问:
How to show mathematically max of uniform is a sufficient statistic?
So I get that $X_i \sim Uniform(0, \theta)$ then $T = max\{X_1,..., X_N\}$ is intuitively aa sufficient statistic.

But how do I prove it mathematically?

Specifically, I learned that if $f(x^n;\theta)$ denotes the joint pdf of a sample $X^n$, then we can factorise it like so:

$$f(x^n;\theta) = g(T(x^n);\theta)h(x^n)$$

Where $g$ is a function of $\theta$ and $T$ and $h$ is not dependant on $\theta$ at all.

Most explanations to prove the max is a sufficient statistic go along these lines:

$$f(x^n;\theta) = \frac{1}{\theta^n} \prod_{i=1}^{n} I_{(0,\theta)} (x_i)$$
$$ f(x^n;\theta) = \frac{1}{\theta^n} I_{(0, \theta)}(t)$$

It seems to me that $g$ here refers to $\frac{1}{\theta^n}I_{(0,\theta)}(t)$, but what does $h$ refer to?

答:
I struggled with a similar issue when I was studying sufficient statistics, in context of showing that the sample mean and variance are sufficient statistics for $\mu$ and $\sigma^2$ in the case of IID random variables with Normal $(\mu, \sigma^2)$ distributions, where both $\mu$ and $\sigma^2$ are unknown.

It turns out that a useful "trick" in this area of mathematics is to define $h(x^n) = 1$. And so you can use the Fisher-Neyman factorisation theorem to show that the the statistic $T(X^n) = \text{max}\{X_1 , ... , X_n\}$ is sufficient.


问:
Question on sequences on banach spaces
I am trying to do this exercise: Suppose $X$ is a Banach Space and $(x_j)_{j=1}^{\infty}$ a sequence in $X$. Suppose $f \in X^{*}$ with $\sum_{j=1}^{\infty}\left|f\left(x_{j}\right)\right|<\infty$. Prove that there is a constant $M>0$ such that $\sup _{k \in \mathrm{N}}\left\|\sum_{j=1}^{k} \epsilon_{j} x_{j}\right\|_{X} \leq M$ for any sequence $\left(\epsilon_{j}\right)_{j=1}^{\infty}\left(\epsilon_{j}=\pm 1\right)$.

I tried to use the fact that
$$\sup _{k \in \mathrm{N}}\left\|\sum_{j=1}^{k} \epsilon_{j} x_{j}\right\|_{X} \leq \sup _{k \in \mathrm{N}}\sum_{j=1}^{k}\left\| \epsilon_{j} x_{j}\right\|_{X} \leq \sup _{k \in \mathrm{N}}\sum_{j=1}^{k}\left\|  x_{j}\right\|_{X},$$
together with defining $\hat{x}_n \in X^{**}$ such that $\hat{x}_n(f)=f(x_n)$ and somehow try to use the uniform boundedness principle to connect both, but I am not sure how to do it, and I am a bit confussed on how to use it, because maybe I can define a sequence of operators using the partial sums or something similar. Can you give me please some hint. Thanks

答:
We wish to show that the set $$S = \left\{\sum_{j=1}^k \varepsilon_jx_j : \varepsilon_j = \pm 1,k \in \Bbb{N}\right\}$$
is bounded. For any $f \in X^*$ we have
$$\left|f\left(\sum_{j=1}^k \varepsilon_jx_j\right)\right| \le \sum_{j=1}^k |\varepsilon_j f(x_j)| = \sum_{j=1}^k |f(x_j)| = \sum_{j=1}^{\infty} |f(x_j)| < +\infty$$
so $\{f(s) : s \in S\}$ is bounded by $\sum_{j=1}^{\infty} |f(x_j)|$ in the field. Since $f\in X^*$ is arbitrary, we conclude that $S$ is weakly bounded. The uniform boundedness principle now implies that $S$ is bounded.

ContentLicense: CC BY-SA 4.0

Title: Question on sequences on banach spaces


问:
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.

I solved it in the following way:

I state that $a$ is the number in front of $2010$, in other words if the number we're looking for is $A$, then $A=10^4\cdot a+2010$.

Hence, $a\cdot10^4\equiv1\bmod2011$.

$1956a\equiv1\bmod2011$

$55a\equiv2010\bmod2011$

So $55a=2010+2011k$ for some positive integer $k$.

So $30+31k\equiv 0 \bmod55$

After doing some trial and error for $k\in[0,54]$

$k_{min}=15$

So $55a=32175$, $a=585$.

Hence the smallest number possible is $5850000$

I solved this question using classical number theory and trial-and-error. However, I realize that this question can be solved using the euclidean algorithm. When I looked it up on the internet the only thing I could find was this: https://en.wikipedia.org/wiki/Euclidean_algorithm which is how to find the gcd using euclid's algorithm. Could you please explain to me how to solve the question using the euclidean algorithm, and also please explain to me what the euclidean algorithm is, how it can be applied and some resources to study from?

Body:
Euclidean Algorithm

$A= 10000K + 2010 = 2011M$ so

$10000K = 2011M -2010 = 2011M-2011 +1 = 2011(M-1) = 1$

$10000K -2011(M-1) = 1$.

Find a small solution to $K, M-1$.

$10000 = 4*2011 + 1956;$ $1956=10000 - 4*2011$

$2011=1956 + 55;$ $55= 2011-1956 = 2011-(10000 -4*2011)=5*2011-10000$.

$1956 = 35*55 + 31;$ $31=1956-35*55= (10000-4*2011)- 35(5*2011-10000)= 36*10000-179*2011$

$55 = 31 + 24;$  $24 = 55-31 = (5*2011-10000)- (36*10000-179*2011)=-37*10000+184*2011$

$31 = 24 + 7;$  $7 =31 -24 =(36*10000-179*2011) -  (-37*10000+184*2011)=73*10000 -363*2011$

$24 = 3*7 + 3;$ $3=24-3*7=(-37*10000+184*2011)-3(73*10000 -363*2011)=-256*10000+1273*2011$

$7 = 2*3 + 1$ $1= 7-2*3 = (73*10000 -363*2011)- 2(-256*10000+1273*2011)=585*10000-2909*2011$

So if we let $K=585$ and $M-1=2909$ we get $5852010 = 2011*2910$


问:
What are some relationships between a matrix and its transpose?
All I can think of are that
If symmetric, they're equivalent

If A is orthogonal, then its transpose is equivalent to its inverse.

They have the same rank and determinant.

Is there any relationship between their images/kernels or even eigenvalues?

答:
The key to understanding the various relations between a matrix and its transpose is to understand what the transpose of a matrix signifies. For simplicity, let's work over the field $\mathbb R$. Recall that for a vector space $V$, its dual, $V^*$, is the vector space of linear functionals: linear transformations $\varphi :V\to \mathbb R$.

Here is an exercise then to explain what transposing a matrix does:

1) Given a linear transformation $T:V\to W$, one obtains a function $T^*:W^*\to V^*$, given by $T^*(\varphi)(v)=\varphi(T(v))$. Actually, $T^*$ is a linear transformation.

2) Fix bases $B_V=(v_1,\cdots, v_n)$ and $B_W=(w_1, \cdots, w_k)$ for $V$ and $W$ respectively (here each $v_i$ is a vector in $V$, each $w_i$ is a vector in $W$). Let $M$ be the matrix representing $T$ in these bases.

3) Recall the dual bases $B^*_{V^*}$ and $B^*_{W^*}$. Prove that the matrix representing $T^*$ with respect to these bases is precisely the transpose of $M$.

Understanding the transpose via the dual spaces explains a lot.

ContentLicense: CC BY-SA 3.0


问:
Why is the divergence of curl expected to be zero?
I was reading the proof of Helmholtz decomposition theorem where I found the relation between the rotational and the irrotational fields are not symmetric. And by that I mean if the divergence of the gradient field is supposed to be zero except where the particle exist aka:

$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$

Then why is the divergence of the curl always zero?

$$\nabla\cdot\nabla\times \bf A = 0$$

and not:
$$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

where $\bf c $ is a vector with speed of light magnitude.

Edit: This question is about asymmetry of electromagnetic potential fields and I have no idea why it was migrated to math forum. Anyways, if mathematicians can see this question my question is:

Suppose there is a vector field $ F=\nabla(1/r)+\nabla \times \bf A $ made out of a scalar potential $1/r$ and a vector potential $\bf A$ where these relations hold:
$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$
and:
$$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

So both potential fields have critical points, considering $\bf F$ should have been sufficiently smooth, can we still apply Helmholtz decomposition theorem?

答:
From the comments you've written, I think that what you're asking is this:

Math tells us that the divergence of a curl is always zero. But suppose we go beyond pure math. What happens when the divergence of the curl is allowed to be nonzero?

And the answer is that it's just not possible. The divergence of the curl is zero, always, everywhere, under all circumstances, in theory and in practice, in the real world and in imaginary worlds.

Asking what happens when the divergence of the curl is nonzero is like asking what happens when two people are both taller than each other. It just can't happen.

OwnerDisplayName: Martin Spinoza


问:
Prove that $x^4 + ax^2 + b^2$ is reducible over $\mathbb{Z}_p$ with $p$ prime.
My work

$$x^4 + ax^2 + b^2 = (x\pm b)^2 + (a \mp 2b)x^2$$

Then if either $(a + 2b) = -r^2$ or $(a - 2b) = -q^2$ with $q,r \in \mathbb{Z}_p$ the result follows.
But I'm not really sure how to prove this last part.

I'd really appreciate any help you could give me.

答:
The key point is to get inspiration from factoring that polynomial over the complex numbers into two quadratic factors (with leading coefficient 1).  Use the quadratic formula to work out $x^2$ and then $x$, so you find all four complex roots $r_1, r_2, r_3, r_4$.  Therefore the quartic in your question factors into two quadratics in three ways:
$$
(x-r_1)(x-r_2) \cdot (x-r_3)(x-r_4), \ \ (x-r_1)(x-r_3) \cdot (x-r_2)(x-r_4), \ \ (x-r_1)(x-r_4) \cdot (x-r_2)(x-r_3).
$$
If you work out what the six possible quadratic factors (such as $(x-r_1)(x-r_2)$ and $(x-r_3)(x-r_4)$) really are, you'll see they involve either $\sqrt{a}$ or $\sqrt{b}$ or $\sqrt{ab}$.  All of this is purely algebraic, and that means that as long as $a$ or $b$ or $ab$ is a square in some field, then the factorization involving that square root will make sense using coefficients in that field.  So we finally reach the key point: for all $a$ and $b$ in the integers modulo a prime $p$, either $a$ or $b$ or $ab$ is a square.


问:
Showing a function $\notin L^1$
Q) Let $F(x)=\frac{1}{x}\int_0^x f(t)dt$ for $x\in (0,\infty)$. If $f>0, f\in L^1$, show that $F\notin L^1$.

My attempt:

Since $f\in L^1$, $\int_0^x f(t)dt<\infty$ for all $x\in (0,\infty)$

$$\int_0^\infty |F(x)|dx = \int_0^\infty \frac{1}{x}\bigg|\int_0^x f(t)dt\bigg| = \int_0^\infty \frac{1}{x}\int_0^x f(t)dt \text{ since $f>0$ } \\
=\int_0^\infty \int_t^\infty \frac{f(t)}{x}dxdt \text{ By Fubini's } \\
=\int_0^\infty f(t)\log x\bigg|_t^\infty dt = \infty
$$

Hence $F\notin L^1$. Is this correct?

答: Alternatively you have $F(x)\underset{+\infty}{\sim}\frac{\|f\|_1}{x}$ which is not integrable.

Title: Showing a function $\notin L^1$


问:
Trigonometric equality $x = 99 \sin (\pi x)$
Find the number of real solutions of $\displaystyle x = 99 \sin (\pi x)$.

I am getting stuck in some trigonometric relations.

答:
There is an obvious root at $x=0$. Since $\sin(-\pi x)=-\sin(\pi x)$, there are just as many negative solutions as there are positive solutions.

So we count the positive solutions, double the result, and add $1$. Now we show how to count the positive solutions. We will use a picture to do the count.

Rewrite the equation as
$$\sin(\pi x)=\frac{x}{99}.$$
For $x\gt 99$, the right-hand side is $\gt 1$, while the left is always $\le 1$. So we will not have to look for solutions beyond $x=99$.

To count the positive solutions $\lt 99$, draw the graph of $y=\sin(\pi x)$. We need to go a moderately long way. Note that $\sin(\pi x)$ has period $2$. So when we go up to $x=99$, there will be $49$ full periods, plus a half period.

Let's take a look at $y=\frac{x}{99}$. This is a line with very shallow slope. It enters the first period of $y=\sin(\pi x)$ and leaves it, giving $2$ intersections, one in the interval $(0,\pi/2)$, the other in the interval $(\pi/2,\pi)$. It enters and leaves the second period ($2$ more solutions). Continue. You will have to think a little about what happens in the final half-period.

Remark: A graphing calculator may help, but you will need to play with the viewing window.


问:
Show that $(x,y)=(0,0)$ is the only solution to the system of equations $ax+by=0$, $cx+dy=0$ iff. $ad-bc \neq 0$
I'm learning Linear Algebra and I can't figure out this problem:

Show that $(x,y)=(0,0)$ is the only solution to the system of equations
$$
ax+by=0
$$
$$
cx+dy=0
$$
if and only if $ad-bc \neq 0$

So far we've studied systems of equations and rank of matrices.

答:
First notice that $ax + by = 0$ and $cx + dy= 0$ is the same as the matrix
$M := \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$
$\begin{pmatrix}
x\\
y
\end{pmatrix}$=
$\begin{pmatrix}
0\\0
\end{pmatrix}
$

"$\Leftarrow$" This direction is easy. If $ad-bc \neq 0$ then the matrix $M$ has an inverse. Consequently
$M^{-1}\begin{pmatrix}
0\\0
\end{pmatrix}
=
\begin{pmatrix}
x\\y
\end{pmatrix}
$
So $x,y=0$.

"$\Rightarrow$" For this direction you can do what Brad S. recommended in the comments. Simple use gaussian elimination (i.e. solving for x,y).


问:
draw the pushdown automaton
How can I draw the pushdown automaton for language $L = \{ bin(n) \text{#} bin(3n)^{R} | n \in  \mathbb{Z}, n \geq 0 \}$ when bin(n) is binary notation of $n$ without leading zeros. Alphabet is $\{ 0, 1, \text{#} \}$

答:
Here is how our PDA works :

if the first symbol is not 1 reject (except if the string is # , see edge cases down)

push all symbols to stack till you read # , do not push # (this step is pushing all symbols of bin(n) to stack)

now using the stack we can read bin(n) from right to left , the input we receive from this point is supposed to be bin(3n)r , meaning we can read also bin(3n) from right to left

Now we consider the symbol at top of stack x , the symbol we read w , and make transitions for valid x & w to states that correspond to the carry

ex 1, after reading # , if x = 0 , 0*3 = 0 , carry = 0 , so if x = 0 , we expect w = 0 , and we head to state carry = 0 , that is 0,0;ε to carry 0

ex 2, if we are in state carry 1 , and x = 0 , 0*3 + 1 = 1 and carry = 0 , so if x = 0 , and we are in carry 1 , we expect w = 1 , and we go to state carry = 0 , that is 1,0;ε to carry 0

Now continue this way and you should be able to construct the PDA , you will find that you will need another state for carry 2 , for each state think about valid x & w to make a transition and what carry will they cause to know which state you will head to , continue this way till you finish the PDA

if we are in state carry 0 , and stack is empty , we non-deterministically guess that input has finished and head to accept state

Hopefully these steps help you build your own PDA , after you try , you can check your answer , here is the PDA :

edge cases :

here we explain how the transition 1,\$;\$ acts like 1,0;ε for ex : if bin(n) = 1 , you see how bin(3n) = 10 , bin(3n)r = 01 , after reading bin(n) the stack will have \$1 after reading 0 from bin(3n)r the stack will be $ and we are in carry 1 , usually we need to pop a 0 to go to carry 0 , but since we read \$ , we know this is the end of bin(n) and so we treat \$ as if it was a leading 0 ( which does not change bin(n) ) , we pop \$ as if we would have popped a 0 , then push it again for the test of emptiness of stack , if you think about it you can make the same transition to the accept state instead of carry 0 , both are the same

in case n = 0 , bin(0) = 0 , since we do not allow leading 0s , the string will end up being just # , so we guess whether n = 0 or not , if n is not 0 we check if b(n) starts with 1

Title: draw the pushdown automaton


问:
Conditional Probability Ball Problem
There are 10 blue balls and 2 green balls inside a bag. Mr. Ali draws a ball inside the bag. If the drawn ball is blue, Mr. Ali will replace it with 4 blue balls inside the bag. If the drawn ball is green, Mr. Ali will replace it with 6 green balls inside the bag. After that, Ms. Bea draws a ball inside the bag. Let A be the event that Mr. Ali draws a green ball and B be the event that Ms. Bea draws a green ball. Determine the following probabilities.

$P(B)$

I am confused if the probability of event B has two values or exactly one value.
Since Mr. Ali will be the first one who will draw a ball in the given scenario, then there must be 2 possible values. Either$P_{A}(B)$ or $P_{\overline{A}}(B)$.

If $P(B)$ has only one value, I am thinking that it is the same as $P(A)$.

$P_B(A)$ Probability of Event A given event B occured

This one is only possible to be solved using the formula for the conditional probability if $P(B)$ was already known. Since I am not sure how many values of $P(B)$ has, I am confused about how to get this value.

Any comments or suggestions will be much appreciated.
Thank you in advance.

答:
Evaluating the situation right before Bea's drawing:

(5/6) of the time, Bea will be faced with 13 blue, 2 green.

(1/6) of the time, Bea will be faced with 10 blue, 7 green.

Therefore

$$p(B) ~=~ \left[\frac{5}{6} \times \frac{2}{15}\right] ~+~
\left[\frac{1}{6} \times \frac{7}{17}\right].$$

I am interpreting the syntax of the second question to be chance of event B occurring assuming that event A has occurred.  This will be

$$\frac{7}{17}.$$

Addendum

Apparently, I misinterpreted the second question.  Instead, it is asking for

$$p(A|B) ~=~ \frac{p(A ~\text{and} ~B ~\text{both occuring})}{p(B)}.$$

Using the work already done at the start of this answer, that evaluates to

$$\frac{\frac{1}{6} \times \frac{7}{17}}{\left[\frac{5}{6} \times \frac{2}{15}\right] ~+~
\left[\frac{1}{6} \times \frac{7}{17}\right]}.$$

Title: Conditional Probability Ball Problem


问:
Why is $y'' = d/dy[(1/2)y'^2]$?
I was wondering why the formula above is correct because if you differentiate $(1/2)y'^2$, shouldn't you get $y'.y''$ by the chain rule?

答:
To me at least, it makes most sense when you use algebraically manipulable differentials.  With these, $$y'' = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
To understand where this comes from, when you take the derivative of $\frac{dy}{dx}$ use the quotient rule.  See also my paper "Extending the Algebraic Manipulability of Differentials."

So, we have:
$$ \frac{d\left(\frac{1}{2}\left(\frac{dy}{dx}\right)^2\right)}{dy} \\
 \frac{2 \cdot \frac{1}{2}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx} - \frac{dy}{dx}\frac{d^2x}{dx}\right)}{dy} \\
 \frac{1}{dy}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx} - \frac{dy}{dx}\frac{d^2x}{dx}\right) \\
\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}
$$

Which is the definition above of the second derivative of $y$.

Title: Why is $y'' = d/dy[(1/2)y'^2]$?


问:
How do I solve this in an understandable and direct way?
For each $i \in \Bbb N$, let $f_i: \Bbb N \mapsto \{0, 1\}$.

Let $A = \{f_i : i \in \Bbb N\}$ and $E = \{n \in \Bbb N : f_n(n) = 0\}$.

Does there exist a $f \in A$ such that $E = \{n \in\Bbb N : f(n) = 1\}$

I think that there does not exist a $f \in A$ such that $E = \{n \in\Bbb N : f(n) = 1\}$. However, I am unsure on how to prove it. How do I show my answer in an understandable way?

答:
No.

Let $f=f_k$ for $k\in\mathbb{N}$.

If $k\in E$, where $E = \{n \in \Bbb N : f_n(n) = 0\}$, then $f_k(k) = 0$. Assume $E=E_2$, where $E_2 = \{n \in\Bbb N : f(n) = 1\}$. Then, $k\in E_2$ as well and $f_k(k)=1$, which contradicts the premise $f_k(k)=0$. It follows, $E\neq E_2$, which contradicts what we wanted to show.

If $k\notin E$, then $f_k(k)=1$. Assume $E=E_2$. Then, $k\notin E_2$ as well and $f_k(k)=0$, which contradicts the premise $f_k(k)=1$. It follows $E\neq E_2$, which contradicts what we wanted to show.


Title: Why is $y'' = d/dy[(1/2)y'^2]$?

Body: I was wondering why the formula above is correct because if you differentiate $(1/2)y'^2$, shouldn't you get $y'.y''$ by the chain rule?

Body: $ d/dy[(1/2)y'^2]=(d/dx [(1/2)y'^2] ) /(dy/dx)=(1/2).2y'.y''. 1/y'=y''$ where $y'=dy/dx.$


问:
Coordinate offset in linear PDE
I have a linear PDE that is not quite traditional of the form $$ F_{x}(x+y,y) - F_{y}(x+y,y) = 0$$ with $F(x+0,0) = f(x).$

I have tried using the method of characteristics in the standard way but it doesn't seem to apply to the present problem. Is this solvable by classical means such as with the method of characteristics and not merely guessing?

答:
Note that
$$\partial_x[F](u,v)=\partial_u F \partial_x u+\partial_v F\partial_x v$$
Similarly
$$\partial_y[F](u,v)=\partial_u F \partial_y u+\partial_v F\partial_y v$$
Using $u=x+y, v=y$
$$\partial_x [F](x+y,y)=\partial_{x+y} [F](x+y,y)$$
$$\partial_y [F](x+y,y)=\partial_{x+y} [F](x+y,y)+\partial_y F$$
So your equation is simply
$$-\partial_y[F](u,y)=0$$
Which hopefully you know how to solve...

Title: Coordinate offset in linear PDE


Body:
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.

I solved it in the following way:

I state that $a$ is the number in front of $2010$, in other words if the number we're looking for is $A$, then $A=10^4\cdot a+2010$.

Hence, $a\cdot10^4\equiv1\bmod2011$.

$1956a\equiv1\bmod2011$

$55a\equiv2010\bmod2011$

So $55a=2010+2011k$ for some positive integer $k$.

So $30+31k\equiv 0 \bmod55$

After doing some trial and error for $k\in[0,54]$

$k_{min}=15$

So $55a=32175$, $a=585$.

Hence the smallest number possible is $5850000$

I solved this question using classical number theory and trial-and-error. However, I realize that this question can be solved using the euclidean algorithm. When I looked it up on the internet the only thing I could find was this: https://en.wikipedia.org/wiki/Euclidean_algorithm which is how to find the gcd using euclid's algorithm. Could you please explain to me how to solve the question using the euclidean algorithm, and also please explain to me what the euclidean algorithm is, how it can be applied and some resources to study from?

Body:
$\!\bmod \color{#0a0}{2011}\!:\,\ 10^4\! = 5(2000)\equiv 5(-11)\,$ so by simple mental arithmetic

$\ \ \dfrac{1}{10^4}\equiv \dfrac{\color{#c00}1}{-5}\dfrac{1}{11} \equiv \dfrac{\color{#c00}{-2010}}{-5}\dfrac{1}{11}\equiv\dfrac{402}{11}\equiv \dfrac{402+\color{#0a0}{2011\cdot \color{#90f}3}}{11} \equiv\, \bbox[5px,border:1px solid #c00]{585}$

We used inverse reciprocity $\bmod 11\!:\ 0\equiv 402+\color{#0a0}{2011\cdot \color{#90f}k}\equiv 6\!-\!2k\!\iff\! \color{#90f}{k\equiv 3}\,$ makes exact the quotient by $11,\,$ and we cast out $11$'s, e.g. $\bmod 11:\ 2011\equiv -2\!+\!0\!-1\!+\!1\equiv-2$.


问:
Find the limit of of such a sequence defined by recurrence
So I need to find the limit of such a sequence defined by recurrence:

$a_1 = \frac{1}{2}$

$a_2=1$

$a_n= \frac{1}{2}a_{n-1} + \sqrt{a_{n-2} } $  for every $n \in \mathbb{N} \geq 3$

I have calculated some elements of that seqence and those are:

$a_1 = \frac{1}{2}$

$a_2=1$

$a_3 = 1+ \frac{\sqrt{2}}{2}$

$a_4= \frac{{3}}{2} + \frac{\sqrt{2}}{4}$

$a_5= \frac{{3}}{4} + \frac{\sqrt{2}}{8} + \sqrt{1+\frac{\sqrt{2}}{2}}$

So I know that the values fo next elements of the sequence grow really slow. But what do I do next?

答:
Your next step should be proving that this increasing sequence has an upper bound. Then we can say that the sequence has a limit because of Completeness of the real numbers. We can prove this by induction.

If $a_{n-2}\leq4$ and $a_{n-1}\leq4$, then $$a_n=\frac{1}2a_{n-1}+\sqrt{a_{n-2}}\leq\frac{1}2\cdot4+\sqrt{4}=2+2=4$$
so $a_n\leq 4$. This holds for base case ($n=3$) since $a_1\leq4$ and $a_2\leq4$.

Hence it is proven that the limit exists. Let $\lim_{n\to\infty}a_n=L$, then the recurrence relation gives you $$L=\frac{1}2L+\sqrt{L}\implies \frac{1}2L=\sqrt{L}\implies L=4$$


问:
GHK = 0, What is the relationship between G and K if H is invertible?
I'm practicing a problem that I came across, it's

$\text{G}$, $\text{H}$, and $\text{K}$ are $4\times4$ matrices. $\text{GHK}=0$ and $\text{H}$ is invertible, does $\text{GK}=0$?

I was trying to think in terms of ranks, if $\text{H}$ is rank $4$, then if $\text{G}$ is rank $3$, and then $\text{GH}$ is rank $3$. If $\text{K}$ happens to be rank $1$ and all in the kernel of $\text{GH}$, then $\text{GHK}=0$. But that doesn't necessarily mean that $\text{K}$ is the kernel of $\text{G}$?

In a more general case, if you a matrix without a kernel, say $\text{A}$, and you multiply it by another matrix without a kernel, you'll always get an invertible matrix? But if you multiply $\text{A}$ by a matrix with a kernel, your result is no longer valid in terms of invertibility?

Thanks!

Body:
$$
 \left(  \begin{array}{rrrr}
  1  &  0  &  0  &  0     \\
  0  &  1  &  0  &  0     \\
  0  &  0  &  0  &  0     \\
  0  &  0  &  0  &  0
\end{array}
  \right)
 \left(  \begin{array}{rrrr}
  0  &  0  &  1  &  0     \\
  0  &  0  &  0  &  1     \\
  1  &  0  &  0  &  0     \\
  0  &  1  &  0  &  0
\end{array}
  \right)
 \left(  \begin{array}{rrrr}
  1  &  0  &  0  &  0     \\
  0  &  1  &  0  &  0     \\
  0  &  0  &  0  &  0     \\
  0  &  0  &  0  &  0
\end{array}
  \right)
  $$
is the zero matrix.


答:
These are two lines passing through origin their slopes are $m_1=-a/b$ and $m_2=-c/d$
if the slopes are equal $(m_1=m_2)$ then $ad-bc=0$, both lines are coincident and there are infinitely mny solutions.

If slopes are unequal $(m_1\ne m_1 \implies ad-bc\ne 0$) they will meet only in the origin $(0,0)$.

Title: Show that $(x,y)=(0,0)$ is the only solution to the system of equations $ax+by=0$, $cx+dy=0$ iff. $ad-bc \neq 0$

Body:
I'm learning Linear Algebra and I can't figure out this problem:

Show that $(x,y)=(0,0)$ is the only solution to the system of equations
$$
ax+by=0
$$
$$
cx+dy=0
$$
if and only if $ad-bc \neq 0$

So far we've studied systems of equations and rank of matrices.


问:
Is this $\epsilon$-$\delta$ limit correct?
I have to show that $\lim_{x\to2}x^2+x+1=7$ with the $\epsilon$-$\delta$ definition of limit. Here is how I have done it: $\left\lvert x^2+x-6\right\rvert=\lvert x−2\rvert\lvert x+3\rvert<\epsilon$ and since we are close to $2$, we can assume that the $\delta$-neighborhood of $c=2$ must be have a radius of max $\delta=1$ which implies that : $\lvert x-2\rvert < \frac{\epsilon}{6}$.
We now choose $\delta = \min\{3,\frac{\epsilon}{6}\}$  and we can conclude that if $\lvert x−2\rvert<\delta$, it follows that $\left\lvert x^2+x+1-7\right\rvert < 6\frac{\epsilon}{6}=\epsilon$. Is this calculation correct? Do I miss something? Or some details?

答:
We put $f(x) = x^2 +x+1$

We can prove that :
$|f(x) - l|<\delta $ $ \Rightarrow $ $ |x-a|<\alpha $

$\alpha , \delta > 0$

$|f(x) - l|= |x^2 +x+1-7|=|x^2 +x-6|=|x-2||x+3|$

$|f(x) - l|<\delta$

$\Rightarrow $ $|x-2||x+3|<\delta$

Suppose $x\in [\frac{3}{2}, \frac{5}{2}] $

$\Rightarrow $$\frac{9}{2}\leq x+3\leq\frac{11}{2}$

$\Rightarrow $$|x+3|\leq \frac {11}{2} $

$\Rightarrow $$|x-2||x+3|\leq \frac{11}{2} |x-2|$

we know that :

$|x-2||x+3|<\delta$

So:

$\frac{11}{2} |x-2|<\delta$

$\Rightarrow $ $|x-2|<\frac{2} {11}\delta$

We put $\alpha=\frac{2\delta} {11}$

Finally :
After the definition of limite we proved $\lim_{x\to 2} f(x) =7$


问:
Show that ${\displaystyle \frac{1}{1-2x^2}} - 1-2x^2 = O(x^4)$ as $x \to 0$.
This is my first time using Big-Oh notation and I just wanted to make sure I got this all correct. Please point out any errors, thank you!

First simplify $f(x) = \frac{1}{1-2x^2} - 1-2x^2$ to $f(x) = \frac{4x^4}{1-2x^2}$.

Observe that since $x\rightarrow 0$ we have that
\begin{equation*}
-1 < x < 1
\end{equation*}
which implies
\begin{equation*}
0 \leq 2x^2 < 2
\end{equation*}
\begin{equation*}
-2 < -2x^2 \leq 0
\end{equation*}
\begin{equation*}
-1 < 1-2x^2 \leq 1
\end{equation*}
\begin{equation*}
0 \leq |1-2x^2| \leq 1
\end{equation*}
We can thus conclude that,
\begin{equation*}
|f(x)| = |\frac{4x^4}{1-2x^2}| = \frac{4x^4}{|1-2x^2|} \leq 4x^4
\end{equation*}

and $f(x) = O(x^4)$.

答: Hint: You want to bound the denominator from below. The denominator equals zero when $x=\pm\frac{\sqrt{2}}{2}$. Set some $c<\frac{\sqrt{2}}{2}$. Then when $|x|\leq c$, it follows that $|1-2x^2|\geq 1-2c^2>0.$


问:
Confusion in Discrete Fourier Transform derivation
In the book "Digital Image Processing" by R. Gonzalez, section 4.4.1 "Obtaining the OFT from the Continuous Transform of a Sampled Function",

they get $$\tilde{F}(\mu)=\sum^{\infty}_{n=-\infty}f_ne^{-j2\pi\mu n\Delta T}$$
where $f_n$ are the sampled points from a continuous function $f$ and $\tilde{F}$ is the fourier transform of the sampled function.

Now, the book says:

"Suppose that we want to obtain $M$ equally spaced samples of $\tilde{F}(\mu)$ taken
over the period $\mu= 0$ to $\mu=\frac1{\Delta T}$. This is accomplished by taking the samples at the following frequencies:
$$\mu=\frac{m}{M\Delta T} \text{     ,                   } m=0,1,2,...,M-1$$
Substituting this result for $\mu$ into <first equation> and letting $F_m$ denote the result yields" $$F_m=\sum^{M-1}_{n=0}f_ne^{-j2\pi \frac{mn}M}\text{   ,   } m=0,1,2,...,M-1$$
Now what I don't understand is that when we are obtaining samples of $\tilde{F}$, then all we have to do is get values of the function at the given points. So how did doing just that, change the limits of the summation? According to me, $$F_m=\sum^{\infty}_{n=-\infty}f_ne^{-j2\pi \frac{mn}M}\text{   ,   } m=0,1,2,...,M-1$$ this should be the correct expression. So where did I go wrong? Please explain.

答: Perhaps you and the author are confusing the term given samples. I think the author means that $M$ values of $f_n$ are given. In this case the finite sum is correct. However you suppose that $M$ values of $F_m$ are given. Here the infinit sum is correct. Note that in both cases the result is not identical.


问:
The determinant of a triangular matrix is the product of its diagonal entries — proof using permutations

Prove that the determinant of a triangular matrix is the product of its diagonal entries.

My attempt:

Suppose $A$ is an $n\times n$ matrix, then:

$$\det(A) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$

The product of diagonal entries is $A_1\sigma_1 \cdots A_n\sigma_n$ when $\sigma_i = i$ for $i = 1, \ldots, n$.

I claim that when the permutation is not an identity, there exist $i$ and $j$ s.t. $\sigma_i < i$ and $\sigma_j > j$.

Let $\sigma_j \neq j$, then $\sigma_k = j$ where either $k > j$ or $k < j$.

If $k > j$, let $\sigma_\ell = k$ where $\ell > k$ or $\ell < k$.

If $\ell < k$, done. Otherwise, $\ell > k$ and suppose for $\sigma_\ell, \ldots, \sigma_n$ that $n>\cdots>\ell$ and follow the above sequence of operations incrementally.

Then, $\sigma_n = n - 1$.

This implies that $\sigma_p = n$ where $p \leq j$.

If $k <j$, let $\sigma_h = k$ where $h > k$ or $h < k$.

Likewise, if $h > k$, done. Otherwise, $h < k$ and suppose for $\sigma_1, \ldots, \sigma_h$ that $1<\ldots<h$ and follow that above sequence of operations incrementally.

Then, $\sigma_1 = 2$.

This implies that $\sigma_q = 1$ where $q \geq j$.

Thus, there is at least one $i$ and $j$ s.t. $\sigma_i < i$ and $\sigma_j > j$.

By definition, if $A$ is upper-triangular, $A_{ij} = 0$ when $i > j$ and so, $A_{i\sigma_i} = 0$. Alternatively, if $A$ is lower triangular, $A_{ij} = 0$ when $i < j$ and $A_{j\sigma_j} = 0$.

So, the determinant of a triangular matrix is indeed the product of diagonal entries.

QED.

I'm not sure if my proof is correct and I wanted someone to go over it. Please let me know if I made any mistake(s) anywhere and if so what I can and should do to correct it.

答:
Suppose $A$ is upper-triangular, so that $A_{ij} = 0$ whenever $i < j$. Then it is clear that the product $A_{1 \sigma(1)} \cdots A_{n \sigma(n)}$ will be zero whenever there is an $i \in \{1, \ldots, n\}$ such that $i < \sigma(i)$. I claim that for every non-identity permutation $\sigma$ we can find an $i$ such that $i < \sigma(i)$, and hence
$$ \det A = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1 \sigma(1)} \cdots A_{n \sigma(n)} = \sum_{\sigma = \operatorname{id}} (\operatorname{sgn} \sigma) A_{1 \sigma(1)} \cdots A_{n \sigma(n)} = A_{1 1} \cdots A_{n n}. $$

Proof of claim: (by contrapositive) suppose $i \geq \sigma(i)$ for all $i \in \{1, \ldots, n\}$. The value $\sigma(1)$ must be an integer between $1$ and $n$, and $1 \geq \sigma(1)$, so we must have $\sigma(1) = 1$. Then the value $\sigma(2)$ must be an integer between $1$ and $n$, but it cannot be that $\sigma(2) = 1$ (since $\sigma(1) = 1$ and $\sigma$ is a bijection), and then $2 \geq \sigma(2)$ forces $\sigma(2) = 2$. Continuing in this way, we see that $\sigma$ must be the identity permutation.


问:
Number of elements of a finite set in at least two subsets.
if three sets $A$, $B$, and $C$ are subsets of a finite set, how many elements are in at least two of the subsets?

currently, I think I am trying to find $|(A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) \ (A ∩ B ∩ C)|$. is this correct so far? or am I missing something incredibly obvious?

答:
Actually, if you think about all the possible subsets of the union $A \cup B \cup C$, they are the followings:

(i) $A - (B \cup C)$

(ii) $(A \cap B) - C$

(iii) $B - (A \cup C)$

(iv) $(A \cap C) - B$

(v) $(A \cap B \cap C)$

(vi) $(B \cap C) - A$

(viii) $C - (A \cup B)$

You need to find element which appears in at least two sets. Thus, you need count (ii) - (iv) - (v) and (vi) in the above list.


问:
Determinant of a triangular matrix
Using the cofactor expansion, explain why the determinant of a triangular matrix is the product of the elements on its diagonal.

Is it the fact that there are $0$'s in the $L$ or $U$ part of the matrix and that somehow comes into play to where only the diagonal is accounted for? I'm not quite sure.

答:
Let ${\rm U}_n$ be an invertible $n \times n$ upper triangular matrix. Let

$${\rm U}_{n+1} := \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix}$$

Using the Schur complement,

$$\det \left( {\rm U}_{n+1} \right) = \det \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix} = \left( u_{n+1} - {\rm 0}_n^\top {\rm U}_n^{-1} {\rm c}_{n+1} \right) \det \left( {\rm U}_{n} \right) = u_{n+1} \det \left( {\rm U}_{n} \right)$$

Let ${\rm U}_{1} =: u_1$. Hence,

$$\begin{aligned} \det \left( {\rm U}_{1} \right) &= u_1\\ \det \left( {\rm U}_{2} \right) &= u_2 \, u_1\\ &\vdots\\ \det \left( {\rm U}_{n} \right) &= \color{blue}{u_n \, u_{n-1}\cdots u_2 \, u_1} \end{aligned}$$

The case where ${\rm U}_n$ is non-invertible is left as an exercise for the reader.

matrices block-matrices schur-complement determinant

Title: Determinant of a triangular matrix


问:
Given value for cosine of an angle calculate cosine of half of an angle
I have a cosine value of an angle $α$. I would like to calculate the cosine value corresponding to half of the angle $α$. For example if $cos(α) = 0.5$, I would like to know what cos(α/2) is equal to. Is this possible to calculate without using the inverse cosine function, diving the resulting angle by two, and calculating the cosine again?

答: Sure, there is a cosine half-angle formula: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Half-angle_formulae</p>


答:
HINT

\begin{align*}
\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b) & \Rightarrow \cos(2x) = \cos^{2}(x) - \sin^{2}(x)\\\\
& \Rightarrow \cos(2x) = 2\cos^{2}(x) - 1\\\\
& \Rightarrow \cos(x) = \pm\sqrt{\frac{\cos(2x)+1}{2}}
\end{align*}

Can you take it from here?

Title: Given value for cosine of an angle calculate cosine of half of an angle

Body: I have a cosine value of an angle $α$. I would like to calculate the cosine value corresponding to half of the angle $α$. For example if $cos(α) = 0.5$, I would like to know what cos(α/2) is equal to. Is this possible to calculate without using the inverse cosine function, diving the resulting angle by two, and calculating the cosine again?


问:
making the domain of $z ↦\tan(z)$ injective
Given the following:

$\sin(z)$ = ($e^i$$^z$ - $e^-$$^i$$^z$)/$2i$

$\cos(z)$ = ($e^i$$^z$ + $e^-$$^i$$^z$)/$2$

$\sin(z)\cos(w) - \cos(z)\sin(w) = \sin(z-w)$

$\sin(z) = 0$ has solution $z = kπ$ for some integer $k$

$\tan(z) = \tan(w)$ has solution $z = w + kπ$ for some integer $k$

The natural domain of the function $z ↦ \tan(z)$ is D = {$z∈C|z ≠ (2m+1)π/2$, m∈ Z}

Question: Use the above to find and sketch a subset $U$ of $D$ so that:

the function $f: U ↦ C$ given by $f(z) = tan(z)$ is injective AND

the set of real numbers {$|z| | z ∈ U$} is not bounded above AND

the set of real numbers {$\arg(z) | z ∈ U$} is equal to [0, π].

--My attempt at this question:

Since $|z|$ is not bounded above, I thought it would be the shaded area outside a circle. Also, since $\arg(z) = [o, π]$ it is only the top half of the circle. I think that covers the last two conditions in the question, however I am not sure. Could someone give me some pointers on how to make $f(z)$ injective? Thanks to anyone that helps. I really appreciate it!

答:
Let's restate the requirements on $U$:

Injectivity:   $U$ does not contain any pair of numbers $z,w$ such that the difference $z-w$ is an integer multiple of $\pi$.

Unboundedness:   $U$ stretches infinitely far away from $0$.

Argument: (a) $U$ is contained in the closed upper halfplane; and (b) $U$ intersects every half-line $\{re^{i\theta}:r\ge 0\}$ with $0\le \theta \le \pi$.

Argument requirements are easy to satisfy: include a half-disk, such as $\{|z|< 1, \mathrm{Im}\,z\ge 0\}$, in  $U$. And do not include anything below the real line.

Concerning unboundedness, the first thing that comes to mind is an infinite horizontal strip. But that would violate injectivity.

A way to satisfy injectivity is to make $U$ "thin" in the horizontal direction. For example, the domain $ \{-1<\mathrm{Re}\,z<1, \ \mathrm{Im}\,z\ge 0\}$ satisfies the injectivity req (as well as the argument req). Imagine it as an infinitely tall tower.

Also, you may want to exclude $0$ from $U$, just to avoid arguing about its argument.


问:
Differentials on hyperelliptic curves
The following is an exercise of Vakil's Foundations of Algebraic Geometry. We consider a hyperelliptic curve $C$ of genus $g$ over an algebraically closed field $k$ of characteristic not $2$. Associated to $C$ is a double cover $\pi \colon C \to \mathbb{P}^1$ of the projective line. This cover has $2g + 2$ ramification points $p_1,\ldots,p_{2g + 2}$ which we may assume to lie on an affine open cover $U = \operatorname{Spec} k[x]$ of $\mathbb{P}^1$.

I am asked to compute the zeros and poles of $\pi^* dx$. I do not know how to proceed. From what I understand, we can think of $C \times_{\mathbb{P}^1} U$ as $\operatorname{Spec} k[x,y] / \big(y^2 - (x - p_1) \cdots (x - p_{2g+2})\big)$, and so on this space we may understand $\Omega^1$ as being the module generated by $dx$ and $dy$ satisfying the relationship
$$2y \,dy = \bigg(\frac{d}{dx}(x-p_1)\cdots(x-p_{2g+2})\bigg)\,dx,$$
and I presume that the $dx$ in this module corresponds to the $\pi^* dx$ we are interested in. But then what? I'm not sure I even know what it means for $\pi^* dx$ to have a zero or a pole anywhere, or how this can be seen from the formula above.

答:
Let $f(x) = \prod_{i=1}^{2g+2} (x - p_i)$ so $C: y^2 = f(x)$. First some intuition: if we set $y=0$ in your equation
\begin{equation}
\label{deriv}
2y \,dy = f'(x) \,dx, \tag{$*$}
\end{equation}
we get $dx = 0$, so the points on $C$ where $y = 0$, namely $(p_1, 0), \ldots, (p_{2g+2},0)$, should be zeroes of $dx$. You might worry that $f'(p_i)$ could be $0$, but this can't happen since $C$ is nonsingular. (See example 21.2.7 (p. 564) in Vakil for an explicit example.)

More formally, the module of differentials $\Omega_{C/k}$ is locally free of rank $1$, so there is an open cover $\{U_i\}_i$ of $C$ such that $\DeclareMathOperator{\O}{\mathscr{O}}\Omega_{C/k}|_{U_i} \cong \O_C|_{U_i}$ for each $i$. A section $s$ of $\Omega_{C/k}$ has a zero or pole at a point $P \in U_i$ exactly when the corresponding function $f_i := \Phi_i(s) \in \O_C(U_i)$ does, where $\Phi_i$ is the isomorphism $\Omega_{C/k}|_{U_i} \overset{\sim}{\to} \O_C|_{U_i}$. (This is true for sections of line bundles in general: to check if a section has a zero at a point, we trivialize the line bundle and then check if the corresponding function does. For another example of computing the zeroes and poles of a section of a line bundle, see this post.)

Since the curve is nonsingular, the distinguished affine opens $D(y)$ and $D(f'(x))$ cover $C$. The equation $(\ref{deriv})$ will allow us to trivialize $\Omega_{C/k}$ over both of these sets. On $D(y)$ we have
$$
dy = \frac{f'(x)}{2y} dx
$$
so $\Omega_{C/k}(D(y))$ is generated by $dx$ and we have the isomorphism
\begin{align*}
\Omega_{C/k}(D(y)) &\overset{\sim}{\to} \O_C(D(y))\\
dx &\mapsto 1\\
dy &\mapsto \frac{f'(x)}{2y} \, .
\end{align*}
The constant function $1$ has no zeroes or poles, so neither does $dx$ on this patch. On $D(f'(x))$ we instead have
$$
dx = \frac{2y}{f'(x)} dy
$$
so $dy$ is a generator on this patch and we have the isomorphism
\begin{align*}
\Omega_{C/k}(D(f'(x))) &\overset{\sim}{\to} \O_C(D(f'(x)))\\
dy &\mapsto 1\\
dx &\mapsto \frac{2y}{f'(x)} \, .
\end{align*}
The function $\frac{2y}{f'(x)}$ has zeroes on $D(f'(x))$ where $y = 0$, which are exactly the Weierstrass points $(p_1, 0), \ldots, (p_{2g+2},0)$.

We've been working in an affine patch, so now we have to check for zeroes and poles on the line at infinity. Here the curve is given in an affine patch by
$$
v^2 = u^{2g+2} f(1/u) = \prod_{i=1}^{2g+2} (1 - p_i u)
$$
where $u = 1/x$ and $v = y/x^{g+1}$. (This can be seen by considering $C$ embedded in the weighted projective space $\mathbb{P}(1,g+1,1)$.) You can try computing the zeroes and poles of $dx$ on this patch and I'll put my computation below.

 Let $\hat{f}(u) = \prod_{i=1}^{2g+2} (1 - p_i u)$ so the curve is given by $v^2 = \hat{f}(u)$. The only points of $C$ on the line at infinity are $(u,v) = (0,1)$ and $(0,-1)$ so we just have to check these two points. These points both lie in the affine open $D(v)$ and differentiating our equation, we have $$2v dv = \hat{f}'(u) du \implies dv = \frac{\hat{f}'(u)}{2v} du$$ so $du$ is a generator for $\Omega_{C/K}$ on this patch. Since $dx = d(1/u) = -\frac{1}{u^2} du$, then $dx$ has a double pole at both $(0,1)$ and $(0,-1)$.

Title: Differentials on hyperelliptic curves


问:
Two homotopy equivalent closed $n$-manifolds are homeomorphic
I heard somewhere that there is a wrong conjecture that states:

Two homotopy equivalent closed $n$-manifolds ($n\geq 3$) are homeomorphic.

and the lens spaces are the first counterexample to this conjecture. I want to know the author of this wrong conjecture, and a reference for it. I am also curious about other counterexamples? (Here is a related question)

答: The right conjecture (due to Armand Borel, not Horowitz) is that any two homotopy-equivalent closed aspherical  manifolds are  homeomorphic, see here. (A manifold is aspherical if it is connected and has contractible universal covering space.) This conjecture is wide-open (in dimensions $\ge 4$) but is expected to be false. The conjecture is known to be true for Riemannian manifolds of negative curvature (very hard work of Farrel and Jones). It is also known to be true for 3-dimensional (aspherical) manifolds, which is primarily due to Perelman's solution of the Geometrization Conjecture.


答:
I'll do an example with $2\times 2$ matrices.  Let
$$G =K= \left(\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}\right) \quad \text{and} \quad H=\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right).$$
We find that $GHK =0$, while
$$GK= \left(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right).$$
What we've done is use the permutation matrix $H$ to move around the eigenvalues of $G$ and $K$.  With appropriate movement, we can change the dimension of the null space.</p>

Title: GHK = 0, What is the relationship between G and K if H is invertible?

Body:
I'm practicing a problem that I came across, it's

$\text{G}$, $\text{H}$, and $\text{K}$ are $4\times4$ matrices. $\text{GHK}=0$ and $\text{H}$ is invertible, does $\text{GK}=0$?

I was trying to think in terms of ranks, if $\text{H}$ is rank $4$, then if $\text{G}$ is rank $3$, and then $\text{GH}$ is rank $3$. If $\text{K}$ happens to be rank $1$ and all in the kernel of $\text{GH}$, then $\text{GHK}=0$. But that doesn't necessarily mean that $\text{K}$ is the kernel of $\text{G}$?

In a more general case, if you a matrix without a kernel, say $\text{A}$, and you multiply it by another matrix without a kernel, you'll always get an invertible matrix? But if you multiply $\text{A}$ by a matrix with a kernel, your result is no longer valid in terms of invertibility?

Thanks!


问:
If f: [2,5] $\rightarrow [4,13]$ is a continuous function prove that there is $c\in [2,5]$ such that $f(c) = 3c-2$
If $f: [2,5] \rightarrow [4,13]$ is a continuous function prove that there is $c\in [2,5]$ such that $f(c)= 3c-2$.

I understand that the extreme value theorem is important in this question and that at most if $c=5$ then $3*5-2=13$, but I am struggling to explain this in a clear and orderly manner. Any advice on how to do Extreme Value Theorem proofs?

答:
In case $f(2)=4+p$ and $f(5)=13-q, p,q>0$, let $g(x)=f(x)-3x+2 \implies g(2)=4+p-4=p, g(5)=13-q-13=-q$, then $g(2)g(5)=-pq<0$. So by IVT it follows that $g(c)=0 \implies f(c)=3c-2$ for some $c\in (2,5)$.

If $pq=0$ the statement satisfied trivially.


答:
Well: $$g(x)=3x-2\implies g(2)=4, g(5)=13$$

and, given $f:[2,5]\to [4,13]$, this means $4\leq f(t)\leq 13$, i.e. $g(2)\leq f(t)\leq g(5)$ for every $t\in[2,5]$

We can infer that, specifically, $f(2)\geq g(2)$ and $f(5)\leq g(5)$, and the only way this is satisfied for a continuous function $f$ is if it has an intersection with $g$ in the range $[2,5]$

The below picture demonstrates. The space $([2,5]\times [4,13])$ on which the function is defined is the green rectangle, and $g$ is the blue line, while $f$ is any continuous curve (with no breaks or jumps) that goes from the left side of the rectangle to the right side without leaving it. It's easy to see it's impossible for this curve to not cross or touch the blue line on the way.

Title: If f: [2,5] $\rightarrow [4,13]$ is a continuous function prove that there is $c\in [2,5]$ such that $f(c) = 3c-2$

Body:
If $f: [2,5] \rightarrow [4,13]$ is a continuous function prove that there is $c\in [2,5]$ such that $f(c)= 3c-2$.

I understand that the extreme value theorem is important in this question and that at most if $c=5$ then $3*5-2=13$, but I am struggling to explain this in a clear and orderly manner. Any advice on how to do Extreme Value Theorem proofs?


问:
What is the codomain of a natural transformation $\alpha : F \Rightarrow G$?
... and (when, if ever) does it matter?

It could be:

the set of objects in the target category of F and G

the set of objects in Im(G)

Im(G)

I think (2) is most intuitive.

Update: There was some ambiguity in the question. In the functor category [C, D] the co-domain of a natural transformation $\alpha$ is just an object $G$. But in implementing $\alpha$ we require a function mapping $FX \to GX$ for all X in the source category. The image of this function of X is clearly the set of all objects in the target category Y for which there exists an $\alpha_x :FX \to Y$. The codomain of this function of X is unclear (and apparently not interesting at all except as a tool for satisfying category axioms of [C,D]).

答:
A morphism only has a codomain in the category in which it is a morphism; there is simply no other answer for the codomain of $\alpha$ but $G$, regardless of how it's implemented.

It looks like you believe that $\alpha$ has codomain $G$ in $[C,D]$, but that it must have some other kind of codomain in some other category; it appears you perhaps want this category to be the category of sets, but it's hard to tell. The point, though, is that perhaps the set theoretic object $X\mapsto\alpha_X$ can be made into a morphism in some other category (presumably you have in mind something other than the function $\mathrm{Ob}(C)\to\mathrm{Mor}(D)$), but there's no hope of answering what its codomain is until you can tell us what category you think it's a morphism in. And if it's not $[C,D]$, then you've answered a different question than "what's the codomain of the natural transformation $\alpha$?"


答:
$$\left(2\cos\frac{\theta}2\right)^2=\left(1+\cos\theta\right)^2+\left(\sqrt{1-\cos^2\theta}\right)^2\\
\Longrightarrow 4\cos^2\frac{\theta}2=2+2\cos\theta\\
\Longrightarrow \cos\frac{\theta}2=\pm\sqrt{\frac{1+\cos\theta}2}$$

If $\quad0\leq\theta\leq\pi\quad$, take positive.

If $\quad\pi\leq\theta\leq2\pi\quad$, take negative.


答:
Drawing out the ven-diagram can help with small numbers of sets.

You want to include all the elements in the green and yellow sections, and none in the red sections. To start with, consider:

$|A \cap B| + |B \cap C| + |A \cap C|$

This gets each green section once, but counts the yellow section three times. So, we need to subtract out the yellow section twice which leaves us with:

$|A \cap B| + |B \cap C| + |A \cap C| - 2 \times |A \cap B \cap C|$

Title: Number of elements of a finite set in at least two subsets.

Body:
if three sets $A$, $B$, and $C$ are subsets of a finite set, how many elements are in at least two of the subsets?

currently, I think I am trying to find $|(A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) \ (A ∩ B ∩ C)|$. is this correct so far? or am I missing something incredibly obvious?


问:
Power Inequality for nonnegative number and power
I am not quite sure the following fundamental inequality:

Let $x$, $y\in \mathbb{R}^+$ and $a$, $b\in \mathbb{R}^+$.  Can I say $$x^a>y^a \Rightarrow x>y$$

and $$x^a>y^b \Rightarrow x>y^{b/a} \text{ ?}$$

Thanks for help.

答:
I think the two relationship is correct  because :

$x^a>y^a$$\Rightarrow (x^{a}) {^{\frac{1}{a}}}>(y^{a}) {^{\frac{1}{a}}} $$\Rightarrow x>y$

And:

$x^a>y^b$$\Rightarrow (x^{a}) {^{\frac{1}{a}}}>(y^{b}) {^{\frac{1}{a}}} $$\Rightarrow x>y^{\frac{b} {a}} $

But that is correct if $( a, b) \in \mathbb{R^{*+}} $and$ (x, y) \in \mathbb{R^{*+}} /({1}) $


问:
Finding the actual vector
In a homework assignment, I have been asked to find the actual vector v when given basis coordinate [v]B = [5 -4 1]. It just wants my reasoning and doesn't want me to solve the problem. B = {[-1 0 1], [0 0 1/2], [3 -2 0]}.

I am seeking to understand how to do this. As of right now, I believe that a zero vector is involved somehow because to find the basis coordinates in another problem, I row reduced the given vector after setting it equal to a zero vector. Could someone guide me in the right direction to answer this question?

答: As I indicated in the comments, we get $\begin{pmatrix}-1&0&3\\0&0&-2\\1&1/2&0\end{pmatrix}\begin{pmatrix}5\\-4\\1\end{pmatrix}=\begin{pmatrix}-2\\-2\\3\end{pmatrix}$.

Title: Finding the actual vector


问:
Integers between $1$ to $1000$ satisfying a condition that $1/2 * {2n \choose n}$ is even
For how many integers $n,$ for $1 \leq n \leq 1000,$ is the number $\frac{1}{2}\left(\begin{array}{c}2 n \\ n\end{array}\right)$ even?

I'm thinking about expanding the top into $(2n \cdot (2n-1) \cdot ... \cdot (n+1))$ and the bottom is simply $n!$ and that expression would be $0 \mod 4$, but I'm not sure how to approach it after that point.

答:
It is simpler to find $n$s where the term is not even.  A quick substitution of $n = 2^i$ for $i = 0, 1, 2, \ldots , 9$ shows this is the case.

If $n = 2^i$ we have $\frac{1}{2} {2 \cdot 2^i \choose 2^i} = \frac{2^i (2^{i+1}-1)(2^{i+1}-2) \cdots (2^{i+1}-2^i)}{2^{i-1}!}$.  The number of factors of $2$ in the numerator are the same as the number of factors of $2$ in the denominator, so the term is not even.

We need not consider any other numbers where there are no factors of $2$, as these simply cannot be even.

Thus the total number is $1000 - 10 = 990$.


问:
Motivation behind the sheaf of relative Kähler differentials
I'm interested in the geometric motivation behind the bundle (or in more general framework the sheaf) of relative differentials $\Omega_{X/Y}$
of a morphism $f: X \to Y$ smooth $k$-varieties. Differential geometry provides following picture of relative differentials:

Let $f: X \to Y$ a equidimensional surjective map between connected manifolds $Y, X$
and moreover assume that every fiber $F:= f^{-1}(y) \subset X$ for $y \in Y$ is also a connected
submanifold of same dimension. We obtain an exact sequence of
tangent spaces

$$  0 \to T_{X/Y} \to T_X \to f^*T_Y \to 0  $$
where $T_{X/Y}$ is the kernel of induced map of tangent bundles. intuitively what is really going on there is that for every $x \in f^{-1}(y)$, the $(T_{X/Y})_x$ is the tangent space of the fiber at $x$. The relative space of Kähler differentials
$\Omega_{X/Y}$ defined as the dual of $T_{X/Y}$ and sits in the sequence which we will
obtain if we dualize the sequence above of tangent spaces:

$$ 0 \to f^*\Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0  $$

Another definition of relative Kähler differentials which is more common to use in modern algebraic geometry works as follows:

Let us embedd $X$ as the image of the diagonal $\Delta: X \to X \times_Y X$
and assume that the ideal sheaf $I \subset O_{X \times_Y X}$ defines the closed image
$\Delta(X) \subset X \times_Y X$.

This gives us another sequence of tagent spaces

$$ 0  \to T_{\Delta(X)} \to T_{X \times_Y X} \to N_{X \times_Y X/X} \to 0 $$

with normal bundle $N_{X \times_Y X/X}$. It's a basic fact that the dual
of $N_{X \times_Y X/X}$ is $I/I^2$ and most books on algebraic
geometry define the sheaf relative Kähler differentials by

$$\Omega_{X/Y} := I/I^2$$

Since this definition not uses that $f$ is a map of smooth maps this
is a far generalization of the old fashion setting from
differential geometry. Now, if there is any justice n this world then these two definitions of relative differentials should coinside if we deal with $X, Y$ and $fY$ nice enough.

Therefore, if $f$ a surjective map between conneted manifolds such that
every fiber is a connected submanifold, why the tangent bundle
$T_{X/Y}$ and the pullback of normal bundle $\Delta^* N_{X \times_Y X/X}$ of
$\Delta(X) \subset X \times_Y X$ are canonically isomorphic?

Can we write down an explicit isomorphism and understand what is geometrically going on there?

答:
Given a linear map $g \colon V \rightarrow W$ between finite dimensional vector spaces, we can form the exact sequence

$$  0 \rightarrow V \xrightarrow{\Delta} V \times_{W} V \rightarrow (V \times_{W} V )/ \operatorname{im}(\Delta) \rightarrow 0 $$

where $V \times_{W} V = \{ (v,v') \, | \, g(v) = g(v') \}$ and $\Delta(v) = (v,v)$ is the diagonal embedding. Then we can see that $(V \times_{W} V) / \operatorname{im}(\Delta)$ is canonically isomorphic to $\ker g \subseteq V$ via the subtraction isomorphism $[(v,v')] \mapsto v - v'$ (whose inverse is given by $v \mapsto [(v,0)]$).

This applies fiberwise to the situation you describe for manifolds. Let's assume that $f$ is a surjective submersion (this is enough to guarantee that the fibered product actually exists in the category of smooth manifolds). Fix $x \in X$ and pull back the short exact sequence via $\Delta$ to obtain
$$ 0 \rightarrow \Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x}
\rightarrow \Delta^{*} T(X \times_{Y} Z)|_{x} \rightarrow \Delta^{*} (N_{\operatorname{im}{\Delta} \hookrightarrow X \times_{Y} X})|_{x} \rightarrow 0 $$

Using the identifications
$$
T(X \times_{Y} X)|_{(x,x')} =  \{ (v,v') \in T_x X \times T_{x'} X, \, | \, df|_{x}(v) = df|_{x'}(v') \}, \\
\Delta^{*} T(X \times_{Y} Z)|_{x} = T_x X \times_{T_{f(x)} Y} T_x X, \\
\Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x} = \{ (v,v) \in T_x X \times T_x X \} \cong T_x X, \\
T_{X/Y}|_{x} = \ker \left( df|_{x} \right)
$$
we see that we are in the algebraic situation described in the beginning of the answer (with $V = T_x X, W = T_{f(x)} Y$ and $g = df|_{x}$).

To see what is going on geometrically, it might be useful to start with the case where $Y = \textrm{pt}$ so that the fibered product is just a regular product $X \times X$ and the relative tangent bundle is the regular tangent bundle of $X$. Take $X = \mathbb{R}$ and identify each equivalence class in a fiber of the normal bundle of $X$ inside $X \times X$ with a slanted line. The isomorphism then identities each such slanted line with its intersection with the (say) $x$ axis (which is the "unslanted, regular" tangent space).


问:
Schema of separation and set of all sets
The schema of separation states (this is probably simplified) states that if $P$ is a property and $X$ is a set, then there exists a set $Y = \{ x \in X : P(x)\}$.

The notes I'm reading say that from this we can conclude that the set of all sets doesn't exist, by applying the schema of separation to the property $x \notin x$.

I'm not seeing how applying the schema of separation to that property shows it.  Can someone clarify this?

答:
This is Russell's paradox.  Starting with the universal set $U$ containing all sets, consider the set

$$ S = \{ x \in U : x \notin x \} $$

So now consider the question: Is $S \in S$?  Well, if $S \in S$, then by its own construction ($S$ is the set of all sets that don't contain themselves) it has to be the case that $S \notin S$, which is impossible.  Otherwise, if $S \notin S$, then by its own construction it must be the case that $S \in S$, a contradiction.  Therefore, $S \in S$ iff $S \notin S$, giving a contradiction.  The only assumption we made (besides the comprehension schema) was that the universal set $U$ exists, and therefore there cannot be a set of all sets.

Hope this helps!


问:
Show that $T$ is a bounded operator and $\left \| T \right \|=1$
Consider in $X=\left ( L^2\left [ 0,1 \right ],\left \| \cdot  \right \|_{2} \right )$ the operator $T:X \rightarrow X$ defined by $\left ( Tf \right )\left ( t \right )=\int_{0}^{t}f\left ( s \right )ds$, where $t \in\left [ 0,1 \right ]$. Show that $T$ is an bounded operator and $\left \| T \right \|=1$.

My attempt:

First,
$$\left \| Tf \right \|_{2}^2=\left \| \left ( Tf \right )\left ( t \right ) \right \|_{2}^2= \int_{0}^{1}\left | \int_{0}^{t} f\left ( s \right )ds\right |^2dt \leq \int_{0}^{1}\left ( \int_{0}^{t} \left | f\left ( s \right ) \right |ds\right )^2 dt \leq \int_{0}^{1}\left ( \int_{0}^{t} \left \| f \right \|_{\infty} \right )^2 dt = \int_{0}^{1} t^2 \left \| f \right \|_{\infty} ^2 dt=\frac{\left \| f \right \|_{\infty} ^2 }{3}.$$

Then,
$$\left \| f \right \|_2^2=\int_{0}^{1}\left | f\left ( s \right ) \right |^2 ds \leq \int_{0}^{1}\left ( \left \| f \right \|_{\infty} \right )^2 ds = \left \| f \right \|_{\infty} ^2.$$

Therefore
$$\frac{\left \| Tf \right \|_{2} }{\left \|f \right \|_{2} } \leq \frac{1}{\sqrt{3}} < 1,$$

and this tell us $T$ is a bounded operator.

Now I'm stuck with the another inequality. What $f$ should I set to get $\left \| T \right \|=1$? Can anyone give me a hint? Thanks in advance.

Body:
We can say $\lvert \int_0^t f(s)\,ds \rvert^2\leq \|1\|_{L^2([0,t])}^2\,\|f\|_{L^2([0,t])}^2=t \|f\|_{L^2([0,t])}^2 \leq t \|f\|_{L^2([0,1])}^2$.

Hence, $\|Tf\|^2=\int_0^1 \lvert \int_0^t f(s)\,ds \rvert^2\,dt \leq \|f\|_{L^2([0,1])}^2 \int_0^1 t\,dt =\frac 12 \|f\|_{L^2([0,1])}^2$. Taking square roots shows
$$
\|Tf\|\leq \frac{\sqrt{2}}{2}\|f|.
$$
What am I missing?


Title: Finding the actual vector

Body:
In a homework assignment, I have been asked to find the actual vector v when given basis coordinate [v]B = [5 -4 1]. It just wants my reasoning and doesn't want me to solve the problem. B = {[-1 0 1], [0 0 1/2], [3 -2 0]}.

I am seeking to understand how to do this. As of right now, I believe that a zero vector is involved somehow because to find the basis coordinates in another problem, I row reduced the given vector after setting it equal to a zero vector. Could someone guide me in the right direction to answer this question?

Body:
Even simpler than Chris's answer,
$$\mathbf{v}=5\mathbf{b}_1-4\mathbf{b}_2+\mathbf{b}_3$$
$$=5(-1,0,1)-4(0,0,1/2)+1(3,-2,0)=(-5+3,-2,5-2)=(-2,-2,3)$$


问:
Why is $y'' = d/dy[(1/2)y'^2]$?
I was wondering why the formula above is correct because if you differentiate $(1/2)y'^2$, shouldn't you get $y'.y''$ by the chain rule?

答:
It may help to think about how to interpret $\displaystyle\frac{1}{2}\frac{\mathrm{d}(y')^2}{\mathrm{d}y}$. We may define

$$ \Delta y:=y(x+\Delta x)-y(x), \qquad \Delta(y')^2:=\big[y'(x+\Delta x)\big]^2-\big[y'(x)\big]^2 $$

which depend on both $x$ and a quantity we call $\Delta x$. Note that $\Delta y$ and $\Delta x$ tend to $0$ together, so

$$ \frac{1}{2}\frac{\mathrm{d}(y')^2}{\mathrm{d}y}:=\frac{1}{2}\lim_{\Delta y\to0} \frac{\Delta (y')^2}{\Delta y} =\frac{1}{2}\lim_{\Delta x\to0} \frac{\Delta(y')^2/\Delta x}{\Delta y/\Delta x} = \frac{1}{2}\frac{\big[(y')^2\big]'}{y'} $$

Of course by chain rule, $\big[(y')^2\big]'=2y'y''$ so you get cancellation.

(This is an elaboration of user159888's answer, really.)

Title: Why is $y'' = d/dy[(1/2)y'^2]$?


答:
Let $V$ be the collection of all sets, and assume that $V$ was a set, let $W=\{x\in V\mid x\notin x\}$, then by the separation schema $W$ is a set.

Now ask yourself, $W\in W$ or $W\notin W$?

If $W\in W$ then $W$ satisfies $W\notin W$, which is a contradiction to the fact that $W\in W$.

If $W\notin W$ then $W$ does not satisfy $W\notin W$, which is again a contradiction!

Therefore $W$ cannot be a set, but this can only mean that $V$ cannot be a set to begin with.

This is also known as Russell's paradox.

Title: Schema of separation and set of all sets

Body:
The schema of separation states (this is probably simplified) states that if $P$ is a property and $X$ is a set, then there exists a set $Y = \{ x \in X : P(x)\}$.

The notes I'm reading say that from this we can conclude that the set of all sets doesn't exist, by applying the schema of separation to the property $x \notin x$.

I'm not seeing how applying the schema of separation to that property shows it.  Can someone clarify this?


问:
Find all sets of positive integers $a, b, c$ satisfying the three congruences :
$a ≡ b \pmod c$ , $b ≡ c \pmod a , c ≡ a \pmod b$ .

What I tried :
I wrote $a ≡ b \pmod c$ as $by + c = a$ , $b ≡ c \pmod a$ as $cz + a = b $, $c ≡ a \pmod b$ as $ax + b = c$ .
Then from these equations I got $ax + by + cz = 0$ ,I don't know what to do further .

答:
WLOG, assume $a\geqslant b \geqslant c$.

Because $a|b-c, a> b-c\geqslant 0$, so $b=c$.

Can you take it from here?


问:
Show that $a=c\mathrm{e}^{\mathrm{i}\beta}+b\mathrm{e}^{-\mathrm{i}\gamma}$ in a triangle $ABC$ with side lengths $a$, $b$ and $c$
The question goes like this:

In triangle $ABC$, the lengths of the sides $BC$, $CA$, $AB$ are $a$, $b$, $c$ respectively, and the radian measures of the angles at $A$, $B$, $C$ are $\alpha$, $\beta$, $\gamma$ respectively. Prove that
$$a=c\mathrm{e}^{\mathrm{i}\beta}+b\mathrm{e}^{-\mathrm{i}\gamma}.$$

My attempt at a solution:

For convenience sake, let $z^{}_1$, $z^{}_2$ and $z^{}_3$ denote the complex numbers represented by the points $A$, $B$ and $C$ respectively in an Argand diagram.

Referring to my drawing above, since $\overrightarrow{CB}$ is $\overrightarrow{CA}$ stretched by a factor of $\dfrac{a}{b}$ and rotated through $\gamma$ radians anticlockwise (about $C$), I conclude that
$$z_2-z_3=\frac{a}{b}e^{i\gamma}(z_1-z_3).$$

Using similar arguments for the directed line segment pairs $\overrightarrow{BC}, \overrightarrow{BA}$ and $\overrightarrow{AB}, \overrightarrow{AC}$, I obtain
$$z_1-z_2=\frac{c}{a}e^{i\beta}(z_3-z_2), \quad z_3-z_1=\frac{b}{c}e^{i\alpha}(z_2-z_1).$$

How do I proceed from here onwards to establish the given result?

答:
By cosine rule we can prove $$b\cos \gamma+c\cos \beta=a$$
Also by sine rule:$$c\sin \beta-b\sin \gamma=0$$
Thius $$a=b\cos \gamma+c\cos \beta+i(c\sin \beta-b\sin \gamma)$$
$$a=ce^{i\beta}+be^{-i\gamma}$$


答:
$y$ is a function of $x$, that is $y=y(x)$; so you must derive with respect of $x$.
Hence when you think of $y'(x)^2$ you have, because of the derivative of a composition, to derive the exterior function (which is the square function) and multiply by the derivative of the interior function (which is $y'(x)$), so you have that
$$\frac{\text{d}}{{\text{d}x}}[y'(x)^2]=2y'(x)y''(x)\iff\frac{1}{2} \frac{\text{d}}{{\text{d}x}} \left[y'(x)^2\right]=y'(x)y''(x)$$
Since $\frac{1}{2}$ is constant you can bring it inside the derivative (remember the linearity of derivative), so
$$\frac{\text{d}}{\text{d}x}\left[\frac{1}{2}y'(x)^2\right]=y'(x)y''(x)$$

Title: Why is $y'' = d/dy[(1/2)y'^2]$?


问:
Proving Lipschitz condition is not satisfied.
I have to prove that this function

$f(x,y) = \frac{x^3y}{x^4+y^2},  \text{ if } (x,y) \neq (0,0)$

$f(x,y) = 0, \text{  if } (x,y) = (0,0)$

in the domain $D = \{ (x,y)\in\mathbb{R}^2:|x|\leq1, |y|\leq2 \}$.

Does not satisfy Lipschitz condition.

I'm using a theorem that says that Lipschitz condition $|f(x,y_1)-f(x,y_2)| \leq L|y_1-y_2|, \ \forall \ (x,y_1), (x,y_2) \in D $, is satisfied by $f(x,y)$ if and only if $\sup\limits_{(x,y)\in D} \left| \frac{\partial}{\partial y} f(x,y) \right| \leq L$.

I don't know how to justify properly that $\sup\limits_{(x,y)\in D} \left| \frac{\partial}{\partial y} f(x,y) \right|$ does not exist because $\frac{\partial}{\partial y} f(x,y)$ is not bounded.

答: If you compute $\frac{\partial f}{\partial y}(x,0)$ you get $\frac{1}{x}$, so you get the unboundness.


答:
Here is my solution with $p$-adic valuation.

I claim $ \frac{1}{2} {2n\choose n}$ is odd if and only if $n = 2^{k}$ for some $k$.

This is true since $v_{2}(2^{k}!) = 2^{k} - 1$, which implies $n = 2^{k}$ as $v_{2}\left({2n\choose n}\right) = v_{2}(2n!) - 2v_{2}(n!)$. Now considering the binary representation of $n$, we see $v_{2}(n!)$ can be written as the sum of terms of the form $v_{2}(2^{i}!)$ from which the result follows.

By complementary counting, the answer is therefore $1000 - \lceil\log_2(1000)\rceil= 990$.

Title: Integers between $1$ to $1000$ satisfying a condition that $1/2 * {2n \choose n}$ is even

Body:
For how many integers $n,$ for $1 \leq n \leq 1000,$ is the number $\frac{1}{2}\left(\begin{array}{c}2 n \\ n\end{array}\right)$ even?

I'm thinking about expanding the top into $(2n \cdot (2n-1) \cdot ... \cdot (n+1))$ and the bottom is simply $n!$ and that expression would be $0 \mod 4$, but I'm not sure how to approach it after that point.


问:
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$
Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$

I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.

I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.

答:
Let
$$y=\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x},$$
then, solving this with respect to $x$, we get
$$x=\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}.$$
So,
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx=\int_0^\infty\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy.$$
The latter integral can be solved by Mathematica and yields $$\pi\log54.$$

Of course, we want to prove that the result returned by Mathematica is correct.

The following statement is provably true, that can be checked directly by taking derivatives of both sides:
$$\int\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy =\\ \frac{1}{2} i \left(2 \text{Li}_2\left(\frac{iy}{8}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{iy}{6}+\frac{1}{3}\right)+2\text{Li}_2\left(\frac{iy}{6}+\frac{2}{3}\right)+\text{Li}_2\left(\frac{iy}{4}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{2i}{y-2 i}\right)-\text{Li}_2\left(-\frac{2 i}{y+2i}\right)-\text{Li}_2\left(-\frac{1}{6} i (y+2i)\right)-\text{Li}_2\left(-\frac{1}{4} i (y+2i)\right)-2 \left(-\text{Li}_2\left(-\frac{2i}{y-4 i}\right)+\text{Li}_2\left(\frac{2 i}{y+4i}\right)+\text{Li}_2\left(-\frac{1}{8} i (y+4i)\right)+\text{Li}_2\left(-\frac{1}{6} i (y+4i)\right)\right)\right)+\pi  \left(\frac{1}{2}\log \left(3 \left(y^2+4\right)\right)+\log\left(\frac{3}{64}\left(y^2+16\right)\right)\right)+\log \left(4\left(y^2+4\right)\right) \arctan\left(\frac{y}{4}\right)-\left(\log576-2\log \left(y^2+16\right)\right) \arctan\left(\frac{4}{y}\right)+\log\left(y^2+4\right) \text{arccot}\left(\frac{6y}{8-y^2}\right)-\arctan\left(\frac{2}{y}\right)\log12 +\arctan\left(\frac{y}{2}\right)\log2$$

The remaining part is to calculate $\lim\limits_{y\to0}$ and $\lim\limits_{y\to\infty}$ of this expression, which I haven't done manually yet, but it looks like a doable task.


问:
Find the limit of such a sequence defined by recurrence (problematic induction because of $0$ in the denominator)
So I need to find the limit of such a sequence defined by recurrence:

$a_1 = \frac{1}{2}$

$a_{n+1}= \frac{\sqrt{1+a_n^2}-1}{a_n}$

I have calculated some elements of that seqence and those are:

$a_1 = 1$

$a_2= \sqrt{2}-1 \approx 0,41$

$a_3 = 1+ \frac{\sqrt{2}}{2} = \frac{\sqrt{4-\sqrt{2}}-1}{\sqrt{2}-1} \approx 0,22$

So I know that the values of next elements of the sequence are falling. Next I wanted to do prove by induction that the limit = $0$ , but I can't put $0$ in the denominator. So what do I do?

答:
Now we prove that there is a lower bound.

Let $a_n>0$, then $$\sqrt{1+a^2_n}-1>\sqrt{1+0^2}-1=0$$ Since both $a_n$ and $\sqrt{1+a_n^2}-1$ are positive, we have $$a_{n+1}=\frac{\sqrt{1+a_n^2}-1}{a_n}>0$$

Hence the limit exists because of Completeness of real numbers. Let $\lim_{n\to\infty}a_n=L$, then the recurrence relation gives you $$L=\frac{\sqrt{1+L^2}-1}{L}$$

If $L\ne0$, then $$\begin{align}L^2=\sqrt{1+L^2}-1&\implies L^2+1=\sqrt{1+L^2}\\&\implies L^4+2L^2+1=L^2+1\\&\implies L^4+L^2=0\\&\implies L^2(L^2+1)=0\\&\implies L=0\end{align}$$
which is a contradiction. Therefore $L=0$.


问:
how to get the vector solution with a system of 3 variables and 4 equations
I'm doing an exercise of solid mechanics in which I got these 4 equations

Own vector associated with eI = 7,391
(i) System of equations:

$1.39116499156263*nx + 0.5*ny + 3.0*nz  = 0$

$0.5*nx - 7.39116499156263*ny + 1.5*nz  = 0$

$3.0*nx + 1.5*ny - 7.39116499156263*nz  = 0$

$nx^2 + ny^2 + nz^2 - 1  = 0$

The solutions is $ \vec{nI} = ( 0.907 ) e_1 + ( 0.142 ) e_2 + ( 0.397 ) e_3.$

I put the system in a Linear Equation Systems Calculator and I got system impossible (and I agree with the solution)

However I can't understand why the vector solution is $\vec{nI} = ( 0.907 ) e_1 + ( 0.142 ) e_2 + ( 0.397 ) e_3$ ... Could someone explain it to me? Shouldn't  the solution be the vector $(0,0,0)$?

Body:
As in comments and answers, admitting a sign error in the first equation, using whole numbers

$$M=\left(
\begin{array}{ccc}
 -\frac{139116499156263}{100000000000000} & \frac{1}{2} & 3 \\
 \frac{1}{2} & -\frac{739116499156263}{100000000000000} & \frac{3}{2} \\
 3 & \frac{3}{2} & -\frac{739116499156263}{100000000000000}
\end{array}
\right)$$

$$|M|=259204707122654630160513716553\times 10^{-42}$$

Using the second and third equations
$$y=\frac{239116499156263 }{1528232998312526}x \qquad z=\frac{100323299831252323990892125169 }{229234949746878900000000000000}x$$ Using the fourth equation
$$x^2=\frac{52548662185454094659180291265210000000000000000000000000000}{638999024282822
   57528322078121921616822143238895937163278561}$$

All of that gives in decimals
$$x=0.9068401383\qquad y=0.1418896460 \qquad z=0.3968731434 $$


问:
Is there a clever solution to Arnold's "merchant problem"?
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

答:
To a first approximation, there is a spoonful of wine in the cup and a spoonful of tea in the barrel.  How much are each of these approximations off by?  Well, there is a bit less than a spoonful of wine in the cup, since a bit of the wine was removed in the second step.  And, there is a bit less than a spoonful of tea in the barrel, since there was a little wine mixed into the spoonful that was put into it.  But these errors are exactly the same: both are the amount of wine that was in the second spoonful.  So the two quantities are the same: both are one spoonful minus the amount of wine that was in the second spoonful.

Or, here's an even slicker way.  Notice that the total volumes of liquid of the cup and barrel have not changed, since the two spoonfuls they exchanged cancelled out.  So, the overall change must be that the barrel exchanged some volume of wine for the same volume of tea from the cup.

Note that your solution is actually wrong--when you compute the amounts of wine and tea in the second spoonful, you are assuming the cup was mixed uniformly after the first spoonful, which the problem tells you not to assume (that's what the "(nonuniform!)" is all about).


问:
A hard 'if and only if' trigonometric identity proof
Prove
$$
\frac{-2+2\tan A+2\cos B\cdot\sin B+\cot^2 A\cdot({\sec^4A-\operatorname{cosec}^2A-2)}}{2+\tan^2A-2\sin^2A} =(\sin A+\cos A)^2
$$

if and only if B is the double angle of A, or $B=2A+2k\pi$, $k=0,1,2,3...$

Advice is welcome as to improve notation and format.

i understand the RHS can be simplified to 1+sin(2A), but that doesn't seen to go anywhere .

Also, I have been able to simplify the left hand side of the expression down a little, but it's not taking me anywhere at all. I have a feeling that the difficulty of this proof lies in using the 2's in the LHS.

again, many thanks,
Yun Fei

答:
The equation can be written as

$\sin 2B =$ (complicated rational function of $\sin A$ and $\cos A$).

The left side is bounded and the right side appears to have an uncancelled 6th-order pole at $\sin A = 0$ from the $\cot^2 A \csc^4 A$ term.  [correction from comments: a 4th order pole from the $\cot^2 A \sec^4 A$ term]


问:
Induction hypothesis for inequality
The solution to a question I found online is in the image below, I'm confused on how did they go from $2^n + 2$ to$ 2^n + 2^n$. Where did they get the extra $2^n$ from? and then they got $2^{n+1}$ from $2^n + 2^n$. It's not multiplication so how did they get $2^{(n+1)}$ suddenly?

Prove that for all integers $n \geq 4,$ we have:

(a) $2 n+1<2^{n}$

Solution. We prove it by induction on $n .$ For $n=4,$ we have $2 \cdot 4+1=9<1$

Now, assume that $2 n+1<2^{n}$. Then,

$$
\begin{aligned}
2(n+1)+1 &=2 n+1+2 \\
&<2^{n}+2
\end{aligned} \quad \begin{array}{l}
\text { [by the IH }]
\end{array}
$$
$$
\begin{array}{ll}
<2^{n}+2^{n} & {\left[\text { as } n \geq 4,2^{n}>2\right]} \\
=2^{n+1}
\end{array}
$$

Title: Induction hypothesis for inequality

Body:
For $n\ge4$, $2<2^n$, so $2^n+2<2^n+2^n$,

and $2^n+2^n=2\cdot2^n=2^{n+1}$.


问:
Intersection of a conic and a line
Suppose $C=\{x_0^2+x_1^2+x_2^2\}\subset \mathbb{P}^2_{x_0,x_1,x_2}$, and $L=\{a_0x_0+a_1x_1+a_2x_2\}\subset \mathbb{P}^2_{x_0,x_1,x_2}$, I was asked to find the set of coefficients $(a_0,a_1,a_2)$ such that $C$ and $L$ do not intersect at two distinct points. Isn't $C$ only contain the origin in this case? How can I find such coefficients?

答: The main idea here is that a conic and a line always meet at two points when counted with multiplicity. In other words, you are looking for triples of coefficients such that the line meets $C$ with multiplicity two, i.e. is tangent to $C$. This is basically a multivariable calculus exercise (hint: you should get a condition of the form $F = 0$, where $F$ is a homogeneous polynomial in $a_0,a_1,$ and $a_2$).

Title: Intersection of a conic and a line


答:
The volume of spoon, $s$, is the conserved quantity. It is also the amount of wine in the cup.
 When you then take some mixture $\mathit{tea}+\mathit{wine} = s$ into the spoon,
$s-\mathit{wine}$ is the amount of wine left in the cup and the amount of tea poured into the wine barrel.

Title: Is there a clever solution to Arnold's "merchant problem"?

Body:
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.


问:
Euclidean Norm Calculation
If ||x||$_2$ = 3, ||4x-5y||$_2$ = 14 and ||5x+2y||$_2$ = 8 find ||y||$_2$

I had attempted to write at the vectors in explicit form and in doing so arrived at a point where I would add (||4x-5y||$_2$)$^2$ = (14)$^2$ and (2)(||5x+2y||$_2$)$^2$= 8$^2$(2) to remove <x,y>. From there I would subtract out 66||x||$_2$ by adding on (-66)(||x||$_2$)$^2$ = (-66)(9). Now I get a negative answer of -270/33. Is it possible to get a negative answer? I was under the impression that Euclidean norms acted like absolute value signs. Any advice and explanation would be appreciated.

答:
Assume $\textbf{x},\textbf{y} \in \mathbb{R}^n$. Recall the definition of the Euclidean norm

$$||\textbf{x}||_2:=\sqrt{\textbf{x}\cdot \textbf{x}}.$$
So the equations you were given really read

\begin{align*}
||\textbf{x}||_2 &=\sqrt{\textbf{x}\cdot \textbf{x}}=3,\\
||4\textbf{x}-5\textbf{y}||_2 &=\sqrt{(4\textbf{x}-5\textbf{y})\cdot (4\textbf{x}-5\textbf{y})}=14,\\
||5\textbf{x}+2\textbf{y}||_2 &=\sqrt{(5\textbf{x}+2\textbf{y})\cdot (5\textbf{x}+2\textbf{y})}=8.
\end{align*}
Squaring both sides to remove the radical gives
\begin{align*}
\textbf{x}\cdot \textbf{x} &=9,\\
(4\textbf{x}-5\textbf{y})\cdot (4\textbf{x}-5\textbf{y})&=196,\\
(5\textbf{x}+2\textbf{y})\cdot (5\textbf{x}+2\textbf{y})&=64.
\end{align*}

Using properties of the dot product, can you solve this system of equations for $\textbf{y}\cdot \textbf{y}$?

Title: Euclidean Norm Calculation


问:
Explanation needed for problem related with differential equation $w''(x)-q^2 w(x)=0.$
I am stuck on the following problem:

Suppose that $q \in \Bbb C.$ Consider the differential equation $w''(x)-q^2 w(x)=0.$ If every solution of this equation satisfies $$\sup_{T >0} \frac{1}{2T} \int_{-T}^{T}|w(t)|dt < \infty$$, then prove that $\text{Re}(q)=0.$

My Attempt:The solution of  the D.E. is given by $w(x)=Ae^{qx}+Be^{-qx},$ where $A,B$ being constants. Now,we compute $\displaystyle \int_{-T}^{T}|w(t)|dt=....=\frac{1}{q}(e^{qT}-e^{-qT})(A-B)$ so that $ \displaystyle \frac{1}{2T} \int_{-T}^{T}|w(t)|dt=\frac{A-B}{qT} \sinh {qT}.$

Now I am not sure how to progress from here. Any idea about how to go with the problem?

答:
Indeed, the general solution is given by $$w=Ae^{qx}+Be^{-qx} \tag1$$ Let's consider what this means: you can pick any real numbers $A$ and $B$ that you want, and (1)  will give you a solution. By assumption, that solution will satisfy

$$
\sup_{T>0} \frac{1}{2T}\int_{-T}^T |w(x)|\,dx <\infty \tag2
$$

That's a lot of freedom you have, picking $A$ and $B$. Use it wisely, making the choices that turn (2) into something tractable.

Here is one such choice: $A=1$, $B=0$. (I set $B$ to zero because I have no desire  to integrate the absolute value of a sum.) With my choice, (2) turns into
$$
\sup_{T>0} \frac{1}{2T}\int_{-T}^T |e^{qx}|\,dx <\infty  \tag3
$$
Now is the time to recall that $|e^{qx}|=e^{(\mathrm{Re}\,q)x}$. Plug this into (3), integrate (noting that antiderivative looks different when $\mathrm{Re}\,q=0$ and when $\mathrm{Re}\,q\ne 0$), and draw your conclusion.


问:
Resonance of a Matrix
I am provided a system of ODEs $$\dot{x} = 5x + xy$$ $$\dot{y} = y + 2x(y-x)$$
for $(x,y) \in \mathbb{R}^2.$

I know the Jacobian Matrix, $Df$, works out to be:
$$Df = \begin{bmatrix} 5+y & x\\ 2y-4x & 1+2x \end{bmatrix} $$
and
$$Df(0) = \begin{bmatrix} 5 & 0\\ 0 & 1 \end{bmatrix} $$
Therefore, the eigenvalues are: $\lambda_1 = 5, \lambda_2 = 1 $

However, I am uncertain whether $Df(0)$ is resonant, given the definition:

We say  $Df(0)$ is resonant if there are non-negative integers $m_i$ with $ \sum m_k \geq 2$ such that

$$(m,\lambda) := \sum^{d}_{k=1}m_{k}\lambda_{k} = \lambda_{s} $$
for some $s \in \{1,...,d\}$.

$|m|$ is the order of the resonance.

答:
Given your definition, $Df(0)$ is indeed resonant.

Take $m_1 = 0, m_2 = 5$. As both are non-negative, these are valid values for them. Then, we get
\begin{equation}
\sum^{d}_{k=1}m_{k}\lambda_{k} = 0\cdot5 + 5\cdot1 = 5 = \lambda_1
\end{equation}

Title: Resonance of a Matrix


问:
What percentage of answers to this question are correct?
I have the following question:

What percentage of answers to this question are correct?

a) 50%

b) 25%

c) 0%

d) 50%

My ideia is the following:

If I say that "a)" is the correct one, thus we have 50% of chance of having the right answer since we have "a)" and "d)" equals to 50%, that is 2 answers out of 4 correct, 50%, which is true.

If  "b)" is correct, then it is already true since "b)" is 1 out of 4, 25%.

"c)" can't be because if it is right, it will be already 25% of the answers.

If "d)" is right, it occurs the same as "a)", true.

Hence, we have 3 possible right answers, thus the percentage is 3/4, 75%.

Is that the case?

答: The intended answer is b.  You are supposed to assume that there is one correct answer out of four, so that is $25\%$.  $50\%$ would only be right if both a and d were correct, but questions are supposed to have only one correct answer unless specified.  c cannot be correct because if it were $25\%$ of the answers would be correct.


问:
Confused how the expression like $ \nabla \cdot (\rho \textbf{v} \otimes \textbf{v}) \cdot \textbf{v} $ is expanded
I'm quite confused to figure out how this expression is expanded:

$
\nabla \cdot (\rho \textbf{v} \otimes \textbf{v}) \cdot \textbf{v}
\stackrel{?}{=}
\nabla \cdot [\rho (\textbf{v} \cdot \textbf{v}) \textbf{v}] -
\rho \textbf{v} \cdot [(\textbf{v} \cdot \nabla) \textbf{v}]
$

$\rho$ is a scalar (density), $\textbf{v}$ is a vector (velocity).

from: The Finite Volume Method in Computational Fluid Dynamics, page 60

Edit: Edited the title expression.

Edit2: Edited the right hand side

答:
This follows from the identity

$$\nabla\cdot(A \otimes B) = (\nabla\cdot A)B + (A\cdot \nabla)B$$

see also: Divergence of a tensor product

OwnerDisplayName: ali b

OwnerDisplayName: IronicalCoffee


答:
This is relatively straightforward to expand out in Cartesian index notation, using the product rule.  We have
\begin{align*}
\nabla \cdot [\rho (\textbf{v} \cdot \textbf{v}) \textbf{v}]
&= \partial_i \left( \rho v_j v_j v_i \right) \\
&= \partial_i \left[ \left(\rho v_j v_i\right) v_j \right] \\
&= \rho v_j v_i \partial_i v_j + \left[\partial_i \left( \rho v_j v_i\right)\right] v_j \\
&= \rho v_j (v_i \partial_i v_j) + \left[\partial_i \left( \rho v_i v_j\right)\right] v_j \\
&= \rho \mathbf{v} \cdot \left[ (\mathbf{v} \cdot \nabla) \mathbf{v}\right] + \left[\nabla \cdot (\rho \mathbf{v} \otimes \mathbf{v}) \right]\cdot \mathbf{v}.
\end{align*}
The requested identity then follows immediately  by rearranging terms to the other side of the equation.

Title: Confused how the expression like $ \nabla \cdot (\rho \textbf{v} \otimes \textbf{v}) \cdot \textbf{v} $ is expanded

Body:
I'm quite confused to figure out how this expression is expanded:

$
\nabla \cdot (\rho \textbf{v} \otimes \textbf{v}) \cdot \textbf{v}
\stackrel{?}{=}
\nabla \cdot [\rho (\textbf{v} \cdot \textbf{v}) \textbf{v}] -
\rho \textbf{v} \cdot [(\textbf{v} \cdot \nabla) \textbf{v}]
$

$\rho$ is a scalar (density), $\textbf{v}$ is a vector (velocity).

from: The Finite Volume Method in Computational Fluid Dynamics, page 60

Edit: Edited the title expression.

Edit2: Edited the right hand side

OwnerDisplayName: Michael Seifert


问:
The problem of the area of a triangle formed by three random points inside another triangular
Thrown three points into a triangle of unit area, find the mathematical expectation of the area of the triangle formed by them

I already found the solution of similar task, but there was a square of unit area. But I really don’t understand what to do with the outer triangle, because it can be anything. I think the outer shape doesn’t matter, all that matters is shape’s area (in this problem it equals 1), but I could be wrong

答:
The answer is $\frac1{12}$, independent of the shape of triangle.

To see the independence, let $T_0$ be a fixed triangle with unit area.
For any triangle $T$ with unit area, let $\eta : \mathbb{R}^2 \to \mathbb{R}^2$ be an affine transformation which sends $T$ to $T_0$.
Recall affine transformation preserves ratio of area of geometric shapes. If $p_1, p_2, p_3$ are $3$ points in $T$, then $\eta(p_1),\eta(p_2),\eta(p_3)$ are three points in $T_0 = \eta(T)$ with
$$\verb/Area/(\triangle_{p_1p_2p_3}) = \verb/Area/(\triangle_{\eta(p_1)\eta(p_2)\eta(p_3)})$$
Furthermore, if $p_k$ are sampled uniformly from $T$, then $\eta(p_k)$ will be uniformly distributed over $T_0$. This implies the expected area of
"uniformly sampled" triangles in $T$ equals to that of $T_0$.

Since this is true for all $T$, the answer is independent of shape of triangle $T$.

This problem of finding the expected area of "uniformly sampled" triangles in a triangle is known as Triangle triangle picking, see the entry on mathworld for more information.

There are various generalizations of this problem. Instead of $3$ points, one can uniformly sampled $n$ points from a triangle of unit area and
askes: What is the expected area of the convex hull formed by those $n$ points?

This is part of the Triangle Point Picking problem, see entry on mathworld again for more information.

The answer of the expected area of convex hull of $n$ points is given by the formula:
$$\Delta_n = 1 - \frac{2}{n+1}\sum_{k=1}^n \frac1k = 1 - \frac{2H_n}{n+1}$$

where $H_n$ is a harmonic number. It is possible to write down a double integral to compute this number. For a derivation, see the second half of my answer to a semi-related question.

In any event, for the problem at hand, the expected area of "uniformly sampled" triangle is

$$\Delta_3 = 1 - \frac{2 H_3}{3 + 1} = 1 - \frac12\left(1 + \frac12 + \frac13\right) = \frac{1}{12}$$


问:
dimension of a quotient space
$V$ is a vector space of polynomials of degree less than or equal to 50 and $W$ is the set of polynomials in $V$ which are divisible by $x^4$.

I have shown that $W$ is a subspace of $V$ and that the dimension of $V/W$ is four.

If $V$ was the space of all polynomials over $\mathbb{R}$, why would the dimension of $V/W$ remain the same?

答:
Let $V$ be the space of polynomials of degree less than or equal to $N$, where $N \ge 3$, and, abusing notation slightly, we allow $N = \infty$ to mean 'all polynomials'.

Suppose $p \in V$, where $p(x) = \sum_{k=0}^n p_k x^k$. Then if $w(x) = \sum_{k=4}^n p_k x^k = x^4 \sum_{k=0}^n p_k x^{k-4}$, we see that $w \in W$. Hence $x \mapsto \sum_{k=0}^3 p_k x^k \in [p] = \{ p \} +W$. Since $p$ was arbitrary, it follows that $\dim V/W \le 4$.

Let $e_k(x) = x^k$ for $k=0,..,3$. Suppose $\sum_{k=0}^3 \alpha_k [e_k] = [0]$, or equivalently, $[\sum_{k=0}^3 \alpha_k e_k] = [0]$. This implies that $\sum_{k=0}^3 \alpha_k e_k = (x \mapsto \sum_{k=0}^3 \alpha_k x_k)  \in W$. However, the only way $\sum_{k=0}^3 \alpha_k e_k$ can be divisible by $x \mapsto x^4$ is if $\alpha_k = 0$. Hence $\{[e_k] \}_{k=0}^3$ are linearly independent, and hence we have $\dim V/W = 4$.

If $N<3$, a slight modification of the above shows that $\dim V/W = \min(4,N+1)$.

Title: dimension of a quotient space


问:
Law of Large Numbers for Martingales
The following question has me stumped: Let $X_n$ be a square integrable martingale with $E((X_n)^2)\leq n$ for all $n$. Prove that $X_n/n$ tends to $0$ almost surely. (this is in a sense a law of large numbers, generalizing the case where $X_n$ is a sum of $n$ iid zero mean random variables.)

Any ideas?

答:
The claim in question is a corollary of a standard SLLN for martingale difference sequences (MDS).

SLLN for MDS

The statement of SLLN for MDS is as follows.
Let $N_t$ be a martingale difference sequence (MDS) such that $\sum\limits_{t=1}^{\infty} \frac{E[N_t^2]}{t^2} < \infty$, then

$$
\frac{1}{n} \sum_{t=1}^n N_t \rightarrow 0 \;\;a.s.
$$

(In this case, the martingale difference sequence $N_t$ is given by differencing the martingale $X_t$: $N_t = X_t - X_{t-1}$.
Then summation by parts gives
\begin{align*}
\sum_{t=1}^n \frac{E[N_t^2]}{t^2} &= \sum_{t=1}^n \frac{E[X_t^2] - E[X_{t-1}^2]}{t^2} \\
                                  &= \frac{E[X_n^2]}{n^2} - \sum_{t = 1}^{n} E[X_{t-1}^2] \left( \frac{1}{t^2} - \frac{1}{(t-1)^2} \right).
\end{align*}

The assumption that $E[X_{t}^2] = O(t)$ implies that
$$
E[X_{t-1}^2] ( \frac{1}{(t-1)^2} - \frac{1}{t^2} ) = O(\frac{1}{t^2}).
$$
Therefore $\sum\limits_{t=1}^{\infty} \frac{E[N_t^2]}{t^2} < \infty$.
)

In turn, the SLLN for MDS can be shown via two arguments. Both are standard devices  for results of this type, one via the martingale convergence theorem and another via Kolmogorov's martingale maximal inequality.

Via Martingale Convergence Theorem

(The previous answer is a variation of this argument.)

If $\sum\limits_{t=1}^{\infty} \frac{E[N_t^2]}{t^2} < \infty$, the martingale $Y_n = \sum\limits_{t = 1}^n \frac{N_t}{t}$, $n \geq 1$, is bounded in $L^2$, therefore converges almost surely (and in $L^2$).
Therefore, by Kronecker's lemma,
$$
\frac{1}{n}\sum_{t = 1}^n N_t    \stackrel{a.s.}{\rightarrow} 0
$$
as $n \rightarrow \infty$.

Via Maximal Inequality

Consider again the $L^2$-martingale $Y_n = \sum\limits_{t = 1}^n \frac{X_t}{t}$, $n \geq 1$.
Let $\sigma^2_t = \frac{E[ X_t^2 ]}{t^2}$.

By the maximal inequality, for all $n > 0$ and for all $\epsilon > 0$,
$$
P( \sup_{m \geq n} | S_m - S_n | \geq \epsilon  ) \leq \frac{K}{\epsilon^2} \sum_{t \geq n} \sigma^2_t
$$
for some constant $K$ independent of $n$.
Therefore
$$
P( \inf_n \sup_{m \geq n} | S_m - S_n | \geq \epsilon  ) = 0
$$
for all $\epsilon > 0$. In other words, the sequence $S_n$, $n \geq 1$, is Cauchy, therefore converges, with probability $1$.
Again by Kronecker's lemma,
$$
\frac{1}{n}\sum_{t = 1}^n N_t
$$
converges to zero as $n \rightarrow \infty$ with probability $1$.

Title: Law of Large Numbers for Martingales


问:
(practical) Codes from sphere packings
I know that good codes allow you to construct good sphere packings in Euclidean space, e.g. the binary Golay code is the key feature of the construction of the Leech lattice.

Going in the other direction, I am wondering whether sphere packings in $\mathbf{R}^n$ actually help with designing good codes (for some definition of "code" that is useful in a practical setting). I know that codes on $n$ bits are sphere packings in Hamming space, but I'm wondering specifically about Euclidean sphere packings.

I know that kissing configurations yield spherical codes (by definition). Are these actually useful in practice? Also, do dense (lattice) packings produce any practically useful codes? From searching on Google I found a small section of a paper of H. Cohn (page 6 of https://arxiv.org/pdf/1003.3053.pdf) which gives a simplified example of how one can do coding with noise using good sphere packings, but I am wondering whether this is actually done in practice.

答:
I believe the question was asked in general terms, so this answer expands the scope of the question to include not only sphere packings, but also sphere coverings. Some sphere coverings based on lattices are simultaneously very good coverings and admit highly efficient nearest neighbor quantization algorithms; see for example

JH Convay and NJA Sloane. Fast quantizing and decoding algorithms for lattice quantizers. IEEE Trans Inform Theory, 28(2):227–232, 1982.

and in particular the beautifully simple E8 lattice fast quantization algorithm.

It seems that variants of this lattice construction and associated quantization algorithm have been used in a standard, see

Lattice Vector Quantization Applied to Speech and Audio Coding, Minjie Xie, ZTE Communications, 2012, Vol. 10, Issue (2): 25-33.

Title: (practical) Codes from sphere packings


问:
QQ, can I find the limit to that sequence with 3 sequence theorem?
As in the title - if it can be done that way, I will manage to get the right answer myself.

Body:
$$\begin{align}\lim_{n\to\infty}a_n&=\lim_{n\to\infty}\sqrt[n]{3^n-\left(2+\frac{1}n\right)^n}\\
&= \lim_{n\to\infty}3\sqrt[n]{1-\left(\frac{2}3+\frac{1}{3n}\right)^n}\\
&= 3\lim_{n\to\infty}\sqrt[n]{1-\left(\frac{2}3+\frac{1}{3n}\right)^n}\end{align}$$

Now,  $$\lim_{n\to\infty}\ln\left(\sqrt[n]{1-\left(\frac{2}3+\frac{1}{3n}\right)^n}\right)=\lim_{n\to\infty}\frac{\ln\left[1-\left(\frac{2}3+\frac{1}{3n}\right)^n\right]}n$$
and note that $\lim_{n\to\infty}\left(\dfrac{2}3+\dfrac{1}{3n}\right)^n=0$. Therefore $$\lim_{n\to\infty}\ln\left(\sqrt[n]{1-\left(\frac{2}3+\frac{1}{3n}\right)^n}\right)=\lim_{n\to\infty}\frac{\ln\left[1-\left(\frac{2}3+\frac{1}{3n}\right)^n\right]}n=\frac{\ln1}\infty=0$$
Therefore $$\lim_{n\to\infty}\sqrt[n]{1-\left(\frac{2}3+\frac{1}{3n}\right)^n}=e^0=1$$ Now we have $$\lim_{n\to\infty}a_n=3\cdot1=3$$


问:
$L=(\{ca^nb^n | n \geq 0\} \cup \{a^nb^{2n } | n \geq 0\})^*$ is not dcfl. Why?

$L=(\{ca^nb^n | n \geq 0\} \cup \{a^nb^{2n } | n \geq 0\})^*$ is not dcfl. Why?

[dcfl - deterministic context free language, cfl - context free language]

How to analyse this? i am not able  to vizualise pda for this? if it is kleane closure of L then i need to loop that L machine between initial and final states right? If it is union then because of dilemma between 2 transition path, we get npda. But both union and kleane closure combined, how to vizualise?

答:
After rereading the question this language is actually a DCFL

We can prove this by constructing a DPDA

first lets describe how the DPDA will work and get an intuition about why it is a DPDA

DPDA works as follows :

if no further input accept

if we read c , then test for canbn , if the string fails reject , else return to 1

if we read a , then test for anb2n , if the strings fails reject , else return to 1

The DPDA works as follows , first since we use * we accept the empty string , if we have an input , we observe the first string , if we read c , then we know we must read a substring canbn , if we read a then we know we must read a substring anb2n , either way we test to see if the input we read matches the substring it should , if not reject , if it matches then we return to accept state , waiting for either an a or c to check again or for the end of input so we accept

Each step in our description follows the other deterministically with no guessing , this should give us a strong basis for our claim that L is a DCFL

Now lets formalise what we stated , to prove L is DCFL we need to provide a DPDA , this DPDA needs to :

follow the formal description of a DPDA , so we know for sure it is a DPDA

accept the language L

First here is the formal description of a DPDA :

The distinguishing part from a PDA is the restrictions imposed on the transition function

Now we provide the DPDA to recognise L:

We can see how the DPDA follows the formal description provided , the transitions from each state follow the description of the transition function in the formal description of the DPDA

We can also see that the DPDA recognises L , as it follows the procedures explained above


问:
Quick question about showing a function is one to one
Let $f : ℕ → P(ℕ)$ be given by $f(n) = {k*n | k ∈ ℕ}$.   (P(ℕ) is the power set of the set A.)

Is f injective?

The answer to this question goes like this:

$f(n) = (n,2n,3n,4n,5n...)$

$min({ n,2n,3n,...})$ is  $n$, as $n<2n<3n<....$

Suppose $f(m)=f(n)$. Then

${ (m,2m,3m,...)}={ (n,2n,3n,...)}$

smallest element of $( m,2m,3m,...)$ = smallest element of $(n,2n,3n,...)$,  $m=n$.

My question is why do we focus on the smallest element of the set?

答: We focus on it because it produces the contradiction we need.  If you claim two sets are equal and I can show the smallest elements are different, it shows you are wrong, because I have identified and element that is in one set and not in the other.  There may be other ways of showing the two sets are different, but this one works.</p>


问:
what is the meaning of this symbol $ ∧?$
I have some confusion in symbolics

what is the meaning of this symbol $\land?$

My attempt : I was googling and got the answer given below

But i didn't  understand the given above  answer

My confusion come from this, pic is given below

答: In this context, $\land$ is the logical AND operator \land.