问:
MATLAB code to find distance and eccentricity in graphs
I was trying to find the distances between vertices in graphs. But as the number of vertices are increasing up to 25 vertices or more, its becoming a tedious job for me to calculate $distance$ and $eccentricity$.

Can I draw graphs in MATLAB and then calculate the $dist$ and $ecc$?

For ex. What is the code if I want to draw $P_n$ and then the distances between vertices.

I am very new to MATLAB. Started few days back only. Can anyone help me here.

Edit : I want the code to draw the graph which are Undirected and non-weighted graphs. And also distance between every two vertices. Can I make the graph if I am taking adjacency matrix. Is it possible to make the graph when entries are like ordered pairs of adjacent vertices .

I would be very thankful. thanks

Coding for any small graph like $P_3$ or $K_4$ will be helpful for me.

答:
You might want to use this http://www.mathworks.com/help/bioinfo/ref/graphshortestpath.html or google "shortest path code in Matlab" and find other options.
This question might also help you.</p>


问:
If $K$ is closed and $C$ is compact then $K\cap C=\emptyset$ implies that there exist a neighborhood of $0$ such that $(K+V)\cap(C+V)=\emptyset$.
What is shown below is a reference from the text Functional Analysis by Walter Rudin.

If $X$ s a vector space over the field $\Bbb F$ then  the following notation well be used
$$
x+A=\{x+a:a\in A\}\\
x-A:=\{x-a:a\in A\}\\
A+B:=\{a+b:a\in A\wedge b\in B\}\\
\lambda A:=\{\lambda a:a\in A\}
$$
for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.

Definition

If $\tau$ is a topology on a vector space $X$ such that

every point of $X$ is a closed set

the vector space operation are continuous with respect to $\tau$

we say that $X$ is a topological vector space equipped with the vector topology $\tau$.

So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.

Lemma

Let be $X$ a topological vector space over the field $F$. So if $W$ is a neighborhood of $0$ in $X$ then there is a neighborhood $U$ of $0$ which is symmetric (that is $U=-U$) and which satisfies $U+U\subseteq W$.

Proof. Omitted.

Now we observe that the proposition can applied to $U$ in place of $W$ so that there exist a new symmetric ceighborhood $U$ of $0$ such that
$$
U+U+U+U\subseteq W
$$
because if there exist a symmetric neighborhood $V$ of $0$ such that $V+V\subseteq U$ then $(V+V)\times (V+V)\subseteq U\times U$ and so
$$
V+V+V+V=s\big((V+V)\times(V+V)\big)\subseteq s(U\times U)=U+U\subseteq W
$$
where $s:X\times X\rightarrow X$ is the vector sum.

So I like to discuss some passages of the following theorem.

Theorem

Suppose $K$ and $C$ are disjoint subsets of a topological vector space X such that K is compact and C is closed. so there exist a symmetric neighborhood $V$ of $0$ such that
$$
(K+V)\cap(C+K)=\emptyset
$$

Proof. If $K$ is empty tehn $K+V$ is too and so the concludion follows immediately. We therefore assume that $K$ is not empty so that we consider a point $x$ of this set. So since $C$ is closed, since $x$ is not in $C$ and since the topology of $X$ is invriant under translations, the preceding proposition shows that $0$ has a symmetric neighborhood $V_x$ such that $x+V_x+V_x+V_x$ does not intersect $C$ and so the symmetry of $V_x$ show then that
$$
(x+V_x+V_x)\cap(C+V_x)=\emptyset
$$
Since $K$ is compact, there are finitely many points $x_1,...,x_n$ in $K$ such that
$$
K\subseteq\bigcup_{i=1}^n\big(x_i+V_{x_i}\big)
$$
so that if we put
$$
V:=\bigcap_{i=1}^n V_{x_i}
$$
then
$$
K+V\subseteq\bigcup_{i=1}^n\big(x_i+V_{x_i}+V\big)\subseteq\bigcup_{i=1}^n\big(x_i+V_{x_i}+V_{x_i}\big)
$$
and no term in this last union intersects $C+V$ by what above shown and so the theorem follows.

So unfortunately I did not understand why $\big(x+V_x+V_x+V_x\big)$ is disjoint from $C$ and then why this implies that $\big(x+V_x+V_x\big)$ and $\big(C+V_x\big)$ are disjoint. Then I like to discuss my arguments below to explain the inclusions of the theorem are correct. Anyway I observe that since $X\setminus C$ is open then $-x+(X\setminus C)$ is too and since $0$ is there then there exist a open symmetric neighborhood $V_x$ of $0$ such that $V_x+V_x\subseteq -x+(X\setminus C)$ and if $s$ is the vector sum this (neighborhood) is such that
$$
x+V_x+V_x=s\big[\{x\}\times(V_x+V_x)\big]\subseteq s\Big[\{x\}\times\big(-x+(X\setminus C)\big)\Big]=X\setminus C
$$
where $s$ is the vector sum but this does not mean that $x+V_x+V_x+V_x$ is disjoint from $C$. Now if $0\in V_x$ for each $x\in K$ then clearly $x\in x+V_x$ which is open and so the collection $\mathfrak V:=\{x+V_x:x\in K\}$ is an open cover of $K$ by which the existece of $x_1,...,x_n$ follows for the compactness of $K$. So by the first inclusion we conclude that
$$
K+V=s[K\times V]\subseteq s\Big[\big(\bigcup_{i=1}^n x_i+V_{x_i}\big)\times V\Big]=s\Big[\bigcup_{i=1}^n\big((x_i+V_{x_i})\times V\Big]=\bigcup_{i=1}^n s[(x_i+V_{x_i})\times V]=\bigcup_{i=1}^n(x_i+V_{x_i}+V)
$$
and finally if $V\subseteq V_{x_i}$ for each $i=1,...,n$ then $(x_i+V_{x_i})\times V\subseteq(x_i+V_{x_i})\times V_{x_i}$ too and so
$$
x_i+V_{x_i}+ V=s[(x_i+V_{x_i})\times V]\subseteq s[(x_i+V_{x_i})\times V_{x_i}]=x_i+V_{x_i}+V_{x_i} $$
for each $i=1,...,n$ and so the last inclusion follows immediately. So could someone help me, please?

答:
The remark after the Lemma shows that $0$ has a symmetric nbhd $V_x$ such that

$$V_x+V_x+V_x+V_x\subseteq(X\setminus C)-x\,,$$

since $(X\setminus C)-x$ is an open nbhd of $0$. And $0\in V_x$, so

$$0+V_x+V_x+V_x\subseteq(X\setminus C)-x\,,$$

and adding $x$ to both sides yields the desired result:

$$x+V_x+V_x+V_x\subseteq X\setminus C\,,$$

i.e.,

$$(x+V_x+V_x+V_x)\cap C=\varnothing\,.\tag{1}$$

Now suppose that $y\in(x+V_x+V_x)\cap(C+V_x)$; then there are $u,v,w\in C_x$ and $z\in C$ such that $x+u+v=z+w$, and hence $x+u+v-w=z$. But $V_x=-V_x$, so $-w\in V_x$, and therefore

$$x+u+v-w=z\in(x+V_x+V_x+V_x)\cap C\,,$$

contradicting $(1)$. It follows that

$$(x+V_x+V_x)\cap(C+V_x)=\varnothing\,.$$

I’m afraid that I can’t follow your subsequent argument; for starters, I don’t understand what $s$ is.

ContentLicense: CC BY-SA 4.0


问:
Prove by induction that...
If $0<a<1$. Prove by induction that $\forall n \in \mathbb N, (1-a)^n < \frac{1}{1+na}$.

I know that induction requires a base case and an induction hybothethis (IH).

For the base case, n = 1:

LHS = (1-a)^1 = 1-a

RHS = $\frac{1}{1+a}$

So $1-a < \frac{1}{1+a}$. By testing out a few numbers, I know this inequality is true. But how do I show the base case formally?

The IH would be: "Assume $(1-a)^k < \frac{1}{1+ka}$ for some natural number k," right?

Then I need to show that $(1-a)^{k+1} < \frac{1}{1+(k+1)a}$? Which I have no idea how to do.

Any help is appreciated. Thanks.

Edit: For the base case, I did
$1-a < \frac{1}{1+a} <=> (1-a)(1+a)<1 <=> 1-a^2<1 <=> a^2 >0$ which is True since a>0, so $a^2>0$

答:
By hypotesis
$$(1-a)^{k+1} =(1-a)^k (1-a)< \frac{1-a}{1+ka}$$
and then
$$\frac{1-a}{1+ka}<\frac{1}{1+(k+1)a} \Leftrightarrow (1-a)(1+(k+1)a)<1+ka \Leftrightarrow \\ \Leftrightarrow -(k+1)a^2<0$$

Title: Prove by induction that...


问:
most general solution for ODE
I am facing some difficulties understanding the difference between a general solution and the most general solution of a 2nd order ODE

for example for homogeneous constant coefficients case with 2 real roots, is $y=C_1 e^{r_1x}+C_2 e^{r_2x}$ a general solution or the most general solution?

and how can I prove that a general solution is the most general one?

答:
There are many kinds of 2nd Order ODEs. Some are (In)homogeneous, some are (Non-)linear. So this question is kind of vague on what you want as a general solution.

Note: I mostly use $\mathbb{R}$ for constants as i'm not always sure whether i can use $\mathbb{C}$ but generally it should work too. All functions here are of the form $y(x)$.
Every equation here for which i provide solutions is second order here. Inhomogeneous means the same as non-homogeneous.

The 'most general' ODE

$$ y''(x)= F(x,y,y') $$

has no general solutions as non-linear equations may never be solved generally. Some can be solved but there is no general rule for it.

The Linear Inhomogeneous ODE((independent) variable coefficient)

$$ (ax^{k}+b)y''+ (cx^{l}+d)y' + (ex^m+g)y = F(x) $$

where $a,b,c,d,e,g \in \mathbb{R} $ and $ k,l,m \in \mathbb{R} $
may or may not have a solution, involves some sort trick to be solved usually and even if there is a solution it's not given like you classically learnt.

When $k=2, l=1, m=2$ and $a=1, b=0, c=1, d=0, e=1, g=-\alpha^2 \hspace{4mm} \alpha \in \mathbb{C}$ and $F(x)=0$
you get the Bessel-equation

$$ x^2y''+xy'+(x^2-\alpha)y=0 $$

whose solution cannot be given with elementary functions and you need a special series with the gamma function to calculate specific values of it.

The Bessel functions of the first kind being $J_\alpha$ are defined as such:

$$ J_{\alpha}(x)=\sum_{n=0}^\infty \dfrac{(-1)^n}{n!\Gamma(n+\alpha+1}\left(\dfrac{x}{2}\right)^{2m+\alpha},$$

where $\Gamma(x)$ is the Gamma-function.

A more 'regular' ODE would be the (Cauchy-)Euler equation whose general form is:

$$ a_nx^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + a_{n-2}x^{n-2}y^{(n-2)} + ... +a_0y=0$$

Here the (n) in the superscript means the n-th order derivative.

The second order version of it let's say is in the form:

$$ ax^2y''+ bxy'+cy=0$$

where $a,b,c \in \mathbb{R} $

The general solution for this kind of equation is reached like this:

First substitute $y=x^m,\;\; y'=mx^{m-1},\;\; y''=m(m-1)x^{m-2} $

Then the transformed equation will look like this:

$$ax^2\cdot m(m-1)x^{m-2} + bx\cdot mx^{m-1} + c x^m =0$$

Notice how the x powers fall out and we can factor out $x^m$.

Thus we get: $$x^m[am(m-1)+bm+c]=0 $$

Divide by $x^m$ which can never be $0$ and we arrive to this simple problem:

$$am^2-(a+b)m+c=0$$

Solve for $m$ and plug your results into this: $$y = c_1x^m_1 + c_2x^m_2$$

If there is a resonance($m_1=m_2$) we multiply our results by $\ln(x)$ so they will be linearly independent(take note how this is similar when this happens with a constant coefficient ODE)

$$ y= c_1x^m\ln(x) + c_2x^m $$

If our solutions are of the complex form $ \alpha \pm \beta i$

$$ y=c_1x^\alpha \cos(\beta\ln(x)) + c_2x^\alpha \sin(\beta\ln(x)) $$

If F(x) is not zero then you need to find the particular solution first, you can find more of this at the bottom of this answer.

Constant Coefficient Linear Homogeneous ODE

$$ ay''+by'+cy=0 $$

where $a, b, c \in \mathbb{C} $

We transform this into an algebraic problem:

$$ ar^2+br+c=0 $$

Solve it for r, get your results in $\alpha$ and $\beta$

the solution of this is as you wrote:

$$ y= c_1e^{\alpha x}+c_2 e^{~beta x}$$

If there is a resonance($\alpha=\beta$)

$$ y= (c_1+c_2x)e^{\alpha x / 2} $$

If the two roots are of the complex form $\alpha \pm \beta i$

$$ y= e^{\alpha x}(c_1\cos(\beta x)+c_2\sin(\beta x)) $$

Constant Coefficient Linear Inhomogeneous ODE

$$ ay''+by'+cy=F(x) $$

When $F(x) \neq 0$, you solve the inhomogeneous term $F(x)$ with Variation of Parameters or Method of undetermined coefficients (some schools also teach the Annihilator Method), then you add this to the homogeneous  general solution thus getting the whole inhomogeneous general solution for that kind of problem.

$$ y_{hom.gen.}+y_{inhom.part.}=y_{inhom.gen.} $$

These are the ones that i had the top off my head. I am not really sure what you could prove about this. Maybe you could (re)formulate your question a bit better.

Title: most general solution for ODE


问:
Restricting split short exact sequence of quasi-coherent sheaves to their coherent sub-sheaves.
Let $X$ be a projective variety, $Z$ a hypersurface section and $U \overset{def}= X \setminus Z$ its complement, an open affine subscheme of $X$. Let $i:U \hookrightarrow X$ be the corresponding open embedding. Given a coherent sheaf $M$ on $U$ we can consider the quasi-coherent sheaf $i_*M$. This seems to have a filtration of coherent sheaves in the following manner: Let's assume we are working on an affine chart and $W=\text{Spec}(R)$ is an affine in this chart and the equation of $Z$ corresponds to the element $f$ in $R$. The quasi-coherent sheaf $i_*M|_{W\cap U}$ on $W$ has this increasing filtration that $M_0=M|_W$, $M_1=\frac 1fM|_W$, $M_2=\frac 1{f^2}M|_W, \ldots$. This filtration seems to glue and give a filtration by coherent sheaves $(i_*M)_0\subset (i_*M)_1 \subset \ldots$ of $i_*M$. (Please correct me if I am wrong). Note that for each $j$, $i^*(i_*M)_j\cong i^*i_*M\cong M$.

Now let's consider a short exact sequence of vector bundles on $X$ like $0\rightarrow E_1 \rightarrow E_2 \rightarrow E_3 \rightarrow 0$. This leads to a short exact sequence of quasi-coherent sheaves $0\rightarrow i_*i^*E_1\rightarrow i_*i^*E_2\rightarrow i_*i^*E_3\rightarrow 0$ which restricts back to the original short exact sequence on $U$. Note that this exact sequence of quasi-coherent sheaves is split. This is because short exact sequence of vector bundles split on $U$. If the contents of the first paragraph is correct, my question is, does this split short exact sequence of quasi-coherent sheaves restrict to a split short exact sequence of coherent sheaves of the form $0\rightarrow (i_*i^*E_1)_j\rightarrow (i_*i^*E_2)_j\rightarrow (i_*i^*E_3)_j\rightarrow 0$ for each $j\geq 0$? (Here $j$ is the same indexing of the filtration defined in the first paragraph.)

答: The answer to the second paragraph is negative. For example the zeroth filtration of the short exact sequence involving $i_*i^*E_k$'s just recovers the original short exact sequence. Every coherent sheaf $M$ such that $M|_U$ is isomorphic to a fixed coherent sheaf $E$, injects into the quasi-coherent sheaf $i_*E$. Applying the filtration corresponding to $M$, just reproduces the coherent sheaves $M, M(1), \ldots , M(n), \ldots$


问:
What is the finite eigenvalue of a controlled system as $\epsilon \rightarrow 0$?
Consider a state space system (A, B) as follows:

$$A = \begin{bmatrix}
    a_1 & a_2 \\
    -a_2 & 0 \\
    \end{bmatrix},
B= \begin{bmatrix}
    0 \\
    1 \\
    \end{bmatrix}$$

where $a_1$ and $a_2$ are real constants and $a_1 \ne 0$. Furthermore, let

$$Q=\begin{bmatrix}
    0 & 0 \\
    0 & 1 \\
    \end{bmatrix},
R = \epsilon^2$$

where $\epsilon > 0$. Let $P(\epsilon)$ be the positive definite solution, assuming it exists, of the algebraic Riccati equation:

$$A^TP + PA - PBR^{-1}B^TP + Q = 0.$$

What is the finite eigenvlue(s) of $A- \epsilon^{-2}BB^TP(\epsilon)$ as $\epsilon \rightarrow 0$?

I have figured out that if a positive definite solution $P(\epsilon)$ exists, then $a_2 \ne 0$. However, I am unable to calculate $P(\epsilon)$. Is there a trick to find the eigenvalues of $A- \epsilon^{-2}BB^TP(\epsilon)$ as $\epsilon \rightarrow 0$?

Thanks in advance!

答:
Since you are dealing with LQR and they ask about the closedloop eigenvalues, the first thing that comes to mind is to use the associated Hamiltonian matrix

$$
H =
\begin{bmatrix}
A & -\epsilon^{-2} B\,B^\top \\ -Q & -A^\top
\end{bmatrix},
$$

which should have negated eigenvalue pairs and the two negative eigenvalues should match the closedloop eigenvalues. Solving directly for the eigenvalues of $H$ as a function of $a_1$, $a_2$ and $\epsilon$ would require solving a quartic equation. However, it is easier to make use of the stated property of $H$, namely from that it follows that its characteristic polynomial would be of the form

$$
\det(\lambda\,I - H) = (\lambda - p_1)\,(\lambda + p_1)\,(\lambda - p_2)\,(\lambda + p_2) = p_1^2\,p_2^2 - (p_1^2 + p_2^2)\,\lambda^2 + \lambda^4.
$$

Matching the actual characteristic polynomial with this form allows one to find expressions for $p_1^2$ and $p_2^2$ relatively easily. Now, when one applies the limit of $\epsilon \to 0$ to these solutions one will remain bounded while the other does not.


问:
Matrix Derivative of $(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta})$
I am trying to calculate the derivative of $$(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta}) $$ where $\mathbf{P}$ is a positive definite matrix. The actual dimensions of each element is not given in the question specification, but since it is for the purposes of minimising $\beta$ for regression analysis, I think $\mathbf{X}$ is mxn, $\mathbf{\beta}\in \mathbf{R}^n$ and $\mathbf{Y}\in \mathbf{R}^m$. First, I expand the expression,

$$(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta}) = (\mathbf{Y^TP-\beta^T\mathbf{X}^TP})(\mathbf{Y-X \beta}) = \mathbf{Y^TPY-Y^TPX\beta -\beta^TX^TPY+\beta^TX^TPX\beta} $$

Now I take the derivative for wrt $\beta$. For the final term, I am using that it is a quadratic form and I think I am assuming $\mathbf{X^TPX}$ is symmetric. I am just using identities on - https://en.wikipedia.org/wiki/Matrix_calculus Anyway I get,

$$\mathbf{-Y^TPX-Y^TPX}+2\mathbf{\beta^TX^TPX} = -2\mathbf{Y^TPX+2\beta^TX^TPX}$$

From here, I can equate to $0$ and take the transpose, to solve for $\beta$ (assuming everything is inversable for now, don't worry).

$$\mathbf{\beta^TX^TPX=Y^TPX}\iff \mathbf{X^TPX\beta=X^TPY} \iff \beta=\mathbf{(X^TPX)^{-1}X^TPY}$$

The solutions solve it slightly differently. They said since, $(\mathbf{Y-X \beta})^T\mathbf{P}(\mathbf{Y-X \beta})$ is already a quadratic form, we can just use this to calculate the derivative as $$\mathbf{-X^T}2\mathbf{P(Y-X\beta})=-2\mathbf{X^TPY} + 2{\mathbf{X^TPX\beta}}$$. As you can see, this is the same as my derivative, but transposed. Of course, once I transpose to solve for $\beta$, this is no longer the case and we get the same final solution. I have 2 questions.

Is the method I have done incorrect, i.e. if the question was just calculate the derivative have I done it incorrectly. If so would you kindly point out where I have made my mistake?

Could anyone recommend some literature/web page that explains the process the solutions took for taking the derivative by spotting it was a quadratic form.

Thank you very much!

答:
The derivative you want is Fréchet derivative (see https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative). Let
$$ \mathbf f(\beta)=(\mathbf{Y-X\beta})^T\mathbf{P}(\mathbf{Y-X \beta}). $$
Then
\begin{eqnarray}
D\mathbf f(\beta)\mathbf h&=&\lim_{t\to0}\frac{\mathbf f(\beta+t\mathbf h)-\mathbf f(\beta)}{t}\\
&=&\lim_{t\to0}\frac{-t(\mathbf{Y}-\mathbf{X}\beta)^T\mathbf{Ph}-t\mathbf{h}^T\mathbf{P}(\mathbf{Y}-\mathbf{X}\beta)+t^2\mathbf{h}^T\mathbf{Ph}}{t}\\
&=&-(\mathbf{Y}-\mathbf{X}\beta)^T\mathbf{Ph}-\mathbf{h}^T\mathbf{P}(\mathbf{Y}-\mathbf{X}\beta).
\end{eqnarray}


问:
Calculating variance of a claim payment
I am very confused about this problem and would appreciate some enlightening.

Question: Losses, $X,$ under an insurance policy are exponentially distributed with mean $10.$ For each loss, the claim payment $Y$ is equal to the amount of the loss in excess of a deductible $d > 0.$ Calculate $\text{Var}(Y).$

My apparently incorrect reasoning:

$Y = X - d,$ and we have this theorem:
"For any constants $a$ and $b,$ $\text{Var}(aX + b) = a^2\text{Var}(X)$"

So $\text{Var}(Y) = \text{Var}(X - d) = \text{Var}(X) = 100$ since $X$ is an exponential RV. Where is my mistake?

The answer given is:
$100[2e^{−0.1d} − e^{−0.2d}].$

答:
Given the memoryless property of the exponential distribution, the distribution of $Y=\max(X-d,0)$ will have the same $\lambda=\frac 1 {10}.$

The expected probability that the insurance company will have to pay at all (claim larger than deductible) is given by the distribution of the exponential evaluated at $x>d$:

$$\Pr(X>d)= e^{-0.1 d}$$

Therefore $Y=0$ with probability $1-e^{-0.1d},$ and follow an exponential distribution with mean $10$ with probability $e^{-0.1 d}.$

The $E(Y)=0\cdot (1-e^{-0.1d}) + 10 \cdot e^{-0.1d}$

We are going to need the formula $\text{Var}=E(Y^2)-E(Y)^2,$ and the latter part after the minus is $100\cdot e^{-0.2d}.$

The $E(Y^2)$ requires some leg work. By LOTUS,

$$\begin{align}
E(Y^2) &= \int_0^\infty y^2 \cdot 0.1 e^{-0.1y}\;dy\\[2ex]
&=\left.y^2 \cdot (-e^{-0.1 y})\right|_{0}^{\infty} - \int_0^{\infty} (-e^{0.1 y}) \cdot 2y \; dy\\[2ex]
&=\frac{2}{0.1}\int_0^{\infty}y\; 0.1e^{0.1 y}  \; dy\\[2ex]
&=\frac{2}{0.1}E(Y) = \frac{2}{0.1}\frac{1}{0.1}=2\cdot 100
\end{align}$$

but before we apply the formula of the variance, we need to go back to the deductible, and the probability that the company actually has to make payments:

$$E(Y^2)=0\cdot (1-e^{-0.1d})+200\cdot e^{-0.1d}.$$

Now we obtained the desired result $100 \left(2 e^{-0.1d}- e^{-0.2d}\right).$

Title: Calculating variance of a claim payment


问:
simplify fraction (stone drops off a cliff)
How do I get from
$$ \frac {dt} {T} = \frac {dx} {gt} \sqrt { \frac {g} {2h} } $$
to
$$ \frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx $$
where $x(t) = \frac {1} {2} gt^2 $ and $ T = \sqrt { \frac {2h} {g}}$.

I'm currently struggling with Griffiths' Introduction to Quantum Mechanics. This is from a worked example on p.11-12 of the 2nd edition.

答:
It’s just algebra. I figured out the steps to get the first expression  for $dt\over T$ from the second, but you can reverse the steps. Using what you are given for $x(t)$ and $T$, and assuming $t\ge0$,

$$\frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx = \frac {1} {{2 \sqrt {h\left({1\over2}gt^2\right)}} } dx = \frac {1} {t\sqrt {2hg} } dx=\frac {1} {gt\sqrt {2h\over g} } dx=\frac {\sqrt {g\over 2h}dx} {gt }.$$


问:
Need help understanding why $f(k+1) \leq \int_k^{k+1} f(x) \, dx \leq f(k)$ in the following context:
I would like to know why the inequality (6.26 in the below pasted solution) $f(k+1) \leq \int_k^{k+1} f(x) \, dx \leq f(k)$ is true. It feels true intuitively since $f$ is a mono decreasing function and the segment length is $1$, but I can't figure out analytically how to show it is true.

The original question is:

And its solution is:

答: $f(k+1)$ is a constant not a function, and $f(k+1) \leq f(x)$ meaning $f(k+1)(k+1 -k) = f(k+1) = \int_{k}^{k+1}f(k+1) \leq \int_{k}^{k+1}f(x)$. The other inequality is the same.


问:
Proving $(AB)^T=B^TA^T$: Can you find the flaw in the reasoning?
Proving $(AB)^T=B^TA^T$ isn't difficult. An example of such a proof can be found here: http://www.math.ucdenver.edu/~esulliva/LinearAlgebra/ABT.pdf

However, if I take a slightly different approach, it doesn't work:

If  $B=(b_{ij})$, then $B^T=(b_{ji})$, and similarly, $A^T=(a_{ji})$.   Thus $B^TA^T=(b_{ji})(a_{ji}) = (\sum_k b_{jk}*a_{ki}).$ But the $(i,j) $ element of $(AB)^T$ is $(\sum_k b_{ki}*a_{jk})=(\sum_k a_{jk}*b_{ki})$, which is generally not equal to $(\sum_k b_{jk}*a_{ki})$.

What is the flaw in the reasoning? Any help is appreciated.

答:
One may also address this problem in the context of inner products, which I find simpler since it avoids the necessity of index bookkeeping, viz:

For any matrix/operator $C$, $C^T$ is the unique operator satisfying

$\langle x, Cy \rangle = \langle C^Tx, y \rangle; \tag 1$

then

$\langle x, ABy \rangle = \langle (AB)^T x, y \rangle; \tag 2$

but

$\langle x, ABy \rangle = \langle A^Tx, By \rangle = \langle B^TA^Tx, y \rangle; \tag 3$

combining (2) and (3) yields

$\langle (AB)^T x, y \rangle = \langle x, ABy \rangle = \langle B^TA^Tx, y \rangle, \tag 4$

which holds for all $x$, $y$; thus

$(AB)^T x = B^TA^Tx \tag 5$

binds for every $x$, whence

$(AB)^T  = B^TA^T. \tag 6$

$OE\Delta$.


问:
Banach fixed point theorem with lip constant possibly attaining $1$
An important condition of Banach fixed point theorem is that the map is strict contraction with Lipschitz constant strictly less than 1. Then the convergence is exponentially fast. But what if we relax the condition to less and equal to 1. Are there still a convergence (albeit not at exponential speed)?</p>

答:
I think you're looking for the following generalization:

If $(S,d)$ is a compact metric space and $f:S\rightarrow S$ is such that $d(f(x),f(y)) < d(x,y)$ for every pair of distinct $x$ and $y$, then there exists some unique fixed point $z$ of $f$ and, additionally, for every $x\in S$ we have $$\lim_{n\rightarrow\infty}f^n(x)=z.$$

The proof of this is harder than the proof of the contraction mapping theorem. I would prove this in two parts.

Part 1: Any two fixed points of $f$ must be equal.

Much as in the original theorem, this just follows by noting that if $f(x)=x$ and $f(y)=y$, then, if distinct, they would have $d(x,y) = d(f(x),f(y)) < d(x,y)$, which is a contradiction.

Part 2: The sequence $f^n(x)$ converges to a fixed point.

This bit is trickier - how I would go about it is to consider the set of limit points $L$ of the sequence. Note that $L$ must satisfy $L=f(L)$ since $f$ is continuous and $S$ is compact and the image of the given sequence under $f$ is just a shift of the original sequence. By compactness, $L$ is non-empty.

Note that $L$ itself is compact, therefore there must be some pair $a,b\in L$ maximizing $d(a,b)$. However, we can then choose $\tilde a$ and $\tilde b$ so that $f(\tilde a)=a$ and $f(\tilde b)=b$ due to the relation $L=f(L)$. Note that, if $a\neq b$, then $d(\tilde a,\tilde b) > d(a,b)$, contradicting maximality - so, to the contrary, we must have $a=b$ - and, by maximality, this means that $d(a',b') \leq d(a,b) = 0$ for every $a',b'\in L$. Otherwise said, $L$ has only one element.

The existence of a unique element of $L$ implies that this element is the limit $\lim_{n\rightarrow\infty}f^n(x)$ and is also a fixed point.

Side note: Another way to prove this theorem is to consider the set
$$I=\bigcap_{n=0}^{\infty}f^n(S).$$
This set also satisfies $I=f(I)$, so can be shown to be a single point similarly. If you do it this way, you can additionally prove that the functions $f^n$ converge uniformly to a constant function.


答: The mean value theorem for integrals implies (for $f$ continuous on $[a,b]$) that $\int_a^b f(t)dt=(b-a)f(c)$ for some $c \in [a,b].$ Replace $a$ by $k$ and $b$ by $k+1$ in this and use monotonicity.

Title: Need help understanding why $f(k+1) \leq \int_k^{k+1} f(x) \, dx \leq f(k)$ in the following context:

Body:
I would like to know why the inequality (6.26 in the below pasted solution) $f(k+1) \leq \int_k^{k+1} f(x) \, dx \leq f(k)$ is true. It feels true intuitively since $f$ is a mono decreasing function and the segment length is $1$, but I can't figure out analytically how to show it is true.

The original question is:

And its solution is:


问:
Non homogenous differential equation

Given the following equations:
$$x'=-x+y-1$$
$$y'=-x-y+3$$
When $t=0$, $x=0$ and $y=3$
Find the general solution for $x$ and $y$ in terms of $t$.

I have already differentiated $x'$ and reduced the equation to find
$$x''+2x'+2x=2$$

I then took the equation to be equal to 0 and used the auxilliary equation and followed through to find the roots to be $$-1 \pm i $$
I then found the PI to be $1$ from $x=2/2$ (unsure about this as there isnt a constant since there is no $x$ involved?)

Therefore leading to the general solution for $x$:
$$x(t)=e^{-t}(A \cos t +B \sin t)+1$$

Can anyone please verify this and help me with how to now find y in terms of t as I am unsure of how to proceed.

Thanks!

答:
Everything you did is correct:
$$x''+2x'+2x=0$$
$$\implies r^2+2r+2=(r+1)^2-i^2=(r+1-i)(r+1+i)=0$$
$$r= -1\pm i$$
Therefore the solution to the homogeenous equation is :
$$x(t)=e^{-t}(A \cos t +B \sin t)$$
And the particular solution is
$$x_p=A \implies A =1$$
$$x(t)=e^{-t}(A \cos t +B \sin t)+1$$
You still need to apply the initial conditions given in the problem.
$$x(0)=0 \implies A=-1$$
$$x(t)=e^{-t}(-\cos t +B \sin t)+1$$
And find $y(t)$:
$$y=x'+x+1$$
$$ye^t=(xe^t)'+e^t$$
$$ye^t=(-\cos t +B \sin t +e^t)'+e^t$$
$$ye^t=\sin t +B \cos t +2e^t$$
$$y(t)=e^{-t}(\sin t +B \cos t) +2$$
Apply intitial condition...
$$ y(0) =3 \implies B=1$$

Title: Non homogenous differential equation


问:
Arithmetic progressions of units in a domain
Let $R$ be a domain with unity, and suppose that $R^\times$ has finite rank as an abelian group.

Can $R^\times$ contain infinitely long arithmetic progressions?

Can $R^\times$ contain arithmetic progressions of arbitrarily long (unbounded) length?

I suspect that the answer to each of these questions is no, but I have been unable to prove either.

A brief note by Morris Newman (PDF) discusses this question for $R$ a number field (for which our question applies by the Dirichlet unit theorem).  For $R$ a number field with degree $n \geq 4$ the lengths of such progressions are bounded by $n$ (and this bound is sharp).  For $n=2,3$ our progressions are instead bounded of length $4$.  Thus each of our numbered claims hold for number fields.

Edit: This edit reflects some recent thoughts I've had on this problem.  Suppose that $R^\times$ is free on $\alpha_1,\ldots, \alpha_m$ (so we've simplified by ignoring torsion elements).  Then an arithmetic progression
$$ \beta_1:=\prod^m \alpha_i^{k_i(1)},\beta_2:=\prod^m \alpha_i^{k_i(2)},\beta_3:=\ldots$$
encodes a number of polynomial relations in the $\alpha_i$, given by
$$p_i(\alpha_1,\ldots, \alpha_m):=\beta_{i+1}-2\beta_i+\beta_{i-1}=0.$$
If the zero sets of these polynomials have no common component, then the fact that $(\alpha_1,\ldots,\alpha_m)$ lies in the intersection of the zero sets should imply a bound on the number of relations that may hold (perhaps $m+1$ such, e.g.).  Because of this tentative connection, I have added the tag "algebraic geometry" to this question.

Edit 2: I've posted a solution below which addresses all aspects of my original questions, save the following (which remains open):

Question: Let $R$ be a domain with characteristic $0$.  If $R^\times$ has finite rank, does there exist a constant $c(R)$ such that any arithmetic progression in $R^\times$ has length at most $c(R)$?

答:
This answer consists of two parts:

I: If $R$ has positive characteristic $n$, then any arithmetic progression has maximal length $n$, for in

$$a,a+b,a+2b,\ldots,a+nb$$

we must have $(a+nb)-a=nb=0$.  This answers questions (1) and (2) in the negative, for $\mathrm{char}(R)>0$.

II: If $\mathrm{char}(R)=0$, suppose that $R^\times$ is finitely generated (as in the original problem statement).  We claim that the Jacobson radical $J_R$ of $R$ vanishes, proven in a series of Lemmas below:

Lemma 1: $R^\times + J_R = R^\times$.

Proof: Let $u \in R^\times$ and $j \in J_R$. If $u+j \notin R^\times$, then $u+j \in \mathfrak{m}$ for some maximal ideal, so that $u \in \mathfrak{m}$ because $j \in J_R \subset \mathfrak{m}$.  This contradicts that $u \in R^\times$, so that $u+j \in R^\times$. //

Lemma 2: If $R^\times$ is finitely generated, then $J_R =(0)$.

Proof: Throughout, we identify $\mathbb{Z}$ with the characteristic subring of $R$.  This proof will proceed in two cases, dependent on whether or not $J_R$ contains non-zero transcendental (over $\mathbb{Q} = \mathrm{Frac}(R)$) elements.  In our first case,  In our first case, take $j \in J_R$ transcendental over $\mathbb{Q}$; then Lemma 1 implies that $1+j\mathbb{Z}[j] \subset R^\times$.  Now, suppose that $\{\pi_i(j)\}_{i=1}^m \subset 1 + j \mathbb{Z}[j]$ is a set of irreducible polynomials.  If $\{\pi_i(j)\}_{i=1}^m$ is not independent (as a multiplicative set), then there exist integers $a_i,b_i \geq 0$ such that

$$\prod \pi_i^{a_i} = \prod \pi_i^{b_i},$$

at which point the fact that $\mathbb{Z}[j]$ is a UFD implies that $a_i=b_i$ for all $i$.  Thus $\{\pi_i(j)\}_{i=1}^m$ is an independent set, whence $R^\times$ has rank at least $m$.  It thus suffices (for the contradiction in our first case) to evince infinitely many irreducible polynomials in $1+j\mathbb{Z}[j]$, and such an example is provided by the cyclotomic polynomials $\Phi_n(j)$ with $n>1$ (or the linear polynomials).

Our second case is just as involved.  Take $j \in J_R$ algebraic; we may assume without loss of generality that $j \in J_R$ is an algebraic integer.  The (multiplicative) abelian group $S \subset R^\times$ generated by $\{1+kj\}_{k \in \mathbb{Z}}$ has finite rank, by assumption on $R^\times$.  The field norm $N: \mathbb{Q}(j) \to \mathbb{Q}$ restricts to a homomorphism $N:S \to \mathbb{Q}^\times$, and the image of this map has finite rank.  Let

$$f(x):= N(1+xj) \in \mathbb{Z}[x].$$

Finiteness of rank implies that the set of prime divisors of $\{f(k) : k \in \mathbb{Z}\}$ is finite.  Yet if $P:= \prod p_i$ represents the product of these primes, it follows that $f(1+kP) \equiv 1 \mod P$, so our hypothesis forces $f(1+kP) = \pm 1$ for all integers $k$.  (This is an extension of Euclid's proof of the infinitude of the primes to a statement about the prime divisors of a polynomial.)  This gives a contradiction, so that no such $j \in J_R$ exists.

Therefore $J_R=(0)$ in either case. //

Now we continue in earnest:

Suppose that $R^\times$ contains the infinite arithmetic progression

$$S:=\{a,a+b,a+2b,\ldots,a+nb,\ldots\}$$

We claim that $b$ fails to generate any given residue domain $R/\mathfrak{p}$.  If not, fix $\mathfrak{p} \in \mathrm{Spec}(R)$ such that $b$ generates $R/\mathfrak{p}$.  Consider the map $\varphi: S \to R/\mathfrak{p}$ given by $a+nb \mapsto (a+nb) \mod \mathfrak{p}$.  This map surjects, hence there exists $a+nb \in S$ such that $a+nb \equiv 0 \mod \mathfrak{p}$.  But $a+nb$ is a unit, which is a contradiction.

When $\mathfrak{p}$ is maximal, $R/\mathfrak{p} \cong \mathbb{Z}/p\mathbb{Z}$ for some rational prime $p$.  Since all non-zero elements generate $\mathbb{Z}/p\mathbb{Z}$, this forces $b \in \mathfrak{p}$, so that $b \in J_R$, the Jacobson radical of $R$.  But $J_R=(0)$ by Lemma 2, so that $b = 0$, a contradiction.  For the case $\mathrm{char}(R)=0$, it follows that the statement of question (1) is universally false.

At this point, I do not know whether or not question (2) can be answered positively under any hypothesis.  As mentioned in my original post, the answer to (2) is "no" when $R$ is a number field of degree $n$, with an explicit upper bound on length given by $\max(n,4)$.  Perhaps it would be possible to look at algebraic subfields of $R$ to reach a similar (if not slightly weaker) conclusion in this more general setting.

Disclaimer: Some of the material above (Lemma 2, in particular) is sourced from my blog, Integer Miscellany.  In that article I consider an ideal that is not quite the Jacobson radical (it behaves the same for these purposes, however), so I've kept this proof self-contained to prevent confusion.

ContentLicense: CC BY-SA 3.0


答:
Let $\phi(x) = \sum_{n\ge 1} f(n) 1_{[n,n+1)}(x)$ and note that
$\phi(x) \ge f(x) \ge \phi(x+1) \ge 0$.

$\int_1^{N+1} \phi(x) dx  = \sum_{n=1}^N f(n) \ge \int_1^{N+1} f(x) dx \ge \int_1^{N+1} \phi(x+1) dx = \int_2^{N+2} \phi(x) dx = \sum_{n=2}^{N+1} f(n)$,
that is
$\sum_{n=1}^N f(n) \ge \int_1^{N+1} f(x) dx \ge \sum_{n=2}^{N+1} f(n)$.

Note that all three quantities are non increasing functions of $N$.

If $\sum_{n=1}^N f(n)$ converges then $\lim_{N \to \infty} \int_1^{N+1} f(x) dx$ exists and if $\int_1^{N+1} f(x) dx$ converges then $\lim_{N \to \infty}  \sum_{n=1}^N f(n)$ exists.


问:
Multiplicative system of a ring and of a category
If A is any category, a class of morphisms $S$ in A is said to be a multiplicative system whether
$(a)$ it's closed by composition, that is: $id_X$ is in $S$ for every $X$ in A and whenever $f$ and $g$ are morphisms in A such that the composition $gf$ makes sense, then $gf$ is in $S$; $(b)$ any diagram of the form $X\overset{f}\longrightarrow Y \overset{s}\longleftarrow Z$ with $s$ in $S$ can be completed as
$\require{AMScd}$
\begin{CD}
W @>g>> Z\\
@VtVV @VVsV\\
X @>>f> Y
\end{CD}
with $t$ in $S$. The same also with all the arrows reversed. Eventually $(c)$ for a pair of morphisms $f,g:X\to Y$ there exists $s$ in $S$ with $sf=sg$ if and only if there exists $t$ in $S$ with $ft=gt$.

My question is: does this definition coincide with the notion of multiplicatively closed set for any ring $R$ if we look at $R$ as an Ab-category with just one object? Certainly condition $(a)$ provides exactly what we desire for a multiplicatively closed set (that is a subset $S\subseteq R$ such that $1\in S$ and $x,y\in S\Rightarrow xy,yx\in S$), and if $R$ is commutative, $(b)$ and $(c)$ become obvious, but in the case of a non-commutative ring I can't find a proof of these conditions.

Could anyone provide a proof or a counterexample? If a counterexample is the answer, is there any profound reason why it happens to works only in the commutative case, or it's the notion of multiplicative system to be designed just to generalize these cases?

答:
Yes, it coincides, but rather trivially (in the commutative case).

See your (commutative unital) ring $R$ as a category as follows. The $R$-module action of $R$ on itself induces a morphism $\iota: R \to \mathrm{Hom}_{\mathbb Z}(R,R)$, so we can consider the category with one object (namely $R$) and the set of morphisms is $\iota(R)$. The fact that this forms an $\mathbf{Ab}$-category is part of the axioms of a ring. You need the ring to be unital for the identity morphism to be present, and commutativity gives you the other axioms. For instance, if you are given
$\require{AMScd}$
\begin{CD}
R @>f>> R @<s<< R
\end{CD}
you are basically given two elements of the original ring $R$. The diagram can easily be completed by assuming that $R$ is commutative since $sf = fs$ leads to the commutative diagram
$\require{AMScd}$
\begin{CD}
R @>f>> R\\
@VsVV @VVsV\\
R @>>f> R
\end{CD}
Statement (c) is proven similarly by taking $t=s$. I don't know about localizing non-commutative rings at subsets $S$ in general, but I would bet that if these ideas made sense, then the localzation $S^{-1}R$ would exist when $R$ is non-commutative in the specific case where those categorical axioms are satisfied, but not in general. I read this to know a bit about non-commutative localization, and it doesn't feel as inspiring as the commutative counterpart.

Hope that helps,


答:
We have
$$
x=\frac{1}{2}gt^2
\quad\Leftrightarrow\quad
gt^2=2x
$$
so
\begin{align*}
\frac{dt}{T}
&
=\frac{dx}{gt}\sqrt{\frac{g}{2h}}
=\sqrt{\frac{g}{g^2t^2\cdot2h}}\,dx
=\sqrt{\frac{1}{gt^2\cdot2h}}\,dx
\\&
=\sqrt{\frac{1}{2x\cdot2h}}\,dx
=\sqrt{\frac{1}{4hx}}\,dx
=\frac{1}{\sqrt{4hx}}\,dx
=\frac{1}{2\sqrt{hx}}\,dx.
\end{align*}

Title: simplify fraction (stone drops off a cliff)

Body:
How do I get from
$$ \frac {dt} {T} = \frac {dx} {gt} \sqrt { \frac {g} {2h} } $$
to
$$ \frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx $$
where $x(t) = \frac {1} {2} gt^2 $ and $ T = \sqrt { \frac {2h} {g}}$.

I'm currently struggling with Griffiths' Introduction to Quantum Mechanics. This is from a worked example on p.11-12 of the 2nd edition.


问:
Two questions on completely regular filters in locales
I'm reading the exposition of the Stone-Čech compactification for locales in Johnstone's book Stone Spaces. In Chapter IV Paragraph 2.2, Johnstone constructs the Stone-Čech compactification of a locale $A$ as the locale $C(A)$ of completely regular ideals in $A$, and in Paragraph 2.3 he shows that the points of $C(A)$ may be identified with the maximal completely regular filters in $A$. There are two points here where I was unable to follow the reasoning. Unfortunately, they're a bit technical.

Let's start with the definitions. Fix a locale $A$.

We say that $a$ is well inside $b$, written $a \eqslantless b$, if there exists $c$ such that $a\land c = 0_A$ and $b\lor c = 1_A$. Equivalently, we could just require that $b \lor \lnot a = 1_A$.

We say that $a$ is really inside $b$, written $a \,\unicode{x02A9B}\, b$ if there is a scale between $a$ and $b$, i.e. if there exist $\{c_q\mid q\in \mathbb{Q}\cap [0,1]\}$ such that $a\leq c_0$, $c_1\leq b$, and $c_p \eqslantless c_q$ for all $p<q$.

We say that $A$ is completely regular if $a = \bigvee_A\{b\mid b\,\unicode{x02A9B}\,a\}$ for all $a$.

We say that an ideal $I\subseteq A$ is completely regular if for all $a\in I$ there exists $b\in I$ with $a\,\unicode{x02A9B}\,b$.

We say that a filter $F\subseteq A$ is completely regular if for all $a\in F$ there exists $b\in F$ with $b\,\unicode{x02A9B}\,a$.

$C(A)$ is the locale of completely regular ideals of $A$. $C(A)$ is a subframe of $\mathrm{Idl}(A)$, so the meet of finitely many completely regular ideals is their intersection and join of some completely regular ideals is the ideal generated by the union. Further, $C(A)$ is compact and completely regular.

Question 1: Lemma 2.3 states that the poset of completely regular filters of $A$ is isomorphic to the poset of completely regular filters of $C(A)$. In the last step, Johnstone writes, "A similar argument shows $\overline{f}_*\overline{f}^*(G) = G$ for any completely regular filter $G$ of $C(A)$."

It doesn't matter what the functions $\overline{f}_*$, $\overline{f}^*$, $f_*$, and $f^*$ are: What matters is that for $J\in C(A)$, $f_*f^*(J) = \{a\in A\mid a \,\unicode{x02A9B}\, \bigvee_A J\}$, and $\overline{f}_*\overline{f}^*(G)$ is the filter of $C(A)$ generated by $f_*f^*(J)$ for all $J\in G$. Since $J$ is completely regular, $J\subseteq f_*f^*(J)$ for all $J$, so $f_*f^*(J) \in G$, and it is clear that $\overline{f}_*\overline{f}^*(G) \subseteq G$.

To prove the converse, we would like to take any completely regular ideal $I\in G$ and find some $J\in G$ such that $f_*f^*(J) \subseteq I$. Since $G$ is completely regular, there is some $J\in G$ with $J\,\unicode{x02A9B}\,I$, so it would be reasonable to try to prove that $J\,\unicode{x02A9B}\,I$ implies $f_*f^*(J)\subseteq I$. But I don't see how to do this.

Question 2: In the second half of Proposition 2.3, $A$ is a compact completely regular locale. We are given a maximal completely regular filter $F$, which we would like to show is prime. If $a\lor b\in F$ and $a\notin F$, we construct the filter $G = \{c\in A\mid c\lor b\in F\}$. It's easy to see that $G$ is a filter and $F\subsetneq G$. But Johnstone writes that $G$ "is easily seen to be completely regular".

So suppose $c\in G$. We need to find some $e\,\unicode{x02A9B}\,c$ with $e\in G$. Well, $c\lor b\in F$, and since $F$ is completely regular there is some $d\in F$ with $d\,\unicode{x02A9B}\,c\lor b$. Since $c\land (c\lor b) = c$, it would be natural to hope that $e = c\land d$ works. But it is not true in general that $x \,\unicode{x02A9B}\,y$ implies $z\land x \,\unicode{x02A9B}\, z\land y$.

答:
$\newcommand{\ri}{\mathbin{\unicode{x02A9B}}}
\newcommand{\wi}{\eqslantless}$
For Question 1, suppose $J\ri I$.  Then in particular $J\wi I$, so there exists $K\in C(A)$ such that $J\wedge K=0$ and $I\vee K=1$.  The latter means there exist $i\in I$ and $k\in K$ such that $i\vee k=1$ and the former then implies that $j\wedge k=0$ for all $j\in J$.

Now let $x=\bigvee_A J$ and suppose $a\ri x$; we wish to show $a\in I$.  Since $j\wedge k=0$ for all $j\in J$, we also have $x\wedge k=0$ and hence $a\wedge k=0$.  This implies $a=a\wedge(i\vee k)=a\wedge i$.  Thus $a\leq i$ and so $a\in I$, as desired.

For Question 2, you have to use complete regularity and compactness of $A$. Take $d\in F$ such that $d\ri c\vee b$; then in particular $d\wi c\vee b$, so there exists $e\in A$ such that $d\wedge e=0$ and $c\vee b\vee e=1$.  Now by complete regularity, we know that $c=\bigvee S$ where $S=\{f\in A:f\ri c\}$.  We then have $\bigvee S\vee b\vee e=1$, so by compactness, there is a finite $T\subseteq S$ such that $\bigvee T\vee b\vee e=1$.  Letting $g=\bigvee T$, we then have $g\ri c$ and $g\vee b\vee e=1$.  We then have $d=d\wedge(g\vee b\vee e)=d\wedge (g\vee b)$ since $d\wedge e=0$.  Thus $d\leq g\vee b$, so $g\vee b\in F$, so $g\in G$ is our desired element.


问:
How is the sample space of a random variable defined?
I was watching this example where the professor said that if we have 2 i.i.d RVs, both of them being binomial random variables, then we can only add them up if their sample spaces are the same. I am wondering what the meaning is of a sample space for a random variable?</p>

答:
The formal definition of a (real) random variable is a function $X:\Omega\to\mathbb R$, where $\Omega$ is the sample space. The sample space is, in informal terms, where the "randomness" of probability comes from.

For example, say you flip a fair coin $5$ times. Then the sample space $\Omega$ is the set of your possible outcomes, e.g. $HHHHH$, $THTTH$ etc.

A random variable $X$ in this space might e.g. be the number of heads you toss. So for example, $X(HHHHH)=5$.

It makes no sense to add two random variables with different sample spaces in the same way it makes no sense to add two functions with different domains. This is because just as with functions, we define $X+Y$ by $$(X+Y)(\omega)=X(\omega)+Y(\omega),$$ for $\omega\in\Omega$. Evidently we require $X$ and $Y$ to have the same sample space.

As an example, consider $X$ and $\Omega$ as defined by the coin tosses, and consider a new random variable $X':\Omega'\to\mathbb R$. The sample space $\Omega'$ is the set of points on Earth, and $X'$ is the current temperature at a point on the Earth. What is $X+X'$?

The situtation can be salvaged though. Suppose $X$, $Y$ are real random variables with sample spaces $\Omega_X$, $\Omega_Y$. Define a random variable $Z:\Omega_X\times\Omega_Y\to\mathbb R$ by $$Z(\omega_X,\omega_Y)=X(\omega_X)+Y(\omega_Y).$$
Then the projections $\bar{X}:(\omega_X,\omega_Y)\mapsto X(\omega_X)$ and $\bar{Y}:(\omega_X,\omega_Y)\mapsto Y(\omega_Y)$ have the same distributions as $X$ and $Y$ respectively, and clearly $\bar{X}+\bar{Y}=Z$ is defined. In fact, $\bar{X}$ and $\bar{Y}$ are even independent.


问:
If ${ x }^{ 4 }+{ y }^{ 2 }=1$ then $x$ and $y$ can be both rational numbers?
Can you give two numbers $x,y\in\mathbb{Q}$ such that  ${ x }^{ 4 }+{ y }^{ 2 }=1$?

I don't know if exists or not. I derived this equation questioning that if $\sin { \alpha  } ={ x }^{ 2 }$ for $x\in \mathbb{Q}$ then for which $\alpha$, $\cos{ \alpha  }=y$ and $y\in \mathbb{Q}$?

Edit: $x,y\neq0$

答:
Consider $\dfrac{p}{q}=x$ and $y=\dfrac{m}{n}$, such that $\gcd (p,q)=(m,n)=1$ and $n \neq 1$, $q \neq 1$

$(\dfrac{p}{q})^4=1-(\dfrac{m}{n})^2=(1-\dfrac{m}{n})(1+\dfrac{m}{n})$

$=(\dfrac{n-m}{n})(\dfrac{n+m}{n})=\dfrac{(n-m)(n+m)}{n^2}$, here $n^2 \nmid (n-m)(n+m)$ since $q \neq 1$.

Let $d$ be a divisor of $n$, note that $d \nmid(n+m)$ and $d \nmid (n-m)$, Using the fact: $\gcd(m,n)=1$

Now this implies $(n-m)(n+m)=p^4, n^2 =q^4$, which isn't possible.(Recalling Area of right angled triangle can't be a square, it doesn't exist as proven by Fermat)


问:
Help to understand the ring of polynomials terminology in $n$ indeterminates
In Hungerford's Algebra, page 150, the author defines a ring of polynomials in "$n$" indeterminates in the following manner:

After the author defines the operations in this ring with a theorem:

saying that the author defines another definition to a ring of polynomials with $n$ indeterminates:

My problem is with the terminology in this theorem which comes after:

If I understood well, the author defines a function $x_i$ which when we applies to $\epsilon_i$, we have the identity in $R$. following this reasoning what is $x_i^k(k\epsilon_i)?$ is this the composition of $x_i$?
I'm very confused with these terminologies, I need help.

Thanks a lot!

答:
First, you shouldn't be frightened by all this; although I haven't read this book I'm pretty sure the author is going to drop all these notations very soon. In fact the theorem 5.4 prepares that dropping.

We are used to writing polynomials as sums of terms, and the author is soon going to do that as well. However he did not define them like that, because expressions are not easy to describe mathematically; one needs to cater for such things as that $x_1+3x_1^2x_2$, $x_1^2x_2+x_1+2x_1^2x_2$ and $x_1(1+3x_1x_2)$ (and many more) are all different expressions but the same polynomial. So instead the mathematical object that represents the polynomial will be the function (not to be confused with a polynomial function) that sends each monomial to its coefficient, which here satisfies $(1,0)\mapsto1$, $(2,1)\mapsto3$, and $(k,l)\mapsto0$ for all other pairs $(k,l)\in\mathbf N^2$.

Once you've understood this, you'll see that the expression $x_i$ stands for the function that maps $\varepsilon_i=(0,\ldots,0,1,0,\ldots,0)$ to $1$ (more precisely to the multiplicative unit $1_R$ of $R$) and everything else to$~0_R$. That's the first part of theorem 5.4. He could then just have said that things like sums, products, and powers are computed using the operations in the polynomial ring he just defined (theorem 5.3(i)), but (maybe since he did not explicitly define powers) he makes explicit what powers (i) and monomials(ii) are, that powers of indeterminates commute among each other (iii) and with coefficients (iv),  and that every polynomial $f$ can be written as $R$-linear combination of distinct monomials (v) (the coefficient $a_{k_1,k_2,\ldots,k_n}$ is going to be nothing else than the value $f(k_1,\ldots,k_n)$ at the tuple labelling the monomial this coefficient is attached to). For point (i), the power $x_i^k$ should map the tuple $(0,\ldots,0,k,0,\ldots,0)\mapsto1_R$ and evertything else to $0_R$; the author writes that tuple as $k\varepsilon_i$, where the multiplication is just the scaling operation in $\mathbf N^n$.


问:
Second order Taylor polynomial at x = 0?
What is the complete second order Taylor polynomial?  at x = 0? I have done solving many times but the results were different.

Let $\epsilon > 0$ and suppose that
$f : (-\epsilon, \epsilon) \to (-1, \infty)$ is a $C^2$ function
which satisfies $f(0) = 0$ and
$$f(x) + \ln(1 + f(x)) = x.$$
Calculate the second order Taylor polynomial of $f$ around the point
$x_0 = 0$

答:
Second order Taylor polynomial of f is:

$$f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2$$

We are evaluating this at $a=0$ so the polynomial becomes the Maclaurin series:

$$f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2$$

Differentiate both sides of the equation you supplied to find $f'$:
\begin{alignat}{3}
&& \frac{d}{dx} (f(x)+\ln(1+f(x)) & = \frac{d}{dx} (x) \\
\implies && \frac{df}{dx} + \frac{\frac{df}{dx}}{1+f(x)} & = 1 \\
\implies && \frac{df}{dx}\left(1 + \frac{1}{1+f(x)}\right) & = 1 \\
\implies && \frac{df}{dx} &= \frac{1+f(x)}{2+f(x)}
\end{alignat}
Evaluate $f'$ at $x=0$:

$$\frac{df}{dx}(0) = \frac{1+f(0)}{2+f(0)} = \frac{1}{2}.$$

Use the quotient rule to solve $f''$:

$$ \frac{d}{dx}(\frac{df}{dx}) = \frac{d}{dx}\left(\frac{1+f(x)}{2+f(x)}\right)$$

$$ \implies \frac{d^2f}{dx^2} = \frac{ (2+f(x))\frac{df}{dx} - (1+f(x))\frac{df}{dx}}{(2+f(x))^2} = \frac{\frac{df}{dx}}{(2+f(x))^2} = \frac{1+f(x)}{(2+f(x))^3}.$$

Evaluate $f''$ at $x=0$:

$$ \frac{d^2f}{dx^2}(0) = \frac{1+f(0)}{(2+f(0))^3} = \frac{1}{8}.$$

Therefore, the Taylor polynomial can be expressed as follows:

$$f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 = \frac{1}{2}x + \frac{1}{16}x^2.$$


问:
Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.

This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?

答:
One can use Lagrange multipliers for:
$$L = \sum_{k=1}^{n} 1/(n-1 + x_{k})  - \lambda(\prod_{k=1}^{n}x_{k} - 1)$$
One has:
$$\frac{dL}{dx_{k}} = -\frac{x_{k}}{(n-1+ x_{k})^{2}} -\lambda\prod_{j\neq k}x_{j}$$
And when this is zero, one can multiply both sides by $x_{k}$ to get:
$$\frac{x_{k}^{2}}{(n-1+ x_{k})^{2}} =\lambda\prod_{k}x_{k}$$
Because of condition on product of $x_{k}'s$ all $x_{k}$ must be the same and equal to 1 at the  maximum.


问:
problems on Lebesgue integral

1) Given a measure space, $f$ a non-negative measurable function and $A$ in the $\sigma$-algebra such that $\mu(A)=0$, prove that $\displaystyle\int_{A} f\;d\mu=0$ .

My try:

$$0\leq \int_{A}f\;d\mu=\int fI_{A}\;d\mu\leq \int \sup{f(A)}I_{A}\;d\mu=\sup{f(A)}\int I_{A}\;d\mu=\sup{f(A)}\mu(A)=0\;,$$
then $\displaystyle\int_{A}f\;d\mu=0\,.$

2) Given a measure space, let $f$ be an integrable function. Show that $\displaystyle \int_{\{|f|>t\}} |f|\;d\mu\rightarrow 0$ as $t\rightarrow\infty$ .

My try: since $f$ is integrable, $|f|\lt \infty$ almost everywhere, $\mu (\{|f|>t\}) = \mu(\Omega)- \mu(\{|f|\leq t\})$, and as $t\rightarrow\infty$,  $\mu (\{|f|>t\})<\epsilon$, then
$$\int_{\{|f|>t\}} |f|\;d\mu=\int |f|I_{\{|f|>t\}}\;d\mu\leq\max\{|f|\}\int I_{\{|f|>t\}}=\max\{|f|\}\;\mu(\{|f|>t\})<\max\{|f|\}\;\epsilon\;.$$
Then it's done; I can take $\epsilon=\frac1t$.

Well, both demonstrations seem right for me. I will be thankful if someone tells me what's wrong on them.

Greetings.

答:
Edit: I missed some things in my first answer and updated my answer accordingly.

Edit 2: Not only that I missed some things in my first answer, I missed the point completely. I leave the answer here for instructional purposes. Tasadduk's answer is definitely the way to go about 2).

I agree with mac that the first solution is almost ok, but some extra care has to be taken (the supremum might be infinite and even if one could argue that $\infty \cdot 0 = 0$ in measure theory, a clear cut argument is certainly preferable to a convention).

I would first prove that for a non-negative and bounded functions $f$ we have $\int_{A} f \,d\mu = 0$, for this you can use your argument. The monotone convergence theorem then implies that it holds for all non-negative measurable functions, simply by approximating $f$ by $f_{n} = \min{\{f,n\}}$ from below.

Remark. The equality $\int_{A} f = 0$ holds true for all measurable $f$, because a general $f$ can be written as $f = f_+ - f_-$ with $f_{+}(x) = \max{\{f(x),0\}}$ and $f_{-}(x) = \max{\{-f(x),0\}}$ and use the definition $\int_{A} f\,d\mu = \int_{A} f_+ \,d\mu - \int_{A} f_-\,d\mu$ provided at least one of the integrals of the right hand side is finite.

As mac also pointed out, there are quite a few problems in the second proof and I don't see how to save it using your idea.

Here's how I would do it. Let $A_{t} = \{|f| \gt t\}$ and note that for $s \lt t$ we have $A_{s} \supset A_t$. Define
$$a_{t} = \int_{\{|f| \gt t\}} |f|\,d\mu = \int_{A_t} |f|\,d\mu.$$
Since $|f| \geq 0$ and $A_{s} \supset A_{t}$ we have $a_{s} \geq a_{t}$ for $s \leq t$. We want to show that $a_{t} \to 0$ as $t \to \infty$. By monotonicity of $t \mapsto a_{t}$, it suffices to show that $a_{n} \to 0$ as $n \to \infty$ runs through the natural numbers. Since $a_{n} \geq 0$ and $a_{n}$ is monotonically decreasing, the limit $a = \lim_{n \to \infty} a_{n}$ exists, and we want to show that $$ a = 0.$$

Now let $f_{n}(x) = \min{\{|f(x)|,n\}}$ and note that $f_{n} \to |f|$ pointwise (and monotonically). By the monotone convergence theorem (or the dominated convergence theorem, applicable since $f_{n} \leq |f|$ and $|f|$ is integrable, if you prefer) we have

$$ \int |f|\,d\mu = \lim_{n \to \infty} \int f_{n}\,d\mu.$$

On the other hand $n \leq |f|$ on $A_{n}$ and $0 \leq f_{n} \leq |f|$ on $\Omega$ imply
$$ \int f_{n}\,d\mu = \int_{A_{n}} n \,d\mu + \int_{\Omega \smallsetminus A_{n}} f_{n}\,d\mu \leq \int_{A_{n}} |f|\, d\mu + \int |f| \,d\mu = a_{n} + \int |f|\,d\mu.$$

Passing to the limit on both sides of the estimate $\int f_{n} \leq a_{n} + \int |f|$ we get
$$\int |f|\,d\mu = a + \int |f|\,d\mu$$
and as $\int |f|\,d\mu \lt \infty$ we conclude $a = 0$, as we wanted.

Title: problems on Lebesgue integral


问:
Proving one function is greater than another
How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?

My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0$ for all $x$. However, $q \neq 0,1$ so $f(x) > g(x)$ for all $x$. However, I'm looking for something more mathematically rigorous. Any suggestions?

Body:
$\begin{align}
f(x)-g(x)
&=(x+1)^2-4qx\\
&=x^2+2x+1-4qx\\
&=x^2+2(1-2q)x+1\\
&=x^2+2(1-2q)+(1-2q)^2-(1-2q)^2+1\\
&=(x+1-2q)^2+1-(1-2q)^2\\
\end{align}
$.

Since $0 <q < 1$,
$-1 < 1-2q < 1$
so $0 < (1-2q)^2 < 1$
so $1 > 1-(1-2q)^2 > 0$
so (finally)
$f(x) > g(x)$.

Another way to get this final inequality is
$1-(1-2q)^2
= (1-(1-2q))(1+(1-2q))
= 2q(2-2q)
= 4q(1-q)
$
and both $1$ and $1-q$ are positive.


问:
We can't show a function is analytic except its singularities with C-R equations?
It's still related to my previous question about proving a function is analytic except at its singularities.

I've found a preposition that said

if $f(z)$ is differentiable at $z$, then it satisfies the
Cauchy-Riemann equations

And it also said the converse isn't true. But i also found another thread said that the converse is also true. Here is the link. I'm confused now. But since I've Sen the counterexample, I believe we can't use Cauchy Riemann to prove this. Here I'm thinking how do we know the function is analytic except the singularity if the function itself I don't know whether it's differentiable except the singularities or not.

By the way, this is the function that I'm talking about:

$$\frac{e^z}{z-2}$$

Anyway, I've been trying to show whether this satisfies Cauchy Riemann equations or not and the answer is yes, it satisfies and of course I'm assuming the function is differentiable except $z=2$.

But again I can't conclude that the function is analytic except at simple pole $2$ since as far as I know the Cauchy Riemann is just a necessary condition for a function to be analytic.

Conclusion: how to prove this problem? Should I show that the function can be written as power series and to support this proof I will also include the fact that that function satisfies Cauchy Riemann? Help me please. I know this is unusual since I'm not talking about how to prove/disprove this function is "differentiable everywhere".

答:
A complex function is complex differentiable at a point if and only if it is real differentiable at that point and satisfies the CR equations at that point. It is analytic on an open set if and only if it is complex differentiable at every point in the set. A function is real differentiable at a point if its partial derivatives exist in a neighborhood and are continuous at the point.

That said, the easiest way to prove that your given function is analytic is by using the fact that sums, differences, products and quotients (as long as the denominator doesn't vanish) of complex differentiable functions are complex differentiable. So your function is complex differentiable everywhere except where the denominator vanishes, since it's composed of everywhere complex differentiable functions. So it's complex differentiable on $\mathbb C\backslash\{2\}$, which is open, so your function is analytic on that open set. CR equations + real differentiability are overkill here.


问:
Monotonically increasing function and supremes
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a monotonically increasing function  and $A \subset \mathbb{R}$ where $A \neq \emptyset$ and boundend.

i) If f is continuous function, prove that $f(\sup (A))= \sup (f(A))$

ii) Find a function $f: \mathbb{R} \rightarrow \mathbb{R}$ wich do not fulfilled i).

For i), I thought because $A$ is boundend it has supreme,

$$ x \le \sup A$$
In other hand, because $f$ is monotonically increasing, if $x \le \sup A$ then $f(x) \le f(\sup A)$ for $x \in A$.

As $f$ is continuous and monotonically  increasing in a boundend set, the set $f(A)$ has a supreme, i.e $f(y) \le \sup f(A)$ for $y \in A$, so $\sup f(A)$ and $f(\sup A)$ are upper bounds.

But I don't know what else I can do.

答: Since $f(x)\leq f(\sup A)$ for all $x\in A$, we should get $$\sup(f(A))\leq f(\sup A).$$ For $\sup A$, you can find an increasing sequence $(a_n)\subseteq A$ converging to $\sup A$. Clearly, for each $n$, we have $$f(a_n)\leq\sup(f(A)).$$ Since $f$ is continuous, the function values $f(a_n)\to f(\sup A)$ and thus $$f(\sup A)\leq\sup(f(A)).$$


问:
which statement is true for $f(x)$ at $x=0$
Let $f(x)$ has continuous second derivative in some neighbourhood of $x=0$ and $\lim _{x\to 0}\frac{f''\left(x\right)}{e^x-1}=1.$ Then $f(x)$ at $x=0$ is

There are extreme values

No extreme values

No turning points

There is an inflection point.

My attempt

Since Let $f(x)$ has continuous second derivative in some neighbourhood of $x=0.$ So, by the property of continuity,$\lim _{x\to 0}\frac{f''\left(x\right)}{e^x-1}=1>0\implies $ $\frac{f''\left(x\right)}{e^x-1}>0$ for some neighborhood of $0$. Let that neighbourhood is $(-\epsilon,\epsilon)$. $x>0 \implies e^x-1>0 \implies f''(x)>0$ and $x<0 \implies e^x-1<0 \implies f''(x)<0.$ By intermediate theorem, $f''(0)=0.$ So, it is an inflection point. Is my logic correct?

答:
Another way to see that $f''(0) = 0$ is to consider the multiplication property of limits:

$$\begin{align*}
f''(x) &= \frac{f''(x)}{e^x-1} \cdot (e^x-1)&&(x\ne 0)\\
\lim_{x\to 0} f''(x)
&= \lim_{x\to 0} \frac{f''(x)}{e^x-1} \cdot \lim_{x\to 0}(e^x-1)\\
&= 1\cdot 0 = 0
\end{align*}$$

In a neighbourhood of $0$, the sign of $f''(x)$ follows the sign of $e^x-1$.

And at $x=0$, since $f''$ is continuous,

$$f''(0) = \lim_{x\to0}f''(x) = 0$$


问:
How can I rearrange the exact sag equation?
I apologize if this is a basic question on a site such as this but my maths skills aren't the best.

The exact sag equation is $s = r - \sqrt{(r^2 - y^2)}$ we are required to know in my module that the rearrangement to find $r$ is $r = \frac{(y^2 + s^2)}{2s}$ but I am struggling to rearrange the equation to find this.
I asked my professor who told me it was fine to just memorize the equation and not understand how we got there but I would like to understand how we rearranged to find this, both to improve my mathematics and because I can bet I will need to know how we got here for the exam.

I searched online and watched many videos on rearrangements and expanding brackets but struggled to find one similar to this. I found a website where somebody goes through how they rearranged it the link is below.
https://www.optiboard.com/forums/showthread.php/15689-Rearranging-the-Sag-formula

They use this as a starting point $r^2 = y^2 + (r - s)^2$ which I understand but then the next step jumps to here $r^2 = y^2 + r^2 - 2rs + s^2$ the step was called after expanding but I am not sure how to expand to get here. I looked at an online calculator that said I should use the perfect square formula and upon researching what this is I became even more confused.

Please help!

Thank you for taking the time to read my question and I will appreciate any response I get! Sorry for the poor formatting I have never used this site and struggled to use the codes to make the squared and square root symbol. Thank you again!

答:
$$s=r-\sqrt{r^2-y^2}\\
\sqrt{r^2-y^2}=s-r\\
r^2-y^2=s^2-2sr+r^2\\
2sr=s^2+y^2\\
r=\frac{s^2+y^2}{2s}$$
As we squared in the third line we may have introduced an extraneous solution, so the solution needs to be checked back into the original equation


答:
$$s = r - \sqrt{r^2 - y^2}$$ can be rearranged to
$$\sqrt{r^2 - y^2} = r - s$$ and squaring both sides (possibly introducing a spurious root)
$${r^2 - y^2} = r ^2 -2rs +s^2$$ and simplifying
$$2rs = y^2+s^2$$ and thus $$r = \frac{y^2+s^2}{2s}$$
and you can check this works correctly by substitution

Title: How can I rearrange the exact sag equation?

Body:
I apologize if this is a basic question on a site such as this but my maths skills aren't the best.

The exact sag equation is $s = r - \sqrt{(r^2 - y^2)}$ we are required to know in my module that the rearrangement to find $r$ is $r = \frac{(y^2 + s^2)}{2s}$ but I am struggling to rearrange the equation to find this.
I asked my professor who told me it was fine to just memorize the equation and not understand how we got there but I would like to understand how we rearranged to find this, both to improve my mathematics and because I can bet I will need to know how we got here for the exam.

I searched online and watched many videos on rearrangements and expanding brackets but struggled to find one similar to this. I found a website where somebody goes through how they rearranged it the link is below.
https://www.optiboard.com/forums/showthread.php/15689-Rearranging-the-Sag-formula

They use this as a starting point $r^2 = y^2 + (r - s)^2$ which I understand but then the next step jumps to here $r^2 = y^2 + r^2 - 2rs + s^2$ the step was called after expanding but I am not sure how to expand to get here. I looked at an online calculator that said I should use the perfect square formula and upon researching what this is I became even more confused.

Please help!

Thank you for taking the time to read my question and I will appreciate any response I get! Sorry for the poor formatting I have never used this site and struggled to use the codes to make the squared and square root symbol. Thank you again!


问:
Does "All additive functions are linear" imply every set of reals is measurable?
Suppose we are working in ZF set theory without choice. An additive function is a function defined over the real line such that $f(x+y)=f(x)+f(y)$. It is known that if every set of reals is measurable, then every additive function is linear. Is the converse true in ZF?

答:
No.

We know that "Every set of reals has the Baire property" will also have the same consequence, and that ZF+"Every set of reals has the Baire property" is actually a weaker(!) theory than ZF+"Every set of reals is Lebesgue measurable", as shown by Shelah.


问:
Non-degenerate conic is projectively equivalent to smooth conic
I'm trying to understand the proof of the proposition:

Every non-degenerate conic $C$ in $\mathbb{P}_{\mathbb{C}}^{2}$ is projectively equivalent to the smooth conic
$$
C_{0}=\left\{\left[x_{0}, x_{1}, x_{2}\right] \in \mathbb{P}_{\mathbb{C}}^{2} \mid x_{1}^{2}+x_{0} x_{2}=0\right\}.
$$

Proof. By a previous result, we may assume that [0,0,1] lies on $C .$ Then $C$ is the zero set of a homogeneous quadratic polynomial of the form
$$
Q\left(x_{0}, x_{1}, x_{2}\right)=a x_{0}^{2}+b x_{1}^{2}+c x_{0} x_{1}+d x_{0} x_{2}+e x_{1} x_{2}
$$
with $a, b, c, d, e \in \mathbb{C} .$ If $d=e=0$ then $Q$ is a product of two linear factors, contradicting the assumption that $C$ is non-degenerate. Thus $d$ or $e$ is non-zero, which means that [0,0,1] is a smooth point of $C .$ By taking a further projective transformation, we may also assume that the projective tangent line of $C$ at $[0,0,1]$ is defined by $x_{0}=0,$ which means that $e=0$ and $d \neq 0 .$ Then $b \neq 0$ because otherwise, $Q$ would be divisible by $x_{0}$ and $C$ would be degenerate. Multiplying $Q$ with $1 / b,$ we can reduce to the case where $b=1$ This leaves us with
$$
Q\left(x_{0}, x_{1}, x_{2}\right)=x_{1}^{2}+x_{0}\left(a x_{0}+c x_{1}+d x_{2}\right)
$$
Now we apply the projective transformation $\Phi_{A}$ where $A$ is the invertible matrix
$$
\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & c & d
\end{array}\right)
$$
This operation transforms $C$ into $\Phi_{A}(C)=C_{0}$.

I had two questions.

How do we know that for every non-degenerate conic, we can choose $4$ points in it such that no three points are collinear? (This is relevant in "By a previous result" part)

In the above proof, we first start with $Q\left(x_{0}, x_{1}, x_{2}\right)=a x_{0}^{2}+b x_{1}^{2}+b' x_{2}^{2}+c x_{0} x_{1}+d x_{0} x_{2}+e x_{1} x_{2}$, and take projective transformations to make $Q$ pass through $[0,0,1]$. So we let $b'=0$. Taking further projective transformations, we let $e=0$. So my question is, how do we know that the form of $Q$ is 'preserved' through these transformations? To make this question more specific, by the first projective transformation, we can let $b'=0$ thus wiping out the $x_2^2$ from $Q$. By the second projective transformation, we can wipe out the $x_1x_2$ factor, but how do we know that the $x_2^2$ doesn't reappear in the polynomial form of $Q$, as we have taken another projective transformation?

答:
Bezout. If three points on your conic were colinear, then that line is an irreducible component of your conic and therefore your conic is degenerate.

Any transformation fixing $[0,0,1]$ keeps $b'=0$. So all you need to do is fix $[0,0,1]$ while moving the tangent line to $x_0=0$ - this is easily accomplished by rotating around that point by the rotation that sends the line $dx+ey=0$ to the line $x=0$. The matrix of this projective transformation is exactly the $3\times 3$ matrix $\begin{pmatrix} R & 0 \\ 0 & 1 \end{pmatrix}$ where $R$ is the $2$-d rotation matrix.


问:
Summation involving subfactorial function
Inspired by this post:

Does the following series converge; if so, to what value does it converge? $$ \sum_{n = 2}^\infty \left|\frac{!n \cdot e}{n!} - 1\right|$$ I am looking for a closed form for the second question.

Note: !n denotes the subfactorial, also known as the number of derangments within $ \mathrm{S}_n $.

答:
Some calculations: We have

$$
\sum_{n=2}^\infty\left|\frac{!n\cdot e}{n!}-1\right|=\sum_{n=2}^\infty\left|\sum_{k=0}^\infty e\frac{(-1)^k}{k!}-1\right|=\sum_{n=2}^\infty\left|\frac{\Gamma(n+1,-1)}{\Gamma(n+1)}-1\right|.
$$
We also have
$$
\frac{\Gamma(n+1,-1)}{\Gamma(n+1)}=\frac{\Gamma(n+1)-\gamma(n+1,-1)}{\Gamma(n+1)}=1+\Gamma(n+1)^{-1}\int_{-1}^0e^{-t}t^n\mathrm{d}t.
$$

EDIT: Oops, I had a factorial too many. Removed.


答:
A small trick

In a case such this one where $f(x)$ is defined by an implicit equation, you can start from where you want to arrive to that is to say
$$f(x)=\sum_{k=1}^n a_k \,x^k+O(x^{n+1})$$ (there is no $a_0$ since $f(0)=0$ and replace.

So, the implicit equation you face is
$$\sum_{k=1}^n a_k \,x^k+\log\left(1+\sum_{k=1}^n a_k \,x^k\right)-x=0\tag 1$$ Using
$$\log(1+\epsilon)=\epsilon-\frac 12 \epsilon^2+\frac 13 \epsilon^3+\cdots$$ we shall make $\epsilon=\sum_{k=1}^n a_k \,x^k$ and use the binomial expansion up to the order which is required.

Let us say that we need the third order. So
$$\epsilon=a_1x+a_2x^2+a_3x^3+O(x^4)$$
$$\epsilon^2=a_1^2 x^2+2 a_1 a_2 x^3+O\left(x^4\right)$$
$$\epsilon^3=a_1^3 x^3+O\left(x^4\right)$$ and now replace in $(1)$ to have
$$a_1x+a_2x^2+a_3x^3+\left(a_1x+a_2x^2+a_3x^3 \right)-\frac 12\left( a_1^2 x^2+2 a_1 a_2 x^3\right)+\frac 13\left(a_1^3 x^3\right)-x=0$$ Group the terms of same power
$$(2 a_1-1) x+\left(2 a_2-\frac{a_1^2}{2}\right)
   x^2+\left(\frac{a_1^3}{3}-a_1 a_2+2 a_3\right)x^3=0$$
Cancel the coefficients one at the time to get
$$a_1=\frac 12 \qquad a_2=\frac 1{16} \qquad a_3=-\frac 1{192} $$

So $$f(x)=\frac 12 x+\frac 1{16}x^2-\frac 1{192}x^3+O(x^4)$$

Title: Second order Taylor polynomial at x = 0?

Body:
What is the complete second order Taylor polynomial?  at x = 0? I have done solving many times but the results were different.

Let $\epsilon > 0$ and suppose that
$f : (-\epsilon, \epsilon) \to (-1, \infty)$ is a $C^2$ function
which satisfies $f(0) = 0$ and
$$f(x) + \ln(1 + f(x)) = x.$$
Calculate the second order Taylor polynomial of $f$ around the point
$x_0 = 0$


问:
Number of ideals of given norm in $\mathbb{Z}[\sqrt[4]{24}]$
Consider the number field $\mathbb{Q}(\sqrt[4]{24})$. We want to find the number of ideals of norm $100$ in $\mathbb{Z}[\alpha]$ where $\alpha=\sqrt[4]{24}$.

We first notice that the ring of integers is given by $\mathbb{Z}[\alpha,\alpha^3/4]$. In here we have unique prime ideal factorization. So if we want to find the ideals of norm $100$ in $\mathcal{O}_{K}$ we can do it in the following way. First notice that if $\mathfrak{p}$ is a prime divisor of such an ideal $I$, then it also divides $[\mathcal{O}_{K}:I]=100$. And therefore it divides $p=2$ or $p=5$. So we only have to factor $p=2$ and $p=5$ in $\mathcal{O}_{K}$ to find all prime ideals above $p=2$ and $p=5$. We find the following factorizations:

$2\mathcal{O}_{K}=(2,\alpha^{3}/4)^4=\mathfrak{p}_{2}^4$ and $5\mathcal{O}_{K}=(5,\alpha^2+2)(5,\alpha^2-2) = \mathfrak{p}_{5}\mathfrak{q}_{5}$.

And the norms are given by $N(\mathfrak{p}_{2})=2$, $N(\mathfrak{p}_{5})=25$ and $N(\mathfrak{q}_{5})=25$. Therefore we find that the only possible ideals of norm $100$ are $\mathfrak{p}_{2}^2\mathfrak{p}_{5}$ and $\mathfrak{p}_{2}^2\mathfrak{q}_{5}$. Since in $\mathcal{O}_{K}$ the norm function is multiplicative.

But how does one determine the ideals of norm $100$ in $\mathbb{Z}[\alpha]$? Since here we have no unique prime ideal factorization and moreover the prime above $2$ is singular. It might be useful to know that $[\mathcal{O}_{K}:\mathbb{Z}[\alpha]]=8$. Any help would be appreciated!

答:
I will  use $a$ for $\sqrt[4]{24}$, easier to type.

Let $I$ be an ideal of $\mathcal O=\Bbb Z[a]=\Bbb Z\langle 1, a, a^2,a^3\rangle$ of norm $100$ in $\mathcal O$, i.e. the quotient
$$R:=\mathcal O/I$$
has cardinality $100$. So the images $\bar 1,\bar a, \bar a^2, \bar a^3$ generate $\Bbb R$ as a $\Bbb Z$-module.

We have than a surjective morphism
$$
\phi:\mathcal O\cong\Bbb Z[x]/(x^4-24)\to R\ ,
$$
parametrized by the image $\bar a\in R$ of $a\in\mathcal O$ (i.e. of $\bar x=x$ modulo $(x^4-24)$). Here $\bar a$ is constrained to, and only to $\bar a^4=24$.

Conversely, each such morphism determines an ideal $I$ of norm $100$.

Which are the possibilities for $R$?

The ring $R/25$ must have $25$ elements as a group with addition, but cannot have the ring structure $\Bbb Z/25$ or $(\Bbb Z/5)^2$, since there is no $\bar a$ (taken modulo $25$) with $\bar a^4=24=-1$. (Since there is no such element modulo $5$.)
So $R/25$ is the field $F=\Bbb F_{25}$ with $25$ elements.
Its unit group is cyclic, $F^\times =\Bbb F_{25}^\times \cong \Bbb Z/24$, generated by some element $u$,
and there are four elements $\bar a\in F$ with $\bar a^4=24=-1=u^{12}$, explicitly
$u^3, u^9, u^{15}, u^{21}$. Over $\Bbb F_5$ we have $(x^4-24)=x^4+1=(x^2-2)(x^2+2)$, so we may realize $\Bbb F_{25}=\Bbb F_5[B]/(B^2-2)=\Bbb F_5[b]$, $b^2=2$, and then $\bar a\in\{b, 2b, 3b, 4b\}$.

The ring $R/4$ has $4$ elements, and from the possibilities $\Bbb Z/4$, $(\Bbb Z/2)^2$, $\Bbb F_4$ only the first one may be realized as surjective image of $\mathcal O/4$, the first one.

This implies that $20=4\cdot 5$ is in the kernel of any surjective $\phi$.
We thus consider
$$
\begin{aligned}
\mathcal O/20
&=\Bbb Z[x]/(20, x^4-24)
\\
&=(\Bbb Z/20)[x]/(x^4-4)
\\
&=(\Bbb Z/4)[x]/(x^4-4) \times (\Bbb Z/5)[x]/(x^4-4)
\\
&=(\Bbb Z/4)[x]/(x^4) \times \Bbb F_5[x]/((x^2-2)(x^2+2))
\\
&=(\Bbb Z/4)[x]/(x^4)
\times
\Bbb F_5[x]/(x^2-2)
\times
\Bbb F_5[x]/(x^2+2)
\ .
\end{aligned}
$$
The first factor has two quotients with $4$ elements,

one is $(\Bbb Z/4)[x]/(x^4)/(x)=(\Bbb Z/4)[x]/(x)=(\Bbb Z/4)$,

the other one is  $(\Bbb Z/4)[x]/(x^4)/(2,x^2)=(\Bbb Z/2)[x]/(x^2)=(\Bbb Z/4)$.
For the first one, one may use the ideal $(5a)$ at the level of $\mathcal O$ as a quotient.
For the second one, one may use $(10, 5a^2)$ at the level of $\mathcal O$.

The last two factors, taken as a whole, as a ring, admit the $4$ morphisms to $\Bbb F_{25}$ mentioned above, two different kernel appear, corresponding to the passage $\Bbb F_5[x]/((x^2-2)(x^2+2))\to\Bbb F_5[x]/((x^2-2)(x^2+2))/(x^2\pm 2)=
\Bbb F_5[x]/(x^2\pm 2)$, and at the level of $\mathcal O$ we may use the ideal
$4(a^2\pm 2)$.

We thus obtain four and only four ideals $I_j$ of $\mathcal O$ with $\mathcal O/I_j$ having $100$ elements, $j=1,2,3,4$. They are:
$$
\begin{aligned}
J_1 &= (20, 5a, 4(a^2+2))\ ,\\
J_2 &= (20, 5a, 4(a^2-2))\ ,\\
J_3 &= (10, 5a^2, 4(a^2+2))\ ,\\
J_4 &= (10, 5a^2, 4(a^2-2))\ .
\end{aligned}
$$

Notes and comments:

The following computations are assisted by sage, and we confirm the inputs from the OP, and the above results:

```<code>sage: R.<x> = PolynomialRing(QQ)
sage: K.<a> = NumberField(x^4-24)
sage: K
Number Field in a with defining polynomial x^4 - 24
sage: a^4
24

sage: K.class_number()
1

sage: Oa = K.order([a])
sage: Oa.index_in(K.OK())
8

sage: factor(K(2))
(a^3 + 3*a^2 + 4*a + 5) * (1/4*a^3 - 1/2*a^2 + a - 2)^4
sage: factor(K(5))
(1/2*a^2 + 1) * (1/2*a^2 - 1)
sage: (a^3 + 3*a^2 + 4*a + 5).norm()
1
sage: (1/4*a^3 - 1/2*a^2 + a - 2).norm()
-2
sage: (1/2*a^2 + 1).norm()
25
sage: (1/2*a^2 - 1).norm()
25
</code>```

For the four ideals $J_1,J_2,J_3,J_4$ of  $\mathcal O$ explicitly written above, if we are considering the corresponding ideals in $\mathcal O_K$ with the same generators, then it is hard to use the information on the norms from the one order to the other. This highly depends on the intersections...

```<code>sage: OK = K.OK()
sage: OK
Maximal Order in Number Field in a with defining polynomial x^4 - 24
sage: a^4
24
sage: JOK1 = OK.ideal([20, 5*a, 4*(a^2+2)])
sage: JOK1.is_principal()
True
sage: JOK1.gens_reduced()
(1/2*a^3 + a^2 + a + 2,)
sage: w = 1/2*a^3 + a^2 + a + 2
sage: w.norm()
-200
sage: JOK1.norm()
200
</code>```

and

```<code>sage: JOK3 = OK.ideal([10, 5*a^2, 4*(a^2+2)])
sage: JOK3.is_principal()
True
sage: JOK3.gens_reduced()
(a^2 + 2,)
sage: 10/(a^2+2)
1/2*a^2 - 1
sage: 5*a^2/(a^2+2)
-1/2*a^2 + 6
sage: (a^2+2).norm()
400
sage: JOK3.norm()
400
</code>```

(A similar functionality is not implemented for non-maximal orders.)

Body:
I will  use $a$ for $\sqrt[4]{24}$, easier to type.

Let $I$ be an ideal of $\mathcal O=\Bbb Z[a]=\Bbb Z\langle 1, a, a^2,a^3\rangle$ of norm $100$ in $\mathcal O$, i.e. the quotient
$$R:=\mathcal O/I$$
has cardinality $100$. So the images $\bar 1,\bar a, \bar a^2, \bar a^3$ generate $\Bbb R$ as a $\Bbb Z$-module.

We have than a surjective morphism
$$
\phi:\mathcal O\cong\Bbb Z[x]/(x^4-24)\to R\ ,
$$
parametrized by the image $\bar a\in R$ of $a\in\mathcal O$ (i.e. of $\bar x=x$ modulo $(x^4-24)$). Here $\bar a$ is constrained to, and only to $\bar a^4=24$.

Conversely, each such morphism determines an ideal $I$ of norm $100$.

Which are the possibilities for $R$?

The ring $R/25$ must have $25$ elements as a group with addition, but cannot have the ring structure $\Bbb Z/25$ or $(\Bbb Z/5)^2$, since there is no $\bar a$ (taken modulo $25$) with $\bar a^4=24=-1$. (Since there is no such element modulo $5$.)
So $R/25$ is the field $F=\Bbb F_{25}$ with $25$ elements.
Its unit group is cyclic, $F^\times =\Bbb F_{25}^\times \cong \Bbb Z/24$, generated by some element $u$,
and there are four elements $\bar a\in F$ with $\bar a^4=24=-1=u^{12}$, explicitly
$u^3, u^9, u^{15}, u^{21}$. Over $\Bbb F_5$ we have $(x^4-24)=x^4+1=(x^2-2)(x^2+2)$, so we may realize $\Bbb F_{25}=\Bbb F_5[B]/(B^2-2)=\Bbb F_5[b]$, $b^2=2$, and then $\bar a\in\{b, 2b, 3b, 4b\}$.

The ring $R/4$ has $4$ elements, and from the possibilities $\Bbb Z/4$, $(\Bbb Z/2)^2$, $\Bbb F_4$ only the first one may be realized as surjective image of $\mathcal O/4$, the first one.

This implies that $20=4\cdot 5$ is in the kernel of any surjective $\phi$.
We thus consider
$$
\begin{aligned}
\mathcal O/20
&=\Bbb Z[x]/(20, x^4-24)
\\
&=(\Bbb Z/20)[x]/(x^4-4)
\\
&=(\Bbb Z/4)[x]/(x^4-4) \times (\Bbb Z/5)[x]/(x^4-4)
\\
&=(\Bbb Z/4)[x]/(x^4) \times \Bbb F_5[x]/((x^2-2)(x^2+2))
\\
&=(\Bbb Z/4)[x]/(x^4)
\times
\Bbb F_5[x]/(x^2-2)
\times
\Bbb F_5[x]/(x^2+2)
\ .
\end{aligned}
$$
The first factor has two quotients with $4$ elements,

one is $(\Bbb Z/4)[x]/(x^4)/(x)=(\Bbb Z/4)[x]/(x)=(\Bbb Z/4)$,

the other one is  $(\Bbb Z/4)[x]/(x^4)/(2,x^2)=(\Bbb Z/2)[x]/(x^2)=(\Bbb Z/4)$.
For the first one, one may use the ideal $(5a)$ at the level of $\mathcal O$ as a quotient.
For the second one, one may use $(10, 5a^2)$ at the level of $\mathcal O$.

The last two factors, taken as a whole, as a ring, admit the $4$ morphisms to $\Bbb F_{25}$ mentioned above, two different kernel appear, corresponding to the passage $\Bbb F_5[x]/((x^2-2)(x^2+2))\to\Bbb F_5[x]/((x^2-2)(x^2+2))/(x^2\pm 2)=
\Bbb F_5[x]/(x^2\pm 2)$, and at the level of $\mathcal O$ we may use the ideal
$4(a^2\pm 2)$.

We thus obtain four and only four ideals $I_j$ of $\mathcal O$ with $\mathcal O/I_j$ having $100$ elements, $j=1,2,3,4$. They are:
$$
\begin{aligned}
J_1 &= (20, 5a, 4(a^2+2))\ ,\\
J_2 &= (20, 5a, 4(a^2-2))\ ,\\
J_3 &= (10, 5a^2, 4(a^2+2))\ ,\\
J_4 &= (10, 5a^2, 4(a^2-2))\ .
\end{aligned}
$$

Notes and comments:

The following computations are assisted by sage, and we confirm the inputs from the OP, and the above results:

sage: R.<x> = PolynomialRing(QQ)
sage: K.<a> = NumberField(x^4-24)
sage: K
Number Field in a with defining polynomial x^4 - 24
sage: a^4
24

sage: K.class_number()
1

sage: Oa = K.order([a])
sage: Oa.index_in(K.OK())
8

sage: factor(K(2))
(a^3 + 3*a^2 + 4*a + 5) * (1/4*a^3 - 1/2*a^2 + a - 2)^4
sage: factor(K(5))
(1/2*a^2 + 1) * (1/2*a^2 - 1)
sage: (a^3 + 3*a^2 + 4*a + 5).norm()
1
sage: (1/4*a^3 - 1/2*a^2 + a - 2).norm()
-2
sage: (1/2*a^2 + 1).norm()
25
sage: (1/2*a^2 - 1).norm()
25

For the four ideals $J_1,J_2,J_3,J_4$ of  $\mathcal O$ explicitly written above, if we are considering the corresponding ideals in $\mathcal O_K$ with the same generators, then it is hard to use the information on the norms from the one order to the other. This highly depends on the intersections...

sage: OK = K.OK()
sage: OK
Maximal Order in Number Field in a with defining polynomial x^4 - 24
sage: a^4
24
sage: JOK1 = OK.ideal([20, 5*a, 4*(a^2+2)])
sage: JOK1.is_principal()
True
sage: JOK1.gens_reduced()
(1/2*a^3 + a^2 + a + 2,)
sage: w = 1/2*a^3 + a^2 + a + 2
sage: w.norm()
-200
sage: JOK1.norm()
200

and

sage: JOK3 = OK.ideal([10, 5*a^2, 4*(a^2+2)])
sage: JOK3.is_principal()
True
sage: JOK3.gens_reduced()
(a^2 + 2,)
sage: 10/(a^2+2)
1/2*a^2 - 1
sage: 5*a^2/(a^2+2)
-1/2*a^2 + 6
sage: (a^2+2).norm()
400
sage: JOK3.norm()
400

(A similar functionality is not implemented for non-maximal orders.)


问:
Expectation inequality for max ratio difference when drawing marbles

There are $n$ marbles in a bag. A strict non-empty subset of these are red, the rest are blue. We draw marbles from the bag one at a time, without replacement. Let $X_i$ be the proportion of red marbles in the bag before we draw the $i$-th marble. Define $$T=\max \{ |X_i-X_j| : 1 \leq i \leq j\leq n\}.$$ Prove that $\mathbb E[T]\leq f(\mathbb E[X_1])$, where $f(x)=-x\log x-(1-x)\log(1-x)$.

I think it might be easier to compute $\mathbb E[\max X_i]$ and $\mathbb E[\min X_i]$. Then $\mathbb E[T]$ is just their difference. But I'm not sure how to evaluate these quantities.

The $X_i$ form a Markov Chain, but I don't think that observation really helps in this case.

答:
OK so here's a pretty nice solution I found.

Let $M=\max X_i$, $m=\min X_i$ and $\mu=\mathbb E[X_1]$. First, note that $X_i$ is martingale, as conditional on $r$ reds and $b$ blues at step $i$,
$$\mathbb  E[X_{i+1}\mid X_i]=\frac{r}{r+b}\cdot\frac{r-1}{r+b-1}+\frac{b}{r+b}\cdot\frac{r}{r+b-1}=\frac{r}{r+b}=X_i.$$
Now the key bound is that $\mathbb P(M\geq x)\leq\frac{\mu}{x}$ when $x\geq\mu$. Indeed, define the stopping time $T$ as the first time $X_i$ either hits $0$ or exceeds $x$. Then by the optional stopping theorem,
$$\mu=\mathbb E[X_T]\geq x\cdot\mathbb P(M\geq x),$$
which implies the claim. Hence
$$\mathbb E[M]=\int_0^1\mathbb P(M\geq x)\mathop{}\!\mathrm{d}x=\mu+\int_\mu^1\mathbb P(M\geq x)\mathop{}\!\mathrm{d}x\leq\mu+\int_\mu^1\frac{\mu}{x}\mathop{}\!\mathrm{d}x=\mu-\mu\log\mu.$$
Applying the above to the ratio of blue marbles, which starts at $1-\mu$ and has maximum $1-m$:
$$\mathbb E[1-m]\leq1-\mu-(1-\mu)\log(1-\mu)\implies\mathbb E[m]\geq\mu+(1-\mu)\log(1-\mu).$$
Hence we get $\mathbb E[M-m]\leq f(\mu)$, as required.


问:
In ZF, does the ring of continuous functions $C([0,1], \mathbb{R})$ have prime ideals which is not maximal?
In ZFC, it is known that the ring of continuous functions $C([0,1], \mathbb{R})$ have prime ideals which is not maximal. But all proofs of this which I saw uses the axiom of choice.

Then, in ZF, does the same statement hold?

If ZF cannot prove this, how strong is $\mathrm{ZF}+(\text{$C([0,1], \mathbb{R})$ has prime ideals which is not maximal})$ as an intermediate between ZF and ZFC?

答:
I haven't found a construction in ZF, but I found this paper which explores the spectrum of $C(X, \mathbb R)$ for $X$ an arbitrary topological space. It gets a bit heavy when the interaction with the Stone-Cech compactification shows up and partially ordered groups/rings appear, and I couldn't find a constructive approach to finding a prime ideal in $C(X,\mathbb R)$ either. But the relationship between
$$
\mathfrak m_x = \{ f \in C(X,\mathbb R) \, | \, f(x) = 0 \}
$$
and
$$
\mathfrak n_x = \{ f \in C(X,\mathbb R) \, | \, \exists V \text{ neighborhood of } x \, \mathrm{s.t.} f|_V = 0 \}
$$
seems to be extensively studied.

I'm curious to see if one could simplify the paper for the case of $C([0,1],\mathbb R)$ (so that the need for the Stone-Cech compactification goes away, simplifying many results) and shed some light on this question that has been triggering me (and probably a lot of other people) for many years. Let me know if you bump into anything interesting.

Hope that helps,


问:
Property of the trace of matrices
Let $A(x,t),B(x,t)$ be matrix-valued functions that are independent of $\xi=x-t$ and satisfy $$A_t-B_x+AB-BA=0$$ where $X_q\equiv \frac{\partial X}{\partial q}$.

Why does it then follow that  $$\frac{d }{d \eta}\textrm{Trace}[(A-B)^n]=0$$ where $n\in \mathbb N$ and $\eta=x+t$?

Is there a neat way to see that this is true?

答:
(This may not be a neat way to prove the assertion, but it's a proof anyway.) Let $\eta=x+t$ and $\nu=x-t$. Then $x=\eta+\nu$ and $t=\eta-\nu$ are functions of $\eta$ and $\nu$, $A=A(\eta+\nu,\eta-\nu)$ and similarly for $B$. As both $A$ and $B$ are independent of $\nu$, by the total derivative formula, we get
\begin{align*}
0 = \frac{dA}{d\nu} &= \frac{\partial A}{\partial x} - \frac{\partial A}{\partial t},\\
0 = \frac{dB}{d\nu} &= \frac{\partial B}{\partial x} - \frac{\partial B}{\partial t}
\end{align*}
and hence
\begin{align*}
\frac{dA}{d\eta} &= \frac{\partial A}{\partial x} + \frac{\partial A}{\partial t} = 2\frac{\partial A}{\partial t},\\
\frac{dB}{d\eta} &= \frac{\partial B}{\partial x} + \frac{\partial B}{\partial t} = 2\frac{\partial B}{\partial x}.
\end{align*}
Therefore
$$
\frac{d(A-B)}{d\eta} = 2\left(\frac{\partial A}{\partial t} - \frac{\partial B}{\partial x}\right) = 2(BA - AB).
$$
and
\begin{align*}
\frac{d}{d\eta}\operatorname{trace}[(A-B)^n]
&= n \operatorname{trace}\left[(A-B)^{n-1}\frac{d(A-B)}{d\eta}\right]\\
&= 2n \operatorname{trace}[(A-B)^{n-1} (BA-AB)].\tag{1}
\end{align*}
Let $\mathcal{P}_m$ denotes the set of products of $m$ matrices from $\{A,B\}$ (e.g. $\mathcal{P}_2=\{AA,AB,BA,BB\}$). Then the function $f:\mathcal{P}_m\to \mathcal{P}_m$ defined by
\begin{cases}
f(B^m) = B^m,\\
f(pAB^k) = B^kAp &\text{ for all } 0\le k<m \text{ and } p\in\mathcal{P}_{m-k-1}
\end{cases}
is a bijection. Since $\operatorname{trace}(AqB)=\operatorname{trace}(Bf(q)A)$ for all $q\in\mathcal{P}_{n-1}$ and the degree of $B$ is preserved by $f$, it follows that $\operatorname{trace}[A(A-B)^{n-1}B]=\operatorname{trace}[B(A-B)^{n-1}A]$. Consequently,
$$\operatorname{trace}[(A-B)^{n-1} (BA-AB)]=0$$
for all square matrices $A,B$ and $(1)$ evaluates to zero.

Title: Property of the trace of matrices


答:
This is about second one.

Consider the function $f$ which is defined as $f(x) = 1$ if $x \ne 0$ and $f(0) = 0$ and $A=[-1,0]$. Clearly we have $sup A = 0$ and $f(sup A) = 0$. Observe that $f(A) =\{0,1\}$ whose supremum is $1$.

Title: Monotonically increasing function and supremes

Body:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a monotonically increasing function  and $A \subset \mathbb{R}$ where $A \neq \emptyset$ and boundend.

i) If f is continuous function, prove that $f(\sup (A))= \sup (f(A))$

ii) Find a function $f: \mathbb{R} \rightarrow \mathbb{R}$ wich do not fulfilled i).

For i), I thought because $A$ is boundend it has supreme,

$$ x \le \sup A$$
In other hand, because $f$ is monotonically increasing, if $x \le \sup A$ then $f(x) \le f(\sup A)$ for $x \in A$.

As $f$ is continuous and monotonically  increasing in a boundend set, the set $f(A)$ has a supreme, i.e $f(y) \le \sup f(A)$ for $y \in A$, so $\sup f(A)$ and $f(\sup A)$ are upper bounds.

But I don't know what else I can do.


问:
Integral $\int_0^{\pi} \frac{\sin^2 x}{x^2} dx.$
It is known that

$$\int_0^{\infty} \frac{\sin^2 x}{x^2} dx=\frac{\pi}{2}.$$

However, it is not so clear how fast $\int_0^{t} \frac{\sin^2 x}{x^2} dx$
converges to $\frac{\pi}{2}$ as $t$ goes to infinity. Also, the following definite integral with trigonometric function often appears in various scientific areas, for example, the analysis on quantum information.

$$\int_0^{\pi} \frac{\sin^2 x}{x^2} dx.$$

Therefore, I am seeking analytical solution of the above definite integral.

答:
$$\int \frac{\sin^2 (x)}{x^2} dx=-\frac 12\int \frac { \cos(2x)-1}{x^2} dx$$Integrate by parts

$$\int \frac { \cos(2x)-1}{x^2} dx=-\frac{ \cos(2x)-1}{x}-2\int\frac {\sin(2x)}{x} dx$$
$$\int\frac {\sin(2x)}{x} dx=\int\frac {\sin(2x)}{2x} d(2x)=\text{Si}(2 x)$$
As a total
$$\int \frac{\sin^2 (x)}{x^2} dx=\text{Si}(2 x)+\frac{ \cos(2x)-1}{2x}+C$$
$$\int_0^\pi \frac{\sin^2 (x)}{x^2} dx=\text{Si}(2 \pi)$$
$$\int_0^\infty \frac{\sin^2( x)}{x^2} dx=\frac \pi 2$$


问:
the ways completion of p-adic integers
I have heard that there are two ways of constructing completion of p-adic numbers:  One gives an equivalence relation on a Cauchy sequence. Unbeknownst to me , the another way  is using inverse limit.....(★)

To begin with, according to the following lecture, the author defines profinite completion.

$G$ be a group and $\mathcal{I}$ be a collection of normal subgroup
$N \triangleleft G $ of finite index. Let's efine $N \le N'$ iff $N'
 \subset N$. Then, this makes directed poset. If $N \subset N'$, then
we have a surjective homomorphism $G/N' \to G/N$ and the inverse limit
$\underset{\longleftarrow}{\lim}G/N =:\hat{G} $ is called profinite
completion.

Here is my question: Is the above definition really related to the completion as we know?
For example, (This is also an example within the following lecture )

Example )  p-adic integer ${\mathbb{Z}}_p$ is a profinite completion of $\mathbb{Z}$. i.e, ${\mathbb{Z}}_p=\underset{\longleftarrow}{\lim}\mathbb{Z}/p^n
\mathbb{Z}$

In the sense of completion, it looks nonsense for me because, unlike $\mathbb{Q}$, $\mathbb{Z}$ is already a complete set. Thus, when comparing such an Example  to constructing $\mathbb{Q}_p$, I think that profinite completion is not totally related to the concept of completion. In other words, the completion of a ring is just an algebraic concept that is not related to the completion. Though, my clumsy conclusion conflicts with the description that I have heard, (★).

答:
Here is a completely (ha-ha) different way of describing the $p$-adic integers.  Start with the ring of formal power series $\mathbf Z[[x]]$.  Inside of here is the ideal $(x-p)$ of multiples of $x-p$.  Take the quotient ring $R = \mathbf Z[[x]]/(x-p)$, a purely algebraic construction.  This is isomorphic to the ring $\mathbf Z_p$!

The intuition here is that every $p$-adic integer is a power series in $p$ with coefficients in $\{0,1,\ldots,p-1\}$, so we could think about it as being a formal power series with restricted coefficients being evaluated at $p$.  In $\mathbf Z_p$ an infinite series $\sum_{n \geq 0} a_np^n$ with $a_n \in \mathbf Z$ and no further restrictions at all will converge since the $n$th term tends to $0$ in $\mathbf Z_p$.  Therefore numbers in $\mathbf Z_p$ might occur as a general integer power series evaluated at $p$ too.

Okay, define $\varphi \colon \mathbf Z[[x]] \to \mathbf Z_p$ by evaluation at $p$: $\varphi(f(x)) = f(p)$.  This sends a formal power series $\sum_{n \geq 0} a_nx^n$ to the $p$-adically convergent series $\sum_{n \geq 0} a_np^n$, where the numbers $a_n$ are arbitrary integers.  Properties of adding and multiplying convergent $p$-adic infinite series show $\varphi$ is a ring homomorphism.  And $\varphi$ is surjective because every $p$-adic integer has a standard $p$-adic expansion with digits in $\{0,\ldots,p-1\}$.

It remains to show the kernel of $\varphi$ is the multiples of $x-p$.
The containment $(x-p)\subset \ker \varphi$ is immediate from $x-p \in \ker \varphi$. To prove $\ker \varphi \subset (x-p)$, suppose $f(x) = \sum_{n \geq 0} a_nx^n$ is in $\ker \varphi$, so $f(p) = 0$ in $\mathbf Z_p$. Then $a_0 \equiv 0 \bmod p\mathbf Z_p$, so the integer $a_0$ has to be an ordinary multiple of $p$: $a_0 = pb_0$ for some $b_0 \in \mathbf Z$. Then
$$
0 = f(p) = a_0 + a_1p + a_2p^2 + \cdots \equiv a_0 + a_1p  \equiv pb_0 + a_1p \bmod p^2\mathbf Z_p,
$$
so $b_0 + a_1$ has to be an ordinary multiple of $p$: $b_0 + a_1 = pb_1$, so $a_1 = pb_1 - b_0$ for some integer $b_1$. Then show the condition $f(p) \equiv 0 \bmod p^3\mathbf Z_p$ implies $a_2 = pb_2 - p_1$ for some integer $b_2$. And so on: there is a sequence of integers $\{b_n\}$ such that
$$
a_0 = pb_0, \ \ a_n = pb_n - b_{n-1}
$$
for $n \geq 1$. That makes
$$
f(x) = \sum_{n \geq 0} a_nx^n = pb_0 + \sum_{n \geq 1} (pb_n - b_{n-1})x^n = (p-x)\sum_{n \geq 0} b_nx^n,
$$
so $f(x) \in (x-p)$ in $\mathbf Z[[x]]$.  Thus $\mathbf Z[[x]]/(x-p) \cong \mathbf Z_p$ as rings for each prime $p$.

Remark. For different primes $p$, the rings $\mathbf Z_p$ are not isomorphic, so the rings $\mathbf Z[[x]]/(x-p)$ are nonisomorphic as $p$ varies. Note the contrast with $\mathbf Z[x]$: for each prime $p$, $\mathbf Z[x]/(x-p) \cong \mathbf Z$ as rings.

I did not use anything special above about $p$ being a prime number.  If you've heard of $10$-adic integers, then the same argument shows $\mathbf Z_{10} \cong \mathbf Z[[x]]/(x-10)$.

Using the $x$-adic topology on $\mathbf Z[[x]]$, which makes $\mathbf Z[[x]]$ complete, the ring homomorphism $\varphi \colon \mathbf Z[[x]] \to \mathbf Z_p$ is continuous because the values at $p$ of all series in $x^k\mathbf Z[[x]]$ are very small for large enough $k$ (a group homomorphism of topological groups is continuous if and only if it is continuous at the identity), and continuity of $\varphi$ implies continuity of the induced mapping $\overline{\varphi} \colon \mathbf Z[[x]]/(x-p) \to \mathbf Z_p$ when using the quotient topology. This $\overline{\varphi}$ is also an open mapping (check first that $\varphi$ is an open mapping), so the bijection $\overline{\varphi}$ is a homeomorphism.  Thus $\overline{\varphi}$ identifies $\mathbf Z[[x]]/(x-p)$ with $\mathbf Z_p$ both algebraically and topologically.

Title: the ways completion of p-adic integers


问:
Law of large numbers?

Given random variables $Z_1,Z_2,Z_3,\ldots$, which are uniformly distributed for $[8,10]$:

If $X_k =\min\{Z_1, Z_2,Z_3,\ldots,Z_k\}$, prove convergence in probability and find the constant.

Prove and find that the mean of the first $k$ of these variables converge in probably to some constant.

I think this involves the law of large numbers and possible central limit theorem, but I am not really sure.

答:
Hint:

1.)

Step 1: $\Pr(X_k > a) = \Pr(Z_1 > a, Z_2 > a, \cdots, Z_k > a) = (1 - F_Z(a))^k$

where $F_Z(.)$ is the c.d.f of the random variables.

Step 2: Use the above and let $k \to \infty$ to show that the sequence 'converges in distribution' to a constant (can you guess the constant?).

Step 3: Now use the fact that convergence in distribution to a constant implies 'convergence in probability'

2.)

The mean of the $X_k$ can be calculated using $\mathbb{E}(X_k) = \int_{0}^{\infty} (1 - F_{X_k} (x)) dx $, since $X_k$ is a positive random variable.

Use the Step 1 of the previous part to calculate the mean.

Title: Law of large numbers?


问:
Adding a constant to all elements of a matrix and solving

Will solving
$$\begin{pmatrix}
1 & 1 & 1 & 1 & 1\\
-1 & 0 & 1 & 2 & 3 \\
1 & 0 & 1 & 4 & 9 \\
-1 & 0 & 1 & 8 & 27 \\
1 & 0 & 1 & 16 & 81 \\
\end{pmatrix}
\begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
w_5 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
0 \\
\end{pmatrix},
$$
produce the same values for $w_i$ as solving for
$$\begin{pmatrix}
1+a & 1+a & 1+a & 1+a & 1+a\\
-1+a & 0+a & 1+a & 2+a & 3+a \\
1+a & 0+a & 1+a & 4+a & 9+a \\
-1+a & 0+a & 1+a & 8+a & 27+a \\
1+a & 0+a & 1+a & 16+a & 81+a \\
\end{pmatrix}
\begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
w_5 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}?$$
And if so, by which law of matrices ? Thank you.

答:
This happens to be true for the particular matrix and vector that are given.  Define the following matrix and vectors:
$$
M = \pmatrix{1 & 1 & 1 & 1 & 1\\
-1 & 0 & 1 & 2 & 3 \\
1 & 0 & 1 & 4 & 9 \\
-1 & 0 & 1 & 8 & 27 \\
1 & 0 & 1 & 16 & 81}, \quad
e = \pmatrix{1\\1\\1\\1\\1}, \quad
e_2 = \pmatrix{0\\1\\0\\0\\0}, \quad
w = \pmatrix{w_1\\w_2\\w_3\\w_4\\w_5}.
$$
We are given that $Mw = e_2$. The first entry of $Mw$ is equal to the product $e^Tw$, so from $Mw = e_2$ we see that it must hold that $e^Tw = 0$.

The new equation that you present can be written as
$$
(M + a ee^T)w = e_2.
$$
We can see that this equation must hold because
$$
(M + aee^T)w = Mw + aee^Tw = Mw + ae(e^Tw) = Mw + 0 = Mw = e_2.
$$