问:
I can't see a reason to reject one solution. Parametized surface
Ok so the question is to consider the parametrized surface

x = $u^2 - v^2$,    y = u + v, z= $u^2 + 4v$

It asks for me to find a vector normal to the surface in terms of u and v which is no problem.

(4- 2u, -8u - 4uv, 2u + 2v)

Which I know is correct: then it asks for the tangent plane at ($\frac{-1}{4},\frac{1}{2},2)$

Which I thought would be no problem just dot product of the normal vector at this point with the whole (x - x0) but Im having troubles getting a single u and v I get $\frac{1}{2}$ or $\frac{-9}{2}$ I just don't know anything else to reject one of them it gives the answer as 4x+z =1 which is fantastic I could figure out both and just reject the one that isnt the answer but it doesnt help me get ready for exams.

答:
You have (at your point) $9/4 = 2-(-1/4) = z-x = (u^2+4v)-(u^2-v^2) = v^2+4v$. So $v = -9/2$ or $1/2$. Next, $u=y-v = 1/2-(-9/2)=5$ or $1/2-1/2=0$. So either $u=5$ and $v=-9/2$ or $u=0$ and $v=1/2$.

If $u=5$ and $v=-9/2$, then $x=5^2-(-9/2)^2=25-36/4 = 16$ (which is wrong).

If $u=0$ and $v=1/2$, then $x=0^2-(1/2)^2=-1/4$, $y=0+1/2=1/2$, and $z=0^2+4(1/2)=2$ (which is correct).

So $u=0$ and $v=1/2$ is the (only) solution.

Explanation: When initially finding $u$ and $v$, we used the following equations: $z-x=v$ and $u=y-v$. This amounts to using a combination of the first and third equations ($x=u^2-v^2$ and $z=u^2+4v$) and the second equation ($y=u+v$). We then arrive at our 2 solutions. Thus this (smaller) system of equations does in fact have 2 solutions. But when we plug these back into our (bigger) system of 3 equations, we find out that one of them is a solution of all 3 equations and the other isn't. This often happens when we solve systems of (non-linear) equations. While searching for a solution we pare down the number of equations (eliminating and substituting). Sometimes in the course of throwing out (seemingly irrelevant) information, we allow "new solutions" to creep in (which aren't actually solutions of the original equations).


问:
Find the integer value of a, b such that $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.
Find the integer value of $a$, $b$ such that  $\lim\limits_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.

My work:

I multiplied with $\frac{ax+\sqrt{x}+b}{ax+\sqrt{x}+b}$, and obtained:

$=\lim\limits_{x \rightarrow 4} \frac{(ax+b)^2-x}{(x-4)(ax+b+\sqrt{x})}$

$=\lim\limits_{x \rightarrow 4} \frac{(a^2x^2+2abx+b^2-x)}{(x-4)(ax+b+\sqrt{x})}$

I factored $(x-4)$ from the numerator:

$(x-4)(a^2x+2ab-1+4a^2)+(b^2+8ab-4+16a^2)$

We have the first equation

$(b^2+8ab-4+16a^2)=0$

$=\lim\limits_{x \rightarrow 4} \frac{(x-4)(a^2x+2ab-1+4a^2)}{(x-4)(ax+b+\sqrt{x})}$
$=\frac{4a^2+2ab-1+4a^2}{4a+b+2}=\frac{3}{4}$

So, the second equation is
$16a^2+8ab-4+16a^2=12a+3b+6$

$32a^2-12a+8ab-3b-10=0$

But I can't find the integer value of $a$, $b$.

答:
$(2-\sqrt{x})(2+\sqrt{x})=4-x$ is a great start for partial fraction decomposition.

$a+\frac{1}{ 2 + \sqrt{x}}+\frac{-2 + 4 a + b}{-2 + \sqrt{x}}=\frac{a(x-4)+B(2-\sqrt{x})+C}{x-4}$

with $B=-1$ and $C=-2+4a+b$.

Take into account that

$$\frac{a x- \sqrt{x} + b}{x - 4}=a-\frac{1}{ 2 + \sqrt{x}}+\frac{-2 + 4 a +b}{-2 + \sqrt{x}}$$

This denominator is not compensable anymore by $\sqrt{x}$ in the nominator. So the nominators have to be zero for the existence of the limit.

The last summand has to be eliminated for a limit to exist!

$$\frac{4 a + b-2}{\sqrt{x}-2}$$

That can only the satisfied once for $-2$  or for $y$ but not for both to not singular.

We have $\frac{3}{4}=a-\frac{1}{4}$ and $4a+b=2$ so that $a=1$ so $b=-2$.

The back substition show we only have reals and therefore only are in need of the positive solution at $x=4$. The calculation remains valid.

The solution is for fitting to the given value/limit:

$$\frac{ x - \sqrt{x} -2}{x - 4}$$.

At $x=4$ the nominator has a zero $ 4 - \sqrt{4} -2=0$ that compensates for the zero of the polynomial of first order in $x$ in the denominator. The nominator is not a polynomial in $x$ but a sum of a polynomial and a second order root. Both a steady and differentiable in the singularity of the denominator.

A plot shows the singularity is gone:

The solution is $\frac{1}{2}$ for $x=0$ and $1$ for $x\rightarrow\infty$ and $\frac{3}{4}$ at $x=4 and this is no limit!$.

The solution process is substition of the root. Then partial fraction decompostion and the fitting the parameters of the given functions to the singularity at $x=4$. Finally we analysed the function and displayed it after resubstitution.

We needed a factor in addition to match the given limit. The pair $(a,b)=(1,-2)$ is unique.

ContentLicense: CC BY-SA 4.0


问:
the proof of implicit function theorem (Terence Tao)
Implicit function theorem: Let $E$ be an open subset of $\mathbb{R}^n$, let $f : E \to \mathbb{R}$ be continuously differentiable, and let $y = ( y_1, ... , y_n)$ be a point in $E$ such that $f(y) = 0$ and $\frac{\partial{f}}{\partial{x_n}}(y) \not= 0$. Then there exists an open subset $U$ of $\mathbb{R}^{n-1}$ containing $(y_1, ... , y_{n-1})$, an open subset $V$ of $E$ containing $y$, and a function $g : U \to \mathbb{R}$ such that $g(y_1, ... ,y_{n-1}) = y_n$, and
$$\{(x_1, ... ,x_n) \in V : f(x_1, ... ,x_n) = 0\} = \{(x_1, ... ,x_{n-1}, g(x_1, ... ,x_{n-1}) : (x_1, ... ,x_n) \in U\}.$$
Moreover, $g$ is differentiable at $(y_1, ... ,y_{n-1})$, and we have
$$\frac{\partial{g}}{\partial{x_j}}(y_1, ... , y_{n-1}) = \frac{\partial{f}}{\partial{x_j}}(y)/ \frac{\partial{f}}{\partial{x_n}}(y)$$
for all $1 \le j \le n-1$.

I don't understand some steps in the proof. I highlight with purple lines the step I am struggling with.

The author let $F^{-1} (x) = (h_1(x), ..., h_n(x))$. Since $F$ is invertible, $x = F(h_1(x), ... ,h_n(x))$, but why $h_j(x) = x_j$?

We know that $W$ is open in $\mathbb{R}^n$, but how do we know that $U$ is open in $\mathbb{R}^{n-1}$?

Similarly, we know that $h_n$ is differentiable at $(y_1, ... ,y_{n-1}, 0)$, but how can we prove that $g$ is differentiable at $(y_1, ... ,y_{n-1})$?

I really appreciate if you give some help!

答:
$$(x_1, \ldots, x_j, \ldots, x_n) = x = F(F^{-1}(x)) = F((h_1(x), \ldots, h_j(x), \ldots, h_n(x))) = (h_1(x), \ldots, h_j(x), \ldots, f(h_1(x), \ldots, h_n(x)))$$
Therefore $x_j = h_j(x)$.

Theorem 2.1.5 (c) states that if $f$ is a continuous function then the inverse image/preimage of an open set is also open. We can make a continuous function $f: \mathbb{R}^{n-1} \rightarrow \mathbb{R}^{n}$ that maps $(x_1, \ldots, x_{n-1})$ to $(x_1, \ldots, x_{n-1}, 0)$. By that theorem,
$$f^{-1}(W) = \{(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1} \vert f((x_1, \ldots, x_{n-1})) = (x_1, \ldots, x_{n-1}, 0) \in W\} = U$$
is open as $W$ is open.

The derivative of $g$ evaluated at $(y_1, \ldots, y_{n-1})$ is the derivative of $h_n$ evaluated at $(y_1, \ldots, y_{n-1}, 0)$ which exists.


问:
Create an isosceles triangle from only equilateral triangles. Possible or impossible?
My friend recently asked me a question. And I think I have a nice solution but wanted to share this question here to see how you approach this problem and hopefully with a solid proof.

The question is:

Can you create an isosceles triangle with an infinite number of any sizes of only equilateral triangles?

答:
Start from the 120-30-30 isosceles triangle. We aim to fill this triangle with equilateral triangles in infinite steps.

Draw an equilateral triangle at one-third of the longest side.

Now the triangle is $\frac13$ filled, with two smaller 120-30-30 isosceles triangles created. Repeat the process:

And now $\frac13 + \frac29$ of the triangle is filled. 3rd iteration below.

Why would the initial triangle be filled in infinite steps? The area filled in $n$ steps is equal to:

$$\sum_{k=1}^n \frac13(\frac23)^{k-1}$$

At infinity, this geometric sum evaluates to

$$\sum_{k=1}^\infty \frac13(\frac23)^{k-1} = \frac {1/3}{1-2/3}=1$$

so the whole triangle will be filled, given that we have infinite equilateral triangles of all sizes.

Or, equivalently, considering the fact that $1/3$ of the remaining area is filled in each iteration, we have the area remaining after the $n$th step is $\dfrac1{3^n}$, which $\to 0$ as $n \to \infty$.


问:
Interesting rotating parabola proof
So, I am reading linear algebra and cartesian geometry by Juan de Burgos for the first time, and came across this problem that I just can't figure out. Goes like this:

Consider a parabola, whose distance from the focus to the directrix is a fixed p, and that rotates while keeping its focus fixed.
We also consider a certain fixed direction in the parabola's plane, and consider the tangent to the parabola that has that given direction.
Find the equation of the geometric space that all the tangency points describe.

The original question is in Spanish but I have tried to translate it the best I can. If anyone speaks Spanish I can send you the original question.

Edit: Ill keep working on it and try to write my work in here, sorry that I didnt do from the beginning

Ok, so first off the book gives the soultion to a problem in the back, but without any explanation or none of that. it says:

**Origin at the focal point, << x axis>> in the given direction;

directrix $xcos(\theta)+ysin(\theta)=p$.

Parabola $x^2 sin^2(\theta)+y^2 cos^2(\theta)-2xysin(\theta)cos(\theta)+2p(xcos(\theta)+ysin(\theta))=p^2$

The polar of (0,1,0) is $xsen^2(\theta)-ysin(\theta)cos(\theta)+pcos(\theta)=0$.

Eliminating $\theta$ we get $p^2(x^2+y^2)=4y^4$.**

I get the part of creating a new coordinates taking the focal point to be F(0,0). as so:
see image here
The directrix in a standard parabola where the vertix is in the origin has directrix x=p/2, so because we are taking the focal point it would be now x=p. Because of the rotation described for the parabola, i thought of using rotation of coordinate axes, as so:
see image
And thats why the directrix would be  $xcos(\theta)+ysin(\theta)=p$ as described in the solution given by the book.
(Thanks to a commenter here who clarified to me that  $xcos(\theta)+ysin(\theta)=p$ or  $xcos(\theta)+ysin(\theta)=-p$ would make no difference, I was confused on that too).

Now, how in the world did they get the parabola described in the solution and the stuff that comes after?

I have read the chapter like three times but I dont know why i cant understand this one problem

答:
HINT.

Instead of following the path suggested by the book, you could have the parabola fixed and rotate the direction of the tangent: it is not difficult to find a function $r=f(\phi)$, where $r=FP$ is the distance between focus $F$ and tangency point $P$, while $\phi$ is the angle between line $FP$ and the tangent at $P$.

It may also be useful to remember that the tangent line at $P$ is the bisector of $\angle FPH$, where $H$ is the projection of $P$ on the directrix.

Title: Interesting rotating parabola proof


问:
Can cube be a Cross-polytope?
I was reading about L1-norm where I came across the term Cross-Polytype. In the link for Cross-Polytope, examples are given for Cross-Polytope in different dimensions. What I understood from the examples is that the sides should be formed from equal length line segments. In that case, can a cube be called a Cross-Polytope in 3 dimensions? and in that case can any shape with equal length sides be called a Cross-Polytope? for instance, a square pyramid.</p>

答:
can a cube be called a Cross-Polytope in 3 dimensions?

It's made clear in the Wikipedia article you link that the cross-polytope in $3$ dimensions is not the cube:

In 3 dimensions it is an octahedron

Also from the article, the definition makes clear that there is only one cross-polytope in $n$ dimensions, the object whose

vertices … are all the permutations of $(±1, 0, 0, \ldots, 0)$

Since that doesn't describe a cube, the answer to the question in your title is simply no, and similarly to your other questions: there is only one $n$-cross-polytope for any given $n$.

ContentLicense: CC BY-SA 3.0

Title: Can cube be a Cross-polytope?


问:
Is the triangle equality an equality if we have ordered objects $x \leq y \leq z$?
Let $X$ be a set with an order on it (say a partial order or total order). Suppose also that $X$ is a metric space with metric $d$. Then does $x \leq y \leq z$ imply $d(x,z) = d(x,y) + d(y,z)$?

If this is not necessarily true, what other assumptions might make it true? Does it depend on whether we assume it is a partial order versus assuming it is a total order? Or perhaps some other kind of order?

My motivation for asking this question comes from thinking of, say, the real numbers (with the usual ordering and metric), where it is true. In general it seems that whenever we have something that resembles a “line”, then this should be true.

答: You have to assume some connection between the order and the metric, otherwise the ordering of the elements does not imply anything for the metric. Consider for example the real numbers with the usual order on it and the discrete metric, i.e. $d(x,y)=1$ if $x\neq y$ and $d(x,y)=0$ if $x=y$ then clearly for any $x<y<z$ we have $1=d(x,z)<d(x,y)+d(y,z)=1+1=2$. Although i do not know what exactly would imply such a behaviour as described by you.


问:
Given $f$ is continuous and nonnegative on $[1,11]$ with $\int_{1}^{11}f(x)dx=6$. Show that $\int_{0}^{2}f(-x^3+3x^2+3x+1)dx$ is less or equal to 2.
Given $f$ is continuous and nonnegative on $[1,11]$ with $\int_{1}^{11}f(x)dx=6$. Show that $\int_{0}^{2}f(-x^3+3x^2+3x+1)dx$ is less or equal to 2.

My Approach

My first instinct was to show that if there exists a partition $P$ such that the lower sum $L(f(-x^3+3x^2+3x+1),P )>2 $, then as $−x^3+3x^2+3x+1$ is one-to-one, we can use the same minimum points to create a $P'$ for $f(x)$, with each closed interval "stretched". If I can show that this lower sum is greater than 6, then I would have a contradiction but I am kinda stuck here. Any hints on how to continue or maybe a new direction? I also noticed that the two functions have the same end point and midpoint, but I haven't figured out if I can use this fact.

答:
Let $g(x):=-x^3+3x^2+3x+1$. Then $g'(x)=-3x^2+6x+3$ and it is easy to see that $g'(x)>0$ on $[0,2]$. So $g(x)$ is strictly increasing there and $g([0,2])=[g(0),g(2)]=[1,11]$.
Moreover, It is not dificult to see that $g'(x)\ge 3$ on $[0,2]$ (indeed, $g'(x)$ is a parabola with negative leading coefficient, $g'(0)=g'(2)=3$ so the vertex is on $g(1)=6$).
$$\int_0^2f(-x^3+3x^2+3x+1)dx=\int_0^2 f(g(x))dx \le \frac{1}{3} \int_0^2f(g(x))g'(x)dx$$
Now, making substitution $g(x)=t$, one gets
$$\int_0^2f(-x^3+3x^2+3x+1)dx\le \frac{1}{3} \int_0^2f(g(x))g'(x)dx=\frac{1}{3}\int_1^{11} f(t)dt=2 $$


问:
Finiteness in compact space classification
Let $X$ be a metric space satisfying that every sequence in $X$ has a subsequence that converges to a point in $X$.

Lemma: Any metric space satisfying the above condition has a countable dense subset.

I draw the proof from Stromberg. He begins the proof by claiming that there exists a finite set $A_\epsilon \subseteq X$ satisfying

\begin{equation}
\begin{split}
&\rho(a,b)\geq\epsilon\; {\rm whenever}\; a\neq b\; {\rm in }\; A_\epsilon,\\
&B_\epsilon(x) \cap A_\epsilon \neq \emptyset\; {\rm for\; each}\; x\in X.
\end{split}
\end{equation}

The proof goes by contradicting that no such $A_\epsilon$ exists. Everything is understandable, since if no such $A_\epsilon$ exists, one can construct a sequence that does not have a convergent subsequence. The only question that I have is that why the requierement that $A_\epsilon$ is finite is needed. I don't see in the proof that if the set is infinite, how the contradiction of no convergent subsequence may arise.

The proof that the above set must exists goes as follows

Assume that the above conditions are false. Let $x_1\in X$. When $x_1,\dots,x_n \in X$ have been chosen so that $\rho(x_i,x_j)\geq \epsilon$ if $i\neq j$, we observe that the set $\lbrace x_1,\dots, x_n\rbrace$ satisfies the first condition, so it cannot satisfy the second. Thus, there exists $x_{n+1}\in X$ such that $$B_\epsilon (x_{n+1})\cap\lbrace x_1,\dots,x_n\rbrace = \emptyset;$$ whence $\rho(x_{n+1},x_i)\geq \epsilon$ for $1\geq i \geq n$. In this way we inductively choose a sequence $(x_n)_{n=1}^\infty \subseteq X$ such that $\rho(x_i,x_j)\geq \epsilon$ whenever $i\neq j$ in $\mathbb{N}$. (...)

答:
The construction of that sequence should go as follow:

By contradiction, we assume that there is an $\varepsilon>0$ such that no finite set $A$ satisfies the two conditions together, i.e. whenever $A$ is a finite set with the first property ($d(a,b)\ge\varepsilon\ \forall a,b\in A$) then it fails the second property, so there is an $x\in X$ whose $\varepsilon$-neighborhood is disjoint from $A$.

First choose an arbitrary element $a_1\in X$.

Then we repeatedly apply the above to add the $x$ found above for the finite set $A=\{a_1,a_2,\dots,a_n\}$ as the next element: $a_{n+1}:=x$.

The main observation is that $A$ satisfies the first property, by construction.

And that's it. Since this construction was applied only for finite sets, we can draw the given conclusion.


问:
Galois's group of $\Bbb{Q}(\sqrt2,\sqrt3,\sqrt5)$ and its reticle of subgroups and subextensions
I got an example of $\Bbb{Q}(\sqrt2,\sqrt3)$  but I don't know how to do that with 3 extensions. Any ideas?
Thanks.

答:
As pointed out by Jyrki Lahtonen, your question (even under a more general form) has been asked many times. For 3 radicals ($\sqrt 2, \sqrt 3, \sqrt 5$ here) it is still possible to give a direct simple solution. Consider first the extension $K=\mathbf Q(\sqrt 2, \sqrt 3)$. I don't know how you solved this case, but I'll give an approach which will be used also for $L=\mathbf Q(\sqrt 2, \sqrt 3,\sqrt 5)$. First notice that, for $a,b \in {\mathbf Q^*}$, it is straightforward that the two quadratic fields coincide iff $ab^{-1} \in {\mathbf Q^*}^2$. So here $K/\mathbf Q$ is biquadratic, with Galois group $G\cong C_2 \times C_2$. Since $G$ has exactly 3 subgroups of order, $K$ has exactly 3 quadratic subfields, which are obviously (fill in the details) $\mathbf Q(\sqrt 2, \sqrt 3,\sqrt 6)$. In particular, $L$ will be the compositum of the 2 linearly disjoint extensions $K$ and $\mathbf Q(\sqrt 5)$, hence $L/\mathbf Q$ will be Galois, with group $J \cong C_2 \times G \cong C_2 \times C_2 \times C_2$. From this you can derive easily the reticle of subgroups (resp. of subextensions). One convenient way to vizualize this is to write additively $G$ as $\mathbf F_2 \times \mathbf F_2\times \mathbf F_2$, where $\mathbf F_2$ is the field with 2 elements, so that $J$ can be viewed as an $\mathbf F_2$-vector space of dimension 3, whose strict non trivial subspaces have dimension 2 ("planes") or 1 ("lines"). The corresponding subextensions are obviously the 3 previous quadratic fields, and the 3 biquadratic fields $\mathbf Q(\sqrt 2, \sqrt 3),\mathbf Q(\sqrt 2,\sqrt 6), \mathbf Q(\sqrt 3,\sqrt 6)$.

A generalization of this approach to $\mathbf Q(\sqrt p_1,..., \sqrt p_n)$, where $p_1,..., p_n$ are $n$ distinct primes, can be adapted e.g. from https://math.stackexchange.com/a/3444776/300700


问:
The finite-rank operators of a Banach space are dense in bounded operators space iff the Banach space is finite dimensional
Let $E$ be a Banach space, $Fr(E)$ be the set of all finite-rank operators from $E$ to $E$, $B(E)$ be the set of all continuous linear operators from $E$ to $E$. Then $Fr(E)$ is dense in $B(E)$ iff $dim(E)$ is finite.

I know that when $dim(E)$ is finite, $Fr(E) = B(E)$. The statement is true in this case. But I do not know how to prove the other side of this statement. There is a Hint: Use Riesz's theorem.If we start from the condition $Fr(E)$ is dense in $B(E)$, how can we derive $dim(E)$ is finite by using Riesz's theorem?

答:
The deeper way to do this is to use Riesz Lemma to show that the unit ball is compact if and only if $\dim E<\infty$, and conclude that the identity operator is not compact, and thus not a limit of finite-rank operators.

But there is a much straightforward way. If $E$ is infinite-dimensional, no finite-rank operator can be invertible. This implies that $\|I-T\|\geq1$ for all finite rank $T$, so $I$ is not a limit of finite-rank operators.

The inequality is justified by the usual Neumann series trick, that if $\|I-T\|<1$, then $T$ is invertible. Namely, if $\|I-T\|<1$, then $\sum_{k=0}^\infty (I-T)^n$ is an inverse for $T$.


问:
Vector Field, Curl and Divergence. Simple computing and proving:

So I went through (a) and showed that curl( v x F) = 2v. (doing cross products twice, which comes out pretty simple.)

Now i am stuck at (b), it says use the results from part (a) to compute:
line integral <2,3,-4> x F * dr..

how should i do (b) and (c)?

答:
For (b), use Stokes's Theorem. What is $\text{curl}\,\mathbf F\cdot\mathbf n$?

For (c), look at various formulas you know about div and curl.


问:
Prove that if $\{ f_n\}$ converges to $f$ pointwise on $(0,1)$, then $\{ f_n \}$ converges to $f$ uniformly on $(0,1)$
Let $\{ f_n \}$ be sequence of differentiable functions on the open interval $(0,1)$ such that for any $x \in (0,1)$ and any positive integers $n$

$$|f'_n(x)| \le 1.$$
Prove that if $\{ f_n\}$ converges to $f$ pointwise on $(0,1)$, then $\{ f_n \}$ converges to $f$ uniformly on $(0,1)$.

How can I solve this?

答:
As someone already mentioned in the comment that the easiest way to do this is to first observe that each $f_n$ is Lipschitz continuous and hence can be extended to the closed interval $[0, 1].$ The proof of this fact, if you are not aware of it, provided in the previous answer.

I will provide an alternate proof below that avoids a little bit of this machinery (although morally I think there is not much of a difference in the two approaches).

Let us get back to basic and see what we have. Given $\epsilon>0,$ and $x\in (0, 1)$ we can find $N_x$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n\ge N_x.$ This is the definition of pointwise convergence. To prove that convergence is uniform, we need to choose $N$ that works for all $x.$ As you might imagine, if we had finitely many $N_x,$ we can choose $N$ to be the maximum of them. To get only finitely many $N_x,$ we need to show something like $N_x$ works not only for $x,$ but it also works for $y$ in a small neighborhood of $x.$ To do this, we let us fix $y$ close to $x$ (we'll see how close in a moment) and see what we can say about $|f_n(y)-f(y)|.$ Obviously, one is going to use triangle inequality
$$|f_n(y)-f(y)|\le|f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(y)-f(x)|.$$

This looks good. We already know that $|f_n(x)-f(x)|<\epsilon$ if $n\ge N_x.$ From the condition that $|f_n'(x)|\le 1,$ it is not hard to conlude that $|f_n(y)-f_n(x)|\le |x-y|$ for all $n.$ The last term does not depend upon $n$ at all, but it is about another animal $f,$ and right now we do not know much about it. So, the first thing to do is to figure out if $|f(y)-f(x)|$ can be controlled if $x, y$ are close. This is again not hard to see from the pointwise convergence that $|x-y|\ge|f_n(y)-f_n(x)|\to |f(y)-f(x)|$. Once we have all these pieces in place, the argument is pretty straightforawrd. And, now I will write it in a linear fashion.

Fix $\epsilon>0.$ For each $x\in (0, 1),$ there exists $N_x$ such that $|f_n(x)-f(x)|\le \epsilon/3.$ If $|y-x|\ge\epsilon/3,$ then $|f_n(y)-f(y)|\le \epsilon$ for all $n\ge N_x.$

Choose finitely many $x_i\in (0, 1)$ such that $B(\epsilon/3, x_i)$ cover $(0, 1).$ This can be done explicitly. If you want start with $0<x_0<\epsilon.$ Once $x_0, \ldots, x_n$ are chosen, take $x_{n+1}$ such that $2\epsilon/3>x_{n+1}-x_n\ge \epsilon/3.$ Since $x_{i}-x_{i-1}\ge \epsilon/3,$ it follows that $\le 3/\epsilon$ many balls suffice to cover $(0, 1).$

Now choose $N\ge \max_{i}N_{x_i}.$ To see that this $N$ works--take $x\in (0, 1).$ Then there exists $i$ such that $x\in B(\epsilon/3, x_i).$ Now,
$$|f_n(x)-f(x)|\le |f_n(x)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(x)|\le \epsilon.$$


问:
Bounded Surface in $\mathbb{R}^3$ with infinite surface area
Does there exist a bounded surface embedded in $\mathbb{R}^3$ with infinite surface area? Would prefer something without boundary and compact, but any example works. The one dimensional analogue to this would be a space filling curve, though this is only an immersion

Edit: ideally without boundary, compact, and $C^1$

Edit: If it's C^1 and compact then it will have finite area necessarily

答: For an example, start with a Koch curve in the right half plane $\{(x,y,z) \mid x > 0, y=0\} \subset \mathbb R^3$, and then revolve it around the $z$ axis to get a surface of revolution. You'll get a surface that is homeomorphic to a torus, i.e. the surface of revolution of an ordinary round torus. But, it will have infinite area.


问:
How do we determine the values of $P(A)$, $P(B)$, $P(C)$ and $P(A\cap B\cap C)$?
The events $A$, $B$ and $C$ have to do with the same experiment and for them we know: $P(A\cup B) = 0.9$, $P(A\cup C) = 0.8$, $P(B\cup C) = 0.8$, $P(A\cap B) = 0.3$ , $P(A\cap C) = 0.3$, $P(B\cap C) = 0.3$ and $P(A\cup B\cup C) = 1$. Find $P(A)$, $P(B)$, $P(C)$ and $P(A\cap B\cap C)$.

答:
HINT

As @Michael has pointed out, you have the following system of equations to solve
\begin{align*}
\begin{cases}
P(A) + P(B) - P(A\cap B) = 0.9\\\\
P(A) + P(C) - P(A\cap C) = 0.8\\\\
P(B) + P(C) - P(B\cap C) = 0.8
\end{cases} & \Longleftrightarrow
\begin{cases}
P(A) + P(B) - 0.3 = 0.9\\\\
P(A) + P(C) - 0.3 = 0.8\\\\
P(B) + P(C) - 0.3 = 0.8
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
P(A) + P(B) = 1.2\\\\
P(A) + P(C) = 1.1\\\\
P(B) + P(C) = 1.1
\end{cases}
\end{align*}

Once you find out the values of $P(A)$, $P(B)$ and $P(C)$, you can make use of the following identity:
\begin{align*}
P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap  C) + P(A\cap B\cap C)
\end{align*}

Can you take it from here?


问:
Is measurable function possible in discrete metric space?
The function defined by $f:X{\rightarrow}Y$ is continuous. If $X$ is endowed with discrete metric.

Now,consider characteristic function

$\chi_{A}:X{\rightarrow}R$ defined by

$\mathcal{X}_E(x) =\begin{cases}1, if& x \in E\\
0, & if& x \not\in E \\  \end{cases}$

is continuous if $X$ is $R$ endowed with the discrete metric.

And Now,

let (X, M) is a measurable space ,

where sigma algebra $M =\{\phi, X, \{a\}, \{b\}\}$ and $X=\{a, b\}$.

Then , the characteristic function is measurable.

Let $E\subset R$,

$\mathcal{X}_A^{-1}(E) =\begin{cases}X, & 0,1 \in E \\A, & 1 \in E, 0 \notin E \\A^C, & 1 \notin E, 0 \in E \\\emptyset, & o.w.\end{cases}$

This gives our characteristic function is measurable and continuous.
And according to definition of simple function a complex measurable function s on a measurable space $X$ whose range is only finitely many points is called simple function.
Hence,the characteristic function consider above is continuous simple function.

Thus,giving us a continuous simple function. I am correct or wrong please check and correct my understanding .I have used Rudin book of Real and complex analysis and some online notes .

答: Yes you are correct. Since every set is open and every set is measurable, we have that every function is continuous and measurable. And a characteristic function is also simple as taking only two values.</p>


答:
Okay, I'll bite.  I don't know if the triangles are allowed to overlap, but my argument would work in either case.

Assume that we can choose a triangle formed from equilateral triangles that is isosceles but not equilateral. Then the smallest angle of the isosceles triangle is strictly less than 60 degrees.  Let A be the vertex of that angle.  Among the infinite number of equilateral triangles that cover the triangle, choose one that includes point A.  Let A' be the point on that equilateral triangle that covers A.  Any neighborhood of A' would have to include points B' and C' such that m$\angle B'A'C'\ge60^\circ$. But then either B' or C' would be outside the isosceles triangle, which is a contradiction.

Needless to say, except that this is a puzzle: an equilateral triangle itself is isosceles, and you could construct that from an infinite number of equilateral triangle using the Sierpinski triangle as a guide.

Title: Create an isosceles triangle from only equilateral triangles. Possible or impossible?

Body:
My friend recently asked me a question. And I think I have a nice solution but wanted to share this question here to see how you approach this problem and hopefully with a solid proof.

The question is:

Can you create an isosceles triangle with an infinite number of any sizes of only equilateral triangles?


问:
What do subgroups of $\mathbb{Z}_{p_1}^{\alpha_1} \oplus ... \oplus \mathbb{Z}_{p_t}^{\alpha_t}$ look like?
If $G = \mathbb{Z}_{p_1}^{\alpha_1} \oplus ... \oplus \mathbb{Z}_{p_t}^{\alpha_t}$ is it true that every subgroup of $G$ looks like $H_1 \oplus ... \oplus H_t$, where $H_i \leqslant \mathbb{Z}_{p_i}^{\alpha_i}$?

UPD:

$p_i$ are primes (they may be equal).

"looks like" means that every subgroup is isomorphic to $H_1 \oplus ... \oplus H_t$ (I want to know if I can find all subgroups using combintaions of subgroups in $\mathbb{Z}_{p_i}^{\alpha_i}$).

答:
Hint: Every subgroup of a finite abelian group is itself a finite abelian group.

 Use the Fundamental Theorem of Finite Abelian Groups.


问:
Simple but challenging probability identity
I am reading E.T. Jayne's Probability Theory book, and I am stuck on a "trivial" fact that he mentions at the end of the derivation of the sum rule in Chapter 2 (page 33):

He states that it is a trivial fact that given

$$
\begin{align}
w(AB|C) = w(A|BC)w(B|C) = w(B|AC)w(A|C) \\
w(A\bar{B}|C) = w(A|C)w(\bar{B}|AC)
\end{align}
$$

one can show that

$$
w^{m}(A|C) - w^m(A\bar{B}|C) = w^m(B|C) - w^m(B\bar{A}|C)
$$

where w is a probability measure.

Can someone demonstrate how this result follows?

Note, I think it's clear in the case that $m = 1$, both sides of the final equation are equal to $w(AB|C)$. I just don't see how you can derive this from the given equations for all powers of $m$.

答:
This works for $m=1$. I don't know if this can be extended to $m \neq 1.$

Notice that
$w(A|C) - w(A\bar{B}|C) = w(B|C) - w(B\bar{A}|C) \Rightarrow \\
w(AC)w(C) - w(A\bar{B}C)w(C) = w(BC)w(C) - w(B\bar{A}C)w(C) \Rightarrow \\
w(AC) - w(A\bar{B}C) = w(BC) - w(B\bar{A}C) \Rightarrow \\
w(AC) + w(B\bar{A}C) = w(A\bar{B}C) + w(BC).$

Observe that $AC$ and $B\bar{A}C$ are disjoint sets. Therefore:

$w(AC) + w(B\bar{A}C) = w(AC+ B\bar{A}C)$.

Similarly, $BC$ and $A\bar{B}C$ are disjoint, and:

$w(BC) + w(A\bar{B}C) = w(BC+ A\bar{B}C)$.

Moreover, using DeMorgan theorem:
$$AC + B\bar{A}C = C(A + B\bar{A}) = C\overline{(\bar{A}\overline{B\bar{A}})} = \\
= C\overline{(\bar{A}(\bar{B} + A))} = C\overline{(\bar{A}\bar{B})} = C(A+B).$$

Similarly:
$$BC+ A\bar{B}C = \text{proof left to the reader} = C(A+B).$$


答:
There is actually a slightly subtle point lurking in your question. Do you want each subgroup of $G$ to be isomorphic to a direct sum of some $H_i$? Or do you want it to be equal to the direct sum of some $H_i$?

In the first case, as Shaun says, you can use the fundamental theorem of finite abelian groups to obtain an affirmative answer; I'll let you work out the details.

In the second case, the answer is negative. For instance, consider the diagonal subgroup $\Delta=\{(0,0),(1,1)\}$ of the finite abelian group $\mathbb{Z}/2\oplus\mathbb{Z}/2$. If $\Delta$ were equal to a direct sum of some $H_1,H_2\leqslant\mathbb{Z}/2$, note that each $H_i$ would have to contain both $0$ and $1$ and hence be equal to all of $\mathbb{Z}/2$. However, $\Delta\neq\mathbb{Z}/2\oplus\mathbb{Z}/2$.

Indeed, $\Delta\cong\mathbb{Z}/2$, so for $\Delta$ to be isomorphic to a direct sum of two subgroups of $\mathbb{Z}/2$ we would need one of the two to be the trivial subgroup.

So, if your question is whether you can "find all subgroups using combinations of subgroups in $\mathbb{Z}_{p_i}^{\alpha_i}$", the answer is true if you add the qualifier "up to isomorphism", but false as written.

Title: What do subgroups of $\mathbb{Z}_{p_1}^{\alpha_1} \oplus ... \oplus \mathbb{Z}_{p_t}^{\alpha_t}$ look like?

Body:
If $G = \mathbb{Z}_{p_1}^{\alpha_1} \oplus ... \oplus \mathbb{Z}_{p_t}^{\alpha_t}$ is it true that every subgroup of $G$ looks like $H_1 \oplus ... \oplus H_t$, where $H_i \leqslant \mathbb{Z}_{p_i}^{\alpha_i}$?

UPD:

$p_i$ are primes (they may be equal).

"looks like" means that every subgroup is isomorphic to $H_1 \oplus ... \oplus H_t$ (I want to know if I can find all subgroups using combintaions of subgroups in $\mathbb{Z}_{p_i}^{\alpha_i}$).


问:
Slightly confused about the commutator subgroup
Let's look at the group of all permutations of a set $\left\{ 1,2,3 \right\}$. There are exactly 6 elements in this group: $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 2 & 3  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 3 & 2  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 1 & 3  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 3 & 1  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   3 & 1 & 2  \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 & 2 & 3  \\
   3 & 2 & 1  \\
\end{matrix} \right)$. It is usually called the symmetric group $S_{3}$. Just out of curiosity I tried to find its commutator subgroup $\left[ {{S}_{3}},{{S}_{3}} \right]$. As a home assignment, I've already proven that the commutator subgroup is a normal subgroup, hence $\forall x\in {{S}_{3}}:x\left[ {{S}_{3}},{{S}_{3}} \right]=\left[ {{S}_{3}},{{S}_{3}} \right]x$. Except for the neutral element $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 2 & 3  \\
\end{matrix} \right)$, I couldn't find a single permutation $p$ with the property: $\forall x\in {{S}_{3}}:xp=px$. For instance, $\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 3 & 2  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 1 & 3  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 3 & 1  \\
\end{matrix} \right)\ne \left( \begin{matrix}
   1 & 2 & 3  \\
   3 & 1 & 2  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 2 & 3  \\
   2 & 1 & 3  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & 2 & 3  \\
   1 & 3 & 2  \\
\end{matrix} \right)$.

Am I missing something?

答:
HINT:

$$\forall x\in {{S}_{3}}:x\left[ {{S}_{3}},{{S}_{3}} \right]=\left[ {{S}_{3}},{{S}_{3}} \right]x$$

does not mean that $xp = px$ for $p \in \left[ {{S}_{3}},{{S}_{3}} \right]$. It means, for $p \in \left[ {{S}_{3}},{{S}_{3}} \right]$, we have $xp = qx$ for some $q \in \left[ {{S}_{3}},{{S}_{3}} \right]$, not necessarily $q = p$.


问:
Limits, counter-example
Regarding this question here:

If there exists an $L ∈ R$ such that $\lim_{x→x_0} f(x) = L$ for every $x_0 ∈ R$, then $f (x) = L$ for every $x ∈ R$.

I think this is false, and I am trying to find a counter example for that.

Can someone please give me a hint for how to find the right $f(x)$ that counter this statement?

Thanks a lot!

答:
Take
$$f(x)=\begin{cases}L& x\notin\Bbb Z\\ L+1& x\in\Bbb Z\end{cases}$$

There are easier exmaples, but this one has the maximum amount of dicontiniities.

For fixed $x_0\in\Bbb R$, you can find $\delta>0$ such that $f(x)=L$ for all $0<|x-x_0|<\delta $ (just take, for instance,  $\delta:=\frac{1}{2}{\rm dist}(x_0,\Bbb Z))=\frac{1}{2}\min\{x_0-\lfloor x_0\rfloor, \lceil x_0\rceil -x_0\}$ -where $\lfloor\cdot\rfloor$ denotes the floor function and $\lceil\cdot\rceil$ the ceil function).

So For any $\varepsilon>0$, take exactly THIS $\delta$ so if $0<|x-x_0|<\delta$, then $x\notin\Bbb Z$ and $f(x)=L$, so  $|f(x)-L|=|L-L|=0<\varepsilon$. And this is the definition of $\lim_{x\to x_0}f(x)=L$.

Title: Limits, counter-example


问:
Prove or disprove $(1,3)A_6=A_6(1, 3)$
Let $A_6$ be the alternating group of order 6. Then prove or disprove
$$(1,3)A_6=A_6(1, 3).$$
If above holds, then
$$(1,3)A_6(1, 3)=A_6.$$So, if we prove the last equality, then  the given equality also holds.
It is easy to see that
$$(1,3)A_6(1, 3)\subseteq A_6.$$ How to proceed further? Please help.

答:
$(1,3)A_6(1,3)\subset A_6$ since its elements are product of an even number of transpositions. The two groups are thus equal since they have the same cardinal.

Let $f(x)=(1,3)x(1,3)$. $f$ is an automorphism (the inner conjugation by $(1,3)$. $f(A_6)=(1,3)A_6(1,3)$, we deduce that $(1,3)A_6(1,3)$ has the same cardinal than $A_6$.

Title: Prove or disprove $(1,3)A_6=A_6(1, 3)$


问:
Given $x_1=1,x_2=2,x_{n+2}=3x_{n+1}-x_n\forall n\in\mathbb N$. Find $x_n$.

Given $x_1=1,x_2=2,x_{n+2}=3x_{n+1}-x_n\forall n\in\mathbb N$. Find $x_n$.

I tried to find ways to telescope, but failed. Please help. Thank you.

答:
There is a general way to solve those type of recurrences, through the characteristic polynomial of the relation.
$$
x_{n+2}+ax_{n+1}+bx_{n} = 0 \leadsto X^2+aX+b=0
$$
Here, it'd be $X^2-3X+1$. The roots are $\phi=\frac{3+\sqrt{5}}{2}$ and $\bar{\phi}=\frac{3-\sqrt{5}}{2}$.Any solution $(x_n)_{n\in\mathbb{N}^\ast}$ is of the form $x_n = \alpha \phi^{n-1} + \beta \phi^{n-1}$ (since the two roots ${\phi},\bar{\phi}$ are distinct). Then use the initial conditions $x_1$ and $x_2$ to get $\alpha,\beta$.</p>


问:
Integral $\int_{0}^{\pi/2} \cos{m\phi}\tan{\phi}\ d\phi$
In the field of Aerospace, for a thin surface to generate zero vertical force, the inclination it has to be kept at, can be written in the form,

$$
\alpha_0 = \frac{1}{\pi}\int_{0}^{\pi} y_x(\theta)(1-\cos\theta)d\theta
$$

where $y_x(\theta)$ is the slope of the thin surface.

If the thin surface is represented as a discrete Fourier series, the slope comes out to be,
$$
y_x(\theta) = \frac{2}{\sin\theta}\sum_{n=1}^{N}nA_n\cos{k_n\theta} \quad, \quad n \in \mathbb{N}
$$

I'm trying to obtain a closed form solution, for $n$ terms in $y_x$, by solving the resulting integral.

$$
\alpha_0 = \frac{2}{\pi}\sum_{n=1}^{N} nA_n \int_{0}^{\pi} \cos{k_n\theta} \big(\frac{1-\cos\theta}{\sin\theta}\big) d\theta
$$

With some manipulation, this further simplifies to the form,
$$
I = \int_{0}^{\pi/2} \cos{m\phi}\tan{\phi}\ d\phi
$$

where $m \in \mathbb{R}$

Any effort/hint is hugely appreciated!

答:
$m$ must be an odd integer for the singularity at $\phi=\pi/2$ to disappear.

So, assume $m=2n+1$ is positive (as we may); then $$\int_0^{\pi/2}\cos m\phi\tan\phi\,d\phi\underset{\phi=\pi/2-t}{\phantom{\big[}=\phantom{\big]}}(-1)^n\int_0^{\pi/2}\sin(2n+1)t\cot t\,dt=(-1)^n(I_n+I_{n+1}),$$ where $$I_n=\int_0^{\pi/2}\frac{\sin 2nt}{2\sin t}\,dt=\int_0^{\pi/2}\sum_{k=0}^{n-1}\cos(2k+1)t\,dt=\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}.$$


问:
Sketch the pre-image $f^{-1}(A)$ where $f(x,y) =\cos(x+y)$ on $A = [1,2]$
Question: Explaining your working, sketch the pre-image $f^{-1}(A)$ where $f(x,y) = \cos(x+y)$ on $A = [1,2].$

So far I have:

\begin{align}
& f^{-1}([1,2]) = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) \in [1,2] \} \\[6pt]
= {} & \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) \in [1,2] \} \\[6pt]
= {} & \{ (x,y) \in \mathbb{R}^2 \mid 1 \leq \cos(x+y) \leq 2 \}
\end{align}

I am not sure where to go from here or how to sketch this graph.

How do I sketch this pre-image?

答:
HINT

I would start by noticing that $f(x,y) = \cos(x+y)\in[-1,1]$. Thus we have that
\begin{align*}
f^{-1}(A) & = \{(x,y)\in\mathbb{R}^{2} \mid f(x,y)\in [1,2]\}\\\\
& = \{(x,y)\in\mathbb{R}^{2} \mid f(x,y)\in \{1\}\}\cup\{(x,y)\in\mathbb{R}^{2} \mid f(x,y)\in (1,2]\}\\\\
& = \{(x,y)\in\mathbb{R}^{2} \mid \cos(x+y) = 1\}\cup\varnothing\\\\
& = \{(x,y)\in\mathbb{R}^{2} \mid (x+y = 2k\pi)\wedge(k\in\mathbb{Z})\}
\end{align*}

Can you take it from here?


答: HINT: Since $(1,3) \notin A_6$ but $(1,3) \in S_6$, if we know $A_6 \unlhd S_6$ (any index $2$ subgroup is normal), we know by normal subgroup definition that $\forall g \in S_6$, $gA_6g^{-1} = A_6$.

Title: Prove or disprove $(1,3)A_6=A_6(1, 3)$

Body:
Let $A_6$ be the alternating group of order 6. Then prove or disprove
$$(1,3)A_6=A_6(1, 3).$$
If above holds, then
$$(1,3)A_6(1, 3)=A_6.$$So, if we prove the last equality, then  the given equality also holds.
It is easy to see that
$$(1,3)A_6(1, 3)\subseteq A_6.$$ How to proceed further? Please help.


问:
question about Hausdorff space
A subset A of a topological space $ (X, \tau)$ has $  \gamma$-property if every open cover $\{U_{\alpha}\}_{\alpha\in I}$ of $cl_{\tau}(A)$ there is finite $I_0 \subset I$so that $A  \subseteq \displaystyle\bigcup_{\alpha\in I_0}   cl _{\tau_{_{A}}} ( U_{\alpha} \cap A)  $.

Is a subset with a $  \gamma$-property in a Hausdorff space, closed?

答: No. Every non-closed subset whose closure is compact (such as $(0,1)$ inside $\mathbb{R}$) satisfies this property.

Title: question about Hausdorff space


问:
Create an isosceles triangle from only equilateral triangles. Possible or impossible?
My friend recently asked me a question. And I think I have a nice solution but wanted to share this question here to see how you approach this problem and hopefully with a solid proof.

The question is:

Can you create an isosceles triangle with an infinite number of any sizes of only equilateral triangles?

答:
You can fill the interior of any triangle (or more generally, any plane region) with non-overlapping equilateral triangles.  Consider an equilateral triangular lattice $L$, and start with those lattice triangles that are contained in your region $R$.

Then refine the lattice (adding the midpoints of all edges), and take the triangles of the new lattice that are contained in $R$ but are not already covered.  Repeat ad infinitum.

EDIT: You get something that looks like this


问:
How to approach series (and sums) like those?
Whats the right test to approach series like these, I spend ages but found none that are similar.
$a_k=\frac{(k+1)^k}{(-k)^k}$ and
$a_k=\frac{k+2^k}{k\cdot 2^k}$
those two have me stumped the ratio and root-test yield nothing. How to procced, any tips greatly appreciated!

Thank you both very helpful, can I conclude from this that the sums of those series are divergent?
$ \sum_{n=1}^{\infty} a_k=\frac{(k+1)^k}{(-k)^k} $ and
$ \sum_{n=1}^{\infty} a_k=\frac{k+2^k}{k\cdot 2^k}$ ?

答:
For thew first
$$\left|\frac{(k+1)^k}{(-k)^k}\right|= \left|\frac{k+1}{k}\right|^k\ge 1 $$
so $a_k:=\frac{(k+1)^k}{(-k)^k}\not\to 0$ (recall $a_k\to0\iff|a_k|\to0$) and the series cannot converges.

For the second, observe that
$$a_k:=\frac{k+2^k}{k2^k}=\frac{1}{2^k}+\frac{1}{k}$$
If it were convergent, then $\frac{1}{k}=a_k-\frac{1}{2^k}$ would be a convergent series, since both $a_k$ and $\frac{1}{2^k}$ were. But this is a contradiction. So the series $a_k$ cannot converges.


问:
How do I calculate typical group size?
If I have a set of groups of individuals (e.g. people), and I want to calculate the typical group size (as observed by individuals), how do I do this?

Wikipedia refers to this as "mean crowding" or "Typical Group Size" but doesn't give a formula: http://en.wikipedia.org/wiki/Group_size_measures

I believe the "typical group size" for a set of group sizes $\{a_1, a_2,..., a_n\}$ is simply:

$$\sqrt{\frac{1}{n}\sum_{i=1}^na_i^2 }$$

i.e. the root mean square or quadratic mean.

Is that correct? Is there a better way to characterise typical group size?

答:
The precise definition (according to the Wikipedia link in your question) seems to be that the typical group size is the average size of the group as observed by the individuals of your population. Thus, if you have a population of $N$ individuals, split into $K$ groups with size $n_k$ (thus, $\sum_{k=1}^K n_k = N$), the typical group size would be $$
  T = \frac{1}{N} \sum_{k=1}^K \left(\sum_{i=1}^{n_K} n_k\right) = \frac{1}{N}\sum_{k=1}^K n_k^2
$$
since the $n_k$ individuals in group $k$ all observe a group size of $n_k$.

It's entirely possible that it's for some reason more convenient to define the typical group size as $\sqrt{T}$ instead of plain $T$, but it then ceases to be the average group size as observed by the individuals.

Title: How do I calculate typical group size?


答:
Assuming $x$ and $y$ are real, the cosine cannot be in $[1,2]$ except by being equal to $1.$ And the cosine is equal to $1$ only when its argument is a multiple of $2\pi.$ So the preimage is
$$
\{(x,y)\in\mathbb R^2 : \text{for some } n\in\mathbb Z,\,\,\, x+y = 2\pi n \}.
$$
The graph of $x+y= 2\pi n$ for any particular $n$ is a straight line with slope $-1.$

So the preimage is the union of infinitely many parallel lines.

Title: Sketch the pre-image $f^{-1}(A)$ where $f(x,y) =\cos(x+y)$ on $A = [1,2]$

Body:
Question: Explaining your working, sketch the pre-image $f^{-1}(A)$ where $f(x,y) = \cos(x+y)$ on $A = [1,2].$

So far I have:

\begin{align}
& f^{-1}([1,2]) = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) \in [1,2] \} \\[6pt]
= {} & \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) \in [1,2] \} \\[6pt]
= {} & \{ (x,y) \in \mathbb{R}^2 \mid 1 \leq \cos(x+y) \leq 2 \}
\end{align}

I am not sure where to go from here or how to sketch this graph.

How do I sketch this pre-image?


问:
P.33 of Karen Smith 'An Invitation to Algebraic Geometry': projective coordinates
I do not get how to use the coordinate of the projective line at infinity as page 33 says.

Specifically when $x_{0} =0 \in \langle x_{0},x_{1},x_{2} \rangle$ a point is at $\langle x_{1},x_{2} \rangle$ in the projective line in $\mathbb{P}^{2}$; in the previous page a point where $x_{0} =0 \in \langle x_{0},x_{1} \rangle$ is at $\infty$, shouldn't the point be at $x_{1}$ to correspond with the subsequent page?

答:
The locus $\{x_0 = 0\}$ in $\mathbb P^2$ is a copy of $\mathbb P^1$, called the "line at infinity", where the projective coordinates are the other two coordinates $x_1, x_2$ on $\mathbb P^2$.  Note that this should not be confused with the point $(x_1, x_2) \in \mathbb C^2$, which corresponds to $\langle 1, x_1, x_2 \rangle \in \mathbb P^2$; the line at infinity is precisely the part of $\mathbb P^2$ that isn't contained in $\mathbb C^2$.

The situation in $\mathbb P^1$ is much the same.  The locus $\{x_0 = 0\}$ in $\mathbb P^1$ is a copy of $\mathbb P^0$, called the "point at infinity".  (Remember that coordinates on projective space are only defined up to scaling by a nonzero constant, so we have $\langle x \rangle = \langle y \rangle$ for all nonzero $x$ and $y$; that is, $\mathbb P^0$ does indeed consist of a single point.)  One usually doesn't impose coordinates on the point at infinity (it is, after all, only one point), but it is perfectly correct to say that it has coordinates $\langle x_1 \rangle$ in $\mathbb P^0$.  Again, this is different from the point $x_1 \in \mathbb C$; the point at infinity is precisely the part of $\mathbb P^1$ that isn't contained in $\mathbb C$.


答:
In any triangle, at least one angle is $\ge 60^{\circ}$.

So starting from the corresponding vertex you can draw an equilateral triangle, inscribed in the starting one , and which leaves out two triangles.

For each  of these remaining triangles you can repeat the process and leave out  four. And so on.


问:
Rationale behind truth values
I originaly asked a question on Programmers.SE to know why $0$ was consider $\text{false}$ and all the other [integral] values were considered $\text{true}$. That was a huge debate and many said it was a legacy from Boolean algebra where $0$ is indeed $\text{false}$ and $1$ is $\text{true}$.

Somebody suggested I go further and ask here why this is actually the case in Boolean algebra. So here is the question: what is the rationale for $0$ to be $\text{false}$ and $1$ to be $\text{true}$ and not the other way around in Boolean algebra?

答: I believe this has to do with boolean algebra where or is considered as plus operation and and is considered to be multiplication. If we assign false to 0 and true to 1 then the usual arithmetic works.</p>

Title: Rationale behind truth values


答:
For the first one we can factor out the $k$ in the numerator and denominator like $$\frac{(k+1)^k}{(-k)^k}=\frac{(k(1+\frac{1}{k}))^k}{(k(-1))^k}=\frac{(1+\frac{1}{k})^k}{(-1)^k}$$
It is well known that the series in the numerator converges to $e$ and the one in the denominator diverges thus the whole sequence diverges by the calculation rules for the limit.

For the second one we observe that $$\frac{k+2^k}{k2^k}=\frac{1}{2^k}+\frac{1}{k}$$ again here it is  a well known fact that both these sequences converge to zero as $k$ goes to infinity thus by the rules for the limit the sum converges to zero.

Title: How to approach series (and sums) like those?

Body:
Whats the right test to approach series like these, I spend ages but found none that are similar.
$a_k=\frac{(k+1)^k}{(-k)^k}$ and
$a_k=\frac{k+2^k}{k\cdot 2^k}$
those two have me stumped the ratio and root-test yield nothing. How to procced, any tips greatly appreciated!

Thank you both very helpful, can I conclude from this that the sums of those series are divergent?
$ \sum_{n=1}^{\infty} a_k=\frac{(k+1)^k}{(-k)^k} $ and
$ \sum_{n=1}^{\infty} a_k=\frac{k+2^k}{k\cdot 2^k}$ ?


问:
Finding probability using Var
I'm stuck on the following question:

Suppose you toss three coins: two fair coins and a positively biased coin (more likely to show Heads than Tails). The variance of the total number of Heads in the three tosses is 0.55. What is the probability that the biased coin shows head?

To start off, I made the following expression: Var(2F+B) = 0.55

I then seperated and simplified to 2Var(F) + Var(B) = 0.55

I used the equation of Var for the binomial distribution (which is np(1-p)) to get

2(2*0.5)(1-0.5) + Var(B) = 0.55

so Var(B) = -0.45. I then used the same equation of Var again to get

p(1-p) = -0.45 but after expanding and solving the quadratic equation, I get 2 values for the probabilities that are less than 0 and greater than 1.

Can anyone tell me where I went wrong.

答:
The main problem here is that you assume Var to be liniear, which it is not.

$Var(2F+B) = 2^2Var(F) + Var(B)$

Also note that the Variance can never be negative. Otherwise you are on a good path.

Title: Finding probability using Var


问:
How can it be that the empty set is a subset of every set but not an element of every set?
How can it be that the empty set is a subset of every set but not an element of every set?

I've understood that the empty set must be a subset of every set because if it were not a subset of every set then the statement $\exists x : x \in \emptyset \land x \notin M$, would need to be ture, but since the empty set has no elements this statement is a contradiction.

But how can it be that the empty set is not element of every set, if it is a subset of every set?

I also understood that it cannot be an element of itself, otherwise it wouldn't be the empty set anymore, but why is it not an element of every non-empty set?
If the empty set is subset of every set, than it also would need to be a subset of itself ($\emptyset \subset \emptyset$), how can that be?

答:
There might be versions of set theory where the requirement " the empty set is an element of every set" is satisfied. What I mean is that  it does not seem absurd prima facie. For example, in the set theoretic consruction of natural numbers, number zero ( that is : the empty set) is an element of every ( natural) number greater than 0 , and these numbers are sets. ( for example , $1=\{\emptyset\}= \{0\}, 2= \{\emptyset, \{\emptyset\}\}=\{0,1\} , 3= \{0,1,2\} $.

However, the question " is every set a member of every set ( different from itself)?"  can be settled  as a pure matter of fact. Any counter-example would do; Consider, for example,  the set : $\{1, 2,3\}$.

I think the question is : why does it seem plausible that, if a set is a subset of every set, then it should also be an element of every set? Maybe one could try to reconstruct the reasoning that produces this false appearence :

(1) The empty set s a subset of every set, say, of set S

(2) Therefore, all the elements of $\emptyset$ are also elements of S.

(3) Therefore the totality of the elements of $\emptyset$ is an
element of S.

(4) But this totality is $\emptyset$  itself .

(5) Therefore $\emptyset\in S$.

The mistake is hidden in steps  (3) and (4).

a) as to (3) : though it is true distributively that every element $\emptyset$ is an element of S, it is not true collectively

b) as to (4) the totality of the elements of $\emptyset$ is nothingness ( nil, nothing) , and therefore is not the same thing as the empty set which is something ( namely, a set).


问:
Symplectic Form Preserved by Orthogonal Transformation
I'm trying to prove that the symplectic form

$$\omega = d(\cos\theta) \wedge d\phi$$

is preserved by the action of $SO(3)$ on $S^2$ where $\phi$ and $\theta$ are spherical polars. Now $SO(3)$ simply acts by

$$\theta \mapsto \theta + \epsilon, \ \phi \mapsto \phi + \delta$$

and writing this diffeomorphism as $F:S^2 \to S^2$ I compute

$$F^*(\omega) = d(\cos(\theta + \epsilon))\wedge d(\phi + \delta) = \cos(\epsilon) d(\cos\theta)\wedge d\phi - \sin(\epsilon) d(\sin \theta)\wedge d \phi$$

Is this correct? It doesn't seem like the form is invariant under $SO(3)$. Perhaps it's only meant to be a local symplectomorphism though.

Am I allowed to claim that it is a local symplectomorphism because it gives the right result in the limit as $\epsilon \to 0$? I think that would be right, because it would mean that the Lie derivative vanishes.

Thanks in advance!

答: Here's a hint: Can you see that the standard area 2-form on $S^2$  is $SO(3)$-invariant?</p>


问:
Immersions vs. embeddings. WHY does an immersion fail to be an embedding?
I understand generally that embeddings are injective maps which are homeomorphic to their immage and that immersions are local embeddings but i fail to see the distinction of precisely why this is the case and why a figure of eight linesegment (immersed via the map $t\mapsto(\cos(\pi/2+t),\sin(2t))$ for $0<t<2\pi$) can be immersed into $\mathbb{R}^2$ but not embedded. I haven't been able to find a mathematical proof of why the crossing point is not homeomorphic and secondly from the definiton of immersion: "A surface is immersed if every point of the domain has a neighborhood which is embedded by the map." What defines how big the neighborhood is and does every point in there need to be embedded, excluding the point itself, or just any for it to be an immersion.

As I understand injective, every point of the domain needs to be mapped to a unique point in the co-domain but how does this work for the crossing point as there are two values of $t$ whch map to identical points in $\mathbb{R}^2$ ($\pi/2$ and $3\pi/2$)

答:
Why can a figure eight, immersed in $\Bbb R^2$ via $t\mapsto \left(\cos\left(\frac\pi 2+t\right),\sin(2t)\right)$, be immersed in $\Bbb R^2$ but not embedded?

You need to first state what the figure eight is. To talk about immersions, you should have a map between smooth manifolds, but the figure eight with a crossing is not a smooth manifold. If the figure eight means a copy of $S^1$ with a twist, then this is diffeomorphic to $S^1$ and can certainly be embedded in $\Bbb R^2$. However, this depends entirely on the map used. It does not make sense to ask if something immersed in $\Bbb R^2$ can be embedded in $\Bbb R^2$. You can either ask if one manifold can be embedded in another, or if a given map between manifolds is an embedding.

Again, as a smooth manifold, the figure eight is just $S^1$, so we need to specify a map $f\colon S^1\to \Bbb R^2$. Really, we want to imagine first embedding $S^1$ into $\Bbb R^3$ so that it looks like a figure eight from above. Here is an example, where $S^1$ is shown as a tube for the sake of visualizing.

Now if we project down to the $xy$ plane, we get a map $S^1\to \Bbb R^2$ whose image is shown below.

One can show that this is an immersion, meaning that for any $p\in S^1$, the differential map $df_p\colon T_pS^1\to T_{f(p)}\Bbb R^2$ is injective. However, $f$ itself is not even injective! There are clearly two points in $S^1$ which both land on the origin in $\Bbb R^2$. So $f$ is not a bijection onto its image, and therefore not a homeomorphism onto its image (embedding).

I haven't been able to find a mathematical proof of why the crossing point is not homeomorphic.

This language doesn't make sense, but you probably mean the following: for any open set $U\subset\Bbb R^2$ containing the origin, there is no homeomorphism between $U\cap f(S^1)$ and a neighborhood of $S^1$. This is more difficult to prove; I would probably use algebraic topology.

To summarize, there are distinct questions one could ask here:

Is there an embedding $S^1\to \Bbb R^2$? (Yes.)

Is the map $f\colon S^1\to \Bbb R^2$ an embedding? (No because $f$ is not injective.)

Is there an embedding from $S^1$ to $f(S^1)$ (topologized as a subset of $\Bbb R^2$)? (No but this is less immediate.)


答:
The problem here is that you are mistreating the distribution $F$.  If $F$ is the distribution for the number of Heads resulting from tossing two fair coins then you want to write $$\text{Var}(F+B)=\text{Var}(F)+\text{Var}(B)$$  We get $\text{Var}(F)=.5$ and all is good. We get $$.5+\text{Var}(B)=.55$$ which makes perfect sense.

But in your first expression, you appear to want $F$ to be the distribution for a single coin.  In that case you should have written $$\text{Var}(F_1+F_2+B)=2\text{Var}(F)+\text{Var}(B)$$

Where the $F_i$ are i.i.d. and both equal $F$.  But in this case $\text{Var}(F)=.25$ and again we get $.5+\text{Var}(B)=.55$

In either case, you should not have written $2F$ as no variable is being multiplied by $2$.

Title: Finding probability using Var

Body:
I'm stuck on the following question:

Suppose you toss three coins: two fair coins and a positively biased coin (more likely to show Heads than Tails). The variance of the total number of Heads in the three tosses is 0.55. What is the probability that the biased coin shows head?

To start off, I made the following expression: Var(2F+B) = 0.55

I then seperated and simplified to 2Var(F) + Var(B) = 0.55

I used the equation of Var for the binomial distribution (which is np(1-p)) to get

2(2*0.5)(1-0.5) + Var(B) = 0.55

so Var(B) = -0.45. I then used the same equation of Var again to get

p(1-p) = -0.45 but after expanding and solving the quadratic equation, I get 2 values for the probabilities that are less than 0 and greater than 1.

Can anyone tell me where I went wrong.


问:
Maximising $Tr(PAPB)$ for orthogonal projection matrix $P$ and symmetric $A,B$.
I want to find a vector $u$ that maximises $Tr(PAPB)$ where $P$ is an orthogonal projection over $u$, for symmetric $A,B$. Does this problem have a closed form solution?

Thanks

答:
Suppose $\|u\|=1$, you can write $P=uu^T$, then
\begin{align*}
\text{Tr}(PAPB) &= \text{Tr}(uu^TAuu^TB)\\
&= \text{Tr}(u^TAuu^TBu)\\
&= u^TAuu^TBu
\end{align*}
since it is a scalar. So this is the expression that you want to maximize.

I do not know a closed form expression for that, however so bounds can be given, a trivial one is that since $u$ has unit norm, we have $u^T A u\in[\lambda_1(A),\lambda_n(A)]$ and $u^T B u\in[\lambda_1(B),\lambda_n(B)]$, where $\lambda_k(X)$ is the $k$'th smalest eigenvalue of $X$. If we take $u$ to be the eigenvector associated to the $k$'th eigenvalue of $A$, we obtain that $u^T A u=\lambda_k(A)$ and so $$u^T A u u^T B u = \lambda_k(A)u^T B u\geq \min(\lambda_k(A)\lambda_n(B),\lambda_k(A)\lambda_1(B))$$This means that
\begin{align*}
\max_u u^TAuu^TBu &\geq \max_k \min(\lambda_k(A)\lambda_n(B),\lambda_k(A)\lambda_1(B))\\
&=\max(\min(\lambda_1(A)\lambda_n(B),\lambda_1(A)\lambda_1(B)),\min(\lambda_n(A)\lambda_n(B),\lambda_n(A)\lambda_1(B)))
\end{align*}
You can also inverse the role of $A$ and $B$ to get that
\begin{align*}
\max_u u^TAuu^TBu &\geq \max(\min(\lambda_1(B)\lambda_n(A),\lambda_1(B)\lambda_1(A)),\min(\lambda_n(B)\lambda_n(A),\lambda_n(B)\lambda_1(A)))
\end{align*}

We can find $A$ and $B$ such that this is tight, but the value can also go up to $\max(\lambda_1(A)\lambda_1(B), \lambda_n(A)\lambda_n(B))$ when the same eigenvector achieves the largest (lowest if they are negative) eigenvalues. So in general any value between those two is achievable by some matrices $A$ and $B$.

If $A$ and $B$ are both PSD, then all eigen values are positive and
\begin{align*}
&\max(\min(\lambda_1(A)\lambda_n(B),\lambda_1(A)\lambda_1(B)),\min(\lambda_n(A)\lambda_n(B),\lambda_n(A)\lambda_1(B)))\\
=&\max(\lambda_1(A)\lambda_n(B),\lambda_1(A)\lambda_1(B))\\
=&\lambda_1(A)\lambda_n(B)
\end{align*}
and similarly we are bounded by $\lambda_n(A)\lambda_1(B)$.


问:
Expected number of unique numbers after drawing $n$ balls from $N$ balls with $N$ different numbers $d$ times.

Problem:

We have $N$ balls and each of the balls are numbered $0,..., N-1$.

Each time, we draw $n$ balls from the set the numbers are recorded. The balls are then put back in.

Given $n<N/2$ what is the number of unique numbers after performing the drawing $d$ times?

Example:

$N=6$, $n=2$, and $d=3$

draw $1$ : $[0,1]$ so I recorded $[0,1]$.

draw $2$ : $[0,1]$ because they are already recorded, I didn't record anything.

draw $3$ : $[0,5]$ I recorded $5$ because $0$ was already recorded.

In this example, after $d=3$,  I recorded 3 unique numbers $[0,1,5]$

答:
I will calculate the expected number of unique numbers found after $d$ such draws.

Let $A_i$ be the event "the number $i$ was not drawn" for any $i \in \{1,\ldots,N\}$.

We have $\mathbb{P}(A_i)=\big(\frac{N-n}{N}\big)^d = \big(1-\frac{n}{N}\big)^d$.

Moreover, the number $X$ of unique numbers is just given by $X=\sum\limits_{i=1}^N \chi_{A_i^c}$.

So $\mathbb{E}[X] = \sum\limits_{i=1}^N \mathbb{E}[\chi_{A_i^c}] = N (1-\mathbb{P}(A_i)) = N\Big(1-\big(1-\frac{n}{N}\big)^d\Big)$.

Just a note: if we use $d=N$, and let $N \to +\infty$ (keeping $n$ constant), we get that the expected fraction $\mathbb{E}[X/N]$ goes to $1-e^{-n}$.


问:
For what values of $\alpha >0$ and $\beta >0$ does the following series $\sum_{n=2}^\infty \frac{1}{n^{\alpha}(\log n)^{\beta}}$ converge?
For what values of  $\alpha >0$ and $\beta >0$ does the following series $$\sum_{n=2}^\infty \frac{1}{n^{\alpha}(\log n)^{\beta}}$$ converge?

if $\alpha >1$ ,then limit comparison test  suggest that given series converge for any $\beta >0$

Also when $\alpha=1$ given series converge iff $\beta >1$ by integral test.

Is it correct ? Also what will happen when $0 <\alpha <1$

Any suggestions?Thanks

答:
What you said until using integral test is correct.

If $0 \lt \alpha \lt 1$, then as

$(\log n)^\beta \lt n^{\frac{1-\alpha}{2}}$ for $n $ large enough, you get

$$\frac{1}{n^\alpha(\log n )^\beta} \ge \frac{1}{n^{\frac{1+\alpha}{2}}}.$$

Hence the series diverges as $\frac{1+\alpha}{2}\le 1$.


问:
Limits, counter-example
Regarding this question here:

If there exists an $L ∈ R$ such that $\lim_{x→x_0} f(x) = L$ for every $x_0 ∈ R$, then $f (x) = L$ for every $x ∈ R$.

I think this is false, and I am trying to find a counter example for that.

Can someone please give me a hint for how to find the right $f(x)$ that counter this statement?

Thanks a lot!

答:
This is true if $f$ is continuous at all $x_0$.

So to find a counterexample we need to consider an $x_0$ there $f$ is not continuous at $x=x_0$ and that $\lim_{x\to x_0} f(x) = L \ne f(x_0)$.

Now our concern is that if $f(x_0) \ne L$ does that mean there is some $x_1$ "near" so that $\lim_{x\to x_1} f(x)$ can't be $L$ because $f(x_0)\ne L$ knocks it out of whack?

Well, no, $f$ is not continuous at $x_0$ and as $x$ "goes to" $x_1$ it will "pass by" $x_0$ and go through infinitely many other $x$s and as $f$ is not continuous at $x_0$ the value of $f(x_0)$ doesn't matter in the least.

Our simplest counter example is $f(x) = \begin{cases}L&x\ne x_0\\\text{something that isn't }L&x=x_0\end{cases}$

In that case for and $\epsilon > 0$ then for any positive $\delta$ at all, for all $x$ so that $0 < |x-x_0| < \delta$ we have $x\ne x_0$ and $f(x)=L$ so  $|f(x)-L| = 0< \epsilon$ so $\lim_{x\to x_0}f(x) = L$.

And for any $y \ne x_0$, then for any $\epsilon > 0$ if we let $\delta = |y-x_0|$ then if $0< |y-x| < \delta$ then $x \ne x_0$ and $f(x)=L$ so $|f(x)-L| = 0 < \epsilon$.  So $\lim_{x\to y}f(x) = L$.

So $\lim_{x\to w}f(x)=L$ for every $w \in \mathbb R$ but $f$ is not the function $f(x) = L$ because $f(x_0) \ne L$.

====

For a more specific function consider $f$ so that $f(x) = \begin{cases} 14&\text{if }x= 3.2\\\sqrt 5&\text{if }x \ne 3.2\end{cases}$

Claim 1:  $\lim_{x\to 3.2} f(x) = \sqrt 5$.

Pf: this makes sense because for every $x$ that is near to $3.2$ but not equal to $3.2$ we have $f(x)=\sqrt{5}$.  But to prove it formally:

For any $\epsilon > 0$ let $\delta = 517,295,746.7$ .  Then if $0 < |x-3.2| < 517,295,746.7$, we will have $x\ne 3.2$ and so $f(x) = \sqrt 5$.  So $|f(x) -\sqrt 5| = |\sqrt 5-\sqrt 5| = 0 < \epsilon$.  And that is the definition of $\lim_{x\to 3.2} f(x) = \sqrt 5$.

(Note:  There is utterly no reason to assume $\lim_{x\to 3.2} f(x) =f(3.2)$ unless $f$ is continuous at $3.2$.  It is completely irrelevant what $f(3.2)$ is in actually calculating what $\lim_{x\to 3.2}f(x)$ is (unless $f$ is continuous at $3.2$.)

Claim 2:  If $M\ne 3.2$ then $\lim_{x\to M}f(x) = \sqrt 5$.

Pf:  This makes sense as for all $x$ that are near $M$ but too close to $M$ to be$3.2$ we have $f(x) = \sqrt 5$.  But more formally.

For any $\epsilon > 0$ if we let $\delta = |M-3.2|$ we will notice if $0 < |M-x| < \delta =|M-3.2|$ then $x\ne 3.2$ so $f(x) =\sqrt 5$.  So $|f(x)-\sqrt 5| = |\sqrt 5 -\sqrt 5|=0< \epsilon$.  And that is the definition fo $\lim_{x\to M}f(x) = \sqrt 5$

So $\lim_{x\to x_0}f(x) = \sqrt 5$ for every possible $x_0$.  But $f(x) \ne \sqrt 5$ for every possible $x$.  (Because $f(3.2) = 14$.)

Title: Limits, counter-example


问:
Supremum of a compact set
For a non-empty compact set C ⊂ R (or $R^2$) prove that there is the point x ∈ C such that y <lex x for any y ∈ C\ {x}.

From the analysis, we know that supremum of any subset of R exits. And since C is a compact set (i.e it is bounded and closed) therefore its supremum belongs to C. Let x = sup C. Then there is a sequence of number i.e. $y_1,y_2,..y_n$ ∈ C. So the limit $y_i$ = x, and due to the closeness x ∈ C.
This is what I have tried in case of R but I am not sure if it's correct or do I have to include something else and am unable to do so for $R^2$. Any kind of feedback is appreciated?

Note: '<lex' denotes the lexicographic ordering by their co-ordinate vertices...i.e. X <lex Y if $x_1=y_1, \ x_2=y_2 ... , x_i< y_i$. where X =($x_1,x_2...x_d$) and Y =($y_1,y_2...y_d$) and i ∈ [d]

Thanks in advance.

答: HINT: Let $\pi_1:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$ and $\pi_2:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto y$ be the projection maps to the coordinate axes. Show that $\pi_1[C]$ has a maximum element, say $a$. Then show that $\{y\in\Bbb R:\langle a,y\rangle\in C\}$ has a maximum element, say $b$. Finally, show that $\langle a,b\rangle$ is the maximum element of $C$ in the lexicographic order on $\Bbb R^2$.

Title: Supremum of a compact set


问:
Finding a linear transformation matrix for a set of vectors
I am wondering if we can find a linear transformation matrix $A$ of size $3\times 3$ over the field of two elements $\mathbb{Z}_2$ i.e. a matrix $A$ of zeros and ones s.t.

$$A \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},
A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},
A \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
A \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},
A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix},
A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},
A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$

As can be seen above, the vectors I am looking at are all the possible 3D vectors with zeros and ones. $A$ is supposed to be an invertible transformation such that $A\neq A^{-1}$ because the effect of $A$ is a bijection from the space of all the possible 3D vectors with zeros and ones to that space itself.

I found an invertible $A$ using three of the equations above but that $A$ does not work for other vectors and so am wondering if it is possible to find such an $A$?

答:
The images of the vector $e_1, e_2,e_3$ of the canonical basis of $\mathbb Z_2^3$ defines entirely $A$.

Then if the image of the other vectors are coherent, $A$ exists. If not, there is no $A$ solution.

Applying this in your particular case, you can see that $A$ doesn’t exist.


问:
Number of binary sequences with exactly $n$ distinct subsequences.
Prove that number of binary sequences that have exactly $n$ distinct subsequences (including empty one) is $\varphi(n+1)$(where $\varphi(n)$ is Euler's totient function). For example, There are $\varphi(8)=4$ sequences with $7$ distinct subsequences:
$$\{000000,111111,010,101\}$$
I came up with this question while solving some other problem and wrote a simple program to calculate it for small numbers and when I searched for the sequence on OEIS, to my surprise it was same as Euler's totient function, but I couldn't prove this result.

答:
Here is a sketch of the proof.

First of all, the following construction is well-known: starting from the interval $\begin{bmatrix}\frac{0}{1}&\frac{1}{1}\end{bmatrix}$, we divide it into two (by placing $\frac{0+1}{1+1}=\frac{1}{2}$ in the middle), then divide those two intervals (by placing $\frac{0+1}{1+2}=\frac{1}{3}$ and $\frac{1+1}{1+2}=\frac{2}{3}$ in the middle(s) etc. At each step, the sub-interval bounded by $\frac{p}{q}$ and $\frac{r}{s}$ is split by a midpoint $\frac{p+r}{q+s}$. We would say that the fractions $\frac{p}{q}$ and $\frac{r}{s}$ are "parents" of the fraction $\frac{p+r}{q+s}$.

One can prove that all those fractions are reduced, and that sooner or later they will enumerate all rational numbers between $0$ and $1$.

Here is what the interval will look like after $5$ divisions:

$$\begin{bmatrix}\frac{0}{1}&\frac{1}{6}&\frac{1}{5}&\frac{2}{9}&\frac{1}{4}&\frac{3}{11}&\frac{2}{7}&\frac{3}{10}&\frac{1}{3}&\frac{4}{11}&\frac{3}{8}&\frac{5}{13}&\frac{2}{5}&\frac{5}{12}&\frac{3}{7}&\frac{4}{9}&\frac{1}{2}&\frac{5}{9}&\frac{4}{7}&\frac{7}{12}&\frac{3}{5}&\frac{8}{13}&\frac{5}{8}&\frac{7}{11}&\frac{2}{3}&\frac{7}{10}&\frac{5}{7}&\frac{8}{11}&\frac{3}{4}&\frac{7}{9}&\frac{4}{5}&\frac{5}{6}&\frac{1}{1}\end{bmatrix}$$

What has this got to do with sequences of $0$'s and $1$'s?

We can label the rational numbers on this interval (all except $0/1$ and $1/1$) using the following algorithm: label the midpoints on each "level $n$" (i.e. added at the division step $n$) using all the words of length $n-1$ in lexicographic order, left-to-right. This means, we label $\frac{1}{2}$ with the empty word $\epsilon$, and then we label $\frac{1}{3}$ and $\frac{2}{3}$ with $0$ and $1$, respectively, etc. We get a correspondence like this one:

$$\begin{bmatrix}\frac{0}{1}&\frac{1}{6}&\frac{1}{5}&\frac{2}{9}&\frac{1}{4}&\frac{3}{11}&\frac{2}{7}&\frac{3}{10}&\frac{1}{3}&\frac{4}{11}&\frac{3}{8}&\frac{5}{13}&\frac{2}{5}&\frac{5}{12}&\frac{3}{7}&\frac{4}{9}&\frac{1}{2}&\frac{5}{9}&\frac{4}{7}&\frac{7}{12}&\frac{3}{5}&\frac{8}{13}&\frac{5}{8}&\frac{7}{11}&\frac{2}{3}&\frac{7}{10}&\frac{5}{7}&\frac{8}{11}&\frac{3}{4}&\frac{7}{9}&\frac{4}{5}&\frac{5}{6}&\frac{1}{1}\\\text{n/a}&0000&000&0001&00&0010&001&0011&0&0100&010&0101&01&0110&011&0111&\epsilon&1000&100&1001&10&1010&101&1011&1&1100&110&1101&11&1110&111&1111&\text{n/a}\end{bmatrix}$$

We can extend the "parent" terminology, and for each sequence of $0$'s and $1$'s call the sequences corresponding to the parents of the fraction corresponding to the given sequence - the "parents" of the sequence. For example, the parents of $0101$ (corresponding to $\frac{5}{13}$) are $010$ (corresponding to $\frac{3}{8}$) and $01$ (corresponding to $\frac{2}{5}$). As we did not associate any sequences to the fractions $\frac{0}{1}$ and $\frac{1}{1}$, the left-most sequence ($000\ldots 0$) and the right-most sequence ($111\ldots 1$) has only one parent.

What is, I think, the crucial observation here is that the number of different subsequences exactly matches the denominator minus $1$ (!)

For example, the sequence $0101$ has the following $12$ subsequences: $\epsilon, 0, 1, 00, 01, 10, 11, 001, 010, 011, 101, 0101$. On the other hand, the fraction that corresponds to it is $\frac{5}{13}$ and $12=13-1$.

Let's take it on faith that this correspondence can be proven. The consequence is that, in the "grand" table (with all rational numbers between $0$ and $1$) the number of binary sequences with exactly $n$ different subsequences will be the same as the number of reduced fractions between $0$ and $1$ with denominator $n+1$. However, all reduced fractions will feature in the "grand" table somewhere, and so the number of those equals the number of potential numerators (coprime to $n+1$) - which is $\varphi(n+1)$. $\Box$

Now, what is left is to prove the correspondence. I have a sketch of a proof below, which goes by induction on the "level" of the particular fraction/binary sequence, which is the same as the induction on the length of the binary sequence itself. A hint that it is possible comes from the aforementioned sequence $0101$ which corresponds to the fraction $\frac{5}{13}$. The parents of the fraction $\frac{5}{13}$ are $\frac{3}{8}$ and $\frac{2}{5}$, which correspond to the two parents of our original binary sequences: $010$ and $01$. Note also the denominator $13=5+8$ (as per construction), i.e. $(13-1)=(5-1)+(8-1)+1$. This hints that we should be able to prove that:

The number of different subsequences corresponding to a sequence is the sum of the numbers of different subsequences corresponding the parents of that sequence, plus $1$.

If we could prove that, then the full correspondence would follow by induction.

To demonstrate, look at the different subsequences of $0101$ - again:

$$\begin{array}{l}\epsilon\\\color{red}{0}\\\color{blue}{1}\\0\color{red}{0}\\0\color{blue}{1}\\1\color{red}{0}\\1\color{blue}{1}\\00\color{blue}{1}\\01\color{red}{0}\\01\color{blue}{1}\\10\color{blue}{1}\\010\color{blue}{1}\end{array}$$

and also note the subsequences for $01$: $\epsilon, 0, 1, 01$ and for $010$: $\epsilon, 0, 1, 00, 01, 10, 010$.

What we get is: either $\epsilon$ (which will give us "plus $1$" in the general statement), or a subsequence of $01$ followed by $\color{red}{0}$, or a subsequence of $010$ followed by $\color{blue}{1}$.

Now for the (sketch) proof of the general correspondence. As said before, we only need to prove the inductive step outlined above.

If the sequence is $\underbrace{000\ldots 0}_n$, its only parent is $\underbrace{000\ldots 0}_{n-1}$, and the number of different subsequences in the given sequence is $n+1$ and in the parent it is $n$, so the inductive step is valid. This works similarly for the sequence $\underbrace{111\ldots 1}_n$.

Suppose that, now, $s_1s_2\ldots s_n$ is a sequence which has both $0$'s and $1$'s in it. Here is another observation (which I will still leave without proof, which is why the whole proof here is sketchy; I am, however, convinced enough that it is true):

Observation: If $s=s_1s_2\ldots s_n$ is a binary sequence containing both $0$'s and $1$'s, then its parents are precisely:

$s_1s_2\ldots s_{n-1}$

$s_1s_2\ldots s_{k-1}$ where $s_k$ is the last occurrence of the symbol different from $s_{k+1}=\ldots=s_n$.$\Box$

In other words, to get the parents, you: (a) remove the last symbol; (b) remove the last symbol, and all the symbols preceding it which are the same as it, and additionally remove the last symbol different from it.

Now, one can see that the number of different subsequences of $s$ is given by the sum of the numbers of the different subsequences of the parents, plus $1$. It is because $s$ has the following subsequences:

The empty sequence $\epsilon$

The sequences obtained by augmenting subsequences of $s_1s_2\ldots s_{n-1}$ (the first parent above) with $s_n$

The sequences obtained by augmenting subsequences of $s_1s_2\ldots s_{k-1}$ (the second parent above) with $s_k\ne s_n$ - the last occurence of the symbol $\ne s_n$.

This is true because every nonempty subsequence of $s$ ends either with $s_n$ or with the opposite symbol, and we can always assume that this symbol has actually come from $s_n$ or from $s_k$.$\Box$


问:
Why is the non-negative matrix factorization problem non-convex?
Supposing $\mathbf{X}\in\mathbb{R}_+^{m\times n}$, $\mathbf{Y}\in\mathbb{R}_+^{m\times r}$, $\mathbf{W}\in\mathbb{R}_+^{r\times n}$, the non-negative matrix factorization problem is defined as:
$$\min_{\mathbf{Y},\mathbf{W}}\left\|\mathbf{X}-\mathbf{Y}\mathbf{W}\right\|_F^2$$

Why is this problem non-convex?

答:
Do you have any reason to believe it is convex? In the space of nonlinear problems, convexity is the exception, not the rule. Convexity is something to be proven, not assumed.

Consider the scalar case; that is, $m=n=1$. Then the problem is
$$\min_{y,w\geq 0}(x-yw)^2=\min_{y,w\geq 0}x^2-2xyw+y^2w^2$$

The gradient and Hessian of $\phi_x(y,w)=x^2-2xyw-y^2w^2$ is
$$\nabla\phi_x(y,w)=\begin{bmatrix} 2yw^2 - 2xw \\ 2y^2w - 2xy \end{bmatrix}$$
$$\nabla^2\phi_x(y,w)=\begin{bmatrix} 2w^2 & 4yw - 2x \\ 4yw - 2x & 2y^2 \end{bmatrix}$$
The Hessian is not positive semidefinite for all $x,y,w\geq 0$. For example,
$$\nabla^2\phi_1(2,1)=\begin{bmatrix} 2 & 6 \\ 6 & 8 \end{bmatrix}, \quad
\lambda_{\min}(\nabla^2\phi_1(2,1))=-1.7082$$


问:
Minimization with Norm Constraint
I want to solve this minimization problem with Lagrange multiplier, but I face with one problem that it seems I need one equation more.
(A is a nn matrix and B, u and x are n1 vectors, I need to find u that minimizes equation)

$$ \min (1 + B(Ax+u)) \\
s.t \space \lVert u \rVert \leq 1 $$
For Lagrange function:
$$ B+ \lambda \frac{u^T}{\lVert u \rVert }=0\\
 \lVert u \rVert =1$$
After that:
$$ \lambda = -\frac{B}{u^T}$$
So here I don't have any more equation to find u.

答:
You get $B+\lambda u^T=0$ and $||u||=1$, so
$$u = -\frac{1}{\lambda} B^T$$
so
$$\left\lVert \frac{1}{\lambda} B^T \right\rVert=1$$
so $\lambda = ||B^T||$, from which you can obtain $u$.

Title: Minimization with Norm Constraint


问:
Chebyshev polynomials properties
In Wikipedia, it is mentioned that for Chebyshev polynomials "An important and convenient property of the Tn(x) is that they are orthogonal with respect to the inner product":

$\langle f(x), g(x)\rangle=\int_{-1}^{1} f(x) g(x) \frac{\mathrm{d} x}{\sqrt{1-x^{2}}}$

I don't understand the statement. Usually a function is orthogonal with respect to another function, however, what does it mean when a function is orthogonal to it's inner product?

答:
Think first of orthogonality of standard discrete vectors in Euclidean space, ${\bf a}$ and ${\bf b}$.  The orthogonality condition is simply ${\bf a}^t {\bf b} = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n = 0$.  Now imagine that we have functions, $f(x)$ and $g(x)$.  The natural extension of the orthogonality condition would be:

$$f(x_1) g(x_1) + f(x_2) g(x_2) + \cdots f(x_n) g(x_n) = 0.$$

Now let the number of $x$s go to infinity, i.e., $n \to \infty$, and the difference between sampled steps approaches zero.  In the limit, you have the integral definition stated in your question, $\int f(x) g(x) dx = 0$.

(The other factor is just for normalization, based on the definition of your class of functions.)

Title: Chebyshev polynomials properties


Title: Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$

Body:
Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$

Body:
According to the Vieta's formulas and the proposed relation, one has that
\begin{align*}
\begin{cases}
x_{1} + x_{2} + x_{3} = 3\\\\
2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = k\\\\
2x_{1}x_{2}x_{3} = 1\\\\
x^{2}_{1} + x^{2}_{2} + x^{2}_{3} = 6
\end{cases}
\end{align*}

Consequently, we have that
\begin{align*}
3^{2} = (x_{1} + x_{2} + x_{3})^{2} = x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = 6 + k \Rightarrow \boxed{k = 3}
\end{align*}


问:
Exponential function 1/x = e
Why is the following identity true?

$$\lim_{x\to0}(1+\sin2x)^{1/x}=\lim_{x\to0}\exp\left(\frac{1}{x}\ln(1+\sin2x)\right)$$

答:
(Explanation is below):

$$\begin{align*}
(1+\sin2x)^{\frac{1}{x}} \quad&{\stackrel{1}{=}}\quad e^{\ln(1+\sin2x)^{\frac{1}{x}}}\\
&{\stackrel{2}{=}}\quad e^{\frac{1}{x}\ln(1+\sin2x)}
\end{align*}$$

Explanation:

$a = e^{\ln a} $

$\ln a^b = b\ln a$

Title: Exponential function 1/x = e


答: It's not possible because $A\begin{bmatrix}0\\1\\1\end{bmatrix}\ne A\begin{bmatrix}0\\0\\1\end{bmatrix}+A\begin{bmatrix}0\\1\\0\end{bmatrix}$.

Title: Finding a linear transformation matrix for a set of vectors

Body:
I am wondering if we can find a linear transformation matrix $A$ of size $3\times 3$ over the field of two elements $\mathbb{Z}_2$ i.e. a matrix $A$ of zeros and ones s.t.

$$A \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},
A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},
A \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
A \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},
A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix},
A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},
A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$

As can be seen above, the vectors I am looking at are all the possible 3D vectors with zeros and ones. $A$ is supposed to be an invertible transformation such that $A\neq A^{-1}$ because the effect of $A$ is a bijection from the space of all the possible 3D vectors with zeros and ones to that space itself.

I found an invertible $A$ using three of the equations above but that $A$ does not work for other vectors and so am wondering if it is possible to find such an $A$?


答:
There is no such $A$, invertible or not. For instance your equations lead to
$$
\begin{bmatrix} 1 \\ 0\\ 0\end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}+ \begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix} =A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} +
A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} =
A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}
$$


答:
A linear transformation can be defined by its values on the basis. So consider the standard basis $e_1,e_2,e_3$ of the $3$ dimensional vector space over the integers modulo $2$. Then your $A$ is defined as a linear map by $Ae_1=(1,1,0),Ae_2=(1,0,1),Ae_3=(0,1,1)$ so we can represent $A$ as a matrix by $[A]_{E,E}=\left[\begin{array}{l}1&1&0\\1&0&1\\0&1&1\end{array}\right]$ We can now easily calculate the determinant of $A$ to be $\det(A)=0+0+0-1-1=-2=0\mod 2$ thus any such matrix is not invertible.
Furthermore such a matrix does not satisfy some of your other desired vector mappings, so there is no matrix satisfying all of your desires.


问:
Solve Logical Problem using contradiction rule and Truth Table
There are three containers, one container has diamond in it, other containers are empty. Each
container has label as clue whatever is inside the containers. Those labels are:

Container#1: The diamond is not present

Container #2: The diamond is not present

Container #3: The diamond is present in container#2

Only one statement is true and other are false. Which container has the diamond? Formalize this
problem into propositional statement and find a valid solution using truth table.

答: There are three possibilities. Either the first statement is true, the second is true, or the third is true. If the first is true, then statement 2 and 3 are false, leading to a contradiction. If the third is true, then statement 1 and 2 are false, leading to a contradiction. Therefore statement 2 must be true, meaning statement 1 is false and the diamond is in container 1.</p>


问:
Are these 2 graphs isomorphic?

They meet the requirements of both having an $=$ number of vertices ($7$).

They both have the same number of edges ($9$).

They both have $3$ vertices of degree $2$ and $4$ of degree $3$.

However, graph two has $2$ simple circuits of length $3$ whereas graph one has only $1$ of length $3$.

Is this not a valid method for checking isomorphism?

My guide says that these two figures are isomorphic.

答:
There's an isomorphism $f$ given on vertices by

$$f(A)=7,\: f(B)=4,\: f(C)=3,\: f(D)=6,\: f(E)=5,\: f(F)=2,\: f(G)=1.$$

An illustration of the isomorphism is given below:

Title: Are these 2 graphs isomorphic?


答:
Using Newton’s identities,

$$x_1^2+x_2^2+x_3^2= (x_1+x_2+x_3)^2- 2\sum x_ix_j=3^2-k=6$$

and therefore $k=3$.


问:
Endomorphism ring of elliptic curve over $\overline{\mathbb{F}_p}$
I want to prove that the endomorphism ring of elliptic curve $E$ over $\overline{\mathbb{F}_p}$ is not isomorphism to $\mathbb{Z}$.

We can find a $q=p^n$ such that $E$ is defined over $\mathbb{F}_q$. I have proved that the Frobenius endmorphism $(x,y)\rightarrow (x^q,y^q)$ is not equal to $[m]$ for any $m\in\mathbb{Z}$, thus the injective map $\mathbb{Z}\rightarrow\operatorname{End}E$ given by $m\rightarrow[m]$ is not surjective, thus this map is not an isomorphism, but can there exist other map such that $\mathbb{Z}\rightarrow\operatorname{End}E$ is an isomorphism?

For example, the map $\mathbb{Z}\rightarrow\mathbb{Z}$ given by $x\rightarrow 5x$ is injective and not surjective, but $\mathbb{Z}\cong\mathbb{Z}$.

Thanks.

答:
$E/\Bbb{F}_3 : y^2=x^3+x$ then $\# E(\Bbb{F}_3)= 4$ thus $\phi_3$ is a root of $X^2+3$ so that $\phi_9 = [-3]$.

$End(E/\Bbb{F}_9)=\Bbb{Z}[i,\phi_3]$ (or maybe $\Bbb{Z}[i,\phi_3]$ just has finite index in $End(E/\Bbb{F}_9)$ ?) it is still not $\Bbb{Z}$ but this is not due to the Frobenius $\phi_9$.


问:
How to solve this candy collectors puzzle?
I am trying to solve this Jane Street puzzle with the solution given here.

For convenience I will rewrite the puzzle here:

Five children went trick-or-treating together and decided to randomly split their candy haul at the end of the night.  As it turned out, they got a total of 25 pieces of candy, 5 copies each of 5 different types (they live in a small town).  They distribute the candies by choosing an ordering of the 25 uniformly at random from all shufflings, and then giving the first 5 to the first child, the second 5 to the second, and so on.

What is the probability that each child has one type of candy that they have strictly more of than every other trick-or-treater? Give your (exact!) answer in a lowest terms fraction.

Here is what I have done so far.

Label each candy type as either candy 0, 1, 2, 3, or 4. Label the children as either child A, B, C, D, or E. Suppose each candy is taken randomly out of a bag and placed in order to produce a sequence like this

[[2, 0, 2, 0, 4], [0, 4, 0, 4, 3], [3, 3, 1, 2, 1], [2, 1, 2, 4, 3], [1, 3, 1, 4, 0]],

where the first 5 go to child A, the second 5 go to child B and so on. The number of permutations of this list is
$$\frac{25!}{(5!)^5}.$$

We need to calculate the number of permutations where every child dominates in 1 and only 1 candy type. We can create a table which looks like this
$$\begin{matrix}
  & A & B & C & D & E \\
0 & 2 & 2 & 0 & 0 & 1 \\
1 & 0 & 0 & 2 & 1 & 2 \\
2 & 2 & 0 & 1 & 2 & 0 \\
3 & 0 & 1 & 2 & 1 & 1 \\
4 & 1 & 2 & 0 & 1 & 1 \\
\end{matrix}$$
where each element of the table shows how much of each candy type each child has. For example the sequence above shows that child A has 2 of candy 0, none of candy 1, 2 of candy 2, none of candy 3 and 1 of candy 4.

For a child to dominate a candy type they must have at least 2 of that type. Therefore we can initialise our table like this
$$\begin{matrix}
  & A & B & C & D & E \\
0 & 2 & * & * & * & * \\
1 & * & 2 & * & * & * \\
2 & * & * & 2 & * & * \\
3 & * & * & * & 2 & * \\
4 & * & * & * & * & 2 \\
\end{matrix}$$
Now we need to create some code to solve all the possible ways to fill in the $*$ such that every child domiantes one candy type (and the sum of each row and column equals 5) where the diagonal elements can have values $\in \{2, 3, 4, 5\}$ and the off-diagonal elements can have values $\in \{0, 1, 2\}$. Where we multiply each result by
$$\prod_{i=1}^{5}\frac{5!}{\prod_{j=1}^{5} a_{ij}!},$$
where $a_{ij}$ corresponds to the value in row $i$, column $j$, to account for the different permutations. Finally, we multiply the final result by $5!$ to account for the different permutations associated with different children dominating different candy types.

Is this the right way of approaching the problem? I don't know how I would make such a piece of code I was thinking of using something similar to the backtracking algorithms which are used to solve sodoku problems.

答:
Here is a solution of a very different flavor, which uses exponential generating functions, along with any CAS which can multiply multivariate polynomials, such as Mathematica.

Let $f_1$ be the following polynomial in five variables:
\begin{align}
f_1(a,b,c,d,e) &= \frac{a^5}{5!}+\frac{a^4}{4!}(b+c+d+e)+\\
&\quad\frac{a^3}{3!}\left(\frac{b^2}{2!}+\frac{c^2}{2!}+\frac{d^2}{2!}+\frac{e^2}{2!}+bc+bd+be+cd+ce+de\right)
\\&\quad\frac{a^2}{2!}(bcd+bce+bde+cde)
\end{align}
Roughly the variables represent children, and each monomial represents the ways to distribute candy number $1$ so that child $a$ has more than anyone else. For example, one of the monomials is $\frac{a^3}{3!}cd$, which means child $a$ gets $3$ of candy number one, while $c$ and $d$ each one of candy number one.

Similarly, let $f_2$ represent the number of ways to distribute candy $2$ so that child $b$ has more than anyone else. Concretely, $f_2(a,b,c,d,e)=f_1(b,c,d,e,a)$. Then, do the same for $f_3,f_4,f_5$ for candies $3,4$ and $5$. It turns out that multiplying these polynomials, then multiplying by $5!^5$ represents the number of ways to distribute each of the candies. This is the magic of exponential generating functions; to learn more of why this works, I suggest looking at Chapter 3 of Herbert Wilf's excellent text generatingfunctionology, available for free online from the author's website.

Since we want each child to receive five candies total, the number of valid ways to distribute the candies is the coefficient of $(abcde)^5$ in the product $f_1f_2f_3f_4f_5$. You then multiply by $5!$ one again to account the for other ways to assign which child has the most of which candy, then divide by the total number of candy distributions, $(25!)/(5!)^2$. Here it the Mathematica code which accomplishes this, which you can run online here.

```<code>f[a_,b_,c_,d_,e_] :=
a^5/5! +
a^4/4!*(b + c + d + e) +
a^3/3!*((b^2+c^2+d^2+e^2)/2!+ b c + b d + b e + c d + c e + d e) +
a^2/2!*(b c d + b c e + b d e + c d e);

EGF = f[a,b,c,d,e]*f[b,c,d,e,a]*f[c,d,e,a,b]*f[d,e,a,b,c]*f[e,a,b,c,d];

numerator = 5!^6 * Coefficient[EGF, (a*b*c*d*e)^5];

Print[numerator / (25!/ 5!^5)]
</code>```

Body:
Here is a solution of a very different flavor, which uses exponential generating functions, along with any CAS which can multiply multivariate polynomials, such as Mathematica.

Let $f_1$ be the following polynomial in five variables:
\begin{align}
f_1(a,b,c,d,e) &= \frac{a^5}{5!}+\frac{a^4}{4!}(b+c+d+e)+\\
&\quad\frac{a^3}{3!}\left(\frac{b^2}{2!}+\frac{c^2}{2!}+\frac{d^2}{2!}+\frac{e^2}{2!}+bc+bd+be+cd+ce+de\right)
\\&\quad\frac{a^2}{2!}(bcd+bce+bde+cde)
\end{align}
Roughly the variables represent children, and each monomial represents the ways to distribute candy number $1$ so that child $a$ has more than anyone else. For example, one of the monomials is $\frac{a^3}{3!}cd$, which means child $a$ gets $3$ of candy number one, while $c$ and $d$ each one of candy number one.

Similarly, let $f_2$ represent the number of ways to distribute candy $2$ so that child $b$ has more than anyone else. Concretely, $f_2(a,b,c,d,e)=f_1(b,c,d,e,a)$. Then, do the same for $f_3,f_4,f_5$ for candies $3,4$ and $5$. It turns out that multiplying these polynomials, then multiplying by $5!^5$ represents the number of ways to distribute each of the candies. This is the magic of exponential generating functions; to learn more of why this works, I suggest looking at Chapter 3 of Herbert Wilf's excellent text generatingfunctionology, available for free online from the author's website.

Since we want each child to receive five candies total, the number of valid ways to distribute the candies is the coefficient of $(abcde)^5$ in the product $f_1f_2f_3f_4f_5$. You then multiply by $5!$ one again to account the for other ways to assign which child has the most of which candy, then divide by the total number of candy distributions, $(25!)/(5!)^2$. Here it the Mathematica code which accomplishes this, which you can run online here.

f[a_,b_,c_,d_,e_] :=
a^5/5! +
a^4/4!*(b + c + d + e) +
a^3/3!*((b^2+c^2+d^2+e^2)/2!+ b c + b d + b e + c d + c e + d e) +
a^2/2!*(b c d + b c e + b d e + c d e);

EGF = f[a,b,c,d,e]*f[b,c,d,e,a]*f[c,d,e,a,b]*f[d,e,a,b,c]*f[e,a,b,c,d];

numerator = 5!^6 * Coefficient[EGF, (a*b*c*d*e)^5];

Print[numerator / (25!/ 5!^5)]


问:
Efficient method to evaluate integrals with multiple roots in denominator
$$ \int \frac{x^2-1}{x\sqrt{(x^2 + \beta x +1 )} \sqrt{(x^2 + \alpha x +1 )}} dx $$

I saw this question but I don't think the methods in answers are much applicable here.

My attempt:

$$ \int \frac{x^2-1}{x\sqrt{(x + \frac{\beta}{2})^2  + ( 1 - \frac{\beta^2}{4} )} \sqrt{(x + \frac{\alpha}{2})^2  + ( 1 - \frac{\alpha^2}{4} )}} dx= \frac{1}{\sqrt{ 1- \frac{\beta^2}{4} } \sqrt{1- \frac{\alpha^2}{4} }}\int \frac{x^2-1}{x\sqrt{\frac{(x + \frac{\beta}{2})^2}{ ( 1 - \frac{\beta^2}{4} )}  +1} \sqrt{\frac{(x + \frac{\alpha}{2})^2}{ ( 1 - \frac{\alpha^2}{4} )}  +1}} dx
$$

Not sure what's the best way to go now..

答:
Divide both numerator and denominator by $x^2$

$$\int \frac{1-\frac{1}{x^2}}{\sqrt{(x+\beta +\frac{1}{x})}\sqrt{(x+\alpha +\frac{1}{x})}}dx$$

which immediately suggests using the substitution $t=x+\frac{1}{x}$

$$\int \frac{dt}{\sqrt{\left(t+\frac{\alpha+\beta}{2}\right)^2-\left(\frac{\alpha-\beta}{2}\right)^2}} = \cosh^{-1}\left[\frac{t+\frac{\alpha+\beta}{2}}{\left|\frac{\alpha-\beta}{2}\right|}\right]$$

from a simple hyperbolic substitution. This makes the final answer

$$\cosh^{-1}\left[\frac{x+\frac{1}{x}+\frac{\alpha+\beta}{2}}{\left|\frac{\alpha-\beta}{2}\right|}\right]$$


问:
Exact angles of a 3-4-5 triangle
I am interested in finding exact values for the angles of a 3-4-5 triangle.

In particular, I would like to know the exact value of $\frac{1}{4}\sin^{-1}(\frac{4}{5})+\sin^{-1}(\frac{3}{5})$.

For context, this came up in an integral i was solving, mainly for fun. Here is the integral, in case there is a simpler solution:

$$\frac{1}{2}\int_0^{\frac{2}{5}}1+f(2t)dt+\int_{\frac{2}{5}}^{\frac{1}{2}}f(t-1)dt-\frac{1}{2}\int_0^{\frac{1}{2}}f(2t-1)dt$$

where $f(x)=\sqrt{1-x^2}$.

I've looked at this question: Prove that the ratio of acute angles in a $3:4:5$ triangle is irrational, so I understand if what I'm asking for is not possible.

答:
As noted in this answer to the question "Prove that the ratio of acute angles in a $3:4:5$ triangle is irrational", we know from Niven's Theorem that $p:=\sin^{−1}(4/5)$ and $q:=\sin^{−1}(3/5)$ are not rational-degree angles (aka, rational multiples of $\pi$), because their sines are rationals other than $0$, $\pm1/2$, and $\pm 1$.

Moreover, considering OP's ultimate target value $r:=\frac14p+q$, we can calculate $\sin(4r)=−44/125$. Again by Niven, we conclude that $4r$ is not a rational-degree angle, so that neither is $r$. $\square$

Title: Exact angles of a 3-4-5 triangle


问:
Wanted : for more formulas to find the area of a triangle?
I know some formulas to find a triangle's area, like the ones below.

Is there any reference containing most triangle area formulas?

If you know more, please add them as an answer

$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 &  b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1  & 1\\
1 &  -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$

Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1  &1 \\
 y_2&z_2  &1 \\
y_3 &z_3  &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1  &1 \\
 z_2&x_2  &1 \\
z_3 &x_3  &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1  &1 \\
 x_2&y_2  &1 \\
x_3 &y_3  &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a  &1 \\
x_b &y_b  &1 \\
x_c &y_c  & 1
\end{vmatrix}$$

In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$

答:
Two more expressions, similar to Heron’s formula
for the area of $\triangle ABC$.

Let cevians $AD,AE,BF,BG,CK,CL$
be such that
(see name-for-the-pair-of-constrained-triples-of-cevians
for more info)
\begin{align}
|BD|&=|CE|=ka
,\quad
|CF|=|AG|=kb
,\quad
|AK|=|BL|=kc
\tag{1}
\end{align}

for some $k\in\mathbb{R}$,
and let
\begin{align}
|AD|&=d_{a1} ,\quad |BF|=d_{b1} ,\quad |CK|=d_{c1}
\tag{2}
,\\
|AE|&=d_{a2} ,\quad |BG|=d_{b2} ,\quad |CL|=d_{c2}
\tag{3}
,\\
\delta_1&=\tfrac12\,(d_{a1}+d_{b1}+d_{c1})
\tag{4}
,\\
\delta_2&=\tfrac12\,(d_{a2}+d_{b2}+d_{c2})
\tag{5}
.
\end{align}

Then for any $k\in\mathbb{R}$

\begin{align}
S_{ABC}&=
\frac{\sqrt{\delta_1(\delta_1-d_{a1})(\delta_1-d_{b1})(\delta_1-d_{c1})}}{k^2-k+1}
\tag{6}
,\\
&=
\frac{\sqrt{\delta_2(\delta_2-d_{a2})(\delta_2-d_{b2})(\delta_2-d_{c2})}}{k^2-k+1}
\tag{7}
.
\end{align}

In fact,
expressions (6)-(7) also include
the famous Heron’s formula
as a special case when either $k=0$ of $k=1$,
and the triplet of cevians becomes a triplet of the sides $a,b,c$.


Title: Exponential function 1/x = e

Body:
Why is the following identity true?

$$\lim_{x\to0}(1+\sin2x)^{1/x}=\lim_{x\to0}\exp\left(\frac{1}{x}\ln(1+\sin2x)\right)$$

Body:
$$y=(1+\sin(x))^{1/x}$$
$$\ln(y)=\ln\left[(1+\sin x)^{1/x}\right]=\frac1x\ln(1+\sin x)$$
but:
$$y=e^{\ln(y)}=e^{\frac1x\ln(1+\sin x)}$$


Body:
Rearrange $2x^3 - 6x^2 + kx -1=0$ as $x(2x^2+k)=6x^2+1$, and square both sides to get the equation in $x^2$
$$x^6 +(k-9)x^4+(\frac{k^2}4-3)x^2-\frac14=0$$

Then

$$x_1^2+x_2^2+x_3^2= 6= -(k-9)$$
which yields $k=3$.