问:
Numbers that are the sum of the squares of their prime factors
A number which is equal to the sum of the squares of its prime factors with multiplicity:

$16=2^2+2^2+2^2+2^2$

$27=3^2+3^2+3^2$

Are these the only two such numbers to exist?

There has to be an easy proof for this, but it seems to elude me.

Thanks

Body:
Giorgos Kalogeropoulos has found 3 such numbers, each having more than 100 digits.

You can find these numbers if you follow the links in the comments of OEIS A339062 &
A338093

or here https://www.primepuzzles.net/puzzles/puzz_1019.htm

So, such numbers exist! It is an open question if there are infinitely many of them...


问:
Z-module homomorphisms from nZ to Z[1/2]/Z
I want to find a description of $\text{Hom}_\mathbb{Z}(n\mathbb{Z}, \mathbb{Z}[1/2]/\mathbb{Z})$. Since they are $\mathbb{Z}$-modules, it is enough to just look at group homomorphisms from the former to the latter. Here's what I have tried so far:

We can write $\mathbb{Z}[1/2]/\mathbb{Z} = \{ \frac{m}{2^a} ; m \ \text{odd}, a \geq 0 \}$. Furthermore, $n\mathbb{Z}$ is a cyclic group with only one generator $n$, so to define a homomorphism $\phi: n\mathbb{Z} \to \mathbb{Z}[1/2]/\mathbb{Z}$, it is enough to just specify a value of $\phi$ at $n$. Furthermore, I believe $n$ can be sent to anything since it has infinite order, i.e. $\phi(n)$ can be $\frac{m}{2^a}$ for any odd $m$ and $a \geq 0$. Is this correct? If I am correct, then does this just mean $\text{Hom}_\mathbb{Z}(n\mathbb{Z}, \mathbb{Z}[1/2]/\mathbb{Z}) \cong \mathbb{Z} \times \mathbb{Z}$? Any pointers would be greatly appreciated.

答:
Your problem is that $\frac{m}{2^a}$ does not have infinite order. Instead, if $m$ is odd, it has order $2^a$; more generally, if $m$ is not required to be odd, it has order $2^{a-b}$, where $m = 2^b n$ and $\gcd(n,2)=1$.

Otherwise, you are completely right in asserting that an arbitrary element defines a morphism. Yet, then of course
$$
\operatorname{Hom}_{\mathbb Z}(n\mathbb Z, \mathbb Z[1/2]/\mathbb Z) \cong \mathbb Z[1/2]/\mathbb Z
$$
by the definition of addition of morphisms (pointwise).

ContentLicense: CC BY-SA 4.0


问:
Series of logarithms $\sum\limits_{k=1}^\infty \ln(k)$ (Ramanujan summation?)
I had this question earlier, so to say as a "standalone" problem, but now it pops up in context of an analysis with the lngamma-function. As well as we can convert the question of sums of like powers $Su_p(n)=1^p+2^p+3^p+\cdots+n^p$ in terms of the Hurwitz-zeta $Su_p(n) = \zeta(p,1)-\zeta(p,n+1)$ (and solve using the Bernoulli-polynomials) I try to express this for sums of logarithms (and powers of logarithms) : $$ Sl_p(n) = \ln(1)^p+\ln(2)^p+\cdots+\ln(n)^p $$

I have seemingly proper coefficients for power series which allow that computations/approximations. The result is always found by the difference of $ Sl_p(n)-Sl_p(1) $ . Here the power series has zero as constant term.

However, to have the analogue to the Hurwitz-zeta I should have the constant term the "value" for the infinite sum $ Sl_p(\infty) $ instead. (This cancels properly if I formulate the sum of (powers of) logarithms always as that difference $ Sl_p(n)-Sl_p(1) $) And also, I'd like that this agreed conceptually more to the notion:
$$ \begin{array} {rll} Sl_p(n)-Sl_p(1) &=& T_p(0)-T_p(n) \\\
 & =&  (\ln(1)+\ln(2)+\ln(3)+\cdots) \\\
 & & - (\ln(n+1)+\ln(n+2)+\ln(n+2)+\cdots) \end{array} $$

as this agrees more with the idea of the Hurwitz-zeta-difference.

But this requires, that $T_p(0)$ represents the (infinite) sum of the $p$'th powers of the logarithms, and the constant in the power series of $T_p(0)$ must contain just such a value. Let's talk about the first powers of the logs $p=1 $ first and denote the assumed sum as $L_1$ : $ L_1 = \lim_{n\to \infty} T_1(0) $

[update 2]

I think, thanks to the hint of J.M., I can answer Q1 myself now; only Q2 remains somehow vague - besides a simple empirical heuristic I did still not get the formally correct approach to the constant term/integral-definition in the Ramanujan summation - but this is now only a side problem here (however it would be nice to get help also for this question).

With the help of the knowledge about the derivatives of the zeta at zero the relevant part of the problem could now satisfyingly be solved, so I put it here in an answer to my own question, see that answer below....

Final remark/conclusion: it is interesting, that the power series for the lngamma pops up here "automatically" - we need no other uniqueness criterion for the argument, that the (Eulerian) gamma-function gives "the correct" interpolation for the factorial problem. It is just the result of the construction of an operator for the "indefinite summation" (which was the comceptual goal from where the problem/question arose initially)

In the following I keep the rest of the original question although the approach to the Ramanujan summation contains an error

[end Update 2]

I tried to give sense to that divergent series by replacing the powers of $x$ in the Mercator series for $ \ln(1+x) $ by appropriate zetas at negative integer arguments. If I understand things correctly then this is similar to the method of Ramanujan summation, where the Bernoulli numbers are just replaced by the according zeta-values (appropriately scaled). With this I got then an approximation of $$L_1 \approx -0.0810614667953 $$

which seems a rather "random" value...

By searching in other online sources I got the suggestion (OEIS), that this is also $$ \int_1^2 \ln(\Gamma(t)) \; dt \approx -0.08106146679532725821967026 $$

With this my power series for $T_1(x)$ begins like

$ \qquad \small \begin{array} {l}
 - & 0.0810614667953 \\
 - & 0.577215664902x \\
 + & 0.533859200973x^2 \\
 + & 0.325578788221x^3 \\
 + & 0.125274140308x^4 \\
 + & 0.0337256506589x^5 \\
 + & 0.00685935357296x^6 \\
 + & 0.00117260810356x^7 \\
 + & O(x^8)
 \end{array}
$

The other coefficients occur also in the power series for $f(x)=\ln(\Gamma(\exp(x)) $

Also, $L_1$ seem to satisfy the following expression:
$$ \exp(L_1) = {\sqrt{(2\pi)} \over e} \approx 0.9221370088957891168791517 $$

The questions are:
$\qquad$   Q1: Is that value $-0.0801\ldots$ a meaningful (or even correct) representation for the infinite sum of logarithms?

$\qquad$   Q2: Did I reproduce the Ramanujan summation correctly here?

[update 1]: J.M. mentions the relation to $ y= - \zeta'(0) \approx 0.918938533205 $ where the representation of the derivative of zeta at 0 equals formally just the sum of logarithms. Now my $L_1$ and the $y$ are related by $L_1 = y - 1$. So I expect some error in my derivation which leads just to that unit difference...

答:
To address your last comment on evaluating infinite sums of powers of logarithms:

Letting

$$(-1)^n \zeta^{(n)}(s)=\sum_{k=1}^\infty \frac{(\log k)^n}{k^s}$$

the question can be recast as the evaluation of the quantity $(-1)^n \zeta^{(n)}(0)$ for various $n\in \mathbb N$.

Apostol and Choudhury had given a bunch of formulae for evaluating $\zeta^{(n)}(s)$; in particular, there is the formula

$$(-1)^n \zeta^{(n)}(0)=\frac{\Im((-\log\,2\pi-i\pi/2)^{n+1})}{\pi(n+1)}+\frac1{\pi}\sum_{j=1}^{n-1} a_j j!\binom{n}{j}\Im((-\log\,2\pi-i\pi/2)^{n-j})$$

where

$$a_j=c_{j+1}+\sum_{\ell=0}^j \frac{(-1)^\ell \gamma_\ell}{\ell!} c_{j-\ell}$$

$$c_k=-\frac{\gamma c_{k-1}}{k}-\frac1{k}\sum_{\ell=1}^{k-1} (-1)^\ell \zeta(\ell+1) c_{k-\ell-1},\qquad c_0=1$$

the $\gamma_n$ are the Stieltjes constants, and $\gamma=\gamma_0$ is the Euler-Mascheroni Constant.

Here are a few explicit evaluations of the formula:

$$\zeta^{\prime\prime}(0)=\frac{\gamma^2}{2}-\frac{\pi^2}{24}-\frac{(\log (2\pi))^2}{2}+\gamma_1$$

$$-\zeta^{\prime\prime\prime}(0)=-\gamma^3-\frac32 \gamma ^2 \log(2\pi)+\frac{\pi^2}{8}\log(2\pi)+\frac{(\log(2\pi))^3}{2}-3\gamma\gamma _1-3\gamma_1\log (2\pi)-\frac32\gamma_2+\zeta(3)$$

As can be surmised, the closed forms become rather unwieldy as $n$ gets large...

An extension of these results to noninteger powers would involve having to appropriately define the differintegral of the zeta function; going the series route would now involve terms containing the incomplete gamma function, but I've no knowledge of any closed forms for the resulting sum.

ContentLicense: CC BY-SA 3.0


问:
Solve a polynomial involving geometric progression?
I have had trouble with this question:

"Solve the equation $8x^3 - 38x^2 + 57x -27 = 0$" if the roots are in geometric progression.

Any help would be appreciated.

答:
Let the roots be $a, a\cdot r, a\cdot r^2$

So using Vieta's formula $a+ a\cdot r+ a\cdot r^2=\frac{38}8\implies a(1+r+r^2)=\frac{19}4$

and $a( a\cdot r)+ a\cdot r(a\cdot r^2)+a\cdot r^2(a)=\frac{57}8\implies a^2r(1+r^2+r)=\frac{57}8$

Divide to get $ar=\frac{\frac{57}8}{\frac{19}4}=\frac32$ as $a(1+r+r^2)\ne0$

Put $a=\frac3{2r}$ in the first equation


问:
Basis determining a unique topology
When I read Munkres' Topology, I have a feeling that if we have a basis $\mathscr{B}$ on a set $X$, then the basis determines uniquely a topology on $X$; that is, if we have two topologies $\mathscr{T}_1, \mathscr{T}_2$ with the same basis $\mathscr{B}$, then $\mathscr{T}_1=\mathscr{T}_2$. I'm not sure if I'm right because I can't see this in the definition, which is as follows:

If $X$ is set, a basis for a topology on $X$ is a collection $\mathscr{B}$ of subsets of $X$ (called basis elements) such that for each $x\in X$, there is at least one $B\in \mathscr{B}$ such that $x\in B$ and if $x\in B_1\cap B_2$, where $B_1, B_2\in \mathscr{B}$, then there exists $B_3\in \mathscr{B}$ such that $x\in B_3\subset B_1\cap B_2$.

Moreover, the basis $\mathscr{B}$ generates a topology

$\mathscr{T}_\mathscr{B}=\left\{ U\subset X: \text{for each $x\in U$, there exists $B\in \mathscr{B}$ such that $x\in B\subset U$}\right\}$,

which is the smallest topology containing $\mathscr{B}$. Hence, I guess it's likely that those topologies whose bases are $\mathscr{B}$ should be equal to $\mathscr{T}_\mathscr{B}$.

By the way, I have consulted the article
Uniqueness of Topology and Basis and one of comments (left by Henno) seems to justify my hunch and they mentioned any open set $O$ is a union of the elements of $\mathscr{B}$, so $O$ is already in the topology $\mathscr{T}_\mathscr{B}$, but how could they know $O$ can be written this way just by the definition of a basis? I mean, in Munkres' book, he mentioned in lemme 13.1, from my understanding, that $\mathscr{T}_\mathscr{B}=\left\{\cup_\alpha B_\alpha:B_\alpha \in \mathscr{B}\right\}$, as opposite to saying it holds for any topology with basis $\mathscr{B}$. Perhaps I'm misunderstanding at this point.

Any help is really appreciated!!

答:
We say that topology $\mathcal T$ has basis $\mathcal B$ if $\mathcal T_{\mathcal B}=\mathcal T$.

Thus, it's immediate that if two topologies have the same basis then they coincide.

Saying that for every $x\in U$ there's a $B_x\in\mathcal B$ such that $x\in B_x\subseteq U$ is equivalent to saying that $U$ is the union of elements of $\mathcal B$, specifically $U=\bigcup_{x\in U}B_x$.

What you might be missing is that

A set $\mathcal B$ of subsets of $X$ is a basis for a topology (meaning $\mathcal T_{\mathcal B}=\left\{\bigcup \mathcal D:\mathcal D\subseteq\mathcal B\right\} $ is a topology) if and only if the given conditions hold, i.e. $\forall x\in X\,\exists B\in\mathcal B: x\in B$ and $\forall x\in X\,\forall B_1,B_2\in\mathcal B\ x\in B_1\cap B_2\implies \exists B\in\mathcal B: x\in B\subseteq B_1\cap B_2$.

Title: Basis determining a unique topology


问:
Finding tangent vector for a composite function at a given point
I would want to insert the whole question but it's kinda complex/difficult to type it out :( sorry guysss
Basically the problem looks similar to

$T(x,y,z)=(u,v)$

$u=x+y+z^2$

$v=\cos(x)+\tan(y)-z+3$

vector $r(t)=(t\cdot e,(t^2+2)^t, e^{-s^2})$

At point $t=0$ what's the tangent vector of $T(r(t))$?

Anybody who can help me out with this problem?
Would appreciate sooooo much $x$.

答:
We have the functions $T:\mathbb R^3\to \mathbb R^2$ and $r:\mathbb R\to\mathbb R^3$, so the composition is such that $$T\circ r:\mathbb R\overset{r}{\longrightarrow}\mathbb R^3\overset{T}{\longrightarrow} \mathbb R^2$$
$r(0)=(0,1,1)$ and the jacobian matrix of $T$ in $(0,1,1)$ is
$$J_T(0,1,1)=\begin{pmatrix}\nabla u(0,1,1)\\\nabla v(0,0,1) \end{pmatrix}=\begin{pmatrix} 1&&1&&2z\\-\sin(x)&&\dfrac{1}{\cos^2(x)}&&-1 \end{pmatrix}\Bigg|_{(0,1,1)}=\begin{pmatrix}1&&1&&2\\0&&\cos^{-2}(1)&&-1 \end{pmatrix},$$
while the jacobian of $r$ in $0$ is the vector $\begin{pmatrix}e\\\log(2)\\0 \end{pmatrix}$, then for the rule of composition we get $$J_{T\circ r}(0)=\begin{pmatrix}1&&1&&2\\0&&\cos^{-1}(1)&&-1 \end{pmatrix} \cdot\begin{pmatrix}e\\\log(2)\\0 \end{pmatrix}=\begin{pmatrix}e+\log(2) \\\log(2)\cos^{-2}(1) \end{pmatrix}$$


问:
If $X$ is compact then $X$ is sequentially compact.
My attempt :

We assume that $X$ is a compact metric space and we let $A=(x_n)$ be a sequence. Then $A \subset G_1 \cup G_2.....G_k$.So suppose $G_i$ contains infinitely many points of $(x_n)$.We choose $y \in G_i$ in such a way so that  we can find a ball $B_d(y,\epsilon) $ such that it is contained in $G_i$ and it contains infinitely many points of $(x_n)$.Then let us say that the susbsequence $(x_{n_k})$ is contained in the ball. I am trying to show that the subsequence $(x_{n_k})$ converges to $y$.

How do I go about from here? If someone could suggest a way out instead of suggesting a different answer.

答: One possible approach: suppose $(x_n)$ has no convergent subsequence. Then, for each integer $n$ there is an open neighborhood $U_i\ni x_i$ such that $U_i\cap U_j=\emptyset$ for all $i\neq j.$ Furthermore, $C:=\{x_n\}_{n\in \mathbb N}$ is closed because it has no limit points, hence has them all. Then $\mathscr U=\{C^c\}\cup \{U_i\}_{i\in \mathbb N}$ is an open cover with no finite subcover.


问:
Exercise 6 on p.111 in "Analysis on Manifolds" by James R. Munkres.
I am reading "Analysis on Manifolds" by James R. Munkres.

There is the following exercise on p.111 in this book.

Exercise 6.

Show that Theorem 13.6 holds without the hypothesis that $f$ is continuous on $S$.

My solution is here:

Since Munkres doesn't use the hypothesis that $f$ is continuous on $S$ in his proof, Theorem 13.6 holds without the hypothesis.

My solution is too simple, so I don't have perfect confidence in my solution.

I think Munkres doesn't use the hypothesis that $f$ is continuous on $S$ in his proof at all.

Am I right or not?

答:
To prove that $f_A = f \chi_A$ is Riemann integrable, it is sufficient that  $f_A$ be bounded and continuous almost everywhere in a rectangle $Q \supset S \supset A$.  The hypothesis that $f_S$ is Riemann integrable is enough since it means that $f_S$ is bounded and continuous almost everywhere in $Q$.

Because $|f_A| \leqslant |f_S|$, we immediately have that $f_A$ is bounded.

The argument in Step 1 then proves that $f_A$ is continuous everywhere in $Q$ except possibly on a set of measure zero. Again this follows from the assumption that $f_S$ is Riemann integrable without a stronger assumption that $f$ is continuous.

This is obvious for $x_0 \in A = \mathrm{Int }\, S$ and $x_0 \in \mathrm{Ext }\,\bar{S}$ since there exist neighborhoods of $x_0$ where $f_A = f_S$, and either $f_S$ is continuous at $x_0$ or $x_0$ belongs to a set of measure zero. The argument is more involved but striaghtforward when $x_0$ is a boundary point. Since $f_A(x_0) = 0$, there is a discontinuity at $x_0$ only if $f_A(x) = f_S(x) \not \to 0$ as $x \to x_0$ from the interior, that is with $x \in A$. However, such points are discontinuity points for $f_S$ and must be contiened in a set of measure zero as well.


答: Since the roots are in geometric progression, you have $x_2^2=x_1x_3$. From Theorem of Viet $x_1x_2x_3=27$. Hence $x_2^3=27$ etc.</p>

Title: Solve a polynomial involving geometric progression?

Body:
I have had trouble with this question:

"Solve the equation $8x^3 - 38x^2 + 57x -27 = 0$" if the roots are in geometric progression.

Any help would be appreciated.


答: Suppose $X$ is not sequentially compact, then there is a sequence $(p_n)\subset X$ which admits no subsequence converging to a point in $X$. Then $\forall x\in X,\exists r\gt 0$ such that $p_j\in B_d(x,r)$ for only finitely many $j$. $\{{B_d(x,r)}:x\in X\}$ is an open cover of $X$, so by compactness it admits a finite subcover $\{B_d(x_i,r_i)\}_{i=1}^N$. Since $(p_n)$ have only finitely many terms in each $B_d(x_i,r_i)$, it follows that $(p_n)$ has only finitely many terms, a contradiction.

Title: If $X$ is compact then $X$ is sequentially compact.

Body:
My attempt :

We assume that $X$ is a compact metric space and we let $A=(x_n)$ be a sequence. Then $A \subset G_1 \cup G_2.....G_k$.So suppose $G_i$ contains infinitely many points of $(x_n)$.We choose $y \in G_i$ in such a way so that  we can find a ball $B_d(y,\epsilon) $ such that it is contained in $G_i$ and it contains infinitely many points of $(x_n)$.Then let us say that the susbsequence $(x_{n_k})$ is contained in the ball. I am trying to show that the subsequence $(x_{n_k})$ converges to $y$.

How do I go about from here? If someone could suggest a way out instead of suggesting a different answer.


问:
Convergence of the sum of moments implies convergence of the sum of arbitrary random variables
I'm trying to solve the following

Let $\{X_j\}$ be a sequence of arbitrary random variables such that $$\sum_{j=1}^{\infty}\mathbb{E}[|X_j|^{r_j}]<\infty$$
for some sequence $\{r_j\}$ in $(0,1]$. Then $$\sum_{j=1}^{\infty}X_j$$
converges absolutely a.s.

I tried it using Markov's Inequality since $$\mathbb{P}(S_n \text{ converges })=\lim_m \lim_n \lim_r\mathbb{P}\left(\max_{n\le k \le r}|S_k-S_n|\le \frac{1}{m}\right)\le \lim_m \lim_n \lim_r\mathbb{P}\left(|S_r-S_n|\le\frac{1}{m}\right)$$

So, $$\mathbb{P}\left(|S_r-S_n|\ge\frac{1}{m}\right)=\mathbb{P}\left(|\sum_{j=n+1}^r X_j|\ge\frac{1}{m}\right)\le m^{h_{n}^r}\mathbb{E}[|\sum_{j=n+1}^rX_j|^{h_{n}^r}]$$

where $h_n^r=\min\{r_j: n+1\le j\le r\}$

From here, I'm stuck.

I'll appreciate any help.

答: The assumption $\sum_{j=1}^{\infty}\mathbb{E}[|X_j|^{r_j}]<\infty$ implies that $\sum_{j=1}^{\infty} |X_j|^{r_j}$ is almost surely finite, which implies in turn that $|X_j|^{r_j}\to 0$ almost surely. Since $0<r_j<1$, this implies that $|X_j| \to 0$ almost surely. Now, for the $\omega$ such that $\sum_{j=1}^{\infty} |X_j|^{r_j}(\omega)$ is finite, the convergence of $$\sum_{j=1}^{\infty} |X_j| (\omega)$$ follows from $|X_j(\omega)| = |X_j(\omega)|^{r_j}|X_j(\omega)|^{1-r_j}$ and the fact that $|X_j(\omega)|^{1-r_j}\leqslant 1$ for $j$ large enough.


问:
determining the limit of an integral
Let $f(x):\mathbb{R} \to \mathbb{R}$ be a continuous function, for every $n > 0$ define:

$$a_n = \int_{\frac{-1}{n}}^{\frac{1}{n}} f(x) \,dx$$

determine $\lim\limits_{n\to\infty} a_n$

At first glance it would seem like the limit approaches zero, is that statement true?

答:
Of course it's true! Since $f$ is continuous, its absolute value assumes its maximum on, say, [-1, 1]. Let $M$ be that maximum. Then
$$
\left| \int_{-1/n}^{1/n} f(x) dx \right| \le \int_{-1/n}^{1/n} |f(x)| dx \le \frac{2M}{n}.
$$

Title: determining the limit of an integral


答:
Yes, the  limit is zero.

Hint:

if $F$ is an anti derivative of $f$,then $F$ is continuous at $0$.

Title: determining the limit of an integral

Body:
Let $f(x):\mathbb{R} \to \mathbb{R}$ be a continuous function, for every $n > 0$ define:

$$a_n = \int_{\frac{-1}{n}}^{\frac{1}{n}} f(x) \,dx$$

determine $\lim\limits_{n\to\infty} a_n$

At first glance it would seem like the limit approaches zero, is that statement true?


问:
Find a minimal polynomial of a primitive 5-th root of unity over $\mathbb{Q}_3(\sqrt{-7})$
Let $K=\mathbb{Q}_3(\sqrt{-7})$ be the base field and $L=K(\zeta_5)$ where $\zeta_5$ is a primitive $5$-th root of unity. I know that the extension $L/K$ is unramified of degree $2$.

Now I would like to find a minimal polynomial of $\zeta_5$ over $K$. I know that such a minimal polynomial must divide $x^4 + x^3 + x^2 + x + 1 = \frac{x^5-1}{x-1}$, i.e. it will split into two quadratic polynomials (which both are potential minimal polynomials of $\zeta_5$). But aside from that, I don't know how to compute the actual factors.

Could you please help me with this problem?

答: Given a prime p not dividing n and a p-adic local field $M$ with residual field $\mathbf F_q$, with $q=p^f$, it is classically known that $M(\zeta_n)/M$ is unramified  cyclic, with Galois group generated by the Frobenius automorphism $s$ determined by $s(\zeta_n)= {\zeta_n}^q$ . In particular $[M(\zeta_n):M]=r$, the smallest integer $r\ge 1$ s.t. $q^r \equiv 1$ mod $n$. See e.g. Serre's "Local Fields", chap. IV,§4, prop. 16. Applying this theorem with your notations $K=\mathbf Q_3 (\sqrt {-7})$ and $L=K(\zeta_5)$, and accepting that $K/\mathbf Q_3$ is quadratic unramified, we find here that $\mathbf Q_3(\zeta_5)/\mathbf Q_3$ and $K(\zeta_5)/K$ are unramified (cyclic) of respective degrees degree $4$ and $2$. But a fixed algebraic closure of $\mathbf Q_3$ contains a unique unramified extension of $\mathbf Q_3$ of given degree, so that $\mathbf Q_3(\zeta_5)=K(\zeta_5)$. In particular, $K$ coincides with the unique quadratic subfield of $\mathbf Q_3(\zeta_5)/\mathbf Q_3$, namely $K=\mathbf Q_3(\zeta_5 +{\zeta_5}^{-1})$, and the minimal polynomial of $K(\zeta_5)/K$ is $X^2 -(\zeta_5 +{\zeta_5}^{-1})X + 1$.


问:
How to show subset of $\mathbb R$ is not an element of $\mathbb R$
How do you show that for some $A\subset \mathbb R$, $A\not \in \mathbb R$.

Intuitively it makes sense, but how do you actually prove it?

答:
To prove that none of the subsets $A\subset\mathbb R$ are contained in $\mathbb R$ as elements, one needs to craft the proof tailored to the particular definition of $\mathbb R$. For some definitions of $\mathbb R$ it may be impossible. For example, if you accept that $0\in\mathbb N\subset\mathbb R$ and if you accept the set-theoretic definition of zero ($0:=\emptyset$), then $\emptyset$ is both an element and a subset of $\mathbb R$.

However, it is possible to prove that at least one subset $A\subset\mathbb R$ is not an element of $\mathbb R$. A well-known consequence of the axiom of regularity is that no set is an element of itself. Let $A:=\mathbb R$. As $A\subset\mathbb R$, if every subset of $\mathbb R$ also belonged to $\mathbb R$, we would have $A=\mathbb R\in\mathbb R$ - contradicting the axiom of regularity.

Without use of the axiom of regularity, you can still mount a proof which would use something like Cantor's diagonal argument. Let $X=\{A\subset\mathbb R\mid A\not\in A\}$. If every subset of $\mathbb R$ is an element of $\mathbb R$, then $X\subset\mathbb R$. Now, try to answer the question: does $X\in X$? (You will conclude that $X\in X\iff X\not\in X$ - a contradiction.)


问:
Measurability of the Cartesian sum of measurable functions
I want to prove the following:

Let $X\subset\mathbb{R}^n$ and $Y\subset\mathbb{R}^m$ be arbitrary Lebesgue measurable subsets. Then, for $f\in L^1(X)$ and $g\in L^1(Y)$, the Cartesian sum of $f$ and $g$, $f\oplus g\colon X\times Y\to\mathbb{R}$, $(x,y)\mapsto f(x)+g(y)$, is an element of $L^1(X\times Y)$.

While the integrability of $f\oplus g$ is trivial to see, I'm really confused about its measurability...

The mapping $(r,s)\mapsto r+s$ is continuous on $\mathbb{R}\times\mathbb{R}$ and thus $(\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}))$-$\mathcal{B}(\mathbb{R})$-measurable. If we show the $(\mathcal{B}(X)\otimes\mathcal{B}(Y))$-$(\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}))$-measurability of the mapping $(f\times g)(x,y):=(f(x),g(x))$, we are done. For this purpose, we choose some element $B$ of the generator of $(\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}))$, i.e., $B\in\{\pi_i^{-1}(A)\colon A\in\mathcal{B}(\mathbb{R}),~i=1,2\}$, where $\pi_i$ is the projection from $\mathbb{R}\times\mathbb{R}$ onto the $i$-th compnent. W.l.o.g. we assume that $B\in\{\pi_1^{-1}(A)\colon A\in\mathcal{B}(\mathbb{R})\}$. Then, $B=A\times\mathbb{R}$, with some $A\in\mathcal{B}(\mathbb{R})$, and we calculate $$(f\times g)^{-1}(B)=\{(x,y)\in X\times Y\colon (f(x),g(y))\in A\times\mathbb{R}\}=f^{-1}(A)\times Y=\pi_X^{-1}(f^{-1}(A))$$ and since $f$ is $\mathcal{B}(X)$-$\mathcal{B}(\mathbb{R})$-measurable, this is some element of the generator of $\mathcal{B}(X)\otimes\mathcal{B}(Y)$ and thus an element of the latter.

Is this proof correct? If so, is there a simpler, if not trivial, argument? Is there some way to use the property $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})=\mathcal{B}(\mathbb{R}\times\mathbb{R})$?

答:
Your proof is correct. Maybe you could write it a slightly lighter way. For instance:

If we show the $(\mathcal{B}(X)\otimes\mathcal{B}(Y))$-$(\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}))$-measurability of the mapping $(f\times g)(x,y):=(f(x),g(x))$, we are done. For this purpose, we choose some element $A\times B$ of the generator of $(\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}))$, i.e. a rectangle $A\times B$ where $A, B  \in \mathcal{B}(\mathbb{R})$. Then

$$(f\times g)^{-1}(A \times B)=\{(x,y)\in X\times Y\colon (f(x),g(y))\in A\times B \}=f^{-1}(A)\times g^{-1}(B)$$ and since $f$ is $\mathcal{B}(X)$-$\mathcal{B}(\mathbb{R})$-measurable and $g$ is $\mathcal{B}(Y)$-$\mathcal{B}(\mathbb{R})$-measurable, we have that $f^{-1}(A)\times g^{-1}(B) \in \mathcal{B}(X)\otimes\mathcal{B}(Y)$.


答:
You need to state explicitly that $X$ is a metric space, since the result is not true for topological spaces in general. The beginning of your argument makes no sense as stated: where do $G_1,\ldots,G_k$ come from, and what are they? We might guess that you started with some open cover of $X$ and took a finite subcover, but you need to say so. You also need to say something about the open cover with which you started. For all we know, you started with the cover $\{X\}$, so that $k=1$ and $G_1=X$.

Next, for a given $\epsilon$ how do you know that there is point $y\in G_i$ such that $B_d(y,\epsilon)\subseteq G_i$ and $B_d(y,\epsilon)$ contains infinitely terms of the sequence? This requires proof, which you have not given. Finally, it is simply not true that the subsequence in question necessarily converges to $y$.

At this point you don’t have enough of a proof for it to be salvageable. I strongly recommend that you try instead to prove the result by showing that if $X$ has a sequence with no convergent subsequence, then $X$ is not compact. A direct proof is possible, by showing that every sequence in $X$ has a Cauchy subsequence and then using the fact that compact metric spaces are complete, but that’s a bit harder.


问:
Application of Implicit Function Theorem to the triangle
Well, I observe the triangle with side-lengths $a, b, c$ and angle $\alpha$, the opposite angle to $a$. The implicit relationship between them should be The Law of Cosines : $a^2-b^2-c^2+2bc\cos \alpha=0$

Then if I write $\alpha$ as smooth function of $a, b, c$, namely
$\alpha=\arccos \frac{c^2+b^2-a^2}{2bc} = f(a,b,c)$, under what circumstances I can do that? This problem should be related to the Implicit Function Theorem. Of course, we must have $b,c\neq0$. Does that mean that $\alpha$ can be written as a smooth function of $a, b, c$ everywhere except in the neighbourhood of $(k,0,0)$, when we look from the point of IFT, or under another circumstances?

答:
Put $F(a,b,c,\alpha) = a^2-b^2-c^2+2bc\cos\alpha$. Then $\alpha$ is defined implicitly as a function of $a,b,c$ by the equation $F(a,b,c,\alpha) = 0$. Examining carefully the statement of the Implicit Function Theorem, given $(a_0,b_0,c_0,\alpha_0)$ which satisfy the Law of Cosines, we may write $\alpha = f(a,b,c)$ (in such a way that $F(a,b,c, f(a,b,c)) = 0$) in a neighborhood of $(a_0,b_0,c_0)$ when $F_\alpha(a_0,b_0,c_0,\alpha_0)\neq 0$.

Since $F_\alpha = -2bc\sin\alpha$, we need $b,c\neq 0$ and $\alpha\neq 0$. In other words, we need $b,c>0$ and the triangle to not be degenerating (implying $a>0$ as well). So we are guaranteed invertibility so long as we can draw a valid triangle with positive side lengths $a,b,c$.


问:
Solving limit which can't be (maybe I'm wrong) solved by L'Hospital
I'm unable to solve the following limit:
$$\lim_{x \to 0+}\frac{\sin( x \sqrt x )-x \sqrt
x}{x\sqrt{x+1}}$$

My attempt:
Using L'Hospital rule:
$$\frac{ \cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x }  { \sqrt{x+1} + x \frac{1}{2\sqrt{x+1}} }$$
$$\frac{\sqrt{x+1} (\cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x)}{2x+1}$$
Since $(\sqrt x)'=\frac{1}{2\sqrt x}$  we can see that we will get $\frac{0}{0}$ when $x \to 0$ if my calculations are right.

答:
$$\lim_{x \to 0_+}\frac{\sqrt{x+1} (\cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x)}{2x+1} = \frac{0}{1} = 0$$

Because the denominator resolves to 1.


问:
Dynamical System problem
PLS HELP ME. I have a problem with this exercise given by the professor for home. It's about Lyapunov equation and autonomous systems. Here it is:

Prove that if the state of equilibrium $x^*=0$ $(x^*\in\mathbb{R}^n)$ of the system:

$x(k+1)=e^A x(k)$  with $A\in \mathbb{R}^{n\times n}$

is asymptotically stable then even the equilibrium state $x^{**}=0$ of the system:

$\dot{x}(t)=Ax(t)$

is asymptotically stable.

I thought I could apply the step invariant transformation for an autonomous system and since it should preserve the response it should be asymptotically stable. But i'm not sure...
Hope someone of you can help me, thanks in advice!

答:
I'm not sure what tools you have available, but here is an approach using the definition of asymptotic stability.

Consider trajectories corresponding to the common initial state $x(0) = x_0$.
For any $k = 0,1,2,3,\dots$ and $t \in [k,k+1]$, we have
$$
\begin{align}
\left\|e^{tA} x_0\right\| &= \left\|e^{(t-k)A} \cdot e^{kA} \cdot x_0\right\|
\\ & \leq
\|e^{(t-k)A}\| \cdot \left\|e^{kA} \cdot x_0\right\|
\\ & \leq
e^{\|(t-k)A\|} \cdot \left\|e^{kA} \cdot x_0\right\|
\\ & \leq  e^{\|A\|} \cdot \left\|e^{kA} \cdot x_0\right\| = e^{\|A\|} \cdot \|x(k)\|.
\end{align}
$$
In other words, we have $0 \leq \|e^{tA}x_0\| \leq e^{\|A\|} \cdot \|x(\lfloor t\rfloor )\|$. If $\lim_{k \to \infty}\|x(k)\| = 0$, then it follows from the squeeze theorem that $\lim_{t \to \infty} \|e^{tA}x_0\| = 0$, which is to say that the continuous time system is asymptotically stable.

Title: Dynamical System problem


答:
The Maclaurin expansion of sine is $\sin(y) = y - y^3/3! + O(y^5)$, so substituting $y = x\sqrt x$ we have $$\sin(x^{3/2})  =x^{3/2} - x^{9/2}/6 + O(x^{15/2}).$$ Hence your limit reduces to

$$\lim_{x \to 0^+} \frac{-x^{9/2}}{6x\sqrt{x+1}} = \lim_{x \to 0^+} \frac{-x^{7/2}}{6\sqrt{x+1}} = 0. $$

Title: Solving limit which can't be (maybe I'm wrong) solved by L'Hospital

Body:
I'm unable to solve the following limit:
$$\lim_{x \to 0+}\frac{\sin( x \sqrt x )-x \sqrt
x}{x\sqrt{x+1}}$$

My attempt:
Using L'Hospital rule:
$$\frac{ \cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x }  { \sqrt{x+1} + x \frac{1}{2\sqrt{x+1}} }$$
$$\frac{\sqrt{x+1} (\cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x)}{2x+1}$$
Since $(\sqrt x)'=\frac{1}{2\sqrt x}$  we can see that we will get $\frac{0}{0}$ when $x \to 0$ if my calculations are right.


问:
Series of Fractional Sums
What is the sum of $$\frac{1}{2\times 5} + \frac{1}{5\times 8}+ \frac{1}{8\times 11}+...+ \frac{1}{2009\times 2012}?$$

What is the easiest way to solve this kind of problem?

答:
HINT:

$$\frac1{r(r+3)}=\frac13\left(\frac1r-\frac1{r+3}\right)$$

It leaves behind a nice telescopic sum

Title: Series of Fractional Sums


问:
Cartesian relation $\preceq$ on $N\times N$ where $(a,b)\preceq (c,d)$ if $a\leq c$ and $b\leq d$.
Studding for midterm. Lets define the Cartesian relation $\preceq$ on $N\times N$ where $(a,b)\preceq (c,d)$ if $a\leq c$ and $b\leq d$.

I'm trying to prove that the partially arranged set $\left(N \times N, \preceq\right)$ maintains the minimum condition.

As I understand $\left(A,\leq\right)$ is partially arranged set that maintains the minimum condition if for each $B\subseteq A$, $B\neq \varnothing$ contains a minimal number (at least one).

I understand the definition but how do I use it to prove the statement? Do I need to use induction for that?

答: HINT: Suppose that $A\subseteq\Bbb N\times\Bbb N$ has no $\preceq$-minimal element. Then there must be a sequence $\big\langle\langle a_n,b_n\rangle:n\in\Bbb N\big\rangle$ such that $\langle a_{n+1},b_{n+1}\rangle\prec\langle a_n,b_n\rangle$ for each $n\in\Bbb N$. Now consider the sequence $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ to get a contradiction.


Body:
Use $\sin z \to z-z^3/6$ if $z|<<1$
$$L=\lim_{x \to 0+}\frac{\sin( x \sqrt x )-x \sqrt
x}{x\sqrt{x+1}}= \lim_{x \to 0^+} \frac{x^{3/2}-x^{9/2}/6-x^{3/2}}{x\sqrt{x+1}}.$$
$$\implies L=\lim_{x \to 0^+} \frac{-x^{7/2}/6}{\sqrt{x+1}}=0.$$


Body:
Hint: Expand $\sin(x^{3/2})$ near $0$ and you get:

$$\lim_{x \to 0^+}\frac{\sin( x^{3/2})-x^{3/2}}{x\sqrt{x+1}} = \lim_{x \to 0^+}\frac{x^{3/2} - \frac{1}{6}x^{-9/2} + o(x^{13/2}) - x^{3/2}}{x\sqrt{x+1}} = \dots$$


问:
he Modified Principle of Mathematical Induction.
Suppose that  $S$ is a subset of $\mathbb{N}$ that satisfy

(i) $a \in S$ for some $a \in \mathbb{N}$

(ii) $\operatorname{If} k \geq a, k \in S,$ then $(k+1) \in S$

then $S=\{n \in \mathbb{N} \mid n \geq a\}$

By the way we define $\mathbb{N}$ as the intersection of all the inductive sets which contains the $1$

Im very stuck and I have no idea what to do, any hint?

答:
I assume that you want to show a modified version of the induction principle (since your title is the modified principle of mathematical induction).
Which states something like this:

Let $a\in\mathbb{N}$ and for all $a\leqslant n$ let $A(n)$ be an assertion.
Suppose:

(i) $A(a)$ holds

(ii) For all $a\leqslant n$ be true that if $A(n)$ holds, $A(n+1)$ holds too.

Then $A(n)$ is true for all $a\leqslant n$.

To prove this variant of induction set $N:=\{n\in\mathbb{N}|A(n+a) holds \} $ and use the induction principle.


问:
Isomorphism between non Banach space and Banach space
I know that a non complete normed space cannot be topologically isomorphic to a complete normed space. However there exist bijective and bounded linear mapping from a Banach space onto a non Banach space.

For example, identity mapping from C([a , b]) with sup norm onto itself with L2 norm. Because the topology on the former is strictly finer than the topology on later.

But I am finding it difficult to find a bijective and bounded linear mapping from a non Banach space onto a Banach space. Is there any such example and if not then how to prove it. May be I am missing some simple result. Any help please. Thanks in advance.

答: The following Example is taken from Bühler and Salamon's book "Functional Analysis". You can find it on page 57, Example 2.2.8. It is preceeded by two other examples that show that in the open mapping theorem and the inverse operator theorem for Banach spaces, the assumption that the source and the target are complete, cannot be removed.


答:
I see a lot of answers using Taylor expansions, but this is overkill.

Since $|\sin u|\le |u|$ you get $|\sin u-u|\le 2|u|\quad$ (use for $u=x\sqrt{x}$).

For the denominator $\sqrt{x+1}\ge 1$ when $x\to 0^+$

Therefore
$$\left|\dfrac{\sin(x\sqrt{x})-x\sqrt{x}}{x\sqrt{x+1}}\right|\le\left|\dfrac{2x\sqrt{x}}{x}\right|=2\sqrt{x}\to 0$$

Sometime using just rough inequalities is enough to do the job.


答:
Dividing numerator and denominator by $x\sqrt x$ gives$$
\lim_{x\to0+}\frac{\frac{\sin (x\sqrt x)}{x\sqrt x}-1}{\sqrt{\frac{x+1}x}}=0$$ because $\lim_{x\to0}\frac{\sin x}x=1$


问:
Square-Trangular Numbers Checking Answer
Problem: The first 2 numbers that are both squares and triangles are 1 and 36. Find the next one and if possible, the one after that.

Answer: 1225, 41616

Problem: Can you figure out an efficient way to find triangular-square numbers?

Answer: $s^2 = t(t+1)/2$, where $s,t \epsilon{} \textbf{Z}$.

Problem: Do you think that there are infinitely many?

Answer: Yes. (Can someone give me an justification for my answer?)

答:
The condition for finding such numbers is to use the square-root-of-two approximation.

```<code>   0    1    2    5   12   29   70   169      A   A' = A+B
   1    1    3    7   17   41   99   239      B   B' = A+A'
</code>```

The square roots of the numbers you require to be both square and triangular, are then the product of the two numbers in the same column, ie $AB$.

So, for example, $70*99 = 6930$, $6930*6930 = 48024900$, while $(9800 * 9801)/2$ gives the same result.  So the triangular number of 9800 and the square of 6930 gives the same answer.

One sees that in the values of 9800 and 9801, that $9800=2*70^2$ and $9801=99^2$.  The ratio of $99/70$ is an approximation to $\sqrt{2}$, which the iterative process above does.

The ratio between the numbers in this table settles down to $\sqrt{2}+1$, which is the so-called silver ratio.

It produces them all, and no others.

Body:
The condition for finding such numbers is to use the square-root-of-two approximation.

   0    1    2    5   12   29   70   169      A   A' = A+B
   1    1    3    7   17   41   99   239      B   B' = A+A'

The square roots of the numbers you require to be both square and triangular, are then the product of the two numbers in the same column, ie $AB$.

So, for example, $70*99 = 6930$, $6930*6930 = 48024900$, while $(9800 * 9801)/2$ gives the same result.  So the triangular number of 9800 and the square of 6930 gives the same answer.

One sees that in the values of 9800 and 9801, that $9800=2*70^2$ and $9801=99^2$.  The ratio of $99/70$ is an approximation to $\sqrt{2}$, which the iterative process above does.

The ratio between the numbers in this table settles down to $\sqrt{2}+1$, which is the so-called silver ratio.

It produces them all, and no others.


Body:
$$L(x)=\frac{\sin(x^{3/2})-x^{3/2}}{x\sqrt{x+1}}$$
we know that:
$$\sin(x)=x-\frac{x^3}{6}+...$$
and so:
$$L(x)=\frac{x^{9/2}/6+...}{x\sqrt{x+1}}=\frac{x^{7/2}+...}{6\sqrt{x+1}}$$
now if you take $x\to0$ it is clear that $L(x)\to0$


问:
Is the rule for squaring differences a "degenerate" case of the law of cosines?
Compare $(a-b)^2=a^2+b^2-2ab$ (the square of a difference) with $c^2=a^2+b^2-2ab\cos{C}$ (the law of cosines), and you will notice the similarity.

So, if one has a (degenerate) "triangle" with three sides $a,b,$ and $c$ for which $c=a-b$, does that mean that the measure of the angle opposite the side with length $c$ is zero degrees, since $\cos{0^{\circ}}=1$?

答: Yes (as long as $c$ is the longest side), as $\Bbb R$ is a model of Euclidean geometry, with $ab$ the dot product. If $a,\,b$ have the same sign, $C=0^\circ$; if they have different signs, $C=180^\circ$, but the "$ab$" in the cosine rule really means $|a||b|$, which in normal multiplication terms is $-ab$, so it works out.


问:
Can we map a complete metric space to $\mathbb{Q}$?
The question I am trying to answer is this:

Given a complete metric space $(X,d)$, in which all of the points are accumulation points. Find if there is a one-to-one and onto map $f: X \rightarrow \mathbb{Q}$.

My initial thoughts are:

Given that $(X,d)$ is complete, we know that all Cauchy sequences in $X$ have a limit also in $X$.

We can easily prove that $\mathbb{Q}$ is not complete, since we can easily find a sequence in $\mathbb{Q}$ which has a limit in $\mathbb{R}\setminus\mathbb{Q}$.

So my thought is that there is no such map $f$. But how do I formally prove it?

答:
If $X$ is a complete metric space in which every point is an accumulation point, it  follows from Baire's category theorem that $X$ is uncountable.

Indeed, assume by contradiction that $X=\{x_1,x_2,\dotsc\}$. Then $X$ is the union of the closed sets $X_n=\{x_n\}$. It follows from Baire's category theorem that there is some $n$ such that $X_n$ has nonempty interior in $X$. It follows that $x_n$ is an isolated point of $X$.


答: Take $T \subseteq N \times N$. Define $T_a= \{a \in N : \exists b \  s.t. (a,b) \in T\}$. Since $N$ is well ordered exists $a_{min}=\min T_a$. Set $T_b=\{b \in B: (a_{min},b)\in T\}$.  Since $N$ is well ordered exists $b_{min}=\min T_b$. Then it is easy to see that $(a_{min},b_{min})$ is a minimal element of $T$

Body:
Studding for midterm. Lets define the Cartesian relation $\preceq$ on $N\times N$ where $(a,b)\preceq (c,d)$ if $a\leq c$ and $b\leq d$.

I'm trying to prove that the partially arranged set $\left(N \times N, \preceq\right)$ maintains the minimum condition.

As I understand $\left(A,\leq\right)$ is partially arranged set that maintains the minimum condition if for each $B\subseteq A$, $B\neq \varnothing$ contains a minimal number (at least one).

I understand the definition but how do I use it to prove the statement? Do I need to use induction for that?


问:
Thinking about a uniform probability distribution over the natural numbers
I well know that in standard probability theory there is no way to define a uniform probability distribution over the natural numbers in the strict sense, but nonetheless, I'm trying to form an intuitive understanding of how such a distribution ( here I'm appealing more to an intuitive idea of probability than the formal one )  should behave, i.e. what should I expect to see, in an aggregated and structural sense, if I were to extract independently an infinite sequence of numbers using this distribution.

My intuition says that if I restrict myself to sequences of ordinal length $\omega$ (simple infinite sequences) then I should expect to observe nothing but pure chaos, with no regularity whatsoever. But if I were to keep extracting (independently, again in a not-quite-formal sense) numbers to obtain a sequence of ordinal length $\omega_1$, let's call it $\langle x_\alpha\rangle_{ \alpha < \omega_1}$, then great regularity would emerge. In particular, I would almost surely expect to see every natural number appearing in this uncountable sequence, and moreover, I would expect every natural number to appear cofinely many times (i.e. $\forall n \in \omega \  \forall \alpha < \omega_1 \ \exists \beta \in [\alpha, \omega_1)$ s.t. $x_\beta = n$). Also there would be no way to distinguish two numbers based on their appearances in the sequence (here the uniform character of the distribution).

Hoping that what I wrote is not too vague, my questions are:

Is this intuition reasonable and acceptable in some way?

Have some attempts been made to formalize this intuition?

More in general I'm getting interested in a critical understanding of what are the true reasons (of various nature) behind the fact that we could not develop a rich theory of probability comprising these "pathological" distributions and also other fringe notions like infinitesimal probabilities (I think these two concepts are intimately related). Is there some literature in this regard? I'm not asking for references to the thousands of non-standard probability theories available, but rather for a study of why these non-standard theories, which in general try to formalize some of the notions written above, all have serious and non-overseeable flaws that make them of no "practical" use, as if there are some structural and profound difficulties in formalizing these notions.

Thanks!

EDIT1: I know that by dropping countable additivity we could define a uniform distribution over the naturals, but dropping such an axiom has a serious (I'd dare to say devastating) impact on the mathematical power of the theory (an example of the flaws I was talking about). Also it does not shed any light on the intuition I wrote above.

EDIT2 (in response to Noah Schweber):

Regarding the first comment: I know that bringing the transfinite into the picture introduces serious difficulties (and I have no precise idea how to deal with it) but to me it is not simply another layer of difficulty, rather it is the necessary layer at which we are able to say something relevant about a possible uniform distribution over the naturals. More specifically I would expect that the "measure" on $\mathbb{N}^{\omega_1}$, once we formalize this uniform distribution over $\mathbb{N}$, should stem from a product-measure-like construction, as it is usually done in standard probability theory.

Regarding the second comment: I didn't say that dropping countable additivity is unnatural, what I say is that by dropping it we'd lose a lot of the theory's mathematical power, i.e. we could develop only very small fragments of Lebesgue's integration theory and would not be able to achieve even the most basic results in stochastic process' theory. And even if we were not interested in such theories, I don't see how we could say anything relevant regarding the uniform distribution over the naturals besides giving it a semblance of formal definition, in particular the intuition about the cofinality of all naturals in the $\omega_1$ sequence (which seems to me a critical property of such distribution) would not be even formalizable, let alone assessable.

答:
I don't know if this helps you but although you cannot define a uniform probability measure in a countable set (easy to prove), you can define a uniform finitely additive probability measure on the naturals: Take a set $A\subset \mathbb N$, and the counting sequence of $A$ defined by $p_A^n=\#\{x\in A:a\leq n\}$, the measure would be $\mu(A)=\lim_n\frac{p_A^n}{n}$, you can easily see that this is finitely additive and $\mu(\{n\})=0$ for every $n\in\mathbb N$.

This is also a very intuitive measure, for example finite sets have measure $0$ and $\mu(n\mathbb N)=\frac{1}{n}$


问:
Solution check: group action.
Let $G=D_{1000}$ (dihedral group) acts transitive on some set $M$ such that for some $x \in M$ $St(x) = \langle \frac{\pi}{10}, S \rangle$, where $S$ is a symmetry. Find 1) $|M|$, 2) kernel of this action, 3) number of stabilizers.

Soulution.

It's easy to see that $St(x) \cong D_{20}$, and since the action is transitive, it has only one orbit. Then $|M| = |G(x)| = \frac{|G|}{St(x)} = \frac{2000}{40} = 50$.

$\ker G = \{g \in G | (\forall y \in M) g(y) =y\} = \cap_{y \in M}St(y) = $ (since M is one orbit all stabilizers are conjugated) $= \cap_{g \in G}(St(x))^g =^? \require{cancel} \cancel{St(x)}$ <s>(it seems to me that $D_{20}$ is a normal subgroup of $D_{1000}$).</s>

UPD. $R_1RR_1^{-1} = R, SRS^{-1} = SRS = R^{-1}$, so we should look only at conjugations of symmetries. It's some exercise to show that for even n the conjugacy class of symmetry is all symmetries of the same type (half of the all symmetries). So, $\ker G$ consists of half of all symmetries and $\langle \frac{\pi}{10} \rangle$

Let G acts on its subgroups with conjugations. Then
$\#St = |G(St(x))| = \frac{|G|}{|N_{G}(St(x))|} = \require{cancel} \cancel{1}$, where $ N_{G}(St(x)) $ is the normalizer of $St(x)$ in $G$.

答:
$\newcommand{\T}{\mathcal T}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\St}{\mathrm{St}}$We can safely forget about $M$. Orbit-stabiliser tells you that you considering the action of $G$ on the (right, say) cosets of the stabiliser $\T = \Span{ \dfrac{2 \pi}{20}, S }$.

The kernel of the action is, as you correctly state, the intersection of all conjugates of $T$. This is the largest normal subgroup contained in $\T$. Now you noted that $\T$ is not normal in $G$, so the kernel will be $K = \Span{\dfrac{2 \pi}{20}}$. The latter is normal, because all subgroups of the cyclic group $\Span{\dfrac{2 \pi}{1000}}$ are characteristic in it.

As to the normaliser of $\T$, let $g = \dfrac{2 \pi}{1000}$. Then if $e = 0, 1$ we have
$$
S^{S^{e} g^{i}}
=
g^{-i} S^{e} S S^{e} g^{i}
=
S S g^{-i} S g^{i}
=
S g^{2 i}.
$$
Since $\dfrac{2 \pi}{20} = g^{50}$, we have that $S^{e} g^{i}$ normalises $\T$ iff $i$ is a multiple of $25$. Since $g^{25} = \dfrac{2 \pi}{40}$, it follows that the normaliser is $\Span{\dfrac{2 \pi}{40}, S}$, of order $80$.

Title: Solution check: group action.


问:
When is 2 a cubic residue mod p?
It's equivalent to compute the primes $p$ such that 2 is not a cubic residue mod $p$. Let p be a prime, now the the map $ \rho: \mathbb{F}_p^\times \rightarrow \mathbb{F}_p^\times $ given by $\rho(n)=n^3$, is not a bijection if and only if it has non-trivial kernel, this only happens if $x^2+x+1=0$ has a solution mod $p$. But this is possible only if $ \left( \frac{-3}{p} \right) =1 \Leftrightarrow p= 1 $ mod $3$.

I started computing the primes under $300$ in which $2$ is not a cubic residue and these are: $$7, 13, 19, 37, 61, 67, 73, 79, 97, 103, 139, 151, 163, 181, 193, 199, 211, 241, 271.$$ Approximately $2/3$ of the primes have the desired property, hence I was thinking we needed to use congruences mod $3^n$, for some $n$, but I'm stuck. Does anyone know to find the answer?

答:
The primes that are $1 \pmod 3$ for which $2$ is a cubic residue are all of form
$$  p= x^2 + 27 y^2 $$
with integers $x,y$

The primes that are $1 \pmod 3$ for which $2$ is not a cubic residue are all of form
$$  q= 4u^2 + 2uv + 7 v^2 $$
with integers $u,v,$  and we allow them negative or positive as needed

Title: When is 2 a cubic residue mod p?


问:
Finding the only multiple zero of a quadratic function
I don't know how to figure this problem out. Can someone please help?

For what value of $\;b\;$ will $\;f(x)=x^2 + bx + 400\;$ have $\;-20\;$ as its only zero?

答:
For a quadratic equation having one root, its discriminant is zero, i.e.

$$b^2- 4\cdot 400=0$$

which leads to $b=\pm40$. Check to determine that $b=40$ from the given root $-20$.


问:
General term of a sequence and its limit.
Having a sequence $\;\sqrt[3]5\;,\;\sqrt[3]{5\sqrt[3]5}\;,\;\sqrt[3]{5\sqrt[3]{5\sqrt[3]5}}\;,\;\ldots\;,\;$ what could be the general term and its limit ?

Note that the second term is cube root of $5$ and inside that there is cube root of $5$ again and the third term is cube root of $5$ and two cube root of $5$ inside the outer cube root of $5$.

I tried to define it recursively as

$c_{n+1}=\sqrt[3]{5c_n}\;$ where $\;c_1=\sqrt[3]5\;.$

答:
Observe the pattern that, if the $n$th term of the sequence is denoted $c_n$, then

$$c_n = 5^{1/3 + (1/3)^2 + (1/3)^3 + \cdots + (1/3)^n}$$

To see this, notice that, for instance:

\begin{align*}
c_2 &= \sqrt[3]{ 5 \sqrt[3] 5} = \left( 5 \cdot 5^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} = 5^{1/3 + 1/3^2} \\
c_3 &= \sqrt[3]{ 5 \sqrt[3]{5\sqrt[3]{5}}} = \left( 5 \cdot \left(5 \cdot 5^{1/3}\right)^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} \cdot 5^{1/3^3} = 5^{1/3 + 1/3^2 + 1/3^3} \\
\end{align*}

and so on and so forth. The exponent of $c_n$ is a geometric series with ratio $1/3$ (just starting at $1/3$ and not $1$), and thus

\begin{align*}
\sum_{k=1}^n \left( \frac 1 3 \right)^k
&= \frac{(1/3)^{n+1} - 1}{1/3 - 1} - 1\\
&= \left( -\frac 3 2 \right)\cdot \frac{1}{3^{n+1}} - \left( -\frac 3 2 \right)\cdot1 - 1\\
&= \frac 1 2 \left( 1 - \frac{1}{3^n} \right)
\end{align*}

and so

$$c_n = 5^{(1 - 3^{-n})/2}$$

Taking the limit $n \to \infty$ is trivial enough, and gives the limit $c_n \to 5^{1/2} = \sqrt 5$.


问:
Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$.
Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.

Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sqrt{3n+3}}$$ $$...$$ $$1<0.$$ After embarrassing myself, I looked around and I found this thread. Using induction, we can then easily prove $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}.$$ This gets me to the original problem. But in a problem solving standpoint, how do you think to use $\frac{1}{\sqrt{3n+1}}$? Is there some point in the first induction that leads to this idea? Or is there a better method than the above?

答:
Notice that
$$\frac{1}{\sqrt{an+b}} \cdot \frac{2n+1}{2n+2} \le \frac{1}{\sqrt{a(n+1)+b}} \\
\iff (a(n+1)+b)(2n+1)^2 \le (2n+2)^2 (an+b) \\
\iff an+a-4bn-3b \le 0$$

Hence if $a=3$, then $b=1$ would work. Of course, you need to prove the initial case ($n$=1).

BTW: how amazing it is that the first two answers got $e$ and $\pi$, respectively.


问:
determining the limit of an integral
Let $f(x):\mathbb{R} \to \mathbb{R}$ be a continuous function, for every $n > 0$ define:

$$a_n = \int_{\frac{-1}{n}}^{\frac{1}{n}} f(x) \,dx$$

determine $\lim\limits_{n\to\infty} a_n$

At first glance it would seem like the limit approaches zero, is that statement true?

答:
Here's another proof.
We know that since $f$ is continuous we may apply the mean value theorem for integrals which guarantees for each of the given intervals $[-\frac{1}{n}, \frac{1}{n}]$ there is a number $c_n$ (depending upon $n$) such that
$$f(c_n)(\frac{1}{n}+\frac{1}{n})=\int_{-\frac{1}{n}}^\frac{1}{n} f(x)dx$$
The intervals $[-\frac{1}{n}, \frac{1}{n}]$ will shrink to the point $0$ in the limit (need to justify this by finite intersection property or something else but it's true)
so we see that the sequence $c_n\to 0$ as $n\to \infty$.

Again since $f$ is continuous $\lim_{n\to \infty} f(c_n)=f(\lim_{n\to \infty} c_n)=f(0)$

Therefore $\lim_{n\to \infty} \frac{2f(c_n)}{n}=0$ and we done.

Title: determining the limit of an integral


问:
Derivatives pricing w/ squared and cubed stock prices
I have an assignment in which $S_t$ is a stock price following a geometric Brownian motion. The task is now to show that at time t the risk-neutral price of a derivative on $S_t$ that pays $S_T^3$ at time T is

$$S_t^3 e^{(2r + 3\sigma^2)(T-t)}$$

Next, I need to work out the risk-neutral price of a derivative that pays $(S_T^2 - K)^+$ at time T. This is also in a continuos time setting.

Any help much appreciated.

答:
For an integer $n>1$, let $D_n(t,T)$ be the price at $t\leq T$ of the derivative that pays $S_T^n$ at $T$.

The risk-Neutral distribution with respect of the stock numeraire is given by $\ln\mathcal{N}\left(\ln S_t+\left(r+\frac{1}{2}\sigma^2\right)(T-t),\sigma^2(T-t)\right)$. So now apply the martingale for $\frac{D_n(t,T)}{S_t}$.

$$E(\frac{D_n(T,T)}{S_T}|S_t)=\frac{D_n(t,T)}{S_t}\implies D_n(t,T)=S_tE(S_T^{n-1}|S_t)$$

Use the formula for $E(X^k)=e^{k\mu+\frac{k^2}{2}\sigma^2}$ for the moments of $X\sim \ln\mathcal{N}(\mu,\sigma^2)$.

$D_n(t,T)=S_t^ne^{(n-1)\left(r+\frac{1}{2}\sigma^2\right)(T-t)+\frac{(n-1)^2}{2}\sigma^2(T-t)}=S_t^ne^{(n-1)(r+\frac{n}{2}\sigma^2)(T-t)}$


问:
Help needed in understanding an argument in Theorem 2.47 of topology by Wayne Patty
I am self studying topology from Wayne Patty and I have a question in Theorem 2.48 on page 90. Subsection is Weak topology and product topology.

The question is that I am not able to understand how $\mathscr B$ is countable in 4th and 5th line of the proof. It is clear to me that for all $\alpha \in \Lambda- \Gamma$, $\pi_{\alpha}^{-1} (B_{\alpha}) =X$. But still I am not able to deduce how $\mathscr B$ is countable.

Please help me with that!

答:
(You may already have realized this, but note that there is a typo in the fourth line: it should read ‘where $\alpha\in\Lambda$ and $B_\alpha\in\mathscr{B}_\alpha$’.)

For each $\alpha\in\Lambda$ let

$$\mathscr{S}_\alpha=\left\{\pi_\alpha^{-1}[B]:B\in\mathscr{B}_\alpha\right\}\,,$$

and let

$$\mathscr{S}=\bigcup_{\alpha\in\Lambda}\mathscr{S}_\alpha\,.$$

If $\alpha\in\Lambda\setminus\Gamma$, then $\mathscr{B}_\alpha=\{X_\alpha\}$, so $\mathscr{S}_\alpha=\left\{\pi_\alpha^{-1}[X_\alpha]\right\}=\{X\}$. If $\alpha\in\Gamma$, then $\mathscr{B}_\alpha$ is countable, so $\mathscr{S}_\alpha$ is also countable. Thus,

$$\mathscr{S}=\{X\}\cup\bigcup_{\alpha\in\Gamma}\mathscr{S}_\alpha$$

is the union of countably many countable sets and is therefore countable. Finally, Patty’s $\mathscr{B}$ is the set of all intersections of finite subsets of $\mathscr{S}$. $\mathscr{S}$, being countable, has only countably many finite subsets, so $\mathscr{B}$ is also countable.


问:
Proving a quantified statement wrong
The domain is the natural numbers set.
I want to prove this statement wrong, however, I'm not sure how should I go about that since, theoretically, I would have to go through all y values in order to find at least one that would be fit for every single x value. Can anyone give me a tip?
Here's the statement:

∃y∀x(2x - y = 0)

答: Asserting that$$\exists y\forall x(2x-y)=0$$is false is the same thing as asserting that$$\forall y\exists x(2x-y\ne0)$$is true. And it is indeed true: for any $y$, just take $x=y$ if $y\ne0$ and $x=1$ if $y=0$.

Title: Proving a quantified statement wrong


问:
Calculating a partial sum
can somebody help me to explain why the partial sum of
$$\sum\limits_{j=1}^n \frac{j}{(j+1)!}=1-\frac{1}{(n+1)!}.$$

I really appreciate any help you can provide!

Title: Calculating a partial sum

Body:
$$\sum\limits_{j=1}^n \frac{j}{(j+1)!}=\sum\limits_{j=1}^n \frac{(j+1) - 1}{(j+1)!} = \sum\limits_{j=1}^n \bigl( \frac{1}{j!}
- \frac{1}{(j+1)!} \bigr) = \sum\limits_{j=1}^n (a_j - a_{j+1})$$
Hence
$$\sum\limits_{j=1}^n \frac{j}{(j+1)!} = (a_1 - a_2) + (a_2-a_3) + \ldots + (a_{n} - a_{n+1}) = $$
$$=a_1 - a_{n+1} =1-\frac{1}{(n+1)!}.$$


问:
How to calculate the probability mass function of $X_N$, the number of people getting back their own hat
How do I calculate the pmf of $X_N$, where $X$ is the number of people out of $N$ getting back their own hat after a random hat exchange?

How can I calculate it without listing all the possible outcomes?

答:
Hint: For $0\leq n\leq N$ : $P(X_{N}=n)$ is the probability that
exactly $n$ people got their hat back.

In how many ways this can be done ?

If want to fix some $n$ hats
in their place (in how many ways can we choose them ?) and dearrange
the other $N-n$ hats (in how many ways can you do that ?)


答:
Mean value theorem for integrals :

$a_n=f(t) \int_{-1/n}^{1/n}1dx=(2/n)f(t)$, where $t \in [-1/n,1/n]$.

$\lim_{n \rightarrow \infty}(2/n) f(t)=$

$\lim_{n \rightarrow \infty} (2/n)\lim_{n \rightarrow \infty} $f(t) =0f(0)=0;$

Used: $\lim_{n \rightarrow \infty} f(t) =0$, since

$t \rightarrow 0$, and $f$ is continuos.

Title: determining the limit of an integral


问:
How to prove these two ways give the same numbers?
How to prove these two ways give the same numbers?

Way 1:

```<code>Step 1 : 73 +  1 =  74. Get the odd part of  74, which is 37
Step 2 : 73 + 37 = 110. Get the odd part of 110, which is 55
Step 3 : 73 + 55 = 128. Get the odd part of 128, which is  1
</code>```

Continuing this operation (with 73 + 1) repeats the same steps as above, in a cycle.

Way 2:

```<code>Step 1:  (2^x) * ( 1/73) > 1 (7 is the smallest number for x)
(2^7) * ( 1/73) - 1 = 55/73

Step 2:  (2^x) * (55/73) > 1 (1 is the smallest number for x)
(2^1) * (55/73) - 1 = 37/73

Step 3:  (2^x) * (37/73) > 1 (1 is the smallest number for x)
(2^1) * (37/73) - 1 =  1/73
</code>```

Repeating the steps with the fraction 1/73 goes back to step 1, and repeats them in a cycle.

The two ways have the same numbers $\{1, 37, 55\}$ in the 3 steps. How can we prove that the two ways are equivalent and give the same number of steps?

答:
Let $M=37$ (or any odd prime for that matter).

To formalize your first "way":
You start with an odd number $a_1$ with $1\le a_1<M$ (here specifically: $a_1=1$) and then recursively let $a_{n+1}=u$, where $u$ is the unique odd number such that $M+a_n=2^lu$ with $l\in\mathbb N_0$.
By induction, one finds that $a_n$ is an odd integer and $1\le a_n<M$

To formalize your second "way":
You start with $b_1=\frac c{M}$ where $1\le c<M$ is odd (here specifically: $c=1$) and then recursively let $b_{n+1}=2^kb_n-1$ where $k\in\mathbb N$ is chosen minimally with $2^kb_n>1$.
Clearly, this implies by induction that $0< b_n\le 1$ and $Mb_n$ is an odd integer for all $n$.

Then we have

Proposition. If $a_{m+1}=M b_n$, then $a_m=M b_{n+1}$.

Proof:
Using $b_{n+1}=2^kb_n-1$, $M+a_m=2^la_{m+1}$, and  $a_{m+1}=M b_n$, we find
$$Mb_{n+1}=2^kMb_n-M = 2^ka_{m+1}-M=2^{k-l}(a_m+M)-M.$$
If $k>l$, we obtain that $Mb_{n+1}\ge 2a_m+M>M$, contradicting $b_{n+1}\le 1$.
And if $k<l$, we obtain $Mb_{n+1}\le \frac12 a_m-\frac 12 M<0$, contradicting $b_{n+1}>0$. Therefore $k=l$ and
$$ Mb_{n+1} = a_m$$
as was to be shown. $_\square$

Since there are only finitely many values available for $a_n$ (namely the odd naturals below $M$), the sequence $(a_n)_{n\in \mathbb N}$ must be eventually periodic, that is, there exists $p>0$ and $r\ge1$ such that $a_{n+p}=a_n$ for all $n\ge r$. Let $r$ be the smallest natural making this true. If we assume $r>1$, then  by chosing $c=a_{r-1+p}$ in the definition of the sequenc $(b_n)_{n\in\mathbb N}$ we can enforce $Mb_1=a_{r-1+2p}=a_{r-1+p}$ and with the proposition find $Mb_2=a_{r-1+p}=a_{r-1}$ contradicting minimality of $r$. We conclude that $r=1$, that is the sequence $(a_n)_{n\in\mathbb N}$ is immediately periodic.

Now the proposition implies that the sequence $(b_n)_{n\in\mathbb N}$ is also immediately periodic: Let $a_1=Mb_1$. Then by periodicity of $(a_n)$, we have $Mb_1=a_{1+p}$, by induction $Mb_k=a_{2+p-k}$ for $1\le k\le p+1$.
Especially, $b_{p+1}=b_1$ and hence by induction $b_{n+p}=b_n$ for all $n$.

Finally, we use the fact that $M$ is prime. Therefore the $Mb_n$ are precisely the numerators of the $b_n$. Our results above then show that these numerators are (if we start with $b_1=\frac{a_1}M$) precisely the same periodic sequence as $(a_n)$, but walking backwards. This is precisely what you observed.

EDIT: As remarked by miket, $M$ need only be odd but not necessarily prime. To see that, one must observe that the $a_n$ are always relatively prime to $M$ if one starts with $a_1$ relatively prime to $M$. Consequently, the $Mb_n$ are still the numerators of the $b_n$ (i.e. their denominators are $M$ in shortest terms).


问:
Group algebra $F[G]$
While studying I came across the concept group algebra which is defined as
$$F[G]=\{\sum_{g \in G} \alpha_{g}g : \alpha_{g} \in F \text{ and } g \in G\}.$$

Can anybody tell what are the elements of this ring?

For example $a$ is an element so it is equal to $a= \alpha_{g_{1}}g_{1}+\alpha_{g_{2}}g_{2}.....+\alpha_{g_{n}}g_{n}$,
what is $\alpha_{g_{1}}$? why is the script $g$ used?

and if you can help me with this question:

$V$ is irreducible $C[G]$ module and $a$ is an element of the center of $C[G]$, $a \in Z(C[G])$. I wanna show that for every element of $V$ there exists element $\gamma \in C$ s.t. $v a = \gamma v$ and $C$ is the set of complex numbers.

Here is my attempt:

I define the map $\theta:V\rightarrow V$ which takes the element $v\in V$ to its image $va\in V$.
$\theta$ is an endomorphism  and its inverse is also an endomorphism ($V$ is irreducible.)

so I try to make use of that. Do you think my appraoch is right?

I'm sorry for the bad writing but it is the first time to write using the notations.

答:
The idea is to take only formal (finite) sums of formal products. An example can be useful to grasp the concept: Lets take the ring $\mathbb Z_4 = \{0,1,2,3\}$ and the group $S_3 = \{e,x,x^2,y,xy,x^2y\}$. Then the elements of $R[G] = \mathbb Z_4[S_3]$ are things like $u = 3x+2xy$ or $v = 2x+3x^2y$. How do you add them? like this $u+v = (3+2)x+2xy+3x^2y = x+2xy+3x^2y$ (since $3+2=1$ in $\mathbb Z_4$). How do you multiply them? Distribute, multiplying the coefficients with the product of $\mathbb Z_4$ and the group elements with the group operation: $u v = (3x+2xy)(2x+3x^2y) = 3\cdot 2 x^2+3\cdot 3 x^3y + 2\cdot 2xyx+2\cdot 3xyx^2y = 2x^2+y+2x^2$  (since in $\mathbb Z_4$ we have $3\cdot2 = 2$, $3\cdot3=1$, $2\cdot 2 = 0$ and $2\cdot 3=2$ and in $S_3$ we have $x^3y = y$ and $xyx^2y = x^2$) which simplifies even more because $2x^2+2x^2 = 0$, so $u v= y$.

In a more general setup you represent each element as a sum of elements of $G$ with coefficients of $R$: $$u = \sum_{g\in G}r_g g$$ If $G$ is infinite you restrict to the formal sums that have only a finite number of coefficients not $0$.

You add them and multiply them as in the example.

Title: Group algebra $F[G]$


问:
Finding generalized inverses in the ring of linear transformations of a finite dimensional vector space V over a division ring
Let $L$ be the ring of linear transformations of a finite dimensional vector space $V$ over a division ring $D$. Show that for any $l$ belonging to $L$, then there exists a $u\in L$ such that $lul=l$.  Such $u$ is called a generalized inverse.

A question of Jacobson’s basic algebra p173, something related to Morita context or Wedderburn-Artin theorem. It can be related in languages of matrix, and maybe it can be solved by some matrix work, but I want to know the solution following the hint of several former questions.

答:
The following are equivalent:

For all $a\in R$, there exists an $x$ such that $axa=a$ (such rings are called von Neumann regular)

Every principal right ideal of $R$ is a summand.

Since this ring is isomorphic to a square matrix ring over $D$, it's a semisimple ring: all right ideals are summands.  This is perhaps where the Artin-Wedderburn theorem is relevant. Morita theory is a bit too general to mention here: it's an extension of what the Artin-Wedderburn theorem says.

You can also do it by just brute force. Pick a basis $\beta=\{l(v_1),\ldots, l(v_n)\}$ for $l(V)$, and extend it to a basis of $V$. Define $u$ by declaring that $u(l(v_i))=v_i$ and $l(b)=0$ for the other basis elements.

Then for every $y\in V$,
$$
l(y)=\sum \lambda_il(v_i)
$$

and now applying $u$

$$
ul(y)=\sum\lambda_iul(v_i)=\sum\lambda_iv_i
$$

and applying $l$ again

$$
lul(y)=\sum\lambda_il(v_i)=l(y)
$$


问:
Prove that $1+ \frac{5}{4}x ≤(1+x)^{\frac{5}{4}}≤1+ \frac{5}{4} x+\frac{5}{16} x^2$
How to prove this inequality:
$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$

I tried applying binomial theorem to get this inequality, but I am not able to get it precisely. Please help.

答:
Update: I found that my solution doesn't work if $x<0$. So far only nguyenhuyen_ag has a correct and complete (and beautiful) proof. Others are either plain wrong, or only correct when $x\ge0$.

I found another proof for the RHS.

Denote $g(x) = \frac{(1+x)^{5/4}-1-5x/4}{x^2}, x>-1$. It's not difficult to prove $g(x)$ and $g'(x)$ are both continuous at $0$. Since $g'(x)$ is ugly we instead look at $$h(z)=g(z-1)=\frac{z^{5/4}-1-5(z-1)/4}{(z-1)^2}, z>0$$
Then
$$h'(z)=-\frac{3\sqrt z+ 4 \sqrt[4]z + 3}{4(z^{3/4}+\sqrt z + \sqrt[4] z+1)^3} < 0, \forall z>0$$

Therefore $g(x) \le g(-1) = \frac 14 \implies (1+x)^{5/4} \le 1 + \frac{5}{4} x + \frac{1}{4} x^2$. This the same (and better than OP's) result as shown in nguyenhuyen_ag's proof.

My original proof:

Many inequalities of this type can be easily solved via Taylor's theorem with the mean-value forms of the remainder.

$f(x)=(1+x)^{\frac 54}, f''(x)=\frac{5}{16(1+x)^{3/4}} > 0, f^{(3)}(x) = -\frac{15}{64(1+x)^{7/4}} < 0.$

Therefore
$$
f(x) = 1 +\frac{5}{4}x + \frac{f''(\xi_1)}{2!}x^2 \ge 1+\frac{5}{4} x
$$

$$
f(x) - \left( 1 +\frac{5}{4}x + \frac{5}{16}x^2 \right) = 1 + \frac{5}{4}x + \frac{5}{32}x^2+\frac{f^{(3)}(\xi_2)}{3!}x^3 - 1 - \frac{5}{4}x-\frac{5}{16} x^2
\\
=-\frac{15}{6\cdot 64(1+\xi_2)^{7/4}}x^3-\frac{5}{32}x^2 = -\frac{5x^2}{32} \left( 1+ \frac{x}{4(1+\xi_2)^{7/4}}\right) \le 0
$$
where $\xi_1, \xi_2$ are between $x$ and $0$.

Another recent example here.


问:
Prove if we have a square unitary Matrix $Q$, then $\det(Q) = e^{i\theta}$
Prove if we have a square unitary Matrix $Q$, then $\det(Q) = e^{i\theta}$

Using $\det(Q)\det(\bar{Q}^T) = I$, I get to the stage $\det(\bar{Q})\det(Q)=1$, but can't do much else with it.

Thanks for the help.

答:
It's easy to show that $\det(\bar{A})=\overline{\det(A)}$ for any square matrix. Thus you have

$$
\det(\bar{Q})\det(Q)=\overline{\det(Q)}\det(Q)=1
$$

which means that $|\det(Q)|=1$. Any complex number $z$ with $|z|=1$ can be represented as $z=e^{i\theta}$ for some $\theta\in[0,2\pi)$.


问:
Trying to show that $r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right)$ is an ellipse
I am trying to prove that planets move in ellipses,

I watched this video: https://www.youtube.com/watch?v=DurLVHPc1Iw

and read this: https://arxiv.org/pdf/1009.1738.pdf .

But both sources end up with this as an equation for the paths that planets follow :
$$
 r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right),
 $$
where $C=G\frac{m_{1}+m_{2}}{h^{2}}$ which is just a constant, as is $\omega$, and $e$ is the eccentricity, so this is an ellipse with the focus at the origin as long as $e<1$ and when I put the equation in graphing software it gives me ellipse.

But I don't understand why it is an ellipse? Can anyone show me how it is an ellipse? Either from the normal polar coordinates of a ellipse centred at the origin and shifting it (I couldn't get the algebra to work out when I added ae to the x coordinate) or just by explaining it conceptionally?

答:
We have
$$
\begin{split}
r&= \dfrac{1}{C} \left(\dfrac{1}{1+e \cos(\theta+\omega)}\right) \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \cdot \dfrac{1/C}{1/C} \\
&= \dfrac{1/C}{1+ e\cos(\theta+\omega)}
\end{split}
$$
Write $P= 1/C$. Then we have
$$
\begin{split}
r= \dfrac{P}{1+e \cos(\theta+\omega)} \\
r+er \cos(\theta+\omega)= P \\
r= P - er \cos(\theta+\omega)
\end{split}
$$
Now let $x= r \cos \theta$. Then noting that in our case, $r^2= x^2+y^2$, we have
$$
\begin{split}
r&= P - er \cos(\theta+\omega) \\
r&= P - ex \\
r^2&= P^2 - 2Pex + e^2 x^2 \\
x^2+y^2&= P^2 - 2Pex + e^2 x^2 \\
x^2 - e^2x^2 +2Pex + y^2&= P^2 \\
(1-e^2)x^2 +2Pex + y^2&= P^2 \\
x^2 + \dfrac{2Pex}{1-e^2} + \dfrac{y^2}{1-e^2}&= \dfrac{P^2}{1-e^2}
\end{split}
$$
Now it is just a routine complete the square in $x$ and we have
$$
\left(x+ \dfrac{Pe}{1-e^2} \right)^2 + \dfrac{y^2}{1-e^2}= \dfrac{P^2}{(1-e^2)^2}
$$
Of course, this is the same as
$$
\dfrac{(x+Pe/(1-e^2))^2}{P^2/(1-e^2)^2} + \dfrac{y^2}{P^2/(1-e^2)} = 1
$$
So that we have an ellipse 'centered' at the point $(-Pe/(1-e^2), 0)$, semimajor axis length $P/(1-e^2)$, semiminor axis $P/\sqrt{1-e^2}$, and with foci at distance
$$
\sqrt{A^2 - B^2}= \sqrt{\dfrac{P^2}{(1-e^2)^2} - \dfrac{P^2}{1-e^2}}= \dfrac{P|e|}{1-e^2}
$$


问:
$dy/dt$ = 0.8y with y(0) = -0.8 , why does the result of this question end up being so different from the formula?
So I'm given a dynamic system $dy/dt$ = 0.8t with y$_0$ = -0.8.

So the Theorem in the book I'm studying says $dx/dt$ = kx
With initial value x$_0$ and k being an ordinary constant.

x is just an arbitrary variable, surely I can bring this principle to 'y' as well, right?

The solution is then $e^{kt}$$x_0$

The understanding is that x will grow or decay exponentially.

The answer I originally got with this understanding was ${-0.8}$$e^{0.8t}$.
Seems simple enough, right? Except that was not the answer. I went online and found the answer, but I do not understand how or why we come to that conclusion and ditch the theorem.

The answer I found was actually $0.4t^2$+C

C being -0.8.
And the final answer being $0.4t^2$ - 0.8.

Why do we end up doing this instead of the theorem I brought up earlier? And with this being used as a tool obviously being Integration. When do I draw the line between using an Integration or an Exponential?

Can I use Integration instead of Exponentials?

答:
You need to be careful with the variables and functions.

The differential equation you which to solve is given by

$$\frac{dy(t)}{dt} = 0.8 y(t)$$

and has solution given by (with $y_0 = y(0) = -0.8$)

$$y(t) = -0.8 e^{0.8t}$$

The solution you propose is, however (ignoring initial conditions)

$$y(t) = 0.4t^2 + C$$

Your solution makes sense if the differential equation is

$$\frac{dy(t)}{dt} = 0.8 t$$

You need to be very careful about the difference between the two.


问:
Any commutative associative operation can be extended to a function on nonempty finite sets
This is a fact we use very frequently in general mathematics when we write such notations as $1+2+3+4$: since we know that $+$ is commutative and associative, we can just "drop the parentheses" and not worry about order of operations. Of course I believe this, but how does one prove this in full generality? Even stating it is giving me trouble. Here's my attempt:

Assume an operation $\oplus:S\times S\to S$ is provided satisfying $x\oplus y=y\oplus x$ and $x\oplus(y\oplus z)=(x\oplus y)\oplus z$ for all $x,y,z\in S$.

Claim: Given any finite set $\emptyset\subset A\subseteq S$, there exists a unique $z\in S$ such that for any function $f:{\cal P}(A)\to{\cal P}(A)$ which satisfies $\emptyset \subset f(B)\subset B$ for all $B\subseteq A$ with $|B|\ge 2$ and any function $g:{\cal P}(A)\to S$ which satisfies $g(\{x\})=x$ for all $x\in S$ and $g(B)=g(f(B))\oplus g(B-f(B))$ for all $|B|\ge 2$, $g(A)=z$.

The operation $\oplus$ does not necessarily have an identity element, so we do not attempt to define an empty sum. Intuitively, this element $z$ represents the finite sum of the elements in $A$, so if $A=\{1,2,3\}$ and $f(\{1,2,3\})=\{1\}$ and $f(\{2,3\})=\{3\}$, then
$$z=g(\{1,2,3\})=g(\{1\})\oplus g(\{2,3\})=g(\{1\})\oplus (g(\{3\})\oplus g(\{2\}))=1\oplus(3\oplus 2).$$
There has got to be a better way to say that, but this is the only way I can think of to capture all the possibilities of parenthesization, and still be amenable to a formal proof. And now that I've stated it, how should I prove it? I suppose I should induct on something, but I've no idea what.

Edit: The goal here is to be able to define an operation $F$ such that $F(\{x_1,\dots,x_n\})=x_1\oplus\cdots\oplus x_n$ and be assured that the operation is well defined and satisfies $F(A\cup B)=F(A)\oplus F(B)$, when $A$ and $B$ are disjoint finite nonempty subsets of $S$.

答:
This is the equivalence between commutative semigroups in the usual sense and their monadic description. It follows, for example, from Beck's monadicity theorem. Here is a down-to-earth formulation: If $S$ is a commutative semigroup, then we have a map which associates to every function $\lambda : S \to \mathbb{N}$ with the property that $\lambda(s)>0$ for at least one $s$, an element $\sum_{x \in S} \lambda(x) \cdot x$ of $S$ (abbreviated by $\sum \lambda$), with the following properties:

If $E=\{s\}$ for some $s \in S$ and $\lambda(s)=1$, then $$\sum_{x \in E} \lambda(x) \cdot x = s.$$

If $\mu$ is a function which associated to every $\lambda$ as above a natural number (with the property that there is some $\lambda$ with $\mu(\lambda)>0$), then
$$\sum_{x \in S} \left(\sum_{\lambda} \mu(\lambda) \cdot \lambda(x)\right) \cdot x = \sum_{x \in S} \left(\sum_{\sum \lambda = s} \mu(\lambda) \right) \cdot x$$

The converse is also true. Whenever $S$ is a set equipped with a sum operation as above, then this endows $S$ with the structure of a commutative semigroup. The point is that the set of functions $\lambda$ equipped with the pointwise addition is the free commutative semigroup on the set $S$. See here for the corresponding monadic description of abelian groups.


问:
Evaluating $\lim_{n\to\infty}( f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0) )$
I'm doing an exercise in Real Analysis and I would like to have my proof check since I'm not 100% sure of it. The problem says to consider a function $f:\mathbb{R}\rightarrow\mathbb{R}$ differentiable at $x_0$ and with $f(x_0)>0$ (which I guess it's useless at this point but the exercise had other points in which it was necessary so I put it anyway). Then, we are asked to evaluate the following

$$\lim_{n\to\infty} \left(f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0)\right).$$

Here is how I would do it:

$f$ is differentiable and hence continuous at $x_0$;

We apply Taylor's Theorem at $x_0$, so that there is a neighborhood of $x_0$, $I=(x_0-\delta,x_0+\delta)$, and a function $h$ such that, for all $x\in I$,
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+h(x)(x-x_0)$$
and $h(x)\rightarrow 0$ as $x\rightarrow x_0$.

Now, if $n>\frac 1{\delta}$ then each of the $n$ points $x_0+\frac 1{n^2},\dots,x_0+\frac 1n$ are in $I$ so we can apply the previous point to get

$$f\left(x_0+\frac k{n^2}\right)=f(x_0)+f'(x_0)\frac k{n^2}+h\left(x_0+\frac k{n^2}\right)\frac k{n^2}.$$

Summing from $k=1$ to $k=n$ we get

$$f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)=nf(x_0)+\frac {f'(x_0)}{n^2}(1+\dots+n)+\frac 1{n^2}\left(h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right)$$

Using the fact that $1+\dots+n=\frac {n(n+1)}2$ our original expression simplifies to

$$f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0)=\frac {f'(x_0)}{n^2}\frac {n(n+1)}2+\frac 1{n^2}\left(h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right).$$

Fix $\epsilon>0$, since $\frac {n(n+1)}{2n^2}\rightarrow \frac 12$ there is an $\bar{n}$ such that $n\geq \bar{n}$ implies that $\left|\frac {n(n+1)}{2n^2}-\frac 12\right|<\epsilon$. On the other hand, there is a $\delta'$ such that $|h(x)|<\epsilon$ on $(x_0-\delta',x_0+\delta')$. Since $x_0+\frac k{n^2}\rightarrow x_0$ and $\frac k{n^2}\leq \frac 1n$ we have that for $n>\frac 1{\delta'}$

$$\left|h\left(x_0+\frac k{n^2}\right)\right|<\epsilon.$$

By picking $n\geq\max\left\{\frac 1{\delta},\bar{n},\frac 1{\delta'}\right\}$ we obtain that

$$\left|\frac {f'(x_0)}{n^2}\frac {n(n+1)}2+\frac 1{n^2}\left(h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right)-\frac {f'(x_0)}2\right|\leq$$

$$f'(x_0)\left|\frac {n(n+1)}{2n^2}-\frac 12\right|+\frac 1{n^2}\left|h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right|<$$

$$f'(x_0)\epsilon+\frac 1{n^2}\left|h\left(x_0+\frac 1{n^2}\right)\right|+\dots+\frac n{n^2}\left|h\left(x_0+\frac 1n\right)\right|<f'(x_0)\epsilon+\epsilon(1+\dots+n)=$$

$$=f'(x_0)\epsilon+\frac {\epsilon}{n^2}\frac {n(n+1)}2<f'(x_0)\epsilon+\epsilon\left(\frac 12+\epsilon\right).$$

Since $\epsilon$ is arbitrary, we can conclude that

$$\lim_{n\to\infty} \left(f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0)\right) = \frac {f'(x_0)}2.$$

Is my proof correct? Is there any smarter/shorter way to do that? Thanks is advance for any help!

答:
Something you could use is the fact that $\lim_{k\to 0}\frac{h(x_0+k)}{k}=0$ so there exists an $n_0$ such that $h(x_0+k)<k$ for $k<1/n_0$, since the limit is taken over $n$ and $\frac{k}{n^2}\leq\frac{1}{n}$ we have that there exists an $n_0$ such that $$\frac{1}{n^2}\sum_{i=1}^n ih\left(x_0+\frac{i}{n^2}\right)<\frac{1}{n^2}\sum_{i=1}^n\frac{i^2}{n^2}<\frac{1}{n^2}\sum_{i=1}^n1=\frac{1}{n}$$
For all $n\geq n_0$, that sequence converges to $0$, so $\lim_n\frac{1}{n^2}\sum_{i=1}^{n}ih(x_0+\frac{i}{n^2})=0$


问:
Find x, y coordinates of a curve by knowing only 3 points?
I was wondering if it's possible to find other coordinates of a curve if I know only 3 points? The points that I know are (1,5), (5.5, 25), (8,50) The curve is exponential I think. It's actually the formula for making a tea. The narrator of a book says that the bigger the container, the less ingredients I will need. In this case, ingredients is x and container size is y

What if I want to find x when the container size = 10 (x, 10)? Is there a formula for this?

答:
Sometimes a figure is worth 1000 words:

If you know the functional form is $a + b e^{c t}$, then it is a simple exercise in nonlinear model fitting to get:  $$-5.2 + 8.01 e^{0.24 t}$$


问:
Osculating circle given parametric coordinates in polar form
I have the following parametric equation for a curve in polar form:
\begin{align}
r(t) &= r_0-kt\\
\phi(t)&=h\ln(r_0-k_0t)+a
\end{align}
where $(r_0,h,a,k_0,k)\in\mathbb{R}$
what is the radius of the osculating circle $R(t)$?
Thanks in advance.

答:
This is the Mathematica output, using the parameter names given.

Let me elaborate it a bit. I just started with $r'$ and $\phi'$, then rotated the vector $(r',r \phi')$ by an angle of $\phi$ to put them in XY coordinates. The norm of the derivative of the normalized resulting vector is the curvature, i.e., the inverse of the radius.


问:
Explanation of marginal likelihood in Gaussian process
I am new to GP/non-parametric regression. I am reading Rasmussen's book on Gaussian processes. In Eq. 2.28, 2.29 (Page 19) and in the subsequent passage he writes the marginal likelihood as the integral of likelihood times prior.

$$
p(\textbf{y}|X) = \int p(\textbf{y}|\textbf{f},X)p(\textbf{f}|X)\,d\textbf{f}
$$

I understand the prior is the GP-prior (the model). $p(\textbf{f}|X)$ part is clear to me. This is assumed by us to be a multivariate Gaussian $\mathcal{N}(0,\textbf{K})$.
But the book immediately says after this that the likelihood is a factorized Gaussian of the form $\textbf{y|f} \sim \mathcal{N}(\textbf{f},\sigma_{n}^{2}\textbf{I})$. I don't get how we can make this jump directly. How did this understanding come into play??

I know the regression model looks like $y = f(x) + \epsilon_{n}$. The model for $f(x)$ is our assumption : The same GP-prior as discussed above - $\mathcal{N}(0,\textbf{K})$. The noise are i.i.d Gaussian which is $\mathcal{N}(0,\sigma_{n}^{2})$. So the target distribution $\textbf{y}$ will be $$\mathcal{N}(0,\textbf{K} + \sigma_{n}^{2}\textbf{I})$$. The way I understand this is $y$ is a sum of 2 independent normal distributions. So we can use the property of sum of 2 normals.

But I don't get the "$\textbf{y|f} \sim \mathcal{N}(\textbf{f},\sigma_{n}^{2}\textbf{I})$" part from the discussion. Could anyone please enlighten me how we can directly assume this. I am sorry if this is a silly question, but I am really new to this.

Thanks in advance.

The link to the book: http://www.gaussianprocess.org/gpml/

答:
The key is the assumption of additive independent identically distributed Gaussian noise $\epsilon_n$, i.e. the assumption that observations are given by $\textbf{y} = \textbf{f} + \epsilon_{n}$ where $\epsilon_n \sim \mathcal{N}(0,\sigma_{n}^{2}I)$ independent from $\textbf{f}$. It should be intuitively clear that if you know the noise-free value $\textbf{f}$ then you should expect the observation $\textbf{y}$ to be a Gaussian centered on $\textbf{f}$ and with the covariance matrix of the noise.

We can show this more rigorously by deriving the conditional distribution $\textbf{y|f}$. First, note the distribution of $\textbf{f}$ and of $\epsilon_n$:

$$
\textbf{f} \sim \mathcal{N}(0, K) \\
\epsilon_n \sim \mathcal{N}(0, \sigma_n^2 I).
$$

We need to know the joint distribution of $\textbf{f}$ and $\textbf{y}$, so we will calculate the covariance matrices of $\textbf{y}$ and of $\textbf{y}$ with $\textbf{f}$. Since $\textbf{y} = \textbf{f} + \epsilon_n$ we have

$$
\textbf{y} \sim \mathcal{N}(0, K + \sigma_n^2 I)
$$

by the properties of idependent Gaussian distributions. Also,

$$
\mathrm{Cov}(\textbf{y}, \textbf{f}) = \mathrm{Cov}(\textbf{f}, \textbf{f}) + \mathrm{Cov}(\epsilon_n, \textbf{f}) = K + 0 = K.
$$

Collect the results above into a statement of the joint distribution of $\textbf{y}$ and $\textbf{f}$

$$
\begin{bmatrix}
\textbf{y} \\
\textbf{f}
\end{bmatrix} \sim \mathcal{N}\left(\begin{bmatrix}
0 \\
0
\end{bmatrix}, \begin{bmatrix}
K + \sigma_n^2 I & K \\
K & K
\end{bmatrix}
\right)
$$

where in the bottom left corner we used the fact that $K = K^T$.

Finally, we find the conditional distribution $\textbf{y|f}$ using identity (A.6) on page 200 in appendix A.2 in Rasmussen (also to be found here)

$$
\textbf{y} | \textbf{f} \sim \mathcal{N}(0 + K K^{-1}(\textbf{f} - 0), K + \sigma_n^2 I - KK^{-1}K^T) \\
\textbf{y} | \textbf{f} \sim \mathcal{N}(\textbf{f}, \sigma_n^2 I)
$$

as expected.


问:
Which is true $A$ is subset of $B$ or $B$ is subset of $A$.

Consider the sets dened by the real solutions of the inequalities $$A=\{(x,y):x^2+y^4\le 1\}$$ and $$B=\{(x,y):x^4+y^6\le 1\}$$Then which is true $A$ is subset of $B$ or $B$ is subset of $A$.

This is Question No. $10.$ of this paper. Please help me.Thanks in advance.

答:
Hints:

$$(x,y)\in A\implies x^2+y^4\le 1\implies |x|,|y|\le 1$$

so

$$ x^4+y^6\le x^2+y^4\implies \ldots\ldots$$


问:
How can I prove that $b^n$ is $ O(n!)$
Hi guys I came upon an exercise from the Big-O chapter that I can not solve.I know that in order to be in $O\bigl(f(n)\bigr)$ has to be smaller or equal to $c$ * $f(n)$ where is c is a constant and is positive but I never solved an exercise where another constant, $b$ in my case, is involved ($b$ is greater than 1 and is real). Any help is highly appreciated, sorry for the writing issues..

答:
Hint:

Use Stirling's formula to prove that yoou have a stronger statement::
$$b^n=o(n!)$$

Title: How can I prove that $b^n$ is $ O(n!)$


问:
Singular points of $\alpha(t)=(t^2\cos(t),t^2\sin(t))$
I have to find the singular points of
$$\alpha(t)=(t^2\cos(t),t^2\sin(t))$$
with $-2\pi\leq t\leq2\pi$. Its first derivative is
$$\alpha'(t)=(2t\cos(t)-t^2\sin(t),2t\sin(t)+t^2\cos(t))$$
and so $t$ is singular if $\alpha'(t)=0$:
$$\left\{\begin{array}{l}2t\cos(t)-t^2\sin(t)=0 \\ 2t\cos(t)+t^2\sin(t)=0 \end{array}\right.$$
which clearly show
$$t=0$$
is a singular point.
But, my question is: in both equations, when I factor them, I got
$$\left\{\begin{array}{l}2\cos(t)-t\sin(t)=0 \\ 2\cos(t)+t\sin(t)=0 \end{array}\right.$$
and so
$$\cos(t)=\frac{t\sin(t)}{2}$$
and then, putting it in the second equation:
$$2\sin(t)+\frac{t^{2}\sin(t)}{2}=0\Rightarrow\sin(t)(4+t^2)=0$$
therefore
$$\sin(t)=0\Rightarrow t=n\pi$$
But I know that the only singular point is $t=0$. What is wrong here?

答:
You have noted that your derivatives are incorrect. They should be the ones below, which is what I will work with.
$$
\begin{cases}
2t\cos(t) - t^2\sin(t)=0 \\
2t\sin(t) + t^2\cos(t)=0
\end{cases}
$$
The first equation is $t(2\cos t - t \sin t)=0$, which implies that either $t=0$ or $2\cos t = t \sin t$. Notice $t=0$ works in the second equation so that $t=0$ is a singular point. Now let's assume that $2 \cos t= t \sin t$. From the second equation, we have $t(2 \sin t + t \cos t)=0$, so either $t=0$ (which we already know is a solution) or $2 \sin t= -t \cos t$. But we know that $2 \cos t = t \sin t$, so that
$$
4 \cos t= 2 \cdot 2\cos t= 2 \cdot t \sin t= t \cdot 2\sin t= -t^2 \cos t
$$
Then we know $t^2 \cos t + 4 \cos t= 0$. This is the same as $\cos t (t^2+4)=0$. Clearly, $t^2+4= 0$ produces no solution so that it must be that $\cos t= 0$. Since you are on $-2\pi \leq t \leq 2\pi$, this means $t= \pm \dfrac{\pi}{2}$. Checking, however, we see that these are extraneous solutions as they satisfy neither equation. Therefore, we are only left with the solution $t=0$.


问:
Does $\lim_{x\to\infty} f(x) = 1$ imply that $f(x) = 1$ when $x = \infty$?
If $\lim_{x\to\infty} f(x) = 1$, then is it correct to state it as $f(x) = 1$ when $x = \infty$?

Can we define a function in this way, if not why? Also if a limit exists at a point $x$ then does the function always has to be defined for that  $x$ with the same value obtained through the limit?

答:
If, say, $f$ is a function from $(0,\infty)$ to $\Bbb R$ and if $\lim_{x\to\infty}f(x)=1$, then, no you cannot say that $f(\infty)=1$ because $f(x)$ is undefined if $x=\infty$; $f(x)$ is defined only when $x\in(0,\infty)$.

And if $f\colon\Bbb R\longrightarrow\Bbb R$ is defined by$$f(x)=\begin{cases}1&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}$$then $\lim_{x\to0}f(x)=1$, but $f(0)=0$.


答:
Remember, a limit is what happens 'as you approach something' not 'once you get there'. Meaning, the function does not even have to be defined there. For example, $\displaystyle \lim_{x \to 0} x^2= 0$ and if $f(x)=x^2$, we have $f(0)=0$. But for $f(x)= \frac{\sin x}{x}$, we have
$$
\lim_{x \to 0} \dfrac{\sin x}{x}= 1
$$
But $f(0)$ is not defined.

Now you undoubtably have seen if a function is failed to be defined at a value but the limit exists there, we have a removable discontinuity that we can 'fill in' by defining it there. So we can define
$$
g(x)=
\begin{cases}
\dfrac{\sin x}{x}, & x \neq 0 \\
1, & x=0
\end{cases}
$$
Now $g(x)$ is continuous. This seems to be what you're thinking about in your question. However, you are working with limits over the real numbers so $x \in \mathbb{R}$, but $\pm \infty \notin \mathbb{R}$. So it makes no sense for you to define a function at $\pm \infty$, instead merely to talk about what happens to a function $f(x)$ as $x \to \pm \infty$.

Title: Does $\lim_{x\to\infty} f(x) = 1$ imply that $f(x) = 1$ when $x = \infty$?

Body:
If $\lim_{x\to\infty} f(x) = 1$, then is it correct to state it as $f(x) = 1$ when $x = \infty$?

Can we define a function in this way, if not why? Also if a limit exists at a point $x$ then does the function always has to be defined for that  $x$ with the same value obtained through the limit?


答:
No.  The notation $\lim_{x\to\infty}f(x) = 1$ has a very precise meaning.  It means that given any $\varepsilon>0$ we can always find a real number $c$ such that for all $x > c$ we have $|f(x) - 1| < \varepsilon.$  Note that nowhere in there do we know what the exact value of $f(x)$ is at a particular point $x$.

Beyond that, $\infty$ is not a number, so it doesn't make sense in this context to define $f(\infty)$ or say $x = \infty$.

To answer your last question, no, we don't require functions to be defined so that $\lim_{x\to a}f(x) = f(a)$.  This means that the function is continuous at $x = a$, which is not true of all functions.


问:
What is the Group of symmetries of Texas Holdem Poker hands?
In Texas Holdem poker, each player is dealt two cards from a standard deck of 52 playing cards. (2,3,4...Queen,King,Ace). What is the group of all the symmetries of all (two-card) poker hands? (Given suits are equivalent, see below)

I like to envisage a poker hand as an abstract mathematical object, and I'd quite like to visualise the group of symmetries. Obviously we have the rank of the card, effectively 1 to 13, but also we have the four suits. In poker, there is no heirarchy to the suits, a spade is no more, or less, important than a club. So Ace-hearts-King-hearts is the same, in effect, as the Ace-diamonds-King-diamonds. Likewise, Ace-spades-Ace-hearts is, in effect, the same hand as Ace-Spaces-Ace-Diamonds. And so on. [From Wikipedia - "There are 169 non-equivalent starting hands in hold 'em, which is the sum total of : 13 pocket pairs, 13 × 12 / 2 = 78 suited hands and 78 unsuited hands (13 + 78 + 78 = 169)."]

I supose that the group of symmetries of suited hands are a sub-group of the group of symmetries of all hands. Also, the group of symmetries of paired hands are another sub-group of the group of all hands.

I'm an amateur with knowlege of the basics, so, if fancy symbols, or elaborate jargon, are used, please explain them to the physics-degree-educated layman who has an understanding of the non-technical basics of group theory. I know the axioms, for example.

Also, what are the representations (matrix?) of the hands, paired hands, and suited hands?

答:
If you're thinking about the symmetries with respect to suites, you have $S_4$, the symmetric group on $4$ letters. Let $\{S,D,C,H\}$ be the suites. Elements of this group look like $\sigma = ($C,D$)$ which permutes clubs and diamonds in a given hand, and leaves the other suites untouched. It is helpful to think of this really in terms of group actions, with the set of suite permutation $S_4$ acting on the set of hands.

We can use information about $S_4$ to re-derive the 169 number you mentioned in an interesting way. We define an $n$-cycle as a single cycle of length $n$. Any element of $S_4$ has what is known as a cycle structure (which you can read more about here: https://en.wikipedia.org/wiki/Cyclic_permutation). For example $(S,D,C,H)$ has cycle structure $4$ since it is composed of just one $4$-cycle, and $(C,S)(D,H)$ has cycle structure $2-2$ because it is composed of two $2$ cycles.

We can easily check that there are only $5$ possible cycle structures in $S_4$:

$0$ cycle: This is just the identity permutation, leaves all suites unchanged usually denoted ()

$2$ cycle: This just swaps any two suites, e.g $(C,D)$ or $(S,D)$

$3$ cycle: Permutes any three of the four suites, e.g. $(S,H,D)$ or $(S,D,H)$

$4$ cycle: Permutes all of the suites, e.g. $(C,S,D,H)$, or $(C,H,S,D)$

$2-2$ cylce: Permutes two pairs of suites, e.g. $(D,S)(C,H)$ or $(S,H)(D,C)$

We can obtain the number of equivalent hands by counting the number of hands which are fixed by each of these cycles types and dividing by the number of elements in the group, $24$. It's easy to see that the identity permutation fixes every hand giving us $1326$.

Any $2$ cycle is going to fix a hand if either the hand contains neither of the permuted suites of which there are $325$, or if both cards are the same rank and one of each of the permuted suites, which yields an additional $13$, of a total of $338$. But there are also $6$ possible $2-2$ cycles, yielding a total of $2028$ fixed hands.

I won't belabor the computation any further as you might find it interesting, but it should come out to $624$ fixed hands under the $3$ cycles, $78$ under the $2-2$ cycles, and $0$ under the $4$ cycles. This sums to $1326+2028+624+78 = 4056$ and dividing by the order of the group gives us $4056/24 = 169$. This is also nice because it is fairly easy to see how you would apply this to other games.

Addendum:

We can think about the general case here for a moment, and while you can certainly apply these ideas here, I don't think they are insightful. Instead of the $S_4$, we can think about the action of $S_4\times S_{13}$ on the set of all hands, where we define the permutation in the first coordinate to be the permutation of the suite and the second coordinate to be the permutation of the rank. One can easily verify that this is indeed a group action.

We can now realize the suite invariance property we explored above as a subgroup $S_4\times S_{13}$, namely the subrgoup composed of elements of the form $(\sigma,())$ where $()$ denotes the identity permutation as above. My understanding is that rank pretty much always matters, so I'm not sure there is much you can do with the second coordinate. I would love for someone to prove me wrong, but I don't think you're going to get a lot more out of this structure. The problem is that while we can use group actions to study the set of hands, there doesn't seem to be a way to actually form an algebraic structure from the set itself. There is no binary operation with which to compose hands into new ones which is an apriori requirement.


问:
Is the sequence defined by $a_{1}=\frac43, a_{n+1} = \sqrt{5a_{n}-6}$ monotonically decreasing?
I want to know where is my mistake in the proof by induction that $(a_n)$ is monotonically decreasing.

Basis of induction: $a_{2} - a_{1} = \sqrt{\dfrac{2}{3}} - \dfrac43 < 0$

Assume $a_{n+1} - a_{n} < 0 $

Then $a_{n+2} - a_{n+1} =  \sqrt{5a_{n+1}-6} -  \sqrt{5a_{n}-6}= ( \sqrt{5a_{n+1}-6}- \sqrt{5a_{n}-6})\dfrac{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}=\dfrac{5a_{n+1}-6-5a_n+6}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}=\dfrac{5(a_{n+1}-a_n)}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}<0$

So $a_{n+1} - a_n < 0 \;\;\forall n \in \Bbb{N}$, hence this sequence is monotonically decreasing.

But how can this be true if $a_3 \notin \Bbb{R}$?

答:
With appropriate initial values, you can have $a_n \in \mathbb R, \forall n \in \mathbb N$. In this case you can do better than writing all those ugly squareroots.

You can prove easily that $a_n \ge 2 \implies a_n \in \mathbb R, \forall n \in \mathbb N$ and

If $a_1=2$ or $3$ then $a_n \equiv a_1$.

If $2 < a_1 < 3$, then $2 < a_n < 3$.

If $a_1 > 3$, then $a_n > 3$.

The trick is to square your equation and subtract the square of one of the stationary points, $2$ or $3$. Let's do $2$.

$$a_{n+1}^2-4 = 5a_n-6-4 \implies (a_{n+1}+2)(a_{n+1}-2)=5(a_n-2) $$

Assume $a_1 > 2$ (thus $a_n>2, \forall n \in \mathbb N$). Then

$$a_1 < 3 \implies a_n < 3 \\ \implies 5(a_n-2) = (a_{n+1}+2)(a_{n+1}-2)< 5(a_{n+1}-2)\\ \implies a_n < a_{n+1}$$

$$
a_1 > 3 \implies a_n < 3 \\ \implies 5(a_n-2) = (a_{n+1}+2)(a_{n+1}-2) > 5(a_{n+1}-2)\\ \implies a_n > a_{n+1}$$


问:
how to find the density function of 2D random variable?
Let $(X,Y)$ be $2$ dimensional random variable with uniform distribution  on the domain $$D = \left\{\left.(x,y)\in\mathbb{R}^2 \ \right| \ x\geq0 \ , \ y\geq0 \ , \ x+y\leq1 \right\}$$ for
$y\in(0,1)$ find the density function $f_{X|Y}(x|y)$

My main problem is that I don't understand how to write $f_{X,Y}(x,y)$ I understand how does it look in one dimensional but when it became 2 or higher I dont understand how to write an expression for it

答: I assume familiarity with indicator functions & Iverson brackets. The locus $\{x|(x,\,y)\in D\}$ is the line segment $[0,\,1-y]\times\{y\}$, so $f_{X|Y}(x|y)=\frac{1}{1-y}1_{[0,\,1-y]}(x)$. Since $D$ is a triangle of area $\tfrac12$, $f_{X,\,Y}(x,\,y)=2[(x,\,y)\in D]$.


问:
Adjoint of the linear transformation over a vector space over complex numbers.
Consider the vector space $V:=P_2(\mathbb{C})$ consisting of polynomials of degree at most $2$ with complex coefficients together with the following inner product $$\langle f, g\rangle=\int_{-1}^1f(t)\overline{g(t)}\, dt$$
Evaluate the adjoint of the linear transformation $T:V\to V$ by $$T(f)=if'+2f.$$
My effort:
I know that $T^*$ is a linear transformation which satisfies
$$\langle Tf, g\rangle=\langle f, T^*g\rangle.$$
Above implies that $$i\int_{-1}^1f'(t)\overline{g(t)}\, dt+2\int_{-1}^1f(t)\overline{g(t)}\, dt=\langle f, T^*g\rangle.$$
Please tell how to proceed next?

答:
The approach suggested in the comments is great for some problems like this, especially where the space in question involves some kind of "zero boundary condition". However, I find that for this problem it is more practical to proceed as follows.

To begin, apply the Gram Schmidt process to $\{1,t,t^2\}$ to get an orthonormal basis for $P_2(\Bbb C)$ relative to this inner product. We end up with
$$
u_1(t) = \frac 1{\sqrt{2}}, \quad u_2(t) = \sqrt{\frac 32} \,t, \quad u_3(t) = k\cdot[3t^2 - 1]
$$
for some (real) $k$ such that $\langle u_3,u_3\rangle = 1$.

Relative to the basis $\mathcal B = \{u_1,u_2,u_3\}$, we find that $T$ has the matrix
$$
[T]_{\mathcal B} =
\pmatrix{2&i\sqrt{3}&0\\
0&2&i\sqrt{\frac 23}\cdot \frac 1{6k}\\
0&0&2}.
$$
The matrix of $T^*$ relative to this same basis is the conjugate-transpose, which is to say that
$$
M = [T^*]_{\mathcal B} =
\pmatrix{2&0&0\\
-i\sqrt{3} & 2 & 0\\
0 & -i\sqrt{\frac 23}\cdot \frac 1{6k} & 2}.
$$
This matrix implicitly gives us a formula for $T^*$. In particular,
$$
[T^*p](t) = \sum_{j=0}^2 \left[\sum_{k=1}^3 M_{jk}\langle p,u_k\rangle\right] u_j(t).
$$
I see no nice way to interpret the operator that we get in this case in terms of derivatives and integrals.