问:
A property of sum of independent binomial random variables
Define independent random variables $$X = B(n_1,p), Y = B(n_2,p), Z = B(n_1+n_2+1,p).$$ Given that $$Pr[X \geq a] > c, Pr[Y \geq b] > c,$$ where a and b are positive integers, and c is a constant in (0,1), can we show that $$Pr[Z \geq a+b] > c?$$ I've tried using convolution, but the resulting expression is too hard to simplify and produce the desired result.

答:
The answer is no. If $n_1=n_2=2, p=\frac 12, a=b=2, c=0.24$ then

$$
\Pr(X\geqslant a)=\Pr(Y\geqslant b)=\frac 1 4>c.
$$

But
$$
\Pr(Z\geqslant a+b)=\frac{6}{32} <c.
$$


问:
Can $p$ divide both $a_1 − a_2$ and $a_1 + a_2$?
If $a_1^2 ≡ a_2^2 \pmod p$, then $p$ divides $a_1^2−a_2^2$, so $p$ divides the product $(a_1 − a_2)(a_1 + a_2)$.

I read in a chapter related to  quadratic residues and nonresidues that

unique prime factorization now tells us that $p$ divides $a_1 − a_2$ or $p$ divides $a_1 + a_2$, and so either $a_1 ≡ a_2 \pmod p$ or $a_1 ≡ −a_2 \pmod p$.

Why? It could be the case where both $p$ divides $a_1 − a_2$ and  $p$ divides $a_1 + a_2$.

答: Yes, this could be the case (for instance if $a_1=a_2=p$). But also, they have not said that this cannot be the case. That's how "either ... or ..." is conventionally interpreted in mathematics: unless explicitly stated, that phrase does not exclude the possibility of both.

ContentLicense: CC BY-SA 4.0


问:
PGF of negative multinomial expansion
I have found the formula for the Probability Generating Function of negative multinomial distribution in Definition 8.1 of this chapter (https://onlinelibrary.wiley.com/doi/abs/10.1002/9781118445112.stat01250). It is as follows. $$G_x(\boldsymbol{t}) = \left(\mathcal{Q} - \sum^k_{i=1} P_{it_i}\right)^{-n}$$, where $n > 0$, $P_i > 0(i = i, \ldots, k)$, and $\mathcal{Q} - \sum^k_{i=1} P_{i} = 1$. I am still confused with the following:

What do $\mathcal{Q}$ and $P_i$ mean?

How can I use this to compute the coefficient of the terms in the series expansion of $\left(x_1, x_2, \ldots, x_m \right)^{-n}$, where, $m > 0$?

答: I am using this space to post the content of Section 2.4 which builds up to the equation I have used in the question.

Title: PGF of negative multinomial expansion


问:
Evaluating $\lim_\limits{x\to 0}(e^{5/x}-6x)^{x/2}$. Is my method correct?
I have to solve the following limit which is in indeterminate form. \begin{equation}\lim_\limits{x\to 0}(e^{\frac{5}x}-6x)^{\frac{x}{2}} \end{equation}
So to solve this limit I attempted to do the following, I ended up making use of a substitution, however, I am not sure I did it right... Please let me know what you think (by the way my TI-nspire CX CAS already gave me the solution of $e^{\frac52}$.)

Work

\begin{align} L &=e^{\lim_\limits{x\to 0^+}(\frac{1}2 x\ln(e^{\frac{5}x-6x}))} \\ u&=\frac{1}x \\ L&=e^{\lim_\limits{u\to+\infty} \frac{1}{2}\frac{\ln(e^{5u}-6\frac{1}u)}{u}} \end{align}
From here I then did the following:
\begin{align} L&=e^{\lim_\limits{u\to+\infty} \frac12 \frac{5\frac{e^{5u}-\frac{6}{u^2}}{e^{5u}-\frac{6}{u}}}1}\end{align}
From Here I multiplied and divided by $\frac{\frac{1}{e^{5u}}}{\frac{1}{e^{5u}}}$
Leading me to this:
\begin{equation}L=e^{\frac{5}{2}\lim_\limits{u\to+\infty}\frac{1-\frac{6}{u^2e^{5u}}}{1-\frac{6}{ue^{5u}}}} \end{equation}
From Here I deduced that the answer was $L=e^{\frac{5}2}$, were my steps wrong, is infinity times infinity indeterminate and it requires further calculation please let me know.

答:
Assuming $x \to 0^+$, your way is a little bit long and can be simplified but it is fine indeed at the end we obtain

$$\frac{1-\frac{6}{u^2 \ e^{5u}}}{1-\frac{6}{u \ e^{5u}}} \to 1$$

since $\frac{6}{u^2 \ e^{5u}}\to 0$ and $\frac{6}{u \ e^{5u}} \to 0$.

As an alternative, more directly we have

$$(e^{\frac{5}x}-6x)^{\frac{x}{2}}=(e^{\frac{5}x})^{\frac{x}{2}}\left(1-\frac{6x}{e^{\frac{5}x}}\right)^{\frac{x}{2}} =e^\frac52\left(1-\frac{6x}{e^{\frac{5}x}}\right)^{\frac{x}{2}} \to e^\frac52 \cdot 1 = e^\frac52$$

where it is important to note that the following

$$\lim_\limits{x\to 0^+}\left(1-\frac{6x}{e^{\frac{5}x}}\right)^{\frac{x}{2}}=1^0=1$$

is not an indeterminate form.

Note that for $x\to 0^-$ by $y=-x \to 0^+$ we obtain

$$\lim_\limits{x\to 0^-}(e^{\frac{5}x}-6x)^{\frac{x}{2}}=\lim_\limits{y\to 0^+}\frac1{\left(6y+\frac1{e^{\frac{5}y}}\right)^{\frac{y}{2}}}=1$$

since

$$\left(6y+\frac1{e^{\frac{5}y}}\right)^{\frac{y}{2}}=\left(6y\right)^{\frac{y}{2}}\left(1+\frac1{6ye^{\frac{5}y}}\right)^{\frac{y}{2}} \to 1\cdot(1+0)^0=1$$


问:
Cannot find the "event" in a probability problem that asks for "at least"
The unemployment of a city is of 10%. If we take 100 people what is the probability of at leat 15 be unemployed. I am doing my homework and I am having difficulty in this one. I am asked to find the "sample", "success", "failure" and "event". Those are the terms we use in portuguese transtalted to english. I hope they are the same. Using the binomial formula, the sample would be 100, the success would be 0,1 and the failure would be 0,9. I am having a hard time to figure out the "event". What the event would be? I am using this calculator

calculator

I put 100 on the first input, 10 on the second, 90 on the third and the fourth for the event and don't know. I get wrong results. I know it because my homework has the answers. I just have to find the variables. The answer is 7,26

答:
Manually, boring but not impossible,  you can calculate

$$1-\sum_{k=0}^{14}\binom{100}{k}0.1^k\cdot 0.9^{100-k}\approx 7.26\%$$

If you want a quick approximation you can use the Central limit Therorem finding immediately

$$\mathbb{P}[X\geq 15]=1-\Phi\Bigg[\frac{14.5-10}{\sqrt{9}}\Bigg]=1-\Phi(1.5)\approx 6.68\%$$


答:
You are correct, another way of writing this would be by using equivalences, since $6x=\underset{0}{o}\left( e^{\frac{5}{x}}\right)$, you get :

$$ e^{\frac{5}{x}}-6x \underset{x\rightarrow0}{\sim}e^{\frac{5}{x}}. $$ So you get :

$$(e^{\frac{5}{x}}-6x)^{\frac{x}{2}} \underset{x\rightarrow0}{\sim} e^{\frac{5}{2}},$$
so :
$$\lim_{x\rightarrow0}(e^{\frac{5}{x}}-6x)^{\frac{x}{2}} = e^{\frac{5}{2}}.$$

Title: Evaluating $\lim_\limits{x\to 0}(e^{5/x}-6x)^{x/2}$. Is my method correct?

Body:
I have to solve the following limit which is in indeterminate form. \begin{equation}\lim_\limits{x\to 0}(e^{\frac{5}x}-6x)^{\frac{x}{2}} \end{equation}
So to solve this limit I attempted to do the following, I ended up making use of a substitution, however, I am not sure I did it right... Please let me know what you think (by the way my TI-nspire CX CAS already gave me the solution of $e^{\frac52}$.)

Work

\begin{align} L &=e^{\lim_\limits{x\to 0^+}(\frac{1}2 x\ln(e^{\frac{5}x-6x}))} \\ u&=\frac{1}x \\ L&=e^{\lim_\limits{u\to+\infty} \frac{1}{2}\frac{\ln(e^{5u}-6\frac{1}u)}{u}} \end{align}
From here I then did the following:
\begin{align} L&=e^{\lim_\limits{u\to+\infty} \frac12 \frac{5\frac{e^{5u}-\frac{6}{u^2}}{e^{5u}-\frac{6}{u}}}1}\end{align}
From Here I multiplied and divided by $\frac{\frac{1}{e^{5u}}}{\frac{1}{e^{5u}}}$
Leading me to this:
\begin{equation}L=e^{\frac{5}{2}\lim_\limits{u\to+\infty}\frac{1-\frac{6}{u^2e^{5u}}}{1-\frac{6}{ue^{5u}}}} \end{equation}
From Here I deduced that the answer was $L=e^{\frac{5}2}$, were my steps wrong, is infinity times infinity indeterminate and it requires further calculation please let me know.


问:
definition of degree matrix of directed graph?
im sorry if this is a trivial question. As the title stated, how is the degree matrix of a directed graph defined?
I looked up wikipedia of degree matrix but there was no defitnition of it. Instead they reffered the indegree and outdegree. I was wondering if we can represent a degree matrix of directed graph. Can someone help me with this?</p>

答: You just put either out- or in-degrees (whatever you like/need more) on the main diagonal.  For example in a graph with vertices $V = \{1, 2, 3\}$ and oriented edges $(1, 2)$ and $(1, 3)$ and $(2, 3)$ the out-degree matrix is $$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ and the in-degree matrix is $$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}.$$ Note that you also can treat your graph as undirected (discard the orientation of every edge), in which case the degree matrix will be $$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}.$$ Which one you need depends on the particular problem you are facing, however I'm afraid I can't tell without more context.


问:
Trigonometry - Addition and subtraction theorem
If $\theta$ and $\phi$ are angles between $0°$ and $90°$, and $\sin \theta=3/5$ and $\tan \phi=7/24$, find without the use of a calculator, the value of each of the following:

a. $\sin(\theta−\phi)$

b. $\cos(\theta+\phi)$

c. $\tan(\theta−\phi)$

Please help, homework!

答:
$\sin\theta=\frac35$,$\cos \theta=\frac45$,$\tan\theta=\frac34$ and $\sin\phi=\frac7{25}$,$\cos \phi=\frac{24}{25}$,$\tan\phi=\frac7{24}$
(using Pythagoras therom we can find out any trigonometric ratio.  )

part a.

$$\sin(\theta-\phi)$$
this formula will be useful:$\sin(A-B)=\sin A.\cos B-\cos A.\sin B$
$$\sin(\theta-\phi)$$
$$\sin\theta.\cos\phi-\cos\theta.\sin\phi$$
$$(\frac{3}{5}).(\frac{24}{25})-(\frac{4}{5}).(\frac{7}{25})$$
$$\frac {44}{125}$$

part b.

$$\cos(\theta+\phi)$$
this formula will be useful:$\cos(A+B)=\cos A.\cos B-\sin A.\sin B$
$$\cos(\theta+\phi)$$
$$\cos \theta.\cos \phi-\sin \theta.\sin \phi$$
$$(\frac{4}{5}).(\frac{24}{25})-(\frac{3}{5}).(\frac{7}{25})$$
$$\frac35$$

part c.

$$\tan(\theta-\phi)$$
this formula will be useful:$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$
$$\tan(\theta-\phi)$$
$$\dfrac{\tan \theta-\tan \phi}{1+\tan \theta.\tan \phi}$$
$$\dfrac{\dfrac34-\dfrac{7}{24}}{1+(\dfrac34)(\dfrac{7}{24})}$$
$$\dfrac{\dfrac{11}{24}}{\dfrac{117}{96}}$$
$$(\dfrac{11}{24})(\dfrac{96}{117})$$
$$\dfrac{44}{117}$$

ContentLicense: CC BY-SA 3.0


问:
Double integral with a change of variables where $u = \frac{y}{x^3}$ and $v = \frac{y}{x}$.
Sketch the planar region D which is bounded by the following four curves and located in the positive quadrant: $y = x^3$, $y = 10x^3$, $y = x$ and $y = 2x$.
Use transformation given by $u=\frac{y}{x^3}$ and $v=\frac{y}{x}$.

Evaluate: $\iint_D \frac{2y}{x^5} \,dx \,dy $

So far I have $x = \frac{u^\frac{1}{4}}{v^\frac{1}{4}}$ and $y = v\frac{u^\frac{1}{4}}{v^\frac{1}{4}}$. Using this to calcuate the Jacobian matrix:

$$\begin{equation}
   \begin{vmatrix}
   \frac{1}{4v^\frac{1}{4}u^\frac{3}{4}} & \frac{u\frac{1}{4}}{4v^\frac{3}{4}}  \\
   \frac{v\frac{3}{4}}{4u^\frac{3}{4}u^\frac{3}{4}} & \frac{3u\frac{1}{4}}{4v^\frac{3}{4}}  \\
   \end{vmatrix}
\end{equation}
= \frac{3-v^\frac{1}{2}}{4v^\frac{1}{2}u^\frac{1}{2}}$$

I used my Jacobian to calculate the integral and I also changed the region D with variables $u$ and $v$ however I ended up with a negative answer. Can anyone check if my Jacobian matrix is correct? I think this is where I have gone wrong.

答:
Solving wrt to $u,v$ we get
$$\left\{x=\sqrt{\frac{v}{u}},y=v \sqrt{\frac{v}{u}}\right\}$$
Jacobian is
$$J=\left(
\begin{array}{cc}
 -\frac{\sqrt{\frac{v}{u}}}{2 u} & -\frac{1}{2} \left(\frac{v}{u}\right)^{3/2} \\
 \frac{\sqrt{\frac{v}{u}}}{2 v} & \frac{3 \sqrt{\frac{v}{u}}}{2} \\
\end{array}
\right)$$
and $\det J=-\frac{v}{2 u^2}$

Substituting in the function we get
$$\iint\limits_{D'} \frac{2 u^2}{v}\cdot \left(-\frac{v}{2 u^2}\right)\,dudv=\\=
-\iint\limits_{D'}\,dudv$$


问:
Converting $3t\cos t\sin3t+2\cos t\cos3t-t\sin t\cos3t-6\cos^2t$ to $4\cos^2t\; ( t\sin2t+\cos2t-2)$
How to alignment.
$$f(t)=3t \cos(t)\sin(3t)+2\cos(t)\cos(3t)-t\sin(t)\cos(3t)-6\cos^2(t)$$
are alternate forms
$$f(t)=4 \cos^2(t)\; ( t \sin(2 t)+ \cos(2 t)-2)$$
Please show me the steps.

答:
First, we can use the angle addition identities:

$$\sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta)$$

$$\cos(\theta + \phi) = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi)$$

From this, we get:

$$\sin(3\theta) = \sin(2\theta + \theta) = \sin(2\theta)\cos(\theta) + \sin(\theta)\cos(2\theta) = 2\sin(\theta)\cos^{2}(\theta) + \sin(\theta)\cos^{2}(\theta) - \sin^{3}(\theta)$$

$$\cos(3\theta) = \cos(2\theta + \theta) = \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) = \cos^{3}(\theta) - \sin^{2}(\theta)\cos(\theta) - 2\sin^{2}(\theta)\cos(\theta)$$

$$\sin(3\theta) = 3\sin(\theta)\cos^{2}(\theta) - \sin^{3}(\theta)$$

$$\cos(3\theta) = \cos^{3}(\theta) - 3\sin^{2}(\theta)\cos(\theta)$$

(In deriving these, we use the double angle identities $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ and $\cos(\theta) = \cos^{2}(\theta) - \sin^{2}(\theta)$, which can be found in the same way.) Can you take it from here?

If you're still stuck:

 Using the triple-angle identities:

 $$f(t) = 3t\cos(t)\sin(3t) + 2\cos(t)\cos(3t) - t\sin(t)\cos(3t) - 6\cos^{2}(t)$$

 $$f(t) = 9t\sin(t)\cos^{3}(t) - 3t\sin^{3}(t)\cos(t) + 2\cos^{4}(t) - 6\sin^{2}(t)\cos^{2}(t) - t\sin(t)\cos^{3}(t) + 3t\sin^{3}(t)\cos(t) - 6\cos^{2}(t)$$

 $$f(t) =8t\sin(t)\cos^{3}(t) + 2\cos^{4}(t)- 6\sin^{2}(t)\cos^{2}(t) - 6\cos^{2}(t)$$

 Using $\sin^{2}(t) = 1-\cos^{2}(t)$:

 $$f(t) = 8t\sin(t)\cos^{3}(t) + 2\cos^{4}(t) - 6(1-\cos^{2}(t))\cos^{2}(t) - 6\cos^{2}(t)$$

 $$f(t) = 8t\sin(t)\cos^{3}(t) + 8\cos^{4}(t) - 12\cos^{2}(t)$$

 $$f(t) = 4\cos^{2}(t)(2t\sin(t)\cos(t) + 2\cos^{2}(t) - 3)$$

 Using $\cos^{2}(t) = 1-\sin^{2}(t)$:

 $$f(t) = 4\cos^{2}(t)(2t\sin(t)\cos(t) + \cos^{2}(t) - \sin^{2}(t) + 1 - 3)$$

 Using the double angle identities:

 $$\boxed{f(t) = 4\cos^{2}(t)(t\sin(2t) + \cos(2t) - 2)}$$


问:
Integration of the function $f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$
Let f be the function on $[-\pi,\pi]$ given by $f(0)=9$ and $f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$ for $x\ne 0$. Find the value of $\frac{2}{\pi }\int\limits_{ - \pi }^\pi  {f\left( x \right)dx} $

My approach is as follow. The function is continuous at $x=0$. Hence it is contionuos for $[-\pi,\pi]$

$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}} \times 9 \times \frac{{\left( {\frac{x}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}} = 9$

$\frac{x}{2}=\theta$

$\frac{2}{\pi }\int\limits_{ - \pi }^\pi  {f\left( x \right)dx} $

$\int\limits_{ - \pi }^\pi  {\frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}}dx}  \Rightarrow 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta  \right)}}d\theta }  \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta  \right)}}d\theta }  $

$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta  + \theta } \right)}}{{\sin \left( \theta  \right)}}d\theta }  \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta  \right) + \cos \left( {8\theta } \right)\sin \left( \theta  \right)}}{{\sin \left( \theta  \right)}}d\theta } $

$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta  \right)}}{{\sin \left( \theta  \right)}} + \cos \left( {8\theta } \right)} \right)d\theta } $

$\int\limits_0^{\frac{\pi }{2}} {\cos \left( {8\theta } \right)d\theta }  = 0$

How so I proceed form here as expansion of $sin(8\theta)$ is cumbersome.

Is their any formula of converting $sin(n\theta)$ in terms of $sin(\theta)$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{{2 \over \pi}\int_{-\pi}^{\pi}
{\sin\pars{9x/2} \over \sin\pars{x/2}}\,\dd x} =
{8 \over \pi}\int_{0}^{\pi/2}
{\sin\pars{9x} \over \sin\pars{x}}\,\dd x
\\[5mm] = &\
{8 \over \pi}\,\Im\int_{0}^{\pi/2}
{\expo{9x\ic} - 1\over \sin\pars{x}}\,\dd x
\\[5mm] = &\
\left.{8 \over \pi}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}
{z^{9} - 1 \over \pars{z - 1/z}/\pars{2\ic}}
\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.{16 \over \pi}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\
{1 - z^{9} \over 1 - z^{2}}
\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\end{align}
I'll "close" the integration around a quarter circle, of radius $\ds{1}$, in the complex plane first quadrant:
\begin{align}
&\bbox[5px,#ffd]{{2 \over \pi}\int_{-\pi}^{\pi}
{\sin\pars{9x/2} \over \sin\pars{x/2}}\,\dd x} =
-{16 \over \pi}\,\Im\int_{1}^{0}\
{1 - \ic y^{9} \over 1 + y^{2}}\,\ic\,\dd y
\\[5mm] = &\
{16 \over \pi}\int_{0}^{1}{\dd y \over 1 + y^{2}} =
{16 \over \pi}\,{\pi \over 4} = \bbx{4} \\ &
\end{align}


问:
Limits, if the limit exists or not

For the following graph there are some questions, i did couple of them but some of them are saying wrong, these are the wrong ones.

```<code>lim x→−1 F(x) = 4

lim x→1 F(x) = 3

lim x→3 F(x) = Does not exist

F(−1) = DNE

F(1) = DNE
</code>```

答:
$F(a)$ refers to what the function is doing when $x$ is EXACTLY equal to $a$.

$\lim_{x\to a} F(x)$ refers to how the function behaves at values of $x$ where $x$ is near $a$ but where $x$ does NOT equal $a$.

So.....

ONE and FOUR

at $x=-1$ then function "jumps off its track" and has that dark red value at $4$.    But when $x$ is close to $-1$ but $x$ is not equal to $-1$ we see the function is on a  track and is approaching the value of $3$ so...

$F(-1)= 4$

$\lim_{x\to -1} = 3$.

TWO and FIVE

and $x = 1$ the function takes on the value of $3$. At $x$ near $1$ but $x < 1$ we see the function was approaching $2$ but "ripped itself" and jump to $3$ at $x = 1$.  And the values of $x$ near $1$ but $x > 1$ we see the function is approaching $3$.  As the function is approaching one value for $x$ near but less than $1$, and is  approaching a different value foor $x$ near but more than $1$, there isn't any one consistent value that the function approaches when it is near but not equal to $1$.  So

$\lim_{x\to 1} F(x)$ does not exist.

$F(1) = 3$.

By the way, we don't ever say something "$= $ does not exist".  "Does not exist" is a statement that something doesn't exist. It's not a number or a value.

THREE (and a non-asked for SIX)

At $x=3$ there is no value of the function.  There's a hole in the graph.  So the value $F(3)$ does not exist.  If we look at the values of $x$ near but not equal to $3$ we see the function approaches the value of $2$. So

$\lim_{x\to 3} F(x) = 2$.

$F(3)$ does not exist.

Title: Limits, if the limit exists or not


问:
Win-loss ratios, and averages of them
There are many games where player performance is ranked by ratio of wins to losses.

I mistakenly had it in my head that the average such ratio must be 1. This is not the case, as can be found from a simple example:

```<code>W   L   W:L
-----------
7   2   3.5
3   3   1
3   6   0.5
4   2   2
2   2   1
1   5   0.2
-----------
sum sum avg
20  20  1.37
</code>```

My question is this: assuming that the numbers come from a closed system, and so the total number of wins is always equal to the total number of losses, is there some way to get a result as "obvious" as I was incorrectly expecting? Perhaps there's some way to scale such ratios, whereby by averaging them you'll always get a constant?

Or to get such a result do you always need to know the absolute number of wins and losses per person? For example, I could award +1 point for a win and -1 for a loss, and then the average score is always going to be 0.

答:
The answer to this explicit question is "no"

is there some way to get a result as "obvious" as I was incorrectly
expecting?

if you want to use W:L ratios. To see that, imagine a tournament in which one player wins all their games. Then their W:L ratio has a zero denominator, which is trouble.

If each of $n$ participants plays exactly $g$ games then the average of the winning ratios $W_i/g$ for the $n$ players will be $1/n$. So for example with four players each player wins $25/%$ of their games - on average.

Title: Win-loss ratios, and averages of them


问:
Constructing a polynomial bump function

Proposition: Suppose $f$ is continuous and $\int_a^bf(x)x^ndx = 0$ for all $n$. Then $f$ is zero on $[a,b]$.

This can be proven by uniformly approximating $f$ with polynomials via the Weierstrass theorem. Before I heard this answer, I was trying to prove it by constructing a sort of polynomial bump function that would peak where $f(x)$ was nonzero, and be small outside that region.

Suppose $f \neq 0$ on $[a,b]$ and suppose WLOG that $f(y)>0$. Choose $(c,d)$ such that $|f(x)|>|f(y)|/2$ on $(c,d)$ and choose $(\gamma,\delta)\supseteq (c,d)$ such that $f(x)>0$ on $(\gamma, \delta)$.

I want a polynomial that is $\geq 1$ on $(c,d)$ and uniformly small on $[a,b]\setminus (\gamma, \delta)$. Is such possible? How could I construct it?

答:
This is more of a note than an answer, but not exactly a comment either.  If we expand on @marty cohen's post.  Let $X$ be a convenient space and define the $C^1(X,\mathbb{R})$ bump function as:
$$
f_c(x)\triangleq (\|x\|^2-c)^2(\|x\|^2+c)^2 I_{\|x\|^2<c};
$$
for some $0<c<\infty$.  Then $f_c$ has the added benefit of taking values in $(0,1]$ and being supported in $Ball(x,c)$.  In particular, when $X=\mathbb{R}^d$ then $f_c$ is compactly-supported.

Title: Constructing a polynomial bump function


问:
Equivalent condition for being a Jacobson ring
(Atiyah-Macdonald, Ex. 5.25)

Let $A$ be a ring. Show that the following are equivalent:

  i) $A$ is a Jacobson ring;

  ii) Every finitely generated $A$-algebra $B$ which is a field is finite over $A$.

I'm trying to solve $i)\Rightarrow ii)$. Almost done, but I need some help. My trial:

Let $f:A\to B$, $B$ a finitely generated $A$-algebra which is a field. Since $f(A)$ is Jacobson and $A$-algebra/module actually means that $f(A)$-algebra/module, we can assume that $A \subseteq B$. Pick $s\in A$ as in Ex. 5.21 (see below). Because $A$ is Jacobson, $J(A)=0$ so we can find a maximal ideal $m$ not containing $s$. Let $k:=A/m$ and $f:A \twoheadrightarrow k \hookrightarrow \bar{k}$, then $f(s) \neq 0$, so $f$ extends to $g:B\to \bar{k}$. Because $B$ is a field, $g$ is either injective or trivial. But $g(s)=f(s) \neq 0$ so $g$ is injective and $B \simeq g(B)$. Suppose $B=A[z_1,\cdots,z_m]$, then $g(B)$ $=$ $g(A)[g(z_1),\cdots,g(z_m)]$ $=$ $k[g(z_1),\cdots,g(z_m)]$. Each $g(z_i)$ is in $\bar{k}$ hence integral over $k$, so $g(B)$ is finitely generated $k$-module.

Now the question:

1. How can I deduce that $B$ is a finitely generated $A$-module?

2. $g(A)=f(A)=k$ but $g$ is injective. How can it be possible? I think that $A$ strictly(?) contains $k$.

(Ex. 5.21) Let $A$ be a subring of an integral domain $B$ such that $B$ is finitely generated over $A$. Show that there exists $s \neq 0$ in $A$ such that, if $\Omega$ is an algebraically closed field and $f:A\to \Omega$ is a homomorphism for which $f(s) \neq 0$, then $f$ can be extended to a homomorphism $B\to \Omega$.

答: Since $A$ is assumed to be a subring of $B$, we have $A\hookrightarrow B \hookrightarrow \bar{k}$, so the composition is injective. But the composition is nothing but $A\twoheadrightarrow k$. It follows that $A=k$ and the maximal ideal $m$ is $0$. Now $B=k(z_1, \ldots, z_m)$ with all $z_i$ algebraic over $k$, hence a finite extension of $k=A$.  </p>


问:
Regular expression for all strings with even length and not containing “ab” and “ba”
I have a problem with this question:

we have a language with alphabet {a, b, c}, all strings in this language have even length and does not contain any substring "ab" and "ba" for example these strings acceptable: "accb", "aa", "bb", "bcac", and these strings not acceptable "ccab", "abca", "cabc" and so on.. actually I think this is not regular language but I can't prove it, anyone can help me by give a regular expression for this language or prove that this language is not regular? or any help that I can figure out how I must think about it.

i tryed (aa)(cc)(bb)+(bb)(cc)(aa)+(ac+ca)+(bc+cb) and ((a+c)(a+c))+((b+c)(b+c))+(ac+bc)(ca+cb)+(ca+cb)(ac+bc) but this not work

答:
We need $7$ non-terminals.  $S$ is the initial state, $T,\ U,$ and $V$ all indicate that an odd number of characters have been read, and that the last character read was $a$, $b$, or $c$, respectively.  Similarly, $X,\ Y$, and $Z$ indicate that an even number of characters have been read, and that the last character read was $a$, $b$, or $c$, respectively. I use $\lambda$ for the empty string.  $S,\ X,\ Y$, and $Z$ are accepting states, so we have the productions $$S\to\lambda\\  X\to\lambda\\ Y\to\lambda\\ Z\to\lambda\\ $$
Also we, have $$S\to aT\\S\to bU\\S\to cV\\T\to aX\\T\to cZ\\U\to bY\\U\to cZ\\V\to aX\\V\to bY\\V\to cZ\\$$

I trust you see how I arrived at these.  If we get a $b$ in state $T$ or an $a$ in state $U$ then we've seen a forbidden substring, so there are no productions corresponding to these possibilities.

It remains to determine the productions with $X,\ Y$ or $z$ on the left-hand side.  I leave it to you to supply those.

EDIT Reading this over, I see that I've slipped into the language of automata at times.  I hope it's obvious what I mean.  If not, ping me.

EDIT

When I wrote this, I though you were mainly concerned with whether or not the language is regular, but it's apparent from your comments that you're mostly concerned with getting a regex.  Remember that the complement of a regular language is regular.  So we want the complement of the union of

```<code>-set of strings containing <span class="math-container">$ab$</span>
-set of strings containing <span class="math-container">$ba$</span>
-set of strings with an odd number of characters
</code>```

So, if you can make regexes for these three languages, you can combine them to construct the regex you seek.

EDIT

The previous edit is wrong.  The discussion prior to theorem $4.5$ in Hopcroft, Motwani and Ullman says that you can't do it that way.  You have to convert the regex to a DFA, construct a DFA that recognizes the complement, and then convert the new DFA to a regex.  I think any regex for this language will be exceedingly long and complicated.

Body:
We need $7$ non-terminals.  $S$ is the initial state, $T,\ U,$ and $V$ all indicate that an odd number of characters have been read, and that the last character read was $a$, $b$, or $c$, respectively.  Similarly, $X,\ Y$, and $Z$ indicate that an even number of characters have been read, and that the last character read was $a$, $b$, or $c$, respectively. I use $\lambda$ for the empty string.  $S,\ X,\ Y$, and $Z$ are accepting states, so we have the productions $$S\to\lambda\\  X\to\lambda\\ Y\to\lambda\\ Z\to\lambda\\ $$
Also we, have $$S\to aT\\S\to bU\\S\to cV\\T\to aX\\T\to cZ\\U\to bY\\U\to cZ\\V\to aX\\V\to bY\\V\to cZ\\$$

I trust you see how I arrived at these.  If we get a $b$ in state $T$ or an $a$ in state $U$ then we've seen a forbidden substring, so there are no productions corresponding to these possibilities.

It remains to determine the productions with $X,\ Y$ or $z$ on the left-hand side.  I leave it to you to supply those.

EDIT Reading this over, I see that I've slipped into the language of automata at times.  I hope it's obvious what I mean.  If not, ping me.

EDIT

When I wrote this, I though you were mainly concerned with whether or not the language is regular, but it's apparent from your comments that you're mostly concerned with getting a regex.  Remember that the complement of a regular language is regular.  So we want the complement of the union of

-set of strings containing $ab$
-set of strings containing $ba$
-set of strings with an odd number of characters

So, if you can make regexes for these three languages, you can combine them to construct the regex you seek.

EDIT

The previous edit is wrong.  The discussion prior to theorem $4.5$ in Hopcroft, Motwani and Ullman says that you can't do it that way.  You have to convert the regex to a DFA, construct a DFA that recognizes the complement, and then convert the new DFA to a regex.  I think any regex for this language will be exceedingly long and complicated.


问:
Extension of irreducible ideal of a Noetherian ring is irreducible of rings of fractions
Let $S$ be a multiplicatively closed subset of the commutative Noetherian ring $R$ and $f:R \to {S^{ - 1}}R$ denote the natural ring homomorphism.

Let $I$ be an irreducible ideal of $R$ for which $S \cap I = \emptyset $. Show that $I^e$ is an irreducible ideal of ${S^{ - 1}}R$.

(Where $I^e$ is extension ideal of $I$ and $I$ is irreducible if and only if $I ≠R$ and, whenever $I=I_1∩I_2$, with $I_1$ and $I_2$ ideals of $R$, then $I=I_1$ or $I=I_2$)

I have just proved that if $I$ is an irreducible ideal of ${S^{ - 1}}R$ then $I^c$ is an irreducible ideal of $R$. I don't know how to use the Noetherian condition to prove the opposite side.

答:
Suppose that $I^e = J_1 \cap J_2$ for distinct proper $S^{-1}R$-ideals $J_i$, then $I^{ec} = (J_1 \cap J_2)^c = J_1^c \cap J_2^c$.  Note that $J_1^c$ and $J_2^c$ are distinct proper ideals of $I$ because for localization maps, $J^{ce} = J$ (more generally this is true of flat epimorphisms).

Thus if we demonstrate that $I^{ec} = I$ under the conditions on $R$ and $I$, then we will have showed the desired implication "$I^e$ reducible $\implies$ $I$ reducible."

Noetherian rings have primary decompositions, so $I$ being irreducible implies its being primary.  All that remains is to understand how primary ideals behave under localization.  The answer is basically that they behave analogously to prime ideals, as well as one could hope for.

Lemma Let $I$ be a primary ideal of a commutative ring $R$ and $R \rightarrow S^{-1}R$ localization at a multiplicative set $S$.  Extension and contraction of ideals induces a one-to-one correspondence between primary ideals of $S^{-1}R$ and primary ideals of $R$ disjoint from $S$.  In particular if $I$ is a primary ideal of $R$ disjoint from $S$ then $I^e$ is primary ideal of $S^{-1}R$ and $I^{ec} = I$.

I leave the details of this to you.

Note that we did not need the full power of being Noetherian here.  All we used was the implication "irreducible $\implies$ primary" which holds for Noetherian rings, and more generally Laskerian rings (in which primary decompositions exist).


问:
How to solve $y'+2 \sin \frac{y}{2} \cos \frac{y}{2} +2x \cos^2 \frac{y}{2} = 0 $
I have to solve the following ODE:

$$y'+2 \sin \frac{y}{2} \cos \frac{y}{2} +2x \cos^2 \frac{y}{2} = 0 $$

I'm stuck. I'm trying to get it to a linear form $y'+a(x)y=b(x)$ so that I can solve it. I've managed to get it to $$y' + \sin y - \sin^2 \frac{y}{2} +2x=0$$
using $\sin y = 2 \sin \frac{y}{2} \cos \frac{y}{2}$ and $\cos^2 \frac{y}{2}  = 1-\sin^2 \frac{y}{2} $. But this still is not helping.

Thanks.

Body:
$$y'+2\sin(y/2) \cos(y/2)+2x\cos^2(y/2)=0 \implies \sec^2(y/2) y'+ 2\tan(y/2)=-2x$$
Let $v=\tan (y/2) \implies v'=\sec^@(y/2) (y/2) \implies v'=\frac{1}{2} \sec^2(y/2) y'$,
The we get a linear ODE:
$$v'+v=-x \implies v=-e^{-x} \int e^{x} x+ Ce^{-x}\implies v=(1-x)+C e^{-x}$$
Finally, $$y=2\tan^{-1}[(1-x)+Ce^{-x}].$$


问:
Numbers which are their own inverse are exactly $1$ and $p − 1$?
We begin by listing all the positive integers from $1$ to $p − 1$:
$$1, 2, 3, . . . , p − 1. $$

$p$ is a prime. The numbers on the list that are their own inverse are exactly $1$ and $p − 1$.

For $x, 1\leq x \leq p − 1, x$ is its own inverse if and only if $x^2 ≡ 1 \pmod p$. Because $p$ is prime, this can happen only if $p$ divides either $x − 1$ or $x + 1$.

Then the author writes -

Thus, the numbers on the list that are their own inverse are exactly $1$ and $p − 1$.

I just couldn't see through the last line, how from  $p$ divides either $x − 1$ or $x + 1$ we deduce  the numbers which are their own inverse are exactly $1$ and $p − 1$?

答:
If $p|x-1$, then because $0 \leq x-1 \leq p-2$, the only possibility is $x-1=0$ i.e. $x=1$.

If $p|x+1$, then because $2 \leq x+1 \leq p$, the only possibility is $x+1=p$ i.e. $x=p-1$.


答:
Well, take $x=1+t$, then: $$(1+t)^2=1+2t+t^2\equiv1\pmod p \iff t^2+2t=t(t+2)\equiv 0\pmod p$$ i.e. $p|t\implies x=kp+1$ or $p|(t+2)\implies x=kp-1$ for $k\in\Bbb Z$

In the range $1\leq x\leq p-1$, only $1, p-1$ are of this form

Title: Numbers which are their own inverse are exactly $1$ and $p − 1$?

Body:
We begin by listing all the positive integers from $1$ to $p − 1$:
$$1, 2, 3, . . . , p − 1. $$

$p$ is a prime. The numbers on the list that are their own inverse are exactly $1$ and $p − 1$.

For $x, 1\leq x \leq p − 1, x$ is its own inverse if and only if $x^2 ≡ 1 \pmod p$. Because $p$ is prime, this can happen only if $p$ divides either $x − 1$ or $x + 1$.

Then the author writes -

Thus, the numbers on the list that are their own inverse are exactly $1$ and $p − 1$.

I just couldn't see through the last line, how from  $p$ divides either $x − 1$ or $x + 1$ we deduce  the numbers which are their own inverse are exactly $1$ and $p − 1$?


问:
Find a sequence of random variables
I have a problem with finding an example of a sequence of nonnegative random variables $X_1, \ldots$ with values in integers s.t. $\mathbb{E}X_n \rightarrow \infty$ and

$\limsup_{n\rightarrow\infty}\mathbb{P}(X_n>0)<1$ (first case),

$\limsup_{n\rightarrow\infty}\mathbb{P}(X_n>0)=1$ (second case).

I think in the second case it can be a sequence $X_n=n$. Is it good reasoning? What with the first case?

答: Hint: your answer for the second case is good. Can you modify it to obtain an example such that $P(X_n = 0) = P(X_n > 0) = \frac{1}{2}$ for all $n$ and $\Bbb{E}(X_n) = \frac{n}{2}$?

Title: Find a sequence of random variables


问:
Gaining an intuition for how changes to the inputs of an equation affect the output
I am currently taking General Chemistry 2 and Physics 2. I have been doing very well, but I am not just preparing to do well in these classes. I also want to do well on the Medical School Admissions Test or MCAT. On the MCAT, you cannot use a calculator. I have spoken with a few people who performed very well on the test, and they told me that an intuition for algebraic equations is a major factor in a good score. However, I have noticed that I do not possess this.

What I mean by an intuition for algebraic equations is the ability to look at an algebraic equation and know how the output will be affected as a result of an input changing. An example is using the Gibbs Free Energy Equation. If some inputs are negative or positive it automatically changes the output/answer.

I don't know how to go about learning this skill. Is this just a matter of going back to Algebra and learning it more deeply?

答:
If you really feel like going back to the basics, I'd advise downloading Geogebra, and plotting some functions.

For example, create 4 sliders ($a$, $b$, $c$, and $d$) by doing "$a = 1$", etc, in a given cell. Then create a function $f$ by writing "$cos(x)$" in a fifth cell. Finally, create a function $g$ by writing "$af(cx + d) + b$" in a sixth cell.

By changing the sliders, you'll realize that a change to the input of $f$ (by changing $c$ & $d$) affects things over the $x$-axis/horizontally; while changing the output of $f$ (by changing $a$ & $b$) affects things on the $y$-axis, vertically. You'll also see that additions are translations/displacements, while multiplications are scalings/dilations. Vary your $f$ function by using something else, like $exp$, or $x^3$ whatever.

This will give you a raw understanding for real functions ($\mathbb{R} \to \mathbb{R}$).

Now open Geogebra's 3D calculator. Try to invent some functions that are from $\mathbb{R}^2 \to \mathbb{R}$ by writing something along the lines of "$exp(x) + x*y$". Once you've experimented with that, try to have another function of the same kind. See what addition "$x+y$" and multiplication "$x*y$" look like. Look up things like the monkey saddle, etc.

As for functional equations: the points where $f(u) = g(u)$ correspond to the point of intersection between your surfaces. You can also plot the function "f(u) - g(u)": its points that pass by the xy-plane have output zero, and should correspond to the same inputs which cause the intersection.

For general equations, look up quadric equations: how would you draw a sphere centered around a point $p$, or a light-cone centered in $(0, 0, 0)$ in $\mathbb{R}^3$ ?

Once you've done that, study functions from $\mathbb{R} \to \mathbb{R}^2$. One example I like to give my students is $(x, 0.2 x^2, cos(x))$. By looking straight down the y axis and the z axis, you can clearly see how both 0.2x² and cos(x) combine into a common continuous curve.

Finally, understand that every solution set to a differential equation is a foliation: a partition of the Input*Output space (eg, your $\mathbb{R}^2$ plane for functions $\mathbb{R} \to \mathbb{R}$) into non-intersecting curves. The go-to basic example is the solution to the equation $f'(x) = f(x)$, which gives rise to the family of functions which are exponentials of the form $y_0 e^{x - x_0}$. Make $y_0$ and $x_0$ sliders, and try to discern the underlying foliation expressed by this differential equation. Can you see that the curves are non-intersecting ? Can you see that they cover the whole space ?

Once that's done, I think you'll have the creativity to explore the problems you are more familiar with (and that you're more interested in) using Geogebra.

Hope this helps, and best of luck !


答:
If $x$ is its own inverse then

$x^2 \equiv 1 \pmod p$

$x^2 -1 \equiv 0 \pmod p$

So $p|x^2 - 1$ and as $x^2 -1= (x+1)(x-1)$ the $p|(x+1)(x-1)$.

As $p$ is prime either $p|x+1$ or $p|x-1$.

Case 1:  $p|x+1$.  That means that $x+1 = pk$ for some integer $k$.  So $x+1 \equiv 0 \pmod p$ and $x \equiv -1 \equiv p -1 \pmod p$.

Case 2:  $p|x-1$.  That means $x -1 =pk$ for some integer $k$.  So $x-1\equiv 0 \pmod p$ and $x \equiv 1 \pmod p$.

And that's it.  Because $p$ is prime there are not other cases.

....

Note if $p$ isn't prime that need not be true.

For example:  If $n = 24$ then if $x$ is it's only inverse then $x^2\equiv 1 \pmod n$ and so $24|x^2 -1 = (x+1)(x-1)$.

We could have $24|x+1$ and so $x+1\equiv 0\pmod {24}$ or  $x\equiv 23 \pmod {24}$.  Or we could have $24|x-1$ so $x-1\equiv 0$ and $x\equiv 1\pmod{24}$.

But we could also have $m|x+1$ and $\frac {24}m|x-1$. For say $m=6$ as so $6|x+1$ and $4|x-1$ and $x=5$ or $x=17$ (and $5^2 \equiv 1 \pmod {24}$).  Or $m=4$ so $4|x+1$ and $6|x-1$ and $x =7$ or $x=19$ (and $7^2\equiv 1 \pmod{24}$). Or $m=3$ so $3|x+1$ and $8|x-1$ and $x\equiv 17$. Or $m=8$ so $8|x+1$ and $3|x-1$ so $x=7$.  Or $m=2$ and $2|x+1$ and $12|x-1$ so $x=13$ or $1$. Or $m=12$ and $12|x+1$ and $2|x-1$ and $x =11$ or $x=23$.


问:
diffusion equation with time and space derivatives in boundary conditions
I think the system I'm working with can be modeled with the following initial boundary value problem based on the 1D diffusion equation.

partial differential equation

$u_t - Du_{xx} = 0$

boundary conditions

$u_t(0,t) = \beta D u_x(0,t)$

$u(L,t) = 0$

initial condition

$u(x,0) = \phi(x)$

Where $u$ is the dependent variable, $0<x<L$ is location, $t$ is time, $D$ is the diffusion coefficient, $\beta$ is a constant, and $\phi$ is some arbitrary function of $x$ specifying the initial value of $u$ over the domain.

What solution methods exist for this sort of IBVP? Would the BC at $x=0$ be considered a Neumann boundary condition? The methods I've read about so far don't deal with any scenarios where both time and space derivatives are present in the BC.

答:
This is tractable by separation of variables. Real solutions of the diffusion equation are of the form

$$
u(x,t) = \sum e^{-D\omega_i^2 t}(a_i cos(\omega_i x) + b_i sin(\omega_i x))
$$

The boundary conditions can be satisfied nontrivially only with certain values of $\omega_i$. Specifically, letting $L = 1$, the first BC gives

$$
-a\omega = b\beta \\
b = -\frac{a\omega}{\beta}
$$

By the way, I don't think this BC is any of the usual types (Dirichlet, Neumann, Robin), because it involves the time derivative, which none of these do. The second BC then becomes

$$
cos(\omega) - \frac{\omega}{\beta}sin(\omega) = 0
$$

In general there are a (countably) infinite number of values of $\omega$ that satisfy that condition. You can satisfy yourself of that by plotting $cos(\omega)-\frac{\omega}{\beta}sin(\omega)$ vs $\omega$. These are the values $\omega_i$ can take on in the general solution. I suspect that for general $\beta$ you will need to find them numerically. Then, of course, you need to find the values of $a_i$ such that the initial condition is satisfied:

$$
\phi(x) = \sum a_i\left(cos(\omega_i x) - \frac{\omega_i}{\beta}sin(\omega_i x)\right)
$$

Unless you can find some nice orthogonality condition that these eigenfunctions satisfy, you may be stuck with solving a system of linear equations to approximate the $a_i$.

Addendum:

Following @Claude_Leibovic's observation that the condition on $\omega$ reduces to

$$
\omega\tan(\omega) = \beta \\
\tan(\omega) = \frac{\beta}{\omega},
$$

I note that there is an easily visualized graphic solution to this constraint. Plot $\tan(\omega)$ and $\beta/\omega$ against $\omega$ -- the points where they cross are the desired solutions.

Now, note that $\tan(\omega)$ and $\beta/\omega$ are odd functions of $\omega$. If $\omega$ is a solution, $-\omega$ is one, too. Thus, it suffices to find the positive solutions. Picturing $\tan(\omega)$ and $\beta/\omega$, it is evident that, for positive $\beta$, there is one solution in $(0,\pi)$ and one solution in $(2\pi,3\pi), (4\pi,5\pi),...$, i.e. $(2n\pi,(2n+1)\pi)$ for every non-negative integer $n$. For $\beta < 0$, there is no solution in $(0,\pi)$, and there is one in each of $(\pi,2\pi), (3\pi,4\pi), ...$.

Often one is not really interested in a full solution, but only in knowing how rapidly the initial condition decays to nothing. The long-term behavior of the solution is dominated by the eigenfunction corresponding to the smallest eigenvalue $\omega_{min} := \min(\left|\omega\right|)$. $\omega_{min}$ can be approximated as described in @Claude_Leibovic's answer. This eigenfunction decays with time constant $\tau = \frac{1}{D\omega_{min}^2}$.


问:
Better proof that vector fields on submanifolds extend globally iff submanifold is closed
I just finished my own proof of one of the problems in Lee's Smooth Manfolds, 2nd ed., but I wonder if anyone knows a better (less messy) solution. It's problem 8-15, the "Extension Lemma for Vector Fields on Submanifolds." Here's the part of the exercise which I had some trouble with:

Let $S$ be an embedded submanifold with or without boundary of a manifold $M$. Every vector field $X \in \mathfrak{X}(S)$ extends to all of $M$ if and only if $S$ is properly embedded.

I proved that if $S$ is properly embedded, then all vector fields extend globally easily. It's the reverse direction I had trouble with. My idea was that if $S$ is not closed in $M$, then take some point $p \in \overline{S} \setminus S$. I showed that one can find a smooth path $\gamma: (-\epsilon, \epsilon) \to M$ such that $\gamma$ is an embedding, $\gamma(0) = p$, $\gamma(-\epsilon, 0) \subset S$, and in some coordinate neighborhood of $p$, $|\gamma'(t)| = 1$. Then I defined the function $f: \gamma(-\epsilon, 0) \to \mathbb{R}$ as $f(x) = -1/\gamma^{-1}(x)$ (the idea is that this is a smooth function which blows up as $x \to p$). Finally I defined the vector field $f(x) \gamma'(\gamma^{-1}(x))$ on $\gamma[-\epsilon/2, 0)$. This is a smooth vector field on a closed subset of $S$, hence by the extension lemma proved earlier in the text, it extends to a smooth vector field on $S$. Clearly it does not extend to a smooth vector field on all of $M$ though, because in the local coordinate chart near $p$, its norm blows up.

The above argument feels very messy to me. I don't like the process I went through to find a smooth function on $S$ which blows up near $p$; surely there's a simpler way to do this?

答:
I know this is an old question but I recently came across this problem. I think this may be a reasonable answer, but I'm really not too sure if it is valid. If $S$ is not properly embedded, then by problem 5-18, there is some $f\in C^{\infty}(S)$ that does not have a smooth extension to all of $M$. Given that $X$ is some smooth vector field on $S$, since smooth vector fields on $S$ form a $C^{\infty}(S)$-module, we have that $fX$ defined by $fX(x) = f(x)X_x$ is a smooth vector field on $S$ that does not have a smooth extension to all of $M$ since $f$ doesn't.

A nice constructive proof that such an $f$ exists is given here: Extension of Smooth Functions on Embedded Submanifolds


问:
The sum of dihedral angles of an $n-$hedral angle is bigger than $(n-2)\pi$
Given a polyhedral angle $O.A_1A_2\dots A_n$ (where $O$ is the vertex), I need to show that the sum of dihedral angles with sides $OA_1,\dots,OA_n$ is bigger than $(n-2)\pi$.

My try is to draw a plane meeting $OA_i$ at $B_i$ and show that the sum of dihedral angles is bigger than the sum of the angles of polygon $B_1B_2\dots B_n$. But it seems that each dihedral angle is not bigger than the corresponding angle of the polygon and I do not know how to continue.

Can someone help me? Thanks a lot!

答: Hint: pick a point $P$ inside the polyhedral angle. Let $P_1. P_2, \ldots, P_n$ be the perpendicular projections of $P$ onto planes $OA_1A_2, OA_2A_3, \ldots, OA_nA_1$, respectively. For every $i$ the dihedral angle at $OA_i$ is equal to $\pi - \angle P_{i-1}PP_i$. The problem boils down to proving that $\sum_{i=1}^n \angle P_iPP_{i+1} > 2\pi$.


问:
Is it normal that u-substitution leaves behind a constant besides C?
Hello in the following problem $\int{(x^2+2)(x^3+6x+1)^5}$, using u-substitution we get the following: $\frac{(x^3+6x+1)^6}{18} + C$, but when we simplify this in the constant end we get $1/18$ alongside $C$, however if we do the integral of the question but fully simplified($x^{17}+.....+2$), we do not get the $1/18$, in fact we don't get any constant. Why are these integrals different? Is there some limitations of u-sub I should know about?

答:
in fact we don't get any constant.

How is that? You should still have the constant of integration and in fact because of this, the difference you think is there, actually isn't...

Why are these integrals different?

When we write
$$\int \left( x+1 \right) \,\mbox{d}x = \color{green}{\frac{x^2}{2}+x}+\color{blue}{C} \quad\quad \left(C \in \mathbb{R}\right)$$
the " $+C$ " makes sure we didn't write an antiderivative of $x+1$, but (all) the antiderivatives.

Now compare with using a (simple) $u$-substitution ($u=x+1$) for the same problem:
$$\int \left( x+1 \right) \,\mbox{d}x =\int u \,\mbox{d}u = \frac{u^2}{2}+C= \frac{\left( x+1 \right)^2}{2}+C= \color{green}{\frac{x^2}{2}+x}+\color{red}{\frac{1}{2}+C}  \quad\quad \left(C \in \mathbb{R}\right)$$

It may not look the same, but we get the exact same (set of) antiderivatives.

To get one specific antiderivative, the values of the "blue $C$" and "red $C$" will differ (by $1/2$).


答:
The answers in both the cases are same. As you can always club the constant of integration with the constant value.

$\frac{1}{18} + C$ can be called a new constant $C_1$.

Title: Is it normal that u-substitution leaves behind a constant besides C?

Body: Hello in the following problem $\int{(x^2+2)(x^3+6x+1)^5}$, using u-substitution we get the following: $\frac{(x^3+6x+1)^6}{18} + C$, but when we simplify this in the constant end we get $1/18$ alongside $C$, however if we do the integral of the question but fully simplified($x^{17}+.....+2$), we do not get the $1/18$, in fact we don't get any constant. Why are these integrals different? Is there some limitations of u-sub I should know about?


问:
A proof problem in mathematical statistics

Let $X_{1}, \ldots, X_{n}$ be independent and identically distributed random variables, with uniform distribution in $(0,1)$. Consider $Y_{1}, \ldots, Y_{n}$ the associated order statistics and define $$V_{i} = \frac{Y_{i}}{ Y_{i + 1}}, i = 1, ..., n -1 \quad \text{and} \quad V_{n} = Y_{n}$$Show that $V_{1}, \ldots, V_{n}$ are independent and $V_{i}\sim  \textbf{Beta}(i, 1), i = 1, \ldots, n$.

How could you prove this result?

My approach: We need two prove two parts,  for to prove that $V_{i}$ are independent, we need to show something like $$f_{V_{1},V_{2},\ldots,V_{n}}(v_{1},v_{2},\ldots,v_{n})=f(v_{1})f(v_{2})\cdots f(v_{n})$$
and for the second part we need to show that $V_{i}\sim \textbf{Beta}(i,1)$, so we need to show that something like $$f(v_{i})=\frac{v_{i}^{i-1}(1-v_{i})^{1-1}}{B(i,1)}$$ is PDF of beta distribution.

Also, I think that I can prove that the joint distribution of order statistics for uniform distribution is also a uniform distribution and maybe use that result in the solution of problem.

I think I can use fundamental transformation theorem, but I don't sure how can I approach this problem.

答:
I have a very unrigourous, but simple answer: To show pairwise independence, it should be sufficient to show that knowing the ratio $Y_i/Y_{i+1}\leq x_i$ for some $i=1,\dots,n+1$ and $x_i\in(0,1)$ (here we let $Y_{n+1}\equiv 1$ for simplicity) does not tell you whether the ratio $Y_l/Y_{l+1}\leq x_l$ for some other $l\neq i,l=1,\dots,n+1$ and $x_l\in(0,1)$.

First of all, we can rewrite both equations to $Y_l\leq x_l Y_{l+1}$ and $Y_i\leq x_i Y_{i+1}$ since both RVs are positive with probability one. Now if $i<l$, the only thing this tells us is that there is some lower bound for $Y_l$. If $i>l$, we have an upper bound for $Y_{l+1}$. Either way the information is useless. This implies pairwise independence. The same goes if we have several of these equations, so we even have independence.

EDIT: I'm pretty sure the argument above does not work at all. I just let it there in the case that it helps someone find an argument that works.

Once you have independence, the rest follows easily from $\textbf{P}(Y_n\leq x_n)=x_n^n$ and $\textbf{P}(Y_{i}\leq x_i Y_{i+1})=\textbf{E}[\textbf{P}(Y_i\leq x_i Y_{i+1}|Y_{i+1})]=\textbf{E}[F_{Y_i}(x_iY_{i+1})]$ (the last equality follows from independence) where $F_{Y_i}$ is the CDF of $Y_i$.

Disclaimer: I'm really unsure about this solution, but it makes a lot of sense to me. It would be great if someone could double check.


问:
Proof of "Induction proof method"
So I have been proving various logical statements using induction method (like structural induction , strong induction , weak induction etc ).I was wondering If there is a proof of this "Induction proof method" . So far , I came to this ,

Induction $\rightarrow$ Well ordering principal $\rightarrow$ Axiom of choice $\rightarrow$ ZFC $\rightarrow$ First-order logic theory

So now I wonder , Is there a way to prove (or show equivalence of) this method of proof using just Logic and no Set theory.Also point out if there is a flaw in my reasoning

Edit:It seems like structural induction doesn't do induction over numbers of any kind , it does in on structures .So I can't use peanos axioms to formulate it .I need ZFC .But ZFC is just a kind of first order logic.So structural induction comes from this particular first order logic . But there are some General theorems (which probably don't necessarily belong to ZFC) in Propositional Calculus which I have to prove using structural Induction .But Structural Induction can only be used inside ZFC , not outside of it.I am confused.In a simpler way , The following general theorem I will show at the end of my question is outside of set theory . And  I need structural induction to prove it . But structural Induction can only prove things inside set theory. Because Structural Induction is a axiom of Axiomatic set theory.

I will give just a example of one of these general theorem.

"Assume $A$$1$ $\equiv$  $A$$2$ . Show that for any formula $C$-containing $A$$1$ as a part , if we replace one of more occurences of the part $A$$1$ by $A$$2$ , then the resulting formula is logically equivalent to $C$."

答:
The Principle of Mathematical Induction boils down to a fact known about the natural (counting) numbers for thousands of years:

Every natural number but the "first" (1 or 0), can be reached by a
process of repeated succession starting at the "first" number.

More formally, induction can be shown to hold on any set $N$ (possibly finite) with $x_0\in N$ and function $S: N \to N$ such that:

$~~~~~~N = \{ x_0,~ S(x_0), ~S(S(x_0)), ~\cdots ~\} $

My formal proof using a form of natural deduction is here. Yes, it does use some very basic set theory, but only an axiom schema for arbitrary subsets (equivalent of specification in ZFC).

Also on this topic, see my blog postings:

Daddy, where do numbers come from?

What is a number again?

Title: Proof of "Induction proof method"


问:
Span of a finite set of linearly dependent vectors.
Show that a linear dependent set of $n$ vectors does not span $V$, where $\dim(V)=n$.

For example, show that $\{(1,0,0),(0,1,0),(2,0,0)\}$ does not span $\mathbb{R}^3$.

答:
Theorem. A linear dependent set of n vectors does not span V, where dim(V) = n.

Proof. Given a linear dependent set of n vectors $(\vec{v_1},\vec{v_2},...,\vec{v_n})$ and a vector space $V$ with $dim(V) = n$, we need to show that there exists at least one $\vec{x}\in{V}$ where the elements in $\vec{x}$ is given by $x_{i}\in{R}$ is not an element of the $span{(\vec{v_1},\vec{v_2},...,\vec{v_n})}$. Note that the $span{(\vec{v_1},\vec{v_2},...,\vec{v_n})} = \{\lambda_1\vec{v_1}+\lambda_2\vec{v_2}+...+\lambda_n\vec{v_n} : \lambda_i\in{R}\} = $$
    \begin{bmatrix}
    \lambda_{11} & \cdots & \lambda_{1n} \\
    \vdots & \ddots & \vdots \\
    \lambda_{n1} & \cdots & \lambda_{nn} \\
    \end{bmatrix} = A.
$
If we can show that a vector in $V$ is not in the span of the linear dependent set, then the set does not span $V$.

To show this, first take the augmented matrix: $$
    \begin{bmatrix}
    \lambda_{11} & \cdots & \lambda_{1n} \\
    \vdots & \ddots & \vdots \\
    \lambda_{n1} & \cdots & \lambda_{nn} \\
    \end{bmatrix}
    \begin{bmatrix}
      x_1 \\
    \vdots \\
    x_n \\
    \end{bmatrix}
$$
Since the $det(A) = 0$ there is a row-equivalent matrix such that there exists a row of all zeros. So, $$
    \begin{bmatrix}
    \lambda_{11} & \cdots & \lambda_{1n} \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & 0 \\
    \end{bmatrix}
    \begin{bmatrix}
      x_1 \\
    \vdots \\
    x_n \\
    \end{bmatrix}
$$
Letting $x_n \neq 0$, then there is no solution to the system and therefore there exists at least one vector in $V$ that is not in the span. Q.E.D


问:
Can $p$ divide both $a_1 − a_2$ and $a_1 + a_2$?
If $a_1^2 ≡ a_2^2 \pmod p$, then $p$ divides $a_1^2−a_2^2$, so $p$ divides the product $(a_1 − a_2)(a_1 + a_2)$.

I read in a chapter related to  quadratic residues and nonresidues that

unique prime factorization now tells us that $p$ divides $a_1 − a_2$ or $p$ divides $a_1 + a_2$, and so either $a_1 ≡ a_2 \pmod p$ or $a_1 ≡ −a_2 \pmod p$.

Why? It could be the case where both $p$ divides $a_1 − a_2$ and  $p$ divides $a_1 + a_2$.

答: A longish comment: If $p\mid(a_1-a_2)$ and $p\mid(a_1+a_2)$, then $p$ divides their sums and differences, which are $2a_1,2a_2$. This occurs when $p=2$, or when $p$ divides both of $a_1,a_2$, or when both statements are true, such as when $p=2$ and $a_1,a_2$ are both even.


问:
find $\lim_{n\to \infty}(\log(1+1/n))^{1/n}$
Question is to find the limit of following as n tends to infinity :

$\lim_\limits{n\to \infty}(\log(1+1/n))^{1/n}$

my attempt:

took expansion of  $\log(1+x)$
so

$\lim_{n\to \infty}(\frac{1}{n^{1/n}}(1-\frac{1}{2n}+\frac{1}{3n^2}....)^{1/n}) $ is of form $1^0$ which is not indeterminate and hence limit is 1.

is it right?

答:
We have that

$$\left(\log\left(1+\frac{1}{n}\right)\right)^{\frac{1}{n}}=\left(\frac1n\right)^{\frac{1}{n}}\left(\frac{\log\left(1+\frac{1}{n}\right)}{\frac1n}\right)^{\frac{1}{n}} \to 0 \cdot 1^0=0$$

indeed

$$\left(\frac1n\right)^{\frac{1}{n}}=e^{-\frac{\log n}{n}} \to e^0=1$$

and by standard limits

$$\frac{\log\left(1+\frac{1}{n}\right)}{\frac1n} \to 1$$


问:
Marking the prime points on a circle
If you travel around a circle and mark all the points on the circle where the distance you travelled is a prime number, where you would go through many rotations*, do you end up marking the entire circumference of the circle as the number of points you pick goes to infinity?

Or do you leave  "holes" in the circle by virtue of only marking the prime points of the circumference?

*I understand that you would have to define the circumference of your circle. Define it as a finite number, prime or not. How does the circumference you pick affect the answer/whether the circumference is prime or not?

Please let me know how I can explain this question further or whether this question does not make sense.

答:
EDIT: In view of the edit made to the question, it would appear that the following was not what OP had in mind. </p>

<p>If you have a finite number, $n$, of points arrayed on a circle, and you go around picking every $m$th point, where $m$ is any number relatively prime to $n$ (doesn't matter whether $m$ is prime or not), you will pick up every one of the $n$ points, eventually. Otherwise, not. </p>

Title: Marking the prime points on a circle

Body:
EDIT: In view of the edit made to the question, it would appear that the following was not what OP had in mind.

If you have a finite number, $n$, of points arrayed on a circle, and you go around picking every $m$th point, where $m$ is any number relatively prime to $n$ (doesn't matter whether $m$ is prime or not), you will pick up every one of the $n$ points, eventually. Otherwise, not.


问:
Some topological properties of "countable lines with one origin"
Let the countable lines with one origin to be a quotient space $CL = ([0, \infty) \times \mathbb N) / \sim$, where $[0, \infty) \times \mathbb N$ has a subspace topology of $\mathbb R^2$ and $0 \times n \sim 0 \times m$ for any $n, m \in \mathbb N$.

Then, is $CL$ Hausdorff, first countable, and/or locally compact?

My attempts:

Hausdorff:

Take for any $x$ and $y$, if both are not $[0 \times n]$, take a disjoint neighborhoods of their correspondents in the original space, and pass it to the quotient space. Otherwise, if $x$ is $[0 \times n]$ and $y = [a \times b]$ is not, I can take something like $\pi([0, \frac{y}{2}) \times b)$ and $\pi((\frac{y}{2}, 2 y) \times b)$, although I am not completely sure if they'd be open.

First countable:

Let $\pi : [0, \infty) \times \mathbb N \to CL$ be a quotient map. If there exists a neighborhood basis $\{ U_n \}$ around $[0 \times 1] \in CL$, I believe I can take something like $U = \bigcup \limits_{n = 1}^\infty \pi([0, \frac{r_n}{2}) \times n)$ where $[0, r_n) \times n \subset \pi^{-1} (U_n) \cap [0, \infty) \times n$, so that it is not properly contained in any of the neighborhood basis, but not sure if it is open again :(.

Locally compact:

Not sure...

答:
Let $p : [0,\infty) \times \mathbb N \to CL$ denote the quotient map and $* \in CL$ denote the common equivalence class of the points $(0,n)$ with $n \in \mathbb N$. Clearly $p^{-1}(*) = \{0\} \times \mathbb N$ which is closed in $[0,\infty) \times \mathbb N$. Thus $\{*\}$ is closed in $CL$ and $CL' = CL \setminus \{*\}$ is open in $CL$.

$CL$ is Hausdorff.

The restriction $p' : (0,\infty) \times \mathbb N \to CL'$ of $p$ is a continuous bijection and it is easy to see that it is an open map. Hence $p'$ is a homeomrphism. This shows that any two distinct $x,y \in CL'$ have disjoint open neigborhoods in $CL$. Now let $x = *$ and $y \in CL'$. We can write $y = p(t,n)$ for some $t > 0$ and $n \in \mathbb N$. Then $U = p((t/2,\infty) \times \{n\})$ is an open neigborhood of $y$ and $V = p([0,t/2) \times \mathbb N)$ is an open neigborhood of $*$. We have $U \cap V  =\emptyset$.

$CL$ is not first countable.

Of course each $x \in CL'$ has a countable neigborhood base. The exceptional case is $*$. Let $\{U_k\}$ be any countable family of open neigborhoods of $*$ in $CL$. Then $p^{-1}(U_k) = \bigcup_{n= 1}^\infty U_k^n \times \{n\}$ with open neighborhoods $U_k^n$ of $0$ in $[0,\infty)$. There are $t_k^n > 0$ such that $[0,t_k^n) \subset U_k^n$. Define $s_n = \dfrac{t_n^n}{2}$ (this is a typical diagonal construction). Then $V = \bigcup_{n= 1}^\infty [0,s_n) \times \{n\}$ is open in $[0,\infty) \times \mathbb N$. We have $p^{-1}(p(V)) = V$, thus $U = p(V)$ is an open neigborhood of $*$. For each $n$ we have $p^{-1}(U_n) \not\subset p^{-1}(U)$ because $[0,t_n^n) \times \{n\} \not\subset [0,s_n) \times \{n\}$. This means $U_n \not\subset U$, thus $\{U_k\}$ cannot be a neighborhood base of $*$.

$CL$ is not locally compact.

Of course each $x \in CL'$ has a compact neigborhood. The exceptional case is again $*$. Let $U$ be any open neigborhood of $*$. We shall show that $\overline U$ is not compact. We have $p^{-1}(U) = \bigcup_{n= 1}^\infty U_n \times \{n\}$ with open neighborhoods $U_n$ of $0$ in $[0,\infty)$. There are $t_n > 0$ such that $[0,t_n] \subset U_n$. The set $V = \bigcup_{n= 1}^\infty [0,t_n) \times \{n\}$ is open in $[0,\infty) \times \mathbb N$. We have $p^{-1}(p(V)) = V$, thus $W = p(V)$ is an open neigborhood of $*$. The sets $W_n = p((\frac{t_n}{2},\infty) \times \{n\})$ are open in $CL$ and we have $W \cup \bigcup_{n=1}^\infty W_n = CL \supset \overline U$. For each $m$ we have $p(t_m,m) \notin W  \cup \bigcup_{n \ne m} W_n$, and since $(t_m,m) \in U_m \times \{m\} \subset p^{-1}(U)$, i.e. $p(t_m,m) \in U \subset \overline U$, we see that $\{W\} \cup \{W_n\}$ does not contain a finite subcover of $\overline U$.


问:
Prove or disprove that the set S is countable
Define set $S$ as follows $$S = \left\{ f \in \{0,1\}^{\mathbb{N}} \middle| \forall x \in \mathbb{N} \ \exists y \in \mathbb{N}: x < y \land f(x) = f(y) \right\},$$ where $\{0,1\}^\mathbb{N}$ denotes the set of boolean functions defined on $\mathbb{N}$.

Prove or disprove that the set $S$ is countable.

I know that the first part before the |-symbol itself is uncountable, but I don't understand the whole $x$, $y$, $f(x)$, and $f(y)$ part, and how it would change the fact that it's already not countable. Could it not just be ignored? I'd be happy about any help.

答:
This $\mid$ symbol you talk about translates to English as "such that", i.e. you want all boolean functions defined on the set $\mathbb{N}$ such that for all $x$ there exists $y$ greater than $x$ with $f(x) = f(y)$. This "such that" part and subsequent condition shrinks the initial set of all boolean functions defined on $\mathbb{N}$, potentially making it countable.

Namely, this condition says that your functions attain every value either zero or infinitely many times, meaning that some functions are excluded from the set, with one possible example being $$f(x) = \begin{cases} 1, & x = 1, \\ 0, & x > 1, \end{cases}$$ or the one suggested by @atticus-stonestrom

This is a partial answer not answering the title question, but I think that you'll be able to complete from here on. Hint: present all excluded functions as a countable union of countable sets, depending on how many times they attain the "wrong" value.


答:
We will define an injection $\varphi:\{0, 1\}^\mathbb{N}\rightarrow S$. Because $\{0, 1\}^\mathbb{N}$ is uncountable, as you have noted, this will be enough to show that $S$ is uncountable. So, define $\varphi(f)$ by $\varphi(f)(n)=f(n/3)$ if $n\equiv 0\text{ (mod 3)}$, $\varphi(f)(n)=0$ if $n\equiv 1\text{ (mod 3)}$, and $\varphi(f)(n)=1$ if $n\equiv 2\text{ (mod 3)}$. Can you show that $\varphi(f)\in S$ and that $\varphi$ is injective? (Answer given below, but try to do it yourself first!)

To show $\varphi(f)\in S$, let $x\in \mathbb{N}$. We wish to show there is $y>x$ such that $\varphi(f)(x)=\varphi(f)(y)$. If $\varphi(f)(x)=0$, let $y=3x+1$, and if $\varphi(f)(x)=1$, let $y=3x+2$.

To show injectivity, suppose that $f\neq g\in\{0, 1\}^\mathbb{N}$. Then there is some $n\in\mathbb{N}$ such that $f(n)\neq g(n)$, we have $\varphi(f)(3n)=f(n)\neq g(n)=\varphi(g)(3n)$, so $\varphi(f)\neq\varphi(g)$ as desired.

Title: Prove or disprove that the set S is countable

Body:
Define set $S$ as follows $$S = \left\{ f \in \{0,1\}^{\mathbb{N}} \middle| \forall x \in \mathbb{N} \ \exists y \in \mathbb{N}: x < y \land f(x) = f(y) \right\},$$ where $\{0,1\}^\mathbb{N}$ denotes the set of boolean functions defined on $\mathbb{N}$.

Prove or disprove that the set $S$ is countable.

I know that the first part before the |-symbol itself is uncountable, but I don't understand the whole $x$, $y$, $f(x)$, and $f(y)$ part, and how it would change the fact that it's already not countable. Could it not just be ignored? I'd be happy about any help.


问:
Problem related to an analytic function
I came across the following problem :

Let $g \colon \Bbb C \to \Bbb C$ be analytic.Suppose that $|g(\frac{1}{n})| \le \frac{1}{n^{3/2}}$ for each $n \in N.$ Then prove that $\{n^2g(\frac{1}{n})\}$ is bounded.

$|g(\frac{1}{n})| \le \frac{1}{n^{3/2}} \implies |n^2g(\frac{1}{n})| \le \sqrt n$....Now,I am not sure which way to go.

答:
Since $g$ is continuous, $g(0)=0$. Show that $g'(0)$ also vanishes.

If that is the case, then there is an entire function $h$ such $g(z)=z^2h(z)$, and then $n^2g(\tfrac1n)=n^2\tfrac{1}{n^2}h(\tfrac1n)=h(\tfrac1n)$ so that the set $\{n^2g(\tfrac1n):n\in\mathbb N\}$ is obviously bounded, because $h$ is bounded on the unit disc.

Title: Problem related to an analytic function


问:
Is this limit correct: $\lim_{x \to+\infty} \frac{\log_{2}(x-1)}{x} = 0$?

Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right]
\end{align*}$.

After some minutes around this limit I did it this way:

$\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$

So,$\space x=2^y+1$.

When $x \to +\infty$,$\space y \to +\infty$ also. By substitution:

$\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}$

$\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}}  \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0
\end{align*}$

Is this correct?Are there any other easy way to find this limit?Thanks

答:
Your way is fine, as an alternative we have that

$$ \frac{\log_{2}(x-1)}{x}= \frac{\log_{2}(x-1)}{x-1}\frac{x-1}{x}\to 0\cdot 1 =0$$

indeed eventually $\log_{2}(x-1) \le \sqrt{x-1}$ and then

$$\frac{\log_{2}(x-1)}{x-1} \le \frac{\sqrt{x-1}}{x-1} \to 0$$


问:
Doubts in $\Bbb Z_2 \times \Bbb Z_2$
My question asks me to prove that whether $\Bbb Z_2 \times\Bbb Z_2$ a subgroup of $S_4$ or not.

I really cant get a clear picture but the only thing that I know is that if we can let $\Bbb Z_2 $ isomorphic to $H_1=\langle(a b)\rangle$ and $\Bbb Z_2$ isomorphic to $H_2=\langle(c d)\rangle$
.Then maybe I can show some isomorphism with some subgroup of $S_4$.[I am having a problem to explicitly write this]

There is also one thing that I know is that if I take $\Bbb Z_2$ as isomorphic to a the group $H_1=\langle(12)(34)\rangle$ and $H_2=\langle(23)(14)\rangle$ Then maybe some sort of isomorphism is possible. Can someone help out please by writing the proof or giving the general structure of the proof.

答:
By Cayley's theorem, any group $G$ embeds into $S_G$ by left multiplication, say $\lambda$. If furtherly $G$ is finite, then $S_G\cong S_{|G|}$ via the isomorphism $\alpha\mapsto f^{-1}\alpha f$, where $f\colon\{1,\dots,|G|\}\to G$ is any bijection. So, yes, $G=\Bbb Z_2\times \Bbb Z_2$, like any other group of order $4$ (other, there is just the cyclic one, $\Bbb Z_4$), is isomorphic to a subgroup of $S_{|\Bbb Z_2\times \Bbb Z_2|}=S_4$. For instance, for:

\begin{alignat}{1}
f(1)&=(0,0) \\
f(2)&=(0,1) \\
f(3)&=(1,0) \\
f(4)&=(1,1) \\
\end{alignat}

we get:

\begin{alignat}{1}
f^{-1}\lambda(0,0)f&=() \\
f^{-1}\lambda(0,1)f&=(12)(34) \\
f^{-1}\lambda(1,0)f&=(13)(24) \\
f^{-1}\lambda(1,1)f&=(14)(23) \\
\end{alignat}

i.e. $\Bbb Z_2\times \Bbb Z_2\cong \{(),(12)(34),(13)(24),(14)(23)\}$. And similarly for other choices of $f$.

Title: Doubts in $\Bbb Z_2 \times \Bbb Z_2$


答:
I think that a way to see this problem is that, with your condition, a sequence isn't in S if 0 or 1 appear a finite amount if time.

So I think you can caracterize $\{0,1\}^{\mathbb{N}}/S=\left(\{0,1\}^*\cup \{0\}^{\mathbb{N}} \right) \cup \left(\{0,1\}^*\cup\{1\}^{\mathbb{N}} \right)$

Where $ \{0,1\}^*$ reprsents all finite sequence of $\{0,1\}$ (which is countable), and $\{0\}^{\mathbb{N}} $ and $\{1\}^{\mathbb{N}} $ are countable (carinal $\aleph_0 $), So you get that $\{0,1\}^{\mathbb{N}}/S$ is countable.

However, $\{0,1\}^{\mathbb{N}}$ isn't countable, so you must have $S$ uncountable


问:
Prove that elementary tensor is non-zero in proof of non-flatness
This is a follow-up question to this question (and the answer there by René Schipperus) about proving that $k[t]$ is non-flat as $k[t^2,t^3]$-module.

I have reduced this to showing that $t\otimes t$ is non-zero in $k[t]/(t^2)\otimes_{k[t^2,t^3]} k[t]$. To show that an elementary tensor is non-zero, I have to find a $k[t^2,t^3]$-bilinear map from $k[t]/(t^2) \times k[t]$ to an abelian group such that the image of $(t,t)$ is non-zero.

Does someone have a hint?

答:
Let $A=k[t]/(t^2)$ and $u=[t]\in A$ satisfying $u^2=0$.

The action is given by $f\cdot(a+bu)=f(u)\cdot(a+bu)$ for $f\in k[t]$, in particular $f=t^2$ annihilates all of $A$.

Then consider $\varphi:A\times k[t]\to A$ mapping
$$(a+bu,\,f)\,\mapsto\,b\cdot f(u)$$
We get $\varphi((a+bu)c,\,f)=\varphi(a+bu,\,cf)$ for any $c\in k$, and the actions of $t^2,t^3$ make both sides $0$, so it's $k[t^2,t^3]$-bilinear, and it maps
$$\varphi([t],\,t)=\varphi(u,t)=1\cdot u=u\,\ne 0\,.$$


答:
$$S = \left\{ f \in \{0,1\}^{\mathbb{N}} \middle| (\forall x \in \mathbb{N})( \ \exists y \in \mathbb{N}): y > x \land f(y) = f(x) \right\}$$

The function gives as result any sequence of $0,1$ these are the binary representation of the real numbers, for instance in the interval $[0,1]$ therefore the set $S$ is more than countable.


问:
L'Hôpital's Rule and Infinite Limits
I was wondering if anyone could help me with computing a limit using L'Hôpital's Rule.

Using L'Hôpital Rule for the following limit, I get the following result:

\begin{equation}
\lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \therefore \lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \stackrel{L'H}{=} \lim_{x \to 0} \frac{e^x-2x}{4x^3+3x^2+2x}
\end{equation}

(1/0) which means that the limit is either $+\infty$, $-\infty$ or it does not exist.

If I want to find out the answer, I should take the side limits - but here lies my question: the side mimits of what? The original function or the function after L'Hôpital rule? Or any of the two?

Should I do this

\begin{equation}
\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty
\end{equation}
\begin{equation}
\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = +\infty
\end{equation}

Or this?

\begin{equation}
\lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty
\end{equation}
\begin{equation}
\lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = +\infty
\end{equation}

Coincidentally (or not) both limits (from the right of both equations and from the left of both equations) give the same answer. Therefore, the limit does not exist.

For the other function below, the same thing happens:

\begin{equation}
\lim_{x \to 0} \frac{sin(x)}{x^4+x^3}
\end{equation}

So what I dind't find out is which function I should use, or if it really doens't matter.

Thank you

答:
The right way to proceed is the following from what you have found

$$\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = \infty \implies \lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = \infty$$

$$\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty \implies \lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty$$

where the LHS are evaluated by l'Hospital's rule and the RHS are derived form this result.

Therefore in this case we conclude that limit doesn't exist.

For the other one we can also proceed as follows

$$\frac{\sin(x)}{x^4+x^3}=\frac{\sin(x)}{x}\frac{x}{x^4+x^3}=\frac{\sin(x)}{x}\frac{1}{x^3+x^2} \to 1 \cdot \infty=\infty$$

Title: L'Hôpital's Rule and Infinite Limits


问:
What is the difference between a random variable and an event?
In layman terms, what is the difference between a random variable and an event? To my understanding a random variable is a function outputting a real number. And an event is an outcome or a set of outcomes.</p>

答:
We have a large (and maybe not well describable) space $\Omega$ of possible outcomes. For example, $\Omega$ might be the set of possible weathers tomorrow at the village where I live.

An event is a decent  subset of $\Omega$, say the set $R\subset\Omega$ of all weathers where it rains at time 10:00 tomorrow. When a probability measure ${\tt P}$ has been defined on $\Omega$ it makes sense to talk of the probability ${\tt P}(R)$ of the event $R$, and the well known rules about the probabilities of unions, etc., of events hold.

A random variable is a given real-valued function defined on $\Omega$, say, the temperature at the graveyard at time 12:00. This is a nice function, and there is nothing random about it. When fate has chosen a weather $\omega\in\Omega$  the function value $f(\omega)$ is determined. This value is considered "random", before one has really observed it.

Usually the probability that $f$ has a particular value, say $21.35^\circ$ Celsius, is zero. This particular event is just too special. But one can easily define  events using $f$ that have an interesting probability, say the event  $Z$, consisting all weathers $\omega$ with $f(\omega)<0^\circ$ Celsius. We see here that a random variable can be used to create lots of events that are of interest in the momentary discussion.

On the other hand an arbitrary event, say the $R$ from above, can be redesigned as a random variable, when we want to argue in the language of random variables: The characteristic function of $R$, being $\equiv1$ on $R$ and $\equiv0$ outside of $R$ is a perfect random variable, albeit one with just two values.


问:
continuous functions problem

I dont understand, isn't this function already continuous on (−∞,∞)

答:
I dont understand, isn't this function already continuous on (−∞,∞)

It is indeed continuous everywhere, except possibly at $x=-5$. There, the continuity depends on the value of (the parameter) $a$.

Now the question is: find the value of (this parameter) $a$ so that the function is indeed continuous at $x=-5$ (and thus, everywhere).

Now rely on the definition and/or relevant properties of continuity you know (think limits!).

Title: continuous functions problem


问:
Can all possible angles on a rational triangle be represented as a rational multiplied by Pi?
The reason I ask: I was wondering if it was possible to find the angle of a rational triangle by only using the lengths of its sides and knowledge of $\pi$ (that is, no inverse trig functions).

So, that probably isn't possible unless every angle $\theta$ such that $\sin (\theta)$ is a rational number can be represented as $\theta = a \pi$ with $a$ is some rational number.

Clarification edit: A rational triangle is a triangle where every ratio of sides is rational.

答: I don't know what you mean by a "rational triangle", but it is extremely unusual for $\sin\theta$ to be rational for $\theta$ a rational multiple of $\pi$; indeed, the only examples are $a=0,1/6,1/2$ and the ones you can obtain from these by symmetries of the sine function. </p>


问:
How to find first term, common difference, and sum of an arithmetic progression?
Given that an arithmetic progression is such that the 8th term is twice the second term, and the 11th term is 18.
Find:
1) The first term and common difference.
2) The sum of the first 26 terms.
3) The smallest of the progression whose values exceed 126?

How on earth am I meant to solve this? I'm guessing you try and find a formula for the nth term, but I have no clue how to get there.
Any suggestions?

Body:
From $$\begin{cases}a+7d=2(a+d),\\a+10d=18\end{cases}$$

$$d=\frac65,a=6.$$

$$S=26\frac{6+\left(6+25\,\dfrac65\right)}2.$$

$$6+(n-1)\frac65>126\implies n\ge102.$$


问:
Is the space of left cosets of a topological group completely regular?
Let $G$ be a (Hausdorff) topological group and $H$ its closed subgroup. Then it's known that $G/H$ is a regular topological space. Is it always completely regular?

A reference is good too.

In the case that $H$ is compact, we know that the quotient map $\pi:G\to G/H$ is perfect and open, therefore $G/H$ is completely regular (since it is an open-and-closed map; image of a completely regular space by a map that is both open and closed is completely regular; see Chaber J. "Remarks on open-and-closed mappings").

答: This is an exercise 3.3.F in the book "Topological Groups and Related Structures" by Arhangel'skii and Tkachenko.</p>


问:
How to express $\cos(20^\circ)$ with radicals of rational numbers?
In showing that the trisection of an angle with ruler and compass is not possible in general one shows that $\cos(20^\circ)$ cannot be constructed (thus the angle $60^\circ$ cannot be trisected) by determining its minimal polynomial, which is $8x^3-6x-1$.

Solving $8x^3-6x-1=0$ yields a solution $x_1=\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}+\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$. Expressing $\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}$ and $\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$ in polar form yields $x_1=\cos(20^\circ)$.

Is it possible to express $\cos(20^\circ)$ with radicals without complex numbers?

答:
The problem arises in the fact that the polynomial you got has 3 real roots and is irreducible over the rationals. It is an example of casus irreducibilis, which essentially shows that certain algebraic numbers whose minimal polynomial is of degree 3 or higher are real themselves but require complex numbers in order to be expressed in radicals.

This is pretty much as simple as you can get:

$$
\cos\left(20^{\circ}\right)=\frac{\sqrt[3]{1-i\sqrt{3}}+\sqrt[3]{1+i\sqrt{3}}}{2\sqrt[3]{2}}
$$

...In which the imaginary parts simply cancel out leaving a real number. Again, you cannot remove the complex numbers from this expression.


问:
How to find if a function reaches its maximum values?
How can I find if a function reaches its maximum values?

For example, for the function $f: (-1, 1) \to \mathbb R, f(x) = \ln\frac{1-x}{1+x}$

To reach its maximum values, there should be an $x_0\in (-1, 1)$ so that $f(x_0) = \inf A = -1$ and an $x_1 \in (-1, 1)$ so that $f(x_1) = \sup A = 1$.

And for the function $ f : [0, 1]\to \mathbb R$, $f(x) = x$ if $x ∈ (0,1] $ and $ f(x) = \frac{1}{2} $ if $ x = 0$

To reach its maximum values, there should be an $x_0 \in [0, 1] $ so that $f(x_0) = \inf A = 0 $ and an $x_1 \in [0,1]$ so that $f(x_1) = \sup A = 1$.

I thought about proving that the first function is continuous, but I can't do the same thing for the second function.

答:
For the first one :
$$f(x)= \ln\left(\frac{1-x}{1+x} \right) = \ln \left(\frac{2}{x+1}-1\right)$$

so $f$ is strictly decreasing. Because its domain is the open interval $(-1,1)$, then $f$ cannot attain its bounds (neither its $\sup$, nor its $\inf$).

For the second one, you can see that for all $x \in [0,1]$, $f(x) \leq 1$, and $f(1)=1$, so $f$ attains its $\sup$. But $f$ does not attains its $\inf$ : indeed, $f(x) \geq 0$ for all $x$, and $\lim_{x \rightarrow 0 \\ x > 0} f(x) = 0$, so $\inf f = 0$, but $0$ is not in the image of $f$.


问:
How to encode the "statement to be proven" for the existence of the cartesian product in FOL.
One of the exercises I am trying to carry out is proving the existence of the cartesian product $S \times T$ for any two arbitrary sets $S$ and $T$. My question is not about how to construct this set (I feel fairly good-to-go with that).

I am trying to develop my ability to document what exactly it is that I am proving. I have two questions about the First Order Logic syntax of the overall statement I am trying to prove.

Firstly: is the overarching statement effectively something along the lines of: $$\forall S,T \  \exists \ N \varphi(N)$$ where $N$ is the "$S \times T$" and $\varphi(N)$ refers to the properties we associate with a cartesian product?

Secondly: when I am trying to demonstrate that this can be done for any arbitrary sets $x$ and $y$ (which would then generalize to any two sets), it seems like I am trying to prove an implication that looks something like (written in pseudo-FOL / English):

$$ x \ \text{and}\  y \ \text{are sets that can be constructed in our domain of discourse} \rightarrow x \times y \  \text{is a set that can be constructed in...etc }$$

I am a little uncertain about my proposed answer to the second question because I am unsure of how to encode the property "can be constructed in our domain of discourse" using FOL.

I've previously seen the following syntax to establish that "something exists": $\exists x (x=x)$.

So maybe the implication is better written as:

$$\exists x (x=x) \land \exists y (y=y) \rightarrow \exists x \times y ( x \times y = x \times y)$$

You can probably see that this question generalizes to the construction of any sets (not just the particular instance of the cartesian product set), so please feel free to speak more generally.

Looking forward to the input!
Thanks~

Edit:

Using the proper syntax provided by Mauro Allegranza below...

We set out to prove the following statement:

$\forall S \forall T \exists C \forall z [z \in C \leftrightarrow z \in \mathcal P( \mathcal P (S \cup T)) \land \exists x \exists y (x \in S \land y \in T \land z=(x,y))].$

Now, the way I would approach this is the following:

Consider two arbitrary sets $N$ and $M$.

Prove the following:

$\exists C \forall z [z \in C \leftrightarrow z \in \mathcal P( \mathcal P (N \cup M)) \land \exists x \exists y (x \in N \land y \in M \land z=(x,y))]$

If I can prove the above statement for arbitrary sets $N$ and $M$, then I have proven it for any two sets.

My question is...what am I "doing" (in the context of FOL) when I assert - "Consider two arbitrary sets $N$ and $M$." What is this statement? Does it have a truth value / interpretation? Is it an "English abbreviation" for some sort of FOL syntax?

答:
We have that: $S \times T = \{ (x,y) \mid x \in S \text {  and  } y \in T \}$.

Thus, a suitable formula defining it would be: $\exists C \forall z [z \in C \leftrightarrow \exists x \exists y (x \in S \land y \in T \land z=(x,y))]$.

But this is not enough, because in order to asserts that set $C$ exists we have to apply Separation.

In order to do this, we have to find a suitable set $A$ from which "separate" it.

If we adopt the common Kuratowski's encoding of $(x,y)$ as $\{ \{ x \}, \{ x,y \} \}$ we have, for $x \in S$ and $y \in T$, that:

$\{ x \} \in \mathcal P(S)$ and $\{ x,y \} \in \mathcal P (S \cup T)$.

Thus, $(x,y) \in \mathcal P( \mathcal P (S \cup T))$.

In conclusion, the correct instance of the axiom will be:

$\forall S \forall T \exists C \forall z [z \in C \leftrightarrow z \in \mathcal P( \mathcal P (S \cup T)) \land \exists x \exists y (x \in S \land y \in T \land z=(x,y))].$

The formula says that, for every pair of sets $S$ and $T$ their cartesian product $S \times T$ exists.


问:
Solve $\sin(z) = z$ in complex numbers
Show that $\sin(z) = z$ has infinitely many solutions in complex numbers.
Little Picard theorem should help, but using big Picard theorem is undesirable.
Thanks a lot!</p>

答:
The function $\sin z - z$ is entire and of order $1$, so if it had finitely-many zeros then the Hadamard factorization theorem would say that

$$
\sin z - z = e^{az+b} p(z),
$$

where $p(z)$ is some polynomial and $a$ and $b$ are some complex numbers.

This can't be true.  We know that $\sin z - z \sim e^{-iz}$ as $z\to +i\infty$ and $\sin z - z \sim e^{iz}$ as $z \to -i\infty$ for $\operatorname{Re} z = 0$, and when taken together these properties imply the absurd conclusions that, not only is $p(z)$ constant, but $\operatorname{Im} a < 0$ and $\operatorname{Im} a > 0$.

Title: Solve $\sin(z) = z$ in complex numbers


问:
Dimension over $\mathbb{Q}$ of infinite sums of rational functions
Let $P(n)=(n+r_1)(n+r_2)...(n+r_k)$ be a polynomial with simple, rational, negative roots (i.e. $r_i>0$) and degree $k\geq 2$ (I stick with negative roots as I don't have to worry about dividing by $0$). My question is: What can we say about the dimension of the set

$$S=\text{span}\left\{\sum_{n=0}^\infty \frac{1}{P(n)},\sum_{n=0}^\infty \frac{n}{P(n)},...,\sum_{n=0}^\infty \frac{n^{k-2}}{P(n)}\right\}$$

over $\mathbb{Q}$ in terms of the roots $r_1,r_2,...,r_k$?

Work so far: For the case $k=2$, it is not hard to show that

$$\sum_{n=0}^\infty \frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}$$

(where $\psi(x)$ is the digamma function). Since this is an increasing function for $x>0$ we know

$$\sum_{n=0}^\infty\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\neq 0$$

and therefore the dimension can only be $1$. In fact, we can use this fact to prove the dimension is never $0$. Assume by way of contradiction that there exists a polynomial $P(n)=(n+r_1)(n+r_2)+...+(n+r_k)$ such that

$$\dim\left(\text{span}\left\{\sum_{n=0}^\infty \frac{1}{P(n)},\sum_{n=0}^\infty \frac{n}{P(n)},...,\sum_{n=0}^\infty \frac{n^{k-2}}{P(n)}\right\}\right)=0$$

(i.e. all of the elements are $0$). Then pick rationals $a_i$ such that

$$Q(n)=\sum_{i=0}^{k-2}a_i n^i=(n+r_3)(n+r_4)+...+(n+r_k)=\frac{P(n)}{(n+r_1)(n+r_2)}$$

Since the dimension of the space is $0$, we know

$$0=\sum_{n=0}^\infty\frac{1}{P(n)}\sum_{i=0}^{k-2}a_in^i=\sum_{n=0}^\infty\frac{Q(n)}{P(n)}=\sum_{n=0}^\infty\frac{1}{(n+r_1)(n+r_2)}\neq 0$$

(which follows from the $k=2$ case). And finally, for the $k=3$ case, a paper by Saradha and Tijdeman proves the dimension is $1$ if and only if all the roots are integer lengths apart (otherwise it is $2$). This pattern extends to higher $k$ as well: For $k\geq 3$, the dimension of $S$ is $1$ if and only if all roots of $P(n)$ are integer lengths apart (this takes more effort to prove but it is indeed true).

For all other $k\geq 4$, the question seems much more difficult.

Motivation: This question is deeply connected to a previous (false) conjecture that I had (found here) which concerned the rationality of infinite sums of rational functions. This question could be thought of as a modification of that original question as I tried to nail down the important behavior of these objects.

答:
No idea for the dimension but each of your series has a closed form. Doing the partial fraction decomposition of $\frac{x^d}{P(x)}$ which has only simple poles and vanishes at $\infty$ we get

$$\sum_{n=1}^\infty \frac{n^d}{P(n)} = \sum_{n=1}^\infty\sum_{m=1}^k \frac{r_m^d/P'(r_m)}{n+r_m}= \sum_{n=1}^\infty\sum_{m=1}^k \frac{B r_m^d/P'(r_m)}{B n+B r_m}$$ $$= -\sum_{m=1}^k \sum_{n,1\le Bn+Br_m\le  Br_m }\frac{B r_m^d/P'(r_m)}{B n+B r_m}+
\sum_{n=1}^\infty\sum_{l=1}^{B-1} c_{l,d} \frac{ e^{2i\pi ln/B}}{ n}$$ $$=-\sum_{m=1}^k \sum_{n,1\le  Bn+Br_m\le Br_m}\frac{B r_m^d/P'(r_m)}{B n+B r_m}-
\sum_{l=1}^{B-1} c_{l,d}\log(1-e^{2i\pi l/ B})$$
where $B$ is the product of the denominators of the $r_m$, $\sum_{m=1}^k Br_m^k/P'(r_m) =0$ and $$c_{l,d} = \frac1B \sum_{m=1}^k e^{-2i\pi l r_m} \frac{B r_m^d}{P'(r_m)} $$ are the FFT coefficients.


问:
Homotopy on $S^{n-1}$ extends to homotopy on $D^n$
I have the following problem:

I have $\beta:D^n\to Z$ such that $\beta\circ i\simeq h$, where $i:S^{n-1}\hookrightarrow D^n$ is the inclusion and $h:S^{n-1}\to Z$ some map. Then is there a $\beta':D^n\to Z$ with $\beta\simeq\beta'$ such that $\beta'\circ i=h$?

I've looked for the answer a bit online, and I've understood that this condition is being a cofibration, but I didn't explicitly see the proof that $i:S^{n-1}\hookrightarrow D^n$ is a cofibration.

答:
It clearly suffices to show that there exists a retraction $r : D^n \times I \to D^n \times \{0\} \cup S^{n-1} \times I$. In fact, the map $\beta$ and the homotopy $\varphi: \beta \circ i \simeq h$ can be pasted to a continuous $\phi : D^n \times \{0\} \cup S^{n-1} \times I \to Z$ and $\phi \circ r$ is what you desire.

Define
$$r(x,t) = \begin{cases} (\frac{2x}{2-t},0)  & \lVert x \rVert \le \frac{2-t}{2} \\ (\frac{x}{\lVert x \rVert},2 - \frac{2-t}{\lVert x \rVert}) & \lVert x \rVert \ge \frac{2-t}{2} \end{cases} $$
This is a well-defined continuous map because

$\lVert \frac{2x}{2-t} \rVert \le 1$ for $\lVert x \rVert \le \frac{2-t}{2}$

$1 \ge t \ge 2 - \frac{2-t}{\lVert x \rVert} \ge 0$ for $1 \ge \lVert x \rVert \ge \frac{2-t}{2}$

For $\lVert x \rVert = \frac{2-t}{2}$ both parts yield the same value.

For $t=0$ we get $\frac{2-t}{2} = 1$ and thus $r(x,0) = (x,0)$ for all $x \in D^n$. For $x \in S^n$  we have $r(x,t) = (x,t)$. Thus $r$ is a retraction.

There is a nice geometric interpretation of $r$: For each $(x,t)$ consider the line $L_{x,t}$ through $(0,2)$ and $(x,t)$. This line intersects $D^n \times \{0\} \cup S^{n-1} \times I$ in a single point $r(x,t)$.

Remark:

It is very easy to see that for a closed subspace $A \subset X$ the inclusion $i : A \to X$ is a cofibration if and only if $X \times \{0\} \cup A \times I$ is a retract of $X \times I$. See for example Retraction induces Cofibration . This is also true for non-closed $A \subset X$, but the proof is sophisticated. See

Strøm, Arne. "Note on cofibrations II." Mathematica Scandinavica 22.1 (1969): 130-142.


问:
Factorizable vector
How I will be able to prove that:

If $|\psi\rangle$ is not factorizable (entagled with two qbits) then the unitary operator $U_1\otimes U_2$ apply on $|\psi\rangle$ is not factorizable?

*factorizable is when a $|\psi\rangle$ vector can be express than $|\psi\rangle = (a|0\rangle + b|1\rangle)\otimes(c|0\rangle + d|1\rangle)$

答:
I assume you're familiar with the concept of a vector space: a collection of elements called vectors that forms an additive abelian group, with multiplication by 'scalars' coming from some field, in many cases the scalar field is the reals $\bf R$ or the complex numbers $\bf C$. Any complex vector space of finite dimension $n$ is isomorphic to ${\bf C}^n$, the space of $n$-tuples with components from $\bf C$.

The tensor product is a way to combine two vector spaces (it can be more general than this, of course, but that is outside the present scope of discussion). If $V$ and $W$ are complex vector spaces, then the tensor product $V\otimes W$ is a vector space spanned by symbols of the form $v\otimes w$ with $v\in V$ and $w\in W$, subject to the following linearity relations for all $v,v_1,v_2\in V$, $w,w_1,w_2\in W$, $\lambda\in\bf C$:

$$\begin{cases}(v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w \\ v\otimes (w_1+w_2)=v\otimes w_1+v\otimes w_2 \\ (\lambda v)\otimes w=\lambda(v\otimes w)=v\otimes (\lambda w).\end{cases}$$

It is noteworthy that not every vector in $V\otimes W$ can be written as "$v\otimes w$"; these are called pure tensors, or in your terms, factorizable. But every vector in $V\otimes W$ can be written as a linear combination of pure tensors. In particular, any pair of bases ${\cal B}_V$ and ${\cal B}_W$ for $V$ and $W$ respectively induce a basis ${\cal B}_{V\otimes W}=\{r\otimes s:r\in{\cal B}_V, s\in{\cal B}_W\}$ for the tensor product $V\otimes W$.

Suppose $V$ to be the $2$-dimensional free vector space over $\bf C$ spanned by the symbols $|0\rangle,|1\rangle$, which we may interpret as pure states of a physical system. $V$ is itself interpreted as a space of possible states for a qubit. Explicitly, ever vector in $V$ is of the form $\alpha|0\rangle+\beta|1\rangle$ for some $\alpha,\beta\in\bf C$.

To represent two qubits, we look at the tensor product $V^{\otimes 2}=V\otimes V$. Since $\{|0\rangle,|1\rangle\}$ is a basis for just $V$, a basis for $V^{\otimes 2}$ is $\{|0\rangle\otimes|0\rangle,|0\rangle\otimes|1\rangle,|1\rangle\otimes|0\rangle,|1\rangle\otimes|1\rangle\}$, so every vector looks like

$$\alpha\,|0\rangle\otimes|0\rangle+\beta\,|0\rangle\otimes|1\rangle+\gamma\,|1\rangle\otimes|0\rangle+\delta\,|1\rangle\otimes|1\rangle.$$

The vector space is isomorphic to ${\bf C}^4$, but I wouldn't call it that because ${\bf C}^4$ is defined to be a space spanned by $4$-tuples, things that look like $(\alpha,\beta,\gamma,\delta)$, whereas $V^{\otimes 2}$ is spanned by tensors and comes with additional information about how vector can be represented in more than one way via to the linearity relations.

Your problem arises in a bit more generality, but for convenience sake we stick to $V\otimes W$ rather than something $V_1\otimes V_2\otimes \cdots \otimes V_l$. Let $A\in{\rm GL}(V)$ and $B\in{\rm GL}(W)$ be any invertible linear maps on $V$ and $W$ respectively. I claim that there is a (unique) linear map $V^{\otimes 2}\to V^{\otimes 2}$ with the property that $v\otimes w$ is always sent to $A v\otimes B w$. To define a linear map it suffices to define where the elements of a given basis should be sent. Pick any bases ${\cal B}_V$ and ${\cal B}_W$, and define $A\otimes B$ to be the linear map that sends $r\otimes s$ to $Ar\otimes Bs$ specifically for $r\in{\cal B}_V$ and $s\in{\cal B}_W$. Then for any two elements $v\in V$ and $w\in W$ we can write $v=\sum_{r\in{\cal B}_V} a_rr$ and $w=\sum_{s\in{\cal B}_W}b_s s$, so

$$\begin{array}{ll} (A\otimes B)(v\otimes w) & =(A\otimes B)\left[\left(\sum_ra_rr\right)\otimes\left(\sum_sb_ss\right)\right] \\ & = (A\otimes B)\left[\sum_{r,s}a_rb_s(r\otimes s)\right] \\ & = \sum_{r,s} a_r b_s (A\otimes B)(r\otimes s) \\ & = \sum_{r,s} a_r b_s (Ar\otimes Bs) \\ & = \left(\sum_r a_r Ar\right)\otimes \left(\sum_s b_s Bs\right) \\ & = \left(A\sum_ra_rr\right)\otimes \left(B\sum_sb_ss\right) \\ & = Av\otimes Bw.\end{array}$$

This is a very nice property, it makes the world a simpler place.

The adjoint of $U_1\otimes U_2$ has a simple representation: $(U_1\otimes U_2)^*=U_1^*\otimes U_2^*$. To show this, one needs to check that $U_1^*\otimes U_2^*$ satisfies the definition for being an adjoint of $U_1\otimes U_2$; note that the inner product on $V\otimes W$ for these purposes is assumed to be induced from the inner products on $V$ and $W$. Towards this end we need only check the defining condition for the pure tensors, since pure tensors span the space, and this is fairly immediate (it amounts pretty much to juggling around parentheses). This is a nice exercise in the way of initiation to tensors, and I leave it at that.

I also leave the following as exercises (consider everything over $\bf C$):

${\rm Id}_V\otimes {\rm Id}_W={\rm Id}_{V\otimes W}$;

$(A\otimes B)(C\otimes D)=AC\otimes BD$ for any $A,C\in{\rm GL}(V)$, $B,D\in{\rm GL}(W)$;

$(A\otimes B)^{-1}=A^{-1}\otimes B^{-1}$;

if $A\otimes B=C\otimes D$, then $A=C$ and $B=D$.

Thus, if $U_1\otimes U_2$ is unitary, it means that

$${\rm Id}_V\otimes {\rm Id}_W=(U_1\otimes U_2)(U_1\otimes U_2)^*=(U_1\otimes U_2)(U_1^*\otimes U_2^*)=U_1U_1^*\otimes U_2U_2^* $$

hence $U_1U_1^*={\rm Id}_V$ and $U_2U_2^*={\rm Id}_W$, i.e. $U_1$ and $U_2$ are unitary and in particular invertible maps. After all of these preliminaries I will finally address your problem.

We wish to show in $V\otimes W$ that, for any $A\in{\rm GL}(V)$ and $B\in{\rm GL}(W)$,

$$|\psi\rangle~{\rm factorizable}\iff (A\otimes B)|\psi\rangle~{\rm factorizable}.$$

First we show $\Rightarrow$. Suppose $|\psi\rangle=|\phi_1\rangle\otimes|\phi_2\rangle$ for some $|\phi_1\rangle\in V$ and $|\phi_2\rangle\in W$. Then by the previous work we know that $(A\otimes B)|\psi\rangle=(A\otimes B)(|\phi_1\rangle\otimes|\phi_2\rangle)=(A|\phi_1\rangle)\otimes(B|\phi_2\rangle)$, which is clearly factorizable (it's already 'factored'!). For the converse, the $\Leftarrow$ direction, we can actually apply the first direction. That is, if $(A\otimes B)|\psi\rangle$ is factorizable, then

$$|\psi\rangle=(A\otimes B)^{-1}(A\otimes B)|\psi\rangle=(A^{-1}\otimes B^{-1})\big((A\otimes B)|\psi\rangle\big)$$

is factorizable, and this is the statement of the reverse direction.

Title: Factorizable vector