问:
Why is $ \frac{1}{4\pi} \int \int_S dx \, dy\, \vec{n} \cdot ( \partial_x \vec{n} \times \partial_y \vec{n} ) = 1 $?
I am trying to understand this notion of "topological charge" from a paper in Physics Review Letters.  They talk about skyrmions and give this integral (here $\vec{n}$ is a vector field on the sphere, from unit vectors in 3-space, which is also a sphere.) :
$$ \frac{1}{4\pi} \int \int_S dx \, dy\, \Big[ \vec{n} \cdot ( \partial_x \vec{n} \times \partial_y \vec{n} ) \Big]  = 1  \in \mathbb{Z}$$
In this case, this is reduced to a vector calculus problem, there's lots of questions that this result was a whole number.  The winding number discusses this notion for a curve in the plane winding around a point or since this an integral over $S^2$ we can talk about "degree".  One other possible statement is:
$$ \pi_3(S^3) = \mathbb{Z} $$
which is relatd to the Hopf fibration.  In this case, there's no more integral.  Discussions of K-theory are very technical, there's is also article on vector bundles.

答:
Your wording is a bit vague and the physicists are being sloppy, as usual. $\vec n$ is simultaneously the position vector on the unit sphere and the unit outward normal field. You cannot parametrize the entire sphere by $(x,y)$, but what the integrand is supposed to represent (if you had a surface that could be parametrized by a region in the $xy$-plane) is the surface area element of the sphere, pulled back by the $(x,y)$-parametrization. Taking all these grains of salt into account, the integral is supposed to represent the surface area of the entire sphere, which, of course, is $4\pi$.

In general, if $\mathbf x(u,v)$ is parametrizing a surface $S\subset\Bbb R^3$ as $(u,v)$ varies in a region $D$ in the $uv$-plane, then for a vector field $\vec F$ on $S$, the integral
$$\int_D \vec F(\mathbf x(u,v))\cdot \big(\frac{\partial\mathbf x}{\partial u}\times \frac{\partial\mathbf x}{\partial v}\big)du\,dv$$
gives the flux of $\vec F$ across $S$. The flux of the position vector on the unit sphere gives the surface area of the sphere.


问:
Find all units, zero divisors, nilpotent elements and idempotent elements in $\Bbb Z[x]/(x^2-1)$
I have no clue on how to do unit elements (inversible), zero divisors (might not exist),  I know only for stuff like $\Bbb Z_n$, and I guessed that nilpotents you get from $(ax+b)^n=0$ and we have $a=0, b=0$ so only $0$ is nilpotent (like $b^n=0$ so b=0 and $a^nx^n=0$ so $a=0$ as well) and for idempotents, you just solve $(ax+b)^2=ax+b$ and you find $a,b$ which have to be something like $0$ and $1$?

Is this true for any case, like what has my function, $x^2-1$ have to do with any of the nilpotent/idempotent stuff?

Maybe I need like $x=\pm 1$, from the function and I put in the equation for idempotents that $x^2=1$? So this is literally the only information I have in all seminars, tutorials etc. I can't find anything related to this stuff, also I'm very bad at math, simple stuff is good.

答:
When you make a quotient ring out of a ring $R$ and one of its ideals $I$, the elements present in that quotient are subsets of $R$ of the form $a + I := \{ a + b \colon b \in I \}$, where $a \in R$. It forms a ring under the operations $(a + I) + (b + I) := ((a+b) + I)$ and $(a + I)(b + I):= (ab +I)$.

In this ring, it is easy to check that its zero (the addition identity) is $0 + I$. So, what you're looking for in that ring are (in the nilpotent case) elements $(a +I) \in R/I$ with $(a + I)^n = (a^n +I) = (0 + I)$.

This in turn is equivalent to stating that $a^n \in I$. Now you can check from the definition of an ideal that in this case $I := (x^2 -1) = \{ g(x)(x^2-1) \colon g(x)\in \mathbb{Z}[x] \}$.

So for an element of $R/I$ to be nilpotent, it must be of the form $q(x) + I$ with $q(x)^n = g(x)(x^2 -1)$ for some $n \in \mathbb{N}$ and some $g(x) \in \mathbb{Z}$.

To give you an example, $((x-1) + I)((x+1) + I) = ((x^2 - 1) + I) = (0 + I)$ since $x^2 -1 \in I$. Note also that $(x-1), (x+1) \notin I$, so $((x-1) + I), ((x+1) + I) \neq (0 + I)$, that is, they are not zero in the ring $\mathbb{Z}[x]/(x^2-1)$

Hope this helps :)

ContentLicense: CC BY-SA 4.0


问:
What is wrong with my derivation of the general formula of the power series for $\cos^{2}(x)$
I have arduously obtained the series for $\cos^2(x)$ by using long multiplication method (multiplying twice the Taylor series of $\cos(x)$ which is $\sum_{n=0}^{+\infty} \frac{(-1)^nx^{2n}}{(2n)!}$
$$\cos^2(x)=1-x^2+\dfrac{1}{3}x^4-\dfrac{2}{45}x^6+\dfrac{1}{315}x^8+...$$

I wish to manipulate with the sigma notation to reduce the cost of labour, so I square the above general formula for $\cos(x)$:

$$\cos^2(x)=\sum_{n=1}^{+\infty} \left(\frac{(-1)^nx^{2n}}{(2n!)}\right)^2=\sum_{n=1}^{+\infty} \dfrac{(-1)^{2n}x^{4n}}{[(2n)!]^2}$$

But this is plain wrong, the formula that I obtain is not the general formula for the series of $\cos^2(x)$.

So what is wrong with my manipulation with the sigma notation of this series. The power of $x$ and $-1$ is too high, it should be $2n$. Also, is my denominator correct?

How do you obtain the general formulae for this series?

答:
Since your error was already explained in other answers let us see what one obtains if the series multiplication in your example is applied correctly. We have:

$$\begin{align}\cos^2 x&=\left[\sum_{i\ge0}(-1)^i\frac{x^{2i}}{(2i)!}\right]^2\\
&=\sum_{j\ge0}\sum_{i\ge0}(-1)^{i+j}\frac{x^{2i+2j}}{(2i)!(2j)!}\\
&=\sum_{k\ge0}(-1)^k \frac{x^{2k}}{(2k)!}\sum_{i\ge0}\frac{(2k)!}{(2i)!(2k-2i)!}\\
&=\frac12+\frac12\sum_{k\ge0}(-1)^k\frac{(2x)^{2k}}{(2k)!}\\
&=\frac{1+\cos2x}2,
\end{align}$$
where the proof of
$$
\sum_{i=0}^k\frac{(2k)!}{(2i)!(2k-2i)!}
=\begin{cases}1,& k=0,\\
2^{2k-1},& k>0
\end{cases}
$$
is left to you.


问:
Is there a Reason why this Iteration Always Lead to a Prime?
Let

$T(n) = n - lpf + 1$

where $n∈\mathbb{N}$,  $n\geq2$ and $lpf =$ least prime factor of $n$.

Let $T^i(n)$ be the output of the $i$th iteration of $T$ on $n$.
Why is it that for all $n$, there exists some $i$ such that $T^i(n)$ is prime?

For example, if $n=95$ we have $95 \to 91\to 85\to81\to79$.

答: Note that the least prime factor of $n$ is always between $2$ and $n$, so $T(n)$ is between $1$ and $n-1$.  In particular, $T(n)$ is always smaller than $n$, so if we iterate $T$ over and over starting with some integer $n\geq 2$, we'll eventually reach $T^i(n)=1$ for some $i$.  (More formally, you can prove this by strong induction on $n$: either $T(n)=1$, or else $2\leq T(n)<n$ and the induction hypothesis says we'll eventually reach $1$ from $T(n)$.)  But now note that $T(m)=1$ means $m$ is equal to its least prime factor, so $m$ is prime.  Thus $T^i(n)=1$ implies $T^{i-1}(n)$ is prime.


问:
Finding the minimum possible size of the edge cut of a complete graph.
I want to find the minimum size of edge cut that separates $K_{3n}$ into three components. I am thinking for the three components after separation, they will all be complete, with orders $x,y$ and $z$. The relations are $x+y+z=3n$, and finding the $\max ((x-1)x/2+(y-1)y/2+(z-1)z/2)$ will lead to the answer. But I am stuck with finding the maximum.

What should I do to continue? Or am I in a wrong direction.

答:
If you have done the two-component version, then you have seen that the minimum edge cut there separates a $1$-vertex component from an $n-1$ vertex component using $n-1$ edges. So it makes sense to conjecture that the optimal solution here will have two $1$-vertex components and an $(n-2)$-vertex component.

This has $2n-3$ edges, which is pretty low. Can we do better?

Well, it takes $k(n-k)$ edges to separate a component of order $k$ from the rest of the graph. This is already more than $2n-3$, unless $k \in \{1, 2, n-2, n-1\}$. (You can check this part with calculus, if you like.) So each of our components must have one of these orders.

The only way to get three components of these orders to add up to $n$ is to take $1$, $1$, and $n-2$, provided that $n > 6$.


问:
An interesting question in number theory
Let $ x_1, x_2, x_3, ..., x_n $ be a sequence consecutive of composite integers. Show that $ x_n $ admits a strict divisor $ d $ such that $ d> n $.
Exemple: 24, 25, 26, 27, 28 is a sequence consecutive of composite integers.

答: Consider the case where $x_k=k$ for $1 \le k \le n.$ This is a sequence of consecutive integers for which $x_n=n.$ But here $x_n$ has only divisors each at most $n$, none $>n.$


问:
Does such $A,B$ exist?
true/false test: there're $n\times n$ matrices $A,\ B$ with real entries such that $(I-(AB-BA))^n=0$

I'm cluesless to begin.

答:
Let $C = AB - BA$ and assume that $(C - I)^n = 0$.

Then the minimal polynomial $m(x)$ of $C$ divides $(x - 1)^{n}$. Thus $m(x) = (x-1)^{k}$ for some $1 \leq k \leq n$. This shows that the only eigenvalue of $C$ (considered in $\Bbb{C}$) is $1$, hence the characteristic polynomial of $C$ is $(x-1)^{n}$. In particular, $\operatorname{tr}(C) = n > 0$. This contradicts the fact that $\operatorname{tr}(C) = 0$. Therefore no such matrices $A$ and $B$ exist.

ContentLicense: CC BY-SA 3.0

Title: Does such $A,B$ exist?


问:
Proof verification: If $\lim_{n\to\infty} |a_n| = 0$, then $\lim_{n\to\infty}a_n = 0$.
Hey I was just working on this problem,

Prove: If $\lim_{n\to\infty} |a_n| = 0$, then $\lim_{n\to\infty}a_n = 0$.

I am not quite sure how to go about it. I was wondering if anyone had any tips on starting this proof?

EDIT: I think I got the answer but I wanted to check it.

Proof: Let $ > 0$ and $\lim_{n\to\infty} |a_n| = 0$.

Then, by definition, $||a_n|-0| \leq  $ for $n \geq N$. Since $||a_n|-0| = |a_n-0|$ we can say $|a_n-0| \leq $ for $n \geq N$. Therefore, by definition $\lim_{n\to\infty}a_n = 0$.

Does this look alright?

答:
Yes. Your newly added proof is correct. Although, it may be better if you reword it slightly.

Suppose $\lim_{n\to\infty}|a_n|=0$ and let $\epsilon>0$. By definition, there exists a positive integer $N$, such that for every $n\ge N$,
$
||a_n|-0|\le\epsilon
$
But $||a_n|-0|=|a_n-0|$ for each $n$. So it follows that
$$
|a_n-0|\le\epsilon
$$
whenever $n\ge N$, which implies by definition that $\lim_{n\to\infty}a_n=0$.

The key step in your proof is the observation that $||a_n|-0|=|a_n-0|$.

Alternatively, you may observe that $-|a_n|\le a_n\le |a_n|$ and apply the Squeeze Theorem.


问:
Prove that $\phi$ is injective by showing the kernel is trivial.

Let $r$ and $s$ be the generator $Q_8$ described in Section $5$. Let $\phi$ be the map from $Q_8$ to $\operatorname{GL}_2(\mathbb{C})$ defined on generators by
$$
\phi(r) = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}
\text{ and }
\phi(s) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
$$
which extends to a homomorphism. Prove that $\phi$ is injective.

Typically I would let $a,b\in Q_8$, such that $\phi(a)=\phi(b)$, and then show that $a=b$. However, I was reading only that you can prove injectivity by showing the $\ker\phi=e_{\operatorname{GL}_2(\mathbb{C})}=I$. Now I don't typically use the kernel so can someone give me some pointers on how to go about this question and future ones of this kind?

答:
What you have read is correct as a general approach, but because $Q_8$ is such a small group, it's better to simply list elements of the form $\phi(q)$ for $q \in Q_8$.

In fact, I recommend the following: compute
$$
\phi(r^2)= \phi(r)^2 = \pmatrix{-1&0\\0&-1}, \\
\phi(r^3) = \phi(r^2) \cdot \phi(r) = \pmatrix{i & 0\\0&-i},\\
\phi(s^3) = \phi(s^2) \cdot \phi(s) = \phi(r^2) \cdot \phi(s) = \pmatrix{0&1\\-1&0},\\
\phi(rs) = \phi(r) \cdot \phi(s) = \pmatrix{0&-i\\i & 0},\\
\phi((rs)^3) = \phi((rs)^2) \cdot \phi(rs) = \phi(r^2) \cdot \phi(rs) = \pmatrix{0&i\\-i&0}.
$$
Because $\phi$ is a homomorphism, $\phi(1)$ must be the identity matrix. With this, you will have shown that $\phi$ maps the elements of $Q_8$ to distinct elements and is therefore indeed injective.

Proof that $(rs)^2 = r^2$: note that
$$
(rs)^2 = (rsr)s = ss = s^2 = r^2.
$$


问:
Digit Sums: A Math Project
I'm doing a Math project on Digits Sums, and I don't really want to try to figure out all the 4-digit numbers that add up to 2, 3, 4, 5, and 6. That's partly because it would be hard, and also, because I don't want to get anything wrong, or it would throw off my entire project... Can anyone help me?</p>

答: Instead of going through a list, think about how you can make these sums.  For example, $4$ can only be $4+0+0+0, 3+1+0+0, 2+1+1+0, 2+2+0+0, 1+1+1+1$.  Now think about the ways to permute them, remembering that $0$ can't be first.  You can compute that and check that you have the right number.  For example, $2+1+1+0$ you can choose the location of the $0$ three ways, then the location of the $2$ three ways, for nine in all. </p>

Title: Digit Sums: A Math Project


问:
Prove the following functions is a Darboux function but discontinuous everywhere
Definition Let $X$ and $Y$ be topological spaces. We say $f: X \rightarrow Y$ is Darboux provided that $F[C]$ is connected for every connected C $\subset$ X.

Problem Consider the Cantor set $C$ in $[0,1]$. if $(a,b)$ is an interval contiguous to $C$ we define $f(x)=\frac{2(x-a)}{b-a}-1$ for $x \in [a,b]$; otherwise put $f(x)=0$. Then $f$ is Darboux and is discontinuous at each point of $C$

My idea to prove f is not continuous, take $b \in [a,b]$ then $$lim_{b^{-} \rightarrow }f(x)=1 \neq 0=lim_{b^{+} \rightarrow }f(x)$$

And I have problems with prove Darboux function, if we denote $(a_{1},b_{1}), (a_{2},b_{2})$ with $b_{1} < a_{2}$ and $0 < a_{1}, b_{2}<1$ intervals contiguous to $C$, we can define the interval $[b_{1}, \frac{b_1+a_2}{2}]$ is a connected subset of $[0,1]$ then

$f[[b_{1}, \frac{b_1+a_2}{2}]]=\{1\} \cup \{0\}$ is not a connected set, is it?

答:
The problem is the following. You can check that if $b_1$ is an the endpoint of an interval contiguous to $C$, then there are infinite intervals $(a_n,b_n)$ contiguous to $C$ converging to $b_1$ from the right. So, $[b_1,\frac{b_1+a_2}{2}]$ will contain some of these intervals, and its image will be the whole $[-1,1]$.

To prove that $f$ is Darboux, take and interval $I$ and let´s see that its image is continuous. If the interval is contained in some interval $[a_n,b_n]$ contiguous to the Cantor set, it is obvious. If not, it has to have a point of the Cantor set in its interior. And for every point of the Cantor set there is a convergent sequence of intervals contiguous to the Cantor set converging to it. So, the image will be $[-1,1]$, which is connected.


问:
Is the following set polyhedral at all of its points?
I'm trying to understand the following definition:

Let $x \in A \subset \mathbb R^n$. We say that $A$ is polyhedral at $x$ iff there is a neighborhood $U$ of $x$ and a polyhedron $B$ such that $A \cap U = A \cap B$.

Let $A = \{x \in \mathbb R^n: \sum_{i=1}^n x_i = 1, \ x_i \geq 0\}$.

Is $A$ polyhedral at each of its points?

Intuitively, it seems to me like the answer should be yes: $A$ is a convex polytope, which is just a special kind of polyhedron, and polyhedra should be polyhedral at each of their points (right?). I'm not really sure how to formally prove this from the definition though.

答:
Edited (seems the in the equality of the question was actually $A\cap U=A\cap B$, although that seems strange to me).

Your set $A$ is a polyhedron, because it is the intersection of $\sum_{i=1}^n x_i \geq 1$, $\sum_{i=1}^n x_i \leq 1$, and $x_i\geq0$.

Thus, for every point $x$ of $A$ and nhood $U$ of $x$, you can pick $U=\mathbb{R}^n$ and $B=A$ and you have $A\cap U=B\cap A$, that is, $A$ is polyhedral at $x$.


问:
Numerically evaluate a series involving Stirling numbers of the second kind
How can I numerically evaluate the following for $n$ above one million:
$$ \sum_{k>=n} \sum_{i=0}^{n-1} (-1)^i {{n-1}\choose{i}} \left( \frac{n-1-i}{n} \right)^{k-1} k (k+1)$$?

I have tried changing the order of summation and the getting closed form for the geometric series in $k$ but the terms are large (even for $n = 100$) and I run into precision problems.

答:
Recall that $${n \brace k}=\frac{1}{k!}\sum _{i=0}^k(-1)^i\binom{k}{i}(k-i)^n,$$
so your expression is equal to
$$\sum _{k\geq n}\frac{(n-1)!k(k+1)}{n^{k-1}}{k-1\brace n-1}=2(n-1)!\sum _{k\geq n}\frac{1}{n^{k-1}}\binom{k+1}{2}{k-1\brace n-1},$$
if $n$ is large, then $k$ is large, so ${k-1\brace n-1}\sim \frac{(n-1)^{k-1}}{(n-1)!}$ and so
$$2(n-1)!\sum _{k\geq n}\frac{(n-1)^{k-1}}{n^{k-1}(n-1)!}\binom{k+1}{2}=2\sum _{k\geq n}\left (\frac{n-1}{n}\right )^{k-1}\binom{k+1}{2}$$
now take $x=(n-1)/n,$ we complete the sum to $\infty$ and take out the finite sum getting
$$2\left (\sum _{k=0}^{\infty}\binom{k+1}{2}x^{k-1}-\sum _{k=0}^{n-1}\binom{k+1}{2}x^{k-1}\right )=2\left (\frac{1}{(1-x)^3}-\sum _{k=0}^{n-1}\binom{k+1}{2}x^{k-1}\right ),$$
recall too that taking the second derivative of  $\sum _{k=1}^{n}x^{k}$ gives you
$\sum _{k=1}^{n-1}(k+1)k\cdot x^{k-1}$ but that sum is $\frac{x^{n+1}-1}{x-1}$ and so we get
$$\frac{2}{(1-x)^3}-\frac{(n + 1)n\cdot x^{n-1}}{x-1} + \frac{2 (n+1) x^n}{(x-1)^2} - \frac{2 (-1 + x^{1 + n})}{(x-1)^3},$$
taking $x=(n-1)/n$ back we get
$$2n^3+(n + 1)n^2\cdot \left (1-\frac{1}{n}\right )^{n-1} + 2 n^2(n+1) \left(1-\frac{1}{n}\right )^n - 2 n^3(-1 + \left (1-\frac{1}{n}\right )^{n+1}).$$
For $n$ as big, if I am not mistaken, your expression should be close to
$$(4-\frac{2}{e})n^3+\frac{3}{e}(n+1)n^2$$


问:
Wald's equation confusion with simple random walk
Suppose we have a simple random walk $S_n$ starting at 20 - i.e. $S_n = X_0+...+X_n$ where $P(X_i = 1)=P(X_i=-1) =1/2$ for all $i$, and $S_0 = 20$. Then $S_n$ is a martingale. Let $\tau$ be the first time that $S_n$ reaches 0 or 100. Then $\tau$ is a stopping time and Wald's equation gives

$$E(S_{\tau})=E(\tau)E(X_0)$$

but we know that $E(S_{\tau}) = E(S_0)=20$, we know that the expectation of the stopping time is finite from other methods, and we know $E(X_0)=0$! These observations aren't consistent with Wald's equation - could anyone point to the mistake here?

答:
If you are starting at $S_0 = 20$, then you cannot say that $S_n = X_0 + \dots + X_n$ where $X_i$ is randomly chosen from $\{-1,1\}$. You could, instead, say that $S_n = 20 + X_1 + \dots + X_n$.

Then Wald's equation gives us $\mathbb E[S_\tau - 20] = \mathbb E[\tau] \mathbb E[X_1]$. Here, it is perfectly reasonable to use $\mathbb E[X_1] = 0$ to conclude $\mathbb E[S_\tau - 20] = 0$: this implies that $S_{\tau} - 20$ is $-20$ with probability $\frac45$ and $80$ with probability $\frac15$.


问:
Prove that $2<\int^2_1ln(x^2+1)+x<3$
To show that $2<\int^2_1ln(x^2+1)+x$ , I took the derivative of the integrand and got that it's positive always in $[1,2]$, so its strictly increasing and the minimum in $[1,2]$ is at $x=1$, so $ln(2)+1<ln(x^2+1)+x$, and $ln(2)<ln(e)=1$ so: $ln(2)+1<2< ln(x^2+1)+x$ in $(1,2]$.

EDIT: I just realized that the lower bound doesn't work too.

About the upper bound I tried to substitute $2$ but it becomes $ln(5)+2>3$, so it doesn't really help.

I would appreciate any help.
 Thanks in advance.

答:
I began to solve this before Bacha's solution appeared, so I did not delete.

Hope I'm correct.
$$\int\ln(x^2+1)dx=(*)\\let\:f=\ln(x^2+1),g=x\\then\:df=\frac{2x}{x^2+1}dx,dg=dx\\\text{using integrating by parts where }\int fdg=dg-\int gdf\text{ we get}\\(*)=x\ln(x^2+1)-\int\frac{2x^2}{x^2+1}dx=x\ln(x^2+1)-2\int \left(1-\frac{1}{x^2+1}\right)dx=\\=x\ln(x^2+1)-2x+2\arctan(x)+C$$
And now calculate
$$\int\limits_1^2(\ln(x^2+1)+x)dx=\\=2\ln(2^2+1)-2*2+2\arctan(2)-\ln(1^2+1)+2*1-2\arctan(1)+\frac{1}{2}2^2-\frac{1}{2}1^2=\\=\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)$$
If we use calculator then $\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)\approx2.6692$. But if you want a more "honest" solution then we can try to estimate this from left and from right but it would hard... Nevertheless let's see how estimating would look.

From left:
$$\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)>\ln(e^2)-\frac{1}{2}-\frac{\pi}{2}+2\arctan(\sqrt{3})=2-\frac{1}{2}-\frac{\pi}{2}+\frac{2\pi}{3}=\\=1.5+\frac{\pi}{6}> 2$$
Ok.

From right:
$$\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)<\\<\left(\ln(e^2)+(\frac{25}{2}-e^2)*\frac{1}{e^2}\right)^{[2]}-\frac{1}{2}-\frac{\pi}{2}+2*\left(\arctan(\sqrt 3)+(2-\sqrt 3)*\frac{1}{\sqrt 3^2+1}\right)^{[1]}<\\<\left(2+5.5*\frac{1}{7}\right)-\frac{1}{2}-\frac{\pi}{2}+\left(\frac{2\pi}{3}+(2-\sqrt 3)*\frac{1}{2}\right)<\\<2.8-\frac{1}{2}-\frac{\pi}{2}+\frac{2\pi}{3}+(2-\sqrt 3)*\frac{1}{2}=2.3+\frac{\pi}{6}+(2-\sqrt 3)*\frac{1}{2}<2.3+\frac{\pi}{6}+0.3*\frac{1}{2}=\\=2.45+\frac{\pi}{6}<3$$
[1] $\forall x\geq 0:\:\arctan(x) + (x+\delta x)*\arctan'(x)\geq\arctan(x+\delta x)$ because $\forall y>x\geq 0:\: \arctan'(x)>\arctan'(y)$.

[2] Same as [1].

Ok, but too much work.

Title: Prove that $2<\int^2_1ln(x^2+1)+x<3$


问:
Finding tangent/ normal space of given manifold
Here it is: $M = \left\{(x,y,z)\in\mathbb{R^3}\big\vert\:\underbrace{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}}_{f} = 1\right\}$, $(a,b,c) \geq 0$

This seems ludicrous but tempting: is the tangent space just given by $T_xM = \frac{2\,x}{a^2}+\frac{2\,y}{b^2}+\frac{2\,z}{c^2} -1 = 0$, since it's ker$(Df)$ ?

Likewise: Is it right to claim the normal space is set by: $T^\perp M = \nabla f$ ?

答: You have$$\nabla f(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2}\right)\tag1$$and therefore the tangent plane at $(x_0,y_0,z_0)$ is the plane passing through $(x_0,y_0,z_0)$ which is orthogonal to $(1)$, which is the plane$$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)=0;$$in other words, it's the plane$$\frac{x_0x}{a^2}+\frac{y_0y}{b^2}+\frac{z_0z}{c^2}=\frac{x_0^{\,2}}{a^2}+\frac{y_0^{\,2}}{b^2}+\frac{z_0^{\,2}}{c^2}.$$And the normal line is the line$$\left\{(x_0,y_0,z_0)+k\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2}\right)\,\middle|\,k\in\Bbb R\right\}.$$


问:
Solution verification: probability of getting tails given you got $2$ on a die in a coin-die experiment
$\newcommand{\P}{\mathbb{P}}$The Problem: Consider an experiment where one first flips a fair coin. If it lands on heads, then the experimenter rolls a fair six-sided die. If it lands on tails, instead, the experimenter rolls an unfair four-sided die. (The probability of face $n$ appearing is $(5-n)/10$ on this die.)

Given that the experimenter has rolled a $2$, what is the probability that they flipped the coin and got tails just beforehand?

My Attempt: We first consider the probability of getting a $2$. This is given by
\begin{align*}
\P(\text{getting a 2})
&= \P(\text{getting heads, then a 2}) + \P(\text{getting tails, then a 2}) \\[4pt]
&= \P(\text{getting heads}) \cdot \P(\text{getting 2 on the fair die}) \\[4pt]
&\;\;\; + \P(\text{getting tails}) \cdot \P(\text{getting 2 on the unfair die}) \\[6pt]
&= \frac 1 2 \cdot \frac 1 6 + \frac 1 2 \cdot \frac{3}{10} \\
&= \frac{1}{12} + \frac{3}{20} = \frac{7}{30}
\end{align*}
Now, we note that
$$
\P(\text{getting tails} \mid \text{you got a 2}) = \frac{\P(\text{getting tails and a 2})}{\P(\text{getting a 2})}
$$
Clearly, as seen earlier, $\P(\text{getting tails and a 2}) = 3/20$. Thus,
$$
\P(\text{getting tails} \mid \text{you got a 2}) = \frac{3/20}{7/30} = \frac{9}{14} \approx 64.29\%
$$
is the probability we desire.

My Question: Basically, is this approach correct? I'm a bit rusty on my probability, so I just wanted to be sure.

My main issue for a lack of confidence is the final step here:

\begin{align*}
\P(\text{getting a 2})
&= \P(\text{getting heads, then a 2}) + \P(\text{getting tails, then a 2}) \\[4pt]
&= \P(\text{getting heads}) \cdot \P(\text{getting 2 on the fair die}) \\[4pt]
&\;\;\; + \P(\text{getting tails}) \cdot \P(\text{getting 2 on the unfair die})
\end{align*}

Is it justified to do this, i.e. to assume that the coin facing and the die value are independent and thus can be multiplied? If so, why? It feels like the two depend on each other. If not, then I'm just not sure how else to approach this.

答:
Recall that $P(A\cap B)=P(A|B)P(B)$.

Let $X$ be the flip of the coin and $Y$ be the number obtained on the second move.

Since $X=H$ and $X=T$ form a partition of the sample space, by the law of total probability $P(Y=2)=P(X=H)P(Y=2|X=H)+P(X=T)P(Y=2|X=T)$.

Define new random variables $Z$ to be the roll of the fair die, and $W$ to be the roll of the rigged die. Then $Y=Z1\{X=H\}+W1\{X=T\}$. This way, you won't get confused even if you happen to roll both die, because you know which one to choose.

Then the expression 2 paragraphs prior becomes $P(X=H)P(Z1\{X=H\}+W1\{X=T\}=2|X=H)+P(X=T)P(Z1\{X=H\}+W1\{X=T\}=2|X=T)=P(X=H)P(Z=2)+P(X=T)P(W=2)$, where I "plugged in" the value of X. Now it is clearer, if you will, that this equals $\frac 1 2 \frac 1 6+\frac 1 2\frac 3 {10}$.

Recall that the definition of independence of two events is that $P(A)=P(A|B)$, i.e. learning about the occurrence of one event has no influence on the probability of the other. Here, we found that $P(Y=2)=\frac 7 {30}$, but $P(Y=2|H=1)=\frac 1 6$. Thus the events $Y=2$ and $H=1$ are not independent. In general, $Y$ and $X$ are sample spaces and perhaps you could talk about events in the sample space as being independent.

Finally, since $X$ is defined and $Y|X$ is defined, then $X,Y$ is defined and so are $Y, X|Y$. So you could rephrase the experiment as rolling a a rigged die with probabilities $\{\frac {17}{60},\frac{7}{30},\frac{11}{60},\frac{2}{15},\frac{1}{12},\frac{1}{12}\}$, followed by rigged coins with probabilities $\{\frac{5}{17},\frac{12}{17}\},\{\frac{5}{14},\frac{9}{14}\},\{\frac{5}{11},\frac{6}{11}\},\{\frac{5}{8},\frac{3}{8}\}, \{1\}, \{1\}$, where the last two coins have heads with probability 1.


问:
prove $\text{abs}(a + b) + \text{abs}(a - b) = \text{max}(a, b) \cdot 2$
after playing around with absolute value for a little, i noticed that $|a + b| + |a - b|$ always appeared to be the larger of $a$ and $b$ multiplied by $2$, whenever both $a$ and $b$ were positive.

$$
|a + b| + |a - b| =
\begin{cases}
2a, & \text{if $a \ge b$} \\
2b, & \text{if $a < b$}
\end{cases}
$$

can this be proven to always be the case, and if so, how do i prove it?

答:
Answer posted to provide industrial strength approach for new students.

To prove:

$|a + b| + |a - b| = 2\max(a,b).$

Either $a \geq b,$ or $a < b.$

Further, either $(a+b) \geq 0$ or $(a+b) < 0.$

Case 1:

$a \geq b$ and $(a+b) \geq 0.$

Then $|a + b| + |a - b| = (a+b) + (a-b) = 2a = 2\max(a,b).$

The assertion is always true in this case.

Case 2:

$a \geq b$ and $(a+b) < 0.$

Then $|a + b| + |a - b| = -(a + b) + (a - b) = -2b.$

$2a = 2\max(a,b).$

The assertion is not always true in this case.

Case 3:

$a < b$ and $(a+b) \geq 0.$

Then $|a + b| + |a - b| = (a+b) + (b-a) = 2b = 2\max(a,b).$

The assertion is always true in this case.

Case 4:

$a < b$ and $(a+b) < 0.$

Then $|a + b| + |a - b| = -(a+b) + (b - a) = -2a.$

$2b = 2\max(a,b).$

The assertion is not always true in this case.


答:
Denote $F$ the cantor set. If $C=(a,b)\subset[0,1]$ (or $[a,b],[a,b),(a,b]$) contains no cantor points, then $f$ is increasing and continuous on $(a,b)$, so $f((a,b))=(f(a),f(b))=(-1,1)$ (or $[-1,1],[-1,1),(-1,1]$), which is connected.

Cantor set has no isolated points, so any connected $C\subset[0,1]$ containing a cantor point $c\in F$ must contain another point $d\in F$ on the cantor set. Since the cantor set is nowhere dense there must exist a point $x\in(c,d)\setminus F$ outside of the Cantor Set. Since the Cantor Set is closed, $[0,1]\setminus F$ is open, and as $x\in[0,1]\setminus F$, there must exist $a,b\in[c,d]$ such that $x\in(a,b)\subset[0,1]\setminus F$, and both $a$ and $b$ are members of the cantor set. As we've already proved, $f([a,b])=[-1,1]$. Then, as $[a,b]\subset[c,d]\subset C$, $f([a,b])=[-1,1]\subset f(C)$. Also, by definition $f(C)\subset[-1,1]$, which completes the proof.

To prove $f$ is discontinuous on every $c\in [0,1]\setminus F$, its enough to use the previous. Let $\epsilon\in(0,\frac{1}{2})$. Then, for any $\delta\in\mathbb{R}^+$, $B(c,\delta)$ is connected; so as we proved before, $f(B(c,\delta))=[-1,1]$. In other words, whatever $f(c)$ is between $-1$ and $1$, there exists $x\in B(c,\delta)$ such that $|f(c)-f(x)|>\frac{1}{2}=\epsilon$.

Title: Prove the following functions is a Darboux function but discontinuous everywhere

Body:
Definition Let $X$ and $Y$ be topological spaces. We say $f: X \rightarrow Y$ is Darboux provided that $F[C]$ is connected for every connected C $\subset$ X.

Problem Consider the Cantor set $C$ in $[0,1]$. if $(a,b)$ is an interval contiguous to $C$ we define $f(x)=\frac{2(x-a)}{b-a}-1$ for $x \in [a,b]$; otherwise put $f(x)=0$. Then $f$ is Darboux and is discontinuous at each point of $C$

My idea to prove f is not continuous, take $b \in [a,b]$ then $$lim_{b^{-} \rightarrow }f(x)=1 \neq 0=lim_{b^{+} \rightarrow }f(x)$$

And I have problems with prove Darboux function, if we denote $(a_{1},b_{1}), (a_{2},b_{2})$ with $b_{1} < a_{2}$ and $0 < a_{1}, b_{2}<1$ intervals contiguous to $C$, we can define the interval $[b_{1}, \frac{b_1+a_2}{2}]$ is a connected subset of $[0,1]$ then

$f[[b_{1}, \frac{b_1+a_2}{2}]]=\{1\} \cup \{0\}$ is not a connected set, is it?


问:
Prove algebraically that $n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$
Prove algebraically that
$n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$ where $n$ and $k$ are integers.

I start with the fact that $(x+1)^n=\sum_{k=0}^{n}\dbinom{n}{k}x^k$ and differentiate both sides of the equation twice, with respect to $x$.

The second derivative (with respect to $x$) of the above equation yields: $n(n-1)(x+1)^{n-2}=\sum_{k=0}^{n}k^2-k\dbinom{n}{k}x^{k-2}$.

Now, I am not sure how to proceed. I know I'm supposed to set $x = 1$. But even then, I am missing the $(n+1)$ on the left hand side and there is still a $-k$ on the right hand side.

Any help would be appreciated. Thank you.

答:
Hint:
$$\begin{align}
\sum_{k=1}^{n}k^2\dbinom{n}{k}&=\sum_{k=1}^{n}k^2\frac nk \dbinom{n-1}{k-1}\\
&=n\sum_{k=1}^{n}k\left[\dbinom{n}{k}-\dbinom{n-1}{k}\right]\\
&=n\sum_{k=1}^{n}\left[n\dbinom{n-1}{k-1}-(n-1)\dbinom{n-2}{k-1}\right].
\end{align}$$


问:
Why does a power series with binomial coefficients reduce to a Taylor series?
I got this surprising result while numerically modeling a differentiable function $f(x)$ with a power series on the interval $[0,t]$. I couldn't find any references to a connection between power series with binomial coefficients and Taylor/Maclaurin series.

$f(t)= \displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^{n}{n \choose k} \left( \frac{t}{n}\right)^kf^{(k)}(0)= \sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}\lim_{n\rightarrow \infty}\frac{\frac{n!}{(n-k)!}}{n^k}=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}$

I obtained the expression on the left by first approximating an $(n-1)th$ derivative as a linear sequence using Euler's method, $\displaystyle f^{(n-1)}\left( \frac{tk}{n} \right)=f^{(n-1)}(0)+\frac{tk}{n}f^{(n)}(0)$ for natural numbers k such that $0\leq k \leq n$. I then numerically integrated to obtain each of the antiderivatives until I got $f(t)$, which revealed binomial coefficients.  I'm betting this connection is well-known, so please share any insights you may have about the relationship here between Taylor series and binomial coefficients. Thank you.

答:
I derived the same result you did while attempting to derive the Taylor series. I think this is basically the same thing you did, but here's how I did it if it might help someone else. It actually involves the binomial series in more than one way.
If we know $f(a)$ and it's derivatives at $a$ and we want to find $f$ just a little ways away at $x$, i.e. $f(x)$, where $x = a + w$ and $w$ is the width of the interval between $a$ and $x$, from the definition of derivative we have our basic
$$f(x) \approx f(a) + wf^{(1)}(a)$$
or to make it clear we're starting at or centered at $a$
$$f(a+w) \approx f(a) + wf^{(1)}(a) \tag{1}\label{eq1}$$
The geometric interpretation is straightforward, we take $f$ at $a$ and then add on one output segment.

To make this more accurate, let's divide the distance $x-a$ into two intervals, so now the width, $w$, of each interval is $w = \dfrac{(x-a)}{2}$, we have $$f(x) \approx f(a) + wf^{(1)}(a) + wf^{(1)}(a+w)$$
We just start at $f(a)$ and add on two segments. We're basically starting at $a$ and integrating over to x.
We can make this as accurate as we like by dividing $(x-a)$ into more and more intervals. So generally, where $w=\dfrac{(x-a)}{n}$ and $n$ is the number of intervals

\begin{align}
f(x) \approx f(a) &+ wf^{(1)}(a) + wf^{(1)}(a+w) + wf^{(1)}(a+2w)  \\&\quad+\ldots+wf^{(1)}(a+(n-1)w) \tag{2}\label{eq2}\end{align}
$f(a)$ is the starting point and $wf^{(1)}(a)$ is the first segment, $wf^{(1)}(a+w)$ is the second segment, etc.
But now the problem is we're not just taking the derivative at $a$, we're also taking the derivative at every interval point between $a$ and $x$ (at $a$, $a+w$, $a+2w$, etc). So we have to perform reduction steps on each segment until they only rely on derivatives of $f$ at $a$. We use the following rule:
$$f^{(k)}(a+mw) = f^{(k)}(a+(m-1)w)+wf^{(k+1)}(a+(m-1)n) \tag{3}\label{eq3}$$
Which looks messy but it's really the same as our original equation \eqref{eq1}.
E.g to reduce the 3rd segment, $wf^{(1)}(a+2w)$, we have:

\begin{align}
wf^{(1)}(a+2w)
&= w[f^{(1)}(a+w)+wf^{(2)}(a+w)]\\
&= wf^{(1)}(a+w)+w^2f^{(2)}(a+w)
\end{align}

We've reduced it from $a+2w$ to $a+w$, so we have to apply the reduction a 2nd time and then collect like terms.
\begin{align}
wf^{(1)}(a+2w)
&= w[f^{(1)}(a) + wf^{(2)}(a)]+w^2[f^{(2)}(a) + wf^{(3)}(a)]\\
&= wf^{(1)}(a) + w^2f^{(2)}(a)+w^2f^{(2)}(a) + w^3f^{(3)}(a)\\
&= wf^{(1)}(a) + 2w^2f^{(2)}(a) + w^3f^{(3)}(a)
\end{align}
So we have the coefficients $(1, 2, 1)$ which are the binomial coefficients! We can write this as a summation and generalize it to any n, giving us the output segment for any interval, $i$:
$$\sum_{k=1}^{i}{i-1 \choose k-1}w^kf^{(k)}(a) \tag{4}\label{eq4}$$
Where $n$ is the number of intervals and $w=\frac{(x-a)}{n}$ and $i$ is the $i^{th}$ interval (e.g, an $i$ of $3$ is the third interval, which starts at $a+2w$)

Substituting \eqref{eq4} into \eqref{eq2}, we arrive at:
$$f(x) \approx f(a) + wf^{(1)}(a)  + \sum_{k=1}^{2}{2-1 \choose k-1}w^kf^{(k)}(a)  + ... +  \sum_{k=1}^{n}{n-1 \choose k-1}w^kf^{(k)}(a)\tag{5}\label{eq5} $$

So each segment we're adding on is using binomial coefficients, but the really interesting part (for me at least) is that when we add up all the segments and collect like terms, our final result is also binomial coefficients! I won't prove it but if you solve the problem using 3 or 4 segments you can easily convince yourself that is indeed the case (it'd just be a proof by induction showing that this mechanical procedure we're doing here holds at each step of the way). Here's an example with 3 intervals. I've arranged it in a grid to more easily illustrate this where each row is one of the segments we're adding on. Each row is the summation for each segment from \eqref{eq4} and then we sum the columns to produce the final result from \eqref{eq5}. Summing the columns is simply collecting like terms.

\begin{array}{rrrrrr|r}
              & f^{(0)} & wf^{(1)} & w^2f^{(2)}& w^3f^{(3)} & w^4f^{(4)} &    \\ \hline
f(a)          & 1 &   &   &   &  &  f(a) \\
wf^{(1)}(a)   &   & 1 &   &   &  &  wf^{(1)}(a) \\
wf^{(1)}(a+w) &   & 1 & 1 &   &  &  wf^{(1)}(a) + w^2f^{(2)}(a) \\
wf^{(1)}(a+2) &   & 1 & 2 & 1 &  &  wf^{(1)}(a) + 2w^2f^{(2)}(a) + 1w^3f^{(3)}(a) \\ \hline
              & 1 & 3 & 3 & 1 &  &  f(a) + 3wf^{(1)}(a) + 3w^2f^{(2)}(a) + w^3f^{(3)}(a)
\end{array}

Each segment has binomial coefficients and they sum together (along with the base $f(a)$) to more binomial coefficients! The above result ends up being a consequence of the hockey stick identity. So the final approximation of $f(x)$ when we divide $a$ to $x$ into 3 interval is
$$f(x) \approx f(a) + 3wf^{(1)}(a) + 3w^2f^{(2)}(a) + w^3f^{(3)}(a)$$
Which of course we can rewrite as a summation and for the general case is:
$$
f(x) \approx \sum_{k=1}^{n}{n \choose k}w^kf^{(k)}(a)
$$
As always, $n$ is the number of intervals and $w=\frac{(x-a)}{n}$.
This approximation becomes exact as the number of intervals tends to infinity
$$
f(x) = \lim_{n\rightarrow \infty} \sum_{k=1}^{n}{n \choose k}w^kf^{(k)}(a)
$$
Substituting $w=\frac{(x-a)}{n}$:
$$
f(x) = \lim_{n\rightarrow \infty} \sum_{k=1}^{n}{n \choose k}\frac{1}{n^k}(x-a)^kf^{(k)}(a)  \tag{6}\label{eq6}
$$
Which I think is the Newton series for finite differences or at least related. This is as far as I got, so I failed to derive the Taylor series, but it's basically the same form as the Taylor series, which is:
$$
f(x) = \sum_{n}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \tag{Taylor Series}\label{Taylor series}$$
Since $f(x) = f(x)$ and we have 2 infinite sums in $(x-a)$, the coefficients have to be equal to each other, so I inadvertently showed:
$$
\lim_{n\rightarrow \infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!} \tag{7}\label{eq7}
$$

I do not understand Qiaochu Yuan's remark that \eqref{eq6} is a generalization of
$e^t = \lim_{n \to \infty} \left( 1 + \frac{t}{n} \right)^n = \sum_{n \ge 0} \frac{t^n}{n!}$ but it sounds interesting and is probably worth a look. So if someone wants to expound on that, that'd be cool!


答:
Using your idea of differentiating the binomial formula & then setting $x=1$ ... gives\begin{eqnarray*}
n2^{n-1} &=& \sum_{k=0}^{n} k \binom{n}{k} \\
n(n-1)2^{n-2}&=& \sum_{k=0}^{n} k(k-1) \binom{n}{k} \\
\end{eqnarray*}
now add these & you will have your result.

Body:
Prove algebraically that
$n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$ where $n$ and $k$ are integers.

I start with the fact that $(x+1)^n=\sum_{k=0}^{n}\dbinom{n}{k}x^k$ and differentiate both sides of the equation twice, with respect to $x$.

The second derivative (with respect to $x$) of the above equation yields: $n(n-1)(x+1)^{n-2}=\sum_{k=0}^{n}k^2-k\dbinom{n}{k}x^{k-2}$.

Now, I am not sure how to proceed. I know I'm supposed to set $x = 1$. But even then, I am missing the $(n+1)$ on the left hand side and there is still a $-k$ on the right hand side.

Any help would be appreciated. Thank you.


问:
Finding planar curve $\beta(s)$ given curvature $k(s) = \frac{1}{s}$, $\beta(1) = (0, 0)$, $\beta'(1) = (1, 0)$
Find the curve $\beta(s) = (x(s),y(s))$ with $|\beta'(s)|= 1$ for all $s >0$ such that $\beta(1) = (0,0)$, $\beta'(1) = (1,0)$ , and curvature $k(s) =\frac{1}{s}$ for all $s >0$.

I am guessing some sort of integral is required since we are given 'initial' conditions, but I do not see how to begin.

答:
EDIT1:

Curve behavior

The curve is natural or intrinsic. A rigid bent curved arc of this shape or can be moved anywhere in the plane merely by varying IC.

$$ \phi'=\frac{1}{s},\;\phi= \log\frac{s}{a}\;; \dfrac{s}{a}= e ^{\phi}$$

where $a$ is an arbitrary constant and $\phi$ is curve rotation with respect to x-axis.

$$ x=\int\cos \phi \;ds=\int\cos\log\frac{s}{a} ; y=\int\sin \phi \; ds=\int\sin \log\frac{s}{a} ds;$$

Curve's  analytical parametrization

$$\boxed{x= \frac{s}{2} [{\sin (\log (s))+\cos (\log (s));\;y=\sin (\log (s))-\cos (\log (s))}] } $$

Calculated by numerical integration. BC  $  (x=1,y=0) $

At start infinite curvature or Cusp occurs. Numerically,  $s=0 $ cusp is avoided, a small value given for x to avoid a singularity.

As $s\to \infty $ the curve becomes flatter.

I have not found a reference to this curve in 2dCurves, Lockwood, McTutor or Mathworld


问:
Is it coincidental that matrices are useful for solving systems of equations or is there a deeper (geometric) connection?
Let's say we have this simple system of equations:

$$-x + y = 1$$
$$ x + y = 3$$

They intersect at the point $(1,2)$ and this is the graph:

One way of going about and solving this system of equations is to convert it into matrix form:

$$\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix} * \begin{bmatrix}
x\\
y
\end{bmatrix} = \begin{bmatrix}
1\\
3
\end{bmatrix}$$

When my brain sees this matrix form, it has the following geometric meaning: If we apply the linear transformation encoded by $\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}$ to vector $\begin{bmatrix}
x\\
y
\end{bmatrix}$, then the output vector is $\begin{bmatrix}
1\\
3
\end{bmatrix}$.
We can then use Gaussian Elimination to determine $\begin{bmatrix}
x\\
y
\end{bmatrix}$.

My question: Is the matrix form of the system of equations simply a useful trick to algorithmically solve these systems of equations use Elementary Row Operations, etc. Or is there a deeper geometric connection between the intersection of two lines and the linear transformation of the vector that describes said intersection.

答:
Ok, I think I've got a potential explanation. The key is my realization that an equation can be thought of geometrically as a linear transformation of space.

Let's take our first equation: $$ -x + y = 1$$
This can be rewritten in "linear algebra" form:

$$\begin{bmatrix}
-1 & 1
\end{bmatrix} *\begin{bmatrix}
x  \\
y
\end{bmatrix} = 1$$

This form is saying that if we apply the
linear transformation encoded by $\begin{bmatrix}
-1 & 1
\end{bmatrix}$ to a 2D vector $\begin{bmatrix}
x\\
y
\end{bmatrix}$, then the 1D output "vector" is $\begin{bmatrix}
1
\end{bmatrix}$.

As it turns out, there are an infinite amount of 2D vectors that will satisfy the above expression. These vectors form the line described by our first equation.

Now when we add the second equation into our system, the question becomes: Is there a vector that satisfies BOTH linear transformations encoded by their respective equations? Indeed, the graphical intersection of these lines is that very vector.

Correct me if I'm wrong, but thinking of equations as linear transformations of space is a beautiful example of duality in mathematics.


问:
Moving value inside of sigma, out of it
I am currently reading the book "Introduction to modern cryptography" by katz et al.
In this book there is a proof that looks something like this:

enter image description here

And written by myself it looks like this, even though the picture is way better:

\begin{align}
Pr[C=c] &= \sum_{m\in \mathcal{M}} \Pr[C=c_j\mid M=m] \Pr[M=m] \\
&= \sum_{m\in \mathcal{M}} \gamma  \cdot \Pr[M=m] \\
&= \gamma \cdot \sum_{m\in \mathcal{M}}    \Pr[M=m] \\
&=  \Pr[C = c|M = m]\\
\end{align}

What I don't understand is the line where we just move the gamma outside of the sigma notation? surely the value the summed value of all c's times gamma, is not the same as every value times gamma summed?

Is this allowed? to just move a value outside of sigma like this?

答: The second part of the proof is trying to show that if the conditional distribution of $C$ given $M$ is constant with respect to the outcome of $M$, then the probability $\Pr[C = c \mid M = m] = \Pr[C = c]$.  We then call $C$ independent of $M$.  This is why they are able to "move" the $\gamma$ term outside of the sum, because the term $\Pr[C = c \mid M = m]$ is the same $\gamma$ for each $m \in \mathcal M$, these terms can be factored out of the summation, leaving behind $$\sum_{m \in \mathcal M} \Pr[M = m] = 1.$$  It is like saying $$\sum_{k=1}^n 2k = 2 \sum_{k=1}^n k = 2 \cdot \frac{n(n+1)}{2} = n(n+1).$$  We just factor out the $2$ because it doesn't depend on the index of summation $k$.  In the proof, it is by construction as one of the conditions of the proof that for all $m_0, m_1 \in \mathcal M$, any distribution $M$, every $c \in \mathcal C$, that $\Pr[C = c \mid M = m_0] = \Pr[C = c \mid M = m_1]$.  This means for every $m \in \mathcal M$, $\Pr[C = c \mid M = m]$ equals some constant, which they call $\gamma$, that is the same for any choice of $m \in \mathcal M$.  The ability to do this is certainly not true in general, such as when the probability of observing some outcome of $C$ does depend on the outcome of $M$.  But the proof is addressing the specific case where it does not.

Title: Moving value inside of sigma, out of it


问:
A counterexample of normal subgroup with cyclic Sylow 2-subgroup
We know that when a group $G$ has order $2^k m$, where $m$ is an odd integer, $G$ should have a normal subgroup with order $m$ from here. When $k=1$, this implies the index of the normal subgroup is $2$. However, when $k=2$, $m$ is a prime, can we also find a normal subgroup which has index $2$?

That is to say, when $|G|=4p$, where $p$ is a prime congruent to $1$ or $3$ modulo $4$, does a normal subgroup of index $2$ always exists? I don't know how to construct the contourexample, any idea or solutions are welcomed, thanks!

答:
If a group $G$ has a normal subgroup $N$ of index $2^k$, then $G/N$ is a group of order $2^k$ and so has normal subgroups of index $2^i$ for all $i=0,1,\ldots,k$, and in particular of index 2. The lattice isomorphism theorem is that the subgroups of $G/N$ are exactly the $H/N$ for subgroups $H$ containing $N$, and that $H/N$ is normal in $G/N$ if and only if $H$ is normal in $G$.

If $G$ is a finite group with a cyclic Sylow 2-subgroup, then $G$ has a normal 2-complement, a subgroup whose index is equal to the order of the Sylow 2-subgroup. By the previous, $G$ has normal subgroups of index $2^i$ for all $2^i$ dividing the order of the Sylow 2-subgroup. As long as $G$ has even order, it must have a normal subgroup of index 2.

If the Sylow 2-subgroup is not cyclic, then $G$ need not have any normal subgroups of index 2, as the alternating group on four points shows.


问:
"Drawing a Picture" of $\operatorname{Spec}(\mathbb{Z})$
Atiyah/Macdonald's commutative algebra book asks the reader to draw pictures of the prime spectrum of $\mathbb{Z}$ in exercise 1.16. I worked through it on my own, figured out what the space looks like as a set and where the open/closed sets are, but when I did some research to check my work I found the following picture:

What exactly is the squiggly line supposed to represent? is it meant to be the ideal $(0)$? That makes a vague bit of sense to me, as the closure of (0) is the entire space, but it contradicts the fact that (0) is a single point in the prime spectrum. Another source said to draw $\operatorname{Spec}(\mathbb{Z})$ as a curve passing through the ideals $(2),(3),(5),\ldots$ and a separate point off to the side to represent $(0)$, but again I don't get how drawing a curve helps clarify the structure of the space. Why not draw it as a discrete set of points and leave out the curve?

Edit: Here's another guess I just came up with: If we put the finite-complement topology on the real line and then create a quotient space by identifying every point that is not a prime positive integer as one point and call it (0), do we get $\operatorname{Spec}(\mathbb{Z})$? If this is the case it would definitely help justify the drawing.

答:
$(0)$ is the generic point. It is a single point, in some sense, "nowhere in particular" (quoting Vakil). A way to think of it is that closed points are actual points, and generic points are "fat/fuzzy" points (quoting here)

The generic point is drawn as going through all of the other points because the closure of the generic point is the entire space.


问:
Alternatives to find the expectancy of a random variable
For actuarial mathematics there is an important identity to calculate the life expectancy for X defined on $[0,\infty)$:
identity
$$
E[X]=\int_0^\infty S_X[x]\,\mathrm{d}x
$$
where $S_{x}(x)$(survival function) is equal to $1-F_{x}(x)$, where $F_{x}$ is the CDF, which theorem or lemma exists so i can understand this identity?

答:
I will elaborate a little on the general measure-theoretical context here, which does not rely on the existence of a density $f_X$ (which would simply be the Radon-Nikodym derivative of the pushforward $X_\ast \mathbb{P}$ with respect to the Lebesgue measure $\lambda$).

I will use the following notation: $(\Omega, \mathcal{B}, \mathbb{P})$ for the probability space on which $X \in L^0(\Omega)$ is defined. Then by definition we have that $\mathbb{E}[X] = \int_\Omega X d\mathbb{P} = \int_{[0, \infty)} x \, dX_\ast\mathbb{P},$ where I used the fact that your variable takes values in $[0, \infty)$ and applied the change of variables formula in the measure-theoretical context at play here. Note that the pushforward $X_\ast\mathbb{P} = \mathbb{P} \circ X^{-1}$ is just the cummulative distribution function of $X.$ But we also have that $x = \int_0^x d\lambda(y),$ so $\mathbb{E}[X] = \int_0^\infty \int_0^x d\lambda(y) \, d X_\ast \mathbb{P}(x).$ Note that we may indeed apply Fubini's Theorem here, finally yielding $\mathbb{E}[X] = \int_0^\infty \int_y^\infty dX_\ast\mathbb{P}(x) \, d\lambda(y) = \int_0^\infty (1 - F_X(y)) \, d \lambda(y) = \int_0^\infty S_X(y) d\lambda(y),$ which is the desired equality.

I hope this is clear, but if not do let me know!


问:
Inconclusive inequality result: show $(id - f)^{-1}$ is bounded
I have the following statement to prove:

Let $V$ be a normed vector space (so not necessarily complete or finite dimensional) over $\mathbb{R}$. Take any norm $\Vert \cdot \Vert_V$ on $V$ and let $\Vert \cdot \Vert_{V\rightarrow V}$ be the induced norm. Suppose $f : V \rightarrow V$ is bounded and $\Vert f \Vert_{V\rightarrow V} \leq \beta < 1$. If $(id - f)^{-1} $ exists, show that $\Vert (id - f)^{-1} \Vert_{V\rightarrow V} \leq \frac{1}{1-\beta}$, where $id :V \rightarrow V$ is the identity operator.

My attempt:

First by reverse triangle inequality, we get
$$
\Vert id - f \Vert_{V \rightarrow V} \geq \Vert id \Vert_{V \rightarrow V} - \Vert f\Vert_{V \rightarrow V} \geq 1 - \beta \implies \frac{1}{\Vert id - f \Vert_{V \rightarrow V}} \leq \frac{1}{1 - \beta}.
$$

Then by submultiplicity of the induced norm, we get
$$
1 = \Vert(id - f)(id - f)^{-1} \Vert_{V \rightarrow V}
\leq \Vert id - f \Vert_{V \rightarrow V}\Vert (id - f)^{-1} \Vert_{V \rightarrow V},
$$
which implies
$$
\Vert (id - f)^{-1} \Vert_{V \rightarrow V} \geq \frac{1}{\Vert  id - f \Vert_{V \rightarrow V}} \leq \frac{1}{1-\beta}.
$$

So this attempt ended up with an inconclusive result. I'm guessing I made a silly mistake somewhere in the inequalities... Any hint on what I did wrong here?

答:
More simply, if $y = (I-f)^{-1}(x)$, then
\begin{equation}
(I - f)(y) = x \Rightarrow y = x + f(y)\Rightarrow \|y\|\le\|x\| + \beta  \|y\|
\end{equation}
Hence
\begin{equation}
(1-\beta)\|y\|\le\|x\|\quad\Rightarrow\quad \|(I-f)^{-1}(x)\| = \|y\|\le\frac{1}{1-\beta}\|x\|
\end{equation}


问:
If $b ∈ \Bbb Z$, then there is no integer a such that $b < a < b + 1$
If $b ∈ \Bbb Z$, then there is no integer $a$ such that $b < a < b + 1.$

I am doing an introductory course for mathematical proofs and this question is an exercise related to inequalities and natural numbers.

I am having trouble articulating a proof for this proposition. I know how to prove that $b < b+1$, but I am not sure how I can prove the statement above by showing that $a$ does not exist.

Should I be doing a proof by contradiction? If I assume $a$ exists, then how do I show that this will result in contradiction? Am I supposed to try to show that $a$ is not a natural number?

I am just looking for suggestions or some instructions on how to write the proof, not an actual full written proof. I would like to be able to figure at least some of it out myself, instead of copying off of a posted answer. So please just provide helpful hints as answers.

Thank you

答:
Assume that $a \in \mathbb{N}$. Since your inequalities give $a \gt b$, then with $b \in \mathbb{N}$, as you state in your comment, we have $a - b \in \mathbb{N}$. Since you also state in your comment $0$ is not an element of $\mathbb{N}$, then if you know or can assume that $1$ in the smallest natural number, we also get

$$a - b \ge 1 \tag{1}\label{eq1A}$$

However, with your inequality, we would like to try to get something to compare \eqref{eq1A} against, ideally to get a contradiction. With that in mind, try subtracting $b$ from both sides of $a \lt b + 1$ to see what the result is.


问:
Minimizing trigonometric expression
How can one get the minimum of a trigonometric expression if the first-order derivative is not in a nice form?

The expression is this:

$$
\left[\cos ^{2}\left(\dfrac{\theta }{2}\right) +
\,\sqrt{\,{1 - \gamma}\,}\,\sin^{2}\left(\dfrac{\theta }{2}\right)\right] ^{2} + \dfrac{\gamma }{4}\,\sin^{2}\left(\theta\right)\quad
\mbox{where}\quad 0 \le \gamma \le 1.
$$
I have tried to solve this but I reach a term which is not easily solvable.

答:
Note $$\cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2}, \quad \sin^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{2},$$ and $$\sin^2 \theta = 1 - \cos^2 \theta.$$ So if we let $x = \cos \theta$, we get the function $$\begin{align}f(x) &= \left( \frac{1 + x}{2} + \sqrt{1-\gamma} \frac{1 - x}{2} \right)^2 + \frac{\gamma}{4} (1 - x^2) \\
&= ax^2 + bx + c,
\end{align}$$ where $$\begin{align} a &= \frac{1 - \gamma - \sqrt{1-\gamma}}{2}, \\ b &= \frac{\gamma}{2}, \\ c &= \frac{1 + \sqrt{1-\gamma}}{2}. \end{align}$$  Now compute the critical point(s) of $f$ with respect to $x$ and consider where these have absolute value less than or equal to $1$, since we require $|x| = |\cos \theta| \le 1$.

Title: Minimizing trigonometric expression


问:
Probability of an event happening after X our of Y attempts
I have a work related question. Based on observation of a system, I determine that I have a 23% chance that when I run a test, the result will be above a certain threshold. Now, let's say that I repeat this same test 19 times, and each test is independent and has this same 23% chance of being above the threshold. I want to calculate the resulting probability of having 10 out of 19 events being above the threshold. It seems like the probability of having 10 out of 19 results above my threshold is much less than 23%, but I don't know how to quantify it.</p>

答: You might want to have a look at the wikipedia article on the binomial distribution, as it is exactly what you are describing and how Brian got the answer he commented before. I hope this clears the air. :)


问:
Why $(\alpha\frac{e}{t})^t e^{-\alpha}$ is an approximation for $P(X > t\alpha)$ for Poisson distribution $\frac{\alpha^ke^{k}}{k!}$?
I am reading Section 3.4 of Algorithms, 4th Edition. Page 466 is a proof of the following proposition:

In a separate-chaining hash table with $M$ lists and $N$ keys, the probability (under Assumption J) that the number of keys in a list is within a small constant factor of $N/M$ is extremely close to 1.

The proof:

The probability that a given list will contain exactly k keys is given by the binomial distribution

$\binom{N}{k}(\frac{1}{M})^k(\frac{M-1}{M})^{N-k}$

When $N$ is large and $p=\frac{1}{M}$ is small, binomial distribution can be approximated by Poisson distribution (The book uses $\alpha$ for $\lambda$):

$\frac{\alpha^ke^{k}}{k!}$

It follows that the probability that a list has more than $t\alpha$ keys on it is bounded by the quantity $(\alpha\frac{e}{t})^t e^{-\alpha}$.

In other words, the probability that $P(X > t\alpha)$ is bounded by the quantity $(\alpha\frac{e}{t})^t e^{-\alpha}$.

I searched the web and browsed https://en.wikipedia.org/wiki/Poisson_distribution. But I couldn't find a reference for the formula $(\alpha\frac{e}{t})^t e^{-\alpha}$.

Here is a candidate answer combining @Did's and @ClementC's comments:

From the background in the book, we can assume that $\alpha \geqslant 10$ and $t \geqslant 2$.

For $s > 0$, $\{X \geqslant t\alpha \} = \{sX \geqslant st\alpha\} = \{e^{sX} \geqslant e^{st\alpha}\}$, so

$P(X > t\alpha ) \leqslant P(X \geqslant t\alpha) \leqslant \frac{E[e^{sX}]}{e^{st\alpha}}$

by Markov's inequality. Since $E[e^{sX}] = e^{\alpha(e^{s}-1)}$, we have:

$P(X > t\alpha ) \leqslant \frac{e^{\alpha(e^{s}-1)}}{e^{st\alpha}} = e^{\alpha(e^{s}-1)-st\alpha}$

Maximizing this bound is equivalent to maximizing $\alpha(e^{s}-1)-st\alpha$. And it is obvious that maximizing $\alpha(e^{s}-1)-st\alpha$ is equivalent to maximizing $f(s) = e^{s}-st$ since $\alpha > 0$.

$f\prime(s) = e^{s} - t$

$f\prime\prime(s) = e^{s}$

Solve $f\prime(s) = 0$, we have $s = \ln{t}$. And $f\prime\prime(\ln{t}) = t > 0$. So $s = \ln{t}$ maximizes $f(s)$.

$P(X > t\alpha ) \leqslant \frac{e^{\alpha(t-1)}}{t^{t\alpha}} = (\frac{e^{\alpha}}{t^{\alpha}})^t e^{-\alpha} $

Now let's compare $\frac{e^{\alpha}}{t^{\alpha}}$ with $\alpha \frac{e}{t}$. When $t \geqslant 3$, we have $\frac{e^{\alpha}}{t^{\alpha}} < \alpha \frac{e}{t}$.

When $t= 2$, we have $(\frac{e}{2})^\alpha$ and $\alpha \frac{e}{2}$. When $\alpha = 10$, we have $(\frac{e}{2})^\alpha = 21.51022050274092$ and
$\alpha \frac{e}{2} = 13.591409142295225 $. When $\alpha = 20$, we have
$(\frac{e}{2})^\alpha = 462.68958607653593$ and $\alpha \frac{e}{2} = 27.18281828459045$. So when $t = 2$, $\frac{e^{\alpha}}{t^{\alpha}} < \alpha \frac{e}{t}$ is not true.

By the above discussion, we can prove $P(X > t\alpha ) \leqslant (\alpha \frac{e}{t})^{t} e^{-\alpha}$ for $t \geqslant 3$. But we can't prove the $t = 2$ case.

答:
I've been spending hours thinking about the same problem!

It turns out that the formula $(\alpha e/t)^t e^{-\alpha}$ is incorrect. The correct formula is $(e/t)^{\alpha t}e^{-\alpha}$.

The correction is posted on the textbook website: Errata.


问:
Strange use of Fubini for infinite sum
Let $k(n)$ be a sequence of natural numbers and $y_m$ a sequence of non-negative numbers. I encountered this line I don't understand:

$$ \sum_{n=1}^{\infty} \frac{1}{k(n)^2} \sum_{m=1}^{k(n)} y_m = \sum_{m=1}^{\infty} y_m \sum_{k(n)\geq m} \frac{1}{k(n)^2}$$

justified as an application of Fubini's theorem. I am aware of this analogue of Fubini but I don't see how this equality could be true because the "integration interval" changes too. Can someone help me see how this would work out?

答:
This is also known as ... "inverting the order the sums" ...

\begin{eqnarray*}
\{ (m,n) \mid  n=1,2, \cdots \infty , m=1,\cdots k(n) \} = \{ (m,n) \mid  m=1,2, \cdots \infty , n=k(m),\cdots \infty \}.
\end{eqnarray*}
Thus
\begin{eqnarray*}
\sum_{n=1}^{ \infty} \sum_{  m=1}^{ k(n) }\cdots = \sum_{m=1}^{ \infty }\sum_{n=k(m)}^{ \infty }\cdots.
\end{eqnarray*}

Title: Strange use of Fubini for infinite sum


问:
Passcode permutations
In how many ways can a passcode of length $7$ be generated from the set $\{0, 1, ..., 9, a, b, ..., z, A, B, ..., Z\}$ with exactly $4$ entries being letters if:

(i) Repetition is allowed?

(ii) Repetition isn't allowed?

My attempt:

(i) There are $52^4$ ways to arrange $4$ letters together. Then, there are $62^4$ ways to arrange the entire passcode. So, there are $52^4 \cdot 62^4$ ways.

(ii) There are $52P4$ ways to arrange $4$ letters together and $62P4$ ways to arrange the whole passcode. So, we have $52P4 \cdot 62P4$ ways in total.

Is my solution correct? I often struggle with permutations so any assistance is appreciated.

答:
In how many ways can a passcode of length $7$ be generated from the set $\{0, 1, \ldots, 9, a, b, \ldots, z, A, B, \ldots Z\}$ with exactly $4$ entries being letters if repetition is allowed?

Choose which four of the seven positions will be filled with letters.  For each of these four positions, there are $52$ choices.  The remaining three positions must be filled with digits.  For each of these three positions, there are $10$ choices.

  There are $$\binom{7}{4}52^410^3$$ passcodes of length $7$ which can be formed if exactly four of the entries are letters and repetition is allowed.

In how many ways can a passcode of length $7$ be generated from the set $\{0, 1, \ldots, 9, a, b, \ldots, z, A, B, \ldots Z\}$ with exactly $4$ entries being letters if repetition is not allowed?

Choose which four of the seven positions will be filled with letters.  The first of these four positions can be filled with a letter in $52$ ways, the second in $51$ ways, the third in $50$ ways, and the fourth in $49$ ways. The remaining three positions must be filled with digits.  The first of these three positions can be filled with a digit in $10$ ways, the second in $9$ ways, and the third in eight ways.

  There are $$\binom{7}{4}P(52, 4)P(10, 3)$$ ways to form a passcode of length $7$ if exactly four of the entries must be letters and repetition is not permitted.

Why is your solution incorrect?

As discussed in the comments, you did not take into account the fact that exactly four of the entries are letters.  The strategy you discussed in the comments is designed for a block of four consecutive letters.  Suppose you wanted to count the number of passcodes of length $7$ with exactly four letters, all of which are consecutive, if repetition is permitted.  You would have four objects to arrange, the block of four letters and the three digits.  The objects could be arranged in $4!$ ways.  There would be $10$ choices for each digit and $52$ choices for each letter within the block, giving you $$4!10^326^4$ such passcodes.

Title: Passcode permutations


问:
Extension and property FH (fixed point for actions on Hilbert spaces)
Remind that a group $G$ has property FH if any $G$ action on a real Hilbert space has a fixed point. For $\sigma$-compact, locally compact groups this is equivalent to the celebrated Kazhdan's property (T).

Is property FH stable under extensions? i.e. if $$
is an extension of a group with property FH by a group with property FH, does it follow that $$ has property FH?

The similar statement concerning property (T) is true for locally compact groups, cf p. 69 of Kazhdan's Property (T) by Bekka, de la Harpe, Valette, but I am really interested in property FH.

答:
I assume that by an "extension" of $N$ by $Q$ you mean a short exact sequence
$$
1\to N\to G\to Q\to 1.
$$
Thus, $N$ can be regarded as a normal subgroup of $G$ and
$G/N\cong Q$ (as a topological group).

Suppose that we have a continuous isometric affine action of $G$ on a Hilbert space $H$. By the FH property of $N$, there exists a nonempty closed affine  subspace $A\subset H$, the fixed point set of $N$. Hence, $A$ is also isometric to a Hilbert space. Since $N$ is normal in $G$, the subspace $A$ is $G$-invariant. The group $Q$ then acts continuously and isometrically on $A$ (via the action of $G$, since $N$ acts trivially on $A$). Hence, by the property FH of $Q$, the group $Q$ has a fixed point $a\in A$. This point is fixed by $G$, of course.


问:
Geometry In Complex Numbers
The line $T$ is tangent to the circumcircle of acute triangle $ABC$ at $B.$ Let
$K$ be the projection of the orthocenter of triangle ABC onto line $T$.
Let $L$ be the midpoint of side $AC$. Show that the triangle $BKL$ is isosceles.

We are supposed to do this using complex numbers. We can easily compute $H, B, L$ But how can we compute $K$? I tried some approaches but it doesn't seem to work or I can't see how to go further. Any help will be appreciated.

答:
We may and do assume that the circumcenter $\odot (ABC)$ is the unit circle, centered in the origin $O$, and that $B$ has the affix $b=1\in \Bbb C$. Let $a,c\in\Bbb C$ be the (affix) complex representations of $A,C$, so $|a|^2=a\bar a=1$, and $|c|^2=c\bar c=1$.

We use this convention further, lower case letters are affixes for the capitalized points. So $o$, the affix of $O$ is zero. The centroid $G$ has affix $g$ with
$3g=a+b+c$.

Vectorially, $GH=-2GO$, since $G$ is placed on the segment $HO$ as it is placed on the median $BL$, so vectorially $GH+2GO=0$, passing to complex numbers
$$
(h-g)+2(o-g)=0\ ,\qquad\text{ i.e. }\qquad
h+2o=3g=a+b+c\ .
$$
In our case, $o=0$, $b=1$, so $h=a+1+c$. The line $T$ is the line parametrized by $1+it$. The point $H,K$ have the same projection on the imaginary axis, parallel to $T$, so we know $k$ if we extract and adapt the imaginary part of $h$. It is
$$
\begin{aligned}
k
&= 1+i\operatorname{Im} h  \\
&= 1+i\operatorname{Im} (a+1+c)\\
&= 1+i\operatorname{Im} (a+c)  \\
&= 1+2i\operatorname{Im} l  \ .
\end{aligned}
$$
Projecting on $T$, we see that $L$ projects to the mid point of $BK$, so
$\Delta LBK$ isosceles.

$\square$

Title: Geometry In Complex Numbers


问:
Conditional probability of picking balls
Suppose that a bag contains $11$ black balls and $9$ white balls. A child takes out a ball $6$ consecutive times, each time replacing the one taken (with a ball of the same color). Find the probability of that he takes out $5$ black balls given that exactly $2$ of the first $3$ times he takes out balls, they were black?

I have absolutely no idea how to approach this problem. It mentions that replacement is involved which implies that binomial distribution may be used. Any assistance is much appreciated.

答:
Hints:

The verbiage about replacement simply means that just before each draw of a golf ball, there are exactly 11 black and 9 white balls.

The given condition that 2 of the 1st 3 draws are black reduces the problem to asking what are the chances that the next three draws are all black.  In other words, the original problem + the assumed event equate to the alternative problem of : what are the chances that the first 3 draws are all black.


问:
How do I solve this solution-mixing problem?
A chemist has a 55% acid solution and a 40% acid solution. How many liters of each should be
mixed in order to produce 100 liters of a 46% acid solution?</p>

答: Let $X_1$ denote the total amount of acid solution 1. Let $X_2$ denote the total amount of acid solution 2. Now, the total amount of solution is $X_1 + X_2$. So we want $$X_1 + X_2 = 100.$$ Now for the acid content. Here, we'll have $0.55X_1$ liters of acid from the first solution, and $0.40X_2$ liters of acid from the second solution. Our second condition says we want $$0.55X_1 + 0.40X_2 = 0.46*100 = 46.$$ So we have a system with two equations. Solve it and get your answers!</p>


问:
Functions in Group Theory
I am reading a math book and in it, it says, "Let $V$ be the set of all functions $f: \mathbb{Z^n_2} \rightarrow \mathbb{R}.$ I know that $\mathbb{Z^n_2}$ is just the cyclic group of order $2$ taken to the $n$th power, but I don't get what the actual function means.

How would such a function, $f$, work? Could someone please give an example?

答:
A function is by definition a mapping between sets. So we have to look at the underlying sets. For simplicity, we write $\Bbb Z_2=\{0,1\}$. Then $\Bbb Z_2^n=\{0,1\}^n$. A map $f\colon\{0,1\}^n\to\Bbb R$ now is an assigment which takes an element of the form $(a_1,\dots,a_n)$ and maps it to some real number $r$ depending on all $a_i\in\{0,1\}$. For example, we can trivially define $f(a_1,\dots,a_n)=\pi$ for all $n$-tupel $(a_1,\dots,a_n)\in\Bbb Z_2^n$. Formally, we write

$$\Bbb R^{\Bbb Z_2^n}=\{f\colon\{0,1\}^n\to\Bbb R\}$$

where we do not enforce any further conditions on $f\in\Bbb R^{\Bbb Z_2^n}$ (like, being a group homomorphism, being linear, etc. which we could ask for; but then we would get a notably different result).

Naturally, this set is equipped with a $\Bbb R$-vector space structure.

Indeed, define the pointwise sum of two elements $f,g\in\Bbb R^{\Bbb Z_2^n}$ by

$$(f+g)(x):=f(x)+g(x)\quad\text{where}\quad x\in\Bbb Z_2^n\,.$$

This is again just a function with domain $\Bbb Z_2^n$ taking values in $\Bbb R$ , hence an element of $\Bbb R^{\Bbb Z_2^n}$. We can define a trivial function sending everything to $0\in\Bbb R$ as neutral element and $g(x)=-f(x)$ is an additive inverse. Thus, this set has an (abelian) group structure inherited from $\Bbb R$ as one can check.

Also, as the functions are valued in $\Bbb R$, we can define a new function $(\lambda\cdot f)(x)=\lambda\cdot f(x)$ for every $\lambda\in\Bbb R$. One can check that the usual compability axioms hold making this set into a $\Bbb R$-vector space.

The functions $f_u(v)=\delta_{uv}$ (where $u,v\in\Bbb Z_2^n$) form a basis for $V=\Bbb R^{\Bbb Z_2^n}$ as $\Bbb R$-vector space as a function is uniquely and completely determined on its values at any point in its domain. The domain happens to be $\Bbb Z_2^n=\{0,1\}^n$ which has $2^n$ elements $u$. Every one of them gives us a unique function $f_u$ as defined earlier and we can decompose a function $f\in\Bbb R^{\Bbb Z_2^n}$ into how it acts on every element, giving us the needed linear combination of the form
$$f=\sum_{u\in\Bbb Z_2^n} a_uf_u,\quad a_u\in\Bbb R$$

Title: Functions in Group Theory


问:
CLT for $P\{X_k = \pm 2^k\} = \frac{1}{2}$
I want to show that the following family of random variables either satisfy or don't satisfy the central limit theorem $P\{X_k = \pm 2^k\} = \frac{1}{2}$.

For the CLT I know that I want to use the Lindeberg condition. However, I'm having trouble understanding the definition given in equation 5.6 of Feller's book volume 1.

答: The moment generating function for $\bar{X}_n$ is $$M_{\bar{X}_n}(t)=\prod_{k=1}^n\cosh\Big(\frac{t2^n}{n}\Big)$$ Note for any $t\neq 0$ we have $$M_{\bar{X}_n}(t)\geq\cosh\Big(\frac{t2^n}{n}\Big)\longrightarrow \infty$$ as $n \longrightarrow \infty$ so the sequence of random variables $\{\bar{X}_n\}_{n=1}^{\infty}$ doesn't converge to any normal distribution.


问:
Are summations the same as a definite integral?
So I was looking at my teacher's solution to a problem, and it starts with a summation, which she converted to a definite integral:

$$\sum_{x=0}^{x=n} \sum_{y=0}^{y=n} |x-y| = \int_{0}^{n}\int_{0}^{n} |x-y| dy dx$$
$$= \int_{0}^{n} \Biggl( \int_{0}^{x} (x-y) dy +  \int_{x}^{n} (y-x) dy \Biggr) dx$$
$$=\int_{0}^{n} x^2 - nx + \frac{n^2}{2} dx$$
$$=\frac{n^3}{3}$$

However, I tried to convert the integral back to a summation and do the calculation that way...just to see if the answer would be the same, and it ended up been differen:
$$\int_{0}^{n} \Biggl( \int_{0}^{x} (x-y) dy +  \int_{x}^{n} (y-x) dy \Biggr) dx$$
$$\sum_{x=0}^{x=n} \Biggl( \sum_{y=0}^{y=x} (x-y) + \sum_{y=x}^{y=n} (y-x) \Biggr) $$
$$\sum_{x=0}^{x=n} \Biggl( x^2 -nx+ \frac{n^2}{2}+\frac{n}{2} \Biggr)$$
$$= \frac{n^3}{3} + n^2 + \frac{2n}{3}$$

I thought because you could make the original summation in to a integral, you could turn the integral back in to a summation midway, but is that not correct?

答:
It may be shown that $$\sum_{x=0}^n \sum_{y=0}^n |x-y| = \frac{n(n+1)(n+2)}{3}$$

and $$\int_{0}^{n}\int_{0}^{n} |x-y| \,dy\, dx = \frac{n^3}{3}$$

The integral may be used as an approximation to the discrete summation. However, strictly speaking, they are not equal.

You calculated the exact discrete sum already!


问:
Finding an equation of a parabola
I was asked to show that a$\pm\frac{cx}{b}=\sqrt{x^2+y^2}$ is a parabola if $b=c$
$$\text{If }b=c\rightarrow a\pm x=\sqrt{x^2+y^2}\\
 \left(a\pm x\right)^2=x^2+y^2\\
 \left(a\pm x\right)^2-x^2=y^2\\
 \pm\sqrt{a^2+2ax+x^2-x^2}=y \text{ or } \pm\sqrt{a^2-2ax+x^2-x^2}=y\\
\pm\sqrt{a^2\pm2ax}=y$$

Can I go from here to a standard form of a parabola? How?

How can I show that if $b>c$ it's an ellipse and if $b<c$ it's a hyperbola?

*Edit: I am also attaching my work for $b>c$. $$\text{If } b>c\rightarrow a\pm\frac{cx}{b}=\sqrt{x^2+y^2}\\
 \left(a\pm\frac{cx}{b}\right)^2=x^2+y^2\\
 a^2\pm2a\frac{cx}{b}+\frac{\left(cx\right)^2}{b^2}=x^2+y^2\\
 \pm2\frac{cx}{ab}+\frac{\left(cx\right)^2}{\left(ab\right)^2}=\frac{x^2}{a^2}+\frac{y^2}{a^2}\\
\frac{cx}{ab}\left(\pm2+\frac{cx}{ab}\right)=\frac{x^2}{a^2}+\frac{y^2}{a^2}\\
\pm2+\frac{cx}{ab}=\frac{b}{cx}\left(\frac{x^2}{a}+\frac{y^2}{a}\right)\\
\pm2=\frac{b}{cx}\left(\frac{x^2}{a}+\frac{y^2}{a}\right)-\frac{cx}{ab}\\
\pm2=\frac{b^2x^2+b^2y^2}{abcx}-\frac{\left(cx\right)^2}{abcx}\\
\pm2=\frac{b^2x^2+b^2y^2-c^2x^2}{abcx}\\
\pm1=\frac{\left(b^2-c^2\right)x^2+b^2y^2}{abcx}$$

答:
All conic sections follow the following general form

$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $, with $A, B, C, D, E, F$ all real and $A,B,C$ non-zero.

(see: https://en.wikipedia.org/wiki/Conic_section#cite_ref-Protter_1970_326_15-0):

The discriminant $B^2 - 4AC$ determines ellipse, circle, parabola, hyperbola, etc.

(1) if $B^2 - 4AC = 0$, the equation is a parabola.

(2) if $B^2 - 4AC < 0$, the equation is an ellipse, and so on.

From $\left(a\pm \frac{cx}{b}\right)^2=x^2 + y^2$,

$\big(\frac{c^2 - b^2}{b^2}\big)x^2 - y^2 \pm2\frac{ac}{b}x + a^2 = 0$,

Then, $A = \big(\frac{c^2 - b^2}{b^2}\big)$, $B = 0$, $C = -1$, $D =\pm2\frac{ac}{b}$, $E = 0$, $F = a^2$.

If $b = c$, then $B^2 - 4AC = 0$, so it is a parabola.

For the ellipse, the discriminant is negative if $b > c$, and similarly for hyperbola.

Alternatively,

$ y^2 = a^2 \pm 2ax$ from your last step,
then, use a substitution $x = x' - \frac{a}{2}$:

$ y^2 = a^2 \pm 2a\big(x' - \frac{a}{2}\big) = \pm 2ax'$, which is an equation of parabola.

Title: Finding an equation of a parabola


问:
Represent the following propositions using quantifiers
Domain: x and y are in the domain of all people
V(x, y): "x has venmoed money to y"

a) "Jake has not venmoed money to Lauren"

b) "There are at least two kids who have not venmoed each other"

c) "Everyone has venmoed the same person" (the person is not themselves)

My answers:

a) ∀x∀y[~V(Jake, Lauren)]

b) ∃x∃y[~V(x,y)] | (x != y)

c) ∀x∃y[V(x,y)] | (x != y)

I am really unsure about my first answer, because I have to use logic and predicates to modify the domain. I am not allowed to actually manipulate the domain.

Are my answers correct?

答:
a) "Jake has not venmoed money to Lauren"

a) ∀x∀y[~V(Jake, Lauren)]

Nope. The only variables in this statement are the constants.  You do not need quantifiers. $$\lnot\mathsf V(\textrm{Jake}, \textrm{Lauren})$$

b) "There are at least two kids who have not venmoed each other"

b) ∃x∃y[~V(x,y)] | (x != y)

N... Well... the $\mid$ symbol is typically used for 'or', while $\&$ is used for 'and'. You need to say that there are two who are not the same and one of them did not venmo the other, so check your symbols.

$$\exists x\exists y~(x\neq y\mathop{\&} \lnot\mathsf V(x,y))$$

$\tiny\textrm{Well the immediate translation is $\exists x\exists y~(x\neq y\mathop{\&} (\lnot\mathsf V(x,y))\mid \lnot\mathsf V(y,x)))$ however, the above is an equivalent rephrasing.}$

c) "Everyone has venmoed the same person" (the person is not themselves)

c) ∀x∃y[V(x,y)] | (x != y)

"There is someone whom everyone else had venmoed."  To refer to the same existance, the existential must be bound before the universal.

$$\exists y~\forall x~(x\neq y\to\mathsf V(x,y))$$


问:
Showing $\sqrt{S^2}$ is a consistent estimator of standard deviation $\sigma$.
Given $X_1,\dots,X_n\stackrel{iid}{f}(x;\mu,\sigma^2)$ with population mean $\mu$ and population standard deviation $\sigma$. I want to show $\sqrt{S^2}$ is a consistent estimator for $\sigma$. So if I read my book correctly, we want to show
$$\lim_{n\rightarrow\infty}P(|\sqrt{S_n^2}-\sigma|>\epsilon)=0\quad\forall\epsilon >0 $$

So invoking Chebyshev's inequality we have

$$
\begin{align*}
P((\sqrt{S_n^2}-\sigma)^2>\epsilon^2)\leq&\frac{\mathbb{E}(\sqrt{S_n^2}-\sigma)}{\epsilon^2}
\\
={}&\frac{\mathrm{Var}(\sqrt{S_n^2})}{\epsilon^2}\\
={}&\frac{\mathbb{E}[(\sqrt{S_n^2})^2]-[\mathbb{E}(\sqrt{S_n^2})]^2}{\epsilon^2}\\
={}&\frac{\mathbb{E}(S_n^2)-[\mathbb{E}(\sqrt{S_n^2})]^2}{\epsilon^2}\\
={}&\frac{\sigma^2-[\mathbb{E}(\sqrt{S_n^2})]^2}{\epsilon^2}\text{ (Since $S^2$ is an unbiased estimator)}
\end{align*}
$$
This is where I get stuck because I know nothing about the distribution besides it being $\textit{iid}$, so I have no idea what to do with the $[\mathbb{E}(\sqrt{S^2})]^2$ term. I know I need to have a term with $n$ to take a limit, but I am unsure if I have ever taken the expected value of a sample standard deviation or if I even know how to properly. Any help or guidance would be appreciated. Just to save time Jensen's inequality is something that I have not learned in the course yet.

答:
Henry said it all, but I will elaborate here. The sample variance is
\begin{equation}
S^2 = \frac{1}{n-1} \sum_{i=1}^n \left(  X_i - \overline{X} \right)^2,
\end{equation}
and since $X_i \sim f(x;\mu,\sigma)$, the population variance is
\begin{equation}
\sigma^2 = \frac{1}{n} \sum_{i=1}^n \left(  X_i - \mu \right)^2.
\end{equation}
It is obvious that $\mathbb{E}(\overline{X}) = \mu$ since $X_i \sim f(x;\mu,\sigma)$,
\begin{equation}
\lim_{n\to\infty} \frac{\sigma^2}{S^2} = \lim_{n\to\infty} \frac{n-1}{n} = 1.
\end{equation}


问:
How to project the surface of a hypersphere into the full volume of a sphere?
The game I mentioned in "Navigating though the surface of a hypersphere in a computer game" is taking shape in here. The world is a 3-sphere where everything belongs. In Euclidean coordinates, for every point $x^2 + y^2 + z^2 + w^2 = 1$. We move the objects by applying orthornormal transformations to them.

We got to a stage where we want to draw a mini-map to help the user navigate through the space, and that is the subject of this question. Like we use stereographic projection to get the 3-sphere into 3-D Euclidean space, how can we project the points of the 3-sphere into the full volume of a "common" 2-sphere? This way, all world objects could be seen in a cool sphere drawn in the corner of the screen, unlike it is with the stereographic projection, which is infinite and some points may be missing in the cropping.

Considering every point $P = (x, y, z, w) \; | \; x^2 + y^2 + z^2 + w^2 = 1$, we need a function $t$ where $t(P) = (X, Y, Z) \; | \; X^2 + Y^2 + Z^2 \leq 1$ and $t((0,0,0,-1)) = (0,0,0)$ (the center of the stereographic projection). Every point of the 3-sphere except $(0,0,0,1)$ must be defined. What is this function? I am not sure on how to look for it...

答:
It's a cool game idea. However, the map you're looking for cannot be called a stereographic projection. Stereographic projection projects a sphere onto a whole infinite plane or, in this case, infinite 3-dimensional space $R^3$.

You want to compactify the 3-space to a 3-ball. However, that's clearly not possible if you want to preserve the angles - which is what the stereographic projection does. A 3-sphere is simply not conformally equivalent to a 3-ball because the latter has a boundary.

Imagine that you have such a map $t(P)$ and you are thinking what points $M$ of the 3-sphere get mapped onto - or near - the boundary of the 3-ball, the 2-sphere. Clearly, $M$ must be some singular manifold inside the $S^3$ which is diffeomorphic to an $S^2$. The points of the 3-sphere on the opposite side of the $M$ cannot be found in the 3-ball, as long as the map is one-to-one.

What you could do - if you want to preserve the angles - would be to divide the $S^3$ into the upper and lower hemisphere, $w>0$ or $w<0$, for example, and these two hemispheres are then very close to and naturally mapped to a 3-ball. In this case, you may use the standard stereographic projection and because of the removal of the wrong hemisphere, the infinite 3-plane will shrink to a 3-ball. But you would get two such 3-balls, not one.

Alternatively, you may sacrifice the conservation of the angles and you may "compactify" the 3-plane from the normal stereographic projection by mapping (using polar coordinates)
$$(r,\theta,\phi) \to (\arctan r,\theta,\phi).$$
I used arctan because it maps $(0,\infty)$ to a finite interval; $\tanh r$ would do the job, too. And there are others (and simpler) such as $r/(1+r)$. As expected, the angles will be severely distorted near the boundary of the 3-ball which will correspond to the vicinity of the removed pole of the 3-sphere.


问:
Integral of $\sqrt{4-x^2}$ by Riemann sum
I was trying to compute $\int_{0}^{2}\sqrt{4-x^2}\,dx$ by Riemann sum like this

$$\sum_{i=1}^{n}\sqrt{4-x_i^2}\,\Delta x_i$$

where $x_i = \frac{2i}{n}$ and $\Delta x_i = \frac{2}{n}$, but I couldn't get something useful for the answer. I think it should be with another kind of partition, but I don't know which kind of partition to use.

答: Note that the partition you mentioned is not in fact a partition of the interval $[0, 2].$ I think that what you would want to do involves considering the partition $P = (\xi_i)_{1 \leq i \leq n},$ where $\xi_i := \frac{2i}{n}$ for all $1 \leq i \leq n$ (and, of course, $\Delta \xi_i = \frac{2}{n}$).


问:
Teaching linear algebra. Is this equation a "linear equation"?
Consider the equation $x+3=3(x+1)-2x$. Is this equation a linear equation?

I would say to my students that an equation in 1 variable is linear when it can be simplified to the form $ax=b$ where $a\neq0$. Also, I would mention that a linear equation in 1 variable has only 1 solution. Therefore, I would say that the equation $x+3=3(x+1)-2x$ is not linear because it can be reduced to $0x=0$ which has infinite solutions.

However, a colleague whose research area is algebra says that a linear equation is one that only involves polynomials of degree 1. Therefore, the equation $x+3=3(x+1)-2x$ is linear. Then, according to the definition of my colleague, the equation $x-x=0$ is linear as well.

I was thinking that maybe a third possibility is that according to the definition of "linear equation" only equations of the form "expression$=0$" can be classified as linear or non-linear.
In this case, the equation $x+3=3(x+1)-2x$ is not linear nor non-linear, but the equation $x-x=0$ would be linear.

What should be the definition for "linear equation" in 1 variable suitable for a math course?

答:
However, a colleague whose research area is algebra says that a linear equation is one that only involves polynomials of degree 1. Therefore, the equation $0=0$(on simplification)  is linear. Then, according to the definition of my colleague, the equation  is linear as well.

The problem is "The degree of a zero polynomial is not defined or $-\infty$". So saying that zero polynomial has degree $1$ is little vague as $0\equiv 0x\equiv 0x^2\equiv 0x^3\equiv.....$

What should be the definition for "linear equation" in 1 variable suitable for a math course?

A linear polynomial $p(x)=ax+b, a\ne 0$ when equated to a real number $\alpha(say)$ i.e. $p(x)=\alpha$ is called a linear equation.


问:
Showing $e^{-x^2}$ small when $x$ is large just from series definition
This is a sort of open ended question. Suppose
$$F(x):=\sum_{L=0}^{\infty} (-1)^L\frac{x^{2L}}{L!}.$$
One can show that the series is absolutely convergent for every fixed $x$, by noting that $x^{2L} \le (L-2)!$ for all large enough $L$. So, $F(x)$ is well defined. I want to show $F(x)$ is very small for $x$ large enough.

I DON'T want to use the fact that $F(x)$ is actually $e^{-x^2}$. Actually, in my research, I am dealing with a similar problem where I have an alternating series quite similar to exponential (but no closed form), where I need to show its small when $x$ is large. So, I was trying with this toy problem.

答:
I like to do unnecessarily complex manipulations on power series for fun, so I thought about this occasionally over the last couple of weeks.  Below I discuss two different solution paths.  The first one is basically done but is unsatisfactory to me.  The second is just some thoughts about an approach that I would find satisfactory.  I'm only including the second part in the hopes of inspiring the OP or someone else to try this approach, since it is nontrivial and (due to the pandemic) there is essentially zero chance that I will have the time to figure it out anytime soon.

Below I'll use $e^{-x}$ instead of $e^{-x^2}$ for simplicity.  I define
$$ f(x) = \sum_n \frac{(-1)^n x^n}{n!} $$
and want to show that $f(x)$ is small when $x$ is large and positive, but without using the fact that $f(x)=e^{-x}$.

Approach 1: This is different than the other answer, and provides a better bound, but has very little chance of working for a different power series.

It's easy to show straight from the power series that $f(x + y) = f(x) f(y)$.  Below, I introduce the index $k=n-m$ to substitute out $n$, at which point the two sums become independent.  (I'm implicitly assuming that we can swap summation order, which I'm not proving, but it's okay because of the convergence properties of the sum.)
$$ \begin{align}
f(x + y) &= \sum_n \frac{(-1)^n (x + y)^n}{n!} \\
&= \sum_n \frac{(-1)^n}{n!} \sum_{m=0}^{n} \binom{n}{m} x^m y^{n-m} \\
&= \sum_n (-1)^n \sum_{m=0}^{n} \frac{x^m y^{n-m}}{m!(n-m)!} \\
&= \sum_m \sum_k \frac{(-1)^{k+m} x^m y^k}{k! m!} \\
&= \left( \sum_m \frac{(-1)^m x^m}{m!} \right) \left( \sum_k \frac{(-1)^k y^k}{k!} \right) \\
&= f(x) f(y)
\end{align} $$

From this, it is obvious that if $n$ is a nonnegative integer, then $f(n) = f(1)^n$.

Since the terms in the series for $f(1)$ are decreasing in magnitude, then we can show that
$$ \frac{1}{3} < f(1) < \frac{1}{2} $$
by just terminating after the third or fourth term in the sum.

Combining these two results, we get that
$$ \frac{1}{3^n} < f(n) < \frac{1}{2^n} $$
for positive integers $n$.

At this point, we want to extend these results to non-integer arguments.  I won't go into details, and there are probably multiple ways of doing this, but one of them would just be to use the same bounds for the derivative, or to place a separate bound on $f(x) - f(n)$ for $x \in (n, n+1)$ using similar means.

As I said, the approach above works, and it gives a reasonably good bound on $f(x)$, but it is extremely unlikely to extend to other series, even approximately.  I feel like it should be possible to show approximate cancellation from the series directly, using some approximations.  I would consider this a "good" solution.

One argument in favor of this, which might be formalizable, is the following.  Consider all of the positive and negative terms in $f(x)$ separately (i.e., the series for $\cosh x$ and $\sinh x$)  The positive ones are $x^{2n}/(2n)!$.  Other than a constant factor, this is the probability of finding value $2n$ in a Poisson distribution with mean $x$; the sum is the probability of finding any even value.  The odd terms are similarly the probability of finding an odd value in the same Poisson distribution.  If the mean is large, then you should be approximately equally likely to find an odd or even value in the Poisson distribution, so basically there should be a lot of cancellation and a small result.

(One way of showing this might be to view the odd and even terms as two different Riemann sums for the integral of the continuous version of the distribution.  But since the distribution is sharply peaked and contains important contributions from many orders of $x$, then it's not obvious to quantify just how small the result should be.)

As another very minor argument in favor of this, it's quite tantalizing that when $x$ is an integer, the two largest-magnitude terms in the series, where $n=x$ and $n=x-1$, cancel exactly.  It's hard to extend this.

I think a general approach, which I have thought about but not tried numerically, is to split the $n^\text{th}$ term in the series into a number of other terms, and rearrange terms in such a way that the result cancels.  (We can do this without fear because the series converges absolutely.)  One way to write this is to say $s_n(x)=(-1)^n x^n/n!$ and write
$$ f(x) = \sum_{n=0}^N s_n(x) + R_N(x) $$
Then we can write the sum as a matrix multiplication,
$$ f(x) = 1^T A s(x) + R_N(x) $$
where $A$ has two properties:

$\sum_j A_{ij} s_j(x) \approx 0$

$\sum_i A_{ij} = 1$

The first property is basically saying that we weight the different terms in the sum in such a way that each weighted "subseries" cancels exactly, and the second ensures that we get the original sum back when we group the terms back together.  Then the bound on the whole sum would be given by $R_N(x)$ and how good the approximation is in part (1).

There are a number of identities involving the binomial coefficients and/or Stirling numbers that might be useful in showing cancellation of each subseries.  The fact that the exponential is its own derivative suggests that considering finite differences of various orders might be helpful.  I wasn't able to get anything to work, but I'm hoping someone else will.


问:
Random variable with infinite value
Consider random variable $X$ which distribution is defined in the following way. Variable $X$ can take any non-negative integer value and $P(X=k)=1/2^{k+2}$ for any $k>0$. Find $P(X=0)$

答:
Since $X$ can take any nonnegative integer value, the distribution is defined for all values in the support of $X$ other than 0.

Then, we can just do the following:

$$ P[X = 0] = 1 - P[X>0] $$

$$ = 1 - P[X=1 \ \cup \ X=2 \ \cup \ ...] $$

$$ = 1 - \sum_{k=1}^\infty P[X=k] $$

$$ = 1 - \sum_{k=1}^\infty \frac{1}{2^{k+2}} $$

This sum is geometric with initial value $\frac{1}{8}$ and common ratio $\frac{1}{2}$, so it sums to:

$$ \sum_{k=1}^\infty \frac{1}{2^{k+2}} = \frac{\frac{1}{8}}{1-\frac{1}{2}} = \frac{1}{4} $$

$$\Rightarrow P[X=0] = \frac{3}{4}$$

Title: Random variable with infinite value


答:
This is nice to illustrate Lindeberg-Feller but you can also solve it with your bare hands.

Consider a the following operation in probability distributions on R:

Starting with a random variable with mean zero and unit variance, add to it an independent random variable equal to ( \pm \sqrt{3} ) with probability 1/2 each, and divide the result by two to get again unit variance.

Check that the Gaussian is not a fixed point of this operation.

This will imply that your sequence does not satisfy the CLT.

Title: CLT for $P\{X_k = \pm 2^k\} = \frac{1}{2}$

Body:
I want to show that the following family of random variables either satisfy or don't satisfy the central limit theorem $P\{X_k = \pm 2^k\} = \frac{1}{2}$.

For the CLT I know that I want to use the Lindeberg condition. However, I'm having trouble understanding the definition given in equation 5.6 of Feller's book volume 1.


问:
If $b ∈ \Bbb Z$, then there is no integer a such that $b < a < b + 1$
If $b ∈ \Bbb Z$, then there is no integer $a$ such that $b < a < b + 1.$

I am doing an introductory course for mathematical proofs and this question is an exercise related to inequalities and natural numbers.

I am having trouble articulating a proof for this proposition. I know how to prove that $b < b+1$, but I am not sure how I can prove the statement above by showing that $a$ does not exist.

Should I be doing a proof by contradiction? If I assume $a$ exists, then how do I show that this will result in contradiction? Am I supposed to try to show that $a$ is not a natural number?

I am just looking for suggestions or some instructions on how to write the proof, not an actual full written proof. I would like to be able to figure at least some of it out myself, instead of copying off of a posted answer. So please just provide helpful hints as answers.

Thank you

答:
The book you mentioned in a  comment under your post, The Art of Proof1, introduces the set of natural numbers in a way that is quite different from the usual one that defines the integers out of natural numbers.

To summarize,

it first defines the integers $(\mathbf{Z},+,\cdot)$ using essentially the "ring" axioms in the first chapter.

Then it defines the natural numbers $\mathbf{N}$ as a subset of $\mathbf{Z}$ that satisfies some properties; in particular, $0\not\in\mathbf{N}$.

Then it defines the order on $\mathbf{Z}$ using the natural numbers $\mathbf{N}$.

And then it introduces the axiom of mathematical induction; this is the section where your exercise is in.

Your book (p.20) suggests that you should first prove the following:

Proposition 2.21. There exists no integer $x$ such that $0<x<1$.

Proof.

Suppose there exists an integer $x$ such that $0<x<1$. Proposition 2.20 says that for all $k\in\mathbf{N}$, $k\ge 1$. Therefore we must have $x\not\in\mathbf{N}$.

On the other hand, by the definition of the order of integers (section 2.2, page 15),
$$
x=x-0\in\mathbf{N}
$$
which contradicts to $x\not\in\mathbf{N}$.

Now we prove

Corollary 2.22 Let $n\in\mathbf{Z}$, There exists no integer $x$ such that $n<x<n+1$.

Let $P(k)$ be the statement that there exists no integer $x$ with $k-1<x<k$.
Then $P(1)$ is true by Proposition 2.21. Let $n\in\mathbf{N}$ and assume that $P(n)$ is true: there is no integer $x$ with $n-1<x<n$.

For sake of argument, suppose there exists an integer $x$ with $n<x<n+1$. Then
$$
n-1=n+(-1)<x+(-1)<n
$$
which contradicts the induction hypothesis. By the induction axiom 2.15, $P(k)$ is true for all $k\in\mathbf{N}$.

So far, we have shown Corollary 2.22 for all integer $n\ge 0$. Now suppose $n<0$. If there exists $x\in\mathbf{Z}$ with $n<x<n+1$, then
$$
-(n+1)<x<-n
$$
which contradicts what we have shown since $-n\in\mathbf{N}$ and $-(n+1)\in\mathbf{N}$ or $-(n+1)=0$.

1 The authors should be Beck and Geoghegan, not Allistair Savage, who is the one wrote his lecture notes based on the book.


问:
$2^{q-1}\equiv 1\pmod{q}.$

The question is asking to show that $q$ must be prime given
$$ 2^{q-1}\equiv 1\pmod{q}. $$

答:
This the counter example of converse of FERMAT'S theorem,

For converse we have to show that if $a^{p-1} \equiv1\pmod p  $ then p must be prime.

Note that $341=31\cdot11$

By Fermat's theorem,

$2^{10} \equiv1\pmod {11}$

$2^{30} \equiv1\pmod{11}$

$2^{31} \equiv2\pmod {11}$

$2^{(31)(11)} \equiv2^{11}\pmod{11}\big(\because a\equiv b\pmod{n} \ \text{then} \ a^n \equiv b^n\pmod{n} \big)$

$2^{341}\equiv2\pmod {11}$($\because$ By Fermat's theorem)

Also,

$2^{10} \equiv1024\pmod{31}$

$2^{10} \equiv1\pmod{31}$

$2^{11} \equiv2\pmod{31}$

$2^{(11)(31)} \equiv2^{31}\pmod{31}$

$2^{341}\equiv2\pmod {31}$($\because$ By Fermat's theorem)

Since gcd$(11,31)=1$

$\big(\text{Here i used the result If} \ \ a \equiv b\pmod{n_1} and \ \ a \equiv b\pmod{n_2} \text{and gcd} \ \ (n_1,n_2)=1 \text{then}  \ \ \  a \equiv b\pmod{n_1\cdot n_2} \big)$

$2^{341} \equiv2\pmod{11\cdot 31}$

$2^{341} \equiv2\pmod{341}$

But $341$ is not a prime number.

Title: $2^{q-1}\equiv 1\pmod{q}.$


问:
Harmonic functions $u$ and $v$ such that $\nabla u = a \nabla v$ for some function $a$.
Let $u,v\colon\Omega\to\mathbb R$ be harmonic functions on the open set $\Omega \subset\mathbb R^2$ such that:

Exists function $a\colon\Omega\to\mathbb R$ such that $\nabla u \left(x,y\right) = a \left(x,y\right) \nabla v\left(x,y\right)$ for every $\left(x,y\right)\in \Omega$.

$\nabla v \ne \mathbf 0$ for every $\left(x,y\right) \in\Omega$.

Prove that there exist $b,c\in\mathbb R$ such that $u\left(x,y\right)=b\,v\left(x,y\right)+c$ for every $\left(x,y\right)\in\Omega$.

答:
Since $a^2=(u_x^2+u_y^2)/(v_x^2+v_y^2)$, the function $a^2$ is differentiable on $\Omega$. Let $B$ be any open sub-rectangular region of $\Omega$ on which $a$ does not change its sign. Hence $a$ is differentiable on $B$. We claim that $a$ is a constant on $B$.

Differentiate $u_x=av_x$ with respect to $x$ to get $u_{xx}=a_xv_x+av_{xx}$. Similarly, $u_{yy}=a_yv_y+av_{yy}$. By comparing the two equations, we have $$a_xv_x=-a_yv_y.~~~~(1)$$ By differentiating $u_x=av_x$ with respect to $y$, we also $u_{xy}=a_yv_x+av_{xy}$, and similarly $u_{yx}=a_xv_y+av_{yx}$. Therefore,
$$a_xv_y=a_yv_x.~~~~~~(2)$$ Square equations (1) and (2) and add. Since $v_x^2+v_y^2 \neq 0$, one has $a_x^2=a_y^2$. If $a$ is not constant on $B$, then $|a_x|=|a_y|\neq 0$ at some point. If at that point $a_x=a_y$, then $v_x=-v_y$ and $v_y=v_x$ hence $v_x=v_y=0$ a contradiction. Similarly, $a_x=-a_y$ leads to a contradiction. Hence $a$ is constant on $B$. By integrating $u_x=av_x$ on $B$, it follows that $u(x,y)=bv(x,y)+c$ on $B$ for some constants $b,c$. On any connected subset of $\Omega$ the constants $b$ and $c$ do not depend on $B$.


问:
Beginner question in first-order logic: Why does it make it difference, whether the quantifier is inside or outside the brackets?
thanks in advance for your help, I feel like I am losing my mind over this. I can't understand the difference between two formulas:

$\exists x (Px \rightarrow \forall y Py)$

$(\exists x Px \rightarrow \forall y Py)$

Again, I don't understand why it makes a difference, whether the existential quantifier is outside the brackets or not. 1. is a tautology and the second is not. I can think of a counterexample for 2. but I feel like it should be possible to get a counterexample for the first. Sorry for my studipity.

答:
The difference is that in the first statement the proposition is that there exists an $x$ such that the statement $(Px\rightarrow \forall yPy)$ is true. This is indeed tautological since, if $\forall yPy$ the implication is true for each $x$. If not $\forall yPy$ there exists an $x$ such that $\neg Px$. implying that for given $x$ the statement $Px\rightarrow\forall yPy$ holds true.

In the second proposition, this is an implication. The implication that $\exists xPx$ implies $\forall yPy$. You could write it with the following parentheses $(\exists xPx)\rightarrow(\forall yPy)$. A counterexample to this being tautological, is for a statement $P$ that is true for some but not all $x$. (For example $x^2 = 1$ in the theory of integer arithmetic.) In this case the premise is true $\exists xPx$, but the implication does not follow $\forall xPx$.


问:
Norm inequality for linear isomorphism
I want to show that if $L:\mathbb{R}^d\rightarrow\mathbb{R}^d$ is a linear isomorphism and $V_1,V_2$ are linear subspaces of $\mathbb{R}^d$ then:

$$\frac{1}{||L||\space||L^{-1}||}\leq\frac{|sin\measuredangle(L(V_1),L(V_2))|}{|sin\measuredangle(V_1,V_2)|}\leq||L||\space||L^{-1}||$$
where the angle $\measuredangle (V,W)$ between two subspaces $V$ and $W$ of $\mathbb {R}^d $ is
the smallest angle between non-zero vectors $v\in V$ and $w\in W$.

I tried to solve the first inequality using the relation between the sinus and the vector product but I think that it is not the correct way.

Can someone help me? Thank you.

答:
First notice that one inequality implies the other by applying $L^{-1}$ instead of $L$, so we will prove only the right one.

Secondly, if the angle between two given vectors $u \in V_1$ and $v \in V_2$ is greater than $\frac{\pi}{2}$, then the angle between $-u$ and $v$ is less than $\frac{\pi}{2}$, so there is no loss of generality by assuming that all vectors we consider have angle less than $\frac{\pi}{2}$. The angle between $u$ and $v$ is the same as the angle between $\frac{u}{|u|}$ and $\frac{v}{|v|}$, so we will also only consider unit vectors. Now take sequences $u_n \in V_1$ and $v_n \in V_2$ such that the angle between $u_n$ and $v_n$ converge to the angle between $V_1$ and $V_2$, since the first sequence is contained in $\mathbb{S}^{d-1} \cap V_1$ and the second one in $\mathbb{S}^{d-1} \cap V_2$, both of these sets being compact sets, we may take subsequences if necessary and suppose that $\lim u_n = u$ and $\lim v_n = v$, that is, the smallest angle is obtained by a pair of vectors, we will show the inequality for this pair of vectors.

Also, composing by an isometry if necessary, we may suppose without loss of generality that $Lu$ is colinear with $u$ and the subspace $V$ generated by $u,v$ is invariant by $L$. Consider an othornomal basis in $V$ where $u$ is the first element of that basis. In such basis, the matrix associated with $L$ in V is written as

\begin{pmatrix}
  a & b \\
  0 & d
 \end{pmatrix}

If $v$ is orthogonal to $u$ then it's obvious since $\|L\|\|L^{-1}\| \geq 1$, therefore, writing $u = (1,0)$ we take a representative of $v$ in the form $v = (1,y)$ which implies $Av = (a+yb,yd)$ and
$$
 \frac{\sin\angle(LV_1,LV_2)}{\sin\angle(V_1,V_2)} \leq \frac{\sin(\min(\angle(Lu,Lv),\angle(Lu,-Lv))}{\sin\angle(u,v)} =
 \frac{|y||d|}{|Lv|}\frac{|v|}{|y|} =\\= \left|L\left(\frac{v}{|v|}\right)\right|^{-1}|d| \leq \|L^{-1}\| \|L\|
 $$

Title: Norm inequality for linear isomorphism