问:
surjectivity of a piece-wise function defined as follows
Define a function $f:\Bbb Z \to \Bbb Z$ by
\begin{equation*}
f(x) =
\begin{cases}
x, & \text{if $x$ is even}\\
x+1, & \text{if $x$ is odd}
\end{cases}
\end{equation*}
for all $x \in \Bbb Z$.
Find a right inverse of $f$ if it exists.

Attempt:
$f$ has a right inverse iff $f$ is surjective, so we must check that $f$ is whether surjective or not.
Let $y=f(x)$. Then,
\begin{equation*}
x =
\begin{cases}
y, & \text{if $y$ is even}\\
y-1, & \text{if $y$ is odd}
\end{cases}
\end{equation*}
for all $x,y \in \Bbb Z$. Now, consider
\begin{align*}
f(x) &=
\begin{cases}
f(y), & \text{if $y$ is even}\\
f(y-1), & \text{if $y$ is odd}
\end{cases}\\
&=
\begin{cases}
y, & \text{if $y$ is even}\\
(y-1)+1,& \text{if $y$ is odd}
\end{cases}
\end{align*}
Hence, we see that $f(x) = y$ for all $x \in \Bbb Z$. Thus, $f$ is surjective and therefore, $f$ has a right inverse.

Am I true?

答:
No; the image of $f:\Bbb Z\to\Bbb Z$ as you have defined it consists of only even numbers --

$y=f(x)$ is always even --

so $f$ is not surjective -- odd numbers do not have pre-images in $\Bbb Z$ --

so $f$ does not have a right inverse.

By the way, $f$ is not injective either, because $f(1)=f(2)$, but $1\ne2$.


问:
Numerical Methods - Relative Error
In our lectures, we've been looking at relative error and in our recent problem sheet we were asked to prove the following:

Let $\hat{x} \neq 0$ be an approximation of a non-zero quantity $x$. Consider the relative error $ e = \frac{|x- \hat{x}|}{|x|}$ and the error normalised using the approximation $\hat{x}$, $\hat{e} = \frac{|x-\hat{x}|}{|\hat{x}|}$.

Assuming that $\hat{e} < 1$, show that there exist quantities $f_1(\hat{e})$ and $f_2(\hat{e})$ such that $f_1(\hat{e}) ≤ e ≤ f_2(\hat{e})$.

I'm honestly not sure where to begin with this, especially since $f_1$ and $f_2$ are functions of $\hat{e}$. Any suggestions would be helpful.

答:
We have
$$e = \frac{|x-\hat{x}|}{|x|} = \frac{|x-\hat{x}|}{|\hat{x}|}\frac{|\hat{x}|}{|x|} \leq \hat{e}\left(1 + \frac{|\hat{x}-x|}{|x|}\right) = \hat{e}(1+e).$$
This implies $e(1-\hat{e}) \leq \hat{e}$ and therefore $e \leq \frac{\hat{e}}{1-\hat{e}}$ assuming $\hat{e} < 1$.

Can you show the other inequality now?

ContentLicense: CC BY-SA 4.0

Title: Numerical Methods - Relative Error


问:
Boundaries in 2D space
Find the volume of figure that is bounded with the three following surfaces:$z=y^2,y=x^2,z=4$.
I don't think this has anything to do with polar coordinates etc. So I tried to put this problem in $OXY$ plane and i get parabola $y=x^2$ and $y=0$ but not sure how to read boundaries, because here(in $OXY$) we don't have that some area is closed, only that is bounded with parabola and x-axis, but not sure how to know boundaries for x,y from that. If someone knows how to proceed following example in this way I would be happy to know.

答:
$z=4$ and $z=y^2$ intersection projection gives lines $y=\pm 2$ on $OXY$ which together with $y=x^2$ gives figure on $OXY$. Then integral is
$$\int\limits_{-\sqrt{2}}^{\sqrt{2}}\int\limits_{x^2}^{2}\int\limits_{y^2}^{4}dzdydx$$

Title: Boundaries in 2D space


问:
What is the probability that the next call come in $3$ minutes provided that in the past $2$ minutes there have been no calls?
At a certain customer service line the waiting time for the next call follows exponential distribution with expected time $3$ minutes. What is the probability that the next call come in $3$ minutes provided that in the past $2$ minutes there have been no calls?

From the given datils we know $\lambda=\dfrac{1}{3}$ so the distribution

$P(X=x)=\frac{1}{3}e^-\frac{x}{3}$

We have to find $P(0<x<3/x>2)=\dfrac{P(2<x<3)}{p(x>2)}=\frac{\int_{2}^3 \frac{1}{3}e^-\frac{x}{3}\,dx}{\int_{2}^\infty \frac{1}{3}e^-\frac{x}{3}\,dx}$

but I am not sure this answer I think we have to find $P(2<x<3)$
Can someone help me to understand?

答:
From the way you phrased the question, I'm assuming that if the current time is 2pm, you are asking for the probability that a call will come before 2:03 given that no call has been received since 1:58.

The key here is that the exponential distribution is memoryless, i.e.
$$\mathbb P(X>a+b\mid X\geq b)=\mathbb P(X>a).$$
So in this case, $$\mathbb P(X>3+2\mid X\geq2)=\mathbb P(X>3)=e^{-\frac{1}{3}\times3}=e^{-1}.$$
So the probability that a customer will arrive in the next three minutes is $1-e^{-1}$. The memoryless property means that the fact we haven't received a call in the past two minutes doesn't affect the probability.

If instead the interpretation is that the current time is 2pm, no call has been received since 1:58, what is the probability a call will come before 2:01, then the memoryless property implies that the answer is $\mathbb P(X\leq1)=1-e^{-1/3}$.


答:
At any given $z$, your surface will be bound by:

$-\sqrt y \leq x \leq \sqrt y \,$ (by parabolic cylinder $y = x^2$)

$0 \leq y \leq \sqrt z \,$ (by parabolic cylinder $z = y^2$).

Bound of $z$ is $0 \leq z \leq 4$.

Integrate in order $dx$ first, $dy$ next and $dz$ last.

EDIT: Here is the integral

$\displaystyle \int_0^4 \bigg [\int_0^{\sqrt {z}} \bigg[\int_{-\sqrt {y}}^{\sqrt{y}} dx\bigg] \, dy \bigg]\, dz$

And the sketch at any given $z \,$:

You can also change the order of integration by writing the limits differently.

Title: Boundaries in 2D space

Body:
Find the volume of figure that is bounded with the three following surfaces:$z=y^2,y=x^2,z=4$.
I don't think this has anything to do with polar coordinates etc. So I tried to put this problem in $OXY$ plane and i get parabola $y=x^2$ and $y=0$ but not sure how to read boundaries, because here(in $OXY$) we don't have that some area is closed, only that is bounded with parabola and x-axis, but not sure how to know boundaries for x,y from that. If someone knows how to proceed following example in this way I would be happy to know.


问:
$\cos x − 2\sin x = 1$. How to solve using $R\cos(x+a)$?
I have this question: $\cos x − 2\sin x = 1$. You have to solve for $x$, the answer being 4.069. I've tried answering but I just keep getting 2.214. This is my method:

$\cos x − 2\sin x = 1$

$R\cos(x+a)= 1$

$R = \sqrt{1^2 + (-2)^2}$

$a = \arctan(-2/1)$

therefore $\sqrt5 \cos(x - 1.107) = 1$

$\cos(x - 1.107) = \sqrt5/5$

$\arccos(\sqrt5/5) = x - 1.707$

$1.707 = x - 1.707$

therefore $x = 2.214$

Can anyone give me a hint where I'm going wrong? Thank you

答:
$$\begin{align}
\cos x-2\sin x&=R\cos(x+\alpha)\\[1ex]
&=R\cos x\cos\alpha-R\sin x\sin\alpha
\end{align}$$

$$\implies\begin{cases}R\cos\alpha=1\\R\sin\alpha=2\end{cases}$$

Dividing the expressions in the second equation by those in the first gives

$$\frac{R\sin\alpha}{R\cos\alpha}=\tan\alpha=2\implies \alpha=\tan^{-1}(2)+n\pi$$

(where $n\in\mathbb Z$)

Taking the squares and summing them gives

$$(R\cos\alpha)^2+(R\sin\alpha)^2=1^2+2^2\implies R^2=5\implies R=\pm\sqrt5$$

Take $R=\sqrt5$ and $n=0$, so that

$$\cos x-2\sin x=\sqrt 5\cos(x+\tan^{-1}(2))=1$$

Can you take it from here?


问:
Solve the Lagrangian dual problem associated to $\max_{x_1,x_2\ge0}(3x_1+4x_2)$ such that $x_1^2+x_2^2\le25$
Consider the (non-linear) optimization problem ($P$)

$$max \quad3x_1 + 4x_2$$

$$s.t. \quad x_1^2 + x_2^2 \leq 25$$

$$ \quad x_1,x_2 \geq 0$$

Solve the Lagrangian dual problem.

I don't have a clue how the dual problem is even obtained since the constraint is nonlinear. Could anyone please help me?

答:
Here is a first couple of steps. See http://en.wikipedia.org/wiki/Duality_(optimization) for more details on Lagrangian duals.

To convert the original problem into a minimization problem, we minimize $-3x_1-4x_2$.
Your Lagrangian is
$$
\Lambda(x_1, x_2, \lambda) = -3x_1 - 4x_2 + \lambda (x_1^2 + x_2^2-25).
$$
The dual problem is to maximimze $\Lambda(x_1, x_2, \lambda)$ over the region $x_1,x_2 \geq 0$.

Your problem is convex, so the min should occur where $\nabla \Lambda = \vec{0}$...

ContentLicense: CC BY-SA 3.0


问:
Given, $f(x+y) =xf(x) +2f(y)-x-y$. If $f(2019)=\frac{a}{b}$, then $a+b=?$

Question:

  $\text{Given, } f(x+y) =xf(x) +2f(y)-x-y\text{. If } f(2019)=\frac{a}{b}, \text{then } a+b\text{ = ?} $

I started solving by replacing $y\to-x$, which yields
$$f(0) = xf(x) +2f(-x)$$
then, replace $x\to-x$
$$f(0)=-xf(-x)+2f(x)$$
on solving, I get,
$$(x+2)f(0)=(x^2+4)f(x)$$

But, I'm stuck here. How to move on?

答:
If $x=0$ we get $f(y)=y$ for all $y$. Pluging this in to starting equation we get $$x+y=x^2+2y-x-y$$

which should be valid for all $x,y$. Since $x^2=2x$ ony for $x=2$ and $x=0$ this FE has no solution unlesss $f$ is defined on the set $\{0,2\}$.


问:
How is this not an elementary anti derivative
i was computing integrals, and was thinking of an anti-derivative of $\int e^{-x^2} dx $. See, i know for a fact that this function has no elementary anti-derivative, but i was curious and tried around a bit. For this i used the known fact, that $e^x=\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}$, so i replaced $x$ by $-x^2$. This gives the following : $\int e^{-x^2}dx=\int \sum\limits_{n=0}^{\infty} \frac{-x^{2^n}}{n!} dx$.
Now, if integrated, term by term, wouldnt this be a countable sum of elementary functions, thus making it elemetary? I would appreciate if anyone could point out my flaw(s) here. Maybe i just cant substitute $x$ for $-x^2$?

答:
I should note that its actually $(-x^2)^{n} = (-1)^{n}x^{2n}$.

There's no problem substituting $x$, in fact it helps define the error-function.

Anyway, the function you can get from this substitution is not elementary, simply because it's an infinite sum of polynomials!

See the definition of an Elementary Function here.

Edit: Thank you Matthew Towers for correcting me - yes, an infinite polynomial can be an elementary function - for example, $\sum_{n \geqslant 0} \frac{x^n}{n!}$. Although some infinite polynomials are elementary functions - some aren't, for example your integral (or the integral of $e^{x^2}, x^x$ and a bunch more). Proving this isn't very easy - it's actually quite difficult.

Unfortunately, as much as I hate saying this (especially in mathematics), you just have to agree that the integral you presented is not elementary, since proving this requires a lot of extra knowledge - which unfortunately I don't know. If you still want to read some more about this, here are a few more questions on Math SE which talk about this:

How can you prove that a function has no closed form integral?

Why can't erf be expressed in terms of elementary functions?

Proving that special functions do not have closed-form expression

Integration of $e^{-x^2}$


问:
An interesting Olympiad Geometry problem.
The point P lies in the interior of $\Delta ABC$. A line is drawn through P, parallel to each side of a triangle. The line divides BC into three parts of length $a , a', a''$ (in that order); AB into three parts, lengths  $c, c', c''$ (in that order); CA into three parts length $b, b', b''$ (in that order) . Prove that $abc = a'b'c' = a''b''c''$ .
This is one of those types of question which I have never encountered before.

I was not able to use Ceva's theorem as there were no Cevians, I could not understand how to apply a similarity argument here, as that is what I think might work.

Any solution is highly appreciated, Thanks!

答:
Quadrilateral $AFPE$, $CHPG$ and $BDPI$ are parallelogram. Sides $PE$, $PF$, $PI$, $PD$, $PG$ and $PH$ are labelled accordingly.

$\triangle DPG\sim \triangle PEH\sim \triangle IFP$.

$\Rightarrow  \frac{a'}{a''}=\frac{b}{b'}=\frac{c''}{c}$ and $\frac{a''}{a}=\frac{b'}{b''}=\frac{c}{c'}$

$\Rightarrow  \frac{a'}{a''}=\frac{b}{b'}   \Rightarrow  a'b'=a''b$------------$(1)$

Also, $\frac{a''}{a}=\frac{c}{c'}       \Rightarrow c'=\frac{ac}{a''}$-----------$(2)$

Multiplying $(1)$ by $(2)$ gives, $a'b'c'=abc$-----------------$(3)$

$\frac{a''}{a}=\frac{b'}{b''}    \Rightarrow a''b''=ab' $-------------------$(4)$

Also, $\frac{b}{b'}=\frac{c''}{c}    \Rightarrow c''=\frac{bc}{b'}$----------------$(5)$

Multiplyng $(4)$ by $(5)$ gives,  $a''b''c''=abc$-------------------$(6)$

From $(3)$ and $(6)$, we get, $\boxed {abc=a'b'c'=a''b''c''}$


问:
Find the dual problem of $\min_x\{||x-a_1||+||x-a_2||+||x-a_3||,a_i\in\mathbb{R}^n\}$

Find the dual problem of

$$\min_x\{||x-a_1||+||x-a_2||+||x-a_3||,a_i\in\mathbb{R}^n\}$$

I've tried the following but got stuck
$$\min_x\{||x-a_1||+||x-a_2||+||x-a_3||,a_i\in\mathbb{R}^n\}=\min_{x,z_i}\{||z_1||+||z_2||+||z_3||,a_i\in\mathbb{R}^n\,z_i=x-a_i\}$$

and the Lagrangian is:\
$L(x,z,\lambda)=  ||z_1||+||z_2||+||z_3||+\sum_{i=1}^3\lambda_i(z_i-x+a_i)=\sum_{i=1}^{3}z_i\lambda_i+||z_i||-x\sum\lambda_i+\sum\lambda_ia_i$

from $-x\sum\lambda_i$ we can deduce that $\sum \lambda_i=0$ else we get $-\infty$.
Now when I try to solve for each $z_i$ I get $\frac{\partial L}{\partial z_i}=\lambda_i+\frac{z_i}{||z_i||}=0$
and I'm not sure how to continue from here.

Any hint?

答:
Based on your arguments, the remainder is to pay attention to the following quantity,
\begin{align}
F(\lambda)=\inf_{z\in \mathbb R^n} (\langle z, \lambda \rangle +||z||).
\end{align}

This quantity can be quantified using the notion of conjugate function. Recall that the conjugate function of $f$, denoted by $f^*$, is defined by
\begin{align}
f^*(y)=\sup_{x \in \mathbb R^n} (\langle x, y \rangle -f(x)).
\end{align}

Now, $F(\lambda)$ can be rewritten in term of conjugate function of $f(x)=||x||$ as follows
\begin{align}
F(\lambda) & =\inf_{z\in \mathbb R^n} (-\langle z, -\lambda \rangle +f(z))\\
           & = - \sup_{z\in \mathbb R^n} (\langle z, -\lambda \rangle - f(z))\\
           & = - f^*(-\lambda).
\end{align}
Here, in the case $f(x)=||x||$, $f^*$ is
\begin{align}
f^*(\lambda)=
\begin{cases}
0, & \text{ if } ||\lambda||_*\leq 1.\\
+\infty, & \text{ if } ||\lambda||_*> 1.
\end{cases}
\end{align}
Where $||\cdot||_*$ is the dual norm of $||\cdot||$. To see how to obtain such formula for $f^*$, see the first answer here.


问:
Independence of random variables for jointly distributed random variables
In our class, our professor told us that, the jointly distributed random variables are said to be independent if the following relation holds:

$f_{X,Y}(x,y) = f_{X}(x)f_{Y}(y)$

In addition to this, he said that when we define the region of integration we should be able to do it in a way such that x and y are not dependent. For example:

$f_{X,Y}(x,y) = 6e^{x}e^{y}, x,y > 0$ are independent. But $f_{X,Y}(x,y) = 6e^{x}e^{y}, x+y \le 1$ are not.

I am unable to understand why this is the case?

If I draw out the regions on a plane, they just look like two ordinary and similar regions to me,the first one being defined on a different range as compared to the next.

答: When $x+y\le 1$, we have $y\le 1-x$. Given a value of $x$, the value of $y$ depends on $x$ with the constraint being $y\le 1-x$. So $y$ is in some way restricted by $x$ in this case, which makes it dependent on $x$.


问:
3b1b Fourier Calculation
At the end of this video, 3b1b gives an exercise asking the viewer to show how the notion of a complex Fourier series presented in the video is equivalent to this alternative real-valued formulation:

So what he showed in the video was that the Fourier series of a function (with real domain $[0,1]$) could be computed as: $$f(t)=c_{0}+\sum_{n=1}^{\infty}(c_{n}e^{n2\pi{i}t}+c_{-n}e^{-n2\pi{i}t})$$
where the $n$th coefficient $c_{n}$ is given by: $$c_{n}=\int_{0}^{1}f(t)e^{-n2\pi{i}t}dt$$
Using the hints he gives, here is what I did: $$f(t)=\frac{a_{0}}{2}+\frac{b_{0}}{2}i+\sum_{n=1}^{\infty}\biggr((\frac{a_{n}}{2}+\frac{b_{n}}{2}i)(\cos(n2\pi{t})+i\sin(n2\pi{t}))+(\frac{a_{-n}}{2}+\frac{b_{-n}}{2}i)(\cos(-n2\pi{t})+i\sin(-n2\pi{t}))\biggr)$$
We may observe that $b_{0}=0$, $a_{n}=a_{-n}$, and $-b_{n}=b_{-n}$, so the equation can be simplified to: $$f(t)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos(n2\pi{t})-b_{n}\sin(n2\pi{t}))$$
which is almost what the video shows, except for that minus sign where there should be a plus sign...where did I go wrong?

答:
If you jump back to the explanation at around 12 minutes in, for a purely real-valued function, $c_{-n} = c_{n}^*$, where $^*$ represents the complex conjugate (if $z = a + bi$, then $z^* = a - bi$). I don't know if it was stated in these terms explicitly, I can only find him talking about the little circles going in the opposite direction.

Using this, the definitions of the coefficients given in the video, and the definition of the complex exponential, we can find that it isn't actually technically $c_n = \frac{a_n}{2} + \frac{b_n}{2} i$ but it's really $c_{-n} = \frac{a_n}{2} + \frac{b_n}{2} i$, so actually $c_n = \frac{a_n}{2} - \frac{b_n}{2} i$

\begin{align*}
  c_n &= \int_0^1 f(t) \mathrm{e}^{-2\pi i n t} \, \mathrm{d}t \\
  &= \int_0^1 f(t) (\cos(-2\pi n t) + i \sin(-2\pi n t)) \, \mathrm{d}t \\
  &= \int_0^1 f(t) (\cos(2\pi n t) - i \sin(2\pi n t)) \, \mathrm{d}t \\
  &= \int_0^1 f(t) \cos(2\pi n t) \, \mathrm{d}t - i \int_0^1 f(t) \sin(2\pi n t) \, \mathrm{d}t \\
  &= \frac{a_n}{2} - \frac{b_n}{2} i
\end{align*}

It's a small mistake in the video.  Excellent question, and good work actually attempting the challenges, and catching the errata.

Title: 3b1b Fourier Calculation


问:
Equivalence of the Heine-Borel theorem for the set of rationals
From the Heine-Borel theorem we can say that a subset $S$ of Euclidean space $\mathbb{R}^n$, the following two statements are equivalent:

$S$ is closed and bounded.

$S$ is compact, that is, every open cover of S has a finite subcover.

Would these statements still be valid in the set of rationals ($\mathbb{Q}$), or would there be any case that contradicts the statements?

答: We need completeness for closed and bounded sets to be compact. Consider the set $\{p\in\mathbb Q: 0\le p^2\le 2\}$ as a counter-example.


问:
Coefficient $x^{62}$ in $(x^3-\frac{1}{x^5})^{50}$
Find the coefficient for $x^{62}$ in $(x^3-\frac{1}{x^5})^{50}$.

Binomial coefficients gives:
$$(x^3)^n (-\frac{1}{x^5})^{n-k}=x^{62}\\
3n-5(n-k)=62\Rightarrow 0=2n+5k-62\\
or\\\
3n+5(n-k)=62\Rightarrow 0=-8n+5k+62\\$$

When I put in the values, none satisfy the equations.

How can i solve this without trial and error?

答:
There's an error in your equations: it should be simply
$$3k-5(50-k)=62\iff 8k=312.$$
Can you take it from there?


Title: Coefficient $x^{62}$ in $(x^3-\frac{1}{x^5})^{50}$

Body:
Find the coefficient for $x^{62}$ in $(x^3-\frac{1}{x^5})^{50}$.

Binomial coefficients gives:
$$(x^3)^n (-\frac{1}{x^5})^{n-k}=x^{62}\\
3n-5(n-k)=62\Rightarrow 0=2n+5k-62\\
or\\\
3n+5(n-k)=62\Rightarrow 0=-8n+5k+62\\$$

When I put in the values, none satisfy the equations.

How can i solve this without trial and error?

Body:
Note that $$\left(x^3-\frac{1}{x^5}\right)^{50}=\sum_{k=0}^{50}\binom{50}{k}(x^3)^k(-x^{-5})^{50-k}=\sum_{k=0}^{50}(-1)^k\binom{50}{k}x^{8k-250}$$
We want the coefficient for $x^{62}$, hence we solve $$8k-250=62\longrightarrow k=39$$
The term corresponding to $k=39$ is given by $$(-1)^{39}\binom{50}{39}x^{62}$$
So the coefficient for $x^{62}$ is $$(-1)^{39}\binom{50}{39}=-37353738800$$


问:
If a number is a square modulo $n$, then it is also a square modulo any of $n$'s factors
Say we have $a \equiv x^2 \bmod n$. How would we prove that this implies:

$$\forall \text{ prime }p_i\text{ such that }\,p_i\mid n,\;\exists y\,\text{ such that }\, a \equiv y^2 \bmod p_i$$

答:
Hint: $ $ note that $\ a\ $ is a square mod $\, n\iff\ f(x) = x^2-a\ $ has a root mod $\, n.\, $

But a root mod $\, n\, $ of a polynomial $\ f(x)\in \Bbb Z[x]\ $ persists as a root mod $\, k\,$ for  $\, k\mid n\,$

because $\ \ f(b)\equiv 0\pmod{n}\ \Rightarrow\ n = jk\mid f(b)\ \Rightarrow\ k\mid f(b)\ \Rightarrow\ f(b)\equiv 0\pmod{k} $

As another example, the linear analog of square preservation is fraction preservation. For example, if $\ ab = 2k- 1\ $ then, mod $\, ab,\,$ we have $\, 2k\equiv 1\ $ so $\, 1/2 \equiv k,\,$ and this congruence remains true both mod $\,a\,$ and mod $\,b.$

More generally, any solution of an equation in an equational algebraic structure remains a solution in any modular (homomorphic) image of the structure.


问:
Proof if $x+y=0$, then $xy \leq 0$.
For all $x,y ∈ \mathbb R$, if $x + y = 0$ then $xy \leq 0$. Decide whether this is true or false.

I assume this is true since one of $x$ or $y$ needs to be positive while the other is negative, so their product is negative, but how would would i go about proving this? Would this be right:

Let $x = 2k$ and $y = -2k$, such that $k \in \mathbb N$. Then $x + y = 0$.

So $xy = -4k^2$ which is less than $0$, hence this is true.

答:
You complicated more then needed and it is not false, but incomplete because your case just applies for even numbers, you should also consider the case x=2k+1(odds)

A more simpler way is:

if x+y=0, then x=-y, therefore $xy=(-y)(y)=-y^2 \leq 0$ for all real.

Title: Proof if $x+y=0$, then $xy \leq 0$.


问:
Prove function f is surjective.
f: $\mathbb{R}$ $\to$ $\mathbb{R}$ is a continuous function and $\overline{f(\mathbb{N})} = \mathbb{R}$. Prove that f is surjective.
I've thought alot about this but i still haven't find a clue. I would really appreciate any help or a hint.

答: The equation $\overline{f(\mathbb{N})} = \mathbb{R}$ means every real number is the limit of a sequence of points in $f(\mathbb N)$. Since $f$ is continuous, you can apply the intermediate value theorem on every closed interval. Can you complete the proof?

Title: Prove function f is surjective.


问:
Possible Values of # of Ordered Factorization of Integers
I was trying to find the number of ordered factorizations of integers into parts $\gt 1$. Starting with $0$, the number of ordered factorizations of $n$ forms this sequence. Interestingly, and in accordance with my program, the values seem to repeat themselves infinitely. While some values repeat, others don't appear at all. It seems that the sequence of possible values of the number of ordered factorizations of an integer appears here, forming the sequence: $$1, 2, 3, 4, 8, 13, 16, 20, 26, 32, 44, 48...$$

In this OEIS entry, they are named "Gozinta numbers". It's not surprising to see that these values solely depend on the exponents of the prime factors of $n$. In this paper, $H(n)$ counts the number of such ordered factorizations, closed-form formula as well as recurrence are given:

$$H(n) = \sum_{j=1}^q \sum_{i=0}^{j-1} (-1)^i \binom {j}{i} \prod_{b=1}^k \binom {r_b +j - i - 1}{r_b} = \sum_{d|n}H(d) , H(1) = 1$$

with $q = \sum_{i=1}^k r_i$, and $n = p_1^{r_1}p_2^{r_1}p_3^{r_3}...p_k^{r_k}$

Can someone shed light on how the sequence of possible values of $H(n)$ is generated? How can I generate these values myself, perhaps using the formulas I mentioned or any other way.

答:
The first thing to say is $$\sum_n H(n)n^{-s}=\sum_{k\ge 0} (\zeta(s)-1)^k= \frac1{2-\zeta(s)}$$

$H(n)$ is the Dirichlet inverse of $2_{n=1}-1$.

$\frac{\zeta(s)}{2-\zeta(s)}=2\frac1{2-\zeta(s)}-1$ gives that
$$\sum_{d| n} H(d)=2 H(n)-1_{n=1}$$

The Tauberian theorem gives that $$\sum_{n\le x} H(n) \sim \frac{x^r}{-\zeta'(r) r}$$ where $r>1$ is the unique solution of $\zeta(r)=2$.

About your question of enumerating in increasing order the elements of $H(\Bbb{N})$ : There is a natural preorder on the exponents of integers, ie. on the sequence $e$ of non-negative integers such that $n=\prod_j p_j^{e_j}$:

$e\ge e'$ if there is a permutation of the primes such that $e_j-e'_{\sigma(j)} \ge 0$ for all $j$. The class of $e$ is obtained by taking a permutation of the primes such that $e_\sigma$ is decreasing.

Then $e\ge e'\implies H(\prod_j p_j^{e_j})\ge H(\prod_j p_j^{e_j'})$.

Given $e$ you can enumerate all the $H(n)$ such that $H(n)<H(e)$: find $k$ such that $H(2^k)>H(n)$, find $J$ such that $H(\prod_{j\le J} p_j)>H(n)$, then enumerate the classes such that $\sum_j e_j' < \max(k,J)$.


问:
Solving rank equations
Suppose we are given a (finite) matrix $A$ with entries $a_{ij} \in k[w_1, \ldots, w_n]$ which are polynomial expressions of $w_1, \ldots, w_n$ over some field $k$. For a fixed integer $m$, is there a way to determine all points $p=(w_1, \ldots, w_n) \in \mathbb A^n(k) = k^n$ such that
$$\operatorname{rank} A=m,$$ i.e. which $p\in k^n$ makes the rank of $A$ to be $m$?

答:
I have reservations about this, Maxima, solution; although it corresponds to some other approaches I tried

Rewriting the particular problem to satisfy my pattern-seeking.

Using Maxima

$ M=\left[\begin{array}{cccc}
d_{1} & d_{2} & 0 & 0\\
-sd_{2} & d_{1} & 0 & 0\\
a_{1} & a_{2} & b_{1} & b_{2}\\
-sa_{2} & a_{1} & -sb_{2} & b_{1}\\
-a_{2}d_{2} & 0 & -d_{2}b_{2} & 0\\
0 & 0 & d_{1} & d_{2}\\
0 & 0 & -sd_{2} & d_{1}
\end{array}\right]$

factor(triangularize(M));

$\left[\begin{array}{cccc}
d_{1} & d_{2} & 0 & 0\\
0 & d_{2}^{2}\cdot s+d_{1}^{2} & 0 & 0\\
0 & 0 & -b_{2}\cdot d_{2}\cdot(d_{2}^{2}\cdot s+d_{1}^{2})\\
0 & 0 & 0 & -\left(b_{2}\right)^{2}\cdot d_{2}\cdot(d_{2}^{2}\cdot s+d_{1}^{2})\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]$

The rank() command also gives these terms.

The only thing wrong with this is the $d_{2}$ in the top row. It should have been eliminated? I left it in because I might have missed something, but it seems wrong (a bug in Maxima?)

Analysis:

$(d_{2}^{2}\cdot s+d_{1}^{2}) =0 $ yields rank 1

$b_{2}=0$ yields rank 2

$d_{2}=0$ yields rank 2

$d_{1}=0$ yields rank 3

Comment: I prefer to use exterior/alternating products for this type of analysis.

That is because it automatically points down to the elementary definition of dependence/independence of linear conditions and fits any dimension/size. It starts out with vector, cross-product, triple product; but these are specialized and actually more complicated than the general alternating products.

But Maxima didn't have specialized routines for your type of problem and the standard function yielded too many terms to be useful. Sage didn't seem to have a function that didn't drag in extra concerns.

Do not use the “echelon” form because I think it will make incorrect decisions when you want to later set terms to zero or not; it normalizes.

Constructive comments welcome :)

Title: Solving rank equations


问:
How calculate the De Rham cohomology group of $3$-torus: $T^3$?
How do I calculate the De Rham cohomology group of the $3$-torus $T^3$? Here $T^3=S^1 \times S^1 \times S^1 $.

Using the Mayer-Vietoris sequence, I can show that $\dim H_3(T^3)=\dim H_0(T^3)=1$. But I don't know how to find $H_1(T^3)$ and $H_2(T^3)$.

答:
A linear basis for the vector space $H^\bullet(T^3, \mathbb{R})$ is given by $$ \{1, \alpha, \beta, \gamma, \alpha\wedge \beta, \alpha\wedge \gamma, \beta\wedge \gamma, \alpha\wedge\beta\wedge \gamma\}.$$

The product structure on the de Rham cohomology algebra is given by wedge product in the exterior algebra of differential forms. Hopefully this gives some intuition behind the $\binom{n}{k}$ dimensions.


问:
Probability of 20 consecutive success in 100 runs.
Suppose a chess player have a win rate equal 90%, what is the chance to have 20 consecutive wins (successes) playing 100 games? Consider that lose/draw = fail.

I've studied basic statistics in college and it seems like a binomial distribution problem (right?), but honestly I can't figure out a way to solve this problem considering "consecutive" successes.

Is there a statistical distribution for this kinda problem?

Thanks very much! I really appreciate any thoughts!

答:
To support @awkward, I have cut and paste a Matlab code that computes the probability values; it is consistent in solution with inputs p = 0.9, r = 20, and n = 100 with the solution provided by awkward. I removed my help menu so the formatting does not interfere with the paste.

function pr = fellerconsecPdf(p,r,n)

d   = [size(p(:),1), size(r(:),1), size(n(:),1)];

id  = setdiff( unique(d), 1);

if(numel(id) > 3/2)

```<code>erorr('MATLAB:fellerconsecPdf.m: non-scalar inputs must match in size');
</code>```

end

m   = max( d );

p   = repmat(p(:), m - (size(p(:),1)-1), 1);
r   = repmat(r(:), m - (size(r(:),1)-1), 1);
n   = repmat(n(:), m - (size(n(:),1)-1), 1);

if(isscalar(p))

```<code>q       = (1 - p);
c       = zeros(r+2,1);
c(1)    = q.*(1 - q).^r;
c(r+1)  = -1;
c(r+2)  = 1;
xsol    = roots(c);
xsol    = xsol( abs(imag(xsol)) < eps );
xsol    = xsol(xsol>0);
[~,ii]  = max( abs(xsol - 1./p) );
x       = xsol(ii);
</code>```

else

```<code>q       = (1 - p);
x       = [];

for k = 1:length(r)

    c           = zeros(r(k)+2,1);
    c(1)        = q(k).*(1 - q(k)).^r(k);
    c(r(k)+1)   = -1;
    c(r(k)+2)   = 1;
    xsol        = roots(c);
    xsol        = xsol( abs(imag(xsol)) < eps );
    xsol        = xsol(xsol>0);
    [~,ii]      = max( abs(xsol - 1./p(k)) );
    x           = cat(1,x,xsol(ii));

end
</code>```

end

pr  = ( (1 - p.*x)./((r + 1 - r.x).(1-p)) ).x.^(-1(n + 1));

return;

Please not that star-dot-star does not show up in the text correctly, but it should be clear from context that that I intend point-wise multiplication of vectors.

Body:
To support @awkward, I have cut and paste a Matlab code that computes the probability values; it is consistent in solution with inputs p = 0.9, r = 20, and n = 100 with the solution provided by awkward. I removed my help menu so the formatting does not interfere with the paste.

function pr = fellerconsecPdf(p,r,n)

d   = [size(p(:),1), size(r(:),1), size(n(:),1)];

id  = setdiff( unique(d), 1);

if(numel(id) > 3/2)

erorr('MATLAB:fellerconsecPdf.m: non-scalar inputs must match in size');

end

m   = max( d );

p   = repmat(p(:), m - (size(p(:),1)-1), 1);
r   = repmat(r(:), m - (size(r(:),1)-1), 1);
n   = repmat(n(:), m - (size(n(:),1)-1), 1);

if(isscalar(p))

q       = (1 - p);
c       = zeros(r+2,1);
c(1)    = q.*(1 - q).^r;
c(r+1)  = -1;
c(r+2)  = 1;
xsol    = roots(c);
xsol    = xsol( abs(imag(xsol)) < eps );
xsol    = xsol(xsol>0);
[~,ii]  = max( abs(xsol - 1./p) );
x       = xsol(ii);

else

q       = (1 - p);
x       = [];

for k = 1:length(r)

    c           = zeros(r(k)+2,1);
    c(1)        = q(k).*(1 - q(k)).^r(k);
    c(r(k)+1)   = -1;
    c(r(k)+2)   = 1;
    xsol        = roots(c);
    xsol        = xsol( abs(imag(xsol)) < eps );
    xsol        = xsol(xsol>0);
    [~,ii]      = max( abs(xsol - 1./p(k)) );
    x           = cat(1,x,xsol(ii));

end

end

pr  = ( (1 - p.*x)./((r + 1 - r.x).(1-p)) ).x.^(-1(n + 1));

return;

Please not that star-dot-star does not show up in the text correctly, but it should be clear from context that that I intend point-wise multiplication of vectors.


问:
Show that if $f: X \to Y$ is injective and $y \in Y$, then $f^{-1}(y)$ contains at most one point

Show that if $f: X \to Y$ is injective and $y \in Y$, then $f^{-1}(y)$ contains at most one point

This seemed very straightforward, but I cannot seem to be able to show this rigorously... If I pick $p,q\in X$ I have that $f^{-1}(f(p)) = \{p\}$ right? So I would get that $$f(p)=f(q) \\ f^{-1}(f(p)) = f^{-1}(f(q)) \\ \{p\} = \{q\} \\ p=q$$ is this anywhere in the direction I would need to go? It seems that I don't want to show that the function is actually injective here since that's given to me, I would need to show that the cardinality of $f^{-1}(y)$ is $1$ right?

答: Let $y\in Y$, then if $f^{-1}(y)$ contains more than one point we have that $p,q \in f^{-1}(y), p\not= q \implies f(p) = y = f(q)$ that is a contradiction because $f$ is an injective function. Then $f^{-1}(y)$ contains one point or none


问:
Is $\lim (\text{max}\{ f \}) = \text{max}\{(\lim f)\}$?
I'm trying to prove some statement from my differential equations class, and for some part of the proof I used that, being each $f_n(x,y)$ a continuous function defined over a compact set $\Omega$,

$$ \lim_{n \rightarrow \infty} \text{max}\{|f_n(x,y)| : (x,y)\in\Omega\} = \text{max}\{|\lim_{n\rightarrow\infty} f_n(x,y)|: (x,y)\in\Omega\ \}.$$
I thought this is true, but I don't know how to justify it, or if it's only true when $f$ is continuous or anything. Is this true in general? If not, under what conditions is it true? Any help will be appreciated, thanks in advance.

答:
Take $f_n(x)=n\,x^n(1-x)$ on $[0,1]$ then $f_n$ converge simply to $0$ while the other limit is $\frac 1e$

(i.e. max is reached in $f_n(\frac n{n+1}$)) : https://www.desmos.com/calculator/ckapsmstdn

Adding another variable $y$ does not change the deal.

Since max is involved or equivalently $||\cdot||_\infty$ norm, then uniform convergence for $f_n$ is most probably required.


问:
Prove: $G/(N_1 \cap N_2)$ is isomorphic with a subgroup of $(G/N_1) \times (G/N_2)$
I have to solve this exercise for my math study, but don't know how to do it. It's keeping me busy for 2 days now.

Let $G$ be a group and $N_{1}, N_{2}$ normal subgroups of $G$. Let $f: G \rightarrow (G/N_{1}) \times (G/N_{2})$ with $f(a) = (aN_{1}, aN_{2})$be a homomorphism with $Ker(f) = N_{1}\cap N_{2}$.

Prove that $G/N_{1}\cap N_{2}$ is isomorphic to a subgroup of $(G/N_{1}) \times (G/N_{1})$.

So far, I have done this:

$f$ is an homomorphism and $Ker(f) = N_{1}\cap N_{2}$, so $N_{1}\cap N_{2}$ is a normal subgroup of $G$. I also think that I have to use the fundamental homomorphism theorem, but I don't know how.

Could you please tell me how to prove this? I have really tried to solve this for a very long time, but I don't see what I have to do.

Thanks in advance!

答:
$\textbf{Def:}$ Let $A,B$ be groups. Then, $A \times B=\{(a,b)|a \in A, b \in B\}$. $A \times B$ is a group by defining $(a_1,b_1) \cdot (a_2,b_2) \,\, \text{for} \,\, a_1,a_2 \in A \,\, \text{and} \,\, b_1,b_2 \in B$ called the $\textbf{Direct Product}$ of $A$ and $B$. (You can check that this is a group by checking closure, identity, associativity, and inverse).

$\textbf{Proof:}$ Define $\varphi:G \rightarrow G/N_1 \times G/N_2$ by $\varphi(G)=(N_1g,N_2g) \,\, \forall g \in G$.

So, $\text{ker}(\varphi)=\{g \in G | \varphi(G)=1_{G/N_1 \times G/N_2}\}=\{g \in G | \varphi(G)=(N_1,N_2)\}=\{g \in G | (N_1g,N_2g)=(N_1,N_2)\}$.

We know that $N_1g=N_1$ if and only if $g \in N$.

Thus, $\text{ker}(\varphi)=\{g \in G | g \in N_1 and g \in N_2\} = \{ g \in G | g \in N_1 \cap N_2\} = N_1 \cap N_2$.

Then, by the First Isomorphism Theorem,
$G/N_1 \cap N_2 \cong \varphi(G)$.
$\,\, \blacksquare$

(To be more rigorous, you can show that $\varphi$ is a homomorphism).


问:
Optimal control input for nonlinear dynamics
This is from Chapter 1 of the book, "Robust Nonlinear Control Design" by Freeman and Kokotovic.

Consider the system
$$\dot{x} = -x^3 + u + wx,$$
where $u$ is an unconstrained input and $w$ is known to take values in the interval $[-1,1]$.

If we consider the cost functional,
$$
J = \int_0^{\infty} (x^2 + u^2) dt,
$$
then the optimal feedback law that minimizes the cost functional (in the worst case) is claimed to be given by
$$
u = x^3 - x - x\sqrt{x^4 -2x^2 + 2}.
$$

Question:

How is the optimal feedback law obtained?

Attempt #1:
I tried using Pontryagin Maximum Principle, and I got the following. The Hamiltonian is given as
$$
H = x^2 + u^2 + \lambda(-x^3 + u + wx).
$$
By taking derivatives with respect to $x$ and $u$ and solving them,
$$
\dot{\lambda} = -\frac{\partial H}{\partial x} = -2x+ 3\lambda x^2 -\lambda wx \Rightarrow \lambda = ke^{(3x^2-w)t}+\frac{2x}{w-3x^2},
$$
$$
\frac{\partial H}{\partial u} = 0 \Rightarrow u = -\frac{\lambda}{2}.
$$
Substituting $\lambda$ in the first expression into the second expression,
$$
u = -\frac{k}{2}e^{(3x^2-w)t}+\frac{2x}{w-3x^2}
$$

The initial or final conditions for the ODE in $\lambda$ are not stated so I cannot solve for $k$, but the form of the solution that I obtained is quite different from what is given.

Does anyone has any ideas or alternative approaches?

Attempt #2 (added on Jan 5 2021):
I also tried to work directly with the given optimal control input, $u$ and the nonlinear dynamics.

Substituting $u$ into the nonlinear dynamics $\dot{x}$ and using the worst-case $w$ of $1$ gives
$$
\dot{x}^2 = x^2 (x^4 - 2x^2 +2) = x^6 - 2x^4 + 2x^2
$$

Also note that we have the following expression for $u$ from the nonlinear dynamics,
$$
\dot{x} = -x^3 + u + x \Rightarrow u = \dot{x} + x^3 - x
$$
This is independent of the given optimal control input.

With this expression of $u$, if $\dot{x}$ can be assumed to be $0$ (may not be a valid assumption), then the following can be said of the cost functional,

$$
\begin{aligned}
J = \int_0^{\infty} (x^2 + u^2) dt &= \int_0^{\infty} (x^2 + (\dot{x} + x^3 - x)^2) dt\\
&= \int_0^{\infty} (x^2 + x^6 - 2x^4 + x^2) dt\\
&= \int_0^{\infty} (x^6 - 2x^4 + 2x^2) dt\\
&= \int_0^{\infty} \dot{x}^2 dt = 0,
\end{aligned}
$$
which is the minimum, since the cost functional is nonnegative.

答:
First of all, your solution for $\lambda$ as a function of time does not look correct. Namely, when one takes the time derivative of it yields

\begin{align}
\dot{\lambda} &= \frac{d}{dt}\left(k\,e^{(3\,x^2 - w)\,t} + \frac{2\,x}{3\,x^2 - w}\right), \\
&= \left(3\,x^2 - w + 6\,x\,\dot{x}\,t\right) k\,e^{(3\,x^2 - w)\,t} +
\frac{2\,(3\,x^2 + w)}{(3\,x^2 - w)^2} \dot{x}, \\
&= - 2\,x + 3\,x^2\,\lambda - w\,\lambda +
\left(6\,x\,t\,\left(\lambda - \frac{2\,x}{3\,x^2 - w}\right) + \frac{2\,(3\,x^2 + w)}{(3\,x^2 - w)^2}\right) \dot{x},
\end{align}

which would only match the desired derivative when $\dot{x} = 0$.

Instead one could solve it in a similar way as one could go about solving infinite time horizon LQR using PMP. Namely, by only considering only the initial conditions that would bring both state to the origin as time goes to infinity. Namely, if $x$ or $u$ (and thus $\lambda$) does not go to zero as time goes to infinity then the cost function integral would become infinite. Once near the origin one could use a linearization of the dynamics. The dynamics after substituting in the expression for $u$ can be written as

$$
\begin{bmatrix}
\dot{x} \\ \dot{\lambda}
\end{bmatrix} = f(x,\lambda),
$$

with

$$
f(x,\lambda) =
\begin{bmatrix}
-x^3 - 1/2\,\lambda + w\,x \\
-2\,x + 3\,x^2 \lambda - w\,\lambda
\end{bmatrix}.
$$

This dynamics in $z=\begin{bmatrix}x & \lambda\end{bmatrix}^\top$ can be linearized near an equilibrium point, where $\dot{z}=0$ and whose solution is denoted with $\bar{z}$. This linearization can be obtained by using a Taylor expansion

$$
\dot{z} = f(\bar{z}) + \left.\frac{\partial\,f}{\partial z}\right|_{z=\bar{z}} (z - \bar{z}) + \text{h.o.t.},
$$

with $\text{h.o.t.}$ short for higher order terms. Applying this to the given dynamics at the equilibrium point $(x,\lambda)=(0,0)$ yields

$$
\begin{bmatrix}
\dot{x} \\ \dot{\lambda}
\end{bmatrix} = \underbrace{
\begin{bmatrix}
w & -1/2 \\
-2 & -w
\end{bmatrix}}_A
\begin{bmatrix}
x \\ \lambda
\end{bmatrix} + \text{h.o.t.}.
$$

It can be noted that the matrix $A$ has one positive/unstable and one negative/stable eigenvalue. So the linearized system would only converge to the origin if its initial condition would lie on the stable manifold, which can be shown to be $\lambda = 2\,(w + \sqrt{w^2 + 1})\,x$. However, this manifold would for the nonlinear system only work very close to the origin. The manifold for the nonlinear system could maybe be found by starting infinitesimally close to the origin, but on the manifold for the linearized system, and solve it backwards in time. However, this is not an easy problem to solve analytically.

Because there are no constraints on the control input $u$ it follows that the Hamiltonian should remain constant (the choice for $\dot{\lambda}$ and $u$ assures that $\dot{H}=0$). The value of the constant value of $H$ can be obtained by substituting any state we know lies on the manifold, which would be $(x,\lambda)=(0,0)$, which gives $H=0$. First, simplifying the Hamiltonian after substituting in the expression $u=-\lambda/2$ yields

$$
H = x^2 + \lambda^2/4 + \lambda\,(-x^3 - \lambda/2 + w\,x).
$$

From this expression it should hopefully also be clear why $H=0$ when $(x,\lambda)=(0,0)$. Using the expression for the Hamiltonian with the expression for $u$ substituted in, then the equation $H=0$ can also be written as the following quadratic equation in $\lambda$

$$
\lambda^2 + 4\,(x^3 - w\,x)\,\lambda - 4\,x^2 = 0,
$$

which has the following two solutions

$$
\lambda = 2\,w\,x - 2\,x^3 \pm 2\,x \sqrt{(x^2 - w)^2 + 1}.
$$

However, only one of these two solutions lies on the manifold for the linearized system near to origin. Namely

$$
\lambda = 2\,w\,x - 2\,x^3 + 2\,x \sqrt{(x^2 - w)^2 + 1}.
$$

Substituting this into the expression for the control input yields the following solution for any $w$

$$
u = x^3 - w\,x - x\sqrt{(x^2 - w)^2 + 1}.
$$

This matches the given control law when $w = 1$. This should also be expected, since for $w=1$ the origin is the most unstable (when $u=0$).


问:
Rings with zero divisors
Is there a ring  $~R~$ with non-trivial multiplication (i.e. $~\exists a,b\in R ~~~ ab\neq 0$) such that each non-zero element of $~R~$ is a zero-divisor? </p>

答: The simplest example might be $R=2\mathbb Z/8\mathbb Z$.  Then $2\cdot 2\neq 0$ but $a\cdot 4=0$ for all $a\in R$.</p>

Title: Rings with zero divisors


问:
Chromatic number of union of graphs
I have two graphs, $G$ and $G'$ on the same vertex set $V$. I have already disproven that $\chi (G\cup G')\leq \chi G + \chi G'$.

Now I have to prove or disprove that $\chi (G\cup G')\leq \chi G \cdot \chi G'$. I think that it is true, but I don't know how to start proving this. Can someone help me? Thanks!

答:
Hint: Suppose that $f:V \to \{1,\dots,\chi G\}$ and $g: V \to \{1,\dots,\chi G'\}$ are valid colorings. Then the function $h:V \to \{1,\dots,\chi G\} \times \{1,\dots,\chi G'\}$ defined by
$$
h(v) = (f(v),g(v))
$$
is a valid coloring of $V$ since we have $h(v) \neq h(w)$ whenever $v,w$ are adjacent in $G \cup G'$.

Title: Chromatic number of union of graphs


问:
Symmetric distribution of points on a sphere
Background:

The VSEPR theory provides a useful guide to predicting molecular geometries. One implication of the theory is that the orientation of ligands in three-dimensional space is such that the seperation between each pair of them is maximum.

This is analogous to the mathematical problem of symmetrically distributing points on the surface of a sphere. With 2, 3, 4, or 6 points, the solutions are trivial. The required distributions would be linear, triangular, tetrahedral, and octahedral respectively. An issue arises when we attempt to distribute 5 points on the surface of a sphere.

We could try to arrange the points such that they lie on the vertices of a regular polyhedron inscribed in the sphere. However, from my knowledge of Euclidean geometry, I believe that such a shape with 5 vertices simply cannot exist.

Chemists suggest an approximately symmetric distribution such as placing the points on the vertices of a square pyramid. But from a mathematical perspective, there ought to be a better solution.

Question: What is the nearest approximation to a symmetric distribution which we can achieve? Can the approximation be generalized to any other number of points which cannot be distributed at the vertices of a platonic solid?

The definition of a symmetrical distribution here is that any non-labelled point on the sphere is indistinguishable from another such point. A nearly symmetrical distribution implies the least variation in the relative distances between the points (distances measured along the surface).

The ambiguity in the definition of symmetry makes the question more difficult to answer. Any additional insights or numerical methods to solve the problem are also welcome.

答:
The answer depends on your definition of best approximation of a symmetric distribution. Personally, I will vote for a "triangular bi-pyramid" because

The optimal packing of 5 equal spheres inside another sphere is degenerate. In one of the degenerate configurations, the centers are arranged in a triangular bi-pyramid. (see wiki entry of sphere packing in a sphere and refs there, please note that the maximum radius of inner spheres is the same for $n = 5$ and $6$).

It is the unique configuration which minimize the 5 electron case of Thomson's problem. (R.E. Schawrtz, 2012, preprint on arxiv).

It is also the configuration which maximizes the volume of corresponding inscribed polyhedron (see refs. in this answer).


问:
Linearization of a function in 1D
I sincerely hope you can provide me with some help. I am currently studying for my advanced Calculus and I got stuck on this question.

The question in regard is: Calculate the linearizations of φ and ψ around 2.

With the functions being given by,
$φ(x)^2$ + $sin(φ(x)ψ(x))$ = 2 − x,
and
$φ(x) + ψ(x)$ = $x^2$

What I tried
Well, what I have tried to do was to create both functions where the right-hand side would be zero and then differentiate them both with respect to x from this you get a system of equations and then just solve them for both.

I see that I have to use the chain rule ( and product rule for sin(phi) to get them but when I differentiate $φ(x)^2$  I get$ 2φ(x)_x$ whereas the answer gives me 2φ(x) I have not tried differentiating $sin(φ(x)ψ(x))$ yet but I expect it to become something like $φ_x(x)ψ(x)cos(φ(x)ψ(x))$+$φ(x)ψ_x(x)cos(φ(x)ψ(x))$

If somebody could help me it would be greatly appreciated.
Picture attached with answers from the professor.enter image description here

答:
You want to write both functions as

$$\phi(x) \approx \phi(2) + \phi'(2) \, (x-2)$$

$$\psi(x) \approx \psi(2) + \psi'(2) \, (x-2)$$

So you need to calculate $\phi(2)$, $\psi(2)$, $\phi'(2)$ and $\psi'(2)$.

$$$$

(a) $\phi(2)$ and $\psi(2)$

$$\begin{matrix}
\phi^2(x) + \sin(\,\phi(x)\,\psi(x)\,) = 2 − x & \rightarrow & \phi^2(2) + \sin(\,\phi(2)\,\psi(2)\,) = 0\\
\phi(x) + \psi(x) = x^2                           & \rightarrow & \phi(2) + \psi(2) = 4
\end{matrix}$$

This system is transcendental and may have more than one solution. But $\phi(2) = 0$ and $\psi(2) = 4$ is clearly one of them.

$$$$

(b) $\phi'(2)$ and $\psi'(2)$

Differentiate each side of both equations to get:

$$\begin{matrix}
2\,\phi(x)\,\phi'(x) + \cos(\,\phi(x)\,\psi(x)\,) \,  [\,\phi'(x)\,\psi(x)+\phi(x)\,\psi'(x)\,] = -1 & \rightarrow & 2\,\phi(2)\,\phi'(2) + \cos(\,\phi(2)\,\psi(2)\,) \,  [\,\phi'(2)\,\psi(2)+\phi(2)\,\psi'(2)\,] = -1\\
\phi'(x) + \psi'(x) = 2x                           & \rightarrow & \phi'(2) + \psi'(2) = 4
\end{matrix}$$

$$
\left[\begin{matrix}
2\,\phi(2) + \cos(\,\phi(2)\,\psi(2)\,) \, \psi(2) & \cos(\,\phi(2)\,\psi(2)\,) \, \phi(2)\\
1 & 1
\end{matrix}\right]
\left[\begin{matrix}
\phi'(2)\\
\psi'(2)
\end{matrix}\right]
=
\left[\begin{matrix}
-1\\
4
\end{matrix}\right]
$$

For $\phi(2) = 0$ and $\psi(2) = 4$, we get:

$$
\left[\begin{matrix}
4 & 0\\
1 & 1
\end{matrix}\right]
\left[\begin{matrix}
\phi'(2)\\
\psi'(2)
\end{matrix}\right]
=
\left[\begin{matrix}
-1\\
4
\end{matrix}\right]
$$

And finally, $\phi'(2) = -1/4$ and $\psi'(2) = 17/4$.

Title: Linearization of a function in 1D


问:
Calculation of $\omega (V_1, V_2)$, when given the two-form and vectors
There is a two form such that $\omega = 6dx \wedge dy+3dy \wedge dz + 2 dz \wedge dx$ and two vectors such that $V_1 = (2, 3, 1), V_2 = (2, 3, 2)$. So I have to find $\omega(V_1, V_2)$ for that I tried to split $\omega$ into some $\omega_1 \wedge \omega_2$ but I got 3 linear equations in 6 variables, which I have never done before.

$\omega_1 = a\cdot dx+b\cdot dy + c \cdot dz$ and $\omega_2 = a'\cdot dx+b'\cdot dy + c' \cdot dz$

$-ba'+ab'=6$, $-cb'+bc'=3$, $ca'-ac'=2$

It would be helpful if someone can give a step by step instruction.

and is there another way of doing this directly without splitting $\omega$?

答:
Notice that $$(dx\wedge dy)(V_1,V_2)=dx(V_1)dy(V_2)-dx(V_2)dy(V_1)=2\cdot 3-2\cdot 3=0$$
Now do the same for the other summands in $\omega$ to get $\omega(V_1,V_2)$


问:
calculating a probability using markov's inequality
How can I calculate $P(X>1)$ for $Mx(t)=1/(1-t)$ using  markov's inequality?

Thank you

答:
Assume $X$ is a non-negative r.v. (which should be true as the mgf looks like the one for an $\text{Exp}(1)$ distribution). Note that Markov's inequality only gives a upper bound for the probability instead of an exact probability. We find the first moment of $X:$ \begin{align*}
M'_X(t)&=\frac{d}{dt}\frac{1}{1-t}=\frac{1}{(1-t)^2}\\ \mathbb E(X)&=M'_X(0)=\frac{1}{(1-0)^2}=1
\end{align*}
By Markov's inequality, $${\displaystyle \mathbb P (X\geq a)\leq {\frac {\mathbb E(X)}{a}}\longrightarrow \mathbb P(X>1)=\mathbb P(X\ge 1)\le \frac{1}{1}=1}$$
(it might sound like a meaningless conclusion, but that's all Markov's inequality can tell you in this case.)


问:
When does the isomorphism $G\simeq \ker(\phi)\times \operatorname{im}(\phi)$? hold?
Suppose you have a group isomorphism given by the first isomorphism theorem:

$$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$

What can we say about the group $\ker(\phi)\times \operatorname{im}(\phi)$? In particular, when does the following hold:

$$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$

I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{>0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/  SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{>0}$, using the surjective homomorphism $\det(M)$. I tried something with semidirect products but I got stuck.

答:
Let $N = ker(\phi)$ and $K = im(\phi)$, then you're asking when, given an exact sequence $1 \to N \to G \to K \to 1$ is trivial.

First you need the extension to be split, that is, there must exist a morphism $s : K \to G$ such that the composition $\phi \circ s$ is the identity. In this case $G \simeq N \rtimes K$, the semidirect product of $N$ and $K$ : this is the splitting lemma (for non-abelian groups).

Now you want this semidirect product to be direct; this is true iff $K$ is also normal in $G$, or equivalently that there exists a morphism $G \to N$ which is the identity on $N$.

I don't include proofs here, as they're found in any basic group theory notes.

In fact you can get away with the first condition. Indeed, if there exists a map $p : G \to N$ which is the identity on $N$, then a section of $\phi$ automatically exists and the isomorphism $G \cong \operatorname{im}(\phi) \times \operatorname{ker}(\phi) = K \times N$ holds. The required isomorphism is $(\phi, p) : G \to K \times N$ (it's not hard to check that this is in fact an isomorphism).


问:
Show that [x] is denumerable

The problem 1 is that: I am not sure about the relation symbol (corner of the $[x]$) means.

Problem 2: I think that if $ x-y \in \mathbb{Q} $, then $x$ and $y$ must be in $\mathbb{Q}$. So all $y$s that make $ x-y \in \mathbb{Q} $ are in $\mathbb{Q}$. Thus they form a subset of $\mathbb{Q}$. Since $\mathbb{Q}$ is denumerable, $[x]$ is denumerable.

Hope I am clear, thank you !

答:
Let me use $[x]$ for the equivalence class of $x.$ A real number $y$ belongs to $[x]$ if and only if $x-y$ is a rational number. This happens if and only if $y=x+q$ for some rational number $q.$ Hence, the number of elements in the equivalence class of $x$ is the same as the number of rationals, that is, $[x]$ is denumerable (by which I'm assuming you mean countably infinite).

For your third question, consider that all rationals belong to the same equivalence class. Next, let $x$ and $y$ be distinct irrationals for which $x-y$ is not rational. What can you say about $[x]$ and $[y]?$ Does that suggest a strategy?

Title: Show that [x] is denumerable


问:
Solution Veriification: Describe the quotient Topology on Y induced by f
Let Y={1,2,3} and define f:$\mathbb{R} \to Y$ by f(x)=1 if x<0 , f(x)= 2 if x=0 and f(x) =3 if x>0 . Describe the quotient topology on Y induced by f.

I think the  topology will be this set ={{1},{3},{13},$\phi$}.

Am I right?

If I am wrong then kindly tell the right topology and why it is so.

Thanks!

答:
The open sets in the quotient are exactly those whose preimages are open.

If $Y$ is finite, then you can find the topology by simply checking the preimages of all the subsets of $Y$.

The subsets of $Y$ which you have not included are $\{2\}, \{1,2\}, \{2,3\}, \{1,2,3\}$. One of them has open preimage.


问:
Proof if $x+y=0$, then $xy \leq 0$.
For all $x,y ∈ \mathbb R$, if $x + y = 0$ then $xy \leq 0$. Decide whether this is true or false.

I assume this is true since one of $x$ or $y$ needs to be positive while the other is negative, so their product is negative, but how would would i go about proving this? Would this be right:

Let $x = 2k$ and $y = -2k$, such that $k \in \mathbb N$. Then $x + y = 0$.

So $xy = -4k^2$ which is less than $0$, hence this is true.

答:
Assume $xy>0;$

Then $x,y >0$, or $x,y <0.$

In in either case $x+y \not =0,$ which contradicts the assumption.

Title: Proof if $x+y=0$, then $xy \leq 0$.


问:
Compact embedding of $l^p$ to $c_0$
if $c_0$ is a space of real sequences that converges to zero with sup norm. I can show that $l^{p} \hookrightarrow c_{0}$ embedding is not compact with the sup norm. But I want something more intresting than that. I want to find norm on $c_0$ such that this emdedding be compact! But until now I can't find such norm. So if one have any idea about this ....

答:
Let $b=\{b_k\}\subset\ell^q$ with $\|b\|_q=1$ and $b_k\geq0$ for all $k$. Define
$$
\|x\|_c=\sum_{k=1}^\infty b_k\,|x_k|
$$
and $$T_nx=\sum_{k=1}^n x_ne_n,$$ where $\{e_n\}$ denotes the canonical basis in $c_0$. Then
$$
\|x-T_nx\|_c=\sum_{k>n}b_k\,|x_k|\leq\Big(\sum_{k>n}|b_k|^q\Big)^{1/q}\,\|x\|_p.
$$
Then
$$
\|I-T_n\|\leq\Big(\sum_{k>n}|b_k|^q\Big)^{1/q}\xrightarrow[n\to\infty]{}0.
$$
Thus $I$ is a limit of finite-rank operators and thus compact.

Title: Compact embedding of $l^p$ to $c_0$


问:
Series Solution of ODE at a Regular Singular Point
I am given the ODE:
$$ (\sin{x})y''+2(\cos{x})y'-(\sin{x})y=0 $$
and asked to:

Show $x=0$ is a regular, singular point

Show both linearly independent solutions can be written as a power series about $x=0$

Find the first two non-zero terms in each power series

I have no problem with the first part and was able to find that $p_0=2$ and $q_0=0$ (the values of the limits of $x*Q/P$ and $x^2*R/P$ as $x \rightarrow 0$ respectively). Using this I was able to find my indicial equation as:
$$ r^2+r=0 \implies r=0,-1 $$
From here, I am not sure how to proceed. I believe my solutions should be of the form:
$$ y_1(x)=\sum\limits_{n=0}^\infty a_nx^n \quad\quad\quad\text{and}\quad\quad\quad  y_2(x)=\sum\limits_{n=0}^\infty a_nx^{n-1} $$
but I'm not entirely sure why. Is this where you have to check the continuity and differentiability of $P, Q, R$? Further, for the last part how do I find the coefficientss of the two solutions?

Any hints and general advice in solving ODEs using the Frobenius Method would be much appreciated!

EDIT: When trying to solve for the coefficients, should I expand $\sin{x}$ and $\cos{x}$ with Taylor Series?

答:
Just for the sake of curiosity, I am going to explain @LutzLehmann's claim:

\begin{align*}
\sin(x)y'' + 2\cos(x)y' - \sin(x)y = 0 & \Longleftrightarrow (\sin(x)y'' + \cos(x)y') + (\cos(x)y' - \sin(x)y) = 0\\\\
& \Longleftrightarrow (\sin(x)y'' + (\sin(x))'y') + (\cos(x)y' + (\cos(x))'y) = 0\\\\
& \Longleftrightarrow (\sin(x)y')' + (\cos(x)y)' = 0\\\\
& \Longleftrightarrow \sin(x)y' + \cos(x)y = c_{0}\\\\
& \Longleftrightarrow (\sin(x)y' + (\sin(x))'y)= c_{0}\\\\
& \Longleftrightarrow (\sin(x)y)' = c_{0}\\\\
& \Longleftrightarrow \sin(x)y = c_{0}x + c_{1}\\\\
& \Longleftrightarrow y(x) = \csc(x)(c_{0}x + c_{1})
\end{align*}

Hopefully this helps!


问:
Basic concepts in finite fields
I need some help with clearing up some some basic concepts in finite fields.
I understand that $\mathbb{F}_p = GF(p)$ where $p$ is a prime is a finite field, which is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. However, I get quite confused with $GF(p^n)$.

Since $GF(p)$ is a finite field, is $GF(p^n)$ a vector space of dimension $n$ over $GF(p)$ ($n$-tuples of elements from $GF(p)$) or another finite field with more ($p^n$) elements? Can I understand this intuitively as like when $\mathbb{R}$ is the field then $\mathbb{R}^n$ is an $n$-dimensional vector space over $\mathbb{R}$?

Another thing is the $\mathbb{F}_p^n $. Is this the same as $GF(p^n)$? But what's $\mathbb F_{p^n}$ then? Are they just two different notations of the same thing? It seems more logical to me that the first one means something like $GF(p)^n$ but is it something meaningful?

Finally, when someone states "Let $V$ be an $n$-dimensional vector space over a finite field..." - does he explicitly mean $GF(p^n)$ for some $p$ prime or something else?

答:
If $q = p^n$ for some prime $p$, then the notations $\rm{GF}(q)$ and $\mathbb{F}_q$ mean the same, namely the (unique up to isomorphism) field with $q$ elements.

This is indeed a vector space over $\rm{GF}(p)$ of dimension $n$.

$\rm{GF}(p)^n$ and $\mathbb{F}_p^n$ also mean the same, namely the set of $n$-tuples of elements from the field $\rm{GF}(p)$, which is also a vector space of dimension $n$ over $\rm{GF}(p)$.

Since these two vector spaces have the same dimension, they are isomorphic as vector spaces. However, they are not isomorphic as rings, since the first is a field, and the other has non-trivial zero-divisors (take for example en element like $(1,0)$ in $\rm{GF}(p)^2$).

When someone says "a vector space of dimension $n$ over $\rm{GF}(p)$" he means just that. You can take any of the two above and it will not matter, since as vector spaces, we cannot distinguish them.

Title: Basic concepts in finite fields


问:
Determining the Distribution of an uncorrelated Random Variable by using Dirac's Delta
Choosing random variables $X \sim U(-1,1)$ and $Y = X^2$, it is possible to show that X and Y are uncorrelated, yet not independent.

I was wondering what the probability distribution of $Y$ now looks like.
We should have
$$P_Y(y) = \int_{-1}^{1}dx~~\frac{1}{2} \delta(x^2-y),$$
where $\delta$ denotes Dirac's delta distribution.

My first intuition would be to go for a substitution via $u = x^2$, but this is bad for the domain of integration.
So to circumvent this inconvenience, I would take the integration domain into an indicator function $\chi_{[-1,1]}(x)$.
Concatenating those ideas, I have
$$ \frac{1}{2}\int \frac{du}{2 \sqrt{u}}~~\biggr(\chi_{[-1,0]}(- \sqrt{u}) + \chi_{[0,1]}( \sqrt{u})\biggr) \delta(u-y)~=~\frac{1}{4\sqrt{y}}\biggr(\chi_{[-1,0]}(- \sqrt{y}) + \chi_{[0,1]}( \sqrt{y})\biggr).$$

Are parts of this solution correct?
How should one proceed to arrive at the correct solution?

答:
You forgot to do the change of variables $du=  2x dx$.

In any case, working with the CDF of $Y$ avoids dealing with the Dirac delta and the degenerate joint distribution.

One can show
$$P(Y \le y) = P(X^2 \le y) = \begin{cases} 0 & y \le 0 \\ \sqrt{y} & 0 < y < 1 \\ 1 & y \ge 1\end{cases}$$
since in the case $y \in (0,1)$ we have $P(X^2 \le y) = P(-\sqrt{y} \le X \le \sqrt{y})$.

Now that you have the CDF of $Y$, you can find the PDF by differentiating.


答:
Provided that @Ben's answer is better (i.e. it gives more insight of what is going on, at least in my opinion), you can also prove your claim as a corollary of your previous result.

Suppose that $\chi(G)=1$, then $G$ necessarily has no edges. Hence, $\chi(G\cup G')=\chi(G')=\chi(G)\chi(G')$. By symmetry, we can then assume that both $\chi(G)$ and $\chi(G')$ are larger than $1$. But under these assumptions, $\chi(G)\chi(G')\geq\chi(G)+\chi(G')$, and hence the claim follows.

Title: Chromatic number of union of graphs

Body:
I have two graphs, $G$ and $G'$ on the same vertex set $V$. I have already disproven that $\chi (G\cup G')\leq \chi G + \chi G'$.

Now I have to prove or disprove that $\chi (G\cup G')\leq \chi G \cdot \chi G'$. I think that it is true, but I don't know how to start proving this. Can someone help me? Thanks!


问:
What distinguishes measure theory and probability theory?
It is clear that the theory of probability works primarily with limited measures on measurable spaces. On the other hand, there is a folklore that says that what distinguishes measure theory and probability theory is the conditional probability and conditional expectation.

But conditional probabilities and conditional expectations are derived from the Radon–Nikodym theorem and a measure with respect to another measure. And the Radon–Nikodym theorem is a typical result of measure theory.

Question 1: So we could see the theory of probability as a subdiscipline of measure theory?

Question 2: Would be possible to give another basis for the theory of probability rather than through measure theory?

答:
Salomon Bochner's words in the introduction to his article entitled 'Stochastic Processes' published in 1947 in volume 48 of the Annals of Matematics:

The link to full text.


问:
Lakers and Bucks arrangements
I know some of them must be trivial but I have pretty poor knowledge of combinatorics:
We have 5 Lakers players and 7 Bucks and we want to

(a) have them seated in a row. My answer: $(7+5)! = 12!$

(b) In a row but with all Lakers to be seated together: We consider the Lakers as one block, so we have 8 "items":  $8!5!$.

(c) Be seated in a circle: It is $(12-1)! = 11!$

(d) In a circle but with at least 2 Lakers players to be seated together: I am not sure about this one: We treat the pair as one, so we have $10!$?

(e) In a row, with two Lakers in the first and last position and no two Lakers in consecutive places. I don't know how to calculate this one. Clearly we need to find the arrangement of 10, with no Lakers in starting and ending positions, and no Lakers in consecutive places, multiplied by the number of ways to select 2 out of 5 (which is 20?)

(f) In a circle but with no two or more Lakers sitting next to each other. I don't know how to calculate this one

(g) In how many ways can we make a subgroup of any 3 players: It is $\binom {12}{12-3}$

(h) a subgroup of size 3 but with exactly 1 Laker player: $\binom {7}{2} \binom {5}{1}$

(i) a subgroup of size 3 but with at least 1 Laker player: $\binom {12}{12-3} - \binom {7}{3}$??

Can you please help me with the ones I've got wrong or haven't answered?

Thank you!

答:
I've checked a, b, c, g, h, i and they look correct.

Hints for the remaining ones:

(e) You are right about the 20 ways to select the two Lakers on the ends. There are $7!$ ways to arrange the Bucks players relative to each other, and $3!$ ways to arrange the remaining Lakers players relative to each other. It remains to count the number of ways to arrange these $7$ Bucks and $3$ Lakers. There are four "gaps" between/around the three Lakers, and there must be at least one Buck in each of these gaps. This is the number of compositions of $7$ into $4$ positive numbers. You can count this manually, i.e. $4+1+1+1$, $3+2+1+1$, $1+2+2+2$, etc.

(f) A similar argument can be used here. There are $4!$ ways to arrange the Lakers in a circle. There are $5$ gaps between them that each must contain at least one Bucks player. You can manually count the number of compositions of $7$ into $5$ positive numbers; they are the five permutations of $2+1+1+1+1$. Once you have decided where the Bucks will seat, there are $7!$ ways to fill the seats.

(d) This is the complement of (f) among all circular seatings

Title: Lakers and Bucks arrangements


问:
support of addition of function
Let $\phi_j:X\to \Bbb{R}$ be collection of smooth function $\phi_i\ge 0$ which support are locally finite.

Given function $\rho(x) = \sum\phi_j$ which is again well defined and smooth due to locally finite.

There is a relation between support that  I can't work it out.

$$\text{supp}(\rho) = \overline{\bigcup\text{supp}(\phi_i)}$$

I know in general $\text{supp}(f+g) \subset \text{supp}(g) \bigcup\text{supp}(f)$ why they are equal even for closure here?

答:
Let $s(f) = \{x \in X \mid f(x) \ne 0\}$. Then $\operatorname{supp}(f)  = \overline{s(f)}$.

Since all $\phi_j \ge 0$, we have $\rho \ge \phi_i \ge 0$ for all $i$ and
$$s(\phi_i) = \{x \in X \mid \phi_i(x) > 0\} \tag{1} ,$$
$$s(\rho) = \{x \in X \mid \rho(x) > 0\} \tag{2} ,$$
$$s(\phi_i) \subset s(\rho) \tag{3} ,$$
$$\bigcup s(\phi_i) = s(\rho)  \tag{4} .$$
$(3)$ implies $\operatorname{supp}(\phi_i) \subset \operatorname{supp}(\rho)$ and we get
$$\overline{\bigcup s(\phi_i)} \subset \overline{\bigcup\operatorname{supp}(\phi_i)} \subset \operatorname{supp}(\rho) \tag{5} .$$
Now let $x \in \operatorname{supp}(\rho)$ and $U$ be an open neigborhood of $x$. There exists $x_U \in U \cap s(\rho)$. By $(4)$ we have $x_U \in \bigcup s(\phi_i)$. Thus $x \in \overline{\bigcup s(\phi_i)}$ and therefore
$$\overline{\bigcup s(\phi_i)} = \overline{\bigcup\operatorname{supp}(\phi_i)} = \operatorname{supp}(\rho) \tag{6} .$$
This is true without any assumptions on the family $\{ \operatorname{supp}(\phi_i) \}$.

Title: support of addition of function


问:
Is $(a_n)_n$ with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ a Cauchy sequence?
Let $0 < q < 1$ and $(a_n)_n$ be a squence with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ for all $n ∈ ℕ$.

I need to show that this is a Cauchy sequence. I'm not sure how to start this one, as we just had cauchy sequences in lecture.
Can I maybe prove this by using induction?
How do start on a problem like this and where do I need to get to?

Body:
Hint:

$|a_{n+1}-a_n|\leq q|a_{n}-a_{n-1}|\leq q^2|a_{n-1}-a_{n-2}|\leq\ldots\leq q^{n-1}\dfrac{1}{1-q}|a_{2}-a_{1}|$

$|a_{n+2}-a_n|\leq |a_{n+2}-a_{n+1}|+ |a_{n+1}-a_{n}|\leq (q+1)|a_{n+1}-a_{n}|\\ \leq q(q+1)|a_{n-1}-a_{n-2}|\leq\ldots\leq q^{n-2}(q+1)|a_{2}-a_{1}|\leq q^{n-2}\dfrac{1}{1-q}|a_{2}-a_{1}|$

$|a_{n+3}-a_n|\leq |a_{n+3}-a_{n+2}|+|a_{n+2}-a_{n+1}|+ |a_{n+1}-a_{n}|\leq\\ (q^2+q+1)|a_{n+1}-a_{n}|\leq q(q^2+q+1)|a_{n-1}-a_{n-2}|\leq\ldots\leq \\
q^{n-2}(q^2+q+1)|a_{2}-a_{1}|\leq q^{n-2}\dfrac{1}{1-q}|a_{2}-a_{1}|$

Now generalize that to show
$$|a_{m}-a_n|\leq\ldots\leq q^{n-2}(q^{m-n}+\cdots+q^2+q+1)|a_{2}-a_{1}| \leq \\ q^{n-2}\dfrac{1}{1-q}|a_{2}-a_{1}|$$
and therefore $(a_n)$ is Cauchy (because $\displaystyle\lim_{n\to\infty}q^n=0$).


问:
How is the Rodrigues formula $L_n^k(x)=\frac{e^x x^{-k}}{n!}\frac{d^n}{dx^n}(e^{-x}x^{n+k})$ derived?
I am trying to deduce the Rodrigues formula for generalized Laguerre polynomials $$L_n^k(x)=\frac{e^x x^{-k}}{n!}\frac{d^n}{dx^n}(e^{-x}x^{n+k})$$ but I have reached a point where I do not know how to proceed, my procedure was as follows:

I started from the following two equalities:

$L_n^k(x)=(-1)^{k}\cfrac{d^k}{dx^k}L_{n+k}(x)$

$L_n(x)=\cfrac{e^x}{n!}\cfrac{d^n}{dx^n}(x^ne^{-x})$

Then substituting $ n + k $ for $ n $ in equation (2) we obtain that:

$$L_{n+k}(x)=\frac{e^x}{(n+k)!}\frac{d^{n+k}}{dx^{n+k}}(x^{n+k}e^{-x})$$

And therefore, substituting the value of $ L_ {n + k} $ in equation (1) we find that:

$$L_n^k(x)=(-1)^{k}\frac{d^k}{dx^k}\cfrac{e^x}{(n+k)!}\frac{d^{n+k}}{dx^{n+k}}(x^{n+k}e^{-x})$$

But I no longer know how to continue.
I would really appreciate your help.

答:
Given:

$$ L^k_n(x) := (-1)^k D^k[ L_{n+k}(x)] \tag{1} $$

using the linear differential operator $\,D[f(x)]:=f'(x)\,$ and where

$$ L_n(x) := e^x/n! D^n[ x^n/e^x ]. \tag{2} $$

Rewrite equation $(2)$ as

$$ L_n(x) = T^n[ x^n ]/n! \tag{3} $$

where the linear differential operator $\,T\,$ is defined as

$$ T[f(x)] := f'(x)-f(x) = (D-I)[f(x)] \tag{4} $$

and because of the identity

$$ T[ f(x) ] / e^x = D[ f(x)/e^x ]. \tag{5} $$

Combine equations $(3)$ and $(1)$ to get

$$ L^k_n(x) = (-1)^k D^k[ T^{n+k}[ x^{n+k} ] / (n+k)! ]. \tag{6} $$

The differential operators $\,D\,$ and $\,T\,$ commute, implying

$$ L^k_n(x) = (-1)^k T^{n+k}[ D^k[ x^{n+k} ] / (n+k)! ]
  = (-1)^k T^{n+k}[ x^n/n! ]. \tag{7} $$

Due to the binomial expansions of $\,T^n\,$ and $\,T^{n+k}\,$ we get

$$ T^n[ x^{n+k} ] = (-1)^k T^{n+k}[ x^n] x^k \tag{8} $$

by expanding both sides into polynomials and
comparing their coefficients.

Substitute this equation $(8)$ into equation
$(7)$ to get

$$ L^k_n(x) = T^n[ x^{n+k}]/(n!x^k). \tag{9} $$

Rewrite this equation $(9)$ using equation $(5)$
to get

$$ L^k_n(x) = \frac{e^x}{n!x^k} D^n[x^{n+k}/e^x]. \tag{10} $$


问:
Prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$
I am trying to prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$. So far I know that $\text{dist}\left({x,y}\right) \leq d_1\left({x,y}\right)$ where $\text{dist}\left({x,y}\right)$ is the standard distance function in k-dimensional Euclidian space. I also know that $d_1\left({x,y}\right) \leq \text{dist}\left({x,y}\right) \leq \sqrt{k}\cdot d_1\left({x,y}\right)$. I have already proven that $d_1\left({x,y}\right) \geq 0$, that $d_1\left({x,y}\right) = 0 \iff x=y$, and that $d_1\left({x,y}\right) = d_1\left({x,y}\right)$, but I am stuck on the triangle inequality property of this particular metric. Any hints or suggestions would be greatly appreciated. </p>

答: Let $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$. Since $d_1\left({x,y}\right)=\left|{x_j-y_j}\right|$ for some $j$ it is obvious that $d_1\left({x,y}\right) \geq 0$, that $d_1\left({x,y}\right) = d_1\left({y,x}\right)$, and that $d_1\left({x,y}\right)=0\iff{x=y}$. Now, by the definition of $d_1\left({x,y}\right)$ we have $\left|{x_j-y_j}\right| \leq d_1\left({x,y}\right)$ for any $j$. It follows that for any $j$ we have $$\left|{x_j-y_j}\right| \leq \left|{x_j-z_j}\right| + \left|{z_j-y_j}\right| \leq \max\{\left|{x_j-z_j}\right|:j=1,2,...,k\}+\max\{\left|{z_j-y_j}\right|:j=1,2,...k\}$$ Thus $$\max\{\left|{x_j-y_j}\right|:j=1,2,...,k\} \leq \max\{\left|{x_j-z_j}\right|:j=1,2,...,k\} + \max\{\left|{z_j-y_j}\right|:j=1,2,...,k\} $$ Or $d_1\left({x,y}\right) \leq d_1\left({x,z}\right) + d_1\left({z,y}\right)$, completing the proof that $d_1\left({x,y}\right)$ is a metric on $\mathbb{R}^k$. </p>


问:
Limit of probability of uniform distribution
Let $U_1,U_2,\ldots,U_n$ be independent and identically distributed random variables each following a uniform distribution $U(0, 1).$ How do I find $\lim_{n\to\infty} P(U_1+U_2+\cdots+U_n\le 3n/4)$?

答:
\begin{align}
& \operatorname E(U_1+\cdots+U_n) = \frac n2. \\[8pt]
& \operatorname{sd}(U_1+\cdots+U_n) = \text{constant} \times \sqrt n. \\
& \quad\text{(where “constant” means not changing as $n$ changes).}
\end{align}
Ask yourself how many standard deviations above the mean $3n/4$ is, and what Chebyshev's inequality says about the probability of being that many standard deviations above the mean.


答: Hint: This is a consequence of the law of large numbers after a suitable rewriting of your expression.</p>

Title: Limit of probability of uniform distribution

Body: Let $U_1,U_2,\ldots,U_n$ be independent and identically distributed random variables each following a uniform distribution $U(0, 1).$ How do I find $\lim_{n\to\infty} P(U_1+U_2+\cdots+U_n\le 3n/4)$?


问:
Why does the column space equal $\mathbb{R}^m$ (where $m$ is the number of rows) when the rank $= m$?
Let $m$ be the number of rows and $n$ be the number of columns. I read that if the rank of some matrix $A$ is equal to $m$, then we know that $Ax=b$ is solvable for every $b$. I believe this means that the column space spans all of $\mathbb{R}^m$.

But I have not been able to come up with a proof to show that rank$(A) =m\Rightarrow C(A)=\mathbb{R}^m$, where $C(A)$ denotes the column space of $A$. Can someone give me the intuition behind why this is be true? The rank is defined based off the rows, so how can it can say anything about the columns?

答:
I will show you a sketch of how we proved equality of the row rank and the column rank of any matrix on a course of linear algebra I studied a long time ago. Your mileage may vary. (For example, in your course some steps may not work, because they are underpinned by theorems which, in your course, you prove assuming the equality of the row and the column rank.)

In the proof, we used elementary operations on rows/columns of a matrix. I believe these are usually taken to be: (1) Add a row/column multiplied by a scalar $\lambda$ to another row/column; (2) Multiply a row/column by a scalar $\lambda\ne 0$ and (3) Swap two rows/two columns.

Lemma 1: Elementary operations on rows (columns) of a matrix don't change the row (column) rank of that matrix.

Proof (sketch): Prove that the span of the resulting rows (columns) is contained in the span of the original rows (columns). However, as all elementary operations are invertible, then the opposite inclusion holds. Consequently, the span of the resulting rows (columns) is the same as the span of the original rows (columns). $\blacksquare$

Lemma 2: Elementary operations on columns (rows) of a matrix don't change the row (column) rank of that matrix.

Proof (sketch): Prove that, if $A_1,\ldots,A_m$ are rows of the matrix $A$, and you choose some of them (say, $A_{i_1},\ldots,A_{i_k}$), then $A_{i_1},\ldots,A_{i_k}$ are linearly independent if and only if the resulting rows (after any elementary operation on columns) are linearly independent. (This is then proven in a pedestrian way for each type of elementary operations.) By symmetry, the statement is now also valid for columns and elementary operations on rows. $\blacksquare$

Lemma 3: Using elementary operations on rows and columns, every $m\times n$ matrix can be transformed into a canonical matrix of the form:

$$\begin{bmatrix}1&0&0&\cdots&0\\0&1&0&\cdots&0\\0&0&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&0\end{bmatrix}$$

with $r$ "ones" on the diagonal ($r\le\min(m,n)$).

Proof: Gaussian elimination.$\blacksquare$

Theorem: The row rank of any matrix is equal to the column rank of the same matrix, and both are equal to the number of "ones" on the diagonal of any matrix of the above canonical form (obtained from the original matrix via elementary operations), making that number $r$ uniquely determined, and making this canonical form unique.

Proof: By Lemma 3, the original matrix can be converted to the above canonical form. By Lemma 1 and 2, the row rank and the column rank of the original matrix are equal to the row rank and column rank of this canonical matrix, respectively. However, obviously the row rank and the column rank of the canonical matrix are the same, and both are equal to $r$. Thus, the original matrix's row rank and column rank are the same, and are both equal to $r$.$\blacksquare$

The last Theorem lets us introduce the name "rank of a matrix" as the name for both the row rank and the column rank.


问:
Find the derivative of the function $F(t) = \int_{t}^{2} \sqrt{tx} \sin{\left(tx \right)}\, dx$
I try to find the derivative of following function:

$$F(t) = \int_{t}^{2} \sqrt{tx} \sin{\left(tx \right)}\, dx$$

But in the end I get an expression that contains the Fresnel integral. Is it possible to get a simpler solution?
I use the following formula:

答:
Hint

For $ t> 0 $,

Put $ \;  xt=u\;  $ to get

$$F(t)=\frac 1t\int_{t^2}^{2t}\sqrt{u}\sin(u)du$$

Thus, by FTC and chain rule,
$$F'(t)=\frac{- 1}{t}F(t)+$$
$$\frac 1t\Bigl(2\sqrt{2t}\sin(2t)-2t\sqrt{t^2}\sin(t^2)\Bigr)$$


答:
We show for $n\geq 2$ the validity of $F(n)=1$ with the help of generating functions. It is somewhat more convenient to consider $F(n+2)$ instead of $F(n)$ and we claim:

The following is valid for $n\geq 0$:
\begin{align*}
\color{blue}{\sum_{k\geq 1}\left(\binom{n+1}{2k-2}-(n+2)\binom{n}{2k-1}-\binom{n}{2k}\right)(2k-1)!!=1}\tag{1}
\end{align*}

The sum (1) is finite since $\binom{p}{q}=0$ for $p,q$ non-negative integers with $q>p$. We use the coefficient of operator $[z^q]$ to denote the coefficient of $z^q$ in a series. We consider each of the parts at the left-hand side of (1) separately and put them together at the end.

First part: We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}}&\color{blue}{\binom{n+1}{2k-2}(2k-1)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}\frac{(n+1)!}{(2k-2)!(n+3-2k)!}\,\frac{(2k )!}{(2k)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}\frac{(n+1)!}{(n+3-2k)!}\,\frac{2k(2k-1)}{2^kk!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+3}{2}\right\rfloor}\frac{(n+1)!}{(n+3-2k)!}\,\frac{2k-1}{2^{k-1}(k-1)!}\\
&=\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{2k+1}{2^kk!}\tag{2}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{1}{2^{k-1}(k-1)!}
+\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{1}{2^{k}k!}\tag{3}\\
&=\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n-1-2k)!}\,\frac{1}{2^{k}k!}
+\sum_{k=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{(n+1)!}{(n+1-2k)!}\,\frac{1}{2^{k}k!}\tag{4}\\
&\,\,\color{blue}{=(n+1)!\left([z^{n-1}]+[z^{n+1}]\right)\exp\left(z+\frac{z^2}{2}\right)}\tag{5}
\end{align*}

Comment:

In (2) we shift the index $k$ by one and start with $k=0$.

In (3) we split the sums according to $2k+1$ and cancel at the left-hand sum. Since the term with $k=0$ is zero at the left-hand sum, we start with $k=1$.

In (4) we shift the index $k$ of the left-hand sum by one.

In (5) we observe we have the coefficients $[z^{n-1}]$ and $[z^{n+1}]$ of $\exp\left(z\right)\exp\left(\frac{z^2}{2}\right)=\exp\left(z+\frac{z^2}{2}\right)$. The derivation for it is similar to that of the third part and is shown there.

Second part: We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}}&\color{blue}{\binom{n}{2k-1}(2k-1)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{n!}{(2k-1)!(n+1-2k)!}\,\frac{(2k )!}{(2k)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{n!}{(n+1-2k)!}\,\frac{2k}{2^kk!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}\frac{n!}{(n+1-2k)!}\,\frac{1}{2^{k-1}(k-1)!}\\
&=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\frac{n!}{(n-1-2k)!}\,\frac{1}{2^kk!}\\
&\,\,\color{blue}{=n![z^{n-1}]\exp\left(z+\frac{z^2}{2}\right)}\tag{6}
\end{align*}

Comment:

In (6) we observe we have a coefficient of $[z^{n-1}]$ of $\exp\left(z+\frac{z^2}{2}\right)$. The derivation for it is similar to that of the third part and is shown there.

Third part: The third part is very similar to the parts before, but it has additionally a nice twist which provides us with the constant $1$. We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}}&\color{blue}{\binom{n}{2k}(2k-1)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(2k)!(n-2k)!}\,\frac{(2k )!}{(2k)!!}\\
&=\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(n-2k)!}\,\frac{1}{2^kk!}\tag{7}\\
&\,\,\color{blue}{=n![z^{n}]\left(\exp\left(z+\frac{z^2}{2}\right)-\exp(z)\right)}\tag{8}
\end{align*}

Comment:

In (8) we have the funny intermediate result:
\begin{align*}
n![z^{n}]\exp(z)=n![z^n]\sum_{j=0}^\infty \frac{z^j}{j!}\color{blue}{=1}
\end{align*}

and the further procedure is clear: We expect that the sum of the coefficients of $\exp\left(z+\frac{z^2}{2}\right)$ cancel away leaving the constant $n![z^{n}]\exp(z)=1$. But at first we have to show the validity of (8).

Derivation of (8): We get
\begin{align*}
\color{blue}{\exp}&\color{blue}{\left(z+\frac{z^2}{2}\right)-\exp(z)}\\
&=\exp(z)\left(\exp\left(\frac{z^2}{2}\right)-1\right)\\
&=\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=1}^\infty\frac{z^{2k}}{2^kk!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{j+2k=n}\atop{j\geq 0, k\geq 1}}\frac{1}{j!}\,\frac{1}{2^kk!}\right)z^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(n-2k)!}\,\frac{1}{2^kk!}\right)\frac{z^n}{n!}}
\end{align*}
and the claim (8) follows.

Now it's (nearly) time to harvest. Let's collect the representation of the parts (5),(6) and (8) we have so far:

Final step: We obtain from (1), (5), (6) and (8)
\begin{align*}
\color{blue}{\sum_{k\geq 1}}&\color{blue}{\left(\binom{n+1}{2k-2}-(n+2)\binom{n}{2k-1}-\binom{n}{2k}\right)(2k-1)!!}\\
&=(n+1)!\left([z^{n-1}]+[z^{n+1}]\right)\exp\left(z+\frac{z^2}{2}\right)\\
&\qquad-(n+2)n![z^{n-1}]\exp\left(z+\frac{z^2}{2}\right)\\
&\qquad-n![z^{n}]\left(\exp\left(z+\frac{z^2}{2}\right)-\exp(z)\right)\\
&=\left((n+1)![z^{n+1}]-n![z^n]-n![z^{n-1}]\right)\exp\left(z+\frac{z^2}{2}\right)\color{blue}{+1}\tag{9}
\end{align*}

We want to show the coefficients in (9) cancel away. In order to do so, we consider
\begin{align*}
A(z)&=\sum_{n=0}^\infty a_nz^n=\exp\left(z+\frac{z^2}{2}\right)\\
\\
\frac{d}{dz}A(z)&=\sum_{n=1}^\infty na_nz^{n-1}\tag{10}\\
\frac{d}{dz}\exp\left(z+\frac{z^2}{2}\right)&=(1+z)\exp\left(z+\frac{z^2}{2}\right)\tag{11}\\
\end{align*}

Coefficient comparison of (10) and (11) gives:
\begin{align*}
(n+1)a_{n+1}&=\color{blue}{(n+1)[z^{n+1}]A(z)}\\
&=[z^n](1+z)\exp\left(z+\frac{z^2}{2}\right)\\
&\,\,\color{blue}{=\left([z^n]+[z^{n-1}]\right)A(z)}
\end{align*}
which shows that in (9)
\begin{align*}
\left((n+1)![z^{n+1}]-n![z^n]-n![z^{n-1}]\right)\exp\left(z+\frac{z^2}{2}\right)=0
\end{align*}
and the claim (1) follows.

Title: How prove this sum $\sum_{k=1}^{+\infty}(2k-1)!!\left(\binom{n-1}{2k-2}-n\binom{n-2}{2k-1}-\binom{n-2}{2k}\right)=1$

Body:
For any  positive integer $n>1$ show that this sum
$$F(n)=\sum_{k=1}^{+\infty}(2k-1)!!\left(\binom{n-1}{2k-2}-n\binom{n-2}{2k-1}-\binom{n-2}{2k}\right)=1$$

where if $n>m$ then $\binom{m}{n}=0$

My try:
it is clear $$F(2)=1!!(\binom{1}{1}-0-0)=1$$
and
$$F(3)=1!!(\binom{2}{0}-3\binom{1}{1}-0)+3!!(\binom{2}{2}-0-0)=1$$
$$F(4)=1!!(\binom{3}{0}-4\binom{2}{1}-\binom{2}{2})+3!!(\binom{3}{2}-0-0)=1$$
and $F(5)=1,F(6)=1$,But How to $F(n)=1,\forall n\in N^{+}$,I have note
$$(n-1)\ge 2k-2,n-2\ge 2k-1,n-2\ge 2k$$,so we have $k\le [\dfrac{n+1}{2}],k\le [\dfrac{n-1}{2}],k\le [\dfrac{n-2}{2}]$


问:
Difference between 'A open subset of B' and 'A open relative to B'
If $A\subset B$ and $B$ is a metric space. What is the difference between "$A$ open subset of $B$" and "$A$ open relative to $B$"?   </p>

答:
I just came across this definition in Principles of Mathematical Analysis and perhaps this might be helpful to someone as I had to read it over a couple of times and even though this is minor, it helped me.

A small notational adjustment to the definition of the open ball:

Definition: Let $X$ be a metric space and $p \in X$ then let $N_r^X(p) := \{q \in X \mid d(p,q) < r\}$.

Now, using that $E$ is open if every point of $E$ is an interior point of $E$ which is Definition 2.18, I prefer this way of defining openness:

Definition: A set $E \subset X$ is open in $X$ if for each $p \in E$ there is an associated $r > 0$ such that $N_r^X(p) \subset E.$

So using the above we now get:

Definition: A set $E \subset Y$ is open relative to Y if for every $p \in E$ there is an associated $r >0$ such that $N_r^Y(p) \subset E.$

Why do I find this helpful? Well, using the above and working an example you can show that an open ball isn't necessarily a "ball" anymore and I think it's very helpful to realize that quickly as it tends to highlight how our usual notation of open/closed sets are relative.

For example, take $X = \mathbb{R}, Y = [0,1] \text{ and } E = [0,\frac{1}{2})$. The open ball $N_r^Y(0)$ turns out to be $[0,r)$ and is no longer $(-r,r)$ as we tend to expect it to be. So in this specific example we can readily see that by setting $r =\frac{1}{4}$ that $N_{1/4}^Y(0) = [0,\frac{1}{4}) \subset [0,\frac{1}{2}) = E $. So the "problematic point" turns out not to be a problem at all.

Hope this may help someone!


问:
Is there a category-theoretic generalisation of function application?
(I'll be using functional programming jargon, because my category theory is not too good).

Is the notion of function application, i.e. the application f a of a function f :: a -> b to a value a :: a to yield a value b :: b, peculiar to functions, or is there a more general category-theoretic concept to generalise it?

答:
(This should be rather a comment, but I will post it as an answer as it is too long)

Besides exponential objects as mention in the comments there is another way of 'evaluating' functions/morphisms:

In a general category we replace the elements (in the set-theoretic sense) with morphisms:

Let $\mathcal{C}$ be a category and $X$ an object in $\mathcal{C}$. We will call a morphism $T\to X$ a "generalized element of shape $T$" or (in algebraic geometry) "$T$-valued point of $X$". Now if you have a morphism $f:X\to Y$ and a generalized element $a$ of shape $T$ of $X$ we get an element $f(a)$ of the same shape of $Y$, i.e. $$f(a):=f\circ a$$
In the category of sets we know that two maps (with the same source and target) are equal iff they agree on elements (in the set-theoretic sense). Now this carries over to these generalized elements. Two morphisms $f,g:X\to Y$ are equal iff $f(a)=g(a)$ for all objects $T$ and generalized elements $a$ of shape $T$ of $X$.

As an example: In the category of sets we can identify the usual elements with these generalized elements of a certain shape $T$, in this case take $T$ to be your favorite set with one element. It is similar in the category of groups: here the usual elements of a group $G$ correspond exactly to the morphisms $\mathbb{Z}\to G$, that is to the generalized elements of shape $\mathbb{Z}$ of $G$.


问:
Why is addition and multiplication associative but not exponentiation and beyond?
The most similar question I found on math.se was concerning commutativity (Why are addition and multiplication commutative, but not exponentiation?). If this ostensibly simple (but not obvious to me) question has already been discussed, please post the link and I apologize for duplicate. I am wondering too how other rules of arithmetic in basic and other algebras (distributivity, transitivity, additive and multiplicative inverses, etc) apply or don't apply to operators and any rhyme to their properties' applicability or hierarchy.

associativity of addition: (α+β)+γ ≡ (α+β+γ) := α+β+γ ≡ α+(β+γ)

associativity of multiplication: (a×b)×c := (ab)⋅c = abc ≡ a(bc)

non-associativity of exponentiation: A^B^C := (A^B)^C := $(A^B)^C$ = ${A^B}^C$ ≢ $A^{B^C}$ = $A^{(B^C)}$ := A^(B^C)

答:
The real question is why $\times$ would associate; properties like that aren't free.

As with the question of commutativity, that $\times$ inherits the property from $+$, but exponentiation doesn't repeat this, can be explained in terms of $\times$ on $\Bbb R^+$ being isomorphic to $+$ on $\Bbb R$. (Distributivity of $\times$ over $+$ extends this to $\times$ on the whole of $\Bbb R$ being commutative & associative.)


问:
Is completeness over a normed space invariant under linear homeomorphisms?
The question is contained in the title. I know that for metric spaces completeness is not invariant under homeomorphisms. What happens if we consider the same problem in the category of normed spaces with continuous linear maps as morphisms?</p>

答: If $X$ and $Y$ are normed spaces and $T : X \to Y$ is a linear homeomorphism then both $T$ and $T^{-1}$ are uniformly continuous, and hence $X$ is complete if and only if $Y$ is complete.


问:
Calculate period doubling bifurcation points
For a given logistic family $f_{\mu}(x)= \mu*x*(1-x),$ where $\mu \in [0, 4]$ and $x \in [0,1].$ This family undergoes the period doubling bifurcation. Let $\mu_{n}$ denote the value of $\mu$ where a $2^n$-cycle first appears. It is quite difficult to calculate $\mu_{n}$ for large $n$. Can someone please tell me how the values of $\mu_{n}$ will be evaluated by Matlab or Mathematica.
P.S.: If anyone has a Code/program to calculate "Bifurcation points", please provide the same.
Thank you!!!

答:
Follows a MATHEMATICA script to plot the bifurcations.

```<code>Logistic = Compile[{{\[Mu], _Real}}, ({\[Mu], #} &) /@Union[Drop[NestList[\[Mu] # (1 - #) &, .2, 300], 100]]];
f = Table[Logistic[\[Mu]], {\[Mu], 0, 4, .0006}];
ListPlot[Flatten[f, 1], PlotStyle -> {Black,Opacity[0.02], AbsolutePointSize[.001]}]
</code>```

Body:
Follows a MATHEMATICA script to plot the bifurcations.

Logistic = Compile[{{\[Mu], _Real}}, ({\[Mu], #} &) /@Union[Drop[NestList[\[Mu] # (1 - #) &, .2, 300], 100]]];
f = Table[Logistic[\[Mu]], {\[Mu], 0, 4, .0006}];
ListPlot[Flatten[f, 1], PlotStyle -> {Black,Opacity[0.02], AbsolutePointSize[.001]}]


问:
Classification groups of order 6 and 10
I have this approach to classify groups of order $6$ and $10$. Let $G$ be a group with $|G|=n$

If $n=6$, then by Sylow there is an element $g$ of order $3$. Hence $N:=\langle g\rangle$ has index $2$ and is normal. Let $x \in G-N$ such that $G=N \cup xN$. Then $x^2\in N$ (otherwise $x^2 \in xN \Rightarrow x \in N $). If $x^2 \neq e$, then $\operatorname{ord}(x)=6$ since $x^3 \in xN$ and we find the cyclic group. If $x^2=e$, then the consider conjugation map $\gamma_x :N\longrightarrow N$ with $\gamma_x(g)=xgx^{-1}$. Now the step I'm going to ask about. Since $N=\{e,g,g^2\}$ it's either $g=xgx^{-1}$ or $g^2=xgx^{-1}$. In the first case $G$ is abelian (and cyclic by the Chinese remainder theorem) and in the second case isomorphic to the dihedral group.

What if $n=10$? I can proceed exactly as in the case $n=6$, but the conjugation map could now have many different images:
i) $\gamma_x(g)=xgx^{-1}=g$ (in this case $G$ is abelian)
ii) $\gamma_x(g)=xgx^{-1}=g^2$
iii) $\gamma_x(g)=xgx^{-1}=g^3$
iv) $\gamma_x(g)=xgx^{-1}=g^4$ (in this case $G\cong D_5$)

I don't see how one can lead ii) and iii) to any contradiction.

答:
The contradiction is as follows: as $x^2=e$, then $\gamma_{x^2}=\gamma_e$ should be identity. However, it's not:

$$g=ege=x^2gx^{-2}=x(xgx^{-1})x^{-1}=xg^2x^{-1}=xg(x^{-1}x)gx^{-1}=(xgx^{-1})(xgx^{-1})=g^2g^2=g^4\ne g$$

(in case (ii), and similarly in case (iii)).

Title: Classification groups of order 6 and 10


问:
Exponential Waiting Times for Bus Arrivals
I thought I understood the Memoryless property for Exponential Distributions but I am unable to get the intuition behind the answer to this question:

You are waiting for a bus at a bus station. The buses arrive at the station according to a Poisson process with an average arrival time of 10 mins. If the buses have been running for a long time and you arrive at the bus station at a random time, what is your expected waiting time. On average, how many minutes ago did the last bus leave?

Clearly expected waiting time is 10 mins. However, I dont get this part: "On average, how many minutes ago did the last bus leave". The explanation in the book is that if you look back in time, the memoryless property implies on average the last bus arrived 10 mins ago as well. I'm basically struggling to see how the memoryless property implies this above statement.

Thanks

答:
The way I understand it is the following:

You are standing on the time point $t_1$, and let us denote the last time that a bus came to be $t_0$. (If $t_1=t_0$, it will be a trivial case, i.e. a bus is coming right now! So we can ignore that in the following discussion and assume there's no bus coming for $t\in (t_0, t_1]$)

Let \Delta t be the time since the left of the last bus, i.e. $\Delta t = t_1 - t_0$. We want to know what is the probability distribution of it. This is equivalent to asking "starting from $t_0$, what is the probability that no bus shows up up to $t=t_1$" or "starting from $t_0$,the first arrival time of a bus would be at least $\Delta t$, i.e. $\tau \geq \Delta t$". So this $\Delta t$ is exponentially distributed with the same parameter.


问:
Zero divisors in a ring $Q^2$
I was asked to show that the ring $\langle\mathbb{Q^2},\oplus,\otimes _m \rangle$, where $m \in \mathbb{N}$ and $(a,b)\oplus(c,d)=(a+c,b+d)$, and $(a,b)\otimes _m (c,d)= (ad+mbc, ac+ bd)$ contains zero divisors $\Leftrightarrow$ when $m$ equals to the square of a rational number.

As I understood I need to show that $ab = 0$ or $ ba=0,\; $  for some $a\in \mathbb{Q^2}$ ,where $b \in \mathbb{Q^2} \setminus \{0\}$

This is what I have done by far. Concluding $m=-(\frac{a}{b})^2$

<img src="https://i.stack.imgur.com/Mu2RJ.jpg" alt="This is what I have done by far. Concluding $m=-(\frac{a}{b})^2$." />

Should I operate with the modulus? or just with the coordinates according under the operations allowed in this ring?.

I want to know what I am supposed to do with the
$(a,b)\oplus(c,d)=(a+c,b+d)$ information. I got confused with the
vector $\mathbb{Q^2}$ information and vector:

答: Hint: Your ring is isomorphic to ${\mathbf Q}[x]/(x^2-m)$, which has no zero divisors if and only if $x^2-m$ is irreducible over $\mathbf Q$.

Title: Zero divisors in a ring $Q^2$


答:
Theorem: Let $X$ and $Y$ be metric spaces and let $f: X \to Y$ be a uniformly continuous homeomorphism. If $Y$ is complete then $X$ is complete.

Proof sketch: Let $x_n$ be a Cauchy sequence in $X$. By uniform continuity $f(x_n)$ is a Cauchy sequence in $Y$. By completeness, $f(x_n)$ has a limit $y$. Let $x = f^{-1}(y)$, then by continuity of $f$, $x$ is a limit of $x_n$. $\square$

Corollary: The affirmative answer to the question follows from the above theorem and the fact that every linear homeomorphism and its inverse are uniformly continuous.

Title: Is completeness over a normed space invariant under linear homeomorphisms?

Body: The question is contained in the title. I know that for metric spaces completeness is not invariant under homeomorphisms. What happens if we consider the same problem in the category of normed spaces with continuous linear maps as morphisms?</p>


问:
Show each interval (a,b) is open in the basis B
I am using Topology Without Tears by Morris

Let B={[a,b)$\in R$:a<b} B is a basis for some topology $\tau$ on R,but  not the Euclidean one.

Regardless show for each interval (a,b)is open  in (R $\tau$)

Attempted proof

Let B be a base for $\tau$.

If for each x$\in$ (a,b)$\subseteq$ U

Let $I_n=[a,b-1/n) $

Then the union of all intervals (a,b)=$\bigcup_{n\in N}I_n$

for some
index n$\in $N and $I_n\in$ B

But x $\in $U implies x$\in I_n$ for n$ \in N$
then [x,b)$\in $(a,b) and [x,b)$\in $B

Any help would be appreciated?

答: Since $a<b$, there is some $N\in\Bbb N$ such that $a+\frac1N<b$. Then$$\bigcup_{n\geqslant N}\left[a+\frac1n,b\right)=(a,b).$$


答:
Why don't you just say that

$$(a,b)=\bigcup_{n=1}^{\infty}\left[a+\frac{b-a}{n+1}, b\right)$$

Title: Show each interval (a,b) is open in the basis B

Body:
I am using Topology Without Tears by Morris

Let B={[a,b)$\in R$:a<b} B is a basis for some topology $\tau$ on R,but  not the Euclidean one.

Regardless show for each interval (a,b)is open  in (R $\tau$)

Attempted proof

Let B be a base for $\tau$.

If for each x$\in$ (a,b)$\subseteq$ U

Let $I_n=[a,b-1/n) $

Then the union of all intervals (a,b)=$\bigcup_{n\in N}I_n$

for some
index n$\in $N and $I_n\in$ B

But x $\in $U implies x$\in I_n$ for n$ \in N$
then [x,b)$\in $(a,b) and [x,b)$\in $B

Any help would be appreciated?


问:
The algebraic possibilities of the (topological) procedure of the compactification of a space
If $X$ is locally compact $K$-vector space, then $X\cup \{\infty\}$ is via the Alexandroff-compactification a compact space.

But this purely topological procedure tells me nothing about the algebraic relationship of $\infty$ to rest of the $x\in X$.    What can be said about this relationship ?

 What kind of definitions for $x+\infty$ and $\lambda\cdot \infty$ are meaningful ? And for which $x\in X$ and $\lambda \in K$ ?

What can be said about the algebraic structure of $X\cup \{\infty\}$ ?

答:
For the one-point compactification $\overline{\mathbb{C}}$ of $\mathbb{C}$, i.e. $\overline{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$, you don't even get that $(\overline{\mathbb{C}},+)$ is a group because $\infty$ has no additive inverse.

So very little of the algebraic structure of a topological space transferrs to its one-point compactification.


问:
Greatest common divisor 2

Find all integers m such that $0 < m < 1008$ and $\gcd(m, 2016-m) = 1$.

The following is my solution:

Note that $1 = \gcd(m, 2016-m) = \gcd(m, 2016)$.

Since $2016$ is even and divisible by $3, 7, 9, 21, 63$ and $\gcd(m, 2016) = 1$, so $m$ is odd and $m\neq 3k, 7l$, where $k, l\in \mathbb Z,$ and $k, l$ are odd numbers.

Since $0 < m< 1008,$ we deduce that $1\leq k\leq 335, \ 1\leq l\leq 143$ and $k, l$ are odd.

We set $$A = \{3k|\ k\in\mathbb Z,\ 1\leq k\leq 335,\ k \mbox{ odd}\}$$
and $$B = \{7l|\ l\in\mathbb Z,\ 1\leq l\leq 143,\ l \mbox{ odd}\}.$$
Then, we have $|(A\cup B)\setminus (A\cap B)| = 168 + 72 - 48 = 192.$

On the other hand, from $1$ to $1008$ we have $\frac{1007-1}{2} + 1 = 504$ odd numbers. Combining these data, we conclude that there are $504 -192 -24 = 288$ positive integers $m$.

答:
Commenting on your solution process, I think you are making life a little more difficult than it needs to be in this instance by separating out odd numbers. You also end up with a wrong answer.

Here it makes more sense to simply take the three prime factors of $2016$ - $2,3,$ and $7$ - and count all the multiples of these numbers up to $1008$, and then by subtraction find how many numbers in the range are coprime to $2016$.

It's true here that $\gcd(m, 2016{-}m) = \gcd(m, 2016)$, but as observed it might be worth another line of justification for a well-paved proof. For solving the challenge it's not so important.

You correctly identify that we need to avoid multiples of $2,3,$ and $7$ in the range

You miss that you have twice subtracted numbers that are multiples of $3\cdot 7 = 21$, so your solution comes out low.

It's perhaps easier to leave $1008$ in the range because it will be eliminated anyway as not coprime to $2016$.

As a hand-worked alternative method of getting the count, you could note that the numbers coprime to $(2,3,7)$ will repeat on a block of $2\cdot3\cdot 7=42$ and in fact that pattern is itself symmetric about $21$.

Title: Greatest common divisor 2