问:
Matrix inequality involving the sum of elements

Let $M$ a $n \times n$ matrix that has all elements either $1$ or
$-1$. Prove that the sum of the elements of $M^2$ is at least
$\dfrac{n(5-6n)}{3}$.

My trial: I will denote the trace of a matrix by $\text{Tr }(...)$. Let $U$ the $n\times n$ matrix that has all the elements $1$. Then the sum of the elements of $M^2$ is $\text{Tr }(UM^2)$. I also thought it would work if we let $M=[b_{ij}]_{1\leq i,j\leq n }$ where $|b_{ij}|=1$. Then, by considering $L_i$ the sum of all the elements of the line $i$ of $M$ and $C_i$ the same but for columns, then the inequality becomes
$$L_1C_1+\dots+L_nC_n\geq\frac{n(5-6n)}{3}.$$

It is also clear that $L_1+\dots+L_n=C_1+\dots+C_n$.

Please help me find a proof to this! Thank you in advance!

答:
The assertion is wrong. Let $M$ be a matrix where the upper right triangular block, including the diagonal itself, has elements $+1$, and the  lower left triangular block has elements $-1$. Then, as introduced by OP,  we consider row sums and column sums. The $k$th row has sum  $L_k = n+2-2\cdot k$, the $k$th column has sum  $C_k = -n+2\cdot k$. This gives
$$
L_1C_1+\dots+L_nC_n = \sum_{k=1}^n (n+2-2\cdot k)(-n+2\cdot k) = \frac{n (4 - n^2)}{3}
$$
Now we see that for $n\ge 6$ we have $\frac{n(5-6n)}{3} - \frac{n (4 - n^2)}{3} > 0 $, so indeed for  $n\ge 6$ the sum of all elements of $M^2$ in this case becomes smaller than the claimed $\frac{n(5-6n)}{3}$. $\qquad \Box$

It is hypothesized that for the presented triangular matrix structure, indeed the sum of elements of $M^2$ is smallest amongst all matrix realisations. Computer simulations for the first $n$ confirm this, note that there are $2^{n^2}$ many realisations of the matrix $M$.

More importantly, there is a "greedy" argument which supports this hypothesis. We want to minimize the sum $L_1C_1+\dots+L_nC_n$. Proceed iteratively. Start with the first term  $L_1C_1$. This becomes minimal if  $L_1$ is maximally positive and $C_1$ is maximally negative (or the other way around). This is achieved if the first row is all positive, and if the first column is all negative. The latter is not possible, since the first row and the first column have the element $M_{1,1}$ in common, hence the best choice is to have all further elements of the first column negative. Changing the sign of $M_{1,1}$ makes no difference.

Now proceed with the second term
$L_2C_2$. The same arguments hold, however $M_{2,1}$ and $M_{1,2}$ are already set in the first step. Let the rest of the second row/column be positive / negative. The first two terms together then give $n\cdot(2-n) + (n-2)\cdot(4-n) = -2(n-2)^2$. Now consider changing the sign and let $M_{1,2} = -1$. The first two terms together then give $(n-2)\cdot(2-n) + (n-2)\cdot(2-n) = -2(n-2)^2$. So indeed nothing changes.

This can be continued with change of  sign of $M_{1,2}$, or with both of $M_{1,2}$ and  $M_{2,1}$, with no effect. Hence minimality is preserved. Further, continue in the same vein with each further term in $L_3C_3 \cdots L_nC_n$. Once completed, this gives exactly the proposed matrix structure (and equivalent ones, applying  sign changes which are ineffective for the minimality of the sum).


问:
Expanding $P(Z \cap X \mid Y)$
I can't see exactly why $P(Z \cap X \mid Y) = P(Z \mid X \cap Y) \mathop{P}(X \mid Y)$. I have been told (and it is also my intuition) that this has to do with the definition of conditional probability, but I can't see how this works in detail. Can someone spell this out for me?

答:
This is simply because
\begin{align*}
\Pr(Z \cap X \mid Y)
&= \frac{\Pr(Z \cap X \cap Y)}{\Pr(Y)} \\
&= \frac{\Pr(Z \cap X \cap Y)}{\Pr(X \cap Y)} \frac{\Pr(X \cap Y)}{\Pr(Y)} \\
&= \Pr(Z \mid X \cap Y) \Pr(X \mid Y).
\end{align*}

ContentLicense: CC BY-SA 4.0

Title: Expanding $P(Z \cap X \mid Y)$


问:
How to understand the exact sequence of differential sheaf on projective space?
I'm reading Harteshorne chapter 2.8. In theorem 8.13, he shows there is a short exact sequence:
$$
0\rightarrow \Omega_{X/A}\rightarrow \mathcal O_{X}(-1)^{n+1}\rightarrow\mathcal O_X\rightarrow 0
$$
where $X = \mathbb P_A^n$ is the projective space over $A$.

I guess I can understand the proof line by line. But I have no idea what's going on in this proof. Do these computations have any geometric interpretations?

Also, this short exact sequence seems to be very important. Is there any initiative way to understand this exact sequence? Does it have geometric meaning? Thank you in advance.

答:
Let $S:=Spec(A)$ be an affine scheme and let $E:=A\{e_0,..,e_n\}$ be the free left $A$-module on the $n+1$ elements $e_i$, and $E^*:=A\{x_0,..,x_n\}$, with $x_i$ the dual basis. Let $\mathcal{E}$ be the sheafication of the $A$-module $E$. There is by Hartshornes book (Chapter II.7) a relative projective space bundle $\mathbb{P}(\mathcal{E}^*)$, and a canonical surjective morphism $\pi: \mathbb{P}(\mathcal{E}^*) \rightarrow S$ with projective spaces as fibers. For any point $s \in S$ it follows the fiber $\pi^{-1}(s) \cong \mathbb{P}^n_{\kappa(s)}$, where $\kappa(s)$ is the residue field of the point $s$ and where $\mathbb{P}^n_{\kappa(s)}$ is projective $n$-space over the field $\kappa(s)$.
The morphism $\pi$ is locally trivial in the Zariski topology. The tautological sequence reflects the fact that we may define the projective space bundle using representable functors: The projective space bundle represents the following functor.
Define the functor $F:Sch(S) \rightarrow Set$ from the category of schemes over $S$ to the category of sets as follows: Let $F(X, g)$ be the
set of pairs $(\mathcal{L}, \phi)/\cong$. Where $\phi: g^*\mathcal{E} \rightarrow \mathcal{L}$ is a surjective map onto an invertible sheaf $\mathcal{L}$ on $X$, modulo the following equivalence relation: Two pairs $(\mathcal{L}, \phi)$ and $(\mathcal{L}', \phi')$ are equivalent iff there is an isomorphism of sheaves $\psi: \mathcal{L} \rightarrow \mathcal{L}'$ making the obvious diagram commute. We may prove that there is an isomorphism of functore $Hom_S(-, \mathbb{P}(\mathcal{E}^*) ) \cong F(-)$ (this is proved in Hartshorne Prop II.7.12). Hence the functor $F$ is represented by the projective space bundle $\mathbb{P}(\mathcal{E}^*)$. If $S:=Spec(k)$ is the spectrum of a field $k$, it follows $\mathbb{P}(\mathcal{E}^*)(k):=Hom_S(S, \mathbb{P}(\mathcal{E}^*) = F(S, id)$, and from this it follows the $k$-rational points $\mathbb{P}(\mathcal{E}^*)(k)$ of the projective space bundle $\mathbb{P}(\mathcal{E}^*)$ equals the set of lines in the $k$-vector space $E$. Hence the tautological sequence reflects the fact that the projective space bundle parametrizes "lines" in $E$.

More generaly if $\mathcal{E}$ is a coherent $\mathcal{O}_S$-module, the EGA book series defines the projective space bundle $\mathbb{P}(\mathcal{E}^*)$ and a surjective morphism $\pi: \mathbb{P}(\mathcal{E}^*) \rightarrow S$ with projective spaces as fibers. If $i: U\rightarrow S$ is an open subscheme where $\mathcal{E}$ is locally trivial it follows the fiber product $U\times_S \mathbb{P}(\mathcal{E}) \cong \mathbb{P}(i^*\mathcal{E}^*) \cong \mathbb{P}(\mathcal{E}_U^*)$ defines a fibration $\pi_U: \mathbb{P}(\mathcal{E}_U^*) \rightarrow U$ which is a locally trivial fibration in the Zariski topology with projective spaces as fibers. A good exercise is to fill in all details of HH, Prop II.7.12 and then try to generalize to the grassmannian bundle. One may construct the grassmannian bundle $\pi: \mathbb{G}(k,\mathcal{E}) \rightarrow S$ of a rank $d$ locally trivial $\mathcal{O}_S$-module $\mathcal{E}$. The map $\pi$ is a surjective map with the property that for any point $s \in S$ it follows the fiber $\pi^{-1}(s)$ is isomorphic to the classical grassmannian $\mathbb{G}(k,\mathcal{E}(s))$,
where $\mathcal{E}(s)$ is the fiber at $s$. It follows the fiber $\pi^{-1}(s)$ parametrize $k$-dimensional $\kappa(s)$-sub spaces of the fiber $\mathcal{E}(s)$. The map $\pi$ is locally trivial in the Zariski topology.
The following book gives all details on this construction. It also defines and studies flag bundles.

A. Grothendieck, Éléments de géométrie algébrique. I. (English) Zbl 0203.23301 Die Grundlehren der mathematischen Wissenschaften. 166. Berlin-Heidelberg-New York: Springer-Verlag. IX, 466 p. (1971).


问:
Taylor polynomial for $\sqrt x$
I need to find the kinds of Taylor polynomial for $\sqrt x$ at the point $a = 1$, $x$ lies in $<1,2>$.

I calculated the second derivative: $-\frac{1}{4}x^\frac{-3}{2}$
and the third: $\frac{3}{8}x^\frac{-5}{2}$

I calculated Taylor's polynomial:
$$ {1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2}$$

I need advice on estimating the error.
I calculated something like:

$$R \leq \frac{3/(8\cdot2^{5/2})}{3!}\cdot{(x-1)^3}$$

But I'm not sure if that's right.

答:
Using Newton's_generalized_binomial_theorem for real power $\frac{1}{2}$  we can obtain

$$\sqrt{x}=(1+ [x-1])^{\frac{1}{2}} = \sum\limits_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(x-1)^k$$

where $\binom{n}{k}=\frac{n(n-1)\cdots (n-k+1)}{k!}$

Title: Taylor polynomial for $\sqrt x$


问:
Throwing 6 sided cube 12 times. Probability of not getting 6 twice
Throwing 6 sided cube 12 times.

What is the probability of rolling no more than two 6's?

My thoughts so far:
There are 6 outcomes for each throw, so 6¹² is the total number of outcomes.
(5/6)¹² is the number of outcomes that I don't get a 6.
(5/6)¹⁰ is that I don't get a 6 in 10 throws, so the last 2 don't matter.
Is my thinking correct here, or am I missing something ?
Can someone help me with understanding the solution. Thanks.

答:
Hint

Binomial distribution:

$$P(X=x)={n\choose x}p^x q^{n-x}$$

Where
$$n=\text{Number of trials}\\
x=\text{Number of succeses}\\
p=\text{Probability of succeding in one trial}\\
q=1-p$$


问:
Complex number polar forms
$(1-i\sqrt{3})^{50}$ in the form x + iy

I have tried this out but seem to be missing something

答:
The polar form of a complex number $z=a+bi$ is $z=r(\cos\theta+i\sin\theta)$.

$$\left(1-i\sqrt{3}\right)^{50}=2^{50} \cdot\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)^{50}$$

$$=2^{50}\left(\cos \left(\frac{-\pi}{3}\right)+i\sin \left(\frac{-\pi}{3}\right)\right)^{50}$$

Using De moivre's formula

$$=2^{50}\left(\cos \left(\frac{-50\pi}{3}\right)+i\sin \left(\frac{-50\pi}{3}\right)\right)$$

In x+iy format, Alpha gets $-562949953421312 - 562949953421312 i \sqrt{3}$

Title: Complex number polar forms


答:
An elementary definition of the determinant of the matrix $A=(a_{i,j})$ is given by the expression
$$
\det(A)=\sum_{\pi}s(\pi)a_{1,\pi(1)}\cdots a_{n,\pi(n)}
$$
where the sum is extended over all permutations of the set $\{1,...,n\}$ and $s(\pi)=\pm1$ is the sign of the permutation $\pi$.

Since each summand is the product of exactly $n$ of the entries of the matrix, if you reverse all the signs each summand--and so the whole sum--will be altered by a factor of $(-1)^n$.

ContentLicense: CC BY-SA 3.0

Title: Why is $\det(-A)=(-1)^n\det(A)$?

Body: Why is $\det(-A)=(-1)^n\det(A)$?</p>


问:
Lebesgue measure coincides on boxes, with measure $m'$ that is translation invariant, and that holds the following equation: $m′([0,1]^d)=1$
I saw this question in measure theory course:

Let $m′:L(R^d)→[0,∞]$ be a measure on the Lebesgue measurable sets, satisfying:

• Unit box measure: $m′([0,1]^d)=1$.

• Translation invariance: $m′(x+A)=m′(A)$ for all $A∈L(R^d)$ and $x∈R^d$.

Show that $m′$ agrees with the Lebesgue measure on boxes, i.e., $m′(B)=m(B)$ for any box $B⊆R^d$.

I was able to show that for any box with integer length edges, I just used the 2 properties of $m'$ and the additivity property of a measure. But how can I show it for general boxes?

答:
Suppose for the moment that $m'([0,1)^d) = m'([0,1]^d) = 1$. Then we could divide the unit half-open hypercube into a disjoint collection of such cubes that look like $[k_1/n, (k_1+1)/n)\times\dotsm\times [k_d/n, (k_d+1)/n)$ where $k_1,\dotsc, k_d \in \{0,\dotsc, n-1\}$. If $Q_k$ denotes the half-open hypercube with $k=(k_1,\dotsc, k_d)$, then
$$\sum_k m'(Q_k) = m'([0,1)^d) = 1,$$
yet by translation invariance all the $m'(Q_k)$ are equal, and since there are $n^d$ of them we get $m'(Q) = 1/n^d$ for any half-open hypercube with side lengths $1/n$.

Piecing together such half-open cubes one can show $m'([0,s_1)\times\dotsm[0,s_d)) = s_1\dotsm s_d$ for rational $s_1,\dotsc, s_d$. Then any half-open box $B$ can be written as the limit from below of an increasing sequence of rational side-length half-open boxes; continuity of measure from below allows us to conclude $m'(B)=m(B)$.

The only thing left to check is that the deleted faces are $m'$-measure zero. To this end, suppose to the contrary that there is a face of the closed unit hypercube with nonzero $m'$-measure $c$ with $c>0$. Say without loss of generality it is $\{1\}\times[0,1]^{d-1}$. This cannot happen though by translation invariance. Indeed, pick $n$ so large that $cn>1$, and also pick $n$ distinct points $x_1,\dotsc, x_n\in (0,1]$. We have
$$E:=\bigcup_{k=1}^n \{x_k\}\times [0,1]^{d-1}\subset [0,1]^d$$
yet $m'(E) = cn$ while $m'([0,1]^d) = 1$, a contradiction. Thus any face of the unit hypercube (and, by extension, any face of any box) has $m'$-measure zero, and our work above suffices for the original closed box problem.


Title: Throwing 6 sided cube 12 times. Probability of not getting 6 twice

Body:
Throwing 6 sided cube 12 times.

What is the probability of rolling no more than two 6's?

My thoughts so far:
There are 6 outcomes for each throw, so 6¹² is the total number of outcomes.
(5/6)¹² is the number of outcomes that I don't get a 6.
(5/6)¹⁰ is that I don't get a 6 in 10 throws, so the last 2 don't matter.
Is my thinking correct here, or am I missing something ?
Can someone help me with understanding the solution. Thanks.

Body:
$$\binom{12}{0}(\frac{1}{6})^0(\frac{5}{6})^{12}+\binom{12}{1}(\frac{1}{6})^1(\frac{5}{6})^{11}+\binom{12}{2}(\frac{1}{6})^2(\frac{5}{6})^{10}$$

$\binom{12}{0}(\frac{1}{6})^0(\frac{5}{6})^{12}$ is the probability of no any 6

$\binom{12}{1}(\frac{1}{6})^1(\frac{5}{6})^{11}$ is the probability of one 6

$\binom{12}{2}(\frac{1}{6})^2(\frac{5}{6})^{10}$ is the probability of two 6's


答:
Rolling no more than two sixes would mean we either roll no six, $1$ six or we roll $2$ sixes in $12$ times. If random variable $X$ is the number of sixes then we have,

$P(X = 0) \, = (\frac{5}{6})^{12}$

$P(X = 1) \, = {12 \choose 1} \times \frac{1}{6} \times (\frac{5}{6})^{11}$

$P(X = 2) \, = {12 \choose 2} \times (\frac{1}{6})^2 \times (\frac{5}{6})^{10}$

Desired probability $ P (X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$


问:
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$
$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

Can we write it as following

$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$

Let's see what happens:-

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$

In the same way for further terms, we will get $0$

Let's also confirm for general term

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$

So the whole expression $E$ will be zero

But actual answer is $1$

Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$

Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?

I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.

But I got the following counter thought:-

$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$

As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$

$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$

$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$

Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.

But surprisingly $\dfrac{1}{3}$ is the correct answer.

I am feeling very confused in these two things. Please help me.

答:
You can use the Squeeze Theorem, with

$$ n\cdot \frac{n}{n^2 + n} \leq a_n \leq n\cdot \frac{n}{n^2 + 1} $$

And as $ n \to \infty$, you get $  1 \leq \lim_{n \to \infty} a_n \leq 1 $, So, using the Squeeze Theorem, $a_n \to 1$.


问:
Is solution of a PDE by separation of variables unique or does it exclude some other solutions?
Using "separation of variables" the PDE
$x \cdot u(x,t)_x = u(x,t)_t$

has solutions of the form

$u(x,t) = C\cdot e^{\lambda\cdot t} x^\lambda $

A more general solution could be found by superposition of this, taking different values for k:

$u(x,t) = \sum_i C_i\cdot e^{\lambda_i\cdot t} x^\lambda_i $

However, another solution would be:
$u(x,t) = C(t+\ln x)$

I wonder now, how those different types of solutions correspond to each other.

Neither of them can be a complete solution, because of the other one.

If I find a solution by separation of variables, how can I trust to have the most general solution at all and not to exclude something due to the special form of the separation $u(x,t)=v(x)\cdot w(t)$?

答:
Thanks to the method of characteristics one can find the general solution :
$$u(x,t)=F(x\:e^t)$$
where $F$ is an arbitrary function.

For example a particular solution in case of $F(X)=C\ln(X)$ is $u=C(t+\ln(x))$.

This is also consistant with the particular cases of $F(X)=CX^\lambda$ thus $u=Cx^\lambda e^{\lambda t}$.

The above general solution includes the cases of fonctions $F(X)$ expressed on form of series such as the series that you found thanks to the method of separation of variables.

Of course, infinity many other forms of particular solutions can be derived, for example with $F(X)=\sin(X)$ thus $u=\sin(xe^t)$.

If some valid condition is added to the wording of the problem, possibly the function $F(X)$ can be determined and the solution can be a unique particular solution.


问:
How does the following change of variables occur?
While reading a book, the author goes from writing that $$I=\int_0^1 \dots \int_0^1(u_1\dots u_s)^{-1+1/k}\frac{\sin 2\pi \lambda(u_1+\dots+u_s-1)}{\pi(u_1+\dots u_s-1)}\mathrm{d}u_1\dots \mathrm{d}u_s$$ to
$$I=\int_0^s \phi(u)\frac{\sin 2\pi \lambda (u-1)}{\pi(u-1)} \mathrm{d}u \dots(\star)$$ where
$$\phi(u)=\int_0^1 \dots \int_0^1 \Bigg(u_1\dots u_{s-1}(u-u_1-\dots-u_{s-1})\Bigg)^{-1+1/k} \mathrm{d}u_1\dots \mathrm{d}u_{s-1}$$
and the region of integration is over $(u_1,\dots, u_s)$ such that  $u-1<u_1+\dots u_{s-1}<u$.

I don't really understand what theorem is being used to do this change of variables. Clearly, they are taking $u=u_1+\dots u_s$. Then $u$ can be between $0$ and $(1+\dots + 1)=s$. We have the following change of coordinates function $(u_1,\dots,u_{s-1}, u_s)\mapsto (u_1,\dots, u_{s-1}, u-u_1-\dots - u_{s-1})$. But then our Jacobian matrix is given by

\begin{pmatrix}
 1 & 0 &\dots 0 & 0\\
 0 & 1  & \dots 0 & 0\\
 0 & \ddots & 1 & 0\\
 -1 & -1 & \dots -1 & 0
\end{pmatrix}

After that, I am not sure what's going on here and how the author gets to $(\star)$. Also, it seems so weird that now this is a integral of one variable (sort of) and I am quite confused. Thanks, any help is appreciated!

答:
The only variable changed is $u_s$, which is replaced with $u-u_1-u_2-\dots-u_{s-1}$. The transformation involved is thus
$$T(u_1,u_2,\dots u_{s-1},u)=(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})$$
and:
$$DT(u_1,u_2,\dots u_{s-1},u)=
\begin{pmatrix}
 1 & 0 &\dots 0 & 0\\
 0 & 1  & \dots 0 & 0\\
 0 & \ddots & 1 & 0\\
 -1 & -1 & \dots -1 & 1
\end{pmatrix}$$
whose determinat is equal to $1$. Changing variables yields:
$$\int_{(0,1)^s}f=\int_{T^{-1}((0,1)^s)}(f\circ T)\cdot |\det DT|$$
where $f(u_1,\dots ,u_s)=(u_1\dots u_s)^{-1+1/k}\frac{\sin 2\pi \lambda(u_1+\dots+u_s-1)}{\pi(u_1+\dots u_s-1)}$.
Therefore, setting $E=T^{-1}((0,1)^s)$ we get:
$$I=\int_E f(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})\textrm{d}u_1\dots\textrm{d}u_{s-1}\textrm{d}u$$
What we need to apply now is Fubini's theorem (by the way, Fubini's theorem and also the change of variables theorem hold under suitable hypoteses, which should be checked first). Since $E\subset\{(u_1,\dots,u_{s-1},u): 0<u<s\} $, thanks to Fubini's theorem we have:
$$I=\int_0^s \bigg(\int_{E^*(u)} f(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})\textrm{d}u_1\dots\textrm{d}u_{s-1}\bigg)\textrm{d}u$$
where
$$E^*(t)=\{(u_1,\dots,u_{s-1}) : (u_1,\dots,u_{s-1},t)\in E\}\subset \Bbb R^{s-1}$$ for ever real $t$.
Now, for $0<u<s$:
$$
\begin{align}
&\int_{E^*(u)} f(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})\textrm{d}u_1\dots\textrm{d}u_{s-1}\\
=&\int_{E^*(u)} \Bigg(u_1\dots u_{s-1}(u-u_1-\dots-u_{s-1})\Bigg)^{-1+1/k}\frac{\sin 2\pi \lambda (u-1)}{\pi(u-1)}\textrm{d}u_1\dots\textrm{d}u_{s-1}\\
=&\frac{\sin 2\pi \lambda (u-1)}{\pi(u-1)}\int_{E^*(u)} \Bigg(u_1\dots u_{s-1}(u-u_1-\dots-u_{s-1})\Bigg)^{-1+1/k}\textrm{d}u_1\dots\textrm{d}u_{s-1}
\end{align}
$$
Finally, observe that:
$$E^*(u)=\{(u_1,\dots,u_{s-1})\in(0,1)^{s-1}: u-1<u_1+\dots u_{s-1}<u\} $$


问:
Help Simplifying Function with Bessel Functions
I'm trying to calculate a function that includes Bessel functions and am running into precision issues, generally due to a small difference between large numbers.

The function is

$$F=\frac{J_0(K\phi)Y_1(\phi)-J_1(\phi)Y_0(K\phi)}{J_0(\phi)Y_1(\phi)-J_1(\phi)Y_0(\phi)} $$

where $J$ and $Y$ are the first and second Bessel functions. The overall function is on the order of unity, but both numerator and denominator tend to zero for the values I'm interested in.  I've tried various rearrangements (including using a normalized Bessel function) but to no avail. I feel there should be some simplification that can be made to help with precision.

As an example, for k=1.05 and $\phi$= -40 000 + 20i, with double precision I get INF because the denominator is zero, whereas when using quadruple precision I get a reasonable results of -0.5533 + 1.0666i.

Any ideas?

edit:
Thanks Paul Enta for the Wronskian link (denominator  = $-2/(\pi\phi)$).  I think that solves half of the problem.  However, the numerator is still problematic.  If I calculate it as shown above I get 0 (quad precision gives -8.8151e-06 + 1.6972e-05i).  If I rewrite it in terms of ratios:
$$J_0(k\phi)Y_1(\phi)\left[1-\frac{J_1(\phi)}{Y_1(\phi)}\frac{Y_0(k\phi)}{J_0(k\phi)}\right]$$
then I also get poor results as the ratios are close to unity.

答:
Evaluating the numerator for small values of $\phi$ and $k\phi$ should not pose numerical problems. For large values, we can use the asymptotic expansions of $J_\nu$ and $Y_\nu$. After simplification of the trigonometric expressions and  by keeping two terms, one obtains of the following asymptotic expansion
\begin{equation}
 J_0(K\phi)Y_1(\phi)-J_1(\phi)Y_0(K\phi)\simeq -\frac{2}{\pi\phi\sqrt{K}}\left[\cos\left( \phi(K-1) \right)+(3K+1)\frac{\sin\left( \phi(K-1) \right)}{8\phi K}\right]
\end{equation}
Then, for large arguments, taking into account the simplification of the denominator which is a Wronskian,
\begin{equation}
 F\simeq\frac{1}{\sqrt{K}}\left[\cos\left( \phi(K-1) \right)+(3K+1)\frac{\sin\left( \phi(K-1) \right)}{8\phi K}\right]
\end{equation}


问:
Quadratic variation of ito integral
Is it true in general that if $f$ is a deterministic function, and $W$ is brownian motion, then the quadratic variation of $\int_0^t f(W_s) dW_s$ is $\int_0^t f^2(W_s) ds$?

Is it also true in general that the quadratic variation of $\int_0^t f(s)dW_s$ is $\int_0^t f^2(s)ds$?

Also, is the quadratic variation of $\int_0^t f(W_s) ds =0$?

I have been using these formulas in my work as they seem to be generally true, but I haven't been able to prove them (I struggle to work with quadratic variation from the definition) and I would love to see a proof.

My definition of quadratic variation of a continuous local martingale $M$ is the unique, continuous, increasing and adapted process $\langle M\rangle$ with $\langle M\rangle _0
=0 $ such that $M^2 - \langle M \rangle$ is a continuous local martingale.

Thanks very much!

答:
Assume that $E \int_0^t f^2(W_u)du < \infty.$ Define
\begin{align}
I_t &:= \int_0^t f(W_u)dW_u, 0\leq t \leq T;\\
X_t &:= \int_0^t f^2(W_u)du, 0\leq t \leq T.
\end{align}
Then
\begin{align}
E[I_t^2-X_t | \mathscr F_s] &= E\left[\left( \int_0^t f(W_u)dW_u \right)^2-\left. \int_0^t f^2(W_u)du \right| \mathscr F_s \right] \\
&= E\left[I_s^2 + \left( \int_s^t f(W_u)dW_u \right)^2 +2I_s \int_s^t f(W_u)dW_u -X_s - \left. \int_s^tf^2(W_u)du \right | \mathscr F_s \right]\\
&= I_s^2 + E\left[ \left(\int_s^t f(W_u)dW_u \right)^2 \right]+2I_s \underbrace{E\left[ \int_s^t f(W_u)dW_u \right]}_{=0} -X_s \\
&- E\left[ \int_s^t f^2(W_u)du \right] \\
&=I_s^2 - X_s,
\end{align}
where in the last step I have used the Ito isometry. Thus the process $I^2 -X$ is a martingale and therefore a local martingale and by definition of quadratic vatiation (your definition in the question), X is the quadratic variation process of I.

The other question, the process $\int_0^t f(W_u)du, 0 \leq t \leq T$ is of finite variation so its quadratic variation is zero.

Title: Quadratic variation of ito integral


问:
Infinite sums and intersections of subsets of $\mathbb{R}^2$
In this exercise I am asked to find $\bigcup_a \bigcap_b X_{a,b}$ (as well as other variations of sums and intersections of $X_{a,b}$) where $X_{a,b}=\{(x,y)\in \mathbb{R}^2 \vert y \leq ax(x-b)\} a,b \in \mathbb{R}, a,b >0 $. I drew the graph of $y =ax(x-b)$ but it's really hard for me to visualize what $\bigcup_a \bigcap_b X_{a,b}$ is going to look like by sheer looking at the graph. I also plugged the formula into desmos to see how it behaves when $a$ and $b$ are changing, but this didn't help either as the results I got were different from the answers provided by the teacher. Is there any systematic approach to a problem like that? Working on exercises of this type becomes dreadful without any good method as the sets get more complicated - what about this monstrosity for instance:
$X_{a,b}=\{(x,y)\in \mathbb{R}^2 \vert ax^2 < y \leq \sqrt[3]{ab^2} a,b \in \mathbb{R}^2 a,b>0\}$?!

答:
Denote $X_{a,b}^+ = X_{a,b} \cap \{(x,y) \in \mathbb R^2 \mid x \gt 0\}$ and $X_{a,b}^- = X_{a,b} \cap \{(x,y) \in \mathbb R^2 \mid x \le 0\}$. Obviously $X_{a,b}= X_{a,b}^+ \cup X_{a,b}^-$.

For $a,b >0$ $f_{a,b}(x) = ax(x-b)$ represents a parabola whose minimum is equal to $-\frac{a b^2}{4}$ and attained at $x_{a,b}=b/2$.

Now take a fix $a > 0$ and $0 \lt b_1 \le b_2$. Looking at the sign of $f_{a,b_1}(x) - f_{a,b_2}(x)$, you get that $X_{a,b_2}^+ \subseteq X_{a,b_1}^+$ and $X_{a,b_1}^- \subseteq X_{a,b_2}^-$.

From above inclusions, considering $(x,y)$ with $x>0$ and noticing that $\lim\limits_{b \to \infty} f_{a,b}(x) = -\infty$ we obtain $\bigcap_b X_{a,b}^+ = \emptyset$. While for $x \le 0$ we notice that $\lim\limits_{b \to \infty} f_{a,b}(x) = a x^2$ and therefore $\bigcap_b X_{a,b}^- = \{(x,y) \in \mathbb R^2 \mid x \le 0\  \mathrm{and} \ y \le ax^2\}.$

From this we get

$$\bigcup_a\bigcap_b X_{a,b}^+ = \emptyset,\, \bigcup_a\bigcap_b X_{a,b}^- = \{(x,y) \in \mathbb R^2 \mid x \le 0\} \setminus \{(0,y) \mid y \gt 0\}$$ and finally

$$\bigcup_a\bigcap_b X_{a,b} = \{(x,y) \in \mathbb R^2 \mid x \le 0\} \setminus \{(0,y) \mid y \gt 0\}.$$

Now providing you with global hints to find those results is not so easy... I would say:

First try to find for $a$ fixed what $\bigcap_b X_{a,b}$ is.

For that take two distinct values $b_1,b_2$ and try to find inclusions between the $X_{a,b_1},X_{a,b_2}$.

This is what I did when I noticed the inclusions $X_{a,b_2}^+ \subseteq X_{a,b_1}^+$ and $X_{a,b_1}^- \subseteq X_{a,b_2}^-$.

From such inclusions, it is easier to find $\bigcap_b X_{a,b}$ as a set limit.


问:
How do I prove that $2^n=O(n!)$?
How do I prove that $2^n=O(n!)$?

Is this a valid argument?

```<code> 2<=2,
 2<3,
 2<4,....
 2<n if n>2

 therefore 2.2.2....n times < 1.2. ... n
 so,2^n <n!
</code>```

Thanks in advance.

答: The argument is very informal and has a small hole in it, but the basic idea is correct. The hole lies in the fact that $2^n$ is actually larger than $n!$ for $n=1,2,3$. Properly you should show by induction on $n$ that $2^n\le n!$ for $n\ge 4$ and conclude immediately that $2^n$ is $O(n!)$.</p>

Title: How do I prove that $2^n=O(n!)$?


问:
Properties with matrices $A$, $B$, and $A-B$ being positive definite.
Let $A, B, W \in \mathbb{R}^{n \times n}$ be three positive definite matrices, and assume $A - B$ is positive definite as well. Let $\{\lambda_{i}\}_{i=1}^{n}$ and $\{u_{i}\}_{i=1}^{n}$ be the eigenvlaues of matrices $A$ and $B$, respectively, sorted in decreasing order.

I wish to prove the three properties on top of page 109 of the convex optimization textbook:

$tr(WA) \geq tr(WB)$,

$tr(A^{-1}) = \sum_{i=1}^{n}\frac{1}{\lambda_i} \leq \sum_{i=1}^{n}\frac{1}{u_i} = tr(B^{-1})$,

$det(A) = \Pi_{i=1}^{n} \lambda_i \geq \Pi_{i=1}^{n} u_i\geq det(B)$.

I have no idea how to prove 1. For 2 and 3, I feel we need to prove $\lambda_i \geq u_i$ for any $i \leq n$? Any ideas?

答:
Property 1

First note that the trace is invariant under cyclic permutations and so
$$
\mathrm{Tr}[WA] = \mathrm{Tr}[W^{1/2}A W^{1/2}].
$$
Secondly note that if $X > 0$ then $Y^T X  Y \geq 0$ for any square matrix $Y$ acting on $\mathbb{R}^n$. Thus we have $W^{1/2} (A-B) W^{1/2} \geq 0$. Therefore as the trace of a positive semidefinite matrix is positive we get
$$
\begin{aligned}
\mathrm{Tr}[WA] - \mathrm{Tr}[WB] &= \mathrm{Tr}[W(A-B)] \\
&= \mathrm{Tr}[W^{1/2}(A-B)W^{1/2}] \\
&\geq 0
\end{aligned}
$$

Property 2

The equalities in property 2 follow from the fact that the trace is equal to the sum of the eigenvalues and that the eigenvalues of $X^{-1}$ are the inverses of the eigenvalues of $X$ (where $X$ is a positive definite matrix). For the inequality, if $A > B$ then $A^{-1} < B^{-1}$, you can find a proof of this statement here.

Property 3

This has been answered before here. Essentially you can use the identity that $A > B > 0 \iff B^{-1/2}BB^{-1/2} > B^{-1/2}AB^{-1/2} > 0$ together with the multiplicative property of determinants and the fact that the determinant of a positive matrix is positive.


问:
When a complex equation has no solutions?
I have to find $a,b,c$ that $az^2 + bz + c = 0$ where $z,a,b,c \in \mathbb{C}$ has no solution. But I don't know how! Would you please help?

答:
This was proved to be false. A polynomial will always have at least one complex solution.

Otherwise, the only scenario where you have no solutions is when

$$ a=b=0, c\ne 0$$

since you get $c=0$, but $c \ne 0$


答:
The equation has no solution in $z$ if and only if $a=b=0\land c\ne0$.

If $a\ne 0$, then the equation has solutions $z=\frac{-b-\rho}{2a}\lor z=\frac{-b+\rho}{2a}$, where $\rho$ is any complex number such that $\rho^2=b^2-4ac$.

If $a=0\land b\ne 0$, then the equation has solution $z=-\frac cb$.

Of course, if $a=b=c=0$, then any $z\in\Bbb C$ satisfies the equation.

Title: When a complex equation has no solutions?

Body: I have to find $a,b,c$ that $az^2 + bz + c = 0$ where $z,a,b,c \in \mathbb{C}$ has no solution. But I don't know how! Would you please help?


问:
Computing a probability of 5 or more same values when rolling 21 fair dice 12 sides each
The way I'd like to approach the problem is to calculate separate probabilities (P(1), P(2), P(3), P(4)) and then calculate P(>=5) = 1 - (P(1) + P(2) + P(3) + P(4)) where:

P(1) is the probability of having exactly one side of the dice represented and 20 others different. For example: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 (1 is repeated exactly once)

P(2) is the probability of having one side of the dice represented exactly 2 times and 19 others different. For example: 5 1 2 3 4 2 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 (5 is repeated 2 times)

P(3) is the probability of having one side of the dice represented exactly 3 times and 18 others different. For example: 6 6 1 6 2 3 4 5 7 8 9 10 11 12 1 2 3 4 5 7 8 (6 is repeated 3 times)

P(4) is the probability of having one side of the dice represented exactly 4 times and 17 others different. For example: 7 1 2 7 3 4 7 5 6 7 8 9 10 11 12 1 2 3 4 5 6 (7 is repeated 4 times)

My issue is that I having trouble calculating the number of combinations for each example yet calculating the overall number of combinations is trivial (${12^{21}}$).
What I came up with so far is as follows:

P(1) = $\binom{21}{1}⋅12⋅{11^{20}}$

P(2) = $\binom{21}{2}⋅{12^2}⋅{11^{19}}$

P(3) = $\binom{21}{3}⋅{12^3}⋅{11^{18}}$

P(4) = $\binom{21}{4}⋅{12^4}⋅{11^{17}}$

I know duplicates aren't ruled out in these formulas but I am not sure I can come up with calculating the number of these duplicates.
To illustrate the point, suppose we have the following combination: 1 5 6 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9.
We can see that it matches the case of P(1) for 1, 5, 6, 7 and 9 which means that combination could have been repeated at least 4 times, however, the number of repetitions varies depending on the combination.
The question is similar to this one, however, I can't properly eliminate duplicates.
Hints and points would be greatly appreciated.

答:
Using the generating functions the probability in question reads:

$$
1-\frac{21![x^{21}]\left(1+\frac x{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)^{12}}{12^{21}},\tag1
$$
where the operator $[x^{n}]$ extracts the coefficient at $x^n$ in the subsequent expression.

UPDATE:

To understand what is going on in (1) substitute:
$$
\left(1+\frac x{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)^{12}\to\prod_{i=1}^{12}\left(1+\frac {x_i}{1!}+\frac{x_i^2}{2!}+\frac{x_i^3}{3!}+\frac{x_i^4}{4!}\right)
$$
After expanding the last expression you will obtain:
$$
\sum_{0\le k_1,\dots, k_{12}\le4}\prod_{i=1}^{12}\frac{x_i^{k_i}}{k_i!}
=\sum_{K=0}^{48}\frac1{K!}\sum_{0\le k_1,\dots, k_{12}\le4}^{\sum k_i=K}K!\prod_{i=1}^{12}\frac{x_i^{k_i}}{k_i!}.
$$

Now observe that the multinomial coefficient $\frac{K!}{\prod_{i=1}^{12}k_i!}$ is exactly the number of ways to permute $K$ objects with the number of objects of type $i$ being $k_i$. If you now set $x_1=x_2=\cdots=x_{12}=x$ all the coefficients appear to be at $x^K$ and summing them you end up with the coefficient at $x^K$ in
$$K!\left(1+\frac x{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)^{12}.\tag2
$$

If you do not have an appropriate computer program for extracting the coefficient, you can either expand the expresson (2) by hand (doing this you may drop all intermediate terms with power of $x$ greater than 21) or resort to the method proposed in the other answer.


答:
Lots of ways to see this, mathematical and geometrical. Here are some mathsy ways first...

Determinants are linear in every row or column. Since you have $n$ of them which you multiply by $-1$ you get this result.

More explicitly, writing out a typical definition with an alternating tensor $\epsilon$, there are $n$ $A$  matrix elements.

You could verify this by computing $\det (-I)$ as Dennis did.

For diagonalizable matrices, determinants are products of eigenvalues, and you're changing $n$ of them in sign.

Most interesting are geometrical approaches. For instance, as a change of basis for $A$ invertible (the other case being trivial due to vanishing determinants), the matrix determinant has an interpretation as the change in volume form. A negative Jacobian factor corresponds to inversion or handedness changes.

In this way of looking at it, we see the result is equivalent to the observation that in even numbers of dimensions, changing the sign of everything can be achieved just by spinning everything around, whereas in odd numbers of dimensions it cannot be smoothy achieved without pushing a basis vector through zero (i.e. a non invertible matrix) where the $\det$ is zero.

Title: Why is $\det(-A)=(-1)^n\det(A)$?

Body: Why is $\det(-A)=(-1)^n\det(A)$?</p>


问:
Prove that a language is not regular.
I want to ask how to prove the following language is not regular using closure properties. I tried to use pumping lemma but I find the proof itself shaky. I'd appreciate if you can help.

The language is $ L = \{0^{i}1^j0^k, where\ i+j \neq k\}$.

When I applied pumping lemma. I picked $s=1^k0^{k+1}$. Then I pump up, let s=wxy, then $wx^iy$ must be in L. But the length $|x|$ would have to be 1 for this proof to work, which I think is shaky...

I'd appreciate your help!

答:
I think the modified pumping lemma is still the easiest for this case.

Let p be the pumping length

$ S = {0^{p}1^p0^{2p!} } $

Now this string can be written as $ wyz $ , where $ wy^{n}z$ belongs to the regular language L. Now y can only be $ 0^{l}$ or $ 1^{l}$, otherwise pumping it would break the whole structure of the string. Now you just need to realize that such a y when pumped would break the condition for $ i + j \neq k $ for any value of l, since any value of l would be a factor of $2p!$ and can be pumped to ensure that it equals $2p!$.

Title: Prove that a language is not regular.


答:
As another contributor has commented, you can use the binomial theorem by writing
$$\sqrt x = (1+(x-1))^{1/2}.$$  You might also note that for $1 \leqslant x \leqslant 2$ the infinite series converges and the terms decrease with alternating sign so the error is always less or equal to the the first neglected term.  Your attempt using derivatives is also perfectly legitimate and should give the same result (but note you have mis-typed the powers of $x$ in your calculation of the derivatives).

So you have obtained
$$
\sqrt{x} = 1 + \frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3+\cdots.
$$
whence truncating after the $(x-1)^2$ term, $$\lvert R \rvert \leqslant \frac{1}{16}(x-1)^3 \leqslant  \frac{1}{16} \quad (x \in [1,2]).$$
If you are interested in the error over the full range ($0 < x \leqslant 2$) then you will need to use one of the forms of the Taylor series error, for example the Lagrange estimate, which should give the same conclusion for $x \geqslant 1$.

Title: Taylor polynomial for $\sqrt x$

Body:
I need to find the kinds of Taylor polynomial for $\sqrt x$ at the point $a = 1$, $x$ lies in $<1,2>$.

I calculated the second derivative: $-\frac{1}{4}x^\frac{-3}{2}$
and the third: $\frac{3}{8}x^\frac{-5}{2}$

I calculated Taylor's polynomial:
$$ {1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2}$$

I need advice on estimating the error.
I calculated something like:

$$R \leq \frac{3/(8\cdot2^{5/2})}{3!}\cdot{(x-1)^3}$$

But I'm not sure if that's right.


问:
A question of Roots of unity

By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
 \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.

I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.

答:
The roots of unity for $z^9=1$ are $e^{i \frac{2\pi k}9},\>k=0, \pm1,\pm2,\pm3,\pm4$ and their sum is equal to zero, i.e.
$$0=1+ e^{i\frac{2\pi k}9} + e^{-i\frac{2\pi k}9}
 + e^{i\frac{4\pi k}9} + e^{-i\frac{4\pi k}9}
 + e^{i\frac{6\pi k}9} + e^{-i\frac{6\pi k}9}
 + e^{i\frac{8\pi k}9} + e^{-i\frac{8\pi k}9}
$$
which leads to
$$\cos\frac{2\pi}{9}+
 \cos\frac{4\pi}{9}+ \cos\frac{6\pi}{9} + \cos\frac{8\pi}{9} =- \frac{1}{2}$$

Title: A question of Roots of unity


问:
Prove with induction that $b_{n+1} \geq ({3 \over 2})^{n-1}$
I need to prove with induction that $b_{n+1} \geq ({3 \over 2})^{n-1} $ $∀ n ∈ \mathbb N$

Sequence is given as $b_{1}=b_{2}=1$ and for every $n \gt 2$ $,b_{n}=b_{n-1} + b_{n-2}$ so basically fibonacci sequence

I tried to prove it with strong mathematical induction

Base:$\quad n=1: 1\geq ({3 \over 2})^{1-1}$ applies

$\quad\quad\quad  n=2: 2\geq ({3 \over 2})^{2-1}$ applies

Step: Let $b_{k+1} \geq ({3 \over 2})^{k-1} $ be correct for every k ∈ $ {1,...,m}$

We have to show that $b_{m+2} \geq ({3 \over 2})^{k} $ is also correct.

So we get that $b_{m-1} + b_{m} \geq ({3 \over 2})^{m} $

Solving this gives me $({3 \over 2})^{m} \geq 0$

So for every natural number this inequality holds

Is this correct?

Any help is appreciated!

答:
As already stated, the base case for the induction must be done for $n = 2$ which is easily verified to be true.

Then from $b_{n+1} \geq (\frac{3}{2})^{n-1}$ we get $b_{n}+b_{n+1}=b_{n+2} \,\color{#c00}\geq \frac {3}{2}b_{n+1} \geq (\frac{3}{2})^{n}$ where $\,\color{#c00}\geq$
proves the required induction step.

$b_{n}+b_{n+1} \,\color{#c00} \geq \frac {3}{2}b_{n+1} \Longleftrightarrow b_{n} \geq \frac{1}{2}b_{n+1} \Longleftrightarrow2b_{n} \geq b_{n+1} \Longleftrightarrow b_{n}+b_{n} \geq b_{n-1}+b_{n} \Longleftrightarrow b_{n}\geq b_{n-1} \Longleftrightarrow b_{n-2}+b_{n-1} \geq b_{n-1} \Longleftrightarrow b_{n-2} \color{green}\geq 0 $

for $n> 2$.

To be clear, although $\color{green} \geq $ is trivial, it still must be shown. This follows immediately using strong induction.


问:
Two persons A and B toss a die. The person who first throws $5$ wins. What is the probability of A’s winning?
It is to prove that if A begins, then

$$P(A_{wins})= \frac{6}{11}$$
Similarly, if B begins,

$$P(A_{wins})= \frac{5}{11}$$
Now, for the given problem, P= P(A wins)= P(A begins and wins or B begins and A wins)
But (A begins and wins) and (B begins and A wins) are mutually exclusive events.

So P= P(A begins and wins) + P(B begins and A wins)$= \frac{6}{11}+ \frac{5}{11}= 1.$

This means that A will surely win and B will surely lose, which is clearly incorrect. So what's the mistake in all of this?

答:
A begins and probability of A wins, it will be an infinity geometric series such that:

$$\frac{1}{6}+(\frac{5}{6})^2\frac{1}{6}+(\frac{5}{6})^4\frac{1}{6}+...=\frac{1}{6}(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)=\frac{1}{6}(\frac{1}{1-(\frac{5}{6})^2})=\frac{6}{11}$$

B begins and probability of A wins, it will be an infinity geometric series such that:

$$(\frac{5}{6})(\frac{1}{6})+(\frac{5}{6})^3\frac{1}{6}+(\frac{5}{6})^5\frac{1}{6}+...=\frac{5}{36}(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)=\frac{5}{36}(\frac{1}{1-(\frac{5}{6})^2})=\frac{5}{11}$$

$\frac{6}{11}+\frac{5}{11}=1$ does not mean A always wins.

We have an event and two possibilities

$$P(A_{\text{first and wins}})=1- P(B_{\text{second and loses}})=\frac{6}{11}$$

$$P(A_{\text{second and wins}})=1- P(B_{\text{first and loses}})=\frac{5}{11}$$

One of them always wins.


问:
Is $f(W)$ closed if $W$ is closed?
Suppose a function $f : R^n \mapsto R^n$ is $C^1$ and the $Jacobian(f) \not= 0$ $\forall x $ in $R^n$

If $W$ is a closed set of $R^n$, is $f(W)$ closed?

I am not sure how to use the Jacobian to comment about the closure?

Do I use the inverse function theorem to talk about its invertibility? But that would only talk about local invertibility.

Can I think in the direction of continuity? But closed set does not necessarily map to a closed set in case of a continuous function.

How do I relate Jacobian to the notion of closed sets?

Please suggest a detailed explanation. Especially on the Jacobian part and  similarly, what other insights I can gain from the Jacobian of a function if its non zero at all points in a domain.

答: No. Consider $f(x)=\arctan(x)$ and $W=\mathbb{R}$.

Title: Is $f(W)$ closed if $W$ is closed?


问:
Am I linearizing this non-linear system properly?
I have the following system:

$$x'=x(1-x)-xy$$
$$y'=y(1-\frac{y}{x})$$

I am trying to linearize it, and from my understanding, if the equilibrium points are hyperbolic, that is, its eigenvalues all have a non-zero real part, I can drop the higher order terms. If I compute $DF_{X_0}$, I get:

$$\begin{pmatrix} 1-2x-y & -x \\ \frac{y^2}{x^2} & 1-\frac{2y}{x} \end{pmatrix}$$

When I drop the higher order terms, I am left with:

$$\begin{pmatrix} 1-2x-y & -x \\ 0 & 1 \end{pmatrix}$$

Am I linearizing this correctly?

答: No, you have misunderstood this, you should not drop higher order terms in the Jacobian like you are doing! What you should to is to substitute the equilibrium that you want to investigate into the Jacobian. That will give you a constant matrix containing the coefficients of the first-order terms in the Taylor expansion around that point, so the “dropping of terms” is automatically taken care of.


答: $f(x)=exp(x)$ defined on $\mathbb{R}$, its image is $(0,+\infty)$, $\mathbb{R}$ is closed, but not its image.

Title: Is $f(W)$ closed if $W$ is closed?

Body:
Suppose a function $f : R^n \mapsto R^n$ is $C^1$ and the $Jacobian(f) \not= 0$ $\forall x $ in $R^n$

If $W$ is a closed set of $R^n$, is $f(W)$ closed?

I am not sure how to use the Jacobian to comment about the closure?

Do I use the inverse function theorem to talk about its invertibility? But that would only talk about local invertibility.

Can I think in the direction of continuity? But closed set does not necessarily map to a closed set in case of a continuous function.

How do I relate Jacobian to the notion of closed sets?

Please suggest a detailed explanation. Especially on the Jacobian part and  similarly, what other insights I can gain from the Jacobian of a function if its non zero at all points in a domain.


答: A simple counterexample in case $n=1$ and $W=\mathbb R$ is $\arctan$.

Title: Is $f(W)$ closed if $W$ is closed?


问:
David Marker's exercise 5.5.2
I want to prove this exercise from David Marker's model theory book. Let $L = {E}$, where $E$ is a binary relation symbol, and let $T$ be the theory of an equivalence relation with infinitely many classes each of which is infinite. Show that in any $M |= T$ we can find infinite sets of indiscernibles $I_0$ and $I_1$ such that $tp(I_0)\neq tp(I_1)$, but if $J$ is any other infinite set of indiscernibles, then $tp(J) = tp(I_i)$ for $i = 0$ or $1$.

I don't really understand what the difference between $I_0, I_1$ and $J$ is. I mean they're all infinite set of indiscernibles. Also, any hint for the solution is appreciated.

答:
Let $M\models T$. Take $I_0$ to consist of $\aleph_0$-many elements from a single equivalence class, and take $I_1$ to consist of $\aleph_0$-many elements from pairwise distinct equivalence classes.

Certainly $\operatorname{tp}(I_0)\neq\operatorname{tp}(I_1)$, as the former contains the formula $E(v_i,v_j)$ for any $i\neq j\in\omega$ and the latter contains the formula $\neg E(v_i,v_j)$ for any $i\neq j\in\omega$. Furthermore, the theory of an equivalence relation with infinitely many infinite equivalence classes has quantifier elimination, so showing that $I_0$ and $I_1$ are indiscernible sets amounts to considering atomic formulas, all of which are the form $E(v,w)$ (or $v=w$), whence by the first sentence of this paragraph $I_0$ and $I_1$ are indeed indiscernible.

Finally, let $J$ be any other infinite set of indiscernibles. To be pedantic, for $J$ to have the same type as $I_0$ or $I_1$ it needs to have the same number of variables as $I_0$ and $I_1$, so we assume this is the case. If any two elements of $J$ come from pairwise distinct equivalence classes, then all elements of $J$ do, in which case $\operatorname{tp}(J)=\operatorname{tp}(I_1)$. Similarly, if any two elements of $J$ come from the same equivalence class, then all elements of $J$ do, in which case $\operatorname{tp}(J)=\operatorname{tp}(I_0)$.

Title: David Marker's exercise 5.5.2


问:
Prove sequence doesn't converge to zero
I need to prove that for every "c" parameter, the following sequence doesn't converge to zero:
$$a_1=c$$
$$a_{n+1}=\begin{cases}
  \frac{a_n}{3} - 1 & a_n\ge0\,\\
  -a_n & a_n<0\,\\
  \end{cases}$$

I really can't think of any solution, and when asking around someone suggested I use the integral test, but we didn't learn it as of now.

Any help would be greatly appreciated!

答:
Suppose it converges towards $0$, there exists $N$ such that $n>N$ implies that $|a_n/3|<{1\over 6}$. Let $n>N$, if $a_n\geq 0$, it implies that $a_{n+1}={a_n}/3-1$, this implies that $|a_{n+1}|\geq 1-{1\over 6}>{1\over 3}$ contradiction.

If $a_n<0,=a_{n+1}=-a_n>0$ and the previous step shows that $|a_{n+2}|>{1\over 3}$. Contradiction.

Title: Prove sequence doesn't converge to zero


答:
Since $e^2=\sum_{n=0}^{+\infty}\frac{2^n}{n!}$ converges (by ratio test if you want), the general term tends to $0$, whence, actually, $
2^n=o(n!)$ and not only $O(n!)$.</p>

Title: How do I prove that $2^n=O(n!)$?

Body:
How do I prove that $2^n=O(n!)$?

Is this a valid argument?

```<code> 2<=2,
 2<3,
 2<4,....
 2<n if n>2

 therefore 2.2.2....n times < 1.2. ... n
 so,2^n <n!
</code>```

Thanks in advance.


问:
Given the growth rate of one component, how much is the total growth rate?
$x = a + b + c + d$

If I increase $a$ by a $5\%$, Can I know by how much $x$ will increase keeping all constant?

答:
It cannot be known. You have that $x=a+b+c+d$ and wonder about how much bigger $x^\prime=1.05 a + b+c+d$ is. The change in $x$ is $\Delta x=x^\prime-x=0.05a$, but the percentage increase in $x$ is

$$ \frac{\Delta x}{x}=\frac{0.05a}{a+b+c+d},$$

which depends on $a+b+c+d$.


问:
Making single product into double product
I have this product: $\prod_{i > j} ( i^2 - j^2 ) $ and I am trying to turn it into either a double product with one for $j$ and the other for $i$ or keep as a single product and rearrange $j$ in terms of $i$. (Note the $i,j$ are for matrix indexes).

I first thought $\prod_{i = 1} ( i^2 - (i-1)^2 ) $ however testing this a few times shows that it does not work... can anyone help me? (The products are up to $n$)

Note I am trying to remove $i>j$ from my product by splitting into two or some other way

答:
First way
$$
\eqalign{
  & P(n) = \prod\limits_{0\, \le \,j\, < \,i\, \le \,n}
 {\left( {i^{\,2}  - j^{\,2} } \right)}  = \prod\limits_{0\, \le \,j\, \le \,n - 1} {\prod\limits_{j + 1\, \le \,i\, \le \,n} {\left( {i^{\,2}  - j^{\,2} } \right)} }  =   \cr
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\prod\limits_{j + 1\, \le \,i\, \le \,n} {\left( {i + j} \right)\left( {i - j} \right)} }  =   \cr
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\prod\limits_{1\, \le \,i - j\, \le \,n - j}
 {\left( {i - j + 2j} \right)\left( {i - j} \right)} }  =   \cr
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\prod\limits_{1\, \le \,k\, \le \,n - j} {\left( {k + 2j} \right)k} }  =   \cr
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\left( {n - j} \right)!\left( {1 + 2j} \right)^{\,\overline {\,n - j\,} } }  =   \cr
  &  = \quad  \cdots  \cr}
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial

Second way
$$
\eqalign{
  & P(n) = \prod\limits_{0\, \le \,j\, < \,i\, \le \,n} {\left( {i^{\,2}  - j^{\,2} } \right)}  =   \cr
  &  = \prod\limits_{1\, \le \,i\, \le \,n}
 {\prod\limits_{0 \le j\, \le \,i - 1} {\left( {i + j} \right)\left( {i - j} \right)} }  =   \cr
  &  = \prod\limits_{1\, \le \,i\, \le \,n}
 {i^{\,\overline {\,i\,} } i^{\,\underline {\,i\,} } }
  = \prod\limits_{1\, \le \,i\, \le \,n} {i!{{\Gamma (2i)} \over {\Gamma (i)}}}  =   \cr
  &  = \prod\limits_{1\, \le \,i\, \le \,n}
 {\Gamma (i + 1){{\Gamma (2i)} \over {\Gamma (i)}}}
  = \prod\limits_{1\, \le \,i\, \le \,n} {i\,\Gamma (2i)}  \cr}
$$

And introducing the Duplication formula for the Gamma function
$$
\Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi  }}\Gamma \left( z \right)
\Gamma \left( {z + 1/2} \right)
$$
we get
$$
\eqalign{
  & P(n) = \prod\limits_{1\, \le \,i\, \le \,n} {{{4^{\,i} } \over {2\sqrt \pi  }}
\Gamma (i + 1)\Gamma \left( {i + 1/2} \right)}  =   \cr
  &  = \left( {{{2^{\,n} } \over {\sqrt \pi  }}} \right)^{\,n}
 \prod\limits_{1\, \le \,i\, \le \,n} {\Gamma (i + 1)\Gamma \left( {i + 1/2} \right)}  \cr}
$$


答:
$$\begin{align*}&(1)\;\;\text{Look at the positive series}\;\sum_{n=1}^\infty\frac{2^n}{n!}\\{}\\
&(2)\;\;\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}=\frac2{n+1}\xrightarrow[n\to\infty]{}0 \;,\;\text{thus}\\{}\\
&(3)\;\;\text{The series in (1) converges}\\{}\\
&(4)\;\;a_n=\frac{2^n}{n!}\xrightarrow[n\to\infty]{}0\end{align*}$$

Thus, for some

$$N\in\Bbb N\;\;\text{and}\;\;\forall n>N\;,\;\;\frac{2^n}{n!}<1\implies 2^n=\mathcal O(n!)$$

Title: How do I prove that $2^n=O(n!)$?


答:
Assuming that "keeping all constant" means $b$, $c$ and $d$ don't change, the  new value of $x$ is
$$
x' = 1.05a + b + c + d = 0.05a + x
$$
so
$$
\frac{x'}{x} = \frac{0.05a}{x} + 1
$$
so the relative increase in $x$ is $\frac{0.05a}{x}$,
which you can write as a percent.

Intuitively, the change in $x$ caused by a change in $a$ is reduced by the fraction   $a$  contributes to $x$.

Title: Given the growth rate of one component, how much is the total growth rate?

Body:
$x = a + b + c + d$

If I increase $a$ by a $5\%$, Can I know by how much $x$ will increase keeping all constant?


答:
P= P(A begins and wins) + P(B begins and A wins)

This is not true. Either A throws it first or B throws it first. You have to assume one of them happens. If not, meaning you don't know who throws first, then you need to assign prior probabilities. Let's say $\pi (\text{A begins first}) = q, \pi(\text{B begins first})=1-q$, then

$$\Pr(\text{A wins}) = \frac{6}{11} q+\frac{5}{11}(1-q)$$

If it's completely random, $q=\frac 12$ then the above is also equal to $\frac 12.$

Title: Two persons A and B toss a die. The person who first throws $5$ wins. What is the probability of A’s winning?

Body:
It is to prove that if A begins, then

$$P(A_{wins})= \frac{6}{11}$$
Similarly, if B begins,

$$P(A_{wins})= \frac{5}{11}$$
Now, for the given problem, P= P(A wins)= P(A begins and wins or B begins and A wins)
But (A begins and wins) and (B begins and A wins) are mutually exclusive events.

So P= P(A begins and wins) + P(B begins and A wins)$= \frac{6}{11}+ \frac{5}{11}= 1.$

This means that A will surely win and B will surely lose, which is clearly incorrect. So what's the mistake in all of this?


问:
How to compute for the coordinates of polynomial function?
This is a multi-part question and im now in the part where coordinates are involved but i'm not sure how to proceed with this.

I was already able to answer up to 2a but not sure how to proceed with 2b. Could anybody point me in the right direction?

答:
First Calculate $L(5t^2-t+2) = 5t+5 = p1$.

You have $B = ${$p2=t-2, p3=3t$}

Now you are looking for $α$ and $β$ that will hold:

$$p1=α*p2+β*p3$$
And then $[p1]_B = (α,β)$

You can see that:
$$p1=-2\frac12*p2+2\frac12*p3$$
And these are the coordinates.


问:
Convergence of a series with sin term
I need to show that the series

$\sum{\frac{1}{n}\sin(\frac{n\pi}{2}}) $

is convergent.

I know that the sine term is $0$ for even $n,$ $1$ when $n$ is $1,5,9, \ldots$ and $-1$ when $n$ is $3,7,11, \ldots$ so it's probably a good idea to use the Alternating Series Test, but I'm struggling to find away to translate this into the problem

答: What you did shows that your series is equal to the series $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}$. And now, yes, you can apply the alternating series test.

Title: Convergence of a series with sin term


问:
What is $\lim_{n \rightarrow \infty}(1-c^n)^\frac{n}{2}$?
What is $\lim_{n \rightarrow \infty}(1-c^n)^\frac{n}{2}$? Where $c \in [0, 0.5)$, and $n$ is an integer. How do I calculate it?

Thank you!!!

答:
Hint. Let $|c|<1$. One may write
$$
(1-c^n)^\frac{n}{2}=e^{\frac{n}{2}\ln\left(1-c^n \right)}
$$ noticing that, as $n \to \infty$, $c^n \to 0$ and
$$
\frac{\ln\left(1-c^n \right)}{-c^n}\to 1.
$$


问:
How do I prove that $2^n=O(n!)$?
How do I prove that $2^n=O(n!)$?

Is this a valid argument?

```<code> 2<=2,
 2<3,
 2<4,....
 2<n if n>2

 therefore 2.2.2....n times < 1.2. ... n
 so,2^n <n!
</code>```

Thanks in advance.

Title: How do I prove that $2^n=O(n!)$?

Body:
You can use induction for $n > 3$.

Assume $ p(k):  2 ^ k < k!,  k > 3 $

Prove that $p(k +1 ): 2 ^ {k + 1} < (k + 1)!$

$ 2 ^ {k + 1} = 2 ^ k * 2$

$ (k + 1)! = k! * k$

$ k > 2 \implies 2 ^ {k + 1} < (k + 1)! $


问:
Definition of normal linear functional

I am confused about the definition of normal linear functional.

In mamy reference books, when we define normal linear functionals , we require that the linear functional is positive. But in Kadison's book, the normal linear functional may not be positive.(See the theorem in the screenshot).

I don't understand why we need the positivity of linear functional when we defining the normality in many cases.

答:
Kadison-Ringrose are very clear on this, in the section you are quoting. In previous sections, "normal" is only applied to states.

But in section 7.4, they state:

7.4.1. DEFINITION. If $\mathscr R$ is a von Neumann algebra acting on the Hilbert space $\mathscr H$, we denote by $\mathscr R_\#$ the linear space of linear functionals on $\mathscr R$ that are weak-operator continuous on the unit ball of $\mathscr R$ (the "normal" linear functionals on $\mathscr R$). We refer to $\mathscr R_\#$ are predual of $\mathscr R$.

And in the page right before the one you quoted:

Applied to ultraweakly continuous states, Proposition 7.4.5 constitutes an additional condition to Theorem 7.1.12, The terms "normal" and "ultraweakly continuous" are now equally applicable to functionals on $\mathscr R$.

Title: Definition of normal linear functional


问:
If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad)
I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:

If
  $$x \cos\theta+y\sin\theta=a$$
  $$x\sin\theta-y\cos\theta=b$$
  then $$\tan\theta=\frac{bx+ay}{ax-by}$$

I've spent many hours trying.
Thanks in advance.

答:
$$x\cos \theta+y\sin\theta=a$$

$$\implies x+y\tan\theta=a\sec\theta$$

$$\implies \sec \theta=\frac{x+y\tan\theta}{a} ...(1)$$

$$x\sin\theta-y\cos\theta=b$$

$$\implies x\tan\theta-y=b\sec\theta$$

$$\implies \sec\theta=\frac{x\tan\theta-y}{b}...(2)$$
From $1$ and $2$,

$$\frac{x+y\tan\theta}{a}=\frac{x\tan\theta-y}{b}$$

$$\implies bx+by\tan\theta=ax\tan\theta-ay$$

$$\implies \tan\theta(ax-by)=bx+ay$$

$$\tan\theta=\frac{bx+ay}{ax-by}$$


问:
Prove that $\sqrt[m]{k}$ is irrational whenever k is not an mth power of some integer.
Let the coefficients of the polynomial $a_0 + a_1x + a_2x^2 + ... + a_{m-1}x^{m-1} + x^m$ be integers. Then any real root of the polynomial is either integral or irrational.

(a) Explain why the Lemma immediately implies that $\sqrt[m]{k}$ is irrational whenever
k is not an mth power of some integer.

(b) Carefully prove the Lemma.

This is problem 1.19 of Mathematics for Computer Science (MIT 6.042j). I am self studying the subject and this question tripped me up.

How is the implication in part (a) related to the polynomial?

I tried messing with sum / product of roots of a polynomial but that seems like a dead end. Also, I already proved previously that square root of a prime number is irrational.

答:
The lemma is the rational root theorem for monic polynomials. I'll sketch the solution:

(a) Consider $x^m-k$.

(b) If $x=p/q$ for coprime $p\in\Bbb Z,\,q\in\Bbb N$ then $\sum_{i=0}^ma_ip^iq^{n-i}=0$ with $a_m:=1$. All but one term is obviously a multiple of $p$, so $p|a_0$; all but one term is obviously a multiple of $q$, so $q|a_m$, i.e. $q=1$.


答:
Let $\epsilon > 0$.  Then for large $n$ we have
$$
0 < c^n < \frac{\epsilon}{n}
\\
1 > (1-c^n) > 1-\frac{\epsilon}{n}
\\
1 > (1-c^n)^{n/2} > \left(1-\frac{\epsilon}{n}\right)^{n/2}
$$
Now $\left(1-\frac{\epsilon}{n}\right)^{n/2} \to e^{-\epsilon}$
so
$$
1 \ge \limsup_n(1-c^n)^{n/2} \ge \liminf_n(1-c^n)^{n/2}
\ge e^{-\epsilon}
$$
Next, $\lim_{\epsilon \to 0} e^{-\epsilon} = 1$, so finally
$$
1 \ge \limsup_n(1-c^n)^{n/2} \ge \liminf_n(1-c^n)^{n/2} \ge 1
$$
and therefore $\lim_n(1-c^n)^{n/2}$ exists and equals $1$.

Body:
What is $\lim_{n \rightarrow \infty}(1-c^n)^\frac{n}{2}$? Where $c \in [0, 0.5)$, and $n$ is an integer. How do I calculate it?

Thank you!!!


问:
Inequality in a bounded real sequence
Prove or disprove that for any bounded real sequence $\{x_n\}_{n\in\mathbb{N}}$ there exist two distinct natural numbers $u,v$ such that:
$$|x_u-x_v|\cdot|u-v|\leq 1.$$</p>

答:
I don't have any solid proof, but I do have some evidence to suggest that the proposition might be false: a proof that if there are always such $u,v$ pairs, then they occur frequently in any sequence, and second, a computed example of a sequence that has no such $u,v$ pairs, yet appears to be bounded.

Take a bound $[a,b]$.
Consider $S_n$, the set of finite sequences such that $|x_u-x_v|\cdot|u-v|<1$ for some $u,v$. Then, $|x_u-x_v|=\frac{1}{|u-v|}-\epsilon$ for some $\epsilon>0$, and so the set $$
[a,b]\times [a,b]\times\cdots\times(x_u-\frac{\epsilon}{2},x_u+\frac{\epsilon}{2})\times[a,b]\times\cdots\times[a,b]\times(x_v-\frac{\epsilon}{2},x_v+\frac{\epsilon}{2})\times[a,b]\times\cdots\times[a,b]
$$ is open in $[a,b]^n$ and is a subset of $S_n$, meaning that $S_n$ is open in $[a,b]^n$. It follows that, since $[0,b]$ is compact, by Tychonoff's theorem, $[a,b]^n$ is compact, and therefore, $\overline{S_n}$ is closed and thus compact. Let $C_n$ be the extension of $\overline{S_n}$ to $[a,b]^\omega$ by appending copies of $[a,b]$, so that by Tychonoff's Theorem, $C_n$ is compact. However, $\{C_n\}$ is a sequence of decreasing compact sets, whose intersection $\bigcap_{n\in N} C_n$ is nonempty if each $C_n$ is. An element of $\bigcap_{n\in N} C_n$ would be a sequence such that $|x_u-x_v|\cdot|u-v|\ge1$ for all $u,v$.

Thus, for a sequence bounded by $[a,b]$, if we can always find some finite sequence of length $n$ such that $|x_u-x_v|\cdot|u-v|\ge1$ for all $u,v$, we can find an infinite sequence violating the proposition. This means that, if the proposition is true, pairs $u,v$ will appear "commonly", in the sense that there is some $n$ at which all finite sequences of length $n$ must have a $u,v$ pair with $|x_u-x_v|\cdot|u-v|<1$, and so $u,v$ pairs will take up at least $\frac{2}{n}$ of the sequence.

A simple strategy of setting $x_0=0$ and $x_n$ to the minimum positive number such that $\{x_0,\cdots,x_n\}$ has $|x_u-x_v|\cdot|u-v|\ge1$ seems to work well, as demonstrated by the computer program below. The output of the program will have the bound on $\{x_n\}$ taper off to just above $2.5$ all the way into a sequence of length $250,000$, which seems to indicate that this naive sequence is bounded.

```<code>#include <stdbool.h>
#include <stdio.h>
#include <float.h>

#define EPS 0.0000000000000000002
#define NUM 250000
long double data[NUM];

static bool within(long double x, long double y, int n){
        long double test = (x-y)*n;
        return test >= (long double)(-1.0) && test <= (long double)(1.0);
}

main(){
        long double max = 0.0;
        int i;
        data[0] = 0.0;
        for(i=1; i<NUM; i++){
                long double test = ((long double)1.0)/i;
                int j;
                for(j=i-1; j>0; j--){
                        if(within(test,data[j],i-j)){
                                test = data[j]+((long double)1.0)/(i-j)+EPS;
                                j=i;
                        }
                }
                data[i] = test;
                if(test > max){
                        max = test;
                }
                if(i%100==0) printf("Got %d/%d -- max=%.15Lf\n",i,NUM,max);
        }
}
</code>```

Title: Inequality in a bounded real sequence

Body:
I don't have any solid proof, but I do have some evidence to suggest that the proposition might be false: a proof that if there are always such $u,v$ pairs, then they occur frequently in any sequence, and second, a computed example of a sequence that has no such $u,v$ pairs, yet appears to be bounded.

Take a bound $[a,b]$.
Consider $S_n$, the set of finite sequences such that $|x_u-x_v|\cdot|u-v|<1$ for some $u,v$. Then, $|x_u-x_v|=\frac{1}{|u-v|}-\epsilon$ for some $\epsilon>0$, and so the set $$
[a,b]\times [a,b]\times\cdots\times(x_u-\frac{\epsilon}{2},x_u+\frac{\epsilon}{2})\times[a,b]\times\cdots\times[a,b]\times(x_v-\frac{\epsilon}{2},x_v+\frac{\epsilon}{2})\times[a,b]\times\cdots\times[a,b]
$$ is open in $[a,b]^n$ and is a subset of $S_n$, meaning that $S_n$ is open in $[a,b]^n$. It follows that, since $[0,b]$ is compact, by Tychonoff's theorem, $[a,b]^n$ is compact, and therefore, $\overline{S_n}$ is closed and thus compact. Let $C_n$ be the extension of $\overline{S_n}$ to $[a,b]^\omega$ by appending copies of $[a,b]$, so that by Tychonoff's Theorem, $C_n$ is compact. However, $\{C_n\}$ is a sequence of decreasing compact sets, whose intersection $\bigcap_{n\in N} C_n$ is nonempty if each $C_n$ is. An element of $\bigcap_{n\in N} C_n$ would be a sequence such that $|x_u-x_v|\cdot|u-v|\ge1$ for all $u,v$.

Thus, for a sequence bounded by $[a,b]$, if we can always find some finite sequence of length $n$ such that $|x_u-x_v|\cdot|u-v|\ge1$ for all $u,v$, we can find an infinite sequence violating the proposition. This means that, if the proposition is true, pairs $u,v$ will appear "commonly", in the sense that there is some $n$ at which all finite sequences of length $n$ must have a $u,v$ pair with $|x_u-x_v|\cdot|u-v|<1$, and so $u,v$ pairs will take up at least $\frac{2}{n}$ of the sequence.

A simple strategy of setting $x_0=0$ and $x_n$ to the minimum positive number such that $\{x_0,\cdots,x_n\}$ has $|x_u-x_v|\cdot|u-v|\ge1$ seems to work well, as demonstrated by the computer program below. The output of the program will have the bound on $\{x_n\}$ taper off to just above $2.5$ all the way into a sequence of length $250,000$, which seems to indicate that this naive sequence is bounded.

#include <stdbool.h>
#include <stdio.h>
#include <float.h>

#define EPS 0.0000000000000000002
#define NUM 250000
long double data[NUM];

static bool within(long double x, long double y, int n){
        long double test = (x-y)*n;
        return test >= (long double)(-1.0) && test <= (long double)(1.0);
}

main(){
        long double max = 0.0;
        int i;
        data[0] = 0.0;
        for(i=1; i<NUM; i++){
                long double test = ((long double)1.0)/i;
                int j;
                for(j=i-1; j>0; j--){
                        if(within(test,data[j],i-j)){
                                test = data[j]+((long double)1.0)/(i-j)+EPS;
                                j=i;
                        }
                }
                data[i] = test;
                if(test > max){
                        max = test;
                }
                if(i%100==0) printf("Got %d/%d -- max=%.15Lf\n",i,NUM,max);
        }
}


问:
How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$?
I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.

I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.

Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$

Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$

Body:
This answer is Community wiki and takes a 'deep dive' into algorithmic solution design.

The 'elemental' exponent reducing identity in ${\displaystyle (\mathbb {Z} /31\mathbb {Z} )^{\times}}$ is

$\quad \large 2^1\cdot3^1\cdot5^1 \equiv 2^0\cdot3^0\cdot5^0 \cdot (-1) \pmod{31}$

and this can be expanded into a 'closed exponent reducing presentation system',

$\quad \large 2^4 \equiv 3^1\cdot5^1\cdot (-1) \pmod{31}$

$\quad \large 3^3 \equiv 2^2 \cdot (-1) \pmod{31}$

$\quad \large 5^2 \equiv 2^1\cdot3^1\cdot (-1) \pmod{31}$

Part 1: Calculate $50^{44} \pmod{31}$

Since

$\quad 50 \equiv 19 \pmod{31}$

$\quad 19 \equiv 2^2 \cdot 3 \cdot (-1) \pmod{31}$

our 'not too smart' algorithm can be applied:

\begin{align}
50^{44}  &\equiv 19^{44}  \pmod{31} \\
&\equiv 2^{88} \cdot 3^{44} \cdot 5^0 \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{66} \cdot 5^{22} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{44} \cdot 3^{0} \cdot 5^{22} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{11} \cdot 5^{33} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{11} \cdot 5^{33} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{16} \cdot 3^{27} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{34} \cdot 3^{0} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{2} \cdot 3^{8} \cdot 5^{9} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{6} \cdot 3^{12} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{14} \cdot 3^{0} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{2} \cdot 3^{3} \cdot 5^{4} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{4} \cdot 3^{5} \cdot 5^{0} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{6} \cdot 3^{2} \cdot 5^{0} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{2} \cdot 3^{3} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{4} \cdot 3^{0} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{1} \cdot 5^{2} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{1} \cdot 3^{2} \cdot 5^{0} \cdot (-1)^{0} \pmod{31} \\
&\equiv 18 \pmod{31}
\end{align}

Part 2: Calculate $44^{50} \pmod{31}$

Since

$\quad 44 \equiv 13 \pmod{31}$

and

$\quad 13 \equiv 2  \cdot 3^2 \cdot (-1)  \pmod{31}$

our 'not too smart' algorithm can be applied:

\begin{align}
44^{50}  &\equiv 13^{50}  \pmod{31} \\
&\equiv 2^{50} \cdot 3^{100} \cdot 5^0 \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{116} \cdot 3^{1} \cdot 5^{0} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{30} \cdot 5^{29} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{20} \cdot 3^{0} \cdot 5^{29} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{34} \cdot 3^{14} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{2} \cdot 3^{22} \cdot 5^{9} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{16} \cdot 3^{1} \cdot 5^{9} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{5} \cdot 5^{13} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{6} \cdot 3^{11} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{12} \cdot 3^{2} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{5} \cdot 5^{4} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{2} \cdot 3^{2} \cdot 5^{4} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{4} \cdot 3^{4} \cdot 5^{0} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{0} \cdot 3^{5} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\
&\equiv 2^{2} \cdot 3^{2} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\
&\equiv 2^{1} \cdot 3^{1} \cdot 5^{0} \cdot (-1)^{1} \pmod{31} \\
&\equiv 25 \pmod{31}
\end{align}


问:
Convergence of sum given convergence of sequence

Let $\{a_n\}_{n\in\mathbb{N}}$ the sequence defined by $a_1 = \tfrac{1}{2}$ and $a_{n+1}=\sqrt{\tfrac{1}{2}+a_n}$ for $n\geq 1$. Prove that the sequence converges and, if $L$ is the limit of the sequence, prove that $$\sum_{n=1}^\infty |a_n - L|$$
converges.

I have proven that the limit $L$ of the sequence $\{a_n\}_{n\in\mathbb{N}}$ exists checking that is an increasing sequence and that is bounded (in particular, $a_n\in [0,2]$ for any $n\geq 1$). Then, for every $\varepsilon >0$, there exists $n_0\in\mathbb{N}$ such that for any $n\geq n_0$ then $|a_n - L|<\varepsilon$.

Using this in the sum, for every $\varepsilon >0$, there exists $n_0\in\mathbb{N}$ such that
$$\sum_{n=1}^\infty |a_n - L| = \sum_{n=1}^{n_0} |a_n - L| + \sum_{n=n_0+ 1}^\infty |a_n - L| < \sum_{n=1}^{n_0} |a_n - L| + \sum_{n=n_0+ 1}^\infty \varepsilon$$

However, I don't know how to finish the proof. Also, I would be interested if the convergence of the sum is true for any convergent sequence $\{a_n\}_{n\in\mathbb{N}}$.

Any help would be appreciated.

Body:
We know $L^2=\frac 12+L, L=\frac{1+\sqrt 3}{2}>1$ and $a_n > 0$. Then
$$
a_{n+1}^2-L^2=\frac 12+a_n-L^2=\frac 12 + a_n - \left(\frac 12 +L\right) = a_n-L\\
\implies |a_{n+1}-L|=\frac{|a_n-L|}{a_{n+1}+L} < \frac{|a_n-L|}{L}.
$$

Can you end it now?


问:
Prove that $a^{(p-1)/2} $$\equiv$-1 (modp). Deduce that if $a, b$ are primitive roots modp, then $a\times b$ is NOT a primitive root mod p.

Let $a$ be a primitive root mod the odd prime p.  Prove that $a^{(p-1)/2} $$\equiv$-1 (modp).  Deduce that if $a, b$ are primitive roots modp, then $a\times b$ is NOT a primitive root mod p.

Here we define that, $a$ is a primitive root of $n$ if $(a,n)=$1 and $a$ has order exactly $\phi(n)$  (where the order refers to the least such $k$ where $a^k$ $\equiv$ 1 (modn), $k \geq 1$) (and $\phi$ denotes Euler's totient function.)

I have started by noting that $a^{(p-1)/2}$ $\equiv \pm 1 $ mod $p$ by Fermats Little Theorem and equating $y=a^{(p-1)/2}$

Also Eulers criterion tells us that $(\frac{a}{p})$ is congruent to  $a^{(p-1)/2}$ mod p

Also, Legendre tells us that $(\frac{a}{p})$ =-1 if $a$ is a quadratic non-residue of $p$

So putting this together, am I to show that $a$ being a primitive root mod p implies that it is a quadratic non-residue mod p?

Can't quite get my head round this problem.  Help appreciated,  thanks in advance.

答:
How to derive this expression $ r ^{ (p-1)/2} \equiv -1 \pmod p$ for primitive root of an odd prime $p$. Gives an answer for your first part. For the second part use the facts of first one i.e:

$a^{\frac{p-1}{2}} \equiv -1(\mod p) $ and $b^{\frac{p-1}{2}} \equiv -1(\mod p)$

$(ab)^{\frac{p-1}{2}} \equiv1(\mod p) \implies ab$ $\text{is not the Primitive root of $p$}$


问:
Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ integrally closed?
I know that $\mathbb{Z}[\sqrt{2}]$ is integrally closed, also $F(\mathbb{Z}[\sqrt{2},\sqrt{3}])/F(\mathbb{Z}[\sqrt{2}])=\mathbb{Q}[\sqrt{2},\sqrt{3}]/\mathbb{Q}[\sqrt{2}]$ is an algebraic field extension, so by a theorem from commutative algebra any element $\alpha$ of $F(\mathbb{Z}[\sqrt{2},\sqrt{3}])$ is integral over $\mathbb{Z}[\sqrt{2}]$ if and only if $\mu_{\alpha,F(\mathbb{Z}[\sqrt{2}])}(X)\in\mathbb{Z}[\sqrt{2}][X]$. Since for $\alpha=a+b\sqrt{3}\in F(\mathbb{Z}[\sqrt{2},\sqrt{3}])=\mathbb{Q}[\sqrt{2},\sqrt{3}]$ with $a,b\in\mathbb{Q}[\sqrt{2}]$ $\mu_{\alpha,F(\mathbb{Z}[\sqrt{2}])}(X)=X^2-2aX+a^2-3b^2$ this is equivalent to $2a,a^2-3b^2\in\mathbb{Z}[\sqrt{2}]$. Trying to derive further conditions for this to show that then $a,b\in\mathbb{Z}[\sqrt{2}]$(since then i would receive the result that $\overline{\mathbb{Z}[\sqrt{2}]}^{\mathbb{Q}[\sqrt{2},\sqrt{3}]}=\mathbb{Z}[\sqrt{2},\sqrt{3}]$ and hence that $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ is integrally closed) i stumble upon ugly equations that seem very tedious to work through, so i was wondering if anyone knows a better approach to this problem or if there is a simple counterexample that i am overlooking.

Thanks in advance!

答: Hint: $\frac{\sqrt{2}+\sqrt{6}}{2}$ is not in this ring, although it is integral over it. Hence the ring is not integrally closed.


答:
$x \cos\theta + y\sin\theta = a, \,\, x \sin\theta - y\cos\theta = b$. Therefore:

$$\dfrac{x \cos\theta + y\sin\theta}{x \sin\theta - y\cos\theta} = \dfrac{a}{b}$$
$$\implies \dfrac{x + y\tan\theta}{x \tan\theta - y} = \dfrac{a}{b}$$
$$\implies bx + by\tan\theta = ax \tan\theta - ay$$
$$\implies (ax - by) \tan\theta = bx + ay$$
$$\implies \tan\theta = \dfrac{bx + ay}{ax - by}$$

Title: If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad)

Body:
I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:

If
  $$x \cos\theta+y\sin\theta=a$$
  $$x\sin\theta-y\cos\theta=b$$
  then $$\tan\theta=\frac{bx+ay}{ax-by}$$

I've spent many hours trying.
Thanks in advance.


问:
What is the definition of this quotient ring: $Z[x]/(3, x^{n}-1)$?
While reading a paper I came up with the following quotient ring definition but I am not sure what it does exactly mean:

$R=Z[x]/(x^{n}-1)$ ---> I know this one, since it has one quotient parameter which is $x^{n}-1$

$R/3=Z[x]/(3,x^{n}-1)$---> But this one has more than one qutient parameter, one is 3 and the other one $x^{n}-1$. What does this exactly mean?

Does it mean that we take the modulo of the elements of $Z[x]$ according to $x^{n}-1$ and then take the modulo again according to 3? If so, does that mean  $Z[x]/(3,x^{n}-1)=Z_3[x]/(x^{n}-1)$?

In other words do we take modulo $x^{n}-1$ of the elements of $Z_3[x]$?

Edit: Does it make any difference if i change the polynomial $x^{n}-1$ to $x^{n-1}+ x^{n-2}+...+1$  or to something else?

答:
The question you have doesn't seem to be "What is $\mathbb{Z}[x]/(3, x^{n-1})?$" so much as it seems to be "What is $(3, x^{n-1})$? To put it shortly, it is the smallest ideal of $\mathbb{Z}[x]$ that contains $3$ and $x^{n-1}$.

To put it a little bit more clearly, it is the set of all polynomials $p(x)$ such that $p(x) = 3q(x) + x^{n-1}r(x)$. To put it even more explicitly, it's the set of all polynomials where the first $x^{n - 2}$ terms have a coefficient divisible by $3$.

Put together, and skipping a lot of machinery in the background, you are right. $\mathbb{Z}[x]/(3, x^{n-1}) \simeq \mathbb{Z}_3[x]/(x^{n - 1})$.


问:
minimization problem on differential equations - optimal control
I am trying to minimize an time-integral of a linear function with respect to differential equations. The problem is formally defined as follows:

Given $\lambda< \mu_1, \mu_2$ fixed parameters(thus $x(s)$ will hit 0 at some point in time and stay there) and $c_1, c_2$ time independent, fixed cost coefficients :

$v(x_1, x_2)   = \min_{u,p} \int\limits_T \ [c_1 X_1(s) + c_2 X_2(s)]ds$ \

subject to

$\dot{X_1}(s) = p(s)\lambda  - u(s)\mu_1$;

$\dot{X_2}(s) = (1- p(s))\lambda  - (1-u(s))\mu_2$;

$ X_i(s) = x_i$

$0   \leq u(s) \leq 1$

$0   \leq p(s) \leq 1$

Thus $u(s), p(s)$ are the controls or decision variables. What would be the methodology to follow in this case?

If I understood properly, for this system to reach optimality HJB equation must be satisfied :

$\min_{u,p} (p(s)\lambda  - u(s)\mu_1)\frac{\partial v}{\partial x1} + (1- p(s))\lambda  - (1-u(s))\frac{\partial v}{\partial x2} + c_1 x_1 + c_2 x_2 = 0 $

But without knowing the exact form of $v$ I can not check whether the equation above holds or not. I guess one way to progress is to guess optimal $v$, but then again how such an argument would work without being circular.

Any help, references, comments appreciated... Thanks in advance..

答:
Use PMP (Pontryagen maximum principle). Pontryagen function (hamiltonian) in your problem is

$$
  H = (c_1x_1+c_2x_2) + \phi_1(\lambda p + \mu_1 u) + p_2(\lambda (1-p) + \mu_2 (1-q))
$$

Here $\phi_1$ and $\phi_2$ are the adjoint variables to $x_1$ and $x_2$. So

$$
\dot\phi_i = -\frac{\partial}{\partial x_i}H
$$

$$
 \dot\phi_1 = c_1\ \ \mbox{ and }\ \ \dot \phi_2=c_2
$$

and control variables $u$ and $p$ can be found from maximum principle:

$$
H\to\max_{u,p}
$$

i.e. $p=sign(\phi_1-\phi_2)$ and $u=sing(\mu_1\phi_1 - \mu_2\phi_2)$.As a result you will have simple ODE system with 4 variables $x_1$, $x_2$, $\phi_1$ and $\phi_2$. It can be solved easely by standard methods


问:
Incompleteness theorem
Correct me if I am wrong at any point!

Godel's incompleteness theorem allows us to express "PA is consistent" in the language of Peano arithmetic, and shows that this is not provable in PA. Let's call this sentence $P$.

Godel's completeness theorem tells us that $P$ does not hold in every model of PA; it also tells us that is holds in some models of PA, since otherwise PA would prove $\neg P$ and so PA would be inconsistent.

However, we have a "standard model" of PA, i.e. $\mathbb{N}$, which we can prove things about using ZFC or second-order arithmetic. So my question is, does $P$, thought of as a statement about natural numbers, hold in $\mathbb{N}$? Or have I misunderstood a subtlety (or not-so-subtlety) somewhere?

答:
Converting and expanding from my comment so that the question is marked as answered.

Godel shows that any sufficiently strong (but not too strong) axiomatization of arithmetic is incomplete, meaning for every such axiomatization there is a sentence in its language that the axiomatization neither proves or refutes (proves the negation of). PA is an axiomatization that is sufficiently strong, and hence is incomplete.

In your third paragraph you say

Godel's completeness theorem tells us that P does not hold in every
model of PA; it also tells us that is holds in some models of PA,
since otherwise PA would prove ¬P and so PA would be inconsistent.

This is not quite correct. The completeness theorem does imply that there are models of PA that are also models of $P$ as well as there being other models that model $\neg P$, however it is not for the reason you give: PA + $P$ and $PA + \neg P$ are both consistent (assuming PA was consistent to start with). If one of them was inconsistent, then PA proves the "other" sentence.

Now to your question. You ask whether $\mathbb{N}$, the standard model of arithmetic, is a model for either $P$ or $\neg P$. It must be a model for exactly one of these sentences by the definition of truth. The sentence that Godel's construction produces is true in the standard model. This is because the sentence $P$ (more or less) says that there is no proof of $P$ from the axioms of PA. More precisely, it says that there is no number that codes a list of numbers each of which encodes a sentence in the language of arithmetic that constitute a proof of P from the axioms in PA.

Now, we know, because of the first incompleteness theorem, that there really is no proof of $P$ from PA, which means that there really is no encoding of such a proof by a natural number. This means that each natural number really fails to encode such a proof, but then, by the definition of truth in a model, $\mathbb{N}$ is a model of the sentence "there is no number that codes a proof of $P$ from PA". But this is exactly the sentence $P$, so $\mathbb{N}$ models $P$.

Title: Incompleteness theorem


问:
Finding eigenvalues of a 3x3 matrix given determinant and trace
Suppose a $3×3$ matrix A has only two distinct eigenvalues. Suppose that $\operatorname{tr}(A)=−1$ and $\det(A)=45$. Find the eigenvalues of $A$.

I have solved a similar problem with a 2x2 matrix by using the properties of trace and determinant (trace = a + d and det = ad-bc). I tried to take the same approach for the 3x3 matrix to no success, as expressing the characteristic polynomial is much more complex. Is there any other approach I could take?

答:
Suppose your eigenvalues are $x$ and $y$.
your matrix $A$ is similar to a diagonal matrix $B$ which has it's eigenvalues on its diagonal.

Now, similar matrices have the same determinant and the same trace, thus we can get to the following equations:
$$2x+y = -1$$
$$x^2y=45$$
The first one is the sum of the diagonal (we know that there are 2 unique eigenvalues thus, one of them will show up 2 times on the diagonal).

The second one is the product of the diagonal (determinant of diagonal matrix).

$$... y=\frac{45}{x^2}$$
$$... x=-3 \space\space\space$$

if $x=-3 => y=5$

$x^2y=45$ and $2x+y=-1$.
And that's our answer :)


问:
Non-standard quadratic matrix equation
I have an equation that looks like the following:

$$
A\cdot\mathrm{diag}(x)\cdot x + B\cdot x + c = 0
$$

where $A, B, C \in \mathbb{R}^{n \times n}$ and $x, c \in \mathbb{R}^n$. $ x $ is unknown. The $\mathrm{diag}(x)\cdot x$ is a trick to get element-wise multiplication.

It is different from the Quadratic Matrix Equation (well-studied and solvable)

$$
Q(X) = AXX + BX + C = 0
$$

where $ A, B, C, X \in \mathbb{R}^{n\times n}$, $ X $ is unknown.

It's also different from the Quadratic Eigenvalue Problem (QEP) (also well-studied, solvable, and related to the Quadratic Matrix Equation).
$$
Q( \lambda ) x = (\lambda^2 A + \lambda B + C)\cdot x = 0
$$

where $ A, B, C \in \mathbb{R}^{n\times n}$, $x \in \mathbb{R}^n$ and $ \lambda \in \mathbb{R} $. $ \lambda$ and $ x $ are unknown.

I feel like there should be a reduction from my problem to either of the problems above, but I'm not well-versed enough in linear algebra to figure it out. Numerical approaches are fine too.

An alternative formulation is
$$
A\cdot\mathrm{diag}(x)\cdot \mathrm{diag}(x) + B\cdot \mathrm{diag}(x) + \mathrm{diag}(c) = 0
$$

This would be in the QME form, but I would have to choose a subset of solutions that fit $X = \mathrm{diag}(x)$.

From research, I think I could also use a Gröbner basis to solve this as a system of multivariate polynomial equations, but it seems non-trivial from first glance.

答:
Treat this as a nonlinear equation, i.e. $\,f(x)=0,\,$ and apply Newton's method.
$$\eqalign{
 f &= A(x\odot x) + Bx + c \\
df &= A(2x\odot dx) + B\;dx \\
   &= \Big(2A\cdot{\rm Diag}(x)+B\Big)\,dx \\
   &= \big(2AX+B\big)\,dx \\
\frac{\partial f}{\partial x} &= (2AX+B) \;\doteq\; J \qquad&\big({\rm Jacobian\,of\,}f\big) \\
x_+ &= x - J^{-1}f&\big({\rm Newton\;iteration}\big) \\
}$$
This iteration will converge to a solution near the initial guess $x=x_0$

Since this is a simple quadratic function, the Newton method should converge quickly.

If the rate of convergence isn't satisfactory, perform a line-search at each iteration.
$$\eqalign{
\lambda &= \arg\min_\lambda\;
  \Big\|\,f\Big(x-\lambda\,J^{-1}f(x)\Big)\,\Big\|_2^2 \\
x_+ &= x - \lambda\,J^{-1}f \\
}$$

Title: Non-standard quadratic matrix equation


问:
Symmetrically splitting an octagon into quadrilaterals
I'm wondering whether it is possible to split an octagon into a finite number of quadrilaterals, such that the result is symmetric from all 8 directions (sides or points). There is one condition — any new points that are created must be on the inside of the octagon (i.e. no new points on the edges of the octagon).

The figure below illustrates 4 of my attempts. Most of them (the $1^{st}$, $2^{nd}$ and $4^{th}$) are not symmetric from all 8 directions, and are therefore no valid solution. The $3^{rd}$ one is symmetric, but creates points on the edges.

I'm curious how one could prove that there is (or isn't) a solution to such a problem?

答:
I'm assuming that you want rotational symmetry of degree 8. (If what you want is reflection along the 4 different axis, then let me know.) Such a configuration is not possible.

Recall that the angles in a octagon sum to $180^\circ \times 6$, and the angles in a quadrilateral sum to $180^\circ \times 2 $.

Since we can only add internal points, note that adding a point would increase the sum of angles by $360^\circ$. If we add a point that is not in the center, then by rotational symmetry, we must add all 8 symmetric points.

By double counting the angles, this shows that we must have $8k+3$ or $8k+4$ quadrilaterals.

However, by rotational symmetry of the figure, there must be $8l$ of $8l+1$ quadrilaterals.


问:
Paradox in Theory of probability of transition to polar coordinates
Let there are two independent random variables $X$,$Y$ with normal distribution. Vector $(X, Y)$ can be considered as a random point on the plane. Let $R$ and $\phi$ polar coordinates of this point.
assuming that $X =Y $ we get that the distribution $R^2 = 2X^2$ coincides with the distribution of the square of a random quantity multiplied by 2.
At the same time, provided that $\phi = \pi/4$ or $\phi = 5\pi/4$ distribution of random variables $R^2 = X^2 + Y^2$ is the same as the distribution of the sum of squares of two independent standard normal values. Therefore, we got different distributions when $X=Y$ and when $\phi = \pi/4$ or $\phi = 5\pi/4$ and it's a paradox. What's the catch?

答: The catch is that as soon as you assumed that $X=Y$, they are not independent anymore.


问:
To prove non-trivial functions are primitive recursive...
Im having struggle with these little things. I have got that the main idea is to prove it by mere induction but I haven't figured how to apply it in more complicated cases. The prove that the function given by $f$ where $f(0)=1$ and $f(n+1)=2^{f(0)} 3^{f(1)} \cdots p_n^{f(n)}$ where $p_n$ is the $n$-th prime number is primitive recursive. And the function given by $f$ where $f(m,n)= m-n$ iff $m \geq n$ and $f(m,n)=0$ otherwise is primitive recursive. Thank you for your time and your answer will be appreciated.

答:
In most cases "brute force" (directly expressing our function using base operations) is the simplest way to go.

Your second function is known as "limited subtraction" and denoted as $m \dot - n$. It is primitive recursive, as it can be defined as $m \dot - 0 = m$, $m \dot - S(n) = P(m \dot - n)$, where $P(0) = 0$, $P(S(n)) = n$ (also primitive recursive function).

The first function is more interesting. We need to show that $a \cdot b$, $a^b$ and $p_n$ ($n$-th prime) are primitive recursive, then we can write $f(S(n)) = g(n, f(n))$ where $g(x, y) = p_x^y \cdot y$.

First, $a + b$ is primitive recursive: $a + 0 = a$ and $a + S(b) = S(a + b)$.

$a \cdot b$ is similarly primitive recursive: $a\cdot 0 = 0$ and $a \cdot S(b) = a\cdot b + b$.

Repeating once more, we get $a^b$: $a^0 = 1$, $a^{S(b)} = a^b \cdot a$.

Proving that $p_n$ is primitive recursive is significantly harder, check, for example, proof wiki.


问:
Prove that $f(0)+f(1)\leq f(x)+f(1-x)$
It's a very hard problem that I've worked on for two months:

Let $0\leq x\leq 1$ then define :

$$f(x)=\operatorname{W}\left(x^{\alpha x +1}+1\right)$$

Where $\alpha$ checking: $$2f\left(\frac{1}{2}\right)=\operatorname{W}(1)+\operatorname{W}(2)$$

Then we have :

$$f(0)+f(1)\leq f(x)+f(1-x)\quad(1)$$

To prove it I have tried something like :

$$f(0)\leq f(x)$$

But it really doesn't work to prove $(1)$ !

I have tried also to use Lagrange inversion theorem on the Lambert's function which gives $x\geq 0$:

$$\operatorname{W}(x)=\sum_{n=1}^{\infty}\frac{(-n)^{n-1}}{n!}x^n$$

Without success until now !

Question :

How to show $(1)$ properly ?

Thanks !

答:
With $$f(x)=\operatorname{W}\left(x^{\alpha x +1}+1\right)\qquad \text{and} \qquad 0 \leq x \leq 1$$ because of the symmetry
$$f(x)+f(1-x)-f(0)-f(1) \geq 0$$ can only be true if
$$2 f\left(\frac{1}{2}\right)-f(0)-f(1) \geq 0$$ that is to say if
$$2 W\left(2^{-\frac{\alpha }{2}-1}+1\right)-W(2)-W(1)\geq 0$$
The solution of the equation
$$2 W\left(2^{-\frac{\alpha }{2}-1}+1\right)-W(2)-W(1)=0$$is given by
$$\alpha_* =-\frac{2 \log \left(e^{\frac{1}{2} (W(1)+W(2))} (W(1)+W(2))-2\right)}{\log (2)}$$ which is $\sim 0.34470677834793192536$.

If $\alpha > \alpha_*$ this is not true for all $x$.

The value of $\alpha_*$ is not recognized by inverse symbolic calculators.

Edit

In order the inequality holds, the first condition is that its first derivative at $x=0$ must be positive. This already gives the condition
$$\alpha < \frac{2 W(1)-W(2)+W(1) W(2)}{(W(1)+1) W(2)}\sim  0.572711$$

Title: Prove that $f(0)+f(1)\leq f(x)+f(1-x)$


问:
Collinearity in n dimensions
What is the best way to check if $m$ points are collinear in $n$ dimensions?

I mean I have

$p_1=(3, 4, 5, 2),\quad p_2=(6, 3, 4, 2),\quad p_3=(5, 3, 5,  6),\quad p_4=(4, 2, 7, 4)$

or

$p_1=(1, 6, 2, 7, 7),\quad p_2=(1, 3, 6, 7, 2),\quad p_3=(6, 2, 5, 2, 8),\quad p_4=(8, 5, 3, 2)$

Are they collinear? What is universal way to do it?

答:
Take any pair of input vectors which are not identical. Their difference is the direction of the line, and either of these two is a start point.

For all other points, subtract the start point and check whether the result is an exact multiple of the direction vector.

Title: Collinearity in n dimensions


答:
There holds for a matrix $A$ that
$$
\sum_i \lambda_i = \operatorname{tr}(A),
\quad
\prod_i \lambda_i = \det(A)
$$
Since you have one eigenvalue twice (i assume $\lambda_1$) this results in:
$$
2 \lambda_1 + \lambda_2 = -1,
\quad
\lambda_1^2 \cdot \lambda_2 = 45
$$

// Edit: corrected result:
You can solve this and get to:

$\lambda_1 = -3, \quad \lambda_2 = 5$

Title: Finding eigenvalues of a 3x3 matrix given determinant and trace

Body:
Suppose a $3×3$ matrix A has only two distinct eigenvalues. Suppose that $\operatorname{tr}(A)=−1$ and $\det(A)=45$. Find the eigenvalues of $A$.

I have solved a similar problem with a 2x2 matrix by using the properties of trace and determinant (trace = a + d and det = ad-bc). I tried to take the same approach for the 3x3 matrix to no success, as expressing the characteristic polynomial is much more complex. Is there any other approach I could take?


问:
Let X and Y be independent standard normal random variables. Find the density function of Z = max(X, Y) and use it to find the expectation of Z.
If $X \& Y$ are independent standard normal variables:
$P(X\cap Y)=P(X)P(Y)$ and their respective densities are = $\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}x^2}$ and $\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}y^2}$.
How do I go about converting this into the density of $max(X,Y)?$ I'm not too sure how to progress my working into the new function.

答:
I'm not too sure how to progress my working into the new function.

$$f_Z(z)=2\phi(z)\Phi(z)$$

where $\phi$ is the standard normal density and $\Phi$ is its CDF


问:
Using Rouche's theorem to find location of roots of the polynomial $z^4+ 4z-1$
This question is from ponnusamy and silvermann Complex variables with applications ( Subsection: Argument principle) and I was unable to solve it.

Show that   $z^4+ 4z-1$ has 1 root in the disk |z|<1/3 and remaining three roots in annulus 1/3 <|z|<2.

Attempt: I have proved the existence of 1 root in the disk |z|<1/3 using rouche's theorem and polynomial 1-4z but I am unable to choose 2nd function to use rouche's theorem whose roots lie in 1/3<|z|<2.

Kindly tell me which polynomial to choose.

答: Let $\varepsilon(z)=4z$ and let $f(z)=z^4-1$. Then $z^4+4z-1=f(z)+\varepsilon(z)$. Furthermore, if $|z|=2$, then\begin{align}\bigl|\varepsilon(z)\bigr|&=|4z|\\&=8\\&<15\\&\leqslant|z^4-1|\\&=\bigl|f(z)\bigr|.\end{align}So, your function has as many zeros on $D(0,2)$ as $f$, which has $4$ zeros there.


问:
Reference for quotient stacks
I would like to understand stacks, but the general theory seems pretty abstract to me and I would be happy to limit myself to the case of a quotient stack $[X/G]$.

Are there introductory notes that focus on such stacks ?

答:
You may look at the lecture notes on the website of Michael Groechenig.</p>
```<code>http://individual.utoronto.ca/groechenig/stacks.html
</code>```

Title: Reference for quotient stacks

Body:
You may look at the lecture notes on the website of Michael Groechenig.

http://individual.utoronto.ca/groechenig/stacks.html


问:
Open balls in $\mathbb{R}^2$
Let $(x,y)\in \mathbb{R}^2$ (equipped with usual euclidian metric)

How can I show for $t>0$ that $B((x,y),t) \subseteq (x-t,x+t)\times(y-t,y+t)$ ?

It is pretty clear to me visually but proving it with the Euclidean metric I don't know. Is it safe to assume that if $(a,b)\in B((x,y),t)$ that $|a-x|<t$ and $|b-y|<t$. That might help but I don't know if I can assume that.

答: Suppose that $(a,b)\in B\bigl((x,y),t\bigr)$. Then $(x-a)^2+(y-b)^2<t^2$ and therefore each of the numbers $(x-a)^2$ and $(y-b)^2$ are smaller than $t^2$. In other words,$$|x-a|<t\quad\text{and}\quad|y-b|<t.$$And this is the same thing as asserting that$$a\in(x-t,x+t)\quad\text{and}\quad b\in(y-t,y+t).$$

Title: Open balls in $\mathbb{R}^2$


问:
Where does this 1 come from when balancing this integral equation?
$$ \int e^{ax}\cos(bx)\,\mathrm dx = \frac1{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,\mathrm dx$$

$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\cos(bx)\,\mathrm dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C$$

Where does the $1$ in $\displaystyle \left(1 + \frac{b^2}{a^2}\right)$ come from?

答: Yeah so they have taken the quantity $-b^{2}/a^{2}$ to the Left hand side. So you have the LHS as $$1 \cdot \int e^{ax}\cos(bx) \rm{dx} + \frac{b^{2}}{a^{2}} \int e^{ax}\cos(bx) \ \text{dx} = \Bigl(1+\frac{b^{2}}{a^{2}}\Bigr)\int e^{ax}\cos(bx) \ \text{dx}$$


问:
Defining a subset
The question I have to answer is as following in Swedish:

```<code>Hur många mängder X uppfyller {a, b, c} ⊆ X ⊆ {a, b, c, d, e}?
</code>```

Loosely translated (I do not know mathematical terminology well in English):

```<code>How many sets X satisfy {a, b, c} ⊆ X ⊆ {a, b, c, d, e}?
</code>```

I do not know how to solve this question. For instance, would {{a,b}, a} be two valid subsets of {a, b, c, d, e}?

Any insight into the main question is greatly appreciated.

答:
It is fairly simple to list them. Sets have no duplicate elements by default. Sets do not include elements of different types or pairs. So you can't have pairs and elements. Pairs are generated through Cartesian products.

{a, b, c, d, e}

{a, b, c, d}

{a, b, c, e}

{a, b, c}

So the answer is 4.

Title: Defining a subset


答:
Hint

If $(a,b)\in B((x,y),t)$ then $|x-a|\leq \sqrt{|x-a|^2+|y-b|^2}<t$. Same for $|y-b|$.

Title: Open balls in $\mathbb{R}^2$

Body:
Let $(x,y)\in \mathbb{R}^2$ (equipped with usual euclidian metric)

How can I show for $t>0$ that $B((x,y),t) \subseteq (x-t,x+t)\times(y-t,y+t)$ ?

It is pretty clear to me visually but proving it with the Euclidean metric I don't know. Is it safe to assume that if $(a,b)\in B((x,y),t)$ that $|a-x|<t$ and $|b-y|<t$. That might help but I don't know if I can assume that.


问:
Prove using the definition of limit $\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 14}{7 - xy}=2$
I've been trying for a while to prove $$\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 14}{7 - xy}=2$$, but I don't even know how to continue after $|\lim_{(x, y)\to (0, 0)}\frac{4x^3y - x^4y^5 + 2xy}{7 - xy}|$. Any hints or solutions?

答:
Subtract 2 to get rid of the 14.

Then, for $|x|, |y| < 1$, $|7-xy| > 6$
so the denominator is bounded away from zero.

Finally, if $|x|, |y| < c$,
see what bounds you can get on the expression.