问:
Discrete Math - Counting Integer Solutions to Inequality
I'm trying to find the number of integer solutions to

$x + y + z + w \le 20$ where $x \ge 1$, $y \ge 2$, $z \ge 3$ and $w \ge 4$.

So far I did $x-1 + y-2 + z-3 + w-4 \le 10$. I believe if it were to be equal to $10$ then there would be ${13\choose10}=286$ solutions but I'm not sure if I did that correctly or how to count with the inequality without overcounting. Any help would be greatly appreciated!

答:
Nicely done. You are right, for $10$ is how you are saying. Now, what about for $9$? $8$? In general, just add
$$\sum _{i=0}^{10}\left ( \text{# of solutions to }a+b+c+d=i\right ),$$
where $a=x-1,b=y-2,\cdots$ This is equivalent to construct a variable $e\geq 0$ such that
$a+b+c+d+e=10$ as hinted by Jaap Scherphuis in the comments.

Both results will be the same and the connection of the formulas comes from the Hockey-Stick identity.


问:
how to prove a matrix is symmetric
Suppose square matrix B has nxn dimensions, has $Null(B) = Col(B)^\perp$ and rank(B)=1, how can we prove the matrix B is equal to its transpose?

I'm not sure how to get started I think I have to start with the eigen vectors and then somehow prove its symmetry but I'm not sure

答:
Since $rank(B)=1$, all the columns of $B$ are scalar multiples of the same $n\times 1$ vector $b$, that is, $$B=bx^\intercal$$ for some $n\times 1$ vector $x$.
By the properties of the four fundamental subspaces (https://www.statlect.com/matrix-algebra/four-fundamental-subspaces) and the stated assumption, we have $$Col\left(B^\intercal\right)=Null\left(B\right)^\perp=\left(Col\left(B\right)^\perp\right)^\perp=Col\left(B\right)$$
The transpose of $B$ is a matrix whose columns are all scalar multiples of $x$:  $$B^\intercal=xb^\intercal$$
But since the column spaces of $B$ and $B^\intercal$ coincide (as just demonstrated), it must be that $x=kb$ for some scalar $k$. Therefore, $$B=kbb^\intercal$$which is clearly symmetric.

ContentLicense: CC BY-SA 4.0

Title: how to prove a matrix is symmetric


问:
Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ be the midpoint of $AO$ and $BC$, respectively.

Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ be the midpoint of $AO$ and $BC$, respectively. Suppose $\angle CBA = 4\angle OPQ$ and $\angle ACB = 6\angle OPQ$ . FiNd $\angle OPQ$ .

What I Tried: Here is a picture :-

You can see what I did, I extended some lines and joined them and got a pair of congruent triangles there, and marked some of the angles, and this is my only progress.

Now can anyone help? Thank You.

答:
In your figure, the angle with $4x$ is greater than the angle with $6x$, so $PQ$ is at the wrong side of $O$. Here is a redrawn figure:

The circle is there to aid visualization of the theorem: angle subtended to the circumference is half the angle at center. With this we have:

$$\angle ABC = 4x \implies \angle AOC = 8x, \quad \angle ACB = 6x \implies \angle AOB = 12x$$

Now as $\angle QOB = \angle QOC$, each of them is equal to $\frac12 (360^\circ - 12x - 8x) = 180^\circ - 10x$.

This gives $\angle POQ = 180^\circ - 10x + 8x = 180^\circ - 2x$.

Hence $\angle OPQ = \angle OQP = x$, and $OQ = OP = \frac12OA = \frac12OB$.

By considering $\triangle BOQ$, we see that $\angle BOQ = 60^\circ$. (Why?) Now we can solve for $x$.


问:
Law of Iterated Expectations and Law of Total Variance
If X is an exponentially distributed random variable, denoted as X ∼ Exp (2).

And E[Y|X] = 15 − 5X + 0.5X^2

Var[Y|X] = 3 + 0.5X

What would be E[Y] and Var[Y] separating the “between” and the “within” components?

For the first part (E[Y]), I got the answer attached, but I don't get a closed-form solution rather than a function. So I'm not sure if it's right or not.

答:
There's a very easy way to proceed...

Observe that the simple moments of your $Exp(2)$ rv are the following

$$\mathbb{E}[X^n]=\int_0^{\infty} x^n 2 e^{-2x}dx=\frac{1}{2^n}\int_0^{\infty}(2x)^n e^{-2x}d(2x)=\frac{n!}{2^n}$$

This result has been obtained using Gamma Function

Let's calculate $\mathbb{E}[Y]$

$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}\Bigg[15-5X+\frac{X^2}{2}\Bigg]=15-\frac{5}{2}+\frac{1}{4}=\frac{51}{4}$$

3.

$$\mathbb{V}[Y]=\mathbb{E}[\mathbb{V}(Y|X)]+\mathbb{V}[\mathbb{E}(Y|X)]=$$

$$=\mathbb{E}\Bigg[3+\frac{X}{2}\Bigg]+\mathbb{V}\Bigg[15-5X+\frac{X^2}{2}\Bigg]=3+\frac{1}{4}+25\mathbb{V}[X]+\frac{1}{4}\mathbb{V}[X^2]-5Cov(X,X^2)$$

...being

$$\mathbb{V}[X^2]=\mathbb{E}[X^4]-(\mathbb{E}[X^2])^2$$

$$Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$$

The result is

$$\mathbb{V}[Y]=3+\frac{1}{4}+\frac{25}{4}+\frac{1}{4}\Bigg[\frac{4!}{16}-\frac{1}{4}\Bigg]-5\Bigg[\frac{3!}{2^3}-\frac{1}{2}\cdot\frac{1}{2}\Bigg]=\frac{117}{16}$$


问:
Proving or disproving basic facts about sequences in Real Analysis
I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.

\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):

(a) sequences $(x_n)$ and $(y_n)$, which both diverge, but whose sum $(x_n + y_n)$ converges;

(b) sequences $(x_n)$ and $(y_n)$, where $(x_n)$ converges, $(y_n)$ diverges, and $(x_n + y_n)$ converges;

(c) a convergent sequence $(b_n)$ with $b_n \ne 0$ for all $n$ such that $1/b_n$ diverges;

(d) an unbounded sequence $(a_n)$ and a convergent sequence $(b_n)$ with $(a_n - b_n)$ bounded

(e) two sequences $(a_n)$ and $(b_n)$, where $(a_n b_n)$ and $a_n$ converge but $(b_n)$ does not.

Solution.

(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.

Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.

(b) This request is impossible. If the $(x_n + y_n)$ is to be convergent, it implies we are able to make the distance $\vert{(x_n + y_n) - (x + y)}\vert$ as small as like. However, we cannot make $y_n$ to lie eventually in a set $(y - \epsilon, y + \epsilon)$. Hence, the sum cannot be convergent.

(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.

(d) This request is impossible. The key here is to show that, assuming here $(a_n)$ is bounded leads to the contradiction, leading to their difference also being bounded.

If $(a_n)$ is a bounded sequence, there exists a large number $M > 0$, such that $\vert{a_n}\vert < M$ for all $n \in \mathbf{N}$. If $(b_n)$ is a bounded sequence, there exists a large number $N > 0$, such that $\vert{a_n}\vert < N$ for all $n \in \mathbf{N}$.

Thus,

\begin{align*}
    \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\
    &\le \vert{a_n}\vert + \vert{-b_n}\vert\\
    &< M + N
\end{align*}

(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not.

答:
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

```<code>I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.
</code>```

(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.

Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.

Short proof.

Consider $a_n = \sqrt{n + 1} - \sqrt{n}$.

Observe that,

\begin{align*}
\sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
&=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\
&< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}}
\end{align*}

Pick $\epsilon > 0$. We can choose $N > \frac{4}{\epsilon^2}$. To show that this choice of $N$ indeed works, we prove that, that for all $n \ge N$,

\begin{align*}
\absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\
&< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon
\end{align*}

Thus, $(\sqrt{n+1} - \sqrt{n}) \to 0$.

Consider $b_n = n - \sqrt{n^2 + 2n}$

Observe that:

\begin{align*}
n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\
&= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\
&< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n}
\end{align*}

Pick an arbitrary $\epsilon > 0$. We choose an $N > \frac{1}{2\epsilon}$. To show that choice of $N$ indeed works, we find that:

\begin{align*}
\absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\
&< \frac{1}{2} \cdot (2\epsilon) = \epsilon
\end{align*}

Thus, $(n - \sqrt{n^2 + 2n}) \to -1$.

(b) This request is impossible. Alternatively, we believe that if $(x_n)$ converges and $(x_n + y_n)$ converges, then $(y_n)$ converges. Let us prove this fact rigorously. Suppose $\lim x_n = a$ and $ \lim x_n + y_n = b$. We shall prove that $\lim y_n = b - a$.

Observe that,

\begin{align*}
\absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\
&= \absval{(x_n + y_n - b) - (x_n - a)}\\
&\le \absval{x_n + y_n - b} + \absval{x_n - a}
\end{align*}

Pick an $\epsilon > 0$. Since $(x_n) \to a$, we can make the distance $\absval{x - a}$ as small as we like. There exists an $N_1$ such that

\begin{align*}
\absval{x_n - a} < \frac{\epsilon}{2}
\end{align*}

for all $n \ge N_1$.

Since $(x_n + y_n) \to b$, we can make the distance $\absval{x_n + y_n - b}$ as small as we like. There exists an $N_2$ such that,

\begin{align*}
\absval{x_n + y_n - b} < \frac{\epsilon}{2}
\end{align*}

Let $N = \max \{N_1,N_2 \}$. To show that this $N$ indeed works, we find that:

\begin{align*}
\absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\
&< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon
\end{align*}

(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.

Consider $(b_n) = \frac{1}{n}$. Pick an arbitrary $\epsilon > 0$. We can choose $N > \frac{1}{\epsilon}$. To show that this choice of $N$ indeed works, we find that:

\begin{align*}
\absval{\frac{1}{n}} < \epsilon
\end{align*}

for all $n \ge N$. Consequently, $(1/n) \to 0$.

(d) This request is impossible.

(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not. Let us prove $\frac{\sin n}{n}$ converges to $0$.

Observe that,

\begin{align*}
\absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\
&\le \frac{1}{n}
\end{align*}

Let $\epsilon > 0$ be an arbitary small but fixed positive real number. We choose an $N > \frac{1}{\epsilon}$. To prove that this choice of $N$ indeed works, we have,

\begin{align*}
\absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\
&< \frac{1}{(1/\epsilon)} = \epsilon
\end{align*}

Consequently, $\frac{\sin n}{n} \to 0$.

Body:
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.

(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.

Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.

Short proof.

Consider $a_n = \sqrt{n + 1} - \sqrt{n}$.

Observe that,

\begin{align*}
\sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
&=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\
&< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}}
\end{align*}

Pick $\epsilon > 0$. We can choose $N > \frac{4}{\epsilon^2}$. To show that this choice of $N$ indeed works, we prove that, that for all $n \ge N$,

\begin{align*}
\absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\
&< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon
\end{align*}

Thus, $(\sqrt{n+1} - \sqrt{n}) \to 0$.

Consider $b_n = n - \sqrt{n^2 + 2n}$

Observe that:

\begin{align*}
n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\
&= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\
&< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n}
\end{align*}

Pick an arbitrary $\epsilon > 0$. We choose an $N > \frac{1}{2\epsilon}$. To show that choice of $N$ indeed works, we find that:

\begin{align*}
\absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\
&< \frac{1}{2} \cdot (2\epsilon) = \epsilon
\end{align*}

Thus, $(n - \sqrt{n^2 + 2n}) \to -1$.

(b) This request is impossible. Alternatively, we believe that if $(x_n)$ converges and $(x_n + y_n)$ converges, then $(y_n)$ converges. Let us prove this fact rigorously. Suppose $\lim x_n = a$ and $ \lim x_n + y_n = b$. We shall prove that $\lim y_n = b - a$.

Observe that,

\begin{align*}
\absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\
&= \absval{(x_n + y_n - b) - (x_n - a)}\\
&\le \absval{x_n + y_n - b} + \absval{x_n - a}
\end{align*}

Pick an $\epsilon > 0$. Since $(x_n) \to a$, we can make the distance $\absval{x - a}$ as small as we like. There exists an $N_1$ such that

\begin{align*}
\absval{x_n - a} < \frac{\epsilon}{2}
\end{align*}

for all $n \ge N_1$.

Since $(x_n + y_n) \to b$, we can make the distance $\absval{x_n + y_n - b}$ as small as we like. There exists an $N_2$ such that,

\begin{align*}
\absval{x_n + y_n - b} < \frac{\epsilon}{2}
\end{align*}

Let $N = \max \{N_1,N_2 \}$. To show that this $N$ indeed works, we find that:

\begin{align*}
\absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\
&< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon
\end{align*}

(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.

Consider $(b_n) = \frac{1}{n}$. Pick an arbitrary $\epsilon > 0$. We can choose $N > \frac{1}{\epsilon}$. To show that this choice of $N$ indeed works, we find that:

\begin{align*}
\absval{\frac{1}{n}} < \epsilon
\end{align*}

for all $n \ge N$. Consequently, $(1/n) \to 0$.

(d) This request is impossible.

(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not. Let us prove $\frac{\sin n}{n}$ converges to $0$.

Observe that,

\begin{align*}
\absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\
&\le \frac{1}{n}
\end{align*}

Let $\epsilon > 0$ be an arbitary small but fixed positive real number. We choose an $N > \frac{1}{\epsilon}$. To prove that this choice of $N$ indeed works, we have,

\begin{align*}
\absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\
&< \frac{1}{(1/\epsilon)} = \epsilon
\end{align*}

Consequently, $\frac{\sin n}{n} \to 0$.


问:
Need help converting 79 to base 5
I am struggling with the topic of converting a number to a base 5....

Tried looking some stuff up but still confused

The one example I have on my homework is 79 to base 5

答:
The basic procedure to convert a number a to base b is by continuous division.

```<code>While quotient is not 0:
remainder = a mod b
display remainder as the digit
qoutient = floor(a/b)
assign a the value of quotient
</code>```

We display each remainder from right to left.

```<code>79 mod5 = 4
floor(79/5) = 15
15 mod 5 = 0
floor(15/5) = 3
3 mod 5 = 3
floor(3/5) = 0
</code>```

End of process.

Displaying the remainders from bottom to top:

$79_{base 10} = 304_{base 5}$

Title: Need help converting 79 to base 5

Body:
The basic procedure to convert a number a to base b is by continuous division.

While quotient is not 0:
remainder = a mod b
display remainder as the digit
qoutient = floor(a/b)
assign a the value of quotient

We display each remainder from right to left.

79 mod5 = 4
floor(79/5) = 15
15 mod 5 = 0
floor(15/5) = 3
3 mod 5 = 3
floor(3/5) = 0

End of process.

Displaying the remainders from bottom to top:

$79_{base 10} = 304_{base 5}$


问:
Work out the percentage of a single unit in a scale
1st question on this forum, and I know it will be an easy-peasy one for someone out there!!  I'm not even sure if the title of the question is accurate for the question I am asking.....I'm also not sure what tags to put it under!

Anyway, here goes....

I have a line (l) which is of an arbitrary length, which represents an arbitrary range of minutes (r).

For example, the line is 900 wide and represents 720 minutes.

How do I work out, as a percentage of l, what one minute is?

Thanks,

ETFairfax

答: The percentage is independent of the length of the line segment. If there are 720 minutes in total, 1 minute is 1/720th of this, or 100/720%.</p>

ContentLicense: CC BY-SA 3.0


问:
Proof about that operator is self-adjoint
In Linear Algebra:

Suppose I have a matrix $A$ that represents the linear product (the Gramian matrix)
Suppose I have a matrix $B$ that represents a linear transformation from Vector Space $V$ to itself.

I'm trying to prove that $T$ cannot be self adjoint.
I'm reading through this explanation but cannot seem to understand the logic behind the second last step. Why would $B^TA = AB$ ?

Why is this true:

If $\langle T(\mathbb{E}x), \mathbb{E}y \rangle = \langle \mathbb{E}x, T(\mathbb{E}y)\rangle \implies x^T(B^TA)y = x^T(AB)y$

Thanks!

答:
It follows from $$
  x^T (B^TA) y = x^T (AB) y \quad\text{for all $x,y$.}
$$

Assume that $B^TA \neq A B$. Then there'd be a vector $y$ with $B^TAy \neq ABy$. Let $y_1 := B^TAy$ and $y_2 := ABy$. Now, since $y_1 \neq y_2$ you can find an $x$ such that $x^T y_1 \neq x^Ty_2$. But then you'd have $$
  x^T(B^TA)y \neq x^T(AB)y
$$
which contradicts the initial assumption.


答:
Just find the remainder of division by 5, taking the positive or negative remainder, at each steps :

$79 = 15*5 + 4$

$15 = 3*5 + 0$

$3 = 0*5 + 3$

Thus in base 5, 79 is 304

As $3*5^2+0*5^1+4*5^0 = 79$

Title: Need help converting 79 to base 5

Body:
I am struggling with the topic of converting a number to a base 5....

Tried looking some stuff up but still confused

The one example I have on my homework is 79 to base 5


问:
Showing that $f(x, y)=xy$ is continuous on $S^1$
This is question from studying general topology.

Let $f : S^1 \times S^1 \rightarrow S^1$ defined by $f(x, y)=xy$

How can I show that $f$ is continuous?

My attempt : Let $U : $ {$e^{i\theta}$ | $\alpha < \theta < \beta $ } : open set in $S^1$.

Then, $f^{-1}(U)$={$(e^{i \theta _1}, e^{i \theta _2})$ | $\alpha + 2n\pi < \theta_1 + \theta_2 < \beta + 2n\pi$} , which is homeomorphic to $V$ = {$(x, y)$ | $\alpha + 2n\pi < x+y < \beta + 2n\pi$} .

Since $V$ is open, $f$ is continuous.

答: $f(e^{it}, e^{ir})  = e^{i (t+r)} $ is a composition of the two continuous functions $(r,t) \mapsto t+r$ and $e^{i x}$.


问:
Example of a topology that requires infinite unions
The standard topology on $\mathbb{R}$ has the property that any open set $U$ can be written as the countable union of open intervals, which are basis elements for the standard topology on $\mathbb{R}$. Thus, I believe, in the standard topology on $\mathbb{R}$, one may replace the axiom of topology that "the union of arbitrarily many open sets is open" with the alternative axiom that "the countable union of open sets is open.''

Are there examples of a topology on $\mathbb{R}$ where this axiom replacement is not possible.  That is, is there a topology $\mathscr{T}$ with basis $\mathscr{B}$ and open set $U\in\mathscr{T}$ such that

$$ \bigcup_{i=1}^\infty B_i \neq U$$

for any choice of $B_i$ in $\mathscr{B}$?

答:
Fixed point topology on $\Bbb R$ where an open set is defined as empty or a set that contains the fixed point $x_0\in\Bbb R$. We take the basis $B=\{\{x_0,y\}:y\in\Bbb R\}$. Then any uncountable set containing $x_0$ can't be written as a countable union of basis elements.

Note also that for the same topology, we might get different results for different bases. If we take the basis as the topology itself, "axiom replacement" is always possible.


问:
Generation of Hermite polynomials with Gram-Schmidt procedure
I want to use the Gram-Schmidt procedure to generate the first three Hermite polynomials. Given the set of linearly independent vectors $\{1,x,x^2,...\}$ in the Hilbert space $L^2(R,e^{-x^2}dx)$, I apply the orthogonalisation procedure as follows:

$u_{0}(x)=1$

$u_{1}(x)=x-\frac{\langle u_{0}|x \rangle}{\langle u_{0}|u_{0}\rangle} \cdot u_{0}=x-\frac{\langle 1|x\rangle}{\langle 1|1\rangle}\cdot 1 =x $

$u_{2}(x)=x^2-\frac{\langle u_{0}|x^2 \rangle}{\langle u_{0}|u_{0}\rangle} \cdot u_{0}-\frac{\langle u_{1}|x^2 \rangle}{\langle u_{1}|u_{1}\rangle} \cdot u_{1}=x^2-\frac{\langle 1|x^2\rangle}{\langle 1|1\rangle}\cdot 1-\frac{\langle x|x^2\rangle}{\langle x|x\rangle}\cdot x =x^2 -\frac{1}{2} $

The inner product I use is the one of $L^2(R,e^{-x^2}dx)$, namely:

$\langle u|v \rangle:=\int_{R}\overline{u(x)}v(x)e^{-x^2}dx$

Then I divide the functions I found by their norm in order to get orthonomal vectors:

$\tilde{H_0}(x)=\frac{1}{\sqrt{\langle u_{0}|u_{0} \rangle}}u_{0}(x)=\frac{1}{\pi^{1/4}}$

$\tilde{H_1}(x)=\frac{1}{\sqrt{\langle u_{1}|u_{1} \rangle}}u_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}x$

$\tilde{H_2}(x)=\frac{1}{\sqrt{\langle u_{2}|u_{2} \rangle}}u_{2}(x)=\frac{\sqrt{2}}{\pi^{1/4}}(x^2-\frac{1}{2})$

I can't understand why the coefficients I get are different from the ones that can be found on Wikipedia (https://en.wikipedia.org/wiki/Hermite_polynomials) or in every list of Hermite's polynomials. Where is the mistake?

答: You are generating the Legendre polynomials with this procedures, i.e., if you take $\{1,x,x^2, x^3, ...\}$ and apply Gram-Schmidt on it you'll end up with Legendre polynomials.


答:
To convert a number $n$ to any base divide $n$ by the base to get both the quotient and remainder. The remainder becomes the rightmost digit in the representation and if the quotient is greater than or equal to the base we divide again to get the next digit, etc.

Let's convert $100$ to base $7$ as an example. If we divide $100$ by $7$ we get a quotient of $14$ and a remainder of $2$. This makes $2$ the right most digit.

Is $14\lt 7$? No so we divide $14$ by $7$ to get a quotient of $2$ and a remainder of $0$. The $0$ becomes the next digit (to the left of our first one) and we again evaluate the quotient.

Is $2\lt 7$? Yes so $2$ becomes our last digit and we get $100=202_7$.

Title: Need help converting 79 to base 5


问:
Calculating polar decomposition
How do I compute the left and right polar decompositions of a matrix by hand? I understand the definition of the decomposition but not how it is calculated.

For instance are the left and right polar decompositions of this matrix different, and how do I compute them?

$$
\begin{bmatrix}
2 & -3\\
1 & 6\\
\end{bmatrix}
$$

答:
A brute force (numerical) method for obtaining the right polar decomposition has been presented in
this answer at MSE.
When applied to the OP's problem (Test4) we get an outcome like:
$$
\begin{bmatrix}
    2.000000 &  -3.000000 \\
    1.000000 &  6.000000
\end{bmatrix} = \\
\begin{bmatrix}
    0.894427 & -0.447214 \\
    0.447214 &  0.894427
\end{bmatrix}
\begin{bmatrix}
    2.236068 &  0.000000 \\
    0.000000 &  6.708204
\end{bmatrix}
$$
However, when searching for the number $0.447214$ on the internet, we find that it's approximately equal to
What is 1 over square root of 5?.
So let's do an educated guess:
$$
\begin{bmatrix}
 2 & -3 \\
 1 & 6
\end{bmatrix} =
\begin{bmatrix}
 2/\sqrt{5} & -1/\sqrt{5} \\
 1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
 \sqrt{5} & 0 \\
 0 & 3\sqrt{5}
\end{bmatrix}
$$
It is noticed that the above question is similar to another one.

EDIT. But the above is, of course, overkill.

Here comes a much simpler solution. First form the transpose times the original matrix:
$$
\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}^T\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} =
\begin{bmatrix} 2 & 1 \\ -3 & 6 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} =
\begin{bmatrix} 5 & 0 \\ 0 & 45 \end{bmatrix}
$$
Then make the following Ansatz:
$$
\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}^2 =
\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix} =
\begin{bmatrix} a^2 & 0 \\ 0 & c^2 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 45 \end{bmatrix}
$$
So the square root of a diagonal matrix is extremely simple; just take the square roots of the diagonal elements:
$$
\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix} = \begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix}
$$
At last, the orthogonal matrix is found with:
$$
\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix}^{-1} = \\
\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & 0 \\ 0 & 1/(3\sqrt{5}) \end{bmatrix} =
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
Same conclusion:
$$
\begin{bmatrix}
 2 & -3 \\
 1 & 6
\end{bmatrix} =
\begin{bmatrix}
 2/\sqrt{5} & -1/\sqrt{5} \\
 1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
 \sqrt{5} & 0 \\
 0 & 3\sqrt{5}
\end{bmatrix}
$$

Title: Calculating polar decomposition


问:
Naturality of Bockstein Homomorphism with respect to Boundary Homomorphism
Let $A \subset X$ be topological spaces and let $0 \to G' \overset{h}{\to} G \overset{k}{\to} G'' \to 0$ be a short exact sequence of abelian groups. We get a long exact sequence of homology groups:
$$\cdots \overset{\beta}{\to} H_n(X,A;G') \overset{h_\sharp}{\to} H_n(X,A;G) \overset{k_\sharp}{\to} H_n(X,A;G'') \overset{\beta}{\to} H_{n-1}(X,A;G') \to \cdots.$$
Here $h_\sharp$ and $k_\sharp$ are coefficient homomorphisms and $\beta$ is the Bockstein operator. I'm trying to show that $\beta$ is natural with respect to the boundary homomorphism of the pair $(X,A)$; that is, $\partial_*\beta = \beta\partial_* : H_n(X,A;G'') \to H_{n-2}(A;G')$. This should be a straightforward exercise in unpacking definitions, but I'm runnning into some trouble. Here's my attempt:

It suffices to verify that the maps commute on the generating set of elements of the form $[u \otimes x]$, where $u$ is a relative $n$-chain in $C_n(X,A)$ and $x \in G''$. Applying $\partial_*$, this element maps to $[\partial u \otimes x]$, where $\partial u \in C_{n-1}(A)$. Now we apply $\beta$: First we pick an element of $C_{n-1}(A;G)$ which maps to $\partial u \otimes x$ under $1 \otimes k$; the element $\partial u \otimes y$ will do the trick, where $y$ is any element such that $k(y) = x$. Now we apply the boundary map $\partial \otimes 1$ to this element, and here's where I run into trouble: doesn't this give $\partial^2u \otimes y = 0$? I tried performing $\beta$ and $\partial_*$ in the opposite order and did not get zero, so something's definitely wrong here. Where did I mess up?

答:
You seem to be using $\partial$ to mean both the connecting homomorphism and the boundary map of chain complexes.  While related, they are not exactly the same.  There are also several connecting homomorphisms in play here associated to the short exact sequences of complexes
\begin{equation}
C_\bullet(A; ?) \to C_\bullet(X; ?) \to C_\bullet(X,A; ?)
\end{equation}
\begin{equation}
C_\bullet(??; G') \to C_\bullet(??; G) \to C_\bullet(??; G''),
\end{equation}
where $? \in \{G', G, G''\}$ in the first line and $?? \in \{A, X, (X,A)\}$ in the second, and you should take care not to confuse them.

Anyway, here is a hands-on way of showing that the Bockstein commutes with the connecting homomorphism of pairs of spaces.  Assuming you've already shown that the chain level definitions are well-defined, you want to show that the following diagram "commutes":
\begin{equation}
\begin{array}{ccccccc}
C_n(X,A;G'') & \leftarrow & C_n(X,A;G) & \rightarrow & C_{n-1}(X,A;G) & \leftarrow & C_{n-1}(X,A;G') \\
\uparrow & & & & & & \uparrow \\
C_n(X;G'') & & & & & & C_{n-1}(X;G') \\
\downarrow & & & & & & \downarrow \\
C_{n-1}(X;G'') & & & & & & C_{n-2}(X;G') \\
\uparrow & & & & & & \uparrow \\
C_{n-1}(A;G'') & \leftarrow & C_{n-1}(A;G) & \rightarrow & C_{n-2}(A;G) & \leftarrow & C_{n-2}(A;G')
\end{array}
\end{equation}

For example, the top row should be read as "take a cycle in $C_n(X,A;G'')$, lift it to a chain in $C_n(X,A;G)$ using the short exact sequence of groups, apply the boundary map to get an element in $C_{n-1}(X,A;G)$, then use the short exact sequence of groups again to get an cycle in $C_{n-1}(X,A;G')$".  Upon passing to homology, this is the definition of the Bockstein homomorphism.  The other sides of the diagram should be interpreted similarly.

To show that the diagram "commutes", simply note that it is built out of smaller squares:
\begin{equation}
\begin{array}{ccccccc}
C_n(X,A;G'') & \leftarrow & C_n(X,A;G) & \rightarrow & C_{n-1}(X,A;G) & \leftarrow & C_{n-1}(X,A;G') \\
\uparrow & & \uparrow & & \uparrow & & \uparrow \\
C_n(X;G'') & \leftarrow & C_n(X;G) & \rightarrow & C_{n-1}(X;G) & \leftarrow & C_{n-1}(X;G') \\
\downarrow & & \downarrow & & \downarrow & & \downarrow \\
C_{n-1}(X;G'') & \leftarrow & C_{n-1}(X;G) & \rightarrow & C_{n-2}(X;G) & \leftarrow & C_{n-2}(X;G') \\
\uparrow & & \uparrow & & \uparrow & & \uparrow \\
C_{n-1}(A;G'') & \leftarrow & C_{n-1}(A;G) & \rightarrow & C_{n-2}(A;G) & \leftarrow & C_{n-2}(A;G')
\end{array}
\end{equation}

You should check that each of the smaller squares commute.  Therefore, the large outer square "commutes" too, and this shows that the Bockstein and the connecting homomorphism commutes.

EDIT: Following the discussion in the comments, here is a simplified toy example that might make things clearer.  Consider the map of chain complexes:
\begin{equation}
\begin{array}{ccc}
0 & \rightarrow & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \xrightarrow{f} & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \rightarrow & 0
\end{array}
\end{equation}
where the maps $\mathbb{Z} \to \mathbb{Z}$ are all identities and everything else is $0$.  It's obvious that $\partial^2 = 0$ for both the chain complexes on the left and right, but if you start at the upper right corner, apply $\partial$, then $f^{-1}$, and then $\partial$, you would get the identity map, not zero.


答:
The first thing to understand when converting a number from base 10 ($x_{10}$) to any given base $b$ is that the number in base 10 is equal to a weighted sum of powers of the base, $x_{10}=a_0b^{0}+a_1b^{1}+\dots+a_nb^{n}$.
So to convert $79_{10}$ to base 5 you need to find the $a_i$. You can find them by dividing the number in base 10 by the base over and over again, until division do not result in a integer quotient and then taking the remainder and the last quotient in the inverse order.

So
\begin{array}{r|r}
79&5\\
\hline
\color{red}4&15
\end{array}
\begin{array}{r|r}
15&5\\
\hline
\color{red} 0&\color{red}3
\end{array}
$79_{10}=304_{5}=4 \cdot 5^0+0\cdot5^1+3\cdot5^2=4+3\cdot25=4+75=79$

Title: Need help converting 79 to base 5


问:
$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$
Prove:

$$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$$

where $H_n=\sum\limits_{k=1}^n\frac1k$

For the $x^k$ i tried  $x^k=((x-1)+1)^k$ and decompose it into binomial expansion but I got dual sum at the one side one sum at the other side

For the second method:

$$\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}-\sum\limits_{k=1}^n \frac{x^k}{k}=H_n$$

and try to show

$$\binom{n}{k}(x-1)^k-x^k=1$$

Either the problem is wrong or I cannot do it.

答:
I think the identity does not hold. Using
$$ x^k=[1+(x-1)]^k=\sum_{i=0}^k\binom{k}{i}(1-x)^i=1+\sum_{i=1}^k\binom{k}{i}(1-x)^i $$
one has
\begin{eqnarray}
&&\sum\limits_{k=1}^n \frac{x^k}{k}\\
&=&\sum\limits_{k=1}^n\frac{1}{k}\bigg(1+\sum_{i=1}^k\binom{k}{i}(1-x)^i\bigg)\\
&=&H_n+\sum\limits_{k=1}^n\sum_{i=1}^k \binom{k}{i} \frac{(x-1)^i}{k}\\
&\neq&H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}.
\end{eqnarray}


问:
Given upper bound of the width, which closed convex curve has the largest area?
The "width" represents the distance between two parallel supporting lines.

I guess it must be the constant width curves, among which the circle has the maximum area. But I cannot prove this.

答:
The maximum width of a bounded planar set $S$ is just equal to its diameter $d$ (the maximum distance between any two points), because for any two parallel lines touching $S$, the distance between their points of contact is at least the width (and hence the width is $\le d$), and by taking the lines to be perpendicular to a diameter of $S$ we can attain this bound.

So your question amounts to "Given an upper bound on the diameter, which convex curve has the largest area?". For this, we can use the isodiametric inequality: $4A\le \pi d^2$, or equivalently, the disc has the largest volume among sets of a fixed diameter in the plane. So your guess is correct.

This PDF goes into more detail about a proof of this fact.

(In fact, everything in this answer holds for arbitrary sets (no convexity necessary) in $\mathbb{R}^n$, if we replace "parallel lines" with "parallel hyperplanes of dimension $n-1$" and "disc" with "$n$-dimensional ball".)


答:
First ask your self what is the highest power of $5$ that is less than or equal to $79$.

$5^0 = 1 < 79, 5^1=5 < 79, 5^2 = 25 < 79; 5^3 = 125 > 79$.

So $5^2 < 79 < 5^3$.

So that's it.  $79 = a\times 5^2 +b\times 5 + c$ for some values of $a,b,c$ and $79$ will have $3$ digits in base $5$.

Figure out what times $25$ will equal $79$ with what remainder.

$79\div 25 = 3$ with $4$ remainder.  That is $79 = 3*25 + 4$.

So the first digit is $3$.  And we have $79 = 3*5^2 + b*5 + c$.  We need to find the next two digits.

Do the so for the next lower power of $5$.  We found out the digit for $5^2$ place so now we must do the some for the $5^1$ place.  So we take the remainder of $4$ and ask ourselves:

What times $5=5^0$ will equal $4$, and what will be the remainder?

The answer to that, and this is probably what is confusing you, $4< 5$ so we must have $0* 5^1$.  In other words $4\div 5 = 0$ with $4$ remainder.  So the second digit is $0$. And we have $79 = 3*5^2 + 0*5^1 + c$.  We must find the last digit.

So we do this for the last power of $5$.  That is $5^0$.  What times $1$ will equal $4$ and what is the remainder.

Well, this is trivial.  $4\div 1 = 4$ with zero remainder.  So last digit is $4$ and we have $79 = 3*5^2 + 0*5^1 + 4$.

and that's it  $79 = 304_5$.

The verify $304_5 = 3*5^2 + 0*5^1 + 4*5^0  = 3*25 + 0*5 + 4*1 = 75+4 = 79$.

That's it.

Title: Need help converting 79 to base 5


问:
Interesting physical phemomena to model
I'm a high-school student looking for some interesting physical problems to model. Some examples include the shape a trampoline makes when you jump on it, the equations describing heat transfer, collisions of particles (something like 3Blue1Brown's video on "How many collisions?" where Pi pops up unexpectedly).

I'm particularly interested in how differential equations can model these scenarios, such as heat transfer. Do you have any good problems you can share?

答:
The Furuta pendulum is a classic example in control theory and it can be interesting if you want to model physics problems

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

Title: Interesting physical phemomena to model


问:
When is 2 a quadratic residue mod p
I'm trying to prove that 2 is a QR when $p = \pm \ 1 \text{ mod } 8.$ I know how to prove via a "factorial" proof, butI'm given the hint to use the fact that if $ \zeta $ is a primitive 8-root of unity, then $(\zeta + \bar{\zeta})^2 = 2$, and would like to solve the exercise using the hint. This is where I got so far: $$ \text{ I can use the fact that } 2 \text{ is a quadratic residue if and only if } 2^{\frac{p-1}{2}}=1 \text{ mod p}$$
$$ 2^{\frac{p-1}{2}} = (\zeta + \bar{\zeta})^{p-1} = \sum_{k = 0} ^{p-1} {p-1 \choose k} \bar\zeta^k \zeta^{p-1-k} = \sum_{k = 0} ^{p-1} {p-1 \choose k}\zeta^{p-1+6k}$$ Now I guess I should use the fact that $ \zeta^4 = -1$, and thus the symmetry of the sum cancels many factors, but I'm stuck at this point, and even computing the sum for small p didn't help much. Can someone help me out? Thanks!

答:
Here's another proof that also uses the fact that $(\zeta+\overline{\zeta})^2=2$.

From the equation $2=(\zeta+\overline{\zeta})^2=(\zeta+\zeta^{-1})^2$, we see that $\Bbb Q(\zeta)$ contains $\Bbb Q(\sqrt{2})$.

It is well-known that the map $(\Bbb Z/8\Bbb Z)^\times \to \mathrm{Gal}(\Bbb Q(\zeta)/\Bbb Q)$ which sends $\overline{a}$ to the unique automorphism $\varphi_a$ that satisfies $\varphi_a(\zeta)=\zeta^a$ is an isomorphism of groups.

Now using that $(\zeta+\zeta^{-1})^2=2$ one sees that under the isomorphism above the subgroup $\mathrm{Gal}(\Bbb Q(\zeta)/\Bbb Q(\sqrt{2})) \subset \mathrm{Gal}(\Bbb Q(\zeta)/\Bbb Q)$ corresponds to the subgroup $\{\overline{\pm 1\}}\subset (\Bbb Z/8\Bbb Z)^\times$.

Now every odd prime $p$ is unramified in $\Bbb Q(\zeta)$, so we have a Frobenius element $\left(\frac{\Bbb Q(\zeta)/\Bbb Q}{p}\right)$ in the Galois group which is just given by $\varphi_p:\zeta \mapsto \zeta^p$. We have that this restricts to a Frobenius element in the Galois group $\mathrm{Gal}(\Bbb Q(\sqrt{2})/\Bbb Q)$ (directly by definition of Frobenius element) $\left(\frac{\Bbb Q(\zeta)/\Bbb Q}{p} \right)|_{\Bbb Q(\sqrt{2})}=\left(\frac{\Bbb Q(\sqrt{2})/\Bbb Q}{p} \right)$.

Now $p$ splits in $\Bbb Q(\sqrt{2})$ iff $\left(\frac{\Bbb Q(\sqrt{2})/\Bbb Q}{p} \right)=\mathrm{id}$. This can be seen easily by the definition of Frobenius element: take $\mathfrak{P}$ above $p$ in $\Bbb Z[\sqrt{2}]$, then the residue field extension $\Bbb Z[\sqrt{2}]/\mathfrak{P}$ over $\Bbb Z/(p)$ has degree $1$ iff $p$ splits.
$p$ splitting in $\Bbb Z[\sqrt{2}]$ is equivalent to the splitting of the polynomial $x^2-2$ mod $p$ which splits iff $2$ is a quadratic residue mod $p$.

In summary, we get that $2$ is a quadratic residue mod $p$ iff $p$ splits in $\Bbb Q(\sqrt{2})$ iff the Frobenius element $\left(\frac{\Bbb Q(\sqrt{2})/\Bbb Q}{p} \right)$ is the identity.

But we have seen that the Frobenius element $\left(\frac{\Bbb Q(\sqrt{2})/\Bbb Q}{p} \right)$ is just the restriction of $\varphi_p$ to $\Bbb Q(\sqrt{2})$. This restriction is the identity iff $\varphi_p \in \mathrm{Gal}(\Bbb Q(\zeta)/\Bbb Q(\sqrt{2}))$. But using the isomorphism $\mathrm{Gal}(\Bbb Q(\zeta)/ \Bbb Q) \cong (\Bbb Z/8\Bbb Z)^\times$, $\varphi_p$ is sent to $\overline{p}$ and $\mathrm{Gal}(\Bbb Q(\zeta)/\Bbb Q(\sqrt{2}))$ is sent to $\{\overline{\pm 1}\}$. Thus $\varphi_p |_{\Bbb Q(\sqrt{2})}=\mathrm{id} \Leftrightarrow p \equiv \pm 1 \pmod{8}$. This finishes the proof.

To summarize the whole argument, we have

$2$ is a quadratic residue mod $p$ $\Leftrightarrow$ $p$ splits in $\Bbb Q(\sqrt{2})$ $\Leftrightarrow$ $\mathrm{id}=\left(\frac{\Bbb Q(\sqrt{2})/\Bbb Q}{p}\right) = \left(\frac{\Bbb Q(\zeta)/\Bbb Q}{p}\right)|_{\Bbb Q(\sqrt{2})}=(\varphi_p)|_{\Bbb Q(\sqrt{2})}$ $\Leftrightarrow$ $\varphi_p \in \mathrm{Gal}(\Bbb Q(\zeta)/\Bbb Q(\sqrt{2}))$ $\Leftrightarrow$ $p \equiv \pm 1 \pmod{8}$

This argument is more advanced because it uses Galois theory and basic algebraic number theory, but it feels more conceptual as it avoids explicit calculations. Given the tags "Galois theory" and "algebraic number theory", it seems appropriate.

Title: When is 2 a quadratic residue mod p


问:
Find the limit $\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$
I have tried simplifying in every way I can think of, but I just can't seem to figure this out. I am asked to find the following limit:
$$\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$$

答:
Elementary rearranging gives:

\begin{eqnarray*}&\,\,&\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\\&=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{\frac1n}\\&=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{\frac1n-\frac1{n^3}}\left(1-\frac1{n^2}\right)\\
&=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{\frac1n-\frac1{n^3}}\lim_{n\to \infty}\left(1-\frac1{n^2}\right)\\
&=&\lim_{n\to \infty} \frac{\sqrt[3]{1+\frac1n}-\sqrt[3]{1+\frac1{n^3}}}{(1+\frac1n)-(1+\frac1{n^3})}
\end{eqnarray*}

Let $f(t)=\sqrt[3](t)$, and let $x_n=1+\frac1n, x'_n=1+\frac1{n^3}$.

Our limit is then:$$
\lim_{n\to \infty} \frac{f(x_n)-f(x'_n)}{x_n-x'_n}=f'(1).$$

Fianlly note $f'(1)=\frac13$.


问:
chebyshev Inequality, estimator only asyomptotically unbiased
Can I use the chebyshev inequality if my estimator is only asymptotically unbiased?
I am trying to show that my estimator converges in probability to $\mu$ if it is asymptotically unbiased and its asymptotical variance is $0$.

Many thanks!

答:
The short answer is yes!

For more detailed answer we denote the estimator of $\mu$ by $\widehat{\mu}$ then in order to show the convergence in probability write

$$
\mathbb{P}(|\widehat{\mu} - \mu| > \varepsilon) \to 0, \text{ as } n \to \infty.
$$

Denote the bias term $b_n = \mathbb{E}\widehat{\mu} - \mu$, then using Markov inequality (or Chebyshev, if you wish) we get

\begin{align}
\mathbb{P}(|\widehat{\mu} - \mu| > \varepsilon) &\le \frac{\mathbb{E}(\widehat{\mu} - \mu)^2} {\varepsilon^2} = \frac{\mathbb{E}(\widehat{\mu} - (\mu + b_n) + b_n)^2} {\varepsilon^2} \\ &= \frac{\mathbb{E}(\widehat{\mu} - (\mu + b_n))^2 + \mathbb{E} b_n^2} {\varepsilon^2} = \frac{\mathbb{V}ar(\widehat{\mu}) + b_n^2}{\varepsilon^2} \to 0,
\end{align}
since the variance tends to $0$ and the bias term $b_n$ tends to $0$ as $n \to \infty$.


问:
Relationship between the sum of the digits of an integer in different bases
Let $n \in \mathbb{Z}$ and let $a, b \in \mathbb{N}$ be two bases in which $n$ can be represented. Then:

$$n = \sum_{i=0}^k \alpha_i a^i = \sum_{j=0}^l \beta_j b^j$$.

I am specifically interested in the quantities $S_a = \sum_{i=0}^k \alpha_i$ and $S_b = \sum_{j=0}^l \beta_j$.  In particular, if I know $a$, $b$, and one of $S_a$ or $S_b$, can I determine the other sum without having to know $n$?  if so, how?

答:
If your bases are prime numbers, and if the representations are standard—that is, $\alpha_i\in\{0,1,\ldots,a-1\}$ and $\beta_j\in\{0,1,\ldots,b-1\}$, then Legendre’s formula for evaluating the the greatest power $v_p(n!)$ of a prime $p$, dividing $n!$ is given by
$$v_p(n!)=\frac{n-S_p}{p-1}\,.$$
This immediately leads to the identity
$$S_a+(a-1)v_a(n!)=S_b+(b-1)v_b(n!)\,.$$
This seems to demand that knowledge of $n$ is necessary to obtain say $S_b$ if you only knew $a,b$ and $S_a$.


答:
You could use binomial series:

$\lim\limits_{n\to \infty}\left[ \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\right]$

$=\lim\limits_{n\to\infty}\left[n\left(1+\dfrac1n\right)^{1/3}-n\left(1+\dfrac1{n^3}\right)^{1/3}\right]$

$=\lim\limits_{n\to\infty}\left[n\left(1+\dfrac{\color{blue}1}{\color{blue}3n}+\cdots\right)-n\left(1+\dfrac1{3n^3}+\cdots\right)\right]=\color{blue}{\dfrac13}$

Title: Find the limit $\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$

Body:
I have tried simplifying in every way I can think of, but I just can't seem to figure this out. I am asked to find the following limit:
$$\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$$


问:
If C is any simple closed piecewise-smooth curve in $\mathbb{R}^3$, what is the value of the line integral?
If C is any simple closed piecewise-smooth curve in $\mathbb{R}^3$, what is the value of the line integral $\oint\limits_{C} x^4 dx +y^4 dy+z^4 dz$

What I know so far:

I think the answer is $0$ because the integral around any close path is $0$.

答:
Consider the function $f(x,y,z):={1\over5}(x^5+y^5+z^5)$ with gradient
$$\nabla f(x,y,z)=(x^4,y^4,z^4)\ .$$
It follows that  for any smooth curve $\gamma$ that connects the point ${\bf r}_0$ with ${\bf r}_1$ one has
$$\int_\gamma x^4\>dx+y^4\>dy+z^4\>dz=\int_\gamma\nabla f({\bf r})\cdot d{\bf r}=f({\bf r}_1)-f({\bf r}_0)\ .$$
A piecewise smooth closed curve $C$ is a finite sum of such $\gamma$, with pairwise matching start and end points. It follows that
$$\int_C x^4\>dx+y^4\>dy+z^4\>dz=0\ .$$


问:
If $g$ is a polynomial and real numbers $a<c<b$ are three roots of $g$, then $g''-g$ has a root in $[a,b]$.
There is a sub-question before this, asking me to prove that $g'+g$ has a root in $[a,b]$ (call this proposition 1), which I did.

My thought is to let $f=g'+g$ and $h=f'-f$ so that $h=g''-g$. Then by proposition 1, $f$ has a root in $[a,c]$ and $[c,b]$. If $r_1\in [a,c)$ and $r_2\in (c, b]$ are two roots of $f$, then I would be able to show that $h$ has a root in $[r_1, r_2]$ by a technique similar to the one used to prove proposition 1.

But what if $f$ has only one root, namely $c$, in $[a,b]$? Is this possible? I'm stuck.

答:
You can show that $f= g'+g$ has a root in each of the open intervals $(a, c)$ and $(c, b)$, by applying Rolle's theorem to $e^x g(x)$.

That gives you two distinct roots $r_1$ and $r_2$ of $f$, and it follows that $f' = g''-g$ has a root in the open interval $(r_1, r_2)$.

Note that $g$ need not be a polynomial, it suffices that $g$ is continuous on $[a, b]$ and differentiable on $(a, b)$.


问:
cardinality problem concerning inclusion of sets
if $A\subseteq B$ then $A\le B$ , i,e if A is a subset of B then A is less than or equinumerous with B
How to do that?
IF A is a subset of B then every element in A is an element in B, so A cannot have more elements than B has right? </p>

答: The meaning of $A\le B$, in this context, is "there exists an injective function $f:A\to B$". Since $A\subseteq B$, one such injective function would be $f:A\to B$, f(a)=a. This proves that $A\le B$. </p>


问:
Detecting closed walks that visit a given edge
Is it possible to indicate whether an edge is part of a closed walk in an undirected graph only by looking at its adjacency matrix and using only matrix powers and linear algebra?

My approach:

Given adjacency matrix $A$ for graph $G$, we denote the edge between nodes $v_i$ and $v_j$ as $\varepsilon_{i, j}$. For $\varepsilon_{i, j}$ not to be in a closed loop, we need to have
$$
A_{i,j}=1
$$
and
$$
A^2_{i,j}+A^3_{i,j}+\dots+A^n_{i,j}=0
$$
where $n$ is the size of graph. However, that includes paths that contain $\varepsilon_{i, j}$. To that end, we can use Roberts approach below:

Let $B$ be obtained from the adjacency matrix $A$ by setting  the
$(i,j)$ and $(j,i)$ entries to $0$.

Then I need to compute a new matrix $B$ for each element $(i,j)$ in A, compute $\sum_{k=1}^{n-1} B^k$. However, that is way too expensive. Is there a method that is computationally less expensive, at least to be able to compute it without n^3 loops?

答: Let $B$ be obtained from the adjacency matrix $A$ by setting  the $(i,j)$ and $(j,i)$ entries to $0$ (i. e. the adjacency matrix of the graph obtained by deleting edge $ij$). The edge $ij$ is part of a closed walk iff $\sum_{k=1}^{n-1} (B^k)_{i,j} = 0$, where the graph has $n$ vertices.


答:
The notation

$$f^{-1}(U)=\{(e^{i \theta _1}, e^{i \theta _2}) \mid \alpha + 2n\pi < \theta_1 + \theta_2 < \beta + 2n\pi\}$$ is fuzzy. What is $n$ here? Also when you state that it is homeomorphic to

$$V= \{(x, y) \mid \alpha + 2n\pi < x+y < \beta + 2n\pi\}$$

this deserves some argument as well as to precise to which set the $(x,y)$ belong to.

This question looks simple... but formulating a precise answer is not so easy (at least for me!).

I would suggest the following:

Take $(x,y) = (e^{i\alpha}, e^{i\beta})\in S^1 \times S^1$. We have to prove that $f$ is continuous at $(x,y)$.

By considering the continuous map $(u,v) \mapsto (ue^{-i\alpha},ve^{-i\beta})$ defined on $S^1 \times S^1$, we can without loss of generality suppose that $(x,y) = (1,1)$.

The open $S^1_+=\{z \in S^1 \mid \Re z >0\}$ is homeomorphic to $(-\frac{\pi}{2},\frac{\pi}{2})$ under the standard homeomorphism $\varphi$ and $(1,1) \in S^1_+ \times S^1_+$.

$f = \overline{\varphi} \circ S \circ \Phi$ where

$$\begin{array}{l|rcl}
\Phi : & S^1_+  \times S^1_+& \longrightarrow & (-\frac{\pi}{2},\frac{\pi}{2}) \times (-\frac{\pi}{2},\frac{\pi}{2}) \\
    & (u, v) & \longmapsto & (\varphi^{-1}(u),\varphi^{-1}(v)) \end{array}$$

$$\begin{array}{l|rcl}
S: &  (-\frac{\pi}{2},\frac{\pi}{2}) \times (-\frac{\pi}{2},\frac{\pi}{2})& \longrightarrow & (-\pi,\pi) \\
    & (r,s) & \longmapsto & r+s \end{array}$$

$$\begin{array}{l|rcl}
\overline{\varphi}: &  (-\pi,\pi)& \longrightarrow & S^1 \\
    & t & \longmapsto & e^{it} \end{array}$$

are all continuous map.

Title: Showing that $f(x, y)=xy$ is continuous on $S^1$

Body:
This is question from studying general topology.

Let $f : S^1 \times S^1 \rightarrow S^1$ defined by $f(x, y)=xy$

How can I show that $f$ is continuous?

My attempt : Let $U : $ {$e^{i\theta}$ | $\alpha < \theta < \beta $ } : open set in $S^1$.

Then, $f^{-1}(U)$={$(e^{i \theta _1}, e^{i \theta _2})$ | $\alpha + 2n\pi < \theta_1 + \theta_2 < \beta + 2n\pi$} , which is homeomorphic to $V$ = {$(x, y)$ | $\alpha + 2n\pi < x+y < \beta + 2n\pi$} .

Since $V$ is open, $f$ is continuous.


问:
Is zero a limit point of the sequence $(\sqrt n \sin n)$
Is zero a limit point of the sequence $(\sqrt n \sin n)_n$?</p>

答: By Dirichlet's approximation theorem, $\pi$ can be approximated by rationals $p/n$ so that $|n\pi-p|<\frac{1}{n}$. Since sine is 1-Lipschitz,  we get$|\sin(p)|<\frac{1}{n}$ from which the result is immediate.</p>


问:
Do I need to verify solutions to functional equations?
I am studying the (basics of) solving functional equations. My teacher stipulates that we check any solutions obtained by substitution. Similar guidelines are given in this IMO training material.

For a typical example, consider $f:\mathbb{R} \setminus \{0\} \mapsto \mathbb{R}$

$$ 2f(x) + f \left( \frac{1}{x} \right) = x.$$

The substitutions $x \mapsto y$ and $x \mapsto \frac{1}{y}$ yield (after some algebra) the solution $f(y) = \frac{2y}{3} - \frac{1}{3y}$. Verifying, we see this is indeed a solution.

My question: In algebra we have some well-known possibilities how extraneous solutions can appear. Are there additional ways of obtaining extraneous solutions in functional equations?

答:
Let's address solutions in algebra first and then move to functional equations. Suppose you have an equation $F_1(x) = 0$ where you are solving for $x$. Then you perform some transformation(s) to yield $F_2(x) = 0$.

In general, there are 3 cases:

$F_1(x) = 0 \implies F_2(x) = 0$. This may introduce extraneous solutions.

$F_1(x) = 0 \impliedby F_2(x) = 0$. This may lose solutions.

$F_1(x) = 0 \iff F_2(x) = 0$. This neither adds nor loses solutions.

For example, $f(x) = g(x) \implies f^2 (x) = g^2(x)$, but the inverse does not always hold. Hence extraneous solutions can arise and you must check your final solution against $f(x) = g(x)$.

The logic is similar for functional equations. Note that substitution, the most common tool for solving functional equations, is not an equivalence transformation. Say we are solving the functional equation $F(x, f) = 0 \ \forall x \in \mathbb{R}$ for $f: \mathbb{R} \to \mathbb{R}$, and we substitute $x \mapsto g(x)$ for some function $g: \mathbb{R} \to \mathbb{R}$. Clearly, $$ \left( F(x, f) = 0 \ \forall x \in \mathbb{R} \right) \implies \left( F(g(x), f) = 0 \ \forall x \in \mathbb{R} \right), \tag{1}$$ but the inverse does not always hold.

$(1)$ becomes an equivalence if $g$ is a bijection, but this is rarely the case. E.g., if you substitute $x \mapsto a$ for some number $a \in \mathbb{R}$, then it is not bijective and extraneous solutions may arise. This is the reason why in practically all applications the final solution must be checked and it is not worth checking every substitution for bijectivity just so that you wouldn't need to check the final solution.


问:
Bolzano-Weierstrass for the sequence ${a_n}=(-1)^n$
I've been studying about sequences and I started to read about the Bolzano–Weierstrass theorem as stated:

"Every bounded sequence {a$_n$} has a convergent subsequence"

So I have a question, if I have the sequence: ${a_n}=(-1)^n $, that is, ${a_n}=-1, 1, -1, 1, ...$

It is clearly bounded within the interval: $[-1, 1]$

Since a subsequence is monotone. How can I find a subsequence for it if a term is greater than the previous one, and then the following is lower than the previous one?

答: A convergent subsequence of your sequence is $1,1,1,1,...$


Body:
As hinted in the comments above:
$$\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}=\\
\lim_{n\to \infty} \frac{\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}}{1}\cdot \frac{(\sqrt[3]{n^3+n^2})^2+\sqrt[3]{n^3+n}\cdot \sqrt[3]{n^3+1}+(\sqrt[3]{n^3+1})^2}{(\sqrt[3]{n^3+n^2})^2+\sqrt[3]{n^3+n}\cdot \sqrt[3]{n^3+1}+(\sqrt[3]{n^3+1})^2}=\\
\lim_{n\to \infty} \frac{(n^3+n^2)-(n^3+1)}{(\sqrt[3]{n^3+n^2})^2+\sqrt[3]{n^3+n}\cdot \sqrt[3]{n^3+1}+(\sqrt[3]{n^3+1})^2}=\\
\lim_{n\to\infty} \frac{1}{\sqrt[3]{\frac{(n^3+n^2)^2}{(n^2-1)^3}}+\sqrt[3]{\frac{(n^3+n)(n^3+1)}{(n^2-1)^3}}+\sqrt[3]{\frac{(n^3+1)^2}{(n^2-1)^3}}}=\frac1{1+1+1}=\frac13.$$


问:
Painting chess board
The task is to paint each of the $64$ squares on a chess board either blue or red.
I need to find the number of distinct ways this can be done given that any $2\times 2$ square on the board has two red and two blue squares.

I've tried solving it for a $4\times 4$ board, but I am getting no where.
Would appreciate any help

答:
For an $m\times n$ chessboard there are $2^m+2^n-2$ ways.

Case I. There are two horizontally adjacent squares of the same color: $2^m-2$ ways.

Case II. There are two vertically adjacent squares of the same color: $2^n-2$ ways.

Case III. None of the above: $2$ ways.

Hint for Case I: There are $2^m-2$ ways to color one row so that two adjacent squares have the same color. The rest of the coloring is determined from that; colors must alternate in each column. (Note, therefore, that Cases I and II do not overlap.)

Title: Painting chess board


问:
Divergent Stopping Times
I have a question regarding a sequence of stopping times in general.

Consider in the context of gambling where $X_n$ denotes your cumulative wealth after the $n^{th}$ gamble.

I recall the intuitive notion of a stopping time with respect to the stochastic process $X$ is a rule that informs you when to quit.

Now, when I think of a sequence of stopping times, it could be some betting strategy that depends on how my wealth evolves over time. For example, I could quit betting when I hit a total wealth of \$0 or \$100. But I could also have a re-enter-the-casino rule as a function of the stopping time. For example, I could set up a re-join rule where I bet on a roulette after 5 cycles once I hit \$100 (e.g. I feel more confident betting after rounds of wins) .

The notion I am a bit unfamiliar is:

$$\lim_{n\rightarrow\infty} \tau_n=\infty.$$

In my example, does this mean as I bet infinitely many times, I will never seize an opportunity to actually stop? If I go bust, this would be true, right? An example would be the gambler's ruin and her stopping time?

$\textbf{My question is why and where is this used?}$

An example I see is when defining something as an approximation in a continuous set-up. Consider the definition of a local martingale.

$X_t$ is a local martingale if there exists a sequence of stopping times $\lim\limits_{n\rightarrow\infty} \tau_n=\infty$ such that $X_{t\wedge\tau_n}$ is a martingale for all $n$.

Are there more intuitive examples where a sequence of stopping times is used to describe something in a continuous set-up to approximate a process? Is this one of the major roles of the sequence?

答:
Your last sentence is basically correct: the sequence of stopping times is often used to approximate a process.  One of the most common applications is to set $\tau_n := \inf_t \{ |X_t| \ge n \}$, i.e. $\tau_n$ is the first time the process $X$ hits $n$.

In the context of betting strategies you discussed, you could consider each $\tau_n$ to be a separate quitting rule that one could follow.  I'm not sure I understand what you mean about "re-entering the casino;" you generally only consider the process stopped at a single $\tau_n$.  In your example, if we consider $X_t$ to be the gambler's wealth at time $t$ (without stopping) and set $\tau_n := \inf_t \{ |X_t| \ge n \}$ then $X^{\tau_n}_t := X_{\tau_n \wedge t}$ is the wealth of a gambler who leaves after reaching a wealth of $n$ dollars (or stays forever if that never happens).  The condition that $\lim_{n \rightarrow \infty} \tau_n = \infty$ basically says that the sequence of stopping times should converge to never stopping, so in this example stopping once your wealth hits $n$ becomes almost the same as never stopping when $n$ is large.

Title: Divergent Stopping Times


问:
How does $|fg|$ compare to $|f||g|$ for two measurable functions $f$ and $g$ on $\mathbb{R}^d$?
Suppose I have two measurable functions $f$ and $g$ on $\mathbb{R}^d$.

I believe that we have $|fg| \leq |f||g|$. However, if $f$ and $g$ are both non-negative, then $|fg| = |f||g|$.

Is this correct ?

Thanks!

答:
$|ab|=|a||b|$ for all real numbers $a,b$

Therefore

$|ab|=|a||b|$ for all measurable functions $a,b$ on $\mathbb R^d$


问:
A category which direct limits but no general colimits
I am looking for a (at best, real life) category that has direct limits, but no general small colimits, or a category that has inverse limits, but no general small limits. Are there any interesting examples that are not too obviously made to be an example for this?

I am asking this because I wonder why many lectures include the construction of direct/inverse limits as an exercise, instead of general small (co)limits. In particular, I wonder this for topology and algebra classes.

答:
This is not really an answer, as I don’t know examples, but I think I might have a reason for why they are considering them separately.

Filtered colimits (I always get confused over directed / inverse) are particularly nice in concrete categories like $\mathsf{Set}, \mathsf{Ab}, \mathsf{Mod}_R, \mathsf{Top}$ and alike. There is an explicit formula for computing and dealing with them in $\mathsf{Set}$, which lifts to similar formulas in other concrete categories. From this formula one can deduce for example that filtered colimits commute with finite products (only for good categories!), which does not hold for arbitrary colimits! They may have even more special properties.

Long story short, often we are not interested in dealing with arbitrary shapes of colimits but only want to work with nice ones like coproducts, quotients, pushouts, gluing constructions or filtered colimits, of which we might know more than just „they are colimits“.

Part of the reason might be as well that most people don’t want to be bothered with abstract nonsense, but rather like to work with these things implicitly...


问:
How to prove a function is computable?
I'm reading Computation Complexity: A Modern Approach and one of the exercises is:

Prove that [the addition function is] computable by writing down a full description (including the states, alphabet, and transition function) of the corresponding Turing [machine].

I successfully wrote this description but I don't think it is a proof the addition function is computable.

According to this question, the definition of a computable function is:

If $f:\Sigma^* \to \Sigma^*$ is function, and $\exists$ a Turing machine which on the input $w\in\Sigma^*$ writes $f(w)$, $\forall w\in\Sigma^*$, then we call $f$ as computable function.

With the Turing machine I wrote, I can prove the machine is able to compute $f(w)$ for some inputs but how can I prove it can do the same for every inputs $w\in\Sigma^*$? In other words, how can I prove $f$ is computable?

答:
If you can prove that for any two integers $m,n$ the output of your machine is $m+n$, you are done.

You are in a situation like this one: prove a theorem. If you find a proof, obviously your proof may be wrong and someone maybe able to correct your error.

Your Turing program may be wrong, but validating it in Mathematical logic won't go through testing it with various values... but by convincing you (and the others) that your Turing program is the right one.

Additional elements

According to Church-Turing thesis, every computable function can be computed by a Turing machine and there are ways to convert a $\mu$-recursive function into a Turing machine. Based on those methods (that you have to validate!) you can convert the addition which is primitive recursive into a Turing machine.

Title: How to prove a function is computable?


答: If $A\subseteq B$ then you have the inclusion mapping $i\colon A\longrightarrow B$ which sends $a\in A$ to itself ($i(a):=a$), seen as an element of $B$. Such a function is trivially injective and then $A\leq B$.</p>

Title: cardinality problem concerning inclusion of sets

Body:
if $A\subseteq B$ then $A\le B$ , i,e if A is a subset of B then A is less than or equinumerous with B
How to do that?
IF A is a subset of B then every element in A is an element in B, so A cannot have more elements than B has right? </p>


问:
Inequality for the matrix infinity norm
Consider the matrix $\ell_{\infty} \to \ell_{\infty}$ operator norm for some
matrix $A \in \mathbb{R}^{m \times n}$, given by
$$
\| A \|_{\infty} := \sup_{x: \| x \|_{\infty} = 1} \| A x \|_{\infty} :=
\max_{j \in [m]} \| A_{i, :} \|_1.
$$

Question: Prove (or disprove via counterexample) the following inequality:
  $$
\left\| V \begin{bmatrix} I_{k_1} & 0 \\ 0 & -I_{k_2} \end{bmatrix}V^\top \right\|_{\infty} \leq C \| V V^\top \|_{\infty},
$$
  for $V$ satisfying $V^\top V = I$ and $V \in \mathbb{R}^{n \times k}$, with
  $k_1 + k_2 = k$.

This is not a homework problem, and I have been unable to come up with a counterexample (for randomly generated $V$, I find $C < 2$). I'm looking for
some $C$ which is ideally in the range $o(\sqrt{k})$.

答:
Let $V:=[v_1,\ldots,v_k]$ written as column vectors. Then $V^\top V=I$ means that $v_i$ are orthonormal.

Now, for any vector $x$,\begin{align}
\|VJV^\top x\|_\infty&\le\|VJV^\top x\|_2\\
&=\|(v_1\cdot x)v_1+\cdots-(v_k\cdot x)v_k\|_2\\
&=\sqrt{\sum_{i=1}^k(v_i\cdot x)^2}\\
&=\|(v_1\cdot x)v_1+\cdots+(v_k\cdot x)v_k\|_2\\
&=\|VV^\top x\|_2\\
&\le\sqrt{n}\,\|VV^\top x\|_\infty\\
&\le\sqrt{n}\,\|VV^\top\|_\infty\|x\|_\infty
\end{align}

Hence $\|VJV^\top\|_\infty\le\sqrt{n}\,\|VV^\top\|_\infty$.

That $\sqrt{n}$ cannot be improved upon much can be seen from this example:

$V:=[v_1,v_2]$, $v_1=(a,b,\ldots,b)$, $v_2=(-a,b,\ldots,b)$. The condition $V^\top V=I$, i.e., $v_1,v_2$ are orthonormal, then forces $a=\frac{1}{\sqrt{2}}$, $b=\frac{1}{\sqrt{2(n-1)}}$.

With $J=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $$\|VV^\top\|_\infty=\left\|\begin{pmatrix}2a^2&0&\cdots&0\\0&2b^2&\cdots&2b^2\\\vdots\end{pmatrix}\right\|_\infty=2\max(a^2,(n-1)b^2)=1$$
$$\|VJV^\top\|_\infty=\left\|\begin{pmatrix}0&2ab&\cdots&2ab\\2ab&0&\cdots&0\\\vdots\end{pmatrix}\right\|_\infty=2(n-1)ab=\sqrt{n-1}$$

Title: Inequality for the matrix infinity norm


答:
Note that for a sub-sequence to be valid, its indexes must be increasing - not the values. So, the sequence of indexes $1,3,5$ is valid, but $1,1,3$ or $1,3,2$ aren't.

So, a converging sub-sequence to your original one will be $$1,1,1,1,\ldots \to 1$$
or
$$-1,-1,-1,-1, \ldots \to -1$$

Title: Bolzano-Weierstrass for the sequence ${a_n}=(-1)^n$

Body:
I've been studying about sequences and I started to read about the Bolzano–Weierstrass theorem as stated:

"Every bounded sequence {a$_n$} has a convergent subsequence"

So I have a question, if I have the sequence: ${a_n}=(-1)^n $, that is, ${a_n}=-1, 1, -1, 1, ...$

It is clearly bounded within the interval: $[-1, 1]$

Since a subsequence is monotone. How can I find a subsequence for it if a term is greater than the previous one, and then the following is lower than the previous one?


问:
Is there a nice proof that $[A,B][A,[C,B]]$ is traceless for real symmetric matrices?
I found a nice identity involving the trace of some commutators of real symmetric matrices:
$$
\operatorname*{Tr}\, [A,B][A,[C,B]] = 0.
$$
Here $A$,$B$, and $C$ are real symmetric $n\times n$ matrices, $[A,B]=AB-BA$, and $\operatorname*{Tr}$ is the trace.  I am simply asking for a nice proof of this identity, which hopefully generalizes to prove
$$
\operatorname*{Tr}\, [A,[A,B]][A,[A,[C,B]]] = 0.
$$
So far I only have a messy proof of the first identity, and an empirical verification of the second (for some random matrices).

A summary of the messy proof

The proof I found is just to expand the commutators out, use the cyclic property of the trace, and rearrange the terms to obtain the identity
$$
\operatorname*{Tr}\, [A,B][A,[C,B]]
 = 2\operatorname*{Tr}\,ABA [C,B] + \operatorname*{Tr} A^2 [B^2,C].
$$
Then I use the fact that the matrices $A$,$B$, and $C$ are symmetric to see that $ABA$ and $A^2$ are symmetric while $[C,B]$ and $[B^2,C]$ are antisymmetric. Then we use the fact that the trace of the product of a symmetric matrix with an antisymmetric matrix is zero.

答:
Since $[A,B]^T=[B^T,A^T]$, the commutator of two symmetric or antisymmetric matrices is antisymmetric, and the commutator of a symmetric and an antisymmetric matrix is symmetric.

As a result, if $A,B,C$ are symmetric, then $[A,B]$ is antisymmetric, $[A,[C,B]]$ is symmetric, and the third one $[A,[A,[C,B]]]$ is again antisymmetric. Thus, you can use the fact the trace of a product of symmetric and antisymmetric matrix is zero without having to expand commutators and use the trace's cyclic property.


答:
Consider the category with two objects and only identity arrows. Or more generally, any poset which has least upper bounds for all chains, but not arbitrary joins (like the disjoint union of two copies of $\mathbb{R}\cup\{\infty\}$).

If you insist that these are not "real life" categories, you might be more satisfied with the example of the category of fields, which has directed colimits but does not have coproducts or an initial object.

Title: A category which direct limits but no general colimits

Body:
I am looking for a (at best, real life) category that has direct limits, but no general small colimits, or a category that has inverse limits, but no general small limits. Are there any interesting examples that are not too obviously made to be an example for this?

I am asking this because I wonder why many lectures include the construction of direct/inverse limits as an exercise, instead of general small (co)limits. In particular, I wonder this for topology and algebra classes.


答:
I'll just give some hints that will allow you to easily deduce the formula that user14111 gave in a comment. Call any pair of adjacent squares a domino, which can be horizontal or vertical. Call two dominos neighbours if their union is a $2\times2$ square. Call a domino monochromatic for a colouring if its squares have the same colour.

If a colouring has some monochromatic domino, then so is any neighbour of it (and it has the opposite colour).

If there is any monochromatic hoizontal domino, then there is one in each row (in the same pair of columns)

If there is any monochromatic vertical domino, then there is one in each column (in the same pair of rows).

Both these conditions cannot be met simultaneously.

Therefore it suffices to count the following cases, and add up the results:

There is at least one monochromatic horizontal domino

There is at least one monochromatic vertical domino

There are no monochromatic dominoes.

A solution for case 1. is completely determined by the colouring of its first row, a solution for case 2. is completely determined by the colouring of its first column, and a solution for case 3. is completely determined by the colouring of its top-left corner square.

Title: Painting chess board

Body:
The task is to paint each of the $64$ squares on a chess board either blue or red.
I need to find the number of distinct ways this can be done given that any $2\times 2$ square on the board has two red and two blue squares.

I've tried solving it for a $4\times 4$ board, but I am getting no where.
Would appreciate any help


问:
Showing that $f(x, y)=xy$ is continuous on $S^1$
This is question from studying general topology.

Let $f : S^1 \times S^1 \rightarrow S^1$ defined by $f(x, y)=xy$

How can I show that $f$ is continuous?

My attempt : Let $U : $ {$e^{i\theta}$ | $\alpha < \theta < \beta $ } : open set in $S^1$.

Then, $f^{-1}(U)$={$(e^{i \theta _1}, e^{i \theta _2})$ | $\alpha + 2n\pi < \theta_1 + \theta_2 < \beta + 2n\pi$} , which is homeomorphic to $V$ = {$(x, y)$ | $\alpha + 2n\pi < x+y < \beta + 2n\pi$} .

Since $V$ is open, $f$ is continuous.

答: Let $\iota\colon S^1 \to \Bbb R$ be the inclusion. By definition of subspace topology, $\iota$ is continuous. So $\iota\times \iota\colon S^1\times S^1 \to \Bbb R \times \Bbb R = \Bbb R^2$ is also continuous. Define $\widetilde{f}\colon \Bbb R^2 \to \Bbb R$ by $\widetilde{f}(x,y) = xy$. This $\widetilde{f}$ is obviously continuous, as it is a polynomial. Then $$f = \widetilde{f}\circ (\iota\times \iota)$$is the composition of continuous maps, hence continuous.


问:
Relationship between multiplicity and order
Here we have

Riemann surface $X$, connected but not necessarily compact,

meromorphic, possibly constant but with no removable singularities, $f: X \to \mathbb C$, i.e. holomorphic $f: X \setminus \text{poles}(f)\to \mathbb C$, corresponding to

holomorphic $F: X \to \mathbb C_{\infty}$, where $F(x):=f(x), x \in \text{Domain}(f)$, $F^{-1}(\infty):=\text{poles}(f)$ and where $F$ is not identically $\infty$

From Rick Miranda - Algebraic Curves and Riemann surfaces

I believe I understand (though not so deeply) the statement of Lemma II.4.7, but I'm really confused about the proof.

Question: How does Lemma II.4.4 give us $mult_p F = ord_p(f-f(p))$, when $p$ is not a pole of $f$ (i.e. $f$ is holomorphic at $p$, since I guess $f$ has no removable singularities)? In particular, I want to know the relationship of the 'h' s.t. $mult_pF = 1 + ord_{z_o}(\frac{dh}{dz})$ and the 'f' that $F$ corresponds to

答:
If $g$ is a holomorphic function on an open subset of $\mathbb{C}$ vanishing at $z_0$, then the order of $z_0$ as a zero of $g$ is the lowest integer $m$ such that $g(z) = (z - z_0)^m \times g_0(z)$ with $g_0$ is a holomorphic function with $g_0(z_0) \neq 0$. Differentiating the previous equality shows that if $z_0$ is a zero of $g$
\begin{align}
\mathrm{ord}_{z_0}(g) = 1 + \mathrm{ord}_{z_0}(g')
\end{align}

Note that if $C$ is a constant and if $f$ is a holomorphic function, then $(f+c)' =f' $.

Now, if $h$ is holomorphic near $z_0$, then $h-h(z_0)$ is holomorphic, vanishes at $z_0$, and
\begin{align}
\mathrm{ord}_{z_0}(h-h(z_0)) &= 1 + \mathrm{ord}_{z_0}\left((h-h(z_0))'\right) \\
&= 1 + \mathrm{ord}_{z_0} (h')
\end{align}

Now suppose $f : X \to \mathbb{C}$ is meromorphic on the Riemann surface $X$. Let $p\in X$ be a point that is not a pole of $f$ and $\phi : U \subset X \to \phi(U)\subset \mathbb{C}$ is an holomorphic chart. Then $h = f\circ \phi^{-1} : \phi(U) \to \mathbb{C}$ is an holomorphic map on an open subset of $\mathbb{C}$, and if $z_0=\phi(p)$,
\begin{align}
\mathrm{mult}_pF &= \mathrm{ord}_p(f-f(p))\\
 &= \mathrm{ord}_{z_0} (h-h(z_0)) & \text{because } \phi \text{ is invertible} \\
&= 1 + \mathrm{ord}_{z_0}\left((h-h(z_0))'\right) \\
&= 1 + \mathrm{ord}_{z_0}(h') \\
&= 1 + \mathrm{ord}_{p}(f')
\end{align}
because $h' = (f' \circ\phi^{-1})\times (\phi^{-1})'$, and as $\phi$ is a diffeomorphism, $h'$ has the same order of annihilation at $z_0$ than $f'$ at $p$.


问:
Showing that the split-step method for PDEs can be written in terms of Fourier transforms
I am trying to learn about the split-step Fourier method for solving partial differential equations, in terms of the Schrodinger equation:

$$
\frac{\partial\Psi(x,t)}{\partial t} = -\frac{i}{\hbar}\hat{H}\;\Psi(x,t) \tag{1}
$$
where
$$
\hat{H} = \hat{T} + \hat{V} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x). \tag{2}
$$

The solution of (1) for advancing the wavefunction by a time step $\Delta t$ should be given by
$$
\begin{align}
\Psi(x,t+\Delta t) &= \textrm{exp}\Big(-\frac{i}{\hbar}\hat{H}\;\Delta t\Big)\Psi(x,t)\\
&\approx \textrm{exp}\Big(-\frac{i}{2\hbar}\hat{V}\;\Delta t\Big)\;\textrm{exp}\Big(-\frac{i}{\hbar}\hat{T}\;\Delta t\Big)\;\textrm{exp}\Big(-\frac{i}{2\hbar}\hat{V}\;\Delta t\Big)\;\Psi(x,t)\\
&=\textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\;\Delta t\Big)\;\textrm{exp}\Big(\frac{i\hbar}{2m}\frac{\partial^2}{\partial x^2}\;\Delta t\Big)\;\textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\;\Delta t\Big)\;\Psi(x,t)\tag{3}
\end{align}
$$
where I have used the symmetrised split-step approximation.

Now from reading, it seems like Eq. (3) can be written in terms of Fourier transforms:
$$
\Psi(x,t+\Delta t) = \textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\Delta t\Big)\;\mathcal{F}^{-1}\Bigg\{\textrm{exp}\Big(-\frac{i\hbar k^2}{2m}\Delta t\Big)\;\mathcal{F}\Big\{\textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\Delta t\Big)\;\Psi(x,t)\Big\}\Bigg\} \tag{4}
$$
but how can I show this explicitly?

Note, I am aware that a second derivative can be taken spectrally like so
$$
\mathcal{F}\Bigg\{\frac{\textrm{d}^2y}{\textrm{d}x^2}\Bigg\} = -k^2 \mathcal{F}\big\{y\big\},\tag{5}
$$
but I am not sure how to handle the second derivative inside the exponential in Eq. (3). It seems I will need to evaluate something of the form
$$
\mathcal{F}^{-1}\Bigg\{ \mathcal{F}\Big\{ \textrm{exp}\bigg(\frac{\partial^2}{\partial x^2}\bigg) \cdot A \Big\} \Bigg\} = \mathcal{F}^{-1}\Bigg\{  \textrm{exp}\big(-k^2\big) \cdot\mathcal{F}\big\{ A \big\} \Bigg\}
$$

Thank you.

答:
Because you want to use the Strang splitting method, you first need to solve the potential term for $\tau \in [t, t + \Delta t/2]$ with timestep $\Delta t/2$ and initial condition $\psi_{0}$

$$\dot{\psi} = -\frac{i}{{\hbar}} V \psi \implies \hat{\psi}^{*} = \mathcal{F}\left\{\exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}\right\} \tag{1}$$

where the hat denotes the Fourier transform. Note that at this iterate we could have left the solution as

$$\psi = \exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}$$

However, because at the next iterate (solving the diffusion part) you want to solve the problem using the FT, and the solution to the potential part acts as an initial condition for this next iterate, the initial condition needs to be FTed as well.

You then solve the diffusion problem for $\tau \in [t, t + \Delta t]$ with initial condition $\hat{\psi}^{*}$ and timestep $\Delta t$

\begin{align}
\dot{\psi} = \frac{i \hbar}{2m} \psi_{xx} \implies \hat{\dot{\psi}} &= -\frac{i \hbar k^2}{2m} \hat{\psi} \\
\implies \hat{\psi} &= \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right)\hat{\psi}^{*} \\
&= \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right) \mathcal{F}\left\{\exp\left( -\frac{i}{{\hbar}} V \Delta t\right) \psi_{0}\right\} \\
\implies \psi^{**} &= \mathcal{F}^{-1}\left\{ \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right) \mathcal{F}\left\{\exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}\right\}\right\} \tag{2} \end{align}

where we have inverted the FT now so that the solution in is physical space because the next iteration is the last in the procedure. Finally, you want to solve the potential term for $\tau \in [t + \Delta t/2, t + \Delta t]$ with timestep $\Delta t/2$ and initial condition $\psi^{**}$

\begin{align}
\dot{\psi} = -\frac{i}{{\hbar}} V \psi \implies \psi &= \exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi^{**} \\
&= \exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \mathcal{F}^{-1}\left\{ \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right) \mathcal{F}\left\{\exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}\right\}\right\}
\end{align}

which is the result you were after.

If you are looking for more information, I have also written a solution that uses splitting methods to solve the KdV problem here.


问:
Sum of kth roots ($\sum\sqrt[k]{m}$)
I'm trying to find an asymptotic to $$S(n) = \sum_{k=1}^n\sqrt[k]{m}$$ From computational tests, it seems to grow nearly as slowly as $n$. However even $$\sum_{k=1}^\infty\sqrt[k]{m}-1$$ diverges (for $m\neq1$) by the comparison test.

I'm thinking it might be something like $n\log{\log n}$, but I don't know how to show it.

Update: So it turns out to be closer to $n\sqrt[n]{m}$. Does anybody know if there is a nice formula?

Title: Sum of kth roots ($\sum\sqrt[k]{m}$)

Body:
You have :
$$ M^{1/k} = 1 + \frac{1}{k} \ln M + \frac{1}{2} \ln^2 M \frac{1}{k^2} + O(1/k^3)$$
Since
$$\sum_{k=1}^n \frac{1}{k} = \ln n + \gamma + \frac{1}{2n} + O(1/n^2)$$
$$\sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6} - \frac{1}{n} + O(1/n^2)$$
we deduce :
$$\sum_{k=1}^n M^{1/k} = n + \ln M. \ln n + C + (\ln M + \ln^2 M) \frac{1}{2n} + O(1/n^2),$$
for some constant $C$.


问:
How do I verify if $2^{35} \equiv 1 \pmod{71}$ is true or not?
I need to know if $2^{35} \equiv 1\pmod{71}$ is true. I tried using Euler and Fermat little theorem and I got stuck. There is probably something trivial I'm not seeing so I appreciate any help, thanks.

Body:
$$2^{35} = 2^{10} \times 2^{10} \times 2^{10} \times 2^5 \\
= 1024 \times 1024 \times 1024 \times 32 \\
\equiv 30 \times 30 \times 30 \times 32 \equiv 1 \mod 71.$$


答:
According to Euler's criterion, $2^{35}\equiv\left(\dfrac2{71}\right)\bmod71$.

Furthermore, $\left(\dfrac2{71}\right)=1$, because $71\equiv-1\bmod8$.

Title: How do I verify if $2^{35} \equiv 1 \pmod{71}$ is true or not?

Body: I need to know if $2^{35} \equiv 1\pmod{71}$ is true. I tried using Euler and Fermat little theorem and I got stuck. There is probably something trivial I'm not seeing so I appreciate any help, thanks.


问:
The function $\phi\colon t \mapsto \int_{0}^{1}f(x,t)\, dx$ is Borel
Suppose $f\colon[0,1]\times [0,1] \rightarrow \mathbb{R} $. Functions $x \mapsto f(x,t)$ are integrable, functions $t \mapsto f(x,t)$ are continuous. I need to prove that the following function is Borel: $$\phi\colon t \mapsto \int_{0}^{1}f(x,t)\,dx.$$

I tried to use monotone class theorem for $E=\{B \in \mathcal{B}(\mathbb{R})\mid \phi^{-1}(B) \in \mathcal{B}([0,1]) \} \subset \mathcal{B}(\mathbb{R})$. I need to show that $E=\mathcal{B}(\mathbb{R})$. The only thing I managed to prove is that $E$ is an algebra. So, the idea seems wrong.

Please give me any ideas how to solve the problem. Thank you for any hints!

答:
The condition is so loose that it had me confused in many ways( it also made me question my understanding of measurability )

Your desired measurability ( of integrals) is already covered in (and also plays a part of ) the demonstration of Fubini's theorem.

Moreover, as you know, the continuity property is still too much for only proving measurability.

If you want to reproduce that theorem, you have realized the right way to it, that is, to use the monotone class theorem( but in a more general setting). Let

$$S:=\{ A \in \mathcal{B}([0,1])\otimes \mathcal{B}([0,1]) | f:= \mathbb{1}_{A} \text{ sasisties the respective }\phi \text{ is measurable}\}$$
then, this family $S$ is a Dynkin system, containing all product $1_{B\times C}$(where $B,C \in \mathcal{B}([0,1])$).

From then you can prove your desired results with integrable positive functions.

Etc.

Discussion

given your constraint, if we suppose further that $f$ is bounded (1), you can even prove the continuity of $\phi$ by using dominated convergence theorem.

The boundedness (1) perhaps can be relaxed, but more elaborate care should be put in.


问:
Dual space of a finite dimensional Hilbert space
I have a statement given that the normed vector space $(L(\mathcal H), ||.||_{\text{1}})$ is the dual space of $(L(\mathcal H), ||.||_{\text{op}})$. Here $L(\mathcal H) := L(\mathcal H, \mathcal H)$ denotes the vector space of linear operators from a vector space $\mathcal H$ to a vector space $\mathcal H$. The $||.||_1$ norm describes the trace norm and $||.||_{\text{op}}$ denotes the operator/spectral norm.

I'm pretty new to the topic and don't know how to show that this statement holds.

答:
I will denote $\mathcal B :=  (L(\mathcal H),\|\cdot\|_{\mathrm{op}})$, $\mathcal B_1 := (L(\mathcal H),\|\cdot\|_1)$, $\mathcal B_2 := (L(\mathcal H),\|\cdot\|_2)$.

First of all, note that every operator $A$ with finite trace norm leads to a corresponding element of $\mathcal B^*$, since we can define $\phi_A(B) = \langle A,B \rangle$ (and make use of the inequality $|\langle A,B\rangle| \leq \|A\|_1 \cdot \|B\|_{\mathrm{op}}$). It remains to be see that for every $\phi \in \mathcal B^*$, there is an $A \in \mathcal B_1$ with $\phi = \phi_A$.

First, we note that restricting $\phi$ (with $\|\phi\|_{\mathrm{op}}$ finite) to the elements of $\mathcal B$ that are in $\mathcal B_2$ gives us a bounded linear map over $\mathcal B_2$ since
$$
|\phi(B)| \leq \|\phi\|_{\mathrm{op}} \|B\|_{\mathrm{op}} \leq \|\phi\|_{\mathrm{op}}\|B\|_2.
$$
Because $\mathcal B_2$ is a Hilbert space, there necessarily exists a unique $A \in \mathcal B_2$ for which $\phi = \phi_A$, but it remains to be seen that $A \in \mathcal B_1$. One argument that this holds is as follows. Let $A = U|A|$ be a polar decomposition.
For every projection operator $P$ of finite rank, we have $\|P\|_{\operatorname{op}} = 1$ and hence
$$
\|\phi_A\|_{\mathrm{op}} \geq |\phi_A(PU^*)| = \langle A,PU^*\rangle =
\operatorname{tr}(P|A|).
$$
It follows that we must have $\|\phi_A\|_{\mathrm{op}} \geq \operatorname{tr}(|A|) = \|A\|_1$, which is what we wanted.


问:
Visual proof for $[\triangle ABC] = rs$?
I'm teaching a high school geometry class and we just got to the part where we proved that the area of a triangle $[\triangle ABC]$ is equal to the inradius $r$ times the semiperimeter $s$.

A student pointed out to me that it's really the same thing as $[\triangle ABC] = \frac{1}{2}rp$ where $p$ is the perimeter. But this representation looks like the regular formula $\frac{1}{2}bh$.

So I was wondering if there's a way to take the original $\triangle ABC$ and construct a new triangle that will have a base of $p$ and a height of $r$ and show that the original and newly constructed triangle have the same area?

答:
It is possible. In fact, in $\Delta ABC$, extending $BC$ to $A'$ and then to $B'$ such that
$$ CA'=AC, A'B'=AB $$
one has
\begin{eqnarray}
S_{\Delta ABC}&=&S_{\Delta OBC}+S_{\Delta OAC}+S_{\Delta OBC}=\frac12ra+\frac12rb+\frac12rc\\
&=&S_{\Delta OBC}+S_{\Delta OCA'}+S_{\Delta OA'B'}=S_{\Delta OBB'}.
\end{eqnarray}


问:
What is the meaning of the third derivative of a function at a point
(Originally asked on MO by AJAY.)

What is the geometric, physical, or other meaning of the third derivative of a function at a point?

If you have interesting things to say about the meaning of the first and second derivatives, please do so.

答:
I've found the time, so I deleted one of my original comments in the OP and decided to expand it into a full answer.

One way to geometrically interpret the third derivative is in the notion of the osculating parabola. In much the same way that the first derivative enters into the defining equation for the tangent line (the line that best approximates your curve in the vicinity of a given point), and that the second derivative is involved in the expression for the osculating circle (the circle that best approximates your curve in the vicinity of a given point), the third derivative is required for expressing the osculating parabola, which is the parabola that best approximates... oh, you catch on quick. ;)

More specifically, if you remember the fact that four points uniquely determine a parabola, you can think of the osculating parabola as the limiting case of the parabola through four neighboring points of a given curve when those four points coalesce, or come together. The so-called aberrancy (a translation of the French "déviation") is the tangent of the angle the axis of the osculating parabola makes with the normal line, and is given by the formula

$$\tan\,\delta=\frac1{3\varrho}\frac{\mathrm d\varrho}{\mathrm d\phi}=\frac{\mathrm d y}{\mathrm d x}-\frac{1+\left(\frac{\mathrm d y}{\mathrm d x}\right)^2}{3\left(\frac{\mathrm d^2 y}{\mathrm d x^2}\right)^2}\frac{\mathrm d^3 y}{\mathrm d x^3}$$

where $\varrho$ is the radius of curvature and $\phi$ is the tangential angle.

From these considerations, one could derive an expression for the osculating parabola: given a curve represented parametrically as $(f(t)\quad g(t))^T$, the parametric equations for the osculating parabola of the curve at $t=t_0$ are

$$\begin{pmatrix}f(t_0)\\g(t_0)\end{pmatrix}+\frac{\varrho\;\cos^4\delta}{2}\begin{pmatrix}\cos\,\phi&-\sin\,\phi\\\sin\,\phi&\cos\,\phi\end{pmatrix}\cdot\begin{pmatrix}(u^2-2)\tan\,\delta-2u-\tan^3\,\delta\\(u+\tan\,\delta)^2\end{pmatrix}$$

Here for instance is the cardioid $(2\cos\,t+\cos\,2t\quad 2\sin\,t+\sin\,2t)^T$ and its osculating parabola at $t=2\pi/3$:

and an animation of the various osculating parabolas for the curve $(3\cos\,t-2\cos\,3t\quad 3\sin\,t-2\sin\,3t)^T$:

Further, one could also give a geometric interpretation for the fourth derivative; what one now considers is the osculating conic (the limiting conic through five neighboring points of a curve when those five points coalesce), and one could classify points of a plane curve as elliptic, parabolic or hyperbolic depending on the nature of the osculating conic. In this respect, the discriminant of the osculating conic depends on the first four derivatives.

A lot more information is in these two articles by Steven Schot (who also wrote a nice article on the "jerk"), and the references therein.


答:
After Ralph's answer I'd like to detail this a bit.  [update](upps, after I've posted this it seems user10670 was 30 sec's ahead)

After some standard-manipulation with the Carleman-matrices for $$ m \to  \sqrt[x]{1+m}-1 =\exp( \log(1+m) / x)-1 $$

I come to the following asymptotic.

We need also the expression for the sum of the first consecutive reciprocals
$$ s_p(a,n) = \sum_{k=a}^n \frac 1{k^p} $$
then we get
$$S_m(n) = \sum_{k=0}^\infty {\log(m)^k s_k(2,n) \over k!}  $$
and clearly this contains a $\zeta(1)-1$-expression in $s_1(2,n)$ if n is assumed to go to infinity.

unchecked remark: To get this possibly converging I think you should not only reduce by 1 but also by the next term, so  something like $$S_m(n) = \sum_{k=1}^n \sqrt[k]{m}-1+m/k $$ to eliminate the $\zeta(1)$-expression

Title: Sum of kth roots ($\sum\sqrt[k]{m}$)

Body:
I'm trying to find an asymptotic to $$S(n) = \sum_{k=1}^n\sqrt[k]{m}$$ From computational tests, it seems to grow nearly as slowly as $n$. However even $$\sum_{k=1}^\infty\sqrt[k]{m}-1$$ diverges (for $m\neq1$) by the comparison test.

I'm thinking it might be something like $n\log{\log n}$, but I don't know how to show it.

Update: So it turns out to be closer to $n\sqrt[n]{m}$. Does anybody know if there is a nice formula?


答:
Since $71\equiv3$ mod $4$, $k$ is a quadratic residue if and only if $71-k$ is a nonresidue. In particular, $70=71-1$ and $35=71-36$ are nonresidues. But since $70=2\cdot35$, we can conclude that $2$ is a quadratic residue, i.e., $2\equiv a^2$ mod $71$ for some $a$, in which case $2^{35}\equiv a^{70}\equiv1$ mod $71$ by Fermat's little theorem.

Remark: This approach shows that $2$ is the square of something without explicitly finding what it's the square of. In fact, as Keith Bachman points out in comments, $2\equiv12^2$ mod $71$.


问:
The group of symmetries of the $d$ dimensional regular simplex is isomorphic to $S_{d+1}$

The group of symmetries of the $d$ dimensional regular simplex is
isomorphic to $S_{d+1}$

($S_n$ is the symmetric group of $n$ elements = the group of permutations)

I was able to prove the above proposition for $d=3$ the following way:

Let $G$ be the group of the symmetries of the $3$-dimensional regular simplex (regular tetrahedron).

There is an obvious homomorphism $\varphi \colon G \to S_4$, sending a symmetry to the corresponding permutation of the vertices.

If a symmetry fixes all the vertices, it must be the identity symmetry, so $\ker \varphi = \{ \text{id} \} \Rightarrow \varphi $ is injective.

Every transposition of neighbours is in the image $\text{Im} \varphi$.

Since $\varphi$ is a group homomorphism $\text{Im}\varphi$ is a subgroup of $S_4$.

$S_4$ is generated by these transpositions $\Rightarrow \text{Im}\varphi = S_4 \Rightarrow \varphi$ is surjective.

So $\varphi$ is a bijective homomorphism $\Rightarrow \varphi$ is an isomorphism $ \Rightarrow G \cong S_4 $

But I don't know how to generalize this proof for dimension $d$. My main problem is that I think that not every transposition lies in Im$\varphi$. Am I right? If I'm wrong about it, can someone show me that all the transpositions are in the image of the homomorphism?
If I'm right about this, can someone help me to prove the theorem in dimension $d$?

Thanks in advance!

答:
Every transposition is still in the image of $\varphi$ in the general case. If you position the simplex in $\mathbb{R}^d$ and fix two points $P_i,P_j$ out of vertices $P_1,\ldots,P_{n+1}$ of the simplex, then the hyperplane determined by all $P_k$ for $k\neq i,j$ along with the midpoint of  $\overline{P_iP_j}$ is of dimension $d-1$, and the reflection across this hyperplane exchanges $P_i$ and $P_j$ while fixing all other vertices (and preserving the shape of the simplex).

Everything else in your post continues to hold in the $d$-dimensional case, so the proof generalizes.


答: Consider any nontrivial group as a 1-object category.  Then it has all filtered (co)limits (exercise: if all the morphisms in a filtered diagram are isomorphisms, then any object in the diagram is a (co)limit by taking an appropriate composition of the isomorphisms and their inverses).  However, it does not have a (co)equalizer of any two distinct morphisms, or a (co)product of any number of copies of the unique object besides 1.</p>


问:
Each alphabet of KANGAROO is replaced with number by $2$ people; which alphabet is replaced with the same number?
In the word KANGAROO Bill and Bob replace the letters by digits, so that the resulting numbers are multiples of $11$. They each replace different letters by different digits and the same letters by the same digits ($K \not = 0$). Bill obtains the largest possible such number and Bob the smallest.
In both cases one of the letters is replaced by the same digit. Which digit is this?

Please give me a hint on the answer or ways to approach this question.

答:
To finish off what Tarlo_x has started:

The maximum is 98,758,066 and
the minimum is 10,250,933

Both, Bill and Bob, assign the 5 to G.


问:
Calculate the area of the part of the cone $x^2+y^2=z^2$ with $z \geq 0$ that is inside of the sphere $x^2+y^2+z^2=2Rz$ where $R>0$

Calculate the area of the part of the cone $x^2+y^2=z^2$ with $z \geq 0$ that is inside of the sphere $x^2+y^2+z^2=2Rz$ where $R>0$

Attempt
Notice that we should apply the formula
$$\int_{S}f \dot dS=\int_{D} ||T_{u} \times T_{v}|| dudv$$ where $T_{u}$ and $T_{v}$ are the tangent vectors to our surface when $S$ is a parametric surface.\
We need find where is the intersection of the cone with the sphere, to can get superior límit and integrate the area of the cone from $0$ to the intersection point namely $P$.\
$$x^2+y^2+z^2=2Rz$$ completing the square
$$x^2+y^2+(z-R)^2=R^2$$
This meaning is our sphere have center in $(0,0,R)$ with radius $R$.\
Now using the fact that $x^2+y^2=z^2$ and substitute it in $x^2+y^2+(z-R)^2=R^2$
$$z^2+(z-R)^2=R^2$$
$$(z^2-R^2)+(z-R)^2=0$$
$$(z-R)[(z+R)+(z-R)]=0$$
from here the intersection are
$z=R$ and $z=0$ but since $z \geq 0$  and
get that the intersection are in the plane $z=R$ that is $p=(0,0,R)$, and $Q=(0,0,0)$

Now parametrizing the surface $S$ given by the intersection we get
$$x=rcos \theta$$
$$y=rsin \theta$$
$$z=r$$
where $r\in[0,R]$ and $\theta \in [0, 2\pi]$ and calculatin the tangent vectors
$$T_{r}=(cos\theta)i+(\sin \theta)j+k$$
$$T_{\theta}=(-r \sin \theta )i+ (r \cos \theta )k+0k$$
$$T_{r}\times T_{\theta}=(-r \cos \theta)i+(-r\sin \theta )k+(r)k$$
$$||T_{r}\times T_{\theta}||=\sqrt{2}|r|$$
finally getting  the integral given by the formula we get
$$ \int_{0}^{2\pi} \int_{0}^{R}\sqrt{2}r dr d\theta=\sqrt{2}\pi R^2$$
Is my answer right or i make a mistake in the problem interpretation.

答:
Your answer is correct but your intersection points are not, due to a typo you made I think.

$z^2+(z-R)^2=R^2 \implies (z^2-R^2) + (z-R)^2 = R^2$

$(z^2-R^2) + (z-R)^2 = R^2 \implies (z-R)(z + R + z - R) = 0$

$2z(z-R) = 0 \,$ gives you intersection points $z = 0, R$.

This makes sense as the cone is an inverted cone with vertex at $(0, 0, 0)$ and the sphere is of radius $R$ with center at $(0, 0, R)$ and so it touches the point $(0, 0, 0)$. It never goes to $z = - R$.

The cone is inside the sphere starting at $z = 0$ and comes out of the sphere at $z = R$ which is exactly at the hemisphere.

So, $0 \leq z \leq R$. Since you discarded $z = - R$ and took the integral from $z = 0$, everything eventually worked out.


问:
Prove that $f(x, y)=xy$ is continuous on $S^1$ without moving to $\mathbb R$
I provided an answer to Showing that $f(x, y)=xy$ is continuous on $S^1$.

However, it was through a cumbersome proof moving back and forth from $S^1$ to $\mathbb R$.

Is there a way to to provide a proof that would avoid that move to $\mathbb R$?

I.e. using the manifold structure of $S^1$?

答:
The multiplication map $\mu : \mathbb C \times \mathbb C \to \mathbb C, \mu(z,w) = z \cdot w$, is obviously continuous because
$$\lvert z \cdot w - z_0 \cdot w_0 \rvert = \lvert z \cdot w -  z \cdot w_0 + z \cdot w_0  -z_0 \cdot w_0 \lvert  \le \lvert z \rvert \cdot \lvert w -  w_0 \rvert + \lvert z -  z_0 \rvert \cdot \lvert w_0 \rvert \\ \le (\lvert z_0 \rvert + \lvert z -  z_0 \rvert) \cdot \lvert w -  w_0 \rvert + \lvert z -  z_0 \rvert \cdot \lvert w_0 \rvert \\ = \lvert z_0 \rvert \cdot \lvert w -  w_0 \rvert + \lvert z -  z_0 \rvert \cdot \lvert w_0 \rvert + \lvert z -  z_0 \rvert \cdot \lvert w -  w_0 \rvert .$$

Your map $f$ is the restriction of $\phi$.


问:
How to calculate the Mahalanobis distance
I am really stuck on calculating the Mahalanobis distance. I have two vectors, and I want to find the Mahalanobis distance between them. Wikipedia gives me the formula of
$$
    d\left(\vec{x}, \vec{y}\right) = \sqrt{\left(\vec{x}-\vec{y}\right)^\top S^{-1} \left(\vec{x}-\vec{y}\right) }
$$

Suppose my $\vec{y}$ is $(1,9,10)$ and my $\vec{x}$ is $(17, 8, 26)$ (These are just random), well $\vec{x}-\vec{y}$ is really easy to calculate (it would be $(16, -1, 16)$), but how in the world do I calculate $S^{-1}$. I understand that it is a covariance matrix, but that is about it. I see an "X" everywhere, but that only seems like one vector to me. What do I do with the Y vector? Can someone briefly walk me through this one example? Or any other example involving two vectors.

答:
Maybe a simple geometric interpretation will help:  Suppose you have a large set of vectors, in your case sets of 'triplets'. Each point can be represented in a 3 dimensional space, and the distance between them is the Euclidean distance.

But suppose when you look at your cloud of 3d points, you see that a two dimensional plane describes the cloud pretty well. So project all your points perpendicularly onto this 2d plane, and now look at the 'distances' between them.

Or maybe a single vector can be defined from the origin through 3d space. Similaly, project ALL your points onto this 1d vector, and look at their 'distances'.

These 1d and 2d 'distances' are called Mahalanobis distances in the reduced 1d and 2d spaces.

I like mental 'pictures' to explain math... hope that helps.


答:
In case you're also interested in a rectangle, I made a diagram.

Rotate one in each pair of the congruent right triangles to obtain a rectangle.

Indeed its length is $$ \color{blue}{(s-a)} + \color{red}{(s-b)} + \color{green}{(s-c)} = s$$

and clearly, $$ rs = \triangle$$

Title: Visual proof for $[\triangle ABC] = rs$?

Body:
I'm teaching a high school geometry class and we just got to the part where we proved that the area of a triangle $[\triangle ABC]$ is equal to the inradius $r$ times the semiperimeter $s$.

A student pointed out to me that it's really the same thing as $[\triangle ABC] = \frac{1}{2}rp$ where $p$ is the perimeter. But this representation looks like the regular formula $\frac{1}{2}bh$.

So I was wondering if there's a way to take the original $\triangle ABC$ and construct a new triangle that will have a base of $p$ and a height of $r$ and show that the original and newly constructed triangle have the same area?


问:
Regarding a counterexample to the reflexivity of a divisibility relation on $\Bbb Z$
I'm asked to either prove or disprove if the relation on all integers $\Bbb Z$ is an equivalence relation for $x \sim y$ if and only if $x+y$ is divisible by $3$. From my understanding, the relation can't be reflexive as if $x=1$ then $x+x=2$ which is not divisible by $3$. Is giving counter examples a correct way to disprove an equivalence relation and is the given counter example appropriate for this question?

答: Yes, giving a counter-example as you have done is a correct way to disprove the assertion that this is an equivalence relation. Your counter-example is fine.</p>


问:
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:

$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$

There is a slight difference in case of $\sin$, which is:
$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$

How do we prove the above two identities?

答:
Small comment:

If we have one of the identities then we can derive the other!

Consider:

$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$

Take the derivative on both sides with $a$ while keeping everything else constant:

$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$