答:
Don't just pick any points $x_n$ in $B_{\frac 1 n} (x_0)$. That will yield  a proof.

We know that $f(U \cap E)$ is not contained in $V$ where $U=B_{\frac 1 n} (x_0)$. So there exists $x_n \in U \cap E$ such that $f(x_n) \notin V$. Now $x_n \to x_0$ so $f(x_n) \to L$. Also $V$ is open and $L \in V$. By defintion of limit this implies that $f(x_n)$ must be in $V$ for $n$ sufficiently large  but $f(x_n) \notin V$ for any $n$. This contradiction proves that a) implies b).

Title: Equivalent definitions of limiting point of function.

Body:
Let $(X,d_X)$, $(Y,d_Y)$ be metric spaces, let $E$ be a subset of $X$ and let $f:X\to Y$ be a function. Let $x_0\in X$ be an adherent point of $E$ and $L\in Y$. Prove the following are equivalent:

a. For every sequence $(x^{(n)})_{n=1}^\infty$ in $E$, converging to $x_0$, the sequence $f((x^{(n)}))_{n=1}^\infty$ converges to $L$.

b. For every open set $V\subseteq Y$ which contains $L$, there exists an open set $U\subseteq X$ containing $x_0$ such that $f(U\cap E)\subseteq V$

Assuming a. I want to prove by contradiction,

Soo suppose $L\in V$ which is open in $Y$ and assume there is no open set $U\subseteq X$ containing $x_0$ such that $f(U\cap E)\subseteq V$

Define the sequence $(x_n)_{n=1}^\infty$ by $x_n\in B_{\frac{1}{n}}(x_0)$

By archimedean property for any $\epsilon >0$ there will exist an $n$ large enough that $d_X(x^{(n)},x+0)<\epsilon$, so this sequence converges to $x_0$.

Since $V$ is open and $L\in V$, there exists an open ball $B_r(L)\subseteq V$

By assuming $a.$ however we know that $f((x^{(n)}))_{n=1}^\infty$ converges thus there exists an $m\in \mathbb{N}$ such that if $n\geq m$, $d_Y(f(x^{(n)}),L)<r$, so it is in $B_r(L)$, or in other words, I know there is a ball small enough $B_{\frac{1}{m}}(x_0)$ such that all point of my sequence within this ball have image in $B_r(L)$.

How do I go from saying I know all the points which belong to my sequence which are in $B_{\frac{1}{n}}(x_0)$ have image in $B_r(L)$ to know the entire open ball is, regardless of whether the point is in my sequence or not?


问:
What is the remainder when $1^n + 2^n + 3^n + \ldots + 99^n$ is divided by $1 + 2 + 3 + \ldots + 99$?
My first idea on how to approach was to create a polynomial expansion for the first few terms and then try to find a pattern for the rest but this became cumbersome and I don't think that this is a right approach as this would quickly become a problem that involves factorials and I have not covered modular arithmetic, is there a way to approach this problem in such a way that does not involve using factorials with mods.</p>

答:
$$\text{We have}\quad \sum_{x=1}^n x^2 =\frac{n(n+1)(2n+1)}{6}\quad \text{and}\quad \sum_{x=1}^n x=\frac{n(n+1)}{2}$$

$$\text{If we divide the first by the second, we can readily see terms cancel so}\quad \frac{\sum_{x=1}^n x^2}{\sum_{x=1}^n x}=\frac{(2n+1)}{3}$$
Plugging in the value $99$ we get $R=2n+1\mod 3 \implies \frac{199}{3}=66\space R1 $

$$\text{For cubes we have}\quad \sum_{x=1}^n x^3=\frac{n^2(n+1)^2}{4}\implies R=n(n+1)\mod 2 \\
\implies \frac{9900}{2}=4950\space R0$$
Sums of odd powers are divisible by $\frac{n(n+1)}{2}$

Sums of even powers are divisible by $\frac{n(n+1)(2n+1)}{6}
=\frac{n(n+1)}{2}\times\frac{2n+1}{3}\\$

I have in the past worked out summation formulas for powers up to $x^{31}$ the process is tedious and I would have trouble locating them right now. I could have gone farther but for lack of arbitrary precision. Any formula can be generated with a set of simultaneous equations represented in matrix form where the coefficients come from a modified Pascal's triangle or the formula's may be generated directly with selected Bernoulli numbers but that is beyond the scope of this question, Here is one more sample what we are looking for.

$$\sum_{x=1}^n x^4=\frac{n(n+1)}{2}\cdot \frac{(2n+1)(3n^2+3n-2)}{15}\implies R=(2n+1)(3n^2+3n-2))\mod 15$$
$$\implies \frac{5909902}{15}=393993\space R7$$

I disagree with the sequence suggested by one of the comments because it appears that, for $x^2, R=1$ and for
$x^3, R=0$ and for $x^4, R=7$ (do check my math)

I can't see a general solution but you might play with the formulas for $x^1$ thru $x^{10}$ shown here.

$\textbf{Update: }$ I thought I had something but perhaps not or my math is bad in the last example below.

$R=n(n+1)(2n^2+2n-1) \mod 6 \implies \frac{196010100}{6} = 32668350\space R0$

$R = (2n+1)(3n^4+6n^3-3n+1) \mod 21\implies \frac{58506059899}{21} =2786002852\space R 7$

$R= n(n+1)(3n^4+6n^3-n^2-4n+2)  \mod 12 \implies \frac{2910504979800}{12}. = 242542081650\space R0$

$R=  (2n+1)(5n^6+15n^5+5n^4-15n^3-n^2+9n-1) \mod 45 \\  \qquad\qquad\qquad  \implies \frac{965252482830701}{45} = 21450055174015\space R26$

$R= n(n+1)(n^2+n-1)(2n^4-4n^3-n^2-3n+3)\mod 10\\ \qquad\qquad\qquad
\implies \frac{18446354100791100}{4950} = 3726536181978\space R0$

$R= (2n+1)(n^2+n-1)(3n^6+9n^5+2n^4-11n^3+3n^2+10n-5) \mod 11\\
\qquad\qquad\qquad \implies
\frac{5732827616228279495}{11}= 173722048976614530 \space  R5$

$\\$

The pattern so far is $1,0,7,0,7,0,26$ but, as indicated above, the last ones discovered may be math errors and the most likely "continuing" sequece is $0,7,0,7,0,7\cdots$

ContentLicense: CC BY-SA 4.0


问:
What is the Point, and Meaning, of the Mean Value of a Function?
Ok, so I know that the mean value of a function, $f(x)$, on the interval $[a,b]$ is given by (or defined by?)
$$\frac{1}{b-a}\int_a^bf(x)~dx$$
but I have $2$ basic questions about this:

$1$: From a purely mathematical point of view, does this have any practical use? Does it give us any extra weapon to add to our mathematical arsenal?

$2$: How excatly can I interpret this as a mean value? I've seen the geometrical interpretation (looking at the area under the graph within the given interval), but I still don't understand how it links to a mean value. For a finite set of values I divide the sum of the values by however many values I have to obtain the mean, but in this case there are infinitely many values, so how is this a mean value?

Thanks for your help.

答:
You wrote:

For a finite set of values I divide the sum of the values by however many values I have to obtain the mean...

Let's look at a Riemann sum for the expression $\frac{1}{b-a} \int_a^b f(x) \, dx$, namely:
$$\frac{1}{b-a} \sum_{i=1}^n f(x_i) \, \Delta x = \frac{1}{b-a} \sum_{i=1}^n f(x_i) \, \frac{b-a}{n} = \frac{\sum_{i=1}^n f(x_i)}{n}
$$
Surprise! It's a finite sum of values divided by the number of values.

Now the amazing thing is that if you study what happens to the numerical values of these finite mean expressions as $n \to \infty$, the absolutely amazing thing is that it approaches a limit (assuming $f : [a,b] \to \mathbb R$ is continuous). That limit has its own special notation
$$\frac{1}{b-a} \left(\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \, \Delta x \right) = \frac{1}{b-a} \int_a^b f(x) \, dx
$$
And for this reason, that last expression is called the mean value of $f$ itself, rather than just the mean value of a selected finite number of values of $f$.


问:
What wrong with pointwise convergent
It is true that the uniform limit of continuous functions is continuous as it has proved as a theorem, but what is wrong with pointwise limit? I mean why this theorem it doesn't work for if the convergence is pointwise?

Can any one clarify this to me may be using epsilon definition if possible?

Thank you.

答:
You can view this as an instance of the question $$
  \lim_{n\to\infty} \lim_{m\to\infty} g(n,m) \overset{?}= \lim_{m\to\infty} \lim_{n\to\infty} g(n,m) \text{,}
$$
i.e. of the question whether two limit-taking operations can be swapped. If you have a pointwise converging sequence of continuous function $f_n$, simply pick an $x$ and an arbitrary sequence $x_m \to x$, and set $$
  g(n,m) = f_n(x_m) \text{.}
$$
The questionable identity above then becomes \begin{align}
  \underbrace{\lim_{n\to\infty} \underbrace{\lim_{m\to\infty} f_n(x_m)}_{=f_n(x) \:\text{($f_n$ continuous!)}}}_{=f(x)\:\text{($f_n\to f$ pointwise!)}}
  &\overset{?}=
  \lim_{m\to\infty} \underbrace{\lim_{n\to\infty} f_n(x_m)}_{=f(x_m) \:\text{($f_n\to f$ pointwise!)}} &\Leftrightarrow \\
  f(x) &\overset{?}= \lim_{m\to\infty} f(x_m) \text{,}
\end{align}
which simply asserts that $f$ is continuous at $x$.

So one possible answer to why this fails is because "limits cannot, in general, be swapped".

ContentLicense: CC BY-SA 3.0

Title: What wrong with pointwise convergent


问:
Is there a name for this figure with alternating colors in the complementary set of Sierpinski's triangle?
Start constructing Sierpinski triangle. On odd step - remove paint center white. On even step paint center blue. This figure has areas with arbitrary small patches of color, yet, area of both white and blue color is greater than zero. Lebesgue measures of white and blue colors are probably 4/7 and 3/7 respectively.

The resulting figure (first 6 iterations) is shown below.

What is the name of this object? Has it been studied?

答:
Your question adresses the way to "access" (with its Computer Science meaning: the way to reach data) triangles in the complementary set of Sierpinski's triangle.

Here is an answer to your issue; I haven't gone into a full description, but I think that all the ideas are there.

It is based on two concepts:

barycentric coordinates,

their binary decompositions.

Let $ABC$ be the filled triangle in which the Sierpinski set is defined.

Let $$\begin{cases}u&=&u_0.u_1u_2u_3\cdots,\\ v&=&v_0.v_1v_2v_3 \cdots,\\ w&=&w_0.w_1w_2w_3\cdots\end{cases}$$

be the barycentric coordinates of a generic point $M$, expressed in bicimals (binary digits). Then

(a) $u_0,v_0,w_0$ are equal to $1$ iff $M=A, B$ or $C$. In such cases, all the other $u_i,v_i,w_i$ are zero.

(b) $M$ belongs to Sierpinski's triangle if and only if $u_i + v_i + w_i = 1$ for all $i$.

Property (a) is evident. Property (b) is presented in this Wikipedia article).

Let us consider for example a recursion level of $6$, the center of each elementary triangle can be described by the placement of zeros in the sequence:

$$R=\begin{pmatrix}u_1&u_2&u_3&u_4&u_5&u_6\\
v_1&v_2&v_3&v_4&v_5&v_6\\ w_1&w_2&w_3&w_4&w_5&w_6\end{pmatrix}$$

In the following graphical representation, an "approximation" of Sierpinski triangle, is represented by small red triangles.

With the example given in its legend, I think it will not be difficult to deduce a general description of the set of triangles you are interested in.

(see below the Matlab program which has generated the figure).

Fig. 1: The yellow triangle delimitated by the black lines is completely characterized by the following set of conditions on the barycentric coordinates of its points: $0<u<1/4, \ 0<v<1/4, \ 1/2<w<3/4$, which can be expressed on the binary decompositions of $u,v,w$ under the form $u_1=0, \ u_2=0, \ v_1=0, $$ \ v_2=0, \ w_1=1, \ w_2=0$.

I leave you the final task. i.e., defining which arrangements of bicimals give precisely the triangles you want to describe, being understood that placing a constraint on the bicimals situated for example on the right of array $R$ will have an incidence on the smallest triangles.

Matlab program for this figure:

```<code>clear all;close all;hold on;axis equal;
v=exp(i*2*pi/3*(0:2)); % cubic roots of 1
p=6;e=2^p;
for a=0:e;
    for b=0:e-a-1;
        c=e-a-b-1;
        % (a/e,b/e,c/e) = barycentric coordinates.
        R=[dec2base(a,2,p)-48;
           dec2base(b,2,p)-48;
           dec2base(c,2,p)-48];
        % Column R(:,k) contains "bicimals" uk,vk,wk
        t=sum(sum(R));
        if t==p;col='r';end; % (red) Sierpinski triangle
        if R(:,1)'==[0,0,0];col='g';end; % green tr.
        if sum(R(:,1))==1;col='y';end; % yellow tr.
        if sum(R(:,2))==1;col='b';end; % blue tr.
        z=a*v(1)+b*v(2)+c*v(3);
        fill(real(z+v),imag(z+v),col,'edgecolor','none');
    end;
end;
</code>```

Body:
Your question adresses the way to "access" (with its Computer Science meaning: the way to reach data) triangles in the complementary set of Sierpinski's triangle.

Here is an answer to your issue; I haven't gone into a full description, but I think that all the ideas are there.

It is based on two concepts:

barycentric coordinates,

their binary decompositions.

Let $ABC$ be the filled triangle in which the Sierpinski set is defined.

Let $$\begin{cases}u&=&u_0.u_1u_2u_3\cdots,\\ v&=&v_0.v_1v_2v_3 \cdots,\\ w&=&w_0.w_1w_2w_3\cdots\end{cases}$$

be the barycentric coordinates of a generic point $M$, expressed in bicimals (binary digits). Then

(a) $u_0,v_0,w_0$ are equal to $1$ iff $M=A, B$ or $C$. In such cases, all the other $u_i,v_i,w_i$ are zero.

(b) $M$ belongs to Sierpinski's triangle if and only if $u_i + v_i + w_i = 1$ for all $i$.

Property (a) is evident. Property (b) is presented in this Wikipedia article).

Let us consider for example a recursion level of $6$, the center of each elementary triangle can be described by the placement of zeros in the sequence:

$$R=\begin{pmatrix}u_1&u_2&u_3&u_4&u_5&u_6\\
v_1&v_2&v_3&v_4&v_5&v_6\\ w_1&w_2&w_3&w_4&w_5&w_6\end{pmatrix}$$

In the following graphical representation, an "approximation" of Sierpinski triangle, is represented by small red triangles.

With the example given in its legend, I think it will not be difficult to deduce a general description of the set of triangles you are interested in.

(see below the Matlab program which has generated the figure).

Fig. 1: The yellow triangle delimitated by the black lines is completely characterized by the following set of conditions on the barycentric coordinates of its points: $0<u<1/4, \ 0<v<1/4, \ 1/2<w<3/4$, which can be expressed on the binary decompositions of $u,v,w$ under the form $u_1=0, \ u_2=0, \ v_1=0, $$ \ v_2=0, \ w_1=1, \ w_2=0$.

I leave you the final task. i.e., defining which arrangements of bicimals give precisely the triangles you want to describe, being understood that placing a constraint on the bicimals situated for example on the right of array $R$ will have an incidence on the smallest triangles.

Matlab program for this figure:

clear all;close all;hold on;axis equal;
v=exp(i*2*pi/3*(0:2)); % cubic roots of 1
p=6;e=2^p;
for a=0:e;
    for b=0:e-a-1;
        c=e-a-b-1;
        % (a/e,b/e,c/e) = barycentric coordinates.
        R=[dec2base(a,2,p)-48;
           dec2base(b,2,p)-48;
           dec2base(c,2,p)-48];
        % Column R(:,k) contains "bicimals" uk,vk,wk
        t=sum(sum(R));
        if t==p;col='r';end; % (red) Sierpinski triangle
        if R(:,1)'==[0,0,0];col='g';end; % green tr.
        if sum(R(:,1))==1;col='y';end; % yellow tr.
        if sum(R(:,2))==1;col='b';end; % blue tr.
        z=a*v(1)+b*v(2)+c*v(3);
        fill(real(z+v),imag(z+v),col,'edgecolor','none');
    end;
end;


问:
sum of binomial coefficients expansion to prove equation
I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$

I already expanded

$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?

Body: $$\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}=\frac{1\times 2}{2}+\frac{2\times 3}{2}+\frac{3\times 4}{2}+...+\frac{(n-1)\times n}{2}=\frac{1}{2}(1\times 2+2\times 3+3\times 4+...+(n-1)\times n)=\frac {1}{2}\times \frac{(n-1)n(n+1)}{3}=\frac{(n-1)n(n+1)}{6}$$


问:
If we anti-differentiate the Laurent series $ \sum_{n=2}^{\infty} a_n(z - a)^{-n}$, how is the radius of convergence affected?
Let $a, a_n \in \mathbb{C}$ ($n \ge 2$). Suppose we are given the Laurent series

$$ \sum_{n=2}^{\infty} a_n(z - a)^{-n}.$$

I know that we can calculate the set on which this expression converges using Hadamard's formula. That is, we have that the series converges for $|z - a| > \frac{1}{R}$, where $\frac{1}{R} = \limsup \sqrt[n]{|a_n|}$. In case $\limsup \sqrt[n]{|a_n|} = 0$, the series converges for $|z-a|> 0$. And in case $\limsup \sqrt[n]{|a_n|} = \infty$, the series is nowhere convergent.

To continue, suppose that we formally anti-differentiate our series to obtain

$$ \sum_{n=1}^{\infty} \frac{-a_{n+1}}{n}(z - a)^{-n}.$$

My question is, how does the radius of convergence of this new series relate to the old one? Maybe the new radius of convergence is potentially larger? I see that the question boils down to just being smart about Hadamard's formula. We need to compare $\limsup \sqrt[n]{|a_n|}$ with $\limsup \sqrt[n]{\frac{|a_{n+1}|}{n}}$, but this seems like a tricky bit of analysis. It's already clear to me that $\limsup \sqrt[n]{\frac{|a_{n+1}|}{n}} \le \limsup \sqrt[n]{|a_{n+1}|}$, so my hope is that I can somehow compare $\limsup \sqrt[n]{|a_{n+1}|}$ with $\limsup \sqrt[n]{|a_n|}$.

Hints or solutions are greatly appreciated.

答:
Derivatives of Taylor series and Laurent series keep the same radius, simply because they always converge where they exist in their respect Ball or Annulus. That is, derivatives do not introduce new poles. </p>

<p>Thus antidifferentiating a Taylor series or Laurent series cannot yield a new radius, because if you differentiate the antiderivative you should have the same radius as your antiderivative. </p>

Body:
Derivatives of Taylor series and Laurent series keep the same radius, simply because they always converge where they exist in their respect Ball or Annulus. That is, derivatives do not introduce new poles.

Thus antidifferentiating a Taylor series or Laurent series cannot yield a new radius, because if you differentiate the antiderivative you should have the same radius as your antiderivative.


问:
Using indicator random variables on random number generation
Suppose a computer is generating a number at random from 1 to n, and can not generate the same number twice.

A) You try to guess the number, without knowing that it can not generate the same number twice (So you are just guessing a number from 1 to n). What is the expected number of predictions that you get correct? Prove your answer is correct using indicator random variables.

Answer: $E[X]= E\Big[\sum_{i=1}^n X_i\Big] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n 1/n = 1$

B) You try to guess the number, this time having a record of all numbers generated previously (So you are not guessing a number already generated). What is the expected number of predictions that you get correct? Prove your answer is correct using indicator random variables.

This is the question I am struggling on, I don't know if I am not setting something up right, or missing some obvious mathematical answer. I have:

$E[X]= E\Big[\sum_{i=1}^n X_i\Big] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n 1/(n-i) = $

and I have no idea where to go next. My best guess would be using something like so:

$= \sum_{i=1}^n 1/n - \sum_{i=1}^n 1/i = 1 - \sum_{i=1}^n 1/i = 1 - H_n$

Which can't be right because it is less then A, when you should have a better chance in B.

答:
For the question in A) you need to take in account that the numbers doesn't repeat, then the probability to correctly guess the $j$-th time is given by $\Pr [X_j=P_j]$
where $P_j$ is your $j$-th guessing and $X_j$ is the r.v. of possible values at time $j$. Then, as you dont know that the numbers doesn't repeat, $P_j$ is distributed uniformly in $\{1,\ldots ,n\}$ but $X_j$ is distributed uniformly in some random subset $S_j\subset \{1,\ldots ,n\}$ with $|S_j|=n-(j-1)$, thus
$$
\Pr [X_j=P_j]=\Pr [X_j=P_j|P_j\in S_j]\Pr [P_j\in S_j]+\Pr [X_j=P_j|P_j\notin S_j]\Pr [P_j\notin S_j]\\
=\frac1{n-(j-1)}\cdot \frac{n-(j-1)}{n}+0\cdot \frac{j-1}{n}=\frac1{n}
$$
And the expected number of correct predictions is given by
$$
\operatorname{E}\left[\sum_{j=1}^n\mathbf{1}_{X_j=P_j}\right]=\sum_{j=1}^n \operatorname{E}[\mathbf{1}_{X_j=P_j}]=\sum_{j=1}^n \Pr [X_j=P_j]=1
$$

Now for part B) this time $P_j$ is uniformly distributed in $S_j$ for each $j$, so $\Pr [X_j=P_j]=\frac1{n-(j-1)}$, therefore as you guessed correctly you have that
$$
\operatorname{E}\left[\sum_{j=1}^n \mathbf{1}_{X_j=P_j}\right]=\sum_{j=1}^n \operatorname{E}[\mathbf{1}_{X_j=P_j}]=\sum_{j=1}^n\frac1{n-(j-1)}=\sum_{j=1}^n\frac1{j}=H_n
$$
There $H_n$ stay for the $n$-th harmonic number (it is not less than the result in A) as you supposed, note that for $j=1$ you have that $\tfrac1j=1$).


问:
Zero vector in a vector space
I know that every vector space contains a zero element so that u + 0 = 0 + u = u but I am not sure if every vector space contains the literal vector with all zero elements or if it is just a vector that acts as a zero vector?

I'm especially wondering this question because a fact about subspaces is that every vector space has at least two subspaces: itself and the zero vector {0}. But in this case, I am not sure if this refers to the vector with all zero elements or if it refers to the vector in the vector space that just acts like a zero vector

答:
Here's a nice vector space: The set of all functions, which takes real numbers to real numbers.

This is a vector space over $\mathbb{R}$: Adding two functions together gives another function, multiplying a function with a scalar still gives a function, addition is associative, ect.

The zero vector in this example is the constant function, $z(t)=0$ for all $t$.

You can check for yourself that if you take any other function and add it to $z$, that you will get the same function back. Also, given any function $f$, you can define $g:\mathbb{R}\rightarrow\mathbb{R}$ by $g(t)=-f(t)$, and you will get that $f+g=z$.  However, the function $z$ is not "the vector with all zero elements", like you would see for vectors in $\mathbb{R}^n$. In general, vector spaces do not look like lists of elements from the base field.

Title: Zero vector in a vector space


问:
Relationship between second order derivatives and cross derivative of smooth surfaces
Probably a silly question, but I wonder if $z=f(x,y)$ is a smooth surface, and the values of its two second order derivatives $\frac{\partial^2f}{\partial x^2}$ and $\frac{\partial^2f}{\partial y^2}$ are known to be both positive throughout a region, say $0<x<1$, $0<y<1$, dose this place a restriction on the possible values of the cross derivative, i. e. $\frac{\partial^2f}{\partial x\partial y}$ ?

The naive geometrical idea behind this is: if I cut a surface horizontally (parallel to the $x$-axis) or vertically (parallel to the $y$-axis) and always find an exponential (or other) curve, does this makes a random cut obliquely more likely to be a exponential-like (or other-like) curve ?

答:
I think your intuition is a bit off. The quantity $f_{xx}$  tells you about the shapes of cross-section curves lying in planes $y = \text{constant}$, as you said. Similarly, $f_{yy}$ tells you about cross-sections in a perpendicular direction. It seems that this ought to tell you something about oblique sections, I agree. But the flaw in your reasoning is to associate $f_{xy}$ with these oblique cross-sections. In fact, $f_{xy}$ tells you how the slopes of cross-section curves change as you move the sectioning plane. In other words, it tells you about variation of cross-sections, not about the cross-sections themselves.

There is a related theorem that might be of interest: Euler's formula tells you how the curvatures of planar sections vary as you rotate the sectioning plane around a surface normal. In fact, the curvature in any direction can be calculated very simply from the curvatures in two specific directions (called the principal directions) that happen to be perpendicular to each other. Not exactly what you had in mind, but in the same spirit, I think.


问:
Uniqueness of a universal covering manifold
In my question, all the domain and codomain of any covering map is assumed to be connected and locally path-connected.

I'm trying to prove corollary 4.43 (whose proof is left as an exercise) of Lee's "Introduction to Smooth Manifolds" 2nd edition:

Corollary 4.43 (Existence of a Universal Covering Manifold). If $M$ is a connected smooth manifold, there exists a simply connected smooth manifold $\widetilde{M}$, called the universal covering manifold of $M$, and a smooth covering map $\pi:\widetilde{M}\to M$. The universal covering manifold is unique in the following sense: if $\widetilde{M}'$ is any other simply connected smooth manifold that admits a smooth covering map $\pi':\widetilde{M}'\to M$, then there exists a diffeomorphism $\phi:\widetilde{M}\to\widetilde{M}'$ such that $\pi'\circ\phi=\pi$.

I proved existence, put how do I prove uniqueness (the second sentence of the above corollary)? I know the following two theorems.

Theorem 80.3. (from Munkres's "Topology" 2nd ed.) Let $p:E\to B$ be a covering map, with $E$ simply connected. Given any covering map $r:Y\to B$, there is a covering map $q:E\to Y$ such that $r\circ q=p$.

Proposition 4.40 (Covering Spaces of Smooth Manifolds). (from Lee's "Introduction to Smooth Manifolds" 2nd ed.) Suppose $M$ is a connected smooth $n$-manifold, and $\pi:E\to M$ is a topological covering map. Then $E$ is a topological $n$-manifold, and has a unique smooth structure such that $\pi$ is a smooth covering map.

By theorem 80.3, I see that there are topological covering maps $q$ and $q'$ such that $\pi'\circ q=\pi$ and $\pi\circ q'=\pi'$. By proposition 4.40, I know there is a unique smooth structure for each $\widetilde{M}$ and $\widetilde{M}'$ which makes $q$ and $q'$ smooth covering maps. But I don't know if those smooth structures are the same as the original smooth structures. Also, I don't know how to get a diffeomorphism from $q$ and $q'$.

答:
Disclaimer: I am not as familiar with differential topology as I am with algebraic topology.

I believe your question can be answered via the classification of covering spaces. I think it can also be done by invoking various lifting/uniqueness criteria regarding covers, but this way is less technical.

(I am being sloppy and ignoring basepoints/additional conditions)

Two covers $p_1 : X_1 \to X$, $p_2:X_2 \to X$ are said to be isomorphic if there is some homeomorphism $f:X_1 \to X_2$ such that $p_2 \circ f = p_1$. This agrees with the intuitive notion of two covers being "essentially the same".

The classification of covering spaces states (among other things) that there is a bijection

$$\{\text{isomorphism classes of path connected covers of }X  \} \leftrightarrow \{\text{conjugacy classes of subgroups of } \pi_1(X,x_0) \}$$

Where an isomorphism class of a covering space $p:X_1 \to X$ is associated with the image of its fundamental group under the induced map $p_\ast$.

Furthermore, if $H \subseteq G \subseteq \pi_1(X,x_0)$, with corresponding covers $p_H:X_H \to X$, $p_G:X_G\to X$, then there is some cover $f:X_H \to X_G$ with $p_G \circ f = p_H$.

Let $p:E \to B$ be some cover. If $E$ is simply connected, then the image of its fundamental group is trivial (note that this makes 80.3 a special case of the classification). The trivial subgroup is of course its own conjugacy class, so it is associated to a single isomorphism class of covers. From this we can conclude that any two covers with simply connected domains must be isomorphic, and in particular your $q,q'$ are homeomorphisms.

Does this solve your problem?


问:
Let $p$ be prime and suppose $P,Q$ are both $p$-subgroups of a group $G$. Show that $Q\leq N_G(P)$ implies that $QP$ is a $p$-subgroup of $N_G(P).$
Here is the problem statement.

Let $p$ be a prime number and suppose that $P$ and $Q$ are both $p$-subgroups of some group $G$. Show that $Q\leq N_G (P)$ implies that $QP$ is a $p$-subgroup of $N_G (P).$

My idea would be let $h\in QP$ so that $h=qr$ for some $q\in Q$ and $r\in P$. Then $|q|=p^m$ and $|r|=p^k$ for some $m$, $k\in \mathbb{N}$. Now, my intuition tells me that $|h|=p^{m+k}$ but that would require $q$ and $r$ to commute. Also, I wouldn't have used the given fact that $Q\leq N_G(P)$ which I am assuming is there for a reason.

Any hints in the right direction would be great.

答:
Now, my intuition tells me that $|h|=p^{m+k}$ but that would require $q$ and $r$ to commute

Even if they did commute it wouldn't imply $|h|$ is that big. For one thing, there's no assumption that $Q$ and $P$ intersect trivially, so if (for instance) $r=q^{-1}$ then $qr$ has order $1$!

The first thing you want to prove is that $QP=\{qr\mid q\in Q,r\in P\}$ is a subgroup of $G$ in the first place. To do that you will need to use the fact that $Q\le N_G(P)$. (Note for this part the fact that $P$ and $Q$ are $p$-groups is irrelevant.) Second you want to show $QP$ is a subset of $N_G(P)$.

After that you can show $QP$ has $p$-power order. Note its order is the product of $[QP:P]$ and $|P|$, and there is a little trick $[QP:P]=[Q:P\cap Q]$...


答:
A vector space $V$ over the real numbers is a set $V$ together with a map $+\colon V \times V \to V$ (called "addition") and a map $\cdot\colon \mathbb{R} \times V \to V$ (called "scalar multiplication") satisfying a certain list of axioms. One of those axioms is that there is an element $0 \in V$ such that $u + 0 = 0 + u = u$ for all $u \in V$. This element is conventionally called the "zero vector" and labelled with the symbol $0$.

A key thing to note here is that this says nothing about what the elements of $V$ (either the zero vector or the other elements) are, just how they act under the given addition and scalar multiplication operations. What the elements are is completely irrelevant except to the extent that this is used in defining the vector space operations. In other words, the thing that "acts like a zero vector" is the zero vector, by definition.

In particular, there's no notion of "coordinates/entries of a vector" for elements of a general vector space. That notion arises when we choose a basis for a vector space; a choice of basis gives a one-to-one correspondence between elements of the vector space and lists of real numbers (indexed by the basis elements). In the finite-dimensional case, this gives the familiar representation of a vector as a finite list of real numbers. And once you choose a basis, you can indeed prove that the representation of the zero vector in any basis has all coordinates zero.

Title: Zero vector in a vector space

Body:
I know that every vector space contains a zero element so that u + 0 = 0 + u = u but I am not sure if every vector space contains the literal vector with all zero elements or if it is just a vector that acts as a zero vector?

I'm especially wondering this question because a fact about subspaces is that every vector space has at least two subspaces: itself and the zero vector {0}. But in this case, I am not sure if this refers to the vector with all zero elements or if it refers to the vector in the vector space that just acts like a zero vector


问:
Find capacity of water tank with a hole

Suppose I have a water tank with unknown capacity. The water drains proportionality to how full it is, and when it is full it drains at 1 gallon/min.

If the tank is half empty in 15 minutes, how much does the tank hold?

I'm not sure how to set this problem up. Do I model it as a diff. eq. That is, if $C$ is the capacity, then

$$\frac{dC}{dt} = C.$$

Solving this gives:

$$\int \frac{1}{C}dC = \int dt \quad\Rightarrow\quad C(t) = e^{t + k},$$

where $k$ is a constant. But I don't think this is right?

答:
Let $C$ be the full capacity, and let $f(t)$ be how much the tank is filled at time $t$ minutes. We have:

$$f'(t) = -\frac{f(t)}{C}$$

$$f(0) = C$$

$$f(15) = \frac{C}{2}$$

Make sure you see why this models the situation. (For instance, why the minus sign? Or the division by $C$?) The first equation has general solution:

$$f(t) = ae^{-\frac{t - b}{C}}$$

Where $a$ and $b$ are any real numbers. Then, from the boundary conditions we must have:

$$(1)\ ae^{\frac{b}{C}} = C$$

$$(2)\ ae^{\frac{b}{C}-\frac{15}{C}}=\frac{C}{2}$$

Dividing $(1)$ by $(2)$, we get $e^{\frac{15}{C}} = 2$. Thus, $\frac{15}{C} = \ln(2)$, so $C = \boxed{\frac{15}{\ln(2)}\approx 21.640\text{ gal.}}$

Title: Find capacity of water tank with a hole


问:
If $A,B\subseteq \mathbb{R}$ are open and dense subsets in $\mathbb{R}$, then $A\cap B$ is open and dense in $\mathbb{R}$.
I would like some feedback or any tips to improve my proof-writing.

If $A,B\subseteq \mathbb{R}$ are open and dense subsets in $\mathbb{R}$, then $A\cap B$ is open and dense in $\mathbb{R}$.

Proof.

We consider $c\in A\cap B \iff c\in A\land c\in B$. As $c\in A$, then because $A$ is open, there exists $\varepsilon_A>0$ such that $c\in (c-\varepsilon_A,c+\varepsilon_A)\subset A$, and as $c\in B$, then there exists $\varepsilon_B$ such that $c\in (c-\varepsilon_B,c+\varepsilon_B)\subset B$. We take $\alpha = \max\{c-\varepsilon_A,c-\varepsilon_B\}$ and $\beta=\min\{c+\varepsilon_A,c+\varepsilon_B\}$, and see that $c\in (\alpha,\beta)=(c-\varepsilon_A,c+\varepsilon_A)\cap (c-\varepsilon_B,c+\varepsilon_B)\subset A\cap B$. Therefore, whatever may be $c \in A\cap B$, it is an interior point of $A\cap B$, and therefore $A\cap B$ is open.

Again, as $A$ and $B$ are open, there exists $\varepsilon_A>0$ such that $a\in (a-\varepsilon_A,a+\varepsilon_A)\subset A$ for all $a\in A$, and also exists $\varepsilon_B$ such that $b\in (b-\varepsilon_B,b+\varepsilon_B)\subset B$ for all $b\in B$. Well, but the subsets $A,B$ are, not only open, but also dense, then we can expand this formulation such that whatever may be $x\in \mathbb{R}$ and $\varepsilon_A>0$, there will be some $a\in A$ such that $a\in (x-\varepsilon,x+\varepsilon)$, and the same argument can be applied to the subset open and dense $B$. We see, therefore, that whatever may be $x\in\mathbb{R}$ and $\varepsilon>0$, we have that $(x-\varepsilon_A,x+\varepsilon_A)\cap A\neq \varnothing$. We then verify that for whatever value for $x\in\mathbb{R}$ and $\varepsilon>0$ we have $(x-\varepsilon_B,x+\varepsilon_B)\cap B \neq \varnothing$ because $B$ is also dense. But if $A$ is dense, then exists $a\in A$ such that $a\in (x-\varepsilon_B,x+\varepsilon_B)\cap B$ and if $B$ is dense then exists $b\in B$ such that $b\in(x-\varepsilon_A,x+\varepsilon_A)\cap A$. We take $c=a=b$, and therefore $c\in A$ and $c\in B$, we see that exists $c\in A\cap B$ such that $c\in (x-\varepsilon_A,x+\varepsilon_A)\cap (x-\varepsilon_B,x+\varepsilon_B)\cap A\cap B\neq \varnothing$. As this is true for all $x\in \mathbb{R}$, it follows that $A\cap B$ is dense.

答:
Your first paragraph is correct but almost certainly unnecessary, as you surely already know that the intersection of two open sets is always open. There are several problems with your second paragraph, however.

First, your notation implies that there is a single $\epsilon_A>0$ such that for each $a\in A$, $(a-\epsilon_A,a+\epsilon_A)\subseteq A$; this, however, is false. For each $a\in A$ there is an $\epsilon_a>0$ such that $(a-\epsilon_a,a+\epsilon_a)\subseteq A$, but $\epsilon_a$ depends on $a$. The same comments apply, mutatis mutandis, to $\epsilon_B$. This could be bad writing on your part rather than a failure of understanding, but there is a clear error later in the paragraph when you say that we can take $c=a=b$: at that point in the argument you have no reason to think that it is possible to choose $a$ and $b$ to be the same point. Indeed, that is essentially what you are trying to prove.

Specifically, you want to prove that for any $x\in\Bbb R$ and any $\epsilon>0$,

$$(x-\epsilon,x+\epsilon)\cap(A\cap B)\ne\varnothing\,.$$

$A$ is dense in $\Bbb R$, so we know that $$(x-\epsilon,x+\epsilon)\cap A\ne\varnothing\,;$$ let $a\in(x-\epsilon,x+\epsilon)\cap A$. Since $(x-\epsilon,x+\epsilon)\cap A$ is open there is a $\delta>0$ such that

$$(a-\delta,a+\delta)\subseteq(x-\epsilon,x+\epsilon)\cap A\;.$$

And $B$ is dense in $\Bbb R$, so we know that

$$(a-\delta,a+\delta)\cap B\ne\varnothing\,.$$

Let $b\in(a-\delta,a+\delta)\cap B$. Then

$$\begin{align*}
b&\in(a-\delta,a+\delta)\cap B\\
&\subseteq\big((x-\epsilon,x+\epsilon)\cap A\big)\cap B\\
&=(x-\epsilon,x+\epsilon)\cap(A\cap B)\,,
\end{align*}$$

i.e., $(x-\epsilon,x+\epsilon)\cap(A\cap B)\ne\varnothing$, and therefore $A\cap B$ is dense in $\Bbb R$.


问:
Combinations and Permutations Question
I am sitting my A level exams (MEI) and I was looking through the S1 past papers and I encountered this question in the January 2010 paper:

Three prizes, one for English, one for French and one for Spanish, are to be awarded in a class of
  20 students.
  Find the number of different ways in which the three prizes can be awarded if

(i) no student may win more than 1 prize,

(ii) no student may win all 3 prizes.

I tried to do part 1 using combinations, ie: 20 choose 3, but that didn't work.
I couldn't even attempt part 2.

The right answers are:

(i) 20 × 19 × 18 = 6840

(i) (20^3) – 20 = 7980

Could someone please explain to me how these answers have been reached?
The paper is at http://www.mei.org.uk/files/papers/s110ja_4766rev.pdf in case you want to see it. Mark scheme is at the end.

答:
The prize for English may be awarded in any of 20 ways: any student may get it. After the English prize has been awarded, the French prize may be awarded to any of the 19 students who have not yet won the price. After both the English and French prize are awarded, the Spanish prize may be awarded to any one of the 18 remaining students who have not won either of the other two prizes. Since all three prizes are awarded, we multiply the total possibilities: $20\times 19\times 18$.

You may feel a little queasy about why I selected to award the prizes in that order: it doesn't matter in which order you award them. The "first prize" can go to any of 20 students; the "second prize" to any of the remaining 19; and the "third prize" to any of the remaining 18.

First we will count all the cases we want, but add some cases we don't want (because it makes counting easier). Then, we will "throw away" (by subtracting) the cases we don't want.

Forget the restriction first: the English prize can be awarded to any of 20 students; the same for French; the same for Spanish. So there are a total of $20\times 20\times 20 = 20^3$ ways of distributing all the prizes.

But this includes cases in which we award all three prizes to the same student, which we aren't supposed to do. How many of these distributions award all prizes to the same student? Well, we can award all prizes to the first student; or all prizes to the second;...; or all prizes to the 20th student. So in 20 of the cases, we award all prizes to the same student. Since there are a total of $20^3$ ways of distributing the prizes, but $20$ of them are "bad", the total number of "good" ways of distributing the prizes is $20^3 - 20$ (total ways minus bad ways).

Title: Combinations and Permutations Question


问:
Proving A Series is not convergent with little information
For a series Xn, that for all n, `X2n≥n. I have to prove it is not convergent, with only this info. I believe that best way is by contradiction, but cannot quite figure out how. I know I assume an arbitrary L s.t. the abs value(xn - L)< epsilon > 0. Beyond this, I cannot really understand the logic

答:
The following is a way to prove by it by contradiction from the definition of convergence.

If we suppose $(x_n)$ does converge to some $0 \leq x\in \mathbb{R}$, one can disprove this by showing that for some $\epsilon > 0$, we have that for any $N \in \mathbb{N}$, there exists an $n> N$ such that $|x_{n}-x| \geq \epsilon$.

Choose $\epsilon := 1$ and consider an arbitrary $N$.

Choose $n=2(N+x) > N$ which by hypothesis we have $x_{2(N+x)} \geq N+x \implies x_{2(N+x)}-x \geq N$ so that $|x_{n} - x| = |x_{2(N+x)}-x| \geq  N\geq 1=\epsilon$. There were some absolute value considerations I was unable to do deal with along the way, but I am sure a similar argument would work for the case where $x < 0$. Then you would have that there does not exist a real number that $(x_{n})$ converges to.


问:
Construct the splitting field for the polynomial of $x^4-3$ over $\mathbb{Q}$?
I've been given this problem. I think I must proceed in the following way:

Given $a=\sqrt[4]{3}$ and $b=\cos\frac{2\pi}{4}+ \sin\frac{2\pi}{4}i$, the roots of $x^4-3$ are $a$, $ab$, $ab^2$, $ab^3$.

We have $\mathbb{Q}\subset \mathbb{Q}[a]\subset\mathbb{Q}[ab]$ so, is $\mathbb{Q}[ab]$ the desired spliting field?

Given a previous example in the book, it seems it is this way it should be done. However, we extend a previous field in various steps, it's not clear to me when this process "ends", for example: I could extend further and find $\mathbb{Q}[ab^2]$ but in a previous example, it stops before adjoining all the roots, I think the same applies here.

答:
The splitting field of a polynomial is the smallest field extension such that the polynomial decompose into linear factors.

I think you probably mean $K=\mathbb{Q}(a,b)$ obtained by adjoining $a=\sqrt[4]{3}$ and $b=i$ to $\mathbb{Q}$.

According to the definition the splitting the smallest field extension in which $x^4-3$ factors into linear. So once you hit the smallest one, the process stop even you find a bigger one that contains $K$.


问:
Which one is true?
In order to assess the number $N$ of individuals of a species of animal living on an island, we propose to adopt the capture-recapture method. For this we capture $800$ individuals. These individuals, which correspond to the proportion $p$ of the total number of individuals living on the islands, are marked and then released. $1000$ animals are subsequently captured, of which there are $250$ animals marked during the first capture. Which of the following statements are true?

We have $0.20$ as estimation of $p$

$p = \frac{800}{N}$

We have $3200$ as estimation of $N$

I am not sure how to deal with this. Can you help?

EDIT:

I know that two of them are right, but I can say which ones?

答:
The assumptions are that in each case, the animals captured constitute a random sample of the animals on the island,  that whether an animal is captured in the second sample is independent of whether that animal is captured in the first sample, and that no animals are born or die between the two samplings.
If there were $N$ animals in all, the expected number of marked animals in
the second sampling would be $1000 \times 800/N$.  Thus if the actual number is
$X$, you might take $800000/X$ as an estimator of $N$.

Title: Which one is true?


问:
Smooth surface and parametrization for $\varphi: U → \mathbb{R}^3$

Let $U = B(0, 1)$, $\varphi: U → \mathbb{R}^3$, $\varphi(u, v) = (2uv, u^2 − v^2, u^2 + v^2)$ and $S = \varphi(U)$. Determine if $S$ is a smooth surface and if $\varphi$ is the parametrization for it.

So I have that $x=2uv$, $y=u^2-v^2$ and $z = u^2+v^2$. Now $z$ can be expressed as $z=(u-v)^2+2uv =(u-v)^2+x$. Also $y = (u+v)(u-v)$ so $u-v = \frac{y}{u+v}$.

I would get that $z = (\frac{y}{u+v})^2 +x$, thus $\varphi$ would be the correct parametrization?

For the part where they ask if $S$ is smooth I'm not sure what to do. Our course material is pretty awful and I couldn't find similar problems online. I think our course material stated that $S$ would be smooth if it's homeomorphic? So if $\varphi$ is bijective, continuous and has a continuous inverse $\varphi^{-1}$?

答:
Write the equation in vector form: $\vec r(u,v)=(2uv, u^2 − v^2, u^2 + v^2).$ So $\vec r$ maps the disk onto some set of points (that we hope will look like a surface) in $\mathbb R^3.$ By definition, $\vec r$ is $smooth$ if each of the coordinate function partial derivatives is continous, and the normal vector $\vec n=\vec r_u\times \vec r_v$ is non-zero on the interior of $B(0,1).$

The partials are obviously continuous, so all we need to do is check the normal vector:

$\vec r_u(u,v)=2v\vec i+2u\vec j+2u\vec k;\quad \vec r_v(u,v)=2u\vec i-2v\vec j+2v\vec k\Rightarrow$

$\vec n(u,v)=8uv\vec i+4(u^2-v^2)-4(u^2+v^2)\vec k$,

which zero at $(0,0)$ so it is not smooth on $B$. On the other hand, if we avoid this pathological point, and pick another, say $(u_0,v_0)$ and some open ball containing it but not containing $(0,0)$ then $\vec r$ is smooth and we can give an intuitive, geometric reason that $\vec r$ is required to satisfy the above conditions. Suppose that the image of $\vec r $ is indeed a surface. Then, if we note that $(u-u_0)\vec r_u(u_0,v_0)$ and $(v-v_0)\vec r_v(u_0,v_0)$ are vectors tangent to the surface, then (the norm of) their cross-product will be the area of the parallelogram built on these two vectors, and this area will be an approximation of the patch of area on the surface "close" to $(u_0,v_0).$

The idea is then to add up all these areas to get an approximation to the area of the entire surface, and then, by a limiting process (an integral), get the exact area of the surface. But this method fais if the cross-product is zero, because in this case, it may not be possible to define a unique tangent plane (and hence calculate an approximate area) at $(u_0,v_0).$


问:
Prove that there is an $x_0\in(0,1]$ s.t. $f(x_0)=0$ and $f(x)>0$ for $0\leq x<x_0$.
Suppose that a function $f:[0,1]\rightarrow\mathbb{R}$ is continuous $f(0)>0$ and $f(1)=0$. Prove that there is a number $x_0\in(0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\leq x<x_0$.

Intuitively, if I think about this function on a graph, we can take a negative cubic function that has a y-intercept greater than 0 and an x-intercept equal to 1 with a local minimum on the x-axis between 0 and 1. However, I am not sure how to prove this for some general function $f$.

答: $f^{-1}(0)$ is a closed subset of the compact set $[0,1]$ it is compact, let $x_0$ be its minimum.


答: Hint: Let $x_0=\inf \{x: f(x)=0\}$. Use continuity to show that $f(x_0)=0$. It is obvious by definition that $f(x) >0$ for $x <x_0$. [Certainly $f(x) \neq 0$ for $x<x_0$. If $0<x<x_0$ and $f(x) <0$ then by IVP there must be some point $y \in (0,x)$ such that $f(y)=0$ but that would contradict the definition of $x_0$].

Title: Prove that there is an $x_0\in(0,1]$ s.t. $f(x_0)=0$ and $f(x)>0$ for $0\leq x<x_0$.

Body:
Suppose that a function $f:[0,1]\rightarrow\mathbb{R}$ is continuous $f(0)>0$ and $f(1)=0$. Prove that there is a number $x_0\in(0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\leq x<x_0$.

Intuitively, if I think about this function on a graph, we can take a negative cubic function that has a y-intercept greater than 0 and an x-intercept equal to 1 with a local minimum on the x-axis between 0 and 1. However, I am not sure how to prove this for some general function $f$.


问:
Problem involving gradient
Let $T(x,y,z)$ be the temperature, in Celsius, inside a room at the point $(x,y,z)$, such that $x,y,z$ are given in meters.

It's given that
$$T(5,4,2)=30º,\quad\frac{\partial T}{\partial x}(5,4,2)=3º/m,\quad\frac{\partial T}{\partial y}(5,4,2)=-1º/m,\quad\frac{\partial T}{\partial z}(5,4,2)=1º/m$$

A fly flies in the room and its path is given by
$$x=t^2+1,\quad y=2t,\quad z=10-t^3$$
at the instant $t$ given in seconds.

I have to find the rate of the temperature subject to this path at $t=2$ seconds.

What I made was: the position of the fly is given by the function
$$s(t)=(t^2+1,2t,10-t^3)$$
and then,
$$s(2)=(5,4,2)$$
and in this point, the partial derivatives are given.

Since
$$s'(t)=(2t,2,-3t^2)$$
so
$$s'(2)=(4,2,-12)$$
Therefore, the rate I'm looking for is
$$s'(2)\cdot\nabla T(5,4,2)=-2$$

My question is: is it right or should I normalize the vector $\vec{u}=(4,2,-12)$?

答:
Related: Why isn't the directional derivative generally scaled down to the unit vector?

In short, by not scaling, you are incorporating the fly's velocity into the answer. If the fly were moving in the same direction, but super slowly the instantaneous rate of change would be smaller, whereas if the fly were moving super quickly in the same direction, the instantaneous rate of change of temperature would be larger.

Title: Problem involving gradient


问:
Gamma distribution PDF
For the following PDF of the Gamma distribution,
$ f(x)= \frac{1}{\Gamma{(k) \theta^k}} x^{k-1} e^{-\frac{x}{\theta}}$

with mean= $k \theta$  and variance = $k \theta^2$ , and if I assume  $k \rightarrow \infty$,

then the random variable x should have approximately a normal distribution with the same mean and variance. Is this correct?

Can anyone please check whether the PDF below is the appropriate representation for the case of $k \rightarrow \infty$? Does it look like the following PDF?

$ f(x)= \frac{1}{\theta \sqrt{2\pi}} e^{ - {\frac{1}{2}} ( \frac{x- \theta}{\theta})^2}$

Thank you so much.

答:
then the random variable x should have approximately a normal distribution with the same mean and variance. Is this correct?

Yes, same variance and mean, say $N(k\theta;k\theta^2)$

Thus

$$f_X(x)=\frac{1}{\theta\sqrt{2\pi k}}e^{-\frac{1}{2k}(\frac{x-k\theta}{\theta})^2  }$$

It is an application of CLT because $Gamma(k;\theta)$ is the sum of $k$ iid $Exp(\theta)$ rv's. Note that, for example $(k=30; \theta=0,5)=\infty$

Take a look here at the densities' graphics when k increases... with $k=9$ the shape is already symmetric around $k\theta$

Title: Gamma distribution PDF


问:
A bag contains 7 blue balls and 3 green balls. What is the probability of the following events?
A bag   contains    7   blue    balls   and 3   green   balls.

I am not sure about my answer but please allow me to share.

If 5 balls are drawn in succession at random without replacement, what is the probability of drawing 1 green, 1 blue, 1 green, and 1 blue in order?

There are 10 balls in the bag initially, 7 of which are blue and 3 balls are green. All balls are equally likely to be drawn from the bag.

Since the order is important, we need to compute the probability of getting 1 green, 1 blue, 1 green, and 1 blue in order.

Hence,

$GBGB = (\frac{3}{10})(\frac{7}{9})(\frac{2}{8})(\frac{6}{7}) = \frac{1}{20}$

The probability of getting 1 green, 1 blue, 1 green, and 1 blue in order is $\frac{1}{20}$.

If 3 balls are drawn at the same time, what is the probability that all the 3 balls drawn are blue?

$BBB = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = (\frac{7}{24})$

If 5 balls are drawn at the same time, what is the probability that 3 balls are blue and 2 balls are green?

Work it out first where the order they are drawn is important.

$BBBGG = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8})(\frac{3}{7})(\frac{2}{6}) = (\frac{1}{24})$

Then work out possible different orders they could be drawn, there are 20 permutations $P(5,2)$. so the answer is

$20 × (\frac{1}{24}) = (\frac{5}{6})$

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

答:
If five balls are drawn in succession at random without replacement, what is the probability of drawing one green, one blue, one green, one blue in order?

You have correctly calculated that the probability that the sequence GBGB occurs in the first four draws is
$$\Pr(GBGB) = \frac{3}{10} \cdot \frac{7}{9} \cdot \frac{2}{8} \cdot \frac{6}{7}$$
However, it could also happen in the second four draws after being preceded by either a green or a blue ball.
$$\Pr(GGBGB) = \frac{3}{10} \cdot \frac{2}{9} \cdot \frac{7}{8} \cdot \frac{1}{7} \cdot \frac{6}{6}$$
and
$$\Pr(BGBGB) = \frac{7}{10} \cdot \frac{3}{9} \cdot \frac{6}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}$$
Since these three cases are mutually exclusive and exhaustive, the probability that the sequence GBGB appears in the first five draws is found by adding the above probabilities.

If three balls are drawn at the same time, what is the probability that all three balls drawn are blue?

Your answer is correct.

An alternate method is to observe that we can select three of the ten balls in
$$\binom{10}{3}$$
ways.  Of these, we can select three blue balls in
$$\binom{7}{3}$$
ways.  Hence, the probability that a random selection of three balls from a bag containing seven blue and three green balls results in the selection of three blue balls is
$$\Pr(\text{three blue}) = \frac{\dbinom{7}{3}}{\dbinom{10}{3}}$$

If five balls are drawn at the same time, what is the probability that three balls are blue and two balls are green.

There are
$$\binom{10}{5}$$
ways to select five of the ten balls.  Of these, we can select three of the seven blue balls and two of the three green balls in
$$\binom{7}{3}\binom{3}{2}$$
ways.  Hence, the probability that a random selection of five balls from a bag with seven blue and three green balls results in the selection of three blue and two green balls is
$$\Pr(\text{three blue and two green balls}) = \frac{\dbinom{7}{3}\dbinom{3}{2}}{\dbinom{10}{5}}$$
Your answer resulted in too large a probability since the number of ordered sequences of three blue and two green balls is
$$\binom{5}{2}$$
since we must choose two of the five positions in the sequence for the green balls.  Notice that
$$\binom{5}{2}\left(\frac{7}{10}\right)\left(\frac{6}{9}\right)\left(\frac{5}{8}\right)\left(\frac{3}{7}\right)\left(\frac{2}{6}\right) = \frac{\dbinom{7}{3}\dbinom{3}{2}}{\dbinom{10}{5}}$$


答:
Problem 1.

Your answer to problem 1. is wrong, because you have computed the probability
of getting G-B-G-B in the first four drawings.  The problem specifically
asked for the probability of the satisfying sequence occuring anywhere
within the first five drawings.

The actual problem is much more complicated, and is best attacked by the
following approach:  the answer will be

$$\frac{N\text{(umerator)}}{D\text{(enoninator)}}$$

where

$$D = 10 \times 9 \times 8 \times 7 \times 6.$$

Because of the constraints of the problem, the easiest approach is to identify
each of the 4 possible satisfying sequences, and enumerate each possibility.

$N_1~: G-B-G-B-G ~:~N_1 = 3 \times 7 \times 2 \times 6 \times 1.$

$N_2~: G-B-G-B-B ~:~N_2 = 3 \times 7 \times 2 \times 6 \times 5.$

$N_3~: G-G-B-G-B ~:~N_3 = 3 \times 2 \times 7 \times 1 \times 6.$

$N_4~: B-G-B-G-B ~:~N_4 = 7 \times 3 \times 6 \times 2 \times 5.$

Final answer:

$$ \frac{N_1 + N_2 + N_3 + N_4}{D}.$$

Problem 2.

Your answer is correct.

Problem 3.

Your answer is almost right.  You overlooked overcounting.

There are only $\binom{5}{2} = 10$ ways of selecting which slots the 2 blue
balls will go in.  So you need to apply a scaling factor of (1/2) to your answer.

Title: A bag contains 7 blue balls and 3 green balls. What is the probability of the following events?

Body:
A bag   contains    7   blue    balls   and 3   green   balls.

I am not sure about my answer but please allow me to share.

If 5 balls are drawn in succession at random without replacement, what is the probability of drawing 1 green, 1 blue, 1 green, and 1 blue in order?

There are 10 balls in the bag initially, 7 of which are blue and 3 balls are green. All balls are equally likely to be drawn from the bag.

Since the order is important, we need to compute the probability of getting 1 green, 1 blue, 1 green, and 1 blue in order.

Hence,

$GBGB = (\frac{3}{10})(\frac{7}{9})(\frac{2}{8})(\frac{6}{7}) = \frac{1}{20}$

The probability of getting 1 green, 1 blue, 1 green, and 1 blue in order is $\frac{1}{20}$.

If 3 balls are drawn at the same time, what is the probability that all the 3 balls drawn are blue?

$BBB = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = (\frac{7}{24})$

If 5 balls are drawn at the same time, what is the probability that 3 balls are blue and 2 balls are green?

Work it out first where the order they are drawn is important.

$BBBGG = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8})(\frac{3}{7})(\frac{2}{6}) = (\frac{1}{24})$

Then work out possible different orders they could be drawn, there are 20 permutations $P(5,2)$. so the answer is

$20 × (\frac{1}{24}) = (\frac{5}{6})$

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


问:
The existence of a finite group with a given order and Sylow numbers?
Is there a finite group $G$ with order $6048=2^5.3^3.7$ satisfying the following three conditions:

The number $n_2(G)$ of Sylow $2$-subgroups is $189=3^3.7$;

The number $n_3(G)$ of Sylow $3$-subgroups is $112=2^4.7$;

The number $n_7(G)$ of Sylow $7$-subgroups is $288=2^5.3^2$?

I have tried to construct a subgroup $H$ of $G$ with small index, but I failed. Does someone have any methods to do this?

答:
I used the grpconst package in GAP to construct the groups of order 6048. There are 36021 isomorphism classes, of which 59 are not solvable.

None of the groups has $n_2=189$ and $n_3=112$.
Exactly one group, $PSU_3(3)$ (as @DerekHolt noted) has $n_2=189$ and $n_7=228$.

540 of the groups have $n_3=112$ (but none of them have $n_2=189$ or $n_7=228$).

Thus the answer to the question is ``no''.


问:
Examples of metrics that are topologically equivalent but have different Cauchy sequences.
I'm using the following definition for topological equivalence.

Definition: Let $X$ be a set and $d_1$ and $d_2$ be two metrics on $X$. Then $d_1$ and $d_2$ are topologically equivalent if for every $x \in X$, for every $r > 0$, there exists $s = s(r,x)$ such that $B_s^{d_1}(x) \subset B_r^{d_2}(x)$ and $B_s^{d_2}(x) \subset B_r^{d_1}(x)$.

Let $d_1 = |x-y|$ be the euclidean metric on $\mathbb{R}$. I want to find a metric $d_2$ that is topologically equivalent to $d_1$. However, I need a sequence that is Cauchy in $d_1$ but not in $d_2$.

I'm trying to find a simple enough example so that I can prove topological equivalence using the definition. I know we can take $d_2 = |\arctan(x) - \arctan(y)|$ and take the sequence $(x_n) = n$, but I can't prove that this $d_2$ is topologically equivalent to $d_1$ using the definition of topological equivalence above. \

So I'm wondering if there is some sort of easier pair of metrics that I'm overlooking.

答:
It is not difficult to prove that $d_1, d_2$ are topologically equivalent.

The function $\arctan : \mathbb R \to \mathbb R$ is continuous, thus for each $x \in \mathbb R$ and each $r > 0$ there exists $s_1 > 0$ such that $\lvert \arctan(y) -  \arctan(x) \lvert < r$ for all $y \in \mathbb R$ with $\lvert y - x \rvert < s_1$. This means $B_{s_1}^{d_1}(x) \subset B_r^{d_2}(x)$.

Similarly, the function $\tan : (-\pi/2,\pi/2) \to \mathbb R$ is continuous. Thus for for each $\xi \in  (-\pi/2,\pi/2)$ and each $r > 0$ there exists $s_2 > 0$ such that $\lvert \tan(\eta) -  \tan(\xi) \lvert < r$ for for all $\eta \in (-\pi/2,\pi/2)$ with $\lvert \eta - \xi \rvert < s_2$. Thus for each $x \in \mathbb R$ and all $y \in \mathbb R$ with we have $\lvert y - x \rvert = \lvert \tan(\arctan(y)) -  \tan(\arctan(x)) \lvert < r$ for $\lvert \arctan(y) -  \arctan(x) \lvert < s_2$. Thus $B_{s_2}^{d_2}(x) \subset B_r^{d_1}(x)$.

Now take $s = \min(s_1,s_2)$.


问:
If $A \in \mathbb{R}^{n \times n}$ then $|a_{ij}| \leq ||A||.$
Let $A \in \mathbb{R}^{n \times n}$ and $||\cdot||$ be an arbitrary matrx norm. I want to show that $|a_{ij}| \leq ||A||.$

I've been looking over the matrix norm properties, and I'm just not seeing a way to compare these two values. I've come across specific matrix norms, but this is an arbitrary one, and I don't know what to do with it.

Does anyone have a hint to point me in the right direction? Since I don't have any input myself, I'm not necessarily looking for a full answer.

答:
This statement is not true. As an example, consider the matrix
$$
A = \pmatrix{0 & 1\\0 & 0},
$$
and let $\|\cdot\|_k$ denote the matrix norm defined by
$$
\|A\|_k = \|S_kAS_k^{-1}\|, \quad S = \pmatrix{k & 0\\ 0 & 1},
$$
where $\|A\|$ denotes the induced Euclidean norm (maximal singular value, AKA spectral norm) of $A$. We find that for every $k>0$, $\|\cdot \|_k$ is indeed a norm. However, $\|A\|_k = k$. So, if we select $k<1$, then we find that $|a_{12}| > \|A\|_k$, contradicting the claim.


问:
Space curves- Vector
enter preformatted text here

I have attached both the problem and approach to the solution. But I am confused with the steps given. I am not able to understand them. Kindly help in providing a simplified version of the solution.

PS- I tried pasting the problem and solution from a word file, but I don't understand why the editor is not accepting my text. That's why I have attached the images.

答:
Let $\alpha(s)$ be an arc-length parametrized curve lying in the sphere of radius $a$ centered at $C \in \Bbb R^3$; then

$(\alpha(s) - C ) \cdot (\alpha(s) - C) = a^2; \tag 1$

thus, differentiating with respect to $s$ we find

$\dot \alpha(s) \cdot (\alpha(s) - C) = 0, \tag 2$

since $a^2$ is constant.  Now

$\dot \alpha(s) = T(s), \tag 3$

the unit tangent vector to $\alpha$; thus (2) becomes

$T(s) \cdot (\alpha(s) - C) = 0; \tag 4$

this shows that the component of $\alpha(s) - C$ along $T(s)$ is zero; differentiating again with respect to $s$ yields

$\dot T(s) \cdot (\alpha(s) - C) + T(s) \cdot \dot \alpha(s) = 0, \tag 5$

which by virtue of (3) is transformed into

$\dot T(s) \cdot (\alpha(s) - C) + T(s) \cdot T(s) = 0, \tag 6$

which, since $T(s)$ is a unit vector, i. e.,

$T(s) \cdot T(s) = 1, \tag 7$

may be written

$\dot T(s) \cdot (\alpha(s) - C) + 1 = 0; \tag 8$

note this implies that

$\dot T(s) \ne 0; \tag{8.1}$

thus we may use the first Frenet-Serret equation

$\dot T(s) = \kappa(s) N(s), \tag 9$

where $N(s)$ is the unit normal vector field to $\alpha(s)$ and

$\kappa(s) > 0 \tag{10}$

is the curvature of $\alpha(s)$; from (8) we obtain

$\kappa(s) N(s) \cdot (\alpha(s) - C) + 1 = 0; \tag{11}$

applying a little ordinary algebra we find

$N(s) \cdot (\alpha(s) - C) = -\dfrac{1}{\kappa(s)}, \tag{11.5}$

which expresses the $N(s)$ component of $\alpha(s) - C$; taking yet another $s$-derivative, this time of (11):

$\dot N(s) \cdot (\alpha(s) - C) + N(s) \cdot \dot \alpha(s) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}; \tag{12}$

via (3),

$\dot N(s) \cdot (\alpha(s) - C) + N(s) \cdot T(s) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{13}$

and with

$N(s) \cdot T(s) = 0, \tag{14}$

$\dot N(s) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}; \tag{15}$

the second Frenet-Serret equation

$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag{16}$

in which $\tau(s)$ is the torsion and

$B(s) = T(s) \times N(s) \tag{17}$

is the unit binormal vector to $\alpha(s)$, may now be used to transform (15) into

$(-\kappa(s) T(s) + \tau(s) B(s)) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{18}$

or

$- \kappa(s) T(s) \cdot (\alpha(s) - C) + \tau(s) B(s) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{19}$

apply (4):

$\tau(s) B(s) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{20}$

a little more algebraic grease and this yields

$B(s) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)}, \tag{21}$

the component of $\alpha(s) - C$ along $B(s)$.  We may expand $\alpha(s) - C$ in the basis comprised of $T(s)$, $N(s)$, and $B(s)$:

$\alpha(s) - C =$
$((\alpha(s) - C) \cdot T(s)) T(s) + ((\alpha(s) - C) \cdot N(s))N(s) + ((\alpha(s) - C) \cdot B(s))B(s); \tag{22}$

in light of (4), (11.5), and (19) we find

$\alpha(s) - C =  -\dfrac{1}{\kappa(s)}N(s) + \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)}B(s); \tag{23}$

finally, we insert this into (1) and, recalling that $N(s)$ and $B(s)$ are orthonormal, obtain

$a^2 = \dfrac{1}{\kappa^2(s)} + \left ( \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)} \right )^2, \tag{24}$

the desired result.  $OE\Delta$.

Title: Space curves- Vector


问:
Understanding Hecke Characters as Extension of Dirichlet Characters
I understand the concept of a Dirichlet character, and am interested in its generalizations to arbitrary number fields. I have heard that this generalization is called a Hecke character. However, I am not familiar with adeles or ideles, so I don't understand that definition. I know there is a "classical" definition of Hecke characters, but I am having trouble finding an easy to understand source about this definition. Could anyone provide a reference on the classical definition of Hecke characters that covers some of the preliminary knowledge needed to understand this definition?</p>

答:
The group of fractional ideals (coprime with some ideal $J$..) is a free abelian group generated by the prime ideals $I_{K,J}\cong \prod_{P\not\ni J}' P^\Bbb{Z}$

So it is easy to construct all the homomorphisms $\psi: I_{K,J}\to \Bbb{C}^\times$.

The Hecke characters are those whose restriction to the principal ideals is defined in term of reduction $\bmod J$ and complex embeddings $\sigma_j$, ie. $$\psi(aO_K)=\chi(a) = \phi(a)\prod_j \sigma_j(a)^{r_j}|\sigma_j(a)|^{s_j}$$
where $\phi$ is an homomorphism $O_K/J^\times\to \Bbb{C}^\times$

It is easy to generate all the possible $\chi$, the big restriction is that we need it to be trivial on $O_K^\times$ so that $\psi(aO_K)=\chi(a)$ is well-defined.

Finally we extend $\psi$ by defining $\psi(I_l)$ for the finitely generators of the class group (of fractional ideals coprime with $J$..) :

$Cl_J(K)\cong \prod C_{n_l}$,

$I_l$ is a generator of $C_{n_l}$, so $I_l^{n_l}=(a_l)$ is principal, we choose  $\psi(I_l)$ such that $\psi(I_l)^{n_l}=\psi(a_l O_K)$.

Try with $K$ a quadratic imaginary field, $O_K^\times$ is finite thus it is much easier to construct all the possible $\chi$.

Try also with $\psi(I)=\varphi(I\cap O)$ where $\varphi$ is a character of the class group of $O=\Bbb{Z}[mi]$ with $m\ne 1$.

The point of this construction is that $\sum \psi(I) N(I)^{-s}$ (in addition to its Euler product) is the Mellin transform of some kind of theta function similar to $\sum_n e^{-\pi n^2 x}$ which will give the analytic continuation and functional equation.


问:
property of a normal subgroup of a transively group acting on a set $S$

I'm trying to solve the part (c), but i can't. Please someone may give me a skill or key, i'll appreciate so much.

答: Suppose $y$ is some other point in $S$. By assumption, there exists some $h\in G$ such that $hx = y$. Now if $g\in G_x$, then $hgh^{-1}y = hgx = hx = y$, so $hgh^{-1}\in G_y$ (the converse holds as well). Use this property to show that if $N$ is normal and every element of $N$ fixes $x$, then every element of $N$ in fact fixes $y$ for every other $y\in S$.


问:
Finding the Posterior Probability
Suppose that the time until a bus arrives is modeled as being equally likely to be $\exp \sim(1.1)$, $\exp \sim(1.2)$, and $\exp \sim (2)$.  The bus actually arrives are time $0.4$.  What are the posterior probabilities for each model?

I figured out that the prior probabilities are $p \sim Unif({1.1,1.2,2})$, but I am not sure how to finish the problem.

答:
From Bayes' Theorem:

$$P(M_i | x) = \frac{p(x|M_i) P(M_i)}{p(x)}$$

where $P(M_i) = 1/3$ and $p(x) = \sum\limits_{i=1}^3 p(x|M_i) P(M_i)$ and $p(x|M_i)$ is given by an exponential with the rate constants you state.

Here we want to calculate $P(M_i|0.4)$.

I find $p(.4) = 0.783213$, which must then be plugged into the above equations.

Title: Finding the Posterior Probability


问:
What is the probability of getting exactly two tails after tossing $x$ number of fair coins ($x\ge2$)?
What is the probability of getting exactly two tails after tossing $x\ge2$ number of fair coins?

I am not sure about my answer, but please allow me to share it here.

Let $A$ be the number of times we get a tail after randomly tossing an x number of different coins, where x is greater than or equal to 2. The probability of getting a tail after randomly tossing a single coin is $\frac{1}{2}$ (H / H, T). The probability of getting exactly two tails by tossing $x$ coins is

$P(A = 2) =  {x \choose 2} (\frac{1}{2})^x$, $x\ge2$

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

答:
Your answer is correct.

A valid outcome corresponds to choosing which two of the $x$ coins you want to be tails, this can be done in $\binom x2$ ways.

The total number of possible outcomes is $2^x$.

Thus the probability is $\binom x2/2^x$.

To feel more confident about my answers of problems like this, I always like to think of them in terms of sets. We have the possibility space $\Omega = \{H,T\}^x$, i.e., the set of all possible outcomes is the set of tuples of length $x$ containing either an $H$ or a $T$. An example of a member of this set is $(H,H,T,H,T,T,H)$, if $x=7$ (for instance). We have $\#\Omega = \#(\{H,T\}^x) = (\#\{H,T\})^x = 2^x$.

Now we want the subset $A\subseteq\Omega$ containing all tuples with precisely two $T$'s. To populate this set, we pick the two out of $x$ positions to place the $T$'s, and fill the rest with $H$'s. This can be done in $\binom x2$ ways, so $\#A=\binom x2$.

Thus our probability is $$\mathsf P(A)=\frac{\#A}{\#\Omega}=\frac{\binom x2}{2^x} = \frac{x(x-1)}{2^{x+1}}.$$


问:
Example of a linear map over $\mathbb R(X)$ which isn't a linear map over $C^1$
Over $\mathbb R$, the only linear maps are those of the form $ax$.

If we discuss rational functions over $\mathbb R$, this extra structure would allow us to describe a wider variety of linear maps.

But the obvious maps such as limits, differentiation, summation, doing $f(x)\mapsto f(x+k)$ seem too easy. Is there a linear map which specifically takes advantage of the fact that we have a rational function?

For instance, something like "double the coefficient of $x$ of the numerator and triple that of $x$ in the denominator". I know this is not a good example because it isn't linear, but it illustrates what I mean by "taking advantage of the fact that we have a rational function", i.e., something which doesn't generalise easily to a wider class of functions (say $C^1$).

答:
Any linear map $T$ of vector spaces $X \to Y$  can be "generalized" to a map of
$Z \to Y$, where $Z$ is any vector space that contains $X$.  Namely, let $P$ be
a projection from $Z$ onto $X$, and take the linear map $T \circ P$.  Of course,
you might not be able to obtain $P$ explicitly: you might need the Axiom of Choice.


问:
show this language is undecidable
$L = \{ \langle G \rangle \mid G$ is context-free, $L(G)$ contains a palindrome $\}$

Reduce the post correspondence problem to $L$.

So I have to show that if I could decide L then I could decide PCP.
So I have to show a translation of any given PCP-problem into some form that makes it appearent that it actually is or can be viewed as a problem of determining whether or not a certain context-free grammar contains a palindrome.

As far as I understand PCP is about lining up blocks, each of them with an upper and a lower string, in a way that the concatenated upper and lower strings are the same.

A context-free grammar is a way of producing strings. Grammars that produce palindromes it seems need rules of the kind $T \rightarrow aTa$ (or something similar).

Given a PCP-instance as a bunch of 'building blocks' with an upper and a lower side, one might try to understand the possibility of arranging them in any order as an application of grammar rules that write a 'left' (corresponding to the upper strings, say) and a 'right' word (corresponding to the lower strings). But there are so many ways to put them together! How to describe this by a few rules? You can even use any building block as many times as you want. So basically there must be a rule for 'everything that can happen'. Also, the grammar needs to spell one of the words (the left or the right one) in reverse order. It should be possible to build a palindrome in that way whenever the blocks can be put together to give the same string upper and lower side.

So let there $k$ blocks be given, with upper words $a_1,...a_k$ and lower words $b_1...,b_k$.
I can start with any of the blocks, so let's introduce a grammar with every rule $$S \rightarrow a_iS\overleftarrow{b_i} $$ for every $i=1,..k$. This rule should suffice whenever I'd like to add block number $i$. Also I think we need $S \rightarrow \varepsilon$ to stop the procedure.

I'm not sure whether we need any further rules.

Now some alignment of blocks, say $i, j, l$, that is an upper string $a_i,a_j,a_l$ and a lower string $b_i,b_j,b_l$ correspond to the use of the rules $S \rightarrow a_iS\overleftarrow{b_i}, S \rightarrow a_jS\overleftarrow{b_j}, S \rightarrow a_lS\overleftarrow{b_l} $, producing the string $a_i a_j a_l \overleftarrow{b_l} \overleftarrow{b_j} \overleftarrow{b_i}$. And this should be a palindrome whenever the $a$-string and the $b$-string match.

Is this it? Am I done? It looks I used a lot of informal language, should this be more formally? Although more or less convincing it doesn't seem too rigourous, as if I could be missing something...

答:
There is a helpful construction for this type of problems

This problem asks you to use this construction to show that  AMBIGcfg is undecidable

Hopefully you can see that T generates a string from the upper strings and B generates a string from the lower strings

The terminals at the end encode the sequence of "dominos" used , so if T generates a string (assume that the alphabet for this PCP Instance is {a,b}) : aaab12 and B generates a string aaab12 , then a match was found , that is because we formed the match aaab using the same sequence of dominos 12 (here we use domino 2 then 1 think about it !)  , this encoding guarantees that both T which uses the upper strings and B which uses the lower strings have used dominos in the same order when they formed the match

A further example to emphasis the idea , assume we use
T --> t1 T a1 , then use T --> t2 a2 we generate the string
t1 t2 a2 a1 , the string generated by dominos is t1 t2 and the sequence of dominos used is a1 then a2 , once again ,if  B generates a string b1 b2 a2 a1, where b1 b2 = t1 t2 , then we found a match , since strings are the same with same sequence of dominos used , hopefully by now you understand how the construction works and are convinced that T and B generate the same string only if we have a match

In this problem , if both T and B generate the same string , the grammer is ambiguous and we accept , else we reject , deciding the PCP

Now can you tweek this construction for our problem ?

We replace S --> T|B by S -->TB , and replace every rule
B --> bi B ai by  B --> ai B bi , and replace every rule
B --> bi  ai by  B --> ai  bi ( reversing the strings on right side of rules which have B on left side )

If a match w exists , then T would generate w , and B would generate w^R , then TB is ww^R which is palindrome , ex : if w is aaab12 then T generates aaab12 and B generates 21baaa , so S is aaab1221baaa , which is a palindrome

I leave it you you to convince your self that the only way for a palindrome to exist is to have a match , if you consider that a match is represented in our grammer by the string followed by a unique sequence of dominos used for this string , then it should be easy to see how it works

So construct G with rules provided above , we assume that a decider R exists for L , we run L on < G> ,

If a match exists G generates some palindrome and L accepts

If a match doesn't exist , G generates no palindromes ,  and L rejects

Title: show this language is undecidable


答: If $X$ represents the number of marked inividuals we observe out of our sample of $1000$ individuals, then for $x\in\{0,1,.\ldots,800\}$ we have $$P(X=x|N=n)=\frac{{800 \choose x}{n-800 \choose 1000-x}}{{n \choose 1000}}$$ Here, $N$ represents the number of individuals in the entire population. Define the function $f$ on the domain $\{1550,1551,\ldots\}$ by $$f(n)=P(X=250|N=n)$$ I would use the value of $n$ that maximizes $f$ as an estimator of $N$.

Title: Which one is true?

Body:
In order to assess the number $N$ of individuals of a species of animal living on an island, we propose to adopt the capture-recapture method. For this we capture $800$ individuals. These individuals, which correspond to the proportion $p$ of the total number of individuals living on the islands, are marked and then released. $1000$ animals are subsequently captured, of which there are $250$ animals marked during the first capture. Which of the following statements are true?

We have $0.20$ as estimation of $p$

$p = \frac{800}{N}$

We have $3200$ as estimation of $N$

I am not sure how to deal with this. Can you help?

EDIT:

I know that two of them are right, but I can say which ones?


问:
Can a function contain a variable?
For example:

f is function

x is a variable

john is a constant

Would $ f(x, john) $ be well formed? Or can a function only map constants to an object in a domain?

答: Sure.  Let's say $f$ is a function that takes two inputs, $x$ and $y$, and gives us some output $z$, i.e. $z = f(x,y)$.  Then if you fix $y = \text{john}$ then you can think of $z = f(x,\text{john})$ as a function itself with one input, $x$, made by looking at a the graph of $z = f(x,y)$ along the plane $y = \text{john}$.  If you wanted to you could give this new function it's own name, say $g(x) = f(x,\text{john})$.

Title: Can a function contain a variable?


答:
Yes, function terms can contain variables. The definition of well-formed expressions of predicate logic reads

...

If $x$ is a variable, then $x$ is a term.

...

If $f$ is an $n$-ary function symbol and $t_1, \ldots, t_n$ are terms, then
$f(t_1, \ldots, t_n)$ is a term.

...

Since function symbols operate on terms, and variables are terms, this is permitted.

BTW, careful about terminology: "valid" has a specialized meaning in logic, meaning that a formula is true in all interpretations; to say that an expression is "grammatical", better say "well-formed" to avoid misunderstandings.

Title: Can a function contain a variable?

Body:
For example:

f is function

x is a variable

john is a constant

Would $ f(x, john) $ be well formed? Or can a function only map constants to an object in a domain?


答:
Using partial fractions and the fact that $\mathbb{R}[X]$ is a PID, it's possible to write every element of $\mathbb{R}(X)$ uniquely as a sum $\sum_{p<\mathbb{R}[X],n\in\mathbb{N}} \frac{f_{p,n}}{p^n}$, where $p$ ranges over the primes of $\mathbb{R}[X]$ and $f_{p,n}$ has smaller degree than $p$.

This expresses $\mathbb{R}(X)$ as an infinite direct sum of finite dimensional vector spaces (namely vector spaces of dimension $1$ for the linear primes and dimension $2$ for indecomposable quadratics), so we can apply any linear maps we like independently to each piece of the decomposition. Most of these maps will be strange and have no obvious generalization to arbitrary continuous functions.

Title: Example of a linear map over $\mathbb R(X)$ which isn't a linear map over $C^1$

Body:
Over $\mathbb R$, the only linear maps are those of the form $ax$.

If we discuss rational functions over $\mathbb R$, this extra structure would allow us to describe a wider variety of linear maps.

But the obvious maps such as limits, differentiation, summation, doing $f(x)\mapsto f(x+k)$ seem too easy. Is there a linear map which specifically takes advantage of the fact that we have a rational function?

For instance, something like "double the coefficient of $x$ of the numerator and triple that of $x$ in the denominator". I know this is not a good example because it isn't linear, but it illustrates what I mean by "taking advantage of the fact that we have a rational function", i.e., something which doesn't generalise easily to a wider class of functions (say $C^1$).


答: A example of a linear functional on $\mathbb R(X)$ ... write it in partial fraction form, and take the coefficient of $1/(X-3)^2$.  (Of course $0$ if that terms does not appear.)


问:
What is the missing number from this pattern?
I tried every combination I could think of, but I just can't figure out this pattern.
A.21  B.9  C.0  D.2

答:
Big hint: $$\frac{2+2+3+3}{2}=5.$$

This pattern holds for the other two, see if you can calculate the blank.


问:
How would I convert a long decimal to binary without using a calculator?
I am aware that you can keep dividing a decimal number by two, finding the remainders in the process, in order to convert the number to binary.

However, when I am working with a long decimal number such as 2147483647, is there any easier way of converting this to binary? Manually dividing this number by two repeatedly would take an entire page of long division to do, and human error is likely.

Thanks in advance.

EDIT: SOLUTION. Thanks to the help I received from joriki and pjs36 I discovered that instead of dividing by 2 I can divide by 16. Where dividing by two will give you binary as a result, dividing by 16 will give you hex as a result. This hex value can then be converted easily to binary , and is is a much shorter process that dividing by 2.

答:
I felt using the remainder method is quite easy for me.
if u got a remainder with decimal when divided with 2 it should be considered a 1 in bitstring and if the remainder doesn't have a decimal it is a 0 bitstring.

then u read from top to bottom as the right to left.
i.e 110011001

409/2 = 204.5 - 1

204/2 = 102   - 0

102/2 = 51    - 0

51/2  = 25.5  - 1

25/2  = 12.5  - 1

12/2  = 6     - 0

6/2   = 3     - 0

3/2   = 1.5   - 1

1/2   = 0.5   - 1


问:
What is the Point, and Meaning, of the Mean Value of a Function?
Ok, so I know that the mean value of a function, $f(x)$, on the interval $[a,b]$ is given by (or defined by?)
$$\frac{1}{b-a}\int_a^bf(x)~dx$$
but I have $2$ basic questions about this:

$1$: From a purely mathematical point of view, does this have any practical use? Does it give us any extra weapon to add to our mathematical arsenal?

$2$: How excatly can I interpret this as a mean value? I've seen the geometrical interpretation (looking at the area under the graph within the given interval), but I still don't understand how it links to a mean value. For a finite set of values I divide the sum of the values by however many values I have to obtain the mean, but in this case there are infinitely many values, so how is this a mean value?

Thanks for your help.

答:
I'd like to complement the answers from @Lee Mosher and @Ivo Terek  with a data compression angle. That's the kind of angle that helped me develop an intuition behind the formal concept.

Your function $f$ typically assumes several values on $[a,b]$, but if you wanted to summarize it with ${\bf one}$ number $v$, then what number would you pick?

There are several possibilities, and surely some numbers are better than others (for instance,  you wouldn't pick a negative number if your function is known to assume positive values). Intuitively, you'd want a value that's right in the middle of the values assumed by $f$. Among all possibilities, the mean value is a really good choice. The reason is that, among all the other possible values you could have picked, the mean minimizes the (quadratic) distance with the values taken by the function:
$$v = \arg_u\min \int_a^b |f(x)-u|^2dx$$
In other words, it is the value that is closest to all values of $f$ on that interval.

So, if you view your function as a signal, then an obvious way to compress (summarize) it would be to take the mean value. There are other interpretations for the mean value. In physics, this mean value represents the center of mass of a solid. In signal processing, it is the first mode of vibration of a complex wave (first Fourier mode). In statistics and probabilities, if you keep drawing a random variable (e.g. rolling a dice) and you look at the average of all the numbers you get, it will converge towards the mean of the distribution (law of large numbers). In data mining, the mean value of a data set is its centroid. And the list goes on and on...


问:
Can flux be proportional to $r^2$ in divergence theorem?
The motivation for the divergence being interpreted as the flux of stuff used the following:

$$\text{div} F(a) = \lim_{r\to0}\frac{3}{4\pi r^3}\int_{|x-a|=r} F\cdot n dA$$

Without the $r^3$ in the denominator the right hand side would converge to $0$. But in the context of the divergence theorem what happens if it is $r^2$ instead, can the limit be nonzero?

答:
$\mathbf F=1 \mathbf e_r$. Then the integral is just $4\pi r^2$ and the limit is 3.

I'm not clear on what you mean by "in the context of the divergence theorem", though. In three dimensions, the size of the sphere of radius $r$ is always proportional to $r^3$, and there is no way you would get the above formula with $r^2$ instead.

Of course, in two dimensions, this is exactly what one does find.

$\nabla\cdot \mathbf F=\lim \frac 1 {\pi r^2} \int_{\text{circle}} \mathbf {F\cdot dn}$

Try $\mathbf F = r \mathbf e_r$ for a simple example. Both sides should give 2.


问:
Expectation and Variance of Binary String
I've been working on the following problem:

Assume a computer generates binary string of length 14 at random. Let $K$ be the random variable that is equal to the number of 0’s in the string plus twice the number of
ones in the string. Calculate $E(K)$ and $Var(K)$.

Now since each individual bit has a 50% chance of either being a 1 or 0, I found the expected number of 0's to be $14*1/2=7$. I'm assuming this problem involves the fact that $E[X+2Y] = E[X] + 2E[Y]$.

But if $X$ represents 0's and $Y$ represents 1's, I get that: $E[X + 2Y] = 7 + 2*7$ = $21$.

However, the length of the string is only $14$, which is why I am lost calculating the expected value.

Any help, hints, and advice would be much appreciated!

答:
Let's do the general case, with length $n$ instead of just length $14$.

We let $X$ be the random variable which counts the number of $0's$ in the string and, of course, $K$ is the random variable defined in the problem.

We remark that the number of $1's$ must be $n-X$ so we have $$K=X+2(n-X)=2n-X$$

Thus we immediately see that $$E[K]=2n-E[X]=2n-\frac n2=\frac {3n}2$$

As to the variance, shifting each value by a constant does not change the variance so $$\text {Var}(K)=\text {Var}(-X)=\text {Var}(X)=\frac n4$$


问:
Show that the following two metrics are topologically equivalent.
This is the definition of topological equivalence I'm working with.

Definition: Let $X$ be a set and $d_1$ and $d_2$ be two metrics on $X$. Then $d_1$ and $d_2$ are topologically equivalent if for every $x \in X$, for every $r > 0$, there exists $s = s(r,x)$ such that $B_s^{d_1}(x) \subset B_r^{d_2}(x)$ and $B_s^{d_2}(x) \subset B_r^{d_1}(x)$.

Note, this definition uses the same $s$. I know in other textbooks there are usually two different $s_1$ and $s_2$.

I am trying to prove that $d_1(x,y) = |x-y|$ the euclidean norm is topologically equivalent to $d_2(x,y) = |e^x - e^y|$ using the above definition.

I'm having trouble with bounding and finding a suitable $s$ since there are exponentials involved.

Any help would be appreciated.

答:
Consider the map $f:\mathbb{R}\to (0,+\infty)$ given by $f(x)=e^x$. This is a continuous bijection with continuous inverse $f^{-1}:(0,+\infty)\to \mathbb{R}$ given by $f^{-1}(x) = \log(x)$.

Now fix $x\in \mathbb{R}$ and $r>0$. By continuity of $f$ there is some $\delta_1 > 0$ such that for all $y\in \mathbb{R}$ the following holds:
$$
|x-y|< \delta_1 \Rightarrow |e^x - e^y| < r.
$$
Similarly, by continuity of $f^{-1}$ at $u=e^x$ there is some $\delta_2 > 0$ such that for all $v\in (0,+\infty)$ we have that
$$
|u-v| < \delta_1 \Rightarrow |\log(u)-\log(v)| < r.
$$
Write $v = e^y$ for some (unique) $y\in \mathbb{R}$, then we obtain that for all $y\in\mathbb{R}$
$$
|e^x - e^y| < \delta_1 \Rightarrow |x-y| < r.
$$
Now take $s=\min\{\delta_1,\delta_2\}$. The inclusions between the balls are just rewritings of the above implications.


问:
Optimization question on a function $(, , , ) = \min\{, 2\} + \max \{3, 4\}$
I have the following utility function

$$(, , , ) = \min\{, 2\} + \max \{3, 4\}$$

I want to find its demand function.

For that

$$\operatorname{Max}(, , , ) = \min\{, 2\} + \max \{3, 4\}$$

S.t.

$$p_1 x+p_2 y + p_3 z +p_4 w \le  I$$

But I cannot proceed from here. Please suggest a solution. Thanks a lot

答:
Introduce a variable $v$ to represent $\min\{x,2y\}$ and solve two linear programming problems:

Maximize $v+3z$ subject to
\begin{align}
v &\le x \\
v &\le 2y \\
3z &\ge 4w \\
p_1 x + p_2 y + p_3 z + p_4 w &\le I \\
\end{align}

Maximize $v+4w$ subject to
\begin{align}
v &\le x \\
v &\le 2y \\
3z &\le 4w \\
p_1 x + p_2 y + p_3 z + p_4 w &\le I \\
\end{align}

Take whichever solution yields the larger of the two resulting objective values.


问:
A curve has the property that the normal line through any point on the curve passes through $(2,0)$. Find its equation.
A curve has the property that the normal line through any point on the curve passes through $(2,0)$.  If the curve contains the point $(2,3)$ find its equation.

My attempt:

Assume $(a,b)$ is a point on the function. The normal line has a slope $m=\frac{b}{a-2}$ and hence the tangent line has a slope $m_T = \frac{2-a}{b}$. Hence, this is the slope we can integrate to get the equation:

$$F(x) = \int \frac{2-a}{b}x dx = \frac{2-a}{2b}x^2+C$$

Since we know that $(2,3)$ is a point we can isolate for $C$ to get:

$$F(x) = \frac{2-a}{2b}x^2+\frac{3b-4+2a}{b}$$

I am getting stuck here because I am not sure how to get values for a $a$ and $b$...

答:
A circle has the property that the normal through any point on the curve will pass through its center.  As such, we can deduce that the center of the circle is $(2,0)$ and the circle contains the point $(2,3)$.  Thus, we have the equation $$(x-2)^2+y^2=9.$$

Taking an approach using calculus instead of geometry:

Define $-\dfrac{dx}{dy}$ as the slope of the normal to the curve $y(x)$.  The equation of the normal to the curve that contains $(x_1,y_1)$ would be $$y-y_1=-\frac{dx}{dy}(x-x_1).$$
Since the normal must contain $(2,0)$, we can substitute and solve a differential equation:
\begin{align}
y-0&=-\frac{dx}{dy}(x-2)\\
y\,dy&=(2-x)\,dx\\
\int y\,dy&=\int2-x\,dx\\
\frac12y^2&=-\frac12(2-x)^2+C_1\\
(x-2)^2+y^2&=C_2
\end{align}

Since this curve must contain $(2,3)$, it follows that $C_2=9$.  Therefore, the curve must have the equation $$(x-2)^2+y^2=9.$$


问:
How to find the discontinuity set?
What is the discontinuity set of the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$,
  $ f(x,y) := \sup \{ \sin (tx) + \sin (ty) : t \in \mathbb{R} \} ?$ </p>

答:
Clearly, $f(0,0)=0$, $f(0,y)=1$ for $y\ne0$, $f(x,0)=1$ for $x\ne0$.

If $x\ne0\ne y$ and $\alpha:=\frac xy$ is rational, say $\alpha=\frac nm$ with $n\in\mathbb Z, m\in\mathbb N$, $\gcd(n,m)=1$, then
$$\tag1f(x,y)=\max_{t\in[0,2\pi]}\sin nt+\sin mt.$$
Especially, $f(x,-x)=0$. For all rationals $\alpha\ne -1$, the value $(1)$ is positive. If $n\equiv m\pmod 4$, the value in $(1)$ is $2$ and for all other rational $\alpha $, it is somewhere inbetween (because the maxima of $\sin nt$ don't match with those of $\sin mt$).

Finally, for irrational $\alpha$, one can readily see that $f(x,y)=2$.

By these results, $f$ is clearly discontinuos at all points $(x,y)$ with $x=0$ or $y=0$ or $\alpha=\frac nm$ rational with $n\not\equiv m\pmod 4$.
At all other points, i.e. whenever $f(x,y)=2$, $f$ is continuous: For $\epsilon>0$, there exists $r>0$ such that $f_r(x,y):=\max\{\sin tx+\sin ty\mid -r\le t\le r\}>f(x,y)-\frac12\epsilon$. The function $f_r$ is continuous, hence $f(u,v)\ge f_r(u,v)>f(x,y)-\epsilon$ for $(u,v)$ close enough to $(x,y)$.

Title: How to find the discontinuity set?


答: Let's consider $f(x)=\frac{2x}{(x^2+1)^2}$. Then $f'(x)=\frac{2(1-3x^2)}{(1+x^2)^3}$, so we have maximum in $x=\frac{1}{\sqrt{3}}$ which is $\leqslant 1$. So $x^2+1 \geqslant \dfrac{2x}{x^2+1}$ for $\forall x \in \mathbb{R}$ - @J. W. Tanner's correction is correct.

Title: If $x$ $\in \mathbb{R}$, then $(x^{2} + 1)^{2}\geq 2x$

Body:
If  $x \in \mathbb{R}$, then:

$x^2+1 \geq \dfrac{2x}{x^2+1}$?


问:
Is this a valid technique for this linear algebra proof?
Let {$A$v$_1$....$A$v$_n$} be a linearly independent set. If $A$ is invertible, prove that {v$_1$....v$_n$} is linearly independent.

PROOF:

Because {$A$v$_1$....$A$v$_n$} is a linearly independent set,

the equation $c_1A$v$_1$+....+$c_nA$v$_n=0$ implies that $c_1$....$c_n$ must all equal $0$.

multiplying the equation by $A^{-1}$ we obtain: $c_1$v$_1$+....+$c_n$v$_n=0$, where $c_1$....$c_n$ must still must all be $0$.

We can therefore conclude that {v$_1$....v$_n$} is linearly independent

答:
I see a bit of a problem with the argument. To prove that a set is linearly independent, you have to allow $c_1...c_n$ to be an arbitrary tuple that yields the zero vector when used as coefficients for the set, then prove they're all zero. That is, you need $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0 \rightarrow c_1...c_n = 0$. You have instead proved $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n =\textbf 0 \rightarrow c_1...c_n = 0$. Now, it certainly is true that if $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n = \textbf 0$, then $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0$, so if you have all $c_1...c_n$ for   which $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n = \textbf 0$, then all of your tuples are tuples that you are looking for. But "everything I find is a tuple I'm looking for" and "all I'm looking for is a tuple I find" are different statement.   If there is any set of coefficients for which $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0$ but $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n \neq \textbf 0$, then you haven't proven that $c_1...c_n=0$ for all $c_1...c_n$ for which $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0$. There aren't actually any such coefficients, and this fact is relatively trivial, but it's about the same level of triviality as what you're being asked to prove, so it isn't really valid to assume it.

When doing proofs, it's important to follow quantifiers exactly.


问:
About a sum of binomial coefficients
I need to calculate the following sum: $\sum_{k=0}^{[\frac{n}{2}]}$ $n\choose 2k$ $\frac{1}{4^{2k}}$.
I guess it has to do something with identity $\sum_{k=0}^{n}$ $n \choose 2k $ $=2^{n-1}$, but I am not sure should I use some substitution for counter of this sum or I am on wrong way?

Title: About a sum of binomial coefficients

Body:
Note that $$\sum_{k \ge 0} a_{2k} = \sum_{k \ge 0} \frac{1+(-1)^k}{2} a_k.$$
Taking $a_k = \binom{n}{k}/4^k$ yields
\begin{align}
\sum_{k \ge 0} \binom{n}{2k}\frac{1}{4^{2k}}
&= \sum_{k \ge 0} \frac{1+(-1)^k}{2} \binom{n}{k}\frac{1}{4^k} \\
&= \frac{1}{2} \sum_{k \ge 0} \binom{n}{k} \left(\frac{1}{4}\right)^k
+ \frac{1}{2} \sum_{k \ge 0} \binom{n}{k} \left(\frac{-1}{4}\right)^k \\
&= \frac{1}{2} \left(1+\frac{1}{4}\right)^n
+ \frac{1}{2} \left(1-\frac{1}{4}\right)^n \\
&= \frac{(5/4)^n+(3/4)^n}{2} \\
&= \frac{5^n+3^n}{2^{2n+1}}
\end{align}


问:
How do you prove by mathematical induction that $n(n+2)$ is divisible by $8$, if n is even and greater than 1
Everything is ok except for when I try to substitute $n = k + 1$. I get $8J + 4k + 8$ (where J is an integer) and cannot further factorize out an 8 to prove it is divisible. Is this even a correct substitution?

答:
If $n$ is even,
then $n = 2m$ for some $m$.

Then $n(n+2)
=2m(2m+2)
=4m(m+1)$.

Now do your induction on $m$,
not $n$.


问:
Translating Polar Functions
How do I rewrite a polar function (expressed in a polar coordinate system $r = F(\theta)$ so the entire curve is shifted right or left $h$ units and up or down $k$ units?</p>

答:
A simple trick is to get $r=F(\theta)$ in parametric equations, and then apply $\sqrt{x^2+y^2}=r$.

If you move the function to right in $\Delta x$ and to up in $\Delta y$, then
$$
\begin{cases}
x_{new}(\theta)=r\cos(\theta)-\Delta x\\
y_{new}(\theta)=r\sin(\theta)-\Delta y
\end{cases}
$$
Then $r_{new}$ can be expressed as
$$
r_{new}=\sqrt{(r\cos(\theta)-\Delta x)^2+(r\sin(\theta)-\Delta y)^2}
$$

Title: Translating Polar Functions


问:
Find quotient space on $\mathbb{N} $
On $\mathbb{N}$ is given equivalence relation R with $nRm \iff 4|n-m$. Topology on $\mathbb{N}$ is defined with $\tau=\{\emptyset\}\cup\{U\subseteq\mathbb{N}|n\in U \wedge m|n \implies m\in U\}$.

I need to find quotient space $(\mathbb{N}/R,\tau_{R})$.

I have solution: $\tau_{R}=\{\emptyset,\mathbb{N}/R,\{[1],[2],[3]\}\}$ where $\mathbb{N}/R=\{[1],[2],[3],[4]\}$.
But I have no idea how to prove that $p^{-1}[\{[1],[2],[3]\}]=\cup_{k\in \mathbb{N}_0}\{4k+1,4k+2,4k+3\}$, where $p$ is quotient mapping, contains all divisors of its elements.
(for other set it's easy to find element whose divisor is not in set)

答:
Note that $p^{-1}(\{[1],[2],[3]\}) = \mathbb{N}\backslash 4\mathbb{N}$.
We need to prove that $n\in\mathbb{N}\backslash 4\mathbb{N}$ and $m|n$ implies $m\in\mathbb{N}\backslash 4\mathbb{N}$ and we do this by contraposition.

Suppose $m\notin\mathbb{N}\backslash 4\mathbb{N}$, then $m = 4k$ and thus $m|n$ implies $n = pm = 4pk$, hence $n\notin\mathbb{N}\backslash 4\mathbb{N}$ concluding the argument.

Title: Find quotient space on $\mathbb{N} $


问:
show that : $\int \sqrt{a^2-x^2}dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin(\frac{x}{a}) + C \ (a > 0)$
How to show that : $\int \sqrt{a^2-x^2}dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin(\frac{x}{a}) + C \ (a > 0)$

My first idea is to take the derivative of both sides. The right side will yield $\sqrt{a^2-x^2}$, the left side is a bit harder to deal with. If $a > 0$ and $a > x$ in order to avoid a complex result, then I can safely say that $x > 0$. Hence we can create a new function for the integral:

$$F(x) = \int_{0}^{x} \sqrt{a^2-x^2}dx$$

By the fundamental theorem of calculus, the derivative of this function is $\sqrt{a^2-x^2}$. Hence showing that the two results are equal on both sides.

Is this solution correct?

答:
Here it is another way to solve it. Let $x = a\sin(\theta)$. Thus we have
\begin{align*}
\int\sqrt{a^{2} - x^{2}}\mathrm{d}x & = a\int\sqrt{a^{2} - a^{2}\sin^{2}(\theta)}\cos(\theta)\mathrm{d}\theta\\\\
& = a^{2}\int\sqrt{1-\sin^{2}(\theta)}\cos(\theta)\mathrm{d}\theta\\\\
& = a^{2}\int\cos^{2}(\theta)\mathrm{d}\theta\\\\
& = a^{2}\int\frac{\cos(2\theta) +1}{2}\mathrm{d}\theta\\\\
& = a^{2}\left[\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right] + c
\end{align*}

Given that $x = a\sin(\theta)$, this implies that
\begin{align*}
\begin{cases}
\displaystyle\theta = \arcsin\left(\frac{x}{a}\right)\\\\
\displaystyle\sin(2\theta) = 2\sin(\theta)\cos(\theta) = \frac{2x\sqrt{a^{2} - x^{2}}}{a^{2}}
\end{cases}
\end{align*}
and we are done.

Hopefully this helps!