问:
Finding the maximum using inequalities
Given that $(a+1)(b+1)(c+1)(d+1)=81$ and $a,b,c,d>0$, find the maximum of $abcd$.

I know the answer should occur when $a=b=c=d$ but how do I get there using inequalities? This feels like AM-GM.

答:
Yeah your intuition is correct about AM-GM. But one can err badly if one naively just applies it on each factor:
$$\frac{a+1}{2}\geq \sqrt{a\cdot1}=\sqrt{a}\implies (a+1)\geq 2\sqrt{a}$$
Similarly for the other three factors and we get:
$$81 = (a+1)(b+1)(c+1)(d+1) \geq 16\sqrt{abcd}$$
$$\implies abcd \leq \bigg(\frac{81}{16}\bigg)^2\approx25.63$$
The issue is that while this bound is correct, it will never be achieved because all these equalities will never be simultaneously satisfied under the problem constraint. For each equality to be achieved we would need $a=b=c=d=1$ which violates the precondition.

But this motivates the cleverness in application of the same technique differently:
$$\frac{\frac{a}{2}+\frac{a}{2}+1}{3}\geq\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a}{2}\cdot\frac{a}{2}\cdot 1}}=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a^2}{4}}}\implies (a+1)\geq 3\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a^2}{4}}}$$
Similarly for the other three factors and we get:
$$81=(a+1)(b+1)(c+1)(d+1) \geq 81\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a^2b^2c^2d^2}{4^4}}}$$
$$a^2b^2c^2d^2 \leq 4^4 \implies abcd\leq16$$
All the equalities of AM-GM are achieved simultaneously iff $\frac{a}{2}=\frac{b}{2}=\frac{c}{2}=\frac{d}{2}=1$ equivalently iff $a=b=c=d=2$, which this time is consistent with the given constraint and thus $16$ is a tight limit, i.e. it is achieved and is the (optimal) maximum value that $abcd$ takes.

Title: Finding the maximum using inequalities


问:
Solving recurrence using EGF
How do I solve the recurrence relation $a_n = na_{n - 1}+ (n - 1)!$ using exponential generating function?

Any help would be highly appreciated. Thanks.

答:
Welcome to MSE!

Let $A = \sum \frac{a_n}{n!} x^n$ be the exponential generating function of $a_n$.

Dividing both sides of the recurrence by $n!$ gives

$$
\frac{a_n}{n!}
= \frac{na_{n-1}}{n!} + \frac{(n-1)!}{n!}
= \frac{a_{n-1}}{(n-1)!} + \frac{1}{n}
$$

Then if we multiply both sides by $x^n$ we get

$$
\frac{a_n}{n!}x^n = \frac{a_{n-1}}{(n-1)!}x^n + \frac{1}{n}x^n
$$

and summing over all $n \geq 1$ (being careful of the $a_0$ boundary term) gives

$$
A - a_0 = xA + \sum_{n \geq 1} \frac{1}{n} x^n
$$

We can solve this for $A$:

$$
A = \frac{1}{1-x} \left ( a_0 + \sum \frac{1}{n} x^n \right )
$$

But we know the $x^n$ coefficient of $\frac{a_0}{1-x}$: it's exactly $a_0$.

We also know the $x^n$ coefficeint of $\frac{1}{1-x} \sum \frac{1}{n} x^n$, at least in theory. It turns out that

$$\frac{1}{1-x} \sum_{n \geq 0} c_n x^n = \sum_{n \geq 0} \left ( \sum_{0 \leq k \leq n} c_k \right ) x^n$$

that is, the $x^n$ coefficient is exactly the $n$th partial sum of the original series. You can see this mentioned here or in the excellent book generatingfunctionology (rule 5 on page $37$).

So then by comparing coefficients on both sides, we find

$$\frac{a_n}{n!} = a_0 + \sum_{k \leq n} \frac{1}{n}$$

In other words,

$$a_n = n! (a_0 + H_n)$$

where $H_n$ is the $n$th Harmonic Number.

In case you're feeling doubtful, it's easy to ask
sage to double check that this is true.
Here I've set $a_0 = 1$:

I hope this helps ^_^

ContentLicense: CC BY-SA 4.0

Title: Solving recurrence using EGF


问:
To prove a sufficient and necessary condition of a quadratic form to equal to zero
The question is the following:

Let X a vector space and $B: V \times V \rightarrow K$ a positive semi-definite form. Show that $q_B(y)=0 \Longleftrightarrow B(x,y)=0$, $\forall$ $x$ $\in$ $V$, where $\ q_B(y)\ $ is the quadratic form $\ B(y,y)\ $.

Is it because $q_B = \sum_{i,j=1}^n a_{ij}x_i \bar{x_j}$? Similar type proof as to proving linear independence?

答:
$\ B(x,y)=0\ \forall x\in V\implies q_B(y)=0\ $ is trivial (put $\ x=y\ $ in the antecedent of the implication)

$\  q_B(y)=0\implies B(x,y)=0\ \forall x\in V\ $ follows from the Cauchy-Schwarz inequality: $$|B(x,y)|\le\sqrt{q_B(x)q_B(y)}\ ,$$ which holds for any positive
semi-definite bilinear form. The proof is essentially the same as for
any inner product.


问:
If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$

Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.

This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at this Math.SE problem, and if this could be proven, then the solution would follow immediately.

答: If $\gcd(a,b)=1$ then there exist integers $s$ and $t$ such that $a s +b t =1$. Hence $c a s +c b  t =c$. Now since $a \vert c$ and $b \vert c$ there exist integers $m$ and $n$ such that $c = m a$ and $c= n b$. Hence $n b a s  + m a b  t =c$. Since $a b $ divides the entire left hand side, it must also divide the right hand side.</p>

ContentLicense: CC BY-SA 3.0


问:
How do you find $B^c$ when $B$ is the solution set of $\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$?
The problem is as follows:

Assuming $B$ is the solution set of the equation from below:

$$\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$$

Find $B^c$.

The given choices in my book are as follows:

$\begin{array}{ll}
1.&\varnothing\\
2.&\mathbb{R}\\
3.&[2,+\infty\rangle\\
4.&[2,3]\\
\end{array}$

I've found this problem in my precalculus workbook but I have no idea how to solve it.

What I'm about to post below is the official solution from my book but it is some unclear to me. The reason for this is that it does not give any further details on the explanation how the vertical pipes are eliminated. Thus I really appreciate someone could explain to me the rationale behind this problem, and more importantly what is the meaning of that $C$ on $B^c$

$\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$

Notice:

$|-x+x^2+3|$ the  $\Delta<0$

Therefore:

$\left|x^2-3x+\sqrt{x^2+2x-3}+3-x^2+x-3\right|+3=-x$

$\left|\sqrt{x^2+2x-3}-2x\right|=-x-3$

and

$-x-3\geq 0$

$\left|\sqrt{x^2+2x-3}-2x\right|=-x-3$

$\sqrt{x^2+2x-3}-2x=-x-3$ and $x\leq -3$

$\sqrt{x^2+2x-3}=x-3$ and $x\leq -3$

Therefore:

The solution set is $\varnothing$

hence the $B=\varnothing$ and $B^c=\mathbb{R}$

Therefore the answer is choice 2.

That's it where it ends the official solution. But as I'm presenting it here I'm just clueless on how did it arrived to that conclusion. There might be some typo in the question regarding who's $C$ but I wonder who could it be so that the answer is all real.

Can someone help me with this?. It would really help me a lot if the answer would include some foreword or an explanation on when the vertical pipes can be erased or just skipped from the analysis I think the latter part is more important to my understanding because this is what I'm lacking the most.

Thus could someone attend these doubts which I have?. Please try not to skip any step and if you could explain each line because I'm lost here.

Regarding the notations which I have been asked before. The closed brackets $[$ means that the left boundary on the interval is closed, conversely $]$ closed in the right boundary and $\rangle$ means that the right boundary on the interval is open.

答:
Choice $1$ implies that the solution is $\mathbb{R}$, which is not true.

Choices $3,4$ imply that, for example, $0$ is a solution, which is not true.

That leaves choice $2$.


问:
If the series $a_n$ is convergent, is the series $(-1)^n \sqrt{a_n}$ convergent?
Given that series $a_n$ converge (which is positive), I have to figure out if the series $(-1)^n \sqrt{a_n}$ is convergent. I would like to know if my proof is correct:

First, I am looking at the series $\sqrt{a_n}$. Since it is always true that $\sqrt{a_n} \leq a_n$, by Basic Comparison Test the series $\sqrt{a_n}$ is convergent, and this implies that the series $(-1)^n \sqrt{a_n}$ is also convergent by absolute convergence test. Does this make sense?

Body: Consider $$a_n=\begin{cases}\frac1{k^2}&n=2k\\\frac1{4k^2}&n=2k-1\end{cases}$$


问:
Fundamental group of the $\mathbb{R}^3$ minus the trefoil knot
I just read a proof stating that $\pi_1(\mathbb{R}^3-K)\cong \langle x,y: x^3=y^2\rangle$ but I am having trouble with some statements.

The proof goes as follows : By using van-Kampen we have that $\pi(S^3-K)\cong \pi_1(\mathbb{R}^3-K)$. Now view $S^3=\partial(D^2\times D^2)=S^1\times D^2\cup D^2\times S^1$ and denote $A:=S^1\times D^2$ and $B:=D^2\times S^1$. We have that $K\subset A\cap B=S^1\times S^1$ and so $A-K$ and $B-K$ are two closed subsets that cover $S^3-K$.  They intersect  at $A\cap B-K$ which is an annulus . (Why is this true ? I can't seem to visualize it , I would guess the idea would be to flatten what is left of the torus but I am not sure this works).The inclusions $A\cap B-K\rightarrow A-K,B-K$ are cofibrations and so we can apply van kampen's theorem and we get that $\pi_1(S^3-K)$ is isomorphic to the pushout of $\mathbb{Z}\leftarrow^{\times 2} \mathbb{Z}\rightarrow ^{\times 3} \mathbb{Z} $.

Now I don't really understand why we have that $\pi_1(A-K)\cong \pi_1(B-K)\cong \mathbb{Z}$, I guess it would follow that we would make a deformation retract into an annulus .
Also I am not sure why we get those induced maps in the fundamental groups. I get that the map that gives the knows is $f(z)=(z^2,z^3)$ but since we are removing I don't see how we can come in with it's influence.

Any help with this is appreciated. Thanks in advance.

答:
To see why the complement of $K$ in $S^1\times S^1$ is a copy of $S^1\times I$, think about what $K$ looks like in the description of the torus as a square with its edges identified - the complement will be easier to visualize there (remember you need to identify edges again to see it properly), and it should be clear that it's an annulus.

As for why $\pi_1(A-K)\cong\pi_1(B-K)\cong\mathbb{Z}$, note that we are now considering the complement of a trefoil knot in a solid torus: the above shows that $(S^1\times S^1)-K$ retracts to an annulus, so have a think about what $(S^1\times D^2)-K$ will retract to.

Finally, for your question regarding the effect on the fundamental groups, it may help to think of it this way: we can think of $S^3$ as the union of two solid tori, as you described in your question. Your curve $K$ is homotopically non-trivial on $S^1\times S^1$, but when we include it into $S^1\times D^2$ (where we think of $K$ as lying first on a hollow donut, then on a solid donut), all the meridional components become nullhomotopic, so the effect on the fundamental group is determined by the longitudinal components. The effect on the "outer" solid torus is the opposite.


问:
How do I model the Expected Value in a trial and error until find some value?
The problem I was thinking is: Supose you have m pieces of cake and  n of them are spoiled, I start picking one piece, if it's normal I take it out and continue doing this until I find one spoiled.

What is the expected value of cakes I catch before the spoiled? I thought it could be something geometric but the probabilities change. How can I attack this problem?

答:
Let $\{G_1,\ldots,G_{m-n}\}$ denote the $m-n$ "good" pieces of cake and $\{B_1,\ldots,B_n\}$ denote the $n$ "bad" pieces of cake. Define $X$ the be the random variable which counts the number of pieces of cake you sample until you obtain your very first "bad" piece of cake. Note that $X$ is supported on the set $\{1,\ldots,m-n+1\}$.

Regarding $\{G_1G_3B_4\}$ and $\{G_3G_1B_1\}$ as two distinct outcomes corresponding to the event that $X=3$ we see that $$\mathbb{P}(X=x)=\frac{^{m-n}P_{x-1}\times n}{^m P _x}$$Therefore $$\mathbb{E}(X)=\sum_{x=1}^{m-n+1}x\mathbb{P}(X=x)=\frac{m+1}{n+1}$$


问:
If $a,b,c>0 $ then $\frac{9}{a+b+c} \leq 2\left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)$
Let $a,b,c>0$ then show taht $\frac{9}{a+b+c} \leq 2\left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\ \right)\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

For show the RHS I do the following.
\begin{equation}
\frac{\frac{1}{a}+\frac{1}{b}}{2} \geq \frac{1}{\sqrt{ab}}\geq \frac{1}{\frac{a+b}{2}}=\frac{2}{a+b}
\end{equation}
Now notice that we can do the same for the other terms, now we have the following.
\begin{equation}
\frac{\frac{1}{b}+\frac{1}{c}}{2} \geq \frac{1}{\sqrt{bc}}\geq \frac{1}{\frac{b+c}{2}}=\frac{2}{b+c}
\end{equation}
And

\begin{equation}
\frac{\frac{1}{a}+\frac{1}{c}}{2} \geq \frac{1}{\sqrt{ac}} \geq \frac{1}{\frac{a+c}{2}}=\frac{2}{a+c}
\end{equation}
and sume the LHS and RHS of those inequalitys we get

\begin{equation}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)
\end{equation}
Therefore we have the RHS of the inequality Holds.
Now for the LHS I don´t know how I should apply the inequialitys, I tryed a lot of things for get a lower bound of $a+b+c$ and get a upper  bound of $\frac{1}{a+b+c}$ but I don´t get it.

I try look the expression like $$\frac{3}{\frac{a+b+c}{3}}$$ and apply the GM-AM inequality but it doesn´t work.Still I try use the fact that
$$a+b+c\geq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}} \geq \frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ac}{a+c} $$
And take inverses, but I don´t get the LHS.

Any advice or hint of where I should apply the GM-AM inequality or how Prove the LHS was very useful.

答:
To prove $\frac{9}{a+b+c} \leq 2\left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)$, we can observe that
$$a+b+c = \frac 12((a+b)+(a+c)+(b+c))$$
denote $x = (a+b),y = (b+c),z=(a+c)$ and we get what we want to prove after simplify
$$\frac {9}{\frac 12 (x+y+z)} \le 2(\frac 1x + \frac 1y +\frac 1z)$$
Use Cauchy Schwarz Inequality directly as follows
$$
(\frac 1x + \frac 1y +\frac 1z)(x+y+z) \ge (1+1+1)^2 =9
$$

Or

we can use the GM-AM inequality only$ \frac {\frac 1x + \frac 1y +\frac 1z}{3} \ge (\frac {1}{xyz})^{\frac 13}$ and $\frac {x+y+z}{3} \ge (xyz)^{\frac 13}$
,we can again get the above result.

Or

we can use the GM-AM inequality $\frac {\frac 1x + \frac 1y +\frac 1z}{3} \ge (\frac {1}{xyz})^{\frac 13}$, then ,
just to prove
$$
\frac {3}{x+y+z} \le (\frac {1}{xyz})^{\frac 13}
$$
which can be simplified as follow
$$
(x+y+z)^3 \ge 27xyz
$$
This has been prooved in this link

I am very grateful to River Li. He helped me a lot here


问:
Given $\triangle ABC$ with $AB=BC$ and $\angle ABC=80^\circ$, and $O$ such that $\angle OAC=30^\circ$ and $\angle OCA=10^\circ$, find $\angle BOC$.
I encountered the following problem in a math problem book, in a section, entitled Geometry.

In an $\Delta ABC$ with $AB=BC$ and $\angle ABC = 80^\circ$, the point $O$ is chosen so that $\angle OAC = 30^\circ$ and $\angle OCA = 10^\circ$. Find $\angle BOC$.

I struggled repeatedly trying to solve this using some reasonable selection of methods, and failed utterly. I can construct a solution using analytic geometry, but that is not a fair choice of methods for this section.

From simple geometry, I keep writing down systems of equations using various triangles in the picture, and always keep missing one constraint, as it seems there are infinite amount of solutions to the system, which I know is wrong since the analytic geometry approach yields a unique solution.

I am happy to add my infinite solution approach using the angle constraints, if you find it helpful. Please let me know if that's the case.

I would like to solve this using methods from geometry, without using more advanced techniques. Thank you very much, I really appreciate your help.

答:
Extend $AO$ to meet $BC$ at $M$ . Note that $\widehat{AMB}=80^o$ (why?) and therefore $\widehat{MOC} = 40^o$ (why?).

Draw $BP$ , bisector of $\widehat B$ . Note that $\triangle ABC$ is symmetric with respect to $BP$ . Let $CN$ be the mirror image of $AM$ with respect to $BP$. Then
$$\widehat{NCB} = \widehat{MAB} = 20^o$$
$$\widehat{NCO} = \widehat{NCA} - \widehat{OCA} = 30^o - 10^o = 20^o $$
Therefore we have:
$$\triangle IOC = \triangle IBC$$
It follows that $BC = OC$, which means $\triangle COB$ is isosceles, hence:
$$\widehat{BOC} = \widehat{OBC} = \frac12 (180 - \widehat{OCB}) = 70^o$$
Et voila!

Fun fact: the segments shown in cyan ($AM,CN,CO,AB,CB$) are of equal length.


问:
Good books and lecture notes about category theory.
What are the best books and lecture notes on category theory?</p>

答:
A relatively new source tha I believe hasn't been mentioned: Notes on Category Theory with examples from basic mathematics by Paolo Perrone.

These notes were originally developed as lecture notes for a category
theory course. They should be well-suited to anyone that wants to
learn category theory from scratch and has a scientific mind. There is
no need to know advanced mathematics, nor any of the disciplines where
category theory is traditionally applied, such as algebraic geometry
or theoretical computer science. The only knowledge that is assumed
from the reader is linear algebra. All concepts are explained by
giving concrete examples from different, non-specialized areas of
mathematics (such as basic group theory, graph theory, and
probability). Not every example is helpful for every reader, but
hopefully every reader can find at least one helpful example per
concept. The reader is encouraged to read all the examples, this way
they may even learn something new about a different field.

Particular emphasis is given to the Yoneda lemma and its significance,
with both intuitive explanations, detailed proofs, and specific
examples. Another common theme in these notes is the relationship
between categories and directed multigraphs, which is treated in
detail. From the applied point of view, this shows why categorical
thinking can help whenever some process is taking place on a graph.
From the pure math point of view, this can be seen as the
1-dimensional first step into the theory of simplicial sets. Finally,
monads and comonads are treated on an equal footing, differently to
most literature in which comonads are often overlooked as "just the
dual to monads". Theorems, interpretations and concrete examples are
given for monads as well as for comonads.


问:
Proof for simple planar graphs using girth
Question 3 For an $n$-vertex simple planar graph with girth $k$, prove that there are most $(n-2)\frac k{k-2}$ edges.

I have been having trouble with this question for a while. All I can think of is that if we know how many faces we have using the $v+f-e = 2$ formula then we can assume that every cycle in the graph is of length k in the worst case, and somehow go from there? But I'm just not able to put the pieces together. Any help would be greatly appreciated as I'm really struggling with this unit.

答: We can bound the number of edges using the girth. Let our graph have $e$ edges, $f$ faces, and $n$ vertices. Each of the graph's $f$ faces must have at least $k$ edges. Since each edge is contained in exactly $2$ faces, we have $2e \geq kf$. By Euler's formula, this is equivalent to $2e \geq k(2 + e - n)$. Some algebra gives us $$(k-2)e\leq (n-2)k \Rightarrow e \leq (n-2)\frac{k}{k-2},$$ as desired.


答:
Pick a planar embedding of $G$ (which we may assume connected). If in this, $f_r$ is the number of faces with $r$ edges, then we have  $2e=\sum rf_r\ge\sum kf_r=kf$. Using Euler,
$$ ke+2k=kn+kf\le kn+2e$$
or,
$$(k-2)e\le k(n-2), $$
whence the claim.

Title: Proof for simple planar graphs using girth

Body:
Question 3 For an $n$-vertex simple planar graph with girth $k$, prove that there are most $(n-2)\frac k{k-2}$ edges.

I have been having trouble with this question for a while. All I can think of is that if we know how many faces we have using the $v+f-e = 2$ formula then we can assume that every cycle in the graph is of length k in the worst case, and somehow go from there? But I'm just not able to put the pieces together. Any help would be greatly appreciated as I'm really struggling with this unit.


答:
Let's look at the first absolute value sign, $|-x+x^2+3|$, the discriminant for $-x+x^2+3$ is negative and the curve is convex, hence $-x+x^2+3$ is nonnegative and hence we can remove the absolute value.

When $x \le -3$, $-2x \ge 0$, hence $\sqrt{x^2+2x-3}-2x \ge 0$, hence we can remove the absolute value.

$B^c$ means the complement of $B$.

Now, given that this is a multiple choice question, let's quickly rule out some options. Let $x=0$,  the left hand side is positive but the right hand side is zero. Hence $0 \notin B$. $0 \in B^c$. We rule out option $1,3,4$ immediately.

Title: How do you find $B^c$ when $B$ is the solution set of $\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$?

Body:
The problem is as follows:

Assuming $B$ is the solution set of the equation from below:

$$\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$$

Find $B^c$.

The given choices in my book are as follows:

$\begin{array}{ll}
1.&\varnothing\\
2.&\mathbb{R}\\
3.&[2,+\infty\rangle\\
4.&[2,3]\\
\end{array}$

I've found this problem in my precalculus workbook but I have no idea how to solve it.

What I'm about to post below is the official solution from my book but it is some unclear to me. The reason for this is that it does not give any further details on the explanation how the vertical pipes are eliminated. Thus I really appreciate someone could explain to me the rationale behind this problem, and more importantly what is the meaning of that $C$ on $B^c$

$\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$

Notice:

$|-x+x^2+3|$ the  $\Delta<0$

Therefore:

$\left|x^2-3x+\sqrt{x^2+2x-3}+3-x^2+x-3\right|+3=-x$

$\left|\sqrt{x^2+2x-3}-2x\right|=-x-3$

and

$-x-3\geq 0$

$\left|\sqrt{x^2+2x-3}-2x\right|=-x-3$

$\sqrt{x^2+2x-3}-2x=-x-3$ and $x\leq -3$

$\sqrt{x^2+2x-3}=x-3$ and $x\leq -3$

Therefore:

The solution set is $\varnothing$

hence the $B=\varnothing$ and $B^c=\mathbb{R}$

Therefore the answer is choice 2.

That's it where it ends the official solution. But as I'm presenting it here I'm just clueless on how did it arrived to that conclusion. There might be some typo in the question regarding who's $C$ but I wonder who could it be so that the answer is all real.

Can someone help me with this?. It would really help me a lot if the answer would include some foreword or an explanation on when the vertical pipes can be erased or just skipped from the analysis I think the latter part is more important to my understanding because this is what I'm lacking the most.

Thus could someone attend these doubts which I have?. Please try not to skip any step and if you could explain each line because I'm lost here.

Regarding the notations which I have been asked before. The closed brackets $[$ means that the left boundary on the interval is closed, conversely $]$ closed in the right boundary and $\rangle$ means that the right boundary on the interval is open.


问:
Prove that if $x\in\{a,b\}^*$ and $abx=xab$, then $x = (ab)^n$ for some $n\in\Bbb N$
This is an exercise in text "Computability, Complexity, and Languages" by Davis et al.:

enter image description here

I can prove it easily by mathematical induction, but this exercise appears after a section titled "Proof by Contradiction", so it is supposed to be proved by deriving a contradiction.

Following is my attempt. From condition $abx=xab$ I can say $x$ begins with and end in $ab$, so I write $x$ as $abuab$ for some string $u$. After plugging it into $abx=xab$ and cancelling $ab$, I got $abu=uab$, which seems to imply nothing contradictory. If I proceed this way by writing $u$ as $abvab$, the same equation results again and again except the variable name. Then I tried to deny the conclusion, but I have difficulty in expressing $x$ in a way easy to proceed when $x\neq(ab)^n$. Can you please give me some hint (you don't have to follow my unsuccessful attempt)? Thank you.

答:
You really already have what I suspect is the intended argument; you just have to look at it properly. Assume that there is a counterexample; then there is a shortest counterexample. Your argument now shows how to get a shorter counterexample and hence a contradiction.

As far as I’m concerned this is really just another way of thinking about induction, but it is technically a proof by contradiction.


问:
Support of a generalized function
Let $F \in D'(\mathbb{R})$. If there exists $f \in D(\mathbb{R})$ which vanishes on $supp(F)$, but $F(f)\neq0$, then how can one prove that $supp(F)$ contains an isolated point?

For example, if $F=\delta'$, and $f(x)=x$, than $F(f)=-1$, $supp(F)=\{0\}$ and $f$ vanishes on it.

If $supp(F)=[0,1]$ or even the Cantor set (!), then such function $f$ does not exist.

答:
If $f(x_n)=0$ for a monotone sequence $x_n\to x_\infty$ the mean value theorem gives you $y_n$ between $x_n$ and $x_{n+1}$ such that $f'(y_n)=0$. The continuity
of $f'$ then implies $f'(x_\infty)=0$. Repeating this argument implies
$f^{(n)}(x_\infty)=0$ for all $n$. If $supp(F)$ has no isolated points, all derivatives vanish there for $f\in D(\mathbb R)$ which vanishes on $supp(F)$. This is enough to conclude $F(f)=0$ (as can be seen in most text books about distribution theory).</p>

Title: Support of a generalized function


问:
Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$
Prove if $x > 0$,  $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.

I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.

So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.

I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.

答:
We’ll let $a = x + 1$

Transform the inequality by substitute $x$ with $a - 1$.

After some calculations we the get

$$ -a^2 + 6a + 3 \leq 8\sqrt{a} $$

Square both sides and we get

$$ a^4 - 12a^3 + 30a^2 - 28a + 9 \leq 0$$

Which is just $(a-1)^3(a - 9) \leq 0$.

Since $a > 1$ , this statement is true for all $a \in [1,9]$.

Thus, we are left to prove this statement for all $a > 9$.

Since $a = x + 1$ , $x > 8$.

Consider $(x - 2)^2 + 12 > 0$

After some calculations we then get $\frac{1}{2}x < 3  - 1 + \frac{1}{8}x^2$.

But since $x > 8$ we then get $1 + \frac{1}{2}x - \frac{1}{8}x^2 < \sqrt{x + 1}$.

Hence, this inequality is true for all $x > 0$.


问:
Use two points and a derivative to approximate a function
I'm wondering How to show the following:

$$
f(x)=-\frac{(x-x_1)(x-2x_0+x_1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)+R(x)
$$

where $R(x)=\frac{1}{6}(x-x_0)^2(x-x_1)f'''(\xi)$

I want to use Taylor to show this, but how can I kill f''(x_1),f''(x_0), it's impossible to do this

so could anyone give me some help? thanks

答:
These sorts of questions are usually solved using the mean value theorem several times. Let $x^*$ be any fixed $x\in(x_0,x_1)$. Then, to start, let

$$G(x) = f(x) -\left(-\frac{(x-x_1)(x-2x_0+x1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}+\frac{(x-x_0)^2}{x_1-x_0}\right)-\lambda(x-x_0)^2(x-x_1)$$

Since $(x^*-x_0)^2(x^*-x_1)\neq 0$, we can find $\lambda$ such that $G(x^*)=0$.

Now, we know that $G(x)$ has zeros at $x_0$, $x^*$, and $x_1$. Thus by the mean value theorem there exists some point in each of $(x_0,x^*)$ and $(x^*,x_1)$ such that $G'(x)=0$.

Now, we know that $G'(x)=0$ at $x_0$ two other points, just call them $\bar{x}_1$ and $\bar{x}_2$. Again applying the meal value theorem, there exists some $\bar{\bar{x}}_1 \in (x_0,\bar{x}_1)$ and some $\bar{\bar{x}}_2 \in (\bar{x}_1,\bar{x}_2)$ such that $G''(\bar{\bar{x}}_1)=G''(\bar{\bar{x}}_2)=0$.

Finally, we can now apply the mean value theorem one more time to say there exists $\xi \in (\bar{\bar{x}}_1,\bar{\bar{x}}_2)$ such that $G'''(\xi)=0$. Turning our attention back to the definition of $G(x)$, we can see that $0=f'''(\xi)-6\lambda$. (To see this, take the derivative of both sides three times. Note: This is easier than it looks initially because the part inside the large parenthesises is quadradic and will vanish when differentiated three times.)

And thus $\lambda = \frac{1}{6}f'''(\xi)$. Since this process can be repeated for any $x$, we are done.


问:
What is a parallel vector space and how do I show it is isomorphic to the solution space?
How can I create an isomorphism between the solution space and a parallel vector space. I'm not sure how to define the vector space and the isomorphism.

$$ \begin{bmatrix}
        -2 & 4 \\
        4 & -8\\
        \end{bmatrix} * \begin{bmatrix}
        x  \\
        y  \\
        \end{bmatrix} =  \begin{bmatrix}
        -8  \\
        16  \\
        \end{bmatrix}  $$

答:
The solution space is the line $y=x/2-2$. Since it does not pass through the origin $(0,0)$, it is not a vector subspace of $\mathbb R^2$. Rather, it is an affine subspace, which by definition is a set of the form $V+u$ where $V$ is a vector subspace and $u$ is a fixed vector. This $V$ is the parallel vector space that you are looking for. The map $x\mapsto x+u$ is a desired isomorphism. The space $V$ is unique for a given affine subspace, but $u$ generally is not.

Since we are talking about lines on a plane, things are not very complicated. Two lines are parallel if they have the slope. So, $y=x/2$ is the parallel vector space. An example of isomorphism between them is $(x,x/2)\mapsto (x,x/2-2)$.

In general, the solution space of the system $Ax=b$ has parallel vector space described by the system $Ax=0$.


问:
Which probability formula do I use?
If the probability of discovering a new planet within the next 1,000 years is 0.99, what is the probability that we will discover a new planet within the next 200 years?

Note: that each discovery of a planet is independent of each other.

I want to use the Poisson process formula, but I don't think that is correct since the probability is given out of 1000 trials, and we are solving for 200 trials.

答: Assuming you meant at least $1$ in $1000$ years, then you could use an exponential distribution for finding none, which  is $0.01$ for $1000$ years. Therefore for $200$ years the probability for finding none is the fifth root, which is equal to $0.3981,$ so the probability of finding at least one in $200$ years is $0.6019.$

Title: Which probability formula do I use?


问:
Comparing the formulas of trigonometric and hyperbolic functions
I just tried to memorize hyperbolic formulas, and realized that there are lots of similarity between formulas for trigonometric and hyperbolic functions. for example:
$$\cos^2x+\sin^2x=1\quad\quad\quad\quad\quad\quad\quad1+\tan^2x=\sec^2x$$
$$\cosh^2x-\sinh^2x=1\quad\quad\quad\quad\quad\quad\quad1-\tanh^2x=\text{sech}^2 x$$
And so on. When I look at it closer I see they are almost the same formulas but the sign comes before a $\sinh$ or $\tanh$ are different than the sign of $\sin$ ,$\tan$ in trig functions. To justify that, I considered the fact that parametric points $(\cos\theta,\sin\theta)$ placed on the circle $x^2+y^2=1$ and points $(\cosh \theta,\sinh \theta)$ are placed on hyperbola $x^2-y^2=1$, so we see the different signs comes before $y^2$ and because $\sin$ and $\sinh$  represents the vertical distance, these signs are different in all formulas.

Is my justification right?

Is there better way to justify this different in the formulas?

答:
It is probably better to think of the difference in algebraic sign as what distinguishes trigonometry from hyperbolic trigonometry.  Consider what happens when $\theta$ is close to zero.  What happens to $\sin\theta$ and  $\sinh\theta$?  Same question for $\cos\theta$ and  $\cosh\theta$.  Those are the properties that you should consider the most basic.

If you understand complex numbers and the natural logarithm, it will be helpful to look at Euler's formula https://brilliant.org/wiki/hyperbolic-trigonometric-functions/ Notice the expressions for $\cosh$ and $\sinh$ are simpler than those for $\cos$ and $\sin$.


问:
Finding the covariance of a mixed pair of r.v.'s given one's distribution and a conditional distribution
$
\newcommand{\E}{\mathbb{E}}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\al}[1]{\begin{align*} #1 \end{align*}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\para}[1]{\left( #1 \right)}
\newcommand{\br}[1]{\left[ #1 \right]}
\newcommand{\B}[1]{\Big[ #1 \Big]}
\newcommand{\var}{\mathrm{var}}
\newcommand{\Beta}{\mathrm{B}}
\newcommand{\b}[1]{\big[ #1 \big]}
\newcommand{\cov}{\mathrm{cov}}
$
The Problem: Let $X,Y$ be jointly distributed r.v.'s such that
\begin{align*}
Y \mid X = x &\sim \mathrm{Binomial}(n,x) \\
X &\sim \mathrm{Uniform}(0,1)
\end{align*}
Among other things, I seek to calculate $\mathrm{cov}(X,Y)$, without using Bayesian techniques.

(This comes to me in the form of a homework assignment, exercise $35$ in Chapter $2$ of Allan Gut's An Intermediate Course in Probability, second edition.)

My Work So Far: Note that, with the given info, we have
$$\al{
f_{Y \mid X = x}(k) &= \binom n k x^k (1-x)^{n-k}, \; k \in \set{0,1,\cdots,n} \\
f_X(x) &= 1, \; x \in (0,1)\\
\E\br{Y \mid X = x} &= nx \\
\E\br{X} &= \frac 1 2\\
\var \para{Y \mid X} &= nx(1-x) \\
\var \para{X} &= \frac{1}{12}
}$$

We'll need to calculate $Y$'s expectation as well:
$$\al{
\E \b{Y}
&= \E \B{ \E \b{Y \mid X} } \tag{known result} \\
&= \E \br{ nX } \tag{known value} \\
&= n \cdot \E \br X \tag{linearity}\\
&= \frac n 2 \tag{known value}
}$$

We can calculate the covariance as so:
$$\al{
\cov(X,Y)
&= \E \B{ \big( X - \E[X] \big) \big( Y - \E[Y] \big) } \tag{definition} \\
&= \E \br{ \para{ X - \frac 1 2 } \para{ Y - \frac n 2 } } \tag{known values} \\
&= \E \br{ XY - \frac n 2 X - \frac 1 2 Y +\frac n 4 }  \tag{expand} \\
&= \E \br{XY} - \frac n 2 \E \br{X} - \frac 1 2 \E \br{Y} + \E \br{ \frac n 4 } \tag{linearity}
}$$

Everything except one thing works to me: how would one find $\E \br{XY}$? We could note that...
$$
f_{Y \mid X = x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)} \implies f_{X,Y}(x,y) = f_{Y \mid X = x}(y) \cdot f_X(x)
$$
Then I think we have this:
$$\al{
\E \br{XY}
&= \sum_{y=0}^n \int_0^1 xy \binom n y x^y (1-x)^{n-y} \, \dd x\\
&= \sum_{y=0}^n y \binom n y \int_0^1 x^{y+1} (1-x)^{n-y} \, \dd x\\
&= \sum_{y=0}^n y \binom n y \Beta(y+2,n-y+1)
}$$
where $\Beta(\cdot,\cdot)$ denotes the usual beta function. And I think this form follows since, essentially, covariance is like
$$
\cov(X,Y) = \kern-30pt \sum \limits_{(x,y) \,\in\, \mathrm{supp}(X) \times \mathrm{supp}(Y)} \kern-30pt  xy \, f_{X,Y}(x,y)
$$
(though with a continuous variable like $X$ we would interpret the respective sum as an integral, though we use sum for discrete r.v.'s like $Y \mid X$).

My Question: Simply put, is this correct?

It doesn't feel like it is at all. I'm struggling with the fact that $Y \mid X$ and $X$ are different "kinds" (discrete versus continuous) of distributions, and how to handle them in a clean way. And the final answer just feels too complicated to be right.

Does anyone have any insights they can give me on this problem? Particularly with respect to finding $\E[XY]$ within the given parameters.

Edit: (3/24/2021)

The suggestion of using that
$$
\E \br{XY} = \E \B{\E \br{XY \mid X}} = \E \B{X \cdot \E \br{Y \mid Y}}
$$
(per StubbornAtom in the comments) has helped in achieving the right answer, namely
$$
\cov(X,Y) = \frac{n}{12}
$$
However, I'm not totally sure if it's kosher to use this in the class, at least without proof; I'm pretty sure this identity hasn't come up at any event. Of course, one can prove it, and some links have been provided hinting at how to do so. But of course this begs the question -- how might one achieve the desired answer, without this identity?

Of course, perhaps the end result is "you just get that nasty expression you derived," but I figured it'd be worth asking to see some alternative approaches. Or maybe my answer can be somehow easily shown to get the same answer anyways?

答:
You can in fact evaluate the sum directly; it is just a bit tedious compared to using iterated expectations.

$$\begin{align}
\operatorname{E}[XY]
&= \sum_{y=0}^n \int_{x=0}^1 xy \binom{n}{y} x^y (1-x)^{n-y} \, dx \\
&= \sum_{y=0}^n y \binom{n}{y} \int_{x=0}^1 x^{y+1} (1-x)^{n-y} \, dx \\
&= \sum_{y=0}^n y \binom{n}{y} \frac{\Gamma(y+2)\Gamma(n-y+1)}{\Gamma(n+3)} \\
&= \sum_{y=0}^n y \frac{n!}{y!(n-y)!} \frac{(y+1)!(n-y)!}{(n+2)!} \\
&= \sum_{y=0}^n \frac{y(y+1)}{(n+1)(n+2)}. \\
\end{align}$$

At this point, we can evaluate the sum directly, using the familiar formulas $\sum y = n(n+1)/2$ and $\sum y^2 = n(n+1)(2n+1)/6$, or we can use the hockey stick identity

$$\sum_{y=0}^n \frac{y(y+1)}{(n+1)(n+2)} = \frac{2}{(n+1)(n+2)} \sum_{y=1}^n \binom{y+1}{2} = \frac{2}{(n+1)(n+2)} \binom{n+2}{3} = \frac{n}{3}.$$

Combining this with the other terms in the covariance expression, we get $$\operatorname{Cov}[X,Y] = \frac{n}{3} - \frac{n}{4} = \frac{n}{12}$$ as claimed.

As an exercise, can you evaluate $\operatorname{E}[XY]$ with the summation and integration order exchanged; i.e., how would we proceed with $$\int_{x=0}^1 \sum_{y=0}^n xy \binom{n}{y} x^y (1-x)^{n-y} \, dx?$$  In fact, is this a more natural way to evaluate this?  Does it mimic the iterated expectation evaluation $$\operatorname{E}[XY] = \operatorname{E}[\operatorname{E}[XY \mid X]]?$$


问:
Fold a paper once and fold it every time along the existing crease. When can we have the crease formed at the center?
Suppose you are folding a piece of paper along the vertical axis. If the top edge (breadth) of the paper is initially $1$ and the first fold reduces the breadth of the paper by $x \in (0,1)$ units, then every fold hereon will be along the existing vertical axis. The existing vertical axis is basically the line/crease you get after each fold.

If we are to mathematically write it, we can recursively track it as :

$x_0 = \{x,1-x\}$

$x_{i+1} = \{\min(x_0), \max(x_0)-\min(x_0)\}$

Can we conclude from here that at some point we will have the crease at the center iff $x \in \mathbb{Q} \backslash \{0\}$?

Here are my observations :

Let $x$ be a non-zero rational number. Then since $x_i$ and $x_j$ for $i \neq j$ will have the same denominator, after finitely many steps, we will have $(\max - \min)(x_n) = 0$ for some $n$. Can we find this $n$?

If $x$ is irrational, then $x_i$ will comprise of numbers of the form $a + bx$ where $a,b \in \mathbb{Z}$. This can never be $0$.

The two observations lead to my conclusion. Can someone verify if my proof is correct?

答:
I could not understand the folding picture by your description alone, but the numerical description let me figure it out. At any point, there will be a left edge, a right edge, and a folding line. Initially, the left edge is at $0$, the right edge is at $1$, and the folding line is at $x$. At each stage, you do one of two things. If the left edge is closer to the folding line than the right edge, you fold the left edge over the folding line, then the left edge and folding line switch roles. If instead the right edge is closer, do the same with the right edge.

You are correct that when $x\in \mathbb Q$, the process eventually stops. Write $x=\frac{p}{q}$, with $\gcd(p,q)=1$. The two distances to the folding line after the $n^{th}$ fold are $a_n/q$ and $b_n/q$, with $a_0=p$ and $b_0=q-p$. The sequence $(a_0,b_0),(a_1,b_1),\dots$ is just the result of applying Euclid's algorithm to $(a_0,b_0)$. Since $\gcd(a_0,b_0)=1$, this eventually reaches $(a_n,b_n)=(1,0)$, wherein the process stops with a paper of width $1/q$.

There is no formula for the number of iterations it takes. The "worst case" is when $a_0$ and $b_0$ are consecutive Fibonacci numbers; if $a_0=F_n$ and $b_0=F_{n-1}$, then the process terminates in $n-1$ steps.

For the reason you said, the process never terminates when $x$ is irrational.


问:
When does this least squares analytical solution based on zeros of partial derivatives start providing more than one solution?
If I want to fit a quadratic function of two variables to some data, I can use

$$f(x, y) = c_1 x^2 + c_2 xy + c_3 y^2 + c_4 x + c_5 y + c_6$$

$$\frac{\partial}{\partial c_i} \sum_j\left( z_j - f(x_j, y_j) \right)^2 = 0$$

to obtain six equations, and then endeavor to solve them.

I've done it for one variable not two, but I'm guessing the process is straightforward.

If I extend this to more variables, and to higher order than 2, when will the analytical expressions be at risk for having multiple solutions?

答:
As @ClaudeLeibovici quickly pointed out, for least squares it will always be linear.

This is one reason why least squares is/was so convenient/popular back when folks were "writing with feathers using light from burning animal fat".

For example:

$$\frac{\partial}{\partial c_1} \sum_j\left( z_j - f(x_j, y_j) \right)^2 = 2 \sum_j \left( z_j - f(x_j, y_j) \right) \frac{\partial f}{\partial c_1}$$ and since $\frac{\partial f}{\partial c_1}$ is now constant and $\left( z_j - f(x_j, y_j) \right)$ is linear in $c_1$, there will always be one solution.

The number of variables and the power to which they are raised is unimportant. It's only the power of 2 in "least squares" that we need to notice here.


问:
Proof Involving Fourier Basis Functions

Given the equations $$u(x,t) = \sum_{k \in \mathbb{Z}} = \hat{u}(k,t) \exp(i \pi k x), \hspace{5mm} \hat{u}(k,t) = \frac{1}{2} \int_{-1}^{1}u(x,t) \exp(-i \pi k x)dx,$$ where $i$ is the imaginary unit. Due to the orthogonality and completeness of the Fourier basis functions, we have $$u_t(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}_t(k,t) \exp(i \pi k x) = \sum_{k \in \mathbb{Z}} \exp(-i \pi k x(1 + \pi^2) t + i \pi^3 k^3t) \hat{u}(k,t) \exp(i \pi k x),$$ and $$ u(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) $$ Prove that for any time $ t > 0$, $$\lVert u(x,t) \rVert := \sqrt{\frac{1}{2} \int_{-1}^1 |u(x,t)|^2 dx} = \sqrt{\frac{1}{2} \int_{-1}^1 |u(x,0)|^2 dx} $$

My attempt: I begin by trying to show that $|u(x,t)|^2 = |u(x,0)|^2$ by using that $$|u(x,t)|^2 = u(x,t)^2 = \bigg ( \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) \bigg )^2$$

Then inserting for $\hat{u}(k,0)$ gives $$u(x,t)^2 = \bigg ( \sum_{k \in \mathbb{Z}} \frac{1}{2} \int_{-1}^1 u(x,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) \exp(-i \pi k x) \bigg ) ^2$$

$$= \bigg ( \sum_{k \in \mathbb{Z}} \frac{1}{2} \int_{-1}^1 u(x,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \bigg )^2$$

But here Im stuck and can't really progress further. Any help would be appreciated.

答:
The assumption that
$$ u(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) $$
implies that for all times $t$ and integer $k$,
$$|\hat{u}(k,t)|=|\hat{u}(k,0)| \,.$$
The desired conclusion now follows from
https://en.wikipedia.org/wiki/Parseval%27s_theorem

Title: Proof Involving Fourier Basis Functions


答:
Here is an "out of triangle thought" for this question. Law of sine can be applied twice to reveal a  hidden equilateral triangle $\triangle BOC$ if O is allowed to be outside $\triangle ABC$. (There might be a simple geometric solution out there someone can find.)

Set BC=1

For  $\triangle ABC$ , AC=$\frac{\sin 80^\circ}{\sin 50^\circ}$

For  $\triangle AOC$, CO=$\frac{AC\sin 30^\circ}{\sin 140^\circ}=\frac{\sin 80^\circ\sin 30^\circ}{\sin40^\circ\sin 50^\circ}=\frac{\sin 80^\circ}{2\sin40^\circ\cos40^\circ}$=1

Therefore,  $\triangle BOC$ is an isosceles triangle, with $\angle BCO=60^\circ$$\to$ $\angle BOC=60^\circ$

Title: Given $\triangle ABC$ with $AB=BC$ and $\angle ABC=80^\circ$, and $O$ such that $\angle OAC=30^\circ$ and $\angle OCA=10^\circ$, find $\angle BOC$.

Body:
I encountered the following problem in a math problem book, in a section, entitled Geometry.

In an $\Delta ABC$ with $AB=BC$ and $\angle ABC = 80^\circ$, the point $O$ is chosen so that $\angle OAC = 30^\circ$ and $\angle OCA = 10^\circ$. Find $\angle BOC$.

I struggled repeatedly trying to solve this using some reasonable selection of methods, and failed utterly. I can construct a solution using analytic geometry, but that is not a fair choice of methods for this section.

From simple geometry, I keep writing down systems of equations using various triangles in the picture, and always keep missing one constraint, as it seems there are infinite amount of solutions to the system, which I know is wrong since the analytic geometry approach yields a unique solution.

I am happy to add my infinite solution approach using the angle constraints, if you find it helpful. Please let me know if that's the case.

I would like to solve this using methods from geometry, without using more advanced techniques. Thank you very much, I really appreciate your help.


问:
Using Ramsey's Theorem to prove that a graph with $\binom{k+l}{k}$ vertices contains $K_{k+1}$ or $\overline K_{l+1}$
I've been presented with the following problem:

Prove that a graph with $\binom{k+l}{k}$ vertices contains $K_{k+1}$ or $\overline K_{l+1}$

It seems to me that a good way to prove this would be using Ramsey's Theorem. But I don't know where to start. Maybe showing that $\binom{k+l}{k}$ is an upper bound for Ramsey numbers? If so what should it bound: $R(k)$ or $R(l)$? And how should I go about proving this bound? I'm stumped.

答:
Lemma. $R(r, s) \leq R(r-1, s) + R(r, s-1)$.

Proof. Let $N = R(r-1, s) + R(r, s-1)$, and let the edges of $K_N$ have a $2$-coloring in, say, red and blue. Fix a vertex $V$ in $K_N$, and partition the other $(N-1)$ vertices of $K_N$ into sets $R$, which contains all vertices with a red edge to $V$, and $B$, which contains all vertices with a blue edge to $V$.

By the pigeonhole principle, either $R$ has at least $R(r-1, s)$ vertices, or $B$ has at least $R(r, s-1)$ vertices. In the former case, $R$ must either contain a red $K_{r-1}$ or a blue $K_s$. If $R$ contains a red $K_{r-1}$, then the graph formed by $K_{r-1}$ and $V$ is a red $K_r$. If $R$ has a blue $K_s$, we are done. Similarly, in the latter case, $B$ contains either a red $K_r$ or a blue $K_{s-1}$. A blue $K_{s-1}$, including $V$, forms a blue $K_s$. Thus, $N$ must contain either a red $K_r$ or blue $K_s$, so $R(r, s) \leq N$.

Now, we will show by induction that $$ R(r, s) \leq \binom{r+s-2}{r-1}. $$

The base case holds, since $R(2, 2) = 2$. Then, by our lemma, $$ R(r, s) \leq R(r - 1, s) + R(r, s - 1) \leq \binom{r + s - 3}{r - 2} + \binom{r + s - 3}{r - 1} = \binom{r + s - 2}{r - 1}, $$ where the last step holds by Pascal's identity.


问:
If there's two eigenvalues for two $n \times n$ matrices with respect to the same eigenvector, then does $ABv = BAv$?
If $A$ and $B$ are both $n\times n$ matrices, and $\lambda_A$ and $\lambda_B$ are the eigenvalues for those matrices with respect to the same eigenvector $v$, then does $ABv = BAv$?

答:
Since scalar multiplication commutes with matrix multiplication, this is not too hard to show. Always go back to the definitions. In this case, $\lambda, v$ form an eigenvector-eigenvalue pair of a matrix $X$ exactly if $Xv=\lambda v$.

Take the left hand side:

$$A(Bv)=A\lambda_Bv=\lambda_B Av = \lambda_B\lambda_A v$$

Taking the right hand side:

$$B(Av)=B\lambda_Av=\lambda_A Bv = \lambda_A\lambda_B v$$

And now of course, $\lambda_A\lambda_B v = \lambda_B\lambda_A v$.


问:
Space of probability density function over compact set is compact or not
Suppose $\Omega\subset\mathbb{R}^n$ is compact. Is the set of probability density functions over $\Omega$ compact with respect to weak convergence or $L^1(\Omega)$ convergence? Thank you!

答: If $L^1(\Omega)$ is assumed infinite dimensional -i.e. $\Omega$ is not finite, this is not true with respect to the weak or strong topology on $L^1(\Omega)$. The set of probability density functions is just the unit sphere in $L^1(\Omega)$, but the unit sphere is never compact in an infinite dimensional space.


问:
Solving $52\sin\left(\frac{2\pi}{365}t\right)+728=750$
I have seen the same formula posted, but the questions are all slightly different from what I am looking for.

Here is the equation I am trying to solve from a Khan Academy Exercise.

$$52\sin\left(\frac{2\pi}{365}t\right)+728=750$$

I can simplify the equation to the following but then get the incorrect result?

$$t = \frac{365}{2\pi}(0.4368+2\pi n) $$

My answer:

$$250 + 365n \quad\text{(incorrect)}$$

Expected answer:

$$25 + 365n \quad\text{(correct)}$$

I have been stuck trying to figure out what I am doing wrong here. If anyone has any ideas please let me know!

Body:
It seems that

$$\frac{750-728}{52} = \frac{22}{52} = 0.423 $$

So the solution is

$$\frac{2\pi}{365}t +2n\pi = 0.4367$$

Now you can rewrite this as

$$t+365n = \frac{0.4367}{2\pi}\cdot 365 = 25.37$$

$$t = 25.37 -365n$$


问:
Is there a connected $n$-sheeted covering space of a closed orientable surface for any positive integer $n$?
Let $S$ be a closed orientable surface. For any positive integer $n$, is there a connected $n$-sheeted covering space of $S$? This is certainly not true if $S=S^2$ because $S^2$ is simply-connected. The result is true for $S=T^2$, since $\pi_1(T^2)=\Bbb Z^2$ has an index $n$ subgroup for each $n$. However I cannot handle the other cases. I know that this is equivalent to finding an index $n$ subgroup of $\pi_1(S)$ but the $\pi_1$ for the other surfaces are quite complicated, I think. Any hints?

答:
Yes. See the picture for an example that a surface of genus 3 got a 5-sheeted covering:

This lay out works for any surfaces of genus $\geq 1$ and any $n$.


问:
Changing variable in a supremum integral inequality
So I'm trying to prove that this operator
$$
T_n(f)=\int_0^t \cfrac{(t-x)^n}{(n-1)!}f(x)dx
$$ is continuous and then calculate its operator norm, provided that

the operator is a map $E\to E$ with $E=C([0,1],\Bbb R)$, and

the norm considered on $E$ is the infinite norm.

To prove the continuity I tried to find a $M>0$ such that $\|T_n(f)\|_\infty  < M\|f\|_\infty $.

After developing  I got to this inequality :
$$
||T_n(f)||_\infty < \int_0^t\sup_{t\in[0,1]} \cfrac{(t-x)^n}{(n-1)!}f(x)dx
$$

What I'm trying to do is change the $t$ of the supremum to $x$ so that I could sort the $f(x)$ out and integrate the quantity $\cfrac{(t-x)^n}{(n-1)!}$ to get the required $M$ for the continuity.

Any tips on how to proceed?

答:
Observe that
$$
 \int_0^t \cfrac{(t-x)^n}{(n-1)!} dx= \cfrac{t^n}{n!} \,.
$$
Therefore

$$\|T_n(f)\|_\infty  \le \cfrac{1}{n!} \|f\|_\infty \,,\quad (*)$$
with equality for $f \equiv 1$, so the operator norm of $T_n$ is $\cfrac{1}{n!}$.


问:
Given that $\sum_{n}^{\infty} a_n$ is convergent, is $\sum_{n}^{\infty} n^n a_n$ convergent?
Given that $\sum_{n}^{\infty} a_n$ is convergent, non-negative series, I have to find out if $\sum_{n}^{\infty} n^n a_n$ is convergent or divergent.

Since $\sum_{n}^{\infty} a_n$ is a "non-negative" series, $a_n \geq 0$ for all $n$ and I cannot use ratio test (since it requires all terms to be >0).

I am considering using limit comparison test but am struggling since all the series I took eventually has a limit 0 and I cannot conclude anything.

答:
Counter-example for 1): $a_n=\frac 1 {n^{n+1}}$. Note that $a_n \leq \frac 1 {n^{2}}$.

Counter-example  for 3): $a_n=0$ for $n$ odd and $a_n=\frac  1 {n^{2}}$ for $n$ even.

The series in 2) is convergent: Use the fact that $\ln (1+x) \sim  x$ to compare the series with $\sum \frac {a_{n+1}-a_n} {2+a_n}$. This series is absolutely convergent since $2+a_n \to 2$ and $|a_{n+1}-a_n| \leq a_{n+1}+a_n$.


问:
How do I show $f$ is one-one on $\mathbb{R}^2$?
Let $f$ be a differentiable map (but not necessarily continuous differentiable) with component functions $f_1$ and $f_2$, that is $f(x_1,x_2)=(f_1(x_1,x_2),f_2(x_1,x_2))$ for all $(x_1,x_2) \in \mathbb{R}^2$. For all $(x_1,x_2) \in \mathbb{R}^2$, one has

$|\frac{\partial f_1}{\partial x_1}-2|+ |\frac{\partial f_1}{\partial x_2}|+|\frac{\partial f_2}{\partial x_1}|+|\frac{\partial f_2}{\partial x_2}-2| \leq \frac{1}{2}$.

Prove that $f$ is one-one on $\mathbb{R}^2$?

I cannot use inverse function theorem to prove this question. Can anyone suggest some direction to solve this question?

答:
Preliminaries: Assume $a,b\in \mathbb R^2.$ Then

i) one of $|b_1-a_1|,|b_2-a_2|$ is at least $|b-a|/\sqrt 2.$

ii) Suppose $g:\mathbb R^2\to \mathbb R$ is differentiable. Then

$$g(b)-g(a) = \nabla g(c)\cdot (b-a)$$

for some $c$ on the line segment from $a$ to $b.$

iii) The hypotheses imply $\partial f_1/\partial x_1 \ge 3/2,$ $|\partial f_1/\partial x_2| \le 1/2,$ similarly for the derivatives of $f_2.$

In our problem assume WLOG $a\ne b$ and $|b_1-a_1|\ge |b-a|/\sqrt 2.$ From ii) we get

$$f_1(b)-f_1(a) = \nabla f_1(c)\cdot (b-a)$$ $$ = [\partial f_1/\partial x_1(c)](b_1-a_1) + [\partial f_1/\partial x_2(c)](b_2-a_2).$$

Taking absolute values and using iii) gives

$$|f_1(b)-f_1(a)| \ge (3/2)|b_1-a_1|-(1/2)|b_2-a_2| $$ $$\ge (3/2)|b-a|/\sqrt 2 - (1/2)|b-a|.$$

It's easy to verify that the last expression is greater than $0,$ which implies $f$ is one-to-one.


问:
Is there a Kubota-Leopoldt $p$-adic zeta function implementation in SageMath?
Exactly as in the title. I am learning the $\mathbb{Z}_p^\times$ measure-theoretic construction of $p$-adic $L$-functions and was wondering if there was an `easy' way produce some example computations with these in the specific case of the interpolated $p$-adic Riemann zeta function.

Any code for this or suggestions for how one might algorithmically implement this would also be very welcomed!

Thank you in advance!

答:
According to the p-Adics section of the reference manual there is nothing I could find to do with zeta functions nor integration.

At any rate, here's a suggestion on algorithmically getting an arbitrary level of accuracy on your integral. In order to be called a p-adic measure we require it to be a bounded p-adic distribution, and this is enough to imply that the limit of Riemann sums of a continuous function is independent of the choices of representatives. It can then be easily shown that we can approximate our integral by an integral of a locally constant function that approximates our function.

By continuity we know there exists a $M$ so that $|x-a|\le p^{-M}$ that gets us $|f(x)-f(a)|\le p^{-N}$, and so in particular the most convenient choice is for $a \in \{0,1,..., p^M-1\}$ define the locally constant function $g(x)=f(a)$ for $|x-a|\le p^{-M}$.

Specifically, let's suppose your measure is bounded $|\mu(U)|_p\le p^{-K}$ for all compact open $U \in \mathbb{Z}_p^\times$ then $$\left|\int f \mu -\int g \mu\right|\le p^{-K-M}$$ and we have the exact expression for a sum that approximates $\int f \mu$ which shouldn't be difficult to code,

$$\int g \mu = \sum_{\substack{0\le a <p^N \\ a+p^N \mathbb{Z}_p \subseteq \mathbb{Z}_p^\times}} g(a)\mu(a+p^N \mathbb{Z}_p) = \sum_{a_0=1}^{p-1} \sum_{a=0}^{p^{N-1}-1} f(a_0+ap) \mu(a_0+ap+p^N \mathbb{Z}_p) $$

I can elaborate on more details here, but I highly recommend you see chapter 2, sections 3 and 5 of Koblitz's p-adic Numbers, p-adic Analysis, and Zeta-Functions for more information and proofs on the theorems above.


问:
Tiling a rectangle with tetris pieces of T-shape
I know that Walkup published a paper stating that $m\times n$ table can be tiled with T-tetrominoes iff $4\mid m$ and $4\mid n$.  The converse is clearly true because we can tile a $4\times 4$ table with 4 T-tetrominoes.  I am not able to prove the direction $\implies$.  I can prove that $8\mid mn$.

Here is my argument.  If $m\times n$ can be tiled, clearly $mn$ is divisible by $4$.  Then we color each $1\times 1$ squares in the checkerboard pattern.  A T-tetromino can occupy $1$ white square and $3$ black squares (and we call it type-A), or $3$ white squares and $1$ black square (and we call it type-B).  If there are $a$ type-A pieces and $b$ type-B pieces, then the number of white squares is $\frac{mn}{2}=a+3b$ and the number of black squares is $\frac{mn}{2}=3a+b$.  Therefore $a=b$ and $mn=4(a+b)=8a$.

But I tried to find Walkup's paper to read online.  I couldn't find a pdf available to read.  I'm only a highschool student and dont have access to https://www.jstor.org/stable/2313337?seq=1.  If you know Walkup's proof, can you post it/give me a sketch?  Thank you.

答:
I answered my question so this question is answered and accepted.  (unless other users post something here, i'll accept my answer when I'm allowed to)  Walkup's argument was the following.

Let the table be given by $\{0,1,2,...,m\}\times \{0,1,2,...,n\}$.  The idea was to prove that points of the form $(2r,2s)$ where $r\not\equiv s\pmod2$ can't be a corner of any T-tetromino and any edge with an endpoint of the form $(2r,2s)$ where $r\equiv s\pmod2$ has to be a boundary of a T-tetromino.  Walkup proved this by strong induction claiming the statement is true for all such points $(x,y)$ with $x+y\le 4\lambda$ and use $\lambda$ to do math induction.


问:
Prove that $\mathbb{Q} {/ \mathbb{Z}}$ contains elements of every possible finite order.
I was able to prove that every element of $\mathbb{Q} {/ \mathbb{Z}}$ has finite order. Now I'm not sure how to prove this one. I usually start my proofs by writing down an equation or end statement of what I have to prove. Any hints on where to start?

答: The element $\frac{1}{n}$ has order $n$

Body: The element $\frac{1}{n}$ has order $n$


问:
Question on the restricted product topology
Let $I$ be an index set and $G_i$ a collection of compact topological groups.  Suppose that for each $i \in I$ there are compact (proper) subgroups $H_i \subseteq G_i$.  We say that $G =\{(a_i) \in \prod G_i \,:\, a_i \in H_i \text{ for almost all $i \in I$}\}$ is the restricted product of $G_i$ with respect to the $H_i$.

We can give $G$ a topology by declaring the collection $\mathcal{B} = \{B \,:\, B = \prod U_i \text{ with $U_i$ open in $G_i$ and $U_i \neq H_i$ only finitely many times}\}$ as a basis.

I must be missing something really basic here, but I can't see why this is closed under arbitrary unions.  For example, consider the sets
\begin{align*}
W_i = G_i \times \prod_{j\neq i} H_j.
\end{align*}
Then $W_i \in \mathcal{B}$, but $W = \bigcup_{i\in I} W_i = \prod_iG_i \not\subseteq G$.  What is going wrong here?

答:
$W$ is not $\prod_{i\in I}G_i$: if $x\in W$, there is at most one index $i\in I$ such that $x_i\notin H_i$. In more detail, $x$ must be in $W_i$ for some $i\in I$, and therefore $x_j\in H_j$ for each $j\in I\setminus\{i\}$.

Moreover, a base for a topology need not be closed under arbitrary unions: one gets the topology by closing it under arbitrary unions.


问:
Conditional expectation of exponential distribution
We were learning about the memoryless property of the exponential distribution, when I came across these seemingly easy problems:

If $X$ ~ $Exp(\lambda)$, find:

$E[X|X>t]$

For this, I used the tail sum formula. $$P(X>s+t|X>t) = P(X>s)$$ $$\int_0^{\infty}P(X>s+t|X>t)ds = \int_0^{\infty}P(X>s)ds$$

I'm not too sure about this step, because I'm not entirely sure of tail sum expectation working for conditional expectations the way I've written it.

If it is correct, it gives the unsurprising result that $E[X|X>t] = E[X]$

$E[X^2|X>t]$

This seemed much harder to tackle for me, since I couldn't easily use the memoryless properly as above. I feel like this should also maybe come out cleanly as $E[X^2]$, but I'm not sure how to go about this.

Thanks in advance!

答:
$E(X|X>t)>t$ so we cannot have $E(X|X>t)=EX$ for all $t$. Memoryless property is not useful in evaluating this.

$E(X|X>t)=\frac {\int_t^{\infty} x\lambda e^{-\lambda x} dx} {\int_t^{\infty} \lambda e^{-\lambda x} dx}$ and $E(X^2|X>t)=\frac {\int_t^{\infty} x^2\lambda e^{-\lambda x} dx} {\int_t^{\infty} \lambda e^{-\lambda x} dx}$. Use integration by parts to evaluate the numertors. Both of these numbers  depend on $t$.


问:
prove $V = Span(S)$
Let $S= \{v_1,\dots,v_p\} \subset \mathbb F^n$ and $V$ be the subspace $\operatorname{span}(S)$. Suppose $v$ is in $S$. Then prove that $V = \operatorname{span}(S \setminus \{v\})$ if and only if $v$ can be written as a linear combination of some vectors in $S \setminus \{v\}$.

So, I felt that the phrasing of the question was confusing, but I tried proving the backwards statement first. Given that $v$ can be written as a linear combination of some of the vectors, then the set $S$ is linearly dependent, by definition. By using this property, we can say that the set $S$ is linearly dependent. After that, how would I relate it to $V$, and how would I prove the forwards statement as well?

答:
Welcome to MSE!

Suppose $\vec{v}$ is a linear combination of some of the vectors in $S\setminus\{\vec{v}\}$ then a basis $B$ for $S$ does not contain $\vec{v}$ and $V=\text{span}(S)=\text{span}(B)=\text{span}(S\setminus \{\vec{v}\})$ where $B \subseteq S\setminus \{\vec{v}\}\subseteq S$.

If $\text{span}(S)=\text{span}(S\setminus \{\vec{v}\})$ then because $\vec{v} \in S$, $\vec{v} \in \text{span}(S)=\text{span}(S\setminus \{\vec{v}\})$ and because $\text{span}(S\setminus \{\vec{v}\})$ is the set of all linear combinations of the vector in $S\setminus \{\vec{v}\}$, $\vec{v}$ must be a linear combination of the vectors in $S\setminus \{\vec{v}\}$.

Hope this helps~!

Title: prove $V = Span(S)$


问:
Use two points and a derivative to approximate a function
I'm wondering How to show the following:

$$
f(x)=-\frac{(x-x_1)(x-2x_0+x_1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)+R(x)
$$

where $R(x)=\frac{1}{6}(x-x_0)^2(x-x_1)f'''(\xi)$

I want to use Taylor to show this, but how can I kill f''(x_1),f''(x_0), it's impossible to do this

so could anyone give me some help? thanks

答:
This problem is estimate the trunction error of Hermite interplotion , Taylor series is the particular instance of Hermite interplotation . I solve this problem with the method by defining an Auxiliary function and Rolle's theorem.

Let
$$ P_2(x) =-\frac{(x-x_1)(x-2x_0+x_1)}{(x_1-x_0)^2}f(x_0)+\frac{(x-x_0)(x-x_1)}{x_0-x_1}f'(x_0)+\frac{(x-x_0)^2}{(x_1-x_0)^2}f(x_1)
$$
we can think $P_2(x)$ is an polynomail of 2th-degree approximation or interplotation of function $f(x)$ with totally three conditions $f(x_0)=P_2(x_0), f(x_1)=P_2(x_1), f'(x_0)=P_2(x_0)$ , two at point $x_0$ (Guarantee better smoothness) and one  at $x_1$

Construct the auxiliary function as follows
$$
\phi(t) = f(t)-P_2(t) - \frac{(t-x_0)^2(t-x_1)}{(x-x_0)^2(x-x_1)}\left(f(x)-P_2(x)\right)
$$

we obsevered that $\phi(x) = \phi(x_0) = \phi(x_1)=\phi'(x_0) = 0$ .
Then repeat the Rolle's Theorem , we can conclude that $\phi'''(t) $ must has one root $\xi \in (\min(x,x_0,x_1),\max(x,x_0,x_1))$ such that $\phi'''(\xi) = 0 $ and note that
$P_2'''(t) \equiv 0 $ , then we get following
$$
0=\phi'''(\xi) = f'''(\xi) - \frac{6}{(x-x_0)^2(x-x_1)}\left(f(x)-P_2(x)\right)
$$
thus we conclude the truction error $R(x) = f(x) - P_2(x) = \frac 16 f'''(\xi)(x-x_0)^2(x-x_1) $


问:
Sum of Self-adjoint elements is self-adjoint
Let $A$ be a commutative $C^*$- algebra. Let $a_1,a_2\in A$, where $a_1=a_1^*$ and $a_2=a_2^*$.

Let $a = a_1+a_2$. Can we say that $a=a^*$?

Thank you in advance!

答: In $C^*$algebra, we have $(a_1+a_2)^* = a_1^* + a_2^*$ which is $a_1+a_2$. We don't need commutativity.


问:
'Obvious' theorems that are actually false
It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true".

Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem).

But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false
(i.e. counter-intuitively-false theorems)?

The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$
).

I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$.

I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself.

Thanks!

答:
Any continuous function is differentiable at least somewhere, right?

False, the Wierstrass function is a famous counterexample

https://en.wikipedia.org/wiki/Weierstrass_function

It is continous everywhere, yet it is differentiable nowhere.


问:
$f:\mathbb R \rightarrow \mathbb R$ be differentiable function such that $f'(x)$ is continuous and $f(x+1)=f(x)+1$ for all $x \in\mathbb R$.
I was thinking about the problem:

Let $f:\mathbb R \rightarrow \mathbb R$ be differentiable function such that $f'(x)$ is continuous and $f(x+1)=f(x)+1$ for all $x \in\mathbb R$. Then which of the following options is correct?

(a) $f'(x)$ must be bounded,

(b) $f(x)$ must be bounded,

(c) Both $f(x)$ and $f'(x)$ must be bounded,

(d) Both $f(x)$ and $f'(x)$ must be unbounded.

My attempts: If I take $f(x)$ to be of the simplest form that is $f(x)=x,$ so that the given condition is satisfied then we see that $f'(x)=1,$ which is bounded. So I think that choice (a) is the correct option. Am I going in the right direction? I want a proof in a more generalized way. Please help. Thanks in advance for your time.

答: $f(x)=x$ is a possibility. So you may remove $(b), (c), (d)$ at once.</p>


答:
Using the first isomorphism theorem, this quotient ring $\mathbb{Q} {/ \mathbb{Z}}$ is isomorphic to the multiplicative group $\{z\in \mathbb{z}:z^n=1, for~ some~ n \in \mathbb{N}\}$, this group contains all the primitive $n-th$ roots of unity for any $n$  which are of the order $n$.

from direct computation, as Jorge mentioned, the coset $\frac{1}{n}+\mathbb{Q}$ will be of the order $n$

Title: Prove that $\mathbb{Q} {/ \mathbb{Z}}$ contains elements of every possible finite order.

Body: I was able to prove that every element of $\mathbb{Q} {/ \mathbb{Z}}$ has finite order. Now I'm not sure how to prove this one. I usually start my proofs by writing down an equation or end statement of what I have to prove. Any hints on where to start?


问:
How to find the range or domain of a function?
This is a general question I'm asking, I really need it explained. Here's an example of what I mean:

The functions $f$ and $g$ are defined by

$f( x)= x^3 + 1$,     $0 ≤ x ≤ 3$

$g(x)= x + 5$,         $x \in \mathbb R$.

And I was asked to find the range of $g(f(x))$??
Which I got to be  $= x^3 + 6$.

But idk how to find the range and if  I was asked to state the domain in another case I wouldn't know how to.

答:
You can find the range of composition of two functions by following the composition step by step. Generally, the range consists of all numbers that the function can produce, given $x$ within a specified interval. First, you find what numbers can be produced by $f$ when given $x$ in $[0,3]$. Then you are to feed the output of $f$ into $g$ and see what $g$ outputs.

It helps to pay attention to the intervals where each function is increasing or decreasing. As a matter of fact, $x^3+1$ is increasing for all $x\ge 0$. Thus, its range on the interval $[0,3]$ is the interval $[f(0),f(3)]$. Which is $[1,28]$.

Next step is to find the range of  $g$ on the interval $[1,28]$. Again, it helps that $g$ is increasing. For this reason, the range is $[g(1),g(28)]$. Which is $[6,33]$. This is the range of the composite function as well.


问:
Uniform convergence of $(f_n)_{n \in \Bbb{N}}$ on $D_1 \cup D_2$
I want to show using the Cauchy criterion that if $(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on both $D_1$ and $D_2$, then it must be uniformly convergent on $D_1 \cup D_2$

Here's what I have:

$(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on $D_1$, therefore, $\forall \varepsilon \gt 0, \exists  n_0$ such that if $ n,m \ge n_0$ then $\vert f_n(x) - f_m(x) \vert \lt \frac{\varepsilon}2$.

Similarly, $(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on $D_2$, therefore, $\forall \varepsilon \gt 0, \exists  k_0$ such that if $ p,q \ge k_0$ then $\vert f_p(x) - f_q(x) \vert \lt \frac{\varepsilon}2$.

Hence letting all $n, m, p, q \ge max\{n_0, k_0\}$:

$$\vert \vert f_n(x)- f_m(x)\vert -\vert f_p(x) - f_q(x)\vert \vert \le \vert f_n(x)- f_m(x)\vert + \vert f_p(x) - f_q(x)\vert \lt \frac{\varepsilon}2 + \frac{\varepsilon}2 = \varepsilon$$

Then it would follow that $$\vert \vert f_n(x)- f_m(x)\vert -\vert f_p(x) - f_q(x)\vert \vert \lt \varepsilon $$

Which I want to think implies that $(f_n)_{n \in \Bbb{n}}$ is convergent on $D_1 \cup D_2$

I don't know if what I did was correct, or if I made a mistake somewhere, so I would appreciate any feedback on my take and some hints on what direction my proof should go assuming it went wrong.

答: Let $N=\max \{n_0,k_0\}$, take any $x \in D_1 \cup D_2$ then we will have $|f_{n_1}(x)-f_{n_2}(x)|< \frac{\epsilon}{2}$ for all $n_1, n_2 \geq N$, because if $x \in D_1$ then the fact that $f$ is uniformly convergent on $D_1$ is used (the inequality you've written in the 3rd paragraph),  similar if $x \in D_2$.


问:
More questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$
This question is a follow-up to an older one about a proof of Rickard that the stable module category can be described as a triangulated Verdier quotient. In fact it's a complemented in that this question asks for clarification of every point that the previous question didn't touch upon.

Setup:

Let $A$ be a self-injective algebra (so projective = injective for modules) and let $D^b(A)$ and $K^b(A)$ be the bounded derived category and the full subcategory consisting of the perfect complexes (complexes whose modules are all projective).  The singularity category is defined to be the quotient $D^b_\mathtt{sg}(A) = D^b(A)/K^b(A)$.

Next let $\mathtt{stmod}(A)$ be the stable module category.  This is the category whose objects are homomorphisms and whose maps are module homomorphisms modulo those that factor through a projective.

Question 1 (Resolved in comments):

In his paper "Derived categories and stable equivalence" Jeremy Rickard proves that the stable module category and the singularity category are equivalent.  The functor
$$F\colon\mathtt{stmod}(A) \to D^b_\mathtt{sg}(A)$$
sends a module $M$ to the complex which is $M$ in degree $0$ and $0$'s in all other degrees.

Rickard proves in two sentences that this functor is an exact functor.

A distinguished triangle in $\mathtt{stmod}(A)$ is a triangle
$$X \to Y \to Z \to X[1]$$
coming from a pushout diagram of modules
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \ra{} & X & \ra{} & I & \ra{} & \Omega^{-1} X & \ra{} & 0 \\
 & & \da{} & & \da{} & & \da{} & &  & \\
0 & \ras{} & Y & \ras{} & Z & \ras{} & \Omega^{-1} X & \ras{} & 0 \\
\end{array}$$
where $X \to I$ is the embedding of $X$ into its injective hull. Since short exact sequences of modules give distinguished triangles in the derived category, and since $FI = 0$,
$$FX \to FY \to FZ \to FX[1]$$
is a distinguished triangle.

Could someone clarify what is going on here? I can understand that $X \to I \to \Omega^{-1} X \to X[1]$ and $Y \to Z \to \Omega^{-1} X \to Y[1]$ are distinguished triangles in $D^b(A)$ and that $FI = 0$, but I entirely fail to see how this gets to the conclusion.

Update: This question is now resolved. One uses the defining property of $D^b(A)$ to infer that we have distinguished triangles $X \to I \to \Omega^{-1} X \to X[1]$ and $Y \to Z \to \Omega^{-1} X \to Y[1]$. Passing to the singularity category, $I$ gets killed, along us to deduce an isomorphism between $\Omega^{-1} X$ and $X[1]$, and hence an isomorphism of triangles between $Y \to Z \to \Omega^{-1} X \to Y[1]$ and $Y \to Z \to X[1] \to Y[1]$. The latter is therefore distinguished, and so by shifting, $X \to Y \to Z \to X[1]$ is distinguished too.

Question 2 (Resolved in comments):

Next, it's proved that the functor is full, where he claims the following.

... no non-projective $A$-module is isomorphic in the derived category to an object of $K^b(A)$.

But this seems unlikely to me. If $M$ is a non-projective $A$-module then $M$ admits a projective resolution $P^* \to M$ and if $P^*$ can be made finite then it's an object of $K^b(A)$ quasi-isomorphic to $M$.

Question 3 (Resolved in answer):

Later on, Rickard proves that the functor is essentially surjective. Every object $X$ in $D^b(A) / K^b(A)$ is isomorphic to a complex of projectives $P^* = \cdots \to P^r \to P^{r+1} \to \cdots P^s \to 0$, where $r < 0$ and $P^*$ has zero homology in degrees less than $r$. The previous question already touched upon the fact that $P^*$ is, in fact, not in $D^b(A)$, so I'll ignore this. The natural map from $P^*$ to $\widetilde{P}^* = \cdots \to P^{r-1} \to P^r \to 0 \to \cdots$ is an isomorphism in $D^b(A) / K^b(A)$ since its mapping cone is a bounded complex of projectives. At this point Rickard writes

... and there is a complex
$$Q^* = \cdots \to P^{r-1} P^r \to Q^{r+1} \to \cdots \to Q^0 \to 0$$
which is the projective resolution of some module $M$ and whose natural map to $\widetilde{P}^*$ is an isomorphism in $D^b(A) / K^b(A)$. Thus $P^* \cong FM$.

This strikes me as simply rephrasing what has to be proved, in terms of projective resolutions rather than in terms of modules. Could someone clarify what is being done here?

答:
The first two questions have already been settled in the comments. For the last one:

First, let's pretend we're working in $\textbf{D}(A)/\textbf{K}^b(A)$ instead of  $\textbf{D}^b(A)/\textbf{K}^b(A)$. Then, the argument shows that within  $\textbf{D}(A)/\textbf{K}^b(A)$, any object $X$ of  $\textbf{D}^b(A)/\textbf{K}^b(A)$ is isomorphic to a complex $\ldots\to P^{r-1}\to P^r\to 0$ with no cohomology below $r$. In other words, it's isomorphic to a stalk complex $M[-r]$ concentrated in degree $r$, for some module $M$. This stalk complex is isomorphic to the degre-$0$ talk complex $(\Omega^{-r} M) [0]$ in $\textbf{D}^b(A)/\textbf{K}^b(A)$ which you see by taking a projective resolution $0\to M\to Q^{r+1}\to \ldots\to Q^0\to \Omega^{-r} M\to 0$ to the right. This is, I think, the argument Rickard presents.

You can eliminate the transit through  $\textbf{D}(A)/\textbf{K}^b(A)$ as follows: Given $X\in \textbf{D}^b(A)$, consider a bicomplex $T$ consisting of projective resolutions of the components of $X$. The totalization of this (unbounded!) bicomplex $T$ is an unbounded complex of projectives quasi-isomorphic to $X$, which you then go on to truncate as described. You can avoid expand+truncate step by already truncating the vertical projective resolutions in $T$ in such a way that the totalization of $T$ is projective except for the lowest non-zero entry (this will require truncating the projective resolutions at different places). This will give you a bounded complex $0\to M\to Q^r\to\ldots\to Q^s\to 0$ quasi-isomorphic to $X$, and hence $X\cong M[-r]$ in $\textbf{D}^b(A)/\textbf{K}^b(A)$ as before.

This is a bit sketchy - let me know if you need mor details.


问:
What is this operator ::== called and how does it work?
I'm learning program analysis and separation logic to create formulas expressing correctness of computer programs and I encountered these equations:

Is it wrong to interpret ::== as "Membership assignment"?

答: The symbol you are talking about is almost equal to the symbol $:=$ in mathematics. It means “is defined as”. You can see a further explanation of the concept behind here Stack overflow question on BNF. It means that the items on the LHS are defined as what is on the RHS for a programming language. More specifically, the term | then works as a comma, so for example you will have that objects such as basic proposition could be either A (atomic propositions in your image), or !A


问:
Can you pick a random natural number? And a random real number?
Is it possible to pick a random natural number? How about a random real number? Is the axiom of choice involved in this? </p>

答:
YES! (sort of)

Here is a construction (note it uses the axiom of choice):

Pick your favorite irrational number and call it $\alpha$. Now, the ergodic circle rotation  by $\alpha$ starting at $x$ is the set $\{x + n \alpha \pmod{1} : n \in \mathbb{Z}\}$. That is, you add (or subtract) $\alpha$ a bunch of times and then keep the fractional part. What's convenient about this is that the orbits form equivalence classes and never repeat, so each element in an equivalence class (with the starting point $x$ given) can be indexed by the value of $n \in \mathbb{Z}$.

OK, so, partition the inteval $[0, 1]$ (or $[0, 1)$ if that makes your life easier, whatever) into orbits of ergodic circle rotations, each with a given starting point $x$ (that is, a representative of the equivalence class of each orbit). We can do this if we assume the axiom of choice.

Now, generating a uniform random integer is easy!

Start by taking a random real number $y$ from $[0, 1]$. It must be in one of the orbits, since they partition the interval. Since you have a fixed starting value $x$ for the orbit, you have a well defined $n \in \mathbb{Z}$ such that $y = x + n \alpha$. So, your random integer is this $n$.

OK, so you might wonder whether this is "uniform" and with good reason. What is the probability of, say, picking the number $5$? Well, it doesn't exist. The set of real numbers $y$ corresponding to any fixed $n$ is not a measurable subset of $\mathbb{R}$, so, technically, no uniform distribution was involved; there weren't even any probabilities!

However, if "uniform" means to you (like it does to me) without bias toward any outcome (or without carrying any information), then this certainly fits the bill! It doesn't favor any integer over another.

And of course this can be extended by a bijection to any countable set, such as the natural numbers. And certainly you can now form a "random" real number by taking a random integer $x \in \mathbb{Z}$ and a random real $y \in [0, 1)$ and adding them together.


问:
Why does this Laurent Series Vanish?
Let $\Phi_{l}(z;q)$ denote the Lerch Transcendent and $Li_{l}(z)$ denote the Polylogarithm function. For $l \in \mathbb{Z}_{\geq 0}$, we have the following Laurent expansion
\begin{align}
\Phi_{-l}(z;0) + (-1)^{l} Li_{-l}(1/z) = \sum_{k \geq 0} k^{l} z^{k} + \sum_{k \leq -1} k^{l} z^{k}.
\end{align}
For $l \in \mathbb{Z}_{\geq 0}$, the left side will always be a sum of two rational functions which cancel. For example, $l = 0$ and $l = 1$ gives
\begin{align}
\frac{1}{1-z} + \frac{1}{(1-1/z)z} = 0 \quad \text{and} \quad \frac{z}{1-z} - \frac{1}{(1-1/z)^{2} z} = 0.
\end{align}
Looking at it another way, I wonder if there is some simple reason why the right side is zero, which can be written (as the meaningless sum)
$$ \sum_{k \in \mathbb{Z}} k^{l} z^{k} = 0 \quad \text{for}\quad l \in \mathbb{Z}_{\geq 0}.$$ Is there some way to interpret this expression and show that its value is zero?

(Compare this to the meaningless expression $\tfrac{1}{12} + \sum_{n \geq 1} n = 0$.)

答:
For nonnegative integers $\ell$ we have

$\Phi_{-\ell}(z;0) = \sum_{n=0}^\infty n^\ell z^n$

converging for $|z| < 1$
while

$(-1)^\ell {\rm Li}_{-\ell}(1/z) = \sum_{n=-\infty}^{-1} n^\ell z^n$

converging for $|z|>1$.  The combined Laurent series $\sum_{n=-\infty}^\infty n^\ell z^n$ diverges everywhere, so this part doesn't make sense.  However,
as for the rational functions

$\Phi_{-\ell}(z;0) = ((\frac{\partial}{\partial t})^\ell \frac{1}{1- e^{t} z})|_{t=0} $

$(-1)^\ell {\rm Li}_{-\ell}(1/z) = ((\frac{\partial}{\partial t})^\ell \frac{1}{e^t z - 1})|_{t=0}$

and of course $\frac{1}{1-e^tz} + \frac{1}{e^tz - 1} = 0$.

Title: Why does this Laurent Series Vanish?


问:
Definite integral - taking a constant term out added a power to it
We have $\int_{0}^{\infty}xe^{-\lambda x}dx$ = $\frac{1}{\lambda^2}\int_{0}^{\infty}ye^{-y}dy $
[Set $y=\lambda x$]

I have been so confused about the $\frac{1}{\lambda^2}$ term. Shouldn't it just be $\frac{1}{\lambda}$?

Because setting $y=\lambda x$, we have:

$\int_{0}^{\infty}\frac{y}{\lambda}e^{-y}dy $.

Taking the constant term $\frac{1}{\lambda}$ out, which would now give:

$\frac{1}{\lambda}\int_{0}^{\infty}ye^{-y}dy $

Am I misunderstanding something?

答: If you note that $\mathrm{d}y = \lambda\mathrm{d}x$, you will get the right answer


答: If $y=\lambda x$, then $dy=\lambda dx$ so you have $$\int_0^{\infty}xe^{-\lambda x}dx=\int_0^{\infty}\frac{y}{\lambda}e^{-y}\left(\frac{dy}{\lambda}\right)$$

Title: Definite integral - taking a constant term out added a power to it

Body:
We have $\int_{0}^{\infty}xe^{-\lambda x}dx$ = $\frac{1}{\lambda^2}\int_{0}^{\infty}ye^{-y}dy $
[Set $y=\lambda x$]

I have been so confused about the $\frac{1}{\lambda^2}$ term. Shouldn't it just be $\frac{1}{\lambda}$?

Because setting $y=\lambda x$, we have:

$\int_{0}^{\infty}\frac{y}{\lambda}e^{-y}dy $.

Taking the constant term $\frac{1}{\lambda}$ out, which would now give:

$\frac{1}{\lambda}\int_{0}^{\infty}ye^{-y}dy $

Am I misunderstanding something?


问:
If $a,b,c>0 $ then $\frac{9}{a+b+c} \leq 2\left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)$
Let $a,b,c>0$ then show taht $\frac{9}{a+b+c} \leq 2\left( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\ \right)\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

For show the RHS I do the following.
\begin{equation}
\frac{\frac{1}{a}+\frac{1}{b}}{2} \geq \frac{1}{\sqrt{ab}}\geq \frac{1}{\frac{a+b}{2}}=\frac{2}{a+b}
\end{equation}
Now notice that we can do the same for the other terms, now we have the following.
\begin{equation}
\frac{\frac{1}{b}+\frac{1}{c}}{2} \geq \frac{1}{\sqrt{bc}}\geq \frac{1}{\frac{b+c}{2}}=\frac{2}{b+c}
\end{equation}
And

\begin{equation}
\frac{\frac{1}{a}+\frac{1}{c}}{2} \geq \frac{1}{\sqrt{ac}} \geq \frac{1}{\frac{a+c}{2}}=\frac{2}{a+c}
\end{equation}
and sume the LHS and RHS of those inequalitys we get

\begin{equation}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)
\end{equation}
Therefore we have the RHS of the inequality Holds.
Now for the LHS I don´t know how I should apply the inequialitys, I tryed a lot of things for get a lower bound of $a+b+c$ and get a upper  bound of $\frac{1}{a+b+c}$ but I don´t get it.

I try look the expression like $$\frac{3}{\frac{a+b+c}{3}}$$ and apply the GM-AM inequality but it doesn´t work.Still I try use the fact that
$$a+b+c\geq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}} \geq \frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ac}{a+c} $$
And take inverses, but I don´t get the LHS.

Any advice or hint of where I should apply the GM-AM inequality or how Prove the LHS was very useful.

答: Just another way, if you're familiar with Karamata's inequality, the  inequality in question is just the majorization $(a, b, c) \succ \left(\frac{a+b}2,\frac{b+c}2, \frac{c+a}2 \right) \succ \left(\frac{a+b+c}3,\frac{a+b+c}3, \frac{a+b+c}3 \right)$ applied to the convex function $t \mapsto \frac1t$.


问:
Evaluate Fourier Series
Fourier series is quite a new concept for me and I'm not sure if I understand them correctly.

I am given a function $f:[-\pi, \pi] \to \mathbb{R} $ defined by

$$
x\mapsto
\begin{cases}
0  & \text{if $-\pi \le x \le -\pi /2$} \\
1 & \text{if $-\pi/2 \lt x \le \pi /2$} \\
0 & \text{if $\pi/2 \lt x \le \pi$}
\end{cases}
$$

the function is extended periodically to the whole of $ \mathbb{R} $, and has the Fourier series, $S_f(x)$. And I am supposed to evaluate $S_f(\frac{\pi}{2})$, and this is the part I am struggling with. So far I have found that the Fourier series for this function is (I might be wrong though):
$$S_f(x)= \frac{1}{2}+\sum_{n=1}^\infty \frac{2}{\pi n} \sin\Big(\frac{\pi n}{2}\Big) \cdot \sin(nx) $$ but after this part, I am stuck.

答:
By Dirichlet's theorem, the answer is $\frac{f\left( \left(\frac{\pi}{2}\right)^+\right) + f\left( \left(\frac{\pi}{2}\right)^-\right)}{2}=\frac12$.

But let's verify it, you made a minor mistake in your series  expression, rather than sines, it should be cosines.

$$S_f(x)= \frac{1}{2}+\sum_{n=1}^\infty \frac{2}{\pi n} \sin\Big(\frac{\pi n}{2}\Big) \cdot \color{red}\cos(nx) $$

Hence

$$S_f\left(\frac{\pi}{2}\right)=\frac{1}{2}+\sum_{n=1}^\infty \frac{2}{\pi n} \sin\Big(\frac{\pi n}{2}\Big) \cdot \cos\left(\frac{n\pi}2\right) =\frac{1}{2}+\sum_{n=1}^\infty \frac{\sin\left(n\pi\right)}{\pi n} =\frac12 + 0=\frac12$$

Title: Evaluate Fourier Series


问:
$T_p(E)/p^nT_p(E)$ is isomorphic to $E[p^n]$
In the case of an elliptic curve $E$ defined over a field $K$, I know that there is a good definition of the so-called Tate module, for every prime $p$, which is the $\mathbb{Z}_p$-module $T_p(E)=\underset{\underset{n}{\longleftarrow}}{\lim}E[p^n]$, where $E[p^n]$ is the $p$-torsion subgroup of $E$.

I would like to prove .

This seems analogy of $\mathbb{Z}_{p}/p^n\mathbb{Z}_{p}$ is isomorphic to $\mathbb{Z}
/p^n\mathbb{Z}$.

Is there any better generalization to module?

Could you tell me the proof of the title, and generalization of that proposition?

Thank you in advance.

答:
An element of the Tate module is an infinite vector $(P_n)$ of $p^n$ torsion points, such that $pP_n = P_{n-1}$.

Map $T_p(E) \to E[p^n]$ by projecting on the $n$th component. The kernel of this map is the set of vectors where the first $n$ coordinates are zero. I claim this is $p^n T_p(E)$.

Certainly $p^n T_p(E)$ has the property that the first $n$ coordinates are zero (the first $n$ coordinates of anything in $T_p(E)$ are killed by $p^n$).

Conversely, given an element of the form $(0, 0, 0, \ldots, 0, P_{n+1}, P_{n+2}, \ldots)$, we note that $P_{n+i}$ is really a $p^i$ torsion point (since the compatiability condition forces $pP^{n+1} = 0$. Therefore this element is $p^n(P_{n+1}, P_{n+2}, \ldots)$.

As a final note, underlying abelian group of the Tate module is (not naturally) isomorphic to the $p$-adics, hence the generalization you noted.

Note that it takes some care to write down the isomorphism, since the transition maps in the inverse limit defining the Tate module are multiplication by $p$ and the transition maps for the $p$-adics are projections. One way to get the isomorphism is to use the isomorphism of $\mathbb{Z}/p^n\mathbb{Z}$ with
$$
\frac{\frac{1}{p^n} \mathbb{Z}}{\mathbb{Z}},
$$
and now the transition maps become multiplication by $p$ on this side too.


问:
Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$
Prove if $x > 0$,  $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.

I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.

So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.

I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.

答: The inequality can be proved using elementary algebra. If $\dfrac{1}{8}x^2 < 1+\dfrac{1}{2}x\implies 2-2\sqrt{3} <x < 2+2\sqrt{3} < 8$, then both sides are positive, and square both of them: $\left(1+\dfrac{1}{2}x-\dfrac{1}{8}x^2\right)^2 < 1+x\iff \left(1+\dfrac{1}{2}x\right)^2-2\left(1+\dfrac{1}{2}x\right)\left(\dfrac{1}{8}x^2\right)+\left(\dfrac{1}{8}x^2\right)^2< 1+x\iff -x+\dfrac{1}{8}x^2 < 0\iff \dfrac{1}{8}x^2 < x\iff x^2 - 8x < 0\iff x(x-8) < 0\iff 0 < x < 8$, and this is true since we've shown that $8 > x > 0$, and if $\dfrac{1}{8}x^2 \ge 1+\dfrac{1}{2}x$, then it is obviously true since the left side is non-positive and the right side is non-negative. You're done !


问:
Evaluating $\lim_{n\to\infty}n\sin^{2n+1}(x)\cos(x)$

I have to find
$$\lim_{n\to\infty}n\sin^{2n+1}(x)\cos(x)$$

Answer given is $0$.

I tried applying L'Hospital's rule and this is what I got:
$$((2n+1)\sin^2n(x)-2\sin^{2n+2}(x)(n+1))/(-1/n^2)$$

How should I proceed from here?

Thanks in advance

答:
The limit is $0$, for any $x\in\Bbb R$:

If $x\in\frac\pi2+\pi\Bbb Z$, then $\cos(x)=0$, and therefore your limit is $0$.

Otherwise, $|\sin(x)|<1$, and therefore your limit is $0$ too, since$$|r|<1\implies\lim_{n\to\infty}nr^{2n+1}=0.$$


问:
Given that $\sum_{n}^{\infty} a_n$ converge, does $\sum_{n}^{\infty} a_n f(\sin n)$ converge?
Assume that $\sum_{n}^{\infty} a_n$ is a non-negative, convergent series. Let $f$ be a continuous function with domain $\mathbb{R}$. I have to figure out if the series

$\sum_{n}^{\infty} a_n f(\sin n)$

converge

diverge

not enough information to decide

The only possible test that is applicable here is I think basic comparison test, somehow using the fact that sin is bounded, but I have no idea how to proceed. Could anyone help me out?

答:
Hint: $$\{f(\sin n)|n\in\mathbb N\}\subseteq f([-1,1])$$

now you can apply one of many useful properties of continuous functions on closed sets.


问:
Joint Probability Mass Function: Wiki definition question
Do the following definitions for the joint probability mass function appear correct?  I put this together from Wiki's definitions.

where $(X_n = x_n) = \{s \in \Omega : X(s) = x_n\}$

$$
f_{X_1, ..., X_n}(a_1, ..., a_n) = P(X_1 = a_1, ... , X_n = a_n) = P(\{s \in \Omega : X_1(s) = a_1 \land \, ... \,  \land X_n(s) = a_n\}) = P(\{s \in \Omega : X_1(s) = a_1\} \cap ... \cap \{s \in \Omega : X_n(s) = a_n\}
$$

答:
Yes, this is correct. This is just a straightforward application of the definition. Note that, if $X_1, X_2$ are two independent Bernoulli random variables of parameter $\frac12$ (which I suspect was what you meant at first), you have, because of the independence,
that knowing that $s \in \Omega$ is such that $X_1(s) = 1$ tells you nothing about $X_2(s)$. Basically in this case you can see your $\Omega$ as $\{(s_i,t_i), i\in \{1,2\}$ with $X_1((s_i,t_i)) = \delta_{s_i,s_1}$ and $X_2((s_i,t_i)) = \delta_{t_i,t_1}$.
Hence
\begin{equation} \mathbb{P}(\{s \in \Omega: X_1(s) = 1\} \cap \{ s \in \Omega: X_2(s) = 1\}) = \mathbb{P}(\{s \in \Omega: X_1(s) = 1\}) \mathbb{P}(\{s \in \Omega: X_2(s) = 1\}).
\end{equation}
and you can check this manually using the particular form of $\Omega$ that I provided above.


问:
Differentiating $x^{x^{x^{...}}}$

How do I differentiate $$ x^{x^{x^{...}}}$$ with respect to $x$? (Note that $x$ is raised infinitely many times.)

My attempt: Let $y = x^{x^{x^{...}}}$. Taking logarithm of both sides we get $\ln y = y \ln x$ and let $f = y \ln x - \ln y$. Now $$\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = \frac{y^2}{x(1 - y \ln x)}$$
Is this approach correct? If not how do I proceed ?

Help would be appreciated. Thanks.

答:
The infinite tetration $y=x^{x^{x^{...}}}$ corresponds to
$$y=-\frac{W(-\log (x))}{\log (x)}$$ where $W(.)$ is Lambert function. Use the chain rule knowing that
$$\frac d{dt} W(t)=\frac{W(t)}{t (W(t)+1)}$$

Title: Differentiating $x^{x^{x^{...}}}$


问:
$z^3+z^2+z+k=0$, $z_1=-i$ what is the other complex root stating whether it is real or complex?
This question I have been stuck on for a while I figured out what $k$ is but I cant find the $z_2$  and $z_3$ this is my working out. Any help is much appreciated thank you. My work so far.The answer is $z_2=i$ and $_3=−1$ but I don't know how to get to it.

答: Once you have $z^3+z^2+z+1 = 0$, this is equivalent to $\frac{z^4 - 1}{z - 1} = 0$ as it is a geometric progression with first term $1$ and common ratio $z$. Note that $z = 1$ is not a solution.


问:
If every point of a topological space $X$ has a closed neighbourhood that is $T_2$ (as subspace), then $X$ is $T_2$.
I have tried the problem in the following way-

Let $x,y$ be two distinct points in $X$ and let $x\in A$ and $y\in B$ where $A,B$ are closed neighbourhoods of $x,y$ respectively and $T_2$. Let $U_1$ and $U_2$ be open sets such that $x\in U_1\subset A$ and $y\in U_2\subset B$.

Case 1: If $y\notin A$, then take $V=A^c$ and $U=U_1$. Then $U\cap V=\emptyset $ and $y\in V$ and $x\in U$.

Case 2: If $x\notin B$ then take $U=B^c$, $V=V_1$. Then $U\cap V=\emptyset$ and $x\in U$, $y\in V$.

But I'm stuck with the case $x\in B$ and $y\in A$ i.e. $x,y\in A\cap B$.

Can anyone help me in this regard? Thanks for your help in advance.

答:
In that last case: $A \cap B$ is Hausdorff, so there are open (in $A \cap B$) sets $O_1, O_2$ containing $x$ resp. $y$ that are disjoint.

So $O_i = (A \cap B ) \cap O'_i$, $O'_i$ open in $X$, $i=1,2$.
But then $U_1 \cap O'_1$ and $U_2 \cap O'_2$ disjoint open neighbourhoods of $x,y$ are as required.


答:
Using Complex conjugate root theorem

$i$ will be another root

If the third root is $a$ by Vieta's formula $$-\dfrac11=a+i+(-i)\iff a=-1$$

$$-\dfrac k1=i(-i)a=?$$

Title: $z^3+z^2+z+k=0$, $z_1=-i$ what is the other complex root stating whether it is real or complex?

Body: This question I have been stuck on for a while I figured out what $k$ is but I cant find the $z_2$  and $z_3$ this is my working out. Any help is much appreciated thank you. My work so far.The answer is $z_2=i$ and $_3=−1$ but I don't know how to get to it.


问:
Proving Continuity by Showing Open Pre-image
I am trying to show that $f(a,b,c):R^3\to R\;$  defined by $f(a,b,c) = a^2+b^2+c^2+1$ is contunious by showing that for every open set, $U$, in $R$ has an open preimage - $f^{-1}(U)$ is open in $R^3$.

All I really have so far is that as $U$ is open, we know that:

$\forall a \in U, \; \exists \varepsilon>0$ such that $N(a,\varepsilon)\subset U$

From here I'm super stuck. I think I want to get to:

$\forall (x,y,z) \in f^{-1}(U) \; \; \exists \varepsilon^{\prime}$ such that $N((x,y,z),\varepsilon^{\prime})\subset f^{-1}(U)$

And I think to do so, I will need to use the function itself, but I am not sure how to do so.

Any help would be greatly appreciated. Thank you!

答:
Take $U = (q-\epsilon,q + \epsilon)$. Then, $f^{-1}(U) = \{ (a,b,c) : q - \epsilon < a^2 + b^2 + c^2 + 1 < q + \epsilon \}$. That condition becomes
$$
q - 1 - \epsilon < a^2 + b^2 + c^2 < q -1 + \epsilon.
$$
This set describes either an open ball ($q - 1 - \epsilon < 0$) or an open ball minus a closed ball ($q - 1 + \epsilon \geq 0$). Either way, we obtain that $f^{-1}(U)$ is open.

Since it's possible to write any open subset of $\mathbb{R}$ as a union over sets of the above form, you get that $f^{-1}(U)$ is open for any $U \subseteq \mathbb{R}$ open.


问:
Binary $[31,22,5]$ code does not exist
Consider a binary $[31,22,5]$ linear code. There are $2^{31-22}=2^9$ cosets and we have $t=\frac{5-1}{2}=2$.

I am looking to find a contradiction, by considering the number of coset leaders with weights $0,1$ and $2$, and considering, for each coset, an upper bound for the number of words of weight $3$ contained in it.

Any help on finding this two values? Thanks in advance!

答:
You say you want to count the weight 3 codewords in each coset; $\mathcal{C}$ contains no weight 3 codeword, and also no weight 1 coset can contain a weight 3 word (I leave it to you to show this).

For a weight 2 coset of the form $u + \mathcal{C}$ with $\mathrm{wt}(u) = 2$, the only way to get a weight 3 word is from a $c \in \mathcal{C}$ with $\mathrm{wt}(c) =5$ and $\mathrm{wt}(u \cap c) = 2$. So if $\{i,j\}$ are the coordinates where $u_{i}=u_{j}=1$, you want to consider how many weight 5 codewords $c$ there can possibly be with $c_{i}=c_{j}=1$, all at distance at least 5 apart. To count these, notice that these codewords cannot be at distance exactly 5, the distance between them must be even since they have the same weight.

Title: Binary $[31,22,5]$ code does not exist


问:
How can I calculate win percentage needed based on risk/reward ratio?
Let's say I have calculated the probability of a win in two games to be $60\%$.
I have $2$ bets I am looking at.

For the first bet I need to risk $165$  to win $85$.

For the second bet I need to risk $263$  to win $237$.

Risk/Reward for first bet is $1.94$ and for second $1.1$.

If I calculate first bet to be $(0.6 * 85) - (0.4 * 165) = -15$ it is losing.

If I calculate second bet to be $(0.6 * 237) - (0.4 * 263) = 37$ it is winning.

But how do I get the minimum win percentage I need for a bet based of money I am risking?

I get confused because the risk is more than reward. All calculations I have seen assume risk is less than reward.

答:
I feel like in your case you would win the amount over the bet. So you would do something like this $win\implies 165+85$ and $lose\implies -165$ so you’re expected value should be something like this assuming that $p=P(win)$ $$\mathbb{E} (bet)= p(165+85)-(1-p)(-165)>0 $$
Solving this for p will give you the interval such that you can have an expected positive value of money after betting. Please note that however this is only the probability, people might have different preferences associated to how much they’d like to risk.


问:
Prove $(AB)^{k} = A^{k}B^{k}$
Apologies if this question was posed already.

I need to prove that $(AB)^{k} = A^{k}B^{k}$ holds if $AB=BA$. After trying it myself, I looked at the solution and found it quite strange.

"Use induction: $(AB)^{k} = A^{k}B^{k}$ is true for $k=1$ since since $AB=BA$. Assume $(AB)^{k-1}=(BA)^{k-1}$ and prove it true for k:
$$ (AB)^{k} = (AB)^{k-1} AB $$
$$  = A^{k-1}B^{k-1} AB $$
$$  = A^{k-1}A B^{k-1} B $$
$$  = A^{k} B^{k} $$"

I don't understand how this really proves anything. We had to assume $(AB)^{k-1}=(BA)^{k-1}$, how does if follow then that $(AB)^{k} = A^{k}B^{k}$  holds?

答:
$$(AB)^k$$
$$=ABAB\dots AB$$
$$=AABBAB\dots AB$$
$$= AABABB\dots AB$$
$$\vdots$$
$$=AA\dots ABB\dots B$$
$$=A^kB^k$$

Title: Prove $(AB)^{k} = A^{k}B^{k}$


答:
As @lab bhattacharjee already answered, the problem is simple if $k$ is a real number and you have the three solutions.

Now, if $k$ is complex, use the long division to have the quadratic
$$\frac{z^3+z^2+z+k} {z+i}=(i k+(1-i)) z^2+(k-i) z-i k$$ for which
$$\Delta=-3 k^2+(4+2 i) k-1$$ Let $k=a+ib$ to make
$$\Delta=(-3 a^2+4 a+3 b^2-2 b-1)+2 (a+2 b-3ab)i$$ If the imaginary part is equal to $0$, then $b=\frac{a}{3 a-2}$ and then
$$\Delta=\frac{-27 a^4+72 a^3-72 a^2+32 a-4}{(2-3 a)^2}$$and the quartic write
$$-9\left(a-\frac{2+\sqrt{2}}{3} \right) \left(a-\frac{2-\sqrt{2}}{3}
   \right)(3 a^2-4 a+2)$$ So, continue looking at the sign of
$$-\left(a-\frac{2+\sqrt{2}}{3} \right) \left(a-\frac{2-\sqrt{2}}{3}
   \right)$$


答: @Exodd has already noted a typo in the statement of the inductive step's inductive hypothesis. Another approach to the inductive step, one that minimizes the rearrangements, is$$(AB)^k=A^kB^k\implies(AB)^{k+1}=A(BA)^kB=A(AB)^kB=AA^kB^kB=A^{k+1}B^{k+1}.$$

Title: Prove $(AB)^{k} = A^{k}B^{k}$

Body:
Apologies if this question was posed already.

I need to prove that $(AB)^{k} = A^{k}B^{k}$ holds if $AB=BA$. After trying it myself, I looked at the solution and found it quite strange.

"Use induction: $(AB)^{k} = A^{k}B^{k}$ is true for $k=1$ since since $AB=BA$. Assume $(AB)^{k-1}=(BA)^{k-1}$ and prove it true for k:
$$ (AB)^{k} = (AB)^{k-1} AB $$
$$  = A^{k-1}B^{k-1} AB $$
$$  = A^{k-1}A B^{k-1} B $$
$$  = A^{k} B^{k} $$"

I don't understand how this really proves anything. We had to assume $(AB)^{k-1}=(BA)^{k-1}$, how does if follow then that $(AB)^{k} = A^{k}B^{k}$  holds?