问:
Does $y = f(x) = ax+b$ actually have two mappings inside it?
I’m just a high school student, so I may be somewhat logically flawed in understanding this.

According to wikipedia, the definition of function requires an input $x$ with its domain $X$ and an output $y$ with its domain $Y$, and the function $f$ maps $x$ to $y$.

But how about $f(x)$? I often see syntaxes such as $f(1) = 0$  in my text book. Doesn’t that mean it is $f(x)$ being first assigned a value and then transfer the value into $y$? So, there must be two transitions/mappings between the input $x$ and the output $y$ right?

My conceptual model of function is like this: A definition of function requires an input $x$ with its domain $X$, a forwarder $f(x)$ with its domain $F$ and an output $y$ with its domain $Y$. The function $f$ first maps $x$ to $f(x)$ then maps $f(x)$ to $y$.

These two definitions are not quite the same.

On 2022.6.29: The picture below had solved my confusion.

答:
We talk about domains and codomains of a function, not of the variables of a function.  You might not have come across the term codomain before, but I think that it's the best for what you're trying to describe.

So, the domain of $f$ is $X$ and the codomain of $f$ is $Y$.  Often, we write this as $f:X\rightarrow Y$ to indicate that the valid inputs to $f$ are points in $X$ and every output of $f$ is a point in $Y$.

In your example, $x$ is a point in $X$, which is the domain of $f$, and $y$ is a point in the codomain of $f$, which is $Y$.


问:
If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.
If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.

I don't think I can find $x$ because using $a/(1-r)$ is only one equation for two variables.

Looking it up on wolfram alpha, it says $x/(x-1)^2$ when $|x| < 1$. Can someone explain to me please?

ContentLicense: CC BY-SA 4.0

Body:
Hint

$$\sum_{k=1}^{\infty}kx^k=x\sum_{k=1}^{\infty}kx^{k-1}=x\left(\sum_{k=1}^{\infty}x^{k}\right)'$$


答:
You can observe what happens to the summation when you expand the first terms:

$$
\sum^{\infty}_{k=1} kx^k  =
1(x) + 2(x^2) + 3(x^3) + \dots
$$

This can be rearranged in a way that will yield in a sum of geometric series:
$$
1(x) + 2(x^2) + 3(x^3) ... =
x + (x^2 + x^2) + (x^3 + x^3 + x^3) + \dots
$$

Now observe what taking the first element of each term, you have a geometric series that starts a $x^1$.

But starting from the second term, taking the second element from all terms you also have a geometric series that starts on $x^2$.

So you actually have a sum of series, each one starting at plus one higher coefficient.

$$
\sum^{\infty}_{k=1} kx^k = \sum^{\infty}_{k=1}\sum^{\infty}_{j=k} x^j
$$

Now you can solve the first summation: $a_0 = x^k$, then:
$$
\sum^{\infty}_{j=k} x^j = \frac{x^k}{1-x}
$$

Now solve the second summation:

$$
\sum^{\infty}_{k=1}\frac{x^k}{1-x} = \frac{1}{1-x}\sum^{\infty}_{k=1}{x^k} = \frac{x}{(1-x)^2}
$$

Body:
If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.

I don't think I can find $x$ because using $a/(1-r)$ is only one equation for two variables.

Looking it up on wolfram alpha, it says $x/(x-1)^2$ when $|x| < 1$. Can someone explain to me please?


问:
Magma : Load file using variable file name
I want to load an external Magma file within another Magma file. (Both files are saved in the same directory.) I want to be able to quickly change which external file is being loaded, ideally at the beginning of the file making the load call, so that I can easily run the same code with various inputs.

(The external file contains computations, whose ultimate result is used by the file making the load call. These computations vary depending on the object being analyzed.)

I tried creating a string-type variable that stores the external file's name, then using Magma's load command with this variable. For example,

```<code>fileName := "externalMagmaFile.txt";
load fileName;
</code>```

However, this results in the error

User error: Could not open file "fileName" (No such file or directory)

The same error results when I include double quotes around the external file name:

```<code>fileName := "\"externalMagmaFile.txt\"";
load fileName;
</code>```

It seems that, for the load command, Magma interprets the variable name as the string specifying the file name, instead of first evaluating the variable, then executing load.

(I am using Magma V2.23-1 on MacOS Version 10.15.5.)

Can I use a variable with the load command in Magma? If yes, how?

答:
Another solution (which is also a bit of a work around) is to set the Magma path based on which file you want to load using the SetPath() function. The Magma path indicates which directories are searched for loading Magma files.

To clarify, if you have 2 different files by the same name in separate directories (or can set it up so that you do) and want to choose which one to load, you can set filepath := "directory1" or filepath := "directory2" and then SetPath(filepath) before loading the file.

If you are concerned about changing the path, you can use GetPath() to save off the previous one and restore afterwards. Or add the new path to the end of the current path before calling SetPath().

Side note: I would have added this as a comment rather than an answer since it is a workaround and it is not the ideal solution for every scenario. But adding comments requires reputation, and I came upon this question because I was in a similar situation and this answer would have helped me. In my case I wanted to load several files from the same path (where the path is a variable determined at runtime) so this worked better for me than trying to do a dummyloadfile for each. Also, sadly, the true answer to your question (afaik) is that there is no direct way to load a file using a variable name.


问:
Find $\frac{\mathrm{d} }{\mathrm{d} x}[f(f(x))-f(x^2)])(3)$
find $\frac{\mathrm{d} }{\mathrm{d} x}[f(f(x))-f(x^2)])(3)$

if $f(3)=3,  \frac{\mathrm{d f} }{\mathrm{d} x}(3)=2,   \frac{\mathrm{d f} }{\mathrm{d} x}(\sqrt{3})=1$ y $\frac{\mathrm{d f} }{\mathrm{d} x}(9)=1$

pls I need a hint

Body:
$\frac {d}{dx} [f(f(x)) + f(x^2)] = f'(f(x))f'(x) + 2xf'(x^2)$

Evaluated at 3.

$f'(f(3))f'(3) + 6f'(9)$

Plug in the values you know... $f(3) = 3, f'(3) = 2, f'(9)  = 1$

$2f'(3) + 6$

And again...

$4 + 6 = 10$


问:
Summation of a series evaluation using fourier series
Is there any procedure to simplify summation of a series, such as,

$$\begin{align}S &=1-\frac 1 3+\frac 1 {10}-\frac 1 {42}+\frac 1 {210}-\frac 1 {1320}+.....=\sum_{n=0}^\infty\frac{(-1)^n}{n!(2n+1)}\end{align}$$

using fourier series expansion, or any criteria should also be given?

If this type of questions has already been asked, please send the link.

Any help will be appreciated, Thank you.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{%
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!\pars{2n + 1}}} =
\sum_{n = 0}^{\infty}
{\pars{-1}^{n} \over n!}\int_{0}^{1}x^{2n}\,\dd x
\\[5mm] = &\
\int_{0}^{1}\sum_{n = 0}^{\infty}
{\pars{-x^{2}}^{n} \over n!}\,\dd x
=
\int_{0}^{1}\expo{-x^{2}}\,\dd x
\\[5mm] = &\
\bbx{{\root{\pi} \over 2}\on{erf}\pars{1}}
\approx 0.7468 \\ &
\end{align}

$\ds{\on{erf}}$ is the Error Function.


问:
How to prove that the first quadrant of the plane is closed.
How can I prove, in topology, that:

$$X = \{(x,y)\in\mathbb{R}^2;y<0;x\in\mathbb{R}\}\cup\{(x,y)\in\mathbb{R}^2;x<0,y\geq0\}$$
is open in $\mathbb{R}^2$?

I tried doing by its complement, but how can I prove that
$$Y= \mathbb{R}^2\setminus X=\{(x,y)\in\mathbb{R}^2;x,y\geq0\}$$
is closed in $\mathbb{R}^2$?

I tried to use, as well, this statement:

$Y$ is closed if and only if for each $y\in \mathbb{R}^2\setminus Y$ there is a neighbourhood $N$ of $y$ such that $N\subseteq \mathbb{R}^2\setminus Y$.

答: Since $X=(\mathbb R\times\mathbb R^-)\cup(\mathbb R^-\times\mathbb R)$, you have that $X$ is the union of two obviously open sets.


问:
$\int \sqrt{x^4+1}\ dx$
I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.

答:
By the reduction and transformation formulas mentioned in my Elliptic Integrals and Functions monograph
$$\int_0^y\sqrt{x^4+1}\,dx=\int_0^y\frac{x^4+1}{\sqrt{x^4+1}}\,dx=\frac{y\sqrt{y^4+1}}3+\frac23\int_0^y\frac1{\sqrt{x^4+1}}\,dx$$
$$=\frac{y\sqrt{y^4+1}}3+\frac23F\left(\sin^{-1}\sqrt{y^2+1-\sqrt{y^4+1}},m=\frac12\right)=\frac{y\sqrt{y^4+1}+F(2\tan^{-1}y,1/2)}3$$
In hypergeometric terms the integral is
$$y{_2F_1}(-1/2,1/4;5/4;-y^4)$$

Title: $\int \sqrt{x^4+1}\ dx$


问:
Evaluating $\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$ without de l'Hospital rule
Is it possible to evaluate this limit without de l'Hopital's rule?
$$\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$$

答:
Why not to use
$$1-x^a=-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)$$
$$\frac {1-x^a}{1-x^b}=\frac{-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)}{-b (x-1)+\frac{1}{2} \left(b-b^2\right) (x-1)^2+O\left((x-1)^3\right)}$$
$$\frac {1-x^a}{1-x^b}=\frac{a}{b}+\frac{a  (a-b)}{2 b}(x-1)+O\left((x-1)^2\right)$$ which shows the limit and also how it is approached.


问:
Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.

That's what I've tried:

Let a Cauchy-Schwarz Inequality be :

\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2 )\left(\left(\cfrac{ab}{\sqrt{a}}\right)^2 +\left(\cfrac{ac} {\sqrt{b}}\right)^2+\left(\cfrac{bc}{\sqrt{c}}\right)^2\right)& \ge (ab+ac+bc)^2 \\
(a+b+c)(ab^2+\cfrac{a^2 c^2}{b}+b^2 c) &\ge (ab+ac+bc)^2 \\
\end{array}

However how should I now factorize the left hand side of the inequality as $(a^2+b^2+c^2) \cdot Something $ ?

I think I've made the problem harder than it needs to be .

答: I have given a detailed proof using an easier method. Please click on this link to see the proof

ContentLicense: CC BY-SA 3.0


问:
Monotone Convergence Property $\iff$ Order Completeness in an arbitrary ordered field.
I have found an answer here but having some doubts about the first answer(highest rep.) posted by Jonas Meyer.

There in Claim 1. He states for all $\epsilon>0$, $U-\epsilon:=\{u- \epsilon: u \in U\}$ is not contained in $U$.

He proved by method of contradiction, he writes "If $U-\epsilon \subseteq U$, then $U-n\epsilon \subseteq U$" for all $n$.

My question is how can we conclude this statement?

Edit: I have a few more questions:

After proving claim 2, he moves on to construct an increasing sequence.
He writes "There exist $n_2>n_1$ such that $x_2 \notin U-\frac{1}{n_1}$" How

can we obtain that $n_2>n_1$? Can it not be smaller than $n_1$?

After creating the monotone increasing sequence $(x_k)$ it is written "(x_k)" is bounded above by each element of U. How can we conclude this?

In the second last line it is written "$\bigcap_{j=1}^\infty U-\frac{1}{n_j}=\bigcap_{n=1}^\infty U-\frac{1}{n}$". How can we conclude this?

答:
By induction on $n$. Suppose that $U-\epsilon\subseteq U$. If $U-n\epsilon\subseteq U$ for some $n\in\Bbb N$, then

$$U-(n+1)\epsilon=(U-\epsilon)-n\epsilon\subseteq U-n\epsilon\subseteq U\,.$$

That’s the induction step, and the claim follows immediately.

Added to answer the additional questions.

For the first question, note that $U-\frac1n\subseteq U-\frac1m$ whenever $n\ge m$.1 Thus, if $x\notin U-\frac1m$, then $x\notin U-\frac1n$ for $n\ge m$. There is certainly some $m$ such that $x_2\notin U-\frac1m$, and we’ve just shown that in that case $x_2\notin U-\frac1n$ for all $n\ge m$, so we can simply take $$n_2=\max\{m,n_1+1\}$$ to ensure that $n_2>n_1$.

1 To see this, suppose that $n\ge m$, and let $x\in U-\frac1n$. Then $$x+\frac1m\ge x+\frac1n\in U\,,$$ so $x+\frac1m\in U$, and hence $x\in U-\frac1m$.

For your second question, note that if $u\in U$, and $v>u$, then $v\in U$. If the sequence is not bounded above by each element of $U$, there are a $k\in\Bbb Z^+$ and a $u\in U$, such that $u<x_k$. But then $x_k\in U$, contradicting the fact that every point of the sequence was chosen not to be in $U$.

For your third question, the fact that

$$\bigcap_{j=1}^\infty\left(U-\frac1{n_j}\right)=\bigcap_{n=1}^\infty\left(U-\frac1n\right)$$

is an immediate consequence of the fact, proved above, that $$U-\frac1n\subseteq U-\frac1m$$ whenever $n\ge m$:

$$U-1\supseteq U-\frac12\supseteq U-\frac13\supseteq\ldots\;.$$

If a point is not in some $U-\frac1n$, then it isn’t in $U-\frac1{n_j}$ for any $n_j\ge n$, and there is certainly an $n_j\ge n$. (In fact the intersection of any infinite subsequence of a decreasing sequence of sets is the same as the intersection of the entire sequence.)


问:
What is a direct proof, formally?
In my logic class I took in college, I sometimes would get wrong marks because I proved a statement indirectly instead of directly. Informally, a direct proof of a conditional $P \implies Q$, is one where you assume $P$ and try to deduce $Q$ "directly", without using the contrapositive of that conditional. But no one has ever given me a formal, rigorous definition of direct proof. Can someone give me one, and with one, can someone give examples of statements which can't be proven directly, only indirectly?

答:
A formal notion of a proof is:

Given a set of axioms $\Delta$ and a formula $\alpha$,
a proof of $\alpha$ is a finite sequence $(x_i)$, $i \in \{1,...,n\}$, such that:

$x_n = \alpha$;

Either $x_j$ , $1 \le j \lt n$, is a axiom or a tautology;

Or $x_j$ , $1 \lt j \lt n$, is deduced through an inference rule from the set of formulas $\{x_k\}$, $k \lt j$.

A standard, and frequently unique, inference rule is the Modus Ponens:
$$\{(\alpha \to \beta), \alpha\} \vdash \beta$$

This notion of proof is from a view of formal systems. As a bibliographic reference, in "Metalogic: An Introduction to the Metatheory of Standard First-Order Logic", Geoffrey Hunter shortly exposes this notion. But this is a wide use notion of proof, far from restricted to this book.

Now to the question of what is a direct proof.

 As indicatet earlier by Some Guy (https://math.stackexchange.com/users/730299/some-guy) from Wikipedia, https://en.wikipedia.org/wiki/Direct_proof, "In order to directly prove a conditional statement of the form 'If p, then q', it suffices to consider the situations in which the statement p is true.". It seems that when someone says that a proof is a indirect proof, it is because it is used reductio ad absurdum or the modus tollens. And that is right, in these cases we use more than just to consider the hypothesis is true, we do negate the conclusion. Lets have a closer look:

modus tollens: $\{(\alpha \to \beta), \lnot\beta\} \vdash \lnot\alpha$

reductio ad absurdum: $\{(\lnot \alpha \to \lnot \beta), (\lnot \alpha \to \beta\} \vdash \alpha$

Both are, actually, valid inference rules in classical logic. They can be deduced just from tautologies and the modus ponens. Then they use can be reduced to the use of modus ponens, but with the crucial use also from the right tautologies. These tautologies are tautologies because of our definitions of the formal classical operators, e.g. $\lnot, \land, \lor, \to,$ etc.

Then why do some peoples desire much more a direct proof or do not trust fully in indirect proofs? The answer lies, I believe, in the credibility of the classical logic. The intuitionists don't believe in the excluded middle, therefore don't they don't hold that the tautology that implies in the validation of the reductio ad absurdum as a inference rule. For these mathematicians, without a direct proof that something is true or that something is false, there has nothing to be said.

I hope this was fruitfull and helpfull, also that there are no mistakens.

Title: What is a direct proof, formally?


问:
How do you find the maximum integer value of $(3\cos \theta-5)^2$ without derivatives?
The problem is as follows:

The owner of a bakery invests $3450\,USD$ in order to prepare $150$
cakes of $1\,kg.$ for his shop. Assuming that he would sell each for
$(B+2)\,USD$, where $B$ is the maximum integer value which it can take
the expression from below:

$$B=(3\cos \theta-5)^2$$

It is known that angle $\theta$ is acute.

Assuming that at the end of the month he is able to sell all his
cakes. Find the profit which it was obtained from the sale of all the
cakes.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{300 USD}\\
2.&\textrm{400 USD}\\
3.&\textrm{350 USD}\\
4.&\textrm{450 USD}\\
\end{array}$

The thing here I'm assuming is that what it is being requested is in short:
Where $P$ is the profit.

$$P=150(B+2)\times 30-3450$$

Thus in the end the question here is how to get the maximum integer value of $B$, as this was specifically stated in the problem.

$B=(3\cos \theta-5)^2$

The only thing which I can remember is that to maximize you could use derivatives. So for that previous equation become into:

$B'=2(3\cos \theta-5)(-3\sin\theta)=-6\sin\theta(3\cos \theta-5)$

From solving this and equating it to zero:

$B=-6\sin\theta(3\cos \theta-5)$

$\sin\theta= 0$

which reduces to

$\theta = 0^{\circ}$

and

$3\cos \theta-5=0$, but this would mean:

$\cos\theta=\frac{5}{3}$

But this solution is not possible in the real set of numbers because $-1<\cos\theta<1$.

In this situation the maximum value attained for that function would be when $\theta = 0^{\circ}$.

This would mean:

$B=(3\cos 0^{\circ}-5)^2=(3-5)^2=4$

But this doesn't seem to be the right approach because the result will at the end be negative?.

Why the approach using derivatives isn't working? Can this be done using another method in precalculus?.

Upon further inspecting the equation. I'm getting the idea that to maximize $B$ I would want $\cos\theta$ to be the less as possible.

Since it mentions that it want an integer value maybe it would mean?

$\cos\theta=\frac{1}{3} \approx 70.53^{\circ}$

Hence:

$B=(1-5)^2=16$

Hence $16\times 150$

Another thing which it came to my mind was:

$0<\cos \theta < 1$

Thus

$-5 < 3\cos \theta-5<-2$

$25 < (3\cos \theta -5)^2< 4$

Although this doesn't seems logical, but would it meant that

the maximum integer value is 24. Hence

$P=150(24+2)-3450=450$

Thus the answer would it be that number?.

Again I'm confused. Can someone help me here?. I'm confused because the sitation doesn't seem to work with derivatives and if using a definition of the boundaries for the cosine function it seems to cause a contradiction. What's wrong here?. Please take the time to read all my effort because it wasn't easy for me and what it would help me the most is an explanation where did my strategy failed?

答: Since we are told that $\theta$ is acute, we have $0 < \theta < \pi/2$.  On this interval, $0 < \cos \theta < 1$, so $3 \cos \theta < 5$.  Hence $$B= (3 \cos \theta - 5)^2 = 9 \left(\frac{5}{3} - \cos \theta\right)^2$$ which shows that as $\cos \theta \to \pi/2$, $B$ increases; specifically, at $\theta = 0$, $B = 25$.  But since $\theta = 0$ is not attainable, the maximum integer value of $B$ must be $B = 24$, for which we can backsolve and get $\theta = \arccos \frac{5 - 2\sqrt{6}}{3}$.  We don't actually need to solve for this specific value of $\theta$, only know that it exists.  The profit is therefore $150(26)-3450 = 450.$


答:
Since $\theta$ is acute

$$0 \lt \cos \theta \leq 1$$

$$\implies 0 \lt 3\cos\theta \leq3$$

$$\implies -5 \lt 3\cos\theta -5 \leq 2$$

$$(3\cos\theta-5)^2 \lt 25$$

Therefore $B = 24$

Yes, the final approach that you made is right - there would be some acute theta for which $B=24$, you aren't really concerned in finding the angle, just it's final value in terms of B

Title: How do you find the maximum integer value of $(3\cos \theta-5)^2$ without derivatives?

Body:
The problem is as follows:

The owner of a bakery invests $3450\,USD$ in order to prepare $150$
cakes of $1\,kg.$ for his shop. Assuming that he would sell each for
$(B+2)\,USD$, where $B$ is the maximum integer value which it can take
the expression from below:

$$B=(3\cos \theta-5)^2$$

It is known that angle $\theta$ is acute.

Assuming that at the end of the month he is able to sell all his
cakes. Find the profit which it was obtained from the sale of all the
cakes.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{300 USD}\\
2.&\textrm{400 USD}\\
3.&\textrm{350 USD}\\
4.&\textrm{450 USD}\\
\end{array}$

The thing here I'm assuming is that what it is being requested is in short:
Where $P$ is the profit.

$$P=150(B+2)\times 30-3450$$

Thus in the end the question here is how to get the maximum integer value of $B$, as this was specifically stated in the problem.

$B=(3\cos \theta-5)^2$

The only thing which I can remember is that to maximize you could use derivatives. So for that previous equation become into:

$B'=2(3\cos \theta-5)(-3\sin\theta)=-6\sin\theta(3\cos \theta-5)$

From solving this and equating it to zero:

$B=-6\sin\theta(3\cos \theta-5)$

$\sin\theta= 0$

which reduces to

$\theta = 0^{\circ}$

and

$3\cos \theta-5=0$, but this would mean:

$\cos\theta=\frac{5}{3}$

But this solution is not possible in the real set of numbers because $-1<\cos\theta<1$.

In this situation the maximum value attained for that function would be when $\theta = 0^{\circ}$.

This would mean:

$B=(3\cos 0^{\circ}-5)^2=(3-5)^2=4$

But this doesn't seem to be the right approach because the result will at the end be negative?.

Why the approach using derivatives isn't working? Can this be done using another method in precalculus?.

Upon further inspecting the equation. I'm getting the idea that to maximize $B$ I would want $\cos\theta$ to be the less as possible.

Since it mentions that it want an integer value maybe it would mean?

$\cos\theta=\frac{1}{3} \approx 70.53^{\circ}$

Hence:

$B=(1-5)^2=16$

Hence $16\times 150$

Another thing which it came to my mind was:

$0<\cos \theta < 1$

Thus

$-5 < 3\cos \theta-5<-2$

$25 < (3\cos \theta -5)^2< 4$

Although this doesn't seems logical, but would it meant that

the maximum integer value is 24. Hence

$P=150(24+2)-3450=450$

Thus the answer would it be that number?.

Again I'm confused. Can someone help me here?. I'm confused because the sitation doesn't seem to work with derivatives and if using a definition of the boundaries for the cosine function it seems to cause a contradiction. What's wrong here?. Please take the time to read all my effort because it wasn't easy for me and what it would help me the most is an explanation where did my strategy failed?


问:
Diagonalizable matrices in $M_{2\times 2}(\mathbb{F}_2)$
List all diagonalizable $2\times 2$ matrices over the a field $F$ consisting of two elements $0$ and $1$.

I want to try and do this using C++, but perhaps this isn't the place to ask. I have an idea as to how I'd do it.

答:
there are only finitely many (and a small number too) of matrices to consider so, of course, you can just write some code that will attempt to diagonlize a $2\times 2$ matrix, and just run over all the relevant matrices.

But in this case it's a lot easier (and more instructive) to figure it out using a bit of linear algebra. For instance, if the characteristic polynomial of a matrix factorizes as the product of distinct linear factors, then the matrix is diagonalizable. Any $2\times 2$ matrix will have characteristic polynomial of degree $2$, and over the field with two elements there are very few such polynomials. Just check them all to see which ones do not factorize as the product of distinct linear factors to find those matrices that might not be diagonalizable. There will not be too many of those, and you can then check them individually.

It will probably take you less time going about this problem mathematically than it would to write and debug your program.


问:
Symmetrizability of conservation laws via mathematical entropy
Let
$$
\frac{\partial \boldsymbol{u}}{\partial t}
+
\sum_{j=1}^d \frac{\partial}{\partial x_j} \boldsymbol{f}_j(\boldsymbol{u})
=
\boldsymbol{0},
$$
be a system of conservation laws and let us assume that we are able to find an additional conservation law
$$
\frac{\partial U(\boldsymbol{u})}{\partial t}
+
\sum_{j=1}^d \frac{\partial}{\partial x_j} F_j(\boldsymbol{u})
=
0,
$$
where $U$ is a strictly convex function. (This is common for many physical system.) Can we now say that the function $U$ is a strictly convex mathematical entropy and conclude that the original system of conservation laws is symmetrizable (and thus locally well-posed)?

In Godlewski, Raviart: Numerical Approximation of Hyperbolic Systems of Conservation Laws (1996) the authors say that a convex function $U$ is a mathematical entropy if there exist functions $F_j$ (entropy fluxes) such that
$$
U'(\boldsymbol{u})\boldsymbol{f}_j'(\boldsymbol{u})
=
F_j'(\boldsymbol{u}).
$$
If this condition holds then, of course, the additional conservation law is satisfied. However, this does not work the other way around. And yet, in Examples 3.1 and 3.2 the authors only check whether the additional conservation law, and not the condition above, holds. Could someone clear this up for me? Is it always sufficient to find an additional conservation law to conclude symmetrizability?

答:
The two conditions are equivalent.

Using the chain rule we can write,
$$
\frac{\partial u}{\partial t} + \sum_{j=1}^d \frac{\partial}{\partial x_j} f_j(u) = \frac{\partial u}{\partial t} + \sum_{j=1}^d f_j'(u) \frac{\partial u}{\partial x_j}.
$$
Then multiply by $U'(u)$ to get
$$
U'(u) \frac{\partial u}{\partial t} + \sum_{j=1}^d U'(u)f_j'(u) \frac{\partial u}{\partial x_j}.
$$
and notice that the entropy equality will hold only when $U'(u)f'_j(u) = F'_j(u)$.
This is discussed in the book "Shock Waves and Reaction-Diffusion Equations" by Joel Smoller (2nd Edition) (See Chapter 20 Part B).


问:
$A^TAx = A^Tb$ is always consistent, independent of $Ax=b$?
Suppose $A$ is a $m\times n$ matrix, $x$ is a vector of unknowns and $b$ is a vector of know entries. Whether $Ax=b$ is consistent or not ,But $A^TAx=B$ is always consistent.

Is that above statement correct? or Am I missing anything?

答:
I will assume you mean $A^Tb$ by "B". Whether $Ax = b$ is asking if b is in the image of the linear map which is multiplication by A. If b is zero, asking for the solutions to $Ax = b$ is the same as asking what x's are in the kernel of "multiplication by A" (we call this linear map $L_A$ by the way). By multiplying both sides by $A^T$ and asking what x's solve the equation $A^TAx = A^Tb$, we should assume A is noninvertible (otherwise it is the same question as what the solutions to Ax = b are). You can also think of this as the kernel of the linear map $(L_A)^TL_A(x) - A^Tb$ (where $(L_A)^T = L_{A^T}$).Additionally,

if $A = \begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $b = \begin{pmatrix} b_1\\b_2\end{pmatrix}$, and $x = \begin{pmatrix} x_1\\x_2\end{pmatrix}$, $A^TA = A^Tb$ iff $Ax = b$ (you should check that). So the relationship between the two equations can be more or less whatever, other than the fact that we must have $Ax = b$ implies $A^TAx = A^Tb$ (by just multiplying both sides of the first equation by $A^T$).

(also I am assuming the entries are in a subfield of the complex numbers).


答:
I suppose the underlying field be $\mathbb{R}$.
It seems that you are a beginner, so this approach may be more friendly.

First is to observe that $A^TAX=0 \iff AX = 0$. (I shall leave this bit of work to you.) So if a linearly independent set $\langle X_1,X_2,\cdots,X_r\rangle$, where $r=\mathrm{rank}\;A$, spans the column space of the matrix $A$, then $\langle AX_1,AX_2,\cdots,AX_r\rangle$ spans the column space of $A^T$ as well. Since $A^Tb$ is always a vector in the column space of $A^T$, it can be derived by a linear combination of $\langle X_1,X_2,\cdots,X_r\rangle$.

Title: $A^TAx = A^Tb$ is always consistent, independent of $Ax=b$?

Body:
Suppose $A$ is a $m\times n$ matrix, $x$ is a vector of unknowns and $b$ is a vector of know entries. Whether $Ax=b$ is consistent or not ,But $A^TAx=B$ is always consistent.

Is that above statement correct? or Am I missing anything?


问:
This Geometry problem seems like it should be easier.
Two ladders are located in and alley between two buildings.  They are leaned up agains the buildings such that they take up the full width of the alleyway and cross at a point 12 ft above the ground.  One ladder is 40 ft long.  The other ladder is 30 ft long.

How wide is the alley?

I was able to find a quartic equation that gave the height above the ground where the 30' ladder meets the building.

$x^4 - 24x^3 + 700x^2 - 16800 x + 100800 = 0$

I did not see an easy way to solve this, and plugged it into wolfram alpha.  WA gives a solution to this quartic of 19.000.
But, 19 can't be a root by the rational root theorem... The root is very nearly, but not exactly 19.

And then we can use the Pythagorean theorem to find that $w\approx \sqrt {539}$

On the face of it this problem seems like it should not require quartic equations.  I am curious to know if anyone has a simpler solution, and even better an exact solution.

答:
You say that you don't think that this problem should require quartic equations, but you also seem to be saying that the answer is a quartic number (i.e. an algebraic number of degree four over the rationals): You say it's not exactly equal to a rational number, and also that it's not the square root of any rational number.

I didn't check your equations, but if the previous assertions are true then how could the answer not involve a quartic equation?

Of course potentially you might be able to write the answer in terms of some trig functions instead of in terms of polynomial equations, but typically we consider an expression in radicals to be "better" than an expression in terms of transcendental functions.


答:
This is the old crossed ladders problem, and the resolution of a quartic is unavoidable in the generic case.

$\frac w{10}$ solves
$$x^8-\frac{1106}{25}x^6+697x^4-\frac{114264}{25}x^2+\frac{6543504}{625}=0$$
which is irreducible and has generic Galois group ($S_4$ with respect to $x^2$), so it cannot be simplified beyond what the quartic formulas provide.

More generally, suppose the ladder lengths are $a$ and $b$, they cross at height $h$ and the alley width is $w$. Let $A=a^2$, $B=b^2$, $H=h^2$, $W=w^2$, $S=A+B$ and $P=AB$. Then $W$ satisfies the (homogeneous in $A,B,H,W$ and symmetric in $A,B$) quartic
$$\begin{align}
&W^4\\
+(4H-2S)&W^3\\
+(S^2+2P-6HS)&W^2\\
+(2H(S^2+2P)-2SP)&W\\
+(H^2(S^2-4P)-2HSP+P^2)&=0
\end{align}$$
For $a=30,b=40,h=12$ this yields $w=23.21673746900\dots$

Title: This Geometry problem seems like it should be easier.

Body:
Two ladders are located in and alley between two buildings.  They are leaned up agains the buildings such that they take up the full width of the alleyway and cross at a point 12 ft above the ground.  One ladder is 40 ft long.  The other ladder is 30 ft long.

How wide is the alley?

I was able to find a quartic equation that gave the height above the ground where the 30' ladder meets the building.

$x^4 - 24x^3 + 700x^2 - 16800 x + 100800 = 0$

I did not see an easy way to solve this, and plugged it into wolfram alpha.  WA gives a solution to this quartic of 19.000.
But, 19 can't be a root by the rational root theorem... The root is very nearly, but not exactly 19.

And then we can use the Pythagorean theorem to find that $w\approx \sqrt {539}$

On the face of it this problem seems like it should not require quartic equations.  I am curious to know if anyone has a simpler solution, and even better an exact solution.


问:
Calculate angle of view based on object sizes in image
I'm trying to find a way to automatically calculate the angle of view of an image by comparing the size and position of two identical objects in the image.

For example, the two bottles below have identical dimensions in reality, but depending on the lens of the camera, the blue bottle may appear to shrink at different rates in the image the further its distance from the camera.

I'm wondering if it is possible to calculate the angle of view of the below image, given the following:

The two objects have the same dimensions in reality (both humans are equally tall and wide). We can assume that these are 180cm tall.

The real-world distance to the objects are not known.

The image dimensions are known.

Both bounding box dimensions are known.

Both bounding box positions relative to the image top-left origin are known.

Both objects are standing on flat ground (their bottom points intersect a common plane).

Question: Is it possible to calculate the angle of view of the image based entirely on how two (equally-sized) objects are represented in the image as they are moved closer or futher away from the camera?

EDIT: See image with bbox coordinates, here: https://i.imgur.com/HhF8IhH.png

答:
For the benefit of anyone coming across this question who is not familiar with the term angle of view, it is described here.
It is a geometric angle that can be computed for a lens of some known focal length set to focus at a known distance and projecting onto a sensor of known dimensions.

For simplicity, since we're dealing with the vertical heights of objects in the frame, let's just consider the vertical angle of view.

Let's try running the problem in reverse.

Suppose we are given a camera with a known angle of view
and a pair of identical objects of known dimensions
which we are required to place on a flat, level surface.
Suppose we are required to point the camera in a horizontal direction while doing this (as was apparently done in the street-scene photograph in the question).

To simplify the problem a little (temporarily), suppose the objects we are photographing are rectangular pieces of cardboard
braced behind by some objects
so that we can stand the rectangles upright on the surface,
and assume we will place the rectangles so they are parallel to the image sensor.
This means that the ratio of height to width of the bounding box of a rectangle in the image will be the same as the ratio of height to width of the actual rectangle.

Given the known height of a rectangle and the known vertical angle of view, you can compute the percentage of the height of the picture frame that the  rectangle's image will occupy at any given distance.
Conversely, you can compute the distance at which to place the  rectangle so that it will occupy a given percentage of the height of the picture frame.
You can achieve any percentage from nearly zero to more than $100$ by moving the  rectangle away from the camera or toward it.

So if we are given two bounding box heights, we can place the first rectangle at a distance such that its bounding box's height is the first given height and place the second rectangle at a distance such that its bounding box's height is the second given height.

Now suppose we are told to put the upper left corner of the first bounding box at some particular coordinates in the frame.
We can achieve the desired horizontal coordinate just by moving the rectangle left or right.
Together with the height of the bounding box, the vertical coordinate of the corner determines a horizontal line on the rectangle (or in the rectangle's plane) that must be imaged exactly halfway between the top and bottom of the picture frame.
We raise or lower the camera to the height of that line, and now the vertical coordinate is correct.

This does not work in practice if the surface is opaque and our chosen coordinates put the bottom of the bounding box above the center of the picture frame,
because this would require us to take the photograph from below the surface.
But let's assume we never require the bottom of the box to be above the center of the frame.

We can achieve any desired horizontal coordinate for the second bounding box, but the constraints of the setup determine the vertical coordinate:
each bounding box must have exactly the same percentage of its height above the center of the frame, just as both rectangles must have the same percentage of their heights above a line at the height of the camera.

We therefore have five degrees of freedom in specifying how we would like the bounding boxes to appear: the height of the first box (which determines the width), the height of the second box (which determines the width),
the horizontal and vertical coordinates of the first box's upper left corner
(which together with the previous data determine the vertical coordinate of the second box's upper left corner)
and the horizontal coordinate of the second box's upper left corner.

So let's say we set up this photograph for a particular set of dimensions and coordinates of the bounding boxes using a lens with a certain angle of view.
Now we change to a lens with a different angle of view.
We will have to move the rectangles forward or backward and left or right so that they again show up with the same dimensions and coordinates of bounding boxes,
but we can do so.

Since two camera setups with different angles of view took identical pictures (as far as you can tell from the bounding boxes of the two known objects), there is no way you can deduce the angle of view from the sizes and positions of the bounding boxes.

If you have some way of measuring the distance between the objects in the two bounding boxes, however, you can determine the distance to the camera.
Using that distance, the portion of the image frame occupied by the bounding box, and the known dimension of the object, you can estimate the angle of view.
In the photograph of the two men in the question, if the paving stones are all the same size and you know what size they are, you might be able to estimate the distance reasonably well.

Things get a little more complicated when dealing with three-dimensional objects in the real world since the points that determine the width of the bounding box may be at a different distance from the camera than the points that determine the height of the bounding box. The width of the bounding box also is not necessarily determined by the points at the maximum width of the object, but rather by the points that subtend the maximum horizontal angle at the camera.
Because of all these complications, the ratio of height to width might vary slightly as the object is moved closer to or further away from the camera,
and if the object holds its shape rigidly enough, if you can measure the bounding box accurately enough, and if you know enough about the shape of the object you can (in principle) tell how far away the object is merely from the ratio of height to width in the picture.
But I think those conditions would be hard to find in practice.


问:
Why don't $A^TAx = A^Tb$ and $Ax=b$ have the same solution?
Suppose $A$ is a $m\times n$ matrix, $x$ is a vector of unkowns and $b$ is a vector of know entries.

Why don't $$Ax=b$$ and $$A^TAx = A^Tb$$ have the same solution ($x$)?

It seems to me that I could get from the first equation to the second equation by simply multiplying both sides by $A^T$, so in my conception this wouldn't change the solutions.

Am I missing something?

答: Let A = 0. Then $A^T = 0$ and $A^TAx = A^Tb$ always and $Ax = b$ iff b = 0.


问:
Write the integral $\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$ in terms of the Euler Beta Function?
How can I write the integral $$\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$$
with $\alpha, \beta >0$, in terms of the Euler Beta Function? For which $\alpha, \beta$ does the integral actually converge? Maybe some change of variable can do the trick?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{-1 < \Re\pars{\beta} < 2\,\Re\pars{\alpha} - 1}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
\pars{1 \over 1 + x^{2}}^{\alpha}x^{\beta}\,\dd x}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}
{x^{\beta/2 - 1/2} \over \pars{1 + x}^{\alpha}}\,\dd x
\end{align}
Lets $\ds{x \equiv {1 \over t} - 1 \implies
t = {1 \over 1 + x}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
\pars{1 \over 1 + x^{2}}^{\alpha}x^{\beta}\,\dd x} =
{1 \over 2}\int_{1}^{0}
{\pars{1/t - 1}^{\beta/2 - 1/2} \over t^{-\alpha}}\,\pars{-\,{1 \over t^{2}}}\dd t
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}
t^{-\beta/2 + \alpha - 3/2}\,\,\,\,\,\pars{1 - t}^{\beta/2 - 1/2}\,\,\,\dd t
\\[5mm] = &\
\bbx{{1 \over 2}\on{B}\pars{\alpha - {\beta + 1 \over 2},{\beta + 1 \over 2}}} \\ &
\end{align}
$\ds{\on{B}}$: Beta Function.


问:
prove function in product space is measurable
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function, define a function $F: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ by $(x,y) \rightarrow f(x-y)$, how do I show that this function is measurable? What I have tried is:

$F^{-1}(-\infty,\alpha) = \{(x,y) \in \mathbb{R} \times \mathbb{R}|  f(x-y) <  \alpha \} = \bigcup\limits_{x \in \mathbb{R}} \{x\} \times \{y \in \mathbb{R} | f(x-y) < \alpha\}$. Since Lebesgue measure is translation invariant, the function $f_{x}: y \rightarrow f(x-y)$ is also measurable, so the set $\{y \in \mathbb{R} | f(x-y) < \alpha\}$ is measurable. The problem is that the union is not over countable set, and I am not sure how to proceed.

答: Since $f$ is measurable by Lusin's theorem, there is a sequence of continuous functions $f_n\to f$ a.e. Since $F_n:\mathbb R\times\mathbb R\to\mathbb R/F_n(x,y)=f_n(x-y)$ is continuous, $F_n$ is measurable and $F_n\to F$ a.e. so $F$ is measurable since the limit of measurable functions is measurable.


问:
Integral residue $\int_{0}^{\infty} {\frac{\cos\left(x\right)}{x^{2} + 2x + 4}}\,{\rm d}x. $
$$
\mbox{I have the integral:}\quad
\int_{0}^{\infty}
{\frac{\cos\left(x\right)}{x^{2} + 2x + 4}}\,{\rm d}x.
$$
I tried to use the residue theorem and semicircle and calculate the integral form minus infinity to infinity, but my function is not even. Do you have any hints $?$.

答:
I suppose you are calculating
$$\int_{-\infty }^{\infty } \frac{\cos (z)}{z^2+2 z+4} \, dz=\frac{e^{-\sqrt{3}} \pi  \cos (1)}{\sqrt{3}}$$
Which can be done by integrating $\frac{e^{i z}}{z^2+2 z+4}$ along large upper semicircular contour, using Jordan's lemma, calculating residue at $-1+\sqrt{3} i$ and taking real part. For your original question,
$$\small \int_{0}^{\infty } \frac{\cos (z)}{z^2+2 z+4} \, dz=-\frac{i \text{Ci}\left(i \sqrt{3}+1\right) \cos \left(1+i \sqrt{3}\right)}{2 \sqrt{3}}+\frac{i \text{Ci}\left(1-i \sqrt{3}\right) \cos \left(1-i \sqrt{3}\right)}{2 \sqrt{3}}-\frac{i \text{Si}\left(i \sqrt{3}+1\right) \sin \left(1+i \sqrt{3}\right)}{2 \sqrt{3}}+\frac{i \text{Si}\left(1-i \sqrt{3}\right) \sin \left(1-i \sqrt{3}\right)}{2 \sqrt{3}}-\frac{\pi  \cos (1) \sinh \left(\sqrt{3}\right)}{2 \sqrt{3}}$$
Which can be done by shifting $z\to z-1$, expanding $\cos(z-1)$ into $\sin/\cos$ and factorizing $\frac{1}{z^2+3}$ into $\frac{1}{z\pm \sqrt{3}i}$ (for partial fractions), using N-L formula based on elementary formula that $\int \frac{\cos (z)}{z-a} \, dz=\cos (a) \text{Ci}(z-a)+\sin (a) \text{Si}(a-z)$ and its sine counterparts. No essential simplification seems to exist for this case, though.


问:
In a polynomial, if $p(x)$ is divisible by $x$, why is the constant $0$?
Here's the problem.

$p(x) = 3(x^2 + 10x + 5) − 5(x − k)$

In the polynomial $p(x)$ defined above, $k$ is a constant. If $p(x)$ is divisible by $x$, what is the value of $k$?

In the answer it explained that because "$p(x)$ is divisible by $x$" we know that the constant is $0$. But why is that true and what is the constant?

答:
Very helpful explanations, here is my twist on describing the logic...

If something can be nicely divisible by a quantity then that means the remainder must be zero.  Example: 21 can be nicely divisible by 3, 21/3 = 7 with a remainder of 0 ("7 R 0", or 7 + 0/3).  Contrary, 22 cannot be nicely divisible by 3, 22/3 does not have a remainder of zero, 22/3 = "7 R 1", or 7 + 1/3.

Go back to the problem and use this reasoning as a template for the solution.

Since we know that p(x) is divisible by x (nicely), then we know that the quantity that remains in terms of x must be equal to zero.  Use the expanded version of p(x) then divide it by x.  Notice that only one component of this result is left with "x" in the denominator, (15+5k)/x .  In other words, because (15+5k)/x is the only component with "x" in the denominator, it is not "nice" and is the remainder.

The problem tell us that the remainder must be "0" (because p(x) is divisible by x).

So, the simple equation, (15+5k)/x, must be equal to 0.  Now just solve for k.

I hope this long explanation helps... it helped with my son's understanding.


问:
Laurent expansion problem
Expand the function $$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)} $$ on the ring $$ 1 < |z| < 2 $$

I used partial fractions to get the following $$f(z)=\frac{1}{(z-2)} +\frac{-2}{(z^2+1)}  $$
then
 $$ \frac{1}{z-2} = \frac{-1}{2(1-z/2)} = \frac{-1}{2} \left[1+z/2 + (z/2)^2 + (z/2)^3 +\cdots\right]
 $$

but I'm stuck with $$ \frac{-2} {z^2+1} $$ how can I expand it?

Title: Laurent expansion problem

Body:
Hint:

$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty (-(-i)^n-i^n) z^n,\quad\mid z\mid<1$$

and:

$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty  (-1+z)^n (-1+i) (2^{-1-n} ((-1-i)^n+i (-1+i)^n))$$


问:
Computing the expectation $E[2^{X1+X2+X3}]$
Suppose that $X_1$, $X_2$ and $X_3$ are three independent random variables satisfying

$$ P(X_i=1)=P(X_i=−1)=\frac{1}{2} \qquad \text{for} \ i = 1,2,3 $$

I am asked to find $E[2^{X1+X2+X3}]$.

So far I have worked out $E[X_i]=0$ from symmetry and tried to separate using Fubini's:

$$ E[2^{X_1}]E[2^{X_2}]E[2^{X_3}] $$

however I am stuck on how to proceed or whether this is the right route.

Body:
Let's just follow the definition

$E(s^X)=s^1 \cdot P(X=1)+s^{-1} \cdot P(X=-1)=s \cdot \dfrac{1}{2}+s^{-1}\dfrac{1}{2}$

In our case, we have $s=2$

so,

$E(2^X)=(2) \cdot \dfrac{1}{2}+(2)^{-1}\dfrac{1}{2}=1+\dfrac{1}{4}=\dfrac{5}{4}$

but we have

$E(2^{X_{1}}2^{X_{2}}2^{X_{3}})=E(2^{X_{1}})E(2^{X_{2}})E(2^{X_{3}})=\dfrac{5}{4}\cdot\dfrac{5}{4}\cdot\dfrac{5}{4}$

$E(ABC)=E(A)E(B)E(C)$ is possible because of independence.


问:
How big must the union of a group's Sylow p-subgroups be?
For various orders $n$ it's a common exercise to prove that a finite group $G$ of order $n$ can't be simple by using the Sylow theorems to show that there is some prime $p \mid n$ such that the number $n_p$ of Sylow $p$-subgroups equals $1$, so the unique Sylow $p$-subgroup is normal. One way these proofs can go is that you show that if $n_p$ isn't equal to $1$, then because $n_p \equiv 1 \bmod p$ it must be very large, so large that there isn't enough room in $G$ for all of its Sylow $p$-subgroups together plus the other Sylow subgroups.

I know how to run this argument if the exponent $a$ of $p$ in $n$ is $1$ and we can show that $n_p = \frac{n}{p}$; in this case the Sylow $p$-subgroups are cyclic, so intersect only in the identity, which means that $G$ has at least $\frac{n}{p}(p - 1)$ elements of order $p$, and hence only room for $\frac{n}{p}$ elements of other orders.

However, I don't know how to run this argument if $a \ge 2$; this came up when I was trying to answer this question and specifically trying to show that a group of order $|G| = 3 \cdot 5 \cdot 7^2 = 735$ can't be simple. The Sylow theorems give that if $n_7 \neq 1$ then $n_7 = 15$, so $G$ has the maximum possible number of Sylow $7$-subgroups. The specific question this gave rise to is:

Specific question: What is the sharpest lower bound on the size of the union of these $15$ Sylow $7$-subgroups?

I wanted to use the Bonferroni inequalities to address this question, using the fact that any two Sylow $7$-subgroups intersect in at most $7$ elements, but something very funny happened: if I apply Bonferroni to all $15$ subgroups I get a lower bound of

$$15 \cdot 49 - {15 \choose 2} \cdot 7 = 0.$$

The problem is that there are too many pairwise intersections between $15$ subgroups. If I instead apply Bonferroni with only $k$ of the $15$ subgroups I get a lower bound of

$$49k - 7 {k \choose 2}$$

which turns out to be maximized when $k = 8$, giving a lower bound of $210$. Is it possible to do better than this? I'm ignoring $7$ of the Sylows!

So the general question is:

General question: What is the sharpest lower bound on the size of the union of the Sylow $p$-subgroups of a finite group $G$ which can be written as a function of the size $p^a$ of such a subgroup and the number $n_p$ of such subgroups? What if $G$ is assumed to be simple?

When $a = 1$ the union has size exactly $(p - 1) n_p + 1$. In general any two Sylows intersect in at most $p^{a-1}$ elements, so Bonferroni with $k$ of the Sylows gives a lower bound of

$$k p^a - {k \choose 2} p^{a-1} = k p^{a-1} \left( p - \frac{k-1}{2} \right)$$

which is maximized when $k \approx p$ as above (or $k = n_p$, if $p$ is more than a little larger than $n_p$). But the smaller $p$ is compared to $n_p$ the less helpful of a bound this will be.

答:
To answer your specific question, the union of the $15$ Sylow $7$-subgroups must have cardinality $637$ (assuming that such a group existed).

By this question, if a group $G$ has fewer than $p^2$ Sylow $p$-subgroups then the Sylow $p$-subgroups have a common intersection of index $p$. In the case of this specific question, this gives
$$7+15\cdot(49-7)=637$$
elements in total.

In general, the number of Sylow $p$-subgroups is of the form $kp+1$, and the cardinality of the union of the Sylow $p$-subgroups is of the form $p^n(t(p-1)+1)$. In the above example, $k=2$ and $t=2$.

There are some simple relationships between $k$ and $t$:

$t=0\iff k=0$

$t=1\iff k=1$

$t\geq2\iff k\geq2$

The case of pairwise trivial intersection gives the upper bound $t\leq\frac{(p^n-1)}{p^{n-1}(p-1)}k<\frac{p}{p-1}k$

If $k<p$ then $t=k$ (this follows from the linked MO question)

Unfortunately, $t$ can be substantially smaller than $k$ (see this answer).


问:
How do you find the equation of the parabola of this envelope created using two opposing sides of a pentagon?
I can kind of understand how this would work with an envelope of a family of lines based on the x and y-axis themselves, but I just can't figure out how that could translate to this situation where the lines connect from the x-axis and a "random" positive slope that passes through the origin.

Or you could check here, if the image link doesn't work: https://www.desmos.com/calculator/eizf3uxuaq

答:
I will just sketch the way to obtain your solution. First, translate and rotate the pentagon such that the vertices at $(2.6,0)$ and $(2.1,1.6)$ are on the $x$ axis, symmetric around the origin. Then these new points are at $(\pm x_0,0)$. Also, for simplicity, make the vertex of the parabola to be below the x-axis. Then the equation of the parabola becomes $$y=ax^2-b$$with $a,b>0$. You need to find out $a$ and $b$. Notice first that $$ax_0^2-b=0$$
Also, the slope of the diagonals (which you can easily calculate as $\pm m$) are tangent to the parabola at $(\pm x_0,0)$.
So $m=2ax_0$. From here on is just plug in some numbers.


问:
subgroup of $\mathbb{Z}_9 \times \mathbb{Z}_{15}$

For a group $G$ and an element $a\in G,$ let $\langle a\rangle$ be the intersection of all subgroups of $G$ containing $a.$ If $G = \langle a\rangle$ for some $a \in G,$ then $G$ is cyclic. Determine with proof the number of cyclic subgroups of $\mathbb{Z}_9\times \mathbb{Z}_{15}$. Find, with justification, a non-cyclic proper subgroup of $\mathbb{Z}_9\times \mathbb{Z}_{15}.$

By Lagrange's Theorem, the order of each subgroup should divide $135,$ which has $4\cdot 2 = 8$ factors. I tried considering the number of cyclic subgroups of each possible order and deduced from considering the orders of elements of $\mathbb{Z}_9\times \mathbb{Z}_{15}$ that the orders for cyclic subgroups are $1, 3, 5, 9, 15, 45.$ The only subgroup with order $1$ is the trivial subgroup, and this is also a cyclic subgroup. The cyclic subgroups of order $3$ seem to be $\langle (0,5)\rangle = \{(0,0), (0,5), (0,10)\}, \langle (3,0)\rangle, \langle (3,5)\rangle, \langle (3,10)\rangle, \langle (6,5)\rangle, \langle (6,10)\rangle.$ However, I'm not sure how to find the number of cyclic subgroups of larger orders.

As for a non-cyclic subgroup, I think it suffices to find a subgroup of order $27$, since no cyclic subgroup has this order. Or there might be a simpler approach.

答: To be sure, Cauchy's theorem will guarantee that we have a cyclic subgroup with the order of any given prime divisor of 135. Some examples, if (j,9) = 1, then <(j,0)> has order 9 and if (m,15) = 1, then <(0,m)> has order 15 (if 15 divides km, then 15 divides k and the smallest such k is 15, same argument for 9). Also, the order of <(1,1)> is lcm(9,15) = 45 (if (k,k) = (0,0), k = 9R = 15S. Then 9 and 15 both divide k). So the order of <(a,b)> is completely contingent on the gcd's (9,a) and (15,b), so I would use that to get the rest of the cyclic subgroups of other orders dividing 135.</p>


问:
Magic dice question.
A pair of dice are magic. When they are rolled together, then half the time, they behave like a normal pair of dice, but the other half of the time they miraculously roll a double ( equally likely to be any of the six possible doubles).  What is the expected value of the product of the two scores when the pair of dice are rolled?</p>

答:
HINT

There are 2 cases, on average with equal weight:

Magic happens: then there are 6 outcomes with equal probability, can you compute the expected value for this case?

No magic happens: then you have regular dice, can you compute the expected value for this case?

Average the expectations for (1) and (2).

Title: Magic dice question.


问:
$\lim_{ x \to 0^+ } \frac{(1+(\sin x)^2)^{\frac{1}{4}}-1}{((1+\sin x)^{(\frac{-1}{x})}-e^{-1}) \ln (1+ \sqrt{1-e^{-x^2}})}$
I have to evaluate this limit:
$$\lim_{ x \to 0^+ }  \frac{(1+(\sin x)^2)^{\frac{1}{4}}-1}{((1+\sin x)^{(\frac{-1}{x})}-e^{-1}) \ln (1+ \sqrt{1-e^{-x^2}})}$$

Applying the L'Hôpital rule. I'm really lost in the calculation and the only simplification that I managed to do is: $$ \ln (1+ \sqrt{1-e^{-x^2}}) \approx \sqrt{1-e^{-x^2}}$$
Are there any other simplifications that I can apply? In my book the suggested solution is $\frac{e}{2}$ or ~$1.359$

答:
Too long for comments.

As usual @Robert Z gave a good solution working with pieces to make life easier.

We could do similar things composing Taylor series one piece at the time and obtain much more than the limit itself. Let me try
$$A=\sqrt[4]{\sin ^2(x)+1}\implies \log(A)=\frac 14 \log \left(\sin ^2(x)+1\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}+O\left(x^8\right)$$
$$\log \left(\sin ^2(x)+1\right)=x^2-\frac{5 x^4}{6}+\frac{32 x^6}{45}+O\left(x^8\right)$$
$$\log(A)=\frac{x^2}{4}-\frac{5 x^4}{24}+\frac{8 x^6}{45}+O\left(x^8\right)$$
$$A=e^{\log(A)}=1+\frac{x^2}{4}-\frac{17 x^4}{96}+\frac{739 x^6}{5760}+O\left(x^8\right)$$

If you continue in the same spirit with the others pieces
$$y=\frac{(1+(\sin x)^2)^{\frac{1}{4}}-1}{((1+\sin x)^{-\frac{1}{x}}-e^{-1}) \log (1+ \sqrt{1-e^{-x^2}})}$$ would be
$$y=\frac e 2\left(1+\frac{7 x}{12}-\frac{77 x^2}{144}+O\left(x^3\right) \right)$$ which, for sure, gives the limit but also how it is approached.


问:
Question on Lebesgue Integration (from old qual)
This problem is from an old real analysis qual...I am trying to review some material I just finished concerning Lebesgue Integration. Below is the problem followed by my attempted (unfinished) solution. Any help towards showing the problem holds would be great.

Problem:

Show that for all integrable functions $f$ over $\mathbb R$ that

$$\lim_{\delta \to 1}\int\left\vert f(\delta x)- f(x) \right\vert dx = 0.$$

(My attmpted) Solution:

Let $\delta_n\in\mathbb{R}$ be a sequence such that $\delta_n\to 1$. Consider the sequence of functions $f_{n}\colon \mathbb{R}\to\mathbb{R}$ defined by,

$$f_{n}:= |f(\delta_nx)-f(x)|.$$

This is where I get stuck and I am unsure if I should go down this route. However, if this route is correct, then the following would be the final step.

Thus,

$$\lim_{\delta \to 1}\int\left\vert f(\delta x)- f(x) \right\vert dx =
\lim_{n \to \infty}\int\left\vert f(\delta_n x)- f(x) \right\vert dx =
\int\lim_{n\to\infty}\left\vert f(\delta_n x)- f(x) \right\vert dx =
\int\left\vert f(1\cdot x)- f(x) \right\vert dx =
\int 0 = 0$$

答:
Here is a proof which does not require approximation by continuous functions:

Let $A \Delta B$ denote the symmetric difference between $A$ and $B$: $A \Delta B=(A\setminus B) \cup (B\setminus A)$. Consider the collection of all Borel sets $A$ such that $m((tA)\Delta A)\to 0$ as $ t \to 1$.  [$tA=\{ta: a \in A\}$].

It is easy to check that this collection contains all intervals.

Use the fact that $(\bigcup_n A_n) \Delta (\bigcup_n B_n)  \subseteq \bigcup_n (A_n\Delta B_n)$ and $A \Delta B=A^{c} \Delta B^{c}$  to verify that our collection is a sigma algebra.

It follows that every Borel set belongs to this collection.

This can be written as $\int |f(\delta_n x)-f(x)|dx \to 0$ when $f=\chi_A$. Extend this to simple functions and then to all integrable functions $f$.


问:
Find derivative of Taylor series
How would one differentiate a function in this form?

$$f(x) = f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}$$

I'm sorry if this is something obvious, I'm not great with maths. I tried searching for the answer with no luck.

答:
The derivative is really simple: it is identical to the original, but the first term is deleted (because it is a constant), and all the other coefficients are shifted to the left (except for the factorial denominator)

$$f(x) = f(a)+{\frac {\color{Blue}{f'(a)}}{1!}}(x-a)+{ \frac {\color{Red}{ f''(a)}}{2!}}(x-a)^{2}+{\frac {\color{Magenta}{f'''(a)}}{3!}}(x-a)^{3}
\\
\\
f'(x) ={\color{Blue}{f'(a)}}+ { \frac {\color{Red}{f''(a)}}{1!}}(x-a)+{\frac {\color{Magenta}{f'''(a)}}{2!}}(x-a)^{2}+{\frac {f''''(a)}{3!}}(x-a)^{3}$$

You can check it, by deriving each term of the summation.

Title: Find derivative of Taylor series


问:
How does the least upper bound property imply the Archimedean property?
This post gives the incomplete proof of the following theorem:
In any ordered field F, the following are equivalent:

F has the least upper bound property

F is complete and has the Archimedean property.

I did not understand how (1) implies the Archimedean property of F. What if we have a metric space that does not have any real numbers? In that case, the Archimedean property can not be applied.

答:
The part of the answer you are referencing that proves that the l.u.b. property implies the Archimedean property is this:

Archimedean Property: For every real $x$ there is an integer $n$ such that $n>x$

Proof: If this is not the case, integers would be bounded by some $x$.

But it's also worth noting that before this part, the answerer wrote:

To simplify discussion, we assume that the order field is $\mathbb{R}$.

The original answerer is suggesting here that integers being bounded above by some $x$ is a contradiction in any ordered field, they just use $\mathbb R$ to keep their answer tidier (basically, the same argument can be made in any ordered field, with $x$ instead just being a member of that field).

The result seems pretty intuitive, but the reliance on the l.u.b. property comes from the fact that if the set of integers, $\langle 1\rangle\subset F$ ($F$ denotes the whole ordered field) is bounded above by some $x$, then $\langle 1\rangle$ (being nonempty as it contains, e.g. $0$) has an l.u.b., $u$. The reason this is a contradiction is that by the property of l.u.b.s, we should be able to choose some integer $n\in \langle 1\rangle$ that is $>u-1$ (specifically, as $F$ is an ordered field, $u-1$ is $<u$. As $u$ is the l.u.b. of $\langle 1\rangle$, it follows that $u-1$ cannot be an upper bound of $\langle 1\rangle$, hence we may choose this $n$). But now (as $F$ is an ordered field), we have $n+1>u$. Clearly $n+1$ is also in $\langle 1\rangle$, and now the contradiction should be clear ($u$ is both an upper bound of $\langle 1\rangle$ and not an upper bound of $\langle 1\rangle$).

Note that in the previous paragraph no mention of the 'metric' (if there even is one) of $F$ was made. The only properties used were those $F$ possessed as an ordered field.


问:
In a CCC with the initial object, can I have a morphism A -> 0?
I'm in a cartesian-closed category with the initial object $0$. What are the consequences of having a morphism from some object $A$ to $0$? Does it mean that there is also a global element of $0$, that is $1\to 0$ ?  Does it mean that we have a zero object, $1 \cong 0$ ? Does it mean that all objects are isomorphic?

Assume we don't have coproducts.

答:
In any cartesian closed category, product $X \times (-)$ distributes over colimits, and in particular over the initial object, so we have $X \times 0 = 0$. If $A$ is any object it follows that

$$\text{Hom}(A, X) \times \text{Hom}(A, 0) \cong \text{Hom}(A, 0)$$

in the sense that the projection map from the LHS to the RHS is an isomorphism, hence that if $\text{Hom}(A, 0)$ is non-empty then $\text{Hom}(A, X) = 1$ for all $X$, so $A = 0$. (Here I am using "equality" to mean that there is a unique map in one direction or another and that unique map is an isomorphism.)

Edit: In more detail, since this was apparently confusing. By $X \times 0 = 0$ I mean that the unique map $0 \to X \times 0$ is an isomorphism. Since this implies $X \times 0$ is the initial object, there is also a unique map $X \times 0 \to 0$, which must be the projection $\pi_0$, and which must be an isomorphism. Now we have the general observation that in any category with finite products, if

$$\pi_Q : P \times Q \to Q$$

is the projection then the induced map

$$\text{Hom}(R, P \times Q) \cong \text{Hom}(R, P) \times \text{Hom}(R, Q) \to \text{Hom}(R, Q)$$

is the projection $\pi_{\text{Hom}(R, Q)}$. So, from the fact that the projection $\pi_0: X \times 0 \to 0$ is an isomorphism, we deduce that for any pair of objects $A, X$ the projection

$$\pi_{\text{Hom}(A, 0)} : \text{Hom}(A, X) \times \text{Hom}(A, 0) \to \text{Hom}(A, 0)$$

is an isomorphism. If $\text{Hom}(A, 0)$ is empty then both sides are empty and we don't learn anything. But if $\text{Hom}(A, 0)$ is non-empty, or equivalently if $A$ admits a morphism to $0$, then the fibers of this map can all be identified with $\text{Hom}(A, X)$, so it follows that if $\text{Hom}(A, 0)$ is non-empty then $\text{Hom}(A, X) = 1$ for all $X$, so $A = 0$.

I had forgotten this argument but apparently I knew it in 2015, although in that answer I wrote it in the opposite category. (We don't use the full strength of the assumption of cartesian closure, only that finite products exist and distribute over initial objects, so this argument continues to work in a distributive category.)

If $A = 1$, or more generally if $A$ has a global point, it follows that the category has a zero object, and now the isomorphisms

$$\text{Hom}(X, Y) \cong \text{Hom}(1, Y^X) \cong \text{Hom}(0, Y^X) \cong 1$$

show that there is a unique morphism between any pair of objects, so the category is contractible.


问:
Why is linear map injectivity defined solely using the null-space?
The thing I'm wondering about is as follows:

Can we have a non-injective linear map $f:M\to N$, with an element $n\in N$, and elements $m_1,m_2 \in M$ such that, $f(m_1)=n$ and $f(m_2)=n$, without $m_1, m_2$ being necessarily in the null-space of $f$ defined as $\mathcal{N}(f)=\{m\in M: f(m)=0\}$? or said differently, without $n$ necessarily being $0$.

If not why?

答:
(while corresponding statements all apply to groups and modules in general, I will focus on vector spaces).

Theorem. Let k be a subfield of $\mathbb{C}$ (so k can be the real numbers or the complex numbers for instance), and $f:U\to V$ be a linear map of k-vector spaces. Then if $u,w\in U$, $u-w\in ker(f):= \{m\in U : f(m) = 0\}$ (typically we use ker(f) and call it the "kernel of f" in this context) if and only if f(u) = f(w).

proof. =>: So $f(u-w) = 0$. f is additive ($f(x+y) = f(x) + f(y)$) so f(-w) = -f(w) and $f(u - w) = f(u) + f(-w) = f(u) - f(w) = 0$ and thus $f(u) = f(w)$.

<=: $f(u) = f(w)$ gives that $f(u) - f(w) = 0$. Again f is additive, so $f(u - w) = f(u) - f(w)$, which we know is zero. So $f(u-w) = 0$, which is what it means to say that $u - w\in ker(f)$.


问:
Given a surjection of rings with nilpotent kernel, the root of a polynomial lifts to a unique root.
Let $ \phi : R_1 \rightarrow R_2$ be a surjection of rings such that the kernel is nilpotent, let $f \in R_1[T]$ be such that its image under the map induced by $\phi, \ $ namely $g \in R_2[T], \ $  has a root, $\overline{\alpha}$, and $g'(\overline{\alpha}) \in R_2^\times$. I'm trying to show that there's a unique $\alpha \in R_1$ such that: $\phi(\alpha)=\overline\alpha \ $ and $ \ f(\alpha)=0.$

Uniqueness

if $ \ \alpha, \beta \in R_1$ are such that $f(\alpha)=f(\beta)=0 \ $ and $ \ \phi(\alpha)=\phi(\beta)=\overline{\alpha},\ $ then $g$ has a double root at $\overline{\alpha}$, thus $g'(\overline{\alpha}) \notin R_2^\times \ \  \unicode{x21af}$.

Existence

I was thinking that we should show that if there's a solution in $\frac{R_1}{(ker \phi)^k}, \ $ then it lifts to a solution in $\frac{R_1}{(ker \phi)^{k+1}},$ and since $$ \exists \ n \text{ with } (\ ker \ \phi \ )^n= (0), \text{ we would get a solution in } \frac{R_1}{(0)} \cong R_1 \text{ as requested}$$  but I've got no clue on how to prove the existence of a lift to $ \frac{R_1}{(ker \phi)^{k+1}}, \  $ any help?

答:
To start with, we can make an (unimportant) reduction to simplify things. Namely, with $R=R_1$, $I=\ker \phi$, we have a polynomial $f\in R[T]$ whose reduction $\bar f\in (R/I)[T]$ has a root $\bar \alpha$, with $\bar f'(\bar \alpha)$ invertible. The goal is to show that, given that $I$ is nilpotent, that $\bar \alpha$ has a unique lift to a root of $f$. If $I^n = 0$, we can consider the filtration $$R = R/I^n \to R/I^{n-1}\to \cdots \to R/I$$ so it suffices to consider the case where $n=2$, since this allows us to lift at each stage.

So suppose $I^2=0$ and $\bar f$ has root $\bar \alpha$ with $\bar f'(\bar \alpha)$ invertible. Let $\alpha_0$ be any lift of $\bar \alpha$; we want to find a unique $a\in I$ for which $f(\alpha_0+a) = 0$. Let $f(x) = a_nx^n+\cdots+a_0$. Since $a^2=0$, we see that for any $m$, we have $(\alpha_0+a)^m = \alpha_0^m + m\alpha_0^{m-1}a$, so in particular we have $f(\alpha_0+a) = f(\alpha_0)+f'(\alpha_0)a$. Since $\bar f'(\bar \alpha)$ is invertible in $R/I$, there is some $b\in R$ and $y\in I$ for which $f'(\alpha_0)b = 1+y$. But then $f'(\alpha_0)$ is invertible in $R$, since $$f'(\alpha_0)b\cdot (1-y) = 1-y^2 = 1.$$ In particular we are forced to choose $a = -f(\alpha_0)f'(\alpha_0)^{-1}$, giving both existence and uniqueness of a lift.

This strategy allows you to even lift a root of $R/I$ uniquely to the $I$-adic completion of $R$, which, in fancier terms, is saying that for any ring $R$ and ideal $I$, the pair $(\hat R_I,I)$ is a henselian pair. This, sometimes in more or less general forms, is called "Hensel's lemma." Your particular question is simply a special case, where $\hat R_I = R$ since $I$ is nilpotent.


问:
Find $\lim_{n \to \infty} \sqrt[n]{n!}$.
I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered $\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}$. I have made the claim that $\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.$ I have begun the proof and I think it is on the right track but some verification would be nice.

Proof:
We can see that for n=1, this obviously holds, $\left(1!\right)^1\geq 1.$ Now suppose that this is the case for some $n=k$, then we have $\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}$. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks!

答:
For an elementary approach
which gives more information,
you can start with
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$.
This is proved by showing that
$(1+1/n)^n$ is an
increasing function of $n$
and
$(1+1/n)^{n+1}$ is an
decreasing function of $n$
and that they have a common limit,
good old $e$.

Using this and induction,
you can show that
$(n/e)^n \lt n! < (n/e)^{n+1}
$
so that
$n/e \lt (n!)^{1/n} < (n/e)(n/e)^{1/n}
$.

Since both $n^{1/n}$ and $a^{1/n}$
approach $1$ as $n \to \infty$,
$\dfrac{(n!)^{1/n}}{n}
\to \dfrac1{e}
$.

Here are the details.

Here's the proof that
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$.

Let
$a_n = (1+1/n)^n$
and $b_n = (1+1/n)^{n+1}
$.
We will prove that
$a_n$ is an increasing sequence
and
$b_n$ is an decreasing sequence.
Since $a_n < b_n$,
this implies,
for any positive integers $n$ and $m$
with $m < n$ that
$a_m < a_n < b_n < b_m$.

We use the very ingenious proof in [1],
which uses the
arithmetic-geometric mean inequality
(AGMI),
which we will use in the form
$((v_1+v_2+...v_n)/n)^n
\gt v_1v_2...v_n
$ (all $v_i$ positive)
with equality if and only if
all the $v_i$ are equal.

For $a_n$,
consider $n$ values of $1+1/n$
and $1$ value of $1$.
By the AGMI,
$((n+2)/(n+1))^{n+1} > (1+1/n)^n
$,
or $(1+1/(n+1))^{n+1} > (1+1/n)^n
$,
or $a_{n+1}
\gt a_n
$.

For $b_n$,
consider $n$ values of $1-1/n$
and $1$ value of $1$.
By the AGMI,
$(n/(n+1))^{n+1}
\gt (1-1/n)^n$
or
$(1+1/n)^{n+1} < (1+1/(n-1))^n
$,
or
$b_n < b_{n-1}
$
so
$b_{n+1} < b_{n}
$.

Since
$b_n-a_n
=(1+1/n)^{n+1}-(1+1/n)^n
=(1+1/n)^{n}(1+1/n-1)
=(1+1/n)^{n}/n
\to 0
$,
$a_n$ and $b_n$
have a common limit,
which I will call $e$.

Here's the proof that
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$
implies that
$(n/e)^n
\lt n!
\lt (n/e)^{n+1}
$.

Suppose
$n! > (\dfrac{n}{e})^n
$.
We want to show that
$(n+1)! > (\dfrac{n+1}{e})^{n+1}
$.

Then

$\begin{array}\\
(n+1)!
&=(n+1)n!\\
&>(n+1) (\dfrac{n}{e})^n\\
&=\dfrac{(n+1)n^n}{e^n}\\
\end{array}
$

so we want
$\dfrac{(n+1)n^n}{e^n}
\gt \dfrac{(n+1)^{n+1}}{e^{n+1}}
$
or
$e \gt \dfrac{(n+1)^n}{n^n}
=(1+1/n)^n
$
which is shown above.

Similarly,
suppose
$n! < (\dfrac{n}{e})^{n+1}
$.
We want to show that
$(n+1)! < (\dfrac{n+1}{e})^{n+2}
$.

Then

$\begin{array}\\
(n+1)!
&=(n+1)n!\\
&<(n+1) (\dfrac{n}{e})^{n+1}\\
&=\dfrac{(n+1)n^{n+1}}{e^{n+1}}\\
\end{array}
$

so we want
$\dfrac{(n+1)n^{n+1}}{e^{n+1}}
\lt \dfrac{(n+1)^{n+2}}{e^{n+2}}
$
or
$e \lt \dfrac{(n+1)^{n+1}}{n^{n+1}}
=(1+1/n)^{n+1}
$
which is shown above.

[1] N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563.


问:
Approximate distribution of $\frac{1}{n}\sum_{i=1}^{n}(X_{2i}-X_{2i-1})$
Question:
Let $\{X_{n}\}_{n \ge 1}$ be a sequence of i.i.d. random variables such that $E(X_i)=1$ and $Var(Xi)=1;,i=1,2,3....$Then the approximate distribution of $\frac{1}{n}\sum_{i=1}^{n}(X_{2i}-X_{2i-1})$ for large n is:

$(A) ܰN(0,1)$

$(B)N(0,2)$ ܰ

$(C)N(0,0.5)$

$(D)N(0,0.25)$

$$\text{My attempt}$$

$\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(X_{2i}-X_{2i-1})=\frac{1}{\sqrt{n}}((X_{2}-X_{1})+(X_{4}-X_{3})+(X_{6}-X_{5})+....)=\frac{1}{\sqrt{n}}(-X_{1}+X_{2}-X_{3}+X_{4}-X_{5}...) $

Seeing options I didn't calculate the mean. So trying to calculate variance.

$\frac{1}{n}V(-X_{1}+X_{2}-X_{3}+X_{4}-X_{5}...)=\frac{1}{n}\bigg(V(X_{1})+V(X_{2})+V(X_{3})+V(X_{4})+V(X_{5}...)\bigg)=\frac{n}{n}=1$

So it follows $N(0,1)$

But the answer is B. I don't know what's my mistake here plus different ways of solving this problem if you could tell me that will wide my way of thinking to solve these kinds of problems.

Thanks.

答: Actually, $X_{2i}-X_{2i-1}$ are independent so it is easy to find the variance of $\frac 1  {\sqrt n} (\sum\limits_{i=1}^{n} X_{2i}-X_{2i-1})$: It is $\frac 1 n (n\, var (X_{2i}-X_{2i-1}))$ or $2$.


问:
Equation of Partial derivatives
I am currently learning multivariable calculus and the following was done in my lecture to prove one of the properties of Jacobian

The property:

($JJ'=1$), where $J$ is $\frac{\partial (x,y)}{\partial(u,v)}$ and $J'$ is $\frac{\partial(u,v)}{\partial (x,y)}$

The equation in the proof:

$\partial u = \frac{\partial u}{\partial x} \partial x +\frac{\partial u}{\partial y} {\partial y}$

Then the professoor proceeded to divide by $\partial u$ on both sides so you end up with $\frac{\partial u}{\partial u}$ on the left hand side which is essential 1. He then did this for $v$ also.

My question: $\partial u\ , \partial x\ , \partial y$ on their own makes no sense. And since partial derivative is defined for suppose u with respect to some other variable, dividing or multiplying by $\partial (some \ variable)$ should be wrong. What is the correct theory ?

答:
First of all, it is important to distinguish between expressions such as $dx$ and $\partial x$.

The differential was first introduced via an intuitive or heuristic definition by Leibniz, who thought of the differential $dy$ as an infinitely small change in the value $y$ of the function, corresponding to an infinitely small change $dx$ in the function's argument $x$.

Later Cauchy's approach was a significant logical improvement over the infinitesimal approach of Leibniz because, instead of invoking the metaphysical notion of infinitesimals. That is, one was free to define the differential $dy$ by an expression

$$dy := f'(x) dx $$

in which $dy$, $dx$ and $f'(x)$ are simply new variables taking finite real values, not fixed infinitesimals as they had been for Leibniz.

Summarizing this: Total differentials such as $dx$, $dy$ and partial derivatives $f'(x) \equiv \frac{\partial y}{\partial x}$ must be thought as mathematical objects and can be manipulated in exactly the same manner as any other real quantities in a meaningful way. Note that this does not hold for $\partial x$ or $\partial y$ alone, which have no meaning in Cauchy's approach.

This might help you:

Differential of a function

Title: Equation of Partial derivatives


问:
Is the product of two Gaussian random variables also a Gaussian?
Say I have $X \sim \mathcal N(a, b)$ and $Y\sim \mathcal N(c, d)$.  Is $XY$ also normally distributed?

Is the answer any different if we know that $X$ and $Y$ are independent?

答:
The product of two Gaussian random variables is distributed, in general, as a linear combination of two Chi-square random variables:

$$ XY \,=\, \frac{1}{4} (X+Y)^2 - \frac{1}{4}(X-Y)^2$$

Now, $X+Y$ and $X-Y$ are Gaussian random variables, so that $(X+Y)^2$ and $(X-Y)^2$ are Chi-square distributed with 1 degree of freedom.

If $X$ and $Y$ are both zero-mean, then

$$ XY \sim c_1 Q - c_2 R$$

where $c_1=\frac{Var(X+Y)}{4}$, $c_2 = \frac{Var(X-Y)}{4}$ and $Q, R \sim \chi^2_1$ are central.

The variables $Q$ and $R$ are independent if and only if $Var(X) = Var(Y)$.

In general, $Q$ and $R$ are noncentral and dependent.


答:
My preference is to define injectivity of linear transformations the same way we define it for all functions: a linear transformation $T$ is injective if and only if $T(m_1)=T(m_2)$ implies $m_1=m_2$.

Now, it is clear that if the linear transformation is injective, then since we know that by virtue of being linear we must have $T(\mathbf{0})=\mathbf{0}$, then by virtue of being injective we will also have that if $T(m)=\mathbf{0}$, then $m=\mathbf{0}$; that is, the nullspace is trivial. This condition is certainly necessary.

But your question is: how come it is sufficient as well? That’s a good question: it doesn’t happen in Calculus, for sure. We can have a function that is not injective, but where there is only one number going to $0$ (for example, $f(x)=x^2$).

The key is the linearity. Essentially, because we are working with a vector space, we can shift everything down to $\mathbf{0}$. Just like we can picture vectors as arrows that either start at the origin or somewhere else, we can “shift” the check to $0$.

Say we want to know whether $T(m_1)=T(m_2)$. Well, we can check instead whether $T(m_1)-T(m_2)=\mathbf{0}$, of course; but because $T$ is linear, this is the same as $T(m_1-m_2)=\mathbf{0}$. That is, we can check whether $T$ takes the same value at $m_1$ and at $m_2$ by checking whether $T$ takes the value $\mathbf{0}$ at $m_1-m_2$, by shifting everything by $-m_2$.

Since $T(m_1)=T(m_2)$ if and only if $T(m_1-m_2)=\mathbf{0}$ (all the processes above can be reversed), and since $m_1=m_2$ if and only if $m_1-m_2=\mathbf{0}$, we get a nice chain of implications:
$$
\begin{align*}
T(m_1)=T(m_2) &\iff T(m_1)-T(m_2)=\mathbf{0}\\
&\iff T(m_1-m_2)=\mathbf{0}\\
&\iff m_1-m_2\in \mathcal{N}(T).
\end{align*}$$
And
$$m_1=m_2\iff m_1-m_2=\mathbf{0}.$$
Putting the two together, we can replace the conditions in the definition of “injective.” Rather than asking that “if $T(m_1)=T(m_2)$ then $m_1=m_2$”, we can ask “if $m_1-m_2\in\mathcal{N}(T)$, then $m_1-m_2=\mathbf{0}$.”

But now... we don’t actually need to write the vector $m_1-m_2$ as a difference. The only thing that matters is that we are asking that “if a vector is mapped to $\mathbf{0}$, then that vector is the zero vector.”

So we get:

Theorem. If $T$ is a linear transformation, then the following are equivalent:

If $T(m_1)=T(m_2)$, then $m_1=m_2$.

If $T(x)=\mathbf{0}$, then $x=\mathbf{0}$.

$T$ is injective.

$\mathcal{N}(T)=\{\mathbf{0}\}$.

So the fact that we can check just the nullspace is a feature. It makes checking injectivity easier.

As to your question... if $T$ is the zero map (sends every vector to $\mathbf{0}$), then it is not injective (unless you are working in the zero vector space), then of course you cannot find $v\neq\mathbf{0}$ and $m_1\neq m_2$ with $T(m_1)=v=T(m_2)$. But in every other case, you can.

If $T$ is not the zero map, then there exists $x$ with $T(x)=v\neq\mathbf{0}$. Note that $x\neq\mathbf{0}$ (because $\mathbf{0}$ always maps to $\mathbf{0}$). If there exists $y\neq \mathbf{0}$ with $T(y)=\mathbf{0}$, then let $m_1=x$ and $m_2=x+y$. Then $m_2\neq m_1$, and $$T(m_2)=T(x+y)=T(x)+T(y) = T(m_1)+\mathbf{0}=T(m_1).$$

But now the problem is that if you pick your $v$ badly, then it could be that $\{x\in V\mid T(x)=v\}$ is empty! And then you won’t be able to detect injectivity at all.

So you could try defining a kind of “nullspace” with other vectors. But (i) It may be empty; and so (ii) useless. The only kind of single “target” you can use that you are guaranteed will work to check injectivity is the usual nullspace. No other “target” will do.

Title: Why is linear map injectivity defined solely using the null-space?

Body:
The thing I'm wondering about is as follows:

Can we have a non-injective linear map $f:M\to N$, with an element $n\in N$, and elements $m_1,m_2 \in M$ such that, $f(m_1)=n$ and $f(m_2)=n$, without $m_1, m_2$ being necessarily in the null-space of $f$ defined as $\mathcal{N}(f)=\{m\in M: f(m)=0\}$? or said differently, without $n$ necessarily being $0$.

If not why?


答:
The only chances for cyclic subgroups are for those of order dividing the order of $G$.  So, as one way of going at this, we could count the number of elements of order $1,3,5,9,15,27$ and $45$.

To do that, we can use the fact that the order of any element of $(a,b)\in G$ is $\rm{lcm}(|a|,|b|)$.

So, to get you started, of course we only have one element of order $1$.  We have $\varphi(3)=2$ elements of order $3$ in $\Bbb Z_9$ and also, by the same reasoning, $2$ in the cyclic $\Bbb Z_{15}$.  Forming ordered pairs, and pairing each element of order $3$ up with an element whose order divides $3$,  we get a total of $8$ elements of $G$ of order $3$.  Since, noting once again that each cyclic group of order $3$ has two elements of order $3$, we get a grand total of $4$ subgroups of order $3$.

We now have five more factors to do this for.  Better you than me.  Just kidding. Let's look at $5$.  Since $\varphi(5)=4$, there are four elements of order five in $\Bbb Z_{15}$.  Since we have to pair these with the identity in the other factor, we get a total of $4$ elements of order $5$, hence only one subgroup of $G$ of order $5$.

Now for $9$.  We have $6$ elements of order nine in $\Bbb Z_9$.  We cam pair each of them with anything of order dividing nine in the other factor, $\Bbb Z_{15}$.  Those are the elements of order dividing three, of course, and there are three.  So we get $18$ elements of order $9$.  But each cyclic subgroup of order nine has $\varphi(9)=6$ elements of order nine.  That makes $18/6=3$ cyclic subgroups of order nine.

For $15$, we have two elements of order $3$ in the first factor, and four elements of order $5$ in the second.  Thus we have $8$ elements of order $15$, plus eight in the second factor times three elements of order dividing three in the first, to give $24$ more, for a total of $32$, just enough, $32/\varphi(15)=32/8=4$, to form four cyclic subgroups of order $15$.

I think I'll leave the last two, $27$ and $45$, for you.

Finally, for part two, it's fairly easy to see that you have a subgroup isomorphic to $\Bbb Z_3\times\Bbb Z_3$, or $\Bbb Z_9\times\Bbb Z_3$, say.

Title: subgroup of $\mathbb{Z}_9 \times \mathbb{Z}_{15}$

Body:
For a group $G$ and an element $a\in G,$ let $\langle a\rangle$ be the intersection of all subgroups of $G$ containing $a.$ If $G = \langle a\rangle$ for some $a \in G,$ then $G$ is cyclic. Determine with proof the number of cyclic subgroups of $\mathbb{Z}_9\times \mathbb{Z}_{15}$. Find, with justification, a non-cyclic proper subgroup of $\mathbb{Z}_9\times \mathbb{Z}_{15}.$

By Lagrange's Theorem, the order of each subgroup should divide $135,$ which has $4\cdot 2 = 8$ factors. I tried considering the number of cyclic subgroups of each possible order and deduced from considering the orders of elements of $\mathbb{Z}_9\times \mathbb{Z}_{15}$ that the orders for cyclic subgroups are $1, 3, 5, 9, 15, 45.$ The only subgroup with order $1$ is the trivial subgroup, and this is also a cyclic subgroup. The cyclic subgroups of order $3$ seem to be $\langle (0,5)\rangle = \{(0,0), (0,5), (0,10)\}, \langle (3,0)\rangle, \langle (3,5)\rangle, \langle (3,10)\rangle, \langle (6,5)\rangle, \langle (6,10)\rangle.$ However, I'm not sure how to find the number of cyclic subgroups of larger orders.

As for a non-cyclic subgroup, I think it suffices to find a subgroup of order $27$, since no cyclic subgroup has this order. Or there might be a simpler approach.


问:
Do exponents in tropical polynomials have to be integers?
Recently I've been learning about tropical geometry, and every time I see a definition of a tropical polynomial in e.g. $k$ variables $x_1,x_2,...,x_k$ such as $\bigoplus_{i=1}^n a_i x_1^{b_{i1}}x_2^{b_{i2}}...x_k^{b_{ik}}$, it usually defines the exponents to be integral. Would it be incorrect to define a tropical polynomial with say rational or real valued exponents? Would the object we get by doing so lose any of the nice properties of tropical polynomials? As far as I can tell the definitions would still work out fine.

答:
The "max-plus" operations (replacing addition with maximum and multiplication with addition) really come from a limiting process. For an expression $f(x_1,\dots,x_n)$ in $n$ variables, its tropicalization is the limit:

$$ \lim_{t \to \infty} \frac{1}{t} \, \log f(e^{tx_1},\dots,e^{tx_n}) $$

If you take $f(x,y) = x+y$, then this limit evaluates to $\mathrm{max}(x,y)$, and if $f(x,y) = xy$, then the limit is $x+y$. Both limits are very easy to compute using L'Hospital's rule.

Exponents become multiplication: if you take $f(x) = x^p$, then

$$ \lim_{t \to \infty} \frac{1}{t} \, \log \Big( (e^{tx})^p \Big) = px $$

This clearly does not depend on $p$ being an integer, so the "tropicalization" of non-integer exponents makes sense in the same way.


问:
Cylinder between two circles in $\mathbb{R}^3$
Let $\mathbf{p}, \mathbf{q} \in \mathbb{R}^3$, $r_p, r_q \in \mathbb{R}^+$ and $\mathbf{n}_p,\mathbf{n}_q \in S^2$. Further let $\mathbf{n} =  \frac{\mathbf{p}-\mathbf{q}}{||\mathbf{p}-\mathbf{q}||}$.

Suppose I have two circles $(\mathbf{p}, r_p,\mathbf{n}_p)$ and $(\mathbf{q}, r_q,\mathbf{n}_q)$ sufficiently far from each other. I can join them up to create a cylindrical shape. The volume enclosed by this tube is the convex hull of the two circles.

Is there a natural way to parametrize this surface (or the solid)? A "natural" map $S^1 \times (0,1) \to \mathbb{R}^3$?

Is there a name for this shape? Or perhaps a linear map to "straighten" it into a more regular shape?

What happens if the surface is extended? Does it self-intersect in a single point (cone) or does something more complicated happen?

How can I calculate its sections with planes $\{ \mathbf{r} : \mathbf{n} \cdot \mathbf{r} = a\}$? Are they ellipses?

答:
I will attempt to answer these in order 2,1,4,3.
Disclaimer: I don't have that deep knowledge in algebraic geometry but since noone has answered yet I will take a stab at it.

The convex hull is a nontrivial developable surface + the two end caps of the circles. In the below, I will ignore the end caps and focus on the developable surface.

2: This surface, let's call it teres surface (I like the suggested name), is a ruled surface:
https://en.wikipedia.org/wiki/Ruled_surface. Some people would call it a scroll.

A ruled surface $S$ is such that through every point of $S$ there is a straight line that lies on $S$.

I think there are nice maps in higher dimensions to "straighten out" such a surface but not with a linear map directly in 3D.

1: If the circles are parameterized with, say angle $u$, then a ruled surface (not the actual teres) connecting the two circles can be expressed as:
$\mathbf{x}(u,v)=(1-v)\mathbf{c}(u)+v\mathbf{d}(u)$ where $\mathbf{c}(u)$ and $\mathbf{d}(u)$ are the parametric equations for the circles (directrices) and $v$ is the parameter running along the generator lines.

To get the convex hull, or rather developable surface (the interesting non-flat part, and not the end cap surfaces), you can then rotate and scale (nonlinearly) the parametrization of one of the circles to align it in such a way that the resulting surface is a developable surface (and hence part of the convex hull). The actually parameterization you would need to use might be highly nontrivial to derive, see an example further below. For most parameterizations you don't get a convex hull but a surface with negative Gaussian curvature. In addition, there is no guarantee that there exists a rational parameterization of this surface, again see below for a non-rational parameterization.

4: The only surfaces of algebraic degree 2, or quadric surfaces, that are ruled surfaces are circular or elliptical cones and cylinders, hyperbolic paraboloids, or hyperboloids of one sheet.
https://en.wikipedia.org/wiki/Quadric

I am ignoring all degenerate cases.

The hyperbolic paraboloid and hyperboloid of one sheet actually have the property that for every point of $S$ there are two straight lines that lies on $S$.
The teres is definitely not one of these two.

Could the teres be an elliptical cone? Indeed, elliptical cones contain circles:
https://en.wikipedia.org/wiki/Circular_section
and we could truncate an elliptical cone in two circles to get an oblique circular cone. However, circular sections of an elliptical cone only exist for certain configurations of the intersection planes so the teres cannot be an elliptical cone in the general case. However, it can be a cone in special cases.

This means the surface is, in general, of algebraic degree 3 or higher, see below.

$S$ is said to be torsal if its line generators have a unique tangent plane (the same tangent plane to the surface all along the line). Some ruled surfaces have only a finite number of torsal lines. I have attached a visualization of the Gaussian curvature of a particular attempt of me to produce an approximate teres surface (by various numerically methods).

Here, blue represents low values (in this case negative values of approximately -0.001, so everything is close to zero here) and red represents high. The Gaussian curvature of a ruled surface is nonpositive and in the figure deep red is zero. In the figure one gets a hint that there might be two or three torsal lines among the generating lines. Note that the edge curves are actually circles lying in two different planes that are not parallel. However, due to the perspective they will inevitably look like ellipses.
However, this is just a near-teres surface. A real teres surface, which by the question is part of the convex hull surface, needs to have zero Gaussian curvature, for an exact teres surface see further below. See also:
https://mathoverflow.net/questions/103982/when-is-the-hull-of-a-space-curve-composed-of-developable-patches
In other words, the true teres surface is a developable surface.

However, the above near-teres comes very close.

The generic planar sections are not ellipses for a general teres surface since its algebraic degree is higher than 2. I have attached a picture of an orthographic projection of a generic section of the above near-teres suface and one can kind of see that it is not an ellipse. Note that since the section curve might have cubic degree it may not be rationally parameterizable (this is a theorem for elliptic curves).

The viewing angle doesn't matter since an ellipse will always look like an ellipse under projection. To find a nice equation for a planar section intersection curve is probably hopeless and would need to be handled numerically.

This reference as well as others:
https://www.researchgate.net/publication/257314770_Developable_Surfaces_Their_History_and_Application
states that any developable surface of this kind might be a generalized cone or a tangent developable. If it is a generalized cone, then this means that although the section curves can be complicated they will all intersect at a point. Think of a line with that singular point as a fulcrum tracing out some interesting shape. The teres surface is characterized by, among other things, having (exactly?) the desired two circles and the surface cannot be a generalized cone  for two circles in general position in space. This means that the teres surface is a tangent developable surface for some curve. The curve is then to be found nontrivially somewhere along the surface, outside of the region between the circles, beyond the smaller circle if the circles have different radii. For an example, see below.

Updated Information:

I worked a bit more on this fascinating problem (applied algebraic geometry).
For a counter example showing that in general the intersection is not a point, but rather a curve belonging to a tangent developable surface, consider the following example: The developable surface beweteen a circle of radius $1$ centered at $x=y=z=0$ and rotated by $\phi=\pi/6$ (30 deg) around the y-axis and a circle of radius $2$ centered at $x=y=0$ and $z=5$ and parallel with the xy-plane.

To find a parameterization of this is a bit tricky and I used the computer. Start by parameterizing the cone correponding to the first circle parallel with the xy-plane  ($\phi=0$).

We then have:

$\mathbf{x}(u,v)=(1-v)\mathbf{c}(u)+v\mathbf{d}(u)$

with:

$\mathbf{c}(u)=(c_x(u),c_y(u),c_z(u))$

$c_x(u)=\frac{1-u^2}{1+u^2}$

$c_y(u)=\frac{2u}{1+u^2}$

$c_z(u)=0$

and

$\mathbf{d}(u)=(d_x(u),d_y(u),d_z(u))$

$d_x(u)=2\frac{1-u^2}{1+u^2}$

$d_y(u)=2\frac{2u}{1+u^2}$

$d_z(u)=5$

Now, to get a more general ruled surface (not developable yet), rotate the first circle around the y-axis using the matrix (with $\phi=\pi/6$):

R=$\begin{bmatrix}
    cos(\phi) & 0 & sin(\phi)\\
    0 & 1 & 0\\
    -sin(\phi) & 0 & cos(\phi)\\
\end{bmatrix}$

This gives

$c_x(u)=cos(\phi) \frac{1-u^2}{1+u^2}$

$c_y(u)=\frac{2u}{1+u^2}$

$c_z(u)=-sin(\phi) \frac{1-u^2}{1+u^2}$

Combine into the ruled surface:

$x(u,v)=(1-v) cos(\phi) \frac{1-u^2}{1+u^2} + 2 v \frac{1-u^2}{1+u^2}$

$y(u,v)=(1-v) \frac{2u}{1+u^2} + 2 v \frac{2u}{1+u^2}$

$z(u,v)=(1-v) (-sin(\phi) \frac{1-u^2}{1+u^2}) + 5 v$

Now this will only give us a ruled surface with negative Gaussian curvature everywhere. This surface can be implicitized, which gives a surface of degree 4 as follows:

$99 x^4 + (196 \sqrt{3} - 388) x^3 z + (1000 \sqrt{3} - 2020) x^3 + (-200 \sqrt{3} + 574) x^2 y^2 + (-364 \sqrt{3} + 630) x^2 z^2 + (-4000 \sqrt{3} + 6960) x^2 z + (-10500 \sqrt{3} + 18354) x^2 + (1748 \sqrt{3} - 3092) x y^2 z + (-6760 \sqrt{3} + 11500) x y^2 + (-116 \sqrt{3} + 204) x z^3 + (-360 \sqrt{3} + 660) x z^2 + (2108 \sqrt{3} - 3516) x z + (5040 \sqrt{3} - 8580) x + (-200 \sqrt{3} + 475) y^4 + (-3292 \sqrt{3} + 5706) y^2 z^2 + (10080 \sqrt{3} - 17440) y^2 z + (-6900 \sqrt{3} + 11950) y^2 + (132 \sqrt{3} - 229) z^4 + (920 \sqrt{3} - 1600) z^3 + (-416 \sqrt{3} + 678) z^2 + (-7160 \sqrt{3} + 12280) z + 7100 \sqrt{3} - 12425=0$

Now, to "twist" this into a developable surface, consider $w$ as a function of $u$ and set $w=w(u)$ and set:

$\mathbf{d}(w)=\mathbf{d}(w(u))$

We have that for a surface on the form:

$\mathbf{x}(u,v)=\mathbf{\gamma}(u)+v\mathbf{\zeta}(u)$

it is a developable surface if and only if

$det(\dot{\gamma}(u),\zeta(u),\dot{\zeta}(u))=0$

For our surface we have:

$\gamma(u)=\mathbf{c}(u)$

$\zeta(u)=\mathbf{d}(w)-\mathbf{c}(u)$

with $w=w(u)$ ($w$ is a function of $u$)

The determinant condition where we use the chain rule for $w=w(u)$

gives

$-8 \frac{d w(u)}{du} \frac{5 w(u)^2 \sqrt{3} u - 2 w(u)^2 u - 9 w(u) u^2 - 5 \sqrt{3} u + 11 w(u) - 2 u}{(u^2 + 1)^2 (w(u)^2 + 1)^2}$

from which we can see that for the determinant to vanish we need:

$5 w(u)^2 \sqrt{3} u - 2 w(u)^2 u - 9 w(u) u^2 - 5 \sqrt{3} u + 11 w(u) - 2 u$

to vanish.

Solving for $u$ now gives, as one of two roots:

$w=\frac{45 \sqrt{3} u^2 + 18 u^2 - 55 \sqrt{3} - 22 + \sqrt{6399 u^4 + 1620 \sqrt{3} u^4 + 6794 u^2 + 1720 \sqrt{3} u^2 + 9559 + 2420 \sqrt{3}}}{142 u}$

And substituting back into:

$\mathbf{d}(w)=\mathbf{d}(w(u))$

gives us the developable surface.

This is not a rational parameterization but a rational parameterization in radicals. There may still be some way of rationally parameterizing this surface but my guess is that this is not possible in general. See also:
http://www.imus.us.es/2cji/archivos/pdf/2cji-MatComp-Sevilla.pdf

Although it is not rational, the radical parameterization means there is an algebraic relationship between $x,y,z,u$, and $v$ and that the surface is algebraic. To see this, just manipulate $w=w(u)$ to eliminate the radicals and get an algebraic relationship between $w$ and $u$:

$5 w^2 \sqrt{3} u- 2 w^2 u- 9 w u^2- 5 \sqrt{3} u+11 w-2 u=0$

This hints that the degree of the corresponding developable surface is probably higher than the previously implicitized ruled surface's $4$. Implicitation of this surface (to get the surface on the form $g(x,y,z)=0$) is difficult. It would be interesting to know its degree. Perhaps someone out there knows a good way to determine the degree.
Trying to implicitized the section curve at $z=5$ hints that the degree is $8$. The fact that the oloid is an $8$th degree surface, https://mathcurve.com/surfaces.gb/orthobicycle/orthobicycle.shtml,  means the teres is almost certainly an $8$th degree surface. Note that the above argument should hold for any types of conic section curve pairs (any end-curves) due to "projective equivalence".

Some visualizations:

The tangent developable surface between the two circles (magenta). Isoparametric lines are shown (blue) as well as the cuspidal curve (green). This particular parameterization is quite nonuniform and doesn't cover the entire algebraic surface (the Zariski closure of the surface), hence the "gap". The parameterization becomes singular along a line and the "gap" is surrounding this line for visualization-numerical reasons:

The cuspidal curve. The wanted surface is swept out by the tangents of this curve:

A zoomed in view of the isoparametric net of original ruled surface and the net of the developable surface. The ruled net is red and the developable is blue. The "rotation" of the isoparametric curves is visible. The original ruled surface and developable surface are almost indistinguishable. For a more general positioning of the circles the difference would probably be greater:

A visualization of the ruled surface (green) and the developable surface (gray)]. They are almost overlapping. However, the Gaussian curvature of the ruled surface is going as low as $-0.005$ whereas numerical evaluation of the developable surface shows the magnitude of the Gaussian curvature to smaller than $10^{-10}$. However, the surface was symbolically derived so we know it to be exactly zero:

Transparent close up view of the interesting area:


问:
$A^TAx = A^Tb$ is always consistent, independent of $Ax=b$?
Suppose $A$ is a $m\times n$ matrix, $x$ is a vector of unknowns and $b$ is a vector of know entries. Whether $Ax=b$ is consistent or not ,But $A^TAx=B$ is always consistent.

Is that above statement correct? or Am I missing anything?

答:
If you are dealing with the reals (complex is similar) we have
$\mathbb{R}^m = {\cal R}A \oplus \ker A^T$.

In particular, we can decompose $b$ (uniquely) as $b=b_1+b_2$ where
$b_1 \in {\cal R}A, b_2 \in\ker A^T$.

Hence there is some $x$ such that $Ax = b_1$, and $Ax-b_1 = b_2$ and so
$A^T(Ax-b) = 0$.