Title: Solving $2^{16^x} = 16^{2^x}$

Body:
Find $x$ if $2^{16^x} = 16^{2^x}$.

If $x = 0$ we have $2=16$.

$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.

$256$ or $2^8$ difference. I can't see anything useful in figuring this out.

I have a feeling $x$ must be a fraction. I know only know about logarithm. I haven't learned anything about the number $e$ or $ln$ yet.

Body:
$16^{2^x}=(2^4)^{2^x}=2^{4\cdot 2^x}=2^{2^2\cdot 2^x}=2^{2^{x+2}}$. So
$$2^{16^x}=16^{2^x}\iff 2^{16^x}=2^{2^{x+2}}\iff$$ $$\iff 16^x=2^{x+2}\iff 2^{4x}=2^{x+2} \iff 4x=x+2 \iff x=\frac{2}{3}.
$$


问:
Failure of group definition with weaker axioms
In "A First Course in Abstract Algebra", Edition 7, p.43, Fraleigh writes that

It is possible to give axioms for a group $\left<G,*\right>$ that seem at first glance to be weaker, namely:

The binary operation $*$ on $G$ is associative.

There exists a left identity element $e$ in $G$ such that $e*x = x$ for all $x \in G$.

For each $a \in G$ there exists a left inverse $a'$ in $G$ such that $a' * a = e$

From this one-sided definition, one can prove that the left identity element is also a right identity element, and a left inverse is also a right inverse for the same element. Thus these axioms should not be weaker, since they result in exactly the same structures being called groups. It is conceivable that it might be easier in some cases to check these left axioms that to check our two-sided axioms. Of course, by symmetry it is clear that there are also right axioms for a group.

Does the above few axioms assume that the right identity element exists in the first place?

Consider $a * b = \left|a\right|b$. There are at least two possible solutions for a "right identity element", namely:

$$-1 * x = -1 \implies x = -1$$
$$1 * x = 1 \implies x = 1$$

Even though the group with operation $*$ satisfies axiom 2 of the weaker definition, it seems that $\left<G,*\right>$ cannot be a group, because using the original axioms of a group, there is no identity element $e$ such that $e*a = a * e = a$ for all $a \in G$.

How then do the above axioms "result in exactly the same structures being called groups"?

答:
Also worth mentioning that the left inverse exists for every element of the group. You can not prove that the left inverse is a right inverse without explicitly using that you can keep taking inverses.

Your -1 element does not have a left inverse, it only has 1 as a right inverse. So it is a regular semigroup but does not satisfy the weak group axioms. Every element except one can have a left inverse and that is enough to fail to be a group.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Showing that, if $|u_{n+1}|<\frac12|u_n|$ for $n>12$, the alternating series $\sum_{n>12}u_n$ converges to a non-zero number, and ...
This question was asked in my msters exam for which I am preparing and I was unable to solve two options and so asking for help here.

Let $(u_n)_{n\geq1}$ be a sequence real numbers satisfying following:

(i) $(-1)^n u_n \geq 0$ for all $n>0$.

(ii) $|u_{n+1}|<\frac12|u_n|$ for all $n>12$.

Prove that $\sum_{n>12} u_n$ converges to a non-zero real number, and that if $|u_{n-1}| < \frac12|u_n|$ for all $1<n<12$, then $\sum_{n\geq 1} u_n$ is a negative real number.

(I am writing only the options I couldn't solve.)

I am not able to prove these two assertions and would be really thankful for help.

答:
Hint

A simple induction on the 2nd condition yields
$$
{|u_n|<\left({1\over 2}\right)^{n-13}|u_{13}|\quad,\quad n\ge 14
\\
|u_n|>2^{n-1}|u_1|\quad,\quad 1<n<12
}
$$
Now try to find the maximal and/or minimal values of $\sum_{i=1}^{12}u_n$ and $\sum_{i=13}^{\infty}u_n$ using the 1st condition to conclude the 2nd part of the question.


问:
Can a function contain a variable?
For example:

f is function

x is a variable

john is a constant

Would $ f(x, john) $ be well formed? Or can a function only map constants to an object in a domain?

答:
Technically, a function doesn't "contain" a variable. A function can, however, be defined by an expression that contains a variable, and its arguments can include a variable. Your question is rather odd: if a function is defined by an expression that doesn't contain any variables, then it's a constant valued function, which technically is a function, but isn't what people generally mean by a "function". For instance, "True" can be considered a function, in that something that sends all inputs to True is a function, but it would normally be considered a constant.

The arguments of a function being variables is what's normal, and them being constants is what's unusual. Often, if a function is defined in terms of a constant, it is treated as a parameter/subscript/index: $f_{John}(x)$.

Title: Can a function contain a variable?


问:
Reformulating an expression to use Gauss's test
I have seen the following two cases for convergence testing using Gauss's test-

$\frac{u_j}{u_{j+1}} = \frac{(2j+1)(2j+2)}{2j(2j+1)-\lambda}$

For large j, my textbook reformulates the RHS expression to-

$\frac{2j+2}{2j} + \frac{B(j)}{j^2} = 1 + \frac{1}{j} + \frac{b(j)}{j^2}$

In a different question, I need to reformulate the following expression -

$\frac{n^2 + a_{1}n + a_0}{n^2+b_1n+b_0}$

What is the right way to reformulate these expressions?

P.S: I do not have the reformulated expression for the second expression.

答:
Just factor out the $n^2$ and use the usual $\dfrac1{1+x}=1-x+x^2+o(x^2)$ for $x\approx 0$. This is a technique you should master. So you get
\begin{align*}
\frac{n^2+a_1n+a_0}{n^2+b_1n+b_0} &= \frac{1+\frac{a_1}n+\frac{a_0}{n^2}}{1+\frac{b_1}n+\frac{b_0}{n^2}} \\ &= \big(1+\frac{a_1}n+\frac{a_0}{n^2}\big)\big(1-\frac{b_1}n+\frac{b_1^2-b_0}{n^2}+o(\tfrac1{n^2})\big).
\end{align*}
Can you finish? (You probably do not need this many terms, but you should learn the technique regardless.)


问:
Proving by contradiction

Consider the statement:  For all $x, y, z ∈ Z$.  At least one of
$x−y$, $x−z$ and $y−z$ is even. Prove this statement by contradiction.

So the contradictory statement would be that neither one of $x−y$, $x−z$ and $y−z$ is even, eg they're all odd. I know how to do the proof, but with proof by contradiction, would I be allowed to let $x$,$y$, and $z$ equal actual integers, eg $1,2,3$ or would I have to make them equal eg. $2k+3$, $2m+6$ and $2n+8$ to complete this proof?

答:
With regards to the question "Would I be able to let $x,y,z$ equal actual integers?":

No. One example of the converse being right is sufficient to disprove a claim, however one example of the converse being wrong is not sufficient to prove it.

You must assume that $x-z, y-z, x-y$ are arbitrary odd numbers, and find a way to show that this makes no sense.

Title: Proving by contradiction


问:
Solving $2^{16^x} = 16^{2^x}$

Find $x$ if $2^{16^x} = 16^{2^x}$.

If $x = 0$ we have $2=16$.

$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.

$256$ or $2^8$ difference. I can't see anything useful in figuring this out.

I have a feeling $x$ must be a fraction. I know only know about logarithm. I haven't learned anything about the number $e$ or $ln$ yet.

答:
If you take the $\log_2$ of both sides you end up with
\begin{align*}
\log_2(2^{16^x}) &= 16^x \log_2(2) = 16^x = 2^x\cdot8^x\\
\log_2(16^{2^x}) &= 2^x\log_2{16} = 2^x\cdot4
\end{align*}
From this you get that $2^x\cdot 4 = 2^x \cdot 8^x \implies 8^x = 4 \implies x =\log_{8}(4) = 2/3.$

($\log_8(4) = \log_8(2^2) = 2\log_8(2) = 2\cdot1/3$)

Title: Solving $2^{16^x} = 16^{2^x}$


问:
A Problem on Euler's totient function.
Find all m,n such that φ(mn)=φ(n)

here φ is the euler's totient function.

My attempt:
So i split this up into two cases, note: { (m,n) = gcd(m,n) }

Case 1: (m,n)=1 , Then we'd get φ(m)φ(n)=φ(n) which would imply m=1 and n is any integer or m=2 and n is an odd integer. Simple enough,

Case 2: (m,n) is not 1. Well here, things get tricky. I could not figure out much so i checked a lot of examples and came up with φ(m)φ(n) = $φ(mn) / (m,n)$ if (m,n) is a prime. I am not even sure if this result is correct or not. And if (m,n) was composite, well then i could't understand what to even do.

Any help/hint would be appreciated, thanks!

答:
There is no solution in case 2.

If there is a prime number $p$ that divides both $m$ and $n$, then we would have $\varphi(pn)\mid \varphi(mn)$, and also $\varphi(pn) = p\varphi(n)$.

This means that $\varphi(mn)$ cannot be equal to $\varphi(n)$ in that case.

Title: A Problem on Euler's totient function.


问:
solving complex roots for a particular equation
I am reading this book. I don't understand how did he get the second answer $1 + i$.

Solve the quadratic equation $z^2 - 3z + 3 + i = 0$.

Body: The roots are$$\frac{3+(1-2i)}2\quad\text{and}\quad\frac{3-(1-2i)}2$$and$$\frac{3+(1-2i)}2=2-i\quad\text{and}\quad\frac{3-(1-2i)}2=1+i.$$


问:
LCM of 3 numbers is given then what can't GCD be
This question was asked to me by a junior and I was unable to solve it and in fact couldn't even start despite of having studied considerable number theory.

$\text{lcm}$ of three different numbers is $120$ then which of the following can't be it's HCF?

(A) $8$

(B) $12$

(C) $24$

(D) $35$

I have studied elementary number theory from David Burton but I am absolutely clueless about this particular problem.

I could think of $\gcd(a,b) \text{lcm}(a,b)=ab$ , but is this extendable  to more than two variables? and even if it is extendable it doesn't seem to be useful.

Thanks for any help!!

答: It's $(D)$ because $7$ is a factor of $35$ but it's not a factor of $120$


问:
Proving Riordan's identity that $\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1$
In a book titled  "Advances in Problem Solving," authored by Sailesh Shirali, Riordan's identity is mentioned which can also be written as
$$S=\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1~~~~(1)$$
Here, we prove it by the integral representation of $(j+1)!$ as:
$$S=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{(j+1)!}{n^{j+1}}=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{1}{n^{j+1}} \int_{0}^{\infty} x^{j+1} e^{-x} dx$$
$$\implies S=\frac{1}{n^{n}}\int_{0}^{\infty} x ~(n+x)^{n-1} e^{-x} dx=n^{-n}\int_{0}^{\infty} [(x+n)^n-n(x+n)^{n-1}]~ e^{-x} dx$$ $$S=-n^{-n}\left. (x+n)^n~ e^{-x}\right|_{0}^{\infty}=1.$$

It will be interesting to see other approaches for proving (1).

Body:
We seek to show that

$$S= \sum_{k=0}^{n-1} {n-1\choose k} \frac{(k+1)!}{n^{k+1}} = 1.$$

The sum is

$$\frac{(n-1)!}{n}
\sum_{k=0}^{n-1} \frac{k+1}{(n-1-k)!} \frac{1}{n^k}.$$

We get without the factor  in front

$$\sum_{k=0}^{n-1} \frac{-n+k+1}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{n}{(n-1-k)!} \frac{1}{n^k}
\\ = - \sum_{k=0}^{n-2} \frac{n-1-k}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = - \sum_{k=0}^{n-2} \frac{1}{(n-2-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = - \sum_{k=1}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = \frac{1}{(n-1)!} \frac{1}{n^{-1}}.$$

Restoring the factor in front we find

$$\frac{(n-1)!}{n} \times
\frac{1}{(n-1)!} \frac{1}{n^{-1}} = 1$$

as claimed.


问:
Why do we need to find the intersection between these lines?
We have the functions

$$ x = -1 + 2 \cos(t)$$

$$ y = 3 + 2 \sin(t)$$

They give P's orbit

with $t$ on $\left[0, \dfrac{3}{2} \pi\right]$

Find (to 2 decimal places accurate) for which values of t the Point t is left of the line $x=-2.$

So we're allowed to use our calculators, and the answer model says that to find the solution you first need to find the intersection between $\;y=-2\;$ and $\;y=-1 + 2\cos (x)$.

My question is; why does this work?

答:
I agree with Ross Millikan that there is a typo, and the suggestion is to solve the intersection between $x = -2$ and $x = -1 + 2\cos t$.

Solving for $t$ at that intersection will give you the value of $t$ exactly where $x = -2$. Knowing that, you can then determine which values of $t$ must therefore lie to the left of the line $x = -2$.

Essentially, the lines will intersect when $x = -2$ and $x = -1 + 2\cos t$, i.e., when $$-2 = -1 + 2\cos t\iff 2\cos t = -1\iff \cos t = \frac{-1}{2}$$

So, when $\cos t = -\frac 12$, we're on the line $x = -2$. So when $\cos t \lt -\frac 12$, we're to the left of the line $x = -2$.


答:
You know that $\sqrt{-3-4i} = 1-2i$ and that $z = \frac{3\pm\sqrt{-3-4i}}{2}$.

All you then have to do is plug in the value of the square root, i.e.

$$z = \frac{3\pm1-2i}{2}$$
$$z_{1}=\frac{3+1-2i}{2}=2-i$$

$$z_{1}=\frac{3-(1-2i)}{2}=\frac{3-1+2i)}{2}=1+i$$

Hope this answers your question.

Title: solving complex roots for a particular equation

Body:
I am reading this book. I don't understand how did he get the second answer $1 + i$.

Solve the quadratic equation $z^2 - 3z + 3 + i = 0$.


Body:
$$z^2 - 3z + 3 + i = 0\to \\
z=\frac{3\pm\sqrt{9-4(1)(3+i)}}{2}\\
z=\frac{3\pm\sqrt{-3-4i}}{2}\\
z=\frac{3\pm\sqrt{1-4-2(2i)}}{2}\\
z=\frac{3\pm\sqrt{(1-2i)^2}}{2}$$ take a look at $$(1-2i)^2=1^2+4i^2-4i=1-4-4i$$ so
$$z=\frac{3\pm\sqrt{(1-2i)^2}}{2}=\frac{3\pm(1-2i)}{2}$$


问:
Divisibility proof through prime-factorization.
I have to show that the following expression:

$$ \frac {1}{2} (3^{2^{n}}-1) $$

can be divided by $n-1$ different and odd prime numbers for every positive $n$ (I assume that $n \in \mathbb{N^*}$ )

So far I tried to prove it with the Lemma that claims, that every number $n \in \mathbb{N} :n \ge 2 $ has at least one prime factor, which seemed not to be utmost successful.

I am looking for a direction and for a better understanding of such proofs.

Thanks in advance for your help :)

答:
The fact in Tito Eliatron's question comment can be used in solving this problem. As stated there,

$$3^{2^{n}} - 1 = \left(3^{2^{n-1}} - 1\right)\left(3^{2^{n-1}} + 1\right) \tag{1}\label{eq1A}$$

You can repeat this factoring with $3^{2^{n-1}} - 1$ and each additional smaller power of $2$ factor down to $2^0 = 1$ to end up with

$$\begin{equation}\begin{aligned}
\frac{1}{2}(3^{2^{n}} - 1) & = \frac{1}{2}\left(3^{2^{0}} - 1\right)\left(3^{2^{0}} + 1\right)\left(3^{2^{1}} + 1\right)\ldots \left(3^{2^{n-1}} + 1\right) \\
& = 4\prod_{i=1}^{n-1}\left(3^{2^{i}} + 1\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Each factor in the product above has $3$ with an exponent $\ge 2$. Consider any one of them to be just a power of $2$, say for some exponent $2^i = j$. Thus, for some $k \ge 2$, we would have

$$3^{j} + 1 = 2^k \implies 2^k - 3^{j} = 1 \tag{3}\label{eq3A}$$

However, Catalan's conjecture, which was proven by Preda Mihăilescu in $2002$, states the only solution for a difference  of positive integers to a power of at least $2$ to be $1$ are for $3^2 - 2^3 = 1$. Thus, this means \eqref{eq3A} will never occur.

As such, each $3^{2^{i}} + 1$ for $i \ge 1$ must have at least one odd prime factor. Also, all of the odd prime factors in $3^{2^{i}} + 1$ for any $i$ are distinct from those of any other value of $i$. To see this, consider for any $i \ge 1$ an odd prime $p \mid 3^{2^{i}} + 1$. This gives, for any integer $m \gt 0$,

$$\begin{equation}\begin{aligned}
3^{2^{i}} & \equiv -1 \pmod{p} \\
3^{2^{i}\left(2^{m}\right)} & \equiv \left(-1\right)^{2^m} \pmod{p} \\
3^{2^{i+m}} & \equiv 1 \pmod{p}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

However, $p \mid 3^{2^{i+m}} + 1$ requires $3^{2^{i+m}} \equiv -1 \pmod{p}$ which means, from \eqref{eq4A}, that $-1 \equiv 1 \pmod{p}$, but this is never true for any odd prime $p$.

This confirms there's at least one distinct odd prime factor among each of the $n - 1$ factors of $3^{2^{i}} + 1$ in \eqref{eq2A}.


问:
Stacking identical sequences with minimal overlap
I stumbled upon the following problem. As computer scientist, I don't know all the correct terminology, which makes it hard to search for existing discussion surrounding this or related problems. If those exist, I guess a pointer to them would already bring me closer to understanding my problem.

The Problem

Given a sequence of elements ( using [ 1, 2, 3, 4, 5, … ] as example in the following), I want to stack multiple copies of this sequence in such a way, that each element overlaps at most once with another element. Pursuing a greedy strategy (always placing the next sequence as soon as possible), leads to the following construction:

```<code>Seq. 1:  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
Seq. 2:    1 2 3 4 5 6 7 8  9 10 11 12 13 14 ...
Seq. 3:        1 2 3 4 5 6  7  8  9 10 11 12 ...
Seq. 4:                1 2  3  4  5  6  7  8 ...
Seq. 5:                              1  2  3 ...

Offsets: |1|<2>|<--4-->|<-----5----->|<------...
</code>```

I say that two elements overlap if they appear in the same column. Thus, since the 1 of the $2^{nd}$ sequence overlaps with the 2 of the $1^{st}$ sequence, any following sequences' 1 cannot overlap with a 2 of any other sequence, or vica versa. I do think, however, that, if this is solved for one element, it holds for all elements.

I define the offset of the $n^{th}$ sequence as how much further it is indented compared to the $(n-1)^{th}$ sequence. Thus, the offsets for the $2^{nd}$ to $5^{th}$ sequence, as displayed here are [1, 2, 4, 5]. The list for more subsequent offsets looks like this: [1, 2, 4, 5, 8, 10, 14, 21, 15, 16, 26, 25, 34, 22, 48, 38, 71, 40, 74, …]. Here, I honestly expected exponential growth. Instead, the $n^{th}$ offset may be smaller than the $(n-1)^{th}$ offset, which is counterintuitive to me.

My questions

Is this, or a problem this can be reduced to, known under a common name that I could look up?

Is my greedy strategy, to always place the sequence as soon as possible, optimal? Or, does there exist some other strategy, where the sum of all offsets is smaller for the same number of stacked sequences?

Does there exist a formula that computes the list of offsets, without having to exhaustively try all possibilities?

答:
Instead of looking at the offsets between adjacent sequences, I want to look at the offset from the start: in your example, the five sequences have total offsets $0, 1, 3, 7, 12$.

If the first $n$ sequences start at total offsets $a_1, a_2, \dots, a_n$, then the overlap condition essentially says that all the pairwise differences $a_i - a_j$ must be distinct. (That's because the $1$ of the $i^{\text{th}}$ sequence overlaps with $1 - a_i + a_j$ of the $j^{\text{th}}$ sequence.) A finite choice of offsets of this type is called a Golomb ruler.

The greedy strategy is already not optimal for five sequences: total offsets of $0, 1, 4, 9, 11$ also work. (This corresponds to offsets of $1, 3, 4, 2$ in your notation.) In fact, for four sequences, $0, 1, 4, 6$ is better than the greedy solution $0, 1, 3, 7$.

As a lower bound, $a_n \ge \binom n2$, because the $\binom n2$ differences $a_j - a_i$ with $1 \le i < j \le n$ must all be in the range $\{1, 2, \dots, a_n\}$. If we just want to solve the problem for a finite $n$, then the Wikipedia article mentions the Erdős–Turán construction $$a_{k+1} = 2pk+ (k^2 \bmod p) \qquad k=0, \dots, n-1$$ where $p$ is the smallest (odd) prime $\ge n$, and this satisfies $a_n = O(n^2)$.

A repeated difference $a_i - a_j = a_k - a_\ell$ is equivalent to a repeated sum $a_i + a_\ell = a_j + a_k$. (Here, we allow repetition between the indices $i,j,k,\ell$, except for the trivial case $a_i - a_j = a_i - a_j$.) Sequences without such a repeated sum are called Sidon sequences. I am mentioning them because the Wikipedia page quotes some result about infinite Sidon sequences as well as finite one, if that's what you're interested in.


问:
Bound for edge chromatic number given size of maximum matching
Problem: A maximum matching in $G$ contains $m$ edges. Show that $\chi(G)\geq \frac{|E(G)|}{m}$, where $\chi$ denotes the minimum number of edge coloring.

I know that each edge must be assigned a colour. Since assigning colors to a graph corresponds  to a graph matching, the number of colors assigned must be less than or equal to $m$ by assumption.

But i'm not sure how to proceed.I am aware that this requires a combinatorial argument, but since I have not taken an official course in combinatorics and my experience with it is almost zero, please explain your argument in detail.

答:
Pick a coloring using $\chi (G)$ color, then
$$E(G)=\sum _{i=1}^{\chi (G)}\text{# of edges colored with color }i.$$
Notice now that the edges colored with one color form a matching on the graph, but this quantity is bounded by $m,$ so
$$E(G)=\sum _{i=1}^{\chi (G)}\text{# of edges colored with color }i\leq m\cdot \chi (G).$$


问:
Convergence of $\int_0^1 \sqrt[3]{\ln(1/x)} \mathrm{d}x $
Does $$\int_0^1 \sqrt[3]{\ln\left(\frac{1}{x}\right)} \mathrm{d}x$$ converge? WA says  it is equal to $\Gamma(4/3)$, however calculating the antiderivative seems approachless to me and can't compare to other functions. I would appreciate any help. Thanks.</p>

答:
Hint: Substitute $x=e^{-t}$ in the integral.  You should end up with

$$\int_0^{\infty} dt \, e^{-t} \,t^{1/3}$$


问:
Proving the duplication formula
I'm trying to prove the famous Legendre duplication formula:

$$ \Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}}   $$

I have to prove this using these two identities/theorems:

$$ \Gamma(z)= \lim_{n \rightarrow \infty}  \frac{n! n^{z}}{z(z+1)...(z+n)}$$

and

$$ (2n)! \sim \frac{2^{2n} (n!)^2}{\sqrt{\pi n}}   $$ as $n \rightarrow \infty $

I put in $ z = 2z $ in the first one and obtained:

$$ \Gamma(2z) = \lim_{n \rightarrow \infty}  \frac{n! n^{2z}}{2z(2z+1)...(2z+n)} $$.

I then tried multiplying both the numerator and denominator by various factors to try to incorporate the other identity, but it was of no use. Can anyone tell me what direction I'm supposed to go from here?

答:
We have (with all limits for $n\to\infty$):
$$
\begin{aligned}
\Gamma(z)
&=
\lim \frac{n!\; n^{z}}{z(z+1)(z+2)\dots(z+n)}\ ,
\\
\Gamma\left(z+\frac 12\right)
&=
\lim
\frac{n!\; n^{z+\frac 12}}
{\left(z+\frac 12\right)\left(z+\frac 32\right)\left(z+\frac 52\right)\dots\left(z+\frac {2n+1}2\right)}\ ,
\\[3mm]
\Gamma(z)\Gamma\left(z+\frac 12\right)
&=
\lim \frac{n!^2\; n^{2z+\frac 12}}{z\left(z+\frac 12\right)(z+1)\left(z+\frac 32\right)(z+2)\left(z+\frac 52\right)\dots(z+n)\left(z+\frac {2n+1}2\right)}\ ,\\
&=
\lim \frac{n!^2\; n^{2z+\frac 12}\; 2^{2n+2}}{(2z)(2z+1)(2z+2)\dots(2z+2n)(2z+2n+1)}\ ,
\\[3mm]
\Gamma(2z)
&=
\lim \frac{n!\; n^{2z}}{(2z)(2z+1)(2z+2)\dots(2z+n)}\ ,
\\
&=
\lim \frac{(2n)!\; (2n)^{2z}}{(2z)(2z+1)(2z+2)\dots(2z+2n)}\ ,\qquad\text{(subsequence)}
\\[3mm]
\frac
{\Gamma(z)\Gamma\left(z+\frac 12\right)}
{\Gamma(2z)}
&=
\lim\frac
{n!^2\; n^{2z+\frac 12}\; 2^{2n+2}}
{(2n)!\; (2n)^{2z}}
\cdot
\frac{1}{2z+2n+1}\qquad\text{ (now use Stirling)}
\\
&=
\lim\frac
{\displaystyle
 {\color{blue}{\left(\frac ne\right)^n}} \sqrt {2\pi n}\;
 {\color{blue}{\left(\frac ne\right)^n}} \sqrt {2\pi n}\;
{\color{green}{n^{2z}}}\cdot n^{\frac 12}\; \color{red}{2^{2n+\color{navy}2}} }
{\displaystyle
 {\color{blue}{\left(\frac {{\color{red}{2}}n}e\right)^{2n}}}\sqrt {2\pi \;2n}\;
\; 2^{2z}\;
{\color{green}{n^{2z}}}}
\cdot
\frac{1}{2z+2n+1}
\\
&=
\lim\frac{\sqrt {2\pi}\cdot \sqrt {2\pi}}{\sqrt {2\pi}\cdot\sqrt 2}
\cdot
\frac{\sqrt n\cdot\sqrt n\cdot n^{\frac 12}}{\sqrt n(2z+2n+1)}
\cdot
\frac{\color{red}2^{\color{navy}2}}{2^{2z}}
\\
&=\frac{\sqrt \pi}{2^{2z-1}}\ ,
\end{aligned}
$$
hence the formula.

Title: Proving the duplication formula


问:
Reference request for Statistical Learning Theory
I want to learn about Statistical Learning Theory. I have a strong background in measure theory, probability theory and functional analysis. What would be a good resource/book to start learning Statistical Learning Theory in the most general setting?

Thanks a lot in advance.

答:
I think the following books can be very useful:

Shai Shalev-Shwartz and Shai Ben-David. Understanding Machine Learning: From Theory to Algorithms.

Mehryar Mohri, Afshin Rostamizadeh and Amit Talwar. Foundations of Machine Learning

Michael Kearns and Umesh Vazirani. An Introduction to Computational Learning Theory. MIT Press, 1994.

Also, you can find the following courses helpful:

Understanding Machine Learning - Shai Ben-David

Computational Learning Theory by Varun Kanade


问:
Prove there exist a unique $n \in \mathbb{N}$ such that $ n \leq \sqrt{10} \leq n+1$
this is my first year in college and my first lesson in algebra is logic and i have this question as an application of the Existential quantification that demand to  Show that :

$\exists n\in \mathbb{N}$ such that $n \leq \sqrt{10} \leq  n+1 $

we know that $\sqrt{10} >\sqrt{9}=3 $

and $3\in \mathbb{N}$ , so 3 verifies the inéquation

My problem is how should i write the demonstration properly and show the unicity .

答:
$9<10<16 \Rightarrow 3=\sqrt9<\sqrt{10}<\sqrt{16}=4$. So $n=3$ works.

FOr the univity. Assume thare are two of them, say $n$ and $m$. That is $n\le \sqrt{10}\le n+1$ and $m\le\sqrt{10}\le m+1$.

$n\le \sqrt{10}\le n+1$ and $-m-1\le-\sqrt{10}\le -m$. By substracting them, you'll get

$$n-m-1\le 0\le n-m +1 \iff m-1\le n\le m+1$$.

If $n=m-1$ would imply that $\sqrt{10}\in[n,n+1]\cap[n+1,n+2]={n+1}$, so $\sqrt{10}\in\Bbb N$, wich is impossible.

The same occur in the case $n=m+1\iff m=n-1$.

So the only possibility is $n=m$ and you're done.


答:
Let $M$ be a matching such that $|M| = m$ and consider a minimal partition of edge set $$E(G) = M_1 \cup M_2 \cup ... \cup M_{k-1} \cup M$$
where each $M_i$ is a matching. In other words, we are partitioning the edge set of $G$ into matchings (note that there is no unique way to do such thing but there is a way always). Then, we know that $|M_i| \le m,\ \forall i \in \{1,2,...,k-1\}$. Thus, we can see that $|E(G)| \le mk$ since matchings $M_i$ and $M$ have no edges in common and $|M_i| \le m$, $|M| = m$. Therefore, $k \ge \frac{|E(G)|}{m}$. Here, since the matchings have no edges in common, it is clear that $\chi(G) \le k$. So, we must show that $\chi(G) = k$.

Suppose for a contradiction that $\chi(G) < k$. Then, by Pigeonhole Principle, at least two elements of $\{M_1, M_2, ..., M_{k-1}, M\}$ have the same colors for their edges. But then, their union is also a matching (since they are colored the same color, there is no two adjacent edges in their union). But then we obtained a matching partition of $k-1$ elements, contradicting the minimality of our first matching partition.

Title: Bound for edge chromatic number given size of maximum matching

Body:
Problem: A maximum matching in $G$ contains $m$ edges. Show that $\chi(G)\geq \frac{|E(G)|}{m}$, where $\chi$ denotes the minimum number of edge coloring.

I know that each edge must be assigned a colour. Since assigning colors to a graph corresponds  to a graph matching, the number of colors assigned must be less than or equal to $m$ by assumption.

But i'm not sure how to proceed.I am aware that this requires a combinatorial argument, but since I have not taken an official course in combinatorics and my experience with it is almost zero, please explain your argument in detail.


问:
why $V= V_{q_1} \cap ....\cap V_{q_n} ? $ why not $V= V_{q_1} \cup ....\cup V_{q_n} ?$
I have some confusion in Baby  Rudin book

My confusion is given marked in red box

My confusion:  why $V= V_{q_1} \cap ....\cap V_{q_n} ? $ why not  $V= V_{q_1} \cup ....\cup V_{q_n} ?$

My attempt :  I thinks  we can write $V= V_{q_1} \cup ....\cup V_{q_n} $ because aribtrary union of open set are open

答: Because you are after a neighborhood of $p$ which does not intersect$$W=W_{q_1}\cup W_{q_2}\cup\cdots\cup W_{q_n}.$$And, although $V_{q_1}\cup V_{q_2}\cup\cdots\cup V_{q_n}$ is indeed a neighborhood of $p$, there is no reason to suppose that it does not intersect $W$. On the other hand, $V_{q_1}\cap V_{q_2}\cap\cdots\cap V_{q_n}$ is also a neighborhood of $p$ and, furthermore, it is sure that it does not intersect $W$.


问:
contradicting identity theorem?
the identity theorem for holomorphic functions states: given functions $f$ and $g$ holomorphic on a connected open set $D$, if $f = g$ on some open subset of $D$, then $f = g$ on $D$

Let $f(z) = \sin z + \cos z$,
    $g(z) = \cos z$

then $(f-g)(0) = 0$
$\implies$ by the above theorem : $f(z) = g(z)$ on $\left|z\right|<1$
$\implies$ $\sin z + \cos z = \cos z$
$\implies \sin z =$ identically zero in $\left|z\right|<1$ which is not true

Where could I be making a mistake ?

答: To conclude that $f=g$ on $D$, you must show that $f=g$ on some open subset of $D$. You did not show that, you just noticed that $f(0)=g(0)$, but $\{0\}$ is not an open subset of $D$.

Title: contradicting identity theorem?


问:
Finding the limit of a sequence $a_n$
I have the following sequence $$ a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1} $$ I want to find $\lim_{n\to\infty}a_n$, I know it is 0, but I can't seem to find a way to show it. It seems like $\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$ is an (good) approximation of $x^3$, but it isn't a taylor series or anything like it I know of. Since it is ( / it looks like) an appromixation, once I try to apply the sandwich rule, I get large numbers, which don't help me. It also seems that Cauchy criteria doesn't make it easier.

答:
HINT:

From Bernoulli's Inequality,

$$\frac1{1+\frac{n-1}{n^3}}\ge \left(1-\frac1{n^3}\right)^{n-1}\ge 1-\frac{n-1}{n^3}$$

Use these inequalities and apply the squeeze theorem.

Can you finish now?

Title: Finding the limit of a sequence $a_n$


问:
Proving by contradiction

Consider the statement:  For all $x, y, z ∈ Z$.  At least one of
$x−y$, $x−z$ and $y−z$ is even. Prove this statement by contradiction.

So the contradictory statement would be that neither one of $x−y$, $x−z$ and $y−z$ is even, eg they're all odd. I know how to do the proof, but with proof by contradiction, would I be allowed to let $x$,$y$, and $z$ equal actual integers, eg $1,2,3$ or would I have to make them equal eg. $2k+3$, $2m+6$ and $2n+8$ to complete this proof?

答:
Alternative approach : pigeonhole principle.

Given integers, $a,b$ then $2 | (a-b) \iff a,b$ have the same parity (odd or even).

There are only two parities : odd or even.  Given three #'s, $x,y,z$ then two of them must have the same parity.

Title: Proving by contradiction


Title: Finding the limit of a sequence $a_n$

Body: I have the following sequence $$ a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1} $$ I want to find $\lim_{n\to\infty}a_n$, I know it is 0, but I can't seem to find a way to show it. It seems like $\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$ is an (good) approximation of $x^3$, but it isn't a taylor series or anything like it I know of. Since it is ( / it looks like) an appromixation, once I try to apply the sandwich rule, I get large numbers, which don't help me. It also seems that Cauchy criteria doesn't make it easier.

Body: Hint: Rewrite your sequence as $$a_{n} = n^{3}- n^{3}\left(\frac{n^{3}+n-1}{n^{3}-1}\right)\left(\frac{n^{3}-1}{n^{3}}\right)^{n}$$


问:
Weak-* convergence in $L^\infty$ of a certain sequence of functions
Let $a < b$, $0 < \lambda < 1$, $U = (0,1) \subset \mathbb{R}$, and consider the sequence of functions

$$ f_k(x) = \begin{cases} a & \mbox{if } \frac{j}{k} \leq x \leq \frac{j+\lambda}{k}, \quad j = 0,1, \dots, k-1,\\ b & \mbox{otherwise}. \end{cases} $$

I'd like to show that $f_k \overset{*}{\rightharpoonup} f := \lambda a + (1-\lambda)b$ in $L^\infty(U)$. Of course, to this purpose one should prove that for each $g \in L^1(U)$ it holds

$$\int_U f_k g \, dx \to \int_U fg \,dx \quad \mbox{as } k \to \infty.$$

Clearly,

$$\int_U f_k g \, dx = (a - b) \sum_{j=0}^{k-1} \int_{j/k}^{(j+\lambda)/k} g(x) \, dx + b \int_0^1 g(x) \,dx,$$

so that to conclude it suffices to prove $\lim_{k\to \infty}\sum_{j=0}^{k-1} \int_{j/k}^{(j+\lambda)/k} g(x) \, dx = \lambda \int_0^1 g(x)\,dx$, which is intuitively quite evident (take $g = \mbox{const.}$, for instance).

We have of course

$$\sum_{j=0}^{k-1} \int_{j/k}^{(j+\lambda)/k} g(x) \, dx = \sum_{j=0}^{k-1} \int_0^1 \chi_{\left(\frac{j}{k},\frac{j+\lambda}{k}\right)}(x) g(x) \, dx,$$

but now I'm stuck. I do not understand what the trick is to conclude convergence. The dominated (monotone) convergence theorem cannot be applied, because $\{f_k\}$ does not have a pointwise a.e. limit (neither it is monotone). I tried to make changes of variables but this brings $j$ and/or $k$ in the argument of the integrand and I do not clearly see how to pass to the limit, although I guess that the missing point should be something really elementary that I am not able to see right now. So, how to finish the proof?

答:
Here's a trick: If $g$ is continuous you have $\int_0^1 f_kg\to\int_0^1fg$. Next note that your sequence $f_k$ is bounded in $L^\infty$ and that the continuous functions are dense in $L^1$, so if $g\in L^1$ you have for any $\epsilon>0$ a continuous $\tilde g$ with $\|g-\tilde g\|_1≤\epsilon$, whence:

$$\left|\int (f_k -f)g\ \right| ≤\left|\int (f_k -f ) \tilde g\ \right| + \|f_k-f\|_\infty\cdot\|g-\tilde g\|_1$$
The first term on the right goes to $0$ since $\tilde g$ is continuous. The second term is smaller than $\sup_{k}\|f_k-f\|_\infty\cdot \epsilon$, here the first part is some constant and the $\epsilon$ is arbitrary.

So why does $\int f_k g\to \int f g$ for continuous $g$? Note that $[0,1]$ is compact so any continuous $g$ is uniformly continuous. So for an arbitrary $\epsilon>0$ you have a $\delta$ with $|g(x)-g(y)|≤\epsilon$ for any $|x-y|<\delta$. In particular if you take $k>\frac1\delta$ you have that
$$\int_{j/k}^{(j+\lambda)/k}g(x)dx = \frac{\lambda}{k}\,[g(j/k)+\epsilon\, \tau_j]$$
where $\tau_j$ is some number of modulus $≤1$. If you sum up all these $j$ you get a Riemann sum:
$$\lambda\frac1k\sum_{j} g(j/k) + \epsilon\,\lambda\frac1k\sum_j \tau_j$$
the first sum goes to $\lambda \int_0^1 g$ as $k\to \infty$, the second is bounded by $\epsilon\lambda$, where $\epsilon$ is arbitrary.


问:
Proving that an nth degree polynomial eventually strictly increases / decreases.
Problem: Let $f(x)$ be an n-th degree real polynomial. Show that $\exists N \in \mathbb{R}$ such that $f(x)$ is either strictly increasing or decreasing on $[N, \infty)$.

I know that by the Fundamental Theorem of Algebra, this polynomial has at most $n$ roots, so past the rightmost root $k$, the function is either strictly positive or strictly negative, but I have no clue how to prove it is strictly increasing or decreasing  past a certain $N$. I know I could try to prove that the function is one to one past $N$ as well, but I have no clue how to do that either. Any tips/help would be greatly appreciated.

答: Hint : use the fondamental theorem of algebra on the derivative.</p>


问:
Multiplicative modular group is internal direct product of subgroups
Let $U(st)$ be the multiplicative modular group mod$st$ where $s,t$ are relatively prime integers.

Let $U_s(st)=\{x \in U(st)\mid x \pmod{s}=1\}$, $U_t(st)=\{x \in U(st)\mid x \pmod{t}=1\}$

Show that $U(st) =U_s(st) \times U_t(st)$

Attempt:
Since the group is abelian, any subgroups or normal.

I know that because $s,t$ are relatively prime $U_s(st) \cap U_t(st)=\{1\}$ however I do not know how to show this. Letting $x \in U_s(st) \cap U_t(st), x=1\pmod{s},x=1 \pmod{t}\implies x=1$. But I am not very confident why?

答:
Hint:

Use the Chinese Remainder theorem: we have a ring isomorphism
$$\mathbf{Z}/st\mathbf Z\simeq\mathbf Z/s\mathbf Z\times \mathbf Z/t\mathbf Z,$$
hence…


答: These two functions must coïncide on an open subset. In your reasoning you use only one point, which is not an open subset.</p>

Title: contradicting identity theorem?

Body:
the identity theorem for holomorphic functions states: given functions $f$ and $g$ holomorphic on a connected open set $D$, if $f = g$ on some open subset of $D$, then $f = g$ on $D$

Let $f(z) = \sin z + \cos z$,
    $g(z) = \cos z$

then $(f-g)(0) = 0$
$\implies$ by the above theorem : $f(z) = g(z)$ on $\left|z\right|<1$
$\implies$ $\sin z + \cos z = \cos z$
$\implies \sin z =$ identically zero in $\left|z\right|<1$ which is not true

Where could I be making a mistake ?


Title: Finding the limit of a sequence $a_n$

Body:
We can use that $\log(1+x)=x+O(x^2)$ and $e^x=1+x+O(x^2)$ as $x \to 0$ to obtain

$$\left(1-\frac{1}{n^{3}}\right)^{n-1}=e^{(n-1)\log\left(1-\frac{1}{n^{3}}\right)}=e^{-\frac{n-1}{n^3}+O\left(\frac1{n^5}\right)}=1-\frac{n-1}{n^3}+O\left(\frac1{n^5}\right)$$

and therefore

$$n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}=n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{n-1}{n^3}+O\left(\frac1{n^5}\right)\right)$$


问:
On well-definedness of $f : Y \to X/H$ for Hatcher's Algebraic Topology, Exercise 1.3.24(a)
I refer to Exercise 1.3.24(a) of Hatcher's Algebraic Topology:

Given a covering space action of a group $G$ on a path-connected, locally path-connected space $X$, each subgroup $H \subseteq G$ determines a composition of covering spaces $X \to X/H \to X/G$. Show that every path-connected covering space between $X$ and $X/G$ is isomorphic to $X/H$ for
some subgroup $H \subseteq G$.

The common tactic is to define $H \subseteq G$ as the subgroup of deck transformations of the covering map $p : X \to Y$, which induces a second covering map $p' : X \to X/H$, then show that $p$ and $p'$ are isomorphic. Then, define maps $f : Y \to X/H$ and $g : X/H \to Y$ as follows: For $x \in p^{-1}(y)$, define $f(y) = Hx$, and $g(Hx) = p(x)$. We conclude by showing that they are well-defined, continuous, and induces the required isomorphism.

My main issue lies in the well-definedness part. The well-definedness of $g$ is clear, but I struggle to show the same for $f$. Given $x_1,x_2 \in X$ such that $p(x_1) = p(x_2) = y$, we want to show that $Hx_1 = Hx_2$, or equivalently $x_2 = hx_1$ for some $h \in H$. Since $p(x_1) = p(x_2)$, the normality of the covering space $X \to X/G$ ensures the existence of $g \in G$ such that $gx_1 = x_2$.

Now if $g \in H$, then we are done. However, since $p : X \to Y$ may not be normal, I'm not sure how to go about showing this.

Any help is appreciated.

答: Showing that $g\in H$ amounts to showing that for all $x\in X, p(gx) = p(x)$. So fix $x\in X$. Since $X$ is path connected, there is a path $\alpha\colon [0,1]\to X$ from $x_1$ to $x$. Let $q$ denote the covering map $X\to X/G$. Then $q\alpha = qg\alpha$ and both $p\alpha$ and $pg\alpha$ are lifts, both starting at $y$. Since $Y\to X/G$ is a covering, uniqueness of lifts implies that $p(x) = p\alpha(1) = pg\alpha(1) =p(gx)$.


答:
Here is a more analytic proof.

Let $$f(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$$ so the derivative is $$f'(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \dots + 2a_2 x + a_1.$$ We can factor out the first term to obtain $$f'(x) = na_n x^{n-1} \left( 1 + \frac{n-1}{n} \frac{a_{n-1}}{a_n} \frac{1}{x} + \dots + \frac{2}{n} \frac{a_2}{a_{n}} \frac{1}{x^{n-1}} + \frac{1}{n}\frac{a_1}{a_n}\frac{1}{x^{n-1}} \right) =na_n x^{n-1} \left( 1 + g(x) \right), $$
where $g(x) \to 0$ as $x \to \pm \infty$. So, there exists $N$ such that $|x| > N$ implies $g(x) \in (-1,1)$ and thus $1 + g(x) \in (0,2)$, which is positive. This allows us to fix the sign of the derivative for $|x|$ large enough.

Title: Proving that an nth degree polynomial eventually strictly increases / decreases.

Body:
Problem: Let $f(x)$ be an n-th degree real polynomial. Show that $\exists N \in \mathbb{R}$ such that $f(x)$ is either strictly increasing or decreasing on $[N, \infty)$.

I know that by the Fundamental Theorem of Algebra, this polynomial has at most $n$ roots, so past the rightmost root $k$, the function is either strictly positive or strictly negative, but I have no clue how to prove it is strictly increasing or decreasing  past a certain $N$. I know I could try to prove that the function is one to one past $N$ as well, but I have no clue how to do that either. Any tips/help would be greatly appreciated.


问:
What does $\lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}}$ evaluate to?
Problem:

Suppose that $f$ is twice continuously differentiable within a
neighborhood of the point $a$ and $f''(a) = 4$. Compute \begin{align*}
 \lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}} \end{align*}

I have done this work
\begin{align*}
\lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}} &= \lim_{x \to a} \frac{\frac{f(x)-f(a)}{x-a} - f'(a)}{x-a} \\
&= \lim_{x\to a}\frac{f'(a)-f'(a)}{x-a} \\
&= 0
\end{align*}
This makes sense, but I am not confident in my answer because I didn't use the given that $f''(a) = 4$. Any help would be greatly appreciated.

答:
According to Taylor's theorem, if $f$ is twice differentiable in $a$ (most versions require continuous differentiability, but it isn't actually necessary), then there exists a remainder function $r(x)$ with
$$\lim_{x\to a}\frac{r(x)}{(x-a)^2}=0$$
such that
$$f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+r(x).$$
With a little bit of algebra we get that
$$\frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2}=\frac{r(x)}{(x-a)^2}+\frac{1}{2}f''(a).$$
The limit now evaluates to $\frac{1}{2}f''(a)$.


问:
What intuition stands behind implicit differentiation
I'm trying to undestand implicit differentation

Let's take as a an example equation y^2 + x^2 = 1

1. How i think about how the equation works

I think the function as : if x changes then the y term have to hold value of "y^2 + x^2" equal 1. Therefore the equation defines some set of numbers at x cordinates and y cordinates.

2. How i think about how differentate the equation

If i want to know how the equation changes as x changes, i'm taking
derivative with respect to x

$\frac{d}{dx}y^2+\frac{d}{dx}x^2=\frac{d}{dx}1$

We can consider $y$ as a function, $y = f(x)$

Therefore: $\frac{d}{dx}(f(x))^2+\frac{d}{dx}x^2=\frac{d}{dx}1$

We can calculate how (f(x))^2 changes as f(x) changes, using chain rule.

$\frac{df(x)}{dx}\frac{d}{df(x)}(f(x))^2+\frac{d}{dx}x^2=\frac{d}{dx}1$

This is equal: $2f(x)\frac{df(x)}{dx}f(x)+\frac{d}{dx}x^2=\frac{d}{dx}1$

As $x$ changes, $x^2$ changes as $2x$, therefore $2f(x)\frac{df(x)}{dx}f(x)+2x=\frac{d}{dx}1$

As x changes, 1 doesn't changes, therefore it is 0.  $2f(x)\frac{df(x)}{dx}f(x)+2x=0$

We don't know derivative of $f(x)$ but we can solve it

If we solve the derivative, we get $f'(x) = -\frac xy$

3. Questions

My way of thinking is right?

What does mean the final answer? It looks strange, it doesn't tell me nothing comparing to norma, explicit derivative of a function.

There is a difference between $\frac{dy}{dx}$ and $\frac{d}{dx}y$ ?

Why i want to know ? Because i want to know how to interpretate steps and solution, not only algorithmically solve some book's problems.

PS. I'm barely after highschool - Therefore i don't know yet set theorem and other high level math things. I'm learning calculus on my own.

答:
You can think of it as nudge both side of the equation and do the derivative.

Example:
For function,

$$y^q = x^p$$

If you increase $x \to x + \Delta x $ then $y \to y+\Delta y $

Or,

$$ (y+ \Delta   y)^q = ( x+ \Delta x )^p$$

Now substract this from original eqn

$$ (y+ \Delta y  )^q - y^q =
(x+ \Delta x)^p - x^p$$
divide both side by $\Delta x $

$$ \frac{ (y+ \Delta y )^q - y^q }{ \Delta x}=\frac{ (x+ \Delta x)^p - x^p}{\Delta x}$$

Now on left side,

$$ \frac{ (y+ \Delta x  )^q - y^q }{ \Delta x} =  \frac{ (y+ \Delta x  )^q - y^q }{ \Delta y} \frac{\Delta y}{\Delta x}$$

Now taking limit on both side of equation as the deltas drop down to zero. You get derivatives!

i.e:

$$ qy^{q-1} \frac{dy}{dx} = px^{p-1}$$


问:
Compute the chromatic polynomial of the $n$-cycle graph, i.e., the only connected graph on $n$ vertices each of which has degree $2$.
Compute the chromatic polynomial of the $n$-cycle graph, i.e., the only connected
graph on $n$ vertices each of which has degree $2$.

I can only think of how to compute this recursively. That is, using the fact that
$$p_G(k)=p_{G-e}(k)-p_{G/e}(k)$$
where $p_G{k}$ is the chromatic polynomial of $G,$ who has some edge $e$, $G-e$ is the graph obtained by the deletion of $e$, and $G/e$ is the graph obtained by the contraction of $e$, we can see that for some $n$-cycle
$$p_{G-e}(k)=\text{some tree on }n \text{ vertices}\\p_{G/e}(k)=\{\text{a }(n-1) \text{-cycle}\}$$
We know that a tree on $n$ vertices has chromatic polynomial
$$p_{G-e}(k)=k(k-1)^{n-1}$$
we can then use this same trick for $p_{G/e}(k)$, and then again and again, eventually recursively building our desired polynomial. I'm not sure this is what was intended for this exercise, as my professor made it seem like we should have some closed formula. Any help would be amazing!

答:
If you want a solution that does not involve a sum, note for some simple cases of $n$ what the chromatic polynomial is. For instance,
$$n=2 \implies p_G(k)=k(k-1)=(k-1)^2+(k-1)\\n=3\implies p_G(k)=k(k-1)(k-2)=(k-1)^3-(k-1)\\n=4 \implies p_G(k)=k(k-1)(k^2-3k+3)=(k-1)^4+(k-1)$$
you can then prove by induction on $n$ that the chromatic polynomial of any $n$-cycle is$$p_G(k)=(k-1)^n+(-1)^n(k-1)$$


问:
Locally finite or not

I am tryıng to learn locally finite and can you give an explanation for my green writing please thank you

答:
I answered the first one here: it’s not even point-finite, because $0$ is in all of the sets.

The same holds true for the cover $\{[0,n]:n\in\Bbb Z^+\}$ of $[1,\to)$: $0$ is in every one of the sets $[0,n]$, so this cover is not even point-finite. In fact, every non-negative real number is in infinitely many of these sets: if $x\ge 0$, there is an integer $n\ge x$, and then $x\in[0,k]$ for every $k\ge n$.

Title: Locally finite or not


答: It's not $0$ in general. Your fallacy was in neglecting the fact that $\frac{f(x)-f(a)}{x-a}-f^\prime(a)\in O(x-a)$. By L'Hôpital's rule, the limit is$$\lim_{x\to a}\frac{f^\prime(x)-f^\prime(a)}{2(x-a)}=\tfrac12f^{\prime\prime}(a).$$

Title: What does $\lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}}$ evaluate to?

Body:
Problem:

Suppose that $f$ is twice continuously differentiable within a
neighborhood of the point $a$ and $f''(a) = 4$. Compute \begin{align*}
 \lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}} \end{align*}

I have done this work
\begin{align*}
\lim_{x \to a} \frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^{2}} &= \lim_{x \to a} \frac{\frac{f(x)-f(a)}{x-a} - f'(a)}{x-a} \\
&= \lim_{x\to a}\frac{f'(a)-f'(a)}{x-a} \\
&= 0
\end{align*}
This makes sense, but I am not confident in my answer because I didn't use the given that $f''(a) = 4$. Any help would be greatly appreciated.


问:
What does $\sum \mu(d)\lfloor{\frac{x}{d}}\rfloor$ count?
Where the sum is over all $d$, where $d$ only has prime factors $\leq \sqrt{x}$. I'm trying to determine, in plain English, precisely what this sum is counting.

I know the Möbius function will be 0 if $p^2|x$ when $p|x$, and $(-1)^{k}$ when $x$ is the product of $k$ distinct primes. Further, I understand that $\lfloor{\frac{x}{d}}\rfloor$ determines the number of multiples of $d\leq x$, I'm just struggling to put these functions together in my head. I know that the only such $d$'s that will contribute to the sum will be the product of distinct primes, to the exponent 1, where the primes are less than or equal to $\sqrt{x}$.

I know that my question is related to Legendre's formula ($\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$,), but I'd like to understand it more combinatorially.

答:
The sum counts the positive integers $\leqslant x$ that are not divisible by any prime $\leqslant \sqrt{x}$. Those integers are $1$ (not divisible by any prime whatsoever) and the primes $\sqrt{x} < p \leqslant x$ (for every composite $1 < n \leqslant x$ is divisible by a prime $q \leqslant \sqrt{n} \leqslant \sqrt{x}$). Accepting this assertion for the moment without proof, we see that
$$\sum_{d \mid P} \mu(d)\biggl\lfloor \frac{x}{d}\biggr\rfloor = 1 + \bigl(\pi(x) - \pi(\sqrt{x})\bigr)$$
where $P$ is the product of all primes $\leqslant \sqrt{x}$.

To see that the sum is indeed counting what I claimed, let's begin with taking a look at the inclusion-exclusion principle. If we have a finite set $U$ and a finite family $\{A_{\kappa} : \kappa \in K\}$ of subsets of $U$, the IE-principle gives us a formula for the cardinality of
$$U \setminus \bigcup_{\kappa \in K} A_{\kappa}\,.$$
For a more compact notation, let us define
$$V_J = \bigcap_{\kappa \in J} A_{\kappa}$$
for all $J \subseteq K$. Since $U$ is our universe of discourse, $V_{\varnothing} = U$ (this is a convention, but like the conventions for empty sums and empty products, it's a convention supported by good practical reasons).

Then the IE-principle says
$$\Biggl\lvert\, U \setminus \bigcup_{\kappa \in K} A_{\kappa}\Biggr\rvert = \sum_{J \subseteq K} (-1)^{\lvert J\rvert}\cdot \lvert V_J\rvert = \sum_{r = 0}^{\lvert K\rvert} (-1)^r\sum_{\substack{J \subseteq K \\ \lvert J\rvert = r}} \lvert V_J\rvert\,.$$
The counting on the right hand side goes: First we count the elements of $U$ ($r = 0$). Then we subtract the sizes of all the $A_{\kappa}$ ($r = 1$). But the elements belonging to two $A_{\kappa}$ have been subtracted twice, so we must re-add their number once ($r = 2$). But the elements belonging to three $A_{\kappa}$ have now been subtracted thrice, and re-added thrice, so effectively haven't been subtracted at all. Therefore we subtract their count ($r = 3$). But the elements belonging to four $A_{\kappa}$ have been subtracted $\binom{4}{1}$ times, then re-added $\binom{4}{2}$ times, and subtraced again $\binom{4}{3}$ times. Thus overall they've been subtracted $4 - 6 + 4 = 2$ times, and we must re-add them once ($r = 4$). And so on.

In total, an element belonging to exactly $m$ of the $A_{\kappa}$ is counted
$$\sum_{r = 0}^{m} \binom{m}{r} (-1)^r = (1 - 1)^m = \begin{cases}1 &\text{if } m = 0 \\ 0 &\text{if } m > 0 \end{cases}$$
times in this formula.

Here, our universe is the set of positive integers not exceeding $x$, thus $U = \{ n \in \mathbb{N} : 1 \leqslant n \leqslant x\}$. The index set is the set of primes not exceeding $\sqrt{x}$, i.e. $K = \{ p \in \mathbb{N} : p \leqslant \sqrt{x} \text{ and } p \text{ is prime}\}$, and $A_p$ is the set of multiples of $p$ in $U$. For $J \subseteq K$ we have
$$V_J = \{n \in U : p \in J \implies p \mid n\}\,,$$
and since distinct primes are pairwise coprime we can also write this as
$$V_J = \{n \in U : d_J \mid n\}$$
where
$$d_J = \prod_{p \in J} p\,.$$
In this form it is easy to see
$$\lvert V_J\rvert = \biggl\lfloor \frac{x}{d_J}\biggr\rfloor$$
and the inclusion-exclusion formula gives us
$$1 + \bigl(\pi(x) - \pi(\sqrt{x})\bigr) = \sum_{J \subseteq K} (-1)^{\lvert J\rvert} \lvert V_J\rvert = \sum_{J \subseteq K} (-1)^{\lvert J\rvert}\biggl\lfloor \frac{x}{d_J}\biggr\rfloor\,.$$
Now it only remains to observe that $J \mapsto d_J$ is a bijection between the subsets of $K$ and the divisors of $d_K$, and that for each divisor we have $(-1)^{\lvert J\rvert} = \mu(d_J)$. With this observation, it is clear that
$$\sum_{d \mid P} \mu(d)\biggl\lfloor \frac{x}{d}\biggr\rfloor$$
is exactly the sum in the inclusion-exclusion formula counting the positive integers $\leqslant x$ not divisible by any prime $\leqslant \sqrt{x}$.


问:
Weighted projective space and projective space are isomorphic
From Vakil's book: Exercise 8.2.N

Show that the weighted projective space $\mathbb{P}(m, n) = Proj(k[x, y])$ (where $x$ and $y$ have degrees $m$ and $n$ respectively) is isomorphic to $\mathbb{P}^1$.

Can anyone give an outline of the solution to this?

Edit:
For convenience I identify $k[x, y]$ with $k[u^m, v^n]$. I tried to see if there is any graded isomorphism between $k[u^m, v^n]$ and the graded ring  ${k[w, z]}^{(d)}$ for some $d$ that divides the gcd of $m$ and $n$. Then it would follow from the fact $\mathbb{P}^1 = Proj(k[w, z]) \cong Proj({k[w, z]}^{(d)})$. But I see that it isn't the case.

答:
Let $d$ be the greatest common divisor of $m$ and $n$. We write $m=m_{1}d,n=n_{1}d,l=m_{1}n_{1}d$.

Consider the morphism $\mathrm{Proj}(k[x,y]_{l})\rightarrow\mathbb{P}^{1}=\mathrm{Proj}(S,T)$ induced by the map of graded rings $S\mapsto x^{n_{1}}, T\mapsto y^{m_{1}}$, where $k[x,y]_{k}$ means the $l$th Veronese subring of $k[x,y]$ (Vakil.6.4.4). It's easily checked that this is an isomorphism. Since $\mathrm{Proj}(k[x,y]_{l})\simeq \mathrm{Proj}(k[x,y])$, we have $\mathbb{P}(m,n)\simeq \mathbb{P}^{1}$.


问:
What does it mean to divide?
So I'm in year 11 (10th grade for americans) and I have a really stupid question..

What does it mean to divide in maths?

I recently looked it up and it said "To divide means to slit the number up into equal groups". However, other sites said "To divide means to see how many times the divisor fits into the dividend"

Could someone please tell me which one it is?

Thanks, sorry for such a stupid question

答: To ''divide'' is to split or place in pieces, think of it like this if you have a chocolate bar with 12 connected pieces and you break off 6 of them you can say you divided them by half because half of 12 is 6 so 12 dived by 2 (which is half) Is six.</p>

Title: What does it mean to divide?


答:
Alternative approach.  Assume that there exists $m \in \mathbb{Z^+}$ such that $m \neq 3$ and $m \leq \sqrt{10} \leq (m+1).$

$\underline{\text{case 1:}}$

$m < 3 \implies \sqrt{10} \leq (m+1) \leq 3 \implies 10 \leq (3^2) = 9 ~ \otimes.$

$\underline{\text{case 2:}}$

$m > 3 \implies \sqrt{10} \geq m \geq 4 \implies 10 \geq (4^2) = 16 ~ \otimes.$

Title: Prove there exist a unique $n \in \mathbb{N}$ such that $ n \leq \sqrt{10} \leq n+1$

Body:
this is my first year in college and my first lesson in algebra is logic and i have this question as an application of the Existential quantification that demand to  Show that :

$\exists n\in \mathbb{N}$ such that $n \leq \sqrt{10} \leq  n+1 $

we know that $\sqrt{10} >\sqrt{9}=3 $

and $3\in \mathbb{N}$ , so 3 verifies the inéquation

My problem is how should i write the demonstration properly and show the unicity .


问:
What is the meaning of $\sin(0.876726)$ here?
one of the solution of $x^2-\sin(x)=0$ is $x\approx 0.876726$  according to Wolfram alpha

Therefor $0.876726^2-\sin(0.876726)\approx0$

But my question is what is the meaning of $\sin(0.876726)$ and how to calculate its value?

I guess it means $0.876726$ radian. If it is true I don't understand why root of $f(x)=x^2-\sin(x)$ should be presented as radian.

答:
When talking about the $\sin(x)$, $x$ is always in radians.

Whenever you have something like $\sin(30^\circ)$ it actually means $\sin(\frac{30^\circ\pi}{180^\circ})$.

To calculate the value of $x$ in the equation:

$$x^2 - \sin(x) = 0$$ you would have to use a numerical method, such as Newton-Raphson.

In this case, you would consider an initial guess $x_0$ such as $x_0 =1$ and then apply the following iteration function.

$$x_{i+1}=x_i - \frac{f(x_i)}{f'(x_i)}$$

Where $f(x)=x^2-\sin(x)$ and $f'(x)=2x-\cos(x)$

For our choice of $x_0$ we get:
$$x_0=1$$
$$x_1=0.891395$$
$$x_2=0.876984$$
$$x_3=0.876726$$
$$x_4=0.876726$$


问:
$n$ lines in a plane, proper coloring of intersection points with just 3 colors
Draw $n$ lines in a plane so that there are no parallel lines and there are no three lines passing through the same point.

Each intersection point is colored red, green or blue. Prove that it is possible to color all intersection points in a “proper” way, so that any two adjacent points (like $A_i, A_j$) have different colors.

This also means that if you "travel" along an arbitrary line, you will cross $n-1$ intersection points, constantly changing colors from one intersection point to another.

My first (and last) try was to use induction. Obviously for two or three lines, we have one or three intersection points and with three colors available we have the base of induction proved. However, the induction step is more difficult. I was able to prove the induction step if in every possible arrangement of lines it was possible to find a line that divides the plane into two halves with one half having no intersection points. However, I was able to construct counter-examples where such line does not exist.

The last time I tried to solve a similar red-green-blue problem I discovered Ramsey theory. I wonder what I will discover this time :)

答:
One can prove this using the Szekeres-Wilf Theorem: For a graph $G$, $$\chi(G) \le 1 + \max_{H\subseteq G} \delta(H).$$

Indeed, it can be shown that any subgraph of the given graph has a vertex of degree at most $2$ (look at the "right-most" vertex.)

For reference, see West's Introduction to Graph Theory Section $5.1$ Pg $196$ for a proof of the Szekeres-Wilf Theorem.


问:
Normalizing a Gaussian function
I want to find the normalization constant for the function $f(x)=e^{-\frac{x^{2}}{2}}$.

I start by calculating the norm of $f(x)$ over $\mathbb{R}$:
$$ \newcommand{\dx}{\text{d}x}||f||^{2} = \int\limits_{-\infty}^{\infty}\left[e^{-\frac{x^2}{2}}\right]^{2} \dx = \int\limits_{-\infty}^{\infty}e^{\frac{-2x^{2}}{2}}\dx = \int\limits_{-\infty}^{\infty}e^{-x^{2}}\dx = \sqrt{\pi}.$$

However, I know that the answer should be $||f||=\sqrt{2\pi}$, and thus $||f||^{2}=2\pi$.

What am I missing here?..

答: Normalization ensures $\int f(x)\ dx = 1$, not that the integral of the square of the functions equals 1.

Title: Normalizing a Gaussian function


问:
A question about Lebesgue measure and integration.
Let $(X,A,\mu)$ be a measurable space.
$f\colon X\to [0,\infty]$ measurable.

Denote $S=\lbrace x\in X : f(x)<1\rbrace$.

Prove

a. $\mu(S)=\lim_{n\to \infty} \int_S e^{-f^n}\,d\mu$

b. Assume $X=S$ , prove that:

$$\sum_{n=1}^{\infty} \int_X f^n\, d\mu= \int_{X} \frac{f}{1-f}\,d\mu$$

I have difficulty where I'm supposed to move between integrals.
We can define the measure in terms of integral. So in a I know that it is the start but it is not clear to me how to connect between the set $S$ to the integral.

In b, this is what I did:
$\sum_{n=1}^{\infty} \int_X f^n \,d\mu$= (by a theory int of sum =sum of int for a sequence of positive measurable functions)
=$\int_X (\sum_{n=1}^{\infty}f^n)\, d\mu$=(sum of a geometry column with $f(x)<1$)= $\int_X \frac{f}{1-f}\, d\mu$.

答:
Hint:

For (a) -

We want to write $\mu(S)$ as the limit of some integrals... So it might be useful to write $\mu(S)$ as an integral. Of course, we know how to do this: $\mu(S) = \int_S 1 \mathrm{d}\mu$.

Notice this already looks kind of like the limit we're meant to compute. If we can show

$$\lim_{n \to \infty} \int_S e^{-f^n} \mathrm{d} \mu = \int_S 1 \mathrm{d} \mu$$

then we're done!

If you play around with this, you might notice $f^n \to 0$ pointwise on $S$ (since $0 \leq f(x) < 1$ for each $x \in S$). Then $e^{-f^n} \to e^{-0} = 1$.

So the stuff we're integrating limits to $1$, which is what we need it to be!

If only we knew how to relate the limit of a sequence of integrals to the integral of the limit of the integrands...

For (b) -

You're on the exact right track! Do you know how to justify the fact that $\sum \int \text{stuff} = \int \sum \text{stuff}$? Since infinite sums are simply limits of finite sums, you'll want to use the same technique of swapping limits and integrals from part (a).

I hope this helps ^_^


答:
Fix $x,y,z\in\mathbb{Z}$ and suppose all of $x-y, x-z,y-z$ were all not even. Then they must all be odd. We know that an odd integer plus or minus an odd integer returns an even integer, so we will check some cases.

$(x-y)-(x-z)=x-y-x+z=z-y$ which is even. However, because we know $-(y-z)=z-y$, then we have that an even integer is equal to an odd integer, a contradiction.

You can check through the other possible cases to show that none of the configurations will hold.

Title: Proving by contradiction

Body:
Consider the statement:  For all $x, y, z ∈ Z$.  At least one of
$x−y$, $x−z$ and $y−z$ is even. Prove this statement by contradiction.

So the contradictory statement would be that neither one of $x−y$, $x−z$ and $y−z$ is even, eg they're all odd. I know how to do the proof, but with proof by contradiction, would I be allowed to let $x$,$y$, and $z$ equal actual integers, eg $1,2,3$ or would I have to make them equal eg. $2k+3$, $2m+6$ and $2n+8$ to complete this proof?


问:
Is this $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}$ a bounded sequence with $F_n$ is a Fibonacci sequence?
I'm interesting to know about boundedness of the following sequence $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}$ with $F_n$ is a Fibonacci sequence  such that I have tried to give its bound , we have $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}\geq   \dfrac{F_n+F_{n-1}}{F_n}\geq 1+\dfrac{F_{n-1}}{F_n} $ using limit on both sides of that inequality we come up to $\lim_{n\to \infty} \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}\geq  \lim_{n\to \infty} \dfrac{F_n+F_{n-1}}{F_n}\geq \lim_{n\to \infty} \big(1+\dfrac{F_{n-1}}{F_n}) =1+\dfrac{1}{\phi}$ but this it doesn't give any validity about boundedness of the titled sequence ? any help ?

答:
Note that
$$
\frac{F_n + F_{n-1}}{F_n - F_{n-1}} = \frac{F_{n+1}}{F_{n-2}}.
$$
We find that
$$
\lim_{n \to \infty} \frac{F_n + F_{n-1}}{F_n - F_{n-1}} =
\left(\lim_{n \to \infty} \frac{F_{n+1}}{F_n}\right) \cdot
\left(\lim_{n \to \infty} \frac{F_{n}}{F_{n-1}}\right) \cdot
\left(\lim_{n \to \infty} \frac{F_{n-1}}{F_{n-2}}\right) = \phi^3 = 2 \phi + 1,
$$
where $\phi = \frac 12(1 + \sqrt{5})$.
Because this sequence has a limit, it must be bounded.


问:
$2(u + v)$ and $2(u-v)$ are orthogonal
Let be two vectors $u$ and $v$ of $\Bbb R^n$. We assume that $||u|| = ||v||$.

The vectors $u$ and $v$ are unitary.

Vectors $2(u + v)$ and $2(u-v)$ are orthogonal.

The vectors $u$ and $v$ are dependent.

All the other answers are wrong.

The vectors $u$ and $v$ are orthogonal.

I have the find the one which is true.

I think that only "$2(u + v)$ and $2(u-v)$ are orthogonal" is true, but it is unclear how I can do that. I know that two vectors $x$ and $y$ are orthogonal if and only if $||x+y||^2 = ||x||^2 + ||y||^2$.

答: Two vectors $y$ and $z$ are orthogonal if $y \cdot z$ is 0. Now setting $y=2(u+v)$ and $z=2(u-v)$, it follows that $y \cdot z = 4||u||^2 - 4||v||^2$. If both $||u||$ and $||v||$ are the same then, it follows that $y \cdot z$ is 0, and so $y=2(u+v)$ and $z=2(u-v)$ are orhtogonal.

Title: $2(u + v)$ and $2(u-v)$ are orthogonal


问:
Find the coordinates of the point P on the line d : 2x − y − 5 = 0, for which the sum AP + PB attains its minimum, when A(−7; 1) and B(−5; 5).
I tried to use the Cosine theorem and get that |(PA)^2 + (PB)^2 - (AB)^2|<=|2PAPB|. Can someone explain to me what should I do? Thank you!

Body:
hint

The coordinates of the point $ P $, will have the form $ (x,2x-5)$

So

$$AP=\sqrt{(x+7)^2+(2x-5-1)^2}$$
$$=\sqrt{5(x^2-2x+17)}$$
and

$$BP=\sqrt{(x+5)^2+(2x-5-5)^2}$$
$$=\sqrt{5(x^2-6x+25)}$$


问:
Using the method of undetermined coefficients, find the general solution of the differential equation
$y''-3y'+2y=te^t$ is the equation I need to work on. I have gotten the roots 2,1 using the quadratic formula. The homogenous solution would be $Y_c = c_1e^{2t} + c_2e^{t}$. I then let $y =Ae^{t}$ because $g(t)=2e^{t}$, plug in $Y$ to get $(Ae^{t})''-3(Ae^{t})'+2(Ae^{t})=te^{t}$ I am stuck after this, I believe I just have to derive and solve for $Y$ somehow.

答:
$$y''-3y'+2y=te^t$$
$$\implies y_h=c_1e^t+c_2e^{2t}$$
For the particular solution, your guess should be:
$$y_p=t(At+B)e^t$$


问:
$N$ day contract, with a probability of $P$ of extending the contract by K days indefinitely. Probability of the contract lasting $J$ days.
On any given day there is a likelihood ($P$) of extending the contract a another $K$ days. I know I can expect the contract to last $\frac{N}{1-K\times P}$ days (KP<1), but I have no idea how to calculate the probability of it lasting exactly $J$ days. I attempted turning it into a binary sequence, but the rule prohibiting when the extension can occur (i.e. $J$ could equal 14, but if $N$ is 4 and $K$ is 10, $K$ could not occur after 4 days have passed because the contract would have already ended) is stumping me. Any help appreciated.

答:
Suppose the contract lasts for $J$ days. Let $X$ be the number of days where the contract was extended, and $Y$ be the number of days where it was not. Then
$$
X+Y=J\\
X+Y-KX=N
$$
These two equations imply that $X=(J-N)/K$, and $Y=J-(J-N)/K$. This means every outcome lasting $J$ days consists of some ordering of $X+Y$ days, where $X$ of the days are extensions, and $Y$ are not. The total number of ways to choose this ordering is
$$
\binom{X+Y}{X}=\binom{J}{(J-N)/K}
$$
However, not all orderings are valid, in the sense that some of them will cause the contract to end too early. For example, when $N=2, J=5, K=3$, then we get $X=1$ and $Y=4$, for $\binom{5}1$ possible orderings. However, only the orderings $xyyyy$ are valid $yxyyy$; for example $yyxyy$ would mean the contract already ended on day $2$.

It turns out that counting the number of valid arrangements is solved exactly by the generalized ballot problem; see https://www.researchgate.net/publication/252059419_Four_Proofs_of_the_Ballot_Theorem. Using this, the probability of lasting $J$ days exactly is equal to
$$
\frac{N}J\binom{J}{X}P^X(1-P)^{J-X}
$$
where $X=(J-N)/K$.


问:
Solving simultaneous equations with multiple variables.
If $(x+a)^2$ is a factor of $x^3 + 6px + k$,
show that $k + 2a^3 = 0$

I've tried different ways to solve this, but is it even possible?

答:
Let $p(x)$ denote the 3rd degree polynomial.

Your condition means that $-a$ is a double root of $p$, therefore it is also a root of the derivative $p(x)=3x^2+6p$.

As a consequence, we have :

$$\begin{cases}p(a)&=&-a^3-6pa+k&=&0\\p'(a)&=&3a^2+6p&=&0 \end{cases}$$

from which you easily derive your constraint.


答:
Original answer: using differentiation.

The fact that $p(x)=x^3+6px+k$ is divisible by $x+a$ at all means that $p(-a)=0$, i.e. $-a^3-6pa+k=0$.

The fact that $p(x)$ is divisible by $(x+a)^2$ also means that $p'(x)=3x^2+6p$ is divisible by $x+a$, i.e. $p'(-a)=3a^2+6p=0$.

Thus, from the second equation we get $6p=-3a^2$, and by substituting this in the first, we get $-a^3+3a^2\cdot a+k=0$, i.e. $2a^3+k=0$, q.e.d.

Update: Alternative solution which does not use differentiation.

Let us extract the factor $(x+a)^2=x^2+2ax+a^2$ from the polynomial $x^3+6px+k$. The remainder should be zero:

$$\begin{array}{rcl}x^3+6px+k&=&x(x^2+2ax+a^2)+(-2a)x^2+(-a^2+6p)x+k\\&=&x(x^2+2ax+a^2)+(-2a)(x^2+2ax+a^2)+4a^2x+2a^3+(-a^2+6p)x+k\\&=&(x-2a)(x^2+2ax+a^2)+(3a^2+6p)x+(2a^3+k)\end{array}$$

So, the quotient in division of $x^3+6px+k$ with $(x+a)^2$ is $x-2a$, and the remainder, which must be $0$, is $(3a^2+6p)x+(2a^3+k)$. The last polynomial (of degree $1$) is zero only if its coefficients are zero, which lets us conclude that $3a^2+6p=0$ and, most importantly, $2a^3+k=0$, as desired.

Title: Solving simultaneous equations with multiple variables.

Body:
If $(x+a)^2$ is a factor of $x^3 + 6px + k$,
show that $k + 2a^3 = 0$

I've tried different ways to solve this, but is it even possible?


答:
A nice way to see that is using the Polarization Identity, that is, $\langle x,y\rangle=\dfrac14\left(|x+y|^2-|x-y|^2\right)$. Setting $x=2(u+v)$ and $y=2(u-v)$ follows
$$\langle 2(u+v),2(u-v)\rangle=\dfrac14\left(|4u|^2-|4v|^2\right)=4(|u|^2-|v|^2)=0$$

Title: $2(u + v)$ and $2(u-v)$ are orthogonal

Body:
Let be two vectors $u$ and $v$ of $\Bbb R^n$. We assume that $||u|| = ||v||$.

The vectors $u$ and $v$ are unitary.

Vectors $2(u + v)$ and $2(u-v)$ are orthogonal.

The vectors $u$ and $v$ are dependent.

All the other answers are wrong.

The vectors $u$ and $v$ are orthogonal.

I have the find the one which is true.

I think that only "$2(u + v)$ and $2(u-v)$ are orthogonal" is true, but it is unclear how I can do that. I know that two vectors $x$ and $y$ are orthogonal if and only if $||x+y||^2 = ||x||^2 + ||y||^2$.


答:
We have
$$
\dfrac{F_{n}}{F_{n-1}} \to \phi
$$
and so
$$
\frac{F_n + F_{n-1}}{F_n - F_{n-1}}
= \frac{\dfrac{F_{n}}{F_{n-1}}+1}{\dfrac{F_{n}}{F_{n-1}}-1}
\to \frac{\phi+1}{\phi-1}
=(\phi+1)\phi=\phi^2+\phi=2\phi+1
$$
Thus the sequence converges and so is bounded.

Alternatively, and simpler:

We have
$
0 < \frac{x+1}{x-1} < 5
$
for $x > \frac{3}{2}$. Now $\frac{F_{n}}{F_{n-1}} > \frac{3}{2}$ for $n \ge 5$; this follows easily by induction.

Title: Is this $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}$ a bounded sequence with $F_n$ is a Fibonacci sequence?

Body: I'm interesting to know about boundedness of the following sequence $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}$ with $F_n$ is a Fibonacci sequence  such that I have tried to give its bound , we have $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}\geq   \dfrac{F_n+F_{n-1}}{F_n}\geq 1+\dfrac{F_{n-1}}{F_n} $ using limit on both sides of that inequality we come up to $\lim_{n\to \infty} \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}\geq  \lim_{n\to \infty} \dfrac{F_n+F_{n-1}}{F_n}\geq \lim_{n\to \infty} \big(1+\dfrac{F_{n-1}}{F_n}) =1+\dfrac{1}{\phi}$ but this it doesn't give any validity about boundedness of the titled sequence ? any help ?


问:
How to compute $\int \frac{\text{sgn}(x-t)}{1-t} dt$?
I can tell that I should be getting logarithms because of the $\dfrac1{1-t}$, but the $\text{sgn}(x-t)$ is messing with me, and I don’t know how to proceed.

I tried writing $\text{sgn}(x-t)$ as $\dfrac{\sqrt{(x-t)^2}}{x-t}$, but got the wrong answer. How do you solve $$\int\frac{\text{sgn}(x-t)}{1-t} dt\;?$$

Any help is greatly appreciated!

答:
Since
$$\text{sgn}(x-t)=\begin{cases} 1, \ \text{if} \ x > t \\ 0, \ \text{if} \ x=t \\ -1 \ \text{if} \ x<t\end{cases}$$
Assuming that your interval of integration is $t\in[a,b]$, you have
$$\int_a^b \frac{\text{sgn}(x-t)}{1-t} \text{d}t=\int_a^x \frac{1}{1-t} \text{d}t +\int_x^b \frac{-1}{1-t} \text{d}t$$


问:
Why the emphasis on Projective Space in Algebraic Geometry?
I have no doubt this is a basic question. However, I am working through Miranda's book on Riemann surfaces and algebraic curves, and it has yet to be addressed.

Why does Miranda (and from what little I've seen, algebraic geometers in general) place so much emphasis on projective space when studying algebraic curves? Why is this the natural setting to conduct algebraic geometry in?

Also, projective spaces and curves in them are hard for me to visualize, and in general I don't have any good intuition about these objects. Do working algebraic geometers simply not visualize things as much, or are there some nice interpretations of projective spaces and algebraic curves I am missing that would make them seem more natural and give me more intuition about them?

答:
This is a less elementary answer, but I think it's worth considering.

(1) As someone else put it, the projective space allows "as many solutions as possible". This is made precise in Bézout's theorem very elegantly: if $X,Y\subset\Bbb{P}^n$ are distinct irreducible hypersurfaces, then
$$\sum_{P\in X\cap Y}i(P, X\cap Y)=\deg(X)\cdot \deg(Y),$$

where $i(P, X\cap Y)$ is the intersection index of $X,Y$ at $P$. The intersection indices can be defined in $\Bbb{A}^n$, but the nice equality above only works in $\Bbb{P}^n$.

So Bézout's theorem is only true in a projective space!

(2) One of the tools used in Riemann Surfaces is divisors, which are widely used in algebraic geometry. Although divisors can be defined in affine varieties, things work specially well in projective varieties.

For a simple example, let $C\subset \Bbb{P}^2$ be a smooth projective curve. For every function $f\in k(C)$, the divisor $\text{div}(f)$ consisting of all zeros minus all poles of $f$ has degree zero and if $f$ either has no zeros or no poles, then $f$ is constant. In particular the Riemann-Roch space $\mathcal{L}(0)=\{f\in k(C)\mid \text{div}(f)\geq 0\}$ is the space of constants. We can of course talk about Riemann-Roch theorem, which relates $\dim \mathcal{L}(D), \deg D$ and the genus $g$.

All of the above can only be said because the curve is projective!

(3) Another very important tool in algebraic geometry is sheaf cohomology. If $X$ is an affine scheme and $\mathcal{F}$ is a quasi-coherent sheaf, then Serre's vanishing theorem says that
$$H^i(X,\mathcal{F})=0\text{ for all }i>0,$$

which essentially means affine schemes are "cohomologically trivial".

But there's real action going on in projective schemes. Just to mention a simple example, one can show that
$$h^n(\Bbb{P^n},\mathcal{O}_{\Bbb{P^n}}(d))=\binom{-d-1}{n},\text { for }d\leq -(n+1)$$
$$\chi(\Bbb{P^n},\mathcal{O}_{\Bbb{P^n}}(d))=\frac{(d+1)(d+2)...(d+n)}{n!} \text{ for all }d\in\Bbb{Z}$$

where $\chi$ is the alternating sum of the $h^i(\Bbb{P^n},\mathcal{O}_{\Bbb{P}^n}(d))$, namely Euler's characteristic.

Take also the crucial Serre's Duality Theorem, which says $h^i(X,\mathcal{F})=h^{n-i}(X,\omega_X\otimes\mathcal{F}^*)$. One of the hypothesis of the theorem is that $X$ is a proper scheme, and the most natural example of a proper scheme is -- guess what -- a projective scheme!


问:
Is the composition series $1 \leq A_3 \leq S_3$ in $S_3$?

Is the composition series $1 \leq A_3 \leq S_3$ in $S_3$?

I thought this was because the composition factors are isomorphic to $\mathbb{Z}_2$ and $\mathbb{Z}_3$ and are therefore simple. Also the subgroups are normal. $A_3$ is normal in $S_3$ since it has index $2$?

答:
$1 \le A_3 \le S_3$ is indeed a composition series of $S_3$.

A subnormal series is a composition series if all of its composition factors are simple, and as you noted, the composition factors are isomorphic to $Z_2$ and $Z_3$, and therefore simple groups.

The trivial group is obviously normal in $A_3$, and $A_3$ is normal in $S_3$ because it has a index of 2, as you noted. therefore this is indeed a proper subnormal series.

This proves that this is a composition series.


问:
Limit of $\sin(\alpha n_k)$
Let $n_{k+1}>n_k \in \mathbb{N}$, let $\alpha \in (0,+\infty)\setminus 2\pi\mathbb{Q}$. Prove that $\sin(\alpha n_k)$ doesn't converge.

Context: showing there exists a sequence $f_n \in C[0,1]$ with no uniformly convergent subsequence.

Proof: assume $\exists \lim_k \sin(\alpha n_k)$. Then $\exists \lim_k \sin(\alpha n_k+b)=\lim_k\sin(\alpha n_k)=:l$.

Now, $\sin(\alpha n_k+b)=\sin(\alpha n_k)\cos(b) + \cos(\alpha n_k)\sin(b)$.

Therefore $(\sin(\alpha n_k+b)-\sin(\alpha n_k)\cos(b))^2= \cos(\alpha n_k)^2\sin(b)=(1-\sin(\alpha n_k)^2)\sin(b)^2$.

Taking the limit yields $l^2(1-\cos(b))^2=(1-l^2)\sin(b)^2$.

If $l \neq \pm 1$ then $\displaystyle \frac{l^2}{1-l^2}=\frac{\sin(b)^2}{(1-\cos(b))^2}$, which is not a constant. Otherwise take the reciprocal.

Question: why "$\exists \lim_k \sin(\alpha n_k)\implies\exists \lim_k \sin(\alpha n_k+b)=\lim_k\sin(\alpha n_k)$"?

The above proof is based on this post

答:
Let $\beta :=\alpha /2\pi$, then from the conditions of the exercise we knows that $\beta $ is irrational (and positive). Now, by the Kronecker approximation theorem we knows that the set $\{\{\beta n\}: n\in \mathbb{N}\}$ is dense in $[0,1]$, where $\{\beta n\}:=\beta n-\lfloor \beta n \rfloor$. Therefore $\{2\pi\{\beta n\}:n\in \mathbb{N}\}$ is dense in the set $[0,2\pi]$, so for every $x\in[0,2\pi]$ there exists an strictly increasing sequence $n_1,n_2,\ldots $ of natural numbers such that $\lim_{k\to\infty}2\pi\{\beta n_k\}=x$.

By last note that
$$
\sin (\alpha n_k)=\sin (2\pi\beta n_k)=\sin (2\pi(\lfloor \beta n_k \rfloor+\{\beta n_k\}))=\sin (2\pi\{\beta n_k\})
$$

So the statement that you want to prove is false in general.

Title: Limit of $\sin(\alpha n_k)$


问:
How to take the first derivative of the entropy-regularized Wasserstein distance?
In optimal transportation theory, it is better to regularize the Wasserstein distance with an entropy constraint because it is differentiable, unlike its unregularized counterpart. By being differentiable, it can then be treated as a loss function that is compatible with common optimization algorithms.

The entropy-regularized Wasserstein distance (aka Sinkhorn distance) is:
$$ \text{inf} _{\gamma \in \Pi } \sum \|x - y\| \enspace \gamma(x,y)- \epsilon H(\gamma)$$
where $H(\gamma) = -\sum \gamma \text{ log}(\gamma)$ is the Shannon entropy of the transport plan $\gamma$, and $\epsilon $ is the regularization parameter.

What then is the derivative of the above formula, could someone show how to derive it?

答: This paper talks about the derivative of Sinkhorn distances https://papers.nips.cc/paper/2018/file/3fc2c60b5782f641f76bcefc39fb2392-Paper.pdf</p>


问:
How to find a solution of a second order differential equation with constant RHS?
This may be a really dumb question but I've never come across this problem before, until now. I have a differential equation with constant right-hand side:

$$y''+ 4y = 5.$$
Normally I would find out the homogeneous solution which would be: $$y=c_1\cos(2t)+c_2\sin(2t).$$

But now I am stuck. I need to find the particular solution. Should/can that be just a constant? Or is it zero? Or five? If so why?

I really don't know.

Thanks.

答:
But now I am stuck. I need to find the particular solution. Should/can that be just a constant? Or is it zero? Or five? If so why?

You can answer your questions easily just try.

$$y_p''+4y_p=5$$

If you try $y_p=0$ you have:
$$(0)''+4\times 0=5$$
$$0\ne 5$$
So it's not correct.
If you try $y_p=5$ then:
$$(5)''+4 \times 5 =5$$
$$20 \ne 5$$
So it's not correct. Try a constant $y_p=A$
$$(A)''+4A=5$$
$$4A=5 \implies A=\dfrac 54$$


问:
About the absolute Galois group
I am surprised by the elements of the absolute Galois group. Except for the identity and the conjugation, we can not write other elements explicitly. Also, any automorphism of a Galois extension $K/\Bbb Q$ can be lifted to an element of the absolute Galois group.

But I have ambiguity with these facts: Consider the automorphism $\sigma(a+b\sqrt 3)=a-b\sqrt 3$ of the Galois extension of the field $K:=\Bbb Q (\sqrt 3)$. Lets denote its lift by $\tau$. Can we find an element $\alpha \in \bar{\Bbb Q}\backslash \Bbb Q (\sqrt 3)$, for which we have $\tau(\alpha) \neq \alpha$? The above fact tells us there should be some $\alpha$'s, but I mean can we write it down explicitly?

答: Consider $\sqrt[4]{3}$. If $\tau(\sqrt[4]{3})=\sqrt[4]{3}$, then $\tau(\sqrt{3})=\sqrt{3}$ which is not the case. We actually need to have $\tau(\sqrt[4]{3})=i\sqrt[4]{3}$ or $\tau(\sqrt[4]{3})=-i\sqrt[4]{3}$.

Title: About the absolute Galois group


问:
Calculate $\lim_{n \rightarrow \infty} E[X_n]$
Let's consider the following sequence : $X_1 \sim U[0,1]$. And if $X_1=a$ then $X_2 \sim U[0,a]$. Now if $X_2=b$ then $X_3 \sim U[0,b]$ and so on so on. I want to calculate $$\lim_{n \rightarrow \infty}E[X_n]$$

Hint Use, properties of supermartingales.

My work so far

Proving martingale is very easy :

$$E[X_{n+1}|\mathbb{F}_t]=\frac12 X_n \le X_n$$

Now how to use it ? Let's see

$$E[X_{n+1}|\mathbb{F}_t] \le X_n \Leftrightarrow E\bigl[E[X_{n+1}|\mathbb{F}_t]\bigr]\le E[X_n] \Leftrightarrow X_{n+1} \le E[X_n]$$

But now I stacked. Could you please give me a hand ?

Body:
You've already shown
$$E[X_{n+1}] = E[E[X_{n+1} \mid X_n]] = \frac{1}{2} E[X_n].$$
This implies $E[X_{n+1}] = \frac{1}{2^n} E[X_1] $.


问:
Find finite sum $ \sum_{k = 1}^{n} {k2^k}$
How do you find the finite sum  of $ \sum_{k = 1}^{n} {k2^k}$ .
Any hint?

No integration and differentiation.

答:
Hint: Note that
$$\sum_{k=1}^n k 2^k = \sum_{j=1}^n \sum_{k=j}^n 2^k = \sum_{j=1}^n2^{j}\sum_{k=0}^{n-j} 2^k $$
and you have a formula for
$$\sum_{k=0}^{n-j} 2^k.$$


Title: Find finite sum $ \sum_{k = 1}^{n} {k2^k}$

Body:
How do you find the finite sum  of $ \sum_{k = 1}^{n} {k2^k}$ .
Any hint?

No integration and differentiation.

Body:
Hint

Let $$S_n=\sum_{k=1}^nk2^k$$

then

$$2S_n=\sum_{k=1}^n(k+1-1)2^{k+1}$$

$$=S_n+(n+1)2^{n+1}-\sum_{k=1}^{n+1}2^k$$

Think of geometric sum.


问:
divisibility observation in particular patterns
Please look at the following observation made by trial and error.

Let us take some $2$-digit numbers like $12, 15, 24,\dots$

$12 = 1 *  2 = 2 \implies 2|12 $ ($2$ divides $12$)

$15 = 1 * 5 = 5 \implies 5|15$

for $3$ digit numbers:
$315 = 3 * 15 = 45 \implies 45|315$

for $4$-digit numbers:
$1352 = 13 * 52 = 676 \implies 676|1352$

Like this kind of applicability is applicable only for some rare such numbers.
I would like to know from this observation the following:

1) In what cases the $2$-digit, $3$-digit cases we can write? Is there any formal method to justify this observation?

2) Is there any particular formula or method to generate such pairs?

Thanks in advance!

答:
I will use concatenation to represent concatenation of digits and $\cdot$ for multiplication.  For the two digit numbers, if the number is $ab$, you are asking when $a \cdot b | 10a+b$  We must then have that $b | 10\cdot a$ and $a|b$, which is when $a=b, b=2\cdot a,$ or $b=5\cdot a$

For three digit numbers, do you insist on breaking off the first digit (like your example) so $a\cdot bc|abc$ or would you also be interested in $ab\cdot c|abc$?  For the first, we want $a\cdot bc | 100\cdot a +bc$  Again, we must have $bc | 100 \cdot a$ and $a |bc$  There are a number of possibilities, which you can catalog.


问:
Integration over the normal pdf using the change of variable technique
I am trying to understand how to integrate normal pdfs and it is proving a bit of a challenge. I think the challenge is more on the mathematical bit of integration. I have the following standard normal pdf and am trying to get $P(-2<x<2)$:

$$f(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}$$

From this MIT opencourseware tutorial, I am supposed to use the change of variable technique and transform my coordinates from cartesian to polar in order to integrate the function.

Introducing $y$ and integrating over the region $-2<x<2$ and $-2<y<2$:

$$I^2=\frac1{\sqrt{2\pi}}\iint e^{-\frac{x^2+y^2}2} \, \mathrm dx\,\mathrm dy$$

Transforming to polar coordinates and integrating over the region $0<r<\sqrt8$ and $\frac\pi4<\theta<\frac{7\pi}4$:

$$\frac1{\sqrt{2\pi}}\iint re^{-\frac{r^2}2}\,\mathrm dr\,\mathrm d\theta$$

$$\frac1{\sqrt{2\pi}}\int\frac{3\pi}2re^{-\frac{r^2}2} \, \mathrm dr$$

$$\frac1{\sqrt{2\pi}}\cdot\frac{3\pi}2\left(e^{-4}-1\right)\approx-1.8455$$

Just from this output, I can already tell that I have made a mistake somewhere since the yielded value is negative. I am hence confused on how to integrate using the change of variable technique. Some bit of insight would be highly appreciated on solving the above problem. Also, is there a simpler way of integrating over the normal cdf?

答:
The main problem is that the method advertised in the video won't work for a finite interval. Let

$$I=\frac1{\sqrt{2\pi}}\int_{-2}^2e^{-\frac{x^2}2}\,\mathrm dx$$

Squaring $I$ and introducing another dummy variable $y$ lets us write

$$I^2=\frac1{2\pi}\int_{-2}^2\int_{-2}^2e^{-\frac{x^2+y^2}2}\,\mathrm dx\,\mathrm dy$$

The domain of integration is the rectangle $[-2,2]\times[-2,2]$ (denoted by $R$), which can be expressed in polar coordinates as the union of $4$ component triangular regions $T_1,\ldots,T_4$:

$$T_1=\left\{(r,\theta) \mid 0\le r\le2\sec\theta\text{ and }-\frac\pi4\le\theta\le\frac\pi4 \right\}$$

$$T_2=\left\{(r,\theta) \mid 0\le r\le2\csc\theta\text{ and }\frac\pi4\le\theta\le\frac{3\pi}4 \right\}$$

$$T_3=\left\{(r,\theta) \mid -2\sec\theta\le r\le0\text{ and }\frac{3\pi}4\le\theta\le\frac{5\pi}4 \right\}$$

$$T_4=\left\{(r,\theta) \mid -2\csc\theta\le r\le0\text{ and }\frac{5\pi}4\le\theta\le\frac{7\pi}4 \right\}$$

For instance, the integral over $T_1$ would be

$$\frac1{2\pi}\iint_{T_1}e^{-\frac{x^2+y^2}2}\,\mathrm dx\,\mathrm dy=\frac1{2\pi}\int_{-\frac\pi4}^{\frac\pi4}\int_0^{2\sec\theta}re^{-\frac{r^2}2}\,\mathrm dr\,\mathrm d\theta=\frac1{2\pi}\int_{-\frac\pi4}^{\frac\pi4}\left(1-e^{-2\sec^2\theta}\right)\,\mathrm d\theta$$

but this cannot be resolved using elementary methods. It has an approximate value of $0.227767$, and incidentally so do the integrals over the other $3$ regions, giving a total of approximately $0.91107$, and hence $I\approx0.9545$.

The reason the method does work for integrating $f(x)$ over $\mathbb R$ is that the infinite plane $\mathbb R^2$, i.e.

$$\{(x,y)\mid-\infty<x<\infty\text{ and }-\infty<y<\infty\}$$

remains an infinite plane after converting to polar coordinates to

$$\{(r,\theta)\mid 0\le r<\infty\text{ and } 0\le\theta\le2\pi\}$$

and the limits of integration in polar coordinates are "nice" in the sense that the resulting integral is trivial, whereas a finite square centered at the origin has to get chopped up into $4$ regions with the limits on $r$ depending on $\theta$.


问:
Finding the nullspace of full column rank matrix
Given tall matrix $D \in \mathbb{R}^{m \times n}, m>n $:
$$
D = \begin{bmatrix}B \\\ C\end{bmatrix}
$$
where $B \in \mathbb{R}^{n \times n}$ is invertible and $C \in \mathbb{R}^{k \times n}$ (so $k=m-n$).

I know that $D$ is full column rank and that $D^TD$ is invertible.

How do I got about finding the basis for the nullspace of $D^T$ in terms of $B$ and $C$?

答:
Because $B$ is invertible, we find that
$$
DB^{-1} = \pmatrix{B\\C}B^{-1} = \pmatrix{BB^{-1}\\ CB^{-1}} = \pmatrix{I\\CB^{-1}}.
$$
We note that $[DB^{-1}]^T = B^{-T}D^T$ has the same nullspace as $D^T$. Thus, it suffices to find the nullspace of
$$
[DB^{-1}]^T = \pmatrix{I & [CB^{-1}]^T} = \pmatrix{I & B^{-T}C^T}
$$
We note that for $x \in \Bbb R^n$ and $y \in \Bbb R^k$, the block-vector $(x,y)$ is an element of the nullspace iff
$$
\pmatrix{I & B^{-T}C^T} \pmatrix{x\\y} = 0 \implies x = -B^{-T}C^T y.
$$
We can now obtain the desired basis as follows:

Find a basis $\{y_1,\dots,y_p\} \subset \Bbb R^k$ for the nullspace of $C^T$. For each $y_i$, the block vector $v_i = (0,y_i) \in \Bbb R^m$ is a basis element for our nullspace of $D^T$.

Extend this to a basis $\{y_1,\dots,y_p,y_{p+1},\dots,y_k\}$ of $\Bbb R^k$. For each $y_i$ with $i \geq p+1$, the block vector $v_i = (-B^{-T}C^Ty_i,y_i)$ is a basis element for our nullspace of $D^T$.

With that, the vectors $v_1,\dots,v_k$ give the basis that we're looking for.

If $C$ has full row-rank, then this gets even easier: the columns of the matrix
$$
\pmatrix{B^{-T}C^{T} \\ -I}
$$
form a basis for the nullspace of $D^T$.


答: Borrowing @hamam_Abdallah's notation, call the sum $S_n$. The first few terms suggest$$S_n=(n-1)2^{n+1}+2.$$There's an easy proof by induction. The base step $n=1$ works (or $n=0$, if you prefer); for the inductive step, note$$k2^{k+2}-(k-1)2^{k+1}=(k+1)2^{k+1}.$$


问:
trees are bipartite

I'm trying to prove that trees are bipartite by induction. Below is what I've come up with.

Let $T$ be a tree. If it has no edges, it is bipartite (one can choose the vertex to be in the set $A$, $B$ to be the empty set, and then for every edge in the edge set, the claim is satisfied vacuously, so $(A,B)$ is a bipartition of $T$). Assume that all trees $T$ with fewer than $m$ edges are bipartite, where $m\geq 1.$ Let $T$ be a tree with $m$ edges. If $m=1$, then there are $2$ vertices of $T$ and we can let each one be in a different set of a bipartition. So we may assume $m>2.$ Let $e = uv$ be an edge of $T.$ Then $e$ is a bridge, so $T-e$ has exactly two components, $T_1$ and $T_2,$ both of which have fewer than $m$ edges and thus by the inductive hypothesis are bipartite. Let $(A_i,B_i)$ be bipartitions for $T_i, 1\leq i\leq 2.$ Then $(A_1\cup B_2, A_2\cup B_1)$ is a bipartition of $T.$ Indeed, let $w = ab$ be an edge of $T$ other than $e.$ Then $a$ is in $A_1\cup B_1\cup A_2\cup B_2,$ by the construction of the $A_i$'s and $B_i$'s. If $a\in A_1,$ then $b\not \in B_2$ as otherwise there would be an edge from a vertex in $T_1$ to one in $T_2,$ which contradicts the fact that $T_1$ is connected in $T-e$. Also $b\not \in A_1$ as $(A_1, B_1)$ is a bipartition of $T_1$ so there are no edges with both ends in $A_1$ in $T_1,$ and so there are no edges with both ends in $T$, since $T_1$ is a subgraph of $T$ that contains all the vertices in $A_1$. Similarly, if $a \in B_2, b \not \in A_1,$ if $a\in A_2, b\not \in B_1,$ and if $a \in B_1, b\not \in A_2.$ Thus every edge connects a vertex in $A_1\cup B_2$ to a vertex in $A_2\cup B_1,$ so $(A_1\cup B_2, A_2\cup B_1)$ is a bipartition and $T$ is bipartite.

Is this incorrect?

答:
The general idea will work, even though you forgot to mention why the edge $e$ isn't causing you any problems (for example if $u$ and $v$ are both in $A_1\cup B_2$).

A hint for a different, maybe easier approach - prove this by induction, but instead of using a random bridge, use the fact that every tree has a leaf, and remove it.

Title: trees are bipartite


问:
Check if an integer is present in a linear recurrence
Given the following recurrence relation :

$f(n) = 5f(n-1) - 2f(n-2)$ where $f(0) = 0, f(1) = 1$

I need to find out if an integer $F_n$ is present in the sequence in $O(1)$ time and space.

Solving the equation, there are two distinct real roots.

$\phi = \frac{5 + \sqrt17}2$

$\psi = \frac{5 - \sqrt17}2$

Therefore, $F_n = \frac{\phi^n - \psi^n}{\sqrt17}$

Similar to Binet's rearranged formula, I want to solve for $n$ in terms of $F_n$.

Since, $\psi = \frac{2}{\phi}$

$\sqrt17F_n = \phi^n - \frac{2^n}{\phi^n}$

$Or,$

$\phi^{2n} - \sqrt17F_n\phi^n-2^n = 0$

Here I'm not able to find out a solution to express $n$ purely in terms of $F_n$ so that I can calculate the perfect square just like in Binet's formula.

答:
Notice how small $\psi \approx 0.44$ is. So in the expression

$$F_n = \frac{\phi^n - \psi^n}{\sqrt{17}}$$

$\phi^n$ is doing all of the work. After all, $0 \leq \psi^n \leq \left ( \frac{1}{2} \right )^n$.

So at least eventually, $F_n = \left \lceil \frac{\phi^n}{\sqrt{17}} \right \rceil$. Some quick computation will show whether this is always true, or if you'll need to have finitely many special cases. Of course, you can handle finitely many special cases without breaking $O(1)$ time/space.

This makes your job substantially easier. As now $F_n \approx \frac{\phi^n}{\sqrt{17}}$, and so some quick algebra gives

$$n \approx \log_\phi \left ( \sqrt{17} F_n \right ).$$

So given any $k$, we have a possible value of $n$. Namely, look at $m$ the nearest integer to $\log_\phi \left ( \sqrt{17} k \right )$. Then we can ask if $F_m$ really does equal $k$.

Both of these steps can be done in $O(1)$ space and time (at least if we use the common lie that real-valued arithmetic is constant time), as desired.

(Caveat Lector: I didn't check if you found these $\phi$ and $\psi$ correctly. So the specifics might change based on that. The idea should still work, though).

I hope this helps ^_^


问:
Does every Lebesgue measurable set have the Baire property?
Title says it all, so I'll just repeat it:
Does every Lebesgue measurable set have the Baire property?</p>

答:
Just to make the solution from the comments to GEdgar's answer more explicit: The set of real numbers can be partitioned as $\mathbb R = M \cup N$, where $M$ is meager and $N$ is Lebesgue null.  ($N$ is the dense $G_\delta$ described by t.b. above.) </p>

<p>Now every subset of $N$ is Lebesgue measurable, but (using AC) one can find a subset without the Baire property. </p>

Body:
Just to make the solution from the comments to GEdgar's answer more explicit: The set of real numbers can be partitioned as $\mathbb R = M \cup N$, where $M$ is meager and $N$ is Lebesgue null.  ($N$ is the dense $G_\delta$ described by t.b. above.)

Now every subset of $N$ is Lebesgue measurable, but (using AC) one can find a subset without the Baire property.


问:
Evaluate $ (\forall y)(\exists x)(x \times x = y) $
$\times$ is as you would think, the multiplication operation.

$ (\forall y)(\exists x)(x \times x = y) $

For the domains $\mathbb{N}$ and $\mathbb{Z}$ I found this formula fairly simple to evaluate, false in both cases.

I'm not certain with the following domains:

Domain: $\mathbb{Q}$

With this domain I concluded it evaluated to false, as for y=2, the $\sqrt[]{2}$ is not a rational number.

Is that correct?

Domain: $\mathbb{R}$

Would this evaluate to true, based off every real number $y$ having a real square root $x$?

答:
With the help of the users in the comment section:

"Not every real number has a real square root" - @Henry

I have worked out that indeed this formula would evaluate to false for domains $ \mathbb{Q} $ (take y=2) and $\mathbb{R} $ (take y=-1). In both domains, no such x exists.


问:
$A$ is a vector subspace of $ℝ^4$?

Am I right to say row space of the matrix $A$ is a vector subspace of $ℝ^4$? I think that because $\operatorname{Col}(A)$ is a vector subspace of $\mathbb{R}^4$ and $\dim \operatorname{Col}(A) = \dim \operatorname{Row}(A)$.

答:
The dimension of a vector space is the number of vectors in a basis. This doesn't say anything about what kind of vectors are in the vector space. You can have a one-dimensional subspace of $\mathbb R$, or $\mathbb R^2$, or of $\mathbb R^{100}$.

The row space is a subspace of $\mathbb R^5$, because its elements (linear combinations of the rows) are $5$-dimensional vectors. The column space is a subspace of $\mathbb R^4$, because its elements (linear combinations of the columns) are $4$-dimensional vectors.

For this matrix, the row space is a $2$-dimensional subspace of $\mathbb R^5$, and the column space is a $2$-dimensional subspace of $\mathbb R^4$, and there's nothing wrong with that.

Title: $A$ is a vector subspace of $ℝ^4$?


问:
integral of special derivative
Assume that $S(0)>=0, I_w(0)>=0, I(0)>=0$ and $F(t)=\mu+\beta_1\frac{I_W(t)}{N}+\beta_2 \frac{I(t)}{N}$(where $\mu, \beta_1,\beta_2$ and N are constants).
If we have that:

$\frac{d}{dt}\{S(t)exp(\int_0^t F(s)ds)\}>=0$, can(by integrating) we conclude that $S(t)>=0$ or $S(t)>0$?
We also have followings:

$\frac{dS(t)}{dt}=\lambda_1-\mu S(t)-\beta_1 \frac{S(t)I_W(t)}{N}-\beta_2\frac{S(t)I(t)}{N},\\
\frac{dI_W(t)}{dt}=\lambda_2+\beta_1 \frac{S(t)I_W(t)}{N}+\beta_2\frac{S(t)I(t)}{N}-(\mu+\alpha_1+\alpha_2)I_W(t),\\
\frac{dI(t)}{dt}=\lambda_3+\alpha_1I_W(t)-(\mu+\alpha_3)I(t).
 $

Thanks  for any help.

答:
The first equation of the system can be written $\frac{d S(t)}{d t} + F(t)S(t)= \lambda_1$. Multiplying by $\exp(\int_0^t F(s) d s)$ one gets, assuming $\lambda_1 > 0$,
\begin{equation}
\frac{d}{d t}\left\{S(t)e^{\int_0^t F(s) d s}\right\} = \lambda_1 e^{\int_0^t F(s) d s}> 0
\end{equation}
Hence $\varphi(t) = S(t)e^{\int_0^t F(s) d s}$ increases strictly with time and $\varphi(0) \ge 0$. It follows that $\varphi(t)>0$ for $t>0$, hence also $S(t)> 0$.

Title: integral of special derivative


问:
Does this multivariable limit exist?
Does the limit  $$ \lim_{x,y \to 0,0} \frac {\cos(x+y)+\cos(x-y)-2}{x^{2}+y^{2}} $$ exist? I can't find a way to prove it.

I am having a problem because I can't find two sequences to disapprove the existence of limit by Heine characterization, but also, I failed to bound the limit using squeeze theorem.

答:
I'm sure there is a simpler way to solve the problem but here's a solution anyway. It's based on using the second order taylor polynomial $ 1-1/2 \;x^2$ of $\cos(x)$.

Define
$$ f(x,y) = 2\frac{ \cos(x)\cos(y)-1}{x^2 + y^2}.$$

We want to find the limit of $f$ at $(0,0)$ if it exists.

Observe that for $x$ close to $0$ and $\epsilon > 0$  we have

$$ 1 - \frac{1}{2}x^2 \leq \cos(x) \leq 1- \frac{1}{2+\epsilon}x^2$$

Note

$$ P_\epsilon(x,y) = 2\frac{(1-  \frac 1 {2+\epsilon}x^2)(1- \frac 1 {2+\epsilon} y^2)-1} {x^2 + y^2}.$$

By the above inequalities we have that for $x,y$ close to $0$ and $\epsilon > 0$ we have

$$ P_0(x,y) \leq f(x,y) \leq P_\epsilon(x,y) $$

Using the first inequality we find that

\begin{align*}
f(x,y) &\geq 2 \frac{(1-\frac 1 2 x^2)(1- \frac 1 2 y^2) - 1}{x^2 + y^2} \\
       &= 2\frac{1- \frac 1 2 (x^2 + y^2) + \frac 1 4x^2y^2 - 1 }{x^2+y^2} \\
       &= 2 \frac{- \frac 1 2 (x^2 + y^2) + \frac 1 4x^2y^2}{x^2+y^2} \\
       &= -1 + \frac 1 2 \frac{x^2y^2}{x^2 + y^2}.
\end{align*}

The second inequality gives

\begin{align*}
f(x,y) &\leq 2\frac{(1-  \frac 1 {2+\epsilon}x^2)(1- \frac 1 {2+\epsilon} y^2)-1} {x^2 + y^2} \\
&= 2 \frac{ 1 - \frac 1 {2+\epsilon}(x^2+y^2) + \frac 1 {(2+\epsilon)^2}x^2y^2 -1 }{x^2 + y^2}\\
&= \frac{-2}{2 + \epsilon} + \frac{2}{(2+\epsilon)^2}\frac{x^2y^2}{x^2 + y^2}.
\end{align*}
Knowing this we have

$$  -1 + \frac 1 2 \frac{x^2y^2}{x^2 + y^2}\leq f(x,y) \leq \frac{-2}{2 + \epsilon} + \frac{2}{(2+\epsilon)^2}\frac{x^2y^2}{x^2 + y^2}$$
Taking the limit as $(x,y) \to 0$ and $\epsilon \to 0$ on both sides and using the squeeze theorem we find that

$$ \lim_{(x,y) \to (0,0} f(x,y) = -1$$

I hope this is correct.

Title: Does this multivariable limit exist?


问:
Find the coordinates of the point P on the line d : 2x − y − 5 = 0, for which the sum AP + PB attains its minimum, when A(−7; 1) and B(−5; 5).
I tried to use the Cosine theorem and get that |(PA)^2 + (PB)^2 - (AB)^2|<=|2PAPB|. Can someone explain to me what should I do? Thank you!

答:
Here is a solution with a geometric flavor:

You may know that the set of points $M$ such as $MA+MB=const.$ is an ellipse with focii $A$ and $B$. The larger the constant, the larger the ellipse. Consider the following figure:

If this constant is too small, the ellipse will not intersect the given line. There is a minimum value for which there is a unique point $M$ belonging to an ellipse of the family and to the straight line, and it is not difficult to see that in this point the line is tangent to the ellipse (colored in cyan).

A particular case occurs here: line $AB$ joining the focii is parallel to the given (green) line. Therefore the tangency will occur along the (black) line bissector of $AB$, with equation $y=-\tfrac12x$.

As a consequence, this minimum will occur at the intersection $P$ between the two lines, with coordinates:

$$P(x=2,y=-1)$$


问:
Can we say anything about $\vert f'(x) \vert$ versus $\vert f''(x) \vert$ if $f$ is concave and goes through the origin?
Suppose we have a function $f(x)$ that is concave, upward sloping, and goes through the origin.

Are we able to say anything about how $\vert f'(x)\vert$ compares to $\vert f''(x)\vert$, such as whether one is greater than/less than another?

答:
Alternative approach.

Depending on your definition of concave, "Concave, upward sloping" is contradictory.  This answer assumes that you intend that $f(x)$ is convex everywhere.

Let $f(x) = ax^2 + bx \implies f$ passes through the origin.

$|f'(x)| = |2ax + b|~$ and $~|f''(x)| = |2a|.$

Convex merely implies 2nd derivative positive which implies that $a > 0$.

Consider the following three functions:

$f_1(x) = 1000x^2 + x~ \implies ~|f'_1(x)| = |2000x + 1|,~$
and $~|f''_1(x)| = 2000.$

For $~-(1/2) < x < (1/2), ~~|f'_1(x)| < |f''_1(x)|.$

For $~1 < x, ~~|f'_1(x)| > |f''_1(x)|.$

$f_2(x) = x^2 - x~ \implies ~|f'_2(x)| = |2x - 1|,~$
and $~|f''_2(x)| = 2.$

For $~0 < x < 1, ~~|f'_2(x)| < |f''_2(x)|.$

For $~3 < x, ~~|f'_2(x)| > |f''_2(x)|.$

$f_3(x) = x^2 + 1000x~ \implies ~|f'_3(x)| = |2x + 1000|,~$
and $~|f''_3(x)| = 2.$

For $~-(501) < x < -(500), ~~|f'_3(x)| < |f''_3(x)|.$

For $~0 < x, ~~|f'_3(x)| > |f''_3(x)|.$