问:
Probability of finding a cycle
Let $n,k\in \mathbb{N}$ where $n\geq k\geq 4$ and they are even. Let's create a $K_{n,n}$ bipartite graph with $n$ random edges chosen uniformly. Take random $k$ edges $e_1, e_2, \dots, e_k$ from $K_{n,n}$.
What is the propability that $e_1, e_2, \dots, e_k$ form a $C_k$ cycle?

It's clear that in a $K_{n,n}$ there are at most ${n \choose k/2}^2$ cycles of length $k$. So I think I should find first the probability of having $C_k$ in $K_{n,n}$, because it doesn't necessarily exist.

答:
In a graph with $n$ edges and $x$ cycles of length $k$, the probability that $k$ randomly chosen edges will form a cycle is $x/\binom nk$.

In a random $n$-edge subgraph of $K_{n,n}$, the number of $k$-cycles is itself a random variable $\mathbf X$, so by the law of total probability, the probability that $k$ randomly chosen edges will form a cycle is $\mathbb E[\mathbf X]/\binom nk$.

We can compute $\mathbb E[\mathbf X]$ by linearity of expectation. First, we should find the number of $k$-cycles in $K_{n,n}$, which is I believe $\frac1k (\frac{n!}{(n-k/2)!})^2$ for even $k$. For each $k$-cycle, the probability that it appears in the random $n$-edge subgraph is $\binom{n^2-k}{n-k}/\binom{n^2}{n}$.

Multiply these together, and we have a somewhat terrifying expression for the probability you want.

Title: Probability of finding a cycle


问:
find the maximum value of $\left(\int_{0}^1\sqrt[3]{f(x)}dx\right)\left(\int_{0}^1\sqrt[5]{f(x)}dx\right)\left(\int_{0}^1\sqrt[7]{f(x)}dx\right)$
If $f:[0,1]\rightarrow (0,\infty)$ is a continuous function such that $\int_{0}^1f(x)dx=1$ then find the maximum possible value of $$\left(\int_{0}^1\sqrt[3]{f(x)}dx\right)\left(\int_{0}^1\sqrt[5]{f(x)}dx\right)\left(\int_{0}^1\sqrt[7]{f(x)}dx\right)$$

How should one choose $f(x)$. Or is there a general version of Cauchy Schwarz Bunyavosky's inequality

答:
By H"older's inequality https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality
(with $p=3,5,7$ respectively), each of the integrals in the product is at most 1. So $f=1$ is the maximizer.

ContentLicense: CC BY-SA 4.0


问:
Is it possible to make a regular 3-polygon by selecting $3$ points in $S$
I would appreciate if somebody could help me with the following problem:

Let $\mathbb{Z}$ be the set of all integers and let $ S = \mathbb{Z} \times \mathbb{Z} $.

Question:

1). Is it possible to make a regular 3-polygon by selecting $3$ points in $S$ ?

2). Is it possible to make a regular 5-polygon by selecting $3$ points in $S$ ?

3). Is it possible to make a regular $n$-polygon by selecting $n$ points in $S$ ?

答:
If it is possible to make an $n$-gon, then the center of that $n$-gon has coordinates in $\frac1n\mathbb Z$ (it is the average of $n$ integers). You can scale the original $n$ gon with a factor of $n$. Then the center of this scaled $n$-gon has integer coordinates.
Thus after translating the $n$-gon to move its center to the origin, we see that the vertices are $$\tag1(x_k,y_k)=\left(r\cos(\frac{2\pi}n\cdot k+\alpha),r\sin(\frac{2\pi}n\cdot k+\alpha)\right)$$ with some (possibly irrational) $r>0$, $\alpha\in[0,2\pi)$.
One can even simplify further by considering the $n$-gon with vertices
$$\tag2(x_k',y_k') = (x_kx_1+y_ky_1,x_ky_1-y_kx_1),\quad k=1,\ldots,n$$
(this is just a scaled and rotated version of the original).
Since $(x_1',y_1')=(x_1^2+y_1^2,0)$, we have in effect made the $r$ in $(1)$ an integer and $\alpha=0$.
Therefore, both $\sin\frac{2\pi}n=\frac{y_2}{\sqrt{x^2+y^2}}$ and $\cos\frac{2\pi}n=\frac{x_2}{\sqrt{x^2+y^2}}$ are at most quadratic irrationals.
Then consider the possible minimal polynomials of the complex number $z=\frac{x_2+iy_2}{\sqrt{x^2+y^2}}$.</p>

ContentLicense: CC BY-SA 3.0


问:
Two definitions of the injectivity radius
For a Riemannian manifold $M$, usually, the injectivity radius of $p \in M$ is defined by the supremum of positive numbers $r$ such that $\exp_p|_{B(0, r)}$ is a diffeomorphism onto the image. Here, $B(0,r)$ is the geodesic ball of radius $r$. However, in Klingenberg's book, the injectivity radius is defined by by the supremum of positive numbers $r$ such that $\exp_p|_{B(0, r)}$ is injective. Are these two definitions equivalent?

Considering the global rank theorem, I tried to prove the exponential map at $p$ is of constant rank, but it seems that this try does not work. In fact, Klingenberg frequently says that the exponential map is just injective instead of saying the map is a diffeomorphism. To guarantee that the exponential map at $p$ is a diffeomorphism on a geodesic ball, is it enough to show that it is injective?

Thanks!

答:
It's not a completely straightforward consequence of the inverse function theorem. Here's a proof.

Theorem. Suppose $(M,g)$ is a Riemannian manifold, $p\in M$, and $B_c(0)\subset T_pM$ is a ball on which $\exp_p$ is defined. Then the restriction of $\exp_p$ to $B_c(0)$ is injective if and only if it is a diffeomorphism onto its image.

Proof.
If the restriction of $\exp_p$ is a diffeomorphism onto its image, then clearly it's injective. For the converse, assume it's injective. For each vector $v\in B_c(0)$, let $\gamma_v(t)= \exp_p(tv)$, which is a geodesic defined for all $t$ such that $|tv|<c$.

There are two key facts needed for the proof. Hopefully the following facts are proved in Klingenberg's book, but I don't have a copy of the book here, so I'll give references to my book Introduction to Riemannian Manifolds (2nd ed.) [IRM].

A vector $v\in B_c(0)$ is a critical point of $\exp_p$ if and only if $\gamma_v(1)$ is conjugate to $p$ along $\gamma_v$ [IRM, Proposition 10.20].

No geodesic is minimizing past its first conjugate point [IRM, Theorem 10.26].

Using these facts, the argument goes as follows. It suffices to show that $\exp_p$ has no critical points in $B_c(0)$, for then it's a local diffeomorphism by the inverse function theorem, and an injective local diffeomorphism is a diffeomorphism onto its image.

Suppose for the sake of contradiction that $v\in B_c(0)$ is a critical point. Then by fact 1 above, $\gamma_v(1)$ is conjugate to $p$ along $\gamma_v$. Thus by fact 2, for any number $r$ such that $|v| < |rv| < c$, it follows that $\gamma_v$ restricted to $[0,r]$ not minimizing. Thus there is a shorter geodesic $\sigma: [0,b]\to M$ with $\sigma(0)=p$ and $\sigma(b) = \gamma_v(r)$. Since every geodesic starting at $p$ is the image of a radial line under $\exp_p$, we can parametrize $\sigma$ so that it has parameter domain $[0,1]$ and is of the form $\sigma(t) = \exp_p(tw)$ for some $w\in T_pM$ with $|w| = \text{Len}(\sigma)$, and the fact that $\sigma$ is shorter implies that $|w|<|rv|<c$. Thus $\exp_p(w) = \exp_p(rv)$, contradicting the assumption that $\exp_p$ is injective on $B_c(0)$. $\square$


问:
Solving for x in $\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$

How can I solve for x?
$$\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$$

I have tried a few methods, but to be honest, I do not think I have what it takes to find the solutions. I'm hoping someone has has experience with these sorts of equations and can help me out!

答:
IfI properly understand, what you are asked is for find, for a given value of $y$ the corresponding $x$.

This equation is very highly transcendental and you will need some numerical method and, for sure, a "reasonable" guess of the solution.

To some extent, this problem is a red herring and, to me, the worst thing to do would be to write $y=f(x)$.

For a given $y$, you know the value of $\left(\frac{y+1}{y-1}\right)$. So, to make the problem much nicer, define $k=\log \left(\frac{y+1}{y-1}\right)$ and, taking logarithms, consider that you look for the zero of function
$$g(x)=(1+x) \log (1+x)+(1-x) \log (1-x)-k$$ which is quite nice and very smooth (almost looking like a parabola).

Expanding the rhs as an infinite series
$$k=\sum_{n=1}^\infty \frac{x^{2 n}}{n (2 n-1)}$$ Using series reversion
$$x=\sqrt k \,\sum_{n=1}^\infty a_n k^n$$ The $a_n$'s make the sequence
$$\left\{1,-\frac{1}{12},-\frac{13}{1440},-\frac{97}{40320},-\frac{25307}{29030400},-\frac{56827}{153280512},\cdots\right\}$$ and this is almost the exact solution.


问:
Convolution of Gamma distribution
I have to show
$$\newcommand\dif{\mathop{}\!\mathrm{d}}
  \text{Gam}(\alpha,\beta_1) \ast \text{Gam}(\alpha,\beta_2) = \text{Gam}(\alpha,\beta_1+\beta_2)
$$
where the probability density function for a $\text{Gam}(\alpha,\beta)$-distributed random variable is defined as follows
$$
  f_{\alpha,\beta}(x) = \frac{\alpha^\beta}{\Gamma(\beta)} x^{\beta-1} e^{-\alpha x} \;\textbf{1}_{(0,\infty)}(x)
$$
Also let's write down the definition of the $\text{Gamma}$ function
$$
  \Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \dif{t}
$$

I know this can be proven easily using the probability-generating function but I want to prove it using the defintion of a convolution. This is what I have done so far:

$$
  \begin{align*}
    & \quad \int_0^\infty \left(\frac{\alpha^{\beta_1}}{\Gamma(\beta_1)} y^{\beta_1-1} e^{-\alpha y} \right)
    \left( \frac{\alpha^{\beta_2}}{\Gamma(\beta_2)} (x-y)^{\beta_1-1} e^{-\alpha(x-y)} \right) \dif{y} \\
    &= \frac{\alpha^{\beta_1 + \beta_2}}{\Gamma(\beta_1)\Gamma(\beta_2)} \int_0^\infty \left(y^{\beta_1-1} e^{-\alpha y} \right) \left( (x-y)^{\beta_2-1} e^{-\alpha(x-y)} \right) \dif{y} \\
&= \frac{\alpha^{\beta_1 + \beta_2}}{\Gamma(\beta_1)\Gamma(\beta_2)} \int_0^\infty y^{\beta_1-1} (x-y)^{\beta_2-1} e^{-\alpha x} \dif{y} \\
&= \frac{\alpha^{\beta_1+\beta_2} e^{-\alpha x}}{\Gamma(\beta_1)\Gamma(\beta_2)} \int_0^\infty y^{\beta_1-1} (x-y)^{\beta_2-1} \dif{y}
  \end{align*}
$$

Now, I know I need to have $\Gamma(\beta_1 + \beta_2)$ in the denominator. The last integral almost looks like the $\text{Gamma}$ function but I don't see how I would arrive at the result.

答: Strangely, you include the indicator function for the gamma density in your first equation, but then you ignore it when you perform the convolution.  You will observe that if $y > x$, the convolution is zero:  $$\int_{y=-\infty}^\infty \frac{\alpha^{\beta_1}}{\Gamma(\beta_1)} y^{\beta_1 - 1} e^{-\alpha y} \mathbb 1(y > 0) \frac{\alpha^{\beta_2}}{\Gamma(\beta_2)} (x-y)^{\beta_2 - 1} e^{-\alpha(x-y)} \color{red}{\mathbb 1(x-y > 0)} \, dy \\= \int_{y=0}^x \frac{\alpha^{\beta_1 + \beta_2}}{\Gamma(\beta_1)\Gamma(\beta_2)} e^{-\alpha x} y^{\beta_1 - 1} (x-y)^{\beta_2 - 1} \, dy.$$  Then a simple scaling transformation $$y = xu, \quad dy = x \, du,$$ yields $$\frac{\alpha^{\beta_1 + \beta_2} x^{\beta_1 + \beta_2 - 1} e^{-\alpha x}}{\Gamma(\beta_1) \Gamma(\beta_2)}\int_{u=0}^1 u^{\beta_1 - 1} (1 - u)^{\beta_2 - 1} \, du.$$  The remaining integral is a Beta function integral and equals $$\frac{\Gamma(\beta_1) \Gamma(\beta_2)}{\Gamma(\beta_1 + \beta_2)},$$ from which the desired result follows.

Title: Convolution of Gamma distribution


问:
Degree of vector bundle under extension of scalars.
I have a somewhat of a trivial question, just want to confirm it (or maybe its wrong). Degree of a vector bundle doesn't change when we extend the scalars is that right? So if a vector bundle is semi-stable after extension to algebraic closure, it had to be semi-stable from the beginning, or is there a counter-example?</p>

答: The degree of a bundle, of course, does not change under extensions of scalars and semi-stability is preserved, because the Harder--Narasimhan filtration is canonical, hence Galois-invariant, hence descends from the algebraic closure of the base field.</p>


问:
An inequality using Cauchy inequality
Let $f\in L^2([0,1])$ and $$g = \int_0^1 \frac{f(t)}{|x-t|^{0.5}}dt.\quad 0<x<1.$$ Prove the following inequality holds:
$$||g||_2 \le 2\sqrt{2}||f||_2.$$

I want to use the fact that $||g|| = max ||gh||_1$ for  $h\in L^2([a,b])$ with $||h||_2=1.$ For arbitrary $h\in L^2$ we have $|gh|_1 = \int_0^1|h(x)|dx(|\int_{-1}^1 \frac{f(x-t)}{|t|^{0.5}}\kappa_{0\le x-t\le 1}(t)dt|)\le \int_{-1}^1\frac{1}{|t|^{0.5}}dt(\int_0^1|h(x)||\tilde{f}(x-t)|dt) \le \int_0^1\frac{1}{|t|^{0.5}}||f||_2||h||_2dt = 2||f||_2$ by Tonelli theorem, here $\tilde{f} = f\kappa_{0\le x-t\le 1}(t).$ Thus I found the coefficient can be greater than $2\sqrt{2}$. But I couldn't find the error in my proof. Could anyone help?

答:
$\def\abajo{\\[0.3cm]}$
Your estimate is wrong since with $f(t)=1$ you get $\|g\|_2=\sqrt{4+\pi}>2$. I don't follow exactly what you did, but it looks like you are missing a term. Below is my take, which I think is very similar to your intention, but using the missing term.

Assume without loss of generality that $f=|f|$, $h=|h|$. Write $a_v=1_{[v,1]}$, $b_v=1_{[0,1-v]}$
\begin{align}
\int_0^1|g(x)h(x)|\,dx
&\leq \int_0^1\int_0^1\frac{h(x)f(t)}{|x-t|^{1/2}}\,dt\,dx\abajo
&\textit{(substitution: $v=t-x$)}\abajo
&= \int_0^1\int_{-x}^{1-x}\frac{h(x)f(v+x)}{|v|^{1/2}}\,dt\,dx\abajo
&\textit{(Tonelli; substitution is explained at the end)}\abajo
&=\int_{-1}^0\frac1{(-v)^{1/2}}\int_{-v}^1h(x)f(v+x)\,dx\,dv
+\int_{0}^1\frac1{v^{1/2}}\int_{0}^{1-v}h(x)f(v+x)\,dx\,dv\abajo
&=\int_{0}^1\frac1{v^{1/2}}\bigg[\int_{v}^1h(x)f(x-v)\,dx\,dv
+\int_{0}^{1-v}h(x)f(v+x)\,dx\bigg]\,dv\abajo
&\textit{(Cauchy-Schwarz twice on each term, also explained below)}\abajo
&\leq\int_{0}^1\frac1{v^{1/2}}\Big[\|h\|_2\,\|f\|_2\,(\|a_v\|_2^2+\|b_v\|_2^2)\Big]\,dv\abajo
&=\|f\|_2\,\int_0^1 v^{-1/2}\,2(1-v)\,dv=\frac83\,\|f\|_2.
\end{align}
The constant $8/3$ is slighly less than $2\sqrt2$, so I assume that whoever asked the question had a different, more sleek argument in mind.

The substitution: The region of integration after the first substitution is, with $x$ as the horizontal axis and $v$ as the vertical one,

To exchange $x$ and $v$ requires two different sets of inequalities, which is what gives the two integrals.

Cauchy Schwarz: what's happening is
\begin{align}
\int_v^1 h(x)\,f(v+x)\,dx
&=\int_0^1 h(x)\,a(x)\,f(x+v)\,a(x)\,dx\abajo
&\leq \|h\|_2\,\|a\|_2\,\|f\|_2\,\|a\|_2.
\end{align}
The other integral is the same, but using the function $b$.

Title: An inequality using Cauchy inequality


问:
Why do smaller intervals give more accurate results?
I'm doing a lot of derivatives. But I'm stumped on this one question.

Why is it that a smaller changes in x correspond to a more accurate value of the slope at that point? Or actually, how do we know that smaller changes mean give us the slope at that point?

I have done quite a bit of calculus without really understanding this and I would appreciate it if anyone could help me understand the reasoning behind this.

答:
To me, Taylor series gives the best explanation.

The formula is
$f(x+h)
=f(x)+hf'(x)+h^2f''(x)+...+h^nf^{(n)}(x)/n!+...
$

If we just look at the first two terms as an approximation we get

$f(x+h)
\approx f(x)+hf'(x)
$
with an error described by the next term which is proportional to
$h^2$.
This means that the smaller $h$ gets,
the more accurate the approximation is.


答:
On the graph of some function $y=f(x)$ (e.g. you could take $f(x)=x^2$ to visualise), pick two points $x_0$ and $x_0+h$ and their corresponding $y$ values. So, you have points $A=(x_0,f(x_0))$ and $B=(x_0+h,f(x_0+h))$. Now draw a line through $A$ and $B$.

Recall that the derivative of a function is essentially the gradient of its tangent line at a point. Now, if $h$ is large, the line $AB$ doesn't look like a tangent line at all. But as you shrink $h$, the line gets closer and closer to becoming a tangent line at $A$. If you don't believe me, just draw the picture yourself (or use some software like GeoGebra) and see with your own eyes.

Another way to see this: it can be shown, mathematically, that any "sufficiently nice" function $f(x)$ looks like a straight line when you zoom in enough, and this straight line is the tangent line. In other words, the tangent like to a graph at a point is usually a very good approximation to the function, near the point of tangency. I say "sufficiently nice" because some "not nice" functions like $f(x)=|x|$ don't satisfy this condition, because at $x=0$ there's a little bump. Roughly speaking, you need the function to be smooth and not have these sharp turns---this condition is known as "differentiability" and determines whether or not the derivative is defined. So, for a differentiable function, the points $A,B$ approximately lie on the same tangent line for small $h$, thus the slope of $AB$ is approximately the same as the derivative at $A$.

Title: Why do smaller intervals give more accurate results?

Body:
I'm doing a lot of derivatives. But I'm stumped on this one question.

Why is it that a smaller changes in x correspond to a more accurate value of the slope at that point? Or actually, how do we know that smaller changes mean give us the slope at that point?

I have done quite a bit of calculus without really understanding this and I would appreciate it if anyone could help me understand the reasoning behind this.


问:
Finitely additivity and continuity from above at $\varnothing$ implies countably additivity.
I know that given a measurable space $(\Omega, \mathcal F)$, if a finitely additive set function $\mu$ is finite and continuous from above at $\varnothing$, then $\mu$ is a measure. But what if $\mu (\Omega)=\infty$? Could you please prove it or give a counterexample?

答:
Here is a counter-example.

Let $\Omega = \Bbb N$ and  $\mathcal F= 2^{\Bbb N}$. Define $\mu$ on $\mathcal F$ by:
$\mu(E)= 0$ if $E$ is finite  and $\mu(E)= +\infty$  if $E$ is infinite.

Note that $\mu(\Omega) = +\infty$.

It is easy to see that $\mu$ is finitely additive and $\mu$ is continuous from above at $\emptyset$. BUT $\mu$ is not $\sigma$-additive.

Remark. Recall the definition: A set function $\mu$ is continuous from above at $\emptyset$  if given any decreasing sequence $E_1 \supseteq E_2\supseteq E_3 \supseteq ... $ of sets in $\mathcal F$ such that $\bigcap_n E_n = \emptyset$, if there is at least one $n$ such that $\mu(E_n)< +\infty$, then $\lim_{n \to +\infty} \mu(E_n) =0$.


问:
Cauchy-Riemann equation analogue but for the quaternions
given a function over the quaternions

$$ U(x,y,z,t)+iV(x,y,z,t)+jW(x,y,z,t)+kR(x,y,z,t)=f(x,y,z,t) $$

what are the analogues of the Cauchy Riemann equation for the quaternionic plane so the function defined above is analytic ??

what happens with the Gauss' Theorem ? , so if the function $ f(x,y,z,t) $ is analytic then the integral over a curve in the quaternionic plane is 0 (closed curve)

$$ \oint f(x,y,z,t)ds =0 $$

where is more info about this equation ?? is there a Cauchy's theorem analogue for this integral or Laurent series in the quaternionic plane ??

答:
This question is fairly old but this paper may be of interest to you: https://dougsweetser.github.io/Q/Stuff/pdfs/Quaternionic-analysis-memo.pdf.

It explains exactly what you are looking for, and allow me to paraphrase for simplicity's sake:

Let any quaternion be described as $q= t+ix+jy+kz.$
Allow an analogue to the CR equations for the quaternions to be $$\frac{\partial f}{\partial t}+i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}=0.$$

Call any function $f: \mathbb{H} \to \mathbb{H}$ which obeys this equation regular.

Given a regular and continuously differentiable function $f$ and a 3-dimensional manifold on the Quaternions, $C$, the following is true:

$$\int_{C} f(q) \; D_q=0, \\ D_q = (dx \,dy\, dz-i\,dt\, dy \, dz - j\,dt\, dx \, dz - k\,dt \, dx \, dy).$$

I wish there were a simpler definition, but it works just fine.


答:
I will follow the standard notation where $K_{n, n}$ denotes the complete bipartite graph on $n$ left and $n$ right vertices, rather than what you defined in the question. What you are asking is to fix $n$ and $k$, choose a subset of $n$ edges randomly, then choose a subset of $k$ of those randomly, and you ask, what is the probability they form a cycle?

The step of choosing $n$ edges is unnecessary as this is equivalent to choosing a subset of $k$ edges from $K_{n, n}$ direclty. If you do this, then what you are asking for is
$$
\frac{\# \text{ of $k$-cycles in } K_{n, n}}{\binom{n^2}{k}}, \tag{1}
$$
where the denominator is the number of ways of choosing $k$ edges out of the $n^2$ total edges.

So it remains to calculate the number of $k$-cycles in $K_{n, n}$ (for $k$ even). There are $2n$ ways to choose the first vertex, $n$ ways to choose the second, $n - 1$ to choose each of the third and fourth, $n - 2$ to choose each of the fifth and second, and so on. This gives $2n \cdot n \cdot \left(\frac{(n-1)!}{(n-(k/2))!}\right)^2$, but this overcounts: we count each cycle $k \cdot 2$ times, since for each cycle there are $k$ possible vertices to start from and $2$ directions to go around the cycle. Therefore dividing, we end up with the number of $k$-cycles in $K_{n, n}$ as
$$
\frac{n!^2}{k (n - (k/2))!^2},
$$
agreeing with Misha Lavrov's calculation for this number. Plugging in to (1), our answer is
$$
\frac{(n^2 - k)! \; n!^2 \; (k-1)!}{n^2 ! \; (n - (k/2))!^2}.
$$

Title: Probability of finding a cycle

Body:
Let $n,k\in \mathbb{N}$ where $n\geq k\geq 4$ and they are even. Let's create a $K_{n,n}$ bipartite graph with $n$ random edges chosen uniformly. Take random $k$ edges $e_1, e_2, \dots, e_k$ from $K_{n,n}$.
What is the propability that $e_1, e_2, \dots, e_k$ form a $C_k$ cycle?

It's clear that in a $K_{n,n}$ there are at most ${n \choose k/2}^2$ cycles of length $k$. So I think I should find first the probability of having $C_k$ in $K_{n,n}$, because it doesn't necessarily exist.


问:
Finding mistake in the proof regarding a prime test
My proof is:

Let $N=pq$. If $p=k\sqrt{N}$ where $k\in\mathbb{Q}^+$, then $q=\dfrac{\sqrt{N}}{k}$.

If $k=1$, clearly $\sqrt{N}$ is a factor of $N$.

If $k<1$, then $p<\sqrt{N}$, so there must be at least one factor of $N$ less than $\sqrt{N}$.

If $k>1$, then $q<\sqrt{N}$, so there must be at least one factor of $N$ less than $\sqrt{N}$.

Therefore, there always exists at leastone number less than or equal to $\sqrt{N}$ that divides $N$. So, if there is no such number, $N$ is definitely prime.

Where did I go wrong?

答: $1$ is "at least one number less than or equal to $\sqrt{N}$ that divides $N$".


问:
Prove that the diameter of a $(n,d,\lambda)$-graph is at most $\lceil \log n / \log (d/\lambda) \rceil$
Let $G$ be a $n$-vertex $d$-regular graph. Let $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$ be the eigenvalues of the adjacency matrix $A_G$ of $G$ and $\lambda\geq \max\{|\lambda_2|, |\lambda_n|\}$. (I.e., $G$ is a $(n,d,\lambda)$-graph.)

Prove that the diameter of $G$ is  at most $\lceil \log n / \log (d/\lambda) \rceil$.

I think this problem is related to the proofof Alon-Boppana bound, but I don't have any idea how to prove.

答:
Have a look at "Chung, F. R. K. (1989). Diameters and Eigenvalues. Journal of the American Mathematical Society, 2(2), 187."

Here is the proof (obviously not my work).


答:
A differentiable function is precisely a function where, as you say, smaller changes in $x$ correspond to a more accurate value of the slope at that point.

It is easy to produce examples of functions that are not differentiable, i.e. where smaller changes in $x$ do not produce better approximations to the slope. Any point where a function is not continuous is also a point where the function is not differentiable.

More exotically, there are functions which are continuous everywhere, but differentiable nowhere. These are known as Weierstrass functions. Roughly, the function oscillates too wildly at every point to have a well-defined slope.


问:
What does $p≡3\mod4$ mean, and how do you find a prime matching this pattern?
I am reading How to Generate a Sequence of Unique Random Integers where it says:

What does that $p≡3\mod4$ mean? How do I find an 8-bit prime, and a 16-bit prime match that $p≡3\mod4$ statement?

答:
$p \equiv 3 \pmod 4$ means that $p = 4 k + 3$ for some $k$, or in other words that the remainder when you divide $p$ by $4$ is $3$.

Note that if you take an odd number and divide it by 4, you'll either get 1 or 3 as a remainder, because if you got 0 or 2 as a remainder then the original number would have had to have been even.

So every odd number (and hence every prime other than 2) satisfies either $p \equiv 1 \pmod{4}$ or $p \equiv 3 \pmod{4}$.

There is no dearth of primes congruent to 3 (mod 4), so there is no shortage of ways to find them.  You can use pretty much any technique you'd use to generate primes and then just check the the resulting prime is congruent to 3 (mod 4).

In fact, $2^{16}$ is a small enough number that you can simply calculate every prime with 16 or fewer bits by brute force and then throw out all the ones that are congruent to 1 (mod 4), with the whole process finishing in a fraction of a second.


问:
To estimate $e$ more accurately, how would you divine to define $e = 2^c$?
I understand all the algebra, but not "let $e = 2^c$". This feels like the key step. I've been staring at this the whole day, and I would've never been able to prognosticate this substitution.

James Stewart, Calculus 7th ed. 2011. Not the Early Transcendentals version. Pp 37-8. This isn't the original screenshot.  Stewart defines $f'(0)$ as $k$, but I deleted his $k$ and restored $f'(0)$. I see no point to defining $f'(0)$ as $k$.

答:
If $f(x) = 2^x$, then we can estimate $f'(0)$ by doing a calculation like
$$
   f'(0) \approx \frac{2^{1/N} - 1}{1/N}
$$
for very large $N$, which only requires taking the $N^{\text{th}}$ root of something. If we take $N = 2^k$, then we can compute $2^{1/2^k}$ by taking $k$ iterated square roots. So we can really compute an estimate such as $f'(0) \approx 0.693147$, from which we then estimate $e$ as $\approx 2^{-0.693147}$.

Picking $2^x$ in particular is not crucial, we could take $3^x$ or $10^x$ or anything else just as easily.


问:
Pigeon Hole Principle question, proving it is possible to select a number of points on a plane such that they are all interior to a circle
Thirteen points are given in the plane so that among any three of them there is a pair whose distance apart is less than 1. Prove that it is possible to select seven of the points so that they are all interior to a circle of radius 1.

I've experimented with PHP, but I didn't get much out of it...

Just for the record, this isn't a homework question, I stumbled upon it when I was doing some other PHP questions (no solution was given :( ...)

答:
Hint: Pick any point ($A$) and consider the circle of radius 1 about $A$.

Case 1: All points lie in the circle, we are done.

Case 2: There is some point $B$ that doesn't lie in this circle.

 Claim: All 13 points lie in a circle of radius 1 about A or about B.

Corollary: Hence, by PP, we are done.


问:
Understanding the (Partial) Converse to Cauchy-Riemann
We have that for a function $f$ defined on some open subset $U  \subset \mathbb{C}$ then the following if true:

Suppose $u=\mathrm{Re}(f), v=\mathrm{Im}(f)$ and that all partial derivatives $u_x,u_y,v_x,v_y$ exists and are continuous on $U$. Suppose further that they satisfy the Cauchy-Riemann equations. Then $f$ is holomorphic on $U$.

The proof for this is readily available, though there is a subtlety that I can't understand.

We essentially want to compute $\lim_{h \rightarrow 0} \dfrac{f(z+h)-f(z)}{h}$ where $h=p+qi \in \mathbb{C}$.

We need a relationship like $u(x+p,y+q)-u(x,y)=pu_x(x,y)+qu_y(x,y)+o(|p|+|q|)$. Why is this relationship true?

答:
Denote $h=\pmatrix{p\\q}.$
Since $u(x,\ y)$ is differentiable at the point $(x,\ y),$   increment of $u$ can be represented as
 $$u(x+p,\ y+q)-u(x,\ y)=Du\;h+o(\Vert h\Vert)=pu_x(x,\ y)+qu_y(x,\ y)+o(|p|+|q|),$$
where $Du=\left(\dfrac{\partial{u}}{\partial{x}}, \ \dfrac{\partial{u}}{\partial{y}} \right).$</p>


问:
If n is divisible by 4 and 2 then why it is not divisible by 8?
Well I'm working on some mathematics aptitude problem on divisibility. There are certain divisibility rules for numbers 6, 12, 15 etc. which are

If n is divisible by 2 and 3, it is divisible by 6

If n is divisible by 3 and 4, it is divisible by 12

If n is divisible by 3 and 5, it is divisible by 15

So, I applied the above rule on 8 and thought

If n is divisible by 2 and 4, it is divisible by 8

but I found a counterexample 576484 to my intuition and it bust the bubble of my figment.
It is divisible by 2 and as well by 4, but not by 8.

Why then it is not divisible by 8? If it is simultaneously divisible by 2 and 4.

Doesn't the rule applies in this case as well? Just like they are applied in the cases of 6, 12, & 15. That is, if factors 'a' and 'b' of number 'n' divides a given number 'm', then 'n' also divides 'm'.

答:
In general your rule holds if $n$ and $m$ are coprime. Otherwise the common factor may be only one time in the prime factorisation, but is two times in the product.

In your case: $2$ and $4$ are both divisible by $2$, that means you should take $4$ instead of $8$.

There is a great function which works in all cases, the least common multiple $lcm(x,y)$. It can be computed directly by many algorithms and plays an role in lots of mathematical subjects, for example there is an equivalent formulation of the Riemann Hypothesis on the growth of $lcm$.

Since this has already been mentioned in the comments, you can find internet links there as well.


问:
A problem on weights of an irreducible representation
Let $\mathfrak{g}$ a semisimple Lie algebra with a Cartan subalgebra $\mathfrak{h}$. Let $\phi$ a root system with base $\Delta$ And $V$ an irreducible representation of $\mathfrak{g}$. Let $\Gamma \subset \mathfrak{h}^*$ finite set, such that $V= \bigoplus_{\gamma \in \Gamma} V_{\gamma}$ , where $V_{\gamma} = \left\{ v \in V | hv = \gamma(h)v \right\}$ (a weight space). I want to prove the following:

$\Gamma \subset \gamma + \sum_{i=1}^{r} \mathbb{Z}\alpha_i$, with $\alpha_i  \in \Delta$

if $\lambda \in \Gamma$ is a maximal weight for $V$,i.e. such that $\Gamma \in \lambda - \sum_{i=1}^{r} \mathbb{Z}_{\geq 0}\alpha_i$, prove that $\langle \lambda, \alpha_i \rangle$ is a positive integer for every $i= 1...r$

So far, I have only managed to notice that to prove 2) it may be enough to show that $(\lambda, \alpha_i) \geq 0$ for every $i$. By contradiction, if there exists $i$ such that $(\lambda, \alpha_i) < 0$ I could infer that $\lambda + \alpha_i \in \Gamma$, if $\alpha_i \in \Gamma$. Therefore I may have found a contradiction since $\lambda$ is maximal. I don’t know if that makes sense, but if it does, I still don’t know how I could prove that at $\alpha_i \in \Gamma$. Any hints?

答:
Note that if $v$ is such that $hv = \gamma(h)v$, and $e_\alpha$ is a basis of the root space to $\alpha$ in $\mathfrak g$, then

$$h(e_\alpha v) = [h, e_\alpha]v + e_\alpha(hv) = (\alpha(h)\cdot e_\alpha) v + e_\alpha(\gamma(h)v) = \left(\alpha(h) + \gamma(h)\right) (e_\alpha v)$$

i.e. (and this is how I "knew" this before I did the computation):

$$v \in V_\gamma \implies e_\alpha v \in V_{\alpha+\gamma} \qquad (*)$$

To conclude for no.1 from there, proceed as follows: W.l.o.g. there exists $\gamma \in \Gamma$ such that $V_\gamma \neq 0$. Let $0 \neq v \in V_\gamma$, and let $W :=$ the subrepresentation of $V$ generated by $v$, i.e. all linear combinations of all $xv$ for all $x \in \mathfrak g$. Use $(*)$ and the root space decomposition of $\mathfrak g$ to show that
$$W \subseteq \bigoplus_{\beta \in \sum \mathbb Z \alpha_i} V_{\gamma + \beta}$$

After that, remember we have not used irreducibility yet; do it now.

For no.2, you should first familiarize yourself with how $\langle \cdot, \cdot\rangle$ is defined.


问:
Question on double sum of proof of Countable subadditivity of outer measure
I am reading Measure Theory from Sheldon Axler's Measure, Integration & Real Analysis. Here's the question that I have:

The rearrangement of $\{ I_{j,k} : j,k \in \mathbb{Z}^{+} \}$ (see proof below) is a little confusing to me. I understand that intuitively what the proof is doing is that we know that there is a function $\sigma : \mathbb{N} \to \mathbb{N}\times \mathbb{N}$ such that $\sigma (1)=(1,1)$, $\sigma(2)=(1,2)$, $\sigma(3)=(2,1)$ and so on and then we are claiming that

$$\sum \limits_{k=1}^{\infty} \mathscr{l}(I_{\sigma (k)})=\sum \limits_{k=1}^{\infty} \sum \limits_{j=1}^{\infty} \mathscr{l}(I_{j,k})$$

However I do not see how to prove this. I am not quite sure how such an rearrangement is possible. Hints will be appreciated in direction to prove this.

答:
The manipulation being done here is an application of the following:

Claim: If $a_{j,k}\ge 0$ for every $j,k=1,2,\ldots$, then $$\sum_{k=1}^\infty\sum_{j=1}^\infty a_{j,k}=\sum_{i=2}^\infty\sum_{j+k=i}a_{j,k}.$$

where $\sum_{j+k=i}a_{j,k}$ is shorthand for $a_{1,i-1}+a_{2,i-2}+\cdots+a_{i-1, 1}$.

One way to prove this rigorously is to first argue for each $j$, $k$:

$$a_{j,k}=\sum_{i=2}^\infty a_{j,k}[j+k=i]\tag1$$ where the Iverson bracket $[j+k=i]$ equals $1$ when $j+k=i$, and zero otherwise. (The reason why (2) is true is that every summand is zero except when $i=j+k$.) Now you can write
$$\sum_{k=1}^\infty\sum_{j=1}^\infty a_{j,k}=\sum_{k=1}^\infty\sum_{j=1}^\infty \sum_{i=2}^\infty a_{j,k}[j+k=i].\tag2$$
Since the summands on the RHS of (3) are all non-negative, it is permissible to interchange the order of summation. Do this repeatedly to obtain
$$\sum_{i=2}^\infty\left(\sum_{j=1}^\infty\sum_{k=1}^\infty a_{j,k}[j+k=i]\right)\tag 3.
$$
Finally argue that for each $i$ the quantity in parentheses reduces to a finite double sum over $j$ and $k$ which can be stopped at $i$: $$\sum_{j=1}^i\sum_{k=1}^i a_{j,k}[j+k=i]\tag4$$ This last is equal to $\sum_{j+k=i} a_{j,k}$, by simple enumeration.

(The claim and proof are a bit tidier if we index $j$ and $k$ starting from zero.)


问:
Subdifferential of convex function.
Let $X$ be a Banach space.

For continuous convex function $f:X \longrightarrow \mathbb{R}$. The linear functional $\phi\in X^*$ is called subgradient of function $f$ at a point $x\in X$ if
$$f(y)-f(x) \geq Re\ \phi (y-x) \ \ \forall\ y\in X. $$
The set  $\partial f(x)$  of all subgradients of $f$ at $x$ is called the subdifferential of $f$ at $x$.

Now we can easily prove that $\partial f(x)$ is a closed and convex subset of $X^*.$

But how can we prove that $\partial f(x)$ is a bounded subset of $X^*?$

答:
This is a little bit more difficult. However, since the continuity of $f$ is given, it is not too hard.

The main idea is:

Prove that $f$ is bounded from above and from below on balls $B_R(x)$ via continuity.

Via convexity, it follows that $f$ is Lipschitz on $B_r(x)$ for $r \in (0,R)$.

Conclude.

Title: Subdifferential of convex function.


答:
If $n$ is divisible by $4$, then $n$ is divisible $2$, and the second part of your hypothesis is redundant.

Hence the implication reduces to:

If $n$ is divisible by $4$, then $n$ is divisible by $8$, which is not true since we only need two $2$'s in the prime expansion of $n$ to satisfy the hypothesis.

Title: If n is divisible by 4 and 2 then why it is not divisible by 8?

Body:
Well I'm working on some mathematics aptitude problem on divisibility. There are certain divisibility rules for numbers 6, 12, 15 etc. which are

If n is divisible by 2 and 3, it is divisible by 6

If n is divisible by 3 and 4, it is divisible by 12

If n is divisible by 3 and 5, it is divisible by 15

So, I applied the above rule on 8 and thought

If n is divisible by 2 and 4, it is divisible by 8

but I found a counterexample 576484 to my intuition and it bust the bubble of my figment.
It is divisible by 2 and as well by 4, but not by 8.

Why then it is not divisible by 8? If it is simultaneously divisible by 2 and 4.

Doesn't the rule applies in this case as well? Just like they are applied in the cases of 6, 12, & 15. That is, if factors 'a' and 'b' of number 'n' divides a given number 'm', then 'n' also divides 'm'.


问:
Why did James Stewart regard it "more elegant" if you define the logarithm as an integral, and the exponential function its inverse?
Stewart didn't explain why the "latter method" is "less intuitive", but I think this is because you can define the exponential function first, without integrals, that he does starting on p. 391. And integrals aren't intuitive. Correct?

Personally I like intuition. I just can't imagine how something can be "less intuitive but more elegant". If it's less intuitive, then it must less elegant!

I read What do we mean by an "Elegant Proof"?, but English is not my first language. Perhaps I don't understand the meaning of "elegant here".

There are two possible ways of treating the exponential and logarithmic functions and each
method has its passionate advocates. Because one often finds advocates of both approaches
teaching the same course, I include full treatments of both methods. In Sections 6.2, 6.3,
and 6.4 the exponential function is defined first, followed by the logarithmic function as
its inverse. (Students have seen these functions introduced this way since high school.) In
the alternative approach, presented in Sections 6.2*, 6.3*, and 6.4*, the logarithm is
defined as an integral and the exponential function is its inverse. This latter method is, of
course, less intuitive but more elegant. You can use whichever treatment you prefer.

James Stewart, Calculus 7th ed. 2011. Not the Early Transcendentals version. Page xiv from Preface.

答:
I'll give you a short overview of the process starting from both points:

$\ln$ first:

Define $\ln x:=\int_1^x\frac1t\mathrm dt$. Now the logarithm rules follow from the substitution rule of integrals (easy). Its differentiability and derivative follow from the fundamental theorem of calculus (reasonably easy). The logarithm is obviously monotonous, so it's injective. It's also surjective, since it's continuous by the fundamental theorem of calculus and unbounded above by comparison with the harmonic series and unbounded below by $\ln(1/x)=-\ln x$ (all reasonably easy), so its inverse $\exp$ exists on the reals. According to the inverse function rule, the inverse is differentiable (since $\ln$ is differentiable by the ftc) and has itself as its own derivative (easy). The rules for exponentiation follow from the logarithm rules (easy).

$\exp$ first:

First we must choose a definition. Shall it be the solution to the IVP $f'=f,~f(0)=1$? Then we must prove existence and uniqueness, which is guaranteed by Picard-Lindelöf (nontrivial theorem). But at least we have the rule for the derivative out of the way. Alternatively, define it as a power series. Now we have to prove that power series are differentiable, and to find the derivative we need to show that we can differentiate power series by differentiating each term individually (kinda non-trivial). Or we define it as $\exp x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$. Now we also need to prove that it's differentiable, which not really hard but messy (you have to rearrange some not-so-nice sums with the binomial theorem). The rest is again reasonably easy (exponential rules, existence and differentiability of the inverse, etc.).

So essentially it boils down to the fact that proving the definitions of $\exp$ to yield sensible results is harder than to do the same for the definition of $\ln$.


问:
$A\neq \{0\}$ is a unit commutative ring. If $a\in A-\{0\}$ is such that for all $b\in A$, $a\ast b=0$ or $a\ast b=a$. Then $(1+a)$ is prime ideal
I have this problem for homework

Let $(A, +, \ast)$ be a ring with unit such that $A \neq \{0\}$. Suppose that each element of A is idempotent. Also, suppose that there exists an element $a \in A-\{0\}$ such that for all $b \in A$ we have $a \ast b = 0$ or $a \ast b = a$. Prove that $(1+a)$, i.e. the ideal generated by $1 + a$, is a prime ideal of A.

I tried to use this proposition:

Let $(A, +, \ast)$ be a commutative ring with unit and $A \neq \{0\}$. Then an ideal P is prime if and only if $A/P$ is an itegral domain.

I proved that $(A, +, \ast)$ is commutative. Let $a,b \in A$ be arbitrary, then since every element of A is idempotent, we have $[a+1] = [a+1]^2 = a^2 + 1^2 + 2a = a + 1 + 2a$, whence $2a = 0$. Also $[a+b] = [a+b]^2 = a^2 + b \ast a + a \ast b + b^2 = a + b \ast a + a \ast b + b$, whence $-b \ast a = a \ast b$ and finally, adding $a \ast b$ on both sides,
\begin{align*}
a \ast b - b \ast a &= 2[a \ast b] \\ a \ast b - b \ast a &= 0 \\ a \ast b &= b \ast a.
\end{align*}
Thus, A is commutative and we can apply the stated proposition. So I have to prove that $A/(1+a)$ is and integral domain. To do this I have to prove that $A/(1+a) \neq \{0 + (1+a)\}$ and that it doesn't have cero divisors. The first one is easy, but I have tried the second one and I couldn't prove it.
Please, any suggestion to finish this prove or another way to prove the problem is welcome. Thanks in advance.

答: Your proof of commutativity looks good to me, so let's show that $P:=(1+a)$ is prime. Since maximal ideals are always prime, it suffices to show that $P$ is maximal, so let $c\in A\setminus P$; we wish to show $1\in P+(c)$. By hypothesis, we know that either $ac=0$ or $ac=a$. In the first case, we have $c=(1+a)c$, meaning $c\in P$, a contradiction, and thus we must have $ac=a$. On the other hand, as you have shown, $A$ has characteristic $2$, so $a+a=c+c=0$. In particular, we have $$1=1+a+c+a+c=(1+a)(1+c)+c\in P+(c),$$ as desired. Thus $P$ is maximal and is hence in particular a prime ideal. (As a further exercise, try proving that every prime ideal in $A$ is in fact maximal.)


问:
Factorial sign before a number
Is there a standard definition for $!n$ like $n!$?

The definition of $n!,n!!,n!!!,\ldots$ is very is easy to find. but I can't find anything about $!n$.
for example, $$5=1\cdot2\cdot3\cdot4\cdot5\\9!!=1\cdot3\cdot5\cdot7\cdot9$$
but, does it mean? $$!5=\text{?}\\!3=\text{?}\\ \vdots$$ can someone help me, or show me a reference? Thanks in advance.

答:
Notation of a preceding exclamation point can represent the subfactorial (a.k.a. derangement number).  In this usage,
\begin{align*}
!0 &= 1  \\
!1 &= 0  \\
!n &= (n-1)(!(n-1) + !(n-2)), n > 1  \text{.}
\end{align*}

Title: Factorial sign before a number


问:
Double limit and iterated limit
I was hanging on the following statement.

"If the double limit $\displaystyle \lim_{(x_1, x_2) \rightarrow (a_1, a_2)} f(x_1, x_2)$ and the iterated limit $\lim_{x_1 \rightarrow a_1} \lim_{x_2 \rightarrow a_2} f(x_1, x_2)$ exist, then they will be equal."

Towards this statement I was thinking the function $x \sin \tfrac{1}{y} + y \sin \tfrac{1}{x}$. But for this function, the iterated limit does not exist.

Can you give a counterexample of the statement.

答: As it is in John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 184, the existence of double limit and either of the two iterated limits, finite or infinite, implies the equality of double and that iterated.</p>

Title: Double limit and iterated limit


问:
Does the series $\sum$$(n!)^2/(2n)!$ diverge or converge?
Does the series: $$\sum \frac{(n!)^2}{(2n)!}$$ converge or diverge?

I used the ratio test, and got an end result as $\lim_{n\to\infty}$ $\frac{n+1}{2}$ which would make it divergent but i know it's convergent. Am i using the right test?

答:
Alternatively, consider that for $n\geq2$ we have
$$\frac{(2n)!}{(n!)^2}={2n\choose n}\geq{2n\choose2}=\frac{(2n)(2n-1)}{2}>n^2$$


答:
Suppose $\lim_{x \to \alpha} f(x) = L$ then for any $\epsilon>0$ there is some $\delta>0$ such that if $x$ satisfies $0< \|x-\alpha\| < \delta$ then
$|f(x)-L| < \epsilon$.

Suppose $x_1$ satisfies $0 < |x_1 - \alpha_1 | < {1 \over \sqrt{2}} \delta$, then for $0 < |x_2 - \alpha_2 | < {1 \over \sqrt{2}} \delta$ we have
$0< \|x-\alpha\| < \delta$ (with $x=(x_1,x_2)$) and so $|f(x)-L| < \epsilon$.

In particular, the two limits are equal.

Title: Double limit and iterated limit

Body:
I was hanging on the following statement.

"If the double limit $\displaystyle \lim_{(x_1, x_2) \rightarrow (a_1, a_2)} f(x_1, x_2)$ and the iterated limit $\lim_{x_1 \rightarrow a_1} \lim_{x_2 \rightarrow a_2} f(x_1, x_2)$ exist, then they will be equal."

Towards this statement I was thinking the function $x \sin \tfrac{1}{y} + y \sin \tfrac{1}{x}$. But for this function, the iterated limit does not exist.

Can you give a counterexample of the statement.


问:
Counting the number of zeros yielded by modulo arithmetic functions
I have 4 functions

\begin{align}
f_1(x) &= x \text{ mod } 1\\
f_2(x) &= x \text{ mod } 3 \\
f_3(x) &= x \text{ mod } 4 \\
f_4(x) &= x \text{ mod } 12\\
\end{align}

for $x \in [0,\infty) $

I want to create a function $g(x)$ that counts the number of zeroes at $x$ for those four functions.

For example, $x=6$ would have

\begin{align}
f_1(6) &= 0\\
f_2(6) &= 0 \\
f_3(6) &= 2 \\
f_4(6)& = 6 \\
\end{align}

Since only two of these are zero, then $g(6) = 2$

Another example is $x=24$

\begin{align}
f_1(24) &= 0\\
f_2(24) &= 0 \\
f_3(24) &= 0 \\
f_4(24)& = 0\\
\end{align}

Since all four of these are 0, then $g(24) = 4$

If $x =0.5$, then

\begin{align}
f_1(0.5) &= 0.5 \\
f_2(0.5)& = 0.5 \\
f_3(0.5) &= 0.5\\
f_4(0.5) &= 0.5 \\
\end{align}

Since none of these are zero, then $g(0.5)=0$

How can I define $g(x)$ elegantly?

Inelegant method I've found

One inelegant way I thought of was basically count them for each possible value.

\begin{align}
g(0)& = 4 \\
g(1) &= 1 \\
g(2)& = 1\\
g(3)& = 2 \\
g(4)& = 2 \\
g(5)& = 1\\
g(6)& = 2 \\
g(7)& = 1\\
g(8)& = 2\\
g(9) &= 2 \\
g(10) &= 1\\
g(11)& = 1\\
\end{align}

Then for any $x$, I could do $g(x \text{ mod } 12)$ and it would either match one of these or else $g(x) = 0$.

This gives me the correct result. However, this seems very inelegant and I'm hoping for a nicer solution.

答:
After some reflexion, the best answer I came up with was

$$g(x) = \sum_{i=1}^{4} \mathbb{I}_{\lbrace f_i(x)=0\rbrace}$$

where $\mathbb{I}$ is the indicator function.

I still feel like there must be a more elegant way to do this, but this one at least is simple.


答:
Let $\mathfrak p := (1+a)$, and take $x \in A$.

If $a*x=0$, then $x=(1+a)*x \in \mathfrak p$.

If $a*x=a$, then $x+1 = (1+a)*(x+1+a) \in \mathfrak p$.

Thus we have $x+\mathfrak p=\mathfrak p$ or $x+\mathfrak p=1+\mathfrak p$, and since $x$ was arbitrary, $A/\mathfrak p = \{\mathfrak p,1+\mathfrak p\} \cong \mathbb Z/2\mathbb Z$, meaning that $A/\mathfrak p$ is an integral domain (in fact, a field).

Title: $A\neq \{0\}$ is a unit commutative ring. If $a\in A-\{0\}$ is such that for all $b\in A$, $a\ast b=0$ or $a\ast b=a$. Then $(1+a)$ is prime ideal

Body:
I have this problem for homework

Let $(A, +, \ast)$ be a ring with unit such that $A \neq \{0\}$. Suppose that each element of A is idempotent. Also, suppose that there exists an element $a \in A-\{0\}$ such that for all $b \in A$ we have $a \ast b = 0$ or $a \ast b = a$. Prove that $(1+a)$, i.e. the ideal generated by $1 + a$, is a prime ideal of A.

I tried to use this proposition:

Let $(A, +, \ast)$ be a commutative ring with unit and $A \neq \{0\}$. Then an ideal P is prime if and only if $A/P$ is an itegral domain.

I proved that $(A, +, \ast)$ is commutative. Let $a,b \in A$ be arbitrary, then since every element of A is idempotent, we have $[a+1] = [a+1]^2 = a^2 + 1^2 + 2a = a + 1 + 2a$, whence $2a = 0$. Also $[a+b] = [a+b]^2 = a^2 + b \ast a + a \ast b + b^2 = a + b \ast a + a \ast b + b$, whence $-b \ast a = a \ast b$ and finally, adding $a \ast b$ on both sides,
\begin{align*}
a \ast b - b \ast a &= 2[a \ast b] \\ a \ast b - b \ast a &= 0 \\ a \ast b &= b \ast a.
\end{align*}
Thus, A is commutative and we can apply the stated proposition. So I have to prove that $A/(1+a)$ is and integral domain. To do this I have to prove that $A/(1+a) \neq \{0 + (1+a)\}$ and that it doesn't have cero divisors. The first one is easy, but I have tried the second one and I couldn't prove it.
Please, any suggestion to finish this prove or another way to prove the problem is welcome. Thanks in advance.


问:
precise definition for multiplicity
Multiplicity in graphs is the number of edges between two distinct vertices. But I'm not sure of the precise definition.

From what I've read, we need to consider directed edges for digraphs. for example, if there are two directed edges from a to b, then the multiplicity of (a,b) is 2. but what if there is another edge in the other direction i.e., (b,a), is the multiplicity of (a,b) still 2 or will it be 3?

How is the number of edges considered for undirected graphs? It would be helpful if you could specify using sets.

答:
For an undirected graph $G = (V, E)$, the multiplicity of a vertex $v \in V$ is the number of edges incident to $v$, i.e.
$$d(v) := \#\{ e \in E \; | \; v \in e \}$$

For a directed graph $D = (W, A)$, we have the in-degree
$$d^-(w) := \#\{a \in A \; | \; \exists x \in V. a = (x, v)\}$$
which is the number of arrows going into $w$, and the out-degree
$$d^+(w) := \#\{a \in A \; | \; \exists x \in V. a = (v, x)\}$$
which is the number of arrows leading out of $w$.

These notions are standard.  Sometimes people also define the "degree" of a vertex in a directed graph to be
$$d(w) := d^+(w) + d^-(w),$$
the total number of arrows incident to $w$, but this is less common.

EDIT: The question is about multiplicities, not degrees, so we should consider graphs where $E$ or $A$ is a multiset and then restrict the above definitions to the edges or arrows between two given vertices.

Title: precise definition for multiplicity


问:
Value of series, Partialsum?
given is the following series

$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$

And I need to find its value.

How can I start finding it?

Thanks for all

does the Telescop-Summing work here as well?:

$\sum_{n=1}^\infty \frac{1}{4n^2-1} $
now:
$\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)} = \frac{1}{2} * ( \frac{1}{2n-1} - \frac{1}{2n+1})$

Now I have to "add the sum":
$\sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{2}* [ \sum_{n=1}^\infty \frac{1}{2n-1} -
\sum_{n=1}^\infty \frac{1}{2n+1}] =  \frac{1}{2} - \frac{1}{4n+2} $ And than for $n \to \infty$ it is $\frac{1}{2}$ ??

答:
split it up the numerator to 2n + 2 - 1 then factor to 2(n+1) -1.
By spliting the fraction you can cancel terms.
This should allow you to see this is a finite result.</p>

Title: Value of series, Partialsum?


问:
Want to verify my approach

If repetition of digit is not allowed, then how many five digit numbers which are divisible by $3$ can be formed
using the digits?

My approach is that i made 3 sets of digits according to the remainder they give on dividing by $3$ -
$ A:(0,3,6,9)$ , $B:(1,4,7)$ , $C:(2,5,8)$, Now the following cases are possible:

Any $1$ digit from $A$ $+$ Any $2$ digit from $B$ $+$ Any $2$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 2} \cdot {3 \choose 2}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 1} \cdot {3 \choose 2} \cdot {3 \choose 2}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$

Any $2$ digit from $A$ $+$ Any $3$ digit from $B$ $+$ Any $0$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1\cdot {3 \choose 1} \cdot {3 \choose 3} \cdot {3 \choose 0}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 2} \cdot {3 \choose 3}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$

Any $2$ digit from $A$ $+$ Any $0$ digit from $B$ $+$ Any $3$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 1}\cdot {3 \choose 0} \cdot {3 \choose 3}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 2} \cdot {3 \choose 3}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$

Any $3$ digit from $A$ $+$ Any $1$ digit from $B$ $+$ Any $1$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 2}\cdot {3 \choose 1} \cdot {3 \choose 1}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 3} \cdot {3 \choose 1}\cdot {3 \choose 1}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$.

Total $= 9072$

pls verify my approach and answer, other methods will also be appreciated ,Thanks.

答:
The count can be simplified using probability  as under:

Without zeroes, there will be $\binom9 5 = 126$ digit strings, and the same number with zeroes, at$\,\binom9 4 = 126$

$\frac 1 3$ of each, i.e. $42$ will be divisible by $3$, but since zeroes can only be allowed at $4$ of the $5$ digits, the multiplication factor while combining them won't be $2$ but $1.8$

Thus answer $= 42\times1.8\times5! = 9072\,$, which confirms your answer

Title: Want to verify my approach


问:
Show this integral equation using Itô's formula
Let $B_t$ be any standard Brownian motion. I want to prove the following equation which is stated in my lecture notes to follow from Itô's formula.
$$tB_t = \int_0^t sdB_s + \int_0^t B_s ds $$
I don't understand how the formula was applied here; in my opinion this looks more like integration by parts...

Any help would be appreciated.

答:
This is a direct application of Ito's rule. Recall that the Ito's rule tells you
\begin{align}
d(f(X_t,t)) = f_t(X_t,t)dt+f_x(X_t,t)dX_t+\frac{1}{2}f_{xx}(X_t,t)d\langle X_t\rangle
\end{align}
or, in the integral form,
\begin{align}
f(X_t,t) = \int_0^t f_s(X_s,s)ds+\int_0^tf_x(X_s,s)dX_s+\frac{1}{2}\int_0^tf_{xx}(X_s,s)d\langle X_s\rangle
\end{align}
where $\langle X_s\rangle$ is the quadratic variation of $X_t$. For more about quadratic variation see this excellent post and the references therein link.

In your case, $X_t=B_t$, $f(X_t,t)=tB_t$ with $f_t(X_t,t)=B_t$, $f_x(X_t,t)=t$, and $f_{xx}(X_t,t)=0$. Use these in the integral form above to obtain the desired formula.


问:
Prove the difference of roots is less than or equal to the root of the difference
I am doing a larger proof that requires this to be true:

$|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|}$

I can square both sides to get

$a - 2\sqrt{a}\sqrt{b} + b \leq |a - b|$

Note that a and b are > 0.

I also know that

$|c| - |d| \leq |c - d|$

It seems like a mistake that $+b$ is there on the left...

How can I prove this is true

答:
Late answer, but we can prove a stronger inequality. Let $a,b\in\mathbb{R}$ and assume, without loss of generality, $|a|\geq |b|$.

$$\begin{align}\big|\sqrt{|a|}-\sqrt{|b|}\big|\leq \sqrt{\big||a|-|b|\big|}&\Longleftrightarrow |a|-2\sqrt{|a||b|}+|b|\leq \big||a|-|b|\big|\\&\Longleftrightarrow |a|-2\sqrt{|a||b|}+|b|\leq |a|-|b|\\&\Longleftrightarrow|b|\leq\sqrt{|a||b|}\\&\Longleftrightarrow|b|^2\leq |a||b|\\&\Longleftrightarrow0\leq |b|(|a|-|b|)\end{align} $$

The third and final inequalities are true since $\big||a|-|b|\big|=|a|-|b|$ if and only if  $|a|-|b|\geq 0$ if and only if $|a|\geq |b|$, as assumed.

Recall that $\big||a|-|b|\big|\leq|a-b|$ so $\sqrt{\big||a|-|b|\big|}\leq \sqrt{|a-b|}$, and finally, $$\big|\sqrt{|a|}-\sqrt{|b|}\big|\leq\sqrt{\big||a|-|b|\big|}\leq \sqrt{|a-b|}$$


问:
Identities with quadratic forms
Let $y_1,\dots, y_n$ be vectors in $\mathbb{R}^p$ and let $\mu$ be an additional vector in $\mathbb{R}^p$. Let $M\in\mathbb{R}^{p\times p}$ be a symmetric positive definite matrix. I have some difficulties into prove the following identity
\begin{equation}\sum_{i=1}^n (y_i-\mu)'M^{-1}(y_i-\mu)=n(\bar{y}-\mu)'M^{-1}(\bar{y}-\mu)+\sum_{i=1}^n (y_i-\bar{y})'M^{-1}(y_i-\bar{y})\end{equation}
where
\begin{equation}\bar{y}\triangleq \frac{1}{n}\sum_{i=1}^n y_i\end{equation}
I've tried to expand the generic product $(y_i-\mu)'M^{-1}(y_i-\mu)$ but this seems to me useless to prove the previous identity.

答:
\begin{align}
& (y_i-\mu)'M^{-1}(y_i-\mu) \\[10pt]
= {} & \big((y_i-\overline y) +(\overline y - \mu)\big)' M^{-1} \big((y_i-\overline y) +(\overline y - \mu)\big) \\[10pt]
= {} & (y_i-\overline y)'M^{-1} (y_i-\overline y) + (y_i-\overline y)'M^{-1}(\overline y - \mu) \tag 1 \\[4pt]
& {} + (\overline y - \mu)' M^{-1}(y_i - \overline y) + (\overline y - \mu)'M^{-1} (\overline y - \mu) \tag 2
\end{align}

Consider the second term on line $(1)$: $\quad (y_i-\overline y)'M^{-1} (\overline y-\mu).$ As $i$ goes from $1$ to $n,$ the factor $M^{-1}(\overline y-\mu)$ does not change, so
$$
\sum_{i=1}^n \Big( (y_i-\overline y)'M^{-1} (\overline y-\mu) \Big) = \left( \sum_{i=1}^n (y_i-\overline y)' \right) M^{-1}(\overline y-\mu).
$$
This last sum from $i=1$ to $n$ evaluates to $0.$ The same thing applies to the first of the two terms on line $(2).$

Now look at $(\overline y-\mu)'M^{-1}(\overline y-\mu).$ As $i$ goes from $1$ to $n,$ this term does not change, so the sum is $n(\overline y-\mu)'M^{-1}(\overline y-\mu).$

Title: Identities with quadratic forms


答:
You can take $g(t,x)=t.x$
$$g(t,x)=t.x\\\frac{\partial g}{\partial t}=x,\frac{\partial g}{\partial x}=t,\frac{\partial^2 g}{\partial x^2}=0\\
dg=\frac{\partial g}{\partial t}dt+\frac{\partial g}{\partial x}dx+\frac12\frac{\partial^2 g}{\partial x^2}(dx)^2$$so
$$dg=xdt+tdB_t+\frac12 (0)(dB_t)^2$$left hand side is complete differential , so take both sides integration
$$\int dg=\int xdt+tdB_t\\
\int d(g)=t.x$$ finally put $x\to B_t$

Title: Show this integral equation using Itô's formula

Body:
Let $B_t$ be any standard Brownian motion. I want to prove the following equation which is stated in my lecture notes to follow from Itô's formula.
$$tB_t = \int_0^t sdB_s + \int_0^t B_s ds $$
I don't understand how the formula was applied here; in my opinion this looks more like integration by parts...

Any help would be appreciated.


问:
Question on basis change and endomorphism
This is a question given in the book Mathematics in Machine Learning.

I know how to solve the first part a. of the question, finding the kernel and image of the transformation matrix.
But I don't understand the second part of the question.

I'm not sure what the fact that it's an endomorphism tells me. I understand that endomorphism means that it is mapping to the same vector space and that it is linear. Does that mean it's mapping to a vector subspace of this vector space? (since it's not bijective) But the transformation matrix seems to say otherwise.

I don't understand what the "(with respect to standard basis in R^3)" part means. Does it mean, the 2 basis sets we are mapping are wrt. to the standard basis, or that we are mapping the standard basis to a new basis? Or something else?

Finally I don't understand what the b. part is asking me to do. I don't understand the "with respect to the basis B" part either.

I've written the formation of a transformation matrix step by step, but I don't really understand what the question is even saying to infer anything.

I do know there is a formula for finding the transformation matrix when a basis is changed,

A' = (T^−1)(A)(S) (T and S being the transformation matrices to map old basis with new).

But I don't know how I would apply it for this question.

答:
I'm not sure what the fact that it's an endomorphism tells me. I understand that endomorphism means that it is mapping to the same vector space and that it is linear. Does that mean it's mapping to a vector subspace of this vector space? (since it's not bijective) But the transformation matrix seems to say otherwise.

But in this example, $\Phi$ is bijective (since the rank is 3, and the kernel is $0$). It's exactly as you said: "endomorphism" just means a linear map from a vector space to itself (there is no assumption on whether or not it is bijective...it might be, but it might not).

I don't understand what the "(with respect to standard basis in R^3)" part means. Does it mean, the 2 basis sets we are mapping are wrt. to the standard basis, or that we are mapping the standard basis to a new basis? Or something else?

That's right: it means both the domain and range are written in the standard basis. If we call them
$$e_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}, \quad e_2 = \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \quad e_3 = \begin{pmatrix}0\\0\\1\end{pmatrix} $$
then saying $A_\Phi$ represents $\Phi$ in the standard basis means
$$ \begin {align*} \Phi(e_1) &= e_1+e_2+e_3 \\ \Phi(e_2) &= e_1-e_2+e_3 \\ \Phi(e_3) &= e_3 \end{align*} $$
This is just obtained from looking at the columns of $A_\Phi$ (the coefficients of $e_1,e_2,e_3$ on the right-hand sides are the columns of the matrix).

Finally I don't understand what the b. part is asking me to do. I don't understand the "with respect to the basis B" part either.

Now, if we write the new basis as
$$ v_1 = \begin{pmatrix}1\\1\\1\end{pmatrix}, \quad v_2 = \begin{pmatrix}1\\2\\1 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 1\\0\\0\end{pmatrix} $$
Then the new matrix $\widetilde{A}_\Phi$ for $\Phi$ in the new basis is
$$ \widetilde{A}_\Phi = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$
where these coefficients satisfy
$$ \begin{align*}
\Phi(v_1) &= a_{11}v_1 + a_{21}v_2 + a_{31}v_3 \\
\Phi(v_2) &= a_{12}v_1 + a_{22}v_2 + a_{32} v_3 \\
\Phi(v_3) &= a_{13}v_1 + a_{23}v_2 + a_{33}v_3
\end {align*}
$$

So part $(b)$ is asking you to find these coefficients $a_{ij}$ of this matrix $\widetilde{A}_\Phi$.

The formula you gave is the correct one: $\widetilde{A}_\Phi = T^{-1} A_\Phi T$, where $T$ is the matrix whose columns are $v_1,v_2,v_3$:
$$ T = \begin{pmatrix} 1&1&1\\1&2&0\\1&1&0 \end{pmatrix} $$


问:
Automorphisms of the Petersen graph
I am trying to find out the automorphism group of the Petersen graph. My book carries the hint: "Show that the $\tbinom{5}{2}$ pairs from {1, . . . , 5} can be used to label the vertices in such a way that a simple rule determines when there is an edge. To find the full automorphism group, consider the subgroup that fixes a vertex and its three neighbors."

Accordingly the rule is that there is an edge if 2-sets are disjoint. What I am not getting is to how to use the second sentence of the hint to find the automorphism group. Its not at all clear what the order of such a subgroup will be and how such a subgroup will be the requisite automorphism group.

Also I am not really proficient with group action, so if that is involved I would appreciate a detailed explanation.

答: Prof. Ravi Vakil has given lecture notes on Complex Algebraic Surfaces, and you can find the proof of your result in the second page.</p>

Title: Automorphisms of the Petersen graph


Title: Value of series, Partialsum?

Body:
given is the following series

$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$

And I need to find its value.

How can I start finding it?

Thanks for all

does the Telescop-Summing work here as well?:

$\sum_{n=1}^\infty \frac{1}{4n^2-1} $
now:
$\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)} = \frac{1}{2} * ( \frac{1}{2n-1} - \frac{1}{2n+1})$

Now I have to "add the sum":
$\sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{2}* [ \sum_{n=1}^\infty \frac{1}{2n-1} -
\sum_{n=1}^\infty \frac{1}{2n+1}] =  \frac{1}{2} - \frac{1}{4n+2} $ And than for $n \to \infty$ it is $\frac{1}{2}$ ??

Body:
HINT:
$$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac1{n^2}-\frac1{(n+1)^2}$$

Can you recognize Telescoping Sum?


问:
Why do you set a hyperbola equation to 0 to find the equation of the asymptotes?
I saw this solution to finding the equations of the asymptotes of the hyperbola of form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. To find the asymptotes, you let the right hand side equal 0, then rearrange to get your equation. Like this: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 0$, then $\frac{x^2}{a^2}  =  \frac{y^2}{b^2}$, then $\pm\frac xa = \frac yb$, and thus $ y = \pm \frac bax$. I do not understand why you set it equal to $0$, though. I get that the asymptotes shouldn't be part of the graph, so we set it to zero to find the points not on the hyperbola, but why set it to 0?

答:
An asymptote denotes a tendency, when a variable goes to infinity.

When $x$ or $y$ tends to infinity the finite terms contribute  only insignificantly . Dividing by $y^2$

$$ \frac{x^2}{a^2 y^2} - \frac{1}{b^2} = \frac{1}{y^2} $$

In order to trace to what curve the graph tends to, we need to a priori set the small finite fraction ... including the increasing denominator ...  to zero.


问:
Prove this functional inequality
Given $f: R \to (0,\infty]$, twice differentiable with $f(0) = 0, f'(0) = 1, 1 + f = 1/f''$

Prove that $f(1) < 3/2$

I found some useless (I believe) facts about $f$, but nothing that gives me a light to answer it.

$f''(x) > 0$

$f'(x) \ge 1$

$e^{((f')^2-1)/2} = f+1$ is another way to write the differential equation

....
?

Title: Prove this functional inequality

Body:
Using Taylor you get for a $\xi \in (0,1)$:

$$f(1) =f(0) + f'(0) + \frac{f''(\xi)}{2!}=1+\frac 12\cdot \frac 1{1+f(\xi)}<1+\frac 12$$


问:
Derivative of Arctangent with definition of derivative
I can't find the derivative of arctangent with definition of derivative.
Here's my way:
$\displaystyle\lim_{\Delta x\to\ 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$ is a definition of derivative with limit. If we define $f(x)=\arctan(x)$, then we get:
$$\left[\lim_{\Delta x\to\ 0} \frac{\arctan(x+\Delta x)-\arctan(x)}{\Delta x}\right]=\left[\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x(x+\Delta x)}\right)}{\Delta x}\right].$$ But I don't know how to continue. Thanks.

答:
Switch to the straight function, find derivative and notice that

$$\dfrac{dy}{dx}=\dfrac{1}{dx/dy}=$$

So
$$x= \tan(y)$$

In the way you have obtained d.c. you can also find:

$$ \dfrac{dx}{dy} = \sec^2 y ={1+x^2}$$

$$ \dfrac{dx}{dy} = \dfrac{1}{\sec^2}  =\dfrac{1}{1+x^2}.$$


问:
Coin betting problem.
A coin bet is placed between two friends such that the person who wins four tosses first is the winner. What is total number of possible ways in which the bet can play out?</p>

答: I believe this is just 7 choose 4, or $\frac{7!}{4!3!}=35$. As Greg said, depending on exactly what you mean by "different ways the bet can play out", you might want to double that.

Title: Coin betting problem.


问:
How do solve this integral $\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\arctan\frac{11-6\,x}{4\,\sqrt{21}}\mathrm dx$?
I need to solve the to following integral:
$$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\arctan\frac{11-6\,x}{4\,\sqrt{21}}\mathrm dx.$$

I tried this integral in Mathematica, but it was not able to solve it. An approximate numeric integration gives $1.6449340668482264364724151666460251892189...$ that is close to $\frac{\pi^2}6$. But when I tried to increase the precision above 60 decimal digits, I began to see a tiny difference, which could be interpreted either as a numerical algorithm glitch, or as $\frac{\pi^2}6$ being just an accidentally close value and not the exact answer. Indeed, $\frac{\pi^2}6$ would be a suspiciously nice result for this integral. Anyway, I need your help with this.

答:
I think the most ecological approach to the problem is as follows:

Denote $x=\sin\varphi$ and recall that $\arctan x=\frac{1}{2i}\ln\frac{1+ix}{1-ix}$. One then obtains the integral
$$\frac{1}{2i}\int_{-\pi/2}^{\pi/2}\ln\frac{4\sqrt{21}+i(11-6\sin\varphi)}{4\sqrt{21}-i(11-6\sin\varphi)}d\varphi=\frac{1}{4i}\int_{0}^{2\pi}\ln\frac{4\sqrt{21}+i(11-6\cos\varphi)}{4\sqrt{21}-i(11-6\cos\varphi)}d\varphi,\tag{1}$$
where at the last step we first used the symmetry of sine function to extend the integration to interval $[-\pi,\pi]$, and then made use of periodicity to shift the integration interval and to replace $\sin$ by $\cos$.

There is a well-known integral (see, for example, here)
$$\int_{0}^{2\pi}\ln\left(1+r^2-2r\cos\varphi\right)d\varphi=\begin{cases}
0, &\text{for}\; |r|<1,\\
2\pi\ln r^2, &\text{for}\; |r|>1.
\end{cases}\tag{2}$$
Obviously, our integral above is a difference of two integrals of this type. Some care should be taken over multivaluedness of logarithms. This can be handled by saying that the arguments of $4\sqrt{21}\pm i(11-6\cos\varphi)$ belong to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Now we have
\begin{align}
4\sqrt{21}\pm i(11-6\cos\varphi)=A_{\pm}\left(1+r_{\pm}^2-2r_{\pm}\cos\varphi\right)
\end{align}
with
$$r_{\pm}=\frac{11-4\sqrt7}{3}e^{\pm i\pi/3},\qquad A_{\pm}=(11+4\sqrt7)e^{\pm i\pi /6}.$$
Since $|r_{\pm}|<1$, the integral (1) reduces to
$$\frac{1}{4i}\cdot2\pi\cdot \ln\frac{A_+}{A_-}=\frac{\pi^2}{6}.$$


答:
$$\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x(x+\Delta x)}\right)}{\Delta x}=\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x^2}\right)}{\Delta x}$$
Now let's introduce the variable $u$ by $$\tan u=\dfrac{\Delta x}{1+x^2}$$
Then $$\Delta x=(1+x^2)\tan u$$and your limit is $$\lim_{u\to\ 0}\frac{\arctan(\tan u)}{(1+x^2)\tan u}=\lim_{u\to\ 0}\frac{u}{(1+x^2)\tan u}=\frac1{1+x^2}\frac1{\lim_{u\to\ 0}\frac{\tan u} u}=\frac1{1+x^2}$$

Title: Derivative of Arctangent with definition of derivative

Body:
I can't find the derivative of arctangent with definition of derivative.
Here's my way:
$\displaystyle\lim_{\Delta x\to\ 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$ is a definition of derivative with limit. If we define $f(x)=\arctan(x)$, then we get:
$$\left[\lim_{\Delta x\to\ 0} \frac{\arctan(x+\Delta x)-\arctan(x)}{\Delta x}\right]=\left[\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x(x+\Delta x)}\right)}{\Delta x}\right].$$ But I don't know how to continue. Thanks.


答: Let $y=\arctan x,\,y+\Delta y=\arctan(x+\Delta x)$ so$$\begin{align}\arctan^\prime x&=\lim_{\Delta x\to0}\frac{\arctan (x+\Delta x)-\arctan x}{\Delta x}\\&=\lim_{\Delta x\to0}\frac{\Delta y}{\tan(y+\Delta y)-\tan y}\\&\stackrel{\star}{=}\lim_{\Delta y\to0}\frac{\Delta y}{\tan(y+\Delta y)-\tan y}\\&\stackrel{\dagger}{=}\frac{1}{1+\tan^2y}\\&=\frac{1}{1+x^2},\end{align}$$where $\star$ uses the continuity of $\arctan x$ and $\dagger$ uses $\frac{d}{dy}\tan y=1+\tan^2y$, which can also be proved from first principles:$$\begin{align}\tan^\prime y&=\lim_{\Delta y\to0}\frac{\tan(y+\Delta y)-\tan y}{\Delta y}\\&=\lim_{\Delta y\to0}\frac{\frac{\tan y+\tan\Delta y}{1-\tan y\tan\Delta y}-\tan y}{\Delta y}\\&=\lim_{\Delta y\to0}\frac{(1+\tan^2y)\tan\Delta y}{\Delta y}\frac{1}{1-\tan y\tan\Delta y}\\&=(1+\tan^2y)\underbrace{\lim_{\Delta y\to0}\frac{\tan\Delta y}{\Delta y}}_1\underbrace{\lim_{\Delta y\to0}\frac{1}{1-\tan y\tan\Delta y}}_1,\end{align}$$where we use $\lim_{h\to0}\frac{\tan h}{h}=1,\,\lim_{h\to0}\tan h=0$.


答:
$$
\lim_{\Delta x\to\ 0} \frac{\arctan(x+\Delta x)-\arctan x}{\Delta x} = \lim_{u\,\to\,v} \frac{u-v}{\tan u - \tan v}
$$
Here we have $u = \arctan(x+\Delta x)$ so $\tan u = x+ \Delta x$ and so on.

Why is it that $u\to v$ as $\Delta x\to 0\text{?}$ In effect this says the arctangent function is continuous. That can only follow from properties of the tangent function on the interval $(-\pi/2,+\pi/2).$

Then
$$
\lim_{u\,\to\,v} \frac{u-v}{\tan u - \tan v} = \frac 1 {\sec^2 v} = \frac 1 {1+\tan^2 v} = \frac 1 {1+x^2.}
$$
Here of course you need to know how to differentiate the tangent function before doing this.


问:
Prove: If $p$ is prime and $p \mid a_1a_2\dots a_n$, then $p \mid a_i$ for $i\in\{1, \dots, n\}$.
Euclid's lemma: If $p\mid ab$, then $p\mid a$ or $p\mid b$.

Prove: If $p$ is prime and $p \mid a_1a_2\dots a_n$, then $p \mid a_i$ for some $i = 1, 2, \dots, n$.

Proof by Induction:

P(1): From $p\in\mathbb{P}$ and $p\mid a_1$ it follows that:
\begin{equation*}
p \mid a_i,
\end{equation*}
for $i = 1$. P(1) is true.

P(2): If $p\in\mathbb{P}$ and $p\mid a_1a_2$ it follows from the lemma that if $p \nmid a_1$, then:
\begin{equation*}
p \mid a_2,
\end{equation*}
so
\begin{equation*}
p \mid a_i
\end{equation*}
for some $i = 1, 2$.

On the other hand, if $p\mid a_1$, then the same conclusion is reached. P(2) is true.

Now assume P(k) inductively:
\begin{equation*}
p \mid a_1a_2\dots a_k \rightarrow p \mid a_i
\end{equation*}
for some $i = 1, 2, \dots, k$.

P(k+1) is also true because from $p \mid (a_1a_2\dots a_k)a_{k+1}$ two cases arise:

Case I: $p\nmid a_1\dots a_k$.

Using Euclid's lemma it follows that if $p\nmid a_1\dots a_k$, then:
\begin{equation*}
p\mid a_{k+1}
\end{equation*}
and thus:
\begin{equation*}
p\mid a_i,
\end{equation*}
for some $i = 1, 2, \dots, k+1$.

Case II: $p\mid a_1\dots a_k$.

By the inductive hypothesis, the conclusion may be reached:
\begin{equation*}
p \mid a_i,
\end{equation*}
for some $i = 1, 2, \dots,k, k+1$.

答:
Here is your argument written more concisely/clearly:

For all arguments for $n=1$ case is trivial.

If $p \mid a_1\cdots a_n \implies p \mid a_{i}$ for some $n$ then suppose $ p \mid a_1\cdots a_na_{n+1}=(a_{1}\cdot\cdot\cdot a_{n})a_{n+1}$

By the $n=2$ case we must have that $p$ divides atleast one of them.

Then if $p \mid (a_1\cdots a_n)$ we are done by the inductive hypothesis, and if $p \nmid (a_1\cdots a_n)$,  $p \mid a_{n+1}$, and the $i$ that satisfies the implication is $n+1$.

Your proof does not use the inductive hypothesis because you are using using an additional $\gcd$ result based on properties of primes. This not necessary as you may use the $n=2$ case of the theorem known as Euclid's lemma. Although, it is a similar result.

The issue is, you are having to note basic results and insights when you could have just proven the $n=2$ case and done a standard induction.

Below are my initial proofs:

One proof is to use strong induction on $n$ and once again Euclid's lemma for the $n=2$ case.

Then $p\mid a_1\cdots a_n \implies p\mid(a_1\cdots a_{n-1})a_n$. If $p \nmid a_n, $then $p\mid(a_1\cdots a_{n-1})$.

You may fill in the details, but by repeating this process of moving the paranthesis inductively, one finds that $p\mid a_i$ for some $i$.

Also, consider the proof where we start with $a_{1}\cdot\cdot\cdot a_{n}=pk$ for some integer $k$.

Then by the Fundamental Theorem of Arithmetic, $p$ must be in the prime expansion of the $LHS$, and hence in the prime expansion of at least one of the $a_{i}$'s.


问:
Relationship between the center and commutator subgroup of a group of nilpotency class 3
I am new to the theory of nilpotent groups. I am dealing with the nilpotent groups of class 3. I want to know is there any relationship between the commutator subgroup $[G, G]$ of the group $G$ and the center $Z(G)$ of the group $G$.  I have a strong belief that $[G, G]^{3}\subseteq Z(G)$ when 3 divides the order of the group $G$. But I could not find any literature for that. Can anyone please help me with this. Thank you, in advance.

答:
Your "strong belief" is false. Consider the $4 \times 4$ Heisenberg group

$$H_4(\mathbb{F}_p) = \left[ \begin{array}{cccc} 1 & \mathbb{F}_p & \mathbb{F}_p & \mathbb{F}_p \\ 0 & 1 & \mathbb{F}_p & \mathbb{F}_p \\ 0 & 0 & 1 & \mathbb{F}_p \\ 0 & 0 & 0 & 1 \end{array} \right]$$

over a prime $p \neq 3$, which is nilpotent of class $3$. Since the order of this group is not divisble by $3$, the map $g \mapsto g^3$ is a bijection on every cyclic subgroup and hence a bijection on every subgroup, so $[G, G]^3 \subseteq Z(G)$ iff $[G, G] \subseteq Z(G)$. The center consists of matrices which have zero entries except in the top right and has order $p$. But the commutator subgroup consists of matrices which have zero entries in the superdiagonal and has order $p^3$.


问:
change to polar coordinates in a ordinary differential equation
I don't know what to do.

write the following system into polar coordinates

$$x'=-ay$$
$$y'=ax$$

with $a\in \mathbb R$

any idea?

答:
$\newcommand{\cost}{\cos\theta} \newcommand{\sint}{\sin\theta}$
Let $(x,y) = (r\cos\theta, r\sin\theta)$. Then,
$$
\begin{align}
x' &= r'\cost - r\theta'\sint = -ar\sint\\
x' &= r'\sint + r\theta'\cos = ar\cost
\end{align}
$$
In a matrix form,

$$
\begin{bmatrix}
\cost & -r\sint\\
\sint & r\cost
\end{bmatrix}
\begin{bmatrix}
r'\\ \theta'
\end{bmatrix}
=
ar
\begin{bmatrix}
-\sint\\
\cost
\end{bmatrix}
$$
I think you know what to do after this.

You will get finally, $r' = 0$ and $\theta' = a$, which means that the solution of the ODE draws a circle of radius $r(0)$ with the angular velocity $a$.


问:
Normal distribution: how to find the mean $µ$ and standard deviation $σ$?
A random variable $X$ is normally distributed with mean $µ$ and standard deviation $σ$.

lf $P(X < 25)$ is $0.1082$ and $P(X> 42)$ is $0.1303$ determine:

(i) $P(25 < X < 42)$

(ii) the values of mean $µ$ and standard deviation $σ$

(iii) $P(X>22)$

For part (i), I have done $P(25 < X < 42) = P(X<42) - P(X<25)$ which gives $(1-0.1303)-0.1082 = 0.7615$.

I'm unsure how to do part (ii). I'm thinking it has something to do with the empirical rule, so $68.3\%$ lies within one standard deviation of the mean, etc. But I'm not sure how to apply this, as there doesn't seem to be any concrete "percent rule" that the $0.7615$ applies to. This is from a high school maths textbook. Any hints would be appreciated.

答:
In normal distribution we use the notation $P(X\le a)=\phi(a)$

$P\bigg(\dfrac{x-\mu}{\sigma} < \dfrac {25-\mu}{\sigma}\bigg)=P\bigg(Z< \dfrac {25-\mu}{\sigma}\bigg)=\phi\bigg(\dfrac{25-\mu}{\sigma}\bigg)=0.1082$

Now look at normal table and find $a$ such that $\phi(a)=0.1082$ then,

$a=\dfrac{25-\mu}{\sigma}$

similarly, you have $P(X>42)=.1303$

So you will get two equations two unknows$(\mu$ and $\sigma)$ solve them you will get $
\mu,\sigma$


问:
Integral of a modified Bessel function of the second kind 1st order
in Gradshteyn-Ryzhik see Bessel functions and exponentials 6.611 (9), there is the formula for
$$\int_0^\infty e^{-\alpha x}K_0(\beta x)\,dx$$
My question is: what about
$$\int_0^\infty e^{-\alpha x}K_1(\beta x)\,dx$$

答:
$$I=\int_0^\infty e^{-\alpha x}K_1(\beta x)\,dx$$ does not converge.

Let $ x=\frac t \beta$ and $k=\frac \alpha \beta$
$$I=\frac 1 \beta\int_0^\infty e^{-k t}K_1(t)\,dt$$ Around $t=0$
$$K_1(t)=\frac{1}{t}+\frac{1}{4} t (2 \log (t)+2 \gamma -1-2 \log (2))+O\left(t^3\right)$$
$$J=\int e^{-kt} \Big[\frac{1}{t}+\frac 12 t\log(t)+c\Big]\,dt$$
$$J=\frac{\left(2 k^2+1\right) \text{Ei}(-k t)-e^{-k t} (2 c k+k t \log (t)+\log (t)+1)}{2 k^2}$$ which tends to $-\infty$ when $t\to 0$.


问:
MULTIPLE EIGEN VECTORS
This is the system that I'm trying to solve:

$
\begin{array}{rrrrrr}
   y'_1=&-2y_1&+5y_2&\\
     y'_2=&-y_1&-6y_2&
 \end{array}
$

$
E_1= \left[
\begin{array}{rrrrrr}
1\\
-(2+j)
 \end{array} \right]
$

$E_2= \left[
\begin{array}{rrrrrr}
1\\
-(2-j)
 \end{array} \right]
$

$\mathbf{y}_h=c_1\mathsf{e}^{(-4+i)t} \left[
\begin{array}{rrrrrr}
1\\
-(2+j)
 \end{array} \right]
 +c_2\mathsf{e}^{(-4-i)t}\left[
\begin{array}{rrrrrr}
1\\
-(2-j)
 \end{array} \right]$

$\mathbf{y}_h= \mathsf{e}^{-4t}\left[
    \begin{array}{rrrrrr}
    c_1 \cos(t)+c_2\sin(t)\\
   - 2c_1 \cos(t)-2c_2\sin(t)+c_2\cos(t)-c_1\sin(t)
     \end{array} \right] $

The book provides the correct answer as

$\mathbf{y}_h= \mathsf{e}^{-4t}\left[
    \begin{array}{rrrrrr}
    c_1 \cos(t)+c_2\sin(t)\\
   \Big(- 2c_1 \cos(t)-2c_2\sin(t)+c_2\cos(t)-c_1\sin(t)\Big)/5
     \end{array} \right] $

答:
The two eigenvectors for $-4+i$ are scalar multiples of each other. Multiply the second one by $2-i$ and you get the first one.

All nonzero scalar multiples of an eigenvector are also eigenvectors with the same eigenvalue.

Title: MULTIPLE EIGEN VECTORS


问:
not understanding a question about IVT

Hello, I do not understand why the first answer is correct : in fact the intermediate value theorem only yields that there exists a $u$ such that $\frac{\pi \sqrt{2}}{8}=f'(u)...$

答:
By the mean value theorem, there is $u \in ]0, \pi/4[$ such that

$$\frac{\pi}{4}\frac{\sqrt{2}}{2}=f( \pi/4)-f(0)=f'(u)(\frac{\pi}{4}-0).$$

Title: not understanding a question about IVT


答:
Hint:

See $\#313$ of The elements of coordinate geometry

From the $\#324$ of the same, the equation to the asymptotes only differs from
any hyperbola by a constant.

So, here the equation of the pair of  asymptotes  here will be $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=K$$ where $K$ is any arbitrary constants

Now use What is condition for second degree equation to represent a pair of straight lines?

$\begin{pmatrix} \dfrac1{a^2} & 0&0\\0&\dfrac1{b^2}&0\\0&0&-K\end{pmatrix}=0\implies \dfrac{-K}{a^2b^2}=0$

Title: Why do you set a hyperbola equation to 0 to find the equation of the asymptotes?

Body: I saw this solution to finding the equations of the asymptotes of the hyperbola of form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. To find the asymptotes, you let the right hand side equal 0, then rearrange to get your equation. Like this: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 0$, then $\frac{x^2}{a^2}  =  \frac{y^2}{b^2}$, then $\pm\frac xa = \frac yb$, and thus $ y = \pm \frac bax$. I do not understand why you set it equal to $0$, though. I get that the asymptotes shouldn't be part of the graph, so we set it to zero to find the points not on the hyperbola, but why set it to 0?


问:
Stuck with a proof regarding cardinality
Problem: For any set $A$, finite or infinite, let $B^{A}$ be the set of all functions mapping $A$ into the set $B=\{0,1\}$. Show that the cardinality of $B^{A}$ is the same as the cardinality of the power set of $A$, $\mathscr{P}(A) .\left[\right.$ Hint: Each element of $B^{A}$ determines a subset of $A$ in a natural way. $]$

Starting with finite sets, the cardinality of $\mathscr{P}(A)$, is  $2^{S}$ where $S$ is the cardinality of $A$. So I need to prove that the cardinality of $B^{A}$ is also $2^{S}$. Now lets look at an example set: $A_3=\{1,2,3\}$.

$B^{A_3}=\{  \{(1,0), (2,0), (3,0)\},\\ \{(1,0), (2,0), (3,1)\}, \\ \{(1,0), (2,1), (3,0)\},\\\{(1,1), (2,0), (3,0)\},\\ \{(1,0), (2,1), (3,1)\},\\ \{(1,1), (2,1), (3,0)\},\\\{(1,1), (2,0), (3,1)\},\\\{(1,1), (2,1), (3,01)\} \}$

Each function in $B^{A_3}$ has a cardinality of 3 because the cardinality of $A_3$ is 3. Since $A_3$ is the domain of the functions in $B^{A_3}$, each function has to have 3 elements to satisfy the properties of in the definition of a function. The only thing that sets all the functions in $B^{A_3}$ apart is the ordering and repetitions of $0$ and $1$ in the second entries of the ordered pairs.

Similarly, for any finite set, $A$, each function in $B^A$ will have a cardinality of $S$ and the only thing setting these functions apart will be the $0's$ and the $1's$. Thus, we can ignore the first entries of the ordered pairs in the functions and reduce this problem down to permutations of $0$ and $1$ where the number of times repetition is allowed is equal to $S$ (because of the cardinality of each function). The formula for permutations with repetitions is $n^r$ where where $n$ is the number of things to choose from, and we choose $r$ of them. Thus the cardinality of $B^{A}$ is $2^{S}$.

I think my proof for the finite sets is correct but my problem is with the infinite sets. My idea is that I can show a pairing between each element in $\mathscr{P}(A)$ to a nonnegative binary integer. That works very nicely with the permutations of the $0's$ and $1's$.The problem is that $A$ could be the set of all real numbers. The pairing to the nonnegative binary numbers doesn't work anymore. So I'm kind of lost here. You could have sets that have cardinalities of higher infinities. How am I supposed to prove the cardinality is the same for all infinities?

答:
Hint: Consider the function
$$F:B^A\longrightarrow \mathcal{P}(A)$$
sending $f$ to $\{a\in A\mid f(a)=1\}$.


问:
$\kappa$-tree and cofinal branch
Definitions: Let $\left<T, \leq_T\right>$ be a tree,

(level of an element) The level of $x\in T$ is defined to be $Lev_T(x)= \mathrm{otp}\{y\in T\ |\ y<_Tx\}$.

Let $T_\alpha = \{x\in T\ |\ Lev_T(x)=\alpha\}$, the height of $T$ is $\min\{\alpha\ |\ T_\alpha =\emptyset\}$

($\kappa$-tree) Let $\kappa$ be a regular cardinal $T$ is called a $\kappa$-tree if the height of $T$ is $\kappa$ and $\forall\alpha<\kappa,\ |T_\alpha|<\kappa$.

(cofinal branch) A set $b\subset T$ is a cofinal branch if for every $x,y\in b,\ x\leq_T y\mathrm{\ or\ }y\leq_T x$ and for all $\alpha<\kappa,\ b\cap T_\alpha\neq\emptyset$

Question: Let $T$ be a $\kappa$-tree and assume that for every $A\subseteq T$, $|A|=\kappa$ there exists some $x,y\in A$ s.t $x<_Ty$. Prove that either $\exists z\in T$ s.t

$$ b=\left\{y\in T\ \big|\ z\leq_Ty\vee y \leq_T z\right\}$$

is a cofinal branch, or there is no cofinal branch.

My Proof: Assume that there is some cofinal branch $b\subset T$ and that there is no $z\in T$ s.t
$$ b_z =\left\{y\in T\ \big|\ z\leq_Ty\vee y \leq_T z\right\}$$
is a cofinal branch. We define a set in the following way, first consider $x_\emptyset = \min b$ then $b_{x_\emptyset}$ is not a cofinal branch thus there exists $\emptyset < \alpha_\emptyset < \kappa$ s.t $\exists y\in b_{x_\emptyset}\cap T_{\alpha_\emptyset}$ and $y\notin b$ choose one and call it $y_{\alpha_\emptyset}$.

Assume we have defined $y_{\alpha_\beta}$ let $z\in b\cap T_{\alpha_\beta}$ then $b_z$ is not a cofinal branch and by the same reasoning there exists $\alpha_\beta<\alpha_{\beta+1}<\kappa$ s.t $\exists y \in b_z\cap T_{\alpha_{\beta+1}}$ s.t $y\notin b$ choose one and call it $y_{\alpha_{\beta+1}}$.

Assume we have defined $y_{\alpha_\gamma}$ for $\gamma<\beta$ define $\alpha = \sup_{\gamma<\beta}\alpha_\gamma$ then from regularity $\alpha <\kappa$ and we can define using the above procedure when taking $z\in T_\alpha\cap b$ the $\alpha_\beta$.

Consider the set $A=\{\alpha_\beta\}$, $\sup\alpha_\beta = \kappa$ then again from regularity $|A|=\kappa$ and $A$ contradicts the assumption because each element is on a different branch hence not in relation to any other element.

My Problem: The definition of the set $A$ does not seem right to me I think I'm doing here something I cant (maybe the use of regularity of $\kappa$?)

Any help would be appreciated. Thanks in advance.

答:
It’s written up a bit clumsily, but your basic idea is fine. The only thing missing is a clearer explanation of why $A$ is an antichain (a subset of $T$ in which no two elements are related by the tree order). Here is a slightly neater way to carry out a very similar idea.

Say that a node of $T$ splits if it has more then one immediate successor in $T$.

Suppose that $b$ is a cofinal branch, let $S=\{x\in b:x\text{ splits}\}$, and suppose that $|S|=\kappa$. The definition of $S$ ensures that each $x\in S$ has an immediate successor $y_x\in T\setminus b$; let $A=\{y_x:x\in S\}$. $|A|=\kappa$, so there are $x,z\in S$ such that $y_x<_Ty_z$. Let $\alpha=\operatorname{Lev}_T(x)$, and let $u$ be the unique member of $b$ in $T_{\alpha+1}$. Then $u\le_Tz<_Ty_z$, so $u$ and $y_x$ are distinct predecessors of $y_z$ in $T_{\alpha+1}$, which is impossible. Thus $|S|<\kappa$. The regularity of $\kappa$ now ensures that there is an $x\in b$ such that $y<_Tx$ for each $y\in S$ and hence that $b=b_x$, as desired.

Title: $\kappa$-tree and cofinal branch


问:
How can you prove Euler's phase angle formula for differential equations?
How can you prove this formula: </p>

<p>$C_1 e^{(\alpha + i\beta) t} + C_2 e^{(\alpha - i\beta)t}=Ke^{\alpha t}\cos {(\beta t + \phi)}$ </p>

<p>This gives $x(t)$ in the second-order differential equation for an underdamped system with the characteristic equation $\lambda^2 + A\lambda + B = 0$ which gives $\lambda_{1,2} = \alpha \pm i\beta = \frac {A \pm \sqrt{A^2 - 4B}} 2$, and $C_1$, $C_2$ and $K$ are constants. </p>

答:
$$e^{ix}=\cos(x)+i\sin(x)$$
Leading to:
$$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=e^{\alpha t} (C_1 \cos(\beta t)+iC_1 \sin(\beta t)+C_2 \cos(\beta t)-iC_2 \sin(\beta t))$$

$$=e^{\alpha t} ((C_1+C_2) \cos(\beta t)-(iC_2-iC_1) \sin(\beta t))$$

$$=e^{\alpha t} (A \cos(\beta t)-B \sin(\beta t))$$

Now, any $(A \cos(\beta t)-B \sin(\beta t))$ can be written in the form $K \cos (\beta t+ \phi)$:

$$K \cos (\beta t+ \phi)=(K\cos(\phi)) \cos(\beta t)-(K\sin(\phi)) \sin(\beta t)$$
As cosine and sine are orthogonal, you can equate the coefficients:

$$K\cos(\phi)=A, K\sin(\phi)=B \Rightarrow \phi=\arctan(\frac{B}{A})$$
Has a unique solution $\frac{-\pi}{2}<\phi<\frac{\pi}{2}$, and
$$K^2(\cos(\phi)^2+\sin(\phi)^2)=A^2+B^2 \Rightarrow K=\sqrt{A^2+B^2}$$

So

$$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=e^{\alpha t}\sqrt{A^2+B^2}\cos(\beta t+ \arctan(\frac{B}{A}))$$

With

$$A=C_1+C_2$$
$$B=iC_2-iC_1$$

Finally,

$$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=2e^{\alpha t} \sqrt{C_1C_2} \cos(\beta t+i \tanh^{-1}(\frac{C_2-C_1}{C_2+C_1}))$$


问:
How to find integration without any sort of modification to the current function?
I want to integrate a difficult single variable function but I don't want to modify the given function i.e. factorising, rationalizing, etc.
The question is:
$$f(x)= (x^3-4x^2+4x)(x^2-4)^{-1}$$

The answer turns out to be equals to:
$g(x)= (x^2/2)+4\ln(|x-4|)+C$ after factorising the question. But I don't want this!
If you can see, there is something special about the question itself.
At $x=2$ the $f(x)=0/0$. But when we calculate derivative of $g(x)$ we get,
$h(x)= (x^2-4x+4)(x-4)^{-1}$ but $h(x)$ is not equals to $0/0$ at $x=2$...
So I want a $F(X)$ whose derivative will give "exactly" my original given $f(x)$- no factorization, no rationalization, etc...
(You can use u-substitution but be careful my objective shall be achieved)
and I'm well aware that $h(x)$ is equivalent to $g(x)$...
In short, I want a F(X) such that its derivative is exactly equals to my given f(x)- it doesn't matter what techniques you use! You can even use technique other than integration if possible.

Thank you for spending your precious time
Respected Regards,
Swayam Jha

答:
What you are asking is not really possible.

The antiderivative of $\dfrac xx$ is $x$ for all $x\ne0$, the derivative of which is $1$.

You might try to express as

$$x\frac xx,$$ formally giving the derivative

$$\frac xx+x\frac{x-x}{x^2}$$ and it would be arbitrary to say that this is $\dfrac xx$ rather than $\dfrac xx+\dfrac{x-x}x$ or other forms.

Also trying $$\frac{x^2}x,$$ we have

$$\frac{2x^2-x^2}{x^2}=\,?$$


问:
Example of complete category with no initial object
My original question is this. I found Zhen Lin's answer very useful, but I couldn't think of a category which is complete but has no initial object. The first category that I thought that has no initial object was the category of fields, but it isn't complete either.
Researching for categories whitout initial object I found this math overflow post that shows the following examples

(1) A groupoid has an initial object precisely if it is equivalent to the trivial groupoid.

(2) Any disjoint union of two categories has no initial object.

(3) Any poset without smallest elements has no initial object.

The category of all grupoids is complete, but this isn't what example (1) intended to be (right?). And for (2) and (3) I couldn't decide completeness for both.

Could you help me with this examples, or even give a new example of a category as in the title?

答:
None of the examples of categories with no initial object you describe are complete. It is actually tricky to write down an example here. The difficulty is that by the adjoint functor theorem, a complete category $C$ has an initial object as soon as the terminal functor $T : C \to 1$ has a left adjoint, which is true once the terminal functor satisfies the solution set condition. This will be true in most common examples.

So let's see what it takes for this to not be true. The solution set condition for $T$ asks for there to exist a (small) family $c_i, i \in I$ of objects in $C$ indexed by a set $I$ such that every object $c \in C$ admits a morphism $c_i \to c$ for some $i$. This is a bit weaker than asking that the $c_i$ be a family of generators (which would involve asking for the morphisms to be epimorphisms); such a family $\{ c_i \}$ is a weakly initial family of objects. I believe the way the adjoint functor theorem works from here is that a suitable limit over this family is used to produce the initial object.

So to build a counterexample we need to find a complete category $C$ which does not admit a weakly initial family of objects. Since an initial object is in particular a weakly initial family of objects this will straightforwardly imply that $C$ does not have an initial object, but the point is that this stronger condition tells us where we need to look. We need a category containing objects which are so "large" that for any set-sized family of objects in the category there is some even larger object which is "inaccessible" relative to that family.

Berci's comment gives what seems to me to be two of the simplest examples of such a thing: take the opposite of the "large poset" of ordinals ordered by inclusion. To say that this category is complete but does not have an initial object is equivalent to saying that the large poset of ordinals ordered by inclusion is cocomplete (has all (small) joins / suprema) but does not have a terminal object (a largest element), both of which are true and standard properties of the ordinals. Similarly we can consider the "large poset" of sets ordered by inclusion, which again is cocomplete (equivalently, admits set-sized unions) but does not have a largest element (since there is no set of all sets).

In the setting of "large posets" it's not hard to see that the infimum of a weakly initial family of objects is an initial object, and dually that the supremum of a weakly terminal family of objects is terminal, so the equivalence between the two is clearer here.


问:
Proof matrix eigenvalues inside the unit circle
Given that matrix $A$ is set as:

$A=(I-P)\left(I+(I-QP)^{-1}QP\right)$.

Could this equation be simplified written as: $A=(I-P)(I-QP)^{-1}$ and how could I proof the eigenvalues of $A$ are inside the unit circle?

Conditions for P and Q are:

$P=diag(p_{1},\cdots,p_{n})$ and $p_{i}\in[0,1)$.

$q_{i,j}$ in $Q$ are in interval $[0,1]$ and all the diagonal elements $q_{i,i}=0$

答:
For the first part of the question, $$(I-QP)^{-1}QP=(I-QP)^{-1}(QP-I)+(I-QP)^{-1}=-I+(I-QP)^{-1}$$ $$\therefore\ I+(I-QP)^{-1}QP=(I-QP)^{-1}$$

For the second part, the eigenvalues need not be bounded in the unit circle. Take $P=\alpha I$, $0<\alpha<\frac{1}{2}$, then $$A=(1-\alpha)(I-\alpha Q)^{-1}$$ It is enough to choose $Q$ symmetric such that $Q$ has an eigenvalue greater than one. For example, take $$Q=\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$$ $$\textrm{Eigenvalues of $Q$ are }2,-1,-1$$ $$\textrm{Eigenvalues of $I-\alpha Q$ are }1-2\alpha,1+\alpha$$ $$\textrm{Eigenvalues of $A$ are }\frac{1-\alpha}{1-2\alpha},\frac{1-\alpha}{1+\alpha}$$ The first eigenvalue is greater than one.


问:
How do we solve the equation $3\sec^{2}(x) - 4 = 0$?
I am confused about the answer to this question:

$$3\sec^{2}(x) - 4 = 0$$

I got the answer

$$x = \frac{\pi}{6} + 2n\pi\quad\text{or}\quad x = \frac{5\pi}{6} + 2n\pi$$

However, the book's answer is $\pi/6 + n\pi, 5\pi/6 + n\pi$

Isn't the period of a secant function $2\pi$? If so, why is the answer saying $\pi$?

答:
$$\sec^2x=\dfrac43 \iff\cos^2x=\dfrac1{\sec^2x}=?$$

$$\iff\sin^2x=\cdots=\dfrac14=\sin^2\dfrac\pi6$$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin\left(x-\dfrac\pi6\right)\cdot\sin\left(x+\dfrac\pi6\right)=0$$

If $\sin\left(x-\dfrac\pi6\right)=0, x-\dfrac\pi6=m\pi$

If $\sin\left(x+\dfrac\pi6\right)=0, x=-\dfrac\pi6+m\pi=(m-1)\pi+\left(\pi-\dfrac\pi6\right)$

where $m$ is ay integer


问:
Uniform convergence of $\sum_{n=1}^{\infty} \frac{\sin(n x) \sin(n^2 x)}{n+x^2}$
I'm not sure wether or not the following sum uniformly converge on $\mathbb{R}$ :

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

Can someone help me with it? (I can't use Dirichlet' because of the areas where $x$ is close to $0$)

答:
The series does converge uniformly. For the proof, put $S_n(x) = \sum_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$ for $n\geq 0$. The general idea is to use summation by parts to reduce ourselves to showing that $S_n(x)$ is bounded uniformly, and then to prove that by giving a closed form for $S_n(x)$.

First, the summation by parts (I write $S_n$ in place of $S_n(x)$ for brevity):
$$
\begin{align}
\sum_{n = 1}^N{\sin{(nx)}\sin{(n^2 x)}\over n + x^2} & = \sum_{n = 1}^N {1\over n+x^2}(S_n - S_{n-1}) \\
& = \sum_{n = 1}^N {S_n\over n+x^2} - \sum_{n = 1}^N {S_{n-1}\over n+x^2} \\
& = \sum_{n = 1}^N {S_n\over n+x^2} - \sum_{n = 0}^{N-1} {S_{n}\over n+1+x^2} \\
& = {S_N\over N+x^2} - {S_0\over 1 + x^2} + \sum_{n = 1}^{N-1} S_n\left({1\over n+x^2} - {1\over n+1+x^2}\right) \\
& = {S_N\over N+x^2} + \sum_{n = 1}^{N-1} {S_n\over (n+x^2)(n+1+x^2)}.
\end{align}
$$
From here it is clear that it is enough to prove that $S_n = S_n(x)$ is uniformly bounded in $x$.

To do this, note that
$$
\begin{align}
2\sin{(kx)}\sin{(k^2 x)} &= \cos{\{(k^2 - k)x\}} - \cos{\{(k^2 + k)x\}} \\
& = \cos{\{k(k - 1)x\}} - \cos{\{(k+1)k)x\}},
\end{align}
$$
and therefore that
$$
\begin{align}
2S_n(x) & = \sum_{k = 0}^n 2\sin{(kx)}\sin{(k^2 x)} \\
& = \sum_{k = 0}^n\left(\cos{\{k(k - 1)x\}} - \cos{\{(k+1)k)x\}}\right)
\end{align}
$$
The last sum telescopes, and we are left with
$$
2S_n(x) = 1 - \cos{\{n(n+1)x\}},
$$
which is plainly bounded uniformly in $x$. So we're done.


问:
About the basis for the usual topology on $\mathbb{R}$
I know that the sets of all open balls with rational radius and center form a basis for usual topology on $\mathbb{R}$. That is,
$$
B = \{I \subset \mathbb{R} \mid \exists a,b\in \mathbb{Q}:a<b \land I = (a,b)\}
$$
is a basis.

Then, can we think of a proper subset of $B$ that forms a basis for the topology on $\mathbb{R}$?

答:
If $D\subseteq \mathbb{Q}$ is a dense subset of $\mathbb{Q}$ then

$$B'=\{(a,b)\ |\ a,b\in D\}$$

will be a basis for $\mathbb{R}$ and a subset of $B$. Proper if $D$ is proper.

This gives us lots of examples, infinitely many in fact. You can remove a single point from $\mathbb{Q}$ or even consider $D_k=\{n/k^m\ |n,m\in\mathbb{Z}\}$ for a fixed natural $k\in\mathbb{N}$.


问:
Modulus of $\exp(-(A+iB)^2)$
I know modulus of $e^z$ is just $e^{\Re(z)}$, and that $|A+iB| \geq A^2$. What I want to know is if the following holds true
$$
|e^{-(A+iB)^2}| \leq e^{-A^2}
$$

答:
In the case when $A, B \in \mathbb{R},$ the answer is: "No".
  The given condition holds if $B=0$.
Because, $|e^{-(A+iB)^2}| = |e^{-(A^2-B^2+2iAB)}|  = |e^{(B^2-A^2-2iAB)}| = |e^{(B^2-A^2)}*e^{-2iAB}|= |e^{(B^2-A^2)}|*|e^{-2iAB}|$
 Using $|e^{i*x}|=1$, for real $x,$ therefore $|e^{-2iAB}|=1,$ So $|e^{-(A+iB)^2}| = e^{(B^2-A^2)}. $
$|e^{-i*2AB}| \leq e^{-A^2} $ if and only if  $e^{(B^2-A^2)} \leq e^{-A^2}.$
 Therefore, $B^2-A^2 \leq -A^2,$ Hence $B^2 \leq 0.$
 Since $B  \in \mathbb{R},$ then $B=0.$ So, the given condition $|e^{-(A+iB)^2}|  \leq e^{-A^2} $ is true if $B=0.$

Counter-example has been given by Kavi Rama Murthy in the comment section just below the question.

Title: Modulus of $\exp(-(A+iB)^2)$


问:
Is there a term for this concept involving functions?
If $f_1:X_1\to Y_1$, $\ldots$, $f_n:X_n\to Y_n$ are maps, is there a term for the map $\left(x_1,\ldots,x_n\right)\mapsto\left(f_1\left(x_1\right),\ldots,f_n\left(x_n\right)\right)$ from $X_1\times\cdots\times X_n$ to $Y_1\times\cdots\times Y_n$? Thanks.

答:
This is the Cartesian product of the functions $f_i$ and is denoted $f_1 \times \cdots \times f_n$.
See for example this Wikipedia article.