问:
finding the limit of $\lim_{x \to 0}\frac{x-\tan(x)}{x-\sin(x)}$
I'm calculating the limit for the following equation, and would like some feedback on my solution. The answer is $-2$ although, I cannot seem to derive this whilst using l'hopitals rule.

$\lim_{x \to 0}\frac{x-\tan(x)}{x-\sin(x)}$

my working out:

$$\lim_{x \to 0}\frac{\frac{d}{dx}(x-\tan(x))}{\frac{d}{dx}(x-\sin(x))}=\frac{1-\sec^2(x)}{1-\cos(x)} = \frac{\tan^2(x)(1+\cos(x))}{1-\cos^2(x)}=0$$

答:
Hint:

The Maclaurin series for tan(x) is:

\begin{equation*}
\tan x=x+\frac{x^{3}}{3} +\frac{2x^{5}}{15} +...
\end{equation*}
And that for sin(x) is
\begin{gather*}
\sin x=x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} +...\\
\end{gather*}
Can you use this data to get to the answer?


问:
The axiom of choice for a category
I am currently studying the counterpart of axiom of choice in ETCS which is the axiom that every surjective function has a right inverse.

In category of sets the surjective functions are epimorphsims however, in general categories not every epimorphism has a right inverse. An example of this is the category of partially ordered set. Every arrow in this category is an epimorphism and non identity arrows in this category does not have a right inverse.

I recently saw in an article that we can reformulate the axiom of choice in any category as the following statement:

$\bf{The\ Axiom\ of\ Choice\ for\ a\ Category}$: every epimorphism is split.

The definition i am using for a split epimorphism is as follow:

A ${split\ epimorphism}$ is a morphism $f: X \rightarrow Y$ where there exists a morphism $g : Y \rightarrow X$ such that the composite of $f$ and $g$ is the identity morphism.

But how can the axiom of choice for a category be true when we can prove that not every epimorphism had a right inverse?

答:
All that means is that the Axiom of Choice is not true for every category. It is true for $\bf Set$ and $\bf Vect_{\Bbb F}$, assuming that we work in $\sf ZFC$, but is provably not true for $\bf Ab,Top$, etc.

The key point is that you want the splitting to occur internally. So whereas $\Bbb Q$ maps onto $\Bbb{Q/Z}$, as groups, no splitting of that epimorphism is a group homomorphism: every element of $\Bbb{Q/Z}$ has a finite order, whereas no element of $\Bbb Q$ has a finite order (except $0$).

Therefore, "The Axiom of Choice for $\bf Ab$" is false; but this has no bearing on whether or not a function in $\bf Set$ exists that splits that epimorphism. If we're only interested in a certain type of structure (i.e. a category), then whatever happens outside that structure is irrelevant.

ContentLicense: CC BY-SA 4.0

Title: The axiom of choice for a category


问:
Irreducible Polynomials and ring of integers
I came across an exercise in a book which I do not how to solve.
Let $f(x) = x^5 -80x -2.$ Let $\alpha$ be a zero of this polynomial. Furthermore let $K = \mathbb{Q}(\alpha)$ be a number field. These are the questions that are asked:

1- Is this polynomial irreducible and why?

2- Find an integer $n$ such that $f(n) = 1.$

3- Show that $\alpha - n \in \mathcal O_K$

For part 1, For prime $p = 2$ we observe that $p$ does not divide the coefficient of the leading term but divides the other coefficients namely $-80, -2.$ Also $p^2 = 4$ does not divide $2.$ Therefore $f(x)$ by Eisenstein Irreducibility Criterion, is irreducible. Is this correct?

I am formally trained as a mechanical engineer and during this lockdown I am teching myself algebraic number theory. Please kindly provide as much details as possible as I am a bit clueless. Many thanks.

答:
Your solution to 1) is correct.  For 2), I can promise you that you will solve it with a few minutes of experimentation.  Try things!

For 3), I think you have omitted the important piece of information that $\alpha$ is a root of $f$!  Furthermore, the question as stated is trivial: $\alpha$ is the root of a monic integer polynomial, so it's an algebraic integer, and a regular integer $n$ is certainly an algebraic integer, and your book should have a proof that the sum of two algebraic integers is an algebraic integer.

I think the original problem was actually to show that $\alpha - n \in\mathcal{O}_K^\times$, i.e. that $\alpha-n$ is an algebraic integer whose inverse is also an algebraic integer.

My hint for showing this is to consider the following ratio: $$\frac{f(\alpha)-f(n)}{\alpha-n}$$

You will need the following fact, which holds for any integer polynomial $f$: $$\frac{f(X)-f(Y)}{X-Y}\in \mathbb{Z}[X,Y]$$

(to see this, look at each term of $f$)


问:
Proof Explanation - $\Bbb R$ is connected, i.e. has no non-trivial clopen subsets
I know that $\Bbb R$ is connected iff it has no non-trivial clopen subsets. To show that $\Bbb R$ has no non-trivial clopen subsets, I came across the following solution which I need help in understanding.

Assume for a contradiction that $A$ is a clopen subset of $\mathbb R,\ $ $\emptyset\ne A\ne\mathbb R.$ Choose $a\in A$ and $b\in\mathbb R\setminus A.$ Without loss of generality, we may assume that $a\lt b.$ Let $c=\sup(A\cap[a,b]).$ Since $A$ is closed we have $c\in A,$ and so $a\le c\lt b.$ Now $c\in A$ and $(c,b]\subseteq\mathbb R\setminus A,$ contradicting the assumption that $A$ is open.

(Cited from this answer.)

I have some questions:

What motivated the choice $c=\sup(A\cap[a,b])$?

Since $A$ is closed, we have $c\in A$ - why is that? (Once $c\in A$, we know that $c\neq b$ and in fact $c < b$. Since $c$ is $\sup$ of $A\cap[a,b]$, $c \ge a$ is also clear).

Why is $(c,b]\subseteq\mathbb R\setminus A$, and how does it contradict the assumption that $A$ is open?

Thanks a lot!

答:
What motivated the choice $c=\sup(A \cap [a,b])$?

The proof basically chooses a point in $A$ ($a$) and a point outside it ($b$). The goal of the proof here is to look at what happens as you go from $a$ to $b$, because you have to "get out" of the set $A$ in some sense. The contradiction arises from the fact that the "edge" of the set cannot be both open and closed. (Obviously this explanation is not meant to be rigorous, but I think it motivates the choice nicely)

Since $A$ is closed, we have $c \in A$

Because closed sets contain all of their boundary points, and the set $A \cap [a,b]$ is closed (as it is the intersection of two closed sets), and its supremum is its boundary point

Why is $(c,b] \subseteq \mathbb R \setminus A$

Because $c$ was the defined as the supremum of $[a,b] \cap A$, so if $(c, b]$ contained more points from $A$, then $c$ wouldn't be the supremum

and how does it contradict the assumption that $A$ is open?

We saw that $c$ is in $A$, and $A$ is assumed to be open, meaning that every point in $A$ (including $c$) has an open ball around it that is still within $A$. But you can see that there can be no such ball, because for any $r > 0$, $c + r$ is already not in the set $A$


问:
Center of mass of points in log scale chart
I would like to know what is the way to find the center of mass of some scattered points in a 2D plane where scales are logarithmic?

Is it safe to first convert them to normal scale and then compute X_mean and Y_mean and then convert that to log scale?

答:
In principle, you don't "convert" the coordinates to normal scale, you just read the values using the tick marks, and they are already what you need.

$$\frac{1+2+100}3,\frac{1+10+100}3.$$


问:
Evaluation of a limite
How to evaluate the following limit

$$
\lim_{t\rightarrow \infty} \beta \, t^\alpha \sum_{k=0}^{+\infty}\dfrac{(-a\,t^\alpha)^k}{\Gamma(\alpha\,k+\alpha+1)}
$$
with $\alpha\in]0,\,1[$, $a>0$ and $\beta\in\Bbb{R}\backslash\{0\}$.

答:
Your function can be expressed in terms of the Mittag–Leffler function as follows:
$$
\beta t^\alpha  E_{\alpha ,\alpha  + 1} ( - at^\alpha  ).
$$
Using the known asymptotics of the Mittag–Leffler function along the negative axis (see, e.g., http://dx.doi.org/10.1080/10652469.2017.1399373) yields
$$
\beta t^\alpha  E_{\alpha ,\alpha  + 1} ( - at^\alpha  ) = \frac{\beta }{a} + \mathcal{O}\!\left( {\frac{1}{{t^\alpha  }}} \right), \;\;\; t\to+\infty.
$$
Thus the limit as $t\to+\infty$ is $\beta/a$.

Title: Evaluation of a limite


问:
Characterizing simply connected spaces
A topological space $X$ is simply connected if it is pathwise connected and each closed path $u : I = [0,1] \to X$ is path homotopic to the constant path at $x_0 = u(0) = u(1)$.

Recall that

A closed path $u$ is one with $u(0) = u(1)$. A closed path will also be called a loop and we write $\mathcal L'(X)$ for the set of all loops in $X$. If we want to be more precise, we say $u$ is a loop based at $x_0 \in X$ provided $x_0 = u(0) = u(1)$. By  $\mathcal L'(X,x_0)$ we denote the set of all loops in $X$ which are based at $x_0$. Also note that each constant path is a loop.

Two paths  $u, v : I \to X$ with the same initial point $x_0 = u(0) = v(0)$ and the same terminal point $x_1 = u(1) = v(1)$ are path homotopic, $u \simeq_p v$, if there exists a  homotopy $H : I \times I \to X$ such that $H(t,0) = u(t)$ and $H(t,1) = v(t)$ for all $t$ and $H(0,s) = x_0, H(1,s) = x_1$ for all $s$ (such a homotopy $H$ is stationary on the boundary $\partial I = \{0,1\}$; it is called a path homotopy). It is well-known that $\simeq_p$  is an equivalence relation on the set $\mathcal P(X)$ of all paths in $X$. Clearly $\simeq_p$ restricts to an equivalence relation on $\mathcal L'(X)$ and to an equivalence relation on each $\mathcal L'(X,x_0)$.

The set of equivalence classes $\mathcal L'(X,x_0)/\simeq_p$ is nothing else than the fundamental group $\pi_1(X,x_0)$ with base point $x_0$.

Question: Find a list, as complete as possible, of properties characterizing topological spaces which are simply connected.

Each of these properties can then be taken as an alternative definition of simply connected.

Note that many questions in this forum deal with aspects of simple connectedness. The answer to the present question will hopefully cover most of them.

答:
The concept of a loop plays an essential role to characterize simple connectedness. Let us begin by introducing an equivalence relation for loops $u, v \in \mathcal L'(X)$ which is coarser than path homotopy. Recall that $u \simeq_p v$ requires that $u, v$ are based at the same point $x_0 \in X$ and $H : u \simeq_p v$ is stationary on $\{0,1\}$.

Given loops $u,v \in \mathcal L'(X)$, a loop homotopy is a homotopy $H : I \times I \to X$ from $u$ to $v$ such that for each $s \in I$ the map $H_s : I \to X, H_s(t) = H(t,s)$, is a loop. In that case we say that $u,v$ are loop homotopic,  $u \simeq_l v$. Clearly $\simeq_l$ is an equivalence relation on $\mathcal L'(X)$ and $u \simeq_p v$ implies $u \simeq_l v$.

Let us next give an alternative interpretation of a loop.

Loops can be identified with maps $S^1 \to X$ living on the circle $S^1 = \{ z \in \mathbb C : \lvert z \rvert = 1 \}$. In fact, let $\mathcal L(X)$ denote the set of all these maps (which is denoted as the free loop space of $X$) and let $\exp : I \to S^1, \exp(t) = e^{2\pi i t} = \cos(2\pi t) + i \sin(2\pi t)$. This is a closed surjection, hence a quotient map. For $t, t' \in I$ we have $\exp(t) = \exp(t')$ if and only if $t = t'$ or $\{t, t'\} = \{0 ,1 \}$. Thus it identifies $0,1 \in I$ to a single point. Therefore $\exp^* : \mathcal L(X) \to \mathcal L'(X), \exp^*(f) = f \circ \exp$, is a bijection. To $S^1$ we give $1$ as a basepoint and say that $f : S^1 \to X$ is based at $x_0 \in X$ if $f(1) = x_0$.

Under this bijection loop homotopy in $\mathcal L'(X)$ corresponds to ordinary homotopy of maps in $\mathcal L(X)$. Path homotopy in $\mathcal L'(X)$ corresponds to basepoint-preserving homotopy in $\mathcal L(X)$, where a basepoint-preserving homotopy is required to be stationary on the basepoint $1 \in S^1$. This is true because the map  $\exp \times id_I : I \times I  \to S^1 \times I$ is a closed surjection and thus a quotient map.

Here are our results.

For a path-connected space $X$ the following are equivalent:

Path form
Circle form

1
Each loop $u : I \to X$ is path homotopic to the constant loop based at $x_0 = u(0) = u(1)$.
Each $f : S^1 \to X$ is basepoint-preserving homotopic to the constant map based at $x_0 = f(1)$.

2
Any two loops $u, v : I \to X$ based at the same $x_0 \in X$ are path homotopic.
Any two $f, g : S^1 \to X$ based at the same $x_0 \in X$ are basepoint-preserving homotopic.

3
Any two paths $u, v : I \to X$ with the same endpoints are path homotopic.

4
Each loop $u : I \to X$ is loop homotopic to a constant loop in $X$.
Each $f : S^1 \to X$ is homotopic to a constant map $S^1 \to X$.

5
Each loop $u : I \to X$ and each constant loop in $X$ are loop homotopic.
Each $f : S^1 \to X$ and each constant map $S^1 \to X$ are homotopic.

6

Each $f : S^1 \to X$ has a continuous extension $F : D^2 \to X$.

For an arbitrary space $X$ simple connectedness is equivalent to each of the following properties:

Path form
Circle form

7
Each loop $u : I \to X$ and each constant loop in $X$ are loop homotopic.
Each $f : S^1 \to X$ and each constant map $S^1 \to X$ are homotopic.

8
Any two loops $u, v : I \to X$ are loop homotopic.
Any two $f, g : S^1 \to X$ are homotopic.

It is clear that path and circle form in the same line are equivalent.

(2) $\Rightarrow$ (1) : Obvious, (1) is a special case of (2).

(1) $\Rightarrow$ (2) : This follows from the fact that $\simeq_p$ is an equivalence relation.

(1) $\Rightarrow$ (4) : Obvious, (4) is a special case of (2).

(4) $\Rightarrow$ (5) : Obvious, in a path connected space any two constant maps $S^1 \to X$ are homotopic.

(5) $\Rightarrow$ (4) : Obvious, pick any constant map $S^1 \to X$.

(4) $\Rightarrow$ (6) : Let $H : S^1 \times I  \to X$ be a homotopy such that $H_0 = f$ and $H_1 = c$, where $c$ is a constant map (with value $x_0 \in X$). The map $p : S^1 \times I \to D^2, p(z,t) = (1-t)z$, is a closed surjection, hence a quotient map. It maps $S^1 \times \{1\}$ to $0 \in D^2$. Therefore $H$ induces a unique function $F : D^2 \to X$ such that $F \circ p = H$. Note that $F(z) = H(z/\lvert z \rvert,1 - \lvert z \rvert)$ for $z \ne 0$ and $H'(0) = x_0$. By the universal property of quotient maps, $F$ is continuous. For $z \in S^1$ we have $F(z) = H(z,0) = f(z)$, thus $F$ is the desired continuous extension of $f$.

(6) $\Rightarrow$ (1) : Let $F : D^2 \to X$ be a continuous extension of $f$. Define $H : S^1 \times I \to X, H(z,t) = F((1-t)z +t)$. This is a well-defined continuous map (note that $\lvert (1-t)z +t \rvert \le (1-t)\lvert z  \rvert + t = 1 - t + t = 1$). We have $H(z,0) = F(z) = f(z), H(z,1) = F(1) = f(1) = x_0$ and $H(1,t) = F(1) = f(1) = x_0$.

(3) $\Rightarrow$ (2) : Obvious, (2) is a special case of (3).

(6) $\Rightarrow$ (3) : Let $x_0 = u(0) = v(0)$ and $x_1 = u(1) = v(1)$. Let $S = I \times \{0,1\} \cup \{0,1\} \times I$ denote the boundary of the square $D = I \times I$. Points of $S$ will be written in the form $(t,s)$. Define
$$\phi : S  \to X, \begin{cases} u(t) & s = 0 \\ v(t) & s = 1 \\ x_0 & t = 0 \\ x_1 & t = 1  \end{cases}$$
Our aim is to find a continuous extension $H : D \to X$ of $\phi$; this is the desired path homotopy from $u$ to $v$. To do so, it suffices to find a homeomorphism $h : D \to D^2$ such that $h(S) = S^1$; then (6) can be applied to find a continuous extension $F : D^2 \to X$ of $f = \phi \circ h^{-1}$. Then clearly $H = F \circ h$ is the desired extension of $\phi$. The existence of such $h$ is well-known. We shall nevertheless give a proof in the end of this answer.

Let us now come to the characterizations (7) and (8) of simple connectedness.

If $X$ is simply connected, then it is path connected and (5) shows that (7) is satisfied. Conversely, let (7) be satisfied and let $a, b \in X$. The constant loops based at $a$ and $b$ are loop homotopic via a homotopy $H: I \times I \to X$. Then $u(s) = H(0,s)$ is a path from $a$ to $b$. This means that $X$ is path connected and satisfies (5), hence it also satisfies (1) and is therefore simply connected.

(7) $\Rightarrow$ (8) : Obvious because loop homotopy is an equivalence relation.

(8) $\Rightarrow$ (7) : Obvious, (7) is a special case of (8).

For the sake of completeness we finally construct a homeomorphism $h : D \to D^2$ such that $h(S) = S^1$.

Let $J = [-1,1]$ and $\alpha : I \to J, \alpha(t) = 2t - 1$. This is a homeomorphism; hence $\beta = \alpha \times \alpha : D \to \bar D = J \times J$ is a homeomorphism such that $\beta(S) = \bar S =  J \times \{-1,1\} \cup \{-1,1\} \times J = $ boundary of $\bar D$. Note that $\bar D = \{ (t,s) \in \mathbb R^2 : \lVert (t,s) \rVert_\infty = \max(\lvert t \rvert, \lvert s \rvert) \le 1 \}$ and $\bar S = \{ (t,s) \in \mathbb R^2 : \lVert (t,s) \rVert_\infty = 1 \}$. Also note that $\lvert t \rvert, \lvert s \rvert \le \sqrt{t^2 + s^2}$, thus $\lVert (t,s) \rVert_\infty \le \lVert (t,s) \rVert = $ Euclidean norm of $(t,s)$. Now define
$$\bar h : \bar D  \to D^2, \bar h(t,s) =   \begin{cases} \dfrac{\lVert (t,s) \rVert_\infty}{\lVert (t,s) \rVert_\phantom{\infty}}(t,s) & (t,s) \ne (0,0) \\ (0,0) & (t,s) \ne (0,0) \end{cases}$$
This map is well-defined because $\left\lVert  \dfrac{\lVert(t,s) \rVert_\infty}{\lVert (t,s) \rVert_\phantom{\infty}}(t,s) \right\rVert = \lVert(t,s) \rVert_\infty \le 1$. It is clearly continuous in all $(t,s) \ne (0,0)$. Continuity in $(0,0)$ follows from $\lVert \bar h(t,s) - \bar h(0,0) \rVert = \left\lVert  \dfrac{\lVert(t,s) \rVert_\infty}{\lVert (t,s) \rVert_\phantom{\infty}}(t,s) \right\rVert = \lVert(t,s) \rVert_\infty \le \lVert(t,s) \rVert$. Similarly
$$h^* : D^2  \to \bar D, h^*(t,s) =   \begin{cases} \dfrac{\lVert (t,s) \rVert_\phantom{\infty}}{\lVert (t,s) \rVert_\infty}(t,s) & (t,s) \ne (0,0) \\ (0,0) & (t,s) \ne (0,0) \end{cases}$$
is well-defined and it has the property $h^* \circ \bar h = id , \bar h \circ h^* = id$. Therefore $h$ is a bijection. But continuous bijections between compact Hausdorff spaces are homeomorphisms. Finally note that $\bar h(\bar S) = S^1$. Then $h = \bar h \circ \beta$ is the desired homeomorphism.

Title: Characterizing simply connected spaces


问:
Indefinite integration concept
What does expression of a function which is indefinitely integrated represents like if we have a function $2x$,

Integrating with respect to $x$(independent variable) we get $x^{2}$ , so what does this $x^{2}$ represents?

答:
Suppose that the function to be integrated is $f(x)$, and the antiderivative exists which is $F(x)$.

We can then say that $F$ is the function whose rate of change is $f$. In symbols,
\begin{gather*}\dfrac{\mathrm{d}}{\mathrm{d}x} F(x)= f(x), \\ \int f(x) \, \mathrm{d}x = F(x) + c\end{gather*}

Title: Indefinite integration concept


问:
Analytical solution for a difficult nonlinear PDE
Is it possible to compute the analytical solution for this nonlinear pde? It doesn't seems to work with Sympy but it doesn't i can do it with it. The point is to prove that the convergeance order of the code below is indeed 2 (since it's using a newton method). Since i didn't manage to get the analytical solution at first, i tried to test a function that verify the boundary conditions (u(1) = 1 and u'(0) = 0) but it diverges. The function was uex = -cos(pi*X)

```<code>close all
clear all
clc
clear;
N=51; L=1;
dx=L/(N-1); X=[0:dx:L];
K0=0.01; sigma=1; beta=300;
I=2:N-1;

% uex = -cos(pi*X)
% Q = beta*4*(-cos(pi*X)-1)-0.01*(pi^2)*cos(pi*X)
delta=0.2;
Q=(X<delta)*beta;

K=inline('t.^2');

Un=ones(1,N); dUn=zeros(1,N); Fn=zeros(1,N);

nitmax=1000; epsilon=1.0e-5;
for it=1:nitmax
    Kn12=0.5*(K(Un(1:N-1))+K(Un(2:N)));
    Fn(I)=K0/dx^2*(Kn12(I-1).*(Un(I-1)-Un(I))+Kn12(I).*(Un(I+1)-Un(I)))-sigma*(Un(I).^4-1.0)+Q(I);
    Fn(1)=K0/dx^2*(Kn12(1)*(Un(2)-Un(1))+Kn12(1)*(Un(2)-Un(1)))-sigma*(Un(1).^4-1.0)+Q(1);
    Fn(N)=0;
    A=[K0/dx^2*(2*Kn12(1))+4*sigma*Un(1)^3,K0/dx^2*(Kn12(I-1)+Kn12(I))+4*sigma*Un(I).^3,1];
    B=[-K0/dx^2*(2*Kn12(1)), -K0/dx^2*(Kn12(I)),0];
    C=[0,-K0/dx^2*(Kn12(I-1)),0];
    J = full(gallery('tridiag', B, [A 0], C));
    J(end-1, 1) = J(end-1, end);
    J(1, end-1) = J(end, end-1);
    J = J(1:end-1, 1:end-1);
    dUn=J\Fn.';
    Err=norm(Fn)/sqrt(N);
    Un=Un+dUn;
    if (Err<epsilon) break; end;
end;
Un(:,1)
plot(Un(:,1))

</code>```

答:
Here is a scheme that might be useful, to reduce the problem
to first order equations.

Take the case where $\kappa(u) = \kappa_0 u^2$
I will write primes for $x$ derivatives since
this is not a PDE. Since
$$
 u^2 u' = \tfrac{1}{3}(u^3)'
$$
your equation takes the form
$$
 c_1(u^3)''+\sigma u^4 = q(x)
$$
where $q$ takes just two values on two subintervals of $[0,1]$, and
for simplicity I won't track all the constants.

Substitute $y = u^3$, and multiply the equation by $y'$ to get
$$
 c_2 y'' y' +\sigma y^{4/3} y' = q(x)y'.
$$
This is, on the two subintervals,
$$
 (c_3y'^2+c_4 y^{7/3})' = (q(x)y)'.
$$
Integrate to get
$$
 c_3y'^2+c_4 y^{7/3} = q(x)y + c_5
$$
on the first subinterval and
$$
 c_3y'^2+c_4 y^{7/3} = q(x)y + c_6
$$
on the second.

These first order equations can possibly be solved numerically
the first to match the boundary condition $u_x(0) = 0$ leaving
$c_5$ as a parameter, the second for $u(1) = 1$, and finally
chosing $c_5$ and $c_6$ to attempt to make $u$ continuous
or maybe smooth at $x = 0.2$.


问:
Problem solving approaches in graph theory
From my experience with problems in graph theory, these pose certain obstacles that to me seem to particular for discrete mathematics, among them are

1) A solution might be obvious at first sight, but extremely hard to rigorously formulate/proof.

2) A problem statement is very hard to grasp, such that I don't even know how to start.

3) I rarely see a measure of "progress towards a solution"; either I get nowhere or I know from the beginning how to solve a problem/what theorem to use

4) Most problems seem to be "individual" in a sense that I can't identify a general concept that might be helpful for a problem that I will face in the future. In particular that means that finding too strong hints or even full solutions does not help at all.
In other fields of math, even copying a well worked out solution might help, since it elucidates in what setting one can properly use a theorem/approach.

Do you have any recommendations of how to figure out conceptual elements in graph theory problem solving, i.e. how to figure out a setting and a general strategy that might be applicable to a certain style of problems?

Are there any good reads in graph theory that point out problem solving strategies/ways of thinking, instead of focusing on results?

答:
As an experienced problem solver I allow myself  to highly recommend to you a book “Mathematical discovery: on understanding, learning and teaching” by George Polya. In this book the author sketches the bases of a general method of mathematical problem solving. In graph theory it works too.

Also I can tell you that when your graph intuition will be developed highly enough, sometimes you will ‘see’ the answers. The checking and the proving of them is a technical task. This is why I regularly starts my answers at MSE with the sentence “It seems the following”. :-)

A couple of words about you partial obstacles.

1) You should be sufficiently skilled in techniques, not only in the intuitive vision.

2) Try to see a problem from different points of view, search familiar elements in the problem. Come up to the problem time after time. Gradually your vision of the problem will become clear.

3) This vision of “a measure of "progress towards a solution"” is a mysterious thing for me. Polya claims that such a feeling exists, but it seems that I does not hove it . Of course, when you have a program or a plan of the solution process (see the respective chapter of “Mathematical discovery” for more details) then after the realization of the phases of the plan you are able to estimate the progress, but this is a quite ideal case.

4) This obstacle is related with the experience. I think that thousands of mathematical problems which I solved or saw the solutions, formed my intuition.

ContentLicense: CC BY-SA 3.0


问:
Proof for Derivative of Dot Product
In Taylor's Classical Mechanics, one of the problems is as follows:

(1.9) If $\vec{r}$ and $\vec{s}$ are vectors that depend on time, prove that the product rule for differentiating products applies to $\vec{r}$ $\cdot$ $\vec{s}$, that is, that:

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$

I'm not totally certain that my solution is correct, so if people could give me a hand in checking my work, I'd really appreciate it!

$\vec{r}$ $\cdot$ $\vec{s} = r_xs_x + r_ys_y$

  $\frac{d}{dt}(r_xs_x+r_ys_y) = \frac{d}{dt}(r_xs_x) + \frac{d}{dt}(r_ys_y)$

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x\frac{d}{dt}s_x + r_y\frac{d}{dt}s_y) + (s_x\frac{d}{dt}r_x + s_y\frac{d}{dt}r_y)$

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x+r_y)\frac{d}{dt}(s_x+s_y)+(s_x+s_y)\frac{d}{dt}(r_x+r_y)$

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$

答:
You're assuming the dot product is $x*y = x_0 y_0 + x_1 y_1 + x_2 y_2 + ... + x_n y_n + ...$

The proof can be extended to any kind of dot product defined over any vector space.

$$\dfrac{d}{dt}(\vec{x}*\vec{y})=\lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)*\vec{y}(t+\Delta t)-\vec{x}*\vec{y}}{\Delta t}=\\ \lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)*\vec{y}(t+\Delta t)-\vec{x}(t+\Delta t)*\vec{y}(t)+\vec{x}(t+\Delta t)*\vec{y}(t)-\vec{x}*\vec{y}}{\Delta t}=\\\lim_{\Delta t\to0}\vec{x}(t+\Delta t)*\dfrac{\vec{y}(t+\Delta t)-\vec{y}(t)}{\Delta t}+\lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)-\vec{x}}{\Delta t}*\vec{y}=\\ \vec{x}(t)*\lim_{\Delta t\to0}\dfrac{\vec{y}(t+\Delta t)-\vec{y}(t)}{\Delta t}+\left(\lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)-\vec{x}}{\Delta t}\right)*\vec{y}=\\ \vec{x}*\dfrac{d\vec{y}}{dt}+\dfrac{d\vec{x}}{dt}*\vec{y}$$

Title: Proof for Derivative of Dot Product


问:
Find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$
I want to find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$. It seems that would simply be to calculate the following triple integral:

$\int_0^3 \int_0^{2x} \int_0^{1-y^2} z\,dz\,dy\,dx$

This is pretty straight-forward to do without any variable substitutions etc. which makes me think it's almost too simple (for a home assignment).

Am I missing something or is the above correct?

答:
It's always a good idea to try and figure out how your shape looks, this helps with setting up the right integral bounds.

Note that calculating the volume using an integral can be written as:

$$ \int_{V}dV = \iiint 1\,dx\,dy\,dz$$

or

$$ \int_{A}f(A)\,dA = \iint f(x,y)\,dx\,dy$$

So using $\iiint z\,dz\,dy\,dx$ is not how you calculated the volume.

The volume bounded by these surfaces could be calculated as:

$$\int_{0}^{1}\int_{y/2}^{3}\int_{0}^{1-y^2} 1\,dz\,dx\,dy = \int_{0}^{1}\int_{y/2}^{3}{1-y^2 \,dx\,dy}   $$

Note that if you want to integrate $x$ at the end, you should split up the integrals as such:

$$\int_{0}^{0.5}{\int_{0}^{2x} 1-y^2}\,dy\,dx + \int_{0.5}^{3}{\int_{0}^{1} 1-y^2}\,dy\,dx$$


问:
Second Order Derivatives in Nested Function
I'm trying to understand Becker's seminal paper Crime and Punishment (1968) particularly the parameter of cost of apprehension and conviction and their second order partial derivatives.

The paper states the costs of various activities to apprehend is given by the function $C = C(A)$ where $C' = \frac{dC}{dA}$ and that activities can be approximate by the clearance rate $(p)$ and number of offences $(O)$ by the formula $A ≅ pO$

Now, partially differentiating once with respect to each $(p)$ and $(O)$ gives the following

$$C_p=\frac{∂C(pO)}{∂p} = C'O \quad \textrm{and;} \quad C_O = C'p \quad \textrm{respectively} \quad$$

Becker then says if $pO ≠ 0$ then an increase in either probability of conviction or number of offenses would increase total costs. This is where my confusion begins, when he discusses the marginal cost of increased activity given by second order differentiation where:

$$C_{pp}= C''O^2;\\
C_{OO}= C''p^2\\
C_{pO}=C_{Op}= C''pO + C'$$

Put simply, why is there an added $C'$ for the cross-partial differentiation results? I went through my calculus books and tried to figure out the chain rule application but to no avail.

答:
This is due to the product rule. Your chain rule
$$
 C_p = C'O
$$
resulted in a product of two functions, $C'$ and $O$. By the product
rule
$$
 C_{pO} = C'\frac{\partial O}{\partial O}+\frac{\partial C'}{\partial O}O
 = C'1+(C''p)O = C'+C''pO.
$$


问:
Forward and backward Kolmogorov equations and stationary distribution
This is a follow up to What is the difference between the forward and backward equations in a CTMC?

Let $Q^T$ be the transpose of $Q$.

Why $\pi Q =0$ is a steady state solution to the continuous time Markov chain, accounting for the forward case, but for the backward we get $ Q^T \pi^T =0^T$? Where does the transpose come from?

For the forward we have  $\pi'(t)=\pi_t Q$, and the backward equation is: $\pi'^T(t)=Q \pi^T_t$.

In both cases, letting $t \rightarrow \infty$ we get $\pi Q =0$ and $Q \pi
^T=0^T$, respectively, no? Where is my mistake?

答:
The backward equation is $P'(t) = Q P_t$.

Then, in steady state, $P'(t) = Q P_t = 0$. Indeed, this equation holds! It is a tautology. Although it is not helpful to find the steady state solution, as all its lines are linearly dependent, it is still true.

The forward equation, in contrast, leads to an alternative system of equations, $\pi Q = 0$, which is helpful to find the steady state solution to the system.


问:
Help locating mini mandelbrots
I would like to be able to list the coordinates of all the first level minibrots.

Here is a picture of the mandelbrot set generated by fraqtive:

zooming in to the circled area we see a slightly distorted copy of the original fractal

the julia set associated to the point right in the middle of the main cardoid looks like this

roughly the coordinates are -0.1564 + i(-1.0320)

The idea to find minibrots is based on this:

The julia set associated to the center of the main cardioid of the mandelbrot set is a circle, so I assume the julia sets associated to the main cardioid of a minibrot will be a julia set built of circles (like pictured): This needs proof but it seems like it would be too difficult so I am happy to leave this unproved for now (but I would like to prove it eventually).

if I can just find a way to recognize which julia sets are built out of circles I should be able to list the minibrots. How can I do that?

a second question which might be more easy is

I have approximate coordinates for this circular julia set, is there any way to use the z^2+c formula to find that point exactly?

答:
The baby-brots are all "centered" on $c$ values with super-attractive orbits.  If $f_c(z)=z^2+c$ and $f_c^n$ denotes the $n^\text{th}$ iterate of $f_c$, then the $c$ values with super-attractive orbits that have a period dividing $n$ are exactly the roots of $f_c^n(0)=0$.  Note that this is a polynomial in $c$.  For example, if $n=3$, then
$$f_c^3(z) = \left(\left(z^2+c\right)^2+c\right)^2+c \: \: \text{ so } \: \: f_c^3(0) = \left(c^2+c\right)^2+c.$$
The roots of this polynomial are approximately $c=0.0$, $c=-1.75488$, and $c=-0.122561 \pm 0.744862i$; these are shown as green dots in the Mandelbrot image below.  The smaller red dots show the other super-attracting parameters for all periods up to period five.

We immediately see two difficulties.  First (and easier), the origin appears.  This corresponds to the algebraic fact that one divides three and to the dynamic fact that a fixed point of $f_c$ is also a fixed point of $f_c^3$.  This is really no big deal; we just find some points more than once.

Trickier, is how to deal with the fact that this technique finds the hyperbolic center of periodic bulbs (i.e. the circles attached to the cardioids), as well as the hyperbolic centers of the baby-brots (i.e. the cardioid centers).  In principle, this can be dealt with as follows: Associated with each of these points is a nearby bifurcation.  The bifurcation points for the bulbs attached to the main cardioid are shown in yellow in the figure.  We can distinguish a bulb from a baby-brot using the nature of this bifurcation.  For the bulbs, these are all so-called period doubling bifurcations; or, really, period-tripling or quadrupling or whatever the case may be.  This simply means that, as we pass through the bifurcation point along the line containing both the bifurcation and the hyperbolic center, that the period of the corresponding attractive orbit increases (or decreases) by a multiplicative factor.  Associated with the baby-brots are so-called saddle-node bifurcations in which the  periodic orbit (or at least its attractive nature) disappears alltogether.  Thus, we can simply check the dynamic behavior for a generic point on this line opposite the hyperbolic center.

Of course, for this to work, we must be able to find the bifurcation point.  This is done by solving the system
$$\begin{align}
f_c^n(z)&=z \\
(f_c^n)'(z)&=1.
\end{align}
$$
This can be done fairly easily using a Newton iteration, since we have good initial guesses for $c$ and $z$.

In the image below, we zoom into the top of the first image.  The most prominent baby-brot corresponds to a super-attractive orbit of period 4.  I've also marked the associated saddle-node bifurcation point using the technique described above.  There are a few other baby-brots in this picture not marked, because the correspond to higher order, super-attracting orbits.

Title: Help locating mini mandelbrots


问:
Find a plane that contains two points and parallel to a line
Find All planes that contain the two points P(2,-1,3), Q(1,0,1) and is parallel to the line x(t)=(2,2,-3)+t(-1,1,-2).

I thought I'd approach it by getting the vector PQ, but PQ is the same vector as the one given in the parallel line. So because the two points and the line are also parallel, this means that the plane can basically only rotate around the two points right? Geometrically, I thought that made sense. I just can't figure out how to get a formula. Help is appreciated.

答:
This solution follows the method of 'user' on https://tinyurl.com/twddfdvx

Line p: (2,-1,3)+t(-1,1,-2) is in the plane

Thus :-a+b-2c = 0 and b=a+2c

Using the equation of the plane: ax+(a+2c)y+cz+d = 0 containing (2,-1,3)

2a+(a+2c)(-1)+c(3)+d=0 which gives d=-a-c

ax+(a+2c)y+cz-a-c = 0

Set a=1, c=k

x+(1+2k)y+kz-1-k=0.....(A)

For mutually perpendicular, dot product is zero:

(-1,1,-2).(1,1+2k,k) gives k=0

On putting k=0 in equation 'A' above gives x+y-1 = 0

This equation satisfies the points in the plane (2,-1,3) and (1,0,1)

The line in the plane and the parallel line are both parallel so cross product calculation can't be used. Is there another method?


问:
Example where the product of two integrable functions is not integrable on $[0,1]$?
I tried searching through the site for examples but to not avail. I was thinking of constructing $f$ and $g$ such that $fg= \frac{1}{x}$, but that proved to be more difficult than I thought. Any thoughts?

答: Assuming that you meant Riemann-integrable functions (and not meaning improper integrals), there is no such pair of functions. Both $f$ and $g$ are bounded  and, for each of them, the set of points at whcih it is discontinuous has Lebesgue measure $0$. So, $fg$ als has those two preperties, and therefore it is Riemann-inegrable.


Title: Example where the product of two integrable functions is not integrable on $[0,1]$?

Body: I tried searching through the site for examples but to not avail. I was thinking of constructing $f$ and $g$ such that $fg= \frac{1}{x}$, but that proved to be more difficult than I thought. Any thoughts?

Body: Take $f(x)=g(x)=\frac{1}{\sqrt{x}}\chi_{(0,1]}(x)$


答:
$$\left(\sum r_is_i\right)'=\sum(r_is_i)'=\sum(r_i's_i+r_is_i')=\sum r_i's_i+\sum r_is_i'$$
$$\mathbf r\cdot \mathbf s=\mathbf{r'}\cdot \mathbf s+ \mathbf r\cdot{\mathbf s'}$$
is enough.

Title: Proof for Derivative of Dot Product

Body:
In Taylor's Classical Mechanics, one of the problems is as follows:

(1.9) If $\vec{r}$ and $\vec{s}$ are vectors that depend on time, prove that the product rule for differentiating products applies to $\vec{r}$ $\cdot$ $\vec{s}$, that is, that:

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$

I'm not totally certain that my solution is correct, so if people could give me a hand in checking my work, I'd really appreciate it!

$\vec{r}$ $\cdot$ $\vec{s} = r_xs_x + r_ys_y$

  $\frac{d}{dt}(r_xs_x+r_ys_y) = \frac{d}{dt}(r_xs_x) + \frac{d}{dt}(r_ys_y)$

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x\frac{d}{dt}s_x + r_y\frac{d}{dt}s_y) + (s_x\frac{d}{dt}r_x + s_y\frac{d}{dt}r_y)$

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x+r_y)\frac{d}{dt}(s_x+s_y)+(s_x+s_y)\frac{d}{dt}(r_x+r_y)$

  $\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$


问:
Why do people use "it is easy to prove"?
Math is not generally what I am doing, but I have to read some literature and articles in dynamic systems and complexity theory. What I noticed is that authors tend to use (quite frequently) the phrase "it is easy to see/prove/verify/..." in the manuscripts. But to me, it is usually not easy at all; maybe because I haven't spent much time in the field, maybe not.

My question is: why do people use the phrase "it is easy" in their proofs?

P.S. I hope this question is not too subjective, and has some value for the community.

答: I know of at least one case where someone did this with the intention of intimidating the referees away from calling them on their claim. "It's easy to prove that..." i.e. "I dare you, referee, to say this isn't true! I will make you look like an idiot!"</p>


问:
With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100$ by using Karamata's inequality

Given $n$ real numbers $a_{1}, a_{2}\cdots a_{n}$ so that
$$a_{1}\geq a_{2}\geq\cdots\geq a_{n}, a_{1}+ a_{2}+ \cdots+ a_{n}= 300, a_{1}^{2}+ a_{2}^{2}+ \cdots+ a_{n}^{2}> 10000$$
With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100.$

I wanna create a contractdiction. So we may assume that
$$a_{1}+ a_{2}+ a_{3}< 100\Rightarrow\sum_{k= 1}^{n}a_{k}^{2}\leq a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+ a_{3}\sum_{k= 4}^{n}a_{k}= a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+$$
$$+ a_{3}\left ( 300- a_{1}- a_{2}- a_{3} \right )\leq 100\left ( a_{1}- a_{3} \right )+ 300a_{3}\leq 100\left ( a_{1}+ a_{2}+ a_{3} \right )= 10000$$
I tried to solve this inequality by using Karamata but unsuccessfully, who can help me with a way ?? Thanks a real lot !

答:
You asked for a proof using Karamata's inequality, so here it is. I'll assume that all $a_j$ are non-negative.

Case 1: $3 \le n \le 9$. Then
$$
 300 = a_1 + \cdots + a_n + \underbrace{0 + \cdots + 0}_{9-n \text{ terms}}
\le 3 (a_1+a_2+a_3)
$$
which implies $a_1+a_2+a_3 \ge 100$.

Case 2: $n > 9$. Assume that  $a_1+a_2+a_3 < 100$. Define
$$
b_1, \ldots, b_n = \underbrace{\frac{100}{3}, \ldots, \frac{100}{3}}_{9 \text{ terms}}, \underbrace{0, \ldots, 0}_{n-9 \text{ terms}} \, .
$$
Then
$$
 100 > a_1+a_2+a_3 \ge 3a_1 \implies b_1 \ge a_1\\
 100 > a_1+a_2+a_3 \ge \frac 32 (a_1+a_2) \implies b_1+b_2 \ge a_1+a_2 \\
 100 > a_1+a_2+a_3 \implies b_1+b_2+b_3 \ge a_1+a_2+a_3 \\
\vdots \\
300 = b_1 + \cdots + b_n = a_1 + \cdots + a_n
$$
which shows that $(b_1, \ldots, b_n)$ majorizes $(a_1, \ldots, a_n)$. Therefore one can apply Karamata's inequality with the convex function $f(x) = x^2$. It follows that
$$
 10\,000 = b_1^2+ \cdots + b_n^2 \ge a_1^2+ \cdots + a_n^2
$$
in contradiction to the assumption.

In the same way one can prove the following generalization:

Let $1 \le k \le n$ be integers and $a_1 \ge a_2 \ge \cdots \ge a_n$ be non-negative real numbers. If
$$
 (a_1 + \ldots + a_n)^2 < k^2 (a_1^2 + \ldots + a_n^2)
$$
then
$$
 a_1 + \ldots + a_k \ge \frac{a_1 + \ldots + a_n}{k} \, .
$$


问:
Is this function locally Lipschitz?
Let $\alpha \in \mathbb{R}$ and $f$ be a real function defined by $f(u)=-ue^{\alpha u}\ln(\lvert u \rvert)$ if $u \neq 0$ and $f(0)=0$. Let $u_0 \in \mathbb{R}^+$ for the following problem:

\begin{equation}
\begin{cases}
u'(t)=f(u(t)), t\in \mathbb{R}\\
u(0)=0
\end{cases}
\end{equation}

I'm trying to see under which conditions $f$ is locally Lipschitz. I tried the mean value theorem without much success.

答:
Let us compute $f'(u)$ for $u > 0$:
$$ f'(u) = -e^{\alpha u}\log(u)-\alpha u e^{\alpha u}\log(u) - e^{\alpha u} = -e^{\alpha u}(\log(u)+\alpha u \log(u) +1) $$

This is continuous on $(0,+\infty)$, so it is locally bounded on $(0,+\infty)$. Hence, $f$ is locally Lipschitz on $(0,+\infty)$.

However,
$$ \lim_{u \to 0⁺} f'(u) = +\infty, $$
hence $f$ is not locally Lipschity on $\mathbb{R}$.

If your initial condition $u_0$ is strictly positive, there is local (in time) existence and uniqueness to the Cauchy problem.

Title: Is this function locally Lipschitz?


问:
Find the suitable vector
Consider $\lbrace \vec{B}_{1},\vec{B}_{2},\ldots,\vec{B}_{n}\rbrace$ be a set of $n>2$ three-dimensional non-unit vectors. My question is can I find a unit vector $\hat{A}$ such that,

$$\hat{A}.\vec{B}_{1}=\hat{A}.\vec{B}_{2}=\hat{A}.\vec{B}_{3}=\hat{A}.\vec{B}_{4}=\ldots=\hat{A}.\vec{B}_{n}$$

For $n=3$, I can easily obtain

$$\hat{A}=\frac{\vec{B}_{1}\times\vec{B}_{2}+\vec{B}_{2}\times\vec{B}_{3}+\vec{B}_{3}\times\vec{B}_{1}}{||\vec{B}_{1}\times\vec{B}_{2}+\vec{B}_{2}\times\vec{B}_{3}+\vec{B}_{3}\times\vec{B}_{1}||}$$

However, what I want is a general construction of $\hat{A}$ for arbitrary $n$.

答:
The requirement that $A$ has length 1 can be replaced by the requirement that $A$ is nonzero, because then if you have found a solution you can always divide by the norm, and it will remain a solution.

Define an $(n-1)\times n$ matrix $M$ with elements $M_{ij}=(B_{i+1}-B_i)_j$, $i=1,2,\ldots n-1$, $j=1,2,\ldots n$. Remove the $j_0$'th column to convert $M$ into a square matrix ${M}^{(j_0)}$ of size $(n-1)\times(n-1)$. Which column you remove does not matter, provided that the determinant of ${M}^{(j_0)}$ is nonzero. The elements of the removed column form a vector $v$ of length $n-1$. Solve the set of $n-1$ linear equations for the $n-1$ unknowns $a_j$,
$$\sum_{j=1}^{n-1}M_{ij}^{(j_0)}a_j=v_i,\;\;i=1,2,\ldots n-1.$$
The solution is given by Cramer's rule,
$$a_i=\frac{\det X^{(i)}}{\det M^{(j_0)}},$$
where the matrix $X^{(i)}$ is obtained by replacing the $i$-th column of $M^{(j_0)}$ by the column vector $v$.

Now the desired vector $A$ has elements $a_1,a_2,\ldots a_{j_0-1},-1,a_{j_0},\ldots a_{n-1}$. Divide by the norm to obtain a unit vector, and you're done.

Title: Find the suitable vector


问:
$P(X > cY)$ where $X,Y$ are gaussian distributed
I believe that as long as $X,Y$ have zero mean, then regardless of what their variance is, and for any constant $c$, $P(X > cY)$ should always be $0.5$. It's easier to visualize if you think about the bivariate distribution of $X$ and $Y$ projected onto the $xy$-plane. It'll be either a perfect circle if the two variables have the same variance, or an ellipse with differing variances. Any line with any  slope that passes through the origin will cut the area of the bivariate pdf in half, hence the probability of 0.5.

Is this correct?

Is this true regardless if $X,Y$ are independent or not?

答:
Ignoring extreme cases such as variances of $0$ or correlation of $\pm1$, it is true that $P(X > cY) =\frac12$ if $X$ and $Y$ have a joint bivariate normal distribution, even if they are correlated.  You get something like this, and by symmetry half the distribution is above the red line.

So any counterexample needs to break the symmetry.  Here is an example using R, where only about $37\%$ of the joint distribution is above the red line, achieved by twisting points on the left hand side below the line without changing the marginal distributions.

```<code>set.seed(2021)
cases <- 10^5
X <- rnorm(cases)
Y <- ifelse(X>=0, X, -qnorm(3/2-pnorm(abs(X))))*24/25 + rnorm(cases)*7/25
mean(Y > 2*X)
# 0.36701

plot(X,Y, pch=".")
abline(h=0); abline(v=0); abline(0,2, col="red")
</code>```

and to check the two distributions are indeed normally distributed, the simulated empirical densities for $X$ (in blue) and $Y$ (in red) are shown over the theoretical density (in black)

```<code>curve(dnorm, from=-4, to=4)
points(density(X), col="blue", type="l")
points(density(Y), col="red" , type="l")
</code>```

Body:
Ignoring extreme cases such as variances of $0$ or correlation of $\pm1$, it is true that $P(X > cY) =\frac12$ if $X$ and $Y$ have a joint bivariate normal distribution, even if they are correlated.  You get something like this, and by symmetry half the distribution is above the red line.

So any counterexample needs to break the symmetry.  Here is an example using R, where only about $37\%$ of the joint distribution is above the red line, achieved by twisting points on the left hand side below the line without changing the marginal distributions.

set.seed(2021)
cases <- 10^5
X <- rnorm(cases)
Y <- ifelse(X>=0, X, -qnorm(3/2-pnorm(abs(X))))*24/25 + rnorm(cases)*7/25
mean(Y > 2*X)
# 0.36701

plot(X,Y, pch=".")
abline(h=0); abline(v=0); abline(0,2, col="red")

and to check the two distributions are indeed normally distributed, the simulated empirical densities for $X$ (in blue) and $Y$ (in red) are shown over the theoretical density (in black)

curve(dnorm, from=-4, to=4)
points(density(X), col="blue", type="l")
points(density(Y), col="red" , type="l")


问:
Convergence in probability of random variables
I want to prove that if $\xi_n \xrightarrow {\mathbb{P}} \xi$ and for some Borel function $\mathit{f}$, where $\mathbb{P}\{\omega \in \Omega|\ \mathit{f} \text{ is discontinuous at  } \xi(\omega)\} = 0$ then $$\mathit{f}(\xi_n)\xrightarrow {\mathbb{P}} \mathit{f}(\xi)$$
I don't know where to start. I would be very grateful for help!

答:
First, note that $X_n\xrightarrow{p}X$ iff every subsequence of $\{X_n\}$ has a sub-subsequence converging to $X$ a.s.

Take a subsequence $\{\xi_{n_k}\}$. We know that $\xi_{n_k}\xrightarrow{p}\xi$ and, thus, there is a subsequence $\{\xi_{n_{k_j}}\}$ converging a.s. to $\xi$. Since  $\{f \text{ is discontinuous at }\xi\}$ is a null set, $f(\xi_{n_{k_j}})\to f(\xi)$ a.s.


问:
Find the limit of $\;\lim\limits_{x\to\infty}\left(\frac{x^2-1}{x^2+1}\right)^\frac{x-1}{x+1}$.
Find limit of
$\;\lim\limits_{x\to\infty}\left(\dfrac{x^2-1}{x^2+1}\right)^\frac{x-1}{x+1}$ without using $L'Hopital$

I tried subtracting $1$ for using $lim$ of $e$ but I got $1^\infty$ form and couldn't continue.

Body:
hint

Use

$$\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}$$

and

$$\lim_{x\to+\infty}\left(1-\frac{2}{1+x^2}\right)^{\frac{x^2+1}{2}}=\frac 1e$$


问:
How many zeros does the function $f(z)=z^3+z^2+4z+1$ have in the first quadrant?

How many zeros does the function $f(z)=z^3+z^2+4z+1$ have in the first quadrant?

Using Rouche's theorem with the function $h(z)=1,g(z)=z^3+z^2+4z$, I've been able to show that all the roots are in $B_3(0)$.

Is it possible to show that they are not in $B_3(0)\cap\{Re(z),Im(z)\ge0\}$ using Rouche's theorem?

答:
The answer by @user is clearly how to tackle this specific question.

But if you are interested in a Rouche/Argument Principle way to tackle such problems let me say how I'd do this one.

$f(z)$ a polynomial, so by the Argument Principle the number of zeros in a region inside a simple contour is $2\pi$ times the increase in the argument of $f(z)$ as we go round the countour.

So consider a contour consisting of (i) the straight line from $0$ to $R$, (ii) a circular arc of radius $R$ from $R$ to $R i$, (iii) the straight line from $Ri$ back to the origin.

Along (i) $f(z)$ is real, and so its argument remains $0$.

Along (ii) we have
$$
\arg f(z)=\arg z^3 +\arg (1+\frac{1}{z}+4\frac{1}{z^2}+\frac{1}{z^3})
$$
which simplifies when $z=Re^{i\theta}$ to
$$
\arg f(z)=3\theta +\arg (1+O(\frac{1}{R})).
$$
Hence as we run round the arc the argument increases by $\frac{3\pi}{2}$.

Along (iii) we have $z=iy$ and so
$$
f(z)=-iy^3 -y^2 +4iy +1
$$
whose argument is
$$
\tan^{-1}Y, \text{  where  }Y=\frac{y(y-2)(y+2)}{(y-1)(y+1)}.
$$

We now need to make a little sketch of $Y$ as $y$ varies -easier on a piece of paper than it is here - and what we see is that at $+\infty$ it starts at $+\infty$, it decrease as $y$ decreases, cutting the axis at $y=2$, runs off to $-\infty$ at $y=1+$, comes back at $+\infty$ at $y=1-$, then decreases to $0$ at $y=0$.

Now we have to account for what the argument of $Y$ does. We got at the end of (ii) to $\frac{3\pi}{2}$, so we start there. When we get to $y=2$ the argument is now $\pi$, and by the time we get to $y=1+$ we are down to $\frac{\pi}{2}$; now we hop on to the next branch of tangent, and as we pass from $y=1-$ to $y=0$ the argument runs down to $0$.

Summing up, as we travel round the whole contour the argument has not changed at all. Hence there are no zeroes of $f(z)$ in the first quadrant.


问:
Is the derivative of a periodic function always periodic?

True or False :
The derivative of a periodic functions is always periodic.

I thought it to be true , as everything about a periodic function repeats itself at regular intervals, and so should it's derivative . But , to my surprise it is given false , which suggests that it might be true most of the time but not always , I have given all my thoughts to finding a counter example but I just can't seem to find even one counter example .

One possibility was $\{x\}$ , which is not differentiable at every integer , and I am confused about whether I should call it periodic or not , because it's graph will be a straight line with holes at every integer , so in a sense it is periodically not defined , just like $\tan x$ wich is not defined at every odd multiple of $\pi\over 2$ but still it is said to be periodic .

Could someone please help me find a counter example and clarify about periodicity of derivative of $\{x\}$.

Thanks !

$\{x\}$ is fractional part of x .

答:
Just a shortcut to what @diracdeltafunk said above.

You could directly apply the chain rule on $f(x)=f(x+T):$
\begin{eqnarray*}f(x)=f(x+T),~\forall x\in\mathbb{R}&\Rightarrow&f'(x)=(x+T)'\cdot f'(x+T),~\forall x\in\mathbb{R}\\
 &\Rightarrow&f'(x)=f'(x+T),~\forall x\in\mathbb{R}
\end{eqnarray*}


问:
What is the equation for a 3D line?
Just like we have the equation $y=mx+b$ for $\mathbb{R}^{2}$, what would be a equation for $\mathbb{R}^{3}$? Thanks. </p>

答:
Here are three ways to describe the formula of a line in $3$ dimensions.  Let's assume the line $L$ passes through the point $(x_0,y_0,z_0)$ and is traveling in the direction $(a,b,c)$.

Vector Form

$$(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$$

Here $t$ is a parameter describing a particular point on the line $L$.

Parametric Form

$$x=x_0+ta\\y=y_0+tb\\z=z_0+tc$$

These are basically the equations that result from the three components of vector form.

Symmetric Form

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$

Here we assume $a,b,$ and $c$ are all nonzero.  All we've done is solve the parametric equations for $t$ and set them all equal.

Title: What is the equation for a 3D line?


问:
DTMC problem： Two machine
In a certain manufacturing system, there are 2 machines $M_1$ and $M_2$: $M_1$ is a fast and high precision machine whereas $M_2$ is a slow and low precision machine. $M_2$ is employed only when $M_1$ is down, and it is assumed that $M_2$ does not fail. Assume that the processing time of parts on $M_1$; the processing time of parts on $M_2$; the time to failure of $M_1$; and the repair time of $M_1$ are independent geometric random variables with parameters $p_1，p_2，f$ and $r$; respectively. Identify a suitable state space for the DTMC model of the above system and compute the TPM. Investigate the steady-state behavior of the DTMC.

The question is: in my point of view, it seems like a CTMC problem so I don't know how to include the $p_1，p_2，f$ and $r$ in the DTMC TPM.

Thanks for your help in advance.

答:
Since you haven't said where you are stuck, I will just work the beginning of the problem.

A geometric random variable is discrete, and each attempt is either a success or a failure.  There is always one machine working.  In each time interval, either machine one is broken or working, and the item being worked on is either finished or not.  At the end of each interval, we need to know whether the item got finished or not, and whether $M_1$ is broken or not. Precision doesn't seem to come into it, except to say that $M_1$ is always used if available.  Each attempt takes one time unit, so as soon as $M_1$ is working again, it makes the nexts attempt, with no time wasted.
If $M_1$ is working, it has probability $f$ of switching to "failed" each time.  If it is broken, it has probability $r$ of being repaired each time.  Meanwhile, if $M_1$ is working, the item has a probability $p_1$ of being completed, but if $M_2$ is working instead, it has probability $p_2$ of being completed.
Suppose

$A = M_1$ working, item finished,

$B = M_1$ failed, item finished,

$C = M_1$ working, item not finished, or

$D = M_1$ failed, item not finished.

I am assuming that if M1 fails it does so after finishing an attempt.

Those would appear to be the possible states.

Title: DTMC problem： Two machine


问:
Axioms that are not evident
How can we give a meaning to the three axioms of prepositional logic? As axioms, is not it supposed to be obvious? For example, how is

$(\phi \to (\psi \to \chi)) \to ((\phi \to \psi) \to (\phi \to \chi))$

evident?

答:
It is a way (in addition to Modus Ponens) to "encode" the crucial properties of the conditional needed to prove the Deduction Theorem.

The fact can be more "obvious" (less obscure...) if we write it as a sequent:

$p → (q → r), p →q \vdash p → r$.

In Natural deduction, it is an easy consequence of the rules for $\to$:

if we have a derivation $\mathcal D_1: \Gamma_1, p \vdash (q \to r)$ and a derivation $\mathcal D_2: \Gamma_2, p \vdash q$, we apply $(\to$-Elim) followed by $(\to$-Intro) to get a derivation:

$\mathcal D_3 : \Gamma_1, \Gamma_2 \vdash (p \to r)$.

The axiom $\phi \to (\psi \to \phi)$ expresses the Weakening rule:

we can add an additional premises in a derivation.

Title: Axioms that are not evident


答:
l'Hopital's rule doesn't apply the to the determinate form $\left[ 1^1 \right]$, so you shouldn't think of applying it here.

You ask explicitly about $\left[ 1^\infty \right]$.  Use the inverse relationship between exponentiation and the natural logarithm together with continuity of the exponential function:  $$  \lim_{x \rightarrow \infty} f(x)^{g(x)} = \mathrm{e}^{\lim_{x \rightarrow \infty} g(x) \cdot \ln f(x)}  \text{.}  $$  Now you have $\mathrm{e}^\left[\infty \cdot 0 \right]$

Of relevance to the problem is \begin{align*}
&\lim_{x \rightarrow \infty} \frac{a x^n + b x^{n-1} + \cdots + c}{d x^n + e x^{n-1} + \cdots + f}  \\
    &\quad = \lim_{x \rightarrow \infty} \frac{x^n}{x^n} \cdot \frac{a + b/x + \cdots + c/x^n}{d + e/x + \cdots + f/x^n}  \\
    &\quad = \frac{a + 0 + \cdots + 0}{d + 0 + \cdots + 0}  \text{.}
\end{align*}

So either apply this last observation directly and get the determinate form $\left[ 1^1 \right]$ or use the first to get the determinate form $\mathrm{e}^\left[1 \cdot 0 \right]$.

Title: Find the limit of $\;\lim\limits_{x\to\infty}\left(\frac{x^2-1}{x^2+1}\right)^\frac{x-1}{x+1}$.

Body:
Find limit of
$\;\lim\limits_{x\to\infty}\left(\dfrac{x^2-1}{x^2+1}\right)^\frac{x-1}{x+1}$ without using $L'Hopital$

I tried subtracting $1$ for using $lim$ of $e$ but I got $1^\infty$ form and couldn't continue.


问:
Prove: $\lim_{x \to \infty} f'(x) = 2$ indicate $\lim_{x \to \infty}f(x) = \infty$
This question is from last year exam in calculus.
In the answers they mentioned more than one way to solve this question.
The method they used is to build a new function but it was a bit complicated.
I would like to get some help.</p>

答:
By definition of limit, let $M$ such that $x \geq M \implies f'(x) > 2-\frac{1}{2} = \frac{3}{2}$. Then we have, for $x > M$
$$f(x) - f(M) = \int_x^M f'(t) dt \geq \frac{3}{2} (x-M),$$
that is
$$ f(x) \geq f(M) + \frac{3}{2} (x-M).$$

Now for any number $L$ we can find $N$ big enough so that
$f(M) + \frac{3}{2} (x-M) > L$
when $x > N$, and this proves that $f(x) \to +\infty$ as $x \to +\infty$.


问:
$f(z)=\frac{z}{1-z} \zeta(\frac {1}{1-z})$ belongs to Hardy Space $H^{1/3}(\mathbf{D})$
The Hardy space $H^p(\mathbf{D})$ is the vector space of holomorphic functions $f$ on the open unit disk that satisfy:
$$
\sup_{0< r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right|^p \; \mathrm{d}\theta\right)^\frac{1}{p}<\infty.
$$Consider $$ f(z)=\frac{z}{1-z} \zeta(\frac {1}{1-z})$$
where $\zeta$ denotes the Riemann zeta function.

Balazard Saias and Yor in a paper used that $f\in H^{1/3}(\mathbf{D}) $

Question - Prove that $f\in H^{1/3}(\mathbf{D})$

My Try -
$$ |f(re^{i\theta})|=|\frac{re^{i\theta}}{1-re^{i\theta}} | |\zeta(\frac {1}{1-re^{i\theta}})|$$
$$ |f(re^{i\theta})|=\frac{r}{|1-re^{i\theta}|}  |\zeta(\frac {1}{1-re^{i\theta}})|   \ \ \  ...(1)$$
$$\zeta(s)= s\int_1^\infty \frac{1-x+[x]}{x^{s+1}}dx +\frac{1}{s-1}, \Re(s)>0$$
$$|\zeta(s)|\leq \frac{|s|}{\Re(s)}+\frac{1}{|s-1|} , \ \ \ \ \Re(s)>0\ \ \ \ \ ... (2)$$
$$s=\frac{1}{1-re^{i\theta}} \Rightarrow|s|=\frac{1}{|1-re^{i\theta}|}$$
$$\Re(s) = \Re(\frac{1}{1-re^{i\theta}}) = \frac{1-r \ cos\theta}{|1-re^{i\theta}|^2}$$
$$\frac{1}{|s-1|}= \frac{|1-re^{i\theta}|}{r}$$
By equation (2),

$$|\zeta(\frac {1}{1-re^{i\theta}})| \leq \frac{|1-re^{i\theta}|}{(1-rcos\theta)}+\frac{|1-re^{i\theta}|}{r}$$
Equation (1) gives,
$$|f(re^{i\theta})|\leq \frac{r}{|1-re^{i\theta}|} [\frac{|1-re^{i\theta}|}{(1-rcos\theta)}+\frac{|1-re^{i\theta}|}{r} ]    $$
$$|f(re^{i\theta})|\leq \frac{1+r(1-cos\theta)}{1-rcos\theta}$$
Further $1+r(1-cos\theta) \leq 1+2r$
$$|f(re^{i\theta})|\leq \frac{1+2r}{1-rcos\theta}$$
After this I am stuck. Please solve

Body: On $\Re(s)=1/2$ $$\zeta(s) = \frac{s}{s-1}+ s\int_1^\infty (\lfloor x\rfloor -x)x^{-s-1}dx= O(|s|)$$ therefore $$\int_{|z|=1} |f(z)|^{1/3}|dz|=\int_{|z|=1} O(\frac1{|1-z|^{2/3}})|dz| <\infty$$


答:
The key is to show that $f$ is bounded below by a function which goes to infinity as $x \to \infty$.  It follows from your hypothesis that there exists an $x_0$ such that for all $x \geq x_0$, $f'(x) \geq 1$.

Now let $x$ be any number bigger than $x_0$.  Consider the average rate of change

$$\frac{f(x) - f(x_0)}{x-x_0}$$

between $x$ and $x_0$.  By the mean value theorem, there exists a $c > x_0$ such that

$$\frac{f(x) - f(x_0)}{x-x_0} = f'(c)$$

Since $f'(c) \geq 1$, this implies that there is a constant $K$ such that $f(x) \geq x +K$ for all $x > x_0$.

Title: Prove: $\lim_{x \to \infty} f'(x) = 2$ indicate $\lim_{x \to \infty}f(x) = \infty$

Body:
This question is from last year exam in calculus.
In the answers they mentioned more than one way to solve this question.
The method they used is to build a new function but it was a bit complicated.
I would like to get some help.</p>


答:
You can describe a line in space as the intersection of two planes.  Thus,
$$\{(x,y,z)\in{\mathbb R}^3: a_1x+b_1y+c_1z=d_1 \text{ and } a_2x+b_2y+c_2z=d_2\}.$$
Alternatively, you can use vector notation to describe it as
$$\vec{p}(t) = \vec{p}_0 + \vec{d}t.$$

I used this relationship to generate this picture:

This is largely a topic that you will learn about in a third semester calculus course, at least in the states.

Title: What is the equation for a 3D line?

Body: Just like we have the equation $y=mx+b$ for $\mathbb{R}^{2}$, what would be a equation for $\mathbb{R}^{3}$? Thanks. </p>


问:
Does the sum $\sum_{k=0}^{\infty}\sin \left(\dfrac{2\pi k!}{2000} \right)$ converge?
I am having som difficulties with series that the title suggests; I would really appreciate some guidance/hint how to get forward. My main difficulty to comprehend is the faculty that appears in the sine and furthermore that the terms in certain intervals of $k$ switches sign... The series apparently converges.

答: For $k$ large enough, $k!$ always has a factor $2000$.  All the terms from then on are zero, so the series converges.


答: When $k\geq 2000$, the argument inside the $\sin$ is an integer multiple of $2\pi$, so the series actually consists of less than 2000 non-zero terms. Hence it converges.

Title: Does the sum $\sum_{k=0}^{\infty}\sin \left(\dfrac{2\pi k!}{2000} \right)$ converge?

Body: I am having som difficulties with series that the title suggests; I would really appreciate some guidance/hint how to get forward. My main difficulty to comprehend is the faculty that appears in the sine and furthermore that the terms in certain intervals of $k$ switches sign... The series apparently converges.


答: what can you say about $\sin\left(\dfrac{2\pi k!}{2000}\right) \ \text{if} \ k >200$


问:
How to solve parametric quadratic function?
i have problem to solve following equation:

Find $b$ and $c$ if the graph of the function $y = 2x ^ 2 + bx + c$ touches the axis $Ox$ at the point $(1; 0)$.

Can anyone help me?

答:
$y = 2x ^ 2 + bx + c$ passes through the point $(1,0)$

$2+b+c=0\tag{1}$

Furthermore it is tangent to $x$-axis, so the solution $x=1$ must be double

Which means that the discriminant of $2x ^ 2 + bx + c=0$ is zero, that is

$b^2-8c=0\tag{2}$
From $(1)$ solve $c=-b-2$ and plug into $(2)$

$b^2-8(-b-2)=0\to (b+4)^2=0\to b=-4$ and $c=2$


问:
Power Series without computation
$$\sum_{n=1}^\infty \frac{n^n}{n!}x^n$$

Now obviously I can use the ratio test to prove that the Radius of convergence is $\frac{1}{e}$ but what the question asks is that using little computation prove why the Radius of convergence of this power series satisfies $R\leq1$. I know the question is ambiguous but is there any concept I can use to prove this? This question was on a past quiz and unfortunately, there is no solution that I can look at...

答:
Let $a_n = \frac{n^n}{n!}$. For the radius of convergence $R$ we have
$$ \frac{1}{R} = \limsup_n \sqrt[n]{a_n}. $$

Since clearly $a_n \geq 1$, also $\sqrt[n]{a_n} \geq 1$, so $\frac{1}{R} \geq 1 \implies R \leq 1$.

Title: Power Series without computation


答:
A quadratic that touches the $x$ axis when $x=1$ has the form $y=a(x-1)^2=ax^2-2ax+a.$

Equate the coefficients to those of $2x^2+bx+c$ to find $b$ and $c$.

Title: How to solve parametric quadratic function?

Body:
i have problem to solve following equation:

Find $b$ and $c$ if the graph of the function $y = 2x ^ 2 + bx + c$ touches the axis $Ox$ at the point $(1; 0)$.

Can anyone help me?


答:
How is

$~~~~(\phi \to (\psi \to \chi)) \to ((\phi \to \psi) \to (\phi \to \chi))$

evident?

Don't despair. The axiom you cite is actually a trivial theorem in natural deduction, the axioms of which are much more intuitive IMHO. I'm sure there is some good historical reason that it was chosen to be an axiom in an early formulation of formal propositional logic, but it eludes me.

(Screenshot from my proof checker)

Title: Axioms that are not evident

Body:
How can we give a meaning to the three axioms of prepositional logic? As axioms, is not it supposed to be obvious? For example, how is

$(\phi \to (\psi \to \chi)) \to ((\phi \to \psi) \to (\phi \to \chi))$

evident?


答:
I think your instructor hoped you would notice that $n^n \ge n!$ (actually, greater when $n>1$).

At $x = 1$  the terms in this infinite sum are all at least  $1$   so the series can't converge there.

Title: Power Series without computation

Body:
$$\sum_{n=1}^\infty \frac{n^n}{n!}x^n$$

Now obviously I can use the ratio test to prove that the Radius of convergence is $\frac{1}{e}$ but what the question asks is that using little computation prove why the Radius of convergence of this power series satisfies $R\leq1$. I know the question is ambiguous but is there any concept I can use to prove this? This question was on a past quiz and unfortunately, there is no solution that I can look at...


问:
the collision in Kuramoto model is not occur for the identical case.
The Kuramoto model is following
$$\dot{\theta_i}=\omega_i+K/N\sum_{j=1}^N \sin(\theta_j-\theta_i),t>0,i=1,2,\dots,N.$$

where $\theta_i=\theta_i(t)\in \mathbb{R}$, $w_i=\dot{\theta}(0)$(intrinsic frequeny) K is coupling strength constant, N is the Number of oscillators.

and the definition of collision is following
for $i\neq j$, we say the oscillators $\theta_i$ and $\theta_j$ collide at time $t_*>0$ if
$$\theta_i(t_*)=\theta_j(t_*)+2k\pi$$ for soem integer $k$.

In the paper, for the identical case( $\omega_i=\omega,\ \  \forall i$
),no collision occur because of the nature of the first order autonomous ode. (one can prove it the uniqueness argument.)

I don't know how to prove why no collsion occur in identical case.

Please give me a help if you have idea.

答:
In the case of uniform angular base velocities, you get for angle differences
\begin{align}
\dot \theta_i &=\omega+K/N\sum_{m=1}^N \sin(\theta_m-\theta_i)\\
\dot \theta_i -\dot \theta_i
&=K/N\sum_{m=1}^N [\sin(\theta_m-\theta_i)-\sin(\theta_m-\theta_j)]\\
&=2K/N\sum_{m=1}^N \cos(\theta_m-\tfrac12(\theta_i+\theta_j))\sin(\tfrac12(\theta_j-\theta_i))
\end{align}
So if a collision occurs at some point, the derivative of the difference will be zero. Thus there is a solution with constant difference. By uniqueness this is also the only variant where a collision occurs.


问:
Expressing the statement "The equation $x^2 + 2x = 15$ has a unique solution." in formal language.
Can the statement "The equation $x^2 + 2x = 15$ has a unique solution." be expressed in a form of the formal language as the following:

\begin{align*}
\exists x,\forall y,(((x^2 + 2x = 15)\wedge (y^2 +2x = 15)) \rightarrow (x=y))
\end{align*}

The universe is $\mathbb{N}$ for both $x$ and $y$.

答:
Personally, I read the English sentence "$F(x)=0$ has a unique solution" as synonymous with "$F(x)=0$ has exactly one solution", and therefore I would write your sentence as $$\exists x,(x^2+2x=15\land\forall y,(y^2+2y=15\to x=y))$$

and I would dismiss your answer as wrong because it means that there is some $x$ such that, for all $y$, either $x$ is not a solution or $y$ is not a solution or $x=y$. In other words, it's an unnatural way to state that either the equation has exactly one solution or there is an element that does not solve it.

As user Mauro ALLEGRANZA has pointed out, $$(\exists x, x^2+2x=15)\land(\forall y,\forall z,((y^2+2y=15\land z^2+2z=15)\to x=y))$$ is another reasonable formulation of the same concept of unique existence, which mimics the way several proofs of such statements are organized.


问:
Differentiability of the Cantor Function
I know that the Cantor function is differentiable a.e. but I want to prove it without using the theorem about monotonic functions. I have already proved that $f'(x) = 0$ for all $x \in [0,1] \backslash \mathbb{C}$ where $\mathbb{C}$ is the Cantor set.

But I'm not sure how to go about proving that if $x \in \mathbb{C}$ then $f$ is not differentiable at $x$.

Actually, upon reflection, I think I have already proved differentiability a.e. but I would still like to know how to finish this part.

Also, the definition I am using for the function:
 $$f:[0,1] \to [0,1]$$
Let $x \in [0,1]$ with ternary expansion $0.a_1a_2...$ Let $N$ be the first $n \in \mathbb{N}$ such that $a_n = 1$. If for all $n \in \mathbb{N}$, $a_n \in \{0,2\}$, let $N = \infty$.

Now define $b_n = \frac{a_n}{2}$ for all $n < N$ and $b_N = 1$.
Then $$f(x) = \sum_{i=1}^{N} \frac{b_n}{2^n}.$$

答:
On the other hand, suppose that one uses a fat Cantor set to produce a similar Cantor function. So now the Cantor set has a positive Lebesgue measure. I would conjecture that now, except for the endpoints of the complementary intervals, the derivative would exist and be finite. Then it would be finite. It would likely be infinite at the endpoints. Remember that, for a function of bounded variation, the derivative exists and is finite almost everywhere. Certainly something of this sort must happen. The devil's
staircase</p>


问:
If $T \in L(X,X)$ is a bounded operator and $lim_{n \to \infty} ||T^n|| = 0$ then $\sigma(T) \subset B_r(0)$ for some $r < 1$
Let $X$ be a Hilbert space and let $T: X \to X$ be a bounded operator such that $lim_{n \to \infty} ||T^n|| = 0$.

I need to show that there exists $r < 1$ such that $\sigma(T) \subset \overline{B_r(0)}$.

I managed to show that if $|\lambda| \ge 1$ then $\lambda \not \in \sigma_p(T)$, because otherwise, there is some $x \in X$ with $||x|| = 1$ such that $Tx = \lambda x$ so $T^nx = \lambda ^n x$  and then $||T^n|| \ge ||T^nx|| = |\lambda|^n \ge 1$, which is a contradiction to $lim_{n \to \infty} ||T^n|| = 0$.

However, I'm not sure how to show that $\lambda \not \in \sigma_c(T)$ or $\lambda \not \in \sigma_r(T)$

Help would be appreciated

答:
The spectral radius of an operator $T$ has the property $$\rho(T)\le\|T^n\|^{1/n}\le\|T\|$$
Hence $$\rho(T)^n\le\|T^n\|\to0$$ so $\rho(T)<1$ as required.


问:
Proofs of the structure theorem for finitely generated modules over a PID
I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID.

If possible, a comparison in terms of clarity, elegance or usefulness would be appreciated.

答:
Fred Goodman has a freely downloadable algebra text that contains this result in section 8.5.  Unfortunately, I am not knowledgeable enough to offer a useful comparison to other references.  But hey, it's free!

(I studied draft versions of this material as it was being written.  I was taking a class from the author, and I found it useful.  It became "Edition 2.5" with substantial additions because the 2nd Edition didn't have enough to cover the full first-year graduate level algebra sequence at Iowa.)

ContentLicense: CC BY-SA 2.5


问:
Composite of relation and its opposite
I have a relation set $R = \{(a, b), (a, c), (b, c), (b, d), (c, d), (c, a)\}$ included in set $\{a, b, c, d\}\times\{a, b, c, d\}$.

Its opposite of relation is $R^{−1} = \{(b, a), (c, a), (c, b), (d, b), (d, c), (a, c)\}$.

I'm asked to find the composite of $R\circ R^{−1}$ as a set, and I came up with:

$R\circ R^{−1} = \{(a, a), (a, d), (b, b), (b, c), (c, b), (c, c), (d, c), (d, d), (d, a)\}$.

However, the practice exam's answer says it is:

$R\circ R^{−1} = \{(a, a), (b, b), (c, c), (d, d), (a, d), (d, c)\}$

After that, I'm asked to find $R^{−1}\circ R$, which I came up with to be:

$R^{−1}\circ R = \{(a, c), (a, a), (a, b), (b, a), (b, b), (b, c), (c, c), (c, b)\}$

Whereas again the practice exam's solution tells me:

$R^{−1}\circ R = \{(a, a), (b, b), (c, c), (d, b), (d, c)\}$

Who is correct? I came to my answers by drawing out arrow diagrams and followed the each member of $X$ to its $Y$, but I seem to be getting way more relations than the solution says I should be.

答:
The text appears to be incorrect.

I was unable to verify your answers.  In one case you seem to be missing one ordered pair and in the other you seem to have one ordered pair too many.

Title: Composite of relation and its opposite


答:
Not quite, even correcting for the minor typo. Let's generalise your statement a bit:
$$\exists x, \forall y (P(x) \land P(y) \implies (x = y))$$
This does not say that $P(x)$ is true for a unique $x$. Indeed, this statement is satisfied if there is some $x$ such that $\lnot P(x)$. If there is such an $x$, then $P(x) \land P(y)$ is false for any $y$, and hence the implication is vacuously true.

Title: Expressing the statement "The equation $x^2 + 2x = 15$ has a unique solution." in formal language.

Body:
Can the statement "The equation $x^2 + 2x = 15$ has a unique solution." be expressed in a form of the formal language as the following:

\begin{align*}
\exists x,\forall y,(((x^2 + 2x = 15)\wedge (y^2 +2x = 15)) \rightarrow (x=y))
\end{align*}

The universe is $\mathbb{N}$ for both $x$ and $y$.


问:
Causality in Vaccine versus Placebo Group
I am having a debate with some people on statistical significance of a certain event.

It is based on a real world situation, namely the AstraZeneca vaccine and the incidence of Transverse Myelitis.

As far as I know the trial invovled two groups with about 25,000 people in each group. (50,000 people altogether).

In the vaccine group there were two incidences of Transverse Myelitis, and in the placebo group there was one incidence.

Without going into the maths I can clearly see that there is no statistical significance of Transverse Myelitis in these two groups. The numbers are just too small.

Can anyone give me a basic mathematical breakdown of this situation? I would just like to back up my suspicions with some cold hard maths.

Thank you.

答: Suppose you "randomly" selected from the general population until you got exactly 3 people with TM. Now (randomly) split the group into 2 equal groups. What is the probability that the sick people will be split 1:2? Since there is 0.5 probability for each TM patient to be in either of the groups, the probability to see exactly 2 in a specific group is 3/8. That is not too bad. If you ask what is the probability to have 0:3 or 1:2, the odds increase to exactly 1/2 ( just from symmetry).</p>


问:
Inverse Matrix of a Special Matrix for Optimization
For some portfolio optimization problems, it finally comes to an inverse matrix of a special matrix
$$B=\begin{bmatrix}
A & \mathbf{1} \\
\mathbf{1}' & 0
\end{bmatrix}$$
where $A$ is a symmetric matrix and $\mathbf{1}=(1,\cdots,1)'\in\mathbb{R}^n$. Can anyone help to find the inverse matrix of matrix $B$?

Thanks a lot!

Below is the background of the above problem:

Assume $\omega$ is the weight vector of the portfolio, $\mu$ is the return vector and $\Sigma$ is the covariance matrix. The constraint is $\sum\omega=1$. For different optimization problems, we will have different targets:

For the optimization problem of minimum variance:
$$\min_\omega\omega'\Sigma\omega$$
applying the Lagrange multiplier, we have the matrix equation
$$\begin{bmatrix}
2\Sigma & \mathbf{1} \\
\mathbf{1}' & 0
\end{bmatrix}
\begin{bmatrix}
\omega \\
\lambda
\end{bmatrix}=
\begin{bmatrix}
\mathbf{0} \\
1
\end{bmatrix}$$

For the optimization problem of mean-variance optimization:
$$\max_\omega\omega'\mu-\frac{1}{2}\delta\omega'\Sigma\omega$$
applying the Lagrange multiplier, we have the matrix equation
$$\begin{bmatrix}
-\delta\Sigma & \mathbf{1} \\
\mathbf{1}' & 0
\end{bmatrix}
\begin{bmatrix}
\omega \\
\lambda
\end{bmatrix}=
\begin{bmatrix}
-\mu \\
1
\end{bmatrix}$$

Anyway, it comes to the problem of inverse matrix of
$$B=\begin{bmatrix}
A & \mathbf{1} \\
\mathbf{1}' & 0
\end{bmatrix}$$

答:
We can use the Schur complement of $A$ to get the formula
$$
B^{-1} = \pmatrix{A^{-1} + A^{-1}\mathbf 1 (B/A)^{-1}\mathbf 1' A^{-1} & -A^{-1}\mathbf 1(B/A)^{-1}\\
-(B/A)^{-1}\mathbf 1' A^{-1} & (B/A)^{-1}},
$$
where
$$
B/A = 0 - \mathbf 1' A^{-1} \mathbf 1 = -\mathbf 1' A^{-1} \mathbf 1 \in \Bbb R.
$$
We can rewrite the above in the following arguably simpler form:
$$
B^{-1} = \pmatrix{A^{-1} & 0\\0 & 0} + \frac 1{\mathbf 1' A^{-1} \mathbf 1}\pmatrix{\mathbf{vv}' & - \mathbf v\\ -\mathbf v' & 1}
\\ =
\pmatrix{A^{-1} & 0\\0 & 0} + \frac 1{\mathbf 1' A^{-1} \mathbf 1}
\pmatrix{-\mathbf v\\ 1} \pmatrix{-\mathbf v \\ 1}',
$$
where $\mathbf v = A^{-1} \mathbf 1$.


问:
Find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$
I want to find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$. It seems that would simply be to calculate the following triple integral:

$\int_0^3 \int_0^{2x} \int_0^{1-y^2} z\,dz\,dy\,dx$

This is pretty straight-forward to do without any variable substitutions etc. which makes me think it's almost too simple (for a home assignment).

Am I missing something or is the above correct?

答:
I ended up doing the following (omitting some algebra steps):

$1 - y^2 \geq 0$ when $y \leq 1$ which gives an integration interval $0 \leq y \leq 1$. By dividing the integral into two parts with the intervals $0 \leq x \leq \frac{1}{2}$ and $\frac{1}{2} \leq x \leq 3$ we get the two simpler integrals:

$V =
V_1 + V_2 =
\int_0^{1/2} \int_0^{2x} \int_0^{1-y^2} dz\,dy\,dx +
\int_{1/2}^3 \int_0^1 \int_0^{1-y^2} dz\,dy\,dx$

$V_1 = \int_0^{1/2} \int_0^{2x} \int_0^{1-y^2} dz\,dy\,dx =
\int_0^{1/2} \int_0^{2x} (1-y^2)\,dy\,dx =
\int_0^{1/2} \left(2x - \frac{8}{3}x^3\right) dx =
\frac{5}{24}$

$V_2 = \int_{1/2}^3 \int_0^1 \int_0^{1-y^2} dz\,dy\,dx =
\int_{1/2}^3 \int_0^1 (1-y^2) dy\,dx =
\int_{1/2}^3 \frac{2}{3} dx =
\frac{5}{3}$

$V = V_1 + V_2 = \frac{5}{24} + \frac{5}{3} = \frac{15}{8}$

Makes sense? Thank you for the comments.


问:
What is the equation for a 3D line?
Just like we have the equation $y=mx+b$ for $\mathbb{R}^{2}$, what would be a equation for $\mathbb{R}^{3}$? Thanks. </p>

答: I am giving you an example. Let $A(-2,0,1),~~B(4,5,3)$ be two points in $\mathbb R^3$. And let $C$ be the end point for the vector which is drawn from the origin. In addition, we assume that this vector has the same direction as the vector $AB$. So we have its coordinates is $(4,5,3)-(-2,0,1)=(6,5,2)$. Therefore the equation of the line passing through $A$ and $B$ is $$L_{AB}: x=(-2,0,1)+t(6,5,2)$$

Title: What is the equation for a 3D line?


问:
Given a scalene triangle and its orthocenter, prove that a point on its Euler circle, has a constant line segment
Given a scalene triangle $\triangle ABC$ and $H$ the orthocenter of the triangle. $P$ is a point on the Euler circle of the triangle $\triangle ABC$. The segments $BH, CH$ intesect the opposite sides $AC,AB$ at the points $E,Z$ respectively. The circumscribed circles of the triangles $\triangle EHP$ and $\triangle ZHP$ intersect the lines $CH, BH$ for a second time at the points $K, R$ respectively. Prove that when the point $P$ moves on the Euler circle of the triangle $\triangle ABC$, the line $KP$ goes through a constant point.

On this question I am very much lost. I immediately thought that $EPHK$ is inscrived and hence has all the properties of an inscribed quadrilateral. Moreover, I thought of somehow manipulating the Euler circle and somehow using the fact that it passes through the midpoints of $AB,BC,AC$ to solve the question, but none of this bared crops. Could you please explain to me how to solve this question?

答:
Let $EH$ and $PK$  meet the nine point circle at points $M$ and $D$ respectively (These are different from points $E$ and $P$ respectively. ). $M$ will be the midpoint of $BH$.

Now, observe that,  $MD\parallel HK$ because $\angle MDP=180-\angle MEP=180-\angle HEP=\angle HKP$.
Also, notice that,  $D$ is the midpoint of $BC$ because $BM=MH$ and $MD\parallel HC$.  Hence, $PK$ always passes through the midpoint of $BC$, a constant point.


答:
By L'Hôpital's rule
$$
\mathop {\lim }\limits_{x \to  + \infty } \frac{{f(x)}}{x} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{f'(x)}}{1} = 2.
$$
Thus
$$
\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } x\mathop {\lim }\limits_{x \to  + \infty } \frac{{f(x)}}{x} =  + \infty .
$$


问:
Multiplying the residue classes mod n by a coprime
I noticed an interesting result when playing around with mods. If I take all the residue classes of $n$ and multiply each of them by an integer $q$ such that $n$ and $q$ are coprime, then each residue class maps to a unique new residue class.

For example, if $n=7$ and $q=3$ then the residue classes
${0,1,2,3,4,5,6}$ get mapped to ${(0\cdot3), (1\cdot3), (2\cdot3),(3\cdot3),(4\cdot3),(5\cdot3),(6\cdot3)}$ which in turn becomes ${0,3,6,2,5,1,4}$. This is simply a permutation of $0,1,2,3,4,5,6$ which are the residue classes we started with.

In other words, is there a way to prove that when you multiply $\{0,1,2,\dots, (n-1)\}$ by $q$, no two products will be equivalent mod n?

答:
Let $gcd(q,n)=1$.
Then the mapping $\Bbb Z_n\rightarrow\Bbb Z_n:x\mapsto qx$ is bijective and so permutes the residues $0,1,\ldots,n-1$.

The reason is that in the above case, $q$ is invertible in $\Bbb Z_n$.
Then $qx=qy$ implies by multiplying with $q^{-1}$, $x=y$ and so the mapping is injective.

But a mapping $X\rightarrow X$, where $X$ is finite, is already bijective if it is injective (or surjective).


问:
How many integers are there from 1 to 10000 that are divisible by exactly two of the numbers 4, 5, 6, and 7?
Could someone please help me solve this question:
How many integers are there from 1 to 10000 that are divisible by exactly two of the numbers 4, 5, 6, and 7?
Thanks!! Also, I think I would have to use PIE but I can't figure out how to set it up.</p>

答:
Inclusion and exclusion is the way to do it.  It's the number divisible by two of the factors, minus the number divisible by three of the factors, plus the number divisible by all four.  As Matthew Daly said in a comment, you have to treat all the cases separately.  For example, a number is divisible by all four factors if and only if it is divisible by their greatest common divisor, namely $420$.  The number of multiples of $420$ less than or equal to $10000$ is $$\left\lfloor\frac{10000}{420}\right\rfloor=23$$

I leave the rest to you.


问:
Is there a nice list of spectral sequences that don't come from particular constructions?
When you first learn about rings, it's important to have examples of, say, a PID which is not a Euclidean domain, a UFD which is not a PID, and so forth, to help build intuition and provide test cases.

Well, I'm just starting to learn about spectral sequences.  According to Eisenbud's treatment of spectral sequences, most everyday spectral sequences come from exact couples, most exact couples come from monic endomorphisms of differential modules, and I guess the next steps are differential modules with filtrations and gradings.

Is there a nice list somewhere with examples of, say, a spectral sequence that doesn't come from an exact couple, and so forth?

Of course, feel free to simply provide such examples here as well. :)

答:
From Wikipedia: "In fact, all known spectral sequences can be constructed using exact couples." (In my opinion, the word "known" in this sentence is a bit odd. Are there unknown spectral sequences that do not come from exact couples? If one does exist, it seems like it would be something contrived and not very helpful towards understanding spectral sequences.)

However, there are some spectral sequences that don't require knowledge of exact couples (and probably better understood w/o exact couples). One is the Grothendieck spectral sequence. This gives you information on how two (right) derived functors compose. The setup is this: Suppose $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ are abelian categories and $\mathcal{A}$, $\mathcal{B}$ with enough injectives  (think $\bf{R-mod}$ or $\text{Sheaves}(X)$), $F:\mathcal{B} \rightarrow \mathcal{C}$ and $G:\mathcal{A} \rightarrow \mathcal{B}$ left exact functors. One last thing: assume $G$ takes injective objects to $F$-acyclic objects.  Then there is a convergent first quadrant spectral sequence for each object $A$ of $\mathcal{A}$:

$$E^{pq}_2 = (R^pF)(R^qG)(A) \Rightarrow (R^{p+q}FG)(A)  $$

(There is a dual statement for right exact functors.)

You can use this to study base change with respect to Ext and Tor. The Leary spectral sequence is another example (this looks at what happens to global sections after applying the direct image functor).  Another special case is the Lyndon-Hochschild-Serre spectral sequence. This appears in group cohomology. It allows you to compute the cohomology of $G$ (with coefficients in $A$) knowing the cohomology of $G/H$ with coefficients in $H^*(H,A)$.  This provides a neat way of computing (for example) the cohomology of the dihedral groups or of the discrete Heisenburg group.

I'm not sure if this is exactly what you are looking for, but I think it is at least relevant. My understanding is that spectral sequences in algebraic topology usually involve exact couples, but spectral sequences in  commutative algebra usually do not.

All of the information above is treated in Weibel's An Introduction to Homological Algebra. I really like this book. In his spectral sequence chapter, he develops the theory without use of exact couples. He doesn't even talk about them until the very end of the chapter.


问:
How to demystify the axioms of propositional logic?
How might I go about getting some intuition on the typical axiom schemes given for propositional logic? They seem rather mysterious at first glance.

For example, these are taken from: http://en.wikipedia.org/wiki/Hilbert_system#Logical_axioms

$\phi \to \phi$

$\phi \to \left( \psi \to \phi \right)$

$\left ( \phi \to ( \psi \rightarrow \xi \right)) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$

$\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$

答:
How might I go about getting some intuition on the typical axiom schemes given for propositional logic? They seem rather mysterious at first glance.

IMHO each of these axioms can be derived as a theorem using the more intuitive axioms of natural deduction.  Following are proofs using a simple form of natural deduction.

(Screenshots from my proof checker)

$\phi \to \phi$

$\phi \to \left( \psi \to \phi \right)$

$\left ( \phi \to ( \psi \rightarrow \xi \right)) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$

$\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$


问:
Partial Derivative of Square Root with Matrix Multiplication
for the scalar case $x\in \mathbb{R}$, the partial derivative of $f(x)  = \sqrt{x^2 + \delta^2}$, where $\delta\in \mathbb{R}$ is a constant, is $f'(x) = x(x^2 + \delta^2)^{-1/2}$.

Now we assume that $x\in \mathbb{R}^n$, and $A\in \mathbb{R}^{n\times n}$. Define the function $f(x) = \sum_{i = 1}^n \sqrt{|Ax|_i^2 + \delta^2}$. Now we hope to find the partial derivative w.r.t. $x_i, i = 1,\dots, n$.

It seems that the partial derivative w.r.t. each element involves other locations. Could anyone give some hints about the derivative? Thanks in advance!

答:
$\def\B{\Big}\def\L{\left}\def\R{\right}\def\o{{\tt1}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$Define
the all-ones vector $\o$ and a vector $v$ such that
$$\eqalign{
 v\odot v &= Ax\odot Ax + \delta^2\o \\
 2\,v\odot dv &= 2\,Ax\odot A\,dx \\
dv &= Ax\odot A\,dx\oslash v \\
}$$
where $\odot$ denotes the elementwise/Hadamard product and $\oslash$ Hadamard division.
Note that $\{\o,v,x\}\in{\mathbb R}^{n}$.

Let's use a colon to denote the trace/Frobenius product, i.e.
$$A:B = {\rm Tr}(A^TB)$$
which coincides with the usual dot product when $\{A,B\}$ are vectors.

Write the objective function in terms of the above notation.

Then calculate its differential and gradient.
$$\eqalign{
f &= \o:v \\
df &= \o:dv \\
 &= \o:(Ax\odot A\,dx\oslash v) \\
 &= (Ax \oslash v):(A\,dx) \\
 &= A^T(Ax\oslash v):dx \\
\p{f}{x} &= A^T\B(Ax\oslash v\B) \\
}$$

Many specialized formulas were used in the steps above. First, due to the properties of the underlying trace function, the terms in a Frobenius product can be rearranged in many different ways, e.g.
$$\eqalign{
CA:B &= C:BA^T = A:C^TB \\
}$$
Second, the Frobenius and Hadamard products commute
$$\eqalign{
A:B &= B:A \\
A\odot B &= B\odot A \\
C:(A\odot B) &= (C\odot A):B \\
}$$
Finally, the vector $\o$ is the identity element for Hadamard multiplication
$$\eqalign{v &= \o\odot v}$$


问:
Help with simple integral
I'd like to know why this is wrong:

\begin{align}
\int \operatorname{sen}(x)·\cos(x) \,dx &\underset{\uparrow}{=}\int u\ du=\frac{u^2}{2}+c=\frac{1}{2}\operatorname{sen}^2(x)+c\\&\boxed{CV\\\phantom{d}u=\operatorname{sen}(x)\\du=\cos(x)\,dx}
\end{align}

When checking on WolframAlpha its done this other way:

\begin{align}
\int \operatorname{sen}(x)·\cos(x) \,dx&\underset{\uparrow}{=}-\int u\ du=-\frac{u^2}{2}+c=-\frac{1}{2}\cos^2(x)+c\\&\boxed{CV\\\phantom{d}u=\cos(x)\\du=-\operatorname{sen}(x)\,dx}
\end{align}

Wouldn't both ways of solving it be correct?

答: As was pointed out in the comments, the Pythagorean Identity states that $$\operatorname{sen}^{2}(x) + \cos^{2}(x) = 1.$$  This means that $$\frac{1}{2}\operatorname{sen}^{2}(x) = -\frac{1}{2}\cos^{2}(x) + 1,$$ and so for a constant $c$ we have $$\frac{1}{2}\operatorname{sen}^{2}(x) + c = -\frac{1}{2}\cos^{2}(x) + 1 + c.$$  Writing $C = 1 + c$, we can write $$\frac{1}{2}\operatorname{sen}^{2}(x) + c = -\frac{1}{2}\cos^{2}(x) + C.$$  This tells us that the two integration constants are different, but since the integration constant is arbitrary it doesn't matter which form of the antiderivative we take.  Both are correct.

Title: Help with simple integral


问:
Limit involving multiple factorials

Let $n, k\in\mathbb N_0$ be fixed (where $k\leq n$), and $S_N\in\mathbb N$ be a sequence such that $\lim_{N\to\infty} S_N/N=p$ for some fixed $p\in(0,1)$.
Show that $$\lim_{N\to\infty} \frac{S_N! (N-S_N)!(N-n)!}{(S_N-k)!(N-S_N-n+k)!N!}=p^k(1-p)^{n-k}.$$

Is there any easy way to evaluate this limit? Rewriting the left hand side to binomial coefficients does not seem to help, $$\lim_{N\to\infty} \frac{{S_N\choose k}{N-S_N\choose n-k}}{{N\choose n}{n\choose k}},$$ at least I don't see a way to start from there.

答:
We will use Stirling's Approximation in the form:

$$n! \sim \sqrt{2 \pi }n^{n+\frac12} e^{-n} $$

Consider, for fixed $r$,

$$\lim_{m \to \infty} m^{-r}\binom{m}{r} = \lim_{m \to \infty} m^{-r} \frac{m!}{r!(m-r)!} $$

$$ \sim \frac{1}{r!}\lim_{m \to \infty} m^{-r} \frac{\sqrt{2 \pi }m^{m+\frac12} e^{-m}}{\sqrt{2 \pi }(m-r)^{m-r+\frac12} e^{-(m-r)}} $$

$$ = \frac{e^{-r}}{r!}\lim_{m \to \infty}  \left( \frac{m }{m-r}\right)^m \left( \frac{m }{m-r}\right)^{-r+\frac12}  $$

$$ = \frac{e^{-r}}{r!}\lim_{m \to \infty}  \frac{1}{\left( 1-\frac{r}{m}\right)^m} \left( 1-\frac{r}{m}\right)^{r-\frac12}  $$

$$ = \frac{e^{-r}}{r!} e^r \, 1 = \frac{1}{r!} $$

So now we can use your rewriting in terms of binomial coefficients:

$$\lim_{N\to\infty} \frac{S_N! (N-S_N)!(N-n)!}{(S_N-k)!(N-S_N-n+k)!N!} = \lim_{N\to\infty} \frac{{S_N\choose k}{N-S_N\choose n-k}}{{N\choose n}{n\choose k}}$$

$$ = \lim_{N\to\infty} \frac{\frac{{S_N}^k}{k!}\frac{(N-S_N)^{n-k}}{(n-k)!}}{\frac{N^n}{n!}{n\choose k}}$$

$$ = \lim_{N\to\infty} \frac{{S_N}^k\,(N-S_N)^{n-k}}{N^n}$$

$$ = \lim_{N\to\infty} \left(\frac{S_N}{N}\right)^k\,\left(\frac{N-S_N}{N}\right)^{n-k}$$

$$=p^k(1-p)^{n-k}$$

Title: Limit involving multiple factorials


问:
Quick Telescopic Series Question
So I was given the following prompt when I was studying for a test:

"Determine the sum of the following convergent series: $\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n+2})$"

I'm a bit confused about what the setup would look like here. I understand the setup for the other type of geometric series (where the sum of the series is equal to the first term divided by $1-$ the common ratio), and I also understand that the formula for this series looks like the following: $S_n=b_1-b_{n+1}$, but I'm confused about what that formula would look like in this specific context. Any help would be appreciated!

答:
Write $$\frac{1}{n}-\frac{1}{n+2}=\left (\frac{1}{n}-\frac{1}{n+1}\right)+ \left( \frac{1}{n+1}-\frac{1}{n+2}\right)$$
Now you do telescoping for two togather.

Title: Quick Telescopic Series Question


问:
How to develop $\frac{1}{1-x}$ into $1+\frac{x}{1+x}+\frac{1\cdot2x^2}{(1+x)(1+2x)}+\dots$
I know that, by long division, or binomial formula or Taylor formula that this function can be developed into the geometric series:
$1/(1-x)=1+x+x^2+x^3+x^4+\ldots $ and I thought that this is the only series that represent the function mentioned above. However, recently, I read "Traite Elementaire des Series" by Eugene Catalan and he said, on page 60, that "The same function can admit multiple expansions". He gives the series below:

$$\frac{1}{1-x}=1+\frac{x}{1+x}+\frac{1\cdot2x^2}{(1+x)(1+2x)}+\frac{1\cdot2\cdot3x^3}{(1+x)(1+2x)(1+3x)}+\dots$$

I am just curious how can we derive this second series. So a function may admit multiple series representation? This is something new that I don't know. Maybe a better way to think about this is that two or multiple series can converge to the same function, which is the same as multiple numerical series may converge to the same sum.

答: Partial answer: Lets begin with the second term. Then $n$th term of this series satisfies the recurrence relation $$f(n)=\dfrac{nx}{(1+nx)}f(n-1),\qquad f(1)=\dfrac{x}{1+x}.$$ For summing series, my favorite method is telescoping. So, let me find another sequence $(a_n)$ such that $f(n)=a_{n-1}-a_n$ with the additional relation $a_{n}=\dfrac{nx}{(1+nx)}a_{n-1}.$ Then we have $$f(n)=a_{n-1}-\dfrac{nx}{(1+nx)}a_{n-1}=\dfrac{a_{n-1}}{1+nx},\qquad  a_{n-1}=(1+nx)f(n),$$ and it determines $(a_n)$ uniquely.  After telescoping we get $$\sum_{n=1}^mf(n)=a_0-a_n$$ and substituting for $(a_n)$ simplifies this to $x-(n+1)xf(n).$


问:
Subsets sharing at most $i$ elements
Assume I have a set $S$ of $N$ elements and I create subsets with $k$ elements from it. With no additional property the number of possible such subsets will be $N \choose k$.

Now I want my subsets to satisfy some additional rules $R_i$:

Given a set $X$ of subsets of $S$ with $k$ elements (i.e. $\forall A \in X, A \subset S , \#A=k$), $X$ satisfies $R_i$ if and only if for all $A$ and $B$ in $X$, $A$ and $B$ share at most $i$ elements, i.e. $\forall A , B \in X, \#(A \cap B) \le i$. Let's call $S_i$ the set of all such $X$ satisfying $R_i$.

I want to know how large a set satisfying $R_i$ can be, i.e. what is $M_i = \max_{X \in S_i} \#X $

For example if $N=1000$ and $k=10$, trivially we have $M_9 = {1000 \choose 10}$: the largest ensemble of subsets of $10$ elements that share at most $9$ with one another is exactly the set of all possible combinations of $10$ elements out of $1000$.

Similarly at the other extreme, still for $N=1000$ and $k=10$, we have $M_0 = 1000/10=100$.

Is there a formula or clever (not brute force) algorithm to compute other less trivial cases for $i=1,2,\dots8$?

BTW even if there is a nice formula, I'll ultimately want to implement a way to get a maximal $X$...

答:
What you are looking for is known in the literature as a constant weight binary code (Wikipedia). Specifically, if $A(n,d,w)$ is the maximum number of binary vectors, each with $w$ ones and $n-w$ zeroes, such that the Hamming distance between any two codewords is $d$, then the largest possible size of your set of sets would be $A(N,2(k-i) ,k)$.

Finding the maximal set is an open problem, and the best known known constructions for $X$ and best bounds are listed on Andries E. Brouwer's website, along with known upper bounds on what is possible.

In general, if you want to find a set $X$ which is "good enough," and you have a comptuer you can program, then lexcodes are the way to go. Simply iterate through the list of all subsets of size $k$ in lexicographic order, and when you find a subset $S$ which shares no more than $i$ elements with everything in your current code, then add $S$ to the code.

Title: Subsets sharing at most $i$ elements


问:
Is there a function $f(x)$ for which $\int_a^b f(x)dx$ converges but $\sum_{i=1}^\infty f(i)$ diverges?
I was recently learning about improper integrals, so I am not that advanced, and I asked myself whether there is some function $f(x)$ for which $\int_a^b f(x)dx$ with the limits being $a$ or $b$, it doesn't matter, converges while $\sum_{i=1}^\infty f(i)$ diverges.

I have found similar questions being asked like this one but there are usually the other way around, so the integral is divergent and the sum convergent, like with $f(x)=\sin(\pi x)$.

答:
As written, it is easy because the integral is over a finite interval.  $f(x)=1$ suffices.  It is a little harder if you make the integral $\int_1^\infty$ but not much.  We know that the sum of $\frac 1i$ diverges but we can make the integral converge by making a bump at each natural and making the bumps get narrower and narrower.
$$f(x)=\begin {cases} \frac 1k & x\in [k-\frac 1k,k+\frac 1k],k \in \Bbb N \\0&\text{otherwise} \end {cases}$$

will work.  The area of each rectangle is $\frac 1{k^2}$ and we know the sum of that converges.  As leslie townes pointed out, you could also just make $f(x)=\frac 1k$ at the naturals and $0$ elsewhere, which makes the integral identically $0$.


问:
The solution of a partial differential equation( I can't work out the equation)
let $ y=f(x,t) $ and $F(x,y,t)=0  $  The function is determined
$t=t(x,y)$ what is $\dfrac{dy}{dx}$
This is original question.

答:
We have that $F(x,y,t)=0$ determines $t$ as a function of $(x,y)$, and then $y=f(x,t)$ determines $y$ as a function of $x$. We are asked to find $\frac{dy}{dx}$.

Differentiating we have
$$
F_x dx+ F_y dy +F_t dt=0 \tag{1}
$$ and
$$
dy= f_x dx+ f_t dt \tag{2}
$$
where I have written $F_x$ for $\frac{\partial F}{\partial x}$ and so on.

From (1) we express $dt$ in terms of $dx,dy$ and substitute this into (2) getting
$$
dy= f_x dx + f_t \big(F_x dx +F_y dy\big)\frac{-1}{F_t}
$$
or
$$
(1+f_t \frac{F_y}{F_t})dy
=
(f_x-f_t \frac{F_x}{F_t})dx
$$
so that finally
$$
\frac{dy}{dx}
=
\frac{f_x-f_t \frac{F_x}{F_t}}{1+f_t \frac{F_y}{F_t}}.
$$


问:
Equilibrium point from Differential equation does not match the solution
I have a differential equation $$y'=-2+t-y$$.

I solved this linear differential equation and got $y=t-3$ as the solution.

And when I tried to solve this without having to solve the actual equation. I got the equilibrium point to be $y=t-2$ which does not match my solution. Why is this? Does this mean y=t-2 is not a solution.

When I plotted the direction field I found that y=t-2 is not in the solution. What does this mean?

答:
There is no equilibrium point because the differential equation is not autonomous: the right side has a $t$ in it.

$y = t-2$ is not a solution, as you can easily check by plugging this in to the differential equation.  What it is is a null isocline: a curve on which solutions have derivative $0$.

And, btw, $y = t-3$ is a solution, but it is not the general solution.
The general solution is $y = t-3+c e^{-t}$.


问:
Show that a local ring is equicharacteristic iff it contains a subfield
A local ring $(A,\mathfrak m)$ is equicharacteristic if $\operatorname{char} A=\operatorname{char}  \kappa (m)$.

Need hints to solve the following question:

A local ring is equicharacteristic iff it contains a subfield.

答:
$\kappa$ is often used to mean "residue field". If $R$ is a ring and $P$ is a prime ideal of $R$, the notation
$$\kappa(P)$$
means the fraction field of $R/P$, and is called the residue field at (or of) the prime $P$.