/2$.


问:
Why does this Laurent Series Vanish?
Let $\Phi_{l}(z;q)$ denote the Lerch Transcendent and $Li_{l}(z)$ denote the Polylogarithm function. For $l \in \mathbb{Z}_{\geq 0}$, we have the following Laurent expansion
\begin{align}
\Phi_{-l}(z;0) + (-1)^{l} Li_{-l}(1/z) = \sum_{k \geq 0} k^{l} z^{k} + \sum_{k \leq -1} k^{l} z^{k}.
\end{align}
For $l \in \mathbb{Z}_{\geq 0}$, the left side will always be a sum of two rational functions which cancel. For example, $l = 0$ and $l = 1$ gives
\begin{align}
\frac{1}{1-z} + \frac{1}{(1-1/z)z} = 0 \quad \text{and} \quad \frac{z}{1-z} - \frac{1}{(1-1/z)^{2} z} = 0.
\end{align}
Looking at it another way, I wonder if there is some simple reason why the right side is zero, which can be written (as the meaningless sum)
$$ \sum_{k \in \mathbb{Z}} k^{l} z^{k} = 0 \quad \text{for}\quad l \in \mathbb{Z}_{\geq 0}.$$ Is there some way to interpret this expression and show that its value is zero?

(Compare this to the meaningless expression $\tfrac{1}{12} + \sum_{n \geq 1} n = 0$.)

答:
As several people have already points out, the actual sum $\sum_{n=-\infty}^{\infty} n^k z^n$ is nowhere convergent. However, there is an interesting phenomenon here.

Let $M$ be the group of functions $\phi: \mathbb{Z}^k \to \mathbb{Q}$ where $\mathbb{Z}^k$ can be partitioned into finitely many lattice polytopes, such that $\phi$ is polynomial on each piece.

Theorem (Lawrence): There is a unique linear map $h: M \to \mathbb{Q}(z_1, \ldots, z_k)$ such that, if $\sum \phi(a_1, \ldots, a_k) z_1^{a_1} \cdots z_k^{a_k}$ converges anywhere, then $h(\phi)$ is equal to this sum.

For example, let $\phi : \mathbb{Z} \to \mathbb{Q}$ be $n \mapsto |n|$. Then $\sum \phi(n) z^n$ is nowhere convergent. However, $\phi(n) = \phi_1(n) + \phi_2(n)$, where $\phi_1(n)$ is $n$ if $n > 0$ and $0$ otherwise, and $\phi_2(n)$ is $-n$ if $n<0$ and $0$ otherwise. The sums $\sum \phi_1(n) z^n$ and $\sum \phi_2(n) z^n$ converge to $z/(1-z)^2$ and $z^{-1} /(1-z^{-1})^2$ respectfully, so $h(\phi) = h(\phi_1 + \phi_2) = z/(1-z)^2 + z^{-1} /(1-z^{-1})^2$. The theorem of Lawrence is a precise statement of the fact that one can work with such expressions consistently.

Moreover, define $x_i \phi$ to be the function $(a_1, \ldots, a_n) \mapsto \phi(a_1, \ldots, a_i -1, \ldots, a_n)$. Then one can check that $h(x_i \phi) = x_i \phi$. This formula is obvious whenever the defining sum converges, and it is not too bad to see that it must then hold for all $\phi$.

Once you believe this, there is a nice explanation for your observation. You are interested in $h(n \mapsto n^k)$. We have
$$(1-x_1)^{k+1} h(n \mapsto n^k) = h \left(n \mapsto n^k - k (n-1)^k + \binom{k}{2} (n-2)^k - \cdots \pm (n-k-1)^k \right) = h(0) = 0$$

The middle equality is using the identity that the $k+1$-st difference of a degree $k$ polynomial is $0$. For example, $n^2 - 3 (n-1)^2 + 3 (n-2)^2 - (n-3)^2 =0$.

Now, $(1-x_1)^{k+1}$ is not a zero divisor in $\mathbb{Q}(x_1)$. So we conclude that $h(n \mapsto n^k)=0$, as you observed.

As far as I know, Lawrence's paper is not available online. The reference is Jim Lawrence, Rational-function-valued valuations on polyhedra, in Discrete and computational geometry (New Brunswick, NJ, 1989/1990), 199–208, DIMACS Ser. Discrete Math. Theoret. Comput. Sci. 6, Amer. Math. Soc. (1991). There is also some good discussion of this in Barvinok's lectures.

Title: Why does this Laurent Series Vanish?


问:
Smallest Counter Example Question?
I'm learning about proving using a smallest counter example but I'm unsure about a specific step our professor did in this example:

$\forall n \geq 0 \\ 1 + 2 + ... + n = \frac{(n)(n+1)}{2}$

Base Case: $n = 0$ checks out

By contradiction, we can suppose $\exists x \geq 0$ such that $p(x)$ is False, so
$1 + 2 \ + \ ... + x ≠ \frac{(x)(x+1)}{2}$, and this means that $p(x-1)$ should be True if $p(x)$ is the smallest counter example.

My question is about what we do to prove how $p(x - 1)$ can be in the same form as $p(x)$ such that it creates a contradiction meaning that $p(x)$ is True.

In class, we did

$1 + 2 \ + \ ... + \ x - 1 = \frac{(x - 1)(x - 1 +1)}{2}$ and then decided that we can add $x$ to this to get it to look like what we are trying to prove:

$1 + 2 \ + \ ... + \ x - 1 + x = \frac{(x - 1)(x - 1 +1)}{2} + x$

After simplifying, we can get the right side to look like $\frac{(x)(x +1)}{2}$, meaning that our original claim that $p(x)$ is not True is in fact False. My question is: why are we allowed to do the step where we add $x$ to our step where we have $p(x - 1)$? Aren't we assuming that isn't True? So why are we allowed to use it?

答:
I think your question is essentially, why should this be done as a proof by contradiction?  After all, if we've shown that $\varphi(0)$ is true, and that $\forall{x\in \mathbb{N}}:\varphi(x)\rightarrow \varphi(x+1)$, aren't we done?

Yes, you're done.  Those two facts together imply that $\forall{x\in\mathbb{N}}:\varphi(x)$ (this is the axiom of induction).

But it's logically equivalent to show the contrapositive of the second fact, which is $\forall{x\in\mathbb{N}}:\neg \varphi(x+1)\rightarrow \neg \varphi(x)$, and in some cases this may make for a clearer argument.

Title: Smallest Counter Example Question?


问:
Why is an unbounded subset of $\omega_1$ uncountable?
(This is not a duplicate of the question in the link below. I want some more explanation but the question is made 7 years ago, so I don't expect to be answered if I just make a comment there.)

Let us say a set is an ordinal if it is well-ordered and transitive under the relation $\in$. $\omega_1$ is the first uncountable ordinal. I give the order topology on $\omega_1$. The statement I want to prove is:

For each positive integer $n$, let $K_n$ be a compact subset of $\omega_1$. Assume $K_n$ has uncountably many elements. Then the intersection $K=\bigcap_{n=1}^\infty K_n$ is an uncountable set.

I found a proof for the above proposition. It is in the link:

The proof that $K$ is uncountable(Math stack exchange)

In the answer, azarel says that:

If $\xi<\omega_1$, then there is an ordinal $\gamma$ in the set $K_0 \cap K_1$ such that $\xi<\gamma$.

But I don't see why this is enough. All I can think is that I can choose $\gamma<\omega_1$, so $K_0 \cap K_1$ has at least countably many elements. How to proceed to the uncountability?

答:
Suppose $K_0\cap K_1$ were countable.  Then the ordinal $\xi=\bigcup K_0\cap K_1$ is a countable union of countable sets, so it is countable.  Thus $\xi<\omega_1$, so there exists $\gamma\in K_0\cap K_1$ such that $\xi<\gamma$.  But this is a contradiction, since $\gamma\in K_0\cap K_1$ means $\gamma\subseteq\xi$ so $\gamma\leq \xi$.

The point here is that $\omega_1$ has uncountable cofinality: you can't "reach the end of it" with a countable set, since the supremum of a countable set of countable ordinals will still be a countable ordinal and thus strictly below $\omega_1$.  So, any countable subset of $\omega_1$ is bounded in $\omega_1$.


问:
Progress on a conjecture of Burnside...
Given a group $G $, the set of automorphisms of $G $ also forms a group, $\rm {Aut}(G) $,with composition as the operation.

An inner automorphism is one determined by conjugation by some element $g\in G $.  That's we have the automorphism  $i_g $ given by $i_g (h)=ghg^{-1}\,\forall h $.

I learned from a comment by @LeeMosher on this site that Burnside once conjectured that any class-preserving automorphism is inner.  Does anyone know about any progress on this?

Of course the converse is trivial.

(Btw, the reference here is to conjugacy classes.  Of course the class equation gives the sizes of these classes.  For example,  in the case of an abelian group,  the class equation consists in all ones.)

For reference,  this would have been early in the  $20$-th century.

答:
This is true for free groups and for surface groups. In fact, in the stunningly beautiful paper*, Edna Grossman proved the following:

Theorem. If $G$ is a finitely generated, conjugacy separable group with every class preserving automorphism inner, then $\operatorname{Out}(G)$ is residually finite.

I mentioned that surface groups satisfy this property, while they are also conjugacy separable. Hence, their outer automorphism groups are residually finite. However, their outer automorphism groups are of particular interest: they are mapping class groups. So we have the following.

Corollary. Mapping class groups of compact orientable surfaces are residually finite.

*Grossman, Edna K. "On the residual finiteness of certain mapping class groups." J. Lond. Math. Soc. (2) 9 (1974/75): 160-164. (doi)

Title: Progress on a conjecture of Burnside...


问:
Prove that $u^m \equiv -1\pmod p \iff p \equiv 1\pmod 4.$

Let $p$ be a prime number with $p \neq 2.$ Let $q = \frac {p-1} {2}$ and $u$ be an element of order $q$ modulo $p.$ Then show that there exists an integer $m$ such that $u^m \equiv -1 \pmod p$ if and only if $p \equiv 1\pmod 4.$

I tried it but failed to conclude. What I did is as follows. $$u^m \equiv -1\pmod p \implies u^{2m} \equiv 1\pmod p.$$ Hence $q\mid 2m.$ Which is same as saying $(p-1)\mid 4m.$ Now I got stuck. How do I proceed now? Any help will be warmly appreciated.

Thanks for your time.

答:
In one direction, if $p \equiv 1 \pmod 4$, then we may take $m = \frac q 2 = \frac {p - 1}4$. Since $u^{2m} = u^q \equiv 1\pmod p$, we see that $u^m \equiv \pm 1 \pmod p$. But $u^m \equiv 1 \pmod p$ would contradict the assumption that the order of $u$ is $q$, hence we have $u^m \equiv -1\pmod p$.

In the other direction, if there exists $m$ such that $u^m \equiv -1\pmod p$, then we have $(-1)^q \equiv u^{qm} \equiv 1 \pmod p$, which (as $p\neq 2$) implies that $q = \frac {p - 1}2$ is even, hence $p \equiv 1 \pmod 4$.


问:
Where do I go wrong with Presburger “multiplication”?
This is a speculation about the Presburger arithmetics, the Peano axioms with only addition added, vis-à-vis the same axioms with both addition and multiplication added.

In the first case I understand that any proposition is decidable but not in the second.

My problem is: how is it possible that I can sneak the multiplication in - without any other actions than appealing to the reader’s feel for geometrical figures – and seemingly go from one system to the other?

Let us assume we are equipped with the natural numbers, the Peano axioms and one composition rule: addition. This will allow us to count objects, numbers for example. We first discover the one-to-one correspondence between the natural numbers such as n, and the geometrical concept of rectangles with length n and width 1, by associating the number n with a rectangle of length n consisting of n unit area element.

We now consider the number of unit area elements of a rectangle of length n and width m, we can count the unit areas line by line and we can denote the result $n*m$ (assuming $*$ to be an unused symbol). We can now see, by applying a physical or geometrical principle of relativity, that (imagining the rectangle placed on the floor) if we move about 90 degrees in a circle around the rectangle it will look like another rectangle of $m*n$ units of area. Assuming our movement has not changed the rectangle, we have $n*m=m*n$.

We can also make a cut in a $n*m$ rectangle perpendicular to the length dimension, after k units of length ( 0 < k < n) which will give us $n*m = k*m + (n-k)*m$ so that $a*(b+c) = a*b + b*c$.  I haven’t described all the components of multiplication here but they don’t seem too difficult to “discover”.

Observe that prime numbers – numbers that fail to appear as the number of rectangles - can be found easily. If we wonder about 7 for example we can start by observing that any rectangle with length 7 or higher, and width >1 will correspond to a number >7. We therefore have a finite number of rectangles to consider, with length < 7 and width < 7 since $n*m=m*n$ (interpreted as numbers of unit areas for a rectangle), i.e. not too different from  solving the same problem in Peano arithmetic.

答:
In Presberger arithmetic, the only objects that you have to work with are natural numbers. So the only way to work with a rectangle would be to represent it, somehow, as one or more natural numbers. For example, you could represent a rectangle by a pair of numbers $(a,b)$ representing the length and width.  But, regardless of the representation, you will not be able to define a function $f(a,b)$ which gives the area of that rectangle. In the question, the idea was to define $f$ by counting the area "line by line". However, there is no way to do that within the language of Presberger arithmetic. The formula that defines multiplication would not be able to mention $f$ at all, and cannot use the sort of inductive definition that you have in mind, which says $f(a+1,b) = b+f(a,b)$. </p>

<p>If you were working in a stronger system, it would indeed be possible to define multiplication from addition. In set theories like ZFC it is possible to define both addition and multiplication from just the successor function $s(n) = n+1$. But that is because set theory is much more expressive, which provides stronger ways of defining new functions. The language of Presberger arithmetic is simply too impoverished to define multiplication from addition. </p>

Body:
In Presberger arithmetic, the only objects that you have to work with are natural numbers. So the only way to work with a rectangle would be to represent it, somehow, as one or more natural numbers. For example, you could represent a rectangle by a pair of numbers $(a,b)$ representing the length and width.  But, regardless of the representation, you will not be able to define a function $f(a,b)$ which gives the area of that rectangle. In the question, the idea was to define $f$ by counting the area "line by line". However, there is no way to do that within the language of Presberger arithmetic. The formula that defines multiplication would not be able to mention $f$ at all, and cannot use the sort of inductive definition that you have in mind, which says $f(a+1,b) = b+f(a,b)$.

If you were working in a stronger system, it would indeed be possible to define multiplication from addition. In set theories like ZFC it is possible to define both addition and multiplication from just the successor function $s(n) = n+1$. But that is because set theory is much more expressive, which provides stronger ways of defining new functions. The language of Presberger arithmetic is simply too impoverished to define multiplication from addition.


问:
Space of probability density function over compact set is compact or not
Suppose $\Omega\subset\mathbb{R}^n$ is compact. Is the set of probability density functions over $\Omega$ compact with respect to weak convergence or $L^1(\Omega)$ convergence? Thank you!

答:
In order for the question to make sense, we must assume that $L^1(\Omega )\neq \{0\}$, which is the same as saying that the Lebesgue
measure of $\Omega $ is nonzero.  Otherwise there are no probability densities on $\Omega $,  whatsoever!

The set $\mathcal D$ of probability densities on $\Omega $ (which may be seen as probability measures that are absolutely
continuous with respect to Lebesgue measure) is,  by definition, the  set of all measurable functions
$$
  \varphi :\Omega \to {\mathbb R}
  $$
which are nonnegative a.e., and such that $\displaystyle \int_\Omega \varphi (x)\, dx=1$.

This set can be naturaly identified with a subset of $L^1(\Omega )$, which in turn has quite a few natural topologies.
Besides the norm topology, it may be equipped with the weak topology
$\sigma (L^1(\Omega ),L^\infty (\Omega ))$,
but there are still other weak
topologies of interest, namely
$\sigma (L^1(\Omega ),A)$,
where $A$ can be taken
to be various  natural subspaces of $L^\infty (\Omega )$, such as the set $C(\Omega )$  of all continuous functions on $\Omega $.

Since the weakest of the topologies so far mentioned is $\sigma (L^1(\Omega ),C(\Omega ))$, should we prove that $\mathcal D$ is not
compact relative to this topology, it will follow that $\mathcal D$ is not compact with the other topologies either.

This said,  let us prove that  $\mathcal D$ is not compact relative to  $\sigma (L^1(\Omega ),C(\Omega ))$.

With this topology, $L^1(\Omega )$ may be seen as a topological subspace of the dual space $C^*(\Omega )$, aka the space  $M(\Omega )$  of all
finite, signed,
Borel measures on $\Omega $, equipped with the weak$^*$-topology.

We will in fact prove that $\mathcal D$ is not closed in
$M(\Omega )$, from where it will follow that $\mathcal D$ is not compact.

Denoting by $\lambda $ the restriction of Lebesgue's measure to $\Omega $, let $x_0$ be any point in the support of $\lambda $, so  any
neigborhood of $x_0$ has nonzero measure.

For each positive integer $n$, let
$$
  V_n=\Omega \cap B_{1/n}(x_0),
  $$
and consider the normalized characteristic function
$$
  \varphi _n = \frac1{\lambda (V_n)}\mathbb 1_{V_n}.
  $$

One can now prove that
$$
  \int_\Omega f(x)\varphi _n(x)\, dx \to  f(x_0), \quad \forall f\in  C(\Omega ),
  $$
as $n\to \infty $,  which means that $\varphi _n$ converges to the Dirac measure $\delta _{x_0}$ in the weak$^*$-topology.  Since $\delta _{x_0}$ is not in
$\mathcal D$, we deduce that $\mathcal D$ is not closed, and hence also not compact.


问:
Q: Prove $\mu^*$ is an outer measure
Let $f : X\to [0,+\infty]$ a positive real function. For all $A\subset X$, we define $\mu^*$ as
$$
\mu^*(A) = \sup\left\{\sum_{x \in B}{f(x) : B \subset A}\right\}
$$

How can I show that $\mu ^*$ is an outer measure?

答:
It is clear that $\mu^*(\emptyset)=0$ and that $A\subset B$ implies $\mu^*(A)\leq \mu^*(B)$. The interesting part is showing that if $B_1,B_2,\dots \subset X$, then $\mu^*\left(\bigcup_j B_j\right)\leq \sum_j \mu^*(B_j)$.

To this end, note that this equation is given by

$\sup \left\{\sum_{x\in B} f(x): B\subset \bigcup_j B_j \right\} \leq \sum_j \sup\left\{\sum_{x\in C_j}f(x) : C_j \subset B_j \right\}$.

We can rewrite the sums on the left hand side into

$\sum_{x\in B} f(x) = \sum_j \sum_{x\in B\cap B_j} f(x)$.

So you can see that this is an expression of the type that we take the supremum over on the right hand side with $C_j = B\cap B_j$. This shows that the set that we are taking the supremum over on the left hand side is contained in the one we are taking the supremum over on the right hand side. Therefore $\sum_j \mu^*(B_j)$ must be greater or equal than $\mu^*\left(\bigcup_j B_j\right)$.

Title: Q: Prove $\mu^*$ is an outer measure


问:
Why are combinations used in the expansion of $(a + b)^4$?
The difference between Permutation and Combination is that in Permutation order matters, Combination order doesn't matter.

So in $ABC, ACB$, the permutation is $2$, and combination is $1$.

But, to expand $(a+b)^4$, the $(a^3)b$ term's coefficient is $4\choose {1}$$  = 4$ ways of getting three $a$'s and $1$ $b$ multiplied together,

namely:

$aaab,  aaba,  abaa,   baaa$. But that's the same combination, shouldn't it (combination) equal to 1? Why 4?

$4\choose 1$$ = \frac{4!}{(1!) \cdot (3!)} = 4$

That's permutation isn't it?

I understand that adding the $(a^3)b$s there are $4$ altogether, but why that's combination?

I'm really confused.

答: Think about the sequence length $n$ of 0s and 1s. You have a total of $(1+1)^n = \sum_{k=0}^{n}\binom{n}{k}$, so $\binom{n}{k}$ is the number of ways to select $k$ different positions in the sequence. All these positions are different (indices),  but the order of choice is not important. So $\binom{4}{1}$ is the number of ways to select a position for $a: 1, 2, 3, 4$, which are different.


问:
Which group is this subgroup isomorphic to?
I am working with SageMath on graphs, in particular on this graph:

I have found that its automorphism group is generated by the elements $(1,3,5)$ and $(2,4,6)$: how do I find to which group is this group isomorphic to? I ask both computationally and mathematically.

答:
The permutations $(135)$ and $(246)$ commute, so this automorphism group is just the direct product of the cyclic groups they generate, i.e. is isomorphic to $(\mathbb{Z}/3\mathbb{Z}) \times (\mathbb{Z}/3\mathbb{Z})$.

I don't know a computational way to recognize the isomorphism class of a group from realizing it as a subgroup of $S_n$ (that might even already be an undecidable problem, not sure). It is likely GAP or some such software can attack this problem, though.


答:
There are e.g. $5C2=10$ ways to select $2$ out of $5$:

$XXOOO$

$XOXOO$

$XOOXO$

$\cdots$

$OOOXX$

Here the order of selecting does not matter and we are dealing with combinations.

There are $5P2=20$ ways to select $2$ out of $5$ if order matters:

$X_1X_2OOO$

$X_2X_1OOO$

$X_1OX_2OO$

$X_2OX_1OO$

$\cdots$

$OOOX_1X_2$

$OOOX_2X_1$

Here the order of selecting matters and we are dealing with permutations.

Title: Why are combinations used in the expansion of $(a + b)^4$?

Body:
The difference between Permutation and Combination is that in Permutation order matters, Combination order doesn't matter.

So in $ABC, ACB$, the permutation is $2$, and combination is $1$.

But, to expand $(a+b)^4$, the $(a^3)b$ term's coefficient is $4\choose {1}$$  = 4$ ways of getting three $a$'s and $1$ $b$ multiplied together,

namely:

$aaab,  aaba,  abaa,   baaa$. But that's the same combination, shouldn't it (combination) equal to 1? Why 4?

$4\choose 1$$ = \frac{4!}{(1!) \cdot (3!)} = 4$

That's permutation isn't it?

I understand that adding the $(a^3)b$s there are $4$ altogether, but why that's combination?

I'm really confused.


答:
Since they are disjoint three-cycles, they commute and have order three; there are no other restrictions on the group; so the group is isomorphic to the group given by the presentation

$$\langle a,b\mid a^3, b^3, ab=ba\rangle,$$

which is $\Bbb Z_3\times\Bbb Z_3$.

Computationally, you could use GAP.

If the automorphism group is, instead, generated by $(135)(246)$, then it is cyclic and the order of the group is that of the generating element, which is three; hence it is isomorphic to $\Bbb Z_3$.

Title: Which group is this subgroup isomorphic to?

Body:
I am working with SageMath on graphs, in particular on this graph:

I have found that its automorphism group is generated by the elements $(1,3,5)$ and $(2,4,6)$: how do I find to which group is this group isomorphic to? I ask both computationally and mathematically.


问:
Calculating the exact value of $\sum_{n=0}^{\infty} (-1)^n (\frac{31}{32})^{2n+1} (\frac{1}{2n+1})$
$\sum_{n=0}^{\infty} (-1)^n (\frac{31}{32})^{2n+1} (\frac{1}{2n+1})$

Hey math stack exchange! I've been trying to figure out how to calculate the exact summation of these series for awhile now. I looked at this post: How to find the sum of an alternating series? to get some clues, but I'm still a bit lost because my formula has the term $(\frac{1}{2n+1})$ in it.

Would appreciate some help! Thanks!

答:
We first look at the power series $f(x) = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1}x^{2n + 1}$. It is clear that the question is asking for the value of $f(\frac{31}{32})$.

We now take derivative of $f(x)$ and get $$f'(x) = \sum_{n = 0}^\infty (-1)^nx^{2n} = \frac 1{1 + x^2}.$$ Special attention should be paid to the convergence problem, but I'll omit details here.

Now we integrate and get $f(x) = \arctan(x)$, noting that $f(0) = 0$.


问:
Can a sum and square sum of a multiset of size n be equal to the square sum and sum of [n]
I have a multiset A of $[n]$, $|A| = n$ and $A\not = [n]$ (i.e., A has at least one element repited) whose sum is equal $\frac{n(n+1)}{2}$, i.e.,

$$\sum_{a\in A} a = \frac{n(n+1)}{2}$$

Can it's square sum $\sum_{a\in A}a^2$ be equal to $\frac{n(n+1)(2n+1)}{6}$?

答:
Searching for numbers that can be written in many ways as a sum of a certain number of squares produces results quickly. I imagine there are very many qualifying results.

Using the result for $129$ noted in Determining the number of ways a number can be written as sum of three squares,
$$10^2+5^2+2^2=8^2+8^2+1^2=129$$

and noting $10+5+2=8+8+1$, it's easy to see that your property holds for the multiset  $\{1,1,3,4,6,7,8,8,8,9,11\}$ (by substituting $\{1,8,8\}$ for $\{2,5,10\}$ in $[11]$).

I "went to eleven" here just so that $n$ was included in $A$.

A fairly simple program search for numbers that can be represented as a sum of three positive squares multiple ways with common sums finds the smallest solution that has at least one of the solutions being composed of distinct numbers as $62 = 1^2+ 5^2 +6^2 = 2^2 + 3^2 + 7^2$ so the least $n$ is probably $7$.

results to $200$:

```<code>sq total 62 sum 12 (1, 5, 6) (2, 3, 7)
sq total 89 sum 15 (2, 6, 7) (3, 4, 8)
sq total 101 sum 15 (1, 6, 8) (2, 4, 9)
sq total 118 sum 16 (1, 6, 9) (3, 3, 10)
sq total 122 sum 18 (3, 7, 8) (4, 5, 9)
sq total 129 sum 17 (1, 8, 8) (2, 5, 10)
sq total 134 sum 18 (2, 7, 9) (3, 5, 10)
sq total 146 sum 18 (1, 8, 9) (3, 4, 11)
sq total 150 sum 18 (1, 7, 10) (2, 5, 11)
sq total 153 sum 19 (2, 7, 10) (4, 4, 11)
sq total 161 sum 21 (4, 8, 9) (5, 6, 10)
sq total 166 sum 20 (2, 9, 9) (3, 6, 11)
sq total 173 sum 21 (3, 8, 10) (4, 6, 11)
sq total 185 sum 21 (2, 9, 10) (4, 5, 12)
sq total 189 sum 21 (2, 8, 11) (3, 6, 12)
sq total 194 sum 20 (1, 7, 12) (3, 4, 13)
sq total 194 sum 22 (3, 8, 11) (5, 5, 12)
</code>```

OEIS A282241 is (I think) a subset of these values.

This property becomes common for larger numbers; $6515$ of the numbers up to $10000$ can be expressed as a sum of three squares in two or more ways where the square roots also have a common sum.

Python code (improved to get to higher numbers):

```<code>lim = int(input('Search limit: '))
rlim = int(lim**0.5)
sqval = {i:i*i for i in range(rlim+1)}

# sieve
allz = []
for a in range(1,rlim+1):
    for b in range(a,rlim+1):
        ab = sqval[a]+sqval[b]
        if ab > lim:
            break
        for c in range(b, rlim+1):
            abc = ab+sqval[c]
            if abc > lim:
                break
            allz.append( (abc, a+b+c, a,b,c) )
# print(len(allz))
allz.sort()
allz.append( (0,0,0,0,0) ) # force action on last totals

pM = 0
pt = 0
mz = []
qct = 0
Mql = 0
for M,t,a,b,c in allz:
    if M != pM or t != pt:
        # report previous block if sutiable
        if len(mz) > 1:
            if any(len(set(mzi))==3 for mzi in mz):
                # report suitable results
                Mql = 1
                print("sq total", pM, "sum", pt, *mz)
        mz = []
        pt = t
        if M != pM:
            qct += Mql
            pM = M
            Mql = 0
    mz.append( (a,b,c) )

print (qct, "numbers in range have a suitable representation")
</code>```

Body:
Searching for numbers that can be written in many ways as a sum of a certain number of squares produces results quickly. I imagine there are very many qualifying results.

Using the result for $129$ noted in Determining the number of ways a number can be written as sum of three squares,
$$10^2+5^2+2^2=8^2+8^2+1^2=129$$

and noting $10+5+2=8+8+1$, it's easy to see that your property holds for the multiset  $\{1,1,3,4,6,7,8,8,8,9,11\}$ (by substituting $\{1,8,8\}$ for $\{2,5,10\}$ in $[11]$).

I "went to eleven" here just so that $n$ was included in $A$.

A fairly simple program search for numbers that can be represented as a sum of three positive squares multiple ways with common sums finds the smallest solution that has at least one of the solutions being composed of distinct numbers as $62 = 1^2+ 5^2 +6^2 = 2^2 + 3^2 + 7^2$ so the least $n$ is probably $7$.

results to $200$:

sq total 62 sum 12 (1, 5, 6) (2, 3, 7)
sq total 89 sum 15 (2, 6, 7) (3, 4, 8)
sq total 101 sum 15 (1, 6, 8) (2, 4, 9)
sq total 118 sum 16 (1, 6, 9) (3, 3, 10)
sq total 122 sum 18 (3, 7, 8) (4, 5, 9)
sq total 129 sum 17 (1, 8, 8) (2, 5, 10)
sq total 134 sum 18 (2, 7, 9) (3, 5, 10)
sq total 146 sum 18 (1, 8, 9) (3, 4, 11)
sq total 150 sum 18 (1, 7, 10) (2, 5, 11)
sq total 153 sum 19 (2, 7, 10) (4, 4, 11)
sq total 161 sum 21 (4, 8, 9) (5, 6, 10)
sq total 166 sum 20 (2, 9, 9) (3, 6, 11)
sq total 173 sum 21 (3, 8, 10) (4, 6, 11)
sq total 185 sum 21 (2, 9, 10) (4, 5, 12)
sq total 189 sum 21 (2, 8, 11) (3, 6, 12)
sq total 194 sum 20 (1, 7, 12) (3, 4, 13)
sq total 194 sum 22 (3, 8, 11) (5, 5, 12)

OEIS A282241 is (I think) a subset of these values.

This property becomes common for larger numbers; $6515$ of the numbers up to $10000$ can be expressed as a sum of three squares in two or more ways where the square roots also have a common sum.

Python code (improved to get to higher numbers):

lim = int(input('Search limit: '))
rlim = int(lim**0.5)
sqval = {i:i*i for i in range(rlim+1)}

# sieve
allz = []
for a in range(1,rlim+1):
    for b in range(a,rlim+1):
        ab = sqval[a]+sqval[b]
        if ab > lim:
            break
        for c in range(b, rlim+1):
            abc = ab+sqval[c]
            if abc > lim:
                break
            allz.append( (abc, a+b+c, a,b,c) )
# print(len(allz))
allz.sort()
allz.append( (0,0,0,0,0) ) # force action on last totals

pM = 0
pt = 0
mz = []
qct = 0
Mql = 0
for M,t,a,b,c in allz:
    if M != pM or t != pt:
        # report previous block if sutiable
        if len(mz) > 1:
            if any(len(set(mzi))==3 for mzi in mz):
                # report suitable results
                Mql = 1
                print("sq total", pM, "sum", pt, *mz)
        mz = []
        pt = t
        if M != pM:
            qct += Mql
            pM = M
            Mql = 0
    mz.append( (a,b,c) )

print (qct, "numbers in range have a suitable representation")


问:
Ordered bases, orientation and the "handedness" of Cartesian coordinates
I'm confused as to how orientations and ordered bases relate to the “handedness” of Cartesian coordinates. On page 237 of Tu's An Introduction to Manifolds he's discussing orientations of a vector space and writes, with reference to the following diagram:

On $\mathbb{R}^{3}$ an orientation is either right-handed (Figure
21.3) or left-handed (Figure 21.4). The right-handed orientation of $\mathbb{R}^{3}$ is the choice of a Cartesian coordinate system such
that if you hold out your right hand with the index finger curling
from the vector $e_{1}$ in the $x$-axis to the vector $e_{2}$ in the
$y$-axis, then your thumb points in the direction of of the vector
$e_{3}$ in the $z$-axis.

On my reading, “the choice of a Cartesian coordinate system” implies that Figure 21.4 should therefore show a left-handed Cartesian coordinate system. But that doesn't make sense as then $e_{2}$ would lie on the $x$-axis and $e_{1}$ would lie on the $y$-axis. On reflection (excuse pun) I'm guessing that an ordered basis defines an orientation on $\mathbb{R}^{3}$, which is then graphically represented by choosing either a right-handed or left-handed Cartesian system. So both Tu's diagrams refer to a right-handed coordinate system. Am I on the right track?

EDIT In other words, how do Tu's diagrams relate to these right and left-handed coordinate systems?

EDIT 2 Part of my confusion stems from this answer by @AndrewD.Hwang here:

It's worth pointing out a final subtlety in $\mathbb{R}^{3}$: If
$\left(\mathbf{e}^{i}\right)_{i=1}^{3}$ denotes the standard basis,
the (algebraic) ordering of the vectors fixes an orientation, but we
must still pick a conventional geometric representation, i.e., must
choose to depict the standard basis as "right-handed" or
"left-handed".

答:
On p. 238 Tu gives a formal definiton of orientation as an equivalence class of orderded bases. In  the sense of this definiton any (finite-dimensional) vector space $V \ne \{0\}$ has exactly two orientations.

Note that $V = \{0\}$ is an exception. Formally it has $\emptyset$ as its only base (if we agree that sums over the empty index set are $0$, then each element of $\{0\}$ has a unique representation as a linear combination of basis vectors) and thus it has only one orientation.

Tu's preceding explanations are purely motivational and may in fact be somewhat confusing. What is he doing?

In dimensions $n = 1, 2, 3$ he considers the standard Cartesian coordinate system in $\mathbb R^n$ and assigns two orientations to it. Be aware that such an orientation is an additional component attributed to the coordinate system, it cannot be found in the coordinate system itself. It occurs in the figures in form of rotation arrows  which are assigned in two variants to the depicted Cartesian coordinate system.

Let us focus to $n = 3$. Tu considers the usual $x$-$y$-$z$-coordinate system with standard unit vectors $e_1$ on the $x$-axis, $e_2$ on the $y$-axis and $e_3$ on the $z$-axis. Drawing the $e_i$ as arrows gives a direction to each of the axes (the positive direction). From the viewer's perspective the graphical representation of Cartesian coordinates in 21.3 and 21.4 shows the $x$-axis pointing forward, the $y$-axis to the right and the $z$-axis upwards. This is an arbitrary convention, you could do it as you want. As an example take the figure in your edit. Anyway, if you decided for a particular graphical representation, you can glue the coordinate system to your right hand or your left hand according to the following convention:

Arrange index finger (I), middle finger (M) and thumb (T) so that they are perpendicular to each other. Then the fingers of your right and of your left hand form two ordered $3$-frames, the right-handed frame $(I_R, M_R, T_R)$ and the left-handed frame $(I_L, M_L, T_L)$. No spatial movement of your hands will transform the right-handed into the left-handed frame - they are mirror images. These two frames are the intuitive prototypes of an orientation of $\mathbb R^3$.

Gluing to the right hand ("right-handed orientation"):  Point with $I_R$ in direction $e_1$, with $M_R$ in direction $e_2$ and with $T_T$ in direction $e_3$.

Gluing to the left hand ("left-handed orientation"):  Point with $I_L$ in direction $e_2$, with $M_L$ in direction $e_1$ and with $T_L$ in direction $e_3$.

This attributes a handedness or orientation to the Cartesian coordinate system. Formally this is described via the two bijections
$$\omega_R = \begin{pmatrix} I_R & M_R & T_R \\ e_1 & e_2 & e_3 \end{pmatrix} \quad \quad\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_2 & e_1 & e_3 \end{pmatrix}$$

In my opinion the graphical representation in your edit is more intuitive than Tu's.

In both cases you use assign $e_1$ to $I$, $e_2$ to $M$ and $e_3$ to $T$ which gives you two variants of a Cartesian coordinate system which are optically distinguishable. Here you have
$$\omega_R = \begin{pmatrix} I_R & M_R & T_R \\ e_1 & e_2 & e_3 \end{pmatrix} \quad \quad\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_1 & e_2 & e_3 \end{pmatrix}$$

However, if you take the graphical representation on the right side as the standard representation of a Cartesian coordinate system (this agrees with Tu, just imagine to look at this coordinate system with your eye facing the arrow tip of $x$), then you will admit that the graphical representation on the left side is non-standard. And in fact, if you use your left hand in the standard representation, then you get
$$\omega_L = \begin{pmatrix} I_L & M_L & T_L \\ e_1 & -e_2 & e_3 \end{pmatrix}$$
Working with equivalence classes of ordered bases, you can easily see that the latter representation of $\omega_L$ coincides with Tu's variant.


问:
finding the number of non-appearing integers in three different sequences
English,Math and Science classes started on the very first day of a month.Mathematics class schedule is 1,3,5,7,9. . . . The schedule for English is 1,4,7,10,13. . . And for Science is 1,5,9,13,17. . .In the next three months how many days will there be no classes of the above subjects?

I first tried to find the number of odd integers between 0 and 181 including the latter and argued that the same odd numbers will appear in the next two pattern sequences but of couse there will be a few more odd numbers.I tried to do some tedious calculation and have ended up with nothing.Then I tried to see the pattern between the digits that are not appearing  but that did not help either. A prod in the right direction would help at this stage,I think.

答: Hint: try looking at the sequences in modulo $12$. You will find that certain values $\mod 12$ will not be in any sequence.</p>


问:
Q: How to smooth out a corner line in a piecewise function of 2 variables?
Let \begin{equation}h(x,y)= \begin{cases} 0, x \leq 0  \\ x(10-5y), x>0 \end{cases}\end{equation} be a piecewise smooth funcion, that is continuous everywhere, and smooth everywhere except at $(0,y)$.

My idea is to smooth out this "corner line" possibly by extracting a small domain around it, a stripe of width $2\delta$, for $\delta$ small but positive, and replace this with a function $g$ such that $f$ and $g$ can be glued together into a new function $H$ that is everywhere smooth.

It seems that all similar questions concern single variable functions, so this one seems trickier to me.

Thank you in advance!

答:
I think the following works:

$$\begin{equation}h(x,y)= \begin{cases} 0,\ x \leq -\delta  \\ \frac{(10-5y)(x+\delta)^2}{4\delta},\ |x|<\delta \\ x(10-5y),\ x\ge\delta \end{cases}\end{equation}$$

I simply treated $y$ as a constant, turning the function into a single-variable piecewise linear function. The two straight lines can be connected by a parabola, giving the result above.


问:
$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$ by substitution
I have to solve this integral $$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$$
It seems that the most convenient way to operate is doing the substitution $ \tan x = t$.
Then after some passages the integral becomes:
$$  \int \frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2}\, dx  $$

Now I'm trying the coefficient A,B,C,D,E and F such as:

$$\frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2} = \frac{A}{t-1}+\frac{2B(t-1)+C}{t^2+1}+ \frac {d}{dx} \frac{D(t-1)^2+E(t-1)+F}{(t-1)(t^2+1)}$$

Perhaps I made mistakes in this last passage because I find some coefficients that don't lead to the result suggested by the book.

答:
$$2(\sin x-\cos x)\cos x =\sin2x-2\cos^2x =\sin2x-\cos2x+1=y\text{(say)}$$

Now $1+\sin4x=2-(\cos2x-\sin2x)^2=2-(y-1)^2=?$

Again, $\dfrac{2-(y-1)^2}y=-y+2+\dfrac1y$

Finally $\dfrac1y=\dfrac1{(\sin x-\cos x)\cos x}=\dfrac{\sec^2x}{\tan x-1}$

Should I leave it here ?


问:
Given a circle with cente $O$, radius $R=6$ and diameter $AB$. On the radius $OB$ we take $C$, so that $OC=2$. Find the length of $EZ$.
Given a circle with cente $O$, radius $R=6$ and diameter $AB$. On the radius $OB$ we take $C$, so that $OC=2$. $EZ$ is a chord of the circle which intersects $AB$ at $C$, so that $\angle ECB=60^o$. Find the length of $EZ$.

After drawing it accurately, I believe that $EZ=10$. I tried to work this out but didn't manage. My thoughts are the following:

$\angle ACZ=\angle ECB=60^o$

$OC=2,CB=4,AO=6$

If $E'$ is Es intidiametric point, then $E'E^2+E'Z^2=EZ^2$. We also have that $AC*CB=EC*CZ$ and $AEBZ$ is inscribed. This is all I could think of. Could you please explain to me how to solve this question?

答:
Connect points $E$ and $Z$ to $O$ and draw a perp from $O$ to chord $EZ$. Say it meets $EZ$ at $H$ then $OH = OC \sin 60^0$.

Now use Pythagoras in $\triangle OEH$ to find $EH$ and that leads to $EZ$.

EDIT: If you want to use intersecting chords theorem,

$ \small EC \cdot CZ = (EH-1) \cdot (HZ+1) = 8 \cdot 4$ (as $\small CH = 1$). Now you know $\small CH = HZ$.


问:
Symmetry of the components of a tensor
If I have to prove that $A_{kh}=g_{ij}\frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^h}$ is symmetric in $k$ and $h$, knowing that $g_{ij}=g_{ji}$, for the term $\frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^h}$ it is allowed to rename $i$ with $j$ and viceversa and so can I write the following?
$$g_{ij}\frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^h}=g_{ij}\frac{\partial x^j}{\partial y^k}\frac{\partial x^i}{\partial y^h}$$
I am not so sure that this is only a matter of renaming the indeces, since for instance if I rename the indeces for a very general (no symmetric) matrix $s_{kj}$ (I rename $k$ with $j$ and viceversa) this becomes $s_{jk}$ but this would mean that this $s$ is  symmetric.

So how can I prove really the symmetry of $A_{kh}$?

答: You can rename indices that are summed over ($\sum_i f(i) = \sum_j f(j)$), so your derivation of $A_{ij}=A_{ji}$ is correct. One is not allowed to rename so-called free indices, whose value is fixed -- doing that would lead to the wrong conclusion that every matrix is symmetric, as you pointed out.

Title: Symmetry of the components of a tensor


问:
In triangle $\triangle ABC$, we have $AB=8cm$, $AC=12cm$ and $BC=15cm$, $I$ the incenter. If $ID//AB$ and $IE//AC$ which is the perimeter of $IDE$?
In triangle $\triangle ABC$, we have $AB=8cm$, $AC=12cm$ and $BC=15cm$, $I$ the incenter. If $ID//AB$ and $IE//AC$ which is the perimeter of $\triangle IDE$?

After drawing the shape accurately I have that the perimeter is $15$.

Since $ID//AB$, then from Thales we have that $\frac{ID}{AB}=\frac{IO}{OA}$ where $O$ is the intersection of $AI$ and $CB$. If it were the barycenter then we could use the property that $AI=2OI$, but it's the incenter. Could you please explain to me how to solve this question?

答:
Hint: We have $\triangle IDE \sim \triangle ABC$ (since all of the sides are parallel), so it suffices to find the similarity ratio. It's a bit hard to do this directly by calculating the side lengths, but it's easier to think about altitudes. If $H$ is the foot from $A$ to $BC$ and $J$ is the foot from $I$ to $DE$, then
$$\frac{\text{perimeter}(\triangle IDE)}{\text{perimeter}(\triangle ABC)}=\frac{IJ}{AH};$$
can you calculate $IJ$ and $AH$?


问:
References for discontinuous Galerkin method with one dimensional case?
Could anyone recommend some references on discontinuous Galerkin method for beginners? It is nice to contain the one-dimensional case. Thanks in advance!</p>

答:
There are a selection of books you could look at:

B. Cockburn, G. E. Karniadakis and C.-W. Shu (eds.), Discontinuous Galerkin methods. Theory, computation and applications, Lecture Notes in Computational Science and Engineering, 11. Springer-Verlag, Berlin, 2000.

P. Lesaint, and P. A. Raviart. "On a finite element method for solving the neutron transport equation." Mathematical aspects of finite elements in partial differential equations 33 (1974): 89–123.

D.A. Di Pietro and A. Ern, Mathematical Aspects of Discontinuous Galerkin Methods. Mathématiques et Applications, Vol. 69, Springer-Verlag, Berlin, 2011.

J.S. Hesthaven and T. Warburton, Nodal Discontinuous Galerkin Methods: Algorithms, Analysis, and Applications. Springer Texts in Applied Mathematics 54. Springer Verlag, New York, 2008.

B. Rivière, Discontinuous Galerkin Methods for Solving Elliptic and Parabolic Equations: Theory and Implementation. SIAM Frontiers in Applied Mathematics, 2008.

I would imagine that they contain the 1D case on the whole - I am in the process of looking at them also.


问:
If $V$ is the standard representation of $S_n$, how to show that $\bigwedge^k V$ is irreducible $(0 \leq k \leq n-1)$?
Edit: As Nate pointed out, the hint was wrong. I added the correct answer (from Fulton & Harris) below.

The hint I get: We have a decomposition of regular representation $R$=$U \oplus V$ where $U$ is the trivial representation. Since we have the expansion

$$
\bigwedge^k(U \oplus V) \cong \bigoplus_{a+b=k}\bigwedge^a U \otimes \bigwedge^b V
$$

It suffices to compute $(\chi_{\wedge^k R},\chi_{\wedge^k R})$, and it should be $2$. Then we can deduce that $\bigwedge^k V$ is irreducible whenever $0 \leq k \leq n-1$.

I managed to compute $(\chi_{\wedge^2 R},\chi_{\wedge^2 R})$ when $n=3$. And  it is indeed $2$. But I don't know how to step on. Is there any way to expand $(\chi_{\wedge^k R},\chi_{\wedge^k R})$ with respect to $U$ and $V$ (some massive computation involved I guess?). I know how to compute $\chi_{\wedge^2 R}(g)$ but how to compute $\chi_{\wedge^k R}(g)$? My guess is there is some much simpler approach for regular representation.

Any hint or solution will be sincerely appreciated!

In case people are confused with terminologies:

Standard representation.

Regular representation.

答:
The inner product $(χ_{∧^kR},χ_{∧^kR})$ can be interpreted as the expectation of $(\text{Tr} \phi(g))^2$, where $\phi$ is the representation of $S_n$ on the exterior product space ${∧^k\mathbb{C}^n}$, and $g$ is a random element of $S_n$ chosen uniformly.

If we define random variables $r_a$ ($a$ runs over the $k$-element subsets of $[1, 2, ... ,n]$) as

$$r_a=\left\{ \begin{array}   & 1  & \text{if the action of $g$ maps the set $a$ to $a$ and preserves the orientation} \\-1 & \text{if the action of $g$ maps the set $a$ to $a$ and reverses the orientation} \\ 0 & \text{otherwise}\end{array}
\right .$$

Then $\text{Tr} \phi (g)$ is just the sum of all the $r_a$s, by the definition of exterior product space.

To compute the expectation of $(\text{Tr} \phi(g))^2$, we expand it as $$\mathbb E(\text{Tr} \phi(g) )^2= \sum_{a\in {n\choose k}} \mathbb Er_a^2+\sum_{a, b\in {n\choose k}\\{|a\cap b|=k-1} } \mathbb Er_ar_b+\sum_{a, b\in {n\choose k}\\{|a\cap b| \leq k-2} } \mathbb Er_ar_b$$

where $n \choose k $ denotes all the $k$-element subsets of $[1,2,...,n]$, and the last two sums are over ordered pairs.

It is straightforward that $\mathbb Er_a^2=\frac{k!(n-k)!}{n!}$, so the first sum evaluates to $1$.

To study the second sum, we consider the action of $g$ on $a \cap b$. For the summand to be nonzero, $g$ must map this set to itself and map the elements $a \text{\\} (a \cap b) $ and $b \text{\\} (a \cap b) $ to themselves respectively.

This happens with probability $\frac{(n-k-1)!(k-1)!}{n!}$; and whenever this happens, the value of $r_ar_b$ is $1$, whether the orientation of $a \cap b$ is reversed by $g$ or not. By counting the number of such ordered pairs $(a,b)$, i.e. $\frac{n!}{(n-k-1)!(k-1)!}$, we find out that the second term also evaluates to $1$.

Every summand in the third term equals $0$. To see this, we may assume that $b_1$ and $b_2$ are two elements in $B\text{\\}A$, and let $\tau$ be the element in $S_n$ that exchanges $b_1$ and $b_2$ and fix everything else.

The expectation $\mathbb E r_ar_b$ can be written as $\frac{1}{n!}\sum _{g\in S_n} r_a(g)r_b(g)$, just to emphasize that $r_a$ and $r_b$ depends on $g$.

But $\sum _{g\in S_n} r_a(g)r_b(g)=\sum _{g\in S_n} r_a(\tau g)r_b(\tau g)=-\sum _{g\in S_n} r_a(g)r_b(g)$, as $\tau$ reverses the orientation on $b$ but has no effect on $a$ if either of the summands are nonzero.

So the third sum evaluates to $0$ and we are done.


问:
Show that $A\in\mathscr{B}(\mathscr{H}) and \|A\|=\sup_{n}\|A_{n}\|.$
The following is the exercise 2.13 from Barbara MacCluer's Elementary Functional Analysis.

Suppose $A_{n}\in\mathscr{B}(\mathscr{H}_{n})$ for $n=1,2,3,\dots$, where each $\mathscr{H}_{n}$ is a Hilbert space. Assume further that $\sup_{n}\|A_{n}\|<\infty$. Define $A$ on $\mathscr{H}\equiv\sum\oplus\mathscr{H}_{n}$ by $$A(h) =A(h_{1},h_{2},\dots)=(Ah_{1},Ah_{2},\dots).$$
Show that $A\in\mathscr{B}(\mathscr{H})$ and $\|A\|=\sup_{n}\|A_{n}\|$. We call $A$ the direct sum of the operators $\{A_{n}\}$ and denote it $\sum\oplus A_{n}$.

I think by completeness of Hilbert space, it should simply follow as claimed, but I'm having trouble deductively showing the result.

答: Hint: estimate the norm of $Ah$ using $\| (A_1 h_1 , A_2 h_2 ,...) \|^2 = \sum_j \| A_j h_j \|^2$ and boundedness of each of $A_j$.


问:
Identify subgroup of a semidirect product in GAP
I have created an external semidirect product of two groups $H$ and $K$ in GAP. Now, I want to identify $H$ as a subgroup of this semidirect product, but I am unable to identify $H$. How can I identify $H$ in this semidirect product in GAP? Please help!!! Thank you. Below is the code that I used to construct the semidirect product.

```<code>gap> H:=DihedralGroup(12);
<pc group of size 12 with 3 generators>
gap> aut:=AutomorphismGroup(H);
<group of size 12 with 3 generators>
gap> inn:=InnerAutomorphismsAutomorphismGroup(aut);
<group with 3 generators>
gap> semi:=SemidirectProduct(inn,H);
<pc group with 5 generators>
</code>```

Now, I want to identify inn (InnerAutomorphism group of H) as the subgroup of this semidirect product.

答:
The group constructed with  SemidirectProduct will have two monomorphisms from the constituents. Embedding(semi,1) is the one from inn, Embedding(semi,2) the one from H. Thus its Image will be the group you want, and you can also use it to find the elements corresponding to particular inner automorphisms:

```<code>gap> ea:=Embedding(semi,1);;
gap> Image(ea);
Group([ f1, f2, f2^2 ])
gap> Image(ea,InnerAutomorphism(H,H.1));
f1
</code>```

Body:
The group constructed with  SemidirectProduct will have two monomorphisms from the constituents. Embedding(semi,1) is the one from inn, Embedding(semi,2) the one from H. Thus its Image will be the group you want, and you can also use it to find the elements corresponding to particular inner automorphisms:

gap> ea:=Embedding(semi,1);;
gap> Image(ea);
Group([ f1, f2, f2^2 ])
gap> Image(ea,InnerAutomorphism(H,H.1));
f1


问:
Two estimations in analytic number theory
I want to show the following two estimation:
$$
\sum_{p|n}\frac{\log p}{p}=O(\log\log n)
$$
And
$$
\sum_{p|n}\frac1p=O(\log\log\log n)
$$

First assume $$n=p_1^{\alpha_1}...p_s^{\alpha_s}$$ then the first can be rewritten as:
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}
$$

I think i should bulid some connections between s and n, and could i do the following?
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}=O(\sum_{i=1}^s\frac{1}{i})
$$

If this is true, why?

答:
Take the first as an example: from what we have done, it suffices to estimate:
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}
$$
Classify $p_1,...,p_s$as follows：
$p_1,...,p_{r_0}\leq\log n$，$p_{r_0+1},...,p_s>\log n$, so:
$$
\sum_{i=1}^{r_0}\frac{\log p_i}{p_i}\leq\sum_{p\leq\log n}\frac{\log p}{p}\leq\log\log n+O(1)
$$

We then estimate how large the $r-r_0$ is :
$$
(\log n)^{r-r_0}<p_{r_0+1}...p_s\leq n\to r-r_0\leq\frac{\log n}{\log\log n}
$$
So,
$$
\sum_{i=r_0+1}^s\frac{\log p_i}{p_i}\leq\frac{\log\log n}{\log n}\times\frac{\log n}{\log\log n}=1
$$
I.e
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}=O(\log\log n)
$$


问:
Why are Fibonacci numbers bad for Euclid's Algorithm and how to derive this upper bound on number of steps needed in general?
I want to ask two things.

The first is why are consecutive Fibonacci numbers the worst case for Euclid's algorithm? I keep seeing people say it in passing and I understand that it's really bad, but how do we know it's the worst?

The second question is about an upper bound on the number of applications of the division algorithm needed to compute the GCD of two distinct positive integers. I am given a hint that it will be of the form $c \log b + d$ when $c, d$ are constants and $b$ is the smaller of the two numbers you start with.

I have seen someone mention that it is $\frac{1}{2} \log b + 1$ but again, I don't know how to derive this and would very much appreciate a hint or push in the right direction.

Thank you in advance!

答:
You can prove this using induction on the number of gcd() calls.

To prove: If a pair (a,b) takes n steps to compute the gcd, then

a >= F_(k+2)

b >= F_(k+1)
With one assumption a >= b

Base case n = 1, the lemma is true.

Consider the lemma to be true for n = k - 1, k > 1

Let a pair (a,b) be such that it takes k steps to compute the gcd.
Hence (b, a mod b) will take k - 1 steps.
Thus b >= F_(k+1) ... by hypothesis
And a mod b >= F_(k)

Now, a = b×q + a mod b >= b + a mod b >= F_(k+1) + F_(k) >= F(k+2)

Hence we have proven that consequentive fibonacci numbers are the worst case. More specifically, they are the least of the numbers to take a certain arbitrary number of steps.


问:
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be continuously differentiable. $\forall x,y\in \mathbb{R}^n$, $d\big(f(x),f(y)\big)\geq a\cdot d(x,y)$
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be continuously differentiable. If there exists a real number $a>0$, such that $\forall x,y\in \mathbb{R}^n$, $d\big(f(x),f(y)\big)\geq a\cdot d(x,y)$, please show that $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is a $C^1$-diffeomorphism.

It's clear that $f$ is injective, and the differential $df(x)$ is invertible for any $x\in \mathbb{R}$, since $f(x+v)=f(x)+df(x)(v)+o(v)$, as $v\rightarrow 0$. Then by the inverse function theorem, we have $f$ is a locally diffeomorphism. If $f$ is surjective, then we can show that $f$ is a $C^1$-diffeomorphism.

How can I show that $f$ is surjective?

答:
I hope that you are familiar with some metric topology. The strategy is to prove that the image of $f$ is both open and closed as a subset of $\mathbb{R}^n$. Since $\mathbb{R}^n$ is connected, it is the only non-empty subset of itself having these two properties.

We first show that $U = f(\mathbb{R}^n)$ is closed in $\mathbb{R}^n$. Let a sequence $y_n\in U$ converge to some $y \in \mathbb{R}^n$. Then there exist $x_n$ such that $y_n = f(x_n)$. By the assumption we have $d(x_n,x_m) \leq a\cdot d(y_n,y_m)$, but since $y_n$ converges in $\mathbb{R}^n$ it is a Cauchy sequence, and thus $x_n$ is Cauchy as well and converges to some $x\in \mathbb{R}^n$. It follows that $y = f(x)\in U$.

The openness follows from the inverse function theorem. It states that the function is invertible on a neighborhood (that is, an open subset of $\mathbb{R}^n$) of each point $f(y)$, which implicitly says that the neighborhood is contained in the image (cf. the proof of Theorem 2 here).


问:
How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped?
The problem is as follows:

The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.

Prove this determinant

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$

What I've attempted to do was to use this well known identity on matrices:

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$

Then working in the right side of this equation I'm getting:

(Here I'm using the notation of capital $V$ for Vandermonde's determinant)

$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

Factoring the common terms I'm getting

$V=bc(c-b)-ac(c-a)+ab(b-a)$

But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.

答:
One way to prove this identity might use some properties of matrices:

$V=\left|\begin{matrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|$

Multiply the first row by $a$ and subtract from the second row, anotate this result as the new second row. Use the original second row and multiply it by $a$ and subtract from the third row, anotate this result as the new third row.

$2R-1R\times a, 3R-2R\times a$

$\left|\begin{matrix}1&1&1\\0&b-a&c-a\\0&b^2-ab&c^2-ac\end{matrix}\right|$

Then using the known identity:

$1\left|\begin{matrix}b-a&c-a\\b^2-ab&c^2-ac\end{matrix}\right|-1\left|\begin{matrix}0&c-a\\0&c^2-ac\end{matrix}\right|+1\left|\begin{matrix}0&b-a\\0&b^2-ab\end{matrix}\right|$

This becomes into:

$V=1\left|\begin{matrix}b-a&c-a\\b^2-ab&c^2-ac\end{matrix}\right|$

$V=1\left|\begin{matrix}b-a&c-a\\b(b-a)&c(c-a)\end{matrix}\right|$

$V=1 (b-a)(c-a)\left|\begin{matrix}1&1\\b&c\end{matrix}\right|$

$V= (b-a)(c-a)(c-b)$

Thus that's it how it can proved. This works using the identity of factoring terms from the columns in the Vandermonde's matrix.


问:
if $\lim_{n \to \infty}{x_{2n}} = a$ and $\lim_{n \to \infty}{x_{2n+1}} = a$, show that $\lim_{n \to \infty}{x_{n}} = a$
Given the sequence $\{x_n\}$ such that $\lim_{n \to \infty}{x_{2n}} = a$  and  $\lim_{n \to \infty}{x_{2n+1}} = a$, show that $\lim_{n \to \infty}{x_{n}} = a$

Can I do this by sub sequences?

I need some help to do this.

Any suggestion?

答:
Let us consider an $ε>0$. We have that $x_{2n}\to a$ so there exists $n_1\in\mathbb{N}$ so that for every $n\geq n_1$ we have $|x_{2n}-a|<ε$. Same goes for the sequence $(x_{2n+1})$, there exists $n_2\in\mathbb{N}$ so that for every $n\geq n_2$ we have $|x_{2n+1}-a|<ε$.
Take $n_0=max\{2n_1,2n_2+1\}$. Now for every $n\geq n_0$ show that the relation $|x_n-a|<ε$ holds true.


问:
Radical equation - can I square both sides with more than 1 radical on one side?
I'm familiar with equations like:

$\sqrt{x+1} - \sqrt{x+2} = 0 $

Has no solutions, it's just an example off the top of my head

Just move the negative square root to the other side, square both sides and solve.

$\sqrt{x+1} = \sqrt{x+2}$

$x+1 = x+2$

0 = 1

My question is, if there are two square roots on one side, then can I still square both sides in this way:

$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$

$x+1 - (x+2) = x+3$

$x+1 - (x-2) = x+3$

$x = -4$

Or does squaring both sides cause something strange to happen on the left hand side?

答:
How I explain it to myself is negative if you want to multiply both sides of the equation by the same number, say 3, you can multiply-across through the pluses and minuses by 3, or take it in stages by putting a bracket around both sides, but the 3 at the start of the brackets on both sides and then multiply across in through the brackets(effectively inflating both sides by the same amount)... but you can’t square across, you need to square both sides, and I find using Brackets here are always a big help.
And if you’ve got say, two square roots (radicals) added together (or subtracted) on one side then it’s an easier calculation to slide one over to the other side, change the plus to a minus , and vice versa, and then square both sides eg $\sqrt{2x-1} + \sqrt{x-1} = 5$   next step $\sqrt{2x-1} = 5 – \sqrt{x-1}$


答:
The following result may be useful:

A sequence $\{x_n\}$ converges to a limit $a$ if and only if every subsequence of $\{x_n\}$ has a further subsequence that converges to $a$.

See here.

Now, choose any subsequence, which I will just call it $L$, of $\{x_n\}$. Note that since $\{x_{2n}\}$ and $\{x_{2n+1}\}$ partitions $\{x_n\}$, then $L$ must contain infinitely many of that $\{x_{2n}\}$’s or $\{x_{2n+1}\}$.

In other words, $L$ contains a subsequence of either $\{x_{2n}\}$ or $\{x_{2n+1}\}$, which converges to $a$ either way, since $\{x_{2n}\}$ and $\{x_{2n+1}\}$ converges to $a$. So by the result, $\{x_{n}\}$ converges to $a$.

Title: if $\lim_{n \to \infty}{x_{2n}} = a$ and $\lim_{n \to \infty}{x_{2n+1}} = a$, show that $\lim_{n \to \infty}{x_{n}} = a$

Body:
Given the sequence $\{x_n\}$ such that $\lim_{n \to \infty}{x_{2n}} = a$  and  $\lim_{n \to \infty}{x_{2n+1}} = a$, show that $\lim_{n \to \infty}{x_{n}} = a$

Can I do this by sub sequences?

I need some help to do this.

Any suggestion?


问:
limits of an integral with delta
Having this integral:

$$\int_{x-t}^{x+t}\delta(z)dz$$

the solution is:

$$
  \begin{cases}
                                   1& \text{for $|x| < t$} \\
                                   0 & \text{else}
  \end{cases}$$

Namely, all the domain of $x-t<z<x+t$ will have $1$ in the integral, but why? Shouldn't I check when $z=x+t=0$ and $z=x-t=0$ so the solution should be:  $1$ when$|x|=t$ else zero? why in the solution all the domain $x-t<z<x+t$  gets 1?

答:
The delta function, $\delta(x)$, is defined by
$\int_A \delta(x)dx= 1$ if $0\in A$ and 0 otherwise.

Title: limits of an integral with delta


问:
Meaning of "Jacobian determinant"?
In his book on optimal control, George Leitmann uses a meaning of "Jacobian determinant" I don't understand (and I do understand the standard definition). Can anyone explain it? In particular, which function's Jacobian is being taken? (Also, why? I understand this why bit might be hard without more context.) The relevant excerpts from the book are, in order:

http://imgur.com/2vKMmkF,2WDQM91#0

http://imgur.com/2vKMmkF,2WDQM91#1

(Unfortunately, I don't have access to "Ref 6.1").

EDIT: I'm not requiring a firm answer like "This is definitely what it means." Even conjecture is fine.

答:
In the determinant symbol,

$g_{\alpha}$ is the partial derivative of the function with respect to $\alpha$ and $g_{t\alpha}$ is the partial derivative with respect to $t$ of the function $g_{\alpha}$ and similarly for $\beta$. The determinant has the value

$$ g_{\alpha}g_{t\beta} - g_{\beta}g_{t\alpha} $$.

So $g_{t\beta}$ for example, really means

$$ \frac{\partial^{2}g}{{\partial t}{\partial \beta}} $$

Title: Meaning of "Jacobian determinant"?


问:
How to describe the stalk on fiber $X_y$?
Let $f:X\rightarrow Y$ be a morphism of schemes. (May assume more if necessary) Let $x\in X$ and $y\in Y$ such that $f(x) = y$ and $X_y$ be fiber over $y$.

My question: Is there any relation in general between the stalk $\mathcal O_{X_y,x}$ and $\mathcal O_{X,x}$? If the answer is no, what could we say or how to understand the stalk $\mathcal O_{X_y,x}$?

My question is from the following a proof in Qing Liu's book:

Lemma 4.3.20. Let $f:X\rightarrow Y$ be a morphism of finite type between locally Noetherian schemes. Then $f$ is unramified if and only if for every $y\in Y$, the fiber $X_y$ is finite, reduced, and if $k(x)$ is separable over $k(y)$ for every $x\in X_y$.

The author says: 'Indeed, the quotient $\mathcal O_{X,x}/m_y\mathcal O_{X,x}$ remains unaltered when we pass from $X$ to the fiber $X_y$'. I'm confused about the meaning of this statement.

答:
My question: Is there any relation in general between the stalk $\mathcal O_{X_y,x}$ and $\mathcal O_{X,x}$?

Yes. If $\mathfrak m_y$ denotes the maximal ideal of $\mathcal O_{Y,y}$, then $$\mathcal O_{X_y,x}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\,,$$
where $\mathfrak m_y\mathcal O_{X,x}\subseteq \mathcal O_{X,x}$ denotes the ideal generated by the image of $\mathfrak m_y$ under the ring morphism $\mathcal O_{Y,y}\rightarrow \mathcal O_{X,x}$. To see this, we may assume $X\cong\operatorname{Spec} B$ and $Y\cong\operatorname{Spec} A$ are affine, with $x$ and $y$ corresponding to prime ideals $\mathfrak q\subseteq B$ and $\mathfrak p\subseteq A$. Then $X_y\cong \operatorname{Spec}(B\otimes_Ak(\mathfrak p))\cong\operatorname{Spec}B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$ (here $B_{\mathfrak p}$ denotes the localisation of the $A$-module $B$ at $\mathfrak p$; moreover, $\mathfrak p B_{\mathfrak p}\subseteq B_{\mathfrak p}$ denotes the ideal generated by the image of $\mathfrak p$). Thus the local ring $\mathcal O_{X_y,x}$ is given by the localisation of the ring $B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$ at its prime ideal $\mathfrak q/\mathfrak pB_{\mathfrak p}$. By elementary properties of localisation, this coincides with $B_{\mathfrak q}/\mathfrak p B_{\mathfrak q}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$.

Also note that $\mathcal O_{X_y,x}$ has the same residue field as $\mathcal O_{X,x}$. Indeed, under the isomorphism $\mathcal O_{X_y,x}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$, the maximal ideal of $\mathcal O_{X_y,x}$ corresponds to $\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x}$, and $$(\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x})/(\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x})\cong \mathcal O_{X,x}/\mathfrak m_x\cong k(x)\,.$$

The author says: 'Indeed, the quotient $\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$ remains unaltered when we pass from $X$ to the fiber $X_y$'. I'm confused about the meaning of this statement.

This is somewhat sloppy language and I'll try to rephrase it in a super pedantic way. Let $Y'=\operatorname{Spec} k(y)$ and let $X'=X\times_YY'\cong X_y$. Let $x'\in X'$ denote the unique point lying over $x\in X$ and let $y'\in Y'$ be the unique point (which automatically lies over $y$). Then what the author is trying to say is that $$\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\cong \mathcal O_{X',x'}/\mathfrak m_{y'}\mathcal O_{X',x'}\,.$$
Indeed, the left-hand side equals $\mathcal O_{X',x'}$, as seen above, and on the right-hand side we have $\mathfrak m_{y'}=0$ since $Y'=\operatorname{Spec} k(y)$ is the spectrum of a field.

For the sake of completeness, let me elaborate a bit more on Liu's proof, since I think quite some things get swept under the rug. Assume first that all fibres $X_y$ are finite over the respective residue field $k(y)$. By the structure theory of artinian rings (see [Stacks Project, Tag 00J4]), $X_y$ has only finitely many points $x_1,\dotsc,x_n$, and $X_y\cong \coprod_{i=1}^n\operatorname{Spec} \mathcal O_{X_y,x_i}$.
If $X_y$ is reduced, then each of the artinian local rings $\mathcal O_{X_y,x_i}$ must be reduced. But a reduced artinian local ring must be a field, whence $\mathcal O_{X_y,x_i}\cong k(x_i)$. By the discussion above, this also implies $\mathcal O_{X,x_i}/\mathfrak m_y\mathcal O_{X,x_i}\cong k(x_i)$. In particular, if each $k(x_i)$ is separable over $k(y)$, then $f$ is indeed unramified.

Conversely, assume $f$ is unramified. We must show that each fibre $X_y$ is finite and reduced over $k(y)$ (and that all residue fields $k(x)$ are separable over $k(y)$, but this is automatic from the assumption that $f$ is unramified). Since $f$ is unramified, $\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\cong k(x)$ holds for all $x\in X_y$. Thus, all local rings $\mathcal O_{X_y,x}\cong k(x)$ are fields. In particular, they are all 0-dimensional, which implies that $X_y$ must be zero-dimensional. This implies again that $X_y$ has only finitely many points $x_1,\dotsc,x_n$ and that $X_y\cong \coprod_{i=1}^n\operatorname{Spec} \mathcal O_{X_y,x_i}$ (see [Stacks Project, Tag 0AAX]; the disjoint union must be finite since $X_y$ is quasi-compact). Since all $\mathcal O_{X_y,x_i}\cong k(x_i)$ are reduced and finite-dimensional over $k(y)$, we see that $X_y$ too must be reduced and finite over $k(y)$, as claimed.


问:
Why do we use $\frac{\pi}{180}$ to convert from degrees to radians?
If I want to convert from degrees to radians, I can use the function that takes degree value as an input, multiplies it with $\frac{\pi}{180}$ and returns the radian value: $\operatorname{DtoR}(d)=d \times \frac{\pi}{180}$.

And if I want to go from radians to degrees I need to only go backwards and  divide radian value with $\frac{\pi}{180}$ (e.g. multiply it with $\frac{180}{\pi}$): $\operatorname{RtoD}(r)=r \times \frac{180}{\pi}$.

My question is this: Why does multiplying/dividing with $\frac{\pi}{180}$ converts degrees into radians/radians into degrees? Why exactly that number, not some other? Also, does this work only for unit circle, or for any circle?

答: Since circumference =2pier if we take half circle it would be pier in which rotation of angle is 180 degree and  arc length would be pie*r then we get theta =pie =180 degree


问:
Progress on a conjecture of Burnside...
Given a group $G $, the set of automorphisms of $G $ also forms a group, $\rm {Aut}(G) $,with composition as the operation.

An inner automorphism is one determined by conjugation by some element $g\in G $.  That's we have the automorphism  $i_g $ given by $i_g (h)=ghg^{-1}\,\forall h $.

I learned from a comment by @LeeMosher on this site that Burnside once conjectured that any class-preserving automorphism is inner.  Does anyone know about any progress on this?

Of course the converse is trivial.

(Btw, the reference here is to conjugacy classes.  Of course the class equation gives the sizes of these classes.  For example,  in the case of an abelian group,  the class equation consists in all ones.)

For reference,  this would have been early in the  $20$-th century.

Title: Progress on a conjecture of Burnside...

Body:
Unless I have made a mistake, the group 43 of order 32, isomorphic to
$$
\langle (2,6)(5,8), (2,8)(3,7)(5,6), (1,2)(3,8)(4,6)(5,7) \rangle
$$
is a counterexample of minimal order. The automorphism
$$
[ (2,6)(5,8), (2,8)(3,7)(5,6), (1,2)(3,8)(4,6)(5,7) ]\rightarrow
[ (1,4)(3,7), (1,4)(2,5)(6,8), (1,2)(3,8)(4,6)(5,7) ]
$$
is not inner, but preserves all conjugacy classes.


问:
How to say if a Matrix is singular or non-singular
I have this question, as I used the method of determinants to prove that this matrix was singular (it is not), and I was told that it wasn't about proving it, but establishing it.

I am very confused. How could then one say if a Matrix is singular or not? (this one is not, but how do I establish it?)

This is the matrix in question:

$A=\begin{pmatrix}
1 & 2 &-3 \\
 -1&  2& 3\\
 0& 1 & 1
\end{pmatrix}$

Any help would be very welcomed. And many kind wishes.

答:
I'll preface this by saying that I'm not certain regarding exactly what you are asking.

As was noted in the comments the answer will depend on what definition of singular you are using. It is the case that a matrix $A$ is singular if and only if $\det A = 0$. You can easily see that $A$ is not singular by computing
$$\det A = \det\begin{pmatrix} 2 & 3\\1 & 1 \end{pmatrix} - 2\det\begin{pmatrix} -1 & 3\\0 & 1 \end{pmatrix} - 3\det\begin{pmatrix} -1 & 2\\0 & 1 \end{pmatrix} = -1 + 2 + 3 = 4 \neq 0.$$
However, depending upon the definition of singularity you're using, you may need to prove that $\det A = 0$ is equivalent to $A$ being singular.


问:
Why is $\ln 0\ne-\ln \infty$?
The title of this post is intentionally sensational, but what I am really going to do is to compare the divergent integrals $\int_0^1\frac1xdx$ and $\int_1^\infty\frac1xdx$.

Let's consider the transform $\mathcal{L}_t[t f(t)](x)$. It is notable by the fact that it preserves the area under the curve:
$$\int_0^\infty f(x)dx=\int_0^\infty \mathcal{L}_t[t f(t)](x) dx$$

But more interesting, it also works well with divergent integrals, allowing us to define the equivalence classes of divergent integrals. Particularly, by successfully applying this transform to $\int_1^\infty \frac1xdx=\int_0^\infty\frac{\theta (x-1)}{x}dx$, one can obtain the following equivalence class of divergent integrals (the first one and the third one turn out to be similar up to a shift):

$\int_0^\infty\frac{\theta (x-1)}{x}dx=\int_0^\infty\frac{e^{-x}}{x}dx=\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\frac{e^x x \text{Ei}(-x)+1}{x}dx=\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx$

On the other hand, applying the transform to $\int_0^1\frac1x dx=\int_0^\infty \frac{\theta (1-x)}{x}dx$ one can obtain another set of equal integrals:

$\int_0^\infty\frac{\theta (1-x)}{x}dx=\int_0^\infty\frac{1-e^{-x}}{x}dx=\int_0^\infty\frac{1}{x^2+x}dx=\int_0^\infty-e^x \text{Ei}(-x)dx=\int_0^\infty-\frac{x-x\ln x-1}{(x-1)^2 x}dx$

Now, having postulated equivalence of divergent integrals in each class, we can pick two integrals, one from each class and compare them. Well, it seems, the integrals in the second class are bigger by an Euler's constant:

$\int_0^{\infty } \left(\frac{1-e^{-x}}{x}-\frac{1}{x+1}\right) \, dx=\gamma$

And thus, we can conclude that $\int_0^1\frac1xdx=\gamma+\int_1^\infty \frac1xdx$. Surprising, is not it, given that one would naively expect $\ln 0=-\ln\infty$?

But we also have an identity $\gamma = \lim_{n\to\infty}\left(\sum_{k=1}^n \frac1{k}-\int_1^n\frac1t dt\right)$ and know the regularized value of harmonic series $\operatorname{reg}\sum_{k=1}^n \frac1{k}=\gamma$. Thus, we obtain
$$\int_0^1\frac1xdx=\gamma+\int_1^\infty\frac1xdx=\sum_{k=1}^\infty\frac1k.$$

So, the integrals differ by Euler-Mascheroni constant. Is there any logical explanation for this?

答:
Why is $\ln 0\ne-\ln \infty$?

They are equal: $\ln 0 = -\infty$ and $-\ln \infty = -\infty$.

And thus, we can conclude that $\int_0^1\frac1xdx=\gamma+\int_1^\infty \frac1xdx$. Surprising, is not it, given that one would naively expect $\ln 0=-\ln\infty$?

Not at all. $$-\infty = \gamma - \infty.$$ You see, $\infty$ is infinite. And this sort of thing is exactly what makes "infinite" different from "finite".

And per user's comments, it is also true that $-\infty = 1 - \infty$. That too is part of infinity's shtick.

There is no logical inconsistency here. The only problem is that you are trying to enforce intuition developed from finite behavior on infinite behavior where it does not hold.

If infinity didn't behave like this, we would have just added it to the real numbers, just like we added $0$ and negative numbers in the past, and even added a square root of $-1$, though for that one we were at least willing to admit the result deserved its own separate name.

But we don't add $\infty$ to the real numbers, and even tell people that "infinity doesn't exist" for years, as if we couldn't just define it, exactly because it doesn't play well with other numbers. And it is more of pain to deal with all the exceptional behavior than it is to just kick infinity out of the playground.

Only after someone develops the mathematical chops to recognize how it needs to be handled do we admit to them that yes, infinity does exist after all.

Title: Why is $\ln 0\ne-\ln \infty$?


问:
Find integral limits for step functions
I have an equation of the form

$$f(t) + \int_0^t H(t') dt' = c$$.

For now, assume that $f(t)$ is linear, and $H(t)$ is some arbitrary step function (not necessarily just a standard heaviside step function. Is there any easy/ logical way of solving this to find the integral limit t. It feels like it should be easy, as the integral of the step function is 'just' a piecewise linear function, but I'm really struggling to do this.

While I don't have any specific values in mind, I am most interested in the following:

The values of the n components of H(t) form a non-negative decreasing sequence.

$f(t) = kt$, where $k > 0$.

$c > 0$.

Under these conditions, I think there should be a unique solution (though would be interested to hear if I am wrong in assuming this!).

答:
Proof of existence and  uniqueness:

The solution exists because $g(t)=f(t) + \int_0^{t} H(u)du $ is a continuous function with $g(0)=0$ and $g(\frac{c}{k} +1) > c$, so by the intermediate value theorem there is at least a solution.

Now, say you have two solutions such that $t_2 > t_1$ then

$$f(t_1) + \int_0^{t_1} H(u)du = c$$

$$f(t_2) + \int_0^{t_2} H(u)du = f(t_2) + \int_0^{t_1} H(u)du +\int_{t_1}^{t_2} H(u)du  =  c$$

So subtracting  you get

$$f(t_2)-f(t_1)+\int_{t_1}^{t_2} H(u)du =0$$

But this is impossible because $ f(t_2)-f(t_1) >0$ and $\int_{t_1}^{t_2} H(u)du \geq 0$.

A method to solve the problem would be finding the component of $H(t)$ in which the solution lies and then calculate the solution from the starting point of that component. The algorithm to find the component is fast because $g(t)$ is a strictly increasing function so you can use a binary search with the values of $g(t)$ at the starting point of every component.

After you find the right component, say it starts (or the previous finishes) at $t_{a}$, you can calculate the solution exactly by just considering the difference $c-g(t_{a})$ and finding the right value in that component, which is easy because you need just to solve

$$(k+\lim_{t \to t_{a}^+}H(t))(t-t_{a}) =c-g(t_{a})$$

where $\lim_{t \to t_{a}^+}H(t)$ is the value at the starting point of the component.

So the solution is

$$t = t_a + \frac{c-g(t_{a})}{k+\lim_{t \to t_{a}^+}H(t)}$$

Title: Find integral limits for step functions


问:
Prove for every $1 \leq n$, that $47\mid3^n4^{2n}-1$
I tried to solve with induction, base $n=1$ is correct such that, $\frac{3*16-1}{47}=1$.

Then assuming that for $n$ the statement is correct such that,  $3^n4^{2n} \equiv 0 \pmod {47}$.

We need prove that for $n+1$ is true such that, $\frac{(3*16*48^n)-1}{47}$ .

Then I tried to simplify it to $\frac{48}{47}*48^n-\frac{1}{47}= 1\frac{1}{47}*48^n-\frac{1}{47}=\frac{(3^n4^{2n})+47(3^n4^{2n})}{47}-\frac{1}{47}=\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$

We get that from $\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$, the expression  $\frac{(3^n4^{2n})} {47}-\frac{1}{47}$ is from the assumption so we are left with $3^n4^{2n}$ which is a natural number.

Is that proof enough to show that $47\mid3^n4^{2n}-1$?

答:
Note that $3^n\cdot 4^{2n}=48^n$, by using the binomial theorem you have $$(47+1)^n-1=\sum_{i=1}^{47}\binom{47}{i}47^i$$

And so every term is divisible by $47$.


问:
Calculating the exact value of $\sum_{n=0}^{\infty} (-1)^n (\frac{31}{32})^{2n+1} (\frac{1}{2n+1})$
$\sum_{n=0}^{\infty} (-1)^n (\frac{31}{32})^{2n+1} (\frac{1}{2n+1})$

Hey math stack exchange! I've been trying to figure out how to calculate the exact summation of these series for awhile now. I looked at this post: How to find the sum of an alternating series? to get some clues, but I'm still a bit lost because my formula has the term $(\frac{1}{2n+1})$ in it.

Would appreciate some help! Thanks!

Body:
$$S = \sum_{n=0}^{\infty}{(-1)^{n}\biggl(\frac{31}{32}\biggr)^{2n+1}\biggl(\frac{1}{2n+1}\biggr)}$$
$$= \sum_{n=0}^{\infty}{(-1)^{n}\Biggl(\int_{0}^{\frac{31}{32}}x^{2n}dx\Biggr)}$$
$$= \int_{0}^{\frac{31}{32}}\Biggl(\sum_{n=0}^{\infty}(-x^{2})^{n}\Biggr)dx$$
$$= \int_{0}^{\frac{31}{32}}{\Biggl(\frac{1}{1+x^{2}}\Biggr)}dx$$
$$= \arctan\Biggl({\frac{31}{32}}\Biggr)$$


问:
Expected value and variance for a homogeneous polynomial of random variables
I want to obtain equations for expected value and variance for a homogeneous polynomial of arbitrary order $n$.

For a quadratic form, which is a special case of a homogeneous polynomial, there are known equations for expected value and variance:

$
E[A^T HA] = tr[H\Sigma]+\mu^TH\mu,
$

$
var[A^T HA] = 2tr[H\Sigma H\Sigma]+4\mu^TH\Sigma H\mu,
$

where $A$ is a vector of $M$ random variables, $H$ is a square symmetric matrix with polynomial coefficients and $\mu$ and $\Sigma$ are the expected value and variance matrix of $A$, respectively.

I want to consider a homogeneous polynomial of order $n$:

$ \sum_{i_1=1}^{M} \sum_{i_2=1}^{M} ... \sum_{i_n=1}^{M} h^{(n)}(i_1, i_2, ...,i_n)a_{i_1}a_{i_2} ... a_{i_n},$

where $a_i$ is a random variable from vector $A$ and $h^{(n)}$ are polynomial coefficients. E.g. a 3-rd order polynomial for $M=3$ will look like this:

$
h_1^{(3)} a_1^3 + h_2^{(3)} a_1^2 a_2 + h_3^{(3)} a_1^2 a_3+ h_4^{(3)} a_1 a_2^2 + h_5^{(3)} a_1 a_2 a_3+ h_6^{(3)} a_1 a_3^2 + h_7^{(3)} a_2^3 + h_8^{(3)} a_2^2 a_3 + h_9^{(3)} a_2 a_3^2 + h_{10}^{(3)} a_3^3 .
$

Question: How to derive equations for expected value and variance of such polynomial or where can I look them up, if they are widely known/used?

答:
I don't understand what the polynomials are, but there is a general formula for the mean and variance of a polynomial function of a random variable.

Let $$p(x)=\sum_{i=0}^na_i x^i$$ be a given polynomial.

The square of the polynomial is
$$[p(x)]^2=\sum_{i=0}^{2n} b_i x^i$$
where $$b_i=\sum_{j=0}^{i}a_j a_{i-j}$$

Also, let $X$ be a random variable with moments about the origin $$\mu'_k=E[X^k]$$ for $k=0, 1, 2, ...$

Finally, let $Y=p(X)$.

Then, the mean and variance of $Y$ are: $$E[Y]=\sum_{i=0}^na_i \mu'_i$$ and $$Var[Y]=E[Y^2]-(E[Y])^2=\sum_{i=0}^{2n} b_i \mu'_i-\left(\sum_{i=0}^na_i \mu'_i \right)^2$$


问:
Is $x^x$ differentiable when $x<0$?
The function $f(x)=x^x$ is only defined for negative values of $x$ if $x$ is a rational number $a/b$, where $b$ is odd. Initially, I thought this meant that the limit
$$
f'(a)=\lim_{h\to0}\frac{(a+h)^{a+h}-a^a}{h}
$$
cannot exist if $a$ is a negative number, since every neighbourhood of $a$ would contain values for which $x^x$ does not make any sense at all. However, this post—about $\lim_{x\to0}\frac{\sin(1/x)}{\sin(1/x)}$—suggests that provided every neighbourhood of $a$ contains values for which $x^x$ is defined, there is no issue with taking limits at the point.

So if $a$ is negative number that is in the domain of $f$, then does $f'(a)$ exist? And if so, is there a general formula for $f'(a)$?

答:
In the usual sense of rational exponentiation, if $x=\frac{a}{b}$ where:

$b$ is odd

$a$ and $b$ are coprime

$x<0$

then $x^x=(-1)^a (-x)^x$. Given any $x<0$ there exist numbers of this form with $a$ even arbitrarily close to it, as well as numbers with $a$ odd arbitrarily close to it, so there is no continuous extension. Moreover we can't take limits or derivatives even at points in the domain, because limits along sequences with $a$ even and limits along sequences with $a$ odd will yield different values.

Title: Is $x^x$ differentiable when $x<0$?


问:
Compute the limit of the integral
Compute

$$ \lim_{n \to \infty} \int_{0}^{n} (1 + \frac{x}{2n})^ne^{-x}dx$$

Edit: We have now learned the dominated convergence theorem. I think it is now a good idea to attempt the question that way. There is already an answer here, so I can't really delete the question. I thought I'd edit it and see if anyone wants to jump in!

答:
As we discussed in a previous problem, write

$$\int_{0}^{n} (1 + \frac{x}{2n})^ne^{-x}dx= \int_{0}^{\infty}\chi_{(0,n)}(x) (1 + \frac{x}{2n})^ne^{-x}dx.$$

You're set up to use the MCT.

Title: Compute the limit of the integral


问:
Show that $\sin6\alpha\equiv \sin2\alpha(16\cos^4\alpha-16\cos^2\alpha+3)$
$$\sin6\alpha\equiv \sin2\alpha(16\cos^4\alpha-16\cos^2\alpha+3)$$

Can you help me with De Moivre's theorem and how I would go about tackling this question.

I understand that De Moivre's theorem states that $(\cos\alpha+i\sin\alpha)^n \equiv \cos n\alpha+i\sin n\alpha$.

But I don't see how this would come to use in this question.

答:
From De Moivres
$$
\cos 6\alpha+i\sin 6\alpha=(\cos\alpha+i\sin \alpha)^6=\\(\cos^6\alpha+\ldots)+i\left({6\choose 1}\cos^5\alpha\sin\alpha-{6\choose 3}\cos^3\alpha\sin^3\alpha+{6\choose 5}\cos\alpha\sin^5\alpha\right).
$$
Therefore
$$\sin 6\alpha={6\choose 1}\cos^5\alpha\sin\alpha+{6\choose 3}\cos^3\alpha\sin^3\alpha+{6\choose 5}\cos\alpha\sin^5\alpha=\\
2\cos\alpha\sin\alpha(\ldots)=\ldots$$</p>


问:
In triangle $\triangle ABC$, we have $AB=8cm$, $AC=12cm$ and $BC=15cm$, $I$ the incenter. If $ID//AB$ and $IE//AC$ which is the perimeter of $IDE$?
In triangle $\triangle ABC$, we have $AB=8cm$, $AC=12cm$ and $BC=15cm$, $I$ the incenter. If $ID//AB$ and $IE//AC$ which is the perimeter of $\triangle IDE$?

After drawing the shape accurately I have that the perimeter is $15$.

Since $ID//AB$, then from Thales we have that $\frac{ID}{AB}=\frac{IO}{OA}$ where $O$ is the intersection of $AI$ and $CB$. If it were the barycenter then we could use the property that $AI=2OI$, but it's the incenter. Could you please explain to me how to solve this question?

答:
Notice that in $\triangle IDB$, $\angle BDI = 180^0 - \angle B$.

So $\angle IBD = \angle BID = \frac{\angle B}{2}$. So $ID = DB$. Similarly show, $IE = EC$.

So $ID + DE + EI = BC = 15$.

If you do not notice that, here is another simple way to proceed instead of doing any calculations upfront -

Assume area of $\triangle ABC = A$. We know that $A = r \cdot \frac{P}{2} \ $ where $r$ is inradius and $P$ is the perimeter of $\triangle ABC$.

So $ \ \displaystyle r = \frac{2A}{P}$

Say altitude from vertex $A$ to $BC$ is $h$, then

$\displaystyle A = \frac{1}{2} BC \cdot h \implies h = \frac{2A}{15}$

As $\triangle ABC \sim \triangle IDE$ and also note that $r$ is the altitude from $I$ to $DE$ in $\triangle IDE$.

So, scale factor of $\triangle IDE$ to $\triangle ABC$, $k = \frac{r}{h} = \frac{15}{P}$.

Perimeter of $\triangle IDE = k \times P = 15$


问:
On random variables made up of independent random digits
Some random variables can be expressed as a binary expansion whose digits are chosen independently at random; this is called a convolution. One example of this kind of random variable is the one for an exponential distribution truncated to the interval $[0, 1]$ (Devroye and Gravel 2020).

However, there seem to be limits on how practical this approach is to describe random variables.

The following is part of Kakutani's theorem (Kakutani 1948): Let $a_j$ be the $j$th binary digit probability in a random variable's binary expansion (starting with $j=1$ for the first digit after the point), where the variable is in $[0, 1]$ and each digit is independently set (that is, each digit is an independent Bernoulli random variable). Then the random variable is absolutely continuous if and only if the sum of squares of $(a_j - 1/2)$ converges. In other words, the digits in the binary expansion become less and less biased as they move farther and farther from the binary point.

An absolutely continuous random variable in $[0, 1]$ can thus be built if we can find an infinite sequence $a_j$ that converges to 1/2. Then that variable could be formed by setting the digits after the point in its infinite binary expansion to 1 with probability equal to the corresponding $a_j$ (that is, the variable's digits are independent Bernoulli random variables with parameter equal to the corresponding $a_j$). However, experiments show that the resulting variable will have a discontinuous probability density function (PDF) in general.

I conjecture the following:

For β = 2, the random variable's PDF will be continuous only if—

the probabilities of the first half, interval (0, 1/2), are proportional to those of the second half, interval (1/2, 1), and

the probabilities of each quarter, eighth, etc. Are proportional to those of every other quarter, eighth, etc.

The random variable's PDF will be continuous only if the sequence has the form—$$a_j=\frac{\exp(w/\beta^j)}{1+\exp(w/\beta^j)}, $$ where β = 2 and w is a constant. Special cases of this include the uniform distribution (w = 0), the truncated exponential(1) distribution (w = −1; (Devroye and Gravel 2020)), and the more general truncated exponential(λ) distribution (w = −$\lambda$). We also have the special case of 1 minus a truncated exponential(w) random variable when w > 0, as well as the special case aj = yv/βj/(1 + yv/βj), with w = ln(y)*v where y > 0 and v are constants. As evidence in favor of this conjecture, experiments show that sequences z(j)/(1 + z(j)), other than sequences of the form just given, will generally result in a discontinuous PDF even if z(j) converges to 1 and even if β is other than 2.

A similar behavior to (1) applies for β other than 2 (non-base-2 or non-binary cases) as it does to β = 2 (the base-2 or binary case).

My question is: Is there a proof for these conjectures? My attempts at trying to prove them using SymPy have failed, notably because it can't yet evaluate the infinite products needed to check my tries. (One example is (exp(2**(-j)*(I*x - 1)) + 1)/(1 + exp(-2**(-j))).) And I suspect that proving these conjectures might involve studying the distribution's characteristic function or Fourier transform, as Claude Gravel told me (see also Devroye and Gravel 2020, section 3.8).

Note: All statements about random variables, etc., are with respect to Lebesgue measure.

REFERENCES:

S. Kakutani, "On equivalence of infinite product measures", Annals of Mathematics 1948.

Devroye, L., Gravel, C., "Random variate generation using only finitely many unbiased, independently and identically distributed random bits", arXiv:1502.02539v6 [cs.IT], 2020.

答:
This is an answer for conjecture $2$, not conjecture $3$.

As aschepler says in the comments, conjecture $1$ is true because the binary digits are chosen independently. This answer is entirely dependent on that conjecture. In order for that to be satisfied, you need that $$P\left(\frac{a}{2^n}\le X< x+\frac{a}{2^n}\mid\frac{a}{2^n}\le X<\frac{a+1}{2^n}\right)$$

is equal for all $a$ such that $0 \le a \le 2^n-1$, where $0 < x < \frac{1}{2^n}$ and $X$ is the randomly generated number. Let $F(x)$ be the CDF of $X$. This expression is then equal to

$$\frac{F\left(x+\frac{a}{2^n}\right)-F\left(\frac{a}{2^n}\right)}{F\left(\frac{a+1}{2^n}\right)-F\left(\frac{a}{2^n}\right)}$$

This means that it must be true that

$$\frac{F(x)}{F\left(\frac{1}{2^n}\right)} = \frac{F\left(x+\frac{a}{2^n}\right)-F\left(\frac{a}{2^n}\right)}{F\left(\frac{a+1}{2^n}\right)-F\left(\frac{a}{2^n}\right)} \to \frac{F(x)}{F\left(\frac{1}{2^n}\right)}\left(F\left(\frac{a+1}{2^n}\right)-F\left(\frac{a}{2^n}\right)\right)+F\left(\frac{a}{2^n}\right) = F\left(x+\frac{a}{2^n}\right)$$

In order for the derivative to be continuous at $\frac{a}{2^n}$, it must be true that $$f\left(\frac{a}{2^n}\right) = \frac{f(0)}{F\left(\frac{1}{2^n}\right)}\left(F\left(\frac{a+1}{2^n}\right)-F\left(\frac{a}{2^n}\right)\right)$$

Using the fact that $F\left(\frac{1}{2^n}\right) = (1-a_1)(1-a_2)\cdots(1-a_n)$ and $f\left(\frac{2a}{2^{n+1}}\right) = f\left(\frac{a}{2^n}\right)$, this can be simplified to $$f\left(\frac{2a+1}{2^{n+1}}\right) = f\left(\frac{a}{2^{n}}\right)\left(\frac{1}{1-a_{n+1}}-1\right)$$

which means that $$\frac{f\left(x + \frac{1}{2^{n}}\right)}{f(x)}=c_n$$

Since this must be true for all $x$ of the form $\frac{a}{2^n}$ and all $n$ and because $f(x)$ must be continuous, $f(x)$ must be an exponential function (i.e. $f(x) = c\cdot e^{bx}$) such that $\int_0^1 f(x)dx = 1$. If $f(x)$ didn't have to be continuous, the function could take any value at numbers not of the form $\frac{a}{2^n}$. But because it does have to be continuous, $$f(x) = \frac{b}{e^b-1}e^{bx}$$

This then means that $$c_n= e^{b/2^n} \to a_n = 1-\frac{1}{1+e^{b/2^n}}=\frac{\exp\left(
b/2^n \right)}{1+\exp\left(
b/2^n \right)}$$

which is the same as your result.


问:
How to show "factors of $x$" in Math?
How to show "factors of $x$" in Math? For example, how can I show "$x$ has $n$ factors" in Math?

答:
If $x\in\Bbb N$ the number of positive factors of $x$ is the divisor function of $x$, and is denoted $\sigma_0(x)$, $d(x)$, $\nu(x)$ or $\tau(x)$, so for your purpose you can follow any of those with $=n$.

The set of such factors has no standard symbol, so write it as $\{d\in\Bbb N:d\mid x\}$. If you intend more than just the positive factors, you can adjust these accordingly.

Title: How to show "factors of $x$" in Math?


问:
On the Leibniz rule in the Serre spectral sequence
Consider the $\mathbb{Z}/2$-cohomological Serre spectral sequence associated to the path fibration
$$ K(\mathbb{Z}/2, 1) \simeq \Omega K(\mathbb{Z}/2, 2) \longrightarrow PK(\mathbb{Z}/2, 2) \longrightarrow K(\mathbb{Z}/2, 2). $$

By universal coefficients, we are allowed to write $E^{p, q}_2 \cong H^p(K(\mathbb{Z}/2, 2); \mathbb{Z}/2)) \otimes H^q(\mathbb{RP}^{\infty}; \mathbb{Z}/2)$. Write $x\in H^1(\mathbb{RP}^{\infty}; \mathbb{Z}/2)$ for the generator of the polynomial algebra $H^*(\mathbb{RP}^{\infty}; \mathbb{Z}/2)\cong \mathbb{Z}/2[x]$ and $\iota_2\in H^2(K(\mathbb{Z}/2, 2); \mathbb{Z}/2)$ for the generator.

Since $PK(\mathbb{Z}/2, 2)$ is contractible, no entry on the $E_2$ page except for $E_2^{0, 0}$ can survive to $E_\infty$. Since $H^1(K(\mathbb{Z}/2, 2); \mathbb{Z}/2) = 0$, we have $E_2^{1, 1} = 0$, and hence the only way that $E_2^{0, 3}$, which I know is generated by $\operatorname{Sq}^1(\iota_2)\otimes 1 \neq 0$, can die is that the map
$$d_3\colon E_3^{0, 2}\to E_3^{3, 0}$$
is an isomorphism.

Now it is not hard to see that $d_2\colon E_2^{0, 2}\to E_2^{2, 1}$ must be the zero map, so $E_3^{0, 2}$ is still generated by $1\otimes x^2$. But now we can calculate using the Leibniz rule that (recall that we're working over $\mathbb{Z}/2$, hence signs can be ignored)
\begin{align*}
d_3(1\otimes x^2) &= d_3((1\smile 1)\otimes (x\smile x))\\
&=  d_3((1\otimes x)\cdot (1\otimes x))\\
&= d_3(1\otimes x)\cdot (1\otimes x) + (1\otimes x) \cdot d_3(1\otimes x).
\end{align*}
But $d_3(1\otimes x)\in E_3^{3, -1} = 0$, and so the entire final term is $0$. This is a contradiction because we argued earlier that the map is an isomorphism.

My takeaway is that there must be an error in the explicit calculation of the differential, but I can't find it. Where is it?

答:
You missed $d_2: E^{0,1} \to E^{2,0}$. This must be an isomorphism, as it is the last chance to kill off $E^{0,1}$. One finds that $d_2(1 \otimes x) = \iota_2 \otimes 1$.

The Leibniz rule for $d_3$ states that if $x,y \in E^3$, then $d_3(xy) = d_3(x)y +  (-1)^{|x|} x d_3(y).$

Now that $E_3^{0,1} = 0$ the element $(1 \otimes x)$ does not exist in the $E_3$ page. In particular, your expression $(1 \otimes x)^2 = 1 \otimes x^2$ takes place in the $E_2$ page! You have a Leibniz rule for $d_2$ of this element and indeed as you claim $d_2(1 \otimes x^2) = 0$. But you get nothing for $d_3$.

Indeed $d_3(1 \otimes x^2) = \text{Sq}^1 \iota_2$, just like you say.

It might be interesting to ask what the Leibniz rule ought to say in your situation: for elements $x,y$ which were cycles on previous pages but are no longer. The $\text{Sq}^1$ suggests that this is the home of primary cohomology operations on spectral sequences. I do not know anything about this.