问:
Symmetry in summation
Can someone please explain the motivation behind this equality?

$\sum_{i=1}^{n} (\sum_{j=1}^{i} + \sum_{j=i}^{n})\frac{1}{ij} = 2\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{1}{ij}$

They explain that it is symmetry of some kind.

答:
We can rewrite the equation as

$$\sum_{i=1}^n\sum_{j=1}^i\frac1{ij}+\sum_{i=1}^n\sum_{j=i}^n\frac1{ij}=2\sum_{i=1}^n\sum_{j=i}^n\frac1{ij}\,,$$

so the question is why

$$\sum_{i=1}^n\sum_{j=1}^i\frac1{ij}=\sum_{i=1}^n\sum_{j=i}^n\frac1{ij}\,.\tag{1}$$

$\sum_{j=1}^i\frac1{ij}$ is the sum of the fractions in row $i$ of the following table, as shown in the last column, and the lefthand side of $(1)$ is therefore the sum of all of the entries in the table, as shown in the lower righthand corner.

$$\begin{array}{c|ccccc|c}
i\backslash j&1&2&3&\ldots&n&\\\hline
1&\frac1{1\cdot1}&&&&&\sum\limits_{j=1}^1\frac1{1\cdot j}\\
2&\frac1{2\cdot1}&\frac1{2\cdot2}&&&&\sum\limits_{j=1}^2\frac1{2j}\\
3&\frac1{3\cdot1}&\frac1{3\cdot2}&\frac1{3\cdot3}&&&\sum\limits_{j=1}^3\frac1{3j}\\
\vdots&\vdots&\vdots&\vdots&\ddots&&\vdots\\
n&\frac1{n\cdot1}&\frac1{n\cdot2}&\frac1{n\cdot3}&\ldots&\frac1{n\cdot n}&\sum\limits_{j=1}^n\frac1{nj}\\\hline
&\sum\limits_{i=1}^n\frac1{i\cdot1}&\sum\limits_{i=2}^n\frac1{i\cdot2}&\sum\limits_{i=3}^n\frac1{i\cdot3}&\ldots&\sum\limits_{i=n}^n\frac1{in}&\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}
\end{array}$$

At the bottom I’ve shown the sums of the columns. Of course the sum of the column sums must be the same as the sum of the row sums, since in both cases we’re just getting the sum of all of the fractions in the body of the table, so the sum across the bottom must be the same as the sum down the righthand edge. The sum across the bottom is

$$\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}\,,$$

and the sum down the righthand edge is

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}\,,$$

so

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}\,.\tag{2}$$

The actual names of the index variables aren’t important: clearly

$$\sum\limits_{j=1}^n\sum\limits_{i=j}^n\frac1{ij}=\sum\limits_{\ell=1}^n\sum\limits_{k=\ell}^n\frac1{k\ell}\,,$$

where we’ve renamed $i$ to $k$ and $j$ to $\ell$. Thus, $(2)$ says the same thing as

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{\ell=1}^n\sum\limits_{k=\ell}^n\frac1{k\ell}\,.\tag{3}$$

Similarly, we can rename $k$ to $j$ and $\ell$ to $i$ and see that

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\frac1{ij}=\sum\limits_{\ell=1}^n\sum\limits_{k=\ell}^n\frac1{k\ell}=\sum\limits_{i=1}^n\sum\limits_{j=i}^n\frac1{ij}\,,$$

which is exactly what we wanted back at $(1)$.

Title: Symmetry in summation


问:
How can a vector space $V$ over a field have $V$ as a basis?
I think my question is related to Free vector space over a set, Non-numerical vector space examples and Formal (series/sum/derivative…).

I'm reading a paper, which starts with the following sentence (the original source is "Flag Algebras" by Razborov)

Let $\mathbb{R}F$ be the real vector space with basis comprised of all elements of $F$; in other words, the space of all formal, finite $\mathbb{R}$-linear combinations of elements of $F$.

Let me summarize my understandings first. We can define the set $\mathbb{R}F$ as a formal sum as $\mathbb{R}F=\{\sum_{f \in F} \lambda_f f | \lambda_f \in \mathbb{R}\}$ and construct the structure of a vector space as described in Free vector space over a set.

Now, here is my trouble: If $\mathbb{R}F$ is supposed to be a vector space, then, due to the vector space axioms, the following must be satisfied: Let $a, b \in F$, then $a+b=c \in F, \forall a,b \in F$, without defining the actual operation "+" of our vector space (since we speak of "formal" sum here). But if $F$ is the basis of our vector space, then all elements of $F$ must be linear independent, thus there cannot exist a $c \in F$ for which $a+b=c$ with $a,b \in F$. (my definition for a vector space is taken from "Advanced Linear Algebra" by Steven Roman").

I would suppose that we define the vector space as the set of all formal sums of elements of $F$ with real coefficients and take the axioms for granted. This works fine for associativity, the inverse element, etc., since we do not define "+" and "juxtaposition" anyway, but I don't see how the basis can be the whole set $F$.

My question is, how can the vector space axioms be satisfied? Can someone give me some examples, or clarifications?

答:
You're misunderstanding what the author denotes $\Bbb RF$. It is different from $F$. Actually,what he/she denotes this way (not a very good notation, in my opinion) is usually denoted as
$$\Bbb R^{(F)},$$
which is the set of maps from $F$ into $\Bbb R$ with finite support. A well-known example is the vector space $\Bbb R^{(\Bbb N)}$, which is known, when endowed with the relevant multiplication (Cauchy product), as the ring of polynomials with real coefficients.

More generally, for any set $X$,  one defines the free $K$-vector space with basis $X$ as $K^{(X)}$ (where $K$ is the base field). The construction of the author is further generalised to the situation where the base field is only a ring $R$ (in this case, it is called an $R$-module) and it can be used to prove that any $R$-module is the quotient of a free $R$-module (all modules, contrary to vector spaces, do not have a basis).

ContentLicense: CC BY-SA 4.0


问:
Prove that inclusion is a "universal" partial order
I wanted to prove a theorem in my textbook which states that inclusion can be viewed as a "universal" partial ordering (how cool is that!). The theorem goes like this: Let $(X, \leq)$ be a partially ordered set. Then, there exists a family of sets $A \subseteq P(X)$ such that partially ordered sets $(X, \leq)$ and $(A, \subseteq |_A)$ are isomorphic. Note: $P(A)$ means power set of set $A$ and $\subseteq|_A$ means that the inclusion was restricted to $A$.

Proof. Define $f(x)=\{a \in X:a \leq x\}$ for every $x \in X$. First let's show that $f:X \rightarrow P(X)$ is injective. Assume that $f(x)=f(y)$, where $x,y \in X$. We see that $x \in f(x)$ because $\leq$ is reflexive by definition. Since $f(x)=f(y)$, $x \in f(y)$, so $x \leq y$. Conversely, $y \in f(y)$, but $f(x)=f(y)$, so $y \leq x$. By antisimmetric property of $\leq$, we get that $x \leq y$ and $y \leq x \iff x=y$. Hence, $f$ is indeed injective. Great. Now we would like to show that $f(x) \subseteq f(y) \iff x \leq y$. Let's define $A=f(X)$. Note that $f(x)$ is a set, so $f(X)$ is a set of sets a.k.a. a family of sets. Now we should probably use the definition of subset.
$$f(x) \subseteq f(y) \iff \forall_\alpha (\alpha \in f(x) \Rightarrow \alpha \in f(y)).$$
We see that $\alpha \in f(x) \iff \alpha \leq x$ and $\alpha \in f(y) \iff \alpha \leq y$, thus
$$\forall_\alpha (\alpha \in f(x) \Rightarrow \alpha \in f(y)) \iff \forall_\alpha(\alpha \leq x \Rightarrow \alpha \leq y) $$
I would like to show that this is equivalent to $x \leq y$, but I have no idea how to proceed from this point onward. It makes sense for the "standard" less or equal to relation, but I don't know how to prove it rigorously. I would really like to prove this think cause it's pretty awesome in my opinion.

EDIT: In case anyone was curious how to finish the proof (basing on the kindly provided answers below).
First, show that $$\forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y) \Rightarrow x \leq  y.$$ Assume $\forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y)$ and choose $\alpha = x$. By the associative property, $x \leq x$, so by assumption $x \leq y$.

Second, prove that
$$x \leq  y \Rightarrow \forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y).$$
Suppose that $x \leq y$. Definition of transitivity tells us that $\forall_\alpha (\alpha \leq x \wedge x\leq y \Rightarrow \alpha \leq y)$. Which, by assumption, reduces to $\forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y)$. QED

答: Well done! You're really extremely close at this point. So what we can do is just prove the last equivalence in two steps, one for each direction. To prove that the statement implies that x is less than or equal to y, we just pick alpha to be equal to x (and use the reflexivity property of a partial order). For the other direction we can just apply the transitivity property.</p>


问:
Is $SL(n,F)$ a Kac-Moody group?
Is $SL(n,F)$ a Kac-Moody group? Do Kac-Moody groups generalize the classical Lie groups or are they a different family of groups?

I can't find an exposition of the material that I can understand, and really I'm just looking for a yes/no answer. A Kac-Moody algebra seems to be a generalization of the classical Lie algebras, as a classical Lie algebra could be described by a generalize Cartan matrix, since this is (obviously) just a generalization of a Cartan matrix.

Sorry if this is a dumb question...

答: Every finite-dimensional semisimple Lie algebra is a Kac-Moody algebra and so every finite-dimensional semisimple Lie group is a Kac-Moody group (assuming "Kac-Moody group" means what I think it means, namely "a group whose Lie algebra is a Kac-Moody algebra"). But it's a bit like "every square is a rectangle"; in practice if someone's writing a paper about Kac-Moody groups or algebras they are studying the infinite-dimensional case.</p>

Title: Is $SL(n,F)$ a Kac-Moody group?


问:
What does $\mathcal{F}_x(S)=\{A \subset S: x \notin A\}$ mean?
Pretty much the title really. What is the set $\mathcal{F}_x=\{A \subset S: x \notin A\}$. This is in the context of rings with the power set ring. I'm trying to prove this is an ideal of the power set (under the necessary operations), which I understand, but I don't have a clue what this set means. Is $\mathcal{F}_x(S)$ a field?

Edit: I missed an S in the title - sorry!

答:
The power set of a ground set $S$ can be turned into a ring by considering the symmetric difference operation for addition and the intersection operation operation for product.

In the text we are defining a subset of the power set for each element $x$. $F_x$ is the set whose elements are the subsets of $S$ that do not contain $x$.


答:
Maybe an example can help. Say we have $S = \{a, b, c, d\}$. Then we get that $\mathcal F_a$ is the set of all subsets of $S$ that do not contain $a$. Which is to say $$\mathcal F_a = \{\{b, c, d\}, \{b, c\}, \{b, d\}, \{b\}, \{c, d\}, \{c\}, \{d\}, \varnothing\}$$
If your power set ring is the standard power set ring, where addition is symmetric difference and multiplication is given by intersection, then this is clearly an ideal, as it is an additive subgroup, and intersecting with a set that doesn't contain $a$ again results in a set which doesn't contain $a$.

Title: What does $\mathcal{F}_x(S)=\{A \subset S: x \notin A\}$ mean?

Body:
Pretty much the title really. What is the set $\mathcal{F}_x=\{A \subset S: x \notin A\}$. This is in the context of rings with the power set ring. I'm trying to prove this is an ideal of the power set (under the necessary operations), which I understand, but I don't have a clue what this set means. Is $\mathcal{F}_x(S)$ a field?

Edit: I missed an S in the title - sorry!


问:
How can I count the number of colored combinations in a set of regions?
Let me first start out by saying as you might have guessed, this is a homework problem. Therefore I am not looking for an answer to the question. I am looking for help in how to analyze it (My textbook is horrid).

I have a grid of 9 squares surrounded by a circle.
I can use three colors to color each region with no region touching any other region with the same color (corners don't count). **`

I am asked to find the number of
  possible different colorings.

I drew a set of nodes with lines connecting neighboring regions, but I still have no idea how to figure out the number of combinations.

I also made a guess of three (which I know is wrong) since you can color the regions once with the three colors then map each color to a different one two times after the first coloring.

No blue is touching any other blue regions, no white is touching any other white regions, no red is touching any other red regions.

答: While it is valuable to attack such problems "by hand," here are some generalities.  The general version of this problem is called graph coloring.  (The graph being colored here is the graph whose vertices are the regions of the diagram where there is an edge between two regions if they are adjacent.)  Counting colorings is known to be #P-hard, so it is not all that easy to do in general.  For small graphs there is a reasonable way to compute the number of colorings by any fixed number of colors $n$ by using the deletion-contraction recurrence for the chromatic polynomial. </p>

ContentLicense: CC BY-SA 2.5


问:
Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$

Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$

This should be fairly straightforward but the proof seems to be alluding me.

I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$.  I recognize this shouldn't be too difficult but perhaps finals have fried my brain.

ContentLicense: CC BY-SA 3.0

Body:
You can use the Taylor series of $\sin(x)$ about $x=0$:

$$\sin(x) = \sum_{n=0}^\infty {\frac {(-1)^n x^{2n+1}} {(2n+1)!}}$$

The first few terms are:

$$\sin(x) = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots$$

Thus, set:

$$x - \frac {x^3}{3!} < \sin x\\
x - \frac {x^3}{3!} < x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots\\
0 < \frac{x^5}{5!} - \frac {x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!}+ \dots\\
$$

For $x>0$, this is true.

Similarly, for $\sin(x) < x$:
$$x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots < x\\
\frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots > 0$$


答:
Suppose $$S=\{w,x,y,z\}.$$

Then
\begin{align}
\mathcal F_x & = \{ A\subset S : x\notin A\} \\[8pt]
& = \big\{ \varnothing, \{w\}, \{y\},\{z\},\{w,y\},\{w,z\},\{y,z\}, \{w,y,z\}\},
\end{align}
i.e. it is the set of all subsets of $S$ that do not have $x$ as a member.


答:
To obtain $(1)$, note that
$$\sum_{i=1}^n\sum_{j=1}^i\frac{1}{ij}
= \sum_{1 \le j \le i \le n} \frac{1}{ij}
\overset{\text{(interchange $i$ and $j$)}}{=} \sum_{1 \le i \le j \le n} \frac{1}{ji}
\overset{\text{(commutativity)}}{=} \sum_{1 \le i \le j \le n} \frac{1}{ij}
= \sum_{i=1}^n\sum_{j=i}^n\frac{1}{ij}$$

Title: Symmetry in summation

Body:
Can someone please explain the motivation behind this equality?

$\sum_{i=1}^{n} (\sum_{j=1}^{i} + \sum_{j=i}^{n})\frac{1}{ij} = 2\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{1}{ij}$

They explain that it is symmetry of some kind.


问:
Prove x = y if $x | y^2$ and $y^2 | x^3$ and $x^3 | y^4$ ...
So, I'm having trouble doing this certain problem. As the title suggests I should prove that if this holds:

$$x | y^2 $$
$$y^2 | x^3$$
$$x^3 | y^4$$
$$...$$

Than for each natural x and y, x is equal to y. I've tried writing the numbers x and y in form of:

$$x = p_1^{\alpha_1} + p_2^{\alpha_2} + ... + a_n^{\alpha_n}$$
$$y = P_1^{\beta_1} + p_2^{\beta_2} + ... + p_n^{\beta_n}$$

Now when I try to express the given equations in this form I get:

$$x \le y^2 \le x^3 \le y^4$$
Which for each $i$ yields:

$$\alpha_i \le 2\beta_i \le 3\alpha_i \le 4\beta_i ...$$

I was now wondering what to do next to prove it. Thanks.

答:
It suffice to prove $\nu_p(x) = \nu_p(y)$ for all prime $p$.

We have the following $\nu_p$ information for any prime $p$ and positive integer $k$,

$(2k+1)\nu_p(x) \le (2k+2) \nu_p(y)$ and $(2k+2) \nu_p(y) \le (2k+3)\nu_p(x)$.

Which implies $\frac{2k+1}{2k+2}\nu_p(x)\leq \nu_p(y)\leq \frac{2k+3}{2k+2}\nu_p(x)$.

This implies $|\nu_p(x) - \nu_p(y)|$ is $0$, and so $\nu_p(x) = \nu_p(y)$ as desired.


问:
First and second derivative of a function log*(n) ("log star").
I would like to ask what are the derivative values (first and second) of a function "log star": $f(n) = \log^*(n)$?

I want to calculate some limit and use the De'l Hospital property, so that's why I need the derivative of "log star":
$$\lim_{n \to \infty}\frac{\log_{2}^*(n)}{\log_{2}(n)}$$

More about this function: https://en.wikipedia.org/wiki/Iterated_logarithm

答:
You might try to use the definition of derivative to find your solutions
$$
\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}
$$
and evaluate at the different intervals that are valid for the function.


答: You can use the Taylor expansion for $\sin (x)$ to get a rather quick solution. If you haven't met Taylor yet, then consider the functions $f(x)=x - \sin x$ and $g(x)=\sin (x) -x + \frac{x^3}{3!}$, compute the derivative, and conclude the functions are increasing for suitable $x>0$. Compute $f(0)$ and $g(0)$ to get the result. </p>

Title: Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$

Body:
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$

This should be fairly straightforward but the proof seems to be alluding me.

I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$.  I recognize this shouldn't be too difficult but perhaps finals have fried my brain.


问:
Show that $\mathbb{C}[x,y] / (y^2 + x^3) \cong \mathbb{C}[t^2,t^3]$
I want to show that $\mathbb{C}[x,y] / (y^2 + x^3) \cong \mathbb{C}[t^2,t^3]$. I'm thinking about using an ismorphism theorem.

Define $\phi:\mathbb{C}[x,y] \rightarrow \mathbb{C}[t^2,t^3]$ by $\phi(x) = t^3$ and $\phi(y) = -t^2$.

First, does this map define a homomorphism of rings? If so can someone elaborate how I can prove that.

Second, how can i show that $\ker \phi = (y^2 + x^3)$ and that Im $\phi = \mathbb{C}[t^2,t^3]$.

Thanks!

答:
(The question originally asked to show that ${\mathbb C}[x,y]/(y^2 + x^3)$ and ${\mathbb C}[t]$ were isomorphic)

These rings are not isomorphic: ${\mathbb C}[t]$ is a unique factorization domain, but ${\mathbb C}[x,y]/(y^2 + x^3)$ is not, because $y^2 = -x^3$ in that ring.

Your reasoning breaks down at the last step: $t$ is not in the image of $\phi$.


问:
If $A$ and $B$ anti-commute then $A$ and $B^2 $ commute
Here is this problem, which I tried to solve for long enough, and I want to know how even to start with this kind of problem:

\begin{align}  AB+BA=0 \Rightarrow A,B^2 \text{  commute.} \end{align}

答:
If $AB=-BA$, then (multiplying both sides by $B$ on the right) $AB^2=-BAB$

and (multiplying both sides by $-B$ on the left) $-BAB=B^2A$,

so $AB^2=B^2A$; i.e., $A$ and $B^2$ commute.


Title: If $A$ and $B$ anti-commute then $A$ and $B^2 $ commute

Body:
Here is this problem, which I tried to solve for long enough, and I want to know how even to start with this kind of problem:

\begin{align}  AB+BA=0 \Rightarrow A,B^2 \text{  commute.} \end{align}

Body: Use $[B^2,\,A]=[B,\,\{A,\,B\}]$.


问:
Best approach for Archimedes problem? Linearization or KKT analysis
I am working through the following problem:

In $R^3$, consider a
sphere of radius r. If one makes a longitudinal cut (i.e., perpendicular to a radius) at a
distance r − h from the center, the cut-off portion has height h. Given a fixed surface area
A > 0 of the cut-off portion, find r and h to maximize the volume of the cut-off pieces.

This yields the nonlinear optimization problem:
\begin{align}
  \min_{r,h \in R} \quad  & -\pi h^2 (r-h/3) \\
  st \quad       & \ 2 \pi rh = A \\
  & \ r, h \ge 0.
 \end{align}

I found the Lagrangian:
$$ L(r,h,\mu_1) = -\pi h^2 (r - \frac{h}{3}) + \mu_1(2\pi r h-A)$$
and KKT conditions
$$\frac{\partial L}{\partial r} = -\pi h^2 + 2\mu_1\pi h = -h^2 + 2\mu_1 h=0 $$
$$\frac{\partial L}{\partial h} = -2\pi r h + \pi h^2 + 2\mu_1\pi r = -2rh+h^2-2r\mu_1=0 $$
$$2\pi r h =A$$
$$\mu_1 \ge 0$$

Then I found a KKT point at $\mu_1 = \frac{1}{2}, h=1, r=1$. However, I couldnt find any more KKT points. The problem asks for an optimal solution. The problem is not convex optimization, so we can not say the KKT point is globally optimal. I tried to model the problem in Gurobi but couldn't linearize the objective function. How can I find the optimal solution to this nonlinear problem?

答:
Hint:

The simplest way of solving this problem is to use the constraint $\ 2\pi rh=A\ $ to reduce the dimension of the objective function's domain.  Substituting $\ r= \frac{A}{2\pi h}\ $ into the function $\ \pi h^2\left(r-\frac{h}{3}\right)\ $ reduces it to $ \frac{Ah}{2}-\frac{\pi h^3}{3}\ $.  There is a unique non-negative point at which the derivative of this function vanishes, and you should be able to show that this point is a maximum by investigating the sign of the derivative.


问:
Why " provided $t$ is not in $\theta$ " at the end of the definition of $\forall $ Intro Rule ? ( Shapiro's Entry On Classical Logic in SEP)
Here is the passage :

"The introduction clause for the universal quantifier is a bit more
complicated. Suppose that a sentence θ contains a closed term t, and
that θ has been deduced from a set of premises Γ. If the closed term t
does not occur in any member of Γ, then θ will hold no matter which
object t may denote. That is, ∀vθ

follows.

```<code>(∀I)
</code>```

For any closed term t, if Γ⊢θ(v|t), then Γ⊢∀vθ provided that t is not
in Γ or θ".

First point :
Something I do not understand is that the preliminary remark suggests that $t$ is contained in $\theta$ while the official statement of the rule requires this  not to be the case.

Second point :
Also, should we not have :

If, for any closed term $t$,  $\Gamma |-\theta(v,t) $, then $\Gamma |- \forall v\theta$

instead of

For any closed term t, if Γ⊢θ(v|t), then Γ⊢∀vθ ?

I mean, in order the universal statement to be deduced, should it not be required that the substitution works for any closed term? ( It seems to me that in the official rule,  " for any closed term" applies to the whole conditional, not to the antecedent in particular.)

Does the alternative reading I suggest make any difference? What do I miss?

答:
$\theta$ might be $v=t$. Then $\theta(v|t)$ is $t=t$ and certainly provable. That doesn't mean that $\forall v\, v=t$ is true.

We might have $t=0$ in $\Gamma$. Then if $\theta$ is $v=0$, we have that $\theta(v|t)$ is $t=0$ and we certainly have $\Gamma\vdash \theta(v|t)$, but not $\forall v\,v=0$.


问:
Normal closure of a finite set
I am studying a proof that states that in every finitely presented residually finite group $G=\langle X \mid R \rangle $ the word problem is solvable.

At some point it states that we can enumerate all words equal to $1$ in $G$ (that means we can enumerate all words for the normal closure of $R$ in $F(X)$.) Why is this possible?

The normal closure is the kernel of $\nu : F(X) \to G$ but I don't see why this should be finite.

答:
It certainly isn't finite! But it is enumerable. That is, possibly countable, but with the bonus property that some computer program can tell you if any given word is in the list.

For instance, the even natural numbers are enumerable, as witnessed by the following program

```<code>def isEven(n):
  return n % 2 == 0
</code>```

Similarly, you can check if a given word is in the normal closure of $R$. How? Well every element of the normal closure is a finite product of conjugates of elements of $R$. We can enumerate all the elements of $G$ by looking at words in the generators $X$, and then we can enumerate all the conjugates of words in $R$ since there's finitely many, and then we can enumerate all the finitely many products of conjugates too.

You have to be a bit careful to make sure you actually hit everything, but it's doable with standard tricks. Here's the idea:

```<code>i = 0
while(true):
  ws = [all tuples of length <= i from R] # note there's finitely many
  gs = [all tuples of length <= i from X] # note there's finitely many
  for (r1,...,rn) in ws:
    for g1 ... gn in gs:
      print(r1^g1 r2^g2 ... rn^gn)
  i += 1
</code>```

The "residually finite" condition tells you how to check that a given element doesn't appear on this list. But I'm not entirely sure how right now.

I hope this helps ^_^

Title: Normal closure of a finite set

Body:
It certainly isn't finite! But it is enumerable. That is, possibly countable, but with the bonus property that some computer program can tell you if any given word is in the list.

For instance, the even natural numbers are enumerable, as witnessed by the following program

def isEven(n):
  return n % 2 == 0

Similarly, you can check if a given word is in the normal closure of $R$. How? Well every element of the normal closure is a finite product of conjugates of elements of $R$. We can enumerate all the elements of $G$ by looking at words in the generators $X$, and then we can enumerate all the conjugates of words in $R$ since there's finitely many, and then we can enumerate all the finitely many products of conjugates too.

You have to be a bit careful to make sure you actually hit everything, but it's doable with standard tricks. Here's the idea:

i = 0
while(true):
  ws = [all tuples of length <= i from R] # note there's finitely many
  gs = [all tuples of length <= i from X] # note there's finitely many
  for (r1,...,rn) in ws:
    for g1 ... gn in gs:
      print(r1^g1 r2^g2 ... rn^gn)
  i += 1

The "residually finite" condition tells you how to check that a given element doesn't appear on this list. But I'm not entirely sure how right now.

I hope this helps ^_^


问:
Size of small connected components in Erdös-Rényi supercritical case
We draw a random graph according to Erdös-Rényi scheme $G(n,p)$ where $p=c/n$ for constant $c=1+\varepsilon$, ($\varepsilon>0$).

It is well known that a.a.s. there is a giant component emerging of size $(1-q)n$ where $q$ is the solution to the equation $e^{c(q-1)}=q$ and all other $qn$ vertices are in 'small' components.

Let $v$ be some vertex. I am interested in a distribution of the size of $C(v)$ (the c.c. of $v$ in the graph) conditioned on $v$ not in the giant component.
$$\forall k\in\mathbb{N}:P(k)=\Pr[|C(v)|=k|v \text{ not in giant c.c.}]=\ ?$$

I noticed corollary 2.7 from here in pages 10-11. Applying it to our case is as follows: Let $Z$ a GW process with $Bin(n,p)$ as its offspring distribution. Conditionally on the extinction event, $Z$ is distributed as a sub-critical GW process $\widetilde{Z}$ with offspring distribution given by:
$$\forall k\in\mathbb{N}\cup \{0\}: Q(k)=q^{k-1}\binom{n}{k}p^k(1-p)^{n-k}=\frac{\binom{n}{k}(pq)^k(1-p)^{n-k}}{q}$$

Is it correct to say that $P(k)$ is the distributed like the total progeny of the process $\widetilde{Z}$ a.a.s. when $n\to\infty$? Is something known about this dist. ?

答:
This seems like the right interpretation; I haven't read the entire 40-page document to check.

Alon and Spencer's Probabilistic Method handles this question differently. As $n \to \infty$, the $\text{Binomial}(n-1, \frac cn)$ distribution converges to the $\text{Poisson}(c)$ distribution, so we can simply study the Poisson branching process. Theorem 11.4.2 in this book says that for any $c>0$, if $T$ is the total size of this branching process, then $$\Pr[T = k] = \frac{e^{-ck} (ck)^{k-1}}{k!}.$$ For $c\le 1$, these probabilities sum to $1$, whereas for $c>1$, they sum to the extinction probability $q$.

For $c>1$, if you want to condition on extinction, you should divide by $q$ to get the probability distribution. Then, $\Pr[T=k]/q$ will give the limiting probability that $|C(v)|=k$, conditioned on $v$ not being in the giant component.


问:
find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$
find all polynomials that satisfy this functional equation
$$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$

I write the polynomial
$$P(x)=\sum_{k=0}^n a_k x^k$$
but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be zero, except the fourth coefficient; $P(x)=a x^4$.
But when I re-substitute it into the equation I found that it's inconsistent but it make sense only if $a=0$.

答:
It is easy to see that the only constant solution is $0$.  Now suppose that $P(x)$ is not a constant.  We remark that, if $P(x)$ is a solution, so is $cP(x)$ for any constant $c$ so we may suppose that the leading term of $P(x)$ is $x^n$ for some $n$.  Comparing the leading terms on the two sides we see that $n=4$.

Letting $x=1$ or $x=16$ shows us that $$P(2)=0=P(16)$$  Letting $x=2$ tells us that $P(4)=0$ and letting $x=4$ then tells us that $P(8)=0$.  It then follows that the only candidate for a monic quartic solution would be $$P(x)=(x-2)(x-4)(x-8)(x-16)$$

It is now a simple matter to confirm that this polynomial does satisfy the functional equation, so the general solution is $$P(x)=c(x-2)(x-4)(x-8)(x-16)$$
for some constant $c$.


问:
Comparison test for series $\sum_{n=1}^{\infty}\frac{n}{n^3 - 2n + 1}$
I am trying to prove the convergence of the series $\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ with the simple comparison test. I know it can be done with other tests but this question came up in my homework for the comparison test before the other tests were discussed so I'm being stubborn and want to use that test.

I know it converges but I haven't been able to find a rigorous justification. This is my reasoning so far.

$\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ converges because $\sum_{n=1}^{\infty}\frac{n}{n^3}$ = $\sum_{n=1}^{\infty}\frac{1}{n^2}$ and I know the latter converges by the p-series criterion.

To use the comparison test as my justification I must show $\frac{n}{n^3−2n+1} \leq \frac{n}{n^3} \ \ \forall \  n$. This would be obvious if the denominator on the left were greater than the denominator on the right but that can't be true as on the left I am subtracting from $n^3$ and on the right $n^3$ is unchanged.

So I switch to proving $\frac{n}{n^3−2n+1} \leq \frac{1}{n^2}$ and then I'm stuck.

If the numerators were both 1 then $n^3−2n+1$ is eventually going to be larger than $n^2$ even if we're subtracting a little bit from $n^3$.

That is, at n=1 we'll get $1^3−2+1=0 \leq 1$ but thereafter $n^3− 2n +1  \geq n^2$ is always true.

But the statement of the comparison test says $a_n \leq b_n \forall n$ where $\sum b_n$ is known to converge.

So that's where I am so far. The problem comes down to showing the necessary inequality holds and is it enough the it holds "eventually".

Any ideas?

答:
We need to start the summation somewhere after $n=1$, since the denominator is $0$ at $n=1$. We will show that $\sum_{n=2}^\infty \frac{n}{n^3-2n+1}$ converges.

For large enough $n$, we have $n^3-2n+1\gt \frac{n^3}{2}$. This follows from general considerations.

But more specifically, $n^3-2n+1\gt \frac{n^3}{2}$ for $n \ge 2$. This is because $n^3-2n\ge \frac{n^3}{2}$. To see that, note that the inequality $n^3-2n\ge \frac{n^3}{2}$ is equivalent to $\frac{n^3}{2}\ge 2n$, which is equivalent to $n^2\ge 4$.

So for $n\ge 2$,  we have
$$\frac{n}{n^3-2n+1} \lt \frac{2}{n^2}.$$
And we know that $\sum_2^\infty \frac{2}{n^2}$ converges.

Remark: We used a common trick. If you have any polynomial $P(x)=a_0x^n +a_1x^{n-1}+\cdots$, where $a_0\ne 0$, then for large enough $|x|$, we have $|P(x)|\lt \left|\frac{a_0 x^n}{2}\right|$. The dominant term is $\dots$ dominant.


问:
Show $\sum_{k=1}^n \frac{(n-1)!}{k!(n-k)!(10n)^{k-1}} \leq 1.06$
I was reading Lecture 7 of Afternotes on Numerical Analysis by G. W. Stewart, and he mentions without proof the following result:

If $n \epsilon_M \leq 0.1$ and $\epsilon_i \leq \epsilon_M (i = 1,2,...,n)$, then

$(1+\epsilon_1)(1+\epsilon_2)\cdots(1+\epsilon_n) = 1 + \eta$

where

$\eta \leq 1.06n \epsilon_M$.

I wanted to figure out where this comes from, and, from pretty standard manipulations, I found that it suffices to show that

$\sum_{k=1}^n \frac{(n-1)!}{k!(n-k)!(10n)^{k-1}} \leq 1.06$. I graphed this on Desmos, and it does appear that this inequality is true. However, I have no idea how to go about showing this or where to even really start with this. Any insight would be appreciated!

答:
The answer above covers the computational part of the question more than adequately but it does not answer how to prove that a bound like that can be rigorously obtained.

Here's an outline of a proof using some basic calculus.

Lemma 1: The function $f(x)=(1+\frac{a}{x})^x$ is increasing for all $x>0$ and $\lim_{x\to\infty}(1+\frac{a}{x})^x=e^a$.

Lemma 2: (Obtaining a bound) $\frac{e^x-1}{x}\leq\frac{2+x}{2-x^2}~,~ x\in(0,\sqrt{2})$

Proof of Lemma 1: One can take a derivative of $f$ noting that
$$f'(x)=f(x)\left[\log\frac{x+a}{x}+\frac{x}{x+a}-1\right]$$
Now use the well known inequality $\log t\leq t-1$, which valid for any argument and plug in $t=\frac{x}{x+a}$. This readily establishes that
$$\log\left(\frac{x+a}{x}\right)\geq-\frac{x}{x+a}+1$$ which in turn shows that f is increasing since it's derivative is positive. The second part of the lemma can be proven in a variety of ways but I will skip the proof here. The monotonicity along with the limit at infinity show that for any $x,a>0$

$$f(x)\leq e^a$$

With Lemma 1 proven we establish that with $x=n, a=1/10$: $$10\left[\left(1+\frac{1}{10n}\right)^n-1\right]\leq \frac{e^{1/10}-1}{1/10}$$

Proof of Lemma 2: For any $n$, one can show using integration by parts and the fundamental theorem of calculus that
$$f(x)=f(0)+xf'(0)+...+f^{(n)}(0)\frac{x^n}{n!}+\frac{1}{n!}\int_{0}^x (x-t)^nf^{(n+1)}(t)dt$$
Select $n=2$ and $f(x)=e^x$. Then

$$e^x-1=x+\frac{x^2}{2}+\int_0^x \frac{(x-t)^2}{2}e^t dt\leq x+\frac{x^2}{2}+\frac{x^2}{2}\int_0^x e^t dt=x+\frac{x^2}{2}+\frac{x^2}{2}(e^x-1)$$
Now if we rearrange and restrict $x$ to be in the range $0<x<\sqrt{2}$ which guarantees that the denominator is positive we can conclude that
$$\frac{e^x-1}{x}\leq\frac{1+\frac{x}{2}}{1-\frac{x^2}{2}}$$
Finally, with $x=1/10$, one obtains a tight upper bound
$$\frac{e^{1/10}-1}{1/10}\leq \frac{210}{199}<1.06$$
and the proof is complete.


答:
By plugging in $x=16$, we find $P(16)=0$. So $P$ has some positive roots. If $P$ is not $\equiv 0$, it has a minimal positive root $x_0$.
Plug in $\frac 12x_0$ to find $0=16(\frac12x_0-1)P(\frac12x_0)$ and so $x_0=2$ as $P(\frac12x_0)\ne 0$. By plugging in $x=2$, we then find $P(4)=0$, next $P(8)=0$ and $P(16)=0$. Similar to above, we find that the maximal positive root is $16$.
So
$$P(x)=(x-2)(x-4)(x-8)(x-16)Q(x) $$
and therefore
$$ (x-16)P(2x)=(x-16)(2x-2)(2x-4)(2x-8)(2x-16)Q(2x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(2x)$$
whereas
$$ 16(x-1)P(x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(x).$$
Therefore $Q(2x)=Q(x)$ for all $x$ (except possibly for $x=1,2,4,8,16$). Hence $Q$ is bounded, hence constant.

Title: find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$

Body:
find all polynomials that satisfy this functional equation
$$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$

I write the polynomial
$$P(x)=\sum_{k=0}^n a_k x^k$$
but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be zero, except the fourth coefficient; $P(x)=a x^4$.
But when I re-substitute it into the equation I found that it's inconsistent but it make sense only if $a=0$.


问:
Modular Arithmetic ( using Euler's Totient Theorem )
Please help me find the last two digits of $312^{84}$ ; I am stuck right here: $312^\phi \equiv 1 \pmod{100}$.

答:
$\phi(100)=40$, but you can't say by Euler's theorem that $312^{40}
\equiv1\bmod100$,

because $\gcd(312,100)\ne1$.

Rather, use Euler's theorem modulo $25$ in conjuction with the Chinese remainder theorem.

$312^{84}\equiv12^{84}\equiv12^4\equiv144^2\equiv(-6)^2\equiv36 \bmod25$,

since $12^{20}\equiv1\bmod25$ by Euler's theorem.

$312^{84}\equiv0\bmod4$, so $312^{84}\equiv36\bmod100,$

so the last two digits of $312^{84}$ are $36$.


问:
Fundamental group of $R^3$ minus two disjoint circles
I know the fundamental group of $X=\mathbb{R}^3-(A\cup B)$ should be $\mathbb{Z}*\mathbb{Z}$, where $A=\{(x,y,1)\in\mathbb{R}^3:x^2+y^2=1\}$ and $B=\{(x,y,2)\in\mathbb{R}^3:x^2+y^2=1\}$.

Could someone show me how to use Van Kampen theorem to do the calculation? This is totally new to me, so I really have no idea.

Let $a=(0,0,0)$ be the base point. I know we should find two path-connected open sets $U$ and $V$ s.t. $X=U\cup V$, and $U\cap V$ is also path-connected and contains $a$. But what is a proper choice of $U$ and $V$? I guess $\pi_1(U,a)$ and $\pi_1(V,a)$ should both equal to $\mathbb{Z}$. Not sure if this is true.

答:
Let us first compute the fundamental group of $\mathbb{R}^3 \setminus S^1$. For this we can use the following homotopy equivalences:

An easy application of the van Kampen theorem (or some other result you can use to compute fundamental groups of wedges) thus yields $$\pi_1(\mathbb{R}^3 \setminus S^1) \cong \pi(S^2 \vee S^1) = \mathbb{Z}.$$ Here I was assuming that you know the fundamental groups of spheres. Now consider the case of two missing circles. I will denote that space by $X$ as you did. We choose open sets $U$ and $V$ as follows:

Note that both $U$ and $V$ and their intersection are path-connected. Moreover, the intersection has trivial fundamental group and $U$ and $V$ are both homotopy equivalent to $\mathbb{R}^3 \setminus S^1$. Thus by the van Kampen theorem (and our computation before) we get $$\pi_1(X) \cong \pi_1 (U) \ast_{\pi_1(U \cap V)} \pi_1(V) = \pi_1 (\mathbb{R}^3 \setminus S^1) \ast_{1} \pi_1(\mathbb{R}^3 \setminus S^1) = \mathbb{Z} \ast \mathbb{Z},$$ as wanted.

Edit: I updated the drawings and the text, so that one can actually read the text.


问:
Is integration by substitution always the reverse chain rule?
I understand that integration by substitution can be justified in the following way. If $I=\int f'(g(x))g'(x) \, dx$ then $I=f(g(x))+C$. If we make the substitutions
$$
u=g(x) \text{ and } du=g'(x)dx
$$
then $I$ becomes $\int f'(u)du=f(u)+C=f(g(x))+C$, which is the same. However, I often see substitutions that don't seem to take this form. For example, a common way to evaluate
$$
I=\int \sqrt{1-x^2} \, dx
$$
is by setting $x=\sin u$ (or, to be more precise, $u=\arcsin x$). The derivative of $\arcsin x$ is
$$
\frac{1}{\sqrt{1-x^2}} \, ,
$$
which does not appear in $I$. It doesn't seem like $I$ is of the form $\int f(g(x))g'(x) \, dx$. So why is this substitution justified? Are we still doing the chain rule in reverse, or is something else going on?

答:
The fact that something does not appear does not mean it is not there. You can always multiply and divide; $$\sqrt{1-x^2}=\frac{1-x^2}{\sqrt{1-x^2}},\qquad \lvert x \rvert <1. $$

Let me give some more detail. Note that, for $\lvert x \rvert <1$,
$$\label{1}\tag{1}
\sqrt{1-x^2}=\frac{\cos^2(\arcsin x)}{\sqrt{1-x^2}}, $$
which equals $f(g(x))g'(x)$ for $f(u)=\cos^2 u, g(x)=\arcsin x$. And so,
$$
\int \sqrt{1-x^2}\, dx = \int \frac{\cos^2(\arcsin x)}{\sqrt{1-x^2}}\, dx = \int \cos^2(u)\, du.$$
As you can see, this is exactly the application of the chain rule that you have  mentioned. The usual technique of letting $u=\arcsin x$ and computing $dx$'s and $du$'s is just a practical method to compute \eqref{1}.


问:
Probabilistic classical algorithm on Deutsch-Jozsa problem
Here is the description of Deutsch-Jozsa problem: Let $f: \{0,1\}^n \mapsto \{0,1\}$ be a function promised to be either constant or balanced ('balanced' means that $f$ outputs as many 0's as 1's).

I need to show that a probabilistic classical algorithm making two evaluations of $f$ can with probability at least 2/3 correctly determine whether $f$ is constant or balanced.

There is also a hint: Your guess does not need to be a deterministic function of the results of the two queries. Your result should not assume any particular a priori probabilities of having a constant or balanced function.

I'm a bit lost here.  My thinking is that if the two evaluations are the different, then $f$ is definitely balanced.  Otherwise, $f$ could be either constant or balanced.  But the chance of success would be depending on the probability of $f$ being constant, which is against the given hint.

How should I approach this problem?

答:
I had the same question that you had, and @PSPACEhard's answer helped me a lot to understand the reasoning. So I'm going to provide another answer with a slightly different approach:

At first sight, you would think that with only two evaluations of $f$, your algorithm should always output "balanced", because there are many more balanced functions than there are constant. However, if you design your algorithm in a certain way, you can ignore the probabilities of the function being balanced vs. constant.So below, I will give the reasoning that would allow you to design the algorithm. Notation:
$$
f_b = \text{The event that the function is balanced}\\
f_c = \text{The event that the function is constant}\\
O_b = \text{The event that our algorithm outputs "balanced"}\\
O_c = \text{The event that our algorithm outputs "constant"}\\
Eq = \text{The event that }f(q_1)=f(q_2)\\
Diff = \text{The event that }f(q_1) \neq f(q_2)\\
$$

Now, given an algorithm that does two evaluations, and then output either balanced or constant:
$$
P(\text{answer is correct}) = P(O_b,f_b) + P(O_c,f_c)
$$
The term $P(O_b,f_b)$ can be written as:
$$
P(O_b,f_b) = \\
P(O_b|f_b) \cdot P(f_b) = \\
(P(O_b,Eq|f_b)+P(O_b,Diff|f_b)) \cdot P(f_b) = \\
(P(O_b|Eq,f_b) \cdot P(Eq|f_b) + P(O_b|Diff,f_b) \cdot P(Diff|f_b)) \cdot P(f_b)
$$

Now, since we want an algorithm that makes it's decisions only based on what was observed, we have:
$$P(O_b|Eq, f_b) = P(O_b|Eq)$$
and
$$P(O_b|Diff, f_b) = P(O_b|Diff)$$
Also, any non-stupid algorithm should output balanced with probability 1 in case of Diff, thus $P(O_b|Diff) = 1$. Taking the above observations into account, the expression becomes:
$$
P(O_b,f_b) = \\
(P(O_b|Eq) \cdot P(Eq|f_b) + 1 \cdot P(Diff|f_b)) \cdot P(f_b) = \\
(\frac{1}{2}P(O_b|Eq) + \frac{1}{2}) \cdot P(f_b)
$$
With a similar analysis we find:
$$
P(O_c, f_c) = \\
P(O_c|Eq) \cdot P(f_c)
$$
And so, going back to our original equation:
$$
P(\text{answer is correct}) =
(\frac{1}{2}P(O_b|Eq) + \frac{1}{2}) \cdot P(f_b) + P(O_c|Eq) \cdot P(f_c)
$$
Now, we notice that if choose our values for $P(O_b|Eq)$ and $P(O_c|Eq)$ carefully (which we are free to do, because this is literally what defines our algorithm's behavior), we can make $(\frac{1}{2}P(O_b|Eq) + \frac{1}{2})=P(O_c|Eq)=\alpha$, and the equation above becomes:
$$
P(\text{answer is correct}) = \alpha \cdot (P(f_b) + P(f_c)) = \alpha
$$
which is not dependent any more on $f_b$ or $f_c$!

So let's see what values achieve this design:
$$
\frac{1}{2}P(O_b|Eq) + \frac{1}{2} = P(O_c|Eq)\\
\text{but } P(O_c|Eq) = 1 - P(O_b|Eq)\\
=> P(O_b|Eq) = \frac{1}{3} \text{ and } P(O_c|Eq) = \frac{2}{3}
$$
And $P(\text{answer is correct}) = \alpha = \frac{2}{3}$

And so you get the algorithm described in @PSPACEhard's answer.

Of course, you can reach this conclusion in shorter ways, but that was the pathway that worked for me.


问:
The Irwin-Hall Distribution with a different parameter
Many of you would know of the Irwin-Hall distribution which states that if you have $n$ independent and identically distributed $U(0,1)$ variables, $U_1, U_2, U_3 \cdots$ and let $X = \sum_{i=1}^{n} U_i$, then the probability density of function of $X$ is: $${\displaystyle f_{X}(x;n)={\frac {1}{2(n-1)!}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}(x-k)^{n-1}\operatorname {sgn}(x-k)}.$$

Is there something similar that gives the cumulative density function of $X$? Also, is there a way to change the parameter for the variables to make it so that they are uniformly distributed from $(-1, 1)$?

答:
Refer to Wikipedia for the CDF, which is

$$F_X(x) = \frac{1}{n!} \sum_{k=0}^{\lfloor x \rfloor} (-1)^k \binom{n}{k} (x-k)^n.$$

As for your second question, this is something you can derive yourself via a suitable transformation, namely if $X \sim \operatorname{IrwinHall}(0,n)$, we have $$Y = 2X-n,$$ thus $X = g^{-1}(Y) = \frac{Y+n}{2}$ and $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| = \frac{1}{2} f_X(\tfrac{y+n}{2}),$$ where the density $f_X$ is the formula you provided.


问:
Prove that a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous in lower and upper limit topologies is constant.
Let $f: \mathbb{R} \to \mathbb{R}$ a function that is continuous in the topologies $\mathcal{T}_{u} \to \mathcal{T}_{[,)}$ and $\mathcal{T}_{u} \to \mathcal{T}_{(,]}$, where $\mathcal{T}_{u}$ is the usual topology on $\mathbb{R}$ and $\mathcal{T}_{[,)}$ and $\mathcal{T}_{(,]}$ are the lower and upper limit topologies on $\mathbb{R}$, respectively. Show that $f$ is constant.

I don't actually know how to begin the proof because I can't see an easy way of using some of the most common characterizations of continuity. Maybe it is useful to use the comparison between the limit topologies and the usual one... I've also about arguing by contradiction, but I cannot see an easy argument. Also, I know there is an easy way of proving this using connection concepts, but we are supposed to do it without using them. I would be pleased of any hints. Thanks in advance.

答:
Let $X\in\Bbb R$. Then $U=(\leftarrow,f(x)]$ is an open nbhd of $f(x)$ in the upper limit topology, and $V=[f(x),\to)$ is an open nbhd of $f(x)$ in the lower limit topology, so

$$f^{-1}\big[\{f(x)\}\big]=f^{-1}[U]\cap f^{-1}[V]$$

is open in the usual topology. On the other hand, $\{f(x)\}$ is closed in both the upper limit and the lower limit topology, so $f^{-1}\big[\{f(x)\}\big]$ is closed in the usual topology. What subsets of $\Bbb R$ are clopen (both closed and open) in the usual topology?


问:
How to solve this difference quotient which doesn't have the standard difference quotient formula?
Haven't touched Calculus in a very long time. Going back through it at the moment and this equation has me stumped right now:

$$ f(x)= \frac{1}{x}, \frac{f(x)-f(a)}{x-a} $$
I know how a regular differential quotient works but this is confusing to me as to what values go where. And my textbook only tells me the answer but doesn't show the steps to get to the answer. Any help would be much appreciated. I Googled for a bit looking for calculators that show every step and also looking for "unique differential quotients" and the like. But came out empty-handed. I'm not looking for an answer, I already have it. I just want to understand the process, what values get substituted and what not. Thanks.

Body: $$f'(a)=\lim\limits_{x\to a}\dfrac{\frac1x-\frac1a}{x-a}=\lim\limits_{x\to a}\dfrac{a-x}{ax}\dfrac{1}{x-a}=\lim\limits_{x\to a}\dfrac1{ax}\dfrac{a-x}{x-a}=\lim\limits_{x\to a}\dfrac{-1}{ax}=\dfrac{-1}{a^2}$$


问:
Solve $-\frac{\partial }{\partial x} (-\gamma p^* - \frac{\sigma^2}{2} \frac{\partial }{\partial x} p^* ) = 0$
I am trying to solve the following Fokker plank equation (partial differential equation)

\begin{align*}
-\frac{\partial }{\partial x} (-\gamma p^*(x,t) - \frac{\sigma^2}{2}  \frac{\partial }{\partial x} p^*(x,t) ) = 0
\end{align*}

or by changing the notation slightly by setting $p^* = p^*(x,t)$
\begin{align*}
-\frac{\partial }{\partial x} (-\gamma p^* - \frac{\sigma^2}{2}  \frac{\partial }{\partial x} p^* ) = 0
\end{align*}

so that I have it in terms of $p$ in the form (I am trying to get to this result explicitly)
\begin{align*}
p^* = C e^{\frac{-\gamma x^2}{\sigma^2}}
\end{align*}
where $C$ represents some constant.

My attempt (not complete at the end)

\begin{align*}
-\frac{\partial }{\partial x} (-\gamma p^* - \frac{\sigma^2}{2}  \frac{\partial }{\partial x} p^* ) = 0 \\
\frac{\partial }{\partial x} (-\gamma p^* - \frac{\sigma^2}{2}  \frac{\partial }{\partial x} p^* ) = 0
\end{align*}

Since our function does not depend on x the value should be equal to some constant $C_1$

\begin{align}
 (-\gamma p^* - \frac{\sigma^2}{2}  \frac{\partial }{\partial x} p^* ) = C_1 \\
  - \frac{\sigma^2}{2}  \frac{\partial }{\partial x} p^*  = C_1 + \gamma p^* \\
   \frac{\partial }{\partial x} p^*  = - \frac{2}{\sigma^2}  \left( C_1 + \gamma p^* \right)
\end{align}

Please can you show more explicitly how to get to the correct result $p^* = C e^{\frac{-\gamma x^2}{\sigma^2}}$.

答:
You have two errors at your current stage. First, note that you have already neglected the time derivative in the Fokker-Planck equation which is equivalent to taking the limit $t \rightarrow \infty$ (steady-state conditions). There is no time variable in the resulting problem, so you have an ordinary differential equation.
Second, you are missing a factor of $x$ for the standard Ornstein-Uhlenbeck Process. The correct Fokker-Planck equation for a steady-state Ornstein-Uhlenbeck process is:
$$
-\frac{d}{dx} \Big[-\gamma x p^*(x) - \frac{\sigma^2}{2}  \frac{d }{d x} p^\ast(x) \Big] = 0,
$$
and integrating once gives
$$ -\gamma x p^\ast(x) -   \frac{\sigma^2}{2}  \frac{d }{d x} p^\ast(x)  = C, $$
where $C$ is some constant.

If your distribution $p$ is to be normalizable, you need to require
$$ \lim_{x\rightarrow \pm \infty} p^\ast = 0, $$
which implies $C=0$.

Therefore we have

$$ \frac{dp^\ast}{p^\ast} = -\frac{2 \gamma x}{\sigma^2}  dx$$

which trivially integrates for

$$ p^\ast(x) = A \exp\big[-\gamma x^2/\sigma^2 \big],$$
where $A$ is a constant. You can the normalize with the well-known / extremely important integral

$$ \sqrt{\pi} = \int_{-\infty}^\infty e^{-x^2} dx.$$


问:
Any example of known strong convexity constant?
A continuously differentiable function $f(x)$ is strongly convex on $\mathbb{R}^{n}$ if there exists a positive constant $\mu$ such that for any $x, y \ \in \mathbb{R}^{n}$,
\begin{align}
f(y)\geq f(x) + \langle \nabla f(x), y-x\rangle + \frac{1}{2}\mu \|y-x\|^2
\end{align}
I'm curious about the strong convexity constant $\mu$. I heard that for some limited cases, it is known.

Could anyone tell me what kind of functions have a known value for $\mu$? For example, is it known for a quadratic function? I searched on Google but I haven't found any useful source.

答: For example, if $f(x) = \|x\|^2$, then $\nabla f(x) = 2 x$, and it's easy to show that your inequality becomes an equality if $\mu = 2$.


问:
Inverse limit of (sub)sets
Let $(X_i)_{i\in I}$ be a family of subsets $X_i\subset X$ partially ordered by inclusion. If $X_j\subseteq X_i$ let $\iota_{ij}\colon X_j\hookrightarrow X_i$ be the inclusion and write $i\le j$. This gives rise to a projective system $(X_i,\iota_{ij})_{i,j\in I,i\le j}$ as clearly

the self inclusion $\iota_{ii}\colon X_i\hookrightarrow X_i$ is in fact just the identity on $X_i$ and therefore $\iota_{ii}=\operatorname{id}_{X_i}$

if we have $i\le j\le k$ the composition $\iota_{ij}\circ\iota_{jk}\colon X_k\hookrightarrow X_j\hookrightarrow X_i$ corresponds to the iterated inclusion $X_k\subseteq X_j\subseteq X_i$ which is precisely given by $\iota_{ik}\colon X_k\hookrightarrow X_i$ and hence $\iota_{ik}=\iota_{ij}\circ\iota_{jk}$

We consider the intersection $\bigcap_{i\in I}X_i$ with (canoncial) inclusions $\iota_j\colon\bigcap_{i\in I}X_i\hookrightarrow X_j$. These are compatible with the maps $\iota_{ij}$ as $\iota_{ij}\circ\iota_j$ corresponds, again, to an iterated inclusion and hence $\iota_i=\iota_{ij}\circ\iota_j$.

In fact, we have $\lim\limits_{\substack{\longleftarrow\\\small i\in I}}X_i=\bigcap\limits_{i\in I}X_i$.

It remains to verify the universal property of the inverse limit. For this purpose let $\mathfrak X$ be a set accompanied by maps $f_i\colon\mathfrak X\to X_i$ which are compatible with the maps $\iota_{ij}$ as usual. Consider $x\in\mathfrak X$ and its image(s) $f_j(x)\in X_j$. By the compatibility property we have $f_i(x)=(\iota_{ij}\circ f_j)(x)=f_j(x)$ for all $i\le j$ since the $\iota_{ij}$ are inclusion maps. In particular, there is some index $k\in I$ such that $i,j\le k$ and hence $X_k\subseteq X_i\cap X_j$. But then also $f_k(x)=f_i(x)=f_j(x)$ and hence there is a well-defined element $\overline x\in\bigcap_{i\in I}X_i$ such that we can set $\rho(x)=\overline x$ which is compatible with the maps $\iota_i,f_i$. This guarentees the existence of a map $\rho\colon\mathfrak X\to\bigcap_{i\in I}X_i$. For the uniqueness part consider a map $\rho\colon\mathfrak X\to\bigcap_{i\in I}X_i$ such that $f_i=\iota_i\circ\rho$ for all $i\in I$. An element $x\in\mathfrak X$ determines an element $f_j(x)$ for all $j\in I$ which in turn uniquely determines this element for all indices $i$ such that $i\le j$. This ensures the uniqueness of $\rho(x)$.

I am not really confident about this proof/construction. While the existence parts seems relatively rigorous to me the uniqueness part appears completely handwavy. I am trying to get a better grasp on inverse limits and the simple example in $\sf Set$ should help for intuition.

So..., here are the main questions: Is the given proof correct? Can it be improved; if so, how? I would like to see a precise verification of $\lim\limits_{\substack{\longleftarrow\\\small i\in I}}X_i=\bigcap\limits_{i\in I}X_i$ as I was not able to locate a reference for this fact (as simple as it might be) and the more involved examples (say, in $\sf Grp$ or ${\sf R}$-${\sf Mod}$) did not really help  me to understand this particular scenario.

Bonus: Explain how this case relates to the dual situation of a inductive system where the direct limit is $\lim\limits_{\substack{\longrightarrow\\\small i\in I}}X_i=\bigcup\limits_{i\in I}X_i$ and hence the role of intersection(s) is replaced by the union(s).

Thanks in advance!

答:
I think the idea is good; let me comment on how I would think about it:

In order for the inverse limit $\varprojlim_{i \in I} X_i$ to be defined, you need $I$ to have the structure of a category, and you need $i \mapsto X_i$ to have the structure of a functor. You do this by making $I$ into a preorder when you define the "$\leq"$ relation, but to be super-pedantic, you didn't explicitly check that the relation $\leq$ is reflexive and transitive. Then you do check that $i \mapsto X_i$ is a functor with the action $(i\leq j) \mapsto \iota_{ij}$ on morphisms. To be super-pedantic again, you never explicitly said that $X$ was to be a functor to the category $Set$ (rather than to some other category); I suppose this was understood, but see the second bullet below.

You then define a cone on the functor $X: I \to Set$, with vertex $\cap_{i \in I} X_i$ and legs $\iota_i$, and verify that this is indeed a cone. Great.

Next you verify the universal property, which I think looks good modulo the one significant issue of your post which I come to next.

The only real issue is the following. In the hypotheses, you never stipulated that the family of sets $(X_i)_{i \in I}$ be downward directed in the sense that for every $i,j \in I$ there exists $k$ with $X_k \subseteq X_i \cap X_j$. But you use this assumption in your proof of the universal property, when you produce the index $k$.

So what you've shown is that if $(X_i)_{i \in I}$ is a family of subsets of $X$ which is downward-directed, then $\varprojlim_{i \in I} X_i = \cap_{i \in I} X_i$. This is good. Without the hypothesis that $(X_i)_{i \in I}$ be downward-directed, the statement would be false in general. For example, suppose that $I = \{1,2\}$ and neither $X_1$ nor $X_2$ is contained in the other. Then $I$ is a discrete poset with the $\leq$ order you define, and $\varprojlim_{i \in I} = X_1 \times X_2$, which of course is typically different from $X_1 \cap X_2$.

Some related notes:

In the usage I'm familiar with, the terms "inverse limit" and "projective limit" are ambiguous. Sometimes they mean (1) "limit indexed by a downward-directed poset" (or more generally, "limit indexed by a co-filtered category"), but sometimes they mean more generally (2) "limit indexed by an arbitrary category". I think that at least nowadays most people tend to just say "limit" when they mean (2), so that the usage (1) is more common, but it's something to watch out for.

In general the limit $\varprojlim_{i \in I} X_i$ can change depending what category you regard the assignment $i \mapsto X_i$ as taking values in. Let's go back to the example where $I = \{1,2\}$ is discrete. If you regard the assignment $i \mapsto X_i$ as taking values in the category $Set$, then as remarked above, you get $\varprojlim_{i \in I} X_i = X_1 \times X_2$, the cartesian product. But instead, you might regard $i \mapsto X_i$ as a functor $I \to Set/ X$, where $Set / X$ is the slice category, i.e. the category where an object is a set $Y$ equipped with a map $Y \to X$, and a morphism is a map $Y \to Y'$ making the obvious diagram commute. If you do this, then you do end up getting $\varprojlim_{i \in I} X_i = X_1 \cap X_2$. Similarly, if you regard $i \mapsto X_i$ as taking values in the poset $Sub(X)$ of subsets of $X$ and inclusions (note that this is a full subcategory of $Set/X$), then you again get $\varprojlim_{i \in I} X_i = X_1 \cap X_2$.

Similarly, the limit can change depending on what we regard the morphisms of $I$ as being, so it's good that you were explicit about this. An exception is when we compute limits in a category which is a preorder. In this case, $\varprojlim_{i \in I} X_i$ is the greatest lower bound of the $X_i$'s (either existing if the other does), and it doesn't matter what category structure you regard $I$ as having for this to be true.

Title: Inverse limit of (sub)sets


问:
Find column of inverted matrix without finding inverted matrix
How would I go about finding, say, the third column of an inverted matrix without finding the inverted matrix itself? I can't wrap my head around this, my standard procedure would be to simply calculate the inverted matrix and extract the desired column.

The matrix in question is as follows:
\begin{bmatrix}-2&-7&-9\\2&5&6\\1&3&4\end{bmatrix}

答:
If $\ e_3\ $ is the third column of the inverse, its entries must satisfy the following system of linear equations:
$$
\begin{bmatrix}-2&-7&-9\\2&5&6\\1&3&4\end{bmatrix}e_3=\begin{bmatrix}0\\0\\1\end{bmatrix}\ .
$$


问:
Solution of $1+x<e^x$

Show that $1+x<e^x$ for all $x>0$

My attempt:

Consider the function $f(x)=e^x-x-1$ and fix $x>0$. By the MVT there exists a point $c\in(0,x)$ such that $\frac{f(x)-f(0)}{x-0}=f'(c)~~~(*)$.

Note that $f(0)=e^0-0-1=0$ and $f'(x)=e^x-1$. This implies that $f'(c)=e^c-1$.

The equality $(*)$ can be rewritten as $\frac{e^x-x-1-0}{x}=e^c-1$.

Since $c$ is a positive number, we have that $e^c>1$. So, the difference $e^c-1>0$.

This means that $\frac{e^x-x-1-0}{x}>0$. Then, $e^x>1+x$ for $x>0$.

Is this solution correct?

答:
Consider the function $f(x) = e^{x} - x - 1$. Consider $f'(x) = e^{x} - 1$, and note that this is positive for all $x > 0$ because $f'(0) = 0$ and $e^{x}$ is an increasing function. Thus, $f(x)$ is strictly increasing for all $x>0$. However, since $f(0) = 0$, this means that $f(x)>0$ for all $x > 0$, and thus:

$$\boxed{e^{x} > x + 1\text{ for }x>0}$$

We can extend this to $e^{x} > x + 1$ for $x < 0$ using a similar argument.

Title: Solution of $1+x<e^x$


答:
Your solution is fine, if a bit more complicated than necessary. If you are allowing derivatives, why not just note that $f'(x)=e^x-1\gt0$ for $x\gt0$, so $f$ is strictly increasing, hence $f(x)\gt f(0)=0$ for all $x\gt0$?

Just for fun, there is a non-calculus proof, provided you know that $(1+x/n)^n$ increases to $e^x$ as $n\to\infty$ (for $x\gt0$).  Then the binomial expansion (with $n\gt1$) tells us

$$1+x=1+n\cdot{x\over n}\lt\left(1+{x\over n}\right)^n\lt e^x$$

for all $x\gt0$.

Added later: On further reflection, it occurs to me that your MVT approach actually works nicely as a proof that $e^x\gt1+x$ for all $x\not=0$, not just for positive $x$.  That is, the Mean Value Theorem guarantees a $c$ between $0$ and $x$ such that $f(x)/x=f'(c)=e^c-1$. Now $x$ and $e^c-1$ have the same sign (e.g., if $x\lt c\lt0$, then $e^c\lt1$), so either way $f(x)$ must be positive. The other proofs here have to do something different to handle the case $x\lt0$.

Title: Solution of $1+x<e^x$

Body:
Show that $1+x<e^x$ for all $x>0$

My attempt:

Consider the function $f(x)=e^x-x-1$ and fix $x>0$. By the MVT there exists a point $c\in(0,x)$ such that $\frac{f(x)-f(0)}{x-0}=f'(c)~~~(*)$.

Note that $f(0)=e^0-0-1=0$ and $f'(x)=e^x-1$. This implies that $f'(c)=e^c-1$.

The equality $(*)$ can be rewritten as $\frac{e^x-x-1-0}{x}=e^c-1$.

Since $c$ is a positive number, we have that $e^c>1$. So, the difference $e^c-1>0$.

This means that $\frac{e^x-x-1-0}{x}>0$. Then, $e^x>1+x$ for $x>0$.

Is this solution correct?


问:
limit along ultrafilter is multiplicative
I am trying to prove that the limit of a sequence along a non-principal ultrafilter on $\mathbb{N}$ is multiplicative and I am getting stuck. This is what I have until now.

Let $U$ be a non-principal ultrafilter on $\mathbb{N}$ and let $\boldsymbol{x}=(x_n)_n$ and $\boldsymbol{y}=(y_n)_n$ be two elements in $\ell^{\infty}$. Define a map $f$ such that $f(\boldsymbol{x})=U-\lim\limits_{n}x_n$. Then

$$
f(\boldsymbol{x})\cdot f(\boldsymbol{y})=U-\lim\limits_{n}x_n \cdot U-\lim\limits_{n}y_n=L_1 L_2,
$$
where $L_1,L_2$ are such that for every $\varepsilon >0$ $\{n\in \mathbb{N} : |x_n-L_1|< \varepsilon  \}\subset U$ and $\{n\in \mathbb{N} : |y_n-L_2|< \varepsilon  \}\subset U$.

Now,
$$
f(\boldsymbol{x}\boldsymbol{y})=U-\lim\limits_{n}x_ny_n=L,
$$
where $L$ is such that for every $\varepsilon >0$ $\{n\in \mathbb{N} : |x_ny_n-L|< \varepsilon  \}\subset U$.

How do I conclude that $L_1 L_2 =L$?

答:
HINT: Fix $\epsilon>0$. You know by the continuity of multiplication that there is a $\delta>0$ such that $|x_ny_n-L_1L_2|<\epsilon$ whenever $|x_n-L_1|<\delta$ and $|y_n-L_2|<\delta$. Let

$$u_x=\{n\in\Bbb N:|x_n-L_1|<\delta\}$$

and

$$u_y=\{n\in\Bbb N:|y_n-L_2|<\delta\}\,,$$

and consider $u=u_x\cap u_y$.


问:
Motivation for the integral notation $\int _{I \in \mathbb{I}} D(I)$ for a categorical limit?
What is the motivation for the integral notation $\int _{I \in \mathbb{I}} D(I)$ for a categorical limit, which is otherwise known as $\underset  {\mathbb{I} \leftarrow} {lim} \ D$?

答: As mentioned in the comments, this notation is most commonly used for ends and coends. In fact, limits and colimits are special cases of ends and coends (the case where one takes the end /coend of a functor $F: C \times C^{op} \to D$ which depends only on one of its variables), so this notation is a special case of the end / coend notation. For more discussion of this notation, see here.


问:
Recursion relation for Euler numbers
I am trying to solve the following:

The Euler numbers $E_n$ are defined by the power series expansion

$$\frac{1}{\cos z}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n\text{ for }|z|<\pi/2$$

(a)  Show that $E_n=0$ when $n$  is an odd integer.

(b) Establish a recursion relation for the sequence.

Through research it seems that part a seems well established but I couldn't find a proof.
Part (b) I found a formula, but couldn't find how to establish it. Any help appreciated.

答:
Related problems: (I), (II). It should be

$$\frac{1}{\cosh(z)} =\sum_{n=0}^\infty \frac{E_n}{n!}z^n,\quad|z|<\pi/2. $$ Here is few terms of Taylor series of $\frac{1}{\cosh(z)}$

$$ \frac{1}{\cosh(z)}=  1-{\frac {1}{2}}{z}^{2}+{\frac {5}{24}}{z}^{4}-{\frac {61}{720}}{z}^{
6}+O \left( {z}^{8} \right),$$

which they match your power series.

Added: Here is another representation. It may be of interest to some people

$$\frac{1}{\cosh(z)} =\sum _{n=0}^{\infty }{\frac {\left(
-1 \right)^{n}\left(\psi \left( 2\,n,\frac{3}{4}\right)-\psi \left( 2\,n,\frac{1}{4} \right) \right) {z}^{2\,n}}{ {4}^{n}{\pi }^{2\,n+1}\left( 2\,n \right) !}},$$

where $\psi(z)$ is the digamma function. Writing the OP's identity as

$$ \frac{1}{\cosh (z)}=\sum_{n=0}^\infty \frac{E_{2n}}{(2n)!}z^{2n} ,$$

one can readily see the following relation

$$ E_{2n} = {\frac {\left(
-1 \right)^{n}\left(\psi \left( 2\,n,\frac{3}{4}\right)-\psi \left( 2\,n,\frac{1}{4} \right) \right) }{ {4}^{n}{\pi }^{2\,n+1}}}, \quad n\in \mathbb{N}\cup \{0\}.  $$

Title: Recursion relation for Euler numbers


问:
Parallelizable product of manifolds
This problem comes from Hirsh's book Differential Topology, chapter 4, section 2, question 5.

Let $M$ and $N$ be (paracompact) $C^r$ manifolds, and $\epsilon^n_E$ the trivial vector bundle on $E$. If $TM$ admits a nonvanishing section, and $TM \oplus \epsilon^1_M$ and $TN \oplus \epsilon^1_N$ are trivial, then the manifold $M \times N$ is parallelizable.

I went for studying short exact sequences of vector bundles however I did not see how to use the section in this scenario and although I got morphisms from $N \times M \times \mathbb{R}^l$ to $T(M \times N)$ and vice versa but I never got any isomorphism.

For example, if $TN \oplus \epsilon^1_N \simeq N \times \mathbb{R}^n$ and $TM \oplus \epsilon^1_M \simeq M \times \mathbb{R}^m$  I can get an morphism $N \times M \times \mathbb{R}^{r+m} \rightarrow T(M \times N)$. Indeed by composing product maps and isomorphism I get the maps:
$$\rho : N \times M \times \mathbb{R}^{r+m} \twoheadrightarrow M \times \mathbb{R}^m \simeq TM \oplus \epsilon^1_M \twoheadrightarrow TM $$
$$\phi : N \times M \times \mathbb{R}^{r+m} \twoheadrightarrow N \times \mathbb{R}^n \simeq TN \oplus \epsilon^1_N \twoheadrightarrow TN $$

Then the morphism is $(\rho, \phi)$. In the other way, we get another morphism $T(M \times N) \rightarrow  N \times M \times \mathbb{R}^{r+m}$ through the inverse inclusion maps. I do not see however how to  use the nonvanishing section as it does not guarantee a monomorphism $M\times \mathbb{R}^{dim(M)} \hookrightarrow TM$ since it is just nonvanishing, nor do I see how to get an isomorphism out of this commutative diagram.   I really would appreciate a tip.

Thank you

Update :

I did found that the section gives an isomorphism $TM \simeq \epsilon^1_M \oplus \eta$ with $\eta$ a vector bundle of base $M$. By the first properties we get $\epsilon^2_M \oplus \eta \simeq M \times \mathbb{R}^n \simeq TM\oplus \epsilon_M^1$. Then in this case we get the sequence.

Then we get the right exact sequence : $$  \epsilon^2_M \oplus \eta \twoheadrightarrow \epsilon^1_M \oplus \eta \rightarrow 0 $$ By the definition of the quotient of vector bundles we get a unique bundle $\zeta$ such that $\frac{\epsilon^2_M \oplus \eta}{\zeta} \simeq \epsilon^1_M \oplus \eta \simeq TM$. This $\zeta$ is also the kernel of the morphism $\epsilon^2_M \oplus \eta \twoheadrightarrow \epsilon^1_M \oplus \eta $ as such it is isomorphic to $\epsilon^1_M$. Thus we get $$TM \simeq \frac{\epsilon^2_M \oplus \eta}{\epsilon^1_M} \simeq \frac{TM\oplus \epsilon_M^1}{\epsilon^1_M} \simeq \frac{M \times \mathbb{R}^m}{M \times \mathbb{R}}$$
From there nothing tells us that quotient of trivial bundles are trivial.

答:
To answer the question I'll use the hint given by Jason De Vito in his comment.
First let's show that : $$\pi^*_M(TM) \oplus \pi^*_N(TN) \cong T(M \times N)$$
Indeed $T(M \times N)$ is indeed a bundle, and the whitney sum can be expressed as a pull back bundle of the diagonal morphism between $M \times N$ and the vector bundle $(p, T(M)\times T(N),  M \times N \times M \times N )$ however through the universal property of pullback bundles, since $T(M\times N) \rightarrow T(M) \times T(N)$ which extends the diagonal map provides an isomorphism between $\pi^*_M(TM) \oplus \pi^*_N(TN) \cong T(M \times N)$.
In the same manner, we get a bundle morphism $(\pi_M^*TN) \oplus \epsilon^1_{M \times N}$ to $TN \oplus \epsilon^1_N$ by summing the projection $\epsilon^1_{M \times N} \twoheadrightarrow  \epsilon^1_N$ and the morphism  we got before $(\pi_N^*TN) \rightarrow TN$ That indeed covers the projection on $N$ gives us an isomorphim again by the universal property : $(\pi_N^*TN) \oplus \epsilon^1_{M \times N} \cong TN \oplus \epsilon^1_N$. By the exact same reasoning combined by the fact that $\eta \oplus \epsilon_{M \times N}^1 \cong TM$. We finally get :

$$\pi^*_M(TM) \oplus \pi^*_N(TN) \cong \pi^*_M(\eta) \oplus \epsilon_{M \times N}^1 \oplus \pi_N^*TN \cong \pi^*_M(\eta) \oplus \pi^*_N(TN \oplus \epsilon^1_N ) \cong \pi^*_M(\eta) \oplus \pi^*_N(\epsilon^n_N) \cong \pi^*_M(\eta) \oplus \epsilon^n_{N \times M} \cong_{\text{by induction}} \pi^*_M(\eta \oplus \epsilon^2_M) \oplus \epsilon^{n - 2}_{N \times M} \cong \epsilon^{m + n -2}_{M \times N} $$

Which completes the proof. ($\pi_N(\epsilon^n_N) \cong \epsilon^n_{N \times M}$ comes by taking $TN = \epsilon^1_N$)

The universal property of the bundle pullback I refer to is the following : If $f : M_0 \rightarrow M$ and $\xi$ a vector bundle over $M$ then for every vector bundle $\eta$ over $M_0$ that have a morphism over $f$ to $\xi$ there exist an isomorphism $f^*\xi \cong \eta$

Title: Parallelizable product of manifolds


问:
Let $p$ be prime and $a$ an integer not divisible by $p$. Prove that if $a^{2^{n}} \equiv -1 \pmod p$, then $a$ has order $2^{n+1}$ modulo $p$.
Apparently I already proved this but I am having trouble the second time around. It is important to use only basic properties of congruences and orders of elements  and Euler's generalization of Fermat's theorem (as that is the section this set of exercises is based on).

It is clear that $a^{2^{n+1}} \equiv 1 \pmod p$ here by squaring both sides.

I know that $a^{p-1} \equiv 1 \pmod p$ here since $a$ is not divisible by $p$ .

It seems obvious to me that the only solutions modulo $p$ to $x^{2} \equiv 1 \pmod p$ are $x = 1,p-1$.

It seems this would help because it would mean that if $2^{n+1}$ was not the order of $a$ modulo $p$, then there must be an $x < 2^{n}$ such that $a^{x} \equiv -1 \pmod p$. I wasn't sure if there was some argument by descent here either.

I also figured that by the division algorithm we can do a similar argument for $a^{x} \equiv -1$ instead of $a^{x} \equiv 1$ where if $x < 2^{n}$ is the order then if the quotient in the division algorithm equation is even, $x | 2^{n} \implies x$ is a power of two $\implies p|2 \implies p=2$ and then check that the $p=2$ case is also satisfied. But then the argument obviously doesn't work if the quotient in that equation is odd because then we can't move a $2$ out to the exponent to square the entire expression.

If we assume for some $x < 2^{n+1}$ we have that $a^{x} \equiv 1 \pmod p$ where $x$ is the order of $a$ modulo $p$ then we must have that $x | 2^{n+1} \implies x$ is a power of $2 \implies x = 2^{k}$ Then $a^{2^{k}} \equiv a^{2^{n+1}} \pmod p \implies p|a^{2^{k}}(a^{d}-1)$ where $d_{2^{n+1}} < 2^{n+1}$. Since $p|a^{2{k}}$ results in the contradiction $p|1$, we must have $p|a^{d}-1$ so that $a^{d} \equiv 1 \pmod p $ if $d < x$, a contradiction too However, we don't necessarily have that $d<x$.

There must be a simple way but I really can't see it.

答:
Note that if $p=2$ then the claim is not  true. In that case, we'd have $-1\equiv 1 \pmod 2$ so the assumption wouldn't say much.  Indeed, in that case, if $a$ is non-zero, then the order of $a$ is $1$.  So assume that $p$ is odd.

Squaring shows that  $a^{(2^{n+1})}\equiv 1\pmod p$ from which we see that the order of $a$ divides $2^{n+1}$.  Hence it is of the form $2^b$ for some $b≤n+1$.  If it were $<n+1$ it would divide $2^n$, but if it were to divide $2^n$ we'd have $a^{(2^n)}\equiv 1 \pmod p$ contrary to assumption.


问:
Prove that the Guassian curvature of the given surface is zero

Let $\gamma_1,\gamma_2:I\to\mathbb{R}^3$ be a pair of regular space curves with the same domain, $I$. Define $\sigma:I\times(0,1)\to\mathbb{R}^3$ as $\sigma(t,u) = u\cdot\gamma_1(t)+(1-u)\cdot\gamma_2(t)$. Assume that $\gamma_1,\gamma_2$ are chosen such that $\sigma$ is regular. Prove that $K(t,1/2)=0$ for all $t\in I$.

In the above, $K$ denotes the Gaussian curvature. To solve the above problem,
\begin{align}
\sigma_t &= u\gamma_1'+(1-u)\gamma_2'\\
\sigma_u&= \gamma_1-\gamma_2\\
\sigma_t\times\sigma_u &= (u\gamma_1'+(1-u)\gamma_2')\times (\gamma_1-\gamma_2)\\
&= \frac{1}{2}(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)\\
&\text{in this case ($u=\frac {1}{2}$).}\\
\sigma_{uu} &= 0\\
\sigma_{tu} &= \gamma_1'-\gamma_2'
\end{align}

So that we need to show $<\gamma_1'-\gamma_2',(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)> = 0$ (This is because, $K = \frac{eg-f^2}{EG-F^2}$ where $e,g,f$ from second fundamental form and $E,F,G$ form the first fundamental form. Since $\sigma_{uu} = 0, g = 0$ and now, to show $K = 0$ is equivalent to show $f = 0$.)

Here is where I'm stuck. Why is $<\gamma_1'-\gamma_2',(\gamma_1'+\gamma_2')\times (\gamma_1-\gamma_2)> =0$?

I used the notation in 'Differential geometry in curves and surfaces' by Tapp.

答:
I do not believe the result. Think about joining two parallel congruent circles by lines to form a hyperboloid of one sheet. The midpoints of these line segments actually trace out the waist of the hyperboloid, which gives the points of maximum $|K|$. For the record, this surface has negative curvature everywhere.

This surface is particularly fascinating because it is in fact doubly-ruled (through each point there are two lines contained in the surface). Projectively, it's the only such surface other than a plane.


问:
Express $\mathbb{E}(|X|)$ using characteristic function of $X$
Given a continuous random variable $X$ with finite expectation, I would like to show the following statement:
$$\mathbb{E}(|X|) = \frac{1}{\pi} \int_{\mathbb{R}} \frac{1-Re(\varphi (t))}{t^2} dt, \tag{1} $$
where $\varphi (t) := \int_{\mathbb{R}} e^{itx}p(X=x)dx$ is the characteristic function of $X$.

What I have attempted so far is to first express $p(|X|=x)$ by
$$p(|X|=x) = \frac{d}{dx}[F_{|X|}(x) - F_{|X|}(-x)]  = \frac{d}{dx}\int_{\mathbb{R}} \frac{e^{itx} - e^{-itx}}{2\pi it}\varphi(t)dt = \frac{d}{dx}\int_{\mathbb{R}} \frac{\sin(tx)}{\pi t}\varphi(t)dt,$$
where the second equality follows from the inversion theorem.

Then the expectation of $|X|$ can be computed by
$$\mathbb{E}(|X|) = \int_{\mathbb{R^+}} x (\frac{d}{dx}\int_{\mathbb{R}} \frac{\sin(tx)}{\pi t}\varphi(t)dt)dx. \tag{2}$$

However, I am stuck here and cannot find a way to turn this expression $(2)$ into $(1)$. My question is, is there a way to obtain $(1)$ by swtiching the order of integration?

答:
Your approach is faulty. Your are not dealing with discrete random variables so $p(|X|=x)$ has no relevance here.

$\int \frac {1-Re \phi (t)} {t^{2}}dt=\int \frac {1-Re Ee^{itX}} {t^{2}}dt=\int \frac {1- E\cos (t|X|)} {t^{2}}dt$ since $\cos $ is an even function. You can interchange  the integral and the expectation by Tonelli's Theorem so we get $E(\int \frac {1- \cos (t|X|)} {t^{2}}dt)$.  Make the change of variable $y=t|X|$ and use the fact that $\int \frac {1-cos y} {y^{2}}dy=\pi$ to complete the proof.


问:
Gambler's Ruin Problem - Explanation needed for the start of the proof for the absorbtion probability
I've been reading the textbook called Introduction To Probability by Dimitri P. Bertsekas, John N. Tsitsiklis and I have the following question regarding the proof of the absorbtion probability of the Gambler's ruin problem.

Given

Transition probability to next state is p in the case of a win

Transition probability to previous state is 1-p in the case of a loss

$$a_0=0$$
$$a_i = (1-p)(a_{i-1}) + pa_{i+1} \hphantom{......}  i = 1,..,m-1$$
$$a_m = 1$$

To solve the equations, the book starts off by saying:

Let us write the equations for the $a_{i}$ as :

$$(1-p)(a_i -a_{i-1}) = p(a_{i+1} - a_{i}), \hphantom{...............} i=1,....,m-1$$

My question

What is the logic for the equation above?

Body:
$$
\begin{align}
a_i&= (1-p)(a_{i-1}) + pa_{i+1}\\
a_i-pa_{i+1}&=(1-p)(a_{i-1})\\
a_i-pa_{i+1}-(1-p)a_i&=(1-p)(a_{i-1})-(1-p)(a_i)\\
p(a_i-a_{i+1})&=(1-p)(a_{i-1}-a_i)\\
p(a_{i+1}-a_i)&=(1-p)(a_i-a_{i-1})
\end{align}$$


问:
Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$

Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$

This should be fairly straightforward but the proof seems to be alluding me.

I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$.  I recognize this shouldn't be too difficult but perhaps finals have fried my brain.

答:
$\text{Here is a geometric proof for $\sin(\theta) < \theta$. Consider a unit circle as shown in the figure.}$
$\text{We then have that the area of the sector formed by $BC$ as }$
$$\color{red}{\dfrac12 \times OB^2 \times \theta = \dfrac12 \times 1^2 \times \theta = \dfrac{\theta}2}$$ $\text{which is greater than the area of triangle $OBC$, which is }$
$$\color{red}{\dfrac12 \times OB \times OC \times \sin(\theta) = \dfrac12 \times 1^2 \times \sin(\theta)}$$
$\text{We hence get that}$
$$\color{blue}{\dfrac{\sin(\theta)}2 < \dfrac{\theta}2 \implies\sin (\theta) < \theta}$$
$\text{The figure was drawn using GeoGebra on Ubuntu. Lets see if some one can come up with a}$
$\text{geometric/pictorial proof for $\color{green}{\theta - \dfrac{\theta^3}{3!} < \sin(\theta)}$.}$


问:
What Does it Mean for a Splitting to be Natural?
$\require{AMScd}$
I'm reading through the proof of the Universal Coefficient Theorem (for homology) given in Massey's Singular Homology Theory, in which he claims that the split SES
$$0 \to H_n(K) \otimes G \overset{\alpha}{\to} H_n(K \otimes G) \overset{\beta}{\to} \operatorname{Tor}(H_{n-1}(K),G) \to 0$$ is natural with respect to coefficient homomorphisms but not with respect to chain maps. In the proof the author writes (after showing that the sequence splits), "Using this procedure, it is easy to prove that the direct sum decomposition is natural vis-a-vis coefficient homomorphisms." The procedure he refers to is the construction of a left inverse $\pi$ for the map $\alpha$ above, from which it follows that the sequence splits.

What does it mean for a direct sum decomposition to be natural? I've only seen the term used in reference to maps. Can I interpret this to mean that the map $\pi$ is natural in the ordinary sense, that is, the following diagram commutes for any homomorphism $h : G_1 \to G_2$?
$$\begin{CD}
H_n(K \otimes G_1) @>{\pi_1}>> H_n(K) \otimes G_1 \\
@V{h_\sharp}VV @V{1 \otimes h}VV \\
H_n(K \otimes G_2) @>{\pi_2}>> H_n(K) \otimes G_2
\end{CD}$$

My other guess for the meaning of the term is that if we write $H_n(K \otimes G_1) = B_1 \oplus B_2$ and $H_n(K \otimes G_2) = C_1 \oplus C_2$ as determined canonically by the map $\pi$, then $h_\sharp(B_1) \subset C_1$ and $h_\sharp(B_2) \subset C_2$. Are either of these guesses correct? Are they equivalent? What if we proved the sequence split by finding a right inverse $s$ for $\beta$ instead of a left inverse for $\alpha$? Would naturality then take the form of a commutative diagram with $s$?

EDIT: It might be relevant that the maps $\alpha$ and $\beta$ have already been shown to be natural with respect to coefficient homomorphisms.

答:
$\require{AMScd}$

Your interpretation (based on the diagram in your question) is correct. Also your other guess is correct.

For a short exact sequence of abelian groups
$$0 \to A \overset{\alpha}{\to} B \overset{\beta}{\to} C\to 0$$
we have canonical bijections between the following three sets:

The set of "left splitting homomorphisms" $\pi : B \to A$ (characterized by $\pi \circ \alpha = id_A$).

The set of isomorphisms $\phi: B \to A \oplus C$.

The set of "right splitting homomorphisms" $\iota : C \to B$ (characterized by $\beta  \circ \iota = id_C$).

Note that these sets are non-empty precisely when the sequence splits. Let us give a sketch without going into details of proofs.

(1) Given $\pi$, define $\phi_\pi : B \to A \oplus C$ by $\phi_\pi(b) = (\pi(b),\beta(b))$ and show that is an isomorphism.

(2) Given $\iota$, define $\psi_\iota : A \oplus C \to B$ by $\psi_\iota(a,c) = \alpha(a) + \iota(c)$ and show that it is an isomorphism. Then define $\phi_\iota = \psi_\iota^{-1}$.

(3) Given $\phi$, define $\pi_\phi = p_A \circ \phi$, where $p_A :  A \oplus C \to A$ denotes projection.

(4) Given $\phi$, define $\iota_\phi = \phi^{-1}\circ i_C$, where $i_C :  C \to A \oplus C$ is given by $i_C(c) = (0,c)$.

Let $h_\sharp : H_n(K \otimes G_1) \to H_n(K \otimes G_2)$ and $h_* : \operatorname{Tor}(H_{n-1}(K),G_1) \to \operatorname{Tor}(H_{n-1}(K),G_2)$ be the induced homomorphisms. The commutativity of your diagram is equivalent to the commutativity of
$$\begin{CD}
H_n(K \otimes G_1) @>{\phi_1}>> H_n(K) \otimes G_1 \oplus \operatorname{Tor}(H_{n-1}(K),G_1)\\
@V{h_\sharp}VV @V{1 \otimes h \oplus h_*}VV \\
H_n(K \otimes G_2) @>{\phi_2}>> H_n(K) \otimes G_2 \oplus \operatorname{Tor}(H_{n-1}(K),G_2)
\end{CD}$$
In fact, if your diagram commutes, then
$$((1 \otimes h) \oplus h_*)(\phi_1(b)) = ((1 \otimes h) \oplus h_*)(\pi_1(b),\beta_1(b)) = ((1 \otimes h)(\pi_1(b)), h_*(\beta_1(b))) \\ = (\pi_2( h_\sharp(b)),\beta_2(h_\sharp(b))) = \phi_2(h_\sharp(b)) .$$
Thus the above diagram commutes. Conversely, if the above diagram commutes, then
$$(1 \otimes h)(\pi_1(b)) = (1 \otimes h)((p_1 \circ \phi_1)(b)) = p_2((1 \otimes h) \oplus h_*)(\phi_1(b)) = p_2 \phi_2(h_\sharp(b)) \\ = \pi_2(h_\sharp(b)) .$$
Thus your diagram commutes.

For the diagram with the right splitting homomorphisms it can be shown similarly that its commutavitity is equivalent to the commutativity of the above diagram.


问:
probability when taking candies out of a jar
there are 300 candies in a jar: 150 red, 100 green, 50 yellow. 10 candies are taken out, randomly, one by one without returning them to the jar.

I need to calculate the probability that:

The first yellow candy in taken out at the 5th pick

if the first yellow was taken out 5th, then the 6th pick is also yellow

if out of the 10 candies exactly 2 are yellow, then exactly 4 candies are red

if the first and second are red, then exactly 6 are red

(1) let $Y_i$ be the event that the first yellow was chosen at the pick no. i, so the probability that the first 4 candies weren't yellow is $$P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c) = (1-\frac{50}{300})(1-\frac{50}{299})(1-\frac{50}{298})(1-\frac{50}{297})\approx 0.48$$ and
$$P\{first\space yellow\space was\space 5th\}=P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)=P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c)P(Y_5) = 0.48 * \frac{50}{296} \approx 0.08$$
(2) let E be the event that yellow was picked 6th. $$P\{6th\space pick\space is\space yellow|first\space yellow\space was\space 5th\}=\frac{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5\cap E)}{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)}=\frac{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)\cdot \frac{49}{295}}{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)} = \frac{49}{295} \approx 0.166 $$

but I'm struggling with (3) and (4).

for (3) I know I can look at $$P\{exactly\space 4\space reds | exactly\space 2\space yellows\} = \frac{P(\{exactly\space 4\space reds\}\cap \{exactly\space 2\space yellows\})}{P\{exactly\space 2\space yellows\}}=\frac{P\{exaclt\space 4\space greens\}}{P\{exactly\space 2\space yellows\}}$$ but I don't know how I can calculate those probabilities without a ton of manual labor as the probability for every pick depends on the order of each of the picks and the color of the candies picked. Same for (4), calculating the probability of choosing exactly 4 reds out of the remaining 8 became super complex. would falling back to "basic" combinatorics simplify this?

答:
Probability the first yellow candy is taken out at the fifth pick.

Your answer is correct.  Here is another method.  If the first yellow candy is selected fifth, then the first four candies must be selected from the $300 - 50 = 250$ candies which are not yellow, then a yellow must be selected from the $296$ candies which remain.
$$\Pr(\text{fifth pick is first yellow}) = \frac{\dbinom{250}{4}}{\dbinom{300}{4}} \cdot \frac{50}{296}$$

Probability that the sixth candy is yellow given that the first yellow was selected with the fifth pick.

Your answer is again correct.

Probability that if exactly two of the first ten candies are yellow, then exactly four candies are red.

For exactly two of the first ten candies to be yellow, eight of the other $250$ candies must be selected.  For exactly two of the first ten candies to be yellow and exactly four of the first ten candies to be red, the other four candies must be green.  If we let $R$ be the event that exactly four of the first ten candies are red and $Y$ be the even that exactly two of the first ten candies are yellow, then
$$\Pr(R \mid Y) = \frac{\dbinom{150}{4}\dbinom{100}{4}\dbinom{50}{2}}{\dbinom{50}{2}\dbinom{250}{8}}$$

Probability that the first and second candies are red, then exactly six of the first ten candies are red.

For this to occur, we must first pick two red candies and then select four red candies among the next eight candies.  If the first two candies are red, we are left with $148$ red candies, $100$ green candies, and $50$ yellow candies.  Let $R$ be the event that the first two candies are red; let $E$ is the event that the first two candies are red and exactly six of the first ten candies are red. Given the first two candies are red, event $R$ occurs if we first pick any eight of the remaining $148 + 100 + 50 = 298$ candies and event $E$ occurs if we select exactly four of the remaining $148$ red candies and exactly four of the other $100 + 50 = 150$ candies with the remaining eight selections.
$$\Pr(E \mid R) = \frac{\dbinom{148}{4}\dbinom{150}{4}}{\dbinom{298}{8}} $$


问:
Spring Calculation - find mass

A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

My work:

The restoring force is $-kx$.  Then

$$7.5 = -k(2.5) \\
-\frac{7.5}{2.5} = k \\
ma = -\frac{7.5x}{2.5} \\
my’’ + 9y’ + -3y = 0,\quad y(0) = 2.5, y(5) = 0 \\
\frac{-9 \pm \sqrt{81 + 4(m)(3)}}{2m} \\
-\frac{9}{2m} \pm \frac{\sqrt{81+12m}}{2m} \\
y = Ae^{-(9/2)x}\cos\left(\frac{\sqrt{81+12m}}{2m}x\right) + Be^{-(9/2)x}\sin\left(\frac{\sqrt{81+12m}}{2m}x\right) \\
2.5 = A + B\cdot 0 \\
0 = (2.5)e^{-45/2}\cos\left(\sqrt{81+12m}\frac{5}{2m}\right) + Be^{-45/2}\sin\left(\sqrt{81+12m}\frac{5}{2m}\right)$$

Any help would be appreciated

答:
Critical damping occurs when $(\gamma^2)-4mk=0$

Therefore,

$$(9^2)-4m(7.5/2.5)=0$$

$$81=12m$$

$$m=6.75 kg$$

Bob Shannon also got the same answer, but I wanted to present it this way because I thought it might be more clear to some.

Title: Spring Calculation - find mass


问:
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgusting, I was mixing up in calculations and couldn't finally get answer.

So, Is there any better way to find such a polinomial (divisor of this one with degree 2)? It would be awesome if proposed way would not require super advanced knowledge. Thanks in advance!

答:
No tricks: Let $f=x^{11}+x+1$ and suppose $g,h\in\Bbb{Z}[x]$ are such that $f=gh$ and $\deg g=2$. Then
$$g=ax^2+bx+c,$$
for some integers $a,b,c\in\Bbb{Z}$. Then $f=gh$ shows that $a$ divides the leading term of $f$, and $c$ divides the constant term of $f$, so $a=\pm1$ and $c=\pm1$. Without loss of generality $a=1$, and so
$$g=x^2+bx\pm1.$$
Now if $\alpha$ is a root of $g$ then $g(\alpha)=0$ and $f(\alpha)=0$, meaning that
$$\alpha^2=-b\alpha\mp1\qquad\text{ and }\qquad\alpha^{11}+\alpha+1=0.$$
and from here it's a cumbersome but straightforward bit of algebra to determine $b$.


问:
Integral of a binomial-like function through Integration by parts
I wonder how to calculate the following through integration by parts:
\begin{align*}
\int_{0}^{1}\binom{n}{i}u^{i}(1-u)^{n-i}\, du
\end{align*}
$i=0,1,2,\cdots,n$. I suppose Gamma function might be useful here?

答:
Indeed integration by parts is quite straightforward: Writing $I_i$ for the given integral and setting $U = u^i$ with $dU = i u^{i-1} du$ and $dV = (1-u)^{n-i} du$ with $V = -\frac{1}{n-i+1}(1-u)^{n-i+1}$ (where we assume $i\le n$), we have
$$
I_i = \binom{n}{i} \left[ \left.(\dots) u^i (1-u)^{n-i+1}\right|_0^1 + \frac{i}{n-i+1} \int_0^1 u^{i-1}(1-u)^{n-(i-1)} du \right] = I_{i-1}
$$
The first term is zero at both boundaries for $1 \le i \le n$, and the second term gives $I_{i-1}$, using
$$
\binom{n}{i}\cdot\frac{i}{n-i+1} = \frac{n!}{i!(n-i)!}\cdot\frac{i}{n-(i-1)} = \frac{n!}{(i-1)!(n-(i-1))!} = \binom{n}{i-1}
$$


问:
Solutions for a diophantine system equations
Does there exist any integer solution for the diophantine system of equations below:

$\left\{\begin{array}{l}
x^2=a^2+b^2-ab
\\
y^2=a^2+b^2+ab
\end{array} \right. ,$

Notice: I know that the two equations above are two similar ellipses with $x=y$ axis that transform to eachother by $\frac {\pi}{2}$ rotation angle, but unfortunately I'm not familiar enough with algebric geometry to solve this system.

Here for example I've find a solution for the first equation for $(x,a,b)=(7,8,3)$ by changing it to a simpler form $m(m+k)=n(n+2k)$ where $(x,a,b)=(n+k,m+k,m)$, but this solution doesn't hold in the second equation.

Now it is clear that we are searching for integer solutions of the intersection points of these two ellipses.

答:
$$ \left( x^2 + xy + y^2 \right) \left( x^2 - xy + y^2 \right) =  x^4 + x^2 y^2 + y^4 $$

Page 19 In Mordell's book Diophantine Equations includes a proof that the only solutions  for
$$ x^4 + x^2 y^2 + y^4 = z^2 \; , \; \gcd(x,y) = 1 $$
integers have either $x=0$ or $y=0$  and the other $\pm 1.$ This is formula (7'), given as an example of (7), that being

$$ x^4 + kx^2 y^2 + y^4 = z^2 \; , \; \gcd(x,y) = 1 $$
Chapter 4 is called Quartic Equations with only Trivial Solutions


问:
On showing that a certain level-set defines a regular surface

Let $f: \Bbb R^4 \to \Bbb R, f(x) =2x_1e^{x_2}+2x_2e^{x_3}+2x_3e^{x_4}$. Show that the level-set $S=f^{-1}(2)$ defines a regular surface.

From Wikipedia, I found this definition for regular surfaces.

Computing the gradient I have that $\nabla f(x)=(2e^{x_2}, 2x_1e^{x_2} + 2e^{x_3}, 2x_2e^{x_3}+ 2e^{x_4}, 2x_3e^{x_4})$. It's clear that $f$ is smooth. Now since $\nabla f \ne 0$ I get that by the by the implicit function theorem there exists an smooth function $g : U \to \Bbb R^3$ such that $\varphi(x_2,x_3,x_3) = (g(x_2,x_3,x_4), x_2,x_3,x_3)$.

Now according to the Wikipedia link, it seems that I would need to show that the rank of this parametrization $\varphi$ would be $3$. I'm a bit stuck here, how should I approach this?

答:
We can avoid a messy computation in the application of the implicit function by using the inverse function theorem instead. Consider $f(x,y,z,w) =2xe^{y}+2ye^{z}+2ze^{w}-2.$

The level set we want to analyze is $S=\{(x,y,z,w):f(x,y,z,w)=0\}.$ Now,

$\partial_xf(x,y,z,w)=2e^y;\ \partial_yf(x,y,z,w)=2xe^y+2e^z$

$\partial_zf(x,y,z,w)=2ye^z+2e^w;\ \partial_wf(x,y,z,w)=2ze^w.$

There are no critical points, since $e^y\neq 0$ so every point $p$ on the level set is regular.

Now take the map $(x,y,z,w)\mapsto (f(x,y,z,w),y,z,w)$ and compute its Jacobian:

$\mathcal J=\begin{pmatrix}
2e^y &2xe^y+2e^z  &2ye^z+2e^w  &2ze^w \\
 0&1  &0  &0 \\
0 &0  &1  &0 \\
0 &0  &0  &1
\end{pmatrix}.$

It's easy to see that this is never zero so NOW we apply the inverse function theorem to get an open neighborhood $p\in U_p$ such that $(U_p,(f,y,z,w))$ is a chart in $\mathbb R^4$ with the property that on $U_p\cap S,$ the first coordinate vanishes. And this, by definition, makes $S$ a regular level set.


问:
Godsil & Royle, Theorem 9.5.1: Extension for digraphs?
I was wondering if there is an extension to digraphs for Theorem 9.5.1 in Godsil & Royle's Algebraic Graph Theory.

The Theorem can also be found in Willem Haemers paper Interlacing Eigenvalues and Graphs, http://arno.uvt.nl/show.cgi?fid=2980 , Theorem 2.1

Any leads would be appreciated.

答:
The fundamental difficulty in extending interlacing to directed graphs is that "interlacing" has no meaning for complex numbers, we cannot say that an eigenvalue of a vertex-deleted subgraph $G\setminus u$ of $G$ lies 'between' two eigenvalues of $G$. So the short answer to the question is "no".

If $C$ is a directed cycle on $n$ vertices, then its characteristic polynomial is $t^n-1$ and its zeros are simple and distinct and are regularly distributed on the unit circle. If we delete vertex we get a directed path with characteristic polynomial $t^{n-1}$. This example shows that many consequences of interlacing in the real case cannot be extended in any obvious way.

For graphs, interlacing is equivalent to the fact that if $u\in V(G)$ then the poles of the rational function $\phi(G\setminus u,t)\phi(G,t)$ are simple, and the residues at the poles are positive. (Here $\phi$ denotes the characteristic polynomial.) For directed graphs, if $\phi(G\setminus u,t)\phi(G,t)$ has these properties then every zero of $\phi(G\setminus u,t)$ lies in the convex hull of
the zeros of $\phi(G,t)$, and this can be viewed as an analog of interlacing. (This a form of the Gauss-Lucas theorem.) Unfortunately these properties do not hold in general, although I do not have an example at hand.

In numerical analysis, there is something of a principle that if a matrix is not normal, the relation between its properties and its eigenvalue is much weaker than we expect, based on our experience with symmetric matrices. (Google on Trefethen and pseudospectra.)


问:
What class of Lie groups embed in a general linear group?
It’s a non trivial fact that every compact Lie groups embeds as a subgroup of SO(n), what larger class of Lie groups can one consider that necessarily embeds in GL(n)?
Possibly locally compact or finitely generated+ some condition. I know residual finite + finitely generated is insufficient.</p>

答:
Here is the answer for finitely generated groups (the proof is rather nontrivial):

Definition. A group $\Gamma$ is said to  have a $p$-congruence structure with the bound $c$ if $\Gamma$ has a descending chain of finite index normal subgroups
$$
\Gamma=N_0\supset N_1 \supset N_2 \supset ...
$$
such that

(1) $\bigcap_i N_i=\{1\}$.

(2) $N_1/N_i$ is a finite $p$-group for each $i>1$;

(3) Each quotient group $N_i/N_j$ ($1\le i<j$) is generated by $c$ elements.

Note that part (1) is equivalent to the residual finiteness of $\Gamma$.

Theorem. Let Γ be a finitely generated group. Then the following conditions are equivalent:

(A) Γ is isomorphic to a subgroup of the group $GL_n({\mathbb C})$ for some $n$.

(B) Γ has a $p$-congruence structure with some bound $c$ for some prime $p$.

See:

Alexander Lubotzky, A group theoretic characterization of linear groups, J. Algebra 113, No. 1, 207-214 (1988). ZBL0647.20045.


问:
A hot body cools so that every 24 min its temperature decreases by a factor of two. Deduce the time-constant and determine the time it will

A hot body cools so that every 24 min its temperature decreases by a factor of two. Deduce the time-constant and determine the time it will take the body to reach 1% of its original temperature.

Hint: The temperature function is $T(t)=T_0e^{-t/T}$ and T is the time constant.

What I did is:
$T = 24$

$\frac{T_0}{100}=T_0e^{-t/24}$

$\frac{1}{100}=e^{-t/24}$

$\ln{\frac{1}{100}}=\frac{-t}{24}$

$4.605=\frac{t}{24}$
$t=110.52$

However, the answer says:

What did I do wrong?

Body:
We have:

$$T(24) = \frac{T(0)}{2}$$

$$T_{0}e^{-\frac{24}{\tau}} = \frac{T_{0}}{2}$$

$$e^{-\frac{24}{\tau}} = \frac{1}{2}$$

$$-\frac{24}{\tau} = \ln(\frac{1}{2}) = -\ln(2)$$

$$\tau = \frac{24}{\ln(2)}\approx\boxed{34.625\text{ min}}$$

Now, we want to solve for $t$:

$$T(t) = \frac{T(0)}{100}$$

$$T_{0}e^{-\frac{t}{\tau}} = \frac{T_{0}}{100}$$

$$e^{-\frac{t}{\tau}} = \frac{1}{100}$$

$$-\frac{t}{\tau} = \ln(\frac{1}{100}) = -\ln(100)$$

$$t = \tau\ln(100) = \frac{24\ln(100)}{\ln(2)} \approx\boxed{159.453\text{ min}}$$


问:
Classic exercise of $L^p$ spaces but using uniform boundedness principle
I have the following exercise:
Let $f$ a real-valued Lebesgue measurable function on $[0,1 ]$, if $fg$ is Lebesgue integrable on $[0,1]$ for any function $g\in L^p([0,1],\mathbb{R})$ with $p \in [1,\infty) $, then $f\in L^{\frac{p}{p-1}}([0,1],\mathbb{R})$.

This is an exercise I remember from my Lebesgue Measure course, I think we proved the results for simple functions and then extend the result to $f$. But here I need to use the uniform boundedness principle, so I think I somehow need a sequence of operators, but I really don't know how to do this.

Any hints appreciated!

答: Hints: Define $T_n(f)=\int fg\chi_{\{|f| \leq n\}}$. You have to verify that $\|T_n\|=(\int |f|^{q} \chi_{\{|f| \leq n\}})^{1/q}$ where $q =\frac p {p-1}$. For this choose $g =sgn(f)|f|^{1/(p-1)}$. Once you get $\|T_n\|$ the result follows immediately from Uniform Boundedness Principle.


问:
Show that the monotonicity in the alternating series test is necessary.
We need to show that monotonicity in the alternating series test is necessary.  I know that it is otherwise, examples like the following:

Monotonicity in alternating Series

could be constructed. But how can we rigorously prove this?

答:
To prove monotonicity is necessary, you simply need a counterexample, i.e. a divergent alternating series which has $|a_n|\to 0$, but not monotonically. Try:
$$\frac{1}{\sqrt 2-1}-\frac{1}{\sqrt 2+1}+\frac{1}{\sqrt 3-1}-\frac{1}{\sqrt 3+1}\ldots$$


问:
Is this Cantor-like set closed?
Let $A$ be the set of real numbers that can be written in the form $\sum_{j \in J} 1/5^j$ with $J \subseteq \mathbb{N}.$ Clearly, $A$ is the set of real numbers in $[0, 1)$ whose canonical base-5 representation only contains the digits 0 and 1.

Thus, $A$ can be constructed by dividing $[0, 1)$ into 5 equal subintervals, removing the last 3 of them $[2/5, 1)$ and repeating the process over and over again with each of the remaining subintervals.

This proves that $A$ is a countable union of finite unions of intervals of the form $[a, b)$, which does not allow us to conclude that $A$ is closed or open or neither.

My intuition is that $A$ is closed. Any help proving or disproving it would be appreciated.

答:
Let $\{0,1\}$ have the discrete topology. Since it is compact, the product $P = \prod_{j = 1}^\infty X_n$ with $X_n = \{0,1\}$ is also compact. Define
$$f : P \to \mathbb R, f((x_j)) = \sum_{j=1}^\infty x_j/5^j .$$
This map is continuous. To see this, let $\epsilon > 0$. For sufficiently large $N$ we have $\sum_{j=N+1}^\infty 1/5^j < \epsilon$. Now consider $x = (x_j) \in P$. The set $U = \{x_1\} \times \ldots \times \{x_N\} \times \prod_{j = N + 1}^\infty X_n$ is an open neigborhood of $x$ in $P$. For $y \in U$ we have
$$\lvert y - x \rvert = \lvert \sum_{j=N+1}^\infty (y_n -x_n)/5^j \rvert \le \sum_{j=N+1}^\infty 1/5^j  < \epsilon .$$
We conclude that $f(P) = A$ is compact, hence closed in $\mathbb R$.

Title: Is this Cantor-like set closed?


问:
Given a percentage of people who have a disease, what's the likelihood of catching it by meeting random people?
As you might guess, this is more than a theoretical question at the moment.

I think that calculating this given that $x$% of people have the disease and someone interacts with $y$ people in a day, and if interacting with someone who has the disease, there's a $z$% change of transmission, then there's probably an equation that will tell me the likelihood of a person catching the disease on a given day.

答:
For each person you meet, if the probability is $\frac{x}{100}$ that they have the disease and $\frac{z}{100}$ that they transmit it given that they have it, the probability is $\frac{xz}{10000}$ that they have the disease and will transmit it.

So you have a $1 - \frac{xz}{10000}$ chance of avoiding transmission from a single contact. This means a $(1 - \frac{xz}{10000})^y$ chance of avoiding transmission from $y$ contacts. The probability of catching the disease is the complement of this probability.