问:
Monty Hall: What is the "unconditional probability of success" and "conditional probability of success"?
Can someone explain the highlighted paragraph from Blitzstein, Introduction to Probability (2019 2 edn), p 69? I'm not understanding the subtlety the author is trying to get at. What is the "unconditional probability of success"? How is it not conditional because we're conditioning on what door the car is behind to solve the law of total probability equation. What is the "conditional probability of success"? Are "unconditional probability of success" and "conditional probability of success" the same thing?

For the normal Monty Hall problem, it seems that the "unconditional probability of success" and "conditional probability of success" are the same; they output the same number (2/3). But they're different in op. cit. Exercise 40, p 91.

Consider the Monty Hall problem, except that Monty enjoys opening Door
2 more than he enjoys opening Door 3, and if he has a choice between
opening these two doors, he opens Door 2 with probability p, where
$\frac{1}{2} \leq p \leq 1$.

(a) Find the unconditional probability that the strategy of always switching succeeds (unconditional in the sense that we do not condition on which of Doors 2,3 Monty opens).

(b) Find the probability that the strategy of always switching succeeds, given that Monty opens Door 2.

(c) Find the probability that the strategy of always switching succeeds, given that Monty opens Door 3.

The answer is 2/3 for (a) because $0 \cdot 1/3 + 1 \cdot 1/3 + 1 \cdot 1/3 = 2/3$ by LOTP. Why this answer is the same as the normal Monty Hall problem, where Monty enjoys all three doors equally?

The answer for (b) is $\frac{1}{p+1}$ and for (c) it's $\frac{1}{2-p}$.

答:
The difference between unconditional and conditional is that for unconditional, we are conditioning on C, and for conditional, we are conditioning on M. Simply put, the thing being conditioned on is different.

You start by picking door 1, it seems as the problem describes.

```<code> |
 |
 v
 _   _   _
| | | | | |
| | | | | |
 -   -   -
 1   2   3
</code>```

The unconditional property only depends on C, the event that the car is behind door C, so we have the following, where "car" really means "get car by switching." You'll notice that this has no information about M. That is why it is the same no matter the probability that the host chooses door 2. It's not in the formula. It only depends on where the car was initially placed. Maybe this is not a satisfactory explanation, but let's move on.

$$P(car)=P(car|C_1)P(C_1)+P(car|C_2)P(C_2)+P(car|C_3)P(C_3)\\
=0*\frac 1 3+1*\frac 1 3+1*\frac 1 3=\frac 2 3$$

You want to find the probability that switching succeeds, given that Monty opens door 2. Let's draw a tree. Maybe take a minute or so to figure out why it is.

```<code>                                      prob
                      / car     0     0
          / M2  p
                      \ no_car  1     p/3
/ C1  1/3
                      / car     0     0
          \ M3  1-p
                      \ no_car  1     (1-p)/3

                      / car     1     1/3
- C2  1/3 - M3  1
                      \ no_car  0     0

                      / car     1     1/3
\ C3  1/3 - M2  1
                      \ no_car  0     0
</code>```

You want $P(car|M_2)=\frac{P(car, M_2)}{P(M_2)}=\frac{P(car,M_2,C_1)+P(car,M_2,C_2)+P(car,M_2,C_3)}{P(M_2,C_1)+P(M_2,C_2)+P(M_2,C_2)}=\frac{0+0+1/3}{p/3+0+1/3}=\frac{1}{p+1}$

Similarly, you want $P(car|M_3)=\frac{P(car, M_3)}{P(M_3)}=\frac{P(car,M_3,C_1)+P(car,M_3,C_2)+P(car,M_3,C_3)}{P(M_3,C_1)+P(M_3,C_2)+P(M_3,C_2)}=\frac{0+1/3+0}{(1-p)/3+1/3+0}=\frac{1}{2-p}$

Lastly, we go back to verify that the "conditional probability" of getting the car conditioning on M is the same as the "unconditional probability" of getting a car conditioning on C, which was 2/3 as determined in part 1. We need the law of total probability $P(car)=P(car|M_2)P(M_2)+P(car|M_3)P(M_3)=P(M_2)\frac{1}{p+1}+P(M_3)\frac{1}{2-p}$.

Convince yourself with the following diagram that the $P(M_2)=\frac 1 3+\frac p 3$. That is, again $P(M_2)=P(M_2,C_1)+P(M_2,C_2)+P(M_2,C_3)=P(M_2|C_1)P(C_1)+P(M_2|C_2)P(C_2)+P(M_2|C_3)P(C_3)=1*\frac 1 3+0*\frac 1 3+p*\frac 1 3=\frac 1 3+\frac 1 3 p$. Alternatively, you could have used the huge tree above. Since opening door 3 is the complement to opening door 2, $P(M_3)=1-P(M_2)=1-\left(\frac 1 3+\frac 1 3p\right)$. Now you can finally find the "conditional" probability that you win a car by switching:

$$\left(\frac 1 3+\frac 1 3p\right)\frac{1}{p+1}+\left(1-\left(\frac 1 3+\frac 1 3p\right)\right)\frac{1}{2-p}=\frac 2 3$$

```<code>  _
|[ ]| [ ] [c]
|[ ]| [c] [ ]
|[c]| [ ] [ ]
  -
</code>```

It is the same as the "unconditional" probability conditioning on C instead of M.

Body:
The difference between unconditional and conditional is that for unconditional, we are conditioning on C, and for conditional, we are conditioning on M. Simply put, the thing being conditioned on is different.

You start by picking door 1, it seems as the problem describes.

 |
 |
 v
 _   _   _
| | | | | |
| | | | | |
 -   -   -
 1   2   3

The unconditional property only depends on C, the event that the car is behind door C, so we have the following, where "car" really means "get car by switching." You'll notice that this has no information about M. That is why it is the same no matter the probability that the host chooses door 2. It's not in the formula. It only depends on where the car was initially placed. Maybe this is not a satisfactory explanation, but let's move on.

$$P(car)=P(car|C_1)P(C_1)+P(car|C_2)P(C_2)+P(car|C_3)P(C_3)\\
=0*\frac 1 3+1*\frac 1 3+1*\frac 1 3=\frac 2 3$$

You want to find the probability that switching succeeds, given that Monty opens door 2. Let's draw a tree. Maybe take a minute or so to figure out why it is.

                                      prob
                      / car     0     0
          / M2  p
                      \ no_car  1     p/3
/ C1  1/3
                      / car     0     0
          \ M3  1-p
                      \ no_car  1     (1-p)/3

                      / car     1     1/3
- C2  1/3 - M3  1
                      \ no_car  0     0

                      / car     1     1/3
\ C3  1/3 - M2  1
                      \ no_car  0     0

You want $P(car|M_2)=\frac{P(car, M_2)}{P(M_2)}=\frac{P(car,M_2,C_1)+P(car,M_2,C_2)+P(car,M_2,C_3)}{P(M_2,C_1)+P(M_2,C_2)+P(M_2,C_2)}=\frac{0+0+1/3}{p/3+0+1/3}=\frac{1}{p+1}$

Similarly, you want $P(car|M_3)=\frac{P(car, M_3)}{P(M_3)}=\frac{P(car,M_3,C_1)+P(car,M_3,C_2)+P(car,M_3,C_3)}{P(M_3,C_1)+P(M_3,C_2)+P(M_3,C_2)}=\frac{0+1/3+0}{(1-p)/3+1/3+0}=\frac{1}{2-p}$

Lastly, we go back to verify that the "conditional probability" of getting the car conditioning on M is the same as the "unconditional probability" of getting a car conditioning on C, which was 2/3 as determined in part 1. We need the law of total probability $P(car)=P(car|M_2)P(M_2)+P(car|M_3)P(M_3)=P(M_2)\frac{1}{p+1}+P(M_3)\frac{1}{2-p}$.

Convince yourself with the following diagram that the $P(M_2)=\frac 1 3+\frac p 3$. That is, again $P(M_2)=P(M_2,C_1)+P(M_2,C_2)+P(M_2,C_3)=P(M_2|C_1)P(C_1)+P(M_2|C_2)P(C_2)+P(M_2|C_3)P(C_3)=1*\frac 1 3+0*\frac 1 3+p*\frac 1 3=\frac 1 3+\frac 1 3 p$. Alternatively, you could have used the huge tree above. Since opening door 3 is the complement to opening door 2, $P(M_3)=1-P(M_2)=1-\left(\frac 1 3+\frac 1 3p\right)$. Now you can finally find the "conditional" probability that you win a car by switching:

$$\left(\frac 1 3+\frac 1 3p\right)\frac{1}{p+1}+\left(1-\left(\frac 1 3+\frac 1 3p\right)\right)\frac{1}{2-p}=\frac 2 3$$

  _
|[ ]| [ ] [c]
|[ ]| [c] [ ]
|[c]| [ ] [ ]
  -

It is the same as the "unconditional" probability conditioning on C instead of M.


问:
Prove that for any integer $n>0$ which is a perfect square, $n+4$ is not a perfect square.
I have been told that this proof is incorrect, however, I'm having a hard time seeing the issue.

Prove that for any integer $n > 0$ which is a perfect square, $n+4$ is not a perfect square:

Proof:

$n > 0$

Since $n$ is a perfect square, $a^2=n$

The subsequent perfect square can be written as ${b^2=\left(a+1\right)}^2$

The difference between consecutive squares is then ${|b}^2-a^2|=|2a+1|$

$\forall a\in\mathbb{Z}(|2a+1|\neq 4)$

$\square$

Edit: Thanks everyone, appreciate it :)

答: Your proof/argument shows that if $n$ is a square, then the next square, namely $\ n+2\sqrt{n}+1,\ $ is not equal to $\ n+4\ $. You haven't in any way shown that $\ n+4\ $ is not a square number. You've just shown that it's not the next square number after $\ n$.

ContentLicense: CC BY-SA 4.0


问:
Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$

$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?

If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.

答:
The relationship between $x,y,z$ and the side of the triangle, $a$ is interesting:

$$3(x^4+y^4 + z^4 + a^4) = (x^2+y^2+z^2 + a^2)^2 \tag{1}$$

Please see here, near reference $19$

If you put $z^2 = x^2 + y^2$ in $(1)$, you'll get
$$3(x^4 + y^4 + (x^2+y^2)^2 + a^4) = (2x^2 + 2y^2 + a^2)^2$$

Solving for $a$, you get

$$a = \pm \sqrt{x^2+y^2 \pm\sqrt{3}xy}$$

Consider only the positive roots.


问:
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots

Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.

But I wonder that there is some logical ways we can find intervals that contain roots without guessing?

答:
Given $f(x)=x^5-5x^3+4x-1$, differentiate first:

$$ f'(x) = 5x^4 - 15x^2 +4 . $$

Setting the derivative equal to $0$, we have

$$ 0 = 5x^4 - 15x^2 +4 . $$

This is quadratic in $x^2$, so we can use the quadratic formula:

$$ x^2 = \frac{15 \pm \sqrt{145}}{10} . $$

Since $\sqrt{145}<15$, $x^2>0$, so our solutions are

$$ x = \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}} . $$

So, we have four distinct local extrema for $f$. If there are five distinct roots, each must be on one of the following intervals:

$$ \left(-\infty, -\sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15+\sqrt{145}}{10}}, -\sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15+\sqrt{145}}{10}}, \infty\right) . $$


问:
$\text{Inn}(G)$ cannot be nontrivial cyclic group

Let $G$ be any group, and let $Z$ be its center.

(a) Show that $G/Z\cong \text{Inn}(G)$.

(b) Conclude that $\text{Inn}(G)$ cannot be a nontrivial cyclic group.

I've already gotten part (a) by considering the mapping $\pi:G\rightarrow\text{Inn}(G)$ such that $\pi(g)$ is the automorphism that takes $x$ to $g^{-1}xg$ for all $x\in G$. The mapping $\pi$ is clearly a surjective homomorphism with kernel $Z$, and part (a) follows from the isomorphism theorem.

For part (b), I must prove that $G/Z$ cannot be a nontrivial cyclic group. If it were, the group would equal $\{Z,Zg,Zg^2,\ldots,Zg^{n-1}\}$ for some $g\in G$. Also, $G/Z$ would be an abelian group, and it follows that the commutator subgroup $G'$ belongs to $Z$. I don't see how to derive a contradiction from there.

答: From $G/Z$ cyclic you can get something stronger: Let $a,b \in G$, then $a = z_1g^n$ and $b = z_2g^m$ for some $z_1,z_2 \in Z$. Now $ab = z_1g^n z_2g^m = z_2g^mz_1g^n = ba$. Thus $G$ is abelian, therefore $G = Z$. What now?</p>

ContentLicense: CC BY-SA 3.0


问:
Independence of random variable comparisons
If $X$, $Y$, and $Z$ are independent continuous random variables with equal distributions, are the two events $X\leq Y$ and $X\leq Z$ independent?

I know that $$P(X\leq Y)P(X\leq Z)=\left(\int_{-\infty}^{\infty}{P(X\leq x)f_X(x) dx}\right)^2$$
but I can't figure out how to solve $P(X\leq Y,X\leq Z)$ to check for independence.

答:
Well, $$\mathsf P(X\leq Y)\mathsf P(X\leq Z) =\left(\int_{-\infty}^\infty\int_s^\infty f(s)f(t)\mathrm d t\mathrm d s\right)^2$$

$$\mathsf P(X\leq Y,X\leq Z) =\int_{-\infty}^\infty\int_s^\infty\int_s^\infty f(s)f(t)f(u)\mathrm d u\mathrm d t\mathrm d s$$

But you might have more luck with:

$$\begin{align}\mathsf P(X\leq Y) ~&=~\mathsf P(X\leq Y\leq Z)+\mathsf P(X\leq Z\leq  Y)+\mathsf P(Z\leq X\leq Y)\\\mathsf P(X\leq Z) ~&=~\mathsf P(X\leq Y\leq Z)+\mathsf P(X\leq Z\leq  Y)+\mathsf P(Y\leq X\leq Z)\\\mathsf P(X\leq Y, X\leq Z) ~&=~ \mathsf P(X\leq Y\leq Z)+\mathsf P(X\leq Z\leq Y)\end{align}$$

Also, since the variables $X,Y,Z$ are iid, they are interchangeable, so ...


问:
How do I estimate log10 of log10 of 8 billionth element of A000670?
How do I estimate log10 of log10 of the 8 billionth element of A000670
as an actual number? I believe it's between 10 and 100, but am not
sure why I believe this.

I realize http://oeis.org/A000670 offers some approximations of
A000670 (in terms of n), but nothing that quickly leads to an actual
value of log10(log10(a(n)))

Goal: I'm trying to calculate the number of possible "prejudices" in
the world, and it's the nth element of A000670, where n is the current
world population. I don't need an exact number, just something like
10^(10^r) [to the nearest integer value for r], and I'm estimating n at about 8 billion right now.

Ideally, I'd like a good general estimate of log10(log10(a(n))) but
that may be asking too much.

EDIT: It's easy to show A000670 is bounded by 2^n*n! (aka http://oeis.org/A000165), but this bound is loose and unhelpful.

http://oeis.org/A000670 notes "Unreduced denominators in convergent to log(2) = lim[n->inf, na(n-1)/a(n)].". Does this mean that: n!*log(2)^n would be an estimate, at least in terms of order of magnitude? It seems a bit low. Since log(2) < 1 this can't be right. Maybe

```<code>n!/(log(2)^n)
</code>```

答:
See for instance Wilf's generatingfunctionology, p. 175, Example 5.2.1 "Ordered Bell Numbers" for the asymptotic
$$a(n) = n! \left(\frac{1}{2 \ln(2)^{n+1}} + O(0.16^n)\right).$$

It just arises from looking at the "leading singularity" of the exponential generating function. This approximation is particularly excellent since the first correction grows exponentially while the remaining correction decays exponentially.

Heuristically, Stirling's approximation says $\log_{10} \log_{10} (8 \cdot 10^9)!$ is going to be on the order of the exponent, namely 9--in fact, Mathematica tells me it's 10.879...--and the corrections are only going to contribute perhaps a few percent since even exponential growth is nothing compared to the growth of $n!$, so the nearest integer is definitely going to be 11.

We can be more precise though and actually prove it. Singularity analysis more fully gives

$$a(n) = \frac{n!}{2} \sum_{k=-\infty}^\infty \frac{1}{(\ln 2 + 2 \pi i k)^{n+1}},$$

the $k=0$ term of which gives the asymptotic above. Hence

$$\frac{1}{n!}\left|a(n) - \frac{1}{2 \ln(2)^{n+1}}\right| \leq \sum_{k=1}^\infty |\ln 2 + 2\pi i k|^{-(n+1)} < \sum_{k=1}^\infty (2\pi k)^{-(n+1)} = \frac{\zeta(n+1)}{(2\pi)^{n+1}}$$

where $\zeta$ is the Riemann zeta function. This says

$$\log_{10} \log_{10} a(8 \cdot 10^9) = \log_{10} \log_{10} (8 \cdot 10^9)! \left(\frac{1}{2 \ln(2)^{8 \cdot 10^9+1}} \pm \frac{\zeta(8 \cdot 10^9 + 1)}{(2\pi)^{8 \cdot 10^9}}\right).$$

Mathematica is happy to compute $\log_{10} \log_{10}$ of this expression, say with 40 digits of precision. I've included some related expressions too.

So, yup, the answer is 11.


问:
$\textit{Why}$ the cases in 4 $\nmid n^2 - 3?$ (bases cases in modular induction)
This question has already been answered here. My question is, when we suppose the hypothesis is not true (proof by contradiction), namely that $4 \mid (n^2 - 3)$, we get $n^2 - 3 = 4k$ for some integer $k$ and then we have two cases. But where does the two cases come from? In addition, why does putting them into two cases work to prove the statment? My thought process would be to try to prove this statement by contradiction and somehow figure out a way to produce a contradiction similar to how $\sqrt{2}$ is proved by contradiction. Any help is great!

Let me know if I need to clarify more!

答:
But why the cases, and where do the two cases come from? In addition, why does putting them into two cases work to prove the statment?

Since these specific questions were not addressed in prior answers, we explicitly address them here. In this type of induction (by modular cases) there are generally multiple base cases corresponding to each possible remainder $\bmod n$. Let's bring to the fore the logical essence of matter.

Congruence $\!\bmod 4\,$ is an equivalence relation on $\,\Bbb Z\,$ whose classes partition $\,\Bbb Z\,$ into $\,4\,$ cosets (arithmetic progressions) $\,r_i + 4\Bbb Z$, $\,r_i \in \{0,1,2,3\}.\,$ The naturals in each coset have the same order structure as $\Bbb N,\,$ so we can use induction to prove that a statement is true for all elements in the coset. If we prove that for all $4$ cosets ($4$ case analyses) then that suffices as a proof that the statement is true for all naturals since, the union of the cosets $= \Bbb Z \supset \Bbb N$.

Sometimes we can give a single proof that works uniformly for all cosets, but generally we may need a (brute force) case analysis, examining each of the $\,n\,$ possible cosets, e.g. when $\,n = 2\,$ this amounts to a proof by parity  using two base cases even & odd. A prototypical example, is the Parity Root Test, which shows that an integer coefficient polynomial has no integer roots if it has no roots $\bmod 2$, essentially performing a parity analysis on possible roots, i.e. if it has no even root and no odd root then it has no integer root, e.g. if $\,n\,$ is odd then $\,x^2+x+n\,$ has no integer roots because it is always odd $\,f(0)\equiv 1\equiv f(1)\pmod{\!2}.\,$ The same idea works $\!\bmod n,\,$ i.e. an integer coef polynomial $\,f(x)\,$ has no integer roots if it has no roots $\!\bmod n\,$ i.e. if $\,f(0),f(1),\ldots,f(n-1)\,$ are all $\not\equiv 0.\,$ The OP follows from the special case $\,f(x) = x^2-3,\ n = 4\,$ (the common proof you linked is essentially just a slight optimization of this modular root test).

As in the proof of this root test, in practice said induction in each coset is not usually explicitly done because typically we employ other results that help to simplify the matter - so the induction in the cosets actually ends up being hidden somewhere down the line in another theorem or lemma. For example, in the proof of the root test it boils down to $\,n\equiv (n\bmod 2)\in \{0,1\}\,$ by the Division Algorithm (computing the least element (remainder) in each coset is essentially induction in that coset). [Down the line it also implicitly uses (structural) induction in the proof of the Polynomial Congruence Rule (to essentially decompose the polynomial into compositions of sums and products) but this is not the modular case induction that we are concerned with here].


问:
Cross product query.
In cross product, we do it like: $\mathbf{a} \times \mathbf{b}  = |a||b|\sin(\theta)$

From where does this $\sin(\theta)$ came from? Can someone please derive the cross product and explain it.

答:
Definition

The cross-product $\vec{u}\times \vec{v}$ is defined as the vector normal to the plane spanned by $\vec{u},\vec{v}$ with a magnitude equal to the area of the parallelogram formed by these two vectors (see above image). You may have meant to say $||\vec{a}\times\vec{b}||=ab\sin\theta$ (where $a,b$ are the respective vector magntiudes) as the expression on the RHS is a scalar quantity.

Proof

We know that the area of a parallelogram is $A=bh$ where $b$ is the length of one base and $h$ is the altitude to the other vector endpoint. From the diagram above, we see that $u$ is the length of a base and $v\sin\theta$ is the altitude to $\vec{v}$'s endpoint. We could have done this the other way by swapping $v$ for $u$. Thus, the area of this parallelogram, and therefore the magnitude of $\vec{u}\times\vec{v}$, is $uv\sin\theta$.

Title: Cross product query.


问:
Showing differentiability of the dot product
Let $f: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ be defined as the dot product $f(a,b) = a \cdot b$. During the course of the proof, I have reduced the expression of the argument in the limit definition of differentiability to the following:

\begin{align*}
\frac{\left|{f(a+h_1, b + h_2}) - f(a,b) - Df(a,b)(h_1,h_2)\right|}{\|h\|}
&= \frac{\left|h_1 \cdot h_2 \right|}{||h||}\\
&\leq \frac{\|h_1|| \|h_2||}{\|h\|}
\end{align*}
The goal is to show that the LHS is bounded by something that converges to $0$ as $h$ goes to $0$. However, I can't figure out how to further bound the inequality, or how to show that the presented inequality tends toward $0$. Also, since $h \in \mathbb{R}^n \times \mathbb{R}^n$ is considered a 2-tuple of vectors, how does one interpret the norm of $h$ in the denominator?

Body:
Write
$$h_1 = (h_{1,1},...,h_{1,n})$$
and
$$h_2 = (h_{2,1},...,h_{2,n}).$$
Thus,
$$\|(h_1,h_2)\|^2 = \left(\sqrt{h_{1,1}^2 + \cdots + h_{1,n}^2 + h_{2,1}^2 + \cdots + h_{2,n}^2}\right)^2 = \|h_1\|^2 + \|h_2\|^2,$$
from where
$$\frac{\|h_1\|^2 + \|h_2\|^2}{\|(h_1,h_2)\|} = \|(h_1,h_2)\|.$$


问:
What is the degree of an n-fold branched cover over a trefoil?
The order-2 cyclic branched cover over a trefoil has degree 6, meaning the preimage of any point off the trefoil has cardinality six. (You can find a wonderful video of this here, made by Moritz Sümmermann.) The order-3 cyclic branched cover over a trefoil has degree 24.

What is the formula for the degree of an order-$n$ cyclic branched cover over a trefoil?

(I'm unsure of the proper terminology for this.)

I'm pretty sure this is $|G_n|$, where
$$G_n=\langle x,y\mid xyx=yxy,~x^n=y^n=1\rangle;$$
however, I have no idea how to find the size of this group.

答:
The group
$$
G_n = \langle x,y \mid xyx=yxy,\, x^n=1,\, y^n=1\rangle
$$
is infinite for all $n\geq 6$.

To see this, observe first that the third relation $y^n=1$ follows from the first two, since by the first relation $y=(xy)x(xy)^{-1}$ and hence $y^n=(xy)x^n(xy)^{-1}=1$.  Thus
$$
G_n = \langle x,y\mid xyx=yxy,\, x^n=1\rangle.
$$

Now, it is well-known that the braid group $B_3=\langle x,y\mid xyx=yxy\rangle$ can also be presented as $\langle a,b\mid a^2=b^3\rangle$, where $a=xyx$ and $b=xy$.  Since $x=b^{-1}a$, it follows that
$$
G_n = \big\langle a,b \;\bigl|\; a^2=b^3,\,(b^{-1}a)^n = 1\big\rangle
$$
Now consider the following quotient of $G_n$:
$$
Q_n = \big\langle a,b \;\bigl|\; a^2=b^3=1,\,(b^{-1}a)^n = 1\big\rangle
$$
(It doesn't affect the argument, but $\langle a,b \mid a^2=b^3=1\rangle$ is a presentation for the modular group $\mathrm{PSL}(2,\mathbb{Z})$, which is a quotient of $B_3$.)

What does the Cayley graph of $Q_n$ look like?  If we treat $a$ edges as undirected, then the Cayley graph is the 1‑skeleton of a regular tiling of a simply connected surface by $2n$-gons corresponding to $(b^{-1}a)^n$ and triangles corresponding to $b^3$, with two $2n$-gons and one triangle meeting at every vertex.

If we try to make the polygons regular and Euclidean, then each $2n$-gon has angles of $\pi\bigl(1-\frac{1}{n}\bigr)$ and each triangle has angles of $\pi/3$, so the total angle at each vertex is
$$
2\pi\biggl(1-\frac{1}{n}\biggr) + \frac{\pi}{3} = 2\pi\biggl(\frac{7}{6}-\frac{1}{n}\biggr).
$$
For $n<6$ this sum is less than $2\pi$, so the Cayley graph of $Q_n$ is the 1‑skeleton of a tiling of the sphere, and hence $Q_n$ is finite.  Indeed, the Cayley of $Q_n$ for $n=2$, $n=3$, $n=4$, and $n=5$, are respectively the 1‑skeleta of a triangular prism, a truncated tetrahedron, a truncated cube, and a truncated dodechedron.  Zeno Rogue's computer code shows that $G_n$ is finite in these cases as well.

For $n=6$, the sum is equal to $2\pi$, so the Cayley graph of $Q_n$ is the 1‑skeleton of the truncated hexagonal tiling of the Euclidean plane by equilateral triangles and regular dodecagons.  For $n>6$, the sum is greater than $2\pi$, which means that the Cayley graph of $Q_n$ is the 1‑skeleton of a tiling of the hyperbolic plane.  For example, when $n=7$ this is the truncated heptagonal tiling of the hyperbolic plane. In particular, $Q_n$ is infinite for all $n\geq 6$, and hence $G_n$ is as well.


问:
If $x+y=2$ then show that $x^2y^2(x^2+y^2)≤2$

Let $x$ and $y$ be two positive real numbers such that $x+y=2$. Then show that $$x^2y^2(x^2+y^2)≤2,$$

In  $x+y=2$ implies $x^3y^3(x^3+y^3)≤2$, for positive $x$ and $y$.  we can see that $t=xy$ solves pretty simple. But here we get
$$t^2(4-2t) \leq \Big({t+t+4-2t\over 3}\Big)^3 = {64\over 27}$$
which is greater than $2$.

Edit: Ahh, true it is not that much harder than I thought at first. Since $$0\leq t =xy\leq \Big({x+y\over 2}\Big)^2 =1$$
we have $$(t-1)(t^2-t-1)\geq 0$$
which is equivalent to $$t ^2(4-2t)\leq 2$$

Body:
Let $x=1+k$ and $y=1-k$, for some $k\in(-1,1)$. Then
\begin{align}
x^2y^2(x^2+y^2)&=(1-k^2)^2\times2(1+k^2)\\
&=2(1-k^2)(1-k^4)\\
&\le 2
\end{align}
Further, equality holds iff $k=0$, i.e. $x=y=1$.


问:
Gödel's incompleteness wrt weakend versions of ZFC
Suppose for the sake of argument that we look at ZFC with the axiom of infinity removed.

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#First_incompleteness_theorem

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem

We would then be in a position where the hypotheses of Gödel's theorems are not satisfied, correct? Basically, I want to remove, for the sake of argument, a minimal amount of axioms of ZFC so that ZFC minus some axiom(s) leads to a theory that does not include arithmetical truths and is not capable of expressing elementary arithmetic.

Is it possible that ZFC- (my shorthand for ZFC with some axiom(s) removed) is consistent and complete?

答:
ZF minus the axiom of infinity plus a new axiom negating the axiom of infinity is bi-interpretable with Peano Arithmetic. And the two equivalent theories are both subject to Gödelian incompleteness.

For the equivalence claim, see Richard Kaye and Tin Lok Wong's paper On interpretations of arithmetic and set theory (which incidentally explores some of the complications alluded to in the answer by @zyx).


问:
Convergence or divergence of $\int_0^2 \frac{\sin^2x}{x^p}\,dx$
I am starting to learn a little about analysis and convergence of Riemann integrals. I came across this problem and it stomped me. I want to find the values of $p$ for which the integral converges.

$$\int_0^2 \frac{\sin^2x}{x^p}\,dx$$

My approach was noting that: $\sin^2x \le \vert \sin x\vert$ and that $\vert \frac{\sin x}{x^p}\vert \le \frac1{x^p} $ then $\vert \frac{\sin^2 x}{x^p}\vert \le \frac1{x^p} $

Integration by parts yields:

$$\int_0^2 \frac{\sin^2x}{x^p}\,dx= \left[ \frac{x^{-p}}{2} \left(x-0.5\sin{2x} \right) \right]_0^2+\frac p2 \int_0^2 x^{-p}-\frac {x^{-p-1}}{2}\sin {2x}\,dx $$

$$= \frac1{2^p}\left( 1 - \frac {\sin4}{2}\right)+\frac p2\left[ \frac{2^{-p+1}}{-p+1} \right]_0^2- \frac p4\int_0^2 x^{-p-1}\sin {2x}\,dx$$

$$= \frac1{2^p}\left( 1 - \frac {\sin4}{2}\right) +\frac {p}{2^p\left(-p+1\right)}-\frac p4\int_0^2 x^{-p-1}\sin {2x}\,dx$$
$$$$
But since $\vert x^{-p-1}\sin{2x} \vert \le x^{-p-1}$ and $p+1 \gt 1$, then $\frac p4\int_0^2 x^{-p-1}\sin {2x}\,dx$ is convergent because $\int_0^2 x^{-p}\,dx$ convergent for $p \gt1$

So I arrived at the conclusion that for $p \gt1$, $\int_0^2 \frac{\sin^2x}{x^p}\,dx$ converges, since all of the terms to the right are finite and the integral on the right also convergent. On the other hand I checked with Wolfram Alpha and  it says $\int_0^2 \frac{\sin^2x}{x^3}\,dx$ does not converge, yet $p=3 \gt 1$. So where did I go wrong?

答:
Notice that
\begin{align}
|\sin x | \le |x|
\end{align}
when $|x|$ is small. In particular, we see that
\begin{align}
\int^2_0 \frac{\sin^2 x}{x^p}\ dx \le \int^2_0 \frac{dx}{x^{p-2}}<\infty
\end{align}
provided $p-2<1$ or $p<3$.

Suppose $p\ge 3$. I need to show that the integral is divergent to complete the problem. In fact, it suffices to show that the following integral is divergent
\begin{align}
\int^{\pi/2}_0 \frac{\sin^2 x}{x^{p}}\ dx
\end{align}
since $\pi/2<2$ and the integrand is non-negative.

Let us make the observation that
\begin{align}
\frac{2}{\pi}x \le \sin x
\end{align}
on the interval $[0, \pi/2]$ (draw a picture to convince yourself). This means that
\begin{align}
\int^{\pi/2}_0 \frac{\sin^2 x}{x^p}\ dx \ge \frac{4}{\pi^2}\int^{\pi/2}_0 \frac{dx}{x^{p-2}} = \infty
\end{align}
provided $p\ge 3$.


问:
Tangent planes and why the normal vector pointed in the opposite direction?

These two methods are giving us the same tangent plane but why did we get normal vectors pointed in the opposite direction?

答: It all comes down to the boundary defined (clockwise or counterclockwise) and then you choose the direction.</p>


问:
What happens to an equality constraint in the primal when you translate to the dual?
\begin{align}
\text{maximize}   &\quad   x_1  +3x_2 &     &    \\
\text{subject to} &\quad   x_1  - x_2   =   2 \\
& -2x_1 +3x_2  \ge 5 \\
&  \quad\quad\quad x_2  \ge  0
\end{align}

I was able to find the dual LP, but I am not seeing any connections to the original LP. Can I get some pointers?

Find the dual LP.

I first converted the LP to standard form then introduced a new variable.

Let $x_1=r_1-r_2$
\begin{align}
\text{minimize}   &\quad   2r_1-2r_2  -5x_2 &     &    \\
\text{subject to} &\quad   r1-r_2+2x_2\geq      1 \\
& -r1+r_2-2x_2\geq      -1 \\
&   -r1+r_2-3x_2\geq    3
\end{align}

What happens to an equality constraint in the primal when you translate to the dual?

What happens to a free variable in the primal when you translate to the dual?

答:
An equality constraint in the primal corresponds to a free variable in the dual formulation.

A free variable in the primal corresponds to an equality constraint in the dual formulation.

$$\max x_1+3x_2$$

subject to $$x_1-x_2=2$$
$$-2x_1+3x_5 \ge 5$$
$$x_2 \ge 0$$

We can associate a dual variable $p$ to the first equation and $q$ to the second constraint. $p$ is a free variable.

Since $x_1$ is a free variable in the primal formulation, the corresponding constraint in the dual variable in an equality which is $p-2q=1$.


问:
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

答:
Here is another way:

Define $f(x)=x-\sqrt{x}-\sqrt{x+1}$. We need to show that $f(x)\ge 0$ for all $x\ge 5$.
Since
$$f'(x)=1-\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x+1}} \ge 0, \quad \forall x\ge 1$$
the function is increasing on $[1,\infty)$. As $f(5)\ge 0$, the result follows.

Title: Show that $n \ge \sqrt{n+1}+\sqrt{n}$


问:
Can you always treat variables as constants in partial derivatives/integrals?
Ok so imagine you are given an expression: $xy : x=y$, & you need to take the partial derivative w.r.t. $x$ but you don't simplify (i.e. substitute x=y) --> $d/dx(xy)=y$.

Now you've gotten a different answer than you would have gotten had you simplified/substituted. But in math you should only be able to get 1 correct answer regardless of simplification. So why does standard partial derivative procedure in this trivial example yield 2 seemingly 'correct' answers?

P.S. Ditto question for partial integrals & cases where $x=y + \epsilon: \epsilon\approx 0$.

答:
Consider $f:\mathbb{R}^2\to\mathbb{R}$ with $f(x,y)=xy$. Also consider the function $g:\mathbb{R}\to\mathbb{R}^2$ with $g(x)=(x,x)$. You can then define the composition $h=f\circ g:\mathbb{R}\to\mathbb{R}$
$$
h(x)=f(g(x))=f(x,x)=x^2
$$

While understandable, it is confusing to write $\frac{d}{dx}f(x,x)$. What you actually have is the composition of two functions.

The derivative of the function $h$ is $h'(x)=2x$.

On the other hand, the partial derivative with respect to $x$ for $f$ is
$$
\frac{\partial}{\partial x}f(x,y)=y.
$$


问:
Prove there exists a real number $x$ such that $xy=y$ for all real $y$

Prove: There exists a real number $x$ such that for every real number $y$, we have $xy=y.$

In class I learned that I can prove a statement by:

proving the contrapositive,

proof by contradiction,

or proof by cases.

Can I do something along the lines of $xy=y$
divide both sides by $x$ to get $y=\frac{y}{x}$
which would then be equal to $y=\frac{y}{x}=xy$
divide everything by $y$ to get $1=\frac{1}{x}=x$
so $x=1$?

There exists a real number $x$ such that for every real number $y$, we have $xy=x$

I know I somehow have to prove $x=0$ right? But I am not sure how to go about this.

答: I am pretty sure that you can just pick x=1 in the real numbers. so for every real number y there exists xy=y because x=1 implies y=y which is true for every real number.</p>


问:
prove $ \mathcal F(f) = c_1\delta + c_2 \delta' + c_3\delta'' + T_g $ with $f(x)=|x^2 -1|$
let $f(x)=|x^2 -1|$ be a tempered distribution (i proved it) , and calculated its 3rd derivation (as a distribution) and then this stopped me :

prove that we have :

$$ \mathcal F(f) = c_1\delta + c_2 \delta' + c_3\delta'' + T_g $$

with $T_g$ is a Tempered distribution and $c_1,c_2,c_3 \in \mathbb R$ and $<\delta,\phi>= \phi(0)$ (the dirac distribution) , calculate $c_1, c_2 , c_3 .$

i tried to do the following :

they asked me before this to calculate $(f)'''$ (as a distribution) , and i thought there will be a connection between  $(f)'''$ and the fact that $\mathcal F(f^{(m)})(x)= (2i\pi x)^m. \mathcal F(f(t))$ soo i calculated $(f)'''$ and i got :

$$ f''' = 4 \delta'_{-1} + 4\delta'_1 - 4\delta_{-1} + 4\delta_1$$

soo we have :

$$ \mathcal F(f''')=(2i\pi x)^3 . F(f) $$
$$ \Rightarrow  F(4 \delta'_{-1} + 4\delta'_1 - 4\delta_{-1} + 4\delta_1) = (2i\pi x)^3 \mathcal F(f)$$

and i stopped here

答:
Hint: you can calculate the derivatives directly, without resorting to Fourrier transform. Just apply the following formula:

If $g$ is distribution represented by piece-wise $\mathcal C^1$ function on $\mathbb R$ with jumps at points $x_n<x_{n+1}$, then its derivative is equal to
$f'= \sum_n \{f(x_n)\}\delta_{x_n} + \sum_n \left(f\large|_{(x_n,x_{n+1})}\right)'$,
where $\{f(x_n)\}$ is a jump of $f$ in the point $x_n$.

It is known as formule des sauts (formula of jumps) in French but I don't know its name in English.

For example, if you have a distribution $g = \begin{cases}-x,&x\le 0,\\1,& x>0\end{cases}$, then by this formula $g'= \delta_0  -\mathbf{1}_{x\le 0}$.

In your case, the function is continuous, then its first derivative can be expressed directly

$f'= 2x\,\mathbf{1}_{x<-1} +2x\,\mathbf{1}_{x>1} -2x\,\mathbf{1}_{|x|\le 1}$. You can continue with Leibnitz's rule and this jump formula until you arrite to $f''' = 4(\delta_1'+\delta_1-\delta_{-1}+\delta'_{-1})$.

Now we need to apply the Fourier transform to these functions: it's well-known that $\mathcal F(\delta_a) = e^{2\pi i a\xi}$ and $\mathcal F(\delta'_a) = -2\pi i \xi e^{2\pi i a\xi}$ (signs and constants may vary depending on the conventions in the definition of the transform). Hence we get

$\mathcal F(f''') = 4(e^{2\pi i \xi}-e^{-2\pi i  \xi}-2\pi i \xi e^{2\pi i  \xi} -2\pi i \xi e^{-2\pi i  \xi}) = 8i\sin (2\pi\xi) - 16\pi i\xi \cos(2\pi\xi) = -8\pi^3i\xi^3\mathcal F(f).$

Now, whenever we solve in the sense of distributions the equation $xT=S$, we obtain $T= c\delta_0 + \frac 1x S$ with $c$- an arbitrary constant (it's a good exercise to see why we have this family of solutions). Similarly, $x^3T=S$ would produce $\delta'$ and $\delta''$, because $\{\delta^{(k)}_0\}_{k=0..n-1}$ are linearly independent and generate all solutions of $x^nT=0$.

Hence, we can write $\mathcal F(f) =\frac{  \sin (2\pi\xi) - 2\pi \xi  \cos(2\pi\xi) }{ - \pi^3 \xi^3 }+c_0\delta +c_1\delta'+c_2\delta''$. Apparently, the first term is a continuous, even, and integrable function on $\mathbb R$.

In order to deal with constants $c_i$, we take well-chosen test functions $\varphi$. It's known, that a function and its Fourier transform are even/odd simultaneously. So, if $\varphi$ is odd, then $\mathcal F(\varphi)$ is odd and this function and its second derivative are both zero in $\xi=0$.  We now can write $0=<f,\varphi>=c_1<\delta',\mathcal F(\varphi)>$. This assures that $c_1=0$.

The final step involves the following statement: if $f$ is even, then $\mathcal F(\mathcal F(f))=f$. Hence, $\mathcal F\left(   \frac{  \sin (2\pi\xi) - 2\pi \xi  \cos(2\pi\xi) }{ - \pi^3 \xi^3 }+c_0\delta   +c_2\delta''\right) = g(x)+c_0-4\pi^2 c_2 x^2$ where $g\in \mathcal C\cap\mathcal L^2\cap\mathcal L^\infty$ and $g(x)\to 0$ as $|x|\to\infty$ because it's an image of a $\mathcal C^{\infty}$ function. As we know the behaviour of $f$ on infinity, we conclude that $c_0=-1$ and $c_2 = -\frac{1}{4\pi^2}$.


问:
An example for finding an explicit definition of "structure maps"
By googling, I found only What is meant by a “structure map”?. But I could not understand that for my example as follows: Let $L$ be a Lie algebra over $K$, and view $K$ as a trivial $L$-module (that is, define $x \dot a=0$ for all $x \in L$, $a \in K$). In other words, we have a trivial structure map. What does it mean when that $\rho : L \to K$ is a structure map? Does it mean that $\rho$ is just a Lie algebra homomorphism?

答:
I don't think I've ever seen this in the context of Lie algebras; only like this for associative algebras, where the homomorphism is a ring homomorphism and the map is going the other way. See Wikipedia, but essentially, for an associative $K$-algebra $A$, the $K$-action on $A$ can be defined by a ring homomorphism $\phi:K \to A$ where for $k \in K$ and $a \in A$ the action is $k.a = \phi(k)a$.

See this too for a more general consideration of "structure maps".


问:
Identifying a certain quotient with a familiar group
Let $a,b\geq 2$ be integers and $N$ the image of the homomorphism
$$\mathbb{Z}_a\rightarrow \mathbb{Z}_{ab}, \;\;\;[k]_a\mapsto[kb]_{ab}$$
Identify the quotient group $\mathbb{Z}_{ab}/N$ with a familiar group.

I have done quite a bit of thinking about this problem, and it seems that the best strategy would be to try and find an isomorphism from $\mathbb{Z}_{ab}/N$ to another group such that the kernel is $N$. The other group should be a familiar group. If this is the correct approach, I'm not sure how to proceed. If it isn't, I would greatly appreciate any hints/guidance.

答:
Let $\varphi$ be the homomorphism. We have that
$$\Bbb{Z}_a/\ker \varphi \simeq N.$$
Take some $k \in \Bbb{Z}_a$, then $[k]_a \in \ker\varphi$ iff $[kb]_{ab} = [0]_{ab}$.

Now, finding $\ker\varphi$ is easy to get $N$ and so $\Bbb{Z}_{ab}/N$.


问:
Existence of solutions for $x(t)=\sin(t)+\int_0^t\frac{x(s)}{\sqrt{s}}ds $
Use the fixed point theorem (or it's generalization) to show there is a continuous function  (unique???) $x:[0,1]\to \mathbb{R}$ such that $$x(t)=\sin(t)+\int_0^t\frac{x(s)}{\sqrt{s}}ds $$

Comment:  We usually define an operator on the space $\mathcal{C}[0,1]$, and using the sup norm $\|\cdot\|_{\infty}$ we prove that for $x,y\in\mathcal{C}[0,1]$, we have  $ |Tx(t)-Ty(t)|\leq L\|x-y\|_{\infty}$, if $L<1$ the theorem guarantees the existence of this unique solution, however the term $\frac{1}{\sqrt{s}}$ near to $0$ makes it difficult. Should I use other theorem rather then this?

If I do this $$|Tx(t)-Ty(t)| = \left|\int_0^t\frac{x(t)-y(t)}{\sqrt{s}}ds \right|\leq \|x-y\|_\infty\int_0^t\frac{1}{\sqrt{s}}ds \leq 2\|x-y\|_{\infty}$$

答:
Let us define the mapping $\Phi:C[0, 1]\rightarrow C[0, 1]$ by
\begin{align}
\Phi(u)(t)= \sin t + \int^t_0 \frac{u(s)}{\sqrt{s}}\ ds.
\end{align}

Actually, let us first check that $\Phi$ indeed maps $C[0, 1]$ to $C[0, 1]$. By direct computation, we have that
\begin{align}
|\Phi(u)(t)-\Phi(u)(t')| \le&\ |\sin t-\sin t'|+\int^t_{t'} \frac{|u(s)|}{\sqrt{s}}\ ds \\
\le&\ |t-t'| + \|u\|_\infty \int^t_{t'}\frac{ds}{\sqrt{s}}\\
\le&\ |t-t'| +2 \|u \|_\infty |\sqrt{t}-\sqrt{t'}|.
\end{align}
Hence, it is clear that $\Phi(u)(t)$ is continuous if $u$ is continuous (in fact, it suffices to be bounded).

Let us show that $\Phi$ is a contraction on a smaller interval $[0, T]$. Observe that
\begin{align}
\|\Phi(u)-\Phi(v)\|_\infty\le \sup_{t \in [0, T]}\int^t_0 \frac{\|u-v\|_\infty}{\sqrt{s}}\ ds = 2\sqrt{T}\|u-v\|_\infty.
\end{align}
If $2\sqrt{T}<1$, then $\Phi$ is a contraction. Hence, we have a unique solution on $[0, T]$. Repeat this argument a finite number of times, then you will have the desired result.


问:
Is this answer to the proof equivalent to the one I wrote
I am doing a mathematical proof in which the answer from the solution is different from what I wrote. Is the one that I wrote also valid or is it wrong?

That was the answer from the solution. Now my answer:

$x^2+5x-1 = (2k)^2+5(2k)-1=4k^2+10k-1=2(2k^2+5k)-1 $

Thank you

答:
It all depends on your definition of an odd number.  In my opinion, both of these solutions are incomplete since they do not sufficiently define what an odd number is.

It would appear that the solution implicitly writes that an odd number $x$ can be written as $x=2k+1$ for some integer $k$.  If you were to define an odd number $x$ as $x=2k-1$ for some integer $k$ instead, your solution is fine.


问:
rank function on Spec (help with definition)
one definition of the line bundle over a ring is: a finitely generated projective A-module such that the rank function Spec A → N (positive integers) is constant with value 1. We call A itself the trivial line bundle.

so here i think that spec is equipped with zariski topology and N with the discrete one. Does this mean that in general the rank function is not continuous? Does one know about a basic example of non constant rank function to illustrate the peculiarities implied by the definition above?

Many thanks

答:
I've never seen this described this way, but here is why it is not "peculiar" to me. If you are thinking about Spec A as a topological space, then this A-module is sort of like a vector bundle on the space (but is allowed to be weirder since it is a module).

The rank function is saying look at the "fiber" of this bundle over a specific point. You get a free module isomorphic to $A^n$. Just as in the vector bundle case, the $n$ is the rank here.

Now if your A-module is actually a vector bundle, then the rank is just constant of whatever the rank. Since you have specified that the module is finitely generated and projective it must be locally free (assuming A Noetherian). You can check that a locally free module will have constant rank (as long as the space is connected), and hence the rank function is continuous.

If you are asking for an arbitrary A-module that is not projective for which the rank function is not continuous, try a torsion module over $\mathbb{Z}$.

(Looks like someone beat me to the answer, but it seems we've given different interpretations of how to produce an example).


答:
In addition to Peter Smith's very correct answer, it should be noted that one can also remove the axioms of choice and regularity without any substantial change.

The reason is that we can still define the relevant finite ordinals and their arithmetics needed for the incompleteness theorems (and because arithmetical truths are absolute between inner models with the same ordinals, so they are true in $L$ which is a model of $\sf ZFC$).

Note also that if $T\subseteq\sf ZFC$ is consistent and complete then $\sf ZFC$ is complete as well (it extends a complete theory), but we know it isn't complete, so no smaller theory is complete -- regardless to the provability of the incompleteness theorems.

Title: Gödel's incompleteness wrt weakend versions of ZFC

Body:
Suppose for the sake of argument that we look at ZFC with the axiom of infinity removed.

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#First_incompleteness_theorem

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem

We would then be in a position where the hypotheses of Gödel's theorems are not satisfied, correct? Basically, I want to remove, for the sake of argument, a minimal amount of axioms of ZFC so that ZFC minus some axiom(s) leads to a theory that does not include arithmetical truths and is not capable of expressing elementary arithmetic.

Is it possible that ZFC- (my shorthand for ZFC with some axiom(s) removed) is consistent and complete?


问:
Proving that if $n$ is odd, then $n^2=8k+1 \text{ for some } k\in\mathbb{Z}$
I'm currently studying Lay's Analysis with an introduction to Proofs.

I am trying to prove this statement:

If $n$ is odd, then there exists a $k\in\mathbb{Z}$ such that $n^2=8k+1$

I first started by saying that $n$ is odd, then it can be written as $n=2m+1$ for $m\in \mathbb{N}$. Substituting this in, I find that
$$4m^2+4m+1=8k+1$$
$$m^2+m=2k$$
$$m(m+1)=2k$$
I now just need to show that $m(m+1)\text{ for } m\in\mathbb{N}$ is even, which I am not sure how to rigorously prove.

Is it enough to say: if $m$ is odd, then $m+1$ is even, thus their product is even; if $m$ is even, then $m+1$ is odd, and their product is once again even? Their product is even because multiplying a number that is divisible by $2$ with any integer yields a product that $2$ still divides.

How can I improve my argument?

答: just consider the 2 cases for m. m can be even or m can be odd. so for case 1 let m=2x and plug that into the equation and you get an even number. case 2 let m=2x+1 and plug into the equation and you get an even number again. so for both cases the number is even and therefore proven.</p>


问:
Is that true If I say posterior = prior multiply likelihood?
I see a formula that we should also have marginal distribution as denominator. But some tutorial said posterior = prior multiply likelihood?</p>

答:
It's possible your tutorial was referring to posterior and prior odds rather than posterior and prior distributions, although the equation should then be posterior odds = prior odds multiplied by likelihood ratio.

If $\ A\ $ is the event whose prior and posterior odds you're interested in, and $\ d\ $ is the evidence, or data, on which the posterior odds are to be calculated, then
\begin{align}
\text{posterior odds}&=\frac{P(A|d)}{P(\tilde A\,|d)}\\
&=\frac{P(A\, \&\, d )}{P(\tilde A \,\&\, d)}\\
&=\frac{P(A)}{P(\tilde A)}\frac{P(d|A)}{P(d|\tilde A)}\\
&=\text{prior odds}\times\frac{P(d|A)}{P(d|\tilde A)}\ ,
\end{align}
the quantity $\ \frac{P(d|A)}{P(d|\tilde A)}\ $ being the ratio of the likelihood of $\ A\ $ to the likelihood of its complement.


问:
Show that $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$
I am trying to understand tensor products and I would like to show that $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$. In Lee's book about smooth manifolds is the following proof for the case $V^*\otimes V^*\simeq L(V,V,\mathbb{R})$:

According to the book it follows directly from $V\simeq V^{**}$, since everything is assumed to have finite dimension. I am fine with that. Nevertheless  I was wondering if it couldn't be done as it is done in the above proof.

For $(v,w)\in V\times V$ and $\omega,\eta\in V^*$ I would define $\Phi :V\times V\to L(V^*,V^*,\mathbb{R}),\Phi(v,w)[\omega,\eta]:=\omega(v)\cdot\eta(w)$. I suppose it is a bilinear map. If it is, then it decents to a linear map

$\tilde{\Phi} :V\otimes V\to L(V^*,V^*,\mathbb{R}),  \tilde{\Phi}(v\otimes w)[\omega,\eta]:=\omega(v)\cdot\eta(w)$. Now I am stuck. I am not sure what the basis vectors for $L(V^*,V^*,\mathbb{R})$ are or how to show that it is a bijection. I am also not sure if the apprach is okay or complete nonsense.

Thank you very much in advance!

答:
You just need to show that, $L(V^*,V^*,\mathbb{R})$ satisfies the universal property of tensor product.

Consider the map $\Phi :V\times V\to L(V^*,V^*,\mathbb{R}).$

Let, $W$ be a vector space with a bilinear map $B$ from $V\times V$ to $W.$

Suppose, $\{v_i^*\}$ is a basis of $V^*$ and if $v^*=\sum_{i} a^i v_i^{*},w^*=\sum_{j} b^j v_j^*,$ we have, $B'(v^*,w^*)=\sum_{i,j} a^ib^jB'(v_i^*,v_j^*)$

Then, there is a unique linear map $T :L(V^*,V^*,\mathbb{R})\to W$ defined by, $T(B')=\sum_{i,j}a^ib^jB(v_i,v_j)$ such that the diagram commutes.

Thus, from the universal property of tensor product,
$V\otimes V\simeq L(V^*,V^*,\mathbb{R})$


问:
How to find the nature of elements in a quotient ring ,with ideal nonprincipal $\frac{\Bbb{Z}[X]}{(2,X)}$
while trying to find the nature of elements in the ring $\frac{\Bbb{Z}[X]}{(2, X)}$. I just put $X=0$, and $2=0$ so the all polynomials with even constant term belongs to the ideal $( X,2)$, and I am left with the elements of the form. $1+(X,2)$, I don't know whether it's correct or not. Can anyone help?

答:
This is almost correct. You are just missing the trivial zero element.
A more straightforward way to see this is

$$\frac{\mathbb{Z}[X]}{(2,X)} \cong \frac{\mathbb{Z}}{(2)} = \mathbb{Z}_2 = \{\overline{0}, \overline{1}\}$$

Edit: In case you are confused about my first isomorphism $-$ you want to show $\mathbb{Z}[X]/(X) \cong \mathbb{Z}$, consider the homomorphism $\phi: \mathbb{Z}[X] \to \mathbb{Z}$ such that $\phi(f) = f(0)$. Find the kernel of this map, use the first isomoprhism theorem and fill in all the details to get the desired result.

So in our question, we have essentially written
$$\frac{\mathbb{Z}[X]}{(X,2)} = \frac{\mathbb{Z[X]}/(X)}{(2)} \cong \frac{\mathbb{Z}}{(2)} = \mathbb{Z}_2$$
which was how I got the first isomorphism.

Edit 2: In case you are wondering about the elements themselves of $\mathbb{Z}[x]/(2,X)$; they basically are the equivalence classes of the constant polynomials $f(x) = 0$ and $f(x) = 1$.


问:
Determining depth in perspective drawing
In his Complete Guide to Perspective Drawing (page 27), Craig Attebery proposes a method for determining depth in a line going towards its vanishing point (VP):

VP = Vanishing Point
HL = Horizon Line
MP = Measuring Point

To put it in words, the author assumes the measuring line will intersect the vanishing point line and the picture plane at the same distance: 2 units. Still, this does not seem to make sense. Here's an apparent proof: the measuring line intersects many lines with different vanishing points. As such lines have different angles, the distance of the interception cannot be all the same:

Is Attebery (or my interpretation of his work) wrong? If so, how should depth be measured in perspective drawing?

答:
If I understand what distance is being measured, the distance to the vanishing point from any other point in the figure is infinite.

But the lines through MP are all parallel in the scene, so lines that cut the viewing plane at uniformly spaced points will also cut the line from $0$ to VP at uniformly spaced points.

Any claim that the segment marked "$2$" is the same length as the segment between the marks $0$ and $2$, however, is a claim that those two segments are legs of an isosceles triangle, which is equivalent to a claim about a relationship between the direction to the vanishing point and the direction to the measuring point.
This implies that the choice of measuring point depends on the choice of vanishing point.

Whether this particular technique gives the claimed result or not depends on how the directions to points in the scene map to points on the horizon and on how the technique tells us to decide on the position of the measuring point given the position of the vanishing point.


问:
Why is "for all $x\in\varnothing$, $P(x)$" true, but "there exists $x\in\varnothing$ such that $P(x)$" false?
There exists an $X\in A$ such that $P(X)$. When $A$ is the empty set then this statement is false because there is nothing in $A$ that when plugged in for $X$, makes $P(X)$ come out True.

However, when the quantifier is the universal one, meaning "For all $X\in A$" (which is the empty set), then the statement is true !!

How is that ? .. Because by the reasoning in the first statement, there are not any values of $X$ that makes $P(X)$ come out True.

I'm missing something !!

(I'm studying Electrical Engineering and self studying velleman's how to prove it and the quantifiers are already kicking my butt so simplify the answers for me if possible :))

答:
The negation of the statement "for all $x\in X$, $P(x)$ holds" is "there exists $x\in X$ such that $P(x)$ does not hold". Well, then if $X$ is empty, then there is not a single $x\in X$ such that $P(X)$ is not true, and thus it is indeed true that for all $x\in X$, $P(X)$ is true.

The confusing part is that in general the validity of a statement of the form $\forall x\in X: P(X)$ does not imply the existence of a single $x\in X$ for which $P(X)$ holds. It only states that if $x\in X$ then $P(x)$ holds.

You can also think of it this way: Suppose I claim that everything in a box in front of you is pink. Can you sue me if you open it up and see a banana in there? Sure, the banana is yellow, so the judge will rule in your favor. But, can you sue me if you open up the box and see nothing at all in there? Well, no, you have nothing to complain about: everything in the box is indeed pink since nothing in it is not pink.


问:
Probability that a character kills the boss in a 3 versus 1 fight
In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.

Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

答:
Here is yet an other answer, targeting the exact fractions that give the needed probabilities, it combines simple combinatorics and simple coding (sage is my weapon of choice), based on the fact that there is a (very) limited amount of complete turns $N$ till the total of $1000$ points can be randomly reached.

The player $A$ wins in a situation like $(ABC)^N\;A$, i.e. there are $N$ complete turns, then $A$ plays and the total sum of points so far is for the first time $\ge 1000$. We denote by $P(A,N)$ this probability.

Similarly, $B$ wins in a situation like $(ABC)^N\; AB$, and we denote by $P(N,B)$ the corresponding probability, and $C$ wins with the remained probability, but for the check, let this be in a situation like $(ABC)^N\; ABC$, and we denote by $P(N,C)$ the corresponding probability.

Now we compute. The first one who can win is $C$, since $5(50+70+80)=1000$, but the corresponding probability
$$
P(5,C)=
{\underbrace{\left(\frac 1{26}\right)}_p\ }^5
{\underbrace{\left(\frac 1{41}\right)}_q\ }^5
{\underbrace{\left(\frac 1{79}\right)}_r\ }^5
$$
is tiny. We have denoted by $p=1/26$,$q=1/41$, $r=1/79$ the corresponding probabilities for each individual result for $A$, $B$, respectively $C$. After this warming up, let us start.

Now let $N$ be fixed (between $4$ and $15$). Consider the polynomials (possibly taken modulo $x^{1001}$):
$$
\begin{aligned}
Q_A &= p(x^{25}+\dots+x^{50})\ ,\qquad Q_A(1) =1\ ,\\
Q_B &= q(x^{30}+\dots+x^{70})\ ,\qquad Q_B(1)=1\ ,\\
Q_C &= r(x^{10}+\dots+x^{80})\ ,\qquad Q_C(1)=1\ ,\\[3mm]
%
F_A(N) &= Q_A^N\; Q_B^N\; Q_C^N\ ,\\
F_B(N) &= Q_A^{N+1}Q_B^N\; Q_C^N\ ,\\
F_C(N) &= Q_A^{N+1}\; Q_B^{N+1}\;  Q_C^N\ ,\\[3mm]
%
&\qquad\text{ and isolate relevant coefficients:}\\
F_A(N) &= \dots + a_{950} x^{950} + a_{951} x^{951} + a_{999} x^{999} + \dots\\
F_B(N) &= \dots + b_{930} x^{930} + b_{931} x^{931} + b_{999} x^{999} + \dots\\
F_C(N) &= \dots + c_{920} x^{920} + c_{921} x^{921} + c_{999} x^{999} + \dots\qquad
\ .
\end{aligned}
$$
Above, the $a$, $b$, $c$ - coefficients in some degree $k$ are probabilities to reach $k$ short before being self on fire.
Then the probability to get on or over $1000$ after turn $N$ for $A,B,C$ is respectively
$$
\begin{aligned}
P_A(N) &=
p(a_{950} + 2a_{951}  + 3a_{952} + \dots + 25 a_{974})
\\
&\qquad\qquad+
( a_{975} + a_{976}+\dots+a_{999})\ ,
\\[2mm]
P_B(N) &=
q(b_{930} + 2b_{931}  + 3b_{932} + \dots + 40 b_{969})
\\
&\qquad\qquad+
( b_{970} + b_{971}+\dots+b_{999})\ ,
\\[2mm]
P_C(N) &=
r(c_{920} + 2c_{921}  + 3c_{922} + \dots + 70 a_{974})
\\
&\qquad\qquad+
( c_{990} + c_{991}+\dots+c_{999})\ .
\end{aligned}
$$
These sums can be calculated. Let us get the explicit fractions.

```<code>p, q, r = 1/26, 1/41, 1/71
R.<x> = PowerSeriesRing(QQ, default_prec=1001)
QA = p*sum([x^k for k in [25..50]])
QB = q*sum([x^k for k in [30..70]])
QC = r*sum([x^k for k in [10..80]])
Q = QA * QB * QC + O(x^1000)

def PA(N):
    QAN = Q^N + O(x^1000)
    return p * sum([ (k-949) * QAN.dict().get(k, 0) for k in [950..974]]) + \
               sum([           QAN.dict().get(k, 0) for k in [975..999]])

def PB(N):
    QBN = Q^N * QA + O(x^1000)
    return q * sum([ (k-929) * QBN.dict().get(k, 0) for k in [930..969]]) + \
               sum([           QBN.dict().get(k, 0) for k in [970..999]])

def PC(N):
    QCN = Q^N * QA * QB + O(x^1000)
    return r * sum([ (k-919) * QCN.dict().get(k, 0) for k in [920..989]]) + \
               sum([           QCN.dict().get(k, 0) for k in [990..999]])

pa = sum([PA(N) for N in [0..16]])
pb = sum([PB(N) for N in [0..16]])
pc = sum([PC(N) for N in [0..16]])

print(f"p_A &= {latex(pa)}\\\\&\\approx {pa.n(200)}\\ ,\\\\[2mm]")
print(f"p_B &= {latex(pb)}\\\\&\\approx {pb.n(200)}\\ ,\\\\[2mm]")
print(f"p_C &= {latex(pc)}\\\\&\\approx {pc.n(200)}\\ .")
</code>```

Let us print the obtained values explicitly in an aligned block:
$$
\begin{aligned}
p_A &= \frac{334033901864090379613127803175076008061329503244624264537089105546493539}{1178392443307351590334920499763343901461744782823432598226397490500042752}\\&\approx 0.28346575350277151214602996031038557893379483738585896702069\ ,\\[2mm]
p_B &= \frac{6309830943686662816904113123282509708706840969580727479884347191047819}{16597076666300726624435499996666815513545701166527219693329542119718912}\\&\approx 0.38017724871382680616769939159810617375709726053541809298534\ ,\\[2mm]
p_C &= \frac{604208147014494132197561989078063573296081662711244943014511790227369}{1796329944066084741364208078907536435155098754304013107052435198932992}\\&\approx 0.33635699778340168168627064809150824730910790207872293999396\ .
\end{aligned}
$$

$\square$

A final check:

```<code>sage: pa + pb + pc
1
</code>```

Body:
Here is yet an other answer, targeting the exact fractions that give the needed probabilities, it combines simple combinatorics and simple coding (sage is my weapon of choice), based on the fact that there is a (very) limited amount of complete turns $N$ till the total of $1000$ points can be randomly reached.

The player $A$ wins in a situation like $(ABC)^N\;A$, i.e. there are $N$ complete turns, then $A$ plays and the total sum of points so far is for the first time $\ge 1000$. We denote by $P(A,N)$ this probability.

Similarly, $B$ wins in a situation like $(ABC)^N\; AB$, and we denote by $P(N,B)$ the corresponding probability, and $C$ wins with the remained probability, but for the check, let this be in a situation like $(ABC)^N\; ABC$, and we denote by $P(N,C)$ the corresponding probability.

Now we compute. The first one who can win is $C$, since $5(50+70+80)=1000$, but the corresponding probability
$$
P(5,C)=
{\underbrace{\left(\frac 1{26}\right)}_p\ }^5
{\underbrace{\left(\frac 1{41}\right)}_q\ }^5
{\underbrace{\left(\frac 1{79}\right)}_r\ }^5
$$
is tiny. We have denoted by $p=1/26$,$q=1/41$, $r=1/79$ the corresponding probabilities for each individual result for $A$, $B$, respectively $C$. After this warming up, let us start.

Now let $N$ be fixed (between $4$ and $15$). Consider the polynomials (possibly taken modulo $x^{1001}$):
$$
\begin{aligned}
Q_A &= p(x^{25}+\dots+x^{50})\ ,\qquad Q_A(1) =1\ ,\\
Q_B &= q(x^{30}+\dots+x^{70})\ ,\qquad Q_B(1)=1\ ,\\
Q_C &= r(x^{10}+\dots+x^{80})\ ,\qquad Q_C(1)=1\ ,\\[3mm]
%
F_A(N) &= Q_A^N\; Q_B^N\; Q_C^N\ ,\\
F_B(N) &= Q_A^{N+1}Q_B^N\; Q_C^N\ ,\\
F_C(N) &= Q_A^{N+1}\; Q_B^{N+1}\;  Q_C^N\ ,\\[3mm]
%
&\qquad\text{ and isolate relevant coefficients:}\\
F_A(N) &= \dots + a_{950} x^{950} + a_{951} x^{951} + a_{999} x^{999} + \dots\\
F_B(N) &= \dots + b_{930} x^{930} + b_{931} x^{931} + b_{999} x^{999} + \dots\\
F_C(N) &= \dots + c_{920} x^{920} + c_{921} x^{921} + c_{999} x^{999} + \dots\qquad
\ .
\end{aligned}
$$
Above, the $a$, $b$, $c$ - coefficients in some degree $k$ are probabilities to reach $k$ short before being self on fire.
Then the probability to get on or over $1000$ after turn $N$ for $A,B,C$ is respectively
$$
\begin{aligned}
P_A(N) &=
p(a_{950} + 2a_{951}  + 3a_{952} + \dots + 25 a_{974})
\\
&\qquad\qquad+
( a_{975} + a_{976}+\dots+a_{999})\ ,
\\[2mm]
P_B(N) &=
q(b_{930} + 2b_{931}  + 3b_{932} + \dots + 40 b_{969})
\\
&\qquad\qquad+
( b_{970} + b_{971}+\dots+b_{999})\ ,
\\[2mm]
P_C(N) &=
r(c_{920} + 2c_{921}  + 3c_{922} + \dots + 70 a_{974})
\\
&\qquad\qquad+
( c_{990} + c_{991}+\dots+c_{999})\ .
\end{aligned}
$$
These sums can be calculated. Let us get the explicit fractions.

p, q, r = 1/26, 1/41, 1/71
R.<x> = PowerSeriesRing(QQ, default_prec=1001)
QA = p*sum([x^k for k in [25..50]])
QB = q*sum([x^k for k in [30..70]])
QC = r*sum([x^k for k in [10..80]])
Q = QA * QB * QC + O(x^1000)

def PA(N):
    QAN = Q^N + O(x^1000)
    return p * sum([ (k-949) * QAN.dict().get(k, 0) for k in [950..974]]) + \
               sum([           QAN.dict().get(k, 0) for k in [975..999]])

def PB(N):
    QBN = Q^N * QA + O(x^1000)
    return q * sum([ (k-929) * QBN.dict().get(k, 0) for k in [930..969]]) + \
               sum([           QBN.dict().get(k, 0) for k in [970..999]])

def PC(N):
    QCN = Q^N * QA * QB + O(x^1000)
    return r * sum([ (k-919) * QCN.dict().get(k, 0) for k in [920..989]]) + \
               sum([           QCN.dict().get(k, 0) for k in [990..999]])

pa = sum([PA(N) for N in [0..16]])
pb = sum([PB(N) for N in [0..16]])
pc = sum([PC(N) for N in [0..16]])

print(f"p_A &= {latex(pa)}\\\\&\\approx {pa.n(200)}\\ ,\\\\[2mm]")
print(f"p_B &= {latex(pb)}\\\\&\\approx {pb.n(200)}\\ ,\\\\[2mm]")
print(f"p_C &= {latex(pc)}\\\\&\\approx {pc.n(200)}\\ .")

Let us print the obtained values explicitly in an aligned block:
$$
\begin{aligned}
p_A &= \frac{334033901864090379613127803175076008061329503244624264537089105546493539}{1178392443307351590334920499763343901461744782823432598226397490500042752}\\&\approx 0.28346575350277151214602996031038557893379483738585896702069\ ,\\[2mm]
p_B &= \frac{6309830943686662816904113123282509708706840969580727479884347191047819}{16597076666300726624435499996666815513545701166527219693329542119718912}\\&\approx 0.38017724871382680616769939159810617375709726053541809298534\ ,\\[2mm]
p_C &= \frac{604208147014494132197561989078063573296081662711244943014511790227369}{1796329944066084741364208078907536435155098754304013107052435198932992}\\&\approx 0.33635699778340168168627064809150824730910790207872293999396\ .
\end{aligned}
$$

$\square$

A final check:

sage: pa + pb + pc
1


问:
Does $\|Tx\|=\|T^*x\|$ for all $x$ imply normality for real Hilbert spaces?
The proof for complex Hilbert spaces crucially relies on the fact that $\langle Sx,x\rangle=\langle Tx,x\rangle$ for all x implies that S=T. This fact fails in the case of real Hilbert spaces as early as $\mathbb{R}^2$: let $S$ be rotation by $\frac{\pi}{2}$ and $T$ be rotation by $-\frac{\pi}{2}$.

Is there a proof that $\|Tx\|=\|T^*x\|$ for all $x$ implies $T^*T=TT^*$ which doesn't rely on this fact?

答:
I'm not sure whether I'm allowed to reply to my own questions when I answer it myself, but here's a solution using the polarization identity for real spaces: for an arbitrary $x,y\in H$,
$$\langle Tx,Ty \rangle=\frac{1}{4}(||Tx+Ty||^2-||Tx-Ty||^2).$$
Apply linearity of $T$ and $T^*$ and the $||T(x-y)||=||T^*(x-y)||$ hypothesis to yield
$$\langle Tx,Ty \rangle=\frac{1}{4}(||T^*x+T^*y||^2-||T^*x-T^*y||^2)=\langle T^*x, T^*y\rangle.$$
Then
$$\langle T^*Tx,y \rangle=\langle T^*Tx,y \rangle,$$
and since $x,y$ are arbitrary, $T^*T=TT^*$.


问:
Parametrising $\mathbb{R^3}$ coordinates using a mesh with barycentric coordinates, and signed distance to the mesh
I need to find a way to parametrise $\mathbb{R^3}$ coordinates relative to a mesh, which is a deformed $UV$ sphere with triangular faces. Every vertex on the mesh has a $UV$ coordinate, and I want to associate every $\mathbb{R^3}-$ point with a coordinate ($u,v$, signed distance). So for instance, points on the mesh would simply be assigned $(u,v,0)$. If a point starts on the mesh at a vertex and then moves outwards by $1$ in the direction of the vertex normal, the new point should be assigned $(u,v,1)$. If a point starts on the mesh inside a face, its normal is a barycentric combination of the surrounding vertex normals, then $(u,v,1)$ is defined as above. Moving inwards gives a negative signed distance. It seems intuitive to me that this formulation assigns a coordinate to every point in $\mathbb{R^3}$, since it is like inflating the mesh until you reach any arbitrary point in $\mathbb{R^3}$. However, I have not been able to find an efficient (or any) way to compute this on an arbitrary $\mathbb{R^3}$ point for a given set of vertex positions. Can anyone see such a way, or is there another approach which is easier altogether?

答:
This sounds like a computer graphics problem.

Suppose $P$ is the point you want to parametrize. Let $T$ be a triangle with vertices $x_1,x_2,x_3$, vertex normals $n_1,n_2,n_3$ and UV coordinates $u_1,u_2,u_3$. You want to find a point $X=ax_1+bx_2+(1-a-b)x_3$ with normal $N=an_1+bn_2+(1-a-b)n_3$ such that $N$ is parallel with the line $XP$.  This will be the case if the cross product $C=N\times(P-X)=0$.  The vector $C$ will have three components that are 2nd degree functions in $a$ and $b$ so you have to solve for $a$ and $b$.  This is essentially solving for the intersection of three conics.  Best to find an existing solver, but check out this question and answers: Algorithm: Intersection of two conics

Once you have this information, compute the signed distance $w$.  And $(u,v)=au_1+bu_2+(1-a-b)u_3$, so the parametrization of $P$ with respect to $T$ is $(u,v,w)$.

It remains to choose a triangle $T$.  Depending on how much the sphere is deformed, there may be more than one $(u,v,w)$ coordinate for a point.
But your best bet should be to choose the triangle closest to $P$.


问:
How to find the least number of extractions to make the word WAAY-TV?
The problem is as follows:

A local tv station is doing a small lottery among their employees to
celebrate the anniversary of the news department.

The owner has set five carton boxes which a set of colored plastic
cards which have printed the letters WAAYTV (the callsign of his
station).

The each box is put one next to another making a row, with the letter
W positioned in the left and V on the right and has a label outside
indicating its contents and these are:

Box W: 16 red cards and 15 blue cards

Box A: 16 red cards and 14 blue cards

Box Y: 16 red cards and 14 blue cards

Box T: 14 red cards and 15 blue cards

Box V: 14 red cards and 15 blue cards

How many of the cards need to be drawn one by one without looking in
order to be sure to make the callsign of the station such that all the
cards share the same color?

It is not specifically mentioned in the problem but I believe looking at similar problems from the same author. The intended meaning is that in this context it is drawing with no replacement.

I'm not sure exactly how to tackle this problem. But it seems to my that in order to guarantee to get all the letters sharing the same color has to do assuming the least likely to happen scenario. In other words looking for an option which assumes you get what you are not looking for and after exhausting these options you can be sure to get what you want.

Given this preamble, I'm assuming that, the contestant, stars taking out cards from the left.

If a card is taken out from the box there, we don't know if its either red or blue it could be any. So in order to compensate for this unknown we have to be sure to extract all the rest which aren't known.

I guess 14+15 from V and from T, and 16+14 from Y and 16 from A + 2 additional which ensure that those have the same color.

Adding all these would be

$1+14+15+14+15+16+14+16+2=107$

But is thia a correct assumption?. I don't know if I'm really doing it right, thus an answer which would help me the most is one which could explain what sort of logic orderly should be followed here to solve this. Since I'm slow at catching up the clues, I require that the answer could be step by step.

答:
Here's a strategy that uses just $64$:

Choose $2$ As:  if they match, continue below; if not, choose a third A.  You are guaranteed to have two As of the same color.  (Think about it.)

Now choose 16 Ws, 16 Ys, 15Ts and 15 Vs, and you'll be guaranteed to have all the same color.


问:
Limit carries with e proof
Suppose that $\lim_{n\to\infty} x_n = x$. Prove that $\lim_{x\to\infty} e^{x_n} = e^{\lim_{x\to\infty} x_n} = e^x$.

My attempt:

Begin by considering $|e^{x_n} - e^x|$:

For $k > 0$:

$$|e^{x_n} - e^x| = e^x|e^{x_n - x} - 1| < e^x|e^{k} - 1| < e^{x + k}$$

Let $\epsilon > 0$. Because $\lim_{n\to\infty} x_n = x$, there is an $N \in \mathbb{N}$ s.t. for all $n \geq N$

$$|x_n - x| < \ln(\epsilon) - x$$

$$|e^{x_n} - e^x| = e^x|e^{x_n - x} - 1| < e^x|e^{\ln(\epsilon) - x} - 1|< e^{x + \ln(\epsilon) - x} = \epsilon$$

Then by definition of the limit, $\lim_{x\to\infty} e^{x_n} = e^x$.

Is this proof correct? If not or even if so, can someone show me a formal one? Any assistance much appreciated.

答:
You need to learn a few properties of exponential function. A key inequality satisfied by exponential function is $$e^t\geq 1+t\tag{1}$$ Replace $t$ with $-t$ and assuming $t\in(0,1)$ to get $$e^t\leq \frac{1}{1-t}$$ or $$e^t-1\leq \frac{t} {1-t}<2t\tag{2}$$ if $0<t<1/2$.

Further we have $$|e^{-t} -1|=1-e^{-t}=\frac{e^t-1}{e^t}<e^t-1\tag{3}$$ and thus it follows from $(2),(3)$ that $$|e^t-1|\leq 2|t|\tag{4}$$ if $|t|<1/2$ and equality holds only if $t=0$.

Now consider the expression $$|e^{x_n} - e^x|=e^x|e^{x_n-x} - 1|\leq 2e^x|x_n-x|\tag{5}$$ via $(4)$ provided that $|x_n-x|<1/2$.

Let $\epsilon>0$ be given and then we can find an  integer $N>0$ such that $$|x_n-x|<\min\left(\frac{\epsilon} {2e^x},\frac{1}{2}\right) $$ whenever $n\geq N$. Now from $(5)$ it follows that $|e^{x_n} - e^x|<\epsilon$ whenever $n\geq N$.

Title: Limit carries with e proof


答:
In classical logic, given a particular structure for some language, if something is not false in the structure, then it's true there. Of course we can easily make things complicated by considering a structure whose universe is made of structures of a particular language, and so on and so forth. So let's stay in the context of one particular structure.

Something is vacuously true if it just cannot be made false. That is to say, there's no counterexample. If it's not false, it has to be true.

So saying "for every $x\in\varnothing$ ..." does not have a counterexample, and therefore it is true; but on the other hand "there exists $x\in\varnothing$ ..." is false because we can prove that there isn't such $x$, and therefore conclude a contradiction.

Title: Why is "for all $x\in\varnothing$, $P(x)$" true, but "there exists $x\in\varnothing$ such that $P(x)$" false?

Body:
There exists an $X\in A$ such that $P(X)$. When $A$ is the empty set then this statement is false because there is nothing in $A$ that when plugged in for $X$, makes $P(X)$ come out True.

However, when the quantifier is the universal one, meaning "For all $X\in A$" (which is the empty set), then the statement is true !!

How is that ? .. Because by the reasoning in the first statement, there are not any values of $X$ that makes $P(X)$ come out True.

I'm missing something !!

(I'm studying Electrical Engineering and self studying velleman's how to prove it and the quantifiers are already kicking my butt so simplify the answers for me if possible :))


问:
numbers-based puzzle problem from a Deductive Reasoning section of Intro to Critical Thinking class. What is the optimal way to solve it?
The following puzzle is one of three deductive-reasoning problems (‘brain-teasers’) that were choosable to attempt for an assignment.  Although it posits that "only simple arithmetic and logic" are required to solve it, I suspect that some particular math skills might help greatly.

What is an optimal method to approach it and satisfactorily find all the solution[s] (in a manner showing that the answer is correct with no missing solutions)?

An initial deduction shows that 9 can belong to only one string-combination (9,3,1), which means that it cannot be a string connector and that (9,3,1) exists which leaves only two other odd letters (5, 7) from the set of 9, and writing out the isolated possibilities for all others 8 thru 1 (although doing so is more exhaustive than an efficient method) yields the number of possible respective combinations being 8=two, 7=two, 6=two, 5=three, 4=four, 3=three, 2=three, 1=three (all more than one).  Also observable is that 13 being odd must consist of an odd number of odds (#= 3 or 1) and an even number of evens (#= 0 or 2), which restricts the possibilities of potential configurations of the entire sequence.

答:
Add $(A+B+C)+(C+D+E)+(E+F+G)+(G+H+I)=52$.

$A+B+C+D+E+F+G+H+I=45$ so we have $C+E+G=7$

$C,E,G$ must be $1,2,4$ in some order.

The $1$ must be part of two of $1+4+8$, $1+3+9$, and $1+5+7$.  We will rule out $1+5+7$ in a bit.
As you said, $9$ must only be used once.  If it is part of one of the middle runs, $3$ would be used twice, but that is not allowed, so $9$ is part of an end run.  The other end of the other run from $1$ has to be a turn, so it has to be $2$ or $4$.  It cannot be $1+5+7$.
We can start (or end) with either $3,9$ or $9,3$.  Say we start with $9,3$.  Then the next run is $1,8,4$ so we have a proper turn at the end.  The other turn is $2$.  The number between them must be $7$.

One solution is $9,3,1,8,4,7,2,5,6$.  Each end pair may be interchanged, giving four solutions.  We can reverse any of them, giving a total of eight solutions.


问:
Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$

$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?

If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.

答:
Rotate $\triangle BCP$ counter-clockwise 60$^\circ$ around the point $B$ to $\triangle BAQ$ and connect $PQ$. Then, $BPQ$ is an equilateral triangle and $APQ$ is a right triangle due to $x^2+y^2=z^2$.  Apply the cosine rule to $\triangle BPA$ to obtain the side $s$
\begin{align}
s^2 & = AP^2+BP^2 - 2AP\cdot BP\cos\angle APB \\
&= x^2 + y^2 - 2x y \cos 150^\circ\\
&= x^2 + y^2 +\sqrt3x y
\end{align}


问:
Good reference for values of Ramsey Numbers
I want to know values of Ramsey Numbers $R(4, 4, \ldots, 4; r)$ for $r \geq 2$, where $4$ is repeated $r$ times. It is defined to be the least $n$ such that any complete graph $K_{n}$ colored in $r$-colors a monochromatic $K_{4}$.

I googled a lot and found out that $R(4, 4; 2) = 18$. I would like to know values of $R(4, 4, 4; 3)$ and $R(4, 4, 4, 4; 4)$. Are these values known? The textbook I own on combinatorics is old and does not seem to provide reference on these numbers.

I would be grateful if someone could provide me a reference to some resource that has Ramsey numbers listed (of course, ones that are known).

答:
"Small Ramsey Numbers" by Stanisław Radziszowski

https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS1


问:
Is a homogeneous ideal radical?
$\newcommand{\radical}{\operatorname{radical}}$I am self studying some introductory level algebraic geometry and I want to ask a naive question.

Let $I\subset k[x_0,\ldots,x_n]$ be a homogeneous ideal, then $\radical(I)=I?$

Recall $\radical(I)=\{f\in k[x_0,\ldots,x_n] \mid \exists N \in \mathbb{Z}_{>0}, \text{ s.th. } f^N\in I\}$

Certainly, $\radical(I)\supset I$

So enough to show another inclusion, if we pick any $f=\sum_{i=0}^r a_i x^i \in \radical(I)$

$f^N = (\sum_{i=0}^r a_i x^i)^N=\sum_{j=0}^{rN} b_j x^j \in I$ which is a homogeneous ideal.

Thus it follows that all the homogeneous component is in $I$, i.e. $b_jx^j\in I$ for all $j=0,...,rN$. And hence, all the homogenous component in $f$ are also in $I$ since they only vary up to multiplying a constant. Therefore, $f$ are also in $I$.

答: No, it is not the case that every homogeneous ideal is radical. Here is an easy counterexample: $(x^2)\subset k[x]$. This is homogeneous, being generated by a homogeneous element, but not radical, since $x\cdot x\in (x^2)$ but $x\notin (x^2)$.

Title: Is a homogeneous ideal radical?


问:
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots

Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.

But I wonder that there is some logical ways we can find intervals that contain roots without guessing?

答:
I would start with the rational root theorem, which fails but gets you values at $x=\pm1$.
Because the coefficients are small, the roots are all close to $0$.  Checking values at the integers close to $0$ is a good approach.  Once you go from $-3$ to $+3$ you have found one root in $(2,3)$.  You should be able to convince yourself there are not roots outside $(-3,3)$ because the $x^5$ term dominates.  Now checking at half integers is a good next step and finds the roots.  There was no guessing involved.  The real question comes if you don't find the roots with the half integral evaluations.  Do you trust the problem setter to guarantee there are five roots?  If so, trying quarters is the next logical step.  Or do you worry the question is wrong?  I have no advice here.


问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答:
As former Chinese then French colonists we Vietnamese dropped the Chinese script and took on the Roman alphabets and the Arabic numbers as our own, thank to France. But we didn't take on the French counting system. Ours is very straight forward, and like Chinese, is mostly a monosyllabic language.

một (one),

hai (two),

ba (three),

Bốn (four),

năm (five),

sáu (six),

bảy (seven),

tám (eight),

chín (nine),

mười (ten),

mười một (ten one = eleven),

mười hai (ten two = twelve),

hai mươi (two tens = 2(10) = 20),

hai mươi một (two tens one = 2(10) + 1 = 21),

ba mươi (three tens = 3(10) = 30),

ba mươi một (three tens one = 3(10) + 1 = 31),

một trăm (one hundred),

một trăm một (one hundred one = 1(100) + 1 = 101),

hai trăm (two hundreds).

Title: Why do the French count so strangely?


问:
P-group of nilpotency class $2$ and Schur multiplier
Let $p$ be an odd prime and let $G$ be a p-group of nilpotency class $2$ in which $V=G/G'$ is an elementary abelian p-group. Let $M(V)$ be the Schur multiplier of $V$ which is isomorphic to the exterior square $V\wedge V$.

Is $M(V)$ necessarily isomorphic to $G'$?

If the answer is no, does there exist a p-group $G$ of nilpotency class $2$ in which $V=G/G'$ is an elementary abelian p-group and $M(G)=1$?

Thank you in advance.

答:
The answer to the first question is “no”. Note that if $K=G\times C_p$, where $C_p$ is cyclic of order $p$, then $K’=G’$, and $K^{\rm ab}\cong G^{\rm ab}\times C_p$. In particular, if $G$ is of class two and $G^{\rm ab}$ is elementary abelian, then the same holds for $K$, but the rank of $K^{\rm ab}$ is strictly larger than that of $G$. That means that you can make $V\wedge V$ have arbitrarily large rank while not changing the commutator subgroup, so we cannot have that $M(V)$ is necessarily isomorphic to $G’$.

As for an example of a $p$ group $G$ of class two with $G^{\rm ab}$ elementary abelian and with trivial Schur multiplier, the easiest example is the extra-special $p$-group of order $p^3$ that is not of exponent $p$,
$$G = \langle x,y \mid x^{p^2}=y^p=[x,y,x]=[x,y,y]=1, x^p=[x,y]\rangle.$$
The same will hold for any group that is isoclinic to $G$.

The Schur multipliers of the four non-abelian groups of order $p^3$ are known classically; you can also read off that this group has trivial Schur multiplier from Theorem 49 in Certain homological functors of $2$-generator $p$-groups of class $2$ by A. Magidin and R.F. Morse, in Computational Group Theory and the Theory of Groups II, Contemporary Mathematics no. 511, American Mathematical Society, Providence RI, 2010, pp. 127-166. The group above corresponds to the tuple $(\alpha,\beta,\gamma;\rho,\sigma)=(1,1,1;0,1)$ in the notation of Theorem 1 of that paper.


问:
Equivalent conditions for $M$-genericity
Suppose $G$ is $M$ generic over some model $M$ of $ZFC$, for some poset $P$. We have the following equivalences:

$G$ is $M$-generic, i.e, for every $D \in M$ such that $D$ is dense in $P$, $G \cap D \not = \emptyset$

$G$ meets every maximal antichain in $M$

$G$ meets every predense set in $M$

$G$ meets every open and dense set in $M$

For every $r \in G$, if $D$ is dense below $r$ (i.e, for every $p \leq r$ there is a $q \in D$ such that $q \leq p$), and $D \in M$, $G \cap D \neq \emptyset$

I wanted to prove these equivalences. Most of these are pretty easy, but some of them I've blanked on.

$1 \implies 4$ is obvious. $4 \implies 1$ is: given a dense $D \in M$, take the downward closure $D' = \{q \in P: \exists p \in D q \leq p \}. D' \in M$ because $M$ satisfies $ZFC$ and is transitive and this is a simple definition. $D'$ is then dense and open so $G \cap D' \not = \emptyset$ and then just use the upwards closure of filters.

$3 \implies 1$ is obvious. $1 \implies 3$ is: given any $D \in M$ predense we have that $(D$ is predense$)^M$ and thus $M$ knows that for any $p \in P$ there is a $q \in D$ such that $p$ and $q$ are comparable, i.e, there is $r_{pq} \in P$ such that $r_{pq} \leq p, q$. Now define $D' = \{r_{pq} : p \in P$ and $q \in D$ such that $p, q$ comparable $\}$ and then I was thinking $D' \in M$ since $(Choice)^M$ (so a random $r_{pq}$ ca nbe chosen if there are many. $D'$ is dense so $G$ meets it at which point you just use upwards closure again. I feel less comfortable with this argument.

$5 \implies 1$ clearly, and $3 \implies 2$ clearly (as maximal antichains are predense). So all that's left is $2 \implies 1,4,5$ and $1,3,4 \implies 5$.

Perhaps I'm being an idiot but I've been having trouble with these ones.

Does this work for $1 \implies 5$? Suppose $D \in M$ dense below some $r$. Then define $D' = \{q \in P:q$ is incompatible with $r$, or $q \in D\}$. This is dense: given any $p \in P$, if $p$ incompat. with $r$, $p \in D'$ so assume not. So that means $p$ is compatible with $r$, which means there is $q$ such that $q \leq p, r$. By density below $r$ there is $q' \in D$ such that $q' \leq q \leq p, r$ so $q' \in D'$ and $q' \leq p$; therefore $D'$ is dense. Then $G \cap D \not = \emptyset$, and it is clear that intersection has to be in $p$, due to the directedness of filters. Does this work?

Finally $2 \implies 1,3,4,5$: this one I'm just stuck with. I may be missing something obvious.

答:
There are multiple questions here; let me address the question of applying $(2)$ (which seems the most unclear point).

The key to applying $(2)$ is to note that every dense set contains a maximal antichain. Specifically, fix a dense set $D$ and let $A$ be a set with the following properties:

$A\subseteq D$,

$A$ is an antichain, and

there is no antichain $B\subseteq D$ with $A\subsetneq B$.

The existence of such an $A$ is guaranteed, as usual, by Zorn's Lemma (consider the partial order of antichains which are subsets of $D$). Now I claim that this $A$ is in fact a maximal antichain.

For suppose otherwise. Let $a\not\in A$ such that $A\cup\{a\}$ is an antichain. Then no extension of $a$ can meet $D$, since if $b\le a$ with $b\in D$ then $A\cup\{b\}$ would be an antichain strictly containing $A$ which is a subset of $D$.

The idea, then, is the following: "If $G$ meets every maximal antichain, then $G$ meets every dense set since every dense set contains a maximal antichain." Do you see how to appropriately formulate this to get $(2)\rightarrow(1)$?

Note that unlike the "maximal-antichain-to-dense-open" translation $$A\leadsto\{p: \exists a\in A(p\le a)\},$$ there is in general no canonical way to find a maximal antichain inside a given dense set. Indeed, it is consistent with $\mathsf{ZF}$ that there is a partial order $P$ with top element $\mathbb{1}_P$ which is separative (= nowhere-trivial, from the forcing perspective) but which has no maximal antichains other than $\{\mathbb{1}_P\}$; in such a poset, the dense set $P\setminus\{\mathbb{1}_P\}$ does not contain a maximal antichain.


问:
Convergence or divergence of $\int_0^2 \frac{\sin^2x}{x^p}\,dx$
I am starting to learn a little about analysis and convergence of Riemann integrals. I came across this problem and it stomped me. I want to find the values of $p$ for which the integral converges.

$$\int_0^2 \frac{\sin^2x}{x^p}\,dx$$

My approach was noting that: $\sin^2x \le \vert \sin x\vert$ and that $\vert \frac{\sin x}{x^p}\vert \le \frac1{x^p} $ then $\vert \frac{\sin^2 x}{x^p}\vert \le \frac1{x^p} $

Integration by parts yields:

$$\int_0^2 \frac{\sin^2x}{x^p}\,dx= \left[ \frac{x^{-p}}{2} \left(x-0.5\sin{2x} \right) \right]_0^2+\frac p2 \int_0^2 x^{-p}-\frac {x^{-p-1}}{2}\sin {2x}\,dx $$

$$= \frac1{2^p}\left( 1 - \frac {\sin4}{2}\right)+\frac p2\left[ \frac{2^{-p+1}}{-p+1} \right]_0^2- \frac p4\int_0^2 x^{-p-1}\sin {2x}\,dx$$

$$= \frac1{2^p}\left( 1 - \frac {\sin4}{2}\right) +\frac {p}{2^p\left(-p+1\right)}-\frac p4\int_0^2 x^{-p-1}\sin {2x}\,dx$$
$$$$
But since $\vert x^{-p-1}\sin{2x} \vert \le x^{-p-1}$ and $p+1 \gt 1$, then $\frac p4\int_0^2 x^{-p-1}\sin {2x}\,dx$ is convergent because $\int_0^2 x^{-p}\,dx$ convergent for $p \gt1$

So I arrived at the conclusion that for $p \gt1$, $\int_0^2 \frac{\sin^2x}{x^p}\,dx$ converges, since all of the terms to the right are finite and the integral on the right also convergent. On the other hand I checked with Wolfram Alpha and  it says $\int_0^2 \frac{\sin^2x}{x^3}\,dx$ does not converge, yet $p=3 \gt 1$. So where did I go wrong?

答:
The problem is at the lower bound.

By Taylor, we have
$$\frac{\sin^2(x)}{x^p}=\frac 1{x^p}\sum_{n=1}^\infty (-1)^{n+1}\,\frac {2^{2n-1}} {(2n)!} x^{2n}$$
$$\int\frac{\sin^2(x)}{x^p}\,dx=\sum_{n=1}^\infty (-1)^{n+1}\,\frac {2^{2n-1}} {(2n)!}\int x^{2n-p}\,dx$$
$$\int\frac{\sin^2(x)}{x^p}\,dx=\sum_{n=1}^\infty (-1)^{n+1}\frac{ 2^{2 n-1} }{(2 n)! \,(2 n-p+1)}x^{2 n-p+1}$$ The very firs term being $x^{3-p}$, the requirement for convergence is then $p<3$.

Just for your curiosity, the definite integral is given by
$$\frac{2^{-p} (\cos (4)-1)}{p-1}-\frac{2^{4-p} }
   {(p-3) (p-1)}\,
   _1F_2\left(\frac{3-p}{2};\frac{3}{2},\frac{5-p}{2};-4\right)$$


问:
How to write "x is a composite number" using predicate logic?
I know that a composite number is a number formed by multiplying 2 smaller natural numbers. I think that will definitely be incorporated in the predicate somehow. I am really confused on how to represent this using predicate logic/logical connectives</p>

答:
So I know that a composite number is a number formed by multiplying 2 smaller natural numbers.

In other words, you are saying that "$x$ is a composite number" means exactly that:

"There are two things which are both natural numbers, each smaller than $x$, and whose product equals $x$."

I am really confused on how to represent this using predicate logic/logical connectives

Here are the symbols you will need: $x, n, m, \Bbb N, \exists, \in, <, =, \times, (,)$.

Just place them into a well formed formula that says what you need to say.


答: $$\exists \{ y, z \}\ {\rm s.t.}\ (yz = x) \wedge (y < x \wedge z < x)\ {\rm for}\ \{ x, y, z \} \in \mathbb{N}$$

Title: How to write "x is a composite number" using predicate logic?

Body: I know that a composite number is a number formed by multiplying 2 smaller natural numbers. I think that will definitely be incorporated in the predicate somehow. I am really confused on how to represent this using predicate logic/logical connectives</p>


问:
Do the genus, codimension and projective space weights determine a Fano threefold?
Say I have a complex Fano threefold of some given codimension, some given genus, embedded into some (possibly weighted) projective space.

My question is: do the genus, codimension and weights of the projective space always uniquely determine a Fano threefold, or do I need more data?

For example, according to the Graded Ring Database, if we pick genus $0$ and codimension $1$, then there are $32$ such Fano threefolds. From a quick inspection, it seems that the weights of the projective spaces that they're embedded into seem different for each Fano threefold. So I'd hazard a guess that the genus, codimension and weights are the only data we need. Is this true?

Thank you.

答:
No. For instance
$$
X = \mathbb{P}^1 \times \mathbb{P}^1 \times \mathbb{P}^1
$$
and a smooth $(1,1)$-divisor
$$
Y \subset \mathbb{P}^2 \times \mathbb{P}^2
$$
both a Fano threefolds of genus 25 in codimension 4 with all weights equal to 1.


答:
I'm assuming that the number of picks from each box must be specified in advance.

Choosing 17W, 17Y, 16T, 16V, and 3A yields at least 1 of each color for WYTV and at least 2 of at least one color for A.  The total is 69.

But a better solution is to choose 17W, 1Y, 16T, 16V, and 18A, which yields at least 1 of each color for WATV, and so either color of the Y card suffices.  The total is 68, and this turns out to be optimal.  An alternative optimal solution is 1W, 17Y, 16T, 16V, and 18A.

Title: How to find the least number of extractions to make the word WAAY-TV?

Body:
The problem is as follows:

A local tv station is doing a small lottery among their employees to
celebrate the anniversary of the news department.

The owner has set five carton boxes which a set of colored plastic
cards which have printed the letters WAAYTV (the callsign of his
station).

The each box is put one next to another making a row, with the letter
W positioned in the left and V on the right and has a label outside
indicating its contents and these are:

Box W: 16 red cards and 15 blue cards

Box A: 16 red cards and 14 blue cards

Box Y: 16 red cards and 14 blue cards

Box T: 14 red cards and 15 blue cards

Box V: 14 red cards and 15 blue cards

How many of the cards need to be drawn one by one without looking in
order to be sure to make the callsign of the station such that all the
cards share the same color?

It is not specifically mentioned in the problem but I believe looking at similar problems from the same author. The intended meaning is that in this context it is drawing with no replacement.

I'm not sure exactly how to tackle this problem. But it seems to my that in order to guarantee to get all the letters sharing the same color has to do assuming the least likely to happen scenario. In other words looking for an option which assumes you get what you are not looking for and after exhausting these options you can be sure to get what you want.

Given this preamble, I'm assuming that, the contestant, stars taking out cards from the left.

If a card is taken out from the box there, we don't know if its either red or blue it could be any. So in order to compensate for this unknown we have to be sure to extract all the rest which aren't known.

I guess 14+15 from V and from T, and 16+14 from Y and 16 from A + 2 additional which ensure that those have the same color.

Adding all these would be

$1+14+15+14+15+16+14+16+2=107$

But is thia a correct assumption?. I don't know if I'm really doing it right, thus an answer which would help me the most is one which could explain what sort of logic orderly should be followed here to solve this. Since I'm slow at catching up the clues, I require that the answer could be step by step.


问:
Find the value of $\int_0^1 f(x)\, dx$

Let $f(x)$ be continous on $[0; 1]$, satisfy the following condition:

$$f(1) = 0$$

$$\int_0^1 [f'(x)]^2 \, \mathrm{d}x = 7$$

$$\int_0^{1} x^2f(x) \, \mathrm{d}x = \dfrac{1}{3}$$
Find the value of $\int_0^{1} f(x) \, \mathrm{d}x$

I have tried but the problem is I was unable to get anything from $(2)$. From $(3)$, using integration by parts, I get this:

$$\int_0^{1} (x^3f(x))' \, \mathrm{d}x = 0$$

But this isn't any help at all. Any idea for this problem?

答:
As done here  by user :Fred

By  IBP $$\int_0^1 x^3 f'(x)=0-\int_0^1 3x^2 f(x)$$ So $$\int_0^1 x^3 f'(x)=-1$$ Now by cauchy schwarz $$1={\left(\int_0^1 x^3 f'(x)\right)}^2\le \int_0^1 x^6\int_0^1 f'(x)^2=\frac{1}{7} \cdot 7$$ That means for equality of C-S must occur ie $f'(x)=\eta x^3\implies f(x)=\frac{\eta}{4} x^4 + C$

Now to find $\eta$  substitute it into $\int_0^1 x^2 f(x)=1/3$ after which you can easily find $\int_0^1 f(x)dx$

Title: Find the value of $\int_0^1 f(x)\, dx$


问:
Equivalence class for a determinant
Show that the relation on the set of all 2 x 2 matrices defined by A ~ B if detA=det B is an equivalence relation. Describe the equivalence class.

I have determined that it is an equivalence relation. I am troubled by the equivalence class part. Determinants exist in the set of real numbers, right? How would I indicate the matrices that satisfy this?

答:
You say, matrix is of order $2\times 2 $.

Det can be any complex number.

Fix a complex number $c$.

Now,,take the matrix $$
    \begin{pmatrix}
    a_{11} & a_{12} \\
    a_{21} & a_{22} \\
    \end{pmatrix}
$$

Now , for the equivalence class of $c$ , $a_{11} a_{22} - a_{12} a_{21}= c $

There infinitely many $a_{11} ,a_{22},a_{12} , a_{21} $ satisfying the above equation.

So,, there are uncountably many class and each class contains uncountably many matrices.

Title: Equivalence class for a determinant


问:
For the value iteration function, if the immediate cost is a linear function in state, then is this value iteration function convex?
For the state value function $V(s)$, if the immediate cost is a linear function in the state, for example $r(s,a) = s + a^2$, where $s$ and $a$ are state and action, which are finite. Is this value function convex in state $s$? if yes, how to prove it?

BTW, the Bellman optimal equation is:
$V(s) = \min_{a\in \mathcal{A}} \Big\{r(s,a)
       + \gamma \sum_{s'\in \mathcal{S}} p(s'|s,a) V(s') \Big\}$.
$\gamma \in [0,1]$ is the discount factor.

答:
Possible duplicate: Is linear function convex or concave?

This link should answer your question. If the value function is linear in $s$, it is also convex (and concave!) in $s$.


问:
How to evaluate $\lim_{(x,y)\to (0,0)} \frac{x^4+y^4}{2x-y}$?
Find the limit :

$$\lim_{(x,y)\to (0,0)} \frac{x^4+y^4}{2x-y}$$

I have tried using polar substitution but the problem is the denominator. Denominator can go to $0$ for some value of theta and is not dependent on only $r$.

答: Along the line $y=x$, the limit is zero, but along the line $y=2x$ we always have zero in the denominator, but apart from $(0,0)$, not in the numerator, so the limit does not exist.


问:
If $E$ has positive measure, then prove that there exists $h \in \mathbb{R}$ such that $|(E+h) \cap E| > 0$.
If $E$ has positive measure, then prove that there exists $h \in \mathbb{R}$ such that $|(E+h) \cap E| > 0$. I tried doing this by contradiction but cannot seem to find my way about the argument.

Any hints or help would be appreciated. Thanks.

答:
It suffices to prove the statement for bounded $E$. So suppose $E$ is bounded.

Note $\int\int 1_E(y)1_E(y-h)dydh = \int 1_E(y)\int 1_E(y-h)dhdy = \int 1_E(y)|E|dy = |E|^2 > 0$.

So, there exists $h$ with $\int 1_E(y) 1_E(y-h)dy > 0$. But this integral is $|E\cap (h+E)|$.