问:
Question on compact subsets of finite dimensional normed spaces
I'm trying to prove this fact:

If $X$ is a finite dimensional normed space and $K$ a compact subset of $X$ that satisfies that if $x, y \in K$ then $\frac{x-y}{2} \in K$ then, given $a \in X \setminus K$ there exists an $\varepsilon >1$ such that $a \not \in\varepsilon K$

I know that the distance from $a$ to $K$ is greater than zero and is attained by some point $x \in K$, so I'm trying to prove that is enough to take $1 < \varepsilon < d(a,K)+1$ but I'm not able to reach a contradiction if I suppose that $a \in \varepsilon K$.

Any ideas?

答:
It is sufficient to assume that $K$ is closed in a topological vector space. Namely $K^c$ is open and $1 \cdot a \in K^c$ so by continuity of scalar multiplication, there exist $s,t > 0$ such that
$$\langle 1-s, 1+s\rangle \cdot B(a,t) \subseteq K^c.$$
In particular, for $\varepsilon := 1+\frac{s}2$ we have $\varepsilon a \in K^c$.


问:
Evaluate $\int \dfrac{1}{\sqrt{1-x}}\,dx$
Find $$\int \dfrac{1}{\sqrt{1-x}}\,dx$$

I did this and got $\dfrac23(1-x)^{\frac32} + c$

But a online calculator is telling me it should be $-2(1-x)^{\frac12}$

What one is on the money and if not me why?

ContentLicense: CC BY-SA 3.0

Body:
$$
\int \frac{1}{\sqrt{1-x}}~dx=\int(1-x)^{-1/2}~dx
$$
Let $u=1-x$, $du=-dx$, so
$$
\int(1-x)^{-1/2}~dx=-\int u^{-1/2}~du
$$
Add one to the power of $u$, and divide by the new power
$$
\int \frac{1}{\sqrt{1-x}}~dx=-\int u^{-1/2}~du=-\frac{u^{1/2}}{1/2}+c=-2(1-x)^{1/2}+c
$$

What you have done is integrated $\sqrt{1-x}$ by mistake
$$
\int \sqrt{1-x}~dx=\int (1-x)^{1/2}~dx=-\int u^{1/2}~du=-\frac{2}{3}u^{3/2}+c=-\frac{2}{3}(1-x)^{3/2}+c
$$


问:
Discrete Riccati equation in discrete $h-\infty$ controller design
I want to design an discrete H infinty controller， I get this reference

consider a discrete system with the state space representation like this
 $$ x_k+1 = Ax_K + B1w_k + B2u_k,X_0 = 0$$
   $$ z_k = C1x_k + D12u_k$$
   $$ y_k = C2x_k + D21w_k$$
and for simplicity ,the following assumptions are made:
 $D12^TD12>0$, $C1^TD12 =0$，$B1D21^T =0$,$D21D21^T>0$

there is a Riccati equation ：$$ M =A^TMP^{-1}A +C1^TC1$$  ,
                         $$ P = I+(B2(D12^T D12)^{-1}*B2^T-\gamma^{-2}B1B1^T)M$$

where $A,B1,B2,C1,C2$ are appropriate matrix ,
there is a problem,  it does't like the normal discrete equation we can use dare to solve it,

$$A^TXA−X−A^TXB(B^TXB+R)^{−1}B^TXA+Q=0 $$

I don't know how to solve this kind of riccati equation , besides  how to choose the parameter $\gamma$,
I konw how to solve the reccati equation mentioned above.l afraid that my presentation is not clearly, so I attached the image of the Riccati equation, l hope this will help you understand more clearly,
enter image description hereenter image description hereenter image description here
I alredy have a discrete system:

```<code> clc;clear;

 A = [0 0 -0.06;
    1  0  0.25;
    0  1    1];

B1 = [0; 0; 1]
B2 = [0; 0; 1];

C1 = [sqrt(2) 0 0;
      0    1   0;
      0    0   1;
      0    0  0];

D12 = [0 0 0 1];

C2 = [0 0 1],

D21 =0;

OB = rank(obsv(A,C2))

CO = rank(ctrb(A,B2))
</code>```

so the riccati equation can be simplified with $D12^TD12 =I$:
$$M =A^TM(I+(B2B2^T-\gamma^{-2}B1B1^T)M)^{-1}A +C1^TC1$$
I am learning discrete H infinty control theory, and stuck with this question for a while,l already konw there are other approachs to deal with output feedback discrete H infinity controller design such that  bilinear transformation ,while acording to my homework requirement l have to use this method,

I appreciate any guidance or suggestion , Thanks in advance!

答:
You can convert the stated matrix equation into the standard DARE form by using the Woodbury matrix identity, which states

$$
\left(A + U\,C\,V \right)^{-1} = A^{-1} - A^{-1} U \left(C^{-1} + V\,A^{-1} U \right)^{-1} V\,A^{-1}. \tag{1}
$$

Using $(1)$ together with $A = I$, $U = F(\gamma) = B_2 B_2^\top - \gamma^{-2} B_1 B_1^\top$, $C = I$ and $V = M$  yields the following

$$
\left(I + F(\gamma)\,M\right)^{-1} = I - F(\gamma)\,(I + M\,F(\gamma))^{-1} M. \tag{2}
$$

When assuming that $F(\gamma) \geq 0$ it should also be possible find a $L(\gamma)$ such that $F(\gamma) = L(\gamma)\,L(\gamma)^\top$. One way to find such $L(\gamma)$ would be for example by using the Cholesky decomposition. Substituting this in $F(\gamma)\,(I + M\,F(\gamma))^{-1}$ and partially applying the push-through identity yields

$$
F(\gamma)\,(I + M\,F(\gamma))^{-1} = L(\gamma)\,(I + L(\gamma)^\top M\,L(\gamma))^{-1} L(\gamma)^\top. \tag{3}
$$

Substituting this in $(2)$ and its resulting expression in the original matrix equation yields

$$
M = A^\top M\,A - A^\top M\,L(\gamma)\,(I + L(\gamma)^\top M\,L(\gamma))^{-1} L(\gamma)^\top M\,A + C_1^\top C_1, \tag{4}
$$

which can be solved for $M$ using a DARE solver using $R = I$ and $B = L(\gamma)$.

ContentLicense: CC BY-SA 4.0


问:
Let $G$ be transitive.Then $\beta\in \operatorname{fix}(G_\alpha)$ implies $G_\alpha = G_\beta$
i am new in this forum.

My question is about group actions

We have a transitive action of $G$ and $\beta$ a element in the fixed points of the stabilizer of another element $\alpha$. Then $\alpha$ and $\beta$ have the same stabilizers?

Thank you for the further answers

答:
If it's finite groups you are talking about, then the following argument applies.

Your hypothesis is that $G_{\alpha} \le G_{\beta}$. Since $G$ acts transitively, there is $g \in G$ such that $\alpha g = \beta$, and then $g^{-1} G_{\alpha} g = G_{\beta}$.

If $G$ is finite, this implies that $G_{\alpha}$ and $G_{\beta}$ have the same order, so they are equal.

If $G$ is infinite, though, this isn't necessarily true. There are examples of groups $G$ with a subgroup $H$ and an element $g \in G$ such that $H$ is properly contained in $g^{-1} H g$. If you let $G$ act by right multiplication on the cosets $H a$, and take $\alpha = H$, $\beta = H g$, then $G_{\alpha} = H$ is properly contained in $G_{\beta} = g^{-1} G_{\alpha} g = g^{-1} H g$.


问:
Given a number $n$ in decimal representation, how do we get the $i$'th digit from the right in the equivalent representation in another base $b$?
Let's say we have a number $n$, how do we know the $i$'th digit from the right in the equivalent representation in another base $b$?

For example, if $n = 12_{10}$, and we want to check what the second $(i = 2)$ bit is in the equivalent representation of $12$ in base $3$ (which would be $110_3$), then the answer is $1$.

Edit: I found a formula on stack overflow, which is $$\Bigl \lfloor \frac{n}{b^{i}} \Bigr \rfloor \bmod b$$
How did we get this formula?

答:
There are general algorithms, but for small numbers you can also just 'go to work' on this, looking at it as an 'algebraic puzzle'.

For your example, $12$ to $\text{base-three}$ representation,

$\text{base-three}$ symbols

$\quad D_\text{three} = \{0,1,2\} \text{ and } \beta \text{ is the next integer after } 2 \quad \text{(using the symbol } 3 \text{ is not a good idea here)}$

So, the symbol $2$ means the same thing in both systems and

$\quad 12_{10} = (2 + 2 + 2 + 2 + 2 + 2)_{\beta}$

and since  $(2 + 2)_{\beta} = 11_{\beta}$,

$\quad 12_{10} = (11 + 11 + 11)_{\beta} = (22 + 11)_{\beta}$

and since  $(20 + 10)_{\beta} = 100_{\beta}$,

$\quad 12_{10} = (20 + 2 + 10 + 1)_{3} = (100 + 2 + 1)_{\beta}$

and since $(2+1)_{\beta} = 10_\beta$ we get the answer,

$\quad 12_{10} = (100 + 10)_{3}  = 110_{3}$


问:
Expression for the sum of the roots of a polynomial of degree n to the power of k
I am trying to find a function f(m) that given a power m returns the sum of the roots to the power of m.
$$(x-a_1)(x-a_2)(x-a_3)...(x-a_n)=0$$
$$f(m)=\sum_{k=1}^n{(a_ k)^m}$$
I attempted doing it manually for small values of n however I wasn't able to spot a pattern after m=3. For m=2 using a quadratic in the form of $x^2+bx+c=0$ I got $b^2-2c$ but that was as far as I got.

答:
If $\,p(x):=x-c_1,\,$ then $\,f(0)=1\,$ and $\,f(m)=c_1^m\,$
which satisfies
$\, f(m+1) = c_1f(m).\,$ Next, if $\,p(x):=x^2-c_1x-c_2,\,$ then
$\, f(0)=2,\,f(1)=c_1\,$ and
$\, f(m+2) = c_1f(m+1)+c_2f(m).\,$ This easily generalizes.
All you need is the first few initial values of $\,f(m)\,$
for $\,m=0,\dots,n\,$ where $\,n\,$ is the degree of $\,p(x).\,$


问:
How to find Isomorphism between groups of order 8
Let $A=\{1,2,3\}$ and $2^A$ is all the subsets of A. Then $(2^A,\triangle)$ is a group. How can I show that the group $\big\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\},\emptyset\big\}$
with operation $\triangle$, where $\triangle$ is the symmetric diffrence $C△B=(C \cup B) \backslash (C \cap B)$
is isomorphic to the group

$S= \{(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),(1,1,1)\}= \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ with operation $+$.

I did not manage to find a bijection, but both groups have order $8$ and the order of the elements correspond.

I did do the cayley tables for both groups but I still did not manage.

答:
Hint: Prove that the group is abelian and that all the nonidentity elements have order two.

There are $|\rm {GL}_3(2)|=168$ isomorphisms.


问:
Expectation of a conditioned geometric random variable
So first let $X$ be the number of independent coin tosses until first head; $P(H)=p$. Suppose we have observed that the first coin toss is tails, what is the expectation $E[X-1|X>1]$?

So intuitively this is just equal to $E[X]$ which makes sense to me. I tried to derive this more formally:

$E[X-1|X>1]=\sum_xg(x)p_{X-1|X>1}(x)=\sum_x(x-1)p_{X-1|X>1}(x)$

Since a geometric random variable is memoryless, we have:

$E[X-1|X>1]=\sum_x(x-1)p_{x}(x)=E[X]-1$

I seem to be off by a factor of 1, I suspect it might be something to do with the fact that I am still summing over all $x$ as denoted in $\sum_x$ but I am not quite sure how to fix this.

答:
It seems to me that the issue here is in applying the law of the unconscious statistician with a mismatch in the random variable. Namely, conditioned on $X>1$ the random variable $Y := X-1$ has the same distribution as $X$. This means that $p_Y(k) = p_X(k)$ for all $k\geq 1$, indeed. So
$$
\mathbb{E}[X-1\mid X>1] = \mathbb{E}[Y] = \sum_{k=1}^\infty k p_Y(k)
= \sum_{k=1}^\infty k p_X(k) = \mathbb{E}[X]\,.
$$
Howver, what you wrote is a bit different: what you wrote is
$$
\mathbb{E}[X-1\mid X>1] = \sum_{k=1}^\infty (k-1)p_Y(k)
$$
which is different. The function $g$ you applied the LOTUS to seems OK, but then you ended up with $p_{X-1|X>1}$ (i.e., $p_{g(X)|X>1}$) instead of $p_{X|X>1}$ (this does not sound very formal as written, but that appears to be the crux of the issue).


问:
Set of Unit Ball’s Extreme Points is not always closed
I am trying to show that $\operatorname{Ext}B(X)$ does not have to be closed where $B(X)$ is closed unit ball of normed space $X$ and $\operatorname{Ext}B(X)$ is the set of extreme points of unit ball.

My definition for extreme point : $x \in S(X)$ is an extreme point of $B(X)$ if $x$ cannot be written as arithmetic mean of distinct elements of $B(X)$. ($S(X)$ is unit sphere)

I thought that $\Bbb R^n$ can be an example for this and I tried to use $\Bbb R^3$ but I could not find the proper sequence to show that $\operatorname{Ext}B(\Bbb R^3)$ is not closed. I really stuck.

I appreciate any help.

Thank you

答:
Let $C=\operatorname{co} ((\pm 1, 0, \pm 1) \cup \{ (0,y,z)| y^2+z^2 = 1\})$.

The points $\{ (0,y,z)| y^2+z^2 = 1, |y| \neq 1\}$ are extreme, but the points
$(0,0,\pm 1)$ are not.

$C$ is convex, $0 \in C^\circ$ and $C=-C$. Then the Minkowski functional is a norm.


问:
Finding an alternate solution to a plane-geometry problem.
I'm having some trouble finding an alternate solution to the following problem:

Let us have a right-angled triangle $ABC$ ($A$ being the right angle vertex). We draw an altitude to the hypotenuse creating point $D$ and an angle bisector of the vertex angle C which intersected with $AB$ gives us another point $E$. Prove that $|AE| \cdot|AC|=|BE| \cdot |CD|. $

I do know how to solve the problem with the use of analytic geometry, however I'm curious wether there's some less computationally demanding way. Any help is appreciated.

答:
Let $ \tan \frac{C}{2}  = t$.  Let $AC = 1 $.

Show that

$AE = t$

$BE = \tan C - t = \frac{ 2t } { 1-t^2 }  - t $

$CD = \cos C = \frac{1-t^2}{1+t^2}$

Hence, $ |AE | \times |AC | = |BE| \times |CD |$


问:
Simple equation for simulating different acceleration curves?
For a video game, I would like to simulate different acceleration curves for different types of cars.

I found out about Logistic and Sigmoid functions and while interesting they produce a symmetrical result:

I'm looking for something asymmetrical, like one car that accelerates quickly, another car that accelerates slowly and so on. (shown in red)

For my input, it will be linear from 0.0 to 1.0, as long as joystick button is held down, I increment X. (shown in black)

Can you suggest a simple equation that would allow me to generate such kind of curves by tweaking a few parameters?

答: Maybe B-Splines are of interest to you.</p>

Body: Maybe B-Splines are of interest to you.


问:
Evaluating $\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$
I have been working on the following integral:
$$\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$$
where $t$ is any nonnegative real number. Would anyone be able to provide a hint or provide a solution on how such an integral should be approached? I have a hunch it involves differentiation under the integral sign.

答:
Firstly, if you want to prove Frullani's integral:
$$I(a,b)=\int_0^\infty\frac{f(ax)-f(bx)}{x}dx$$
$$\frac{\partial I}{\partial a}=\int_0^\infty f'(ax)dx=\frac1a\int_0^\infty f'(x)dx$$
$$\frac{\partial I}{\partial b}=\int_0^\infty f'(bx)dx=\frac1b\int_0^\infty f'(x)dx$$
also:
$$dI=\frac{\partial I}{\partial a}da+\frac{\partial I}{\partial b}db$$
and so:
$$\int dI=\int_0^\infty f'(x)dx\left(\int\frac {da}a-\int\frac{db}b\right)$$
and so:
$$\int_0^\infty\frac{f(ax)-f(bx)}{x}=\left[\lim_{x\to\infty }f(x)-\lim_{x\to 0^+}f(x)\right]\left(\ln a-\ln b\right)$$

now that we have this we want to define our function, we can say:
$$f(x)=e^{-x^2}$$
and so this gives us: $a=\sqrt{t},b=1$ and so finally our integral is equal to:
$$I=\left[f(\infty)-f(0)\right]\left(\ln\sqrt{2}-\ln 1\right)$$
$$I=(0-1)(\ln\sqrt{t}-0)=-\ln\sqrt{t}=-\frac12\ln t$$


问:
Convergence of $\int_0^{\infty} s^{2-2\alpha}e^{is^2}ds$
I need to evaluate the following integral

$$\int_0^{\infty} s^{2-2\alpha}e^{is^2}ds$$
where $\alpha\in[0,\,1]$. I shall call this integral Gamma-function like integrals since it can be brought to Gamma function as follows:

One first replace $s=\sqrt{it}$, then the integral transforms to
$$\frac{i^{3/2-\alpha}}{2}\int_0^{-i\infty} t^{1/2-\alpha}e^{-t}dt$$

Finally, one can analytically continue the integral from negative $t$ axis to real $t$ axis to obtain $\frac{i^{3/2-\alpha}}{2}\Gamma(3/2-\alpha)$.

When I throw the two integrals into Mathematica, it does gives me the above result. However, it seems to tells me the integrals only converge on $[1/2,1]$ and diverge on $[0, 1/2]$. Could anybody explain why it is the case?

Update: I think now I understand the convergence by myself. Let us start with the second integral. We denote break the second integral into two parts: $I=I_1+I_2$, where
$$I_1=\int_0^{-i\epsilon} t^{\beta}e^{-t}dt$$
$$I_2=\int_{-i\epsilon}^{-i\infty} t^{\beta}e^{-t}dt$$
and $\beta=1/2-\alpha$

To guarantee $\lim_{\epsilon\to0}I_1=0$, we must have
$$\beta>-1$$

Now we would like to analytically continue the integral for $I_2$ to the real positive axis. But the condition of doing this is that the integrand should not have any singularity on the fourth quadrant, apart from the small quarter-circle of radius $\epsilon$ near the origin.
This analyticity of the integrand is guaranteed only if it vanishes at $t=-i\infty$. So we should require

$$\beta<0$$
which leads to

$$I_2=\int_{\epsilon}^{\infty} t^{\beta}e^{-t}dt$$

When $\beta \in(-1,0)$,i.e., $\alpha\in(1/2,3/2)$, we can express $I$ as
$$I=\int_{0}^{\infty} t^{\beta}e^{-t}dt=\Gamma(\beta+1)=\Gamma(3/2-\alpha)$$

答:
The following is partially analytical but relies on Mathematica for the closed-forms. We start by deforming the contour to the ray $\arg s=\phi$ where $0<\phi<\pi/2$. This is accomplished by the substitution $s=re^{i\phi}$, obtaining

\begin{align}
I(\phi)
&:=\int_0^{\infty e^{i\phi}} s^{2-2\alpha}\exp(i s^2)\,ds\\
&=e^{(1-2\alpha)i\phi}\int_0^\infty r^{2-2\alpha} \exp(i e^{2i\phi}r^2)\,dr\\
&=e^{(1-2\alpha)i\phi}\int_0^\infty r^{2-2\alpha} \exp(i r^2 \cos(2\phi))\exp(-r^2 \sin(2\phi))\,dr.\\
\end{align}

Since $\sin(2\phi)>0$ due to the assumption on $\phi$, the integrand decays like a Gaussian at infinity and so the integral converges. With this in mind, we can ask Mathematica to carry out the integral under this assumption:

Integrate[Exp[I r^2 Exp[2I \[Phi]]]r^(2-2\[Alpha]),{r,0,\[Infinity]},Assumptions->0<\[Phi]<\[Pi]/4]

That is, the integral converges so long as the real part of $\alpha$ doesn't exceed $3/2$. Including the prefactor and taking the limit $\phi\to 0^+$, we obtain

FullSimplify[Limit[ConditionalExpression[ 1/2 (-I E^(2 I \[Phi]))^(-(3/2) + \[Alpha])Gamma[3/2 - \[Alpha]], \[Alpha] + Conjugate[\[Alpha]] < 3],\[Phi]->0,Direction->"FromAbove"]]

Since Mathematica uses $-i=e^{-i\pi/2}$, we obtain $$I(0^+) =\frac12 e^{i(3/2-\alpha)\pi/2}\tag{Re $\alpha<3/2$}$$
which is seemingly valid on the entirety of $\alpha\in[0,1]$.

A warning, though. On the one hand, for $1/2\leq \alpha<1$ the above formula agrees with the original Mathematica calculation. But the new calculation purports to be valid for $\alpha\leq 1/2$ as well. I'm not sure I understand the issue myself, but note that the exponent $e^{i s^2}$ has modulus 1 along the real line. Hence the rate at which the integrand's modulus grows along the real line is entirely controlled by $s^{2-2\alpha}$: For $\alpha>1/2$ the modulus grows slower than linear, whereas for $\alpha<1/2$ it is faster than linear. Evidently this transition is enough to Mathematica to declare the integral divergent.

By contrast, if we deform the contour just above the real line then the modulus of $e^{i s^2}$ decreases exponentially fast and so there's no issue with convergence. But it does raise the question of what exactly we intend by the original integral: Is it okay to regularize the growth in this way, or is it not? That will probably depend on the original context.


问:
Relaxation of valuation axioms?
Let $K$ be a field, and $v:K\rightarrow \mathbb{R}$ be a valuation, i.e. a function such that, for all $x,y\in K$: $v(x)\geq 0$, $v(x+y)\leq v(x)+v(y)$ (if $v$ is Archimedian, alternatively $v(x+y)\geq \min(v(x),v(y))$) and $v(xy)=v(x)v(y)$. My question is, are there meaningful objects that come out of relaxing any of these conditions? I would be particularly interested to know if one can relax to a "super" or "sub" valuation satisfying $v(xy)<v(x)v(y)$ or $v(xy)>v(x)v(y)$.

答:
Another broad generalisation is that of filtered rings (and modules). Some common filtrations are basically valuations except you allow $v(xy) \ge v(y)+v(y)$.

The theory is rich. Concepts you can search for are the associated graded ring (module, algebra), or the Artin-Rees condition, or microlocalisations. Bourbaki has an entire chapter about filtrations and graduations (chapter III in Algèbre Commutative), another bible is https://link.springer.com/book/10.1007/BFb0067331.

Title: Relaxation of valuation axioms?


问:
Help with a probability homework problem
I'm trying to solve the problem in the image but having trouble formulating the problem mathematically:

There are $20$ identical laptops on a trolley, out of which $12$ have
a hard disk with capacity $160$ GB and $8$ with a capacity of $320$
GB. A teacher randomly takes two laptops from the trolley. A student
then takes a laptop from the trolley to complete a project. Given that
the student took a laptop with $160$ GB, find the probability that the
teacher took both laptops with $320$ GB.

so any help will be appreciated.

I've tried the following: let $A$ denote the event of choosing a 160GB laptop, and $B$ denote the event of choosing a 320GB laptop. Then $P(A)= 12/20, P(B)=8/20.$ Next, let $C$ denote the event of choosing 2 laptops, regardless of their memory (160GB or 320GB). Then $P(C)=2/20$ (right?) Then $B \cap C$ will denote the event of choosing two 320GB laptops (right?). But then how do I calculate $P(B \cap C)?$

Also, how do I calculate the probability of $D:=$  choosing two 160GB laptops and $P(A|D)$? I understand that at the end we'll need to calculate $P(D|A)$, and hence we'll need to use the formula $P(D|A)= \frac{P(D \cap A)}{P(A)}= 20/12 * P(D \cap A)=20/12 * P(A|D) P(D).$ Here's where I'm confused: how to I calculate $ P(A|D), P(D)?$

答:
I don't think the student picking a laptop with a 160 GB hard drive after the fact affects the probability of the teacher choosing two laptops with 320 GB hard drives. If the question was what is the probability of the teacher choosing two 320 GB laptops and then a student choosing a 160 GB, that would be a different story. But, let's focus on the question at hand.

Let A represent the event of choosing a laptop with a 320 GB hard drive on the first try and B represent the event of choosing a laptop with a 320 GB hard drive on the second try.

The probability of choosing 320 GB on the first try and 320 GB on the second try is $$P(A\cap B) = P(B|A)P(A).$$

The probability of A is calculated as follows:

$$P(A) = \frac{8}{20} = \frac{2}{5}.$$

Since there are 7 laptops with 320 GB hard drives remaining of 19 total laptops, the probability of B given A is calculated as follows:

$$P(B|A) = \frac{7}{19}$$.

Therefore, the probability of both A and B occurring is:

$$P(A\cap B) = P(B|A)P(A) = (\frac{7}{19})(\frac{2}{5}) = \frac{14}{95}.$$


问:
Infection in a village
Consider the following problem:

Suppose a lonely wanderer infected with a virus came into an isolated village with $M$ villagers and stayed there. Every week each of the infected villagers coughs onto $n$ random other villagers (each of them chosen uniformly and independently among everyone) and then develops antibodies becoming immune to it. All villagers who are coughed upon become infected if they are not immune. Nobody left or entered the village after the arrival of the lonely wanderer. Consider time to be discrete and measured in weeks. We say, that the virus survives as long as someone is infected with it. For what $n$ is the expected time of its survival the longest?

The extremum clearly is not achieved in the border cases here.

Indeed, if $n = 0$ the lonely wanderer becomes immune before being able to infect anyone else, thus the virus will survive only for $1$ week.

If $n \to \infty$ the probability that the lonely wanderer infects everyone in the first week tends to $1$. Thus the expected time of the survival of the virus tends to $2$ in this case.

So, we must look for optimal $n$ somewhere in between. However, I have no idea how to find it (or even its asymptotic for large $M$)…

At the first glance the problem looked to be somewhat similar to two well studied problems: branching processes (villagers infected by a given infected villager - their descendants in terms of branching processes) and coupon collector problem (uninfected villagers as coupons to be collected). However, it is different from both of them (the number of ‘descendants’ changes each turn here, which makes it different from a Galton-Watson branching process, and the number of ‘coupons collected per turn’ depends on the number of ‘coupons collected on the previous turn’, which makes it different from a classical coupon collector) and methods, similar to the ones used to solve them, are unlikely to work here.

答:
Putting that at week $w$ we have the following numbers of infected, susceptible (infectable), and immunized
$$
I(w) = {\rm infected}\quad S(w) = {\rm infectable}\quad H(w) = {\rm immunized}
$$
then, allowing for the numbers to be rationals, we shall have
$$
\left\{ \matrix{
  I(0) = 1\quad S(0) = m\quad H(0) = 0 \hfill \cr
  I(w) + S(w) + H(w) = m + 1 \hfill \cr
  H(w) = I(w - 1) \hfill \cr
  I(w+1) = {n \over {m + 1}}S(w) \hfill \cr}  \right.
$$

Solving for $I$ (excluding the banal case $n=0$)
$$
I(w) + {{m + 1} \over n}I(w + 1) + I(w - 1) = m + 1\quad \left| \matrix{
  \;I(0) = 1 \hfill \cr
  \;I(1) = n{m \over {m + 1}} \hfill \cr}  \right.
$$
and recasting it by including the initial conditions,  so that the recursion is valid for any $0 \le w$, under the acception that $w <0 \; \Rightarrow  \; I(w)=0 $
$$
I(w) + {n \over {m + 1}}I(w - 1) + {n \over {m + 1}}I(w - 2)
 + \left[ {0 = w} \right]\left( {n - 1} \right) = n\quad \left| {\;0 \le w} \right.
$$
We can solve the above via ordinary generating function
$$
F(z) = \sum\limits_{0\, \le \,w} {I(w)\,z^{\,w} }
$$
readily obtaining
$$
\eqalign{
  & F(z) = {{{n \over {1 - z}} - \left( {n - 1} \right)} \over {1 + {n \over {m + 1}}\left( {z + z^{\,2} } \right)}}
 = {{m + 1} \over n}{{1 + \left( {n - 1} \right)z}
 \over {\left( {1 - z} \right)\left( {z^{\,2}  + z + {{m + 1} \over n}} \right)}} =   \cr
  &  = {{m + 1} \over n}{{1 + \left( {n - 1} \right)z}
 \over {\left( {1 - z} \right)\left( {z + {1 \over 2} + {{\sqrt {1 - 4\left( {m + 1} \right)/n} } \over 2}} \right)
\left( {z + {1 \over 2} - {{\sqrt {1 - 4\left( {m + 1} \right)/n} } \over 2}} \right)}} =   \cr
  &  = {A \over {\left( {1 - z} \right)}} + {B \over {\left( {z + {1 \over 2}
 + {{\sqrt {1 - 4\left( {m + 1} \right)/n} } \over 2}} \right)}}
 + {C \over {\left( {z + {1 \over 2} - {{\sqrt {1 - 4\left( {m + 1} \right)/n} } \over 2}} \right)}} \cr}
$$
where, however, $A,B,C$ are rather complicated expressions
of $m,n$.

Normally it would be $n \le m$, so that the square roots will turn into immaginary values.

Title: Infection in a village


问:
What does the phrase "by a polynomial in a finite number of functions of $\mathcal{F}$ means?
In Royden's "Real Analysis" (second edition), there is the following exercise concerning the Stone-Weierstrass Theorem (it can be found on page 175, Chap. 9, Sec. 7):

33. Let $\mathcal{F}$ be a family of continuous real-valued functions on a compact Hausdorff space X, and suppose that $\mathcal{F}$ separates the points of X. Then every continuous real-valued function on X can be uniformly approximated by a polynomial in a finite number of functions of $\mathcal{F}$.

I am trying to understand what does the phrase "a polynomial in a finite number of functions of $\mathcal{F}$" menas in this context, but I am failing to do so. I would greatly appreciate if anyone can enlighten me with this one (not with the solution, just with an explanation of what this phrase means).

Also, this problem appears as 44 on Royden's "Real Analysis" (third edition) in Chap. 9, Sec.9 and as 33 on Royden's "Real Analysis" (fourth edition) in Chap. 12 Sec. 12.3.

答:
A real polynomial in $k$ variables is a function of variables $X_1,X_2,\ldots,X_k$ that can be expressed as a linear combination of products of these variables, with real coefficients.  The general definition is therefore a finite sum of the form
$$
  p(X_1,X_2,\ldots,X_k) = \sum a_{n_1,n_2,\ldots,n_k}X_1^{n_1}X_2^{n_2}\ldots X_k^{n_k},
  $$
where each index (and exponent) $n_1,n_2,\ldots,n_k$ can vary from 0 up.

If $f_1,  f_2,  \ldots , f_k$ are members of your family $\mathcal F$,  then you can compose them with a polynomial such as the one above,
obtaining  a function $F$ defined on  $X$ as follows:
$$
  F(x) =   p\big (f_1(x), f_2(x), \ldots  f_k(x)\big ).
  $$

The function $F$ so defined is what one might call a polynomial in a finite number of functions of $\mathcal{F}$.

Of course,  by choosing other polynomials $p$ and other finite subsets of $\mathcal F$ you will get a whole lot of such
functions.

PS: For the cognoscenti,  I am fully aware that I am adopting a functional interpretation of the concept of a polynomial,  which is
perfectly suited to answer the present question!


问:
Prove an initial value problem blows up in finite time.
Given the initial value problem $$y'(t)=p(t)+y^2(t)$$ $$y(0)=\frac{1}{2}$$ for $t>0$
where $p(t)$ is a continuous function such that  $p(t)>0$.

Prove the initial value problem blows up in finite time.

My attempt, $$y'\geq p(t)$$
So $$y(t) \geq \int_0^t p(s)ds$$
For some $t_0$ I get  $$y(t_0) \geq \int_0^{t_0} p(s)ds$$
Now I have $$\frac{y'}{y^2} \geq 1$$ and by integrating both sides from $t_0$ to $t$ I get $$y(t)\geq \frac{1}{\frac{1}{y_0}-t+t_0}$$
So since the lower bound blows up for $$t=t_0+\frac{1}{y(t_0)}$$ so does $y$.

Is my attempt correct?

答:
Your attempt is not correct as the inequality $\frac{y^\prime}{y}\ge 1$ is not correct.

Let’s consider a solution at the right of $0$. As $p$ is supposed to be positive, $y(t)$ is increasing and we have

$$0\le \frac{y^\prime(t)}{p(t)+y^2(t)}=1\le  \frac{y^\prime(t)}{y^2(t)} $$ integrating the inequalities on $[0,t]$ we get

$$0 \le t \le 2-\frac{1}{y(t)}\le 2 $$ as $y(t) \ge y(0)=\frac{1}{2}$.

A solution can’t exist beyond $t=2$.


问:
Eigenvalues of a block matrix given eigenvalues of the blocks
I want to find the eigenvalues of the following $6\times6$ block matrix:

\begin{pmatrix} A & B \\ B & C
\end{pmatrix}
having zero diagonal terms, where each block is a $3\times3$ matrix of given eigenvalues,
Is their any property I can use?

答:
It is not determined by the eigenvalues of the blocks.  Consider
$$\pmatrix{0 & 1 & 1 &  -2 \sin(t) \cos(t) & 2 \cos^2(t)-1 & \cos(t)-\sin(t)\cr
1 & 0 & 1 & 2 \cos^2(t)-1 & 2 \sin(t) \cos(t) & \sin(t)+\cos(t)\cr
1 & 1 & 0 & \cos(t)-\sin(t) & \sin(t)+\cos(t) & 0\cr
-2 \sin(t) \cos(t) & 2 \cos^2(t)-1 & \cos(t) - \sin(t) & 0 & 1 & 1\cr
2 \cos^2(t)-1 & 2 \sin(t) \cos(t) & \sin(t)+\cos(t) & 1 & 0 & 1\cr
\cos(t)-\sin(t) & \sin(t)+\cos(t) & 0 & 1 & 1 & 0}
$$
where $A,B,C$ all have the same eigenvalues $2, -1, -1$ but the block matrix has characteristic polynomial $\lambda (\lambda + 2)(\lambda -2 - 2 \cos(t)) (\lambda + 2 \cos(t))(\lambda^2 + 4 \cos^2(t) + 4 \cos(t) - 8)$
that depends on $t$.


问:
Series representation of the reciprocal of prime
In the article on unique primes in Wikipedia. It is stated that the representation of the reciprocal of a prime $p$ in the numeral base $b$ is periodic of period $n$ if

$$\displaystyle {\frac {1}{p}}=\sum _{i=1}^{\infty }{\frac {q}{(b^{n})^{i}}}$$ where $q$ is a positive integer smaller than $b^n$

I have no idea how this particular representation is obtained. Kindly provide a proof or point to some resources where I can find this representation of $\frac{1}{p}$.

Thanks in advance !

Update:

Let $p$ be a prime and $n$ be the period of $\displaystyle \frac{1}{p}$ in base $b$ and $b \not\mid p$.

If,  $\displaystyle \frac{1}{p} = (0.\overline{a_1a_2a_3\dots a_n})_b$ where $a_1, a_2, a_3, \dots \in \mathbb{Z}_b$

Then, $\displaystyle \frac{1}{p} = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right) = \sum_{i=1}^{\infty}\frac{q}{(b^n)^i}$, where $ q = a_1a_2a_3 \dots a_n$.

I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds

Body:
I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds

Well, $(0.\overline{a_1a_2a_3\dots a_n})_b=0.a_1a_2a_3\dots a_na_1a_2a_3\dots a_na_1a_2a_3\dots a_n\dots$

$=\dfrac{a_1}{b}+\dfrac{a_2}{b^2}+\cdots\dfrac{a_n}{b^n}+\dfrac{a_1}{b^{n+1}}+\dfrac{a_2}{b^{n+2}}+\cdots\dfrac{a_n}{b^{2n}}+\dfrac{a_1}{b^{2n+1}}+\dfrac{a_2}{b^{2n+2}}+\cdots+\dfrac{a_n}{b^{3n}}+\cdots$

$=\dfrac{a_1b^{n-1}+a_2b^{n-2}\cdots+a_n}{b^n}+\dfrac{a_1b^{n-1}+a_2b^{n-2}+\cdots+a_n}{b^{2n}}+\dfrac{a_1b^{n-1}+a_2b^{n-2}+\cdots+a_n}{b^{3n}}+\cdots$

$=a_1a_2\dots a_n\left(\dfrac1{b^n}+\dfrac1{b^{2n}}+\dfrac1{b^{3n}}+\cdots\right).$


问:
Evaluate $\int \dfrac{1}{\sqrt{1-x}}\,dx$
Find $$\int \dfrac{1}{\sqrt{1-x}}\,dx$$

I did this and got $\dfrac23(1-x)^{\frac32} + c$

But a online calculator is telling me it should be $-2(1-x)^{\frac12}$

What one is on the money and if not me why?

Body: $$\int\frac1{\sqrt x}dx=2\sqrt x+C\implies \int\frac1{\sqrt{ax+b}}dx=\frac2a\sqrt{ax+b}+C\;,\;\;a,b,C\;\text{constants}$$


问:
Proof that $(a,b) = (c,d)$
Show that if the ordered pair $(a,b)$ is defined to be $\{\{a\}, \{a,b\}\}$ then $(a,b) = (c,d)$ if and only if $a=c$ and $b=d$.

Someone attempted to prove that and started by the forward direction $\Rightarrow$ saying that let us assume that $(a,b) = (c,d)$, then by the definition of ordered pair $(a,b) = \{\{a\}, \{a,b\}\}$:

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

So, my question is as to why he assumed that $$\{\{a\}, \{a,b\}\} = \{\{c\}, \{c,d\}\}$$ from the definition of the ordered pair $(a,b) = \{\{a\}, \{a,b\}\}$ given the only thing given was $(a,b) = \{\{a\}, \{a,b\}\}$?

答:
$\Rightarrow$ Assume that $a=c$ and $b=d$. Then $(a,b) = \{\{a\}, \{a,b\}\} = \{\{c\}, \{c,d\}\} = (c,d) \implies (a,b) = (c,d)$

$\Leftarrow$ Assume that $(a,b) = (c,d)$ Then $(a,b) = \{\{a\}, \{a,b\}\} = (c,d) \implies c=a, b=d$ by the given definition of an ordered pair.

As @azif00 said in the comment to your question, you might find it easier to define $(x,y)$ as $\{\{x\}, \{x,y\}\}$ and using this definition, prove $(a,b) = (c,d) \iff a=b, c=d$

Title: Proof that $(a,b) = (c,d)$


问:
Explosion rule in paraconsistent logic
In classical logic the rule of noncontradiction states $\neg(A\wedge\neg A)$
In the Paraconsistent logic, this rule doesn't apply, at least in some case. Now, in the classical logic, if we avoid to state this rule, we can't avoid the 'ex falso sequitur quodlibet', that means:
$$(A\wedge\neg A)\to B$$
So, under what conditions we can accept the noncontradiction rule to be non valid in order to avoid the $(A\wedge\neg A)\to B$? Or, alternatively is there in this logic a rule that allows to accept this 'explosion'?</p>

答: The principle of explosion is not valid in paraconsistent logic. Somewhat loosely, the meaning of an implication $p\to q$ is such that if $p$ is false, then $p\to q$ has not truth value. It's a bit like saying "don't ask me about such nonsense of whether $q$ follows from $p$ with $p$ is not even true". </p>

Title: Explosion rule in paraconsistent logic


问:
From derivatives to gradients in backpropagation
Consider the following explanation of backpropagation from Wikipedia:

Given an input–output pair {\displaystyle (x,y)}(x,y), the loss is:

$ C(y,f^{L}(W^{L}f^{L-1}(W^{L-1}\cdots f^{1}(W^{1}x)\cdots )))$

The derivative of the loss in terms of the inputs is given by the chain rule; note that each term is a total derivative, evaluated at the value of the network (at each node) on the input $x$:

${\frac {dC}{da^{L}}}\cdot {\frac {da^{L}}{dz^{L}}}\cdot {\frac {dz^{L}}{da^{L-1}}}\cdot {\frac {da^{L-1}}{dz^{L-1}}}\cdot {\frac {dz^{L-1}}{da^{L-2}}}\cdots {\frac {da^{1}}{dz^{1}}}\cdot {\frac {\partial z^{1}}{\partial x}}$

These terms are: the derivative of the loss function; the derivatives of the activation functions; and the matrices of weights:

$\displaystyle {\frac {dC}{da^{L}}}\cdot (f^{L})'\cdot W^{L}\cdot (f^{L-1})'\cdot W^{L-1}\cdots
(f^{1})'\cdot W^{1}.$

So far so good. Then they say:

The gradient $\nabla$  is the transpose of the derivative of the output in terms of the input, so the matrices are transposed and the order of multiplication is reversed, but the entries are the same:

$ \nabla _{x}C=(W^{1})^{T}\cdot (f^{1})'\cdots \cdot (W^{L-1})^{T}\cdot (f^{L-1})'\cdot (W^{L})^{T}\cdot (f^{L})'\cdot \nabla _{a^{L}}C.$

Why? i.e. why does converting this to "gradient form" (not sure if that's what they actually did above) requires "transposing matrices and reversing the ordering of multiplication"?

In case it helps, I thought this article here (layout conventions in matrix calculus) may help, but I'm not able to cross-reference hat convention was used above, and why vectors / matrices are transposed, and why they chose to re-arrange the ordering of the terms

答:
This is a quite late, but I also had the same issue while trying to understand back-propagation and implementing it in code. I figure others may run into this as well.

I believe the transpose is there simply to reduce the computational load required to evaluate the first expression, i.e:

$$
{dC \over da^L} \cdot {da^L \over dz^L} \cdots {\partial z^1 \over \partial x}
$$

Since every term after ${dC \over da^L}$ is either a matrix or higher ranked tensor, evaluating it from right to left, as you would, is considerably more expensive than starting with the vector $dC \over da^L$ and having it collapse into more simple vector/matrix operations and then transposing it back at the end; hence, we use the fact that for matrices $A, B, C$ where $ABC$ is defined, $ABC = (C^T B^T A^T)^T$:

$$
ABC = (C^T B^T A^T)^T = (C^T (AB)^T)^T = ABC
$$

And this makes sense because the gradient of a scalar function is, of course, a vector, and for all practical purposes, the transpose of it is insignificant.


问:
Show that $\int |f_n-f|\to 0$

Let $f$ be an integrable function and $(f_n)_{n=1}^\infty$ a sequence of non-negative measurable functions such that $f_n\to f$ and $\int_\Omega f_n\,d\mu\to \int_\Omega f\,d\mu$.
Show that
$
   \int_\Omega |f_n-f|d\mu\to 0.
  $

The naive idea of using $|\int f|\leq \int|f|$ fails because the $\leq$ sign points in the wrong directio.

This question is a follow-up to monotone convergence for decreasing functions, so perhabs we should use that?
I think that since $f_n\to f$ we can take $|f_n-f|$ decreasing, then
$$
\lim_{n\to\infty}\int_\Omega |f_n-f|d\mu = \int_\Omega \lim_{n\to\infty}|f_n-f|d\mu,
$$
and perhabs what's inside vanishes.
I think this is incorrect, because we didn't use the hypothesis $\int f_n \to \int f$.
Any ideas?

答:
The question asks to prove the result:

Let $f$ be an integrable function and $(f_n)_{n=1}^\infty$ a sequence of non-negative measurable functions such that $f_n\to f$ (pointwise) and $\int_\Omega f_n\,d\mu\to \int_\Omega f\,d\mu$. Then $ \int_\Omega |f_n-f|d\mu\to 0$.

We are going to present two ways proofs.

Proof $1$. Here is a proof using Dominated Convergence Theorem.

Since $\{f_n\}$ is a sequence of nonnegative measurable functions and $\{f_n\}$ converges to the function $f$ pointwise, we have that $f\geqslant 0$. So, since $f$ is integrable and $f\geqslant 0$, we have that $\int_\Omega f d\mu=\int_\Omega|f|d\mu<+\infty$

Now consider $f_n \wedge f$ defined by $(f_n \wedge f)(x)=\min\{f_n(x),f(x)\}$, for each $x \in \Omega$.

Since $\{f_n\}$ converges to $f$,
we have that $\{f_n \wedge f\}$ converges to $f$. But we know that, for all $n$, $\vert f_n \wedge f \vert =  f_n \wedge f \leqslant f $ and   $\int_{\Omega }  f d\mu< \infty $.  So we can apply Lebesgue Dominated Convergence Theorem and we have that
$$\lim_{n \to \infty}\int_{\Omega }  f_n \wedge f d\mu = \int_{\Omega }  f d\mu$$
To conclude the proof, note that
$$\vert f_n-f\vert = f_n+f-2(f_n\wedge f)$$
So
$$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } f_n d\mu +\int_{\Omega } f d\mu -2\int_{\Omega } (f_n\wedge f) d\mu $$
And since $\lim_{n \to \infty}\int_{\Omega }  f_n d\mu = \int_{\Omega }  f d\mu$, we have
$$  \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

Proof $2$. Here is a proof using Fatou's lemma.

Since $\{f_n\}$ is a sequence of nonnegative measurable functions and $\{f_n\}$ converges to the function $f $ pointwise, we have that $f\geqslant 0$.

Note that $|f_n -f|\leqslant f_n +f$. So, for each $n$, the function $f_n +f - |f_n -f|$ is non-negative.

Note also that, since $\{f_n\}$ converges to $f $ pointwise, we have $|f_n-f|$ converge to $0$ pointwise, and so
$ 2f= \lim (f_n+f - |f_n-f|) = \lim\inf (f_n+f - |f_n-f|)$.

Since $\lim_{n \to \infty} \int_\Omega f_n d\mu = \int_\Omega f d\mu$, using Fatou's Lemma, we have:
\begin{align}
2 \int_\Omega f d\mu
&=\int_\Omega \lim\inf(f_n +f - |f_n -f|)d\mu \leqslant \lim\inf \int_\Omega (f_n +f - |f_n -f|)d\mu = \\
&=\lim\inf \left (\int_\Omega f_n d\mu +\int_\Omega f d\mu - \int_\Omega|f_n -f|d\mu \right) = \\
&= \left(\lim\inf\int_\Omega f_n d\mu\right) +\int_\Omega f d\mu - \left(\lim\sup\int_\Omega |f_n -f|d\mu\right)  = \\
&=2\int_\Omega f d\mu - \left(\lim\sup\int_\Omega|f_n -f|d\mu\right)
\end{align}
So we have
$$2 \int_\Omega f d\mu \leqslant 2\int_\Omega f d\mu - \left(\lim\sup\int_\Omega |f_n -f|d\mu\right) $$
Since $f$ is integrable and $f \geqslant 0$, we know that $\int_\Omega f d\mu=\int_\Omega|f|d\mu<+\infty$, and so we get
$$\lim\sup\int_\Omega|f_n -f|d\mu \leqslant  0$$
So we can conclude that
$$\lim\int_\Omega|f_n -f|d\mu =  0$$

Title: Show that $\int |f_n-f|\to 0$


问:
Confusion on the definition of accumulation points
I've been trying to learn a bit about limits of sequences and accumulation points to get a better intuition behind the workings for calculus and I got confused on the definitions of limits,limit points and accumulation points of sequences and sets.

My first question is a limit of a sequence the same as the accumulation point and is that the same as the limit point i looked online and its all very vague.
My second confusion is that is the limit of a sequence the same as the limit of a set if not is there some proof or intuitive explanation as to why not?.

I know this is probably a very simple and probably trivial concept for all of you on here but its confused me a lot.
Thanks in advance

答:
A limit point is the same thing as an accumulation point, and its definition is that:

A point $x$ is a limit point of a set $A$ if for every neighbourhood $S$ of $x$ there exists $y \in S$ such that $y \in A$, $y \neq x$.

I strongly prefer the name "accumulation point", because you are not actually taking limits here... it's the other way around! In order to be able to do limits you normally require accumulation points, since the topological definition of a limit requires taking neighbourhoods and computing the function there.

About your second question:

A point $x$ is an accumulation point for a sequence $\{x_n\}$ if any neighbourhood $S$ of $x$ is such that there are infinitely many indices $n$ such that $x_n \in S$.

It is essentially the same definition as above, but you take $A=\{x_n\}_{n \in \mathbb{N}}$. However, a point is a limit point for a sequence if all the indices after a certain $n$ are in any neighbourhood. Formally:

A point $x$ is the limit of a sequence $\{x_n\}$ if any neighbourhood $S$ of $x$ is such that there exists $N \in \mathbb{N}$ such that $x_n \in S$ for all $n>N$.

And this is stronger than simply being an accumulation point: you can see the difference by considering the sequence $x_n = \frac{(-1)^n n}{n+1}$. Any neighbourhood of $1$ contains infinitely many points of this sequence, namely all the $x_{2n}$ after a certain $n$. Similarly, any neighbourhood of $-1$ will contain all the $x_{2n+1}$ after a certain $n$, so both $1$ and $-1$ are cluster points for $x_n$. However, there is no limit (in fact limits are unique, if they exist).


问:
Find the parameter $a$ and the coordinate $x$ of a function so that the function's maximum has a given value
I have implemented in C++ the code for a quasi-Newton method given by

$x_{n+1} = x_n - \dfrac{\delta f(x_n)}{f(x_n+\delta) - f(x_n)}$

for a computational project in which $f$ is a function given by:

$f(x) = 2e^{-ax}(a\sin(3x) + \cos(3x))$

One of the exercises asks to find the parameter $a$ and the coordinate $x$ so that the maximum value of the function $f$ is 3. In the problem is left as a suggestion that we use the above quasi-Newton method implemented to find the value of $x$ and the bisection method to find $a$. Any tip on how to do this?

Thanks,

Patrick

答:
Your Quasi-Newton method is not correct, you need a $-$ after $x_n$. Note also that if $x$ is a local maximum, then $f'(x)=0$ and to solve this with your Quasi-Newton you don't have to provide $f''(x)$ analytically, as it's already estimated "in place" with finite differences ( i.e. $f'(x_0)= \frac{f(x_0+\delta) - f(x_0)}{\delta} + O(\delta)$).

After a quick test, I obtained $x = -0.09679574815$, $a= -6.209538561$. A plot confirms that the local maximum is $3$ at point $x$.

Another found solution is $x=-2.052921249$ and $a=0.1898554466$.

I applied "externally" the bisection method for the parameter $a$ and inside of it I used the quasi-Newton method in order to solve for $x$ the equation  $f_a'(x)=0$ for different parameters $a$. As $(x,a)$ are coupled, you can't split the two methods.

I wrote the following C++ snippet: here a1 and a2 are the leftmost and rightmost point of the starting interval in the bisection routine.

```<code>    #include <cmath>
    #include <iostream>
    #include <iomanip>      // std::setprecision

    constexpr double tol = 1e-12;
    constexpr double delta = 1e-5;
    double f(double x, double a){
        return 2*exp(-a*x)*(a*sin(3*x)+cos(3*x));
    }

    double fprime(double x, double a){
        return exp(-a*x)*(4*a*cos(3*x) - 2*(a*a+3)*sin(3*x));
    }

    double quasi_newton(double f(double ,double),double x0,double a){
        double res =  -(delta*f(x0,a) )/(f(x0+delta,a)-f(x0,a));
        while (std::fabs(res)>tol) {
            x0+=res;
            res = -(delta*f(x0,a))/(f(x0+delta,a)-f(x0,a));
        }
        return res+x0;
    }

    int main(){
        double x0=0.5;
        int it{0};
        double a1{-10.0};
        double a2{10.0};
        double c{0};

        double xc,gc,ga1,ga2,xa1,xa2;
        while (std::fabs(a1-a2)>tol) {
            std::cout<< std::fabs(a1-a2) <<std::endl;
            c =0.5*(a1+a2);
            xc = quasi_newton(fprime,x0,c);
            xa1 = quasi_newton(fprime,x0,a1);
            xa2 = quasi_newton(fprime,x0,a2);

            gc = f(xc,c)-3.0;
            ga1 = f(xa1,a1)-3.0;
            ga2 = f(xa2,a2)-3.0;

            if (gc*ga1<0) {
                a2 = c;
            }
            if (gc*ga2<0) {
                a1=c;
            }

        }
        std::cout <<"Point: " << std::setprecision(10) << xc << ", Value of a: " <<std::setprecision(10)<< c << ", Check:" << std::setprecision(10) << f(xc,c) << std::endl;
        return 0;

    }
</code>```

Body:
Your Quasi-Newton method is not correct, you need a $-$ after $x_n$. Note also that if $x$ is a local maximum, then $f'(x)=0$ and to solve this with your Quasi-Newton you don't have to provide $f''(x)$ analytically, as it's already estimated "in place" with finite differences ( i.e. $f'(x_0)= \frac{f(x_0+\delta) - f(x_0)}{\delta} + O(\delta)$).

After a quick test, I obtained $x = -0.09679574815$, $a= -6.209538561$. A plot confirms that the local maximum is $3$ at point $x$.

Another found solution is $x=-2.052921249$ and $a=0.1898554466$.

I applied "externally" the bisection method for the parameter $a$ and inside of it I used the quasi-Newton method in order to solve for $x$ the equation  $f_a'(x)=0$ for different parameters $a$. As $(x,a)$ are coupled, you can't split the two methods.

I wrote the following C++ snippet: here a1 and a2 are the leftmost and rightmost point of the starting interval in the bisection routine.

    #include <cmath>
    #include <iostream>
    #include <iomanip>      // std::setprecision

    constexpr double tol = 1e-12;
    constexpr double delta = 1e-5;
    double f(double x, double a){
        return 2*exp(-a*x)*(a*sin(3*x)+cos(3*x));
    }

    double fprime(double x, double a){
        return exp(-a*x)*(4*a*cos(3*x) - 2*(a*a+3)*sin(3*x));
    }

    double quasi_newton(double f(double ,double),double x0,double a){
        double res =  -(delta*f(x0,a) )/(f(x0+delta,a)-f(x0,a));
        while (std::fabs(res)>tol) {
            x0+=res;
            res = -(delta*f(x0,a))/(f(x0+delta,a)-f(x0,a));
        }
        return res+x0;
    }

    int main(){
        double x0=0.5;
        int it{0};
        double a1{-10.0};
        double a2{10.0};
        double c{0};

        double xc,gc,ga1,ga2,xa1,xa2;
        while (std::fabs(a1-a2)>tol) {
            std::cout<< std::fabs(a1-a2) <<std::endl;
            c =0.5*(a1+a2);
            xc = quasi_newton(fprime,x0,c);
            xa1 = quasi_newton(fprime,x0,a1);
            xa2 = quasi_newton(fprime,x0,a2);

            gc = f(xc,c)-3.0;
            ga1 = f(xa1,a1)-3.0;
            ga2 = f(xa2,a2)-3.0;

            if (gc*ga1<0) {
                a2 = c;
            }
            if (gc*ga2<0) {
                a1=c;
            }

        }
        std::cout <<"Point: " << std::setprecision(10) << xc << ", Value of a: " <<std::setprecision(10)<< c << ", Check:" << std::setprecision(10) << f(xc,c) << std::endl;
        return 0;

    }


问:
Vector normal to the gradient
I know that the gradient points in the direction of the maximum rate of increase, with the maximum rate given by its magnitude.  Similarly, the maximum rate of decrease is given by the negative of its magnitude.  Does a vector normal to the gradient point in the direction of zero increase?  I.e., if I were standing on a level curve and traveled in a direction normal to the gradient at that point, will I land on a level curve with the same value as when I began?</p>

答:
Yes. We can define the directional derivative of $f$ at the point $\mathbf r$ along a unit vector $\mathbf n$ by $$\mathbf n\cdot\nabla f=\lim_{h\to0}\frac{1}{h}\left[f(\mathbf r+h\mathbf n)-f(\mathbf r)\right].$$
If $\mathbf n$ is perpendicular to $\nabla f$, then the directional derivative is $0$, so $f$ doesn't change when you move in the direction of $\mathbf n$.

Title: Vector normal to the gradient


问:
Show that $E[ E[Z\lvert \mathcal{F}_{T}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]$
Let $T$ and $S$ be stopping times and $Z$ some integrable random variable. I already know that:

$\mathcal{F}_{T\land S}=\mathcal{F}_{T}\cap\mathcal{F}_{S}$ and $\{T>S\},\;\{T\leq S\},\; \{T<S\},\;\{T\geq S\} \in \mathcal{F}_{T\land S}$. And furthermore I have shown that:

$E[Z\lvert \mathcal{F}_{T}]=E[Z\lvert \mathcal{F}_{S\land T}]\; \; \mathbb P$-a.s. on $\{T\leq S\} \; (*)$

I want to show that:

$E[ E[Z\lvert \mathcal{F}_{T}]\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]\; \; \mathbb P$-a.s.

I guess we have to use $(*)$ but I get stuck:

$E[ E[Z\lvert \mathcal{F}_{T}]\lvert \mathcal{F}_{S}]=E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T\leq S\}}\lvert \mathcal{F}_{S}]+E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]= E[Z\lvert \mathcal{F}_{T\land S}]1_{\{T\leq S\}}+E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]$

How can I go about showing that the second term

$$ E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]1_{\{T>S\}}$$ or am I going about this the wrong way?

Reference to Problem 2.17 from Karatzas and Shreve p. 9 Brownian Motion and Stochastic Calculus

答:
Let $Y$ be the random variable $Y=E[Z| \mathcal{F}_{T}]$. I am assuming we work with a complete filtration $(\mathcal F_t)$ on some space $(\Omega, \mathcal F,\Bbb P)$.

Then $(*)$ gives:
$$
\begin{aligned}
E[Z| \mathcal{F}_{T}]
&=E[Z |\mathcal{F}_{T\land S}] \qquad\text{ on } A:=\{T\le S\}\ ,\\
E[Y| \mathcal{F}_{S}]
&=E[Y |\mathcal{F}_{T\land S}] \qquad\text{ on } B:=\{S\le T\}\ ,\text{ so }
\\[2mm]
E[Z\cdot 1_A| \mathcal{F}_{T}]
&=E[Z\cdot 1_A |\mathcal{F}_{T\land S}] \qquad(1)
\\
E[Y\cdot 1_{\bar A}| \mathcal{F}_{S}]
&=E[Y\cdot 1_{\bar A} |\mathcal{F}_{T\land S}]
\qquad(2)\qquad\text{ so }
\\[2mm]
E[\ E[Z\cdot 1_A| \mathcal{F}_{T}]\ | \mathcal{F}_{S}]
&= E[\ E[Z\cdot 1_A |\mathcal{F}_{T\land S}] \ |\mathcal F_S]
   \qquad\text{ by }(1)\\
&= E[Z\cdot 1_A |\mathcal{F}_{T\land S}]\\
&= E[Z |\mathcal{F}_{T\land S}]\cdot 1_A
\\[2mm]
E[\ E[Z\cdot 1_{\bar A}| \mathcal{F}_{T}]\ | \mathcal{F}_{S}]
&=E[Y\cdot 1_{\bar A}| \mathcal{F}_{S}]\\
&=E[Y\cdot 1_{\bar A}| \mathcal{F}_{T\land S}]
   \qquad\text{ by }(2)\\
&=E[Y| \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\
&=E[\ E[Z| \mathcal{F}_{T}]\ | \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\
&=E[Z | \mathcal{F}_{T\land S}]\cdot 1_{\bar A}\\
\end{aligned}
$$
And we add.
(We have used $\bar A\subseteq B$.)

$\square$


问:
Showing two measures are equal up to scalar multiplication.
I've been stuck on the following question and was wondering if it'd be possible to get some help.

Let $\mu, \nu$ be two finite Borel measures on a metric space $X$ s.t. $\mu << \nu$ and let $c>0$ and suppose that $\nu$ is doubling and that $$\lim_{r \to 0} \frac{\mu(B_r(x))}{\nu(B_r(x))} = c$$ for $\nu$ almost every $x\in X$. By using the Vitali Covering theorem to show that $\mu = c\nu$. (We assume the balls are closed)

I've been able to prove this using Radon-Nikodym combined with the lebesgue differentiation theorem but wish to prove it solely using Vital's covering theorem but have been unsuccessful in even finding an appropriate Vitali cover.

Theorem(Vitali Covering Theorem)

Let X be a metric space and let $\nu$ be a doubling measure on $X$ and let $\mathcal{B}$ be a Vitali cover for $S \subset X$ then there exists a countable $\mathcal{B}' \subset \mathcal{B}$ such that all elements of $\mathcal{B}'$ are disjoint and $$\nu \bigg(S \setminus \bigcup_{B' \in \mathcal{B}'}B'\bigg)=0$$

Definition(Vitali Cover)

Let $S \subset X$ then $\mathcal{B}$ a collection of closed balls such that $\forall \epsilon >0 , \forall x \in S$ there exists $B \in \mathcal{B}$ such that $x \in B$ and $\rm{rad}(B) < \epsilon$ then we call $\mathcal{B}$ a Vitali cover for S.

Many Thanks.

答:
Let's prove that for any $\mu$ and $\nu$ measurable $S$ in $X$ we have $\mu(S) = c\nu(S).$ Let $0 < \epsilon < 1.$

For $\mu$-a.e. $x \in S$ there exists $r_1(x) > 0 $ such that for all $r < r_1(x).$

\begin{equation} \mu(S \cap B(x,r)) \geq (1- \epsilon)\mu(B(x,r)). \tag{1} \end{equation}

For $\nu$-a.e. $x \in S$ there exists $r_2(x) > 0$ such that for all $r < r_2(x)$ we have

\begin{equation} \nu(S \cap B(x,r)) \geq (1- \epsilon)\nu(B(x,r)). \tag{2} \end{equation}

Note that (1) and (2) come from Lebesgue's density theorem which we can apply since $\mu$ and $\nu$ are finite doubling measures. Finally for $\nu$-a.e. $x \in X$ there exists $r_3(x) > 0$ such that for all $r < r_3(x)$ we have

$$\frac{c}{1 + \epsilon}\nu(B(x,r)) \leq \mu(B(x,r)) \leq c(1 + \epsilon) \nu(B(x,r)) \tag{3}.$$

Let $R(x) = \min(r_1(x), r_2(x), r_3(x))$ and let

$$\mathcal{B} = \{B(x,r) \mid r < R(x), B(x,r) \ \text{satisfies (1) (2) and (3)}\}.$$

This is an (a.e.) Vitali cover of $S,$ so we can apply the Vitali covering theorem to get a disjoint subset $\{B_1, B_2,...\}$ such that

$$\nu\bigg(S\setminus \bigcup_{i=1}^\infty B_i\bigg) = 0 = \mu\bigg(S\setminus \bigcup_{i=1}^\infty B_i\bigg),$$

with the second inequality following from $\mu \ll \nu.$ Since $S$ is measurable we deduce that

$$\mu(S) = \mu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg)$$

and

$$\nu(S) = \nu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg).$$

Applying all of the above we have

\begin{align} \mu(S) &= \mu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg) \\
&= \sum_{i=1}^\infty \mu(S \cap B_i) \\
&\leq \sum_{i=1}^\infty \mu(B_i) \\
&\leq c(1+\epsilon)\sum_{i=1}^\infty \nu(B_i) \\
&\leq \frac{c(1 + \epsilon)}{(1-\epsilon)}\sum_{i=1}^\infty \nu(S \cap B_i) \\
&=\frac{c(1 + \epsilon)}{(1-\epsilon)}\nu\bigg(\bigcup_{i=1}^\infty(S\cap B_i)\bigg)\\
&= \frac{c(1 + \epsilon)}{(1-\epsilon)} \nu(S).
\end{align}

Since $\epsilon$ was arbitrary it follows that $\mu(S) \leq c\nu(S).$ The reverse inequality is virtually identical giving that $\mu(S) = c\nu(S).$


问:
Proving $\exists C>0$ s.t. $\sup_{t\in[0,T]}\mathbb{E}(X_t^2)\leq C$ for geometric Brownian Motion
I am trying to show that $\exists$ a constant $C>0$ s.t. $\sup_{t\in[0,T]}\mathbb{E}(X_t^2)\leq C$ for geometric BM $\mathrm{d}X_t=r\mathrm{d}t+\sigma\mathrm{d}W_t$ for $r>0$. I thought this would have appeared on this site before but I can’t find any trace of it here. I am not sure what theorems I should be using. So far, here's what I think could be (potentially) useful for proving this.

$X_t>0$ a.s. It. is also a submartingale.

Writing $t\wedge T$ and using $T$ as a (fixed and deterministic) stopping time, and possibly using Fatou's lemma since $X_t$ is non-negative.

Doob's submartingale inequality (although it seems to operates on $\mathbb{E}(\sup_{t\in[0,T]}X_t^2)$ instead).

$f(x)=x^2$ is convex (although Jensen’s inequality goes in the opposite direction of the requested inequality).

I'm not particularly where to begin. Generally speaking, I have trouble proving inequalities when it comes to stochastic processes, does anyone have a good framework to tackle such problems?

答:
Since you wrote 'geometric Brownian motion', I am going to assume that there is a typo and the SDE is
$$dX_t = X_t r dt + X_t \sigma dW_t,  \quad X_0 = x_0 \in \mathbb R \quad \text{ on } [0,T]. \tag{1}$$

where $r$ and $\sigma$ are constants.

The solution to the SDE (1) is $X_t = x_0 e^{(r-\frac{1}{2} \sigma^2)t} e^{\sigma W_t}.$ Then
\begin{align}
E[X_t^2] &= x_0^2 e^{2(r-\frac{1}{2}\sigma^2)t} E[e^{2 \sigma W_t}] \\
&= x_0^2 e^{2(r-\frac{1}{2}\sigma^2)t} e^{2 \sigma^2 t},
\end{align}
where in the last equality I have used the fact that $2\sigma W_t \sim N(0,4\sigma^2t)$ and for a normal random variable $Y \sim N(\mu, \sigma^2)$, we have $E[e^Y]=e^{\mu + \frac{\sigma^2}{2}}.$

Thus, we have two cases. If $(r-\frac{1}{2}\sigma^2) \geq 0,$ then
$$\sup_{t \in [0,T]} E[X_t^2] = x_0^2 e^{2(r-\frac{1}{2}\sigma^2)T}
e^{2 \sigma^2 T},$$
otherwise
$$\sup_{t \in [0,T]} E[X_t^2] = x_0^2 e^{2 \sigma^2 T}.$$


问:
Is there a finitist semantics for transfinite mathematics?
I'm sympathetic to the Aristotelian view that potential infinity makes sense while actual (completed) infinity doesn't. However, I also find transfinite set theory to be fascinating, and I'm under the impression that this domain of mathematics is founded upon the idea of performing calculations with completed infinities. So for a while I felt like I had to "pick sides", so to speak.

Upon reflection, however, this doesn't really make sense. Formalists and fictionalists ignore or deny the reality of mathematical objects without ignoring/denying mathematics as an activity, so why should my ontological reservations matter when it comes to working with "higher infinities"? Clearly there is something interesting and systematic going on in this domain.

It seems like my difficulty with transfinite set theory is merely semantic: I don't think it makes sense to talk about adding one to $\omega$ if the symbol refers to a completed infinity, but there's nothing wrong with adding one to $\omega$ as long as $\omega$ means something else. My question here is basically: what are my options for that "something else"?

(As an example: one thing I'm wondering is if it's possible to understand transfinite mathematics as operating on potential infinities instead of actual infinities.${}^1$ Perhaps we could understand $\omega$ not as a completed object but as an endless process.${}^2$ Is there a "process" interpretation of transfinite set theory?)

Edit: What about coalgebras? Is there a coalgebraic version of transfinite set theory that treats all completed infinities as incomplete processes?

${}^1$ Potentially relevant: Can we formally distinguish between actual and potential infinities?

${}^2$ What would it mean to "add" and "multiply" processes? Can we make sense of the countable/uncountable distinction in a process context?

答:
The hyperreal numbers do what I believe you are trying to do.  While there are many different notations, I prefer the one that uses $\epsilon$ for a standard marker for an infinitesimal, and $\omega$ as a standard marker for an infinity, such that $\epsilon = \frac{1}{\omega}$.

This makes infinities that act semantically basically like uncompleted infinities.  You can do arithmetic with them in the ordinary sense, just treating $\omega$ and $\epsilon$ as if they were variables, but then using the "standard part" function (std(x)) to reduce it to a real if you need.

In this system, $2\omega + 1$ is a different value than $\omega - 1$.  They can also be put into ratio with one another and cancel: $\frac{5\omega}{10\omega} = \frac{1}{2}$.

You can even do things like $\frac{\omega^2 + 2}{\omega^2}$ which yields $1 + \frac{2}{\omega^2} = 1 + 2\epsilon^2$, whose standard part is just $1$.

This also allows you to assign values to certain divergent series, essentially treating them as if they were convergent to a hyperreal value.  For instance, $\frac{1 + 3 + 5 + \ldots}{1 + 2 + 3 + \ldots} \simeq 2$.


问:
Are the reals genuinely a subset of the complex numbers?
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?

答:
Here is slightly abstract point of view: When we say $A\subseteq B$, we don't actually mean that $A$ is literally a subset of $B$. The symbol $\subseteq$ and its cousins $\subset,\supset,\subsetneq,\supsetneq,\dots$ are sensitive to context. If $A$ and $B$ are considered as sets, then the subset symbols mean what you were taught that they mean. But when $A$ and $B$ bear some additional structure, like being groups, vector spaces, metric spaces, fields, etc., then $A\subseteq B$ means:

"There is a natural, i.e., in some sensible way unique, injective homomorphism $i:A\longrightarrow B$, and from now on we mean $i(A)$ when we write $A$."

What a homomorphism is depends on the exact kind of structure we're looking at. If it's groups, we mean a group homomorphism. If it's vector spaces, we mean linear maps. If it's topological fields (fields equipped with a topology such that addition, multiplication and inversion are continuous), then we mean continuous field homomorphism.

Some authors try to get around this overloading of $\subseteq$ by specifying that $\subseteq$ literally means a subset and $\leq$ means what I wrote above. But if you ask me, in the big picture, it's not at all helpful to distinguish between the two cases. I can't think of a single instance where we actually care that, for instance, $\mathbb R$ is not literally a subset of $\mathbb C$ according to the standard construction of $\mathbb C$. We only care about how things behave, not how they look. And $\mathbb C$ contains a unique subset which, if equipped with the restricted field operations, behaves exactly like $\mathbb R$. And it's just cumbersome to always write "$\mathbb C$ contains a subfield isomorphic to $\mathbb R$" instead of just writing "$\mathbb C$ contains $\mathbb R$". So we do the latter.

Also, here's a better, less ambiguous way to talk about the subject: let $F$ be a field. A subfield of $F$ is a pair $(E,i)$, where $E$ is a field and $i:E\to F$ is an injective field homomorphism (field homomorphisms are automatically injective, but injectivity is needed for other structures). "Classical" subfields in the sense of subsets which are also fields are subfields in this sense if we take the natural inclusion mapping as $i$. And if $i$ is clear from context, we just say $E$ to be a subfield of $F$. Take this definition and then just say $\mathbb R$ is a subfield of $\mathbb C$, without using the subset symbol. It's what we care about, anyway: that one is a subfield of the other.


问:
What's the most elementary way to solve this trigonometric problem?

A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.

答:
This is the closest I could come up with to a purely "trigonometric" answer, and has minimal algebra. Firstly note you have a triangle divided into two, so it may be useful to use the standard formulae for a divided triangle. Quoting from my (rather old!) secondary school mathematics textbook:

When D divides AB in the ratio $m:n$ then:
$$(m+n) \cot \theta = n \cot A - m \cot B = m \cot \alpha - n \cot \beta \tag{1}$$

Here the large triangle is $\triangle ABC$ and the Greek-lettered angles are $\alpha = \angle ACD$, $\beta = \angle BCD$ and $\theta = \angle BDC$ and is supplementary to $\angle ADC$. Given your example is about a string attached to a wall, you might not be surprised this is listed in the "Mechanics" section of the textbook, and usually used in school examinations to solve three-force problems. There is a very useful special case listed, which sadly doesn't apply here but I'll include it for completeness:

When $D$ is the midpoint of $AB$, then
$$2 \cot \theta = \cot A - \cot B = \cot \alpha - \cot \beta $$

The textbook has $B$ and $D$ labelled the other way round to your diagram, but other than that it just takes a rotation to sort out. We have $m = 2$, $n = y$, $BC = 1$ and $B$ (using the textbook formula's labels) is a right angle with $\cot B = 0$.

From $\triangle BCD$ we have $\cot \theta = \frac{y}{1} = y$ and $\cot \beta = \frac{1}{y}$, while $\triangle ABC$ gives $\cot A = \frac{2+y}{1} = 2 + y$. Hence $(1)$ becomes:

$$(2 + y) \cdot y = y \cdot (2 + y) - 2 \cdot 0 = 2 \cot \alpha - y \cdot \frac{1}{y}$$

The first equality doesn't tell us anything new, but the second yields (with a little bit of algebra):

$$y^2 + 2y = 2 \cot \alpha - 1 \\
\implies (y+1)^2 = 2 \cot \alpha \tag{2}$$

Secondly, note that since $AC + BC = 4$, we know that $\triangle ABC$ has perimeter 6 and semi-perimeter 3. Hence it is very inviting to use Heron's formula or the closely-related law of cotangents. In fact, since for the one side we know, we also have the altitude (perpendicular height) and hence can calculate the area, this temptation becomes overwhelming. I find the most practically useful formulation of the law of cotangents when solving a triangle is:

$$\frac{\cot \frac{\alpha}{2}}{s-a} = \frac{1}{r} = \frac{s}{T} \tag{3}$$

or for calculator purposes, should we lack a cot button, the reciprocal:

$$(s - a) \tan \frac{\alpha}{2} = r = \frac{T}{s}$$

Here $\alpha$ and $a$ are any opposite angle-side pair (indeed the formula is often stated with equalities also for similar expressions involving $\beta$ and $\gamma$, resembling the law of sines, but I find that form is generally far more useful for trigonometric proof than solving a given triangle), $r$ is the radius of the incircle (not relevant here but sometimes useful), $T$ is the area of the triangle and $s$ the semi-perimeter. In our case $(3)$ becomes:

$$ \frac{\cot \frac{\alpha}{2}}{3-2} = \frac{3}{ \frac{1}{2} \cdot 2 \cdot 1 } \\ \implies \cot \frac{\alpha}{2} = 3 \tag{4} $$

With an electronic calculator, it is now straightforward to calculate $\alpha$ and hence $2 \cot \alpha$, substitute into $(2)$ and solve the simple quadratic for $y$. If an exact or non-calculator solution is required, we can instead use the double-angle formula:

$$ \cot \alpha = \frac{\cot^2 \frac{\alpha}{2} - 1}{2 \cot \frac{\alpha}{2}}
 = \frac {3^2 - 1}{2 \cdot 3} = \frac{4}{3}$$

Then the positive solution of $(y+1)^2 = \frac{8}{3}$ is

$$y = \sqrt{\frac{8}{3}} - 1 = \frac{2\sqrt{6}}{3} - 1 \approx 0.63$$

and corresponds to the bead's vertical level lying beneath the lower attachment point, as drawn. By symmetry, the bead will also be a horizontal distance of one metre from the wall when it is held this far above the vertical level of the upper attachment point. In terms of vertical distance above the lower attachment point, this is then around $2.63$ metres, which — as we might hope —  coincides neatly with the absolute value of the negative root of the quadratic.

Ignore my verbal commentary, and this method gets to the answer in just a couple of lines of working, involves no algebraic rearrangement except solving a simple quadratic, and packs a very hearty dose of trigonometry from a high school textbook. Accept the use of an electronic calculator after $(2)$ and it doesn't even require the angle-sum, double-angle or harmonic addition ("R") formulae. Sadly, I expect it is the case that your local schools' curriculum lacks some of the goodies in my textbook. This is a shame! The first time I saw the divided triangle formula, I recall the sinking feeling that it looked like an awful lot of hard work. What I've discovered in practice, whenever I've had recourse to use it, is just how much hard work it saves me...

Post-script: proof of the divided triangle formulae

Many proofs are available online and in textbooks for the law of cotangents but the divided triangle formula seems more obscure.

Often half the battle with trig is identifying the correct formula to use: which best represents the state of our knowledge? For example, knowing the area, perimeter and one side of a triangle, the law of cotangents is ideal for finding the angle opposite the known side. Here's another rule true for any $\triangle PQR$ and deduced by dropping the perpendicular from $Q$, which a little thought shows is true in both the acute and obtuse case:

$$q = r \cos P + p \cos R \tag{5}$$

where $q = PR$ etc. Formula $(5)$ is useful if we want to write an included side in terms of its including angles and the other two sides. In practice it's rare that we are given four out of five of these quantities, or wish to represent all five in one equation. However, $p$, $r$, $P$ and $R$ are redundant by the law of sines, so we can cut one out:

\begin{align}
q &= r \cos P + \frac{r \sin P}{\sin R} \cdot \cos R \\
  &= r \left( \cos P + \sin P \cot R \right) \\
  &= r \sin P \left( \frac{\cos P}{\sin P} + \cot R \right) \\
q &= r \sin P \left( \cot P + \cot R \right) \tag{6}
\end{align}

Formula $(6)$ lets us write an included side in terms of its including angles and one other side. Like the law of sines, this formula links two sides and two angles. Unlike the law of sines, there is only one opposite angle-side pair!

Returning to the figure in the textbook, the key to representing the situation is that the smaller triangles $\triangle ACD$ and $\triangle BCD$ are linked by a common side $CD$ and the fact $ADB$ is a straight line, so $\angle ADC$ and $\angle BDC$ are supplementary. Hence $CD$, $\angle ADC$ and $\angle BDC$ must appear in our equations if those facts are to be brought to bear. Moreover, we want $AD = m$ and $BD = n$ to appear too, since they were indicated as of particular interest (the ratio $\triangle ABC$ was divided in for the textbook; the lengths marked on the wall in the OP's question). Needing to represent two sides and an included angle in each of the smaller triangles rules out the law of sines; we could apply the law of cosines if  we wanted $AC$ and $BC$ to appear in our equations also. Alternatively we can apply $(4)$ if we prefer two sides and two angles in each equation.

At this stage we have a choice. Treating $AD$ as included by $A$ and $\pi - \theta$, and $BD$ as included by $B$ and $\theta$,

\begin{align}
m &= CD \sin (\pi - \theta) \left(\cot (\pi - \theta) + \cot A \right) = CD \sin \theta \left( -\cot \theta + \cot A \right) \\
n &= CD \sin \theta \left( \cot \theta + \cot B \right) \\
\implies \frac{m}{n} &= \frac{-\cot \theta + \cot A }{\cot \theta + \cot B} \tag{7}
\end{align}

where $(5)$ is obtained by dividing the previous two equations. Alternatively we can treat $CD$ as included by $\alpha$ and $\pi - \theta$ in $\triangle ACD$ and included by $\beta$ and $\theta$ in $\triangle BCD$:

\begin{align}
CD &= m \sin (\pi - \theta) \left(\cot (\pi - \theta) + \cot \alpha \right) = CD \sin \theta \left( -\cot \theta + \cot \alpha \right) \\
CD &= n \sin \theta \left( \cot \theta + \cot \beta \right) \\
\implies 1  &=  \frac{m}{n} \cdot \frac{-\cot \theta + \cot \alpha }{\cot \theta + \cot \beta} \tag{8}
\end{align}

Rearranging $(7)$ and $(8)$ to combine them into the textbook's $(1)$ is straightforward algebra.


问:
Probability of disease transmission to future generations
Assume disease D is transmitted through a father (male) to his children. Let us assume that every family in a society have $C\geq1$ children. Let the father of a family have the disease D. What is the probability that disease D is transmitted to at least one person in the $N$-th generation after him $(N\geq1)$? $N=1$ represents the children of the father, $N=2$ represents the grandchildren, etc.

One can calculate the probability for a given $N$ and $C$, but I was wondering if it possible to find a general closed-form solution.

P.S.: The problem is equivalent to the problem of passing the family-name. In a society where the family-name is passed from father to children, what is the probability of having the family-name of a father in the $N$-th generation after him.

答:
I don't think it's possible.

I assume $C > 0$. I also suppose that repartition of male/female is balanced.

Let $q_N$ be the probability that the disease is not transferred to any person of the $N$-th generation. (This is the complementary of the probability you are looking for.)

Then:

$q_1 = 0$

$q_{N+1} = (0.5 + 0.5 \times q_N)^C$

This sequence converges (because it's bounded and monotonic) and the limit $q_{\infty}$ satisfies:

$$q_{\infty} = (0.5 + 0.5 \times q_{\infty})^C$$

I'm quite sure the solution of this equation has no closed-form in general.

That's not a complete answer of your question and I hope someone will be able to help you more on this.


问:
What is the expected value of X, the number of fans they will gain or lose over the next 3 games?
The All-Star Pigeons lose $10$ fans every time
they lose a game and gain $100$ fans every time they win. Assume they have
a 30% chance of losing each game and that their performance in each game is
independent. What is the expected value of $X$, the number of fans they will
gain or lose over the next 3 games?

My work:
Let $Y$ represent the number of games the All-Star Pigeons win.

$P(Y=0) = 0.3^3 = .027$

$P(Y=1) = 3 \cdot 0.7 \cdot 0.3^2 = .189$

$P(Y=2) = 3 \cdot 0.7^2 \cdot 0.3 = .441$

$P(Y=3) = .7^3 = .343$

So the expected value for $Y$ is $E(Y) = 0 \cdot .027 + 1 \cdot .189 + 2 \cdot .441 + 3 \cdot .343 = 2.1$

So with this logic the expected number of wins is $2.1$ so I believe that $E(X) = 2.1 \cdot 100 = 210$ fans gained

So the expected value of $X$ is $210$.

My question is, is my methodlogy (and answer) for this question correct? I can't think of any other way to calculate $E(X)$. Could someone here let me know if my thinking is correct? Thank You!

答:
You are nearly there: you need to subtract an extra $0.9\times10=9$ for the games they lose.

Here's another approach, which generalises easily to any number $n$ of games.

Let $X_i$ be the number of fans they gain after game $i$ (which is negative for a loss). Then
$$\mathbb E[X_i]=0.7\times100+0.3\times(-10)=67.$$
So by linearity of expectation, $$\mathbb E[X]=\mathbb E[X_1+X_2+X_3]=\mathbb E[X_1]+\mathbb E[X_2]+\mathbb E[X_3]=3\times67=201.$$


问:
Why are rotation matrices given in one dimension higher
Hello Stackexchange :) I am currently reviewing some notes from my linear algebra class and found out that my professor gave the rotation matrices for $\mathbb{R}^2$ with an 3x3 matrix and the one for $\mathbb{R}^3$ with an 4x4 one. Why is this neccessary? The last column and line of those seems superfluous.

Thank you in advance for your answers! :)

Edit: Here is an example for the rotation by the x-Axis

答:
Here is what is going on.  An isometry of $\mathbf R^n$ is a mapping $f: \mathbf R^n \to \mathbf R^n$ that preserves distances.  If you want $f$ to fix the origin, then $f$ can be described by an $n \times n$ orthogonal matrix: $f(\mathbf v) = A\mathbf v$ where $A$ is an $n \times n$ matrix such that $AA^\top = I_n$ (that condition is what captures the "preserving distances" property).

But if you do not want to fix the origin, to truly get all isometries of $\mathbf R^n$, then you have to allow translations as well. The general isometries of $\mathbf R^n$ have the form
$$
f(\mathbf v) = A\mathbf v + \mathbf b
$$
for an $n \times n$ orthogonal matrix and a vector $\mathbf b$ in $\mathbf R^n$.  Note $f(\mathbf 0) = \mathbf b$, so $\mathbf b$ is nothing other than where the origin is sent.  A general isometry of $\mathbf R^n$ can be described as a rotation around the origin followed by a translation in some direction.

Let's call that isometry above $f_{A,\mathbf b}$.  How do two isometries $f_{A,\mathbf b}$ and $f_{A',\mathbf b'}$ compose?  For $\mathbf v$ in $\mathbf R^n$,
$$
(f_{A,\mathbf b} \circ f_{A',\mathbf b'})(\mathbf v) = f_{A,\mathbf b}(A\mathbf v' + \mathbf b) = A(A'\mathbf v + \mathbf b) + \mathbf b' = (AA')\mathbf v + A\mathbf b + \mathbf b'.
$$

So we have the composition rule
$$
f_{A,\mathbf b} \circ f_{A',\mathbf b'} = f_{AA', A\mathbf b + \mathbf b'}.
$$
It turns out that this composition rule can be described by going up $1$ dimension by thinking of $f_{\mathbf A,\mathbf b}$ as the block matrix
$$
\begin{pmatrix}A&\mathbf b\\0&1\end{pmatrix}
$$
acting on vectors $\binom{\mathbf v}{1}$ where $\mathbf v \in \mathbf R^n$.
That is, the effect of that $(n+1) \times (n+1)$ matrix on the vector $\binom{\mathbf v}{1}$ is
$$
\begin{pmatrix}A&\mathbf b\\0&1\end{pmatrix}
\begin{pmatrix}\mathbf v\\1\end{pmatrix}
=
\begin{pmatrix}A\mathbf v + \mathbf b\\ 1\end{pmatrix}
$$

It is not important for $A$ to be orthogonal for this "increase dimension by $1$" algebra to work out. For example, take $n = 1$: the function $f(x) = ax + b$ on $\mathbf R$ could be described as an effect of certain $2 \times 2$ matrices on vectors with bottom component $1$:
$$
\begin{pmatrix}
a& b\\0&1\end{pmatrix}
\begin{pmatrix}x\\1\end{pmatrix}
=
\begin{pmatrix}ax+b\\1\end{pmatrix}
$$

This explains why in the notes you displayed, the vectors on both sides have bottom component $1$ and the orthogonal matrix is in the upper $3 \times 3$ block of the $4 \times 4$ matrix.  The convenience of such notation is it allows the not-entirely-matrix-looking operation $f(\mathbf v) = A\mathbf v + \mathbf b$ with nonzero $\mathbf b$ to be described entirely by a matrix operation up one dimension higher using $(\begin{smallmatrix}A&\mathbf b\\0&1\end{smallmatrix})$.  If the translation part $\mathbf b$ is $\mathbf 0$, meaning the isometry is a rotation around the origin, then the $(n+1) \times (n+1)$ matrix will have zeros along the right column except in the bottom right entry, which will be $1$. And if you only want to discuss rotations around the origin, then it is kind of silly (or superfluous, as you wrote) to use the $(n+1) \times (n+1)$ matrices.

A compelling reason to use the larger matrices is the potential to allow translations as part of the transformations under consideration.

Another compelling reason is if you want to allow rotations around a general point of $\mathbf R^n$, not just the origin.  For example, suppose an $n \times n$ orthogonal matrix $A$ describes a rotation around $\mathbf 0$ that you are interested in, but you'd like to apply it to rotate around some other point $\mathbf r$ instead.  The way to do this is

Translate a vector backwards by $\mathbf r$ ("recenter from $\mathbf r$ to origin")

Apply $A$ (desired kind of rotation around origin)

Translate forwards by $\mathbf r$ ("recenter from origin to $\mathbf r$).

That means
$$
f(\mathbf v) = A(\mathbf v - \mathbf r) + \mathbf r =
A\mathbf v + (\mathbf r - A\mathbf r).
$$
Notice $f(\mathbf r) = \mathbf r$, so this is fixing $\mathbf r$ just as you want (a rotation around $\mathbf r$ should not move $\mathbf r$).  This has the form $f(\mathbf v) = A\mathbf v + \mathbf b$ where $\mathbf b = (I_n - A)\mathbf r$.  The upshot is that rotation around a point other than the origin is the same thing as rotation around the origin followed by a certain translation that involves the rotation matrix $A$ and the point $\mathbf r$ you want to be centering everything around.  So even if you are interested only in rotations, including rotations around general points of $\mathbf R^n$, you should be allowing translations too and that's where the nonzero $\mathbf b$'s come from merely from interest in rotations (around general points, not just the origin).


答:
While It's difficult to say without context, it seems that the notes are referring to homogeneous coordinates, a system in which points in $n$-dimensional (projective) space are represented by $n+1$ coordinates, where two coordinate tuples $\vec{x}$ and $\vec{y}$ are considered equivalent if $\vec{x}=c\vec{y}$ for some $c\in\mathbb{R}$. It's common to normalize the last coordinate to $1$ to recover the Euclidean coordinates.

Despite the apparent redundancy, this system has a few advantages, one being that translations can be encoded as matrices just like rotations. As a result, they are used extensively in computer graphics.

Title: Why are rotation matrices given in one dimension higher

Body:
Hello Stackexchange :) I am currently reviewing some notes from my linear algebra class and found out that my professor gave the rotation matrices for $\mathbb{R}^2$ with an 3x3 matrix and the one for $\mathbb{R}^3$ with an 4x4 one. Why is this neccessary? The last column and line of those seems superfluous.

Thank you in advance for your answers! :)

Edit: Here is an example for the rotation by the x-Axis


问:
Cycle in a graph.
Given an non oriented graph $G=(V,E)$, prove that if $|V| \geq 3$ and there exists an unique $v_0 \in V  $ s.t. $d(v_0)=1$ then the graph $G$ contains a cycle.

Using the hand-shaking lemma we have that $ \sum_{v \in V} d(v)=2|E| \rightarrow \sum _{v\in V\setminus{v_0}}d(v) + d(v_0)=\sum _{v\in V\setminus{v_0}}d(v)+1=2|E| $ from where it follows that we have atleast one more vertice of odd degree. How do I prove that $G$ contains a cycle?

答:
I don't think you really need hand - shaking lemma.
From $d(v_0)=1$ there exists $v_1 $connected to it. Now for vertex different from $ v_0$ if you have anny neighbour you have at least two of them. So there exists $v_2$ connected to $v_1$ with degree at least two...

But as graph is finite this has to stop at some $v_n$ from which I believe you can deduce existence of a cycle.

Title: Cycle in a graph.


问:
What is the point of model theory?
Wikipedia defines model theory as ...

... the study of the relationship between formal theories (a collection of sentences in a formal language expressing statements about a mathematical structure), and their models, taken as interpretations that satisfy the sentences of that theory.

Why does this field need to exist? If we have proof theory, then we can discuss any formal system $\Gamma$ and statements within $\Gamma$ like, say, some arbitrarily chosen sentences $\phi$ and $\psi$, and even theorems like $\Gamma \vdash \phi \to \psi$, assuming that $\psi$ is derivable from $\phi$ in $\Gamma$, of course. Why should anyone care whether $\Gamma \models \phi$ if one already knows that $\Gamma \vdash \phi$?

Thanks for any advice.

答:
Model theory is an excellent tool for talking about the semantics of formal languages, particularly first order logic with equality and with constants, functions and non-logical predicates, by building models out of sets. You can do other, fancier things with it like considering models of set theory itself, but I don't understand those applications and can't comment on them.

One interesting fact from model theory is the existence of nonstandard models of arithmetic. I think this is a good motivating example because it shows that axiomatizations of seemingly simple things like arithmetic do not pin down a semantics as much as we would like.

I can try to answer this from the perspective of someone who found out about model theory through the proof theory and programming language theory route. I don't know the subject well; I'm just a beginner. As a meta note, I think Noah Schweber is right and you should phrase your question more politely.

Basically, a model is a structure that assigns interpretations to various symbols and has a domain of discourse. The interpretations are all drawn from our universe of sets in our underlying set theory.

A symbol is an abstract component in a signature. It can be a constant symbol (like a nullary function), a function symbol, or a predicate symbol. Functions and predicates can have different arities. Normally, there's a single domain of discourse, but you can generalize this to multiple sorts if you want to.

So, symbols in signatures can be used in two ways. They can be used in your formal language. For instance, if $f$ is a function symbol, you can express properties about $f$ in your language.

$$ \forall x \mathop. \forall y \mathop. f(x, y) = f(y, x) $$

The function symbol $f$ is also supplied an interpretation in a given structure.

A set of sentences using non-logical symbols taken from a particular signature can be used to define a theory. A structure where all the sentences in a theory are true is called a model of the theory.

But, structures are also made of sets. This means that their domains have cardinalities.

This means that you can talk about the spectrum of a theory, which is the number of models up to isomorphism for each cardinality.

The two results that Noah Schweber mentioned in a comment constrain the shape of spectra of a theory for infinite cardinalities. NOTE: I'm counting models up to isomorphism here, not "raw" models.

The Löwenheim–Skolem theorem says that if your theory has an infinite model, it has at least one model of every infinite cardinality.

Morley's categoricity theorem says that if your theory has exactly one infinite model for some uncountable cardinality $\kappa$, then it has exactly one infinite model for all uncountable cardinalities.

Title: What is the point of model theory?


答:
Let's just follow our nose.

We start with a vertex $v_1$ with $\deg v_1=1$. Let $v_2$ be its neighbour. As $\deg v_2>1$, we know $v_2$ has another neighbour $v_3\neq v_1$.

Likewise, $v_3$ has a neighbour $v_4\neq v_2$. Also, $v_4\neq v_1$, as $v_1$ is a leaf. Now $v_4$ has a neighbour that is not $v_1$ or $v_3$. This neighbour could be $v_2$ (in which case we have a cycle), or it could be another vertex $v_5$.

So we get a path $v_1,v_2,v_3,\dots$, which (if $G$ is finite) has to repeat a vertex eventually. When it does, we get a cycle.

The claim is false when $G$ is infinite, e.g. take your vertices to be $\mathbb N$, and draw an edge between $m$ and $n$ iff $|m-n|=1$.

Title: Cycle in a graph.

Body:
Given an non oriented graph $G=(V,E)$, prove that if $|V| \geq 3$ and there exists an unique $v_0 \in V  $ s.t. $d(v_0)=1$ then the graph $G$ contains a cycle.

Using the hand-shaking lemma we have that $ \sum_{v \in V} d(v)=2|E| \rightarrow \sum _{v\in V\setminus{v_0}}d(v) + d(v_0)=\sum _{v\in V\setminus{v_0}}d(v)+1=2|E| $ from where it follows that we have atleast one more vertice of odd degree. How do I prove that $G$ contains a cycle?


问:
Finding a four-digit number $\overline{abcd}$ such that $(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $
Consider a four digit number $\overline{abcd}$ such that $$(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $$
$\overline{ab}$, $\overline{bc}$ and $\overline{cd}$ are two digit numbers

What is this number ?

What I have tried is

$$(10a+b)^2+(10b+c)^2+(10c+d)^2=\overline{abcd} $$ then I used $(a+b)^2$ formula but this method didn't help.

I need mathematical solution

答:
Denote $S = (\overline{bc})^2+(\overline{cd})^2$. Note that $S < 3500$ — otherwise, we have $a \geq 3$, therefore $\overline{abcd} \geq 3500 + 900 = 4300$, so $a \geq 4$, and if we continue we reach a contradiction.

In particular, this implies $b, c < 6$.

Taking our original equation mod 10, we obtain
$$b^2 + c^2 + d^2 \equiv d \pmod{10}$$
$$b^2 + c^2 \equiv - d(d-1) \pmod{10}$$

Checking each value of $d$ separately, we see that the right side can only have values 0, 4, and 8. Similarily, considering the quadratic residues mod 10 (all of which happen to be reachable with $b, c < 6$), you'll find five possible pairs for $\{b, c\}$:

3, 1

3, 3

3, 5 (which can be eliminated with the $S < 3400$ condition)

2, 0 (which violates the condition that the numbers being squared should have two digits)

2, 4

2, 2

Each of those also has a small subset of possible values of $d$.

While this gets us near the threshold where checking the cases by hand becomes possible, I personally wouldn't want to do that.

I'd try finding a lower bound for $S$, to complement or upper bound of 3400 and help prune some of these possibilities.


问:
When is an integral domain a localization of a proper subring?
Given an integral domain $R$, is $R$ a localization of a proper subring if and only if there is some unit $u \in R^{\times}$ for which $u^{-1} \notin \mathbb{Z}[u]$, where $\mathbb{Z}[u]$ denotes the (unital) subring of $R$ generated by $u$?

One direction clearly holds: If $R$ is a localization of a proper subring $S$, then any for any nonunit $u$ in $S$ that becomes a unit in $R$, one must have $u^{-1} \notin \mathbb{Z}[u]$.

Conversely, suppose that $u$ is a unit in $R$ for which $u^{-1} \notin \mathbb{Z}[u]$. Then, there is clearly at least one subring of $R$ containing $u$ but not $u^{-1}$, namely $\mathbb{Z}[u]$. But of course, $\mathbb{Z}[u,u^{-1}]$, the subring of $R$ generated by $u$ and $u^{-1}$, is not necessarily all of $R$, so $\mathbb{Z}[u]$ is not the subring we want for $S$.

Perhaps, $S$ should be a subring maximal among those that contain $u$ but not $u^{-1}$, but we need to show whether such a subring actually exists (with Zorn's lemma, maybe) and whether for this subring $S$, $S[u^{-1}]=R$.

答:
Let $F=Frac(\Bbb{Z}/char(R))$,

If $u$ is transcendental over $F$,

With the axiom of choice let $X\subset R$ be a transcendence basis of $Frac(R)$ chosen such that $u\in X$.

Note that $F[X]_{(u)}[u^{-1}]=F(X):\ \ $ for $\frac{a(X)}{b(X)}\in F(X)=Frac(F[X])$ there is some $j$ such that $b(X)u^{-j}\in F[X]-u F[X]$ and $u^{-j} \frac{a(X)}{ b(X)u^{-j}}\in F[X]_{(u)}[u^{-1}]$

Any $r\in R$ is algebraic over $F(X)$, it is the root of some $h(t) \in F(X)[t]$,

$u^m h(t) \in F[X]_{(u)}[t]$ whose leading coefficient is $u^{n(r)} a$ with $a\in F[X]_{(u)}^\times$ so $r u^{n(r)}$ is integral over $F[X]_{(u)}$.

Let $$S = \Bbb{Z}/(char(R))[\{ r u^{n(r)} ,r\in R\}]$$

Then $S$ is integral over  $F[X]_{(u)}$ so $u^{-1}\not \in S$ and clearly $R=S[u^{-1}]$.

If $u$ is algebraic over $F$ then $char(R)=0$. Let $O$ be the integral closure of $\Bbb{Z}[u]$ in $\Bbb{Q}(u)$ and $m\subset O$ a maximal ideal  containing $u$ (which exists because $u^{-1}\not \in O$) and $(\pi)$ the maximal ideal of $O_m$.

Take again a transcendence basis $X$. Then $\Bbb{Q}(u)(X) = O_m[X]_{(\pi)}[\pi^{-1}]=O_m[X]_{(\pi)}[u^{-1}]$ and with $n(r)$ constructed as above such that $r\pi^{n(r)}$ thus $r u^{n(r)}$ is integral over $O_m[X]_{(\pi)}$ then $$S = \Bbb{Z}[\{ r u^{n(r)} ,r\in R\}]$$ works (it is integral over $O_m[X]_{(\pi)}$ so $u^{-1}\not \in S$)


问:
Prove that $k \cdot |I| \leq |\bigcup_{i \in I} X_i|$ for each $I \subseteq \{1,...,n\}$
Let ${\cal X} = (X_1,...,X_n)$ be family of subsets and integer $k > 0$. I know that $|\bigcup_{i \in I} X_i| \geq |I|$ for each $I \subseteq \{1,...,n\}$. Moreover, $$k \cdot (\min_{I \subseteq \{1,...,n\}} (|\bigcup_{i \in I} X_i|)) \leq |\bigcup_{i \in I} X_i|, \forall I \subseteq \{1,...,n\}$$

How can to show that $k \cdot |I| \leq |\bigcup_{i \in I} X_i|$ for each $I \subseteq \{1,...,n\}$?

答:
Since $\left|\bigcup_{i\in I}X_i\right|\ge|I|$ for each $I\subseteq[n]$, we know that $|X_i|\ge 1$ for each $i\in I$.

Let $I_0$ be a non-empty subset of $[n]=\{1,\ldots,n\}$ such that

$$\left|\bigcup_{i\in I_0}X_i\right|=\min_{\varnothing\ne I\subseteq[n]}\left|\bigcup_{i\in I}X_i\right|\,.$$

We're told that there is a positive integer $k$ such that

$$k\cdot\min_{\varnothing\ne I\subseteq[n]}\left|\bigcup_{i\in I}X_i\right|\le\left|\bigcup_{i\in I}X_i\right|$$

for each non-empty $I\subseteq[n]$, so in particular

$$k\cdot\min_{\varnothing\ne I\subseteq[n]}\left|\bigcup_{i\in I}X_i\right|\le\left|\bigcup_{i\in I_0}X_i\right|=\min_{\varnothing\ne I\subseteq[n]}\left|\bigcup_{i\in I}X_i\right|\,.$$

What must $k$ be? Then remember that $\left|\bigcup_{i\in I}X_i\right|\ge|I|$ for each non-empty $I\subseteq[n]$.


问:
The multiplicative identity of the vector space formed by the set of all linear transformations
I was asked to prove that the set of all linear transformations from the vector space $V$ to the vector space $W$ is a vector space. I was going through the axioms, and I am stuck on the multiplicative identity part.

The function $T(x)=1$ is not a linear transformation since $T(\alpha x)=1$ whereas $\alpha T(x)=\alpha$.

Which linear transformation is the multiplicative identity then?

I asked this to one of my classmates and he said it is the function $T(x)=x$. If we are taking multiplication to mean composition, I understand why this is true. (In that case, I don't understand the multiplicative associativity. Also, I feel like this should have been stated.)

However, if we are talking about multiplication in the normal sense, I don't understand how this is true. For example, if we let $V=W=\mathbb{R}$, all the linear transformations are of the form $T(x)=ax$ for some $a$. If we multiply this by $x$, we get $ax^2$ which is not even a linear transformation, let alone equal to $T$.

I feel like I am asking a very easy question. If that is the case, I apologize. However, I don't understand what I am missing.

答:
Product of linear maps is defined as composition:

(ST)(u) = S(T(u))

hence multiplicative identity T(x)=x makes sense.

Multiplicative associativity (ST=TS), where S,T are linear maps,  is not a requirement for a vector space. In order for linear maps to form vector space they should only adhere to scalar multiplicative associativity, i.e. (ab)S = a(bS), where a,b are scalars (R, C, etc).


问:
Convergence in distribution of independent random variables
Let $(X_n)_{n\in \mathbb N}$ and $(Y_n)_{n\in \mathbb N}$ be random sequences independent of each other. Further suppose both $X_n$ and $Y_n$ converge in distribution (to some random variable).

I know that if $X'$ and $Y'$ are weak limits of $X_n$ and $Y_n$, then they need not be independent.

My question is, can I (always) find "independent" random variables $X$ and $Y$ that are weak limits of $X_n$ and $Y_n$ respectively?

Thanks and regards.

答:
Let $X'$ & $X$ be $S_1$-valued r.v.s, and $Y'$ & $Y$ be $S_2$-valued ones. We denote the distribution that r.v. $X$ follows as $P_X(\cdot)$ ($= P(X \in \cdot)$, and similarly for other r.v.s.

We can construct a product measure $P_{X'} \otimes P_{Y'}$ on $(S_1 \times S_2, \sigma (\mathcal{P}) )$ that satisfies

$$
\forall A \in \mathcal{B}(S_1), B \in \mathcal{B}(S_2) \quad P_{X'} \otimes P_{Y'}(A \times B) = P_{X'} (A) P_{Y'} (B).
$$

where $\mathcal{B}(\cdot)$ describes the Borel family and $\mathcal{P}$ is the family of "Borel rectangles", i.e.

$$
\mathcal{P} = \{ E \times F \mid E \in \mathcal{B}(S_1), \ F \in \mathcal{B}(S_2) \}.
$$

When you think of a distribution of r.v.s $X$ & $Y$ as

$$
P_{(X, Y)}(A,B) = P_{X'} \otimes P_{Y'}(A,B)
$$

then

\begin{eqnarray*}
&& P_X(A) = P_{(X, Y)}(A,\mathbb{R}) = P_{X'} \otimes P_{Y'}(A \times \mathbb{R}) = P_{X'}(A), \\
&& P_Y(B) = P_{(X, Y)}(\mathbb{R},B) = P_{X'} \otimes P_{Y'}(\mathbb{R} \times B) = P_{Y'}(B).
\end{eqnarray*}

(We used the last equal of the each equation the definition of $P_{X'} \otimes P_{Y'}$.)

We can say $X$ & $Y$ are independent, as

$$
P_{(X, Y)}(A,B) = P_{X'} \otimes P_{Y'}(A \times B) = P_{X'} (A) P_{Y'} (B) = P_X(A) P_Y(B).
$$

I am not sure if this answer is clear enough...