问:
Polynomials Reference Request
I am searching for bibliographical references to polynomials that are a product of sums where each term of the product has one more summand variable than the previous one, i.e. something like $P(x_1,\dots,x_n) = x_1(x_1+x_2)(x_1+x_2+x_3)\dots(x_1+\dots+x_n)$. Do these have a certain name, special properties, or been studied at all in the maths literature?

答:
You can call it the rising factorial of the sequence $\{x_n\},$ and denote it by $\left (x_k\right )^{\overline{n}}$ notice that if $x_i=1$ then you recover the factorial $n!,$ if $x_n = x+n-1$ you recover the normal rising factorial $x^{\overline{n}}.$ If $x_i = q^{i-1},$ you recover the $[n]_q!$ $q-$factorial.

 People study these objects but not in full generalized form because is hard to say anything about them, they are too general! The only think that I can think you can say is that
$$(x_k)^{\overline{n}}=\sum _{\substack{(a_1,\cdots ,a_n)\\1\leq a_j\leq j}}\prod _{i=1}^nx_{a_i}.$$ which is saying not much.

Probably you will enjoy this reference: Generalized j-Factorial Functions, Polynomials, and Applications by Maxie Schmidt.

Title: Polynomials Reference Request


问:
Singular values of product of matrices
I asked this question earlier and a user pointed out my error with a counterexample. However, it seems like if I change my statement slightly, all the counterexamples pointed out are not an issue anymore.

Given two diagonalizable matrices $A$ and $B$ of same dimension, is it true that

$$\sigma_i(AB) \leq \sigma_i(A)\sigma_1(B),$$

where $\sigma_i(\cdot)$ are the singular values ordered in nonincreasing order? How can one show this?

答:
Yes, this is true. One way to prove this is using the min-max characterization of singular values. In particular,
$$
\sigma_i(PQ) = \max_{S: \dim(S) = i} \min_{x \in S, \|x\| = 1} \|PQx\|\\
\leq \max_{S: \dim(S) = i} \min_{x \in S, \|x\| = 1} \|P\| \cdot \|Qx\|\\
= \sigma_1(P) \cdot \max_{S: \dim(S) = i} \min_{x \in S, \|x\| = 1} \|Qx\|
\\ = \sigma_1(P) \sigma_i(Q).
$$
With that, we have $\sigma_i(PQ) \leq \sigma_1(P) \sigma_i(Q)$. It follows that
$$
\sigma_i(AB) = \sigma_i([AB]^T) = \sigma_i(B^TA^T) \leq \sigma_1(B^T) \sigma_i(A^T) = \sigma_i(A) \sigma_1(B).
$$

ContentLicense: CC BY-SA 4.0

Title: Singular values of product of matrices


问:
Sum of sum of elementary s of subsets
Let n ≥ 1 be an integer. For each subset S ⊂ {1,2,...,3n}, let f (S) be the sum of the elements of S, with f  (ø) = 0.
Determine, as a function of n, the sum
∑
f (S) S ⊂{1,2,...,3n }
3|f (S)
where S runs through all subsets of {1,2,...,3n} such that f (S) is a multiple of 3.
Please help me , im really stuck on this one , maybe its induction , but idk how to find the general formula for n , the case n=1,2,3 are trivial</p>

答:
I don't know the answer but here's a approach that might help: Let $A=\{1,...,3n\}$ and let $M = \{S\subset A : f(S) \text{ is a multiple of } 3\}$. Notice that for any $S \subset A$ we have $f(S) + f(S^c) = f(A) = 3n(3n+1)/2$. Hence $S \in M\iff S^c \in M$, so
$$\sum_{S\in M} f(S) = \frac{1}{2}\sum_{S\in M} \left[f(S) + f(S^c)\right]$$
$$= \frac{3n(3n+1)}{4}\big|M\big|$$

So the problem is equivalent to finding the size of the set $M$. I don't know if that's easier.

Title: Sum of sum of elementary s of subsets


问:
Calculate the integral $\int \frac{\exp(kx)}{x}dx$
How can you calculate the integral $$\int_a^b\frac{\exp(kx)}{x}\,\mathrm dx$$for any $k>0$?</p>

答:
I think you should take a look at Exponential integrals. According to wiki, it is not an elementary function.

For $k>0$ one gets
$\int_a^b\frac{\exp(kx)}{x}\,\mathrm dx = \int_{ka}^{kb}\frac{\exp(x)}{x}\,\mathrm dx = \text{Ei}(kb)-\text{Ei}(ka)$.

ContentLicense: CC BY-SA 3.0


问:
Rank of $3 \times 7$ matrix over GF(2)
Consider the following matrix:
\begin{equation}
 \boldsymbol{A} = \begin{bmatrix}
                     1     & 0     & 0     & 1     & 0     & 1     & 1 \\
                     0     & 1     & 0     & 1     & 1     & 1     & 0 \\
                     0     & 0     & 1     & 0     & 1     & 1     & 1
                  \end{bmatrix},
\end{equation}
where the entries are from GF(2). Checking with MATLAB, the rank of matrix $\boldsymbol{A}$ turns out to be equal to 3. However, it is clear from simple observation that the addition of first and third columns is equal to the last column. Hence the number of linearly independent columns in $\boldsymbol{A}$ is equal to 2, so the rank should be equal to 2.

What is the reason for this discrepancy?

答:
You have found one column that can be expressed as a linear combination of the others. Hence, the $7$ columns are not linearly independent and the conclusion you can make, is that the rank is at most $6$. However, since column rank and row rank always agree, it can be at most $3$ anyway, so you didn't gain anything.

Now note that the first three columns are in fact linearly independent, so the rank is equal to $3$.

Title: Rank of $3 \times 7$ matrix over GF(2)


问:
How can we choose the identity element?
Let $\mathbb{R}^{*}=R\setminus \left \{ 0 \right \}$. Consider maps $f_{i}:\mathbb{R}^{*}\rightarrow \mathbb{R}^{*}$:

$f_{1}(x)=x, f_{2}(x)=\frac{1}{x},f_{3}(x)=-x,f_{4}(x)=-\frac{1}{x}$.
There is a binary operation (called $S$) *  satisfied: $f_{i}$*$f_{j}$=$f_{i}.f_{j}$

Prove that,  the maps above is a group and the group is commutative.

I supposed $e=f_{1}$, but $\forall f_{i} $, choose $i=3$ then $e*f_{3}=f_{3}*e=-x^{2}\neq f_{3}$, I also got trouble with $i=1,2,4$. Please help me, thank you very much!

My problem is, with the definition about the set of $f_{i}$, how can we choose the identity element?

答:
Actually its $f_1 * f_3(x) = f_1(f_3(x)) = f_1(-x) = -x = f_3(x)$.

It is not $f_1(x)×f_3(x)$, which is what you did.

In fact $f_1$ is indeed the identity element.

Title: How can we choose the identity element?


问:
Analysis proof about irrational numbers
Prove that every closed interval $[a,b]$ is a subset of $\mathbb{R}$ contains at least one irrational number.</p>

答: If $a$ or $b$ is irrational, we're done. Otherwise (assuming $a \ne b$), $a+(b-a)/\sqrt2$ is irrational.</p>

Title: Analysis proof about irrational numbers


问:
Find the extreme values of $f(x,y)=x^2y$ in $D=\{x^2+8y^2\leq24\}$
Find the extreme values of $f(x,y)=x^2y$ in $D=\{x^2+8y^2\leq24\}$

It was easy to find using Lagrange multipliers the local extreme values on $\partial{D}$ since we have the condition $x^2+8y^2=24.$ I found that $(-4,1),(-4,-1),(4,1)$ are critical points.

Now I have to find the extreme values inside $D$. $\nabla f(x,y)=(2xy,x^2)=(0,0)$ gives me that every $(0,y)$ is a critical point. However, I'm not sure how to get extremes from here, since I cannot use the determinant test. Would be a possibility to use the monotony of $x^2$ depending the sign of $y$?

答: Rescaling any point with $x^2+8y^2<24$ so that $x^2+8y^2=24$ increases $|x^2y|$, so extrema of $x^2y$, which can change its sign under $(x,\,y)\mapsto(x,\,-y)$, satisfy $x^2+8y^2=24$, so we should extremize $8y(3-y^2)$. The first two derivatives of this are $24(1-y^2),\,-48y$, so at $y=1$ ($y=-1$) with $x^2=16$ we have a maximum (minimum) with $f=16$ ($f=-16$).


问:
What property of modular arithmetic lets us conclude than a^(2b) mod n is congruent to a^b mod n * a^b mod n?
After several years of absence, I seem to have lost track of the very basic of number theory. In my head, I know what I am allowed to do, but not why I am allowed to do it. I have been consulting Wikipedia's list of properties of modular arithmetic, but I can't find any property that matches up to the following:

What property of modular arithmetic lets us conclude that $a^{2b}\mod{n} \equiv a^{b}\mod{n} \cdot a^{b}\mod{n}$ for $a,b,n \in \mathbb{Z}$? I worry that I've confused the Computer Science style of "mod as in remainder upon division"  and the mathematics style of "mod as in congruence", but I was sure that this worked for both. For example, I am confident that $5^4\mod{3} = 5^{2}\mod{3} \cdot 5^{2}\mod{3}=(5^{2}\mod{3})^2=(5\mod{3})^4 =(2\mod{3})^4 =(2)^4\mod{3} = 16 \mod{3} = 1\mod{3}$ is completely valid.

答:
Well the multiplication in $\Bbb Z_n$ is defined as $\overline{ab} = \overline a\cdot\overline b$, where $\overline a$ denotes the residue class of $a$ mod $n$.

This amounts to the congruence $ab\mod n = [(a\mod n)\cdot (b\mod n)]\mod n$.

Your congruence $a^2b^2\mod n≡ab\mod n⋅ab\mod n $ is not correct as it requires $\mod n$ at the end: $a^2b^2\mod n ≡ab\mod n⋅ab\mod n \mod n$


问:
Is there an analog of the spectral theorem for symmetric, non-degenerate bilinear forms?
I was thinking of persymmetric matrices after I saw this question. Persymmetric matrices are notable in that they satisfy $(Ax,y) = (x,Ay)$, where $(\cdot,\cdot)$ the bilinear form over $\Bbb R^{n}$ given by
$$
(x,y) = x_1y_n + x_2 y_{n-1} + \cdots + x_{n-1}y_2 + x_n y_1.
$$
I suspect that if $A$ is "self-adjoint" relative to this bilinear form, then $A$ must be diagonalizable. However, the usual proof of the spectral theorem cannot be extended to this context because of the existence of non-zero vectors $x$ for which $(x,x) = 0$.

Is my suspicion correct? Does there exist some generlization of the spectral theorem to this context? Are there further generalizations (e.g. to non-symmetric bilinear forms)?

答:
Counterexample: Take $n=2$. The given form is $$(x,y)=x^* Qy=x^*\begin{pmatrix}0&1\\1&0\end{pmatrix}y$$
Self-adjointness with respect to $Q$ is equivalent to $$x^*A^*Qy=(Ax,y)=(x,Ay)=x^*QAy$$ $$\iff A^*Q=QA$$
Let $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$. Then it is routine to show it is self-adjoint wrt $Q$, yet there is no matrix $P$ such that $P^{-1}AP$ is diagonal.


问:
Proving that $h$ is not a homotopy equivalence.
Suppose $h:S^1\times S^1\times S^1\rightarrow S^1\times S^1\times S^1$ is given by $h(u, v, w)=(u, v^2, w^7)$. I want to show that $h$ is not a homotopy equivalence.

My thought is this: Suppose for contradiction, we assume that $h$ is a homotopy equivalence. Then $h_\ast$ must be an isomorphism. But since $h$ is not injective, it is not an isomorphism, therefore our assumption was wrong?

答:
Suppose that $p:\tilde{X}\rightarrow X$ is a covering map with $\tilde{X}$ connected. Then $p_*\pi_1(\tilde{X},x_0)\subseteq \pi_1(X,\pi(x_0))$ is a subgroup of index $p^{-1}(p(x_0))$. In particular, a covering map $\pi$ with connected covering space induces and epimorphism of fundamental groups if and only if $p^{-1}(p(x_0)) = \{x_0\}$.

Now your $h$ is a covering with a connected covering space, but has fibers of cardinality $2\cdot 7 = 14$ or something like that. Hence $h_*$ is not surjective and thus $h$ is not a homotopy equivalence.


问:
To construct a set with a limit point.
I learned how to construct a Cantor Set, and I am asked to do the following.

"Construct a bounded set with exactly 3 limit points."

Since the Cantor set contains infinitely many points, I don't think something like it will not work.
But this is the only thing that I have learned from the book that tells me anything about constructing a set that has a limit point.

I am also considering the interval $[0,1]$ and constructing a set so that the limit points are $\{0,1/2,1\}$.

If possible, I would like to see more than one simple examples because I am new to analysis and I have no teacher. It's very tough.

答: Construct a set with exactly one limit point and then add two distinct copies of it; let me be more clear: you'll surely agree with the fact that $\{0\} \cup \{{\frac{1}{n} : n \in \mathbb{N} ^{*}}\}$ has only one limit point. Thus if we repeat twice a translation we'll get what we're looking for: $(\{0\} \cup \{{\frac{1}{n} : n \in \mathbb{N} ^{*}}\}) \sqcup (\{2\} \cup \{{\frac{1}{n}+2 : n \in \mathbb{N} ^{*}}\}) \sqcup (\{4\} \cup \{{\frac{1}{n} +4: n \in \mathbb{N} ^{*}}\})$.</p>

Title: To construct a set with a limit point.


问:
Every singleton in Postman Metric Space is open set?
This is a question from Metric Spaces by Jain and Ahmad.
In metric space $(\mathbb R^2, d)$ I need to find if every singleton set is open or not, where

$d(x,y)=|x_1-y_1|+|x_2-y_2|$

for $x=(x_1, x_2)$ and  $y=(y_1, y_2)$ in $\mathbb R^2$

My Attempt I know that the open sphere centered at any $x\in \mathbb R^2$ will be an open square and it won't be contained in the singleton set considered, but the answer says that every singleton would actually be an open set, except {0}. Please help out.

答:
This is a mistake in the book. As you correctly found, for the metric
$$d^{\ast\ast}(x,y) = \lvert x_1 - y_1\rvert + \lvert x_2 - y_2\rvert$$
the open balls are open squares (with the diagonals parallel to the coordinate axes). This metric induces the standard topology on $\mathbb{R}^2$, hence no nonempty finite set is open. Common names for this metric are $\ell^1$-metric (since it is induced by the $\ell^1$-norm), Manhattan metric (since Manhattan is somewhat famous for a more or less rectangular road network, so the distance one has to travel between two points is the sum of the north-south distance and the east-west distance) or Taxicab metric.

They probably intended to give the metric as
$$d^{\ast\ast}(x,y) = \begin{cases}\qquad 0 &\text{if } x = y \\ \lVert x\rVert + \lVert y\rVert &\text{if } x \neq y \end{cases}$$
where $\lVert\,\cdot\,\rVert$ is a norm on $\mathbb{R}^2$ (often the Euclidean, aka $\ell^2$, norm). This metric has the stated property, for $x \neq 0$ and $0 < r < \lVert x\rVert$ the open ball $B_r(x)$ is the singleton $\{x\}$.

This latter metric is - in my opinion unfittingly - also known as the British Rail metric, SNCF metric, or post office metric. These names are unfitting, because if two points lie on the same line, one doesn't need to travel via London or Paris respectively to reach one's destination, whereas the metric says one does have to.

A more fitting name, under which I first encountered this metric, but which unfortunately doesn't seem to have spread, is the metric of Gaul fishmongers. ("Das Meer? Was hat denn das Meer mit meinen Fischen zu tun?" My translation: "The sea? What has the sea to do with my fish?"; because he buys the fish for his shop in Paris rather than fishing in the sea right behind the village as Asterix suggested.)


问:
subtracting exponents with different bases and finding all the prime factors of it
Here is the question:

Find all prime factors of $3^{18}-2^{18}$

No Calculators allowed, and it is a competition problem, fastest answers are the best.

How can I approach this problem? If it was just $3^{18}$, it would just be 3 and likewise for 2. What do I do?

答: You are supposed to see it as a difference of squares and cubes to start.  $3^{18}-2^{18}=(3^9-2^9)(3^9+2^9)$ and you can factor each of these as a sum or difference of cubes.  That gets the numbers down to primes you may know, or you may need to do some trial division.


问:
What is the number of spanning sub_tree $_$ that have exactly $( − 2)$ leaves?
What is the number of labelled spanning trees of $_$ that have exactly $(n-2)$ leaves?(Consider that $n\geq 4$ )

Do not consider isomorphic trees as one.

I know that the number of labelled spanning trees of $K_$ is equal to $n^{n-2}$.

I think it can be proved with the theorem of inclusion and non-inclusion, but it is very long!

Can you help me?

答:
Hint: consider what a tree with $n-2$ leaves looks like. It is one single central edge $uv$, where every vertex is adjacent to either $u$ or $v$. Further, $u$ has at least one neighbor other than $v$, and $v$ has at least one neighbor other than $u$.

Answer: (click spoiler)

 $$\text{number of spanning trees of $K_n$ with $n-2$ leaves} = \binom{n}{2}\cdot \left(2^{n-2}-2\right)$$

Proof: (click spoiler)

 Each spanning tree with $n-2$ leaves is uniquely determined by a choice of edge $uv$ (i.e., the 'middle' of the spanning tree), and a choice of which of the remaining $n-2$ vertices are adjacent to $u$ --- subject to the constraint that the neighborhood of $u$ cannot contain either all or none of these remaining vertices (see the diagram). There are $\binom{n}{2}$ choices for the middle edge, and $(2^{n-2}-2)$ non-empty, non-full subsets of the remaining $n-2$ vertices that can be the set of neighbors of $u$.


Title: subtracting exponents with different bases and finding all the prime factors of it

Body:
Here is the question:

Find all prime factors of $3^{18}-2^{18}$

No Calculators allowed, and it is a competition problem, fastest answers are the best.

How can I approach this problem? If it was just $3^{18}$, it would just be 3 and likewise for 2. What do I do?

Body:
Note that $3^{18}-2^{18} = (3^{9}-2^{9})(3^{9} + 2^{9})$ by Difference of Squares. Using Difference and Sum of Cubes:

$$(3^{9}-2^{9})(3^{9} + 2^{9})$$

$$(3^{3}-2^{3})(3^{6} + 3^{3}2^{3} + 2^{6})(3^{3}+2^{3})(3^{6}-3^{3}2^{3}+2^{6})$$

Now the factors are a bit more manageable:

$$19\cdot 1009\cdot 35\cdot 577$$

Both of the larger numbers are prime (which can be verified by a bit of trial division), so we have:

$$3^{18}-2^{18} = \boxed{5\cdot 7\cdot 19\cdot 577\cdot 1009}$$


问:
How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
How can we prove the following trigonometric identity?

$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$

答:
Another way to solve it using the following theorem found here (author B.Sury):

Let $p$ be an odd prime, $p\equiv -1 \pmod 4$ and let $Q$ be the set of squares in  $\mathbb{Z}_p^*$. Then, $$\sum_{a\in Q}\sin\left(\frac{2a\pi}{p}\right)=\frac{\sqrt{p}}{2}$$

You may also need to use $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$.

ContentLicense: CC BY-SA 2.5


问:
value of this inverse trigonometric expression.
How to evaluate this expression.
$$\sec^2(\tan^{-1} 2)+\csc^2(cot^{-1}(3))$$
I'm stuck on how to process squares, which is on sec and cosec function?.</p>

答: Let $x=\tan^{-1}2$.  Then $\tan x=2$ and $\tan^2x=4$.  If $\tan^2x=4$, what does $\sec^2x$ equal?</p>


问:
If a measure takes infinitely many values then there is a disjoint sequence of sets of positive measure
I want to prove that if a measure takes infinitely many values then there is a disjoint sequence of sets of positive measure. The notes I'm reading says there is a sequence $$0<\mu(Y_1)<\mu(Y_2)<\dots$$ and that we can construct the sequence $(A_n)$ from $(Y_n)$ recursively. The first idea was to define $$A_n=Y_n\setminus(Y_1\cup\dots\cup Y_{n-1})$$ but by considering a particular example I saw that it didn't work.

Edit : From the comments I realized that I cannot extract such a sequence either.

答:
Let $(S,\Sigma,\mu)$ be a measure space.

Let us start with a definition and some remarks.

We say $A$ is an finite atom if $0<\mu(A)<+\infty$ and, for every $B\subseteq A$, $\mu(B)=0$ or $\mu(B)=\mu(A)$

Note that if $A_1$ and $A_2$ are finite atoms, then either $\mu(A_1 \Delta A_2)=0$ or $\mu(A_1 \cap A_2)=0$. (It can be easily proved by considering the  $\mu(A_1 \cap A_2)=0$ or $\mu(A_1 \cap A_2)=\mu(A_i)$, $i=1,2$).

If $A_1$ and $A_2$ are finite atoms and $\mu(A_1 \Delta A_2)=0$  we say that $A_1$ and $A_2$ are equal $a.e.$ (or almost equal). We write $A_1=A_2$ $a.e.$ (the $a.e.$ comes from the fact that the $\chi_{A_1} = \chi_{A_2}$ $a.e.$).

If $A_1$ and $A_2$ are finite atoms and $\mu(A_1 \cap A_2)=0$  we say that $A_1$ and $A_2$ are disjoint $a.e.$ (or almost disjoint).

Now let us prove that

If $\mu$ takes infinitely many values, then there is a disjoint sequence of sets of positive measure.

Let us prove by cases.

Case 1.  There are infinitely many almost disjoint finite atoms. In this case there is a sequence $\{A_i\}_i$ of almost disjoint finite atoms. In this case it is clear that $\mu$ takes infinitely many values (just consider $\mu(A_1)$, $\mu(A_1 \cup A_2)$ , ...).

Let us prove that there is a disjoint sequence of sets of positive measure.

For each $i$, take
$$E_i= A_i \setminus \left ( \bigcup_{j>i} (A_j\cap A_i) \right )$$
It is easy to see $\{E_i\}_i$ is a sequence of disjoint finite atoms. So $\{E_i\}_i$ is a disjoint sequence of sets of positive measure.

Case 2. There are finitely many almost disjoint finite atoms $A_1, \dots A_n$ and, for any $B\in \Sigma$, if $B \subset S\setminus  \bigcup_{i=1}^n A_i$, then $\mu(B)=0$ or $\mu(B)=\infty$. In this case $\mu$ has only finitely many values. The result is vacuously true.

Case 3. There are finitely many almost disjoint finite atoms $A_1, \dots A_n$ and, there is $B\in \Sigma$, $B \subset S\setminus  \bigcup_{i=1}^n A_i$, such that $0<\mu(B)<\infty$. In this case $B$ is not a finite atom and it does not contain any finite atom. Take $B_1=B$. There is $B_2 \subset B_1$  such that $0<\mu(B_2)<\mu(B_1)$. Since $B_2$ is not a finite atom, there is $B_3 \subset B_2$, such that $0<\mu(B_3)<\mu(B_2)<\mu(B_1)$.

So, by induction, we have a sequence $\{B_i\}_i$ such that
$$ B_1 \supset B_2 \supset B_3 \supset \cdots$$
and
$$\infty > \mu(B_1) > \mu(B_2) > \mu(B_3) > \cdots $$

So it is clear that, in this case, $\mu$ takes infinitely many values.

Let us prove that there is a disjoint sequence of sets of positive measure.

Define, for all $i$, $E_i=B_i\setminus B_{i+1}$.  It is clear that $\{E_i\}_i$ is a sequence of disjoint sets. And, for all $i$, we have

$$\mu(E_i) = \mu(B_i\setminus B_{i+1})= \mu(B_i)-\mu(B_{i+1})>0$$

So $\{E_i\}_i$ is a disjoint sequence of sets of positive measure.

This completes the proof.


问:
In a 52 card deck, with 2 white, 2 black, and the rest blue, what is the probability of picking 4 cards at once with 2 white and 2 black?
Cards: 52

Blue: 48

Black: 2

White: 2

What is the probability of picking a 4 card bundle (all cards at once) in which you have 2 black cards, and 2 white, in any order?

I've already tried to solve this, but the fact that we take all cards at once confuses me as I'm not sure whether we subtract cards from the 'total probabilities' side after picking one. Perhaps this is a hypergeometric problem?
I'd appreciate any help. I wanted to give this problem to my friend and realized I could not solve it myself.

答:
There is only one way to have 2B and 2W, thus the probability is

$$\frac{1}{\binom{52}{4}}=\frac{1}{270,725}$$


问:
Showing that a topology is coarser than another using neighborhood basis
I have the following statement that I need to prove:
Let $ \tau_{1} $and $ \tau_{2}$ be two topologies on the set $X$ and for every $x\in X$ let $\mathcal B^{1}(x)$ and $\mathcal B^2(x)$ be bases of a Neighborhood-System of x in respectively $\tau_{1}$ and $\tau{2}$.

Prove that:

$\tau_{1}$ is coarser(weaker) than $\tau{2}$ $\iff \forall x \in X, \space\forall B^1\in \mathcal B^{1}(x) \exists B^2\in\mathcal B^2(x):B^2 \subseteq B^1 $

So I am completely stuck. I understand that for a topology to be coarser it needs to fufill the condition that $\tau_{1} \subseteq \tau_{2}$, and I tried to start with an element of $\tau_{1}$ and tried to find a connection with the right hand side but without success. I know also that these neighborhood bases need to fufill 3 conditions and I know the conditions but I am not certain at all at how that might help me out eventually in this case.

I would be very thankful for any sort of help

答: Hint:  The given condition means anything open in $\tau_1$ is open in $\tau_2$.  That is, let $U\in\tau_1$.  Then let $x\in U$. Then there is $B_1\in\tau_1$ such that $x\in B_1\subset U$.  By the given condition there is $B_2\in\tau_2$ such that $x\in B_2\subset B_1\subset U$.  Thus $U\in\tau_2$.


问:
${\mathrm{GL}}(4,{\mathbb{F}}_q)$ action on the $(2,4)$-Grassmann
Let ${\mathbb{F}}_q$ be the finite field of order $q$ and ${\mathcal{G}}(2,4)$ denote the set of all $2$-dimensional subspaces of ${\mathbb{F}}^4_q$. Let $P$ denote the $(2,2)$-parabolic (which is the subgroup of all matrices of ${\mathrm{GL}}(4,{\mathbb{F}}_q)$ which has lower triangular $2 \times 2$ block zero while written in $2 \times 2$ sub-blocks).

There is a natural transitive left action of ${\mathrm{GL}}(4,{\mathbb{F}}_q)$ on ${\mathcal{G}}(2,4)$: if $W \in {\mathcal{G}}(2,4), g \in {\mathrm{GL}}(4,{\mathbb{F}}_q)$ with $W = {\mathrm{span}} \{ v_1, v_2 \}$, then $g \cdot W := {\mathrm{span}} \{ g \cdot v_1, g \cdot v_2 \}$. Setting $\{ e_1, e_2, e_3, e_4 \}$ as the canonical basis of ${\mathbb{F}}^4_q$, clearly $P$ turns out to be the stabiliser of $W_0 = {\mathrm{span}} \{ e_1, e_2 \}$.

${\mathrm{\bf Question.1:}}$ Show that ${\mathrm{GL}}(4,{\mathbb{F}}_q)/P$ can be identified with ${\mathcal{G}}(2,4)$.

Since $P$ is not a normal subgroup ${\mathrm{GL}}(4,{\mathbb{F}}_q)$, this identification doesn't seems to come from the natural action. What is the best way to see this?

${\mathrm{\bf Question.2:}}$ Under this identification, two elements of ${\mathrm{GL}}(4,{\mathbb{F}}_q)/P$ are in the same orbit of $P$ if and only if the corresponding subspaces intersect in the same dimensional subspaces of ${\mathrm{span}} \{ e_1, e_2 \}$.

What is the meaning of orbit here?

One can make an equivalence relation on ${\mathcal{G}}(2,4)$ satisfying this property to explain the orbit. However, I don't see a natural connection with ${\mathrm{GL}}(4,{\mathbb{F}}_q)/P$. What am I missing?

答:
If $G$ acts on a set $\Omega$, the action is called transitive if for every $\omega,\omega'\in\Omega$ there exists a $g\in G$ for which $\omega'=g\omega$. The stabilizer $\mathrm{Stab}(\omega)\le G$ of a point $\omega\in G$ is the set of all $g\in G$ which fix $\omega$, i.e. $g\omega=\omega$. The orbit $\mathrm{Orb}(\omega)$ of a point $\omega\in\Omega$ is the set of all "translates" $g\omega$ of $\omega$ (for all $g\in G$). Note a $G$-set $\Omega$ is transitive if and only if $\Omega=\mathrm{Orb}(\omega)$ for any (or all) $\omega\in\Omega$, i.e. it's all one orbit. Two $G$-sets $\Omega_1$ and $\Omega_2$ are called equivalent or isomorphic if there is a $G$-equivariant bijection between them. A coset space $G/H$ is itself a $G$-set, with the action by left-multiplication.

The orbit-stabilizer theorem says if $G\curvearrowright\Omega$ is a transitive action then $\Omega$ is equivalent (as a $G$-set) to a coset space $G/\mathrm{Stab}(\omega)$ (for any point $\omega\in\Omega$). Specifically, $g\mathrm{Stab}(\omega)\leftrightarrow g\omega$. Note $\mathrm{Stab}(g\omega)=g\mathrm{Stab}(\omega)g^{-1}$ so this sets up a one-to-one correspondence between isomorphism classes of transitive $G$-sets and conjugacy classes of subgroups $H$.

In particular, $\mathrm{GL}_4\mathbb{F}_2$ acts on $\mathrm{Gr}_2\mathbb{F}_2^4$ transitively, and the stabilizer of the first-two-coordinates-plane is $P$, so we automatically know that $\mathrm{GL}_4\mathbb{F}_2/P$ is equivalent to $\mathrm{Gr}_2\mathbb{F}_2^4$. This answers (1).

For (2) you'll want to explore how $P$ acts on $\mathrm{Gr}_2\mathbb{F}_2^4$...


问:
Formal Definition of differentiable
Given
\begin{align*}
y(x) =\begin{cases} \int^x_0 \frac{e^{e^t}}{e^x} dt \ &\mbox{for} \ x > 0 \\
   a + bx + cx^2                                 &\mbox{for} \ x \leq 0 \end{cases}
\end{align*}

I am trying to use the formal definition of limit,

$f$ is differentiable in $x = x_0$ if the limit
\begin{align*}
f'(c) = \lim_{h\rightarrow 0} \frac{f(x_0+h) - f(x_0)}{h} = m \ \mbox{exist}.
\end{align*}
to show that $f(x)$ is differentiable. So far I've written that y must satisfy \begin{align*}
\lim_{x\rightarrow 0_-} = \lim_{x \rightarrow 0_+} = \frac{y(x) - y(0)}{x - 0}
\end{align*}
and calculated that
\begin{align*}
\lim_{x\rightarrow 0_-} &y(x) = a\\
\lim_{x\rightarrow 0_+} &y(x) \ \mbox{gives} \ \frac{0}{0} \ \mbox{form, so I used L'Hôpitals rule again}
\end{align*}
\begin{align*}
y'(x)= \begin{cases}  \frac{e^{e^x}}{e} \ &\mbox{for} \ x > 0 \ \mbox{(by the fundmental theorem of calculus)}\\
 2cx + b   \ &\mbox{for} \ x \leq 0 \end{cases}\\
\end{align*}

\begin{align*}
\lim_{x\rightarrow 0_+} y(x) = \lim_{x\rightarrow 0_+} \frac{e^{e^x}}{e} = 1    \ ( => a \ \mbox{must be} \ = 1)
\end{align*}

And I have found that $f(x)$ is continous in $x = 0$ because
\begin{align*}
\lim_{x\rightarrow 0} y(x) = \lim_{x\rightarrow 0} a + bx + cx^2 = a = f(0).
\end{align*}
Now, trying to use the formal definition of differentiable:

\begin{align*}
 f'(0) &= \lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h} \\
      &= \lim_{h\rightarrow 0} \frac{\int_0^h \frac{e^{e^t}}{eh} dt - a}{h}  \\
      &= \lim_{h\rightarrow 0} \frac{\int_0^h \frac{e^{e^t}}{eh}dt - 1}{h} |\cdot eh \\
      &= \lim_{h\rightarrow 0}  \frac{\int_0^h e^{e^t}dt - eh}{eh^2}
\end{align*}

But I dont know how to proceed from here. I tried getting rid of the integral, and ended up with
\begin{align*}
f'(0) &= \lim_{h\rightarrow 0} \frac{[e^{e^0} - e^{e^0}] - eh}{eh^2} \ \mbox{(Direct substitution for} \ h = 0) \\
&= \lim_{h\rightarrow 0} \frac{[e^1 - e^1] - eh}{eh^2}\\
&= \lim_{h-\rightarrow 0} \frac{0 - eh}{eh^2}= \frac{-1}{h}
\end{align*}
And I know this doesn't work because I get zero in the denominator. I appreciate all help I can get!

答:
$\lim_{x\to0^+}y(x)=\lim_{x\to0^+}\frac1{e^x}\int_0^xe^{e^t}dt=\lim_{x\to0^+}\int_0^xe^{e^t}dt=0$ since $e^x\to1^+$. It does not have the $0/0$ form. This gives $\color{red}{f(0)=a=0}$ for continuity.

$y'(x)$ is not $e^{e^x}/e$ for $x>0$. $$\begin{align*}y(x)&=e^{-x}\int_0^xe^{e^t}dt\\y'(x)&=-e^{-x}\int_0^xe^{e^t}dt+e^{-x}\times e^{e^x}\\&=e^{-x}\left[e^{e^x}-\int_0^xe^{e^t}dt\right]\\&=e^{e^x-x}-y(x)\end{align*}$$by the product rule.

Left hand derivative at $0$ is $b$. Right hand derivative is given by$$\begin{align*}\lim_{h\to0^+}\frac{y(h)-y(0)}h&=\lim_{h\to0^+}\frac{\int_0^he^{e^t}dt}{he^h}\\&=\lim_{h\to0^+}\frac{e^{e^h}}{e^h(1+h)}\\&=e\end{align*}$$giving $\color{blue}{b=e}$ for differentiability. Note that this limit has the $0/0$ form.

Title: Formal Definition of differentiable


答:
So you have $4$ good cards and $48$ bad cards and $52$ total cards.

You choose $4$ cards.  There are literally "$52$ choose $4$" $= {52 \choose 4}$ ways of doing that.  And only one of them are your $4$ good cards.

So the probability $\frac 1{52 \choose 4}$.

$\frac 1{52\choose 4} = \frac 1{\frac{52!}{48!4!}}= \frac {4!48!}{52!}=\frac {4!}{49*50*51*52}= \frac {1*2*3*4}{49*50*51*52}=\frac 1{49}*\frac 1{25}*\frac 1{17}*\frac 1{13}=\frac 1{270725}$

Title: In a 52 card deck, with 2 white, 2 black, and the rest blue, what is the probability of picking 4 cards at once with 2 white and 2 black?

Body:
Cards: 52

Blue: 48

Black: 2

White: 2

What is the probability of picking a 4 card bundle (all cards at once) in which you have 2 black cards, and 2 white, in any order?

I've already tried to solve this, but the fact that we take all cards at once confuses me as I'm not sure whether we subtract cards from the 'total probabilities' side after picking one. Perhaps this is a hypergeometric problem?
I'd appreciate any help. I wanted to give this problem to my friend and realized I could not solve it myself.


问:
Hint for a question about automorphisms
let $F$ a field and $f$ an automorphism of the polynomials ring $F[x]$. show that exists $g $automorphism of $F$ such that $f(c) = g(c)$ for all $c \in F$ and exists $a,b \in F, a \neq 0$ such that $f(x) = ax + b. $

can anyone send me a hint to solve this?

答:
$f$ must map $0\mapsto 0$ and $1\mapsto 1$ and then also must map $F[x]^\times \to F[x]^\times$, therefore induces an automorphism of $F = F[x]^\times\cup \{0\}$. So let us call this $g$. Then clearly $g(c)=f(c)$ for all $c\in F$.
Now $f(x)$ must be some polynomial in $x$ of some degree $d$.  Observe that the degree of $a_nx^n+\cdots +a_1x+a_0$ (with $a_n\ne 0$) then turns out to be $nd$. By surjectivity of $f$, we must have $d=1$.

Title: Hint for a question about automorphisms


答: Hints: if $p, q \in F[x]$ and $pq = 1$, then$p, q \in F$. If $p \in F$ has $\deg(p) > 1$ or $\deg(p) = 0$, then $x$ does not belong to the ideal generated by $p$.

Title: Hint for a question about automorphisms

Body:
let $F$ a field and $f$ an automorphism of the polynomials ring $F[x]$. show that exists $g $automorphism of $F$ such that $f(c) = g(c)$ for all $c \in F$ and exists $a,b \in F, a \neq 0$ such that $f(x) = ax + b. $

can anyone send me a hint to solve this?


问:
Greedy algorithm , the coin change problem proof
How to proof that a greedy algorithm will work on this set of coins $ S = {1,2,3...K} $.
I haven't  seen a proof on this type of sets but saw on this set $1,5,10$

答: The greedy algorithm will use $\lceil \frac nK\rceil$ coins. Any better method would use $r$ coins for some $r$ with $rK<n$, which is absurd.


问:
Does $[G,G] \trianglelefteq \text{ker}(\Psi)$ hold?
Let $\Psi:G \to H$ be a homomorphism between a group $G$ and an Abelian group $H$,then show :

$$[G,G] \le \text{ker}(\Psi)$$
Where $[G,G]$ is the commutator subgroup of $G$.

I came up with the question: Does $$  [G,G] \trianglelefteq \text{ker}(\Psi)$$

hold ?

My work:

let $x,y \in [G,G]$ then $x=[a,b]$ and $y=[c,d]$ for some $a,b,c,d \in G$,then:

$$\Psi(xy^{-1})=\Psi(x)\Psi(y^{-1})$$

From the fact that $[h,g]^{-1}=[g,h]$ it follows that:

$$\Psi(xy^{-1})=\Psi([a,b])\Psi([d,c])=\Psi(a)\Psi(b)\Psi(a)^{-1}\Psi(b)^{-1}\Psi(d)\Psi(c)\Psi(d)^{-1}\Psi(c)^{-1}$$

Since $\forall g \in G:\Psi(g) \in \Psi(G)=\text{Im}(G) \subseteq H$ and $H$ is Abelian hence:

$$\Psi(xy^{-1})=e_H$$
I conclude that $xy^{-1} \in \text{ker}(G)$ but this does not imply $xy^{-1} \in [G,G]$ ,on the other hand $\text{ker}(G) \trianglelefteq G$ we see that $xy^{-1} \in G$,again this is not useful.

So is my proof  wrong or it's not true to claim that $[G,G] \trianglelefteq \text{ker}(\Psi)$?

答:
Yes because $\Phi(G)\le H$ must be abelian, as a subgroup of an abelian group.  By the first isomorphism theorem $\Phi(G)\cong G/\rm{ker}\Phi$.  And, finally, $G/H$ abelian if and only if $G'\subset H$.

Furthermore, $G'\trianglelefteq G\implies G'\trianglelefteq H$.


答:
By definition, $[G,G]$ is generated by elements of the form $[x,y]=xyx^{-1}y^{-1}$ (and not $xy^{-1}$). It suffices to show that every commutator is in $\ker \psi$ to show that $[G,G]\subset \ker \psi$. But

\begin{align}
\psi(xyx^{-1}y^{-1}) &= \psi(x)\psi(y)\psi(x^{-1})\psi(y^{-1})\\
&= \psi(x)\psi(x)^{-1}\psi(y)\psi(y)^{-1}\\
&= 1
\end{align}
Because $H$ is abelian.

The fact that $[G,G]$ is normal is because every conjugate of a commutator is a commutator :
\begin{align}
z [x,y]z^{-1} &= zxyx^{-1}y^{-1}z^{-1}\\
&=zxz^{-1}zyz^{-1}(zxz^{-1})^{-1}(zyz^{-1})^{-1}\\
&= [zxz^{-1},zyz^{-1}]
\end{align}
Thus, if $g = [x_1,y_1]\cdot [x_2,y_2]\cdots[x_n,y_n] \in [G,G]$, and if $h \in G$ :
\begin{align}
hgh^{-1} &= h\left([x_1,y_1]\cdot [x_2,y_2]\cdots[x_n,y_n]\right)h^{-1}\\
&= h[x_1,y_1]h^{-1}h[x_2,y_2]h^{-1}\cdots h[x_n,y_n]h^{-1}\\
&= [hx_1h^{-1},hy_1h^{-1}]\cdot[hx_2h^{-1},hy_2h^{-1}]\cdots[hx_nh^{-1},hy_nh^{-1}] \in [G,G]
\end{align}

Title: Does $[G,G] \trianglelefteq \text{ker}(\Psi)$ hold?

Body:
Let $\Psi:G \to H$ be a homomorphism between a group $G$ and an Abelian group $H$,then show :

$$[G,G] \le \text{ker}(\Psi)$$
Where $[G,G]$ is the commutator subgroup of $G$.

I came up with the question: Does $$  [G,G] \trianglelefteq \text{ker}(\Psi)$$

hold ?

My work:

let $x,y \in [G,G]$ then $x=[a,b]$ and $y=[c,d]$ for some $a,b,c,d \in G$,then:

$$\Psi(xy^{-1})=\Psi(x)\Psi(y^{-1})$$

From the fact that $[h,g]^{-1}=[g,h]$ it follows that:

$$\Psi(xy^{-1})=\Psi([a,b])\Psi([d,c])=\Psi(a)\Psi(b)\Psi(a)^{-1}\Psi(b)^{-1}\Psi(d)\Psi(c)\Psi(d)^{-1}\Psi(c)^{-1}$$

Since $\forall g \in G:\Psi(g) \in \Psi(G)=\text{Im}(G) \subseteq H$ and $H$ is Abelian hence:

$$\Psi(xy^{-1})=e_H$$
I conclude that $xy^{-1} \in \text{ker}(G)$ but this does not imply $xy^{-1} \in [G,G]$ ,on the other hand $\text{ker}(G) \trianglelefteq G$ we see that $xy^{-1} \in G$,again this is not useful.

So is my proof  wrong or it's not true to claim that $[G,G] \trianglelefteq \text{ker}(\Psi)$?


问:
No covering map from $S^1\vee S^1\vee S^1$ to $S^1\vee S^1$
I want to see why there is no covering map from $S^1\vee  S^1\vee  S^1$ to $S^1\vee  S^1$. The fundamental group of $S^1\vee  S^1\vee  S^1$ is $\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ but the fundamental group of $S^1\vee S^1$ is $\mathbb{Z}\ast\mathbb{Z}$. Since their fundamental group is not isomorphic, there is no covering map. Is my reasoning right?

答:
There are good places to use algebra and there are bad places. You might be tempted to use algebra here because a covering map induces an injective map on fundamental groups, so this implies that the free group on 3 letters injects into the free group on 2 letters, and surely this is false, right?

Well actually this is true, and we do have a covering that witnesses this injection. It just is the case that the domain cannot be the wedge of three circles.

So instead of using algebra, let's just use topology! A covering map is a local homeomorphism. Where do these spaces look different? A small open set around the wedge point of the three circles is clearly not homeomorphic to any open set in the wedge of two circles. Try proving this rigorously!


问:
How to check continuity of this multivariable function
Here's the function:
$$f(x,y)=\begin{cases}{y+{1\over y}\arctan({x^2y})} & y\neq 0\\ 0 & y=0\end{cases}$$
I need to study it's continuity and I've got a hard time understanding exactly how to rigorously formulate my findings (because I don't understand everything that well), so I would like to know if my reasoning is correct.

Firstly what I found was that in $f(x,y)=(0,0)$ the function is continuous in $(0,0)$  because the limit of this function for $(x,y)\to(0,0)$ is indeed $0$.

Secondly, the function isn't continuous in $f(x,y)=(x,0)$ if $x$ isn't zero because the limit of f when $y\to 0 $ and $x$ is "fixed" is $1$.

Is this reasoning correct and complete ?

答:
We have that for $x=0$

$$\lim_{y\to 0}\,\left(y+{1\over y}\arctan({0\cdot y})\right)=\lim_{y\to 0}\,y=0$$

and for $x =x_0\neq 0$

$$\lim_{y\to 0}\left(y+{1\over y}\arctan({x_0^2y})\right)=\lim_{y\to 0}\left(y+\frac{\arctan({x_0^2y})}{x^2y}x_0^2\right)=x_0^2$$

therefore the function is not continuous.


问:
more gluing ideals together
Let $X$ be a scheme and let $U_i= Spec R_i$ be an open cover of $X$ by open affines.  Suppose we have a family of ideals $I_i\subset \mathcal{O}_X(U_i)=R_i$ and let's say I want to know if the family $\{I_i\}$ cuts out a closed subscheme of $X$. Here are two observations:

It is necessary and sufficient to check that any two $I_i,I_j$ generate the same ideal in every $W\subset U_i\cap U_j$ when $W$ is distinguished open affine in both $U_i$ and $U_j$.

By observation 1., it suffices to show that any two $I_i,I_j$ generate the same ideal in every $\mathcal{O}_X(U_i\cap U_j)$ (since the restriction from $  U_i$ to $W$ factors through the restriction from  $U_i$ to $ U_i\cap U_j$)

My question is whether 2. is also necessary. I suspect the answer is no. A counterexample will have to have $X$ non separated since we need $ U_i\cap U_j$ to be non affine.  and I guess I haven't met too many of those

答:
Your suspicion is correct.

Consider an affine plane $\mathbb{A}^2_k$ over some field $k$ and let $o$ be its origin. We glue two copies of $\mathbb{A}^2_k$ along $\mathbb{A}^2_k\setminus \{o\}$. We obtain a $k$-scheme $X$ which is called the plane with double origin. We denote two copies of $\mathbb{A}^2_k$ in $X$ by $U_1,U_2$. Each one of them has its own private origin $o_1,o_2$, respectively. Moreover, $U_1\cap U_2 = \mathbb{A}^2_k\setminus \{o\}$. Note also that restrictions
$$\Gamma(U_1,\mathcal{O}_{X}) \rightarrow \Gamma(U_1\cap U_2,\mathcal{O}_{X}),\,\Gamma(U_2,\mathcal{O}_{X})\rightarrow \Gamma(U_1\cap U_2,\mathcal{O}_{X})$$
are isomorphism, because open subset $U_1\cap U_2$ is of codimension two in normal schemes $U_1,U_2$.

Now we pick $Z = \{o_1\}$ with reduced structure. This is a closed subscheme of $X$. Denote by $I_1,I_2$ ideals of $Z\cap U_1,Z\cap U_2$ in $\Gamma(U_1,\mathcal{O}_{X})$ and $\Gamma(U_2,\mathcal{O}_{X})$, respectively. Then $I_1\subsetneq \Gamma(U_1,\mathcal{O}_{X})$ is some proper ideal. On the other hand $I_2 = \Gamma(U_2,\mathcal{O}_{X})$ due to the fact that $Z\cap U_2 = \emptyset$. Since restrictions
$$\Gamma(U_1,\mathcal{O}_{X}) \rightarrow \Gamma(U_1\cap U_2,\mathcal{O}_{X}),\,\Gamma(U_2,\mathcal{O}_{X})\rightarrow \Gamma(U_1\cap U_2,\mathcal{O}_{X})$$
are isomorphism, we derive that the image of $I_1$ in $\Gamma(U_1\cap U_2,\mathcal{O}_{X})$ is proper ideal and the image of $I_2$ in $\Gamma(U_1\cap U_2,\mathcal{O}_{X})$ is nonproper. Thus these images are not the same.

P.S. For 2 to be neccessary it suffices to assume that $X$ is semiseparated, which means that the diagonal $X\rightarrow X\times X$ is affine or equivalently intersection of any two open, affine subschemes is affine. This is a more general class of schemes than separated schemes. The plane with double origin is the simplest example (that I am aware of) of a scheme which is not semiseparated.

Title: more gluing ideals together


问:
Examine the continuity of $f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$ at $x=0$

Examine the continuity of $$f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$$ at $x=0$

I tried to solve the problem using $$\lim_{x \to0^+}f(x)=\lim_{x \to0^-}f(x)=f(0)$$ and I got both limits to be $0$. I think there is an error somewhere since my book says it is discontinuous.

Can you help me out?

 Thankyou

答:
Note that it's a geometric series, as Haral Hanche-Olsen commented. You can write $f(x)$ as

$$f(x) = x^2 \frac{1-\frac{1}{(1+x^2)^{n+1}}}{1-\frac{1}{1+x^2}}=x^2\frac{(1+x^2)^{n+1}-1}{x^2(1+x^2)^{n}} = \frac{(1+x^2)^{n+1}-1}{(1+x^2)^{n}}$$

This function is continuous at $x=0$ in my sense...


问:
Proving $\forall n \geq10, n^3 < 2^n$
Im asked to prove the following:

$$\forall n \geq10, n^3 < 2^n$$

I'm 100% sure that the easiest way to prove this is via induction but I'm not being able to do so. How can I prove this?

答:
Proof without induction : The function $$f : x \mapsto \frac{2^x}{x^3}$$

is differentiable over $[10, +\infty)$, and $$f'(x)=\frac{2^x\left[\ln(2)x - 3\right]}{x^4} > 0$$

for all $x \geq 10$. Therefore $f$ is strictly increasing over $[10, +\infty)$, so for $n \geq 10$, you have $$f(n) \geq f(10) = \frac{2^{10}}{10^3} = \frac{1024}{1000} > 1$$

i.e. for all $n \geq 10$, you have
$$2^n > n^3$$

Title: Proving $\forall n \geq10, n^3 < 2^n$


问:
An olympiad question on numbers

Consider the sum of $92 + 105$, so the expression is $197$, however if you turn it upside down, you get $501+ 26$ or $527$ :
\begin{align}
105+92 \to 26+501
\end{align}
Now I'm looking for an expression that when you turn it upside down the sum will increase by exactly 2020 units !!!

But for solving this puzzle we have the following rules :

1- We are just allowed to use digits and the signs $+ , - $

2- None of the $2$ numbers (before and after the spin) can start with $0$

3-The value of the phrase must be positive.

Here is my try :
If we pay attention to the example we'll get some information: the difference between two numbers $92$ and $26$ is $66$ and the difference between $105$ and $501$ is $396$ and also the difference between $197$ and $527$ is $330$ now look at this : $ 396 - 66 = 330$; I guess there must be a relation between this and our answer!

Can anyone help ?

答:
To begin, I assume that you are allowed many numbers and many $+$'s and/or $-$'s in your expression.  If you only allow your expression to involve two numbers and a single $+$ or $-$ then more thought is required.

Let the flip-function be labeled as $F$ which takes in an unevaluated expression and outputs the flipped expression which can then be evaluated.

First, let us notice that if the expression passed in to $F$ contains only addition, that the result has the same remainder modulo $3$ before and after.  This is because the digits themselves who get flipped stay the same remainder class modulo $3$ before and after and rearranging the digits in a summation or in a number itself does not change the remainder modulo $3$.  As such, an expression involving only addition minus the flip of the expression will always result in an expression that evaluates to a multiple of $3$ and could never result in $2020$.

It follows then that for such an expression to exist who when subtracting the flip of the expression evaluates to $2020$ there must be subtraction involved.

Well, $92-15=77$ while $F(92-15)=51-26=25$.  We have $(92-15)-F(92-15)=77-25=52$ which has remainder $1$ when considered modulo $3$, the same equivalence class as $2020$ is.

Now, note that:

$9+9+9+\dots + 9= 9k$

$F(9+9+9+\dots+9)=6+6+6+\dots+6=6k$

$9+9+\dots+9-F(9+9+\dots+9)=9k-6k=3k$

So, we could have the following expression:

$$\underbrace{9+9+9+\dots+9}_{k~\text{copies}} + 92 - 15$$

The flip of the expression:

$$51-26+\underbrace{6+6+6+\dots+6}_{k~\text{copies}}$$

The difference of these:

$52 + 3k$ which by setting $k=656$ is precisely equal to $2020$

With regards to only allowing our initial expression to include two numbers and a single operation, we can limit our search noting what we did earlier that it must be a minus and could not have been a plus.  Writing a computer script can yield some results:

For instance $$6969-2296$$

as well as $$6969-2626$$

Indeed, $6969-2296 = 4673$ and $F(6969-2296)=9622-6969=2653$ which is precisely $2020$ off from the earlier evaluation.

Admittedly, I don't see a clean way to have arrived at these numbers by hand without a great deal of trial and error, but knowing the numbers must have been different equivalence classes modulo $3$ and a minus sign must be used helps narrow it down considerably.

 The admittedly very dirty script: flip = function(exp){ exp = exp.replaceAll("6","n").replaceAll("9","x"); exp = exp.replaceAll("n","9").replaceAll("x","6"); ret = ''; for(i=0;i<exp.length; i++){ret+=exp[exp.length-i-1]} return ret} for(a=5000;a<7000;a++){ for(b=1;b<a; b++){ exp = '' + a + '-' + b; if(exp.contains("3")||exp.contains("4")||exp.contains("7")){continue} else{if(eval(exp +'-('+flip(exp)+')')==2020){console.log(exp)}}}} which searched for solutions with $a$ only in the range $5000$ to $7000$ and ignored limitations on leading zeroes.

Title: An olympiad question on numbers


问:
Working with a Cyclic System of Equations
I recently found the following problem:

Find all non-negative reals $(x,y,z,w)$ such that:
$$\begin{cases}x^2 = 3y + 10z + 5w\\
y^2 = 3z + 10w + 5x\\
z^2 = 3w + 10x + 5y\\
w^2 = 3x + 10y + 5z\end{cases}$$

After looking at previous MSE forums regarding cyclic systems, the main piece of advice I got was to try adding everything up. That yields: $x^2 + y^2 + z^2 + w^2 = 18(x+y+z+w)$ but I don't know how to continue on from there.

After entering the problem into WolframAlpha, I figured out that the only solutions that actually work are $(0,0,0,0)$ and $(18,18,18,18).$ That makes me want to find a way to prove that $x=y=z=w$ but I don't really get how to.

Any help would be greatly appreciated!

答:
Just a partial answer :

suppose we have $x \leq y \leq z \leq w  $,

then  $x^2 \leq y^2 \leq z^2 \leq w^2 $ which we can use to write

$$3y + 10z + 5w \leq 3z + 10w + 5x \iff 3(y-z)+10(z-w)+5(w-x) \leq 0 $$
$$ 3z + 10w + 5x \leq 3w + 10x + 5y \iff 3(z-w)+10(w-x)+5(x-y) \leq 0 $$
$$3w + 10x + 5y \leq 3x + 10y + 5z \iff 3(w-x)+10(x-y)+5(y-z) \leq 0$$

Summing all the eq. we obtain :

$$3(y-x)+10(z-y)+5(w-z) \leq 0 $$

but if at least one of the three terms is strictly greater than $0$ (e.g. $y>x$) we can notice that

$$3(y-x)+10(z-y)+5(w-z) > 0 $$

which is a contradiction.

I don't think this method can be extended to other possible cases.


答: To show that $(n+1)^3 < 2n^3$, you can also use the fact that for $n \ge 10$: $$\dfrac {(n+1)^3}{n^3} = \left(1+\dfrac1n\right)^3 \le \left(1+\dfrac1{10}\right)^3 < 2$$

Title: Proving $\forall n \geq10, n^3 < 2^n$

Body:
Im asked to prove the following:

$$\forall n \geq10, n^3 < 2^n$$

I'm 100% sure that the easiest way to prove this is via induction but I'm not being able to do so. How can I prove this?


答:
Suppose there is an algorithm that in some case gives an answer that includes two coins $a$ and $b$ with $a,b<K$. If $a+b\le K$, then the two coins can be replaced with one coin, which would mean the algorithm is not optimal. If $a+b>K$, then you can replace the two coins by a $K$ coin and a $a+b-K$ coin for an equally good solution using more of the value $K$ coins.

Therefore, any optimal algorithm can without loss of generality be assumed to use at most one coin that is not of value $K$. There is only one way to do this for any total value, and this is the same as the solution that the greedy algorithm gives.

Title: Greedy algorithm , the coin change problem proof

Body:
How to proof that a greedy algorithm will work on this set of coins $ S = {1,2,3...K} $.
I haven't  seen a proof on this type of sets but saw on this set $1,5,10$


答:
You are correct that $f$ is continuous at $0$ although I would really like to see your calculation before confirming that your logic is correct.

Note that the limit of $f$ at $(m,0)$ for $m\ne0$ over $y>0$ is $m^2$, not $1$. This is because$$\lim_{(x,y)\to(m,0^+)}\left[y+\frac{\arctan(x^2y)}y\right]=\lim_{y\to 0^+}y+\lim_{(x,y)\to(m,0^+)}\left[\frac{\arctan(x^2y)}{x^2y}\right]\lim_{x\to m}x^2\\=0+1\cdot m^2\ne0$$which concludes that the limit does not exist since it is $0$ along $y=0$.

Title: How to check continuity of this multivariable function

Body:
Here's the function:
$$f(x,y)=\begin{cases}{y+{1\over y}\arctan({x^2y})} & y\neq 0\\ 0 & y=0\end{cases}$$
I need to study it's continuity and I've got a hard time understanding exactly how to rigorously formulate my findings (because I don't understand everything that well), so I would like to know if my reasoning is correct.

Firstly what I found was that in $f(x,y)=(0,0)$ the function is continuous in $(0,0)$  because the limit of this function for $(x,y)\to(0,0)$ is indeed $0$.

Secondly, the function isn't continuous in $f(x,y)=(x,0)$ if $x$ isn't zero because the limit of f when $y\to 0 $ and $x$ is "fixed" is $1$.

Is this reasoning correct and complete ?


问:
Is $[x, x)$ an invalid interval for *all* real values of $x$?
I'm writing some software at work and I believe it isn't a valid interval because it seems contradictory. To me, it reads as both an inclusion and exclusion of $x$ which seems like it would nonsense in all possible cases -- not even the empty interval would satisfy both criteria.

I wanted to ask here because I don't necessarily know enough to say for sure that this interval should be considered invalid. Rather than pulling such a statement out of relatively nowhere and asserting it in code for others, I'd like to hear from those who know more about it.

答: I would call it $\emptyset$, consistently with the notation $[a,b)=\{x\in\Bbb R\,:\, a\le x\land x<b\}$. This is an interval because it is true that $$\forall x\in\emptyset,\forall y\in\emptyset,\forall z\in\Bbb R,(x<z<y\rightarrow z\in\emptyset)$$


问:
Operations that preserve non-convexity
There are a number of operations that may be done for convex functions such that the resulting function is convex as well.

What about the opposite? Do the operations that preserve convexity also preserve non-convexity?

In particular, I am interested whether the following two statements are true:

Vector composition:
Let $g_i(x)\in\mathbb{R},i=1,\dots,n$ be convex and $h(y)\in\mathbb{R},y\in\mathbb{R}^n$ be non-convex. Then, $h(g_1(x),\dots,g_n(x))$ is non-convex.

Positive weighted sum: Let $h_j(y)$ be non-convex. Then $\sum_{j=1}^m \alpha_j h_j(y),\alpha_j>0,\sum_{j=1}^m\alpha_j=1$ is non-convex.

Title: Operations that preserve non-convexity

Body:
(1) fails for $g(x) = x^4$ and $h(x) = \sqrt(|x|)$.

(2) fails for $h_1(x)=\sin(x)$, $h_2(x)=-\sin(x)$, $\alpha_1=\alpha_2=\frac12$.


Title: Proving $\forall n \geq10, n^3 < 2^n$

Body:
For $n=10$, it is obvious that $2^{10}>10^3$

Assume that for $k\geq 10, \space   2^k>k^3$

Prove for $n=k+1$

$$2^{k+1}>(k+1)^3$$

So

$$2^{k+1}=2\times2^k>2k^3=k^3+k^3>k^3+3k^2+3k+1=(k+1)^3$$

Since $$k^3>3k^2+3k+1$$

$$k\geq10>3+3+1$$

$$k^3>3k^2+3k^2+k^2>3k^2+3k+1$$


答:
Let $P(n):\ n^3<2^n$.

For the base case $n=10$, we have $P(10):\ 1000<1024$ which is true.

Let us assume that $P(k)$ is true for some $\mathbb{N}\ni k\geq 10$. Then,
$$k^3<2^k\quad (*)$$
Now, we shall prove that $P(k+1)$ is also true.
Note that $2^{k+1}=2\cdot 2^k$. Using $(*)$, we get
$$2^{k+1}>k^3+k^3\quad (**)$$
The key here is to note that for $n\geq 10$, $n^3\geq 1000$ grows much faster than $3k²+3k+1\geq 331$. If required, this can be proved easily.

Now, using the inequality above, $(**)$ becomes
$$2^{k+1}>k^3+3k^2+3k+1$$
$$\implies 2^{k+1}>(k+1)^3$$
Thus, we have proved that $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction, we conclude that $P(n)$ is true for all $\mathbb{N}\ni k\geq 10$.

Hope this helps :)

Title: Proving $\forall n \geq10, n^3 < 2^n$


问:
Asymptotic expansion of multiple integral at $\infty$
I am attempting to figure out the first term of the asymptotic expansion at $t \to \infty$ of the following triple integral:

$$
I(t) = \int_0^1 \int_0^\pi \int_0^\pi e^{[r\cdot(\cos(x)-\cos(y))+2 \cos(y)]\cdot t} \sin^2(x) \sin^2(y) \mathrm d x \mathrm dy \mathrm dr
$$

Intuitively, I am expecting to find something like $I(t)\sim \gamma e^{2t} t^\alpha$ with some constants $\alpha, \gamma$, as the exponential is most certainly going to dominate for such values (for instance when $y=0, r=0$)

As for solving it, I am considering using Laplace's method - or at least a variant of it. Now, it's a bit puzzling because the inward term in the exponential function : $f(r,x,y) = r\cdot(\cos(x)-\cos(y))+2 \cos(y)$ has some issues: its maximum value 2 is reached for all values for which we have $y=0$ and either $x=0$ or $r=0$ - so that means potentially two segments on which the exponential is maximal (while the link provides inputs for a single point). Secondly, these points are on the border of the integration surface.

Would you know another variant of this method to solve this properly?

Thank you in advance!

答:
As remarked by @user2673062 result in the comments, the triple integral simplifies. By using the integral representation for the modified Bessel function
$$I_{\nu}\left(z\right)=\frac{(\frac{1}{2}z)^{\nu}}{\pi^{\frac{1}{2}}\Gamma\left(\nu+\frac{1}{2}\right)}\int_{0}^{\pi}e^{\pm z\cos\theta}(\sin\theta)^{2\nu}\mathrm{d}\theta$$
for $\nu>-1/2$, we can perform the integrations over $x$ and $y$ with $\nu=1$ to obtain:
\begin{align} I(t) &= \int_0^1 \int_0^\pi \int_0^\pi e^{[r\cdot(\cos(x)-\cos(y))+2 \cos(y)]\cdot t} \sin^2(x) \sin^2(y) \mathrm d x \mathrm dy \mathrm dr\\
&=\frac{\pi^2}{t^2}\int_0^1\frac{I_1(tr)}r\frac{I_1\left(t(2-r)\right)}{2-r}\,dr
\end{align}
By changing $r\to 2-r$, this integral can be written as
\begin{align}I(t)&= \frac{\pi^2}{2t^2}\int_0^2\frac{I_1(tr)}r\frac{I_1\left(t(2-r)\right)}{2-r}\,dr\\
&=\frac{\pi^2}{2t^2}\Phi(2)
\end{align}
where
$$\Phi(\theta)=\int_0^\theta\frac{I_1(tr)}r\frac{I_1\left(t(\theta-r)\right)}{\theta-r}\,dr$$
can be considered as a convolution integral. Its Laplace transform is then
\begin{align}\mathcal L[\Phi] &=\left\lbrace \mathcal L\left[\frac{I_1(tx)}x\right]\right\rbrace^2\\
&=\frac1{t^2}(u-\sqrt{u^2-t^2})^2
\end{align}
By inversion, we find
$$ \Phi(\theta)=\frac2\theta I_0(\theta t)-\frac4{\theta^2 t}I_1(\theta t)$$
These integral transforms were found using a CAS but could certainly be found from classical tables of Laplace transforms (Ederlyi TI, for example).
Finally,
$$ I(t)=\frac{\pi^2}{2t^2}\left[I_0(2 t)-\frac1{ t}I_1(2t)\right]$$
which seems to be numerically correct.
Using the asymptotic expansions for the Bessel functions, we find
$$ I(t)\sim \frac{\pi^{3/2}}{4}\frac{e^{2t}}{t^{5/2}}$$ which is not very far from @Raffaele estimation.


问:
Modulo of an arithmetic sequence
I have an arithmetic sequence, $a+dn$, where $a$ and $d$ are constants and $n$ is the term number. How would I efficiently calculate solutions to the following equation,

$$
(a+dn)\%b=0
$$

where $b$ is also a constant. (and $\%$ means modulo)

I attempted the question myself and found out that I really just need to find the smallest $n$ that satisfies the equation and find the rest by adding ${kb\over GCD(d,b)}$ to $n$, where $k$ is any integer. I also noticed that the smallest solution will always be less than $b$ and if no solution exist within this range, then there is no solution. So I tried to find the solution by brute forcing from 0 to $b$, until I got a solution. But this method is not nearly as efficient as I need it to be.

答:
If $a + dn \equiv 0(\mod{b})$, $$dn \equiv -a (\mod{b}) \implies dn \equiv b - a (\mod{b})$$ (assuming $b > a$, and of course the remainder should be less than the divisor as per Euclid's Division Lemma, and remainders should be positive)
$$\implies n \equiv \dfrac{b - a}d (\mod{b}) \space \text{if} \space gcd(b,d) = 1 \space \text{and} \space d \mid (b - a)$$

Therefore, $n = bx + \dfrac{b - a}d$ is the kind of $n$ you might want.

But there may arise cases where this alone won't work, so I'd suggest attending a course on some advanced topics in modular arithmetic (even I am not an expert and I used what I knew to answer this question).

Refer the suggestions in @J.W.Tanner's comments beneath your question for more methods.

Title: Modulo of an arithmetic sequence


答:
Using the representation of the integral in terms of modified Bessel functions we have:

$$
I(t)=\frac{\pi^2}{t^2}\int_0^1dr \frac{I_1(tr)I_1(t(2-r))}{r(2-r)}
$$

Observation: $t(2-r)\gg1$ for $r\in(0,1)$ so we won't do much harm if we just replace $$
I_1(t(2-r))\approx\frac{e^{2t-r}}{\sqrt{2t\pi(2-r)}}$$
(I am ignoring $O$'s for the moment but all statements can be made precise with a bit of extra work),

We get
$$
I(t)=\underbrace{\frac{\pi^{3/2} e^{2t}}{\sqrt{2}t^{5/2}}}_{C(t)}\underbrace{\int_0^1dr \frac{I_1(tr)e^{-tr}}{r(2-r)^{3/2}}}_{\mathcal{J(t)}}
$$

We now split this integral at $1/t \ll\delta \ll 1$:
$$
\mathcal{J(t)}=\mathcal{J_1(t)}+\mathcal{J_2(t)}
$$
We get (Lemma 1 plus large argument asymptotics):
$$
\mathcal{J_1(t)}\approx \frac{1}{2^{3/2}}\int_0^{\delta} I_1(tr)e^{-tr}/r=\frac{1}{2^{3/2}}((1+O(\sqrt{\delta t}^{-1}))
$$

for $\mathcal{J_2(t)}$ we can use again large argument asymptotics for $I_1(rt)$:
$$\mathcal{J_2(t)}\approx\frac{1}{\sqrt{2\pi t}}\int_{\delta}^1\frac{dr}{r^{3/2}(2-r)^{3/2}}=\frac{\eta}{\sqrt{2\pi t}}-O(\delta^{5/2})
$$

where $\eta$ is a constant independet of $t$.
Finally since $\mathcal{J_2(t)}\ll\mathcal{J_1(t)}$

$$
I(t)\approx C(t)\mathcal{J_1(t)} \approx \frac{C(t)}{2^{3/2}}\\
=\frac{e^{2t}\pi^{3/2}}{4t^{5/2}}
$$

Lemma 1:
$$ \int \frac{dz}ze^{-z}I_1(z)=Const-e^{-z}(I_1(z)+I_0(z))
$$

Proof:

Differentiate w.r.t. $z$ and use $I_0'(z)=I_1(z), \,\, 2 I_1'(z)=I_0(z)+I_2(z)$ as well as $I_0(z)-I_2(z)=I_1(z)/(2z)$

Title: Asymptotic expansion of multiple integral at $\infty$

Body:
I am attempting to figure out the first term of the asymptotic expansion at $t \to \infty$ of the following triple integral:

$$
I(t) = \int_0^1 \int_0^\pi \int_0^\pi e^{[r\cdot(\cos(x)-\cos(y))+2 \cos(y)]\cdot t} \sin^2(x) \sin^2(y) \mathrm d x \mathrm dy \mathrm dr
$$

Intuitively, I am expecting to find something like $I(t)\sim \gamma e^{2t} t^\alpha$ with some constants $\alpha, \gamma$, as the exponential is most certainly going to dominate for such values (for instance when $y=0, r=0$)

As for solving it, I am considering using Laplace's method - or at least a variant of it. Now, it's a bit puzzling because the inward term in the exponential function : $f(r,x,y) = r\cdot(\cos(x)-\cos(y))+2 \cos(y)$ has some issues: its maximum value 2 is reached for all values for which we have $y=0$ and either $x=0$ or $r=0$ - so that means potentially two segments on which the exponential is maximal (while the link provides inputs for a single point). Secondly, these points are on the border of the integration surface.

Would you know another variant of this method to solve this properly?

Thank you in advance!


问:
Proof of Volume of sphere
I want a simple proof for the formula of volume of sphere. Does the proof/explanation without integration possible?</p>

答:
A brute proof:

One can "transform" sphere to some cone/pyramid:

$$
V_{sphere} = V_{cone/pyramid} = \frac{1}{3}HS = \frac{1}{3}R\cdot 4\pi R^2 = \frac{4}{3}\pi R^3.
$$
(You need to know here that sphere surface area is $4\pi R^2$.)

The same way one can "prove" that circle area is $\pi R^2$.
Just "transform" circle to triangle:

$$
S_{circle} = S_{triangle} = \frac{1}{2}HL=\frac{1}{2}R\cdot 2\pi R = \pi R^2.
$$
(You need to know here that circle's circumference length is $2\pi R$.)

Title: Proof of Volume of sphere


问:
Having trouble understanding (intuitively) a combinatorics problem
I was given the following puzzle:

You manage a bank and you have 5 employees. With an infinite number of locks (and keys), you need to create a system that enables each triplet of workers open the vault of the bank, but a pair of employees should never be able to open together the vault.

Intuitively I immediately asked myself how many possible triplets of employees are there, and I know the answer from the combinations formula ($n!/r!(n-r)!$), which I also understand intuitively.

So we can represent all of our cases with this table:

https://upload.wikimedia.org/wikipedia/commons/6/65/Combinations_without_repetition%3B_5_choose_3.svg

We can say each case (or triplet) is represented by a row, and each worker is represented by a column. Obviously, when I see the table I immediately understand that the rows can also represent keys, but I can’t explain to myself in my head, logically, why this isomorphism (if I use the term correctly) is possible here. Perhaps I would have been able to get to this table myself, but it is not clear to me how (or why) the transformation from a table of triplets or cases, to a table of keys, occur. I try to imagine it somehow in my head but can’t. We can create 10 (unique) triplets (without repetition) from a set of 5, but why this necessarily implies that in order for each set of (only) three employees to be able to open a vault, we need 10 locks, and to give each employees 6 keys? Can you explain the isomorphism between the possible triplets and the number of locks, and the number of keys per employee?

I hope my question is making sense.

答:
For every doubleton of employees, there has to be at least one lock they cannot open.  However, if two different doubletons of employees (which might be a total of either three or four employees) couldn't open the same lock, then there would be a tripleton that couldn't open the vault.  Therefore, there have to be at least ${5\choose2} =10$ locks.

Let us see if ten locks are sufficient.  As noted before, there is a correspondence between the locks and doubletons of employees, so for convenience let us call the employees A, B, C, D, E, and then the locks will be AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.  No employee can have the key for a lock with his name in it.  Can an employee be missing a key without his name in it?  No.  Without loss of generality, let us assume that A does not have key BC.  Then {A, B, C} cannot open the vault.  Therefore, each employee must have keys for the six locks that don't have his name on it.

Let's verify that this solution works.  As before, we imagine that each lock is "named" for the two people that don't have its key, so therefore each lock has three keys held by the three people the lock isn't named for.  Imagine any two people: they cannot open the lock with both their names on it, so they cannot open the vault.  But for any three people, at least one of them isn't named on every lock, so they will be able to open every lock of the vault.


问:
Throwing a biased coin $m$ times
We throw a biased coin (probability for heads is $\frac{1}{n}$) $m$ times. It is known that in first two throws we got at least one tail. Calculate the probability that we got $k$ heads.

So, let's define:

$A - $ at least one tail in two first throws

$B \ - $ $k$ heads

I need to calculate $\mathbb P(B|A)=\frac{\mathbb P(B\cap A)}{\mathbb P(A)}$, so:

$\mathbb P(A)= 1-\mathbb P($first are two heads$ )=1-(\frac{1}{n})^2$

$\mathbb P(B|A)=\mathbb P(B\cap A)= \frac{1}{n}\cdot\frac{n-1}{n}{m-2 \choose k-1}(\frac{1}{n})^{k-1}(1-\frac{1}{n})^{m-k-1}+\frac{n-1}{n}\cdot\frac{1}{n}{m-2 \choose k-1}(\frac{1}{n})^{k-1}(1-\frac{1}{n})^{m-k-1}+\frac{n-1}{n}\cdot\frac{n-1}{n}{m-2 \choose k}(\frac{1}{n})^k(1-\frac{1}{n})^{m-k-2}$

Am I correct?

答:
you can use the Total probability Theorem: it results to me

$$\mathbb{P}[B]=\Bigg(\frac{1}{n}\Bigg)^k\Bigg(1-\frac{1}{n}\Bigg)^{m-k}\Bigg[\binom{m-2}{k}+2\binom{m-2}{k-1}\Bigg]$$

In fact, the number of H is always $k$ and the number of T always $(m-k)$. What changes is  only the way to combine the k H in $m$ tosses.

These combinations are

$\binom{m-2}{k}$ or $\binom{m-2}{k-1}$, it depends if you have zero or one H in the first two tosses, respectively

The binomial coefficient $\binom{m-2}{k-1}$ has to be multiplied by 2 because we can have $TH$ or $HT$ in the first two tosses

Title: Throwing a biased coin $m$ times


问:
Proving that $h$ is not a homotopy equivalence.
Suppose $h:S^1\times S^1\times S^1\rightarrow S^1\times S^1\times S^1$ is given by $h(u, v, w)=(u, v^2, w^7)$. I want to show that $h$ is not a homotopy equivalence.

My thought is this: Suppose for contradiction, we assume that $h$ is a homotopy equivalence. Then $h_\ast$ must be an isomorphism. But since $h$ is not injective, it is not an isomorphism, therefore our assumption was wrong?

答: If you don't want to use covering spaces, use that the fundamental group of the three-fold product is the product of the fundamental groups, and check that the induced map is of the form $(a,b,c) \mapsto (a,2b,7c)$. This is not an isomorphism of groups.


问:
Is 30 a 2 sf number or a 1 sf number?
Something about significant figures really confuses me, for instance:

If we are counting, say, apples and we say there are 30 apples, there are exactly 30, should 30 be a 2 significant figures number or a 1 significant figure.

And If I am in a phyiscs exams and a quantity like 30 kg is given, this number should be a 2sf numbers but in the same time if say it was 31 kg and I rounded it to 1 sf it would still be 30 and I would say it's a 1 sf number so I'm really confused.

答:
This is a confusion, that's why scientists prefer the scientific notation, which clearly indicates the number of significant figures.
If your number is $3.0\times 10^1$, then it has two significant figures; if it is $3\times 10^1$, then it has one significant figure.

As for your question, I would say that $30$ in the case of $30$ apples has 2 significant figures, because here the $0$ is significantly counted.

If this sounds confusing, consider this, you have a metre scale (marked up to $30\ \mathrm{cm}$), then if you measure a length of $30\ \mathrm{cm}$, then it has two significant figures as you counted the $0$ significantly, you are able to count $31$ or $32$ as well! However, if you say, you have measured $3000\ \mu\mathrm{m}$ ($\textit{i.e.}\ 3\ \mathrm{mm}$) then none of the zeroes is significant because the $3$ here is significantly counted, the zeroes are not! Any other person could have measured the same and could have said $3500$! That's why only $3$ is significant!

Significant errors count all the digits measured accurately as well as one additional digit to count the errors. In the above case, the $3$ is the additional digit itself!

Hope this is clear :)

Title: Is 30 a 2 sf number or a 1 sf number?


问:
Prove by induction that $n^{+}+m=n+m^{+}$ for all $m,n \in \omega$
Just to make sure we use same symbols: $\omega= \mathbb{N}_{0}$, $n^{+}$ is successor of number $n$.
And we define addition by recursion:

$+:\omega \times \omega \rightarrow\omega$

$m+0=m$

$m+n^+=(m+n)^+$

I'm trying to prove: $n^++m=n+m^+$ by induction.

Proof:

Let $m \in \omega$ be fixed and $S=\{n\in \omega| n^++m=n+m^+\}$.

Base: $0\in S$ (proof by induction)

Let $S'=\{m\in \omega| 0^++m=0+m^+\}$ (now I'm using this fixed $m$ as variable in induction (not sure if that's ok?))

$0 \in S$ because  $0^++0=0^+=1$ and $0+0^+=(0+0)^+=0^+=1$

 Let $m\in S$ which means $0^++m=0+m^+$.

 Then $0^++m^+=(0^++m)^+=(0+m^+)^+=0+(m^{+})^{+}$. So, $S'=\omega.$

We proved: $0\in S$.

Hypothesis: $n \in S$ which means  $n^++m=n+m^+$

And now I'm stuck. I don't know how to prove step of induction, and I'm not even sure that begining where I start with fixed $m$ is wnough to prove that statement is true for all $m \in \omega$.

答:
You started with the wrong $S$ and you inducted on the wrong variable.

Since the induction is done of the second argument of $+$ you need to induct on $m$.
So if you want to prove $\forall n \forall m. m+n^+ = (m+n)^+$, you have to pick $S = \{m / \forall n. n^+ + m = n + m^+\}$


答:
No, you have $P(A)=P(\text{there are $k$ heads})$ and this is not true. I would argue that $P(A)=1-P(\text{first two are heads})=1-(\frac{1}{n})^2.$

So you will hve
$$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(A\cap B)}{1-(\frac{1}{n})^2},$$
your computation of $P(A\cap B)$ seems correct to me.

Title: Throwing a biased coin $m$ times

Body:
We throw a biased coin (probability for heads is $\frac{1}{n}$) $m$ times. It is known that in first two throws we got at least one tail. Calculate the probability that we got $k$ heads.

So, let's define:

$A - $ at least one tail in two first throws

$B \ - $ $k$ heads

I need to calculate $\mathbb P(B|A)=\frac{\mathbb P(B\cap A)}{\mathbb P(A)}$, so:

$\mathbb P(A)= 1-\mathbb P($first are two heads$ )=1-(\frac{1}{n})^2$

$\mathbb P(B|A)=\mathbb P(B\cap A)= \frac{1}{n}\cdot\frac{n-1}{n}{m-2 \choose k-1}(\frac{1}{n})^{k-1}(1-\frac{1}{n})^{m-k-1}+\frac{n-1}{n}\cdot\frac{1}{n}{m-2 \choose k-1}(\frac{1}{n})^{k-1}(1-\frac{1}{n})^{m-k-1}+\frac{n-1}{n}\cdot\frac{n-1}{n}{m-2 \choose k}(\frac{1}{n})^k(1-\frac{1}{n})^{m-k-2}$

Am I correct?


问:
<< notation on measures
If $\nu$ and $\mu$ are two signed measures of finite variation, what does the notation $\nu << \mu$ means?

答:
If $\mu$ is positive, then we write that $ \nu \ll \mu $ iff $\mu(S)=0\implies\nu(S)=0$ for every measurable set $S$. We say that $\nu$ is absolutely continuous with respect to $\mu.$

If $\mu$ is signed, then we replace $\mu$ by $|\mu|$ in the above.

Title: << notation on measures


问:
$\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ is an index two subgroup of $\mathbb{Z}\ast\mathbb{Z}$
I want to prove $\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ is an index two subgroup of $\mathbb{Z}\ast\mathbb{Z}$. Can I use covering map to prove this? Any ideas and suggestions are greatly appreciated.

答:
Directly:
Define $\color{red}{\Bbb Z}*\color{green}{\Bbb Z}*\color{blue}{\Bbb Z}\to \color{red}{\Bbb Z}*\color{green}{\Bbb Z} $ by $\color{red}1\mapsto \color{red}2$, $\color{green}1\mapsto \color{green}2$, $\color{blue}1\mapsto \color{red}1\cdot\color{green}1$.

If you use $a$, $b$, and  $c$ as the standard generators of $\mathbb{Z\ast Z\ast Z}$, and $x$ and $y$ for $\mathbb{Z\ast Z}$, then this map is given by

\begin{align}
a &\mapsto x^2 \\
b &\mapsto y^2 \\
c &\mapsto xy
\end{align}


问:
How was the following Integral identity calculated?
$$\frac{1}{2\pi} \int_{0}^{2\cdot\pi}\left |1+e^{j\cdot k\cdot d\cdot \cos(\theta )}  \right |^{2}d\theta =2(1+ J_{0}\left ( k d \right ))$$
Where:
$$j=\sqrt{-1}$$
and $J_{0}$
is the Bessel function of the first kind of order zero.
k,d are constants.

答:
$$\frac{1}{2\pi}\int_{0}^{2\cdot\pi}|1+e^{j\cdot k\cdot d\cdot \cos(\theta )}|^{2}d\theta =
\frac{1}{2\pi}\int_{0}^{2\cdot\pi}(1+e^{j\cdot k\cdot d\cdot \cos(\theta )})(1+e^{-j\cdot k\cdot d\cdot \cos(\theta )})d\theta $$
$$=\frac{1}{2\pi}\int_{0}^{2\cdot\pi}(2(1+cos(cos(k\cdot d\cdot cos(\theta)))d\theta
$$
Using the following identity:
$$\frac{1}{2\pi}\int_{0}^{2\cdot\pi}cos(a\cdot cos(\theta)d\theta =J_{0}(a)$$
We get:
$$\frac{1}{2\pi}\int_{0}^{2\cdot\pi}(2(1+cos(cos(k\cdot d\cdot cos(\theta)))d\theta=2(1+J_{0}(k\cdot d))$$


答:
Yes, you can use coverings. In fact, you can use coverings to prove that every subgroup of index two is isomorphic to $\mathbb{Z\ast Z\ast Z}$.

To use coverings, note that your group is the fundamental group of the graph with a single vertex and two loop edges.

A two-sheeted cover of this graph is a connected graph with two vertices, each with degree four. Up to isomorphism there are 2 such graphs (two edges between the two vertices, or four). In each case, the graph has fundamental group $\mathbb{Z\ast Z\ast Z}$, as required.

Body: I want to prove $\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ is an index two subgroup of $\mathbb{Z}\ast\mathbb{Z}$. Can I use covering map to prove this? Any ideas and suggestions are greatly appreciated.


问:
Integer solutions using stars and bars with bounds on integers
In this page of brilliant , an example problem is as follows

$$\sum_{i=0}^6 a_i \leq 100$$ How many positive integers $(a_1,a_2...,a_6)$ with $ i \leq a_i$ for $i=1,2,3,4,5,6$ satisfy the sum?

In the solution they start of with introducing an extra variable $c$ into the equation making it:

$$ c+ \sum_{i=0}^6 a_i  \leq 100$$

With, $ 0 \leq c$

But I don't understand why?  They call this addition of 'c' is some kind of bijection but I'm not sure what purpose it is serving..

答:
The term $c$ has been introduced to the expression as a dummy variable, which saves you a lot of work in the long run.

Via the introduction of this new dummy variable '$c$', the below problem has been transformed from
$$\sum_{i=0}^6 a_i  \leq 100$$
to
$$\sum_{i=0}^6 a_i + c = 100$$

Why this works out so well is that each non-negative value taken up by this dummy variable $c$ corresponds to a particular case of the inequality. For example, when $c=10$, $\sum_{i=0}^6 a_i = 90$. To reiterate, taking this extra variable saves you the effort of having to calculate so many different cases.

The number of positive integral solutions of the above equation can be calculated easily via the well known formula for the same. I think you can take it from here!


问:
Two proofs involving Harmonic Mean
If H is the harmonic mean between $a$ and $b$,then show that
$$\frac{1}{H-a}+\frac{1}{H-b} = \frac{1}{a} + \frac{1}{b}$$ and $$\frac{H+a}{H-a}+\frac{H+b}{H-b} = 2$$

I substituted  $\displaystyle H = \frac{2ab}{a+b}$, then tried some algebraic manipulation, but I am not getting there.

Are they even valid? If yes, could somebody give me some ideas how to approach these?

答:
I don't know why you aren't getting this. You have
\begin{align*}
\frac{1}{H-a} + \frac{1}{H-b} &= \frac{1}{\frac{2ab}{a+b} -a} + \frac{1}{\frac{2ab}{a+b}-b} \\ &= \frac{a+b}{ab-a^{2}} + \frac{a+b}{ab-b^{2}} \\ &= (a+b) \cdot \biggl[ \frac{1}{a(b-a)} + \frac{1}{b(a-b)}\biggr] \\
\end{align*}

I think this should help you out. Try taking the factor $(a-b)$ common. For the next problem also again substitute the value of $H$ and try doing manipulations. Be patient, and be careful, with your calculations, you shall arrive at the result.

Title: Two proofs involving Harmonic Mean


问:
Joint cdf of guaranteed event
I need someone to sanity check my maths here, I think I'm right but need some help!
Let X be a continuous random variable taking values in R+ and let Y be a discrete random variable taking values in {0,1} (to simplify the equation). X and Y are not independent.

Are the following statements true:

i) $F_{X,Y}(x,1) = P(X \leq x, Y \leq 1) = P(X \leq x)$

ii) $f_{X,Y}(x,1) = f_X(x)$

where $f_{X,Y}$ is the joint 'density' defined as in this wiki page.

i) is just the marginal as $Y \leq 1$. I think ii) follows as this is just simple differentiation/subtraction of each case (continuous and discrete)?

Thanks for the help in advance!
By the way if independence makes a difference to the answer I'd appreciate if someone could explain why!

EDIT: Simplfied question and key points.

答:
You cannot differentiate with respect to $y$ so there is not a density $f_{X,Y}(x,y)$ in the standard sense.

$f_{X}(x)$ may be meaningful as the derivative of $F_X(x)=\mathbb P(X\le x)=F_{X,Y}(x,1)$.  Is this what you want?

Or are you looking for something like $f_{X\mid Y=1}(x)$, which should integrate to $1$,

or like $f_{X\mid Y=1}(x) \mathbb P(Y=1)$, which should integrate to $\mathbb P(Y=1)$?

Title: Joint cdf of guaranteed event


答:
To me, it reads as both an inclusion and exclusion of x which seems like it would nonsense in all possible cases -- not even the empty interval would satisfy both criteria.

Oh!  I see what you are asking.  It took me a few readings to get your concern.

The definition of $[x, y)$ is $\{w\in \mathbb R| x \le w < y\}$.  By that definition $[x,x) = \{w\in \mathbb R|x\le w < x\} = \emptyset$.  There is no problem with that.

But I think you are assuming the following propositions:

Proposition 1:  $x\in [x,y)$.

Proposition 2: $y\not \in [x,y)$

If so, then we can't have $[x,x)=\emptyset$ as that would imply $x \in \emptyset$.  But we can't have $[x,x)$ be any set as we need $x \in [x,x)$ and $x \not \in [x,x)$ and that's impossible.

....

But..... proposition 1: is not true.  It just isn't.

.....

By definiton:   $[x,y)=\{w|x\le w < y\}$.  This  is non-empty if $y > x$ (and in this case $x\in \color{blue}{[x,y)}$ because $\color{blue}{x \le} x\color{blue}{< y}$) and $[x,y)$ is empty if $y \le x$ (and in these cases $x\not \in \color{blue}{[x,}\color{red}{y)}$ because $\color{blue}{x \le} x \color{red}{\not < y}$.... and because.... it's empty... we just said that....).

So a proper stating of Prop 1 would be:

Prop 1':  If $[x,y)$ is non-empty, then $x\in[x,y)$. Furthermore $[x,y)$ is non-empty if and only if $y >x$.

And it this case $[x,x) = \emptyset$.

And $\emptyset=[x,x)$ satisfies both Prop 1' and Prop 2. Prop 1': If $[x,x)$ were non-empty the $x$ would be in it, but $[x,x)$ is empty and $x$ isn't in it.  Prop 2:  $x \not\in [x,x)=\emptyset$.

Title: Is $[x, x)$ an invalid interval for *all* real values of $x$?

Body:
I'm writing some software at work and I believe it isn't a valid interval because it seems contradictory. To me, it reads as both an inclusion and exclusion of $x$ which seems like it would nonsense in all possible cases -- not even the empty interval would satisfy both criteria.

I wanted to ask here because I don't necessarily know enough to say for sure that this interval should be considered invalid. Rather than pulling such a statement out of relatively nowhere and asserting it in code for others, I'd like to hear from those who know more about it.


问:
Are non-constant meromorphic functions on compact Riemann surjective?
Let $f$ be meromorphic on compact Riemann surface $X$ (Assume $X$ if connected if need be), i.e. we have $f: X \setminus poles(f) \to \mathbb C$ holomorphic.

This corresponds to holormorphic $F: X \to \mathbb C_{\infty}$, where $F(x):=f(x)$, $x \in Domain(f)$ and $F^{-1}(\infty):=poles(f)$.

We have $f$ non-constant if and only if $F$ non-constant.

If $F$ is non-constant, then $F$ is surjective.

$F$ is surjective if and only if $f$ is surjective.

Therefore, if $f$ is non-constant, then by (2),(3) and (4), $f$ is surjective.

Question 1: Is this correct?

Question 2: Is this incorrect if $X$ is not connected?

Remark: I believe for the case of $X$ compact and connected: If question 1 is answered affirmatively, then I believe $f$ has a pole because '$f$ has a pole' is equivalent to '$F^{-1}(\infty) \ne \emptyset$'.

答:
The argument is correct for connected $X$. The crucial step is 3, and that works since a non-constant holomorphic mapping on a connected Riemann surface is an open mapping. Thus $F(X)$ is an open subset of the Riemann sphere. Also, since $X$ is compact $F(X)$ is compact, thus closed. And since the Riemann sphere is connected, it follows that $F(X) = \mathbb{C}\cup \{\infty\}$ (using the fact that $X \neq \varnothing$, for otherwise $F$ couldn't be non-constant).

This breaks down if $X$ is not connected. If say $X$ has two connected components $X_1$ and $X_2$ the meromorphic (even holomorphic) function $f(z) = j$ if $z \in X_j, j=1,2$ is non-constant, but not surjective.

We get surjectivity if [and only if] there is (at least) one component of $X$ on which $F$ is non-constant.


问:
Prime elements of $\mathbb{Z}[i\sqrt5]$.
I was studying the Gaussian integers and I proved that every composite number in $\mathbb{N}$ is not a prime in $\mathbb{Z}[i]$. This is true because this ring is an Euclidean domain, and if $n=ab$ is composite then
$$\lambda(n)=\lambda(a)\lambda(b),$$
and $\lambda(a),\lambda(b)>1$ where $\lambda(\alpha)=\alpha\overline{\alpha}$. Basically I used the property that every irreducible is prime.

This same property is not valid in $\mathbb{Z}[i\sqrt5]$ because for example $2$ is irreducible but is not prime, as $2$ divides $6=(1+i\sqrt5)(1-i\sqrt5)$ but $2$ does not divide either of the factors.

How could I prove that there is some composite number of $\mathbb{N}$ which is prime in $\mathbb{Z}[i\sqrt5]$? Is it possible to list the primes of this ring?

答:
Because $\Bbb{N}\subset\Bbb{Z}[i\sqrt{5}]$, any factorization of $n$ in $\Bbb{N}$ is also a factorization of $n$ in $\Bbb{Z}[i\sqrt{5}]$.

This relies on the fact that the ring extension $\Bbb{Z}\subset\Bbb{Z}[i\sqrt{5}]$ does not introduce any new units, as noted in the comments.


问:
Let $W$ be a Wiener process and $X(t):=W^{2}(t)$ for $t\geq 0.$ Calculate $\operatorname{Cov}(X(s), X(t))$.
Let $W$ be a Wiener process. If $X(t):=W^2(t)$ for $t\geq 0$, calculate $\operatorname{Cov}(X(s),X(t))$</p>

答:
Assume WLOG that $t\geqslant s$. The main task is to compute $E(X_sX_t)$.

Write $X_t=(W_t-W_s+W_s)^2=(W_t-W_s)^2+2W_t(W_t-W_s)+W_s^2$, and use the fact that if $U$ and $V$ are independent random variables, so are $U^2$ and $V^2$.


问:
subsequences of sin(n)
It is well known that $( \sin(n) )_{n\ge 1}$ is dense in $[-1,1]$.  Often one falls in exercise calculation on subsequences of this one: like $\sin(n^2)$ for example; let us generalize right away to $\sin(n^\alpha)$ for $\alpha>0$. These sequences should be dense as well, but that is not clear right away.

For $\sin(n^2)$ for example, Weyl's argument would require to show that  $\sum_{k=1}^n \exp( i k^2 m)  = o(n)$ which is not clear to me, since I cannot calculate explicitly this sum. Is there some nice trick to handle the density of $\sin(n^\alpha)$  ?

答:
Some answers, I eventually found.

In this book, https://web.maths.unsw.edu.au/%7Ejosefdick/preprints/KuipersNied_book.pdf we have a nice Theorem 3.2 stating that for any polynomial with real coefficients and at least one irrational coefficient, p(n) is unif. distributed mod 1.  In my setting, $\sin(n^2) =\sin(\pi n^2/\pi)$ the polynomial p(n) = n^2/\pi$ satisfies the criterion.

Another situation is $\sin(n^n)$ that seems to be treated  here https://www.jstor.org/stable/43678733.

Title: subsequences of sin(n)


问:
Power series with cubic factorials
I am looking for a starting point to obtain the closed form of
$$ \sum_{k=0}^{\infty} x^k \frac{(2k)!}{(k!)^3}. $$
I have tried transforming the ratio of factorials into combinations, but this lead me nowhere. I also computed several values of the series on Wolfram and I expect the answer to look like $e^{2x} I_0(2x)$ for $I_0$ the $0$-th modified Bessel function.
Does anyone has an idea on how to properly obtain the result ?

答: Yes, that is the correct formula for the exponential generating function of the central binomial coefficients $\binom{2k}{k}$.  See https://oeis.org/A000984

Title: Power series with cubic factorials


问:
What happens to the Stone-Cech compactification if you change "compact Hausdorff" to "compact"?
Here if in the diagram below

Universal property and functoriality

we take a modification that any continuous map $f:X\to K$ where
$K$ is compact but not necessarily Hausdorff what would be our $\beta X$ (if $X$ is discrete) in this case instead of all ultrafilters over $X$? Would it always exist?

答:
No, if you drop the Hausdorff condition when talking about the Stone-Cech compactification, then it never exists for any non-compact space.  Indeed, suppose $X$ is not compact and suppose there existed an initial continuous map $f:X\to Y$ to a compact space $Y$.  Consider the space $K$ obtained by adjoining two points $a,b$ to $X$ and declaring that a set is open in $K$ iff it is either an open subset of $X$ or is equal to all of $K$.  Then $K$ is compact and the inclusion map $i:X\to K$ would be continuous, so there would have to be a unique continuous $g:Y\to K$ such that $gf=i$.  In particular, this means that the image of $g$ contains all of $X$, and therefore must also contain at least one of $a$ and $b$ since the image of $g$ must be compact and $X$ is not compact.  But now define $g':Y\to K$ by $g'(y)=g(y)$ if $g(y)\in X$, $g'(y)=b$ if $g(y)=a$, and $g'(y)=a$ if $g(y)=b$.  This $g'$ is still continuous, since the open sets containing $a$ are the same as the open sets containing $b$.  Also, for any $x\in X$, $g(f(x))=x\in X$ so $g'(f(x))=x$ as well.  That is, $g'f=i$.  This contradicts the uniqueness of $g$.

From a categorical perspective, what's going on here is that compact spaces (unlike compact Hausdorff spaces) are not closed under limits in the category of topological spaces, and thus are not a reflective subcategory.  The issue is with equalizers: the equalizer of two maps between Hausdorff spaces is closed in the domain, and thus compact if the domain is compact.  However, the equalizer of two maps between compact spaces need not be compact.


问:
Tell me which of $691$ and $697$ is prime (no calculator)
For two numbers, $691$ and $697$, tell me which (or both) is prime.

Without using calculator, what is the strategy for tackling this? I tried all the numbers up to 8 which are easy to test. How do you deal with better numbers?

答:
You should eliminate single digit prime factors first: $2, 3, 5, 7$.

By observation, both are odd. Eliminate $2$.

By sum of digits rule, both are not divisible by $3$.

By observation of last digit, both are
not divisible by $5$.

For 7, you can apply this rule and check
that both are not divisible by $7$

Let us also eliminate 11 since it has an easy to check divisibility rule.

For 11, you can apply the divisibility rule (Form the alternating sum of the digits. The result must be divisible by 11) and check that both are not divisible by $11$

Therefore one of the factor is at least 2 digits.

Answer #1: Use the divisibility rules given here for primes up to 30. This works since the square root of 697 lies between 26 and 27.

Answer #2: Using digital roots

Since the numbers themselves are 3 digits and no single digit factor exists, both factors must be 2 digits (for both numbers).

We have, $p = 10a + b, q = 10c+d$

Further steps for 691:

$pq = (10a + b)(10c + d) = 691$

The digital root of an integer $n$ is denoted by $dr(n)$.

The digital root of $691$ is $dr(691) = 7$.

Using this Vedic square, also shown below, we see that:

$$\{(dr(p),dr(q))\} \equiv \{(1,7), (2,8), (4,4), (5,5), (7,1), (8,2)\}$$

are the possibilities. Due to commutativity of multiplication, we can consider the reduced set:

$$\{(1,7), (2,8), (4,4), (5,5)\}$$

Check the primes $11 < p,q \lt 60$ i.e., $13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$
Their respective digital roots are:        $4,  8,  1,  5,  2,  4,  1,  5,  7,  2,  8,  5$

So, $\{(p,q)\} \in \{(19,43), (37,43), (29,17), (29,53), (47,17), (47,53), (13,13), (13,31), (31,13), (31,31), (23,23), (23,59),(59,23), (59,59)\}$

We can eliminate redundant pairs using commutativity:

So, $\{(p,q)\} \in \{(19,43), (37,43), (29,17), (29,53), (47,17), (47,53), (13,13), (13,31), (31,31), (23,23), (23,59), (59,59)\}$

Using $691 \mod 10 = 1$, we can eliminate pairs leaving

$${(p,q)\} \in {(37,43), (47,53), (31,31), (59,59)}$$

All these products are greater than 691 which can be checked quickly. For eg: $30 \times 40 = 1200, 40 \times 50 = 2000$ and so on.

Therefore, $691$ must be prime.

Further steps for 697:

The digital root of 697 is $dr(697) = 4$.

Using this the Vedic square again, we see that:

$$\{(dr(p),dr(q))\} \equiv \{(1,4), (2,2), (4,1), (5,8), (7,7), (8,5)\}$$

are the possibilities. Due to commutativity of multiplication, we can consider the reduced set:

$$\{(dr(p),dr(q))\} \equiv \{(1,4), (2,2), (5,8), (7,7)\}$$

Check the primes $11 < p,q \lt 60$ i.e., $13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$
Their respective digital roots are:          $4,  8,  1,  5,  2,  4,  1,  5,  7,  2,  8,  5$

So, $\{(p,q)\} \in \{(19,13), (37,13), (19,31), (37,31), (29,29), (29,47), (41,17), (47,29), (47,47), (23,17), (23,53), (43,17), (43,53), (59,17), (59,53)\}$

We can eliminate redundant pairs using commutativity:

So, $\{(p,q)\} \in \{(19,13), (37,13), (19,31), (37,31), (29,29), (29,47), (41,17), (47,47), (23,17), (23,53), (43,17), (43,53), (59,17), (59,53)\}$

Using $697 \mod 10 = 7$, we can eliminate pairs leaving only those marked by the green cells in the figure below:

Of these 15, we can eliminate all whose products are lesser than 697 or greater than 697 by using the product of numbers rounded to nearest 10. For eg: $19 \times 13 < 300$, $29 \times 43 > 800$ and so on. You need to check only 3 products $(667, 697 and 767)$ by hand to find out that $697 = 17 \times 41$

This works for small numbers where we can calculate by hand.

You can also force a condition that $p \le q$, but if we do that we should not do the reduction using the commutativity rule.

For larger integers, I have a conjecture that we may be able to extend this digital roots technique to speed up the recovery of divisors. It is still an evolving idea. See my related question: An old multiplication technique and its reverse for Integer Factoring


问:
How can I show that $\begin{cases} x = 2 \pmod 6 \\ x = 5 \pmod 8 \end{cases}$ does not have a solution?
I thought about using the Chinese Remainder Theorem here as $\gcd(6,8)=2$ and not $1$, but

$$\begin{cases} x = 2 \pmod 6 \\ x = 6 \pmod 8 \end{cases}$$

has indeed a solution.

But right now, I'm stuck with

$$\begin{cases} x = 2 \pmod 6 \\ x = 5 \pmod 8 \end{cases}$$

Which obviously after some calculations does not seem to have an integer solution, but how can I show it more "elegantly", like by using some theorem?

答:
If you want to have a more general perspective on these examples, then you can use the following precise form of the Chinese Remainder Theorem (for the case of two equations):

Given integers $a,b$ and $m,n>0$, there is a solution to
\begin{align*}
x &\equiv a~(\text{mod}~m)\\
x &\equiv b~(\text{mod}~n)
\end{align*}
if and only if $a\equiv b~(\text{mod}~\gcd(m,n))$.

Note in the usual form of CRT one assumes $\gcd(m,n)=1$, and of course any two $a$ and $b$ are equivalent mod $1$.

However, in your case $\gcd(m,n)=\gcd(6,8)=2$. In the first set of equations you have a solution since $2$ and $6$ are equivalent mod $2$. But in the second set, $2$ and $5$ are not equivalent mod $2$, and there is no solution.

This generalizes to arbitrarily many equations. See: Chinese Remainder theorem with non-pairwise coprime moduli


问:
How to find parallel tangents for a parametric equation

For the function
$$ \vec x(t) = \begin{pmatrix}
    2t + 3 \\
    2 - t \\
    t^3 - 2t^2 + t \\
    \end{pmatrix}
 t ≥ 0$$
Are there 2 points $\vec x(t_1)$, $\vec x(t_2)$, such that the function’s tangent vectors
at these points are parallel to each other? Find such points, or show that none
exist.

I know that the derivative of the function is

$$ \vec x'(t) = \begin{pmatrix}
    2 \\
    -1 \\
    3t^2 - 4t + 1 \\
    \end{pmatrix} $$

and that in order for the tangent vectors to be parallel to each other the functions will equal the same value.  However, I am not sure how to go about finding the values.

答: Notice that $x'(t)=(2,-1,3t^2-4t+1)$ and $g(t)=3t^2-4t+1$ is not injective for $t\geq0$. In particular the equation $g(t)=0$ has two positive distinct solutions.


问:
CLT for non identically distributed bounded random variables
Let X1, X2, ... be independent, zero-mean, bounded random variables. Let $S_n$ = $X_1 +... + X_n$ with variance $s^2_n$ = Var($S_n$), $s^2_n \rightarrow \infty$. Show that $S_n/s_n$ has a central limit.

I believe that I am supposed to apply either Lindeberg or Lyapunov here but I am struggling on how to go about that

答:
In order to check Lindeberg's condition, note that the term
$$
\mathbb E\left[X_j^2\mathbf{1}_{\{\lvert X_j\rvert/s_n>\varepsilon\}}\right]
$$
equals to zero if $\lVert X_j\rVert_\infty/s_n\leqslant \varepsilon$.


问:
Checking derivative of norm squared of a linear transformation

Given $f: L(\mathbb{R}^n, \mathbb{R}^k) \to \mathbb{R}$ where $f(A)=|Ax|^2$ and fix $x\in \mathbb{R}^n$ then find $\partial_H f(A)$ where $H\in L(\mathbb{R}^n, \mathbb{R}^k)$.

My attempt:
$$
\partial_H f(A) = \lim_{t\to 0}\left(\frac{|(A+t\cdot H)x|^2 - |Ax|^2)}{t}\right)\\
= \lim_{t\to 0}\left(\frac{\sum_{j=1}^k\left(\sum_{i=1}^n (a_{ji} + t\cdot h_{ji})x_i\right)^2-\sum_{j=1}^k\left(\sum_{i=1}^n a_{ji}x_i\right)^2}{t}\right)\\
= \lim_{t\to 0}\left(\frac{\sum_{j=1}^k\left(\sum_{i=1}^n a_{ji}x_i\right)^2 + 2t\sum_{j=1}^k\left(\sum_{i=1}^n (a_{ji}x_i) \sum_{i=1}^n(h_{ji}x_i)\right) + t^2\sum_{j=1}^k\left(\sum_{i=1}^n h_{ji} x_i\right)^2-\sum_{j=1}^k\left(\sum_{i=1}^n a_{ji}x_i\right)^2}{t}\right)\\
= \left(2\sum_{j=1}^k\left(\sum_{i=1}^n (a_{ji}x_i) \sum_{i=1}^n(h_{ji}x_i)\right)\right)\\
$$
however I can't seem to simplify this further.

答:
Assume for the moment that the Frechet derivative $f'(A)$ exists. Then,

$\tag1f(A + v) = f(A) + f'(A)(v) + r(v)\  \text{where}\  \left|\frac{r(v)}{v}\right|\to 0\ \text{as}\ v\to 0.$

We have then

$\tag2\frac{f(A+tu)-f(A)}{t}=\frac{f'(A)(tu)+r(tu)}{t}\to f'(A)(u)\ \text{as}\ t\to 0.$

It follows that

$\tag3(Df_A)(u)=f'(A)(u)$

so it suffices to calculate $f'(A)(u).$

Now, $f$ is a compostion $A\overset  g\mapsto Ax\overset h\mapsto |Ax|^2$ and so $f'(A)$ does exist and is given by

$\tag4 f'(A)(u)=h'(g(A))(g'(A))(u)=h'(g(A))(g(u))$

because $g$ is linear. Then,

$\tag5 h'(g(A))(g(u))=h'(Ax)(ux)=2\langle Ax,ux\rangle.$


问:
How to formulate weighted distance?
Given we have 2 trees and their coordinates on xy plane is : $(-2,1)$, $(0,2)$ and their weights are $3$ and $4$ respectively. We want to plant a new tree with the condition that the weighted sum of the absolute distances from the existing trees will be minimized. Without including the weights I formulate as below:

$$\min.  (|x1 + 2| + |x2 -1|) + (|x1| + |x2 - 2|) $$

where $(x1,x2)$ are the coordinates of the tree that will be planted.

How could I include the weights of the existing trees in the formula?

One way I can think of is as below:

$$\min.  3(|x1 + 2| + |x2 -1|) + 4(|x1| + |x2 - 2|)$$

Although I can't intuitively get how this would be the logical way.

答: Yes, that is the correct formula for weighted sum of absolute distances.  The optimal $(x_1,x_2)$ turns out to be $(0,2)$, with objective value $3(2+1)+4(0)=9$.

Title: How to formulate weighted distance?


问:
Uniform Convergence of a serie
Prove that

$$\eta(z) = \sum_{n=1}^{\infty} \frac{1}{n^{z}} $$

does not converge uniformly on A = { z | Re $z$ > 1 }

I don't know how to proceed with this exercise. Could anyone explain me why I cannot use the Weierstrass M test?

Thanks in advance.

答:
You cannot use the Weierstrass $M$-test because $\sup_{z\in A}\left|\frac1{n^z}\right|=\frac1n$ and the harmonic series diverges.

And the convergence is not uniform because when a series of bounded functions converges, its sum is bounded too. But $\displaystyle\lim_{z\to1^+}\sum_{n=1}^\infty\frac1{n^z}=\infty$.

Title: Uniform Convergence of a serie


问:
Proving $\forall n \geq10, n^3 < 2^n$
Im asked to prove the following:

$$\forall n \geq10, n^3 < 2^n$$

I'm 100% sure that the easiest way to prove this is via induction but I'm not being able to do so. How can I prove this?

答:
We have that

$$n^3 < 2^n \iff \sqrt[n] n < \sqrt[3] 2$$

and we can check the cases $n=10, 11$ by inspection and then using that for $n\ge 12$

$$\sqrt[n] n =e^{\frac{\log n}{n}} \le1+\frac{\log n}{n}+\left(\frac{\log n}{n}\right)^2\le 1+\frac{\log 12}{12}+\left(\frac{\log 12}{12}\right)^2 <\sqrt[3] 2$$

Title: Proving $\forall n \geq10, n^3 < 2^n$


问:
How to find basis for an image of a subset under a transformation?
I was watching the khan acadamey videos on linear algebra.

Sal shows that the transformation $T$ of a subspace $S$ is another subspace $T(S)$. Since $S$ and $T(S)$ can have different dimensions, how do I find the basis for $T(S)$ if know the matrix $\mathbf A$ such that $T(\mathbf x) = \mathbf A\mathbf x$ and I know the basis for $S$?

For example if $S$ is a line in $\mathbb R^2$, and $T:\mathbb R^2 \rightarrow \mathbb R^3$; does the transformation of the basis of $S$ form a basis of $T(S)$? Is it possible $T(S)$ be a plane in $\mathbb R^3$ or must it be a line or a point?

答:
Always $dim(T(S))\le dim(S)$, so the answer for the question in the example is a line through the origin or the origin (both are subspaces).
 To form a basis for $T(S)$ form a matrix $V$ w/columns consisting of the basis vectors for $S$, then find the largest square submatrix w/non-zero determinant in the product matrix $AV$, the corresponding column vectors form a basis for $T(S)$.
 SVD decomposition of $A$ may give an alternative algorithm.</p>