问:
Asymptote for $y=\frac {x^2}{kx+1}$
We want to find the asymptote of
$$y=\frac {x^2}{kx+1}$$
as $x\to \infty$.
By synthetic division or binomial expansion, we arrive at $$y=\frac 1k \left(x-\frac 1k\right)$$
which is the correct answer.

However we could also have divided top and bottom by $x$ to get:
$$y=\frac x{k+\frac 1x}$$
and conclude that as $x\to\infty$, $\frac 1x \to 0$, hence $y\to \dfrac kx$ which is close but not the right answer. How do we reconcile this method with the ones mentioned earlier?

Thanks.

答:
Aim:

Find the best linear approximation $y=mx+b$ to the function $y=f(x)$ for large $x: $

1)The second option gives

$y=mx$, where $m=(1/k)$, which is a linear approximarion for large $x$

2)A better linear approximation of

$y=(1/k)x(1+1/(kx))^{-1}=$

$(1/k)x(1-1/(kx)+O(1/x^2))$

is given by

$y=(1/k)(x-1/k)$, where you retain the constant term.

Title: Asymptote for $y=\frac {x^2}{kx+1}$


问:
How to find domain of definition of solution?
For the given initial-value problem:
$$\dfrac{dy}{dt}=\dfrac{1}{\left(y+2\right)^{2}}, \quad y(0)=1$$
we are asked to solve it then to state the domain of definition of the solution. So first of all I separated the variables and applied the initial condition and obtained:
$$y(t)=\left(3t+27\right)^{1/3}-2$$

but now I can't figure out why does the solution exist only when $t>-9$? I am having trouble understanding also what does this have to do with $\dfrac{dy}{dt}$ not being defined at $y=-2$. Why are we letting
$$3t+27 >0 $$

what if $t=-10$? how does this make $y$ undefined?

答:
It is perhaps not stated often enough outright, but in standard ODE theory, where the "O" stands for "ordinary", the domain of the ODE is supposed to be an open set that is chosen so that the right side exists, and is continuous in all variables. A solution is then a continuously differentiable function with a graph inside that set.

In this case this means that the line $y=-2$ is a boundary of the domain, and as the initial condition is $y(0)=1$, the domain is the upper part $\{(x,y):y>-2\}$. As you computed, the solution only stays inside this domain for $t>-9$. There can be no more solution in the given context.

Now it is of course quite possible to consider limits of the solution at the boundary, and solutions in the second domain $\{(x,y):y<-2\}$, pairs of solutions that can be joined together at the boundary, or result from the same algebraic expression. But these are extended solution concepts, which need to be requested explicitly.

ContentLicense: CC BY-SA 4.0


问:
Simple example to apply the SLLN
I'm trying to find concrete examples of the SLLN theorem. Before, let's see the statement of this theorem precisely from this book, page 81:

Definition: We say that $X_n$ converges almost surely to $X$, written $X_n\xrightarrow{a.s.} X$, if

$$\mathbb{P}(\{s:X_n(s)\to X(s)\})=1$$

Theorem (The Strong Law of Large Numbers). Let $X_1,\ldots X_n$ be IID. If $\mu=\mathbb E|X_1|\lt \infty$ then

$$\bar X_n\xrightarrow{a.s.}\mu$$

I'm thinking about for example the $X\sim Bern(1/2)$, such as the coin experiment where we can define as $X(T)=0$ and $X(H)=1$.

The way I understand the theorem is it states for every $s$ in the outcome from the experiment (except the ones with $P(s)=0$), we have $\bar X_n(s)\to \mu$.

I can't see this in my example. Let's start with $s=H$. Since we already know $\mu =1/2$, we have:

$$\bar X_n(H)=\frac{X_1(H)+\ldots+X_n(H)}{n}=\frac{1+\ldots+1}{n}=1\nrightarrow 1/2$$

as $n$ goes to infinity.

Where am I wrong?

答:
Given that the proof of SLLN is well known and you can find it in a lot of books (Here, for example, you can find a very basic proof) I would like to make you think about an intuitive reasoning:

SLLN in your example can be stated in the following way: For $n$ large, calculate the following probability

$$\mathbb{P}\Bigg[\lim\limits_{n}\overline{X}_n=\frac{1}{2}\Bigg]=?$$

Say: toss the fair coin $10,000,000$ times: what is the probability you observe a numbers of H $\rightarrow 5,000,000$?

...I think it is intuitive that this probability is 1.

SLLN provides a proof of this intuition

Title: Simple example to apply the SLLN


问:
What is the fastest way to solve this very difficult polynomial question?
If $x^3 +3px^2 + 3qx + r = 0 $ has a double root, show that the double roots must be $\frac{pq - r}{2(q-p)^2}$

Now the solutions provide an extremely complicated proof, that involves letting the double root be $\alpha$, and then proceeding to do various substitutions and equating. Does anyone know of a method that is actually reasonable and simpler?

答: Hint;if $a$ is the double roots of $f(x)$ then $$f(a)=0 ,f'(a)=0$$


问:
Using $\mathbb{E}(1/X)$ and $1/\mathbb{E}(X)$ to prove $X=Y$ a.s.?
Consider two r.v.s $X,Y\geq1$ defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ which may not necessarily be integrable. Let $\mathbb{E}[X\mid \mathcal{G}]:=\lim_{n\to\infty}E[X\wedge n\mid\mathcal{G}]$ where $\mathcal{G}$ is any sub-$\sigma$-algebra of $\mathcal{F}$. I have $\mathbb{E}(X\mid Y)=Y$ and  $\mathbb{E}(Y\mid X)=X$. I need to prove the following: i) $\mathbb{E}\left(\frac{X}{Y}\right)=\mathbb{E}\left(\frac{Y}{X}\right)=1$; and ii) $X=Y$ a.s. by comparing $\mathbb{E}(1/X)$ and $1/\mathbb{E}(X)$. I also need to somehow use these to show that if instead $X,Y\geq-1$, $X=Y$ a.s.

I have read through the proofs outlined in here but the conditions for this case are not the same as the one here (since we cannot assume integrability anyway). What should I do in this case? I don’t really know to go about both cases. Thanks!

答:
Suppose that $\mathbb E\left[\frac{X}{Y}\right]=\mathbb E\left[\frac{Y}{X}\right]=1$.

Since $U=\frac{X}{Y}> 0$ is integrable and $x\rightarrow \frac{1}{x}$ is strictly convex on $]0,+\infty[$, Jensen inequality says that
\begin{align*}
\mathbb E\left[\frac{1}{U}\right]&\geq\frac{1}{\mathbb E[U]}
\end{align*}
with equality if and only if $U$ is almost surely constant (and so $U=\mathbb E[U]~a.s.$). By the previous inequality, we know that $\mathbb E[U]=1$ and $\mathbb E\left[\frac{1}{U}\right]=\mathbb E\left[\frac{Y}{X}\right]=1$ so we have equality. This means that $\frac{X}{Y}=\mathbb E[U]$ almost surely and so we indeed get $X=Y$ almost surely.

We can compute
\begin{align*}
\mathbb E\left[\frac{X}{Y}\middle| Y\right]
&= \lim_{n\to \infty} \mathbb E\left[\frac{X}{Y}\land n\middle| Y\right]\\
&= \frac{1}{Y}\lim_{n\to \infty} \mathbb E\left[X\land (Yn)\middle| Y\right]\\
&= \frac{1}{Y}\lim_{x\to \infty} \mathbb E\left[X\land x\middle| Y\right]\\
&= \frac{1}{Y}\mathbb E[X|Y]\\
&= 1
\end{align*}
almost surely. We change the limit from $n$ to $x$ by multiplying by $Y\geq 1$, observe that $X\land (Yn)$ and $X\land x$ will converge to the same limit almost surely by conditional dominated convergence theorem (actuelly they converge to $X$). However there may be some integrability conditions I missed for the conditional dominated convergence theorem.


问:
Integral with Bessel function and exponential
I have a very direct question.

Is there an explicit solution for this integral:

$$\int_a^b e^{-\frac{x^2}{2\sigma^2}}I_0\left(\frac{x\cdot z}{\sigma^2}\right)\mathrm{d}x$$

where $z,\sigma^2\in\mathbb{R}^+$ and $I_0$ is the 0-order modified Bessel function of first kind?

Thanks in advance.

Body:
$$J=\int e^{-\frac{x^2}{2\sigma^2}}\,I_0\left(\frac{x\, z}{\sigma^2}\right)\,dx$$

Let $x=\frac{\sigma ^2 y}{z}$ and $k=\frac{\sigma ^2}{2 z^2}$ to make
$$J=\frac {\sigma ^2 }{z} \int e^{-k y^2}\,I_0(y)\,dy$$

Now using the infinite series
$$I_0(y)=\sum_{n=0}^\infty \frac{2^{-2 n}}{(n!)^2}\, y^{2n}$$

we face a series of integrals
$$J_n= \int e^{-k y^2}\,y^{2n}\,dy=-\frac{1}{2} k^{-\frac{2n+1}{2}} \Gamma \left(n+\frac{1}{2},k y^2\right)$$

For example, let $z=\frac 12$ and $\sigma=\frac 1{\sqrt 2}$ to make the simple
$$J=\int_0^1 e^{-x^2}\, I_0(x)\,dx=\sum_{n=0}^\infty \frac{ \Gamma \left(n+\frac{1}{2},0\right)-\Gamma
   \left(n+\frac{1}{2},1\right)}{2^{2 n+1}\,(n!)^2}$$ which converge very fast. For the partial sums up to $n=p$
$$\left(
\begin{array}{cc}
 p  & \text{approximation} \\
 0 & 0.7468241328 \\
 1 & 0.7941922193 \\
 2 & 0.7957589192 \\
 3 & 0.7957878829 \\
 4 & 0.7957882194 \\
 5 & 0.7957882221 \\
 6 & 0.7957882221
\end{array}
\right)$$


问:
Why irreduciblilty and not division leads to relative primeness in a PID
I'm unable to understand reasoning behind following conclusion: If $p$ is irreducible element of PID $R$ and $p$ doesn't divide $a$ of R. Then $p$ and $a$ are relatively prime.

The definition of a irreducible element $p$ is

It's non-invertible

If $p=ef$ then one of $e$ and $f$ is invertible

My problem is following reasoning $p\not|a$ doesn't imply $e\not|a$ and if $e$ does divide $a$ and since it divides $p$ (from def). Wouldn't then the GCD of $p$ and $a$ be $e$ instead of $1$? So they won't be relatively prime.

答:
Recall that - generally - in a domain a gcd is defined only up to associates, i.e. up to unit factors, In particular: $\,p,a\,$ are coprime means $\,\gcd(p,a)\approx 1,\,$ i.e. $\,d\mid p,a\iff d\mid 1\iff d\,$ is a unit.

The sought result is a bit more intuitive when viewed slightly more generally. By the Lemma below, in a PID, $\,p\,$ is irreducible $\iff (p)\,$ is a maximal ideal. Generally if $P$ is a maximal ideal we have

$$\begin{align} P\,\ {\rm max},\ P\not\supset A\,&\Rightarrow\, P\!+\!A = (1)\\[.2em]
{\rm so}\ \ p\,\ {\rm irred,}\ \ \ p\,\nmid\, a \,&\Rightarrow\, (p,\,a) = (1),\ \text{for $\,P\!=\!(p),\ A\!=\!(a)\,$ in a PID}\end{align}\qquad$$

therefore we conclude that $\ (1) = (p,a) = (\gcd(p,a)),\,$ so $\,p,a\,$ are coprime.

Lemma $ $ For principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e.  $\,\color{#0a0}{(a)\supseteq (b)}\iff \color{#c00}{a\mid b},\,$   thus having no proper $\rm\color{#0a0}{containing}$ ideal (maximal) is the same as having no proper $\rm\color{#c00}{divisor}$ (irreducible), $ $ i.e.

$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal}
&\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\
&\iff&\  p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\
&\iff&\  p\ \ \text{ is irreducible}\\
\end{eqnarray}$

Remark $ $ See also Euclid's Lemma in Bezout form, gcd form, and ideal form.


问:
Evaluating a limit with variable in the exponent
For
$$\lim_{x \to \infty} \left(1- \frac{2}{x}\right)^{\dfrac{x}{2}}$$
I have to use the L'Hospital"s rule, right? So I get:
$$\lim_{x \to \infty}\frac{x}{2} \log\left(1- \frac{2}{x}\right)$$<br>
And what now? I need to take the derivative of the log, is it:
$\dfrac{1}{1-\dfrac{2}{x}}$  but since there is x, I need to use the chain rule  multiply by the derivative of $\dfrac{x}{2}$  ? </p>

答:
We know, $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$

Put $n=-\frac x2, \left(\lim_{x\to\infty}\left(1-\frac 2x\right)^{-\frac x2}\right)^{-1}=e^{-1}$

ContentLicense: CC BY-SA 3.0


问:
Can we define a set containing all sets within ZFC and then prove that it does not exist?
I am currently reading "Elements of Set Theory" by Herbert B. Enderton. The first theorem proven in the book states:

There is no set to which every set belongs.

But isn't it the case that phrase "set to which every set belongs" cannot even be formulated within a theory. According axiom schema of specification (also called the axiom schema of separation or of restricted comprehension) a new set can be defined only using the following formula:

$\forall z \forall w_1 \forall w_2\ldots \forall w_n \exists y \forall x [x \in y \Leftrightarrow (( x \in z )\land \phi )]$

In other words, any new set $y$ can be defined only as a subset of some other set ($z$ in the formula). So, how can we define a set containing all the other sets under this restriction?

So, my concern not about the prove that "set containing all sets" does not exist but about the fact that we prove existence (non-existence) of something that cannot be formulated in terms of the language of theory. It looks like we try to prove within ZFC that "cats" exist. ZFC does not know what is a "cat".

答:
The natural-language statement "there is no set of all sets" is formalized in set theory as $$\neg\exists x\forall y(y\in x),$$ or perhaps more readably $$\forall x\exists y(y\not\in x).$$

(Keep in mind that in $\mathsf{ZFC}$ all things are sets, so "is a set" is just a dummy phrase.)

You seem to be confused about the role of the axioms, e.g. when you write

According axiom schema of specification (also called the axiom schema of separation or of restricted comprehension) a new set can be defined only using the following formula [...].

Axioms are not creative, they are descriptive: they don't in any sense generate the objects that exist, they merely tell us how those objects must behave. The separation axioms don't say that every set has to be created in a certain way, they just describe certain types of set which are guaranteed to exist. And the axioms certainly don't affect what we can define - that's just a language matter, it has nothing to do with the axioms at play.

The following might help:

We can express the property of being a universal set in the language of set theory by the formula $$\upsilon(x):\quad \forall y(y\in x).$$ The $\mathsf{ZFC}$ axioms subsequently imply that no universal set exists, that is, we have $$\mathsf{ZFC}\vdash\forall x\neg\upsilon(x).$$ Different axiom systems may yield different specific results - e.g. the alternative set theory $\mathsf{NFU}$ proves that there is a universal set, and $\mathsf{NFU}$ has strictly weaker consistency strength than $\mathsf{ZFC}$.


问:
When the function $y(x)=f(x)^{g(x)}$ is defined?
Suppose we have the function $y(x)=f(x)^{g(x)}$. should $f(x)>0$ in order to $y(x)$ being defined? and why?

For example if we have $y=(\sin x)^x$ what happens when $x\in(-1,0)$ or $x\in(\pi,\frac{3\pi}{2})$ ?

I am confused.

答:
If working strictly in the reals:
$$f^g=e^{g\ln f}$$
and then we see that $f(x)>0$ is necessary for the exponentiation to be defined in any interval larger than isolated points ($(-1)^x$ is only defined at integral $\mathbb Z$ for example).

If complex numbers are allowed, such a restriction disappears, though the result may be multivalued.


问:
Integrating left to right versus right to left.
OK, I understand that when integration is done left to right with respect to x increasing left to right (dx is positive), that the answer is positive, and vice versa when integrating right to left. But, as a reality check, there is something that I have never understood (or, understanding, I find so counterintuitive as to want to question it): Interpreting the result as area, one and the same area is either positive or negative not depending on what it is, but on how it is counted. I recognize that it is a logical extension of, say, -1 versus +1 on the number line, but at least -1 and +1 occupy different positions! Whenever I've brought this up, I tend to get blank looks, like, "how could you be so stupid", or "huh", but, clearly it does not keep me from being bothered by it, so I'll ask here ... once again! To me, apart from what the mechanics of integration require, a more logical "positioning" of the negative area would be its reflection across the x-axis.</p>

答:
I will try to explain it for the Riemann-Intgral:
If you look at the definition of R-Integral you will notice that the term $(a-b)$ plays an important role. Namely $(a-b)/N$ gives you the length of the intervalls over wich the values of the function are avaluated and then summed. Here we are integrating over $[a,b]$. If we invert the direction of integration the previous expression changes to $(b-a)/N$ wich is equal to $-(a-b)/N$. Informally the Integral is given as  the value taken in the interval $(a-b)/N$ times the Value that it takes in a precise point inside that interval. Hence if we invert $a$ and $b$ these values will be multiplied by $-(a-b)/N$ and hence the Integral with  $a$ and $b$ inverted becomes negative</p>


问:
Help in understanding this, convergence in measure.
Let $(X,A,\mu)$ be a measurable space.
$g,f_1,f_2\ldots \in L^1(\mu)$ , $g\geq 0$.
$f:X\to C$ measurable.
Suppose $f_n$ converges to $f$ in measure ($\mu$) and $\forall n\in N |f_n|\leq g$.

Prove that :

$lim_{n\to \infty} \int_{X} f_n d\mu = \int_X f d\mu$.

I'm trying to show this with using the dominated convergence theorem of
Lebesgue since we have almost all the conditions .
But we have to show that $f_n$ converges to $f$ almost everywhere.
From the given information we can conclude that $f_n$ is a couchy sequence So by another theorem there is a sub sequence $f_{n_j}$ such that $f_{n_j}$ converges to $f$ almost everywhere.
I don't know how to finish it.Though I know that there is a theory in calculus which states:
$a_n$ converges iff  every subsequent of $a_n$ has a subsequent that converges.
But I don't know how to use it.

Can you help in this, and explain how it can be showed.

答:
Prove it by contradiction. Suppose it is not true that $\int f_n d\mu \to \int f d\mu$. Then there exists $\epsilon >0$ and integers $1 \leq n_1<n_2<....$ such that $|\int f_{n_k} d\mu- \int f d\mu| \geq \epsilon$ for all $k$ $\,\,\, \cdots(1)$.

Now $(f_{n_k})$ converges to $f$ is measure because $f_n \to f$ in measure. Hence $(f_{n_k})$ has  a subsequence $(f_{n_{k_i}})$ converging almost everywhere to $f$. By DCT we get $\int f_{n_{k_i}} d\mu \to \int fd\mu$ But this contradicts (1).


答:
Assuming that we're restricting ourselves to the reals:

It is not quite necessary to have $f(x) > 0$, but close. In particular:

If $g(x) \in \mathbb{Z}\setminus \{0\}$ for all $x$, then $f(x)^{g(x)} = f(x)$ is well-defined, regardless of what $f(x)$ is.

If $g(x) \in \mathbb{Z}$ for all $x$, then $f(x)^{g(x)}$ is well-defined unless there is $x$ such that $f(x) = g(x) = 0$.

These are essentially the only exceptions, for the reason that the other answer gives.

Title: When the function $y(x)=f(x)^{g(x)}$ is defined?

Body:
Suppose we have the function $y(x)=f(x)^{g(x)}$. should $f(x)>0$ in order to $y(x)$ being defined? and why?

For example if we have $y=(\sin x)^x$ what happens when $x\in(-1,0)$ or $x\in(\pi,\frac{3\pi}{2})$ ?

I am confused.


问:
What does this bar notation mean?
I have a problem in which I need to solve for $x$ in the question (in image) below, but I don't even understand it. Is this suppose to be a matrix?
$$\begin{vmatrix}\ln(2x+1)&2-\ln x\\1&1\end{vmatrix}=0$$

答: The bar notation means taking the determinant of the matrix enclosed. This is a shorthand for $\det M$.

Title: What does this bar notation mean?


问:
$\mathbb{Z}[2\sqrt2]$ is not a UFD.
Dummit and Foote q9.3.1

Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $\text{Prove that if a(x)} \not \in \text{R[x] then R is not a U.F.D.}$ Deduce that $\mathbb{Z}[2\sqrt2]$ is not a U.F.D.

How to do the last part?

$\mathbb{Z}[2\sqrt2]$ is integral domain. $F$ be its quotient field. To find $p(x)\in \mathbb{Z}[2\sqrt2][x](=\mathbb{Z}[2\sqrt2, x])$,    $a(x) \not \in \mathbb{Z}[2\sqrt2, x]$,       $a(x), b(x) \in F[x]$ of smaller degree than $p(x)$ s.t. $p(x) = a(x)b(x)$. $deg_x(p(x)) \geq 2$ because it can't be $0$ or $1$.

Please give a hint how to find $p(x), a(x), b(x)$. Please do not give solution.

Thanks!

This is a direct solution which is maybe not expected here.

答:
Hint: Think about $x^2+2\sqrt 2 x + 2$.
Full solution:

 Let $p(x)= x^2+2\sqrt 2x +2$

 Then, $a(x)=b(x)=x+\sqrt 2 \not \in \mathbb Z[2\sqrt 2,x]$.

Title: $\mathbb{Z}[2\sqrt2]$ is not a UFD.


答:
Recall the limit:
$$\lim_{y \to \infty} \left(1+\dfrac{a}y\right)^y = e^a$$
I trust you can finish it from here, by an appropriate choice of $a$ and $y$.</p>

Title: Evaluating a limit with variable in the exponent

Body:
For
$$\lim_{x \to \infty} \left(1- \frac{2}{x}\right)^{\dfrac{x}{2}}$$
I have to use the L'Hospital"s rule, right? So I get:
$$\lim_{x \to \infty}\frac{x}{2} \log\left(1- \frac{2}{x}\right)$$<br>
And what now? I need to take the derivative of the log, is it:
$\dfrac{1}{1-\dfrac{2}{x}}$  but since there is x, I need to use the chain rule  multiply by the derivative of $\dfrac{x}{2}$  ? </p>


答:
As you've mentioned, a sequence $(x_n)$ in a topological space X converges to a point $x\in X$ if and only if any subsequence of $(x_n)$ has a further subsequence that converges to $x$.

Now, let $(f_n)_{n \in M \subset \mathbb N}$ be a subsequence of $ (f_n)_{n \in \mathbb N}$. We wish to show that there exists a further subsequence $(f_n)_{n \in L\subset M}$ such that
$$ \left (\int f_{n}  \right)_{n \in L} \to \int f .$$
Evidently, $(f_n)_{n \in M} $ is dominated by $g$ and it converges to $f$ in measure. Thus, there exists a subsequence $(f_n)_{n \in L \subset M}$ such that $(f_n(x))_{ n \in L}  \to f(x)$ for almost all x. By L.D.C theorem, we get the desired.

Title: Help in understanding this, convergence in measure.

Body:
Let $(X,A,\mu)$ be a measurable space.
$g,f_1,f_2\ldots \in L^1(\mu)$ , $g\geq 0$.
$f:X\to C$ measurable.
Suppose $f_n$ converges to $f$ in measure ($\mu$) and $\forall n\in N |f_n|\leq g$.

Prove that :

$lim_{n\to \infty} \int_{X} f_n d\mu = \int_X f d\mu$.

I'm trying to show this with using the dominated convergence theorem of
Lebesgue since we have almost all the conditions .
But we have to show that $f_n$ converges to $f$ almost everywhere.
From the given information we can conclude that $f_n$ is a couchy sequence So by another theorem there is a sub sequence $f_{n_j}$ such that $f_{n_j}$ converges to $f$ almost everywhere.
I don't know how to finish it.Though I know that there is a theory in calculus which states:
$a_n$ converges iff  every subsequent of $a_n$ has a subsequent that converges.
But I don't know how to use it.

Can you help in this, and explain how it can be showed.


问:
Convergence in probability implies convergence in $L^2 $
Let $(X_n)_n$ a sequence of gaussian r.v. such that for any $m$ and $ n$
$(X_n − X_m)$ is gaussian.
If $X_n \overset{P}{\longrightarrow X}$ then $X_n \overset{L^2}{\longrightarrow}X$ (more precisely in $ L^p ,\ p\in [1, +∞)$).

I tried to do this: convergence in probability implies convergence in law , and convergence in law of gaussians implies (Paul-Levy) that $ X \sim N(m,\sigma^2) $ with $m=\lim\limits_{n \to \infty} m_n $ and $\sigma^2=\lim\limits_{n \to \infty} \sigma^2_n$.

Convergence in probability also implies that the limit is a.s. unique.

Because $L^2$ is complete it is enough to prove that the sequence is Cauchy. Using the hypothesis :

$\int \lvert X_n-X_m \rvert^2 dP= \mathbb{E}[(X_n-X_m)^2]= (m_n-m_m)^2+(\sigma_n^2+\sigma_m^2)$

And i can find, fixed $\epsilon>0 , \hspace{0,3cm} n^* \in \mathbb{N} \hspace{0,3cm} $such that $\hspace{0,1cm} \forall m,n \geq n^*\hspace{0,4cm}  (m_n-m_m)^2+(\sigma_n^2+\sigma_m^2)<\epsilon \hspace{0,2cm}$ because of the limits above.

I would like to know where is the mistake and how i could fix it. I think i'm mistaken because the absolute moment equals the moment only if p is even, moreover it seems like i'm using convergence in probability only for the a.s. uniqueness of the limit. Any help would be appreciated.

答:
There is a mistake in your computation of $\mathbb E(X_n-X_m)^{2}$. It is not given that $X_n$ and $X_m$ are independent so it is not possible to compute this in terms of $m_n,m_m$ and $\sigma_n, \sigma_m$ at all!.

Here is a valid proof:   $X_n-X=\lim_{m \to \infty} (X_n-X_m)$ in probability. This implies that $X_n-X$ is Gaussian for each $n$. This, together with $X_n -X \to 0$ in probability implies that $\mathbb E(X_n-X)^{2} \to 0$.


问:
How to integrate $\int \sqrt{x^2+a^2}dx$
$a$ is a parameter. I have no idea where to start</p>

答:
I will give you a proof of how they can get the formula above. As a heads up, it is quite difficult and long, so most people use the formula usually written in the back of the text, but I was able to prove it so here goes.

The idea is to, of course, do trig-substitution.
Since $$\sqrt{a^2+x^2} $$ suggests that $x=a\tan(\theta)$ would be a good one because the expression simplifies to $$a\sec(\theta)$$

We can also observe that $dx$ will become $\sec^2(\theta)d\theta$.
Therefore$$\int \sqrt{a^2+x^2}dx = a^2\int \sec^3(\theta)d\theta$$

Now there are two big things that we are going to do.
One is to do integration by parts to simplify this expression so that it looks a little better, and later we need to be able to integrate $\int \sec(\theta)d\theta$.

So the first step is this. It is well known and natural to let $u=\sec(\theta)$ and $dv=\sec^2(\theta)d\theta$ because the latter integrates to simply, $\tan(\theta)$.

Letting $A = \int \sec^3(\theta)d\theta$,you will get the following $$A = \sec(\theta)\tan(\theta) - \int{\sec(\theta)\tan^2(\theta)d\theta}$$

$$=\sec(\theta)\tan(\theta) - \int{\sec(\theta)d\theta - \int\sec^3(\theta)d\theta}$$

therefore,$$2A = \sec(\theta)\tan(\theta)-\int \sec(\theta)d\theta$$

Dividing both sides give you $$A = 1/2[\sec(\theta)\tan(\theta)-\int \sec(\theta)d\theta]$$

I hope you see now why all we need to be able to do is to integrate $\sec(\theta)$.

The chance that you know how is rather high because you are solving this particular problem, but let's just go through it for the hell of it.

This is a very common trick in integration using trig, but remember the fact that $\sec^2(\theta)$ and $\sec(\theta)\tan(\theta)$ are derivatives of $\tan(\theta)$ and $\sec(\theta)$, respectively. So this is what we do.

$$\int \sec(\theta)d\theta = \int {{\sec(\theta)(\sec(\theta)+\tan(\theta))} \over {\sec(\theta)+\tan(\theta)}} d\theta$$

Letting $w = \sec(\theta)+\tan(\theta)$,

$$= \int {dw \over w} = \ln|w|$$

So, long story short,

$$\int \sqrt{a^2+x^2}dx = a^2/2[\sec(\theta)\tan(\theta) - \ln|\sec(\theta)+\tan(\theta)|]$$

$$= {x\sqrt{a^2+x^2}\over 2} + {{a^2\ln|x+\sqrt{a^2+x^2}|}\over 2} + C$$


问:
eigen value of the $n \times n $ Matrix $x y^T$
Let $x,y$ be two non zero $n \times 1$ vectors. If $y^T$ denotes the transpose of the matrix $y$, then what are the eigen value of the $n \times n$ matrix $x y^T$?

Attempt: Denoting $x= \big(x_1,\cdots,x_n \big)^T, y = (y_1,\cdots,y_n)^T$

Then, To find the eigen value of $xy^T,$ we solve the system : $xy^T v = \lambda v~:~v=\big(v_1,\cdots,v_n)^T \in \Bbb R^n$

Expanding, we obtain the following equation system:

$$x_1(y_1v_1+\cdots+y_nv_n)=\lambda v_1 \\x_2(y_1v_1+\cdots+y_nv_n)=\lambda v_2 \\.\\.\\.\\x_n(y_1v_1+\cdots+y_nv_n)=\lambda v_n \\$$

How does one move ahead and solve this system?

答:
The case $n=1$ is trivial. Let's see what happens for $n=2$ just to get a flavor:
$$
\begin{cases}
x_1(y_1v_1+y_2v_2)=\lambda v_1 \\[6px]
x_2(y_1v_1+y_2v_2)=\lambda v_2
\end{cases}
$$
We want to identify a nonzero eigenvalue $\lambda$. The system becomes
$$
\begin{cases}
(x_1y_1-\lambda)v_1+x_1y_2v_2=0 \\[6px]
x_2y_1v_1+(x_2y_2-\lambda)v_2=0
\end{cases}
$$
and we know from elementary arguments that we need
$$
(x_1y_1-\lambda)(x_2y_2-\lambda)-(x_2y_1)(x_1y_2)=0
$$
which easily yields
$$
\lambda^2-(x_1y_1+x_2y_2)\lambda=0
$$
So a nonzero eigenvalue, if any, has to be $\lambda=x_1y_1+x_2y_2$.

Now the extension should be evident, isn't it?

If we take $v=x$, then your system is satisfied with $\lambda=x_1y_1+x_2y_2+\dots+x_ny_n$. Since $x\ne0$, we're done.

Can there be other eigenvalues? Certainly so, for $n>1$, but with the restriction that they're $0$, because $xy^T$ has rank $1$ and so nullity $n-1$.

In a faster way: for three column vectors $u,v,w$, it holds $(uv^T)w=(v^Tw)u$, where $v^Tw$ is considered as a scalar. This identity is proved easily by direct computation. Take $u=w=x$ and $v=y$.


问:
Integrate $\log(x+i)/(x^2+1)$ from $-\infty$ to $\infty$
In the first part of the question I was asked to find a branch of $log(z+i)$ which is holomorphic on the complex numbers except the imaginary axis below $-i$. I think this is $\log(z+i)=\ln(z+i)+i\text{arg}(z+i), \text{ for } -\pi/2<\text{arg}(z+i)<3\pi/2$. That said I am having trouble with the concept of branch cuts so this may be wrong.

The hint is to integrate $\log(z+i)/(z^2+1)$. I set this equal to $\log(w)/w(w-2i)$ where $w=z+i$. Then taking an integral around $[-R,-k]\cup [k,R] \cup z=k e^{2\pi i \theta} \cup z=Re^{2\pi i \theta}$ for some very small $k$. Then the only pole I need to consider is $w=2i$. Can I use Residue Theorem here and let $k\to 0$ to do this?

Is this in any way the right approach? Any help very much appreciated!

答: Think I've got it - integrate over the semicircle in the positive plane, use the estimation lemma to show the integral over the arc = 0.</p>


问:
Show that the set of the given vectors form a basis in $\mathbb R^3$ and represent the standard basis as a linear combination of these vectors

Show that the vectors $\langle 1,0,-1\rangle$,$\langle 1,2,1\rangle$,$\langle 0,-3,2\rangle$ from a basis in $\mathbb R^3$.

Represent standard basis in $\mathbb R^3$ as a linear combination of the given vectors.

By the definition a nonempty  linearly independent subset of vectors $S=\langle v_1,v_2,...,v_n \rangle$  from a vector space $V$ is a basis of $V$ if it spans $V$, in other words if every element of $V$ can be represented as a linear combination of the vectors in $S$.

We've given three vectors in $\mathbb R^3$,so we are allowed to use the fact that  The columns of an invertible  matrix are linearly independent ,thus the matrix is given by:

\begin{pmatrix}
1 & 1 & 0 \\
0& 2 & -3 \\
-1 & 1  & 2
\end{pmatrix}

With a nonzero determinant,so it follows that the vectors are linearly independent.

Now it's needed to show that every vector $\langle x,y,z \rangle$ in $\mathbb R^3$ can be defined as a linear combination of the given vectors,let $\lambda_1,\lambda_2,\lambda_3$ be some scalars,then :

$$  \begin{bmatrix} 1 & 1 & 0 \\
0& 2 & -3 \\
-1 & 1  & 2 \\  \end{bmatrix}  \left[ \begin{array}{c}  \lambda_1  \\
  \lambda_2  \\
 \lambda_3 \end{array} \right]=
\left[ \begin{array}{c}   x   \\
   y \\
   z  \end{array} \right] $$

Reduce it by elementary row operations gives:

$$  \begin{bmatrix} 1 & 0 & 0 \\
0& 1& 0 \\
0 & 0  & 1 \\  \end{bmatrix}  \left[ \begin{array}{c}  \lambda_1  \\
  \lambda_2  \\
 \lambda_3 \end{array} \right]=
\left[ \begin{array}{c}   x'   \\
   y' \\
   z'  \end{array} \right] $$

Where $x',y',z'$ are expressed as $x,y,z$.

Thus for every vector $\langle x,y,z \rangle$ in $\mathbb R^3$ it's possible to find scalars for which the vector can be defined as a linear combination of the given vectors which implies that the vectors span $\mathbb R^3$ and so the set containing them is a basis.

Another way would be using the fact that The set of three linearly independent vectors in $\mathbb R^3$ is a spanning set.

For the second part I thought that we need to find scalars and solve an equation ,but I guess for each one of the three standard basis we need to solve one system of equations which takes much time,so what else can I do? Does there exist a faster method?

答:
Let's call your matrix $A$.

Showing $\det A\ne 0$ is indeed enough to get linear independence of the column vectors, and since they are three, they form a basis of $\Bbb R^3$.

For the second part, you're basically looking for the inverse of $A$, as its $i$th column contains the coordinates of the $i$th standard basis vector in the given basis, as we can read it from the matrix product $AA^{-1}=I$.

If you already performed a full row reduction to obtain $I$ from $A$, then the exact same steps in the same order, but starting out from $I$ will produce $A^{-1}$.

This is because every elementary row operation can be expressed by a left multiplication of a corresponding elementary matrix $E_i$, and thus we have $E_k\dots E_2E_1A=I$, so $E_k\dots E_2E_1I=A^{-1}$.


问:
This is a question in number theory. It seemed I can't start the proof.
How can this be solved?

I knew that the concept of an order of an integer is needed, but rather than that I think I'm really lost with the proofs.

Body: Hint: exploit innate symmetry: $\ \ \underbrace{\overbrace{-a^4 = 1^{\phantom{|}}_{\phantom{.}}\!\!}^{\!\!\!\!\textstyle \color{#0a0}{(-a^3)}a\!=\! 1}}_{\!\!\!\!\!\!\textstyle(\color{#c00}{-a^2})a^2\! =\! 1}\Rightarrow\ (a+\overbrace{a^{-1}}^{\!\!\textstyle \color{#0a0}{-a^3}})^2 = a^2+\underbrace{a^{-2}}_{\!\textstyle \color{#c00}{-a^2}}+ 2\, =\, 2$


问:
Can we use sine and cosine graphs for varying temperatures?
I have been researching for a while looking for what to include more than just deriving the equation of Newton’s law of cooling and a website gave me inspiration by stating what if the temperature of the environment is changing (± C). In conclusion, I wasn’t sure if that person carried out the mathematical process in the right way and the only place I could ask for accurate information is here so please let me know what you think about the mathematical process he went for and if it is right or wrong.

Thanks a lot.

Website: http://calculus7.com/sitebuildercontent/sitebuilderfiles/nwtnslawcling1.pdf

答:
Assuming that you refer to the equation
$$y'(t)=\frac{3}{20} \left(55+10 \sin \left(\frac{\pi  }{12}t\right)-y(t)\right)$$ we can integrate it.
$$y(t)=55+C \,e^{-3 t/20}+\frac{15 \left(162 \sin \left(\frac{\pi  }{12}t\right)-90 \pi
   \cos \left(\frac{\pi  }{12}t\right)\right)}{243+75 \pi ^2}$$
Using the initial condition $y(0)=45$, we have
$$C=\frac{450 \pi }{81+25 \pi ^2}-10$$


问:
Question related to diagonalization of a matrix.
Suppose we are given that A=\begin{bmatrix}4&11&20&30\\ 0&4&11&20\\ 0&0&4&11\\ 0&0&0&4\end{bmatrix}
I need to show that if F is a Fourier matrix then
$$F^{-1}AF=D$$ where D is a diagonal matrix containing eigen values of A.
I am doubtful before starting solving this problem. Clearly the matrix A is not diagonalizable.  Is this scenario possible?

答:
Circulant matrices are ensured to be diagonalizable by the Fourier matrix.

The matrix you have is not circulant, but it is Toeplitz.

You can follow this answer for how to construct a circulant matrix from a Toeplitz one.

https://math.stackexchange.com/a/3415198/213607

Depending on what you want to do, this might help you.


问:
Computing Conditional Probability: rolling a die
I am facing a problem like this:

We roll a die with 6 equiprobable outcomes. Compute the conditional probability:
Pr[outcome is odd ∣ outcome is even]

So by the question, I need to calculate the probability of outcome is odd under the condition outcome is even.

So,

outcome is odd is: 1, 3, 5

outcome is even is: 2, 4, 6

But, as I understand, the outcome is odd and the outcome is even are two events that don't relate to each other(you can only get outcome is odd or outcome is even in each roll).

Do I understand this question correctly? If not then please explain it for me. Thank you very much.
Sorry for bad English, I'm from non-English speaking country.

答:
the probability is obviously zero.

The possible outcomes are the following

$$\Omega=\{2;4;6\}$$

because "we know an even number has been drawn"

As you can see, there are no odd numbers available in your sample space.


问:
How to find the $b$ of a hyperbola equation?

It is easy to find the answer is either A or B, knowing the center coordinate $(2,2)$ and the value of $a$, which is $1$. However, I don't know how to calculate $b$ to find the equation.

Without finding $b$, though, I can still  get the right answer A by plugging in the origin that is passed by the hyperbola.

But I still hope someone can teach me how to derive $b$.

答:
Currently you have found $$\dfrac{(x-2)^2}{1^2}-\dfrac{(y-2)^2}{b^2}=1$$

Now use the condition that hyperbola passes through origin. Subbing $(x,y)=(0,0)$ in above will give value of $b$.

This is the standard way. Without using this condition, there is not enough information in the question to fix the hyperbola to be unique. There are infinitely many hyperbolas with vertices at the two points $(1,2)$ and $(3,2)$ only.

One needs minimum five points to uniquely determine a conic. Here we have $(0,0)$ and its reflections in the axes to fix our required hyperbola.


问:
Question about the theory of independent unary predicates and categoricity
I am studying the theory of independent unary predicates in a model theory textbook, and I am struggling to solve a problem that is left open in the notes. By the theory of independent unary predicates I mean the theory $T_n$ in language $L_n$ with $n$ unary disjoint predicates, each $P_i$ is satisfied by infinite number of elements from the domain, and also there are infinitely many elements in the domain that do not satisfy each $P_i$. Each $P_i$ is disjoint from all other $P_j$'s. I have shown that $T_n$ is $\aleph_0$-categorical, and from here, using the Los-Vaught Test, concluded that $T_n$ is complete. I now want to answer the following question: How many models of cardinality $\aleph_1$(up to isomorphism) does $T_2$ have?

I am really stuck on where to start with this, and any hints and help would be really appreciated. Thank you.

答:
The theory you are describing is not the theory of independent unary predicates, but rather, the theory of disjoint unary predicates (although it does not change the properties of the theory very much, as long as there are only finitely many of them).

To find the number of models, show that if $M, N$ are models of $T$, then they are isomorphic if and only if all the atoms in the Boolean algebra generated by the predicates have the same number of elements in $M$ and $N$.

For $n$ disjoint infinite predicates with infinite complement, this leads to the conclusion that there are $2^{n+1}$ models of cardinality $\aleph_1$.


问:
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$?
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$?
Please help!

Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint?

答: Since $f$ is supposed to be entire, it is continuous at $0$.  Let $n \to \infty$. Then we must have $f(0) = 0$.  Now let $n \to -\infty$. What is $f(0)$ now?</p>


问:
Show that the function $f(z)=\frac{1-\cos z}{z^n}$ has no anti-derivative on $\mathbb C\setminus\{0\}$ if $n$ is an odd integer.
Show that the function $f(z)=\frac{1-\cos z}{z^n}$ has no anti-derivative on $\mathbb C\setminus\{0\}$ if $n$ is a positive odd integer.

I think, If we can able to prove $\int_S f(z) dz \neq 0$, on the unit circle $S$, then by fundamental theorem of algebra gives the desired one. But, how can I use the definition $\int_\gamma f(z)dz=\int_a^b f(\gamma(t))\gamma'(t)dt$ for a continuous function $f(z)$ on a piecewise smooth contour $\gamma:=\gamma(t),~a \leq \gamma \leq b$ ?

答: Let $g(z)=1-\cos z$. Then\begin{align}\int_Sf(z)\,\mathrm dz&=\int_S\frac{g(z)}{z^n}\,\mathrm dz\\&=2\pi i\frac{g^{(n-1)}(0)}{(n-1)!}\end{align}which is $0$ if and only if $n$ is even.


问:
Sums of the form $S_k=\sum_{n\geq 1}\frac{1}{\sinh^{2k}(n \pi)}$ and the residue theorem.
Let us define the sums

$$S_k=\sum_{n\geq 1}\frac{1}{\sinh^{2k}(n \pi)}$$

A few days ago, i was able to give a surprisingly simple derivation of the fact
that
$$
S_1=\sum_{n\geq 1}\frac{1}{\sinh^2(\pi n)}=\frac{1}{6}-\frac{1}{2\pi}
$$
as an answer to this nice old question in terms of contour integration.
Being curious i immediatly tried to generalize this result to higher orders of $\sinh$, which as it turned out will not work that easily. For example, considering the case of $k=2$ it turns out that

$$\text{Res}\left(\frac{
\cot(\pi z)}{ \sinh^4(\pi z)},z=z_k\right)=
\begin{cases}
\frac{1}{\pi\sinh^4(\pi k)} \quad\text {if}\,\,k \in \mathbb{Z}/0,\\
-\frac{1}{\pi\sinh^4(\pi k)} -\frac{4}{3\pi\sinh^2(\pi k)} \quad\text {if}\,\,ik \in \mathbb{Z}/0
\end{cases}
$$

So integrating over a large quadratic contour gives us absolutly no information about $S_2$, since the residues cancel (For details of the exact procedure please have a look at the post linked above). This is in fact a simple consequence of the transformation properties of $\sinh^{2k}(z)$ along the imaginary axis: $\sinh^{2k}(i y)=(-1)^k\sin^{2k}(y)$

Looking at $S_3$ with the same approach, it turns out that it contains $S_2$ as well as $S_1$ so it is clear that we need this value to make any progress in a generalization at all.

I also want to mention that a Laplace/Mellintransform approach seem to suffer from the same cancellations then what i tried above, so this  is maybe also not the way to go...:/

My questions are

1.) Is there any chance to proof the result for $S_2$ (and $S_k$ in general if possible) using an an approach which is close to the one i used to calculate $S_1$ ?

2.) What is a general approach to derive calculate such sums using a minimum of knowledge about the realm of elliptic integrals, Jacobi theta functions etc. (where i'm sure they can be derived but i have far too less knowledge in this field to do this by my own)

PS:
It turns out that a closed form is, according to mathematica, given by

$$
S_2=-\frac{11}{90}+\frac{1}{3\pi}+\frac{\bar{\omega}^4}{7680
\pi^4}
$$
where $\bar{\omega}$ is the Lemniscate constant.

答:
($1$) IMHO there is little chance to compute $S_2$ using contour integration due to existence of the lemniscate constant in all $S_n, n\ge 2$ (see below).

($2$) Here is a general approach. The normalized Eisenstein series can be written as
$$G_{2k}(i)=\sum_{m,n\in\mathbb R, mn\not=0}\ \frac{1}{(m+ni)^{2k}}=2\zeta(2k)+2\sum_{n>0}\sum_{m\in\mathbb R}\frac{1}{(m+ni)^{2k}}$$
Where the inner sum can be computed through differentiating Mittag-Leffler expansion of $\cot$:
$$\sum_{m\in\mathbb R}\frac{1}{(m+ni)^{2k}}=\sum_{j=1}^{k}a_{k,j}\text{csch}^{2j}(n\pi)$$
Thus the following recurrence
$$G_{2k}(i)=2\zeta(2k)+2\sum_{j=1}^{k}a_{k,j}S_k$$
with well-known initial value $S_1=\frac{1}{6}-\frac{1}{2\pi}$ (see for instance J. Borwein's monograph Pi and AGM) allows us to compute all $S_n$ in terms of $G_{2k} (k=2,\cdots,n)$, all of which are expressible via products of $G_4(i)=\frac{\Gamma \left(\frac{1}{4}\right)^8}{960 \pi ^2}, G_6(i)=0$ (due to theory of modular forms). Therefore we conclude that $S_n$ lie in the algebra of $\mathbb Q$ generated by $\frac{1}{\pi}, \Gamma\left(\frac14\right)$, for instance
$$S_5=\frac{14797}{187110}-\frac{64}{315 \pi }-\frac{\Gamma \left(\frac{1}{4}\right)^{16}}{2580480 \pi ^{12}}-\frac{41 \Gamma \left(\frac{1}{4}\right)^8}{90720 \pi ^6}$$


问:
Law of Clavius explained
Law of Clavius states  $ \sim P \Rightarrow  P \vdash P$

And the only explanation I sort of understand is

"If we ought to philosophise, then we ought to philosophise; and if we ought not to philosophise, then we ought to philosophise (i.e. in order to justify this view); in any case, therefore, we ought to philosophise."

which to me reads as $(P \Rightarrow  P)\wedge (\sim P \Rightarrow P ) \vdash P   $

where P is "we ought to philosophise"
Which is not quite the same.

I am aware of the following relationship

$(\sim P \Rightarrow P ) \Leftrightarrow \; (\sim(\sim P)\; \vee \: P)$

and combining this with the above would basically be a form of "Or Elimination" or proof by cases. But this is just an explanation of how its proof would be worked out.

Anyway the above is more of a digression, but perhaps it may have some weight on my question. Really what I am interested in is an intuitive understanding of Clavius' law. What are some common examples in everyday logic?

答:
Clavius Law aka Consequentia Mirabilis is an instance of the more common Proof by Cases. Proof by Cases is the axiom schema:

/* Proof by Cases */

(¬ B → A) → ((B → A) → A)

If you now instantiate B = A, you get:

/* Clavius Law aka Consequentia Mirabilis */

(¬ A → A) → A

Interestingly you can go backwards and derive Proof by Cases from Consequentia Mirabilis as well. The interderivability
holds over minimal logic. The instantiation B = A is also mentioned here:

Aristotle and the Consequentia Mirabilis

William Kneale - 1957

https://www.jstor.org/stable/628635?seq=1

Title: Law of Clavius explained


问:
What does this bar notation mean?
I have a problem in which I need to solve for $x$ in the question (in image) below, but I don't even understand it. Is this suppose to be a matrix?
$$\begin{vmatrix}\ln(2x+1)&2-\ln x\\1&1\end{vmatrix}=0$$

答:
The bar notation is used to indicate the determinant of a matrix. It converts a matrix into a scalar. For example,
$$\begin{align}\begin{vmatrix} a & b\\c & d \end{vmatrix}=ad - bc \end{align}$$
Other common notations are $\det A$ and $\det (A)$.

In particular, for your case, the mentioned determinant becomes
$$\begin{align}\begin{vmatrix} \ln (2x+1) & 2-\ln x\\1 & 1 \end{vmatrix}=\ln (2x+1)-2+\ln x = \ln (2x^2+x)-2 \end{align}$$

Hope this helps :)

Title: What does this bar notation mean?


问:
Maximum interval of definition of $\frac{dy}{dt} = f(t, y) $, where $f \in C^1$, $y(t_0) = y_0 $
Given the equation
$$\frac{dy}{dt} = f(t, y) $$where $f \in C^1$, $y(t_0) = y_0 $ and assuming that $|y - y_0| \leq L|t - t_0|$ (1)  and  $|L| \leq  1$

By using the Picard Iterations and by saying that $f(t, y)$ is locally Lipschitz, could I say that the maximum interval of definition of the solution of the IVP is $\mathbb{R}$ ?

I have some difficulties in understanding the meaning of the condition 1. Could someone explain me this too?

Thanks in advance.

答:
This claim should have nothing to do with Picard iterations, only with the theorem about the local solution itself.

The corollaries about the maximal solution/maximal domain tell that the maximal domain is an open set and that the graph of the maximal solution leaves every bounded set inside the domain of the ODE (to both sides, "past" and "future"). So if you consider the box $[t_0-a,t_0+a]\times[y_0-La,y_0+La]$, then the solution can only leave this box to the sides, meaning that the maximal domain is larger than this interval. As this is true for any $a>0$, the maximal domain is indeed the whole of $\Bbb R$.

The condition $L\le1$ is not necessary.


问:
If $f_k \to f$ a.e. and the $L^p$ norms converge, then $f_k \to f$ in $L^p$
Let $1\leq p < \infty$. Suppose that

$\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite),

$f_k \to f$ almost everywhere, and

$\|f_k\|_{L^p} \to \|f\|_{L^p}$.

Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$

A statement in the other direction (i.e. $\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}$ ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though.

答:
Consider $g_k = 2^p(|f_k|^p + |f|^p) - |f_k - f|^p$.

Since $g_k \geq 0$ (why?), and $g_k \to 2^{p+1}|f|^p$ a.e., we can apply Fatou's Lemma:
$$\int \liminf g_k \leq \liminf \int g_k$$
so that
$$\int 2^{p+1}|f|^p \leq \liminf \left(\int 2^p |f_k|^p + \int 2^p |f|^p - \int |f_k - f|^p \right),$$
and I'll let you take it from here.


问:
Group extension reference request
I'm looking for a reference for the following "well known" result

Let  $C$ be an abelian group and $G$ a finite group, and  let $$0 \rightarrow C \rightarrow W \rightarrow G \rightarrow 0$$ be a group extension of $G$ by $C$ corresponding to the fundamental class in $H^2(C,G)$. Now the result I have says that if we have a homomorphism $f$ from $C$ to some abelian group  $A$ and an exact sequence of groups $$0 \rightarrow A \rightarrow A \rtimes G  \rightarrow G \rightarrow 0, $$ then you can extend $f$ to some map $$f': W \rightarrow A' \rtimes G$$ such that both these sequences fit into a commutative diagram with the first vertical arrow given by $f$ and the third given by the identity map if and only if $f$ is $G$ invariant and it "sends" (I don't quite get what this means) the fundamental class in $H^2(C,G)$ to zero.

The place I got this result from just gives a reference to Serre's Local fields but not a page number or anything that might narrow where this result is. I can imagine it is pretty standard but I cant seem to find anything of this sort.

Thank you

答:
We had better suppose that $A$ is abelian, and so a $G$-module; otherwise I'm not sure the question makes sense (because the $H^2$ with coeffs. in $A$ would not be defined).

Group cohom. is functorial in the coefficients, so $C \to A$ induces
$f_*: H^2(G,C) \to H^2(G,A)$.  This latter map has an interpretation in terms of
extensions: it sends the extension $0 \to C \to W \to G \to 0$ to the
extension $0 \to A \to W' \to G \to 0$ obtained by "pushing out" along the
morphism $C \to A$.  One way to think about $W'$ is that it is the unique extension of $G$ by $A$ compatible with given action of $G$ on $A$, and which
receives a map from $W$ that restricts to $f$ on $C$ and induces the identity on $G$.

Now by definition the $0$ class in $H^2$ corresponds to the split extension,
i.e. the semi-direct product.  Thus $f_*$ sends the class of $W$ (an element of $H^2G,C)$) to zero in $H^2(G,A)$
if and only if $W'$ is a semidirect product.

Now if you think about the above description of the push-out, you will see that $W'$ equals
the semidirect product if and only if $W$ admits a map to the semidirect product, compatibly with the given map $f: C \to A$ and the identity on $G$.

Title: Group extension reference request


问:
Show $ M_1 \# M_2= U_1 \cup U_2 $ for some open subsets $U_i$ of manifolds $M_i$ such that the following properties hold.
This is problem 4-19 of Lee's "Introdunction to topological manifolds" 2ed.

I want to show that for two connected $n$-manifolds $M_1$ and $M_2$, there are open subsets $U_1$ and $U_2$ of the connected sum $M_1 \# M_2$ and points $p_i \in M_i$ such that $U_i \cong M_i\setminus \{p_i\}$, $U_1 \cap U_2 \cong \mathbb R^n \setminus\{0\}$, and $U_1 \cup U_2 = M_1 \# M_2$ where an equivalence means homeomophism.

What I have tried:

By definition, $M_1 \# M_2$ is formed by attaching $M_1 \setminus B_1$ to $M_2 \setminus B_2$ along the boundary. I tried choosing $p_i$'s to be the center of the coordinate balls $B_i$. Then if the equivalences were homotopy equivalences, the statement seems to be true.

Any help is appreciated.

答:
Set up notation:

Let $B_i\subseteq M_i$ be open discs and $N_i$ normal neighbourhoods of the boundaries. Let $\varphi:N_1\to N_2$ a diffeomorphism that induces the gluing map (ie $\varphi$ maps the inward collar of $N_1$ to the outward collar of $N_2$ and vice versa). Let $V_i$ be the points of the open discs not in the normal neighbourhoods, note $V_i$ is a closed disk.

Then $M_1\#M_2$ is equal to $M_1 - V_1$ and $M_2-V_2$ glued along the normal collars, ie $M_1\#M_2$ is defined as the quotient space
$$(M_1- V_1) \cup_\varphi (M_2- V_2):= [(M_1-V_1)\cup (M_2-V_2)]/\{x\sim \varphi(x)\}$$

Let $U_i$ then be the image of $M_i- V_i$ in the quotient space. This is open since the quotient map necessarily is an open map. Further it is diffeomorphic to $M_i-V_i$ since this quotient map is a local diffeomorphism and it is injective on this set. The intersection $U_1\cap U_2$ is then the image of both normal neighbourhoods $N_1, N_2$ under the quotient mapping - infact these two neighbourhoods have the same image so it suffice to just take the image of $N_1$. But the quotient map is a local diffeomorphism and injective on $N_1$, hence $U_1\cap U_2\cong N_1$.

Now prove:

For any connected manifold $M$ if you remove a closed disk the result is diffeomorphic to $M-\{p\}$ for a point $p$.

A normal neighbourhood of the boundary of an $n$-dimensional ball is diffeomorphic to $\Bbb R^n - \{0\}$.


问:
Minimum specificity for a diagnostic test
I'm hoping somebody here can help with an argument me and my father have been having recently regarding Covid-19 and the "false positive paradox".

My father has been making probability trees similar to the guy in this video.

https://youtu.be/6WuTNMleuQI

My father has been trying to work out how accurate the Covid-19 PCR tests are if you return a positive test.

My father has been using 94.9% specificity which I argue is not mathematically possible in the U.K.

My argument is that with 42,000,000 tests and 1,600,000 positives returned, the minimum specificity of the test must be at least 96.2% and is likely to be much higher because that would assume all of the tests returned are incorrect.

He's arguing that these aren't "real numbers" and that they're only "probabilities" I think my problem is that I don't have the necessary mathematical knowledge to explain to him where I think he's mistaken.

Who is correct here? Or are we both mistaken?

Thanks in advance.

答:
Let's set up a frequency table.

$$\begin{array}{|c|c|c|c|}
\hline
& \text{Infected} & \text{Not Infected} & \text{Total} \\
\hline
\text{Test Positive} & \text{True Positive} = p_1 - n_1 & \text{False Positive} = n_1 & p_1\\
\hline
\text{Test Negative} & \text{False Negative} = p_2 - n_2 & \text{True Negative} = n_2 & p_2\\
\hline
\text{Total} & (p_1 + p_2) - (n_1 + n_2) & n_1 + n_2 & p_1 + p_2 \\
\hline
\end{array}$$

All variables are counts.  The specificity is the true negatives divided by the total of those not infected; i.e., $$\text{Specificity} = S = \frac{n_2}{n_1 + n_2}.$$
We are given $p_1 = 1.6$ (in millions) and $p_1 + p_2 = 42$. Thus $p_2 = 42 - 1.6 = 40.4$.  We want to determine the smallest $S$ that allows the table to be filled correctly.  This means maximizing $n_1$ subject to the above constraints.  To see why, say $n_1 < p_1$.  Then for any fixed $n_2$, we have $$n_1 + n_2 < p_1 + n_2,$$ which in turn implies $$\frac{1}{n_1 + n_2} > \frac{1}{p_1 + n_2},$$ and $$\frac{n_2}{n_1 + n_2} > \frac{n_2}{p_1 + n_2},$$ thus showing that choosing $n_1 = p_1$ minimizes $S$ for a fixed $n_2$.  This is simply the formal mathematical way to state what you already stated:  the specificity is least under the condition that all of the positive results are assumed to be false positives.

Next, we observe that minimizing $n_2$ under the given constraints once $n_1 = p_1$ is selected, gives the minimum $S$.  If every negative result were false, that is to say, $n_2 = 0$, then the specificity is $0$.  The given data does not restrict this choice because we can fill in the table as

$$\begin{array}{|c|c|c|c|}
\hline
& \text{Infected} & \text{Not Infected} & \text{Total} \\
\hline
\text{Test Positive} & \text{True Positive} = 0 & \text{False Positive} = 1.6 & 1.6 \\
\hline
\text{Test Negative} & \text{False Negative} = 40.4 & \text{True Negative} = 0 & 40.4 \\
\hline
\text{Total} & 40.4 & 1.6 & 42 \\
\hline
\end{array}$$

Of course, this is hardly a realistic result.  It would mean that the test has perfectly incorrect determination--if you test positive, you're actually uninfected; and if you test negative, you're actually infected.  If among the negative results half are true and half are false, then

$$\begin{array}{|c|c|c|c|}
\hline
& \text{Infected} & \text{Not Infected} & \text{Total} \\
\hline
\text{Test Positive} & \text{True Positive} = 0 & \text{False Positive} = 1.6 & 1.6 \\
\hline
\text{Test Negative} & \text{False Negative} = 20.2 & \text{True Negative} = 20.2 & 40.4 \\
\hline
\text{Total} & 20.2 & 21.8 & 42 \\
\hline
\end{array}$$

and the specificity would be $S = 0.926606$ assuming that all positive results are false.

What is the proportion of negative results that are false (i.e. failed to detect infection) for a specificity of $94.9\%$, assuming that all of the positives are false?  We would need to first solve for $$0.949 = \frac{n_2}{1.6 + n_2}$$ which gives $n_2 = 29.77$.  Then the detection failure rate is $$\frac{p_2 - n_2}{p_2} = 1 - \frac{29.77}{40.4} = 0.263056.$$  This means, in order for the specificity to be as low as what your father claimed, all of the positive results would need to be false positives, and at least $26.3\%$ of the negative results would need to be false.  This suggests that the test specificity should be substantially higher.

However, let's visit your father's assertion, which amounts to claiming that the observed frequencies are basically random realizations of an underlying probability distribution with some true underlying sensitivity and specificity.  While theoretically valid, due to the extremely large sample size, this point is inadequate to address the claimed specificity as I have illustrated.  If we consider the observed data as a point estimate for the true specificity, it remains highly unlikely that the true specificity is that low even with unreasonable assumptions that all of the positive results and over a quarter of the negative results are all false.


问:
volume between two surfaces
In my textbook they solve this problem like this, Find volume between two surfaces $\quad$  $x^2+y^2-1\leq z \leq 2+\frac{2}{3}(x^2+y^2), \ \ z\geq0$

They have called the first region $D_1=x^2+y^2-1=0$, and second region $D_2=x^2+y^2=9$ and then they solve this problem with two integral
$$\iint _{D_1}   \left(2+\frac{2}{3}(x^2+y^2)-0\right) dxdy + \iint_{D_2} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right)dxdy$$

My question is that why they counted the first integral? and why they divided the region into two parts?

Thanks in advance

答:
Please see a $2D$ sketch of the region (projection on $XZ$ or $YZ$ plane) and what each integral is finding.

What you see shaded on the left is what the first integral is finding. What you see shaded on the right is what the second integral is finding.

Since it needs to find the volume between solids above $XY$ plane ($z \geq 0$), it should have set up the first integral for $x^2 + y^2 - 1$ instead of for $(2 + \frac {2}{3}(x^2+y^2))$ to remove the volume below $XY$ plane.

So the correct integral would be -

$\displaystyle \iint _{D_1} (x^2 + y^2 - 1) \, dx \, dy + \iint_{D_2} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right) \, dx \, dy$

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (x^2 + y^2 - 1) \, dx \, dy = -\frac{\pi}{2}$

$\displaystyle \int_{-3}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right) \, dx \, dy = \frac{27 \pi}{2}$

Adding them gives you $13 \pi$ as the volume of the desired region.

EDIT:

As per the integral setup you sent, this is what the textbook is doing.

Title: volume between two surfaces


问:
Constructing Distribution By Coin Flipping
I am interested in any example of construction distribution by coin flipping.

Actually I want to show the process of construction a random variable $X$ with distribution $(p_1,...,p_n)$ by coin flipping  and to prove expectation of the number of coin flipping is at least entropy $H(X)$.

So far I have no idea where to start from. Will appreciate for examples and hits.

答:
I'm not sure exactly what you're trying to ask here.  You'll probably be able to find this answer in a textbook somewhere.  If you were asking something else it would be helpful if you gave a little more detail in your question.

Suppose $p_i< 2^{-k}$.  If I toss a coin $k$ times I have $2^k$ possible events each of probability $2^{-k}$.  So I cannot generate an event of probability $p_i$ with $k$ coin flips.

Therefore if my generator returns $X=i$ it must have flipped the coin at least $-\log_2(p_i)$ times.

So if $K$ is the number of coin flips before my algorithm returns a value we must have
$$\mathbb E(K|X=i) \geq \log_2(p_i).$$

Now we just add these up
$$\begin{array}{rcl}\mathbb E(K) &=& \sum_{i=1}^n \mathbb E(K|X=i)\mathbb P(X = i) \\ &\geq& \sum_{i=1}^n -\log_2(p_i)p_i \\ &=&\mathcal H(X).\end{array}$$

In reality it can take much longer.  If I have a random variable with $p_1 = 0.50001$ $p_2 = 0.49999$ I need a very large number of coin flips to simulate that exactly, even though the entropy is pretty much one.

The easiest way to generate your random number is to think of an infinite string of coin flips as a random number.

If you replace every Head with a $1$ and every Tail with a $0$, Then a sequence of coinflips
$THTHHT\dots$ would represent a binary number $0.010110\dots\equiv 0.17187$.

Let's make this rigorous.

Set $$\begin{array}{rcl}
d_k &=& \left\{\begin{array}{rl} 2^{-k} &\text{if the $k$th toss is a head} \\0 &\text{if the $k$th toss is a tail}\end{array}\right. \\
X_i &=& \sum_{k=1}^i d_i \\
X &=& \sum_{k=1}^\infty d_i \\
q_i&=& \sum_{k=1}^i p_i \\
&=& \mathbb P(M\leq i).
\end{array}$$

Now we want to set $m=i$ whenever $p_{i-1} < X < p_{i}$
So if we toss the coin $k$ times we know that $X_k < X < X_k + 2^{-k}$ so we stop tossing the coin the first time $k$ such that $p_{i-1} <X_k <  X_k + 2^{-k} < p_{i}$ for some $i$.


问:
soft question: solution manual for Demidovich Calculus
Context: I'm taking multivariable calculus this semester and my professor takes exercises from Problems in Mathematical Analysis by Demidovich - a book from a few decades ago, written by russian mathematicians. The problems in this book are of much higher difficulty than, say, Thomas Calculus (probably the most used textbook for calculus courses). I'm finding the course pretty difficult and it would be useful to have a solutions manual. However, there isn't any official one, so I thought I'd ask here if anyone has worked on the book and has solutions to the problems.

Thanks

答: A Spanish version is circulating over there. You may try this: [https://www.academia.edu/29687679/Solucionario_demidovich_Analisis_Matematico_1_ByPriale].</p>


问:
Question on Subring test and Subfield test
The question goes like this:

$P = \{x - 2^{1/3}y - 2^{2/3}z : x, y, z \in \mathbb{Z}\}$

Part(i) prove that $P$ is a subring of the ring of real numbers.

Part(ii) prove that $P$ is a subfield of the field of real numbers.

My attempt:

I use the subring test for part (i), I had already shown that addition and multiplication binary operations are both closed in $P$.

Now I am confused by the remaining criteria(s) I need to show for completing the subring test.

Some abstract algebra books said that I only need to show that subtraction and multiplication is closed in $P$.

While some youtuber said I need to show the additive identity of the ring of real number is in $P$.

While still some others said that I have to show that $P$ is an abelian subgroup and show both associate and distributive laws exists in $P$.

So I don't know exactly what I need to show for subring test.

For part (ii) for showing subfield, I have no clue yet, please help.

Thank you very much!

答:
To show that $P$ is a subring, we need to show that it is a ring. That is: we need to show that:

Addition in $P$ is associative.

Addition in $P$ is commutative.

Multiplication in $P$ is associative.

Multiplication is left and right distributive over addition.

For all $x, y \in P$, there is $x + y \in P$.

For all $x, y \in P$, there is $xy \in P$.

$0 \in P$

For all $x \in P$, there is $-x \in P$.

$1 \in P$ (see note).

Note: opinions vary on whether rings must have multiplicative identities. I like them to, so I've included it.

I've listed them in that slightly odd order for a reason: the first four are all immediate: they're all true in all of $\mathbb{R}$, and don't have any existential quantification going on, so they're true in $P$ as well.

So we actually only have those last five things to check, and we can, if we like, just go through and check them all separately. However, we can be lazy:

If $P$ is closed under subtraction and nonempty, then we have some $x \in P$, so we have $0 = x - x \in P$ (so we don't need to check $0 \in P$ separately), and also $0 - x = -x \in P$ (so we don't need to check that $P$ has additive inverses separately), and finally for any $y \in P$, we have $y - (-x) = y + x = x + y \in P$ (so we don't need to check closure under addition separately. However, $0$ is very often the easiest thing to show that $P$ contains, so very often we'll do that, then subtraction, and have all of the addition axioms sorted.

So to finish the job off, we only need to check that $1 \in P$ (if we like our rings to have identities) and $xy \in P$ for all $x, y \in P$.

However, we can be even lazier: we can look ahead, and check that it's a subfield, which will then immediately imply that it's a subring.

We could check that it's a subfield by just doing the above, then checking the extra field axioms (commutativity of multiplication is immediate as with the first four above, we now definitely need $1 \in P$ if we didn't already, we need to check multiplicative inverses, and we need $1 \neq 0$, but that's also immediate from $\mathbb{R}$).

We could, again, just check both of the new things that need checking. However, we can now do the same thing as we did with addition above, and just check that $P$ is closed under non-zero division and that $P$ contains something other than $0$, and get the rest for free (we then have $x \in P \setminus \{0\}$, and so $1 = xx^{-1}$ and $x^{-1} = 1x^{-1}$, and finally $xy = y(x^{-1})^{-1}$ for all $y \in P$).


问:
Arithmetic progression topology
I'm being asked to prove that $\mathcal{A}=\left\lbrace A_{a,b}:a,b\in\mathbb{Z}\right\rbrace$ is a basis for a topology on $\mathbb{Z}$, being:
$$A_{a,b}=\left\lbrace a+nb:n\in\mathbb{Z}\right\rbrace=\left\lbrace ...,a-2b,a-b,a,a+b,a+2b,...\right\rbrace$$

Obviously every $a\in \mathbb{Z}$ belongs to a basis element, for example $A_{a,b}$, for any $b$. I'm having difficulties though, proving that if two of those sets $A_{a,b}$ and $A_{a',b'}$ share an element $x$, then there exists another $A_{c,d}$, such that $x\in A_{c,d}$ and $A_{c,d}\subset A_{a,b},A_{a',b'}$.
Any hint?

答:
To avoid the space becoming discrete, we must not allow $b = 0$. Also, $A_{a,b} = A_{a,-b}$. Hence it is sufficient to take $b \in \mathbb{Z}^+$. Finally, if $a > b$, then by taking $c = a \mod b$, we see that $A_{a,b} = A_{c,b}$. Thus the basic open sets are of the form $A_{a,b}$, $(b \in \mathbb{Z}^+, 0 \le a < b)$. Or in other words, the basic open sets are of the form $b\mathbb{Z} + a$.

Next given $A_{a,b}, A_{a',b'}$, we have to find $A_{c,d} \subset  A_{a,b} \cap A_{a',b'}$. Let $z,w \in A_{c,d}$. Then $z,w \in A_{a,b} \Rightarrow b \mid (z-w)$ and similarly, $ b' \mid (z-w)$. Thus $lcm(b,b') \mid (z-w)$. So, we take $d = lcm(b,b')$.

Next we have to find $c$. As noted we shall find $c$ such that $0 \le c <d$. Let $x$ be the smallest nonnegative integer in $A_{c,d}$. Then $x \in A_{a,b} \Rightarrow b \mid (x - a)$. This gives $x \cong a \mod b$. Similarly, $x \cong a' \mod b'$. Using the Chinese remainder theorem, we get $c$. Note that in Chinese remainder theorem, either there is no solution or there are infinitely many solutions. Here our requirement of proving that the given sets form an open base, forces that we have to find the solution, only when the two sets intersect. Thus the solution is guaranteed.

Title: Arithmetic progression topology


问:
Evaluating a limit with variable in the exponent
For
$$\lim_{x \to \infty} \left(1- \frac{2}{x}\right)^{\dfrac{x}{2}}$$
I have to use the L'Hospital"s rule, right? So I get:
$$\lim_{x \to \infty}\frac{x}{2} \log\left(1- \frac{2}{x}\right)$$<br>
And what now? I need to take the derivative of the log, is it:
$\dfrac{1}{1-\dfrac{2}{x}}$  but since there is x, I need to use the chain rule  multiply by the derivative of $\dfrac{x}{2}$  ? </p>

答:
If you insist in working with the l'Hospital:

$$\lim_{x\to\infty}\frac x2\log\left(1-\frac2x\right)=\lim_{x\to\infty}\frac{\log\left(1-\frac2x\right)}{\frac2x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac2{x^2}\frac1{1-\frac2x}}{-\frac2{x^2}}=-\lim_{x\to\infty}\frac1{1-\frac2x}=-1$$

Thus, the limit is $\,e^{-1}\,$ ...


问:
Big-O notation and polynomials
In my text, I am given that the sum of the first n positive integers can be understood in terms of big-O notation.

''Since each of the integers in the sum of the first $n$ positive integers does not exceed $n$'', we can write:

$$1 + 2 + \cdots + n \leq n + n \cdots + n= n^2$$

Why does $n + n +\cdots + n = n^2$ ?

答:
1+2+3+...+n is always less than n+n+n...+n n times. you can rewrite this n+n+..+n as n*n.

f(n) = O(g(n)) if there exists a positive integer n0 and a positive
constant c, such that f(n) ≤ c * g(n) ∀ n ≥ n0

since Big-Oh represents the upper bound of the function, where the function f(n) is the sum of natural numbers up to n.

now, talking about time complexity, for small numbers, the addition should be of a constant amount of work. but the size of n could be humongous; you can't deny that probability.

adding integers can take linear amount of time when n is really large.. So you can say that addition is O(n) operation and you're adding n items. so that alone would make it O(n^2). of course, it will not always take n^2 time, but it's the worst-case when n is really large. (upper bound, remember?)

Now, let's say you directly try to achieve it using n(n+1)/2. Just one multiplication and one division, this should be a constant operation, no?
No.

using a natural size metric of number of digits, the time complexity of multiplying two n-digit numbers using long multiplication is Θ(n^2). When implemented in software, long multiplication algorithms must deal with overflow during additions, which can be expensive. Wikipedia

That again leaves us to O(n^2).

Title: Big-O notation and polynomials


问:
Show that the image of a finite ring is also finite, when the map is a surjective ring homomorphism
Can you help me solve iv) in the following problem?

I believe I'm supposed to use the results from i), ii) and iii).

But why can't I just say that $R/I$ is a finite ring from ii), and since $\varphi$ is surjective we have $\varphi(R/I)=R/J$ which must be finite ring, when $R/I $ is finite?

答:
To elaborate on my comment. We have an ideal $J \subset R = \mathbb{Z}[i]$, and we wish to construct another ideal $I \subset J$ of the form $I = nR$ for some $n \in \mathbb{Z}$.

We know $\mathbb{Z} \cap J$ is a nonzero ideal of $\mathbb{Z}$, hence is of the form $(n)$ for some $n \neq 0$.

By the above $n\mathbb{Z} \subset J$ so we know that $I := nR \subset J$ (you should convince yourself of this using the axioms of an ideal). But then we can use the fact that $R/I$ is finite and surjects onto $R/J$.

Note the difference with your comment - $J \cap \mathbb{Z}$ is not an ideal of $R$, you have to throw in some extra stuff.


Title: Prove $\gcd\left((a^{2m}−1)/(a+1),a+1\right)=\gcd(a+1,2m)$

Body:
Show or prove that
$$
\gcd\left(\frac{a^{2m}−1}{a+1},a+1\right)=\gcd(a+1,2m),
$$
and that
$$
\gcd\left(\frac{a^{2m+1}+1}{a+1},a+1\right)=\gcd(a+1,2m+1).
$$</p>

Body:
Hint

$$\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+...a^2+1)$$
$$=(a+1)(a^{2m-2}+a^{2m-4}+...a^2+1)-2(a^{2m-2}+a^{2m-4}+...a^2+1)$$
$$=(a+1)(a^{2m-2}+a^{2m-4}+...a^2+1)-2(a^{2m-2}-1+a^{2m-4}-1+...a^2-1+2m)$$
$$=(a+1)(a^{2m-2}+a^{2m-4}+...a^2+1)-2[(a^{2m-2}-1)+(a^{2m-4}-1)+...(a^2-1)]+2m$$

Hint 2 Each of $(a^{2m-2}-1),(a^{2m-4}-1),...,(a^2-1)$ is divisible by $a+1$.


问:
Convergence in sequence
in $\mathbb{R^2} $ $(a_n, b_n)\rightarrow(a, b) \Leftrightarrow a_n\rightarrow a, b_n \rightarrow b$

Show that in $d_2$.

How can I show that?

答:
Try to show that, for $N_2$ your norm in $\mathbb R^2$ and $N_1$ your norm in $\mathbb R$, there is a constant $C$ such that
$$C^{-1}(N_1(a) + N_1(b)) \le N_2((a,b)) \le C(N_1(a) + N_1(b))$$
and use that to show the equivalence.

Title: Convergence in sequence


问:
Solve $x''-(\tan t)x'+2x=0$

Solve $x''-(\tan t)x'+2x=0$ knowing that $x=\sin t$ is one of the solutions

I really have no clue how to do this one. The fact that one of the solutions is given makes it look a bit similiar to Riccati's equation, but I dont think thats the right track.

答:
Let $x=(\sin t)u$ ,

Then $x'=(\sin t)u'+(\cos t)u$

$x''=(\sin t)u''+(\cos t)u'+(\cos t)u'-(\sin t)u=(\sin t)u''+2(\cos t)u'-(\sin t)u$

$\therefore(\sin t)u''+2(\cos t)u'-(\sin t)u-(\tan t)((\sin t)u'+(\cos t)u)+2(\sin t)u=0$

$(\sin t)u''+2(\cos t)u'-(\sin t)u-\dfrac{\sin^2t}{\cos t}u'-(\sin t)u+2(\sin t)u=0$

$(\sin t)u''+\left(2\cos t-\dfrac{\sin^2t}{\cos t}\right)u'=0$

$(\sin t)u''=\left(\dfrac{\sin^2t}{\cos t}-2\cos t\right)u'$

$\dfrac{u''}{u'}=\tan t-2\cot t$

$\ln u'=-2\ln\sin t-\ln\cos t+c$

$u'=\dfrac{C_1}{\sin^2t\cos t}$

$u=C_1(\tanh^{-1}\sin t-\csc t)+C_2$

$\therefore x=C_1(\sin t\tanh^{-1}\sin t-1)+C_2\sin t$

Title: Solve $x''-(\tan t)x'+2x=0$


问:
Asymptote for $y=\frac {x^2}{kx+1}$
We want to find the asymptote of
$$y=\frac {x^2}{kx+1}$$
as $x\to \infty$.
By synthetic division or binomial expansion, we arrive at $$y=\frac 1k \left(x-\frac 1k\right)$$
which is the correct answer.

However we could also have divided top and bottom by $x$ to get:
$$y=\frac x{k+\frac 1x}$$
and conclude that as $x\to\infty$, $\frac 1x \to 0$, hence $y\to \dfrac kx$ which is close but not the right answer. How do we reconcile this method with the ones mentioned earlier?

Thanks.

答:
Your second analysis merely hints that the curve behaves like a straight line at infinity -- it doesn't give you its exact form. This is as a result of the fact that the analysis is quite crude.

For exact analysis, write the expression as $$\frac{x^2}{kx}\left(\frac{1}{1+1/kx}\right)=\frac xk\left(1-\frac{1}{kx}+\frac{1}{k^2x^2}-\cdots\right),$$ from where the exact form of the linear asymptote becomes clear.

Title: Asymptote for $y=\frac {x^2}{kx+1}$


问:
Conjecture: $a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}\leq 1$
Hi it's a conjecture inspired by a work of Vasile Cirtoaje :

Claim :

Let $0.5\geq a \geq b \geq c\geq 0$ such that $a+b+c=1$ and $n\geq 1$ a natural number then we have :

$$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}\leq 1$$

To prove it I have tried Bernoulli's inequality as we have :

$$a^{(2(1-a))^{\frac{1}{n}}}\leq 1-(1-a)(2(1-a))^{\frac{1}{n}}$$

But it's not enough and wrong.

I have tested for $n=1,2,3,4,5$.Moreover the equality case is $a=b=0.5$.

Case $n=1$ :

As in my previous posts we have the inequality $x\in[0,0.5]$ :

$$x^{2(1-x)}\leq 2^{2x+1}x^2(1-x)$$ applying this for each variable we want to show :

$$2^{2a+1}a^2(1-a)+2^{2b+1}b^2(1-b)+2^{2(1-(a+b))+1}(1-(a+b))^2(a+b)+1-(a+b)\leq 1$$

Now by Bernoulli's inequality we have:

$$2^{2x+1}\leq 2(1+2x)$$

Remains to show :

$$2(1+2a)a^2(1-a)+2(1+2b)b^2(1-b)+2(1+2(1-(a+b)))(1-(a+b))^2(a+b)+1-(a+b)\leq 1$$

Wich is wrong because we have an equality case at $b=c=0.25$ and $a=0.5$!

My question :

Have you a counter-example or later (if it's true) a proof ?

Thanks in advance !

Body:
if you pick $a=b=c=\frac{1}{3}$ and $n=1$ then
$$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}=(\frac{1}{3})^{\frac{4}{3}} +(\frac{1}{3})^{\frac{4}{3}}+(\frac{1}{3})^{\frac{4}{3}}+\frac{1}{3}\geq \frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}>1$$
And even if you delete the last term $\frac{c}{n}$ in this case it would be also bigger than 1.
It is also valid for $n=2$ because you will have:
$$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}=3\cdot(\frac{1}{3})^{\sqrt{\frac{4}{3}}} +\frac{1}{6}=3^{1-\sqrt{\frac{4}{3}}}+\frac{1}{6}=1.010368679... $$


问:
Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$
If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$

So $x+y+z=0$

We need to minimize $$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{z}\right)^{4}+\left(\frac{z}{x}\right)^{4}$$
$\implies$
$$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{x+y}\right)^{4}+\left(\frac{x+y}{x}\right)^4$$
Letting $\frac{x}{y}=t$ we get:
$$E=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)^4}{t^4}$$
Now i tried using derivatives, but very tedious.

Body:
My second solution (using derivative):

Since $\lim_{t\to \infty} E(t) = \infty$, $\lim_{t\to - \infty} E(t) = \infty$,
$\lim_{t\to 0} E(t) = \infty$ and $\lim_{t\to -1} E(t) = \infty$, the minimum of $E(t)$
is attained at the points with $E'(t) = 0$.

We have
\begin{align}
E'(t) &=
\frac{4(t^2+t+1)^2}{(t+1)^5t^5}(t^9+3t^8+t^7-3t^6+t^5+2t^4-10t^3-13t^2-6t-1)\\
&= \frac{4(t^2+t+1)^2}{(t+1)^5t^5}F(t)[t^3+(c+3)t^2+ct-1]
\end{align}
where
$$c = -\frac{1}{6}\sqrt[3]{108+12\sqrt{849}} + \frac{1}{6}\sqrt[3]{-108 + 12\sqrt{849}}-2,$$
and
\begin{align}
F(t) &= t^6-ct^5+(c+1)^2 t^4+(-c^3-4c^2-7c-5)t^3\\
&\quad +(c^2+3c+4)^2t^2+(-c^5-8c^4-31c^3-67c^2-81c-45)t+1.
\end{align}

It is not difficult to prove that $F(t) > 0$ for all $t$ in $\mathbb{R}$.
Thus, all real roots of $E'(t) = 0$ are the real roots of $t^3+(c+3)t^2+ct-1 = 0$.

By using $t^3+(c+3)t^2+ct-1 = 0$, we have
\begin{align}
E(t) &= \frac{t^8(t+1)^4+t^4+(t+1)^8}{(t+1)^4t^4}\\
&= \frac{\mathrm{rem}(t^8(t+1)^4+t^4+(t+1)^8,\ t^3+(c+3)t^2+ct-1)}{\mathrm{rem}((t+1)^4t^4,\ t^3+(c+3)t^2+ct-1)}\\
&= c^4+8c^3+32c^2+68c+69\\
&= \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}}
+ \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}
\end{align}
where $\mathrm{rem}(p(x),q(x))$ is the remainder of $p(x)$ divided by $q(x)$ for two polynomials $p(x), q(x)$.

Thus, the minimum of $E = E(t) = t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}, \ t \ne 0, -1$ is given by
$$E_{\min} = \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}}
+ \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}.$$
Note: $E_{\min} \approx 12.5037895937134016209875697987512$.


问:
Olympiad-type combinatorial problem on convex polyhedra
Consider a convex polyhedron $P$ in three-dimensional Euclidean space. Let $V$ be the set of vertices and $E$ be the set of edges of $P$.

I use the following ad-hoc definition of a mapping: A mapping on P is any function from $E$ to the set $\{-1, +1\}$.

An additional ad-hoc definition: given a polyhedron $P$ as above, with vertices $V$ and edges $E$, and a mapping $m$ (as just defined), for every vertex $v \in V$ let the product at $v$ be the product of the $-1$ and $+1$ values associated by $m$ to the edges that have $v$ as one of the endpoints.

With these definitions, prove the following:

For every polyhedron $P$ with 2019 vertices, and for every mapping on $P$, there is always at least one vertex $v$ so that the product at $v$ is $+1$.

For every polyhedron $P$ with 2020 vertices, there is at least one mapping on $P$ so that the product at $v$ is $-1$ for all vertices $v \in V$.

Of these, 1. is relatively easy (and relatively standard for an olympiad problem). The interesting problem, really, is 2.

Make sure you understand 2. correctly. It is not about finding one counterexample to 1. on some polyhedron with 2020 vertices; that's very easy. Rather, it is to show that 1. fails for all polyhedra with 2020 vertices, no matter what their (combinatorial) shape.

BACKGROUND: A very long time ago I created this problem and we used it in one of the examinations to select our national team for the IMO. There was no Internet then (at least, certainly, not in the East European country I'm from), so I doubt it very strongly that this problem has ever made it to the Internet. Obviously, I have a solution (my own!) but I am offering this as a fun challenge to math enthusiasts on SE. It will be fun to see what other solutions are being offered.

答:
$$\prod_{v \in V} \text{product at } v = \prod_{v} \prod_{v \in e } P(e) = \prod_{e} P(e)^2 = 1.$$

Hence the number of vertices that have product of -1 is even, so an odd number (in particular, at least 1) of vertices have a product of 1.

2.

Claim: For any connected graph with an even number $n=2m$ of vertices, we can find a subset of edges such that the degree at each vertex is odd.

Proof: This is an olympiad problem, of which there are several (pretty distinct) solutions.
The solution that I like takes the symmetric difference of $m$ paths, connecting distinct vertices. Adding a path changes the parity of the endpoints and nothing else, thus the resultant parity of each vertices degree is odd.

Corollary: Giving those edges $-1$ will lead to each vertex having a product of $-1$.

Note:

We don't need "convex polytope" graph. No special features of it were used, other than connectedness in part 2. Specifically,

Any graph with an odd number of vertices will have an odd number of vertices with a product of 1.

Any connected graph with an even number of vertices will yield a mapping such that all vertices have a product of -1.

The olympiad problem I referenced is essentially identical to yours, just with a different skin. I don't know which was published first.


答:
I'll assume you have already done (i) and (ii).

If $S$ is a subring of $R$ and $J$ is an ideal of $R$, then $J\cap S$ is an ideal of $S$.

In your case $S=\mathbb{Z}$, so $J\cap\mathbb{Z}=n\mathbb{Z}$ for some $n\ge0$.

Since $J\ne0$, take $a+bi\in J$, $a+bi\ne0$; then $(a+bi)(a-bi)=a^2+b^2\in J\cap\mathbb{Z}$ and $a^2+b^2\ne0$. Hence $n\ne0$. Let $I=nR$. Then $n\in I\cap(J\cap\mathbb{Z})$, so $n\in J$, which implies $I=nR\subseteq J$.

Now apply (ii) and (i).

Title: Show that the image of a finite ring is also finite, when the map is a surjective ring homomorphism

Body:
Can you help me solve iv) in the following problem?

I believe I'm supposed to use the results from i), ii) and iii).

But why can't I just say that $R/I$ is a finite ring from ii), and since $\varphi$ is surjective we have $\varphi(R/I)=R/J$ which must be finite ring, when $R/I $ is finite?


问:
Second order linear ODE question
I am working on this equation: $ x^4y''+2x^3y'+y=0$ and i need a little help. Should I substitute y=exp(integrate(u)dx), and transform given equation into Riccati's: $ u'=-u^2-2u/x-1/x^4 $ (but i dont know know how to solve this either) or is there any other way? Thanks for tips and help</p>

答:
Let $t=\dfrac{1}{x}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{dt}=-t^2\dfrac{dy}{dt}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2y}{dt^2}-2t\dfrac{dy}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2y}{dt^2}-2t\dfrac{dy}{dt}\right)(-t^2)=t^4\dfrac{d^2y}{dt^2}+2t^3\dfrac{dy}{dt}$

$\therefore\dfrac{1}{t^4}\left(t^4\dfrac{d^2y}{dt^2}+2t^3\dfrac{dy}{dt}\right)-\dfrac{2}{t}\dfrac{dy}{dt}+y=0$

$\dfrac{d^2y}{dt^2}+\dfrac{2}{t}\dfrac{dy}{dt}-\dfrac{2}{t}\dfrac{dy}{dt}+y=0$

$\dfrac{d^2y}{dt^2}+y=0$

$y=C_1\sin t+C_2\cos t$

$y=C_1\sin\dfrac{1}{x}+C_2\cos\dfrac{1}{x}$

Title: Second order linear ODE question


问:
Type of singularity of the Laurent series $\frac{1}{z^{2n}}$
What is the type of singularity of the Laurent series $\sum_2^{\infty} 1/z^{2n}$ at point $z=0$? is it an essential singularity because the series has infinitely many terms or is it a pole because $\sum_2^{\infty} 1/z^{2n}=1/(z^2(z^2-1))$

答:
$\sum \frac1 {z^{2n}}$ is divergent for $|z|\leq 1$. So it is not even defined in any neighborhood of $0$ and the type of singularity at $0$ is not defined.

However $\frac 1 {z^{2}(z^{2}-1)}$ has  a pole of order $2$ at $0$.


问:
Consider set of integers between 1 000 and 9 999 inclusive. How many integers in this set: are divisible by 2? divisible by 3? divisible by 2 & 3?
Consider the set of integers between $1000$ and $9 999$ inclusive.
How many integers in this set:

(i) are divisible by $2$?

(ii) are divisible by $3$?

(iii) are divisible by $2$ and $3$?

What I've tried so far:

(i) $\frac{9999-1000}{2} = 4499.5$

i.e. $4499$ (but since it's inclusive of $1000$), $4499 + 1 = 4500$.

(ii)$\frac{9999-1000}{3} = 2999.66\dots$

i.e. $2999$ (but since it's inclusive of $9999$), $2999+1 = 3000.$

I'm sure this logic/method of solving this question is wrong, and have come across many variations of this problem in exams before but do not know the reasoning behind solving it.

What are the steps for approaching such problems? Thanks!

答:
Your logic is correct except for one part.

Because the set of numbers includes both $1,000$ and $9,999$, its cardinality will be $9,999-1,000+1=9,000$.

So the answers are:

(i) $\frac{9,999-1,000+1}{2}=4,500$ (This works because $1,000$ is a multiple of $2$.)

(ii) $\frac{9,999-1,000+1}{3}=3,000$ (This works because $9,999$ is a multiple of $3$.)

(iii) $\frac{9,999-1,000+1}{6}-1=1,499$ (This works because neither $1,000$ or $9,999$ is a multiple of $6$.)


问:
Union of union of elements of a basis $B $ of a topology $\tau$
I'm trying to show that in a topological space $(X, \tau)$ then a set $B$ is a basis iff $B$ has the following properties:

X = $\cup^{}_{A \in B}A$

for any ${A_1}, {A_2} \in B, {A_1} \cap{A_2}$ is a union of elements in $B$.

I proved the left to right implication and now I'm trying to work out the other one.

Let $\tau=${ union of elements of $B$}, we need to see that $\tau$ is indeed a topology.
I already showed that $X$ and $ \emptyset $ are in $\tau$. My problem is when I need to show that if ${A_1} $and ${A_2} \in \tau$ then so is ${A_1}\cap{A_2}$. Here's where I am at:

If ${A_1} $and ${A_2} \in \tau$ then they are union of elements of $B$. Write them as:

${A_1}= \cup^{}_{i \in I}U_i$ for I an index set.

${A_2}= \cup^{}_{j \in J}U_j$ for J an index set.

Then ${A_1}\cap{A_2} = (\cup^{}_{i \in I}U_i) \cap (\cup^{}_{j \in J}U_j) = \cup^{}_{i \in I, j \in J}U_i \cap U_j$ . By hypothesis, $U_i \cap U_j$ is a union of members of $B$.

Why can we say that the union of the $U_i \cap U_j$ is a union of members of $B$? A union of elements of a set is not the same as a union of a union of members of said set.

Sorry if it's not very well written or explained.

答:
When you speak about unions of elements in $B$ you mean that you have a family of elements $U_r$, where $r$ belongs to a set of indices $R$, and take $\bigcup _{r \in R} U_r$. You should be aware that your indexed family of sets $U_r$ formally is a function $\iota : R \to B, \iota(r) = U_r$. Note that this allows $U_r = U_{r'}$ for $r \ne r'$, i.e. your indexed family of sets $U_r$ may have multiple occurences of the same set in $B$.

We could also confine ourselves to taking only subsets $S \subset B$ als index sets in which case $\bigcup_{U \in S} U$ is the union of all members of $S$. Doing so, we avoid multiple occurences of the same set. The first approach is more flexible, but in some sense it is a matter of taste.

Concerning condition 2. let me remark that we must allow the empty set as an index set when $A_1 \cap A_2 = \emptyset$. No member of  $B$ is indexed by an element of $\emptyset$, but it is standard to define the union of this "empty family of sets" to be $\emptyset$.

We have $U_i \cap U_j = \bigcup_{k \in K(i,j)} U^{ij}_k$ with a suitable index set $K(i,j)$. Therefore

$$\bigcup_{i \in I, j \in J} U_i \cap U_j = \bigcup_{(i,j)\in I\times J} U_i \cap U_j =  \bigcup_{(i,j)\in I\times J}\bigcup_{k \in K(i,j)} U^{ij}_k = \bigcup_{(i,j,k) \in M} U^{ij}_k$$
with
$$ M = \{(i,j,k)  \mid i \in I, j \in J,  k \in K(i,j)\} .$$

Let me close with a remark concerning a general misunderstanding on your part:

In a topological space $(X, \tau)$ a set $B$ is a basis iff $B$ has the following properties:

X = $\cup^{}_{A \in B}A$

for any ${A_1}, {A_2} \in B, {A_1} \cap{A_2}$ is a union of elements in $B$.

Properties 1. and 2. guarantee that $B$ is the basis for some topology on $X$, but this topology may differ from $\tau$. It thus would be better to say that $B$ is the basis for a topology on the set $X$. To assure that $B$ is a basis for $\tau$ you must add that $B \subset \tau$. In that case it would be easier to replace 1. and 2. by

Each $U \in\tau$ is a union of elements in $B$.


问:
Proving two equations involving the greatest common divisor

Show or prove that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m),$$
  and that
  $$\gcd \left(\frac{a^{2m + 1}+1}{a+1} , a + 1\right)=\gcd(a + 1 , 2m + 1).$$

Body:
I will show that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m)$$

First note that $$a^{2m}-1=(a^2)^m-1=(a^2-1)(a^{2(m-1)}+
a^{2(m-2)}+\cdots+1)$$

so that $$\frac{a^{2m}-1}{a+1}=(a-1)(a^{2(m-1)}+\cdots+1)$$

But $a\equiv -1 \mod a+1$ so we have

$$\frac{a^{2m}-1}{a+1}\equiv -2((-1)^{2(m-1)}+\cdots+1)=-2m$$

which means $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m)$$
Can you do something similar with the other?


问:
For every $a, b, x \in \mathbb{R}$ such that $a \le x \le b$, we have $|x| \le \max\{|a| , |b|\}$
I was given an assigmnent in my Infintesiaml math course, and I'm having trouble proving the statement below:

For every $a, b, x  \in \mathbb{R}$ such that $a \le x \le b$, we have $|x| \le \max\{|a| , |b|\}$

Hint: denote $r = \max \{|a| , |b|\}$.

Could it be false? I have tried using the triangle inequality, with and without the mentioned hint, and also tried isolating $x$ and $a$, or $x$ and $b$, but i think there is something important that I am missing.

I saw a similar question in a different post but the answers there are not explained very well.

答:
Hint:

Remember  that for any $u$, $|u|=\max(u,-u)$. Therefore if $r=\max\bigl(|a|,|b|\bigr)$, we have $\;-r \le a,\,b,-a,-b\le r$. What can you deduce for $\max(x,-x)$?


问:
Question about linear operator defined in C[0,1].
I'm trying to solve the following exercise:

Let be $X=C([0,1])$. We define $T:X \longrightarrow X$ as $T(f)(x)=\displaystyle \int_{0}^{x} f(t^2)dt$. Prove that $T^{-1} : T(X) \longrightarrow X$ is not continuous.

I have proved that $T$ is a linear operator, continuous and injective, but I don't know how to prove that $T^{-1}$ is not continuous. In other exercises, people show that $||T^{-1}||$ is not bounded, but I think this is not useful in this exercise.

In X, I am considering the uniform norm (the max-norm).

答:
Solve $Tf=g$ by differentiating both sides to get $T^{-1}(g)=g'(\sqrt x)$. [This is defined on the range of $T$ which is the whole of $C[0,1]$).
Suppose there exists $C \in [0,\infty)$ such that $|T^{-1} g(x) | \leq C \|g\|$ for all $g$ for all $x$. Take $g_n(x)=\frac {x^{n}} {\sqrt n}$ to get a contradiction. [The inequality fails when $x=1$ and $n$ is large].


问:
Does anyone know of a much faster way to solve this polynomial question?
A monic cubic $p(x) $ is divided by $(x^2 +x +1)$ and the remainder is $(2x+3)$. When $p(x)$ is divided by $x(x+3)$, the remainder is $5(x+1)$. Find $p(x)$

So the way I solved it was through using long division - as you can imagine very lengthy, and not very efficient in exam situations. I then equated the remainders with the given remainders. However, in an exam situation, would there be an even faster way? I am open to suggestions.

答:
This may not be a quicker way, but it is systematic and will work in all cases where the two moduli have no common factors. Hopefully, it also might make the use of the Extended Euclidean Algorithm with the Chinese Remainder Theorem more transparent.

We want
$$
\begin{align}
p(x)&\equiv2x+3&&\pmod{x^2+x+1}\tag{1a}\\
p(x)&\equiv5x+5&&\pmod{x^2+3x}\tag{1b}
\end{align}
$$
so we apply the Extended Euclidean Algorithm (as implemented in this answer and adapted to polynomials) to $x^2+3x$ and $x^2+x+1$:
$$
\begin{array}{c|c|c|c}
x^2+3x&1&0\\
x^2+x+1&0&1\\
2x-1&1&-1&1\\
\frac74&-\frac{2x+3}4&\frac{2x+7}4&\frac{2x+3}4
\end{array}\tag2
$$
Thus, we get
$$
(2x+7)\left(x^2+x+1\right)-(2x+3)\left(x^2+3x\right)=7\tag3
$$
So we have
$$
-\frac{(2x+3)\left(x^2+3x\right)}7\equiv\left\{
\begin{align}
&1&&\pmod{x^2+x+1}\\
&0&&\pmod{x^2+3x}
\end{align}
\right.\tag4
$$
and
$$
\frac{(2x+7)\left(x^2+x+1\right)}7\equiv\left\{
\begin{align}
&0&&\pmod{x^2+x+1}\\
&1&&\pmod{x^2+3x}\\
\end{align}
\right.\tag5
$$
Adding $2x+3$ times $(4)$ to $5x+5$ times $(5)$ yields one solution to $(1)$:
$$
\begin{align}
&\overbrace{(5x+5)\frac{(2x+7)\left(x^2+x+1\right)}7}^{\substack{0&\pmod{x^2+x+1}\\5x+5&\pmod{x^2+3x}}}\overbrace{-(2x+3)\frac{(2x+3)\left(x^2+3x\right)}7}^{\substack{2x+3&\pmod{x^2+x+1}\\0&\pmod{x^2+3x}}}\\
&=\frac{6x^4+31x^3+45x^2+53x+35}7\tag6
\end{align}
$$
The Chinese Remainder Theorem says that this is the solution mod $\left(x^2+x+1\right)\left(x^2+3x\right)$, so we can subtract $\frac67$ of the modulus to get
$$
\begin{align}
p(x)
&=\frac{6x^4+31x^3+45x^2+53x+35}7-\frac67\left(x^2+x+1\right)\left(x^2+3x\right)\\
&=\bbox[5px,border:2px solid #C0A000]{x^3+3x^2+5x+5}\tag7
\end{align}
$$


问:
How does one go about evaluating the limits of nonelementary antiderivatives of functions?
We obviously know the following to be equal:

$$\lim_{z \to \pm\infty} \text{erf}(z) = \pm1$$

But as far as I'm aware, the only way we know how to calculate any given value of the error function is to approximate it. How do we know the limits at infinities without any way to evaluate it algabraically?

答:
$$\lim_{x\to \pm \infty}\text{erf}(x)=\lim_{x\to \pm \infty}\frac{2}{\sqrt \pi}\int_0^x e^{-t^2}dt = \frac{2}{\sqrt \pi} \int_0^{\pm\infty}e^{-t^2}$$
using the fact $e^{-t^2}$ is an even function:
$$ \pm \frac{2}{\sqrt \pi} \int_0^\infty e^{-t^2}$$
The integral is known as the gaussian integral(half of it) and is known to equal $\frac{\sqrt \pi}{2}$ and thus, $$\lim_{z \to \pm\infty} \text{erf}(z) = \pm1$$


问:
Conditions on finite groups such that there exists some $g$ for which $g \mapsto g^n$ is invertible.
I am interested in the following question about finite groups. Consider a finite group $G$ and a function which takes group elements and returns their $n$-th power, equivalently $g \mapsto g^n$. For certain finite groups, e.g., cyclic groups of prime order $p$ and $n \perp p$ this map is invertible on all elements, while for the alternating group on $4$ elements and $m = 2$, the map is only invertible on cycle types of the form $(1, 0, 1, 0)$ (e.g., $(1,2,3)(4)$).

What are methods to determine the size of the pre-image for a given $h = g^m$, and conditions under which this pre-image is uniquely $g$?

答:
Proposition Let $G$ be a finite group, $n$ a non-zero integer, then the map $f: G \rightarrow G$ defined by $f(g)=g^n$ is a bijection if and only if gcd$(n,|G|)=1$.

Proof Assume $f$ is bijection and let $p$ be a prime dividing both $|G|$ and $n$. By Cauchy's Theorem, one can find a $g \in G$ of order $p$. But $f(g)=g^n=(g^p)^{\frac{n}{p}}=1$, so $g=1$ as $f$ is bijective, a contradiction.

If $|G|$ and $n$ are relatively prime, then by Bézout's Theorem, we can find  $a,b \in \mathbb{Z}$ with $1=a|G|+bn$. Hence for each $g \in G$, $g=g^1=g^{a|G|+bn}=(g^{|G|})^a \cdot (g^b)^n=(g^b)^n$, showing that $f$ is surjective. Since $G$ is finite, it follows that $f$ is injective.

If $n$ and $|G|$ are not relatively prime, then one can use character theory of finite groups over $\mathbb{C}$ to say something about "$\sqrt[n]{g}$". One can define, assuming $n$ is positive, $\vartheta_n(g)=\#\{h \in G: h^n=g\}$. This is a class function on $G$ (constant on conjugacy classes) and hence one can write $\vartheta_n=\sum_{\chi \in Irr(G)}\nu_n(\chi)\chi$, where $\nu_n(\chi) \in \mathbb{C}$ are uniquely determined. So if you know these values of $\nu_n$ for each $\chi \in Irr(G)$, you know the numbers $\vartheta_n$. More on this theory can be found in many textbooks, for example that of I.M. Isaacs, Character Theory of Finite Groups, Chapter $4$, Products of Characters, page $49$ and following. Or see Y.G. Berkovich, L.S. Kazarin, E.M. Zhmud Characters of Finite Groups, page $152$, IV.5-7.


问:
$a_n = u^n+v^n+(-1)^n$
Anyone has solutions for this problem?

Let $u, v$ are distinct roots of equation:
$$x^2-tx+1=0 (t \in \mathbb{N}, t>2)$$
and sequence $$(a_n): a_n = u^n+v^n+(-1)^n$$
Prove that: $$a_{3^m}\mid a_{3^mn} \forall m,n \in \mathbb{N} ,  gcd(n,3)=1$$
Notes

$S_n=\frac{a_{3^mn}}{a_{3^m}}\\$

$a=u^{3^m}$ then $\frac{1}{a}=v^{3^m}$

+If $n$ is odd

$$S_n = \frac{a^{2n}-a^n+1}{a^{n-1}(a^2-a+1)}$$

Because $gcd(n,6)=1$, polynomial $a^{2n}-a^n+1$ is divided by $a^2-a+1$, then $\frac{a^{2n}-a^n+1}{a^2-a+1}$ is a symmetric polynomial with degree $2(n-1)$

So $S_n=\sum\limits_{i=0}^{n-1}s_i(a^i+\frac{1}{a^i})$

+If $n$ is even, then $n=2^tn'$

$S_n = S_{n'}.a_{\frac{n}{2}}a_{\frac{n}{4}}...a_{\frac{n}{2^t}}$

答:
By either using Vieta's formulas or just expanding $(x - u)(x - v) = x^2 - tx + 1$ and comparing coefficients, we get

$$\begin{equation}\begin{aligned}
u + v & = t \\
uv & = 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

For non-negative integers $i$, define $f_i(t)$ to be $u^i + v^i$ expressed as a function of $t$. Thus, $f_0(t) = u^0 + v^0 = 2$ and $f_1(t) = u + v = t$. For any $i \ge 2$, using \eqref{eq1A}, we get

$$\begin{equation}\begin{aligned}
(u^{i-1} + v^{i-1})(u + v) & = u^{i} + u^{i-1}(v) + v^{i-1}(u) + v^{i} \\
t(f_{i-1}(t)) & = u^{i} + v^{i} + uv(u^{i-2} + v^{i-2}) \\
t(f_{i-1}(t)) & = f_{i}(t) + f_{i-2}(t) \\
f_{i}(t) & = t(f_{i-1}(t)) - f_{i-2}(t)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

This confirms $u^{i} + v^{i}$, for all $i \ge 0$, can be expressed as a polynomial in $t$ with integral coefficients and, thus, is an integer. Next, for any odd positive integer $j$, your sequence gives

$$a_j = u^j + v^j - 1 \tag{3}\label{eq3A}$$

For conciseness, with any integer $k \ge 0$, let

$$w_k = u^{kj} + v^{kj} \tag{4}\label{eq4A}$$

With $k = 0$ and $k = 1$, we get

$$\begin{equation}\begin{aligned}
& w_0 = u^{0} + v^{0} = 1 + 1 \equiv 2 \pmod{a_j} \\
& w_1 = u^{j} + v^{j} \equiv 1 \pmod{a_j}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$

For $k \ge 2$,

$$\begin{equation}\begin{aligned}
(u^{(k-1)j} + v^{(k-1)j})((u^j + v^j) - 1) & \equiv 0 \pmod{a_j} \\
u^{kj} + u^{(k-1)j}(v^{j}) + v^{(k-1)j}(u^j) + v^{kj} & \equiv u^{(k-1)j} + v^{(k-1)j} \pmod{a_j} \\
u^{kj} + v^{kj} + (uv)^{j}(u^{(k-2)j} + v^{(k-2)j}) & \equiv w_{k-1} \pmod{a_j} \\
w_{k} + (u^{(k-2)j} + v^{(k-2)j}) & \equiv w_{k-1} \pmod{a_j} \\
w_k & \equiv w_{k-1} - w_{k-2} \pmod{a_j}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$

This therefore gives

$$\begin{equation}\begin{aligned}
& w_2 \equiv w_1 - w_0 \equiv 1 - 2 \equiv -1 \pmod{a_j} \\
& w_3 \equiv w_2 - w_1 \equiv -1 - 1 \equiv -2 \pmod{a_j} \\
& w_4 \equiv w_3 - w_2 \equiv -2 - (-1) \equiv -1 \pmod{a_j} \\
& w_5 \equiv w_4 - w_3 \equiv -1 - (-2) \equiv 1 \pmod{a_j} \\
& w_6 \equiv w_5 - w_4 \equiv 1 - (-1) \equiv 2 \pmod{a_j} \\
& w_7 \equiv w_6 - w_5 \equiv 2 - 1 \equiv 1 \pmod{a_j}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Due to \eqref{eq6A} showing $w_k$ depends only on the previous $2$ smaller indice values, then since $w_6 \equiv w_0 \pmod{a_j}$ and $w_7 \equiv w_1 \pmod{a_j}$, the congruence values will repeat cyclically with a period of $6$. Thus, using \eqref{eq5A} and \eqref{eq7A}, for all non-negative integers $r$,

$$\begin{equation}\begin{aligned}
& a_{(6r)j} \equiv w_0 + 1 \equiv 3 \pmod{a_j} \\
& a_{(6r + 1)j} \equiv w_1 - 1 \equiv 0 \pmod{a_j} \\
& a_{(6r + 2)j} \equiv w_2 + 1 \equiv 0 \pmod{a_j} \\
& a_{(6r + 3)j} \equiv w_3 - 1 \equiv -3 \pmod{a_j} \\
& a_{(6r + 4)j} \equiv w_4 + 1 \equiv 0 \pmod{a_j} \\
& a_{(6r + 5)j} \equiv w_5 - 1 \equiv 0 \pmod{a_j}
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$

For all $n \in \mathbb{N}$ with $\gcd(n, 3) = 1$, \eqref{eq8A} shows $a_{nj} \equiv 0 \pmod{a_j} \implies a_j \mid a_{nj}$. Since $j$ was any odd positive integer, this includes your question's specific case of $j = 3^m$. In addition, apart from the one case of $t = 4$ and $j = 1$, since in all other cases $a_j \gt 1$ and $a_j \neq 3$, we also have for $\gcd(n, 3) = 3$ that $a_{j} \not\mid a_{nj}$.

Title: $a_n = u^n+v^n+(-1)^n$