问:
Feynman on indefinite integral?
Consider:

$$ \int \frac{ \ln (t-1)}{t} dt$$

And then,

$$ I( \alpha) = \int \frac{ \ln [\alpha (t-1)] }{t} dt$$

D.w.r.t. $\alpha$

$$ \frac{dI}{d \alpha} = \int \frac{ 1}{ \alpha t} dt$$

$$ \frac{dI}{d \alpha} = \frac{1}{\alpha} \ln t +C \tag{1}$$

Integrating both side with $\alpha$,

$$ I =  \ln \alpha \ln t + C \alpha$$

And putting $ \alpha =1$

I get a weird result...:

$$ I =  C$$

What did I do wrong..? Also is  the constant in (1) a function of $ \alpha$ ..? (generally speaking)

答:
Ok, the differentiation is where I messed up.

$$ I_{\alpha} = \int \frac{\partial}{\partial \alpha} \frac{ \ln (\alpha (t-1) )}{t} dt$$

Or,

$$I_{\alpha}  = \int \frac{1}{t} dt$$

Hence,

$$ I = \ln t \alpha +C$$

Hence,

$$ \int \frac{ \ln [\alpha (t-1)] }{t} dt  = \alpha \ln t  + C$$

Too bad there is no easy points to evaluate this at >:(

Title: Feynman on indefinite integral?


问:
What are some good resources to learn arithmetic combinatorics?
I have been very intrigued by the theory of arithmetic combinatorics after the result regarding primes in arithmetic progressions proven by Tao and Green.
PAPERS ARE OK. I notice that most people usually skip out on papers, but papers are what got me into the theory. I have found a book, (additive combinatorics-Tao), which so far seems quite nice. But for instance, are there any notes, or "paper-notes" or EVEN lectures based around, or even on this topic? Even some other books talking about this theory, or things related to that?
(I have watched the Ben Green lectures on higher-order Fourier analysis, and I am aware of Tao's notes on this subject).</p>

答:
By no means am I an expert on this, so I don't know how good these resources are. However, these are other resources you might find helpful:

https://www.amazon.com/Additive-Combinatorics-Proceedings-Lecture-Notes/dp/0821843516

https://math.stanford.edu/~ksound/Notes.pdf

http://people.maths.ox.ac.uk/greenbj/notes.html

Additive Combinatorics is also known as Discrete Harmonic Analysis in some circles (no pun intended). There is a book on Discrete Harmonic Analysis here:

https://www.maa.org/press/maa-reviews/discrete-harmonic-analysis

If you are interested in a Theoretical Computer Science perspective, there are a couple good starting points:

Daniel Stefankovic's Master's Thesis surveying applications of the Fourier transform in Combinatorics and TCS. (https://people.cs.uchicago.edu/~laci/students/stefankovic-masters.pdf)

Ryan O'Donnell's book on Analysis of Boolean Functions (https://www.amazon.com/Analysis-Boolean-Functions-Ryan-ODonnell/dp/1107038324#reader_1107038324).

Ryan O'Donnell's course webpage in this same area (https://www.cs.cmu.edu/~odonnell/boolean-analysis/).

I hope this is helpful!

ContentLicense: CC BY-SA 4.0


问:
Prove that $6|n(n+1)(2n+1)$, $n\in\mathbb{Z}$
I am trying to solve this question without modular arithmetic.

Let $n\in\mathbb{Z}$. Prove that $6$ divides $n(n+1)(2n+1)$.

Using induction:

$\text{For }n=1
\\P(1) = 1(1+1)(2(1)+1)=6
\\\text{For }n=k
\\P(k)=k(k+1)(2k+1)
$

$P(k)$ is divisible by $6$ so $k(k+1)(2k+1)=6h, h\in\mathbb{Z}$. Once by the induction hypothesis, $P(k)$ and $P(k+1)-P(k)$ being divisible by $6$ so $P(k+1)$ will be divisible by $6$. as well.

$P(k+1)= (k+1)(k+1+1)(2(k+1)+1) = (k+1)(k+2)(2k+3)$

So,

\begin{align}
P(k+1)-P(k) & =(k+1)(k+2)(2k+3)-k(k+1)(2k+1) \\
& = 2k^3+9k^2+13k+6 - (2k^3+3k^2+k) \\
& = 6k^2+12k+6 \\
& = 6(k^2+2k+1) \\
& = 6(k+1)^2
\end{align}
Therefore, $P(k+1)$ is true and $6$ divides $n(n+1)(2n+1)$.

But once $n\in\mathbb{Z}$ the proof is not complete. I was wondering about considering $6|(-n)(-n+1)(-2n+1)$, $n\in\mathbb{Z}$ but I am not really sure how to proceed. It doesn't seem to be the right approach. How can I keep doing that or what would be a good proof? Proofs without modular arithmetic are preferable, but if you got a nice proof it would be welcomed. I've seen similar questions, but they considered $n\in\mathbb{N}$.

答:
For $n$ such that $(n,6)=1$, we could apply Euler's theorem, with $\varphi(6)=2$.  We get $n^2\equiv1\bmod6$.  Then look at $n(n+1)(2n+1)\equiv2n^3+3n^2+n\equiv3n+3\equiv3(n+1)\equiv0\bmod6$, since $n$ is odd.

When $n\equiv2;3;4;0\bmod6$, we get $n(n+1)(2n+1)\equiv0\bmod6$ by calculation.


问:
How to find $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$ using known result for $\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx$
I need to find integral of $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$. From
this Riemann_zeta_function , it gives $\zeta\left(  s\right)  =\frac{1}{\Gamma\left(
s\right)  }\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx$. Hence for $s=3$ this gives

\begin{align}
\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx  & =\zeta\left(  3\right)
\Gamma\left(  3\right)  \nonumber\\
& =2\zeta\left(  3\right)  \tag{1}%
\end{align}

My question is, how to use the above result to find integral of $\int%
_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$? Mathematica gives

\begin{equation}
\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx=\frac{3}{2}\zeta\left(  3\right)
\tag{2}
\end{equation}

I tried to see if there is change of variable or other trick, but I do not
see one. I do know about a direct method given in this answer here, but I want
to see if it is possible, with some trick, to use the result from (1) to find (2).

答:
Well using the fact that:
$$(a+1)(a-1)=a^2-1$$
we can say that:
$$\frac{1}{e^{2x}-1}=\frac12\frac{1}{e^x-1}-\frac12\frac{1}{e^x+1}$$
$$\therefore \frac{1}{e^x+1}=\frac{1}{e^x-1}-\frac{2}{e^{2x}-1}$$
so In term of your integral we can say:
$$I=\int_0^\infty\frac{x^2}{e^x+1}dx=\int_0^\infty\frac{x^2}{e^{x}-1}dx-2\int_0^\infty\frac{x^2}{e^{2x}-1}dx\tag{1}$$
$$I_1=\int_0^\infty\frac{x^2}{e^x-1}dx\tag{2}$$
Now, the first integral you know the value of so lets look at the second, using the substitution $u=2x\Rightarrow dx=\frac{du}2$ we can say the following:
$$\int_0^\infty\frac{x^2}{e^{2x}-1}dx=\int_0^\infty\frac{(u/2)^2}{e^u-1}\frac{du}2=\frac18\int_0^\infty\frac{u^2}{e^u-1}du=\frac18I_1\tag{3}$$

Now from this we can say:
$$I=I_1-\frac28I_1=\frac34I_1\tag{4}$$
which is the same as saying:
$$\int_0^\infty\frac{x^2}{e^x+1}dx=\frac34\int_0^\infty\frac{x^2}{e^x-1}dx\tag{5}$$


问:
Range such that $g(\alpha)$=100
This is a problem which I am having a really hard time solving this problem. Any help would be appreciated. Also sorry for my poor English.

For some real number $0<\alpha<\pi$, $a_n$ is defined as

$$a_n=(2n-1)\alpha$$
There exists a function $f(x)$, being differentiable in all real number and such that f(0)=0. $f'(x)$ is given as this.
$$f'(x) = \begin{cases}\cos x & x < a_1 \\ cos(x-2n\alpha) & a_n\leq x<a_{n+1}\end{cases}$$
Let $g(\alpha)$ be function of the number of positive roots of $f(x)=1$. The range of $\alpha$ such that $g(\alpha)=100$ is $m<\alpha<n$. Find the value of $\frac{\sin m}{\sin n}.$

I am aware that the integration constant is different for each n and have so far integrated $f(x)$ and tried to graph it, but it didn't go well...Any hints or solutions please?

答:
$f'(x)$ just repeats the graph of $\cos x$ from $-\alpha$ to $\alpha$ periodically with a period of $2\alpha$. Because $\cos(-\alpha) = \cos \alpha, f'(x)$ is continuous. If $\alpha \le \frac \pi2$, then $f'\ge 0$ everywhere and only $= 0$ at isolated points, so $f(x)$ is strictly increasing, and $f(x) = 1$ would have only one root.

Conversely, if $\alpha$ were allowed to be $\pi$, then $f'(x) = \cos x$ everywhere, and because the area between the $x$-axis and $\cos x$ is the same below the axis as above, integrating $\cos x$ keeps adding and removing the same amount of area, which is why the integral $\sin x$ is also periodic. However, for lower values of $\alpha > \frac \pi2$, some portion of the cosine curve below the $x$-axis is removed, while the portion above the axis remains, so when integrating, with each repetition of the period, more positive area is added than negative area, with the result that $f(x)$ will have an overall upward trend. But since $f'(x)$ is periodic with period $2\alpha$, integrating it over an interval of width $2\alpha$ will pick up the same area every time. So there is some constant $A$ such that $f(x + 2\alpha) = f(x) + A$ for all $x$ (like $f, A$ depends on $\alpha$ but not $x$).

Note also that for $\alpha > \frac \pi2$ that we know exactly where the local minima and maxima of $f$ occur. The minima occur at $-\frac\pi2 + 2k\alpha$, while the maxima occur at $\frac\pi2 + 2k\alpha$ for $k \in \Bbb N$.

Put those last two facts together (in particular, figure out what $A$ is), and the solution should be in sight.

Title: Range such that $g(\alpha)$=100


问:
What is the image of y>0 under the map $w= e^{z}$
What is the image of y>0 under the map  $w= e^{z}$

It is and entire circunference of radius 1? or only half circunference
of radius 1?

I see this video
https://www.youtube.com/watch?v=d18M390rMaI

And, in minute 4, he only graphics half of the circunference...
And my question is why??? If y>0 is not to supossed to be an entire
circunference of radius 1?

Thanks you for help

答: They only drew the half circumference because they stoped at drawing on the left at $y = \pi$. If there were to keep going upward on the $y$ axis, the point in the right picture would go round and around the unit circle (making a full trip for each interval $y = k 2\pi i$ and $y = (k + 1) 2\pi i$, $k$ an integer)


问:
Given that $x,y,z$ are positive reals such that $xyz=1$.What is the minimum value of $(x^2+4)(y^2+4)(z^2+4)$.
Help me to prove that minimum value of expression $ (x^2+4)(y^2+4)(z^2+4) $ is 125 when $ xyz = 1$. I think that minimum value achieved when $x=y=z=1$. But I can prove only $ (x^2+4)(y^2+4)(z^2+4) = xyz(x+\frac{4}{x})(y+\frac{4}{y})(z+\frac{4}{z}) = (x+\frac{4}{x})(y+\frac{4}{y})(z+\frac{4}{z}) \ge 2\sqrt{x \frac{4}{x}} \cdot 2\sqrt{y \frac{4}{y}} \cdot 2\sqrt{z \frac{4}{z}} = 2^6=64$

Body: By Holder $$(x^4+4)(y^4+4)(z^4+4)\ge {\left(\sqrt[3]{x^4y^4z^4}+\sqrt[3]{4\cdot 4\cdot 4}\right)}^3=125$$


问:
How to prove that each edge of tree is a bridge?

How to prove that each edge of tree is a bridge?

My attempt:

Tree is a connected graph which has no cycle, and in a connected graph, bridge is an edge whose removal disconnects the graph.

Let G be a tree, and each edge of G is not a bridge.

I should find a contradiction from my assumption.

But, now I can't go proceed. I think by this way I can prove, and I can't express it.

How can I go further?

答: Let $G$ be a tree, and suppose that one of its edges $\{v,w\}$ is not a bridge. This implies that removing this edge does not disconnect $G$, i.e., if $G'$ is the graph $G$ with $\{v,w\}$ removed, $G'$ is connected. Hence, there exists a path connecting vertices $v$ and $w$ in $G'$. This path together with $\{v,w\}$ forms a cycle in $G$, which contradicts the assumption that $G$ is a tree. Therefore, every edge in $G$ must be a bridge.

ContentLicense: CC BY-SA 3.0


问:
The following vectors are linearly independent if and only if k≠
Does anybody know how to solve this Question ?

https://imgur.com/gallery/gJkyaPg

I managed to get K= to a number so that there is a solution, but apparently it is wrong and I have to find that k!= something..

答: Hint:  Find what linear combination $(3,-9)$ is of $(-4,7)$ and $(2,-4)$.  You'll get $(3,-9)=a(-4,7)+b(2,-4)$ for certain $a,b$.  Then we get that $11a+6b+kb=-12$.  Solve for $k$.


问:
Epsilon-N method: proving recursive sequence
How do you prove an iterative sequence like
$$x_{n+1} = ax_n + b$$
$$ 0< |a|< 1   $$
by using epsilon-N method?

I tried create a closed form of the sequence
$$a^{n+1}x_0 + b\sum_{i=0}^n a^i$$.

And then I set an inequality that
$$|a^{n+1}x_0 + b\sum_{i=0}^n a^i - b(1-a)^-1| < \epsilon $$
where
$$\epsilon > 0$$
$$n> N$$
Which is the lowerbound for n.

I don't know the next step of the proof. How can I rearrange the inequality so that only n stands on one side of the inequality?

答:
How can I rearrange the inequality so that only $n$ stands on one side of the inequality?

Like this, maybe?

$$|a^{n+1}x_0 + b\sum_{i=0}^n a^i - b(1-a)^{-1}| < \epsilon$$
$$|a^{n+1}x_0 + b\frac{a^{n+1}-1}{a-1} - b(1-a)^{-1}| < \epsilon$$
$$\left|\frac{1}{a-1}\right|\cdot|a^{n+1}(a-1)x_0 + b(a^{n+1}-1) + b| < \epsilon$$
$$\left|\frac{1}{a-1}\right|\cdot|a^{n+1}(a-1)x_0 + b\cdot a^{n+1}| < \epsilon$$
$$\left|\frac{(a-1)x_0+b}{a-1}\right|\cdot|a^{n+1}| < \epsilon$$
Now you can take $\log$ (unless $(a-1)x_0+b=0$, and be careful with signs, $|a|<1$) and separate $n$.


问:
Prove that every edge in a tree is bridge
I have found this link

Which apparently ask same question like me right now.  But, the answers don't provide me information to solve the problem.  Once again, I have read many times about what was discussed in the comments,  but still I don't understand. This one:
@JAEMTO Ok. Suppose there is an edge AB which is not a bridge. Then after removing it there is a path from A to B. That path cannot involve the edge AB because you have just removed it. So how does that give you a contradiction?

The theorem may be simple to solve, but somehow I can't proceed how to  prove it.

Please help me.  Regards.

答:
Don't need induction:
Let $G$ be a tree, and suppose that one of its edges $\{v,w\}$ is not a bridge. This implies that removing this edge does not disconnect $G$, i.e., if $G'$ is the graph $G$ with $\{v,w\}$ removed, $G'$ is connected. Hence, there exists a path connecting vertices $v$ and $w$ in $G'$. This path together with $\{v,w\}$ forms a cycle in $G$, which contradicts the assumption that $G$ is a tree. Therefore, every edge in $G$ must be a bridge.


问:
The ring $\Bbb Z\left [\frac{-1+\sqrt{-19}}{2}\right ]$ is not a Euclidean domain
Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:

Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.

Now the exercise asks:

Prove that $A$ is not a Euclidean Domain.

Everything hints to an argument by contradiction: let $(A, d)$ be a ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.

How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!

I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this?

答: Here is a solution that may turn out to be helpful: http://www.maths.qmul.ac.uk/~raw/MTH5100/PIDnotED.pdf</p>


问:
Sequents: if-introduction and discharging assumptions
I am reading through "Mathematical Logic by Ian Chiswell & Wilfred Hodges"(amazon, and publisher)

for context I am reading through this for self-study, so I don't have the normal support of a classroom environment - and the lack of exercise solutions makes it hard to check my understanding.

On page 21 there is Exercise 2.4.4.f which asks "write out a derivation to prove the following sequent"

$\{(\phi \rightarrow (\psi \rightarrow \chi))\} \vdash ((\phi \wedge \psi) \rightarrow \chi)$

So far we have covered $\rightarrow$-Introduction and discharging in the following form

$$\begin{array}{lr}
\phi & \psi\\
\hline
 & (\phi \rightarrow \psi) \\
\end{array}
$$

which will discharge $\phi$ using $\rightarrow$-Introduction

My solution to 2.4.4.f so far is

(f) derivation of sequent $ \{ (\phi \rightarrow (\psi \rightarrow \chi)) \} \vdash ((\phi \wedge \psi) \rightarrow \chi) $

$$
\begin{array}{rrr}
\phi & & (\phi \rightarrow (\psi \rightarrow \chi)) \\
\hline
 \psi & & (\psi \rightarrow \chi) \\
\hline
 & & \chi
\end{array}
$$

which now leaves us with the undischarged assumptions
$ \{\phi, \psi, (\phi \rightarrow (\psi \rightarrow \chi)) \} $

from this I have $\chi$ so I can then (using $\rightarrow$-introduction)

$$
\begin{array}{r}
\chi \\
\hline
((\phi \wedge \psi) \rightarrow \chi) \\
\end{array}
$$

I don't have to discharge $(\phi \rightarrow (\psi \rightarrow \chi))$ as it is captured in the LHS of the sequent - but I do have to discharge both $\phi$ and $\psi$

In my $\rightarrow$-introduction above this means I can then discharge $(\phi \wedge \psi)$, but by the rules given so far it doesn't mean I can discharge $\phi$ and $\psi$ even though I know they are logically equivalent - that is if we can assume $(\phi \wedge \psi)$ then it is easy to show this entails both $\phi$ and $\psi$

Is it allowable to use $\rightarrow$-introduction of $(\phi \wedge \psi)$ to then discharge the assumptions of both $\phi$ and $\psi$ ?

If so, what is the generalisation of this ?

If not, how can I correctly form a derivation to prove this sequent?

答:
It is helpful to indicate what assumptions are being discharged and where; usually by indices and either canceling or boxing off the assumption.   So conditional introduction should be presented similar to: $$\require{cancel}\dfrac{\cancel{~\phi~}^n\quad\psi}{\phi\to\psi}{\small{\to}\mathsf I^n}\qquad\dfrac{[\phi]^n\quad\psi}{\phi\to\psi}{\small{\to}\mathsf I^n}\\~\\\tiny\text{NB: I prefer boxing, as it is cleaner.}$$

Why is this helpful? Well, the assumption may not appear immediately above the line in which it is discharged; and may appear multiple times.

In this case, you seek to discharge $\phi\land\psi$ so to introduce the conditional; further, the only assumption that should remain undischarged at the end of the proof is the premise $\phi\to(\psi\to\chi)$.  So you cannot leave the $\phi$ and $\psi$ naked; they must be covered by the discharged assumption; ie: you must derive them using conjunction elimination.

$$\dfrac{\dfrac{\dfrac{\lower{1.5ex}{\phi\to(\psi\to\chi)}\quad\dfrac{[\phi\land\psi]^1}{\phi}{\small\land\mathsf E}}{\psi\to\chi}{\small{\to}\mathsf E}\quad\dfrac{[\phi\land\psi]^1}{\psi}{\small\land\mathsf E}~}{\chi}{\small{\to}\mathsf E}}{(\phi\land\psi)\to\chi}{\small{\to}\mathsf I^1}$$

In Fitch style, this proof would look like$$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~~\phi\to(\psi\to\chi)}{\fitch{~~2.~~\phi\land\psi}{~~3.~~\psi\hspace{12ex}{\land}\mathsf E~2\\~~4.~~\phi\hspace{12ex}{\land}\mathsf E~2\\~~5.~~\psi\to\chi\hspace{7ex}{\to}\mathsf E~4,1\\~~6.~~\chi\hspace{12ex}{\to}\mathsf E~3,5}\\~~7.~~(\phi\land\psi)\to\chi\hspace{4ex}{\to}\mathsf I~2{-}6}$$

So the assumption is raised in line 2, used twice (in lines 3 and 4), and discharged for line 7


问:
Is the zero subspace 0 independent with any other subspace (of vector space) (Linear algebra hoffman)
Recall Definition: (in sec 6.6 on 209p)

Let $W_1,...,W_k$ be subspaces of a vector space $V$. We say that $W_1,...,W_k$ are independent if $\alpha_1+...+\alpha_k=0$, $\alpha_i \in W_i$ implies that each $\alpha_i$ is $0$.

Question:

By this definition, I think the zero subspace $0$ is independent of all subspace of $V$. Am I right? Or is there a problem?

答:
$\newcommand{\set}[1]{\{ #1 \}}$Yes.

Let $\set{W_i}_{i=1}^n$ be an arbitrary finite family of subspaces of $V$ which are independent. Consider along with them the trivial space $Z := \set{\vec 0}$. I claim that $\set Z \cup \set{W_i}_{i=1}^n$ is an independent family of spaces. Indeed, take $w_i \in W_i$ and $z \in Z$, and consider the combination

$$S := z + w_1 + w_2 + \cdots + w_n$$

Since the $W_i$ are independent, and $z \in Z \iff z = \vec 0$, it immediately follows $S \equiv 0$ always.

In particular, if the family in question is a singleton, i.e. it only consists of a single space $\set{W}$, then $W$ is independent from $Z$ and $\set{W,Z}$ are a family of independent subspaces of the vector space $V$.


问:
Finding common ratio of a geometric series from a cubic
The question is asking to find the sum of a geometric series up to term n, where the first value of r is 1.
The expression is $$ \sum_{r=1}^n \ 4r^3-3r^2+r$$.
I am aware of the general formula for the sum of a geometric series, but don't know what to use as the common ratio - is there any way to figure out the common ratio from the equation?

Body:
First, we use the identities:

$$\begin{align}&\sum_{i = 1}^{n}i^{3} = \bigg(\frac{n(n +1)}{2}\bigg)^{2}
\\&\sum_{i = 1}^{n}i^{2} = \frac{n(n + 1)(2n + 1)}{6}
\\&\sum_{i = 1}^{n}i = \frac{n(n + 1)}{2}\end{align}$$

These can be easily proved by induction. Then, for your sum we have:

$$\begin{align}\sum_{r=1}^{n}4r^{3}-3r^{2} + r&= 4\sum_{r = 1}^{n}r^{3}-3\sum_{r=1}^{n}r^{2}+\sum_{r=1}^{n}r
\\&= n^{2}(n+1)^{2}-\frac{n(n+1)(2n+1)}{2}+\frac{n(n + 1)}{2}
\\&=n^{2}(n + 1)^{2}-\frac{n(n + 1)(2n + 1 - 1)}{2}
\\&= n^{2}(n + 1)^{2} - n^{2}(n + 1)
\\&= \boxed{n^{4} + n^{3}}\end{align}$$


问:
For what $a$ and $b$ is  $9x^4-12x^3+28x^2+ax+b$ a perfect square?
If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)</p>

答:
Even though it doesn't affect the final outcome in this problem, I think it's better long-term practice not to assume that the leading term is positive. This emphasizes that there are two square roots to a square polynomial, differing by sign, just like with integers.

Suppose $9x^4-12x^3+28x^3+ax+b=g(x)^2$ for some polynomial $g(x)$. Because the degree of $9x^4-12x^3+28x^3+ax+b$ is 4, we have $4=\deg(g(x)^2)=2\deg(g(x))$, so that $g$ is a degree  2 polynomial.

Write $g(x)=cx^2+dx+e$. Then
$$9x^4-12x^3+28x^3+ax+b=(cx^2+dx+e)^2=(c^2)x^4+(2cd)x^3+(2ce+d^2)x^2+(2de)x+e^2.$$
Thus $c^2=9$, $2cd=-12$, $2ce+d^2=28$, $2de=a$, and $e^2=b$.

$c^2=9$ implies $c=3$ or $c=-3$.

Combined with $2cd=-12$, this implies that $d=2$ or $d=-2$.

Combined with $2ce+d^2=28$, this implies that $e=4$ or $e=-4$.

If $c=3$, then $d=-2$ and $e=4$. If $c=-3$, then $d=2$ and $e=-4$. Thus,

$$g(x)=\pm(3x^2-2x+4),$$

and $a=2de=-16$ and $b=e^2=16$.


问:
If $f$ is complex differentiable then so is $\bar{f}$?

Suppose $f$ is complex differentiable on some open set $U$, is $\bar{f}$ also differentiable?

Assuming $f'(z)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists does $\overline{f(z)}=\lim_{z\to z_0}\frac{\overline{f(z)}-\overline{f(z_0)}}{z-z_0}$ also exist?

Is this just properties of limits?

$\overline{f}'(z)=\lim_{z\to z_0}\frac{\overline{f(z)}-\overline{f(z_0)}}{z-z_0}=\frac{\lim_{z\to z_0}\overline{f(z)}-\overline{f(z_0)}}{\lim_{z\to z_0} z-z_0}$

Then let $\epsilon >0$ by assumtion there exists $\delta >0$ s.t if $\vert z-z_0\vert <\delta$ then $\vert f(z)-f(z_0)\vert <\epsilon$

and $\vert f(z)-f(z_0)\vert=\vert \overline{f(z)}-\overline{f(z_0)}\vert<\epsilon$

so $\frac{\lim_{z\to z_0}\overline{f(z)}-\overline{f(z_0)}}{\lim_{z\to z_0} z-z_0}$ exists and $f'(z)=\overline{f'(z)}$?

答:
$\overline{f}'(z)=\lim_{z\to z_0}\frac{\overline{f(z)}-\overline{f(z_0)}}{z-z_0}=\frac{\lim_{z\to z_0}\overline{f(z)}-\overline{f(z_0)}}{\lim_{z\to z_0} z-z_0}$

You can't split the limits like that.

The property of limits says $$\lim_{z\to z_0}\frac{f(z)}{g(z)} = \frac{\displaystyle \lim_{z\to z_0} f(z)}{\displaystyle \lim_{z\to z_0} g(z)}$$ as long as both limits on the RHS exists and  $\displaystyle\lim_{z\to z_0} g(z) \ne 0$

Then let $\epsilon >0$ by assumtion there exists $\delta >0$ s.t if $\vert z-z_0\vert <\delta$ then $\vert f(z)-f(z_0)\vert <\epsilon$
and $\vert f(z)-f(z_0)\vert=\vert \overline{f(z)}-\overline{f(z_0)}\vert<\epsilon$

This proves that if $f$ is continuous so is $\overline{f}$.


问:
Can we show that the result of an integral of a function of polynomial order is always one degree higher?
Is it true that if $$ \int f(x)\,\mathrm dx = g(x) $$

for $f\in\mathcal O(x^n)$, then $g$ must be in $\mathcal O(x^{n+1})$ where $\mathcal O$ denotes big-O notation and $ n \ge 1 $ is an integer?

If it is true, how can we prove it?

If it is false, can we have a counterexample as the answer?

答:
(I've decided to delete my old answer since I had made a step that was false. I decided that the proof was unsalvageable and wrote this proof instead)

For some $n>-1$ and $f(x)\in O\left(x^n\right)$, suppose that $$g(x)=\int f(x)dx\notin O\left(x^{n+1}\right)$$

Now this implies that there does not exist real $k$ such that, provided $x>x_0$ for some real $x_0$, $$g(x)\leq kx^{n+1}$$

Or alternatively, for every real $k$ and $x>x_0$ we have $$g(x)\leq kx^{n+1}$$

Taking the derivative:

$$\frac d{dx}g(x)>\frac d{dx}kx^{n+1},x>x_0$$

$$f(x)>k(n+1)x^n,x>x_0$$

However, since $f(x)\in O\left(x^n\right)$, this implies that there exists $M$ and real $x_1$ such that $$f(x)\leq Mx^n,x>x_1$$

Let $k=\frac M{n+1}$. The equation we obtained by taking the derivative becomes $$f(x)>Mx^n,x>x_0$$

But when $x>x_0$ and $x>x_1$ then we have

$$Mx^n\geq f(x)>Mx^n$$

Clearly a contradiction.

$\therefore$ $g(x)\in O\left(x^{n+1}\right)$


问:
How to prove that Limit $\lim_{n\to\infty} \cos(n\pi)$ doesn´t exist, or exist?
I believe that when n tends to infinity, the expression $\cos(n\pi)$ doesn´t converge, because the oscilation values. But, how can we formally prove that?
Thank you!!

答: If $n$ is even $cos(n\pi) = 1$, and if $n$ is odd $cos(n\pi) = -1$. So, there is a subsequence ($cos(2n\pi)$) that converges to $1$ and another subsequence ($cos((2n+1)\pi)$) that converges to $-1$. So, the sequence does not converge.


答:
Just do the definition.

If $\lim_{n\to \infty} \cos (n\pi) = a$ then any $\epsilon > 0$ the there is an $N$ so that if $n > N$ then $|\cos(n\pi)-a| < \epsilon$.

Well if $n$ is even than $\cos (n\pi) = 1$ and if $n$ is odd then $\cos (n\pi)=-1$.

So both  $|1- a| < \epsilon$ and $|-1 -a | < \epsilon$ which can be shown to be impossible for $\epsilon \le 1$ by the triangle inequality.

$2= |-1-1|=|(-1-a) +(a-1)|\le |-1 + a|+|a-1|=|-1+a|+|1-a|< \epsilon + \epsilon = 2\epsilon < 2$.

A contradiction.  so $\lim_{n\to \infty}\cos(n\pi) \ne a$ for and $a\in \mathbb R$.

I'll leave it to you to show $\lim_{n\to \infty}\cos(n\pi) \ne \pm \infty$.

($\lim_{n\to \infty}\cos(n\pi) =\infty$ would imply for any $M\in \mathbb R$ there is an $N$ so that $n > N$ would mean $\cos(n\pi) >M$  but $\cos (n\pi) \le 1$.)

Title: How to prove that Limit $\lim_{n\to\infty} \cos(n\pi)$ doesn´t exist, or exist?

Body:
I believe that when n tends to infinity, the expression $\cos(n\pi)$ doesn´t converge, because the oscilation values. But, how can we formally prove that?
Thank you!!


答: If the limit exists, then the sequence $(x_n)$ given by $x_n:=\cos(\pi n), n\in\Bbb N$ converges as well.  The sequence isn't Cauchy though, because of the oscillation.  Successive terms differ by at least $2$.  More precisely, given $\epsilon\lt2$, there is no $N$ such that $n,m\ge N\implies|x_n-x_m|\lt\epsilon$.


问:
Solve $4 \log_2 (n) \leq \frac{n}{2}$
I have to find the $n \in \mathbb{N}$ whenever $4 \log_2 (n) \leq \frac{n}{2}$. So

$$4 \log_2 (n) \leq \frac{n}{2}$$
$$ \iff \log_2 (n) \leq \frac{n}{8}$$
$$ \iff \frac{1}{n}\log_2 (n) \leq \frac{1}{8}$$
$$ \iff \log_2 (n^\frac{1}{n}) \leq \frac{1}{8}$$
$$ \iff n^\frac{1}{n} \leq 2^\frac{1}{8}$$

I am not sure how I can use that to find the appropriate $n$. Is there a better technique to find the right $n$?

EDIT

Not sure I can use induction here. I have to find the minimum $n_0 \in \mathbb{N}$ where $4 \log_2 (n) \leq \frac{n}{2}$ for $n \geq n_0$. How to find that $n_0$?

答:
A hint:

You can use the fact that $2^{x}$ is an increasing function.

$$8log_{2}(n)\leq n$$

whenever

$$2^{8log_{2}(n)} \leq 2^{n} $$

simplifying

$$n^{8} \leq 2^{n} $$

Whenever

$$0 \leq 2^{n} - n^{8} $$

Either solve the equation 2^{x} - x^{8} = 0 and take the next largest integer value which follows the largest solution. If this is within the methods and techniques that you have been taught. The solution looks quite messy so it may be too complex in your course to find the exact value of such x.

Otherwise try solutions of the form $n=2^{k}$ find the first such $k > 0$ which satisfies the inequality, which is $k=6$ and you can look for $2^{5}=32<n<64=2^{6}$. Plotting the graph also helps narrow down where to look.

$n=44$ is the first such value.

Maybe there is some slick way to finish the computation that I am missing.

Title: Solve $4 \log_2 (n) \leq \frac{n}{2}$


问:
Why to obtain the coordinates of vectors in the basis that themselves belong?

Let $ \space T: \mathbb{R^2} \to \mathbb{R^3}$ a linear transformation defined as $ \space T(x,y)=(3x+2y,x+y,-2x-y)$, where $\beta=\{(1,-1),(0,1)\}$ is a basis of $\mathbb{R^2}$. Is not specify a basis for $\mathbb{R^3}$.

By the expression of $ \space T(x,y)$ one known that,

$$T(1,-1)=(1,0,-1)$$
$$T(0,1)=(2,1,-1)$$

The matrix will have $3$ lines by $2$ columns. Since there wasn't a specific basis for $\mathbb{R^3}$ I assumed the canonic basis. So the linear combination of each $T(1,-1)$ and $T(0,1)$ in the canonic $\mathbb{R^3}$ basis are themselves.

The matrix will be,

$\begin{bmatrix}1 & 2 \\ 0 & 1 \\ -1 & -1 \end{bmatrix}$

But,

$\begin{bmatrix}1 & 2 \\ 0 & 1 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}1 \\ -1 \end{bmatrix}\neq$$\begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}$

My doubt is, why it's need to write each vector of $\beta$ as a linear combination of themselves to get their coordinates in the basis that they belong? Thanks

答:
Since that $(1,-1)$ is the first vector of $\beta$ it has coordinates $(1,0)$.

When you use matrices to apply the linear transformation you are using the coordinates of the vector on a fixed basis.

So you should write $$\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}.$$

If $\{v_1,\ldots,v_n\}$ is a basis and $v=\sum_{i=1}^{n} \alpha_iv_i$ then you can say that the coordinates of $v$ on this basis is $(\alpha_1,\ldots,\alpha_n)$.

So what is the coordinate of $v_i$ on this basis? That is $(\underbrace{0,\ldots,0}_{i-1},1,0,\ldots,0)$.


问:
Inverse fourier transform of function with support on [0,1]
Consider the following function $$F(x)=x,~ for~0<x<1 ~; 0~otherwise.$$

I was trying to compute the inverse fourier transformation as (by using wolframalpha) $$f(y)=\int_{R} F(x)e^{2 \pi i x y}dx=\int_0^1 xe^{2 \pi i x y}dx= \frac{-1+e^{2 \pi i x}(1-2 \pi i x)}{4 \pi ^2 x^2}.$$
Now, can I conclude that $$f(x)=\frac{-1+e^{2 \pi i x}(1-2 \pi i x)}{4 \pi ^2 x^2}$$ (here, I am not sure, if I should define $f$ by $0$ outside $(0,1)$ or not) and that FT of $f$ is $F$ (the FT of the result expresion differs from the origin)?

答:
Yes.  Just perform the integration by parts

$$\begin{align*} f(y) &= \int_{-\infty}^\infty F(x)e^{2\pi i yx} dx \\
\\
&=  \int_0^1 xe^{2\pi i yx} dx \\
\\
&= \dfrac{xe^{2\pi i yx}}{2\pi i y}\biggr{\rvert}_0^1 - \int_0^1 \dfrac{e^{2\pi i yx}}{2\pi i y} dx\\
\\
&= \dfrac{e^{2\pi i y}}{2\pi i y} -  \dfrac{e^{2\pi i yx}}{\left(2\pi i y\right)^2}\biggr{\rvert}_0^1 \\
\\
&= -\dfrac{2\pi i y e^{2\pi i y}}{4\pi^2 y^2} + \dfrac{e^{2\pi i y}}{4\pi^2y^2} - \dfrac{1}{4\pi^2y^2} \\
\\
&= \dfrac{-1 + e^{2\pi i y}\left(1-2\pi i y \right)}{4\pi^2y^2}\\
\end{align*}$$

No, the resulting function is not limited in domain; $y \in (-\infty, \infty)$

It is worth noting that $f(y)$ has a singularity at $y=0$.  You'll have to use L'Hopital's rule to check if the singularity is removable or not.


问:
Show that $\sum_{k=0}^{n} {n \choose k}2^k=3^n$ using mathematical induction.
Ok so I'm not looking for someone to just straight up tell me the answer but if you could nudge me in some direction approaching a valid proof that'd be great.

$\sum_{k=0}^{n} {n \choose k}2^k=3^n \ \ \ \ \ \ \ \ \ \ \ (eq.1)$

(1) for $n=1$:
$$ \sum_{k=0}^{1}{1 \choose k}2^k = {1 \choose 0}2^0+{1\choose1}2^1 = 1+2 = 3 = 3^1 $$

so eq.1 is true for n=1.

(2) Assume eq.1 holds for $n=p$. That is,
$$\sum_{k=0}^{p} {p \choose k}2^k=3^p $$
is a true statement.

(3)  Show that (2) necessitates eq.1 be true for $n=p+1$. That is, show
$$\sum_{k=0}^{p+1} {p+1 \choose k}2^k=3^{p+1}.$$

So anyhow, I've been attempting to start (3) by splitting the sum from k=0 to p then having the last term separate like so :
$$\sum_{k=0}^{p+1} {p+1 \choose k}2^k=\sum_{k=0}^{p} {p+1 \choose k}2^k+2^{p+1}$$

This is the method the prof suggested but I'm getting nowhere with it.  And yes I must use induction.

答:
(maybe is cheating) You can always use that $$\sum_{k=0}^n {n \choose k}1^{n-k}2^k=
 (1+2)^n$$


问:
Tail bounds for maximum of sub-Gaussian random variables
I have a question similar to this one, but am considering sub-Guassian random variables instead of Gaussian. Let $X_1,\ldots,X_n$ be centered $1$-sub-Gaussian random variables (i.e. $\mathbb{E} e^{\lambda X_i} \le e^{\lambda^2 /2}$), not necessarily independent. I am familiar with the bound $\mathbb{E} \max_i |X_i| \le \sqrt{2 \log (2n)}$, but am looking for an outline of a tail bound for the maximum.

A union bound would give
$$\mathbb{P}(\max_i |X_i| > t) \le \sum_i \mathbb{P}(|X_i| > t) \le 2n e^{-t^2/2},$$
but I am looking for a proof of something of the form
$$\mathbb{P}(\max_i |X_i| > \sqrt{2 \log (2n)} + t)
\le \mathbb{P}(\max_i |X_i| > \mathbb{E} \max_i |X_i| + t)
\le 2e^{-t^2/2}.$$
Does anyone have any hints?

答:
I needed something along those lines recently and didn't have a specific reference to cite, so here is a proof of a self-contained statement implying yours.

Theorem. Let $X_1,...,X_n$ be independent $\sigma^2$-subgaussian random variables. Then
$$
\mathbb{E}[\max_{1\leq i\leq n} X_i] \leq \sqrt{2\sigma^2\log n} \tag{1}
$$
and, for every $t>0$,
$$
\mathbb{P}\!\left\{\max_{1\leq i\leq n} X_i \geq \sqrt{2\sigma^2(\log n + t)}\right\} \leq e^{-t}\,. \tag{2}
$$

Proof.
The first part is quite standard: by Jensen's inequality, monotonicity of $\exp$, and $\sigma^2$-subgaussianity, we have, for every $\lambda \in \mathbb{R}$,
$$
e^{\lambda \mathbb{E}[\max_{1\leq i\leq n} X_i]}
\leq \mathbb{E}e^{\lambda \max_{1\leq i\leq n} X_i}
= \max_{1\leq i\leq n}\mathbb{E}e^{\lambda X_i}
\leq \sum_{i=1}^n\mathbb{E}e^{\lambda X_i}
\leq n e^{\frac{\sigma^2\lambda^2}{2}}
$$
so, taking logarithms and reorganizing, we have
$$
\mathbb{E}[\max_{1\leq i\leq n} X_i] \leq \frac{1}{\lambda}\ln n + \frac{\lambda \sigma^2}{2}\,.
$$
Choosing $\lambda := \sqrt{\frac{2\ln n}{\sigma^2}}$ proves (1).
${}$
Turning to (2), let $u := \sqrt{2\sigma^2(\log n + t)}$. We have
$$
\mathbb{P}\{ \max_{1\leq i\leq n} X_i \geq u \}
= \mathbb{P}\{ \exists i,\; X_i \geq u \}
\leq \sum_{i=1}^n \mathbb{P}\{ X_i \geq u \}
\leq n e^{-\frac{u^2}{2\sigma^2}}
 = e^{-t}
$$
the last equality recalling our setting of $u$.
$\square$

Here is now an immediate corollary:

Corollary. Let $X_1,...,X_n$ be independent $\sigma^2$-subgaussian random variables. Then, for every $u>0$,
$$
\mathbb{P}\!\left\{\max_{1\leq i\leq n} X_i \geq \sqrt{2\sigma^2\log n}+ u \right\} \leq e^{-\frac{u^2}{2\sigma^2}}\,. \tag{3}
$$

Proof. For any $u>0$,
$$
\mathbb{P}\!\left\{\max_{1\leq i\leq n} X_i \geq \sqrt{2\sigma^2\log n} + u \right\}
\leq e^{-\frac{u^2}{2\sigma^2} - u\sqrt{\frac{2\log n}{\sigma^2}}}
\leq e^{-\frac{u^2}{2\sigma^2}},
$$
the inequality by choosing $t := \frac{u^2}{2\sigma^2} + u\sqrt{\frac{2\log n}{\sigma^2}}$ and using (2).       $\square$

More: the slides of some lecture notes by John Duchi.


问:
Compare the number of comparisons used by the insertion sort to sort the list $ 7, 4, 3, 8, 1, 5, 4, 2$.
Compare the number of comparisons used by the insertion sort to sort the list $ 7, 4, 3, 8, 1, 5, 4, 2$? I found that based on the following pseudocode for the algorithm:

```<code>/* Function to sort an array using insertion sort*/
void insertionSort(int arr[], int n)
{
    int i, key, j;
    for (i = 1; i < n; i++)
    {
        key = arr[i];
        j = i - 1;

        /* Move elements of arr[0..i-1], that are
        greater than key, to one position ahead
        of their current position */
        while (j >= 0 && arr[j] > key)
        {
            arr[j + 1] = arr[j];
            j = j - 1;
        }
        arr[j + 1] = key;
    }
}
</code>```

The total should be starting from phase 1 that begins with 4 and 7 is: $1+2+0+4+3+4+7$, but the correct answer should be $1+1+3+1+4+3+2=15$, can you please explain why we need 15?

My solution:

Phase 1: compare 4 with 7 needs 1.

Phase 2: compare 3 with 7 with 3 with 4 needs 2.

Phase 3: compare 8 with 7 needs 1.

Phase 4: compare 1 with 8 with 7 with 4 with 3 needs 4.

...etc

答: For your answer $1+2+0+4+3+4+7$, I think that $0$ should be a $1$. Other than that, your answer looks good to me. It might be worthwhile to discuss with your instructor.


问:
show that : $\mathcal{S}\cap\mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra on $\mathbb{R}$ making the function $x \mapsto x^2$ measurable.
Let $\mathcal{S}$ be the set of symmetrical subset of $\mathbb{R}$ $\big($ i.e. $\mathcal{S}:= \{A\subset\mathbb{R}:\forall x \in\mathbb{R}~~,~~ x \in A \Leftrightarrow -x \in A\}$$\big)$.

Problem:

I want to show that : $\mathcal{S}\cap\mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra $\mathcal{A}$ on $\mathbb{R}$ making the function $f:x \mapsto x^2$ from $\mathbb{R}$ to $\mathbb{R}$, $(\mathcal{A}, \mathcal{B} (\mathbb{R}))$ - measurable.

My effort

We have $$\sigma \big(f^{-1}(\mathcal{B} (\mathbb{R})\big)=\sigma\bigg(f^{-1}\bigg(\sigma\bigg(\big\{]-\infty,q]:q\in\mathbb{Q}\big\}\bigg)\bigg)\bigg)=\sigma\bigg(\bigg\{f^{-1}\big(]-\infty,q]\big):q\in\mathbb{Q}\big\}\bigg)\bigg)\bigg)=\sigma(\{[-q,q]:q\in\mathbb{Q}\})$$

Then $\mathcal{A}=\sigma(\{[-q,q]:q\in\mathbb{Q}\})$ is the smallest $\sigma$-algebra on $\mathbb{R}$ making the function $f:x \mapsto x^2$ from $\mathbb{R}$ to $\mathbb{R}$, $(\mathcal{A}, \mathcal{B} (\mathbb{R}))$ - measurable.

Is what I wrote true? If so, you can help me show that $\sigma(\{[-q,q]:q\in\mathbb{Q}\})=\mathcal{S}\cap\mathcal{B}(\mathbb{R})$?

An idea please.

答:
If $f(x) = x^2$ we see that $\{x | f(x) < \alpha \} = (-\sqrt{\alpha},\sqrt{\alpha})$ if $\alpha >0$ and $\emptyset$ otherwise. Hence the smallest $\sigma$ algebra
that makes $f$ (Borel) measurable is $\Sigma = \sigma(\{(-x,x)\})$.

We want to show that $\Sigma = \mathcal{S} \cap \mathcal{B}$. (I have dropped the $\mathbb{R}$ notation.)

Note that $\mathcal{S}$ is a $\sigma$ algebra and so $\mathcal{S} \cap \mathcal{B}$ is a $\sigma$ algebra.
Since $(-x,x) \in \mathcal{S} \cap \mathcal{B}$ we see that
$\Sigma \subset \mathcal{S} \cap \mathcal{B}$.

To show that $\mathcal{S} \cap \mathcal{B} \subset \Sigma$, let $\mathcal{C} = \{ B \in \mathcal{B} | \{B\} \cap \mathcal{S} \subset \Sigma
 \}$. It is straightforward to
see that $\mathcal{C}$ is a $\sigma$ algebra and that $(a,b) \in \mathcal{C}$ and
hence $\mathcal{C} = \mathcal{B}$.


答:
As a real, the solution of $$4 \log_2 (n) = \frac{n}{2}$$ is given in terms of Lambert function
$$n=-\frac{8 }{\log (2)}W_{-1}\left(-\frac{\log (2)}{8}\right)\sim 43.5593$$ so $\lceil n \rceil=44$.

Checking
$$4 \log_2 (43) - \frac{43}{2} \sim +0.205059$$
$$4 \log_2 (44) - \frac{44}{2} \sim -0.162274$$

Edit

If you do not know about Lambert function (or if you cannot use it), consider the function and derivatives
$$f(n)=\frac{4 \log (n)}{\log (2)}-\frac{n}{2}$$
$$f'(n)=\frac{4}{n \log (2)}-\frac{1}{2}$$
$$f''(n)=-\frac{4}{n^2 \log (2)} \quad < 0 \quad \forall n$$

The first derivative cancels at
$$n_*=\frac{8}{\log (2)}\implies f(n_*)=-\frac{4 \left(1+\log \left(\frac{\log (2)}{8}\right)\right)}{\log (2)} <0$$ Develop $f(n)$ as a Taylor series around $n_*$ to get
$$f(n)=f(n_*)+\frac 1 2 f''(n_*)(n-n_*)^2+O((n-n_*)^3$$
Solve the quadratic to get a starting point
$$n_0=n_*+\sqrt{-2\frac{f(n_*)}{f''(n_*)}}\sim 31.1687$$ Now, start Newton method and get the following iterates
$$\left(
\begin{array}{cc}
k & n_k \\
 0 & 31.1687 \\
 1 & 44.7107 \\
 2 & 43.5645 \\
 3 & 43.5593
\end{array}
\right)$$

Title: Solve $4 \log_2 (n) \leq \frac{n}{2}$

Body:
I have to find the $n \in \mathbb{N}$ whenever $4 \log_2 (n) \leq \frac{n}{2}$. So

$$4 \log_2 (n) \leq \frac{n}{2}$$
$$ \iff \log_2 (n) \leq \frac{n}{8}$$
$$ \iff \frac{1}{n}\log_2 (n) \leq \frac{1}{8}$$
$$ \iff \log_2 (n^\frac{1}{n}) \leq \frac{1}{8}$$
$$ \iff n^\frac{1}{n} \leq 2^\frac{1}{8}$$

I am not sure how I can use that to find the appropriate $n$. Is there a better technique to find the right $n$?

EDIT

Not sure I can use induction here. I have to find the minimum $n_0 \in \mathbb{N}$ where $4 \log_2 (n) \leq \frac{n}{2}$ for $n \geq n_0$. How to find that $n_0$?


Title: Solve $4 \log_2 (n) \leq \frac{n}{2}$

Body: Iterate: $$k_0=1, k_{n+1}=8\log_2(k_n)\implies \lim_{j\to\infty} k_j\approx 43.55926044$$


问:
Probability of a specific order of outcomes
Suppose there is a student who wanders into a classroom to take an exam that he is not prepared for. There are four multiple choice questions on the exam, each with four given answers to select from. There is only one correct solution for each question. What is the probability that the student gets the first and third questions correct, but misses the second and fourth questions?

This problem is nested in a binomial distribution question, but the binomial distribution doesn't seem to apply here since it only tells you the probability of a certain number of successes.

My initial thought is to just multiply the respective probability for each question, i.e., $\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}$, but that would only tell you the probability that the student gets any two questions correctly. My second thought is to divide that quantity by 7 since there are seven possible ways for the student to get two questions correctly (assuming that order doesn't matter).

Any thoughts or insight would be appreciated. Thanks!

答:
My initial thought is to just multiply the respective probability for each question, i.e., $\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}$, but ...

No buts.   That is the answer.   That is the probability for obtaining: first is correct, second is incorrect, third is correct, and fourth is incorrect.   Ie: two correct and two incorect results in the specified order.


问:
Prove that expected value of X is greater than Y, if given that $P(X\ge Y)=1$
I have to prove that $E(X)$ (Expected Value of a random variable X), is greater than $E(Y)$, if given that $P(X\ge Y)=1$.

my thoughts so far:

I know from the $P(X\ge Y)=1$ statement, that the values that X "receives" are always greater than the values that Y "receives", and because the expected value of a random variable is always a weighted mean of its valid values, than of course $E(X) \ge E(Y)$. But how do I prove it formally?

答: Take a new random variable, say Z, where Z = X - Y. Then you have $P(Z \geq 0) = 1$. Now try to show $E(Z) \geq 0$.</p>


问:
Calculate Geometric Progression with logarithm
I'm trying to find the following summation but cant seem to find the answer. Would really appreciate any help from here!

$$\sum_{i=0}^{k}(4/3)^{i}\log(n/3^i)$$

答:
$$(4/3)^i\log(n/3^i)=(4/3)^i(\log n-\log(3^i))=(4/3)^i(\log n-i\log 3)$$

So, your sum is,

$$\log n\left[\sum_{i=0}^k (4/3)^i\right]-\log 3\left[\sum_{i=0}^k i(4/3)^i\right]$$

The first summand has a finite geometric series and the second summand has a finite arithmetico-geometric series. Do you know how to compute them?


问:
Can I use a z test to compare two samples if there is a significant difference in the size?
I have to compare two unequal size samples from a population of 53k. One has a proportion of 56% and size of 50k and the other has a proportion of 50% and size of 3k. Can I do a two proportion z test to compare to make a case that the proportion of sample 1 is greater than sample 2?
Does the difference in the sample size pose an issue?

Some more info:
Here is the situation: Lets say I had 53k people and 50k received treatment A and 3k received treatment B. From the 3k, 50% showed positive results and from the 50k, 56% showed positive effect. How can I conclude that my treatment A was more effective than treatment B? I am looking for a way to make that case.

答:
Yes. In R, the procedure prop.test uses sizes of the two
samples (which may differ) and counts of 'successes' in
the two samples.

Then the test is as shown below:
The very small P-value indicates a highly significant
difference between the proportions $\hat p_1 =0.56$ and
$\hat p_2 =   0.50.$ [Because of the large sample sizes
I have not used the continuity correction cor=F.]

```<code>p1 = .56; n1 = 50000; x1 = round(p1*n1)
p2 = .50; n2 = 3000;  x2 = round(p2*n2)
prop.test(c(x1,x2), c(n1,n2), cor=F)

    2-sample test for equality of proportions
    without continuity correction

data:  c(x1, x2) out of c(n1, n2)
X-squared = 41.284, df = 1, p-value = 1.317e-10
alternative hypothesis: two.sided
95 percent confidence interval:
 0.04158663 0.07841337
sample estimates:
 prop 1 prop 2
   0.56   0.50
</code>```

Notes: (1) The X-squared statistic shown in the
output is the square of the normal $Z$-statistic.

(2) The version of the test used here is the one in the (yellow) note at the end of
this page
so that the confidence interval 'matches' the test. [If the 95% CI does not include $0,$ then the two proportions are significantly different
at the 5% level.]

Body:
Yes. In R, the procedure prop.test uses sizes of the two
samples (which may differ) and counts of 'successes' in
the two samples.

Then the test is as shown below:
The very small P-value indicates a highly significant
difference between the proportions $\hat p_1 =0.56$ and
$\hat p_2 =   0.50.$ [Because of the large sample sizes
I have not used the continuity correction cor=F.]

p1 = .56; n1 = 50000; x1 = round(p1*n1)
p2 = .50; n2 = 3000;  x2 = round(p2*n2)
prop.test(c(x1,x2), c(n1,n2), cor=F)

    2-sample test for equality of proportions
    without continuity correction

data:  c(x1, x2) out of c(n1, n2)
X-squared = 41.284, df = 1, p-value = 1.317e-10
alternative hypothesis: two.sided
95 percent confidence interval:
 0.04158663 0.07841337
sample estimates:
 prop 1 prop 2
   0.56   0.50

Notes: (1) The X-squared statistic shown in the
output is the square of the normal $Z$-statistic.

(2) The version of the test used here is the one in the (yellow) note at the end of
this page
so that the confidence interval 'matches' the test. [If the 95% CI does not include $0,$ then the two proportions are significantly different
at the 5% level.]


问:
Determining the contour to use during contour integration
Let us say we want to integrate

$$\int_{-\infty}^{\infty} \frac{dx}{1+x^4}$$

We do this by c contour integration of the form:

$$\oint_{-\infty}^{\infty} \frac{dz}{1+z^4}$$
However my question concerns the type of contour we use.
Now everywhere that I have seen do this online do it via taking the semicircle counter.
However, we know we have poles at $$z   = 1,-1,i,-i$$

So my question is since we have a singularity at -1, 1 then surely these would lie on the path integral if we choose a semicircle. Surely a more appropriate contour would be a semi circle with 2 small semi circles of very small (tend to 0) radius $\epsilon$ as shown below.

答:
The poles are not at $z \in \{1,-1,i,-i\}$. Note that, for all of these, $z^4 = 1$ and thus the denominator is nonzero.

Rather, the poles are at

$$z \in \left\{\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; \frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i\right\}$$


问:
Forward Euler is a method of order 1 using Taylor expansion
I am reading MIT's numerical analysis notes. On page 6 they say it is a good exercise to show the explicit Euler method is of order 1. I thought this would be a good exercise because I am not familiar.

Given a numerical scheme $$y_{n+1} = \Psi(h,y_n,t_n)$$ they defined it to have order $p$ if the local error $e_{n+1}(h)=\Psi(h,y_n,t_n)-y(t_{n+1})$ is of order $O(h^{p+1})$. If we plug the forward Euler into this definition we get (note that $t_n = nh$ for some step $h$)

$$e_{n+1}(h)=y_n+hf(t_n,y_n)-y(t_{n+1})$$
They suggest we use 'elementary' Taylor expansion to show that $\Psi$ is of order 1. I dont know what they mean by elementary, other than a first order expansion. They also dont say what to take a Taylor expansion of, but I assume it is of $y(t_{n+1})$. Also not sure where to take the centre of the expansion, so I leave it as $t$ for now. Then
$$
y(t_{n+1}) \approx y(t) + y^{\prime}(t) t_{n+1}
$$
We can use the initial value problem $y'(t) = f(t,y(t))$ to re-write the prime term to get
$$
y(t_{n+1}) \approx y(t) + f(t,y(t)) t_{n+1}
$$I thought that I would probably need to put this into the local error formula and do some algebra to get a $h^2$ term, my attempt goes as follows

\begin{align*}
e_{n+1}(h) &\approx y_n + hf(t_n,y_n) - y(t) -f(t,y(t)) t_{n+1}\\
&= y_n - y(t) + h(f(t_n,y_n) - (n+1) f(t,y(t)))
\end{align*}
from here I cant see how to argue that $e_{n+1}$ is of order $O(h^2)$.

答:
Suppose your step size is $h$, so $t_{n+1}=t_n+h$. Then Taylor expanding $y(t_n+h)$ gives
\begin{align*}
y(t_{n+1})=y(t_n+h)&=y(t_n)+hy'(t_n)+O(h^2)\\
&=y_n+hf(t_n,y_n)+O(h^2)\\
&=y_{n+1}+O(h^2),
\end{align*}
which tells you that the local truncation error is $O(h^2)$. Remember that when computing the local truncation error, you implicitly assume that you got your previous steps right, i.e. $y_n=y(t_n)$.


问:
Maximum area for minimum perimeter of a triangle
Does the fact that a triangle has the maximum area for a given perimeter when it is an equilateral triangle (Isoperimetric Property of Equilateral Triangles) imply that the ratio of a triangle's area to its perimeter will have a maximum when all side lengths are equal? I found the derivative of Heron's formula for area divided be perimeter for an isosceles triangle and this is not the case. The ratio is at a maximum when the third side is $\sqrt5 - 1$ times the length of the two equal sides, which interestingly enough comes out to be $2(\phi - 1).$

答:
This is a demo to show that the interpretation of Heron is wrong. I have not had good luck with him either so perhaps misreading him is just a common thing. Note: D is used for area because it is the fourth term of a cubic equation that often results in area computations and it avoids confusion with sides or vertices A,B,C of any triangles used.

Given a triangle with $P=240$ we will let  $(\sqrt{5}-1)x$ be the base.

$$\big(\sqrt5 - 1\big)x+2x=240\implies x=\frac{240}{\sqrt{5}+1}\approx 74.16\\
\implies (\sqrt{5}-1)x\approx 91.67
\implies h=\sqrt{74.16^2-\bigg(\frac{91.67}{2}\bigg)^2}\approx 58.2997...\\
\implies D=area=\bigg(\frac{91.67}{2}\bigg)\cdot 58.2997\approx 2672.17$$

In comparison, an equilateral triangle where $P=240$ has sides of $80$.
$$x=80\implies h=\sqrt{80^2-40^2}\approx 69.28\\
\implies D=area= 40\cdot 69.28\approx 2771.2$$

The equilateral triangle has the larger area for the same perimeter and the larger D/P ratio.
$$
\frac{D_1}{P}=\frac{2672.17}{240}\approx 11.134\\
\frac{D_2}{P}=\frac{2771.2}{240}\approx 11.546$$


问:
Gap between composite numbers that aren't divisible by 2,3 or 5
It is easy to determine if a number is divisible by 2,3 or 5, so in my experiment i am concentrating on a consecutive list of composite numbers that are not divisible by 2,3 or 5.

My question is if there is a maximum gap limit between such numbers, and if so what is it?

Please review the image as an example:

答:
These numbers are prime

1418575498609,  1418575498607,  1418575498603,  1418575498601,  1418575498597,  1418575498591,  1418575498589,  1418575498583,  1418575498579,  1418575498577,  1418575498573

which gives a gap of 42 between 1418575498571 and 1418575498613

They are 614101948*2310 - 1260 -[11,13,17,19,23,29,31,37,41,43,47]

Edit
Suppose we look at numbers near $N$, which might be $10^{13}$.

The chance a number $x$ is prime is $1/\ln(N)$.  Prime factors up to 11 are ruled out by the 2310 formula, so the chance of being prime is increased to

$$P=\frac{\frac21\frac32\frac54\frac76\frac{11}{10}}{\ln N}$$
The chance all eleven numbers are prime is $P^{11}$.

The 2310 formula gives two opportunities every 2310, namely $2310k+1260+[11,...]$ and $2310k-1260-[11,...]$

So an estimate for the number of solutions within $N/2$ of $N$, so between $N/2$ and $3N/2$, is

$$\frac{P^{11}N}{1155}$$
That suggests about 4 near $N=10^{12}$, but I have only found one.


问:
Burnside's Lemma bracelet with beads in 2 colours (uneven cardinality)
How many bracelets can be made with 5 beads in one colour and nine in the other (rotated and flipped bracelets are considered the same).

I'm trying to use Burnside's Lemma yet I'm little confused. So far I've got:

Let $X=\{\text{bracelets of length 14 beads = 5 red + 9 blue}\}$, $|X|=\frac{14!}{5!9!}=2002$.
Now G in my case is $G=D_{14}$, $|G|=id + \text{13 rotations + symmetries(diagonal and perpendicular)}=14+14+14(?)$
Yet I've found that $D_{14}$ is order 28. What I'm missing here? Am I counting some actions twice?(I did)

Another step I am not sure is applying Burnside's Lemma with specified number of beads.

I should count fixed points for each $g\in G$.

Identity $g_{r_{14}}$ would is trivial and sends any $v$ back to itself $|fix(g_{r_{14}})|=|X|=2002$

First I can consider rotating bracelet by $\frac{360}{14}^\circ $ . For given colouring there are none, since permutation induced by $g_{r_{1}}$ maps initial set to one cycle(orbit) that should be in one colour: $r_1 \rightarrow (1\,2\,3\,4\,5\,6\,7\,8\,9\,10\,11\,12\,13\,14)$ (If we choose one vertex and colour it, we need all of them in the same colour).

Next $g_{r_{2}}$. For given colouring again there are none, since permutation induced by $g_{r_{2}}$ maps initial set to two cycles, both should be monochromatic: $r_2 \rightarrow (1\,3\,5\,7\,9\,11\,13)(2\,4\,6\,8\,10\,12\,14)$ The orbits are order 7 and there are only 5 beads in one colour and 9 in the other.
I've found that $r_n$ has $gcd(n,14)$ orbits of equal size.

For $r_3$ I've found one: $r_3 \rightarrow (1\,4\,7\,10\,13\,2\,5\,8\,11\,14\,3\,6\,9\,12)$

And $r_4$ I've found again 2: $r_4 \rightarrow (1\,5\,9\,13\,3\,7\,11)(2\,6\,10\,14\,4\,8\,12)$

Both intuitively cannot result in proper colouring. How to avoid writing all of them?
Can I note that $r_5$,$r_7$,$r_9$,$r_{11}$ and $r_{13}$ are coprime with 14 and do not allow for proper colouring. Same with $r_6$,$r_8$,$r_{10}$ and $r_{12}$, there would be only 2 orbits of even size and thus uncolouruable with given beads.
$r_{7}$ would result in 7 orbits of size 2. Since they should be monochromatic it's impossible to find colouring.

And what about flips?
Do I correctly assume that 14 $s_d$ flips has 6 orbits of size 2 (vertices on both sizes of symmetry axis) and 2 orbits of size 1 (vertices that are immutable). And another 14 $s_p$ flips has 7 orbits of size 2 each. Eg. $s_d$ along 1-8 axis:
$$s_{d_1} \rightarrow (1)(2\,14)(3\,13)(4\,12)(5\,11)(6\,10)(7\,9)(8)$$
Which allows us to make such colouring with fixed vertices being in two colours and assigning the colour to rest of them is possible because now, we have even number of each group of beads. It gives us 2 choices in first step and 15 ways of choosing 2 out of 6 orbits, and for following 4 and the last one is only one option. So $2\cdot 15\cdot1^5=30$

and $s_{p_{1}}$ on edges $1-2$ and $8-9$:
$$s_{p_1}\rightarrow(1\,2)(3\,13)(4\,13)(5\,12)(6\,11)(7\,10)(8\,9)$$
And in this case there are non, since all orbits are even.

$$N=\frac{1}{\left|G\right|}\sum_{g\in G} \left|fix(g)\right|$$
here it would be
$\frac{1}{28}\sum_{g\in G} \left |fix(g)\right|=\frac{1}{28}\cdot(14\cdot 15\cdot2+2002) =$ 74  (?)

答:
Yet I've found that D14 is order 28. What I'm missing here?

Your count should be $(1+13)+(7+7)$. There are $7$ diagonal and $7$ perpendicular reflections.

Both intuitively cannot result in proper colouring. How to avoid writing all of them? Can I note that [...] Since they should be monochromatic it's impossible to find colouring.

Correct: instead of computing the fixed points for every $g$, you can group the $g$ according to cycle type (or conjugacy class, or by other similarity considerations in other problems) and simply compute the fixed points of one representative of each group.

It gives us 2 choices in first step and 15 ways of choosing 2 out of 6 orbits

Correct, $2\cdot\binom{6}{2}=30$ is how many fixed points a diagonal reflection has.

And in this case there are non, since all orbits are even.

Correct, perpendicular reflections have only $2$-cycles so can only have an even number of beads of each color for a fixed bead configuration. (Also keep in mind an orbit of the cyclic subgroup $\langle g\rangle$ is called a cycle so it doesn't get confused with an orbit of the whole group $G$.)

The corrected calculation should then be

$$ \frac{1}{28}(2002+7\cdot30)=79. $$


问:
Are the following sets closed in $C[0,1]$?
I am trying to figure out if the following two sets are closed in $C[0,1]$ with $||f||_{\infty}$ and $||f||_1$.

$A = \{f \in C[0,1]: f(\frac{1}{2}) = 0\}$

$B = \{f \in C[0,1]: f(x) \geq 0, \forall x \in [0,1] \}$

Here $C[0,1]$ denotes the set of all real-valued continuous functions over the interval $[0,1]$. For $A$, I do think it is closed with respect to $||f||_{\infty}$. The reason is that since $C[0,1]$ is Banach with the infinity norm, then any subset of it is closed if and only if it is a complete metric space, with the metric induced the norm of $C[0,1]$. However, I am not too sure if $A$ is closed with respect to $||f||_1$. I want to say it is not closed but I do not know how to form my argument. I cannot use the contrapositive of the statement I mentioned above since $C[0,1]$ with $||f||_1$ is not Banach. I think $B$ is closed with respect to the infinity norm, but not the 1-norm.

Any help would be appreciated!

Thanks guys.

答:
$A$ is closed in $\| \cdot\|_\infty$  (since uniform convergence implies pointwise convergence) but it is not closed in $\| \cdot\|_1$: Consider functions $f_n$ that satisfy
$f_n(0)=f_n(1/2-1/n)=1$ and $f_n(1/2)=0$ and $f_n(1/2+1/n)=f_n(1)=1$ and  interpolate linearly between the specified points. These $f_n$ converge in  $\| \cdot\|_1$ to the constant function $1$.

$B$ is closed in $\| \cdot\|_\infty$ and in  $\| \cdot\|_1$. It suffices to prove that $B^c$ is open in $\| \cdot\|_1$.
Suppose $f \notin B$. Then for some $\epsilon>0$ and $x_0 \in [0,1]$, we have $f(x_0)<-2\epsilon$, so by continuity, there exists $\delta>0$ such that $f(x)<-\epsilon$ whenever $|x-x_0|<\delta$.
This implies that the $\epsilon \delta$ ball centered at $f$ in the norm $\| \cdot\|_1$ is disjoint from $B$.


答:
A is closed in $∥\cdot∥_{\infty}$ because of the uniform convergence theorem.

If you know about bump functions and how you can get smooth functions very close to the indicator functions, I can layout a way to find non-continuous examples in the closure for the $||\cdot||_{1}$.

Let I(A) be an indicator function for a set A.

You can consider the sequence of functions $g_{n} = I([0,\frac{1}{n}])$ each function $g_{n}$ is in the closure of both of the sets under $||\cdot||_{1}$ because of bump functions. So the discontinuous function $g(x)=0$ for $x\neq 0$ and $g(0)=1$ is in the closure of both the sets.

If you don't know about bump functions then we can make this construction $h_{n}(x) = 1$ for $x\in [0,\frac{1}{n}]$. $h_{n}(x) = 1 - n(x - \frac{1}{n})$ for $x\in [\frac{1}{n},\frac{2}{n}]$. $h_{n}(x) = 0$ for $x\in [\frac{2}{n}, 1]$. And consider the sequence for  $n>3$.

Title: Are the following sets closed in $C[0,1]$?

Body:
I am trying to figure out if the following two sets are closed in $C[0,1]$ with $||f||_{\infty}$ and $||f||_1$.

$A = \{f \in C[0,1]: f(\frac{1}{2}) = 0\}$

$B = \{f \in C[0,1]: f(x) \geq 0, \forall x \in [0,1] \}$

Here $C[0,1]$ denotes the set of all real-valued continuous functions over the interval $[0,1]$. For $A$, I do think it is closed with respect to $||f||_{\infty}$. The reason is that since $C[0,1]$ is Banach with the infinity norm, then any subset of it is closed if and only if it is a complete metric space, with the metric induced the norm of $C[0,1]$. However, I am not too sure if $A$ is closed with respect to $||f||_1$. I want to say it is not closed but I do not know how to form my argument. I cannot use the contrapositive of the statement I mentioned above since $C[0,1]$ with $||f||_1$ is not Banach. I think $B$ is closed with respect to the infinity norm, but not the 1-norm.

Any help would be appreciated!

Thanks guys.


问:
Given a single die, what is the probability it takes an even number of rolls to get a 4?
So I've decomposed the probability as meaning we failed $2k-1$ times and succeeded on the $2k^{th}$ roll. I set up a  series for this: $\sum_{k=1}^{\infty}\frac{1}{6}\cdot\left(\frac{5}{6}\right)^{2k-1}$. The problem is that my calculus is very shaky and I don't know how to simplify this and get a fraction as as if it were k-1 instead of 2k-1. What's my next step here?


问:
multiply and sum in Finite Field
$F_q$ is a finite field of $q$ (which equals $p^n$，p is a prime number). $F_q^*$ denotes the elements in $F_q$ which has inverse.
Prove that:

The multiply of all the elements in $F_q^*$ equals -1.

$p＞2$,The sum of all elements in $F_q$ equal 0.

Here is my proof,and I want to konw wehther it is true?

Proof: $F_q$ is a finite field,then $F_q^*$ is a cyclic group generated by $u$ (which $u^{q-1}=1$)  with respect to mulitplication in $F_q$ . $F_q=\{0,1,u,…,u^{p^n-2}\}$.

1 .Let  $S=1×u×u^2×…×u^{p^n-2}$,then $S=u^{1+2+…+(p^n-2)}=
u^{\frac{{(p^n-2)(p^n-1)}}{2}}$

Case (1). $p=2$,then $p^n-2$ could divided by 2,then $S=1$ and $1=-1$ (Does this hold in $F_{2^n}$?) in $F_q$ . So $S=-1$

Case (2). $p＞2$ , then $p$ is odd,$S=u^{\frac{(p^n-2)(p^n-1)}{2}}=u^{\frac{-(p^n-1)}{2}}$,we know $S^2=1$ .Thus $(S+1)(S-1)=0$. $S≠1$ because  the order of $u$ is $p^n-1$ (Is this right or wrong?).Then $S+1=0$(in the the field )

2.This is directly by the order of $u$: Let $T=0+1+u+…+u^{p^n-2}$.$u≠1$ in this case. Thus $T=\frac{1-u^{p^n-1}}{1-u}=0$

答:
(1) $p=2$ case: yes, $v=u^{(p^n-2)/2}$ makes sense and $v^{p^n-1}=1$ because $\mathbb{F}_{2^n}^\times$ has order $p^n-1$, and indeed $1=-1$ in any field of characteritic two.

(1) $p>2$ case: you have a typo, should be $S=u^{-(p^n-1)/2}$. And yes, we know $S\ne1$ since $u$ has order $p^n-1$, so you can conclude $S=-1$ from $S^2=1$.

(2) The proof you edited in is correct, and even works for $p=2$ as long as $n>1$.

You can also say that if $p>2$ then $x\ne -x$ for all $x\in\mathbb{F}_{p^n}$, which means each element will cancel with its additive inverse in the sum.

Title: multiply and sum in Finite Field


问:
Embedded Lp spaces
Let $L^\infty(Ω,F,P)$ be the vector space of bounded random variables $(X ∈ L^\infty (Ω,F,P)$ means that there exists a constant C such that $|X(ω)|≤C$, a.s.$)$. Show that $$L^\infty(Ω,F,P)⊂L^2(Ω,F,P)⊂L^1(Ω,F,P)$$

答:
It is a consequence of Holder inequality
$$
E[|XY|]\leq E[|X|^p]^{1/p}E[|Y|^q]^{1/q}
$$</p>

Title: Embedded Lp spaces


问:
Factorize polynomial of degree 4 given statements
If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?

Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$
Can I solve $a,b,c,d$ from there? How can I do so?

答:
Let $$f(x) = x^4-x^3-13x^2+26x-8$$

Observe that $f(2) = 0$ and $f(-4) = 0$

Hence $(x-2)$ and $(x+4)$ are two factors of $f(x)$

Also $f(x)$ is divisible by $(x-2)(x+4) = x^2 + 2x - 8$

Now divide $f(x)$ by $x^2 + 2x - 8$ to obtain the other quadratic factor $x^2 - 3x +1$

If you solve $x^2 - 3x +1 = 0$, you'll find $x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$

Finally
$$f(x) = \left(x - \left(\frac{3}{2} + \frac{\sqrt{5}}{2}\right) \right) \left(x - \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right) \right) (x + 4)(x-2)$$


问:
How to integrate the function $\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}$
How to integrate the following family of functions? I really do not have too many ideas. The context of this is to find moment generating functions but the context is not too important here.

$$I_n:=\int_{0}^{\infty}n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$

I wanted to try this inductively/iteratively but I got nowhere (by the way I don't have to do this iteratively).  I should also stress that $\lambda>t$ so that I think $I_n$ converges for each $n$. Could someone please given me  a hint or two? I know how to do $n=1$ btw.

Many thanks!

答:
Using hypergeometric functions$$J=\int\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$
$$J=\frac{\lambda  e^{t x} \left(1-e^{-\lambda  x}\right)^n \left(1-e^{\lambda
   x}\right)^{-n} \, _2F_1\left(1-n,\frac{t}{\lambda }-n;-n+\frac{t}{\lambda
   }+1;e^{x \lambda }\right)}{\lambda  n-t}$$ if $\Re(t)<\Re(\lambda )\land \Re(\lambda )>0\land \Re(n)>0$
$$K=\int_0^\infty\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx=\frac{\Gamma (n) \Gamma \left(1-\frac{t}{\lambda }\right)}{\Gamma
   \left(n-\frac{t}{\lambda }+1\right)}$$
$$I_n=n\frac{\Gamma (n) \Gamma \left(1-\frac{t}{\lambda }\right)}{\Gamma
   \left(n-\frac{t}{\lambda }+1\right)}$$


问:
Prove this formula $\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$
I am trying to use prove, by just simple algebraic manipulation, to prove the equality of this formula.
$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$$

I have been given hints and instructions from this thread

Write using Euler’s identity $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Factor the denominator

Find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again).

This is how I have done:

RHS: $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{1-2r(\dfrac{e^{ix}+e^{-ix}}{2})+r^{2}}=\dfrac{1-r(\dfrac{e^{ix}-e^{-ix}}{2})}{1-re^{ix}-re^{-ix}+r^2}$$

Then from this thread, I have learnt how to factorize the denominator (See Jack D'Aurizio answer, first answer of the thread, first line). The trick is to write $1=e^0$.

$$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(re^{ix})(r-e^{-ix})}$$

The third step is to decompose the fraction:

$$\dfrac{A}{r-e^{ix}}+\dfrac{B}{r-e^{-ix}}=\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(r-e^{ix})(r-e^{-ix})}$$

$$A(r-e^{-ix})+B(r-e^{ix})=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I notice that let $x=0$, we will have

$$A(r-1)+B(r-1)=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I don't know to find $A$ and $B$

Also, could you provide in details how to finish the part "expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again)".

My symbolic manipulation skill is not very good, so a detailed answer is great!

Thanks!

Body:
I might derive the series from
$$
\frac{1-r \cos (x)}{r^2-2 r \cos (x)+1}
$$
by setting $\cos x = (e^{ix}+e^{-ix})/2 = (z+z^{-1})/2$ with $z=e^{ix}$.
Then
$$
\eqalign{
  \frac{1-r \cos (x)}{r^2-2 r \cos (x)+1} &=
  \frac{r z^2+r-2 z}{2 (r-z) (r z-1)} =
  \frac{1}{2}\left(1+\frac{r/z}{(1-r/z)}+\frac{1}{(1-r z)}\right) \cr
 &= \frac{1}{2}\left(2+\frac{r/z}{(1-r/z)}+\frac{r/z}{(1-r z)}\right) =
  \frac{1}{2}\left(2+\sum_{n=1}^{\infty}{(r/z)^n}+\sum_{n=1}^{\infty}{(r z)^n}\right) \cr
 &= 1+\sum_{n=1}^{\infty}{z^n+1/z^n\over2} {r^n} =
  1+\sum_{n=1}^\infty r^n \cos nx \ \ \text{or}\ \sum_{n=0}^\infty r^n \cos nx \,. \cr
}
$$


问:
A spinner is made with red and blue sections. The desired probability for RED is 0.75. What angle measure is required for the BLUE section
question: A spinner is made with red and blue sections. The desired probability for RED is 0.75. What angle measure is required for the BLUE section.

Hi, I saw these qs in my practice questions and that's how I answered them.

as we have 100-75=25
25% of the circle would be 90 degrees
The angle measure for blue would be 90.
I'm not sure if that how should do it because my answer looks so basic.
I would appreciate it if you tell me that I have answered it right.

答: Your answer looks correct. The only way that answer would be wrong is if there are, say, 8 total sections of equal size. In such a case we would have 2 blue sections and they would be 45 degrees each. Otherwise, I see no issue with your solution.</p>


问:
Meaning of a formula for the $y$-intercept?
Suppose a line goes through the points $A = (x_1, y_1)$ and $B = (x_2, y_2)$. One can easily check (as I did today while doodling) that
$$
b = - \frac{x_1 y_2 - x_2 y_1}{x_2 - x_1}
$$
where $b$ is the y-coordinate of the y-intercept. This can be written more suggestively as
$$
b = -\frac{\det(A, B)}{\Delta x} \label{b}\tag{1}
$$
The presence of $\det(A,B)$ suggests a geometrical interpretation, but I couldn't think of one.
This reminds me of Cramer's Rule, but I couldn't make that connection explicit either.

Can anyone give an interpretation for equation \ref{b}?

答:
Basically you are using the fact that the area enclosed by three collinear points is zero

$A(x_1, y_1), B(x_2, y_x), C(0,b)$ are the three points, then

$$\begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\0  & b & 1\end{vmatrix} = 0$$

Now expand along the bottom row to get the equation you have


问:
Why does $PV\int_{0}^{\infty} dx \frac{e^{-x^2}}{(x-1)^2}$ have a finite answer?
Mathematica gives the result of the above integral in terms of a Meijer G function, or numerically

$$ PV\int_{0}^{\infty} dx \frac{e^{-x^2}}{(x-1)^2} \approx -0.168 - 2.31 i $$

Ideally I would like to be able to approximate this answer using some numerical integration. But I simply can't understand how there is a finite answer for this. Around the singularity we have

$$ \frac{e^{-x^2}}{(x-1)^2}\Bigg|_{x=1-\epsilon} \approx \frac{1}{e}\Big(\frac{1}{\epsilon^2} - \frac{2}{\epsilon}+1+O(\epsilon) \Big)$$
$$ \frac{e^{-x^2}}{(x-1)^2}\Bigg|_{x=1+\epsilon} \approx \frac{1}{e}\Big(\frac{1}{\epsilon^2} + \frac{2}{\epsilon}+1+O(\epsilon) \Big)$$

I can understand how the first order terms will 'cancel' on the left and right sides of the singularity in a principal value sense, but the leading terms both go to infinity in the same direction.

I would appreciate any guidance with understanding this. Thanks!

Edit: The exact expression from Mathematica is

```<code>-(MeijerG[{{0, 1/2, 1}, {}}, {{1/2, 1}, {}}, -1, 1/2]/\[Pi])
</code>```

which I believe is
$$ -\frac{1}{\pi} G_{3,2}^{2,3}\Big(-1,\frac{1}{2}\Big| \begin{matrix}0,1/2,1\\1/2,1\end{matrix}  \Big) $$

答:
I was doing the same when @Felix Marin's  comment appeared and the result is
$$\frac{\pi \, \text{erfi}(1)+\text{Ei}(1)-e \left(1+\sqrt{\pi }\right)}{e}\sim-0.167837 $$


答:
Re-write the equation as
$b (x_2 - x_1)= (x_2 y_1) – (x_1 y_2)$

$x_2 y_1 = (2)+(3)+(4)+(5)+(6)$

$x_1 y_2 = (1) + (2)+(3)+(4)$

Added: A is a point on the diagonal of the rectangle involving (1), (2), (3), (5), and the part in white. Point A, together with the added lines, will divide the whole rectangle into four parts (namely (1), (2) + (3), (5) and the part in white). A theorem in geometry (its proof is skipped) says (1) is equal to (5) in area.

RHS$ = (5)+(6) – (1)= (6) = b(x_2 - x_1) = $ LHS

Title: Meaning of a formula for the $y$-intercept?

Body:
Suppose a line goes through the points $A = (x_1, y_1)$ and $B = (x_2, y_2)$. One can easily check (as I did today while doodling) that
$$
b = - \frac{x_1 y_2 - x_2 y_1}{x_2 - x_1}
$$
where $b$ is the y-coordinate of the y-intercept. This can be written more suggestively as
$$
b = -\frac{\det(A, B)}{\Delta x} \label{b}\tag{1}
$$
The presence of $\det(A,B)$ suggests a geometrical interpretation, but I couldn't think of one.
This reminds me of Cramer's Rule, but I couldn't make that connection explicit either.

Can anyone give an interpretation for equation \ref{b}?


问:
What would one consider $\lim_{x \to \infty}(\frac{1}{x}+0^{\frac{1}{x}})$ given $0^0=1$? Is it continuous for $x\gt0$?
I had a curious thought cross my mind earlier. Given for the sake of conjecture that $0^0=1$, what would be the value of:

$$\lim_{x \to \infty}{\left(\frac{1}{x}+0^{\frac{1}{x}}\right)}$$

Say this value is equal to $1$. What would be the value of:

$$\lim_{x \to \infty^-}{\left(\frac{1}{x}+0^{\frac{1}{x}}\right)}$$

I don't know if $\lim_{x \to \infty^-}$ is any sort of accepted notation, and if it is used for something else, let me know, but it is the best way I could think of to describe what I'm wondering. Would this limit be equal to $0$ as opposed to the first limit being equal to $1$ (given that it is)?

Another question about this is whether we could consider the equation $f(x)=\frac{1}{x}+0^{\frac{1}{x}}$ to still be continuous for $x\gt0$, or whether this would be considered to have a "discontinuity at infinity".

I just wanted to spark curiosity with what sparked some for me, and also learn a little more about any nuances limits could have like this.

答:
The definition of a limit to infinity: $\lim_{x \rightarrow \infty} f(x) = L$ means that for all $\epsilon > 0$, there exists a $x_0 \in \mathbb{R}$ such that for all $x > x_0$, $|f(x) - L| < \epsilon$.

Now, $\lim_{x \rightarrow \infty } 0^{1/x} = 0$, since for any $\epsilon > 0$, we can choose $x_0$ to be anything, (I'll choose $x_0 = 1$ arbitrarily). Then, for all $x \in \mathbb{R} , x > 1$, we have $0^{1/x} = 0$.

Then apply the same reasoning for your original problem.


问:
Prove/disprove the set of positive reals is a field under a different definition of addition and multiplication.
Define addition and multiplication as

$a \oplus b = ab \text{ and } a \otimes b = a^{\log{b}}.$

Given that the set is commutative since $a^{\log{b}}=b^{\log{a}},$ is it a field?
I have found that the identity element is 10 since $a \otimes 10 = a^{\log{10}} = a,$ but I can't find a solution $x$ in the positive reals for the equation $a \otimes x = 10.$

I'm just beginning Abstract Algebra so my foundation is the subject of rings is still shaky. Sorry if I've made any obvious mistakes.

答: It's isomorphic to $\Bbb R$ with the usual field structure.  The isomorphism is given by $\varphi: \Bbb R^+ \to \Bbb R$ with $\varphi(x)=\log x$.


答:
You know that $abcd = -8$ and $cd = -8.$ Therefore $ab= 1.$

From the coefficient of $x$ in the polynomial, you know that $abc + acd + abd + bcd = -26$ (note: not $+26$).

But since $ab=1$ and $cd=-8$, you can see that
$abc + acd + abd + bcd = c - 8a + d - 8b.$
Therefore

$$ -8a - 8b + c + d = -26. \tag1 $$

But you also know that

$$ a + b + c + d = 1. \tag2 $$

Subtract Equation $(1)$ from Equation $(2)$:

$$ 9a + 9b = 27. $$

That is, $a + b = 3.$  But $b = \frac1a,$ so we have

\begin{align}
a + \frac1a &= 3, \\
a^2 + 1 &= 3a, \\
a^2 - 3a + 1 &= 0. \\
\end{align}

Apply the quadratic formula to solve $y^2 - 3y + 1 = 0.$
Note by symmetry that $a$ and $b$ both are solutions of this equation.
But you are given that $a > b$, so you can see how to match $a$ and $b$ with the two solutions of the quadratic formula.

For $c$ and $d,$ multiply Equation $(2)$ by $8$ and add the result to Equation $(1)$.
You get $9c + 9d = -18.$ But also $d = -\frac8c.$ Again you can get a quadratic equation out of this and solve it, then use the information that $c < d$ to know which root is which.

No guesswork is required, although if you do guess cleverly you can shorten the path a little.

Title: Factorize polynomial of degree 4 given statements

Body:
If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?

Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$
Can I solve $a,b,c,d$ from there? How can I do so?


问:
Regarding $\lim_{n \to \infty} n^{\frac{1}{n}}$
Suppose $\lim_{n \to \infty} n^{\frac{1}{n}} = l \in \mathbb{R}$. The function $f(x) = x^n$ is continuous, then

$$l^n=\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right ) =\lim_{n \to \infty} n = \infty.$$

Then it follows that $l = \infty $. This is false, but I couldn't find which step is wrong. Could you please help me? Thank you for your time.

答: Why would $\lim a_n = L $ imply $\lim a_n^n = \lim L^n$?</p>


问:
A lemma about Radon-Nikodym theorem
We use a lemma to prove Radon-Nikodym theorem.

Lemma Let $\mu$ and $\nu$ be finite positive measures on a measurable space $(X, \mathcal{A}) .$ Either $\mu \perp \nu$ or else there exists $\varepsilon>0$ and $G \in \mathcal{A}$ such that $\mu(G)>0$ and $G$ is a positive set for $\nu-\varepsilon \mu .$

However if $\mu \perp \nu$ , $\mu$ is concentrated on $E$ , $\nu$ is concentrated on $F$ and $E\cap F=\emptyset,$ is F a positive set for $\nu-\varepsilon\mu(\forall \varepsilon)$?

For all $\varepsilon$,take a subset $G$ of $F$ then $\nu(G)-\varepsilon\mu(G)=\nu(G)\geq0.$

If this is right,then in any case,we always find a $G$ with such property.

Am I right?Any help will be thanked.

答: $\mu$ is concentrated in $E$  and $E \cap F$ is empty. So $\mu (G) \leq \mu (F)=0$. So you cannot take $G \subset F$.

Title: A lemma about Radon-Nikodym theorem


问:
If $n^n|m^m$, is it true that $n|m$?
I was solving some number theory problems and couldn't solve this one. I've already proved that if $a^n|b^n$, then $a|b$ (since positive integer power of a rational number is an integer implies that the rational number is an integer). But in this case, the exponents are different so that result is not applicable. Also, since $x^x$ is an increasing function, so $n<m$ in this case, so the exponent of the denominator is lesser (if it would be higher, we could've reduced the exponent of the denominator and then the previous result would be applicable.)

Please help.

答:
Note that $$
4^4 | 10^{10}
$$

but $4 \nmid 10$. The basic reason why this doesn't work is that if we consider a prime $p$ and look at $\nu_p(k)$, the maximum power of $p$ which divides $k$, then $n^n | m^m$ implies that $n\nu_p(n) \leq m \nu_p(m)$ i.e. $\nu_p(n) \leq \frac{m}{n} \nu_p(m)$ for every prime $p$, while $n|m$ implies $\nu_p(n) \leq \nu_p(m)$ for every prime $p$. You can see that going from the weaker to the stronger statement is a jump in an inequality that could be violated for small primes.

In fact, as pointed out in a deleted answer, $n=4$ and $m=2k$ for any odd $k>4$ works, and you can try to verify this yourself. I've picked out the smallest counterexample.

EDIT : Suppose that $n= p^2$ for a prime $p$ then $n^n | m^m$ for $m>p^2$ a multiple of $p$ but not $p^2$. However, $n \nmid m$. I thank Jyrki in the comments for this example. This can be extended to $n$ being a multiple of $p^2$ and $m$ being a multiple of $\frac np$ without being a multiple of $p^2$ for any prime $p$.

Essentially, one creates contradictions in the given statement by realizing that for every prime $p$, $\nu_p(n) \leq \frac{m}{n} \nu_p(m)$ can be ensured by raising $m$ so that $\frac mn$ becomes large enough. This can be done, even if $\nu_p(m)$ is just conditioned to stay non-zero and below $\nu_p(n)$, in which  case $n \nmid m$.

Title: If $n^n|m^m$, is it true that $n|m$?


问:
Why is the direction of angle theta in circular motion towards and inwards?
Why is the direction of angle $\theta$ in circular motion towards and inwards of plane x-y axis?

I am not getting this concept at all.As the angel theta is changing ; the arc length (s in diagram ) is also changing;

We know about vector cross product of two vectors a and b whose direction is perpendicular to the plane of paper( or perpendicular xy axis on my paper i.e xz axis).So my sir tells me that to find direction of angle θ (it is on a 2D plane, X-Y axis), then its direction is on X-Z plane. (You can interpret the direction of vector cross product by right hand thumb rule ).

Also, that z axis is drawn from the centre of circle. Only from there only.

答:
The direction of $\theta$ being positive moving counterclockwise in the $xy$-plane is typically an assumption that we use along with the right-hand rule to determine the direction of the normal vector for the surface enclosed by a closed path, a circle specifically in this question.

Given a circle in the $xy$-plane, with the positive direction of $\theta$ being counterclockwise along that circle, we could theoretically define the direction of the normal vector either in the positive $z$ direction or the negative $z$ direction. Either way, the vector would be normal since both directions are orthogonal to any vector in the $xy$-plane.

By convention, we define the positive $z$-direction using the right-hand rule, and we use the same convention to define the positive direction of the normal vector for our surface. This is also the same convention that gives us $\mathbf{\hat{i}} \times \mathbf{\hat{j}} = \mathbf{\hat{k}}$, where $\mathbf{\hat{i}}$ is the unit vector in the $x$-direction, $\mathbf{\hat{j}}$ is the unit vector in the $y$-direction, and $\mathbf{\hat{k}}$ is the unit vector in the $z$-direction. This is all to ensure that "positive" means the same thing in slightly different contexts (though the underlying mathematics is the same).


问:
Show a series in $\ell^{\infty}$ converges to a point in $c_0$
Let $x =(x_1, x_2, ...) \in \ell^{\infty}$

Suppose $\displaystyle\sum_{n=1}^{\infty}x_n e_n$ converges to $x$ with respect to $\| \cdot\|_{\infty}$.
Show that $\displaystyle\lim_{n \rightarrow \infty}x_n=0$.

Things I have tried:

Since $\sum_{n=1}^{\infty}x_n e_n$ converges to $x$, there exists some $N$ such that for all $k>N$,

$$\bigg\|x- \sum_{n=1}^k x_n e_n \bigg\|_{\infty} = \bigg\| \sum_{n=k+1}^{\infty} x_n e_n \bigg\|_\infty < \infty.$$

But now I'm confused about the notation. Not sure how to finish this.

Thanks in advance!

答:
Choose $k_0$ such that $\|x- \sum_{n=1}^k x_n e_n \|_{\infty}<\epsilon$ for all $k \geq k_0$. Then $ \| \sum_{n=k_0+1}^{\infty} x_n e_n \| =\sup \{|x_n|: n >k_0\}$ by definition of the norm in $\ell^{\infty}$. Hence, $|x_n| <\epsilon $ for all $n >k_0$.

[The norm in $\ell^{\infty}$: $\|\sum\limits_{k=1}^{\infty}c_ke_k\|=\|(c_1,c_2,\cdots)\|=\sup \{|c_k|: k=1,2,\cdots\}$].


问:
Proving Least Common Multiple
So I completed the first part of the proof where I prove $\frac{xy}{gcd(x,y)}$ is a multiple of $x$ and $y$. Now I need to prove for the second part that for any integer $m$ such that $x | m$ and $y | m$, $xy | (m*gcd(x,y))$ and then use these parts to prove the final part that $lcm(x,y) = \frac{xy}{gcd(x,y)}$. I'm confused on how to approach the second part and I also don't see how to connect the two parts when I do. Any help is appreciated

答:
Let $k=\gcd(x,y)$ and write $x=ak, y=bk$. You need to show $abk|m$ so that $abk^2|mk$.

To do this you write $m=pak=qbk$ and simplify to $pa=qb$. Then use $\gcd(a,b)=1$ to infer $b|p$. Therefore $p=rb$ and $m=rabk$.

Title: Proving Least Common Multiple


问:
This is regarding Chi square test
A chi square test is conducted to check whether a person's ability in Mathematics has an impact on his/her interest in Statistic. The test statistic is 13.277 under the tested null hypothesis.
write a recommended null hypothesis and an alternative hypothesis. Briefly describe your conclusion on this test at the 0.01 significance level.</p>

答:
Alternative hypothesis (what the study is trying to find out): interest in statistics is not the same on average for people of all mathematical abilities
Null hypothesis (negation of the alternative hypothesis): interest in statistics is the same on average for people of all mathematical abilities

If there are at least four degrees of freedom, then the test is significant at the 0.001 level. The p-value can be found in R using the command pchisq(13.277,k) where k is the number of degrees of freedom. Alternatively, the p-value is given by

$1-\frac{1}{\Gamma(\frac{k}{2})}\gamma(\frac{k}{2},\frac{x}{2})$

where $\gamma(s,x)=\int_0^xt^{s-1}e^{-t}dt$

Title: This is regarding Chi square test


答:
Note that $a \oplus b = \exp(\log(a)+\log(b))$  and  $a \otimes b = \exp(\log(a)\log{(b)})$. So the structure of $\Bbb R$ as a field is ''transportet'' by $\exp$ to $(\Bbb R^+,\oplus,\otimes). $ In other words: Let $f:=\log$, Then
$a\oplus b=f^{-1}(f(a)+f(b))$ and $a\otimes b=f^{-1}(f(a)\cdot f(b))$.

Title: Prove/disprove the set of positive reals is a field under a different definition of addition and multiplication.

Body:
Define addition and multiplication as

$a \oplus b = ab \text{ and } a \otimes b = a^{\log{b}}.$

Given that the set is commutative since $a^{\log{b}}=b^{\log{a}},$ is it a field?
I have found that the identity element is 10 since $a \otimes 10 = a^{\log{10}} = a,$ but I can't find a solution $x$ in the positive reals for the equation $a \otimes x = 10.$

I'm just beginning Abstract Algebra so my foundation is the subject of rings is still shaky. Sorry if I've made any obvious mistakes.


问:
Why does the principle of superpositions not seem to work in this case? Please help me understand this better without a proof.
I just learned about the principle of superpositions and I tried it with a simple example from physics which led to questionable results...

$$y''(t) = g = 9.8\text{ m/s}^2\iff y'(t) = gt + c_1\iff y(t) = \frac12 gt^2 + c_1 t + c_2$$

If I take two specific solutions; say for $y_1$, $c_1 = 2$, $c_2 = 0$ and for $y_2$, $c1 = 3, c2 = 4$; when I add $Ay_1 + By_2$ (where A, B are constants), I end up with a term $(A + B)(\frac12 gt^2)$. This would imply that one solution to $y''(t) = g$ involves a term where the coefficient to $t^2$ is not $1/2$! This seems like a contradiction because I know that the constant in front of $t^2$ for the position with constant acceleration function is always $\frac12 g$. Where did I go wrong? Am I misunderstanding the principle of superposition?

I may just have tunnel vision and/or have made a mistake, but I can't seem to find it. Sorry in advance!

答:
To restate the principle as you learned it:

If $y_1(x)$ and $y_2(x)$ are both solutions of the linear homogeneous equation
$$ P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=0, \tag1$$
and $c_1,c_2$ are any constants, then the function $y(x)=c_1y_1(x)+c_2y_2(x)$ is also a solution.

So let's now consider the differential equation
$$ y''(t) = g. \tag2$$
Since $y''(t)$ is just another notation for $\frac{d^2y}{dx^2},$
or in the format of Equation $(1),$
another way to write Equation $(2)$ is

$$ \frac{d^2y}{dx^2} = g. \tag3$$

So let's compare this to Equation $(1)$. We see the coefficient of $\frac{d^2y}{dx^2}$ is just $1,$ so we must have $P(x)=1.$ The $\frac{dy}{dx}$ and $y$ terms are missing, so we must have $Q(x)=R(x) = 0.$

Now the only remaining terms to match up are the $0$ on the right-hand side of $(1)$ and the $g$ on the right-hand side of $(3)$. If $g = 0$ then $(3)$ is an equation in the same form as $(1)$, the Principle of Superposition applies, and any linear combination of solutions of $(3)$ is a solution of $(3).$

If $g \neq 0$ then you don't have an equation in the form of Equation $(1)$ and you cannot use the principle.

A familiar example from physics that does work is a mass suspended from a perfect Hooke's Law spring with constant $k$:

$$  y''(t) + k y(t) = 0. $$

Try the Principle of Superposition on a couple of solutions of that equation.


问:
Sorting out prime and composite numbers

this is a question I was asked to solve:

The mathematician wrote a three-digit whole number on the whiteboard and asked the students to decide whether that is a prime number or not. Big-brain Billy divided it from 2 to 31 but didn’t find any factors. There he stopped and stated that the given number is a prime.

a) Was he right? Why?

b) Is 2021 a prime number?

How will I even begin with this question!

答:
If a positive integer $n$ is composite then $n=ab$ where $a$ and $b$ are positive integers with $a>1$ and $b>1.$ It cannot be true that $n=ab$ with $a>\sqrt n$ and $b>\sqrt n,$ otherwise we would have $n=ab>a\sqrt n>(\sqrt n)(\sqrt n)=n$. So if $n$ is composite then $n$ has a factor ($a$ or $b$) that is greater than $1$ but less than or equal to $\sqrt n. $

We have $32^2=2^{10}=1024>999.$ So if $100\le n\le 999$ then $\sqrt n <32.$

Big-brain Billy could do better. If $a$ is a factor of $n$ with $1<a\le \sqrt n$ then $a$ has a prime factor $p$ with $p\le a\le \sqrt n,$ so $p$ is  a $prime$ factor of $n$ with $p\le \sqrt n.$ So if $100\le n\le 999$ and $n$ is not divisible by any of $2,3,5,7,11,13,17,19,23,29,31$ then $n$ is prime.

There are some short-cuts for finding whether $n$ is divisible by $a.$ The notation $a|n$ means $a$ divides $n.$ In base ten:

$3|n$ iff $3$ divides the sum of the digits of $n$ (...iff $3$ divides the sum of the digits of the sum of the digits of $n,$ etc.).$

$5|n$ iff the last digit of $n$ is $0$ or $5.$

$11|n$ iff $11$ divides the alternating $\pm$-digit  sum of $n.$ E.g. $11$ does not divide  $+2-0+2-1=3$ so $11$ does not divide $2021$.

If $k$ is an  integer then $a|n$ iff $a|(n- ka).$ Choose $k$ so that the last digit of $ka$ is the last digit of $n$.

E.g. $29|2021$ iff $29|(2021+29)=2050$ iff $29|205$. Also $29|205$ iff $29|(2)(205)=410$ iff $29|41.$ Therefore $29$ does not divide $2021$ .

Similarly $31|2021$ iff $31|(2021-31)=1990$ iff $31|199$ iff $31|(199+31)=230$ iff $31|23$. Therefore $31$ does not divide $2021.$

Title: Sorting out prime and composite numbers


问:
Solve using the Principle of Inclusion and Exclusion:How many ways can 5 boys and 4 girls be arranged such that no two girls stand next to each other?
I've been trying to solve this problem with the principle of inclusion and exclusion (it's easy using the standard technique of seating the boys first and then the girls afterwards but I'm stumped using this particular technique (PIE)). Could someone please help me out?

This is what I've tried: 9! - 4C2×8!×2! + 4C3×7!×3! - 4C4×6!×4! = -17280
This gives a negative result which is obviously impossible.
Why doesn't this work? Surprisingly, if there are only 3 girls, the method above works like a charm. It breaks down though with 4.

Edit: By arrangements I mean linear arrangements. The 9 people are to stand in a straight line and then impose this condition.

答:
PIE is trickier here, because the "bad" situations are indexed by pairs of girls. I agree with initial two terms, $9!-\binom{4}2 8!\cdot 2!$. However, when adding back in the doubly subtracted terms, you would need to add back in all arrangements where each pair of pair of girls are both together. These pairs of pairs come in two types, either disjoint pairs or overlapping pairs, and would need to be counted separately. Things only get worse when you look at triple intersection, then all the way up to six-way intersections (EDIT: Actually, it's not that bad at all, see the other wonderful answer!).

Here is an easier way, still with inclusion-exclusion. First, count the arrangements where all boys are identical, as well as all girls. Then, multiply by $5!\cdot 4!$ at the end.

We are arranging $5$ B's and $4$ G's in a line. Let $E_1$ be the set of arrangements where the leftmost G has another G to its right, let $E_2$ be the arrangements where the second G from the left has a G to its right, and $E_3$ where the third G from the left has a G to its right. Using PIE, the number of valid arrangements of identical symbols is (where $AB$ is shorthand for $A\cap B$)
$$
\binom{9}4-|E_1|-|E_2|-|E_3|+|E_1 E_2|+|E_1 E_3|+|E_2 E_3|-|E_1 E_2 E_3|
$$
You can then show (how?) this is
$$
\binom94-\binom31\binom{8}{3}+\binom{3}2\binom{7}2-\binom{3}{3}\binom{6}1=15
$$
so the final answer is $5!\cdot 4!\cdot 15$.


问:
Can I use a z test to compare two samples if there is a significant difference in the size?
I have to compare two unequal size samples from a population of 53k. One has a proportion of 56% and size of 50k and the other has a proportion of 50% and size of 3k. Can I do a two proportion z test to compare to make a case that the proportion of sample 1 is greater than sample 2?
Does the difference in the sample size pose an issue?

Some more info:
Here is the situation: Lets say I had 53k people and 50k received treatment A and 3k received treatment B. From the 3k, 50% showed positive results and from the 50k, 56% showed positive effect. How can I conclude that my treatment A was more effective than treatment B? I am looking for a way to make that case.

答:
The R method provided above is correct for this specific case.

The statistics is a Z statistics, but you have to be careful on how you calculate.

You can find a formula in https://en.wikipedia.org/wiki/Test_statistic, and searching for 'Two-proportion z-test'. The page also provides when you can use Z statistics.

The above web page provides formulas for many statistical tests: so you can use for other analyses.


问:
Show $\int_{-1}^{1} T_{n}(x)\,dx = \frac{(-1)^{n}+1}{1-n^{2}}$
$T_{n}$ being

$$T_{n}(\cos (\theta)) = \cos(n\theta)$$

Could anyone explain me why

$$\int_{-1}^{1} T_{n}(x)\,dx = \frac{(-1)^{n}+1}{1-n^{2}}$$

if $n \neq 1$ and $0$ if $n=1$?

With the integral I came as far as the integral evaluates to

$$ \frac{\cos(n+1)}{n+1}-\frac{\cos(n-1)}{n-1}$$

but from here I am not sure how to get the results with given boundaries.

Body:
Pretty straightforward proof.
$$\int_{-1}^1 T_n(x)\mathrm{d}x=\int_0^\pi \cos(n\theta)\sin(\theta)\mathrm{d}\theta$$
Now we use the identity
$$\cos(a)\sin(b)=\frac{1}{2}(\sin(a+b)-\sin(a-b))$$
So
$$\int_0^\pi \cos(n\theta)\sin(\theta)\mathrm{d}\theta=\frac{1}{2}\int_{0}^{\pi} \sin\big((n+1)\theta\big)-\sin\big((n-1)\theta\big)~\mathrm{d}\theta$$
$$=\frac{1}{2}\bigg[\frac{-1}{n+1}\cos\big((n+1)\theta\big)\bigg]^\pi_0-\frac{1}{2}\bigg[\frac{-1}{n-1}\cos\big((n-1)\theta\big)\bigg]^\pi_0$$
$$\frac{-1}{2(n+1)}\big((-1)^{n+1}-1\big)+\frac{1}{2(n-1)}\big((-1)^{n-1}-1\big)$$
$$=\frac{\big((-1)^{n-1}-1\big)(n+1)-\big((-1)^{n+1}-1\big)(n-1)}{2(n^2-1)}$$
$$=\frac{(-1)^{n-1}n+(-1)^{n-1}-n-1-(-1)^{n+1}n+(-1)^{n+1}+n-1}{2(n^2-1)}$$
$$=\frac{(-1)^{n-1}n-(-1)^2(-1)^{n-1}n+(-1)^{n-1}+(-1)^2(-1)^{n-1}-2}{2(n^2-1)}$$
$$=\frac{2(-1)^{n-1}-2}{2(n^2-1)}=\frac{(-1)^n+1}{1-n^2}.$$
Done.


问:
Proving the existence/non-existence of prime elements
No answers, please. Hints only.

Given that $n$ is a positive integer and $x=(n+1)!+2$, I want to prove whether or not the sequence $x, x+1, \cdots, x+(n-1)$ contains no primes.

I did a lot of algebra to find that $x+(n-1)$ is no more than just $(n+1)[n!+1]$. But I'm not sure what to do with this, I'm still unsure what this says about the conclusion.

答: Note that $(n+1)!$ is the product of the first $n+1$ positive integers. Can you show that $2$ divides $(n+1)!+2$? Can you generalize from here?


答:
You cannot swap the $n$ around, that is, inside and outside the limit. The step $$\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right )$$ is wrong.

The $n$ outside the LHS is fixed, while the $n$ inside, in the RHS, varies like the other $n$. The result would be correct, if you said, say,

$$\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^3=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^3 \right )$$

As Thomas pointed out, $n$ is a dummy variable, that is $$\lim_{n\to\infty}n^{1/n}=\lim_{k\to\infty}k^{1/k}=\lim_{\sigma \to\infty}\sigma^{1/\sigma}$$

so you cannot really force $n$ to "match up" in the limit.

Title: Regarding $\lim_{n \to \infty} n^{\frac{1}{n}}$

Body:
Suppose $\lim_{n \to \infty} n^{\frac{1}{n}} = l \in \mathbb{R}$. The function $f(x) = x^n$ is continuous, then

$$l^n=\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right ) =\lim_{n \to \infty} n = \infty.$$

Then it follows that $l = \infty $. This is false, but I couldn't find which step is wrong. Could you please help me? Thank you for your time.


问:
Trying to prove infinite primes by infinite coprime sequences.
Rules

Infinite primes do not exist.

Conjecture

There exists infinite coprime sequences of the form

$a_n = \{(k, n-k) \,\  |\  \, 1 \le k < n,\  \gcd(k, n-k) = 1\}$

proving infinite primes exist.

Examples

$a_2\ =\{(1,1)\}$

$a_3\ =\{(1, 2), (2, 1)\}$

$a_5\ =\{(1, 4), (2, 3), (3, 2), (4, 1)\}$

$a_7\ =\{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$

$a_{11}=\{(1,10), (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2), (10,1)\}$

...

$a_n\  =\small \{(1,n-1),...,(\left \lfloor{\frac{n}2-2}\right \rfloor ,\left \lceil{\frac{n}2+2}\right \rceil),,(\left \lfloor{\frac{n}2}\right \rfloor ,\left \lceil{\frac{n}2}\right \rceil),(\left \lceil{\frac{n}2}\right \rceil ,\left \lfloor{\frac{n}2}\right \rfloor),,(\left \lceil{\frac{n}2+2}\right \rceil ,\left \lfloor{\frac{n}2-2}\right \rfloor),...,(n-1,1)\}$

Notes

The $n^{th}$ sequence always has several known coprimes as shown.

答:
If a sequence $a_n$ satisfies your requirements, then we have:
$$\gcd(k,n-k)=1$$
for all $1 \leqslant k <n$. By the Euclidean Algorithm, this means $\gcd(k,n)=1$ for all $1 \leqslant k < n$. Since $n>1$, and since $\gcd(k,n)=1$ for all $k<n$, this is equivalent to saying that $n$ is prime. Clearly, our requirement holds when $n$ is prime. Moreover, $n$ cannot be composite, since if $k$ is a divisor of $n$ that is neither $1$ or $n$, then $\gcd(k,n)=k \neq 1$. Hence, $a_n$ satisfies your requirements iff $n$ is prime. Hence:
$$\text{infinite $a_n$ $\iff$ infinite primes }$$


答:
Here is another solution using the in-and-out formula:

$$9!-4\cdot3\cdot8!+\left(4\cdot3\cdot2+\frac{4\cdot3\cdot2\cdot1}2\right)7!-4\cdot3\cdot2\cdot1\cdot6!$$
$$=362880-483840+181440-17280=\boxed{43200}$$
Explanation. $9!$ is the total number of arrangements, from which I subtract the number of bad arrangements, which are classified into $12$ overlapping types.

$N(AB)=8!$ is the number of arrangements in which Ann and Betty sit together in that order, with Ann on the left; so the second term is $-4\cdot3\cdot N(AB)=-4\cdot3\cdot8!$.

$N(ABC)=N(AB\cap BC)=7!$ is the number of arrangements in which Ann, Betty, and Carol sit together in that order from left to right; and $N(AB\cap CD)=7!$ is the number of arrangements where Ann and Betty sit together, and Carol and Dorothy sit together, with Ann on the left of Betty, and Carol on the left of Dorothy; so the third term is $4\cdot3\cdot2\cdot N(ABC)+\frac{4\cdot3\cdot2\cdot1}2\cdot N(AB\cap CD)$.

The fourth term is $-4\cdot3\cdot2\cdot1\cdot N(ABCD)$ where $N(ABCD)=6!$ is the number of arrangements where all four girls sit together in alphabetical order from left to right.

All other terms of the in-and-out formula are zero.

Title: Solve using the Principle of Inclusion and Exclusion:How many ways can 5 boys and 4 girls be arranged such that no two girls stand next to each other?

Body:
I've been trying to solve this problem with the principle of inclusion and exclusion (it's easy using the standard technique of seating the boys first and then the girls afterwards but I'm stumped using this particular technique (PIE)). Could someone please help me out?

This is what I've tried: 9! - 4C2×8!×2! + 4C3×7!×3! - 4C4×6!×4! = -17280
This gives a negative result which is obviously impossible.
Why doesn't this work? Surprisingly, if there are only 3 girls, the method above works like a charm. It breaks down though with 4.

Edit: By arrangements I mean linear arrangements. The 9 people are to stand in a straight line and then impose this condition.