问:
Which of the following lines offers the better fit to the given points in the least-squares?
I have two similar exercises:

Which of the following lines, $y = 1 - x$ or $y = 4 - x$, offers the better fit to the points $(1,2),(2,1), (3,1)$ in the least-squares?
Justify?

My thought process (haven't solved it yet) was to solve for $A^tAx=A^tb$ and find the line that best approximates it, and then for each point find the error... but now I see that doesn't make much sense. I might not have to solve that system because on this other exercise:

Determine and justify, without solving any system of equations, which
line $y = -1 -x$ or $y = 1-x$ offers the better fit, in the least squares,
the points $(-2,3),(-1,2),(0,1),(1,0)$.

How do I solve this? I remember vaguely some formula that involved a sum of the residue or error or something, and squares, but I don't remember it well.

答:
The first one can be solved without computation at all.

Clearly $4-x$ is the better fit in the least-square sense.  Maybe we'd rather have a formula.  For each point, compute the square of the difference between the height of the point and the height of the line.

$1-x$: $((1-\overbrace{1}^{x_1})-\overbrace{2}^{y_1})^2 + ((1-2)-1)^2 + ((1-3)-1)^2 = 4 + 4 + 9 = 17$.

$4-x$: $((4-1)-2)^2 + ((4-2)-1)^2 + ((4-3)-1)^2 = 1 + 1 + 0 = 2$.

This computation confirms that $4-x$ is a better fit to the data in the least squares sense.

As a formula, the quantity to minimize is
$$  \sum_{i=1}^n (f(x_i) - y_i)^2  \text{,}  $$
where $\{(x_1, y_1), \dots, (x_n, y_n)\}$ is the set of points being fit and $f$ is (in these examples) the line.

The second one is even easier -- all the point lie on one of the lines.


问:
Find the coefficients of this polynomial.

Given $f(x)=x^4 +ax^3 +bx^2 -3x +5$ , $  a,b\in \mathbb{R}$ and knowing that $i$ is a root of this polynomial. what are $a,b$?

My work until now: since all the coefficients are real, I now know that $i$ and $-i$ are roots of this polynomial, there must be two more roots. I tried to use vietta but it didn't really help, I get that $x_3+x_4=-a$ and $i(-i)x_3x_4=5$. What am I missing?

Appreciate any help.

答:
Alternative (but poor) solution:

As we know that $\pm i$ are roots, $x^2+1$ is a factor. Then by long division we obtain the remainder

$$-(a+3)x+(-b+6).$$

But this remainder must vanish, and $a=-3,b=6$.

ContentLicense: CC BY-SA 4.0


问:
What is rigorous notation for functions?
I have seen many ways to denote a function: $f(x)=x^2, y=x^2, f: x\mapsto x^2$ and so on. What is exact notation for functions? Please include lethal doses of rigor, set theory, and of course notational exactness.

Note: I am very familiar with functions in general. I just know that a lot of mathematical literature abuses notation when it comes to functions.

答:
Using ZF a function $f$ is a subset $f \subseteq \mathcal P(X \times Y)$ for some fixed sets $X,Y$ such that $\forall x \in X \ \exists ! y \in Y \ (x,y) \in f$. We denote with $f(x) \in Y$ the unique value in $Y$ such that $(x,f(x)) \in f$.

Now, what is allowed to describe for a given $x \in X$ its image $f(x) \in Y$, heavily depends on the structure of $Y$ and we may introduce notation such as $f: X \rightarrow Y, \ x \mapsto f(x)$ for our own convenience...

ContentLicense: CC BY-SA 3.0


问:
Trying to find a function from line length
I was wondering if it was possible to find a function $f(x)$ whose line length is equal to its integral, thinking about it logically the function $f(x)=1$ satisfies these conditions as the line length is just equal to $x$ and so is the antiderivative, However I was wondering if there could be any more functions that satisfy this, that is to say:
$$\int\sqrt{1+f'^2}\,dx=\int f\,dx$$
Which I think is fair to change to:
$$\sqrt{1+f'^2}=f$$
$$1+f'^2(x)=f^2(x)$$
Any suggestions?

答:
$$
\frac{df}{dx} = \pm \sqrt{f^2-1},\\
\frac{df}{\sqrt{f^2-1}} = \pm dx,\qquad\text{if $f\neq1$}\\
\frac{d(\cosh u)}{\sqrt{\cosh^2 u-1}} = \pm dx,\qquad f = \cosh u, \\
|u|=C\pm x,\\
f=\cosh(C\pm x) = \cosh(x+C).
$$

So there are two solution $f_1=\cosh(x+C)$ and $f_2=1$. They can transform into each other at points where $f_1=f_2=1$, which for function $f_1$ happens only at one point $x=-C$, so a general solution is:

$$
f(x) = \begin{cases}\cosh(x-a), &x\le a,\\
1,&a<x<b,\\
\cosh(x-b),& x\ge b\end{cases}
$$

where $b\ge a$ are constants.


问:
Help with reasoning on how to set up a recurrence relation
I'm working with the following problem:

An alphabet $\Sigma$ consist of the four numeric characters 1, 2, 3, 4, and the seven alphabetic characters a,b, c, d, e, f, g. Find and solve a recurrence relation for the number of words of length n (in $\Sigma$), where there are no consecutive (identical or distinct alphabetic characters.

My problem is how to reason when I'm setting up the recurrence relation. My reasoning goes like this:

If there's a numeric character last, then the character before it can be either an numeric or an alphabetic character which there exist 4+7=11 characters of. So $11a_{n-1}$

If there's a alphabetic character last, then the character before it can only be an numeric character which there exist 4 characters of. So $4a_{n-2}$

Which gives me the recurrence relation: $a_n = 4a_{n-1} + 11a_{n-2}$

But I don't think my reasoning is sound. please help me out

答:
I usually find it less error-prone to break problems like this into cases explicitly.

Let $s_n$ be the number of valid $n-$character strings than end in an alphabetic character, and let $t_n$ be the number of valid $n-$character strings than end in a numeric character, so that $a_n=s_n+t_n$.

We have $$\begin{align}
s_n&=7t_{n-1}\\t_n&=4(s_{n-1}+t_{n-1})
\end{align}$$

Substituting the first equation into the second gives $$\frac{s_{n+1}}7=4\left(s_{n-1}+\frac{s_n}7\right)$$ which we can solve in the usual way.  Then we can appeal to the first equation again to get $t_n$, and finally $a_n$.

Alternatively, the last equation can be rewritten as $$s_{n+1}=28s_{n-1}+4s_n,$$ and the first equation shows that $t_n$ must satisfy the same recurrence
$$t_{n+1}=28t_{n-1}+4t_n.$$  Adding these two gives $$a_{n+1}=28a_{n-1}+4a_n.$$


问:
What's the difference between 'for any' and 'for all'?
There are lots of expressions like, for all x, for any x, for some x, etc.

It may be because of the fact that I am not a native English speaker, but I can't tell the difference between 'for all' and 'for any'.

(I think 'for some x' means 'there exists at least one point', right?)

Could you explain it? thank you!

答:
The term "any" is troublesome, because in natural usage it could mean "all" or "at least one", depending on the context. Here are examples to consider.

(1) For any $a > 0$ there is an $x > 0$ such that $x^2 = a$.

(2) Does the equation $x^3 + y^3 + z^3 = 33$ have any integral solution?

(3) Have you solved any of those problems?

(4) Using this new technique, I can solve any of the problems from that list.

In the first example, "any" = "all".  In the second one, "have any" is asking about existence. In the third, "any" means "at least one" (existence). In the fourth, "any" means "all".

I have known weak math students who are native English speakers and think (1) is proved by showing it works when $a = 1$, even though that way of interpreting (1) makes it into a trivial statement. In other words, they interpret "For any" in (1) as meaning "For some", and hence turn (1) into an existence claim instead of a universal claim. Such usage of "any" is present in non-mathematical English (see the third example), and I think this is the basis for the student's misunderstanding (comparable to having to learn the different meaning of "or" in mathematical English compared to non-technical English). I don't think any native English speaker would  misunderstand the different senses of "any" in (3) and (4).

I would advise someone who is not a native English speaker to avoid using "any" in mathematical statements.  You can convey what you need with other choices of words.


Title: Find the coefficients of this polynomial.

Body:
Given $f(x)=x^4 +ax^3 +bx^2 -3x +5$ , $  a,b\in \mathbb{R}$ and knowing that $i$ is a root of this polynomial. what are $a,b$?

My work until now: since all the coefficients are real, I now know that $i$ and $-i$ are roots of this polynomial, there must be two more roots. I tried to use vietta but it didn't really help, I get that $x_3+x_4=-a$ and $i(-i)x_3x_4=5$. What am I missing?

Appreciate any help.

Body:
By long division,
$$
\require{enclose}
\begin{array}{r}
x^2+ax+(b-1)\\[-4pt]
x^2+1\enclose{longdiv}{x^4+ax^3+bx^2-3x+5}\\[-4pt]
\underline{x^4\phantom{\ +ax^3}+x^2\ }\phantom{-3x+5\ \ }\\[-2pt]
ax^3+(b-1)x^2-3x\phantom{\,+5\ \ }\\[-4pt]
\underline{ax^3\phantom{\,\,+(b-1)x^2}+ax}\phantom{\ +5}\\[-2pt]
(b-1)x^2-(a+3)x\ +5\\[-4pt]
\underline{(b-1)x^2\phantom{\ \qquad}+(b-1)}\\[-2pt]
-(a+3)x+(6-b)
\end{array}
$$
We must have that $-(a+3)x+(6-b)=0$. Therefore, $b=6\text{ and }a=-3$. Thus,
$$
\left(x^2+1\right)\left(x^2-3x+5\right)=x^4-3x^3+6x^2-3x+5
$$


问:
Is the converse of divisibility of $a\mid b \land a\mid c \implies a\mid (b+c)$ always true?
If $a\mid b$ and $a\mid c$ then $a\mid(b+c)$ where $a$ and $b$ are non zero and belong to $Z$.

But does the converse hold true?

E.g. $2\mid 8$ but it does not divide $5$ or $3$. Of course we can express $8$ differently i.e. $4+4$ and then it holds.

But I was wondering in the generality if $a\mid x$ then can $x$ always be expressed in sums of numbers that are divisible by $a$? Of course in case that $x$ is not prime.

Update (thanks to @EricTowers)

Basically the question is if it can be proven that for all ,∈ ℤ, if |, then there exist ,∈ℤ such that =+, |, and |.

答:
Let me clarify this a bit. Let $a,b$, and $x$ be integers.

Original statement (true): if $x|a$ and $x|b$, then $x|(a+b)$.

Converse statement (false): if $x|(a+b)$, then $x|a$ and $a|b$.

The first statement is true of course, but the second statement is false. In fact, you already gave a counterexample: take $x=2$, $a=5$, $b=3$.

The other statement you hinted at would be the following, which is different from either of the first two.

New statement: if $x|y$, then we can find integers $a$ and $b$ such that $y=a+b$, $x|a$, and $x|b$.

This is true because we can take $a=y$ and $b=0$.


问:
Expected average gain from implementing a policy via majority vote
Suppose a country with $n$ citizens decides whether to implement a new policy. The value of implementing that policy for each individual citizen is known only to them and is uniformly distributed on $[-1; 1]$ (it is positive, when they would gain from this policy being implemented and negative if they will lose). If the policy is not implemented nobody will either gain anything or loose anything. Whether to implement the policy is decided via a majority vote: the policy will be implemented if and only if at least half of the citizens votes for it.

A weakly dominant strategy for each citizen is to vote for the policy if their gain from it is positive and against it, if their gain from it is negative. What is the average expected gain of citizens if all of them follow this strategy?

The policy will be selected only if and only if more citizens gain from it than lose from it. Suppose $A$ is the set of all citizens with positive gain. Than conditional expectation for fixed $A$ will be $\frac{2|A|-n}{n 2^{n+1}}$. That means that total expected gain will be:

$\sum_{k=\lceil \frac{n}{2} \rceil }^n \frac{C_n^k (2k-n)}{n 2^{n+1}}$

It is quite obvious, that this value is bounded by 1 from above. It is also monotonously non-decreasing with growing $n$, because:

$\frac{C_{n+1}^k (2k-n-1)}{(n+1) 2^{n+2}} - \frac{C_{n}^k (2k-n)}{n 2^{n+1}} = \frac{C_{n+1}^k (2k-n-1)n - C_{n}^k (2k-n)(n+1)}{n (n+1) 2^{n+1}} = \frac{C_n^k}{{n 2^{n+1}}} (\frac{n(2k-1-n)}{n+1-k} - 2k - n) \geq 0$ when $k \geq \frac{n}{2}$

That means, it converges to some number on $[0;1]$. I would like to find that number, but I don't know how.

答:
Let $X$ be the number of voters whose gain is positive. Provided $X\ge n/2$, each of the $X$ voters gains $1/2$ on average, and each of the $n-X$ other voters gains $-1/2$ on average, so the average total gain is
$$
E[\text{total gain}]
=E\left[\frac12(X-(n-X))\cdot \def\1{{\bf 1}}\1(X\ge n/2)\right]
=E\Big[(X-n/2)\cdot \1(X\ge n/2)\Big]
$$
Using the equality $z\cdot \1(z\ge 0)=\frac12(|z|+z)$, this is equivalent to $$E[\text{total gain}]=E\left[\frac12(|X-n/2|+(X-n/2))\right]=\frac12 E|X-n/2|.$$

To compute this we use the following Lemma, first proved by de Moivre (see this excellent explanation of the history by [Diaconis and Zabell], or this question on cross-validated).

Lemma: (de Moivre) If $Y$ is binomially distributed with $n$ trials and probability $p$, then
$$
E|Y-np|=2(1-p)\cdot (\lfloor np\rfloor +1)\cdot P(Y= \lfloor np\rfloor +1).
$$

In this case, we have
$$
E[\text{total gain}]=\frac12(\lfloor n/2\rfloor +1)\cdot 2^{-n}\binom{n}{\lfloor n/2\rfloor +1}=\boxed{2^{-(n+1)}\lceil n/2\rceil \cdot \binom{n}{\lfloor n/2\rfloor}.}
$$

Diaconis, Persi; Zabell, Sandy. Closed Form Summation for Classical Distributions: Variations on a Theme of De Moivre. Statist. Sci. 6 (1991), no. 3, 284--302. doi:10.1214/ss/1177011699. https://projecteuclid.org/euclid.ss/1177011699


问:
Substitution with Möbius Transformation and polynomial whose leading coefficient disappears after substitution
Let $K = \mathbb{Q_3}$ and $L=K(\alpha)$ where $f := \min_K(\alpha) = x^4 - 3x^2 + 18$.

The roots of $f$ are given by $\pm \alpha$ and $\pm \alpha'$ where
$$
\alpha' = \frac{(2 \alpha^2 -3) u}{\alpha}
$$
and $u = \sqrt{-2/7} \in K$.

Consider the Möbius Transformation
$$
\phi(x) = \frac{a x + b}{c x + d}
$$
with coefficients $a,b,c,d \in L$ such that $\phi(0) = \alpha$, $\phi(1) = -\alpha$ and $\phi(\infty) = \alpha'$.

In fact, this is possible for $a = \alpha' c$, $c = -\frac{2\alpha d}{\alpha+\alpha'}$, $b = \alpha d$ and $d \in L$ arbitrary.

I was told that because $\alpha'$ is a root of $f$ and $\phi(\infty) = \alpha'$, we obtain
$$
f(\phi(x)) = \frac{1}{(c x + d)^4} f_1(x)
$$
where $f_1$ is a polynomial of degree $3$ (and not $4$), i.e. the coefficient corresponding to the monomial $x^4$ disappears!

Question: Without computing that by hand, can someone explain why this happens?

I could compute all of this by hand but this would not give me any insight on how the Möbius Transformation works here.

答:
Let $g(x)=f(\phi(x))$.  We have $g(\infty)=f(\phi(\infty))=f(\alpha')=0$.  If a rational function vanishes at $\infty$, that means its numerator has lower degree than its denominator.  How you prove this depends on how you define evaluation at $\infty$. For instance, one way of defining it is that $g(\infty)=h(0)$ where $h$ is the rational function given by $h(x)=g(1/x)$.  If $g(x)=p(x)/q(x)$ for polynomials $p$ and $q$ of degree $m$ and $n$, we then have $$h(x)=\frac{p(1/x)}{q(1/x)}=\frac{x^{-m}r(x)}{x^{-n}s(x)}=x^{n-m}\frac{r(x)}{s(x)}$$ where $r(x)$ and $s(x)$ are polynomials not divisible by $x$.  So, $h(0)=0$ iff $n>m$.

In your case, since the denominator has degree $4$, the numerator must have degree at most $3$.  (And in this case, the numerator actually has degree exactly $3$, since $g$ has $3$ zeroes at finite points, namely at $\phi^{-1}(\alpha)$, $\phi^{-1}(-\alpha)$, and $\phi^{-1}(-\alpha')$.)


问:
Solving NH 2nd order ODE using Frobenius method
Quite stumped with this one so far. I have the following non-homogeneous ODE:

$$2x^2y''+3xy'-xy=x^2+2x$$

And I need to find a solution for $x_0<0$ using Frobenius. Obviously we can center the solution about $x_0=0$, and we can see it is a regular singular point based on the coefficients of $y'$ and $y$. But I searched quite a bit and couldn't find anything about the non-homogeneous part of the equation (every example that I find on here either is homogeneous, or the NH part is a constant). I know for $p(x)$ and $q(x)$ we check for RSP such that:

$$xp=x\left(\frac{3x}{2x^2}\right)=3/2$$
$$x^2q=x^2\left(\frac{-x}{2x^2}\right)=-x/2$$

Which are analytical for $x_0=0$, so no problem there. But I have no clue what to do with the non-homogeneous part of the equation to verify that it is really a RSP. Of course the problem would easily be solved simply with a regular power series, but I absolutely have to use Frobenius without using series to check for analyticity. Any pointers?

Apologies if the terminology is a bit off, english is not my mother tongue. Thanks in advance!

答:
It would be convenient if there was just a low degree polynomial solution, so let's see if there is by plugging in a few monomials into the LHS:

$$L(1)=-x \\
L(x)=3x-x^2.$$

This is already enough play to get what you want:

$$L(bx+a)=-ax+3bx-bx^2$$

and so you want

$$-b=1 \\
-a+3b=2.$$

Note that this would have failed if the RHS had a constant term.

Now $z=y-(bx+a)$ solves the homogeneous equation, and $bx+a$ is analytic, so the analytical behavior of $y$ and $z$ is the same, after adjusting the initial data appropriately.


问:
$\sum_{n=1}^{\infty}(A_n+B_n)=\sum_{n=1}^{\infty}A_n+\sum_{n=1}^{\infty}B_n $
If $\sum_{n=1}^{\infty}A_n$ is convergent and $\sum_{n=1}^{\infty}B_n$ is divergent then can we write $\sum_{n=1}^{\infty}(A_n+B_n)=\sum_{n=1}^{\infty}A_n+\sum_{n=1}^{\infty}B_n $?

答: If $\sum_{n=1}^\infty B_n$ is divergent, it is not clear what is meant by $\sum_{n=1}^\infty B_n$. It might be $-\infty$, $+\infty$ or maybe even something else (consider, for example, $\sum_{n=1}^\infty(-1)^n$).


答:
If at any two of the series $\sum_{n=1}^{\infty}(A_n+B_n)$, $\sum_{n=1}^{\infty}A_n$, $\sum_{n=1}^{\infty}B_n$ converge, then so does the third and
$$
\sum_{n=1}^{\infty}(A_n+B_n)=\sum_{n=1}^{\infty}A_n+\sum_{n=1}^{\infty}B_n
$$
So in your case, we can only conclude that $\sum_{n=1}^{\infty}(A_n+B_n)$ diverges.  Normally, when two series both diverge, we do not say that they are equal.  (Nor do we say that they are unequal.)

Body: If $\sum_{n=1}^{\infty}A_n$ is convergent and $\sum_{n=1}^{\infty}B_n$ is divergent then can we write $\sum_{n=1}^{\infty}(A_n+B_n)=\sum_{n=1}^{\infty}A_n+\sum_{n=1}^{\infty}B_n $?


问:
Is there a positive series of rational numbers that converge to $\pi$?
I'm looking for a series of rational numbers that approach $\pi$ that are all non-negative. In addition, I also want this series to have a general rule. So something like $\dfrac31+\dfrac1{10}+\dfrac4{100}+\ldots$ is not gonna cut it.

Does such a series exist?

Body:
We have

$$\pi = 3 + \sum_{k=0}^\infty \frac{24}{(4 k + 2) (4 k + 3) (4 k + 5) (4 k + 6)}$$

$$ = 3 +\frac{2}{15}+\frac{2}{315}+\frac{6}{5005}+\frac{2}{5355}+\frac{2}{13167} + \frac{6}{82225} + ...$$
which is equivalent to the integral

$$ 2 \int_0^1 \frac{x(1-x)^2}{1+x^2} dx = \pi - 3 $$


问:
Does a closed walk necessarily contain a cycle?
[HOMEWORK]

I asked my professor and he said that a counter example would be two nodes, by which the pathw ould go from one node and back. this would be a closed path but does not contain a cycle. But I am confused after looking at this again. Why is this not a cycle? Need there be another node?

答:
I guess the answer depends on the exact definition of cycle. If it is as you wrote in your comment - a closed walk that starts and ends in the same vertex, and no vertex repeats on the walk (except for the start and end), then your example with two nodes is a cycle.

However, a definition of a cycle usually contains a condition of non-triviality stating that a cycle has at least three vertices. So a graph with two vertices is not a cycle according to this definition.


问:
How does this "non-standard definition" of a tangent vector relates to the definitions given in Lee's book on smooth manifolds?
I am trying to understand the following definition of a tangent vector to a manifold which I came across in some lecture notes and how it relates to the three "standard definitions" as they can be found in for example Lee's book on smooth manifolds. The definition is

That $X:\mathcal{C}^{\infty}(M)\to\mathbb{R}$ is a linear map and it satisfies $Xf=(f\circ\gamma)'(0)$ made me guess that it probably also satisfies the Leibniz rule. Hence, we would have $X_{p}\in\{\text{Derivations at p}\}$.Would't we then have $T_{p}M\subset\{\text{Derivations at p}\}$. I would also like to have that or rather my guess is that $\{\text{Derivations at p}\}\subset T_{p}M$. I think this guess is due to the following proposition in Tu's book on smooth manifolds:

One of the things I don't get my head around too is that it also reminds me of the definition in terms of equivalence classes of curves, but apperently one can get away without them.

I would really appreciate some hints and tips. Many thanks in advance!

答:
Your guess is correct.

First note that we have to interpret Definition 2.1 in the sense that there exists $\gamma$ such that for all $f$ one has $Xf = (f \circ \gamma)'(0)$, i.e. $\gamma$ does not depend on $f$.

Given $f,g \in C^\infty(M)$, we get $(f \cdot g) \circ \gamma = (f \circ \gamma) \cdot (g \circ \gamma)$ and therefore by the product rule of elementary calculus (applied to functions $(-\epsilon, \epsilon) \to \mathbb R$)
$$X(f \cdot g) = ((f \cdot g) \circ \gamma)'(0) = (f \circ \gamma)'(0) \cdot (g \circ \gamma)(0) + (f \circ \gamma)(0) \cdot (g \circ \gamma)'(0)\\ = (Xf)\cdot g(p) + f(p) \cdot (Xg)$$
which means that $X$ is a derivation at $p$.

That each derivation $X$ at $p$ has the property in Definition 2.1 follows indeed from Tu's Proposition 8.17. There exists $c$ such that $c'(0) = X$, and by Tu's Proposition we have
$$Xf = (f \circ c)'(0) .$$
Note that Tu's Proposition says that this is true for any such $c$.


问:
Prove that A’C + A’B + BC equals C (A XOR B)’ + A’B
Prove
$$A’C + A’B + BC = C (A \oplus B)’ + A’B$$

Here is what i have tried:
$$A’C + A’B + BC = C(A’+B) + A’B$$
Now, need to prove $(A \oplus B)’ = A’+B$

$$(A \oplus B)’ = A'B'+AB = \text{now what??}$$

答:
Please note that $(A \oplus B)’ \not = A’+B$

Your mistake is that just because $$C (A \oplus B)’ + A’B = C(A’+B) + A’B$$

does not mean that $(A \oplus B)’ = A’+B$

That is: just because two larger statements are equivalent does not mean that two components of those larger statements are the same, even if the rest is the same. As another example:

Just because $P + P'Q = P + Q$ does not mean that $P'Q=Q$

So, your approach is not going to work.

Instead, I would start with the right side, so you can first rewrite the $\oplus$ into the more familiar boolean operations (e.g. use $A \oplus B = AB'+A'B$) so you can then use the more familiar laws for those more familiar operators to try and end up with the left hand side.


问:
Determine with proof if the following integral converges
Determine if the following integral converge.
$$\int^\infty_2 \ln\left(\sec\left(\pi\over x\right)\right)\ dx$$

I have a function to compare the above over the interval [2, 3], but am struggling to find one that will compare nicely over [3, $\infty$]. $1\over x^{1.1}$ can be used to show that it does converge using comparison, but is having trouble showing the function is greater than the one above over [3, $\infty$]. Any help would be appreciated!

答:
At $z=0$ we have the expansion
$$\sec z=1+\frac{z^2}{2}+O\left(z^3\right)$$
so as $z\to 0$
$$\log\sec z \sim \log \left(1+\frac{z^2}{2}\right)\sim \frac{z^2}{2}$$
thus as $x\to\infty$
$$\log\sec\frac{\pi}{x}\sim \frac{\pi ^2}{2 x^2}$$
therefore for the comparison test the integral converges at $+\infty$


问:
Is there any way to evaluate/ estimate this integral apart from wolfram?
I am looking for a way to evaluate (if not, then estimate) the following $$\int_0^x \dfrac {1}{\sqrt{\cos t - \cos x}}dt, \ 0 < x \leq \pi$$

Wolfram tells me that it blows up at $x = \pi$ and for lower values, it does not. (I tried $\pi/2$ and $\pi/3$.)

Edit 1

I have tried to make use of comparing the integrand with some other function, whose behaviour is known. For $\cos t - \cos x$, I am trying with $t-x$, but once it goes inside the root, then the range no longer stays real.

答:
Maple does this as an elliptic integral.  If $0 < x < \pi$,
$$
\int_{0}^{x}\!{\frac {1}{\sqrt {\cos \left( t \right) -\cos \left( x
 \right) }}}\,{\rm d}t=\sqrt {2}\;{\rm F} \left({\frac {2\sin
 \left( x/2 \right) }{\sqrt {2-2\,\cos \left( x \right) }}},
\frac{\sqrt {2-2\,\cos \left( x \right) }}{2} \right)
$$
The graph:

There is a vertical aspymptote as $x \to \pi^-$.


问:
Fact Check: Is any maximum clique also a maximal one?
According to Wikipedia:

A maximal clique is a clique that is not included in a larger clique.

and

A maximum clique is a clique that includes the largest possible number of vertices.

答: If it has the largest possible number of vertices, a larger clique can't exist, so ...</p>


问:
finding inverse map
Could someone help me find the inverse of $\psi(u,v)= (\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1})$ I tried $u=x,\frac{2u}{u^2+v^2+1}=x $ ect, but I can't find it. any hints?

答:
Avoid coordinates whenever possible.  Regard $\psi$ as a map from $\mathbb R^3$ to itself identifying $(u,v)$ with $(u,v,0)$.  Define $n=(0,0,1)$ and $q=(u,v,0)$.  We have
$$\psi(q)=\frac{1}{\|q\|^2+1}\bigr(2q+n(\|q\|^2-1)\bigr).$$
Geometrically, $\psi(q)$ is the intersection of a point $q$, which is perpendicular to $n$ and the point $n$ on the unit sphere.

Now the inverse mapping maps a point $p$ on the unit share -- different from $n$ -- to the plane orthogonal to $n$. To determine $\psi^{-1}(p)$ calculate the point on the straight line through $n$ and $p$,
$$n+t(p-n),$$
which is orthogonal to $n$, that is
$$\langle n+t(p-n),n\rangle=0\iff t=\frac{1}{1-\langle n,p\rangle}.$$
Finally,
$$\psi^{-1}(p)=\frac{p-\langle n,p\rangle n}{1-\langle n,p\rangle}.$$
You may feel free to write the equation with coordinates ...

Title: finding inverse map


问:
Urn model: closed formula for: at least a given number of blue balls
I was discussing the following question with a friend but we are stuck, this is probably obvious for someone with a mathematical background.

Say we have an urn with $N$ balls of which a $M$ are blue. We now take $n$ balls out of the urn without putting them back. What is the probability of having at least $k$ blue balls.

Wikipedia tells me the probability of getting exaktly k blue balls is

$ P(X=k)=\frac{{n\choose k}{N-n\choose M-k}}{{N\choose M}}$

Is there a closed form solution for the probability for drawing i or more blue balls?

$\sum_i^N P(X=i)$

答: Yes and no. You're asking for $$\sum\limits_{i=k}^n P(X=i)=1-\sum\limits_{i=0}^{k-1}P(X=i)=1-P(X\leq k-1),$$which is the cif of the hypergeometric distribution (see. wikipedia). If k=0, so you're asking the probability of having zero or more blue balls, the answer is obviously 1.


问:
Expectation Value with exponential function
I've got a short question about the existence of the expectation value.

Let $X$ be a random variable without further knowledge about it (e.g. if it has a density or not).
If I know $\mathbb{E}(e^X)<\infty$ can I conclude that $\mathbb{E}(|X|)<\infty$, hence the expectation value of X exists?

Thanks!

答:
Suppose $X=-2,-2^2, -2^3,-2^4,\ldots$ with probabilities  $\frac12,\frac1{2^2},\frac1{2^3},\frac1{{2^4}},\ldots$ respectively

then  $\mathbb{E}(e^X)$  is certainly finite and between $0$ and $e^{-2}$, in fact about $0.07228849$,

but $\mathbb{E}(X)=-\infty$ and $\mathbb{E}(|X|)=\infty$.

What is true is that $\mathbb{E}(e^{|X|})\lt \infty \implies \mathbb{E}(|X|)\lt \infty$.


问:
If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$.
$x,y,z \in \mathbb R$

My attempt:

Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.

Maybe I’m wrong.

答:
I guess you guys are too fast and / or assume too much prior knowledge. Let's digest it.

We need to be careful since all numbers are reals. Hence AM-GM and square root  methods don't work.

First, note that if $xyz < 0$, then there is nothing to prove, since in this case $x^2 + y^2 + z^2 \geq 0 \geq xyz$ holds in all cases. So we need to consider  $xyz \ge  0$ only.

We want to establish the following train of inequalities:
$$
(x^2+y^2+z^2)^2 \ge (xy+yz+zx)^2 \ge {3xyz(x+y+z)} \ge  {3}  (xyz)^2 = 3 |xyz|^2
$$
The first one follows from
$$
 x^2+y^2+z^2 - ( xy+yz+zx)  = \frac12 \Big[ (x-y)^2+(y-z)^2+ (z-x)^2 \Big] \ge 0
$$
The second one follows from
$$
(xy+yz+zx)^2 -  3xyz(x+y+z) =  \frac12 \Big[(xy-yz)^2+(zx-xy)^2+(yz-zx)^2 \Big] \ge 0
$$
For the third one, for $xyz > 0$, then this follows from the question's condition: $x + y+ z \ge xyz$.

So indeed it holds for all reals:
$$
x^2+y^2+z^2 \ge \sqrt{3} |xyz| \ge \sqrt{3} \cdot xyz
$$

This is also the tightest inequality, since we observe that equality holds for the question's condition and for this solution at $x=y=z = \sqrt{3}$. (River Li asked for that constant $k = \sqrt{3}$).


答:
First, we prove for all positive $x,y,z$.

If all of $x,y,z\geq1$ we have

$$x^2\geq x ,\\ y^2\geq y \\ z^2\geq z$$ $$⇒x^2+y^2+z^2\geq x+y+z \geq xyz\\$$

If at least one of $x,y,z$ is less than $1$ then take (WLOG) $x\geq y\geq z>0$ & $1>z>0$

$$\frac{x}{y}\geq1 $$

$$⇒\frac{x}{yz}\geq1 $$
$$⇒\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\geq 1$$
$$⇒x^2+y^2+z^2\geq xyz$$

If two of $x,y,z$ is negative (say $x$ and $y$)
we have $$-x-y+z ≥ x+y+z ≥xyz=(-x)(-y)z$$
$$⇒-x-y+z ≥(-x)(-y)z$$ Here $-x,-y,z$ are positive numbers. So from the earlier result we can say,
$$(-x)^2+(-y)^2+z^2\geq (-x)(-y)z$$ $$⇒x^2+y^2+z^2\geq xyz$$

The inequality is obvious when one of $x,y,z$ is $0$ or if one of $x,y,z$ is negative or if all of $x,y,z$ is negative.

Title: If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$.

Body:
$x,y,z \in \mathbb R$

My attempt:

Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.

Maybe I’m wrong.


问:
Is every fundamental group of a mapping torus of a surface homeomorphism a CAT(0) group?
TLDR; Is every fundamental group of a mapping torus of a surface homeomorphism a CAT(0) group?

Let $S$ be a compact surface of negative Euler characteristic and let $f :S\to S$ be a homeomorphism. Write $M_f$ to denote the mapping torus of $f$,  $$M_f = \frac{S \times [0,1]}{(0,f(x)) \sim (1, x)}.$$

Thurston shows that:

If $f$ is pseudo-Anosov then $M_f$ admits a complete hyperbolic structure of finite volume.

If $f$ is finite order then $M_f$ admits a complete $\mathbb{H}^2 \times \mathbb{R}$ structure.

If $f$ is reducible then $M_f$ admits an embedded incompressible torus, not parallel to the boundary.

If $S$ has empty boundary, then in cases 1 and 2 we can conclude that $M_f$ acts properly and cocompactly on $\mathbb{H}^{3}$ and $\mathbb{H}^2 \times \mathbb{R}$, respectively, and thus that $\pi_1(M_f)$ is CAT(0).

My question: What can be said in the case when $S$ (and thus $M_f$) has non-empty boundary? What can be said in case 3?

答:
Yes, there are  examples of mapping tori which are not CAT(0). The simplest one is when $f$ is a single Dehn twist.

See Theorem 3.7 in

Kapovich, Michael; Leeb, Bernhard, Actions of discrete groups on nonpositively curved spaces, Math. Ann. 306, No. 2, 341-352 (1996). ZBL0856.20024.

You would also need Theorem 2.4 in their paper.

The ultimate result is contained in

Buyalo, S.; Svetlov, P., Topological and geometric properties of graph-manifolds, St. Petersbg. Math. J. 16, No. 2, 297-340 (2005); translation from Algebra Anal. 16, No. 2, 3-68 (2004). ZBL1082.57015.

In this paper you will find a combinatorial criterion for the mapping torus of a completely reducible homeomorphism of a surface to admit a metric of nonpositive curvature. (I find their condition to be too complicated to be discussed here.) See also references there to the earlier papers by Buyalo and Kobelsky on this subject. One of the more surprising results they prove in the paper (it is actually due to Svetov) is that the fundamental group of a closed graph-manifold is CAT(0) if and only if the manifold virtually fibers over the circle.


问:
Why in the axiom of comprehension in von Neumann-Gödel-Bernays set theory only predicative formulas are considered?
For predicative formula I mean a formula where all the quantifiers are restricted to sets.

In Zermelo-Fraenkel set theory the axiom of comprehension is replaced by the axiom of separation that doesn't lead to inconsistencies like Russel paradox.

In NGB the idea to reduce the power of axiom of comprehension is accomplished considering predicative formulas only. Otherwise, I think it might be some formulas that allows a class to belong to another class, that is absurd.

Can someone express a clearer reason for this choice?

Thanks in advance :)

答:
Class-quantifier-comprehension does not in fact force (proper) classes to be elements of classes: remember that, regardless of the type of variable being quantified over, the object produced by a comprehension axiom is a priori a collection of sets. So you won't run into this problem here.

In fact, allowing class quantifiers in the comprehension axioms is relatively safe. Working in $\mathsf{ZFC}$, suppose $\kappa$ is inaccessible and consider the two-sorted structure $\mathfrak{M}=(V_\kappa, V_{\kappa+1})$ (here I'm following the Bernays approach where each set "appears twice," once as a set and again as a class). It's easy to check that $\mathfrak{M}$ satisfies all the NGB axioms together with the stronger comprehension principle allowing quantification over "classes" (= elements of $V_{\kappa+1}$), the point being that every collection of $\mathfrak{M}$-sets (= every subset of $V_\kappa$) is an $\mathfrak{M}$-class (= element of $V_{\kappa+1}$) due to the nature of the $V$-hierarchy.

Put another way, $\mathsf{ZFC}$ + "There is an inaccessible cardinal" proves the consistency of this strengthening of $\mathsf{NGB}$. And as usual, this can be optimized by being a bit more careful about the details. So allowing quantification over classes in the comprehension axioms for $\mathsf{NGB}$ will not actually result in an inconsistency, barring a gigantic surprise.

So why does $\mathsf{NGB}$ not allow quantification over classes in the comprehension axiom? Well, I think this comes down to the philosophical motivation for $\mathsf{NGB}$: a certain amount of predicativity is desired for its own sake, not merely to avoid contradictions. Not being an historian I don't have much to say here, but that's my understanding of the situation.

One important point here is that the quantifier restriction lets us prove that $\mathsf{NBG}$ is, appropriately construed, a conservative extension of $\mathsf{ZFC}$. However, I don't think this motivated that axiomatic choice, since I believe the conservativity result was only observed after $\mathsf{NBG}$ was codified. But I could be wrong here, too - the early set-theoretic literature is sometimes a bit out of whack in terms of what was known when vs. what was published when.

EDIT: And let me add to that observation the dual, which is that strengthening comprehension to allow quantification over classes does increase the consistency strength. See this summary of Hamkins, with the caveat that Hamkins' $\mathsf{GBC}$ is what we've been calling $\mathsf{NGB}$ (and thanks to Jason Zesheng Chen for reminding me of it in a comment above).


问:
Show that $2n+1$ is a factor of the sum of positive odd powers of the first $2n$ integers
This question comes from a high school math book. We are currently learning about binomial coefficients and combinatorial proofs, which we should use for this according to our teacher, but I don't see how that can be used for this particular problem.

The question is:

Show that for all $a,n \in \mathbb{N}$, $2a+1$ is a factor of the sum $\sum_{i=1}^{2a} i^{2n+1}$.

I know that for $n=0$, this follows from the summation formula, which is $\frac{2a(2a+1)}{2} = a(2a+1)$. But I don't know how to proceed.

答: Hint. $i^{2n+1}+(2a+1-i)^{2n+1}\equiv 0\pmod{2a+1}$ since we are considering odd powers. Pair the summands up, and the conclusion should follow.


答:
Hint 1: use the fact that $(x+y) | (x^{2k+1}+y^{2k+1})$.

Hint 2: try grouping the numbers so you get the factor you are looking for.

Title: Show that $2n+1$ is a factor of the sum of positive odd powers of the first $2n$ integers

Body:
This question comes from a high school math book. We are currently learning about binomial coefficients and combinatorial proofs, which we should use for this according to our teacher, but I don't see how that can be used for this particular problem.

The question is:

Show that for all $a,n \in \mathbb{N}$, $2a+1$ is a factor of the sum $\sum_{i=1}^{2a} i^{2n+1}$.

I know that for $n=0$, this follows from the summation formula, which is $\frac{2a(2a+1)}{2} = a(2a+1)$. But I don't know how to proceed.


问:
Question about Do Carmo's notation
Let M be a differentiable manifold. On page 26 of Riemannian Geometry by Do Carmo, he makes this statement:

Observe that if $\varphi:M\longrightarrow M$ is a diffeomorphism, $v\in T_pM$ and $f$ is a differentiable function in a neighborhood of $\varphi(p)$, we have
$$(d\varphi(v)f)\varphi(p)=v(f\circ\varphi)(p).$$
Indeed, let $\alpha:(-\varepsilon,\varepsilon)\longrightarrow M$ be a differentiable curve with $\alpha'(0)=v$, $\alpha(0)=p$. Then
$$(d\varphi(v)f)\varphi(p)=\left.\frac{d}{dt}(f\circ\varphi\circ\alpha)\right|_{t=0}=v(f\circ\varphi)(p).$$

I understand what he means by the right hand side. Namely, a vector $v$ in the tangent space at $p$ acting as a derivation on the differentiable function $f\circ\varphi$ and then evaluating the resultant function at $p$. However, I'm unsure if I'm understanding what the object on the left hand side is supposed to be. You take the differential of the map $\varphi$ (not at any point in particular?, I assume there's supposed to be a subscript $p$?) and then act this Jacobian on the vector $v$. This is fine, and it should give you a vector in the tangent space at $\varphi(p)$. You then treat this as a derivation and so it maps $f$ to some other differentiable function which you then evaluate at $\varphi(p)$. Is this correct? If so, I still can't see where the middle step comes from.

答:
Let $g = f\circ \varphi : M \to M $. Then $g$ is differentiable and the chain rules says
$$
\mathrm{d}g (p) = \mathrm{d}(f\circ \varphi)(p) = \mathrm{d}f(\varphi(p))\circ\mathrm{d}\varphi(p).
$$
If $\alpha$ is a path in $M$ with $\alpha(0)=p$ and $\alpha'(0)=v$, then by definition
$$
(g\circ\alpha)'(0) = \mathrm{d}g(\alpha(0))\alpha'(0)=\mathrm{d}(f\circ \varphi)(p)v = v\cdot \left( f\circ \varphi\right),
$$
where the last equality is just a notation for the vector $v$ acting on the function $f\circ \varphi$ by evaluating its differential.
But if you write $\beta = \varphi\circ \alpha$, which is a path with $\beta(0) = \varphi(p)$ and $\beta'(0) = \mathrm{d}\varphi(p)v$
$$
(f\circ\beta)'(0) = \mathrm{d}f(\beta(0))\beta'(0) = \mathrm{d}f\left(\varphi(p)\right) \left(\mathrm{d}\varphi(p)v\right) = \left(\mathrm{d}\varphi(p)v\right)\cdot f,
$$
where the last equality is the vector $\mathrm{d}\varphi(p)v$ acting on the function $f$.

The equality, $g \circ \alpha = f \circ \beta$ shows that
$$
v\cdot\left(f\circ \varphi \right) = \left(\mathrm{d}\varphi(p)v \right)\cdot f.
$$

Title: Question about Do Carmo's notation


问:
Find the area lying inside the cardioid $r=1+\cos\theta$ and outside the parabola $r(1+ \cos\theta)=1$
I need to find the area lying inside the cardioid $r=1+ \cos\theta$ and outside the parabola $r(1+ \cos\theta)=1$.

ATTEMPT

First I found the intersection point of two curves which comes out to be $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

The integral setup will be $$\int_{\theta =\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{r=\frac{1}{1+\cos\theta}}^{1+\cos \theta}\ dr\ d\theta.$$On integrating this, I got the answer as $\pi$ but answer was given to be $\frac{3\pi}{4}-\frac 43$.

Can anybody check my integral setup?

答:
Actually, it's neither $\pi$ nor $\frac{3\pi}4-\frac43$.

You are right when you claim that the possible values for $\theta$ go from $-\frac\pi2$ to $\frac\pi2$; see the picture below. But you seem to have forgotten that, when you compute an area in polar coordinates, the function that you should be integrating is $r$, not $1$. So, you should compute$$\int_{-\pi/2}^{\pi/2}\int_{1/(1+\cos(\theta))}^{1+\cos(\theta)}r\,\mathrm dr\,\mathrm d\theta,\tag1$$which is equal to $\frac{3\pi}4+\frac43$. However, if, in $(1)$, you forget the $r$ inside the integral, then the integral will be equal to $\pi$.

Besides, you can see from the picture below that the area that you are trying to compute is more than half of the area of the region bounded by the cardioid. But the area of the region bounded by the cardioid is $\frac{3\pi}2$, and therefore that area that you're after must be greater than $\frac{3\pi}4$.


问:
Proving $r(m + n) = rm + rn$ for a type of scalar multiplication.
Here is the question I want to answer:

Let $R \subset S$ be commutative rings and let $M$ be an $R$-module. Then $S \otimes_R M$ is an $R$-module generated by $\{s \otimes m \mid s \in S, m \in M\}.$ Define scalar multiplication by elements $t \in S$ as follows:

$$t. \sum_i r_i(s_i \otimes m_i) = \sum_i r_i(ts_i \otimes m_i)$$
where $r_i \in R, s_i \in S, m_i \in M.$ Show that this gives $S \otimes_R M$ the structure of an $S$-module.

Here is my trial:

Proof:

Let $R \subset S$ be commutative rings and let $M$ be an $R$-module. Since $S \otimes_R M$ is an $R$-module then it is an abelian group with respect to $+,$ i.e. $(S \otimes_R M, +)$ is an abelian group.  Define scalar multiplication by elements $t \in S$ as a function $S \times (S \otimes_R M) \rightarrow S \otimes_R M$ as follows:

$$t. \sum_i r_i(s_i \otimes m_i) = \sum_i r_i(ts_i \otimes m_i)$$
where $r_i \in R, s_i \in S, m_i \in M \,\,\,\, \forall i = 1, \dots ,n$

Now, for all $t,t' \in S$ and $\sum_{i=1}^ n r_i(s_i \otimes m_i) , \sum_{j=1}^m r_j'(s_j' \otimes m_j') \in S \otimes_R M:$ we have:

$(i)$ $t. (\sum_{i = 1}^n r_i(s_i \otimes m_i) + \sum_{j=1}^m r_j'(s_j' \otimes m_j')) = t.\sum_{i = 1}^n r_i(s_i \otimes m_i) + t.\sum_{j=1}^m r_j'(s_j' \otimes m_j')) $

Proof:

I do not know how to prove this, could anyone help me in doing so?

$(ii)$ $(t + t')\sum_{i = 1}^n r_i(s_i \otimes m_i) = \sum_{i = 1}^n r_i((t + t') s_i \otimes m_i) = \sum_{i = 1}^n r_i((ts_i + t's_i)  \otimes m_i) = \sum_{i = 1}^n r_i((ts_i \otimes m_i) + (t's_i \otimes m_i)) = \sum_{i = 1}^n r_i(ts_i \otimes m_i) + \sum_{i = 1}^n r_i(t's_i \otimes m_i) = t.\sum_{i = 1}^n r_i(s_i \otimes m_i) + t'.\sum_{i = 1}^n r_i(s_i \otimes m_i).$ as required.

Where the first equality by the given definition of scalar multiplication by elements in $S$, the second equality is because $S$ is a ring, the third equality by the biadditivity of tensor product, the fourth equality because $S \otimes_R M$ is an $R$-module and finally the fifth equality by the given definition of scalar multiplication by elements in $S.$

$(iii)$ $(tt').\sum_{i = 1}^n r_i(s_i \otimes m_i) = \sum_{i = 1}^n r_i((tt')s_i \otimes m_i) = \sum_{i = 1}^n r_i(t(t's_i) \otimes m_i) = t.\sum_{i = 1}^n r_i((t's_i) \otimes m_i) = t.( t'.\sum_{i = 1}^n r_i(s_i \otimes m_i))$ as required.

Where the first equality is by the given definition of scalar multiplication by elements in $S$, second equality is by associativity of multiplication in the ring $S$ , the third equality by the biadditivity of tensor product, the fourth and fifth equality is by the given definition of scalar multiplication by elements in $S.$

$(iv)$ $ 1.\sum_{i = 1}^n r_i(s_i \otimes m_i) = \sum_{i = 1}^n r_i(1.s_i \otimes m_i) = \sum_{i = 1}^n r_i(s_i \otimes m_i)$

Where the first equality by the given definition of scalar multiplication by elements in $S$, the second equality is because $S$ is a ring.

Thus the given definition of scalar multiplication by elements in $S$ i.e.,  $S \times (S \otimes_R M) \rightarrow S \otimes_R M$ gives $S \otimes_R M$ the structure of an $S$-module.

My questions are:

1- How can I prove $(i)$?

2- Why the sum should be finite?

3- Do I have to prove that the given definition of scalar multiplication by elements in $S$ is well defined as given in Dummit and Foote (3rd edition) on pgs. 360 & 361?

4- What is the importance of the given that the 2 rings are commutative? is it because I will only care about proving that it is $S$-module and not about right and left $S$-module?

5- Is my solution correct?

Could someone demonstrate these points to me please?

答:
(2.) By definition, every element of the $R$-module $S \otimes_R M$ is a finite sum of so-called elementary tensors $s \otimes_R m$ for some elements $s \in S$ and $m \in M.$ (This is precisely your second line: "generated by" means that every element is a finite sum of the generators.) (3.) Unfortunately, it is not true in general that the elementary tensors in the finite sum are unique, hence an element of $S \otimes_R M$ can have two different representations $\sum_{i = 1}^k (s_i \otimes_R m_i)$ and $\sum_{j = 1}^\ell (s_i' \otimes_R m_i').$ Ultimately, this requires us to check that the prescribed action is well-defined, i.e., if we assume that $\sum_{i = 1}^k (s_i \otimes_R m_i) = \sum_{j = 1}^\ell (s_i' \otimes_R m_i'),$ then we must have that $$s \cdot \sum_{i = 1}^k (s_i \otimes_R m_i) = s \cdot \sum_{j = 1}^\ell (s_i' \otimes_R m_i').$$ Of course, there are a few ways to accomplish this, but this answer expounds upon the technique used in Dummit and Foote, so it might be worth a look.

(4.) If $S$ is a commutative ring, then any subring of $R$ of $S$ is commutative, so we need only assume that $S$ is commutative. But also, any left $S$-module is automatically a right $S$-module: indeed, if $s \cdot m$ is a left-action of $S$ on an $S$-module $M,$ then one can define a right-action $m * s = s \cdot m$ that satisfies all of the necessary properties. (For more, check this answer.)

(1.) Last, we have that $s \cdot (s_1 \otimes_R m_1 + s_2 \otimes_R m_2) = s s_1 \otimes_R m_1 + s s_2 \otimes_R m_2$ by definition, and this is precisely $s \cdot (s_1 \otimes_R m_1) + s \cdot (s_2 \otimes_R m_2).$ Consequently, your first property follows. (Expand the sums into one sum, and use the fact from the previous sentence.)


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: All eigenvalues of a Hermitian matrix A are real. No immediate intuition as to why it must be true. If we think of the Riemann hypothesis "All non-trivial zeros of the Riemann zeta function are of the form $ \frac{1}{2}+zi $ where z is real" and try to think of Hermitian matrices having real eigenvalues in equal awe, the wonderment increases. The two ideas are also  related.</p>

LastEditorDisplayName: anon


问:
Why the $\nabla f(x)$ in the direction orthogonal to $f(x)$?
Why the $\nabla f$ in the direction orthogonal to $f$?

I can't understand it intuitively.

Is it because of the definition of gradient?

答:
The correct statement is that the gradient   vector $\nabla f(x_0,y_0)=\left(\frac{\partial}{\partial x} f(x_0,y_0),\frac{\partial}{\partial y} f(x_0,y_0) \right)$ is orthogonal to any level curve $\mathbf{\alpha(t)=(x(t),y(t))}$ of the function $\mathbf{ f} $  in point $\alpha(t_0)=(x_0,y_0)$.

That is, the gradient vector $\nabla f(x_0,y_0)=\left(\frac{\partial}{\partial x} f(x_0,y_0),\frac{\partial}{\partial y} f(x_0,y_0) \right)$  of $f$ in point $(x_0,y_0)$ is orthogonal to the tangent vector $\alpha'(t_0)=\big(x'(t_0),y'(t_0)\big)$.

Let us explain this in a more detailed way. This in fact is the statement of a theorem of calculus in several variables.

Theorem. Let $f:\Omega\subset\mathbb{R}^2\to \mathbb{R}$ be a differentiable function in a open $\Omega\subset \mathbb{R}^2$ and suppose that all partial derivatives  of $f$ are continuous. Consider a  curve $\alpha:I\to \mathbb{R}^2$ differentiable in interval $I=(a,b)$ and living in

  $$
f^{-1}(c)=\{ (x,y)\in\Omega :  f(x,y)=c\}.
$$
  A curve with this property is called a level curve in $c$.
  Suppose that the curve $\alpha(t)=(x(t),y(t))$ have coordinates $x(t)$ and $y(t)$  whose derivatives $x'(t)$ and $y'(t)$ are continuous. For all $t_0\in I$ and $\alpha(t_0)=(x_0,y_0)$,
  $$
\langle \nabla f(x_0,y_0), \alpha'(t_0)\rangle = \frac{\partial}{\partial x} f(x_0,y_0)\cdot x'(t_0)+\frac{\partial}{\partial y} f(x_0,y_0)\cdot y'(t_0)=0
$$

Proof. The function $(a,b)\ni t\mapsto f\circ \alpha(t)= f\big(x(t),y(t)\big)\in\mathbb{R}$ is constant equal to $c$, that is, $f\circ \alpha(t)= f\big(x(t),y(t)\big)=c$. Looking $f\circ \alpha(t)$ as a function in the variable $t$ we know that
$$
\frac{d}{dt}(f\circ \alpha)(t)= \frac{d}{dt} f\big(x(t),y(t)\big)=0.
$$
On the other hand, by the chain rule and gradient property we have that
$$
\left.\frac{d}{dt}(f\circ \alpha)(t)\right|_{t=t_0}
=
\frac{\partial}{\partial \vec{v}}f(x,y)
=
\langle \nabla f(x_0,y_0), \vec{v}\rangle
=
\langle \nabla f(\alpha(t_0)), \alpha'(t_0)\rangle
$$
for $\alpha'(t_0)=(x'(t_0),y'(t_0))=(v_1,v_2)=\vec{v}$.
Then
$$
\langle \nabla f(x_0,y_0), \alpha'(t_0)\rangle = \frac{\partial}{\partial x} f(x_0,y_0)\cdot x'(t_0)+\frac{\partial}{\partial y} f(x_0,y_0)\cdot y'(t_0)=0
$$


问:
Countable Subsets of Irrational Sets
If A is an uncountable subset of the reals such that all elements of A are irrational, is there a countable subset Z of A such that for all x,y in A, there is a z in Z such that x>z>y?

In general, I want to claim that for any uncountable subset B of the reals, there is always a countable subset Y such that $\forall b,b' \in B$ such that $b>b'$, $\exists y\in Y$ such that $b>y>b'$.

答:
As noted in the comments, this is not always possible, because $A$ can contain points that adjacent in $A$. However, it cannot contain too many of them.

Let $A\subseteq\Bbb R$ be uncountable. Let

$$S=\left\{x\in A:\exists y\in A\big(x<y\text{ and }(x,y)\cap A=\varnothing\big)\right\}\,,$$

the set of elements of $A$ that have immediate successors in $A$. For $s\in S$ let $s^+$ be the immediate successor of $s$ in $S$. The intervals $(s,s^+)$ for $s\in S$ must contain distinct rationals, so $S$ is countable.

$\Bbb R$ is hereditarily separable, so $A$ has a countable subset $D$ that is dense in $A$ in the subspace topology on $A$. Suppose that $x,y\in A$ and $x<y$. If $(x,y)\cap A=\varnothing$, then $x\in S$, and $y=s^+$. Otherwise, $(x,y)\cap A$ is a non-empty open subset of $A$ in the subspace topology on $A$, so there is a $z\in D$ such that $x<z<y$.

This is the best that you can do along these lines unless you begin by requiring that $A$ be densely ordered.

Title: Countable Subsets of Irrational Sets


问:
Given a category $\mathcal{C}$, is $\mathcal{C} = (\mathcal{C}^{OP})^{OP}$ true?
The way I understand the definition of $\mathcal{C}^{OP}$, it is the case that $\mathcal{C}(X, Y) = \mathcal{C}^{OP}(Y,X)$ for every pair of objects $X, Y$.

Consider the category $(\mathcal{C}^{OP})^{OP}$, it has the same objects as $\mathcal{C}$, and for every pair of objects $X, Y$, $\mathcal{C}(X, Y)$ and $(\mathcal{C}^{OP})^{OP}(X, Y)$ coincide, having the same morphism and composition rules. I take $\mathcal{C}$ and $(\mathcal{C}^{OP})^{OP}$ to be the same.

But does $\mathcal{C} = (\mathcal{C}^{OP})^{OP}$ implies that every functor is covariant and contravariant at the same time? Does it imply that every covariant functor can also be viewed as a contravariant functor?

答:
Yes, $(\mathcal{C}^\mathrm{op})^\mathrm{op} = \mathcal{C}$.  This is a check of the definition.

Thus a covariant functor $F: \mathcal{C} \to \mathcal{D}$ may be regarded as a contravariant one $F: \mathcal{C}^\mathrm{op} \to \mathcal{D}$ and vice versa.  So, I suppose, yes, every functor may be viewed as contra- or covariant, but that requires a change of the domain, so this is not the "same" functor.


问:
Uniqueness of LDL decomposition of positive semi-definite matrices
Consider a real and positive semi-definite matrix $A \in \mathbb{R}^{n \times n}$ with rank $p < n$.

It is known from Wiki that $A$ admits a decomposition as
$$
A = L D L^\top,       \quad \quad\quad                      (1)
$$
where $L$ is lower unit triangular; $D$ is diagonal and the number of its non-zero diagonal entries equals the rank of $A$.

The question is that given $A$, if the $D$ matrix in the above decomposition is unique.

I think the decomposition, and thus the pair $(L,D)$, is not unique, because the columns of $L$ that correspond to the zero diagonal entires in $D$ can be modified freely without violating $(1)$. However, my interest here is the uniqueniess of $D$ only.

答:
you rarely have uniqueness in factorizations involving singular matrices, and your argument about $L$ not being unique is correct.  It is true, however, that $D$ is unique.

suppose

$A=L_1D_1L_1^* = L_2D_2L_2^*$

where $D_1$, $D_2$ are diagonal matrices and $L_1$, $L_2$ are lower triangular with 1's on the diagonal.

$D_2 = L_2^{-1}L_1D_1L_1^*(L_2^*)^{-1}=  \big(L_2^{-1}L_1\big)D_1\big(L_2^{-1}L_1\big)^{*} = R^*D_1R$

with $R:=\big(L_2^{-1}L_1\big)^{*}$ which is upper triangular with ones on the diagonal

$\implies D_2R^{-1}=R^*D_1$

the LHS is uppper triangular while the RHS is lower triangular, which in fact implies each side is diagonal.  And since $R$ has ones on the diagonal:

$\implies D_2= D_2R^{-1}=R^*D_1 =D_1$

alternatively, looking at $D_2R^{-1}=R^*D_1$ and examining component (k,k) of each side

$\implies d^{(2)}_{k,k} \cdot 1 =d^{(1)}_{k,k}\cdot 1 \implies  D_2 = D_1$


问:
What is the residue of a even function?
Suppose that the function $f(z)$ has an isolated singularity  at point $a\in \Bbb C$. Now suppose that $f(z)$ is even  in $(z-a)$, i.e.

$$
f(z-a) = f(-(z-a))
$$
Claim: the residue of $f(z)$ at $a$ is zero.

How i can do this ?? Any help is highly appreciated. Please help and give  a  descriptive  one.

答:
For simplicity, take $a=0.$ Then $f$ is analytic in $U=\{0<|z|<r\}$ for some $r>0,$ and we have $f(-z)=f(z)$ there. Let $\sum_{-\infty}^{\infty} a_nz_n$ be the Laurent expansion of $f$ in $U.$ Then

$$f(-z)=\sum_{-\infty}^{\infty} a_n(-1)^nz^n = f(z) = \sum_{-\infty}^{\infty} a_nz_n,\,\, z\in U.$$

By the uniqueness of the Laurent expansion, $a_n=a_n(-1)^n$ for all $n.$ In particular, $a_{-1}= a_{-1}(-1)^{-1} = -a_{-1}.$ Therefore $a_{-1}=0,$ which is the desired result.

Title: What is the residue of a even function?


问:
Compositions of equal a.e. functions with injective function need not be equal a.e.
Apparently there are measurable functions $f,g,h$ on $\mathbb{R}$ such that $f=g$ almost everywhere, $h$ is injective, but we do not have $f \circ h = g \circ h$ almost everywhere. What is an example of such functions?

答:
Prasiortle! My answer will rely upon some basic facts regarding the Cantor set $C.$

Consider $f := \chi_C,$ the indicator function of the Cantor set (since the Cantor set is measurable, so is $f$) and $g := 0.$ Since the measure of the Cantor set is zero, it follows that $f = g$ almost everywhere. Moreover, since the Cantor set is uncountable, there is some injective function $h: \mathbb{R} \rightarrow C,$ which can of course be viewed as a function $h: \mathbb{R} \rightarrow \mathbb{R}$ (by composing it with the natural inclusion of the Cantor set into the reals). Then it is clear that $g \circ h = 0.$ On the other hand, $f \circ h (x) = \chi_C(h(x)) = 1$ for all $x \in \mathbb{R}.$ It follows that it is not true that $f \circ h = g \circ h$ almost everywhere, so this is an example of the behaviour you were looking for. Hope you will find this helpful.


问:
$\sum_{n=1}^{\infty}(A_n+B_n)=\sum_{n=1}^{\infty}A_n+\sum_{n=1}^{\infty}B_n $
If $\sum_{n=1}^{\infty}A_n$ is convergent and $\sum_{n=1}^{\infty}B_n$ is divergent then can we write $\sum_{n=1}^{\infty}(A_n+B_n)=\sum_{n=1}^{\infty}A_n+\sum_{n=1}^{\infty}B_n $?

答:
The statement is false in the extended real line, by example suppose that $A_n:=n$ and $B_n:=(-1)^n$, then $\sum_{n\geqslant 0}(A_n+B_n)=\infty $ however $\sum_{n\geqslant 0}A_n+\sum_{n\geqslant 0}B_n$ is not defined as $\sum_{n\geqslant 0}B_n$ doesnt converge to something in the extended real line, therefore the equality would be false.

In the standard real line when both sides of an equality doesn't exists then the equality doesn't make sense, that is, the statement is as meaningless as if you write 172345g1j3kg123hu1263rt. You "can" write it (that is, you already had written it!!!) but it have no mathematical meaning, in short: it is not mathematics.


问:
$\lim \arcsin\bigl(f(x)\bigr)/f(x)$ for $f(x)$ that goes to $0$
this might be a trivial question for some of you but I'm really not sure about it.
(I can't use latex sorry)

For $f(x)\to >0$, $\lim \arcsin\bigl(f(x)\bigr)/f(x) = 1$? And mostly, why/why not?

Thank you all very much, have a nice day :)

答:
It's quite simple, with equivalence: near $0$, we have $\arcsin u\sim u$ which means exactly that $$\lim_{u\to 0}\frac{\arcsin u}u=1,$$
therefore, by substitution,
$$\lim_{f(x)\to 0}\frac{\arcsin\bigl(f(x)\bigr)}{f(x)}=1.$$


问:
Non-standard five-point formula for second derivative used in Tracker
Tracker is an open-source program used to analyze object trajectories from video. The typical data sets Tracker produces are time series of object positions at various times, i.e., data pairs (t_i,x_i) for $i=1$ to $N$. The time samples are assumed to have uniform time step $t_{i}-t_{i-1}=\Delta t$. Tracker then uses this data to generates estimates for the first and second derivatives, corresponding to the velocity $v=dx/dt$ and acceleration $a=d^2x/dt^2$ respectively. By default, it does this using the following finite difference schemes:

$$v_i = \frac{x_{i+1}-x_{i-1}}{\Delta t},\qquad  a_i = \frac{1}{7(\Delta t)^2}(2x_{i+2} - x_{i+1} - 2x_i - x_{i-1} + 2x_{i-2}) $$

The first is a standard two-point formula which requires no comment. However, the standard 5-point formula for the second derivative is

$$f''(x)\approx \frac{-f(x+2h)+16 f(x+h)-30 f(x)+16f(x-h)-f(x-2h)}{12h^2} \tag{1}$$
which has error of order $O(h^4)$. (Note that this formula is exact for quartic polynomials but not quintic.) The formula used by Tracker, by contrast, would correspond to $$f''(x)\approx \frac{2f(x+2h)- f(x+h)-2 f(x)-f(x-h)+2f(x-2h)}{12h^2}. \tag{2}$$ Both sides do match in the limit $h\to 0$, but the error is instead $O(h^2)$. So Tracker uses a scheme which, on the face of it, is not as precise.

Tracker's documentation doesn't address this point in detail, but does state the following: "Note: there are many other finite difference algorithms. Tracker's algorithms define the velocity for a step to
be the average velocity over a 2-step interval, and the acceleration to be the second derivative of a parabolic fit over a 4-step interval, with the step at the center. Tracker's acceleration algorithm is less sensitive to position uncertainties than others." The last sentence is especially notable. So my question as a whole is: What's the precise motivation behind Tracker's use of approximation (2), and in what sense (if any) is it "less sensitive to position uncertainties?"

答:
Let's try and fit a parabola
$$
x_i=c_0+c_1t_i+c_2t_i^2+\varepsilon_i
$$
and WLOG assume $\Delta t=1$ and $t_0=0$ for easier calculation.  Using ordinary least square regression (i.e., minimizing $\sum\varepsilon_i^2$), we need to solve the system
\begin{align*}
5 c_0 + 10 c_2 &=\sum_{i=-2}^2 x_i\\
10 c_1 &=\sum_{i=-2}^2 ix_i\\
10 c_0 + 34 c_2 &=\sum_{i=-2}^2 i^2x_i
\end{align*}
i.e.,
$$
\begin{pmatrix}
c_0\\c_1\\c_2
\end{pmatrix}
=
\begin{pmatrix}
\dots\\
\frac1{10}\sum_{i=-2}^2 ix_i\\
\frac1{14}\sum_{i=-2}^2 (i^2-2)x_i
\end{pmatrix}
$$
and the acceleration being $2c_2$ by the usual 'suvat' equations.  This gives the formula for $a$ as the Tracker doc claims.

Curiously, note that the velocity estimate $c_1$ is different here.  If we kept the two-point estimate of velocity in our fit, it would deliver a different estimate for acceleration.

Edit: After thinking more about it, "less sensitive to position uncertainties" could also mean the individual coefficients are smallest possible: $\lVert\alpha\rVert=\max_i\lvert\alpha_i\rvert$ is smallest over all $a=\sum_i \alpha_i x_i$ while still accurate for the case of uniform acceleration (so the error of a single reading contributes as little as possible).  Clearly we only need to consider $\alpha_i=\alpha_{-i}$ (otherwise we take the average of $\alpha$ and its reverse yielding something with no larger $\lVert\cdot\rVert$).  So $\alpha$ is an affine combination of $(\frac14,0,-\frac12,0,\frac14)$ and $(0,1,-2,1,0)$
\begin{align*}
\alpha&=(\frac14,0,-\frac12,0,\frac14)\lambda + (0,1,-2,1,0)(1-\lambda)\\
&=(\frac14\lambda,1-\lambda,\frac32\lambda-2,1-\lambda,\frac14\lambda)
\end{align*}
and so we must be either $\pm\frac14\lambda=1-\lambda$, $\pm\frac14\lambda=\frac32\lambda-2$ or $\pm(1-\lambda)=\frac32\lambda-2$ and it is easy to check these cases.


问:
On the degree of $F(ab)$ if degrees of $F(a)$ and $F(b)$ are relatively primes
Let $a,b$ be two algebraic elements over $F$. Suppose $[F(a):F]=m$ and $[F(b):F]=n$, with $m,n$ relatively prime. I think $[F(ab):F]=mn$, but I don't know how to prove it (I have $[F(ab):F]\leq mn$).

Any hint to prove the another inequality?

答:
The claim is false.

Let $F=\mathbb Q$.

Let $a$ be a root of $x^2+x+1$, so that $a$ is a primitive cubic root of unity.

Let $b=\sqrt[3]{2}$.

Then $[F(a):F]=2$ and $[F(b):F]=3$ but $[F(ab):F]=3$, not $6$.


问:
probability of non of the urns empty
So I have

$10$ balls, and $6$ urns

All of the balls and urns are numbered/labeled. This tells me that everything is unique.

The question is, what is the probability that none of the urns are empty.

Attempt

$Pr($None are empty$)= Pr($all are full$)=Pr($I not empty$)*Pr($II not empty$)*...Pr($VI not empty)
In a previous problem I found the probability of Urn $I$ to be empty as $\frac{5^{10}}{6^{10}}$ So the probability of Urn $I$ not empty is 1-$\frac{5^{10}}{6^{10}}$. Since the probability of any one urn to be empty is the same. its 6*(1-$\frac{5^{10}}{6^{10}}$). This doesn't feel right based on the answer solutions in this question appears to be similar however I am not sure how to extend that here.

I am not sure how to extend the answer solution there to this case either. Ball 1 has only 1 choices, urn $I$. Ball 2: has two choices($I, II) =>$ ($0.5$), ... Ball 6: =>$\frac{1}{6}$. However ball 7, 8, 9, 10 now has 6 choices. So does this mean that the $Pr($none are empty)=($\frac{1}{6!*6^3}$)

答:
By inclusion exclusion principle the probability in question is:
$$
\frac1{6^{10}}\sum_{i=0}^6(-1)^i\binom{6}i (6-i)^{10}\equiv\frac{6!}{6^{10}}{10\brace 6}\approx0.27,
$$
where ${m \brace n} $ is the Stirling number of the second kind.

Title: probability of non of the urns empty


问:
Find the limit $\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$
Find the limit if it exists

$$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$$

答:
Let's see the limit in the plane $z=0$. So,
\begin{align*}
     \lim_{(x, y, z)\rightarrow (0,0,0)} \frac{xy + yz + xz}{x^2 + y^2 + z^2} &= \lim_{(x, y) \rightarrow (0,0)} \frac{xy} {x^2 + y^2}
\end{align*}
Now analyzing by paths, we have to take the path $y=mx$, with $m>0$.
\begin{align*}
\require{cancel}
     \lim_{(x, y)\rightarrow (0,0)} \frac{xy}{x^2 + y^2} &= \lim_{x\rightarrow 0} \frac{mx^2}{x^2 + m^2x^2}\\
     & = \lim_{x\rightarrow 0} \frac{mx^2}{x^2(1 + m^2)}\\
     & = \lim_{x\rightarrow 0} \frac{m\cancel{x^2}}{\cancel{x^2}(1 + m^2)}\\
     & = \frac{m}{1 + m^2}.
\end{align*}
The above tells us that the limit is not unique, because it varies according to the value that $m$ takes. Therefore the limit $\displaystyle \lim_{(x, y, z) \rightarrow (0,0,0)}\frac{xy + yz + xz}{x^2 + y^2 + z^2}$ does not exist.


问:
Solving NH 2nd order ODE using Frobenius method
Quite stumped with this one so far. I have the following non-homogeneous ODE:

$$2x^2y''+3xy'-xy=x^2+2x$$

And I need to find a solution for $x_0<0$ using Frobenius. Obviously we can center the solution about $x_0=0$, and we can see it is a regular singular point based on the coefficients of $y'$ and $y$. But I searched quite a bit and couldn't find anything about the non-homogeneous part of the equation (every example that I find on here either is homogeneous, or the NH part is a constant). I know for $p(x)$ and $q(x)$ we check for RSP such that:

$$xp=x\left(\frac{3x}{2x^2}\right)=3/2$$
$$x^2q=x^2\left(\frac{-x}{2x^2}\right)=-x/2$$

Which are analytical for $x_0=0$, so no problem there. But I have no clue what to do with the non-homogeneous part of the equation to verify that it is really a RSP. Of course the problem would easily be solved simply with a regular power series, but I absolutely have to use Frobenius without using series to check for analyticity. Any pointers?

Apologies if the terminology is a bit off, english is not my mother tongue. Thanks in advance!

答:
Suppose that the particular solution is a polynomial
$$y_p=ax^2+bx+c$$
We have
$$y'=2ax+b;\;y''=2a$$
so the DE becomes
$$4ax^2+3 x (2 a x+b)-x \left(a x^2+b x+c\right)\equiv x^2+2 x$$
that is
$$-a x^3+x^2 (10 a-b)+x (3 b-c)\equiv x^2+2x$$
this means
$$
\begin{cases}
-a=0\\
10a-b=1\\
3b-c=2\\
\end{cases}
$$
which gives $a=0;b=-1;c=-5$
so the particular solution is $y_p=-x-5$

Added to the general solution we have
$$y=\frac{C_1 e^{\sqrt{2 x}}-C_2 e^{-\sqrt{2 x}}}{\sqrt{2 x}}-x-5$$


问:
Given nondegenerate symmetric bilinear form $f$, how is $L_f(x) = f(x,-)$ defined for $x = 0$?
Let $f$ be a nondegenerate symmetric bilinear form over an $n$-dimensional vector space $X$. For each nonzero vector $x \in X$, define the dual vector
$$f(x,-) \in X^{\prime}, \hspace{3mm} f(x,-): y \mapsto f(x,y).$$
Now we define $L_f:X \rightarrow X^{\prime}$ by $L_f: x\mapsto f(x,-)$.

Questions. Since $L_f$ is defined for all vectors in $X$, then what does $L_f$ do with $x = 0$? The only way I understand the definition of $L_f$ is that it maps nonzero vectors to $f(x,-)$, which makes sense because $f(x,-)$ was constructed using nonzero vectors. Is it somehow implied that $L_f$ maps $0$ by $0$?

答:
The map $L_f:X\to X'$ is linear, so it maps the zero vector of $X$ to the zero vector of $X'$, that is the zero linear map.

Another way to see this is to use linearity in $x$: For all $y\in X$, we have $f(0,y)=f(0+0,y)=f(0,y)+f(0,y)$. Substracting $f(0,y)$ on both sides yields $f(0,y)=0$.

PS. Your definition of $L_f$ works perfectly for $x=0$, I don't understand what is puzzling you.


问:
How can letters abcdef can be arranged so that a appears before b, c appears before d and e appears before f?
How can letters abcdef can be arranged so that a appears before b, c appears before d and e appears before f? If a general formula can be derived with this?

Note: a may be somewhere before b or c may be somewhere before d or e may be somewhere before f.....it need not to be immediate but will include cases where they are immediate before

答:
Place $a,b$ in a line with $a$ before $b$. There are now three places to put $c$ and $d$ (to the left of $a$, between $a$ and $b$ or to the right of $b$).

They can go into $1$ or $2$ of these places which can be chosen in
$3+\binom{3}{2}=6$ ways.

There are now five places for $e$ and $f$ which can be chosen in $5+\binom{5}{2}=15$ ways.

This gives us a total of $90$ possibilities.

I'm sure you can now see how to extend this to letters $g,h$ etc.


问:
Prove that a given function is not integrable over any interval $[a,b]$.
Consider the following function:

$$
f(x)=\begin{cases}
      0 & x\text{ is a rational number}, \\
      1 & x\text{ is an irrational number}.
   \end{cases}
\
$$

Show that the function $f(x)$ is not integrable over any interval $[a,b]$.

答: The function does not have left-hand and right-hand limits at each point, hence it is not regulated, hence it is not Riemann-integrable.</p>


答:
You can prove that $[F(ab):F]$ divides $nm$. Indeed, we have that $[F(a,b):F]=nm$, since
$$\begin{align*}
[F(a,b):F]&=[F(a,b):F(a)][F(a):F] = m[F(a,b):F(a)]\\
&\leq m[F(b):F] = mn,\\
[F(a,b):F]&=[F(a,b):F(b)][F(b):F] = n[F(a,b):F(b)]\\
&\leq n[F(a):F] = mn.
\end{align*}$$
So $n$ and $m$ both divide $[F(a,b):F]$, and since $\gcd(n,m)=1$, then $nm$ divides $[F(a,b):F]$.  Since $[F(a,b):F]\leq nm$, and is a multiple of $nm$, then $[F(a,b):F]=nm$.

Now we have that $ab\in F(a,b)$, hence $nm=[F(a,b):F] = [F(a,b):F(ab)][F(ab):F]$,
so $[F(ab):F]$ divides $nm$. But it need not be equal, as  lhf's example shows.

Title: On the degree of $F(ab)$ if degrees of $F(a)$ and $F(b)$ are relatively primes

Body:
Let $a,b$ be two algebraic elements over $F$. Suppose $[F(a):F]=m$ and $[F(b):F]=n$, with $m,n$ relatively prime. I think $[F(ab):F]=mn$, but I don't know how to prove it (I have $[F(ab):F]\leq mn$).

Any hint to prove the another inequality?


问:
Why define vector spaces over fields instead of a PID?
In my few years of studying abstract algebra I've always seen vector spaces over fields, rather than other weaker structures. What are the differences of having a vector space (or whatever the analogous structure is called[module?]) defined over a principal ideal domain which is not also a field? What properties of vector spaces break down when defining it over a PID? Please give an example (apart from letting the vector space be the field/PID itself) where the vector space over a field has properties that the module over a PID does not.</p>

答:
There is a very nice structure theory for finitely generated modules over a PID, which is frequently discussed here at MSE.

However, the general theory is much more complicated already in the case of (not finitely generated) abelian groups, i.e., $\Bbb{Z}$-modules.

The first thing that's missing is that not all modules over a PID will have a base. Take for instance the $\Bbb{Z}$-module $\Bbb{Z}/2\Bbb{Z}$. When dealing with vector spaces, the only invariant is the dimension. With modules over a PID, this is not the case anymore..


问:
8.8.32 - Mary L Boas

Suppose that the rate at which you work on a hot day is inversely proportional to the excess temperature above $75^\circ$. One day the temperature is rising steadily, and you start studying at $2$ p.m. You cover $20$ pages the first hour and $10$ pages the second hour. At what time was the temperature $75^\circ$?

My initial model for a differential equation is:
$$\frac{dw}{dt}=\frac{k}{\frac{dT}{dt}+75},$$
where $w$ goes for work. But I can't find the correct way to develop it to the right answer. Please help.

答:
For time $t$ hours after $2$pm, let the excess temperature above $75^0$ be $T+at$. (In other words $T$ is the excess temperature at $2$pm)

Then $$\frac{dw}{dt}=\frac{k}{T+at}$$
Integrating $$w=\frac{k}{a}\ln (T+at)+C.$$

Then $$20=\frac{k}{a}\ln (\frac{T+a}{T})\text { and } 30=\frac{k}{a}\ln (\frac{T+2a}{T})$$
and therefore $$(1+\frac {a}{T})^3=(1+\frac {2a}{T})^2.$$

Multiplying out and solving gives $\frac{a}{T}=1.618$, the Golden ratio.

We now know that the temperature above $75^0$ is $2.618T$ at 3pm and $4.236T$ at 4pm. In other words, a change of $1.618T$ takes an hour. So, for the temperature to rise from an excess of $0$ to an excess of $2.618T$ takes
$$\frac{2.618}{1.618}= 1.618\text { hours}.$$

The temperature was $75^0$ at a time of $1$ hours $37$ minutes before $3$pm i.e. at $1.23$pm.

Title: 8.8.32 - Mary L Boas


问:
Find $\cos (x)$ and $\tan (x)$ if $\sin (x) = 3/5$ over the interval $[\pi/2,\pi]$
Need help with this please. I seem to be making a mistake somewhere because my answer is different from the text book's. Mine are:

$\cos(x) = \frac{4}{5}$

$\tan(x) = \frac{3}{4}$

But in the text book it's:
$\cos(x) = \frac{-4}{5}$

$\tan(x) = \frac{-3}{4}$

I'm not sure where I went wrong.

My approach was to treat the ratio of $\sin(x)$ as the sides of a right angled triangle and solve for the final side using Pythagoras' theorem. But I don't know where the negative values are coming from. Please help if you can.

答: If $\theta\in\left(\frac\pi2,\pi\right]$, both $\cos\theta$ and $\tan\theta$ are smaller than or equal to $0$. So, you cannot possibly have $\cos\theta=\frac45$ and $\tan\theta=\frac34$ there.


问:
If a subset of the square grid can be tiled by $1\times n$ rectangles for every $n$, can it be tiled by infinite rays?
Suppose that we have some set $S$ of grid-aligned squares in the plane; equivalently, we can think of our set as $S\subset \mathbb{Z}^2$. Suppose that for every positive integer $n$, $S$ can be tiled by disjoint copies of a $1\times n$ rectangle (possibly rotated). Is it necessarily the case that $S$ can be tiled by congruent copies of an infinite "ray" of squares, i.e., a region congruent to the set $\{(k,0)\ |\ k\ge0\}$?

For example, the region given by removing from $\mathbb{Z}^2$ the points $(0,0), (1,1)$, and $(-1,2)$ can be tiled by every $1\times n$ rectangle, and also by infinite rays, as shown:

I'd like to say that this follows from some kind of compactness argument, but my attempts to apply such reasoning ran into trouble (I can't rule out that the $1\times n$ tilings have their endpoints move away from the origin without a well-defined limiting behavior). Any thoughts on how to prove this (or a counterexample, if my intuition is badly wrong)?

I believe the same statement should hold in $\mathbb{Z}^n$ with $1\times1\times\ldots\times1\times n$ tiles and the same kind of rays; if the original statement proves easy to show, I would be interested in this generalization.

As a remark, the converse is obviously true, since the infinite ray can be tiled by any $1\times n$ rectangle.

答:
Shortly after reading Edward H's answer, I realized that there is a self-contained compactness argument, which I missed the first time around; I thought I would present it here.

Let $T$ be an infinite set of polyominoes. Say that an infinite "polyomino" $P$ is an extension of $T$ if, for every finite region in the plane, its possible intersections with that region could also be an intersection of $t$, for infinitely many $t\in T$. (Intuitively, we're saying that local patches of $P$ look like part of a generic large element of $T$ - if you have a finite field of view, you can't distinguish it from infinitely many possibilities in $T$.)

Then, we claim that for every region $R$ which can be tiled by every element of $T$, $R$ has a tiling using extensions of $T$. (As a trivial consequence, at least one such extension must exist.)

In the original example, $T$ is the set of all $1\times n$ rectangles, and the extensions are $\{\text{infinite ray},\text{infinite line}\}$ - but since infinite lines can be tiled by infinite rays, we can just use the rays for everything.

Proof: For every positive integer $N$, let $S_N$ denote the square $[-N,N]\times[-N,N]$. Consider the restrictions of the tilings of $R$ to $R\cap S_n$ for each tile $t\in T$. Since there are finitely many ways to partition $R\cap S_n$ into tiles, but infinitely many tilings of $R$ by some $t\in T$, at least one of these partitions of $R\cap S_n$ must extend to a $t$-tiling of $R$ for infinitely many $t\in T$.

Now, fix our partition of $S_N\cap R$, and consider ways of extending it to a partition of $S_{N+1}\cap R$. Again, there are finitely many ways to do this, so again, of the infinitely many ways to extend to a $t$-tiling of $R$, infinitely many $t$ must cluster around one such partition. (The wording of this suggests a canonical choice of tiling for each $t\in T$, but we don't need to make this selection.)

Extending this construction inductively, we end up with a tiling of $R$ with tiles that have the property that every restriction of them to a finite square resembles (a part of) infinitely many tiles in $T$. Hence, all our tiles are extensions of $T$.

To show that this generality is actually good for something, here are some other special cases of the above theorem:

A region tileable by arbitrarily large squares can be tiled by infinite quadrants. (The possible extensions are quadrants, half-planes, and the whole plane, but all of these can be tiled by quadrants.)

A region tileable by arbitrarily long "zigzag" shapes can be tiled by an infinite staircase ray, i.e. the points $\{(x,y)\ |\ x,y\ge 0, x-y\in \{0,1\}\}$.

A region tileable by all "hooks" ($1\times n$ rectangles with an extra cell on the side of one end, e.g. :.......) can be tiled using an infinite ray and an infinite ray with a cell on the side of one end.

The same proof extends to most reasonable sort of grids - $\mathbb{Z}^n$, polyiamonds on the triangular grid, etc.


答:
Another approach:
$$2\pi i Res_a{f} = \int_{\partial B(a,r)}f(z) \\
=  \int_{\partial B(0,r)}f(z-a) \\
= \int_{\partial B(0,r)}f(-(z-a)) \\
= -\int_{\partial B(a,r)}f(z) \\
= -2\pi i Res_a{f}
$$

Title: What is the residue of a even function?

Body:
Suppose that the function $f(z)$ has an isolated singularity  at point $a\in \Bbb C$. Now suppose that $f(z)$ is even  in $(z-a)$, i.e.

$$
f(z-a) = f(-(z-a))
$$
Claim: the residue of $f(z)$ at $a$ is zero.

How i can do this ?? Any help is highly appreciated. Please help and give  a  descriptive  one.


问:
Lie Algebra of $SL_n(\mathbb H)$
The Lie algebra of $SL_n(\mathbb C)$ are the matrices where the trace is $0$. But what is the Lie algebra of $SL_n(\mathbb H)$ where $\mathbb H$ is the quaternions?</p>

答:
The Lie algebra of $\mathrm{GL}_n(\mathbb H)$ is just the full matrix space $M_n(\mathbb H)$ with commutator bracket. Viewed as Lie algebra this way, it's called $\mathfrak{gl}_n(\mathbb H)$).

One can show that this is a reductive Lie algebra. Since its centre $\mathfrak z$ is easily seen to be exactly the real scalar matrices, i.e. $\mathrm{diag}(a,a, ..., a)$ with $a\in \mathbb R$, its derived subalgebra

$$\mathfrak{sl}_n(\mathbb H):= [\mathfrak{gl}_n(\mathbb H), \mathfrak{gl}_n(\mathbb H)]$$

consists of those matrices where the real part of the trace is $0$. (Remember a quaternion $q = a +bi+ cj +dk$ has a real part $\mathfrak{Re}(x) := a$ and a pure quaternion part $bi+cj+dk$).

This has real codimension $1$ in $\mathfrak{gl}_n(\mathbb H)$ and corresponds to the Lie group $SL_n(\mathbb H)$ defined as the kernel of the Dieudonné determinant (1, 2).

This obviously avoids the problem in Mariano Suárez-Álvarez' answer. For example, the computation  $[ \begin{pmatrix}i&0\\0&-i\end{pmatrix}, \begin{pmatrix}j&0\\0&-j\end{pmatrix}] = \begin{pmatrix}2k&0\\0&2k\end{pmatrix}$ happens without problem in $\mathfrak{sl}_2(\mathbb H)$. (While the commutator of two matrices with trace $0$ need not have trace $0$ necessarily, the commutator of two matrices whose traces have real part $0$ again has trace with real part $0$, and that's all we need.)

This also corresponds to the construction in the Wikipedia article cited in John Baez' answer, but with the correction alluded to in my comment: When we write a matrix $A \in M_n(\mathbb H)$ as matrix
$$\tilde{A}= \pmatrix{X & -\overline{Y}\\Y & \overline{X}} \in M_{2n}(\mathbb C)$$ (with $X, Y \in M_n(\mathbb C)$), then the condition to be in $\mathfrak{sl}_n(\mathbb H)$ is not $Tr(X) = 0$ as claimed there, but rather $\mathfrak{Re}(Tr(X))=0$ or equivalently $Tr(\tilde{A})=0$.

Note 1: It's worthwhile to check all this already in the non-trivial basic case $n=1$. We have $\mathrm{GL}_1(\mathbb H) = \mathbb H^*$, $\mathrm{SL}_1(\mathbb H) =$ the norm-1-quaternions (which is a real Lie group $\simeq SU_2$), and $\mathfrak{gl}_n(\mathbb H) = \mathbb H$ which splits, stable under commutators, as

$$ \mathbb H  = \underbrace{\mathbb R}_{\text{centre}} \oplus \underbrace{\text{ pure quaternions }}_{\text{real Lie algebra } \simeq \mathfrak{su}_2 (\simeq \mathfrak{so}_3)}$$

And when written as complex matrices $\pmatrix{\alpha & -\bar{\beta}\\\beta& \bar{\alpha}}$, the subspace of pure quaternions is indeed that with $\alpha + \bar\alpha = 0$ or equivalently $\mathfrak{Re}(\alpha)=0$, not just the ones with $\alpha=0$.

Note 2: All this generalizes in the best way possible to a base field $k$ (to be safe, $char(k)=0$ although much goes through at least if for $char(k)$ does not divide $n$) and a $k$-central division algebra $D$ (of degree $d:= \sqrt{\dim_k D}$ say). Then again one defines the Lie algebra $\mathfrak{sl}_n(D)$ as the derived subalgebra of the full matrix algebra $\mathfrak{gl}_n(D)$, which is of $k$-codimension $1$. It turns out to be an absolutely simple $k$-Lie algebra of type $A_m$, where $m = dn-1$, of $k$-rational rank $n-1$, with Satake-Tits diagram

For more details, see sections 4.4 and 4.5.2 of my thesis. (Note the errata for page 93.)

It matches the group case as presented in Tits, Jacques (1966), "Classification of algebraic semisimple groups", Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965), Providence, R.I.: American Mathematical Society, pp. 33–62

And of course for the Hamilton quaternions, we just have $d=2$ and the above collapses to the second row in Onishchik / Vinberg's table 4 (they just exclude the compact form $\mathfrak{sl}_1(\mathbb H)$, which in their notation would be $p=0$ and as said above is just $\simeq \mathfrak{su}_2$, with diagram one black dot):

Title: Lie Algebra of $SL_n(\mathbb H)$


问:
Relation between Ax=0 and Ax=b

I'm not sure how to solve this. My teacher told me that if $A\vec{x}=0$ has one unique solution (the trivial) then $A\vec{x}=\vec{b}$ has only one unique solution. But I don't know how to prove this.

答:
Hint

Case of a square matrix $A$

By the rank nullity theorem $A$ injective if and only if it’s kernel is reduced to the zero vector if and only if $A$ is surjective.

Case where $Ax=b$ has no solution

Take
$$A=\begin{pmatrix}
1\\
1\end{pmatrix} \text{ and }b=\begin{pmatrix}
1\\
2\end{pmatrix}$$

Title: Relation between Ax=0 and Ax=b


答:
It is not just one way. Modules over $\mathbb Z$ are abelian groups, and have a rich structure theory, involving the existence of finite modules, and prime factorisation.

Passing to $\mathbb Q$ as field of fractions we find that every scalar is a unit and every non-trivial vector space is infinite.

So modules can have properties which vector spaces don't share.

Title: Why define vector spaces over fields instead of a PID?

Body: In my few years of studying abstract algebra I've always seen vector spaces over fields, rather than other weaker structures. What are the differences of having a vector space (or whatever the analogous structure is called[module?]) defined over a principal ideal domain which is not also a field? What properties of vector spaces break down when defining it over a PID? Please give an example (apart from letting the vector space be the field/PID itself) where the vector space over a field has properties that the module over a PID does not.</p>


答:
We know that

$ sin(x) ^2+cos(x)^2=1$$\Rightarrow$ $cos(x) =\sqrt{1-sin(x)^2} $or $- \sqrt{1-sin(x)^2}$

So

$cos(x) =\frac{4}{5}$ or $- \frac{4}{5}$

We know that $x\in [\pi, \frac{\pi} {2}]$ $\Rightarrow$$ cos(x) =<0 $

so $cos(x) =- \frac{4}{5}$

$Tan(x) =-\frac{3}{4}$

Title: Find $\cos (x)$ and $\tan (x)$ if $\sin (x) = 3/5$ over the interval $[\pi/2,\pi]$

Body:
Need help with this please. I seem to be making a mistake somewhere because my answer is different from the text book's. Mine are:

$\cos(x) = \frac{4}{5}$

$\tan(x) = \frac{3}{4}$

But in the text book it's:
$\cos(x) = \frac{-4}{5}$

$\tan(x) = \frac{-3}{4}$

I'm not sure where I went wrong.

My approach was to treat the ratio of $\sin(x)$ as the sides of a right angled triangle and solve for the final side using Pythagoras' theorem. But I don't know where the negative values are coming from. Please help if you can.


答:
A useful result on linear systems is the following:

Let $K$ be a field,  $A$ the matrix of a linear map from $K^n$ to $K^p$, $\:b\in K^p$.

The linear equation $Ax=b$ ($x\in K^n$) has a solution if & only if the matrix $A$ and the augmented matrix $[A|b]$ have the same rank.
 Furthermore, this common rank is the codimension of the affine subspace of solutions.

Title: Relation between Ax=0 and Ax=b

Body: I'm not sure how to solve this. My teacher told me that if $A\vec{x}=0$ has one unique solution (the trivial) then $A\vec{x}=\vec{b}$ has only one unique solution. But I don't know how to prove this.


问:
Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

I did this by creating 3 sets:

$S\,\,=\{a,b,c,d\}$

$T\,\,=\{c,d,e,f\}$

$T'=\{a,b,c,d\}$

And then I performed the operations and checked they are the same. Is there something else I should do or only this will suffice?

答:
A different approach: let $x$ be any element in the universe. The following equivalences hold:

$$\begin{align} x\in S\cap (T\cup T') &\iff x\in S\land x\in T\cup T'\\
&\iff x\in S\land (x\in T\lor x\in T')\\
&\iff (x\in S\land x\in T)\lor (x\in S\land x\in T')\\
&\iff x\in S\cap T\lor x\in S\cap T'\\
&\iff x\in (S\cap T)\cup (S\cap T').\end{align}$$


问:
What is the probability of drawing 2 cards and having one card be higher or lower than the other?
I'm creating a Casino game for math class. My game's rules are, the dealer deals one "community" card and one house card. Players can bet on the community card being higher, lower, or the same as the house card but I'm having trouble finding the probability of the cards being higher or lower.

This is the equation I used to come up with an answer:

$$ P(\text{Same value}) = \frac{3}{52} = 5.7\% $$

$$ P(\text{Higher}) = \frac{(100 - P(\text{Same value}))}{2} = 47.15\% $$

When I think about it though, I'm not sure whether or not drawing the first card would affect the outcome.

答:
First, as Math Lover noted, the probability for drawing the same value should be $\frac{3}{51}$, since you've already drawn the first card. The numerator decreases to 3, since 3 of that first card remain in the deck, and the denominator decreases to 51, since there are 51 total cards remaining in the deck.

As for the second question, your approach works because there's an equal probability for each possible first card drawn. The probability of higher/lower will change if you know what card is drawn, but if you don't know, then the probabilities average out because you're just as likely to draw a high card as a low card first.

Let $X_1$ represent the first card drawn, and $X_2$ the second, and let $x$ represent the observed value of $X_1$, where $x = 1$ is the lowest possible card value (A or 2, typically), and $x = 13$ the highest possible card value (K or A, typically).

Now, once we've observed $x$, we also know that there are $4(x-1)$ cards with a lower value and $4(13-x)$ cards with a higher value. Then, we have the following:

$$ P[X_2 > x] = \frac{4(13-x)}{51} , $$
$$ P[X_2 < x] = \frac{4(x-1)}{51} . $$

These two values obviously aren't the same, so the probability of higher/lower depends on that first draw. If $x$ is a low card, then the probability of a higher card on the second draw is higher. Similarly, if $x$ is a high card, then the probability of a higher card on the second draw is lower.

Now, prior to that first draw, we have to allow for all possible values it could be. We do this by summing over all possible outcomes for that first draw (using the Law of Total Probability):

$$ P[X_2 < X_1] = \sum_{x=1}^{13} {\left(P[X_2 < X_1 | X_1 = x]*P[X_1 = x]\right)} $$
$$ = \sum_{x=1}^{13} {\left(P[X_2 < x]*P[X_1 = x]\right)} $$
$$ = \sum_{x=1}^{13} {\left(\frac{4(x-1)}{51}*\frac{4}{52}\right)} $$
$$ = \frac{4}{51*13}\sum_{x=1}^{13} {(x-1)} $$
$$ = \frac{4*78}{51*13} $$
$$ = \frac{24}{51} $$

Similarly,

$$ P[X_2 > X_1] = \sum_{x=1}^{13} {\left(P[X_2 > X_1 | X_1 = x]*P[X_1 = x]\right)} $$
$$ = \sum_{x=1}^{13} {\left(P[X_2 > x]*P[X_1 = x]\right)} $$
$$ = \sum_{x=1}^{13} {\left(\frac{4(13-x)}{51}*\frac{4}{52}\right)} $$
$$ = \frac{24}{51} . $$

So, before you know whether the first card is high or low, there's an equal probability that you'll end up with a higher second card as there is with a lower second card - this stems from the fact that the first card itself could be high or low. Once you've seen that first drawn card, the probabilities change because you have more information.


问:
How many 10-digit numbers have at least 2 equal digits
Please help go solve this question. I tried it solving, by various methods. But nothing gave a positive result.</p>

答:
A ten digit number without any equal digits is a permutation of $0,1,2,3,4,5,6,7,8,9$.

Assuming the first digit cannot be zero, there are $9\times 9!$ such permutations.

In total there are $9\times 10^9$ ten digit numbers so you can easily calculate the probability.


问:
Fourier Transform in polar coordinates of 1
Like in the table of transforms https://en.wikipedia.org/wiki/Fourier_transform#Distributions,_one-dimensional the FT (Fourier transform) of $\delta$ is 1 and the FT of 1 is $\delta$, but in polar coordinates the expressions for transformations is quite different, but like in section 4 of this review https://www.researchgate.net/publication/241069465_Two-Dimensional_Fourier_Transforms_in_Polar_Coordinates
the Dirac delta for polar coordinates is $\frac{1}{r}\delta(r-r_0)\delta(\theta-\theta_0)$ and your FT is $e^{-i\vec{w}.\vec{r}_0}$, I am interested in Dirac delta function at the origin (subsection 4.1 of the review), this delta is $\frac{\delta(r)}{2\pi r}$ and your FT is 1, but the FT of 1 is $\frac{\delta(r)}{2\pi r}$?

My question arose because the FT in polar coordinates for radially symmetric functions is
$$\mathbb{F}(f(r))=2\pi\int^{\infty}_0f(r)J_0(\rho r)dr=2\pi\mathbb{H}(f(r))$$
where $\mathbb{H}$ is the Hankel transform.

But, when f(r)=1, the integral above have solution only for finite intervals.

The table https://en.wikipedia.org/wiki/Hankel_transform#Some_Hankel_transform_pairs have the Hankel transform of 1 and it is $\frac{\delta(r)}{r}$, this is the only place where I had some answer, but I am not convinced. So my questions are, the FT in polar coordinates of 1 is $\frac{\delta(r)}{2\pi r}$? How prove it?

答:
This was mostly proved in my answer here.

One has to be a bit careful when manipulating distributions (i.e. generalized functions). The first remark is that in general, the multiplication of two distributions is not defined (one can only define the multiplication of distributions with smooth functions). However, one can still look case by case. Here for example, one can see the distribution $f(r) := \,"\frac{\delta_0}{2πr}"$ as the solution of the equation
$$
2\pi\, r\, f(r) = \delta_0.
$$
However, this equation have several solutions (see again here). If we choose the only homogeneous distribution that is solution, we find (see again here for proof)
$$
f(r) = -\frac{1}{2\pi}\,\delta_0'(r)
$$
where $\delta_0'$ is the derivative of the Dirac delta distribution. Again in the same post, I also proved that $f(r)$ is the radial representation of the Dirac delta $\delta_0$, an so since the Fourier transform of $FT(1)=\delta_0$ (in cartesian coordinates), its (homogeneous) radial representation is $f(r)$.


问:
Is trace( inv(A) B ) a distance?
Given two symmetric p.d. matrices of dimension $(p \times p)$, $\mathbf{A}$ and $\mathbf{B}$, can we interpret
$$
d = \operatorname{trace}(\mathbf{A}^{-1} \mathbf{B})
$$
as a distance between the two?

For this it would obviously have to be the case that $d$ is minimized if $\mathbf{A} = \mathbf{B} \Rightarrow d = p$. Is this the case?

答:
The answer is no. Consider the case $A,B$ are diagonal positive definite matrices. It is easy to make $d$ arbitrarily small/arbitrarily close/equal to $p$ without the matrices being close/equal.

To make this clear, let $x_1,..., x_p$ be arbitrary positive numbers with $x_1+...+x_p=p$. Set $A=I, B= \mbox{diag}(x_1,...,x_p)$.

Then $d(A,B)=p$.

Title: Is trace( inv(A) B ) a distance?


问:
A possible proof of the existence of a prime in the interval [n,2n].
Consider the theta function, $\theta(x)$=$\sum_{p\leq x} \log p$.

Lemma: $\theta(x)$ is strictly monotone increasing.

We thus have that $\theta(x)$<$\theta(2x)$. Note also that $\theta(2x)$=$\theta(x)$+$\sum_{p<q\leq2x} \log p$. So there is a q in between x and 2x. Assume there did not exist such a q. Then by the monotone condition, one has $\theta(x)$<$\theta(2x)$=$\theta(x)$+$\sum_{p<q\leq2x} \log p$. But by assumption, $\sum_{p<q\leq2x} \log p$=0. Thus we have that $\theta(x)$<$\theta(x)$. A contradiction. Thus we have a prime q, such that x<q$\leq$2x.

My only dilemma is the proof of the monotone condition. I can't seem to find it. But once I do, then this would imply Bertrand's postulate. I am sorry if I am not very good with these types of things. I just tried my best to find the most elementary proof of this conjecture.

So I would like to know if; 1) my proof is correct, 2) help with the proof of the lemma.

EDIT: The above lemma is false. Let me restate it:

Lemma: $\theta(x)$<$\theta(2x)$.
Now I need help with this lemma. :/

答:
The correct formula is $$\theta(2x) = \theta(x) + \sum_{x<p\le 2x}\log p$$
Then $\theta(2n) > \theta(n)$ is equivalent to say that there is a prime in $(n,2n]$. However, $\theta(x)$ is NOT strictly increasing: $\theta(7)=\theta(8)=\theta(9)=\theta(10)$.


问:
What specifically is wrong with this naive approach to solve Hilbert's 10th problem over the rationals?
By MDPR, c.e. sets are the same as Diophantine sets. An existential definition of $\Bbb Z$ in $\Bbb Q$ would solve Hilbert's 10th problem over $\Bbb Q$.

The idea is not so crazy, you just input a pair $(a,b)$ of integers (I guess a pair of pairs of naturals, but that's a detail) and if $b \vert a$ you halt and if it doesn't you run forever. So the pairs $(a,b)$ for which the algorithm halts are rational numbers that are actually integers and if it doesn't they're rational numbers that are not integers.

N.B.
This is so simple I know I cannot be correct. Many people much more experienced than me have tried and failed with many more complicated methods. This is more of a request to explain why this is wrong.

What I am guessing could be my incorrect assumption:

You can just give an algorithm that distinguishes between integers and rationals and by MDPR the c.e. set this algorithm produces is a Diophantine set and hence a Diophantine definition of $\Bbb Z$ in $\Bbb Q$.

You can sneak the integers into my algorithm by defining rationals as pairs of integers and not worry about circularity.

答:
If I understand your idea correctly, you're not actually defining $\mathbb{Z}$ in $\mathbb{Q}$. You're defining $\mathbb{Z}$ in a structure $\tilde{\mathbb{Q}}$, which intuitively consists of $\mathbb{Q}$ together with the details of its construction via ordered pairs of integers. This is unavoidable when you try to check whether $a\vert b$ in a rational $q=(a,b)$: you need to be able to first "unpack" the rational into an ordered pair, and second check whether there is an integer (not a rational) which when multiplied by the left coordinate yields the right coordinate.

Now this absolutely works, but it doesn't help define $\mathbb{Z}$ in $\mathbb{Q}$ at all, let alone in a $\Sigma_1$ way. The structure $\tilde{\mathbb{Q}}$ is, when we dive into the details, $\mathbb{Z}$ itself (or perhaps more accurately, $\mathbb{Z}$ together with some inessentialy "syntactic sugar")! And an existential definition of $\mathbb{Z}$ in $\mathbb{Z}$ isn't very surprising.

It may help at this point to look at a context where we know such an idea must fail - for example, the field $\overline{\mathbb{Q}}\cap\mathbb{R}$ of algebraic real numbers. As a real closed field this is decidable, and so cannot define $\mathbb{Z}$ at all, but it can be implemented in $\mathbb{Z}$ in a manner similar to the implementation of rationals as (equivalence classes of) ordered pairs of integers.

I think the source of this confusion is a reflexive identification of computations with $\Sigma_1$-definitions. The fact that we can do this over $\mathbb{Z}$ (or similar) is a highly nontrivial property of $\mathbb{Z}$, and we're not guaranteed anything similar about $\mathbb{Q}$; in fact, this is essentially what you're trying to prove in the first place!

For this reason it's important to grapple with $\Sigma_1$ definitions themselves, not intuitive analogues. A $\Sigma_1$ definition of $\mathbb{Z}$ in $\mathbb{Q}$ can't "unpack" an element of $\mathbb{Q}$ into some explicit implementation in a different context, nor can it in any way "search" over natural numbers or similar objects.


问:
Maximum matching for general graph
I am studying the maximum matching problem and I was trying to understand why the classical augmenting path algorithm does not work for the general graph (i.e. for non bipartite graph) and you must recur to the blossom algorithm.

The explanations that I found (1. here, 2. and here) are based on the fact that the direction that I choose to start exploring an odd loop in the Depth First Search (DFS) could make me miss an augmented path.

But it seems to me that this issue can be solved by considering the same vertex distinct during the DFS if I am entering it through a matched edge or an unmatched one, why isn't this enough?

答:
I posted the question on the computer science sub as well (is that legal?) and ending finding the solution at the end:
https://cs.stackexchange.com/questions/135278/maximum-matching-for-general-graph/135336#135336</p>

Title: Maximum matching for general graph