问:
Absolute continuity in terms of integral
Below is the question and my attempted solution. I'm curious if it's correct, and if not can it be fixed? Or is it completely wrong?

Let $\left(X,\mathcal{M},\mu\right)$ be a measure space, and let $\{\mu_{\alpha}\}_{\alpha\in[0,1]}$ be a collection of measures on the measurable space $\left(X,\mathcal{M}\right)$. Define,
$$\nu(E)=\int_0^1\mu_\alpha(E)dx, \quad E\in \mathcal{M}.$$
Show that, if $\mu_{\alpha}\ll \mu$-a.e $x$ in $[0,1]$, then $\nu\ll \mu$.

This problem seems almost too easy, so I figured I may be doing something wrong. Below is my solution:

Let $E_0\subseteq X$ be such that $\mu(E_0)=0$. As $\mu_{\alpha}\ll \mu$, it follows by definition $\mu_{\alpha}(E_0)=0$. By direct calculation,

$$\nu(E_0)=\int_0^1\mu_\alpha(E_0)dx = \int_0^10dx = 0\cdot1=0.$$

Therefore, by definition $v\ll \mu$.

答:
Seems okay. Just note that $\mu_\alpha \ll \mu$ only almost everywhere w.r.t. $\alpha$. Thus $\mu_\alpha(E_0) = 0$ for almost every $\alpha$. But that's enough to conclude that $\nu(E_0) = 0$ anyway.

This question looks like a fancier version of the usual $f = 0\ \text{a.e.} \implies \int f = 0$, maybe that's the nontrivial part.


问:
Confusion about a proof on Mycielski construction and chromatic number
Theorem 10.10 of the textbook "A First Course in Graph Theory (2012)" by Gary Chartrand and Ping Zhang is as follows:

For every integer $k \ge 3$, there exists a triangle-free graph with chromatic number $k$.

This is proved by induction on $k$ and in the inductive step it actually shows that

From a $(k-1)$-chromatic triangle-free graph $F$, Mycielski's construction produces a $k$-chromatic triangle-free graph $G$.

The Mycielski's construction (wiki) of $F$ is obtained from $F$ by first adding, for each vertex of $F$, a new vertex $v'$, called the shadow vertex of $v$, and joining $v'$ to the neighbors of $v$ in $F$ and then adding a new vertex $z$ and joining $z$ to all the shadow vertices.

For the part the correctness of $\chi(G) = k$, it proceeds as follows:

Assume, to the contrary, that $\chi(G) = k-1$. Let there be a $(k-1)$-coloring $c$ of $G$, say with colors $1, 2, \ldots, k-1$. We may assume that $c(z) = k-1$. Since $z$ is adjacent to every shadow vertex in $G$, it follows that the shadow vertices are colored with the colors $1, 2, \ldots, k-2$. For every shadow vertex $x'$ of $G$, the color $c(x')$ is different from the colors assigned to the neighbors of $x$. Therefore, if for each vetex $y$ of $G$ belonging to $F$, the color $c(y)$ is replaced by $c(y')$, we have a $(k-2)$-coloring of $F$. This is impossible, however, since $\chi(F) = k - 1$.

My Problem: The argument in bold does not seem clear to me. How to make sure that the resulting coloring is a proper coloring? In other words, how to show that for each pair of adjacent vertices $u,v$ in $F$, $c(u') \neq c(v')$?

答:
I think the proof in that book might be flawed.

If $x$ and $y$ are adjacent vertices of $F$ such that their shadow vertices have the same color (that is $c(x')=c(y')$), then replacing the colors as in the proof will not give a proper coloring.

There is no (obvious to me) reason why this cannot happen.

Both answers ignore this issue by assuming that if a vertex changes colour, then none of its neighbours does.

To fix this, only recolor vertices of $F$ that have the same colour as $z$. (And change their color to the one of their shadow, as in the book.) One can check this is a proper colouring of $F$.

ContentLicense: CC BY-SA 4.0


问:
Proof that $\dim\big(\mathcal{L}(V,W)\big)=\dim V *\dim W$, when $V, W$ are finite dimensional.
I've been looking a few proofs of this, and I have a problem with the standard one, which uses the concept of isomorphism. Suppose $\dim V=n$ and $\dim W=M$. The proof uses the function $M$ from $\mathcal{L}(V,W)$ (the linear maps from $V$ to $W$), to $\text{Mat}_{m\times n}(F)$ (matrices of $m\times n$), which assigns to each linear map $T$ it's matrix $M(T)$. To do this, they fix first some basis $(v_1,....v_n)$ of $V$ and $(w_1,...w_m)$ of $W$. My problem is this fixing of the basis. Because then the function $M$ is not a function, at least not from $L(V,W)$, because each linear map has many matrices depending of the basis. How can we then state that $\dim\big(\mathcal{L}(V,W)\big)=(\dim V)(\dim W)$ ?</p>

答: Suppose, m = dim($V$) and n = dim($W$). Since $L(V,W)$ and $\mathbb{F}^{m,n}$ are isomorphic, dim $(L(V,W))$ = dim $\mathbb{F}^{m,n}$ where dim $\mathbb{F}^{m,n} = mn$ = dim($V$) dim($W$).

ContentLicense: CC BY-SA 3.0


问:
Understanding subset notation
So in this answer, user326210 used the notation, $f(W) \equiv \{ f(w) : w\in W\} \subseteq E$.
Maybe I am misunderstanding this but this part $\{ f(w) : w\in W\}$ is defined as the set of workers. How can the set of workers be a subset of E (which is itself the set of farms)..aren't they different?

答:
$W$ is the set of workers.

$E$ is the set of farms.

And if $w \in W$ is a worker, $f(w)$ is the farm where $w$ works.

$f:W\to E$ and $\{f(w)| w \in W\}$ is NOT defined to be a set of workers.  $\{f(w)|w\in W\}$ is defined to be the set of farms where all the workers work.

The elements in $\{f(w)| w \in W\}$ are all the $f(w)$s; not the $w$s.  And every $f(w)$ is a farm; not a worker.

Title: Understanding subset notation


问:
Why is square root by long division found so?
We were taught the long division method of finding square root in junior classes. The  logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.

答:
This is based on my comments to the question.

Consider that you have evaluate the square root of $a$. Let $x$ be the square root and $h$ be the least significant digit of the square root and $b$ be the number formed from $x$ by removing this least significant digit $h$. Then we have $x=10b+h$. (for example consider $a=144,x=12,b=1,h=2$).

The process of long division allows you to find $h$ given $a, b$. Thus we can find the next digit of square root if previous digits are known.

Our job is to find $h$ such that $(10b+h)^2=a$ or $$(2\cdot 10b+h)h=a-(10b)^2$$ the subtraction in right hand side is done by placing digits accordingly (for $a=144,b=1$ you place the digit $1$ of $b$ below $1$ of $a$ and get $44$ so that this is essentially doing $a-(10b)^2$).

Now analyze the left hand side. $2\cdot 10b$ involves doubling the number $b$ and multiplying further by $10$ to make space for next digit $h$. We have to choose next digit $h$ so that the new number $(20b+h)$ when multiplied by $h$ equals the number obtained after subtraction on right hand side. This is exactly what we do in the long division process (try to see for the example $a=144,b=1$ and you can easily guess that $21\cdot 1=21,22\cdot 2=44$).

Consider the same in terms of polynomials and roots. Let us solve $x^2-a=0$ digit by digit. We guess the root $b$ (which is the approximate square root of first two digits of $a$) and find an equation whose roots are $t=x-b$. This equation is clearly $$(t+b) ^2-a=0$$ Next multiply the roots by $10$ so that $z=10t$ gives $$(z+10b)^2-100a=0$$ or $$z(20b+z)=100a - 100b^2$$

Let us see that $100a$ is essentially about getting two more digits from $a$ and then subtracting $100b^2$ from it means that we subtract $b^2$ from first two digits of $a$ and bring in next two digits from $a$. The guess for $z$ is based on $z(20b+z)$ which is doubling $b$, creating next digit $z$ and multiplying it by $z$.

We can continue the process by subtracting $z$ and multiplying by $10$ and get another equation. The pattern remains same and can be continued till any desired number of digits.

The mysterious part in the long division process is doubling the divisor at each step and creating space for quotient as next digit. I have tried to explain that part in both the approaches above.


问:
Uniformiser in book by Miranda
I am doing Exercise V.2D in Miranda's book Algebraic Curves and Riemann Surfaces.

Let $X$ be the projective plane curve defined by the equation $y^2 z=x^3-xz^2$. Let $p_0=[0:1:0]$, $p_1=[0:0:1]$, $p_2=[1:0:1]$ and $p_3=[-1:0:1]$. Show that $p_1+p_2+p_3\sim 3p_0$.

Taking $f=Y/Z$, the points where it has zeros and poles are precisely the $p_i$ above.

When I want to calculate the order of vanishing at $[0:0:1]$, I go to the affine chart $Z\neq 0$, then the point becomes $(0,0)$ and I have to calculate $ord_{(0,0)}(y)$. I believe this is one, because there is no power in the $y$. However, I know from the theory of uniformisers in algebraic geometry that this is not necessarily the case. For example, when I do the same trick for $
[0:1:0]$  then the order of vanishing should be -3 but is $ord_{(0,0)}(1/z)=-1$ when I reason like this.

I can not find anywhere in Miranda's book the theorem that says that a function that vanishes at a point but is not the tangent line is a uniformiser. Could someone help me out ??
Also, in Miranda's book it seems that the order of vanishing is defined by the lowest non-zero term in the Taylor expansion, but isn't this $y$ ?

答:
Miranda's book has more of a complex geometry flavor, so he defines order of vanishing using holomorphic charts and local coordinates. (It's really not that different from using uniformizers and local rings, though.) The analogue of the result you're looking for is the passage preceding Theorem 2.3 of Chapter I (p. 11), which says more or less the following.

Proposition. Let $C$ be a nonsingular affine plane curve given by the equation $f(x,y)=0$, where $f \in \mathbb{C}[x,y]$ is irreducible. Given $P \in C$, if $\frac{\partial f}{\partial y}(P) \neq 0$ then $x$ is a local coordinate on some neighborhood $U$ of $P$. In other words $(U, \pi_x)$ is a holomorphic chart containing $P$, where $\pi_x : (x,y) \mapsto x$.

(And similarly if $\frac{\partial f}{\partial x}(P) \neq 0$, then $y$ is a local coordinate near $P$.) The proof is basically the Implicit Function Theorem (Theorem I.2.1). Since the tangent line at the point $P = (a,b)$ is $\frac{\partial f}{\partial x}(P) (x-a) + \frac{\partial f}{\partial y}(P) (y-b)$, this agrees with the criterion you stated in terms of tangent lines.

Okay, so let's apply the proposition to your example. Let $E$ be the projective curve given by $Y^2 Z = X^3 - X Z^2$. (Forgive me for changing the notation, but I prefer to use lower case letters for affine coordinates.) In the affine open where $Z \neq 0$, $E$ is given by $y^2 = x^3 - x$, where $x = X/Z$ and $y = Y/Z$, so let $f = y^2 - (x^3 - x)$. Here $p_2 = (1,0)$, and since $f_x(p_2) = -(3(1)^2 - 1) \neq 0$, then $y$ is a local coordinate near $p_2$, so $\DeclareMathOperator{\ord}{ord} \ord_{p_2}(y) = 1$. (As you said, its Taylor series is simply $y$!)

To compute the order of vanshing at $p_0 = [0:1:0]$, we move to the affine open where $Y \neq 0$. Letting $w = X/Y$ and $z = Z/Y$, then $E$ is given by $z = w^3 - w z^2$, and $p_0 = (0,0)$. From this equation we can already see that $z$ can't be a uniformizer at $p_0$ since $z = w^3 - w z^2$ and both terms on the righthand side vanish to at least order $3$. Letting $h = z - (w^3 - wz^2)$, then
$$
h_z(p_0) = 1 - (-2wz)\big{|}_{(w,z) = (0,0)} = 1 \neq 0
$$
so $w$ is a local coordinate near $p_0$. To compute $\ord_{p_0}(z)$ we need to express $z$ as a Taylor series in $w$. One way to do this is by repeatedly substituting $z = w^3 - w z^2$ into the righthand side of the equation:
\begin{align*}
z &= w^3 - w z^2 = w^3 - w (w^3 - w z^2)^2 = w^3 - w^7 + 2 w^5 z^2 - w^3 z^4\\
&= w^3 - w^7 + 2 w^{11} - 5 w^{15} + \cdots
\end{align*}

If you don't like that approach, the Implicit Function Theorem implies that $z = g(w)$ for some function $g$ holomorphic near $w=0$. Substituting this into the equation for the curve yields
$$
g(w) = w^3 - w \, g(w)^2
$$
and we can compute as many Taylor coefficients as we like by repeatedly differentiating and substituting $w = 0$.

In any case, we find that
$$
\ord_{p_0}(y) = \ord_{p_0}(1/z) = -\ord_{p_0}(z) = -3 \, .
$$

Title: Uniformiser in book by Miranda


问:
Prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$
Q: If $M$ is a positive number less than 1. Prove the terms in $\displaystyle \left\{\dfrac{n}{n + 1}\right\}_{n = 1}^{\infty}$ exceed $M$ for sufficiently large $n$; that is, prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$.

Not exactly sure how to prove this but here is my attempt so far.

By the definition of the limit, $\displaystyle \lim_{n \to \infty} \dfrac{n}{n + 1} > M$, we have $\forall M\in(0, 1)\exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{1cm} \rightarrow \hspace{1cm} a_n > M
\end{equation*}

If $a_n = \dfrac{n}{n + 1}$, then we want to show that it exceeds $M$ as $n$ gets larger.

This is the part where I am kind of lost. I am not exactly sure how to proceed from here. Some tips or advice would be useful. Thanks

答:
$\frac n{n+1} > M\iff n > M(n+1)\iff $

$n> Mn + M \iff n-Mn = n(1-M) > M$.

As $M < 1$ then $1- M > 0$ so

$n(1-M) > M \iff n \ge \frac M{1-M}$

So if $N$ is any integer $\ge \frac M{1-M}$ you get your result.

$n > N \ge \frac 1{1-M} \implies \frac n{n+1} > M$.


问:
How can I compute an expression for $P(X_1>X_2>X_3>X_4)$ if $X_1,X_2,X_3,X_4$ are normal and mututally independent?
How can I compute an expression for $P(X_1>X_2>X_3>X_4)$ if $X_1,X_2,X_3,X_4$ are normal and mututally independent?

I managed to compute $P(X_1>X_2>X_3)$ as follows but the same technique does not seems to be working for 4 variables since I cannot change variables in an efficient way. For the case with 3 random variables I can show that

$P(X_1>X_2>X_3)=\int \int \int_{x_1>x_2>x_3} f_1(x_1)f_2(x_2)f_3(x_3)dx_1dx_2dx_3 =\int_{-\infty}^{\infty}dx_3 \int_{x_3}^{\infty}dx_2 \int_{x_2}^{\infty}f_1(x_1)f_2(x_2)f_3(x_3)=\int_{-\infty}^{\infty}dx_3 \int_{x_3}^{\infty}f_2(x_2)f_3(x_3)[1-F_1(x_2)]dx_2=\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{x_2}f_2(x_2)f_3(x_3)[1-F_1(x_2)]dx_3=\int_{-\infty}^{\infty} F_3(x_2)[1-F_1(x_2)]f_2(x_2)dx_2=\int_{-\infty}^{\infty}F_3(y)[1-F_1(y)]f_2(y)dy$

In the case of 4 variables I am unable to change the variables in a nice way to get a simplifies expression as above. Can somebody help me here?

答: Interestingly, if $X_1,X_2,\ldots,X_n$ are normal and mutually independent, you can show that $$\mathbf P(X_1 > X_2>X_3 > \ldots > X_n) = \frac{1}{n!}$$ using essentially the same approach as in @hypernova's answer.


问:
Let $f(x,y,z)$ be a convex function. Is the reciprocal convex?
Suppose that $f(x,y,z)$ be a convex function. Prove $\frac{1}{f(x,y,z)}$ is convex. Or give an example of $f$ where $1/f$ is not convex.

For example, I know that $f(x,y,z)=(1+x^2)\sqrt{1+y^2+z^2}$ is convex on $\mathbb R^3.$  $g=1/f$  is not convex  on $\mathbb R^3? $

$f$ convex implies the inequality
$f(x, y+v, z+w) - f(x,y,z) \geq f_y(x,y,z) + f_z(x,y,z)$ for all $x,y,z,v,w \in \mathbb R$.

However $g=1/f$ does not satisfy the inequality  $g(x, y+v, z+w) - g(x,y,z) \geq g_y(x,y,z) + g_z(x,y,z)$ for x=0, y=1, z=1, v= -2, w=-2 . Thus it is not convex in $\mathbb R^3$.

Therefore in general $1/f$ is not convex. Any comments?

答:
This is false in general, even in one dimension.  A smooth function $f : \mathbb{R} \to \mathbb{R}$ is convex iff $f''(x) > 0$, but
$$ \left(\frac{1}{f}\right)'' = \left(-\frac{f'}{f^2} \right)' = -\frac{f''}{f^2} + \frac{2f'^2}{f^3}$$
Thus, for example, if $f' = 0$, and $f'' > 0$, $f$ is convex but its reciprocal is not.</p>


问:
For similar triangles ADG, BDF, and CDE, find EF given AB, BC, and DE
I was tutoring a student today and one of the practice problems we were working on asked the question posed in the title, including an image like the one below. The quadrilaterals each represent a car in a train in a perspective sketch with the triangle CDE representing all of the following cars to the vanishing point on the horizon.
The triangles in question

After looking at the problem, neither of us could see how to find the length of EF with the given information and we moved on. Is there a solution that we missed or is this problem just missing information?

答:
We need to apply the the intercept theorem (a.k.a Thales' Theorem) three times. First,
$$\frac{1.9}{8.4}=\frac{EF}{FG}=\frac{1.9k}{8.4k}$$
Now, focus on $\bigtriangleup DBF$ to find CD in terms of $K$.
$$\frac{5.4}{1.9k}=\frac{CD}{BC}=\frac{\frac{5.4}{k}}{1.9} \implies CD=\frac{5.4}{k}$$
Finally, focus on $\bigtriangleup ADG$ to find $k$.

$$\frac{\frac{5.4}{k}+1.9}{8.4}=\frac{5.4+1.9k}{8.4k}$$

I believe $k$ turns out to be exactly $1$ and hence, $EF=1.9$ units.


问:
Probability of n heads in 2n tosses, given at least one head in n tosses
Suppose we toss a fair coin $2n$ times. Conditioned on the fact that the coin came up heads at least once in the first $n$ tosses, what is the probability that we get precisely a total of $n$ heads over all $2n$ heads?

答:
By the $P(B \mid A) = P(A \cap B) / P(A)$ definition of conditional probability, you need to compute

the probability of getting $n$ heads over all tosses, with at least one of the heads occurring in the first $n$ tosses, and

the probability of getting at least one head in the first $n$ tosses.

For the second term, it is easier to compute the probability of the complement.

For the first term, it might be helpful to consider the probability of getting $n$ heads over all tosses, with no heads occurring in the first $n$ tosses.


问:
The minimum of $f(x)=1+x+\cdots+x^{2n}$
For the function $$
f(x)=\sum_{k=0}^{2n}{x^k}
$$
I think this function has no zeros, and if ${f(x_0)}$ is the minimum point, then ${x_0\in \left[-1,0 \right] }$.

For $1+x+x^2+x^3+x^4$ and $1+x+x^2$, I can get the minimum by calculating $f'(x)$ and then using the root fomula. But this method doesn't work for $n\geqslant 3$ since there is no root fomula for equations higher than the fifth degree.

I have no idea for the situation where  $n\geqslant 3$, so any advice could be helpful

Title: The minimum of $f(x)=1+x+\cdots+x^{2n}$

Body:
It seems possible to generate an estimate of the value of $x$ where the function
$$f(x)=\sum_{k=0}^{2n}{x^k}=\frac{x^{2 n+1}-1}{x-1}$$ is minimum.
As shown in other answers, the problem is to approximate the zero of function
$$g(x)=2nx^{2n+1}-(2n+1)x^{2n}+1$$
This have been done using the first iteration of  an high order iterative method starting with $x=-1$ as an initial guess. The result is given by
$$x_0=-\frac{256 n^6+10240 n^5+75296 n^4+212560 n^3+273764 n^2+159400 n+33199}{2 (2 n+3)
   \left(64 n^5+2688 n^4+20224 n^3+53604 n^2+56320 n+20001\right)}$$

Now, one single interation of Newton method is supposed to sufficiently polish the solution (it would not be any problem to use Halley or Householder methods in place of Newton).

Below are reported some results ($x_1$ by Newton).
$$\left(
\begin{array}{cccc}
n & x_0 & x_1 & \text{solution} \\
 1 & -0.500137 & -0.500000 &  -0.500000 \\
 2 & -0.605614 & -0.605830 &  -0.605830 \\
 3 & -0.669716 & -0.670334 &  -0.670332 \\
 4 & -0.713566 & -0.714543 &  -0.714538 \\
 5 & -0.745773 & -0.747064 &  -0.747054 \\
 6 & -0.770592 & -0.772160 &  -0.772142 \\
 7 & -0.790396 & -0.792205 &  -0.792178 \\
 8 & -0.806624 & -0.808643 &  -0.808605 \\
 9  & -0.820203 & -0.822403 &  -0.822353 \\
 10 & -0.831760 & -0.834116 &  -0.834053 \\
 11 & -0.841736 & -0.844223 &  -0.844148 \\
 12 & -0.850449 & -0.853047 &  -0.852958 \\
 13 & -0.858136 & -0.860826 &  -0.860724 \\
 14 & -0.864977 & -0.867742 &  -0.867627 \\
 15 & -0.871111 & -0.873936 &  -0.873809 \\
 16 & -0.876648 & -0.879520 &  -0.879381 \\
 17 & -0.881675 & -0.884583 &  -0.884433 \\
 18 & -0.886264 & -0.889197 &  -0.889037 \\
 19 & -0.890473 & -0.893421 &  -0.893252 \\
 20 & -0.894349 & -0.897304 &  -0.897127 \\
 21 & -0.897932 & -0.900888 &  -0.900703 \\
 22 & -0.901257 & -0.904206 &  -0.904015 \\
 23 & -0.904351 & -0.907289 &  -0.907091 \\
 24 & -0.907239 & -0.910161 &  -0.909958 \\
 25 & -0.909942 & -0.912843 &  -0.912636
\end{array}
\right)$$

It is possible to do better for $x_0$ at the price of an higher order method.

Edit

For the fun of it, I used the same procedure with two more orders. The result write, just as before,
$$x_0=-\frac{\sum_{k=0}^8 a_k\,n^k}{\sum_{k=0}^8 b_k\,n^k}$$ and the coefficients are
$$\left(
\begin{array}{ccc}
k & a_k & b_k \\
 0 & 2300106 & 8278851 \\
 1 & 13284828 & 37098288 \\
 2 & 30007592 & 67149620 \\
 3 & 34417264 & 63318640 \\
 4 & 21546112 & 33284272 \\
 5 & 7272832 & 9611392 \\
 6 & 1199488 & 1382080 \\
 7 & 73216 & 75520 \\
 8 & 512 & 512
\end{array}
\right)$$

Just to give an idea, for $n=25$ this produces for $x_0$ a value of $-0.912088$.

Update

Up to now, we just focused on the value of $x$ for which $f(x)$ is minimum. What is interesting if that when $f'(x_0)=0$ by elimination we have the simple
$$f(x_0)=\frac{2 n+1}{1+2 n(1-x_0)}$$

Using the last set of coefficients given in the edit, we have
$$\left(
\begin{array}{ccccc}
n & f_{\text{min}}^{\text{est}} & f_{\text{min}}^{\text{calc}}  & x_0^{\text{calc}}& x_0^{\text{est}}\\
 1 & 0.749996 & 0.750000& -0.500000 & -0.500011 \\
 2 & 0.673570 & 0.673553& -0.605830 & -0.605782 \\
 3 & 0.635115 & 0.635094& -0.670332 & -0.670271 \\
 4 & 0.611584 & 0.611567& -0.714538 & -0.714486 \\
 5 & 0.595555 & 0.595543& -0.747054 & -0.747017 \\
 6 & 0.583866 & 0.583858& -0.772142 & -0.772117 \\
 7 & 0.574928 & 0.574922& -0.792178 & -0.792159 \\
 8 & 0.567852 & 0.567846& -0.808605 & -0.808586 \\
  9 & 0.562099 & 0.562091& -0.822353 & -0.822327 \\
 10 & 0.557321 & 0.557309& -0.834053 & -0.834013 \\
 11 & 0.553284 & 0.553267& -0.844148 & -0.844089 \\
 12 & 0.549825 & 0.549801& -0.852958 & -0.852876 \\
 13 & 0.546825 & 0.546793& -0.860724 & -0.860614 \\
 14 & 0.544196 & 0.544156& -0.867627 & -0.867487 \\
 15 & 0.541872 & 0.541823& -0.873809 & -0.873636 \\
 16 & 0.539802 & 0.539743& -0.879381 & -0.879173 \\
 17 & 0.537945 & 0.537876& -0.884433 & -0.884189 \\
 18 & 0.536269 & 0.536190& -0.889037 & -0.888755 \\
 19 & 0.534749 & 0.534660& -0.893252 & -0.892932 \\
 20 & 0.533363 & 0.533263& -0.897127 & -0.896769 \\
 21 & 0.532094 & 0.531984& -0.900703 & -0.900306 \\
 22 & 0.530926 & 0.530807& -0.904015 & -0.903580 \\
 23 & 0.529850 & 0.529719& -0.907091 & -0.906618 \\
 24 & 0.528852 & 0.528712& -0.909958 & -0.909447 \\
 25 & 0.527926 & 0.527777& -0.912636 & -0.912088
\end{array}
\right)$$

Update

After this question (which looks quite similar), I thought that it could be interesting to look for the zero of function
$$h(x)=\log \left((2 n+1) x^{2 n}-2 n x^{2 n+1}\right)$$ Expanding it as a series around $x=-1$ and using series reversion, we end with
$$x=-1+z+\sum_{k=2}^\infty (-1)^{k+1}\frac {P_k(n) } { 2^{k-1}\, k! \,(4 n+1)^{k-1} }\,z^k$$ with
$$z=\frac{(4 n+1) \log (4 n+1)}{4 n (2 n+1)}$$
The first $P_k(n)$ are
$$\left(
\begin{array}{cc}
 2 & 8 n+1 \\
 3 & 64 n^2+8 n-1 \\
 4 & 512 n^3+32 n^2-24 n-1 \\
 5 & 4096 n^4+256 n^3+32 n^2+104 n+13
\end{array}
\right)$$


问:
Could $(2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) $ be an integer?
Let $ a,b  \in \mathbb{R}_{>0} \backslash \left \{ \tfrac{2N-1}{2M-1} : N,M \in \mathbb{N} \right \} $ and $n \in \mathbb{N}$.

I want to prove that the quantity
\begin{equation}\label{key}
 (2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right)
\end{equation}
cannot be an integer.

I got the following:

Assume that it is an integer, $N$, say. Then
\begin{equation}
 \dfrac{Nab}{a-b}
\end{equation}
is odd.
However, I get stuck here.

答:
None of the expressions expressed need to be integer.

We can have $(2n-1)(\frac 1a - \frac 1b) = M\in \mathbb Z$

if $\frac 1a - \frac 1b = \frac M{2n-1}$

Nos $a,b$ can be any positive reals that are not a ration of two odd integers.

So we can have $b$ but any irrationa and then $a = \frac 1{\frac M{2n-1}-\frac 1b}$ will not be rational and that will work.

Example $b = \frac 1{\pi}$ and $a = \frac 1{2-\pi}$ and

$(2n-1)(\frac 1a -\frac 1b) = (2n-1)(2-\pi + \pi) = (2n-1)\cdot 2$ is an integer.

If we want $a$ and $b$ to be rational we can have $a$ and $b$ but have an even components.  Let $b = \frac 12$ and $a =\frac 14$.  Thn

$(2n-1)(\frac 1{\frac 14} - \frac 1{\frac 12})= (2n-1)(4-2) = 2(2n-1)$ is an integer.


问:
Multiplicative Group of a Field
The multiplicative group $F^{\times}=F\setminus \{0\}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $\mathbb{Q},\mathbb{R}$ and $\mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $\mathbb{C}^{\times}$. The question I would like to ask is

Question: Is there an infinite field $F$ such that $F^{\times}$ is isomorphic to the Prüfer $p$-group?

答:
Nice question! The answer is no.

If $F$ has characteristic zero it contains a copy of $\mathbb{Q}$, so $F^{\times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{\times}$ is isomorphic to a Prüfer $\ell$-group for some prime $\ell$, then $F$ contains all $\ell$-power roots of unity, and $\mathbb{F}_p^{\times}$ must consist of $\ell$-power roots of unity, so in particular $p \neq \ell$.

Let $k$ be a positive integer. Since $F$ contains all $\ell^k$-th roots of unity, $F$ contains $\mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $\ell^k | p^n - 1$. Then $F^{\times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $\ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.

Title: Multiplicative Group of a Field


答:
If $a$ and $b$ are both odd integers, then $ab$ will be an odd integer. Then

$n= \frac{ab+1}{2}$ is an integer.

$\frac{2n-1}{a} = b$

$\frac{2n-1}{b} = a$

$(2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) = b - a \in \Bbb Z$

Title: Could $(2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) $ be an integer?

Body:
Let $ a,b  \in \mathbb{R}_{>0} \backslash \left \{ \tfrac{2N-1}{2M-1} : N,M \in \mathbb{N} \right \} $ and $n \in \mathbb{N}$.

I want to prove that the quantity
\begin{equation}\label{key}
 (2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right)
\end{equation}
cannot be an integer.

I got the following:

Assume that it is an integer, $N$, say. Then
\begin{equation}
 \dfrac{Nab}{a-b}
\end{equation}
is odd.
However, I get stuck here.


问:
Basis of neighbourhoods on cofinite topology
Let $(E,\tau)$ be a topological space where $\tau=\{ \emptyset \}\cup \{G\subset E, |E\setminus G|<\infty\}$

How I can write $W_x$ the basis of neighbourhoods  of $x \in E$?

Thank you

答: You can write $W_x=\{E \setminus A: A \subseteq E, |A|<\infty, x \notin A\}$. If $U$ is any open neighborhood $x$ then let $E \setminus U \subseteq B$ where $x \notin B$ and $|B|<\infty$ this implies $x \in E \setminus B \subseteq U$.


问:
How can I compute probability P(E|D) from Bayes Net
How can I compute probability P(E|D) from Bayes Net
Please refer to the image</p>

答:
The network encodes the factorisation:$$\def\P{\operatorname{\mathsf P}}\P(A,B,C,D,E,F)=\P(A)\P(B)\P(C\mid A,B)\P(D\mid C)\P(E\mid A)\P(F\mid B)$$

And therefore by the law of total probability: $$\require{cancel}\begin{align}\P(A,B,C,D,E)&=\P(A)\P(B)\P(C\mid A,B)\P(D\mid C)\P(E\mid A)\color{blue}{\cancelto 1{\color{silver}{\sum_{\dot F}\P(\dot F\mid B)}}}\\[2ex]\P(A,B,C,D)&=\P(A)\P(B)\P(C\mid A,B)\P(D\mid C)\color{blue}{\cancelto 1{\color{silver}{\sum_{\dot E}\P(\dot E\mid A)}}}\end{align}$$

So applying the Law of Total Probability (again):

$$\def\tilda#1{\overset\sim #1}\begin{align}\P(D,E)&=\sum_{\dot A,\dot B,\dot C}\P(\dot A)\P(E\mid\dot A)\P(\dot B)\P(\dot C\mid\dot A,\dot B)\P(D\mid\dot C)\\[2ex]\P(D)&=\sum_{\dot A,\dot B,\dot C}\P(\dot A)\P(\dot B)\P(\dot C\mid\dot A,\dot B)\P(D\mid\dot C)\end{align}$$

(With the usual understanding that $\small\raise{1.5ex}{\sum\limits_{\dot X}}$ represents $\small\raise{1.5ex}{\sum\limits_{\dot X\in\{X,{\tilda{X}}\}}}$ )

And so...


问:
Does Lyapunov theorem extend to dynamical system that have "equilibrium sets"?
Suppose that I have some dynamical system

$$\dot x = f(x)$$
$f$ is locally Lipschitz, etc. I know that $f(x) = 0$ whenever $x \in \Gamma$, where $\Gamma$ is some closed (and possibly bounded) set in $\mathbb{R}^n$, could be a small disc, a line, a plane. In other words, not just an equilibrium.

How can dynamical system theory deal with this case?

Normally the approach is through Lyapunov theorem, but it is usually with respect to a single point.

Is it possible to extend the Lyapunov theorem by defining a Lyapunov function $V$ such that it is $V(\Gamma) = 0$ and $V(x) > 0$ elsewhere and proceed as usual by showing $\dot V(\Gamma) = 0$ and $\dot V(x) < 0$ elsewhere?

Is this possible? Are there any catch for using the above? I have went through several books but could not find this extension.

答:
You can use https://en.wikipedia.org/wiki/LaSalle%27s_invariance_principle, though I don't like the statement off of Wikipedia. The following comes from Nonlinear Systems, Third Edition by Khalil.

Consider the autonomous system

\begin{align}
\dot{x} = f(x) \qquad (1)
\end{align}

where $f: D\to \mathbb{R}^{n}$ is a locally Lipschitz map from a domain $D\subset \mathbb{R}^{n}$ into $\mathbb{R}^{n}$.

Theorem 4.4 in Nonlinear Systems, Third Edition by Khalil:

Let $\Omega \subset D$ be a compact set that is positively invariant with respect to $(1)$. Let $V: D \to \mathbb{R}$ be a continuously differentiable function such that $\dot{V} \leq 0$ in $\Omega$. Let $E$ be the set of all points in $\Omega$ where $\dot{V}=0$. Let $M$ be the largest invariant set in $E$. Then every solution starting in $\Omega$ approaches $M$ as $t\to\infty$.

Note that this theorem is more general than the extension you mention, $V$ doesn't need to be positive definite. As long as you can show that $\Omega$ is a compact set and positively invariant. If $V$ is positive definite then $\Omega$ can be obtained as the sublevel set of $V$, i.e., $\Omega := \{ x\ |\ V(x) \leq c\}$ for some $c$ and showing that $\dot{V}\leq 0$ for all elements in $\Omega$. If you can find a Lyapunov function such that your set $\Gamma = M$ then you can use LaSalles to show that you converge to the set $\Gamma$.


问:
prove $\left\{\begin{array}{l}X^{\prime \prime}+\lambda X=0,0<x<1 \\ X(0)=0, X(1)+X^{\prime}(1)=0\end{array}\right.$ has $\lambda >0$ without solving
Can you help me some idea to prove it or some similar question about it. I know prove $\lambda>0$ is easy when solving it, it's just like normal differential equation.

答:
The idea is that the differential equation either has sines and cosines for solutions (if $\lambda > 0$) or exponentials (if $\lambda < 0$) or linear ($\lambda = 0$).

The second boundary condition looks a bit strange.  If it had been $X(1) = 0$, that would have been enough to conclude it must be trig functions right there.

The $\lambda = 0$ case can be ruled out because $X(1) + X'(1) = 0$ means that there must be a positive slope to a negative value, or vice versa, which doesn't work starting from the origin except with a zero solution.


问:
Multivariable piecewise continuity + notation for "x ≠ y except when x = y = 1/2"
In the problem, I have to find the subset of the domain Z = {(x, y) $\in$ $\mathbb{R}^2$ | x, y > 0} for which the following function is continuous:

\begin{equation*}
h(x,y) = \left\{
        \begin{array}{ll}
            f(x,y) & \text{if x $\neq$ y} \\
            g(x,y) & \text{if $x=y$}
        \end{array}
    \right.
\end{equation*}

I have proved that the $\lim_{(x,y) \to (n,n)} f(x,y)$ = g(n,n) when n =$\frac12$

I took this to mean that h(x,y) is continuous for {x>0, y>0, x $\neq$ y except when x=y=$\frac12$}

Could I write this as D = [f(x, y) $\in$ $\mathbb{R}^2$ | x>0, y>0, x $\neq$ y \ {x=y=$\frac12$}]?

I might have completely missed the mark, but any help is greatly appreciated, thanks!

答:
I would write this as
$$\{(x, y) \in (0, \infty) \times (0, \infty) \mid x \neq y\} \cup \left\{\left(\frac{1}{2}, \frac{1}{2}\right)\right\}.$$
Just as a word of caution, you shouldn't use the set difference when dealing with equations. Writing $x \neq y \setminus \{x = y = \frac{1}{2}\}$ doesn't make sense. I understand what you're going for (you're reading \ as "except for"), but this is not how the notation works. It's really only used for sets.


问:
Minimize the perimeter of a quadrilateral
Given ABCD is a square. E is a point on BC (not overlapping with B or C), and F is a point on CD (not overlapping with C or D).  $\angle$EAF=$45^o$, P is the mid point of EF.  AP=x, BE+DF=y.

Question: Are there two points M,N on AB and AD, respectively, so that quadrilateral MEFN has minimum perimeter? If yes, express this perimeter with x and y.

I started to explore the relationship among different segments and found y= EB+DF=EF, but do not know how to continue from this discovery.

答: As a hint, I offer this figure:

Body: As a hint, I offer this figure:


问:
Prove that the integral of a monotone function is between a given area interval
So I have to prove the following:

If $f$ is monotone in an interval $I:\{a\leq x\leq b\}$, $\int_a^bf(x)dx$ is between $f(a)(b-a)$ and $f(b)(b-a)$.

I clearly see the geometric interpretation of this. The previous statement is saying that the area represented by the integral is "between" the area of a rectangle with height $f(a)$ and base $(b-a)$, and a rectangle with height $f(b)$ and base $(b-a)$. Obviously the function could be strictly increasing, decreasing or just being a constant (as far as I know that's the definition of a monotone function), as such I can visualize that the statement holds, however I don't seem to be able to put this into math notation and actually prove it.

I have tried to use a Riemann sum to solve it, I've also tried to use upper and lower integrals but I didn't really know what to do with those. Any help or insight on how to tackle this would be truly appreciated.

EDIT 1:

So, from the comment I came up with the following (for the case in which we have a strictly increasing function):

If $f(b)>f(a)$, then $\int_a^b[f(b)-f(a)dx]\geq 0$, since the integrand is positive, while $\int_a^b[f(a)-f(x)]dx\leq 0$, because the integrand is negative. From this we get the following inequality:

$\int_a^b[f(a)-f(x)]dx\leq 0\leq\int_a^b[f(b)-f(a)]dx$

By using properties of the integrals, we simplify as:
$\int_a^bf(a)-\int_a^bf(x)dx\leq 0\leq\int_a^bf(b)dx-\int_a^bf(x)$

It is then obvious,

$\int_a^bf(a)\leq\int_a^bf(x)\leq\int_a^bf(b)$

For this last part I am not really sure how to convert the integrals of the function evaluated at the lower and upper bound of the integral into $f(a)(b-a)$ and $f(b)(b-a)$. I can see that the height will always be the same and the base is just going to be $(b-a)$. Maybe a Riemann sum of the previous integral would help? I have tried doing this:

$\int_a^bf(a)=\lim_{x\to \infty}\sum_{k=1}^nf(\xi)\Delta_kx$. Since we have that for every $\xi$, $f(\xi)$ will be the same as $f(a)$, then by properties of limits and sums (because f(a) is a constant), $\lim_{n\to \infty}\sum_{k=1}^nf(a)\Delta_kx=f(a)\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$, and since the sum of all the little sub-intervals that compose $\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$ is $(b-a)$, we finally have $f(a)(b-a)$. We then do the same for $f(b)$. This leads us to,

$f(a)(b-a)\leq\int_a^bf(x)\leq f(b)(b-a)$

答: If $f(b)> f(a),$ then $\int_a^b (f(b)-f(x)) d x \geq 0,$ since the integrand is positive, while $\int_a^b (f(s)-f(x)) d x \leq 0,$ since the integrand is negative.


问:
Show that if $X^*AX\in \mathbb{R}$, then $A$ is hermitian
As the title says the goal is to prove: If $X^*AX\in \mathbb{R}$ for all $X$, then $A$ is Hermitian.

This is my attempt:

$X^*AX=P\in \mathbb{R}$ then $(X^*AX)^*=(AX)^*X=X^*A^*X=P^*=P^t$. Furthermore, let $X$ be invertible, then $A=(X^*)^{-1}PX^{-1}$ and $A^*=(X^*)^{-1}P^tX^{-1}$. If I can show that $P=P^t$, then I am done, but I haven't been able to do that. I suspect one has to choose the right $X$ (since the above it is true for all invertible $X$.

I would appreciate if someone can give me a hint on how to go on or a different approach to prove the statement. Ideally, I don't want to invoke theorems. Thanks!

答:
First we show that if $\langle AX, X \rangle =0 ~~~\forall ~X$ then $A\equiv0$. It suffices to show that $\langle AX, Y \rangle =0~~~\forall ~X,Y$ , because \begin{align}
&\langle AX, Y \rangle =0~~~\forall~ Y \\
&\Rightarrow AX=0 ~~~\forall~ X \\
&\Rightarrow A \equiv0
\end{align}
Let $\alpha, \beta \in \mathbb{C}$ be arbitrary, by hypothesis $\forall ~X,Y$
\begin{align}
&\langle A(\alpha X+\beta Y),\alpha X+\beta Y \rangle =0\\
&\Rightarrow |\alpha|^2 \langle AX,X \rangle + |\beta|^2 \langle AY,Y\rangle+\alpha\bar\beta\langle AX,Y\rangle+ \bar\alpha\beta\langle AY,X \rangle=0 \\
&\Rightarrow \alpha\bar\beta\langle AX,Y\rangle+ \bar\alpha\beta\langle AY,X \rangle=0
\end{align}
Now putting $\alpha=\beta=1$ then $\alpha=i,~\beta=1$ in the above equation we get two equations in two unknowns, solving we get $\langle AX,Y \rangle=0~~~\forall~X,Y$ this implies $A\equiv 0$. Thus we get $\langle AX, X \rangle =0~~~\forall~X \Rightarrow A\equiv0$.

Note that $\overline{\langle X, AX \rangle}=\langle X, AX \rangle$.
\begin{align}
&\overline{\langle X, AX \rangle}=\langle AX, X\rangle ~~~\forall ~X\\
&\Rightarrow \langle X, AX \rangle =\langle X, A^*X \rangle ~~~\forall ~X\\
&\Rightarrow \langle X, (A-A^*)X \rangle =0 ~~~\forall ~X \\
&\Rightarrow (A-A^*)X=0 ~~~\forall ~X \\
&\Rightarrow A=A^*
\end{align}


问:
Is it true that if $E(X)=0$, then $E(X\mid\mathcal{G})=0$?
If $\mathcal{G}\subset \mathcal{F}$ is a sub sigma-algebra and $X$ is $\mathcal{F}$-measurable r.v. such that $E(X)=0$. Is it correct that $E(X\mid\mathcal{G})=0$? I know the converse is true.

答: No. Let $\Omega$ be some finite set, $\mathcal F = \mathcal G=2^\Omega,$ and $X$ be some nonconstant random variable with expectation zero.


问:
Indices confusion
Consider the equation $3^{y} = 9^{x}$

It follows that $3^{y} = 3^{2x}$

But $3^{2x} \equiv (3^{x})^{2} \equiv (3^{2})^{x}$ (I think? Since e.g. $(x^{2})^{3} \equiv x^{2 \cdot 3} \equiv (x^{3})^{2}$ right?)

So which of the following is correct? $y = 2x$ or $y = x^2$ or $y = 2^x$?

Thanks!

答: $y=2x$ is the correct option. Because if $$x^{a}=x^{b} \Longrightarrow a=b$$</p>

Title: Indices confusion


问:
Prove that the group, under multiplication, of all nonzero elements in a finite field must be a cyclic group.

Prove that the group, under multiplication, of all nonzero elements in a finite field must be a cyclic group.

This is what I did, but I'm not sure if it's right:

First, we look at the additive group G of the field in order to determine the elements in the field. Any finite additive group in a field must be of prime characteristic. Otherwise, if n (where n is not prime) was the characteristic of G, then $n=uv=0 \implies$ either u or v must be zero (since all fields are integral domains), which contridicts the fact that n is the smallest element that equals zero.

If we have a finite additive group of characteristic p in the field, then it must be cyclic (since it is a group of prime order). Now if we want to look at the multiplictative group of the field, we must only look at the invertible elements, which is precisely the group $\Bbb{Z}^{\times}_p \cong \Bbb{Z}_{p-1}$.

Do you think my proof is correct?

Thank you in advance

答:
One way to do this is to look at the group's invariant factor decomposition.

Consider the invariant factor decomposition $\Bbb{F}_{p^n}^{\times} \cong \Bbb{Z}/k_1\Bbb{Z} \oplus \Bbb{Z}/k_2\Bbb{Z} \oplus ... \oplus \Bbb{Z}/k_r\Bbb{Z}$, where $k_{i+1} | k_i$ for all $i$. We wish to show that $\Bbb{F}_{p^n}^{\times}$ is cyclic, or that $r=1$.

The polynomial $x^{k_1} = 1$ must have at most $k_1$ roots. Since, for all $i$, $k_i | k_1$, $x^{k_1} = 1$ for every element $x \in \Bbb{F}_{p^n}^{\times}$. That should be enough for you to find a contradiction if $r>1$.

One possible source of confusion here is that I am writing all groups multiplicatively, using $1$ for the identity, juxtaposition for the group operation and exponents, while the groups $\Bbb{Z}/k\Bbb{Z}$ are usually written additively.


问:
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$. Is it possible to find a closed form for this sum? I was trying to upper bound this sum by another series, but could not find one.
Can anyone help?

Body:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{a^{\frac{n}{2}}}=\text{Li}_{-\frac{1}{2}}\left(\frac{1}{\sqrt{a}}\right)$$ where appears the polylogarithm function.

For $a=2$, the result is
$$4.1450444026566332049672175338821574521236562661674935536$$


答:
The magic words are: Bayes' rule.

$$P(A|B) = \frac{P(B|A) P(B)}{P(A)}.$$

In this case: $A$ is the event of getting exactly $n$ heads, so $P(A) = \frac{\binom{2n}{n}}{2^{2n}}.$ $P(B)$ is the probability of getting at least one head in the first $n$ tosses: $1-\frac1{2^n}.$
$1-P(B|A) = \frac1{2^{2n}}$ (since this is the event of getting all tails in the first $n$ tosses and all heads in the last $n$ tosses).

Title: Probability of n heads in 2n tosses, given at least one head in n tosses

Body: Suppose we toss a fair coin $2n$ times. Conditioned on the fact that the coin came up heads at least once in the first $n$ tosses, what is the probability that we get precisely a total of $n$ heads over all $2n$ heads?


答:
Suppose you perform $2n$ coin flips. There are $2^{2n}$ possible outcomes, because each of the $2n$ flips can be either heads or tails.

Suppose you perform $2n$ coin flips and got exactly $n$ heads. Combinatorically, this involves picking $n$ out of $2n$ coins to come up heads; there are ${2n \choose n}$ such outcomes (out of $2^{2n}$ possible outcomes).

Suppose you perform $2n$ coin flips and get $n$ tails followed by $n$ heads: TTT...TTT HHH...HHH.  This is the one and only outcome where you have exactly $n$ heads but none of them occur within the first $n$ tosses.

Hence by (1) and (2), there are exactly $\left[{2n \choose n} - 1\right]$ outcomes that have exactly $n$ heads and also at least one head in the first $n$ flips.

Because the outcomes are all equally likely by assumption, the probability of this event is $ \left[{2n \choose n} - 1\right] / 2^{2n}$. This is the probability of getting exactly $n$ heads, at least one of which occurs within the first $n$ tosses.

Let $X$ denote the event "You flip a coin $2n$ times". Let $A$ denote the event "...and get exactly $n$ heads". Let $B$ denote the event "...and there are no heads in the first $n$ tosses." Then:

The quantity you want to compute is "The probability of getting exactly $n$ heads, conditioned on getting at least one head in the first $n$ tosses." That is, $P(A|\overline B) = \frac{P(A\cap \overline B)}{P(\overline B)}$.

By (3), the probability $P(A\cap \overline B)$ is $\left[{2n \choose n} - 1\right] / 2^{2n}$.

The probability of $B$ is the probability of getting no heads in the first $n$ tosses. How many outcomes are like this? : they have $n$ tails for the first $n$ tosses, and either heads or tails for the  next $n$ tosses; hence $2^n$ possibilities overall.  Because each outcome is equally likely, their combined probability is $P(B) = 2^n / 2^{2n} = \frac{1}{2^n}$. Therefore, the quantity we really want, $P(\overline B)$, is equal to $1-2^{-n}$.

Hence, completing the division, the probability is $$P(A\cap\overline B)/P(\overline B) = \frac{\left[{2n \choose n} -1\right]}{2^{2n}}\left[\frac{1}{1-2^{-n}}\right].$$


问:
The coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$
To find the coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$.

I saw a solution to this problem in which the above expression was differentiated 2 times and they put x=0. I really can't understand the significance behind these steps.

答:
We know that, whatever the expansion of $\left(x^3 + 2x^2 + x +4 \right)^{15}$ is, it will look something like
$$
\left(x^3 + 2x^2 + x +4 \right)^{15} = c_0 + c_1 x + c_2 x^2 +c_3x^3+ ... + c_{45} x^{45}
$$
where $c_i$'s are some constants possibly equal to $0$. Now, in the above equation what you want to know is the value of $c_2$. So now what happens if we differentiate the above equation twice? We get
\begin{align*}
\frac{d^2}{dx^2}\left(x^3 + 2x^2 + x +4 \right)^{15} & = \frac{d^2}{dx^2}\left(c_0 + c_1 x + c_2 x^2 + c_3 x^3 + ... + c_{45} x^{45}\right)\\
& = \frac{d}{dx}\left(c_1  + 2c_2 x + 3c_3 x^2 + ... + 45c_{45} x^{44}\right)\\
& =  2c_2  + (3\cdot 2c_3) x + ... + (45 \cdot 44c_{45}) x^{43}
\end{align*}
Plugging in $x=0$ in the R.H.S. of the above makes all the terms after $2c_2$ go to $0$, so we get
$$
\frac{d^2}{dx^2}\left(x^3 + 2x^2 + x +4 \right)^{15} \Bigg\vert_{x=0} = 2 c_2
$$
from where, by using the chain rule on the L.H.S., it's easy to solve for $c_2$.


问:
Finding the matrix from the null space
If I had a kernel as span of a line (4,7,0), how do I find the matrix from this kernel. Assume no rows and columns have been given for the matrix. I have been at this for days and am completely clueless.</p>

答: If all you know is the kernel, then you can't determine the matrix.  All you know is that each row is orthogonal to the kernel.</p>

Title: Finding the matrix from the null space


问:
Does gradient descent converge in convex optimization problems? If so, how?
Sorry in advance if this question sounds too broad or a little bit too obvious.

I know for sure that gradient descent, i.e., the update equation

$x_{k+1} = x_k - \epsilon_k \nabla f(x_k)$

converge to the unique minimizer of $f$ with domain $\text{dom}(f) = \mathbb{R}^n$ whenever $f$ is strictly or strongly convex.

However, I could not remember if it converges to a minimizer in convex functions, and how it achieves this convergence.

What is bothering me is that

I've seen some conflicting results where instead of $x_k$, an averaged sequence $\hat x_{k} = \frac{1}{K} \sum_k x_k$ converges.

I've also seen conflicting results where the step size is decreasing $o(1/k)$ vs it is constant.

There is also the issue of weak vs strong convergence. I'm not sure what this means exactly.

I have know some results but they are for quadratic functions, not for convex functions in general.

Can someone chime in on what this basic result in optimization look like?

答:
In https://stanford.edu/~boyd/papers/pdf/monotone_primer.pdf, section 5, subsection "Gradient method" or "Forward step method" shows a proof for functions that are strongly convex with Lipschitz gradient, using constant step sizes.

If the function is convex with Lipschitz gradient then https://www.stat.cmu.edu/~ryantibs/convexopt-F13/scribes/lec6.pdf
shows a proof of convergence with constant step sizes.

For strictly convex functions (not Lipschitz), gradient descent will not converge with constant step sizes. Try this very simple example, let $f(x) = x^{4}$. You will see that there is no constant step size for gradient descent that will converge to $0$ (for any initial condition). In this case, people use diminishing step sizes. I've never seen results about convergence of the average but it might be what happens when you use a constant step size on functions that are strictly convex and don't have Lipschitz gradient.

Unless $x$ lives in an infinite-dimensional space, weak and strong convergence is the same and I wouldn't worry about it. Here is additional reading if you are interesting,
https://people.maths.bris.ac.uk/~mazag/fa17/lecture9.pdf


答:
Let $f(x)=(x^3+2x^2+x+4)^{15}=\sum_{k=0}^{45}a_kx^k$. Now $f''(x)=\sum_{k=2}^{45}k(k-1)a_kx^{k-2}$ and on plugging zero we get $f''(0)=2(2-1)a_2$.

An alternative is to note that to get a term with $x^2$, we either select the $2x^2$ term from any one of the $15$ factors and the constant term from the remaining factors, or the $x$ term from any two of the $15$ factors and constants from the rest. The required coefficient thus should be$$\binom{15}14^{14}2+\binom{15}24^{13}$$

Title: The coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$

Body:
To find the coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$.

I saw a solution to this problem in which the above expression was differentiated 2 times and they put x=0. I really can't understand the significance behind these steps.


问:
How to determine where a function is complex differentiable
I know the definition of complex differentiability and also am aware that $f$ is complex differentiable at $z_0$ iff it is real differentiable at $z_0$ and that the partial derivatives satisfy the Cauchy-Riemann Equation.

In general, what are the steps required to show that $f$ is (not) complex differentiable at a point $z\in\mathbb C$?

EDIT: Made a typo on $f(z)$.
I was also given an example of a $f(z)=exp(-z^{-4})$ to show that $f$ satisfies the Cauchy Riemann equations at $0$ but is still not complex differentiable at $0$.

Does it mean that whenever we want to show whether $f$ is not differentiable at $z\in\mathbb C$, the Cauchy-Riemann Equation doesn't give us any information on it and we just have to work with the definition of complex differentiability and find sequences that tend to different limits?

答:
Two comments suffice for the specific function:

$f(z)=\exp(−z^4)$ is not even defined at $z=0$, so of course it does not satisfy CR in the first place. I assume that you considered the (non-continuous) extension obtained from additionally defining the value $f(0)=0$.

– Hagen von Eitzen May 1 at 13:19

For this new $f(z)$, try the points $z_n=1/(e^{i\pi/4}n)$. Can the function be continuously extended at $0$? Not even speaking of differentiability.

– julien May 1 at 13:24

As for the general question:

Does it mean that whenever we want to show whether $f$ is not differentiable at $z\in \mathbb C$, the Cauchy-Riemann Equation doesn't give us any information on it and we just have to work with the definition of complex differentiability and find sequences that tend to different limits?

That depends on what the function is. If you are given $f(x+iy)=2x+iy$, you'll naturally point to the failure of the  Cauchy-Riemann equations. Working with difference quotients $(f(z)-f(z))/(z-a)$, while possible, would be unnecessarily awkward.

When you inspect a function, you should be able to tell if it looks okay in the real sense, or if there are some quirks. If it looks real differentiable, then the CR equations are the thing to consider. Otherwise you try to disprove real differentiability in some way.


问:
Law of the Excluded Middle for Mere Propositions
In the Homotopy Type Theory book, they define the (non-contradictory) law of the excluded middle as
$$\operatorname{LEM}:\prod_{A:\mathcal U}\big(\operatorname{isProp}(A)\to(A+\lnot A)\big).\tag{3.4.1}$$
Here, $\operatorname{isProp}(A)$ is the assertion that all elements of $A$ are propositionally equal. The book goes on to treat this as an axiom; I'm guessing this means that $\operatorname{LEM}$ does not follow from univalence but would like confirmation.

Additionally, how would we go about showing that we cannot exhibit $\operatorname{LEM}$?

答: As Mark says, univalence does not imply LEM, and one way to prove this is to exhibit a model satisfying univalence but not LEM.  To my knowledge the first such model was constructed in my paper Univalence for inverse diagrams and homotopy canonicity, and is not that complicated to describe: given essentially any model of HoTT, one can construct a new model in which the types are pairs $(A_0,A_1)$ where $A_0$ is a type in the old model and $A_1$ is an $A_0$-indexed family of types in the old model.  This new model, called the "Sierpinski topos" over the old model, violates LEM, because it has three different closed propositions: $0=(0,0)$, $1=(1,1)$, but also $(1,0)$.  It's based on the "simplest Kripke model" for set-based constructive mathematics.  Now we know there are many other models too, including models in all higher toposes and also cubical set models that are computational.


问:
Nature and number of solutions to $xy=x+y$

Find all solutions to $$xy=x+y$$ Initially the given condition was $x,y\in \Bbb{Z}$.

$$$$In this case, I just guessed that the solutions were $(0,0)$ and $(2,2)$. As far as I can see, these are the only 2 integral solutions possible. However, i'm quite surprised as usually a Diophantine Equation has infinite solutions. Could somebody please show me how to actually $solve$ this equation instead of just guessing the values? Is there any way to $show$ that there are just 2 sets of solutions?

$$$$Secondly, what if $x,y\in \Bbb{R}$? In that case how could the equation be solved?
$$$$For both these conditions on the values of $x,y$, is there any way to use Coordinate Geometry to achieve an answer?
$$$$
Many thanks in anticipation!

Edit: I forgot to mention that I know the solution using factoring. I was hoping to find a solution using Coordinate Geometry. I was told by a friend that the non integral solutions lie on 2 unique lines, and am hence particularly interested in a geometrical solution.

答: To solve $x+y=xy$ in integers, note that it implies that $x\mid y$ and $y\mid x$.  Therefore, $y=\pm x$.  Can you show from there that $(0,0)$ and $(2,2)$ are the only integer solutions?


问:
element of fraction of DVR
Let us now assume that $A$ is a discrete valuation ring. Any element $\pi \in A$ for which $v(\pi) = 1$ is called a uniformizer. Such a uniformizer necessarily exists, since $v$ maps $A$ surjectively onto $\mathbb Z_{\ge 0}$.

If we fix a uniformizer $\pi$, why every $x ∈ k^\times$ can be written uniquely as
$x = u\pi^n$
where $n = v(x)$ and  $u = x/\pi^n ∈ A^\times$ ?

I think this is fundamental fact of DVR, but how to prove formally from definition?

Thank you in advance.

答: Well, just define $u=x/\pi^n$.  Then $v(u)=0$, so $u\in A$.  Also, $v(u^{-1})=0$ so $u^{-1}\in A$ as well, so $u$ is a unit of $A$.

Title: element of fraction of DVR


问:
Few topological operations under which the classes of CDH-spaces
It's known that $\mathbb{R}$ is a CDH space. Does anyone have any idea why $\omega^{\omega}$ is a CDH space? Then, Does anyone have any suggestion why $\mathbb{R}\times\omega^{\omega}$ is not a CDH space?

Also, the Cantor set $K$ in $[0,1]$ is CDH, and any compact metrizable space is continuous image of $K$. Then why this implies that continuous images of CDH-spaces need not be CDH?

Finally, note that $\mathbb{Q}$ is not CDH, so the property of being CDH is not hereditary.

Thanks

答: Following the ideas of @Brian Scott we have the following


答:
To build such a matrix, first note that the nullspace of our matrix has dimension $1$ (it's spanned by a single non-zero vector), and the number of columns must be $3$ (since this single non-zero vector lives in $\Bbb{R}^3$). By the rank-nullity theorem, the rank of the matrix must be $3 - 1 = 2$. This implies that our matrix must contain at least $2$ rows (and only $2$ linearly independent rows).

Further, as Robert Israel points out, these rows must be orthogonal to $(4, 7, 0)$. The way matrix multiplication works is by a series of dot products, and multiplying a column vector to a matrix simply dot products this column with each row, forming a new column vector. Since multiplying the matrix with $(4, 7, 0)$ should produce the zero vector, each of these dot products must be zero.

So, let's build a $2 \times 3$ matrix, with two linearly independent rows (so that the rank is $2$), each of which is orthogonal to $(4, 7, 0)$ (so that the kernel contains $(4, 7, 0)$). By the rank-nullity theorem, the kernel of this matrix will be only $1$ dimension, and hence will be exactly equal to $\operatorname{span}(4, 7, 0)$.

Our two vectors can be $(7, -4, 0)$ and $(0, 0, 1)$, for example. This gives us a matrix

$$\begin{pmatrix}
7 & -4 & 0 \\
0 & 0 & 1
\end{pmatrix}.$$

By the reasoning above, I'm confident that this matrix satisfies our conditions. But, I would suggest computing the general solution, and verifying that it comes to $\operatorname{span}(4, 7, 0)$.

Title: Finding the matrix from the null space

Body: If I had a kernel as span of a line (4,7,0), how do I find the matrix from this kernel. Assume no rows and columns have been given for the matrix. I have been at this for days and am completely clueless.</p>


问:
How many numerical multiplications for the product of three matrices?
For example you have the matrices $A, (5 \times 8), B, (8 \times 4), C (4 \times 10). $
The question wants you to find the number of multiplications if you were to multiply these matrices like $(A\times B)\times C$.

The answer is $5 \times 4 \times 8 + 5 \times 10 \times 4 = 160 + 200 = 360$ multiplications.

答:
For a matrix mutiplication of order $(m\times n)$ and $(n\times p)$.

No of mutiplications involved are $= m\cdot n\cdot p$.

No Of additions involved are $= m\cdot n\cdot (p-1)$

so here for

first mutiplication: $5\cdot 8\cdot 4= 160$

Second mutiplication: $5\cdot 4\cdot 10 = 200$

Total $= 360$

Similarly Total addition would be $=300.$


问:
Whether the given series are convergent or divergent:
Whether the given series is absolutely convergent, conditionally convergent or divergent:

$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.

$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.

$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$

Am i right?

答:
For (b), $\frac{n}{n+2} \to 1$,
so the series is absolutely
fairly violently divergent.

The sum of consecutive even-odd terms is
$\frac{2n}{2n+2} - \frac{2n+1}{2n+3}
= \frac{2n(2n+3)-(2n+2)(2n+1)}{(2n+2)(2n+3)}
= \frac{4n^2+6n-(4n^2+3n+2)}{(2n+2)(2n+3)}
= \frac{3n-2}{(2n+2)(2n+3)}
$,
and the sum of consecutive odd-even terms is
$-\frac{2n+1}{2n+3}+\frac{2n+2}{2n+4} -
= \frac{-(2n+1)(2n+4)+(2n+2)(2n+3)-}{(2n+3)(2n+4)}
= \frac{-(4n^2+7n+4)+(4n^2+6n+6)}{(2n+3)(2n+4)}
= \frac{-n+2}{(2n+3)(2n+4)}
$,
so the series is not even conditionally convergent.


问:
Why is square root by long division found so?
We were taught the long division method of finding square root in junior classes. The  logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.

答:
I’ll just add the following intuition to the other answers.

Suppose you are extracting the square root of $x$. First you guess, say, $x_0$. If $x_0^2\neq x$, then you must correct your guess by adding some new term $x_1$ to your initial guess $x_0$. Your guess for $\sqrt{x}$ is thus updated to $x_0+x_1$, whose square is $x_0^2+\color{red}{2x_0x_1+x_1^2}$.

The key point of the algorithm is that incrementing your guess by $x_1$ increments the square of your guess by the amount in red, which can be written $\color{red}{x_1(2x_0+x_1)}$. This fact helps you determine by how much to increment your guess.

Now apply this idea to $\sqrt{2704}$. First we guess $50$, since $50^2=2500$ gets us pretty close. The gap between the squares is $204$. If we increment our guess of $50$ by some amount $a$, we’ll increment the square of our guess by $a(2\cdot50+a)$. We’d like this to be as close to $204$ as possible. Notice $a=1$ increments by $101$, but $a=2$ increments by $204$: perfect! So we increment our initial guess of $50$ by $2$, and since that fills the gap perfectly we’re done! We’ve discovered $2704$ is a perfect square, and we iteratively arrived at $52$ as the square root.

It all boils down to noting how the square of your estimate changes when you change your estimate.


问:
How do we find the prime ideals of a ring of integers of a number field?
How do we find the prime ideals of a ring of integers of a number field?

For example, for $F=\mathbb Q(\sqrt{-5})$ the ring of integers of $F$ is $\mathbb Z[\sqrt{-5}]$ (since $-5\equiv3 \pmod 4$). How can we determine the prime ideals of this ring? Another problem is the corresponding valuations.

答:
A starter would be the observation that prime ideals of an extension field are always above prime ideals of the ground field. In this case, the extension field is $F$, and the ground field is $\mathbb Q$, where all prime ideals are principal ideals generated by prime numbers. Therefore we are reduced to looking for the prime ideal factorisations of primes in the extension $F$.

By the theorem of Dedekind, we can find the prime ideal factorisation of $p\in \mathbb Z$ by examining the factorisation of the polynomial $f=x^2+5$ modulo $p$. If $p=5$, then $(p)$ ramifies in $F$, and is equal to $\mathfrak p^2$ for some prime ideal $\mathfrak p$ of $F$. In fact, as $5=-\sqrt{-5}^2$, we know that $(5)=(\sqrt{-5})^2$. And if $p=2$, then $x^2+5\equiv(x+1)^2\pmod2$ shows that $(2)$ ramifies as well: $(2)=(1+\sqrt{-5})^2$.

Now, if $p\ne2,5$, then $(p)$ is unramified in $F$, so either $p$ is inert, i.e. $(p)$ is still a prime ideal in $F$, or $p$ splits, meaning that $(p)=\mathfrak p_1\mathfrak p_2$. And a prime $p$ splits if and only if $x^2+5\equiv 0\pmod p$ has solutions in integers. Namely, if and only $(\dfrac{-5}{p})=1$. By quadratic reciprocity, we find that this is thus equivalent with $p\equiv 1, 3, 7, 9\pmod{20}$.

In summary, $\begin{cases}(p)=(\sqrt{-5})^2 \text{ if }p=5 \\ (p)=\text{ still prime if }p\equiv 11, 13, 17, 19\pmod{20}\\ (p)=\mathfrak p_1\mathfrak p_2 \text{ if } p\equiv1, 3, 7, 9\pmod{20}\end{cases}.$

So all prime ideals of $F$ are classified now.

As to the second question, I cannot understand properly the content of the question, as we know that the valuations correspond to the prime ideals, and hence information about one is information about the other. As a further detail, consider the localisations at various primes: this should make clear what the corresponding valuations are.

Please, inform me of errors.

Thanks for the attention.


问:
show that $y^2 = x^5 - x +2 $ has no integer solutions
I considered everything$\mod 3$:

$x,y \equiv 0 \mod 3 ,\;\;\;\; y^2 \equiv 0 \mod 3 ,\;\;\;\; x^5 \equiv 0 \equiv x \mod 3$

$x,y \equiv 1 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 1 \equiv x \mod 3$

$x,y \equiv 2 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 2 \equiv x \mod 3$

thus in particular $x^5 - x \equiv 0 \mod 3, \;\;\; \forall x \in \mathbb{Z},$ and furthermore
$x^5 - x + 2 \equiv 2 \mod 3, \;\;\; \forall x \in \mathbb{Z}$.

But, $y^2 \equiv 0,1 \mod 3$, and if we were considering that there would be an integer solution then both sides should have the same remainder modulo $3$, but LHS and RHS never have the same remainder thus there are no integer solutions?

This is my first time doing a proof of this form, I'm familiar with modular arithmetic, however I've never used it to prove that a polynomial equation didn't have integer solutions.

答:
Your reasoning is correct.

Here a similar way using little Fermat for $p=5$:

$$x^5\equiv x \mod 5$$

Hence,

$$y^2\equiv 2 \mod 5$$

but the quadratic residues $\mod 5$ are $0,1,4$.

So, no solution in integers exists.


问:
Prove that among any five distinct real numbers there are two, $a$ and $b$, such that $|ab+1|\gt|a-b|$. Solution without trigonometry?

Prove that among any five distinct real numbers there are two, $a$ and $b$, such that $\lvert ab+1\lvert
 \ \gt \ \vert a-b \vert$.

Solution: I have solved the above problem using Trigonometrical substitution by writing $-90^\circ <\tan\ x_k<90^\circ$, $k=1,2,3,4,5$,  and considering the intervals $(-90^\circ,-45^\circ)$; $(-45^\circ,0^\circ)$; $(0^\circ,45^\circ)$; $(45^\circ,90^\circ)$. Then by Pigeonhole Principle , at least two of $x_1,x_2,x_3,x_4,x_5$ will lie in the same interval. Let those two be $(x_i,x_j)$ as $\vert x_i-x_j\vert\lt 45^\circ$ setting $a=\tan x_i$ and $b=\tan x_j$ we get the desired inequality. $\square$

However, I was wondering whether we could approach this problem in an algebraic way because this question is in an Algebraic Book which I have.

Any help would be appreciated.
Thanks!

答:
Your use of the tangent function is a good inspiration from your knowledge of trigonometry, but you can avoid it just by making the intervals be for the $x_k$ instead of their tangents and making the intervals $(-\infty,-1], (-1,0],(0,1],(1,\infty)$.  Now show that if two of the points are in the same interval the desired inequality is satisfied.  For example, if $a,b \in (0,1]$ we have
$$|ab+1|=ab+1\gt 1 \gt |a-b|$$

You can prove it for the other three intervals as well.  Someone who does not know any trig might think you had a miraculous inspiration to choose these intervals, but $0$ and $\pm 1$ are often special when multiplication is involved.


答:
Essentially the same argument can be made algebraically: at least one of the intervals $(\leftarrow,-1)$, $[-1,0)$, $[0,1)$, and $[1,\to)$ must contain two of the numbers, say $a$ and $b$. We may assume that $a<b$ and let $d=b-a$. (Note that your four intervals are not quite exhaustive, since all of them are open.) Taking the intervals in reverse order:

if $1\le a<b$, then $$|ab+1|=a^2+ad+1\ge d+2>d=|a-b|\,;$$

if $0\le a<b<1$, then $$|ab+1|=a^2+ad+1\ge 1>d=|a-b|\,;$$

if $-1\le a<b<0$, then $$|ab+1|=ab+1>1>d=|a-b|\,;\text{ and}$$

if $a<b<-1$, then $$|ab+1|=b^2-bd+1>2-bd=2+|b|d>d=|a-b|\,.$$

Title: Prove that among any five distinct real numbers there are two, $a$ and $b$, such that $|ab+1|\gt|a-b|$. Solution without trigonometry?

Body:
Prove that among any five distinct real numbers there are two, $a$ and $b$, such that $\lvert ab+1\lvert
 \ \gt \ \vert a-b \vert$.

Solution: I have solved the above problem using Trigonometrical substitution by writing $-90^\circ <\tan\ x_k<90^\circ$, $k=1,2,3,4,5$,  and considering the intervals $(-90^\circ,-45^\circ)$; $(-45^\circ,0^\circ)$; $(0^\circ,45^\circ)$; $(45^\circ,90^\circ)$. Then by Pigeonhole Principle , at least two of $x_1,x_2,x_3,x_4,x_5$ will lie in the same interval. Let those two be $(x_i,x_j)$ as $\vert x_i-x_j\vert\lt 45^\circ$ setting $a=\tan x_i$ and $b=\tan x_j$ we get the desired inequality. $\square$

However, I was wondering whether we could approach this problem in an algebraic way because this question is in an Algebraic Book which I have.

Any help would be appreciated.
Thanks!


答:
You don't have to speak of $5$ distinct real numbers, since you're not using the remaining three
But instead I suggest, $2$ distinct number
$$\lvert ab+1\lvert
 \ \gt \ \vert a-b \vert$$
$$ (ab+1)^2 \gt (a-b)^2 $$
$$ a^2b^2+2ab+1 \gt a^2-2ab+b^2$$
$$ a^2b^2+2ab+1+4ab \gt a^2-2ab+b^2+4ab$$
$$a^2b^2+6ab+1 \gt a^2+2ab+b^2$$
$$a^2b^2+6ab+1 \gt ( a+b)^2 $$
$$ab( ab+6)+1 \gt (a+b)^2 $$
It's easy to know that $ ab \gt a+b$, for $a,b \in \mathbb{Z}$
$$ab \cdot (ab+6) +1 \gt (a+b) \cdot (a+b) $$
Then the identity is true, or are you trying to proof it with trigonometry alone


问:
Proof verification: $\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$
While exploring the various applications of integral reduction formulae, I stumbled into what I believe to be a beautiful elementary proof of the equality

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$

I'd greatly appreciate it if someone took the time to review it!

To arrive at the desired result, we'll need to establish the following:

Lemma: For all $n\geq 1$,$$\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)dx=(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)$$

We use mathematical induction to do this. The base case $n=1$ can be proven as follows:

\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^2(x)\text{ }dx &= \int_{0}^{\frac{\pi}{4}}\left[\sec^2(x)-1\right]dx\\
&= \left[\tan(x)-x\right]_{0}^{\frac{\pi}{4}}\\
&= 1-\frac{\pi}{4}\\
&= -\left(\frac{\pi}{4}-1\right)\\
&= (-1)^1\left(\frac{\pi}{4}-\sum_{k=0}^{0}\frac{(-1)^k}{2k+1}\right)\\
&= (-1)^1\left(\frac{\pi}{4}-\sum_{k=0}^{1-1}\frac{(-1)^k}{2k+1}\right)
\end{align*}

Thus, the equality holds for $n=1$. If it holds for some $m\in\mathbb{N}$, then it follows that

\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{2(m+1)}(x)\text{ }dx &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+2}(x)\text{ }dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\tan^2(x)\text{ }dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\left[\sec^2(x)-1\right]dx\\
&= \int_{0}^{\frac{\pi}{4}}\left[\tan^{2m}(x)\sec^2(x)-\tan^{2m}(x)\right]dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\sec^2(x)\text{ }dx-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)
\end{align*}

Here, we make the substitution $u=\tan(x)$. This gives

\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\sec^2(x)\text{ }dx-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right) &= \int_{0}^{1}u^{2m}\text{ }du-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\
&= \frac{1}{2m+1}-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\
&= \frac{(-1)^m(-1)^m}{2m+1}-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\
&= (-1)^m\left(\frac{(-1)^m}{2m+1}-\frac{\pi}{4}+\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\
&= (-1)^m\left(-\frac{\pi}{4}+\sum_{k=0}^{m}\frac{(-1)^k}{2k+1}\right)\\
&= (-1)^{m+1}\left(\frac{\pi}{4}-\sum_{k=0}^{(m+1)-1}\frac{(-1)^k}{2k+1}\right)
\end{align*}

This shows that the equality is true for $n=m+1$. Thus, mathematical induction implies that it is true for all $n\geq 1$.

With that settled, notice that

\begin{align*}
\left|\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right| &= \left|(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)\right|\\
&= \left|\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx\right|\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx
\end{align*}

because $\tan^{2n}(x)\geq 0$. We now obtain an upper bound for the integral $\int_{0}^{\pi/4}\tan^{2n}(x)dx$. Making the substitution $x=\tan^{-1}(u)$ gives

$$\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx=\int_{0}^{1}\frac{u^{2n}}{1+u^2}du$$

It's clear that $u^{2n}/(1+u^2)< u^{2n}$ except when $u=0$, so

$$\int_{0}^{1}\frac{u^{2n}}{1+u^2}du< \int_{0}^{1}u^{2n}\text{ }du=\frac{1}{2n+1}$$

Thus, we have that

$$\left|\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right|<\frac{1}{2n+1}$$

and since $1/(2n+1)\to 0$ as $n\to\infty$, it immediately follows from the squeeze theorem that

$$\lim_{n\to\infty}\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)=0$$

which is equivalent to $\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$. $\blacksquare$

Let me know what you think!

Edit: This is harmless, but should still be addressed: the inequality $u^{2n}/(1+u^2)<u^{2n}$ should have been $u^{2n}/(1+u^2)\leq u^{2n}$, since $u^{2n}/(1+u^2)$ and $u^{2n}$ are equal if and only if $u=0$. The inequality of integrals is still strict, though.

答:
This is not an answer - just writen for your curiosity

In this question, was given for
$$I_n=\int_0^{\frac \pi 4} \tan^n (x)\, dx$$ the result
$$I_{n+1}-I_{n-1}=-\frac 1 n+\frac{1}{2} \left(\psi \left(\frac{n}{4}+1\right)-\psi   \left(\frac{n}{4}+\frac{1}{2}\right)\right)$$ and using the same approach, it is simple to show that
$$I_{n+1}+I_{n-1}=\frac 1 n$$ Adding them gives
$$I_{n+1}=\frac{1}{4} \left(\psi \left(\frac{n}{4}+1\right)-\psi   \left(\frac{n}{4}+\frac{1}{2}\right)\right)$$

So
$$J_n=\int_0^{\frac \pi 4} \tan^{2n} (x)\, dx=\frac{1}{4} \left(\psi \left(\frac{n}{2}+\frac{3}{4}\right)-\psi
   \left(\frac{n}{2}+\frac{1}{4}\right)\right)$$


问:
Why if the columns of $Q$ are an orthonormal basis for $F^n$ then $Q$ is unitary?
Why if the columns of $Q$ are an orthonormal basis for $F^n$ then $Q$ is unitary?

There exist any available theorems to prove it?

答:
With $Q=(q_{i,j})_{i,j=1}^n$ the equation $Q^*Q=I$ becomes explicitly
$$
  \sum_{k=1}^n \overline{q_{k,i}}q_{k,j}=\delta_{i,j}
 \qquad\text{for all $i,j\in\{1,\ldots,n\}$,}
$$
which just says that columns $i,j$ of $Q$ are orthogonal for $i\neq j$, and that their scalar product is$~1$ when $i=j$.</p>


问:
Free variables in first order logic

Let $\sigma = \{\subseteq\}$ be a signature with the binary relation symbol $\subseteq$. For a set $S$, we define the $\sigma-$structure $\mathcal{S} = (\mathcal{P(S), \subseteq^\mathcal{S}})$  with $\mathcal{P(S)}$ and $\subseteq^\mathcal{S}$ being the powerset of $S$ and the ordinary subset relation respectively. Give two formulas $\varphi_1, \varphi_2 \in \text{FO}[\sigma]$ such that:

$ \varphi_1(\mathcal{S}) = \emptyset.$

$\varphi_2(\mathcal{S}) = \{ x \in \mathcal{P(S)} \mid \text{ } \mid x \mid = 1 \}$.

What I understand is that our universe is a set all subsets of a set $S$. For $\varphi_1$ I have:

$$\varphi_1(x) = \forall y (x \subseteq y)$$

Since the empty set is a subset of every set including itself.

For $\varphi_2$:

$$\varphi_2(x) = \forall y (y \subseteq x \rightarrow x \subseteq y \lor \varphi_1(y))$$

Here I'm basically saying that $y$ and $x$ have to be the same set (with one element).

Can you please check my work?

答:
$\varphi_1$ is fine, but $\varphi_2$ actually defines $\{x\in\wp(\mathcal{S}):|x|\color{red}{\le}1\}$, since $\varnothing$ satisfies it. That’s very easily fixed:

$$\varphi_2(x)=\neg\varphi_1(x)\land\forall y\big(y\subseteq x\to x\subseteq y\lor\varphi_1(y)\big)$$

Title: Free variables in first order logic


问:
Is it true that if $E(X)=0$, then $E(X\mid\mathcal{G})=0$?
If $\mathcal{G}\subset \mathcal{F}$ is a sub sigma-algebra and $X$ is $\mathcal{F}$-measurable r.v. such that $E(X)=0$. Is it correct that $E(X\mid\mathcal{G})=0$? I know the converse is true.

答:
The answer and comment above use counter-examples where $X$ is $\mathcal{G}$-measurable.

For the sake of argument, let's do an example where $X$ is not $\mathcal{G}$-measurable (in particular, $\mathcal{F} \neq \mathcal{G}$).

Let $\Omega = \{1,2,3,4\}$, $\mathcal{F}$ be the power set of $\Omega$, and $\mathcal{G} = \{\phi, \{1,2\}, \{3,4\}, \Omega\}$.  Put the uniform measure on $(\Omega,\mathcal{F})$, that is, $\mathbb{P}(A) = \frac{1}{4} (\# A)$.

Define $X$ by $X(1) = -2$, $X(2) = -1$, $X(3) = 1$, and $X(4) = 2$.  A computation shows that $\mathbb{E}(X) = 0$.  On the other hand, $Y = \mathbb{E}(X \mid \mathcal{G})$ is given by $Y(1) = Y(2) = -\frac{3}{2}$ and $Y(3) = Y(4) = \frac{3}{2}$.  In particular, $Y$ is non-constant, even though $\mathbb{E}(Y) = \mathbb{E}(X) = 0$.


问:
Game theory problem about a 5x5 table with 25 entries and 2 players
There is a 5x5 table with 25 cells, each of which has a different positive integer inside. Person A and B play a game.
They both receive a copy of the table.

Person A first picks the largest number that has not been crossed out, then crosses out the number, and all numbers in its row and column, then repeats this 5 times (until all numbers are crossed out).

Person B does the same, but with the smallest number available.

Can the table be filled in such a way that the sum of the 5 numbers person B picked is larger than the sum of the 5 numbers person A picked?

I tried to prove that the answer is no by changing the order, as in the last number person A picks is at least equal to the first number person B picks, the second last number person A picks is equal to the second number person B picks, and so on, but I don't know how to finish the proof.

答:
That's a helpful start. If the cells chosen by A in order are $A_1, \ldots, A_5$ and the cells chosen by B in order are $B_1, \ldots, B_5$, and the number in a cell $C$ is $N(C)$, it is in fact possible to show

$$N(A_{6-k}) \geq N(B_k)$$

for every $k$ in $1, \ldots, 5$.

We know that $B_k$ was chosen from the set of cells not in the same row or column as any of $B_1, \ldots, B_{k-1}$. $A_{6-k}$ was chosen from the set of cells not in the same row or column as any of $A_1, \ldots, A_{5-k}$. Consider these restrictions together, as though on a fresh copy of the board we cross out the rows and columns containing cells $B_1, \ldots, B_{k-1}$ AND the rows and columns containing cells $A_1, \ldots, A_{5-k}$. This is at most $(k-1)+(5-k)=4$ different rows and at most $4$ different columns, so at least one cell $C$ remains. That is, cell $C$ was not crossed out on A's board when $A_{6-k}$ was chosen, and was not crossed out on B's board when $B_k$ was chosen. Since A always chooses a cell with largest value and B always chooses a cell with smallest value,

$$ N(A_{6-k}) \geq N(C) \geq N(B_k) $$

Nothing about the number $5$ was really used here, so the same proof works for a more general game with any size square table.


问:
Help make Wonder Woman's box big.

Wonder Woman wants a box for her lasso. It is to be built from a rectangular piece of steel measuring 25 cm by 40 cm by cutting out a square from each corner and then bending up the sides. Find the size of the corner square which will produce a container that will hold the most.

Per suggestion by user Mark S, rewrote question to accept answer.

答:
I am making this an answer rather than a comment,
because I find the question itself upsetting.

There is enough misogyny in the world and
on the internet
that I object to the phrasing of this question
in terms of "Wonder Woman's box" objectionable.

There are many different ways this exact same problem
could be phrased, and, in my opinion,
this is one of the worst.

Title: Help make Wonder Woman's box big.


问:
Explanation about connected graph
Assume the function $h\colon \Bbb R\to \Bbb R$ with dense graph such that $h\restriction C$ connected for any connected set $C\subset\Bbb R.$ let $f\colon (0,1)\to \Bbb R$ be a homeomorphism.

Now define a function $g\colon \Bbb R\to \Bbb R$ as $g=h\circ f$ on $(0,1)$ and 0 on its complement. I want to check if the graph of $g$ will be connected or not. It is clearly that the graph  will be as union of three connected pieces.

I need to check if the whole graph will be connected or not. I think it will be problem in the end points. Any help will be helpful.

答:
To avoid the question of whether a function $h$ satisfying your assumptions exists, let's just consider a dense, connected set $H\in\Bbb R^2$.

If $f_1:\Bbb R\to(0,1)$ is a homeomorphism, then so is $f:(x,y)\mapsto(f_1(x),y)$, so $K=f(H)$ is connected and dense in $(0,1)\times\Bbb R$. The point $(0,0)$ is a limit point of $K$, because every open neighborhood of $(0,0)$ contains an open subset of $(0,1)\times\Bbb R$, which must contain an element of $K$ by density.

Let $R=\{(x,0):x\le0\}$. You're interested in whether $K\cup R$ is connected. Since $K$ and $R$ are each connected, the only potential disconnection to check is $\{K,R\}$. But $R$ is not open because it contains $(0,0)$ and every open set containing $(0,0)$ includes elements of $K$. So $K\cup R$ is indeed connected.

Title: Explanation about connected graph


问:
How to solve the equation $k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$
I've been struggling a little on this problem for a while. I'm not sure what I'm missing. It goes like this:

$$k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$$

Here $k, a$ are constants and $v, t$ represent velocity and time respectively. I want to know $v$ as a function of $t$. Is it possible to solve this? Any help would be appreciated!

答:
If $v$ is velocity and is given by $v=\frac{ds}{dt}$ then

$$\int_0^tv(\tau)d\tau=\int_0^t\frac{ds}{d\tau}d\tau$$

By the fundamental theorem of calculus and Leibniz's rule:

$$\int_0^t\frac{ds}{dt}d\tau=\frac{d}{dt}\int_0^tsd\tau=\frac{d}{dt}st=s$$

Note: We are only allowed to do this because these specific limits of integration(zero at the left part, variable/parameter on the right part)

Substituting back in the original equation:

Which is 1st order in $v$

$$k=\frac{dv}{dt}(a+2s)^2$$

But because $v$ depends on $s$ we need to integrate it wrt. to $s$, then it's a 2nd order linear ODE:

$$k=\frac{d^2s}{dt^2}(a+2s)^2$$

$$\iint kdtdt=\iint(a+2s)^2dsds$$

$$k\frac{t^2}{2}=\int \frac{(a+2s)^3}{3a}ds+C=\frac{(a+2s)^4}{4a^2}+C$$

Now solve for $s$ and differentiate wrt. $t$ get back your solution to $v$.

Why doesn't distance depend on time?

To OP: try going through each step, i'll leave you to differentiate it on your own. Tell me if you got stuck.


问:
Plot $2|z|=|z-2|$ over the complex plain
I was given a home work to Sketch $2|z|=|z-2|$ on the complex plane.I tried to simplify and found out that the last expression was $3x^2+3y^2+4x-4$. However I don't know how to sketch this on the complex plane and how I can describe the set of points

答:
Note that $\phi(z) = {z-2 \over 2z}$ is a Möbius transformation and you are looking for $\{ z | |\phi(z)|=1 \} = \phi^{-1}(\partial B(0,1))$.

Since the inverse of a Möbius transformation is a Möbius transformation, and Möbius transformations transform circles into circles, we see that the set in question is a circle.

It is simple to verify that $-2, {2 \over 3}$ and ${2 \over \sqrt{3}}i$ are in this locus and from these three points we can easily plot the set.


答:
The equation $2|z| = |z - 2|$ represents the locus of points in the complex plane such that the distance from $(2,0) = 2 + 0i$ is twice the distance from the origin.  So one obvious example of such a point would be $z = 2/3$, since the distance from the origin is $2/3$ and the distance from $2$ is $2 - 2/3 = 4/3$.

Another (slightly less) obvious example would be $z = -2$, since its distance to the origin is $2$, but the distance to $2$ is $4$.

Since you found an equation $3x^2 + 3y^2 + 4x - 4$, where $x$ is the real part of $z$ and $y$ the imaginary part (i.e., $z = x + iy$), what happens when you plug in $(x,y) = (2/3, 0)$?  What happens when you plug in $(x,y) = (-2, 0)$?  Why do both of these choices give you $0$ for this expression?

Now, what does the equation $$3x^2 + 3y^2 + 4x - 4 = 0$$ correspond to in the Cartesian coordinate plane?  What curve does it look like?

Title: Plot $2|z|=|z-2|$ over the complex plain

Body: I was given a home work to Sketch $2|z|=|z-2|$ on the complex plane.I tried to simplify and found out that the last expression was $3x^2+3y^2+4x-4$. However I don't know how to sketch this on the complex plane and how I can describe the set of points


问:
Reduction formula for $\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $
I'm struggling to go any further. Anyone have any hints?

$$ I _ n = \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $$

I used integration by parts where I differentiated $ x ^ n $, but it resulted in an $ \arcsin x $ term which didn't get me anywhere.

$$ u = x ^ n $$
$$ v = \arcsin x $$
$$ \mathrm d v = \frac { \mathrm d x } { \sqrt { 1 - x ^ 2 } } $$
$$ I _ n = \int u \ \mathrm d v = u v - \int v \ \mathrm d u = x ^ n \arcsin x - ( n - 1 ) \int x ^ { n - 1 } \arcsin x \ \mathrm d x $$

答:
$$I_n=\int\frac{x^n}{\sqrt{1-x^2}}dx$$

Let $x=\sin y\implies dx=\cos y \,dy$

$$I_n=\int\sin^ny \,dy$$

which has a well known reduction formula


问:
How can i solve this LCM puzzle?
I have this puzzle "A woman went to the market and a horse broke the eggs she had in her basket. The owner of the horse offered her to pay for the damage caused. He asked her how many eggs were broken by his horse. She said she didnt knew , but remebered that when she arrenged them in pairs, there was one left. Same thing happened when she arrenged them in groups of 3, 4, 5 and 6. But when she arrenged them in groups of 7 there was none left. what is the minimum quantity of eggs that were in the basket?" I tried showing the equations

2a+1=x

3b+1=x

4c+1=x

5d+1=x

6e+1=x

7f=x

But I really dont know what to do with them

答:
Hint

Let $m$ be the least common multiple of $2,3,4,5,6$.

Find $x$ such that $x \equiv +1 \pmod{m}$ and $x \equiv 0\pmod{7}.$

If the Chinese Remainder Theorem is not your thing...

if $a \equiv +1 \pmod{m}$, then so is $\{a + m, a+2m, a + 3m, \cdots\}.$

Find the element in that set that is a multiple of $7$.

Edit

If $b \equiv m\pmod{7}$, then $(1 + kb) \equiv (1 + km) \pmod{7}.$

Title: How can i solve this LCM puzzle?


问:
Computing stable and unstable points of dynamical system
For a dynamical system governed by the equation $$\frac{dx}{dt} = 2\sqrt{1-x^2}$$

where $|x|\leq 1$. I want to compute its stable points. The question was asked in one of the entrance exam for PG course.

My approach I am completely new to this topic dynamical system. However, going through web searches I found the following way to solve this problem. Kindly correct me if I am wrong.

Let us take $f(x) = 2\sqrt{1-x^2}$. Then its equilibrium points are given by setting $f(x) = 0$ which are $x^* = -1, 1$.
Further, if $f′(x^*)<0$, the equilibrium $x(t)=x^*$ is stable, and
if $f′(x^*)>0$, the equilibrium $x(t)=x^*$ is unstable.

Since in this case $f^\prime(x) = \frac{-4x}{\sqrt{1-x^2}}$, we see that $f^\prime(x^*)$ goes to infinity so the points $x^* = -1, 1$ are unstable equilibrium points.

Thanks in advance

答: Well, one goes to $-\infty$ and the other goes to $+\infty$, so you can expect the former to be stable and the latter to be unstable. In point of fact, notice that any solution $x(t)$ is strictly increasing as long as $x(t)\in(-1,1)$, therefore $1$ is asymptotically stable and $-1$ is unstable.


问:
What is the coefficient of the second term in $ xyz+(x+1)(y)+4x$?
The definition I found for a coefficient is 'the number multiplied by a variable.'
For the expression
$$
xyz + \left(x+1\right)\left(y\right) + 4x
$$
are both the factors $\left(x + 1\right)$ and $y$ considered variables?

In this case, I think the coefficient of the second term is $1$. Am I right $?$.

答:
I think a more appropriate definition for coefficient might be any multiplicative factor, as wikipedia suggests here. We usually use it to refer to the number that's multiplied to a variable in an expression. For example, in the expression $4x$, $4$ is typically referred to as a coefficient, and $x$ is a variable.

At the same time, variables can also be coefficients. For example, in the general form of a quadratic expression $ax^2 + bx + c$, we typically call $a$, $b$, and $c$ the coefficients, even though they are unknown. In some types of problems, they can actually be the unknown - for example, when fitting a parabola through a set of known points. Though, just as a caveat, it would be strange to refer to the independent variable ($x$ in this case) as a coefficient.

Also notice that in the quadratic above, we refer to $a$ as a coefficient, even though it's not multiplied to a variable - it's multiplied to $x^2$, another, slightly more complex expression.

My point is, a coefficient doesn't have to be limited to a number that's multiplied to a variable.

Going back to the problem at hand, in the expression $xyz+(x+1)(y)+4x$, there are three terms. The first is $xyz$, the second is $(x+1)(y)$, and the third is $4x$. In the same way that we could think of $x^2$ in the quadratic example above as a variable expression, and then think of $a$ as the coefficient on it, we can use the same type of logic here with respect to $xyz$ and $(x+1)(y)$. We can think of $xyz$ as a variable expression with coefficient $1$, and $(x+1)(y)$ as a variable expression with coefficient $1$. So, the coefficient on the second term is the $1$ that's the coefficient on $(x+1)(y)$.


问:
Is the graph of this problem statement incorrect?

Here is the problem statement:

According to the parabola in the figure of the function $f(x) = ax^2 + bx + c$, it is given that $a + b + c = 0$. The points $A$ and $B$ are the x-intersects, and the point $C$ is the y-intersect of the parabola. Let the origin be the point $O$, and determine the area of the triangle $COA$.

Here are my solving steps and why I think the plot doesn't seem right.

From the figure, it is apparent that $C = 3$, which leaves me with $a + b = -3$, from which I can write $b = - a - 3$.

I find the roots of the parabola, which turn out to be $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Substituting $b = - a - 3$ and $c = 3$ the roots are $x_1 = 1$ and $x_2 = \frac{3}{a}$.

However, since the value from under the square root on step 2 above is a positive value, I can assume that $x_1$ > $x_2$, simply because I used $+$ of the $\pm$ for $x_1$, and the $-$ for $x_2$. But the problem's answer accepts 1 to be the x coordinate of the point $A$ and proceeds on calculating the area from there with $\frac{3\times1}{2}=1.5$. I'm intending to appeal for the problem, but can't make sure whether or not I'm missing something.

答:
We have $a+b+c=f(1)=0$, hence clearly $1$ is a root. The question is whether it is the smallest root.

Also $f(0)=c=3$.

From vieta's formula, $\frac{c}{a}=1\cdot x_2$, that the other  root is $\frac{c}{a}=\frac3a$.

Hence if $\frac3a > 1$, then the answer is $\frac32$.

However, it is also possible that $\frac3a < 1$ as well. Then the answer is $\frac{3}{2}\cdot \frac{3}{a}=\frac{9}{2a}$ which is a function of $a$.

For example, consider the graph of $$f(x)=6x^2-9x+3=3(2x^2-3x+1)=3(2x-1)(x-1).$$


问:
Zero-ary cartesian product
Let $A=(A_1,...,A_n)$ be a vector of sets. The Cartesian product in $A$ could be defined as:

$\times_A :=\left\{(a_1,..,a_n)\mid \forall a_i \in A_i\right\}$

Which one is correct?

a) $\times_{\emptyset}  = \emptyset$ (empty set)

b) $\times_{\emptyset}  = \left\{(\ )\right\}$ (set containing the empty vector)

I think b) is correct because the empty $\forall$-operator is always true.

答: b) is the correct answer. The all-operator evaluates to true of empty.</p>

Title: Zero-ary cartesian product


问:
What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$?
I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I have to integrate it. I would think that the $-x$ would equal to $\frac{-x^2}{2}$ but the answer on the back of my book is: $\displaystyle \frac{y^2}{2} - \frac{2}{5} y^{5/2} =\frac{x^3}{3}-x-\frac{54}{5}$

What is the particular solution given $y=4$ and $x=3$?

My Solution
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}\\
y(1-\sqrt y)\frac{dy}{dx}=\frac{1}{x}\cdot (x^3-x)\\
\int (y-y^\frac{3}{2}) \frac{dy}{dx}=\int{x^2-x}\\=\frac{y^2}{2}-\frac{2y^{5/2}}{5}=\frac{x^3}{3}-\frac{x^2}{2}+c\\
\frac{y^2}{2}-\frac{2y^{5/2}}{5}-\frac{x^3}{3}+\frac{x^2}{2}=c$$

答:
The ODE is $$\displaystyle \small xy\frac{dy}{dx} = \frac{x^3-x}{1- \sqrt y},$$ which can be rewritten as

$$\displaystyle \small (y-y^{3/2}) \ dy  = (x^2-1) \ dx. \ $$ This is the step where you have a mistake! By integrating both sides we get:

$$\displaystyle \small \frac{y^2}{2} - \frac{2}{5} y^{5/2} = \frac{x^3}{3} - x + C,$$

Now, plug in $\small y = 4, x = 3$ and so you'll get $\small C = \displaystyle - \frac{54}{5}$.


问:
Étale cohomology vs flat cohomology
I'm recently reading the Milne's Etale Cohomology (1980 Princeton University Press). When I read the part of etale cohomology vs flat cohomology in page 115, I have trouble understanding the first step in the proof. Let $X$ be affine scheme, $X_0$ be closed sub-scheme of $X$ defined by an ideal $\mathfrak{I}$ of square zero, $G$ be smooth (quasi-projective) commutative group scheme over $X$. Define sheaf $\mathcal{N}$ (or just functor) by $Y\mapsto\ker(G(Y)\rightarrow G(Y_0))$ where $Y_0=X_0\times_XY$. $X'$ is another affine $X$-scheme which is finite and faithfully flat over $X$.

Milne tried to show cech cohomology of $\mathcal{N}$ for small finite flat topology is trivial, alternatively, $\check{H}(X'/X,\mathcal{N})=0$ which is crucial in Step 2 in the proof. His strategy is to claim $\mathcal{N}$ is representable by quasi-coherent $\mathcal{O}_X$-Module (say, $\mathcal{M}=\mathcal{Hom}_{\mathcal{O}_X}(e^*(\Omega_{G/X}),\mathfrak{I})$ where $e$ is neutral section $X\rightarrow G$) and apply Prop. 3.7 in page 114.

The question arises in the following way : I know $\mathcal N$ is a torsor of $\mathcal M$ and therefore, they coincide as sets. When we calculate cech cohomology, group structure have to be involved, however, group structure of $\mathcal{N}$ is induced from commutative group $G$ while group structure of $\mathcal{M}$ is induced from modules. How can we say the cech cohomology of $\mathcal{M}$ is trivial will imply triviality of cech cohomology of $\mathcal{N}$?

I found the similar question arised in the proof of Grothendieck's Le Groupe de Brauer III : Exemples et Complements, page 177. Their references are all just telling the coincidence of sets, which confuse me for a long while. Any help will be welcomed.

答: As you say, $\mathcal{N}$ is a torsor for $\mathcal{M}$. But $\mathcal{N}$ has a canonical point, namely the identity element, and hence is canonically isomorphic to $\mathcal{M}$.

Title: Étale cohomology vs flat cohomology


答:
Okay, so you have to solve the equation:
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
The first step is to move things around, which you did properly.
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}$$

Now we can move the $dx$ to the other side and write the equation as,

$$y(1-\sqrt y)dy=\frac{1}{x}\cdot (x^3-x)dx$$
Simplifying and adding in the integration,
$$\int \left(y-y^\frac{3}{2}\right)dy= \int(x^2-1)dx$$

So, integrating both sides, we have,
$$\frac{y^{1+1}}{(1+1)}-\frac{y^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}=\frac{x^{2+1}}{(2+1)}-x+C\\
\frac{1}{2}y^2-\frac{2}{5}y^{\frac{5}{2}}=\frac{1}{3}x^3-x+C$$

Finally, plugging in $y=4$ and $x=3$, we have,
$$\frac{1}{2}(4)^2-\frac{2}{5}(4)^{\frac{5}{2}}=\frac{1}{3}(3)^3-3+C\\
-\frac{24}{5}=6+C\\ C=-\frac{54}{5}$$ or, we can make it a little cleaner by saying that there is some other constant, say $C_1$, such that, $$C_{1}=\frac{54}{5}$$

Title: What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$?

Body:
I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I have to integrate it. I would think that the $-x$ would equal to $\frac{-x^2}{2}$ but the answer on the back of my book is: $\displaystyle \frac{y^2}{2} - \frac{2}{5} y^{5/2} =\frac{x^3}{3}-x-\frac{54}{5}$

What is the particular solution given $y=4$ and $x=3$?

My Solution
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}\\
y(1-\sqrt y)\frac{dy}{dx}=\frac{1}{x}\cdot (x^3-x)\\
\int (y-y^\frac{3}{2}) \frac{dy}{dx}=\int{x^2-x}\\=\frac{y^2}{2}-\frac{2y^{5/2}}{5}=\frac{x^3}{3}-\frac{x^2}{2}+c\\
\frac{y^2}{2}-\frac{2y^{5/2}}{5}-\frac{x^3}{3}+\frac{x^2}{2}=c$$


问:
Evaluating $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}$ (and another)

Compute
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$ and
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}\sin{\frac{2a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$
where $a$ is a positive real number.

I think the second limit may be solved by using the comparison criteria(limit comparison test), putting $-1 \le \sin{\frac{2a}{3^k}} \le 1$ and using the first limit. I suspect the first limit is $0$, so the second one should be $0$ too, but I might be wrong. I'm not sure how would you do the first one - I tried using the comparison test on the first one, but I didn't succeed. Can you help?

答: Neither limit can be zero, because the summand is not a function of $n$.  Only the index of summation is a function of the limiting variable; as such, it is necessary that each limit is bounded below by even just the first term in the sum, in as much as $$\sum_{k=1}^\infty \frac{1}{3^k} > \frac{1}{3}.$$  I suspect there is no simple closed form solution for these sums, unless there is some nice telescoping identity I have missed.


问:
show that $y^2 = x^5 - x +2 $ has no integer solutions
I considered everything$\mod 3$:

$x,y \equiv 0 \mod 3 ,\;\;\;\; y^2 \equiv 0 \mod 3 ,\;\;\;\; x^5 \equiv 0 \equiv x \mod 3$

$x,y \equiv 1 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 1 \equiv x \mod 3$

$x,y \equiv 2 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 2 \equiv x \mod 3$

thus in particular $x^5 - x \equiv 0 \mod 3, \;\;\; \forall x \in \mathbb{Z},$ and furthermore
$x^5 - x + 2 \equiv 2 \mod 3, \;\;\; \forall x \in \mathbb{Z}$.

But, $y^2 \equiv 0,1 \mod 3$, and if we were considering that there would be an integer solution then both sides should have the same remainder modulo $3$, but LHS and RHS never have the same remainder thus there are no integer solutions?

This is my first time doing a proof of this form, I'm familiar with modular arithmetic, however I've never used it to prove that a polynomial equation didn't have integer solutions.

答:
You can also use $\bmod 3$:

$$y^2-2=x(x^2-1)(x^2+1)$$

RHS: $x^2-1\equiv 0\bmod 3\Rightarrow RHS \equiv 0\bmod 3$

But :

LHS: $$y^2-1-1\equiv (0\bmod 3-1\bmod 3)\equiv -1\bmod 3$$


问:
prove $\left\{\begin{array}{l}X^{\prime \prime}+\lambda X=0,0<x<1 \\ X(0)=0, X(1)+X^{\prime}(1)=0\end{array}\right.$ has $\lambda >0$ without solving
Can you help me some idea to prove it or some similar question about it. I know prove $\lambda>0$ is easy when solving it, it's just like normal differential equation.

答:
We compute the Rayleigh quotient. Multiply your ODE by $X$ (assuming $X \ne 0$ is non-trivial) and integrate over the domain $x \in [0, 1]$

\begin{align}
- \lambda X^{2} &= X X'' \\
\implies - \lambda \int_{0}^{1} X^{2} dx &= \int_{0}^{1} X X'' dx \\
&= XX' \bigg \lvert_{0}^{1} - \int_{0}^{1} X'^{2} dx \\
&= - X(1)^{2} - \int_{0}^{1} X'^{2} dx
\end{align}

where we have used integration by parts and the boundary conditions to get the last line. Hence

$$\lambda = \frac{X(1)^{2} + \int_{0}^{1} X'^{2} dx}{\int_{0}^{1} X^{2} dx} > 0$$

as the integral of a non-negative function is non-negative and the numerator is always positive.