问:
Confusion in one of the combinatorics/probability problem, with my approach for its solution.

The only contents of a container are 10 disks that are each numbered
with a different positive integer from 1 through 10, inclusive. If 4
disks are to be selected one after the other, with each disk selected
at random and without replacement, what is the probability that the
range of the numbers on the disks selected is 7?

My approach: (Using the concept of conditional probability and combinatorics)

Possible ranges: 1-8, 2-9, 7-10 (Total 3)

Now, taking one range at a time, (say 1-8):
Hence, among the 4 disks to be selected, the probability of selecting the first disk will 1/10 (for 1).
Similarly, 1 is selected, hence second place could be filled by 8. Prob(second place is 1/9).
Now, since out of range 1-8 1 and 8 are selected, therefore we have only 6 choices (out of the remaining 8) for third place. Henceforth, only 5 choices (out of 7 choices) for fourth place.
Combining the probability of all 4 places, (1/10)(1/9)(6/8)(5/7).

Now, the number in four places can be arranged in 4! ways. Also, apart from the (1-8) range, there is also other two range of numbers.
Therefore, combining all, the required probability should be (3) x (1/10)(1/9)(6/8)(5/7) x (4!).
But my answer is wrong. My answer is double of actual answer.

答:
Re the glitch in your answer, (3) x (1/10)(1/9)(6/8)(5/7) x (4!)

(6/8)(5/7) implies that the two discs within the limiters are already permuted, so the (4!) needs to be replaced by (4!/2!)

If you had written (2/10)(1/9) for the limiters, the final factor would become (4!)/(2!2!)

It is because of such hidden snags that I prefer to use the safer route of combinations,  as @saulspatz has done.


问:
Is function $f(x,y)=2x+3y$ linear?
Is function

$f(x,y)=2x+3y$

linear?

If it is linear, how can I prove that this is linear(superposition, homogeneity)?

I don't know how to prove linearity of more than two variable function.

ContentLicense: CC BY-SA 4.0

Title: Is function $f(x,y)=2x+3y$ linear?

Body:
You have to show that

$$f((x_1,y_1)+(x_2,y_2))=f(x_1,y_1)+f(x_2,y_2)$$

and

$$f(c(x,y))=cf(x,y).$$

($c$ is a scalar.)


问:
How to solve a simple ODE using Green's function
How do I find the solution to

    $\ddot{x} = f(t) \equiv\begin{cases} a, 0\leq t \leq T, \\ 0, T<t<\infty \end{cases}$
where $a=$ const,
subject to $x(0)=0$, $\dot{x}(0)=0$ using a Green's function? This should be a very straightforward problem but I am stuck.
So far I have found (using the boundary + jump conditions + continuity) that the Green's function is given by: $G(t,T) = \begin{cases} G_1(t,T) = 0, 0\leq t \leq T, \\ G_2(t,T) = t-T, t>T \end{cases}$.
The solution is then \begin{align} x(t) &= \int_0^{\infty} G(t,T) f(T)dT \\
&=\int_0^{t}G_2(t,T)f(T)dT + \int_t^{\infty}G_1(t,T)f(T)dT \\
&=\int_0^t(t-T)f(T)dT \end{align}
as $G_1(t,T) \equiv 0$.

 But now $f(T) =0$ for $T<t$, so isn't the integral above just zero? I'm confused.

答:
Using  $y$ as the variable of integration instead of $T$
we have \begin{align*} x(t) &= \int_0^\infty G(t,y)f(y)dy\\ &= \int_0^tG_2(t,y)f(y)dy + \int_t^\infty G_1(t,y)f(y)dy \\&=\int_0^t(t-y)f(y)dy, \end{align*}
since $G_1 \equiv 0$. If $t\geq T$,
\begin{align*} x(t) &= \int_0^Ta(t-y)dy + \int_T^t0\cdot(t-y)dy \\ &=aT(t-T/2)\end{align*}
If $t<T$, we get \begin{align*}x(t) &= \int_0^ta(t-y)dy \\&= \frac{at^2}{2}\end{align*}
Hence the solution is $x(t)$ = \begin{cases} \frac{at^2}{2} & \mbox{if }t<T \\ aT(t-T/2) & \mbox{if } t\geq T \end{cases}


问:
Is $K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi}$
Working on this conjecture, I found its corollary, which is also supported by numeric calculations up to at least $10^5$ decimal digits:
$$K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi},$$
where $K(x)$ is the complete elliptic integral of the 1st kind. I did not find this specific value at MathWorld, Wolfram Functions Site, Wikipedia or DLMF.

Is it a known value?

答:
See here: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html and also here: http://mathworld.wolfram.com/EllipticIntegralSingularValuek3.html

Your value is actually
$$ \sin \frac\pi{12} = \frac{\sqrt{2-\sqrt3}}{2}, $$
and according to MathWorld, it is known as the third singular value $k_3$. It satisfies
$$ K(\sqrt{1-k_3^2}) = \sqrt{3}K(k_3) $$
and
$$ K(k_3) = \frac{\sqrt{\pi}\Gamma(1/6)}{2\cdot 3^{3/4}\Gamma(2/3)}. $$
Mathematica says that the two closed forms are equal.

ContentLicense: CC BY-SA 3.0


问:
Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.
I am reading "An Introduction to Analysis vol. 2" (in Japanese) by Kazuo Matsuzaka.

There is the following example in this book:

Let $f_n(x) := \frac{\sin n^2 x}{n}$ for $n = 1,2,\dots$.

Then $f(x):=\lim_{n\to\infty} f_n(x)=0$, so $f'(x) = 0$.

On the other hand, $f_n^{'}(x) = n \cos n^2 x$ and $f_n^{'}(x)$ doesn't converge as $n\to\infty$.

So, $f_n\to f$ and, $f_n$ and $f$ are differentiable, but $f_n^{'}\to f^{'}$ doesn't hold.

The author wrote $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

But I cannot prove that $n\cos n^2x$ doesn't converge as $n\to\infty$ even for $x=1$.

Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

答:
As @user20672 said, if $\lim_{n\to\infty} n\cos n^2 x\in\mathbb{R}$ then $\lim_{n\to\infty}\cos n^2 x=0$.

Assume that $\lim_{n\to\infty} n\cos n^2 x=A\in\mathbb{R}$ for some $x\in\mathbb{R}$.

Let $a_n := \cos n^2x$.

Then $\lim_{n\to\infty} a_n =\lim_{n\to\infty} \frac{1}{n} \times n\cos n^2x= 0 \times A=0$.

Since $\{a_{2n}\}$ is a subsequence of $\{a_n\}$, so $\lim_{n\to\infty} a_{2n}=0$.

On the other hand, $a_{2n}=\cos(2n)^2x=\cos4n^2x=2(\cos2n^2x)^2-1=2(2(\cos n^2x)^2-1)^2-1=2(2a_n^2-1)^2-1$.

So, $\lim_{n\to\infty}a_{2n}=\lim_{n\to\infty}2(2a_n^2-1)^2-1=2(2\times0^2-1)^2-1=2-1=1$.

This is a contradiction. So $\lim_{n\to\infty} n\cos n^2 x$ diverges.


问:
The equivalence condition for a linear map $T:V\longrightarrow Z$ with $W\subset \ker(T)$ between normed vector space to be continuous.
Let $V$ be a normed vector space and $W$ be a closed linear subspace. The quotient norm on $V/W$ is defined by $$\|[x]\|_{V/W}:=\inf_{y\in W}\|x-y\|_{V}.$$ (As $W$ is closed, the "norm" is indeed a norm)

We also define the quotient map $Q:V\longrightarrow V/W$. I am learning several properties of the map $Q$, so far, I have proved that

$\|Q\|_{op}=1$, bounded (hence continuous).

$Q:V\longrightarrow V/W$ is an open map, by proving $Q(B_{r}^{V}(x))=B_{r}^{V/W}([x]).$

$A\subset V/W$ is open if and only if $Q^{-1}(A)$ is open in V.

Now, I want to use 1 and 2 to prove the following statement:

For every vector space $Z$, a linear map $T:V\longrightarrow Z$ with $W\subset \ker(T)$ is continuous if and only if the unique map $\tilde{T}$ such that $T=\tilde{T}\circ Q$ is continuous.

By the description of $\tilde{T}$, it should be the map $\tilde{T}:V/W\longrightarrow Z$. But then I don't know where to start. How is this related to $\ker(T)$?

What should I do? Thank you!

答:
The map $\overline{T}:V/W\to Z$ should be defined by

$$ \overline{T}\Big( v+W \Big):= T(v)  $$

as all maps from the quotient space are.

Since composing continuous maps is continuous, one implication should be obvious. So you only have to show that $\overline{T}$ is also continuous if $T$ is.

So take an open subset $U\subset Z$ and we want to show that $\overline{T}^{-1}[U]$ is open in $V/W$. ${T}^{-1}[U]$ is open by assumption in $V$. Since $Q$ is open, you know that $ Q\circ T^{-1}[U] $ is open in $V/W$. Now just verify that $Q\circ T^{-1}[U]=\overline{T}^{-1}[U]$, and you're done.


问:
Lagrange constraints for points on a triangle and points on a square
If I have a function

$P = \frac{q}{\sqrt{(x_1-x_2)^2 + (y_1+y_2)^2}} + \frac{q}{\sqrt{(x_1-x_3)^2 + (y_1+y_3)^2}} + \frac{q}{\sqrt{(x_3-x_2)^2 + (y_3+y_2)^2}}$

where the points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ are subject to the constraint that they must lie on an equilateral triangle with side length $1$ and with vertices at $(-\frac{1}{2}, 0)$, $(\frac{1}{2},0)$, $(0, \frac{\sqrt{3}}{2})$, what would the constraints look like if I'm using a Lagrange multiplier? I know for circle with radius $1$ centered around the origin we would get three different constraint functions

$g(x_1, y_1) = x_1^2 + y_1^2 - 1$,

$h(x_2, y_2) = x_2^2 + y_2^2 - 1$,

$j(x_3, y_3) = x_3^2 + y_3^2 - 1$,

but I'm not sure how to set up the constraints for a triangle.

And what about for a square with side length $1$ with vertices at $(-1,0)$, $(-1, 1)$, $(1, 1)$, $(1, 0)$?

答:
Given the points $p_1,p_2,p_3$ characterizing the vertices of an equilateral triangle with known locations, the sides can be described by three parameters $\lambda,\mu,\nu$ such that

$$
\cases{
s_{12}(\lambda)= \lambda p_1+(1-\lambda)p_2,\ \ 0\le\lambda\le 1\\
s_{23}(\mu)= \mu p_2+(1-\mu)p_3\ \ 0\le\mu\le 1\\
s_{31}(\nu)=\nu p_3+(1-\nu)p_1\ \ 0\le\nu\le 1\\
}
$$

and the problem could be transformed into

$$
\min(\max)_{\lambda,\mu,\nu}\left(\frac{q}{\|s_{12}(\lambda)\|}+\frac{q}{\|s_{23}(\mu)\|}+\frac{q}{\|s_{31}(\nu)\|}\right)\ \ \text{s. t.}\ \ \cases{0\le\lambda\le 1\\  0\le\mu\le 1\\ 0\le\nu\le 1}
$$

NOTE

The restrictions can be eliminated by doing

$$
\cases{
\lambda = \frac 12(\sin u+1)\\
\mu = \frac 12(\sin v + 1)\\
\nu = \frac 12(\sin w + 1)
}
$$


问:
Show that series $\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$ is divergent
Show that series $$\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$$

is divergent . What inequality can we use here. i tried various method but none of these give any result. Any hint please .

答:
By the integral test:
$$
\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}\sim \int_2^\infty\frac{dx}{(\log x)^{\log(\log x)}}\stackrel{\log x\mapsto u}=
\int_{\log2}^\infty\frac{e^udu}{u^{\log(u)}}\stackrel{\log u\mapsto t}=
\int_{\log(\log2)}^\infty e^{e^t-t^2+1}d t.
$$
which clearly diverges.


问:
Is a sequence with this property convergent or at least bounded?
Let $(u_k)_{k=0}^\infty, (v_k)_{k=1}^\infty$ and $(w_k)_{k=1}^\infty$ be nonnegative sequences that satisfy

$u_k^2-u_{k-1}^2\le v_k u_{k-1}+w_k$ for all $k \geq 1$

$\sum_{k=1}^\infty v_k < +\infty$ and $\sum_{k=1}^\infty w_k < +\infty$

I've come across this question while doing stuffs, and I suspect that the sequence $(u_k)$ is convergent. But I still haven't been able to either prove it or find a counter example. I don't know if there is any result regarding this available? Any idea/ answer would highly be appreciated.

答:
Let $\displaystyle V=\sum^\infty_{k=1}v_k$, $\displaystyle W=\sum^\infty_{k=1}w_k$ and $\displaystyle m_n=\max_{1\le k\le n}u_k$. Then,
$$
u^2_k-u^2_0=\sum^k_{i=1}(u^2_i-u^2_{i-1})\le m_{k-1}\,\sum^k_{i=1}v_i+\sum^k_{i=1}w_i\le V\,m_{k-1}+W\le V\,m_n+W
$$
for $1\le k\le n$, so we must have
$$m^2_n\le V\,m_n+W+u^2_0,$$ and that means
$$u_n\le m_n\le V/2+\sqrt{V^2/4+W+u^2_0}.$$


问:
Convergence of series (factorial)
I cant think of a way to approach

$$\sum_n\frac{1}{n\ln(n!)}$$

I tried Cauchy condensation test, but then i don't know how to deal with

$$\frac{1}{\ln((2^n)!)}$$

Any hints are welcome, thanks for your time

Title: Convergence of series (factorial)

Body:
Hint: $n!=1\cdot 2\cdot 3\cdots\cdot n\ge 1\cdot 2\cdots\cdot 2=2^{n-1}$ so:

$$\frac{1}{n\ln(n!)}\le\frac{1}{n\ln(2^{n-1})}=\frac{1}{n(n-1)\ln 2}$$


问:
Kernels and Cokernels of differentials on a bounded long exact sequence.
Given a bounded long exact sequence on a smooth projective variety where objects are direct sums of $\mathcal{O}(n)$ for different integer $n$'s i.e. $\bigoplus_i\mathcal{O}(n_i)$. Is there anything we can tell about the kernels and cokernels of the differentials? Like are they necessarily generated by extensions of line bundles $\mathcal{O}(n)$ or they can be very complicated vector bundles?

答:
Question: "Like are they necessarily generated by extensions of line bundles O(n) or they can be very complicated vector bundles?"

Answer: Hartshorne Corr. II.5.18 says: If $X \subseteq \mathbb{P}^n_A$ is projective with $A$ noetherian, it follows any coherent sheaf $F$ is a quotient

$$ f_1:\oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$

If $A$ is noetherian it follows $X:=Proj(A[x_1,..,x_n]/I)$ with $B:=A[x_i]/I$ noetherian. Hence if $F$ is a coherent module it follow any submodule $E \subseteq F$ is coherent. It follows $\oplus \mathcal{O}(n(1)_i)$ is coherent hence $ker(f_1)$ is a coherent module. Hence there is a surjection

$$ f_2: \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow ker(f_1)$$

and you get a long exact sequence of $\mathcal{O}_X$-modules

$$S1.\text{   }\cdots \rightarrow  \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow  \oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$

If $E$ is a finite rank locally free sheaf, it follows $E^*$ is coherent, hence there is a surjection

$$f: \oplus_{i=1}^m \mathcal{O}(n_i) \rightarrow E^* \rightarrow 0$$

Dualize to get

$$0 \rightarrow E \cong E^{**} \rightarrow \oplus_i \mathcal{O}(-n_i) \rightarrow \cdots .$$

Hence any finite rank locally free module can be realized as a submodule of a direct sum $\oplus_i \mathcal{O}(-n_i)$.

The "syzygy theorem" says that such a sequence is finite when $X$ has an open cover $U_i:=Spec(A_i)$ where $A_i$ is a polynomial ring over $k$.

In such cases we get an exact sequence  with a finite number of terms

$$0 \rightarrow E_{k+1} \rightarrow \oplus_j \mathcal{O}(n(k)_j) \rightarrow \cdots \rightarrow E_1 \rightarrow 0.$$

Example: Projective space is such a variety: $\mathbb{P}_k^N$ has an open $D(x_i) \cong \mathbb{A}^N_k$. Hence on projective space it follows any coherent sheaf has a finite resolution of direct sums of invertible sheaves. A similar result holds for the grassmannian variety.

What is true is that by HH.II.5.18 for any coherent sheaf $F$ you always get a long exact sequence where the terms in the sequence are finite direct sums of  linebundles. Note that for the grothendieck group $K_0(X)$ if there is a long exact sequence

$$ 0 \rightarrow F_k \rightarrow \cdots \rightarrow F_2 \rightarrow F_1 \rightarrow 0$$ you get a relation in $K_0(X)$:

$$[F_1]=[F_2]-[F_3]+ \cdots +(-1)^k[F_k].$$

We get

$$[\mathcal{O}(n)]=[\mathcal{O}(1)^{\otimes n}]=[\mathcal{O}(1)]^{ n}$$

hence if the $F_2,F_3...$ are direct sums of linebundles it follows

$$[F_1]\in  \{ \sum_i [\mathcal{O}(n_i)]:\text{$n_i\in \mathbb{Z}$} \} \subseteq K_0(X).$$

If $F$ has a finite resolution of invertible sheaves for any finite rank vector bundle $F$ on $X$ it follows $K_0(X)$ is generated by the invertible sheaves $\mathcal{O}(n)$ for $n \in \mathbb{Z}$. Hence there is a surjection

$$ \rho: \mathbb{Z}[t, t^{-}] \rightarrow K_0(X)$$

defined by $\rho(f(t^i)):=f([\mathcal{O}(i)]), i=1,-1$, and an isomorphism

$$I1.\text{  }K_0(X) \cong \mathbb{Z}[t, t^{-1}]/(ker(\rho)).$$

There is an isomorphism $K_0(X)_{\mathbb{Q}}\cong CH^*(X)_{\mathbb{Q}}$

where $CH^*(X)$ is the Chow group of $X$ and $CH^*(X)$ is "infinite dimensional". But this does not immediately contradict the isomorphism I1.

Question: "Is there anything we can tell about the kernels and cokernels of the differentials? Like are they necessarily generated by extensions of line bundles O(n) or they can be very complicated vector bundles?"

Answer: As shown above: Any coherent module is the cokernel of such a map. The kernel of such a map is always coherent. If $E$ is finite rank and locally free, it follows $E$ can be realized as a subsheaf of such a finite direct sum. In the case of projective space it follows any coherent sheaf $E$ has a finite resolution by finite direct sums of invertible sheaves and you get an equality

$$ K_0(\mathbb{P}^N_k)\cong \mathbb{Z}\{[\mathcal{O}(n)]:n \in \mathbb{Z} \}.$$

The projective bundle formula confirms this calculation: It gives

$$K_0(\mathbb{P}^N_k)\cong \mathbb{Z}[t]/(t^{N+1})$$

where $t:=1-[\mathcal{O}(-1)]$.

https://en.wikipedia.org/wiki/Hilbert%27s_syzygy_theorem

Comment: "This doesn't answer the question as it is not known whether this resolution will end or not. If you re-read the question it is about a bounded long exact sequence on such objects."

Answer: It is well known the grothendieck group $K_0(X)$ is not generated by invertible sheaves in general, hence the answer to your question is: Not in general. For projective space it is true as shown above.


问:
Characterization of pretty compact spaces
I believe that the following problem have already been considered by some sophisticated topologist.

Definition 1. A non-compact Hausdorff topological space $X$ is called almost compact if its Stone-Cech compactification coincides with its one point compactification.

An example of almost compact space is $[0,\omega_1)$ for first uncountable ordinal $\omega_1$.

Definition 2. A compact Hausdorff space $X$ is called pretty compact if $X\setminus\{p\}$ is almost compact for all non-isolated points $p\in X$.

I would like to hear answers to any of the following questions.

Questions:

What are examples of pretty compact spaces?

Is it true that pretty compact spaces are extremally disconnected?

Is it true that pretty compact spaces contain dense extremally disconnected subspace?

Does there exist any characterization of pretty compact spaces?

答: The question was partially answered on mathoverflow.net. It seems like there is no short clear description for such spaces since they include essentially different classes of topological spaces.</p>


问:
Ace Of Spades Right After An Ace and Two of Clubs Right After An Ace
I came across this question

A Deck of 52 playing cards is shuffled and the cards are turned up and
kept on the table. What is the probability that ace of spades comes right after an ace and 2 of
clubs comes right after an ace?

I tried the following approach:

Sample space consists of $52!$ orderings of the cards.

Let an event A be all the orderings such that ace of spades is right next to an ace and two of clubs is right next to an ace.

lets denote the set of events described below as $B$. We visualize the 3 aces and denote their positions as $i, j, k (1 \leq i, j, k \leq 52)$

We can choose 3 slots in ${51\choose 3}$ ways (51 because we should not choose the $52^{nd}$ position as we need slots that have an empty slot right next to it. We can permute the 3 aces in these 3 slots in $3!$ ways. Out of these 3 slots, we can choose any 2 slots(for ace of spades and two of clubs) in ${3\choose 2} \cdot 2!(=P(3, 2))$. And the remaining 47 cards can be arranged in $47!$ ways.

Therefore, $$|B| = {51\choose 3} \cdot P(3, 2) \cdot 3! \cdot 47!$$

lets denote the set of events described below as $C$. We visualize the 3 aces with one ace in
the $52^{nd}$ position. We can choose the remaining 2 slots in ${51\choose 2}$ ways. We can permute the 3 aces in these 3 slots in $3!$ ways. Out of these 2 slots, we can choose any 2 slots(for ace of spades and two of clubs) in $P(2, 2)$ ways.  And the remaining 47 cards can be arranged in $47!$ ways.
Therefore. $$|C| = {51\choose 2} \cdot P(2, 2) \cdot 3! \cdot 47!$$

Lets denote the set of events described below as $D$. We can choose from 50 positions for the 3 aces( we cannot choose the 51 and 52 positions because we need 2 empty slots right next to the ace), that is ${50 \choose 3}$ ways. The pair (ace of spades, 2 of clubs) can sit right next to either of these 3 slots in 3 ways. The remaining 47 cards can be arranged in $47!$ ways.
Therefore, $$|D| = {50 \choose 3} \cdot 3! \cdot 3 \cdot 47!$$

$B$ and $C$ are mutually exclusive. Also, $|A| = |B| + |C| + |D|$.
Therefore $$P(A) = {\dfrac{{51\choose 3} \cdot P(3, 2) \cdot 3! \cdot 47! + {51\choose 2} \cdot 3! \cdot P(2, 2) \cdot 47! + {50 \choose 3} \cdot 3! \cdot 3 \cdot 47!} {52!}} \approx 0.002404$$

Is my approach correct? If it is, is there a simpler approach? If not, please guide me to the correct approach via hints/solution. Thank you!

Note

The problem statement previous to this edit was

A Deck of 52 playing cards is shuffled and the cards are turned up and
kept on the table. What is the probability that ace of spades and 2 of
clubs come right after an ace?

The problem statement turned out to allow for multiple interpretations and @Math Lover's solution is a very elegant one for one of them. I am sorry for the inconvenience caused and will be careful to check that the problem statement is unambiguous before posting it.

答:
Here is how I interpret it -

Question says ace of spades and $2$ of clubs come right after an ace. Number of arrangements where that happens is $2 \cdot 3 \cdot 50!$

Explanation: We need to consider $\{A \ \textbf{A} \ 2 \}$ as one combined card - that is arrangements of $50$ cards. Now the first ace can be any of the $3$ non-spade ace. Also, $2$ of club and ace of spade can swap places as question does not state their internal order.

So the probability should be $ \displaystyle \frac{2 \cdot 3 \cdot 50!}{52!}$


问:
Projectivity of projective bundles over algebraic varieties.
In Kodaira's famous 1954 paper 《On Kähler Varieties of Restricted Type 》  https://www.jstor.org/stable/pdf/1969701.pdf?refreqid=excelsior%3A3c883bd969da3e5252b449ceb0e723b1, the author proved that "the projective bundles over a projective manifold is a projective manifold", the full statement is:

Let $B$ be an analytic fibre bundle over a complex analytic variety $V$ whose fibre is a complex projective space and whose structure group is the group of projective transformations. If the base space $V$ is a non-singular algebraic variety imbedded in a projective space, then the bundle $B$ is also a non-singular algebraic variety imbedded in a projective space.

and my question is: what if we remove the condition "whose structure group is the group of projective transformations"? or if we only require the fiber is projective space, is the statement still remained true?

Here, for the fiber space $\pi:B\to V$, if we only require that all the fibers are projective spaces, not requiring anything about the structure group or transition functions, can we say that this fiber space is a projective bundle?

And I think this is a similar question, for complex manifolds $E,X$, and the holomorphic map $\pi:E\to X$, if we only ruquire that all the fibers of $\pi^{-1}(t)$ are $\mathbb C^n$, not requiring there exist local trivalisation $\pi^{-1}(U_i)\cong U_i\times \mathbb C^n$, can we still conclude that $E$ is a vector bundle?

答:
Question: "..if we only ruquire that all the fibers of $π^{−1}(t)$ are $\mathbb C^n$, not requiring there exist local trivalisation $π^{−1}(U_i)≅U_i×\mathbb C^n$, can we still conclude that $E$ is a vector bundle?"

Answer: I believe the answer is no. A vector bundle has transition maps linear maps. There are locally trivial fibrations with fiber $\mathbb{C}^n$ where the transition maps are not linear (affine bundles): For any quasi projective variety $X \subseteq \mathbb{P}^n_{\mathbb{C}}$ there is a "torsor"  $\pi: Y \rightarrow X$. The map $\pi$ is surjective and locally trivial with fiber $\mathbb{C}^n$ but the transition maps are affine maps. The variety $Y$ is an affine variety. The construction is explicit, and you should try yourself to construct non-trivial examples. I would guess you get non-trivial examples for $\mathbb{P}^1$ or $\mathbb{P}^n$:

https://en.wikipedia.org/wiki/Jouanolou%27s_trick

Question: "Here, for the fiber space π:B→V, if we only require that all the fibers are projective spaces, not requiring anything about the structure group or transition functions, can we say that this fiber space is a projective bundle?"

Answer: If $\pi: B \rightarrow X$ is a surjective map with fibers $\pi^{-1}(x)\cong \mathbb{P}^n_{\kappa(x)}$ for all $x\in X$, it is not immediate that this implies that $\pi$ is locally trivial with transition maps being linear.  For real smooth manifolds there is the "Ehresmann fibration lemma" and for complex manifolds there is the "period mapping":

https://en.wikipedia.org/wiki/Ehresmann%27s_lemma

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


问:
Solving two simultaneous equations in x and y
During solving a complex numbers question, after comparing the real and imaginary parts, it finally came down to these 2 equations-

$$x^3-3xy^2 = y \quad \text{ and }\quad  y^3-3yx^2= -x.$$

I am unable to solve further as I can't think of any factorisation or manipulation which can solve the above.

Does anyone have any approach which can help solve them?

P.S. The orignal equation was : z^3 = i* conjugate(z)

答:
$z^3=i\overline{z}$ implies $z^4=i|z|^2$ (multiply both sides by $z$), so $|z|^4=|z|^2$ (take the modulus of both sides). This gives you $|z|=0$ or $|z|=1$.

If $|z|=0$ then $z=0$

If $|z|=1$ then solve $z^4=i|z|^2=i$, i.e. your solutions are fourth roots of $i$: $z=\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}$, $z=\cos\frac{5\pi}{8}+i\sin\frac{5\pi}{8}$, $z=\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8}$, $z=\cos\frac{13\pi}{8}+i\sin\frac{13\pi}{8}$

Of course, as we used implication method to solve the equation, one is advised to check that those are all actual solutions (which they do turn out to be).


答:
From $z^3= i \cdot \bar z$ it follows
$$|z|^3= |z|$$ and therefore you have either $z=0$ or $|z|=1$.
In the latter case $\bar z= z^{-1}$ and the original equation becomes $z^4= i$, which has the following 4 solutions
$$ e^{i \frac \pi 8}, e^{i \frac {5\pi} 8}, e^{i \frac {9\pi} 8}, e^{i \frac {13\pi} 8}. $$

Title: Solving two simultaneous equations in x and y

Body:
During solving a complex numbers question, after comparing the real and imaginary parts, it finally came down to these 2 equations-

$$x^3-3xy^2 = y \quad \text{ and }\quad  y^3-3yx^2= -x.$$

I am unable to solve further as I can't think of any factorisation or manipulation which can solve the above.

Does anyone have any approach which can help solve them?

P.S. The orignal equation was : z^3 = i* conjugate(z)


问:
Generalizing the Borsuk problem: How much can we shrink a planar set of diameter 1 by cutting it into $k$ pieces?
Borsuk's problem asks whether a bounded set in $\mathbb{R}^n$ can be split into $n+1$ sets of strictly smaller diameter. While true when $n=1,2,3$, it fails in dimension $64$ and higher; I believe all other $n$ are open as of this writing.

However, it turns out that at least in the $n=2$ case we can be more precise than "strictly smaller diameter"; if the original set has diameter 1, we can ensure that each piece has diameter at most $\frac{\sqrt{3}}{2}\approx 0.866$, a bound attained by the circle of diameter $1$. To see that this holds, we note that the regular hexagon of width $1$ is a solution to Lebesgue's universal covering problem, and can be split into three sets of diameter $\frac{\sqrt{3}}2$ as well:

I am interested in putting bounds on such dissections with more than $3$ pieces: what is the minimum diameter one can ensure when cutting a planar set of unit diameter into $k$ pieces?

Using the same approach as above (finding specific sets with a lower bound, and dissecting a universal cover for sets of diameter 1), I have some bounds for higher $k$ as well, though only for $k=3,4,7$ are they exact:

(Extending this table beyond $k=7$ would be difficult, as working out optimal dissections for the circle would become much more complicated.)

Edit: By taking spokes at $72^\circ$ angles on a regular hexagon (with one spoke meeting the hexagon at the midpoint of a side), I think I can get a slightly better upper bound of around $0.6434$ for the case $k=5$. Optimizing spoke placement further (so that the distances between spoke endpoints are equal) gets me around $0.6223$.

In the limit, I think the diameter of each piece is asymptotic to $\sqrt{\frac{2\pi}{3\sqrt{3}k}}\approx \frac{1.1}{\sqrt{k}}$ by tiling with regular hexagons. Certainly one can do no better than $1/\sqrt{k}$ when dividing the circle, using the isodiametric inequality (if the pieces were any smaller, they would have too little area). Using a trivial dissection of the square, one also has an upper bound of $\frac{\sqrt{2}}{\lceil\sqrt{k}\rceil}$.

Some questions I have about this problem:

Has this question been investigated before in the literature? If so, what is known?

Are there any $k$ for which the circle does not present the worst-case scenario for dissection?

Can the $k=5,6$ upper bounds be substantially improved? I think using Pal's slightly smaller solution to the universal covering problem would allow for a few adjustments when $k=6$, but haven't worked out the details.

答:
Here V. P. Filimonov, Covering planar sets, Mat. Sb., 2010, Volume 201, Number 8, 127–160 some more results about this problem, but the article is in Russian.
Here link this article on English, But I don't know how to access it.

In any case, you can see pictures of the partitions of the universal covering set to prove the upper bounds of $d_n$(in his article $d_n$ is $\delta_2(n)$).
On the last page, you can see the improved results at that time. The first and last columns show improved scores from the bottom and top, respectively


问:
Constrained optimization using lagrange multipliers
I was given the following question:

Find the maximum and minimum of the function $f(x,y)=x^2y^2$ subject to $x^2+4y^2=24$.

I tried using lagrange multipliers and got the following:

I was assuming that I would get different answers, and that the biggest one would be the maximum and the smallest would be the minimum. This did not happen. Where did I go wrong, and what steps can I take to solve this problem?

答:
Your function $ \ f(x,y) \ = \ x^2y^2 \ $ has four-fold symmetry about the origin, in that changing the sign of $ \ x \ $ or $ \ y \ $ or both does not alter the value of the function.  As the constraint curve, $ \ x^2 \ + \ 4y^2 \ = \ 24 \   $  (an ellipse centered on the origin) also has four-fold symmetry, we should expect to see critical points of a particular type "mirrored" in all four quadrants.  Also, since $ \ f(x,y) \ \ge \ 0 \ $ everywhere, we already expect the function to have minima wherever it is equal to zero.

When your Lagrange equations have all of the terms taken to one side of each equation and are factored, we find

$$ 2xy^2 \ - \ 2 \lambda x \ \ = \ \ 2x · (y^2 \ - \ \lambda) \ \ = \ \ 0 \ \ \ , \ \ \ 2x^2y  \ - \ 8 \lambda y \ \ = \ \ 2y · (x^2 \ - \ 4\lambda) \ \ = \ \ 0 \  \ . $$

Your calculation for $ \ \lambda \ $ correctly led you to the four absolute maximal points on the ellipse, one in each quadrant, with $ \ f(\pm 2 \sqrt{3} \ , \ \pm  \sqrt{3}) \ = \ 36 \ $ .

The other factors in the equations, giving $ \ x \ = \ 0 \ $ or $ \ y \ = \ 0 \ $ then tell us that the absolute minima on the ellipse [ $ \ f(x,y) \ = \ 0 \ $ ] occur at the endpoints of its major and minor axes, $ \ (\pm 2 \sqrt{6} \ , \ 0 ) \ $ and $ \ (0 \ , \ \pm \sqrt{6}) \ . $

$$ $$

(Although it is not asked for here, if we had been asked about critical points in the interior of the ellipse, the first partial derivatives indicate that there is one at the origin.  When we work with the second derivatives, however, we receive an inconclusive result (for example, the Hessian determinant is zero there).  The problem here is that $ \ f(x,y) \ = \ 0 \ $ everywhere on the coordinate axes, so our usual devices for investigating critical points fail us.  We instead need to look at the properties of the function to understand its behavior near the origin.)


问:
Math equation a little hard for me
I got this equation to solve.

Im not very good at equation.

I got this equation

$( A\times \cos(B) ) \times C = D$

I want this form

$( A \times ? ) \times \cos(B \times ?) = D$

I want to do something like merge C in A and B for giving the result D

答:
If we have the equation: $(A\cdot \cos{B}) \cdot C = D$

Then by commutativity/associativity we can rewrite it as: $(A \cdot C) \cdot \cos{B} = D$

We can NOT "distribute" the $C$ in the way that you have suggested. That is, in general:

$(X \cdot Y) \cdot Z \ne (X \cdot Z) \cdot (Y \cdot Z)$

$(\cos{X}) \cdot Y \ne \cos(X \cdot Y)$

Title: Math equation a little hard for me


问:
Torsion-free Lie groups
Out of curiosity, I'm looking for some examples of Lie groups that are torsion-free.

For some reason (and perhaps there is a good reason), most torsion-free groups I've heard of are discrete. One Lie example that comes to mind is $\mathbb{R}^n$. Also, if $G$ is a Lie group and its torsion subgroup, call it $\Gamma$, is both normal and discrete, then $G/\Gamma$ provides another source of examples.

Question 1: What are some more examples of torsion-free Lie groups?

Question 2: Are there any examples that have non-trivial compact Lie subgroups?

答:
The group of real upper triangular matrices with $>0$ diagonal entries is a good example of a group without elemnt of finite order.

If a Lie group contains a compact group , this subgroup is a compact Lie group, as every closed subgroup of a Lie group is a Lie group (Chevalley). But a compact Lie group is either finite (hence torsion) or contains a maximal torus, hence $T^1$, the group of complex number of modulus 1, and therefore contains elemnets of order $n$ for every $n$.

Title: Torsion-free Lie groups


问:
prove that if $f$ is decreasing and bijective, then $f$ and $f^{-1}$ are strictly decreasing.
I have done the first part proving f is strictly decreasing, but it is so confusing to prove the inverse function is also strictly decreasing. This is what I have so far, I am not sure if it's correct or not because of so many inequality signs. Here's my proof.

Let $A, B$ the domain and codomain of f. We need to prove that for any $x, y \in A$ and $u,v \in B$, if $x < y$ and $u < v$, we have $f(x) >  f(y)$ and $f^{-1}(u) > f^{-1}(v)$. Assume $x < y$. Since f is decreasing, we have $f(x) \ge f(y)$. Since $x \neq y$ and f is bijective, $f(x) \neq f(y)$, we must have $f(x) > f(y)$. Next, we need to prove that $f^{-1}$ is strictly decreasing, so assume $u < v$. set $a = f^{-1}(u)$ and $b = f^{-1}(v)$, so we claim that $a > b$. If not, then $a \ge b$. Since f is decreasing, we have $u = f(a) \ge f(b) = v$. This is a contradiction since by our choice, $u < v$. Hence $f^{-1}$ is decreasing.

I am not sure about the part which is $u = f(a) \ge f(b) = v$ is right or not. Because we have $a > b$, $f^{-1}(u) \ge f^{-1}(v)$ which is $a>b$. Then, we take inverse on both side which becomes to $u \ge v$. Am I understanding the definition of decreasing function right?

答:
Define $B$ as the range of $f$, not codomain. There is also an error in the second part of the proof since $a\ge b$ does not imply $f(a)\ge f(b)$. Anyhow the negation of $a>b$ is $a\le b$ which needs to be proven false.

Suppose $a\color{red}{\le}b$. Then due to the decreasing nature of $f$ we have $u=f(a)\ge f(b)=v$ which is a contradiction. Thus $a>b$.


答:
Direct resolution:

$x=y=0$ is an obvious solution.

Then $x$ time the first equation plus $y$ times the second give

$$x^4-6x^2y^2+y^4=0,$$

which factors as

$$(x^2-(3-\sqrt8)y^2)(x^2-(3+\sqrt8)y^2)=0$$

then

$$\left(x-\sqrt{3-\sqrt8}\,y\right)\left(x+\sqrt{3-\sqrt8}\,y\right)\left(x-\sqrt{3+\sqrt8}\,y\right)\left(x+\sqrt{3+\sqrt8}\,y\right)=0.$$

Now, plugging $y=ax$ in the first equation,

$$x^3-3x^3a^2=ax$$ gives

$$x=\pm\sqrt{\frac{a}{1-3a^2}}.$$

As only two among the four $a'$s are positive, this gives us four distinct solutions.


问:
Find an integrable $g(x,y) \ge |e^{-xy}\sin x|$
I want to use Fubini theorem on $$\int_0^{A} \int_0^{\infty} e^{-xy}\sin x dy dx=\int_0^{\infty} \int_0^{A}e^{-xy}\sin x dx dy$$

Must verify that $\int_M |f|d(\mu \times \nu) < \infty$. I'm using the Lebesgue theorem, so far I've come up with $g(x,y)=e^{-y}$ but am not sure whether it's correct.

My argument is that if $x\in (0,1)$ then the $\sin x$ part is going to ensure that the inequality holds.

答: Try $g(x,y)=x\mathrm e^{-xy}$, then $|\mathrm e^{-xy}\sin x|\leqslant g(x,y)$ for every nonnegative $x$ and $y$. Furthermore, $\int\limits_0^\infty g(x,y)\mathrm dy=1$ for every $x\gt0$ hence $\int\limits_0^A\int\limits_0^\infty g(x,y)\mathrm dy\mathrm dx=A$, which is finite.</p>


答:
The sum is $$\displaystyle \sum_{t=0}^{1000}(-1)^t C(2000,2t)=\sum_{t=0}^{1000}(-1)^t\binom{2000}{2t}$$ Now consider the binomial expansion of $$(1+x)^{2000}=\binom{2000}{0}+\binom{2000}{1}x+\binom{2000}{2}x^2+\cdots+\binom{2000}{2000}x^{2000}$$

Now, plug $x=i$ in it and the real part of that expansion is your expansion. Hence, we evaluate the real part of the LHS, we can write $$1+i=\sqrt{2}\left(\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)\right)=\sqrt{2}e^{i\frac{\pi}{4}}$$ This raised to $2000$th power is $2^{1000}e^{i500\pi}=2^{1000}(\cos(500\pi)+i\sin(500\pi)$ which equals $2^{1000}$.

Title: Evaluating $\sum_{t=0}^{1000}(-1)^t \binom{2000}{2t}$

Body:
Calculate the sum $$\sum_{t=0}^{1000}C_{2000}^{2 t}(-1)^{t}$$
Please kindly help me or show me how to do this.
$$\sum_{t=0}^{1000}C_{2000}^{2 t}(-1)^{t}=\sum_{t=0}^{1000}C_{2000}^{2 t}(i)^{2t}$$
$$\Rightarrow\sum_{t=0}^{1000}C_{2000}^{2 t}(-1)^{t}=(1+i)^{2000}=2^{1000}$$
Please kindly check my answer.
I mean $C_{n}^{r}=C(n,r)$, $r$ is smaller or equal than $n$
Thank in advance!


问:
Let $X,Y$ be two independent random variables and $P_X$, $P_Y$ are induced probability measure. Prove $P(X+Y\in B)=\int_{\Bbb{R}} P_Y(B-x)\ dP_X(x)$
Let $X,Y$  be two independent random variables and $P_X$, $P_Y$ are induced probability measure. Prove $P(X+Y\in B)=\int_{\Bbb{R}} P_Y(B-x)\ dP_X(x)$

Here $P_X, P_Y$ are probability distribution of $X,Y$ respectively defined on $\mathcal{B}(\Bbb{R})$ as $P_X(B)=P(X\in B)$ and $P_Y(B)=P(Y\in B)$ for all $B\in\mathcal{B}(\Bbb{R})$.

My thoughts:

$\{\omega|\ X(\omega)+Y(\omega)\in B\}=\bigcup\limits_{x\in\Bbb{R}} \{\omega|\ Y(\omega)\in B-x\}\cap \{\omega|\ X(\omega)=x\}$

As $X, Y$ are independent, we have $P(\{\omega|\ Y(\omega)\in B-x\}\cap \{\omega|\ X(\omega)=x\})=P(Y\in B-x)=P(X=x)P(Y\in B-x, X=x)=P_Y(B-y)P_X(x)$

As $P(X+Y\in B)=P(\{\omega|\ X(\omega)+Y(\omega)\in B\})=P\left(\bigcup\limits_{x\in\Bbb{R}} \{\omega|\ Y(\omega)\in B-x\}\cap \{\omega|\ X(\omega)=x\}\right)$

Intuition says it has to be $P(X+Y\in B)=\int_{\Bbb{R}} P_Y(B-x)\ dP_X(x)$ but I'm unable to write it rigorously. If $X$ is discrete then the union $\bigcup\limits_{x\in\Bbb{R}}$ will turn out to be a countable disjoint union and we can use the property of measure to writing it as a summation.

Can anyone give any idea to prove the statement rigorously? Thanks for your help in advance.

答:
Only discetre random variable can be handled the way you have done. You have to use Fubini's Theorem to prove this result.

$P(X+Y \in B)=\int \int I_{\{(x,y): x+y \in B\}} dP_Y(y)dP_X(x)=\int P(Y \in B-x)dP_X(x)$  since $\int I_{\{(x,y): x+y \in B\}} dP_Y(y)=P(Y \in B-x)$.


问:
Find all functions satisfying $ f \left( m ^ 2 + f ( n ) \right) = f ( m ) ^ 2 + n $, for all $ m , n \in \mathbb N $
Find all functions $f:\mathbb{N}\to\mathbb{N}$ such that
$$f(m^{2}+f(n))=f(m)^{2}+n\text {, for all }m,n\in \mathbb N$$

I was initially unable to solve this problem, so I referred to a hint. The hint is as follows:

Let $f(1)=k$. From the given equation, we get $f(m^{2}+k)=1+f(m)^{2}$ and $1+k^{2}=f(1+k)$. Use this to prove that
$$f(f(m))^{2}-f(k)^{2}=m^{2}-1 \ \ \ \ \ \ (*)$$
Taking $m=2$, conclude that $f(k)=1 \ \ \ (*)$.
 Obtain $f(f(m))=m$ for all $m\in \mathbb{N}$. Use this to prove that $$f(n+1)=f(n)+k^{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$
By induction $f(n+1)=f(1)+nk^{2}=k+nk^{2}$. Conclude that $k=1$ and hence $f(n)=n$ for all $n\in \mathbb{N}$

This solution is very unclear to me. I have marked all unclear steps with $(*)$. Please breakdown this solution and provide well explained arguments for the marked steps. Please try to use the same notation I have provided in the solution for ease of understanding. Thanks for all help :)

答:
You've been given all of the steps, and just need to creatively figure out why they are true by playing around enough, using what was just discovered/stated.

I've given you additional hints to each $(*)$, and encourage you to think about how to make it true before reading the spoiler.

Let the statement be $P(m, n): f(m^2 + f(n)) = f(m)^2 + n$.

Consider $P(m,1)$ and $P(1,1)$ to get the first 2 equations.

Setting $ m = f(m)$ is the cleanest way to get $ f(f(m))^2$. However, $P(f(m), n), P(f(m),1), P(f(m),k) $ all do not have an easily manipulated LHS. We need to massage it to get another expression of the form $f(m^2 + f(n))$.

$P(f(m), 1+k): f(f(m)^2 + 1 + k^2) = f(f(m))^2 + 1 + k $.

 - $P(k, m^2+k): f( k^2 + 1 + f(m)^2 ) = f(k)^ 2 + m^2 + k$.

Taking the difference gives $f(f(m))^2 - f(k)^2 = m^2 - 1$, which is the first $(*)$.

Taking $m = 2$, $(f(f(2)) - f(k) ) ( f(f(2))+f(k)) = 3 =  1 \times 3$.

 -  which forces $f(f(2)) = 2, f(k) = 1$, which is the second $(*)$.

Hence $ f(f(m))^2 = m^2$, so $f(f(m)) = m$.

$P(k, n): f(k^2+f(n)) =  f(k)^2 + n = 1 + n$.

 - Taking $f$ on both sides, $k^2+f(n) = f \left( f\left( k^2 + f(n)\right)\right)  = f(1+n)$,which is the third $(*)$.

By induction, $f(n) = k+(n-1)k^2$.

Substituting into the original equation, $k + ( m^2 + k +(n-1)k^2 - 1) k^2 = (k+(m-1)k^2)^2 + n $.

Hence $ k = 1$.

Hence $ f(n) = n$.

Note: There are several different ways of approaching this problem.

The key seems to be getting to $f(f(n))=n$.


Body:
You are using non-standard notation. I expect you mean to calculate:
$$\sum_{t=0}^{1000}\binom{2000}{2 t}(-1)^{t}$$

Consider
$$(1+1)^{2000}=\sum_{t=0}^{2000}\binom{2000}{t}1^{2000-t}1^{t}$$
$$(1-1)^{2000}=\sum_{t=0}^{2000}\binom{2000}{t}1^{2000-t}(-1)^{t}$$
Add:
$$2^{2000}+0=2\sum_{t=0 \text{, even}}^{2000}\binom{2000}{t}=2\sum_{t=0 }^{1000}\binom{2000}{2t}$$
$$\sum_{t=0 }^{1000}\binom{2000}{2t}=2^{1999}$$


问:
Find all irreducible representations of $A_4$ over $\mathbb{F}_p$ where p is an odd prime
I have no issue with finding representations over $\mathbb{C}$ but I'm not quite sure how to find them over finite fields, particularly when that field might not be algebraically closed or the characteristic of the field divides the order of the group (both of which are possible in the above question). Thank you in advance for your help.</p>

答:
Here is a quick answer. We get the deleted permutation module of dimension 3, which is irreducible over all fields of odd characteristic.

This is the only faithul irreducible representation. The others all have the subgroup of order 4 in their kernels, so they are effectively representations of the cyclic group of order 3.

We always have the trivial representation, and that is the only other one when when $p=3$.

When $3|p-1$, there are two faithful 1-dimensional representations. (Remember that ${\rm GL}_1(p)$ is just the cyclic group of order $p-1$.)

Otherwise, we have a single irreducible but not absolutely irreducible 2-dimensional representation, which decomposes into two 1-dimensionals over the field of order $p^2$. In this case, you can think of the cyclic group of order 3 as a subgroup of ${\rm GL}_1(p^2)$.


问:
How to compute the operator norm of an integral?
I have to compute the operator norm of $T$ of the following integral operator:

$Tf(x) = \int_0^1\frac{f(y)}{1+x^2+y^2}dy$ equipped with the norm $\|f\|_\infty = \sup_{x\in[0,1]}|f(x)|$

This is what I tried, I was also thinking of using the Holder inequality.

$\Vert Tf(x)\Vert
=\sup\limits_{x\in[0,1]}\left|\int_0^1 \frac{1}{1+x^2+y^2}f(y)\,dy\right|
\leq\sup\limits_{x\in[0,1]}\int_0^1 |\frac{1}{1+x^2+y^2}||f(y)|\,dy
\leq\sup\limits_{x\in[0,1]}\int_0^1 1\cdot\sup\limits_{y\in[0,1]}|f(y)|\,dy
\\=\sup\limits_{x\in[0,1]}\int_0^1 \Vert f(y)\Vert\, dy
=\sup\limits_{x\in[0,1]}\Vert f(y)\Vert
=\Vert f(y)\Vert$

Hence, $\|T\| \leq 1$.

I also tried the following:

$\Vert Tf(x)\Vert
=\sup\limits_{x\in[0,1]}\left|\int_0^1 \frac{1}{1+x^2+y^2}f(y)\,dy\right|
\leq\sup\limits_{x\in[0,1]}\int_0^1 \bigl|\frac{1}{1+x^2+y^2}\bigr|\,|f(y)|\,dy
\leq\sup\limits_{x\in[0,1]}\int_0^1 \bigl|\frac{1}{1+y^2}\bigr|\,|f(y)|\,dy$

and then I know that the antiderivative of $\frac{1}{1+y^2} = \arctan(y)$ combining that with the boundaries, one would get $\frac{\pi}{4}$.

Any help would be grateful. I am really struggling finding a correct answer and I am not sure one of the two ways are the right ones. Thanks in advance.

答:
You have already proved that $\|T\|\leq \frac{\pi } 4$. Now obsetve that if $f(x)=1$ for all  $x$ then $\|Tf\| \geq Tf(0)=\int_0^{1} \frac 1 {1+y^{2}} dy=\frac{\pi } 4$. Since $\|f\|=1$ this proves that $\|T\| \geq \frac {\pi } 4$. Hence, $\|T\| = \frac{\pi } 4$.

[$\|T\| \leq 1$ is also correct  but that does not give you the actual value of the norm].


问:
Reformulating homology theory from pairs to single spaces
I want to convert chapter IX of Eilenberg-Steenrod book to an exposition in single spaces, reminiscent to usual introductory expositions of simplicial or singular homology, but for Cech homology. This is done in Hocking-Young but not in an inverse system and only done fully for finite polytopes.

I was wondering if, given a homology theory, there is any general way to go back from pairs to single spaces, such that at least induced homomorphisms make sense? Or perhaps, where could it fail? I apologize for such an open ended question, I would greatly appreciate any pointers to references. I am not sure how to proceed in doing this myself, and I worry there is a good reason why Hocking-Young did it on such elementary level. Thanks!

Added after the accepted answer: the question was not phrased well, of course the point a comparison of constructions of homology rather than functors.

答:
Your question is misleading, it seems to ask the following:

If we are given homology functors for pairs $H_*(X,A)$, how do we get homology functors for single spaces $H_*(X)$?

The answer is obvious: Take $H_*(X) = H_*(X,\emptyset)$.

But your comment makes clear that this is not what you want to know. You are not interested in the homology functors as the final outcome of some construction, but in the construction itself.

There is a plethora of homology theories satisfying the Eilenberg-Steenrod axioms, and some of them are based on the following type of construction:

Define two functors associating to pairs and single spaces chain complexes $C_*(X,A)$ and $C_*(X)$.

Take $H_n(X,A) = H_n(C_*(X,A))$ and $H_n(X) = H_n(C_*(X))$.

This works for singular homology and we moreover only need the "absolute" chain complex $C_*(X)$ and define the "relative chain" complex as $C_*(X,A) = C_*(X)/C_*(A)$.

For Cech homology such an approach does not work. Eilenberg and Steenrod start with coverings of pairs, then form the nerves of these coverings and finally take the direct limit of the (simplicial) homology groups of these nerves. This is a more sophisticated construction than that of singular homology.

Here are some quotations:

p. 234: One should be careful not to confuse the sets $Cov(X)$ and $Cov(X,\emptyset)$.
However one may regard $Cov(X)$ as the subset of $Cov(X,\emptyset)$ consisting
of coverings $\alpha$ indexed by $(V_\alpha, \emptyset)$.

p. 236: Beginning with 2.4 the discussion of this chapter was limited to
coverings of pairs. However the definitions and results may be duplicated
for coverings of spaces (without a distinguished subset) and to
indexed families of sets which are not necessarily coverings.

p. 245: The Cech homology and cohomology groups of $(X,A)$, $A$, and $X$ are
defined as limits of suitable systems of groups defined over the directed
sets $Cov(X,A)$, $Cov(A,\emptyset)$, and $Cov(X,\emptyset)$ respectively.

This shows that Eilenberg and Steenrod define indeed $H^{Cech}_n(X) = H^{Cech}_n(X,\emptyset)$; they do not separately consider pairs and single spaces. But of course one could perform an alternative construction based on $Cov(X)$ to get functors $H^{Cech'}_n(X)$. Eilenberg and Steenrod don't do it, but you can repeat everything for $Cov(X)$ instead of $Cov(X,\emptyset)$. This is somewhat tedious and I doubt you learn much if you are performing it explicitly.

Anyway, if you have done this, you can use the fact that $Cov(X)$ is cofinal in $Cov(X,\emptyset)$ (recall Definitions VIII 2.1 and IX 2.8 plus Corollary VIII 3.16.) to show that $H^{Cech'}_n(X) \approx H^{Cech}_n(X,\emptyset) = H^{Cech}_n(X)$ canonically.


问:
$\frac{1}{\cosh{x}}+\log\left ( \frac{\cosh{x}}{1+\cosh{x}} \right )$ for $x \rightarrow \pm \infty$ has a limit
Show from the definition of a limit that $$\frac{1}{\cosh{x}}+\log\left ( \frac{\cosh{x}}{1+\cosh{x}} \right )$$ for $x \rightarrow \pm \infty$ has a limit.

My attempt

This one is really tough for me. I know the limit is 0 and I know that $f(x) \rightarrow b$ for $x \rightarrow \infty$ if $\forall \epsilon>0 \exists M >0: \left \| f(x)-b \right \|<\epsilon$ for all $x \in A, x>M$ ($A$ is the domain of the map $f$ and $A \subset \mathbb{R}^m$). From this I should be able to prove the following but have no idea on how to do it and where to apply my inequalities. I am pretty sure I have to use the fact that $\frac{x}{1-x}\leq \log(1-x)\leq x$ and $0<\frac{1}{\cosh{x}}\geq 2e^{-x}$. When I get the result for $x\rightarrow \infty$ I should get it for $x\rightarrow -\infty$ because of symmetri around the $y$-axis.

答:
First solution: Re-write your expression as
$$
\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right).
$$
Then
$$
\left| {\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right)} \right| \le \frac{1}{{\cosh x}} + \log \left( {1 + \frac{1}{{\cosh x}}} \right) \le \frac{2}{{\cosh x}}
$$
for all real $x$. Let $\varepsilon>0$ be arbitrary. Then if $\left| x \right| > \log \left( {\frac{4}{\varepsilon }} \right)$,
$$
\frac{2}{{\cosh x}} \le 4e^{ - |x|}  < \varepsilon .
$$
Second solution: A more optimal solution may be given as follows. Note that
$$
\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right) = \frac{1}{{\cosh ^2 x}}\int_0^1 {\frac{t}{{1 + \frac{t}{{\cosh x}}}}dt} .
$$
Then
$$
\left| {\frac{1}{{\cosh x}} - \log \left( {1 + \frac{1}{{\cosh x}}} \right)} \right| \le \frac{1}{{\cosh ^2 x}}\int_0^1 {tdt}  = \frac{1}{2}\frac{1}{{\cosh ^2 x}}.
$$
Let $\varepsilon>0$ be arbitrary. Then if $\left| x \right| > \frac{1}{2}\log \left( {\frac{2}{\varepsilon }} \right)$,
$$
\frac{1}{2}\frac{1}{{\cosh ^2 x}} \le 2e^{ - 2\left| x \right|}  < \varepsilon .
$$


Body:
If you mean
$$
\begin{align}
\sum_{k=0}^{1000}(-1)^k\binom{2000}{2k}
&=\sum_{k=0}^{1000}i^{2k}\binom{2000}{2k}\tag1\\
&=\frac12\left(\sum_{j=0}^{2000}i^j\binom{2000}{j}+\sum_{j=0}^{2000}(-i)^j\binom{2000}{j}\right)\tag2\\
&=\frac12\left((1+i)^{2000}+(1-i)^{2000}\right)\tag3\\[6pt]
&=\frac12\left(\left(\sqrt2e^{\pi i/4}\right)^{2000}+\left(\sqrt2e^{-\pi i/4}\right)^{2000}\right)\tag4\\[9pt]
&=2^{1000}\tag5
\end{align}
$$
Explanation:

$(1)$: $i^{2k}=(-1)^k$

$(2)$: when $j$ is even, $j=2k$ and $i^j+(-i)^j=2i^{2k}$

$\phantom{\text{(2):}}$ when $j$ is odd, $i^j+(-i)^j=0$

$(3)$: Binomial Theorem

$(4)$: $1+i=\sqrt2e^{\pi i/4}$ and $1-i=\sqrt2e^{-\pi i/4}$

$(5)$: $\sqrt2^{2000}=2^{1000}$ and $\left(e^{\pm\pi i/4}\right)^{2000}=1$


问:
Extrema homework — maximizing the viewing angle of a picture on a wall
I have hit a problem in my homework and don't know how to solve it.
Here it is:

"A picture with height of 1.4 meters hangs on the wall, so that the bottom edge of the picture is 1.8 meters from the viewers eye. How far does the viewer have to stand in order to have the best viewing angle? "

I understand that the viewing angle has to be maximum, but I have no idea how to begin solving it.
Hints and tips would really help!
Thank you in advance!

答:
Sketch the problem:

Write down the cosine rule:

$${ x }^{ 2 }+{ y }^{ 2 }-2xy\cos { \alpha  } ={ 1.4 }^{ 2 }\\\alpha =\arccos { \left( \frac { { x }^{ 2 }+{ y }^{ 2 }-{ 1.4 }^{ 2 } }{ 2xy }  \right)  } $$

Now find $y$ in terms of $x$ using Pythagorean theorem:

$${ x }^{ 2 }-{ 3.2 }^{ 2 }={ y }^{ 2 }-{ 1.8 }^{ 2 }$$

Maximize $\alpha$ and find $x$. The distance is:
$$d=\sqrt { { x }^{ 2 }-{ 3.2 }^{ 2 } } $$

Also see:
Regiomontanus' angle maximization problem


问:
Understanding Hatcher's Proposition 1.26
First, let me state the proposition in Hatcher's textbook

(a) If $Y$ is obtained from $X$ by attaching 2-cells as described above, then the inclusion $X\hookrightarrow Y$ induces a surjection $\pi_1(X,x_0)\to\pi_1(Y,x_0)$ whose kernel is $N$. Thus $\pi_1(Y)\simeq\pi_1(X)/N$

Note that $N$ is a normal subgroup of $\pi_1(X,x_0)$ generated by all the loops $\gamma_\alpha\varphi_\alpha\gamma_\alpha^{-1}$. I'm wondering the orientation $\varphi_\alpha$ goes. Let me just write an example:

If I want to attach single 2-cell to below 1-skeleton...

...like this

Then there are several ways of attaching 2-cell. For example, attaching 2-cell  along $aba^{-1}b^{-1}cbc^{-1}$ or $aba^{-1}b^{-1}cb^{-1}c^{-1}$ or $cbc^{-1}aba^{-1}b^{-1}$...etc. Is this attaching orientation(?) matters? I mean even attached orientations are different but still, those spaces are homeomorphic? Or at least the fundamental groups are always isomorphic?

答:
I think it's not true in general that the orientation of attaching cells is irrelevant to the resulting attached space.

Consider the following example:

The following is 1-skeleton

I'm going to attach 2-cell as the following image.

There're essentially two ways to attach 2-cell. One is along $abab^{-1}cac^{-1}$ and the other is $aba^{-1}b^{-1}ca^{-1}c^{-1}$. Let $X_1$ be $X^2$ whose 2-cell is attached by the former way and $X_2$ be $X^2$ whose 2-cell is attached by the latter way.

Then, $\pi_1(X_1)=\langle a,b,c|abab^{-1}cac^{-1}\rangle$ and $\pi_1(X_2)= \langle a,b,c|aba^{-1}b^{-1}ca^{-1}c^{-1}\rangle$. Then $\pi_1(X_1)^{\text{ab}}=\Bbb Z_3\oplus\Bbb Z\oplus\Bbb Z$ and $\pi_1(X_2)^{\text{ab}} = \Bbb Z\oplus\Bbb Z$ so that $X_1$ and $X_2$ are not homeomorphic (even in the level of fundamental group).


问:
What is $\lim_{h\to0}\int_{-h}^h \delta(i+x)f(x)dx$?
This answer says that

$\int_{-\infty}^\infty \delta(i+x)f(x)dx=f(-i)$
(for well-behaved functions).

But what happens if we make the limits of integration to go to zero? What is $$\lim_{h\to0}\int_{-h}^h \delta(i+x)f(x)dx$$?

Is the answer the same?

答:
When you operate with distributions that are not proper functions, you cannot simply change the limits of integration. In a sense $$ \int_{-\infty}^\infty\delta(x+i) (\cdot) dx$$
is a single symbol and it has a mathematical meaning only as a whole, as the linear map
$$ \int_{-\infty}^\infty\delta(x+i) (\cdot) dx : f \mapsto f(-i)$$
defined on a set of functions continuous at $-i$.

Before you ask about the limit $h \to 0$ you need to first properly define what
$$ \int_{-h}^h\delta(x+i) (\cdot) dx$$
and it's not obvious how it should be defined. If it was a real number $a$ in place of $i$ one would usually say that
$$ \int_{-h}^h\delta(x+a) (\cdot) dx : f \mapsto f(-a){\bf 1}_{[-h,h]}(-a) $$
and is defined on functions $f$ such that $f(x){\bf 1}_{[-h,h]}(x)$ is continuous at $x=-a$. You cannot however use this for $\delta(x+i)$, because ${\bf 1}_{[-h,h]}(x)$ is only defined for real arguments. Therefore, to get an answer you must first define what $\int_{-h}^h\delta(x+i) (\cdot) dx$ means.


问:
Real analysis when solving a complex number problem?
We are to solve $(x-1)^3+8=0$ in the complex set to do so we bring 8 to the RHS and cube root on both sides $$(x-1)=2(-1)^{1/3}$$

The RHS is of three different forms $(-1,\omega,\omega^2)$

But the LHS is just written in a single form why do we do that ?

I'm nothing close to a mathematician so forgive me for my mathmatical illiteracy .

答:
Let's pick another, less complex example. When solving $x^2 = 1$, we conclude $x = \pm 1$. So, why does the left hand side have a single value, whereas the right has two values?

When you solve an equation like $x^2 = 1$, you are finding all possible values of $x$ that satisfy this equation. You start with the assumption that $x$ is a single value that satisfies the equation $x^2 = 1$. We can then conclude, logically, from this assumption, that $x = 1$ or $x = -1$. It's not that $x$ has both values, it has one value or the other (but not both).

Now, in your example, there is an issue where complex exponentiation can produce multiple answers. In real numbers, we understand that $x^{1/3}$ refers to the one and only cube root of $x$, i.e. the unique number $y \in \Bbb{R}$ such that $y^3 = x$. In the complex numbers, there is no unique $y \in \Bbb{C}$ such that $y^3 = x$ (for $x \neq 0$). As such, you should interpret $2(-1)^{1/3}$ as one of three possible complex numbers, much like you would treat $\pm 1$ as one of two real numbers.


问:
If $A$ and $B$ are $n \times n$ matrices invertible matrices, can the equation $A+B = AB$ be true?
We just need to find one example to back this up right?

as answered by @peek-a-boo ,
taking A= B = 2I

$2I + 2I = (2I )(2I)$

$4I = 4I$(TRUE)
Hence, it can be true.

答:
Let us rearrange the equation as $(A-I)(B-I)=I$.  Then as long as $1$ is not an eigenvalue of $A$, then we can find a $B$ that solves the equation.

If you further want $AB=BA=I$, then $A+A^{-1}=I$, so $A^2-A+I=0$, and so if $\lambda$ is an eigenvalue of $A$, then $\lambda^2-\lambda +1=0$.  Further, $A$ satisfies an equation with no repeated roots, so $A$ must be diagonalizable.  Therefore, the equation is solvable if and only if $A=PDP^{-1}$ where $D$ is diagonal with diagonal entries equal to $\frac{1\pm \sqrt{-3}}{2}$


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
One might suppose that there needs to be some profound insight or conception needed.  That is, one either has to deconstruct the notation, or see something in a different light.

For example, a recent question on whether other sets of weights, other than $1,3,9,27,81$ might yield all numbers from $1$ to $121$.  One might be tempted to solve this by seeking the largest of the weights, as one does filling a backpack.  The actual proof involves hunting down the smallest weight.

When one devises notation, such as my Polygloss, the notation gives rise to insights that were not apparent without it.

One thing, for example, is that it's not apparently obvious that a dot-product of vectors in an oblique coordinate system could be done relatively easily.  But this can indeed be done by the use of a matrix, which normally does not appear in the process.  One multiplies one of the vectors by a matrix, and takes the dot product of this and the second vector.

I have had many profound insights into many different things, and the chief stumbling block is that one has to deconstruct the current notions and objections.  For example, the notion that a base consists of digits $0$ to $b-1$ goes, when one deals with alternating arithmetic.


问:
Converting recursive formula into non-recursive
I am currently playing around with recursive formulars. The following problem is not taken froma book or course.

Denote by $R_k$ a matrix of dimension $k \times k$ and by $r_k$ a vector of dimension $1 \times k$. Furthermore we are given a lower unittriangular matrix $R_2$ and the recursive formula $R_{k+1} = \begin{pmatrix} R_k & 0 \\ r_kR_k & 1 \end{pmatrix}$. It is easy to compute the matrix $R_3, R_4, R_5...$ one after another, but does there exist a closed formula such that given an integer $n$ one can compute the matrix $R_n$ without computing $R_3, ..., R_{n-1}$?

I already tried computing the sequence and find an explicit description but the therms become very fast very complex: Denote by $r_k^i$ the i-th entry of the vector $r_k$:

$R_2=\begin{pmatrix} 1 & 0 \\
r_1^1 & 1 \end{pmatrix}$

$R_3 =\begin{pmatrix} 1 & 0 & 0 \\
r_1^1 & 1 & 0 \\
r_2^1+r_2^2r_1^1 & r_2^2 & 1
\end{pmatrix}$

$R_4 =\begin{pmatrix} 1 & 0 & 0 & 0 \\
r_1^1 & 1 & 0 & 0 \\
r_2^1+r_2^2r_1^1 & r_2^2 & 1 & 0\\
r_3^1+r_1^1r_3^2+r_3^3r_2^1+r_3^3r_2^2r_1^1 & r_3^2+r_3^3r_2^2 & r_3^3 & 1
\end{pmatrix}$

So far I only see the pattern at the subdiagonals.

Thank you in advance for your help.

答:
WLOG, $R_0=(1)$. Then

$$R_1=\begin{pmatrix}1&0\\r&1\end{pmatrix}$$

$$R_2=\begin{pmatrix}1&0&0\\r&1&0\\s+rs'&s'&1\end{pmatrix}$$
$$R_3=\begin{pmatrix}1&0&0&0\\r&1&0&0\\s+rs'&s'&1&0\\t+rt'+(s+rs')t''&t_1+s't''&t''&1\end{pmatrix}$$

does not seem to lead us anywhere.


问:
Show that series $\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$ is divergent
Show that series $$\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$$

is divergent . What inequality can we use here. i tried various method but none of these give any result. Any hint please .

答:
Note that for all $n\geq 2$,
$$
(\log n)^{\log \log n}  = e^{(\log \log n)^2 }  \le e^{4e^{ - 2} \log n}  < e^{\log n}  = n.
$$
Consequently
$$
\sum\limits_{n = 2}^\infty  {\frac{1}{{(\log n)^{\log \log n} }}}  > \sum\limits_{n = 2}^\infty  {\frac{1}{n}} ,
$$
and we are done since the harmonic series is divergent.


问:
Integration By parts for Lebesgue integrals
Given bounded distribution functions $F$ and $G$, on $[a,b]$, such that $F$ and $G$ do not share a common point of discontinuity, following is true.

$$\int_{(a,b]} FdG +\int_{(a,b]}GdF\\ = F(b)G(b)-F(a)G(a)$$

I was able argue it using Fubini's theorem, what I don't understand is where does the condition "no common points of discontinuity" play into all this.

答:
In fact, a more general result is true (I'm going off memory, so I'd highly recommend you verify this closely). Given $F,G:\Bbb{R}\to\Bbb{R}$ right continuous and of bounded variation, define
\begin{align}
F_-(x):=\lim_{y\to x^-}F(y)\quad \text{and} \quad \Delta F(x):= F(x)-F_-(x),
\end{align}
and likewise for $G$. Then, we have
\begin{align}
\int_{(a,b]}F_-\,dG + \int_{(a,b]}G_-\,dF &= (FG)(b)-(FG)(a)-\int_{(a,b]}\Delta F\, dG\\
&=(FG)(b)-(FG)(a)-\sum_{n=1}^{\infty}\Delta F(x_n)\cdot \Delta G(x_n),
\end{align}
where $\{x_n\}_{n=1}^{\infty}$ is the set of discontinuities of $F$. The proof is almost identical to what you claim: apply Fubini to a carefully chosen subset of the plane (you just have to be more careful). So, from this you can see that if $F$ and $G$ are both discontinuous at a common point then you'll pick up an extra term.

Sorry this is slightly indirect; but to get a more direct answer, you'll have to show us precisely the proof you've seen so that we can point out exactly where the assumption is being used. It all boils down to the fact that the Stieltjes measures of singleton sets are non-zero at points of discontinuity:
\begin{align}
\mu_F(\{x\})&=\Delta F(x).
\end{align}


问:
Inequality expectations
Suppose $X$ is a nonnegative random variable and $a$ and $b$ are nonnegative constants. I am trying to prove that

$\begin{equation}
\mathbb{E}[((a-X+b)^+-X)^+] \leq \mathbb{E}[((a-X)^++b-X)^+],
\end{equation}$

where $x^+=\max(x,0)$. So far without any progress. Any thoughts?

答:
As $b$ is nonnegative you have (by exhaustion)
$$(a-x+b)^+\leq(a-x)^+ +b$$
and since $x\mapsto\max(x,0)$ is a monotone increasing function you receive
$$((a-x+b)^+-x)^+\leq((a-x)^+ +b-x)^+.$$
Now you just have to use the monotonicity of the expected value. You see by this proof that the assumption "$a$ is nonnegative" is not requiered.

Title: Inequality expectations


问:
Modular Arithmetic/Congruence
Hello I was wondering if anyone could help better my understanding of finding remainders using congruence and modular arithmetics as I cannot wrap my head around it.

The question I have been presented is to find the remainder when $4444^{4444}$ is divided by 9.

Now I have started by working out $4444$ (mod 9)

$4444\equiv16\equiv-2 \pmod 9$

but for the remaining part I am unsure how to work out.

I have been told through the answers that it should be done by the following:

$4444^{4444} \equiv-2^{4444}\equiv 2^{3*1481+1} \equiv -2 \equiv 7$

I follow the first step as I am aware from a Theorem that;

$a \equiv b \pmod m$ implies $a^{k} \equiv b^{k}\pmod m$ for any integer $k\geq 0$

But I don't understand why we split up the power and how to integer 2 flips between positive and negative.

Thanks in advance for any help

答: The point is that $(-2)^3\equiv 1\pmod 9$, and therefore $(-2)^{3k}\equiv1^k=1$ for any natural number $k$. So you can rewrite $(-2)^{4444}$ as $(-2)^1\times(-2)^{3\times 1481}$, and the second factor is $1\pmod 9$.

Title: Modular Arithmetic/Congruence


问:
Solving $\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2 $
How can I solve the equation  $$\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2$$
I know that it has two roots: $x=1$ and $x=3$. </p>

答:
Substituting $x = \sqrt{t^3+1}$ and twice squaring, we arrive to the equation
$$   36t^6-24t^5-95t^4+8t^3+4t^2-48t=0.$$ Its real roots are $t=0$ and $t=2$ (the latter root is found in the form $\pm \text{divisor}(48)/\text{divisor}(36)$), therefore $x=1$ and $x=3$.</p>


问:
Can this differential equation be solved manually?
$ x\;\ddot{y} - 2\dot{x} + 9y + \frac{2}{3} = 0$

I tried making use of wolfram alpha but it was unable to provide me with a solution.
I also wanted to apply the Laplace transform method, however I am not sure how to apply it given the first term of the equation.

答:
There are two unknowns with only one equation, it is undetermined and wont take off and cannot be solved.

But as an aid to insight, a second arbitrary supporting ode function .. here an accompanying Sine function is assumed with the following BC:

$ (x',x)=(-0.25,-2), (y,y') = (0,0.25) \;;$

$$ x''+ \dfrac{2 \pi x }{\lambda} =0 ;\quad \lambda =0.2 ;\text{
Numerical solution looks like: } $$

EDIT1:

With coupled DE s the wavelength is seen $  \lambda\approx 1$ whereas given uncoupled wave length is only $0.2 ;$ it makes the arbitrary function choice important too.


答:
For problems related to remainders it is often convenient and possible to get the remainder as $1$. In this case $4444 \equiv 7\equiv-2\mod{9}$ also $7^3 \equiv 1\mod9$. This means that $4444^3\equiv 1 \mod9$. Using this you split the exponent in a multiple of 3 and a remainder $4444^{4444}=4444^{3\times1481+1}$.

This is common and convenient method to solve this problem.

Title: Modular Arithmetic/Congruence

Body:
Hello I was wondering if anyone could help better my understanding of finding remainders using congruence and modular arithmetics as I cannot wrap my head around it.

The question I have been presented is to find the remainder when $4444^{4444}$ is divided by 9.

Now I have started by working out $4444$ (mod 9)

$4444\equiv16\equiv-2 \pmod 9$

but for the remaining part I am unsure how to work out.

I have been told through the answers that it should be done by the following:

$4444^{4444} \equiv-2^{4444}\equiv 2^{3*1481+1} \equiv -2 \equiv 7$

I follow the first step as I am aware from a Theorem that;

$a \equiv b \pmod m$ implies $a^{k} \equiv b^{k}\pmod m$ for any integer $k\geq 0$

But I don't understand why we split up the power and how to integer 2 flips between positive and negative.

Thanks in advance for any help


问:
Remove element conflicts to keep most elements possible
I originally wanted to post this to Stack Overflow, but then I realized it's not really code, but just an algorithm, so I decided to switch here. I hope I understood the purpose of this site correctly.

I have a set of things, which can "conflict" with some of the others. What I'd like to do is to eliminate some of them to remove all conflicts, while keeping as many items as possible.

I start with a half matrix of which items conflict (marked with #, item can't conflict with itself):

(e.g. C conflicts with B and D)

```<code>   A  B  C  D  E  F
A
B  #
C  .  #
D  #  #  #
E  #  .  .  .
F  .  .  .  #  .
</code>```

Now my idea was to create an array of current number of collisions for each item, and iteratively keep removing those with most collisions. In first round, it is item D, so I mark it deleted (null), and then I also decrease the number of collisions for A, B, C, F, which it was colliding with.

Full procedure of this algorithm:

```<code> A     B  C     D     E  F
[3,    3, 2,    4,    1, 1] - remove D, decrement A,B,C,F
[2,    2, 1, null,    1, 0] - now there are 2 options, I choose to remove B (decr. A,C)
[1, null, 0, null,    1, 0] - again options, choose to remove E
[0, null, 0, null, null, 0] - done
</code>```

However, I'm not sure if this always gives the best possible solution (the most items remain). So: does this always keep as many elements as possible, and if not, is there any algorithm that guarantees that? Should there be just some additional logic in the case there are more options?

Some programming stuff:

I don't care about performance that much, it's going to run on sets of at most around 100 items. What I care about is result that really includes as many items as possible.

I would love to hook some additional logic when there are multiple possibilities: the elements have sizes, so I'd like to keep biggest total sum. However, since the item sizes will be relatively close to each other and quite big (for any 3 element sizes, $x + y \gg z$), the number of remained elements is the more important part

答:
Let's formulate this with a graph $(V,E)$:

your "things" will be the nodes $V$

whenever there's a conflict between two things, we put an edge $e \in E$ between the corresponding nodes.

What you're trying to find is a set $S \subset V$ of nodes, so that when removing it, your graph has no edge anymore. In other terms, every edge in the first graph has to have at least one end that is part of $S$.

This problem is known as the vertex cover problem, which is one of the classic NP-complete problems. This means that there is no known optimal algorithm that runs in polynomial time. So even your set size of $100$ might be a problem for exponential algorithms (yet, you can still try to run one ; sometimes it actually runs fast). But for this problem, there are approximation algorithms, which are guaranteed to give a solution within a certain bound of the optimal solution. You can look in this direction.


问:
Convergent subsequence in $C^1$
We consider two closed interval in $\mathbb{R}$ : $[a,b]$ and $[c,d]$. Moreover, let us consider a sequence $\{f_{n}\}_{n = 1}^{\infty}$ uniformly bounded in $C^1([a,b]\times [c,d])$. My question is: Is there exist a subsequence $\{f_{n_{j}}\}_{j = 1}^{\infty}$ and $f \in C^{1}([a,b]\times [c,d])$ such that
$f_{n_{j}}$ converges to $f$ in the  weak our $\mbox{weak}^{*}$ sense?

答:
You need the Eberlein-Smuljan Theorem. It states the following:

Let X be a Banach space. Then every bounded sequence in X has a weakly convergent subsequence.

The space $C^1([a,b]\times[c,d])$ is a Banach space, so you get the weakly convergent subsequence directly from the theorem. As the space is banach, the limit $f$ obviously is an element of your space.

Title: Convergent subsequence in $C^1$


问:
How to represent combinations of plus and minus
Just wondering if there is a shortcut for representing the following:
$$x=1+i,1-i,-1+i,-1-i$$
I know that you can do:
$$x = 1\pm i,-1\pm i$$
But can you use something like:
$$x = \pm 1 \pm i$$
NOTE: I know the example above just gives $x = 1+i, -1-i$, but I just wanted to provide an example

答: Do you think $$(-1)^x \left( 1 \pm i \right) \\ x\in\{0,1\}$$ is a nice idea?


答:
Suggestion: let $z=1+i$ and

$$x \in \{z , \bar z, -z , - \bar z \}.$$

Title: How to represent combinations of plus and minus

Body:
Just wondering if there is a shortcut for representing the following:
$$x=1+i,1-i,-1+i,-1-i$$
I know that you can do:
$$x = 1\pm i,-1\pm i$$
But can you use something like:
$$x = \pm 1 \pm i$$
NOTE: I know the example above just gives $x = 1+i, -1-i$, but I just wanted to provide an example


问:
Derive $u$ for an intersection between sphere and a segment of line $P1 + u(P2-P1)$
If we have a segment defined by points $P1$ and $P2$ and a sphere defined by origin $P3$ and radius $r$. And the segment is part of a line defined by $P1 + u(P2-P1)$. Then if the segment and the sphere intersect, the intersection point lies between $u >= 0$ and $u <=1$. How do we derive the formula for u? The formula below I tried to derive is incorrect.

$$u = \frac{(P3-P1)(P2-P1)}{(P2-P1)(P2-P1)}$$

Do we need to extend the endpoints of the segment distance $r$ along the direction of the line and set the new endpoints in the equation above?

Body:
Calling $s_{12}(\lambda) = \lambda p_1 + (1-\lambda) p_2$ and given the sphere as $\|p-p_3\|^2 = r^2$ we have

$$
\|s_{12}(\lambda)-p_3\|^2 = \lambda^2\|p_1\|^2+2\lambda(1-\lambda)p_1\cdot p_2+(1-\lambda)^2\|p_2\|^2 - 2\lambda p_1\cdot p_3-2(1-\lambda)p_2\cdot p_3+\|p_3\|^2= r^2
$$

solving for $\lambda$. If $\ \ 0\le \lambda\le 1$ the segment intersects the sphere.

NOTE

$$
\lambda=\frac{\|p_2\|^2-p_1p_2+p_1p_3-p_2p_3\pm\sqrt{(\|p_2\|^2-p_1p_2+p_1p_3-p_2p_3)^2-\|p_1-p_2\|^2(\|p_3-p_2\|^2-r^2)}}{\|p_1-p_2\|^2}
$$

Attached a plot showing the solution for

$$
\cases{
p_1 = (2,1,1)\\
p_2 = (5,6,4)\\
p_3 = (1,0,0)\\
r=3
}
$$

giving $\lambda = \frac{1}{43} \left(54-\sqrt{379}\right)$ and after substitution into $\lambda p_1+(1-\lambda)p_2$...


问:
Identity on the sum $\sum_{i=1}^{n-2}{i-1 \choose k}{n-2-i \choose k-1}={n-2 \choose 2k}$
I need to evaluate the first summation for $k\geq1$ and $n\geq2$.
Some computer calculations allow the identification with the expression ${n-2 \choose 2k}$
above but, how can it be derived? Are there different methods or any
combinatorial interpretation?

答: Hint: From the set $\{1, 2, \dots, n-2\}$, choose $2k$ elements and let $i$ be the $(k+1)$-th element chosen, using the standard ordering. The first $k$ elements are chosen from $\{1, 2, \dots, i-1\}$ and the last $k - 1$ elements are chosen from $\{i+1, i+2, \dots, n-2\}$.


问:
Is a mean of 12 unusual for a sample of size 30 from Exp(1/10)?
I am doing an assignment in my statistics course and have trouble with the following:

Maria claims that she has drawn a random sample of size 30 from the exponential distribution with $\lambda = 1/10$. The mean of her sample is $12$.

c) Is a mean of $12$ unusual for a sample of size $30$ from $Exp(1/10)$?

Is there any formal way to check this? I can't find any formal definition of "unusual" in my book. I am reading "Mathematical Statistics with Resampling and R" Laura Chichara.

TIA for any help.

答:
No, it is not unusual.

"Usually" do define an "unusual" value one refers to the fact that this events happens in less than $5\%$ of the times.

In your exercise, before getting the distribution of the sample mean, first observe that

$$E(\overline{X}_{30})=10$$

$$V(\overline{X}_{30})=\frac{10}{3}$$

thus we have a sigma of $\sigma=\sqrt{3.33}\approx 1.83$

in a Gaussian approx, unusual values are over $2\sigma$, that is in your 30 size sample to have a value greater than $10+1.83\times2=13.66$

Additionally, your size 30 sample from an exponential is good to be approximated with a gaussian.

Of course, if you want to have a more precise result, you can derive the distribution of your sample mean, that is

$$\overline{X}_{30}\sim \text{Gamma}(30;3)$$

and thus

$$6\cdot\overline{X}_{30}\sim \chi_{(60)}^2$$

Using the tables, you get that the 95% quantile is 13.18 thus 12 is perfectly in the expected range; In fact, your 12 is included in the 86.21% quantile


问:
If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.

I think it is trivial because they are distinct.

So I wonder just saying "Since they are distinct" is enough to prove it?

Of course there could be several more detailed versions but I just want to know that reasoning is true or not.

Body:
HINT:
$$a^3+b^3+c^3-3abc$$

$$=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab(a+b+c)$$

$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$

$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2-3ab\}$$

$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$=(a+b+c)\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2$$

which will be $>=<0 $ according as $a+b+c>=<0$ as for real distinct $a,b,c,$ each of $(a-b)^2,(b-c)^2,(c-a)^2>0$


问:
Distance from vector to a subspace

Determine the distance of $v = (1,2,3)^T$ from the subspace $V =\operatorname{span}(v_1,v_2,v_3)$ where $v_1=(1,1,0)^T, v_2=(0,1,1)^T, v_3=(2,4,2)^T.$

Using Gram-Schmidt I obtain the basis $\left[ \begin{array}{c} \frac{\sqrt{5}}{5} \\\\ 0 \\\\ \frac{2 \sqrt{5}}{5} \end{array} \right], \left[ \begin{array}{c} - \frac{4 \sqrt{5}}{15} \\\\ \frac{\sqrt{5}}{3} \\\\ \frac{2 \sqrt{5}}{15} \end{array} \right], \left[ \begin{array}{c} 0\\\\ 0 \\\\0 \end{array} \right]$ for $V.$ Now the projection from $v$ to the subspace $V$ $$\hat{v} =  \frac{v \cdot v_1}{v_1 \cdot v_1}\cdot v_1 + \frac{v \cdot v_2}{v_2 \cdot v_2}\cdot v_2 + \frac{v \cdot v_3}{v_3 \cdot v_3}\cdot v_3 = \begin{bmatrix}\frac{1}{3}\\ \frac{4}{3}\\ \frac{10}{3}\end{bmatrix}$$ so the distance would be $\|v-\hat{v}\|= 1$, but this seems not to be correct. What might I be doing wrong? The orthonormal basis should be enough to measure the distance just from the projection vector?

答: Simple test of row reduction will show that subspace V has dimension 2 which is a plane. $v_1,v_2$ can be taken a basis, take a cross-product $v_1\times v_2 = (1,-1,1)$ which is the normal('s direction ratios). Equation of plane that represents $V$ is given by $x-y+z=0$ and using distance formula $$d=\left|\frac{ax_1+by_1+cz_1-d}{\sqrt{a^2+b^2+c^2}} \right| = \frac{1-2+3}{\sqrt{3}}=\frac{2}{\sqrt3}$$

Title: Distance from vector to a subspace


问:
Modular form Fourier series calculations
Let $f(z)$ be a modular form of weight $k$ for $\text{SL}_{2}(\mathbb{Z})$ and set $g(z) := f(Nz)$ for some $N>1$, $g$ is a modular form of weight $k$ for $\Gamma_{0}(N)$. I am trying to evaluate $g$ at the cusp $0$. Suppose $f$ has some Fourier expansion $\displaystyle f(z) = \sum_{n=0}^{\infty}a_{n} e^{2\pi i nz}$, then in essence I'm trying to compute the Fourier expansion of $z^{-k}g(-\frac{1}{z})$.

So far the only reasonable way I can see of doing this is to use the modularity of $f$ as follows:

$$ z^{-k}g\left(-\frac{1}{z}\right) = z^{-k}f\left(-\frac{N}{z}\right)$$

well if $z \in \mathbb{H}$, then $-\frac{1}{z} \in \mathbb{H}$ and thus $-\frac{N}{z} \in \mathbb{H}$. Since $f$ is modular of weight $k$, we have that for any $z\in \mathbb{H}$ that

$$ f\left(-\frac{1}{z} \right) = z^{k}f(z) \Rightarrow f\left(-\frac{N}{z} \right) = \frac{z^{k}}{N^{k}}f\left( \frac{z}{N}\right) = \frac{z^{k}}{N^{k}}\sum_{n=0}^{\infty}a_{n} e^{2\pi i n \frac{z}{N}}.$$

So it would appear that

$$ z^{-k}g\left(-\frac{1}{z}\right) = \frac{1}{N^{k}}\sum_{n=0}^{\infty}a_{n} e^{2\pi i n \frac{z}{N}}$$

So I guess my question first of all: is this right, or at least does it look right?

From this, allegedly one can deduce modular forms in $\Gamma_{0}(N)$ which take the value of $0$ at $\infty$ and $1$ at $0$ and vice versa. However, the leading constant coefficients in both the expansions of $g(z)$ and $z^{-k}g\left(-\frac{1}{z}\right)$ must both be zero if one of them is. So I feel the above must be wrong.

Any help appreciated!

答:
Yes it is correct.

If $g\in M_k(\Gamma_0(N))$ then the value at the cusp $\Gamma_0(N)i\infty$ can be different to the value at the cusp $\Gamma_0(N)0$ (ie. $g(i\infty)\ne z^{-k}g(-1/z)(i\infty)$

But here you are taking $g$ in the subspace such that $g(z/N) = (z/N)^{-k}g(-1/z)$ so obviously it vanishes at one cusp iff it vanishes at the other.

Try with $f(z)-f(Nz)$ for $f\in M_k(SL_2(\Bbb{Z}))$ not a cusp form, it vanishes at $\Gamma_0(N)i\infty$ but not at $\Gamma_0(N)0$.


问:
Ellipse centre from two points and radius
I might need some help on this problem. I searched online and didn’t find any way to solve it.

Given two points, I need to find the center of the two ellipses of given radius $rx, ry$ that intersects at those two points.

I know the two A and B point coordinates, as well as ellipses radius $rx$ and $ry$ (on $X-$axis and $Y-$ axis respectively).

Is there a way, from those $4$ variables, to calculate the center $C$ and $C'$ of each ellipse?

Here is a little diagram I made to make things clearer.

答:
Scale the $y$-axis until $rx=ry$.

Fit two circles.

Undo the scaling.


问:
How to expand $\langle \textbf{x}-\textbf{y},\textbf{x}-\textbf{y}\rangle$ using the Cauchy–Schwarz inequality

The Question

\begin{align}||\textbf{x}+\textbf{y}||^2+||\textbf{x}-\textbf{y}||^2&=\langle\textbf{x}+\textbf{y},\textbf{x}+\textbf{y} \rangle+\langle\textbf{x}-\textbf{y},\textbf{x}-\textbf{y} \rangle\\&=\langle\textbf{x},\textbf{x}+\textbf{y}\rangle+\langle\textbf{y},\textbf{x}+\textbf{y}\rangle+\langle\textbf{x},\textbf{x}-\textbf{y}\rangle-\langle\textbf{y},\textbf{x}-\textbf{y}\rangle\\&=\langle\textbf{x},\textbf{x}\rangle+\langle\textbf{x},\textbf{y}\rangle+\langle\textbf{y},\textbf{x}\rangle+\langle\textbf{y},\textbf{y}\rangle+\langle\textbf{x},\textbf{x}\rangle-\langle\textbf{x},\textbf{y}\rangle-\langle\textbf{y},\textbf{x}\rangle+\langle\textbf{y},\textbf{y}\rangle\\&=2\langle\textbf{x},\textbf{x}\rangle+2\langle\textbf{y},\textbf{y}\rangle\\&=2||\textbf{x}||^2+2||\textbf{y}||^2\end{align}

My Understanding

I understood how $\langle \textbf{x}+\textbf{y},\textbf{x}+\textbf{y}\rangle$ was expanded. I believe you take the $\textbf{x}$'s from each side and make them into a vector. Then the $\textbf{y}$'s. And then the $\textbf{x}$ from the first vector and $\textbf{y}$ from the second. And lastly the $\textbf{y}$ from the first vector and $x$ from the second. I repeated this same process for $\langle \textbf{x}-\textbf{y},\textbf{x}-\textbf{y}\rangle$ and got: $\langle \textbf{x},\textbf{x}\rangle+\langle -\textbf{y},-\textbf{y}\rangle+\langle \textbf{x},-\textbf{y}\rangle+\langle -\textbf{y},\textbf{x}\rangle$. How did they get a negative in front of the $\textbf{x}$? Also, why does $\langle -\textbf{y},-\textbf{y}\rangle$ become $\langle \textbf{y},\textbf{y}\rangle$?

答:
This is due to two properties of the dot product:

linearity in the first argument $\langle\alpha{x} + \beta{y},z\rangle=\alpha\langle{x}, z\rangle+\beta\langle{y}, z\rangle$, where $\alpha$ and $\beta$ are complex numbers, $x$, $y$ and $z$ are vectors;

$\langle x, y\rangle=\langle y, x\rangle^{*}$, where the symbol $*$ is a complex conjugation.

So, $\langle x, \alpha y + \beta z\rangle=\langle \alpha y + \beta z, x\rangle^{*} = \alpha^{*}\langle y, x\rangle^{*} + \beta^{*}\langle z, x\rangle^{*} = \alpha^{*}\langle x, y\rangle + \beta^{*}\langle x, z\rangle$.


答:
Hint

The Cauchy–Schwarz inequality is used to prove that the inner product is a continuous function with respect to the topology induced by the inner product itself. Therefore, using the dot product, the following hold for any vectors $\textbf{x}, \textbf{y}$, and $\textbf{z}$ in $\mathbb{R}^n$, and scalars $c \in \mathbb{R}$

$\textbf{x}\cdot\textbf{y}=\textbf{y}\cdot\textbf{x}.$

$\textbf{x}\cdot\textbf{x}=||\textbf{x}||^2.$

$\textbf{x}\cdot(\textbf{y}+\textbf{z})=\textbf{x}\cdot\textbf{y}+\textbf{x}\cdot\textbf{z}$ and $(\textbf{x}+\textbf{y})\cdot\textbf{z}=\textbf{x}\cdot\textbf{z}+\textbf{y}\cdot\textbf{z}.$

$(c\textbf{x})\cdot\textbf{y}=\textbf{x}\cdot c\textbf{y}=c(\textbf{x}\cdot\textbf{y}).$

Title: How to expand $\langle \textbf{x}-\textbf{y},\textbf{x}-\textbf{y}\rangle$ using the Cauchy–Schwarz inequality

Body:
The Question

\begin{align}||\textbf{x}+\textbf{y}||^2+||\textbf{x}-\textbf{y}||^2&=\langle\textbf{x}+\textbf{y},\textbf{x}+\textbf{y} \rangle+\langle\textbf{x}-\textbf{y},\textbf{x}-\textbf{y} \rangle\\&=\langle\textbf{x},\textbf{x}+\textbf{y}\rangle+\langle\textbf{y},\textbf{x}+\textbf{y}\rangle+\langle\textbf{x},\textbf{x}-\textbf{y}\rangle-\langle\textbf{y},\textbf{x}-\textbf{y}\rangle\\&=\langle\textbf{x},\textbf{x}\rangle+\langle\textbf{x},\textbf{y}\rangle+\langle\textbf{y},\textbf{x}\rangle+\langle\textbf{y},\textbf{y}\rangle+\langle\textbf{x},\textbf{x}\rangle-\langle\textbf{x},\textbf{y}\rangle-\langle\textbf{y},\textbf{x}\rangle+\langle\textbf{y},\textbf{y}\rangle\\&=2\langle\textbf{x},\textbf{x}\rangle+2\langle\textbf{y},\textbf{y}\rangle\\&=2||\textbf{x}||^2+2||\textbf{y}||^2\end{align}

My Understanding

I understood how $\langle \textbf{x}+\textbf{y},\textbf{x}+\textbf{y}\rangle$ was expanded. I believe you take the $\textbf{x}$'s from each side and make them into a vector. Then the $\textbf{y}$'s. And then the $\textbf{x}$ from the first vector and $\textbf{y}$ from the second. And lastly the $\textbf{y}$ from the first vector and $x$ from the second. I repeated this same process for $\langle \textbf{x}-\textbf{y},\textbf{x}-\textbf{y}\rangle$ and got: $\langle \textbf{x},\textbf{x}\rangle+\langle -\textbf{y},-\textbf{y}\rangle+\langle \textbf{x},-\textbf{y}\rangle+\langle -\textbf{y},\textbf{x}\rangle$. How did they get a negative in front of the $\textbf{x}$? Also, why does $\langle -\textbf{y},-\textbf{y}\rangle$ become $\langle \textbf{y},\textbf{y}\rangle$?


答:
Note: I failed to notice that the problem as stated allows $a,b$, and $c$ to be arbitrary real numbers. As others have noted, in this generality the result is false. It is, however, true if $a,b,c\ge 0$. If $-a^3-b^3-c^3+3abc=0$, then

$$\frac13\left(a^3+b^3+c^3\right)=abc=\sqrt[3]{a^3b^3c^3}\;,$$

i.e., the arithmetic mean of $a^3,b^3$, and $c^3$ is the same as their geometric mean. Now use the full form of the AM-GM inequality.

I’ll leave this up for now in case it turns out that the problem was actually supposed to include the hypothesis that $a,b,c\ge 0$.

Body:
If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.

I think it is trivial because they are distinct.

So I wonder just saying "Since they are distinct" is enough to prove it?

Of course there could be several more detailed versions but I just want to know that reasoning is true or not.


问:
If a two variable smooth function has two global minima, will it necessarily have a third critical point?
Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a  $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$, and then imagine an endless rain pouring with water level rising steadily on the entire plane.

Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima.

Those two points are connected by a compact line segment $K$. As a continuous function, $f$ attains a maximum value $M$ on the set $K$. This means that when the water level has reached $M$, the two lakes will have been merged.

The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$.

It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$. But, can we prove that one always exists?

Follow-ups:

Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ($S^1\times S^1$) or the surface of a sphere ($S^2$). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima?

Similarly, what if we have local minima instead of global?

If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded - not sure that would be at all relevant, but who knows).

答:
Suppose we have a compact orientable manifold $M$ with boundary $\partial M$. Suppose that we have a Morse function $f$ whose gradient flow is transverse to the boundary $\partial M$. Then the boundary splits into two components (which are themselves not necessarily connected) $\partial M=\partial_- M^{}+\partial_+M$ where $-\nabla f$ points outward on $\partial_-M$ and inward along $\partial_+M$. Then we have relative Morse inequalities. Define $b_k=\mathrm{rank} H_k(M,\partial M)$ and $c_k$ the number of critical points of $f$ of index $k$. Define the relative Poincare Polynomial and Morse polynomial
$$
P_t(M,\partial_- M)=\sum_{k=0}^n b_k \qquad M_t(f)=\sum_{k=0}^n c_k.
$$
The Morse relations state that there exists a polynomial $Q_t$ with non-negative coefficients $Q_t$ such that
$$
M_t(f)=P_t(M,\partial_- M)+(1+t)Q_t
$$
An immediate consequence is that the number of critical points of index $k$ must be larger than the $k$-th Betti number of $(M,\partial_-M)$. And that the alternating sum of the number of critical points of index $k$ must equal the relative euler characteristic of $(M,\partial_-M)$ (evauate the expression above at $t=-1$). In general the Morse relations are stronger than the corollaries I just outlined.

For non-compact manifolds the relation is a bit more complicated, and growth conditions on $f$ should be given. (But the statement for compact manifolds with boundary should give you an idea what to expect). But there are weaker conditions available than demanding that $\lim_{|x|\rightarrow \infty}f(x)=\infty$ (A keyword is isolation or Palais-Smale). You can also make sense of this for non-Morse functions, or even more general vector fields (beyond the gradient of function). This is known as Conley theory. You can have a look at my phd thesis where I discuss some of these matters.


问:
Whether the following is a Dedekind cut?
I am able to verify that

$$ T= \{t \in \mathbb{Q} : t^2\lt{2} \text{ or } t\lt{0}   \} $$

is a Dedekind cut.

However i have some confusion in verifying if

$$ U= \{t \in \mathbb{Q} : t^2\le{2} \text{ or } t\lt{0} \} $$

is a Dedekind cut.

I think that the set $ U $ has no largest element in $\mathbb{Q} $. Can someone show me a proof that $U$ has no largest element?

答: We have $U=T$. So since one is a cut, the other is a cut as well. It is, however, marginally more immediate that $T$ is a cut, so that one is generally preferred.


问:
What is the mathematical term and symbol for division without remainders?
What is the mathematical term and symbol for division without remainders?

Take for example 322 / 100.

I know 322 / 100 is division and the result is 3.22.
I know 322 % 100 is a modulo and the result is 22.

But what is the proper term and symbol for division with no remainder?

Eg, 322 SYMBOL 100 would give 3.

答: The term is 'integer division'. The symbol used is $\lfloor \frac{322}{100} \rfloor$ or sometimes $[\frac{322}{100}]$ and $100\,|\,300$ if one is a divisor of the other.


问:
Evaluate integral in terms of Gamma function
Need help evaluating the following integral in terms of the gamma function where gamma function is: $$\Gamma(x)=\int_0^\infty e^{-t}t^{x-1}dt.$$  The integral is the following: $$\large{\int_0^\infty t^{x-1}e^{-\lambda t\cos(\theta)}\cos(\lambda t \sin(\theta))dt}$$</p>

答:
You just rewrite the cosine in terms of the complex exponential:
$$ \cos(\lambda t \sin\theta) = \frac 12\left[\exp(i\lambda t \sin\theta)+\exp(-i\lambda t \sin \theta)\right] $$
Your integral is then
$$ \frac 12\int_0^\infty dt\,t^{x-1}
\left(
\exp(-t(\lambda\cos\theta-i\lambda \sin\theta))+
\exp(-t(\lambda\cos\theta+i\lambda \sin\theta))
\right)$$
That's a sum of two similar terms. In each of them, one only rescales $t$ by the factor of
$$\lambda\cos\theta\mp i\lambda \sin\theta = \lambda\exp(\mp i \theta) $$
so that the exponentials become $\exp(-T)$ while a factor of the coefficient above to the $(-x)$-th power is picked from $dt\,t^{x-1}$. So the result is
$$ \dots = \frac {\Gamma(x)}{2\lambda^x} (\exp(ix\theta)+\exp(-ix\theta  )) = \frac{\Gamma(x)}{\lambda^x}\cos(x\theta) $$
I am going to fix the minor mistakes now. Note that the limits of the definite integral aren't really changed even though $t$ was rescaled by a complex factor – as long as the phase is within some limits.</p>


问:
Is this set a basis of the vector space $H$?
When $H=\{(x_1,x_2,x_3)^T\in\Bbb{R^3}:3x_1-2x_2+6x_3=0\}.$ I claim that the set $V=\{v_1 = (2,0,3), v_2 = (2,3,0), v_3 = (0,3,1)\}$ is a basis. My logic is this. This set is linearly independent, and it spans the entirety of $\Bbb{R^3}$. Thus it must also span each subspace, including $H$. Is this enough to $V$ is a basis? Are there other methods we could use?

答: No, it is not enough. For a set to be a basis of, let's say, $V$, it has to be a linearly independent set whose span is $V$. For example, in $\mathbb{R}^2$, a line through the origin with a slope of 1: a basis would be $\{(1,1)\}$, not $\{(1,0), (0,1)\}$, since the span of the second set is all $\mathbb{R}^2$, not just the line itself.


问:
Determining the quotient group $(\mathbb{Z} \times\mathbb{Z})/{\langle(n,n)\rangle}$
I am trying to classify, $\frac{\mathbb{Z} \times \mathbb{Z}}{\langle(n,n)\rangle}$, $n=1,2,3,\cdots, $ using fundamental theorem of finitely generated abelian groups.

Inspire by @user134824 in Classifying the factor group $(\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle$, I construct $\varphi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}_n$ by $(k,l) \mapsto (k-l,l)$. Then its kernel is $\langle(n,n)\rangle$ and from the 1st isomorphism theorem, $\mathbb{Z} \times \mathbb{Z} \simeq \mathbb{Z}_n \times \mathbb{Z}$.

But how I can understand this from the fundamental theorem of finitely generated abelian groups?

答:
I don't see what you mean. You have an automorphism of $\Bbb{Z}^2$ $$(a,b)\to (a-b,b)$$ which sends $(n,n)$ to $(0,n)$ and $\Bbb{Z}^2/\langle (0,n)\rangle$ is not too hard to understand.

This generalizes with the Smith normal form.


问:
Can this differential equation be solved manually?
$ x\;\ddot{y} - 2\dot{x} + 9y + \frac{2}{3} = 0$

I tried making use of wolfram alpha but it was unable to provide me with a solution.
I also wanted to apply the Laplace transform method, however I am not sure how to apply it given the first term of the equation.

答:
$$xy''-2x'+9y+\frac23=0 \tag 1$$
$y''=\frac{d^2y}{dt^2}$ and $x'=\frac{dx}{dt}$ .

Let $x(t)=u(t)e^{y'/2}\quad\implies\quad x'=u'e^{y'/2}+\frac12 ue^{y'/2}y'' $

$$(ue^{y'/2})y''-2(u'e^{y'/2}+\frac12 ue^{y'/2}y'')+9y+\frac23=0$$
$$-2u'e^{y'/2}+9y+\frac23=0$$
$$u'=\left(\frac92 y+\frac13\right)e^{-y'/2}$$
$$u(t)= \int \left(\frac92 y+\frac13\right)e^{-y'/2}dt+c$$
$c=$ arbitrary constant.
$$\boxed{x(t)=c\:e^{y'/2}+e^{y'/2}\int \left(\frac92 y+\frac13\right)e^{-y'/2}dt}$$
This is the general solution of equation $(1)$ for any arbitrary function $y(t)$ , insofar supposed integrable.


答: You statement is wrong, see $(0,-1,1)$.</p>

Title: If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.

Body: You statement is wrong, see $(0,-1,1)$.


问:
Irreducible polynomials of degree $n>3$
I don't quite understand how I know if a polynomial is irreducible, and I would like to know if someone could help me understand that.

For example, if I have the following polynomial: $x^5+x^2+x^3+1$, is this polynomial irreducible in $\mathbb{Z}_{2}[x]$? I know that $\mathbb Z_{2}[x]=\{x^2, x^2+1, x^2+x, x^2+x+1\}$ but I will not continue from there to answer that question

答:
It is easy to see that the set of  irreducible polynomials of degree $2$
over $\Bbb F_2$ also contains $x+1$ and $x^2+x+1$. Dividing
$x^5+x^3+x^2+1$ by $x+1$, or $x^2+x+1$ we see that
$$
x^5+x^3+x^2+1=(x^2 + x + 1)(x + 1)^3.
$$

Title: Irreducible polynomials of degree $n>3$


答:
The most immediate issue is that $v_1 \notin V$, which violates the fact that bases must be subsets of their spaces. This is important for all sorts of reasons, for example, we cannot say $\operatorname{span} \{v_1, v_2, v_3\} = V$, as $v_1$ lies in the left hand side, but not the right.

More intuitively, this should hopefully not make sense to you. If we have a basis of three vectors, then all bases for $V$ have three vectors, and $V$ should be a $3$-dimensional space. The definition of $V$ makes it clear that it's a plane, specifically the plane $3x - 2y + 6z = 0$ in general form. Planes are two-dimensional objects, so a basis ought to only contain $2$ vectors.

Instead, observe that $\{v_2, v_3\}$ is a linearly independent subset of $V$, hence $\dim V \ge 2$. Since $V$ is a proper subspace of $\Bbb{R}^3$, which has dimension $3$, then $\dim V \le 2$. Thus, $\dim V = 2$, and hence any linearly independent $2$-element subset of $V$ must be a basis.

Title: Is this set a basis of the vector space $H$?

Body: When $H=\{(x_1,x_2,x_3)^T\in\Bbb{R^3}:3x_1-2x_2+6x_3=0\}.$ I claim that the set $V=\{v_1 = (2,0,3), v_2 = (2,3,0), v_3 = (0,3,1)\}$ is a basis. My logic is this. This set is linearly independent, and it spans the entirety of $\Bbb{R^3}$. Thus it must also span each subspace, including $H$. Is this enough to $V$ is a basis? Are there other methods we could use?


答:
An Extension of Vandermonde's Identity
$$
\begin{align}
\sum_{\substack{k\\n-k\ge a\\m+k\ge b}}\binom{n-k}{a}\binom{m+k}{b}
&=\sum_{k}\binom{n-k}{n-k-a}\binom{m+k}{m+k-b}\tag1\\
&=\sum_{k}(-1)^{n-k-a}\binom{-a-1}{n-k-a}(-1)^{m+k-b}\binom{-b-1}{m+k-b}\tag2\\[9pt]
&=(-1)^{n+m-a-b}\binom{-a-b-2}{n+m-a-b}\tag3\\[9pt]
&=\binom{n+m+1}{n+m-a-b}\tag4\\[9pt]
&=\binom{n+m+1}{a+b+1}\tag5
\end{align}
$$
Explanation:

$(1)$: symmetry of Pascal's Triangle

$(2)$: negative binomial coefficients

$(3)$: Vandermonde's Identity

$(4)$: negative binomial coefficients

$(5)$: symmetry of Pascal's Triangle

Applying $(5)$ gives
$$
\sum_{i=1}^{n-2}\binom{i-1}{k}\binom{n-2-i}{k-1}=\binom{n-2}{2k}\tag6
$$

Title: Identity on the sum $\sum_{i=1}^{n-2}{i-1 \choose k}{n-2-i \choose k-1}={n-2 \choose 2k}$

Body:
I need to evaluate the first summation for $k\geq1$ and $n\geq2$.
Some computer calculations allow the identification with the expression ${n-2 \choose 2k}$
above but, how can it be derived? Are there different methods or any
combinatorial interpretation?


问:
How to simplify dot product of cross product by itself $(a \times b) \cdot (a \times b)$
Given $a$ and $b$ are vectors in $R3$. Is it possible to simplify this? I could try to simplify it myself but if the simplification result is already widely known I wanted to ask it here.

答:
We have

$$(a \times b) \cdot (a \times b)= ||a \times b||^2= ||a||^2 \cdot ||b||^2-(a \cdot b)^2.$$

Lagrange's identity.