问:
Euler's proof of Fermat's Last Theorem for cubes
I'm working through Richard Freidburg's An Adventurer's Guide to Number Theory, and on p. 149, chapter 7, he presents Euler's proof of Fermat's Last Theorem for cubes.

He starts with $z^3 = x^3 + y^3$, where $x, y, z$ have no factors in common. This implies that $z$ is even and that $x$ and $y$ are odd.

He doesn't state explicitly, but I think his proof assumes $x > y$.

Because $x$ and $y$ are both odd, $x + y$ is even and $x - y$ is also even.

He defines $p = (x + y)/2$, $q = (x - y)/2$, therefore $x = p + q$ and $y = p - q$.

By substitution, binomial expansion and factoring, he converts $z^3 = x^3 + y^3$ to $z^3 = 2p(p^2+3q^2)$

He then presents 4 'remarks', that the reader should prove. I think I've got 1, 2, and 4 figured out & need help with 3:

Remark 1: $p$ and $q$ have no common factor

Proof: Suppose $p$ and $q$ have a common factor $a$, then $x = a(p\prime + q\prime)$ and $y = a(p\prime - q\prime)$, therefore $a$ is a common factor of both $x$ and $y$, but $x$ and $y$ have no common factors

Remark 2: $p^2 + 3q^2$ is odd

Proof: $x = p + q$, $x$ is odd, therefore either $p$ is odd or $q$ is odd. Both cannot be odd, nor can both be even. If $p$ is odd then $q$ is even and $p^2$ is odd, $q^2$ is even and thus $3q^2$ is also even, and $p^2 + 3q^2$ is an odd number plus an even number, hence an odd number. Conversely, if $p$ is even, then $p^2$ is also even, and $q$ is odd, as is $q^2$, thus $3q^2$ is also odd. In this case $p^2 + 3q^2$ is an even number plus an odd number, hence an odd number.

Remark 3: $q$ is not divisible by 3

(Hints provided by the author: For this you must use the fact proven earlier in this chapter, that every cube is congruent to 0, 1 or -1 modulo 9. If $q$ is divisible by 3, then $x \equiv y$ modulo 3, and hence $x^3 \equiv y^3$ modulo 9.)

Remark 4: $p$ and $p^2 + 3q^2$ have no prime factor in common, except possibly 3.

Proof: Follows from remark 1. $p$ and $q$ have no common factors, therefore $p^2$ and $q^2$ have no common factors. However if 3 is a factor of $p$, then $p$ and $3q^2$ share 3 as a common factor.

答: Well, if $x^3$ and $y^3$ are congruent to $0$ mod $9$, then so is $z^3$, which contradicts "no common factors". If they are both $1$ (resp $-1$) then $z^3$ is $2$ (resp $-2$) which contradicts that "cube should be congruent to $0,1,-1$".


问:
Why is this integral diverging? $\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$
$$\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$$

I can easily prove that this integral is diverging by taking the limit over the following proper integral,

$$\lim_{R_1,R_2\to \infty} \int\limits^{R_2}_{-R_1} \,\frac{x}{x^2+1} dx$$

Mathematically, this makes sense to me, but intuitively I am not able to absorb this.

If we observe the function $y=\frac{x}{x^2+1}$, it is clearly an odd function. And since, integrals return signed areas, an integral to an odd function having limits that are the negatives of each other, should evaluate to zero.

As clearly seen by the following plot.

Can someone please clarify this?

Thank you.

答: You have to look more carefully at the definition of $\int_{- \infty}^{\infty}.$  Your intuition is treating it as though it means $\lim_{x \to \infty} \int_{-x}^x$, but that's not correct.  It's actually $\lim_{x \to - \infty}(\lim_{y \to \infty} \int_x^y)$ (assuming the integrand is in fact defined everywhere), and for this function the inner limit does not exist for any $y$.

ContentLicense: CC BY-SA 4.0


问:
Approximating $\int_2^4 e^{\frac{1}{x}} dx$

Estimate the integral precisely up to the error of $0.001$:$$\int_2^4 e^{\frac{1}{x}} dx$$

I started by dividing the segment $[2,4]$ into some pieces $(n=5)$ as in the following figure:

$\qquad\qquad\quad$

Now, I have $$\Delta x = \frac{4-2}{5} = 0.4$$ and therefore, $$S \approx \sum_{i=1}^5 f(x_i)(x_i - x_{i-1}) = \Delta x \sum_{i=1}^5 e^{\frac{1}{x_i}}$$ where $x_i = 2 + 0.4i$.

Then, approximating the area boils down to approximating the exponential function at $5$ points. What is the best way to proceed? And how do I know that the final result will give the error at most $0.001$?

答:
The Midpoint Rule error is given by

$$\tag 1 |E_n| \le \left|\dfrac{(b-a)^3}{24 n^2}\right | |\max_{2 \le x \le 4}f''(x^*)|$$

The second derivative of $f(x) = e^{1/x}$ is

$$f''(x)= \dfrac{e^{1/x} (2 x + 1)}{x^4}$$

So, we want to find the max of $|f''(x)|$ over the interval $[2,4]$

$$\displaystyle \max_{2 \le x \le 4}\left| \dfrac{e^{1/x} (2 x + 1)}{x^4}\right| = \dfrac{5 \sqrt{e}}{16} ~ \mbox{at}~ x = 2$$

Substituting everything we have into $(1)$

$$|E_n| \le \dfrac{(4-2)^3}{24 n^2} \left(\dfrac{5 \sqrt{e}}{16}\right) \le 0.001$$

Can you continue by solving that for $n$? You should find

$$n \ge 13.105$$

We would choose an integer at $n = 14$.

Note that that is an upper bound and can be way off.

In practice, using trial and error gives eight intervals for an absolute error $< 0.001$

$$2.83456$$

Using WA, the closed form is ugly, and the numerical result is

$$\displaystyle \int_2^4 e^{1/x} dx \approx 2.835422294875796952$$

Update If you want to see the intermediate Midpoint Rule steps above, use this calculator.


答:
This integral is not absolutely convergent and thus you cannot "easily prove that this integral is diverging"; such expression is usually regarded as ill-defined.

This is similar to $\int_{-\infty}^{\infty}xdx$, the "meaning" of which depends on how you interpret the symbol $\int_{-\infty}^\infty$.

If you take it as "a Cauchy principal value"
$$
\lim_{N\to\infty}\int_{-N}^N xdx
$$
then it is zero because the function is odd.

However, if you interpret it as an iterated limit:
$$
\lim_{a\to-\infty}\left(\lim_{b\to\infty}\int_{a}^b xdx\right)
$$
you already have a problem with the inner limit.

On the other hand, if a function is absolutely convergent,  then all the following "interpretations" of $\int_{-\infty}^\infty$ are the same:

$\int_{-\infty}^\infty:=\lim_{a\to\infty}\int_{-a}^a$;

$\int_{-\infty}^\infty:=\lim_{a\to-\infty}\lim_{b\to\infty}\int_a^b$;

$\int_{-\infty}^\infty:=\lim_{b\to\infty}\lim_{a\to-\infty}\int_a^b$;

$\int_{-\infty}^\infty:=\int_0^\infty+\int_{-\infty}^0$

Title: Why is this integral diverging? $\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$

Body:
$$\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$$

I can easily prove that this integral is diverging by taking the limit over the following proper integral,

$$\lim_{R_1,R_2\to \infty} \int\limits^{R_2}_{-R_1} \,\frac{x}{x^2+1} dx$$

Mathematically, this makes sense to me, but intuitively I am not able to absorb this.

If we observe the function $y=\frac{x}{x^2+1}$, it is clearly an odd function. And since, integrals return signed areas, an integral to an odd function having limits that are the negatives of each other, should evaluate to zero.

As clearly seen by the following plot.

Can someone please clarify this?

Thank you.


问:
Matrix determinant where $a_{ij} = i + j$
So I'm studying for my course of linear algebra and the following problem was inside the book with exercises. "Given a matrix $A \in \mathbb{R}^{n \times n}$, where $a_{ij} = i + j$, calculate it's determinant".

The first thing I did was make a small visualization of what this sort of matrix would look like. I came up with this:

$$ A = \begin{pmatrix}
1 + 1 & 1 + 2 & ... & 1 + n\\
2 + 1 & 2 + 2 & ... & 2 + n\\
... & ... & & ...\\
n + 1 & n + 2 & ... & n + n
\end{pmatrix}$$

I set of to set up an expression for the determinant using cofactor expansion (I don't know if this is the correct English terminology for it, the formula is given below, I also found on wikipedia this is called "Laplace Expansion"?) to the first row (thus making $j = 1$ in the expression):

$$ \sum^{n}_{i = 1}(-1)^{(i + 1)}a_{i1}det(M_{i1})$$

where $det(M_{i1})$ is the minor for the $i1$-th element. This is the point where I am stuck and could definitely use some help. I've tried writing $A$ in terms of summations to maybe see something in it, but this evolved into some very complex expression which I'm not even going to bother typing down (it's really, really long).

I have a feeling that I can bring the $a_{i1}$ out of the summation but I'm not sure how I would go on doing that.

Thanks in advance

答:
The determinant is undefined if $m\neq n$, so consider matrices $A^{n\times n}$ with $a_{i,j}=i+j$.

For $n\ge3$, the determinant is $0$. The second row can be written as a linear combination of the first and third rows:

$$\begin{align}
\begin{bmatrix}2&3&4&\cdots&1+n\end{bmatrix}^\top+\begin{bmatrix}4&5&6&\cdots&3+n\end{bmatrix}^\top&=\begin{bmatrix}6&8&10&\cdots&4+2n\end{bmatrix}^\top\\[1ex]
&=2\begin{bmatrix}3&4&5&\cdots&2+n\end{bmatrix}^\top
\end{align}$$

The cases of $n<3$ are trivial (but non-zero!).


答:
Let $ f $ be locally integrable at $ (-\infty,+\infty)$

By définition, we say that the integral $$\int_{-\infty}^{+\infty}f(x)dx$$
is convergent if only if there exists $ c\in \Bbb R $ such that, Both integrals
$$\int_{-\infty}^cf(x)dx\text{ and } \int_c^{+\infty}f(x)dx$$
Are convergent.

If ine of them is divergent, the integral will be divergent.

In your case, the integral $$\int_0^{+\infty}\frac{xdx}{1+x^2}$$ is divergent.


问:
Prove a coin game is fair

Suppose $2n$ identical, equal valued fair coins are flipped. If no more than half land head, you win all the heads. If more than half land head, you lose all the heads. Prove the game is fair with combinatorial argument only.

This is derived from a recently closed question, which was linked to Prove that $\sum\limits_{i=1}^n 2i\binom{2n}{n-i}= n\binom{2n}{n}$.

If the game is fair then
$$\sum_{i=0}^{n} i \binom{2n}{i} = \sum_{j=n+1}^{2n} j \binom{2n}{j}\\
\implies n\binom{2n}{n} + \sum_{i=0}^{n-1} i \binom{2n}{i} = \sum_{i=0}^{n-1} (2n-i) \binom{2n}{2n-i}\\
\implies n\binom{2n}{n} = \sum_{i=0}^{n-1} 2(n-i)\binom{2n}{i} = \sum_{i=0}^{n-1} 2i \binom{2n}{n-i}$$

I found that it reduced to proving this simple symmetry:

$$(n-i) \binom{2n}{n-i} = (n+i+1) \binom{2n}{n+i+1}, \forall 0 \le i \le n-1 \tag 1$$
i.e., your gain with $2n$ heads is offset by your loss with $2n+1$ heads, gain with $2n-1$ heads is offset by loss with $2n+2$ heads, and so on, and when all land tail you breakeven.

Algebraically $(1)$ is trivially true but is there a pure combinatorial argument? Or better yet, an argument to prove the game is fair without breaking it down?

答:
Based on Mike Earnest's answer I have an argument to prove the following:

$$\sum_{i=0}^{n} i \binom{2n}{i} = \sum_{j=n+1}^{2n} j \binom{2n}{j}$$

Now the question is: if there are no more than half heads, paint one of them red; otherwise paint one of more than half heads green. Show that there are same number of ways of getting a red head vs. a green head.

Now we pick a particular coin. We look at each case where it is the red coin. That means there are $k$ other heads and $2n-1-k$ tails where $0 \le k \le 2n-1$. If we flip all other $2n-1$ coins then we have $2n-k$ heads where $n+1 \le 2n-k\le 2n$ which corresponds to a case where this particular coin is a green head. And this map is one-to-one, therefore we have the same number of cases where this particular coin is a red head vs. a green head. And this is true for each and every one of the $2n$ coins. We are done.

Title: Prove a coin game is fair


问:
Notation Confusion (Specifically: "Such That", & Multiple $\exists$)
There are a couple of notation-related matters that I find myself clumsily working around quite often when writing proofs, and I was hoping for some help with dealing with them:

Let's say that $L$, $L_1$ and $L_2$ are regular languages.

I claim:

$w \in (L_1 \cup L_2) \cdot L \iff \exists w_1 \in (L_1 \cup L_2), w_2 \in L$ such that $w=w_1\cdot w_2$.

So I have two questions about this:

How should I handle that "there exist several $w_i$ such that..."? That is to say: is it okay to separate them with a comma as I did ($\exists w_1 \in ..., w_2 \in ...$), or would it be better to add a new $\exists$ per $w_i$ (e.g. $\exists w_1 \in ... \wedge \exists w_2 \in ...$)?

Is there any kind of standard notation to use as a substitute for "such that"? I gather that $|$ and $:$ are used in set-builder notation and not in cases such as this one; so it'd be off to write something like: "$w \in (L_1 \cup L_2) \cdot L \iff \exists w_1 \in (L_1 \cup L_2), w_2 \in L: w=w_1\cdot w_2$". Is that correct?

Looking at some questions & answers on SE and on Google, my understanding was that the common consensus seems to be that writing "such that" is the preferred method, usually, but my native language is written RTL, and that makes it a bit awkward sometimes when I want to add "such that" in the middle of a LTR mathematical statement (I usually find myself adding line breaks in unfortunate places to deal with it), so I was hoping for some clear, creative solution based on standard notation for this.

Does it look okay if I write something like: "$\exists w_1, w_2$ such that $w_1\in ... \wedge w_2 \in ... \wedge w = w_1 \cdot w_2$"? That is, to concentrate all the "exists" in one place, and all their "descriptions" in another?

I get a strong sense that I'm overthinking it all, but I dislike the way a lot of my notation usages feel so clumsy to me. If you can see some sort of root cause (maybe a mindset issue or something more general like that) for my confusion & are able to point it out, that'd be great, too.

答:
If you absolutely insist on using symbols, I would write

$$w\in(L_1\cup L_2)\cdot L\leftrightarrow\exists w_1\in L_1\cup L_2\,\exists w_2\in L(w=w_1\cdot w_2)\,.$$

But if I were actually doing it myself, I’d use intelligible English:

$w\in(L_1\cup L_2)\cdot L$ iff there are $w_1\in L_1\cup L_2$ and $w_2\in L$ such that $w=w_1\cdot w_2$.

Or on occasion even:

A word $w$ is in $(L_1\cup L_2)\cdot L$ if and only if there are words $w_1\in L_1\cup L_2$ and $w_2\in L$ such that $w=w_1\cdot w_2$.


问:
What function gives this sequence?
Recently, I have been trying to find what function $f(n)$, for natural $n$, gives the following sequence:

$$1, 5, 105, 4725, 363825, 42567525, \cdots $$

It turns out that $\frac{d^{2n}}{dy^{2n}} (1-y^2)^{-5/2}|_{y=0}$, generates it. However, is there any function $f(n)$ that gives the same?

So far, it has been kind of hard to spot a pattern:

$$1, 5, \underbrace{3 \cdot 5 \cdot 7}_{105}, \underbrace{3^3 \cdot 5^2 \cdot 7}_{4725}, \underbrace{3^3 \cdot 5^2 \cdot 7^2 \cdot 11}_{363825}, \underbrace{3^5 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13}_{42567525}$$

Any suggestions would be very much appreciated! Thanks

答:
As mentioned by others, this is an example of binomial series.
$$ (1-x^2)^{-X/2} = 1 + X\frac{x^2}{2!} + 3!! X(2+X)\frac{x^4}{4!} + \\
5!! X(2+X)(4+X)\frac{x^6}{6!} + \cdots.  $$
This implies that the exponential power series
coefficient is
$$ (2n\!-\!1)!! \prod_{k=0}^{n-1}
 (2k\!+\!X)= (2n\!-\!1)!!
\frac{(X\!+\!2n\!-\!2)!!} {(X\!-\!2)!!}. $$
where the $\,(...)!!\,$ denotes a
double factorial. When $\,X=5\,$
OEIS sequence A051577 is the double factorial quotient.

Title: What function gives this sequence?


问:
Does the function $g\in\mathbb{C}^\infty(\mathbb{R}^2)$ exist if...
Does the function $g\in\mathbb{C}^\infty(\mathbb{R}^2)$ exist if
$\{g > 1\}$ is simply connected and $\{ g > -1 \}$ is not simply connected. I think it does not exist, but I can't find any counterexample of its existence.

答: Take $g:\Bbb R^2\to\Bbb R$ to be defined as $g(x)=-\frac{1}{\|x\|^2+1}$. Then $g^{-1}((1,\infty))=\emptyset$ and $g^{-1}((-1,\infty))=\Bbb R^2\setminus\{0\}$.


Body: $$\lim_{l\to\infty}\lim_{u\to\infty}\int_l^u f(t)\,dt\ne\lim_{u\to\infty}\int_{-u}^u f(t)\,dt.$$


问:
Ordered triple condition
Compute the number of ordered triples $(a,b,c)$ of positive integers such that
$$a2^b + b2^c + c2^a = 2000.$$

I tried to make some bijection but no luck.
I have tried setting some variables to see if stars and bars (or balls and urns) works, as well as casework counting.

答:
On investigation numerically, there are no positive integer triples $(a,b,c)$ for which the function $ f(a,b,c) := a2^b + b2^c + c2^a$ fulfils the condition
$$  f(a,b,c) = 2\cdot 10^k $$
for any value of $k\in \mathbb N$ up to $600$.

Allowing $a,b,c$ to also be zero (but forbidding any of $2^a, 2^b, 2^c$ to be greater than the target value) gives us effectively one solution in that space, $f(1,4,0) = 20$, so also $(4,0,1)$ and $(0,1,4)$.

The closest approach to the original problem (where $k=3$) is $f(5,8,6)=1984$

I might have become a little obsessed with the form $a2^b + b2^c + c2^a$ ...

Anyway the original format above has no solutions out to $2\cdot 10^{10000}$.

I also had a quick look at variations in the form $a2^b + b2^c + c2^a = N^k$ (just for fun) and

$4\cdot 2^5 +5\cdot 2^3 +3\cdot 2^4 = 6^3$  (quite pretty, all low numbers)

$13\cdot 2^{17}+ 17\cdot 2^{14} + 14\cdot 2^{13} = 8^7$, weirdly

$3\cdot 2^9 +  9\cdot 2^7 +  7\cdot 2^3 = 14^3$ and

$2\cdot 2^7 + 7\cdot 2^3 + 3\cdot 2^2 = 18^2$

Title: Ordered triple condition


问:
Divisibility Rule for 21 similar to rule for 12
I already understand the divisibility rule for 12. $$10\equiv -2 \pmod{12} \implies 10^n\equiv (-2)^n  \pmod{12}$$

Then for some number $n = abcd = 1000a + 100b + 10c + d,  abcd\equiv1000a + 100b + 10c + d \pmod{12}$

Then  $abcd\equiv (-2)^3a + (-2)^2b + (-2)c + d$

Example:  $7368\equiv -8*7 + 4*3 -2*6 + 8\equiv -48\equiv 0 \pmod{12}$

But why does it work for 21 as well?

If I switch the order that the power rises in, so that now it is $a + (-2)b + (-2)^2c + (-2)^3d$ now, all of a sudden, it works for mod 21

Example: $2331\equiv 2 + (-2)*3 + (-2)^2*3 + (-2)^3*1 \equiv 0 \pmod{21}$

It's obvious why this works for numbers like $48$ and $84$ since with both you are multiplying $4 * -2$ and $8 * 1$ and their sum is then $0$ for both.

But why does it still work for numbers like $105$.

$501$ is obviously not divisible by $12$ and yet: $105\equiv 1 + (-2)*0 + (-2)^2*5\equiv 21 \equiv 0 \pmod{21}$

I'm also fairly sure this is part of the reason that $f$ works in Proof of general divisibility rule

答:
For $21$, the reason this works is that $10\times(-2)=1\pmod{21}$, so $10\equiv(-2)^{-1}\pmod{21}$. Hence
$$1000a+100b+10c+d\equiv(-2)^{-3}a+(-2)^{-2}b+(-2)^{-1}c+(-2)^{0}d\pmod{21},$$
so multiplying the RHS by $(-2)^3$ implies the result.

You will observe a similar result for all teen numbers $\overline{1m}=10+m$, because
$$10\times(-m)\equiv-\overline{m0}\equiv1\pmod{\overline{m1}}.$$


问:
reduction of order when one solution is given
I'm currently working on this differential equation, where one solution is already known, however, I'm having difficulties:

The equation:

$(x+1)^2y''-4(x+1)y'+6y=0$

where:

$y_1{(x})=(x+1)^2;$ where $y_2{(x)}=u_xy_1{(x)}$

Then differentiating to obtain $y_2'$and $y_2''$, I get:

$u_x(x+1)^2+2u_x'(x+1)+2u_x''$

substituting this in the differential equation:

$2u_x''(x+1)^2-8u_x'(x+1)^2+6u(x+1)^2=0$

when expanding the bracket:

$2u_x''x^2+4u_x''x+2u_x''-8u_x'x^2-16u_x'x-8u_x'+6ux^2+12ux+6u=0$

I'm stuck on what to do next?

Body:
$$(x+1)^2y''-4(x+1)y'+6y=0$$
For simplicity set $z=x+1$
$$z^2y''-4zy'+6y=0$$
Then $y=z^2 v$
$$z^2(v'z^2+2zv)'-4z(v'z^2+2zv)+6z^2v=0$$
$$z^2(v''z^2+4zv'+2v)-4v'z^3-2z^2v=0$$
$$z^2(v''z^2+4zv')-4v'z^3=0$$
$$v''=0$$
The order is reduced.
$$v=c_1z+c_2$$
$$y=vz^2=c_1z^3+c_2z^2$$
$$y(x)=c_1(x+1)^3+c_2(x+1)^2$$


问:
On splitting of the tensor product.
Let ${\Bbb C}$ be a complex number field. We shall define the two-variable polynomial ring $R$ as follows$\colon$
\begin{equation*}
R \colon = {\Bbb C}[X,Y] = {\Bbb C}[X] \otimes_{\Bbb C}{\Bbb C}[Y] \quad \cdots\cdots \quad (\lozenge).
\end{equation*}
Let us denote by $R_{{\frak m}}$ the localisation of $R$ at the maximal ideal ${\frak m} \colon= (X, Y)$ of $R$. $R_{\frak m}$ is a two-dimensional regular local ring with the unique maximal ideal ${\frak m}$.

Now we consider the quotient ring $S \colon= R_{\frak m}/(XY)$, which is a one-dimensional local ring with the unique maximal ideal $\overline{\frak m}$ and two generic points $(X)$ and $(Y)$. The complement $U \colon= {\mathrm{Spec}}\,S \setminus {\overline{\frak m}}$ consists of two generic points.

Q. How is the affine ring ${\cal O}_U$ of $U$ written explicitly in terms of the presentations $(\lozenge)$?

This might be elementary commutative algebra, but I got confused when trying to write down the explicit product ring structure of ${\cal O}_U$ with respect to two presentations of $R$ at $(\lozenge)$.

答:
You can think of $\mathcal O_U$ as the localization of $S$ by the multiplicative system $Q$ generated by $X+Y$, that is, $Q=\{1,X+Y,(X+Y)^2,...\}$. The point here is that $X+Y\in\mathfrak m$ but $X+Y\notin(X)\cup(Y)$. Instead of $X+Y$ you can use any other polynomial belonging to $\mathfrak m\setminus((X)\cup(Y))$; the resulting localization will be canonically isomorphic to $S_Q$. The space $Spec\ S_Q$ embeds naturally into $Spec\ S$ and its image is $(X)\cup(Y)$.

The ring $\mathcal O_U$ is isomorphic to the direct product of two fields: $\mathcal O_U\cong\mathbb C(X)\times\mathbb C(Y)$.

Title: On splitting of the tensor product.


问:
Let $\mathcal{P}$ be the set of all polynomials. Let $D : \mathcal{P} \to \mathcal{P}$ be a derivative mapping, show that $\text{im}D= \mathcal{P.}$

Let $\mathcal{P}$ be the set of all polynomials $p : \mathbb{R} \to \mathbb{R}$, $x\mapsto a_n x^n + \cdots + a_1 x + a_0$ defined with addition and scalar multiplication. Let $D : \mathcal{P} \to \mathcal{P}$ be a derivative mapping, show that $\operatorname{im}D= \mathcal{P.}$

I've previously posted a similar question regarding the set of polynomials at most degree $n$. Now, this seems to be a generalization for the set of all polynomials?

For the image of $D$ I have that $\operatorname{im}D=\{Dp(x) \mid p(x) \in \mathcal{P} \}$ right? What I don't understand here is that the derivative operator applied to $p(x)$ outputs a polynomial of degree $n-1$ so how can $\operatorname{im}D = \mathcal{P}$ when it seems that $\operatorname{im}D = \mathcal{P_{n-1}}$?

答:
Note that

$\mathcal P = \Bbb R[x], \tag 0$

and hence contains all polynomials of every degree.  Thus:

For any

$b \in \Bbb R\tag 1$

and any

$\displaystyle \sum_0^n a_ix^i \in \Bbb R[x] = \mathcal P \tag 2$

we have

$D \left ( \displaystyle \sum_0^n \dfrac{a_i}{i + 1}x^{i + 1} + b \right ) =  \displaystyle \sum_0^n a_ix^i, \tag 3$

and we see that

$D: \mathcal P \to \mathcal P \tag 4$

is surjective; that is

$\text{im} \; D = \mathcal P = \Bbb R[x]. \tag 5$


问:
Prove that if $\nu(A)=\int_A fd\mu$ and $\mu$ is $\sigma$-finite, then $\nu$ is $\sigma$-finite.
Let $\nu$ be the measure defined on a $\sigma$ finite measure space with measure $\mu$ and a non negative, measurable, real valued function $f$ such that $\nu=f*\mu$ so that $\nu(A)=\int_Afd\mu$.I know that $\nu$ is a measure. My attempt has been to construct a nested sequence of measurable sets $A_i$ so that $A_i\subset A_{i+1}$ and $\Sigma_{i\in \mathbb{N}}=\Omega$. Now because $\mu$ is $\sigma$- measurable $\mu(A_i)<\infty$ $\forall A_i$. Now if $\nu(A_N)=\infty$ it follows from the linearity of the integral that $\nu(A_{i>N})=\infty$. However I haven't been able to produce a contradiction from this assumption.

答:
I could be wrong, but the following was what I had in mind when I suggested to approximate $f$ with simple functions.

Let $E_n$ be the sequences of sets through which $\mu$ is $\sigma$ finite.

If $f=\chi_A$, then $v(E_n)<\infty $. Now if $v_1$ and $v_2$ are $\sigma$ finite measures (with sequences $I_n$ and $J_m$), then $v_1+v_2$ is a $\sigma$ finite measure as well:

$$(v_1+v_2)(I_n\cap J_m)<\infty$$

and $$\bigcup_{n=1}^{\infty}\bigcup_{m=1}^{\infty} I_n \cap J_n = X$$

It doesn't take much to conclude that linear combinations of $\sigma$ finite measures is also a $\sigma$ finite measure. Let $s_n$ be a simple function such that $|\int f - s_n|<\epsilon$,  then $$\int f <\int s_n +\epsilon$$

Notice that, $\int s_n = s_n * \mu$ and $s_n$ is a linear combination of characteristic functions. So $s_n * \mu$ is a $\sigma$ finite measure for some sequence of sets $F_n$. As a result, $\int f$ is a sigma finite measure for some sequence of sets $F_n$.


问:
Solving for $x$: $a \sin(bx + c) = \sin(x)$
Ok, I scoured the internet for more than a few months for this one (whenever I had the time). But just because of this question I've created my first Stack Exchange account.

I have thrown this at all Computer Algebra System software I was accessible to. (not Maple or Mathematica though)

I also tried complex transformation, no luck there (not too much experience there)

I know this could be easily numerically solved because it is guaranteed that there is exactly one solution for every interval when: $a \sin(bx + c) \le 1$. In my case, the general solution is crucial because I'm applying this to an optimization where I can't afford to have an additional dimension to be optimized.

This problem can be visualized as the intersection of two sine waves.

A simplified version of the problem (when a = 1) could be easily solved with the sum-to-product identity:

$\sin(bx + c) = \sin(x)$

$0 = \sin(bx + c) - \sin(x)$

$0 = 2 \cos(\frac{x(b + 1) + c}{2})\sin(\frac{x(b - 1) + c}{2})$

$0 = \cos(\frac{x(b + 1) + c}{2})$ and $0 = \sin(\frac{x(b - 1) + c}{2})$ etc...

But for the time being, the only valid assumption in my case is that b is rational

答:
This is a supplement to the answer of @Piquito. Recalling the identity theorem for power series, we make a coefficient comparison of the series expansion
\begin{align*}
a\sin(c)\sum_{k=0}^\infty (-1)^k\frac{\left(bx\right)^{2k}}{(2k)!}
+a\cos(c)\sum_{k=0}^\infty (-1)^k\frac{\left(bx\right)^{2k+1}}{(2k+1)!}
=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}
\end{align*}
In the following we denote with $[x^n]$ the coefficient of $x^n$ of a series.

We consider even and odd coefficients separately and obtain for $n\geq 0$:
\begin{align*}
&[x^{2n}]:\quad &a\sin(c)\frac{(-1)^nb^{2n}}{(2n)!}&=0&\\
&&a\sin(c)b^{2n}&=0&\tag{1}\\
\\
&[x^{2n+1}]:\quad &a\cos(c)\frac{(-1)^nb^{2n+1}}{(2n+1)!}&=\frac{(-1 )^n}{(2n+1)!}&\\
&&a\cos(c)b^{2n+1}&=1&\tag{2}\\
\end{align*}

From (1) we conclude that at least one factor has to be equal to zero. Looking at (2) we see  that $a,b\ne 0$ and it follows
\begin{align*}
\sin(c)=0\qquad\rightarrow\qquad c=j\pi\qquad j\in\mathbb{Z}\tag{3}
\end{align*}
Putting $c=j\pi$ in (2) we obtain
\begin{align*}
a(-1)^jb^{2n+1}=1\qquad\qquad n\geq 0,\ j\in\mathbb{Z}\tag{4}
\end{align*}
From (4) we see that $j\in2\mathbb{Z}$. Since (4) has to be valid for all $n\geq 0$ we conclude that $b=1$ and finally that $a=1$.

Conclusion: The equation
\begin{align*}
a\sin(bx+c)=\sin(x)\qquad\qquad a,b,c\in\mathbb{C}
\end{align*}
has a solution if and only if $a=b=1$ and $b\in 2j\pi, j\in\mathbb{Z}$. In this case we have the identity
\begin{align*}
\sin(2j\pi x)=\sin(x)\qquad\qquad x\in\mathbb{C}
\end{align*}


问:
Solve non-homogeneous recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$
I'm trying to solve this recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$, but the solution of the homogeneous is $a_n=b+cn$, and I don't know how to change it to find the non-homogeneous solution.

答:
If the non-homogeneous part is constant you can always take the difference and make it homogeneous.  In your case,

$$(\mathbb E-1)^2 x_n = 8 \implies (\mathbb E-1)^3 x_n = (\mathbb E - 1)8=0$$

And you get your characteristic function immediately which is $(\lambda-1)^3=1$.

Here $\mathbb E$ is the forward shift operator, $\mathbb E^i x_n = x_{n+i}, \forall i \in \mathbb N$.

For a non-constant $f(x)$ example, check out this answer.


问:
prove that $\{A_1,...,A_m\}$ cannot span $M_{nn}$ if $\vec{0} \neq y \in \mathbb{R}$ and $A_1y=A_2y=...=A_my=0$
Let $A_1, A_2,...,A_m$ denote n by n matrices if $\vec{0} \neq y \in \mathbb{R}$ and $A_1y=A_2y=...=A_my=0$, prove that $\{A_1,...,A_m\}$ cannot span $M_{nn}$

My Proof (Edited)

Let $B \in M_{nn}$ and invertible. Now assume that $M_{nn} = span\{A_1, A_2,...,A_m\}$

$$B = c_1A_1 + ...+c_mA_m$$
$$By = [c_1A_1 + ...+c_mA_m]y$$
$$By = c_1A_1y + ...+c_mA_my = c_10+...+c_m0=0$$

Hence $By=0$. Since it is invertible $y = B^{-1}0 = 0$ which is a contradiction.

Is this approach (now) correct?

答:
"Hence...either..." is false.

In general, $Ax=0$ does not imply that either $A=0$ or $x=0$.

Hint for a proof.

Let $B$ be an invertiable matrix of $M_{nn}$ and assume that $B$ is a linear combination of the matrices $A_k$. Calculate $By$.


问:
Inequality for a determinant
Let $A=(a_{i,j})$ an $n\times n$ symmetric definite positive matrix i.e. for all $n\times1$ vector $x\ne0$ we have
$$x^T A x>0.$$
We write the matrix $A$ by blocks:
$$A=\begin{pmatrix}A_1 &y^T\\
y& a_{n,n}\end{pmatrix}$$
and the question is to prove that $\det(A)\le a_{n,n}\det(A_1)$. Honestly I have no idea how to get this inequality but at least I have proved that $a_{n,n}>0$ and in fact $a_{i,i}>0$ for all $i$.

答:
$A$ is real symmetric PD.  Running Cholesky Factorization:
$A=LL^T$ which we partition as

$L=\begin{pmatrix}L_1 &0\\
\mathbf m^T& l_{n,n}\end{pmatrix}$

$\begin{pmatrix}A_1 &\mathbf y\\
\mathbf y^T& a_{n,n}\end{pmatrix} =LL^T=\begin{pmatrix}L_1 &0\\
\mathbf m^T& l_{n,n}\end{pmatrix}\begin{pmatrix}L_1^T &\mathbf m\\
0& l_{n,n}\end{pmatrix}$

$\implies A_1 =L_1L_1^T$

$\implies a_{n,n} =\mathbf m^T\mathbf m +l_{n,n}^2\geq l_{n,n}^2\gt 0$

using the multiplicativity of determinants and the fact that the determinant of a triangular matrix is the product of its diagonals, we have

$\det\big(A\big)$

$=\det\big(L\big)\det\big(L^T\big)$

$=\det\big(L^2\big)$

$=\big(\prod_{k=1}^{n-1}l_{k,k}^2\big)\cdot l_{n,n}^2$

$=\det\big(L_1^2\big)\cdot l_{n,n}^2$

$\leq \det\big(L_1\big)^2\cdot a_{n,n}$

$= \det\big(L_1L_1^T\big)\cdot a_{n,n}$

$=\det\big(A_1\big)\cdot a_{n,n}$

Title: Inequality for a determinant


答:
$$z^22y''−4zy'+6y=0,~~z=x+1,$$
is an Euler-Cauchy equation with characteristic polynomial
$$
0=r(r-1)-4r+6=(r-2)(r-3)
$$
This gives $(z^2,z^3)$ or $\bigl((x+1)^2,(x+1)^3\bigr)$ as basis solutions.

In the reduction approach, it pays to be careful in the derivatives computation. With $y_2(x)=u(x)(x+1)^2$ the derivatives are
$$
y_2'(x)=(x+1)^2u'(x)+2(x+1)u(x)\\
y_2''(x)=(x+1)^2u''(x)+4(x+1)u'(x)+2u
$$
where you can see that you reverted somehow the place of the derivatives.

In general, the terms without derivative in $u$ have to cancel, else you have an error in the prior computations.

Title: reduction of order when one solution is given

Body:
I'm currently working on this differential equation, where one solution is already known, however, I'm having difficulties:

The equation:

$(x+1)^2y''-4(x+1)y'+6y=0$

where:

$y_1{(x})=(x+1)^2;$ where $y_2{(x)}=u_xy_1{(x)}$

Then differentiating to obtain $y_2'$and $y_2''$, I get:

$u_x(x+1)^2+2u_x'(x+1)+2u_x''$

substituting this in the differential equation:

$2u_x''(x+1)^2-8u_x'(x+1)^2+6u(x+1)^2=0$

when expanding the bracket:

$2u_x''x^2+4u_x''x+2u_x''-8u_x'x^2-16u_x'x-8u_x'+6ux^2+12ux+6u=0$

I'm stuck on what to do next?


问:
Solving a system of equation:
Solve for $x,y$:
\begin{align}
x^3 + y^3&=2\\
x^2 +x + 9y - 3y^2&=8 \\
\end{align}

I can find $x=y=1$ by guessing. Please help me solve it without using computer.

Thanks

Edited, sorry, I have to change $x^2$ to $x^3$ to have a nice solution, but still can't done it by hand.

答: Substracting equation $2$ from $1$ we obtain $y^3 - x - 9y + 3y^2 = -6 \iff y^3 - \sqrt{2-y^3} -9y + 3y^2 = -6$ Now you have an expression that only consists of $y$'s. Try to see if you can manipulate this such that you get your desired result. I am not sure though if you cand find it algebraically. If you can, chances are that long division may be needed.</p>

ContentLicense: CC BY-SA 3.0

Title: Solving a system of equation:


问:
Number of pairs of subsets that have no elements in common

A set $M$ consists of $n$ elements. Determine the number of pairs of subsets of $M$ which have no elements in common (don't forget to account for the empty set).

If we choose a subset of one element, then there are $(n-1)+1$ corresponding different subsets of the same size, hence the total is ${n\choose1}+\frac12{{n-1}\choose 1}{n\choose1}$ when we shuffle through each subset and repeat the same operation.

Then, if we choose one with two elements, then there are $1+{{n-2}\choose1}+{{n-2}\choose2}$ corresponding different subsets of the same size and of size = 2-1, which gives us a total of ${n\choose2}+{{n-2}\choose1}{{n}\choose2}+\frac12{{n-2}\choose2}{n\choose2}$.

With three, we get for the one selection ${n\choose3}+{{n-3}\choose1}{n\choose3}+{{n-3}\choose2}{n\choose3}+\frac12{{n-3}\choose3}{n\choose3}$ for the same size, size-1 and size-2.

If I am to do this for a subset with $k$ elements, then the total is $s_k={n\choose k}+\frac12{{n-k}\choose k}{n\choose k}+\sum_{i=1}^{k-1}{{n-k}\choose i}{n\choose k}$.

The big total is when I shuffle through all possible values of $k$, so $\sum_{k=1}^{n-1}s_k$.

Is there any flaw in my reasoning? Thank you for your time!

答:
The question asks for pairs of subsets that share no elements. When I read "pairs", I typically assume that means an ordered pair of sets. This would mean that the question is asking for the number of pairs of subsets $(A, B)$ where $A\cap B=\varnothing$. Your answer works if the question wants you to count the number of sets $\{A, B\}$ where $A\cap B=\varnothing$ (though you could consider the case that $A=B=\varnothing$, in which case you're off by $1$).  I'll give an answer for the ordered case, which is what I assume is intended.

One approach is to revise your argument to treat the subsets as being distinct. We let $t_k$ count the number of pairs of subsets $(A, B)$ such that $A$ and $B$ have no elements in common, and $|A|=k$. Then $$t_k={n\choose k}\sum_{i=0}^k {n-k\choose i}.$$ That is, you first choose a subset $A$ of $k$ elements, then you choose a subset $B$ from the remaining $n-k$ elements of arbitrary size. Hence, the number of such pairs can be counted by $$\sum_{k=0}^n t_k.$$

However, the nice thing about the ordered case is that there's a much easier way of counting such pairs. Each element $x$ of $M$ has three choices. Either

$x\in A$ and $x\notin B$,

$x\notin A$ and $x\in B$, or

$x\notin A$ and $x\notin B$.

For each of these $n$ elements, you choose one of these three options, so there are $3^n$ possible pairs.


答:
Note: Like you, I take pairs to mean unordered pairs. If ordered pairs are intended, the calculation is a bit simpler, both via your approach and via mine.

The explanation could be a good bit clearer, but it appears to be right. However, it results in a very complicated expression that can be greatly simplified. It’s easier, however, to adopt a different approach from the start.

We can choose disjoint subsets of $M$ by first choosing a set $C\subseteq M$ and then partitioning $C$ into two sets. There are $\binom{n}k$ ways to choose a $C\subseteq M$ of cardinality $k$. $C$ has $2^k$ subsets, and if $k>0$, these subsets come in $2^{k-1}$ complementary pairs. Thus, $C$ can be split into two disjoint subsets in $2^{k-1}$ ways if $k>0$. If $k=0$, $C=\varnothing$, which is the union of two disjoint subsets in only one way: $\varnothing=\varnothing\cup\varnothing$. Altogether, then there are

$$\begin{align*}
1+\sum_{k=1}^n\binom{n}k2^{k-1}&=1+\frac12\sum_{k=1}^n\binom{n}k2^k\\
&=1+\frac12\left(\sum_{k=0}^n\binom{n}k2^k-\binom{n}02^0\right)\\
&\overset{*}=1+\frac12\left(3^n-1\right)\\
&=\frac12\left(3^n+1\right)\,,
\end{align*}$$

where the starred step uses the binomial theorem applied to $(2+1)^n$.

Title: Number of pairs of subsets that have no elements in common

Body:
A set $M$ consists of $n$ elements. Determine the number of pairs of subsets of $M$ which have no elements in common (don't forget to account for the empty set).

If we choose a subset of one element, then there are $(n-1)+1$ corresponding different subsets of the same size, hence the total is ${n\choose1}+\frac12{{n-1}\choose 1}{n\choose1}$ when we shuffle through each subset and repeat the same operation.

Then, if we choose one with two elements, then there are $1+{{n-2}\choose1}+{{n-2}\choose2}$ corresponding different subsets of the same size and of size = 2-1, which gives us a total of ${n\choose2}+{{n-2}\choose1}{{n}\choose2}+\frac12{{n-2}\choose2}{n\choose2}$.

With three, we get for the one selection ${n\choose3}+{{n-3}\choose1}{n\choose3}+{{n-3}\choose2}{n\choose3}+\frac12{{n-3}\choose3}{n\choose3}$ for the same size, size-1 and size-2.

If I am to do this for a subset with $k$ elements, then the total is $s_k={n\choose k}+\frac12{{n-k}\choose k}{n\choose k}+\sum_{i=1}^{k-1}{{n-k}\choose i}{n\choose k}$.

The big total is when I shuffle through all possible values of $k$, so $\sum_{k=1}^{n-1}s_k$.

Is there any flaw in my reasoning? Thank you for your time!


问:
interior product equality
im having a hard time proving that for 2 differential forms $\omega\in \Omega^m(V)$ and $\eta \in \Omega^n(V)$ and a smooth vector field $v$ on $V$, that we have $j_v(\omega \wedge \eta)= j_v\omega\wedge \eta +(-1)^m\omega \wedge j_v\eta$.  This looks extremely similar to this equation $d(\omega \wedge \eta)=d\omega \wedge \eta +(-1)^m\omega \wedge d\eta$, which i already proved. Where exactly is the connection between those two, and what exactly do i need to modify in my prove for the second equality to obtain the first one? ( Also im sorry for not making it more readable, because i dont know how to make new lines). Thanks in advance !

答:
Here is one way to prove it: Let $\alpha^1,\dots,\alpha^k\in\Omega^1(V)$ be one-forms. Then we have $$\iota_v(\alpha^1\wedge\dots\wedge\alpha^k)=\sum_{i=1}^k(-1)^{i+1}\alpha^i(v)\alpha^1\wedge\dots\wedge\widehat{\alpha^i}\wedge\dots\wedge\alpha^k$$
where the hat over $\alpha^i$ means that this term is omitted. This is just the Leibniz expansion of determinants in the following formula $$(\alpha^1\wedge\dots\wedge\alpha^k)(v_1,\dots,v_k)=\det(\alpha^i(v_j))_{ij}$$
By linearity of $\iota_v$ we are reduced to show that the claim holds in the case where $\omega,\eta$ are of the form \begin{align*}
\omega&=\alpha^1\wedge\dots\wedge\alpha^m\\
\eta&=\beta^1\wedge\dots\wedge\beta^n
\end{align*}
for one-forms $\alpha^i,\beta^j$. So we get: \begin{align*}
\iota_v(\omega\wedge\eta)&=\iota_v(\alpha^1\wedge\dots\wedge\alpha^m\wedge\beta^1\wedge\dots\wedge\beta^n)\\
&=\sum_{i=1}^m(-1)^{i+1}\alpha^i(v)\alpha^1\wedge\dots\wedge\widehat{\alpha^i}\wedge\dots\wedge\alpha^m\wedge\beta^1\wedge\dots\wedge\beta^n\\&\hspace{0.7cm}+\sum_{j=1}^{n}(-1)^{j+1+m}\beta^i(v)\alpha^1\wedge\dots\alpha^n\wedge\beta ^1\wedge\dots\wedge\widehat{\beta ^j}\wedge\dots\wedge\beta ^m\\
&=\iota_v(\omega)\wedge\eta+(-1)^m\omega\wedge\iota_v(\eta)
\end{align*}

Title: interior product equality


问:
Finding matrix determinant based on Trace, Spectrum and Eigenspace
Given the Matrix $A$ $\in$ $\mathbb {R}^{4\times4}$ with the following known properties:

$|Spec(A)| = 3$

$Tr(A) = 92$

$Tr(A^2) = 4060$

$Rank(A - E)= 2$

I need to determine the determinants of this matrix. So far I'm gathering that since $|Spec(A)|$ = 3 I have 3 eigenvalues and because the determinant is the product of all the eigenvalues I have $ det(A) =\lambda_1\lambda_2\lambda_3 $

From 4. I gather that $Rank(A-E) = 2 < 4 \iff \lambda = 1$ is an eigenvalue of $A$

The relationship between determinant and trace as well as the trace of $A^2$ is not clear to me aside from one of the coefficients of the characteristic of the equation being written in the form of the trace but not for an $n$ this high.

Thank you all in advance!

答:
The fact that $\operatorname{rank}(A - E) = 2$ tells you not only that $1$ is an eigenvalue but also that its algebraic multiplicity is at least $2$. Because $A$ has $3$ distinct eigenvalues, the eigenvalues must be $1,1,\lambda_1,\lambda_2$ for some distinct $\lambda_1,\lambda_2$ not equal to $1$.

The trace of a matrix is the sum of its eigenvalues.  Thus, we have
$$
\operatorname{tr}(A) = 2 \cdot 1 + \lambda_1 + \lambda_2 = 92,\\
\operatorname{tr}(A^2) = 2 \cdot 1^2 + \lambda_1^2 + \lambda_2^2 = 4060.
$$
The rest is algebra.


Body:
Let $t:=x+1$. We solve

$$t^2y''-4ty'+6y=0,$$ knowing the solution $y_1=t^2$.

With $y_2=vy_1$,

$$t^2(v''t^2+4v't+2v)-4t(v't^2+2vt)+6vt^2=v''t^4=0$$

and

$$v(t)=at+b.$$

Hence,

$$y(t)=(cx+d)(x+1)^2.$$


问:
Solving PDE resembling the Airy equation
Hey I am trying to solve following PDE:

$\frac{\partial \Psi}{\partial t}+Ax_2\frac{\partial \Psi}{\partial x_1}=\nu (\frac{\partial^2 \Psi}{\partial x_1^2}+\frac{\partial^2 \Psi}{\partial x_2^2}+\frac{\partial^2 \Psi}{\partial x_3^2})$

I tried using the seperation of variables as $Psi(x_1,x_2,x_3,t)=F_1(x_1)F_2(x_2)F_3(x_3)T(t)$

However I do find it diffult to solve this, since in the PDE it seems the dependency between $x_1,x_2$ is very linked... MAPLE also could not really help.

Any advice or someone who could provide me with an approach to solve this PDE?

I would appreciate any kind of help.

答:
Using separation of variables, and by assuming $F_1(x_1)=e^{r_1x_1}$, $F_3(x_3)=e^{r_3x_3}$, $T(t)=e^{r_4t}$, you're left an ODE describing $F_2(x_2)$,
\begin{align}
\frac{\mathrm d^2F_2}{\mathrm dx_2^2}+\left(r_1^2+r_3^2-\frac{r_4+Ar_1y}{\nu}\right)F_2=0,
\end{align}
whose solution is some sort of Airy function.
https://www.wolframalpha.com/input/?i=f%27%27%2B%28a%2Bby%29f%28y%29%3D0


问:
calculate derivative for standard inner product
$L\colon\mathbb{R}^n\to \mathbb{R}$ $L_y(x)=\langle x,y\rangle$ for some inner product, $DL$ be the derivative of $L$. Its a Linear map so I know derivative will be itself only.
 I want to calculate explicitly the derivative for standard inner product, could any one help me?
$y=(y_1,\dots,y_n)$ and $x=(x_1,\dots,x_n)$
Then $L_y(x)=x_1y_1+\dots+x_ny_n$ using standard inner product.
Now what I have to do?

In particular $DL(u)=DL(v)\quad \forall u,v\in\mathbb{R}^n$?

答:
The function $L_y: \mathbb{R}^n \rightarrow \mathbb{R}$ is differentiable if there exists a linear transformation $\lambda: \mathbb{R}^n \rightarrow \mathbb{R}$ such that,

$$\lim_{h\to 0} \frac{|L_y(x+h) - L_y(x) - \lambda(h)|}{|h|} = 0.$$

In this case we know it's differentiable so $DL_y = \lambda$.

I won't work it out because if you substitute for $L_y$ above and expand the inner products you'll immediately see what $\lambda(h)$ must be for the limit to be $0.$

Hope this helps.


问:
How to prove that a local extrema is an Absolute extrema on an open interval?
So I know how to find the Absolute extrema of a continuous function on a Closed interval by using the Extreme Value theorem. Additionally, I know how to find the set of local extrema by using the first and second derivative tests.

Since all Absolute extrema are local, How would I prove that a local extrema is an Absolute extrema on an open interval?

For example, If I would like to find the Absolute Max of a function over it's Domain. I'd find the critical points and check how the sign changes at the critical points to determine whether a local max is attained at f(c) where is c is a critical point. If we only have one local max is must the the absolute maximum.

But what would I do if we attain multiple local maxima? Would the Absolute Maximum just be the largest value of the set of local maxima? Or is there a whole other method of finding Absolute extrema on an open interval?

Any help would be appreciated.

答:
Absolute maximum is indeed the largest value of local maxima, if it exists.

Unlike closed interval, function on open interval (or any non-compact set for this matter) doesn't necessary have an absolute maximum. Simple example is unbounded function ($\frac{1}{x}$ on $(0, 1)$), more interesting is $(1 - x) \cdot \sin \frac{1}{x}$ on $(0, 1)$ - it has values arbitrarily close to $1$, but in no point is it equal (or greater) than $1$.


问:
Given a function and compute the definite integral of its inverse
So I have a function
$$ f(x)=x^{3}-12x^{2}+69x+6 $$

The question askes me to find the relationship between $$\int_{1}^{2} f(x)dx$$ $$and$$ $$\int_{64}^{104} f^{-1}(x)dx$$

And compute the value of $\int_{64}^{104} f^{-1}(x)dx$

I found out that $f(1) = 64$ and $f(2) = 104 $

But not sure how do I proceed.

I could not find how this is related to the course material presented this week, and am completely lost to be honest. Any help would be appreciated

答:
Hint.

By Laisant's formula, if $f(a)=c$ and $f(b)=d$, then
$$
\int_{c}^df^{-1}(x)dx+\int_a^bf(x)dx=bd-ac
$$

Notes.

When $f$ is differentiable, one can prove the result by change of variables:
$$
\int_c^d f^{-1}(y)dy=\int_a^b f^{-1}(f(x))f'(x)dx=\int_a^b xf'(x)dx=xf(x)|_{a}^b-\int_a^bf(x)dx
$$

A proof without words:


问:
Implicit function theorem and derivative
The implicit function theorem (in the $2$ dimensional case): Let $F\colon D \subset \Bbb R^2  \to \Bbb R^2$ and let $(x_0, y_0)$ be an interior point of $D$ with $F(x_0, y_0) = 0$. Suppose both first order partial derivatives of $F$ exist in $D$ and are continuous at $(x_0, y_0)$ with $F_y(x_0, y_0)= 0$. Then there is an interval $I \subset \Bbb R$, with $x_0$ an interior point of $I$, and a function $φ \colon I \to \Bbb R$, such that $φ$ is differentiable on $I$, $φ(x_0) = y_0$ and $F(x, φ(x)) = 0$ for each $x \in I$.

I don't see how I can get from the chain rule that $$φ'(x)=\frac{-F_x(x,φ(x))}{F_y(x,φ(x))}.$$
I know that it should follow from the chain rule, but why?

(I stated here the theorem for the $2$ dimensional case, but of course we look at it in the $n$-dimensional case).

答:
Given that for all $x\in I$ we have $F(x,\varphi(x))=0$, we differentiate with respect to $x$:
$$0 = \frac{d}{dx}F(x,\varphi(x))=F_{x}(x,\varphi(x)) + F_{y}(x,\varphi(x))\varphi'(x).$$  Solving for $\varphi'(x)$ gives us the desired result: $$\varphi'(x) = \frac{-F_{x}(x,\varphi(x))}{F_{y}(x,\varphi(x))}.$$


问:
Continuity, constant functions
I am completely stuck and would be delighted if someone could explain an approach. Thanks!

Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function with $f (x) = f (x^2)$ for all $x \in \mathbb{R}$.
Show that $f(y^{(2^{-n})})=f(y)$ for all $n ∈ \mathbb{N} \cup\{0\}$ (the natural numbers including zero) and $y ≥ 0$ and conclude that $f$ is constant.

答:
First observe that $f(x)=f(x^2)=f(-x)$ for all $x\in\Bbb R$, because $(-x)^2=x^2$.

Let $x>0$, then $x^{\frac1{2^n}}\to x^0=1$ if $n\to\infty$, so the constant sequence $f(x)=f(x^{\frac12})=f(x^{\frac14})=\dots$ converges to $f(1)$ by continuity, so we must have $f(x)=f(1)$.

Finally, continuity at $0$ implies $f(0)=f(1)$.

Title: Continuity, constant functions


答:
By chain rule:
$$D (F\circ(\operatorname{Id},\varphi))(x)=DF(x,\varphi(x))\cdot\begin{pmatrix}1 \\ \varphi'(x)\end{pmatrix}=D_xF(x,\varphi(x))+D_y F(x,\varphi(x))\cdot\varphi'(x)$$
Also, $D(F\circ(\operatorname{Id},\varphi))(x)=0$ as $F(x,\varphi(x))=0$ for all $x\in I$, so $F\circ(\operatorname{Id},\varphi)$ is constant. Then you get the result.

Title: Implicit function theorem and derivative

Body:
The implicit function theorem (in the $2$ dimensional case): Let $F\colon D \subset \Bbb R^2  \to \Bbb R^2$ and let $(x_0, y_0)$ be an interior point of $D$ with $F(x_0, y_0) = 0$. Suppose both first order partial derivatives of $F$ exist in $D$ and are continuous at $(x_0, y_0)$ with $F_y(x_0, y_0)= 0$. Then there is an interval $I \subset \Bbb R$, with $x_0$ an interior point of $I$, and a function $φ \colon I \to \Bbb R$, such that $φ$ is differentiable on $I$, $φ(x_0) = y_0$ and $F(x, φ(x)) = 0$ for each $x \in I$.

I don't see how I can get from the chain rule that $$φ'(x)=\frac{-F_x(x,φ(x))}{F_y(x,φ(x))}.$$
I know that it should follow from the chain rule, but why?

(I stated here the theorem for the $2$ dimensional case, but of course we look at it in the $n$-dimensional case).


问:
Find two functions f,g such that $f(n)∼g(n)$ and $f(n)^n=o(g(n)^n)$
I am trying to find two functions $f(n)$ and $g(n)$ such that $f(n) ∼ g(n)$ and $f(n)^n=o(g(n)^n)$ where $f,g > 1$.

I tried it to solve it by using $\lim$ formula but unable to get $f(n)$ and $g(n)$ satisfied by both of them. Please help.

答: Take $f,g$ to be functions that go to $1$ at different speeds, e.g. $f(n)=1+\frac{1}{n}, g(n)=1+\frac{1}{\sqrt{n}}$. Then it is clear that $f\sim g$, but $f(n)^n\to e$ and $g(n)^n\to\infty$, so $f(n)^n=o(g(n)^n)$.


问:
Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross section
looking for some assistance with a problem I've been exploring with a friend for kicks.  I haven't taken uni level calculus or algebra in a couple of decades so I'm rusty!

I'm trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through  covering every cross section throughout the height of the 'cone' is constant.  Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Any takers?

答:
Such a solid of revolution cannot exist. Let the solid be generated by the rotation around $x$-axis of the curve with equation $y=f(x)$. Then its volume, from $x=0$ to $x=X$ is
$$
V(X)=\int_0^X\pi f^2(x)\,dx
$$
while the area of the cross section at $x=X$ is $A(X)=\pi f^2(X)$. If $A(X)=kV(X)$ for some constant $k$ and any $X\ge0$, then $A'(X)=kV'(X)$, that is:
$$
2\pi f(X)f'(X)=k\pi f^2(X).
$$
Hence, either $f(X)=0$ or $f'(X)=(k/2)f(X)$, that is: $f(X)=Ae^{kX/2}$.
But that solution doesn't work, because it gives:
$$
V(X)={\pi A^2\over k}\big(e^{kX}-1\big)
\quad\text{and}\quad
A(X)=\pi A^2 e^{kX}.
$$


问:
Line Integral - Independence of the way (Why these conditions)
Given the following differential form
$$ydx - x^2dy$$
It is desired to obtain the value of the line integral along the segment going from (0,0) to (1,1). In addition, it is stated that for any parameterization $x = y = f (t), 0 \leq t \leq 1, f'(t) > 0, f(0) = 0, f(1) = 1$ the value of the integral will be the same.

Well, the first part I already got, which is the value of the integral equal to 1/6.

However, in the second one, which is to verify the statements, I thought about using the independence of the path. For that, I checked if the field is conservative.

$$\frac{dP}{dy} = \frac{dQ}{dx}$$

But, equality is not satisfied. Implying that I cannot use the independence of the path to verify the claims. I'm wrong? What is the correct way to proceed. Furthermore why the need for the conditions mentioned ($x = y = f (t), 0 \leq t \leq 1, f'(t) > 0, f(0) = 0, f(1) = 1$)?

Thanks in advance!

答:
"Not depending on parametrizations (of one path)" is different from "independent of paths".

The former says that the notion of the line integral using one particular parameterization is well-defined.

The latter is a property of the underlying vector field in the integral.


问:
Interpretation of Hellinger distance
Given to discrete probability distribution $\mathbf{p}:=(p_1,p_2,\dots,p_n)$ and $\mathbf{q}:=(q_1,q_2,\dots,q_n)$, the Hellinger distance between $\mathbf{p}$ and $\mathbf{q}$ is defined as:
$$
d_H(\mathbf{p},\mathbf{q}):=\frac{1}{\sqrt{2}}\left\|\mathbf{p}^{1/2}-\mathbf{q}^{1/2}\right\|_2=\frac{1}{\sqrt{2}}\left(\sum_{i=1}^n \left(\sqrt{p_i}-\sqrt{q_i}\right)^2\right)^{1/2},
$$

Why is this distance extensively exploited in statistics and probability? What is the geometrical/statistical interpretation of this distance? Assuming that $\mathbf{p},\mathbf{q}$ represent vectors and not probability distributions, has this distance been studied in other areas different from statistics?

My questions are not technical, but I was not able to find references which clearly address them.

Thank you for your help.

答:
I guess ultimately the geometric interpretation comes from the fact that the Hellinger distance is the 2-norm distance of the square roots of the point wise probability vectors. And as such, it corresponds to a scalar product:

$$\frac{d_H(p, q)}{\sqrt{2}} =\frac{||\sqrt{p} - \sqrt{q}||_2}{\sqrt{2}} = \sqrt{1 - \langle\sqrt{p}|\sqrt{q}\rangle}$$

Some notes that are obvious but still helpfull for the interpretation:

the 1-norm resp. the TV distance does not correspond to a scalar product (through the parallelogram inequ.) but the 2-norm does

probability vectors may have a vanishing (as a fct of the dimension) 2-norm, so this is not a good norm for measuring differences

the 2-norm of the square root however is the 1-norm of the original vector. So for $p$ a probability vector $\sqrt{p}$ has 2-norm equal to 1

it can be shown that the 1-norm or the TV distance is upper and lower bounded by the Hellinger distance somewhat as $c_1 d_H^2(p,q)\leq TV(p,q) \leq c_2 d_H(p,q)$ for some constants $c_1$ and $c_2$.

So the Hellinger distance has a geometric interpretation (in terms of a vector space with scalar product and hence angles, where each point in the positiv unit sphere corresponds 1:1 to a probability vector) AND taking the TV distance as the 'natural' distance, the Hellinger distance can be used almost equivalently in the extrem cases ($d(p,q)\rightarrow 0$ or $1$)

Title: Interpretation of Hellinger distance


问:
In the case where ⊕ is involved as a composite system M1 ⊕ M2, what kind of equation does ⊕ denote?
I am reading about relations between equations using kronecker products as well as the use of ⊕ in situations such as M1 ⊕ M2, M ⊕ N, S ⊕ S, S⊕S⊕…n and others besides. In these papers they are called composite systems.

I understand that in a kronecker product a multiplication is performed on each element of the first matrix by every element in the second and it forms a block matrix of the two systems. Such as that Amxn ⊗ Bpxq is the matrix pm x nq. With situations such as a11 x b11, a11xb12, etc...

But what kind of equation is the followed out by the ⊕ symbol in the case of the above equations?

答:
You are working in the category of vector spaces over some field $k$. Let us denote that category by $C$.

If $M,N\in \text{obj}(C)$ then $M\oplus N$ denotes the direct sum of spaces and $M\otimes N$ denotes the tensor product of two spaces. These are both objects of this category.

Suppose now that you have two morphisms $f_1:M_1\to N_1$ and $f_2:M_2\to N_2$. The direct sum will give you the direct sum of morphisms $f_1\oplus f_2:(M_1\oplus M_2)\to (N_1\oplus N_2)$ and the tensor product will give a morphism $f_1\otimes f_2:(M_1\otimes M_2)\to (N_1\otimes N_2)$.

Since $f_1\oplus f_2$ and $f_1\otimes f_2$ are morphisms in the category $C$ it means they can be represented by matrix multiplication. If $A$ is the matrix that represents $f_1$ and $B$ is the matrix that represents $f_2$ then the "direct sum of matrices" is just the corresponding matrix for $f_1\oplus f_2$ which is formed by sticking the two matrices diagonally. Likewise for $f_1\otimes f_2$ you get the "Kronecker product" for the two matrices.


问:
isomorphism of two localized ring
Let $K$ be an algebraic closed field, $K[X,Y]$ be it's polynomial ring, $K[X,Y]/\langle XY\rangle=A$ be a quotient ring. Let $f=[X]\in A$, define $S=\{f^{n}|n\in\mathbb{N_{0}}\}$. Also define $T=\{X^{n}|n\in\mathbb{N_{0}}\}$, then show that $S^{-1}A$ isomorphic to $T^{-1}K[X]$.

I have problem finding the isomorphism. $S^{-1}A$ spanned by $Y^{n}X^{-m},n,m\in\mathbb{N}$ and $X^{p}, p\in\mathbb{N_{0}}$, while $T^{-1}K[X]$ spanned by $X^{q}, q\in\mathbb{Z}$. The degree of freedom isn't the same, how can it be possible to be isomorphic?

答:
You want to prove that $\{[x]^n\}^{-1}A\cong\{x^n\}^{-1}K[x]$.

Consider the natural morphism given by mapping $\frac{[f(x,y)]}{[x]^n}$ to $\frac{f(x,0)}{x^n}$, then try to prove this map is well-defined and is an isomorphism of rings. This is not difficult.

Title: isomorphism of two localized ring


答:
The connection lies in the Lie derivative.
For all vector fields $X$, the Lie derivative, as the name suggests, is an operator on forms $\mathcal L_X : \Omega^k(M)\to\Omega^k(M)$ defined as
$$\mathcal L_X(\alpha):=\frac{d}{dt}\Bigg|_{t=0}(\varphi^{(X)*}_t \alpha),$$
where $\varphi^{(X)}_t$ is the flux associated to $X$. Here, the definition is unimportant: what matters is that
$\mathcal L_X$ is a differential operator, and as such it satisfies the product rule
$$
\mathcal{L}_{X}(\omega \wedge \eta)=\left(\mathcal{L}_{X} \omega\right) \wedge \eta+\omega \wedge\left(\mathcal{L}_{X} \eta\right); \tag1
$$
also, Cartan's "magic" formula connects it with $j_X$ and $d$ in an interesting way:
$$\mathcal L_X\omega = j_X(d\omega)+d(j_X\omega). \tag2$$
This entails
$$\begin{split}
\mathcal L_X(\omega \wedge \eta)&=j_Xd(\omega\wedge\eta)+dj_X(\omega\wedge \eta) \\
(\mathcal L_X \omega)\wedge \eta + \omega\wedge(\mathcal L_X\eta)&=j_X((d\omega)\wedge \eta)+ (-1)^mj_X(\omega\wedge(d\eta)) \\
&\quad+d((j_X\omega) \wedge \eta + (-1)^m \omega \wedge (j_X\eta)) \\
0 &= (-1)^{m+1}(d\omega)\wedge(j_X\eta) + (-1)^m(j_X\omega)\wedge(d\eta)\\
&\quad+(-1)^{m}(d\omega)\wedge(j_X\eta) + (-1)^{m-1}(j_X\omega)\wedge(d\eta) \\
0&=0,
\end{split}$$
by applying the magic formula on the LHS as well, and the graded product rules you stated on the RHS (twice), thus proving their consistency (at least if you trust $(1)$ and $(2)$!).

Title: interior product equality

Body: im having a hard time proving that for 2 differential forms $\omega\in \Omega^m(V)$ and $\eta \in \Omega^n(V)$ and a smooth vector field $v$ on $V$, that we have $j_v(\omega \wedge \eta)= j_v\omega\wedge \eta +(-1)^m\omega \wedge j_v\eta$.  This looks extremely similar to this equation $d(\omega \wedge \eta)=d\omega \wedge \eta +(-1)^m\omega \wedge d\eta$, which i already proved. Where exactly is the connection between those two, and what exactly do i need to modify in my prove for the second equality to obtain the first one? ( Also im sorry for not making it more readable, because i dont know how to make new lines). Thanks in advance !


答:
Number of ordered pairs: since each element is either in subset $A$, or subset $B$, or neither, it's $3^n$.

Number of unordered pairs: if $A \neq B$, the pair is counted twice in the ordered case. Only when they are both empty are they the same, which was counted once. Therefore the total number of unordered pairs is $1+(3^n-1)/2=(3^n+1)/2$.


问:
There are 7 people (two of which are John and Andy) in a group. The group must consist of 4, 5, or 6 people.
There are 7 people (two of which are John and Andy) in a group. The group must consist of 4, 5, or 6 people. Find the probability that John is in the group if the probability for each group size is the same.

So I have found the probability for each group size, for 4-people it is 4/7, for 5-people it is 5/7 and for 6 people it is 6/7.

My question is, do I multiply the probabilities here, or what should my next step be?

答:
You say "the probability for each group size is the same" so, since there are three possible group sizes, this may be $\frac13$ for each size

That makes the overall probability $$\frac13 \times \frac47 + \frac13 \times \frac57 + \frac13 \times \frac67 =\frac57$$


问:
If $G$ is a finite group such that every maximal subgroup is nilpotent, are any two maximal subgroups of $G$ conjugate?
I’m trying to prove that if every maximal subgroup of a finite group $G$ is nilpotent, then $G$ is solvable. I know already that if any two maximal subgroups of a finite group are conjugate, then it is cyclic, so it is solvable. So my question is:

If $G$ is a finite group such that every maximal subgroup is nilpotent, are any two maximal subgroups of $G$ conjugate?

I’m using the following definition of “nilpotent group”: $G$ has a lower central series terminating in the trivial subgroup after finitely many steps.

答:
As a counterexample, let $G$ be any non-cyclic finite abelian group.

Since $G$ is abelian, every maximal subgroup of $G$ is abelian, hence nilpotent.

Since $G$ is a non-cyclic finite abelian group, $G$ has more than one maximal subgroup, no two of which are conjugate (in an abelian group, no two distinct subgroups are conjugate to each other).


问:
$U\subseteq V$ is $T$ invartiant $\Rightarrow$ $\left(T\,|_{U}\right)^{*}=\left(P \circ T^{*}\right)\bigl|_{U}$
I want to proove that given $T\in\mathcal{L}\left(V,V\right)$ ($V$ is a finite dimensional inner product space)  and a subspace $U\subseteq V$ which is $T$ invariant that

$\left(T\,|_{U}\right)^{*}=\left(P\circ T^{*}\right)\bigl|_{U}$

Where $P$ is the projection on $U$.

My attempt:
I've tried working with the inner product but  but with no luck because I'm unsure how can I "defuse" all the $^*$. I've all so thought about finding some orthonormal basis of $U$ $\mathcal{B}=\left\{ f_{1},\dots,f_{k}\right\}$ but I didn't know how to express those linear transforms with the basis.

I do understand intuitively why restricting $T^{*}$ is the same as using $T^*$ first and then just projection the result to $U$ but I'm struggling writhing a correct rigorous proof.

I'd be happy to provide any additional information.

答:
For $u, v \in U$ we have (denoting by $(,)$ the inner product on $V$ and by $(,)_U$ its restriction to $U$):
\begin{align*}
  (T|_Uu, v)_U &= (Tu, v)\\
               &= (u, T^*v)\\
               &= (u, PT^*v) + (u, (I - P)T^*v)\\
               &= (u, PT^*v)\\
               &= (u, (PT^*)|_Uv)_U
\end{align*}
So $(T|_U)^* = (PT^*)|_U$. In the forth step we used that$\mathrm{ran}(I-P) = U^\bot$ (i. e.  the range of $I-P$ is the orthogonal complement of $U$, as $I-P$ is the projection onto $U^\bot$), hence $(u, (I-P)x) = 0$ for all $x \in V$.</p>


答:
I would recommend that you not work in coordinates like this. Think of the space of $3\times 3$ matrices as a vector space $V$ and compute derivatives from the definition. In particular, since $f(A) = AA^\top$ is smooth, we know that for any $B\in V$, we have
\begin{align*}
df_A(B) &= \lim_{t\to 0}\frac{f(A+tB)-f(A)}t = \lim_{t\to 0}\frac{t(AB^\top + BA^\top +tBB^\top)}t \\ &= AB^\top + BA^\top.
\end{align*}
Now, you already noticed that $f$ maps $V$ to the vector space $W$ of symmetric $3\times 3$ matrices. We want to show that $df_A\colon V\to W$ is surjective for any $A\in O(n)$. So, for an arbitrary $C\in W$, your charge is to figure out $B\in V$ so that $df_A(B)=C$.

Title: Prove $SO(3)$ is a smooth regular surface in $\mathbb R^{3\times 3}$

Body:
I proved this for $O(3)$. To do that, I defined $O(3)$ implicitly by $f(A) = AA^t - I = 0$. Identifying $\mathbb R^{3\times 3}$ with $\mathbb R^9$, I got $$f (a_{11}, ..., a_{33}) = \cases {
a_{11}^2 + a_{12}^2 + a_{12}^2 - 1\\\\
a_{21}^2 + a_{22}^2 + a_{23}^2 - 1\\\\
a_{31}^2 + a_{32}^2 + a_{33}^2 - 1\\\\
a_{11}a_{21} + a_{12}a_{22} + a_{13}a_{23}\\\\
a_{11}a_{31} + a_{12}a_{32} + a_{13}a_{33}\\\\
a_{21}a_{31} + a_{22}a_{32} + a_{23}a_{33},}$$ and then showed that rank of $(\partial{f_i}/\partial{x^j})$ is maximal on $O(3)$. I tried to do it similarly for $SO(3)$. In this case $f(A) = \det A(AA^t) - I= 0$ should work, but then $(\partial{f_i}/\partial{x^j})$ seems to look like
$$\small\begin{pmatrix}
d_{11} + 2a_{11} & -d_{12} + 2a_{12} & d_{13} + 2a_{13} & -d_{21} & d_{22} & -d_{23} & d_{31} & -d_{32} & d_{33}\\
d_{11} & -d_{12} & d_{13} & -d_{21} + 2a_{21} & d_{22} + 2a_{22} & -d_{23} + 2a_{23} & d_{31} & -d_{32} & d_{33}\\
d_{11} & -d_{12} & d_{13} & -d_{21} & d_{22} & -d_{23} & d_{31} + 2a_{31} & -d_{32} + 2a_{32} & d_{33} + 2a_{33}\\
a_{21} & a_{22} & a_{23} & a_{11} & a_{12} & a_{13} & 0 & 0 & 0\\
a_{31} & a_{32} & a_{33}  & 0 & 0 & 0 & a_{11} & a_{12} & a_{13}\\
0 & 0 & 0 & a_{31} & a_{32} & a_{33} & a_{21} & a_{22} & a_{23}\\
\end{pmatrix},$$
where $d_{ij}$ are corresponding minors of $A$. I'm not sure how to show its rank is maximal. Can someone help with this, or should I use some other approach?


答:
EDIT2:

(After OP comments on 26 Jan 2021)

Let $X=Z = r,Y=y $ symbol $r$  used for cross section radius.Surface area and volume of revolution are found. The ratio is a constant $c/2$, say

$$ \dfrac{\int\pi r^2 dy}{\pi r^2}= c/2  \tag1 $$

Cancel $\pi$ and differentiate with respect to $y$ using Chain Rule

$$ \dfrac{r^2}{2r \frac{dr}{dy}} = c/2$$

$$ \dfrac{dy}{c}=\dfrac{dr}{r} \to \dfrac{dr}{dy}= \dfrac{r}{c} \tag2 $$

Integrate with BC $y=0, r=a\; $

$$ \dfrac{y}{c}=\log \dfrac{r}{a}  \tag 3$$

Or

$$ r=a \;e^{y/c} \tag 4 $$

So it is an exponential curve rotated around $Y$ axis.

To verify the obtained result plug (2) into (1) and simplify

$$\dfrac{\text{Volume}}{\text{Cross-Section Area}}=\dfrac{\pi a^2 \int e^{2y/c} dy} {\pi(a^2 e^{y/c})^2} =\dfrac{ e^{2y/c} \cdot c/2} {(e^{y/c})^2}= \dfrac{c}{2} \tag5 $$

which is a constant. So the profile obtained by rotating around y-axis is an exponential curve. It is not a paraboloid of revolution, nor a cone. The ice-cream cup has bottom  diameter $2a$ with an upward flare, tallies OK. A rough sketch:

EDIT3:

( EDIT1: First post before 26 Jan 2021)

To me looks that you mean  just a surface of revolution not any parabolic cone.

If a surface of revolution between two z- limits is considered, a single area for comparison.. am not able to figure out what it is. So I assumed that you refer to its  swept out surface area. With z-axis as symmetry axis, volume/ area ratio is

$$ \dfrac{\pi\int r^{2} dz} {2 \pi\int r \sqrt{1+r^{'2}} dz} =\text{constant}$$

where $r'=\tan \phi,$ the slope. Applying Quotient Rule reduces it to

$$ r\cdot \cos \phi = c$$

which on integration yields as solution the catenoid of revolution.

With initial condition $z=0, \phi=0$,

$$ r= c\;\cosh(z/c)$$

Title: Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross section

Body:
looking for some assistance with a problem I've been exploring with a friend for kicks.  I haven't taken uni level calculus or algebra in a couple of decades so I'm rusty!

I'm trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through  covering every cross section throughout the height of the 'cone' is constant.  Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Any takers?


问:
Definition of inner product
I'm studying the inner product part in linear algebra and it's a bit tricky to understand what inner product really means. It indicates the length of vectors? So if we calculate $<x,y>$ we have to make it as a scalar?

Here is an example. If $x$ and $y$ are column vector of $2*2$ matrix and $F=R$ then we calculate $\langle x,y\rangle=x^T y$. I wonder it is equivalent to $x^T y=y^T x$ (because it results in the same value)? Seriously, there is no difference?

Also, the example says if $x$ and $y$ are  column vector of $2*2$ matrix and $F=R$ then $\langle x,y\rangle=y^* x$ where $*$ means conjugate transpose. I can't understand why...

答:
The inner (or dot) product of two real-valued vectors is defined by
$$\sum_ix_iy_i$$
where $x_i$ and $y_i$ are the components of the two vectors. This is why $x^Ty=y^Tx$. If the vectors are complex-valued the inner product is defined by
$$\sum_ix_iy_i^{*}$$
This results in $x^{*}x=||x||^2$ (real-valued norm), which is what you want, also for complex vectors.</p>

Title: Definition of inner product


问:
Finding a specific member of an ascending list of whole number powers sequence.
I've been tasked with writing a program to find a $a_k$ member of a sequence:

$
a_0=4=2^2, a_1=8=2^3, a_2=9=3^2, a_3=16=4^2=2^4, a_4=25=5^2, a_5=27=3^3, a_6=32=2^5
$

and so on. I've coded up a brute-force solution, which takes advantage of $m^n$s logarithmic contours for $m>2$ and $n>2$:

The checking for each following member goes right-to-left through the table below, checking max $m$ for every $n$ and filtering out the repeated values (i.e. $64=8^2=4^3=2^6$). However, the method is too slow for me and I was wondering if there was a way to get to that $a_k$ element ($1<=k<=10^6$) by expressing this sequence as a formula? Thanks in advance.

答:
This problem seems difficult, see the information provided in OEIS. However, brute force approaches give some results.

One may for instance list all numbers satisfying your conditions, until a given maximal value, and then sort them. The maximal value must be large enough to reach $k$ values (it seems from experiments and OEIS that it should be close tp $k^2$).

The following Python code makes the job, if I am correct:

```<code>sqrt_max_val = 10**6
max_val = sqrt_max_val**2
values = set()

for i in range(2,sqrt_max_val):
        val = i*i
        while val<max_val:
                values.add(val)
                val *= i

print("nb values: "+str(len(values)))
for x in sorted(values):
        print(x)
</code>```

Notice that it uses a set to avoid duplicates, and that a full answer would have to find sqrt_max_val automatically (by increasing it and adapting the range, typically). Many optimizations are still possible.

Body:
This problem seems difficult, see the information provided in OEIS. However, brute force approaches give some results.

One may for instance list all numbers satisfying your conditions, until a given maximal value, and then sort them. The maximal value must be large enough to reach $k$ values (it seems from experiments and OEIS that it should be close tp $k^2$).

The following Python code makes the job, if I am correct:

sqrt_max_val = 10**6
max_val = sqrt_max_val**2
values = set()

for i in range(2,sqrt_max_val):
        val = i*i
        while val<max_val:
                values.add(val)
                val *= i

print("nb values: "+str(len(values)))
for x in sorted(values):
        print(x)

Notice that it uses a set to avoid duplicates, and that a full answer would have to find sqrt_max_val automatically (by increasing it and adapting the range, typically). Many optimizations are still possible.


问:
Determine whether or not the topological spaces are homeomorphic or not
• $(X, τ_X)$ where $X = \{(x, y) ∈ \mathbb{R}^2: x^2 +y^2 = 1\}$ with the topology $τ_X$ induced by
the Euclidean metric on $\mathbb{R}^2$
;

• $(Y, τ_Y )$ where $Y = \{(x, y) ∈ \mathbb{R}^2 : x^2 −y^2 = 1\}$ with the topology $τ_Y$ induced by
the Euclidean metric on $\mathbb{R}^2$
.

答: Hint: Look at connectedness. Two topological spaces that are homeomorphic are either both connected or both disconnected.


问:
Defining a real number
No answers, please, hints only.
I want to express every nonzero $x\in \mathbb R$ as a product of two numbers that are not rational.

My attempt is $x=(a_1+b_1i)(a_2+b_2i)$ where $a_1, b_2\in \mathbb R$

Am I correct or do I need a hint?

答:
Hints:

Why is addressing the case with positive $x\ne1$ the hard part?

For such $x$, consider rational and irrational $x$ separately, and think about how $n$th roots would help.

Title: Defining a real number


答:
Hint. Play with examples.

$2=\sqrt{2}\cdot \sqrt{2}$

$-4=\frac{-4}{\sqrt{3}}\cdot \sqrt{3}$.

$\frac{7}{5}=\sqrt{\frac{7}{5}}\cdot\sqrt{\frac{7}{5}}$

$\sqrt{2}=\frac{\sqrt{2}}{2}\cdot 2\sqrt{2}$

etc.

Now generalize...

Title: Defining a real number

Body:
No answers, please, hints only.
I want to express every nonzero $x\in \mathbb R$ as a product of two numbers that are not rational.

My attempt is $x=(a_1+b_1i)(a_2+b_2i)$ where $a_1, b_2\in \mathbb R$

Am I correct or do I need a hint?


问:
Showing that $f(U) \subset W$ is a subspace and that $f^{-1}(X)$ is a subspace

Let $V$ and $W$ be vector spaces and $U \subset V$ and $X \subset W$ be subspaces. Let $f: V\to W$ be a linear transformation. Show that $f(U) \subset W$ is a subspace and that $f^{-1}(X)$ is a subspace.

In order to show that $f(U)$ is a subspace I have to show that the zero vector is in $f(U)$ and that it's closed under addition and scalar multiplication or that any linear combination of two elements in $f(U)$ is also in $f(U)$?

Let $w_1,w_2 \in f(U)$ and $c_1,c_2 \in \mathbb{R}$. Now since $w_1,w_2 \in f(U)$ there exists $u_1$ and $u_2$ $\in U$ such that $f(u_1)=w_1$ and $f(u_2) = w_2$. Considering $c_1w_1+c_2w_2$ I have that $$c_1w_1+c_2w_2=c_1f(u_1)+c_2f(u_2)= f(c_1u_1+c_2u_2) \in f(U)$$

and so $f(U)$ would be a subspace?

For the second part I couldn't came up with much... how should I get started?

答:
The first part of your proof is absolutely correct. The second part is quite similar. Let $a,b\in f^{-1}(X),$ which is to say that $f(a)\in X$ and $f(b) \in X$. Consider $c_1 a + c_2 b$. Is $f(c_1 a + c_2 b)\in X$? Well, this is just $c_1 f(a) +c_2 f(b)$, which is a linear combination of two elements of $X$. Since $X$ is a subspace, it is true that $c_1 f(a) +c_2 f(b)\in X$, which shows that $f^{-1} (X)$ is closed under linear combinations.
You are right that for both parts, you also need to argue that $0$ is part of the subset in question. This part shouldn't be too hard.


答: Since the probability for each of the group sizes is equal, we can take the average of the probabilities 4/7, 5/7, and 6/7 to get a probability of 5/7.</p>

Body:
There are 7 people (two of which are John and Andy) in a group. The group must consist of 4, 5, or 6 people. Find the probability that John is in the group if the probability for each group size is the same.

So I have found the probability for each group size, for 4-people it is 4/7, for 5-people it is 5/7 and for 6 people it is 6/7.

My question is, do I multiply the probabilities here, or what should my next step be?


问:
How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$

The answer is
 $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

Please help me out; thanks!

答:
Given

$$A: x^2 + x - 2$$

you're trying to do the 'magic' in your head in order to get backwards to

$$B: (x+2)(x-1)$$

What is it that you are trying to do backwards. It's the original multiplication of $(x+2)(x-1)$. Note that

the -2 in $A$ comes from multiplying the +2 and -1 in $B$

the +1 (it's kind of invisible it's the coefficient of $x$ ) in  $B$ comes from:

$x$ in the first part times -1 in the second, plus

+2 in the first part times $x$ in the second

or $(-1)+2 = +1$.

So that's how the multiplication works going forward. Now you have to think of that to go backwards. In $x^2 + x - 2$:

where does the -2 come from? From two things that multiply to get -2. What could those possibly be? Usually we assume integers so the only possibilities are the two pairs 2, -1, and -2, 1.

of those two pairs, they have to -add- to the coefficient for $x$ or just plain positive 1. So the answer has to be the pair 2 and -1.

Another example might help: given

$$x^2-5x+6$$

what does this factor to? (that is, find $(x-a)(x-b)$ which equals $x^2 -5x + 6$).

So the steps are:

what are the factors of 6? (you should get 2 pairs, all negative.

for those pairs, which pair adds up to -5?

The main difficulty is keeping track in your head of what is multiplying, what is adding, and what is positive and negative.

The pattern for any sort of problem solving skill like this that seems like magic (but really is not) is to:

Do more examples to get a speedier feel for it.

Check your work. Since you're going backwards, once you get a possible answer, you can do the non-magic (multiplying) to see if you can get the original item in the question.

Title: How to factor quadratic $ax^2+bx+c$?


问:
initial value problem: y'' + 4y = f(t) , y(0)= y'(0)=0. f(t) = { 0 if t <3; t if t >3}
Solve the initial value problem:

$$y'' + 4y = f(t) , y(0)= y'(0)=0. $$
where
$$ f(t) = \begin{cases} 0 &t < 3 \\ t & t > 3\end{cases} $$

I've solved for the homogeneous equation, $y'' + 4y = 0,$ to get the general solution,
$y = A\cos(2t) + B \sin(2t)$ for some arbitrary constants $A$ and $B$.

But, now where do I go next?

Thanks!

答:
Again, as referenced above, I would Laplace transform.  The LT of the RHS is not hard:

$$\int_0^{\infty} dt \, f(t) e^{-s t} = \int_3^{\infty} dt \, t\, e^{-s t} = \frac{(3 s+1) e^{-3 s}}{s^2}$$

The equation for the LT $Y(s)$ is then

$$Y(s) = \frac{(3 s+1) e^{-3 s}}{s^2 (s^2+4)}$$

The ILT $y(t)$ may be found by using the theory of residues on

$$y(t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+ i\infty} ds \frac{(3 s+1)}{s^2 (s^2+4)} e^{s (t-3)}$$

That is, for $t>3$, we close the loop to the left and sum the residues of the poles at $s=0$ and $s=\pm2 i$.  For $t<3$, we close the loop to the right and enclose no poles, so the ILT is zero there.

I will leave the details of evaluating the residues to the reader; I get

$$f(t) = \left [ \frac{t}{4} - \frac{3}{4} \cos{[2 (t-3)]} - \frac18 \sin{[2(t-3)]}\right] \theta(t-3)$$

where $\theta$ is the Heaviside step function.


问:
Show that if $A\neq B$ then $C^A\cap C^B=\emptyset$.
Definition

A set $R$ is a (binary) relation if any its element is a ordered pair, that is $z\in R$ if and only if there exist $x$ and $y$ such that $z=(x,y)$. In particular if $R$ is contained in the cartesian product of two set $A$ and $B$ we say that $R$ is a relation of between $A$ and $B$ and if $B$ is equal to $A$ we say that $R$ is a relation in $A$.

Definition

If $R$ is a relation we call domain of $R$ that set whose elements are a first coordinate of some pair of the relation, that is
$$
\text{dom}\,R:=\{x:\exists\,y\,\text{such that}\,(x,y)\in R\}
$$
and analogously the range of $R$ is that set whose elements are the second coordinate of the relation, that is
$$
\text{rank}\,R:=\{y:\exists\,x\,\text{such that}\,(x,y)\in R\}
$$

So it is possible to prove that the above two defined set exist using the $ZFC$'s formalism but this now has not matter so we proceed to give the following well know definition.

Definition

A function $f$ is a relation such that if $(x,y)$ and $(x,z)$ are such that $(x,y),(x,z)\in f$ then $y=z$.

Definition

Let $A$ and $B$ sets. So the set whose element are functions from $A$ to $B$ is denoted by the symbol $B^A$.

Again it is possible to prove that the above defined set exist using $ZFC$'s formalism but now this again has not matter.

So with the previous definitino I ask to prove that when is true that if $A$ and $B$ are two different set then the set $C^A$ and $C^B$ for any other set $C$ are disjoint. So could someone help me, please?

答:
Prove by contradiction. Suppose the contrary that $C^{A}\cap C^{B}\neq\emptyset$.
Choose $f\in C^{A}\cap C^{B}.$ Since $A\neq B$, there exists $x$
such that $x\in A\setminus B$ or $x\in B\setminus A$. Case 1: $x\in A\setminus B$.
Note that $f:A\rightarrow C$ is a function and $x\in A$, so there exists (uniquely)
$y\in C$ such that $(x,y)\in f$. On the other hand, that $f:B\rightarrow C$
is a function and $(x,y)\in f$ implies that $x\in\mbox{dom}(f)=B$.
This is a contradiction.

Case 2: $x\in B\setminus A$ can be treated similarly.


答:
If $f\in C^A\cap C^B$ then
$$ A=\operatorname{dom} f = B $$

Title: Show that if $A\neq B$ then $C^A\cap C^B=\emptyset$.

Body:
Definition

A set $R$ is a (binary) relation if any its element is a ordered pair, that is $z\in R$ if and only if there exist $x$ and $y$ such that $z=(x,y)$. In particular if $R$ is contained in the cartesian product of two set $A$ and $B$ we say that $R$ is a relation of between $A$ and $B$ and if $B$ is equal to $A$ we say that $R$ is a relation in $A$.

Definition

If $R$ is a relation we call domain of $R$ that set whose elements are a first coordinate of some pair of the relation, that is
$$
\text{dom}\,R:=\{x:\exists\,y\,\text{such that}\,(x,y)\in R\}
$$
and analogously the range of $R$ is that set whose elements are the second coordinate of the relation, that is
$$
\text{rank}\,R:=\{y:\exists\,x\,\text{such that}\,(x,y)\in R\}
$$

So it is possible to prove that the above two defined set exist using the $ZFC$'s formalism but this now has not matter so we proceed to give the following well know definition.

Definition

A function $f$ is a relation such that if $(x,y)$ and $(x,z)$ are such that $(x,y),(x,z)\in f$ then $y=z$.

Definition

Let $A$ and $B$ sets. So the set whose element are functions from $A$ to $B$ is denoted by the symbol $B^A$.

Again it is possible to prove that the above defined set exist using $ZFC$'s formalism but now this again has not matter.

So with the previous definitino I ask to prove that when is true that if $A$ and $B$ are two different set then the set $C^A$ and $C^B$ for any other set $C$ are disjoint. So could someone help me, please?


答:
If $\frac{x}{\sqrt 2}$ is irrational we are done;

If $\frac{x}{\sqrt 2}$ is rational. Let $x = \sqrt 2 \cdot \frac mn$ where $\gcd(m,n)=1$, and $$y = \frac{\sqrt 2}{\sqrt{4n^2+4n-1}}, z = \frac{m}{n}\cdot \sqrt{4n^2+4n-1} $$ Then $x=yz$ where $y$, $z$ are both irrational. In fact $4n^2+4n-1$ can be replaced with any non-square integer coprime to both $2$ and $n$, for example any prime number greater than both $n$ and $2$.

Title: Defining a real number


答:
How many ways are there to express $x=ab$ as product of two reals? How many of these ways have rational $a$? How many have rational $b$?

More constructively: Can you achieve $a=\sqrt 2$? Can you achieve $a=\sqrt 3$? Could it happen that $b$ is rational in both cases?

Title: Defining a real number


问:
$2 \int_a^b [f(x) \int_x^b f(y) dy] dx= [ \int_a^b f(x) dx] ^2$?
A hint on proving $2 \int_a^b [f(x)  \int_x^b f(y) dy] dx= [ \int_a^b f(x) dx] ^2$ would be much appreciated.

PS I couldn't made any progress so to write my attempt...

答: Write $F(x)$ for an anti-derivative of $f(x)$. By the fundamental theorem, the left hand side becomes $2\int_a^b f(x) (F(b)-F(x))\thinspace dy$. You can expand this to be $2\left(F(b)\int_a^bf(x)\thinspace dx -\int_a^b F(x)f(x)\thinspace dx \right)$. The remaining trick is to recognize that an antiderivative for $F(x)f(x)$ is $\frac{1}{2}F(x)^2$. Hope this helps.


答:
The map $f\colon \mathbb{R}\to X$, $f(t)=(\cos t,\sin t)$ is continuous and surjective, so $X$ is …

Consider $Y=(Y\cap Z_+)\cup(Y\cap Z_-)$, where
$$
Z_+=\{(x,y)\in\mathbb{R}^2:x>0\},
\quad
Z_-=\{(x,y)\in\mathbb{R}^2:x<0\}
$$
so $Y$ is not …

Title: Determine whether or not the topological spaces are homeomorphic or not

Body:
• $(X, τ_X)$ where $X = \{(x, y) ∈ \mathbb{R}^2: x^2 +y^2 = 1\}$ with the topology $τ_X$ induced by
the Euclidean metric on $\mathbb{R}^2$
;

• $(Y, τ_Y )$ where $Y = \{(x, y) ∈ \mathbb{R}^2 : x^2 −y^2 = 1\}$ with the topology $τ_Y$ induced by
the Euclidean metric on $\mathbb{R}^2$
.


答:
Similar to the other answer but lets say $F'(x)=f(x)$ then:
$$2\int_a^bf(x)\int_x^bf(y)dydx=2\int_a^bf(x)[F(b)-F(x)]dx$$
If we break this up:
$$\int_a^bf(x)F(b)dx=F(b)[F(b)-F(a)]$$
now notice for the second integral we have something of the form $uu'$ so:
$$\int_a^bf(x)F(x)dx=\frac{F^2(b)-F^2(a)}{2}$$
now combine the two:
$$F(b)[F(b)-F(a)]-\frac{F^2(b)-F^2(a)}{2}=\frac{F^2(b)+F(a)[F(a)-2F(b)]}{2}$$
and multiplying by two gives us:
$$F^2(b)+F(a)[F(a)-2F(b)]=F^2(a)-2F(a)F(b)+F^2(a)$$
now lets have a look at the second part:
$$\left(\int_a^bf(x)dx\right)^2=\left(F(b)-F(a)\right)^2=F^2(b)-2F(a)F(b)+F^2(a)$$
so its true

Title: $2 \int_a^b [f(x) \int_x^b f(y) dy] dx= [ \int_a^b f(x) dx] ^2$?

Body:
A hint on proving $2 \int_a^b [f(x)  \int_x^b f(y) dy] dx= [ \int_a^b f(x) dx] ^2$ would be much appreciated.

PS I couldn't made any progress so to write my attempt...


问:
Dual of axiom of choice
The (internal version of) axiom of choice, stated in the language of category theory says that every epimorphism splits - i.e., every epimorphism has a section.

Now, just like any other category theoretic statement, it has a dual statement which can be produced by inverting all of the arrows. Thus, dual of axiom of choice, we which we dub axiom of co-choice in this question, states that every monomorphism is split -- i.e. has a retraction.

Now, without any assumptions beyond ZF, this holds in the topos Set. However, like axiom of choice, it fails in Top. Off the top of my head, I don't know it failing in any other relevant category.

However, despite looking for some more information on it, there seems to be no literature studying axiom of co-choice. So, my question are these:

Is the axiom of co-choice not mathematically interesting? Is there any philosophical or mathematical justification for the dual of axiom of choice - a very interesting problem on its own - being boring? If it is truly a boring or trivial result, what classes of categories do satisfy it - i.e. what are some statements or requirements which are equivalent to it?

答:
"Every epimorphism splits" and "every monomorphism splits" turn out to be equivalent for abelian categories: both conditions are equivalent to the condition that every short exact sequence splits. You might call such an abelian category semisimple, although the discussion at this MO question suggests "split" instead. It is equivalent to the condition that $\text{Ext}^n = 0$ (as a functor) for $n \ge 1$. For example, the category $\text{Mod}(R)$ of $R$-modules satisfies this condition iff $R$ is semisimple.

So it is an interesting and natural condition in this setting, but 1) it's equivalent to its opposite and 2) nobody would call it "the axiom of cochoice" here.

It's also an interesting question in various settings to ask whether specific epimorphisms or monomorphisms split. In $\text{Top}$, for example, "does this epimorphism split?" specializes to asking whether a bundle has a section, and in $\text{CRing}$, "does this monomorphism split?" specializes to asking when a Diophantine equation has a solution. I wrote a blog post Topological Diophantine equations which discusses the analogy between these two cases in some more detail.

As another example, in abelian categories an object $P$ is projective iff every epimorphism with target $P$ splits, and dually an object $I$ is injective iff every monomorphism with source $I$ splits.

Title: Dual of axiom of choice


问:
How to solve $a^2 - 2b^2 - 3 c^2 + 6d^2 = 1$ over $\mathbb{Z}$?
I would like to find the integral elements int the quaternion algebra $D_{2,3}(\mathbb{Q})$.  These are $2 \times 2$ matrices with elements in $\mathbb{Z}[\sqrt{2}]$ with determinant $1$.  Observe these matrices are closed under multiplication.

$$ \det \left| \begin{array}{cc} x_1 + \sqrt{2} x_2 & x_3 + \sqrt{2} x_4 \\
3(x_3 - \sqrt{2} x_4) & x_1 - \sqrt{2} x_2\end{array}  \right| = x_1^2 - 2x_2^2 - 3x_3^2 + 6 x_4^2 = 1 $$
Books as recent as 2003, such on John Conway's On Quaternions and Octonions they collect fact about integral quaternions.

The usual mistake here is that I could have written an equation with no solutions, though I suspect that's unlikely.  This is similar to Pell's equation and we're looking for integer points on a conic section, a hyperboloid.

By exhaustive search (a standard problem-solving heuristic) I was able to show that:

$$ \det \left| \begin{array}{cc} 4 + 3 \sqrt{2} & 7 + 5 \sqrt{2}
 \\
3(7-5\sqrt{2}) & 4 - 3 \sqrt{2}
\end{array}\right| = 4^2 - 2 \times 3^2 - 3 \times 7^2 + 6 \times 5^2 = 1$$
These number theory problems are notorious for their brevity. In fact, they will get arbitrarily complex.  We also get an element of $D_{2,3}(\mathbb{Z})$.

There are books or papers on integer solutions to quadratic forms.  Cassels' book proves outlines and theory of qudratic forms over $\mathbb{Q}$.  For example,  there's no complete list and even writing down generators of this group could be complicated.

答:
You are looking for units of the order $\Lambda$ formed by the matrices, denote them $M(x_1,x_2,x_3,x_4)$ with $x_i,i=1,2,3,4,$ ranging over $\Bbb{Z}$. Obviously Conway can say more about this than I can dream of ever learning. I just wanted to make a comment explaining, why I think this is difficult. But that comment is way too long, so an answer it is.

I once had a need to find the units of a maximal order of another division algebra, but a simple computer search proved that the structure of the group is more complicated than in the case of maximal orders of number fields, and I gave up on the idea. A colleague searched for more information. A number of general facts are known, but IIRC in general the answer is rather messy.
Furthermore, I have not checked whether $\Lambda$ is a maximal order of $D$. That may not be relevant you (and may not have too much of an impact on the answer), so let's ignore that question.

The main comment I want to make is that there are copies of orders of infinitely many quadratic number fields $K, [K:\Bbb{Q}]=2$ inside $\Lambda$. Consider the following.
Any matrix $A=M(x_1,x_2,x_3,x_4)$ satisfies its characteristic polynomial $\chi_A(T)$. The coefficients of that polynomial are the reduced trace and norm of the element $A$ respectively. Because we are in an order we can conclude that $\chi_A(T)\in\Bbb{Z}[T]$.
Furthermore, unless $A$ is in the center (= $\Bbb{Q}$), $\chi_A(T)$ cannot have rational roots. For if $\chi_A(q)=0$, $q\in\Bbb{Q}$, then $qI_2-A$ would be non-invertible contradicting the fact we have a division algebra.

It follows that the set
$$
\Bbb{Q}(A)=\{q_1+q_2A\mid q_1,q_2\in \Bbb{Q}\}
$$
is a splitting field of $\chi_A(T)$, call it $K_A$. That is, it is isomorphic to a quadratic number field. Furthermore, the set
$$\Bbb{Z}[A]=\{n_1+n_2A\mid n_1,n_2\in\Bbb{Z}\}$$
is an order of $K_A$.

By basic facts about quadratic number fields we know that whenever $K_A$ is real (or, equivalently, the discriminant of $\chi_A(T)$ is positive), the order $\Bbb{Z}[A]$
has infinitely many units.

A few examples:
$$
A=M(0,1,1,1)=\left(
\begin{array}{cc}
 \sqrt{2} & 1+\sqrt{2} \\
 3-3 \sqrt{2} & -\sqrt{2} \\
\end{array}
\right)$$
has $\chi_A(T)=T^2+1$, implying that $A$ is itself a unit of order four and $K_A\simeq \Bbb{Q}(i)$. Observe that
$$
-2\cdot1^2-3\cdot1^2+6\cdot1^2=1.
$$

If, instead, we look at the element
$$
A=M(1,1,2,0)=\left(
\begin{array}{cc}
 1+\sqrt{2} & 2 \\
 6 & 1-\sqrt{2} \\
\end{array}
\right)
$$
we get $\chi_A(T)=T^2-2T-13$. The zeros of that polynomial are $1\pm\sqrt{14}$ implying that
$\Bbb{Z}[A]\simeq \Bbb{Z}[\sqrt{14}]$. In that ring $15+4\sqrt{14}$ is a unit (solve a Pell equation to find it). The isomorphism has $A=1+\sqrt{14}$, so
$$
15+4\sqrt{14}=11I_2+4A=M(15,4,8,0)
$$
should also be a unit.
Indeed,
$$
15^2-2\cdot 4^2-3\cdot8^2=1.
$$

Further facts/observations/complications:

There are several quadratic number fields inside your division algebra. By the so called double centralizer theorem each $K_A$ is its own centralizer in $D$. So if $u\notin K_A$ is a unit of $\Lambda$, then the conjugate subfield $uK_Au^{-1}$ is isomorphic to $K_A$ but intersects it only at $\Bbb{Q}$.

But there exists a method for describing exactly which quadratic number fields occur as (maximal) subfields of $d$. That high powered theory may be overkill for our purposes given that we easily verify
$$
\chi_{M(a,b,c,d)}(T)=T^2-2aT+(a^2-2b^2-3c^2+6d^2)
$$
(should have spelled this out earlier).
And the discriminant of that polynomial is
$$
\Delta=4(2b^2+3c^2-6d^2)
$$
implying that $K_A\simeq\Bbb{Q}(\sqrt{2b^2+3c^2-6d^2})$. Each and every such quadratic field produces its share of units to $\Lambda$. Together with the conjugates.

Skolem-Noether -theorem tells us that whenever $K_A$ and $K_{A'}$ are isomorphic fields, they are conjugates in the division algebra $D$. Meaning that the units of a given quadratic number field appear as units of $\Lambda$ in many distinct copies gotten by conjugating with other units.

Nothing I can say will make my claim that the group of units is "messy" more explicit. Hopefully a more knowledgable user shows up and can shed additional light to your question.


问:
Prove $\vert z\vert\leq \vert \mathfrak{Re}(z)\vert+ \vert \mathfrak{Im}(z)\vert$ where $z$ is a complex number
Prove $\vert z\vert\leq \vert \mathfrak{Re}(z)\vert+ \vert \mathfrak{Im}(z)\vert$ where $z$ is a complex number

Seems quite simple but I don't know how to.

答:
$$ |z| = \sqrt{Re(z)^2 + Im(z)^2} $$
$$     \leq \sqrt{Re(z)^2} + \sqrt{Im(z)^2}$$
$$ = |Re(z)| + |Im(z)|$$

The inequality follows since in general for any non-negative $x$ and $y$ we have $\sqrt{x+y} \leq \sqrt{x} + \sqrt{y}$ (consider the square of both sides)