问:
A limit of a given sequence
Let $(a_n)_{n\ge1}$ a sequence of real positive numbers such that $a_n+\frac{n}{a_{n+1}^2} \le a_{n+1} \le \frac{a_{n-1}^2}{a_n}+\frac{n+1}{a_n^2}$ for any $n \ge 1$. Prove that $b_n = \frac{a_n}{\sqrt[3]{n^2}}$ is a convergent sequence and find its limit.

First of all, it's clear that $a_n$ is a increasing sequence from the first inequality given. So $a_n$ has a limit and it is either a finite number or infinity. If we suppose that $a_n$ is convergent and has a finite limit (let it be $l$), by using limit in the first inequality we obtain: $l+\infty \le l$, which is false and it means that $\lim_{n \to \infty} a_n = \infty$, so we may apply Cesaro-Stolz for finding the limit of $b_n$, but here I didn't know how to continue. Can you help me?

Title: A limit of a given sequence

Body:
We have $a_n$ increasing and positive, and
$$
a_{n} \leq a_{n+1} - \frac{n}{a_{n+1}^2},
$$
Thus $a_{n+1}^3 \geq n$, and
$$
\frac{a_{n-1}^2}{a_n} + \frac{n+1}{a_n^2} \leq a_n + \frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2},
$$
and using  $a_{n}^3 \geq (n-1)$
$$
\frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2} \leq \frac{(n-1)^2}{a_n^5} \left(1-\frac{n-3}{(n-1)^2}a_n^3 \right) \leq 2  \frac{(n-1)}{a_n^5} < 2\frac{1}{(n-1)^{2/3}}
$$
so altogether
$$
a_n + \frac{n}{a_{n+1}^2}\leq a_{n+1} \leq a_n  + \frac{2}{(n-1)^{2/3}} .
$$
So
$$
\sum_{k=2}^n a_{k+1} -a_{k} \leq 2\sum_{k=2}^n k^{-2/3} \leq 2 \int_{1}^{n-1} \frac{1}{t^{2/3}} dt = 6 n^{1/3}-6,
$$
So we have obtained that
$$
(n-1)^{1/3}\leq a_n \leq 6(n-1)^{1/3}+C,
$$
and the conclusion is clear.


问:
Probability problem regarding distributions

Here which distribution should I use? I am confused between hypergeometric and multinomial distributions

答:
The multinomial is for an independent series of experiments that all have the same finitely many outcomes, and we want to count the number of experiments with cretain outcomes.

The hypergeometric is exactly for what you problem states: one selection without replacement from a finite number of options, that are divided into subclasses.

You can also just use combinatorial reasoning: there is a total number of $\binom{180}{50}$ draws that are possible from $180$ pieces of candy when we pick $50$ of them at once. ("all outcomes")

There are $\binom{100}{15}\binom{80}{35}$ ways to pick both $15$ from the hundred orange-flavoured ones and $35$ from the eighty strawberry-flavoured ones. (the "good" outcomes).

Their quotient will be the desired probability.

ContentLicense: CC BY-SA 4.0


问:
Calculate limit of $\Gamma$ function for special values.
I would like to calculate the limit without any software but have no idea how to do it.

$$f(n) = \lim_{c \rightarrow 0} \frac{\Gamma(-n + c) + \Gamma(-n - c))}{2}$$
$$n = 0, 1, 2, ...$$

Wolfram in some way claculates it, for example:
$$f(0) = - \gamma$$
$$f(1) = \gamma - 1$$
$$f(2) = \frac{3 - 2 \gamma}{4}$$
$$f(3) = \frac{6 \gamma - 11}{36}$$
$$(...)$$

It seems that the solutions will be somethink like that:
$$f(n) = \frac{(a - \gamma)(-1)^{n}}{b}$$

答:
Here's another way to do it. Using the reflection formula for the gamma function write
$$
\begin{aligned}
f(n)
&=\frac{1}{2}\lim_{\epsilon\to 0}(\Gamma(-n+\epsilon)+\Gamma(-n-\epsilon))\\
&=\frac{\pi}{2}\lim_{\epsilon\to 0}\left(\frac{1}{\Gamma(1+n-\epsilon)\sin\pi(-n+\epsilon)}+\frac{1}{\Gamma(1+n+\epsilon)\sin\pi(-n-\epsilon)}\right)\\
&=\frac{(-1)^n\pi}{2}\lim_{\epsilon\to 0}\left(\frac{1}{\Gamma(1+n-\epsilon)\sin\pi\epsilon}-\frac{1}{\Gamma(1+n+\epsilon)\sin\pi\epsilon}\right)\\
&=\frac{(-1)^n\pi}{2\Gamma(1+n)^2}\lim_{\epsilon\to 0}\left(\frac{\Gamma(1+n+\epsilon)-\Gamma(1+n-\epsilon)}{\underbrace{\sin\pi\epsilon}_{\sim\pi\epsilon}}\right)\\
&=\frac{(-1)^n}{2(n!)^2}\lim_{\epsilon\to 0}\left(\frac{\Gamma(1+n+\epsilon)-\Gamma(1+n-\epsilon)}{\epsilon}\right),\quad^\ast\text{L'Hôpital}\\
&=\frac{(-1)^n}{2(n!)^2}\lim_{\epsilon\to 0}\left(\psi(1+n+\epsilon)\Gamma(1+n+\epsilon)+\psi(1+n-\epsilon)\Gamma(1+n-\epsilon)\right)\\
&=\frac{(-1)^n}{n!}\psi(n+1).
\end{aligned}
$$
Then again use the specialized values for $\psi(n+1)$.


问:
Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$
Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?</p>

答:
few thoughts on the first one:

$$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$

This is just a Riemann sum, and thus its limit is $$\int_{0}^1 \ln(\sin(x)) dx$$

This is an improper integral though, so the RS approach might not be best, but I think it is a convergent improper integral, since $\int_0^1 \ln(x)dx $ is convergent..

Maybe someone can take over....

ContentLicense: CC BY-SA 3.0


问:
Putting colored balls into colored urns

There are six balls and urns each. They are equally distributed in three colors: R, W, and B. Place the balls randomly in the six urns, arranged linearly. What is the probability that no ball is in an urn of the same color?

We can think of this as a matching problem, consider $X = \{r_1,r_2,b_1,b_2,w_1,w_2\}$. We want to find the number of bijections $f:X\to X$ such that:

$f(r_i) \ne r_j$ for $1\le i,j\le 2$

$f(b_i) \ne b_j$ for $1\le i,j\le 2$

$f(w_i) \ne w_j$ for $1\le i,j\le 2$

The total number of bijections is $6! = 720$. Define a few sets:

$R = \{f: X\to X \text{ is a bijection }: f(r_i)\ne r_j, 1\le i,j\le 2\}$

$B = \{f: X\to X \text{ is a bijection }: f(b_i)\ne b_j, 1\le i,j\le 2\}$

$W = \{f: X\to X \text{ is a bijection }: f(w_i)\ne w_j, 1\le i,j\le 2\}$

The answer we require then is $1 - \frac{N}{720}$, where $N = |(R\cap B\cap W)^c| = |R^c\cup B^c\cup W^c|$. Using inclusion-exclusion principle, we know that, $$|R^c\cup B^c\cup W^c| = |R^c| + |B^c| + |W^c| - |R^c\cap B^c| - |R^c\cap W^c| - |W^c\cap B^c| + |R^c \cap B^c\cap W^c|$$

Any other ways to approach the problem? Thanks a lot!

答:
There are $6!$ possible orderings of the balls in the urns, all of which we assume are equally likely.  We will show by the method of Rook Polynomials, which is an extension of the Principle of Inclusion / Exclusion, that the number of arrangements in which no ball is placed in an urn of the same color is $80$, so the probability of such  an arrangement is $80/6! = 1/9$.

Number the urns from $1$ to $6$, with urns $1-2$ all one color, $3-4$ the second color, and $5-6$ the third color.  Then the number of acceptable arrangements is the same as the number of ways to place $6$ non-attacking, distinct rooks on the following $6$ by $6$ chessboard, where an X marks a forbidden square:

```<code>    X X . . . .
    X X . . . .
    . . X X . .
    . . X X . .
    . . . . X X
    . . . . X X
</code>```

We want to find the rook polynomial $R(x)$ of the forbidden sub-board.  By definition
$$R(x) = \sum r_i x^i$$
where $r_i$ is the number of ways to place $i$ identical non-attacking rooks on the forbidden squares, for $0 \le i \le 6$.  We define the number of ways to place zero rooks as $1$, so $r_0=1$.

For a start, let's consider the forbidden $2$ by $2$ sub-board in the upper left hand-corner:

```<code>    X X
    X X
</code>```

By inspection, the rook polynomial of this sub-board is
$$1 + 4x + 2x^2$$

Since the larger forbidden area consists of three such $2$ by $2$ similar areas with no rows or columns in common among the three areas, the rook polynomial of the forbidden sub-board is
$$R(x) = (1 + 4x + 2x^2)^3$$
On expansion,
$$R(x) =1+12 x+54 x^2+112 x^3+108 x^4+48 x^5+8 x^6$$

Now we can find the number of ways to place $6$ distinguishable non-attacking rooks on the allowable area of the $6$ by $6$ board.  By inclusion / exclusion, that number is
$$1(6!) -12 (5!)+54 (4!)-112 (3!)+108 (2!)-48(1!)+8 (0!) = \boxed{80}$$

Title: Putting colored balls into colored urns

Body:
There are $6!$ possible orderings of the balls in the urns, all of which we assume are equally likely.  We will show by the method of Rook Polynomials, which is an extension of the Principle of Inclusion / Exclusion, that the number of arrangements in which no ball is placed in an urn of the same color is $80$, so the probability of such  an arrangement is $80/6! = 1/9$.

Number the urns from $1$ to $6$, with urns $1-2$ all one color, $3-4$ the second color, and $5-6$ the third color.  Then the number of acceptable arrangements is the same as the number of ways to place $6$ non-attacking, distinct rooks on the following $6$ by $6$ chessboard, where an X marks a forbidden square:

    X X . . . .
    X X . . . .
    . . X X . .
    . . X X . .
    . . . . X X
    . . . . X X

We want to find the rook polynomial $R(x)$ of the forbidden sub-board.  By definition
$$R(x) = \sum r_i x^i$$
where $r_i$ is the number of ways to place $i$ identical non-attacking rooks on the forbidden squares, for $0 \le i \le 6$.  We define the number of ways to place zero rooks as $1$, so $r_0=1$.

For a start, let's consider the forbidden $2$ by $2$ sub-board in the upper left hand-corner:

    X X
    X X

By inspection, the rook polynomial of this sub-board is
$$1 + 4x + 2x^2$$

Since the larger forbidden area consists of three such $2$ by $2$ similar areas with no rows or columns in common among the three areas, the rook polynomial of the forbidden sub-board is
$$R(x) = (1 + 4x + 2x^2)^3$$
On expansion,
$$R(x) =1+12 x+54 x^2+112 x^3+108 x^4+48 x^5+8 x^6$$

Now we can find the number of ways to place $6$ distinguishable non-attacking rooks on the allowable area of the $6$ by $6$ board.  By inclusion / exclusion, that number is
$$1(6!) -12 (5!)+54 (4!)-112 (3!)+108 (2!)-48(1!)+8 (0!) = \boxed{80}$$


问:
Let $A$ be a vector subspace of the normed space $E$. Let $x \in E, y \in A$ and $c >0$. Show that $d(y+cx,A) =cd(x,F).$

Let $A$ be a vector subspace of the normed space $E$. Let $x \in E, y \in A$ and $c >0$. Show that $d(y+cx,A) =cd(x,A).$

I'm very confused with the notation here. The distance function takes in points of a space/set as an input, how can they pass the whole set there? Are they looking at the distance of a point from the whole set or what is happening here?

答:
First of all $d(x,A) = \inf \{d(x,y)\mid y \in A\}$

We can also use that a metric from a norm is translation invariant: $d(x+z,y+z)= d(x,y)$ for all $x,y,z$, and also $|c|d(x,y)=d(cx,cy)$ for all scalars $c$. Both are obvious from $d(x,y)=\|x-y\|$ and properties of the addition and the norm.


问:
probability distribution of a random variable that is uniformly distributed in [-1, 1]
If $X$ is a random variable that is uniformly distributed between $-1$ and $1$, find the PDF of $\sqrt{\vert X\vert}$ and the PDF of $-\ln\vert X\vert$.
Solution or an explantion of approach wooulf work.

thank you in advance for your help

答:
$X \sim U[-1, 1]$, $Y = \sqrt{|X|} \Rightarrow$

$$
F_Y(y) = \text{Pr}\left(Y \leq y\right), \quad y \geq 0,
$$

is a cumulative distribution function (c.d.f.) of $Y$ (by definition).

$$
\begin{aligned}
\text{Pr}\left(Y \leq y\right) &= \text{Pr}\left(\sqrt{|X|} \leq y\right) = \text{Pr}\left(|X| \leq y^2\right) = \text{Pr}\left(-y^2 \leq X \leq y^2\right) = \\
&= \left|\begin{aligned}
&\text{since }X\sim U[-1, 1] \Leftrightarrow \text{c.d.f. of } X\text{ is }\\
&F_X(x) =
\left\{
\begin{array}{rl}
0, & x \leq -1 \\
\frac{x-(-1)}{1-(-1)} = \frac{x+1}{2}, & -1 < x \leq 1 \\
1, & x > 1
\end{array}
\right.
\end{aligned}\right| = \\
&= F_X(y^2) - F_X(-y^2) =
\left\{
\begin{array}{rl}
\frac{y^2+1}{2}-\frac{-y^2+1}{2} = y^2, & 0 \leq y^2 \leq 1 \Leftrightarrow 0 \leq y \leq 1\\
1, & y^2 > 1  \Leftrightarrow y > 1
\end{array}
\right.
\end{aligned}
$$

So, the c.d.f. of $Y$, $F_Y(y) = \text{Pr}(Y \leq y)$, $\quad y \geq 0$, is equal to

$$
F_Y(y) =
\left\{
\begin{array}{rl}
y^2, & 0 \leq y \leq 1\\
1, & y > 1
\end{array}
\right.
$$

Then, the probability density function (p.d.f.) of $Y$ equals to

$$
f_Y(y) = F'_Y(y) =
\left\{
\begin{array}{rl}
2y, & 0 \leq y \leq 1\\
0, & y > 1
\end{array}
\right.
$$


问:
Is it true that for any $n\geq 1$, $(\ln(x))^n < x$?
I try to graph it on Desmos but it does not seem to be the case?</p>

答: We know that $\mathrm e<3$. Now take $x=\mathrm e^2$ and $n=4$. Then $(\ln x)^4=2^4=16$, while $x<3^2=9$.


问:
set of sequence and it limits in a Hausdorff space (not metric)Is closed?
I want to prove that for $K=\{x_n, n\in\mathbb{N}\}\cup\{\ell\}$ we have $\overline{K}\subset K$ in a Hausdorff topological space.

let $x\in \overline{K}$ that is. any open V such that $x\in V$: $V\cap A\neq \emptyset$

it means that there exists $y\in V\cap A$ , if $y=x$ we are done.

But if $x\neq y$, $V\setminus \{x\}$ is an open containing $y$.

Here I don't know how to continue?

Thanking you

答: Take $x\notin K$; you want to prove that $x\notin\overline K$. Since $x\notin K$, $x\ne l$. Let $V_x$ and $V_l$ be neighborhoods of $x$ and of $l$ such that $V_x\cap V_l=\emptyset$. Since $l=\lim_{n\to\infty}x_n$, there is some $N\in\Bbb N$ such that $n\geqslant N\implies x_n\in V_l$. So, $n\geqslant N\implies x_n\notin V_x$. For each $n\in\{1,2,\ldots,N-1\}$, let $V_n$ be a neighborhhod of $x$ such that $x_n\notin V_x$. Then $V_1\cap V_2\cap\ldots\cap V_{N-1}\cap V_x$ is a neighborhood of $x$ which contains no element of $K$. Therefore, $x\notin\overline K$.


答:
If $x_n \to l$ in any space, $K:= \{x_n: n \in \Bbb N\} \cup \{l\}$ is compact (easily seen by using the open cover definition: any open set containing $l$ already contains all but finitely many $x_n$ by convergence etc.)

In a Hausdorff space a compact subset is closed.

Title: set of sequence and it limits in a Hausdorff space (not metric)Is closed?

Body:
I want to prove that for $K=\{x_n, n\in\mathbb{N}\}\cup\{\ell\}$ we have $\overline{K}\subset K$ in a Hausdorff topological space.

let $x\in \overline{K}$ that is. any open V such that $x\in V$: $V\cap A\neq \emptyset$

it means that there exists $y\in V\cap A$ , if $y=x$ we are done.

But if $x\neq y$, $V\setminus \{x\}$ is an open containing $y$.

Here I don't know how to continue?

Thanking you


问:
Find a line with maximum points from N points
You are given N points and you want to draw a line such that maximum points lie on the line. What is the efficient way to find the number of maximum points?</p>

答:
Follows a script in MATHEMATICA which implements a procedure to obtain the line passing through more points.

NOTE

This script admits many improvements regarding computational efficiency. It was released in this basic form with the proposal to show the geometric ideas involved in the algorithm. Calling $p_i, i=1,\cdots n$ the set of points the first step is to form a matrix $V$ such that $V_{i,j} = p_i-p_j$. After, we construct a list $\mathcal{L}$ (list0) with the element indexes in $V$ such that for $i\ne j\ne k$ we have $V_{i,k}\times V_{j,k}=0$. The last operation consists in joining elements on $\mathcal{L}=\{\mathcal{l}_1,\cdots,\mathcal{l}_p,\cdots,\mathcal{l}_q,\cdots,\mathcal{l}_m\}$ such that $\mathcal{l}_p,\mathcal{l}_q$ have at least two indexes (points) in common. The biggest resulting elements in this union are the solutions.

```<code>n = 30;
SeedRandom[5]
list = {};
points = Union[RandomInteger[{0, 10}, {n, 2}]];
n = Length[points]
V = Table[Join[points[[i]] - points[[j]], {0}], {i, 1, n}, {j, 1, n}];
For[i = 1, i <= n, i++,
 For[j = 1, j <= n, j++,
  For[k = 1, k <= n, k++,
   If[i != k && j != k && i != j,
    cross = Cross[V[[i, k]], V[[j, k]]];
    If[cross == {0, 0, 0}, AppendTo[list, Sort[{i, j, k}]]]]
  ]
 ]
]
list0 = Union[list];

lines = {};
npts = 0;
For[i = 1, i <= Length[list0], i++,
 lini = list0[[i]];
 line = lini;
 For[j = 1, j < i, j++,
  linj = list0[[j]];
  If[Length[Intersection[lini, linj]] == 2, line = Union[line,linj]]
 ];
 nl = Length[line];
 If[nl >= npts,
  AppendTo[lines, line];
  npts = nl
 ]
]
nlin = Length[lines]
LINES = {};
For[k = 1, k <= nlin, k++,
 If[Length[lines[[k]]] == npts, AppendTo[LINES, lines[[k]]]]]
LINES = Union[LINES]
NL = Length[LINES];
gr1 = ListPlot[points];
gr2 = Flatten[Table[Graphics[{Red, PointSize[0.02], Point[points[[list0[[j, k]]]]]}], {k, 1, 3}, {j, 1, Length[list0]}]];
gr2a = Flatten[Table[Table[ParametricPlot[lambda points[[LINES[[j, k]]]] + (1 - lambda) points[[LINES[[j, k + 1]]]], {lambda, 0, 1}], {k, 1, Length[LINES[[j]]] - 1}], {j, 1, NL}]];
Show[gr1, gr2a, gr2, PlotRange -> All, Axes -> False, AspectRatio -> 1]
</code>```

with SeedRandom[51]

and with SeedRandom[52]

Body:
Follows a script in MATHEMATICA which implements a procedure to obtain the line passing through more points.

NOTE

This script admits many improvements regarding computational efficiency. It was released in this basic form with the proposal to show the geometric ideas involved in the algorithm. Calling $p_i, i=1,\cdots n$ the set of points the first step is to form a matrix $V$ such that $V_{i,j} = p_i-p_j$. After, we construct a list $\mathcal{L}$ (list0) with the element indexes in $V$ such that for $i\ne j\ne k$ we have $V_{i,k}\times V_{j,k}=0$. The last operation consists in joining elements on $\mathcal{L}=\{\mathcal{l}_1,\cdots,\mathcal{l}_p,\cdots,\mathcal{l}_q,\cdots,\mathcal{l}_m\}$ such that $\mathcal{l}_p,\mathcal{l}_q$ have at least two indexes (points) in common. The biggest resulting elements in this union are the solutions.

n = 30;
SeedRandom[5]
list = {};
points = Union[RandomInteger[{0, 10}, {n, 2}]];
n = Length[points]
V = Table[Join[points[[i]] - points[[j]], {0}], {i, 1, n}, {j, 1, n}];
For[i = 1, i <= n, i++,
 For[j = 1, j <= n, j++,
  For[k = 1, k <= n, k++,
   If[i != k && j != k && i != j,
    cross = Cross[V[[i, k]], V[[j, k]]];
    If[cross == {0, 0, 0}, AppendTo[list, Sort[{i, j, k}]]]]
  ]
 ]
]
list0 = Union[list];

lines = {};
npts = 0;
For[i = 1, i <= Length[list0], i++,
 lini = list0[[i]];
 line = lini;
 For[j = 1, j < i, j++,
  linj = list0[[j]];
  If[Length[Intersection[lini, linj]] == 2, line = Union[line,linj]]
 ];
 nl = Length[line];
 If[nl >= npts,
  AppendTo[lines, line];
  npts = nl
 ]
]
nlin = Length[lines]
LINES = {};
For[k = 1, k <= nlin, k++,
 If[Length[lines[[k]]] == npts, AppendTo[LINES, lines[[k]]]]]
LINES = Union[LINES]
NL = Length[LINES];
gr1 = ListPlot[points];
gr2 = Flatten[Table[Graphics[{Red, PointSize[0.02], Point[points[[list0[[j, k]]]]]}], {k, 1, 3}, {j, 1, Length[list0]}]];
gr2a = Flatten[Table[Table[ParametricPlot[lambda points[[LINES[[j, k]]]] + (1 - lambda) points[[LINES[[j, k + 1]]]], {lambda, 0, 1}], {k, 1, Length[LINES[[j]]] - 1}], {j, 1, NL}]];
Show[gr1, gr2a, gr2, PlotRange -> All, Axes -> False, AspectRatio -> 1]

with SeedRandom[51]

and with SeedRandom[52]


问:
How do I proceed with this trigonometric integral $∫\sec^3θ\cot θ\; dθ$
I tried to perform trig sub on the equation below and what I got was
$$∫\sec^3θ\cot θ\; dθ$$
I'm not sure if this is correct and I'm also really not sure how to proceed from here.

$$\int{{\sqrt{\ln^2 x+1}\over{x\ln{x}}}\; dx}$$

My solution

答:
Note

\begin{align}
\int \sec^3t\cot  t\ dt
&= \int \frac{\sec t}{\tan t} (1+\tan^2t)dt
= \int \left(\frac1{\sin x} + \sec t \tan t\right)dt\\
&=\ln\tan \frac t2 + \sec t+C
\end{align}


问:
$f'(x)=f(x)$ and $f(0)=0$ implies that $f(x)=0$ formal proof
How can I prove that if a function is such that $f'(x)=f(x)$ and also $f(0)=0$ then $f(x)=0$ for every $x$. I have an idea but it's too long, I want to know if there is a simple way to do it. Thanks!
Obviously in a formal way.</p>

答:
Below I show how the standard proof (e.g. egreg's answer) is a special case of more general results on uniqueness theorems and Wronskians. Let's rewrite the proof in slightly more general form. Given that $\,f\,$ and $\,g = e^x\,$ are  solutions of $\,y' = y\,$ we deduce

$$\begin{eqnarray}\color{#c00}{f'=f}\\ \color{#0a0}{g'=g}\end{eqnarray}\ \ \Rightarrow\ \ \left(\dfrac{f}g\right)' = \dfrac{\color{#c00}{f'}g-f\color{#0a0}{g'}}{g^2} = \dfrac{\color{#c00}fg-f\color{#0a0}g}{g^2} = 0\ \ \Rightarrow\  \dfrac{f}g = c\ \ \rm constant$$

Therefore $\ f = cg = ce^x\ $ so $\ 0 = f(0) = c,\ $ thus $\, f = 0.\, $ This generalizes. The same proof shows that, assuming appropriate differentiability/continuity conditions, if $\,f,g\,$ are solutions of $\, y' = h y\,$ on an interval $\,I\,$ where the Wronskian $\,W(f,g) := f'g - fg' = 0\,$ on $\,I,\,$ then $\,f,g\,$ are linearly dependent on $\,I,\,$ i.e. $\,c_1 f = c_2 g \,$ on $\,I\,$ for some constants $\,c_i.\,$

Thus, by a very simple proof, we've deduce a uniqueness theorem for solutions of linear first-order differential equations of said type. These ideas extend to analogous higher-order linear differential equations. See here for a proof of the second-order case (which generalizes to n'th order), using variation of parameters, and see here for the discrete analog of difference equations (recurrences). See also the classical result below on Wronskians and linear dependence.

Theorem $\ \ $  Suppose  $\rm\:f_1,\ldots,f_n\:$   are $\rm\:n-1\:$ times differentiable on interval $\rm\:I\subset \mathbb R\:$
and suppose they have Wronskian $\rm\: W(f_1,\ldots,f_n)\:$ vanishing at all points in $\rm\:I\:.\:$ Then  $\rm\:f_1,\ldots,f_n\:$  are linearly dependent on some subinterval of $\rm\:I\:.$

Proof $\ $ See this answer.


问:
How to show $\prod_{n=1}^{\infty}\frac{1-x^{2 n}}{1-x^{n}} = \prod_{n=1}^{\infty}\frac{1}{1-x^{2 n-1}}$?
I've tried many different ways of rearranging the expression $\frac{1-x^{2 n}}{1-x^{n}}=\frac{1}{1-x^{2 n-1}}$ , but I can not seem to prove that the left hand side equals the right hand side.

Am I missing something?

This is part of a larger formula in this paper on proving the Euler partition identity with generating functions. https://people.clas.ufl.edu/alladik/files/alladipaperpaule60-1.pdf

答: The terms in the numerator and the terms with even $n $ in the denominator cancel, so that only terms with odd $n $ in the denominator survive. I assume that $|x|<1$ so that the product converges.


问:
Comparison function for the Rossler system
we are studying the chaotic Rössler system:
$$
\begin{align}
& \dot{x}_1 = - x_2 - x_3\\
& \dot{x}_2 = x_1 +\alpha x_2\\
& \dot{x}_3 = \beta + x_3(x_1-\gamma)
\end{align}
$$
With $\alpha = \beta = 0.1$ and $\gamma = 14$, and $x_3(0)>0$, such that $x_3>0$ $\forall$ $t\geq0$.

we have found that for the comparison function $W=x_1^2+x_2^2+2x_3$ the following inequality holds:
$$
\dot{W} \leq 2\alpha W + 2\beta
$$

This implies that the solutions of this system are well defined on the infinite time interval $[0,\infty)$, and implies that solutions can not escape to infinity in finite time.

However, we are not able to derive this last conclusion, if anyone could point us in the right direction, it would greatly be appreciated. Thank you!

答:
One can easily confirm that $x_3>0\implies w(t)=W(x(t))>0$ and thus
$$
αw(t)+β\le (αw(0)+β)e^{2αt}
\implies w(t)\le \bar w(t)=w(0)e^{2αt}+\frac{β}{α}(e^{2αt}-1).
$$
This means that over any finite time interval $[0,T]$ the solution is bounded by $$W(x(t))\le \bar w(T).$$ So if one considers the region $W(x)<2\bar w(T)$, it is bounded, and has $x(T)$ as inner point. Thus the ODE function has bound and a Lipschitz constant there, thus the IVP with the IC at $x(T)$ can be solved locally in forwar direction and the solution $x$ thus continued.

Note that the movement forward in time is essential for the claim that $x_3$ stays positive, backwards in time $x_3$ can become negative and thus unbounded by $\bar w$.


问:
Null space and range of generic projection matrix
Let $v \in \Bbb R^2$ that is nonzero. I've already proved that $R(P_v) = Sp\{v\}$, but I need to find the null space of $P_v$, as well as the dimensions of the range and null space. How would I go about doing that?
($P_v = \frac{vv^T}{||v||^2}$).

I know the null space of $P_v$ = $\{x: P_vx = 0\} $, so it's $\{x: vv^Tx = 0\}$? I'm not sure where to go from here.

I also need to do the same problem but with $v \in \Bbb R^n$.

Thanks :)

答:
As you have shown that the range of $P_v$ is the span of $v$, i.e. the line through the origin parallell to $v$, you know that the dimension of its range is $1$ ($v$ is a basis). This will be the case for $\mathbb{R}^n$. The range of $P_v$ is the column space $\text{Col}(P_v)$. Since $P_v = \frac{1}{\lVert v \rVert^2}vv^T$, its easy to see that all columns in $P_v$ are linear combinations of $v$, so $\text{Col}(P_v) = \mathbb{R}v$. Since you know the dimension of the column space, you know the dimension of the null space by the Rank-Nullity Theorem. So $\dim \text{Nul} (P_v) = n-1$.

You can also see this directly. Choose an orthogonal basis $\{v_1, \ldots v_n \}$ for $\mathbb{R}^n$ with $v_1 = v$. We claim that $\{v_2, \ldots v_n \} \subseteq \text{Nul} (P_v)$. This is easily varified:

$$P_v v_i = \frac{1}{\lVert v \rVert^2} (vv^T)v_i = \frac{1}{\lVert v \rVert^2} v(v^Tv_i) = \frac{v \cdot v_i}{\lVert v \rVert^2} v = 0$$

since $v_i$ is orthogonal to $v$ for $2 \leq i \leq n$. Hence, $\{ v_2, \ldots, v_n \}$ forms a basis for the null space of $P_v$.


问:
Continuity of $f(t)$ from $\frac{df}{dt}+kf(t)=\delta (t)$
Consider the differential equation
$$\frac{df}{dt}+kf=\delta (t)$$
where $k$ is constant. I'm asked to give statement about the continuity of $f(t)$ and  $f'(t)$ at $t=0$.

I tried solving the equation using Laplace transform seeing the $\delta $ function, it seems much convenient.
$$\mathcal{L}\left(\frac{df}{dt}+kf\right)=\mathcal{L}(\delta(t))$$
$$s\mathcal{L}(f)-f(0^+)+k\mathcal{L}(f)=1$$
$$\mathcal{L}(f)=\frac{1+f(0^+)}{s+k}$$
$$f(t)=(1+f(0^+))\mathcal{L^{-1}}\left(\frac{1}{s+k}\right)=(1+f(0^+))e^{-kt}$$
But looking on the function, I don't How can I say about it anything. For example
$$\lim_{t\rightarrow 0^+}f(t)=f(0^+)=(1+f(0^+))e^0\Rightarrow 1=0$$
which is complete non-sense. I don't understand what I'm missing. Can any help me figure it out?

答:
The direct answer without solving anything is that the primary source for the singularity on the right side is the highest derivative on the left. So for $f'$ to give $\delta$ then $f$ must have a jump of size $1$ at $t=0$ and consequently $f'$ a jump of $-k$.

To quickly solve this, apply the integrating factor and sifting property of delta to get
$$
(e^{kt}f(t))'=e^{k·0}δ(t)
$$
Now use that delta integrates to the unit jump to get
$$
e^{kt}f(t)-f(0^+)=u(t)-u(0^+),\\
f(t)=(u(t)-1+f(0^+))e^{-kt}.
$$
To ensure that $f(t)=0$ for $t<0$ one needs that $f(0^+)=1$.

The Laplace transform is mostly concerned with things that happen at $t>0$. Using singularities like jumps or Dirac deltas at $t=0$ is just another form of encoding initial conditions. The usual treatment is to integrate a short way formally
$$
f(ε)-f(-ε)+k\int_{-ε}^εf(s)\,ds=1\implies f(0^+)=1+f(0^-).
$$
One could also just shift the Dirac delta some small distance $ε$. If shifting to the left, it leaves the scope of the Laplace transform and one is left with a transformed equation
$$
sF(s)-f(0^+)+kF(s)=0.
$$
Shifting the Dirac pulse to the right to some position $ε$ results in the transform
$$
sF(s)-f(0^-)+kF(s)=e^{-εs},~~~ε\to 0
$$
Both interpretations lead to consistent results.


问:
How to make an injective function $ f $ such that $ f : \left[0,1\right]\rightarrow2^\mathbb{N} $ using binary notation?
$ 2^\mathbb{N} $ is a power set of $ \mathbb{N} $.

By using binary notation, we can take $ x\in\left[0,1\right]$ and $ x=0.a_1a_2a_3\cdots\left(a_n\in\left\{0,1\right\}\right) $

Then we can make a set $ A=f(x) $ as an element of $ 2^\mathbb{N} $ following next rules:

$n\in A $ when $ a_n=1 $ and $ n\notin A $ when $ a_n=0 $

For example,

$$\begin{align*}
0.00000\cdots&\longmapsto\emptyset\\[5pt]
0.10000\cdots&\longmapsto\left\{1\right\}\\[5pt]
0.01000\cdots&\longmapsto\left\{2\right\} \\[5pt]
0.01111\cdots&\longmapsto\left\{2,3,4,\cdots\right\} \\[5pt]
0.10111\cdots&\longmapsto\left\{1,3,4,5,\ \cdots\right\} \\[5pt]
\end{align*}$$

This seems nice. However, since $ 0.10000\cdots=0.01111\cdots $, there exist two $ f(x) $ for same $ x$. I can't fix this problem and I've been thinking of this all night.. How can I make the injective function $ f $ with solving this?

答:
Here's an indirect way to produce the $1-1$ function using the hint that I dropped earlier in comment.

First we show that $P(\mathbb N)$ and $T=\{f| f:\mathbb N\to \{0,1\}\}$ are similar.

Let $A\subseteq \mathbb N$ (i.e, $A\in P(\mathbb N)$). Let's define a function $\chi_A: \mathbb N\to \{0,1\}$ such that $\chi_A(x)= 1,$ if $x\in A$ else $0$. Clearly, $\chi_A\in T$.

Define $\theta: P(\mathbb N)\to T$ by $\theta (A)=\chi_A$. Verify that $\theta$ is a bijection.

Now, every $r$ in $[0,1]$ has a binary representation which can be restricted to "unique binary representation" by choosing the one which has forever repeating $0$s at the end in case of two binary representations (the other being with forever repeating 1s at the end) for some $r$. It's an easy exercise to show that $[0,1]$ and $T$ are similar. Hence there exists a bijection $\phi:[0,1]\to T$.

Now consider the composition $\theta^{-1} o\phi:[0,1]\to P(\mathbb N) $. This is clearly $1-1$.


问:
A linear Equation in 4D space corresponds to the whole 3D space
This equation is said to be a plane in 4D space $u+ v+ w+ z=6$

Does that plane correspond to the whole real 3D space?

I think so. Below is my attempt

Say we think of the three real space axis as given by $u ,v, w$ ,then fixing $z=d$ ($d $ is some arbitrary number ),gives me $u+ v+ w=6-d$ which is a plane in 3D space. This plane is a set of solution.

Now I change  value of $z$ and I again get another plane as a solution , continuing this way I can fill all of 3D space with the solution planes.

Therefore we can say that a plane in 4D is the whole 3D space.

答:
Consider just two variables, $x$ and $y$. The way to describe the 1D number line, i.e. the $x$ axis, is with the equation $y=0$.

If you take the equation $x+y=6$, this describes a different line in the plane. $y$ cannot have freedom to vary.

So for your four variables, $u$, $v$, $w$, and $z$: If you want to describe the $uvw$ "plane", the only value $z$ can have is $0$. So the equation should just be $z=0$. Your equation describes a different 3D space.


问:
How to write $P = \{ [x] \mid x \in S \}$ in first-order logic?
I  am trying to write $P =  \{ [x] \mid x \in S \}$  in first-order logic. My intuition is that it is written like this:

$\forall y (y \in P \Leftrightarrow \exists x (y = [x]) \land  x \in S))$

Is this correct?

答: $[x]$ is not in first-order-logic. And the definition of equivalence class need a equivalence relation. Let the relation be $R$. Then the correct answer is that $\forall y(y\in P \leftrightarrow \exists x(x\in S\wedge \forall z(z\in y\leftrightarrow Rxz)))$


问:
What does point-wise convergence in the sup norm mean?
I am starting out with functional analysis and learning about point-wise and uniform convergence. In particular, I need help with my proof that the space $$ \mathcal{B} = \left\{ f:X\rightarrow \mathbb{R} \, \middle| \, f \, \text{bounded} \right\}$$
of bounded real functions on an arbitrary space $X$ is complete in the sup norm $$ \lVert f \rVert_{\infty} := \sup_{x\in X} \lvert f\left(x\right) \rvert. $$

My book (Intro to Topology and Modern Analysis by Simmons) poses this proof as a problem before any discussion on convergence of function sequences, so I will not use terms like "point-wise" or "uniform" convergence in the proof itself.

My Proof

Let $\left(f_n\right)_{n\in \mathbb{N}} \subset \mathcal{B}$ be a Cauchy sequence.

Letting $\epsilon >0$, there is an $N\in \mathbb{N}$ such that $$ m,n\geq N \implies \lVert f_m - f_n\rVert_{\infty} < \epsilon. $$ Thus, by definition of the sup norm we have for every $x \in X$, $$ m,n \geq N \implies \lvert f_m\left(x\right) -f_n\left(x\right) \rvert \leq \lVert f_m-f_n\rVert_{\infty} < \epsilon $$ This shows that, for fixed $x$, the real sequence $\left(f_m\left(x\right)\right)_{m\in \mathbb{N}}$ is Cauchy and converges since $\mathbb{R}$ is complete. Hence, we define $f:X\rightarrow \mathbb{R}$ by $$ f\left(x\right) = \lim_{m\rightarrow \infty} f_m\left(x\right). $$

So far we have merely constructed a function $f$ to which our sequence $\left(f_m\right)_{m\in \mathbb{N}}$ may converge (see discussion after proof); it remains to show that it does converge as a bona fide sequence in the sup norm.

To this end, let $\epsilon >0$ and take $N$ in the Cauchy hypothesis so that for any $x\in X$and $m,n> N$ we have $\lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}$. Then for this same $n>N$ we have $$ \lvert f\left(x\right) - f_n\left(x\right)\rvert = \lvert \lim_{m\rightarrow \infty} f_m\left(x\right) - f_n\left(x\right) \rvert = \lim_{m\rightarrow \infty} \lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}  $$
(pulling the limit out of the absolute value required a small side lemma). Thus, we have $$  n>N \implies \lVert f-f_n\rVert_{\infty} = \sup_{x\in X} \lvert f\left(x\right) -f_n\left(x\right)\rvert \leq \frac{\epsilon}{2} < \epsilon $$ so that $\left(f_n\right)_{n\in \mathbb{N}}$ converges to our $f$ in the sup norm.

Proving the boundedness of the limit function $f$ is a whole other can of worms which is irrelevant to my confusion, so I will avoid typing it out here. Once the limit function is bounded, we know that the Cauchy sequence converges in $\mathcal{B}$ and so $\mathcal{B}$ is complete.

My Confusion

I was happy with this proof until I learned more about point-wise versus uniform convergence. My proof shows that the sequence $\left(f_n\right)_{n\in\mathbb{N}}$ converges in the sup norm to the $f$ we constructed. In doing so, I see in retrospect that I also proved that this is uniform convergence. This uniform convergence seemed to come naturally out of the proof since we used the sup norm: we found an $N\in \mathbb{N}$ for which the $\epsilon$ property holds for all $x\in X$. However, I didn't set out to prove uniform convergence, only convergence in the norm we had defined on the function space, so it looks like this convergence in this norm might imply uniform convergence.

My question is: where does the notion of point-wise convergence enter in? The construction of my limit function $f$ is done in the usual "point-wise" manner, and so I assume that by construction I can say that "$\left(f_n\right)_{n \in \mathbb{N}}$ converges point-wise to $f$," but I am bothered by when I specialize to a point $x\in X$, my "point-wise convergence" ends up using only the standard norm on $\mathbb{R}$ and not the sup norm.

Hence, I can imagine we might have function sequences which converge "point-wise" but not uniformly; how are we justified in saying that they "converge," then, since they won't converge as sequences in $\mathcal{B}$, where the norm is the sup norm?

答:
After focusing a bit more on the basic definitions, I think I've made sense of this on my own. I'll leave my notes here in this answer to help anyone else who hits upon this.

I will generalize the definitions of convergence that I was using to more general spaces. Let $\left(X, d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces and $f$ and $\left\{f_n\right\}_{n\in\mathbb{N}}$ be functions from $X$ to $Y$.

We say that "$\left(f_n\right)_{n\in\mathbb{N}}$ converges pointwise to $f$" if $\forall \epsilon>0$ and $x\in X$ we can find some $N \in \mathbb{N}$ so that $$ n>N \implies d_{Y}\left(f_n\left(x\right), f\left(x\right) \right) < \epsilon. $$ Notice that we don't use the metric on $X$ anywhere in this definition, since we are concerned with a single point in the domain; the metric on $Y$ allows us to quantify the distance between the function values (in my original problem, this was simply the usual metric on $\mathbb{R}$ since my functions were real-valued).

We say that "$\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly to $f$" if $\forall \epsilon >0$ we can find some $N\in \mathbb{N}$ so that for all $x\in X$ we have $$ n>N \implies d_Y\left(f_n\left(x\right), f\left(x\right) \right) < \epsilon. $$ Again we do not use the metric on $X$.

These two notions of convergence of function sequences are certainly not equivalent, though uniform convergence implies pointwise. My earlier proof proceeded by finding a pointwise limit of the sequence and showing that --- by the grace of its Cauchy-ness --- this convergence was also uniform. My confusion started when trying to consider these function sequences as sequences in a function space equipped with the uniform metric. My proof below shows that uniform convergence of a function sequence is equivalent to convergence of the sequence in a function space equipped with the uniform norm. Thus, I am happily convinced that these notions of convergence coincide and it makes sense to consider uniform convergence as true convergence of a sequence of elements (functions) in a (function) space.

Proof

First let $\left(f_n\right)_{n\in \mathbb{N}}$ converge uniformly to $f$, all functions from metric space $\left(X, d_X\right)$ to $\left(Y, d_Y\right)$ belonging to some function space, call it $\mathcal{C}$. Define the uniform metric on $\mathcal{C}$ to be the usual $$ d_{\mathcal{C}}\left(f,g\right) = \lVert f - g\rVert_{\infty} := \sup_{x\in X} d_Y\left(f\left(x\right), g\left(x\right)\right). $$

Let $\epsilon >0 $ arbitrary. Then by uniform convergence there exists some $N \in \mathbb{N}$ so that $$ \forall x\in X, n > N \implies d_Y\left(f_n\left(x\right), f\left(x\right) \right) < \frac{\epsilon}{2}. $$ Now let $$B = \left\{ d_Y\left(f_{n'}\left(x'\right), f\left(x'\right)\right) \, \middle| \, x' \in X, n' > N\right\}. $$ We have just shown that this set $B\subset \mathbb{R}$ is bounded, and as a subset of $\mathbb{R}$ it has a supremum $\beta = \sup B.$ But then $\beta \leq \frac{\epsilon}{2} < \epsilon$. Now, notice that since for a fixed n > N, $$ \left\{ d_Y\left(f_n\left(x\right), f\left(x\right)\right)\, \middle| \, x\in X  \right\} \subseteq B, $$ we thus have $$ n>N \implies \lVert f_n - f\rVert_{\infty} = \sup_{x\in X} d_Y\left(f_n\left(x\right), f\left(x\right)\right) \leq \beta < \epsilon. $$ That is, $\left(f_n\right)_{n\in \mathbb{N}}$ converges to $f$ in the uniform norm on $\mathcal{C}$.

Conversely, let $\left(f_n\right)_{n\in \mathbb{N}}$ converge to $f$ in the uniform norm on $\mathcal{C}$. Then let $\epsilon >0 $ and take $N \in \mathbb{N}$ so that $$ n>N \implies \lVert f_n - f\rVert_{\infty} = \sup_{x\in X} d_Y\left( f_n\left(x\right), f\left(x\right)\right) < \epsilon.$$ But then for arbitrary $x\in X$ we have $$ n>N \implies d_Y\left(f_n\left(x\right), f\left(x\right)\right) \leq \lVert f_n - f\rVert_{\infty} < \epsilon,$$ so that $\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly to $f$. $\blacksquare$


问:
Proving a recursive definition about decreases in permutations
Definition

A permutation $\pi = a_1 a_2 \cdots a_n \in S_n \; \; i \in \{1,\cdots,(n-1)\}$ is called a decrease if $a_i > a_{i+1}$. For $k \geq 1$, let $A(n,k)$ be the number of permutations of $[n]$ having $k-1$ decreases. We set $A(0,0) = 1$ and $A(0,k) = 0$ for $k \geq 1$.

Excercise

Prove that for $n \geq 0, k \geq 1$:
$$A(n,k) = k \cdot A(n-1,k) + (n-k+1) \cdot A(n-1,k-1)$$

I first tried to write down the permutations and their decreases for n = 2,3,4:

$
\begin{array}{l|r}
\text{Permutation} & \text{#} \\
12 & 0 \\
21 & 1 \\
\end{array}
$

$
\begin{array}{l|r|l|r}
 \text{Permutation} & \text{#} & \text{Permutation} & \text{#} \\
123 & 0 & 132 & 1 \\
213 & 1 & 231 & 1 \\
312 & 1 & 321 & 2
\end{array}
$

$
\begin{array}{l|r|l|r|l|r}
 \text{Permutation} & \text{#} & \text{Permutation} & \text{#} & \text{Permutation} & \text{#}  \\
1234 & 0 & 1243 & 1 & 1324 & 1 \\
1342 & 1 & 1432 & 2 & 1423 & 1 \\ \hline
2134 & 1 & 2143 & 2 & 2314 & 1 \\
2341 & 1 & 2413 & 1 & 2431 & 2 \\ \hline
3124 & 1 & 3142 & 2 & 3214 & 2 \\
3241 & 2 & 3412 & 1 & 3421 & 2 \\ \hline
4123 & 1 & 4132 & 2 & 4231 & 2 \\
4213 & 2 & 4312 & 2 & 4321 & 3
\end{array}
$

I played around with this for quite a while, but looking for example at

$\begin{array}{lllll}
A(4,1) & = 1  \cdot A(3,1) & + 4 \cdot A(3,0) & =  1 \cdot 1  + 4 \cdot 0 & = 1 \\
A(4,2) & = 2 \cdot A(3,2) & + 3 \cdot A(3,1) & = 2 \cdot 4 + 3 \cdot 1 & = 11 \\
A(4,3) & = 3 \cdot A(3,3) & + 2 \cdot A(3,2) & = 3 \cdot 1 + 2 \cdot 4 & = 11 \\
A(4,4) & = 4 \cdot A(3,4) & + 1 \cdot A(3,3) & = 4 \cdot 0 + 1 \cdot 1 & = 1
\end{array}$

didn't make me see anything that helped me understanding and proving that recursive definition.

Could you please help me a bit?

Thanks in advance!

答:
The first summand comes from the following construction: take a $n-1$-permutation with $k-1$ decreases; now you can add a new element, $n$ (which is the largest) without inducing new decreases by either entering it directly before any given decrease ($k-1$ places) or the end (1 additional place).

The second summand is similar, only now we consider the places where adding $n$ will create a new decrease (namely, exactly those where a decrease didn't happen before, and not the end).


问:
zeros of exponential polynomials
Let $\exp[n;z]$ denote the $n$th Taylor polynomial for the exponential function.

In the 1920's Szegő initiated the study of the asymptotic properties of the zeros (rescaled by dividing by $n$) of this family of polynomials and one consequence of his results is that they can approach arbitrarily closely to the imaginary
axis.  This prompts the following question:

Is it possible for $\exp[n;z]$ to have a root which lies precisely on the imaginary axis?

答:
This is equivalent to asking if there is a simultaneous real zero of the two polynomials $\cos[n,z]$ and $\sin[n,z]$. But for any $n$, one of these two polynomials is the derivative of the other, so they are only simultaneously zero at a repeated root of the higher-degree one.

So the question is equivalent to asking if the Taylor polynomial centered at 0 of $\sin$ or $\cos$ ever has a repeated real root.

Edit: Apparently this blog has been following our discussion and states that their are no repeated roots.

Title: zeros of exponential polynomials


问:
Calculate the measure in degrees of the acute angle formed by $ NMA $?
Consider the $PAT$ triangle such that the angle $ \angle P = 36 $ degrees, $ \angle A = 56 $ degrees and $ PA = 10 $. Knowing that the points $ U $ and $ G $ belong, respectively, to the sides $ TP $ and $ TA $, so that $ PU = AG = 1 $. If $ M $ and $ N $ are the midpoints of the $ PA $ and $ UG $ segments, respectively. What is the measure in degree of the acute angle formed by $ NMA $?

Can anyone give me any tips?

答:
Connect $MN$ and $MU$. Rotate $\bigtriangleup MPU$ $\;180°$ about $M$ and name the exterior point $U'$.

Connect $U'G$. Observe that $\bigtriangleup AGU'$ is isosceles and $U'G \; \parallel MN$.

The rest is simple angle chase.


问:
Convex Hull of Unit Circles
I know that if we're trying to get the convex hull of $n$ unit circles, we can simply shrink the circles down to their centers and consider the convex hull of their centers, but I'm trying to prove some intermediate steps towards that.

a) Show that boundary of the convex hull consists of only straight line segments and parts of circles.

b) Show that each circle an appear at most once in the boundary of the convex hull.

(This is from de Berg's Computational Geometry book.)

I sort of have an intuition of why these are true, but my problem is that whenever I try to come up with a solution, I end up considering lots of cases, and I feel like it's not rigorous or elegant enough because of so many cases. Is there a neat way to prove these?

(Note: I know that this has been posted before in Convex Hull Algorithms, but I'm not satisfied with the answers there, and I'm trying to go for something more rigorous.)

答:
just a clarifying comment.
a) the boundary of the convex hull of a family K of unit circles is the union of common outer tangential segments to pairs of circles and circular arcs connecting them
b*) the supporting lines to K are tangents to a circle with a point on the boundary of K or a line containing a common outer tangential segment of two circles of K

Title: Convex Hull of Unit Circles


问:
Show that $\|x\| = \sum_{n=1}^{\infty}2^{-n}(\min\{|x_n|,1\})$ satisfies the triangle inequality

Let $$\|x\| = \sum_{n=1}^{\infty}2^{-n}(\min\{|x_n|,1\})$$ where $x=(x_1,\dots) \in A$. And  $A$ is the vector space formed by $F(\mathbb{N}, \mathbb{R}).$ Show that $\|x\|$ satsifies the triangle inequality,

I'm trying to determine wheter this sum satisfies the triangle inequality, but the minimum function inside the sum is giving me hard time I have that $$\|x+y\| = 2^{-1}(\min\{|x_1+y_1|,1\}) + 2^{-2}(\min\{|x_2+y_2|,1\}) + \dots + 2^{-n}(\min\{|x_n+y_n|,1\})$$ but I'm not sure how to go further. How can I bound this?

答: First of all, note that the map $\eta(x)=\min\{|x|,1\}$  satisfies the triangle inequality. In fact,\begin{align}\eta(x+y)&=\min\{|x+y|,1\}\\&\leqslant\min\{|x|+|y|,1\}\\&\leqslant\min\{|x|,1\}+\min\{|y|,1\}.\end{align}But then $2^{-n}\eta$ also staisfies the triangle inequality. So, since your function is a sum of functions for which the triangle inequality holds, then it also holds for your function.


问:
Find the maximum rate change.
Find all points from the domain of
$$
f(x,y)=e^{x^2-xy-1}
$$
in which the function f reaches the maximum rate change (I mean gain/increase) in the direction of x-axis

the domain: (am I right?)
$$
x \in R, y \in R
$$
so the domain is real numbers>0.

I thought about directional derivative and it's property:
$$
[e^{x^2-xy-1}(2x-y), e^{x^2-xy-1}(-x)]*[cos\alpha, cos\beta]
$$
where
$$
cos\alpha=1, cos\beta=0
$$
so it's:
$$
e^{x^2-xy-1}(2x-y)
$$
and what should I do now?

答: You now have a function (call this g(x,y)) that gives the rate of change of f in the direction of the x-axis for every point (x,y). The question remains when is g(x,y) a maximum? i.e. when is the rate of change of f in the direction of the x-axis a maximum?</p>

Title: Find the maximum rate change.


问:
Guaranteeing that three straight lines form a triangle in the third quadrant
The points of intersection of the following three straight lines form a triangle with all of the vertices in the third quadrant.

$x-2y=1$

$y-4x=3$

$3y+2x=-19$

How can we guarantee that any three lines will intersect to form a triangle with all of its vertices in the third quadrant?

I have explored looking at the gradients, $x$-intercepts and $y$-intercepts, as well as looking at the forming the equations of the lines given the points, but I'm wondering if there is an easier way.

答:
Choose minimum  values ( say -6 each ) for coordinates  $x,y$  ( domains/range) as minimum to include all desired vertices.

$$ 0<x<x_{min}  \text{; and } 0<y<y_{min}; $$

All coordinates points $x,y$ .should be non-positive.


问:
Value of $a+a^a+a^{a^a}+a^{a^{a^a}}+... $
Is there any closed form expression of $a+a^a+a^{a^a}+a^{a^{a^a}}+...$ up to $n$ terms? I am mostly interested in $a=2$

My try:

$a+a^a+a^{a^a}+a^{a^{a^a}}+... =a(1+a^{a-1}+a^{{a^a}-1}+a^{{a^{a^a}}-1}+... )$, but I do not know how to continue.

答:
With $a=2$, the first terms make the sequence
$$\{2,6,22,65558,\cdots\}$$ The problem is that the next one is
$$\sim 2.0035\times 10^{19728}$$

Notice that, just by itself,
$$2^{65558} \sim 8.4034\times 10^{19724}$$


问:
Finding probabilities given continous random variable
Given the random variable $Y_1 \sim \mathcal{N}(μ, σ^2 )$

Where $μ=350$ and $σ^2 =12365$

How can one calculate probability that the random variable $Y_1$ is max 275

Likewise how would you find the probability that the random variable $Y_1$ is at least 425 ?

Im not sure how one would solve this problem using the Φ function

答:
$$P(Y_1\leq 275)=\Phi\left(\frac{275-350}{\sqrt{12365}}\right)\approx 0.25$$

and similarly the second question.


问:
How to prove that $\vdash \neg \neg (p \vee \neg p)$? (Natural Deduction)
I do know that double negation and LEM are equivalent, but can we prove
$$\vdash \neg \neg (p \vee \neg p)$$
without using either of them, in a Fitch-style proof?

答:
Yes, we can.

Hint: assume $\lnot (p \lor \lnot p)$ and derive a contradiction.

The conclusion follows by $(\lnot \text I)$, which is intuitionistically valid.


问:
Hpergeometric Reduction with Mathematica
We know that:

$$\,_2F_1(a,b,b,z)=(1-z)^{-a}$$

But putting $b<0$ with $b$ integer Mathematica does not use the previous formula but generates the hypergeometric polynomial. For instance:

$$\,_2F_1(a,-2,-2,z)= \frac{1}{2} (2 + 2 a z + a z^2 + a^2 z^2)$$

I don't understand why this happens because the 2 pochammer should cancel in the hypergeometric definition also if they are negative. Any Idea?

答:
This comes down to how Mathematica interprets these types of special cases for the Hypergeometric function. Specifically, it is using this definition so that your hypergeometric function is interpreted as
$$
\begin{aligned}
F(a,-2;-2;z)
&=\lim_{c\to-2}\lim_{b\to-2}F(a,b;c;z)\\
&=\lim_{c\to-2}\frac{a^2 z^2-2 a c z+a z^2-2 a z+c^2+c}{c (c+1)}\\
&=\frac{1}{2} (2 + 2 a z + a z^2 + a^2 z^2)\\
&=1 +a z +\frac{a(1+a)z^2}{2}\\
&=\sum_{k=0}^2 \frac{(a)_k}{k!}z^k.
\end{aligned}
$$

Here the Mathematica code: Limit[Hypergeometric2F1[a, -2, c, z], c -> -2].

So we see that the $(-2)_k$ terms do cancel out; however, the series also truncates to only the first three terms and thus does not gives the binomial result you expected.


问:
Preservation of closed sets under linear transformation.
Let $f:\mathbb{R^n} \rightarrow \mathbb{R^m}$ and let $A$ be a closed set in $\mathbb{R^n}$. I would like to know if $f(A)$ is a closed set.

I know this question is pretty much the same as this one: A linear transform of a closed set is closed. But the answer given in that question is wrong as the image of $E$ given is said to not be closed when I would claim that it actually is closed. Just take the compliment of $f(E)$ and find that it is open.

As for the correct answer, my intuition would tell me that linear transformations do in fact preserve closedness. To be clear: I am looking for a better counterexample to disprove this fact or indeed a proof for it.

答:
For posterity, a necessary and sufficient when $A$ is an arbitrary subset of $\mathbb{R}^n$ and $f: \mathbb{R}^n \to \mathbb{R}^m$ is linear, is:
$$
f(A) \textrm{ is closed} \iff A + \textrm{ker}(f) \textrm{ is closed.}
$$
This is a corollary of the infinite dimensional version which can be found, for example, in Holmes' Geometric Functional Analysis and Its Applications (Lemma 17.H).


问:
How Do You Expand and Simplify?
How do you expand and simplify $(x + 2y)^2$ or $(7x + 5y)^2$. I don't understand, how would you multiply that by itself?

I mean I can simplify and expand two brackets; e.g. $(x + 3) (x + 5)$ but I'm not sure how to do one bracket with a squared.

Title: How Do You Expand and Simplify?

Body:
Another example:

$$(7x + 5y)^2 \;\; = \;\;(7x + 5y)(7x + 5y)$$

So just expand (distribute) as you normally would:

$$\begin{align}(7x + 5y)^2 & = (7x + 5y)(7x + 5y)\\ \\
& = 7x(7x + 5y) + 5y(7x + 5y) \\ \\
& = 49x^2 + 35 xy + 35 xy + 25 y^2 \\ \\
& = 49x^2 + 70 xy + 25 y^2
\end{align}$$


问:
How can I find all integers $x≠3$ such that $x−3|x^3−3$
How can I find all integers $x≠3$ such that $x−3|x^3−3$?

I tried expand $x^3−3$ as a sum but I couldn't find a way after that.

Body:
$$x - 3 | x^3 - 3$$
$$\frac{x^3 - 3}{x - 3} = \frac{x^3 - 3 + 27 - 27}{x - 3}$$

$$ = \frac{x^3 - 27 + 24}{x - 3}$$

$$ = \frac{x^3 - 3^3 + 24}{x - 3}$$

$$ = \frac{(x - 3) \times (x^2 + 3x + 9) + 24}{x - 3}$$

$$ = x^2 + 3x + 9 + \frac{24}{x - 3}$$

The result is integer when $x - 3$ is a factor of $24$ and $x \ne 3$


问:
What is a transcedental Equation?
I was studying equations, and came across the term Transcedental being use for equations of the form,
$$ \sin(x) -e^{x} + x^{2 }= 0 $$
From what I understand, the equations involving terms like exp, sin are called transcedental. I also googled and found that  equations of the form  $\space a_{n}x^{n} + a_{n-1}x^{n-1} ... + a_{0} = 0 \space$ (the lhs is a polynomial) are called algebraic.
Finally, then what are equations of the form $ \space x^{\frac{3}{2}} - 2x + x^{\frac{5}{3}} = 0 \space $ called?

NOTE: These are just random examples.

答: A transcendental equation is an equation that is not algebraic.  The equation $x^{\frac{3}{2}} - 2x + x^{\frac{5}{3}} = 0$ is algebraic, since its solutions are solutions of some polynomial equation (which takes some manipulation to find).

Title: What is a transcedental Equation?


问:
Proving positive definiteness for a symmetric matrix
Let A be an $n\times p$ matrix with rank $p$. I want to prove that $X = A^TA$ is a positive definite matrix.

I know that for any symmetric matrix to be positive definite, all of its eigenvalues must be strictly positive.

So I need a way to show that $X$'s eigenvalues are all positive, or perhaps any other method to proof that $X$ is positive definite.

答:
As suggested above by phdmba7of12, we consider the quadratic form

$$x^T X x = x^T (A^T A)x = (Ax)^T (Ax) = (Ax, Ax) = \lVert Ax \rVert^2 \geq 0.$$

Since $A$ has rank $p$, the only solution of $Ax = 0$ is the trivial solution. Hence,

$$x^T X x = 0 \iff x = 0.$$

Hence, we have varified the definition of positive definiteness.


问:
Simple non-closed geodesic.
In torus there exists simple non-closed geodesic. One example is take the irrational slope curves in $\mathbb{R}^2$ and project it down to torus. Is this thing can happen in closed hyperbolic surface i.e.

Does there exists a simple non-closed geodesic in a closed (i.e. compact without boundary) hyperbolic surface?

答:
Let's call $\Sigma$ a closed hyperbolic surface. Each free homotopy class of curves on a hyperbolic surface has exactly one geodesic representative. Conversely, a closed geodesic gives a free homotopy class. Thus (up to basepoint) there is a bijective correspondence
$$\bigg\{ \mbox{ closed geodesics }\bigg\} \longleftrightarrow \pi_1(\Sigma).$$

Since $\pi_1(\Sigma)$ is finitely generated, it is a countable set. In particular, it lifts to a countable set of geodesics in $\mathbb{H}^2$. Any and every geodesic in $\Sigma$ can be obtained by pushing a geodesic of $\mathbb{H}^2$ forward under the covering map, so the set of all geodesics of $\Sigma$ lifts to the set of geodesics of $\mathbb{H}^2$, which is uncountable. So there are many more open geodesics than closed geodesics on $\Sigma$.

Here's a simple example on a non-closed hyperbolic cylinder that illustrates the phenomenon. Let $\Sigma = \mathbb{H}^2/\mathbb{Z}$, where the $\mathbb{Z}$ action is given by $z\mapsto \lambda z$, for some $\lambda > 1$. The single closed geodesic is the image (under quotient) of the $y$-axis. Take any other geodesic which limits to $0$ and push it down under the covering map; it will spiral in from one side of the cylinder and limit to the closed geodesic.

Title: Simple non-closed geodesic.


问:
What do vectors over a ring look like?
I read the following on this Wikipedia page:

Let ${\displaystyle \mathbb {Z} _{q}}$ denote the ring of integers modulo ${\displaystyle q}$ and let ${\displaystyle \mathbb {Z} _{q}^{n}}$ denote the set of ${\displaystyle n}$-vectors over${\displaystyle \mathbb {Z} _{q}}$.

What I don't get (being a noob) is what the members of ${\displaystyle \mathbb {Z} _{q}^{n}}$ might be like:

Say ${\displaystyle q}$ is $5$, then will the members of ${\displaystyle \mathbb {Z} _{q}^{n}}$  be of the form such that for every $x \in \displaystyle \mathbb {Z} _{q}^{n}$, there is a one-to-one correspondence to a member in $\displaystyle \mathbb {Z} _{q}$, for example a binary encoding of the integers $0, 1,... 4$? Or does it simply mean that the $n$-vector members are such that every element in the vector is in ${\displaystyle \mathbb {Z} _{q}}$.

答:
Your last sentence is the right interpretation.

$$\mathbb{Z}_p^n = \mathbb{Z}_p \times \cdots \times \mathbb{Z}_p  = \{(x_1 , \ldots, x_n) \, | \, x_1, \ldots x_n \in \mathbb{Z}_p \}$$

$\mathbb{Z}_p^n$ may be viewed as a ring with the usual addition and multiplication $\mod p$ defined componentwisely.

Example: Two elements in $\mathbb{Z}_5^3$ are $(1, 2, 3)$ and $(2, 4, 3)$. We have

$(1, 2, 3) + (2, 4, 3) = (3, 1, 1) \quad \text{and} \quad (1, 2, 3)(2, 4, 3) = (2, 3, 4)$

Title: What do vectors over a ring look like?


问:
why is the gradient vector to rotated ellipse wrong?
A formula for the rotated ellipse can be found by taking the standard ellipse
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad\qquad\qquad\qquad {\bf{x}}(t) = (a\cos t, b\sin t) \qquad \text{ for } t\in [0, 2\pi]
$$
and rotating it counter-clockwise using a rotation matrix
$$
R{\bf{x}}(t) =
\begin{pmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
    a\cos t \\
    b\sin t
\end{pmatrix}
=
\begin{pmatrix}
    a\cos t\cos\theta - b\sin t\sin \theta\\
    a\cos t\sin \theta + b \sin t \cos \theta
\end{pmatrix}
=
{\bf{x}}_r(t)
$$
I can easily find the gradient of the standard ellipse (non-rotated) by taking the derivative of the first formula.
$$
g =
\begin{pmatrix}
    \frac{2x}{a^2},  &
    \frac{2y}{b^2}
\end{pmatrix}^\top
$$

Now I tried to methods to compute the gradient at a point on the rotated ellipse.

Method 1: I take the gradient of
$$
\frac{(x\cos\theta - y\sin\theta)^2}{a^2} + \frac{(x\sin\theta + y\cos\theta)^2}{b^2} = 1
$$
obtaining
$$
g_{r1} =
\begin{pmatrix}
    \frac{2(x\cos\theta - y\sin\theta)\cos\theta}{a^2} + \frac{2(x\sin\theta + y\cos\theta)\sin\theta}{b^2} \\
    \frac{-2(x\cos\theta - y\sin\theta)\sin\theta}{a^2} + \frac{2(x\sin\theta + y\cos\theta)\cos\theta}{b^2}
\end{pmatrix}
$$

Method 2: Rotating the gradient of the non-rotated ellipse counterclockwise
$$
g_{r2} = \begin{pmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
    \frac{2x}{a^2} \\
    \frac{2y}{b^2}
\end{pmatrix}
 =
\begin{pmatrix}
    \frac{2x\cos\theta}{a^2} - \frac{2y\sin\theta}{b^2}\\
    \frac{2x\sin\theta}{a^2} + \frac{2y\cos\theta}{b^2}
\end{pmatrix}
$$

To me, method 1 should work while method 2 should be wrong. However, as you can see in the plot below, $g_{r1}$ seems to actually be totally wrong while $g_{r2}$ is right. How come $g_{r2}$ seems to be right but not $g_{r1}$?

答:
Method 1 is wrong because the equation
$$
\frac{(x\cos\theta - y\sin\theta)^2}{a^2} + \frac{(x\sin\theta + y\cos\theta)^2}{b^2} = 1
$$

expresses the fact that the result of rotating the point $(x,y)$ belongs to the ellipse $x^2/a^2+y^2/b^2=1$, but this is not what you need.


问:
Roots of polynomials that are the product of 2 polynomials
Given $p, q  \in \mathbb{F}[X]$, $deg(p) \land deg(q) \geq 1$ with no roots in $\mathbb{F}$. How could we show that $p.q$ has no roots as well?
Any hint would be appreciated.

答: The magic words are "division with remainder". If $a$ is a root of the product, then $(x-a)$ divides $p q,$ but we know that $x-a$ does not divide either $p$ or $q.$ So $p = s (x-a) + r_1, q = t (x-a) + r_2,$ $pq = (x-a) v +. r_1 r_2.$ $r_1 r_2 \neq 0$ (since $\mathbb{K}$ is a field and has no zero divisors.)


问:
Find the orthogonal proyection of $x^4$.
A doubt.  $V=Span(1,x,x^3)$ generates $L^2([0,1])$?

I ask it because I am trying to find an orthonormal basis from V and then calculate the orthogonal projection of x^4 but I got that doubt.

I reached the next orthonormal set

$V'=Span(v_1,v_2,v_3)$ with $v_1=1$, $v_2=(x-1/2)/|v_2|_2$ and $v_3=(x^3-1/4-1/10(x-1/2))/|v_3|_2 $ and like uses that $x^4=\sum_{k=1}^{3} a_kv_k$ for find the orthogonal proyection of $x^4$ in $V'$...but I am somewhat lost with what arose in doubt

Thanks.

答:
No. The span of $\{1,x,x^2\}$ does not span $L^2([0,1])$, because it is is infinite dimensional, so it cannot be spanned by only three vectors.
The main claim of this answer is that $\{1,x,x^2\}$ does not span $L^2([0,1])$, and not as may be suggested otherwise by some comments.


问:
Rudin's proof that the Cantor set has no segments
In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(\alpha,\beta)$ contains a segment of the form $(\frac{3k+1}{3^{n}},\frac{3k+2}{3^{n}})$ if $3^{-n}\lt\frac{\beta-\alpha}{6}$. I don't know how $\frac{\beta-\alpha}{6}$ is obtained.

答:
The condition $3^{-m} < \frac{\beta - \alpha}{6}$ is equivalent to  $3^{-(m-1)} < \frac{\beta - \alpha}{2}$.

Now if $3^{-(m-1)} < \frac{\beta - \alpha}{2}$, then $[\alpha, \alpha+\frac{\beta - \alpha}{2}]$ contains $\frac{k}{3^{m-1}}$ for some integer $m\geq 2$ and $k\geq 0$.

We can conclude that $[\alpha, \beta] \supset [\frac{k}{3^{m-1}}, \frac{k+1}{3^{m-1}}]$, which is equivalent to $[\frac{3k}{3^{m}}, \frac{3k+3}{3^{m}}]$. Then clearly $(\alpha, \beta) \supset (\frac{3k+1}{3^{m}}, \frac{3k+2}{3^{m}})$.

What I have shown is that the condition you asked, $3^{-m} < \frac{\beta - \alpha}{6}$, is sufficient to conclude that every segment $(\alpha, \beta)$ contains a segment of the form $(\frac{3k+1}{3^{m}}, \frac{3k+2}{3^{m}})$.


问:
Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$

Let $x,y,z\in \mathbb R^+$ prove that:

$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz \le RHS$. How to prove above inequality ? It is too hard for me.

答:
$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{AM-GM}{\ge} 3xyz$$

DOES help, since you can also prove:

$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

Note that this is the same as

$$2^6(x^2y+ y^2z+z^2x)^3(y^2x+z^2y+x^2z)^3 \geq 3^6x^2y^2z^2(x^2+yz)^2(y^2+xz)^2(z^2+xy)^2 \,.$$

which after long computations and many tears can be reduced to this all junk being positive. If you look to the few negative terms , you can get easely rid of them by invoking the AM-GM inequality.

Now, all this suggests that the inequality

$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

is true, all you have to do is find a neater solution :)


问:
Expressions $x^a+1$ and $x^b+1$, when $a$, $b$ are odd positive integers
Let $a\geq 3$ and $b\geq 3$ be odd integers. Let $x\geq 2$ be an integer.  I would like to know if we can state that $x^a+1|x^b+1 \Longleftrightarrow a|b$. I know how to prove that $a|b$ implies $x^a+1|x^b+1$. Is the other implication also true? I'm especially interested in the case when $x$ is a prime number, but I don't think that this additional assumption matters so much. I've checked the result for particular primes $x$ and it seems that it holds. Still, I would like to know how to prove it or, if it isn't true, which would be a counterexample?

There is a similar result which states: $x^a-1|x^b-1 \Longleftrightarrow a|b$. By following the proof for this similar result, I've tried to apply the same argument (the quotient remainder theorem) to show the implication above, but I did not manage to finish the proof. Any hints are appreciated.

答:
Suppose $x^a+1\mid x^b+1$ (where $x$ is either an unknown and we work in polynomials $\Bbb Z[x]$, or $x$ is an integer $\ge 2$ and we work in $\Bbb Z$ itself).

Suppose $x^a+1\mid x^b+1$ where $a,b$ are odd positive integers and $x$ is an integer $\ge 2$.
Then necessarily $b\ge a$ and also
$$ x^a+1\mid (x^a+1)x^{b-a}-(x^b+1)=x^{b-a}-1.$$
If $b\le 2a$, the right hand side is $<x^a+1$ hence must be $=0$, i.e. $a=b$ and of course $a\mid b$.
So far, we have proved $a\mid b$  for $b\le 2a$. Now we can treat the rest by induction on $b$ because we have
$$ x^a+1\mid (x^b+1)-(x^a+1)(x^a-1)x^{b-2a}=x^{b-2a}+1$$
where $b-2a$ is an odd positive integer $<a$ so that by induction hypotheseis, $a\mid b-2a$, which then implies $a\mid b$.


问:
How many subsets of S contains A?
Given a set $S$ of size $n$ and $A \subseteq S$ of size $k$, how many subsets of $S$ contain $A$?

I think it should be $2^{n} - 2^{n-k}$ ? Please tell is it correct or not ?

答: The answer is: the number of subsets of $S\backslash A.$ You can figure out the number from this.

Title: How many subsets of S contains A?


问:
Find the maximum rate change.
Find all points from the domain of
$$
f(x,y)=e^{x^2-xy-1}
$$
in which the function f reaches the maximum rate change (I mean gain/increase) in the direction of x-axis

the domain: (am I right?)
$$
x \in R, y \in R
$$
so the domain is real numbers>0.

I thought about directional derivative and it's property:
$$
[e^{x^2-xy-1}(2x-y), e^{x^2-xy-1}(-x)]*[cos\alpha, cos\beta]
$$
where
$$
cos\alpha=1, cos\beta=0
$$
so it's:
$$
e^{x^2-xy-1}(2x-y)
$$
and what should I do now?

答: In general you should take the derivative and set it equal to zero to find extrema, to find out if it is a (local) minimum or maximum you should take a look at the second derivative.</p>

Title: Find the maximum rate change.


问:
Sequence involving $\sqrt[n]{n}$ and $\text{log}$
I know that the sequence $\sqrt[n]{n}$ converges to 1 and that $\text{log}(\sqrt[n]{n})$ thus converges to 0 as $n\to\infty$ since the logarithmic function is continuous. But how can I calculate the limits as $n\to\infty$ of the following sequences?

(a) $\frac{\sqrt[n]{n^2}-1}{\text{log}(\sqrt[n]{n})}$

(b) $\frac{\sqrt[n]{1/n}-1}{\text{log}(\sqrt[n]{n})}$

Any help would be much appreciated!

答:
(a) Write this quotient as

$$\frac{e^{(1/n)\ln n^2}-1}{(1/n)\ln n} = \frac{e^{2(1/n)\ln n}-1}{(1/n)\ln n}.$$

As $n\to \infty,$ $(1/n)\ln n\to 0.$ So the limit of the above has the form

$$\lim_{u\to 0}\frac{e^{2u}-1}{u}.$$

That should look familiar.


问:
Disc vs Shell Method, getting different answers AP calc
Can someone please check my work.

$R$ is the region in the first quadrant bounded by $y=1/x$, $y=1$ and $x=e$

Find the volume of the solid generated when $R$ is revolved about the line $y=1$

Disk:

$$V= \int_{1}^{e} \pi \left(1-\frac{1}{x}\right)^2 \,dx=\pi \left(e-\frac{1}{e}-2\right)$$

Shell:

\begin{align*}
V&=-2\pi \int_{1/e}^{1} -\left(1-\frac{1}{y}\right)\left(e-\frac{1}{y}\right)\, dy\\[5pt]
&=-2\pi\left[ey-(1+e)\ln(y)-\frac{1}{y}\right]\Bigg|_{1/e}^{1}\\[5pt]
&=-2\pi\bigl[(e-(1+e)\cdot 0-1)-(1-(1+e)(-1)-e)\bigr]\\[5pt]
&=-2\pi\bigl[(e-1)-2\bigr]\\[5pt]
&=-2\pi(e-3)
\end{align*}

答:
Your working for the disk method is correct. In shell method, the integral should be,

$\displaystyle 2 \pi \int_{1/e}^{1}  \color {blue} {(1-y)} (e-\frac{1}{y}) \ dy$

The mistake that you made was in the radial distance of the shell from $y=1$. Instead of $(1 - \frac{1}{y})$, it should be $(1-y)$ as the shell is at distance $y$ from the x-axis.


问:
Logarithm of a matrix with zero diagonal elements.
I want to calculate the natural logarithm of this matrix$$A=\begin{bmatrix}0&0&0&0\\0&\frac{1}{2}&\frac{i}{2}&0\\0&-\frac{i}{2}&\frac{1}{2}&0\\0&0&0&0\end{bmatrix}$$

After calculating the eigenvalues and eigenvectors I find: $\lambda_1=1,\lambda_2=0,\lambda_3=0,\lambda_4=0$. These values are the diagonal elements in  D of $A=M D M^{-1}$. Here $M$ is the modal matrix and $D$ the diagonal matrix looking like:$$D=diag(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$$

Now taking the natural logarithm of the matrix would require me to take the natural logarithm of the elements of $D$. So what is my result? I should just say that three of the elements are undefined or that the logarithm of this matrix doesn't exists/diverges? Is there some pre requirement for A that I did not take into account?

答:
There is no answer in real or complex matrices.

However you can get answer depending on what extension of complex numbers you allow.

On the projective complex numbers the answer is

$$\left(
\begin{array}{cccc}
 \infty  & 0 & 0 & 0 \\
 0 &\infty  &  \infty  & 0 \\
 0 &  \infty  & \infty  & 0 \\
 0 & 0 & 0 & \infty  \\
\end{array}
\right)$$

On oriented projective complex numbers the answer is

$$\left(
\begin{array}{cccc}
 -\infty  & 0 & 0 & 0 \\
 0 & -\infty  & i \infty  & 0 \\
 0 & -i \infty  & -\infty  & 0 \\
 0 & 0 & 0 & -\infty  \\
\end{array}
\right)$$

In terms of divergent integrals, the answer is

$$\left(
\begin{array}{cccc}
 -\int_0^1 \frac{dx}x  & 0 & 0 & 0 \\
 0 & -\int_0^1 \frac{dx}x  & i \int_0^1 \frac{dx}x & 0 \\
 0 & -i \int_0^1 \frac{dx}x  & -\int_0^1 \frac{dx}x  & 0 \\
 0 & 0 & 0 & -\int_0^1 \frac{dx}x  \\
\end{array}
\right)$$


问:
Every finite group $G$ such that $p\mid |G|$ has a maximal p-subgroup
I'm reading the book "An Introduction to the Theory of Groups" by Joseph Rotman. Maximal $p$-subgroup is defined as: $P$ is a maximal $p$-subgroup of a group $G$ if for every $Q\le G$, $Q$ is a $p$-subgroup and $P\le Q$, then $P=Q$.

1º Wouldn't a subgroup, $M$, of order $m>1$ such that $\gcd(m,p) = 1$ be a maximal $p$-subgroup, since there is no $p$-subgroup which contains $M$ as subgroup?

The author then proceeds to prove the Sylow Theorems, in the proof he considers the set $X=\{P_1,P_2...,P_r\}$ the family of all conjugates of $P$, a maximal $p$-subgroup.

2ºHow can we be sure that this set is not empty? In other words how can we be sure that every finite group $G$, such that $p\mid |G|$ has a maximal $p$-subgroup?

答:
No, because $M$ is not a $p$-subgroup.

By Cauchy theorem, $G$ has an element of order $p$, hence a cyclic sugroup of order $p$. Thus, the set of $p$-subgroups of $G$ is not empty. Take any $p$-subgroup $P$ of $G$ with maximal order (which does exist since $G$ is finite. Note that such a $P$ is not necessarily unique). Now, if $Q$ is a $p$-subgroup containing $P$, $\vert P\vert\leq\vert Q\vert $. By maximality of the order of $P$ among the orders of $p$-subgroups, $\vert Q\vert\leq\vert P\vert.$ Hence $P$ and $Q$ have same order, and since $Q$ contains $P$, $Q=P$.  Hence, $P$ is a maximal $p$-subgroup.


问:
A question about the statement of Bertini's theorem.
Bertini's theorem states that if $X$ is a closed non-singular subvariety of $\mathbb{P}^n$ over an algebraically closed field, there exists some hyperplane $H$ not containing $X$ such that $H\cap X$ is non-singular.

My question is this: does $H\cap X$ mean the scheme-theoretic intersection or the reduced closed sub-scheme structure on $H\cap X$?

答:
This is a community-wiki post recording the answer from the comments so that this question may be marked as answered.

Scheme theoretic. – Mohan


问:
integration by parts for numerical definite integrals
I am new to calculus, and I am trying to learn integration by parts.
So I have got a question here, and I have no idea what I am doing wrong (i have been looking at it for about 4 hours now..)

The method as I understand it, is as follows: $\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x$

Now i have a definite integral, evaluated from 0 to 2, and which looks as follows:

$\int 0.5 x \cdot 0.375 x^{2} d x$

I know the answer to be: $\left.0.046875 x^{4}\right|_{0} ^{2}=0.75$, but nomatter how i twist and turn this thing, my answer is wrong.

My last attemt looks like this:

$\begin{array}{l}
f(x)=0.375 x^{2} \\
f^{\prime}(x)=0.75 x \\
g(x)=0.25 x^{2} \\
g^{\prime}(x)=0.5 x
\end{array}$

$0.375 x^{2} \cdot 0.25 x^{2}-\int 0.75 x \cdot 0.25 x^{2}$

$0.375 x^{2} \cdot 0.25 x^{2}- \frac{0.75 x^{2}}{2} \cdot \frac{0.25 x^{3}}{3}$

Which, evaluated at 0 and 2, comes out to be 0.5.

Where am i going wrong?

答:
You evaluated the integral
$$\int 0.75 x \cdot 0.25 x^{2}=\frac{0.75 x^{2}}{2} \cdot \frac{0.25 x^{3}}{3}.$$
This is wrong. The right solution is
$$\frac{3x^4}{64}\: (+C).$$

Check:

Here $$0.375 x^{2} \cdot 0.25 x^{2},$$
if $x=2$ is equal to $0.75$ and equal to $0$ if $x=0$. So the final solution is
$$0.75-0=0.75.$$


答:
I'm assuming you were told to use integration by parts here, because you showed that you know the answer via just multiplying the integrand. The error occurs when you take the integral $\int 0.75x \cdot 0.25x^2$.

You can't integrate them separately as you do.

Usually when you're doing integration by parts, your goal is to pick $f$ and $g'$ so that when you apply integration by parts $\int f' g$ becomes simpler. So in your case, you would probably want to chose $f = .5x$ so that when you apply the integration by parts formula $\int f' g $ is very simple because only $g$ still contains a term with $x$ in it. Try this out and see if it makes the problem easier!

This is suggestive of how you apply integration by parts in more complicated examples where it is required. For example $\int x e^x dx$. Here you wouldn't want to choose $f = e^x$ because this doesn't get any simpler when you differentiate.

Title: integration by parts for numerical definite integrals

Body:
I am new to calculus, and I am trying to learn integration by parts.
So I have got a question here, and I have no idea what I am doing wrong (i have been looking at it for about 4 hours now..)

The method as I understand it, is as follows: $\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x$

Now i have a definite integral, evaluated from 0 to 2, and which looks as follows:

$\int 0.5 x \cdot 0.375 x^{2} d x$

I know the answer to be: $\left.0.046875 x^{4}\right|_{0} ^{2}=0.75$, but nomatter how i twist and turn this thing, my answer is wrong.

My last attemt looks like this:

$\begin{array}{l}
f(x)=0.375 x^{2} \\
f^{\prime}(x)=0.75 x \\
g(x)=0.25 x^{2} \\
g^{\prime}(x)=0.5 x
\end{array}$

$0.375 x^{2} \cdot 0.25 x^{2}-\int 0.75 x \cdot 0.25 x^{2}$

$0.375 x^{2} \cdot 0.25 x^{2}- \frac{0.75 x^{2}}{2} \cdot \frac{0.25 x^{3}}{3}$

Which, evaluated at 0 and 2, comes out to be 0.5.

Where am i going wrong?


问:
Unconstrained integer quadratic programming
What are the approaches for finding the global optimum of an unconstrained quadratic optimization problem where all variables are integers? When the integrality constraint is relaxed, the problem is convex. The variables can be binary as well.</p>

答: For unconstrained integer convex optimization, you can solve the continuous relaxation by any method, and an integer optimal solution is obtained by rounding each component up or down.  Explicitly, if the continuous optimal solution is $(x_1^*, \dots,x_n^*)$, an integer optimal solution is among the $$\prod_{j=1}^n (\lceil x_j^* \rceil - \lfloor x_j^* \rfloor + 1)\le 2^n$$ points of the form $(y_1^*, \dots, y_n^*)$, where $y_j^* \in \{\lfloor x_j^* \rfloor, \lceil x_j^* \rceil\}$.


问:
Bertanlaffy growth - differential equation
Suppose a tumor of mass $m$ satisfies the differential equation $\frac{dm}{dt}=am^{\frac{3}{4}}-bm-cm^{\frac{2}{3}}$, where $c$ increases slowly from zero. Show that as $c$ increases, the tumour eventually completely disappears.

To do this, I tried sketching a graph of $\frac{dm}{dt}$ but this didn't really yield any results for me.

答:
Hint

Let $m=u^{12}$ to make
$$12\,u'\,u^3+b\,u^4-a\,u=-c$$ Switch variables to get $t(u)$ and compute $\partial t/\partial c$ and analyze.


问:
Differentiability of continuous continuation
Let $f$ be a strictly convex, differentiable function on $[0,\infty)$ with $f(0)=0$ and $f(x_0)=0$ for a $x_0>0$.
Thus $f<0$ on $(0,x_0)$ and $f>0$ on $(x_0,\infty)$.

Let $g(x) = f(x)/(x-x_0)$ on $x\neq x_0$ and assume, that there is a $y_0 \in [0,\infty)$ with
$g(x) \rightarrow y_0$ for $x\rightarrow x_0$.
Define $gx_0 = y_0$.

$g$ is differentiable on $x\neq x_0$ as a composition of differentiable functions.

Is $g$ also differentiable at $x_0$ ? Waht if $y_0 = f'(x_0)$ ?

答:
You are assuming that $f$ is differentiable so $g$ admits a limit at $x_0$ that is $f'(x_0)$.

For the differentiation of $g$ at $x_0$, we have that $g$ is differentiable at $x_0$ if and only if for some $l \in \mathbb{R}$,
$$ \frac{g(x) - g(x_0)}{x-x_0} \underset{x \rightarrow x_0}{\longrightarrow} l$$
and
\begin{align}
\frac{g(x) - g(x_0)}{x-x_0} \underset{x \rightarrow x_0}{\longrightarrow} l &\Longleftrightarrow g(x) \underset{x_0}{=} g(x_0) + l(x-x_0) + o(x-x_0) \\
&\Longleftrightarrow \frac{f(x)}{x-x_0} \underset{x_0}{=} f'(x_0) + l(x-x_0) + o(x-x_0) \\
&\Longleftrightarrow f(x) \underset{x_0}{=} f'(x_0)(x-x_0) + l(x-x_0)^2 + o((x-x_0)^2)
\end{align}
that is to say $f$ admits a development at order $2$ at $x_0$. But in general a strictly convex function that is merely differentiable doesn't admit a development at order $2$ at all points. For instance take $f$ defined by $f(x) = e^x$ if $x \leq 0$ and $f(x) = 1+x+x^2$ if $x \geq 0$, you can check that $f$ is differentiable and strictly convex on $\mathbb{R}$ but doesn't admit a development at order $2$ at $0$. So in general we cannot conclude on the fact that $g$ is differentiable at $x_0$.


答:
For any continuous function $f(x)$ such that $f(x)\rightarrow 1$ as $x\rightarrow a$ we can say that $\lim_{x\rightarrow a} \frac{\ln (f(x))}{f(x)-1}=1$, i.e., $\ln(f(x))$ starts approaching $f(x)-1$. Make the same replacements in your question as

(i) $\frac{n^{2/n}-1}{\ln{n^{1/n}}}\rightarrow\frac{n^{2/n}-1}{n^{1/n}-1}\rightarrow n^{1/n}+1\rightarrow 2$ as $n\rightarrow \infty$.

(ii) Using the above replacement for $\ln$ we can write it as $\frac{1-n^{1/n}}{n^{1/n}(n^{1/n}-1)}\rightarrow \frac{-1}{n^{1/n}}\rightarrow -1$ as $ n\rightarrow\infty$.

Title: Sequence involving $\sqrt[n]{n}$ and $\text{log}$

Body:
I know that the sequence $\sqrt[n]{n}$ converges to 1 and that $\text{log}(\sqrt[n]{n})$ thus converges to 0 as $n\to\infty$ since the logarithmic function is continuous. But how can I calculate the limits as $n\to\infty$ of the following sequences?

(a) $\frac{\sqrt[n]{n^2}-1}{\text{log}(\sqrt[n]{n})}$

(b) $\frac{\sqrt[n]{1/n}-1}{\text{log}(\sqrt[n]{n})}$

Any help would be much appreciated!


问:
Derivative and inverse
Is there a function whos derivative of its inverse equals the inverse of its derivative? This may also hold on a specific interval only.
$$\frac{d}{dx}f^{\langle-1\rangle}(x)=\left(\frac{d}{dx}f(x)\right)^{\langle-1\rangle}$$
Considering that $f(x)=x$ is its own inverse and $e^x$ is its own derivative, neither of them satisfies the equation, but both are the only functions with the respective properties (over the reels). Is this already a proof that there is no solution to the above equation?

Also, I dont know much about complex analysis but if a complex-valued function meets the criteria I'd be happy to know too :)

答:
I foud this video about complex solutions to the problem of the form $Cx^N$:

$$\frac{d}{dx} f^{\langle -1 \rangle}(x) = \left(\frac{d}{dx} f(x) \right)^{\langle -1 \rangle} \quad \to \;
\frac{ x^\frac{1-N}{N}}{NC^\frac{1}{N}} = \left(\frac{x}{CN} \right)^\frac{1}{N-1}$$

This only holds when $\quad \frac{1-N}{N} = \frac{1}{N-1} \qquad N^2-N+1=0 \qquad
N = \frac{1 \, \pm \, i \sqrt{3}}{2} = e^{\pm \frac{i \pi}{3}}$

and $\quad NC^\frac{1}{N} = (CN)^\frac{1}{N-1} \qquad C = N^{(N-2)N}=
\pm \, ie^{\pm \frac{\pi}{2\sqrt{3}}} \;$ or $\; e^{\pm \frac{\pi}{\sqrt{3}}}$

Title: Derivative and inverse


问:
Little $o$ and big $O$

The sum of a little $o$ and a big $O$ is given by the little $o$?

The sum of two big $O$ is a little $o$?

In studying I have encountered an implication that seems to imply these facts but I am not sure. If it is, someone can explain me why this holds? Thanks in advance.

答:
Simplify $f(x)=O(g(x))+O(h(x))$.

$$f(x)=O(\max\{g,h\}).$$
Here is a proof from MSE.

Simplify $f(x)=o(g(x)+O(h(x)).$

Note that $o(f)\subseteq{O}(f)$. That means that if something is $o(f)$, it's also $O(f)$. (The reverse does not always hold). So
$$f(x)=o(g(x))+O(h(x))=O(g(x))+O(h(x))=O(\max\{g,h\}).$$

Edit:
The statement $O(h^2)+O(h^2)=o(h)$ is incorrect:
Note that
$$O(h^2)+O(h^2)=O(\max\{h^2,h^2\})=O(h^2).$$

In other words you are saying that $O(h^2)=o(h)$. Choose $h(x)=x$ to show an counterexample.

Observe that $x^{1+1/2}=O(x^2)$. So that would mean by $O(x^2)=o(x)$ that
$$x^{1+1/2}=o(x),$$
which is absolutely wrong.

A note: As @user6725906 correctly pointed out in the comments, it depends on the limit. But if you write expressions like $\sim,o,O,\Omega,$ etc. without further information, this is (always) understood as $x\to\infty$. If you want to specify the limit (for example for $a$) you should write something like
$$f(x)=o(g(x)),\qquad (x\to a).$$

Title: Little $o$ and big $O$


问:
Linear bounded operators in $l^2$
I'd appreciate some assistant to solve the following problem:

Let $H=l^2(\mathbb{N})$. If $N\geq1$ and $f:H\rightarrow\mathbb{C}$ defined by $f(x)=f((x_j)_j)=\alpha x_N+x_{N+1}$, where $\alpha$ is a given complex scalar.

i. Prove that $f∈H'$, where $H'$ represents the set of all bounded linear operators in $H$.

ii. Find $h_0∈H$ such that $f(x)=<x,h_0>$ for any $x∈H$. Is $h_0$ unique?

I already proved part i. But I'm not sure how to proceed on part ii.

Thank you.

答:
Let $g(x):=x_1=\langle x,e_1\rangle$ and let $L,R$ be the left/right-shift operators.

Then $f=g\circ(\alpha L^N+L^{N+1})$ is the composition of linear continuous operators, hence itself a bounded functional. Moreover, if $\phi^*$ is the vector associated with the functional $\phi$, then $$f(x)=\langle x,f^*\rangle=\langle x,(gT)^*\rangle=\langle x,T^*g^*\rangle$$ so

$$h_0=f^*=(\alpha I+L)^*(L^*)^N\circ g^*=(\bar{\alpha}I+R)R^Ne_1=\bar{\alpha}e_N+e_{N+1}$$

Title: Linear bounded operators in $l^2$


问:
If $\frac{y}{x}=\frac{3x+y-w}{6x-y-z}=\frac{y+w}{7y-x+z}$ with $x,y,z,w>0$, $6x\ne y+z$, $7y\ne x-z$, then what is the value of $\frac{x}{y}$?
If $\frac{y}{x}=\frac{3x+y-w}{6x-y-z}=\frac{y+w}{7y-x+z}$ with $x,y,z,w>0$, $6x\ne y+z$, $7y\ne x-z$, then what is the value of $\frac{x}{y}$?

I tried to solve it as follows:

From these equations we have the following equations:

$\frac{y}{x}=\frac{3x+y-w}{6x-y-z}$, $\frac{y}{x}=\frac{y+w}{7y-x+z}$, $\frac{3x+y-w}{6x-y-z}=\frac{y+w}{7y-x+z}$

Hence we have the following:

$y(6x-y-z)=x(3x+y-w)$, $y(7y-x+z)=x(y+w)$, $(3x+y-w)(7y-x+z)=(y+w)(6x-y-z)$

$6xy-y^2-yz=3x^2+xy-xw$, $7y^2-yx+zy=xy+xw$, $21xy-3x^2+3xz+7y^2-yx+yz-7yw+wx-wz=6xy-y^2-yz+6xw-wy-wz$

And I continued along these lines, but it didn't lead me anywhere. Could you please explain to me how to solve this question?

答:
Hint :

Using $$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$$

we get

$$\frac{y}{x}=\frac{3x+2y}{\ldots}$$

a quadratic in $x,y$ which can be solved for positive $x,y$.


问:
Roots of polynomials that are the product of 2 polynomials
Given $p, q  \in \mathbb{F}[X]$, $deg(p) \land deg(q) \geq 1$ with no roots in $\mathbb{F}$. How could we show that $p.q$ has no roots as well?
Any hint would be appreciated.

答: $p, q  \in \mathbb{K}[X] \land \forall a \in \mathbb{K}:p,q \neq 0 \implies pq \neq 0$ because $ \mathbb{K} $ is a field and thus has no  zero divisors (a product of nonzero elements can only be zero if at least one of them is zero).


问:
If $E_r = \{x \in X \mid f(x) < r \}$ is open for all $r \in \mathbb{Q}$, then show that $E_r$ is open for all $r \in \mathbb{R}$.

If $X$ is a metric space and $f: X \to \mathbb{R}$ a map such that $E_r = \{x \in X \mid f(x) < r \}$ is open for all $r \in \mathbb{Q}$, then show that $E_r$ is open for all $r \in \mathbb{R}$.

If $E_r$ is open for all $r \in \mathbb{Q}$ it means that for all $x \in E_r$ there is an $\varepsilon >0$ such that $B(x, \varepsilon) \subset E_r$. However I also have that $\mathbb{R} \subset \mathbb{Q}$ doesn't this immediately imply that $E_r$ is open for all $r \in \mathbb{R}$?

答:
For every $r\in\mathbb{R}$ there exists a sequence $q_n,\in\mathbb{Q}$ with $lim_nq_n=r, q_n<r$. We have $\cup_nE_{q_n}=E_r$.
Firstly remark that since $q_n<r, E_{q_n}\subset E_r$,

Let $x\in E_r, f(x)<r$, there exists $q_n$ with $f(x)<q_n<r$ implies that $x\in E_{q_n}$ and $E_r\subset \cup_n E_{q_n}$.

We deduce that $E_r$ is open since it is the union of open sets.


问:
Solving an exponential recurrence relation
Let $f(t)$ be a sequence s.t. $e^{f(t)}\geq e^{\frac{2}{n}\log n}+e^{\frac{2}{n}f(t-1)}$ where $f(0)=e^{\frac{2}{n}\log n}, t\in \mathbb{N}, t\geq 1$. I am trying to find a lower bound on $f(t)$ as a function of $n,t$. So I was trying to think of this as a recurrence inequality but am not sure how to solve an exponential recurrence relation because I'm used to linear recurrence relations. Any ideas?

If we look at a few terms, it looks as follows:

\begin{align*}
e^{f(t)} &\geq e^{\frac{2}{n}\log n} + e^{\frac{2}{n}f(t-1)} \\
&= e^{\frac{2}{n}\log n} + \left( e^{f(t-1)} \right)^\frac{2}{n} \\
&\geq e^{\frac{2}{n}\log n} + \left( e^{\frac{2}{n}\log n} + e^{\frac{2}{n}f(t-2)} \right)^{\frac{2}{n}}\\
&\geq \cdots
\end{align*}

答:
For any $t$, the exponential being positive on $\mathbb{R}$,
$$ e^{f(t)} \geq e^{\frac{2}{n}\ln(n)} + e^{\frac{2}{n}f(t-1)} \geq e^{\frac{2}{n}\ln(n)} $$
and taking the logarithm of the last inequality we have
$$ f(t) \geq \frac{2}{n}\ln(n). $$


问:
How to take algebraic Bianchi identity into account when calculating the number of independent components of Riemann curvature tensor
I understant that the Riemann curvature tensor has the following symmetry properties:

Antisymmetry in last two indices:
$$R_{a b c d}=-R_{a b d c}$$

Symmetry in first pair of indices with second pair of indices:
$$R_{a b c d}= R_{c d a b}$$

Algebraic Bianchi identity:
$$R_{a b c d}+R_{c a b d} +R_{b c a d}=0$$

In $N$ dimensions, I can have ${ }^{N} C_{2}$
choices for $c$ & $d$ (they cannot be equal otherwise $R_{abcd}$ will be zero).

By combining the first & second symmetry property, we can say that $R_{abcd}$ is antisymmetric in $a$ & $b$ too. So there are ${ }^{N} C_{2}$ choices for the first pair of indices.

Based on the second symmetry property:

We can have ${ }^{N} C_{2}$ choices for the first pair of indicies. We can then have ${ }^{N} C_{2}-1$ different than the first one choices for the second pair. Their is pair swap symmetry, so in total we have $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}-1)$ independent components when the first & second pair differs. We have ${ }^{N} C_{2}$ choices where they are the same. So, not taking tha algebraic Bianchi identity into account we have $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1)$ independent compontent for the Riemann curvature tensor.

In the algebraic Bianchi identity, all indicies have to be different. There are then ${ }^{N} C_{4}$ different choices we have. This answer just subtracts this number from the earlier result, and gets the right answer:

$$\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1) - { }^{N} C_{4} = \frac{n^{2}\left(n^{2}-1\right)}{12}$$

The equality can be established after a few lines of algebra, or plotting the RHS & LHS and concluding that they are the same. Here is such a plot.

Why can we just subtract this number from the earlier result? This is the part I don't get.

答:
Your thoughts are right about the first part, the two first conditions of the curvature tensor gives you $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1)$ choices for $a,b,c,d$. Denote by $\mathfrak{B}(V^*)$ the linear subspace of all curvature tensor satisfying your first two conditions, and not the Bianchi identity, and by $\mathfrak{T}(V^*)$ the space of all covariant tensors $(0,4)$.

This means that by your argument, $\mathfrak{B}(V^*)$ has dimension $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1)$. Now consider the linear map $\pi: \mathfrak{B}(V^*) \rightarrow \mathfrak{T}(V^*)$ given by
$$
\pi(T)(w,x,y,z)=\frac{1}{3}(T(w,x,y,z)+T(x,y,w,z)+T(y,w,x,z)).
$$
Hence, you have that the space of all curvature tensors (it is, the space of $(0,4)$ satisfying all the properties you write) is the kernel of this map. So by classical linear algebra theorem, you only have to compute the dimension of the image of that map, which will be ${ }^{N} C_{4} $ (because is the set of alternating $4$-covariant tensors), and substract it to the number you compute to get the result. That is the formal reason why you cand substract that number.

This argument I wrote is part of the general proof you can find in Introduction to Riemannian Manifolds by John M. Lee, proposition 7.21, page 212.

For an easier way to remember this, I use this (NON FORMAL) trick:
I have $n^4$ possibilities, but by the antisymmetry of last two indices, I may substract $n^2$ choices, so I have $n^4-n^2$ in total. But by the second restriction, I must divide this number by two due to they are related $2\leftrightarrow2$, so I have
$$
\frac{n^4-n^2}{2}
$$
and then, by the Bianchi identity, I only have to put off the sum of cyclic permutations of $3$ indixes, so I divide by 6. REMEMBER this is only a trick I use remember the formula, nothing formal.


问:
Are continuous bijections between compact and topological spaces homeomorphisms?
We know that continuous bijections between compact and Hausdorff spaces are homeomorphisms. I.e. if $(X,\tau_{X})$ is a compact topological space, $(Y,\tau_{Y})$ is a Hausdorff space and $f:(X,\tau_{X})\rightarrow(Y,\tau_{Y})$ is a continuous bijective function, then $f$ is a homeomorphism.

However, does this hold if we relax the assumption that $(Y,\tau_{Y})$ is a Hausdorff space, i.e. only a topological space? If not, provide a counterexample.

答:
Take $X$ to be a finite set ($|X|>1$) with discrete topology and $X'$ be the same set with indiscrete topology.

Consider $i : X \to X'$, the identity function.

Clearly, $X$ is compact (as it is finite). $i$ is clearly bijective and continuous (you cannot demand more open sets in the left space).

But, obviously, this is not a homeomorphism.


问:
Random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $Z$ be the random variable denoting the distance of the point $B$ to its neighbour in the anti-clockwise direction.

Find the pdf of $Z$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_Z(z)= P(Z \le z)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

A.3. Then $Z$'s image is also $(0,1)$ (or $[0,1)$ or whatever). (more details in part D.)

B. Answer (that I found online):

Edit: it says $Z$ is uniform, but I think this is wrong.

C. My intuition for the answer (but I would like to know precisely please):

Edit: Nvm. I think the answer is wrong.

For parts D,E,F,G: Edit: nvm this is wrong.

H. New section: What exactly is the formula of $Z$ in relation to $A,B,C$? Supposedly $f_Z(z)=2(1-z)1_{(0,1)}(z)$. But what exactly is $Z$? $Z=|A-C|$? $Z=\min\{?,?\}$

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
The three points are independently and uniformly distributed on the circle

So given the location of $B$, the anticlockwise arc distances from $B$ of each of the other two points are independently uniformly distributed on $[0,1)$.

If $Z$ is the minimum of the anticlockwise arc distances from $B$ of each of the other two points, and $0 \le z \lt 1$, then:

the probability a particular one of these arc distances is greater than $z$ is $(1-z)$, given its uniform distribution

the probability both of these arc distances are greater than $z$ is $\mathbb P(Z\gt z)=(1-z)^2$, since they are independent

and the probability the nearest is less than or equal to $z$ is $F_Z(z)=\mathbb P(Z\le z)=1-(1-z)^2=2z-z^2$, as the complementary probability

making the probability density for the nearest $f_Z(z)=2-2z$, by taking the derivative of the absolutely continuous cumulative distribution function

You are correct in saying $Z$ does not have a uniform distribution.  It has a triangular distribution on $[0,1)$ with the mode at $0$.

$Z$ is not independent of the positions of $A$, $B$ and $C$ (for example it is exactly equal to the anticlockwise arc distance between two of them), though it is pairwise independent of the position of any one of them.

Title: Random points on a circle


问:
How to Factorise these expressions?
I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$

答:
Remember, and you must learn it by heart:
$$(a{+}b)^2=a^2+2ab+b^2$$
and
$$(a-b)^2=a^2-2ab+b^2$$
So try to write each of two expressions in one of this two forms</p>

Title: How to Factorise these expressions?


问:
Is my understanding of backward stability correct?
An algorithm $\tilde{f}$ of a problem $f$ is said to be backward stable if for any input $x$,
$$ \tilde{f}(x) = f(\tilde{x}) \ \text{ for some } \tilde{x} \text{ such that }
\frac{||\tilde{x} - x||}{||x||} = O(\epsilon) $$
as $ \epsilon \to 0 $, where $\epsilon$ is the machine epsilon.

The first time I saw this definition I was very confused: where does the $ \epsilon $ come from? There is no $ \epsilon $ in the definition of $ f, \tilde{f}, x, \tilde{x} \cdots $ etc.

After some thought, I realized that the algorithm $\tilde{f}$ is meant to be implemented on a family of idealized machines for which machine epsilon $\epsilon \to 0$. Thus, the algorithm is actually a family of functions $ \{ \tilde{f_\epsilon} \}_{\epsilon >0} $. Each $ \tilde{f_\epsilon} $ corresponds to different machine epsilon $\epsilon$.

Here is my understanding of backward stability:

An algorithm $ \{ \tilde{f_\epsilon} \}_{\epsilon >0} $ is backward stable if there exists $C, \delta>0$ such that for all input $x$ and $ 0 < \epsilon < \delta$, there exists $\tilde{x}$ such that
$$ \tilde{f_\epsilon}(x) = f(\tilde{x}) \ \text{ and }
\ ||\tilde{x} - x|| \leq C \epsilon ||x|| $$

Is my understanding correct? This definition seems to be too complicated.

答:
Your understanding is very close to correct, I would say. The difficulty with the original definition that you quote stems from confusing an algorithm with its output when implemented on a variety of machines with inexact arithmetic.

I hesitate to give a general definition of what an algorithm is, since I am not a computer scientist.  But commonly we need to talk about implementing the same algorithm on different machines. On any given machine $m$, this determines an output $\tilde f_m(x)$ from any given input $x$.

To analyze the machine-dependent error, we suppose that the relative error in inexact arithmetic operations can be bounded in terms of machine $\epsilon$ in a standard way, that, for example, IEEE floating-point arithmetic should satisfy.
(Different machines with the same machine $\epsilon$ can produce different outputs, as I know from bitter experience: Around 1990, a certain British PC maker refused to fix an overflow-on-underflow bug in their Intel 8088 software floating-point emulation that I ran into and told them about.)

Under these assumptions on machine arithmetic, we can hope to prove that, as you say, there exists $C,\delta>0$ such that if $0<\epsilon<\delta$, for each input $x$ there exists $\tilde x$ such that
$$ \tilde f_m(x)=\tilde f_0(\tilde x) \quad\mbox{and } \|\tilde x-x\|\le C\epsilon\|x\|,
$$
where $\tilde f_0(\tilde x)$ is the (idealized) output of the algorithm run in exact arithmetic with input $\tilde x$. We may prefer here to replace $\tilde f_0$ by the quantity $f(\tilde x)$ that the exact algorithm should output if it could be run exactly, without worrying about whether this is possible or whether it is defined differently. In this sense, evaluating $f(x)$ is the original "problem."

As to why this is fairly complicated, it has to do with distinguishing error generated by inexact arithmetic from error due to problem sensitivity. There is a  possibility that the values of $f$ may depend very sensitively on the inputs, so that $f(x)$ may differ very greatly from $\tilde f_m(x)$, but it is not the algorithm's fault: It computes the exact answer $f(\tilde x)$ to a problem with input $\tilde x$ as close to $x$ as we can reasonably expect, given finite-precision arithmetic.  See answers to this question for more on this and further references.


问:
Proving positive definiteness for a symmetric matrix
Let A be an $n\times p$ matrix with rank $p$. I want to prove that $X = A^TA$ is a positive definite matrix.

I know that for any symmetric matrix to be positive definite, all of its eigenvalues must be strictly positive.

So I need a way to show that $X$'s eigenvalues are all positive, or perhaps any other method to proof that $X$ is positive definite.

答:
It is straightforward to show that $A^TA$ is positive semi-definite by showing $x^TA^TAx \geq 0$ for all $x \in \mathbb{R}^{p}$, therefore $A^TA$'s all eigenvalues are non-negative.

Now using the fact $\mathrm{rank}(A^TA) = \mathrm{rank}(A) = p$, hence as an order $p$ matrix, $A^TA$ is non-singular. Therefore $A^TA$ doesn't have zero eigenvalues.

Combining these two arguments together, we see that all eigenvalues of $A^TA$ are positive, thus $A^TA$ is positive definite.


问:
Show that if $T\in\mathcal{L}(V,W)$ injective then there exists $S \in \mathcal{L}(W,V)$ such that $ST$ is the identity map on $V$

Let $W$ be a finite-dimensional and $T\in \mathcal{L}(V,W)$. Show that if $T\in\mathcal{L}(V,W)$ injective then there exists $S \in \mathcal{L}(W,V)$ such that $ST$ is the identity map on $V$. I saw some posts on this statement but I didn't find the same approach as mine. If someone could give a feedback on my proof, I would really appreciate it. Thank you in advance.

As $T\in \mathcal{L}(V,W)$ is injective, then by fundamental theorem of linear maps we have that $\dim V=\dim \text{null}\ T+\dim \text{range}\ T=0+\dim \text{range}\ T\le \dim W$. So, $\dim V\le \dim W$.

Let $(v_1,...,v_n)$ be a basis of $V$ and $(w_1,...,w_m)$ be a basis of $W$ (so $n\le m$). Define $T$ and $S$ as the following:

$T(v_i)=w_i, S(w_i)=v_i$ with $i=1,...,n$.

Both $T$ and $S$ are well defined as $n\le m$. Thus, $ST(v_i)=S(w_i)=v_i$which is exactly an identity map as each $v\in V$ can be expressed uniquely.

答:
Unfortunately, you don't get to define $T$, it's given to you in the problem. Here's similar approach to yours:

Let $\{ v_i \mid i \in I \}$ be a basis for $V$. Then because $T$ is injective, $\{ T(v_i) \mid i \in I \}$ is a linearly independent set in $W$. Extend it to a basis $\{ T(v_i) \mid i \in I \} \cup \{ w_j \mid j \in J \}$ of $W$, and define $S : W \to V$ on this basis by sending $S(T(v_i)) = v_i$ for all $i \in I$, and $S(w_j) = 0$ for all $j \in J$. Then $S \circ T = \text{id}_V$ since these two maps $V \to V$ agree on the basis $\{v_i\}$.

Note that this does not require either $V$ or $W$ to be finite-dimensional!