问:
To check $N\preccurlyeq^+ M$, does it suffice to consider single-variable formulas?
Let $R$ any unital ring, and let $\mathcal{L}_R=\{+,-,0,r\}_{r\in R}$ be the language of left $R$-modules, where each $r$ is a unary function symbol. Recall that a positive primitive (pp) $\mathcal{L}_R$-formula is one of the form $\exists\overline{w}\bigwedge_{i=1}^k\overline{r_i}\bullet\overline{v}=\overline{s_i}\bullet\overline{w}$, where each $\overline{r_i}$ and $\overline{s_i}$ is a tuple of elements of $R$, and $\bullet$ denotes the "dot product". (So, eg, $\overline{r_i}\bullet\overline{v}=\sum_{j=1}^mr_{ij}\cdot v_j$.) It is well-known that, modulo the complete theory of any $R$-module, every $\mathcal{L}_R$-formula is equivalent to a boolean combination of pp-formulas. Given a pair of modules $N\leqslant M$, we say that $N$ is a pure submodule of $M$, denoted $N\preccurlyeq^+M$, if $$N\models\psi(\overline{a})\iff M\models\psi(\overline{a})$$ for any pp-formula $\psi$ and any tuple $\overline{a}\in N$. (Note that the forward implication always holds, so the content of purity is the backwards implication.)

Question: When checking that $N$ is a pure submodule of $M$, does it suffice to consider pp-formulas in a single variable? $\square$

This seems like a very natural and straightforward question, but the answer is unclear to me. Certainly if $\psi$ is of the form $\exists\overline{w}(\overline{r}\bullet\overline{v}=\overline{s}\bullet\overline{w})$ then the desired implications hold. Indeed, then the sentence $\psi(\overline{a})$ for a tuple $\overline{a}\in N$ is equivalent to the sentence $\psi'(a^1)$, where $a^1=\overline{r}\bullet\overline{a}\in N$ and $\psi'$ is the single-variable formula $\exists \overline{w}(v=\overline{s}\bullet\overline{w})$. However, things seem less straightforward if $\psi$ has multiple conjuncts. By a similar argument as above (considering each $\overline{r_i}\bullet\overline{a}$ as a single element of $N$), it suffices to consider formulas of the form $\exists\overline{w}\bigwedge_{i=1}^kv_i=\overline{s_i}\bullet\overline{w}$. Then saying that $N$ is pure in $M$ with respect to formulas in one variable amounts to saying that each equation $a_i=\overline{s_i}\bullet\overline{w}$ has a solution in $N$ if and only if it has a solution in $M$, for every $a_i\in N$. But I don't immediately see how we can "stitch" these solutions together in the way that we want, and I'm not even sure if we can in general. Does anyone have any thoughts?

A counterexample, or a proof for a particularly nice class of ring (PIDs perhaps?), would also be much appreciated. For instance, I believe the result holds if $R$ is a local PID. Sketch: suppose $R$ has maximal ideal $(\pi)$. Then every element of $R$ is associate with a power of $\pi$, so we may assume that the length of the tuple $\overline{w}$ in the formula above is $1$, and that each $s_i$ is of the form $\pi^{n_i}$ for some $n_i$. Then, assuming wlog that $n_1$ is minimal among the $n_i$, we have $N\models\exists w\bigwedge_{i=1}^ka_i=\pi^{n_i}w$ if and only if
$$\text{(i) }a_1\text{ is divisible by }\pi^{n_1}\text{ in }N\ \ \ \ \ \ \ \text{  and }\ \ \ \ \ \ \ \text{(ii) }a_i=\pi^{n_i-n_1}a_1\text{ for each }i.$$ Clearly (ii) holds in $N$ if and only if it holds in $M$, and (i) is covered by the pp-criterion for single variable formulas, and so we are done.

答:
Okay, I think I've found a proof in the case that $R$ is a PID. Throughout all modules will be taken over $R$. For convenience, we will call a submodule $N\leqslant M$ "weakly pure" if $N$ and $M$ agree on all pp-sentences with a single parameter from $N$. In other words, $N$ is weakly pure in $M$ if $$N\models\varphi(a)\iff M\models\varphi(a)$$ for any single-variable pp-formula $\varphi(v)$ and any $a\in N$. (Comment: the lemma below is based on a fact proved in Kaplansky's Infinite Abelian Groups.)

Lemma: If $N$ is weakly pure in $O$, and $O\big/N$ is finitely generated, then $N$ is a direct summand of $O$.

Proof: Since $R$ is a PID, the structure theorem for finitely generated modules over a PID tells us that $O\big/N$ is a direct sum of cyclic modules. Say $O\big/N=R\overline{o_1}\oplus\dots\oplus R\overline{o_n}$ for some $o_i\in O$.

Claim: We may assume that $\operatorname{ann}\overline{o_i}=\operatorname{ann}{o_i}$ for each $i$. To see this, suppose $\operatorname{ann}\overline{o_i}=(r)$ for some $r\in R$, recalling that $R$ is a PID. Then $ro_i\in N$, and clearly $O\models\exists w(ro_i=rw)$, so by weak purity $N\models\exists w(ro_i=rw)$ and thus there is $n\in N$ with $ro_i=rn$. Then $\overline{o_i-n}=\overline{o_i}$ mod $N$, and $$(r)=\operatorname{ann}\overline{o_i}\supseteq\operatorname{ann}(o_i-n)\supseteq (r),$$ so $o_i-n$ gives the desired generator of $R\overline{o_i}$. $\dashv$

In particular, $N\cap Ro_i=\{0\}$; indeed, if $n=so_i$ for some $n\in N$ and $s\in R$, then $s\overline{o_i}=\overline{0}$. But this means $s\in\operatorname{ann}\overline{o_i}=\operatorname{ann}o_i$, so $n=so_i=0$, as desired. We claim $O=N\oplus \bigoplus_{i=1}^nRo_i$. Indeed, if $$n+r_1o_1+\dots+r_no_n=0$$ for some $n\in N$ and $r_i\in R$, then $\sum_{i=1}^nr_i\overline{o_i}=\overline{0}\in O\big/N$, and so (since $O\big/N$ is a direct sum of the $\overline{o_i}$) we have $r_1\overline{o_1}=\dots=r_n\overline{o_n}=\overline{0}$. This means each $r_io_i\in N$, and hence, since $N\cap Ro_i=\{0\}$, we have $r_io_i=0$. Thus $n=0$ too, as desired. $\square$

Theorem: If $N$ is weakly pure in $M$, then $N\preccurlyeq^+M$.

Proof: Let $\psi(\overline{v})\equiv\exists\overline{w}\bigwedge_{i=1}^k\overline{r_i}\bullet\overline{v}=\overline{s_i}\bullet\overline{w}$ be an arbitrary pp-formula and suppose that $M\models\psi(\overline{a})$ for some tuple $\overline{a}\in N$. Then there is a tuple $\overline{c}\in M$ such that $\overline{r_i}\bullet\overline{a}=\overline{s_i}\bullet\overline{c}$ for each $i\in\{1,\dots,k\}$, and we want to show that such a tuple also exists in $N$. To see this, consider the submodule $$O:=N+Rc_1+\dots+Rc_n$$ of $M$. Clearly $O\big/N$ is finitely generated, by the $\overline{c_i}$. Also, we claim that $N$ is weakly pure in $O$. Indeed, if $O\models\varphi(n)$ for a single-variable pp-formula $\varphi$ and an element $n\in N$, then $M\models\varphi(n)$ with the same witness, and so, since $N$ is weakly pure in $M$, $N\models\varphi(n)$ too, as desired. Thus the hypotheses of our lemma apply, and so $N$ is a direct summand of $O$. Denoting by $\pi:O\to N$ the canonical projection map, we then have $\overline{r_i}\bullet \overline{a}=\pi(\overline{r_i}\bullet\overline{a})=\pi(\overline{s_i}\bullet\overline{c})=\overline{s_i}\bullet\pi(\overline{c})$ for each $i$, since $\pi$ commutes with scalar multiplication. In particular, the $\pi(c_i)$ provide a witness for $\psi(\overline{a})$ in $N$, and so we are done. $\square$

Note that this proof relies heavily on the structure of PIDs, so I am now skeptical that the desired result holds in the general case, but I'm still struggling to think of a counterexample. Any thoughts would be appreciated!


问:
I can't wrap my head around it as no roots are given.
If $x^4+2x^3+px^2+qx+9=0$ is a complete square. $p$ and $q$ are positive. Find the value of $p$ and $q$.

答:
If it is a complete square, then

$$x^4+2x^3+px^2+qx+9=(x^2+ax+b)^2=x^4+2ax^3+(2b+a^2)x^2+2abx+b^2$$

(we have the coefficient in front of $x^2$ is $1$ since the coefficient in front of $x^4$ is $1$). Clearly, $b=3$ and $a=1$ which gives us

$$x^4+2x^3+px^2+qx+9=(x^2+x+3)^2=x^4+2x^3+7x^2+6x+9$$

$$px^2+qx=7x^2+6x$$

So take $p=7$ and $6=q$.

ContentLicense: CC BY-SA 4.0


问:
Find the dual problem to the following $\min$ $\alpha$ s.t $Aq=\alpha f$
I came across this question while studying duality

$\min$ $\alpha$

s.t $Aq=\alpha f$

$||q||^2\leq\epsilon$

where $\alpha\in\mathbb{R},q\in\mathbb{R}^n$ are the variables and $A\in\mathbb{R}^{m\times n},f\in\mathbb{R}^m,\epsilon\in\mathbb{R}_{++}$ and the rows of A are linearly independent.
Find the dual problem to and don't assign a lagrange multiplier to the quadratic constraint.

The lagragian is :

$$\underset{||q||^2\leq\epsilon}{L(\alpha,q,\lambda)}=\alpha(1-\lambda^Tf)+\lambda^TAq$$
Solving first for $\alpha$ I get that $\underset{\alpha}{\min} \alpha(1-\lambda^Tf)=\left\{\begin{array}{rcl}0&1-\lambda^Tf=0\\-\infty&else\end{array}\right.$

And for $q$ i get the following $\underset{||q||^2\leq\epsilon}{\min}\lambda^TAq$ and I'm not sure how to continue from here. I think that I need to demand that $\lambda^TA=0$ or else I $-\infty$ but not sure.

any hint?

答:
First, rewrite for simplicity

$$
L'(q, \lambda) := \min_{\alpha} L(\alpha, q, \lambda) = \delta_{C}(\lambda) + \langle A^* \lambda, q \rangle,
$$

where $\delta_{C}(\lambda)$ is $0$ if $1 - \lambda^* f = 0$ and $+\infty$ otherwise.
From there, you can minimize over $q$ since

$$
\min_{q \in \sqrt{\varepsilon} B} \left\{ \delta_{C}(\lambda) + \langle A^* \lambda, q \rangle \right\} = \delta_{C}(\lambda) + \min_{q \in \sqrt{\varepsilon} B} \langle A^* \lambda, q \rangle \\
= \delta_{C}(\lambda) + \frac{1}{\sqrt{\varepsilon}} \min_{z \in B} \langle A^* \lambda, q \rangle \overset{(*)}{=}
\delta_{C}(\lambda) - \frac{\| A^* \lambda \|_2}{\sqrt{\varepsilon}},
$$

where in $(*)$ we made use of the fact that the minimum of $x \mapsto \langle y, x \rangle$ over the unit ball is $- \| x \|_2$.


问:
Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$
Hopefully I am reading the correct line from LMFDB.

Let $a$ be the algebraic number solving $a^4 - a^3  - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3  - x^2 + x + 1)$.

Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.

答:
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$

divide both sides by $a^2$ to find

$$0=a^2-a-1+\dfrac1a+\dfrac1{a^2}=\left(a-\dfrac1a\right)^2-\left(a-\dfrac1a\right)+1=0$$

$\implies a-\dfrac1a=-w$ where $w$ is a complex cube root of unity

$\implies a^2=1-wa$

$\implies a^3=a(1-wa)=a-wa^2=a-w(1-wa)=a(1+w^2)-w=-wa-w$

Compare the values of $wa$

$$1-a^2=a^3-w$$

$$\iff a^2-a^3=1+w=-w^2\ne1$$

$$\implies(a^2-a^3)^2=?\ne1$$

$$\implies(a^2-a^3)^3=?\ne1$$

$$\implies(a^2-a^3)^6=?$$


问:
Extending the Spectral Theorem of Unbounded Self-Adjoint Operators on Infinite-Dimensional Hilbert Spaces
I'm a physics student trying to do the maths of the Hilbert space in quantum mechanics with a bit more rigour than I'm accustomed to. I am trying to find ways to extend the spectral theorem for unbounded self-adjoint operators on an infinite dimensional Hilbert space to other operators. The form of the spectral theorem I'm working with is from Hall, which says that a self-adjoint operator is unitarily equivalent to a multiplication operator on some other Hilbert space. (Page 207, Theorem 10.10.) Specifically, I have an operator $\hat{H}$ on the infinite dimensional Hilbert space, which is not itself self-adjoint but is related to self-adjoint operators by at least two simple relations. First, both

$$
      \hat{H}\hat{\gamma}^0 \; \text{and} \; \hat{\gamma}^0\hat{H}
$$

are self-adjoint, where $\hat{\gamma}^0$ can be written in block-diagonal form as

$$
    \hat{\gamma}^0 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
$$

It therefore has all sorts of useful properties, such as being bounded, Hermitian, invertible, unitary, and $(\hat{\gamma^0})^2=1$. I also have the knowledge that

$$
     (\hat{H})^2
$$

is self-adjoint. Is there any way of proving that $\hat{H}$ must be isomorphic (though presumably not in general by a unitary transformation) to a multiplication operator on a Hilbert space, from the fact that $\hat{H}^2$ and/or $\hat{H}\hat{\gamma}^0$ are? It feels like there might or might not be something obvious I'm missing. Would any further properties of $\hat{H}$ be needed?

答: I'd suggest checking Frederic Schuller's Youtube Channel. He covers those kinds of topics from a physics perspective.</p>


问:
Everyone is passed everything exactly once, but never from the same person
Say 4 or more people are sitting around a table. Each has a sheet of paper. Devise an algorithm to pass these papers between these people that guarantees:

Each person passes and signs every piece of paper exactly once, no more, no less

Every time the papers are passed, each person receives a paper from a person they have not received it from before

The papers must all be passed the exact same number of times as there are people (i think this follows from 1, just making sure)

People can only have one paper at a time; in other words, no person can be signing two papers at the same time.

Questions:

Is there a mathematical name for this problem? Or can anyone find internet pages where this problem has been discussed before?

Is this possible for every number of players?

If so, is there an algorithm to generate this?

If not, is there an algorithm that can at least minimize the number of times people receive a paper from someone they have received it from before?

This has real world implications!!! I would like to use this theoretical algorithm in my game Drawphone, more info here about how it fits in. Thanks! :)

答:
W1hat you are looking for is equivalent to a row-complete Latin square. A Latin square is an $n\times n$ table filled with the numbers $1$ to $n$ so that each row and column has no repeats. A Latin square is further called row-complete if the $n\times (n-1)$ ordered pairs of numbers which occur horizontally adjacently in the square are also all distinct.

To apply this to your problem, given a row-complete Latin square, numbered the papers from $1$ to $n$. The rows represent people, the columns represent time, and the label at any row and column represents the paper that that person signs at the time. The Latin square conditions imply everyone has a different paper, and that everyone signs all papers. The row-complete condition ensures everyone receives their papers from a different person each time.

It is easy to solve the problem when $n$ is even. First, everyone passes their paper one spot to the left, then two spots to the right, then three spots to the left, then four spots to the right, and so on, ending with passing $(n-1)$ spots to the left.

It is proven in [Higgam] that when $n$ is odd and composite, a row complete Latin square of order $n$ exists. The construction uses finite fields. The construction is provided in that paper, and is also available in this survey by Otis, available online at https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS10, in section 3.6.

All that remains are odd primes. No row-complete Latin square has been found for any odd prime order. Furthermore, exhaustive searches have shown there are no squares of orders $3,5$ and $7$. It is an open problem for higher odd primes.

In the odd prime case, where no row-complete Latin square is known to exist, the following method ensures that there is at most one repeat among the people who pass you the papers. Let $n=2m+1$. The algorithm differs slightly when $m$ is even vs odd.

In either case, the first $m$ steps are the same: start with passing one left, two right, three left, four right, and so on, ending at passing $m$ in a direction depending on the parity of $m$.

Next, if $m$ is odd...

Pass one spot to the left (this is the repeated pass). Then, pass $(m-1)$ spots left, $(m-2)$ right, $(m-3)$ left, $(m-4)$ right, and so on, down to passing one spot right.

If instead $m$ is even...

Start pass one spot to the right. Then, pass $(m-1)$ spots right, $(m-2)$ spots left, $(m-3)$ right, $(m-4)$ left, and so on down to $1$ spot right (this is the repeated step).

Examples

Approximate solution for $n=7$: (note the each explanation is for each column despite being written next to the row).

```<code>1 2 7 3 4 6 5   The first column is the initial paper distribution.
2 3 1 4 5 7 6   The second column is attained from the first by shifting up one.
3 4 2 5 6 1 7   The third is attained by the second by shifting down two.
4 5 3 6 7 2 1   The fourth from the third by shifting up three.
5 6 4 7 1 3 2   Next, you shift up one (again).
6 7 5 1 2 4 3   Then, shift up two.
7 1 6 2 3 5 4   Finally, shift down one.
</code>```

Approximate solution for $n=9$:

```<code>1 2 9 3 8 7 4 6 5
2 3 1 4 9 8 5 7 6    up one
3 4 2 5 1 9 6 8 7    down two
4 5 3 6 2 1 7 9 8    up three
5 6 4 7 3 2 8 1 9    down four
6 7 5 8 4 3 9 2 1    down one
7 8 6 9 5 4 1 3 2    down three
8 9 7 1 6 5 2 4 3    up two
9 1 8 2 7 6 3 5 4    down one
</code>```

Exact solution for $n=9$:

```<code>1 4 7 2 6 8 5 9 3
4 2 1 8 7 3 9 5 6   For the explanation, you will have to read the cited sources.
7 8 6 5 2 4 3 1 9
2 5 8 3 4 9 6 7 1
5 3 2 9 8 1 7 6 4
8 9 4 6 3 5 1 2 7
3 6 9 1 5 7 4 8 2
6 1 3 7 9 2 8 4 5
9 7 5 4 1 6 2 3 8
</code>```

Higham, J. (1998), Row‐complete latin squares of every composite order exist. J. Combin. Designs, 6: 63-77. https://doi.org/10.1002/(SICI)1520-6610(1998)6:1<63::AID-JCD5>3.0.CO;2-U

Body:
W1hat you are looking for is equivalent to a row-complete Latin square. A Latin square is an $n\times n$ table filled with the numbers $1$ to $n$ so that each row and column has no repeats. A Latin square is further called row-complete if the $n\times (n-1)$ ordered pairs of numbers which occur horizontally adjacently in the square are also all distinct.

To apply this to your problem, given a row-complete Latin square, numbered the papers from $1$ to $n$. The rows represent people, the columns represent time, and the label at any row and column represents the paper that that person signs at the time. The Latin square conditions imply everyone has a different paper, and that everyone signs all papers. The row-complete condition ensures everyone receives their papers from a different person each time.

It is easy to solve the problem when $n$ is even. First, everyone passes their paper one spot to the left, then two spots to the right, then three spots to the left, then four spots to the right, and so on, ending with passing $(n-1)$ spots to the left.

It is proven in [Higgam] that when $n$ is odd and composite, a row complete Latin square of order $n$ exists. The construction uses finite fields. The construction is provided in that paper, and is also available in this survey by Otis, available online at https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS10, in section 3.6.

All that remains are odd primes. No row-complete Latin square has been found for any odd prime order. Furthermore, exhaustive searches have shown there are no squares of orders $3,5$ and $7$. It is an open problem for higher odd primes.

In the odd prime case, where no row-complete Latin square is known to exist, the following method ensures that there is at most one repeat among the people who pass you the papers. Let $n=2m+1$. The algorithm differs slightly when $m$ is even vs odd.

In either case, the first $m$ steps are the same: start with passing one left, two right, three left, four right, and so on, ending at passing $m$ in a direction depending on the parity of $m$.

Next, if $m$ is odd...

Pass one spot to the left (this is the repeated pass). Then, pass $(m-1)$ spots left, $(m-2)$ right, $(m-3)$ left, $(m-4)$ right, and so on, down to passing one spot right.

If instead $m$ is even...

Start pass one spot to the right. Then, pass $(m-1)$ spots right, $(m-2)$ spots left, $(m-3)$ right, $(m-4)$ left, and so on down to $1$ spot right (this is the repeated step).

Examples

Approximate solution for $n=7$: (note the each explanation is for each column despite being written next to the row).

1 2 7 3 4 6 5   The first column is the initial paper distribution.
2 3 1 4 5 7 6   The second column is attained from the first by shifting up one.
3 4 2 5 6 1 7   The third is attained by the second by shifting down two.
4 5 3 6 7 2 1   The fourth from the third by shifting up three.
5 6 4 7 1 3 2   Next, you shift up one (again).
6 7 5 1 2 4 3   Then, shift up two.
7 1 6 2 3 5 4   Finally, shift down one.

Approximate solution for $n=9$:

1 2 9 3 8 7 4 6 5
2 3 1 4 9 8 5 7 6    up one
3 4 2 5 1 9 6 8 7    down two
4 5 3 6 2 1 7 9 8    up three
5 6 4 7 3 2 8 1 9    down four
6 7 5 8 4 3 9 2 1    down one
7 8 6 9 5 4 1 3 2    down three
8 9 7 1 6 5 2 4 3    up two
9 1 8 2 7 6 3 5 4    down one

Exact solution for $n=9$:

1 4 7 2 6 8 5 9 3
4 2 1 8 7 3 9 5 6   For the explanation, you will have to read the cited sources.
7 8 6 5 2 4 3 1 9
2 5 8 3 4 9 6 7 1
5 3 2 9 8 1 7 6 4
8 9 4 6 3 5 1 2 7
3 6 9 1 5 7 4 8 2
6 1 3 7 9 2 8 4 5
9 7 5 4 1 6 2 3 8

Higham, J. (1998), Row‐complete latin squares of every composite order exist. J. Combin. Designs, 6: 63-77. https://doi.org/10.1002/(SICI)1520-6610(1998)6:1<63::AID-JCD5>3.0.CO;2-U


问:
Is my algebra alright, despite obtaining a different output from Matlab
I am going through previous algebra exercises that I had performed from a textbook in Matlab.

However, although I am certain that my results are correct, Matlab is throwing up different answers to mine.

Examples of some of these questions are:

$$ 16x^2 - 40xy + 25y^2 = 0$$

For this, I get factors of $$ (4x - 5y)^2 = 0$$

$$ 5x^2 + 13x + 6 = 0$$

For this, I get factors of $$ (5x + 3)(x + 2) = 0$$

$$ 2x^2 - 11x + 12 = 0$$

For that, I get factors of $$ (2x - 3)(x - 4) = 0$$

And the results produced by Matlab are shown below:

Ultimately, what I am seeking is just confirmation that my answers are correct and that Matlab is just displaying other possible solutions!

答:
Your algebra is simply wrong for part 1.

$$
(3x-4)(2y+5) = 6xy - 8y + 15x - 20
$$
which is not at all the same as $$16^2−40+25^2$$.

In fact, $16^2−40xy+25^2 = 0$ actually factors as
$$
(4x - 5y)^2 = 0
$$
for which any pair $(x,y) = (a, -\frac{5}{4}a)$ is a solution.

I don't understand Matlab's answer that $y$ is $0$ or $0$, however.

Part 2: Your algebra is right here, but Matlab's answer is baffling. No positive value of $x$ can yield a root of that equation, because all three terms are positive, and hence cannot sum to zero. Perhaps "solve" doesn't work the way you (and I) may think it does. Then again, when I tried running it, Matlab gave the answers $-2, -3/5$. Did you remember to declare x symbolic with sym x?


问:
How to show that a set is not in PCP (Post Correspondence Problem)?
In my lecture slides I have two examples of sets of tuples, where the first one is in PCP while the other one is not. The first is:

$$PCP:=\{<(x_1,y_1)=(1,101), (x_2,y_2)=(10,00), (x_3,y_3)=(011,11)>\} \in PCP$$

The solution for this is $x_1 \cdot  x_3 \cdot x_2 \cdot x_3=y_1 \cdot  y_3 \cdot y_2 \cdot y_3$. Since the set only has $3$ elements you can use a simple trial and error method and figure out a solution. However, what if the number of tuples is larger, let's say 9 tuples? How can you show that or rather proof that no matter the arrangement of the tuples, the set is not in PCP?

PS - Since PCP is not in NP, that means that one could write a program that could check whether a given set is in PCP. However, I am looking for a more mathematical solution.

答:
Well, the PCP is undecidable. I.e., there is no general algorithm to decide if a PCP has a solution or not.

You could use some heuristics, or simply trial and error.


问:
$X\geq0$ and $X$ a random variable, show that $\lim_{n\to\infty} \frac{1}n\cdot E(\frac1X\cdot 1_{X>\frac1n})$=0
The only result I got so far is, that the limit has to be smaller than 1. This I derived via monotonicity and $\frac1X$<n. I have no clue how to continue and a hint would be appreciated.

答: Hint: Lebesgue dominated convergence theorem.</p>


问:
Show through $\varphi (z) = \frac{z-i}{z+1}$ diffeomorphism $\mathbb{R}^{2}_{+}\longrightarrow U$, defined w.r.t. the hyperbolic metric is an isometry

$\mathbb{R}^2_+$ and $U$ are subsets of $\mathbb{C}$ with:

$\mathbb{R}^2_+ = \left \{ (x,y) \in \mathbb{R}^2 : y > 0  \right \}$

$U = \left \{ (u,v) \in \mathbb{R}^2 : u^2 +v^2 < 1 \right \}$

Show that through $\varphi (z) = \frac{z-i}{z+1}$ a Diffeomorphism $\mathbb{R}^2_+ \rightarrow U$ is
defined, with respect to the respective hyperbolic metric is an isometry.

I calculated the metric tensors:
$g_{ij}(U) = \begin{pmatrix}
\frac{4}{(1-u^2-v^2)^2} &0 \\
0 & \frac{4}{(1-u^2-v^2)^2}
\end{pmatrix}$
and $g_{ij}(\mathbb{R}^2_+) = \begin{pmatrix}
\frac{1}{y^2} &0 \\
0 & \frac{1}{y^2}
\end{pmatrix}$

However, I dont know how to show that its an isometry.

答:
The map that you wrote is not an isometry, I'll give you the correct definition. It's a very long calculation but not so difficult.

Recall that a smooth map $f:(X,g)\rightarrow (Y,h)$ between riemannian manifold is an isometry if it is a diffeomorphism and $f^*h=g$, where $\forall u,v\in T_pX$ one has $f^*h(u,v):=h(d_pf(u),d_pf(v))$ and $d_pf: T_pX\rightarrow T_{f(p)}Y$ is the differential of $f$ at $p$.

Step 1:

Let $\mathbb{H}=\{(u,v)\in\mathbb{R}^2 | v>0 \}$ with metric $g=\frac{du^2+dv^2}{v^2}$ and let $\mathbb{B}=\{(x,y) | \sqrt{x^2+y^2}<1\}$ with metric $h=\frac{4}{1-x^2-y^2}(dx^2+dy^2)$. Define the map $\displaystyle f:\mathbb{H}\rightarrow\mathbb{B}$ to be $f(z):=\frac{z-i}{z+i}$ where I'm identifying $z=x+iy$.
The first thing one has to show is that Im$(f)\subset\mathbb{B}$, i.e. that for all $z\in\mathbb{H}$ then $f(z)\in\mathbb{B}$. This is equivalent to prove that if $\Im(z)>0$ then $|z-i|<|z+i|$ and you can check it by yourself.

Step 2:

$f$ is a diffeomorphism (notice that $-i\notin\mathbb{H}$).

Check that the map $\sigma:\mathbb{B}\rightarrow\mathbb{H}$ given by $\sigma(w):=\frac{w+1}{i(w-1)}$ is a well defined smooth map and inverse of $f$.

Step 3:

$f^*h=g$ which implies that $f$ is an isometry between the hyperbolic plane and the Poincarè disk.

To do this just write $h$ and $g$ in terms of complex numbers $z=x+iy$ and $w=u+iv$ and then use the definition of the pullback metric above.

Try to do this by your own it's a very useful exercise!


问:
How to "invent" a function?
I'm currently making a game. I was given some numbers the leader wants me to stick close to. They are used for leveling; level 1, for example, needs 100 experience points, level 2 needs 250 experience points, etc.

Is there any way, maybe with a computer program, to calculate a function that will do something like this, instead of having to hard-code it?

Thanks in advance!

Simon.

答:
I think interpolation is not the way to go, as levels are discrete in any way and especially the lagrange interpolation might really not do what you want (most answers suggest to use it to interpolate the few first values, but then the function can even be negative for higher levels although it fits what you originally wanted, and if you interpolate over all values you could directly use a table, I see no advantage here).

The experience function is something that should be carefully designed as it is one of the main goals for the player to level up (and the game should not get too boring after a few levels, at the same time it shouldn't be too fast to level up). Also you have to consider that the experience you gain from defeating enemys should maybe also increase over the time.

Therefore there are a few things you should consider in your formula:

The level of the character (of course, this will be the main parameter)

The average amount of experience the character gets by defeating an enemy

A difficulty factor?

As an example you can learn from the experience function of this popular game. I can assure you that they thought a lot about balance and especially experience. For example in patch 2.3 they added a different factor to speed up the level process in certain levels because it was too difficult to level up. I also recommend this. I would recommend you to use a similar approach and maybe you should use a hardcoded table.

ContentLicense: CC BY-SA 3.0

Title: How to "invent" a function?


问:
Plane geometry tough question
$\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.</p>

答:
Hint:
You can solve following these steps:

Title: Plane geometry tough question


问:
Matrix to represent a map
Consider the map $f:\mathbb{R}^3 \rightarrow  \mathbb{R}^3  $ defined by $$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=(x_1-x_2+x_3)b_1 + (x_1-x_2-x_3)b_2 + (x_1+x_2-x_3)b_3$$

where $b_1 = \begin{pmatrix}
1\\
0\\
1
\end{pmatrix},b_2 = \begin{pmatrix}
0\\
1\\
1
\end{pmatrix},b_3 = \begin{pmatrix}
1\\
1\\
0
\end{pmatrix}$ is a basis for $\mathbb{R}^3 $ (this has been shown).

Problem

Determine the matrix $C$ that is represented by the given $f$ with respect to the basis $(b_1,b_2,b_3)$ for both domain and codomain. Furthermore, determine the coordinates of $f(b_1) \in \mathbb{R}^3$ with respect to the basis $(b_1,b_2,b_3)$.

Try

First off this is a pratice problem. I would think that $C=BP$ where $P$ would just be the matrix of the $j$'s basis vectors as columns and $B$ would be matrix representing the standard basis evaluated in the map $f$ and thus give us the $j$'s standard basis would give the $j$'s column in the matrix. Is this correct or is there an easier way since it is a endomorphism?

答: Note that$$f(b_1)=2b_1,\ f(b_2)=-2b_2\text{, and }f(b_3)=2b_3.$$So,$$C=\begin{bmatrix}2&0&0\\0&-2&0\\0&0&2\end{bmatrix}.$$I think that this is the simplest way.

Title: Matrix to represent a map


问:
Quantitative Freivalds' algorithm
I recently became aware of Freivalds' algorithm: Given $n\times n$ matrices $A, B, C$, we can test whether $A\cdot B = C$ by simply multiplying either side with $k$  random vectors and check whether the results coincide. Since the failure rate of the test  decreases exponentially as $\mathcal O(2^{-k})$, in practice we only need to test $k\ll n$ vectors.

Now there are 2 issues with this:

Due to round-off errors, even if $AB=C$ we shouldn't expect $ABv=Cv$ exactly when computing with floating point numbers

If the test fails, we only know $AB\neq C$, but not how unequal they likely are, e.g. measured in terms of a matrix norm

So is there a quantitative version of Freivalds' algorithm, that can tell me something about $\|AB-C\|$, given $\|ABv_i - Cv_i\|$ for random vectors $(v_i)_{i=1\ldots k}$?
E.g. an error bound of the following form would be nice:

If $\|ABv_i - Cv_i\|\le \epsilon$ for random vectors $(v_i)_{i=1\ldots k}$ then $\|AB-C\|\le \mathcal O(g(\epsilon))$ with probability at least $p_k$

PS: Should there be a tag for probabilistic algorithms / probabilistic numerics?

答:
The HMT error bound (see Lemma 4.1) accomplishes this goal. For $E = AB - C$,

$$
\|E\|\le \alpha \sqrt{\frac{2}{\pi}} \max_{i=1,2,\ldots,k} \|E v_i\| \mbox{ with probability } \ge 1-\alpha^k
$$

where $v_1,\ldots,v_k$ are independent vectors with IID standard normal entries and $\|\cdot\|$ the spectral (operator $2$-) norm. This is precisely the resume you as for with $g(\epsilon) = \alpha \sqrt{2/\pi}$ and $p_k = 1-\alpha^k$.

Title: Quantitative Freivalds' algorithm


问:
Scale Invariance of the Einstein-Hilbert Functional
Define the normalised Einstein Hilbert functional for an n-dimensional Psuedo-Riemannian manifold ($M, g$) as

$\mathcal{E}(g) = \frac{{\int_{\mathcal{M}}S_g \Omega_g}}{\left({\int_{\mathcal{M}} \Omega_g}\right)^{\frac{n-2}{n}}}$.

where $S_g$ is the scalar curvature and $\Omega_g$ is the Riemannian volume form for $g$. I'd like to show that this is scale invariant, ie. $\mathcal{E}(g) = \mathcal{E}(cg)$ for $c \in \mathbb{R}$.

Defining $\Omega_{cg}$ and $S_{cg}$ in the obvious way and using the rule on scalar curvature $S_{cg}=\frac{1}{c}S_g$ I get

$\frac{\mathcal{E}(g)}{{\mathcal{E}(cg)}} = \frac{{\int_{\mathcal{M}}S_g \Omega_g}}{{\int_{\mathcal{M}}S_{cg} \Omega_{cg}}}$  $\frac{{\left({\int_{\mathcal{M}} \Omega_{cg}}\right)^{\frac{n-2}{n}}}}{{\left({\int_{\mathcal{M}} \Omega_g}\right)^{\frac{n-2}{n}}}}$ = $ \frac{S_g(vol)_g}{\frac{S_g}{c}(vol)_{cg}} \left(\frac{(vol)_c}{(vol)_{cg}}\right)^{\frac{n-2}{n}}$.

But $(vol)_{cg} = c^{n}(vol)_g$ leaving us with

$\frac{\mathcal{E}(g)}{{\mathcal{E}(cg)}} = \frac{c}{c^n} \frac{1}{c^{n-2}} = c^{-1} \neq 1$.

I'm wondering what I have done wrong here?

答:
You've made two mistakes.

The first mistake is that $\int_M S_g\Omega_g \neq S_g\operatorname{vol}_g$; the left hand side is a number, while the right hand side is a function unless $S_g$ is constant, in which case you do get an equality.

Secondly, $\Omega_{cg} = c^{\frac{n}{2}}\Omega_g$ not $c^n$ (think about what happens when you replace $g$ by $cg$ in this local expression).

With these things in mind, note that

$$\int_M S_{cg}\Omega_{cg} = \int_M c^{-1}S_g c^{\frac{n}{2}}\Omega_g = c^{\frac{n}{2}-1}\int_MS_g\Omega_g = c^{\frac{n-2}{n}}\int_M S_g\Omega_g$$

and

$$\left(\int_M\Omega_{cg}\right)^{\frac{n-2}{n}} = \left(\int_Mc^{\frac{n}{2}}\Omega_g\right) = \left(c^{\frac{n}{2}}\int_M\Omega_g\right)^{\frac{n-2}{n}} = c^{\frac{n-2}{2}}\left(\int_M\Omega_g\right)^{\frac{n-2}{n}}.$$

Therefore

$$\mathcal{E}(cg) = \dfrac{\displaystyle\int_M S_{cg}\Omega_{cg}}{\left(\displaystyle\int_M\Omega_{cg}\right)^{\frac{n-2}{n}}} = \dfrac{c^{\frac{n-2}{n}}\displaystyle\int_M S_g\Omega_g}{c^{\frac{n-2}{2}}\left(\displaystyle\int_M\Omega_g\right)^{\frac{n-2}{n}}} = \dfrac{\displaystyle\int_M S_g\Omega_g}{\left(\displaystyle\int_M\Omega_g\right)^{\frac{n-2}{n}}} = \mathcal{E}(g).$$


问:
How to compute: $\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$
This question was asked in a masters of mathematics exam for which I am preparing.

Compute $\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$.

I could only think of substituting $y^3 = x$ and that does not change much.

Could somebody post a solution using residues or in ways besides here in this link?:
How to compute the integral $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$

Edit : I am interested in the answers which use contour integration and residue calculus.

答:
Substitution
$$x=y^{3/2}$$
allows to write
$$I=\int\limits_0^\infty\dfrac{\sqrt[\large3]x\,\text dx}{1+x^2}
= \dfrac32\int\limits_0^\infty\dfrac{y\,\text dy}{1+y^3}.$$
At the same time, substitution
$$y=\dfrac1z$$
gives
$$I= \dfrac32\int\limits_0^\infty\dfrac{\text dz}{1+z^3}
= \dfrac32\int\limits_0^\infty\dfrac{\text dy}{1+y^3}.$$
Therefore,
$$I= \dfrac34\int\limits_0^\infty\dfrac{(1+y)\,\text dy}{1+y^3}
= \dfrac34\int\limits_0^\infty\dfrac{\text dy}{1-y+y^2}
=3\int\limits_0^\infty\dfrac{\text dy}{3+(2y-1)^2}.\tag1$$
Integral $(1)$ does not require the residue approach:
$$I=\dfrac{\sqrt3}2\arctan\dfrac{2y-1}{\sqrt3}\bigg|_0^\infty
 = \dfrac{\sqrt3}2\left(\dfrac\pi2+\dfrac\pi6\right),$$
$$\color{brown}{\mathbf{I=\dfrac\pi{\sqrt3}.}}$$


答:
Okay, I'm assuming you mean that the polynomial $f(x) = x^4 + 2x^3 + px^2 + qx + 9$ is a square, that is there is some other polynomial $g(x)$ with $g(x) * g(x) = f(x)$.

So what could this polynomial $g(x)$ be?  Well, the highest-degree term in $f(x)$ is $x^4$, so that means the highest-degree term in $g(x)$ should be $x^2$ ($x^2 * x^2 = x^4$).

So $g(x) = x^2 + ax + b$, for some constants $a$ and $b$.  Now lets look at what happens when we multiply it out.  We get $f(x) = g(x) * g(x) = x^4 + 2ax^3 + (2b + a^2) x^2 + 2abx + b^2$.  This tells us a few things.  First of all, by looking at the $x^3$ term we see $2a = 2$, or $a = 1$.  Second, by looking at the constant term, we see $b^2 = 9$, so $b = \pm 3$.  So that gives us two possibilities for $g(x)$. $g(x) = x^2 + x + 3$ or $g(x) = x^2 + x - 3$.

Now lets look at what happens with $p$ and $q$ in both cases. Remember, by looking at the multiplication we have $p = (2b + a^2)$ and $q = 2ab$.  So if $p$ and $q$ are positive, we have to take the $+3$ option.  In which case $p = 7$ and $q = 6$.

Title: I can't wrap my head around it as no roots are given.

Body: If $x^4+2x^3+px^2+qx+9=0$ is a complete square. $p$ and $q$ are positive. Find the value of $p$ and $q$.


问:
Ellipse curvature

Let $a,b>0$ and let $\alpha$ be the ellipse
$\alpha(t)=\big(x(t),y(t)\big)=\big( a\cos t, b\sin t\big)$.

Let $(u(t),0)$ be the point where the line through $\alpha(t)$
perpendicular to $\alpha'(t)$ meets the $x$-axis, and let $d(t)$ be
the distance between $\alpha(t)$ and $(u(t),0)$.
Find a formula for $\kappa(t)$ in terms of $a$, $b$ and $d(t)$.

The curvature of the ellipse has been discussed thoroughly in this question: how to calculate the curvature of an ellipse

However, I can't find a way to use $d(t)$ in expressing curvature.

I tried to sketch the graph and realized that I can inscribe a circle inside the ellipse. But that is as far as I can go.

I also tried to explicitly write out $d(t)$, but it gives me an ugly expression.

答:
It is well known that (see figure below, $F$ and $G$ are the foci):

$$
\kappa={\cos\alpha\over2}\left({1\over p}+{1\over q}\right)
={a\cos\alpha\over pq}.
$$
On the other hand, from the formula for the length of a bisector we get:
$$
d^2={b^2\over a^2}pq
$$
while from the cosine rule applied to triangle $FPG$ we obtain
$$
\cos^2\alpha={b^2\over pq}.
$$
Inserting these results into the formula for $\kappa$ we finally get:
$$
\kappa={b^4\over a^2d^3}.
$$

Title: Ellipse curvature


问:
Knowing that $f(n)=n$, $g(n)=n\sqrt3$. Find $h(n)$ that contains all elements of $f(n)$ and $g(n)$ in ascending order.
Knowing that:

$n$ is an integer going from zero to infinity,

$f(n)=n$,

$g(n)=n\sqrt3$.

I need a formula for $h(n)$ that can generate the series: $0, 1, \sqrt3, 2, 3, 2\sqrt3, 4, 5, 3\sqrt3, 6, 4\sqrt3 ... $ as a function of $n$.

$h(n)$ is a series that contains all elements of $f(n)$ and $g(n)$ in ascending order.

答:
You don't want the formula; it is going to be ugly.

.

.

.

.

.

.

Well, you've been warned. Have it, then:
$$h(n) = \min\left(\sqrt3\left\lceil{n\over1+\sqrt3}\right\rceil,\left\lceil{n\sqrt3\over1+\sqrt3}\right\rceil\right)$$

where $\lceil\cdot\rceil$ is of course the ceiling function.


答:
Add up the the sides of $\triangle ADB$ and $\triangle ADC$ and you get their perimeters and then equate them:

$x+\frac{100}{3}-y+\frac{80}{3}=x+y+20$

Solve for $y$ to get $y=20$

and then for $x$ using the cosine rule we get

$x=\sqrt{20^2+20^2-2(20)(20)\cos\left(\tan^{-1}\frac{80}{20\times3}\right)}$

$x=8\sqrt{5}$

Title: Plane geometry tough question

Body: $\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.</p>


问:
Is this process a Brownian motion?
I have some problem with deciding whether the process $Y_t$ is a Brownian motion. Here is the structure:

$X_t = \int_0^tW_s^2ds$ where $\{W_t\}_{t≥0}$ is a Brownian motion, and $Y_t=\frac{X_t-\frac{1}{2}t^2}{\frac{1}{2}t^{3/2}}$.

So here are what I did:

I calculated the mean and variance for $X$ and $Y$: $\mathbb{E}X_t=\frac{1}{2}t^2$, $\text{Var}X_t= \frac{1}{3}t^4$, $\mathbb{E}Y_t=0$, $\text{Var}Y_t= \frac{4}{3}t$.

I didn’t find the wrong places in my calculations. I found the variance of $Y_t$ is not $t$, so I wanted to say that $Y_t$ is not a Brownian motion. But I feel this argument is not that convincing. I am wondering there might be something wrong but I didn’t notice. Could you give me a convincing proof on whether $Y_t $ is a Brownian motion or not?

Thank you so much!

答:
Note : From Danny's observation, we already know that $Y_t$ cannot be a Brownian motion as it is pathwise differentiable. However, I wish to derive the mean and variance of $Y_t$ for the sake of computation, as OP has demanded this of me.

To calculate the mean and variance of $Y_t$ we need to find the respective quantities for $X_t$.

Now, $E[X_t] = E\left[\int_0^t W_s^2 ds\right]$. We can exchange the integral and expectation since the integrand is non-negative hence Tonelli's theorem applies. Thus, $E[X_t] =  \int_0^t E[W_s^2]ds  = \int_0^t sds = \frac{t^2}{2}$.

Now, $E[X_t^2] = E\left[\left(\int_0^t W_s^2ds\right)^2\right] = E[\int_0^t W_s^2ds \int_{0}^t W_u^2du] = E\left[\int_0^t\int_0^t W_s^2W_u^2\ du\ ds\right]$. Once again, we have Tonelli applying here, and we get :
$$
E[X_t^2] = \int_0^t \int_0^t E[W_s^2W_u^2] duds
$$

Now, we will calculate the inner expectation for the case that $s > u$, and then for $u>s$ we can just switch the roles of $u$ and $s$. The advantage of doing that is that we may then write:
$$
W_s^2W_u^2 = ((W_s - W_u)+W_u)^2W_u^2 = (W_s-W_u)^2W_u^2 + 2(W_s-W_u)W_u^3 + W_u^4
$$

and therefore by independence and stationarity of increments :
$$
E[W_s^2W_u^2] = E[(W_s - W_u)^2]E[W_u^2] + 2E[(W_s - W_u)]E[W_u^3] + E[W_u^4] \\ = (s-u)u + 3u^2 = 2u^2 + us
$$

For $u>s$ we have $2s^2+su$. So we have :
$$
E[X_t^2] = \int_0^t \left(\int_0^s (2u^2+us)du + \int_s^t (2s^2+us)du\right) ds \\
= \int_0^t \left[\frac{7s^3}{6} + \frac {s}{2}(-5s^2+4st+t^2)\right]ds\\ =\left[\frac 16\left(-2s^4+4s^3t+\frac{3s^2t^2}{2}\right)\right]^t_0
\\ = \frac{7t^4}{12}
$$

and therefore $\mbox{Var}[X_t] = E[X_t^2] - E[X_t]^2 = \frac{7t^4}{12} - \frac{t^4}{4} = \frac{1}{3}t^4$, which matches the correct answer.

Now, $E[Y_t] = 2t^{-\frac 32} (E[X_t] -\frac 12 t^2) = 0$.

Then $\mbox{Var}[Y_t] = 4t^{-3}\mbox{Var}(X_t) = \frac {4t}3$, which doesn't match with the variance of a Brownian motion $B_t$ at time $t$.

Title: Is this process a Brownian motion?


答: This is not an answer to the question, but I did want to add a fun fact about the sequence. If you center a circle on a point in a triangular lattice, then this sequence describes the length of all possible radii in which the boundary of the circle touches only six points.</p>

Title: Knowing that $f(n)=n$, $g(n)=n\sqrt3$. Find $h(n)$ that contains all elements of $f(n)$ and $g(n)$ in ascending order.

Body:
Knowing that:

$n$ is an integer going from zero to infinity,

$f(n)=n$,

$g(n)=n\sqrt3$.

I need a formula for $h(n)$ that can generate the series: $0, 1, \sqrt3, 2, 3, 2\sqrt3, 4, 5, 3\sqrt3, 6, 4\sqrt3 ... $ as a function of $n$.

$h(n)$ is a series that contains all elements of $f(n)$ and $g(n)$ in ascending order.


问:
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. if one of its diagonal is 8 cm long, find the length of another diagonal.
If I solve for diagonal by using area = 1/2 (D1 x D2) and area = base x height and if I solve for diagonal by using Pythagoras theorem, why the answers are different?</p>

答: The thing is, if a side of a rhombus is $6$ and one of the diagonals is $8$, the area cannot be $24$. The area will consist of four right triangles with hypothenuse $6$ and one of the legs $4$. $A=4\cdot(2\sqrt{36-16})=16\sqrt{5}\approx 35.8$


问:
Polynomial vs Integer Rings
I have a more conceptual question on the differences between polynomial rings and integer rings. I know they are both Euclidean domains, but is one a subset of the other? More precisely are all polynomial rings integer rings or all integer rings polynomial rings?

Herstein's Abstract Algebra suggests that the notion of roots separates integers compared to polynomials, but it didn't seem obvious to me why this is important,

答: Absolutely no. For instance, the ring $\mathbb{Z}[X]$ of polynomials in the variable $X$, then $\mathbb{Z}[X]$ is indeed an UFD (unique factorization domain, Gauss Lemma) but it's not a PID thus not even an Euclidean Ring. What you actually meant is that $K[X]$ is euclidean whenever $K$ is a field, now it is more precise. Indeed $K[X]$ and $\mathbb{Z}$ do enjoy some similiarities but they are quite different: notice that $K\subsetneqq K[X]$ and $\mathbb{Z}$ does not contain any non trivial field. If you deepen your studies over quotient rings then you will see even more features which holds only on $K[X]$

Title: Polynomial vs Integer Rings


答:
The indices when roots appear seems to be oeis.org/A054088

Based on this I propose the following:

$\phi(n)=n+\lfloor(n\sqrt{3})\rfloor\qquad$ the indices of $x\sqrt{3}$ numbers

$\Phi(n)=\lceil\frac n{1+\sqrt{3}}\rceil\qquad$ the $x$ before $\sqrt{3}$, i.e. $\phi^{-1}$

$\delta(n)=\Phi(n+1)-\Phi(n)\in\{0,1\}\qquad$ indicates that this $n$ is a root

$N(n)=n+1-\Phi(n)\qquad$ the natural numbers mingled between the roots

The resulting formula:
$$h(n):\begin{cases}N(n)&\quad&\delta(n)=0\\\Phi(n)\sqrt{3}&&\delta(n)=1\end{cases}$$

Or without piecewise selection $h(n)=\delta(n)\Phi(n)\sqrt{3}+(1-\delta(n))N(n)$


答:
You noted that $\frac{1}{nX} 1_{X>\frac{1}{n}} \le 1$ almost surely. Since $E[1] < \infty$, this satisfies the condition in the dominated convergence theorem, so you may push the limit into the expectation.

$$\lim_{n \to \infty} E[\frac{1}{nX} 1_{X>\frac{1}{n}} \le 1]
= E[\lim_{n \to \infty} \frac{1}{nX} 1_{X>\frac{1}{n}} \le 1 ] = E[0] = 0.$$

Title: $X\geq0$ and $X$ a random variable, show that $\lim_{n\to\infty} \frac{1}n\cdot E(\frac1X\cdot 1_{X>\frac1n})$=0

Body: The only result I got so far is, that the limit has to be smaller than 1. This I derived via monotonicity and $\frac1X$<n. I have no clue how to continue and a hint would be appreciated.


问:
Is $\mathcal B(\mathbb R) \times \mathcal B(\mathbb R)$ closed under finite intersection?
Can anyone help me show that the Cartesian product of the Borel-sigma-algebra is closed under finite intersection?</p>

答: Hint: Use that if $A,B,C,D$ are subsets of $\mathbb{R}$, then $(A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D)$.


答:
I agree with you

Using area = 1/2 (D1 x D2)
we get $$area = 4 * D_2$$

and with area = base x height

$$area = 6 * 4 = 24$$

so $$D_2 = 24/4 = 6$$

Now Pythagorus' Theorem:
taking $$ a = 4 b = D_2/2 c = 6$$

we get $$D_2 = 4*\sqrt(6^2-4^2)$$
so $$D_2 = 4*\sqrt(5)$$

I believe There may be a typo in the original question. If you take side length to be 5 then the answers agree. A rhombus with the dimensions you give doesn't exist

Title: Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. if one of its diagonal is 8 cm long, find the length of another diagonal.

Body: If I solve for diagonal by using area = 1/2 (D1 x D2) and area = base x height and if I solve for diagonal by using Pythagoras theorem, why the answers are different?</p>


问:
Probability and statistics expected value
The number of car accidents during one week on a busy motorway is modelled by random variable $X$ with distribution $\mathbb P(X=k) = e^{-\lambda} \cdot \frac{\lambda^k}{k!}$ for $k=0,1,2,\ldots$, where $\lambda=0.5$.

How can I solve how many car accidents are expected to happen in a week?

答: Use the formula for expected value: $E(X)=\displaystyle \sum_{k=0}^\infty P(X=k)k=e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^k}{k!}k$


问:
How to solve this Riccati Differential Equation
I need to solve following Differential Equation

$\dfrac{dy}{dx} = \dfrac{1}{x^2} - y^2$, $x >0$

I know  this is a Riccati Equation, but I am unable to see the substitution required to convert this into a Linear Differential Equation.

Any hints will be helpful.

Thank You.

答:
Letting $y_p=c/x$, we find for $c$

$$c^2-c-1=0$$

Substituting $y=y_p+u$ into the original equation, we find the differential equation for $u$

$$ u'+u^2=- \frac{2c u}{x}$$

Now let $v=1/u$ to find

$$ v'=\frac{2cv}{x}+1$$

Which is easily solvable, as the nonlinear term has been eliminated. The homogeneous solution is

$$ v(x)_H=Ax^{2c}$$

A particular solution is

$$ v_P(x)=Bx$$

For which we find $B=\frac{1}{1-2c}$. Our solution for $v$ is

$$ v(x)=\frac{x}{1-2c}+Ax^{2c}$$

Hence, the solution for $y$ is

$$ y(x)=\frac{c}{x}+\left(\frac{x}{1-2c}+Ax^{2c} \right)^{-1} $$

$A$ is your constant of integration, and $c$ has values $ \frac{1\pm\sqrt{5}}{2}$


问:
How do you show that $7a^2 - 12b^2 = 8c^2$ has no integer solutions?
How do you show that $7a^2 - 12b^2 = 8c^2$ has no integer solutions

When $x = a/c$ and $y = b/c$ then $\gcd(a,b,c) = 1$ I believe.

If we use mod5, and neither $a,b,c$ is divisible by 5 then $7a^2 - 12b^2$ can have remainder 1 or 4 and $8c^2$ can only have remainder 2 or 3. But if either $a$ or $b$ is divisible by 5 then $7a^2 - 12b^2$ will have remainder 2 or 3.

I've considered using mod7 so $-12b^2$ would have remainder either $1,2,$ or $4.$ But $8c^2$ can have remainder 1,2 or 4 as well.

I'm not really sure how to go about this, any help would be appreciated thanks.

答:
Hint

WLOG $(a,c)=1$

Take modulo $\pmod3$

$$a^2\equiv2c^2\pmod3$$

But for any integer $d,$

$$d^2\equiv0,1\pmod3\not\equiv2$$


问:
Optimality conditions in linear programming problems with lower- and upper-bounded variables
Background:

According to a paper by Andersen and Andersen (https://www.davi.ws/doc/Andersen.pdf), any linear programming problem can be formulated as:

min $c^Tx$,

s. t. $Ax = b$,

$l \le x \le u$,

$ ...(1)$

where $x, c, l, u \in \Bbb R^n$, $b \in \Bbb R^m$ and $A \in \Bbb R^{m \times n} $, and some of the simple bounds $l_j$ and $u_j$ may be $-\infty$ and  $+\infty$ respectively. With $L = \{j: l_j \gt -\infty \}$ and $U = \{j: u_j \gt +\infty \}$, then the optimality conditions to 1) can be stated as follows:

$Ax^* = b$,

$A^Ty^* + z^* = c$,

$(x^*_j - l_j) z^*_j = 0$ $\forall j \in L $,

$(u_j - x^*_j) z^*_j = 0$ $\forall j \in U $,

$z^*_j \ge 0$ $\forall j \in \{j \in L: x^*_j = l_j$ $\land$ $l_j \le u_j  \} $,

$z^*_j \le 0$ $\forall j \in \{j \in U: x^*_j = u_j$ $\land$ $l_j \le u_j  \} $,

$z^*_j = 0$ $\forall j \notin L\cup U $,

$l \le x^* \le u$,

$...(2)$

where $y \in \Bbb R^m$ and $z \in \Bbb R^n$. $y$ and $z$ are the Lagrange multipliers corresponding to the linear constraints and linear inequalities in (1), respectively.

My problem:

I am having difficulty seeing whether the optimality conditions (2) came from. I have attempted to derive them as follows:

Introduce a new variable $x'= x - l$ and a new (slack) variable $x'' = u - x'$. Then the primal problem can be cast as:

min $c^T(x' + l)$

s. t. $A(x' + l) = b$

$x' + x'' = u - l$

$x', x'' \ge 0$.

That is:

min $c^T x' + c^T l$

s. t. $\bigl( \begin{smallmatrix} A & 0 \\ 1 & 1 \end{smallmatrix} \bigr)$  $\bigl( \begin{smallmatrix} x' \\ x'' \end{smallmatrix} \bigr)$ $=$  $\bigl( \begin{smallmatrix} b - A l \\ u -l \end{smallmatrix} \bigr)$,

$ x', x'' \ge 0$.

Then, with dual variables $y'$ and $y''$, complementary slackness gives, for optimality:

$x' (c - (y'A + y'')) = 0$, and $x''(-y'') = 0$

$...(3)$

Also, for dual feasibility:

$c \ge y'A + y''$, and $0 \ge y''$

$...(4)$

Is my working correct so far, and if it is, how can I derive (2) from (3) and (4)? Or is there a simpler way to prove (2)?

I would be very grateful for any help.

答:
It occurred to me after posting that the proof can be made much simpler if inequality primal constraints are used, i.e.

$
    \begin{pmatrix}
    A \\
   -A \\
    1  \\
   -1 \\
    \end{pmatrix} x \ge \begin{pmatrix}
   b \\
   -b \\
   l  \\
   -u \\
    \end{pmatrix}
$

Then with dual variables $y1, ..., y4 \ge 0$, setting $y = y1-y2$ and $z = y3-y4$ yields the desired result directly.


答:
Using mod $4$, a is divisible by $2$

Let $a = 2k$

$7k^2 - 3b^2 = 2c^2$

HINT: GO mod 3 and use that $t^2 \equiv 0,1 \pmod 3$ for all positive integers t

Title: How do you show that $7a^2 - 12b^2 = 8c^2$ has no integer solutions?

Body:
How do you show that $7a^2 - 12b^2 = 8c^2$ has no integer solutions

When $x = a/c$ and $y = b/c$ then $\gcd(a,b,c) = 1$ I believe.

If we use mod5, and neither $a,b,c$ is divisible by 5 then $7a^2 - 12b^2$ can have remainder 1 or 4 and $8c^2$ can only have remainder 2 or 3. But if either $a$ or $b$ is divisible by 5 then $7a^2 - 12b^2$ will have remainder 2 or 3.

I've considered using mod7 so $-12b^2$ would have remainder either $1,2,$ or $4.$ But $8c^2$ can have remainder 1,2 or 4 as well.

I'm not really sure how to go about this, any help would be appreciated thanks.


答:
WLOG, assume that $\gcd (a,b,c) = 1$ as you did.

Taking modulo $3$ gives $a^2 \equiv 2c^2 \pmod 3$.

As $n^2 \equiv 0,1 \pmod 3$, we must have $a \equiv c \equiv 0 \pmod 3$.

Hence $3 \nmid b$, and $a^2 \equiv c^2 \equiv 0 \pmod 9$.

Now taking modulo $9$, we have:

$$-3b^2 \equiv 0 \pmod 9$$

suggesting that $3 \mid b$, which is a contradiction.


问:
What does the null hypothesis $H_0:\beta_2=0$ mean?
For the model $$Y=\beta_1+\beta_2X_2+\beta_3X_3+\epsilon$$ what does the null hypothesis $H_0:\beta_2=0$ mean?

I think that then $X_2$ would not affect the expected value of $Y$ since the coefficient os then $0$.

But is that correct?

答: Yes that is correct. If $\beta_3$ is 0, the variable has no predictive value for the response. No matter what value $X_2$ has, it does not alter the predicted value of the response.


问:
Get the equation of a circle when given 3 points
Get the equation of a circle through the points $(1,1), (2,4), (5,3) $.

I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?

Body:
As a programmer this was the best solution for me  as there is no division by zero:

For given (x1,y1) , (x2,y2) and (x3,y3) first form A,B,C:

$$A=x_1(y_2-y_3) - y_1(x_2-x_3)+x_2y_3 -x_3y_2$$
$$B=(x_1^2 + y_1^2)(y_3-y_2)+(x_2^2+y_2^2)(y_1-y_3)+(x_3^2+y_3^2)(y_2-y_1)$$
$$C=(x_1^2+y_1^2)(x_2-x_3)+(x_2^2+y_2^2)(x_3-x_1)+(x_3^2+y_3^2)(x_1-x_2)$$
$$D=({x_1^2}+{y_1^2})({x_3}{y_2} - {x_2}{y_3})+(x_2^2 + y_2^2)(x_1y_3-x_3y_1)+(x_3^2+y_3^2)(x_2y_1 - x_1y_2)$$

If A=0 then the points are colinear elasewhere using A,B,C you can find center and radius of circle:

$$x_c=-B/2A$$
$$y_c=-C/2A$$

$$r= \sqrt {\frac {B^2+C^2-4AD}{4A^2}}$$

Finally the formula of the circle is:

$$(x-x_c)^2+(y-y_c)^2=R^2$$


问:
for which real values of $a$ the limit is finite and different from $0$ $\lim_{ x \to 0 } \frac{ \cos (\pi \cdot \frac{1-\cos ax}{x^2})}{x^2}$
I have to find for which real values of $a$ this limit is finite and different from $0$:
$$\lim_{ x \to 0 }  \frac{ \cos (\pi \frac{1-\cos ax}{x^2})}{x^2}$$

I've tried to change the form of cos but with no success.
According to my book the final result should be for $a^2=2k+1 (k \in N)$ for which $\lim=\frac{(-1)^k (2k+1)^2 \pi}{24}$.

答:
Hint

Use

$$\lim\limits_{x \to 0} \frac{1-\cos ax}{x^2}= a^2\lim\limits_{x \to 0} \frac{1-\cos ax}{(ax)^2}=\frac{a^2}{2}$$


问:
How can I prove that A is the associated Matrix to g(x)
What are nice ways to prove that

$$A= I_{n}+w v^{T}$$ is the associated matrix of the following function
$$g (x)=x+\left(v^{T} x\right) w$$ and $v,w \in \mathbb R^n$

My idea is to show that A is just the picture of the unit vectors but somehow I am stuck.
Thanks for your help.

答:
Hint:

You only have to check that the matrix $B=w{\,}^{\mathrm t \mkern-1.5mu}v$ is the matrix of the linear map
$$h(x)=({}^{\mathrm t \mkern-1.5mu}v\,x)w.$$
So determine $B$  and the images of the canonical basis: $h(e_1),h(e_2),\dots,h(e_n)$, and compare with the columns of $B$.


问:
Explanation of a proof of $\lim_{x \to 0}\frac{\sin(x)}{x}=1$ using the epsilon-delta definition of limits
I am reading this proof of $\lim _{x\to \:0}\left(\frac{\sin \left(x\right)}{x}\right)=1$ by using epsilon delta technique, directly applied from the epsilon delta definition of limit.

All the steps are understood except the circle red one.

I don't understand where does the $x^2$ come from? I am bit stumped, perhaps the proof leaves out some steps that I need to trace.

Thank you!

答:
That comes from $|\sin(x)|<|x|$ on the second line of your "solution" and also
$$
1+\cos x>1
$$
since $\cos x>0$ on $(-\pi/2,\pi/2)$.


问:
Example of a finitely generated flat module which is not free
I couldn't come up with an example of a finitely generated flat module which is not free. I know that over local rings, freeness and flatness are equivalent. So the ring cannot be a local ring. </p>

答:
Let $R$ be a semisimple ring which isn't a division ring, and take an idempotent $e\notin \{0,1\}$. Then we have that $R=eR\oplus(1-e)R$ is a nontrivial decomposition of $R$, and both pieces are cyclic and projective (since they are summands of $R$) hence flat.

Actually neither piece is a free module, but we'll argue here that at least one of them isn't free to simplify things. If they were both free, that would imply that $R\cong R^n$ for some $n>1$ as modules. (Each factor contributes at least one $R$, you see.) But since $R$ has the IBN property, this is impossible.

Thus at least one of the pieces is flat but not free.

If you really want to be concrete, you can use a finite semisimple ring to make things obvious: let's try $R=\Bbb F_2\times\Bbb F_2$. $\Bbb F_2$ is denoting the field of two elements.

The module $I=\Bbb F_2\times \{0\}$ is a direct summand of $R$, but it can't be free: a free module would have to have at least four elements, and this module only has two elements!


问:
Efficiently beating this game
Say that I choose an integer between 1 and 100 with uniform probability.
Your goal is to determine the number I have chosen by guessing. Once you guess, I will tell you if that guess is too high or too low. You may make as many guesses as you wish, but may only guess too high twice.

What is the most efficient way to beat this game, and how do I prove that this method is the most efficient?

An obvious strategy is to guess 1, then 2, then 3, then ... until the nth guess is too high. Then n is the number.

A better strategy would be to guess 10, 20, 30, ... until you get a number too high (e.g. 30 is too high), then guess 21, 22, 23, ... until you are too high again, giving the answer.

Apparently, the best approach is to use the fact that 105 is the first triangular number over 100, and then guess 14, 14+13, 14+13+12, ..., until too high. This guarantees that in the worst possible case you make 14 guesses to win.

However, I can't prove to myself why this is the best approach.

My idea was to count how many guesses it would take to reach every number between 1 and 100 with a given strategy. The optimal strategy is then the strategy which minimizes the sum of all the guesses.

Eg with the triangular number setup:

n
Number of guesses to guess n

1
3

2
4

3
5

...
...

12
14

13
14

14
3

...
...

You can show quite easily that 14 is maximum number of trials you will have to run here.

Anyone know the name of this game? It's quite practical, e.g. minimise the number of trials to find the wattage that blows a lightbulb, you only have 2 lightbulbs to test with.

Even better, can anyone prove why a certain strategy is best?

答:
EDIT: I know realize now you wanted to optimize the average number of drops, not worst case, so this answer is not relevant.

Here is a slick proof of optimality I got from this answer on Puzzling stack exchange.

Suppose you have a strategy which allows you to succeed in $n$ guesses. When you apply this strategy to a given chosen number, you will be told a sequence of $n$ responses which are all "too high" (H) or "too low" (L), where there are at most two H's. Furthermore, the collection of all possible response sequences must be distinct; if there were two chosen numbers which caused the same response sequence, then your strategy could not distinguish those two numbers.

Therefore, we need the number of possible response sequences to be at least $100$, one for each possible unknown number. The number of possible response sequences with $n$ guesses is
$$
1+n+\binom{n}2,
$$
corresponding to the cases where there are zero, one, and two responses which are H. You can then check that the smallest value of $n$ for which that quantity is at least $100$ is $n=14$.

The same logic applies to when you can hear "too high" at most $k$ times, in which case you would instead have $1+n+\binom{n}2+\dots+\binom{n}k$ for the quantity which must meet or exceed the number of unknown numbers.

Title: Efficiently beating this game


问:
Field extensions and unions
If $F$ is a subfield of a field $E$ and if $S = S_1 \cup S_2$, the $F(S) = F(S_1)(S_2)$.

I understand why this has to be so, but I can't seem to prove it rigorously.

Could someone help?

答:
Notice $S_1\subseteq F(S_1)(S_2)$ and $S_2\subseteq F(S_1)(S_2)$, so $S\subseteq F(S_1)(S_2)$; of course also $F\subseteq F(S_1)(S_2)$ so since $F(S)$ is the smallest subfield of $E$ containing $F$ and $S$ we deduce $F(S)\subseteq F(S_1)(S_2)$.

The reverse inclusion is similar.

Title: Field extensions and unions


答: Observe that : $\cos x > 0, \forall x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ,and $0 \le |\sin x | \le |x|$ as well. This implies $\dfrac{\sin^2 x}{1+\cos x} \le \sin^2 x \le x^2$.

Title: Explanation of a proof of $\lim_{x \to 0}\frac{\sin(x)}{x}=1$ using the epsilon-delta definition of limits

Body:
I am reading this proof of $\lim _{x\to \:0}\left(\frac{\sin \left(x\right)}{x}\right)=1$ by using epsilon delta technique, directly applied from the epsilon delta definition of limit.

All the steps are understood except the circle red one.

I don't understand where does the $x^2$ come from? I am bit stumped, perhaps the proof leaves out some steps that I need to trace.

Thank you!


答:
You always have $\vert \sin x \vert \le \vert x\vert$.

And for $x$ close to zero $\cos x \ge \frac{1}{2}$. Therefore

$$0 \le \frac{1}{1 +\cos x} \le \frac{2}{3}.$$


答: $|\sin x| \le |x|$ implies $\sin^2 x \le x^2$, and on $x \in [-\pi/2, \pi/2]$, $\cos x \ge 0$ hence $1 + \cos x \ge 1$ and $$\frac{1}{1 + \cos x} \le 1.$$  Combining these results yields $$\frac{\sin^2 x}{1 + \cos x} \le \frac{x^2}{1 + \cos x} \le x^2.$$


问:
Finding the limits of integration for the volume of a region inside a cube

Premise

We are restricted to the region $x, y, z \in [f, 1]$ where $0 \leq f < 1$. The surface $y^2=4xz$ divides this cube into two regions. We are interested in finding the volume of the region where $y^2 - 4xz \geq 0$.

My Attempt

Let $t=\frac{y^2}{4}$ for brevity.

When $f=0$, it is relatively easy to figure out the limits. We can separate the region into two where $x \leq t$ and $x > t$. In the first case, we have $y^2 - 4xz \geq 0$ for all $z \in [0, 1]$ because $xz \leq x\cdot 1 \leq t$. Similarly, in the second case, we have $y^2 - 4xz \geq 0$ for all $z \in [0, \frac{t}{x}]$ because $xz \leq x\cdot \frac{t}{x} = t$. Applying these limits, we have

$$
\begin{align}
V &= \int_{0}^{1}{dy\int_{0}^{t}{dx\int_{0}^{1}{dz}}} + \int_{0}^{1}{dy\int_{0}^{t}{dx \int_{0}^{\frac{t}{x}}{dz}}}\\
  &= \int_{0}^{1}{dy\int_{0}^{t}{dx\cdot (1)}} + \int_{0}^{1}{dy\int_{0}^{t}{dx\cdot \frac{t}{x}}}\\
  &= \frac{1}{12} + \frac{1 + 3\log{2}}{18} \\
  &\approx 0.254
\end{align}
$$

However, when $f > 0$, I cannot seem to understand the proper limits for $z$. Is it $f$ to $\frac{t}{x}$ or is the lower limit $f^2$? What happens when $f$ is less than or equal to $\frac{y^2}{4}$?

I tried a few times, but the results ended up being negative, which is clearly wrong as volumes are nonnegative.

Specific Questions

Primary objective

I would like help in figuring out the limits in the case where $f>0$.

Secondary objectives

Is this approach sensible? Perhaps there is a better way to solve this problem?

Are there any glaring oversights?

What books/courses/reference should I follow to be able to solve these types of problems?

答:
We need to find the volume of the cube formed by $x, y, z \in [f, 1], 0 \leq f \lt 1$, with the condition $y^2 \geq 4xz$ (outside the cone).

For the cube, the upper limit for $x, y, z$ is $1$.

For the cone, please note when $x = z = 1, y = 2$ (I am only considering first octant given our cube is in the first octant) but as we are restricted by $y = 1$ for the cube, we have $4xz \leq 1 \implies x, z \leq  \min (\frac{1}{4f}, 1)$ for $f \ne 0$.

At the same time for the points outside the cone, $y^2 \geq 4f^2 \implies y \geq 2f$ for $x = z = f$ and $y \geq 2 \sqrt f \ \gt f $ as $f \lt 1$ for $x = 1, z = f$ or $z = 1, x = f$.

These are important observations to find the right limits for our volume integral. What it indicates is that if $f \leq 0.25, $ we have following $3$ vertices of the cube outside the cone and rest $5$ are inside -

$(f,1,f),(f,1,1), (1,1,f)$

In fact at $f = 0.25$, $(f,1,1)$ and $(1,1,f)$ are on the cone and only $(f,1,f)$ is outside.

Also note that at $f = 0.5$, $(f,1,f)$ is on the cone and rest are inside and so for $f \geq 0.5$, we have not volume bound (as all points of the cube are inside the cone, $y^2 \leq 4xz$).

So here is the integral to find the desired volume

i) $0 \leq f \leq 0.25$ (first integral is zero at $f = 0.25$)

$\displaystyle  \int_{f}^{0.25} \int_{f}^{1} \int_{2\sqrt{xz}}^{1} dy \ dx \ dz +
\int_{0.25}^{1} \int_{f}^{1/(4z)} \int_{2\sqrt{xz}}^{1} dy \ dx \ dz$

And your answer is indeed correct for $f = 0$.

ii) $0.25 \leq f \leq 0.5$ (zero for $f = 0.5$)

$\displaystyle \int_{f}^{1/(4f)} \int_{f}^{1/(4z)} \int_{2\sqrt{xz}}^{1} \ dy \ dx \ dz$

To your other specific questions, no there does not seem anything wrong with your approach. But your approach did not extend to other cases, probably because you did not visualize the surface well enough. A $3D$ sketch using an online tool helps. I literally had to keep a cube in front of me while answering the question.


问:
Almost sure convergence on integral of Poisson process
Let $N_t$ be a Poisson process with intensity $\lambda$.
Then how to show that
$$
  \lim_{t\rightarrow 1} (1-t) \int_0^t \frac{2}{(1-s)^2}dN_s = 0
$$
almost surely?

What I have done.

Since for some $\varepsilon>0$, the probability $\mathbb{P}\left\{ N_{1-\varepsilon,1}=\infty \right\}$ or $\mathbb{P}\{dN_1 > 0\}$ is zero, so my intuition says that the above equality should hold.
First, I tried to apply Borel-Cantelli Lemma:
$$
  \sum_{n=1}^{\infty} \mathbb{P}(E_n)<\infty \implies \mathbb{P}(\lim_{n\rightarrow \infty} E_n) = 0.
$$
So I define $E_n = \left\{(1-t_n) \int_0^{t_n} \frac{2}{(1-s)^2} dN_s > \varepsilon \right\}$ for some $\epsilon>0$ and $t_n \rightarrow 1$ as $n\rightarrow \infty$, and using Markov's inequality:
$$
 \mathbb{P}(E_n) \le \frac{\mathbb{E}\left[(1-t_n)\int_0^{t_n}\frac{2}{(1-s)^2}dV_s\right]}{\varepsilon} \\
= \frac{1-t_n}{\varepsilon}\int_0^{t_n}\frac{2}{(1-s)^2}\lambda ds \\
= \frac{2}{\varepsilon}-2\frac{1-t_n}{\varepsilon}
$$
is not tight enough to bound the sum $\sum_{n=1}^{\infty} \mathbb{P}(E_n)$.
This is same for the higher moment bounds.
Then is it possible to show the above equality?

Thanks.

答:
As you observe, $N$ has no jump at time $1$, a.s. Let $T\in[0,1)$ be the last jump of $N$ before time $1$. ($T=0$ means there are no such jumps.) Then because, $s\mapsto(1-s)^{-2}$ is increasing,
$$
\int_0^t{2\over (1-s)^2}dN_s\le 2N(t)\cdot(1-T)^{-2}\le 2N(1)\cdot (1-T)^{-2}.
$$


问:
Limit to $e^2$.
I have to show
$$
\lim_{x \to \infty} {\left(\frac{x^2 + 1}{1 - x^2}\right)}^{x^2} = e^2,
$$
but I don't get the trick to see it, I suppose I can use something like
$$
{\left(\lim_{x \to \infty} {\left(1 + \frac1x\right)}^x\right)}^2 = e^2,
$$
but I do not see how I can apply to the problem. On the other hand,
$$
\frac{x^2 + 1}{1 - x^2} = 1 + \frac{2 x^2}{1 - x^2}.
$$
Any hint?

答:
If the limit of interest is $\lim_{x\to\infty}\left(\frac{x^2+1}{x^2-1}\right)^{x^2}$, then we can write

$$\begin{align}
\left(\frac{x^2+1}{x^2-1}\right)^{x^2}&=\left(\frac{x^2-1+2}{x^2-1}\right)^{x^2}\\\\
&=\left(\left(1+\frac{2}{x^2-1}\right)^{x^2-1}\right)^{x^2/(x^2-1)}
\end{align}$$

Now, use $\lim_{y\to\infty}\left(1+\frac{t}{y}\right)^y=e^t$ with $y=x^2-1$.  Can you wrap this up now?

Title: Limit to $e^2$.


Title: for which real values of $a$ the limit is finite and different from $0$ $\lim_{ x \to 0 } \frac{ \cos (\pi \cdot \frac{1-\cos ax}{x^2})}{x^2}$

Body:
I have to find for which real values of $a$ this limit is finite and different from $0$:
$$\lim_{ x \to 0 }  \frac{ \cos (\pi \frac{1-\cos ax}{x^2})}{x^2}$$

I've tried to change the form of cos but with no success.
According to my book the final result should be for $a^2=2k+1 (k \in N)$ for which $\lim=\frac{(-1)^k (2k+1)^2 \pi}{24}$.

Body:
We start from

$$\cos(ax)=1-\frac{a^2x^2}{2}+\frac{a^4x^4}{4!}(1+\epsilon(x))$$

to get

$$\frac{1-\cos(ax)}{x^2}=\frac{a^2}{2}-\frac{a^4x^2}{24}(1+\epsilon(x))$$

If The limit is finite, then necessarily

$$\cos(\pi.\frac{a^2}{2})=0$$

or

$$a^2=1+2k , k\ge 0$$

In this case

$$A=\pi(\frac{a^2}{2}-\frac{a^4x^2}{24}(1+\epsilon(x))=$$
$$\frac{\pi}{2}+k\pi-\frac{a^4x^2\pi}{24}(1+\epsilon(x))$$

and

$$\cos(A)=-(-1)^k\sin(\frac{a^4x^2\pi}{24}(1+\epsilon(x)($$

$$\sim -(-1)^k\frac{a^2x^2\pi}{24}$$

the limit will be
$$-(-1)^k\frac{a^4\pi}{24}=-(-1)^k\frac{(1+2k)^2\pi}{24}$$


问:
Are spaces of finite sequences nuclear?
Let $I$ be some index set and $c_{00}$ the set of functions $c$ from $I$ to $\mathbb{C}$ such that $c(i) \neq 0$ for only finitely many $i \in I$.

Let this space carry the locally conves topology induced by the seminorms
$$ \|c\|_f := \sum_{i \in I} c(i) f(i)$$
where $f \in \mathbb{C}^I$ is any function.

Is this space nuclear?

答:
I think you mean $\|c\|_f= \sum_i |c(i)f(i)|$. Then the space is nuclear if and only if $I$ is countable.

If $I=\mathbb N$ and $f$ is a weight function you put $g(i)=2^i f(i)$ to obtain rather easily the nuclearity of the canonical map $X_g \to X_f$, where $X_f$ is the (Hausdorff-) completion of $(c_{00},\|\cdot\|_f)$ which is the weighted space
$\ell_1(f)=\lbrace (x_i)_{i\in I}: \sum_i |x(i)f(i)|<\infty\rbrace$.

If $I$ is uncountable, put $f(i)=1$ and assume that there is $g\ge f$ such that
the map $J:X_g\to X_f$ is nuclear with a nuclear representation
$J=\sum_{n\in\mathbb N} \phi_n \otimes y_n$. Each $y_n \in \ell_1(f)$ has countable support $S_n$ so that $J(x)$ has support in the countable set $S=\bigcup_n S_n$ for every $x\in X_g$
(because $\lbrace y\in\ell_1(f):$ support$(y)\subseteq S\rbrace$ is closed).
For $x=\delta_i$ with $i\in I\setminus S$ you get a contradiction.

Title: Are spaces of finite sequences nuclear?


问:
Union of all finite fields of order $p^i$, for some prime number $p$ is algebraically closed.
Let $p$ be a prime number. Consider for all $k,r\in\mathbb{N}_0$ with $k|r$ the field $\mathbb{F}_{p^k}$ as a subfield of $\mathbb{F}_{p^r}$. Define

$$ \overline{\mathbb{F}_p}:=\bigcup_{i\in\mathbb{N}_0}\mathbb{F}_{p^i}.$$

I want to show that $\overline{\mathbb{F}_p}$ is the algebraic closure of $\mathbb{F}_p$. I already managed to show that $\overline{\mathbb{F}_p}$ is a field and that it is algebraic over $\mathbb{F}_p$, but I can't seem to show that it is also algebraically closed.

I know that for every $i\in\mathbb{N}_0$, $\mathbb{F}_{p^i}$ is a decomposition field of the polynomial $X^{p^i}-X$. Which means this polynomial can be expressed as
$$X^{p^i}-X=\prod_{a\in\mathbb{F}_{p^i}}(X-a),$$

and I thought that this could help me prove that every irreducibel polynomial in $\overline{\mathbb{F}_p}[X]$ has degree $1$ but so far it hasn't.

Any help would be appreciated.

答:
If $q(x) \in \mathbb{F}_{p^k}[x]$ is of degree $n$ and irreducible, $q$ has a non zero root in the rupture field $\mathbb{F}_{p^k}[x]/(q)$ which is isomorphic to $\mathbb{F}_{(p^k)^n}= \mathbb{F}_{p^{kn}}$.

By induction, you can prove that $q$ splits in one of the $\mathbb{F}_{p^m}$. Which implies the desired result.


问:
Is this process a Brownian motion?
I have some problem with deciding whether the process $Y_t$ is a Brownian motion. Here is the structure:

$X_t = \int_0^tW_s^2ds$ where $\{W_t\}_{t≥0}$ is a Brownian motion, and $Y_t=\frac{X_t-\frac{1}{2}t^2}{\frac{1}{2}t^{3/2}}$.

So here are what I did:

I calculated the mean and variance for $X$ and $Y$: $\mathbb{E}X_t=\frac{1}{2}t^2$, $\text{Var}X_t= \frac{1}{3}t^4$, $\mathbb{E}Y_t=0$, $\text{Var}Y_t= \frac{4}{3}t$.

I didn’t find the wrong places in my calculations. I found the variance of $Y_t$ is not $t$, so I wanted to say that $Y_t$ is not a Brownian motion. But I feel this argument is not that convincing. I am wondering there might be something wrong but I didn’t notice. Could you give me a convincing proof on whether $Y_t $ is a Brownian motion or not?

Thank you so much!

答: $Y$ is not a Brownian motion because the sample path $t\mapsto Y(t,\omega)$ is differentiable for each $\omega\in\Omega$.

Title: Is this process a Brownian motion?


问:
Prove that this logical statement is a tautology
$$\left( A \land \neg B \right) \lor \left[ \left( \left( \neg C \land D \right) \land \left( A \lor \neg E \right) \land \left( F \lor \neg G \right) \right) \implies \left( D \lor \neg E \right) \right]  $$

I started by rewriting the implication:
$$\left( A \land \neg B \right) \lor \left[ \neg\left( \left( \neg C \land D \right) \land \left( A \lor \neg E \right) \land \left( F \lor \neg G \right) \right) \land \left( D \lor \neg E \right) \right]$$

Simplify:

$$\left( A \land \neg B \right) \lor \left[ \left( \neg \left( \neg C \land D \right) \lor \neg \left( A \lor \neg E \right) \lor \neg \left( F \lor \neg G \right) \right) \land \left( D \lor \neg E \right) \right]$$

$$\left( A \land \neg B \right) \lor \left[ \left( C \lor \neg D \lor  \left( \neg A \land  E \right) \lor  \left( \neg F \land G \right) \right) \land \left( D \lor \neg E \right) \right]$$

Expand:

$$\left( A \land \neg B \right) \lor  \left(  C \land \left( D \lor \neg E \right) \right) \lor \left( \neg D \land \left( D \lor \neg E \right) \right) \lor  \left( \left( \neg A \land  E \right) \land \left( D \lor \neg E \right) \right) \lor \left( \left( \neg F \land G  \right) \land \left( D \lor \neg E \right) \right)$$

$  C \land \left( D \lor \neg E \right) \Leftrightarrow (C \land D) \lor (C \land \neg E)$

$\neg D \land \left( D \lor \neg E \right) \Leftrightarrow \neg D \land \neg E$

$\left( \neg A \land  E \right) \land \left( D \lor \neg E \right) \Leftrightarrow \neg A \land D \land E$

$\left( \neg F \land G  \right) \land \left( D \lor \neg E \right) \Leftrightarrow (D \land \neg F \land G) \lor (\neg E \land \neg F \land G)$

$$ (A \land \neg B) \lor ((C \land D) \lor (C \land \neg E)) \lor (\neg D \land \neg E) \lor (\neg A \land D \land E) \lor ((D \land \neg F \land G) \lor (\neg E \land \neg F \land G))$$

I feel like this can be simplified even further. How do I continue from here?

Thanks

答:
Firstly, material implication should be $A \implies B \equiv \neg A \lor B$, so you made a mistake when rewriting the implication.

Secondly, the intuition should be that the implication must be true. So we review the premises $\neg C \land D$, $A \lor \neg E$, $F\lor \neg G$ to see how $D \lor \neg E$ can be derived. Indeed from $\neg C \land D$ we can derive $D$, so we write:

$$\left( A \land \neg B \right) \lor \left[ \left( \left( \neg C \land D \right) \land \left( A \lor \neg E \right) \land \left( F \lor \neg G \right) \right) \implies \left( D \lor \neg E \right) \right]  $$

$$\equiv\left( A \land \neg B \right) \lor \left[ \neg  \left( \neg C \land D \right) \lor \neg \left( A \lor \neg E \right) \lor \neg \left( F \lor \neg G \right) \lor \left( D \lor \neg E \right) \right]  $$

$$\equiv\left( A \land \neg B \right) \lor \left[ \left(C \lor \neg D \right) \lor \neg \left( A \lor \neg E \right) \lor \neg \left( F \lor \neg G \right) \lor \left( D \lor \neg E \right) \right]  $$

$$\equiv\left( A \land \neg B \right) \lor \left[ \left(D \lor \neg D \right) \lor \neg \left( A \lor \neg E \right) \lor \neg \left( F \lor \neg G \right) \lor \left(C \lor \neg E \right) \right]  $$

$$\equiv\left( A \land \neg B \right) \lor \left[ \left(T \right) \lor \neg \left( A \lor \neg E \right) \lor \neg \left( F \lor \neg G \right) \lor \left(C \lor \neg E \right) \right]  $$

$$\equiv T$$

if the Law of Excluded Middle is allowed. To derive the tautology at the end, we observe that the tautology from LEM is conjuncted with other statements, so it must be a tautology as well.


问:
Boyd & Vandenberghe, page 27 — intuition and proof for how a ray is convex rather than affine
Can someone provide intuition & proof for why a ray is convex, I don't see the sum to $1$ constraints for theta :

A ray of the form $\left\{x_{0}+\theta v \mid \theta \geq 0\right\}$, where $v$ != $0$ is convex, but not affine. It is a convex cone if its base $x_{0}$ is $0.$

From Boyd & Vandenberghe Convex Optimization pg. 27

答:
I will prove the case where $x_0 = 0$, since convexity is invariant under translation.

According to the definition, your ray is of the form, for a fixed $\mathbf{x} \in \mathbb{R}^d$,

$$R= \{\lambda\mathbf{x} : \lambda \geqslant 0\}$$

Now let $\mathbf{y},\mathbf{z} \in R$, so $\mathbf{y} = y\mathbf{x}$ and $\mathbf{z} = z\mathbf{x}$ for some $y,z\geqslant 0$ by definition of the ray. As this is a subset of a $2$-dimensional linear (affine, if $x_0 \neq 0$) subspace, all convex combinations will require at most two terms without any redundancy (it is not hard to show for many-termed sums, I will leave this as an exercise to help understand the proof). So for all $t \in [0,1]$, it suffices to show:

$$t\mathbf{y} + (1-t)\mathbf{z} \in R$$

The terms $t\mathbf{y}, (1-t)\mathbf{z}$ are the same as $ty\mathbf{x}$ and $(1-t)z\mathbf{x}$ by the result above, and as $t$ and $(1-t)$ are both at least $0$, $t\mathbf{y}$ and $(1-t)\mathbf{z}$ are in the ray. As such, the above sum is just:

$$(ty + (1-t)z)\mathbf{x}$$

which is in the ray, so the ray is convex.

Generally, proving shapes convex involve showing the conditional:

$$\{\mathbf{x}\}_i \in X \Longrightarrow \sum_i \lambda_i \mathbf{x}_i \in X$$

for $\sum \lambda_i =1$ and each $\lambda_i \geqslant 0$, holds true. You can see how the above proof is essentially just this conditional.

Geometrically and intuitively, a shape is convex when the line between any two points is in the shape. I dare you to find two points in a ray where the line between them is not in the ray.


问:
Recurrence relation $ T(n)=5^n+3T(\lfloor n^\frac{2}{5}\rfloor) $
I need to prove that the following recurrence relation is $O(5^n)$:
$$ T(n)=5^n+3T(\lfloor n^\frac{2}{5}\rfloor) $$
And $T(n)=\Theta(1)$ for $n\le 9$.

I am trying induction, and proving that there exists $c$ such that $T(n)\le c\cdot5^n$, but I am stuck at the induction step:

$T(n)=5^n+3T(\big\lfloor n^\frac{2}{5}\big\rfloor)\le5^n+3c\cdot 5^{\lfloor n^\frac{2}{5}\rfloor}\le5^n+3c \cdot 5^{n^{\frac{2}{5}}}=\bigg(1+\frac{3c}{5^{n- n^\frac{2}{5}}}\bigg)\cdot 5^n$.

How can I continue in the induction (turn the parenthesis into the constant $c$ / is there a different way to prove it?

答:
Considering the recurrence

$$
T(n) = 5^n+3T\left(n^a\right)
$$

$$
T\left(a^{\log_a n}\right) = 5^n+3T\left(a^{\log_a n^a}\right)
$$

now making $z = \log_a n,\ \ \mathcal{T}(\cdot) = T\left(a^{(\cdot)}\right)$ we follow with

$$
\mathcal{T}(z) = 5^{a^z}+3\mathcal{T}(a z)
$$

or

$$
\mathcal{T}(a^{\log_a z}) = 5^{a^z}+3\mathcal{T}(a^{\log_a (a z)})
$$

making now $u = \log_a z,\ \ \mathbb{T}(\cdot) = \mathcal{T}\left(a^{(\cdot)}\right)$ we follow with

$$
\mathbb{T}(u) = 5^{a^{a^u}}+3\mathbb{T}(u+1)
$$

This recurrence has the solution

$$
\mathbb{T}(u) =3^{1-u}\left(c_1 - \sum_{k=0}^{u-1}3^{k-1} 5^{a^{a^k}}\right)
$$

and going backwards

$$
T(n) = 3^{1-\frac{\ln \left(\frac{\ln (n)}{\ln (a)}\right)}{\ln (a)}}\left(c_1-\sum _{k=0}^{\frac{\ln \left(\frac{\ln (n)}{\ln(a)}\right)}{\ln (a)}-1}3^{k-1} 5^{a^{a^k}}\right)
$$

NOTE

With $a = \frac 25$ and $k = \frac{\ln \left(\frac{\ln (n)}{\ln(a)}\right)}{\ln (a)}-1$

$$
 5^{a^{a^k}} = 5^{n^{\frac 52}}
$$


问:
Question about continuity of linear transformations

Assume $ W,V $ are two normed vector spaces over $ \mathbb{C} $ and let $ T:V\to W $ be a linear transformation.

Prove that $ T $ is continuous if and only if for any sequence of vectors $ (v_n)_n $ such that $v_n \to \vec{0}$, we have
$$ \sup_{n}\|Tv_{n}\|<\infty $$

The first direction is easy, if $T$ is continuous then for any sequence of vectors that converges to $\vec{0}$ we have
$$
\lim_{n}\|Tv_{n}\|
 =\|T(0)\|
 = 0
$$
and thus the sequence $\|Tv_{n}\|$ converges and also bounded.

I'm struggling with the other direction. Say for any sequence of vectors that converges to $\vec{0}$ we have that $\sup_{n} \|Tv_{n}\|<\infty$. How can I use that to prove that $T$ is continuous in $0$ ? (This would imply that $T$ is continuous in the whole space).

Alternatively, if I'll find a way to show that the given detail implies that $\|T\|<\infty$, this would also be enough to conclude that $T$ is continuous.

Thanks in advance.

答: Suppose that $T$ is discontinuous at some $x_0\in X$. Then, since $T$ is linear, it is also discontinuous at $0$. So, there's some sequence $(v_n)_{n\in\Bbb N}$ such that $\lim_{n\to\infty}v_n=0$ and that you don't have $\lim_{n\to\infty}Tv_n=0$. So, for some $\varepsilon>0$, there are infinitely many $n$'s such that $\|Tv_n\|\geqslant\varepsilon$. You can assume without loss of generality that this occurs for every $n$. But then $\lim_{n\to\infty}\frac{v_n}{\sqrt{\|v_n\|}}=0$ too, $$\left\|T\frac{v_n}{\sqrt{\|v_n\|}}\right\|\geqslant\frac\varepsilon{\sqrt{\|v_n\|}}$$and, since $\lim_{n\to\infty}\sqrt{\|v_n\|}=0$,$$\lim_{n\to\infty}\left\|T\frac{v_n}{\sqrt{\|v_n\|}}\right\|=\infty.$$


问:
The matrix of $T(x_1,x_2)=(x_1+x_2,x_1-x_2)$ with respect to a basis
Consider the linear map $T:\mathbb{R}^2\to \mathbb{R}^2, T(x_1, x_2)=(x_1+x_2,x_1-x_2)$. Let $B_1$ be the canonical base of $\mathbb{R}^2$ and consider another basis $B_2=\{f_1,f_2\}$, where $f_1=(1,1)$ and $f_2=(1,2)$.

So, according to my computations, the matrix of $T$ with respect to $B_1$ is $\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}$ and the matrix of $T$ with respect to $B_2$ is $\begin{pmatrix}
2 & 3 \\
0 & -1
\end{pmatrix}$. However, I tried computing the transition matrices. I got that the transition matrix from $B_1$ to $B_2$ is $\begin{pmatrix}
1 & 1\\
1 & 2
\end{pmatrix}$ and the transition matrix from $B_2$ to $B_1$ is $\begin{pmatrix}
2 & -1\\
-1 & 1
\end{pmatrix}$. I should have that $\begin{pmatrix}
2 & 3 \\
0 & -1
\end{pmatrix}= \begin{pmatrix}
2 & -1\\
-1 & 1
\end{pmatrix}\cdot \begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}\cdot \begin{pmatrix}
1 & 1\\
1 & 2
\end{pmatrix}$, but this is not true. Where did I go wrong?

答:
The second column of your matrix of $T$ with respect to $B_2$ is wrong.

Indeed, $T(f_2) = T(1,2) = (3,-1) \neq  3f_1 - f_2 = (2,1).$


问:
Gradient, Hessian, and minimum of a vector function?
Given the function $f(\vec{x})=(\frac{1}{2})\vec{x}^TP^TP\vec{x}+q^T\vec{x}+r$, where $\vec{x}$ and $q \in \mathbb{R}^n$, and $P \in \mathbb{R}^{n x n}$ is full rank and $r \in \mathbb{R}$, I am trying to find the gradient, Hessian, and what the minimum of this function is.

For the sake of simplicity, I am going to say $n=2$ and define the following as:

$\vec{x}= \begin{bmatrix}
x_1\\
x_2
\end{bmatrix}$

$ P = \begin{bmatrix}
p_{11} & p_{12}\\
p_{21} & p_{22}
\end{bmatrix}$

$\vec{q}= \begin{bmatrix}
q_1\\
q_2
\end{bmatrix}$

This is what I computed for the gradient:

$\frac{\partial f(\vec{x})}{\partial x_1} = x_1p^2_{11} + x_1p^2_{21} + x_2p_{11}p_{12} + x_2p_{21}p_{22} + q_1$

$\frac{\partial f(\vec{x})}{\partial x_2} = x_1p_{11}p_{12} + x_1p_{21}p_{22} + x_2p^2_{12} + x_2p^2_{22} + q_2$

For the Hessian, I got the following: $\begin{bmatrix}
p^2_{11} + p^2_{21} & p_{11}p_{12} + p_{21}p_{22}\\
p_{11}p_{12} + p_{21}p_{22} & p^2_{11} + p^2_{21}
\end{bmatrix}$

Are my computations of the gradient and Hessian correct? From here, how do I find the minimum of this function?

答:
What is is interesting is at least sufficient condition to have that extremum:

If $a$ is a critical point

$f$ has a strict minimum in $a$ if $H(f)(a)$ is defined positive (resp. defined negative for a maximum)

Recalling that defined positive is that

$$  \forall x \in R^2, x \neq \vec{0,  }\, x^tH(f)(a)x>0 $$

Begin by determining the critical point (annulation of gradient is necessary)


问:
Computing the moments of a triangle
Assume you have a triangle in the plane defined by its three vertices at $(x_0,y_0)$, $(x_1,y_1)$, and $(x_2,y_2)$. Is there a general expression for the moments of the triangle, where the moments are defined as averages by the following expression:

$$\langle x^n y^m \rangle = \frac{1}{A}\int_\triangle x^n y^m \, dA $$

Integrating manually to compute the first few moments, one can obtain:

$$\langle x \rangle = \frac{1}{3} \left(x_0+x_1+x_2\right) $$

$$\langle x^2 \rangle = \frac{1}{6} \left(x_0^2+x_1^2+x_2^2+x_0 x_1+x_0 x_2+x_1 x_2\right) $$

$$\langle x y \rangle = \frac{1}{12}\left(2 x_0 y_0+x_1 y_0+x_2 y_0+x_0 y_1+2 x_1 y_1+x_2 y_1+x_0 y_2+x_1 y_2+2 x_2 y_2 \right)$$

Is it possible to obtain a closed form expression for general $n$ and $m$ using combinations of the coordinates?

EDIT:

Can the same be done for line segments defined by their endpoints $(x_0,y_0)$ and $(x_1,y_1)$?
Again the first few moments in this case are:

$$\langle x \rangle = \frac{1}{2} \left(x_0+x_1\right) $$
$$ \langle x^2 \rangle = \frac{1}{3} \left(x_0^2+x_0 x_1+x_1^2\right) $$
$$ \langle x^3 \rangle = \frac{1}{4} \left(x_0^3+x_0^2 x_1+x_0 x_1^2+x_1^3\right) $$
$$ \langle x y \rangle = \frac{1}{6} \left(2 x_0 y_0+x_1 y_0+x_0 y_1+2 x_1 y_1\right) $$
$$ \langle x^2 y \rangle  = \frac{1}{12} \left( 3 x_0^2 y_0+2 x_0 x_1 y_0+x_1^2 y_0+x_0^2 y_1+2 x_0 x_1 y_1+3 x_1^2 y_1\right)$$

答:
If you don't mind a quadruple sum as a closed form (it was the best simplification of the integral I gave in the comments that I could manage, using repeated applications of the binomial theorem), then

$\langle x^p y^q\rangle=\displaystyle 2\sum_{k=0}^p\sum_{\ell=0}^q \frac{(-1)^{k+\ell}\binom{p}{k}\binom{q}{\ell}\nu_{k\ell}}{k+\ell+2} x_0^{p-k} y_0^{q-\ell}$

where

$\displaystyle \nu_{k\ell}=\sum_{i=0}^k\sum_{j=0}^\ell \frac{\binom{k}{i}\binom{\ell}{j}}{k-i+\ell-j+1}\left(x_0-x_1\right)^i \left(x_1-x_2\right)^{k-i} \left(y_0-y_1\right)^j \left(y_1-y_2\right)^{\ell-j}$

is an explicit expression for $\langle x^p y^q\rangle$.

FWIW, Mathematica is unable to evaluate this (as usual with such complicated multiple sums).

For the problem of finding the moments of a line segment with endpoints $\left(x_0, y_0\right)$ and $\left(x_1, y_1\right)$, the integral can be derived by noting that a set of parametric equations for the line passing through the two given points is

$\displaystyle \begin{align*}x&=x_0+t(x_1-x_0)\\y&=y_0+t(y_1-y_0)\end{align*}\qquad 0\leq t\leq1$

Substituting the two equations into the expression $x^p y^q$ and integrating over the parameter range gives

$\displaystyle \int_0^1\left(x_0+t\left(x_1-x_0\right)\right)^p\left(y_0+t\left(y_1-y_0\right)\right)^q\mathrm{d}t$

which can be expanded into a double sum (again by using the binomial theorem):

$\displaystyle \sum_{j=0}^p\sum_{k=0}^q\frac{\binom{p}{j} \binom{q}{k}}{j+k+1}\left(x_1-x_0\right)^j \left(y_1-y_0\right)^k x_0^{p-j} y_0^{q-k}$

which can be "simplified" into the Appell hypergeometric function:

$x_0^p y_0^q F_1\left(1;-p,-q;2;1-\frac{x_1}{x_0},1-\frac{y_1}{y_0}\right)$

ContentLicense: CC BY-SA 2.5

Title: Computing the moments of a triangle


问:
How to prove the inequality $\Theta(x,y)\le \Theta(x,z)+\Theta(z,y)$?
Let $x, y$ be two complex vectors, $$\cos\Theta(x,y):=\operatorname{Re} \frac{y^*x}{\|x\|\|y\|} .$$ Then I want to prove that $$\Theta(x,y)\le \Theta(x,z)+\Theta(z,y) .$$  </p>

答:
The inequality simply says that the "angle between $\vec x$ and $\vec z$ plus the angle between $\vec z$ and $\vec y$ is never smaller than the angle between $\vec x$ and $\vec y$". That's obvious in the Euclidean space. First, as ncmathsadist pointed out, the angles don't change if you normalize $\vec x,\vec y,\vec z$ to unit vectors.

Then the angles (in radians) between two vectors are measured just as distances measured along the surface of the unit sphere and the inequality says that there the crow-fly, straight distance between $x$ and $y$ can't be made shorter by inserting a detour to $z$. Well, I hope that it's not hard to see how to prove that the crow-fly distance is the shortest path between the two vectors on the sphere.


问:
Hartog's extension theorem for codimension 2
I need to use a version of Hartog's extension theorem, that I did not find by googling around. However I think I found a solution myself, and wanted to ask if this is correct. Any feedback would be appreciated :)

Most versions of Hartog's extension theorem are similar to this one from Wikipedia:

Let $f$ be a holomorphic function on a set $G \setminus K$, where $G$ is an open subset of $\mathbb C^n (n ≥ 2)$ and $K$ is a compact subset of $G$. If the complement $G \setminus K$ is connected, then $f$ can be extended to a unique holomorphic function on $G$.

However, I've got a holomorphic function defined on an open set  $U \setminus Z$, where $U \subset \mathbb C^n$ is open, and $Z$ is an analytic set of complex codimension $\geq 2$. So in general $Z$ will not be compact.

My idea is to proceed by induction on $n$. If $n = 2$, $Z$ is a discrete set of points, so this case is covered by the version of Hartog's theorem cited above.

For $n > 2$, I would like to choose some hyperplane
$H \subset \mathbb C^n$, such that for some fixed $v \in \mathbb C^n \setminus H$,
\begin{equation} \tag{$*$}
\dim (H + z \cdot v) \cap Z = \dim Z - 1, \quad \forall z \in \mathbb C.\end{equation} This happens if and only if $H + z \cdot v$ does not contain any irreducible component of $Z$. Choosing such an $H$ is always possible, since each irreducible component is contained in a Zariski closed subset of hyperplanes.

So by induction we can extend $f$ to any of the slices $U \cap (H + z \cdot v)$, and this gives a well-defined function
$$ f : U \to \mathbb C. $$
To see that this extension is holomorphic, suppose $p \in H+z \cdot v$, and $z_1, \dotsc, z_{n-1}$ are coordinates on $H + z \cdot v$. Then by Cauchy's formula on $H + z \cdot v$ one has
$$f(p) = \int_{|w_1| = r} \frac{f(w_1, z_2(p), \dotsc, z_{n-1}(p), z(p))}{w_1 - z_1} dw_1$$
And since $\frac{\partial f}{\partial \overline z} = 0$ outside of $Z$, the same is true for the integral on the right hand side. So $f$ is holomorphic everywhere.

答:
I am not sure about all the details, but since you assume to deal with complex analytic sets whose codimension is at least 2, your result is similar to Hassler Whitney's version ([1] §2.5, pp. 50-51, theorem 5B) of  Hartogs' theorem on separate analyticity:

Theorem. Let $H\subset \Bbb C^n$ be a open set and let $V$ be an analytic set of codimension $\ge 2$ in $H$. Then any holomorphic function $f$ in $H\setminus V$ has a unique holomorphic extension to the whole $H$.

His proof does not uses directly Cauchy's formula, but he uses other results he proved earlier by using it. This perhaps is not a direct answer to your question, but it may help yourself checking it.

References

[1] Hassler Whitney, Complex analytic varieties, (English)
Addison-Wesley Series in Mathematics. Reading, Massachussets-Menlo Park, California-Lomndon-Don Mills, Ontario: Addison-Wesley Publishing Company. XII, 399 p. (1972), MR0387634, Zbl 0265.32008.


问:
Without calculating the partial derivatives, explain why sin x cosh y and cos x sinh y are harmonic functions in C.
It could be simple if I was allowed to use partial Derivatives.
Without calculating partial derivative I don't know how to prove that it's harmonic. Can someone please amswer this question?</p>

答: Note that, if $x,y\in\Bbb R$,\begin{align}\sin(x+yi)&=\sin(x)\cos(yi)+\cos(x)\sin(yi)\\&=\sin(x)\cosh(y)+\cos(x)\sinh(y)i\end{align}and therefore your functions are the real part and the imaginary part of a holomorphic function. Therefore, they are harmonic.


问:
Composition of inverse trigonometric functions with trigonometric functions
I found in this Wikipedia article a useful table showing the algebraic expressions for the composition of trigonometric functions with inverse trigonometric functions, along with a picture explaining where they come from. Among others,

$$\begin{align}
\sin({\arccos{x}})=\sqrt{1-x^2}, \quad \sin({\arctan{x}})=\frac{x}{\sqrt{1+x^2}}\\
\cos({\arcsin{x}})=\sqrt{1-x^2}, \quad \cos({\arctan{x}})=\frac{1}{\sqrt{1+x^2}}\\
\tan({\arcsin{x}})=\frac{x}{\sqrt{1-x^2}}, \quad \tan({\arccos{x}})=\frac{\sqrt{1-x^2}}{x}\\
\end{align}$$

What would be, instead, the algebraic expressions for the composition of inverse trigonometric functions with trigonometric functions, such as the followings? Could someone provide them or give a source where they can be obtained?

$$\begin{align}
\arcsin({\cos{x}})=?, \quad \arcsin({\tan{x}})=?\\
\arccos({\sin{x}})=?, \quad \arccos({\tan{x}})=?\\
\arctan({\sin{x}})=?, \quad \arctan({\cos{x}})=?\\
\end{align}$$

答:
Hint

Let $$ t=\arcsin(\cos(x))$$

then

$$-\frac{\pi}{2}\le t\le \frac{\pi}{2}$$

and

$$\sin(t)=\cos(x)=\sin(\frac{\pi}{2}+x)$$

thus

$$t=\arcsin(\cos(x))=x+\frac{\pi}{2}+2k\pi$$
or
$$t=\frac{\pi}{2}-x+2k\pi$$

we choose the the value in $ [-\frac{\pi}{2},\frac{\pi}{2}]$.


问:
The nature of variables within proofs
I am confused about how best to view certain kinds of variables that show up in the course of an elementary proof. For example, to prove that the square of every odd integer is odd, I would write:

Let $m \in \mathbf{Z}$ be arbitrary. Suppose $m$ is odd. Choose $k \in \mathbf{Z}$ such that $m = 2k + 1$. Then $m^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1$, so since $2k^2 + 2k \in \mathbf{Z}$, it follows that $m^2$ is odd.

The symbol $m$ is introduced as part of the technique of universal generalization; Velleman [1] describes this step as "introducing a new variable [$m$] ... to stand for an arbitrary object." The symbol $k$ is introduced as part of the technique of existential instantiation; [1] describes this as "introducing a new variable [$k$] to stand for an object for which [the predicate $m = 2k + 1$] is true."

In what sense are $m$ and $k$ variables? They do not seem to be bound variables, as they are in the scope of no quantifier. They also can't be free variables, because if they were, then, for example, the sentence being proved true after the introduction of arbitrary integer $m$ -- namely that if $m$ is odd then $m^2$ is odd -- would be a predicate and not even have a truth value; I am also obviously not free to substitute values into $m$ and $k$, if for no other reason than $m$ and $k$ are dependent.

Stoll [2] says a variable is free if it is not bound, so there is no third option.

Is it more correct to think of the phrases "let $m \in \mathbf{Z}$ be arbitrary" and "choose $k \in \mathbf{Z}$ such that ..." as introducing new constants rather than variables, as their values are fixed despite being unspecified ($m$) or unknown ($k$)? More generally, the idea of a symbol that stands for a particular but arbitrary element of some set seems incompatible with the typical description of a variable as a placeholder as seen in, e.g., Epp [3].

[1] Velleman, D. (2006). How to Prove It: A Structured Approach (2nd ed.). Cambridge: Cambridge University Press.

[2] Stoll, Robert Roth (1963). Set Theory and Logic. W.H. Freeman.

[3] Epp, Susanna S. Variables in Mathematics Education. https://condor.depaul.edu/~sepp/VariablesInMathEd.pdf

答:
You say:  "They also can't be free variables, because if they were, then, for example, the sentence being proved true after the introduction of arbitrary integer $m$ -- namely that if $m$ is odd then $m^2$ is odd -- would be a predicate and not even have a truth value."  I would say this a little differently.  A predicate does not have a truth value until values are assigned to the free variables.  But if you assign values to the free variables, then a predicate has a truth value.

Now, let me suggest a slightly different version of Lereau's answer.  I like the idea of distinguishing between context and goal, but I would say that the context consists of two things: a specification of which variables are assumed to have values assigned to them, and a specification of which statements are assumed to be true (these are called "givens" in my book).  With this modified version of Lereau's idea of context, let me go through the proof.  We start with:

Context:  empty; that is, no variables are assumed to have values and no givens are assumed to be true

Goal:  $\forall m(m \text{ is odd} \Rightarrow m^2 \text{ is odd})$

When you write "Let $m \in \mathbb{Z}$ be arbitrary," you change the situation to:

Context: variables: $m$; givens: none

Goal:  $m \text{ is odd} \Rightarrow m^2 \text{ is odd}$

So now you are assuming that $m$ stands for something (so the goal, which contains $m$ as a free variable, has a truth value), but you are making no assumption about what $m$ stands for, so it could be anything.  Now you assume $m$ is odd:

Context: variables: $m$; givens: $m$ is odd

Goal:  $m^2$ is odd

Writing out the definition of "odd", this means:

Context: variables: $m$; givens: $\exists k \in \mathbb{Z}(m = 2k+1)$

Goal:  $\exists j \in \mathbb{Z}(m^2 = 2j+1)$

Now since the given is an existential statement, we can introduce the variable $k$  to stand for an integer such that $m = 2k+1$:

Context:  variables: $m$, $k$; givens:  $m = 2k+1$

Goal:  $\exists j \in \mathbb{Z}(m^2 = 2j+1)$

Now $m$ and $k$ are both assumed to have values assigned to them, and the statement $m = 2k+1$ is assumed to be true with those values assigned.  Now it is just simple algebra to come up with the value of $j$ that demonstrates that the goal is true.

Title: The nature of variables within proofs


问:
What are the units of the variance/rate parameter in a 1D continuous Brownian diffusion process?
In a 1-dimensional continuous Brownian diffusion process following the SDE:
$$
dx(t) = \sigma dW(t)
$$
where $W(t)$ represents the Wiener process, what are the units of $\sigma$?

Given that the probability density of, say, $x(s) = 0$ at time $s>0$, given that at time $0$ $x(0) = 0$, is $0 \sim N(0, \sigma^2 s)$, is $\sigma^2$ in $\frac{x^2}{unit \: time}$?

答:
Involving irrational function in $\mathcal{N}(0,\sigma^2t)$,

We must have in terms of dimension :

$$ [\sigma^2t]=1$$

so

$$ [\sigma]=T^{-1/2}$$

Conclusion

$W(t)$ Wiener process is of a dimension of $[L.T^{1/2}]$ unit of $m.t^{-1/2}$ ISU and $\sigma$ is of a dimension of $[T^{-1/2}]$ unit of $t^{-1/2}$ ISU.

ISU means International System of Unit.