问:
Let $f(x)$ be continuous everywhere and consider the function $g(x)$. Prove that $g(x)$ is continuous everywhere.

Let $f(x)$ be continuous everywhere and consider the function:
$$g(x)=\begin{cases}f(x) &\text{if}\ f(x)>0\\0 &\text{if}\ f(x)\le0\end{cases}$$
Prove that $g(x)$ is continuous everywhere.

Hello, can anyone please help me with this one? I have a feeling this has to do with composition but cannot figure out the correlation between $f(x)$ and $g(x)$.
Thank you!

答:
You want to use the fact that $g(x) = \max \{f(x), 0\}$. The continuity of the functions involved is key.

Hope that helps,


问:
when is the region bounded by a Jordan curve "skinny"?
How can I formalize and prove the following intuition?:

Picture a very skinny rectangle, one with base length 1 and sides length $\epsilon$.  Or imagine a very flattened ellipse.  The interiors of these objects are "skinny" (or we might say $\epsilon$-skinny)  in the sense that each point in the interior is very close (or within $\epsilon$) to a point on the boundary. This notion seems easy to formalize.

Now think of the sides of the rectangle, or the boundary of the ellipse, as the image of a Jordan curve $f : S^1 \rightarrow \mathbb{R}^2$ . It's intuitively "obvious" (though not at all obvious) that what makes the bounded region skinny (in the sense of the above paragraph) is the fact that: For every $\theta \in S^1$ there exists $\theta ' \in S^1$ such that $f (\theta )$ is close to $f (\theta ' )$ but $\theta$ is not very close to $\theta '$.

I have been thinking about the right way to formalize these ideas and I'm a bit stuck.  I'd like to formulate them in the $C^0$ setting, that is to say without assuming tangent vectors or any calculus-related things.   Do you have any thoughts?  Thanks!

Also: This is motivated by the completely well-defined question here:why is an annulus close to it's boundary when it's boundary curves are close?

答: How about: a region is "skinny" if the ratio of its area to its perimeter is small.</p>

ContentLicense: CC BY-SA 3.0


问:
determine the convergence of this infinity serie
determine the convergence of
$$ \sum^{\infty}_{n=1}n^{\frac{p^2 -2}{p^2 +2p -3}}$$
how do I prove this?

答:
To avoid confusion, I will call your $p$ in my answer $r$.

The series $\sum_{n=1}^{\infty}n^{-p}$ is called $p$-series and known to be convergent iff $p>1$. Thus, your series is convergent if and only if $-\frac{r^2-2}{r^2+2r-3}>1$, which holds for all $r \in (-3,\frac{-1-\sqrt{11}}{2}) \cup (1,\frac{-1+\sqrt{11}}{2})$.

Further research

If you are interested in $p$-series, you should also take a look at the Riemann zeta function, which is the extension to complex $s$ instead of $p$.

EDIT: Does that answer your question?

ContentLicense: CC BY-SA 4.0


问:
How to prove (analytically) that $\sin(\tan^{-1}(\sqrt{2}))=\sqrt{\frac{2}{3}}$?
I calculated $\sin(\tan^{-1}(\sqrt{2}))$ using Wolfram and got it as a result $\sqrt{\frac{2}{3}}$. However, I was unable to find an identity to obtain this result.

Is there any way to calculate this value without using Wolfram?

答:
Let $t=\tan^{-1} \sqrt2$. Then, $ \tan t=\sqrt2$ and
$$\sin t = \tan t\cos t= \frac {\tan t}{\sqrt{1+\tan^2t}}=\sqrt{\frac23}
$$


问:
How does family of curves formula exactly work?
I am familiar with the following equations

$$S_1 + \lambda S_2=0$$ for two circles

$$L_1 + \lambda L_2=0$$ for two lines
$$L_1 + \lambda S_1=0$$ for circle and line

But I never really understood how they worked. Now this is a sample questions

A circles touches the parabola $y^2=2x$ at P $(\frac 12, 1)$ and cuts parabola at vertex V. If centre of circle is a Q, find the radius of circle

The formula used here was

$$(x-\frac 12)^2 + (y-1)^2 +\lambda (2x-2y+1)=0$$

Now it’s easy to see that equation is basically hinting at a curve passing through $(1/2, 1)$ and tangent to $(2x-2y+1)$, but how exactly was the form mat determined? How can we tell if this will us gives a circle? Why was the distance formula used in the first part of the equation? Basically I want to know the process of writing such equations .

答:
If $S_1(x,y)=0$ and $S_2(x,y)=0$ are the equations of two curves, the equation

$$S_1(x,y)+\lambda S_2(x,y)=0$$ describes a family of curves that contains $S_1$ (when $\lambda=0$) and $S_2$ (when $\lambda=\infty$).

The type of the intermediate curves depend on the formula of the two curves. In particular, if the extreme curves are conics, so are all curves in the pencil. We can also say that if $S_1,S_2$ have common points (intersections), then any curve of the family passes through these points. (See why ?)


问:
How to do Fourier series, deduce question?

Hello,

This is a question using Fourier series to show that $t^2$ = $f(t)$ in Fourier series.

I could do the Fourier series part but I am not sure how to do the deduce = $\pi^2/3$ part of the question?

Could somebody help me out?

答:
Hint:

Can you find two expressions for $f(0)$?

We have $f(t)=t^2$ and $f(t)=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2}\cos(nt)$ and they give the same value for $t=0$. Then after equating these two expression, re-arrange and deduce that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}.$


问:
Question about definition of symmetric affine connection from do Carmo
I have been self studying Riemannian Geometry using do Carmo, and am a bit stuck in the section regarding the introduction of a symmetric affine connection.

According to do Carmo, and affine connection $\nabla$ on a smooth manifold is symmetric when $$\nabla_X Y -\nabla_YX = [X,Y] \text{ for all }X,Y \in \chi(M)$$.

Although I think I understand this bit, the next part where he speaks about the result in coordinates confuses me. He uses the fact that the Lie Bracket is always 0 to conclude the symmetry of the lower indices of the Christoffel symbols, but how does he know the Lie bracket is always 0? Wouldn't that just imply that the Lie bracket of every pair of vector fields is zero? I realize that this is probably a result of the symmetric condition, but because the Lie bracket is used in its definition I dont understand how we can use it to conclude the Lie brackets are always zero.

Any help understanding this would be greatly appreciated!

答: Having $[\partial_i,\partial_j]=0$ (which holds simply because second order partial derivatives in Euclidean space commute) for all $i$ and $j$ does not imply that $[X,Y]=0$ for all fields $X$ and $Y$, because the Lie bracket is not bilinear over smooth functions, only over real scalars (that is, it is not a tensor). The derivation character of vector fields implies that $$[fX,gY]=fg[X,Y]+fX(g)Y-gY(f)X$$holds for all fields $X$ and $Y$, and smooth functions $f$ and $g$. So $$\left[\sum_iX^i\partial_i,\sum_jY^j\partial_j \right]= \sum_jX(Y^j)\partial_j -\sum_i Y(X^i)\partial_i.$$


问:
Neighborhood shuffle problem
Suppose there is a street of houses lined up horizontally, numbered $1$ through $N$.

My family lives in house $k$. My neighbor's family gets along with mine. Without loss of generality, let's assume the neighbor's family lives in house $k + 1$.

One morning, every family wakes up at a house of the $N$ houses of the street. (You may assume some sort of uniform selection mechanism.)

I get along with my neighbor's family, so ideally, I would want to live as close to my neighbor's family as possible. After this house shuffle happened, what is the expected distance between my neighbor's family's house and my house?

(Assume if that we live in adjacent houses after the shuffle, that is considered a distance of $1$.)

My attempt at this problem

Suppose we have houses $H = \{1, \dots, N\}$.

Without loss of generality, suppose the pair consisting of both me and my neighbor's famillies is $(k, k+1)$.

Let $\sigma: H \to H$ be a permutation (bijection).

Then I wish to calculate
$$\mathbb{E}\left[\left|\sigma(k)-\sigma(k+1) \right|\right]\text{.}$$

What I am unsure of is how to calculate this expectation without brute force calculation (i.e., getting all possible permutations), as well as somehow trying to take into account the uniform distribution of the shuffle (if that's even what it is). The difficulty that I am having a hard time wrapping my head around is that we're not looking at a uniform distribution over $H$; what we're looking at is a uniform distribution conditioned on prior houses already being selected for families.

I found a Wikipedia page and learned that this is a concept known as a random permutation, but this idea is well beyond my expertise.

答:
So, assume $|\sigma (k)-\sigma (k+1)|=i,$ in how many ways can you do this? Well you have $2$ options to choose the sign(which one is bigger) and then you can place the smallest one at $1,$ the biggest one at $i+1$ and take them to the right until the biggest one hits $N.$ So, you have $2(N-i)$ ways to do this. Also, you really do not care about what happens on the other houses, so you just pick a random permutation for the $N-2$ choices. You can pick this permutation in $(N-2)!$ ways.

Recall that the expectation is just $$\sum _{i=1}^{N-1}i\cdot P(|\sigma (k)-\sigma (k+1)|=i),$$
but then you now know that this probability, should be
$$\frac{2(N-i)(N-2)!}{N!}.$$ This last is just because we counted the possibilities and they are uniform, so each permutation has $1$ out of total permutations chance to appear, that means $\frac{1}{N!}.$

Can you finish?

Title: Neighborhood shuffle problem


问:
Expected value of two random variables and an increasing function.
Let $X,Y$ two real random variables defined over the same probabilistic space with $X(\omega),Y(\omega)\geq 0$ for every $\omega\in\Omega$ and such that $\text{E}[X]\leq\text{E}[Y]$. Let $g : [0,\infty)\rightarrow [0,\infty)$ be an increasing function. I wonder to know if $\text{E}[g(X)]\leq\text{E}[g(Y)]$ or find a counterexample. I was thinking in the function $g(x) = \sqrt{x}$, but i'm pretty lost.

Thanks in advance.

答:
Not necessarily.

Let X be 1,100 or 200 with probability 0.1,0.8,0.1 so E[X]=100.1

and Y be 1,100 or 200 with probability 0.2,0.6,0.2 so E[Y]=100.2

But, $E[log(X)]\approx 4.2>E[log(Y)]\approx 3.8$

Also, $E[\sqrt{X}]\approx 9.5>E[\sqrt{Y}]\approx 9.0$


问:
Given $\lim_{x\to0}\frac{\sin^2x}{x}=0$, evaluate $\lim_{x\to0}\frac{1-\cos x}{x}$

Given that $\lim_{x\to0}\frac{\left(\sin x\right)^2}{x}=0$, find the limit $\lim_{x\to0}\frac{1-\cos x}{x}$.

a. $1\quad$ b. $0\quad$ c. $\frac12\quad$ d. The limit does not exist

New to Calculus, and these trig questions really trip me up. Thanks

答:
hint:
$$
(1-\cos x)(1+\cos x) = 1-\cos^2(x) = \sin^2(x)
$$

more hints:

 \begin{align*} \frac{1-\cos x}{x} = \frac{\sin^2(x)}{x(1+\cos x)} \end{align*}

 $\cos$ is continuous in $0$.

solution:

 \begin{align*} \lim_{x\to 0} \frac{\sin^2(x)}{x(1+\cos x)} = \underbrace{\lim_{x\to 0} \frac{\sin^2(x)}{x}}_{=0} \cdot \underbrace{\frac{1}{1+\cos 0}}_{=\frac{1}{2}} = 0\end{align*}

edit:
The most important properties (and much more) are listed for example at the wikipedia article about sine and cosine. As for what I used: The first equations just follows from the fact $(a-b)(a+b) = a^2-b^2$ (which has nothing to do with trig). The second equation follows from $\sin^2 x + \cos^2 x =1$. While this might look surprising at first, this is something you are probably familiar with but just don't recognize. You can view $\cos x$ and $\sin x$ as the lengths of the sides of a right triangle with a hypotenuse length of $1$. Then it follows from the Pythagorean theorem. But this is also stated on wikipedia, there is also a nice visualization of sine and cosine on the unit circle.


答:
Hint:

Use the mid-school formula:
$$\sin^2\theta=\frac{\tan^2\theta}{1+\tan^2\theta}$$
and use that $\sin\theta$ and $\tan\theta$ have the same sign on $\bigl(-\frac\pi 2,\frac\pi 2\bigr)$.

Title: How to prove (analytically) that $\sin(\tan^{-1}(\sqrt{2}))=\sqrt{\frac{2}{3}}$?

Body:
I calculated $\sin(\tan^{-1}(\sqrt{2}))$ using Wolfram and got it as a result $\sqrt{\frac{2}{3}}$. However, I was unable to find an identity to obtain this result.

Is there any way to calculate this value without using Wolfram?


问:
Solving $7^{2020} \equiv x \operatorname{mod} 10000$ using CRT
I want to compute the last four digits of $7^{2020}$ using the Chinese Remainder Theorem (CRT).

Here is what I've got so far:

Obviously
\begin{align}
10000 = 10^4 = 2^4 \cdot 5^4,
\end{align}
so due to CRT we know
\begin{align}
\mathbb{Z}/10000\mathbb{Z} \cong \mathbb{Z}/16\mathbb{Z} \times \mathbb{Z}/625\mathbb{Z}.
\end{align}
I also do know that
\begin{align}
\varphi(16) = 8 \quad \text{and} \quad \varphi(625) = 500
\end{align}
and therefore
\begin{align}
7^8 \equiv 1 \operatorname{mod} 16 \quad \text{and} \quad 7^{500} \equiv 1 \operatorname{mod} 625.
\end{align}

(I know that one can find a lot of almost similar questions here on MSE, but I am only interested in solutions, which use CRT. That's due to the fact that I struggle to see how CRT works for those kind of problems. Up to now I used CRT only to compute a solution for multiple congruences. Hence, I would appreciate solutions/hints which straight forward show how to apply CRT here (without any fancy tricks etc.).)

答:
Continuing with what you have already.

$7^{2020} = 7^{2016}7^{4}\equiv 7^{4}\pmod {16}\\
7^{2020} = 7^{2000}7^{20}\equiv 7^{20}\pmod {625}$

But we can do better.

$7^2 = 49 \equiv 1\pmod {16}\\
7^{2020} \equiv 1\pmod {16}$

$7^4 = 2401 \equiv -99 \pmod {625}\\
7^{20} \equiv (1-100)^5 \pmod {625}$

Every term in the binomial expansion of $(1-100)^5$ after the first two be  multiples of $10,000$ that can be ignored.

$(1-100)^5 \equiv -499\equiv 126 \pmod {625}$

Now you can use the Chinese remainder theorem.

$7^{2020} \equiv 2001 \pmod {10,000}$

In fact, $7^{20} \equiv 2001 \pmod {10,000}$ and you can stop there.


答:
Hint:

$$\lim\limits_{x\to0}\dfrac{1-\cos x}x=\lim\limits_{x\to0}\dfrac{\sin^2\left(\dfrac x2\right)}{x/2}$$

Title: Given $\lim_{x\to0}\frac{\sin^2x}{x}=0$, evaluate $\lim_{x\to0}\frac{1-\cos x}{x}$

Body:
Given that $\lim_{x\to0}\frac{\left(\sin x\right)^2}{x}=0$, find the limit $\lim_{x\to0}\frac{1-\cos x}{x}$.

a. $1\quad$ b. $0\quad$ c. $\frac12\quad$ d. The limit does not exist

New to Calculus, and these trig questions really trip me up. Thanks


问:
Weak topology and separation
Let $X$ have the weak topology induced from the collection of functions $f_\alpha\colon X\to X_\alpha$, for $\alpha\in A$. I am trying to show that $x,y\in X$ s.t. $x \neq y$, $\exists\alpha\in A: f_{\alpha}(x)\neq f_{\alpha}(y)$ does not imply whenever $B$ is closed in $X$ and $x\notin B$, then for $\alpha\in A$, $f_{\alpha}(x)\notin \overline{f_\alpha[B]}$.

I consider first the subspace topology via the inclusion map, however, it satisfies both properties.

答: HINT: Let $X=\prod_{\alpha\in A}X_\alpha$, and let $f_\alpha:X\to X_\alpha$ be the projection map. You can even take $A=\{0,1\}$ and $X_0=X_1=\Bbb R$.

Title: Weak topology and separation


答:
$\lim_{x\to0}\frac{(sinx)^2}{x}=0$
This is universally true.

Also, $$\lim_{x\to0}\frac{(1-cosx)}{x}= \lim_{x\to0}\frac{(1-cosx)}{x^2}*x$$
$$=\lim_{x\to0}\frac{1}{2}*x=0$$

This is because of the standard limit:
$$\lim_{x\to0}\frac{(1-cosx)}{x^2}=\frac{1}{2}$$


问:
Name (or proof) of $\frac{a}{b} - 1 = \frac{a-b}{b}$
Can someone tell me the name of the equation (or the proof) which I show here:

The equation is always valid:

$$\frac{a}{b} - 1 = \frac{a-b}{b}$$

I used $(a-b)/b$ to calculate the difference in percentage between $a$ and $b$. But it seems, that I can also use $(a/b) - 1$ (which return always the same result).

答: Its just $\frac{a}{b}-1=\frac{a}{b}-\frac{b}{b}=\frac{a-b}{b}$ as @Randall mentioned.


问:
Convergence of Taylor series expansion
The Taylor series expansion of $\frac{1}{(x^2 -1)^2}$ at $x=-1$ is given by

$\frac{1}{(x^2 -1)^2} =\frac{1}{4(x+1)^{2}}+\frac{1}{4(x+1)}$+$\frac{3}{16}$+$\frac{(x+1)}{8}$ $\ldots$

Is this series convergent? If so, how do I prove it?

答:
Hint.

To find the Laurent series at $z=-1$, note that
$$
\frac{1}{(z^2-1)^2}=
\frac14
\left(\frac{1}{(z+1)^2}
+\frac{1}{(z+1)}
-\frac{1}{z-1}+\frac{1}{(z-1)^2}
\right)
$$

So you need to work with the last two terms.

$$
\frac{1}{z-1}=\frac{1}{z+1-2}=-\frac12\frac{1}{1-\frac{z+1}{2}}
$$

You can then find a series for $z$ with $|z+1|<2$. Taking the derivative term by term, you can find the series for $\frac{1}{(z-1)^2}$

Title: Convergence of Taylor series expansion


问:
My first proof that epsilon delta $\displaystyle\lim_{x \to 5}3x+2=17$
Prove: $\displaystyle\lim_{x \to 5}3x+2=17$

General format $\displaystyle\lim_{x \to a}f(x)=L$ (included to verify various parameters)

What needs to be proved: $\forall\epsilon>0, \exists\delta>0$, such that $0<|x-a|<\delta|\implies|f(x)-L|<\epsilon$

Let $\epsilon>0$, $\epsilon$ be arbitrary, pick $\delta=\color{red}{\epsilon/3}\implies3\delta=\epsilon$ (Picked later)

Hypothesis: $\exists\delta>0$ (I am afraid of saying for every $x \in\mathbb{R}$), there exists a $x$ such that $0<|x-5|<\delta\implies|3x+2-17|<\epsilon$

Proof:

Now, $|3x+2-7|=|3x-15|=3|x-5|$

$|x-5|<\delta\implies3|x-5|<3\delta=\epsilon/3\implies|x-5|<\epsilon$ (Q.E.D)

This proof is based on https://www.youtube.com/watch?v=dXr6aoJ1nVI

Remark:

I redo to see if I can write a proof that follows logical order.

This proof somehow doesn't seem to me to spell out that there exists such an $x$, while approaching $a$ will make the distance between $|f(x)-L|<\epsilon$. Did I prove it correctly?

Also with this kind of proof, I find that you usually pick a $\delta$ in such a way that eventually will give you a $\epsilon$. Also this kind of excercise is easy to do because $|x-a|$ can be written as $N|f(x)-a|$, with $N$ being a constant factor out to make $|f(x)-L|=|x-a|$

Do you think doing the proof as shown in the video with a double inequality is better?

答:
The point of an $\epsilon,\delta$ proof of a limit is to find $\delta$ given $\epsilon$, because we are after all proving that for any $\epsilon$, there exists a $\delta$ ...

With this in mind, let's start the proof.

Proof: Let $\epsilon>0$. We must find $\delta>0$ such that
$$|x-5|<\delta\Rightarrow |3x+2-17|<\epsilon.$$
Keep in mind that our choice of $\delta$ depends only on the value of $\epsilon$.
We may notice that $|3x+2-17|<\epsilon$ is equivalent to the condition $-\epsilon<3x-15<\epsilon$, so
$$-\epsilon/3<x-5<\epsilon/3.$$
However, our desired value $\delta$ is defined to satisfy $-\delta<x-5<\delta$. Thus, $\delta=\epsilon/3$ is a perfect choice!

The proof is now complete, because we were given any value $\epsilon>0$, and from that we were able to prove the existence of (in other words, find) a value $\delta>0$ such that
$$|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon.$$


问:
Properties of Map: Morphisms, isomorphism and Zariski continuous
I am self studying Commutative algebra out of Kemper's text. I came across an exercise in his text that is meant to better solidify the idea of mappings between affine varieties. As I am very new to the subject, I was hoping for some help in navigating through the question. It goes as follows:

Let $$X = \{ ( x_1 , x_2) \in \mathbb{C}^2 \: | \: x_1x_2 = 1 \}$$

Which of the following maps $\varphi_i : X \rightarrow X$ are morphisms, isomorphisms, or Zariski
continuous?

(a) $\varphi_1(x_1,x_2) = (x_1^{-1} , x_2^{-1})$

(b) $\varphi_2(x_1,x_2) = (x_1^{2} , x_2^{2})$

(c) $\varphi_2(x_1,x_2) = (\bar{x_1} , \bar{x_2})$ (complex conjugation).

Here are some definitions to work with.

Let $K$ be a field and let $X \subseteq K^m$ and $Y \subseteq K^n$ be affine
varieties. A map $f: X → Y $ is called a morphism (of varieties) if there
exist polynomials $f_1,...,f_n \in K[x_1,...,x_m]$ such that $f$ is given by
$f(P)=(f_1(P),...,f_n(P))$ for $P \in X$.

A morphism that has an inverse which is also a morphism is an isomorphism.

My question here is how do i approach these? Is there a technique I should think about. Any help would be appreciated.

答:
I was not able to solve c), but I hope a) and b) will help you in your studies:

a)

First note, that neither $\{0\}\times\mathbb{C}$, nor $\mathbb{C}\times\{0\}$ is a subset of $X$, since the condition $x_1x_2=1$ is not fufilled. Therefore

$$\varphi_1:X\rightarrow X,\quad \varphi_1(x_1,x_2)=(x_1^{-1},x_2^{-1})$$

is well-defined. (Neither $x_1$, nor $x_2$ is equal to $0$.)

Using the condition of $X$, we can calculate as follows:

$$\varphi_1(x_1,x_2)=(x_1^{-1},x_2^{-1})=(1 x_1^{-1},1 x_2^{-1})=(x_1x_2x_1^{-1},x_1x_2x_2^{-1})=(x_2,x_1)=(f_1(x_1,x_2),f_2(x_1,x_2))$$

with polynomials $f_1,f_2\in\mathbb{C}[x_1,x_2]$, defined by

$$f_1(x_1,x_2)=x_2,\quad f_2(x_1,x_2)=x_1.$$

Therefore $\varphi_1$ is a morphism. Since $\varphi_1\circ\varphi_1=id_X$, it is even an isomorphism.

b)

$$\varphi_2:X\rightarrow X,\quad \varphi_2(x_1,x_2)=(x_1^2,x_2^2)$$

is well-defined, since any complex number can be squared. Again, both component functions are functions of $\mathbb{C}[x_1,x_2]$. Therefore $\varphi_2$ is a morphism. This is not an isomorphism, since for all $f\in \mathbb{C}[x_1,x_2]$ and all $(x_1,x_2)\in X$, there exists a $n\ge 0$:

$$f(\varphi_2(x_1,x_2))=f(x_1^2,x_2^2)=\sum_{k=0}^n a_k x_1^{2k}+b_k x_2^{2k}$$

This holds because for any $n_1\ge n_2\ge 0$, we find

$$(x_1^2)^{n_1}(x_2^2)^{n_2}=(x_1^2)^{n_1-n_2}$$

again by using the property of $X$. Especially we cannot find $f_1$ or $f_2$, such that

$$f_1(\varphi_2(x_1,x_2))=x_1\quad\text{and}\quad f_2(\varphi_2(x_1,x_2))=x_2$$

$$$$

c)

$$\varphi_3:X\rightarrow X,\quad \varphi_2(x_1,x_2)=(\overline{x_1},\overline{x_2})$$

is well-defined, since for any complex number exists a complex conjugate.


问:
Prove that exists a vector n-dimensional $x \neq 0$ such that $(A^2+B^2)x=0$ for two different square matrices $n\times n$.
Let $A$ and $B$ be two different square matrices $n\times n$. If $A^3=B^3$ and $A^2B=B^2A$. Prove that exists a vector n-dimensional $x \neq 0$ such that $(A^2+B^2)x=0$.

My first thought was that $(A^2+B^2)$ should be equal to $0$ as $x \neq 0$, but then I realize that with matrices this is not true. For example: if we have the matrix $$C=\begin{bmatrix}1&2&3\\1&2&3\\1&2&3\end{bmatrix}$$
Then we should find a vector $x \neq 0$ such that $Cx=0$. $$\begin{bmatrix}1&2&3\\1&2&3\\1&2&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=x_1\begin{bmatrix}1\\1\\1\end{bmatrix}+x_2\begin{bmatrix}2\\2\\2\end{bmatrix}+x_3\begin{bmatrix}3\\3\\3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ So a possible answer might be $$x=\begin{bmatrix}3\\3\\-3\end{bmatrix}$$ proving that the matrix is not necessarily equal to $0$.

I am not sure how I should use the fact that $A^3=B^3$ and $A^2B=B^2A$.

答: First, prove that since $A$ and $B$ are different there must be a vector $\mathbf{x}$ with $A\mathbf{x} \neq B\mathbf{x}$.  This means that $\mathbf{v} = A\mathbf{x} - B\mathbf{x} \neq \mathbf{0}$. Now show that $(A^2+B^2)\mathbf{v} = \mathbf{0}$.  For this, you will need the hypotheses on $A$ and $B$.


问:
Reference for Exercise 5.3.1 in [Probability: Theory and Examples, 4th edition, Durrett, 2010]
Exercise 5.3.1 on page 205 of [Probability: Theory and Examples, 4th edition, Durrett, 2010] reads:

Exercise 5.3.1. Let $X_n$, $n \ge 0$, be a submartingale with $\sup X_n < \infty$. Let $\xi_n = X_n − X_{n−1}$ and suppose $\mathbb{E}(\sup \xi_n^+) < \infty$. Show that $X_n$ converges a.s.

In [Probability: Theory and Examples, 5th edition, Durrett, 2019], the same question appears as Exercise 4.2.4. on page 223.

I would like to cite this result, but I hesitate to cite an exercise. Could anyone kindly provide a reference for this conclusion, or for a variant similar to the following?

Theorem. Let $X_n$ be a submartingale with  $\mathbb{E}(\sup (X_n-X_{n-1})^+) < \infty$ (or more strongly, $\sup|X_n-X_{n-1}|\le c <\infty$, which is adequate for my application).
Then
$$
\mathbb{P}\left(\{X_n \text{ converges}\}\bigcup \{\sup X_n = \infty\}\right) = 1
$$

I would be surprised if this kind of result has never been mentioned in any other monographs/textbooks, although I have found nothing after searching.

Any comments or criticism will be appreciated. Thank you.
(I am looking for a reference. Nevertheless, if you would like to post a proof, it is also very welcome.)

答:
You can essentially follow the same proof as Theorem $5.3.1$, which appears right above this exercise. Briefly, let $0<K<\infty$, and let $N:=\inf\{n:X_n \ge K\}$. Then $\{X_{n\wedge N}\}$ is a submartingale, and $X_{n\wedge N}^+ \le K + \xi_N^+$. It follows by the martingale convergence theorem that $\lim_{n\to\infty}X_{n\wedge N}$ exists a.s., and in particular that $\lim_{n\to\infty}X_n$ exists on $\{N=\infty\}$. Letting $K\to\infty$, and using the fact that $\mathbb P(\limsup_{n\to\infty}X_n<\infty) = 1$, we see that $\lim_{n\to\infty}X_n$ exists a.s.

In terms of citation, you could probably get away with either citing the exercise, or Theorem $5.3.1$ itself. You might even do both, i.e. "by a straightforward generalization of [Durrett, Theorem $5.3.1$] (see [Durrett, Exercise $5.3.1$]), one has..." But you could also just prove it directly, since it wouldn't take more than a couple of lines.


问:
Connecting the rank with the determinant of a skew symmetric matrix.
We need to find the rank of the matrix:

\begin{bmatrix}
0&c_{1}&-b_{1}&a_{2}\\
-c_{1}&0&a_{1}&b_{2}\\
b_{1}&-a_{1}&0&c_{2}\\
-a_{2}&-b_{2}&-c_{2}&0
\end{bmatrix}

Where all $a_{i},b_{i},c_{i}$ are non zero.

The only thing I could observe was the given matrix is skew-symmetric, and the determinant is $(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2})^2.$However I could not proceed any further than that. The rank is ofcourse = the number of non zero rows in the row echelon form of the matrix, but taking that route seems to be too tedious , and I believe the purpose of giving a skew symmetric matrix was to exploit the (skewed)symmetry somehow.

The solution given in my book is :

Rank=$4$,if $det(A) \neq 0$. (This is straightforward)

Rank=$2$ otherwise, since "all determinants of the third order have the value zero".

Questions:

What does the statement in quotes mean, and how is it true?

What connection does it have to finding the rank?

答:
There's basically two ways to address why a skew symmetric matrix $A$ never has odd rank.

1.) The most general uses bilinear forms and holds over any field (with extra careful definition of skewness in characteristic 2).  This seems against the spirit of "connecting the rank with the determinant of a skew symmetric matrix" and the details are outside the scope.

2.) Here is a proof that hold over any field $\mathbb F$ with $\text{char}\big(\mathbb F\big) \neq 2$.

Suppose $\text{rank}\big(A\big)=r$ and $r\%2=1$ -- i.e. $r$ is odd.

Then $A$ has an $r\times r$ principal submatrix, $C_{r\times r}$ that is invertible-- i.e. $ \det\big( C_{r\times r} \big)\neq 0$

proof: Prove the existence of a principal submatrix of order $r$ in $M\in\Bbb F^{n\times n}, M=-M^T,\ \operatorname{rank}(M)=r$

But since $C_{r\times r}$ is a principal matrix, it is skew symmetric, which tells us

$\det\big(C_{r\times r}\big)=\det\big(C_{r\times r}^T\big)=\det\big(-I_r C_{r\times r} \big)=(-1)^r\cdot \det\big( C_{r\times r} \big)=-1\cdot \det\big( C_{r\times r} \big)$

$\implies 0 = 2\cdot \det\big( C_{r\times r} \big)$
$\implies 0 = \det\big( C_{r\times r} \big)\neq 0$

which is a contradiction

addendum

approach (2) actually works when $\text{char}\big(\mathbb F\big) = 2$ with a little bit of cleverness. First prove the lemma, working over the ring $\mathbb Z[\mathbf x]$ that for any $r\times r$ skew symmetric matrix matrix $S$ with zeros on the diagonal, and off diagonal elements are $x_{i,j} \in \mathbb Z[\mathbf x]$ then $\det\big(S\big) = 0$.  One can do this by considering an arbitrary substitution homomorphisms of $x_{i,j}\to a_{i,j}\in \mathbb R$, and the determinant is zero for every possible substitution by the above argument-- i.e. in reals we see it implies 2 times the determinant is 0 hence the determinant is 0.  By principle of permanence of identities the polynomial given by taking the $\det\big(S\big)\in \mathbb Z[\mathbf x]$  is equal to $ 0$.  Now apply a different substitution homomorphism mapping the $x_{i,j}$ to elements in the desired field $\mathbb F$ of characteristic 2, and said polynomial is still zero -- i.e. we've proven that for odd r, any $r\times r$ skew symmetric matrix is singular (with explicit constraint of zeros on diagonal, which is the right definition of skew symmetry in char 2, and is redundant in other characteristics). From here the original argument (2) applies showing such a matrix cannot have odd rank.


问:
Side-limit of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$
The problem wants the value of left hand limit of the derivative of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ when $x\to3^-$.I graphed the function and the answer is $\frac{3}{2}$ but what's wrong with my method:

For any value $x\to3^-$ we have $[x]=2$ , $|x-1|=x-1$ so if $h\to0^-$, then $f(3+h)=\frac{2(3+h)^2}{2+h}=\frac{2h^2+12h+18}{2+h}$. So the value of the derivative is equal to:
$$\lim_{h\to0^-}\dfrac{f(3+h)-f(3)}{h}=\lim_{h\to0^-}\frac{\frac{2h^2+12h+18}{2+h}-\frac{27}{2}}{h}$$ which is infinity and clearly wrong with or without the graph. I'm not new into this but I can't figure out what's wrong here despite I thought about the method for one hour straight.

答:
Alright I got the problem. In the following figure the green graph is $f(x)$ and the red one is $f'(x)$. Now if you calculate the left hand limit of the derivative using $\displaystyle\lim_{h\to0^-}\dfrac{f(x+h)-f(x)}{h}$, you're actually calculating the slope of the line that approaches to the segment $AB$ which is infinity so we're not wrong on this point.

The major problem about this method is that we actually want the slope of the line tangent to the graph of $f(x)$ when we're still before $x=3$. For doing that simply note that for $x\in(2,3)$ which contains the left neighborhoods of $x=3$, $f(x)=\frac{2x^2}{x-1}$ and this function's derivative in $x=3$ is exactly equal to $\frac{2}{3}$.

Furthermore, if you insist on solving this with the definition of derivative one can modify the formula like the following: (the following form is clearly a more general definition that will also work for the discontinuity points)
$$\displaystyle {f'_\pm} (x)=\lim_{h\to0^{\pm}}\bigg(\lim_{t\to0^{\pm}}\dfrac{f(x+h+t)-f(x+t)}{h}\bigg)$$

In which we shift the points in the wanted direction first, and then do the rest.

*P.S. The segment $AB$ is not a part of the graph. I added it myself


问:
How many three digit strings can be formed from numbers in a set such that the number represented by the string is divisible by three?
Consider the set of numbers $\{0,1,2,3,...,9 \}$ , now we want the number of ways to find three digit strings such that the string is divisible by three (with repetition).

For example,

012 is a good because twelve is divisible by three

009 is good because it is also divisible by three

031 is not because it is not divisible by three

My thought was to use divisibility rules, we can find that the problem can be found equivalent to finding the number of solutions to the equation:

$$ a+b+c = 3n$$
Where $n$ is some integer and $(a,b,c)$ represents the number we picked from the set. For example, $(0,1,2)$ would mean the number 012 and that's divisible by three because

$$ 0 + 1 + 2 = 3n$$

Which is true when $n=1$

Now, a way to solve this would be to use brute force counting but I'm hoping for a better way to count using combinatronics

答: I realized a much more direct observation later, the set of strings is the set of all multiple of three between $ 0$ to $1000$


问:
How many vectors are linearly independent?
A sequence of vectors $\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}$ from a vector space $V$ is said to be linearly dependent, if there exist scalars $a_{1}, a_{2}, \ldots, a_{k},$ not all zero, such that
\begin{equation*}
a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+\cdots+a_{k} \mathbf{v}_{k}=\mathbf{0}
\end{equation*}
where $\mathbf{0}$ denotes the zero vector.

Suppose we have $\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}$ such that $a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+a_{3} \mathbf{v}_{3}=\mathbf{0}$ has solution $a_3=-2a_2 + 4a_1$. How many linearly independent vectors is there in the set $\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\}$ ? I feel like if the solution was $a_3=-2a_2$, $a_1 =0$, the answer would be $2$ ($\mathbf{v}_{1}$ and another one). Is it still $2$ for the above case ? Or 1 ?

答:
There could be:

0 linearly independent vectors

```<code>all vectors are 0
</code>```

1 linearly independent vector

```<code>1 non zero vector and 2 zero vectors
2 non zero vectors that lie on the same line and 1 zero vector
3 non zero vectors that lie on the same line
</code>```

2 linearly independent vectors

```<code>1 non zero vector, 1 linearly independent vector, 1 zero vector
2 non zero vectors that lie on the same line and 1 non zero vector not on the same line
1 nonzero vector, 1 nonzero vector that doesn't lie on the same line, and 1 nonzero
     vector that lies in the plane but not on the same line as either of those vectors
</code>```

3 linearly independent vectors

```<code>the dimension of the vector space must be at least 3, and the vectors are linearly
     independent (the coefficients of them are all 0)
</code>```

Body:
There could be:

0 linearly independent vectors

all vectors are 0

1 linearly independent vector

1 non zero vector and 2 zero vectors
2 non zero vectors that lie on the same line and 1 zero vector
3 non zero vectors that lie on the same line

2 linearly independent vectors

1 non zero vector, 1 linearly independent vector, 1 zero vector
2 non zero vectors that lie on the same line and 1 non zero vector not on the same line
1 nonzero vector, 1 nonzero vector that doesn't lie on the same line, and 1 nonzero
     vector that lies in the plane but not on the same line as either of those vectors

3 linearly independent vectors

the dimension of the vector space must be at least 3, and the vectors are linearly
     independent (the coefficients of them are all 0)


问:
Differentiability of $f$ at $z_0$ implies $f$ is bounded in open neighbourhood of $z_0$

If $f(z)$ is differentiable at $z = z_0 \in \mathbb{C}$, then $f(z)$ is bounded in a neighbourhood of $z_0$.

My attempt: Since $f$ is differentiable at $z_0$, it is continuous at $z_0$. For any $\varepsilon > 0$, we can find a $\delta$-neighbourhood of $z_0$ such that

\begin{align*}
|z - z_0| < \delta &\implies |f(z) - f(z_0)| < \varepsilon \\
&\implies \left| |f(z)| - |f(z_0)| \right| \leq |f(z) - f(z_0)| < \varepsilon\\
&\implies |f(z_0)| - \varepsilon < |f(z)| < |f(z_0)| + \varepsilon
\end{align*}

Taking $\varepsilon = 1$, since $|f(z_0)|$ is finite, we have that $|f(z)|$ is bounded, and hence $f(z)$ is bounded in some neighbourhood of $z_0$.

My question: is the proof above correct? Also, is $f$ being continuous at $z_0$ sufficient to show that $f$ is bounded in a neighbourhood of $z_0$?

Thanks!

答: Yes, it is correct. And, yes, continuity is enough. A slightly shorter proof would be: take $\delta>0$ such that $|z-z_0|<\delta\implies\bigl|f(z)-f(z_0)\bigr|<1$, and then, if $|z-z_0|<\delta$,\begin{align}\bigl|f(z)\bigr|&=\bigl|f(z)-f(z_0)+f(z_0)\bigr|\\&\leqslant\bigl|f(z)-f(z_0)\bigr|+\bigl|f(z_0)\bigr|\\&<1+\bigl|f(z_0)\bigr|.\end{align}


问:
$f$ nonnegative, continuous in $[a,b]$ and such that $\int_a^b f(x)dx=0$ implies $f(x)=0$ in $[a,b]$
My book proves the following statement: "Let $f$ be nonnegative and continuous in $[a,b]$ and such that
$$\int_a^b f(x)dx=0$$
then $f$ is identically $0$ in $[a,b]$." using an argument by contradiction.

However, I've tried to prove it in a direct way (which takes way more effort, but I would like to know if there is a flaw in my reasoning or not), can someone check if my work is okay please?

My try: by hypothesis $f$ is nonnegative in $[a,b]$: if $f(x)=0$, the theorem is true.

So let's suppose $f(x)>0$: let $x_0 \in [a,b]$, by hypothesis $f$ is continuous in $[a,b]$, so in particular it is continuous in $x_0$. Since $f$ is positive in this case, I can chooose $\epsilon=\frac{1}{2}f(x_0)$ in the definition of continuity to get the inequality $\frac{1}{2}f(x_0) \leq f(x) \leq \frac{3}{2}f(x_0)$ if $x_0-\delta_\epsilon < x < x_0+\delta_\epsilon$. Since by hypothesis $f$ is continuous in $[a,b]$, it is integrable in $[a,b]$ as well, so by monotonicity of integral it is
$$\int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} \frac{1}{2}f(x_0)\,dx \leq \int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} f(x)\,dx \leq \int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} \frac{3}{2} f(x_0)\,dx$$
By hypothesis $\int_a^b f(x)\,dx=0$, so in particular $\int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} f(x)\,dx=0$; hence from the last inequality it follows that $\delta_\epsilon f(x_0) \leq 0 \leq 3\delta_\epsilon f(x_0)$. Since $\delta_\epsilon>0$, This means that $f(x_0)$ is both $\geq0$ and $\leq 0$, the only possibility is that $f(x_0)=0$. Since $x_0 \in[a,b]$ is arbitrary, this holds for all $x\in[a,b]$ and so $f(x)=0$ for all $x\in[a,b]$.

Is this correct? If it is maybe I should've put more attention in the intervals of the kind $[a,x_0+\delta_\epsilon]$ and $[x_0-\delta_\epsilon,b]$ because since $\delta_\epsilon$ if $x_0$ is too much near the boundary I risk to "go out" of $[a,b]$, but I'm not sure if this is a problem or not since if $\delta_\epsilon$ works a lesser $\delta'_\epsilon$ would work too and so I can restrict it as much as I need.

答:
You must pick some $x_0$ such that $f(x_0)>0$, you can't just pick $x_0$ completely arbitrarily. It can easily happen that $f$ is only positive on, say, some small interval.

Other than that, yes, once $f(x_0)>0$, there is a nondegenerate interval containing $x_0$ on which $f(x)>f(x_0)/2$, and so the integral on that interval is positive. Meanwhile the integral elsewhere is nonnegative. So you conclude that if $f$ is nonnegative, continuous, and not identically zero, then its integral is nonzero, which is a contrapositive of the desired statement.

I would call this general type of argument either contradiction or contraposition, though, so I'm a bit surprised that you think this is especially direct.


问:
Expectation as unique linear functional
A textbook on Probability theory by A. Obretenov I'm reading (in Bulgarian) has the following:

Let $(\Omega, \mathbb{F}, \mathbb{P})$ be a probability space.

For a simple random variable $\xi(\omega) = \sum_{i=1}^n x_i I[A_i]$ (where $I[A_i]$ is the indicator of event $A_i$), they define
$$\mathbb{E}\xi = \sum_{i=1}^n x_i \mathbb{P}I[A_i]$$

Then they state the following:

Lemma. The expectation operator defined above is the unique linear functional of the space of simple random variables, which satisfies:

$\mathbb{E}[I[\Omega]] = 1$

If $\xi \ge 0$ then $\mathbb{E}[\xi] \ge 0$

If $\xi_n(\omega) \downarrow 0$ for each $\omega$, then $\mathbb{E}\xi_n \downarrow 0$ ($\downarrow$ denotes monotonically decreasing convergent sequence).

I'm trying to understand in what sense can we talk about uniqueness here, since clearly changing the probability measure will change the expectation without violating the conditions.

答: The expectation operator is the unique linear operator satisfying the conditions you have given which extends the definition you have given on simple functions. Obviously, if you use a different probability measure, you will have a different expectation of some simple function.</p>

Title: Expectation as unique linear functional


问:
Help me solve the invariant measure of $Q$
My $Q$ matrix is given by:

\begin{bmatrix}
-\lambda &0  &\lambda  &0  &0  &... \\
 \mu&-(\lambda+\mu) &0  &\lambda  &0  &... \\
 0&\mu  &-(\lambda+\mu)  &0  &\lambda  &... \\
 0&0  &\mu  &-(\lambda+\mu)  &0  &... \\
 ...& ... & 0 &\mu  &-(\lambda+\mu)  &... \\
 ...& ...&...  &...  &...  &...
\end{bmatrix}

This is an infinite matrix, hopefully the pattern is clear.  The first row is unique, but each row after the second row is just the second row shifted over to the right by one more.  I'm trying to find its invariant measure, and I get the following equations:

$-\lambda\pi_1+\mu\pi_2=0$

$-(\lambda+\mu)\pi_2+\mu\pi_3=0$

$\lambda\pi_i-(\lambda+\mu)\pi_{i+2}+\mu\pi_{i+3}=0$ for $i=1,2,3,...$

But I'm finding these equations intractable, is there another way to find the invariant measure?  Or can someone help me resolve these equations.

答:
The stationary distribution $\pi$ exists if and only if the chain is positive recurrent. Since the drift at $i$, for every $i\geqslant1$, is $2\lambda-\mu$, this happens when $2\lambda\lt\mu$. Then, the standard approach to compute every $\pi_i$ is to look for the generating function $g(x)=\sum\limits_{i\geqslant1}\pi_ix^i$. The relations between the coefficients $\pi_i$ given above yield readily
$$
g(x)=\frac{\mu\pi_1x}{\mu-\lambda x-\lambda x^2}.
$$
The fact that $g(1)=1$ yields $\mu\pi_1=\mu-2\lambda$, hence
$$
g(x)=\frac{(\mu-2\lambda)x}{\mu-\lambda x-\lambda x^2}.
$$
To go further, one decomposes the denominator as
$$
\mu-\lambda x-\lambda x^2=\mu(1-\alpha x)(1+\beta x),
$$
for some suitable positive $\alpha$ and $\beta$, and one rewrites $g(x)$ as
$$
g(x)=\frac{\mu-2\lambda}{\mu(\alpha+\beta)}\left(\frac1{1-\alpha x}-\frac1{1+\beta x}\right),
$$
which yields finally, for every $i\geqslant1$,
$$
\pi_i=\frac{(\mu-2\lambda)(\alpha^i-(-1)^i\beta^i)}{\mu(\alpha+\beta)}.
$$</p>


问:
Find the maximum value of $2x-4x^2 > 0$
$x \in \mathbb{R}$.

My attempt:

We’re gonna check the case where: $2x-4x^2 >0$.

And by some basic algebra you will find that : $$ x \in ]0,\frac{1}{2}[$$
But since we are in Real numbers we can’t just plug numbers.

答:
hint

Use the canonical form ;
$$f(x)=-(4x^2-2x)$$
$$=\frac 14-(2x-\frac 12)^2$$

$$\le \frac 14$$

Title: Find the maximum value of $2x-4x^2 > 0$


答:
Your concerns are in the right place, as the "solution" $a_3=-2a_2+4a_1$ lack some information. If by this terminology you also say that $v_1,v_2$ are lineary independent (and hence there is need to state $a_2$ AND $a_1$, then the answer is as you said, $2$.
But there is the option that for example $v_1 = \alpha v_2$ which means that actualy $v3 = (-2+4\alpha)v1$ and hence there is one independent vector in the set

Title: How many vectors are linearly independent?

Body:
A sequence of vectors $\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}$ from a vector space $V$ is said to be linearly dependent, if there exist scalars $a_{1}, a_{2}, \ldots, a_{k},$ not all zero, such that
\begin{equation*}
a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+\cdots+a_{k} \mathbf{v}_{k}=\mathbf{0}
\end{equation*}
where $\mathbf{0}$ denotes the zero vector.

Suppose we have $\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}$ such that $a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+a_{3} \mathbf{v}_{3}=\mathbf{0}$ has solution $a_3=-2a_2 + 4a_1$. How many linearly independent vectors is there in the set $\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\}$ ? I feel like if the solution was $a_3=-2a_2$, $a_1 =0$, the answer would be $2$ ($\mathbf{v}_{1}$ and another one). Is it still $2$ for the above case ? Or 1 ?


问:
Maximum number of members in a club
Some people form $n=36$ clubs. There are no two clubs with same set of club members. No two people are in the same set of clubs.

If a person is not in a club, he/she must be friend with one person in that club.

However, nobody is friend with another person in the same club (yes, a weird bunch of people).

What's the maximum possible size of a club in this setup?

This is similar to the odd town/even town problem and the Fisher inequality. I suspect we could approach in a similar linear algebra setup: https://www.cs.utexas.edu/~panni/lec20.pdf

答:
There can be at most $2^{n-1} = 2^{35}$ people in any given club $C$. That's because there are $35$ clubs other than $C$, and for every subset of those $35$ clubs (including the empty subset), at most one person can be in exactly those clubs, and also in club $C$.

As pointed out in hinkypunk's answer, we can achieve this bound. Suppose that there are $2^{36}$ people: for every subset of the $36$ clubs, there is a person exactly in those clubs. Furthermore, suppose that every person is friends with only one other person: whoever is in exactly all the clubs they are not in. Then all the conditions in the problem are satisfied, and every club contains the maximum possible amount of $2^{35}$ people.

Title: Maximum number of members in a club


答:
Let $f(x) = 2x-4x^{2} \implies f'(x)=2-8x$.

Note that $f(x)$ is represented graphically as a downward facing parabola because of the sign of the coefficient of the $x^{2}$ term.

Hence, finding when the slope is $0$ we have the maximum when the value is valid.

Setting $f'(x) =0$ we have $8x=2 \implies x =
\frac{1}{4}$ which is in the interval you found.

Title: Find the maximum value of $2x-4x^2 > 0$

Body:
$x \in \mathbb{R}$.

My attempt:

We’re gonna check the case where: $2x-4x^2 >0$.

And by some basic algebra you will find that : $$ x \in ]0,\frac{1}{2}[$$
But since we are in Real numbers we can’t just plug numbers.


问:
Banach's Fixed Point theorem extra condition
My question doesn't concern Banach's fixed point directly but rather an extra condition that I have seen pop up at different times, but I can't remember where.

Let $T:X\rightarrow X$ be the contraction from Banach's fixed point theorem where we assume $X$ is also a normed space with the norm ||.||.

Then the extra condition for $f\in X$:$||Tf||\leq||f||$ is often shown.

Of course if one verifies this condition one obtains that $T$ indeed maps $X$ into $X$, however if we already know this what does this condition ensure?

I'm asking from a differential equations viewpoint. If I remember correctly this condition helped with achieving a continuation of a solution in time, i.e. rule out blow-ups.

I am sorry if this question is vague, but I was just wandering what I had previously used this condition for?

答:
This question is probably motivated by a typical application of the contraction mapping principle to PDEs. Now I am going to just make up a toy model, to illustrate the main points of this argument; these points are marked by blackboard font. The answer to the question lies in the last point.

Consider the initial value problem
$$\tag{1}
\begin{cases}
\partial_t u = \Delta u +u^2, & t\in (0, T), x\in \mathbb R^n, \\
u(0, x)=u_0(x).
\end{cases}
$$
We want to establish that, if $T$ is sufficiently small, then there is a solution $u\in C(0, T; L^{\infty}(\mathbb R^n))$. This is the Banach space of functions $u=u(t,x)$ that are bounded in $x$ for all $t\in (0, T)$, and such that the map
$$
t\in (0, T)\mapsto u(t, \cdot)\in L^\infty(\mathbb R^n)$$
is continuous.

This is not the only evolution space, but it certainly is the most natural; we can afford $u(t, x)$ to be defined for almost all $x$, but we must require it to be defined for all $t$, for the initial condition $u(0, x)=u_0$ to make sense. Thus, a space such as $L^p(0, T; L^{\infty}(\mathbb R^n))$ is not good, because its elements are defined only for almost all $t$. This is why we chose a space of continuous functions of $t$. With this argument, we have chosen the ambient Banach space.

Now we need to choose the candidate contraction mapping. Here, the input comes from the classical theory of ODEs; we define
$$
\Phi_{u_0}(u) = e^{t\Delta}u_0 +\int_0^t e^{(t-s)\Delta}(u^2(s, \cdot))\,ds,$$
because, if $v(t, x)$ is an arbitrary function, then $\Phi_{u_0}(v)$ solves the initial value problem
$$
\begin{cases}
\partial_t u = \Delta u + v^2, \\
u(0, x)=u_0(x).
\end{cases}
$$
We have given a formal definition of our candidate contraction mapping.

To finish the setup of the functional framework, we need one more input. We gave a formal definition of $\Phi_{u_0}$, we established that it must be defined in some closed subset $X$ of the Banach space $E:=C(0, T;L^\infty)$, but we still haven’t specified such subset. To do so, we observe that, since $\lVert e^{t\Delta}f\rVert_\infty\le \lVert f\rVert_\infty$, then
$$\tag{2}
\lVert \Phi_{u_0}(u)\rVert_E \le \lVert u_0\rVert_\infty + T\lVert u\rVert_E^2.$$
And this suggests the definition of $X$. We let

$$
X=\{v\in E\ :\ \lVert v\rVert_E\le R\}.$$

I can now, finally, answer the question. Why is this a good choice for $X$? First, it is a complete metric space. Second, by (2) we see that
$$
\lVert\Phi_{u_0}(u)\rVert_E \le \lVert u_0\rVert_\infty +TR^2,\quad \forall u\in X, $$
thus if $R=2\lVert u_0\rVert_\infty$, then for all
$$
T\le \frac{1}{4\lVert u_0\rVert_\infty}, $$
we have that $\Phi_{u_0}$ maps $X$ into $X$. Notice that what we did with (2) is exactly what the question asks; controlling
$$
\lVert \Phi_{u_0}(u)\rVert_E \text{ in terms of }\lVert u\rVert_E.$$
We have found the correct domain for the mapping $\Phi_{u_0}$.

We are not done. We still need to prove that $\Phi_{u_0}$ is a contraction mapping. I am not going to do so; it may be that this requires taking a smaller $T$. If somebody is interested then they should look into some book in parabolic PDEs, such as the one of Quittner and Souplet, which is the one I used to write this answer. (P. Quittner, P. Souplet, “Superlinear parabolic problems”, Theorem 15.2, https://www.springer.com/gp/book/9783764384425).


答:
The major points here are the right idea but this isn't necessarily a "direct proof" (if by direct you mean showing $p$ implies $q$) since you are saying that $f(x) > 0$ cannot happen by some contradiction.
One thing to note is that not every proof has a "direct way" to prove it as even some reasoning used in the proof could actually by due to contradiction or contrapositive

For example, let's consider another possible argument which is simply a modification of what you wrote here.
Say you take arbitrary $x_0 \in [a,b]$ and then show (using similar reasoning here) that for all $\epsilon > 0$, we have $f(x_0) < \epsilon$.
Then you would conclude that since $f(x_0) \geq 0$ that $f(x_0) = 0$.
However, the very fact that given a number $r \geq 0$, if $r \leq \epsilon$ for all positive $\epsilon$, then $r =0$ is proven necessarily with contradiction, so it can't always be avoided. This is part of the nature of real analysis.

Title: $f$ nonnegative, continuous in $[a,b]$ and such that $\int_a^b f(x)dx=0$ implies $f(x)=0$ in $[a,b]$

Body:
My book proves the following statement: "Let $f$ be nonnegative and continuous in $[a,b]$ and such that
$$\int_a^b f(x)dx=0$$
then $f$ is identically $0$ in $[a,b]$." using an argument by contradiction.

However, I've tried to prove it in a direct way (which takes way more effort, but I would like to know if there is a flaw in my reasoning or not), can someone check if my work is okay please?

My try: by hypothesis $f$ is nonnegative in $[a,b]$: if $f(x)=0$, the theorem is true.

So let's suppose $f(x)>0$: let $x_0 \in [a,b]$, by hypothesis $f$ is continuous in $[a,b]$, so in particular it is continuous in $x_0$. Since $f$ is positive in this case, I can chooose $\epsilon=\frac{1}{2}f(x_0)$ in the definition of continuity to get the inequality $\frac{1}{2}f(x_0) \leq f(x) \leq \frac{3}{2}f(x_0)$ if $x_0-\delta_\epsilon < x < x_0+\delta_\epsilon$. Since by hypothesis $f$ is continuous in $[a,b]$, it is integrable in $[a,b]$ as well, so by monotonicity of integral it is
$$\int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} \frac{1}{2}f(x_0)\,dx \leq \int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} f(x)\,dx \leq \int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} \frac{3}{2} f(x_0)\,dx$$
By hypothesis $\int_a^b f(x)\,dx=0$, so in particular $\int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} f(x)\,dx=0$; hence from the last inequality it follows that $\delta_\epsilon f(x_0) \leq 0 \leq 3\delta_\epsilon f(x_0)$. Since $\delta_\epsilon>0$, This means that $f(x_0)$ is both $\geq0$ and $\leq 0$, the only possibility is that $f(x_0)=0$. Since $x_0 \in[a,b]$ is arbitrary, this holds for all $x\in[a,b]$ and so $f(x)=0$ for all $x\in[a,b]$.

Is this correct? If it is maybe I should've put more attention in the intervals of the kind $[a,x_0+\delta_\epsilon]$ and $[x_0-\delta_\epsilon,b]$ because since $\delta_\epsilon$ if $x_0$ is too much near the boundary I risk to "go out" of $[a,b]$, but I'm not sure if this is a problem or not since if $\delta_\epsilon$ works a lesser $\delta'_\epsilon$ would work too and so I can restrict it as much as I need.


问:
How to verify if vectors of $\mathbb{R}^{2 \times 2}$ span $\mathbb{R}^{2 \times 2}$
If I have three square matrices as following:
$$\begin{pmatrix} 1 & 0 \\ 0&0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0&1\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1&0\end{pmatrix}$$

How can I verify if they span $\mathbb{R}^{2 \times 2}$ ?

So far I have only learned the fact that if vectors span $\mathbb{R}^n$, the matrix A composing of the vectors in their column form will have n pivots

My main problem is that I am not sure how can I put them into column form as they are square matrices. I have only dealt with vectors of $n \times 1$. In these cases, it is easy for me to verify if they span a particular space.

答: How would you get $\begin{pmatrix} 0 & 0 \\ 1&0\end{pmatrix}$?


问:
How to define my moments of a fake dice?
Say I have a dice.

Say I suspect that the dice is unfair with probability 75 % with probabilities $p_1, ..., p_6$ that are NOT all equal to 1/6.

However, with 25 % probability, the dice is actually fair and the probabilities are indeed 1/6.

I have two questions:

What is the mean and variance of this dice's outcomes?

Are there different ways of defining what I meant by "fake with 75 % probability" and do those different ways lead to different answers? For example, maybe the die is EITHER fake OR real, and I don't know which one it is .... that is one way to define it. OR maybe the die is BOTH fake AND real, and it switches between these two states every time I throw it. This definition is different, but does it give the same answer? Are there other definitions that make sense?

答:
You'd have to calculate it based on $p_i$, for instance on the average one would do:
$$E(X) = \sum_{k=1}^6 k \cdot \left(\frac{3}{4}p_1 + \frac{1}{4}\cdot\frac{1}{6}\right)  = \frac{3}{4}E(X|\text{fake}) + \frac{1}{4}E(X|\text{real}) $$
The equivalent way to see this question that would more intuitively work is saying you randomly choose a dice out of 4, 3 of which are identical but unfair with probabilities $p_i$, and $1$ is a fair dice, if the chance of an unfair dice were lower one could estimate the mean and variance well enough, but with a $\frac{3} {4}$ that a die can go from rolling straight sixes to rolling straight 1 means the range there is too big to get anything meaningful, in fact:
$$\frac{13}{8}\le E(X)\le \frac{43}{8}$$


问:
Is it true that bounded metric can never be induced by norm.
Let $(X, d)$ be a metric space where, $d$ is metric on $X$.

We know that metric space $X$ is called bounded if there exists some number $r$, such that $d(x,y) ≤ r$ for all $x$and $y$ in $X$.

I want to know is it true that a bounded metric doesn't satisfy homogeneity condition of norm. And thus could we conclude that bounded metric $d$ defined on metric space $X$ can never be induced by any norm.

Thanks

答:
If $\|\cdot\|$ is a norm (of a real vector space), and $\|v\|=r$ then for every $n\in\Bbb N$ we have $\|n\cdot v-0\|=\|n\cdot v\|=n\cdot r$.

It follows that $\|\cdot\|$ cannot be bounded.


问:
$\phi(n)=\frac{2n}{5}$ how do I find the number of solutions?
I'm not really all too sure how to approach this problem, any help would be much appreciated. (Specifically talking about #4)

答:
I assume, it is Euler Totient.

$$\frac{2n}{5}=n(1-\frac 12)(1-\frac 15)$$

A solution will be of the form  $$n= 2^p\times5^q$$

There are many many solutions.


问:
Intuition behind "ideal"
To briefly put forward my question, can anyone beautifully explain me in your own view, what was the main intuition behind inventing the ideal of a ring? I want a clarified explanations in these points:

Why is the name "ideal" coined?. In English 'ideal' means "One that is regarded as a standard or model of perfection or excellence." Why did people gave the name of ideal to such group?

And why are the ideals not present in the case of groups?

And give me a very fantastic intuition and motivation behind the ideals and what are the roles served by them in advanced mathematics.

Thanks a lot, to every one.

答:
Let $ R $ be a commutative ring with unity (Like in Atiyah-Macdonald, from now "ring" would mean "commutative ring with unity"). We can ask ourselves :

$\textbf{Q}$) In how many ways can we put an equivalence relation $ \sim $ on $ R $ (i.e. partition set $ R $), and put binary operations $ \oplus,\odot $ on set $ R/{\sim} $, such that "Equations in $ R $ give corresponding equations between equivalence classes in $ R/{\sim} $" that is "($a+b=c$ in $R$ implies $[a] \oplus [b] = [c]$ in $ R/{\sim} $) and ($ab=c$ in $ R$ implies $ [a] \odot [b] = [c]$ in $R/{\sim} $)" ?

[Under such $ \sim, \oplus, \odot $, notice $ (R/{\sim}, \oplus, \odot) $ automatically becomes a ring]

$\underline{\textbf{Part-1}}$ (Looking at the potential candidates for $ \sim, \oplus, \odot$)

Let $ \sim, \oplus, \odot $ be as needed. Unwrapping the constraints one by one...

$$ (1) \text{ } \sim \text{ is an equivalence relation} $$
and
$$ (2) \text{ } [a]=[a'], [b]=[b'] \implies [a]\oplus [b] = [a'] \oplus [b'], [a]\odot[b] = [a']\odot [b'] $$
and
$$ (3) \text{ } [a]\oplus [b] = [a+b], [a]\odot [b] = [ab]. $$

Using (3), we see (2) modifies as $ a \sim a', b \sim b' \implies a+b \sim a' + b' , ab \sim a' b'.$

Notice $ a \sim b \iff a-b \in [0] $

($ \implies $: As $ a \sim b $ and $ (-b) \sim (-b) $, we have $ a + (-b) \sim b + (-b) = 0 $.

$ \impliedby $: As $ a - b \sim 0 $ and $ b \sim b $, we have $ (a-b)+b \sim 0+b $, that is $ a \sim b $ )

Using this, constraint (1) can be rewritten as {$a - a \in [0]$; $a - b \in [0] $ implies $b-a \in [0] $; $ a-b, b-c \in [0] $ implies $ a-c \in [0] $}, which just means "$ [0] $ is a subgroup of $ (R, +) $".

Similarly constraint (2) becomes "$a-a', b-b' \in [0] $ implies $ (a+b)-(a'+b') , ab-a'b' \in [0]$", that is "$ a - a', b - b' \in [0] $ implies $ ab - a'b' \in [0] $", that is "$ x,y \in [0] $ implies $ (a' + x)(b' + y) - a'b' \in [0] $", that is "$x,y \in [0] $ implies $ a' y + b' x + xy \in [0] $", that is "$ t \in [0] $ implies $ at \in [0]$".

Constraint (3) remains the same.

So to summarise, in any such $ \sim, \oplus, \odot $ we have :

$ a \sim b \iff a - b \in I $, where $ I $ is a subgroup of $ (R, +) $ satisfying $ a I \subseteq I $ for all $ a \in R $

[Above condition gives $ [a] = a+I $] The operations $ \oplus, \odot $ satisfy $ (a+I) \oplus (b+I) = (a+b+I) $ and $ (a+I) \odot (b+I) = (ab + I) $.

$ \underline{\textbf{Part-2}} $ (That all such $ \sim, \oplus, \odot $ work)

Let $ I $ be a subgroup of $ R $ with $ aI \subseteq I $ for all $ a \in R $. We can readily verify $ a \sim b \overset{\text{def}}{\iff} a - b \in I $, and $ (a+I) \oplus (b+I) \overset{\text{def}}{=} (a+b+I) $, $ (a+I) \odot (b+I) \overset{\text{def}}{=} (ab+I) $ satisfy the constraints in question.

To summarise the entire discussion, equivalence relations $ a \sim b \iff a - b \in I $, arising from subgroups $ I $ of $ (R, +) $ satisfying $ a I \subseteq I $ for all $ a \in R $, are precisely the ones we were looking for. The subgroups here are traditionally called "Ideals".

Title: Intuition behind "ideal"


问:
sum of floor function simplification
Let $n \in \mathbb{N}$.

I'm trying to show $\sum_{i=1}^{\infty} \left(\left \lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}}\right \rfloor \right) = 2^k-1$.

I know that when $2^i > n-1$, the right floor function will become zero.

I'm not sure how I cancel out the $n$'s.

I also tried solving the sum of each floor function separately, but didn't get anywhere.

Any suggestions will be appreciated!

Thanks.

答:
The point is that the LHS is a geometric series, because extraneous parts of the sums cancel. However, the result doesn't seem quite right, because you get some extra terms.

$$
\begin{align}
&\sum_{i=1}^{\infty}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)\\
&= \sum_{i=1}^{k}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)
+\sum_{i=k+1}^{\infty}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)\\
&= \sum_{i=1}^{k}
\left( \frac{2^k}{2^i} +\left\lfloor{\frac{n-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)
+\sum_{i=k+1}^{\infty}
\mathbb{1}(n-1<2^i, n+2^k-1\ge 2^i)\\
&= \sum_{i=1}^{k}2^{k-i}+ \text{something}\\
&= \sum_{j=0}^{k-1}2^j + \text{something}\\
&= 2^k-1 + \text{something}
\end{align}
$$

Title: sum of floor function simplification


问:
Find the number of positive integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=x_1\cdot x_2\cdot x_3\cdot x_4\cdot x_5$
Find the number of positive integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=x_1\cdot x_2\cdot x_3\cdot x_4\cdot x_5$

This is one of the questions from the Indian Institution of Technology Joint Entrance Examination question. I have been struck on this problem for hours.

I was solving some intriguing maths problems on combinatorics. Then I came across this wonderful questions and I was surprised looking at it. Firstly what I thought to give it a try was to Give it a BIJECTION of Distinct Balls distribution in Identical Boxes. For example $x_1+x_2+x_3=20$ can have $(20+3-1)C(3-1)$ solutions (including zero) but how does it work here?

答:
$$ 0=  x_1 + x_2 + x_3 + x_4  +x_5 - x_1 x_2 x_3 x_4 x_5 \tag{1}$$

An immediate observation I see is that the the above expression is symmetric , so finding one solution will lead to 5! other solutions. Hence, we search unique solution after putting an ordering on our variables:

$$ 1 \leq x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5$$

From the comment by lulu(*), we can deduce that:

$$ x_1 x_2  x_3 x_4 x_5 \leq 5x_5$$

Or,

$$ x_1 x_2 x_3 x_4 \leq  5 \tag{2}$$

For the max case consider putting $x_4= 5$ , this leads to the following ordering:

$$ x_1 \leq x_2 \leq x_3 \leq  x_4 \leq 5 \tag{3}$$

Since the cases are small , we can find them by directly counting as the four number lists which satisfy (2) and (3) :

$$ (x_1,x_2, x_3 , x_4 ) = \{ (1,1,1,1), (1,1,1,2) ,(1,1,1,3), (1,1,1,4) , (1,1,2,2) \} \tag{4}$$

Now consider $(1)$ and rearrange it such that we write $x_5$ as a function of othervariables:

$$ 0 = x_1 + x_2 + x_3 + x_4 + x_5(1-x_1 x_2 x_3 x_4)$$

Or,
$$ \frac{x_4 + x_2 + x_3 + x_1 }{x_3 x_4 x_2 x_1 - 1} = x_5$$

Plug in the number lists to the above equation, this will lead us the following set of $(x_1,x_2,x_3,x_4,x_5)$ quintuples : $\{(1,1,1,2,5),(1,1,1,3,2) , (1,1,2,2,3) \}$ (As checked using a Pascal program by @Raffaele )[ I have neglected the cases where plugging in the list gave me negative/ undefined number to satisfy the conditions of the question]

Now, with the solutions, I'll leave it to you to permute them and find the total :^)

(*): If the values of $x_1,x_2,x_3,x_4,x_5$ are positive integers, then after putting the ordering, it must be that the sum of rest of numbers is less than five times the sum of the largest number. For example,

$$ 1 +2 +3 +4 +5 \leq 5(5)$$

The sum of first five integers is definitely less than five times the largest integer


问:
Particular proof of the fundamental theorem of equivalence relations
I have some trouble understanding the following proof. After doing some research, I realized that this is (probably) the Fundamental Theorem of equivalence relations. I didn't understand the proofs I found online and I struggled specifically with the following proof:

Theorem: Let $R$ be an equivalence relation on the set $X$. Then every element $a \in X$ belongs to exactly one equivalence class and for any two equivalence classes $A, A'$ we either have $A=A'$ or $A \cap A' = \emptyset$.

Proof: For a fixed $a \in X$ we define $A := \{x \in X : x \sim a\}$. We show that $A$ is an equivalence class that contains $a$. Because of $a \sim a$, we have $a \in A$ and therefore $A \neq \emptyset$. For $x,y \in A$, we have $x \sim a$ and $y \sim a$ and therfore also $x \sim y$.

If $x \in A, y \in X$ and $x \sim y$, then we have $x \sim a$ and $y \sim a$ and therefore $y \in A$. Now we have shown that $a$ is contained in at least one equivalence class[*].

Now we need to show that two equivalence classes $A$ and $A'$ are either the same or disjoint. Let $A \cap A' \neq \emptyset$ and $a \in A \cap A'$. If $x\in A$, then we have $x \sim a$ and because of $a \in A'$ we also have $x \in A'$. Therefore, we have $A \subset A'$. We can prove $A' \subset A$ analougsly. Therefore, $A = A'$.

[*] This is the step I am having trouble with. Why did we assume that $y \in X$ and not $y \in A$ as we have done before? First we let $x,y \in A$ to show that $x \sim y$ and then we set $y \in X$ to show that $y \in A$? Don't we use $x \sim y$ to show that point and $x \sim y$ dependes on $x,y \in A$? And why does it follow that $a$ is contained in at least one equivalence class?

答:
Let $\sim$ denote an equivalence relation on set $X$.

That means that the relation is reflexive, symmetric and transitive.

If we define for every $a\in X$ the set:$$[a]:=\{x\in X\mid  a\sim x\}$$ then the reflexivity of $\sim$ tells us immediately that $a\in[a]$.

Transitivity tells us that:$$c\in[a]\text{ and }x\in[c]\implies x\in[a]$$
or equivalently:$$c\in[a]\implies[c]\subseteq[a]\tag1$$

By symmetry we find:$$c\in[a]\iff a\sim c\iff c\sim a\iff a\in[c]$$This tells us that $(1)$ can be replaced by the stronger statement:$$c\in[a]\iff[c]=[a]\tag2$$

So if $[a]$ and $[b]$ have an element $c$ in common then $(2)$ tells us immediately that $[a]=[c]=[b]$.

This shows that for sets $[a]$ and $[b]$ there are only two options: they are disjoint or they coincide.

The sets of form $[a]$ are by definition the equivalence classes induced by $\sim$.

If $A\subseteq X$ is such an equivalence class then $A=[a]$ for every $a\in A$ and an element of $A$ is a so-called representative of the class.

As shown above these sets form a partition of $X$.


答:
Using the AM-GM inequality, given $2x-4x^2>0$,

$\sqrt{2x-4x^2 }=\sqrt{2x(1-2x)}\le\dfrac{(2x)+(1-2x)}2=\dfrac12$.

Now square both sides.

Title: Find the maximum value of $2x-4x^2 > 0$


答:
You could also view it as follows: A matrix in $\Bbb R^{2\times 2}$ consists of 4 real values. This means $\Bbb R^{2\times 2}$ is basically $\Bbb R^4$, since you get an isomorphism of vector spaces via
$$
\Bbb R^{2\times 2} \to \Bbb R^4, \begin{pmatrix} a & b \\ c & d\end{pmatrix} \mapsto \begin{pmatrix} a\\b\\c\\d \end{pmatrix}.
$$
But $\Bbb R^4$ has dimension 4 (for example $\{(1,0,0,0)^T, (0,1,0,0)^T, (0,0,1,0)^T , (0,0,0,1)^T\}$ is a basis), so $\Bbb R^{2\times 2}$ must have dimension 4, thus $3$ elements can't span the whole vector space.

Title: How to verify if vectors of $\mathbb{R}^{2 \times 2}$ span $\mathbb{R}^{2 \times 2}$

Body:
If I have three square matrices as following:
$$\begin{pmatrix} 1 & 0 \\ 0&0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0&1\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1&0\end{pmatrix}$$

How can I verify if they span $\mathbb{R}^{2 \times 2}$ ?

So far I have only learned the fact that if vectors span $\mathbb{R}^n$, the matrix A composing of the vectors in their column form will have n pivots

My main problem is that I am not sure how can I put them into column form as they are square matrices. I have only dealt with vectors of $n \times 1$. In these cases, it is easy for me to verify if they span a particular space.


问:
Lie Algebra of $SL_n(\mathbb H)$
The Lie algebra of $SL_n(\mathbb C)$ are the matrices where the trace is $0$. But what is the Lie algebra of $SL_n(\mathbb H)$ where $\mathbb H$ is the quaternions?</p>

答:
The group $\mathrm{SL}(n, \mathbb{H})$ is defined here:

Wikipedia, GL(n,H) and SL(n,H).

It works like this:

The algebra of quaternions is a subalgebra of the $2 \times 2$ complex matrices in a standard way. This lets us regard $n \times n$ quaternionic matrices as certain special $2n \times 2n$ complex matrices.  This in turn lets us define a complex-valued determinant of an $n \times n$ quaternionic matrix, called the Study determinant, and also a complex-valued trace.

The $n \times n$ quaternionic matrices for which the Study determinant is 1 form the group $\mathrm{SL}(n,\mathbb{H})$.   The Lie algebra of $\mathrm{SL}(n,\mathbb{H})$ then consists of $n \times n$ quaternionic matrices for which the complex-valued trace is 0.

The Study determinant is the square of the Dieudonné determinant, as shown here:

Helmer Aslaksen, Quaternionic determinants, The Mathematical Intelligencer 18 (1996), 57–65.

But the Dieudonné determinant is nonnegative, so in fact the Study determinant also takes only nonnegative real values, and $\mathrm{SL}(n,\mathbb{H})$ is also the group of quaternionic $n \times n$ matrices with Dieudonné determinant 1.

Title: Lie Algebra of $SL_n(\mathbb H)$


问:
How do we solve the equation $4^{x} + 2^{x} = 20$?
It is easy that $x = 2$ below, but can I solve it with a general formula?

$$4^x + 2^x = 20$$

答: Do the substitution $t=2^x$, then solve the quadratic equation you get. Undo the subsitution, you will have to decide whether one answer works or not.


问:
How is "n+n/2+n/4....1" equal to "2n-1" using the formula for geometric series?
I never knew not having good knowledge of basic maths will be so crippling!! So please help me out this time. I'll be working on my maths from today on.

I was discussing about complexity of an algorithm on StackOverflow and I was told that the series $n+n/2+n/4 + \dots + 1$ evaluates to $2n-1$ and I was linked to the following formula on Wikipedia:

Even after trying hard, I regret to say I still don't get it how using this formula I can conclude that my series evaluates to $2n-1$. Please help me out as I am sure it will take only a few seconds for you.

https://stackoverflow.com/questions/16748454/complexity-for-nested-loops-diving-by-2?noredirect=1#comment24124682_16748454

答:
If you have knowledge of progression than this is a geometric series with common ratio of $\frac{1}{2}$ and formula you have written is sum to n terms of geometric progression.

But you  require to calculate the sum of this series upto $2n-1$ term, so you can replace $n$ with $(2n-1)$ and then proceed. Here , $a = n$ (which is your first term).

So, by putting these values we get : $\frac{a(1-r^{2n-1})}{1-r}$ =

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

I hope from here you  will get your answer....


问:
Side-limit of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$
The problem wants the value of left hand limit of the derivative of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ when $x\to3^-$.I graphed the function and the answer is $\frac{3}{2}$ but what's wrong with my method:

For any value $x\to3^-$ we have $[x]=2$ , $|x-1|=x-1$ so if $h\to0^-$, then $f(3+h)=\frac{2(3+h)^2}{2+h}=\frac{2h^2+12h+18}{2+h}$. So the value of the derivative is equal to:
$$\lim_{h\to0^-}\dfrac{f(3+h)-f(3)}{h}=\lim_{h\to0^-}\frac{\frac{2h^2+12h+18}{2+h}-\frac{27}{2}}{h}$$ which is infinity and clearly wrong with or without the graph. I'm not new into this but I can't figure out what's wrong here despite I thought about the method for one hour straight.

答:
$$f(x) = \frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$$

Assume $0 \le \delta < 1$. Then

\begin{align}
   f(3)&= 13.5\\
   f(3-\delta) &= 9 \\
\hline
   f'(3^-)
   &= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &= \lim_{\delta \to 0^+}\dfrac{4.5}{\delta} \\
   &= \infty
\end{align}

You did it correctly. Why? Because the floor function is discontinuous from below.

$f(3^-) = 9$ and $f(3) = f(3^+) = 13.5$


问:
Proof by induction: base case for a limit as x goes to infinity?
With induction we always start with a base case; What would the base case for this be?
Choosing 1 seems nonsensical.
Choosing infinity seems wrong as well.

Prove, using induction, that $\lim\limits_{x\to\infty}\dfrac{(\ln x)^k}x=0$.

答:
As I said in a comment, it's a strange idea, but let's give it a try: so we want to prove $$\lim_{x\to\infty} \frac{\ln^k x}x=0$$ for $k=0,1,\ldots$ This is obvious for $k=0$. Defining $$f_k(x)=\frac{\ln^k x}x,$$ it isn't a good idea to use $f_{k+1}(x)=\ln x\,f_k(x)$ for the induction step, but
$$f_{k+1}(x)=2^{k+1}\,f_k(\sqrt{x})\frac{\ln\sqrt{x}}{\sqrt{x}}<2^{k+1}\,f_k(\sqrt{x})$$ will turn the trick.


答:
What you're trying to show is not always true. For example, let $n = 4$, so $n - 1 = 3$, and $k = 1$, so $2^k = 2$. This means $n + 2^k - 1 = 5$. However, using that each summation term is non-negative, the first few terms then become

$$\begin{equation}\begin{aligned}
& \sum_{i=1}^{\infty} \left(\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) \\
& = \left(\left\lfloor\frac{5}{2}\right\rfloor - \left\lfloor\frac{3}{2}\right\rfloor\right) + \left(\left\lfloor\frac{5}{4}\right\rfloor - \left\lfloor\frac{3}{4}\right\rfloor\right) + \ldots \\
& = (2 - 1) + (1 - 0) + \ldots \\
& = 1 + 1 + \ldots \\
& = 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note I've used that all of the remaining terms, i.e., the "$\ldots$" part, are $0$. This is larger than your right side of $2^{k} - 1 = 1$.

Note your statement would be correct with an appropriate restriction between $k$ and $n$. In particular, something like $2^k \gt n - 1$ works. This is because, for $i \le k$, each of the summation terms becomes

$$\begin{equation}\begin{aligned}
\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor & = \left(\left\lfloor{\frac{2^k}{2^i} + \frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\
& = \left(2^{k-i} + \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\
& = 2^{k-i}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

For $i \gt k$, then $2^i \ge 2(2^k) \gt n + 2^k - 1$ and $2^i \gt n - 1$, so both floor terms become $0$. Thus, the total sum would be that of a geometric series of

$$\sum_{i=1}^{k}(2^{k-i}) = 2^k - 1 \tag{3}\label{eq3A}$$

which then matches your right side.

Title: sum of floor function simplification

Body:
Let $n \in \mathbb{N}$.

I'm trying to show $\sum_{i=1}^{\infty} \left(\left \lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}}\right \rfloor \right) = 2^k-1$.

I know that when $2^i > n-1$, the right floor function will become zero.

I'm not sure how I cancel out the $n$'s.

I also tried solving the sum of each floor function separately, but didn't get anywhere.

Any suggestions will be appreciated!

Thanks.


问:
Understanding the relation between multiplication order of covectors and interpretation as linear functionals
I have read over similar discussions in other questions, e.g. here: What is the difference between a vector and its transpose?

But I'm still confused about covectors and their relation to matrix multiplication. Wikipedia says the definition of a covector is an element of the dual. But vectors and covectors are also just ordered sets of $n$ elements from the underlying field. In this way, they are interchangeable. We also multiply vectors on the right (of a matrix) and covectors on the left. Can this fact be deduced from the definition of a covector? Is it in some way natural, or is there a design choice here in the way we write them as rows/columns? Is the transpose of a vector defined as the covector with the same field elements in the same order?

What I'd like to get is one simple definition for a vector and one simple definition for a covector from which the multiplication order and the fact that they're transposes of each other follow naturally.

I'm sorry if there are a lot of questions all in one here. Thanks.

答:
"But vectors and covectors are also just ordered sets of n elements from the underlying field."

No. that's not at all what they are. If you fix a vector space $V$ over a field $F$, then by convention, we call elements of $V$ vectors and elements of $V^*=\text{Hom}(V,F)$ covectors. $V$ could be any abstract vector space, so a-priori, there is no meaning at all in thinking of elements of $V$ as ordered tuples of numbers. Also $V$ and $V^*$ are completely different spaces and in general you should never conflate the two.

Even if you're considering the special case where $V=F^n$ for some positive integer $n$, $(F^n)^*$ and $F^n$ are still completely different spaces, so they're not equal. Furthermore, there is no natural isomorphism between these spaces, so there's no reason at all to think they're the same thing. It is only AFTER you choose a basis $\beta$ for a (finite-dimensional) vector space that you can construct an isomorphism $\Phi_{\beta}:V\to V^*$.

So, in this sense, your entire conception of vectors and covectors seems to be misguided by your thinking in terms of row/column vectors.

So, what is the relationship between vectors (i.e elements of $V$) covectors (elements of $V$) and vertical ordered lists of numbers (column vectors) and horizontal lists of numbers (row vectors)? Well, it all stems from choosing a basis for $V$.

If you have a basis $\beta=\{v_1,\dots, v_n\}$ for $V$, then given any $v\in V$, there exist unique $a_1,\dots, a_n\in F$ such that $v=\sum_{i=1}^na_iv_i$. In this case, we can write
\begin{align}
[v]_{\beta}&=
\begin{pmatrix}
a_1\\\vdots \\a_n
\end{pmatrix}
\end{align}
We call $[v]_{\beta}$ the coordinate representation of $v\in V$ relative to the basis $\beta$.

Next thing we shall discuss is matrix representations of linear transformations. So let $V,W$ be vector spaces over $F$, let $\beta=\{v_1,\dots, v_n\}$ be a basis for $V$, let $\gamma=\{w_1,\dots, w_m\}$ be a basis for $W$, and let $T:V\to W$ be a given linear transformation. In this case, we can associate with $T$, a certain rectangular array of numbers $[T]_{\beta}^{\gamma}$ called the matrix representation of $T$ relative to the ordered basis $\beta$ and $\gamma$. This will be a certain $m\times n$ matrix with entries coming from the field $F$. How is this matrix obtained? Well, for each $v_j\in \beta$, we can consider $T(v_j)\in W$. Since $\gamma=\{w_1,\dots, w_m\}$ is a basis, there exist unique scalars $a_{1j},\dots, a_{mj}\in F$ such that
\begin{align}
T(v_j)= \sum_{i=1}^ma_{ij}w_i
\end{align}
So, we take this whole massive collection of numbers $a_{11},\dots, a_{m1}, a_{12}\dots, a_{m2},\dots, a_{1n},\dots a_{mn}$ and stick it into a rectangular array (a matrix):
\begin{align}
[T]_{\beta}^{\gamma}&:=
\begin{pmatrix}
a_{11} & \dots &a_{1n}\\
\vdots &\ddots &\vdots\\
a_{m1}&\dots & a_{mn}
\end{pmatrix}
\end{align}

I hope you appreciate the difference: $T$ is an element of $\text{Hom}(V,W)$ and it is the thing we are really interested about, while $[T]_{\beta}^{\gamma}\in M_{m\times n}(F)$ is just a rectangular array of numbers, and until you appreciate the significance of the basis-free development of linear algebra, you should not assign any extra significance to this array of numbers other than that it is a convenient short-hand way of keeping track of some information.

Now, what you can prove is that for any $v\in V$, we have
\begin{align}
[T(v)]_{\gamma}&=[T]_{\beta}^{\gamma}\cdot[v]_{\beta}
\end{align}
What is this saying? Well, if you start with a vector $v\in V$ and a linear map $T:V\to W$, then $T(v)\in W$. So, the equation above is telling you that the coordinate representation of $T(v)$ is obtained by multiplying the matrix representation of $T$, with the coordinate representation of $v$.

Now, consider what a covector is: it is an element of $V^*$, i.e it is a linear map $\lambda:V\to F$, so said once again, in the discussion above, we're considering the special case of $W=F$. Again, we are fixing a basis $\beta=\{v_1,\dots, v_n\}$ for $V$. What is a basis for $F$? Well, the most obvious one is $\gamma=\{1\}$ ;i.e it is a basis consisting of only one element, namely the element $1\in F$. So, by the conventioned described above, we can consider the matrix representation of the linear transformation $\lambda$, which gives us a $ (\dim F)\times (\dim V)=1\times n$ matrix $[\lambda]_{\beta}^{\gamma}=[\lambda]_{\beta}^{\{1\}}$.

At this stage a very tempting thing to do is the following: start with a $\lambda\in V^*$, then write out its $1\times n$ matrix representation, say
\begin{align}
[\lambda]_{\beta}^{\{1\}}&=
\begin{pmatrix}
c_1 & \dots &c_n
\end{pmatrix}
\end{align}
Now, you can look at the vector $v=\sum_{i=1}^nc_iv_i$. This has the coordinate representation
\begin{align}
[v]_{\beta}&=
\begin{pmatrix}
c_1\\\vdots\\c_n
\end{pmatrix}
\end{align}

So, now if you just look at the numbers alone, and completely forget everything I've told you about the importance of the distinction between vectors and covectors, you'll (rightfully) say that the only difference between these things is that one list of numbers is given horizontally while the other is given vertically.

What I have just described is a way of associating to each element of $V^*$ a certain element of $V$. You can infact reverse this procedure. What this means is I've described for you an isomorphism $V^*\to V$ which assigns to each $\lambda \in V^*$ an element $v\in V$ according to the procedure described above. But of course, you should keep in mind that $\lambda \neq v$ (of course, how can they be equal if they live in different spaces). The correct relationship is that $[\lambda]_{\beta}^{\{1\}}=([v]_{\beta})^t$.

One final thing I should mention is that transpose of a matrix comes in the following manner: suppose I have a linear transformation $T:V\to W$. Then, this "induces" a very natural linear map $T^*:W^*\to V^*$ defined by $T^*(\phi):= \phi\circ T$. This is called the pull-back/transpose map of $T$. The reason for this terminology is that if you take a basis $\beta$ of $V$, $\gamma$ of $W$, and you consider their dual bases $\beta^*$ of $V^*$ and $\gamma^*$ of $W^*$, then
\begin{align}
[T^*]_{\gamma^*}^{\beta^*}&= \left([T]_{\beta}^{\gamma}\right)^t
\end{align}
The fact that $T:V\to W$ while $T^*:W^*\to V^*$ seems to go in the "opposite direction" is exactly why if you unwind all the definitions carefully, you'll see that the matrix representation of $T^*$ is the transpose of the matrix representation of $T$ (the fact that the linear map "goes the other way" is reflected in the fact that the matrix representation is somehow "flipped", i.e transposed).


问:
Let $V = \{(x, y, z) \in \mathbb{R}^3 : y - z = 0 \}$, and let $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an endomorphism such that ...
Let $V = \{(x, y, z) \in \mathbb{R}^3 : y - z = 0 \}$, and let $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an endomorphism such that $V = \text{ker}(f)$ and $-1$ is an eigenvalue of $f$.

Which of the following is true?

(a) $\text{dim}(\text{Im}(f)) = 2$

(b) The characteristic polynomial of $f$ can be $-t(t + 1)^2$

(c) $f$ has three distinct eigenvalue

(d) $f$ is simple

The correct answer is (d), but I don't know how he arrived at that conclusion. Can you help me?

答:
I also do not know what you mean by f is simple but as mentioned above you can argue that all assumptions a,b,c are false the following way.

a)Since $ker(f)=V$ and $V$ is 2 dimensional than $Im(f)$ has to be 1 dimensional, by the dimensionequation for a linear map.

b) The characteristic polynomial can not be the following because that would mean that the algebraic multiplicity of the eigenvalue $0$ should be $1$, but its geometric multiplicity which is given by $dim(V)=dim(f-0*Id)=2$ is greater than $1$ , leads to a contradiction. Look up the inequality for algebraic and geometric multiplicity.($a(\lambda)<g(\lambda)$)

c)$f$ also can not have three distinct eigenvalues. If one of the eigenvalues were to be $0$ than you can lead the assumption to a contradiction by the above argumentation. If $0$ were no eigenvalue the kernel would than be trivial which is also a contradiction.

d) i am not sure what simple means


问:
Analytic tools in the theory of Galton-Watson processes
The questions basically aims at discussing the relative power of using probability generating functions, moment generating functions and characteristic functions as an example for Galton-Watson-processes (GWP).

Within the theory of GWP I see people extensively use the probability generating function $\mathbb E[s^X], x\in [0,1]$ for a random variable $X$.

So far, I have thought that the characteristic function $\mathbb E[e^{itX}]$ is the most powerful tool out of three mentioned above, because it yields all information about the distribution and has nice analytic properties, in particular always exists.

Is there a good reason why the probability generating function is used for GWP?

答: All three are strictly equivalent, of course, hence each is exactly as powerful as the two others, in particular the generating function is always analytic on $[0,1)$. Some natural properties of branching processes are best expressed using generating functions though, for example the probability of extinction $q$ solves $q=f(q)$ where $f(s)=E[s^X]$ for every $s$ in $[0,1]$ and $X$ denotes the number of descendants of each individual.</p>


问:
Differentiable function and the limit as x goes to infinity
Let $$f: \mathbb{R} \to \mathbb{R}, \lim_{x \to \infty} f'(x) = 0.$$

Prove: $$\lim_{x \to \infty} (f(x + 2) - f(x)) = 0$$

I know I need use the MVT somewhere in here. I know $f$ is differentiable on $(x, x+2)$ so $$f(x+2) - f(x) = 2f'(c)$$ for some $c \in (x,x+2).$ So if we keep looking at the interval as $x \to \infty$ then it seems that's all we need, but I'm not sure how to make it more rigorous.

Edit:
Am I on the right track here:
$$f(x+2) - f(x) = 2f'(c)$$
implies $$\lim_{x \to \infty} f(x+2) - f(x) = \lim_{c \to \infty} 2f'(c) = 0$$

答:
You are on the right track. And let us make it very, very rigorous, i.e. let us use the $\varepsilon$ definition.

So let $\varepsilon > 0$ be arbitrary. Indeed, there is some $c_x \in (x, x+2)$ such that
$$
\lvert f(x+2) - f(x) \rvert = 2\lvert f'(c_x) \rvert.
$$
Since $\displaystyle \lim_{x \rightarrow \infty} f'(x)  = 0$, there exists $M>0$ such that
$$
\lvert f'(y) \rvert < \frac{\varepsilon}{2}
$$
for all $y > M$. So, for all $x > M$, we have $c_x > M$ and then
$$
\lvert f(x+2) - f(x) \rvert = 2\lvert f'(c_x) \rvert < 2 \cdot \frac{\varepsilon}{2} = \varepsilon
$$
for all $x>M$.


问:
Estimate bias of perturbed Bernoulli process
I have a Bernoulli process with parameter $\theta$,

$$
Z_1, Z_2, \dots, Z_n \sim \text{Bernoulli}(\theta),
$$

as well as a perturbed process:

$$
X_1, X_2, \dots, X_n \sim \text{Bernoulli}(\theta + \varepsilon), \qquad \varepsilon \geq 0, \qquad (\theta + \varepsilon) \in [0, 1].
$$

Given realizations $\{z_i\}_{i=1}^n$ and $\{x_i\}_{i=1}^n$, can I estimate $\varepsilon$?

答:
You basically just have to build an estimator for $\varepsilon$. If $X_1,\cdots,X_n, Z_1,\cdots,Z_n$ are all i.i.d., then
$$
\mathbb E\left[ \frac 1n \left( \sum_{i=1}^n (X_i - Z_i) \right) \right] = \frac 1n \left(\sum_{i=1}^n(\theta+\epsilon) - \theta \right) = \varepsilon.
$$
So this is the simplest unbiased estimator you could come up with:
$$
\widehat{\varepsilon} = \frac 1n \sum_{i=1}^n (X_i - Z_i).
$$
It will have a variance, namely since $\mathrm{Var}(Z_i) = \theta(1-\theta)$ and $\mathrm{Var}(X_i) = (\theta + \varepsilon)(1-(\theta + \varepsilon))$, thus
$$
\mathrm{Var}\left(\frac 1n \sum_{i=1}^n (X_i-Z_i) \right) = \frac 1{n^2} \sum_{i=1}^n \mathrm{Var}(X_i) + \mathrm{Var}(Z_i) \\
= \frac 1{n^2}\left( \sum_{i=1}^n (\theta+\varepsilon)(1-\theta-\varepsilon) + \theta(1-\theta) \right) \\
= \frac 1n \left( 2\theta(1-\theta) + \varepsilon(1-\theta) + \theta(1-\varepsilon) - \varepsilon^2 \right).
$$
Hope that helps,


答:
We know that $\text{dim(Im}(f))+\text{dim(Ker}(f))=\text{dim}(\mathbb R^3)$.

$V$ is defined as the set $\{\underline x \in \mathbb R^3:y-z=0\}$, which has dimension equal to $2$ because the system $\begin{cases}x=\alpha&\alpha\in\mathbb R\\y=z=\beta &\beta\in\mathbb R\end{cases}$ has two free parameters.

This fact implies that the geometric multiplicity of the eigenvalue zero, which corresponds to the dimension of the eigenspace relative to the eigenvalue $\lambda= 0$, is equal to $2$ and also that $\text{dim(Im}(f))=3-2=1$.

For these reasons we can state that (a) and (b) are false. The second is false because the characteristic polynomial can't have the $t$ term with algebraic multiplicity one since $\forall \lambda\in\mathbb R$ eigenvalue $mg(\lambda)\le ma(\lambda)$.

Of course the characteristic polynomial $\chi(t)=t^2(t+1)$ doesn't have three distinct roots for the same facts I wrote for (a) and (b).

Title: Let $V = \{(x, y, z) \in \mathbb{R}^3 : y - z = 0 \}$, and let $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an endomorphism such that ...

Body:
Let $V = \{(x, y, z) \in \mathbb{R}^3 : y - z = 0 \}$, and let $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an endomorphism such that $V = \text{ker}(f)$ and $-1$ is an eigenvalue of $f$.

Which of the following is true?

(a) $\text{dim}(\text{Im}(f)) = 2$

(b) The characteristic polynomial of $f$ can be $-t(t + 1)^2$

(c) $f$ has three distinct eigenvalue

(d) $f$ is simple

The correct answer is (d), but I don't know how he arrived at that conclusion. Can you help me?


问:
Show the sequence converges uniformly
Here is another uniform convergence problem I have been struggling with. Show that $e^{\frac{x}{k}}$ converges uniformly on $[0,1]$.

Attempt: I see that the function sequence converges pointwise to $1$.Again, I am having difficulty finding a bound for $|e^\frac{x}{k}-1|$ for all $x \in [0,1]$, so that for any $\epsilon$ I can choose an $N$ for all $x \in[0,1]$ such that $k \geq N \implies |e^\frac{x}{k}-1|<\epsilon$ . What am I supposed to do here?

Edit: So I looked at one of the answers and I decided that this would be sufficient.

Let $\epsilon>0$. Choose $N>\frac{1}{\text{ln}(1+\epsilon)}$ then $k \geq N \implies |e^\frac{x}{k}-1|<|e^\frac{1}{k}-1|<\epsilon$ for all $x \in [0,1]$, can I get a confirmation that this is ok?

答:
hint

For any $ k>0$, the function $$g_k: x\mapsto e^{\frac xk}-1$$

is positive and strictly increasing at $ [0,1]$.

$$M_k=\max_{x\in[0,1]}|g_k(x)|=$$
$$g_k(1)=e^{\frac 1k}-1$$

$$\lim_{k\to +\infty}M_k=1-1=0$$
So, the convergence is uniform at $ [0,1]$.

Title: Show the sequence converges uniformly


问:
Examples of non-complete lattices with enough prime elements
In this nice answer, Eric Wofsey shows that a lattice $L$ is isomorphic to the lattice of open sets of a topological space if, and only if, the following two conditions hold:

$L$ is complete: every subset $S\subseteq L$ has a least upper bound.

$L$ has enough prime elements (elements $p\in L\setminus\{1\}$ such that $a\wedge b\leq p$ implies $a\leq p$ or $b\leq p$): that is, if $a,b\in L$ and $a\not\leq b$, then there exist a prime element $p\in L$ such that $b\leq p$ and $a\not\leq p$.

Although I already know complete lattices without enough prime elements (here), I wonder about examples of non-complete lattices with enough prime elements. Since I am new to this pointless perspective, I don't know whether this question is trivial.

Anyway, I would like to ask for examples of lattices satisfying condition 2 but not condition 1, or references where I can find them.

答: For a very simple example, let $X$ be any infinite set and let $L$ be the lattice of subsets of $X$ that are either finite or cofinite.  Then $L$ is not complete (for instance, if $A\subset X$ is infinite and coinfinite, the collection of finite subsets of $A$ has no join in $L$).  However, $L$ has enough prime elements: the prime elements of $L$ are the complements of singletons, and if $a\not\leq b$ in $L$ then there is some $x\in a$ that is not in $b$ and the prime $p=X\setminus\{x\}$ satisfies $b\leq p$ and $a\not\leq p$.


问:
Volume of revolution around a line
Find the volume of revoulution of the region enclosed by $y = x^3, x = 0, x = 1$ and $y=-1$ around the axis $y = -1$.

This question is most easily done via the disk method, however, I tried cylindrical shells and I have an incorrect answer.

The region is composed of two sections: A square with side 1 in the fourth quadrant and a region enclosed by $y=x^3$ in the first quadrant.  So using shells, I believe I need to split the integral up, since the shells will be varying in height:

For ths lower region, I have $2\pi \int_{-1}^0 y\,dy = \pi$  For the upper region, I have $2 \pi\int_{0}^1 (1+y)(1-y^{1/3})\,dy = \frac{9\pi}{14}$  Evaluating and adding together yields an incorrect answer.

Many thanks.

答:
Note that you made a small typo; the lower region should be set up as:
$$2\pi \int_{-1}^0 -y\,dy = \pi$$

Thus, your total volume would be: $\pi+\dfrac{9\pi}{14}=\dfrac{14\pi}{14}+\dfrac{9\pi}{14}=\boxed{\dfrac{23\pi}{14}}$

Otherwise, your method looks correct. The easier solution via the disk method yields the same answer:
$$\pi \int_0^1 (x^3+1)^2dx = \boxed{\dfrac{23\pi}{14}}$$

Title: Volume of revolution around a line


答:
The definition of uniform convergence may help: you need to verify that
$$sup_{x\in[0,1]}\|e^{\frac{x}{k}}-1\| \to 0 $$ and it may worth to notice where do you get that supermum (which is maximum in this case). from there you can complete the proof

Title: Show the sequence converges uniformly

Body:
Here is another uniform convergence problem I have been struggling with. Show that $e^{\frac{x}{k}}$ converges uniformly on $[0,1]$.

Attempt: I see that the function sequence converges pointwise to $1$.Again, I am having difficulty finding a bound for $|e^\frac{x}{k}-1|$ for all $x \in [0,1]$, so that for any $\epsilon$ I can choose an $N$ for all $x \in[0,1]$ such that $k \geq N \implies |e^\frac{x}{k}-1|<\epsilon$ . What am I supposed to do here?

Edit: So I looked at one of the answers and I decided that this would be sufficient.

Let $\epsilon>0$. Choose $N>\frac{1}{\text{ln}(1+\epsilon)}$ then $k \geq N \implies |e^\frac{x}{k}-1|<|e^\frac{1}{k}-1|<\epsilon$ for all $x \in [0,1]$, can I get a confirmation that this is ok?


问:
how many base $10$ decimal expansions can a real number have?
A somewhat unintuitive result of real analysis is that decimal expansions are not unique. For example,
$$0.99999...=1.$$
So it can be gathered that every real number has at least one base-$10$ decimal expansion, sometimes even two. But is two the maximum? Are there any real numbers with three different base-$10$ decimal expansions?

Intuitively, I would think not, but I have close to no idea how to prove it other than knowing that the easiest method would be a proof by contraction.

It may help to restrict the problem by considering the expansions of numbers only in $[0,1]$. That is because if $a\in[0,1]$ has more than two base-$10$ decimal expansions, then so does $a+x$ for all $x\in\Bbb R$. And likewise, if $x\in\Bbb R\setminus [0,1]$ has more than two base-$10$ expansions, it can be written as $x=\mathrm{sgn}(x)(\lfloor x\rfloor+a)$, where $a\in[0,1)$, and by necessity $a$ has more than two base-$10$ expansions.

To be clear, I should define what I mean by base-$10$ expansion. Let $N\in[0,1]$. Then a decimal expansion of $N$ is a sequence $\delta=(\delta_0,\delta_1,\delta_2,...)$ of integers $0\le \delta_i\le 9$ such that
$$\sigma(\delta):=\sum_{i\ge0}\frac{\delta_i}{10^i}=N.$$
Furthermore, let $\mathcal U=\{(a_0,a_1,a_2,...):0\le a_i\le 9,\, a_i\in\Bbb Z\}$ and let
$$D_N=\{\delta\in\mathcal U :\sigma(\delta)=N\}.$$
Lastly, let
$$\mathcal C_k=\{N\in[0,1]:\#(D_N)\ge k\},\qquad k\in\Bbb N$$
where $\#(S)$ is the number of elements in the set $S$.

So, is $\mathcal C_k$ empty for $k>2$?

答:
$\cal C_k$ is indeed empty for $k > 2$.

Suppose that a number has two distinct decimal expansions $\delta = (\delta_0, \delta_1, \delta_2, \ldots)$ and $\delta' = (\delta_0', \delta'_1, \delta'_2, \ldots)$. Then, we must have

$$0 = \sum_{i \ge 0}\dfrac{\delta_i - \delta'_i}{10^i}.$$

Wlog, we may assume that $\delta_0 \neq \delta'_0$. (There must be some smallest $i$ for which they are unequal. By multiplying by a suitable power of $10$, we may assume that it is $i = 0$.) Furthermore, we may assume that $\delta_0 > \delta'_0$.

Thus, we have
$$1 \le \delta_0 - \delta'_0 = \sum_{i \ge 1}\dfrac{\delta'_i - \delta_i}{10^i}.$$

Taking absolute value on all sides and using triangle inequality for series, we see that

$$1 \le \delta_0 - \delta'_0 \le \sum_{i \ge 1}\dfrac{|\delta'_i - \delta_i|}{10^i} \le \sum_{i \ge 1}\dfrac{9}{10^i} = 1.$$

Thus, we have equality throughout. Note that this means that $|\delta'_i - \delta_i| = 9$ for all $i$. In fact, we see that $\delta'_i - \delta_i$ must have the same sign for all $i$. This sign must, of course, be positive.

Thus, we have $\delta_0 = \delta'_0 + 1$ and $\delta_i = 0$ for all $i \ge 1$, $\delta'_i = 9$ for all $i \ge 1$.

(All the manipulations above were justified as the series converged absolutely.)

The above shows that if a number has two decimal expansions, it must actually just be a variant of $1 = 0.\bar{9}$. In particular, there are at most two decimal expansions.


答:
The functions $f_n(x)=e^{x/n}$ are positive in $\mathbb R$ and $f_n\in\mathcal C^1(\mathbb R)$.

For $x_0\in\mathbb R$ fixed, $\lim_{n\to+\infty}f_n(x_0)=1:=f(x)$, $\forall x\in\mathbb R$. So the convergence is punctual in $\mathbb R.$

Since $|f_n(n)-1|=|e-1|\ne 0$, the convergence is not uniform in $\mathbb R$.

The functions $f_n(x)-1$ have derivative $\dfrac{1}{n}e^{x/n}$, which is always postive. So, for a fixed  positive real number $M>0$, we have
$$\sup_{x\in(-\infty,M]}|f_n(x)-f(x)|\le|f_n(M)-1|\underset{n\to+\infty}\longrightarrow 0$$
and this shows that the convergence is uniform $\forall x\in D_M:=(-\infty,M]$, $M>0.$

Title: Show the sequence converges uniformly


问:
Finding eccentricity of an ellipse from latus rectum
The latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the same as latus rectum of a parabola $y^2=4cx$ . Find eccentricity of the ellipse .</p>

答:
HINT:

from this and this, the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  is $2a(1-e^2)$ and $b^2=a^2(1-e^2)$ where $a$ is Semi major  Axis, $b$ is the Semi-minor Axis and $e$ is the Eccentricity

and the length of the latus rectum of the parabola $y^2=4ax$ is $4a$

Can you utilize the relations to find the value of $e$

Check separately for cases when $a\ge b$ and when $a<b$

EDIT: after a drastic change in the question $y^2=4ax$ to $y^2=4cx$

Now WLOG, we can choose $a\ge b$

As the focus of the parabola is $(c,0)$ and those of the ellipse are $(\pm ae,0)$

$c=ae$

Now, $2a(1-e^2)=4c=4ae\implies 1-e^2=2e\implies e^2+2e-1=0$

$\implies e=\frac{-2\pm{\sqrt{2^2-4(-1)1}}}{2\cdot1}=-1\pm\sqrt2$

As $0\le e<1, e=\sqrt2-1$