问:
Is this step mathematically allowed?
In my signal and systems course, my professor claims that the energy of a signal $x(t)$ over an infinite and continuous time interval is:
$$
E_{\infty}=\lim _{T \rightarrow \infty} \int_{-T}^{T}|x(t)|^{2} d t=\int_{-\infty}^{+\infty}|x(t)|^{2} d t
$$
My concern is about the substitution of $\infty$ in place of $T$ before evaluating the integral. Is this always correct or is it mathematically more correct to evaluate the integral then compute the limit?

答:
Many people write $\int_{-\infty}^{\infty} |x(t)|^2dt$ as $\lim \limits_{T \to \infty} \int_{-T}^{T} |x(t)|^2dt$ because they believe that using infinity as a number is not proper, so they write it as a limit. However, since $$\int_{-\infty}^{\infty} |x(t)|^2dt = \lim \limits_{T \to \infty} \int_{-T}^{T} |x(t)|^2dt$$

is true, that must mean that the converse,
$$\lim \limits_{T \to \infty} \int_{-T}^{T} |x(t)|^2dt =\int_{-\infty}^{\infty} |x(t)|^2dt $$

is true as well.

Title: Is this step mathematically allowed?


问:
Exhaustive events and the law of total probability
Suppose $X$, $Y$, and $Z$ are exhaustive events with $Y$ and $Z$ being mutually exclusive. Can $Y$ and $Z$ be treated as a partition? Also, as a follow-up, can the law of total probability be applied to $X$ this way: $P(^cX) = P(^cX \cap Y) + P(^cX\cap Z)$, where $^cX$ is the complement of $X$?

答:
$Y$ and $Z$ are mutually exclusive, but may not be exhaustive on their own, so therefore they will not form a partition of $X\cup Y\cup Z$, unless $X\smallsetminus(Y\cup Z)$ is empty.

Still, $\{Y, Z, X\smallsetminus(Y\cup Z)\}$ would be a partition over that outcome space, and so therefore...

ContentLicense: CC BY-SA 4.0


问:
Prove that $\forall k$ and for any $g \in G$, the sets $A = \{a \in G : a^k = g\}$, $B = \{b \in H : b ^k = f(g)\}$ have the same number of elements.

Let $f : G \to H$ be a group isomorphism. Show that for any integer $k$ and for any $g \in G$, the sets $A = \{a \in G : a^k = g\}$ and $B = \{b \in H : b^k = f(g)\}$ have the same number of elements.

I've been trying this problem for quite a while but to no avail. I don't understand what the saying.

答:
Note that, for any $a \in A$, $$f(a)^k = f(a^k) = f(g)$$
so $f(a) \in B$, and then $f(A) \subseteq B$. Now, apply the same argument with $f^{-1}$ to show that $f^{-1}(B) \subseteq A$. Thus $f(A) = B$.


问:
Simplify the probability of events
Given three exhaustive events $X, Y$ and $Z$, $X$ and $Y$ are mutually exclusive, and $X$ and $Z$ are independent. If $P(X) = m, P(Y) = n$ and $P(z) = m$, express the following in terms of $m$ and $n$: $P(Z|X\cup Y)$

I'm not sure how to simplify this except using the fundamentals of conditional probability:

$$P(Z|X\cup Y) = \frac{P(Z \cap (X \cup Y))}{P(X \cup Y)}$$

I also know that $P(X\cup Y) = P(X) + P(Y) = m + n$.

Any assistance is much appreciated.

答:
$P(Z\cap(X \cup Y))=P((Z\cap X)\cup (Z \cap Y)) = P(Z \cap X) + P(Z \cap Y)$ because $X$ and $Y$ are mutually exclusive.

$P(Z \cap X) + P(Z \cap Y) = P(Z)P(X) + P(Z \cap Y) = m^2 + P(Z \cap Y)$ as $X$ and $Z$ are independent.

$P(Z \cap Y)=P(Z \cap \bar{X})$, as $X$, $Y$, and $Z$ are exhaustive. But $P(Z \cap \bar{X})=P(Z)P(\bar{X})=m(1-m)$ as $Z$ and $X$ (and therefore $\bar{X}$) are independent.

So, $P(Z|X \cup Y) = \frac{m^2 + m(1-m)}{m + n}=\frac{m}{m+n}$.

Title: Simplify the probability of events


问:
Which numbers of [0,1) have a unique base g expansion?
Good evening,

i know that is question is rather standard, but unfornunately I have not much knowledge of number theory.

Take $2 \leq g\in \mathbb{N}$. I know that every $x \in [0,1)$ can be represented as $x = \sum_{k=1}^\infty \frac{x_k}{g^k}$ with $x_k \in \{0, .. , g-1\}$.

I know that this repesentation is not unique (for example if $x_k= g-1 ~~ \forall k > N_0 \in \mathbb{N}$). I know that there are  $x \in [0,1)$ with a unique representation (for example if $x_k= 0 ~~ \forall k > N_1 \in \mathbb{N}$).

Here is my question: Is there any characterisation of the $x \in [0,1)$ which have a unique base g expansion?

I am very thankful for any reference! With best regards.

答:
Here's a proof that supports the characterization stated in the comments:
The $g$-ary representations for numbers $x\in[0,1)$ are of the form
$$\tag1 x=\sum_{k=1}^\infty x_kg^{-k}$$
with $x_k\in\{0,\ldots,g-1\}$. Let $(x_k)$ and $(y_k)$ be two distinct digit sequences in $\{0,\ldots,g-1\}$ representing the same number and wlog. $(x_k)$ lexically precedes $(y_k)$, say $x_n<y_n$ whereas $x_k=y_k$ for $k<n$.
Then
$$\begin{align}0&=\sum_{k=1}^\infty y_kg^{-k}-\sum_{k=1}^\infty x_kg^{-k}\\&=\sum_{k=1}^\infty (y_k-x_k)g^{-k}\\&=g^{-n}(y_n-x_n)+g^{-n}\sum_{k=1}^\infty (y_{n+k}-x_{n+k})g^{-k}\\&\stackrel{(1)}\ge g^{-n}+g^{-n}\sum_{k=1}^\infty (y_{n+k}-x_{n+k})g^{-k}\\&\stackrel{(2)}\ge g^{-n}+g^{-n}\sum_{k=1}^\infty(0-(g-1))g^{-k}\\
&=g^{-n}-g^{-n}(g-1)\sum_{k=1}^\infty g^{-k}
\\&=g^{-n}-g^{-n}(g-1)\frac1{g-1}=0.\end{align}$$
Since we end up with equality, we must have equality at $(1)$ and $(2)$.
But equality at $(1)$ holds if and only if $y_n=x_n+1$ and at $(2)$ if and only if for all $k\ge1$ we have $y_{n+k}=0$, $x_{n+k}=g-1$.
Note that in this case we additionally obtain
$$\tag2x = \sum_{k=1}^\infty y_kg^{-k} =\sum_{k=1}^n y_kg^{-k}=\frac{\sum_{k=1}^ny_kg^{n-k}}{g^n}.$$
Likewise, if $x=\frac m{g^n}$ in shortest trerms (i.e. with $m$ not divisible by $g$), the $g$-ary integer representation of $m$ ands in a nonzero digit, gives us a terminating representation with $y_n>0$ of $x$ via $(2)$, from which we obtain two $g$-ary representations as above.

This shows that a number $x\in[0,1)$ has two different $g$-ary representations if and only if $x=\frac m{g^n}$ with $0<m<g^n$.

ContentLicense: CC BY-SA 3.0


问:
there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$.
Let $X$ be a reflexive Banach space of infinite dimension.

a) Prove that there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$.

b) Let $x_n$ be a sequence such that $\forall f \in X' \quad \exists \lim\limits_{n\to\infty} f(x_n)<\infty$ .Prove that $x_n$ converges weakly.

c) Find a Banach space non riflexive where b is false.

My idea:
the weak closure of $S^1$ is the closed unit ball, thus $0\in$ the weak closure of $S^1$. But to prove that there exists a sequence in $S^1$ that converges weakly to 0 we must prove that the weak topology is metrizzable.Is it always true in riflexive spaces?

I would appreciate any idea on the solution.Thank you in advance.

答: What you said is not true in general, but there's a way around it. Find a closed subspace $S \subset X$ which is separable (and it is reflexive also by a common result). To do this, pick $\{x_{n}\}_{n \in \mathbb{N}}$ linearly independent vectors in $X$ and consider the subspace spanned by them. This should be separable (why? considering rational scalings of all $\{x_{n}\}_{n \in \mathbb{N}}$). Now since $S$ is separable and reflexive, $S^{*}$ is separable (this is a theorem, not trivial), which shows that $B_{X} \cap S$ is metrizable. Apply your argument to $B_{X} \cap S$.


问:
tensor product of ideal $I$ and $R/I$
Let $R$ be a commutative ring, $I$ an ideal. View $I$ and $R/I$ as $R$-module. Then I want to know why $I \otimes R/I=I/I^{2}$.

I know there is a property in many books that $R/B \otimes C=C/BC$. But when I get down to element level, let $a \in I$, $b \in R/I$, then $a\otimes b=1\otimes ab=1\otimes 0=0$.

Where is the problem in this calculation?

答:
The problem is the step $a \otimes b$ to $1 \otimes ab$, because it's only valid when 1 is an element of $I$.

However if 1 were in $I$, then $R / I$ would be zero, aligning with your calculation.

Title: tensor product of ideal $I$ and $R/I$


答:
As you say, this comes from the fact that $R/I \otimes_R M = M/IM$, applied to $M = I$ as an $R$-module. Concretely, the map
$$
[r] \times i \in R/I \times i \mapsto [r\cdot i] \in I/I^2
$$
is bilinear and $R$-balanced, hence induces a map
$$
[r]\otimes i \in R/I \otimes I \mapsto [ri] \in I/I^2.
$$

It is surjective, since $[a]$ is the image of $[1] \otimes a$. Moreover, if $[ri] = 0$, then $ri = \sum c_jd_j$ with $c_j,d_j \in I$ and

$$
[r] \otimes i = [1] \otimes ri = \sum_{j} [1] \otimes c_jd_j = \sum_j
[c_j] \otimes [d_j] = 0.$$

Hence the map is also mono, thus an iso.

The problem with your calculation is that your are passing coefficients from one side to the other in the wrong way: we have $[r] \otimes si = [sr] \otimes i$ for each $s \in R$. Not every element of $I$ is of the form $si$. We can't write $[r] \otimes a = [ar] \otimes 1$ because $1 \not \in I$, unless $I = R$.

Title: tensor product of ideal $I$ and $R/I$

Body:
Let $R$ be a commutative ring, $I$ an ideal. View $I$ and $R/I$ as $R$-module. Then I want to know why $I \otimes R/I=I/I^{2}$.

I know there is a property in many books that $R/B \otimes C=C/BC$. But when I get down to element level, let $a \in I$, $b \in R/I$, then $a\otimes b=1\otimes ab=1\otimes 0=0$.

Where is the problem in this calculation?


问:
Proving the existence of minimal distance
Let $(X,d)$ be a metric space and $x \in X$. Let $Y$ be a compact subset of $X$. I want to show that there exists a point $\hat{y} \in Y$ such that:

$$d(x,\hat{y}) = \inf_{y\in Y}\{d(x,y)\}$$

not using the sequence characterization of the compactness of $Y$, but rather the topological defintion, that is every open subcover of $Y$ has a finite subcover.

My attempt was to derive a contradiction. Assume such a point $\hat{y}$ does not exits. Then consider the open cover of $Y$:

$$ U_n = \{z \in X| d(z,y) > \inf_{y \in Y} \{d(z,y)\}+\dfrac{1}{n}\}$$

Can we show that it has no finite subcover? Is there any other way to prove the statement using the topological defintion of compactness?

答:
Your way looks like it works. There is another, slicker way to go that becomes more natural once you've learned a bit about compactness. You need a few facts

The function $d(x,-) : X \to \mathbb{R}$ defined by taking the distance to a fixed point $x$ is continuous.

For any continuous function of topological spaces $f:A \to B$ and compact set $K \subset A$, the image $f(K) \subset B$ is compact.

Any compact subset of a metric space is closed and bounded. In particular, a compact subset of $\mathbb{R}$ contains its maximum (and minimum).


问:
Construction of symmedians but different.
Given $2$ lines $l_1$ and $l_2$ intersecting at point $F,$ let $A$ and $B$ on line $l_1,$ construct point $E$ on $l_2$ such that $\angle AEF=\angle (BE,EF) .$

Here's the diagram of  the problem:

For the construction, I thought of taking point $D$ such that $BD\perp DF.$ But couldn't proceed much.

答:
The construction:
Take any point $D$ on $l_2$. Now intersect $AD$ and $BD$ with any line through $F$ to get two points $X$ and $Y$ respectively. Now intersect $AY$ and $BX$ at $P$. Then intersect $DP$ with $l_1$ at $Z$. Draw the circle with diameter $FZ$. Then the intersection of this circle with $l_2$ is our point $E$.

Proof that it works:
The point $Z$ is just the harmonic conjugate of $F$ with respect to segment $AB$, so the circle with diameter $FZ$ is the Apollonian circle through $F$. It's well known that for any point $E'$ on this circle, $\angle AE'B$ is bisected by $E'Z$, and more importantly $E'F$ is the external angle bisector of $\angle AE'B$. So intersecting this with $l_2$ gives the desired point.


问:
Question about null-homotopic function
How to prove that every continuous function $f:M\to X$ from the Möbius strip $M$ into a simply connected space $X$ is null-homotopic? Thanks in advance.</p>

答:
Since $M \simeq S^1$, take WLOG $f:S^1 \to X$. By definition a loop in $X$ is a path $g:I \to X$ such that $g(0)=g(1)$. Notice $S^1=I/\sim, 0\sim1$. So any map $f:S^1 \to X$ is just a loop and induces a loop in $\pi_1(X)$.

$\pi_1(X)=0$ so the loop has to be homotopic to a constant (trivial) loop and hence by definition null-homotopic.

Title: Question about null-homotopic function


问:
How to find $\int\sqrt{(26x-x^2)}dx $
How do I find $\int \sqrt{(26x-x^2)} dx $

Is this an integration by parts question?

Thanks,

--Nick

Title: How to find $\int\sqrt{(26x-x^2)}dx $

Body:
Integrate as follows

\begin{align}
\int \sqrt{26x-x^2} dx &=
\int \frac{\sqrt{26x-x^2}}{2(x-13)}d[(x-13)^2]\\
&=\frac12(x-13)\sqrt{26x-x^2}+ \frac{169}2\int\frac1{ \sqrt{26x-x^2}}dx\\
&= \frac12(x-13)\sqrt{26x-x^2}+\frac{169}2\sin^{-1}{\frac {x-13}{13}}+C\\
\end{align}


问:
Alternatives to the politician theorem
A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R and S are invited to a birthday party. We are told that:

each pair of guests have exactly one common friend

A and G only have one common friend: C

A and G are friends

I and C only have one common friend: A

How many friends does the most popular (in terms of friends) guest have ?

Invoking the politician/friendship theorem: it is straightforward that A has the most friends (18), as he is the common friend to all the guests.

My question is: can anyone think of an alternative solution to this problem ?

答:
I had never encountered this surprising theorem - thanks for posting.

In order to give a concrete example, the question adds some extra conditions. Under these $A$ has at least three friends: $G$ (by the second condition) and $I$ & $C$ (by the third condition). Therefore it is $A$ who is the politician. Note that this implication was invalid until Alex Ravsky added the second condition. Also the first condition ($C$ the common friend of $A$ & $G$) is redundant.

Shorn of these conditions, the ask is for alternative proofs from the one in "The Book in which God keeps the most elegant proof of each mathematical theorem". That's a fair question but a tall ask. Erdős & spectral analysis have both proved very powerful in graph theory, and we are asked to turn our backs on both...

Here's a different approach based on triangles. It doesn't lead to the full result yet, but I will keep playing with it, and maybe it will inspire someone else.

With no assumptions about regularity for now, for any vertex $A$ with degree $d_A$, there are $C(d_A,2)$ unordered pairs of adjacent vertices. We are going to triage each of the $\sum _{A \in G} C(d_A,2)$ cases. Suppose $B$ & $C$ are two vertices adjacent to $A$. There are two possibilities:

(1) If $B$ & $C$ are non-adjacent, then as $A$ is the "sole mutual friend" of $B$ & $C$, there is a one-to-one-correspondence to the set of non-edges in $G$, of which there are $C(n,2) - m$ items.

(2) If $B$ & $C$ are adjacent then, with $A$, the three vertices form a triangle $ K_3$. Because each vertex is the "sole mutual friend" of the other two, each edge in $G$ belongs to exactly one such triangle. And each such triangle is discovered exactly three times in our examination, depending on which of the three vertices is discovered first. So if $m$ is the number of edges in $G$, there are hence $3 \times m/3$ = $m$ items to be counted here.

Adding these two populations: $m + C(n,2) - m = C(n,2)$

So $\sum_{A \in G} C(d_A,2) = C(n,2)$. [Equation *]

If we require also that $G$ is $k$-regular, then we derive:

$n \times C(k,2) = C(n,2)$

=> $n = k^2 - k + 1$, which is indeed an equation from The Book, so we are on the right track.

Equation * is satisfied happily by all the actual (windmill) solutions. If one "politician" vertex $A$ has degree $2p$, and the other $2p$ vertices have degree $2$, for any $p \ge 0$, then $n = 2p+1$. Then:

$C(2p,2) + 2p \times C(2,2) = 2p\times (2p-1)/2 + 2p \times 1 = p(2p-1)+2p = p(2p+1) = C(n,2)$.

Title: Alternatives to the politician theorem


问:
$1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^\times$.
I have seen this problem:

Let $p$ be an odd prime and let $n$ be a positive integer. Use the Binomial Theorem to show that $(1+p)^{p^{n-1}}\equiv 1 \mod p^n$ but $(1+p)^{p^{n-2}}\not\equiv 1 \mod p^n$. Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^\times$.

be posted but no one has given a complete answer for the last part:

Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^\times$.

Can someone give an explanation for how this comes from the previous part?

答: Define $\pi$ and $f$, respectively, on $\Bbb N$ and $\Bbb N\times\Bbb N$ by $f(x,y)=(\pi(x,y),y+1)$ and $$(1+xp^y)^p=1+\pi(x,y)p^{y+1}\;\;\;\;\;;\;\;\;\;\;\pi(x,y)=\sum_{r=1}^p\frac{\binom{p}{r}}{p}x^rp^{y(r-1)}\;\;\;\;\;\;\;$$ Letting $\pi_n(x,y)$ denote the   first coordinate of $f^n(x,y)$ we must have $$(1+xp^y)^{p^n}=1+\pi_n(x,y)\cdot p^{y+n}$$ where $f^n:=f\circ\cdots\circ f\;\;\text{(n copies of $f$)}$ denotes the function composition of $n$ copies of $f$. $$\therefore\;(1+p)^{p^{n-1}}=1+\pi_{n-1}(1,1)\cdot p^n\equiv 1\mod{p^n}$$


问:
Where to start when learning math (again)?
I have a few questions I hope you can help me answer.

First, I'll introduce myself. I'm a finance undergraduate student in Australia, but I'm originally from Norway. Throughout school I always loved math, but I ended up studying finance. The last year or so I have started to realise that I should have done Computer Science or Engineering in stead, as I would like to see myself in a quant role when I finish studying. Even so, I've decided to finish this finance degree.

The last month or so I've started programming in C++, and refreshing up on my maths knowledge (Khan Academy, algebra, precalculus). My problem is I don't know where to go from here, and what order I should be learning the different branches of math.

I've picked up a book on linear algebra today; "Linear algebra and its applications" by Gilbert Strang. Do you think this is an alright place to start? Or do I start with basic calculus? And where can I go next? Differential equations?

I hope some of you can help me in the right direction. Thanks.

答: I'd recommend you check out the Mathematics section here for free online video courses offered by some of the top institutions.</p>


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

So invoking Chebyshev's inequality we have

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

答:
I have done a little more research - here is my answer.

Given $X_1,\dots,X_n\stackrel{iid}{f}(x;\mu,\sigma^2)$, we know that by the weak law of large numbers:

$$\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2\stackrel{p}{\longrightarrow} \sigma^2$$

By the continuous mapping theorem, the square root function preserves continuity, thus

$$\sqrt{\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2}\stackrel{p}{\longrightarrow} \sqrt{\sigma^2}$$

$$\Rightarrow \sqrt{S^2}\stackrel{p}{\longrightarrow}\sigma$$

It was more theory heavy and that is where I am weak, this has been a good learning experience.


答:
Let $P$ and $Q$ be points on the line $l_{2}$ such that $AP\perp l_{2} $ and $BQ\perp l_{2}$. Extend $AP$ and $BQ$ to points $R$ and $S$ such that $AP=PR$ and $BQ=QS$. Now draw $AS$ and $BR$. These two segments will intersect at some point on the line $l_{2}$.That point is our desired point $E$.

Reason : Let $BR$ intersect $l_{2}$ in point $T$.Draw $AT$ and $ST$. Observe that $\triangle ATR$ and $\triangle BTS$ are both isosceles and $\angle ART=\angle SBT$; Thus, $\triangle ATR\sim \triangle BTS$. Hence $\angle ATP=\angle STQ$ and thereafter $AS$ will intersect $l_{2}$ in point $T$.Now $\angle ATP=\angle BTQ$ and hence Point $T$$=$ Point $E$.

Title: Construction of symmedians but different.

Body:
Given $2$ lines $l_1$ and $l_2$ intersecting at point $F,$ let $A$ and $B$ on line $l_1,$ construct point $E$ on $l_2$ such that $\angle AEF=\angle (BE,EF) .$

Here's the diagram of  the problem:

For the construction, I thought of taking point $D$ such that $BD\perp DF.$ But couldn't proceed much.


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

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

答:
Too long for comments.

Starting from the point where @Phicar wrote
$$S_n=2\sum _{k= n}^\infty\left (\frac{n-1}{n}\right )^{k-1}\binom{k+1}{2}$$ we have
$$S_n=\sum_{k= n}^\infty k (k+1) \left(\frac{n-1}{n}\right)^{k-1}=(n-1)^{n-1} n^{3-n} (5 n-3)$$ and for large values of $n$
$$S_n=\frac 1e\left(5 n^3-\frac{n^2}{2}-\frac{n}{24}+\frac{1}{16}+\frac{95}{1152 n}+\frac{917}{11520
   n^2}+O\left(\frac{1}{n^3}\right)\right)$$ which seems to be quite different; however, none of these match the numbers that you provided as examples.


问:
using diagonals of the a trapezoid inside a triangle you can find the midpoint of the bases
In the following triangle, AB is parallel to DE. F is an intersection point of the segments BD, AE and CG. Is it true that AG = GB?

答:
In the following, I have incorrectly assumed D and E are the midpoints of CA and CB respectively. Please jump right to the section labeled  as "CORRECTED version".

We can play withit using areas. The respective regions are labelled as shown.

(1) + (2) = (3) = (6) = (8)+ (7)

∴  (1) + (2) + (3) = (6) + (8)+ (7) = x, say

$\dfrac {x + (4)}{x +(5)} = \dfrac {(4)}{(5)}$

This gives (4) = (5) and the required result follows

CORRECTED version

Let C(F)G cut DE at H. Suppose that DH : HE = p : q.

Then, (1) by similar triangles, AG = kp and GB = kq for some non-zero constant k; (2) by triangles with the same altitude, [FDH] = hp and [FEH] = hq for some non-zero constant h.

Let [AFG] = x and [BFG] = y. Then, by considering areas of similar objects, $x = hq \times (\dfrac {kp}{q})^2$ and $y = hp \times (\dfrac {kq}{p})^2$.

Therefore, $\dfrac {x}{y} = … = (\dfrac {p}{q})^3$

Since, by considering area of triangles with the same altitude, $\dfrac {x}{y} = \dfrac {kp}{kq}$

From $(\dfrac {p}{q})^3 = \dfrac {kp}{kq}$, we have $p = q$ after rejecting the negative quantities.


问:
Are $D_{20}$ and $\mathbb{Z}_{2} \times \mathbb{Z}_{10}$ isomorphic?

Are  $D_{20}$ and $\mathbb{Z}_{2} \times \mathbb{Z}_{10}$ isomorphic?

For $D_{20}$ I am using the notation $D_{2n}$ and $\mathbb{Z}_{2}, \mathbb{Z}_{10}$ are additive groups.

I think not, and my justification is this: In group $\mathbb{Z}_{2} \times \mathbb{Z}_{10}$ we have 2 elements of order 2, namely, $(1,5) $ and $(0,5)$; while in $D_{20}$ there are 10 rotations and 10 flips, where each flip has order 2, therefore they are not isomorphic. Is that so?

答: $D_{20}$ has exactly $10$ reflections but $\Bbb Z_2\times\Bbb Z_{10}$ has only $\it{three}$ elements of order $2$, namely $(0+2\Bbb Z,5+10\Bbb Z),(1+2\Bbb Z,5+10\Bbb Z),(1+2\Bbb Z,0+10\Bbb Z)$, and thus $\require{cancel} D_{20}\;\cancel{\approx}\;\Bbb Z_2\times\Bbb Z_{10}$.


问:
How does one derive the Fourier transform of the Ramp function?
One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $\frac{1}{2} [\delta(t) - \frac{1}{\pi t}  ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $\frac{i}{4\pi} \delta'(t) - \frac{1}{4\pi^2 t^2} $

I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch!

答: If I am right Fourier analysis is applicable only to bounded or finite integral or stable system (Dirichlet Conditions). And ramp function is not a bounded signal. So we can't apply Fourier transform for ramp function.</p>


问:
Subgroups of $PSL_2(\mathbb Z/m\mathbb Z)$
Let $G$ and $H$ be two (finite) groups. Say that $G$ is "involved" in $H$ if $G$ is a quotient of a subgroup of $H$.

The question is the following : prove that, for every $m$, there is a prime $p$ and $r\geq 1$ such that any group involved in $PSL_2(\mathbb Z/m\mathbb Z)$ is involved in $PSL_2(\mathbb Z/p^r\mathbb Z)$.

Of course, it is enough to do it only for $PSL_2(\mathbb Z/m\mathbb Z)$ itself. It's tempting to use the Chinese Remainder Theorem : this group is a product of groups of the form $PSL_2(\mathbb Z/p_i^{r_i}\mathbb Z)$. And then I'm blocked... Am I missing something ?

答:
I think the question is not correctly worded.  As it is currently worded, it is impossible.

Take m = 77, then as you say PSL(2,Z/77Z) ≅ PSL(2,7) × PSL(2,11).  However, the non-abelian simple groups involved in PSL(2,Z/prZ) are at most PSL(2,p) and PSL(2,5).  In particular,  PSL(2,7) × PSL(2,11) is never involved in PSL(2,Z/prZ).


问:
Finding bounds on $\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}$
I am supposed to approximate the value of $\lim_{n\to \infty}\sum_{k=0}^{n}1/\sqrt{n^2+k}$. If it were $k^2$, then it could be rewritten as an integral, which would come out to be $\ln(\sqrt{2}+1)$.

$$\lim_{n\to \infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}\lt \lim_{n\to \infty}\frac{n+1}{n}=1$$

So the limiting sum should lie in the interval $(0,1)$. Is this correct? Thanks.

答:
If you use generalized harmonic numbers
$$S_n=\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}=H_{n^2+n}^{\left(\frac{1}{2}\right)}-H_{n^2-1}^{\left(\frac{1}{2}\right)}$$ Now, using asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2\sqrt{p}}+O\left(\frac{1}{p^{3/2}}\right)$$ apply it twice and continue with Taylor series to obtain
$$S_n=1+\frac{3}{4 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$

For example, the "exact" value of $S_{10}$ is $\sim 1.07386$ while the above truncated series gives $\frac{859}{800}= 1.07375$


问:
Find the sum of the maximum of the trigonometric polynomial coefficient
somedays ago,I have studying and discussing this Maximum Trigonometric polynomial coefficient
problem

Now I want consider a simple problem:

(The sum of the maximum of the trigonometric polynomial coefficient:)  Let $n$ be  give postive integers  If for all real numbers $x$ we have $$f(x)=A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\ge -1$$
Find the maximum value of $A_{1}+A_{2}+\cdots+A_{n}$.

Here is what I tried:use this well kown :
$$\sum_{j=0}^{n-1}\xi^k_{j}=\begin{cases}
n&n|k\\
0&otherwise
\end{cases}$$
where $\xi_{i}$be the $n$th roots of unity
let $x_{k}=\dfrac{2k\pi}{n+1},k=0,1,\cdots,n$,so we have
$$\sum_{k=0}^{n}\cos{(mx_{k})}=0,m=1,2,\cdots,n$$

so $$\sum_{k=0}^{n}f(x_{k})=\sum_{m=1}^{n}\sum_{k=0}^{n}A_{m}\cos{(mx_{k})}=0$$
then we have
$$A_{1}+A_{2}+\cdots+A_{n}=f(0)=-\sum_{k=1}^{n}f(x_{k})\le n$$

I Conjecture the maximum is $n$.But I can't find  example such reach this maximum

I have find example for  $n=2$,then let $A_{1}=\dfrac{4}{3},A_{2}=\dfrac{2}{3},A_{1}+A_{2}=2$,then
$$f(x)=\dfrac{1}{3}(2\cos{x}+1)^2-1\ge -1$$
for all postive $n$,I can't find it.Thanks

答:
Solution due to @quasi:

Let $A_k = 2 - \frac{2k}{n+1}$. We have $A_1 + A_2 + \cdots + A_n = n$.

We have
\begin{align}
&\sum_{k=1}^n A_k \cos(kx) + 1\\
=\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\sum_{k=1}^n k \cos (kx) + 1\\
=\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\left(\sum_{k=1}^n \sin (kx)\right)' + 1\\
=\ & 2\left(\frac{-\sin \frac{x}{2} + \sin\frac{(2n+1)x}{2}}{2\sin \frac{x}{2}}\right)
- \frac{2}{n+1}\left(\frac{\cos \frac{x}{2} - \cos \frac{(2n+1)x}{2}}{2\sin \frac{x}{2}}\right)' + 1\\
=\ & \frac{\sin x \cos(nx) + \cos x \sin (nx) - \sin x + \sin(nx)}{\sin x}\\
&\quad - \frac{2}{n+1}\left(\frac{1 + \cos x - \cos(nx) - \cos x \cos (nx) + \sin x \sin (nx)}{2\sin x}\right)' + 1\\
=\ & \frac{1 - \cos((n+1)x)}{(n+1)(1-\cos x)}\\
\ge\ & 0.
\end{align}

Update: We can find $A_k = 2 - \frac{2k}{n+1}$ through the following steps.

Note that when $A_k$ is a polynomial in $k$, $\sum_{k=1}^n A_k \cos(kx)$ has a closed form.

We first try $A_k = a_n + b_n k$.

From $\sum_{k=1}^n A_k = a_n n + \frac{b_n n (n+1)}{2} = n$, we have $a_n + \frac{b_n (n+1)}{2} = 1$.
Thus, $b_n = \frac{2(1-a_n)}{n+1}$.

We need
\begin{align}
\sum_{k=1}^n \left(a_n + \frac{2(1-a_n)}{n+1}k\right) \cos(kx) + 1\ge 0, \ \forall x \in (0, 2\pi). \tag{1}
\end{align}

When $n=2$, (1) becomes
$$\frac{1 - \cos x}{3}(1 + 2\cos x)\Big(a_2 + \frac{4\cos x - 1}{1 - \cos x}\Big) \ge 0, \ \forall x \in (0, 2\pi). \tag{2}$$
It is easy to prove that (2) holds if and only if $a_2 = 2$.

When $n=3$, (1) becomes
$$2\cos x (1 - \cos^2 x)\Big(a_3 + \frac{3\cos x - 2}{1 - \cos x}\Big) \ge 0,  \ \forall x \in (0, 2\pi). \tag{3}$$
It is easy to prove that (3) holds if and only if $a_3 = 2$.

We guess that when $a_n = 2$, (1) holds. Fortunately, it works.


问:
Proof of De-Morgan's law.
The law states : $(A \cup B)' = A' \cap B' $

Although I can do it using Venn diagrams, I tried it using this method:

Let $x$ be any arbitrary element which exists in the set $(A \cup B)'$

$x \in (A \cup B)'$

$\Longrightarrow  x \notin (A\cup B)$

$\Longrightarrow  x \notin A $  and  $x \notin B $

$\Longrightarrow  x \in A' $  or  $x \in B' $

But now I'm getting or so that means union , so the final proof which I'm getting is :

$(A \cup B)' = (A' \cup B') $

Please tell where I am going wrong.

答:
Your last implication:  Why are you switching from "and" to "or"?  You shouldn't.

Your proof only leaves the job half done.  You have only proved $(A \cup B)' \subseteq A' \cap B'$.  To establish equality, you also need to prove the other containment:  if $x \in A' \cap B'$, then $x \in (A \cup B)'$.

Title: Proof of De-Morgan's law.


答: Welcome to Math Stack Exchange. You correctly had $x\notin A$ and $x\notin B$, which is precisely what it means to say that $x\in A'$ $\textbf{and}$ $x\in B'$. So $x\in A'\cap B'$. If $x\in A'\cap B'$, then $x\notin A$ and $x\notin B$, so $x$ is not in A or B, i.e., $x\notin A\cup B$. So $x\in (A\cup B)'$.

Title: Proof of De-Morgan's law.

Body:
The law states : $(A \cup B)' = A' \cap B' $

Although I can do it using Venn diagrams, I tried it using this method:

Let $x$ be any arbitrary element which exists in the set $(A \cup B)'$

$x \in (A \cup B)'$

$\Longrightarrow  x \notin (A\cup B)$

$\Longrightarrow  x \notin A $  and  $x \notin B $

$\Longrightarrow  x \in A' $  or  $x \in B' $

But now I'm getting or so that means union , so the final proof which I'm getting is :

$(A \cup B)' = (A' \cup B') $

Please tell where I am going wrong.


答:
I think you've gone wrong with your last statement.

I.E.
$\Longrightarrow  x \epsilon A' $ or  $ x \epsilon B'$

Note that $A'$includes $(A \cup B)'$ and $B-A$

And, $B'$includes $(A \cup B)'$ and $A-B$

So, according to your last statement, x can belong to $(A \cup B)'$, as well as $B-A$ or $A-B$. And that is not what you mean by your second last statement, i.e. $\Longrightarrow  x \notin A $ and  $ x \not in B'$

Thus, you shouldn't switch from AND to OR. You'll get the required relation, then.

Title: Proof of De-Morgan's law.


问:
Prove for Lp norm
If

$$f_n = n I\left[\frac{1}{n},\frac{2}{n}\right],$$

show that the $L_p$ norm of $f_n$ is 1.

I am using the definition of $L_p$ norm which is $\left(\int |X|^p dP\right)^{1/p}$. In this case it would be $$\left(\int_{\frac{1}{n}}^{\frac{2}{n}} n^p dx\right)^{\frac{1}{p}} = \left(\frac{n^p}{n}\right)^\frac{1}{p} \ne 1.$$
I am not sure where I am making the mistake. Can any one correct it?

答: The $L_p$ norm is not $1$.  The only mistake in the computation, perhaps a typo, was fixed in Cameron's edit.</p>

Title: Prove for Lp norm


问:
what is the circle with the smallest radius that contains the three points $(0,0)$ , $(1,1)$, and $(2,3)$?
How to write a model that gives us a circle with the smallest radius that contain three points $(0,0)$ , $(1,1)$, and $(2,3)$?

I've tried to model this as following:

if $x, y$  be the location and $r$ is radius:

$ Min \ \ x^2+y^2 = r^2$

$ x^2+y^2 \le 0$

$ (x-1)^2+(y-1)^2 \le 1$

$ (x-2)^2+(y-3)^2 \le \sqrt 13$

Body:
How about:

min $r^2$ subject to:

$(x-0)^2 + (y-0)^2 \leq r^2$

$(x-1)^2 + (y-1)^2 \leq r^2$

$(x-2)^2 + (y-3)^2 \leq r^2$


问:
Modus ponens reasoning: truth of p implies q and why can we say q follows?
I've found Isn't the modus ponens just the definition of what 'if' means? but I don't feel like it is particularly relevant. I also found  Why is modus ponens a q-implication? which I understand if I don't dig too deep ($p$'s truth and $q$'s truth) but I don't feel like I fully understand modus ponents.

Modus ponens states

$$ p \rightarrow q $$

Given $p$

Therefore $q$

I'm trying to approach this by looking at the truth table but it isn't helping me.

$p$
$q$
$p\rightarrow q$

1
1
1

1
0
0

0
1
1

0
0
1

What is modus ponens saying exactly? I'm having a problem interpreting it.

Is it saying that given some $p$ (can this be true or false or must this only be true) and given some implication (can THIS be true or false or must this be true)?

We can determine that $q$ is true?

Assuming $p$ must be true and $p\rightarrow q$ must be true that DOES imply that $q$ must be true but in any other case we cannot determine $q$?

答:
Abstract answer. An inference rule is something of the form (for some $n \geq 0$)

$$\tag{*}
\frac{H_1 \quad \dots \quad H_n}{T}
$$

where $H_1, \dots, H_n, T$ are formulas; more precisely, $H_1, \dots, H_n$ are the hypotheses or premises of $(*)$ and $T$ is the thesis or conclusion of $(*)$.

The inference rule $(*)$ is read as follows: "Given $H_1, \dots, H_n$, then $T$ follows" (or similar expressions). This means that whenever $H_1, \dots, H_n$ hold simultaneously true, then by necessity $T$ holds true too.
Said differently, it is impossible that $H_1, \dots, H_n$ are all simultaneously true and $T$ is not.

Concrete answer. Modus ponens is the following inference rule:

$$
\frac{p \to q \qquad p}{q}
$$

According to the abstract explanation above, modus ponens says that whenever $p \to q$ and $p$ are simultaneously true, then $q$ is necessarily true.
Indeed, this is exactly what happens in the truth table you wrote.
Check the rows where both $p\to q$ and $p$ are simultaneously true (this happens only on the first row): in each of them (i.e. in the first row) it turns out that $q$ is true too.
Said differently, in the truth table you wrote, it is impossible to find a row where $p \to q$ and $p$ are both true and $q$ is false.


问:
Construction of symmedians but different.
Given $2$ lines $l_1$ and $l_2$ intersecting at point $F,$ let $A$ and $B$ on line $l_1,$ construct point $E$ on $l_2$ such that $\angle AEF=\angle (BE,EF) .$

Here's the diagram of  the problem:

For the construction, I thought of taking point $D$ such that $BD\perp DF.$ But couldn't proceed much.

答:
Let $C$ be symmetric to $B$ with respect to $l_2$.

Then the line $AC$ intersects $l_2$ at $E$ you seek.

Proof of correctness:

Let $K$ be the intersection of $l_2$ and $BC$. Then triangles $\triangle EKB$ and $\triangle EKC$ are congruent by the SAS rule (angles at $K$ are both right, hence equal, and $BK=KC$ by symmetry). Consequenlty their angles at $E$ are equal. Finally, the angle $\angle AEF$ is opposite to $\angle CEK$, so congruent to it and to $\angle KEB$, as required.


问:
Is BZC inconsistent with Reinhardt cardinals
A Reinhardt cardinal is defined as the critical point of a nontrivial elementary embedding $j: V\rightarrow V$ from the universe $V$ to itself, and is known to be inconsistent with the axiom of choice when paired with standard ZFC. Are Reinhardt cardinals still inconsistent with the axiom of choice when paired with the weaker BZC, Zermelo set theory with only bounded separation and choice?

答:
Your theory is consistent modulo large cardinal axioms.
You can see that if $\lambda$ witnesses $I_3$ (i.e., there is a non-trivial elementary embedding $j: V_\lambda\to V_\lambda$), then $V_\lambda$ satisfies the followings:

$\mathsf{ZFC}$ (but separation and replacement schemas does not allow $j$ to appear,)

The Wholeness axiom $\mathsf{WA}$ (separation schema for formulas with $j$.)

Hence $V_\lambda$ satisfies $\mathsf{BZC}$ with $j$ allowed to appear in bounded separation.

If you do not require $j$ can appear in bounded separation, then weaker large cardinal axioms suffice to provide the consistency of $\mathsf{BZC}$ with an elementary embedding $j:V\prec V$; for example, if $0^\sharp$ exists and $j:L\to L$ be an elementary embedding which sends the $n$th Silver indiscernibles to $(n+1)$th, and fixes all other indiscernibles, then $L_\iota$ for $\omega$th Silver indiscernible $\iota$ thinks it is a model of $\mathsf{ZFC}$ with an embedding $j:L_\iota\to L_\iota$.


问:
I need help with a word problem .
The length of a rectangle is 1 meter less than twice its width. If the area of the rectangle is 120 square meters, find the dimensions of the rectangle</p>

答:
Let $x$ be the length of the rectangle and $y$ be the width of the rectangle.

Area is given by $A=xy\implies xy=120 \tag{1}$

And according to the question, the length $x$ is $1$ less than twice the width $2y$.

$\therefore x=2y-1 \tag{2}$

Solve $(1)$ and $(2)$ to get $x=15$ and $y=8$.

Title: I need help with a word problem .


问:
Is there a positive integer $n$ such that the prime divisors of $n^3 - 1$ are $2$, $3$ and $7$?
For a positive integer $k$ write $\pi(k)$ for the set of all prime divisors of $k$.  For example, $\pi(24) = \{2,3\}$ and
$\pi(1) = \emptyset$.

Question.  Is there a positive integer $n$ for which $\pi(n^3 - 1) = \{2,3,7\}$?

An alternative point of view asks whether there are positive integers $a$, $b$ and $c$ such that $1 + 2^a3^b7^c$ is a perfect cube (of a positive integer).

To be clear, $n^3 - 1$ must be divisible by all three primes $2$, $3$ and $7$, and not divisible by any other prime.

I think the answer is "no", but have so far been unable to prove it.  I actually only need to find an answer for prime $n$, but it appears it may be true without assuming that $n$ is prime.  Using Maple, I have checked this for all $n \leq 10^9$, and for the first $106000000$-ish primes.

This is actually just one case ($r = 3$) of the problem I am working on which is to find all primes $p$ and integers $r>1$ for which $\pi(p^r - 1) = \{ 2, 3, 7 \}$.  I think I have managed to show that $r = 2$ and $p \in \{ 13, 97, 127 \}$, unless $r = 3$. (And, as you might guess, this whole thing is just one very special case of a more general problem I am looking at.)

答:
Partial? answer Way too long for a comment.

Since $2,3$ must divide $n^3-1$ we have $n$ odd and $3|n-1$. Therefore
$$
n=6k+1
$$
for some positive integer.

Then
$$
n^3-1=(6k+1)^3-1=18k(12k^2+6k+1)
$$

Since $12k^2+6k+1$ is relatively prime to $2,3$ it follows that we must have
$$
12k^2+6k+1=7^l
$$
for some positive integer $ l \geq 1$.

This immediately implies that $7 \nmid k$ and hence there exists non-negative integers $a,b$ such that
$$
k=2^a3^b
$$

This leads to the diophantine equation
$$
2^{2a+2}3^{2b+1}+2^{a+1}3^{b+1}+1=7^l \\
a,b,l \in \mathbb Z \\
a,b \geq 0 \\
 l\geq 1
$$

Here I got stucked.

Note that there are good chances
$$
12k^2+6k+1=7^l
$$
already does not have positive solutions.

Note that if this quadratic in $k$ has a solution
$$
\Delta=36-48(1-7^l)=48 \cdot 7^l-12
$$must be a perfect square. This implies that there exists a positive integer $c$ such that
$$
12 \cdot 7^l-3=c^2
$$

Since $3|c$ there exists some integer $d$ such that $c=3d$ and hence
$$
4 \cdot 7^l=3d^2+1
$$
??


答:
Consider the points pair-wise, call them $A$ and $B$. Let the other
point be $C$

Consider the circle with $AB$ as diameter and check if $C$ is within this circle.

If you consider all three possibilities, you'll find that the line joining $(0, 0)$ and $(2, 3)$ is the diameter of the required circle. The equation of the circle is

$$(x - 1)^2 + (y - \frac{3}{2})^2 = \left(\frac{\sqrt{13}}{2} \right)^2$$

It is easy to verify that the other point $(1,1)$ lies inside the above circle.

Title: what is the circle with the smallest radius that contains the three points $(0,0)$ , $(1,1)$, and $(2,3)$?

Body:
How to write a model that gives us a circle with the smallest radius that contain three points $(0,0)$ , $(1,1)$, and $(2,3)$?

I've tried to model this as following:

if $x, y$  be the location and $r$ is radius:

$ Min \ \ x^2+y^2 = r^2$

$ x^2+y^2 \le 0$

$ (x-1)^2+(y-1)^2 \le 1$

$ (x-2)^2+(y-3)^2 \le \sqrt 13$


问:
Rule of inference in logic
There is a slide in my class notes that mention

$$
\neg p \rightarrow F_0
$$

therefore p.

It then follows up and it says

if we want to establish the validity of the argument

Eqn 1: ($p_0 \wedge p_1... \wedge p_n) $ -> q,

we can establish the validity of logically equivalent argument.

Eqn 2: ($p_0 \wedge p_1... \wedge p_n \wedge \neg q) -> F_0$

My questions:

The implication

$$
\neg p \rightarrow F_0
$$

leads to "therefore p" because p is assumed to be true and the only way p can lead to false is through negation of p? I'm very confused about this

How do we go from the first equation to the second?

答:
Assuming that $F_0$ means $\bot$ (the always False logical constant), we can easily check (with truth table) that $p \to F_0$ is equivalent to $\lnot p$: due to the fact that $F_0$ is always False, we have only two cases: $\text T → \text F$ and $\text F → \text F$, and the result is exactly the opposite of $p$.

Thus, $\lnot p \to F_0$ is $\lnot \lnot p$ that, in classical logic, is equivalent to $p$.

For the same reason, $(p_i \land \lnot q) \to F_0$ is equivalent to $\lnot (p_i \land \lnot q)$ which is in turn equivalent to $p_i \to \lnot \lnot q$.

Title: Rule of inference in logic


问:
How to judge the form of matrix multiplication?(faster ways please)

I wonder if it exists a faster way to know it is B=AC OR CA

答:
If $B$ and $A$ are column vectors, then you have $B=CA$ and $B_i=\sum_{k=1}^n C_{ik}A_k$.

If $B$ and $A$ are row vectors, then $B=AC$ and $B_i=\sum_{k=1}^n C_{ki}A_k$.

So this is the latter.


问:
Given $v, w, z \in \mathbb{R}$, show this equation as true
I am now trying to understand, given $v, w, z \in \mathbb{R^{3}}$, its true that $(v\times w)\times z$=$c_{1}v+c_{2}+c_{3}(v \times w)$. $c_{1}, c_{2}, c_{3} \in \mathbb{R}$. This is cros product by the way.

So this is a true equation. But I want to know how to show $c_{1}=-w\cdot z, c_{2}=v\cdot z, c_{3}=0$.

I first want to prove $c_{3}=0$, I want to let $v$x$w=n$, make the dot product of n and $c_{1}v+c_{2}+c_{3}(v \times w)$ to prove that $c_{3}=0$.

答:
The definition of cross product is $c=a\times b$, is

$c_x=a_yb_z-a_zb_y$

$c_y=a_zb_x-a_xb_z$

$c_z=a_xb_y-a_yb_x$

Now, for your expression, $d=(v\times w)\times z=-z\times (v\times w)$ because the cross product is "anticommutative." By the component wise definition of cross product, $$\begin{split}d_x&=-z_y(v\times w)_z-z_z(v\times w)_y\\&=-z_y(v_xw_y-v_yw_x)+z_z(v_zw_x-v_xw_z)\\&=v_x(-z_yw_y-z_zw_z)+w_x(z_yv_y+z_zv_z)+[v_xz_xw_x-v_xz_xw_x]\\&=-v_x(z_xw_x+z_yw_y+z_zw_z)+w_x(v_xz_x+z_yv_y+z_zv_z)\\&=(z\cdot w)w_x-(z\cdot w)v_x\end{split}$$

Likewise, for the y and z components you get $d_y=(z\cdot v)w_y-(z\cdot w)v_y$ and $d_z=(z\cdot v)w_z-(z\cdot w)v_z$. Putting the components together, you find $d=(z\cdot v)w-(z\cdot w)v$.

Thus $c_1=-z\cdot w, c_2=(z\cdot v)w, c_3=0$.

You could also use directly the vector triple product, which says that $a\times (b \times c)=b(a\cdot c)-c(a\cdot b)$, or $(a\times b)\times c=-c\times(a\times b)=-(c\cdot b)a+(c\cdot a)b$


问:
Intuition on how to exchange finite sums
I am asked to prove the following
$$\sum_{r=1 (r,q)=1}^{q}f(r)=\sum_{d|q}\mu(d)\sum_{s=1}^{q/d}f(ds)$$
I assume this can be proven by exchanging the sums. However, this made me realize I never really understood how to exchange finite sums like this. How I do it is I list all combinations and then I try to notice how to exchange the sum.
For example, take a simple $q$ say $q=p_1p_2$ $p_1<p_2$. Normally the summation is as such
$$(1,1)\cdots (1,p_1)\cdots(1,p_2)\cdots(1,p_1p_2)$$
$$(p_1,1)\cdots (p_1,p_1)\cdots(p_1,p_2)$$
$$(p_2,1)\cdots(p_2,p_1)$$
$$(p_1p_2,1)$$
Now I stare at this untill i can see how to revert the sum.
In this case I think it turns into $$\sum_{s=1}^{q}\sum_{d|q \text{s.t $d<\frac{q}{s}$}}$$
Is this correct? Also, how else should i arrive at this answer without listing the numbers. It takes too long and can be hard for me

EDIT: Perhaps this is a wrong example, as you do not want to exchange the sums but my question still stands, how do I know how to exchange such sums, without writing out an example like I have shown above.

答: Here's how I see the process of "exchanging" (well, rearranging) the sums. The LHS is $\sum_{r\in A}f(r)$ over some set $A$ (assume $q$ fixed), the RHS is $\sum_{(d,s)\in B}\mu(d)f(ds)$ over some set $B$. So, we group the terms on the RHS with $ds=r$, for each possible $r$ (that is, for $1\leqslant r\leqslant q$). What are $(d,s)\in B$ with $ds=r$? We see that $d$ must divide both $q$ and $r$, and each such $d$ determines $s=r/d$ uniquely. This means $$\sum_{d\mid q}\mu(d)\sum_{s=1}^{q/d}f(ds)=\sum_{r=1}^q f(r)\sum_{d\mid(q,r)}\mu(d).$$ (Now recall that $\sum_{d\mid n}\mu(d)$ is $1$ if $n=1$ and is $0$ otherwise.)


问:
pi-character of a compact Hausdorff space does not exceed its tightness
I'm studying parts of "recent progress in topology", the chapter about homogeneous spaces. In one of the corollaries (corollary 5.5) authors use this theorem, and reference to a paper from Šapirovski for the proof, but I can't find the link or anything for it. Does anybody know the proof, or a paper or book or anything I can find it there?

Thanks ahead!

答:
If you're interested in cardinal functions, you should surely look at Juhász' classic book(let)s Cardinal functions in topology and Cardinal functions in topology - ten years later. These made the subject quite popular for a while. They're rather self-contained introductions that contain detailed proofs, also of the underlying set theory combinatorics in some cases.

In the latter I found as theorem 3.14 (a), that for compact Hausdorff spaces we indeed have $\pi\chi(X) \le t(X)$. The proof is too long to reproduce here, but it's based on the characterisation of tightness in compact Hausdorff spaces using so-called free sequences, which is also proved in this second book. He also refers to some original Šapirovski papers (some in Russian, some in English translation ), so going to the original papers should be doable too, if your library has access to these. But the exposition in Juhász is fine, and more self-contained I think.

I believe the original paper is

B. E Šapirovski , $\pi$-character and $\pi$-weight in bicompacta, Dokl. Akad. Nauk SSSR 223 (1975), no. 4, 799–802.


问:
Proving that ${n \choose k}\sim \frac{(n-k)^k}{k!}$ as $n\to \infty$.
Question:

Prove that $${n\choose k} \sim \frac{(n-k)^k}{k!}$$ as $n\to \infty$.

Attempt and Ideas:

I don't have too many ideas but I guess I want to manipulate Stirling's Formula, which is
$$ n! \sim n^n e^{-n}(2\pi n)^{1/2}\quad \text{as }n\to\infty  $$ and is equivalent to $$ \lim_{n\to \infty} \left( \dfrac{n!}{n^n e^{-n}(2\pi n)^{1/2}} \right) = 1.$$ I can get rid of the $e^{-n}$ term (hopefully?) by using the fact that $$ e^{-k} = \lim_{n\to \infty} \left(1-\frac{k}{n}\right)^n. $$
However, I have no idea how to get rid of the $(2\pi n)^{1/2}$ term and so I'm really lost.

I have also tried using the definition $$ {n\choose k} = \dfrac{n!}{k!(n-k)!} $$ and dividing by $ \frac{(n-k)^k}{k!}$ but it quickly proved fruitless as I don't know how I can demonstrate that $$ \frac{n!}{(n-k)!(n-k)^k} \to 1,\quad \text{as }n\to \infty. $$ Any help will be greatly appreciated, thank you.

Body:
$$
\frac{n!}{(n-k)!(n-k)^k}
= \frac{n(n-1) \cdots (n-(k-1))}{(n-k)^k} = \frac n{n-k} \frac{n-1}{n-k} \cdots \frac{n-(k-1)}{n-k}.
$$


问:
Does the series $\sum_{1}^{\infty}\big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big]$ converge?
As title says, does the following series converge? $$\sum_{1}^{\infty}\big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big]$$

It is supposed to converge but I don't know how.

My attempt:

$$\sum_{1}^{\infty}\big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big] = \sum_{1}^{\infty}a_n,$$
where

\begin{align}
a_n &= \big[1 + n\ln\big(\frac{2n-1}{2n+1}\big) \big] \\
    &= 1 + n\big[\ln(2n-1) - \ln(2n+1)\big].
\end{align}

I couldn't come up with any good convergence tests and started listing out a few terms:

$a_1 = 1 + 1\ln(1) - 1\ln(3)$,

$a_2 = 1 + 2\ln(3) - 2\ln(5)$,

$a_3 = 1 + 3\ln(5) -3\ln(7)$,

and so on.

Since some terms cancel, I think for the $N$-th partial sum, we have:

\begin{align}
\sum_{1}^{N}a_n &= N + \ln(2N-1) - N\ln(2N+1) \\
                &= N + \ln\frac{2N-1}{(2N+1)^N}
\end{align}
At this point, I can only see divergence as $N \rightarrow \infty$.

What am I doing wrong? How can I correctly test for convergence in this case?

Body:
Let

$$f(x)=\frac{1}{x}\left(\log\frac{1-\frac x2}{1+\frac x2}\right)+1$$

Then

$$f(x)=\frac1x\log(1-\frac x2)-\frac1x\log(1+\frac x2)+1$$

Use Taylor's formula:

$$\log(1-x)=-x-\frac12x^2-\frac13x^3+O(x^4)$$

$$f(x)=(-\frac12-\frac18x-\frac1{24}x^2+O(x^3))-(\frac12-\frac18x+\frac1{24}x^2+O(x^3))+1$$

$$f(x)=-\frac{1}{12}x^2+O(x^3)$$

Hence $1+n\log \frac{2n-1}{2n+1}=-\frac{1}{12n^2}+O(\frac 1{n^3})$

And your series is convergent.


问:
How many possible combinations of 12 numbers are there, such that the first six numbers are ascending and the last?
Given the numbers 1,2,3,..., 12. How many possible combinations of ordering these numbers are there, such that the first 6 numbers and the last six numberes are ascending?

May someone please give me a hint on how to compute the solution? I know, there are 12! possible combinations of these 12 numbers and $\binom{12}{6}$ possibilities of choosing 6 numbers out of 12. But how do I eliminate those of not ascending order?

答: There are $\binom{12}{6}$ ways of choosing which six numbers will appear first. There is only one way to arrange them to be ascending. The remaining six numbers will appear last, and there is only one way to arrange them to be ascending.


问:
How to solve equations of this form: $x^x = n$?
How would I go about solving equations of this form:

$$
x^x = n
$$

for values of n that do not have obvious solutions through factoring, such as $27$ ($3^3$) or $256$ ($4^4$).

For instance, how would I solve for x in this equation:
$$x^x = 7$$

I am a high school student, and I haven't exactly ventured into "higher mathematics." My first thought to approaching this equation was to convert it into a logarithmic form and go from there, but this didn't yield anything useful in the end.

My apologies if this question has been asked and answered already; I haven't been able to find a concrete answer on the matter.

答:
Can be solved by solving the problem of finding an approximate value.But accuracy ...
$(x_0+\triangle x)^{x_0+\triangle x}=7\\x=2;\triangle x=?\\f(x)=f(x_0)+f'(x_0)\triangle x\\f(x_0)=2^2=4\\f'(x)=x^x(lnx+1)*\triangle x\\7=4+4(ln2+1)*\triangle x\\\triangle x=\frac{3}{4(ln2+1)}\approx0.44\\x=2+0.44\approx2.44\\2.44^{2.44}\approx8.8$


问:
Different ways to define distance and the definition of $\pi$
I saw a post a couple of posts about "proving that $\pi$ = $4$", and I thought of using a naive way to say that it is not impossible.

We know that $\pi$ is commonly  defined like this:

$$\pi = \frac{\text{Circumference}}{\text{Diameter}}.$$

How will you measure the circumference? That is equivalent to saying that "What metric I will choose to measure the length of this curved line?" The important point is that the metric we choose will affect what value of $\pi$ we will get.For example, we will get $\pi = 4$ if we choose the taxicab metric $(|\Delta x|+|\Delta y|)$ and we will get $\pi = 3.1415...$ if we choose the Euclidean metric $(\sqrt{\Delta x^2 + \Delta y^2})$.

In short, we will get different answers based on the underlying assumptions we are taking beforehand. Is my approach wrong?

答:
I'd say that in the end it simply depends on how you define $\pi$. Most people (especially non-mathematicians) know the definition you mention, i.e. $\pi$ is the ratio between the circumference and the diameter of a unit circle. Implicitly we use the euclidean metric to define what a unit circle is (because who wouldn't think of the euclidean metric in everyday life first, right?).

Now I think what you are doing is exactly the 'easiest' way to justify an assessment like $\pi = 4$. But we'll have to deviate from the 'everyday life definition' above in the following way. Let's take a different metric, e.g. $d(x,y) = \max(|x|,|y|)$. What is the unit circle $S$ in this metric? It's

$$S = \{(x,y)\in \mathbb{R}^2 | d(x,y) = \max(|x|,|y|) = 1\} $$

You can easily draw this circle, and you will see that it turns out to be the square of length 2 centered around the origin, which has of course circumference 8. If we now apply our definition of $\pi$ being equal to the circumference divided by the diameter, we get $\pi = \frac{8}{2} = 4$. (Small remark: you get a different result when using the taxicab metric as you proposed. You can also draw the unit circle with respect to the taxicab metric to figure it out!).

So as often: It depends on your definition.

I hope this helps!


问:
Problem related with a linear transformation and its rank,nullity...
I am stuck on the following problem:

Suppose $W$ be  a vector space of dimension $m \geq 2.$ Let $F \colon W \to W$ be a linear transformation such that $F^{n+1}=0$ and $F^n \ne 0$ for some $n \ge 1.$ Then I have to check which of  the following statement(s) is/are true?

1.$\text{trace}(F) \ne 0 $

2.$\text{Rank}(F^n) \le \text{Nullity}(F^n)$

3.$n=m$

4.$F$ is diagonalizable.

I do not know how to progress with the problem. Any kind of hints with some explanation will be helpful.

答:
$2:True:$ $FF^n=0\\\implies\text{Im}(F^n)\subset\text{Kernel}(F)\subset\text{Kernel}(F^n)\\\implies\text{Rank}(F^n)\le\text{Nullity}(F^n)$

$1,3,4: False:$ $F:\mathbb R^3\to\mathbb R^3$ given by $F=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$


问:
Prove that if a set $S$ has an accumulation point at $0$, it contains a sequence of nonzero terms which converge to $0$
I met this problem in exercises in regards to Complex Analysis, and felt very confused.

Our aim to establish that, if a set $S$ has an accumulation point at $0$, it contains a sequence of nonzero terms which converge to $0$. This fact proves a rather important corollary: If a power series equals zero at the points of a set with an accumulation point at the origin, the power series is identically zero.

Since accumulation point is just another way to express limit point, it means the set $S$ has infintely many points surrounding $0$.

Then, we can take $\lvert z_{1}\rvert $ equal to any real number

Take $\lvert z_{2}\rvert = 1/2 * \lvert z_{1}\rvert $ Repeat the steps and we have a sequence converge to $0$.

Now how does the above steps contribute to proving the power series must be identically $0$ if there is an accumulation point at the origin?

答:
Your proof is not correct.  The points you construct are not necessarily in $S$.

Take $\epsilon \in \mathbb{R}$, $\epsilon \gt 0$ Then we have $z_1 \in S$ such that $0 \lt |z_1 - z| \lt \epsilon$.  Now start dividing $\epsilon$ by $2$. You get $z_2, z_3, ... $ etc. (all in $S$) such that

$$0 \lt |z_n - z| \lt \frac{\epsilon}{2^{n-1}}$$ for every $n = 1,2,3, \dots$

This is the sequence you are looking for.

All in all this statement is obvious, it just follows from the definition.

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


问:
Spectrum of a unilateral weighted shift
Let $H$ be an infinite dimensional, separable, complex Hilbert space.

Let $T \in B(H)$ be a unilateral weighted shift. i.e. $T e_n= t_n e_{n+1}$, where $(t_n)_n \in \ell^{\infty}$.

Suppose $\lambda \in \sigma(T)$. Show that $\{z\in \mathbb{C}: |z|= \lambda \} \subset \sigma(T)$.

I showed that if $\lambda=1$, then we have $UTU^{-1}=\lambda T$, where $U$ is a unitary. Not sure if this is related to what we want to show.

Thank you!

答:
As per this answer, for any $\mu=e^{i\theta}$ there is a unitary operator such that $UTU^{-1}=\mu T$.

Applying the spectrum to both sides, $$\sigma(T)=\sigma(UTU^{-1})=\mu\sigma(T)$$ This means that $\sigma(T)$ is rotationally invariant. Hence if $\lambda\in\sigma(T)$ and $|z|=|\lambda|$ then $z=\frac{z}{\lambda}\lambda\in\frac{z}{\lambda}\sigma(T)=\sigma(T)$ as well.

Title: Spectrum of a unilateral weighted shift


问:
Proof that $AA^T$ is invertible if $A$ has independent rows
While looking for the answer for my question I came across this post. It may be a silly idea, but if $A^{t}$ has independent rows can I just transpose it and get $A$ with independent columns and proceed as is shown in the linked solution? Or it is not working this way?

答: For every matrix $A$ with  independent rows, we have $det A \neq 0$ also $det A= det A^t$so both $A$ and $A^t$ are invertible. Then $AA^t$ is invertible if and only if $A$ is.


问:
UMVUE OF $\sum_{k=4}^{\infty}\frac{e^{-\theta}\theta^k}{k!}$

Let $x_1=1, x_2=4$ be the data on a random sample of size $2$ from a
Poisson($\theta$) distribution, where $\theta \in (0,\infty)$. Let T
be the uniformly minimum variance unbiased estimate of T($\theta$)=
$\sum_{k=4}^{\infty}\frac{e^{-\theta}\theta^k}{k!}$ based on the given
data. Then T equals____?

I am using $x$ instead of $k$.

$P(X\ge 4)=\sum_{x=4}^{\infty}\frac{e^{-\theta}\theta^x}{x!}$

$I(X_1)=1$ if $X\ge4$,$0$ otherwise

So using Rao–Blackwell theorem

$E(I(X_1)|T=\sum_{i}X_i)$
=$1\cdot P\left( \dfrac{X_1\ge4,X_1+X_2+...+X_n=T}{T=\sum_{i}X_i} \right)$

How do I solve this? if I was given $X_1=1$ I would've utilized it and then used the fact of independence but I am not sure how to proceed here. Also, tell me if there is any other way possible way to solve this problem.

Body:
Using help from stubbornAtom's comments
$n=2,X_1+X_2=t=5$

$1\cdot P\left( \dfrac{X_1\ge4,X_1+X_2=t}{T=\sum_{i}X_i} \right) =P(X_1\ge4|X_1+X_2=t)=\frac1{2^t}\sum_{j=4}^t \binom{t}{j})=\frac1{2^5}\sum_{j=4}^5 \binom{t}{j})=\frac{1}{2^5}(\binom{5}{4}+\binom{5}{5})=\frac{1}{32}(5+1)=\frac{6}{32}$


问:
Compute the min of $a^Tx+||x||_{\infty}$

Prove that for any $a\in\mathbb{R}^n$ the following holds

$$\min a^Tx+\|x\|_\infty=\left\{\begin{array}{rcl} 0&\|a\|_1\leq1\\-\infty&\text{else}\end{array}\right.$$

I've tried looking at $\|x\|\cdot\|a\| \cos(\theta)+\|x\|_\infty$ and bound it from below by looking at the norm of $a$ but it didn't get me anywhere.
edit:

I've tried taking norm 1 on $a^Tx+\|x\|_\infty\geq-\|a\|_1 \|x\|_\infty + \|x\|_\infty$ from holder's inequality and from here i got $\|x\|_\infty(1-\|a\|_1) > -\infty\iff\|a\|_1\leq1$ but to get that the minimum is zero still stuck

any hint will be welcome :)

答: For the  case $\|a\|_1 \leq 1$ use the fact that $-a^{T}x \leq \sum |a_i||x_i|  \leq \max_i |x_i| \sum |a_i| \leq \|x\|_{\infty}$.  This proves that $a^T x + \|x\|_\infty \geq 0$ for all $x$ so $0$ is a lower bound. This bound is attained when $x=0$ so the minimum value is $0$. For the case $\|a\|_1 > 1$ take $x_i=-n \operatorname{sgn} (a_i)$.


问:
The Cardinality of the Set $\{\lfloor\frac{N}{i^k}\rfloor \ | \ i \in \mathbb{Z}^+\}$ Given $k$
Given a fixed real number $k$, I'd like to ask how to calculate the asymptotic
behavior of the cardinality of the set
$$S_k=\left\{\left\lfloor\frac{N}{i^k}\right\rfloor \ \bigg| \ i \in \mathbb{Z}^+\right\}$$ for sufficiently large $N$, here $\lfloor\cdot\rfloor$ is the floor function.

An example: when $k=1$, it turns out that $|S_k|=\Theta(N^{\frac{1}{2}})$.

It seems that $|S_k|=\Theta(N^{\frac{1}{k+1}})$ asymptotically, but I cannot prove it rigorously.

答:
Write $S_k=S_k^-\cup S_k^+$, where
$$
S_k^+=\{\lfloor N/i^k\rfloor :i\in [1,N^{1/(k+1)})\cap \mathbb Z\}\\
S_k^-=\{\lfloor N/i^k\rfloor :i\in (N^{1/(k+1)},N^{1/k}]\cap \mathbb Z\}
$$

Obviously, $|S_k^+|\in O(N^{1/(k+1)})$, since there $\lfloor N^{1/(k+1)}\rfloor$ possibilities for $i$. With some more thought, the values $\lfloor N/i^k\rfloor$ are distinct for distinct $i$, so $|S_k^+|\in \Theta(N^{1/(k+1)})$.

Furthermore, $|S_k^-|\in O(N^{1/(k+1)})$, since $S_k^-\subseteq [1,N^{1/(k+1)})\cap \mathbb Z$. You can in fact show the other inclusion by proving that for $i\ge N^{1/(k+1)}$, that
$$
\frac{N}{(i+1)^k}\ge \frac{N}{i^k}-1
$$
so that as $i$ increases from $N^{1/(k+1)}$ to $N^{1/k}$, it hits every element of $[1,N^{1/(k+1)})$ in decreasing order.

Combining these gives you want you want.


问:
If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$

If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$

My attempt-

Since $x^5=2,x^6=2x,x^7=2x^2..$and so on

The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$

Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.

Any help will be appreciated.

Note: Please don't use root tables and complex numbers too!

Body:
Working mod $x^5-2$,
$$
\scriptsize\frac{3x^3}{1+x}=x^3\frac{1+x^5}{1+x}=x^3-x^4+x^5-x^6+x^7=2-2x+2x^2+x^3-x^4\tag1
$$
$$
\scriptsize\frac{5x}{1+x^2}=x\frac{1+x^{10}}{1+x^2}=x-x^3+x^5-x^7+x^9=2+x-2x^2-x^3+2x^4\tag2
$$
$$
\scriptsize\frac{9x^4}{1+x^3}=x^4\frac{1+x^{15}}{1+x^3}=x^4-x^7+x^{10}-x^{13}+x^{16}=4+8x-2x^2-4x^3+x^4\tag3
$$
$$
\scriptsize\frac{17x^2}{1+x^4}=x^2\frac{1+x^{20}}{1+x^4}=x^2-x^6+x^{10}-x^{14}+x^{18}=4-2x+x^2+8x^3-4x^4\tag4
$$
Adding $\frac13\text{(1)}+\frac15\text{(2)}+\frac19\text{(3)}+\frac1{17}\text{(4)}$, we get
$$
\scriptsize\frac{x^3}{1+x}+\frac{x}{1+x^2}+\frac{x^4}{1+x^3}+\frac{x^2}{1+x^4}=\frac{1336+233x+79x^2+122x^3-44x^4}{765}\tag5
$$


问:
How to judge the form of matrix multiplication?(faster ways please)

I wonder if it exists a faster way to know it is B=AC OR CA

答:
In this question, it is assumed that the first coordinate represents rows and the second columns.

$a^Tc=\left(\begin{matrix}a_1&...&a_n\end{matrix}\right)\left(\begin{matrix}|&&|\\c_1&...&c_n\\|&&|\end{matrix}\right)=\left(\begin{matrix}a^Tc_1&...&a^Tc_n\end{matrix}\right)=\left(\begin{matrix}a_1c_{11}+...+a_nc_{n1}&...&a_1c_{1n}+...+a_nc_{nn}\end{matrix}\right)$

also $a^Tc=\left(\begin{matrix}a_1&...&a_n\end{matrix}\right)\left(\begin{matrix}-c_1^T-\\...\\-c_n^T-\end{matrix}\right)=a_1c_1^T+...+a_nc_n^T=\left(\begin{matrix}a_1c_{11}+...+a_nc_{n1}&...&a_1c_{1n}+...+a_nc_{nn}\end{matrix}\right)$

On the other hand,

$ca=\left(\begin{matrix}|&&|\\c_1&...&c_n\\|&&|\end{matrix}\right)\left(\begin{matrix}a_1\\...\\a_n\end{matrix}\right)=a_1c_1+....+a_nc_n=\left(\begin{matrix}a_1c_{11}+...+a_nc_{1n}\\...\\a_nc_{n1}+...+a_nc_{nn}\end{matrix}\right)$

$ca=\left(\begin{matrix}-c_1^T-\\...\\-c_n^T-\end{matrix}\right)\left(\begin{matrix}a_1\\...\\a_n\end{matrix}\right)=\left(\begin{matrix}c_1^Ta_1\\...\\c_1^Ta_n\end{matrix}\right)=\left(\begin{matrix}a_1c_{11}+...+a_nc_{1n}\\...\\a_nc_{n1}+...+a_nc_{nn}\end{matrix}\right)$

Thus I conclude if it is the first coordinate that stays still in each $b_i=\sum _jc_{ij}$ (the second coordinate varies) that $b=ca$, i.e. a is a column vector multiplied from the right. If it is $b_j=\sum_ic_{ij}$ (the first coordinate varies), then $b=a^Tc$, i.e. a is a column vector multiplied from the left.

The trick is if it is b_i, then b is a column vector. If b_j, then b is a row vector.


问:
Is there a way to eliminate an extraneous index from this sum?
Fix a natural number $n$.  Suppose we have a triple sum of the form
$$ \sum_{i=1}^n \sum_{j=1}^n \sum_{k=j+1}^{i-1} f(i,k) $$
where the summands only depend on $i$ and $k$.  </p>

<p>Is there a way to rewrite the sum so that it does not include the index $j$?  </p>

答:
This sum can be written as
$$
\sum_{1 \le j < k < i \le n} f(i,k)
$$
(do you see why?)
From this form, it should be clear that for fixed $i,k$ the term $f(i,k)$ appears $k - 1$ times in the sum.  So you can write the sum as
$$
\sum_{2 \le k < i \le n} (k - 1) f(i,k).
$$
In general, a very helpful approach to multiple nested sums is to write them as a single sum over some inequality in the variables, as I have done above.</p>


问:
Definition of a tensor for a manifold
While reading Nakahara's geometry, topology and physics. I came across the following definition of a tensor. </p>

<p>A tensor $T$ of type $(p, q)$ is a multilinear map that maps
$p$ dual vectors and $q$ vectors to $\mathbb{R}$. While generalizing to manifolds, the dual vectors are replaced by elements of the cotangent space, and the vectors by elements of the tangent space. How is this definition equivalent to the definition of a tensor (which I knew before) as a linear multi-dimensional array? When I think of tensors, I think of them as generalizations of matrices to many dimensions. More specifically, how do I show that a linear map between two vector spaces is a tensor of rank (1,1). In the above definition, I can see a map to $\mathbb{R}$, while this is a map between two vector spaces. </p>

答:
More specifically, how do I show that a linear map between two vector
spaces is a tensor of rank $(1,1)$.

Since you are reading Nakahara's book, you are certainly talking about exercise $2.12$:

Let $V$ and $W$ be vector spaces and let $f\colon V\to W$ be a linear
map. Show that $f$ is a tensor of type $(1,1)$.

I consider his definition of tensors unclear, but I guess the best way to solve the exercise is to show that $L(V,W)$ together with a suitable bilinear map $V^*\times W\to L(V,W)$ has the universal property of $V^*\otimes W$, asssuming that $V$ and $W$ are finite-dimensional real vector spaces.

It turns out that
\begin{align}
\phi\colon V^*\times W&\to L(V,W)\\
(v,w)&\mapsto v\otimes w
\end{align}
with
\begin{align}
v\otimes w\colon V&\to W\\
u&\mapsto v(u)\cdot w
\end{align}
has the desired properties. That is, if $X$ is a real vector space and $f\colon V^*\times W\to X$ is a bilinear map, there is exactly one linear map $\tilde{f}\colon L(V,W)\to X$ such that $f(v,w)=\tilde{f}(v\otimes w)$ for all $(v,w)\in V^*\times W$ (the proof is left to the reader). In other words, $L(V,W)$ together with the map $\phi$ has the universal property of $V^*\otimes W$.

Title: Definition of a tensor for a manifold


问:
find the exponential form of a matrix
Let
\begin{equation*}
A =
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}\end{equation*}

\begin{equation*}
B =
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix}\end{equation*}

\begin{equation*}
C =
\begin{pmatrix}
4 & 1 & 0 & -4 & 0 & 0\\
0 & 4 & -1 & 0 & -4 & 2\\
0 & 0 & 6 & 0 & 0 & -6\\
2 & 0 & 0 & -2 & 1 & 0\\
0 & 2 & -1 & 0 & -2 & 2\\
0 & 0 & 3 & 0 & 0 & -3\\
\end{pmatrix}\end{equation*}

Suppose $\exp(z)=e^z$, for $z\in \mathbb{C}$. Find $\exp(C)$.

Hint:

\begin{equation*}
D =
\begin{pmatrix}
2 & 1 \\
1 & 1 \end{pmatrix}\end{equation*}

is invertible.

I'm very confused about how to approach this problem.

It would be nice if someone could walk me through how to solve this. But any hints or ideas will be appreciated!

Thank you!

答:
$$C=\begin{pmatrix}
2B-A & 2A-2B\\ B-A & 2A-B
\end{pmatrix}$$

Let $U=\begin{pmatrix}
-1 & 2\\ -1 & 2
\end{pmatrix}$ and $V=\begin{pmatrix}2 & -2\\ 1 & -1
\end{pmatrix}$.

Then $C=U\otimes A+V\otimes B$, $U^2=U$, $V^2=V$ and $UV=VU=0$.

Hence, using the Kronecker mixed product,

$$C^2=(U\otimes A)^2+(V\otimes B)^2+(U\otimes A)\cdot(V\otimes B)+(V\otimes B)\cdot(U\otimes A)$$

$$C^2=U^2\otimes A^2+V^2\otimes B^2+(UV)\otimes (AB)+(VU)\otimes(BA)$$

$$C^2=U\otimes A^2+V\otimes B^2$$

By induction, we would find that

$$C^n=U\otimes A^n+V\otimes B^n$$

Note that the preceding works also for $C^0$, because $U+V=I_2$, hence
$$U\otimes A^0+V\otimes B^0=U\otimes I_3+V\otimes I_3=(U+V)\otimes I_3=I_6$$

Therefore

$$\exp C=U\otimes \exp A+V\otimes \exp B$$

$A$ is nilpotent and $A^3=0$ hence

$$\exp A=I+A+\frac12A^2=\begin{pmatrix}1 & 1 & \frac12\\
0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$$

$B$ is a Jordan matrix, and the first Jordan block is $\begin{pmatrix}2 & 1 \\ 0 & 2\end{pmatrix}=2\begin{pmatrix}1 & \frac12 \\ 0 & 1\end{pmatrix}$.

And $\begin{pmatrix}1 & \frac12 \\ 0 & 1\end{pmatrix}^n=\begin{pmatrix}1 & \frac n2 \\ 0 & 1\end{pmatrix}$, hence

$$B^n=\begin{pmatrix}2^n & n 2^{n-1} & 0\\
0 & 2^n & 0 \\ 0 & 0 & 3^n\end{pmatrix}$$

And

$$\sum_{n=0}^{\infty} \frac{n2^{n-1}}{n!}=\sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!}=e^2$$

Hence

$$\exp B=\begin{pmatrix}
e^2 & e^2 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & e^3\end{pmatrix}$$

All in all,

$$\exp C=\begin{pmatrix}
-1 & 2\\ -1 & 2
\end{pmatrix}\otimes\begin{pmatrix}1 & 1 & \frac12\\
0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}+
\begin{pmatrix}2 & -2\\ 1 & -1
\end{pmatrix}\otimes\begin{pmatrix}
e^2 & e^2 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & e^3\end{pmatrix}$$

$$\exp C=
\begin{pmatrix}
2e^2-1 & 2e^2-1 & -\frac12 & -2e^2+2 & -2e^2+2 & 1\\
0 & 2e^2-1 & -1 & 0 & -2e^2+2 & 2\\
0 & 0 & 2e^3-1 & 0 & 0 & -2e^3+2\\
e^2-1 & e^2-1 & -\frac12 & -e^2+2 & -e^2+2 & 1\\
0 & e^2-1 & -1 & 0  & -e^2+2 & 2 \\
0 & 0 & e^3-1 & 0 & 0 & -e^3+2\end{pmatrix}$$

And the hint with $D=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$?

Well, we have already noticed that $U$ and $V$ are two commuting projectors, and we have additionally:

$$D^{-1}UD=\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}$$
$$D^{-1}VD=\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$

That is, $U$ projects on the subspace spanned by $(1, 1)^T$, and $V$ projects on the subspace spanned by $(2, 1)^T$.

The eigendecomposition might have helped to discover that $U^2=U$, $V^2=V$ and $UV=VU=0$, which are needed to simplify $C^n$.

Title: find the exponential form of a matrix


答:
Denote $f(x) = a^Tx + \Vert x \Vert_\infty$.

Case $\Vert a \Vert_1 \gt 1$

Suppose that $\Vert a \Vert = \vert a_1 \vert + \dots + \vert a_n \vert \gt 1$ and denote for $\lambda \gt 0$ $x_\lambda = -\lambda(\text{sign}(a_1), \dots, \text{sign}(a_n))$ where $\text{sign}$ denotes the sign function.

You have $f(x_\lambda) = - \lambda \Vert a \Vert_1 + \lambda = \lambda (1 - \Vert a \Vert_1)$ and therefore $\lim\limits_{\lambda \to \infty} f(x_\lambda) = -\infty$.

Case $\Vert a \Vert_1 \le 1$

Then $$ 0 \le -\Vert a \Vert_1 \Vert x \Vert_\infty + \Vert x \Vert_\infty\le f(x)$$ and $f(0) = 0$. Therefore $\min\limits_x f(x) = 0$

Title: Compute the min of $a^Tx+||x||_{\infty}$

Body:
Prove that for any $a\in\mathbb{R}^n$ the following holds

$$\min a^Tx+\|x\|_\infty=\left\{\begin{array}{rcl} 0&\|a\|_1\leq1\\-\infty&\text{else}\end{array}\right.$$

I've tried looking at $\|x\|\cdot\|a\| \cos(\theta)+\|x\|_\infty$ and bound it from below by looking at the norm of $a$ but it didn't get me anywhere.
edit:

I've tried taking norm 1 on $a^Tx+\|x\|_\infty\geq-\|a\|_1 \|x\|_\infty + \|x\|_\infty$ from holder's inequality and from here i got $\|x\|_\infty(1-\|a\|_1) > -\infty\iff\|a\|_1\leq1$ but to get that the minimum is zero still stuck

any hint will be welcome :)


问:
What is the completion of the normed space?
What is the completion of the normed space $C_{0,0}, ||~||_p $ ?

$(1 \leq p < \infty)$ remember that if $\overline{x} = (Xn)^\infty_{n=1}$, then $||\overline{X}||_p = (\sum_{n=1}^{\infty}|Xn|^p)^{1/p}$

We have $C_{0,0}$ is the set of sequences with entries almost all zero except for a finite quantity. Then, we have the completion of $ C_{0,0}, ||~||_p $, is the space  $lp, ||~||_p$.

I think I have to prove that $ C_{0,0} \subset lp , \forall p$ tq $(1 \leq p < \infty)$ and $ C_{0,0}$ is  dense in $lp$

I don't know how can I follow this prove, could you help me, please?

答:
Your sequence space is more commonly denoted by $c_{00} = \{(x_n) \in \Bbb C^{\Bbb N}\mid \{n: x_n \neq 0\} \text{ is finite }\}$ (or take real sequences instead of complex ones, whatever your text does).

First off, it's clear that all such sequences are in $\ell^p$ for any $p\ge 1$ because we only have a finite sum to compute for the norm. So the norm is always well-defined; in short an "eventually $0$-sequence" is always $p$-summable. for any $p$.

The density is also clear because we can, for a given $x:=(x_n)_n \in \ell^p$, define a sequence $(x_1, 0, 0, \ldots), (x_1,x_2, 0,0,\ldots), (x_1,x_2,x_3,0,0,\ldots)$ if vectors in $c_{00}$ that converges to $x$ in the $\|\cdot\|_p$-norm (if $\varepsilon>0$, there is some $N\in \Bbb N$ such that $\sum_{n=N}^\infty |x_n|^p < \varepsilon$ etc.), so that $\overline{c_{00}} = \ell^p$ for every $1 \le p < \infty$.

The completion of $c_{00}$ in $\ell^\infty$ is the space $c_0$ of seqences that converge to $0$ in the sup-norm. The arguments are quite similar.


问:
Convergence of $ \int\limits_{0}^1 f(tx)t^{-1}dt$
Let $f   \in C ^\infty (\mathbb{R})$.

Why this is true :

If f(tx) vanishes for t= 0, then, $F(f)= \int\limits_{0}^1 f(tx)t^{-1}dt$, is well defined .

答: Since $f$ is infinitely differentiable, it is Lipschitz (see https://en.wikipedia.org/wiki/Lipschitz_continuity for why this holds). So in particular, we have the estimate $|f(tx)| \leq K|tx|$, where $K$ is the Lipschitz constant. This is enough to show integrability, since $$\int_0^1 |f(tx)| |t^{-1}|dt \leq K\int_0^1 |tx| |t^{-1}| dt = K |x| < \infty.$$


问:
Bijection between multisets and directed animals?
The number of directed animals (aka polyominoes) of size $n$ (A005773) is enumerated by the generating function $$\frac{1}{2} \left(1+\sqrt\frac{{1+z}}{{1-3 z}}\right).$$
This
generating function also enumerates the number of $n$-element multisets of $\{1,\ldots,n\}$ containing no pair of consecutive integers (e.g. $111, 113, 133, 222, 333$ for $n=3$).

The first few numbers in the sequence are
$$1, 1, 2, 5, 13, 35, 96, 267, 750, 2123, 6046, 17303, 49721.$$
Is there a good bijection between the two combinatorial classes?

This question is now resolved in this paper.

答: This question is now resolved in this paper.</p>


问:
Finding an explicit isomorphism between $\mathbb{Z}^{\times}_n$ and $\mathbb{Z}^{\times}_{2n}$
For an odd integer $n$, find an explicit isomorphism between $\mathbb{Z}^{\times}_n$ and $\mathbb{Z}^{\times}_{2n}$.

How do I do this? I don't really know where to start. I can easily find bijections but am yet to find a structure preserving one.

答:
If $R$ is any ring, let $R^{\times}$ be its group of units.  Then if

$f: R \rightarrow S$

is a ring homomorphism, it restricts to a group homomorphism

$f^{\times}: R^{\times} \rightarrow S^{\times}$

on the unit groups.  Here you have a natural (i.e., quotient) map

$f: \mathbb{Z}/2n \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$

so $f^{\times}$ is a homomorphism between two finite groups that you want to show are isomorphic.  Since they are finite and you say you already know that they have the same order, you are in a good place: it suffices to show either that $f^{\times}$ is injective or that $f^{\times}$ is surjective.  Give it a try.

(But, to be sure, if $f^{\times}$ is in fact an isomorphism, you can presumably show that it is both injective and surjective from scratch, so you don't actually need to know that $\varphi(2n) = \varphi(n)$ for odd $n$ in advance -- not that this is at all hard to see.)

By the way, studying the injectivity/surjectivity of $f_{m,n}^{\times}$ where

$f_{m,n}: \mathbb{Z}/mn \mathbb{Z} \rightarrow \mathbb{Z}/n \mathbb{Z}$

is a good exercise in general -- i.e., for any $m,n \in \mathbb{Z}^+$.  It's a nice mixture of algebra and number theory.


问:
How do you define definition symbol :=?
How is "$:=$" defined formally and why?  "$\iff$", "$=$", ...? </p>

答: There is no need of a formal definition of a definition. A definition is just an abbreviation. For example, in formal arithmetic (number theory) we write $x\mid y$ ($x$ divides $y$) as an abbreviation for $\exists z (y=x\times z)$. Whenever $s\mid t$ is used, it can be replaced in principle by the "long" form.

Title: How do you define definition symbol :=?


问:
What do Gödel's incompleteness theorems actually tell us, and how?
I'm currently reading a book on the famous Gödel incompleteness theorems which, at least as I originally understood it, purport to prove that mathematics itself cannot be axiomatized (that is, there exists no axiomatic system upon which all of math can be constructed). Upon further inspection, however, it seems that the truth is a bit more murky. Wikipedia describes the theorems as "that [which] demonstrate the inherent limitations of every formal axiomatic system capable of modelling basic arithmetic"... a bit vague.

What do Gödel's incompleteness theorems actually tell us, and how do they go about proving it, roughly?

No doubt I will have a deeper understanding after spending more time with this, however I can't help but want to anticipate the conclusion now. I'm looking for an answer in simple terms (I'm familiar with fundamental mathematical proof techniques, logic, e.g. at the undergraduate level). Is it true that math itself somehow cannot be axiomatized?

答:
In simple terms the way I understand it is this. If you pick any consistent set/system $A$ of axioms (which obeys certain conditions), and thus build a (math) theory, there will always be statements $S$ in this theory which you can formulate and which are true, but you cannot prove just by using your set $A$ of axioms.

This is Goedel's 1st incompleteness theorem.

That's why it's called incompleteness theorem. Because any consistent system of axioms is not complete i.e. cannot prove all the statements which can be formulated.

There's also a 2nd incompleteness theorem by Goedel which states that no set/system of axioms can prove its own consistency.

Note that both formulations given here are rough and loose.

See also:

How Goedel's incompleteness theorems work?

Book:

Ernest Nagel, James Newman, "Gödel's Proof", 1958


问:
inhomogeneous elements in vector superspace?
I am currently studying Lie Superalgebra, and having a confusion on basic concepts of superalgebra. From Kac's Lie Superalgebra(p.13), it says that "if deg$a$ appears in a given Superalgebra, then it is assumed that $a$ is homogeneous, and that the expression is extended to the other elements by the linearity". What I am confused is that in any superalgebra, is there a inhomogeneous elements in Superalgebra i.e which is neither even nor odd?
Thanks

答: Almost any element of a superalgebra $\mathfrak g$ is inhomogeneous. In fact, if $\mathfrak g_0$ is the even part of $\mathfrak g$ and if $\mathfrak g_1$ is the odd part, then for every $X\in\mathfrak g_0\setminus\{0\}$ and for every $Y\in\mathfrak g_1\setminus\{0\}$, $X+Y$ is inhomogeneous.


问:
closeness of a set of probability distributions
how can one check if a family of distributions is closed? what does closeness of a set of distributions mean? I understand the concept of closed sets in the context ofEuclidean geometry, but I have difficulty understanding what these concepts mean for a set of probabilities.</p>

答: It all depends (by definition, really) on what topology you put on the set of distributions. There are probably several that are used (weak topology, vague topology IIRC), depending on context. So go find out what that context is and you'll know what closed means. It probably won't be easy, depending on how strong your background in general topology or functional analysis or measure theory is.


问:
Inequality with sides and angles in a triangle

Let $ABC$ be a triangle. Show that
$$ 60^o \le \frac{aA+bB+cC}{a+b+c} \le 90^o$$
where $A$ is the measure in degrees of the angle corresponding to the vertex $A$ and $a$ the side opposite to it ($B$, $b$, $C$ and $c$ are defined similarly).

I saw this problem in a Facebook group and couldn't figure out the solution. It was a long time ago, so there is no way I can find that post again (only took a screenshot of the problem).

I have no idea where to start. My thoughts about the problem:

$a+b+c$ is the perimeter of the triangle.

The quantity $\frac{aA+bB+cC}{a+b+c}$ is the weighted average angle with sides as weights.

The formula is obvious for an equilateral triangle.

Since $\frac{aA+bB+cC}{a+b+c}=60^o$ is obtained for an equilateral triangle, I was thinking using Lagrange multipliers method with the Law of Cosines, Law of Sines and $A+B+C=180^0$ has constraints but the computations seem complicated.

Any idea?

答:
Let's start with the right inequality. Notice that $a,b,c$ satisfy $a+b+c>2a, a+b+c>2b, a+b+c>2c$ due to the triangle inequality. Multiply these inequalities respectively with $\alpha, \beta,\gamma$ $$(\alpha+\beta+\gamma)\cdot (a+b+c)>2\cdot(a\alpha+b\beta+c\gamma)\iff90^\circ=\frac{\alpha+\beta+\gamma}2>\frac{a\alpha+b\beta+c\gamma}{a+b+c}$$

For the second one, suppose wlog $a\geqslant b\geqslant c$. This implies, as it is well-known, that $ \alpha\geqslant \beta\geqslant \gamma$ so you can use Chebyshev's inequality as follows $$\frac{a\alpha+b\beta+c\gamma}3\geqslant \frac{(a+b+c)}3\cdot\frac{(\alpha+\beta+\gamma)}3\iff \frac{a\alpha+b\beta+c\gamma}{a+b+c}\geqslant \frac{\alpha+\beta+\gamma}3=60^\circ$$


问:
general solution of equation and relation
I am interested in learning the below question in some elementary way. Please discuss this problem and help me to get mind free state.

How to get solutions for $x^2 - 10y^2$ = $1$? I would like to learn some easy method. Also, I have seen in some text book $x= 19$ and $y = 6$ is one of the solution and general solution can be prepared by using $x = 19$ and  $y = 6$ in terms of $n$. where $n$ is some positive integer. Please explain this general solution and relation with $x = 19$ and $y =6$.

Advance thanks to all.

答:
This is what is known as Pell's equation.

This is how to find all solutions in an elementary way.

$$x^2-10y^2=1$$ can be factored to give $$(x+\sqrt{10}y)(x-\sqrt{10}y)=1$$

You can find more solutions by taking $(19+\sqrt{10}6)$ to some integer power, collecting like terms and putting $y$ equal to the coefficient of $\sqrt{10}$ and $x$ to the integer part of the result. These will always be solutions because $1^k=1$ and because the binomial theorem confirms that the absolute value of the negative term in the second bracket will always be equal to the value of $\sqrt{10}y$.

Note we have not shown that this is a complete set of solutions. That is much harder to do, and not as elementary as you would probably want. If you do want an explanation however, just leave a comment below.

Why $$x=\frac{(x+\sqrt{10}y)^n+(x-\sqrt{10}y)^n}{2}$$ is always a solution:

We know from the above that $x$ is integer part of $(19+\sqrt{10} \cdot 6)^n$.

Note that in the expansion of $(x+\sqrt{10}y)^n+(x-\sqrt{10}y)^n$ all the irrational terms cancel out (by the binomial theorem http://en.wikipedia.org/wiki/Binomial_theorem). What is left is twice the required coefficient so we divide by 2 to get $x$.


问:
Linear Maps and component functions
Prove or disprove the following.

Let $f$: $\mathbb{R}^2$ $\to$ $\mathbb{R}^2$.

be a map with component functions $g$ and $h$ such that $f(x,y)$ = $(g(x,y),h(x,y))$. If $f$ is linear, then $g$ and $h$ are linear as well.

What's the best method to prove/disprove the following? Would showing f be linear in terms of g and h work?

Title: Linear Maps and component functions

Body: Given $f(k_1(x_1,y_1)+k_2(x_2,y_2))=k_1f(x_1,y_1)+k_2f(x_2,y_2)$, i.e.$$\begin{bmatrix}g(k_1(x_1,y_1)+k_2(x_2,y_2))\\h(k_1(x_1,y_1)+k_2(x_2,y_2))\end{bmatrix}=k_1\begin{bmatrix}g(x_1,y_1)\\h(x_1,y_1)\end{bmatrix}+k_2\begin{bmatrix}g(x_2,y_2)\\h(x_2,y_2)\end{bmatrix}$$which gives $g(k_1(x_1,y_1)+k_2(x_2,y_2))=k_1g(x_1,y_1)+k_2g(x_2,y_2)$ and similarly for $h$.


问:
Derivative of a function and Sobolev space
Let function $f$ belongs to the Sobolev space or order $\beta$ defined by
$$
\mathcal{S}^{\beta}(\mathbb{R}) = \left\{u \in L^2(\mathbb{R}): \int_{\mathbb{R}}(1+|\xi|^2)^{\beta}|\hat u(\xi)|^2d\xi < \infty \right\},
$$
where $\hat u(\xi) = \int_{\mathbb{R}} e^{-ix\xi}u(x)dx$ is the Fourier transform of $u$. Can we deduce directly that the derivative $f'$ of $f$ belongs to $\mathcal{S}^{\beta - 1}(\mathbb{R})$? Thank in advance for any help.

答:
By the answers from this question:

Fourier Transform of Derivative

It follows that $|\hat{u'}(\xi)|^2=\frac{1}{({2\pi|\xi|})^2}|\hat{u}(\xi)|^2$. Inserting this with $f\in\mathcal{S}^\beta(\mathbb{R})$ then brings

\begin{align*}
\int_\mathbb{R} \frac{1}{4\pi^2}(1+|\xi|^2)^{\beta-1}|\hat{u'}(\xi)|^2d\xi\leq \int_\mathbb{R} \frac{(1+|\xi|^2)^{\beta}}{4\pi^2|\xi|^2}|\hat{u'}(\xi)|^2d\xi=\int_\mathbb{R}(1+|\xi|^2)^\beta|\hat{u}(\xi)|^2d\xi <\infty.
\end{align*}


问:
Mathematicall term for connected region the boundary of which is connected
I want to know the mathematical term for connected region the boundary of which is connected, if exists.

Intuitively, I want to define a connected region which does not have any hollow inside. For example, in the figure below, the first one is OK as it is connected and also does not have any hollows inside. The second one is not OK as it is not connected. The third one is not OK as it has a hollow so the boundary of the region is not connected.

答: A region without "holes" is usually called SIMPLY CONNECTED, but this is NOT the same as you ask fo: For exmaple, think on the exterior of a circle: it is a connected region and its boundary is connected, BUT, this is not a simply connected region.</p>


问:
Finding a Power Series Solution for $y''-y'+xy=0$
I'm solving the problem as part of a homework in my second applied analysis course, and I'm having a hard time solving the recurrence. I got

$$
     \begin{cases}
          \begin{align}
               a_1-2a_2&=0\\
               (n+3)(n+2)a_{n+3}-(n+2)a_{n+2}+a_{n}&=0
          \end{align}
     \end{cases}
$$

From which I got

$$
     \begin{align}
          a_{3}=-\frac{1}{3!}a_{0}+\frac{1}{3!}a_{1}\\
          a_{4}=-\frac{1}{4!}a_{0}-\frac{1}{4!}a_{1}\\
          a_{5}=-\frac{1}{5!}a_{0}-\frac{4}{5!}a_{1}\\
          a_{6}=\frac{3}{6!}a_{0}-\frac{8}{6!}a_{1}\\
          a_{7}=\frac{8}{7!}a_{0}-\frac{3}{7!}a_{1}\\
          a_{8}=\frac{14}{8!}a_{0}+\frac{2}{8!}a_{1}\\
     \end{align}
$$.

Any ideas on how to solve it?


问:
Image of continuous Galois representation constructed by Deligne corresponding to a eigenform.
It is a well known theorem that given a eigneform $f$ of weight $k$ and level $N$, Deligne, Serre, Shimura and others have constructed continuous $\ell$-adic Galois representation $\rho_f : \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\to \mathrm{GL}_2(E)$ such that the trace and determinant of the image of Frobenius at prime $p$ different from $\ell$ and not dividing $N$ satisfies nice properties which connects it to the Fourier coefficients of $f$.
But I have seen far too many times that an expert while giving a talk says that we can take the representation $\rho_f : \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}) \to \mathrm{GL}_2(\mathcal{O}_E)$. What is the reason of this switch from $\mathrm{GL}_2(E)$ to $\mathrm{GL}_2(\mathcal{O}_E)$ ? Does it have something to do with the compactness of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ ?

Thanks for the help !!

答:
This follows from the compactness of $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$. Let $G$ be the image of $\rho_f$. The point is that, since $G$ is compact, there is a $G$-stable lattice in $\Lambda \subset E^2$ and $\mathrm{Aut}(\Lambda)\cong \mathrm{GL}_2(\mathcal O_E)$

Indeed, $G$ is a compact subgroup of $\mathrm{GL}_2(E)$ so, in particular, its denominators are bounded.

Consider $\Lambda = \sum_{g\in G}g\mathcal O_E^2$, an $\mathcal O_E$-module preserved by $G$. Then
$$\mathcal O_E^2\subseteq\Lambda\subseteq \pi^{-k}\mathcal O_E^2$$
for some $k$, where $\pi$ is a uniformiser of $E$. It follows that $\Lambda$ is a free $\mathcal O_E$-module and hence there exists $h\in\mathrm{GL}_2(E)$ such that $h\mathcal O_E^2 = \Lambda$. Then $h^{-1} G h\subset \mathrm{GL}_2(\mathcal O_E)$.


问:
How many ways to add an edge to a directed acyclic graph?
Suppose I have a directed acyclic graph (DAG) $G=(V,E)$, how many ways can I add a single edge and still have a DAG?

I saw a similar question here whose solution uses a topological order on the vertices. I thought to count (somehow) the number of edges that could be potentially added while preserving the order. However since the ordering may not be unique you'd have to enumerate over all possible orderings whilst making sure to not count pendent edges twice. But perhaps there's a better way, ideally a closed expression?

答:
Let $n$ and $m$ denote the number of vertices and edges of $G$, respectively. I'll assume that you'd like to know how many directed edges we can add, so that if for $u,v \in V$ we can add either of $uv$ and $vu$, we count that as two ways of adding a new edge. Otherwise, the answer is ${n \choose 2} - m$, since for any pair of non-neighbouring vertices $u,v$, you can add at least one of $uv$ and $vu$ to the graph (e.g. if in a topological ordering $u$ comes before $v$, then you can add $uv$).

Now for any pair of vertices $u,v$, if you can't add the edge $uv$ to the graph, that's either because $uv$ is already an edge, or because adding $uv$ would create a directed cycle, which happens precisely if $u$ is reachable from $v$. Since $G$ is a DAG, these two possibilities are mutually exclusive. This means that if $m'$ denotes the number of ordered pairs $(u,v)$ such that $u$ is reachable from $v$, then the number of directed edges you can add is $n(n-1) - m - m'$. Note that $m'$ is the number of edges of the transitive closure of $G$.


答:
Using that $\hat{u'}(\xi) = i\xi\hat u(\xi)$ and $\lvert\xi\rvert \leq 1+\lvert\xi\rvert^2$:
$$
\int_\mathbb R (1+\lvert\xi\rvert^2)^{\beta-1}\lvert\hat{u'}(\xi)\rvert^2d\xi = \int_\mathbb R \lvert\xi\rvert(1+\lvert\xi\rvert^2)^{\beta-1}\lvert\hat{u}(\xi)\rvert^2d\xi \leq \int_\mathbb R (1+\lvert\xi\rvert^2)^\beta\lvert\hat u(\xi)\rvert^2 d\xi.
$$

Title: Derivative of a function and Sobolev space

Body:
Let function $f$ belongs to the Sobolev space or order $\beta$ defined by
$$
\mathcal{S}^{\beta}(\mathbb{R}) = \left\{u \in L^2(\mathbb{R}): \int_{\mathbb{R}}(1+|\xi|^2)^{\beta}|\hat u(\xi)|^2d\xi < \infty \right\},
$$
where $\hat u(\xi) = \int_{\mathbb{R}} e^{-ix\xi}u(x)dx$ is the Fourier transform of $u$. Can we deduce directly that the derivative $f'$ of $f$ belongs to $\mathcal{S}^{\beta - 1}(\mathbb{R})$? Thank in advance for any help.


问:
Unscented (dual) quaternion kalman filter for pose estimation
This paper Unscented Dual Quaternion Particle Filter for SE(3) Estimation shows a method to use Unscented Kalman Filter (UKF) on dual quaternions in the pose estimation problem.

It proposes a projection/retraction method to project/retract sigma points on the dual-hypersphere to Euclidean space back and forth. The following is the projection equation.

A is a 4x3 matrix, which is also the last 3 columns of Q (4x4) matrix. By simple matrix multiplication, the numerator of tau is 0, which doesn't make sense, because it means all the projected points tau, regardless what x is, is 0. Do I miss something here?

答:
I suspect that the paper might have a typo in those equations. Namely, instead of $\mathcal{A}_{x_r}^{\circ}$ and $\mathcal{Q}_{x_r}^{\circ}$ I think one is supposed to use $\mathcal{A}_{v_r}^{\circ}$ and $\mathcal{Q}_{v_r}^{\circ}$ respectively. However, when using this and substituting $\tau_r = \mathcal{P}_{v_r}\!(x_r)$ in $x_r = \mathcal{U}_{v_r}\!(\tau_r)$ I get that

$$
\mathcal{U}_{v_r}\!(\tau_r) = x_r \frac{|v_r^\top x_r|}{v_r^\top x_r} = \pm x_r.
$$

This would also make more sense for the notation of $\mathcal{U}_{v_r}\!(\tau_r)$, which otherwise would also be directly a function of $x_r$, instead of just $v_r$ and $\tau_r$. It can be noted that negating a unit quaternion represents the same attitude. Therefore, this possible change in sign might not matter for the rest of the algorithm.

This sign issue doesn't happen when one keeps using $\mathcal{A}_{x_r}^{\circ}$ and $\mathcal{Q}_{x_r}^{\circ}$, but that would indeed yield $\tau_r = 0$ which is quite a boring projection. However, in order to be completely sure about whether this is indeed a typo you could try to contact one of the authors.


问:
If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$
Matrix Question

If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!

Thanks for help in advance!

答: $A,B$ belong to the group $GL_n(F),~(F$ being the field over which $A,B$ are defined$).$ So you may apply cancellation on $(AB)^2=A^2B^2.$</p>


问:
Calculating Variance using PDF $2−2x$ , for $x∈(0,1)$ and $0$ otherwise.
$2−2x$ for  $x∈(0,1)$ and $0$ otherwise.

i started by calculating expectation $E(x)$
so integrated $(2−2x)x$ from $0$  to  $1$
and it was $E(x)=1/3$

now to find variance(x) it is $var(x)= E(x-E(x))^2$ and i am getting it zero which is wrong

答:
You have to calculate the second moment $\mathbb{E}[X^2]$ then easy find

$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$

$$\mathbb{E}[X^2]=\int_0^1 x^2 f(x) dx=\dots=\frac{1}{6}$$

thus

$$\mathbb{V}[X]=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}$$


Title: Calculating Variance using PDF $2−2x$ , for $x∈(0,1)$ and $0$ otherwise.

Body:
$2−2x$ for  $x∈(0,1)$ and $0$ otherwise.

i started by calculating expectation $E(x)$
so integrated $(2−2x)x$ from $0$  to  $1$
and it was $E(x)=1/3$

now to find variance(x) it is $var(x)= E(x-E(x))^2$ and i am getting it zero which is wrong

Body:
$$Var(X)=E[X^2]-E[X]^2=2\int_0^1(x^2-x^3)dx-\frac19=2\left(\frac{x^3}{3}-\frac{x^4}{4}\right)_0^1-\frac19=$$$$=2\left(\frac13-\frac14\right)-\frac19=\frac16-\frac19=\frac{1}{18}
$$

Using your method

$$Var(X)=E[(X-E[x])^2)=\int_0^1(x-1/3)^2(2-2x)dx=$$ $$=\frac19\int_0^1\left(-18 x^3+{30 x^2}-{14 x}+{2}\right)dx=\frac19 \left(-\frac{9x^4}{2}+10x^3-7x^2+2x\right)_0^1=\frac19(-9/2+10-7+2)=\frac{1}{18}$$


Body: $Ex^{2}=\int_0^{1} x^{2}(2-2x)dx=\frac 2 3 -\frac 2 4 =\frac 1 6$ so the variance is $\frac  1 2-(\frac  1 3)^{2}=\frac1  {18}$.


问:
Find $y$-Lipschitz constant
$$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$

I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$

I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$

now I don't know what to do.

答: Better to find $\frac{df}{dy}$, then you will get either $2a^3$  or $2a^4$.</p>

Title: Find $y$-Lipschitz constant


问:
Why is $\frac{n-1}{n}$ always in its simplest form?
So I was just messing around on my calculator and I noticed that for every positive integer the formula $\frac{n-1}{n}$ is always a fraction in its simplest form im not quite sure why this happens is there a case where this is not true?

答:
Because $n-(n-1)=1$ and this means that $n$ and $n-1$ are co-prime. Can you see why ?

Any common divisor to $n$ and $n-1$ divides their difference...