问:
A Polynomial-Inequality problem from Vietnam National Olympiad 2021
Abstract

I have just completed 2 days of our National Olympiad. This year's problems are not difficult, yet new and strange to many of the students. Mentioned below will be problem 5 out of 7, the first problem of Day 2.

To anyone who wonders why it is VMO 2021, we always do national olympiad for a year before the previous year ends, so that the result will be published in early 2021

Problem

Given that $P(x)=a_{21}x^{21}+a_{20}x^{20}+a_{19}x^{19}+ \dotsb + a_1x+a_0$ is a polynomial with real coefficients such that $a_i \in [1011,2021]$ for all $ i \in \{ 0,1,2,\dotsb,21\}$ and $c \in \mathbb{R}$ such that $ | a_{k+2}-a_{k} | \le c$ for all $k \in \{ 0,1,2, \dotsb,19 \}$ Given that $P(x)$ has at least one integer root.

Prove that $P(x)$ has exactly one integer root

Prove the following inequality: $\displaystyle\sum_{k=0}^{10} |a_{2k+1}-a_{2k}|^2 \le 440c^2$

My works

For part 1, it is easy to see that since $a_i$ are all positive, then if $P(x)$ has a root, that root must be negative. From the condition $a_i \in [1011,2021]$ it is also easy to see that $ \frac{a_i}{a_j} <2$ for all $0 \le i,j \le 21$, thus if $x$ is an integer root of $P(x)$ then $x \ge (-2) $. Thus $x = (-1)$ is the only integer root of $P(x)$.

For part 2, some of my teammates say that it could be solved using Jensen inequality, but none of us succeeded. I noticed that $ \displaystyle\sum_{k=0}^{10} a_{2k+1} =\displaystyle\sum_{k=0}^{10} a_{2k}$, but how do I finish the problem?

Any help is appreciated.

答:
(This is before the edit.)

The inequality is not true.

Take $ c = 0.1$ (Any $ 0 < c < \frac{3}{22} $ will work.)

Take $ a_{2k} = 1011 + kc $, $a_{2k+1} = 1011 + (10-k) c $.

This satisfies the conditions as 1) $P(-1) = 0$ is an integer root, 2) The difference between terms is exactly $c$.

However,  $ \sum |a_{2k+1} - a_{2k}| = 10c + 8c + \ldots + 0c + 2c + \ldots + 10c = 60c > 440 c^2$.

My suspicion is that $ \sum |a_{2k+1} - a_{2k}| \leq 60c$, though I don't know how to prove it. The equality case is as above.

Claim: If

$\sum_{i=0}^{21} (-1)^i a_i = 0$

$|a_{k+2} - a_k  | \leq |c| $

Then $ \sum_{k=0}^{10} |a_{k+1} - a_k |^2 \leq 440 c^2$.

Equality case is when $ a_{2k} = a + kc $, $a_{2k+1} = a + (10-k) c $.


问:
Generalisations belong to any affine open subset of a scheme
I was reading the proof of Lemma 26.13.2. here and it seems they use the following fact, but I haven't been able to find a proof of it. Could anyone provide a reference/proof?

Let $X$ be a scheme and $x,x'\in X$ points. Suppose that $x\in\overline{\{x'\}}$, this is, $x'$ is a generalisation if $x$. If $U\subseteq X$ is an affine open neighbourhood of $x$, then $x'\in U$.

答:
This is just a general topological result. If $X$ is a topological space and  $Y \subset X$ then $x
\in \bar{Y}$ if and only if we have $U \cap Y \neq \emptyset$ whenever $U$ is an open neighbourhood of $x$.

ContentLicense: CC BY-SA 4.0


问:
Computation of the likelihood of the following two questions with balls in urns

Urn A contains three balls: one black, and two white; urn B contains
three balls: two black, and one white. One of the urns is selected at
random and one ball is drawn. The ball is black. What is the
probability that the selected urn is urn A?

For the above problem I solved the following:

$p(U_a | B) = \frac{p(B|U_a)P(U_a)}{P(B)}$

with $P(B) = \sum_{U} p(B|U_a)P(U_a) + P(B|U_B)P(U_B) = 2/3 \times 0.5 + 1/3 \times 0.5 = 0.5$

So $p(U_a | B) = \frac{1/3 \times 0.5}{0.5} = \frac{1}{3}$

Anyone can confirm if this is in fact right?

Similarly for the second problem below I used the same method

Urn A contains five balls: one black, two white, one green and  one
pink; urn B contains five hundred balls: two hundred black, one
hundred white, 50 yellow, 40 cyan, 30 sienna, 25 green, 25 silver, 20
gold, and 10 purple. [One fifth of A’s balls are black; two-fifths of
B’s are black. One of the urns is selected at random and one ball is
drawn. The ball is black. What is the probability that the urn is urn
A?

$P(U_a|B) = \frac{0.2 * 0.5}{(0.2 \times 0.5 + 0.4 \times 0.5} = \frac{1}{3}$

More unsure of the second answer but would love it if someone could confirm my thinking, thanks

答:
You did fine.

For the first problem there is a more direct solution that avoids the use of the rule of Bayes.

Since both urns have the same number of balls and secondly the choosing of the urn is at random we can conclude that all balls have equal probability to be the drawn ball.

One of the black balls comes from urn A and two from urn B, so the probability that the drawn ball came from urn A equals $\frac13$.


问:
Can complex numbers be visualized

If $z=|x+iy|$ is a complex number satisfying $\;|z-2|-|z-3i|=\sqrt5\;$ then locus of $z$ is $\dots$

On solving this I easily get an equation of part of hyberbola.

My question is: Is there any general result for locus problems like $(x-a)^2-(y-b)^2=c$ is a general equation of hyperbola and can be visualized easily?

Is there any sort of visualization for complex numbers?

答:
Geometrically, $|z_1-z_2|$ is the distance between points which represent complex numbers $z_1$ and $z_2.$

Set the points $M(z),A(2),B(3i).$ The number $z$ satisfies $$|z-2|-|z-3i|=\sqrt5$$ if and only if the difference of distances $$|MA|-|MB|=\sqrt5.$$
This is equation of one branch of hyperbola with foci $A$ and $B.$ The branch is that which is near the focus $B(3i).$

Title: Can complex numbers be visualized


答:
Both of your answers are correct (didn't check the algebra but the reasoning and formula are correct), as a rule of thumb whenever you have a partition of your sample space and a conditional probability you can/should Bayes' theorem.

Since your sample space is partitioned between two urns and you need to get the probability of drawing from a certain urn GIVEN the ball is black then Bayes' theorem is suitable here.

Title: Computation of the likelihood of the following two questions with balls in urns

Body:
Urn A contains three balls: one black, and two white; urn B contains
three balls: two black, and one white. One of the urns is selected at
random and one ball is drawn. The ball is black. What is the
probability that the selected urn is urn A?

For the above problem I solved the following:

$p(U_a | B) = \frac{p(B|U_a)P(U_a)}{P(B)}$

with $P(B) = \sum_{U} p(B|U_a)P(U_a) + P(B|U_B)P(U_B) = 2/3 \times 0.5 + 1/3 \times 0.5 = 0.5$

So $p(U_a | B) = \frac{1/3 \times 0.5}{0.5} = \frac{1}{3}$

Anyone can confirm if this is in fact right?

Similarly for the second problem below I used the same method

Urn A contains five balls: one black, two white, one green and  one
pink; urn B contains five hundred balls: two hundred black, one
hundred white, 50 yellow, 40 cyan, 30 sienna, 25 green, 25 silver, 20
gold, and 10 purple. [One fifth of A’s balls are black; two-fifths of
B’s are black. One of the urns is selected at random and one ball is
drawn. The ball is black. What is the probability that the urn is urn
A?

$P(U_a|B) = \frac{0.2 * 0.5}{(0.2 \times 0.5 + 0.4 \times 0.5} = \frac{1}{3}$

More unsure of the second answer but would love it if someone could confirm my thinking, thanks


问:
Is the map $\Psi:S \to \ell^2$ a bijection?
Let $S$ be the set of all entire functions $f$ such that $|f(z)| \le Ae^{\pi |z|}$ for all complex $z$ and $f \in L^2(-\infty,\infty)$. Let $\ell^2$ be the set of of all complex sequences $\{x_n\}$ such that $\sum_{n=1}^\infty |x_n|^2<\infty$. Take $\Psi:S \to \ell^2$ to be the map $\Psi(f)=\{f(1),f(2),\ldots\}$. Show that $\Psi$ is a bijection. My work is that if $\Psi(f-g)=0$, then I want to show that $f=g$. I can show that $f(n)=g(n)$ for $n=1,2,\ldots$. Combining this and the paley-wiener, we can show that
\begin{equation}
\int_{-\pi}^\pi [F(\theta)-G(\theta)]e^{in\theta}d\theta=0
\end{equation} for some $F,G \in L^2(-\infty,\infty)$, where $n=1,2,\ldots$. Then I stop here. Are my approach of proving injection of $\Psi$ correct?

答:
If you know the Paley-Wiener theorem this result should just fall out by Parseval. I re-read the question. In fact the version above is wrong; it's the correct version that's trivial...$\newcommand\s{\operatorname{sinc}}$ (if $f(z)=\s(\pi(z+1))$ then $\Psi f=0$).

So we change the definition of $\Psi f$ to be the bilateral sequence $$\Psi f=(f(n))_{n\in\Bbb Z}.$$

(Note that at this point it would be wrong to write $\Psi:S\to\ell_2$ as you did; we will see below that in fact $\Psi:S\to\ell_2$, but that's not clear yet.)

For $F\in L^2([-\pi,\pi])$ define $f=\hat F\in H(\Bbb C)$ by $$f(z)=\hat F(z)=\int_{-\pi}^\pi F(t)e^{-izt}\,dt.$$For $n\in\Bbb Z$ define $e_n\in L^2(-\pi,\pi)$ by $$e_n(t)=e^{int},$$so $(e_n)_{n\in\Bbb Z}$ is an orthonormal basis for $L^2(-\pi,\pi)$. Let $$s_n(z)=\hat e_n(z)=\s(\pi(z-n)).$$

Now if we define a norm in $S$ by $$||f||_S=||\hat F||_{S}=||F||_{L^2}$$then $S$ becomes a Hilbert space, with $(s_n)_{n\in\Bbb Z}$ as an orthonormal basis. So for every $f\in S$ there exists a unique sequence $\alpha\in\ell_2(\Bbb Z)$ with $$f=\sum_{\alpha\in\Bbb Z}\alpha_ns_n,$$with convergence in $S$.

Easy Lemma: If $K\subset\Bbb C$ is compact there exists $c$ such that $|\hat F(z)|\le c||F||_{L^2}$ for every $z\in K$ and $F\in L^2$.

So the series $f=\sum \alpha_ns_n$ converges uniformly on compact sets, hence to an entire function. And hence it converges pointwise, so $\alpha_n=f(n)$.

Hence $\langle f, s_n\rangle = f(n)$, so the fact that $(s_n)$ is an orthonormal basis says precisely that $\Psi$ is a bijection from $S$ onto $\ell_2(\Bbb Z)$.


问:
Understanding the Taylor expansion of the magnitude of a vector
I am working on some problems where I encountered the need to find the Taylor expansion of the magnitude cubed of a vector. I have looked at the answer but I cannot understand how the result was obtained. I can place the equation here and I will really appreciate if anyone can show me the explicit working to achieve the result. I have not been able to find any Taylor expansions of the magnitude of a vector anywhere on the internet
$$\frac{1}{\left|\underline{r} \pm \frac{1}{2} \underline{a}\right|^{3}} \approx \frac{1}{r^{3}}\left(1 \mp \frac{3}{2} \cdot \frac{\underline{a}\cdot\underline{r}}{r^{2}}\right)$$

答:
Since $|\vec x\pm \vec y|^2= x^2+y^2\pm 2\vec x\cdot \vec y$, $$\frac{1}{|\vec r+\frac 12 \vec a|^3} = \left(r^2 + \frac 14 a^2 \pm \vec r\cdot \vec a\right)^{-3/2} \\ = r^{-3}  \left(1+\frac{a^2}{4r^2} \pm \frac{\vec r\cdot \vec a}{r^2} \right)^{-3/2} $$
Presumably $a$ is so small to make $\frac{a^2}{4r^2} \approx 0$, in which case using $(1+x)^n \approx 1+nx$ gives $$r^{-3}\left(1\mp \frac 32 \cdot \frac{\vec a \cdot \vec r}{r^2} \right)$$


问:
Product rule of probabilities and conditioning
I have a probably very basic question regarding the product rule for probabilities. If we have a two probability densities say $p(a|b)$ and $p(b|c,d)$, is $p(a|b)p(b|c,d) = p(a,b | c,d)$? What has me slightly confused is that the second term is conditioned on variables $c$ and $d$ and so I am not sure the product rule applies directly here. I would have thought that the first term $p(a|b)$ would need be conditional on $c$ and $d$ explicitly also.

Would someone mind clarifying this for me please?

答:
In terms of events you are actually asking:

Do we have: $$P(A\cap B)\times P(B\cap C\cap D)=P(B)\times P(A\cap B\cap C\cap D)\text{ ?...}$$

Certainly not in general.


问:
Estimation of probability of the distribution of a continuous random variable
I am trying to find the probability distribution of a continuous random variable X which is defined by
(X=x)=(2−x)(x+1),where 0≤=x≤2

Can anybody help me to find x and mean expected value E(x) ?

答:
You can't find $x$, the question doesn't make sense and I suspect you meant find $k$ so I will proceed with that
A continuous random variable must satisfy the property $\int f(x) dx=1$  or in other words: the sum of all probability must equal to $1$.

To find $k$ we need $k\times\int_0^2(2-x)(x+1)dx=1$, we have $\int_0^2(2-x)(x+1)dx=\frac{10}{3}$ (this was done quickly on WolframAlpha as the goal here isn't to teach you integration) so we have $k\times \frac{10}{3}=1$ which gives us $k=\frac{3}{10}$

For your second question, we have $E(X)=\int xf(x)$ so to find the expected value you should evaluate $\frac{3}{10}\int_0^2x(2-x)(x+1) dx$ which gives $\frac{4}{5}$


问:
Finding the limit of $\frac{(2n)!!}{(2n+1)!!}$ without using Stirling’s approximatation
I have proven that the sequence is convergent.

Let $x_n=\frac{(2n)!!}{(2n+1)!!}$

$\frac{x_{n+1}}{x_n}$ $=\frac{2n+2}{2n+3}<1$ Therefore the sequence decreases. On the other hand
$x_n>0$. Therefore the sequence is bounded below.
Since we have a sequence that is bounded below and decreases the sequence must converge.

But from here I don’t know how to find the limit. Could you please help me? And could you please say if what I’ve done is necessary or not?

Thanks in advance!

答:
See that

$$x_n = \prod_{k=1}^n \frac{2k}{2k+1}, $$

so that

$$\ln x_n  = \sum_{k=1}^n \ln \left(\frac{2k}{2k+1}\right). $$

So we work with this series.  First argue that

$$\sum_{k=1}^{\infty} \ln\left(\frac{2k}{2k+1}\right) - \int_{t=1}^{\infty} \ln\left(\frac{2t}{2t+1}\right) \; dt$$

finite.  (If you graph the summmand and the integrand, they differ by a set of little (near) triangles.  Because the functions are monotonic, the triangles can be pushed together and fit in a small rectangle.)

Integration by parts gives

$$ = \int \ln(2t) - \ln(2t+1) \; dt =  t \ln(2t) -\frac{1}{2}\ln(2t+1)(2t+2) + 1/2.$$

And when you evaluate the improper integral you get $-\infty.$   So the limit of the sequence is $e^{-\infty} = 0.$


问:
Intermediate fields of the simple extension $\mathbb{C}(x)$
Let $\mathbb{C}(x)$ be the field of rational functions over $\mathbb{C}$. Of course $\mathbb{C}(x)$ is a field extension of $\mathbb{C}$. My question now is: are there any intermediate fields between $\mathbb{C}$ and $\mathbb{C}(x)$? If so, what can we say about their dimension? Is it always infinite?

答:
A summary of the comments (excluding reuns result that they should post separately!)
Below $K$ stands for an arbitrary intermediae field strictly in-between, $\Bbb{C}\subset K\subset\Bbb{C}(x)$.

Because $\Bbb{C}$ is algebraically closed, it has no algberaic extensions. Hence no finite extensions. Therefore $[K:\Bbb{C}]=\infty$.

On the other hand, if $u=f(x)/g(x)$ is an arbitrary element of $K\setminus\Bbb{C}$, $f,g\in\Bbb{C}[x]$, then $x$ is a zero of the polynomial
$$
P(T):=f(T)-g(T)u\in K[T].
$$
Therefore $x$ is algebraic over $K$. Hence $[K(x):K]<\infty$. But, $K(x)=\Bbb{C}(x)$, so we can conclude that $[\Bbb{C}(x):K]<\infty$. Nothing more can be said, as we easily see that $[\Bbb{C}(x):\Bbb{C}(x^n)]=n$ for every positive integer $n$, so the extension degree can be arbitrarily high.

By Lüroth's theorem every intermediate field $K$ is actually a simple transcendental extension of $\Bbb{C}$. In other words, $K$ is $\Bbb{C}$-isomorphic to $\Bbb{C}(x)$.


问:
Induced topology base
Let $X$ be a topological space, $B$ an open base for $X$ and $Y \subseteq X$. Prove that $B_Y = \{U \cap Y: U \in B\}$ is an open base for $Y$ (considering, over $Y$, the subspace topology).

We must show that given an open $A$ relative to the topology of $Y$, we have to show that there is an element of $B_Y$ such that $ A = \cup (U \cap Y) $. Since $ B $ is a base for X, we have that Y can be written as a union of elements of $B$. That is, there is $W \in B$ such that $ Y = \cup W $. Is this demonstration absurd?

答:
Use the fact that the subspace toplogy $\tau_Y$ is the coarsest one for which the inclusion $i:Y\to X$ is continuous.

If $A$ is open in $Y$ then there is a $V$ open in $X$ such that $i^{-1}(V)=A$. Since $V$ is open in $X$ there is a basis element $B\in \tau_X$   such that $B\subseteq V.$ Then, $i^{-1}(B)=B\cap Y\subseteq A$ and $B\cap Y$ is open (in $Y$) by definition of $\tau_Y.$ It follows that the collection $\{B\cap Y\}_{B\in \tau_X}$ is a base for $\tau_Y.$

Title: Induced topology base


问:
Prove that $\int_0^\infty\frac1{x^x}\, dx<2$

Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$

 Note: This inequality is rather tight. The integral approximates to $1.9955$.

Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.

Expressing the left-hand side as a Frullani integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ means that $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ that satisfies the functional equation. Is there such a function?

 (For context, user371838's post below proves sophomore's dream which I also asked about originally.)

Body:
This approach is based on the intermediate inequalities, which follow from the series representations.

Using  inequality
$$x^{-x}\le1+(\sqrt[e]e -1)ex(2-ex)+ \frac{116-ex}{635} \sqrt{e x} (ex-1)^2,\quad\text{if}\quad x\in[0,1],\tag1$$

one can get
$$I_1 \le \dfrac1e\int\limits_0^e (1 + (\sqrt[e]e - 1) (2 y - y^2) + \dfrac{116-y}{635} \sqrt y\, (y - 1)^2)\,\text dy,$$
$${\small I_1 \le 1 - e + \dfrac{e^2}3 + \sqrt e\left(\dfrac{232}{1905} - \dfrac{466 e}{3175} + \dfrac{236 e^2}{4445} - \dfrac{2 e^3}{5715}\right) + \dfrac{\sqrt[e]e (3e-e^2)}3 < 1.291733}\tag2$$
(see also Wolfram Alpha integration).

If $\;x\in[1,3],\;$ then
$$\begin{align}
&(2-x)\left(1+\dfrac1{120}(x-1)^6\right) + \dfrac1{12}(x-1)^3 (x^2-6x+11)\\[4pt]
&+ \dfrac1{602910}(x-1)^8(470 +381(x-e) -270 (x-e)^2 +115(x-e)^3) -x^{-x} \le 0,
\end{align}\tag3$$

$$\begin{align}
&I_2 = \int\limits_1^3 x^{-x}\,\text dx \le \int\limits_0^2\bigg(1-y)\left(1+\dfrac1{120}y^6\right) + \dfrac1{12}y^3 (y^2-4y+6) + \dfrac1{602910}y^8\\[4pt]
&\times\bigg(470 +381(y+1-e) -270 (y+1-e)^2 +115(y+1-e)^3\bigg)\bigg)\,\text dy = \bigg(\dfrac{y^9}{5426190}\\[4pt]
&\times\left(\dfrac{345}4 y^3 - \dfrac{135(23e - 5)}{11} y^2 + \dfrac{27(62 - 50 e + 115 e^2)}{10} y + 696 - 186 e + 75 e^2 - 115 e^3\right) \\[4pt]
& - y^8/960 + y^7/840 + y^6/72 - y^5/15 + y^4/8 - y^2/2 + y\bigg)\bigg|_0^2,
\end{align}$$
$$I_2 \le \dfrac{2 (61687647 - 11159040 e + 4899840 e^2 - 809600 e^3)}{149220225}
< 0.687550.\tag4$$

At last,
$$I_3 = \int\limits_3^\infty x^{-x}\,\text dx = 3\int\limits_0^\infty (3+3y)^{-3-3y}\,\text dy = 3\int\limits_0^\infty3^{-3-3y}(1+y)^{-3-3y}\,\text dy$$
$$I_3 \le \dfrac19\int\limits_0^\infty \dfrac{27^{-y}}{(1+y)^3}\,\text dy
= \dfrac{1-3\ln3\,(1 + 81\ln3\, \text{Ei}(-3\ln3))}{18} < 0.018865\tag5$$
(see also WA integration)

From $(2),(4),(5)$ should
$$I=I_1+I_2+I_3 < 1.9982 <2.$$

On the other hand,
$$I_1=\sum\limits_{n=1}^\infty \dfrac1{n^n}\le 1+\dfrac14+\dfrac1{27} +\dfrac1{256} + \dfrac1{3125}+\sum\limits_{n=6}^\infty\dfrac1{n^6},$$
$$I_1 \le 1+\dfrac14+\dfrac1{27} +\dfrac1{256} + \dfrac1{3125}+\dfrac{\pi^6}{945} < 1.291302,$$

$$\color{brown}{\mathbf{I< 1.291302 + 0.687550 + 0.018865 < 1.99772}}.$$


答:
Observe that, for any $ n\in\mathbb{N} $ : $$\frac{\left(2n\right)!!}{\left(2n+1\right)!!}=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x} $$

Since : \begin{aligned} \int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}&=\int_{0}^{1}{\left(1+x\right)^{n}\left(1-x\right)^{n}\,\mathrm{d}x}\\&\leq\int_{0}^{1}{\left(1+x\right)^{n}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}

Then : $$ \frac{\left(2n\right)!!}{\left(2n+1\right)!!}\underset{n\to +\infty}{\longrightarrow}0 $$

Title: Finding the limit of $\frac{(2n)!!}{(2n+1)!!}$ without using Stirling’s approximatation

Body:
I have proven that the sequence is convergent.

Let $x_n=\frac{(2n)!!}{(2n+1)!!}$

$\frac{x_{n+1}}{x_n}$ $=\frac{2n+2}{2n+3}<1$ Therefore the sequence decreases. On the other hand
$x_n>0$. Therefore the sequence is bounded below.
Since we have a sequence that is bounded below and decreases the sequence must converge.

But from here I don’t know how to find the limit. Could you please help me? And could you please say if what I’ve done is necessary or not?

Thanks in advance!


问:
Is there a proof system for first order logic that does not use premisses and auxiliary variables?
Before I have started to learn proof theory for first order logic I had the following simple (and maybe naive) expectations:

We can use formal language (first order logic) to formulate expressions (statements).

We formulate several statements that we accept without a proof (axioms).

We have a small set of "derivation rules". A derivation rule is something that can take one or two statements and generate one (or none or several) statements that follow from the input statement(s). One example might be a rule that is based on commutativity of "and  (A and B = B and A). So, we can take an expression containing "and" and change places of two "inputs" of this binary function. A new statement should not be equivalent to the input statements. It just need to follow from input. For example: A and B -> B.

So, within this approach we can start from axioms and by applying different derivation rules we can generate new statements some of which might be useful theorems.

So far I have learned sequent calculus and Hilbert's system. Hilbert's system looks closer to what I expected / wanted but it is still not the same because of several reasons:

In a proof we start from axioms and premisses (assumptions). And I want any proof to use only axioms and statements that have been proved earlier (so, no assumptions).

In a proof we use "introductions" (universal and existential). As a result we get some "intermediate / auxiliary" formula that are not valid statements by themselves. You cannot take these statements and claim that they are some lemmas or theorems because they contain some "variables" (till the end of the proof we need to get rid of them).

So, my question is, if there is a proof system that I have naively imagined (without use of premisses and in which any statement generated in the proof is a "valid" statement that can be used as a theorem).

答:
I think a typical Hilbert system is what you are looking for.

If you see premises being used, then there are probably two reasons for this:

The proof uses a premise that may have been derived as a Lemma in some earlier proof. As such, we can in fact substitute those earlier proofs and get rid of those premises, and obtain a premise-free proof.

The proof is about some domain, e.g. set theory or arithmetic or what have you. Since he theorems in those domains are not logical theorems, they have to rely on premises: axioms for those specific domains. So in that case you cannot get rid of the premises ... but that's only because you are trying to prove something beyond pure logic

Now, you also say that some proofs have lines/statements with free variables that, as such, aren't really claims by themselves. But actually, you can regard those statements as claims: any expression of the logic language requires some interpretation, and you can simply make the variable assignment part of the interpretation. As such, every line is a logically valid statement.

But if you really don't like the use of these variables, note that we can also use constants instead of the variables. The universal introduction rule can take a statement using constants and make it into a universal statement provided the use of that constant in the proof as a whole satisfies certain restrictions ... restrictions that of course make the constant really act like a variable and thus there is really no practical difference here .... but still, maybe you find this more palatable.


问:
Proof involving continuous functions

$\DeclareMathOperator{\Im}{Im}$
We've just started to learn about continuity, and this is one of the questions we were assigned.
As I am new to this topic, my confidence level in the material is quite low and I was wondering whether I am doing okay with my understanding of the material..
My proof is as follows;

Note that f is both continuous and bounded, therefore by Weierstrass'
Theorem, there exists $a∈ \Bbb R$ such that $f(x) ≤ f(a)$ for every $x∈\Bbb R$.
Since $f(a)$ is the maximal point in $\Im(f)$, by definition of a
maximum, $f(a) = \sup\Im(f) = M$.

In the same way there exists $b∈R$ such that $f(b) ≤ f(x)$ for every
$x∈R$. Since $f(b)$ is the minimal point in $\Im(f)$, by definition of
a minimum, $f(b) = \inf\Im(f) = m$ .

Clearly $M,m ⊆ \Im(f)$, and since $f$ is continuous $m ≤ f(x) ≤ M$

I'm especially not sure about the last line, is there anything related to properties of continuous functions that will help me show that the image of $f$ is contained in $[m,M]$?

答:
Your approach is wrong from the start. It is false that, since $f$ is continuous and bounded, there is some $a\in\Bbb R$ such that $(\forall x\in\Bbb R):f(x)\leqslant f(a)$; take the $\arctan$ function, for instance.

Since, by definition, $M=\sup\operatorname{Im}(f)$, it is clear that $(\forall x\in\Bbb R):f(x)\leqslant M$. By a similar argument, $(\forall x\in\Bbb R):f(x)\geqslant m$. So, $\operatorname{Im}(f)\subset[m,M]$.

Now, take $a\in(m,M)$. Then, since $M=\sup\operatorname{Im}(f)>a$, there is some $x_0\in\Bbb R$ such that $f(x_0)>a$. And, since $m=\inf\operatorname{Im}(f)>a$, there is some $x_1\in\Bbb R$ such that $f(x_1)<a$. So, by the intermedite value theorem, there is some $x$ between $x_0$ and $x_1$ such that $f(x)=a$, and so $a\in\operatorname{Im}(f)$. This proves that $\operatorname{Im}(f)\supset(m,M)$.

Title: Proof involving continuous functions


问:
Show that a certain complex power series satisfies $f'(w)=\frac{f(w)}{2w}$
Suppose $$a_{n+1}=\left(\frac 1 2 - n\right)a_n$$ with $a_0=1$ defines a real sequence of numbers.

I am then given that the following complex power series is differentiable on $D(1,1)$:
$$ f(w)= \sum_{n=0}^\infty \frac{a_n}{n!}(w-1)^n.$$
I know how to differentiate complex power series on the disk with radius $1$. Its derivative is given by:
$$f'(w)=\sum_{n=1}^\infty n \frac{a_n}{n!}(w-1)^{n-1}.$$

$$f'(w)=\sum_{n=1}^\infty \frac{a_n}{(n-1)!}(w-1)^{n-1}.$$
An index shift gives us:
$$f'(w)=\sum_{k=0}^\infty  \frac{a_{k+1}}{k!}(w-1)^{k}.$$
Which almost is the original series again, apart from an incremented coefficient.

Now I am asked to show the following relation:

$f'(w)=\frac{f(w)}{2w}$

But I do not see how I can possibly get there from what I have. I tried binomial expansion and using the recursion, but by using the recursion I get two terms and I'm struggling to make it one single series again. Any insights/hints?

答:
Observe that:
$$ f'(w) =\sum_{n =1} ^\infty n \frac{a_n}{n!} (w-1)^{n-1} =\sum_{n=1} ^\infty  \frac{a_n}{(n-1)!} (w-1)^{n-1}.$$
We are now asked to show:
$$ f'(w) = \frac{1}{2w}f(w).$$
This is equivalent to:
$$ 2w f'(w) = f(w).$$
Which can be written as:
$$ 2((w-1) +1) f'(w) = f(w).$$
This is what we will prove instead. We fill in $f '(w)$:
$$2wf'(w) =  2((w-1) +1)\sum_{n =1} ^\infty n \frac{a_n}{n!} (w-1)^{n-1}.
$$
We work out the brackets:
$$2wf'(w)= \sum_{n =1} ^\infty \frac{2 n a_n}{n!} (w-1)^{n} + \sum_{n =1} ^\infty  \frac{2 a_n}{(n-1)!} (w-1)^{n-1}.$$
We apply an index shift in the second series:
$$2wf'(w)= \sum_{n =1} ^\infty \frac{2 n a_n}{n!} (w-1)^{n} + \sum_{n =0} ^\infty \frac{2 a_{n+1}}{n!}(w-1)^{n}.$$
We combine the power series:
$$2wf'(w)=\frac{2 a_1}{0!}+ \sum_{n=1}^\infty \frac{2 na_{n}+2 a_{n+1}}{n!} (w-1)^{n}.$$
Since $a_0=1$ we have $a_1=(1/2-0)a_0=1/2 \cdot 1=1/2.$ So $2a_1=1$, we also see that $a_{n+1}=(1/2-n)a_n \implies a_{n+1}+ n a_n= 1/2 a_n $ and thus $a_n=2na_n + 2a_{n+1}$. This is precisely what appears in our power series. We see that:
$$2wf'(w)=\frac{2 a_1}{0!}+ \sum_{n=1}^\infty \frac{2 na_{n}+2 a_{n+1}}{n!} (w-1)^{n}=1+ \sum_{n=1}^\infty \frac{a_n}{n!} (w-1)^{n}, $$
$$2wf'(w) =  \sum_{n=0}^\infty \frac{a_n}{n!} (w-1)^{n} = f(w)  .$$
This then in turn is equivalent to out desired equality:
$$ f'(w) = \frac{1}{2w} f(w). \quad \square $$


答:
Your question holds a couple of mistakes. First, if you are dealing with continuous random variable then, for any x the probability P(X=x) is equal to 0 by definition. Secondly, the task to find $x$ makes no sense, since $x$ is a name of an argument here.

I would assume that by P(X=x) you mean a probability density function (PDF) $f_X(x)$, which is also defined as 0 everywhere outside $x\in [0,2]$, and will build my answer based on that information.

Find $k$. For a PDF following conditions hold by definition

$f_X(x)\geq0$ for all $x$ (clearly holds);

$\int_{-\infty}^{\infty} f_X(x)dx=1$

Calculating the integral above gives us the equation $\frac{10}{3}\cdot k=1$, thus $k=\frac{3}{10}$.

calculate $E[X]$. By definition,
$$E[X]=\int_{-\infty}^{\infty} xf_X(x)dx = \int_{0}^{2} x\cdot \frac{3}{10}(2-x)(x+1)dx=\frac45$$

Title: Estimation of probability of the distribution of a continuous random variable

Body:
I am trying to find the probability distribution of a continuous random variable X which is defined by
(X=x)=(2−x)(x+1),where 0≤=x≤2

Can anybody help me to find x and mean expected value E(x) ?


问:
$\mathbb C^{2 \times 2}$ be the vector space of $2 \times 2$ matrices ,and $T:\mathbb C^{2 \times 2} \to \mathbb C^{2 \times 2}$ is a linear operator
Let $\mathbb C^{2 \times 2}$ be the vector space of $2 \times 2$ matrices with entries in $\mathbb C$,and assume $T:\mathbb C^{2 \times 2} \to \mathbb C^{2 \times 2}$ is a linear operator such that, for all $A \in \mathbb C^{2 \times 2}$, $T(A)=BA$, where $$B=
\begin{bmatrix}
1& -1  \\
-4 & 4\\
\end{bmatrix}
\quad
$$

What is the rank of $T$?

By rank–nullity theorem I know that :

$${\displaystyle \operatorname {Rank} (T)=\dim \mathbb C^{2 \times 2}}-\operatorname {Nullity} (T)$$
Where ${\displaystyle \operatorname {Nullity} (T):=\dim(\operatorname {Ker} (T)).}$

Kernel of $T$ is the set of all  $A \in \mathbb C^{2 \times 2}$ such that $$BA=0_{\mathbb C^{2 \times 2}}=
\begin{bmatrix}
0& 0  \\
0 & 0\\
\end{bmatrix}
\quad$$

From which I conclude that $$\text{kernel (T)}=\left\{\begin{bmatrix}
x_1& x_2  \\
x_1& x_2\\
\end{bmatrix}
\quad:x_1,x_2 \in \mathbb C\right\}$$

I know that $\dim \mathbb C^{2 \times 2}=4$,so I only left with $\dim \left\{\begin{bmatrix}
x_1& x_2  \\
x_1& x_2\\
\end{bmatrix}
\quad:x_1,x_2 \in \mathbb C\right\}$,and after that I can apply the theorem and determine the rank of the linear operator.

So how can I find $\dim \text{kernel (T)}$?

Besides what is the best way to determine the rank of $T$?

答:
Since$$\ker(T)=\left\{\begin{bmatrix}x_1&x_2\\x_1&x_2\end{bmatrix}\,\middle|\,x_1,x_2\in\Bbb C\right\}$$and since $\left[\begin{smallmatrix}1&0\\1&0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0&1\\0&1\end{smallmatrix}\right]$ are two linearly independent matrices which span $\ker(T)$, $\dim\ker(T)=2$.

So,$$\operatorname{rank}(T)=4-\dim\ker(T)=4-2=2.$$


问:
Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$
Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.

What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?

Body:
The equation's defined for $\;x>1\;\;\text{or}\;\;x\le-1\;$ , so now square the whole thing:

$$x^2-1=(x+5)^2\,\frac{x+1}{x-1}\iff x^3-x^2-x+1=x^3+11x^2+35x+25\iff$$

$$12x^2+36x+24=0\iff12(x+1)(x+2)=0\ldots$$


问:
Nice geometry question to prove tangency
Let $\triangle ABC$ be a scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at a point $P$. Let $AP$ meet $BC$ at a point $K$ and let $M$ be the midpoint of $BC$. Let $X$ be a point on $AC$ such that $AB$ is tangent to the circumcircle of $\triangle BXC$. Let $BX$ meet $AM$ at a point $T$. Let the line through $C$ and parallel to $AM$ meet $TK$ at a point $V$.

Prove that $AV$ is tangent to $\omega$.

I tried the question with $\sqrt{bc}$ inversion with reflection over angle bisector. This approach does not seem to help much. Please provide hint?

答: First, $BX$ is parallel to the tangent to $\omega$ at $A$. So if $L=BX\cap AK$, we have $$-1=A(B,C;AK\cap\omega,A)=(B,X;L,\infty_{BX}).$$ So $L$ is the midpoint of $BX$, hence $L$ lies on $MN$, where $N$ is the midpoint of $AB$. Then $$-1=T(M,L;N,TK\cap MN)=(A,B;N,TK\cap AB),$$ so $TK\parallel AB$. Reflect $A$ over $M$ to $D$, so $CD\parallel AB\parallel TK$. Then $$-1=C(A,D;M,\infty_{AM})=(CA\cap TK,\infty_{TK};K,V)=(AC,AB;AK,AV).$$ This implies the result, as $(AC,AB;AK,AA)=-1$.


问:
Variable substitution in Universal Quantifier
During my math studies , I encountered variable substitutions as shown in the following examples:

$ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \epsilon $

$ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $

Another example:

$  \forall n\in \mathbb{N} \, \exists s \in \mathbb{R} .n \leq s $

$  \forall n\in \mathbb{N} \, \exists s \in \mathbb{R} .n+2 \leq s $

$  \forall n\in \mathbb{N} \, \exists s \in \mathbb{R} .n+4 \leq s $

How do we go from $ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \epsilon $

to:
$ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $
?

My naive thought was just to change the bound variable $ \epsilon $ to $ \frac{\epsilon}{4} $ so as to have:

$ \forall \frac{\epsilon}{4}>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $

But clearly this is incorrect since $ \frac{\epsilon}{4} $ is just a symbol and the 4 in the denominator is just part of the symbol and not really a number.

How do we go from  $  \forall n\in \mathbb{N} \, \exists s \in \mathbb{R} .n \leq s $

to:
$  \forall n\in \mathbb{N} \, \exists s \in \mathbb{R} .n+2 \leq s $?

Can you please give a rigorous explanation? ( I understand the intuition behind these representations, but I can't really concisely explain and/or write on a paper how do we go from one representation to another of these logical statements. )

答:
Think of how you would prove that universal statement $ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $. You would let $\epsilon$ be any real number greater than $0$, and you would have to show that for such an $\epsilon$, we have that $\exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $

OK, but why would the latter be true? It is because if $\epsilon > 0$, then we also have that $\frac{\epsilon}{4} > 0$, and we already know that for any real number (let's call it $\gamma$) it is true that $ \forall \gamma>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \gamma $. Why do we know that? Because we were given that $ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \epsilon $, but the $\epsilon$ is just a 'dummy' variable, i.e. we can use any other variable ... in this case $\gamma$.

So, since that is true for any $\gamma > 0$ that $ \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \gamma $, the latter is specifically true for $\gamma = \frac{\epsilon}{4}$, and thus we know that for any $\epsilon > 0$ it is true that $\exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $.

In other words:  $ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n \in N . | a_n - L | < \frac{\epsilon}{4} $.


问:
Atiyah-Macdonald: Exercise 3.6
Q: Let $A$ be a nonzero ring and let $\Sigma$ be the set of multiplicatively closed subsets of $A$ such that $0 \notin S.$ Show that $\Sigma$ has a maximal element, and that $S \in \Sigma$ is maximal iff $A \setminus S$ is a minimal prime ideal of $A$.

It can be shown using Zorn's Lemma $\Sigma$ has a maximal element, be showing every chain has an upper-bound. (for this $0 \notin S$ condition is not required.)

$\underline{Claim}$: $S$ maximal in $\Sigma$ iff $A \setminus S$ is a minimal prime ideal. I proceeded as follows:

$"\impliedby"$

As $A\setminus S$ is a prime ideal, $S \in \Sigma$.

Now how do I show that $S$ is maximal?

$"\implies"$

$S$: a maximal element in $\Sigma$.

$0 \in A\setminus S$, and if $xy \in A\setminus S$ then either $(x \in A
   \setminus S)$ or $(y \in A \setminus S)$ as $S$ is multiplicatively closed.

Also, $A \setminus S$ does not contain any prime ideal properly.

If it does, let $\mathfrak{p}$ be one such prime ideal. Then $A \setminus
   \mathfrak{p} \in \Sigma$
containing S properly, a contradiction.

It remains to show is $A \setminus S$ is a subgroup of the additive group $A.$ How can I show this?

Edit

Using @SteveD's comment I have completed the above proof, which I have posted as an answer. But the proof is a bit set-theoretic. I would like to see a proof using tools from Commutative Algebra.

Is there any nice application of this result?

答:
This is essentially the same answer as that of the OP's, but written to my liking. I put it here for my own future reference. Kindly ignore.

Lemma 1: Let $S$ be a maximal multiplicative subset of $A$ not containing $0$, then $A\backslash S$ is an ideal.

Proof. Let $\mathcal{I}$ be the set of ideals of $A$ contained in $ A\backslash S$ and ordered by inclusion.

$ 0 \in \mathcal{I}.$

If $\{\mathfrak{a}_i\}$ is a chain of ideals of $\mathcal{I},$ then $ \cup \mathfrak{a}_i\in \mathcal{I}.$

By Zorn's lemma, $\mathcal{I}$ contains a maximal element, say $\mathfrak{m}.$ Assume for contradiction that $\mathfrak{m} \subsetneq A\backslash S,$ then there is $x \notin \mathfrak{m} $ and $x \in A\backslash S,$ i.e. $ x \notin S.$
Consider the ideal $\bar{\mathfrak{m}} =(\mathfrak{m}, x),$ I claim that $\bar{\mathfrak{m}}$ is contained in $A\backslash S.$

By maximality of $S,$ the monoid generated by $S \cup \{x\}$ contains $0,$ i.e. there exists $s \in S$ and $r \in \mathbb{N}$  such that $ x^rs =0.$
Now assume $ m + \alpha x \in \bar{\mathfrak{m}} \cap (A\backslash S)^c =\bar{\mathfrak{m}} \cap S,$ then
$ m x^{r-1}s + \alpha x^rs = mx^{r-1}s \in S,$ but $mx^{r-1}s \in \mathfrak{m},$ thus we have a contradiction. Hence $ \mathfrak{m} = A \backslash S,$ as desired. $\square$

Lemma 2: Let $\mathfrak{p}$ be an ideal, then
$$ \mathfrak{p} \text{ is prime }\iff A\backslash \mathfrak{p} \text{ is multiplicative.} $$

Proof. This the contrapositive of the definition of a prime ideal. $\square$

The claim now follows from the following:

If $S$ maximal, then for a prime ideal $ \mathfrak{p},$
$$ \mathfrak{p} \subsetneq A \backslash S  \iff S \subsetneq A\backslash \mathfrak{p}.$$
If $\mathfrak{p}$ is minimal prime, then for a maximal multiplicative set $S'$

$$ A \backslash S' \subsetneq \mathfrak{p} \iff A \backslash \mathfrak{p} \subsetneq S'.$$

Title: Atiyah-Macdonald: Exercise 3.6


问:
Can complex numbers be visualized

If $z=|x+iy|$ is a complex number satisfying $\;|z-2|-|z-3i|=\sqrt5\;$ then locus of $z$ is $\dots$

On solving this I easily get an equation of part of hyberbola.

My question is: Is there any general result for locus problems like $(x-a)^2-(y-b)^2=c$ is a general equation of hyperbola and can be visualized easily?

Is there any sort of visualization for complex numbers?

答: Represent $x+iy$ as the point $(x,\,y)$ in the plane, so the question is which points are $\sqrt{5}$ more distant (in an additive rather than multiplicative sense) from $2$ than from $3i$. That "the points $\ell$ more distant from $p_1$ than from $p_2$" defines a hyperbola for $\ell>0$ is a theorem of Euclidean geometry in $\Bbb R^2$. Hyperbolae are $e>1$ conic sections; $e<1$ conic sections (ellipses) are the locus of points whose sums of distances to some $p_1,\,p_2$ are some $\ell>0$.

Title: Can complex numbers be visualized


问:
$K(x,t)=-2 \log \left( \sin \left( \frac{1}{2} \left( x-t \right) \right) \right)$ eigenvalues and eigenfunctions

Find the eigenvalues and eigenfunctions for the following kernel:
$$K(x,t)=-2 \log \left( \sin \left( \frac{1}{2} \left( x-t \right) \right) \right) \in \mathcal{L}_2([0,2 \pi ]^2).$$

What I have: $$y(x)=\int_0^{2 \pi} -2 \log \left( \sin \left( \frac{1}{2} \left( x-t \right) \right) \right) y(t)dt = \\ -2\int_0^{2\pi} \left( \sum_{n=1}^\infty \frac{\sin(nx) \sin(nt)}{n}+\sum_{n=1}^\infty \frac{\cos(nx) \cos(nt)}{n} \right)y(t)dt$$ by Fourier series. It sort of seems that eigenfunctions should be in the form $y(x)=A\cos(nx)+B\sin(nx)$ but I don't have any arguments for that. I know that the eigenvalues should be $\lambda_n=- \frac{n}{2 \pi}$.

Also I've found a similar question (Eigenvalues of an integral operator with non-degenerate kernel) but its kernel is much simpler. Not sure if the same technique would work in my case. What corresponding differential equation could I get from this: $y''(x)+\lambda y(x)=0$, $y(0)=y(2 \pi)=0$? However I wouldn't get the same eigenvalues $\lambda_n=- \frac{n}{2 \pi}$. Help?

答:
It is a convolution operator, convolution in $L^2(\Bbb{R/2\pi Z})$, ie. $f$ is $L^2_{loc}$ and $2\pi$-periodic and $$(Tf)(x)= \int_0^{2\pi} f(t) h(x-t)dt, \qquad h(x)=-2 \log \left( \sin \left( \frac{1}{2} \left( x \right) \right) \right)$$

The complex exponentials $e^{i n x},n\in \Bbb{Z}$ are an orthonormal basis of eigenfunctions with eigenvalues $\lambda_n=(T e^{-in x})(0)$, such that $$h = \frac1{2\pi}\sum_n \lambda_n e^{inx}$$ We find the $\lambda_n$ from $$h = \lim_{r\to 1^-}\Re(-2 \log \left(\frac{e^{ix/2}}{2i} (1-r e^{-ix})\right))$$ (convergence in $L^2(\Bbb{R/2\pi Z})$) and the Taylor series of $-\log(1-z)$

Decompose $$f = \sum_n c_n e^{inx}, \qquad Tf=\sum_n \lambda_n c_n e^{inx}$$
(convergence in $L^2(\Bbb{R/2\pi Z})$)

Then $Tf= \lambda f$ iff $c_n\ne 0 \implies\lambda_n=\lambda$.

Since we are keeping only the real part $h =\lim_{r\to 1^-}\Re(-2\log \left(1-r e^{-ix}\right))$ then from the Taylor series of $-\log(1-z)$ we have for $r\in (0,1)$ $$\Re(-2\log \left(1-r e^{-ix}\right))= \Re(2\sum_{n\ge 1} r^n e^{-inx}/n)=\sum_{n\ne 0} r^n e^{inx}/|n|$$ and $$h = \lim_{r\to 1^-} \sum_{n\ne 0} r^n e^{inx}/|n|= \sum_{n\ne 0} e^{inx}/|n|$$ which converges without problem in $L^2[0,2\pi]$. So the $\lambda_n$ are $0$ and $2\pi/|n|$ and the eigenfunctions are the linear combination of complex exponentials with the same eigenvalue ie. $ae^{inx}+be^{-inx}$.


问:
Dimensionality reduction estimation
Assume we have a vector $x\in R^d$ and consider $F(x) = s^Tx$, where $s$ is a $d$-dimensional random vector with entries drawn uniformly and independently from $\{-1,1\}$.
What is the value of $E[F(x)^2]$ ? How can it be formally proved?

答: Since $s_i^2=1$ while $E[s_is_j]=E[s_i]E[s_j]=0^2=0$ for $i\ne j$,$$F^2=(s_ix_i)^2=s_is_jx_ix_j\implies E[F^2]=\delta_{ij}x_ix_j=x^2.$$

Title: Dimensionality reduction estimation


答:
More directly, factorize  the equation as

$$\sqrt{\dfrac{x+1}{x-1}}( |x-1|-(x+5))=0$$

which leads to $x+1=0$ and $(1-x)-(x+5)=0$.

Title: Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$

Body:
Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.

What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?


问:
If $\limsup_{n\to\infty}a_n<\infty$, then $\limsup_{n\to\infty}(-a_n)>-\infty$?
Assume $(a_n)$ is a positive sequence. I stuck to find if the following always true?

If $\limsup_{n\to\infty}a_n<\infty$, then $\limsup_{n\to\infty}(-a_n)>-\infty$.

答:
Yes, it is true, because
\begin{align}
\limsup(-a_n) = -\liminf (a_n) \geq -\limsup(a_n)>-\infty.
\end{align}
The first equality should be clear from definitions, while the other steps follow from $\liminf(a_n)\leq \limsup(a_n)<\infty$. The proof even shows that this is true for all sequences $\{a_n\}_{n=1}^{\infty}$ in $[-\infty,\infty]$; positivity is not required (a little reflection should convince you that your "if then" is even an "if and only if").


问:
$\mathfrak{p}$ a prime ideal, $\varphi: A\rightarrow S^{-1}A$, then $\varphi(\varphi^{-1}(\mathfrak{p}))$
Let $A$ be a commutative ring, $S$ a multiplicatively closed subset. Let $S^{-1}A$ denote the localization. I want to show that for $\mathfrak{p}$ a prime ideal, $\varphi: A\rightarrow S^{-1}A$, the ideal generated by $\varphi(\varphi^{-1}(\mathfrak{p}))$ is indeed equal to $\mathfrak{p}$.

Denote by $q$ the ideal generated by $\varphi(\varphi^{-1}(\mathfrak{p}))$. We want to show that $q\subset p$ and that $p\subset q$.

We know $\varphi: A\rightarrow S^{-1}A$ sends $a$ to $a/1$. So $\varphi^{-1}(a/s)=a$ assuming $a$ and $s$ are coprime. So if $a/s$ is in $\mathfrak{p}$ we have that $a/1\in \varphi(\varphi^{-1}(\mathfrak{p}))$, and hence $1/s\cdot a/1\in \mathfrak{q}$.

For the other direction. If $b/q$ is in $\mathfrak{q}$ then $b/q=a'/q\cdot a/1$ where $a'/q$ is in $S^{-1}A$ and $a/p$ is in $\mathfrak{p}$ for some $p$. I don't see how to conclude that $b/q$ is in $\mathfrak{p}$. For example if $S=\{1,2,3\}$ and $\mathfrak{p}$ consisted of $a/(2^{1+n}3^{m})$ then $a/3$ would not be in $\mathfrak{p}$ but $a/p$ would be for some $p$.

答:
Note that for any function $\varphi: X \to Y$ and $U \subset Y$ we have $\varphi(\varphi^{-1}(U))\subset U$. In particular $\varphi(\varphi^{-1}(\mathfrak{p})) \subset \mathfrak{p}$. It follows that $\mathfrak{q} \subset \mathfrak{p}$ as $\mathfrak{p}$ is an ideal.

I am having some trouble understanding the argument you're trying to make in your last paragraph, but note that your set $S$ is not multiplicative.

For the other inclusion, note that if $\frac{p}{s}\in \mathfrak{p}$, then as $\mathfrak{p}$ is an ideal we have $\frac{p}{s} \frac{s}{1}= \frac{p}{1}\in \mathfrak{p}$. Hence, $\frac{p}{1}\in \varphi(\varphi^{-1}(\mathfrak{p}))$ and $\frac{p}{s} = \frac{p}{1}\frac{1}{s}\in \mathfrak{q}$.


问:
Definition of removable singularity at infinity in Stein's book
In Stein's Complex Analysis book, he wrote  'If $f$ is holomorphic for all large values of $z$, we consider $F(z)=f(1/z)$, which is now holomorphic in a deleted neighborhood of the origin. Similarly, we can speak of $f$ having an essential singularity at infinity, or a removable singularity (hence holomorphic) at infinity in terms of of the corresponding behavior of $F$ at $0$. A meromorphic function in the complex plane that is either holomorphic at infinity or has a pole at infinity is said to be meromorphic in the extended complex plane.

What does the author mean by saying 'hence holomorphic'?

答:
Let $U$ be a neighbourhood of $0$ and $F$ be holomorphic in $U \setminus \{ 0 \}$.
$F$ has a removable singularity at $z=0$ if it can be extended over $0$, i.e. if there is holomorphic function $\tilde F$ on $U$ with $\left. \tilde F \right|_{U \setminus \{ 0 \}} = F$.

In that case one sometimes says (in a slight abuse of language) that “$F$ is holomorphic at $0$” or “$F$ is holomorphic in $U$”, even though $F$ is not defined at $z=0$. In other words, the extension $\tilde F$ is again called $F$.

In the same sense, $f$ is said to be “holomorphic at infinity” if $F(z) = f(1/z)$ is holomorphic at $0$, i.e. if $F$ has a removable singularity at $z=0$.


问:
Finding $a>0$ such that $a^x\geq x^a$ for all positive $x$
I want to find all real numbers $a>0$ such that for all $x>0$ we have
$$a^x\geq x^a.$$
My guess is that the only possible solution is $a=e$. But that's really all I have so far.

答:
Hint:

It amounts to showing the function
$$f(x)=\frac{\ln x} x\quad(x>0),$$
has a unique maximum, attained at a single point.


问:
Deducing a coefficient from a cubic polynomial given a divisor and remainder?
I got this question which I don't understand:

"Suppose $x^3 - 2x^2 + a = (x + 2) Q(x) + 3$ where $Q(x)$ is a polynomial. Find the value of a."

I know the identity: $P(x)=A(x)Q(x)+R(x)$, but I'm not sure how to apply it in this question.

Any help would be appreciated, thanks.

答: Well, when the polynomial is divided by $x+2$ the remainder is $3$, so substitute $-2$ for each $x$ value, and it will equal $3$. Then, you will be able to find $a$.</p>

ContentLicense: CC BY-SA 3.0


问:
How many words are required for What3Words?
This is just curiosity / a personal exercise.

What3Words allocates every 3m x 3m square on the Earth a unique set of 3 words. I tried to work out how many words are required, but got a bit stuck.

$$
Area
    = 510 \times 10^6 km^2
    = 5.1 \times 10^{14} m^2
    => ~ 5.4 \times 10^{14} m^2
$$

(rounding up to make the next step easier!)

And so there are ~ $6\times10^{13}$ 3m x 3m squares.

I assumed I could use the equation to calculate number of combinations to find the number of words needed:

$$
_nC_r = \frac{n!}{r! (n - r)!}
$$

where $r$ is 3, and total number of combinations is the number of squares: $6\times10^{13}$

$$
6\times10^{13} = \frac{n!}{3! (n - 3)!}
$$
$$
6\times10^{13} = \frac{(n)(n-1)(n-2)(n-3)!}{3! (n - 3)!}
$$
$$
n^3 - 3n^2 + 2n - (36\times10^{13}) = 0
$$

... and then, I can't work out the first factor to use to solve the cubic equation, I'm not sure I've ever had to solve a cubic eqtn with a non-integer factor and none of the tutorials I've found have helped.

(And, my stats is also not good enough for me to be convinced this is the correct equation to use anyway!)

Any hints as to the next step would be appreciated.

答:
Almost there! Your equation is not correct tho, cause you need an order in the words. So the right inequality is $$3!\cdot\binom{n}{3}\geq 6\cdot 10^{13},$$
which is $$n(n-1)(n-2)\geq 6\cdot 10^{13}.$$ Now, you can depress it (actual math term) meaning that you can make a change of variable to make your polynomial of the form $x^3+px+q$ by doing the change of variable $x+1=n$ getting
$$(x+1)\cdot x\cdot (x-1)=x^3-x\geq 6\cdot 10^{13},$$
now you can use Cardano's formula.
getting
$$x=\sqrt[3]{3\cdot 10^{13}+\sqrt{9\cdot 10^{26}-\frac{1}{27}}}+\sqrt[3]{3\cdot 10^{13}-\sqrt{9\cdot 10^{26}-\frac{1}{27}}}=39148.67,$$
so $n=39149.67$ so take $n=39150.$


问:
What is the importance of fixing an origin in a coordinate system?
In analytic geometry, a coordinate system is defined as as pair $(O, \mathcal{B})$ consisting of a point $O$ in the space, called the origin, and a basis $\mathcal{B}=(\overrightarrow{b}_1, \overrightarrow{b}_2, \overrightarrow{b}_3)$ for the space of free vectors.

Since $\mathcal{B}$ is a basis for the space of free vector, every free vector $\overrightarrow{v}$ might be written uniquely as:

$$\overrightarrow{v}=v_1\cdot \overrightarrow{b}_1+v_2\cdot \overrightarrow{b}_2+v_3\cdot \overrightarrow{b}_3.$$ In particular, we have coordinates $v_1, v_2$ and $v_3$ for $\overrightarrow{v}$ with respect $\mathcal{B}$, which I write as $$\overrightarrow{v}=(v_1, v_2, v_3)_{\mathcal{B}}.$$

What is really the role of the origin in the coordinate system?

I mean, once we choose a basis we already have a bijection of the space of free vectors to $\mathbb R^3$.

I know that once we fix the origin $O$ we can find the coordinate of any vector $\overrightarrow{AB}$ in terms of the coordinates of $\overrightarrow{OA}$ and $\overrightarrow{OB}$, for:

$$\overrightarrow{OA}=(a_1, a_2, a_3)_{\mathcal{B}}\ \textrm{and}\ \overrightarrow{OB}=(b_1, b_2, b_3)_{\mathcal{B}}\implies \overrightarrow{OA}=\overrightarrow{OB}-\overrightarrow{OA}=(b_1-a_1, b_2-a_2, b_3-a_3)_{\mathcal{B}}.$$ I believe there is a conceptual gap in my understading of the real meaning of a coordinate system. So, what is the theoretical importance of fixing an origin after all?

Maybe what I'm missing is that the idea of the coordinate system is to establish a bijection between $\mathbb E^3$ (the space of euclidean geometry) and $\mathbb R^3$. In order to do that, we could define:

$$\varphi_{O, \mathcal{B}}: \mathbb E^3\rightarrow \mathbb R^3$$ setting

$$\varphi_{O, \mathcal{B}}(P)=(p_1, p_2, p_3)$$ where $\overrightarrow{OP}=(p_1, p_2, p_3)_{\mathcal{B}}$. Is that it? And, in fact, for defining this bijection is necessary to fix the point $O$.

答:
You've hit upon the distinction between vector spaces and affine spaces. If you identify vectors with points, the zero vector is identified with the origin. But we could translate all these identifications to change the origin, thereby preserving the paths between points. An affine space identifies each vector with the set of such paths that involve the same displacement; for example, the zero vector is identified with the set of point-to-same-point, zero-length paths.

You can think of these sets as equivalence classes labelled with vectors. With respect to the equivalence relation thereby induced, our different choices of vector-to-point identifications are all equivalent; the only element of the resulting quotient space is the aforementioned affine space.


问:
Finding Jordan Canonical form but equation for finding the generalized eigenvector becomes inconsistent
I have been given a problem to find the Jordan Canonical form of the following $ 3 \times 3 $ matrix

$$ A=
\begin{bmatrix}
3 & 0 & 1 \\
-4 & 1 & -2 \\
-4 & 0 & -1
\end{bmatrix}
$$

I have been told to do this using the transformation matrix way, wherein $A$ can be written as

$$
A = TJT^{-1}
$$
where $T$ is the modal matrix analogue which contains the eigenvectors and/or the generalized eigenvectors as columns and $J$ is Jordan Canonical form of A. Doing some matrix operations on the above equation, $J$ can be obtained as

$$
J = T^{-1}AT
$$

Coming back to the problem, I have found the characteristic equation and got all three eigenvalues as $1$. I found eigenvectors by

$$
Ax_1 = \lambda x_1
$$

Solving this I got two linearly independent eigenvectors, since I was getting only 1 pivot after row transformation operations on $A$ the above equation must give me two solutions. I got these

$$
x_1 = \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}, \begin{bmatrix}
1 \\ 0 \\ -2
\end{bmatrix}
$$

For the next step, according to my textbook, I should get a generalized eigenvector by solving for $x_2$

$$
Ax_2 = \lambda x_2 + x_1
$$

The textbook example only shows one possible value for $x_1$, so clearly I've got a problem regarding which one of the two eigenvectors I got, should be chosen for the next step.

But even without considering that, I have yet another problem

$$
Ax_2 = \lambda x_2 + x_1 \\
(A - \lambda I)x_2 = x_1
$$

If I find out the matrix $A - \lambda I = A - I$, I get

$$
\begin{bmatrix}
2 & 0 & 1\\
-4 & 0 & -2\\
-4 & 0 & -2
\end{bmatrix}
$$

which forms an inconsistent set of equations to solve, because of the repeated row but different values in the rows of (either) $x_1$. So I can't proceed further to get $x_2$.

Have I done something wrong, or is there some pre-condition to be checked on $A$ that I missed?

答:
You need to choose an eigenvector that is also in the column space of the matrix $A-\lambda I$. In this case, by looking at the matrix you have and at your eigenvector basis, one sees that an eigenvector that will work is $\begin{bmatrix}1\\0\\-2\end{bmatrix}-2\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}1\\-2\\-2\end{bmatrix}$.

Hence, you have two eigenvectors $ v_1=\begin{bmatrix}1\\0\\-2\end{bmatrix}$ and $v_2 = \begin{bmatrix}1\\-2\\-2\end{bmatrix}$. Define $v_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}$ and note that $(A-I)v_3 =v_2$. Now using $T=[v_1,v_2,v_3]$ we will get the Jordan form.

Note that if we just wanted the Jordan form $J$, without caring about what T is, then it is sufficient to know that A has $2$ eigenvectors. For then, as A is a $3\times3$ matrix the only possibility for the matrix $T$ is that it contains $2$ eigenvectors and one generalized eigenvector, which is mapped to one of the eigenvectors under $A-I$. Hence, we will get two Jordan blocks, one of size $1\times 1$ and one of size $2\times 2$. So up to ordering of the blocks $J=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$.


答: $\displaystyle a^x \ge x^a \implies x^{\frac{1}{x}} \le a^{\frac{1}{a}}$, so at which $x$ does the function $f(x)=x^{\frac{1}{x}}$ gets maximized?

Title: Finding $a>0$ such that $a^x\geq x^a$ for all positive $x$

Body:
I want to find all real numbers $a>0$ such that for all $x>0$ we have
$$a^x\geq x^a.$$
My guess is that the only possible solution is $a=e$. But that's really all I have so far.


问:
How we can write 36 in base 37?
i've been imagining that in the base 16 the number 10 is A and 11 is B so in the base 37, 35 will be Z so how we can write 36 ?</p>

答: You could use a different symbol.  Or you could use commas to separate place values in this base.  Something like $5,22,36$ could represent the three-digit base thirty-seven numeral whose value in decimal would be $5\cdot 37^2+22\cdot 37 +36$.

Title: How we can write 36 in base 37?


答:
The quickest way is to evaluate the identity $x^3 - 2x^2 + a = (x + 2) Q(x) + 3$ for $x = -2$.

You get $-8 - 8 + a = (-2 + 2) Q(-2) + 3$, so $a = 19$,

Title: Deducing a coefficient from a cubic polynomial given a divisor and remainder?

Body:
I got this question which I don't understand:

"Suppose $x^3 - 2x^2 + a = (x + 2) Q(x) + 3$ where $Q(x)$ is a polynomial. Find the value of a."

I know the identity: $P(x)=A(x)Q(x)+R(x)$, but I'm not sure how to apply it in this question.

Any help would be appreciated, thanks.


答: As the comments have said, you have to define a symbol for it.  Going to lower case letters is one possibility.  A more general approach is to use two decimal digits for each base $37$ digit so your base $37$ digits range from $00$ through $36$.  This has the advantage of being extendable to any base.

Title: How we can write 36 in base 37?

Body: i've been imagining that in the base 16 the number 10 is A and 11 is B so in the base 37, 35 will be Z so how we can write 36 ?</p>


答:
According to wikipedia and wikiwand this is called Heptatrigesimal, it uses all numbers $0-9$ and all letters of the Spanish alphabet. So people have apparently already thought about this. Here you can also find other clever solutions for all sorts of numeral systems.

This seems to be postponing the problem till we run out of letters so I definitely prefer the other options listed by the others. Though it is fun to find out that other people have at least been naming these systems and enumerating them.

Title: How we can write 36 in base 37?


答:
$a^x \geqslant x^a$ is equivalent to $x \ln a \geqslant a \ln x$.

So it is equivalent to $\frac{\ln x}{x} \geqslant \frac{\ln a}{a}$.

So we have to look at the function $g:x\mapsto \frac{\ln x}{x}$.

By deriving it, we get that $g$ is increasing then decreasing, with a maximum for $x=e$, where $e$ is the unique solution of the equation $\ln x = 1$ (it is unique because this last function is strictly increasing).

So $g$ as a unique maximum, and it is reached for a positive $x$.

So the only $a$ that verifies $a^x \geqslant x^a$ for all $x>0$ is $e$.

Hope it helped.


问:
n choose 2 for maximum possible number of inversions
I was watching this Algorithms Course and section where he mentions n choose 2. I wrote out the maximum number of inversions by hand for a 6-array element like this answer.

However, I am curious how he got n choose 2. I don't see how combinations come from there. My best idea to get the formula would have been start with 1 and do an arithmetic series up to n-1 as the general formula but I don't see how he got n choose 2.

I get that for the maximum number of inversions the array has to be ordered in reverse.

答:
Well, the array has to be ordered in reverse, no? So the first element is the biggest one, that gives you $n-1$ elements in inversion with it. Then the second one has $n-2$ elements in inversion with it. In general then the inversions are
$$\sum _{i=1}^n n-i=\sum _{i=0}^{n-1}i=\frac{(n-1)\cdot n}{2}=\binom{n}{2}.$$
Another way to think about this is that any pair of different elements in the array are in inversion! And so there are $\binom{n}{2}$ ways to choose two elements in the array.

Edit: In general $\binom{n}{k}$ is the number of ways to pick $k$ objects without order out of $n$ objects. In your particular problem say you pick one number first, you have $n$ ways to do this. Then you want to pick a different number so now you have $n-1$ options, then you have $n\cdot (n-1)$ ways to do this by the multiplication principle. The problem is that you are giving them an order (for example, you could have chosen $3,2$ or $2,3$) so every pair of numbers you are picking twice so divide by $2$ to get $\frac{n(n-1)}{2}=\binom{n}{2}.$ The same principle can be extended to $\binom{n}{k}.$


问:
Is a maximal-square-covering unique?
Let X be a shape in 2-dimensional space.

Define a square covering of X as a set of axis-aligned squares, whose union exactly equals X.

Note that some shapes don't have a finite square covering, for example, a circle or a triangle.

Define a maximal square covering of X as a square covering where each square is needed, i.e.:

For every square, there is a point in X that is covered by that square only.

No two squares can be replaced by a larger square that is contained in X.

Is a maximal square covering, defined in this way, unique? (i.e. can there be two different maximal square coverings of the same shape?)

答:
Let $Q(a,b,c,d)$ denote the square having $(a,b)$ and $(c,d)$ as opposite vertices.
Then
$$ \begin{align}X&=Q(-10,1,10,21)\cup Q(-10,-21,10,-1)\cup Q(-2,-1,2,3)\\& = Q(-10,1,10,21)\cup Q(-10,-21,10,-1)\cup Q(-2,-3,2,1)\end{align}$$
shows that uniqueness does not hold.</p>

Title: Is a maximal-square-covering unique?


问:
Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932)
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.

Is there a shorter/simpler solution than the one presented below that I made some months ago? It seems rather ‘lengthy’.

Solution

Let the angles be $\alpha$ and $\beta$.

We have
$$
\left\{
\begin{aligned}
\alpha+\beta&=135°,\\
\tan(\alpha)+\tan(\beta)&=5.
\end{aligned}
\right.
$$

Since
$$
\tan(x)+\tan(y)
=\frac{\sin(x)}{\cos(x)}+\frac{\sin(y)}{\cos(y)}
=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)}
=\frac{\sin(x+y)}{\cos(x)\cos(y)}
$$
we have
$$
5
=\tan(\alpha)+\tan(\beta)
=\frac{\sin(\alpha+\beta)}{\cos(\alpha)\cos(\beta)}
=\frac{\sin(135°)}{\cos(\alpha)\cos(\beta)}
=\frac{\frac{1}{\sqrt{2}}}{\cos(\alpha)\cos(\beta)}
$$
which gives
$$
\cos(\alpha)\cos(\beta)=\tfrac{1}{5\sqrt{2}}.
$$

Further
$$
-\tfrac{1}{\sqrt{2}}
=\cos(135°)
=\cos(\alpha+\beta)
=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}-\sin(\alpha)\sin(\beta)
$$
which gives
$$
\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}+\tfrac{1}{\sqrt{2}}
=\tfrac{6}{5\sqrt{2}}.
$$

Since $\alpha+\beta=135°$ we have
\begin{align*}
\sin(\alpha)\sin(\beta)
&
=\sin(\alpha)\sin(135°-\alpha)
\\&=\sin(\alpha)\bigl(\sin(135°)\cos(\alpha)-\cos(135°)\sin(\alpha)\bigr)
\\&=\sin(\alpha)\bigl(\tfrac{1}{\sqrt{2}}\cos(\alpha)+\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigl(\cos(\alpha)+\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\bigl(\sin(\alpha)\cos(\alpha)+\sin^2(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\Bigl(\tfrac{1}{2}\sin(2\alpha)+\tfrac{1}{2}\bigl(1-\cos(2\alpha)\bigr)\Bigr)
\\&=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\end{align*}
why
\begin{gather*}
\tfrac{6}{5\sqrt{2}}=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\\\quad\Leftrightarrow\quad
\tfrac{12}{5}=\sin(2\alpha)+1-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sin(2\alpha)-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sqrt{2}\sin(2\alpha-\tfrac{\pi}{4})
\\\quad\Leftrightarrow\quad
\tfrac{7}{5\sqrt{2}}=\sin(2\alpha-\tfrac{\pi}{4})
\end{gather*}
which gives
\begin{gather*}
2\alpha-\tfrac{\pi}{4}=
\begin{cases}
\arcsin(\tfrac{7}{5\sqrt{2}})+n_12\pi\\
\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+n_22\pi
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\bigl(\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_12\pi\bigr)\\
\tfrac{1}{2}\bigl(\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_22\pi\bigr)
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{8}+n_1\pi\\
-\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{5\pi}{8}+n_2\pi
\end{cases}
\end{gather*}
where $\beta=135°-\alpha=\frac{3\pi}{4}-\alpha$ and $n_1,n_2\in\mathbb{Z}$, and vice versa since the problem is ‘symmetrical’.

The original exam

答:
Take $\tan$ of both sides of first equation, $$\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$

$$\Rightarrow \tan 135 = -1 = \dfrac{5}{1-\tan \alpha \tan \beta}  $$

$$\Rightarrow \tan \alpha \tan \beta=6$$

Setting $\tan \alpha = x$,  $\tan \beta = y$ we get two equations : $$x+y=5 \quad xy=6$$

I believe you can easily solve from here.


答: Recall that $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. Now, given $\alpha+\beta=135^{\circ}$ and $\tan(\alpha+\beta)=\tan 135^\circ=-1$, you get $\frac{5}{1-\tan\alpha\tan\beta}=-1$, i.e. $\tan\alpha\tan\beta=6$. Now, knowing the sum ($5$) and the product ($6$) of $\tan\alpha$ and $\tan\beta$, using Vieta formulas we conclude that $\tan\alpha$ and $\tan\beta$ are the solutions of the quadratic equation $x^2-5x+6=0$, i.e. $\tan\alpha$ and $\tan\beta$ are $2$ and $3$ in some order. This is now very easy to finish off: $\alpha=\arctan 2+n\cdot 180^{\circ}$ and $\beta=135^{\circ}-\alpha=\arctan 3-n\cdot 180^{\circ}$, or vice versa.

Title: Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932)

Body:
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.

Is there a shorter/simpler solution than the one presented below that I made some months ago? It seems rather ‘lengthy’.

Solution

Let the angles be $\alpha$ and $\beta$.

We have
$$
\left\{
\begin{aligned}
\alpha+\beta&=135°,\\
\tan(\alpha)+\tan(\beta)&=5.
\end{aligned}
\right.
$$

Since
$$
\tan(x)+\tan(y)
=\frac{\sin(x)}{\cos(x)}+\frac{\sin(y)}{\cos(y)}
=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)}
=\frac{\sin(x+y)}{\cos(x)\cos(y)}
$$
we have
$$
5
=\tan(\alpha)+\tan(\beta)
=\frac{\sin(\alpha+\beta)}{\cos(\alpha)\cos(\beta)}
=\frac{\sin(135°)}{\cos(\alpha)\cos(\beta)}
=\frac{\frac{1}{\sqrt{2}}}{\cos(\alpha)\cos(\beta)}
$$
which gives
$$
\cos(\alpha)\cos(\beta)=\tfrac{1}{5\sqrt{2}}.
$$

Further
$$
-\tfrac{1}{\sqrt{2}}
=\cos(135°)
=\cos(\alpha+\beta)
=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}-\sin(\alpha)\sin(\beta)
$$
which gives
$$
\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}+\tfrac{1}{\sqrt{2}}
=\tfrac{6}{5\sqrt{2}}.
$$

Since $\alpha+\beta=135°$ we have
\begin{align*}
\sin(\alpha)\sin(\beta)
&
=\sin(\alpha)\sin(135°-\alpha)
\\&=\sin(\alpha)\bigl(\sin(135°)\cos(\alpha)-\cos(135°)\sin(\alpha)\bigr)
\\&=\sin(\alpha)\bigl(\tfrac{1}{\sqrt{2}}\cos(\alpha)+\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigl(\cos(\alpha)+\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\bigl(\sin(\alpha)\cos(\alpha)+\sin^2(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\Bigl(\tfrac{1}{2}\sin(2\alpha)+\tfrac{1}{2}\bigl(1-\cos(2\alpha)\bigr)\Bigr)
\\&=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\end{align*}
why
\begin{gather*}
\tfrac{6}{5\sqrt{2}}=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\\\quad\Leftrightarrow\quad
\tfrac{12}{5}=\sin(2\alpha)+1-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sin(2\alpha)-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sqrt{2}\sin(2\alpha-\tfrac{\pi}{4})
\\\quad\Leftrightarrow\quad
\tfrac{7}{5\sqrt{2}}=\sin(2\alpha-\tfrac{\pi}{4})
\end{gather*}
which gives
\begin{gather*}
2\alpha-\tfrac{\pi}{4}=
\begin{cases}
\arcsin(\tfrac{7}{5\sqrt{2}})+n_12\pi\\
\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+n_22\pi
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\bigl(\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_12\pi\bigr)\\
\tfrac{1}{2}\bigl(\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_22\pi\bigr)
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{8}+n_1\pi\\
-\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{5\pi}{8}+n_2\pi
\end{cases}
\end{gather*}
where $\beta=135°-\alpha=\frac{3\pi}{4}-\alpha$ and $n_1,n_2\in\mathbb{Z}$, and vice versa since the problem is ‘symmetrical’.

The original exam


问:
Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$
Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.

What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?

答:
$x^2 - 1 = (x+1)(x-1)$.

So you have $\sqrt{(x+1)(x-1)} =(x+5) \sqrt\frac  {x+1}{x-1}$.

Now my first thoughts are dividing both sides by $\sqrt{x+1}$ and multiplying both sides by $\sqrt{x-1}$ to get $x-1 = x+5$.

BUT BE CAREFUL!

First off.  When we divide both sides by $\sqrt{x+1}$ we are assuming that $\sqrt{x+1}\ne 0$.  We must consider the case where $\sqrt{x+1} = 0$ so $x = -1$ as a potential solution.

Indeed, if $x = -1$ we have $\sqrt{x^2 -1} =0$ and $(x+5) \sqrt{\frac {x+1}{x-1}} = 0$.  So $x=-1$ is a solution.

And we could divide both sides by $\sqrt{x+1}$.... except....

$\frac {\sqrt{(x+1)(x-1)}}{\sqrt{x+1}} \ne \sqrt{x+1}$ and $\sqrt{\frac {x+1}{x-1}}{\sqrt{x+1}}\ne \frac 1{\sqrt{x-1}}$ unless both $x+1$ and $x-1$ are positive.  If they are both negative the terms we are refering to simply are not designed.

So we could two cases where we consider both $x+1, x-1$ are positive and a case were they are both negative, in which case we divide both sides by $\sqrt{-(x+1)}$ and get $\sqrt{1-x} = (x+5)\frac 1{\sqrt {1-x}}$.

But more sophisticatedly we shouldn't divide and multiply in two steps.  We should simply multiply both sides by $\sqrt{\frac{x-1}{x+1}}$.

In this case we still must make the case where $x+1 = 0$ for a potential solution.

So if $x = -1$ that is a solution.  We did that above.

If $x \ne -1$ multiply bot sides by  $\sqrt{\frac{x-1}{x+1}}$ to get:

$\sqrt{(x^2-1)\frac {x-1}{x+1}}=(x+5)\sqrt 1$ or

$\sqrt{(x-1)^2} = x+5$ or

$|x-1| = x+ 5$.

(Note my initial thought of $x-1$ being positive was erroneous and we get $|x-1| = x+5$ not $x-1=x+5$)

If $x -1 \ge 0$ we get $x-1=x+5$ which has no solutions.

If $x -1<0$ we get $1-x = x+5$ or $x=-2$.

So our two solutions are $x=-1$ and $x=-2$ both of which fit in the domain that you calculated.


答:
$$\sqrt{x^{2}-1}=\left(x+5\right)\frac{\sqrt{x+1}}{\sqrt{x-1}}$$
Both sides are only defined for $|x|>1$ or $x=-1$

Squaring both sides,
$$x^{2}-1=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
We can factor the LHS using difference of squares,
$$\left(x+1\right)\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
Assuming $|x|>1$, we can divide both sides by $x+1$ which is nonzero (We can check $x=-1$ as a sperate case).
$$\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{1}{x-1}\right)$$
Multiplying both sides by $x-1$,
$$\left(x-1\right)^{2}=\left(x+5\right)^{2}$$
$$\Rightarrow \pm\left(x-1\right)=\pm\left(x+5\right)$$
There won't be any solutions for $(x-1)$=$(x+5)$ or $-(x-1)$=$-(x+5)$, so $(x-1)$ and $(x+5)$ must be additive inverses.
$$(x-1)=-(x+5)$$
$$\Rightarrow x=-2$$
Both sides of the original equation are defined for $x=-2$, so it's a solution.

Earlier we only considered values of x that satisfied $|x|>1$. If $x=-1$, both sides of the equation are $0$, so $x=-1$ is also a solution.


问:
$\pi$ is isomorphism from one Peano system $(N, S, e)$ to another $(N', S', e')$, then $\pi^{-1}$ is isomorphism from $(N', S', e')$ to $(N, S, e)$
This is an exercise from Cunningham's book "Set Theory: A First Course".

Theorem: Let $(N, S, e)$ and $(N', S', e')$ be Peano systems. Let $\pi$ be an isomorphism from $(N, S, e)$ onto $(N', S', e')$. Prove that $\pi^{-1}$ is thus an isomorphism from $(N', S', e')$ onto $(N, S, e)$.

Definition: Isomorphism $\pi$ is the following bijection
$$
\pi =
\begin{cases}
\pi(e) = e' \\
(\forall x \in N)\pi(S(x)) = S'(\pi(x))
\end{cases}
$$
Proof: Let $(N, S, e)$ and $(N', S', e')$ be Peano systems. Let $\pi$ be an isomorphism from $(N, S, e)$ onto
$(N', S', e')$. $\pi$ is a bijection, so is $\pi^{-1}$. To prove that $\pi^{-1}\colon N'\to N$ is an isomorphism we have to show
that:
$$
\pi^{-1} =
\begin{cases}
\pi^{-1}(e') = e \\
(\forall x \in N')(\pi^{-1}(S'(x)) = S(\pi^{-1}(x)))
\end{cases}
$$

Since $\pi(e)=e'$ then $\pi^{-1}(e') = e$. Now we prove the second statement by induction.
Consider a set $A = \{y \in N': \pi^{-1}(S'(y)) = S(\pi^{-1}(y)))\}$, prove that it is closed under
$S'$. First show that $e'\in A$.

$$
\begin{align*}
\pi(S(e)) &= S'(\pi(e)) &\text{def. of $\pi$} \\
\pi(e) &= e' \\
\pi^{-1}(e') &= e \\
\pi^{-1}(S'(e')) &= S(\pi^{-1}(e')) \\
e' &\in A
\end{align*}
$$
For induction hypothesis assume $y \in A$, therefore $\pi^{-1}(S'(y)) = S(\pi^{-1}(y))$. Show that $S'(y)$ is also in $A$ by proving that $\pi^{-1}(S'(S'(y))) = S(\pi^{-1}(S'(y)))$ holds. Since $y \in N'$ and $\pi^{-1}:N' \to N$ is bijection, then there exists unique $x \in N$ such that $x = \pi^{-1}(y)$
$$
\begin{align*}
(\forall x \in N)(\pi(S(x)) &= S'(\pi(x))) &\text{def. of $\pi$}\\
S(x) &\in N \\
\pi(S(S(x)) &= S'(\pi(S(x)))) &\text{from def. of $\pi$} \\
\pi(S(x) &= S'(\pi(x))) &\text{from def. of $\pi$} \\
\pi(S(S(x)) &= S'(S'(\pi(x))) \\
\pi^{-1}(S'(S'(\pi(x)))) &= S(S(x)) \\
\pi^{-1}(S'(S'(\pi(\pi^{-1}(y))))) &= S(S(\pi^{-1}(y)))) &\text{replacing $x$ with $\pi^{-1}(y)$}\\
\pi^{-1}(S'(S'(y)))) &= S(S(\pi^{-1}(y))) \\
\pi^{-1}(S'(S'(y)))) &= S(\pi^{-1}(S'(y))) &\text{by Induction Hypothesis}\\
S'(y) &\in A \\
\text{The set $A$ is closed under $S'$, hence $A=N'$.} \\
(\forall x \in N')(\pi^{-1}(S'(x)) &= S(\pi^{-1}(x)))
\end{align*}
$$
Therefore $\pi^{-1}$ is an isomorphism from $(N', S', e')$ onto $(N, S, e).\qquad\square$

Even though the derivation above seems valid to me it feels too "straight forward", therefore my questions are: is the proof valid and whether more "elegant" solution exists. Cheers.

答:
Induction is not needed here.

For $x \in N'$, let $y = \pi^{-1}(x)$, so $x = \pi(y)$. Since $\pi$ is an isomorphism $$S'(x) = S'(\pi(y)) = \pi(S(y))$$ and $$\pi^{-1}(S'(x)) = S(y) = S (\pi^{-1}(x))$$
as required.

This is a simple theorem. One would expect its proof to be straight-forward.


问:
To show that function is constant
Let $f$ be defined on $\mathbb{R}$ and suppose that |$f(x)$ - $f(y)$| $\leq$ $(x-y)^2$ $x,y \in\mathbb{R}$. Here I have to show that $f$ is a constant function.

I think I have to show that $f'(x)$ = 0 for all $x$.
But I don't know from where to start this. I tried taking it as (|$f(x)$ - $f(y)$|/|$x$-$y$|) $\leq$ |$x$ - $y$|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks

答:
For any $x, y \in \mathbb{R}$, subdivide the line segment between $x$ and $y$ into $N$ pieces, we have:

$$\begin{align}
| f(y) - f(x) | = &\left|\sum_{i=1}^N \left( f(x + \frac{i}{N}(y-x)) - f(x+\frac{i-1}{N}(y-x) \right) \right|\\
\le & \sum_{i=1}^N \left| f(x + \frac{i}{N}(y-x)) - f(x+\frac{i-1}{N}(y-x)) \right|\\
\le & \sum_{i=1}^N \left|\frac{y-x}{N}\right|^2 = \frac{|y-x|^2}{N}
\end{align}$$
Since $N$ can be taken to be arbitrary large, we have:
$$|f(y) - f(x)| \le \liminf_{N\to\infty} \frac{|y-x|^2}{N} = 0\quad\implies\quad f(y) = f(x)$$

Please note that this argument not only works for $f : \mathbb{R} \to \mathbb{R}$, but for any map between normed vector spaces provided the $|\cdot|$ is interpreted properly.

Title: To show that function is constant


问:
Quantile function of Student's T-distribution and Normal distribution
We all know that the t-distribution has a heavier tail than normal distribution (this means for a given percentile level $\alpha$, the quantile of student's T-distribution $q_{t_v}^\alpha$ is greater than the quantile of normal distribution $q_{N(0,1)}^\alpha$, $v$ is the degree of freedom of the t-distribution). But it seems to me that we have this conclusion just because we are comparing apples and oranges, the two distribution don't have the same variance. Why don't compare a normal distribution with a variance of $1$ with a Student's T-distribution with a variance of $10000$ ?

Now, if we compare the quantile of a normal distribution with a Student's T-distribution of same variance, we will have a paradoxical conclusion.

Let $X$ and $Y$ two random variables with mean of $0$ and variance of $1$. $X$ follows the normal distribution $N(0,1)$ while $Y$ follows the Student's T-distribution $t_v$.

As the variance of $Y$ is equal to $1$, $Y$ must equal in distribution to $\sqrt{\frac{v-2}{v}} t_v$ (because the variance of $t_v$ is equal to $\frac{v}{v-2}$ ):
$$Y \overset {d}{=} \sqrt{\frac{v-2}{v}} t_v$$

Now, we calculate the quantile at the level $\alpha$ for both $X$ and $Y$:
$$
P(X \leq q_{X}^\alpha)  = P(N(0,1) \leq q_{X}^\alpha)=\alpha  \Rightarrow   q_{X}^\alpha=F_{N(0,1)}^{-1}(\alpha)
$$
and
$$
P(Y \leq q_{}^\alpha)  = P(\sqrt{\frac{v-2}{v}} t_v \leq q_{Y}^\alpha)=\alpha
$$
$$
\Rightarrow   \sqrt{\frac{v}{v-2}} q_{Y}^\alpha = F_{t_v}^{-1}(\alpha)
$$
$$
\Rightarrow   q_{Y}^\alpha = \sqrt{\frac{v-2}{v}} F_{t_v}^{-1}(\alpha)
$$

We take $\alpha = 0.95$ for example, the quantile $q_{X}^\alpha \approx 1.64 $ is always greater than $q_{Y}^\alpha$, $\forall v$  (for example, if $v = 5$, $q_{Y}^\alpha = \sqrt{\frac{5-2}{2}} F_{t_5}^{-1}(0.95) \approx 1.56$.

In practice, we usually have observations from a random variable $Z$ and we can easily compute the variance and mean of this variable, so the variance and mean are known. From my demonstration, the quantile of $Z$ is smaller if we suppose $Z$ follows the Student's T-distribution rather than Normal distribution.

This conclusion seems contradictory to what we read about the heavier tail of Student's T distribution, doesn't it?

答:
Let's compare any Student's distribution $t_{\nu}$ and any normal distribution $N(a,\sigma^2)$.

Student's distribution $t_{\nu}$ has density $$f_{\nu}(t) \sim C_{
\nu} t^{- \nu -1}, t \to +\infty$$
(see en.wikipedia.org/wiki/Student%27s_t-distribution) and normal distribution $N(a,\sigma^2) $has density
$$f_{a, \sigma^2}(t) = C_{a, \sigma^2} e^{-\frac{(x- a)^2}{2\sigma^2}}.$$

Hence $$P(t_{\nu} > x) >   P(N(a,\sigma^2)  > x) \text{ ${}$(1)}$$
for $x \ge x_0(\mu, a, \sigma)$.

But (1) may be false for $x$ which are not big enough.

For example, $(n+100)^{100}$ grows slowlier, than $e^n$, but $(n+100)^{100}  > e^n$ for $n \le 100$.


答:
From scratch:

Applying the definitions:

$\tag1 \limsup_{n\to\infty}(-a_n)=\underset{k\to \infty}\lim g_k\quad  \text{where}\quad  g_k=\sup_{n\ge k}(-a_n).$

It suffices then to show that

$\tag2 -\sup (-S)=\inf S\  \text{for any set}\  S$

The definition of $\sup$ implies that if $\alpha=\sup (-S)<\infty$, then

$\tag3 \alpha\ge -s\quad  \text{for each}\  s\in S\quad  \text{and if}\quad  \epsilon>0\ \text{there is a}\  -s\in -S\  \text{such that}\  \alpha-(-s)<\epsilon.$

Or equivalently,

$\tag4 -\alpha\le s\quad \text{for all}\quad  s\in S\quad \text{and} -\alpha-s>\epsilon$

The cases $\alpha=\pm \infty$ are even easier.

Title: If $\limsup_{n\to\infty}a_n<\infty$, then $\limsup_{n\to\infty}(-a_n)>-\infty$?

Body:
Assume $(a_n)$ is a positive sequence. I stuck to find if the following always true?

If $\limsup_{n\to\infty}a_n<\infty$, then $\limsup_{n\to\infty}(-a_n)>-\infty$.


问:
Prove that for $f(x) = \frac{x^\textrm{T}Ax}{x^\textrm{T}x}$ and any $v$, $v^\textrm{T}Hv=0$
$f(x) = \frac{x^\textrm{T}Ax} {x^\textrm{T}x}$, where $A$ is a symmetric $n \times n$ matrix and $x \neq 0$. I need to prove that if $H=H(f)(v)$, where $H(f)(v)$ is the Hessian of function $f$ evaluated for vector $v$, $v^\textrm{T}Hv=0$ for any $v$.

I have already proved that:

(a) $\frac{df}{dx}=0$ iff $x$ is an eigenvector of $A$;

(b) $f(cx) = f(x)$ for $c \in R$;

(c) $\lambda_\min \leq f \leq \lambda_\max$ where $\lambda_\max$ and $\lambda_\min$ are the maximal and minimal eigenvalues of $A$.

The hint to the question states the following: “ Hint: do not attempt to calculate the Hessian directly: the algebra is very messy. Use
the intuition gave by (b), that a certain function is constant.”

答:
Let $v_1,v_2,\cdots, v_n$ be a basis of $R^n$ such that they are eigenvectors of $A$. By (a),
$$ \nabla f(x)v_i=\frac{\partial f(x)}{\partial v_i}=0, \forall x\in R^n. $$
So
\begin{eqnarray}
v_j^TH(f)v_i=H(f)(v_i,v_j)=\lim_{t\to0}\frac{\nabla f(x+tv_i)(v_j)-\nabla f(x)(v_j)}{t}=0.
\end{eqnarray}
Hence for $\forall v\in R^n$, $v=\sum_{i=1}^nc_iv_i$, one has
$$ vH(f)v=\sum_{i,j=1}^nc_ic_jv_j^HH(f)v_i=0. $$


问:
Nice geometry question to prove tangency
Let $\triangle ABC$ be a scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at a point $P$. Let $AP$ meet $BC$ at a point $K$ and let $M$ be the midpoint of $BC$. Let $X$ be a point on $AC$ such that $AB$ is tangent to the circumcircle of $\triangle BXC$. Let $BX$ meet $AM$ at a point $T$. Let the line through $C$ and parallel to $AM$ meet $TK$ at a point $V$.

Prove that $AV$ is tangent to $\omega$.

I tried the question with $\sqrt{bc}$ inversion with reflection over angle bisector. This approach does not seem to help much. Please provide hint?

答:
Here you have an approach employing barycentric coordinates. For more information, have a look at this awesome handout by Evan Chen. Following his advice, I will use , when referring to normalized coordinates, and : for homogeneous coordinates.

Let $\triangle ABC$ be the reference triangle, i.e. $A=(1,0,0); B=(0,1,0);C=(0,0,1)$ with side lengths $a=BC, b=CA, c=AB$. Observe that $M=(0:1:1).$ Furthermore, notice that $(\triangle BXC)$ being tangent to $AB$ is equivalent to $$\angle XBA=\angle ACB\implies \triangle ABX\sim \triangle ACB\implies \frac{AX}{AC}=\frac{c^2}{b^2} $$ Therefore $$\overrightarrow{AX}=\frac{c^2}{b^2}\cdot \overrightarrow{AC}=\frac{c^2}{b^2}\cdot (-1, 0, 1)=\left(-\frac{c^2}{b^2}, 0, \frac{c^2}{b^2}\right)\implies X=\left(1-\frac{c^2}{b^2},0, \frac{c^2}{b^2}\right)\equiv \left(b^2-c^2:0:c^2\right)$$
Thus, we infer that $AM: z-y=0$ and $BX: c^2\cdot x-\left(b^2-c^2\right)\cdot z=0$. It follows that $T=\left(b^2-c^2:c^2:c^2\right)$.

On the other hand, looking at the definition of $K$, it is well-known that $AK$ is but the $A$-symmedian. Hence, $$AK:b^2z-c^2y=0, \; BC:x=0\implies AK\cap BC=K=\left(0:b^2:c^2\right)$$ It follows that \begin{align*}TK&: (b^2-c^2)b^2\cdot z+c^4\cdot x-c^2b^2\cdot x-(b^2-c^2)c^2\cdot y=0\\\iff TK&: (b^2-c^2)\cdot (b^2z-c^2y)+c^2x\cdot (c^2-b^2)=0\\\iff TK&: b^2z-c^2y-c^2x=0\end{align*} In order to determine the equation of the line $l$ through $C$ parallel to $AM$, I will use the method described here (see page $76$). This yields $$l: x+2y=0$$ Thus $l\cap TK=V=(-2b^2:b^2:-c^2)$. Besides, the equation of the tangent $t$ to $\omega$ at $A$ is given by $b^2z+c^2y=0$ (cf. Chen's handout, corollary $15$). We are almost there, since $$V\in t\iff b^2\cdot(-c^2)+c^2\cdot b^2=0 $$ Which is clearly true.


问:
Open Sets and Boundary Points

Preliminary Definitions

Definition (Open Sets and Topological Space): Let $X$ be a set. A topology on $X$ is a subset $\mathcal{T}\subseteq \mathcal{P}(X)$ with the following properties:

$\emptyset \in \mathcal{T}$ and $X\in\mathcal{T}$

Unions of elements of $\mathcal{T}$ belong to $\mathcal{T}$.

Intersectiosn of finitely many elements of $\mathcal{T}$ belong to $\mathcal{T}$.

The elements $T_{i}\in\mathcal{T}$ are called open sets and the pair $(X,\mathcal{T})$ is called a topological space.

Definition (Neighbourhood): Let $(X,\mathcal{T})$ be a topological space. A subset $U\subseteq X$ is called a neighbourhood of a point $x\in X$, if $U$ contains an open set $T_{x}\in\mathcal{T}$ containing $x$, i.e. $x\in T_{x}\subseteq U$.

Definition (Interior points and boundary points): Let $(X,\mathcal{T})$ be a topological space and $A\subseteq X$ a subset of $X$.

A point $x\in X$ is called an interior point of $A$, if $A$ is a neighbourhood of $x\in X$. The set of all interior points of $A$ is called the interior of $A$, denoted as $A^{\circ}$.

A point $x\in X$ is called a boundary point of $A$, if every neighbourhood of $x\in X$ contains points of $A$ and points of $X\setminus A$.

Question

Some texts state an open set $A\subseteq \mathcal{T}$ contains all interior points of $A$ but no boundary points of $A$ (which sounds kinda plausible since in the set of real numbers, for example, we call $(a,b)$ an open intervall and it obviously contains all interior points of that intervall but none of the two boundary points).

I tried to prove that statement using the definitions given above without success so far. Any ideas?

答:
All sets contains its interior points by definition, because if $U$ is neighborhood of $x$ then $x\in U$

But if $A$ is open then all its points are interior points.

But interior point can't be boundary point, because if $x \in A^{\circ}$ then is neighborhood of $x$, but $A$ contains no points of $X\setminus A$, so $x$ not boundary for $A$.

Therefore $A$ contains no boundary points.

Title: Open Sets and Boundary Points


问:
Proving or disproving a statement about discontinuity

I'm trying to disprove this statement, but as spend more time on it I'm beginning to think it might actually be true.
How should I approach on proving/disproving this statement?

答:
Counter-example:

Let $f(x)=x$ for every $x.$ Let $x_0=0.$ Let $\epsilon =1/3.$

For any $\delta >0$ there exist $x'$ and $x''$ in $(-\delta +x_0,\,\delta +x_0)$ with $x'\in \Bbb Q$ and $x''\not\in \Bbb Q.$ Then $g(x')=f(D(x'))=f(1)=1$ and $g(x'')=f(D(x''))=f(0)=0.$

So it cannot be true that $|g(x)-g(x_0)|<\epsilon$ for all $x\in (-\delta +x_0,\,\delta +x_0),$ else we would have $1= |g(x')-g(x'')|\le |g(x')-g(x_0)|+|g(x_0)-g(x'')|<2 \epsilon=2/3. $

So $g$ is not continuous at $x_0$ but we still have $f(x_0)=f(0)=0.$


问:
Is magnitude operator |..| linear or non-linear?
I am working on a problem in electrical engineering where I have to take the magnitude square of the following term as given in the picture below

I have to take the magnitude square of the equation no -4 which is complex, so that is going to be a^2 +b^2 + 2 a*b or just a^2 + b^2?

as I read somewhere above picture

答:
I'm not sure if this is what you're asking, but here goes.The RHS of equation 4 is of the form $|z_1 + z_2|^2$, where $z_1, z_2$ are complex numbers. Let $*$ denote complex conjugation. Note that $|z|^2=z^*\cdot z$. Writing it out as follows

$$|z_1 + z_2|^2 = (z_1 + z_2)^* \cdot(z_1 + z_2)= z_1^*z_1+z_1^*z_2+z_2^*z_1+z_2^*z_2$$
$$= |z_1|^2+|z_2|^2+z_1^*z_2+z_2^*z_1$$

After which you plug in the terms $z_1, z_2$


问:
How to correctly apply probability multiplication and addition laws?
I'm trying to apply the probability addition rule to the problem below.

There are 12 different socks in a drawer.
The table below shows the different varieties:

Thickness
chunky (C)  or  thin (T)

Style
stripy (S) or dotty (D) or plain (P)

Colour
red (R) or blue (B)

Thickness
Style
Colour

C
S
R

C
S
B

C
D
R

C
D
B

C
P
R

C
P
B

T
S
R

T
S
B

T
D
R

T
D
B

T
P
R

T
P
B

Based on the table, some simple observations:

Probability that a chunky sock is taken out: 6:12

Probability that a stripy red sock is taken out: 2:12

This is where i'm getting confused based on applying the laws:

Probability that a dotty and red sock is taken out:

probability of dotty sock = 4:12

probability of red sock = 6:12

applying multiplication law, probability of dotty and red sock = 4/12 * 6/12 = 1:6

1:6 seems to correctly reflect observed data in table, so I assume multiplication law is correctly applied in this case?

Probability that a sock that is neither plain nor blue is taken out:

probability of plain sock = 4:12

probability of blue sock = 6:12

applying addition law, probability of plain or blue sock = 4/12 + 6/12 = 10:12

therefore probability of neither plain or blue sock is everything else i.e 2:12 = 1:6

observed data in table suggest this should be 4:12 = 1:3

What might be wrong in my understanding of the problem and/or application of the addition law?

答:
The probability that a dotty and red sock is taken is 1:6 is correct.

The mistake in the second method:

Let A be one event and B be the second event.

Neither A nor B means (not A) and (not B)

The probability that neither A nor B is selected is $P($not $A) \cdot P($not $B)$

In your case,

Probability that a sock that is neither plain nor blue is taken out = $P($not blue$) \cdot P($not plain$)$

P(not blue)= $1 - \frac{6}{12} = \frac{1}{2}$

P(not plain) =  $1 - \frac{4}{12} = \frac{2}{3}$

Probability that a sock that is neither plain nor blue is taken out = $\frac{1}{3}$

Hope this helps

EDIT:

P(A or B) = P(A) + P(B) - P(A and B)

P(A and B) = P(A).P(B) only when A and B are independent. Independent means that an effect on A does not affect B.

Basically

P(neither A nor B) = 1- P(A or B) = 1 - P(A) - P(B) +P(A and B)

Now in this question, A and B are independent so P(A and B) = P(A)P(B)

So,

P(neither A nor B) = 1- P(A or B) = 1 - P(A) - P(B) +P(A)P(B)

$---------------------------------------$
Also,

P(neither A nor B) = not(P(A)) and not (P(B))

So,

P(neither A nor B) = (1 - P(A))(1 - P(B)) = 1 - P(A) - P(B) +P(A)P(B)

You get the same result in both cases.

If you have any more doubts you can ask in the comment


答: Further to the calculations in @BotnakovN's answer, you can always compare a $t$-distribution to the Normal distribution with the same mean and variance, or even compare both distributions' z-scores and obviate any "different variance" objection. You then find the $t$ tails are heavier in the sense their decay is subexponential (indeed, they're fat), while the Gaussian tails decay superexponentially. (See also all definitions here.)

Title: Quantile function of Student's T-distribution and Normal distribution

Body:
We all know that the t-distribution has a heavier tail than normal distribution (this means for a given percentile level $\alpha$, the quantile of student's T-distribution $q_{t_v}^\alpha$ is greater than the quantile of normal distribution $q_{N(0,1)}^\alpha$, $v$ is the degree of freedom of the t-distribution). But it seems to me that we have this conclusion just because we are comparing apples and oranges, the two distribution don't have the same variance. Why don't compare a normal distribution with a variance of $1$ with a Student's T-distribution with a variance of $10000$ ?

Now, if we compare the quantile of a normal distribution with a Student's T-distribution of same variance, we will have a paradoxical conclusion.

Let $X$ and $Y$ two random variables with mean of $0$ and variance of $1$. $X$ follows the normal distribution $N(0,1)$ while $Y$ follows the Student's T-distribution $t_v$.

As the variance of $Y$ is equal to $1$, $Y$ must equal in distribution to $\sqrt{\frac{v-2}{v}} t_v$ (because the variance of $t_v$ is equal to $\frac{v}{v-2}$ ):
$$Y \overset {d}{=} \sqrt{\frac{v-2}{v}} t_v$$

Now, we calculate the quantile at the level $\alpha$ for both $X$ and $Y$:
$$
P(X \leq q_{X}^\alpha)  = P(N(0,1) \leq q_{X}^\alpha)=\alpha  \Rightarrow   q_{X}^\alpha=F_{N(0,1)}^{-1}(\alpha)
$$
and
$$
P(Y \leq q_{}^\alpha)  = P(\sqrt{\frac{v-2}{v}} t_v \leq q_{Y}^\alpha)=\alpha
$$
$$
\Rightarrow   \sqrt{\frac{v}{v-2}} q_{Y}^\alpha = F_{t_v}^{-1}(\alpha)
$$
$$
\Rightarrow   q_{Y}^\alpha = \sqrt{\frac{v-2}{v}} F_{t_v}^{-1}(\alpha)
$$

We take $\alpha = 0.95$ for example, the quantile $q_{X}^\alpha \approx 1.64 $ is always greater than $q_{Y}^\alpha$, $\forall v$  (for example, if $v = 5$, $q_{Y}^\alpha = \sqrt{\frac{5-2}{2}} F_{t_5}^{-1}(0.95) \approx 1.56$.

In practice, we usually have observations from a random variable $Z$ and we can easily compute the variance and mean of this variable, so the variance and mean are known. From my demonstration, the quantile of $Z$ is smaller if we suppose $Z$ follows the Student's T-distribution rather than Normal distribution.

This conclusion seems contradictory to what we read about the heavier tail of Student's T distribution, doesn't it?


问:
How shall I write the derivative of a function $f(x)$: $f_{,x}(x)$ or $f(x)_{,x}$?
I want to write the derivative of a function $f(x)$ wrt $x$. I have two options:  $f_{,x}(x)$ and $f(x)_{,x}$. Which one is right?

答:
If your function is $f(x,y)$, and $x$ is a single variable (i.e. $x\in \mathbb R$) and not a vector, I'd go with the notation $f^{(1,0)}(x)$, which means one partial derivative after the first variable, and no after the second variable.

From your two options though, I'd go with $f_{,x}(x)$, as the $(x)$ is the point being substituted, so you might have long expressions in there, in which the second notation style will be confusing
$$
f_{,x}(\frac{e^{x-2/x}+4x}{3x}) \qquad \text{vs.}\qquad f(\frac{e^{x-2/x}+4x}{3x})_{,x}
$$


答: First one is the more standard notation. It makes sense since you are really having the function $f_{,x}$ and then you evaluate it at the point $x_0$. Then you get $f_{,x}(x_0)$. In your case you are then confusing $x$ and $x_0$.

Title: How shall I write the derivative of a function $f(x)$: $f_{,x}(x)$ or $f(x)_{,x}$?

Body: I want to write the derivative of a function $f(x)$ wrt $x$. I have two options:  $f_{,x}(x)$ and $f(x)_{,x}$. Which one is right?


答:
I have never seen either notation. If $f$ is a function of just one variable then $f'$ is its derivative. If you want to indicate that the derivative is with respect to a variable called "$x$" you can write
$$
\frac{d}{dx}f \quad \text{ or }\quad \frac{df}{dx}
\quad \text{ or }\quad f'(x).
$$


问:
Inequality of a recursive function
Let $x_1=1$ and $x_{n+1}=\tfrac{1}{1+x_n}$.
I want to proof that for every $n,m\in \mathbb{N}$ the inequality $|x_{n+1}-x_{m+1}|\leq \tfrac{1}{2}|x_n-x_m|$ is true. My idea is to somehow exchange $x_{n+1}-x_{m+1}$ with $\tfrac{1}{1+x_n}-\tfrac{1}{1+x_m}$. Maybe one can argue with Cauchy and the Epsilon-criteria?

答:
We can easily prove by induction that for all $n$, $x_n \leqslant 1$.

So for all $n$, $x_{n+1} \geqslant 1/2$, which also means that for all $n\geqslant 2$, $x_n \geqslant 1/2$.

Then if we note $f:x\mapsto \frac{1}{1+x}$, we have $f'(x)=\frac{-1}{(1+x)^2}$, so $|f'(x)|=\frac{1}{(1+x)^2}$ which is decreasing with x (we consider $x\geqslant 0$).
So for $x \geqslant 1/2$, we have $|f'(x)|\leqslant |f'(1/2)| = 4/9 \leqslant \frac{1}{2}$.

By the mean value theorem, we have that :
$\forall (x,y) \in [1/2, +\infty[, |f(x)-f(y)|\leqslant \underset{z\geqslant 1/2}{\text{sup}} |f'(z)| |x-y|$

By replacing $x$ and $y$ by $u_n$ and $u_{n+1}$, we get the answer.

Title: Inequality of a recursive function


问:
Geometry problem - prove that incentre of two triangles coincide

$AB$ and $AC$ are tangent to the circle at $B$ and $C$ respectively. Let $OA$ and $BC$ intersect at M. Let $E$ be an arbitrary point on the circle. Extend $EM$ to meet the circle at F.

(1) Prove that $\Delta AEF$ and $\Delta ABC$ have a common incentre.

(2) Prove that $\Delta AEF$ and $\Delta ABC$ have a common excentre.

I verified (1) on Geogebra and one of the interesting things I found was that $\Delta AEM \sim \Delta AOF$, but I have no idea how to prove it (or will it even be useful).

My Question

I want to know how to solve (1). To do it I tried to show that $OA$ is the angle bisector of $\angle MAF$ first. However, although I was quite sure that it is true, I couldn't come up with a proof of it. Anyone can give me a hint?

答:
Draw $AO$, $OE$, $OF$, $OB$ & $OC$. $OB\perp AB$, $OC\perp AC$ & $AO$ will pass through $M$.

Notice that, $EM\cdot MF=BM\cdot MC$.

Since, $BACO$ is a cyclic quadrilateral, $BM\cdot MC=AM\cdot MO$.

Thus, $EM\cdot MF=AM\cdot MO$.
$\Rightarrow$  $AEOF$ is a cyclic quadrilateral.

$\angle OAE=\angle EFO=\angle FEO$($\triangle OEF$ is isosceles)$=\angle FAO$

$\Rightarrow$  $AO$ bisects $\angle EAF$.

But, $AO$ bisects $\angle BAC$ as well; Which indicates that the incentres of both $\triangle ABC$ & $\triangle FAE$  lie on $AO$.

Assume that, point $I$ is the incentre of $\triangle ABC$ [ We don't necessarily know whether $I$ is the point of intersection of $AO$ with the circle or not ]. Draw $CI$ & $EI$.

$\Rightarrow  \frac{AI}{IM}=\frac{AC}{CM}$[By the angle bisector theorem]$=\frac{CO}{OM}$[$\triangle ACM\sim\triangle COM$]$=\frac{OF}{OM}=\frac{AE}{EM}$[ Since $AEOF$ is a cyclic quadrilateral ]

$\Rightarrow   \frac{AI}{IM}=\frac{AE}{EM}$

Hence, $IE$ is the angle bisector of $\angle AEF$ as well; Which indicates that $\triangle ABC$ and $\triangle AEF$ have the same incentre.


问:
Solving $9x \equiv 21 \bmod 30$

Solving $9x \equiv 21 \bmod 30$

My approach:

Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}

So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$3 = 30-9 \cdot 3 \ \ \ \ \ \ (1)$$

We have to solve this equation to find $x_0$:
$$9x-30y=21 \ \ \ \ \ \ (2)$$

Multiplying $(1)$ by $7$ we get:
$$21=30 \cdot7-9 \cdot 21$$

Rewritten in form of $(2)$:
$$-9 \cdot (21) + 30 \cdot (7) = 21 $$

So $x_0=21$, the incongruent solutions are:
$$21 + 10n, \ \ \ \ \ n = 0,1,2$$

Then $x=21,31,41$. But my answer is wrong, the answer is: $x=9,19,29$. Can you tell me which step is wrong. Thanks for your help!

答: $9x\equiv 21\pmod{30}$ means $9x=30k+21$ or $3x=10k+7$ or $3x\equiv 7\pmod{10}$ or $21x\equiv 49\pmod{10}$ or $\color{red}{x\equiv 9\pmod{10}}$.

Title: Solving $9x \equiv 21 \bmod 30$