问:
$\mathbb Q$ is homeomorphic to a subspace of $\mathbb N^{\mathbb N}$
I'm trying to prove that $\mathbb Q$ is homeomorphic to a subspace of $\mathbb N^{\mathbb N}$.

More precisely $\mathbb Q$ is homeomorphic to the sequence space $S$ of finite support over $\mathbb N$.

Actually I have been reading a proof of Sierpiński's topological characterisation of rational numbers, namely, every countable metric space $\left<X,d\right>$ without isolated points is homeomorphic to $\mathbb Q$, from an article by Krzysztof Chris Ciesielski. Here the author has given a homeomorphism between $X$ and $S$. I am trying to find a homeomorphism between $\mathbb Q$ and $S$. Then we have $X$ is homeomorphic to $\mathbb Q$.

Here is my idea : we can define a metric on $\mathbb N^{\mathbb N}$ so that when it is restricted to the subspace $S$, it gives the metric on $S$.

So first we topologize $\mathbb N^{\mathbb N}$ by defining a metric on it.

We identify each sequence $\left<\alpha_1,\alpha_2,\alpha_3,\ldots\right>$ with the unique real number : $\alpha_1+\dfrac1{\alpha_2+\dfrac1{\alpha_3+\dfrac1{\ddots}}}\tag*{(*)}$

And then we give a metric on $\mathbb N^{\mathbb N}$ as :

$$\text{distance of }\left<a_1,a_2,a_3,\ldots,\ldots\right>, \left<b_1,b_2,b_3,\ldots\ldots\right>=\text{usual distance between the real numbers arose by the identification }(*)$$

Now each rational number can be written in its terminating simple-continued-fraction form, and we can take this representation of rational numbers to be unique.

Then I identify each sequence $\left<\alpha_1,\alpha_2,\ldots,\alpha_k,0,0,\ldots\right>$ in $S$ with a unique rational number like :

$$\alpha_1+\dfrac1{\alpha_2+\dfrac1{\alpha_3+\dfrac1{\ddots+\alpha_{n-1}+\dfrac1{\alpha_n}}}}\tag*{(**)}$$

So if we give a metric to $S$ :

$$\text{distance of }\left<a_1,a_2,\ldots,a_m,0,0,\ldots\right>, \left<b_1,b_2,\ldots,b_n,0,0,\ldots\right>=\text{usual distance of the rational numbers arose by the identification }(**)$$

So this is an isometry and therefore is a homeomorphism.

I would like to know whether my idea is correct, and if not, what is the right way to think about this problem.

And I also like to know if there are any other explicit homeomorphisms between $S$ and $\mathbb Q$.

答:
If you simply want to prove that $S$ is homeomorphic to $\Bbb Q$, there is a much simpler approach. $\Bbb N$ with the discrete topology is metrizable, and the product of countably many metrizable spaces is metrizable, so $\Bbb N^{\Bbb N}$ is metrizable, and hence so is its subspace $S$. $S$ is clearly countable, and it is easy to check that it has no isolated points, so it must be homeomorphic to $\Bbb Q$.

If you want an explicit homeomorphism, your idea works if you can prove that the metric on $\Bbb N^{\Bbb N}$ induced by the continued fraction embedding in $\Bbb R$ induces the product topology on $\Bbb N^{\Bbb N}$; that, however, takes quite a bit of work.


问:
Does a $g$ exists such that $\ker(g\circ f) = \{(x,y,z)\in \mathbb{R}^3: x = z\}$?

Let $f:\Bbb{R}^3 \to \mathbb{R}^3$, $f(x,y,z)=(x-y,y-z,x-z)$ be a linear map. Does it exists a surjective linear map $g:\mathbb R^3 \to \mathbb R^2$ such that: $\ker(g\circ f) = \{(x,y,z)\in \mathbb{R}^3: x = z\}$

Using the Rank–nullity theorem I arrived at the conclusion that $g\circ f$ is not surjective, but because $f$ isn't surjective, this does not allow me to conclude anything about $g$. I don't any different way I can approach this exercise. I Don't know if this helps but did the calculations and discovered that: $\ker(f)=\{(t,t,t)|t\in \mathbb R\}$ and that $f(\mathbb R^3)=\{u(1,0,1) +v(-1,1,0)|u,v\in \mathbb R\}$. How can I solve this?

答:
$\ker f$ is contained in the plane $x = z$. So let us pick a basis for this plane that contains a basis for $\ker f$. That will be $u := (1,1,1)$ and $v :=(0,1,0)$. Remember: we want $g(f(u)) = 0$ and $g(f(v)) = 0$. Well, by choice, $f(u) = 0$ so that's easy. We just need to remember later to choose $g$ such that $g(f(v)) = 0$.

First, complete $\{u, v\}$ to a basis by adding some $w$ (any $w$ not in the plane works). So we have a basis $\{u, v, w\}$ that contains the basis for the kernel we want, and if we apply $f$ to this basis, we get $\{0, f(v), f(w)\}$ where $\{f(v), f(w)\}$ is a basis for the image of $f$. Let us finally complete $\{f(v), f(w)\}$ to a basis $\{f(v), f(w), z\}$ for $\mathbb{R}^3$.

These bases are picked in a way that makes defining $g$ easy. Because the bases include vectors for which we know where they need to go to. Namely, we need $g(f(u)) = 0$ and $g(f(v)) = 0$. So now, given this basis, we can define $g$, by saying what $g(f(v)), g(f(w)), g(z)$ are:

We can define $g(f(v)) = 0$

We can define $g(f(w))$ and $g(z)$ so that $g$ is surjective

It follows from the fact that $g(f(w)) \neq 0$ that $\ker g = \{x = z\}$ (i.e. that it isn't larger)

ContentLicense: CC BY-SA 4.0


问:
Atiyah MacDonald: Proposition 3.5
In Atiyah-Macdonald's Commutative Algebra, on page 39, Proposition 3.5, it says

Let $M$ be an $A$-module. Then the $S^{-1}A$-modules: $S^{-1}M$ and $S^{-1}A \otimes_{A}M$ are isomorphic.

Since $S^{-1}A$ and $M$ are $A$-modules so is $S^{-1}A\otimes_{A}M$.

My question is: how can one tell $S^{-1}A\otimes_{A} M$ is an $S^{-1}A$-module?

Is it true that if $M$ is an $A$-module and $S$ being a multiplicatively closed subset of $A$. Then is $M$ also $S^{-1}A$-module?

答: No, it is not. For instance, $\mathbb{Z}$ is a $\mathbb{Z}$-module, but not an $(\mathbb{Z}\backslash \{ 0\})^{-1} \mathbb{Z}=\mathbb{Q}$ module. However, for any $A$-module $M$, the tensor product $S^{-1}M:=M\otimes_{A} S^{-1}A$ is an $S^{-1}A$-module.

Title: Atiyah MacDonald: Proposition 3.5


问:
Probability to get 5 successes in at most 7 trials
The night before the party, you plan to rehearse the song you are to sing and would like to sing it perfectly 5 times before sleeping. Since you are a great singer, the probability of you singing it perfectly on each attempt is 5/6. Suppose your attempts are mutually independent, what is the probability that you'll rehearse the song at most 7 times?

Here's my approach:
\begin{align*} \text{Probability to get 5 successes in 5 tries} = 4C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^0\dfrac{5}{6} &\approx 0.402\\ \text{6 tries} = 5C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^1\dfrac{5}{6} &\approx 0.334\\ \text{7 tries} = 6C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^2\dfrac{5}{6} &\approx 0.167\\\\ \text{Probability to get 5 successes in at most 7 tries} &= 0.903 \end{align*}

Was my approach correct? May I also know what type of Probability Distribution is suitable to model this problem?

答:
Revised answer

You have been revising your question from time to time $\,!\,$
I take it that your actual question is   "the probability that you will get your fifth head/ fifth perfect take in at most $7$ tries.

The simplest solution comes  by realizing that in $7$ trials, we must have $\leq2$ failures, and applying the binomial distribution formula

$\binom 7 0p^7q^0 + \binom71p^6q^1 + \binom 7 2p^5q^2$

Putting $ p = \frac56, q = \frac 16,$ the answer comes to $\dfrac{3125}{3456}, \approx 0.9042$


问:
Order of standard normal random variables, all independent
Let $X,Y,Z \sim N(0,1)$ and $X \perp Y, Y\perp Z, X \perp Z$

A combinatorial argument gives that $\displaystyle P(Z > Y >X) = \frac{1}{6}$ since the possible orderings are $3! = 6$ and each is just as likely.

I would like to for learning and understanding sake compute this without the use of a combinatorial argument.

We have:

$P(Z > Y > X) = P(Z-X > Y-X > 0)$ and in the RHS we can see that the independence disappears.

Initially one could think to try:

$P(Z > Y > X) =^{?} P(Z-Y-X>0 \mid Y- X > 0)P(Y-X>0) $ but the RHS seems to be $\displaystyle \frac{1}{2}\frac{1}{2}= \frac{1}{4}\neq \frac{1}{6}$

答:
One way you can do this is

$$
P(Z > Y > X) = \iiint_{\mathbb{R}^3} P(Z > Y > X | X = x, Y = y, Z = z) P(X = x, Y = y, Z = z) dV
$$

and so

$$
P(Z > Y > X) = \iiint_{\mathbb{R}^3} P(Z > Y > X | X = x, Y = y) f(x,y,z) dV
$$

hence

$$
P(Z > Y > X) = \frac{1}{\sqrt{2\pi}^3} \int_{-\infty}^{\infty}\int_{-\infty}^{z}\int_{-\infty}^{y} (1) \cdot e^{-\frac{x^{2}}{2}}e^{-\frac{y^{2}}{2}}e^{-\frac{z^{2}}{2}}dxdydz = \frac{1}{6}
$$


问:
triangle, vectors, proving an identity.
I'm trying to prove something but unfortunately I can't.

Let $ABC$ be a triangle and $M$ a point in $[AB]$ where $d(A,M)=d(B,M)$.Let also be
$N$ be a point in $[AC]$ where $d(A,N)=d(B,N)$.

Prove that $\overrightarrow{BC}=2\overrightarrow{MN}$.

Thanks!

答:
$1^{st}$ MODE:

prove the similarity between the 2 triangles ($ABC$,$AMN$) you have, and then calculate the ratio between the sides

to be more exhaustive, hence you proved the similarity (they have the same angles) the ratio between sides is given you by hypothesis (let's call it $k \in \mathbb{R}$) so $k\overrightarrow{MN}=\overrightarrow{BC}$.

Pay attention of what you use in the numerator/denominator of the fraction to compute $k$.

$2^{nd}$ MODE:

following the advices given you in your previous question we can conclude that $$\overrightarrow{AM}-\overrightarrow{AN}=\overrightarrow{NM}$$ $$\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CB}$$

by hypothesis $$\overrightarrow{AB}=2\overrightarrow{AM}$$ $$\overrightarrow{AC}=2\overrightarrow{AN}$$ then the proof follows easily (substitution :) )

ContentLicense: CC BY-SA 3.0

Title: triangle, vectors, proving an identity.


问:
According to what rigorous definition of "polynomial" is $\sum_{k=0}^n a_k D^k$ a polynomial, where $D$ is the derivative operator?
Let $Df$ be the derivative of $f\in C^{\infty}$.
\begin{align}
D^k\colon C^{\infty}&\to C^{\infty}\\
f&\mapsto f^{(k)}=\begin{cases}f&\text{ if }k=0\\
Df^{(k-1)}&\text{ if }1\leq k
\end{cases}
\end{align}
I was told that if $(a_0,\ldots,a_n)\in\mathbf{C}^{n+1}$, $\sum_{k=0}^n a_k D^k$ is a polynomial and can thus be factorized. That is, there is a $(b_1,\ldots,b_n)\in\mathbf{C}^n$ such that
\begin{equation}
\sum_{k=0}^n a_k D^k = a_n \prod_{i=1}^{n}(D^1 - b_iD^0).
\end{equation}

According to what rigorous definition of "polynomial" is that true and where can I find more information?

Unfortunately, I didn't find the Wikipedia article helpful and I'm not sure if the definition as sequences in some ring is what I'm looking for, as I have no background in abstract algebra.

答:
The polynomial ring $\mathbb C[X]$ satisfies the following universal property: If $\iota:\mathbb C\longrightarrow S$ is a commutative ring extension of $\mathbb C$ (so $S$ is commutative and $\iota$ is an injective ring homomorphism) and $s\in S$, then there is exactly one homomorphism $e:\mathbb C[X]\longrightarrow S$ such that $e(X)=s$ and $e(z)=\iota(z)$ for all $z\in\mathbb C$. We call $e$ an evaluation homomorphism, since it evaluates each polynomial at $s$ by plugging in $s$ for $X$. You can find this property on Wikipedia, or really any decent book on abstract algebra. Look for universal property of polynomial rings.

Now here the ring $S$ is the ring of differential operators of the form $\sum_{k=0}^n a_k\mathrm D^k$, and we have $\iota:z\mapsto z\mathrm D^0$. Also, we choose $s=\mathrm D$. We notice that the operator $\sum_{k=0}^n a_k\mathrm D^k$ is exactly the image of the polynomial $\sum_{k=0}^n a_k X^k$ under the evaluation homomorphism. And since the polynomial can be factored, and homomorphisms are multiplicative, its image can be factored in exactly the same way.


问:
Quaternion starting assumptions
Went through the explanation of quaternion rotations in here.

I try to figure out some starting assumptions and I am not sure how to approach them.

If I have a left-hand coordinate system where rotations are made in the order: yaw (Y - up axis) -> pitch (X - right axis) -> roll (Z - forward axis) and moreover they are defined as follows:

yaw axis is defined by the left-hand rule (counter-clockwise along the Up axis)

pitch axis is defined by the right-hand rule (clockwise along the Right axis)

roll axis is defined by the right-hand rule (clockwise along the Forward axis)

Does this orientation thing have effect on the signs of cosines and sines of quaternions?

Are the order of those i, j, k related to XYZ frame or to the order of rotations? So in my case i would be for yaw (first axis) or pitch (X-axis)?

Thanks in advance!

答:
We can write quaternions $w+xi+yj+zk$ as 4D vectors $[w,x,y,z]^T$.

Unit quaternions $q$ represent rotations (so $\lVert q\rVert_2=1$).

Every unit quaternion can be written as
$$ q = \begin{bmatrix}\cos(\theta/2)\\\mathbf{v}\sin(\theta/2)\end{bmatrix},$$
where $\theta$ is the rotation angle and $\mathbf{v}\in\mathbf{R}^3$ with $\lVert\mathbf{v}\rVert_2=1$ is the rotation axis.

This means, if I understood your description correctly:

If you have a unit quaternion $[w,x,y,z]^T$ with $y=z=0$, it represents a rotation around the x axis, so it is a pure pitch.

If you have a unit quaternion $[w,x,y,z]^T$ with $x=z=0$, it represents a rotation around the y axis, so it is a pure yaw.

If you have a unit quaternion $[w,x,y,z]^T$ with $x=y=0$, it represents a rotation around the z axis, so it is a pure roll.

But I have to point out that in a unit quaternion $w+ix+jy+kz = [w,x,y,z]^T$ the $x,y,z$ do not directly correspond to amount pitch, yaw and roll amounts!

Title: Quaternion starting assumptions


问:
Find the first 3 decimal places of a definite integral

$I(x)=\int_0^x \frac{\sin (t)}{t} \, dt$

Find $I(1)$ correct to 3 decimal places

I thought that we just put $1$ instead of $x$ and we get $\frac{\sin 1}{1} = 0.841 $ but the answer is different.
Any ideas what is the correct way to solve it?

答:
You are supposed to use the Maclaurin series of $\sin x$, divide by $x$ and integrate term-by-term:
$$\int_0^1\left(1-t^2/3!+t^4/5!-t^6/7!+\cdots\right)\,dt$$
$$=1-\frac1{3\cdot3!}+\frac1{5\cdot5!}-\frac1{7\cdot7!}+\dots$$
In this case, three terms are sufficient because the fourth is smaller than $\frac1{35000}$ in absolute value, so the answer is $0.946$.


问:
Statistics, Hypothesis and P-value - Check my answers
I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the class hours and can not follow the class examples, I thought someone can help me to solve and learn the subject.

With help of the answers from here I try to give an answer. Still need some improvements but tried to do my best. I still do not have answer for question D and confused about CL(the part C) and Significance level(part B)

My answers:

$N\ =\ 9\ \ \ \ \ \ \ \ Sum\ of\ x\ =\ 3970\ \ \ \ \ \ \ \ Mean,\ µ = 441.1111  \      Variance,σ^2  = 161.1111$
$ \sigma\ =\ \sqrt{161.1111}  = 12.6929$

$t\ =\ \frac{m\ -\ \mu}{s\ /\ \sqrt n}$

$t\ =\ \frac{500\ -\ 441.1111}{12.6929\ /\ \sqrt9} = 13.918545$

We subtract 1 to get degrees free 9 - 1 = 8

Degrees of freedom = n – 1 = 8

$Probability: P( T ≤ 13.918545) = 0.00000069 $ So, this is the p-Value

$We\ will\ reject\ H_0\ at \ \alpha = 1% $ and also any > 1%

$$ (i) 0.10\
The\ information\ from\ the\ first\ question\, the\ critical\ t-value\ for\ α = 0.10\ and\ df = 8,\ t_c=\ 1.86
 \\ CI\ =\ (\bar{X}\ -\ \frac{t_c\ \times\ s}{\sqrt n},\ \bar{\ X}\ +\ \frac{t_c\ \times\ s}{\sqrt n})
 \\CI\ =\ (441.1111\ -\ \frac{1.86\ \times\ 12.6929}{\sqrt9},\ \ 441.1111+\frac{1.86\ \times\ 12.6929}{\sqrt9})
= (441.1111 – 7.868, 441.1111 + 7.868)
= (433.243, 448.979)
 $$

$For\ the\ other\ t_c\ values:$

(ii) 0.05  $t_c=\ 2.306$ CL = (431.354, 450.868) (iii) 0.01 $t_c=\
> 3.355$
= (426.915, 455.308)

Based on the answers in part 2 for (i) = 0.90, (ii) = 0.95, (iii) = 0.99, none of the confidence intervals contain 500.

The Question:

The worker says that the mean purchasing cost is 500 USD. We decide to test this.

For a random sample of 9 purchases drawn from a normally distributed population with unknown variance, the costs are:

```<code>430, 450, 450, 440, 460, 420, 430, 450, 440.
</code>```

A) Conduct a hypothesis test of whether the population mean purchasing equals 500 USD. Include all assumptions, the hypotheses, test statistic, and P-value and interpret the result in context.

B) For which significance levels can you reject $H_0?$ (i) 0.10, (ii) 0.05, or (iii)
0.01.

C) Based on the answers in part B), for which confidence levels would the confidence interval contain 500? (i) 0.90, (ii) 0.95, or (iii) 0.99.

D) Use part B) and part C) to illustrate the correspondence between results
of significance tests and results of confidence intervals.

答:
Notice that all nine of the observations are \$460 and below? Just from common sense, what does that tell you about the claim that average cost is \$500.

You already have a thoughtful Answer from @tommik (+1), but because you ask I will show some additional detail.

Here is a relevant t test from a recent release of Minitab. How much of the output can your find by hand? What parts of the question can you answer from this?

```<code>One-Sample T: x

Test of μ = 500 vs ≠ 500

Variable  N    Mean  StDev  SE Mean       95% CI            T      P
x         9  441.11  12.69     4.23  (431.35, 450.87)  -13.92  0.000

Descriptive Statistics: x

Variable  N    Mean  SE Mean  StDev  Minimum      Q1  Median      Q3  Maximum
x         9  441.11     4.23  12.69   420.00  430.00  440.00  450.00   460.00
</code>```

I don't know whether you are taking this course in a classroom or online. A lot of online courses are using hastily written texts with confusing, useless problems. By contrast, this is a very nice problem carefully written (probably by a real statistician) to encourage your intuitive insight into hypothesis testing and confidence intervals. It will be worth your trouble to do computations, look at results, compare with computer printout, and think carefully about each of your answers.

Below is output from R statistical software for the same problem.

```<code>x = c(430, 450, 450, 440, 460, 420, 430, 450, 440)
summary(x); length(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  420.0   430.0   440.0   441.1   450.0   460.0
[1] 9        # sample size
[1] 12.69296 # sample SD

t.test(x, mu = 500, conf.lev=.99)

        One Sample t-test

data:  x
t = -13.918, df = 8, p-value = 6.874e-07
alternative hypothesis:
  true mean is not equal to 500
99 percent confidence interval:
 426.9145 455.3077
sample estimates:
mean of x
 441.1111

boxplot(x, ylim=c(400,500), col="skyblue2")
 abline(h=500, col="green2")
</code>```

Body:
Notice that all nine of the observations are \$460 and below? Just from common sense, what does that tell you about the claim that average cost is \$500.

You already have a thoughtful Answer from @tommik (+1), but because you ask I will show some additional detail.

Here is a relevant t test from a recent release of Minitab. How much of the output can your find by hand? What parts of the question can you answer from this?

One-Sample T: x

Test of μ = 500 vs ≠ 500

Variable  N    Mean  StDev  SE Mean       95% CI            T      P
x         9  441.11  12.69     4.23  (431.35, 450.87)  -13.92  0.000

Descriptive Statistics: x

Variable  N    Mean  SE Mean  StDev  Minimum      Q1  Median      Q3  Maximum
x         9  441.11     4.23  12.69   420.00  430.00  440.00  450.00   460.00

I don't know whether you are taking this course in a classroom or online. A lot of online courses are using hastily written texts with confusing, useless problems. By contrast, this is a very nice problem carefully written (probably by a real statistician) to encourage your intuitive insight into hypothesis testing and confidence intervals. It will be worth your trouble to do computations, look at results, compare with computer printout, and think carefully about each of your answers.

Below is output from R statistical software for the same problem.

x = c(430, 450, 450, 440, 460, 420, 430, 450, 440)
summary(x); length(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  420.0   430.0   440.0   441.1   450.0   460.0
[1] 9        # sample size
[1] 12.69296 # sample SD

t.test(x, mu = 500, conf.lev=.99)

        One Sample t-test

data:  x
t = -13.918, df = 8, p-value = 6.874e-07
alternative hypothesis:
  true mean is not equal to 500
99 percent confidence interval:
 426.9145 455.3077
sample estimates:
mean of x
 441.1111

boxplot(x, ylim=c(400,500), col="skyblue2")
 abline(h=500, col="green2")


问:
Recursive function that outputs its own code
This problem is probably a rather trivial one, since I have the impression, that it is a textbook-style one, but nonetheless somehow it won't give in. Here it is:

I have to show that there exists a (unary) recursive function, that has code $c$ and also takes the constant value $c$ (i.e., it outputs its own code).

I am pretty sure, I have to use Kleene's (second) recursion theorem, that says that for a given (total) recursive function $f:\mathbb{N} \rightarrow \mathbb{N}$ there is a number/code $c$ such that $\phi_c=\phi_{f(c)}$ (where $\phi_a$ is the partial recursive function that has code $a$), but I can't figure out how...

答:
Let $T_i$ be a Turing Machine with Godel number $i$ and $\phi_i$ the function computed by $T_i$.

We wish to show there exists a number $n_0$ such that $T_{n_0}$ prints its own coding $n_0$.  Let $\phi_0, \phi_1, ...$ be any enumeration (Godel numbering) of all computable functions and let $h$ be any totally computable function.  Then, by the fixed point theorem (as you have suspected) there exists a number $i_0$ such that $\phi_{i_0} = \phi_{h(i_0)}$, i.e. $i_0$ is a fixed point for $h$.

Let us now define the function $h$ to be the function such that for each natural number $n$, we have that $h(n)$ is equal to a machine that prints $n$ and halts, i.e., such that $\phi_{h(n)} = n$.  Let $n_0$ be a fixed point for $h$.  Then $\phi_{n_0} = \phi_{h(n_0)} = n_0$.


问:
Given $f(I) \subseteq I$, prove that there exists $x \in I$ such that $f(x)=x$
Given an interval $I = [0, 1]$ and a continuous transformation $f: I \to\mathbb R$.

a) Given $f(I) \subseteq I$. Prove that there exists $x \in I$ such that $f(x)=x$.

b) Given $I \subseteq f(I)$. Prove that there exists $x \in I$ such that $f(x)=x$.

答:
The question $a)$ is well well-known.

For the second,

$ f $ is continuous at the compact $ I=[0,1] $, thus $$f(I)=[m,M]=[f(a),f(b)]$$

with $ a,b \in I$.

$$I\subset f(I)\implies $$
$$[0,1]\subset [f(a),f(b)] \implies$$
$$f(a)\le 0\;  \wedge \;  f(b)\ge 1\implies$$
$$f(a)\le 0\le a\le b\le 1\le f(b)\implies$$
$$f(a)-a\le 0 \; \wedge \; f(b)-b\ge 0$$

You can finish by IVT.


问:
"Imaginary" part of a quaternion
With complex numbers, we write $Im(z)$ to denote the imaginary part of a complex number $z$. I was wondering what $Im(q)$, where $q$ is a quaternion, would show: would it be the $i$ part of the quaternion or all of the $i$, $j$, and $k$ parts? And if it shows all three parts, is there a notation for showing each of the individual parts?

In case this is unclear, if $q = 1 + 2i + 3j + 4k$, would $Im(q)$ be $2i + 3j + 4k$ (or possibly the vector $(2, 3, 4)$) or just $2$? And how would you the individual parts $3$ and $4$ separately?

答:
Real and imaginary part of quaternions is well-defined. Let's consider a quaternion $w+ix+jy+kz$. Often, we write the quaternion as a 4D vector $q=[w,x,y,z]^T$. Now we define the real part
$$ \operatorname{Re}q = w \qquad\text{and}\qquad \operatorname{Im}q = \begin{bmatrix}x\\y\\z\end{bmatrix}. $$

This is really helpful, since you can represent quaternions $q$ in polar form
$$ q = r\exp\biggl(\begin{bmatrix}0\\\mathbf{v}\end{bmatrix}\biggr), $$
where $r=\lVert q\rVert_2$ is multiplied by the exponential of the purely imaginary quaternion $[0,\mathbf{v}^T]^T$. This is very similar to
$$ z = r\mathrm{e}^{i\phi},$$
where $z$ is a complex number, $r=|z|$ and $i\phi$ is a purely imaginary number.

If you just want the factor in front of one of $i,j,k$, refer to José's answer.

Title: "Imaginary" part of a quaternion


问:
all topological spaces whose all subsets are compact
I have this question on my recent homework problem set, and the only thing I came up with, is to eliminate some "options", like metric spaces or compact spaces. Is there any characteristic answer for that? Is the answer as obviuos as just "finite topological spaces" or there is something non-trivial?</p>

答:
Such spaces are said to be hereditarily compact. In a Hausdorff space all compact subsets are closed, so every subset of a hereditarily compact Hausdorff space $X$ must be closed, and $X$ must therefore have the discrete topology. But a compact space with the discrete topology must be finite, so the hereditarily compact Hausdorff spaces are precisely the finite discrete spaces.

If we don’t require the space to be Hausdorff, however, there are many more hereditarily compact spaces. For instance, let $X$ be an infinite set with the cofinite topology. Of course all finite subsets of $X$ are compact, so suppose that $A$ is an infinite subset of $X$, and let $\mathscr{U}$ be an open cover of $A$. Let $U_0\in\mathscr{U}$. Then $A\setminus U_0\subseteq X\setminus U_0$, so $A\setminus U_0$ is finite, and it’s clear that some finite subset of $\mathscr{U}$ covers $A$. Thus, $X$ is hereditarily compact.

There are also non-$T_1$ spaces that are hereditarily compact. For instance, for each $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k\ge n\}$, and let $\tau=\{\varnothing\}\cup\{U_n:n\in\Bbb N\}$; then $\tau$ is a $T_0$ topology on $U_0$ that is not $T_1$, and you can easily check that $\langle U_0,\tau\rangle$ is hereditarily compact. Alternatively, you can notice that $\tau$ is a subset of the cofinite topology on $\Bbb N$ and prove the following result:

Proposition. Let $\tau_0$ and $\tau_1$ be topologies on $X$ such that $\tau_1\subseteq\tau_0$. If $\langle X,\tau_0\rangle$ is hereditarily compact, so is $\langle X,\tau_1\rangle$.

The space $\langle U_0,\tau\rangle$ is an example of a Noetherian space: a space $X$ is Noetherian if it satisfies the ascending chain condition for open sets, meaning that if $U_n$ is an open set in $X$ for each $n\in\Bbb N$, and

$$U_0\subseteq U_1\subseteq U_2\subseteq\ldots\,,$$

then there is an $n_0\in\Bbb N$ such that $U_n=U_{n_0}$ for all $n\ge n_0$. It’s easy show that every Noetherian space is hereditarily compact. These spaces arise naturally in algebraic geometry.

And if we don’t require any separation axioms at all, the indiscrete (or trivial) topology on any set is hereditarily compact, as is any other finite topology.

Added: As Ulli noted in the comments below, a space is hereditarily compact if and only if it is Noetherian.

In this answer Henno Brandsma proves that a space is Noetherian if and only if every open subset of it is compact. If every subset of $X$ is compact, then certainly every open subset of $X$ is compact, so $X$ is Noetherian.

For the other direction suppose that $X$ is Noetherian, and let $A\subseteq X$. Let $\mathscr{U}$ be a relatively open cover of $A$. For each $U\in\mathscr{U}$ there is an open $V_U$ in $X$ such that $U=V_U\cap A$; let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. Then $\mathscr{V}$ is an open cover of the open set $\bigcup\mathscr{V}$. By the result that I quoted above, $\bigcup\mathscr{V}$ is compact, so there is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ such that $\{V_U:U\in\mathscr{U}_0\}$ covers $\bigcup\mathscr{V}$. But then $\mathscr{U}_0$ covers $A$, so $A$ is compact, and $X$ is hereditarily compact.


问:
If $\vec{\omega}\cdot\vec{n}=0$ for $t=0$ then it holds for all time
I am trying to show that if $\vec{u}$ is a solution of the incompressible Euler equations with zero body force in a bounded domain $D$, and if $\vec{\omega}.\vec{n}=0$ on the boundary of $D$ for $t=0$ then $\vec{\omega}.\vec{n}=0$ also holds for all time $t\geq0$.

I thought that if I can show $\frac{∂\omega}{∂t}=0$ then by time independence of $\omega$ the above is true. Taking the curl of the incompressible Euler equations with $\vec{f}=\vec{0}$ then gives me $\frac{∂\omega}{∂t}=\nabla\times(\vec{u}\times\vec{\omega})$. (Of course after a few lines of vector identities/algebra).

I'm not sure where to go from here.

答:
In inviscid flow, the fluid velocity satisfies the no-flux  condition $\mathbf{u} \cdot \mathbf{n} = 0$ on the boundary $\partial D$, but the tangential component of velocity need not vanish as in viscous flow.

Let $\Sigma(t)$ be a smooth material surface element, bounded by the simple, smooth closed curve $C(t)$. As $\Sigma(t)$  evolves in time,  it contains the same fluid particles present at time $t= 0$.  If $\Sigma(0) \subset \partial D$, then no flux condition ensures that $\Sigma(t) \subset \partial D$ for all $t > 0$.

By Stokes' theorem we have

$$\oint_{C(t)}\mathbf{u} \cdot dl = \int_{\Sigma(t)}(\nabla \times \mathbf{u}) \cdot \mathbf{n} \, dS = \int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS $$

In an inviscid flow with the absence of body forces it follows from Kelvin's circulation theorem that

$$\frac{d}{dt}\oint_{C(t)}\mathbf{u} \cdot dl = \frac{d}{dt}\int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS  =0$$

By a generalization of the Leibniz integral rule we can pass the derivative under the integral on the RHS to obtain

$$\frac{d}{dt}\int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS  = \int_{\Sigma(t)} \left[\frac{\partial\mathbf{\omega}}{\partial t}+ (\nabla \cdot \mathbf{\omega})\mathbf{u}  \right]\cdot \mathbf{n} \, dS- \oint_{C(t)}(\mathbf{u}\times \mathbf{\omega}) \cdot dl$$

Note that $(\nabla \cdot \mathbf{\omega})\, \mathbf{u} \cdot \mathbf{n} = 0$ and $\mathbf{u}\times \mathbf{\omega} = \mathbf{u}\times (\nabla \times \mathbf{u}) = \mathbf{u} \cdot \nabla\mathbf{u} - \mathbf{u} \cdot \nabla\mathbf{u} = \mathbf{0}$.

Thus,

$$\int_{\Sigma(t)} \frac{\partial\mathbf{\omega}}{\partial t}\cdot \mathbf{n} \, dS   = 0$$

Under the usual smoothness assumption for the velocity and vorticity fields, since this holds for any $\Sigma(t) \subset \partial D$, we get $\frac{\partial\mathbf{\omega}}{\partial t}\cdot \mathbf{n} = 0$ for all $t > 0$.


问:
Combination of two random variables
Suppose there are two kinds of stock, and that each of them currently sells for same price in the market. The return on the one share of the first stock is a random variable with E(X)= π and Var(X) = s^2. The return on one share of the second stock also has E(Y) = π and Var(Y) = s^2. Consider the following investment plans: Plan 1: Buy 10 shares of the first stock. Plan 2: Buy 5 shares of each stock.

Compute the average return for each of the two plans and compare.

My approach: I know how to do the algebra with random variables and for both plan I'm getting average = π. As all shares have same mean, so when we averaged them, we will get average as π. Therefore, both plans will have same average return.

Can anyone plz confirm, if my way is correct or not.

答:
The choice depends on the investment plan variance. If you buy 1 share of 10 different stocks (independent) with same perfomance and risk, the total risk is obviouly very lower compared to the investment risk of buying 10 shares of one single stock...one share can gain another can loss and the risk decreases...

Formally the variance passes from $10^2$ to $10$ times

$$\mathbb{V}[10X_1]=10^2\mathbb{V}[X_1]$$

$$\mathbb{V}[X_1+X_2+\dots+X_{10}]=10\mathbb{V}[X_1]$$

Title: Combination of two random variables


问:
Is $O(m \log n + n \log n )$ equal to $O((m+n) \log n )$?
I have a self study question and maybe very simple.

Can we conclude the following is true? Is it a property of asymptotic or log property?

$O(m \log n + n \log n )$ is equal to $O((m+n) \log n )$?

答:
If you look at the expression

$$ \mathcal{O}(m \log(n) + n \log(n) )  = \mathcal{O}((m+n) \log(n)) $$

this is just factoring the expression since

$$ m \log(n) + n \log(n) = (m+n)\log(n) $$

You'd probably be interested in the relationship between $m$ and $n$. Which one is larger?


答:
Write $$G = \pmatrix{a& b&c\\ d&e&f}$$

We see $\ker(GF)$ is generated with $u=(1,0,1)$ and since $Fu = (1,-1,0)$ we get $$0=GFu = (a-b,d-e)\implies a = b\;\;{\text and} \;\; d=e$$   so $$G = \pmatrix{a& a& c\\ d&d & f}$$

So if you take $a=1=c=f$ and $d=0$ you are done since $\dim (Im(G)) = 2$.

Title: Does a $g$ exists such that $\ker(g\circ f) = \{(x,y,z)\in \mathbb{R}^3: x = z\}$?

Body:
Let $f:\Bbb{R}^3 \to \mathbb{R}^3$, $f(x,y,z)=(x-y,y-z,x-z)$ be a linear map. Does it exists a surjective linear map $g:\mathbb R^3 \to \mathbb R^2$ such that: $\ker(g\circ f) = \{(x,y,z)\in \mathbb{R}^3: x = z\}$

Using the Rank–nullity theorem I arrived at the conclusion that $g\circ f$ is not surjective, but because $f$ isn't surjective, this does not allow me to conclude anything about $g$. I don't any different way I can approach this exercise. I Don't know if this helps but did the calculations and discovered that: $\ker(f)=\{(t,t,t)|t\in \mathbb R\}$ and that $f(\mathbb R^3)=\{u(1,0,1) +v(-1,1,0)|u,v\in \mathbb R\}$. How can I solve this?


问:
For $s(t) = a_2t^2 + a_1t + a_0$ and $r(t) = b_2t^2 + b_1t + b_0$ in $P_2[t],$ define $<s(t),r(t)> = 2a_2b_2 + a_1b_1 + 3a_0b_0$
EDIT: I'm still trying to figure out! Will ask for help if I can't answer it still by the end of the day. Thank you :)

Online classes hasn't been as easy as face-to-face classes and with that my professor is quite old and isn't very familiar with the online methods of teaching these days.

He said that the above is an inner product on $P_2[t].$
In attempting to prove other < , > examples, I would like to know how to know and show that < , > are inner products on $P_2[t].$ You may use the example I provided. Thank you so much!

答:
an orthonormal basis is given by
$$    \frac{t^2}{\sqrt 2}, \; \; \; t, \; \; \; \frac{1}{\sqrt 3}  $$

If you have real numbers with $$e^2 + f^2 + g^2 = 1,$$
a unit vector is given by
$$    \frac{ e t^2}{\sqrt 2} \; \; \; +ft + \; \; \; \frac{g}{\sqrt 3}  $$

The quadratic form is given by the norm of your $s(t)$ being
$$ 3 a_0^2 +  a_1^2 + 2 a_2^2  $$
which becomes $g^2 + f^2 + e^2 $ when $a_0 = g/ \sqrt 3, \; \; a_1 = f, \; \; a_2 = e/ \sqrt 2$


问:
Counting Possible combinations
I do not now how to count the number of possibilities. Perhaps you can help.

I have 12 sided dice.
The sides are colored.
4 yellow
3 blue
2 green
1 red
1 black
1 white

If I roll 2 dice there are 16 possible combinations of two yellow sides showing out of a possible 144 combinations.

If I add dice, how do I count the number of possible yy combinations?

To complicate matters, the black dice are sometimes wild.
When they are, then I want to count the number of yellow and black.

If I have two dice, then I want to count the possible combinations of yy, yb, by, and bb. I can make a simple table and observe the number of instances. In this case, there are 25 possible combinations out of 144 possibilities.

When I have more than 2 dice how many yy combinations are possible given x dice?  And if I add the black dice how many yy, by, yb, bb combinations are possible given x dice?

Is there a formula that I can use to count the various combinations?  Tables become impractical when there are more than 2 dice.  I am unable to find formulas that help in counting the number of specific combinations our of the total number of combinations. In most examples, the identification of subset# out of total# seems to be done manually.

答:
In the cases you've described, you just multiply the possibilities.  Just as you got $16=4\cdot4$ YY combinations, with $3$ dice you would have $64=4\cdot4\cdot4$ YYY combinations.  Similarly with four dice, there would be $4^4$ ways of getting all yellow.

For different colors of dice, you again multiply the possibilities.  For blue and yellow, three are $3\cdot4$ possibilities, and for yellow and blue there are $4\cdot3$ possibilities.  Usually, we would only do one multiplication and double the result.

When you have many dice, it would be very arduous to list all the possibilities.  There's a short cut using binomial coefficients, which I gather you are unfamiliar with.  There's a ton of stuff about it on the web.

Title: Counting Possible combinations


问:
Prove whether this is true or false: For every two sets $A$ and $B$, $\{A\oplus B, A\cap B\}$ is a partition of $A \cup B$.
I am having trouble with this question. Does anyone have an idea?</p>

答: If $B = \emptyset$ and $A\ne\emptyset$ then $\{A\oplus B, A\cap B\} = \{A,\emptyset\}$ is NOT a partition of $A\cup B =A$


问:
Quadrilateral $ABCD$ has $∠BDA = ∠CDB = 50^\circ, ∠DAC = 20^\circ$ and $∠CAB = 80^\circ.$ Find angles $∠BCA$ and $∠DBC.$

Quadrilateral $ABCD$ has $∠BDA=∠CDB=50°$, $∠DAC=20°$ and $∠CAB=80°$. Find angles $∠BCA$ and $∠DBC$.

This seems to be a very easy problem, but I am not able to solve it.

I am attaching a diagram which also contains all the simple angle chasing I have done.

I can not see any cyclic quads either or any parellelograms so I am not sure what to do.

答:
In a triangle, that is $\triangle ADC$, bisector of an interior angle, that is $BD$ bisector of $\angle ADC$, and bisector of non-adjacent two exterior angles, they are  $AB$ bisector of $\angle CAB$ and $CB$ bisector of $\angle ACB$ are concurrent.

Hence

$$\angle BCA=60^ {\circ}$$

$$\angle CBD=10^{\circ}$$

Given that point $B$ is the intersection of $DB$ and $AB$

$$\angle ADB = \angle CDB=50^{\circ}$$

$$\angle CAB = \angle BAQ=80^{\circ}$$

where $Q$ is a point on $DA$

Hence point $B$ is the centre of excircle of $\triangle ADC$ then $CB$ is the exterior bisector.

$$\angle ACB= \angle BCP = 60^{\circ}$$

where point P is on $DC$


问:
Integration of $\int_{0}^{\infty}{\frac{1}{(x^2+1)(x^{2019}+1)}dx}$

The original problem is:

The proof of the equation show above is what I want. I found that the first integral equals $\pi/4$.  The changes of 2019 from 1 to any real numbers bigger than 1 cause no change of the final integration.

OwnerDisplayName: Zw L

Body:
A numeric verification

Clear["Global`*"]

The integrand of the lhs can be written as 1/(x^2 + 1) - 1/((x^2 + 1)(x^2019 + 1))

x^2019/((x^2 + 1) (x^2019 + 1)) ==
  (1+x^2019)/((x^2+1)(x^2019+1)) - 1/((x^2+1)(x^2019+1)) ==
  1/(x^2 + 1) - 1/((x^2 + 1) (x^2019 + 1)) // Simplify

(* True *)

lhs1 = Integrate[1/(x^2 + 1), {x, 0, ∞}]

(* π/2 *)

The numeric part of the verification:

lhs2 = π ((NIntegrate[1/((x^2+1)(x^2019 + 1)), {x, 0, ∞}]/π) //
     RootApproximant)

(* π/4 *)

lhs = lhs1 - lhs2

(* π/4 *)

rhs = 1/2 Integrate[1/(x^2 + 1), {x, 0, ∞}]

(* π/4 *)

lhs == rhs

(* True *)

OwnerDisplayName: Bob Hanlon


Title: Integration of $\int_{0}^{\infty}{\frac{1}{(x^2+1)(x^{2019}+1)}dx}$

Body:
The original problem is:

The proof of the equation show above is what I want. I found that the first integral equals $\pi/4$.  The changes of 2019 from 1 to any real numbers bigger than 1 cause no change of the final integration.

Body:
You should change the variable $x\to \frac{1}{x}$

(1/(1 + x^2)*1/(1 + x^2019) /. x -> 1/x)*D[x /. x -> 1/x, x] // Simplify

$$-\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}$$

That is
\begin{align}
&\int_{0}^{\infty} \frac{1}{(1+x^2)(1+x^{2019})}\,\mathrm{d}x\\
=&\int_{\infty}^{0} \frac{1}{(1+(\frac{1}{x})^2)(1+(\frac{1}{x})^{2019})}\,\mathrm{d}\frac{1}{x}\\
 =&\int_{\infty}^0-\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}\,\mathrm{d}x
\end{align}

So we get
$$\int_{0}^{\infty} \frac{1}{(1+x^2)(1+x^{2019})}\,\mathrm{d}x=\int_{0}^{\infty}\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}\,\mathrm{d}x$$

It means that all of them equal to
$$\frac{1}{2}\int_{0}^{\infty} \frac{1}{(1+x^2)(1+x^{2019})}+\frac{x^{2019}}{\left(x^2+1\right) \left(x^{2019}+1\right)}\,\mathrm{d}x$$

Just equal to
$$\frac{1}{2}\int_0^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x=\frac{\pi}{4}$$

1/2 Integrate[1/(1 + x^2), {x, 0, ∞}]

π/4

OwnerDisplayName: cvgmt


答:
Two chances:

(A)

```<code>(x^2019/((1 + x^2) (1 + x^2019)) -
   1/2 (1 + x^2019)/((1 + x^2) (1 + x^2019))) // FullSimplify
</code>```

(* (-1 + x^2019)/(2 (1 + x^2) (1 + x^2019)) *)

(B)

```<code>x^2019/((1 + x^2) (1 + x^2019)) -
 1/2 (1 + x^2019)/((1 + x^2) (1 + x^2019))
</code>```

(* -(1/(2 (1 + x^2))) + x^2019/((1 + x^2) (1 + x^2019)) *)

The [NIntegrate] results must be wrong too. We know for sure from NIntegrate that

```<code>NIntegrate[1/(1 + x^2), {x, 0, \[Infinity]}]
</code>```

$\frac{\pi}{4}$

By the famous majorant criteria for infinite integrals we have ease to estimate that this has to be bigger then the

Take as a reference for example this page: Integral test for convergence

To do a harder attempt:

```<code>NIntegrate[1/(1 + x^2)*1/(1 + x^2019), {x, 0, 1},
 PrecisionGoal -> 100, WorkingPrecision -> 200]
</code>```

(*

```<code>0.78522650738967301568031546551813653513306507248637888350236342703381\
7598418491157836764104246377397390853048300788950931805991775817967566\
46995443684292857095545206581998393114079894683526201949452667
</code>```

*)

That fails and is indeed as expected smaller than $\frac{\pi}{4}$.

```<code>Plot[{1/(1 + x^2) , 1/((1 + x^2) (1 + x^2019)),
  1/((1 + x^2) (1 + x^2020))}, {x, 0.8, 1.2},
 PlotLegends -> "Expressions", PlotRange -> {{0.8, 1.2}, Automatic}]
</code>```

Just for value a little bigger than $1$ the second factor in the denominator explodes to very high values and the curves are really different.

```<code>Table[{1/(1 + x^2) , 1/((1 + x^2) (1 + x^2019)),
   1/((1 + x^2) (1 + x^2020))}, {x, 0.5, 1.5, 0.5}] // TableForm
</code>```

```<code>NIntegrate[1/((1 + x^2) (1 + x^2019)), {x, 0, 1}]
</code>```

(*  0.785398 *)
is a good approximation to the integral with the high requirements by [NIntegrate].

Despite it looks like a sharp jump it is not.

[NIntegrate] is much like Plot it needs the hand of an experienced Mathematician to give good results.

A powerful plot is for the answer of Your question:

```<code>Plot[NIntegrate[1/(1 + x^2)*1/(1 + x^2019), {x, 0, y},
  PrecisionGoal -> 100, WorkingPrecision -> 200], {y, 0, 1.05},
 PlotLegends -> "Expressions", PlotRange -> Full]
</code>```

So the two calculations at the beginning of my answer did prove already the equation does not hold. The integral value is close to $0.785398$. The functions series $f(x,n)=\frac{1}{(1+x^2)(1+x^n))}$ with $n$ Integers is strongly monoton dropping with increasing $n$. $0.785398$ is already close to the limit for $n\rightarrow\infty$.

Body:
Two chances:

(A)

(x^2019/((1 + x^2) (1 + x^2019)) -
   1/2 (1 + x^2019)/((1 + x^2) (1 + x^2019))) // FullSimplify

(* (-1 + x^2019)/(2 (1 + x^2) (1 + x^2019)) *)

(B)

x^2019/((1 + x^2) (1 + x^2019)) -
 1/2 (1 + x^2019)/((1 + x^2) (1 + x^2019))

(* -(1/(2 (1 + x^2))) + x^2019/((1 + x^2) (1 + x^2019)) *)

The [NIntegrate] results must be wrong too. We know for sure from NIntegrate that

NIntegrate[1/(1 + x^2), {x, 0, \[Infinity]}]

$\frac{\pi}{4}$

By the famous majorant criteria for infinite integrals we have ease to estimate that this has to be bigger then the

Take as a reference for example this page: Integral test for convergence

To do a harder attempt:

NIntegrate[1/(1 + x^2)*1/(1 + x^2019), {x, 0, 1},
 PrecisionGoal -> 100, WorkingPrecision -> 200]

(*

0.78522650738967301568031546551813653513306507248637888350236342703381\
7598418491157836764104246377397390853048300788950931805991775817967566\
46995443684292857095545206581998393114079894683526201949452667

*)

That fails and is indeed as expected smaller than $\frac{\pi}{4}$.

Plot[{1/(1 + x^2) , 1/((1 + x^2) (1 + x^2019)),
  1/((1 + x^2) (1 + x^2020))}, {x, 0.8, 1.2},
 PlotLegends -> "Expressions", PlotRange -> {{0.8, 1.2}, Automatic}]

Just for value a little bigger than $1$ the second factor in the denominator explodes to very high values and the curves are really different.

Table[{1/(1 + x^2) , 1/((1 + x^2) (1 + x^2019)),
   1/((1 + x^2) (1 + x^2020))}, {x, 0.5, 1.5, 0.5}] // TableForm

NIntegrate[1/((1 + x^2) (1 + x^2019)), {x, 0, 1}]

(*  0.785398 *)
is a good approximation to the integral with the high requirements by [NIntegrate].

Despite it looks like a sharp jump it is not.

[NIntegrate] is much like Plot it needs the hand of an experienced Mathematician to give good results.

A powerful plot is for the answer of Your question:

Plot[NIntegrate[1/(1 + x^2)*1/(1 + x^2019), {x, 0, y},
  PrecisionGoal -> 100, WorkingPrecision -> 200], {y, 0, 1.05},
 PlotLegends -> "Expressions", PlotRange -> Full]

So the two calculations at the beginning of my answer did prove already the equation does not hold. The integral value is close to $0.785398$. The functions series $f(x,n)=\frac{1}{(1+x^2)(1+x^n))}$ with $n$ Integers is strongly monoton dropping with increasing $n$. $0.785398$ is already close to the limit for $n\rightarrow\infty$.


问:
How do I find the equation of hyperbola passing through a given point and has given two asymptotes
If I know that my hyperbola passes through the point $$(1;1)$$ and has the asymptotes $$2x-y+2=0$$ and $$y=0$$ how can I find the equation of the hyperbola?
Thanks in advance!

答:
When $x$ and $y$ are large the curve approximates the pair of straight lines as given by the equations of the asymptotes.
Therefore the hyperbola will have an equation of the form given by the product of the two equations equal to a constant.
So the hyperbola is $$y(2x-y+2)=c$$

Since $(1,1)$ lies on the curve, $c = 3$


问:
What is $\sup_X \{ \inf_Y \{P(X > Y)\}\}$?
Suppose $X$ and $Y$ are two independent random variables, such that $E[X] = E[Y] = 0$. What is $\sup_X \{ \inf_Y \{P(X > Y)\}\}$?

I managed to derive the following fact:

$\sup_X \{ \inf_Y \{P(X > Y)\}\} \leq \frac{1}{2}$.

One can get this simply by examining the case, when $X$ and $Y$ are i.i.d.

However, I failed to find any $X$ for which this bound is tight (i.e. such $X$, that $\forall Y$ $P(X > Y) \geq \frac{1}{2}$).

答:
It's actually $0$.  Fix any $X$ with $\mathbb{E}[X] = 0$, then there exists $N > 0$ such that $P(X > N) \le \varepsilon$.  Define $Y$ by $P(Y = N) = 1-\varepsilon$ and $P(Y = -K) = \varepsilon$, where $K = \frac{1-\varepsilon}{\varepsilon} N$ is chosen so that $\mathbb{E}[Y] = 0$.  Then
\begin{align*}
P(X > Y) = (1-\varepsilon)P(X > N) + \varepsilon P(X > -K) \le (1-\varepsilon)\varepsilon + \varepsilon \le 2\varepsilon.
\end{align*}

Since $\varepsilon$ was arbitrary, we have $\inf_Y P(X > Y) = 0$ and since $X$ was arbitrary we have $\sup_X \inf_Y P(X > Y) = 0$.


答:
General equation of hyperbola is $$ax^2+bxy+cy^2+dx+ey+f=0$$

Since $ac\ne 0$ we can assume $c=1$. If we express $y$ we get $$y= {-(bx+e)\pm \sqrt{(bx+e)^2-4(ax^2+dx+f)}\over 2}$$

Now, use that:

$\lim_{x\to \infty} y =0$ and

$\lim_{x\to \infty} {y\over x} =2 \;\;$(so $a+2b+4=0$) and

$\lim_{x\to \infty} (y-2x) =2$

And finally, since $(1,1)$ is on it we get $a+b+1+d+e+f=0$.

Title: How do I find the equation of hyperbola passing through a given point and has given two asymptotes

Body:
If I know that my hyperbola passes through the point $$(1;1)$$ and has the asymptotes $$2x-y+2=0$$ and $$y=0$$ how can I find the equation of the hyperbola?
Thanks in advance!


问:
proof with differential equation

How to prove the last equation from the first one. If I integrate both side there is something left side in terms of arc sine and on left side there is t. then how to prove the equation that is given in terms of cos???

答:
Given equation is from standard spring-mass harmonic motion. Energy conservation KE+ PE

$$ \dfrac12 m  \dot x^2 + k \dfrac{x^2}{2} =E  \tag 1$$

which is the same as the given equation

$$\dot x^2 = \dfrac{2E}{m}- \dfrac{k x^2}{m} \tag 2 $$

Differentiate wrt time $t$ gives force equilibrium

$$ m \ddot x + k x= 0 \tag3 $$

having solution $ \omega^2 = k/m $

$$ x = x(t) = A \sin  \omega t  + B \cos \omega t \tag 4 $$

To find phase let

$$  \dfrac{A}{\sqrt{A^2+B^2}} =\sin K ,\dfrac{B}{\sqrt{A^2+B^2}} = \cos K, \tan^{-1}\dfrac{A}{B} =K, x_{max}= {\sqrt{A^2+B^2}} \tag 5 $$

to write

$$ x = x_{max} \cos(\omega t - K)  \tag 6 $$

From(1) at extremity maximum potential energy at $ t=0, x_{max}=\sqrt{ \dfrac{2E}{k}} \tag 7$

so that when phase difference $K$ is included

$$ x =\sqrt{ \dfrac{2E}{k}} \cos(\omega t + K).  \tag 8 $$

The sign of arbitrary constant $K$ is also arbitrary. There is a printo in the given text at 4. , missing radical sign of $ \sqrt{k/m} $.

Title: proof with differential equation


问:
Show contractibility implies single set proof
Problem. Suppose $Y$ is a contractible space. Let $X$ be any space. Then $[X,Y]$, the set of homotopy classes, has only one element.

Proof:

Suppose $Y$ is contractible. So $Y$ is path connected. Let $X$ be any space. Let $[f],[g]\in [X,Y]$. It suffices to show that there exists a homotopy between $f$ and $g$. Since $Y$ is path connected, for each $x\in X$, there exists a continuous map $f_x:I\rightarrow Y$ such that $f_x(0)=f(x)$ and $f_x(1)=g(x)$. Define

$F:X\times I\rightarrow Y$ by $F(x,t)=f_x(t)$.

This map is the desired homotopy.

Is the proof correct?

答:
This is not quite correct, as you have not ensured that $F$ be continuous. The idea is correct, but as stated, you are picking maps $f_x$ for each $x$, and we need these to also depend continuously on $x$.

To be contractible means that the identity map is homotopic to a constant map. So for some fixed point $y_0\in Y$, we have a continuous homotopy $G\colon Y\times I\to Y$ where $g_0(y)=G(y,0)=y$ and $g_1(y)=y_0$ for all $y$.

Now given $f\colon X\to Y$, we can define $F\colon X\times I\to Y\times I$ as $f'(x,t)=(f(x),t)$. The composition $G\circ F$ is now a map $X\times I\to Y$ which is guaranteed to be continuous. At time $0$ this composition is $(x,0)\mapsto (f(x),0)\mapsto f(x)$, and at time $1$ it is the constant map.

This shows that every map $f\colon X\to Y$ is homotopic to the same constant map, so $[X,Y]$ has one point.


问:
The space of absolutely continuous functions is separable with norm $\|F\|_{AC}=\sup |F|+\int_a^b|F'|$
Consider the space of absolutely continuous functions $AC([a,b])$ equipped with norm
$$
\|F\|_{AC}=\sup |F|+\int_a^b|F'|
$$

It can be shown that $AC([a,b])$ is a Banach space under this norm.

Please try to prove that $AC([a,b])$ is separable, i.e. there is a countable dense subset.

答:
One can prove directly that it is separable, which is always a bit of work. A more powerful (but elegant) way to prove it may be the following:

Note that, using a decomposition of absolutely continuous functions, $AC([a,b])$ is isometrically isomorphic to the direct sum of $L^1([a,b])$ and a one-dimensional space. Since separability is stable under isometric isomorphisms and finite direct sums, the result follows from the separability of $L^1([a,b])$ and the one-dimensional space.


问:
Prove that $\operatorname{ack}(3,y)=2^{y+3}-3$.

Prove that:
$$\operatorname{ack}(3,y)=2^{y+3}-3$$
where $\operatorname{ack}$ refers to the Ackermann function.

Here's what I have so far.

This statement can be prove by induction over $y$.

Induction Start: $\operatorname{ack}(3,0)= \operatorname{ack}(2,1)=5=8-3=2^3-3=2^{0+3}-3$

Induction Assumption: The statement $\operatorname{ack}(3,y)=2^{y+3}-3$ holds for all $y \in \mathbb{N}$.

Induction Step: The statement should also hold for $y \to y +1 \in \mathbb{N}$, $\operatorname{ack}(3,y+1) = 2^{(y+1)+3}-3$.

$$\operatorname{ack}(3,y+1) = \operatorname{ack}(2, \operatorname{ack}(3,y+1-1))= \operatorname{ack}(2, \operatorname{ack}(3,y))$$

I know that I can use my induction assumption and set $\operatorname{ack}(3,y) = 2^{y+3}-3$, but I'm not able to see how I can use that to get to the final answer.

答:
Let's use an "operator name" $\operatorname{ack}$ instead of $ack$.

Since $\operatorname{ack}(0,\,y)=y+1$ and $\operatorname{ack}(1,\,0)=\operatorname{ack}(0,\,1)=2$,$$\operatorname{ack}(1,\,y)=\operatorname{ack}(0,\,\operatorname{ack}(1,\,y-1))=\operatorname{ack}(1,\,y-1)+1$$for $y\ge1$, so by induction $\operatorname{ack}(1,\,y)=y+2$. Then $\operatorname{ack}(2,\,0)=\operatorname{ack}(1,\,1)=3$ and$$\operatorname{ack}(2,\,y)=\operatorname{ack}(1,\,\operatorname{ack}(2,\,y-1))=\operatorname{ack}(2,\,y-1)+2$$for $y\ge1$, so by induction $\operatorname{ack}(2,\,y)=2y+3$. Finally, $\operatorname{ack}(3,\,0)=\operatorname{ack}(2,\,1)=5$ and$$\operatorname{ack}(3,\,y)=\operatorname{ack}(2,\,\operatorname{ack}(3,\,y-1))=2\operatorname{ack}(3,\,y-1)+3$$for $y\ge1$, so by induction $\operatorname{ack}(3,\,y)=2^{y+3}-3$. I'll only make the base step explicit in this last proof by induction, as it's the hardest one: $2(2^{k+3}-3)+3=2^{k+4}-3$.


问:
Evaluation Functional
Let $V$ be a vector space over $\mathbb R$. The algebraic dual of $V$, denoted $V^*$, is the space of all linear functions from $V$ to $\mathbb R$. These functions are called functionals. Most textbooks list the following as examples for functionals:

The integration operator $f\mapsto \int f$ is a functional.

The evaluation functional (at the query point $x$) $f\mapsto f(x)$

Obviously, both examples are only meaningful in the case $V$ is a function space of real-valued functions, e.g. $L^p(X,\mathbb R)$, or $C^p(X,\mathbb R)$.

Now, suppose we are looking at spaces of functions which map the domain $X$ to, say, $\mathbb R^k$, or even more general to an arbitrary (Banach) spaces $B$. The integral operator remains a functional as integration, by definition, yields eventually a real number. However, the evaluation functional is no longer a functional as the function in place no longer takes values in $\mathbb R$, right? So I was wondering whether an evaluation "functional" is also defined for such cases? Or do I get something fundamentally wrong?

答:
Evaluation operator $\Phi_x: f \mapsto f(x)$ defined on the space $V$ of functions from $X$ to $\mathbb{R}^k$ for $k > 1$ is not a functional for the reason you stated: its codomain is not the scalar field.

However, you can compose the evaluation operator with a linear functional on $\mathbb{R}^k$ to obtain a linear functional on $V$. For an example, consider a basis $e_i$ of $\mathbb{R}^k$ and the corresponding dual basis $\xi_i$ of $\mathbb{R}^{k*}$. The composition $\xi_i \circ \Phi_x$ is a linear functional on $V$ and hence a member of the dual space $V^*$.

Title: Evaluation Functional


答: The word "functional" is just short for "linear function which codomain is the scalar field". So no, the evaluation is not a functional if the codomain is a vector space different from the field. In that case is just called "linear function".</p>

Title: Evaluation Functional

Body:
Let $V$ be a vector space over $\mathbb R$. The algebraic dual of $V$, denoted $V^*$, is the space of all linear functions from $V$ to $\mathbb R$. These functions are called functionals. Most textbooks list the following as examples for functionals:

The integration operator $f\mapsto \int f$ is a functional.

The evaluation functional (at the query point $x$) $f\mapsto f(x)$

Obviously, both examples are only meaningful in the case $V$ is a function space of real-valued functions, e.g. $L^p(X,\mathbb R)$, or $C^p(X,\mathbb R)$.

Now, suppose we are looking at spaces of functions which map the domain $X$ to, say, $\mathbb R^k$, or even more general to an arbitrary (Banach) spaces $B$. The integral operator remains a functional as integration, by definition, yields eventually a real number. However, the evaluation functional is no longer a functional as the function in place no longer takes values in $\mathbb R$, right? So I was wondering whether an evaluation "functional" is also defined for such cases? Or do I get something fundamentally wrong?


问:
Binomial coefficients and $p$-adic limits
I was reading Schoof's Catalan's Conjecture, and on page 37, while proving Cassels' Theorem it is mentioned that by Exercise 5.7, the denominators of the binomial coefficients $\binom{\frac{p}{q}}{k}$ for $0 \leq k \leq \lfloor \frac{p}{q} \rfloor + 1$, where $p$ and $q$ are odd prime numbers, are powers of $q$.

I found a proof by Keith Conrad, but it uses $p$-adic limits, so I decided to read through Gouvea's $p$-adic Numbers: An Introduction.

I still don't know what is going on, any insight please?

答:
The proof you cited does not really use $p$-adic numbers but just the $p$-adic topology on $\mathbf Q$ and continuity of polynomial functions in the $p$-adic topology.

Suppose $F$ is a field and $|\cdot|$ is an absolute value on $F$. Let $f \colon F \to F$ be a continuous function and $S \subset F$ be a subset such that $|f(s)| \leq 1$ for all $s \in S$. If $x \in F$ is a limit of points in $A$, say $x = \lim_{n \to \infty} s_n$.  By continuity, $f(x) = \lim_{n \to \infty} f(s_n)$.  The absolute value on $F$ is continuous, so $|f(x)| = \lim_{n \to \infty} |f(s_n)|$. Since $|f(s_n)| \leq 1$ for all $n$, we get $|f(x)| \leq 1$.

Now apply this to $F = \mathbf Q$, $|\cdot| = |\cdot|_p$ ($p$-adic absolute value), and $f(x) = \binom{x}{k}$, and $S = \mathbf Z$.  Polynomials are continuous functions since addition and multiplication are continuous.  A rational number whose denominator is not divisible by $p$ is a $p$-adic limit of integers: if $a/b$ is a fraction where $p \nmid b$, then for each positive integer $n$ we can solve $bs_n \equiv a \bmod p^n$ for some integer $s_n$, which means $|bs_n - a|_p \leq 1/p^n$, so $|s_n - a/b|_p = |(bs_n-a)|_p/|b|_p = |bs_n-a|_p \leq 1/p^n$.  Thus $s_n \to a/b$ as $n \to \infty$.
Therefore $|\binom{a/b}{k}|_p \leq 1$ since $|\binom{s_n}{k}|_k \leq 1$ for all $n$. That shows the fraction $\binom{a/b}{k}$ has no $p$ in its denominator when $p \nmid b$.


问:
Valid reason the sequence of functions does not converge uniformly
Is this a valid reason why the sequence of functions $f_n(x)=\frac{e^{nx}-e^{-nx}}{e^{nx}+e^{-nx}}$ does not converge uniformly?

$\text{lim}\frac{d}{dx}f_n(x) \neq \frac{d}{dx}\text{lim}f_n(x)$?

Also I do not think a reason could be $\text{lim} \int_{-1}^{1}f_n \neq \int_{-1}^{1}\text{lim}f_n$

since the latter statement appears untrue.

答:
You see that the sequence of functions you have written is $f_{n}(x)= \tanh(nx)$. If you take the limit for $n \rightarrow \infty$ you can see that it converges pointwise to a discontinuous function, hence it doesn't converge uniformly.

For the interchange of limit and derivative, it is not enough to prove that $f_n$ is not converging uniformly to $f$. You can prove that you have a sequence of functions that is uniformly convergent to $f$ and still the interchange between limit and derivative cannot be performed. I give an example: consider the functions $f_n : [0,2\pi] \rightarrow \mathbb{R}$ be $f_n(x)= n^{-1/2}$sin$(nx)$ and $f(x)=0$. Here by the squeeze thm $f_n$ converges uniformly to $f$ but you cannot interchange lim & derivative. So, from the condition you mentioned you cannot conclude that it is not uniformly converging.


问:
How does the $p$-group fixed point theorem proof make sense?
Let $G$ be a finite $p$-group acting on a finite set $X$, and ${\rm Fix}_G(X)$ the subset of $X$ consisting of fixed points under this action. Then
$$|X| \equiv | {\rm Fix}_G(X)| \pmod{p}.$$

The proof of this consists of saying $X$ is a disjoint union of orbits that all have a cardinality equal to $p^y$ where $y$ is an integer.

But surely this isn't true?

Using the $\Bbb Z_4$ group as an example, let $X = \{ 0,2,3\}$. There is one orbit here of length $4$, but $X$ does not have the element '$1$' in it, so it cannot be called a disjoint union of that orbit.

答:
You seem to be assuming that if $G$ acts on a set $X$ and $Y$ is a subset of $X$, then the action restricts to an action of $G$ on $Y$. This is not the case in general; to make this work you need the hypothesis that $Y$ is what we call $G$-invariant, which means $g\in G$ and $y\in Y$ implies $gy\in Y$.

In the example you've given the set $\{0,2,3\}$ is not a $\mathbb Z_4$-invariant subset of $\{0,1,2,3\}$.


问:
Infinite Closed Discrete Subset of a Normal non-compact Space
I need to prove that given a space $X$ that is normal and not compact, there exists an infinite closed set $S \subseteq X$ that is discrete. (where singletons are open in the subspace topology)

答:
The proposed result is false.

Let $X$ be the space $\omega_1$ of countable ordinals with the order topology. (This space is sometimes denoted by $\Omega$.) I showed here that every linearly ordered space with the order topology is not just normal, but hereditarily normal. To see that $X$ is not compact, for each $\alpha\in X$ let $U_\alpha=\{\xi\in X:\xi<\alpha\}$; then $\{U_\alpha:\alpha\in X\}$ is an open cover of $X$ that has no countable subcover, let alone a finite subcover.

However, $X$ is limit point compact. Let $A$ be an infinite subset of $X$. Then there is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $A$. Let $\alpha=\sup_{n\in\omega}\alpha_n$; then $\alpha$ is a limit point of $\{\alpha_n:n\in\omega\}$ and hence of $A$.


问:
Right-angled isosceles triangles
If a right-angled triangle is isosceles then the other two angles must be equal to $45^\circ$ ?

Is this always the case or are there other possible right-angled isosceles triangles?

答:
From this, we know in isosceles triangles, the angles at the base are equal to each other.

If each is $\theta,2\theta<180^\circ$ as the other angle $>0$

$\implies \theta<90^\circ$

So, the other unequal angle $=90^\circ$ and we have $90^\circ+2\theta=180^\circ$

Title: Right-angled isosceles triangles


问:
Solve Matrix Linear Least Squares with Frobenius Norm Regularization and Linear Equality Constraints
How to solve the following constrained minimization problem:
$$
\arg_S min_\; \frac{1}{2}\left \{ \left \| K_2SK_1^T-M \right \|_F^2 +\lambda \left \| S \right \|_F^2\right \} \\
s.t. \sum_{1}^{col}S=Sum1 \\
\sum_{1}^{row}S=Sum2 \\
$$
where $K_1$,$K_2$,$M$ and $S$ are 2d Matrix, and only $S$ is unknown. In the constraints, $Sum1$ is the sum along the column of $S$, which is a row vector. $Sum2$ is the sum along the row of $S$, which is a column vector.

Here is the data stored in mat format. How to solve this kind of problem?

```<code>    load('matlab.mat');
%　 min norm( K2*X*K1'-M,'fro')^2+lambda*norm(X,'fro')^2
%   s.t. sum(X,1) = Sum1 ;   sum(X,2) = Sum2;
</code>```

答:
The idea is to bring the problem into the form:

$$\begin{aligned}
\arg \min_{ \boldsymbol{s} } \quad & \frac{1}{2} {\left\| K \boldsymbol{s} - \boldsymbol{m} \right\|}_{2}^{2} + \frac{\lambda}{2} {\left\| \boldsymbol{s} \right\|}_{2}^{2} \\
\text{subject to} \quad & A \boldsymbol{s} = \boldsymbol{u} \\
\quad & B \boldsymbol{s} = \boldsymbol{v}
\end{aligned}$$

Using the Kronecker Product we can see that:

$ K = {K}_{1} \otimes {K}_{2} $.

$ \boldsymbol{s} = \operatorname{vec} \left( S \right) $ where $ \operatorname{vec} \left( \cdot \right) $ is the vectorization operator.

$ \boldsymbol{m} = \operatorname{vec} \left( M \right) $.

The matrices $ A $ and $ B $ are just Selectors of the corresponding elements in $ \boldsymbol{s} $.

Remark

Pay attention that if $ A $ and $ B $ represent a matrix which selects each element of $ \boldsymbol{s} $ exactly once then $ \sum_{i} {u}_{i} = \sum_{i} {v}_{i} $ must hold as it represent the sum of $ \boldsymbol{s} $. Namely $ \boldsymbol{1}^{T} A \boldsymbol{s} = \boldsymbol{1}^{T} B \boldsymbol{s} = \sum_{i} {s}_{i} $. This is the case for your constraints. So it must be like that in order to have a feasible solution.

Now the above is a basic Convex problem which can be solved by Projected Gradient Descent where we project onto the intersection of the 2 equality constraints.

You could even do something simpler by concatenate the matrices and vectors:

$$ C \boldsymbol{s} = \begin{bmatrix} A \\ B \end{bmatrix} \boldsymbol{s} = \boldsymbol{w} = \begin{bmatrix} \boldsymbol{u} \\ \boldsymbol{v} \end{bmatrix} $$

Then it is very similar to Linear Least Squares with Equality Constraint.

An interesting resource with that regard is Robert M. Freund - Projection Methods for Linear Equality Constrained Problems.


问:
Nested Interval Property and Infinity
I'm having difficulty understanding some of the properties of real numbers which invoke infinity. For example, the fact that $$\bigcap_{n \in \mathbb{N}}\left(0,\frac{1}{n}\right) = \emptyset$$
What confuses me about this is that it seems that it can be proven to be false for arbitrarly large natural numbers. I considered something like this:

Consider the set $\bigcap_{n=1}^{m}\left(0,\frac{1}{n}\right)$ with $m \in \mathbb{N}$. Due to the desnity of the rationals, for any interval there exists a value c such that $0 < \frac{1}{c} < \frac{1}{n}$. So every interval $(0,\frac{1}{n})$ is non-empty. Now, choose c such that $0<\frac{1}{c}<\frac{1}{m}$. Since $\frac{1}{m} < \frac{1}{m-1} < ... < \frac{1}{1}$, then, since $\frac{1}{c}$ is in every interval, then $\frac{1}{c} \in \bigcap_{n=1}^{m}\left(0,\frac{1}{n}\right)$. Since m was an arbitrarily chosen natural number, this should hold for an arbitrarily large natural number.

But, I know that the intersection over all natural number is the empty set, and that means that something must be different between an arbitrarily large natural number and iterating over the entire set of natural numbers.

This breaks my intuition, and I realize I don't have an accurate and rigorous conception of infinity. For example, here are some central questions that I'm unsure how to answer:

What is wrong with viewing infinity as some arbitrarily large natural number?

How would I prove that while I can always create a smaller fraction than $\frac{1}{n}$ by constructing $\frac{1}{n+1}$, there isn't a fraction between $(0,\frac{1}{n})$ 'at infinity'?

I would appreciate advice on how I should intuitively view infinity, especially regarding the difference between something holding true for an arbitrarily large number but failing 'at infinity.' Thank you for your time!

答:
Since you've shown some confusion about it in the comments, let's start with the meaning of $$\bigcap_{n=0}^\infty A_n = \bigcap_{n\in \Bbb N} A_n$$
The definition of the first expression is the second (not vice versa). And the definition of the second is
$$\bigcap_{n\in \Bbb N} A_n = \{a\mid \forall n \in \Bbb N, a \in A_n\}$$
There is no taking of limits here, and no approximations. It is simple logic: To be in the interesection, it has to be in every set.

The real numbers $\Bbb R$ is the unique (up to isomorphism) complete ordered field. "Ordered field" just means a set with addition and multiplication behaving normally, and an order operation $\le$ that respects the addition and multiplication. By "complete" I mean that the order has the supremum property: If non-empty $A \subset \Bbb R$ is bounded above, then it has a supremum, or least upper bound. Lets play with that:

First, if $\Bbb N$ is bounded above, then it has a supremum $\omega \in \Bbb R$. But then $\omega -1 < \omega$ is not an upper bound, so there must be some $n \in \Bbb N$ with $ n > \omega -1$, and therefore $n + 1 > \omega$. Since $n+1 \in \Bbb N$, this contradicts that $\omega$ is an upper bound. Therefore $\Bbb N$ cannot be bounded. And so for every $x \in \Bbb R$, there is an $n \in \Bbb N$ with $n > x$, the "Archimedean principle".

But then if $\epsilon > 0$, it is invertible, and there is some $n > \frac 1\epsilon$. And that means $\frac 1n < \epsilon$. So for any $\epsilon > 0$, there is an $n \in \Bbb N$ with $\frac 1n < \epsilon$.

If $x \in \bigcap_{n\in \Bbb N} \left(0,\frac 1n\right)$, then $x \in (0,1)$. So $x > 0$. And for all $n, x \in \left(0,\frac 1n\right)$, so for all $n, x < \frac 1n$, in contradiction to the previous result. Hence there cannot be such an $x$, so $$\bigcap_{n\in \Bbb N} \left(0,\frac 1n\right) =\varnothing$$

Title: Nested Interval Property and Infinity


问:
Closest point on a̶ ̶s̶u̶r̶f̶a̶c̶e̶ 3d bell curve to a point
I am developing a game using SDFs and ray marching. High school starts next year so I don't really know any calculus (not even pre calc) or linear algebra.

Here's the problem:

A surface is defined by this function (removed constants, not sure if it changes anything):
$$S(x,y) = e^{-(x^2+y^2)}$$
graph of S

A vector A is on the surface if it satisfies $S(A_x,A_y) = A_z$ .This is the distance function (with constants removed):
$$\mathrm{dist}(a,b) = \sqrt{(a_x - b_x)^2 + (a_y - b_y)^2 + (a_z - b_z)^2}$$

Given any vector A, how can I find a vector B on the surface, such that $\mathrm{dist}(A, B)$ is minimum, or at least approximate it?

答:
as @Raffaele described, the obvious (and accurate) approach is to compute the partial derivatives and iteratively optimize, which is slow, really heavy on the hardware and requires multivariable calculus which I understand absolutely nothing, yet

while playing around a little, i noticed that the surface is symmetrical along the z axis, this can be exploited to perform faster optimization, or a straightforward algebraic approach

instead of $Surface(x,y) = e^{-(x^2+y^2)}$ we can use $Surface(x) = e^{-x^2}$

$m = Polar(A_x, A_y)$

$n = A_z$

$B = intersection( (m, Surface(m)), Slope(m), (m,n), Slope(m)^{-1})$

these functions can be changed to according to the surface:

$Polar(x,y)=|(x,y)|$

$Slope(x) = \partial S / \partial x $ (but $Slope(x)=x*k$ works for me which is faster)

$intersection(A, S_a, B, S_b) = (buf, S_a*(buf-A_x)+A_y)$

where $buf=\frac{(S_a*A_x - A_y - S_b * B_x + B_y)}{(S_a - S_b)}$


问:
Convergence of a rational sequence
Im trying to prove the convergence of a sequence and so far I did not have a whole lot of ideas.

Lets assume we have a convergent sequence $(a_n)_{n \in \mathbb{N}}$ of rational numbers, so lets say $a_n = \frac{r_n}{q_n}$ with $r_n \in \mathbb{Z},q_n \in \mathbb{N}\setminus\{0\}$ and $\mathrm{gcd}(r_n,q_n)=1$ with $\lim_{n \longrightarrow \infty} a_n = x_0 \in \mathbb{R} \setminus \mathbb{Q}$.

I now want to show that $\left( \frac{1}{q_n}\right)_{n\in\mathbb{N}}$ is a convergent sequence with limit $0$.

First of all, Im not entirely sure that this statement is true, since I did not read it anywhere, but it feels like it has to be true, but I dont know why. So far I have tried to prove it using the normal definition without any success.

Any help is appreciated!

答: Obviously for some integer $m$, we have $ x_0 \in (m,m+1) $. Now observe that for every $ M > 0 $ there are only a finite number of proper fractions $r_n / q_n $ with $ q_n < M $ inside $(m,m+1)$. You can use this to show that $ q_n \to \infty $.

Title: Convergence of a rational sequence


问:
Calculating series expansions within a matrix: matrix exponential
I have a $(3 \times 3)$ matrix $$ Y = \begin{pmatrix} 0 & - e^{-i \theta} & 0 \\ e^{i \theta} & 0 & - e^{-i \theta} \\ 0 & e^{i \theta} & 0 \end{pmatrix} $$ for which I would like to calculate the matrix exponential $\exp(t Y) = I + t Y + \frac{t^2 Y^2}{2!} + \ldots $ If I let $z : = e^{i \theta}$, I have $$ Y^2 = \begin{pmatrix} - |z|^2 & 0 & |z|^2 \\ 0 & -2 |z|^2 & 0 \\ |z|^2 & 0 & - |z|^2 \end{pmatrix} \\ Y^3 = \begin{pmatrix} 0 & 2 \overline{z} |z|^2 & 0 \\ |z|^2 (-z - \overline{z}) & 0 & |z|^2 (z + \overline{z}) \\ 0 & -2z |z|^2 & 0 \end{pmatrix} $$ and $$ Y^4 = \begin{pmatrix} - \overline{z} |z|^2 (-z - \overline{z}) & 0 & - \overline{z} |z|^2 (z + \overline{z}) \\ 0 & 4 |z|^4 & 0 \\ z |z|^2 (-z- \overline{z}) & 0 & z |z|^2 (z+ \overline{z}) \end{pmatrix}. $$ Setting $|z| = 1$ and calculating the matrix exponential above up the fifth power $Y^5$, I got $$ \begin{pmatrix} 1 - \frac{t^2}{2!} + \frac{t^4}{4!} \overline{z} (z + \overline{z}) + \ldots & - t \overline{z} + \frac{t^3}{3!} (2 \overline{z}) - \frac{t^5}{5!} 4 \overline{z} + \ldots & \frac{t^2}{2!} - \frac{t^4}{4!} \overline{z} (z + \overline{z}) + \ldots \\ tz - \frac{t^3}{3!} (z + \overline{z}) + \frac{t^5}{5!} 2 (z + \overline{z}) + \ldots & 1 - \frac{2 t^2}{2!} + \frac{t^4}{4!} 4 + \ldots & - t \overline{z} + \frac{t^3}{3!} (z + \overline{z}) - \frac{t^5}{5!} 2 ( z+ \overline{z}) + \ldots \\ \frac{t^2}{2!} - \frac{t^4}{4!} z (z + \overline{z}) + \ldots & tz - \frac{t^3}{3!} 2 z + \frac{t^5}{5!} 4 z + \ldots & 1 - \frac{t^2}{2!} + \frac{t^4}{4!} z (z + \overline{z}) + \ldots
 \end{pmatrix} $$
I think I must be able to rewrite this with the help of $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots$.

For example, if I look at the $a_{22}$ term above, I see that it is almost $\cos(t)$, except the numerical factors that don't work out. Also, the $a_{11}$ term is almost $\cos(t)$, except there appears a term $\overline{z} (z+ z)$ from the fourth power onwards, and the same happens for the $a_{33}$ term with $z$ and $\overline{z}$ switched. The $a_{32}$ term seems to be $z \sin(t)$, but again the numerical coefficients don't work out.

Question: Does anyone recognize the pattern in these entries (i.e. the series) and is able to calculate the matrix exponential $e^{tY}$ in closed form?

Also, what would be matrix exponential $\exp(tZ)$ of the 'generalization' $$Z = \begin{pmatrix} 0 & - \overline{z} & - \overline{z} \\ z & 0 & - \overline{z} \\ z & z & 0 \end{pmatrix} $$ with $z = e^{i \theta}$ again?

答:
Setting $z  = e^{i \theta}$ is a good idea. It becomes a bit clearer if $(- e^{-i \theta})$ is replaced by $-1/z$ instead of $-\overline z$ (and it makes the result correct even for complex $\theta$).

So we have
$$
Y = \begin{pmatrix}
 0 & -1/z & 0 \\
 z &  0   & -1/z \\
 0 &  z   & 0
\end{pmatrix}
$$
and the first powers are
$$
Y^2 = \begin{pmatrix}
 -1 & 0 & 1/z^2 \\
 0 &  -2   & 0 \\
 z^2 &  0   & -1
\end{pmatrix}\, , \,
Y^3 = \begin{pmatrix}
 0 & 2/z & 0 \\
 -2z &  0   & 2/z \\
 0 &  -2z   & 0
\end{pmatrix}\,. \\
$$
One can see that $\boxed{Y^3 = -2Y}$, which allows to compute all powers $Y^n$ in terms of $Y$ or $Y^2$:
$$
 Y^{2k+1} = (-2)^{k} Y \\
 Y^{2k+2} = (-2)^{k} Y^2
$$
for $k \ge 1$. Therefore
$$
\begin{align}
\exp(tY) &= I + \left(t-\frac{2t^3}{3!} + \frac{2^2t^5}{5!} - \frac{2^3t^7}{7!} + \ldots\right)Y
\\ &\quad + \left(\frac{t^2}{2!} - \frac{2t^4}{4!} + \frac{2^2t^6}{6!} - \frac{2^3t^8}{8!} + \ldots \right)Y^2 \\
&= I +  \frac{\sin(\sqrt 2 t)}{\sqrt 2}Y +  \frac 12 \left(1- \cos(\sqrt 2 t)\right)Y^2 \, .
\end{align}
$$

The general case is described in Computing the Matrix Exponential The Cayley-Hamilton Method: If $A$ is an $n$-dimensional square matrix and $\lambda_1, \ldots, \lambda_n$ the zeros of the characteristic equation $\det(\lambda I - A) = 0$, then
$$
 \exp(tA) = \sum_{k_0}^{n-1} \alpha_k A^k
$$
where $\alpha_0, \ldots, \alpha_{n-1}$ are the solutions of the linear equation system
$$
 e^{\lambda_i t} = \sum_{k_0}^{n-1} \alpha_k \lambda_i^k \, , \,
1 \le i \le n \, .
$$

In our case $\det(\lambda I - Y) = \lambda^3 + 2 = 0$ has the zeros $\lambda_1 = 0$, $\lambda_2 = i\sqrt 2$, $\lambda_3 = -i \sqrt 2$. The linear equation system is
$$
\begin{align}
 1 &= \alpha_0 \\
e^{i\sqrt 2 t} &= \alpha_0 + i \sqrt 2 \alpha_1 - 2 \alpha_2 \\
e^{-i\sqrt 2 t} &= \alpha_0 - i \sqrt 2 \alpha_1 - 2 \alpha_2
\end{align} \, .
$$
The solution is
$$
 \alpha_0 = 1, \, \alpha_1 = \frac{\sin(\sqrt 2 t)}{\sqrt 2}, \,
\alpha_2 = \frac 12 \left(1- \cos(\sqrt 2 t)\right)
$$
confirming the result for $\exp(tY)$ that we obtained above.


问:
Limit of a recursive sequence with $u_n$
It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.

The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.

答:
If you are only interested in finding the limit, then here is an algebraic way to find it. Assume

$$\lim_{n\to \infty} u_n = a \implies \lim_{n\to \infty} u_{n+1} = \lim _{n\to \infty }(1+\frac{1}{u_n}) \implies a = 1+\frac{1}{a}\implies a^2-a-1=0  $$

$$ \implies a = \dots\,. $$

Note that, the terms of the sequence are positive.


问:
Merging multiple non linear equations into one
I have 3 nonlinear equations in 3 unknowns,

$\Large A=\frac{P_L}{1-c}\quad B=\frac{P_L}{1-ce^{-10k}}\quad  C=\frac{P_L}{1-ce^{-20k}}$

where $A$, $B$, $C$ are known values and $P_L$, $c$ and $k$ are unknowns. I want to get them into one single equation in one of three unknowns ($P_L$, $c$ or $k$, does not matter) so I can use one of the root finding methods to determine that single unknown.

Body:
You have $$c=1-\frac{P}{A}$$

$$ce^{-10k}=1-\frac{P}{B}$$
$$ce^{-20k}=1-\frac{P}{C}$$

Therefore $$(1-\frac{P}{B})^2=(1-\frac{P}{A})(1-\frac{P}{C})$$


问:
Surprising results about the geometry and topology of Lie groups
I am new to the study of Lie groups, and I would like to know beforehand what are the most surprising and important results in this field of study. For example, as a person especially interested in algebraic topology, I find this post extremely interesting. What other results describe the particular geometry and topology of these objects?</p>

答:
One of my favourite results is a description of the cohomology of a compact Lie group in terms of invariants in the exterior algebra of its lie algebra.

To be more precise:

Let $G$ be a compact Lie Group and let $g$ be its Lie algebra (that can be viewed as the algebra of left translation invariants vector fields on $G$).

Conjugation induces an action of $G$ on $g$.
If you identify $g$ with the tangent plane at $e$ of $G$, this action extends to a degree preserving action on $\Lambda g$, i. e. the space of left invariant differential forms.

A very famous theorem due to Cartan and proved in a more Lie-theoretic flavour by Reeder (see  M.Reeder, "On the cohomology of compact Lie groups", L'Enseignement Math., t. 41, (1995), pages 181-200 )  , asserts that

$$H^*(G) \simeq (\Lambda g )^G$$

as graded vector spaces, where $(\Lambda g )^G$ is the ring of invariant differential forms for the above described action.

Moreover, a beautiful and instructive overview about the XX century topological discoveries concerning Lie groups can be found in the very illuminating

A. Borel "Topology of Lie groups and characteristic classes." Bull. Amer. Math.
Soc.,Vol. 61, number 5 (1955), 397-432


答:
In a right angle triangle two sides which are equal will base and perpendicular because hypotaneous is always greatest side in right angle triangle.
so one angle is $90^\circ$ and sum of other two is also $90^\circ$ since sides are equal so there angle made with other side is also same so each will be $45^\circ$.

Title: Right-angled isosceles triangles

Body:
If a right-angled triangle is isosceles then the other two angles must be equal to $45^\circ$ ?

Is this always the case or are there other possible right-angled isosceles triangles?


问:
Cardinalities in which abelian group theory is categorical?
I want to find all cardinalities in which abelian group theory is categorical. It is clear for some finite cardinalities (how to determine them?) by fundamental theorem of finite abelian groups, but what about infinite ones?</p>

答:
For any infinite set $X$, the groups $(\mathbb{Z}/2\mathbb{Z})^{\oplus{X}}$ and $(\mathbb{Z}/3\mathbb{Z})^{\oplus{X}}$ are two non-isomorphic abelian groups having the same cardinality as $X$.

For a finite cardinality $n$, all abelian groups of order $n$ are isomorphic if and only if $n$ is square-free.


问:
Lie derivative of this 1-differential form
Let M be a compact manifold provided with an action of a compact lie group G. Let $\langle.,, \rangle$ be a G-invariant Riemannian metric on M.

Let $X \in \mathfrak{g}$, let $\theta_X \in \Omega^1(M)$ be a 1-form on M, s.t $\theta_X(Y)= \langle X_M ,Y \rangle, $ $Y \in TM$.

How to prove that $\mathcal{L} (X)\theta  = 0 ?$

I've tried this:

Let $\phi_t: M \rightarrow M, $ $ \phi_t(m) = e^{-tX}.m$. let  $Y \in TM$
$(\mathcal{L} (X)\theta_X)(Y) = {\frac{d}{dt}}_{|t=0} (\phi^*_t(\theta_X))(Y) = {\frac{d}{dt}}_{|t=0} \langle X_M, (\phi_t)_*(Y) \rangle = \langle {\frac{d}{dt}}_{|t=0}  X_M, {\frac{d}{dt}}_{|t=0} (\phi_t)_*(Y) \rangle  = 0 $,

since $X_M$ does not dependent of t.

Is this correct ?

答:
Let $x\in M$, and $Y\in T_xM$, we have:

$\phi_t^*\theta_X(Y)=\theta(d\phi_t.Y)=\langle X_M(\phi_t(x)),d\phi_t.Y\rangle=\langle d\phi_t.X(x),d\phi_t.Y\rangle=\langle X(x),Y\rangle$.

This implies that ${d\over{dt}}_{t=0}\phi_t^*\theta= L_X\theta_X=0$.


问:
Atiyah MacDonald: Proposition 3.5
In Atiyah-Macdonald's Commutative Algebra, on page 39, Proposition 3.5, it says

Let $M$ be an $A$-module. Then the $S^{-1}A$-modules: $S^{-1}M$ and $S^{-1}A \otimes_{A}M$ are isomorphic.

Since $S^{-1}A$ and $M$ are $A$-modules so is $S^{-1}A\otimes_{A}M$.

My question is: how can one tell $S^{-1}A\otimes_{A} M$ is an $S^{-1}A$-module?

Is it true that if $M$ is an $A$-module and $S$ being a multiplicatively closed subset of $A$. Then is $M$ also $S^{-1}A$-module?

答:
Not at all. This is just a particular instance of a general result:

If $A\longrightarrow B$ is a (commutative) $A$-algebra, and $M$ an $A$-module, the $A$-module $B\otimes_AM$ is in a natural way a $B$-module  through the scalar multiplication
$$\beta((b\otimes m) \overset{\text{def}}{=}(\beta b)\otimes m.$$
This $B$-module is said to be ‘obtained by extension of scalars’ from the $A$-module $M$.

Title: Atiyah MacDonald: Proposition 3.5


问:
Noether‘s Theorem and Moment Maps
Noether‘s Theorem says that every continuous symmetry of a physical system (i.e., a Lie group action on phase space ${\bf R}^{2n}$ preserving a Hamiltonian $H$) leads to a conservation law (i.e., a function $I$ satisfying $\left\{I,H\right\}=0$).

In the literature one finds this as an application of the method of moment maps. The general result is that for a group action of a Lie group $G$ on a symplectic manifold $(M,\omega)$ one has the associated moment map $\mu\colon M\to {\mathfrak g}^*$, in particular for fixed $X\in {\mathfrak g}$ a map $\mu^X\colon M\to{\bf R}$, and then $\left\{\mu^X,H\right\}=\omega(X_H,X)$ for the Hamiltonian vector field $X_H$. The right hand side equals $\iota_X dH$, which vanishes because the Lie group action is assumed to preserve $H$.

I do not understand, why the literature makes the assumption that the Lie group preserves the symplectic form and, more seriously, I do not understand why this assumption should be true in the setting of Noether‘s Theorem. That theorem is just about continuous symmetries preserving the given Hamiltonian on ${\bf R}^{2n}$, so why should they preserve the symplectic form $\sum_k dx_kdy_k$?

答:
That theorem is just about continuous symmetries preserving the given Hamiltonian on ${\bf R}^{2n}$, so why should they preserve the symplectic form $\sum_k dx_kdy_k$?

No it is not. In classical mechanics appropriate coordinate changes are the canonical transformations or, in mathematical language, symplectomorphisms, of the phase space.

This is maybe an important point: we want transformations that preserve the nature of the evolution equations (aka Hamilton equations); these turn out to be symplectomorphisms of the phase space.

Such a transformation is a symmetry if it preserves the Hamiltonian.

Now, any diffeomorphism on the configuration space lifts to a canonical transformation.

If your Hamiltonian is of the form "potential + kinetic" and the potential energy $U$ depends only on the position variables, then any diffeo of the configuration space will in fact lift to a symmetry (in the above sense) of the mechanical system. More generally, in the Lagrangian formulation here, perturbing the $q$ coordinates implicitly perturbs "their derivative" $p$ coordinates. If the Lagrangian is "invariant under small perturbations" you will get that the corresponding "perturbation" on the phase space is a symmetry -- i.e. is given by a symplectic vector field (one generating symplectomorphisms) and preserving the Hamiltonian; you will have to go through the Legendre transform from the Lagrangian to the Hamiltonian picture first, though.

On the other hand, there are symplectomorphisms that are not lifts of diffeos on the configuration space, so canonical transformations - aka symplectomorphisms - are more general than that (though not too much more general if you look at things locally, but that is a different story to think about).

Title: Noether‘s Theorem and Moment Maps


问:
Number of paths in Cayley graph of free group / random walk on tree / Cayley graph of free group
I want to calculate number of paths of $n$ "components" with length $k$ which goes in Cayley graph of free group $F_2$, i.e.:

$F_2 =  = <S>$ - free group with 2 generators, how much words $w = g_1 * ... * g_n, g_i \in S$, such that $l(w) = k$, where $l$ - length of word (we reduce our word to shortest and count all letters) exists?

I`d try to google "random walk on Cayler group of free group"/"word lengths in free group" and found nothing

I have next idea:

Denote $p_n(g, h)$ - number of ways from vertex $g$ to vertex $h$. Then I have the following hypothesis (I don't know is it true, it looks correct, but I didn`t proof it yet): $p_n(g, h) = p_n(e, g^{-1}h)$ - depends only on length of $g^{-1}h$. I need that to calculate recurrence relation for $p_n(0, 0)$:

$p_n(0) = 4 * p_{n-2}(0) + p_{n-2}(2)$, where $p_n(k)$ - number of words $g_1 *...* g_n$ of length $k$ (i.e. if we reduce our word, it will have exactly $k$ letters)

And here I stuck:

I didn`t proof the hypothesis yet (it looks like something trivial, but still)

I only find recurrence relation for $p_n(0)$, but I want to find all $p_n(k)$

Any suggections?

答:
Your hypothesis is trivially true: a walk from $g$ to $h$ is found by taking a walk from $0$ to $g^{-1}h$, and left multiplying all steps by $g$. This is bijective, with inverse map of

Finding a recurrence for all $n,k$ was quite tricky to think about, because there is a special case when $k=1$. Here is a recurrence which works for all $n,k$:
$$
p_{n}(k)=\begin{cases}
3p_{n-1}(k-1)+p_{n-1}(k+1) & k\neq 1\\
4p_{n-1}(0)+p_{n-1}(2) & k = 1
\end{cases}
$$
To explain, consider a string $s$ whose un-reduced length is $n$, which reduces to a string $r$ with length $k$. Let $s'$ be the result of deleting the first entry of $s$, and let $r'$ be the reduced version of $s'$.

If the first symbol of $s$ is the opposite of the first symbol of $r'$, then $r'$ must have length $k+1$. Such strings are counted by $p_{n-1}(k+1)$.

If the first symbol of $s$ is the not the opposite of the first symbol of $r'$, then $r'$ must have length $k-1$. There are three choices for the first symbol, so these are counted by $3p_{n-1}(k-1)$.

Finally, if $r'$ is the empty string, then the first symbol of $s$ can be any of four choices.


问:
How can I prove that $b^n$ is $ O(n!)$
Hi guys I came upon an exercise from the Big-O chapter that I can not solve.I know that in order to be in $O\bigl(f(n)\bigr)$ has to be smaller or equal to $c$ * $f(n)$ where is c is a constant and is positive but I never solved an exercise where another constant, $b$ in my case, is involved ($b$ is greater than 1 and is real). Any help is highly appreciated, sorry for the writing issues..

答:
For $b>1$, $m=\lfloor  b  \rfloor$ and enough big $n$
$$ \frac{b^n}{n!} = \underbrace{\frac{b}{1}\cdot \frac{b}{2} \cdots \frac{b}{m}}_{\lfloor b \rfloor}\cdot \underbrace{\frac{b}{m+1}\cdot \frac{b}{m+2} \cdots \frac{b}{n}}_{n-\lfloor  b  \rfloor}<\frac{b^m}{m!} \cdot \left(\frac{b}{m+1} \right)^{n-m} \to 0$$

Title: How can I prove that $b^n$ is $ O(n!)$


问:
Quadrilateral $ABCD$ has $∠BDA = ∠CDB = 50^\circ, ∠DAC = 20^\circ$ and $∠CAB = 80^\circ.$ Find angles $∠BCA$ and $∠DBC.$

Quadrilateral $ABCD$ has $∠BDA=∠CDB=50°$, $∠DAC=20°$ and $∠CAB=80°$. Find angles $∠BCA$ and $∠DBC$.

This seems to be a very easy problem, but I am not able to solve it.

I am attaching a diagram which also contains all the simple angle chasing I have done.

I can not see any cyclic quads either or any parellelograms so I am not sure what to do.

答:
While $ABCD$ is not a cyclic quadrilateral, if we extend $DC$ to point $E$ such that $\angle ABE = 80^0, \, ADEB$ is a cyclic quadrilateral. We then have,

$\angle AEB = \angle ADB = 50^0\,$ and $\angle BAE = \angle BDE = 50^0$.

So, $\triangle ABE$ is isosceles.

As $\angle BAE = 50^0, \angle DAE = 50^0$ so $\angle CAE = 30^0$.

Also, $\angle ACE = \angle ADE + \angle CAD = 120^0$

So, $\angle AED = 30^0 = \angle CAE$.

$\triangle ACE$ is isosceles.

So perpendicular from $B$ and $C$ will both meet $AE$ at its midpoint which means $BC$ is perpendicular to $AE$.

That leads to $\angle CBE = 180^0 - (90^0 + 50^0) = 40^0 \implies \angle DBC = 10^0$.

So we also have in $\triangle ABC, \angle ACB = 60^0$


问:
real analysis and integral
Let $f$ be a continuous function on $[a, b]$ satisfying
$$\int_a^b f(x)g^\prime(x)\,\mathrm{d}x = 0$$
whenever $g$ is a continuously differentiable function on $[a, b]$ satisfying $g(a) =
g(b) = 0$. </p>

<p>Show that $f$ must be constant. </p>

答:
Hint: Choose any two points $c \neq d$ in $(a, b)$.

Let $\varphi(x) = \max \{ 1 - |x|, 0 \} $ a bump near $x = 0$, and let

$$g_{\epsilon}(x) = \int_{a}^{x} \frac{1}{\epsilon} \left( \varphi\left(\frac{t-c}{\epsilon}\right) - \varphi\left(\frac{t-d}{\epsilon}\right) \right) \, dt. $$

For sufficiently small $\epsilon$, $B_{\epsilon}(c), B_{\epsilon}(d) \subset (a, b)$ and we have $g_{\epsilon}(a) = g_{\epsilon}(b) = 0$. Also, it is not hard to show, by continuity of $f$, that

$$ \int_{a}^{b} f(x) g_{\epsilon}'(x) \, dx \to f(c) - f(d) \quad \text{as} \quad \epsilon \to 0. $$

Thus $f(c) = f(d)$ for all such $c, d$ and it follows that $f$ is constant on $[a, b]$.

Title: real analysis and integral


问:
Exact Differential 1-form on $ R^{3}$
Check this link for the question:  total differential means exact in this case

The question want to show $ I\eta$ is not exact ($I\eta \neq dq$, where d is exterior derivative and q is 0-form). I am failing to prove that $I\eta$ is not  $C^{1},C^{2},C^{3}$ otherwise $I\eta$ is exact. Any help would greatly appreciated.

答:
SKETCH: I will switch to the letter $f$ for the function. Suppose $f\eta$ is exact. If so, then it must be closed, so
$$d(f\eta) = df\wedge\eta + f\,d\eta = 0.$$
This implies that
$$d\eta\wedge\eta = 0.$$
(Why?) Does this equation hold?

Title: Exact Differential 1-form on $ R^{3}$


问:
Direct limit over cofinal subset of a preordered, not directed, set is isomorphic to the original one
I have been struggling with this exercise I've been assigned. The difficulty is in the fact that the set of indeces is not necessarily directed. Here's the text:

Let $I$ be a preordered set,
$\mathbb{M}=(M_i,\mu_{ii'})$ a direct system of right modules over a
ring $R$, and $(M,\mu_i)$ the direct limit of $\mathbb{M}$. Let
$J\subseteq I$ be a cofinal subset of $I$, that is, a subset of
$I$ such that for every $i\in I$ there exists $j\in J$ with $i\le j$.
Let $\mathbb{N} =(N_i,\nu_{jj'})$ be the direct system with $N_j=M_j$ for every $j\in J$ and $\nu_{jj'}=\mu_{jj'}$ for every $j,j'\in J$, $j\le j'$. Let $(N,\nu_i)$ be its direct limit. Show that there exist an isomorphism $\varphi\colon M\to N$ such that $\varphi\mu_j=\nu_j$ for every $j\in J$.

Of course, using the universal property of direct limit I have been able to prove that there exist a (unique) morphism $\psi:N\to M$ such that $\mu_j=\psi\nu_j$. Since $J$ is cofinal, it is easy to see that $\psi$ is surjective. Unfortunately, I haven't managed to prove that it is injective. At this point, I'm not even sure that it is true. Can someone help me, please?

答: The smallest non-directed and non-discrete poset already furnishes an example: the colimit of $a\leftarrow b\to c$ is a pushout while the colimit of $a\quad c$ is a coproduct,


问:
Last man standing strategy
Let's imagine a last man standing scenario in a video-game. A last man standing situation is one where each player is for himself and has to eliminate all other players to win. For analyzing the situation, consider the game mechanics of a simple shooter game like CSGO.

There are 3 people in this last man standing type of game, they all have the same level of skill and luckily, you aren't in combat. Yet.

Let's assign each player a letter, you are A and the other two are labelled B and C. B and C are in close combat, while you, on the other hand, have a sniper and have the choice to shoot at number B or C. Number B has a health of 30 and number C has 80 health. A shot from the sniper will do damage of fifty.

Considering you don't miss shots, Who would you shoot? My brother said ''Well I would shoot at B to eliminate him out of the equation.'' But think about it this way: If you shoot at C, he is down to 30 health and when either of them is done with his enemy (I am talking about B and C) he will have 30 health, even in the best scenario. So, when it comes to 1v1, you will have a much better advantage.

I think that the best strategy for last man standing scenarios are to feed and assist the weaker (but not too much) so that they will eat each other to the point where it's a Pyrrhic victory.

Which strategy has the best outcome for Last Man Standing scenarios? Should they 'level' the playing field by reducing the opponent's health or should he go for eliminating whole players out of the game?

答:
What you might be looking for in game theory the games are called Truels (three people) or N-uels (n people) or the simplified version Duals (two people). There are three kinds to consider

Sequential (fixed order): The players fire one at a time in a fixed, repeating sequence, such as A, B, C, A, B, C, A...

Sequential (random order): The first player to fire, and each subsequent player, is chosen at random among the survivors.

Simultaneous: All surviving players fire simultaneously in every round.

If needed a firing order is chosen the when it is the players turn to fire (or they all fire at once) each may fire at either of their co-players. A player's marksmanship is the probability that he hits his chosen target. Throughout the game each player has no information as to the previous strategy choices of his co-players, except through their successful shots. There can also be variants of the rules in that each player will have only $n$ shots available or giving the players the ability to abstain from shooting in a round. All players share the same goal: to be the one survivor. In order to maximize their payoff, players have to chose strategies that maximize their survival probability.

As you get to fire first it looks like a sequential fixed order game without the possibility of abstaining from shots is the closest to what you are describing. A player who is about to fire, whether their turn comes up in a fixed order or is chosen at random, has a choice whom to fire. To make this choice optimally, we assume that a player considers only their own survival probability. Regardless of what the other players do, a player in a sequential truel maximizes their survival probability by firing at the opponent against whom it would less prefer to fight a duel. If a player's shot misses, then it makes no difference who was the target; but if a shot hits the target, the shooter is better off, because their opponent in the next duel is weaker. Thus, a player who follows this stronger opponent strategy fires at the opponent whose marksmanship is highest, trying to eliminate this player so as to set up a duel with the lower marksmanship player.

In you case you give health so you should target the player with the most health as you would prefer to try and set up a dual the weaker player.

Title: Last man standing strategy


问:
At least one of each gender
A class of 22 students consists of 12 girls and 10 boys. The class selects 3 students for student council. What is the probability that at least one of each gender is on student council?</p>

答: number of groups of three with at least one of each gender= total number of groups of three - number of groups of three made up of people with the same gender= $\binom{22}{3}-\binom{12}{3}-\binom{10}{3}$; also the total number of groups of three is $\binom{22}{3}$ thus $P=\frac{\binom{22}{3}-\binom{12}{3}-\binom{10}{3}}{\binom{22}{3}}=\frac{60}{77}\approx 78$%

Title: At least one of each gender


问:
A weird directional derivative, escalar function composed with a real function
Let $f:\mathbb{R}\to\mathbb{R}$ be a diferenciable function. Define $\Phi(x,t):=f(x-t)$. If $f'(0)=6$, calculate the directional derivative of $\Phi$ in the direction of the vector $v=(-19,0)$. (It's not necessary any point.)

答:
$\partial_v\Phi(x,t)=\Phi'(x,t)v$. We have

$$\Phi'(x,t)=(\partial_xf(x-t)\quad \partial_tf(x-t))=(f'(x-t)\quad -f'(x-t))$$
Then the directional derivative is $\partial_v\Phi(x,t)=\Phi'(x,t)v=-19f'(x-t)$. For $x=t$, this is an explicit number, with the information you are given.


答:
Consider the complement - what is the probability that all 3 students are of the same gender?
$P(\textrm{All girls}) = \frac{12}{22}\times\frac{11}{21}\times\frac{10}{20}=\frac{1}{7}$

$P(\textrm{All boys}) = \frac{10}{22}\times\frac{9}{21}\times\frac{8}{20}=\frac{6}{77}$

Thus, the probability is $1-(\frac{1}{7}+\frac{6}{77})=\frac{60}{77}\approx78\%$

Title: At least one of each gender

Body: A class of 22 students consists of 12 girls and 10 boys. The class selects 3 students for student council. What is the probability that at least one of each gender is on student council?</p>


问:
sum of monotonic increasing and monotonic decreasing functions
I have a question regarding sum of monotinic increasing and decreasing functions. Would appreciate very much any help/direction:

Consider an interval $x \in [x_0,x_1]$. Assume there are two functions $f(x)$ and $g(x)$ with $f'(x)\geq 0$ and $g'(x)\leq 0$. We know that $f(x_0)\leq 0$, $f(x_1)\geq 0$, but $g(x)\geq 0$ for all $x \in [x_0,x_1]$. I want to show that $q(x) \equiv f(x)+g(x)$ will cross zero only once. We know that $q(x_0)\leq 0$ and $q(x_1)\geq 0$.

Is there a ready result that shows it or how to proceed to show that? Many thanks!!!

答:
Alas, the answer is no.

$$f(x)=\begin{cases}-4& x\in[0,2]\\ -2& x\in [2,4]\\0& x\in[4,6]\end{cases}$$

$$g(x)=\begin{cases}5 & x\in [0,1]\\3& x\in[1,3]\\1& x\in[3,5]\\ 0 & x\in[5,6]\end{cases}$$

$$q(x)=\begin{cases} 1 & x\in [0,1]\\ -1& x\in[1,2]\\ 1 & x\in[2,3]\\ -1 & x\in[3,4]\\ 1& x\in[4,5]\\ 0 & x\in[5,6]\end{cases}$$

This example could be made continuous and strictly monotone with some tweaking.