问:
Heat loss of a cube insulated on four sides.
I'm looking for a way to make sense of this. A cube, perfectly insulated on four sides is heated to an initial temperature of 100C. The two non-insulated sides will be the top and the bottom of the cube. We place the cube on a plate which maintains a temperature of 0C. The loss of heat through the top is given, and we are left to solve a pretty standard PDE.

Here's the issue. I'm saying the insulated sides correspond to the BCs of the $X(x)$ and $Y(y)$ functions after separation. When I go to compute the Fourier coefficients I end up with 0! The entire problem collapses. I've never encountered a problem where this happens... Am I just supposed to say the eigenvalues must be 0? I'm not sure if this physically makes sense, but would this just be equivalent to an infinite number of thin rods, implying that solution is independent of $x$ and $y$ values?

答:
Looks like you are right. From the physical standpoint, there is no horizontal heat transfer. Consider a thought experiment: if instead of the cube you had an entire thick plane, due to symmetry there will be no horizontal heat transfer. It means that you may cut the plane into cubes and insulate them. Nothing will change.

It is indeed a one-dimensional problem, and your computations confirmed that.


问:
Why is an element in the countable product of real numbers regarded as a sequence?
A sequence of real numbers is a just an enumerated list of real numbers. Now in my lecture notes it is mentioned that an element of the countable product topological space $\mathbb{R}^w$ (Tychonoff topology) can be regarded as a sequence of real numbers. I'm a bit confused by this statement. How can an element of the product topology be a real sequence? Could someone clarify this with a concrete example?

答:
You probably think of $\Bbb R^\omega$ as the set of "$\omega$-tuples" (by analogy with $n$-tuples) with coordinates elements of $\Bbb R$.  But then one has the sequence $(x_n)$, where $x_n:=\text{the n-th  coordinate of x}$ for any element $x\in\Bbb R^\omega$.

But what's also kind of nice is the functional analysis flavor of considering $\Bbb R^\omega$ as the set of functions from $\Bbb N$ to $\Bbb R$.  Then we simply have, for each $f\in\Bbb R^\omega$, the sequence $x_n:=f(n)$.

ContentLicense: CC BY-SA 4.0


问:
Proving/disproving a given set $A$ is open in $\Bbb R^2$ with the usual metric
$
\newcommand{\R}{\mathbb{R}}
$
Let $(\R^2, d)$, where $d$ denotes the usual Euclidean distance, be a metric space. A set $A$ is defined as follows:

$$A = \left\{ (x,y) \in \R^2 \; \middle| \; e^{2x} + \left( \frac y 5 \right)^4 < 1 \right\}$$

I think the set should be open in this metric space.

To prove it, is it correct to show the boundary set of $A$ is a subset of $\R^2\setminus A$?

Let $P_0(x_0,y_0)$ be an arbitrary element belongs to the boundary set.

There exists a converging sequence in $A$ that converge to $P_0$.

Also, there exists a converging sequence in the complement of $A$ that is also converging to $P_0$.

Then we have for all $n>N$ for some $N$,

$$e^{2x_n} + \left( \frac {y_n} 5 \right)^4 \ge 1$$

And since the sequence when $n \to \infty$ is $P_0 (x_0,y_0)$.

So after taking the limit we have

$$e^{2x_0} + \left( \frac {y_0} 5 \right)^4 \ge 1$$

Thus $P_0$ is a element of $\R^2\setminus A$.

Does my proof make sense?

答:
It is a correct approach. In any topological space $X$ and any $A\subseteq X,$ with $A^c= X$ \ $A,$ the sets $int(A),\, int(A^c), \,\partial A$ are pairwise-disjoint and their union is $X.$

Now $int(A)$ is the union of all subsets of $A$ (including possibly $A$ itself) that are open in the space $X,$ so $int (A)$ is open and $int (A)\subseteq A.$

If $A\cap \partial A=\phi$ then $$A=A\cap X=(A \cap int (A))\,\cup\, (A\cap A^c)\,\cup\, (A\cap \partial A)=(int(A))\cup (\phi)\cup (\phi)=int (A)$$ so $A$  is open.


问:
Finding probability in "at least " scenario
Suppose there is a grocery store with 3 branches, with the probability that KoolAid is present in store
A : 0.35
B: 0.55
C : 0.10

How can I find the probability that KoolAid will be present in at least 2 stores?

My idea so far:
Divide into 3 cases:

Present in A and B (not C)

Present in B & C (not A)

Present in A & C (not B)

I'm not sure how to calculate these however or if I'm missing a case?

答:
Hint :

Assuming the probabilities of KoolAid present in each store are independent then you can use the definition of independence on the three subcases you found plus another case that you missed (all store with KoolAid)


问:
Discriminant of integral basis
Let $A=\mathbb{Z}\alpha_1+...+\mathbb{Z}\alpha_n$, $B=\mathbb{Z}\beta_1+...+\mathbb{Z}\beta_n$, be two lattices, s.t. $A\subseteq B$. Then $d(\alpha_1,...,\alpha_n)=c^2d(\beta_1,...,\beta_n)$, where $d$ is the discriminant and $c$ is the index of $A$ in  $B$. Why this is true?

I read a proof of discriminant being square free, makes and integral basis, there I encountered this. Can anyone help?

答:
I assume everything takes place in a number field $K/\Bbb Q$ with $[K:\Bbb Q]=n$. Let $\{\sigma_1, \dots, \sigma_n\}=\mathrm{Hom}(K,\overline{\Bbb Q})$ be the field homomorphisms to a fixed algebraic closure so that $\mathrm{Tr}_{K/\Bbb Q}=\sum_{i=1}^n \sigma_i$.

For a $n$ linearly independent elements $\alpha_1, \dots, \alpha_n \in K$ consider the $n \times n$ matrix $A=(\sigma_i(\alpha_j))_{i,j}$.

Then one has $A^TA=(b_{i,j})_{i,j}$ where $b_{i,j}=\sum_{k=1}^n\sigma_k(\alpha_i) \sigma_k(\alpha_j)=\mathrm{Tr}_{K/\Bbb Q}(\alpha_i\alpha_j)$

We see that $\det(A^TA)=\mathrm{Disc}(\alpha_1, \dots, \alpha_n)$

Now consider another $n$ linearly independent elements $\beta_1, \dots, \beta_n$. Let $T:K \to K$ be the linear transformation determined by $T\beta_j = \alpha_j$ for all $j$. Let $T$ be represented by the matrix $(t_{i,j}) \in \mathrm{GL}_n(\Bbb Q)$ so that we have $\sum_{j,k}t_{j,k}\beta_k=\alpha_j$

Because each $\sigma_i$ is $\Bbb Q$-algebra homomorphism we can apply it to this equation and obtain $\sum_{j,k}t_{j,k}\sigma_i(\beta_k)=\sigma_i(\alpha_j)$.

This means if we consider the matrix $B:=(\sigma_i(\beta_j))_{i,j}$, we have $TB=A$. As before with $A$, we have $\det(B^TB)=\mathrm{Disc}(\beta_1, \dots, \beta_n)$
But $TB=A$, so $\det(A^TA)=\det(T)^2\det(B^TB)$, so $\mathrm{Disc}(\alpha_1, \dots, \alpha_n) = \det(T)^2 \mathrm{Disc}(\beta_1, \dots, \beta_n)$

Now assume that $\Bbb Z\alpha_1+ \dots+\Bbb Z\alpha_n \subset \Bbb Z\beta_1 + \dots +\Bbb Z\beta_n$. Then $T$ has integer entries. We can thus find the Smith normal form of $T$, this means that there are $U,V \in \mathrm{GL}_2(\Bbb Z)$ such that $UTV=\mathrm{diag}(a_1, \dots, a_n)$ for integers $a_1, \dots, a_n$.
We can think of $U$ and $V$ as automorphisms (base-changes over $\Bbb Z$) of the two lattices. We obtain from this that $\Bbb Z\beta_1 + \dots + \Bbb Z\beta_n/\Bbb Z \alpha_1 + \dots + \Bbb Z\alpha_n \cong \Bbb Z/a_1 \Bbb Z \oplus \dots \oplus \Bbb Z/a_n \Bbb Z$, so that the index of $\Bbb Z \alpha_1 + \dots + \Bbb Z\alpha_n$ in $\Bbb Z\beta_1 + \dots + \Bbb Z\beta_n$ is $\prod_{i=1}^n |a_i|=|\det(T)|$ (using that $U$ and $V$ have determinant $\pm 1$)

Plugging this into $\mathrm{Disc}(\alpha_1, \dots, \alpha_n) = \det(T)^2 \mathrm{Disc}(\beta_1, \dots, \beta_n)$ proves the result.

Title: Discriminant of integral basis


问:
Conjugation under Permutation Matrices
This seems like a basic matrix algebra question but I can't seem to find any easy reference/trick for it. If I am given two (diagonal) matrices $D_1$ and $D_2$ and guaranteed that they are related via conjugation of a permutation matrix  i.e. $$D_2=P.D_1.P^{-1}$$ can I find what the permutation matrix P is, completely or partially?

答:
I follow the notation on the Permutation matrix wiki page.

Let $p_1,\dots,p_n$ denote the diagonal entries of $D_1$, and let $q_1,\dots,q_n$ denote the diagonal entries of $D_2$. Suppose that $\pi$ is a permutation for which $q_i = \pi(p_i)$ for $i = 1,\dots,n$. In this case, we find that
$$
D_2 = P_\pi D_1 P_{\pi}^{-1} = P_{\pi}D_1 P_{\pi}^T.
$$

Title: Conjugation under Permutation Matrices


问:
Orthogonality of Legendre polynomials using specific properties
I'm having significant issues with a problem and would appreciate any help at all with it. It is regarding proving the orthogonality of Legendre polynomials using a specific recursion formula and property of Legendre polynomials.

I have been given the following relation $$x P'_n(x) -P'_{n-1}(x) = n P_n (x) $$
And that $$ \int^{1}_{-1} f(x) P_n (x) dx = 0 $$ for any polynomial $f(x)$ of degree less than $n$.

I must then show that for $ n \geq  1$

$$ \int^{1}_{-1} P_n (x)^2 dx = \frac{1}{2n} \int^{1}_{-1} x \frac{d}{dx} (P_n (x)^2)dx $$

And then determine the value of the integral, that being $\frac{2}{2n+1}$. I must use the two stated facts, I'd be grateful for any help with this, thank you guys.

答:
$$\int\limits_{-1}^{1} P_n(x)^2 \;dx = $$
$$\int\limits_{-1}^{1} P_n(x)P_n(x) \;dx = $$
$$\frac{1}{2n}\int\limits_{-1}^{1} 2n P_n(x)P_n(x) \;dx = $$

Substituting your first relation in:
$$\frac{1}{2n}\int\limits_{-1}^{1} 2 \left[x P'_n(x) - P'_{n-1}(x)\right] P_n(x)\;dx = $$

$$\frac{1}{2n}\int\limits_{-1}^{1} 2 x P'_n(x) P_n(x) \; dx - \frac{1}{2n}\int\limits_{-1}^{1} 2 P'_{n-1}(x) P_n(x)\;dx = $$

Given your second property defining orthogonality (since after all $P'$ is of lesser degree than $P$) the entire second integral drops out.

$$\frac{1}{2n}\int\limits_{-1}^{1} x 2 P_n(x) P'_n(x) \; dx  = $$

Reverse of the chain rule:

$$\frac{1}{2n}\int\limits_{-1}^{1} x \frac{d}{dx} P^2_n(x) \; dx $$

QED

And for anyone who asks/wonders, or is inclined to criticize... I answered this question because it was asked, and this is a Q&A forum.  I shouldnt have to justify answering questions on a Q&A forum though, but this is the culture here.


问:
Theory in $(\mathbb{N};0,S,<)$ is complete?
My question is how I show the corollary 32B(a) from the book A mathematical introduction to logic by Herbert Enderton.
Corollary 32B(a) on page 196 says that:

Cn$A_{L}$ is complete.

where $A_{L}=(\mathbb{N};0,S,<)$ and $\text{Cn} A_L$ is the set $\{\sigma \mid A_L \models \sigma\}$ of consequences of $A_L$.

The proof of Corollary 32B(a) simply says:

The argument that followed the proof of Theorem 31G is applicable here also.,

Theorem 31G (on page 191) says that another theory admits quantifiers elimination, however I don't know how to write the proof of Corollary 32B(a) more formally.

答:
You don't have to use Theorem 31G (which states theory of natural numbers with successor admits elimination of quantifiers), but you have to develop an argument similar to the one explained after (the proof of) Theorem 31G on page 192, to show that from the elimination of quantifiers in $\text{Cn}A_L$(Theorem 32A) it follows that $\text{Cn}A_L$ is complete (Corollary 32B.a).

First, a clarification on what we are talking about. In Enderton's book (Chapter 3), $A_L$ is the finite set of axioms $\{S3, L1, L2, L3, L4, L5\}$ below, in the first-order language with a constant symbol $\mathbf{0}$, a unary function symbol $\mathbf{S}$ and a binary relation symbol $\mathbf{<}$.
\begin{align}
(S3)&&& \forall y \, (y \neq \mathbf{0} \to \exists x \, y = \mathbf{S}x) \\
(L1)&&& \forall x\forall y \, (x \mathbin{\mathbf{<}} \mathbf{S}y \leftrightarrow x \mathbin{\mathbf{\leq}} y) \\
(L2) &&& \forall x \ x \not\!\!\mathbin{\mathbf{<}} \mathbf{0} \\
(L3) &&& \forall x \forall y \, (x \mathbin{\mathbf{<}} y \lor y \mathbin{\mathbf{<}} x \lor x = y) \\
(L4) &&& \forall x \forall y \, (x \mathbin{\mathbf{<}} y \to y \mathbin{\not\!\!\mathbf{<}} x) \\
(L5) &&& \forall x \forall y \forall z \, (x \mathbin{\mathbf{<}} y \to y \mathbin{\mathbf{<}} z \to x \mathbin{\mathbf{<}} z)
\end{align}

The structure $\mathfrak{N}_L = (\mathbb{N}; 0, S, <)$ interprets the symbols  $\mathbf{0}$ (zero), $\mathbf{S}$ (successor) and $\mathbf{<}$ (order) as expected in the domain $\mathbb{N}$ of natural numbers.

When you write $A_L = (\mathbb{N}; 0, S, <)$ you are confusing two distinct notions: a finite set of axioms $A_L$ (a syntactic object) and a structure $\mathfrak{N}_L$ (a semantic object).

Clearly, $\mathfrak{N}_L$ is a model of $A_L$ and actually, $A_L$  finitely axiomatizes the theory of $\mathfrak{N}_L$ (i.e. $A_L$ is a finite axiomatization of the theory of natural numbers with successor and natural order), but this is not relevant to answer your question.

Notations: $\text{Cn} A_L$ is the set $\{\sigma \mid A_L \models \sigma\}$ of theorems (sentences) that can be proved starting from the axioms in $A_L$; while
$\text{Th}\,\mathfrak{N}_L$ is the theory of $\mathfrak{N}_L$, i.e. the set of  sentences that are true in the structure $\mathfrak{N}_L$.

Saying that of $\text{Cn} A_L$ is complete means that $A_L \models \sigma$ or $A_L \models \lnot \sigma$ for every sentence $\sigma$ in the first-order language with $\mathbf{0}$, $\mathbf{S}$ and $\mathbf{<}$.

How to prove that $\text{Cn} A_L$ is complete, knowing that $\text{Cn}A_L$ admits elimination of quantifiers? Let $\sigma$ be a sentence. The quantifier elimination procedure gives a quantifier-free sentence $\tau$ such that
$$\tag{1} A_L \models (\sigma \leftrightarrow \tau).$$
Now, I claim that either $A_L \models \tau$ or
or $A_L \models \lnot \tau$. Indeed, $\tau$ is built up from atomic sentences by means of $\lnot$ and $\to$. Any atomic sentence is of the form $\mathbf{S}^i\mathbf{0} = \mathbf{S}^j \mathbf{0}$ or $\mathbf{S}^i\mathbf{0} \mathop{\mathbf{<}} \mathbf{S}^j \mathbf{0}$.
It can be easily proven that $A_L \models \mathbf{S}^i\mathbf{0} = \mathbf{S}^j \mathbf{0}$ if $i = j$, otherwise $A_L \models \lnot (\mathbf{S}^i\mathbf{0} = \mathbf{S}^j \mathbf{0})$ ($*$); similarly, $A_L \models \mathbf{S}^i\mathbf{0} \mathop{\mathbf{<}} \mathbf{S}^j \mathbf{0}$ if $i < j$, otherwise $A_L \models \lnot (\mathbf{S}^i\mathbf{0} \mathop{\mathbf{<}} \mathbf{S}^j \mathbf{0})$ ($**$). Thus, in $A_L$, every atomic sentence can be proved or refuted (i.e. its negation can be proved), so every quantifier-free sentence $\tau$ can be proved or refuted, and hence by $(1)$ every sentence $\sigma$ can be proved or refuted. Therefore, $A_L$ is complete.

($*$) Indeed, Enderton shows (p. 194) that $A_L \models S1$ and $A_L \models S2$ where
\begin{align}
(S1) &&& \forall x \ \mathbf{S}x \neq \mathbf{0} \\
(S2) &&& \forall x \forall y \, (\mathbf{S}x = \mathbf{S}y \to x = y)
\end{align}

From that, it is easy to prove, by induction on $j \in \mathbb{N}$, that $A_L \models \mathbf{S}^i\mathbf{0} = \mathbf{S}^j \mathbf{0}$ if $i = j$, otherwise $A_L \models \lnot (\mathbf{S}^i\mathbf{0} = \mathbf{S}^j \mathbf{0})$.

Base case ($j = 0$ and hence $\mathbf{S}^j\mathbf{0} $ is $\mathbf{0}$). If $i = j = 0$, then $A_L \models \mathbf{0} = \mathbf{0}$ because $=$ is reflexive. Otherwise, $i \neq j = 0$ and hence
$$A_L \models \mathbf{S} \overbrace{\mathbf{S} \dots \mathbf{S}}^{i-1 \text{ times}}\mathbf{0} \neq \mathbf{0}$$
because of $S1$, taking $x = \overbrace{\mathbf{S} \dots \mathbf{S}\mathbf{0}}^{i-1 \text{ times}}$.

Inductive case ($j>0$ and hence $\mathbf{S}^j\mathbf{0}$ is $ \mathbf{S}\overbrace{\mathbf{S} \dots \mathbf{S}}^{j-1 \text{ times}}\mathbf{0}$).
If $i = 0 \neq j$, then $\mathbf{S}^i\mathbf{0}$ is $ \mathbf{0}$ and hence
$$A_L \models \mathbf{0} \neq \mathbf{S} \overbrace{\mathbf{S} \dots \mathbf{S}}^{j-1 \text{ times}}\mathbf{0}$$
because of $S1$, taking $x = \overbrace{\mathbf{S} \dots \mathbf{S}\mathbf{0}}^{j-1 \text{ times}}$, and commutativity of $=$.
Otherwise, $i > 0$ and hence $\mathbf{S}^i\mathbf{0} $ is $ \mathbf{S}\overbrace{\mathbf{S} \dots \mathbf{S}}^{i-1 \text{ times}}\mathbf{0}$. There are two sub-cases:

If $ i = j$, then $\mathbf{S}^i \mathbf{0}$ and $ \mathbf{S}^j \mathbf{0}$ are the same term, and thus $A_L \models \mathbf{S}^i \mathbf{0} = \mathbf{S}^j \mathbf{0}$ by reflexivity of $=$.

If $i \neq j$, then $\mathbf{S}^{i-1} \mathbf{0}$ and $\mathbf{S}^{j-1} \mathbf{0}$ are distinct terms, and $A_L \models \mathbf{S}^{i-1} \mathbf{0} \neq \mathbf{S}^{j-1} \mathbf{0}$ by the inductive hypothesis (since $i-1 \neq j-1$). Therefore,
$$A_L \models \mathbf{S}\mathbf{S}^{i-1} \mathbf{0} \neq \mathbf{S}\mathbf{S}^{j-1} \mathbf{0}$$
according to (the contrapositive of) $S2$, taking $x = \mathbf{S}^{i-1} \mathbf{0}$ and $y = \mathbf{S}^{j-1} \mathbf{0}$.

The proof of ($**$) is analogous to the one for ($*$). You have to use properties similar to $S1$ and $S2$, concerning $\mathbf{<}$ instead of $\mathbf{S}$.

Hint. Axiom $L2$ seems very similar to $S1$.


问:
Quick way to solve computational congruences
The specific problem at hand is $$34x \equiv 60 \bmod{98}$$
I reduced to get $$17x \equiv 30 \bmod{49}$$
and from this I have
$$17x \equiv 30 \bmod{7}$$
which is easy to solve and yields $x \equiv 3 \bmod{7}$. How can I use this fact to help me solve the more complex equation? I know the solution will be of the form $7k + 3$, but how can I utilize this fact?</p>

答:
You can use your fact directly: plug in $7k+3$ for $x$ and continue solving!

More precisely, you can take $x=7k+3$ as the definition of the new variable $k$, and your calculation shows that $k$ is an integer. So you can continue solving congruences to determine the complete solution for $k$, which in turn gives you the complete solution for $x$.

ContentLicense: CC BY-SA 3.0


问:
What is the connection between the norm of a sequence in $l^2(N)$ and the norm of a function in $L^2(X)$? Little $l$ and big $L$ spaces
If I have a a function in $l^2(N)$, can they converge to a function in $L^2$ of some set $X$ subset of R? I am having a little trouble connecting functions of $l^2$ with finite support with functions in $L^2$?

答:
Let $f \in \ell^2(\Bbb N)$, consider it as a function $f:\Bbb N \to \Bbb C$. We can define an extension $F$ of $f$ to $\Bbb R$ as follows: set $F(x)=f(\lfloor x\rfloor)$ if $x\geq 1$ and $F(x)=0$ if $x<1$. Then $F \in L^2(\Bbb R)$ and we have $\|F\|_{L^2}=\|f\|_{\ell^2}$.
The function $\ell^2(\Bbb N) \to L^2(\Bbb R), f \mapsto F$ is an isometric embedding.


问:
$ P (| X |> 1) = P (| X | <1) $
Let $X$ be a random variable following a uniform distribution over the interval $[-a, a]$ where $a$ is a constant greater than 1.

If we have $ P (| X |> 1) = P (| X | <1) $, then a must be equal to a value that is in the interval:

$[0, 1.5[$

$[1.5, 2[$

$[2, 2.5[$

$[2.5, 3[$

None of the preceding intervals contains $a$.

I have computed that $a=1$, but I think my answer is wrong. I think the answer is rather the choice 5. Am I wrong?

答:
We can immediately dismiss the possibility that $a = 1$, because if that were true, then $X \in [-1,1]$, from which it would follow that $\Pr[|X| > 1] = 0$.

Let's translate this into something a bit more intuitive.  $X$ is a number chosen at random between $-a$ and $a$, inclusive, and all values in this range are equally likely to be chosen.

If $a$ is less than $1$, then its impossible to get a value of $X$ whose absolute value will be greater than $1$, because the resulting interval is too small.

If $a$ is very large, say $1000$, then the outcome $|X| < 1$ is extremely unlikely because the range of possible values is $[-1000, 1000]$ of which $[-1,1]$ is only a tiny part.

Another way to understand the question is to think of a number line.  Since we know that $a > 1$, we have the points $-a, -1, 0, 1, a$ in order from left to right.  From $-a$ to $-1$, color the interval red.  From $-1$ to $1$, color the interval blue.  And from $1$ to $a$, color the interval red.  The condition $\Pr[|X| < 1] = \Pr[|X| > 1]$ means that the total length of the red intervals must equal the length of the blue interval.  What would $a$ need to be to make this true?

Title: $ P (| X |> 1) = P (| X | <1) $


问:
Let $m, n ∈ \mathbb{N}$ such that $2m^{2} + m = 2n^{2} + n$ , then prove that $m-n$ is a perfect square .
After simply factorizing the equation given in the question I got $\mathbf{m + n = -0.5}$
. But the question mentioned $m, n ∈ \mathbb{N}$ . Then how $m + n = -0.5$ ?

Did I do some mistake or is the question wrong?

答:
Special case of that below, namely $\, d=2,\ f = 2x^2\!+\!x$.

Theorem $\ $ If $\,f(x)\in\Bbb Z[x]\,$ and $\!\bmod d\!:\ f\equiv bx\!+\!c,\, \color{#c00}{b\not\equiv 0}\,$ then $\,f(m)=f(n)\Rightarrow m = n$

Proof $\,\ 0 = f(m)\!-\!f(n) = (m\!-\!n)\:\!k,\,$ for $\,k = \dfrac{f(m)-f(n)}{m-n}\in\Bbb Z,\,$ thus if $\,m\neq n\,$ then $\,k = 0\,$ thus $\bmod d\!:\,\ \color{#c00}0 = k \equiv \dfrac{bm\!+\!c-(bn\!+\!c)}{m-n}\equiv \color{#c00}b,\,$ contra hypothesis.


问:
Why $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k}) = \textbf{CAlg}$?
Let $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k})$ be the category of monoids in the monoidal category of algebras $\textbf{Alg}$, with product $\otimes_{\mathbb{k}}$ and unit element $\mathbb{k}$. I want to show that $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k}) = \textbf{CAlg}$, but the thing I obtained is $\textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k}) = \textbf{Alg}$.

This is because $X \in \textbf{Mon}(\textbf{Alg},\otimes_{\mathbb{k}},\mathbb{k})$ is an algebra with an associative multiplication $X \otimes X \rightarrow X$ and a unit $u: \mathbb{k} \rightarrow X$. So, the multiplication is only associative, not commutative.

How can it turn out to be commutative in order to obtain $X \in \textbf{CAlg}$?

答:
The multiplication $\mu : X \otimes X \rightarrow X$ is also an algebra homomorphism (since it lives in $\mathbf{Alg}$), not just a linear map as in the usual case, so satisfies the relation $\mu (x*x' \otimes y*y' ) = \mu (x \otimes y) * \mu (x' \otimes y')$ where $*$ denotes the algebra product on $X$. This is not true in general if $\mu = *$ but only true if the algebra is commutative. Just set $x = y' = 1$ so we get $x'*y = y*x'$.

The tricky part is proving that $\mu = *$ which you can do with the Eckmann Hilton argument.


问:
Probability of a letter sequence from a sequence of $n$ random letters
What is the probability of this specific sequence in a sequence of $n$ letters -

$A$, then any three letters, then $B$ and then $C$.

My idea: first treat this sequence of 6 bytes as one block. This could occur in $n-6$ places in n letters.
Then, amongst the 6 letters, the number of ways in which three of them are any letters and three are $A, B$ and $C$  specifically is $26^3$.
Then, we want a specific ordering for 3  of these 6 letters. The three bytes in the middle could be in any order, so the number of ways in which to order these 3 occurring  letters is 3!. The other 3 letters $A, B$ and $C$ occur in a specific order. Namely, $A, x , y , z, B, C$.

The total number of combinations of $n$ letters is $26^n$.

Therefore, this probability should be :

$$  \frac{(n-6)(26^3) 3!}{26^n}     $$

Is this correct?

Edit:

@NickPeterson pointed out that $n=6$ will give me a probability of 0.
Thus, I correct my formula to:

$$  \frac{(n-5)(26^3) 3!}{26^n}     $$

答:
This is not an answer.  I am posting this long-winded comment here only for legibility.

Clearly the answer will take the form $$\frac{T}{(26)^n}$$ where $T$ represents the
total # of ways that there is at least one occurrence of the satisfying sequence "A---BC" anywhere in the $n$-character string.

It is dead wrong to compute $T$ as $$(n-5) \times (26)^3.$$

This is because the occurrences of the satisfying sequence "A---BC" in one of the $(n-5)$ slots is not a mutually exclusive event.

This means (for example) that you could have partial strings that look like either "A-A-BCBC" (two occurrences) or "A---BCA---BC" (two occurrences).

One approach is to let $f(k)$ denote the # of ways that exactly $k$ occurrences of the sequence

"A---BC" can occur.

The idea would then be to compute
$$T = f(1) + f(2) + f(3) + \cdots$$

As has been indicated in the comments, this is very difficult, because the assumption (for example) that the first 6 characters are "A---BC" affects whether another "A---BC" sequence can start in the first 6 characters.

Another approach is to count the # of ways that you can have an $n$-character string where the sequence "A---BC" does not occur.

If I was writing a computer program, I would spin through all 26 characters at each of the $n$ positions.  Then I would discount those situations where the character is a "C", the previous character (if any) is a "B", and the 5th character back (if any) is an "A".

In my opinion, this type of algorithm does no good re manually computing $T$ as a function of $n$.

It is tempting to reason that there are $\binom{n-5}{k}$ ways of distributing k "A"'s among the $(n-5)$ slots.  However, this merely disguises the problem that was broached in the first approach given in this response.  Some of the distributions of k "A"'s will interfere with each other.

I emphasize: this response is not an answer.  The only point to this response is to clarify the challenges in this problem.


问:
Expected value of function of negative binomial
With $X$ representing the total number of trials, and m the fixed number of successes. The pdf is then

$f(x|p)=$${x-1}\choose{m-1}$$p^m (1-p)^{x-m} \ \ \ \ x \ge m$

As a step in something else I'm trying to find $E[\frac{1}{X-1}]$.

Here's what I've got up to now:

$E[\frac{1}{X-1}]=\sum_{x=m}^{\infty}\frac{1}{x-1}$${x-1}\choose{m-1}$$p^m (1-p)^{x-m} \ \ \ \  \ \  \ \ $   #lotus

$E[\frac{1}{X-1}]=(1-p)\sum_{x=m}^{\infty}(x-m)$${x-2}\choose{m-1}$$p^m (1-p)^{x-1-m} \ \ \ \  \ \  \ \ $ #algebra

$E[\frac{1}{X-1}]=(1-p)\sum_{y=m-1}^{\infty}(y+1-m)$${y-1}\choose{m-1}$$p^m (1-p)^{y-m} \ \ \ \  \ \  \ \ $ #let y = x -1

Note the first term of the sum is zero to change the starting index.

$E[\frac{1}{X-1}]=(1-p) \left[ \sum_{y=m}^{\infty}y P(y|p)+ (1-m)\sum_{x=m}^{\infty} P(y|p) \right] \ \ \ \  \ \  \ \ $ #splitting up the sum

$E[\frac{1}{X-1}]=(1-p)\left[ \frac{m}{p} + (1-m) \right] \ \ \ \  \ \  \ \ $ #left = expectation of negative binomial, right = pdf sums to one

Is this correct? Is there a neater way to do it?

Body:
$$
\begin{align}
&\sum_{x=m}^\infty\frac1{x-1}\binom{x-1}{m-1}p^m(1-p)^{x-m}\\
&=\sum_{x=m}^\infty\frac1{m-1}\binom{x-2}{m-2}p^m(1-p)^{x-m}\tag{1}\\
&=\frac{p^m}{m-1}\sum_{x=m}^\infty\binom{x-2}{x-m}(1-p)^{x-m}\tag{2}\\
&=\frac{p^m}{m-1}\sum_{x=m}^\infty\binom{-m+1}{x-m}(-1)^{x-m}(1-p)^{x-m}\tag{3}\\
&=\frac{p^m}{m-1}\sum_{x=0}^\infty\binom{-m+1}{x}(-1)^x(1-p)^x\tag{4}\\
&=\frac{p^m}{m-1}p^{-m+1}\tag{5}\\[6pt]
&=\frac{p}{m-1}\tag{6}
\end{align}
$$
$(1): \binom{x-1}{m-1}=\frac{x-1}{m-1}\binom{x-2}{m-2}$

$(2): \binom{x-2}{m-2}=\binom{x-2}{x-m}$ and bring constants out front

$(3): \binom{x-2}{x-m}=\binom{-m+1}{x-m}(-1)^{x-m}$

$(4): x\mapsto x+m$

$(5):$ binomial theorem

$(6):$ algebra


问:
How to show $\operatorname{ann}(M) = \operatorname{ann}(X)$.
If $R$ is left Artinian, $X$ a subset of a left $R$-module $M$, define $\operatorname{ann}(X) = \{a \in R: ax = 0, \forall x\in X \}.$

(a) Show $\operatorname{ann}(M) = \operatorname{ann}(X)$ for some finite subset $X \subset M$

(b) If $\operatorname{ann}(M) = 0$ show $R$ is isomorphic to a submodule of $M$.

So let me tell you what I thought first.

(a) I think that anything that annhilates $m \in M$, certain annhilates $x \in X$. But since there may be more elements annihilating $M$ (because $M$ is the larger set containing $X$), we should get $\operatorname{ann}(X) \subset \operatorname{ann}(M).$ For the other inclusion, we need to construct a left descending series to use the Artinian property.

(b) I am almost certain that this is just done by the multiplication map $R \to M$ given by $r \to rm$, then $R/\ker = R/\operatorname{ann} M = R/0 = M$

The above is the answer to part (a), but I just don't quite understand it.

答:
The claim of part $(b)$ of the exercise is:

If $R$ is an artinian ring and $M$ is a finitely generated left $R$-module such that $\text{ann}(M)=0$, then $M$ has a submodule isomorphic to $R$.

However the claim, as stated, is false.

Before presenting a counterexample, we state and prove a lemma.

lemma:

If $R$ is an artinian ring, then no proper left ideal of $R$ is isomorphic, as a left $R$-module, to $R$.

Proof of the lemma:

Let $R$ be an artinian ring and suppose $A$ is a left ideal of $R$ which is isomorphic, as a left $R$-module, to $R$.

Our goal is to derive a contradiction.

Let $A_0=A$.

Since $A_0$ is isomorphic, as a left $R$-module, to $R$, it follows that $A_0$ has a proper $R$-submodule, $A_1$ say, which is isomorphic, as a left $R$-module, to $R$.

Note that $A_1$ is also a left ideal of $R$.

Iterating the process, we get a strictly descending infinite chain
$$
A_0\supset A_1\supset A_2\supset\cdots
$$
of left ideals of $R$, contradiction, since $R$ is an artinian ring.

This completes the proof of the lemma.

The counterexample:

Let $R=M_n(K)$ where $K$ is a field and $n\ge 2$.

Regarding $R$ as a vector subspace over $K$, $R$ is finite-dimensional (more precisely, it has dimension $n^2$).

Since every left or right ideal of $R$ is closed under multiplication by elements of $K$, it follows that every left or right ideal of $R$ is vector subspace of $R$.

Since the subspaces of a finite-dimensional vector space satisfy the descending chain condition, it follows that $R$ is an artinian ring.

For $1\le i\le n$, let $a[i]\in R$ be the $n{\times}n$ matrix with all entries in the $i$-th row equal to $1$ and all other entries equal to $0$.

Let $A$ be the left ideal generated by $a[1]$.

Thus, regarding $A$ as an $R$-module, $A$ is finitely generated.

Note that $\det(a[1])=0$.

If $a\in A$, then $a=ra[1]$ for some $r\in R$, hence $\det(a)=\det(r)\det(a[1])=0$.

Thus all elements of $A$ are singular, so the inclusion $A\subset R$ is proper.

Applying the lemma, no $R$-submodule of $A$ is isomorphic to $R$.

It remains to show $\text{ann}(A)=0$.

Let $r\in\text{ann}(A)$.

For $1\le i,j\le n$, let $e[i,j]\in R$ be the $n{\times}n$ matrix with $(i,j)$-th entry equal to $1$ and all other entries equal to $0$.

By definition of $A$, we have $a[1]\in A$.

For $2\le i\le n$, we have $a[i]=e[i,1]a[1]$.

Thus we have $a_1,...,a_n\in A$.

Then from $rA=0$ we get $ra[i]=0$ for all $i\in\{1,...,n\}$.

From $ra[i]=0$ it follows that the $i$-th column of $r$ is zero.

But then all columns of $r$ are zero, so $r=0$.

Thus we have $\text{ann}(A)=0$, so $A$ qualifies as a counterexample to the claim.


问:
Expressing "does not imply''
In ordinary discourse, when we say that "$A$ implies $B$", we shall formalize it by writing the following:
$$A\rightarrow B$$
But when we say that "$A$ does not imply $B$", we cannot formalize it as the following:
$$\neg(A\rightarrow B)$$
Because by material implication, we have the following equivalences for the second formula:
$$\neg(A\rightarrow B)\Longleftrightarrow\neg(\neg A\vee B)\Longleftrightarrow A\wedge\neg B$$
But the ordinary meaning of the sentence "$A$ does not imply $B$" is that $B$ does not follow from $A$: if $A$ is true, $B$ is either false or undecidable. I wonder how "$A$ does not imply $B$" is formally expressed in the object language (not at the meta-level). Thanks!

答:
There is a way to express it in the object language, using modal operators. "$A$ implies $B$" in English often means "$⬜(A⇒B)$", which can be interpreted as "we can justify that $A⇒B$". So "$A$ does not imply $B$" in natural language often means "$¬⬜(A⇒B)$". Note that $¬⬜(A⇒B)$ does not imply $⬜¬(A⇒B)$, because it may be that both $¬⬜(A⇒B)$ and $¬⬜¬(A⇒B)$ hold.

For example, $x,y∈ℝ$ and $x≤y$ do not imply $x<y$, equivalently $¬⬜( x,y∈ℝ∧x≤y ⇒ x<y )$. And this is not equivalent to $⬜¬( x,y∈ℝ∧x≤y ⇒ x<y )$, which can be interpreted as "we can justify that $x,y∈ℝ$ and $x≤y$ and $x≥y$" (which is clearly nonsense).

There is in fact a connection between this formalization of English "implies" and meta-logic: This "$⬜$" can be taken as the provability operator from provability logic (where "$⬜$" is given a higher precedence than other logical operators), and so "$⬜(A⇒B)$" does in fact express "$⊢A⇒B$" internally, and is stronger than "$⬜A⇒⬜B$" (which expresses "$(⊢A)⇒(⊢B)$" internally. A distinguishing example is that PA (Peano Arithmetic) proves $⬜(¬⬜{⊥})⇒⬜{⊥}$ but (we strongly believe) does not prove $⬜(¬⬜{⊥}⇒⊥)$ (where "$⬜Q$" can be expressed in PA using Godel coding).

Title: Expressing "does not imply''


问:
Ignoring exponential terms in asymptotic matching of two point boundary value ODE
So I'm not sure how much background I need to give to set up this question. But in my lecture notes I have that

$e^{-\eta / \epsilon^{1-\alpha}}$

can be ignored where $\epsilon << 1$ and $0 <\alpha <1$ and $\eta$ is fixed. In particular that:

$\lim_{\epsilon \rightarrow 0} exp(-\eta/ \epsilon^{1-\alpha})/\epsilon^k = 0$ for all $k \ge 0$

I tried playing around with this limit with L'Hopitals but I wasn't really sure how to show it rigorously. Applying L'Hopitals it seemed like the limit is equal to

$\lim_{\epsilon \rightarrow 0} \eta(1-\alpha)exp(-\eta/ \epsilon^{1-\alpha})/k\epsilon^{k-1 + (1-\alpha)^2} \le \lim_{\epsilon \rightarrow 0} \eta(1-\alpha)exp(-\eta/ \epsilon^{1-\alpha})/k\epsilon^{k-1}$

So then it seems like when you keep applying L'hopitals, the constants change but you will keep getting the power of the denominator to decrease by 1, so it seems like it will go to zero. But I'm not sure how to make this rigorous. Is this argument even the right track? How would you show this limit? Thanks for your time!

答:
One way is to take the logarithm:
$$-\frac{\eta}{\epsilon^{1-\alpha}} - k\log \epsilon$$
Here $\log\epsilon$ is much smaller than the first term, so the expression tends to $-\infty$. One way to show it from scratch is to consider
$$f(\epsilon)=-\frac12\frac{\eta}{\epsilon^{1-\alpha}} - k\log \epsilon$$
observe that $f'(\epsilon)>0$ for all small $\epsilon$, and conclude that $f$ is bounded from above in a neighborhood of zero. Say: $f\le M$, then
$$-\frac{\eta}{\epsilon^{1-\alpha}} - k\log \epsilon\le -\frac12\frac{\eta}{\epsilon^{1-\alpha}} +M $$
which tends to $-\infty$ as $\epsilon\to 0$.</p>


问:
Expand multilinear functions
From the definition, we know that the multilinear function behaves linearly on one variable when all other variables are fixed. For example, $f(x_1, x_2+x_3, x_4)=f(x_1, x_2, x_4)+f(x_1, x_3, x_4)$.

However, if we know that a function is multilinear and there are more than 1 variables in the calculation, how should we deal with it? For example, if we have $f(x_1, x_2, x_4)+f(x_1, x_4, x_2)$, how should we calculate this as we see both the second and the third are variables? Does this equal $f(x_1 ,x_2+x_4, x_2+x_4)$?

Title: Expand multilinear functions

Body:
Note that
\begin{align}
f(x_1 ,x_2+x_4, x_2+x_4)
&= f(x_1,x_2,x_2+x_4) + f(x_1,x_4,x_2+x_4) \\
&= f(x_1,x_2,x_2) + f(x_1,x_2,x_4) \\
&\qquad +  f(x_1,x_4,x_2) + f(x_1,x_4,x_4)
\end{align}
which, in general, is not the same as $f(x_1, x_2, x_4)+f(x_1, x_4, x_2)$.


问:
Angle between vectors $a$ and $b$
Given $a = (4,5,-3)$ and $b = -i + 2 j + k$.

I am asked to measure the angle between $a - b$ and $-j - k$.
I know how to get the $a-b$ angle but I don't understand what to do for the $-j-k$ angle.

Am I supposed to take just  $a= (5,-3)$ and $b=(2,1)$ and use those?

答:
The first vector is $\vec x = \vec a - \vec b = (5,3,-4)$.

The second vector $\vec y =  -\textbf{j} - \textbf{k} = 0\textbf{i} -1\textbf{j} - 1\textbf{k}=(0,-1,-1)$

Use the dot product. $\vec x \cdot \vec y = |\vec x||\vec y| \cos \theta$, where $\theta$ is the angle you need to find.

This gives you $(5)(0)+ (3)(-1) +(-4)(-1) = \sqrt{5^2 + 3^2 + (-4)^2} \sqrt{0^2 + (-1)^2 + (-1)^2} \cos \theta$ which yields $\theta = \arccos \frac 1{10} \approx 84.26^{\circ}$

Title: Angle between vectors $a$ and $b$


问:
Recurrence relation to find number of ways to cover $2×n$ chessboard with red and blue $2 \times 1$ tiles.
Let an be the number of ways to cover a 2 x n chessboard using 2 x 1 red tiles and 2 x 1 blue tiles. Find a formula for an. Find a recurrence relation for an, with initial conditions.

Solution: given 2 x 1 red and blue tiles

If n =1 (2 x 1) then there are 2 ways to cover the chessboard. 1st: 1 red tile, 2nd 1 blue tile --> a1 = 2

If n = 2 (2 x 2) then there are {r, r}, {b, b}, {r, b}, {b, r} --> a2 = 4

If n = 3 (2 x 3) we have:

For red:

{r, r, r}, {r, b, r}, {r, r, b}, {b, r, r}

For blue:

{b, b, b}, {b, r, b}, {b, b, r}, {r, b, b}

--> a3 = 8

If n = 4 ( 2 x 4) then

For red:

{r, r, r, r}, {r, b, r, r}, {r, r, b, r} {r, r, r, b}, {b, r, r, r}, {r, b, b, r}, {r, b, r, b},

For blue:

{b, b, b, b},{b, r, b, b}, {b, b, r, b} {b, b, b, r}, {r, b, b, b}, {b, r, r, b}, {b, r, b, r}.

--> a4 = 14

At this point, I was unsure of my red and blue order. For example is {b, b, r} and {r ,b, b} the same thing? Am I overcounting the number of solutions for each an? Please help

答:
Let $b_n$ be the number of tilings of $2 \times n$ with uncolored dominoes.  Then
\begin{align}
b_1&=1\\
b_2&=2\\
b_n&=b_{n-1}+b_{n-2} &&\text{for $n\ge 3$}
\end{align}
which you can derive by conditioning on the last column.  Note that $b_n$ satisfies the same recurrence as the Fibonacci numbers.
Now $a_n = 2^n b_n$ because you have two choices for the color of each domino.  So
\begin{align}
a_1&=2 b_1 = 2\\
a_2&=4 b_2 = 8\\
a_n&=2 a_{n-1} + 4 a_{n-2} &&\text{for $n\ge 3$}
\end{align}


问:
Product of nonzero elements of Z/pZ
I was wondering if someone could guide me regarding the following question: "If p is congruent to 1 modulo 4, we may proceed as follows. Recall that Z/pZ is a field. Let
L = 1 × 2 × · · · × (p − 2) × (p − 1),
viewed as product of all the nonzero elements of Z/pZ. Explain why, as an element of Z/pZ,
L = (−1)^(p−1/2) × (1^2) × (2^2) × · · · × ((p − 1)/2)^2"

I understand that p congruent to 1 mod 4 implies (p-1)/2 is even, and that p-1 is congruent to -1 mod p but I'm not sure how to progress from here. Could someone give me some advice on how to get started? Thanks!

答: The inversion map $\iota:t\mapsto t^{-1}$ defines a permutation of the units group $\Bbb F_p^*$ of the field $$\Bbb F_p:=\frac{\Bbb Z}{p\Bbb Z}=\{\bar0,\bar1,\dots,\overline{p-1}\}$$ Except for the $\iota$ orbit $\{-\bar1\}$, the elements of an $\iota$ orbit together have a product of $\bar1$. $$\therefore\;\prod_{t\in\Bbb F_p^*}t=\prod_{X\in\mathcal O(\iota)}\prod_{t\in X}t=\prod_{t\in\{-\bar1\}}t=-\bar1$$ where the $\Bbb F_p^*$ partition $\mathcal O(\iota)$ is the family of all $\iota$ orbits.

Title: Product of nonzero elements of Z/pZ


问:
Expected value of $|H-T|$ in $n$ coin flips
Let $H_n$ be the number of heads in $n$ coin flips.

Let $T_n$ be the number of tails in $n$ coins flips.

Is there a good way to calculate $E_n = E[|H-T|]$ that isn't brute force computation, i.e. directly evaluating $$E[|H-T|] = \frac{1}{2^n} \sum {n \choose r} |n-2r|. $$

I don't see any approach that can simplify this calculation.

答:
$$\sum_r \dbinom{n}r \vert n-2r \vert = \sum_{r < n/2} \dbinom{n}r (n-2r) + \sum_{r>n/2} \dbinom{n}r (2r-n) = 2\sum_{r>n/2} \dbinom{n}r r - \sum_{r<n/2}\dbinom{n}r r$$
I trust you can take it from here. You may need to use the identities
$$r \binom{n}r = n\dbinom{n-1}{r-1}$$
$$\sum_{r=0}^{k-1} \dbinom{2k}r = \dfrac{4^k - \dbinom{2k}k}2$$
$$\sum_{r=0}^{k} \dbinom{2k+1}r = 2^{2k}$$</p>


Title: If $B=\sqrt[3]{x\sqrt[5]{x^4\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$

Body:
What value does $\dfrac{B}{x}$ approach?

Source: Select Problems of "Lumbreras Editors"

If I divide I get that:

$$\dfrac{B}{x}=\sqrt[3]{\dfrac{x}{x}.\dfrac{1}{x^2}\sqrt[5]{x^4\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$$

$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{\dfrac{x^4}{x^4} \dfrac{1}{x^6}\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$$

$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{ 1\sqrt[9]{\dfrac{x^{24}}{x^{24}}\dfrac{1}{x^{30}}\sqrt[17]{x^{240}...}}}}$$

$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{ 1\sqrt[9]{1\sqrt[17]{\dfrac{x^{240}}{x^{240}}\dfrac{1}{x^{270}}...}}}}$$
Can I assume that $\dfrac{B}{x}$ is approximately 1?

Body:
$B = x^\alpha$ where
$$
\begin{align}
\alpha &=
{\small \frac13}(1 +
{\small \frac15}(4 +
{\small \frac19}(24 +
{\small \frac{1}{17}}(240+\cdots ))))\\
&= \frac13 + \frac{4}{3\cdot 5} + \frac{24}{3\cdot 5 \cdot 9} + \frac{240}{3\cdot 5 \cdot 9\cdot 17} + \cdots\\
&= \frac{2}{2\cdot 3}  + \frac{4}{3\cdot 5} + \frac{8}{5\cdot 9} + \frac{16}{9\cdot 17} + \cdots\\
&= \sum_{k=0}^\infty\frac{2^{k+1}}{(2^k+1)(2^{k+1}+1)}\\
&= \sum_{k=0}^\infty \left(\frac{2}{2^k+1} - \frac{2}{2^{k+1}+1}\right)\\
&= \frac{2}{2^0+1}\\ &= 1
\end{align}
$$
This means $\frac{B}{x} = \frac{x}{x} = 1$.


问:
Let $H$ be a subgroup of $G$ and suppose that $Ha=bH$ for $a,b$ elements of $G$. Show that $aH=Hb$.
I am struggling pretty bad with this homework problem, I've been beating my head against the wall for quite some time and still can't arrive at an answer or even know where to start.

Let $H$ be a subgroup of $G$ and suppose that $Ha=bH$ for $a,b$ elements of $G$. Show that $aH=Hb$.

I feel like I'm supposed to apply Lagrange's theorem, but I don't know how...

答:
Since $a\in Ha=bH$, we have $a=bh$ for some $h\in H$. Therefore, $$aH=(bh)H=b(hH)=bH=Ha.$$
Since $ab^{-1}=h\in H$, we have $Ha=Hb$.

Hence we conclude that $aH=Ha=Hb$.


问:
Zero KL-divergence $\Rightarrow$ same distribution?
I am looking at relative entropy=KL-divergence and trying to prove, at least in case of probability mass functions $q,\,p$ with possible states $x\in A$ that if:

$$
D(p \,\lVert\,q)=\sum_{x\in A}p\left(x\right)\,\log\left[\frac{p\left(x\right)}{q\left(x\right)}\right] = 0
$$

then $p\left(x\right)=q\left(x\right)\: \forall x \in A$.

I can see that the converse is true, and my book Cover, Thomas Elements of Information Theory does state that the above result is also valid, but the proof there is, in my opinion, incomplete. The authors use concavity of $\log$ to prove that $D(p \,\lVert\,q)>0$ (happy there). And then state, that the only way to have equality, as opposed to inequality, is to have $,\log\left[\frac{p\left(x\right)}{q\left(x\right)}\right]$ being not strictly concave. Sort of ok with that (not too happy), but then the authors state that the only way to get that is to have $\left[\frac{p\left(x\right)}{q\left(x\right)}\right]=const$. I am not sure how to make that last jump.

Can anyone help? Or suggest an alternative proof? Thanks

答:
"The only way to have equality is to have $\log \frac{p(x)}{q(x)}$ being not strictly concave" is a weird way to say something here, so you may have misread the proof, or it may be badly worded. However, the idea is to figure out the equality case of the inequality we used to prove that KL-divergence is nonnegative.

To begin with, here's a proof that KL-divergence is nonnegative, which is probably essentially the same as the one you read. Let $f(x) = -\log x = \log \frac1x$; this is a strictly convex function on the positive reals. Then by Jensen's inequality
$$
   D(p\,\|\,q) = \sum_{x \in A} p(x) f\left(\frac{q(x)}{p(x)}\right) \ge f\left(\sum_{x \in A} p(x) \frac{q(x)}{p(x)}\right) = f(1) = 0.
$$
For a strictly convex function like $f(x)$, assuming that the weights $p(x)$ are all positive, equality holds if and only if the inputs to $f$ are all equal, which directly implies $\frac{q(x)}{p(x)}$ is constant and therefore $p(x)=q(x)$ for all $x$.

We have to be careful if $p(x)=0$ for some inputs $x$. Such values of $x$ are defined to contribute nothing to the KL-divergence, so essentially we have a sum over a different set $A' = \{x \in A : p(x) > 0\}$. Then on the right-hand side of the inequality, we get
$$
    f\left(\sum_{x \in A'} p(x) \frac{q(x)}{p(x)}\right) = f\left(\sum_{x \in A'} q(x)\right).
$$
But now, if the sum over $A'$ is some probability $q<1$, then $f(q) > 0$, and we conclude $D(p\,\|\,q) > 0$. So the sum of $q(x)$ over $A'$ should be $1$. We conclude that when the KL-divergence is $0$, $p(x)=0 \implies q(x)=0$ for all $x \in A$. In order for the KL-divergence to even be defined, $q(x)=0 \implies p(x)=0$ for all $x\in A$. So we can throw away any values where $p(x)=q(x)=0$, and now we're back in the simple case, where all the weights in Jensen's inequality are positive.


问:
$MSE(θ_1) = MSE(θ_2) = 0.06$?
We offer two estimators for the average concentration $μ$ of lead in the atmosphere of a region of Quebec where factories manufacturing dyes are located. The first estimator $θ_1$ has a bias equal to 0.2 and a variance of 0.02. The second estimator $θ_2$ is unbiased and has a variance equal to 0.06.

Which of the following is true?

$θ_1$ is better than $θ_2$ to estimate $μ$.

$θ_2$ is better than $θ_1$ to estimate $μ$.

None of the above is true

I have some difficulties to find the right one. If I use the mean square error (MSE), then $MSE(θ_1) = MSE(θ_2) = 0.06$. So does the the choice 3 is the right one?

答:
The two estimators have the same MSE but $\theta_2$ is correct while $\theta_1$ is biased

For a moment forget all you studied and answer intuitively: which one do you prefer?

Title: $MSE(θ_1) = MSE(θ_2) = 0.06$?


问:
How to linearize the following optimization exercises with absolute values?
I have a problem with non-linear variables and I have to present a way for linearizing such a situation.

$\min |x_1| - |x_2-1| \quad \mathrm{s.t.} \quad 3x_1 + 2x_2 \geq 1,
    \quad x_2 \geq 1$

$\min -2x_1 + 3.|x_2-10| \quad \mathrm{s.t.} \quad |x_1+2| + |x_2|
    \leq 5$

Could you also please give me another example with absolute values (same difficulty as this ones)? - statement and solution

(ps) I don't want programming language, it's an exercise to write by hand

答:
For #1, you can replace $|x_2-1|$ with $x_2-1$ because $x_2 \ge 1$.  Now introduce $y_1$ to represent $|x_1|$ and two linear constraints $y_1 \ge x_1$ and $y_1 \ge -x_1$ to enforce $y_1 \ge |x_1|$.  Minimizing $y_1-(x_2-1)$ will then guarantee that $y_1 = |x_1|$.

For #2, introduce three new variables and six linear constraints.

For some other examples, see https://or.stackexchange.com/questions/tagged/absolute-value


答: It’s clear that any time that you place a tile horizontally, you must pair it with another horizontal tile in the same two columns. Thus, any tiling consists of some combination of vertical tiles and pairs of horizontal tiles covering a $2\times 2$ square. If we look at just one row of such a tiling, it consists of $1\times 1$ and $1\times 2$ tiles covering a $1\times n$ board. It is well known that there are $F_{n+1}$ such tilings, where $F_n$ is the $n$-th Fibonacci number, and $F_0=0$. The other row will have the same sequence of $1\times 1$ and $1\times 2$ tiles, so ignoring color there are $F_{n+1}$ tilings of the $2\times n$ board with $1\times 2$ tiles. Any tiling of the $2\times n$ board by $1\times 2$ tiles uses $n$ tiles, each of which can be either red or blue, so there are $2^n$ combinations of colors and therefore $2^nF_n$ distinguishable tilings.

Title: Recurrence relation to find number of ways to cover $2×n$ chessboard with red and blue $2 \times 1$ tiles.

Body:
Let an be the number of ways to cover a 2 x n chessboard using 2 x 1 red tiles and 2 x 1 blue tiles. Find a formula for an. Find a recurrence relation for an, with initial conditions.

Solution: given 2 x 1 red and blue tiles

If n =1 (2 x 1) then there are 2 ways to cover the chessboard. 1st: 1 red tile, 2nd 1 blue tile --> a1 = 2

If n = 2 (2 x 2) then there are {r, r}, {b, b}, {r, b}, {b, r} --> a2 = 4

If n = 3 (2 x 3) we have:

For red:

{r, r, r}, {r, b, r}, {r, r, b}, {b, r, r}

For blue:

{b, b, b}, {b, r, b}, {b, b, r}, {r, b, b}

--> a3 = 8

If n = 4 ( 2 x 4) then

For red:

{r, r, r, r}, {r, b, r, r}, {r, r, b, r} {r, r, r, b}, {b, r, r, r}, {r, b, b, r}, {r, b, r, b},

For blue:

{b, b, b, b},{b, r, b, b}, {b, b, r, b} {b, b, b, r}, {r, b, b, b}, {b, r, r, b}, {b, r, b, r}.

--> a4 = 14

At this point, I was unsure of my red and blue order. For example is {b, b, r} and {r ,b, b} the same thing? Am I overcounting the number of solutions for each an? Please help


问:
Showing that a function is a homeomorphism
I've come up with an answer to this problem that I have precisely zero faith in, so a 'yay' or 'nay' would be enormously appreciated with regards to whether I'm an idiot or not. So here it goes:

$(X,\mathcal{T})$ is a topological space, and $\mathbb{I}= \{t \in \mathbb{R} \;\; | \;\; 0 \leq t \leq 1 \}$ has the Euclidean topology. $X \times \mathbb{I}$ is given the product topology, and for $\lambda \in \mathbb{I}$, take the function
$$i_\lambda: X \longrightarrow X \times \mathbb{I}, \quad x \longmapsto (x,\lambda)$$
The image of $i_\lambda$, $\text{im}(i_\lambda)$, has the topology induced by the product topology on $X\times\mathbb{I}$. I need to show that
$$\phi_\lambda: X \longrightarrow \text{im}(i_\lambda),\quad x\longmapsto i_\lambda(x)$$
is a homeomorphism.

OK, so, firstly I decided that $\text{im}(i_\lambda) = X \times \{\lambda\}$.

For injectivity, given $a,b \in X$, $\phi_\lambda(a) = \phi_\lambda(b) \Longleftrightarrow
 (a,\lambda)=(b,\lambda) \Longleftrightarrow a=b$.

For surjectivity, given $(a,\lambda)\in X\times \{\lambda\}$, $\exists b\in X$ such that $\phi_\lambda(b) = (a,\lambda)$, namely $b=a$.

So $\phi_\lambda$ is bijective.

Now I define
$$\zeta_\lambda: X\times \{\lambda\} \longrightarrow X,\quad (x,\lambda)\longmapsto x.$$
I think that $\zeta_\lambda \circ \phi_\lambda = \text{id}_X$ and that $\phi_\lambda \circ \zeta_\lambda = \text{id}_{X\times \{\lambda\}}$, so $\zeta_\lambda = \phi_\lambda^{-1}$. Now it just remains to be shown that $\phi_\lambda$ and $\phi_\lambda^{-1}$ are continuous. To this end, I showed that the topology on $\text{im}(i_\lambda)=X\times\{\lambda\}$ is

$$\mathcal{G} = \{U\times\{\lambda\}\;|\; U \in \mathcal{T}\,\},$$
and so to show that $\phi_\lambda$ is continuous, take $A \in \mathcal{G}$, i.e. $A=U\times\{\lambda\}$ for some $U\in \mathcal{T}$. Then
$$\phi_\lambda^{-1}[A] = \phi_\lambda^{-1}[U\times\{\lambda\}]=U\in\mathcal{T}.$$
Thus $\phi_\lambda$ is continuous. Then for $\phi_\lambda^{-1}$, take $K\in\mathcal{T}$;
$$\left(\phi_\lambda^{-1}\right)^{-1}[K] = \phi_\lambda[K] = K\times\{\lambda\} \in \mathcal{G}.$$
Hence $\phi_\lambda^{-1}$ is continuous.

Therefore $\phi_\lambda$ is a homeomorphism.

Please be as brutally honest as you feel you need to be, as I have no delusions about my mathematical ability. If it makes no sense and I understand nothing, so be it. Thank you for your time.

答:
Your argument seems correct, but slightly overcomplicated.

First of all, a function is always onto its image, so an injective function is automatically a bijection with its image.

As for checking the continuity of $\phi_\lambda$ simply take an open set in $X\times\{\lambda\}$, namely $U\times\{\lambda\}$ and note that its preimage is the open set $U$ and therefore $\phi_\lambda$ is continuous.

Lastly, the product topology is the finest topology under which the projections are continuous.

Therefore $\phi_\lambda^{-1}$ is continuous by that virtue, and it would require no checking.

In case you did not learn that property of the product topology, it is still simple enough to show its continuity as the preimage of $U$ is indeed $U\times\{\lambda\}$.


问:
Orthogonality of Legendre polynomials using specific properties
I'm having significant issues with a problem and would appreciate any help at all with it. It is regarding proving the orthogonality of Legendre polynomials using a specific recursion formula and property of Legendre polynomials.

I have been given the following relation $$x P'_n(x) -P'_{n-1}(x) = n P_n (x) $$
And that $$ \int^{1}_{-1} f(x) P_n (x) dx = 0 $$ for any polynomial $f(x)$ of degree less than $n$.

I must then show that for $ n \geq  1$

$$ \int^{1}_{-1} P_n (x)^2 dx = \frac{1}{2n} \int^{1}_{-1} x \frac{d}{dx} (P_n (x)^2)dx $$

And then determine the value of the integral, that being $\frac{2}{2n+1}$. I must use the two stated facts, I'd be grateful for any help with this, thank you guys.

答:
I immediately thought of integration by parts. Here is my work:

By the product rule of derivative we have
$$ \frac{d}{dx} x\, P_n(x)^2 = P_n(x)^2 + x\, \frac{d}{dx}(P_n(x)^2)
\tag{1} $$

Note that Legendre polynomials have the property
$$ P_n(1) = 1,\quad \text{ and } \quad P_n(-1)=(-1)^n. \tag{2} $$

By integrating equation $(1)$ and using equation $(2)$ we get
$$ 2 = 1 - (-1) = \int_{-1}^1 P_n(x)^2 dx + \int_{-1}^1
  x\, \frac{d}{dx}(P_n(x)^2) dx. \tag{3} $$

Define $\, a_n := \int_{-1}^1 P_n(x)^2 dx.\,$ Equation $(3)$
implies that
$$ 2 - a_n = \int_{-1}^1   x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{4} $$
We are given the property that
$$ x P'_n(x) -P'_{n-1}(x) = n P_n (x). \tag{5} $$
Multiply this equation $(5)$ by $\,P_n(x)\,$ and integrate
to get
$$  \int_{-1}^1\!P_n(x)(xP'_n(x)\!-\!P'_{n-1}(x))dx
\!=\!\int_{-1}^1\!P_n(x)nP_n(x)dx. \tag{6} $$
We are also given the property that
$$ \int_{-1}^1 P_n(x)\,f(x)\,dx = 0 \tag{7} $$
for any polynomial $\,f(x)\,$ of degree less than $\,n.\,$
Now notice that
$$ x\, \frac{d}{dx}(P_n(x)^2) = 2\,x\,P_n(x)P'_n(x). \tag{8} $$
Combining equations $(6)$, $(7)$, and $(8)$ we get that
$$ \frac12(2-a_n) = n\,a_n. \tag{9}$$
Dividing this equation $(9)$ through by $\,n\,$ and using
the definition of $\,a_n\,$ and equation $(4)$ gives the result we wanted:
$$ \int_{-1}^1 P_n(x)^2 dx = \frac{2-a_n}{2n} = \frac1{2n}
\int_{-1}^1   x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{10}$$
Also, solving for $\,a_n\,$ in equation $(9)$ gives
$$ a_n = \frac2{2n+1} \tag{11}$$
which was also requested.

NOTE: The property in equation $(2)$ was not given to
us. It would have to be proven using equations $(5)$
and $(7)$ which are given to us. It can be done, and
would take some effort, but I have not given the
steps required.


问:
How binary operation defined as $\operatorname{gcd}(a,n)=1$ is same as multiplication modulo $n$?

How binary operation defined as $\operatorname{gcd}(a,n)=1$ is same as multiplication modulo $n$?

I have recently started learning group theory on my own. There is a problem that the set of all numbers coprime to $n$ is a group.

To prove this,

$$(a,n)=1 = (b,n) \implies (ab,n) =1$$

If $(ab, n) = d > 1$ then there exists some prime $p$ that divides $d$ and so $p$ divides $a$ or $b$ and $p > 1$. This leads to contradiction and so closure is satisfied

$(1, n) = 1$ and so $1$ is identity element in the set of all numbers coprime to $n$.

But in my textbook while proving inverse, the author suddenly takes binary operation defined by $\operatorname{gcd}$ w.r.t. $n=1$ as multiplicative modulo $n$. How is it possible?

I have following doubts:

If we have a number "$a$" coprime to $n$ then how to show that there exists another number $b$ coprime to $n$ such that a (some binary operation defined by $\operatorname{gcd}$ w.r.t. $n$) $b=$ unity $=1$

How is this binary operation is nothing but multiplication modulo $n$

Kindly clarify.

答:
Q2. Let $S=\{x:1\leq x<n,(x,n)=1\}$ and $\cdot$ is the multiplication modulo $n$. Then $(S,\cdot)$ form a group. Note that the elements in $S$ are the integers less than $n$ which are coprime to $n$. However, I think you misunderstood the operation defined by author. Please check again your reference book.

For example, take $n=8$. Then $S=\{1,3,5,7\}$. In $(S,\cdot),$ we have \begin{align}3\cdot5=7,\\
3\cdot7=5,\\
5\cdot7=3.\end{align}
You can check that the operation is the usual multiplication modulo $8$.

Q1. Let $a\in S$. By how we define $S$, we have $(a,n)=1$. By Bezout Identity, there exist $x,y\in \Bbb{Z}$ such that $$ax+ny=1.$$
Let $b$ be the remainder of $x$ when divided by $n$.

Then we have $ab\equiv ax\equiv 1 (\bmod n)$. In other words, $b\in S$ such that that $a\cdot b=1$ in $(S,\cdot)$.


问:
Continuity of $\int_{x+V} fdm$
Assume $V\subset \mathbb {R}^n$, an open set. $f\in L^1(\mathbb {R}^n)$ and non-negative.

QUESTION: Consider function: $x\mapsto \int_{x+V} fdm (x\in \mathbb {R}^n)$, is it continuous? If not, is it semi-continuous?

I think the answer is on the tip of my tongue, but I cannot state any in a formal math way. Any help will be appreciated!

答:
It is continuous if $V$ is bounded (I am not sure of unbounded $V$). This follows from what is called absolute continuity of integral which says that if $f$ is an $L^1$ function then for any $\epsilon >0$ there exists a $\delta >0$ such that for ANY measurable set $E$ with measure less than $\delta$, $\int_E |f| < \epsilon$. This is not at all an obvious result.

Now in your situation the point is that $x+V$ and $y+V$ differ by a small set provided $x$ and $y$ are close enough. Well, on a second thought, maybe this is not too obvious. The difference seems to do with the boundary of $V$ which can be crazy :(

Title: Continuity of $\int_{x+V} fdm$


问:
swapping limit and series
Consider i.i.d non negative integer valued random variables $Z_t$ with $Y_t= \sum_{k=1}^t Z_k$
and the partial sum
$$s_n:=\sum\limits_{j=c}^{\infty} P(Y_{j+1-c}=j) F^{n+c-j} 1_{\{j \leq c+n\}}$$ with

$$F^{n+c-j}:= E\left[\left(1- \frac{Y_{n+c-j}}{n+c-j}\right)\cdot 1_{\{Y_{n+c-j} \leq n+c-j \}}\right]  $$

I know $$\lim\limits_{n \rightarrow \infty} F^{n+c-j}=1-z$$ for $z<1$ a.s.

I want to prove: $$\lim\limits_{n \rightarrow \infty} s_n = (1-z) \sum\limits_{j=c}^{\infty} P(Y_{j+1-c}=j)$$

答:
Normally, Dominated Convergence Theorem would work fine in your case. Thing is I'm still concerned about the convergence of your limit object, especially the convergence of the series
$$ \sum_{j=c}^{\infty} \mathbb{P}( Y_{m+1-c} =m ) $$
(Though the presence of that convergence is not important to for our limit proof)

Let :
$$G_j(n) = F^{n+c-j} 1_{ j \le n+c}$$

According to the definition $G_m$,  $0 \le G_j \le 1$ and $\lim_m G_m= (1-z) $ pointwise.

Also, for the sake of clarity, we define another object $\mu$ which is a measure on $\mathbb{N}$ such that:
$$\mu(\{m\}) = \mathbb{P}( Y_{j+1-c} =j ) \text{ when }m \ge c$$
So :
$$ s_n= \int G_m d\mu$$
and
$$  \text{your limit object is }\int \lim G_m \mu $$

Now, let's treat your problem in different cases:

If $\mu( \mathbb{N})<+\infty$, then $(1-z) \in L^1(\mu)$ hence by Dominated Convergence Theorem, we imply our desired result.

If $\mu( \mathbb{N})= +\infty$, then by Fatou we have:
$$\liminf s_n \ge \int \liminf G_m d\mu =+\infty  $$
and note that $(1-z)\mu( \mathbb{N})$ is also equal to $+\infty$, so we still have our conclusion.

Discussion It'd be nice if you can expend my knowledge by having me informed of the convergence of the initial series when $\mathbb{E}(Z_1^2)=+\infty$

Title: swapping limit and series


问:
Determine the smallest integer $k$ such that $4^k \equiv 1 \pmod{19}$.
The following is a question I've come across, which states:

Determine the smallest integer $k$ such that $4^k \equiv 1 \pmod{19}$.

We know that, according to Fermat's Little Theorem, $a^{p-1}\equiv 1 \pmod{p}$, and because $19$ is prime, it must be that $4^{19-1}\equiv 1$; in other words, $k=18$.

However, this is not the smallest solution - the smallest solution is $9$. I don't know how to find the smallest solution, though I would assume it has to do with the answer of $k=18$ somehow.

Any help would be appreciated.

答:
Recall that the set

$\quad {\displaystyle {\mathbb {Z} /19\mathbb {Z}}^{\,*} =  \mathbb {Z} /19\mathbb {Z} \setminus \{0\}}$

forms a group under multiplication and contains $18$ elements. It follows from Lagrange's theorem that the order of any element in ${\displaystyle {\mathbb {Z} /19\mathbb {Z}}^{\,*}}$ must divide $18$ and belong to the set $\{1,2,3,6,9,18\}$.

We begin by determining the order of $[2]$.

Since $2^6 \equiv 2^4 \, 2^2 \equiv (-3) (4) \equiv 7 \pmod{19}$, the order of $[2]$ can't be $6$, and since $2 \mid 6$ and $3 \mid 6$, the order of $[2]$ must be either $9$ or $18$.

Since $2^9 \equiv 2^4 \, 2^4 \, 2^1 \equiv (-3) (-3) (2) \equiv 18 \pmod{19}$, the order of $[2]$ must be equal to $18$.

Let $k$ be the order of $[4]$. Then $1 \equiv 4^k \equiv 2^{2k} \pmod{19}$, so $2k \ge 18$ and therefore $k \ge 9$.

Also, $4^9 \equiv 2^{18} \equiv 1 \pmod{19}$ and therefore $k \le 9$.

We conclude that the order of $[4]$ is equal to $9$.


问:
Epsilon of room trick, Tao Measure Theory
In proving a Lemma regarding $m^{*}$, Tao ('An Introduction to Measure Theory') states that 'by definition of the Lebesgue outer measure we can cover $E\cup F$ by a countable family of boxes such that'

$$\sum_{n=1}^{\infty}m(B_{n})\leq m^{*}(E\cup F) +\epsilon$$

Question: What is the reason or proof of this result?
Since elementary sets are just finite unions of boxes, we can write
$$m^{*}(E\cup F)=\inf\left\{\sum_{n=1}^{\infty}m(B_{n}): B_{n}\text{ are boxes that cover $E\cup F$}\right\}$$
The way I intuitively explained it to myself was that since $m^{*}(E\cup F)$ is the infimum, we should be able to choose $B_{n}$'s to arbitrarily well approximate $m^{*}(E\cup F)$ from above so that
$$\sum_{n=1}^{\infty}m(B_{n}) - m^{*}(E\cup F)\leq \epsilon$$
Is this correct? And how would one more rigorously explain this? I know that for real numbers, if $a > b$, then there must exist $\epsilon>0$ so that $a>b+\epsilon$, but I am not sure if that is relevant here.

答: Your argument is rigorous enough. Maybe you have thought about it too hard.


问:
Correlation of the min and max of random numbers in $(0,1)$
Generate three random numbers independently and uniformly distributed in $(0,1)$. Let $X$ be the smallest of these three numbers and $Y$ the biggest.

(a)  Do you expect $X$ and $Y$ are positively correlated, negatively correlated, or uncorrelated? Explain your reasoning intuitively then find the value of the correlation coefficient $\rho(X,Y)$.

(b)  Suppose $n$ (as opposed to three) random numbers are generated. Compute the correlation coefficient $\rho(X,Y)$.

Answer: Am I correct in assuming that $X$ and $Y$ are uncorrelated because they are independent, and thus $\rho(X,Y)$ is 0 in both cases?

答:
I think you are mistaking the independence of the three random numbers, call them $A, B, C$ for the independence (whether or not they actually are independent) of the minimum $X = \min(A,B,C)$ and maximum $Y = \max(A, B, C)$. As you can see, $X$ and $Y$ are both functions of $A, B, C$, so it is not immediate that they would be independent of each other.

To start building your intuition, if I tell you that $Y = 1/2$, then does that give you any information about $X$? Does that suggest anything about the correlation of $X$ and $Y$?


问:
Proving a certain symmetric matrix has positive determinant
This problem comes arises from the Markowitz Portfolio optimization with $n$ risky assets.

Suppose $\mathbf{1} = (1, 1, ..., 1)\in\mathbb{R}^{n}$ and let $\mathbf{z}\in\mathbb{R}^{n}$ be the vector of expected returns and $\Sigma$ be the covariance matrix for the assets. I have the matrix
$$D = \begin{bmatrix}A & B \\ B & C\end{bmatrix}$$
where $$
\begin{align*}
A & = \mathbf{1}^{\top}\Sigma^{-1}\mathbf{1} \\
B & = \mathbf{1}^{\top}\Sigma^{-1}\mathbf{z} \\
C & = \mathbf{z}^{\top}\Sigma^{-1}\mathbf{z}.
\end{align*}
$$
Question: How can I prove that the determinant of $D$ is positive?

My attempt: I know $A,B,C$ are all positive because $\Sigma^{-1}$ is positive definite. Also, $D$ is symmetric, but I do not think I have enough to conclude $D$ is positive definite. I tried finding the eigenvalues, but arrived that the expression $$\lambda = \frac{A+C\pm\sqrt{(A+C)^{2} - 4(AC - B^{2})}}{2},$$
from which I could not get any further.

答:
I found the answer thanks to Michal's hint. Since $\Sigma^{-1}$ is positive definite, $\langle u, v\rangle =u^{\top}\Sigma^{-1}v$ defines an inner product on $\mathbb{R}^{n}\times\mathbb{R}^{n}$. By the Cauchy Schwarz inequality,

$$|\langle 1, z\rangle|^2 = B^{2}\leq |\langle 1, 1\rangle||\langle z, z\rangle|=AC$$

Hence,
$$\Delta = AC-B^2\geq 0$$


答:
Am I correct in assuming that X and Y are uncorrelated because they are independent,

Totally wrong!

Define $U,V$ two independend rv's form Uniform distribution in $[0;1]$

$X$ and $Y$ are not independent but positively correlated. In fact the value of $Y=max(U,V)$ depends on the values of $X$.

If, as an example $X=0.7$ the values of Y are not anymore in $[0;1]$ but greater than 0.7.

Easy example: just as an hint, suppose we generate only 2 random numbers independently, say $U,V$.

Set $X=min(U;V)$ and $Y=max(U;V)$

The correlation between $X$ and $Y$ can be misured with their covariance

$$Cov(X,Y)=E(XY)-E(X)E(Y)=E(UV)-E(X)E(Y)=\frac{1}{2}\cdot\frac{1}{2}-\frac{1}{3}\frac{2}{3}=\frac{1}{36}$$

skipping for a moment case a), for the general case b), after some calculations based on Order Statistics, the covariance results to me

$$Cov(X;Y)=\frac{1}{(n+1)^2(n+2)}$$

Which, for $n=2$ results exactly $\frac{1}{36}$ as derived before and $\frac{1}{80}$ for case a), when $n=3$

This means, as intuitive it is, that the greater is $n$ the lower is the linear dependence between Max and min.

If you want,as requested, you can easy calculate also $\rho(X,Y)$ but covariance is enough to capture linear dependence between the rv's

How to derive the general solution (sketch)

By definition,

$$Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$$

$\mathbb{E}[X]$ and $\mathbb{E}[Y]$ are very easy to calculate using the distribution of min and max, obtaining respectively $\frac{1}{n+1}$ and $\frac{n}{n+1}$

The hardest probelm is to calculate $\mathbb{E}[XY]$. To solve it I used the definition of density of order statistics that you can find in the link above.

$$f_{XY}(x,y)=n(n-1)(y-x)^{n-2}$$

where $x<y$

Thus

$$\mathbb{E}[XY]=\int_0^1 n xdx\Bigg[\int_x^1 (n-1)y(y-x)^{n-2}dy    \Bigg]dx=\dots=\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}$$

When solving the integral, you will arrive at the following

$$\mathbb{E}[XY]=n\int_0^1 x(1-x)^{n-1}dx-\int_0^1 x(1-x)^{n}dx=n\frac{\Gamma(2)\Gamma(n)}{\Gamma(n+2)}-\frac{\Gamma(2)\Gamma(n+1)}{\Gamma(n+3)}=$$

$$=\frac{n(n-1)!}{(n+1)!}-\frac{n!}{(n+2)!}=\frac{1}{n+1}-\frac{1}{(n+2)(n+1)}$$

This results follow using Beta Function

Title: Correlation of the min and max of random numbers in $(0,1)$

Body:
Generate three random numbers independently and uniformly distributed in $(0,1)$. Let $X$ be the smallest of these three numbers and $Y$ the biggest.

(a)  Do you expect $X$ and $Y$ are positively correlated, negatively correlated, or uncorrelated? Explain your reasoning intuitively then find the value of the correlation coefficient $\rho(X,Y)$.

(b)  Suppose $n$ (as opposed to three) random numbers are generated. Compute the correlation coefficient $\rho(X,Y)$.

Answer: Am I correct in assuming that $X$ and $Y$ are uncorrelated because they are independent, and thus $\rho(X,Y)$ is 0 in both cases?


问:
Equivalent condition for $V\subset X$ to be a neighborhood in the inductive topology on $X$.
Let $\{X_i\}$ a family of topologcial vector space(equipped with topology $\mathcal{T_i}$)

let $X = \bigcup_{i = 1}^\infty X_i$ equipped with inductive topology w.r.t $\{(X_i,\mathcal{T}_i,I_i)\}$(the finest topology which makes $I_i:X_i \to X$ continuous where $I_i$ is the identity map into)

Prove if for each $i$ we have $V\bigcap X_i$ is neighborhood near 0 in topological vector space $X_i$

Then $V$ is a neighborhood near 0 in topological space $X$

答:
I think what you're asking comes down to the following result.  It has nothing to do with topological vector spaces.

Proposition: Let $\mathcal T$ be the inductive topology on $X$ coming from the spaces $(X_i, \mathcal T_i)$, and let $\mathcal T'$ be the topology on $X$ given by the condition that $V \subseteq X$ is open if and only if $V \cap X_i$ is open in $X_i$ for each $i$.  Then $\mathcal T = \mathcal T'$.

Proof: By definition, $\mathcal T$ is the largest topology on $X$ for which each inclusion $X_i \subset X$ is continuous.  In other words, if $\mathcal T''$ is a topology on $X$ for which each inclusion $X_i \rightarrow X$ is continuous, then $\mathcal T'' \subseteq \mathcal T$.  Since each inclusion $X_i \rightarrow X$ is clearly continuous for $\mathcal T'$, we have $\mathcal T' \subseteq \mathcal T$.

On the other hand, if $V \subseteq X$ belongs to $\mathcal T$, then each $V \cap X_i$ is open in $X_i$, and hence $V \in \mathcal T'$.  Thus $\mathcal T \subseteq \mathcal T'$.


问:
Integral $E_n(a_1,...,a_n;t)=\int_{-\infty}^{\infty}\frac{x^2\cos(tx)}{\prod_{k=1}^{n}(x^2+a_k^2)}dx$

Evaluate
  $$E_n(a_1,...,a_n;t)=\int_{-\infty}^{\infty}\frac{x^2\cos(tx)}{\prod_{k=1}^{n}(x^2+a_k^2)}dx$$
  for $i\ne j\iff a_i^2\ne a_j^2$, and $a_i\in\Bbb R\setminus \{0\}$.

I've been able to find the similar integral
$$J_n(a_1,...,a_n;t)=\int_{-\infty}^{\infty}\frac{\cos(tx)}{\prod_{k=1}^{n}(x^2+a_k^2)}dx$$
by seeing that we can write
$$\prod_{k=1}^{n}\frac1{z+\alpha_k}=\sum_{k=1}^{n}\frac{1}{z+\alpha_k}\prod_{k\ne j=1}^{n}\frac{1}{\alpha_j-\alpha_k}\ ,$$
And then using $z\equiv x^2$, $\alpha_k\equiv a_k^2$ to get
$$J_n(a_1,...,a_n;t)=\sum_{k=1}^{n}J_1(a_k;t)\prod_{k\ne j=1}^{n}\frac{1}{a_j^2-a_k^2}\ .$$
Then we take the Laplace transform of the remaining integral:
$$\begin{align}
\mathcal{L}\{J_1(q;t)\}(s)&=2\int_0^\infty e^{-st}\int_0^\infty \frac{\cos(tx)}{x^2+q^2}dxdt\\
&=2\int_0^\infty\frac{1}{x^2+q^2}\int_0^\infty \cos(tx)e^{-st}dtdx\\
&=2\int_0^\infty\frac{1}{x^2+q^2}\text{Re }\int_0^\infty e^{-(s-ix)t}dtdx\\
&=2\int_0^\infty\frac{1}{x^2+q^2}\text{Re }\left[\frac1{s-ix}\right]dx\\
&=2s\int_0^\infty\frac{1}{(x^2+q^2)(x^2+s^2)}dx\\
&=\frac{2s}{s^2-q^2}\left[\int_0^\infty\frac{dx}{x^2+q^2}-\int_0^\infty\frac{dx}{x^2+s^2} \right]\\
&=\frac{\pi s}{s^2-q^2}\left[\frac{1}q-\frac{1}s \right]\\
&=\frac{\pi}{q}\left[\mathcal{L}\{\cosh(qt)\}(s)-\mathcal{L}\{\sinh(qt)\}(s)\right]\\
&=\frac{\pi}{q}\mathcal{L}\{e^{-qt}\}(s)\ .
\end{align}$$
Which gives $$J_1(q;t)=\frac{\pi}{q}e^{-qt}$$
and $$J_n(a_1,...,a_n;t)=\pi\sum_{k=1}^{n}\frac{e^{-a_k t}}{a_k}\prod_{k\ne j=1}^{n}\frac{1}{a_j^2-a_k^2}\ ,$$
from which many remarkable results may be derived.

I have the suspicion however, that we may be able to find $E_n$ in a similar way, but I don't know how. One idea:
$$\int_{-\infty}^{\infty}\frac{(x^2+1)\cos(tx)}{(x^2+1)(x^2+\mathbf{a}_n^2)}dx=J_n(\mathbf{a}_n;t)=E_{n+1}(1,\mathbf{a}_n;t)+J_{n+1}(1,\mathbf{a}_n;t)$$
Where we used the shorthand notation $$x^2+\mathbf{a}_n^2\equiv\prod_{k=1}^{n}(x^2+a_k^2)$$
$$\mathbf{a}_n\equiv(a_k)_{k=1,...,n}\ .$$
It should be noted that the above holds for $1\not\in \mathbf{a}_n$.

Any Ideas? Thanks.

Body:
I can't believe I've overlooked this for so long.

Let $$u(x,t)=\frac{\cos(tx)}{\prod_{k=\color{red}{2}}^{n}(x^2+a_n^2)}.$$

We may write
$$E_n(a_1,...,a_n;t)=\int_\Bbb R \frac{x^2+a_1^2-a_1^2}{x^2+a_1^2}u(x,t)dx=\int_\Bbb Ru(x,t)dx-a_1^2\int_{\Bbb R}\frac{u(x,t)dx}{x^2+a_1^2},$$
which is
$$E_n(a_1,...,a_n;t)=J_{n-1}(a_2,...,a_n;t)-a_1^2J_n(a_1,...,a_n;t).$$


问:
Recursive function of $n^2$?
How would you convert $n^2$ into a recursive function?

Like for example, I can say the recursive function of $2^n$ is $2 \cdot 2^{n-1}$, and it can be applied recursively since it requires the previous value.

Title: Recursive function of $n^2$?

Body: $a_{n+2}=2a_{n+1}-a_n+2$, with $a_0=0,a_1=1$


问:
Let $a \in \mathbb R^m$. What is an analytic formula for the maximum value of $\|Xa\|_2$ over all $k \times m$ matrices with $\|X\|_F \le 1$?
Let $a \in \mathbb R^m$.

Question. What is an analytic formula for the maximal value of $\|Xa\|_2$ subject to $X \in \mathbb R^{k \times m}$, $\|X\|_F \le 1$ ?

Trivial ower-bound. Let $r(a)$ be the max value. Observe that $r(a) \ge \|a\|_2$. Indeed, this is the value we get uppon restricting the problem to matrices $k \times m$ matrices $X$ all of whose columns are zero except for the first one.

答:
Using the fact that the operator norm is at most the Frobenius norm, we have

$$\|Xa\|_2 \le \|X\| \cdot \|a\|_2 \le \|X\|_F \cdot \|a\|_2 \le \|a\|_2.$$

You can check that this upper bound is attainable by any matrix of the form $X = \dfrac{1}{\|a\|_2}ua^T$ where $u \in \mathbb{R}^k$ is a unit vector.


问:
Smallest number of vertices in tree with specific degrees.

Find, with proof, the smallest number $r$ of vertices in a tree having two vertices of degree $3$, one vertex of degree $4$, and two vertices of degree $6$. Give an example of such a tree.

Clearly, I need to make use of the degree requirements to impose a lower bound on the number of vertices. Intuitively, I think the higher the average degree, the more vertices there'll be as compared with graphs with lower degree. I know a few properties about trees, including the fact that every tree with at least two vertices has at least $2$ leaves. Trees are connected graphs without cycles. So far, the smallest number I can find is $19,$ and that is by placing the vertex of degree $6$ at the root and by making the other vertices as neighbours. But this approach seems too simple; it's likely that the minimum number is smaller. Clearly it's greater than $7$.

Body:
For $d \ge 1$, let $n_d$ be the number of vertices of degree $d$.  Then we have
\begin{align}
r &=\sum_d n_d \\
2(r-1)&=\sum_d d n_d
\end{align}
By eliminating $r$, we find that
$$2\sum_d n_d = 2 + \sum_d d n_d,$$
equivalently,
$$0 = 2 + \sum_d (d-2) n_d.$$
We are given $n_3 = 2$, $n_4 = 1$, and $n_6 = 2$, so
\begin{align}
0 &= 2 - 1n_1 + 0n_2 + 1n_3 + 2n_4 + 3n_5 + 4n_6 + \sum_{d \ge 7} (d-2) n_d \\
&= 2 -n_1 + 0 + 2 + 2 + 3n_5 + 8 + \sum_{d \ge 7} (d-2) n_d \\
&= 14 -n_1 + 3n_5 + \sum_{d \ge 7} (d-2) n_d.
\end{align}
Hence $$n_1 = 14 + 3n_5 + \sum_{d \ge 7} (d-2) n_d \ge 14,$$
and so $$r \ge n_1 + n_3 + n_4 + n_6 \ge 14 + 2 + 1 + 2 = 19.$$


问:
Why is the Mazur swindle named so?
Often results or techniques in mathematics are called 'theorems'. Sometimes they are called 'tricks'. In no other context have I seen a result called a 'swindle'. Is there a historical reason for this trick being called a swindle?</p>

答:
The earliest use of the term ‘swindle’ for this little ploy, this ruse, appears to be in H. Bass. Big Projective Modules Are Free, Illinois J. Math. 7(1) (1963), 24–31. He writes:

Our method relies on two basic tools. … The second is an elegant little swindle, observed several years ago by Eilenberg, and which might well have sprung from the brow of Barry Mazur.

This is in reference to the 1959 and 1961 papers of B. Mazur, namely On the structure of certain semi-groups of spherical knot classes, Publications Mathématiques de l'IHÉS 3 (1959), 19–27, and On embeddings of spheres, Acta Mathematica 105(1–2) (1961), 1–17.

As for why Bass chose the term ‘swindle’… Well, the trick does appear to be quite the swindle, quite the sham, and maybe almost a scandal indeed! For surely you would feel cheated if someone told you something like
$$1=1+(-1+1)+(-1+1)+\dots=(1-1)+(1-1)+\dots=0.$$

Title: Why is the Mazur swindle named so?


问:
Finding the limit of a sequence $a_n$
I have the following sequence $$ a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1} $$ I want to find $\lim_{n\to\infty}a_n$, I know it is 0, but I can't seem to find a way to show it. It seems like $\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$ is an (good) approximation of $x^3$, but it isn't a taylor series or anything like it I know of. Since it is ( / it looks like) an appromixation, once I try to apply the sandwich rule, I get large numbers, which don't help me. It also seems that Cauchy criteria doesn't make it easier.

答:
$$a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$$

We need to focus on the last term
$$b_n=\left(1-\frac{1}{n^{3}}\right)^{n-1}\implies \log(b_n)=(n-1)\log\left(1-\frac{1}{n^{3}}\right)$$ By Taylor
$$\log\left(1-\frac{1}{n^{3}}\right)=-\frac{1}{n^3}-\frac{1}{2 n^6}+O\left(\frac{1}{n^{9}}\right)$$
$$\log(b_n)=-\frac{1}{n^2}+\frac{1}{n^3}-\frac{1}{2 n^5}+\frac{1}{2
   n^6}+O\left(\frac{1}{n^{8}}\right)$$
$$b_n=e^{\log(b_n)}=1-\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{2 n^4}-\frac{3}{2
   n^5}+\frac{5}{6
   n^6}+\frac{1}{n^7}+O\left(\frac{1}{n^8}\right)$$
All of that makes
$$a_n=\frac{1}{2 n}-\frac{1}{2 n^2}-\frac{1}{3
   n^3}+\frac{1}{n^4}+O\left(\frac{1}{n^5}\right)$$ which shows the limit but also that the limit is approched from above.

Moreover, it allows a fast way to find the approximate value of $n$ for a given value of $a_n$. Suppose $a_n=0.01$ so the first term of the expansion gives $n \sim 50$; using the first and second terms gives $n \sim 49$.

Title: Finding the limit of a sequence $a_n$


问:
Approximate the age of the employees with an histogram

The following histogram represents the age distribution of employees of a textile company. (The absence of gradation on the vertical axis is intentional)

knowing that 225 employees are under 45, there are fewer than 505 employees in the factory.

The approximate average age of employees is greater than 46.

Which are true?

I have no clue how to say which one is true. How can I deal with this question?

EDIT

Intuitively, I think only the number 2 is true. That question is hard for me.

答:
You are expected to extend the ticks on the vertical axis through the bars.  Each rectangle will then represent the same number of employees. There are $9$ rectangles below age $45$ and $11$ above, so $20$ total rectangles.  Each rectangle represents $25$ employees.  That says there are $500$ total employees.  1 is true.

It could be that all the employees in a rectangle are at the minimum age for that rectangle.  I didn't do the arithmetic, but I am sure the average age would then come out below $46$.  This would say that 2 could be false.


问:
A one-point compactification of a non-separable metric space: separable or not?
I'm interested in compactifications of some non-separable metric spaces of functions.  For concreteness, consider $D[0,1]$, the space of right-continuous real-valued functions on $[0,1]$ with left limits, and give it the uniform (supremum) norm $\|f\|_\infty=\sup_t |f(t)|$.

The Alexandroff one-point compactification $D^*$ of $D[0,1]$ is compact, and $(D[0,1],\,\|\cdot\|_\infty)$ is a dense subspace of it. Is $D^*$ separable?

If it isn't, $D^*$ is a somewhat interesting example of a compact space that isn't separable (this question)

If it is, $(D[0,1],\,\|\cdot\|_\infty)$ and $D^*$ are an interesting example of a non-separable subspace of a separable space (this question)

But I have no idea which.

答: The space $D^*$ is not separable, because otherwise $D[0,1]$ would be separable, being an open subspace of a separable space.


问:
If $R$ is an integral domain without $1_R$ and $a=b$, then $ac \neq b$ for every $c$?
As it says in the question, let $a = b$, where $a, b \neq 0$.
Assume by contradiction that there exists some $c \in R$ such that $ac = b$. Then,
$a=ac$, so $a-ac = 0$. I was thinking about using distributive property, but this cannot be applied properly since it forces the existence of $1_R \in R$, i.e. $a(1_R-c) = 0$. Is there an alternate way to prove the statement?

Thank you

答:
If $ac=a\neq 0$ then $acc=ac$ implies $c^2=c$ is a nonzero idempotent.

Then $(x-xc)c=0=c(x-cx)$. Implies $x=cx=xc$ for any $x\in R$. That would mean $c$ is an identity element.

Since you assumed there is no identity, the original assumption that $ac=a$ for some $c$ cannot be true for any $c$.


答:
Here is one more.

Let $a_n = n^2$. Then simple school algebra should convince you that $$a_{n+1} = a_n + 2\sqrt(a_n) + 1$$
with $a_1  = 1$ and $a_0=0$

Title: Recursive function of $n^2$?

Body:
How would you convert $n^2$ into a recursive function?

Like for example, I can say the recursive function of $2^n$ is $2 \cdot 2^{n-1}$, and it can be applied recursively since it requires the previous value.


问:
Proving the duplication formula
I'm trying to prove the famous Legendre duplication formula:

$$ \Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}}   $$

I have to prove this using these two identities/theorems:

$$ \Gamma(z)= \lim_{n \rightarrow \infty}  \frac{n! n^{z}}{z(z+1)...(z+n)}$$

and

$$ (2n)! \sim \frac{2^{2n} (n!)^2}{\sqrt{\pi n}}   $$ as $n \rightarrow \infty $

I put in $ z = 2z $ in the first one and obtained:

$$ \Gamma(2z) = \lim_{n \rightarrow \infty}  \frac{n! n^{2z}}{2z(2z+1)...(2z+n)} $$.

I then tried multiplying both the numerator and denominator by various factors to try to incorporate the other identity, but it was of no use. Can anyone tell me what direction I'm supposed to go from here?

答:
This is not an answer since not using all the identities/theorems.

Showing that $$\Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}}$$ is the same as showing that
$$\log (\Gamma (2 z))-\log \left(\frac{2^{2 z-1} \Gamma (z) \Gamma \left(z+\frac{1}{2}\right)}{\sqrt{\pi  }}\right)=0$$ Expand the rhs in terms of logarithms; use Stirling approximation (up to any order) three times and continue with Taylor expansions for large values of $z$

Title: Proving the duplication formula


问:
Equivalence of statements via geometric inequality on $4$ Points
Let us consider $4$ Points $A,B,C,D$ on a line $g$.

We are to prove:
If $\vert AP\vert+\vert PD\vert >\vert BP\vert+\vert CP\vert $ for all $P$ not on $g$, then $\vert AB\vert=\vert CD\vert$.

I solved the question using the contraposition of the statement and calculus:
We will first show that the inequality can be reversed for a point $P$ on $g$.
Let us consider a point further to the right on $g$ then $D$, and let it be $P$; additionally, let $z>x$.
We define $\vert AB\vert=x, \; \vert BC\vert=y, \;\vert CD\vert=y, \; \vert DT \vert=w$.
Then, if the inequality were still to hold, we must have: $\vert AP\vert+\vert PD\vert >\vert BP\vert+\vert CP\vert$
or, $x+y+z+2t\geq 2z+y+2t$ or $x>z$, a contradiction.

We now choose $P$ such that for its foot $T$ on $g$, we have $\vert DT\vert=w$.
Let $\vert TE\vert=\epsilon$.
Now, using the pythagorean theorem, we must have:
$\vert AP\vert+\vert PD\vert =\sqrt{(x+y+z+w)^2+\epsilon^2}+\sqrt{z^2+\epsilon^2}>\sqrt{(y+z+w)^2+\epsilon^2}+\sqrt{(z+w)^2+\epsilon^2}$
Letting $\epsilon$ approach $0$, by continuity we find a point of equality and thus for the inequality to hold, we must have $x=z$.

While this approach works it is definetly not the most elegant, because the problem was posed in a competition for middle schoolers. What is an elementary solution?

答:
WLOG assume $x<z$. Place P on top of D such that $PD \perp AD$.

Let $\angle PAB = \alpha, \angle PBC=\beta, \angle PCD = \gamma$. Then $0 <\alpha < \beta < \gamma <\frac{\pi}{2}\Rightarrow \cos \alpha > \cos \beta > \cos \gamma.$

Now,
$$
|AP| = \frac{x+y+z}{\cos \alpha}, |BP|=\frac{y+z}{\cos \beta},|CP|=\frac{z}{\cos \alpha}, |DP| = z \tan \alpha.
$$

$$
0 < |AP|+|DP|-|BP|-|CP|=(|AP|-|BP|) - (|CP|-|DP|)\\
= \left(\frac{x+y+z}{\cos \alpha} - \frac{y+z}{\cos \beta}\right)-\left(\frac{z}{\cos \gamma} - \frac{z\sin \gamma}{\cos \gamma}\right)\\
< \left(\frac{x+y+z}{\cos \beta} - \frac{y+z}{\cos \beta}\right)-\frac{z(1-\sin \gamma)}{\cos \gamma}\\
= \frac{x}{\cos \beta} -\frac{z(1-\sin \gamma)}{\cos \gamma}
<\frac{x}{\cos \gamma} -\frac{z(1-\sin \gamma)}{\cos \gamma}
$$

If we choose $\gamma$ very close to 0, such that $x < z(1-\sin \gamma)$ then $0 < \frac{x}{\cos \gamma} -\frac{z(1-\sin \gamma)}{\cos \gamma} < 0, \rightarrow \leftarrow$

Therefore $x=z.\blacksquare$


答:
If you want a homogeneous linear difference equation then
$$
a_{n+3}-3a_{n+2}+3a_{n+1}-a_n=0, \text{ with } a_0=0, a_1=1, a_2=4.
$$

Title: Recursive function of $n^2$?


问:
Open set in irrationals
$A$ is the set of all irrationals in $\mathbb R$. Is there a non empty open set subset of $A$?

My hunch is that the irrationals can not contain an open set because rationals are dense in $\mathbb R$? But I am not able to reconcile to the fact that the rationals are countable and hence are contained in an open set of small radius. Please get me out of the muddled thinking.

答:
Suppose a subset $B$ of $A$ is open. Then all interior points are contained in $B$. So for any point in $B$ , there is a neighborhood of that point which is contained in $B$.

But neighborhood is an open interval and by denseness property it must contain rational numbers which is not possible since $B$ contains only irrational numbers. So $B$ cannot be open.

Title: Open set in irrationals


问:
Why do statisticians like "$n-1$" instead of "$n$"?
Does anyone have an intuitive explanation (no formulas, just words! :D) about the "$n-1$" instead of "$n$" in the unbiased variance estimator
$$S_n^2 = \dfrac{\sum\limits_{i = 1}^n \left(X_i-\bar{X}\right)^2}{n-1}?$$</p>

答: using samples we try to estimate the population mean. But since it is a sample, it will not include the full spectrum of data ; only contains a subset. So squared distance of sample mean from each reading divided by sample size would be lower than that of population. We try to reduce this effect by dividing with a small denominator (N-1) in case of sample variance.</p>


答:
Let $\Bbb P=\Bbb R\setminus\Bbb Q$ be the set of irrationals. Let $U$ be a non-empty open set in $\Bbb R$; then there are $a,b\in\Bbb R$ such that $a<b$ and $(a,b)\subseteq U$. As you say, the rationals are dense in $\Bbb R$, so there is a rational $q\in(a,b)$, and it follows that $$q\in(a,b)\setminus\Bbb P\subseteq U\setminus\Bbb P$$ and hence that $U\nsubseteq\Bbb P$. Thus, as you suspected, the irrationals do not contain any non-empty open subset of $\Bbb R$.

The fact that $\Bbb Q$ is countable and therefore can be covered by an open set of small radius is simply irrelevant: there is no reason to think that in general an open set covering a dense set should cover the whole space. A dense set can even be open. For a very simple example, consider the subspace

$$X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$

of $\Bbb R$: the set $X\setminus\{0\}$ is a dense, open subset of $X$, so it is an open cover of itself that covers only itself, not all of $X$.

Title: Open set in irrationals

Body:
$A$ is the set of all irrationals in $\mathbb R$. Is there a non empty open set subset of $A$?

My hunch is that the irrationals can not contain an open set because rationals are dense in $\mathbb R$? But I am not able to reconcile to the fact that the rationals are countable and hence are contained in an open set of small radius. Please get me out of the muddled thinking.


问:
Must two morphisms into an algebraic space which agree on closed points be the same?
First I apologize if this is elementary. I have just started looking at the basics of stacks and algebraic spaces so my understanding is lacking.

Let's work over an algebraically closed field $k$. Suppose I have an algebraic space $\mathcal{A}$ and two morphisms $f_i:S\rightarrow\mathcal{A}$ such that for every closed point $\operatorname{Spec}(k)\rightarrow S$ the induced maps $f_i|_{\operatorname{Spec}(k)}\rightarrow \mathcal{A}$ agree. Suppose $S$ is a reduced scheme with $\operatorname{Spec}(k)$ points which are dense in $S$. Must $f_1=f_2$? If not is there a condition we can impose on the algebraic space so this holds?

My intuition says since an algebraic space is fibered over sets instead of groupoids this should hold.

Thanks

答:
It was difficult to capture the essence of what I was looking at, so I apologize that I never nailed down the correct hypotheses. This exercise from Hartshorne is what I was looking for an analog of in algebraic spaces:

II.4.2: Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let $Y$ be a separated scheme over $S$. Let $f$ and $g$ be two $S$-morphisms which agree on an open dense subset of $X$. Then $f=g$.


问:
Is there is any example of continuous bounded function on $\mathbb{R}$ which attains 'either' maximum 'or' minimum on $\mathbb{R}$ but not both.
I need to find example of continuous bounded function on $\mathbb{R}$ which attains 'either' maximum 'or' minimum on $\mathbb{R}$ but not both.

My attempt:

First of all, is such function exists? (How)? If it exists then isn't it will contradict extreme value theorem?

Is $f(x)=\frac{1}{x^2+1}$ for $x\in\mathbb{R}$ works?

答:
As pointed out in a comment your example works if you replace maximum/minimum by supremum/infimum.

Otherwise, consider $f(x)=|x|.$ It achieves its minimum on $\mathbb{R}$ but the maximum does not exist. Similarly, $g(x)=\exp(-x^2)$ achieves its maximum on $\mathbb{R}$ but does not achieve the infimum (minimum, of course, does not exist).

If you say that the 'maximum' and 'minimum' should exist, then your question does not make sense. Because maximum/minimum is by definition achieved by the function,


答: Let $S\subset \mathbb{R}\setminus \mathbb{Q}$ be any non empty subset of the irrational . Choose any point $x\in S$, let $(a,b)$ be any open interval in $\mathbb{R}$ containing the point $x$. Then $(a,b)$ cannot be contained inside $S$ because of the fact that every interval contains atleast one rational number. Therefore $x$ cannot be an interior point of $S$, and hence $S$ is not open in $\mathbb{R}$.

Title: Open set in irrationals


问:
How to find the inverse Laplace Transforms?
How do you find the inverse Laplace Transform of the following,

$$\frac{2 (s^2+4 s+5)^2+40}{(s^2+4 s+5)^2}$$

Separating them into complex coefficients is to long.

答:
The problem reduces to $$2+\frac{40}{(s^2+4s+5)^2}$$

Let us denote the Laplace Transform of $f(t)$ as $L\left(f(t)\right)=F(s)$

and the inverse as $L^{-1}\left(F(s)\right)=f(t)$

For $L^{-1}\frac1{(s^2+4s+5)^2},$

$$\text{If }L\left(f(t)\right)=\frac{s-a}{\{(s-a)^2+b^2\}^2},$$

$$L\left(\frac {f(t)}t\right)=\int_s^\infty \frac{s-a}{\{(s-a)^2+b^2\}^2} ds$$
$$=\frac12\int_s^\infty \frac{2(s-a)}{\{(s-a)^2+b^2\}^2} ds=-\frac12\left(\frac1{(s-a)^2+b^2}\right)_s^\infty=\frac12\frac1{(s-a)^2+b^2}$$

$$\implies \frac {2f(t)}t=L^{-1}\left(\frac1{(s-a)^2+b^2}\right)=\frac{e^{at}\sin bt}{b} $$

$$\implies L^{-1}\frac{s-a}{\{(s-a)^2+b^2\}^2} =f(t)=\frac{te^{at}\sin bt}{2b}$$

$$a=0\implies  L^{-1}\frac s {\{s^2+b^2\}^2} =\frac{t\sin bt}{2b}$$

As  $L\{g(t)\}=G(s)\implies \int_0^tg(t)dt=L^{-1}\left(\frac{G(s)}s\right)$

$$L^{-1}\frac1{(s^2+b^2)^2}=L^{-1}\left(\frac1s\cdot \frac s{(s^2+b^2)^2}\right)=\int_0^t\frac{t\sin bt}{2b}dt=\frac{\sin bt-bt\cos bt}{2b^3}$$ (Integrating by parts)

Now using shifting property  $L\{g(t)\}=G(s)\implies L(e^{at}g(t))=G(s-a)$

So, $$L^{-1}\frac1{\{(s-a)^2+b^2\}^2}=\frac{e^{at}(\sin bt-bt\cos bt)}{2b^3}$$

Here $s^2+4s+5=(s+2)^2+1^2\implies a=-2,b=1$


问:
Does the series $\sum\ln(1+\frac{1}{n\ln(n)})$ Converge?
As the title states I wanna prove this series converges or not however I noted that if we fan transform $\ln(1+\frac{1}{n\ln(n)})$ into some known limit identity and use Comparison test I would be done

答:
Consider the series :
$$
\sum b_{n}=\sum_{n=2}^{+\infty}\frac{1}{n\ln(n)}
$$

Applying the limit comparison test yields :
$$
\lim_{n\to+\infty}\frac{\ln\left(1+\frac{1}{n\ln(n)}\right)}{\frac{1}{n\ln(n)}}
$$
Let $\displaystyle u=\frac{1}{n\ln(n)}$, then :
$$
\lim_{u\to0}\frac{\ln(1+u)}{u}=1
$$
Since the series $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln(n)}$ diverges by the integral test, it follows that your main series diverges by the limit comparison test.


问:
Let $H$ be a subgroup of $G$ and suppose that $Ha=bH$ for $a,b$ elements of $G$. Show that $aH=Hb$.
I am struggling pretty bad with this homework problem, I've been beating my head against the wall for quite some time and still can't arrive at an answer or even know where to start.

Let $H$ be a subgroup of $G$ and suppose that $Ha=bH$ for $a,b$ elements of $G$. Show that $aH=Hb$.

I feel like I'm supposed to apply Lagrange's theorem, but I don't know how...

答:
A (left or right)  coset  is generated by any of its elements.

Therefore $Ha=aH$  ( since $= bH$ a left coset, and contains $a$).

Similarly $bH = Hb$.


问:
Properties of orthonormal basis in $\mathbb R^n$
Is the intuition that at least one of the vectors of any $\mathbb{R}^n$ orthonormal basis has all entries $e_i\geq 0$ or $e_i\leq 0$ correct?

答:
For $n = 3$, consider $e_1 = \begin{bmatrix}-1/3 \\ 2/3 \\ 2/3 \end{bmatrix}$, $e_2 = \begin{bmatrix}2/3 \\ -1/3 \\ 2/3 \end{bmatrix}$, and $e_3 = \begin{bmatrix}2/3 \\ 2/3 \\ -1/3 \end{bmatrix}$.

For larger $n$, try to find vectors $\{e_k\}_{k = 1}^{n}$ of the form $(e_k)_j = \begin{cases}a & \text{if} \ j \neq k \\ b & \text{if} \ j = k\end{cases}$ where $a$ and $b$ are constants with opposite signs (which could depend on $n$).


问:
Sequence of positive integers being degree sequence of tree.

Let $k\geq 2$ and $T$ be a tree on $k$ vertices. Let $ D_k = (d_1,\cdots, d_k)$ be a sequence of positive integers. Show that $D_k$ is the degree sequence of $T$ iff $\sum_{i=1}^k d_i = 2k-2.$

For the forward implication, we know that $2|E(T)| = 2(k-1) = 2k-2 = \sum_{i=1}^k d_i$.

For the reverse implication, suppose $\sum_{i=1}^k d_i = 2k-2.$ We want to show that $(d_1,\cdots, d_k)$ is the degree sequence of $k.$ We use induction on $k.$ For $k=2,$ we have $d_1 + d_2 = 2.$ Since both $d_1$ and $d_2$ are positive integers, $d_1 = 1 = d_2,$ and so $(d_1, d_2)$ is the degree sequence of a tree with $k$ vertices. So the base case holds. Now suppose that for all $2\leq k < m, m\geq 3, $ whenever $(d_1,\cdots, d_k)$ is a sequence of positive integers so that $\sum_{i=1}^k d_i = 2k-2,$ $(d_1,\cdots, d_k)$ is the degree sequence of a tree on $k$ vertices. Let $D_{m} = (d_1,\cdots, d_{m})$ be a sequence of $m$ positive integers so that $\sum_{i=1}^m d_i = 2m-2.$ If one $d_i = 2,$ then $D_m[i] := (d_1,\cdots, d_{i-1}, d_{i+1},\cdots, d_m)$ is a sequence of $m - 1$ positive integers with $\sum_{1\leq j\leq n, j\neq i} d_j = 2m-4 = 2(m-1) - 2.$ Thus, by the inductive hypothesis, $D_m[i]$ is the degree sequence of a tree $T_{m-1}$ on $m-1$ vertices. Since  $m-1\geq 2, T_{m-1}$ has at least $1$ leaf $t_1$. Add a new vertex $t'$ to $T_{m-1}$ so that $t_1 t'$ is an edge and let $T_{m-1}'$ be the resulting tree. Then $T_{m-1}'$ is a graph with $m$ vertices, and the only difference between the degree sequences of the two trees is that the degree sequence of $T_{m-1}'$ has one more entry of degree $2$. We claim that $T_{m-1}'$ is a tree. Observe that it has $m-1$ edges, since $T_{m-1}$ has $m-2$ edges, and for any two vertices $u \neq t', v \neq t'\in V(T_{m-1}')$ there is a path in $T_{m-1}'\backslash t' = T_{m-1}$ from $u$ to $v$. Also, we can append the neighbour of $t', t_1,$ to the beginning of a path from $t_1$ to any vertex other than $t_1$ and $t'$ ($t'$ and $t_1$ are joined by a path by definition, so we just need to consider vertices distinct from these two). So $T_{m-1}'$ is connected, and hence a tree. Thus, $(d_1,\cdots, d_m)$ is the degree sequence of the tree $T_{m-1}'.$

However, I'm having a lot of trouble dealing with the case where no $d_i=2$, and I can't fully prove it. Is there some simpler approach?

答:
The induction argument that occurs to me is a bit different. Suppose that the result is true for all sequences shorter than $k$ that satisfy the conditions of the theorem, and let $D_k=\langle d_1,\ldots,d_k\rangle$ be a sequence of positive integers such that $\sum_{i=1}^kd_i=2k-2$.

The idea is to remove all of the $1$ terms from the sequence, so that if $D_k$ really is the degree sequence of a tree $T$, we’re removing the pendant vertices. Of course that would also reduce the total degree of the remaining vertices by the number of pendant vertices, so we have to adjust the remaining terms of $D_k$ downward by a total amount equal to the number of $1$ terms. The trick is to do this in such a way that we get a shorter sequence satisfying the conditions of the theorem, so that we can apply the induction hypothesis to get a tree $T'$ and then add the appropriate leaves to to to get a tree $T$ whose degree sequence is $D_k$, and the induction is complete.

If $d_i\ge 2$ for $i=1\ldots,k$, then $\sum_{i=1}^kd_i\ge 2k$, which is impossible, so there is at least one $i$ such that $d_i=1$. (In fact there are at least two.) We may assume that $d_1\le d_2\le\ldots\le d_k$. Let $\ell=\max\{i\in[k]:d_i=1\}$; then

$$\sum_{i=\ell+1}^kd_i=2k-2-\ell=\big(2(k-\ell)-2\big)+\ell\,.$$

If $\ell=k$, then $k=\sum_{i=1}^k1=2k-2$, so $k=2$, and $\langle 1,1\rangle$ is indeed the degree sequence of the tree on $2$ vertices; otherwise $\sum_{i=\ell+1}^kd_i\ge\ell$.

If $\sum_{i=\ell+1}^kd_i=\ell$, then $2k-2=2\ell$, so $\ell=k-1$, and we have the degree sequence of the tree $K_{1,k-1}$. Otherwise, $\sum_{i=\ell+1}^kd_i>\ell$. And

$$\sum_{i=\ell+1}^k(d_i-1)=2k-2-\ell-(k-\ell)=k-2\,,$$

so $k-2>\ell-(k-\ell)$, $2k-2>2\ell$, $k-1>\ell$, and $\sum_{i=\ell+1}^k(d_i-1)\ge\ell$.

Let $m$ be maximal such that $\sum_{i=\ell+1}^m(d_i-1)\le\ell$. For $i=1\ldots,m-\ell$ let $d_i'=1$, and if $m<k$ let $d_{m-\ell+1}'=\sum_{i=\ell+1}^{m+1}(d_i-1)-\ell+d_{m+1}$. If $m+1<k$ let $d_i'=d_{\ell+i}$ for $i=m-\ell+2,\ldots,k-\ell$. Then

$$\sum_{i=1}^{k-\ell}d_i'=\sum_{i=1}^kd_i-2\ell=2(k-\ell)-2\,,$$

so by the induction hypothesis $\langle d_1',\ldots,d_{k-\ell}'\rangle$ is the degree sequence of a tree $T'$ on $k-\ell$ vertices. Let the vertices of $T'$ be $\{v_1,\ldots,v_{k-\ell}\}$, and let $d(v_i)=d_i'$ for $i=1,\ldots,k-\ell$. For $i=1,\ldots,m-\ell$ attach $d_{\ell+i}-1$ leaves to $v_i$, and attach $\ell-\sum_{i=\ell+1}^m(d_i-1)$ leaves to $v_{m-\ell+1}$, if that vertex exists. The resulting tree has $k$ vertices and degree sequence $\langle d_1,\ldots,d_k\rangle$.


问:
Solve an Integral
Can anyone help me integrate $$\int_0^1\frac{x^2}{e^x+1}dx$$

Title: Solve an Integral

Body:
$$I(x)=\int \frac{x^2}{e^x+1} dx = \int x^2 \frac{e^{-x}}{1+e^{-x}} dx = \int \sum_{k=1}^{\infty} (-1)^{k+1}  x^2 e^{-kx} dx$$
$$I(x) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{2-2kx+k^2x^2}{k^3} e^{kx}$$ $$\implies I(x) = \frac{x^3}{3}-x^2\ln (1+e^{x})-2x \text{Li}_2(-e^{-x})+2\text{Li}_3(-e^{-x})$$


问:
Proving matrix exponential
Can anyone tell me how the following is derived?

where $A$ is a matrix.

答:
Write

$\dot x = Ax \tag 1$

as

$\dot x - Ax = 0; \tag 2$

multiply through by $e^{-At}$:

$e^{-At} \dot x - e^{-At}Ax = 0; \tag 3$

note that

$e^{-At}A = Ae^{-At}, \tag{3.5}$

and re-arrange (3):

$e^{-At} \dot x - Ae^{-At}x = 0; \tag 4$

observe that the Leibniz rule for differentiating products yields

$\dfrac{d}{dt} (e^{-At} x) = -Ae^{-At}x + e^{-At} \dot x = e^{-At} \dot x - Ae^{-At}x = 0; \tag 5$

thus

$e^{-At} x = c, \; \text{a constant vector}, \tag 6$

whence

$x(t) = e^{At}c; \tag 7$

taking $t = 0$ yields

$x(0) = e^{A \cdot 0}c = e^0c = Ic = c; \tag 8$

thus (7) becomes

$x(t) = e^{At}x(0). \tag 9$

Title: Proving matrix exponential


问:
Are the Schatten p-norms induced norms?
The Schatten p-norm of a matrix is given by

$$\|A\|_p = \left[\text{tr}\left((A^*A)^{p/2}\right)\right]^{1/p}$$

Some of them are induced vector norms. For example, $\|A\|_\infty = \max\{\|Au\|: \|u\|= 1\}$ is the operator norm and $\|\cdot\|$ is the Euclidean 2-norm for vectors.

Can all the other Schatten p-norms be written as induced vector norms? More generally, is there any connection between the vector p-norms and the matrix Schatten p-norms?

答: The Schatten. $p$-norm of a matrix $A\in M_n(\mathbb C)$ is the $p$-norm of the vector containing the singular values of $A$. That is, $\|A\|_p=\|\left(\sigma_1(A),\ldots,\sigma_n(A)\right)\|_p$. When $A$ is a square matrix whose size is at least $2\times2$, $\|A\|_p$ is not an induced norm unless $p=\infty$, because $\|I_n\|_p=n^{1/p}>1$ when $p$ is finite.

Title: Are the Schatten p-norms induced norms?