问:
The probability that two integers picked at random are Prime.
It's known that The probability that two integers m and n picked at random are relatively prime is $\frac{6}{\pi^2}$

To see for myself, I ran 100,000,000 random tests of $gcd(m=random(), n=random())$ (random max 32767) and the probability was 60.09% (approx $\dfrac{6}{\pi^2}$).

Also ran a similar test that accepts two "random" integers, then tallies any prime pairs it encounters and the probability was 1.15% (approx $\dfrac{0.11}{\pi^2}$).

Question, is there a known approximation for primes as there is for coprimes?

Just curious.

Thanks.

EDIT

Based upon the comments, "Randomly chosen positive integers less than $n$ would be fine." I'll try playing with increasing values of $n$ (currently set to $32767$) to see how it tends to $0$ as $n$ tends to infinity. Thanks!

答:
It is a nontrivial thing to define a good idea of "randomly chosen integer", as @JMoravitz notes. The normal way to do this is to definite the density of a subset $S \subset \mathbb{Z}_{\geq 1}$ as
$$ \lim_{N \to \infty} \frac{ \#(S \cap \{n | 1\leq n \leq N \} ) }{N}$$

With this idea in mind, the only thing to do is recall the prime number theorem that $\pi(N) \sim N/\log(N)$ (i.e., $\pi(N) = N/\log(N) + O(N)$ ) where $\pi$ is the prime counting function. Then we have the density of prime numbers is
$$\lim_{N \to \infty} \pi(N)/N $$
which tends to $1/\log(N)$ which tends to zero, as $N \to \infty$.

From here I'll leave it to you as an exercise to show that the concurrent primality you are asking about is "rarer" (in our sense of density - which I stress is not a probability, so don't use those rules).


问:
prove that infinite vector spaces are elementarily equivalent
let F be a field. I want to show that any two infinite F-vector spaces are elementarily equivalent using model theory.

Setup: Let $L$ be first order language of vector spaces over F. Non-logical symbols of $L$ are a constant $0$, a binary function symbol $+$, and a unary function symbol $F_a$ for any $a ∈ F$. For an $F$-vector space $V$ , we regard $V$ as an $L$-structure in the obvious way: 0 is interpreted by the identity element of $V$ , + is interpreted by the addition of $V$ , and each $F_a$ is interpreted by the operation of scalar multiplication by $a$.

My attempt is to take infinite F-vector spaces $A \subseteq B$, and show that A is the elementary substructure of $B$, i.e. $A$≺$B$. For this I would have to show that for every finite sequence $a_1,...,a_m \in A$ and $b\in B$, there exists an automorphism of $B$ that fixes the $a_1,...,a_m$ and sends the $b$ into $A$. How is this done rigorously?

答:
You want to show that the theory $T$ of infinite $F$-vector spaces is complete. The easiest way to do this is by the Łoś-Vaught test. For any cardinal $\kappa>|F|$, $T$ is $\kappa$-categorical. Since $T$ also has no finite models, the Łoś-Vaught test tells us that $T$ is complete.

The proof of the Łoś-Vaught test and $\kappa$-categoricity of $T$ are both so easy that we can reformulate this argument as a short direct proof:

Suppose $V$ and $W$ are two infinite $F$-vector spaces. Let $\kappa>|T|$ be a cardinal. By Löwenheim-Skolem, there are $F$-vector spaces $V'$ and $W'$ of cardinality $\kappa$ such that $V\equiv V'$ and $W \equiv W'$. Pick a basis $B_{V'}$ for $V'$ and a basis $B_{W'}$ for $W'$. Since $|V| = |W| = \kappa$ and $|F|<\kappa$, by counting we must have $|B_{V'}| = |B_{W'}| = \kappa$. Any bijection $B_{V'}\to B_{W'}$ extends to an isomorphism $V'\cong W'$. Thus $V\equiv V'\equiv W'\equiv W$ so $V$ and $W$ are elementarily equivalent.

Now what about your attempted proof? You want to show that if $A\subseteq B$ are models of $T$, then $A\preceq B$. Showing this would actually establish that $T$ is model complete. Despite the name, this is a totally different condition: there are theories that are complete but not model complete (e.g. the complete theory of $(\mathbb{N},<)$) and theories that are model complete but not complete (e.g. the theory of algebraically closed fields of arbitrary characteristic).

However, if you show that $T$ is model complete (which it is) and additionally that there is a model $M\models T$ that embeds in any other model (you can use an $F$-vector space of dimension $1$ if $F$ is infinite or an $F$-vector space of dimension $\aleph_0$ if $F$ is finite), then you can conclude that $T$ is complete. This is because for any models $V$ and $W$, there are embeddings $f\colon M\to V$ and $g\colon M\to W$, and by model-completeness $f$ and $g$ are elementary embeddings. So $V\equiv M\equiv W$.

Ok, so let's try to follow your strategy to show that $T$ is model complete. We have infinite $F$-vector spaces $A\subseteq B$, and you want to show that $A\preceq B$ by applying the Tarski-Vaught test. It suffices to show that for every finite sequence $a_1,\dots,a_m\in A$ and $b\in B$, there exists an automorphism of $B$ that fixes the $a_1,\dots,a_m$ and maps $b$ into $A$.

If $A$ is infinite-dimensional (which it must be if $F$ is a finite field), we can do this. If $b\in A$ already, we're done, using the identity automorphism. So assume $b\notin A$. First pick a basis $B_0$ for the subspace $\text{Span}(\{a_1,\dots,a_m\})\subseteq A$. Extend $B_0$ to a basis $B_1$ for $A$. Since $b\notin A$, $B_2 = B_1\cup \{b\}$ is linearly independent. Extend $B_2$ to a basis for $B_3$ for $B$. Now since $B_0$ is finite and $A$ is infinite-dimensional, we can pick some $b'\in B_1\setminus B_0$. Let $\sigma\colon B_3\to B_3$ be any bijection which fixes $B_0$ and maps $b\mapsto b'$. Then $\sigma$ extends to an automorphism of $B$ which fixes $\text{Span}(\{a_1,\dots,a_m\})$ and maps $b$ to $b'\in B_1\subseteq A$.

Unfortunately, there's no hope of doing this if $A$ is finite-dimensional (which it might be, when $F$ is an infinite field). Consider the case when $a_1,\dots,a_m$ is a basis for $A$. Any automorphism of $B$ which fixes $a_1,\dots,a_m$ fixes all of $A$, so it cannot map an element of $B\setminus A$ into $A$.

A better approach to proving model-completeness is to prove directly that $T$ has quantifier elimination.

ContentLicense: CC BY-SA 4.0


问:
Questions about $\mathrm{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$

Consider the extension $\mathbb{Q} \subset\mathbb{Q} (\sqrt{2}, \sqrt{3})$. How many elements are there in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})?$ Describe all elements in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$. Find all subgroups of $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$ and their fixed fields.

I'm using Dummit & Foote Chapter 14 (Galois Theory) if it needs to be referenced. Chapter 13 was a breeze for me so I just feel like I need someone to explain how easy of a connection this is and it'll just click (or maybe it's not and I need to be told that, too). I really just need direction.

答: You know that any automorphism of $\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ fixing $\Bbb{Q}$ has to send $\sqrt{2}$ and $\sqrt{3}$ to another root of their respective minimal polynomials, so that you might want to look at those polynomials/roots and see/count the different (coherent) ways you can permute the roots. Once you do that, you should be able to determine what an arbitrary element does by seeing what it does to the elements of a basis of $\Bbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ (viewed as a $\Bbb Q$ vector space).</p>

ContentLicense: CC BY-SA 3.0


问:
What is the approach to solve the following combinatorial problem?
I would like to request a hint on how to approach the problem below:

The following are axioms for a (hitherto undefined) mathematical structure known as a bureaucracy. This consists of:

a set B of bureaucrats,

a set C of committees,

a relation S between B and C (read serves on), satisfying the following axioms:

(B1) Every bureaucrat serves on at least three different committees.

(B2) Every committee is served on by at least three different bureaucrats.

(B3) Given two distinct committees, exactly one bureaucrat serves on both.

(B4) Given two distinct bureaucrats, there is exactly one committe on which they both serve.

Prove from these axioms that if the number of bureaucrats is finite, so is the number of committees.

Prove that there are always at least seven bureaucrats in a bureaucracy, and find a bureaucracy with exactly seven bureaucrats.

答:
Each committee corresponds to a certain subset of bureaucrats. No two committees can have the exact same subset of bureaucrats (why?). Therefore, if there are $n$ bureaucrats, than there are at most $2^n$ committees.

Consider a particular bureaucrat. They must be on at least three committees, and these committees must each have at least three members. Add up all of the bureaucrats mentioned so far.

This is tricky, I do not know how to give a hint without giving it away. Look up the Fano plane.


问:
For a given real square matrix $A$ what is meant by $e^{kA}$ where $k$ is real.
For a given real square matrix $A$ what is meant by $e^{kA}$ where $k$ is real?

I've problem involving this notion and I wondered if $e^{kA}=(e^{ka_{ij}})$ where $A=(a_{ij}).$

答:
It's the matrix exponential. By definition,

$$e^A = \sum_{i=0}^{\infty}\frac{A^i}{i!}.$$

In your case, the matrix we're exponentiating is $kA$.

Your "intuition" that this can be computed component-wise is generally only true for the diagonal entries of diagonal matrices.


问:
Given a board of size $1 \times n$,in how many this board can be colored with the colors white,yellow,blue assuming the number of blue cells is even
Given a board of size $1 \times n$,in how many this board can be colored with the colors white,yellow and blue assuming the number of blue cells is an even number.

My try:

Given a board of size $1 \times n$,then there are two possibilities for the color of the first cell:

If the first cell is blue then the it's needed to find another cell to color it with the blue color ,this can be done in $n-1$ ways and for each one of these choices there are $b_{n-2}$ ways to color the remaining $1 \times (n-2)$ board with the colors yellow,white and blue assuming the number of blue cells is an even number.

If the first cell is yellow or white then the remaining $1 \times (n-1)$ board can be colored in $b_{n-1}$ ways assuming the restriction on the number of blue cells is satisfied.

Gathering these cases follows:

$$b_n=(n-1)b_{n-2}+2b_{n-1}$$

With $b_0=1$ and $b_1=2$ as initial conditions.

Calling $G(x)=\sum_{k\ge 0}^{} b_k x^k$,omitting some elementary calculations follows:

\begin{align*}
G(x)&=\frac{-1}{\left(n+1\right)^{2}x^{2}+2x-1}\\
&=\frac{-1}{\left(n+1\right)^{2}}\frac{1}{(x-a)(x-b)}\\
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\frac{1}{a-x}-\frac{1}{b-x}\right)\\
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\frac{1}{a}\frac{1}{1-\frac{x}{a}}-\frac{1}{b}\frac{1}{1-\frac{x}{b}}\right)\\
&
\end{align*}

From which we conclude:
\begin{align*}
b_n=\left[x^{n}\right]G\left(x\right)
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\left(\frac{1}{a}\right)^{n+1}-\left(\frac{1}{b}\right)^{n+1}\right)\\
\end{align*}

Where $$a=\frac{-1+\sqrt{n+2}}{n+1},b=\frac{-1-\sqrt{n+2}}{n+1}$$

But the closed form does not coincide with the recurrence that I've found.(I only want the question to be solved using recurrence relations.)

答:
Consider the expansion of $(w+y+b)^n$. Then each term represents a pattern on the board in a natural way.

Now let $w=y=1, b=-1$. Then each pattern with an even number of Blues counts $+1$ and  each pattern with an odd number of Blues counts $-1$.

Since $(1+1-1)^n=1$, there is one more 'even' pattern than 'odd'.

Hence there are $\frac {3^n+1}{2}$ 'even' and   $\frac {3^n-1}{2}$ 'odd'.


问:
Fibonacci variant with small exponent
The standard Fibonacci sequence defined by $F(n) := F(n-1) + F(n-2)$ has exponential asymptotic growth as stated e.g. here (with the approximate base being the golden ratio). What happens if the recurrence equation is altered by adding an exponent to $F(n-2)$? Intuitively, instead of 'add the previous number at every step', we get 'add a constant power of the previous number at every step'. I.e.,

$$ F(n) := F(n-1) + F(n-2)^c  \;\;\;\;\;\text{ where }\;\;\;\; c \in (0,1) $$

What is the asymptotic growth of such a sequence, assuming starting values are still 0 and 1? (Or if this problem is too difficult, what is the asymptotic growth for  $c = \frac12$ specifically?)

答:
This modified sequence with $0<c<1$ grows at a rate somewhere between linear and exponential as illustrated by the graph of some values on a log-y scale:

The smaller the exponent, the less the increment will change at one step compared to the previous. The asymptotic behavior should tend towards (nearly) linear.

All such sequences start $0,1,1,2,3$ - the exponent has no effect applied to $0$ or $1$ of course.

Title: Fibonacci variant with small exponent


问:
Need help with the general formula for the Taylor series of $e^x+\sin(x)$
A while ago, I asked about ways to derive the sigma notation for the infinite series of $e^x+sin(x)$.

$$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$$

In this thread,

Clive Newstead brilliantly gave me two formulas.

There, he said that the pattern of the series, regarding the coefficient of the numerators, is as followed: 1,2,1,0,1,2,1,0

He said that the pattern is "is periodic with period 4"

This comment still puzzles me. I have tried to validate his formula and it is true, they are all correct, but I don't know why he can come up with succinct representation of the pattern:

$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$

How does one realize that the power is related to $4k, 4k+1$ and $4k+2$ and $4k+3$?

This seems to relate to finding the general formula for a sequence or series of number, for example, trivially

$1, 3, 5, 7, ...$ can be represented as $2k+1$

$0, 2, 4, 6, ...$ can be represented as $2k$

with $k$ as the position of the term in the sequence.

So my thought is you to count the terms that have the numerator's coefficient as $1$, $2$ and $0$

The ones that have $1$ is $1^{st}$, $3^{sd}$, $5^{th}$, $7^{th}$, etc.

The ones that have $2$ is $2^{sd}$, $6^{th}$, $10^{th}$, $14^{th}$, etc.

The ones that have $0$ is $4^{th}$, $8^{th}$, $12^{th}$, $16^{th}$, etc.

So the difference between the positions of the terms that contain the coefficient $1$ is $2$

So the difference between the positions of the terms that contain the coefficient $2$ is $4$

So the difference between the positions of the terms that contain the coefficient $0$ is $4$

I find that the general term for terms that contain $1$ is $2k+1$

I find that the general term for terms that contain $2$ is $4k-2$

I find that the general term for terms that contain $0$ is $4k$

But I don't know how to derive these into $4k$, $4k+1$, $4k+2$, $4k+3$

I realize that this is similar to finding patterns of a sequence on an IQ test, but I don't know how to derive it. One my friend said I should look into Lagrange's interpolation, but I don't why this technique has anything to do with writing down the general terms for sequences and series.

Could you help me on this?

Methinks, this is a simple question. I am not good at writing down general term of a sequence yet.

答:
$e^x = 1 + x + \frac {x^2}{2} + \frac {x^3}{3!} + \cdots$

It seem pretty natural to write this as

$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}$

But we could look at pairs of terms.

$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{2n}}{2n!}+\frac {x^{2n+1}}{(2n+1)!}\right)$

Or, even triples of terms.

$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{3n}}{3n!}+\frac {x^{3n+1}}{(3n+1)!}+\frac {x^{3n+2}}{(3n+3)!}\right)$

And do the same thing with $\sin x$

$\sin x = \sum_\limits{n=0}^\infty (-1)^n \frac {x^{2n+1}}{(2n+1)!} = \sum_\limits{n=0}^\infty \left(\frac {x^{4n+1}}{(4n+1)!} - \frac {x^{4n+3}}{(4n+3)!}\right)$

Now find a representation of $e^x$ that plays nicely with our representation of $\sin x$ add them together and you have what you show above.


问:
Inverse of $\sum^n_{i=1} 1/x_i$ convex?
Is this function convex?
$$f(x)=\left(\sum^n_{i=1} 1/x_i \right)^{-1}$$
where $\operatorname{dom} f =\mathbb{R}^n_{++}$.

I having a hard time coming up with a clean Hessian from the partial derivatives.

答:
If I've not made any mistakes in my calculations: if $n \geq 2$, then

$$ \partial_1 f(x) = -\frac{1}{\left( \sum_{i=1}^n x_i^{-1} \right)^2} \cdot \left(- \frac{1}{x_1^2} \right) = \frac{1}{ \left(1 + \sum_{i \neq 1} \tfrac{x_1}{x_i} \right)^2 }, \\
\partial_1^2 f(x) = - \frac{2}{\left(1 + \sum_{i \neq 1} \tfrac{x_1}{x_i} \right)^3} \cdot \left( \sum_{i \neq 1} x_i^{-1} \right) < 0 \text{ on } \mathbb{R}_{++}^n
    $$
so all Hessians will have a negative entry in the upper left corner, and hence they can't be positive (semi-)definite. Hence $f$ is not convex for $n \geq 2$.

EDIT: It becomes even less mystical when considering restrictions of the function at specific values. Let $n = 2$ and $y = 1$, then $f(x,1) = \left(\tfrac{1}{x} +1 \right)^{-1}$ is easily plotted to not be convex.

Title: Inverse of $\sum^n_{i=1} 1/x_i$ convex?


问:
How do I calculate mean of a second variable with just just the mean and standard deviation of the first? (AP stats)
I don know really how to explain but I am currently working on some ap stats homework and I am incredibly confused. I will post the question below.

A certain box of cereal is labeled as 32 ounces, but boxes actually contain, on average, 33.1 ounces of cereal with
a standard deviation of 0.4 ounces. The distribution of weights is approximately Normal. Let X represent the weight
of cereal in a randomly selected box.
a) Let B represent the extra cereal contained in a box of cereal (i.e. the cereal above the advertised 32 ounces). Find
the mean and standard deviation of B.
b) Find the P(B > 2).

Honestly, this is the easiest question on the assignment and I am already lost so I don't have a great feeling about the rest lol.

答:
I've taken AP statistics so I can help you out.

As mentioned in the comments, the extra cereal is $B=X-32$ cuz you have X cereal and you find the extra. $B=X-32\sim N(1.1,.4^2)$ cuz you just shift it. The squiggly means it follows a normal distribution with mean 1.1 and variance .4^2.

So the answer for part a is 1.1 and .4, because standard deviation is the square root of variance.

For part b, you need to calculate a zscore. The zscore is $z=\displaystyle \frac {2-1.1}{.4}$. Calculate this and see what percent of the normal distribution is to the right of it. You will likely need to find the probability to the left of it and subtract that from $1$ by using a table provided in your book or elsewhere.


问:
How to pass this step using the squezee theorem?
$\lim\limits_{n \to \infty} \sqrt[n]{ a_1^n + a_2^n + ... + a_k^n }$, $a_i >0$, $i = 1,2,...,k$

Let $a = max(a_1, a_2, ..., a_k)$

Then we have

$a\le\sqrt[n]{ a_1^n + a_2^n + ... + a_k^n }\le a \sqrt[n]{k}$

How did we get $\sqrt[n]{k}$ i know that there are k elements under the root but i can't grasp how we get then n-th root ?

P.S. Thanks for all the help, i know i worded the question really bad its my first time here if someone could point out how to form these types of questions so i know later on.

答:
We know that $a_i\le a$ by definition of $a$. If you now take the sum $a_1^n+\cdots +a_k^n$ we find the same inequality $k$ times:
$$a_1^n+\cdots+a_k^n\le a^n+\cdots +a^n=ka^n$$
Taking the $n$-th root on both sides, we get
$$\sqrt[n]{a_1^n+\cdots+a_k^n}\le \sqrt[n]{a^n+\cdots +a^n}=\sqrt[n]{ka^n}=a\cdot \sqrt[n]{k}.$$


问:
Beginner Probability Question
This is my second day of stats & probability and I'm a little confused.

Here's the question:

The Abigail Construction Company is determining whether it should
  submit a bid for the construction of a new shopping mall. In the past,
  its main competitor, the Jared Construction Company, has submitted
  bids 60% of the time. When Jared does not submit a bid, the
  probability the Abigail will win the job is 70%. However, when Jared
  does submit a bid, the probability that the Abigail will win the job
  is only 40%. If Abigail wins a job, what is the probability that Jared
  submitted a bid?

What I've solved for thus far (it's not much):
J not - .2
  does - .7

I think the .6 in the beginning is irrelevant, but I could be wrong. How do I move forward?

P.S. If you know of a good place online to learn probability, please don't hesitate to post it!
Thanks!

答:
$A$ will represent Abigail submitting a vote, and $J$ will represent Jared submitting a vote.
We are given the following:

$P(J) = 0.6$

$P(A|J^c) = 0.7$

$P(A|J) = 0.4$

We want to find $P(J|A)$.

This becomes a relatively straightforward problem when you remember that $P(A|B) = \frac{P(A\cap B)}{P(B)}$.

Using this formula, we can expand the second equation:

$$P(A|J^c) = \frac{P(J^c\cap A)}{P(J^c)} = \frac{P(J^c\cap A)}{0.4} = 0.7$$

$$P(J^c\cap A) = 0.28$$

Now the third equation:

$$P(A|J) = \frac{P(J\cap A)}{P(J)} = \frac{P(J\cap A)}{0.6} = 0.4$$

$$P(J\cap A) = 0.24$$

I double checked, and you can prove that $P(J\cap A) + P(J^c\cap A) = P(A)$, and so we can find $P(A)$ to be equal to $0.24 + 0.28 = 0.52$. We then can expand $P(J|A)$ to find your answer:

$$P(J|A)$$

$$= \frac{P(J \cap A)}{P(A)}$$

$$= \frac{0.24}{0.52}$$

Which is approximately $0.46153846153846153846153846153846$.

Title: Beginner Probability Question


问:
Complex Integral proof explanation

Compute $$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$

DonAntonio gave a well-constructed answer here but I have a few confusions on his answer as I am  new to complex integral:

How did we derive $\text{Res}_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\frac{e^{-\alpha}}{2i\cdot 3i\cdot (-i)}=-\frac{e^{-\alpha}}6i\;\;$? (the step when solving the limit)

What is the purpose of using Jordan's Lemma to get
$\lim_{R\to\infty}\int\limits_{\gamma_R}f(z)dz=0$? The result equalling to 0 does not seem to be applied to the next equation.

Should the final answer be $\frac16\left(2e^{-\alpha}-e^{-2\alpha}\right)\pi$ or without $\pi$?

Thanks in advance.

答:
Because\begin{align}\lim_{z\to i}(z-i)f(z)&=\lim_{z\to i}(z-i)\frac{e^{i\alpha z}}{(z+i)(z-i)(z^2+4)}\\&=\lim_{z\to i}\frac{e^{i\alpha z}}{(z+i)(z^2+4)}\\&=\frac{e^{-\alpha}}{(2i)\times3}\\&=-\frac{e^{-\alpha}}6i\end{align}

You need the fact that $\lim_{R\to\infty}\int_{\gamma_R}f(z)\,\mathrm dz=0$ to deduce that$$\int_{-\infty}^\infty f(x)\,\mathrm dx=\lim_{R\to\infty}\int_{C_R}f(z)\,\mathrm dz.$$

It's $2\pi i$ times the sum of the residues at $i$ and at $2i$. So, yes, $\pi$ should be there.

Title: Complex Integral proof explanation


问:
Why is variance formula x^2f(x), not x^2f(x^2)
The variance formula is $E[X^2]-E[X]^2$ and in the continuous case this means $\int x^2f(x)dx-\mu^2$. This means that $E[X^2]=\int x^2f(x)dx$ which is confusing to me because $E[X]=\int xf(x)$, so I would expect $E[X^2]$ to be $\int x^2f(x^2)$. What am I missing here?

答:
$f(x)$ is the so-called PDF (Probability Density Function). However the variable ($X$) changes, the probability density won't change (otherwise you will get another distribution).

Edit: By the way, as the comment points out, the variance formula is $E[X^2]-(E[X])^2$, not the way you wrote it.


问:
Minimal steps to reach a natural number by a signed arithmetic progression
$F:=\big\{f\,\big|\,\text{function }f: \mathbf N\rightarrow \{1,2\}\big\}$.
\begin{align}
&\min_{f\in F,\,k\in\mathbf N} k, \\
&\sum_{i=1}^k (-1)^{f(i)}i = n \in\mathbf N.
\end{align}

Is there an analytical or an asymptotic solution to this problem for general $n$?

For a specifically given $n$, what would a good algorithms to solve this problem? The dynamic programming?

答:
Continuing on @inavda's answer, for a given $n$ we find the minimum number of terms, which is the index of the smallest triangular number not smaller than $n$, and with the same parity. As the parities of the triangular numbers follow the pattern $e,o,o,e,e,o,o,e,e,\cdots$, we can find the smallest triangular number not smaller than $n$, and increment once or twice as needed to reach the desired parity.

Then we repeat this operation recursively on the half of the residue $\dfrac{k(k+1)}2-n$ to obtain the desired correction.

Example: $n=37$

$$37\le T_9=45, \text{half residue }4\to+++++++++$$

$$4\le T_3=6, \text{half residue }1\to---+++++$$

$$1=T_1\to+--++++++$$

The smallest triangular number is obtained by

$$\frac{k(k+1)}2\ge n$$ or

$$k\ge\frac{\sqrt{8n+1}-1}2,$$

$$k=\left\lceil\frac{\sqrt{8n+1}-1}2\right\rceil.$$

Notice that the procedure fails for $n=2$, as the smallest triangular number is $T_3=6$ and the half residue is again $2$. Also for $n=5$, $T_5=5$ and the half residue is $5$. These are the only fixed points. We solve these with $2=1-2+3$ and $5=1-2-3+4+5$.

Notice that the residue is on the order $O(\sqrt n)$, so that the successive residues decrease very quickly, and the recursion depth is of order $O(\log\log n)$.

Update:

Checking the example, it turns out that this procedure is wrong.


答:
Remember that for the series centered around the point b, the coefficient $a_{n} = \frac{f^{(n)}(b)}{n!}$. So if $f^{(4)}(x) = f(x)$ then  $f^{(n)}(b)$ is periodic with not necessarily prime period 4.

This is why you can be sure that the $f^{(n)}(b)$ is periodic with period 4 in your case.

Let $f(x) = e^{x}$, then $f'(x) = e^{x}$, and so continuing to take derivatives $f^{(4)}(x)=e^{x}$.

Let $g(x) = sin(x)$ then $g''(x)=-sin(x)$ and so $g^{(4)}(x) = sin(x)$

So for $h(x)=f(x) + g(x)$ we have $$h^{(4)}(x) = (f(x)+g(x))'''' = f^{(4)}(x) + g^{(4)}(x) = f(x) + g(x) = h(x)$$.

In this way you can see that the $f^{(n)}(b)$ has the period 4 structure.

Title: Need help with the general formula for the Taylor series of $e^x+\sin(x)$

Body:
A while ago, I asked about ways to derive the sigma notation for the infinite series of $e^x+sin(x)$.

$$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$$

In this thread,

Clive Newstead brilliantly gave me two formulas.

There, he said that the pattern of the series, regarding the coefficient of the numerators, is as followed: 1,2,1,0,1,2,1,0

He said that the pattern is "is periodic with period 4"

This comment still puzzles me. I have tried to validate his formula and it is true, they are all correct, but I don't know why he can come up with succinct representation of the pattern:

$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$

How does one realize that the power is related to $4k, 4k+1$ and $4k+2$ and $4k+3$?

This seems to relate to finding the general formula for a sequence or series of number, for example, trivially

$1, 3, 5, 7, ...$ can be represented as $2k+1$

$0, 2, 4, 6, ...$ can be represented as $2k$

with $k$ as the position of the term in the sequence.

So my thought is you to count the terms that have the numerator's coefficient as $1$, $2$ and $0$

The ones that have $1$ is $1^{st}$, $3^{sd}$, $5^{th}$, $7^{th}$, etc.

The ones that have $2$ is $2^{sd}$, $6^{th}$, $10^{th}$, $14^{th}$, etc.

The ones that have $0$ is $4^{th}$, $8^{th}$, $12^{th}$, $16^{th}$, etc.

So the difference between the positions of the terms that contain the coefficient $1$ is $2$

So the difference between the positions of the terms that contain the coefficient $2$ is $4$

So the difference between the positions of the terms that contain the coefficient $0$ is $4$

I find that the general term for terms that contain $1$ is $2k+1$

I find that the general term for terms that contain $2$ is $4k-2$

I find that the general term for terms that contain $0$ is $4k$

But I don't know how to derive these into $4k$, $4k+1$, $4k+2$, $4k+3$

I realize that this is similar to finding patterns of a sequence on an IQ test, but I don't know how to derive it. One my friend said I should look into Lagrange's interpolation, but I don't why this technique has anything to do with writing down the general terms for sequences and series.

Could you help me on this?

Methinks, this is a simple question. I am not good at writing down general term of a sequence yet.


问:
dividing two complex numbers
I'm trying to divide two complex numbers, namely:$$z = 2\left(\cos\left(\frac{3\pi}{11}\right) + i\sin\left(\frac{3\pi}{11}\right)\right)$$ and $$w = \left(\cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right)\right)$$

I'm trying to figure out the following:  $\frac{\overline z}{6w}$.

The way I've done it is figuring the conjugate of $z$ which is $\overline z = 2\left(\cos\left(\frac{3\pi}{11}\right) - i\sin\left(\frac{3\pi}{11}\right)\right)$ if I'm not mistaken.
But now I'm sort of stuck, I tried to divide it by simply dividing $\frac{2}{6}$ and substracting the argument from the $\cos$ and $\sin$ of $\overline z$ but I didn't get anywhere for figuring out $e^{\theta i}$ which would be (according to the answer)  $\frac{2}{6} = r$ and $e^{\theta i} = \frac{73\pi}{55}$.

I would appreciate some help, thanks in advance!

答:
Using Euler's formula, we can write $$z=2e^{\frac{3\pi i}{11}}, w=e^{\frac{2\pi i}{5}}$$ So $$\overline z=2e^{\frac{-3\pi i}{11}}$$ and then we can simplify it to $$\frac{\overline z}{6w}=\frac{2e^{\frac{-3\pi i}{11}}}{6e^{\frac{2\pi i}{5}}}=\frac{e^{i\pi(\frac{-3}{11}-\frac{2}{5})}}{3}=\frac{e^{i\pi(\frac{-37}{55})}}{3}$$

Adding $2\pi$ to the argument yields $\frac{e^{i\pi\frac{73}{55}}}{3}$, so $r=\frac{1}{3}$ and $\theta=\frac{73\pi}{55}$

Title: dividing two complex numbers


问:
When is the independence relation induced by the amalgamation property not monotonic?
Let's consider a Fraïssé class $K$ whose amalgamation property is witnessed functionally: for a diagram $A \leftarrow B \rightarrow C$ of embeddings in $K$, the structure $A\otimes_B C$ is an amalgam of the diagram. Moreover, let's assume that automorphisms on $A$ and $C$ fixing $B$ pointwise extend onto $A\otimes_B C$.  (I don't know if the second part is strictly relevant in my question, but I'm following Tent and Ziegler's On the isometry group of the Urysohn space.  It's a natural requirement, anyway.)

Let $M$ be the limit of $K$.  We can define a ternary relation $A \downarrow_B C$ on the finite subsets of $B$ by letting $A \downarrow_B C$ hold if and only if there is an isomorphism $ABC \to AB \otimes_B BC$ respecting the relevant embeddings (I denote unions by justapositions.  Also, this definition may not make sense in a non-relational language, but it can be modified so it may make sense.)

The said work by Tent and Ziegler considers several properties of the ternary relation and lists those that hold under the aforementioned most general setting.  A property that is excluded from that list is Monotonicity: if $A \downarrow_B CD$, then $A \downarrow_B C$ and $A \downarrow_{BC} D$.

After thinking about this a bit, it seems to me that for many, if not all, natural class $K$ and $\otimes$, the induced ternary relation does satisfy Monotonicity.  For what $(K, \otimes)$ does $\downarrow$ not satisfy monotonicity?  Ideally I would like to have a natural (counteexample.

答:
Certainly it's easy to give ad hoc counterexamples.

For example, let $K$ be the class of finite graphs. If $|B|>1$, define $A\otimes_B C$ to be the free amalgamation (the disjoint union of $A$ and $C$ over $B$, with no edges between vertices of $A\setminus B$ and $C\setminus B$). If $|B|\leq 1$, define $A\otimes_B C$ to be the co-free amalgamation (the disjoint union of $A$ and $C$ over $B$, with an edge between each vertex in $A\setminus B$ and each vertex in $C\setminus B$). To see that this doesn't satisfy monotonicity, consider the graph $\{a,b,c,d\}$ with edges $aEc$ and $aEd$. Then $a\downarrow_b cd$, but $a\not\downarrow_{bc} d$.

Of course, the above definition of $\otimes$ doesn't seem "canonical", because of the distinction between the cases $|B|\leq 1$ and $|B|>1$. I agree that  "canonical" classes $(K,\otimes)$ will tend to satisfy monotonicity. But sometimes there's no way to make "uniform" decisions about how to amalgamate.

A natural counterexample is actually given by one of the key classes in the paper you're reading: $K = $ finite metric spaces with all distances in $\mathbb{Q}$. When $B$ is non-empty, we can define the "free amalgamation" $A\otimes_B C$ by setting, for each pair of points $a\in A\setminus B$ and $c\in C\setminus B$, $$d(a,c) = \min_{b\in B} \,(d(a,b)+d(b,c)).$$ This puts $a$ and $c$ as far apart as possible without violating the triangle inequality. But the formula doesn't work when $B$ is empty, since we cannot put $a$ and $c$ at distance $\infty$. So if we want to extend the definition down to amalgamation over the empty metric space, we need to make some choice of how to define $A\otimes_{\varnothing} C$, e.g. define $D= \max(\max_{a,a'\in A} d(a,a'),\max_{c,c'\in C}d(c,c'))$ and set $d(a,c) = D$ for all $a\in A$ and $c\in C$. But this definition breaks monotonicity.

In fact, there is no way to extend $\otimes$ down to amalgamation over $\varnothing$ while preserving monotonicity. Suppose for contradiction that we have a "good" definition of $\otimes$. Let $A = \{a\}$, $B = \{b\}$, and $C = \{b,c\}$ with $d(b,c) = 1$. Consider $A\otimes_\varnothing C$, i.e. assume $A\downarrow_\varnothing BC$. Since $C$ has an automorphism swapping $b$ and $c$, we must have $d(a,b) = d(a,c)$. But if $A\downarrow_B C$, then we would have $d(a,c) = d(a,b)+d(b,c) = d(a,b)+1$.


问:
Understanding a Hermitian Matrix Derivation in a Mechanics Spring Setting
On pg. 46 of Principles of Quantum Mechanics:

Question: What is going on with the red highlighted portion of (1.8.34)?

It seems like (1.8.34) follows completely from (1.8.33) (i.e. subtracting one side from the other).  Isn't this true? So How does the operator $\frac{d^2}{dt^2}-\Omega$ have anything to do with deriving (1.8.34)?

EDIT: The proof later goes on to state that

How does this follow?

答:
Regarding the first question: begin with (1.8.32),
$$
|x(t) \rangle = |\text{I}\rangle x_\text{I}(t) + |\text{II}\rangle x_{\text{II}}(t).
$$
Applying $(\frac{d^2}{dt^2} - \Omega)$ to the left side gives us
$$
(\frac{d^2}{dt^2} - \Omega)|x(t) \rangle = |\ddot x(t) \rangle - \Omega |x(t)\rangle = |0 \rangle.
$$
Applying the operator to the right side gives us
\begin{align}
\left(\frac{d^2}{dt^2} - \Omega \right)[|\text{I}\rangle x_\text{I}(t) &+ |\text{II}\rangle x_{\text{II}}(t)] =
\left(\frac{d^2}{dt^2} - \Omega \right)|\text{I}\rangle x_I(t)
+ \left(\frac{d^2}{dt^2} - \Omega \right)|\text{II}\rangle x_{\text{II}}(t)
\\ & =
|\text{I}\rangle \ddot x_{\text{I}}(t) - \Omega |\text{I}\rangle x_\text{I}(t) + |\text{II}\rangle \ddot x_{\text{II}}(t) - \Omega |I\rangle x_{\text{II}}(t)
\\ & =
|I\rangle \ddot x_{I}(t) - - \omega_I^2|I\rangle x_I(t) + |\text{II}\rangle \ddot x_{\text{II}}(t) - - \omega_{\text{II}}^2 |I\rangle x_{\text{II}}(t)
\\ & =
|\text{I}\rangle (\ddot x_\text{I}(t) + \omega_\text{I}^2 x_\text{I}(t))
+ |\text{II}\rangle (\ddot x_{\text{II}}(t) + \omega_{II}^2 x_{\text{II}}(t)),
\end{align}
which indeed gives us the right-hand side of (1.8.34).

Regarding the second question: (1.8.31a) tells us that the "ket" $|\text{I}\rangle$ corresponds to $\frac 1{\sqrt{2}}\left( \begin{smallmatrix} 1\\1 \end{smallmatrix}\right)$ in the $1,2$ basis. To get the corresponding "bra", take the conjugate-transpose:
$$
\langle \text{I} | \leftrightarrow \frac 1{\sqrt{2}}\pmatrix{1 & 1}.
$$
Plugging that in yields
$$
\langle \text{I} |x(0) \rangle = \frac 1{\sqrt{2}} \pmatrix{1 & 1} \pmatrix{x_1(0)\\x_2(0)},
$$
and matrix multiplication yields $\frac{x_1(0) + x_2(0)}{\sqrt{2}}$. The second line can be derived similarly.


问:
Composition law for limits of vector valued multivariable function
Let $A\subseteq\mathbb{R}^n$ and $B\subseteq\mathbb{R}^{m}$ be two non empty sets, and

$\bullet \ \ \ f:A\to B; \ \ \ g: B\to\mathbb{R^{s}}$

Suppose that $a\in\mathbb{R}^{n}$ is a limit point for $A$, $l$ a limit point for $B$, and

$\bullet \ \ \ \displaystyle\lim_{x\to a}f(x)=l\in\mathbb{R}^n \ \ \ \mbox{and} \ \ \lim_{y\to l}g(y)=m\in\mathbb{R}^{s}$

What are the hypotheses that guarantee the following equality

$$\displaystyle\lim_{x\to a}g(f(x))=m\, ?$$

Thanks.

答: I guess you are searching for minimum hypothesis. Continuity is more than enough to let this work. Can we actually easen it? Not really, it also depends on the hypotheses you have over $A$ and $B$. In general, if $f$ is not continous then such equality does not need to hold. If for instance $A$ is convex and also $f$ is convex, I think you can make it work but not even these last ones are enough.


问:
What are some applications of apeirogons, apeirohedra, or n-apeirotopes?
I was initially reading about the Apeirion concept in philosophy when I came across this mathematical concept of apeirogons and its higher-dimensional analogues. Being a layman in mathematics, the definition of an apeirogon doesn't seem of much use. I imagine that they must have some applications in higher-level mathematics, but I couldn't come up with any ideas.

What applications of these infinite-sided polygons, polyhedra, and/or n-polytopes do you know of?

答:
They're one case of the classification of abstract polygons (or polyhedra, or polytopes, for the higher-dimensional versions). It's not a very interesting case, but we include it so that the classification is complete.

In hyperbolic space, apeirogons can be more interesting, because they can be closer to our usual definition of a polygon: sides of the same length with the same (nontrivial!) angle at every corner. For example, here is a tiling of the hyperbolic plane by infinitely many apeirogons, which is no different in spirit from tiling the Euclidean plane by infinitely many hexagons.


答:
When we look at expected values we multiply each outcome times the probability of that outcome and sum or integrate over all possibilities.  We can use the frequency distribution $f(x)$ to find the likelihood of all possible values of $x^2$

Lets look at a simple example:

$f(x) = \begin{cases} \frac 12 & x = 0\\\frac 13 & x = 1\\\frac 16 & x = 2\end{cases}$

That is $X = 0$ with probability $\frac 12, 1$ with probability $\frac {1}{3}, 2$ with probability $\frac 16.$

What is the probability $X^2 = 4$?  It is the same as the probability $X = 2$

$E[X^2] = \frac 12 (0^2) + \frac 13 (1^2) + \frac {1}{6}(2^2)$

This example is with discrete probabilities.  As we move from discrete probabilities to continuous probabilities summation is replaced with integration.

Title: Why is variance formula x^2f(x), not x^2f(x^2)

Body: The variance formula is $E[X^2]-E[X]^2$ and in the continuous case this means $\int x^2f(x)dx-\mu^2$. This means that $E[X^2]=\int x^2f(x)dx$ which is confusing to me because $E[X]=\int xf(x)$, so I would expect $E[X^2]$ to be $\int x^2f(x^2)$. What am I missing here?


问:
Why does $\cos(n \pi)=1$ if $n$ is even, but $\cos(n\pi) = -1$ if $n$ is odd?
I am taking Calculus 1 after graduating college several years ago. I have forgotten a good amount of info and am having difficulty searching info for this on my own.

Here is what is causing me confusion:

$\cos{n\pi}=1$ (if $n$ is even)

$\cos{n\pi}=-1$ (if $n$ is odd)

I am seeking an explanation as to why this is. From what I've gathered so far it has to do with radians and trigonometric functions, but the info is very unclear to me.

答:
You have to notice that, being $k\in\mathbb{Z}$, then
$$\cos (x) = \cos (x+2k\pi),$$
(this is because the $\cos x$ fuction has period $2\pi$) so from this it's clear that $\cos(n\pi)=\cos(n\pi+2k\pi)=\cos((n+2k)\pi)$ for all $n,k\in\mathbb{Z}$, and this means that all even values of $n$ will give you the same value (and all odd number will aso give the same value), since $n+2k$ is even if and only if $n$ is even (respectively for $n$ odd). So you conclude that
$$\text{$n$ even: }\ \cos(n\pi)=\cos(0\cdot\pi)=\cos(0)=1$$
$$\text{$n$ odd: }\ \cos(n\pi)=\cos(1\cdot\pi)=\cos(1)=-1$$


答:
The mod $4$ periodicity can perhaps be most easily understood using the formula $e^{ix}=\cos x+i\sin x$, so that $\sin x={e^{ix}-e^{-ix}\over2i}$, which gives

$$e^x+\sin x={1\over2i}\sum_{n=0}^\infty{(2i+i^n-(-i)^n)x^n\over n!}$$

Since $i^4=1$, the coefficients $2i+i^n-(-i)^n$ cycle through the values

$$\begin{align}
2i+i^0-(-i)^0&=2i+1-1=2i\\
2i+i^1-(-i)^1&=2i+i-(-i)=4i\\
2i+i^2-(-i)^2&=2i-1-(-1)=2i\\
2i+i^3-(-i)^3&=2i-i-i=0
\end{align}$$


问:
Period lattice of entire function

Let $f$ be an entire function, and let $\alpha, \beta \in \mathbb{C}$ such that
$$f(z+\alpha) = f(z) \quad f(z+\beta) = f(z) \quad \forall z \in \mathbb{C}$$
Then if $\alpha$ and $\beta$ are linearly independent over $\mathbb{Q}$, $\{n\alpha +m\beta: n,m\in \mathbb{Z}\}$ is dense in $\mathbb{R}$. Prove that $f$ is constant.

I know I have to come up with some kind of sequence converging to any point on $\mathbb{R}$ but I dont know how to do it, and what it has to do with the fact that $\alpha $ and $\beta$ are linearly independent. The second part is easier, I can just use Liouville's theorem, as f is bounded over the real line because $f(0) = f(n \alpha)$ and f is bounded at 0, but I don't know if I can prove that f is bounded just by looking at if on the real line.

Any help will be appreciated.

答:
Case 1 $\alpha, \beta$ are linearly dependent over $\mathbb R$. This means $\beta=s\alpha$ for some $s \in \mathbb R$.

Show first that

$$ \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}=0$$

Indeed, assume by contradiction that
$$
a= \mbox{ inf } \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}>0
$$

First show that $a \in\mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$.

Note that otherwise you could find two elements $b,c \in \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$ with
$$
a<b<c <\frac{3a}{2} \,.
$$
But then $c-b \in\mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$ and $c-b <a$ contradicting the definition of $\mbox{inf}$.

Next, since $a \in \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\} $ it follows
that
$$
\mathbb Z a \alpha \subseteq \{n\alpha +m\beta: n,m\in \mathbb{Z}\}  \,.
$$

Complete the contradiction by observing that the above inclusion must be equality, as otherwise you can create an element $d \in\mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$ with
$$
0 < d< a
$$
which is not possible.

You can now reach the contradiction:
$$
\mathbb Z a \alpha = \{n\alpha +m\beta: n,m\in \mathbb{Z}\}
$$
gives that
$$
\alpha =ka \alpha \\
\beta=la \alpha$$
The first relation gives $a \in \mathbb Q$, and then the second contradicts the linear independence over $\mathbb Q$.

Next, show that $\{n\alpha +m\beta: n,m\in \mathbb{Z}\}$ If $(e,f)$ is dense in the line $\mathbb R \alpha$ . Indeed, if $(e,f) \subseteq \mathbb R$ is any interval, by the above there exists some $m,n$

$$
0<\frac{n\alpha +m\beta}{\alpha}< f-e
$$
Pick the smallest $k \in \mathbb Z$ such that $k\frac{n\alpha +m\beta}{\alpha}>e$ and show that $k(n \alpha+m\beta) \in (e\alpha ,f \alpha)$.

FInally, since $f$ is constant on $\{ n\alpha +m\beta : m,n \in \mathbb Z\}$ and continuous, it is constant on its closure which is $\mathbb R \alpha$. The claim follows.

Case 2  $\alpha, \beta$ are linearly independent over $\mathbb R$.

In this case you can show that there exists some $R>0$ such that for each $z \in \mathbb C$ there exists $m,n \in \mathbb Z$ such that
$$
|z-m\alpha-n \beta| \leq R \,.
$$

Since $f$ is bounded on $\{ w : |w| \leq R \}$, the above implies that $f$ is bounded on $\mathbb C$.

Title: Period lattice of entire function


答:
We solve the problem informally, since it is early in the course. This will be done in two slightly different ways. Later in the course, one might write out a formal conditional probability argument.

Imagine construction project opportunities occurring a large number of times, say $1000$. In about $600$ of these times, Jared submits a bid, and about $400$ times it doesn't.

Ourt of the $400$ times that Jared does not submit a bid, Abigail wins about $70\%$ of the time, so $280$ times.

Out of the $600$ times Jared submits a bid, Abigail wins about $40\%$ of the time (we corrected a presumed typo). So this is $240$ times.

Thus Abigail wins about $520$ times. In $240$ of these, Jared had a bid. So our required probability should be $\frac{240}{520}$.

Another way: (Well, it really is the same way.) Draw a tree diagram. The first branching is Jared does not submit (write a $0.6$), Jared submits (write a $0.4$). From each of the two nodes, draw two branches, one for Abigail wins, the other for she loses. Write the appropriate probabilities along these $4$ branches.

Look at the two paths that lead to an Abigail win. Multiplying probabilities along the branches, we find that the probability Abigail wins is $(0.4)(0.7)+(0.6)(0.4)$. (There are two paths along which Abigail can win.). This probability turns out to be $0.52$.

Now calculate the probability of the path that leads to an Abigail win and a Jared bid. This path has probability $(0.6)(0.4)=0.24$.

It follows that our conditional probability is $\dfrac{0.24}{0.52}$.

Title: Beginner Probability Question

Body:
This is my second day of stats & probability and I'm a little confused.

Here's the question:

The Abigail Construction Company is determining whether it should
  submit a bid for the construction of a new shopping mall. In the past,
  its main competitor, the Jared Construction Company, has submitted
  bids 60% of the time. When Jared does not submit a bid, the
  probability the Abigail will win the job is 70%. However, when Jared
  does submit a bid, the probability that the Abigail will win the job
  is only 40%. If Abigail wins a job, what is the probability that Jared
  submitted a bid?

What I've solved for thus far (it's not much):
J not - .2
  does - .7

I think the .6 in the beginning is irrelevant, but I could be wrong. How do I move forward?

P.S. If you know of a good place online to learn probability, please don't hesitate to post it!
Thanks!


问:
Find all positive integers $x$ such that $x(x+2021)$ is a perfect square
I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you

I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$

答:
For $x(x+2021)$ to be a square there has to be integers $a,b,c$ such that $$x=ab^2,x+2021=ac^2.$$
Subtracting these two equations we obtain $a(c-b)(c+b)=2021=43\times 47$. Since $43$ and $47$ are primes the only possibilities for $a, c-b$ and $c+b$ are
$$(1,43,47),(1,1,2021),(43,1,47),(47,1,43).$$

These give, in turn, $x=ab^2\in \{ 4,1020100,22747,20727\} .$


答:
To compute the expected value of $X^2$, using the definition of expected value rather than the LOTUS rule, what you would need to do is first find a PDF $g$ for the random variable $X^2$. Then, by the definition of expected value, we would have $E(X^2) = \int x g(x) \, dx$. Now the question is, what is $g$? That is not obvious!

The LOTUS rule saves us from the necessity of finding $g$. According to the LOTUS rule (that is, the "law of the unconscious statistician"), we have
$$
E(X^2) = \int x^2 f(x) \, dx
$$
where $f$ is a PDF for $X$.


问:
How to calculate the Rotation Matrix from other known values?
Dear Math Stackexchange community, how can I compute $R$ rotation matrix in this equation ($P=Rp+t$)? (Please see the figure. More detailed explanation is under the figure.)

The equation is for the transformation of $p=p_1,p_2,p_3$ points. $p$ is in Oxyz (local) reference frame.     $p$ consists of 3 points of $p_1,p_2,p_3$. And $p_1,p_2,p_3$ each has 3 coordinates of $x,y,z$. Therefore p is 3x3 matrix. $p=\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}$

$t$ is the distance between Oxyz(local) and OXYZ(global) reference frame.  $t=\begin{bmatrix} t_x \\ t_y \\ t_z \end{bmatrix}$. To be able to do matrix operations, we can take $t$ as $t=\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

$P$ is the coordinates of p_1, p_2, and p_3 in OXYZ global reference frame. $P$ is a 3x3 matrix.
$P=\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix}$

$R$ is the (orthogonal) rotation matrix.
$R=\begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$

Therefore, all in all:
$P=Rp+t$

$\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix} = \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}.\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}+\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

We can derive the following form of the equation of  $P=Rp+t$

$(P-t)=R.p$

$(P-t).p^{-1}=R$

$p$ is an arbitrary matrix, and not always be taken of its inverse. I can't take the inverse of $p$, if last raw of $p$ is zero.

Let's look at an example to make it clear:
$P=Rp+t$

$p_1=(1,1,0)$ , $p_2=(2,4,0)$ , $p_3=(5,10,0)$ is the coordinates of three points. So in matrix form:
$p=\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}$

$t=\begin{bmatrix} 0 \\ 0 \\ 50 \end{bmatrix}$

$\alpha=10^\circ $ (rotation around the x axis)

$\beta=15^\circ $ (rotation around the y axis)

$\gamma=20^\circ $ (rotation around the z axis)

$R_x=\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) \\ 0 & sin(\alpha) & cos(\alpha) \end{bmatrix}$

$R_y=\begin{bmatrix} cos(\beta) & 0 & sin(\beta) \\ 0 & 1 & 0 \\ -sin(\beta) & 0 & cos(\beta) \end{bmatrix}$

$R_x=\begin{bmatrix} cos(\gamma) & -sin(\gamma) & 0 \\ sin(\gamma) & cos(\gamma) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$R=R_z.R_y.R_x$

Therefore, for $\alpha=10^\circ $, $\beta=15^\circ $, $\gamma=20^\circ $, the rotation matrix is :

$R=\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$

$P=R.p+t$

$P=$$\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$$.\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$  $+\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

$P=\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}$

$p_1=(0.6131,1.2712,49.90)$,$p_2=(0.6370,4.4239,50.1533)$,$p_3=(1.5925,11.0597,50.3832)$.

Now let's take opposite example with the same values to find $R$, with the values of $P,p,t$ as known.

$P=R.p+t$

$\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}=R.$ $\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}+$ $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

Could you find $R$ here in the above equation?

To repeat the formulation:

$P=Rp+t$

$(P-t)=Rp$

$(P-t).p^{-1}=R.p.p^{-1}$

$(P-t).p^{-1}=R$

For the inverse of $p$ (where $p$ is $\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$ ), I can't take the inverse of $p$.

Here I am stuck. It is interesting for me, because for the condition that $R,p,t$ is known, $P$ can be calculated(9 unknown elements in 3x3 matrix of $P$). However, for the condition that $P,p,t$ is known, I am having problem to compute $R$ (9 unknown elements in 3x3 matrix of $R$). I should be able to compute the equivalent $R$ matrix for that inverse problem.

Do you have any idea about how to find $R$ rotation matrix? Any suggestion would be  appreciated.

Thanks in advance.

答:
Firstly, I think the extra gymnastics with $t$ is a distraction, so I will state the problem as:

$\mathbf{Q}=\left[\mathbf{q}_1\,\dots,\mathbf{q}_k\right]=\mathbf{P}-\mathbf{t}=\mathbf{R}.\boldsymbol{\mathcal{P}}=\mathbf{R}.\left[\mathbf{p}_1\,\dots\,\mathbf{p}_k\right]$

Where $\mathbf{Q}$ is some $3\times k$ real-valued matrix, which we can also view as $k$ column vectors $\mathbf{q}_{1,\dots k}$. $\mathbf{R}$ is a real-valued rotation matrix, which means it is orthogonal ($\mathbf{R}^T\mathbf{R}=\mathbf{Id}=identity$) and has determinant of $det\left[\mathbf{R}\right]=+1$. Finally, matrix $\boldsymbol{\mathcal{P}}$ is made up from 3d real-valued column vectors $\mathbf{p}_{1,\dots k}$. The aim is to determine the rotation matrix.

Here, for simplicity, I will work with linearly independent sets of $\mathbf{p}$-vectors, so $k=\mbox{rank}\left(\mathbf{Q}\right)=\mbox{rank}\left(\boldsymbol{\mathcal{P}}\right)$

$k=\mbox{rank}\boldsymbol{\mathcal{P}}=3$

If you can pick three linearly independent vectors $\mathbf{p}_{1,2,3}$ you can have a unique solution for $\mathbf{R}$ given $\mathbf{Q}$ (by inverting $\boldsymbol{\mathcal{P}}$). If you cannot such unique solution does not exist.

Use rank function to check whether you have independent $\mathbf{p}$-vectors

$k=\mbox{rank}\boldsymbol{\mathcal{P}}=2$

If you only have two linearly independent vectors, I think the best you can in principle get two possible solutions, of which both fit the data. In some special cases you will still be able to get the unique solution.

Use Gram-Schmidt process to extract two orthonormal vectors $\mathbf{\hat{n}}_{1,2}$ which can be expressed as $\mathbf{\hat{n}}_{1,2}=\alpha_{1,2}\mathbf{p}_1+\beta_{1,2}\mathbf{p}_2$.

There will be two further orthonormal vectors $\mathbf{\hat{q}}_{1,2}=\alpha_{1,2}\mathbf{q}_1+\beta_{1,2}\mathbf{q}_2$.

Now try to find $\psi$ that minimizes: $s=\left|\left(\cos\psi\,\mathbf{\hat{q}_1+\sin\psi\,\mathbf{\hat{q}}_2}\right)-\left(\cos\psi\,\mathbf{\hat{n}_1+\sin\psi\,\mathbf{\hat{n}}_2}\right)\right|$, this will give you the projection of the axis around which matrix $\mathbf{R}$ causes the rotation. Let the actual rotation axis be $\boldsymbol{{\hat{\omega}}}$, and its projection into plane spanned by $\mathbf{\hat{n}}_{1,2}$ be $\boldsymbol{{\omega}}_\mathcal{P}$. The logic here is simple if $\boldsymbol{\hat{\omega}}=\left(\cos\psi\,\mathbf{\hat{n}_1+\sin\psi\,\mathbf{\hat{n}}_2}\right)$ then rotating such vector will not change it at all, by definition.

Two interesting cases here are when $s=const$ for all $\psi$. Then your rotation matrix's axis is perpendicular to the plane spanned by $\mathbf{\hat{n}}_{1,2}$, and you can make the problem 2d (i.e. make $\mathbf{R}$ 2d) and find exact solution.

Another interesting case is when $s=0$ or very close to 0, say 1e-3, for some $\psi$. Then you found a way to express the rotation axis in terms of your orthonormal vectors: $\boldsymbol{\hat{\omega}}=\left(\cos\psi\,\mathbf{\hat{n}_1+\sin\psi\,\mathbf{\hat{n}}_2}\right)$. I think you will be able to work out the rotation angle of $\mathbf{R}$ in this case by looking at what happens with $-\sin\psi\mathbf{\hat{n}}_1+\cos\psi\mathbf{\hat{n}}_2$, i.e. vector that is perpendicular to rotation axis.

In general, if you found $\boldsymbol{\omega}_\mathcal{P}$ you can extend it to full rotation axis by adding to it a component of vector perpendicular to the plane spanned by $\mathbf{\hat{n}}_{1,2}$, $\mathbf{\hat{m}}=\mathbf{\hat{n}}_1\times \mathbf{\hat{n}}_2$. So

$$
\boldsymbol{\hat{\omega}}=\frac{\boldsymbol{\omega}_{\mathcal{P}}+f\mathbf{\hat{n}}_1\times \mathbf{\hat{n}}_2}{\left|\boldsymbol{\omega}_{\mathcal{P}}+f\mathbf{\hat{n}}_1\times \mathbf{\hat{n}}_2\right|}
$$

There will be ambiguity in choosing the sign of $f$ - but that's the non-uniqueness for you. Once you have the rotation axis, you should be able to reduce the problem to 2d again, and solve it exactly (up to sign of $f$)

$k=\mbox{rank}\boldsymbol{\mathcal{P}}=1$

Not sure there is much you can do here.


问:
Identity with a product
I have some difficulties to show the following identity

$$\prod_{i=1}^m 2^{(n-i-1)/2}=2^{(nm/2)-m}$$

My attempt:

$$\prod_{i=1}^m 2^{\frac{n-i-1}{2}}\triangleq 2^k$$
where
$$\begin{aligned}
k&\triangleq\sum_{i=1}^m \frac{n-i-1}{2}=\frac{1}{2}\sum_{i=1}^n (n-i-1)\\
&=\frac{1}{2}\left[(n-1)\sum_{i=1}^m1-\sum_{i=1}^m i\right]\\
&=\frac{1}{2}\left[(n-1)m-\frac{m(m+1)}{2}\right]\\
&=\frac{nm}{2}-\frac{m^2+3m}{4}
\end{aligned}$$

so my conclusion is,

$$\prod_{i=1}^m 2^{(n-i-1)/2}=2^{nm/2-(m^2+3m)/4}$$

Where am I wrong?

Update:

I have inferred the previous equality from the following equation
$$
\prod_{i<j}^m [(2\pi)^{1/2}]\prod_{i=1}^m\left\{2^{(n-i-1)/2}\Gamma\left[\frac{1}{2}(n-i+1)\right]\right\}=\pi^{m(m-1)/4}\prod_{i=1}^m \Gamma\left[\frac{1}{2}(n-i+1)\right]\cdot2^{mn/2-m}
$$

which is written at page 71 in Aspects of Multivariate Statistical Theory by Muirhead.

答:
Due to the bijectivity of $\log_2$ (logarithm base $2$), your identity is equivalent to

$$\sum_{i=1}^m \dfrac12(n-i-1) = \dfrac{nm}{2}-m$$

Let us transform its LHS:

$$\dfrac12 \underbrace{\sum_{i=1}^m  n}_{nm}- \dfrac12\underbrace{\sum_{i=1}^m( i+1)}_{\tfrac{(m+1)(m+2)}{2}-1} = \dfrac{nm}{2}-m$$

A problem: there is no agreement for the second terms in the LHS and RHS !

Title: Identity with a product


答:
Your question seems to show some confusion about the definition of sine and cosine.

We have a circle of radius 1.  Starting at the rightmost point on the circle coordinate (1,0) you proceed to travel around the circle in the counter-clockwise direction $x$ units.  The coordinates of the point at the end of this arc is $(\cos x, \sin x).$  This should be your definition of sine and cosine.  It is a more useful definition than anything based off of right triangles.  There are other ways to define the functions based on complex exponentials, but you are not ready for that.

Suppose you travel exactly half-way around.  How far have you traveled?  The circumference of the circle is $2\pi$ units.  You have covered a half a circle, so you have traveled $\pi$ units.  What are the coordinates of the point half way from the starting point?  $(-1,0)$

$\cos \pi = -1, \sin \pi = 0$

Suppose you travel a full rotation.  You have traveled $2\pi$ units, and you are back at your starting point.

$\cos 2\pi = 1, \sin \pi = 0$

You can make multiple full rotations around your circle, pass go, and continue to your destination.

For any $x, \cos (x+2\pi) = \cos x, \sin (x+2\pi) = \sin x$

I hope this helps, and helps to trigger some memories from pre-calculus.

Title: Why does $\cos(n \pi)=1$ if $n$ is even, but $\cos(n\pi) = -1$ if $n$ is odd?

Body:
I am taking Calculus 1 after graduating college several years ago. I have forgotten a good amount of info and am having difficulty searching info for this on my own.

Here is what is causing me confusion:

$\cos{n\pi}=1$ (if $n$ is even)

$\cos{n\pi}=-1$ (if $n$ is odd)

I am seeking an explanation as to why this is. From what I've gathered so far it has to do with radians and trigonometric functions, but the info is very unclear to me.


问:
No weak derivative in singularity
Let $\Omega := B_{\frac{1}{2}}(0) \subseteq \mathbb{R}^n$ and $n>1$. It is rather well known that $u: \Omega \setminus \lbrace 0 \rbrace \rightarrow \mathbb{R}$ lives in the Sobolev space $W^{1, n}(\Omega)$, where
$$
u(x) := \log \lvert \log \lvert x \rvert \rvert.
$$
This is in spite of the fact that $u$ can not be extended such that it is a function in $C^1(\Omega)$. An approximation argument can be applied to show weak differentiability while using that $u \in C^1(\Omega \setminus \lbrace 0 \rbrace)$.

Now, out of pure interest: Do you know an example of a function $f \in C^1(\Omega \setminus \lbrace 0 \rbrace)$ with $f, \nabla f \in L^p(\Omega)$ (for some $1 \leq p \leq \infty$) but that is not weakly differentiable, i.e. $f \notin W^{1, p}(\Omega)$?

答:
When you write $\nabla f\in L^p(\Omega)$ I have to interpret it as the statement:

the classical gradient of $f$ on $\Omega\setminus\{0\}$ extended (say)
by zero to $\Omega$ is in $L^p(\Omega)$.

Then, in one dimension, consider on $\Omega\setminus\{0\}:=(-1,1)\setminus\{0\}$ the indicator function $\chi$ of $(0,1)$. Clearly $\chi$ is in $L^\infty(-1,1)$ as well as its classical gradient (which is identically zero). But the distributional derivative is a measure (Dirac measure). Therefore it is not in any of the Sobolev spaces $W^{1,p}(-1,1)$.

You can cook up similar examples in any dimension if you increase the dimension of the singular set. In fact, you are essentially asking when $W^{1,p}(\Omega)$ and $W^{1,p}(\Omega\setminus \{0\})$ are different. This should happen only in dimension one as the question is related to the capacity of a set (the capacity of the singleton $\{0\}$ here).

Title: No weak derivative in singularity


问:
a flatness criterion
I'm having trouble with part (b) of Exercise 10.5.25 from Dummit & Foote (the goal of the problem is to prove that $A$ is a flat $R$-module iff $A\otimes_R I\to A\otimes_R R$ is one-to-one for all finitely-generated ideals $I$ of $R$).

Part (b) has 3 parts. The first and third I've done, but I'm stuck on the second. Here're the first two parts of (b):

If $A\otimes_R I\to A\otimes_R R$ is injective for every finitely-generated ideal $I$, prove that $A\otimes_R I\to A\otimes_R R$ is injective for every ideal $I$. Show that if $K$ is any submodule of a finitely generated free module $F$, then $A\otimes_R K\to A\otimes_R F$ is injective.

Here's what I've tried:

Let $e_1,\ldots,e_n$ be a basis for $F$. Let $x\in \ker(1\otimes\iota)$, where $\iota\colon K\to F$ is inclusion. Then
$$x=\sum_{i=1}^m a_i\otimes k_i$$
for some $a_i\in A$, $k_i\in K$. Express each $k_i$ in terms of the basis: $k_i=\sum_j r_{ij}e_j$. Then
$$0 = 1\otimes \iota(x)=\sum_i\sum_j a_i\otimes r_{ij}e_j=\sum_j\left(\sum_i a_i r_{ij}\right)\otimes e_j.$$
So, using the fact that $A\otimes_R F\cong A^n$ via $\sum_j b_j\otimes e_j \mapsto (b_1,\ldots,b_j)$, we get
$$\sum_i a_i r_{ij}=0$$
for all $j$. My issue is that I don't know how to lift this back to $A\otimes_R K$. Also, I think I should be using the first part of (b) somewhere.

Any hints? Am I even on the right track?

答:
Since Jack Schmidt has mentioned direct limits here is how we can prove the first part of (b) using direct limits. Let $I$ be any ideal of $R$; we can write $I = \operatorname{colim} I_\alpha$ where the colimit is over all finitely generated ideals contained in $I$. Now consider the ses

$$0 \to I\to R  \to R/I \to 0$$
from which we get a long exact sequence in Tor

$$\ldots \to 0 \to \text{Tor}_1^R(A,R/I) \to A \otimes_R I\to A \otimes_R R \to A \otimes_R R/I \to 0$$

where the zero on the left appears because $R$ a free module over itself implies $\text{Tor}_1^R(A,R) = 0$. Now I claim that in fact $\text{Tor}_1^R(A,R/I) = 0$. Indeed we have
$$\begin{eqnarray*} \text{Tor}_1^R(A,R/I)&=& \text{Tor}_1^R(A,\left(R/\operatorname{colim}I_\alpha\right)) \\
&=& \text{Tor}_1^R\left(A,\operatorname{colim} (R/I_\alpha)\right) \hspace{1cm} (\text{Colimits are right exact}) \\
&=&  \operatorname{colim} \text{Tor}_1^R\left(A, (R/I_\alpha)\right) \hspace{1cm} (\text{Tor commutes with colimits})\\
&=& 0
\end{eqnarray*}$$

where we get $0$ at last because for all $\alpha$, $\text{Tor}_1^R\left(A, (R/I_\alpha)\right)  = 0$ by assumption. Thus the first part of (b) is proven.

Title: a flatness criterion


问:
Contour integration proof explanation (term goes to 0)

Prove that $$\int_\infty^{-\infty} \frac {cos x} {e^x + e^-x} dx = \frac {\pi} {e^{\pi/2} + e^{-\pi/2}}$$.

I have a question about the proof shown below: why does the third integral (the union of left and right side contour) vanish?

I think I understand intuition wise, i.e. $i\pi$ becomes insignificant as $R$'s in both $R+i\pi$ and $R$ get infinitely large. However, as practice I tried to calculate the third integral myself to show it equal to 0 and couldn't do it.

I am wondering if someone could show me how to calculate the third integral.

答: The third integral is indeed zero but not because of taking $R$ to be infinitely large. It is zero no matter what $R$ we use. I integrate it by taking $z=R+\theta i$ in the upwards path in positive side and taking $z=-R+\theta i$ in the negative downward path. I put a negative in front of the downward integral which make $\theta$ in both integral going from $0$ to $\pi$ . Making two integral into one you will find that the numerator cancel to $0$. Be careful that the denominator is not the same. I guess you have made some careless mistake in your calculation maybe.


问:
Forcing GCH with iteration of $\mathsf{Add}(\kappa,1)$
For a successor cardinal $\kappa^+$, forcing with $\mathsf{Add}(\kappa^+,1)$ forces $2^\kappa=\kappa^+$ in the extension. Can one use an iteration of this kind of forcing to for the GCH? If so, I'd like to know if there is a write-up of this proof somewhere. In particular, what kind of support should it have? Is there a characterization of what cardinals are preserved or collapsed?

答:
It is well-known that iterating $\mathsf{Add}(\kappa^+,1)$ with Easton support iteration (or reverse Easton iteration) forces $\mathsf{GCH}$. Here Easton support iteration means the following:

If $\alpha$ is a singular limit, then take the inverse limit, and

If $\alpha$ is a regular limit, then take the direct limit.

You may find some part of the proof from Sy Friedman's Fine Structure and Class Forcing. I believe there is a full proof somewhere, but I have no idea where it is. Here is the proof I sketched:

We will clarify the definition of the iteration we will use. Let $\langle \mathbb{P}_\alpha \mid \alpha\in\mathsf{Ord}\rangle$ be the iteration of $\langle\dot{\mathbb{Q}}_\alpha\mid\alpha\in\mathsf{Ord}\rangle$, where

If $\alpha$ is a cardinal in $V^{\mathbb{P}_\alpha}$, then $\dot{\mathbb{Q}}_{\alpha}=\mathsf{Add(\alpha^+,1)}$, and

If $\alpha$ is not a cardinal in $V^{\mathbb{P}_\alpha}$, then $\dot{\mathbb{Q}}_{\alpha}$ is a trivial forcing.

Let $\mathbb{P}$ be the Easton support iteration of all $\langle \mathbb{P}_\alpha \mid \alpha\in\mathsf{Ord}\rangle$. Then $\mathbb{P}$ preserves $\mathsf{ZFC}$. (Using the Easton support is necessary to prove it.)
Then we have the following property:

Theorem. Let $\gamma$ be an ordinal such that $V^{\mathbb{P}_\gamma}$ thinks $\gamma$ is regular. Then $V^{\mathbb{P}_\gamma}$ thinks the tail forcing $\dot{\mathbb{P}}_{\gamma,\infty}$ is $(\gamma^+)$-closed.
(Note that $\mathbb{P}\cong \mathbb{P}_\gamma*\dot{\mathbb{P}}_{\gamma,\infty}$).

Now we prove $\mathbb{P}$ forces $\mathsf{GCH}$ as follows:

Proof. Assume that $\mathbb{P}_\beta$ forces $\mathsf{GCH}$ below $\beta$ for all $\beta<\alpha$. Furthremore, assume that $\mathbb{P}_\alpha$ forces $\alpha$ is a cardinal. (We do not need to consider when $\mathbb{P}_\alpha$ forces $\alpha$ is not a cardinal.)

If $\mathbb{P}_\alpha$ forces $\alpha=\beta^+$ for some cardinal $\beta<\alpha$, then $\mathbb{P}_\beta$ forces $\beta$ is a cardinal. Furthermore, $\mathbb{P}_{\beta+1}$ forces $2^\beta=\beta^+$. Hence $\mathbb{P}_\alpha=\mathbb{P}_{\beta+1}$ forces $\mathsf{GCH}$ below $\alpha$.

Now assume that $\mathbb{P}_\alpha$ forces $\alpha$ is a limit ordinal. Fix $\beta<\alpha$ which is a cardinal in $V^{\mathbb{P}_\alpha}$, and take $\gamma$ such that $V^{\mathbb{P}_\alpha}\models \gamma=\beta^+<\alpha$.
Then $\mathbb{P}_\gamma$ forces $\gamma$ is regular (since $\mathbb{P}_\alpha$ does and $\gamma<\alpha$.) Since $\mathbb{P}_\alpha\cong \mathbb{P}_\gamma*\dot{\mathbb{P}}_{\alpha,\gamma}$, $V^{\mathbb{P}_\gamma}\models 2^\beta=\beta^+<\alpha$ and ($V^{\mathbb{P}_\gamma}$ thinks) $\dot{\mathbb{P}}_{\alpha,\gamma}$ is $\gamma^+$-closed, we have $V^{\mathbb{P}_\alpha}\models 2^\beta=\beta^+<\alpha$.

We can see that inaccessible cardinals are preserved.
But some cardinals may collapse; for example, if the ground model satisfies $2^{\aleph_0}=\aleph_2$, then $\mathsf{Add}(\aleph_1^V,1)$ forces $2^{\aleph_0}=\aleph_1$ while preserving all cardinals below $\aleph_1^V$ and above $\aleph_3^V$, so it collapses $\aleph_2^V$.


问:
How do I calculate mean of a second variable with just just the mean and standard deviation of the first? (AP stats)
I don know really how to explain but I am currently working on some ap stats homework and I am incredibly confused. I will post the question below.

A certain box of cereal is labeled as 32 ounces, but boxes actually contain, on average, 33.1 ounces of cereal with
a standard deviation of 0.4 ounces. The distribution of weights is approximately Normal. Let X represent the weight
of cereal in a randomly selected box.
a) Let B represent the extra cereal contained in a box of cereal (i.e. the cereal above the advertised 32 ounces). Find
the mean and standard deviation of B.
b) Find the P(B > 2).

Honestly, this is the easiest question on the assignment and I am already lost so I don't have a great feeling about the rest lol.

答:
Read @Stacker's (+1) Answer first:

You have $B \sim \mathsf{Norm}(\mu=1.1, \sigma = 0.4).$
Then $P(B > 2) = 1 - P(B \le 2).$

From R statistical software, where pnorm is a normal CDF, using the standard deviation:

```<code>1 - pnorm(2, 1.1, .4)
[1] 0.01222447
</code>```

According to @Stacker's advice for printed normal tables,
you will not get that many digits of accuracy, but
the answer from R will give you a clue whether you
got the right numerical answer from the table. Depending on the
format 0f  your table, you might get $1 - 0.9878.$

Body:
Read @Stacker's (+1) Answer first:

You have $B \sim \mathsf{Norm}(\mu=1.1, \sigma = 0.4).$
Then $P(B > 2) = 1 - P(B \le 2).$

From R statistical software, where pnorm is a normal CDF, using the standard deviation:

1 - pnorm(2, 1.1, .4)
[1] 0.01222447

According to @Stacker's advice for printed normal tables,
you will not get that many digits of accuracy, but
the answer from R will give you a clue whether you
got the right numerical answer from the table. Depending on the
format 0f  your table, you might get $1 - 0.9878.$


问:
Intersection curve parametrization
How can we find parametrization of the intersection curve between a pseudo-sphere and sphere ?

Elimination of variables appears not readily amenable/elegant. Thanks for the same.

答:
The $z \gt 0$ half of a pseudosphere with radius $r$ is a surface of revolution (rotated around the $z$ axis) described by
$$\left\lbrace ~ \begin{aligned}
x(t) &= r t - r \tanh t \\
y(t) &= r \operatorname{sech} t = \frac{r}{\cosh t} \\
\end{aligned} \right.$$
where $0 \lt t \in \mathbb{R}$.

This intersects a circle of radius $r$ centered at origin when
$$x(t)^2 + y(t)^2 = r^2$$
i.e.
$$(r t - r \tanh t)^2 + \frac{r^2}{(\cosh t)^2} - r^2 = 0$$
The degenerate sphere case ($r = 0$) is uninteresting, so we can assume $r \gt 0$.  For $t \gt 0$, $\cosh t \gt 0$, so we can divide the equation by $t r^2$ and multiply by $\cosh t$, getting
$$\begin{aligned}
t \cosh t - 2 \sinh t &= 0 \\
t \cosh t &= 2 \sinh t \\
t &= 2 \tanh t \\
\end{aligned}$$
which has only one solution for $0 \lt t \in \mathbb{R}$,
$$t \approx 1.91500804815453748135300306100481565057336256878593...$$

Because both shapes are symmetric around the $z$ axis, the intersection is a circle of radius $y(t)$ perpendicular to the $z$ axis at $z = x(t)$, i.e.
$$\left\lbrace ~ \begin{aligned}
R = y(t) &\approx 0.28841990894499 \, r \\
z = x(t) &\approx 0.95750402407727 \, r \\
\end{aligned}\right.$$

Title: Intersection curve parametrization


问:
In $\mathbb{F}_p[X]$ with $p$ prime and $p = 1 \text{ mod } 3$, prove that there exists an element of order 3.
Let $p$ be prime, and $p = 1 \text{ mod } 3$. Consider $\mathbb{F}_p[X]$.

Prove that in $\mathbb{F}_p$ there exists an element of order 3. Use that to conclude that $X^2 + X + 1$ has a root in $F_p[X]$.

I am preparing for an exam and this is one of the example questions. I've thought about it for a bit, but to be honest I don't really know where to start. Can somebody give me a hint, so that I know where to start looking?

答:
Write $p=3m+1$. Then $x^{p-1}-1=x^{3m}-1=(x^3-1)q(x)$, where $q(x)$ has degree $3(m-1)$.

The  $3m+1$ elements of $\mathbb{F}_p^\times$ are exactly the roots of $x^{p}-x$. They cannot all be the roots of $x(x-1)q(x)$, which has degree $3m-1$. Therefore, there are at least $2$ roots of $x^2+x+1$ in $\mathbb{F}_p^\times$.


问:
Does the integral converge $\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx$ for $b>0$?
I try to calculate integral $$\int_{-\infty}^{+\infty}\frac{\exp(ibx)}{1+e^x}\,dx.$$ Does it converge for $b>0$? Wolfram Mathematica shows contradictory results: for $b=2$ it converges, for $b=1$ and $b=3$ it does not. I need to calculate it.

答:
Probably something along these lines ...

Set $f(z) = e^{i\alpha z}/(1 + e^z)$ for complex $z$ and integrate round the closed contour running along the real axis from $z = -R$ to
$z = R$, then continuing on the semicircle of radius $R$ in the upper half plane:

\begin{align}
I_1(R) &= \int_{-R}^R f(x)\, dx, \\
I_2(R) &= iR \int_0^\pi f(Re^{i\theta}) e^{i\theta}\, d\theta.
\end{align}

We're aiming for using Cauchy's integral theorem. On the semicircle
\begin{align}
f(Re^{i\theta}) &= \frac{\exp(i\alpha R e^{i\theta})}{1 + \exp(Re^{i\theta})} \\
&= \frac{\exp(-\alpha R \sin\theta + i\alpha R \cos\theta)}{1 + \exp(R\cos\theta + iR\sin\theta)}
\end{align}

For $0 < \theta < \pi/2$, both $\cos\theta$ and $\sin\theta$ are positive. The numerator is $O(e^{-\alpha R \sin \theta})$ and the denominator is $O(e^{R\cos\theta})$ so the ratio is $O(e^{-\alpha R \sin\theta - R\cos\theta})$.

When $\pi/2 < \theta < \pi$, $\sin\theta$ remains positive but $\cos\theta$ is negative. The numerator is $O(e^{-\alpha R\sin\theta})$ and the denominator is
$O(1)$, so the ratio is $O(e^{-\alpha R\sin\theta})$.

All of this to justify that $I_2(R) \rightarrow 0$ as $R \rightarrow \infty$. It follows from Cauchy's integral theorem, letting $R \rightarrow \infty$, that
$I_1(\infty)$ is $2\pi i$ times the sum of the residues of $f$ in the upper half-plane. Here the poles of $f$ are simple and occur at $z_k = (2k + 1)\pi i$ for $k = 0, 1, \dots$. Verify that $\text{Res}(f, z_k) = -e^{-\pi\alpha (2k + 1)}$,
which yields:
\begin{align}
I_1(\infty) &= -2\pi i \sum_{k=0}^\infty e^{-\pi\alpha(2k + 1)} \\
&= -2\pi i e^{-\pi \alpha}\sum_{k=0}^\infty e^{-2\pi \alpha k} \\
&= -2\pi i e^{-\pi \alpha}\frac{1}{1 - e^{-2\pi\alpha}} \\
&= -\frac{\pi i}{\sinh(\pi \alpha)}.
\end{align}


问:
Explanation of solution of a question about polynomials
Taken from https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_10

Problem

Given the polynomial $f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n$ with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a, b, c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$, show that there is no integer $k$ such that $f(k)=8$.

Solution

Set $g(x) = f(x) − 5$. Since $a, b, c, d$ are all roots of $g(x)$, we must have

$g(x) = (x − a) (x − b) (x − c) (x − d) h(x)$

for some $h(x) ∈ Z[x].$
...

How do we know that h(x) has integral coefficients here? How is it obvious?

I tried simplifying the $g(x)$ expression and comparing coefficients, and it seems to be true, but cannot prove it. Moreover, how is it so obvious?

答:
If $n<4$ then $g=0$ because $g(x)=0$ when $x$ is any of the 4 distinct values $a,b,c,d.$

If $n\ge 4$ let $j(x)=(x-a)(x-b)(x-c)(x-d).$

There exist $h(x), s(x)\in \Bbb Z[x]$ with $g(x)=h(x)j(x)+s(x)$ and $deg(s)<n$ because $deg (\,g(x)-x^{n-4}j(x)\,)<n.$

So consider $m,$ the least degree of an $s(x)\in \Bbb Z[x]$ such that there exists  $h(x)\in \Bbb Z[x]$ with $g(x)=h(x)j(x)+s(x).$

Suppose $m\ge 4.$ Let  $g(x)=h_1(x)j(x)+s_1(x)$ with $h_1(x)$ and $s_1(x)$ in $\Bbb Z[x]$ and $deg(s_1)=m.$ Let the leading term of $s_1(x)$ be $\sigma x^m.$  Let $s_2(x) =s_1(x)-\sigma x^{m-4}j(x).$ Let $h_2(x)=h_1(x)+\sigma x^{m-4}.$ Then $h_2(x)$ and $s_2(x)$ are in $\Bbb Z[x]$ with $g(x)=h_2(x)j(x)+s_2(x)$ and $deg (s_2)\le m-1,$ contradicting the minimality of $m.$

So by contradiction we have $m<4.$

So let $g(x)=h(x)j(x)+s(x)$ with $h(x)$ and $s(x)$ in $\Bbb Z[x]$ and $deg (s)<4.$ But $s(x)=0$ when $x$ is any of the 4 different values $a,b,c,d$ and $deg (s)<4$. So $s=0$ and $g(x)=h(x)j(x).$


问:
What is this function called?
What is this function called?
Here is a graph:

```<code>          _______
         /
        /
_______/
</code>```

Sort of like:

$f(x, a, b) = 0$, if $x < a; \quad \quad \displaystyle \frac {(x - a)} {(b - a)}$ if $a < x  b$.

Closest name is Sigmoid function, but they all are smooth. Also I remember I saw similar formula for saturation, but not sure about that, googling didn't help.

Edit: Ramp function is closer:

But it is general one, while my is more specific one, used for normalization.

答:
The first two "pieces" of your graph seem to fit the "ramp" function, with $(a,0)$ replacing $(0,0)$ of the "general" example you provide for a ramp function. But the first and third "pieces" comprise what could be described as a step-function (with endpoints joined by the second "piece"/line).

But in the end, since it seems we need to define and discuss three separate "pieces" of the function, depending on the value of x with respect to the parameters $a$ and $b$, perhaps the most accurate (albeit general) description for your function is simply a piecewise-linear function, (consisting of a linear-ramp function which continues/plateaus at its maximum value, $y=1$  for all $x>b$?). I can't seem to find any more concise description that captures all you'd like it to capture.

Title: What is this function called?


问:
inverse laplace transform of $\frac{1}{s(e^s+1)}$
How to compute this inverse Laplace transform ?

$$\displaystyle{ \mathcal{L^{-1}} \left\{  \frac{1}{s(\exp(s)+1)} \right\} }$$

Thanks.

答:
OK, I think I have a systematic way of getting this ILT.  Let's consider a periodic function $f(t)$ with period $T$.  The Laplace transform of such a function is

$$\begin{align}\int_0^{\infty} dt \, f(t) e^{-s t} &= \sum_{k=0}^{\infty} \int_{k T}^{(k+1) T} dt \, f(t) e^{-s t}\\ &= \sum_{k=0}^{\infty} e^{-k s T} \int_0^T du \, f(u) e^{-s u} \\ &= \frac{\hat{f_0}(s)}{1-e^{-s T}} \end{align}$$

where

$$f_0(t) = f(t) \theta(t) \theta(T-t)$$

and $\theta$ is the Heaviside step function.

This is a pretty standard result.  But what if we have another function $g$ that was anti-periodic; that is:

$$g(t) = -g(t+T) = g(t+2 T)$$

Then by similar reasoning as above, we find that

$$\int_0^{\infty} dt \, g(t) e^{-s t} = \sum_{k=0}^{\infty} (-1)^k e^{-k s T} \int_0^T du \, f(u) e^{-s u} = \frac{\hat{g_0}(s)}{1+e^{-s T}}$$

So we consider an anti-periodic function $g$ of period $T=1$ with

$$\hat{g_0}(s) = \frac{e^{-s}}{s}\implies g_0(t) = \theta(t-1)$$
Note that

$$\frac{e^{-s}}{s} \frac{1}{1+e^{-s}} = \frac{1}{s(e^s+1)}$$

Thus, we can conclude from this that the ILT is the anti-periodic function constructed from the anti-periodic version of $g_0(t) = \theta(t-1)$.  That is, when

$$\hat{g}(s) = \frac{1}{s(e^s+1)}$$

then

$$g(t) = \begin{cases} \\ 0 & t \in [4 k, 4 k+1]\\ 1 & t \in [4 k+1,4 k+2]\\0 & t \in [4 k+2,4 k+3]\\-1 & t \in [4 k+3,4 k+4) \end{cases} \forall k \in \mathbb{Z}_+ \cup \{0\}$$


问:
Write the contrapositive of "if $n$ is an even integer such that n + 1 is a square then n is divisible by 8"
For the purpose of clarity, I do not want anybody to prove this statement, I am just looking to get some help translating it into the contrapositive using symbolic logic. Right now I have the following translation:

Original Statement:

$$(\exists k \in \mathbb{Z} \quad n = 2k) \quad \land \quad (\exists m \in \mathbb{Z} \quad n+1 =m^2) \quad \Rightarrow \quad \text{n is divisible by 8}$$

Contrapositive Statement:

$$ \text{n is not divisible by 8} \quad \Rightarrow \quad (\exists k \in \mathbb{Z}  \quad n = 2k+1) \quad \lor \quad (\forall m \in \mathbb{Z}  \quad n+1 \neq m^2) $$

A couple of points of clarification:

I wasn't sure how to write $n$ is divisible by 8. Perhaps it could have been stated as $\exists p \in \mathbb{Z} \quad n = 8p$?

For the negation of $\exists k \in \mathbb{Z} \quad n = 2k$, I know that typically you might want to reverse the $\exists$ to an $\forall$, but I think what I wrote makes sense since the negation of even numbers is odd numbers.

Eventually I will try to prove this statement by contrapositive, but I want to get some practice in with translating mathematical statements into a more formal structure.

My question(s) are following:

Is the original translation correct? How might it be improved?

How could I more formally translate '$n$ is divisible by 8'?

Is the contrapositive stated correctly?

I also am not able to use $mod$ notation, $a|b$ notation, or anything that is unique to Abstract Algebra.

答:
"if $n$ is an even integer such that $n+1$ is square then n is divisible by 8"

The original statement can be written as:

$$ (\forall k \in \mathbb{Z} \quad [n = 2k \quad \land \quad n \neq2k+1]) \quad \land \quad (\exists m\in \mathbb{Z}\quad m^2 = n+1) \quad \Rightarrow \quad 8 |n $$

The contrapositive is then:

$$ 8 \nmid n \quad \Rightarrow \quad (\exists k\in \mathbb{Z} \quad [n \neq 2k \quad \lor \quad n = 2k+1]) \quad \lor \quad (\forall m \in \mathbb{Z} \quad m^2 \neq n +1) $$


问:
Corollary of Kronecker-Weber Theorem (J. Neukirch's ANT)
I have a question about the proof of Corollary (1.9), Chap V page 324 from Jürgen Neukirch's
Algebraic Number Theory:

Claim: Every finite abelian extension of $L \vert \mathbb{Q}_p$ is contained
in a field $\mathbb{Q}_p(\zeta)$, where $\zeta$ is a root of unity. In other words:
The maximal abelian extension $\mathbb{Q}_p^{ab} \vert \mathbb{Q}_p$ is generated
by adjointed roots of unity.

Proof: For suitable $f$ and $n \in \mathbb{N}$, we have
$(p^f) \times U^{(n)}_{\mathbb{Q}_p} \subset N_{L \vert \mathbb{Q}_p} L^*$.
Therefore $L$ is contained in the class field / [the proof continues...]

About used notations: $(p) \subset \mathbb{Z}_p$ is the unique maximal ideal
of $p$-adic integers and $(p^f)$ it's $f$-th power. $U^{(n)} = 1 +(p)^n$ and
$N_{L \vert \mathbb{Q}_p} L^* \subset \mathbb{Q}_p$ is the image of $L^*$ under
norm map.

Question: why such $f, n$ exist and now they can be found or their existence
be showed?

答:
Here $(p)$ is not $p \Bbb{Z}_p$ but $p^\Bbb{Z}$.

With $f = [L:\Bbb{Q}_p]$ and $p^{n-2}$ the largest power of $p$ dividing $f$
$$p^f= N_{L/\Bbb{Q}_p}(p), \qquad 1+p^n \Bbb{Z}_p= (1+p^2 \Bbb{Z}_p)^{p^{n-2}}=(1+p^2 \Bbb{Z}_p)^f=N_{L/\Bbb{Q}_p}(1+p^2\Bbb{Z}_p)$$


问:
Proper notation for corresponding members in sets
I'm an amateur 42 years removed from academia so I don't always know "the language".

In a paper I am writing, I specify search limits like $\;4\le m  \le 9\;$ and then show, in examples, how some members in the search set yield integers for a function: $\;k=f(m).\quad$ In an example where $f(4)=1\land f(8)=7\;$ I have been using the following notation.
$$m\in\{4,8\}\implies k \in\{1,7\} $$

The "sets" can be larger or more complex such as for a cubic where there can be up to $3$ solutions for a given $m$.
$$m\in \{7\}\implies k\in\{5,8,3\}\qquad\land\qquad m\in\{8\}\implies k\in\{7\}$$

Does this notation make sense at all and does it make it clear that $4$ corresponds to $1$ and that $8$ corresponds to $7$? Alternatively, is there a better way to express these "summaries"?

This notation is the briefest I could think of outside of showing ordered pairs such as $\quad(7,5)\;(7.8)\;(7,3)\quad (8,7).\quad$ What I want to emphasize is the the "search" found its objective(s) within the specified limits because the point of the bigger "thing" is that what we are looking for can be found $\textbf{without brute force}$.

$\textbf{Update:}$ In response to where this fits in the middle of a paper, here is one of a dozen or more "problems" as I have written them so far. The question remains about the best way to present the results of the search.

$$mk^3-m^3k+D=0\implies$$
\begin{equation}
k_0=\sqrt{\frac{4m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\cos^{-1}{\biggl(-\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}\\
\quad k_1=\sqrt{\frac{4m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\cos^{-1}{\biggl(\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}\\
\quad k_2=k_1-k_0
\end{equation}

$$\text{ for }\quad\lfloor\sqrt[4]{2D}\rfloor\approx\sqrt[4]{\frac{8D}{3}}
 \le m \\
 \le \frac{\sqrt[3]{\sqrt{3}
  \sqrt{27 D^2 - 4} + 9 D}}{\sqrt[3]{2} \sqrt[3]{3^2}} + \frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{\sqrt{3} \sqrt{27 D^2 - 4} + 9 D}}\approx\big\lceil\sqrt[3]{D}\big\rceil$$

$$D=840\implies \lfloor\sqrt[4]{2(840)}\rfloor=6 \le m \le \lceil\sqrt[3]{840}\rceil=10\quad\text {and we find}$$
$$m\in \{7\}\implies k\in\{5,8,3\}\qquad\land\qquad m\in\{8\}\implies k\in\{7\}$$
$$F(7,5)=(24,70,74)\quad F(7,8)=(-15,112,113)\quad F(7,3)=(40,42,58)\\ F(8,7)=(15,112,113)$$

Not related to the original question. The "approximate" limits shown are what I found heuristically before I was able to determine the more complicated "exact" limits. I'm thinking of trashing the latter because the approximate limits are lower and higher and would not add much to the search. Still, the exact limits $are$ exact and omitting them might detract from the precision expected in a paper for publication.

答:
Without further context to override this, a list or table of pairs would be appropriate. Based on the snippet of the text, I also recommend using a lot more words/sentences to break up what's going on. If you read more papers in your area, you'll get a sense for how much they use words to present things clearly.

Note that if there are many similar calculations, extra sentences would get in the way after the first couple times, if it's completely clear from context what is going on. If that's your situation, representing all of the routine steps of the calculation in a table may be appropriate.

So for the snippet in the question, I would recommend sentence(s) along the lines of:

For $D=840$, the bounds for $m$ are $\lfloor\sqrt[4]{2(840)}\rfloor=6$ and $\lceil\sqrt[4]{2(840)}\rceil=10$. The only pairs of integer solutions $(m,k)$ in that range are then $(7,3)$, $(7,5)$, $(7,8)$, and $(8,7)$.

The ordering by increasing $m$ and then $k$ helps the reader to follow along. For more complicated situations, tables may be appropriate. Perhaps something like $$\begin{matrix}m&k\\ 3&4\\ 3&7\\ \\ 4&3 \\ 4&5 \\ 4&12 \\ \\ 5&4 \\ 5&6 \end{matrix}$$or
$$\begin{matrix}m&k\\ 3&4\\ 3&7\end{matrix}\,\,\,\,\begin{matrix}m&k\\  4&3 \\ 4&5 \\ 4&12 \end{matrix}\,\,\,\,\begin{matrix}m&k\\  5&4 \\ 5&6 \end{matrix}$$


问:
Uniformly Convergent Subsequence of a Pointwise Convergent Sequence of Functions
I've come across a question in my Real Analysis course that I'm having quite a lot of difficulty finding an answer to. This post has gotten me the closest to understanding the problem, which is as follows:

"Provide an example or explain why it is not possible: a sequence of functions $(f_n)$ that converges pointwise (but not uniformly) to a function $f$, with a subsequence $(f_{n_k})$ of $(f_n)$ that converges uniformly to $f$."

In general, what conditions need to be satisfied in order for a sequence of functions to have a uniformly convergent subsequence?

答:
Consider :
$$
f_{n}(x)=1-\frac{1}{x^{n}+1}
$$
Note that $(f_{n}(x))_{n\in\mathbb{N}}$ converges to $f$ point-wise where :
$$
f(x)=\begin{cases}
0,& \text{if $x \in[0,1)$} \\
\frac{1}{2}, &\text{if $x=1$} \\
1,& \text{if $x \in(1,2]$}
\end{cases}
$$
Each sequence of $f_{n}$ is continuous but the limit function is not continuous. Moreover, we know that if $f_{n}$ is continuous and uniform convergent, then it would imply continuity so if we take the contrapositive statement we can see that no subsequence of $f_{n}$ can converge uniformly to $f$ due to Arzelà–Ascoli theorem


答:
You omitted the powers of $2$ arising from the first product:
$$\prod_{i<j}^m 2^{1/2} = 2^{\binom{m}{2}/2} = 2^{m(m-1)/4}$$

Title: Identity with a product

Body:
I have some difficulties to show the following identity

$$\prod_{i=1}^m 2^{(n-i-1)/2}=2^{(nm/2)-m}$$

My attempt:

$$\prod_{i=1}^m 2^{\frac{n-i-1}{2}}\triangleq 2^k$$
where
$$\begin{aligned}
k&\triangleq\sum_{i=1}^m \frac{n-i-1}{2}=\frac{1}{2}\sum_{i=1}^n (n-i-1)\\
&=\frac{1}{2}\left[(n-1)\sum_{i=1}^m1-\sum_{i=1}^m i\right]\\
&=\frac{1}{2}\left[(n-1)m-\frac{m(m+1)}{2}\right]\\
&=\frac{nm}{2}-\frac{m^2+3m}{4}
\end{aligned}$$

so my conclusion is,

$$\prod_{i=1}^m 2^{(n-i-1)/2}=2^{nm/2-(m^2+3m)/4}$$

Where am I wrong?

Update:

I have inferred the previous equality from the following equation
$$
\prod_{i<j}^m [(2\pi)^{1/2}]\prod_{i=1}^m\left\{2^{(n-i-1)/2}\Gamma\left[\frac{1}{2}(n-i+1)\right]\right\}=\pi^{m(m-1)/4}\prod_{i=1}^m \Gamma\left[\frac{1}{2}(n-i+1)\right]\cdot2^{mn/2-m}
$$

which is written at page 71 in Aspects of Multivariate Statistical Theory by Muirhead.


问:
Category-theoretic description of the real numbers
The familiar number sets $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$ all have "natural constructions", which indicate, why they are mathematically interesting.

For example, equipping $\mathbb{N}$ with the usual sucessor function and the constant $0$, it can be described as the initial $(0,1)$-Algebra. Or, if we want it to be an additive monoid, it is the free monoid on some one-point set. Since monoids are something very elementary and can be described purely categorically in many equivalent ways, this gives an idea, why considering the natural numbers might be interesting.

Now the forgetful functor $\mathbf{Groups}\longrightarrow\mathbf{Monoids}$ has a left adjoint sending $\mathbb{N}$ on the additive group of integers $\mathbb{Z}$. Alternatively, $\mathbb{Z}$ is the initial ring. If one questions that considering rings is interesting, we could reply that they are just monoids in the "natural" category of abelian groups with the "natural" tensor product making it a monoidal category.

From $\mathbb{Z}$ to $\mathbb{Q}$ it's not far, since the rationals are the image of $\mathbb{Z}$ under the left adjoint of the embedding $\mathbf{Fields}\longrightarrow\mathbf{Domains}$.

(One could go one step further and move on to the algebraic numbers as the field-theoretic completion of the rationals.)

My question now is: Why do we consider the reals $\mathbb{R}$, from a structural point of view? It is clear to me (or at least I don't feel I have the right to question it) that considering real numbers in physics, finance, etc. is necessary, because it provides good models for our reality.

I rather wonder, whether there are mathematical reasons making the reals interesting. The only description of the real numbers by some universal property that I am aware of is that they are the Cauchy-Completion of the rationals as a metric space - but since the definition of a metric space already depends on some notion of the reals, this is just a cheap trick. (One could of course define a metric space as a set $X$ with a set-function into some archimedian ordered field, satisfying the usual axioms, but this is a little artificial, I think.) Also note that it is almost impossible to do Algebra, Set Theory or Graph Theory without knowing what the natural numbers are, whereas one can prove many results of Algebra and Topology without ever coming across the reals.

I hope you can provide some ideas, showing why the reals numbers considered as a topological space/a field/a group/an ordered set are interesting in conceptual (i.e. category-based) mathematics. Of course, if one questions that the real numbers are interesting, one also has to question complex numbers $\mathbb{C}$ and other concepts based on these notions (like almost all of Analysis, Differential topology, etc.), so I am well aware of the fact, that I should not refuse the real numbers as old-fashioned even in case, I don't receive many answers.

Edit: Of course, my question is implicitly based on my strong belief that mathematical interestingness and categorical interestingness are equivalent concepts. As suggested by some of the comments, I would like to rephrase my question: How can I deduce from the mathematical properties of the real numbers that they are mathematically interesting, and thus that they are the optimal formalization of our intuitions of geometry and infinitesimal operations?

答: Expanding on @bruckner's answer, $\mathbb{R}$ is the unique complete, ordered field, up to isomorphism. Being complete is natural for describing e.g. Euclidean space, because we want "every number to exist", i.e. we don't want lines which don't have a length because that length is irrational. Being ordered is natural because we want to be able to add up enough of any one length to reach another length. We want to be able to multiply and add lengths. I don't have a physical intuition for wanting $\mathbb{R}$ to be a field is, but it gives us a lot of the symmetries of $\mathbb{R}^n$.


问:
A proof for the interior product
How should I approach a proof like the following for the interior product:

Let $K$ be a field and $V$ a $K$-vector space. And $\omega_i \in V^*$ with $V^*$ the dual space to $V$. It is given that $$\iota_v(\omega_1 \otimes \cdots \otimes \omega_p) = \omega_1(v)\omega_2 \otimes \cdots \otimes \omega_p.$$
Now it is to be proven that:
$$\iota_v(\omega_1 \wedge \cdots \wedge \omega_p)=\sum_{i=1}^p (-1)^{i-1}(\iota_v\omega_i) \omega_1\wedge \cdots \wedge \widehat{\omega_i} \wedge \cdots \wedge \omega_p. \tag{1}$$

I tried using the definition of the wedge product through the sum over the permutations of $\operatorname{sgn}(\sigma)$ and the tensor product of the given dual vectors and used the iota function on this definition, because it is given how the tensor product behaves under it. So
$$\iota_v(\omega_1 \wedge \cdots \wedge \omega_p) = \iota_v \bigg(\sum_{\sigma\in S_p} \operatorname{sgn}(\sigma) \omega_{\sigma(1)} \otimes \cdots \otimes \omega_{\sigma(p)} \bigg).$$

But wasn't sure whether or not I was allowed then to change the sum over the permutations to a sum over $1$ to $p$.

答: OK, you've made some serious progress. Now here's a hint to proceed to the next step. Consider dividing $S_p$ up into subsets $S_p(i) = \{\sigma: \sigma(1)=i\}$. You should convince yourself that these are all (right) cosets of $S_{p-1}$. This breaks your sum up into $\sum_{i=1}^p$. Can you finish now?

Title: A proof for the interior product


问:
Solving a Quartic Function
A user named 'Uzdawi' from another post asked a question about how to solve the quartic function of

One of the responses included an answer from the user 'Peđa Terzić', which is as follows:

Could anyone kindly elaborate on the method used to get to the answer please.

And how the same method could be used to solve the following function:

If not the same method used by Peđa Terzić, how else could you approach to solving a quartic function?

Thank you for your time.

答:
What he did there from the first equation to the second was a very convenient rearrangement of the expression, in such a way that the quartic polynomial became factorable. For example, note that he wrote $21x^2$ as $25x^2$ followed by adding a $-4x^2$.

The real "magic" happened from
$$
x^4-10x^3+21x^2+40x-100=0. \tag{1}
$$
to
$$
x^4-10x^3+25x^2-(4x^2-40x+100)=0. \tag{2}
$$
The rest of the solution consists of high school algebraic manipulations.After written in terms of factors, it's easy to get the roots.

There is no algorithmic method for doing what he did from (1) to (2). It takes mathematical maturity and some ingenuity. Another way to solve a quartic equation is numerically, because not all quartic equation will factor out nicely like that.

Title: Solving a Quartic Function


问:
$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series
Find limit without using L'Hospital or Taylor's series:

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = $$

$$= \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{x^5} \cdot \frac{x^2}{5^{x^{2}} - 1} \cdot \frac{ x^3}{\tan^3(3x)}$$

$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} = 1$

$\displaystyle \lim_{x \to 0} \frac{\sin(2x^5)}{x^5} = 2$

$\displaystyle \lim_{x \to 0} \frac{x^2}{5^{x^{2}} - 1} = \frac{1}{\ln5}$

$\displaystyle \lim_{x \to 0} \frac{ x^3}{\tan^3(3x)} = \displaystyle \lim_{x \to 0} \cos^3(3x) \cdot \displaystyle \lim_{x \to 0} \frac{x^3}{\sin^3(3x)} = 1 \cdot \left( \frac{1}{3} \right)^3 = \frac{1}{27}$

We get:

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = 1 * 2 * \frac{1}{\ln5} * \frac{1}{27} = \frac{2}{27 \ln5}$$

All credits to @Bernard. Thank you all, question closed.

答:
You can do it using equivalents and some standard limits: rewrite  the fraction as
$$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}.$$
By substitution, the first factor tends to $1$. Also by substitution, $\:\dfrac{5^{x^2}-1}{x^2}$ tends to the derivative of $5^x$ at $0$, i.e. $\ln 5$.
Last, $\sin(2x^5)\sim_0 2x^5$ and $\tan 3x)\sim_03x$, so $\tan^3(3x)\sim_0(3x)^3=27x^3$, and ultimately we obtain the equivalence
$$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}\sim_0\frac{2\not x^5}{27\not x^3\cdot\ln5\cdot \not x^2}=\frac 2{27\ln 5}.$$


答:
The approach you described is based on clever factorization of the quartic into the product of two quadratics. A process for doing that is described here.

Another approach is to use the formulas for solving quartics. See here.

If you want to write software to solve quartics, ask again, because there are some nasty numerical problems that you'll need to deal with.

Title: Solving a Quartic Function

Body:
A user named 'Uzdawi' from another post asked a question about how to solve the quartic function of

One of the responses included an answer from the user 'Peđa Terzić', which is as follows:

Could anyone kindly elaborate on the method used to get to the answer please.

And how the same method could be used to solve the following function:

If not the same method used by Peđa Terzić, how else could you approach to solving a quartic function?

Thank you for your time.