问:
Cyclotomic units modulo $p$-th powers in the cyclotomic tower
Let $E$ be the unit group of the cyclotomic field $L_n=\mathbb{Q}(\zeta_{p^n})$ where $\zeta_{p^n}$ is a primitive $p^n$-th root of unity. Let $G=Gal(L_n/\mathbb{Q})$ and let $\Delta=Gal(L_1/\mathbb{Q})$. If we let $\omega: (\mathbb{Z}/p\mathbb{Z})^{\times} \to \mu_{p-1}\in \mathbb{Z}_p$ be the Teichmuller character (which we may also regard as a character to $\mathbb{Z}/p^n\mathbb{Z}$ by reducing modulo $p^n$), then we may embed $\Delta$ into $G$ by sending $\sigma_a \mapsto \sigma_{\omega(a)}$. In this way, $E$ is a $\Delta$-module and $E/E^p$ is a $\mathbb{Z}/p\mathbb{Z}[\Delta]$-module. Define the usual idempotents for $i=0,\dots,p-2$
\begin{align*}
\theta_i=-\sum_{a=1}^{p-1}a^i\sigma_{\omega(a)}^{-1}
\end{align*}
and decompose $E/E^p$ as a direct sum of its eigenspaces in the usual way. Now if we further suppose that $p$ is regular, then $p$ does not divide the index of the cyclotomic units in the full unit group so we may take cyclotomic units as generators for each of the subspaces. It is shown in Washington's book that when $n=1$, $\theta_i(E/E^p)=0$ for odd $i \neq 1$, $\theta_1(E/E^p)=\langle \zeta_p \rangle$, $\theta_0(E/E^p)=0$ and $\theta_i(E/E^p)\cong \mathbb{Z}/p\mathbb{Z}$ for even $i\neq 0$. In addition, if $g$  is a primitive root modulo $p$, we have that
\begin{align*}
E_i=\prod_{a=1}^{p-1}\Big(\zeta^{(1-g)/2}\frac{1-\zeta^g}{1-\zeta}\Big)^{a^i\sigma_a^{-1}}
\end{align*}
are generators for $\theta_i(E/E^p)$ and even $i$.
When $n>1$, things are not so simple.
The odd components are the same since every unit in $L_n$ is a product of a root of unity and a real unit and the odd components of real units vanish.
What I am having trouble with is determining the even components. For the $n=1$ case, there was a simple, perhaps naive, guess that turned out to be true: If $E_{+}$ is the group of real units, then $E_{+}/E_{+}^p\cong (\mathbb{Z}/p\mathbb{Z})^{(p-3)/2}$ by the Dirichlet unit theorem, the $0$-th component vanishes since $\theta_0$ is just the inverse of the norm, so the obvious guess is that the remaining $(p-3)/2$ even eigenspaces  are all cyclic of order $p$. For $n>1$, the $0$-th component doesn't necessarily vanish since it isn't a norm to $\mathbb{Q}$ but a norm to the subfield of degree $p^{n-1}$ over $\mathbb{Q}$. Also, $E_{+}/E_{+}^p \cong (\mathbb{Z}/p\mathbb{Z})^{p^{n-1}(p-1)/2-1}$ so unless I'm completely overlooking something it doesn't seem obvious how the sizes of the components should be distributed.

What I do have is a set of generators for $E_{+}/E_{+}^p$: If $g$ is a primitive root modulo $p^n$ then
\begin{align*}
\zeta_{p^n}^{(1-g)/2}\frac{1-\zeta_{p^n}^g}{1-\zeta_{p^n}}
\end{align*}
generates the  real cyclotomic units modulo $\{\pm 1\}$ over $\mathbb{Z}[G]$ so hitting all Galois conjugates of these elements with the idempotents $\theta_i$ will give us a set of generators for the components but it will be redundant.

Has anyone come across an analysis of this decomposition or something relevant?

答:
Perhaps you are not aware of the deep level of your question. A partial answer is known, in close relation with the so-called Iwasawa Main Conjecture (now a theorem of Mazur-Wiles (1984) when the base field is $\mathbf Q$, and Wiles (1990) when it is a totally real number field). Let us stick here to your hypotheses and notations but, for simplification, assume $p$ odd and consider only totally real fields and even parts (so + disappears from the indices). Given an abelian group $B$ of finite type, $B/pB$ is naturally an $\mathbf F_p$-vector space of finite dimension.

Recall that the subgroup $C_n$ of cyclotomic units inside $E_n$ has finite index equal to the class number $h_n$. Applying the snake lemma to the exact sequence $1 \to C_n \to E_n \to E_n/C_n \to 1$, we derive (with obvious notations) an exact sequence of vector spaces $ 1\to (E_n/C_n)[p] \to C_n/C_n^p \to E_n/E_n^p \to (E_n/C_n)/p \to 1$, so that $dim E_n/E_n^p= dim C_n/C_n^p$, and the problem is brought back to the study of $dim C_n/C_n^p$. This remains of course valid when working characterwise, i.e. applying idempotents of $\Delta$. This approach presents at the same time an advantage ($C_n$ is more or less "explicitly" known) and a drawback (the indirect intervention of the $p$-class-groups $A_n$, which  are the main arithmetic mystery). Note that the Vandiver conjecture is known to "propagate" along the cyclotomic tower, hence the  class-group problem vanishes in that case.

Without Vandiver, the point of view of Iwasawa's theory is to study the $A_n$ not one by one, but all together, more precisely to study $structurally$ the projective system $Y_{\infty}:=\varprojlim A_n$ w.r.t. norms. Actually $Y_{\infty}$ is endowed with a canonical structure of noetherian torsion module over the algebra $\Lambda:=\varprojlim \mathbf Z_p[\Gamma_n]$, where $\Gamma_n=Gal(L_{\infty}/\mathbf Q_n)$. The main technical point is that the seemingly complicated algebra $\Lambda$ is isomorphic to the complete power series algebra $\mathbf Z_p [[T]]$, which allows to attach to each noetherian torsion $\Lambda$-module $Z$ a characterisic power series $char_{\Lambda} (Z)$ describing the $\Gamma$-action on $Z$ analogously to the characteristic polynomial attached to a linear map acting on a vector space over a field. To make a lengthy story short, Iwasa introduced an important arithmetic $\Lambda$-module $X_{\infty}$ defined as the Galois group over $L_{\infty}$ of the maximal abelian pro-$p$-extension of $L_{\infty}$ which is unramified outside $p$, and he showed in a celebrated paper (1964) that his MC is equivalent to $char_{\Lambda}(X_{\infty})= char_{\Lambda} (E^1_{\infty}/ C^1_{\infty})$. For simplicity, I don't explain the upper indices $(.)^1$; suffice it to say that they stand for minor modifications of $E_{\infty}$ and $C_{\infty}$ (see the notations at the beginning of [CS]). This was the crucial step just before the final Mazur-Wiles proof of the MC, which expresses $char_{\Lambda}(X_{\infty})$ in terms of the $p$-adic zêta function (see thm. 1.4.3 of [CS]).

To get back to our original problem, let us recall that in the course of his theorem, Iwasawa showed that $C^1_{\infty}$ is $\Lambda $-cyclic, generated by the proj. lim. of an "explicit" system b of cyclotomic units (lemma 4.3 of [CS]). To "go down" to the $n$-th level, we have just to take the $\Gamma_n$ co-invariants of $C^1_{\infty}$ and b. Applying idempotents, we can get in priciple the semi-simple decomposition along $\Delta$ of $C_n/C_n^p$. "In principle" only, because the description of the co-invariants of b, although "explicit", is still not accurate enough without Vandiver. Actually, by the so-called "reflection theorems", the situation is analogous to the semi-simple decomposition of the $\mathbf F_p$-vector spaces $A_n/{A_n}^p$ .

Ref. Instead of Washington's book, in the particular case of the base field  $\mathbf Q$, I prefer the monograph [CS] by J. Coates and R. Sujatha, "Cyclotomic Fields and Zeta Values", Springer Mono. in Math., where the simplifications due to the base field allow to delineate much more clearly the flow of arguments.


问:
Proof by strong Induction for a recurrence
We have a sequence of $a_n = a_{n-3}+a_{n-2}+a_{n-1} \text{ where } a_{1} = a_{2} =a_{3} = 1$ for all $n\ge 4$. Prove $a_n < 2^n$ is true for all positive integers $n$.

What I have done so far is starting with the inductive hypothesis because this is strong induction. So far my inductive hypothesis is: Consider $P(k)$ is true for all $a_k < 2^k$ where $k$ is a positive integer. Now show that $a_{k+1} < 2^{k+1}.$

This is what I have done so far: $$a_{k+1} < 2^{k+1} \implies a_{k+1} < 2^k \cdot 2 \implies 2 \cdot \underbrace{a_k < 2^k}_{IH} \cdot 2 \\ \text{So we see: } a_k < 2^k < 2^{k+1}$$

Now I am stuck because I don't know how to show $a_{k+1} < a_k$.

If my way of solving this looks wrong please let me know.

答:
We show that $P_n : a_n < 2^n$ is true.

for $n=1$, it's ok.

for $n >1$, suppose $P_1,...,P_{n-1}$. We see that
$$
a_n = a_{n-3}+a_{n-2}+a_{n-1} \le 2^{n-3}+2^{n-2} + 2^{n-1}=2^n(\frac{1}{2^3}+\frac{1}{2^2}+\frac{1}{2})<2^n
$$
So $P_n$ is true.

Here are some comments about your proof:

The way you stated your inductive hypothesis is odd (to me). Also here we know that $a_n$ is defined by induction so using strong induction is a good idea. It seems to me that you got kind of lost when applying the hypothesis. You started your proof from the result not the hypothesis...

ContentLicense: CC BY-SA 4.0


问:
System of recurrence relations with different even and odd relations
I know the following:

if n is even, then: $A_{n+1}=A_n+B_n$ and $B_{n+1}=B_n$

if n is odd, then it is the reverse: $A_{n+1}=A_n$ and $B_{n+1}=A_n+B_n$

I am given the initial conditions that $A_1=A_2=B_1=1$ and $B_2=2$

I've solved recurrence relations before, but this has two going at once with each dependent on the other so I'm not sure what to do. I can code it out well enough, but I'm trying to get a closed form instead of a recurrence. I would appreciate a solution if anyone could demonstrate how to tackle this.

答:
To start, let's write the recurrence relations in matrix form.

if n is odd:
$$
\begin{bmatrix}
A_{n + 1}\\
B_{n + 1} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
A_n \\
B_n \\
\end{bmatrix}
$$

if n is even:
$$
\begin{bmatrix}
A_{n + 1}\\
B_{n + 1} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 \\
0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
A_n \\
B_n \\
\end{bmatrix}
$$
Claim: $A_n = F_{n + 1}, B_n = F_{n} : \forall n \in Odd $ and $A_n = F_{n}, B_n = F_{n + 1}: \forall n \in Even$. where F is Fibonacci sequence.

proof: Induction.

step of induction:

if n is Odd, by applying induction hypothesis in the first equation we have:
$$
\begin{bmatrix}
A_{n + 1}\\
B_{n + 1} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
F_{n + 1} \\
F_{n} \\
\end{bmatrix}
=
\begin{bmatrix}
F_{n + 1} \\
F_{n + 2} \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
A_n \\
B_n \\
\end{bmatrix}
=
\begin{bmatrix}
F_n \\
F_{n + 1} \\
\end{bmatrix}
: \forall n \in Even
$$

if n is even, by applying induction hypothesis in the second equation we have:
$$
\begin{bmatrix}
A_{n + 1}\\
B_{n + 1} \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 \\
0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
F_{n} \\
F_{n + 1} \\
\end{bmatrix}
=
\begin{bmatrix}
F_{n + 2} \\
F_{n + 1} \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
A_n \\
B_n \\
\end{bmatrix}
=
\begin{bmatrix}
F_{n + 1} \\
F_n \\
\end{bmatrix}
: \forall n \in Odd
$$
Base of induction:
$$ n = 1 \rightarrow
\begin{bmatrix}
A_1 \\
B_1 \\
\end{bmatrix}
= \begin{bmatrix}
F_2 \\
F_1 \\
\end{bmatrix}
= \begin{bmatrix}
1 \\
1 \\
\end{bmatrix}
$$
$$ n = 2 \rightarrow
\begin{bmatrix}
A_2 \\
B_2 \\
\end{bmatrix}
= \begin{bmatrix}
F_2 \\
F_3 \\
\end{bmatrix}
= \begin{bmatrix}
1 \\
2 \\
\end{bmatrix}
$$

always make good use of matrices in Fibonacci questions.


Title: Proof by strong Induction for a recurrence

Body:
We have a sequence of $a_n = a_{n-3}+a_{n-2}+a_{n-1} \text{ where } a_{1} = a_{2} =a_{3} = 1$ for all $n\ge 4$. Prove $a_n < 2^n$ is true for all positive integers $n$.

What I have done so far is starting with the inductive hypothesis because this is strong induction. So far my inductive hypothesis is: Consider $P(k)$ is true for all $a_k < 2^k$ where $k$ is a positive integer. Now show that $a_{k+1} < 2^{k+1}.$

This is what I have done so far: $$a_{k+1} < 2^{k+1} \implies a_{k+1} < 2^k \cdot 2 \implies 2 \cdot \underbrace{a_k < 2^k}_{IH} \cdot 2 \\ \text{So we see: } a_k < 2^k < 2^{k+1}$$

Now I am stuck because I don't know how to show $a_{k+1} < a_k$.

If my way of solving this looks wrong please let me know.

Body: Our statement is true for $n=1,2,3$. Suppose $P(k):a_k<2^k$ is true for all $k< n\ge4$. Then$$a_n=a_{n-1}+a_{n-2}+a_{n-3}<2^{n-1}+2^{n-2}+2^{n-3}=2^{n-1}(1+1/2+1/4)<2^{n-1}\cdot2$$


问:
Solving differential equation with boundary conditions $y=2$ at $x=0$
I need to find a solution to this differential equation $(1+x^2) $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = x-xy^2$, with boundary condition $y=2$ at $x=0$

I collect all terms on the RHS to get $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{1+x^2}  (1-y^2)$$

I integrate then move the y terms to the left side and integrate both sides and simplify to get $$\frac{1}{\sqrt{1-y^2}} = k\sqrt{1+x^2}$$

I then isolate y in order to get it in a y= form and sub in the y and x value but I don't get the correct answer; I'm not sure where I'm going wrong

答:
$\frac{dy}{1-y^2}=\frac{xdx}{1+x^2}$

Integrate both sides

$\frac{1}{2}\log_e\lvert\frac{1+y}{1-y}\rvert=\frac{1}{2}\log_e (1+x^2) +c$

Can you proceed from here?


问:
Conditional expectation of uniform given normal
Let $X \sim \mathcal{N}(\mu, \sigma^2)$, $Y \sim U(-\alpha, \alpha)$ and $Z = X - Y$.
Assume $Y$ and $Z$ are independent. Note that $X$ and $Y$ are necessarily dependent.

What is $\mathbb{E}[Z|X]$?

What I've tried

$$
\mathbb{E}[Z|X](x) = \frac{1}{p_X(x)} \int z \, p_Y(x-z)p_Z(z) \mathrm{d}z
$$

I don't know what $p_Z(z)$ is. I've tried getting it from the characteristic functions
$$ p_Z(z) = \frac{1}{2\pi}\int e^{-itz} \Phi_X(t)/\Phi_Y(t) \mathrm{d} t $$
but the integral looks intimidating.

It would be nicer to beat the integral for $\mathbb{E}[Z|X]$ out of $\Phi_Z(t)$ without finding a closed form for $p_Z(z)$...

答:
The setup is not possible

We have
$$X = Y+Z$$

hence

$$\Phi_X(t) = \Phi_Y(t) \Phi_Z(t) $$

Because the first factor is a $sinc$ function, the product cannot be a gaussian.


问:
Prove the following integral inequality: $\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$

Let $f$ be a twice continuously differentiable function from $[0,1]$ into $\mathbb R$. Given that
  $$f(0)+2f(\frac{1}{2})+f(1)=0,$$
  show that
  $$\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2.$$

I tried some methods, such as Cauchy-Schwarz inequality
$$\int_{0}^{1}(f''(x))^2dx\cdot\int_{0}^{1}g^2(x)dx\ge\left(\int_{0}^{1}f''(x)g(x)dx\right)^2,$$
where $g(x)$ is polynomial function and $\max{\deg{(g(x))}}\le 2$.

And by parts integral we have
$$\int_{0}^{1}f''(x)g(x)dx=f'(1)g(1)-f'(0)g(0)-f(1)g'(1)+f(0)g'(0)+\int_{0}^{1}f(x)g''(x)dx=
f'(1)g(1)-f'(0)g(0)-f(1)g'(1)+f(0)g'(0)+C\int_{0}^{1}f(x)dx.$$

Body:
Let $g(x)$ be the piecewise differentiable function defined as,
$$g(x)=\begin{cases}
g_1(x),\quad x\in\left[0,\frac{1}{2}\right)\\
g_2(x),\quad x\in\left[\frac{1}{2},1\right]
\end{cases}$$
By integration by parts, we find
\begin{align*}
\int_0^{\frac{1}{2}}g_1(x)f''(x)\mathrm{d}x
&=g_1(x)f'(x)\bigg|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}g_1'(x)f'(x)\mathrm{d}x\\
&=g_1(x)f'(x)\bigg|_0^{\frac{1}{2}}-g_1'(x)f(x)\bigg|_0^{\frac{1}{2}}+\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x
\end{align*}
and
\begin{align*}
\int_{\frac{1}{2}}^1g_2(x)f''(x)\mathrm{d}x
&=g_2(x)f'(x)\bigg|_{\frac{1}{2}}^1-\int_{\frac{1}{2}}^1g_2'(x)f'(x)\mathrm{d}x\\
&=g_2(x)f'(x)\bigg|_{\frac{1}{2}}^1-g_2'(x)f(x)\bigg|_{\frac{1}{2}}^1+\int_{\frac{1}{2}}^1g_2''(x)f(x)\mathrm{d}x
\end{align*}
First, we need
$g_1''(x)=g_2''(x)=\text{constant}$, so, $g_1(x)$ and $g_2(x)$ are quadratic polynomials
$$\begin{cases}
g_1(x)=cx^2+a_1x+a_0,\quad x\in[0,\frac{1}{2})\\
g_2(x)=cx^2+b_1x+b_0,\quad x\in[\frac{1}{2},1]
\end{cases}$$
Next, Let $$g_1(x)f'(x)\bigg|_0^{\frac{1}{2}}+g_2(x)f'(x)\bigg|_{\frac{1}{2}}^1=0$$
we get,
$$g(1)=g(0)=0\Rightarrow a_0=0,c+b_1+b_0=0 \tag{1}$$
and
$$g_1\left(\frac{1}{2}\right)=g_2\left(\frac{1}{2}\right)\Rightarrow \frac{1}{2}a_1=\frac{1}{2}b_1+b_0\tag{2}$$
Next, Let
$$g_2'(x)f(x)\bigg|_{\frac{1}{2}}^1+g_1'(x)f(x)\bigg|_0^{\frac{1}{2}}=g'(1)f(1)+\left(g_1'\left(\frac{1}{2}\right)-g_2'\left(\frac{1}{2}\right)\right)f\left(\frac{1}{2}\right)-g'(0)f(0)=0$$
Since, $f(0)+2f(\frac{1}{2})+f(1)=0$, we get
$$g_1'\left(\frac{1}{2}\right)-g_2'\left(\frac{1}{2}\right)=-2g'(0)=2g'(1)$$
so
$$a_1-b_1=-2a_1=4c+2b_1 \tag{3}$$
from $(1),(2),(3)$, we get
$$a_0=0,\quad a_1=-\frac{1}{2}c,\quad b_1=-\frac{3}{2}c,\quad b_0=\frac{1}{2}c$$
Therefore,
$$\int_0^1g(x)f''(x)\mathrm{d}x
=\int_0^1g''(x)f(x)\mathrm{d}x=2c\int_0^1f(x)\mathrm{d}x$$
by Cauchy-Schwarz inequality, we get
$$\left(2c\int_0^1f(x)\mathrm{d}x\right)^2\leqslant \int_0^1(g(x))^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x$$
Among,
$$\int_0^1(g(x))^2\mathrm{d}x=\int_0^{\frac{1}{2}}(g(x))^2\mathrm{d}x+\int_{\frac{1}{2}}^1(g(x))^2\mathrm{d}x=\frac{c^2}{480}$$
Finally, we get
$$\int_{0}^{1}(f''(x))^2\mathrm{d}x\geqslant 1920\left(\int_{0}^{1}f(x)\mathrm{d}x\right)^2. \tag*{$\Box$}$$


问:
Examples on why isomorphism of categories are "too restrictive"
While I'm reading the Algebra: Chapter 0 by Paolo Aluffi, I've encountered the following paragraph where he discuss that the isomorphism on categories is too restrictive:

When are two categories ‘equivalent’?
Essentially every notion of ‘isomorphism’ encountered so far has boiled down to ‘structure-preserving bijection’,
and the reader may well expect that the natural notion identifying two categories
should be drawn from the same model: a functor matching objects of one category
precisely with objects of another, preserving the structure of morphisms.

One could certainly introduce such a notion, but it would be exceedingly restrictive. Recall that solutions to universal problems, that is, just about every construction we have run across, are only defined up to isomorphism. Requiring solutions in one context to match exactly with solutions in another would be problematic. The structure of an object in a category is adequately carried by its isomorphism class”, and a natural notion of ‘equivalence’
of categories should aim at matching isomorphism classes, rather than individual
objects. The morphisms are a more essential piece of information; the quality of a
functor is first of all measured on how it acts on morphisms.

My problem lies in the part

Requiring solutions in one context to match exactly with solutions in another would be problematic.

while I can understand that the notion of functor isomorphism seems too restrictive on categories, I do want to further convince myself by an actual example. Is there an example that can justify the sentence "requiring solutions in one context to match exactly with solutions in another would be problematic"?

答: Consider the category of sets and the operation of the cartesian product of sets. It is rather straightforward to prove that the cartesian product operation is associative up to canonical isomorphisms. This is simply a matter of moving parenthesis in the syntactic description of the cartesian products as tuples but also follows from the universal property of products. Now, you define the notion of cardinality of a set as follows. Let $C$ be the category of sets and bijections and let $Card$ be a skeleton of it. There is an inclusion $Card \to C$ which is an equivalence but not an isomorphism. Now we wish to define, say, the product of cardinalities. We do so by $Card \times Card \to C\times C \to C \to Card$ using the injection followed by the cartesian product followed by the quasi-inverse of the inclusion. Now we phrase the problem: is the product of cardinalities associative? To answer that we translate the problem back to $C$ where we do not get an equality between the sets representing the problem but we do get an isomorphism. That isomorphism tells us that back in $Card$ we do have an equality. So the solutions in $C$ do not match exactly with the solutions in $Card$. But the equivalence is sufficient.


问:
Does sample mean converge in distribution to population mean?
$X_i$ is i.i.d. random variable.
$\overline{X}=\frac{1}{n}\sum X_i$

Can we say that
$\overline{X} \stackrel{d}{\longrightarrow } E(X)?$

答: As Stefan says, it does follow that $\bar{X} \to E(X)$ in distribution since $\bar{X} \to E(X)$ almost surely by the SLLN and convergence a.s. implies convergence in distribution.  The difference between this and the CLT is that the CLT says $$\sqrt{n} (\bar{X} - E(X)) \to N(0, \sigma^2)$$ in distribution, so we are scaling by $\sqrt{n}$ at each step. While it's tempting to conclude from this that $\bar{X} \to N( E(X), \sigma^2 / n)$ in distribution, it's not valid to move the $\sqrt{n}$ term to the right side because it is not constant with respect to $n$.  So the CLT does not contradict the fact that $\bar{X} \to E(X)$ in distribution.

ContentLicense: CC BY-SA 3.0


问:
If $R$ has zero divisors, does $R[x]$ have invertible polynomials of degree greater than $0$?
In my book, it is claimed without proof that if a ring (commutative with identity) $R$ has zero divisors, then $R[x]$ has invertible polynomials of degree greater than $0$. They give the following example in $\mathbb{Z}_4$: $$(1+2x)(1+2x)=1$$

This seems like a pretty lucky example: in this ring we have a zero divisor whose sum with itself is $0$. In general this doesn't seem like an easy question to tackle. First i thought that maybe could find some $a$ such that $a^2 = 0$ and use the factorization $(1+ax)(1-ax)$ but in general such $a$ does not exist: $\mathbb{Z}_6$ is a counterexample. No polynomial of form $(1+ax)$ has a degree $1$ inverse in $\mathbb{Z}_6[x]$ either. Trying out higher degrees seems like a nightmare.

答: This is false. A polynomial is a unit if it has the form $\sum_{i=0}^k a_i x^i,$ where $a_0$ is a unit and $a_i$ is nilpotent for $i>0.$ (see this problem set). In the ring $R=\mathbb{Z}_6$ there are no nilpotents, so there are no invertible polynomials of degree greater than $0.$


问:
Show that the function $f(x)=2\lfloor x\rfloor-x$ defined from $f:\mathbb{R}\to\mathbb{R}$ is onto
$f:\mathbb{R}\to\mathbb{R}$ is defined by $f(x)=2\lfloor x\rfloor-x$ where $\lfloor \cdot \rfloor$ is greatest integer function.

How to show that it is an onto function without the help of graph?

It is given it is invertible.

And what would be $f^{-1}$?

答:
If the fractional part of $x$ is denoted by $\{x\},$
then $f^{-1}(x) = \lfloor x \rfloor + (1 - \{x\}).$


问:
the difference between big and small asymptotic notation
I saw this example and i am a little confused.

$10n^2-5n+6046$=$ω(n^2)$

isn't that a mistake and it is necessary to use $Ω$?

because $n^2$=$n^2$

答:
There is indeed a mistake in this example. $$\lim_{n\to\infty}\frac{10n^2-5n+6046}{n^2}=10\ne \infty$$
So $10n^2-5n+6046\ne \omega(n^2)$. However we can say that $10n^2-5n+6046=\Omega(n^2)$.


问:
how do determine the distribution of outcomes for a given probability?
For a game I generate various block types given certain odds. Say, there's a $0.001$ chance for the karma block. If a typical game has $600$ blocks, what's the distribution of games that have $0$ karma blocks, $1$ karma blocks, $2$ karma blocks, etc.

I've tried reading the wikipedia page on probability distributions, but I can't figure out how it applies to this.

答:
You want the binomial distribution because the probability of each block being karma or not-karma (whatever that means), that is, the probability of a success or a failure in each trial is constant and independent of the other trials.

In general, the probability that there are k successes out of n trials in a binomial distribution is ${n\choose k} p^k(1-p)^{n-k}$ (where p is the probability of a success). In your case the probability of 'k' "karma blocks" would be ${600 \choose k} (0.001)^k(0.999)^{600-k}$.

What's actually happening is you determine there must be exactly $k$ karma-blocks, and that probability is $0.001^k$. The rest of the 600 have to be failures, hence $0.999^{600-k}$ (because if it's not a success, it's a failure, 1-0.001=0.999). And you don't care about order so you multiply by the number of different orders there could be for k karma blocks in 600 blocks, which is $600 \choose k$


问:
in the diagram, $O$ is the origin of $OC$ and $OD$ find $CD$
While doing some past tests for the IGCSE I encountered this question and I couldn't figure out what's the answer to this question.

I found out using the mark sheet that the answer is:
$CD = OC + OD = 6a - 2b$, However I couldn't figure out why.

Furthermore I know a triangle can't have a side that's equal to the other two side combined, and even more

$(-2a + 3b) + (4a + b) = 2a + 4b. $

But

$(4a + b) - (-2a + 3b) = 6a - 2b$

I also couldn't understand what the other question wanted.

答:
Welcome to MSE

Case 1:If the tail of second vector $\vec {CD}=r_2=(x_2 i+y_2 j)$ is on the head of the first one $\vec{OC}=r_1=(x_1 i+y_1 j)$ , then $\vec{OC}+\vec{CD}=(x_1+x_2) i +(y_1+y_2) j$

Case 2:If the tail of two vectors are coincident, like OC and OD, then the sum is:

$OC+OD=(x_2-x_1)i+(y_1-y_2)j$ , so we have:

$OC+OD=[4-(-2)]a+ (1-3)b=6a-2b$

For second question the position vector of E is  like case 1 and we can write:

$r=OE=OD+DE=(4+1) a +(1-2)b=5 a -b$ or $\vec r=5i-1j$


问:
Strong duality by Slater's condition
In Section 5.3.2 of Boyd, Vandenberghe: Convex Optimization, strong duality is proved under the assumption that ker(A^T)={0} for the linear map describing the equality constraint, though it is remarked that this condition is not necessary. Where can I find a proof of the statement without the assumption on the kernel?</p>

答:
I think that the proof in the book can be very easily modified to work for the rank-deficient case, too, by simply using a better version of the supporting hyperplane theorem: Indeed, the proof for that in the book (and in many other places) gives a trivial supporting hyperplane when the interior of the convex set is empty, namely, one with a normal vector orthogonal to the affine subspace generated by the convex set. However, there also exist non-trivial supporting hyperplanes; one simply has to use the supporting hyperplane theorem in a subspace, and then extend the linear functional to the whole space. I think if we use such a non-trivial supporting hyperplane from the beginning in the proof of Slater's theorem, then $\langle\nu,Ax-b\rangle\ge 0$ for all $x$ in the domain of the problem implies that
$\nu$ has to be perpendicular to $\{Ax-b:\,x\in\mathcal{D}\}$, but then, by the preceding arguments, $(\lambda,\nu,\mu)=(0,\nu,0)$ is perpendicular to $\mathcal{A}$, which contradicts our initial choice of a non-trivial supporting hyperplane.

Title: Strong duality by Slater's condition


问:
Help locating mini mandelbrots
I would like to be able to list the coordinates of all the first level minibrots.

Here is a picture of the mandelbrot set generated by fraqtive:

zooming in to the circled area we see a slightly distorted copy of the original fractal

the julia set associated to the point right in the middle of the main cardoid looks like this

roughly the coordinates are -0.1564 + i(-1.0320)

The idea to find minibrots is based on this:

The julia set associated to the center of the main cardioid of the mandelbrot set is a circle, so I assume the julia sets associated to the main cardioid of a minibrot will be a julia set built of circles (like pictured): This needs proof but it seems like it would be too difficult so I am happy to leave this unproved for now (but I would like to prove it eventually).

if I can just find a way to recognize which julia sets are built out of circles I should be able to list the minibrots. How can I do that?

a second question which might be more easy is

I have approximate coordinates for this circular julia set, is there any way to use the z^2+c formula to find that point exactly?

答:
This is an answer to the question in your very first sentence.

See for instance this paper:

McMullen, Curtis T.,
  The Mandelbrot set is universal. In The Mandelbrot set, theme and variations, 1–17,
  London Math. Soc. Lecture Note Ser., 274, Cambridge Univ. Press, 2000.
  MR1765082 (2002f:37081)

PDF available at the author's site.

See also The significance of the Mandelbrot set.

Title: Help locating mini mandelbrots


问:
Count Number of Slots for Balls with Repetition
I have $3$ slots and $3$ balls. Each ball needs to go into a slot, but each slot does not need to be used. By writing out all of the possibilities, I know there are $27$ possible combinations.

For example,

$$000$$

$$001$$

$$002$$

$$010$$

$$011$$

$$012$$

$$...$$

However, is there some sort of formula that can be used to apply this to larger problems where we can't write all of them out? Just straight permutation and combination doesn't seem to give the correct number, nor does factorial. All of these are assuming you want to use each slot, but in my case I can leave $n-1$ slots empty.

答:
Imagine the balls and urns are numbered. The first ball must go into one of the three urns, hence we have $3$ alternatives. For each one of those alternatives, we have $3$ new alternatives for the second ball, etc. Hence we have $3 \times 3 \times 3 = 3^3$ ways. In general, if we have $B$ balls and $X$ urns we have $X^B$ ways.

More in general, see the rule of product


问:
Ellipse: product of the distance from foci to a tangent is a constant
I am supposed to determine what is the result of said product. Given $P(x_0,y_0)$, I need to calculate the distance from the foci of an ellipse to the tangent line that passes through $P$, and then multiply the distances.

In essence it is quite simply. We take:
$$
\frac{x_0}{a^2}x + \frac{y_0}{b^2}y = 1
$$
as the tangent line. Then we simply calculate its distance to each focus $(c,0)$ and $(-c,0)$, using the formula and then, multiplying.

$$
d=\frac{\frac{x_0c}{a^2}±1}{\sqrt{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}}
$$

$$
\text{Some constant k}=\frac{\frac{x_0^2c^2}{a^4}-1}{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$

I'm having trouble getting things cancelled here. The constant k is $b^2$, but I can't get to it. Help?

Body:
$$
\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac{y_0^2}{b^4}}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac1{b^2}\left(1-\frac{x_0^2}{a^2}\right)}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac1{b^2}\left(1-(a^2-b^2)\frac{x_0^2}{a^4}\right)}
=b^2,
$$
where in the final step we used the identity $a^2-b^2=c^2$.


问:
Domain, codomain, range and image of a function
I am confused by the concept of domain, range, co-domain and image set of a function.

For example, I have $f:[0,5] \ \to [\frac{1}{5}, \infty]$, defined by $f(x) = \frac{1}{x}$.

I would like to ask the questions:

What is the domain of $f$ ? Is it $[0,5]$ or $(0,5]$ ?

Is the range, codomain and image of $f$$: [\frac{1}{5}, \infty]$ or $\mathbb{R}$ ?

What is the condition for 2 functions $f$ and $g$ to be equal ?

Thank you very much!

P/s: This question on mathexchange relates but not fully satisfies my concern

Domain, Codomain, Range, Image and Preimage

答:
Domain D is set of all real values of $x$ for which the function takes real,finite and unique value. Range R is all values taken by the function over all the $x$ values of the domain. A set larger than the the range is co-domain C. Infinity is never included in D and R. So in your example $$D=(0,5], ~~R=[1/5,\infty),~~ C= \Re (Real)$$

Image is $f(a)$, the value of function at $x=a$ when $a \in D$. Set  of all images is nothing but the range R. $x=a$ can also be called pre-image of $f(a)$.


问:
Finding the vertex of a quadratic function
I'm in the process of relearning math and came across a problem when calculating the vertex for $5x^2-20x+15$ function. I took the following steps to find the vertex:

I divided every term by 5 to get a simplified equivalent function $x^2-4x+3$

Put it in the square form $(x^2 -4x +4) -4 +3$ => $(x-2)^2 -1$

From the $(x-2)^2 -1$ form I would say the vertex coordinates are [2;-1], but I'm wrong. Having looked at wolframalpha it looks likes dividing every term by 5 is a problem, but I don't understand why. Shouldn't it be the same thing? Sorry for the trivial question.

答:
Let's say you have a quadratic function $f(x) = ax^2 + bx + c$. Then, you create a new quadratic function by multiplying by a factor of $\lambda$: $g(x) = \lambda f(x) = \lambda ax^2 + \lambda bx + \lambda c$.

Putting $f$ into vertex form, we have:

$$ y = ax^2 + bx + c $$
$$ y = a \left( x^2 + \frac{b}{a}x \right) + c $$
$$ y = a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right) - \frac{b^2}{4a} + c $$
$$ y = a \left( x - \left(-\frac{b}{2a}\right) \right)^2 - \frac{b^2}{4a} + c $$
$$ y - \left( c - \frac{b^2}{4a} \right) = a \left( x - \left(-\frac{b}{2a}\right) \right)^2 $$

Comparing this to vertex form, $y-k = a(h-k)^2$, we have the following coordinates for the vertex of $f$:

$$ (h_f,k_f) = \left( -\frac{b}{2a} , c - \frac{b^2}{4a} \right) . $$

Now, let's consider what happens to the vertex if we scale our quadratic $f$ by a factor of $\lambda$:

$$ g(x) = \lambda f(x) = \lambda ax^2 + \lambda bx + \lambda c . $$

Following the same process of completing the square, we have:

$$ y = \lambda ax^2 + \lambda bx + \lambda c $$
$$ y = \lambda a \left( x^2 + \frac{b}{a}x \right) + \lambda c $$
$$ y = \lambda a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right) - \frac{\lambda b^2}{4a} + \lambda c $$
$$ y = \lambda a \left( x - \left(-\frac{b}{2a}\right) \right)^2 - \frac{\lambda b^2}{4a} + \lambda c $$
$$ y - \lambda \left( c - \frac{b^2}{4a} \right) = \lambda a \left( x - \left(-\frac{b}{2a}\right) \right)^2 $$
$$ y - \lambda h_f = \lambda a \left( x - k_f \right)^2 $$

Notice, the vertex of $g$ is at coordinates

$$ (h_g, k_g) = (\lambda h_f, k_f) . $$

So, scaling our function by a factor of $\lambda$ leaves the $x$-coordinate of the vertex unchanged, but it does scale the $y$-coordinate of the vertex by that same factor.

In this case, that scaling factor was $\lambda = \frac{1}{5}$, so the $x$-coordinate was the same for both:

$$ h_f = h_g = 2 . $$

But, the $y$-coordinate of the scaled function was $\frac{1}{5}$ the $y$-coordinate of the original function:

$$ h_g = -1 = \frac{1}{5} * -5 = \frac{1}{5} * h_f . $$


问:
A detail in the proof that left invariant vector fields are smooth
I am trying to understand a proof of the following proposition:

Let $X$ be a left invariant vector field on the Lie Group $G$. Then, $X$ is smooth.

The proof goes as follows:

"Let $X$ be left invariant. It suffices to prove that$^1$
$$
             Xf: G\rightarrow \mathbb{R}
                   \\ x\mapsto f_*(x)(X(x))
$$
is smooth for any function $f$.

$$
(Xf)(x) = f_*(x)(X(x)) \\
        = f_*(x)(L_{x,*}(e)(X(e)))\\
        = (f\circ L_x)_*(e)(X(e)))
$$
which is clearly smooth."

Question:

$^1$: I thought that the definition of $Xf$ was such that $Xf(x):= X_xf$. And this is not the same as $f_*(x)(X(x))$.

答: I think the problem here is the canonical isomorphism $\mathbb{R}\simeq T_p\mathbb{R}$ for all $p\in\mathbb{R}$. Notice that if $f:G\to\mathbb{R}$, then $f_*|_x : T_x G\to T_{f(x)}\mathbb{R}\simeq \mathbb{R}$, and that is why you can identify $f_*|_x(X|_x)\in T_{f(x)}\mathbb{R}$ with $X|_x f\in \mathbb{R}$.


答:
Let $g(y)=2\lceil y\rceil-y,$ where $\lceil\cdot\rceil$ is the ceiling function.

Then, since $0\leq \lceil y\rceil -y<1$ we have:

$$g(y)=\lceil y\rceil +(\lceil y\rceil-y)$$

So
$$\lfloor g(y)\rfloor =\lceil y\rceil$$

So $$\begin{align}f(g(y))&=2\lfloor g(y)\rfloor -g(y)\\
&=2\lceil y\rceil -(2\lceil y\rceil-y)\\
&=y
\end{align}$$

A similar argument gives $g(f(x))=x.$

This can be seen as related to two bijections $\mathbb Z\times[0,1)\to\mathbb R$:

$$q_+(n,z)=n+z, q_-(n,z)=n-z.$$

Then $$q_+^{-1}(x)=(\lfloor x\rfloor,x-\lfloor x\rfloor)\\
q_-^{-1}(y)=(\lceil y\rceil, \lceil y\rceil-y)$$

Then your $f=q_-\circ q_+^{-1}$ and thus $f^{-1}=q_+\circ q_-^{-1}.$

Title: Show that the function $f(x)=2\lfloor x\rfloor-x$ defined from $f:\mathbb{R}\to\mathbb{R}$ is onto

Body:
$f:\mathbb{R}\to\mathbb{R}$ is defined by $f(x)=2\lfloor x\rfloor-x$ where $\lfloor \cdot \rfloor$ is greatest integer function.

How to show that it is an onto function without the help of graph?

It is given it is invertible.

And what would be $f^{-1}$?


问:
Why does the Cauchy distribution model resonance?
I can see that being the quotient of two independent normal distributions it may model things that spin (although I don't understand how it can be well defined at zero).

But what is the connection to the solution of differential equations modelling resonance?

答:
Given the absence of more knowledgeable answers, I will share what I found on the topic - hopefully better or more more comprehensive answers will come along.

In short, free-induction decay (FID), which results from interactions among dipoles in MRI (magnetic resonance imaging) is influenced by inhomogeneities in the magnetic field which have long tails (not exponential or Gaussian), and can be modelled using a Cauchy distribution:

... The distribution of microscopic inhomogeneous fields is not necessarily Gaussian. In some systems of importance in areas as diverse as oilwell
logging and medical magnetic resonance imaging (MRI) the distribution of inhomogeneous fields is approximated by the Cauchy-Lorentz (C-L, Cauchy in statistics, Lorentz in physics) distribution. This is the
field due to randomly located point dipoles within a sample. Again, the phase distribution width at an effective exchange time $t_{ex}$ is proportional to $\chi \nu t_{ex}.$ However, the distribution width for $n = t/t_{ex}$ random samples is $t/t_{ex}$ times the original width, so the width of
the phase distribution at time $t$ does not depend on $t_{ex}.$ Here, the Fourier transform of the $n$-fold convolution of a C-L distribution is the $n$-fold product of its simple exponential FT, not an exponential with the width squared as in the Gaussian. That is, the FID decay does not depend on $a$ or $D.$ Furthermore, the additional FID decay rate is proportional to the width of the CL distribution of phases, not to the square of the width, as for the Gaussian.

where $\chi\nu$ (susceptibility difference times frequency), $D$ (diffusion coefficient), $a$ (a pore dimension), echo-time $t$ for single echoes and half-echo-spacing $\tau$ for CPMG (Carr-Purcell-Meiboom-Gill (CPMG) spin-echo measurements), $t_{ex}$ for exchange time.

More relevant, and in reference to a more robust correspondence between time and frequency domains in MRI than the FFT (e.g. non-parametric fast Padé transform)

Considered strictly, a general line spectrum is associated only with for purely stationary, stable bound states (e.g. ground states of atoms/molecules) of practically infinite lifetimes. On the other hand, peaks in spectra are resonances that represent non-stationary, transient states of molecules. Such metastable states last for a given lifetime, after which they decay. This is reflected in distributions of frequencies in the spectral lineshapes, instead of the stick spectra. These distributions are mainly the Lorentzian functions that are in mathematics known as the Cauchy distributions. In physics, they are also called Breit-Wigner lineshape profiles from the theory of resonant scattering of particles and photons on general targets. The Cauchy, Lorentz and Breit-Wigner distributions coincide with one of the simplest forms of the Padé approximant (PA), symbolized as [0/2], and given by:

$$\begin{aligned} Y_{\mathrm{A}} \left( \nu \right) =\frac{d_0 }{2\pi }\frac{\varGamma _0 }{\left( {\nu -\nu _0 } \right) ^{2}+\left( {\varGamma _0 /2} \right) ^{2}}\quad \hbox {(Absorptive Lorentzian)} \end{aligned}$$

It is seen that the Padé spectrum $Y_A(\nu)$ depends on $\nu$ in a continuous way. If $Y_A(\nu)$ is written in the general form $P_L(\nu)/Q_K(\nu),$ it would follow that $P_0(\nu)=d_0Γ_0//(2π)$ and $Q_K(\nu)=Q_2(\nu)=(\nu−\nu_0)^2+(Γ_0/2)^2.$ Here, $d_0$ is the amplitude, $\nu_0$ and $Γ_0$ are the resonant frequency and the full width at half-maximum (FWHM), respectively. All three parameters $\nu_0,$ $Γ_0$ and $d_0$ are real-valued quantities, with the understanding that the phase of the generally complex amplitude is equal to zero. The notation [L/K] is the usual symbol for the general PA as a ratio of two polynomials $P_L$ and $Q_K$ of degrees $L$ and $K,$ respectively. The physical meaning of a stick or bar spectrum is that it consists of lines associated with mono-energetic (mono-chromatic) absorption or emission of radiations. These latter processes result from transitions between discrete rotational, vibrational or rovibrational molecular levels for the ground state as well as for excited states. As noted, such line spectra belong to pure bound, stable states of molecules. Metastable states are prone to decay and their spectra are not lines or sticks. Rather, due to the underlying dissipative dynamics (there is a loss of energy), the absorptive radiations cannot be mono-energetic. This implies that the resulting spectra do not have sharp lines, in lieu of which a distribution of energies (or frequencies) appears. The most typical example of such frequency distributions is the absorptive Lorentzian function. In other words, the Lorentzian lineshape centered at $\nu_0$ is a broadened line of breadth or full width $Γ_0.$ These notions are also familiar by reference to a vibrating dipole which radiates energy according to classical physics. As a result, the vibrational amplitudes decrease. In other words, the dipole vibrations are attenuated, and the lineshape of the emitted radiation energy or frequency is the same Lorentzian distribution instead of a simple mono-chromatic phenomenon.

In comparing to the pdf of the Cauchy distribution, the equation for the Absorption Lorentzian above would be encapsulate the density of the random variable resonant frequency, $\nu,$ with scale parameter $\gamma = 2,$ and location $x_0=\nu_):$

$$f_X(x) = \frac 1 {\pi \gamma\left ( 1 + (\frac {x-x_0}{\gamma}) \right)^2}$$

Body:
Given the absence of more knowledgeable answers, I will share <a href="https://pdf.sciencedirectassets.com/271222/1-s2.0-S0730725X00X00148/1-s2.0-S0730725X96001579/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEDAaCXVzLWVhc3QtMSJIMEYCIQDqo6Y43LCAxy%2FZRXWX6FJenZlWu4NQ91Y5RttzsFklqAIhAPwsTjV9A1Ze6sgsUGASOFY0rAb9suR2daADdYx9I6RMKrQDCEkQAxoMMDU5MDAzNTQ2ODY1IgyWuWqwcwU6fHWyrDcqkQOgVuL8XpAc5enKMg7yWJ6CsgY9oIDKn52JBtNp%2FHC1VEEUMnc6ekX8cgslX1z4OpQ1TmjKK%2BUfCO9PH9g6%2FOozF%2Fyp7AOEov%2Bw0O%2F36JBErqY4Vpy8U1AHa8zrUC85T7C3u1R2l1thsWfGEVIXgYcL9%2Bm%2BSsM0xUKjUFHrK87zqbQEy3oA%2BjFt3v4sBpXtn0BwIgfxTOMS0w1K94ddG%2F5T2lcTah8ESr%2FyggsshR9l2%2FGGSid3qyrSYpfrGwtiysFor9pe3UQ4ln8JAWljenl02qChn1PNJKzaU3tqxQsStwVFRWOdAXKILsGwtDCJd3p%2FrGtjxRuFJmqPt7WQTepTyV9rPXh8w%2Fp7Cm%2Byu9Bj2gixLr16KrOSRwPD38kZ2o3PL0rjOVSa%2BqiEL0qiwuPp53FIy%2FQh8CzvvJUSYSEwOYqmejCGtZrCdZbMybdFgR4lWQooZjevcyTBKERCzGS1bE2mr%2FDXOD3wXP4Cxss5RgSKXdcaaN%2F%2Fqt4Pn8J76KgnCb0UkMVg6wwSilRvry4P6DCVt%2BSBBjrqAQks9Ne56CnM9u9T39Tkp6xsyaac4qLEFzXT%2FVBt1XSs5tO0H86qz%2FVRMH7qTOioyEs6bTQsjCx%2FKC2wQcix542jNzEI1GjhDYprfDbODcIuKeHL%2BGYQE4DZX%2BAqrssUDtaFpss2Tl%2Bs5SdaiazWbn2JsNRfZcbXxFV%2B4PDVCTCCGnRQPv%2BDWFX4epbPL0HU55oTMIg2YG3kKXtBPBUFZJ7wl%2FYEICZUe7jGXkxGEGsREVdlyyMQGX6m4KgukJndkap%2FH7LBCagUM9getiDDcS9FzsSFHJy3jEaa2jxv3OdH9yGgRPfikoAiBg%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20210226T163041Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYVQNCBPPG%2F20210226%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=131b5bc85f235bb5f0f811b3456c6d795be767fe4a7f3334c062871175e743e8&hash=7261cb39fa4a1772b0fb7c56bb4725b862656981a0ce7568bb5a4719ec61e54b&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0730725X96001579&tid=spdf-c498c68a-8583-4257-a62d-3d95bae0fe96&sid=d5b6f5d66e406245970873581b9b86762015gxrqa&type=client" rel="nofollow noreferrer">what I found on the topic - hopefully better or more more comprehensive answers will come along.

In short, free-induction decay (FID), which results from interactions among dipoles in MRI (magnetic resonance imaging) is influenced by inhomogeneities in the magnetic field which have long tails (not exponential or Gaussian), and can be modelled using a Cauchy distribution:

... The distribution of microscopic inhomogeneous fields is not necessarily Gaussian. In some systems of importance in areas as diverse as oilwell
logging and medical magnetic resonance imaging (MRI) the distribution of inhomogeneous fields is approximated by the Cauchy-Lorentz (C-L, Cauchy in statistics, Lorentz in physics) distribution. This is the
field due to randomly located point dipoles within a sample. Again, the phase distribution width at an effective exchange time $t_{ex}$ is proportional to $\chi \nu t_{ex}.$ However, the distribution width for $n = t/t_{ex}$ random samples is $t/t_{ex}$ times the original width, so the width of
the phase distribution at time $t$ does not depend on $t_{ex}.$ Here, the Fourier transform of the $n$-fold convolution of a C-L distribution is the $n$-fold product of its simple exponential FT, not an exponential with the width squared as in the Gaussian. That is, the FID decay does not depend on $a$ or $D.$ Furthermore, the additional FID decay rate is proportional to the width of the CL distribution of phases, not to the square of the width, as for the Gaussian.

where $\chi\nu$ (susceptibility difference times frequency), $D$ (diffusion coefficient), $a$ (a pore dimension), echo-time $t$ for single echoes and half-echo-spacing $\tau$ for CPMG (Carr-Purcell-Meiboom-Gill (CPMG) spin-echo measurements), $t_{ex}$ for exchange time.

More relevant, and in reference to a more robust correspondence between time and frequency domains in MRI than the FFT (e.g. non-parametric fast Padé transform)

Considered strictly, a general line spectrum is associated only with for purely stationary, stable bound states (e.g. ground states of atoms/molecules) of practically infinite lifetimes. On the other hand, peaks in spectra are resonances that represent non-stationary, transient states of molecules. Such metastable states last for a given lifetime, after which they decay. This is reflected in distributions of frequencies in the spectral lineshapes, instead of the stick spectra. These distributions are mainly the Lorentzian functions that are in mathematics known as the Cauchy distributions. In physics, they are also called Breit-Wigner lineshape profiles from the theory of resonant scattering of particles and photons on general targets. The Cauchy, Lorentz and Breit-Wigner distributions coincide with one of the simplest forms of the Padé approximant (PA), symbolized as [0/2], and given by:

$$\begin{aligned} Y_{\mathrm{A}} \left( \nu \right) =\frac{d_0 }{2\pi }\frac{\varGamma _0 }{\left( {\nu -\nu _0 } \right) ^{2}+\left( {\varGamma _0 /2} \right) ^{2}}\quad \hbox {(Absorptive Lorentzian)} \end{aligned}$$

It is seen that the Padé spectrum $Y_A(\nu)$ depends on $\nu$ in a continuous way. If $Y_A(\nu)$ is written in the general form $P_L(\nu)/Q_K(\nu),$ it would follow that $P_0(\nu)=d_0Γ_0//(2π)$ and $Q_K(\nu)=Q_2(\nu)=(\nu−\nu_0)^2+(Γ_0/2)^2.$ Here, $d_0$ is the amplitude, $\nu_0$ and $Γ_0$ are the resonant frequency and the full width at half-maximum (FWHM), respectively. All three parameters $\nu_0,$ $Γ_0$ and $d_0$ are real-valued quantities, with the understanding that the phase of the generally complex amplitude is equal to zero. The notation [L/K] is the usual symbol for the general PA as a ratio of two polynomials $P_L$ and $Q_K$ of degrees $L$ and $K,$ respectively. The physical meaning of a stick or bar spectrum is that it consists of lines associated with mono-energetic (mono-chromatic) absorption or emission of radiations. These latter processes result from transitions between discrete rotational, vibrational or rovibrational molecular levels for the ground state as well as for excited states. As noted, such line spectra belong to pure bound, stable states of molecules. Metastable states are prone to decay and their spectra are not lines or sticks. Rather, due to the underlying dissipative dynamics (there is a loss of energy), the absorptive radiations cannot be mono-energetic. This implies that the resulting spectra do not have sharp lines, in lieu of which a distribution of energies (or frequencies) appears. The most typical example of such frequency distributions is the absorptive Lorentzian function. In other words, the Lorentzian lineshape centered at $\nu_0$ is a broadened line of breadth or full width $Γ_0.$ These notions are also familiar by reference to a vibrating dipole which radiates energy according to classical physics. As a result, the vibrational amplitudes decrease. In other words, the dipole vibrations are attenuated, and the lineshape of the emitted radiation energy or frequency is the same Lorentzian distribution instead of a simple mono-chromatic phenomenon.

In comparing to the pdf of the Cauchy distribution, the equation for the Absorption Lorentzian above would be encapsulate the density of the random variable resonant frequency, $\nu,$ with scale parameter $\gamma = 2,$ and location $x_0=\nu_):$

$$f_X(x) = \frac 1 {\pi \gamma\left ( 1 + (\frac {x-x_0}{\gamma}) \right)^2}$$


问:
Approximation property and separability
I have a question regarding separability. Let $Y$ be a normed space, and suppose the identity $I$ is the pointwise limit of finite range operators $F_n$.

Does this imply that $Y$ is separable?

Context: giving conditions under which the approximation property holds

My thoughts: this is true, I think. Indeed, $F_n(Y)$ is finite dimensional hence separable (being homeomorphic to $\mathbb{R}^n$ for some $n$. Pick therefore a countable dense subset $C_n$, and define $C:=\bigcup_n C_n$, which I claim to be dense.

Let $y\in Y$. We know $y=\lim_n F_n(y)$ and $F_n(y)=\lim_j c_j^n$, for $c_j^n \in C_n$. But then $||y-c_j^n||\leq ||y-F_n(y)||+||F_n(y)-c^n_j||$. The first term is small for $n$ large, and fixing $n$, I can surely find $j=j(n)$ such that the second is as small as the first one. So I found $y=\lim_n c_{j(n)}^n$.

Is this correct?

答:
It is correct. I'd write:

If $y \in Y$ let $\varepsilon >0$ be given. Then we know that $F_n(y) \to I(y)=y$ as $n \to \infty$ so we find $N$ such that for all $n \ge N$ we have $\| F_n(y) - y\| < \frac{\varepsilon}{2}$. Then as $C_N$ is dense in $F_N[Y]$, we have $c \in C_N \subseteq C$ such that $\|F_n(y) - c\| < \frac{\varepsilon}{2}$. It follows that $\|y - c\| < \varepsilon$ and we're done.

Title: Approximation property and separability


问:
About the functoriality of the long exact sequence in cohomology
I'm writing some notes on homological algebra and there I proved the long exact sequence in cohomology using the snake lemma. (I can give more details if anyone wants.) The next natural step is to prove that the connecting morphisms arise "functorially". In most books (Iversen, both books by Gelfand and Manin, both books by Kashiwara and Shapira, etc...) this is said to be either "obvious" or is an exercise to the reader.

Using elements (as Weibel does it), this is not too difficult but, since my main example of abelian category is the category of sheaves, I would rather avoid it.

Another possibility is to use the mapping cone to describe "explicitly" the connecting morphism and then the functoriality comes for free.

Because of this, I am thinking about using this (the need for an explicit description of the connecting morphisms) as a first motivation for the mapping cone construction. However, this would be very silly if the functoriality was indeed obvious. That's why I'm here: there is an "obvious" proof of this without using elements nor mapping cones?

(Not really related, but I would also love to know if someone can think of a better motivation for mapping cones, which I'll need to define the triangulated structure of the homotopic and derived categories.)

答:
This is not really an answer to my question but rather my understanding of the idea given by @Mindlack. (Proposition 5.4.2 is the fact that a functor category is abelian if the target is and "the previous theorem" is the existence of the long exact sequence in cohomology.)

I'm still very much interested in understanding better the motivation for the mapping cone.

(I've edited the answer to add the precise statement of the theorem.)


问:
Why is the Legendre polynomials eigenfunctions to this operator?
The operator $$Au = -\frac{d}{dx}((1-x^2)\frac{du}{dx}), \quad -1<x<1$$
supposedly has eigenfunctions
$$P_0 = 1, \quad P_1 = x, \quad P_2 = \frac{3}{2}x^2-\frac{1}{2}x$$
with eigenvalues
$$\lambda_0 = 0, \quad \lambda_1 = 2, \quad \lambda_2 = 6.$$

Controlling this, the first two are ok, but the third one seems wrong,

$$AP_2 = -((1-x^2)(3x-\frac{1}{2}))' = -(3x-3x^3-\frac{1}{2}+\frac{x^2}{2})' = (-3+9x^2 -x) \neq \lambda_2\frac{3}{2}x^2 - \lambda_2\frac{1}{2}x \implies AP_2 \neq \lambda_2P_2.$$

What am I missing?

答: A sanity check: The Legendre polynomials have the same parity as their degree. In particular, the second Legendre polynomial should be an even function (i.e., only contain even powers of $x$). But your $P_2(x)$ is not even, and so cannot be correct.


问:
How many combinations of groups are there where no member of a group has been with another member before?
I found this hard to word in the title, so let me give an example.

I have 16 students, and I want to split them up into 4 groups of 4. However, I want to make sure that every time I have a new combination of groups, that a student has never been with any of their new groupmates before.

I found info on how to separate objects into groups, but not in only unique ways as I've described above.

I think I figured this out the rote way, by assigning each student a number and then finding the unique combinations, and got 3. But I'd like to know the math behind it, and how it could be applied to other configurations (20 students w/ 4 groups of 5?).

Thanks for any help you can give!

答:
If you have $n^2$ students being divided into $n$ groups of size $n$, then a solution which works for $d$ ways is equivalent to a set of $d+2$ mutually orthogonal $n\times n$ Latin squares (MOLS). Some partial results:

For any $n$, you can succeed for at most $n+1$ days.

When $n$ is a prime power, it is possible to achieve the limit of $n+1$ days.

For $n=6$, three days is the max, as there are no two orthogonal $6\times 6$ Latin squares.

For $n=10$, four days is possible, but it is an open problem to determine if five days is possible (see Find three $10\times10$ orthogonal Latin squares.).

There is more data on the maximum known number of MOLS here: http://www.math.stonybrook.edu/~tony/whatsnew/column/latin-squaresII-0901/latinII3.html

In general, if you have $nm$ students to be divided into $n$ groups of size $m$, this is known as the Social Golfer problem. Again, this is a hard open problem in most cases. See

http://www.mathpuzzle.com/MAA/54-Golf%20Tournaments/mathgames_08_14_07.html

for some partial results, and you can find more by searching "Social Golfer problem."


问:
Computing $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$
how can i compute: $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$?

i tried l'hopital's rule  but it's like a loop.

also if it can be done without that rule i'd like to know how.

Thanks.

答:
If you recognize some special limits in this expression, the rest is relatively easy.

$\lim_{x\to \infty} xe^{-x}=0 $

$\lim_{x\to 0} xe^{-x}=0$

Begin with the substitution: $$\frac{1}{\cos^2(x)}=\sec^2(x)=\tan^2(x)+1$$ Rewriting the limit expression: $$\lim_{x\to -\frac{\pi}{2}}(\tan^2(x)+1)\space e^{\tan(x)}=\lim_{x\to -\frac{\pi}{2}} \tan^2(x)\space e^{\tan(x)}\space + \space \lim_{x\to -\frac{\pi}{2}} e^{\tan(x)} $$ If we make the substitution $-\tan(x)=y$, we get: $$\lim_{y\to \infty} y^2 e^{-y}\space +\space \lim_{y\to \infty} e^{-y}=0+0=0 $$


问:
How to effectively learn by self learning.
I'm a self studying student.I live in a place where I've limited resources,to give an example,in winters to study I've to use candles because there is no electricity,no light and so on.
I have no teacher so all that I have is books and sometimes internet,if the internet is not banned .

I'm average in studying but I love to study and try my best to understand. The thing that is the biggest hurdle is as explained below:

I'm going through a text smoothly and then I get stuck.I have a confusion, doubts and questions. I try to think for hours, and then they turn into days.
I can't ask anyone near me and I feel I can't move ahead though the text. I feel I haven't understood the previous paragraphs so my motivation to go ahead is dead.
The only way I know out is to accept the conclusion that's presented and move on.
But I am not able to do it. I have this irresistible  urge to know it before I can move ahead.

Could you please tell me how to proceed in such a situation.

Should I note my doubts and move on and try to clear them when possible?

I will be thankful for even a hint that might help me or any other general tips.
Warmest regards, Thank you.

答:
Here's a collection of suggestions

no internet required

Reading about the same topic from different books.

Study partner, talking about the material with other people.

Keeping notes (of any kind)

Keeping organized notes of theorems, definitions, and examples

Trying to prove results you already know

internet required

Ask (well-structured) questions related to items you do not understand on this forum


问:
Show that the function $f(x)=2\lfloor x\rfloor-x$ defined from $f:\mathbb{R}\to\mathbb{R}$ is onto
$f:\mathbb{R}\to\mathbb{R}$ is defined by $f(x)=2\lfloor x\rfloor-x$ where $\lfloor \cdot \rfloor$ is greatest integer function.

How to show that it is an onto function without the help of graph?

It is given it is invertible.

And what would be $f^{-1}$?

答:
To prove any function is onto you simply show that for any $w \in $ the co-domain that $f(x) = w$ will have a solution.

SO if $w\in \mathbb R$ we have to show that there exist an $x$ so the $2[x] - x= w$.

Now bear in mind $x = [x] + \{x\}$ (where $[x]$ is defined to be an integer so that $n \le x < n+1$ and $\{x\}$ is defined to be $x-[x]$).

So $2[x]-\{x\} = 2[x] - ([x]+\{x\}) = [x]-\{x\}$

So we need to solve $[x]-\{x\} = w$.

Well $w = [w] + \{w\}$ so we need $[x] - \{x\} = [w] + \{w\}$ and $[x]-[w]=\{x\} + \{w\}$.

Now $[x] -[w]$ is an integer.  And $0\le \{x\} +\{w\} < 2$. so we must have $\{x\} + \{w\} = 0$ or $1$.

For $\{x\} +\{w\} =0$ we must have both $\{x\}$ and $\{w\}$ equal to $0$.

For $\{x\} + \{w\} = 1$ well each term is strictly less than $1$ we must have both $\{x\}$ and $\{w\}$ not equal to $0$.

So case 1:  $\{w\} = 0$ and $w = [w] + \{w\}$ is an integer.  The we must have $[x]-[w] = 0$ so $[x] = [w]$ and $\{x\} = 0$ so $x = [x] = [w] = w$.

So if $w \in \mathbb Z$ then $x= w$ and $f(w) = 2[w] -  w= 2w-w = w$.

Case 2:  $\{w\} \ne 0$ so $[w] < w < [w]+1$ and $\{w\} + \{x\} = 1$ so $\{x\} = 1 - \{w\}$.

Ane $[x] -[w] = 1$ so $[x] = [w] + 1$ and so $x = [x] + \{x\} = [w] + 1 + (1-\{w\})= [w] + 2-\{w\}$.

That does it.  If $w\not \in \mathbb Z$ then

$f([w] + 2-\{w\}) = 2[([w]+1)+(1-\{w\})] - ([w]+2-\{w\}) =$

$2([w]+1) - [w] - 2 + \{w\} = [w] + \{w\} = w$.

Either case there will always be a real $x$ so that $f(x) = w$

.....

If we use the notation that $\lceil x \rceil$ is the unique integer $m$ so that $m-1 < x \le m$, and if we use the notation $\}x\{$ = \lceil x\rceil -x$ then

if $w\not \in \mathbb Z$, $\lceil w \rceil =\lfloor w \rfloor + 1$ and $\}w\{ = 1-\{w\}$ so we can write $[w] + 2-\{w\} = [w]+1 + (1-\{w\} =\lceil w\rceil -\}w\{$.

And if $w \in \mathbb Z$ then $\lceil w\rceil =\lfloor w \rfloor = w$ and $\}w\{ = \{w\} = 0$.

And $w =\lceil w\rceil -\}w\{$

So in either case  $x = \lceil w\rceil -\}w\{$ will be a solution to $f(x) = w$


问:
show that $f$ is Riemann integrable on $[0,1]$
Let $f(x)=\sin\left(\frac{1}{x}\right)$ if $0<x\le1$ and $f(x)=0$ if $x=0$. Show that $f$ is Riemann integrable on $[0,1]$ and calculate it's integral on $[0,1]$.

I would like to know if my proof holds, please. And, I would like have a hint on how we can calculate the intergal of this such of functions, please.

My attempt is to pass by Darboux upper and lower sums.

To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$.

Let $0<\epsilon<2$ and consider the following intervals $[0,\epsilon/2]$ and $[\epsilon/2,1]$. First of all, as $f(x)$ is continuous on $[\epsilon/2,1]$($\sin(\frac{1}{x})$ is continuous on $[\epsilon/2,1]$), it is integrable on this interval. We would like to show now that $f$ is intergable on $[0,\epsilon/2]$. Consider the partition $\sigma=\{0,\frac{\epsilon}{2}\}$ on $[0,\epsilon/2]$ We have that

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)=\sum_{i=0}^{0}M_i(x_{i+1}-x_i)-\sum_{i=0}^{0}m_i(x_{i+1}-x_i)$ with $M_i=\sup\{f(x):x\in[x_i,x_{i+1}]\}$ and $m_i=\inf\{f(x):x\in[x_i,x_{i+1}]\}$.

We have that $M_i\le1$ and $m_i\le 0$. Thus,

$\sum_{i=0}^{0}M_i(x_{i+1}-x_i)-\sum_{i=0}^{0}m_i(x_{i+1}-x_i)\le1\cdot \frac{\epsilon}{2}< \epsilon$. Therefore, as $f(x)$ is integrable on $[0,\epsilon/2]$ and on $[\epsilon/2,1]$, we conclude that $f$ is integrable on $[0,1]$.

Now, I would like to calculate it's integral value... If someone could help with it, I would appreciate it. Honestly, I have no idea how to integrate this such of discontinuous functions at finitely many points.

Edit:

As $f(x)$ is continuous on $[\epsilon/3,1]$, it is integrable on this interval. Therefore, $\forall \epsilon>0 \ \exists$ partition $\tau$ on the interval $[\epsilon/3,1]$: $\overline{S}_{\tau}(f)-\underline{S}_{\tau}(f)<\epsilon/3$

To show that $f$ is integrable on $[0,1]$, we have to show that: $\forall \epsilon>0 \ \exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}<\epsilon$

Suppose $0<\epsilon<3$ and let $\sigma=\tau \ \cup \{0\}$ be a partition on the interval $[0,1]$. We have that (for the following inequality I use the same definition for $M_i$ and $m_i$ as above in the post):

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}=\Big(\sum_{i=0}^{0}\underbrace{M_i}_{\le1}(x_{i+1}-x_i)-\sum_{i=0}^{0}\underbrace{m_i}_{\ge -1}(x_{i+1}-x_i)\Big)+\underbrace{\Big(\sum_{i=1}^{n}M_i(x_{i+1}-x_i)-\sum_{i=1}^{n}m_i(x_{i+1}-x_i)\Big)}_{<\epsilon/3}<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$.

Therefore, $f$ is integrable on $[0,1]$.

答:
Integrate by parts
$$\int \sin \left(\frac{1}{x}\right) \, dx=x\sin \left(\frac{1}{x}\right) -\int x \left(-\frac{1}{x^2}\right)\cos \left(\frac{1}{x}\right)\,dx=$$
$$=x \sin \left(\frac{1}{x}\right)-\text{Ci}\left(\frac{1}{x}\right)+C$$
Ci is cosine integral
$$\text{Ci}(x)=-\int_x^{\infty}\frac{\cos x}{x}$$

$$\int_0^1 \sin \left(\frac{1}{x}\right) \, dx=\sin (1)-\text{Ci}(1)-\lim_{a\to 0^+}\left[a \sin \left(\frac{1}{a}\right)-\text{Ci}\left(\frac{1}{a}\right)\right]$$
set $\frac{1}{a}=t$
$$\lim_{t\to\infty}\left[\frac{1}{t} \sin t-\text{Ci}\left(t\right)\right]=0$$
$$\int_0^1 \sin \left(\frac{1}{x}\right) \, dx=\sin (1)-\text{Ci}(1)\approx 0.504$$


问:
How many epsilon numbers $<\omega_1$ are there?
An epsilon number is an ordinal $\epsilon$ such that $\epsilon=\omega^\epsilon.$ What is the cardinality of the set of all epsilon numbers less than $\omega_1$?

I'm asking this because of a proof I've just read that seems to presuppose that there are countably many such ordinals, and it seems to me intuitively that there should be uncountably many (although I don't know how to prove it).

Added. OK, I've just understood that the proof I mentioned is OK even if there are uncountably many such ordinals, but I still don't see how I can find their number.

答: Your intuition is correct: $\omega_1=\epsilon_{\omega_1}$. See the discussion in Wikipedia: the constructions of $\epsilon_{\alpha+1}$ from $\epsilon_\alpha$ and of $\epsilon_\alpha$ for a countable limit ordinal $\alpha$ preserve countability, so there must be $\omega_1$ countable $\epsilon$-numbers.


问:
Demonstrate that the set of orthonormal matrices is closed
I'm trying to demonstrate that the set of orthonormal matrices $\mathcal{O}(n,m) =\{ A | A^TA = I_m\} $ is closed. I am in a metric space $\mathbb{R}^{n*p}$ that uses the measure $d^2(X,Y) = trace((X-Y)^T(X-Y))$. I don't really understand how to interpret the notion of closed when it comes to this example since it's with matrices and not singular values, functions, or open/closed balls.

答:
Let's denote $\mathcal M(n,m)$ the linear space of real matrices of dimension $ n \times m$. The map

$$\begin{array}{l|rcl}
\phi : & \mathcal M(n,m) & \longrightarrow & \mathcal M(m,m) \\
    & A & \longmapsto & A^tA \end{array}$$

is continuous as $A^tA$ entries are polynomial functions of $A$ entries.

The singleton $\{I_m\}$ is a closed subset of the normed vector space $(\mathcal M(m,m), \Vert \cdot \Vert_2)$ where $\Vert \cdot \Vert_2$ stands for the norm defined by

$$\Vert A \Vert_2^2 = \sum_{i,j}  a_{ij}^2.$$ Therefore, $\mathcal O(n,m)$ which is the inverse image of $\{I_m\}$ under $\phi$ is closed for the topology induced by the distance $d$ as

$$d^2(X,Y) = \Vert X-Y \Vert_2^2.$$


问:
Need help with proof on trigonometric integrals
Question:

Let $f:[0,\pi]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^{\pi} f(t)\sin(t)dt = \int\limits_0^{\pi} f(t)\cos(t)dt =0,$$ then the equation $f(x)=0$ admits at least two solutions in $[0,\pi].$

My thoughts:

If $f(x) > 0$, then $f(x)\sin(x) > 0 \hspace{1em}\forall x \in(0, \pi)$,

but $f(x)\cos(x) > 0 \hspace{1em}\forall x \in[0,\frac{\pi}{2})\hspace{1em}$ and $\hspace{1em}f(x)\cos(x) < 0 \hspace{1em}\forall x \in(\frac{\pi}{2},\pi]$

I'm not really sure how to proceed.

答:
As observed in the comments, $f$ must have at least one zero: otherwise $f$ has no zeroes, and by continuity $f$ is always positive, or always negative. Suppose that $f>0$. Recalling that $\sin x \ge 0$ for $x \in [0, \pi ]$ you get
$$\int_0^\pi f(x) \sin x \ \mathrm dx >0$$

Now, suppose by contradiction that $f$ has exactly one zero. Such a zero must be in the interior of $(0, \pi)$, otherwise we would have a contradiction using the same argument above.

Say $f( A)=0$ is the unique zero of $f$, with $A \in (0, \pi)$. Since $f$ is continuous, it changes sign only once. Without loss of generality we can suppose that $f >0$ in $[0, A)$, and $f<0$ in $(A, \pi]$.

Recall that for all $x \in \Bbb R$ you have
$$\sin (x-A) = \sin x \cos A - \cos x \sin A$$

In particular, for all $x \in [0, \pi]$ you can multiply by $f(x)$, and the you can integrate over $[0, \pi ]$ to get
$$\int_0^\pi f(x) \sin (x-A)  \ \mathrm dx = 0$$

However this cannot happen. Indeed for all $x \in [0, A)$ you have
$$\sin (x-A) < 0$$
while for all $x \in (A, \pi]$ you have
$$\sin (x-A) > 0$$
This means that for all $x \neq A$ you have $f(x) \sin (x-A) <0$. Hence
$$0=\int_0^\pi f(x) \sin (x-A)  \ \mathrm dx <0$$
A contradiction.


问:
Tensor product of two elements, one them is zero
Let V and W be two vector spaces over $\mathbb{R}$. I've see this property somewhere  but I'm not sure how to prove it.

Let  $ x \in V $ and $y \in W$, such that $x=0 $ or $y=0, $
then $x \otimes y = 0 \in V \otimes W$.

Help please!

答:
By definition of the tensor product, for any scalar $c \in \mathbb{R}$ and any vectors $x\in V$ and $y\in W$ we have

$$
(cx) \otimes y = x \otimes (cy) = c(x \otimes y).
$$

Suppose $x = 0$, then for any vector $z\in V$ we can write $x = 0 \cdot z$ , so

$$
0 \otimes y = (0\cdot z) \otimes y = 0 \cdot (z \otimes y) = 0 \in V\otimes W
$$

and similarly when $y=0$.


问:
Well-ordered sets two examples
Say I have the following two sets:

$\mathbb{R} \cup \{ 0\}$ under < (less than operator)

$\{(a,b) | a,b \in \mathbb{Z^+} \}$ under $(a,b) \prec (c,d) \iff (a<c) \land (b <d)$.

For $\mathbb{R} \cup \{ 0\}$ I said yes because every subsets will have a least element. For example $\{0\} \subset \mathbb{R} \cup \{ 0\} \text{ and } \{0,1,2\} \subset\mathbb{R} \cup \{ 0\}$, $0$ is the least element in the last subset.

For $\{(a,b) | a,b \in \mathbb{Z^+} \}$ I am saying no because there is not greatest element in each of these subsets.

Looking for some verification to see if my reasoning is correct with these problems.

答:
Note for the first example that $\mathbb{R} \cup\{0\}=\mathbb{R}$, which is known to be non well ordered with the usual order given by <. (The subset $(- \infty,0)$ has no least element).

For $\{(a,b) | a,b \in \mathbb{Z^+} \}$, note that this is not a total ordered set. For instance, there is no order relation between $(1,2)$ and $(2,1)$, so it cannot be well ordered.

Title: Well-ordered sets two examples


问:
calculating $\frac{1}{1-\cos^2 (x)}-\frac{1}{1+\cos^2(x)}$
I'm stuck on simplifying this equation, and would appreciate your support on my approach.

$\frac{1}{1-\cos^2 (x)}-\frac{1}{1+\cos^2(x)}$

My working out:

Using the identity: $\sin^2(x)=1-\cos^2 (x)$ and $\frac{1}{\sin^2 (x)}=\csc^2(x)$

$$\csc^2(x)-\frac{1}{1+\cos^2(x)} = \frac{1-[\csc^2(x)+\csc^2(x)\cos^2(x)]}{1+\cos^2(x)}=\frac{[1-\csc^2(x)]-\csc^2(x)\cos^2(x)]}{1+\cos^2(x)}=\frac{-\cot^2(x)-\csc^2(x)\cos^2(x)}{1+\cos^2(x)}=-\frac{\csc^2(x)}{1+\cos^2(x)}$$

答: Succinctly, it's $\frac{2\cos^2x}{(1-\cos^2x)(1+\cos^2x)}=\frac{2\cot^2x}{1+\cos^2x}$. You've miscalculated; in your display-line equation, the first $=$ has multiplied its RHS by $-1$. Note also $\csc^2x\cos^2x=\cot^2x$.


问:
How to integrate$\int_0^1 \frac{\ln x}{x-1}dx$ without power series expansion
I happen to watch the video here,
which gives a solution to the definite integral below using the power series approach. Then answer is $\frac{\pi^2}{6}$, given by:

$$\int_0^1 \frac{\ln x}{x-1}dx=\int_{-1}^0 \frac{\ln(1+u)}{u}du=\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}=\frac{\pi^2}{6},$$

where the power seires expansion of the function $\ln(1+u)$ is used.

I tried for some time, but could not find another approach. Does anyone know any alternative methods to evaluate above definite integral without using the infinite series expansion?

Any comments, or ideas, are really appreciated.

答:
Here is another way of showing that $\int_0^1 \frac{\log(x)}{1-x}\,dx=\pi^2/6$.  First, enforce the substitution $x\mapsto 1-x$ to obtain

$$I=-\int_0^1 \frac{\log(1-x)}{x}\,dx$$

Then, noting that $\int_0^1 \frac{1}{1-xy}\,dy=-\frac{\log(1-x)}{x}$ we write $I$ as

$$I=\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy\tag1$$

In THIS ANSWER, I used the transformation $x=s+t$, $y=s-t$ to write the double integral in $(1)$ as

$$\begin{align}
\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy&=\int_0^{1/2}\int_{-s}^{s}\frac{2}{(1-s^2)+t^2}\,dt\,ds+\int_{1/2}^{1}\int_{s-1}^{1-s}\frac{2}{(1-s^2)+t^2}\,dt\,ds\\\\
&=\int_0^{1/2}\frac{4}{\sqrt{1-s^2}}\arctan\left(\frac{s}{\sqrt{1-s^2}}\right)\,ds\\\\&+\int_{1/2}^{1}\frac{4}{\sqrt{1-s^2}}\arctan\left(\sqrt{\frac{1-s}{1+s}}\right)\,ds\\\\
&=4\int_0^{1/2}\frac{\arcsin(s)}{\sqrt{1-s^2}}\,ds+4\int_{1/2}^1\frac{\arccos(s)}{2\sqrt{1-s^2}}\,ds\\\\
&=2\arcsin^2(1/2)+\arccos^2(1/2)\\\\
&=2\left(\frac{\pi}{6}\right)^2+\left(\frac{\pi}{3}\right)^2\\\\
&=\frac{\pi^2}{6}
\end{align}$$

And we are done!


问:
Picking out unique solution of $x,y,z$ from two constraints

Let $x,y,z$ be positive real numbers in $\mathbb{R}$ satisfying the conditions:$x+y+z=12$ and $x^3 y^4z^5=(.1)600^3$. Then the value of $x^3 + y^3 +z^3=?$

This is a question from JEE mains, in it I have a doubt of how we are able to uniquely determine an $x,y,z$ triplet from just two equations(upto permutation)! The answer involves observing $x=3, y=4 $ and $z=5$ but how can we make that inference?

答:
For positive reals, we can use AM-GM inequality. This inequality for $12$ terms applies as

$$\frac{3\cdot (x/3)+4\cdot (y/4)+5\cdot (z/5)}{3+4+5} \ge \left(\frac{x^3}{3^3}\cdot \frac{y^4}{4^4}\cdot \frac{z^5}{5^5} \right)^{1/(3+4+5)}$$
$$\Rightarrow x^3\cdot y^4 \cdot z^5 \le 3^3 \cdot 2^8 \cdot 5^5$$
$$\Rightarrow x^3y^4z^5 \le (0.1)(600)^3$$

where equality (maximum) is achieved when each term is equal i.e.,
$$\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{12}{12}=1$$

from which one infers $x=3, y=4, z=5$.


问:
Solving differential equation with boundary conditions $y=2$ at $x=0$
I need to find a solution to this differential equation $(1+x^2) $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = x-xy^2$, with boundary condition $y=2$ at $x=0$

I collect all terms on the RHS to get $$\tfrac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{1+x^2}  (1-y^2)$$

I integrate then move the y terms to the left side and integrate both sides and simplify to get $$\frac{1}{\sqrt{1-y^2}} = k\sqrt{1+x^2}$$

I then isolate y in order to get it in a y= form and sub in the y and x value but I don't get the correct answer; I'm not sure where I'm going wrong

答:
$$\int\frac{dy}{1-y^2}=\int\frac{x\,dx}{1+x^2}$$
now you can do the left a few ways but we can use PFD:
$$\int\frac{dy}{1-y^2}=-\frac12\int\left(\frac{1}{y-1}-\frac{1}{y+1}\right)dy=\frac12\ln\left|\frac{y+1}{y-1}\right|$$
now do RHS:
$$\int\frac{x}{x^2+1}dx=\frac12\ln(x^2+1)+C_1$$
now equate:
$$\ln\left(\frac{y+1}{y-1}\right)=\ln(x^2+1)+C_2$$
exponentiate:
$$\frac{y+1}{y-1}=C_3(x^2+1)\tag{1}$$
now ideally get $y$ on its own, so you can use that:
$$\frac{y+1}{y-1}=\frac{y-1+2}{y-1}=1+\frac{2}{y-1}$$
bring it back into the equation:
$$1+\frac{2}{y-1}=C_3(x^2+1)$$
$$y-1=\frac{2}{C_3(x^2+1)-1}$$
$$y=1+\frac{2}{Cx^2+(C-1)}$$


问:
Two Questions Involving Differential Equations Calculus
Given the equation $x''(t)+k^2x(t)=0$, how do I find a correct solution to the equation? I tried letting $x(t) = c$, where c is a constant, so that $x'(t) = 0$. But the equation won't make sense since $x''(t)$ will also be 0, and there won't be a solution. I consulted with someone else and they said that it should be a sinusoidal equation - which I don't know how to obtain from this.

答:
I'll just answer the first question for now. Your friend said it should be a sinusoid because the equation relates the derivative of a function to the function itself - think of how the second derivative of sine is a constant times sine itself.

More generally, it would be better to use an exponential (because by Euler's formula sines and cosines are linear combinations of exponentials). So assume your function has form $x(t)=ce^{lt}$ for some undetermined $c,l$. Substitute this into your equation to get $cl^2e^{lt}+k^2ce^{lt}=0$. You can cancel the exponentials here and just get an algebraic equation for $l$ in terms of $k$. So $l = \pm ik$. Since linear combinations of solutions to homogenous ODEs are also solutions, we have to add the solution with $l=ik$ and the solution with $l=-ik$. Our solution said nothing about $c$, in order to get this kind of information you would need some initial conditions. Hence your solution is $x(t) = c_1e^{ikt}+c_2e^{-ikt}$.


问:
Dominos ($2 \times 1$ on $2 \times n$ and on $3 \times 2n$)
How many ways are there to tile dominos (with size $2 × 1$) on a grid of $2 × n$?

How about on a grid of $3 × 2n$?

答:
In the case of a $2\times n$ grid, the resulting Fibonacci recurrence is pretty much clear, and you can read up on the solutions on Wikipedia. However, I believe the previous answers have not treated the $3\times n$ case with enough care, so I'd like to fill that gap. Please, let me know of mistakes if there are any.

So we start with a grid such as

To make a first step in the iteration, we ask the question: how can we put a number of dominos on this grid in all different ways such that the first column labeled by $n$ is covered completely?

The possible variations clearly are:

This reveals that apart from the initial $3\times n$ type grid, the number of different tilings for which we can denote as $f_n$, we also need to consider a second grid of $3\times n$ type with one corner missing, the number of tilings for which we call $g_n$. (By symmetry, it does not matter whether an upper or lower corner is missing, the number of tilings will be the same.)

Therefore, the above all possible coverings of the first column labeled by $n$ suggest the recurrence

$$f_n = f_{n-2}+2g_{n-1}$$

Since we have introduced the quantity $g_n$, it seems we need a recurrence relation for it as well. For that end, we consider all possible ways to add dominos to the grid 1) above, such that all squares of column $n-1$ are covered:

Note that in grid 5) we also have added an additional domino occupying columns $n-2$ and $n-3$ at the top, since this is the only possible way to cover the remaining square in column $n-2$ anyway.

This suggests the following recurrence for $g_n$:

$$g_{n-1}=f_{n-2}+g_{n-3}$$

With these two recurrence relations determined, we can proceed to solve for the general solution.

As is usual in such recurrence problems, we can attempt an exponential ansatz for the solution and fix free parameters by demanding consistency with the recurrence relations. For that end, we assume:

$$f_n= q^n\\
g_{n}= a q^n$$

for some unknown $q$, and allowing for a different relative factor $a$ in $g_n$ compared to $f_n$. Plugging the above ansatz into the two recurrence relations, we cancel all trivial overall coefficients and see that the relations reduce to the two equations

$$q^2-2 a q-1= 0 ~~~,~~~ a q^2-q-a =0$$

In total, these equations have four independent solutions:

$$\begin{align*}q_1&=-\sqrt{2-\sqrt{3}}~~~,~&&a_1=\frac{1}{\sqrt{2}}~,\\
q_2&=\sqrt{2-\sqrt{3}}~~~,~&&a_2=-\frac{1}{\sqrt{2}}~,\\
q_3&=-\sqrt{2+\sqrt{3}}~~~,~&&a_3=-\frac{1}{\sqrt{2}}~,\\
q_4&=\sqrt{2+\sqrt{3}}~~~,~&&a_4=\frac{1}{\sqrt{2}}~,\end{align*}$$

where $q_i$ and $a_i$ belong together as a single solution respectively.

Finally, the general solution for $f_n$ is therefore given by

$$f_n=c_1 q_1^n+c_2 q_2^n+c_3 q_3^n+c_4 q_4^n.$$

The four unknown coefficients $c_i$ can be fixed by taking the initial conditions

$$f_0=1~~,~~f_1=0~~,~~f_2=3~~,~~f_3=0~,$$

where we used that the $3\times n$ grid can only be successfully tiled if $n$ is even, and that by convention there is only one way to tile a grid of zero width: place no dominos at all. The case $f_2=3$ can be essentially read off of the grids 1), 2) and 3) above. These initial conditions lead to:

$$c_1=c_2=\frac{1}{12} \left(3-\sqrt{3}\right)~~~,~~~c_3=c_4=\frac{1}{12} \left(3+\sqrt{3}\right)~,$$

and we are done.

As a consistency check, we then observe e.g. $f_4=11$, as was mentioned in the article concerning this problem.


问:
Primitive polynomials of finite fields
there are two primitive polynomials which I can use to construct $GF(2^3)=GF(8)$:

$p_1(x) = x^3+x+1$

$p_2(x) = x^3+x^2+1$

$GF(8)$ created with $p_1(x)$:

0

1

$\alpha$

$\alpha^2$

$\alpha^3 = \alpha + 1$

$\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$

$\alpha^5 = \alpha^4 \cdot \alpha = (\alpha^2+\alpha) \cdot \alpha=\alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1$

$\alpha^6 = \alpha^5 \cdot \alpha=(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha+1+\alpha^2+\alpha=\alpha^2+1$

$GF(8)$ created with $p_2(x)$:

0

1

$\alpha$

$\alpha^2$

$\alpha^3=\alpha^2+1$

$\alpha^4=\alpha \cdot \alpha^3=\alpha \cdot (\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1$

$\alpha^5=\alpha \cdot \alpha^4=\alpha \cdot(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1$

$\alpha^6=\alpha \cdot (\alpha+1)=\alpha^2+\alpha$

So now let's say I want to add $\alpha^2 + \alpha^3$ in both fields. In field 1 I get $\alpha^2 + \alpha + 1$ and in field 2 I get $1$. Multiplication is the same in both fields ($\alpha^i \cdot \alpha^j = \alpha^{i+j\bmod(q-1)}$. So does it work so, that when some $GF(q)$ is constructed with different primitive polynomials then addition tables will vary and multiplication tables will be the same? Or maybe one of presented polynomials ($p_1(x), p_2(x)$) is not valid to construct field (altough both are primitive)?

答: The situation is not so different in a simpler context, the field of 5 elements, also known as the integers modulo 5. Whether $\alpha$ is $2$ or $3$, the field is $0,1,\alpha,\alpha^2,\alpha^3$, but whether $\alpha+\alpha+1=0$ depends on which $\alpha$ you choose. </p>

Title: Primitive polynomials of finite fields


问:
Why $s/(1-s) = 1$ at $s=1$ in bode plot?
Wolfram plot of $\frac{s}{1-s}$ is $\pm\infty$ at $s=1$. But, bode plot of $\frac{s}{1-s}$ results in $1$ at $s=1$. Obviously, this is wrong. Why?</p>

答:
The x-axis in the Bode plot isn't $s$, but $\omega$. Remember that $s=j\omega$, so what you're seeing in the Bode plot is as follows:

$$20\log|H(j1)|=20\log|\frac{j1}{1-j1}|=20\log|\frac{j(1+j)}{2}|$$


问:
How do I compare two expressions in Maple to see if they’re equivalent?
I have to compare two expressions as an assignment but the professor didn’t explain how to do it (I doubt he actually knows how to do the stuff he asks us to do). The two expression pairs in question:

```<code>tg(x) + tg(y) = (sin(x+y))/cos(x)*cos(y)

arcsin(x) + arcsin(y) = pi/2
</code>```

Now I know these are true, but whenever I use evalb or is I invariably get false. I tried many different things but none of them work and I’ve searched the internet for help but I didn’t find anything useful. The most similar thing to mine was this but I tried the solution and it didn’t work. Please help, I’m losing my sanity.

答:
You may have mistyped tg(...) when you actually mean tan(...), where tan is the name of a trigonometric function in Maple.

In that case you seem to be missing brackets in the right hand side of the first equation. Note the difference in the output that the extra pair of brackets makes:

```<code>tan(x) + tan(y) = (sin(x+y))/cos(x)*cos(y);

                     sin(x + y) cos(y)
   tan(x) + tan(y) = -----------------
                          cos(x)

tan(x) + tan(y) = (sin(x+y))/(cos(x)*cos(y));

                      sin(x + y)
   tan(x) + tan(y) = -------------
                     cos(x) cos(y)
</code>```

And now,

```<code>eq1 := tan(x) + tan(y) = (sin(x+y))/(cos(x)*cos(y)):

is(eq1);

             true
</code>```

For your second equation, note that the well-known constant is spelled Pi in Maple. The lowercase name pi has no special meaning to Maple.

You might try the solve command, as one way to reformulate the implied relationship between x and y. Eg,

```<code> eq2 := arcsin(x) + arcsin(y) = Pi/2:

 solve(eq2);

                    2     1/2
     {x = x, y = (-x  + 1)   }
</code>```

This forum is for mathematics, and questions about Maple syntax and programming are better suited to stackoverflow.com or www.mapleprimes.com (the Maplesoft user community).

[edit] You provided a revision to your second equation in a comment to another answer.You gave it as,

```<code>eq2 := arcsin(x) + arccos(x) = Pi/2:
</code>```

Note the conversion,

```<code>convert(arcsin(x), arccos);

     1/2*Pi - arccos(x)
</code>```

And so these all work here.

```<code>convert((lhs-rhs)(eq2),ln);

           0

convert((lhs-rhs)(eq2),arcsin);

           0

convert((lhs-rhs)(eq2),arccos);

           0
</code>```

Body:
You may have mistyped tg(...) when you actually mean tan(...), where tan is the name of a trigonometric function in Maple.

In that case you seem to be missing brackets in the right hand side of the first equation. Note the difference in the output that the extra pair of brackets makes:

tan(x) + tan(y) = (sin(x+y))/cos(x)*cos(y);

                     sin(x + y) cos(y)
   tan(x) + tan(y) = -----------------
                          cos(x)

tan(x) + tan(y) = (sin(x+y))/(cos(x)*cos(y));

                      sin(x + y)
   tan(x) + tan(y) = -------------
                     cos(x) cos(y)

And now,

eq1 := tan(x) + tan(y) = (sin(x+y))/(cos(x)*cos(y)):

is(eq1);

             true

For your second equation, note that the well-known constant is spelled Pi in Maple. The lowercase name pi has no special meaning to Maple.

You might try the solve command, as one way to reformulate the implied relationship between x and y. Eg,

 eq2 := arcsin(x) + arcsin(y) = Pi/2:

 solve(eq2);

                    2     1/2
     {x = x, y = (-x  + 1)   }

This forum is for mathematics, and questions about Maple syntax and programming are better suited to stackoverflow.com or www.mapleprimes.com (the Maplesoft user community).

[edit] You provided a revision to your second equation in a comment to another answer.You gave it as,

eq2 := arcsin(x) + arccos(x) = Pi/2:

Note the conversion,

convert(arcsin(x), arccos);

     1/2*Pi - arccos(x)

And so these all work here.

convert((lhs-rhs)(eq2),ln);

           0

convert((lhs-rhs)(eq2),arcsin);

           0

convert((lhs-rhs)(eq2),arccos);

           0


问:
Need help with proving that $(6n^2 -1/ (2n^2 + 5n))$ is a Cauchy Sequence by using definition

Using the definition of the Cauchy Sequence, show that $\forall\varepsilon\gt 0, \exists n_0\in\mathbb N$ s.t. $\forall m\gt n\ge n_0$, $\vert x_m - x_n\vert\le\varepsilon$.

I have tried to estimate it from above and got $\dfrac{12m^2}{4m^2}$, but in that case, the $m$ terms in the numerator and the denominator get canceled.

Body:
For
$$ x_n = \frac{6n^2-1}{2n^2+5} = \frac{3(2n^2+5)-16}{2n^2+5} = 3 - \frac{16}{2n^2+5}$$
we have, for $m>n\ge n_0$
$$ |x_n-x_m| = \left|\frac{-16}{2n^2+5} - \frac{-16}{2m^2+5}\right| = \frac{32(m^2-n^2)}{(2n^2+5)(2m^2+5)} $$
using
$$ m^2-n^2 < m^2 $$
$$ 2m^2+5 > 2m^2 $$
$$ 2n^2+5 > 2n^2 \ge 2n_0^2 $$
we get then
$$ |x_n-x_m| < \frac{32m^2}{(2n^2+5)2m^2} = \frac{16}{2n^2+5} < \frac{8}{n_0^2}$$
Therefore, for any $\epsilon>0$, if we take $n_0 >\sqrt{8/\epsilon}$ we have
$$ |x_n-x_m| < \frac{8}{(\sqrt{8/\epsilon})^2} = \epsilon$$


问:
"Infinite Sperner's Theorem"
Sperner's theorem gives the size of a maximal antichain in $\mathcal{P}(X)$ where $X$ is a finite set.

In this, "antichain" means a set of mutually incomparable subsets (i.e. subsets such that for any two $A,B$ of them, neither $A\subseteq B$ nor $B\subseteq A$).

There must be a generalization of this to infinite sets: how is the cardinality of the maximal antichain in $\mathcal{P}(X)$ related to the cardinality of $X$? In particular, for infinite $|X|$, is it $2^{|X|}$? Is it a known statement? Is it perhaps trivial?

答:
Assuming the axiom of choice, if $X$ is infinite then $\mathcal{P}(X)$ has an antichain (with respect to inclusion) of size $2^{\vert X\vert}$ (which is obviously as large as possible).

Here's one way to prove this. For $A\subseteq X$, let $$\hat{A}=\{(x,1): x\in A\}\cup\{(x,0): x\not\in A\}.$$ Note that if $A$ and $B$ are distinct subsets of $X$, the sets $\hat{A}$ and $\hat{B}$ are inclusion-incomparable subsets of $X\times\{0,1\}$.

Now via the axiom of choice, since $X$ is infinite there is a bijection $f: X\times\{0,1\}\rightarrow X$. We use $f$ to "push through" the hat construction above to get a family of subsets of $X$. Specifically, for $A\subseteq X$ let $A'=\{f(i): i\in \hat{A}\}.$ It's easy to check that the family of subsets of $X$ $$\{A': A\subseteq X\}\subseteq\mathcal{P}(X)$$ is an antichain with respect to $\subseteq$ and has cardinality $2^{\vert X\vert}$.

Title: "Infinite Sperner's Theorem"


问:
How to effectively learn by self learning.
I'm a self studying student.I live in a place where I've limited resources,to give an example,in winters to study I've to use candles because there is no electricity,no light and so on.
I have no teacher so all that I have is books and sometimes internet,if the internet is not banned .

I'm average in studying but I love to study and try my best to understand. The thing that is the biggest hurdle is as explained below:

I'm going through a text smoothly and then I get stuck.I have a confusion, doubts and questions. I try to think for hours, and then they turn into days.
I can't ask anyone near me and I feel I can't move ahead though the text. I feel I haven't understood the previous paragraphs so my motivation to go ahead is dead.
The only way I know out is to accept the conclusion that's presented and move on.
But I am not able to do it. I have this irresistible  urge to know it before I can move ahead.

Could you please tell me how to proceed in such a situation.

Should I note my doubts and move on and try to clear them when possible?

I will be thankful for even a hint that might help me or any other general tips.
Warmest regards, Thank you.

答:
I'm going through a text smoothly and then I get stuck.I have a confusion, doubts and questions. I try to think for hours, and then they turn into days. I can't ask anyone near me and I feel I can't move ahead though the text. I feel I haven't understood the previous paragraphs so my motivation to go ahead is dead. The only way I know out is to accept the conclusion that's presented and move on. But I am not able to do it. I have this irresistible urge to know it before I can move ahead.

There are around three main reasons that I've identified as to why we get stuck/ have confusions. I've ranked them on the order of likeliness:

Not going over examples/ Illustration on the topics I am doing

Most of the times, the reason I don't get some thing is because I don't have a clear example of the subject I am handling with relate to in my mind. So, either checking back of the book exercises or trying to conjure your own examples of where a theorem can be applied will help.

Lack of some 'prior' knowledge

Sometimes it is like we can't even begin to understand the material we are doing, like we feel a strong disconnect between what we are reading and what we know. This usually means, you've skipped a lot of the basics/ pre - requisites for the subject you are doing. Try find a self contained introduction to the topic you want to study/ do a bit more watered down version before doing the complicated stuff.

The materials explanation is not clear

This is an unlikely case (if you searched up what books are good), but can happen. Sometimes it is that the book is just not clear enough to convey the information to you. In this case, look up on youtube / google. If you can't find answers to your queries in that way, look up questions on MSE/ ask on MSE.

I personally am not a big believer for keeping notes in mathematics and all the notes I tried to take for it ended up being waste paper. I think a better way to do it is to keep thinking of what you are studying, and try to write down your thoughts on a piece of paper. Even if you don't get the intended result that you wanted, just do a rough work and save the paper.

The one exception I have for this is when engaging completely new material, then I take a paper and write down important definitions with some cartoons so I don't have to flip back to see them again if I forgot.

I started writing donw my math thoughts like these on an app called obsidian over the months and it has helped me quite a bit in thinking of new ideas and figuring out good questions to ask about the subject.

The kind of self studying while trying to understand everything at some X level is a hard one but very rewarding. Good luck!


问:
Six ducks in a pond
This puzzle has got me stumped.</p>

<p>$6$ ducks are swimming on a pond radius $5$. Show that at any moment there are two ducks a distance at most $5$ apart. </p>

答:
If one duck sits at the center of the pond we are done. Otherwise each duck has a well defined argument (polar angle), and we can number the docks cyclically according to increasing arguments. In this way we have
$$\arg(D_i)-\arg(D_{i-1})\geq0,\quad \sum_{i=1}^6 \bigl(\arg(D_i)-\arg(D_{i-1})\bigr)=2\pi\ .$$
It follows that $\arg(D_i)-\arg(D_{i-1})\leq{\pi\over3}$ for at least one $i\in[6]$, and since a circular sector of radius $5$ and central angle ${\pi\over3}$ has diameter $5$ it follows that $|D_i-D_{i-1}|\leq5$ for such $i$.</p>

Title: Six ducks in a pond


问:
How common are integer solutions to this problem?
Consider the problem of finding the smallest number $x \in \mathbb{R}$ that satisfies the constraint $x \lfloor x \rfloor \geq S$ where $S$ is an integer (of course, $\lfloor x \rfloor$ simply means $x$ rounded down).

Often, solving this problem will yield an integer solution for $x$. For example, if $S = 100$, then the solution is $x = 10$ (since $x \lfloor x \rfloor = 10\lfloor 10 \rfloor = 10^2 \geq 100$). Similarly, if $S = 99$, then the solution remains at $x = 10$ (clearly, $x = 10$ satisfies the inequality; and if $x < 10$, then $x \lfloor x \rfloor < 10\cdot9 = 90 < 99$ so the inequality is violated). But for other values of $S$, e.g. $S = 101$, the solution is not an integer (it is $x = 10.1$.)

Question: Suppose we solve this problem for every possible value $S = 0, 1, 2, ...$, each time updating the proportion of solutions $p$ that are integer solutions. Is there some way to evaluate the limit of $p$ (as $S \rightarrow \infty$)?

答:
Let $S\geq0$ and let $n:=\lfloor\sqrt{S}\rfloor$ so that
$$n^2\leq S<(n+1)^2.$$
If $S<n(n+1)$ then $x=\frac{S}{n}$ satisfies $\lfloor x\rfloor =n$ and hence $x\lfloor x\rfloor=S$.

If $S\geq n(n+1)$ then $x=n+1$ is the smallest real number satisfying $x\lfloor x\rfloor\geq S$.

In particular this shows that $x$ is an integer if and only if
$$n(n+1)\leq S\leq(n+1)^2,$$ for some positive integer $n$. From here it is not difficult to show that the limit equals $\tfrac12$.


答:
If $n(n-1)\leq S\leq n^2$ for some integer $n\geq 2$ then the answer is an integer (in fact, is the integer $n$). This is because for any $x<n$ we have $\lfloor x\rfloor\leq n-1$ and so $x\lfloor x\rfloor< n(n-1)$.

Conversely, if $(n-1)^2<S<n(n-1)$ the answer is not an integer, since $\lfloor S/(n-1)\rfloor=n-1$ so $x=S/(n-1)$ works (but $n-1$ doesn't).

There are $2n-1$ values of $S$ with $(n-1)^2<S\leq n^2$ for each $n$. Of these, $n+1$ work and $n-2$ don't. These intervals cover all natural numbers. So the limit will be $1/2$.

Title: How common are integer solutions to this problem?

Body:
Consider the problem of finding the smallest number $x \in \mathbb{R}$ that satisfies the constraint $x \lfloor x \rfloor \geq S$ where $S$ is an integer (of course, $\lfloor x \rfloor$ simply means $x$ rounded down).

Often, solving this problem will yield an integer solution for $x$. For example, if $S = 100$, then the solution is $x = 10$ (since $x \lfloor x \rfloor = 10\lfloor 10 \rfloor = 10^2 \geq 100$). Similarly, if $S = 99$, then the solution remains at $x = 10$ (clearly, $x = 10$ satisfies the inequality; and if $x < 10$, then $x \lfloor x \rfloor < 10\cdot9 = 90 < 99$ so the inequality is violated). But for other values of $S$, e.g. $S = 101$, the solution is not an integer (it is $x = 10.1$.)

Question: Suppose we solve this problem for every possible value $S = 0, 1, 2, ...$, each time updating the proportion of solutions $p$ that are integer solutions. Is there some way to evaluate the limit of $p$ (as $S \rightarrow \infty$)?


问:
Limit of an definite integral

Calculate
$$\lim\limits_{n\to\infty}\int_0^1 \frac{x^n(ax^2+ax+1)}{e^x}.$$

a) $0$

b) $a$

c) $2a+1$

d) $\dfrac{2a+1}{e}$

I tried to note separated $\displaystyle\int_0^1 \frac{x^n\cdot x^2}{e^x}$ with $I_{n+2}$ and to integrate it by parts until I have it in the function of $I_{n+1}$ and the Same $I_{n+1}$ in the function of $I_n$ but I got that the limit is $\dfrac{a+1}{e}$ and it is not there.

答:
$\textbf{Hint: }$Let $f_n(x)=\dfrac{x^n(ax^2+ax+1)}{e^x}$, with $x\in[0,1]$.

If $x$ is fixed in $[0,1]$, then we have
$$\lim_{n\to \infty}f_n(x)=\begin{cases}0&&\text{if }x\in[0,1)\\\dfrac{2a+1}e&&\text{if }x=1  \end{cases}$$
so we have pointwise convergence of $f_n$ in $[0,1]$ to $f(x):=\lim_{n\to\infty}f_n$. Does $f_n$ uniformy converge in $[0,1]?$

Is $\lim_{n\to\infty}\underset{x\in I}\sup|f_n-f|=0$?

If the answer is positive, we have
$$\lim_{n\to\infty}\int_0^1f_n(x)dx=\int_0^1\lim_{n\to\infty}f_n(x)dx=\int_0^1f(x)dx$$

Title: Limit of an definite integral


问:
Modulo question with negative
$60-88  \equiv \,\,? \pmod 5$

$60-88 = -28$

Then what do I do?

Please tell me how to answer this question. Thanks.

答:
As Key Ideas pointed in the comments...

$60-88  \equiv \,\,x \pmod 5$

$-28 \equiv \,\,x \pmod 5$

$-28+30 \equiv \,\,x+30 \pmod 5$

$2 \equiv \,\,x \pmod 5$

$x=5n-2$ with $n\in\mathbb{Z}$

Title: Modulo question with negative


问:
Terminology to describe that, eg, $ax^2+bx+c=0$ cannot be solved by simply moving terms around to isolate $x$
For example, a quadratic equation in the form
$$ax^2+bx+c=0$$
cannot be solved by simply "moving" the terms around; we wouldn't be able to reduce $x$ to just one term without introducing new terms.

Is there a way to describe this? Say I'm trying to teach the derivation of the quadratic formula. How would I explain that we must introduce new terms, construct a perfect square trinomial, factor it, etc. rather than simply attempt to solve it via basic algebraic manipulation?

答:
The following older (mostly before 1900) usage may be of interest. Equations of the form $ax^2 + c = 0$ such that $a \neq 0$ used to be called pure quadratic equations (sometimes incomplete quadratic equations) and equations of the form $ax^2 + bx + c = 0$ such that $a \neq 0$ and $b \neq 0$ used to be called affected quadratic equations (sometimes complete quadratic equations). See this google books search.

Footnote * on p. 280 of College Algebra by Edward A. Bowser:

1888 1st edition at google books and 1893 reprint of 1st edition at internet archive

The term adfected, or affected, was introduced by Vieta, about the year 1600, to distinguish equations which involve, or are affected with, different powers of the unknown quantity from those which contain one power only.


问:
How to prove that $\sum_{k=1}^m \cos^2(\frac{2 \pi k}{m}) = \frac{m}{2}$ for $m>2$

How to prove that for $m>2$ $$\sum_{k=1}^m \cos^2\left(\frac{2 \pi k}{m}\right) = \frac{m}{2}.$$

This could be useful for me solving a Differential Geometry problem. And Wolfram Alpha says it's true. But i can only be certain with a proof

答:
Note that $\cos^2 z=\frac{1+\cos2z}{2}$. Substituting gives us
$$\sum_{k=1}^m \cos^2\left(\frac{2 \pi k}{m}\right) = \frac{m}{2}+\frac{1}{2}\sum_{k=1}^m \cos\left(\frac{4 \pi k}{m}\right).$$
Also
$\operatorname{Re}e^{iz}$ is $\cos z$. Your problem can be rewritten to
$$\frac{1}{2}\sum_{k=1}^m \operatorname{Re}\left(\exp\left(\frac{2 \pi k}{m}\right)\right) = \frac{1}{2}\operatorname{Re}\sum_{k=1}^m\exp\left(4 \pi i\frac{k}{m}\right)=\operatorname{Re}\frac{1-\exp\left(4\pi i\right)}{1-\exp\left(4\pi i\frac{k}{m}\right)}=0.$$
Now you may use the geometric progression.


问:
Hi! I'm having a problem with the integral $\int_{n}^{n+1}x^2 \,dx$
Basically, I noticed that when $n\subset  \mathbb{N}$, $\displaystyle\int_{n}^{n+1}{x^2}\,dx$ gives a prime number divided by 3 or a multiplication of prime numbers.
Here's a table i made.

Is this something already known or a "function that generates prime numbers"?
I don't know much about math and I don't have a great academic background so I'm asking this here.

答: All numbers are factorizable into some prime numbers by the prime factorization theorem</p>