问:
A counter example on uniform continuity

Let $f:(0,1]\rightarrow \mathbb{R} $ be a continuous function and bounded is it uniformly continuous?

I know this isn't true, but I can't find a good counter example I was thinking something like this

$f(x)=\min\{3,1/x\}$ can someone give a better example.

答:
You example $f(x)$ doesn't work, i.e. $f(x)$ is uniformly continuous  on $(0,1]$. This is because it extends to a continuous function on $[0,1]$ by defining $f(0)=3$, and all continuous functions on closed and bounded intervals (or compact sets) are automatically uniformly continuous.

As for an actual example, someone else has already suggested $\sin(1/x)$, so see their answer.

Title: A counter example on uniform continuity


问:
Complex integral $\int_{-\infty}^{\infty} e^{-zt^2+2wt}dt$
The goal is to show this integral
$\int_{-\infty}^{\infty} e^{-zt^2+2wt}dt = \sqrt{\frac{\pi}{z}} e^\frac{w^2}{z}$ for $z,w \in \mathbb{C}$ with $\Re(z)>0$ using the identity:

$
\int_{-\infty}^{\infty}e^{-\pi x^2}e^{-2\pi ix \psi}dx = e^{-\pi \psi^2}
$ for $\psi \in \mathbb{R}$

I have proven in a previous far that the function is holomorphic as a function of $z$ on all of $\mathbb{C}$ but I'm not sure how to do this next part. I looked for a substitution that would help but I only know the substitution formula for real number so I cannot just replace the $x$ in the identity with some complex number.

Any help welcome :)

答:
I'm not certain this is correct but it is my best shot:

We complete the square in the exponent:

$zt^2 -2xt = \pi x^2 + 2\pi \psi ix +C\\
(\sqrt{z}t-\frac{w}{\sqrt{z}})^2-\frac{w^2}{z} = \pi x^2 + 2\pi ix \psi + C$

This gives us a value for $C = \frac{-w^2}{z}$, a value for $\psi = 0$ and a substitution $t =\sqrt{\frac{\pi}{z}}x + \frac{w}{z}$ (this is the bit I am unsure of - can I substitute a real number for this complex one????) But this reduces the function down to:

$
f(z) = e^{\frac{w^2}{z}} \sqrt{\frac{\pi}{z}} \int_{\infty}^{\infty}e^{-\pi x^2}dx
$

Using the identity given with $\psi =0$ we know the integral is one and therefore we get the result:

$f(z) = e^{\frac{w^2}{z}}\sqrt{\frac{\pi}{z}}$

ContentLicense: CC BY-SA 4.0


问:
the existence of R-regular and M-regular element over Noetherian local ring

$(R,\mathfrak m,k)$ is a Noetherian local ring, and $M≠0$ is a finitely generated $R$-module. Let $\operatorname{depth}R≧1$ and $\operatorname{depth}M≧1$. Then $\mathfrak m$ contains an element which is both $R$-regular and $M$-regular.

I tried to solve this by a lemma (see below) but couldn't.

Lemma: Let $R$ be a Noetherian ring, $M, N$ finite $R$-modules, and $I=\operatorname{Ann}(N)$. Then $\operatorname{Hom}(M, N)=0$ implies that $I$ contains an $M$-regular element.

答:
Let $\newcommand{\zdv}{\operatorname{Zdv}}\zdv(M)$ denote the set of all zerodivisors on $M$. Suppose that $\mathfrak m\subset \zdv(R)\cup \zdv(M).$ Since $$\zdv(M)=\bigcup_{\mathfrak p\in{\rm Ass}(M)}\mathfrak p,$$
we see that $\mathfrak m$ is contained in a union of prime ideals. Then, by prime avoidance, we obtain that $\mathfrak m\subset \zdv(R)$ or $\mathfrak m\subset \zdv(M)$, a contradiction to the hypothesis. Thus $\mathfrak m\not\subset \zdv(R)\cup \zdv(M)$. This means that $\mathfrak m$ contains an element which is both $R$-regular and $M$-regular.


问:
How to find $\;\lim_{x\to 0} \frac{x}{2-\sqrt{4}-2x}\;$ without using L'Hôpital's rule?
Well, I'm studying calculus and doing some exercises. First of all, from the answers that were given by my teacher the result of this limit should be $4$. I'm beginning my calculus class now but I've already studied calculus by myself before so I know how to apply l'Hôpital's Rule to this limit, following l'Hôpital's I get $-1$, the step-by-step resolution answer from WolframAlpha gives me $-1$ as well.

I thought of all I know about factoring and simplifying equations but I'm not able to actually solve it without l'Hôpital's. Any hint or way about how to do it?

$$\lim\limits_{x\to 0}  \frac{x}{2-\sqrt{4}-2x}$$

答:
It is fairly simple to solve the question. It happens to the best of us but it seems that you have not noticed that you can simplify the denominator $2-\sqrt{4}-2x$ to $-2x$. So then you will have:

$$\lim\limits_{x\to 0}  \frac{x}{-2x}$$

Which then simplifies to

$$\lim\limits_{x\to 0}  \frac{1}{-2}$$

So your limit is:
$$\lim\limits_{x\to 0}  \frac{x}{2-\sqrt{4}-2x} = -\frac{1}{2}$$

ContentLicense: CC BY-SA 3.0


问:
How do interpret the notation $\operatorname{Max}(\mathopen|x\mathclose|,\mathopen|y\mathclose|)$
I just started calc 3 and there are problems with the notation $\operatorname{Max}(a,b)$. I haven't seen it in other courses, for example, calc 2, yet there is nothing in my book about it so I suspect it's supposed to be prerequisite knowledge.

An example question is, sketch the following set in $\mathbb{R}^{2}$: $\{(x,y); \operatorname{Max}(\mathopen|x\mathclose|,\mathopen|y\mathclose|) \leq1\}$.

What do I do/how should I think?

答: $\operatorname{Max}(a,b)$ is just the number that is the maximum of $a$ and $b$.  So, for example, $$\operatorname{Max}(41,12) = 41.$$  As for $$\{(x,y)\in\mathbb{R}^{2}: \operatorname{Max}(\mathopen|x\mathclose|,\mathopen|y\mathclose|) \leq1\},$$ we can think of this as the region of the $(x,y)$ plane made of the regions where either $|y|\le |x|\le 1$ or $|x|\le |y|\le 1$.  See Desmos.


问:
How do I find the intervals in which I have to calculate the integral?
I have to calculate the $\int_{E}1\,dx\,dy$ of the following set $C=\bigl\{(x,y)\in\mathbb{R}:1\leq x^2y \leq8, x\leq y\ \leq 27x\bigr\}$. I have to know in which interval I take  $x$ and in which I take $y$.
I have so many exercises like this, that I have to understand which the boundaries are from a given set, and I am struggling a lot. Sometimes at least one of them is clearer and I set the other as a function of the first one. Please If somebody could explain to me how I can generally do it, any sort of tip.

I have an end exam coming up and I have no idea what I'll do with this sort of exercise.

答:
We can rewrite the inequalities defining $C$ as

$$ \frac 1 {x^2} \leq y \leq \frac{8}{x^2} \quad, \quad   x\leq y \leq 27x$$

Which is the same as

$$ \max\left(\frac{1}{x^2},x \right)\leq y \leq \min\left(\frac{8}{x^2}, 27x\right)$$

Since

$$ \max\left(\frac{1}{x^2},x \right) =
 \begin{cases} 1/x^2 & x \leq 1 \\ x
 & x \geq 1 \end{cases}$$

and
$$ \min\left(\frac{8}{x^2},27x \right) =
 \begin{cases} 27 x & x \leq 2/3 \\ 8/x^2
 & x \geq 2/3 \end{cases}$$

We will split the integral in three parts with

$ 1/x^2 \leq y \leq 27x$  and $x \leq 2/3.$

$ 1/x^2 \leq y \leq 8/x^2$ and $x \in [2/3,1]$.

$ x \leq y \leq 8/x^2$ and $x \geq 1$

Clearly these conditions are the same as

$  1/x^2 \leq y \leq 27x$  and $1/3 \leq x \leq 2/3.$

$ 1/x^2 \leq y \leq 8/x^2$ and $x \in [2/3,1]$.

$ x \leq y \leq 8/x^2$ and $1 \leq x \leq 2$

Of course you could've found this just by making a sketch like the one Bernard provided.


问:
Can't understand this solution to the combinatorics problem
```<code>You have 3 literature books, 2 computer books, and 1 math book.
How many ways can books be sorted on a shelf
if the books of literature and those of mathematics are all close together?
</code>```

The right answer: $3!3!2!$

my answer: $3 \cdot 4!2!$ because I have 4(=3+1) elements to order, so $4!$ and the other 2 books to order so $2!$, all multiplied by 3 because I have 3 ways to sort this books.

$$\_ \ \_ \ I \ I \ I \ I$$
$$\_ \ I \ I \ I \ I \ \_$$
$$I \ I \ I \ I \ \_ \ \_$$
Why I'm wrong?

答:
I believe that the problem meant to say that all the literature books must be in one group, and also all the computer books (rather than the math books) must be in one group. It certainly seems like something like that must have been intended, since -- as others noted in the comments -- the "close together" constraint seems meaningless for a single book.

Under this interpretation, we'd have $3!$ ways to arrange the literature books among themselves and $2!$ ways to arrange the computer books. (If we're being a bit cheeky, we could also observe that there's just $1! = 1$ way to arrange the single math book.) Then we have $3!$ ways to arrange the groups of Literature, Computers, and Math, yielding $3!3!2!$ arrangements, as claimed.


问:
How do I show that some eigenvalues of two square matrices of different dimensions are the same?
I have three matrices in a field $F$:

$X \in F^{a,a}, Y \in F^{b,b}, Z \in F^{a,b}$, where $a,b \in \mathbb{N}, a \geq b$ and $\text{rank}(Z) = b$. The following term describes their relation:

$X \cdot Z = Z \cdot Y$

I want to show that all eigenvalues of $Y$ are eigenvalues of $X$.

答:
Note: Not all of the determinants you wrote down are well-defined, since $XZ$ and $ZY$ are not quadratic if $a \neq b$.

Also, as stated, this does not seem quite correct. Set $a = 2, b = 1$ and

$$X = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad
Y = (1), \quad
Z = \begin{pmatrix} 1 \\ 0 \end{pmatrix},$$
then $a \geq b$, the rank of $Z$ is equal to $b = 1$, and
$$XZ = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = ZY.$$

Then $X$ has the eigenvalues 1 and 2, but $Y$ has the eigenvalue 1. Note that you can replace the 2 in $X$ with any other value.

Generally, with problems like these it's a good idea to start by comparing if you can somehow make eigenvectors of one of the matrices into an eigenvector of the other one, and here this gives us some insight, as well.

If $x \neq 0$ is an eigenvector of $Y$ to the value $\lambda$, then

$$\lambda Zx = Z(\lambda x) = Z Yx = XZx,$$
hence, if $Zx \neq 0$, then it is an eigenvector of $X$ to the value $\lambda$. But the full rank condition on $Z$ and $a \geq b$ means that its kernel is empty, so if $x \neq 0$ then also $Zx \neq 0$. So every eigenvalue of $Y$ is an eigenvalue of $X$.

But, looking at the counterexample above, that's all that you can show from the given requirements. X might have a bunch of other eigenvalues that you do not know anything about from the given data.


问:
Prove that $A^T \cdot A$ equal $I$ for an orthonormal matrix $A$ directly using matrix multiplication
Apparently it is always true that $A \cdot A^T=A^T \cdot A=I$ for an orthonormal matrix $A$. I know this can be proven using theorems regarding the rank and invertibility of matrices, but I would like to show it directly with matrix multiplication.

Suppose that
$A =
\begin{bmatrix}
u_1 &v _1 \\
u_2 &v_2 \\
\end{bmatrix}
$ is an orthonormal matrix.
Then we know that $\begin{bmatrix} u_1\\u_2 \end{bmatrix}^T \begin{bmatrix} u_1\\u_2 \end{bmatrix}=1$
and $\begin{bmatrix} v_1\\v_2 \end{bmatrix}^T \begin{bmatrix} v_1\\v_2 \end{bmatrix}=1$, and also that $\begin{bmatrix} u_1\\u_2 \end{bmatrix}^T \begin{bmatrix} v_1\\v_2 \end{bmatrix}=0$

However, we must have that $A \cdot A^T = \begin{bmatrix}
u_1^2+v_1^2 & u_1u_2+v_1v_2 \\
u_1u_2+v_1v_2  & u_2^2+v_2^2 \\
\end{bmatrix}
$, and I do not see why $u_1^2+v_1^2 =1$ or why $u_1u_2+v_1v_2=0.$ It seems like $A \cdot A^T = I$ is not necessarily true.

Where is my mistake?

答:
I think that the answers/responses given do not exactly answer the question asked. The question seems to have been asked specifically to try to build a concrete understanding of why "orthonormal columns implies orthonormal rows". While this desire probably comes from noble intentions, it is also likely built upon the assumption that the result is obvious-- even without using the powerful tools of linear algebra.

However, even in the 2x2 case (as originally asked) if we tie our hands behind our backs and restrict ourselves only to tools which came before linear algebra it turns out to be a rather complicated thing to show. But, I'll include an uglier and weaker answer than those already provided, to emphasize the value of learning linear algebra and how non-trivial the results from linear algebra are. Even "simple" ones.

To answer the question:

Fix orthonormal vectors $u,v \in \mathbb{R}^{2}$. In particular, $|u| = |v| = 1$. So, we know that there exists $\theta, \alpha$ so that $u = (\cos(\theta), \sin(\theta))$ and $v= (\cos(\alpha), \sin(\alpha))$. Moreover, since $u \cdot v = 0$, without loss of generality (by replacing $u$ with $-u$ if necessary) we also know that $\theta = \alpha \pm \pi/2$.

Using the identity $\cos(x \pm \pi/2) = \mp \sin(x)$ we deduce that $u_{1}^{2} + v_{1}^{2} = (\mp \sin(\alpha))^{2} + \cos^{2}(\alpha) = 1$ as desired. One can show identically that $u_{2}^{2} + v_{2}^{2} =1$.

Since $\theta = \alpha \pm \pi/2$ we have $\alpha = \theta \mp \pi/2$. So, $\cos(\alpha) = (\pm \sin(\theta))$. Hence,
\begin{align*}
u_{1} u_{2} + v_{1} v_{2} &= \cos(\theta) \sin(\theta) + \cos(\alpha) \sin(\alpha) \\
& = (\mp \sin(\alpha)) \sin(\theta) + \cos(\alpha) \sin(\alpha) \\
& = \sin(\alpha) \left( \cos(\alpha) \mp \sin(\theta) \right) \\
& = 0,
\end{align*}
painfully verifying that $A A^{T} =I$.

Good luck generalizing this to any higher-dimensional cases.


问:
How to express $2 \cos X = \sin X$ in terms of $\sin X$?
The Question was:

Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.

I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.

答: Given the equation $$2 \cos(x) = \sin(x)$$ and the instruction to write solely in terms of $\sin(x)$, I would begin by looking for an identity that involves $\cos(x)$, the term we want to transform and $\sin(x)$ the term we want to write everything in. This leaves us with the identity $$\cos(x)^2 + \sin(x)^2 =1.$$ We may then subtract $\sin(x)^2$ from both sides to have have $$\cos(x)^2 = 1 - \sin(x)^2.$$ As $\cos(x)^2$ is non-negative, we may take the square root of both sides and are left with $$\cos(x) = \pm \sqrt{1 - \sin(x)^2}.$$ We then replace the $\cos(x)$ in our equation with $\pm\sqrt{1 - \sin(x)^2}$ and thus we have two equations $$2 \sqrt{1 - \sin(x)^2} = \sin(x)$$ and $$-2 \sqrt{1 - \sin(x)^2} = \sin(x)$$ </p>


问:
Is WolframAlpha wrong on piecewise differentiation at $x=0$?
I'm differentiating the function:

$${ f\left(x\right) = \begin{cases}{x \sin ( \frac{1}{x}) } && {x\>\ne\>0} \\ {0} && {x\>=\>0}\end{cases} }$$
In my understanding it is not differentiable at $x= 0$, as it is not a constant function so we must use the definition of a limit to differentiate at $x=0$.

Wolfram says its derivative is $0$ at $x=0$.

Where am I wrong?

答:
Wolfram is wrong as it is often in similar questions.

$f$ is not differentiable at $0$ as
$$\lim\limits_{x \to 0} \frac{f(x)-f(0)}{x}$$ doesn’t exist.


问:
If sum of LI vectors is Cauchy, do the vectors themselves have to be Cauchy?
Say I have a normed space $X$ and a Cauchy sequence $(x_n) \in X$. We can define $W$ to be an arbitrary subspace of X, with the Hamel Basis $\{e_1, e_2, e_3 ...\}.$

Given that $W$ is a subspace of $X$, we can define a Hamel Basis of $X$ as $$\{e_1, e_2, e_3 ... \}\cup\{e'_1,e'_2,e'_3...\}$$
From this, I think we can define unique sequences $(y_n)$ and $(w_n)$ such that $x_n = y_n + w_n$ for all $n \in \mathbb{N}$, and that $w_n \in W$ and $y_n \in \text{Span}(\{e'_1,e'_2,e'_3...\})$ for all $n$. Is it true then that for all normed spaces $X$, that $(y_n)$ and $(w_n)$ have to be Cauchy? $X$ and $W$ are not necessarily complete and not necessarily of finite dimension.

I think it's true, but in my attempts, I can't think of a way to get to where I need to go without using the assumption that $||y+w||\ge||y||,||w||$ for $y \in \text{Span}(\{e'_1,e'_2,e'_3...\})$ and $w \in W$. I don't think that inequality is true in general.

Either pointers or solutions would both be appreciated.

答:
Consider $c_{00}$, the space of all finitely supported sequences equipped with $\ell^2$-norm and let $W = \operatorname{span}\{e_1\}$ and $Y = \left\{x \in c_{00} : \sum_{k=1}^\infty kx_k = 0\right\}$. Then $c_{00} = W \dotplus Y$. Let
$$x_n = \left(1,\frac12, \frac13, \ldots, \frac1n, 0, 0, \ldots\right).$$
Then $(x_n)_n$ is Cauchy in $c_{00}$. The unique decomposition $x_n = y_n + w_n$ with $y_n \in Y$ and $w_n \in W$ is
$$w_n = ne_1, \quad y_n = x_n - ne_1.$$
Neither sequence $(y_n)_n$ or $(w_n)_n$ is bounded so in particular they aren't Cauchy.


问:
Hölder inequality and interpolation
I was seeking the other day for intuition about the Hölder inequality and more precisely the geometric intuition behind the relation $\frac{1}{p} + \frac{1}{q} = 1$.

I found this interpretation of the Hölder inequality.

It says that we are trying to find an estimate of :

$$ \int f^a g^b$$ with the knowledge of $\int f^p$ and $\int g^q$.

For $(a, b) = (1,1)$, we need the relationship $\frac{1}{p} + \frac{1}{q} = 1$ so that the points : $(p, 0), (1,1), (0, q)$ lie on the same line. We can know use interpolation to have the Hölder inequality which give an estimate at the point $(1, 1)$.

But I feel like an argument is missing here. Why the estimate is using the geometric mean of the  function $f$ and $g$ ?  Why we don't have something like (using arithmetic mean) :
$$
\int \mid fg \mid \leq \frac{1}{p} \int f^p + \frac{1}{q}\int g^q \;\;?
$$
Moreover the article is saying we can generalize this approach to approximate $\int f^ag^b$ but in this case we need three points such that $(a,b)$ is in the convex hull of these three points. Why for $(1, 1)$ two points are needed while for general $(a, b)$ we need three points ?

答:
At first we look at the geometrical situation of the convex hull of three points in general position in $\mathbb{R^2}$.

Convex hull of three points with $(a,b)$ inside:

In OPs referenced article we have a triangle with points $(u_j,v_j),1\leq j\leq 3$ and a point $(a,b)$ inside the triangle as shown in the graphic below:

$\qquad\qquad\quad$

We can write $(a,b)$ as convex linear combination of $(u_j,v_j)$ according the graphic as follows:
\begin{align*}
\binom{a^{\prime}}{b^{\prime}}&=\binom{u_2}{v_2}+\lambda\left[\binom{u_3}{v_3}-\binom{u_2}{v_2}\right]\\
&=(1-\lambda)\binom{u_2}{v_2}+\lambda\binom{u_3}{v_3}\\
\\
\color{blue}{\binom{a}{b}}&=\binom{u_1}{v_1}+\mu\left[\binom{a^{\prime}}{b^{\prime}}-\binom{u_1}{v_1}\right]\\
&=(1-\mu)\binom{u_1}{v_1}+\mu\binom{a^{\prime}}{b^{\prime}}\\
&=(1-\mu)\binom{u_1}{v_1}+\mu\left[(1-\lambda)\binom{u_2}{v_2}+\lambda\binom{u_3}{v_3}\right]\\
&\,\,\color{blue}{=(1-\mu)\binom{u_1}{v_1}+\mu(1-\lambda)\binom{u_2}{v_2}+\lambda\mu\binom{u_3}{v_3}}\tag{1}
\end{align*}

Note we have in (1) a convex combination of the points $(u_j,v_j)$ since
\begin{align*}
(1-\mu)+\mu(1-\lambda)+\lambda\mu=1
\end{align*}

Convex hull of three points with $(a,b)=(1,1)$ at the boundary:

Now we consider the triangle with the three points $(0,0), (p,0), (0,q)$, $p>1, q>1$ and the point $(a,b)=(1,1)$ at the line segment from $(p,0)$ to $(0,q)$.

$\qquad\qquad\qquad$

We calculate similarly as above:
\begin{align*}
\color{blue}{\binom{1}{1}}&=\binom{p}{0}+\nu\left[\binom{0}{q}-\binom{p}{0}\right]\\
&\,\,\color{blue}{=(1-\nu)\binom{p}{0}+\nu\binom{0}{q}}
\end{align*}

We observe in this special case where $(a,b)=(1,1)$ is at the boundary of the convex hull, we need only two points $(p,0)$ and $(0,q)$ to obtain a convex combination of the triangle for $(1,1)$.

Hölder's inequality:

A few words to Hölder's inequality which might be helpful. We have for positive real numbers $x$ and $y$ and $p>1, q>1$ real numbers with $\frac{1}{p}+\frac{1}{q}=1$ the arithmetic-geometric means inequality:
\begin{align*}
x^{\frac{1}{p}}y^{\frac{1}{q}}\leq \frac{1}{p}x+\frac{1}{q}y\tag{2}
\end{align*}
We define for a $p$-integrable measurable function $f$:
\begin{align*}
N_p(f):=\left(\int|f|^p\,d\mu\right)^{\frac{1}{p}}
\end{align*}
and take
\begin{align*}
x:=\left(\frac{|f(\omega)|}{N_p(f)}\right)^p,\quad y:=\left(\frac{|g(\omega)|}{N_q(g)}\right)^q\tag{3}
\end{align*}

We obtain from (2) and (3):
\begin{align*}
\frac{|f\,g|}{N_p(f)N_q(g)}&\leq \frac{1}{p\left(N_p(f)\right)^p}|f|^p+\frac{1}{q\left(N_q(g)\right)^q}|g|^q\tag{4}\\
\frac{1}{N_p(f)N_q(g)}\int|f\,g|\,d\mu
&\leq \frac{1}{p\left(N_p(f)\right)^p}\int|f|^p\,d\mu+\frac{1}{q\left(N_q(g)\right)^q}\int|g|^q\,d\mu\\
&=\frac{1}{p}+\frac{1}{q}\\
&=1\\
\color{blue}{N_1(fg)}&\color{blue}{\leq N_p(f)N_q(g) }\tag{5}
\end{align*}
and (5) gives Hölder's inequality.

Note the transformation from the arithmetic mean in (4) to the multiplication of $N_p(f)$ with $N_q(g)$ in (5).

Title: Hölder inequality and interpolation


问:
An integral that can be solved by trigonometric substitution must be solved by algebraic substitution
EDIT:THANK YOU FOR THE HELP, I am sorry I did not include my progress so far I will learn from all of your feedback

I am new to calculus 2 and I would appreciate some help, our teacher gave us this question:

$$\int \frac{\sqrt{z^2-49}}{z}\,dz$$

He showed us the answer which is

$$-7\sec^{-1}\frac{z}{7}+\sqrt{z^2+49}+C$$

But the catch is we have to solve it using algebraic substitution, not trigonometric substitution. I am really stumped I can't seem to get that answer,  I would really be grateful for the help, I apologize if this question seems easy to you

答:
Integrate by parts and then apply $(\csc t)’= -\frac1{t\sqrt{t^2-1}}$ with $t=\frac x7$

\begin{align}
\int \frac{\sqrt{z^2-49}}{z}\,dz=& \sqrt{z^2-49}-\int \frac {49}{z\sqrt{z^2-49}}\,dz\\
=& \sqrt{z^2-49}+7 \csc^{-1}\frac z7 +C
\end{align}


问:
How to show the associativity in this group?
The group law is given by $a \oplus b = \frac{a+b}{1+a\,b}$ for $a,b \in G = (-1,1)$

For associativity I guess I have to show $(a \oplus b) \oplus c = a \oplus (b \oplus c)$, hence:

$$\begin{equation}\frac{\frac{a+b}{1+a\,b}+c}{1+\frac{a+b}{1+a\,b}\,c}= \frac{\frac{c+b}{1+c\,b}+a}{1+\frac{c+b}{1+c\,b}\,a}\end{equation}$$

but I got a hard time doing so. Is this actually the task?

答: Yes, it is. And in the expression$$\frac{\frac{a+b}{1+ab}+c}{1+\frac{a+b}{1+ab}c},$$you multiply both the numerator and the denominator by $1+ab$, then you will get$$\frac{a+b+c+abc}{1+ab+ac+bc}.$$Now, do a similar thing to the RHS.


问:
Uniqueness of endpoints of non-empty open bounded intervals in a linear order
This is a follow-up to my previous question on uniqueness of endpoints of closed bounded intervals. Suppose $(L, \leq)$ is a linear order, and suppose $a,b,c,d$ are elements of $L$ such that $a < b$ and $c <d$. Suppose further that the open intervals $(a,b)$ and $(c,d)$ are both non-empty. If $(a,b)=(c,d)$, must $a=c$ and $b=d$? I would be very surprised if there was a counterexample in a linear order. I know there are counterexamples in general partial orders, but what about linear orders?

答:
It's not true if the open intervals are empty. For example, take $a=1,\ b=2,\ c=3,\ d=4$ in $\Bbb N$.

On the other hand, if we suppose that $(a,b)=(c,d)$ is nonempty, the result is true.

First, by definition of intervals, both $b$ and $d$ must be strict upper bounds of the nonempty set $(a,b)=(c,d)$, similarly both $a$ and $c$ are strict lower bounds, hence in particular $c<x< b$ for any $x\in(a,b)=(c,d)$, so $c<b$ follows. Similarly, $a<d$.

In a linear order we either have $a<c,\ a>c$ or $a=c$.

In the first case we would have $c\in (a,b)$ whereas $c\notin (c,d)$, and similarly in the second case we would have $a\in (c,d)$ whereas $a\notin (a,b)$. So we must have $a=c$.

Analogously, $b=d$ also follows.


答: This is a known bug.  You can get WolframAlpha to return the correct result if you replace D[Piecewise[{{xsin(1/x), x != 0}, {0, x = 0}}], x] with D[Piecewise[{{xsin(1/x), x < 0 || x > 0 }, {0, x = 0}}], x].  Using the input x < 0 || x > 0  signals to WolframAlpha that $x$ is real, and so the algorithm it uses to differentiate is more robust.  See here for the correct result in WolframAlpha.

Title: Is WolframAlpha wrong on piecewise differentiation at $x=0$?

Body:
I'm differentiating the function:

$${ f\left(x\right) = \begin{cases}{x \sin ( \frac{1}{x}) } && {x\>\ne\>0} \\ {0} && {x\>=\>0}\end{cases} }$$
In my understanding it is not differentiable at $x= 0$, as it is not a constant function so we must use the definition of a limit to differentiate at $x=0$.

Wolfram says its derivative is $0$ at $x=0$.

Where am I wrong?


问:
Proving Bijections in $\mathbb{R}^n$
I have a question that may seem trivial or silly, however I'll try to make my point clear. I'm a student of Mathematical Physics and unfortunately my college doesn't offer a set theory course, so that we are not taught to prove bijections between arbitrary sets. The single variable calculus course we have teaches us to prove bijections by looking for one inverse function in a very clear way: we write $y = f(x)$, try to solve for $x$ and then create the function desired (it's of course more formal than that, first we prove injectivity showing that $f(x)=f(y)$ implies $x=y$ and then we look for the inverse to prove surjectivity).

The multivariable calculus course on the other hand doesn't talk about proving bijections from subsets of $\mathbb{R}^n$ to subsets of $\mathbb{R}^m$. On the other hand we have the analysis course, the topology course and the differential geometry course that all presumes we know how to prove those bijections and so on.

Now, I'm confused, because none of our courses teaches to prove that kind of thing. If we have some function $f: \mathbb{R}^n \to \mathbb{R}^m$, proving injectivity is the easiest part, but for surjectivity, seeking for some inverse seems kind of painful. By now I'm studying differential geometry of curves and surfaces (it uses Do Carmo's book), and the only thing I'm needing is this basic notion on how to prove bijections in these cases (for instance, when we come to show that something is a chart, proving continuity is rather easy, we have been taught on how to do it, but proving the bijection is complicated).

Is there any general guidelines as there is in the case of functions from $\mathbb{R}$ to $\mathbb{R}$? Is there any reference that can be given that treats this in a good way? Again, I know this is basic and that we all should know before coming to things like differential geometry, but since the course doesn't teach this, I came to ask here to fill this gap.

Thanks very much in advance, and sorry if this question is silly.

答:
For example, $\tan$ gives a (smooth) bijection $(-\frac\pi2,\,\frac\pi2)\to \Bbb R$.
 By linear maps and shifting we can get bijections between any two open intervals.

Similarly, there is a bijection from the open $n$ dimensional (unit) ball to $\Bbb R^n$, using the previous bijection on each ray. These bijections (and their inverses) are also continuous, so they are homeomorphims.

An $n$ dimensional manifold is a topological space that has an $n$ dimensional ball around each of its point.

To formulate it, we need the bijections to the balls (or either to $\Bbb R^n$).

Title: Proving Bijections in $\mathbb{R}^n$


答:
Assume that the columns of $A$ are orthonormal, meaning
$$\sum_{i=1}^n A_{ij}A_{ik} = \delta_{jk}$$
for all $1 \le j,k \le n$.

Then for all $1 \le j,k \le n$ we have
$$(A^TA)_{jk}= \sum_{i=1}^n (A^T)_{ji}A_{ik}= \sum_{i=1}^n A_{ij}A_{ik} = \delta_{jk} = I_{jk}$$
so $A^TA = I$. By general matrix theory this implies $AA^T = I$ as well, so for all $1 \le j ,k \le 1$ we have
$$\delta_{jk}=I_{jk} = (AA^T)_{jk}= \sum_{i=1}^n A_{ji}(A^T)_{ik}= \sum_{i=1}^n A_{ji}A_{ki}$$
meaning that the rows are orthonormal.

Title: Prove that $A^T \cdot A$ equal $I$ for an orthonormal matrix $A$ directly using matrix multiplication

Body:
Apparently it is always true that $A \cdot A^T=A^T \cdot A=I$ for an orthonormal matrix $A$. I know this can be proven using theorems regarding the rank and invertibility of matrices, but I would like to show it directly with matrix multiplication.

Suppose that
$A =
\begin{bmatrix}
u_1 &v _1 \\
u_2 &v_2 \\
\end{bmatrix}
$ is an orthonormal matrix.
Then we know that $\begin{bmatrix} u_1\\u_2 \end{bmatrix}^T \begin{bmatrix} u_1\\u_2 \end{bmatrix}=1$
and $\begin{bmatrix} v_1\\v_2 \end{bmatrix}^T \begin{bmatrix} v_1\\v_2 \end{bmatrix}=1$, and also that $\begin{bmatrix} u_1\\u_2 \end{bmatrix}^T \begin{bmatrix} v_1\\v_2 \end{bmatrix}=0$

However, we must have that $A \cdot A^T = \begin{bmatrix}
u_1^2+v_1^2 & u_1u_2+v_1v_2 \\
u_1u_2+v_1v_2  & u_2^2+v_2^2 \\
\end{bmatrix}
$, and I do not see why $u_1^2+v_1^2 =1$ or why $u_1u_2+v_1v_2=0.$ It seems like $A \cdot A^T = I$ is not necessarily true.

Where is my mistake?


问:
Limit of $f: \mathbb{R}^3 \rightarrow \mathbb{R}$
Given a function
$$ f\left(x{,}\ y{,}\ z\right)=\frac{\alpha_1x+\alpha_2y+\alpha_3z}{\beta_1x+\beta_2y+\beta_3z},$$
does the limit $$ \lim_{\left(x{,}\ y{,}\ z\right)\rightarrow\left(0{,}\ 0{,}\ 0\right)}f\left(x{,}\ y{,}\ z\right) $$ exist?

So far I have solved for one case, that if all the coefficients $$ \alpha_1, \alpha_2, \alpha_3,  \beta_1, \beta_2, \beta_3 $$are non-zero, then the limit exists, if vectors $$\left(\alpha_1{,}\ \alpha_2{,}\ \alpha_3\right)\ \mathrm{and}\ \left(\beta_1{,}\ \beta_2{,}\ \beta_3\right) $$  are linearly dependent.

What about other cases, if I can't assume that all coefficients are non-zero? I may however assume, that there exists at least one non-zero coefficient β (for the the divisor not to be equal to zero) , and in this particular case which I am interested in, that there exists at least one non-zero coefficient α.

答:
Suppose $f(x,y,z)$ converges to some number $\lambda$ as $(x,y,z)\to(0,0,0)$.

Then taking $y=0=z$ and $x\to0$, we get $$\lambda=\lim_{x\to0}\frac{\alpha_1x}{\beta_1x}=\frac{\alpha_1}{\beta_1}$$

Similarly, taking $y\to0$ and $z\to0$ keeping the other coordinates zero, we get $$\lambda=\frac{\alpha_2}{\beta_2}=\frac{\alpha_3}{\beta_3}$$

Hence $\alpha_i=\lambda\beta_i$.

Conversely, if $(\alpha_1,\alpha_2,\alpha_3)=\lambda(\beta_1,\beta_2,\beta_3)$ then $f(x,y,z)=\lambda$ is constant, so it obviously converges as $(x,y,z)\to0$.


问:
How to use method of moments to estimate mean?
I have given the density function
$$f(\epsilon,\mu) = (1-\epsilon)\frac{1}{3}e^{-x/3}+\epsilon\frac{1}{\mu}e^{-x/\mu}.$$
And would like to estimate $\mu$ using the method of moments and then numerically
maximize the likelihood as a function of $\epsilon$. I know that mgf is
$$M_n(x) = E[(X-EX)^n],$$
but I don't know how to calculate this to use method of moemnts in thus task. Will appreciate a hint, thank you!

答:
First, I'm rusty.  Second, I haven't seen all the notation.  But I'm a little suspicious of your bit above on moment-generating functions, my primary concern being that the RHS looks a lot like the $n$th central moment.

Anyway.

If $a_1, a_2, \ldots a_n$ are a sequence of non-negative numbers summing to $1$ and $f_1, f_2, \ldots f_n$ are pdfs, then we can define a new pdf by $ f(x) = \sum_{i \le n} a_i f_i (x) $.  The linearity of the definite integral means that we'll also have, for example, $EX^n = \sum_{i\le n} a_i E_i[X^n]$, where each $E_i$ is the expectation taken with respect to $f_i$.

Assuming the support for your distribution is non-negative values of $x$, your pdf is this type of mixture, specifically one comprising two exponential distributions, one with mean $3$ and another with unknown mean $\mu$.  If $W \sim \exp(\lambda)$, then $EW = \lambda$ and $EW^2 = 2\lambda^2 $.  Therefore, we have

\begin{aligned} \
\\ EX &= (1 - \epsilon)\cdot 3 + \epsilon\cdot\mu \
\\ EX^2 &= (1 - \epsilon)\cdot 18 + \epsilon\cdot 2\mu^2.
\end{aligned}

Using the method of moments, we'd set the above expressions equal to $\bar{x}$ and $\frac{1}{n}\sum x_i^2$.  So then

\begin{aligned} \
\\ \bar{x} &= 3(1 - \epsilon) + \epsilon\mu \
\\ \frac{1}{n}\sum x_i^2 &= 18(1 - \epsilon) + 2\epsilon\mu^2.
\end{aligned}

From here, we can solve for both $\mu$ and $\epsilon$ in terms of the known values of $\bar{x}$ and $\frac{1}{n}\sum x_i^2$.


问:
Making sense of GLM's problem setting
I'm totally new (and on my own) into generalized linear models (GLM's), and I'm having trouble understanding the problem setting, sad, I know. This is what I figured out trying to read the book "Foundations of Linear and Generalized Linear Models" by Alan Agresti. For simplicity I will just assume nothing about the distribution of $Y$ and I will consider the identity link function.

We have a random variable $Y$ (response variable) and $p$ random variables $X_1,\dotsc,X_p$ (explicative variables), so let's define $\mathbf{X}:=(X_1,\dotsc,X_n)$. We state that there exists a vector $\beta\in\mathbb{R}^p$ such that $E(Y|\mathbf{X}=\mathbf{x})=\mathbf{x}^\top\beta$.

Well, this seems like a fairly reasonable idea, we are trying to find a functional relation between $Y$ and $\mathbf{X}$, that is, $Y=h(\mathbf{X})$ for some measurable function $h$, but, sadly, $h\circ\mathbf{X}\in\sigma(\mathbf{X})$ and, probably, $Y\not\in\sigma(\mathbf{X})$. However, we know that $E(Y|\mathbf{X})$ is the closest (in the $L^2$ sense) $\sigma(\mathbf{X})$ measurable function to $Y$. So we are just assuming that a linear function of $\mathbf{X}$ is the best we can get.

Our assumption is cool, but, back to the real world, we don't have a clue about who $\beta$ is, so we can take two different (equivalent?) approaches.

Take $n$ independent observations of the random vector $(Y,\mathbf{X})$

Pre-select certain values for $\mathbf{X}$, observe $Y$, and repeat this process $n$ times independently.

In any case, as a result, we have a vector $\mathbf{y}\in\mathbb{R}^n$, which corresponds to the $n$ independent observations of $Y$ and a matrix $\mathtt{X}\in\mathfrak{M}_{n\times p}(\mathbb{R})$ which corresponds to the $n$ independent observations (or the $n$ selections) of $\mathbf{X}$. Now, we want to estimate $\beta$ from $\mathbf{y}$ and $\mathtt{X}$, there are plenty of methods and heuristics to do this (and plenty of statistical theory to evaluate the "goodness" of the estimation), just to mention one of them, we may find the least squares solution $\hat{\beta}$ of the system of equations $\mathbf{y}=\mathtt{X}\beta$, and use it as our estimation.

Probably, everything you have read above is totally wrong, but please, help me, did I get something right?

答:
I'll try to give a valuable answer. I believe that your main question (i.e., method 1 vs. method 2 for data collection) is the difference between controlled experiment vs. observational data. Namely, in a control experiment you know what are your explanatory variables $X$ are because you choose them apriori and sometimes even manipulate them. In such a case, for $n$ independent "subjects", you observe $n$ realizations of $(Y, X)$ or $Y|X=x$.

However, for observational data, you basically don't know what are your explanatory variables $X$. So, although you still observe $(Y, X)$, but $Y$ and $X$ maybe independent.

Practically, assume that you are interested in the association of height $X$ and weight $Y$. So you observe $n$ realizations of $(Y, X)$. You can even switch between them. This is the first case. For the second case, you are interested in finding what determines weight $Y$. Now, you collect everything that you can from every "subject", thus your $X$ has many variables, some are relevant, other probably not. You still observe $n$ independent realizations of $(X,Y)$ since you cannot randomize your $X$s and $Y$, as every subject $i$ "brings" $(y_i, x_{1i}, ..., x_{pi})$. If you collect randomly $X$, and then heights $Y$, not from the same individuals, your model will have no meaning (whatever estimation method you use).

Title: Making sense of GLM's problem setting


问:
Given a board of size $1 \times n$,in how many this board can be colored with the colors white,yellow,blue assuming the number of blue cells is even
Given a board of size $1 \times n$,in how many this board can be colored with the colors white,yellow and blue assuming the number of blue cells is an even number.

My try:

Given a board of size $1 \times n$,then there are two possibilities for the color of the first cell:

If the first cell is blue then the it's needed to find another cell to color it with the blue color ,this can be done in $n-1$ ways and for each one of these choices there are $b_{n-2}$ ways to color the remaining $1 \times (n-2)$ board with the colors yellow,white and blue assuming the number of blue cells is an even number.

If the first cell is yellow or white then the remaining $1 \times (n-1)$ board can be colored in $b_{n-1}$ ways assuming the restriction on the number of blue cells is satisfied.

Gathering these cases follows:

$$b_n=(n-1)b_{n-2}+2b_{n-1}$$

With $b_0=1$ and $b_1=2$ as initial conditions.

Calling $G(x)=\sum_{k\ge 0}^{} b_k x^k$,omitting some elementary calculations follows:

\begin{align*}
G(x)&=\frac{-1}{\left(n+1\right)^{2}x^{2}+2x-1}\\
&=\frac{-1}{\left(n+1\right)^{2}}\frac{1}{(x-a)(x-b)}\\
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\frac{1}{a-x}-\frac{1}{b-x}\right)\\
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\frac{1}{a}\frac{1}{1-\frac{x}{a}}-\frac{1}{b}\frac{1}{1-\frac{x}{b}}\right)\\
&
\end{align*}

From which we conclude:
\begin{align*}
b_n=\left[x^{n}\right]G\left(x\right)
&=\frac{1}{\left(n+1\right)^{2}(a-b)}\left(\left(\frac{1}{a}\right)^{n+1}-\left(\frac{1}{b}\right)^{n+1}\right)\\
\end{align*}

Where $$a=\frac{-1+\sqrt{n+2}}{n+1},b=\frac{-1-\sqrt{n+2}}{n+1}$$

But the closed form does not coincide with the recurrence that I've found.(I only want the question to be solved using recurrence relations.)

答:
Roughly, half of the colorings have an even number of blue cells. To make this precise, for each coloring, $C$, define the "mate" of $C$ to be the coloring you get when you find the leftmost cell of $C$ which is white or blue, and change it to be blue or white. Every coloring is given a mate, except for the single all-yellow coloring which has no blue or white cells. However, the remaining $3^n-1$ colorings are divided into pairs of mates. Furthermore, in each pair of mates, exactly one coloring has an even number of blues. Therefore, the number of colorings with an even number of blues is
$$
\frac{3^n-1}2+1.
$$


问:
Proof Verification: Prove that $AB-BA=A$ is false, given $A$ is invertible
Steps taken:

Multiply both sides by $A^{-1}$, given $A$ is invertible.

$A^{-1}AB - A^{-1}BA = A^{-1}A $

Thus,  $B - A^{-1}BA = I$

Take the trace of both sides.

$\operatorname{Tr}(B-A^{-1}BA) = \operatorname{Tr}(I)$

Utilize the linear mapping property of trace.

$\operatorname{Tr}(B)-\operatorname{tr}(A^{-1}BA) = \operatorname{Tr}(I)$

Utilize the cyclic property of trace.

$\operatorname{Tr}(B)-\operatorname{tr}(BAA^{-1}) = \operatorname{Tr}(I)$

$\operatorname{Tr}(B)-\operatorname{tr}(BI) = \operatorname{Tr}(I)$

$\operatorname{Tr}(B)-\operatorname{tr}(B) = \operatorname{Tr}(I)$

Reaching a contradiction

$0 = \operatorname{Tr}(I)$

Is my usage of the trace properties correct?

答:
It is correct.

Here is a different, more high level proof. Consider the linear map $\delta : M_n \to M_n$ given by
$$\delta(X):= XB-BX.$$

Then we can show by induction that for all $k \in \Bbb{N}$ holds $\delta(A^k) = kA^k$. Indeed, for $k=1$ the claim is true by assumption. Assuming it holds for some $k$, we have
\begin{align}
\delta(A^{k+1}) &= A^{k+1}B-BA^{k+1} \\
&= A^k(AB-BA) + (A^kB - BA^k)A \\
&= A^k\delta(A) + \delta(A^k)A \\
&= (k+1)A^{k+1}
\end{align}

so it holds for $k+1$ as well.

If all $A^k$ were nonzero, we would get that $k$ is an eigenvalue of $\delta$ for all $k \in \Bbb{N}$. This is impossible since there are at most $n^2$ different eigenvalues of $\delta$ so it has to be $A^k =0$ for some $k \in \Bbb{N}$. This is a contradiction since $A$ is invertible.


问:
Finding the locus of $b^2-2x^2=2xy+y^2$
I was asked to determine the locus of the equation
$$b^2-2x^2=2xy+y^2$$

This is my work:

Add $x^2$ to both sides:
$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
 b^2-x^2 &=\left(x+y\right)^2
\end{align}$$

I see that this is similar to the equation of a circle. How can I find the locus of this expression?

答:
The matrix associated to the conic is
$$\begin{pmatrix}-2 & -1 & 0\\-1 & -1 & 0\\0 & 0 & b^2 \end{pmatrix}\implies\Bigg|\begin{pmatrix}-2 & -1\\-1 & -1 \end{pmatrix}\Bigg|=2-1=1>0$$
and this shows that the conic is an ellipse.

Title: Finding the locus of $b^2-2x^2=2xy+y^2$


问:
Generate all possible combinations of 3 digits without repetition
It's possible to generate all possible combinations of 3 digits by counting up from 000 to 999, but this produces some combinations of digits that contain duplicates of the same digit (for example, 099). I could obtain all combinations of 3 digits without repetition by counting up from 000 to 999, and then removing all numbers with duplicate digits, but this seems inefficient. Is there a more efficient way to accomplish the same task?</p>

答:
Yes, there does exist such a way.  First you select a digit d from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.  Then you select a digit e from ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d).  Then you select a digit f from (({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d)-e).  You first select 0 for d, then 1, and so on until you get to 7.  And, you always select the least digit first for e and f also, with the additional condition that d < e < f.  List out the first sequence, 012, 013, 014, 015, 016, 017, 018, 019.  Then list all the other numbers beneath them with the condition that for all numbers e and f, and with d held constant, the digits for e and f follow the natural number sequence down the column.  Partition each set of sequences by d.  The column rule only applies within each partition.  (this description might come as incomplete or could use some revision).

The list thus goes:

```<code>012, 013, ...,         019
023, 024, ...,     029
034, 035, ..., 039
.
.
.
089

123, 124, ...,      129
134, 135, ..., 139
.
.
.
189

.
.
.

789
</code>```

I'll clarify the last part here:

```<code>567, 568, 569

578, 579

589

678, 679

689

789
</code>```

Perhaps better, say we try to do the same thing in base 4.  The entire sequence goes

```<code>012, 013

023

123
</code>```

In base 5 the process yields:

```<code>012, 013, 014

023, 024

034

123, 124

134

234
</code>```

Thus, in base 10 the sum of the first 8 triangular numbers gives us the number of such combinations: +(1, 3, 6, 10, 15, 21, 28, 36)=120.  Similarly, it should logically follow that for x digit numbers in base z, where x < z, or x=z, there exist +[T$_1$, ..., T$_ (z-(x+1))$] such combinations, where T$_n$ indicates the nth triangular number.

I guess the underlying idea I've used here lies in following the natural number sequence across the rows, and down the columns for the digits e and f also.

Body:
Yes, there does exist such a way.  First you select a digit d from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.  Then you select a digit e from ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d).  Then you select a digit f from (({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d)-e).  You first select 0 for d, then 1, and so on until you get to 7.  And, you always select the least digit first for e and f also, with the additional condition that d < e < f.  List out the first sequence, 012, 013, 014, 015, 016, 017, 018, 019.  Then list all the other numbers beneath them with the condition that for all numbers e and f, and with d held constant, the digits for e and f follow the natural number sequence down the column.  Partition each set of sequences by d.  The column rule only applies within each partition.  (this description might come as incomplete or could use some revision).

The list thus goes:

012, 013, ...,         019
023, 024, ...,     029
034, 035, ..., 039
.
.
.
089

123, 124, ...,      129
134, 135, ..., 139
.
.
.
189

.
.
.

789

I'll clarify the last part here:

567, 568, 569

578, 579

589

678, 679

689

789

Perhaps better, say we try to do the same thing in base 4.  The entire sequence goes

012, 013

023

123

In base 5 the process yields:

012, 013, 014

023, 024

034

123, 124

134

234

Thus, in base 10 the sum of the first 8 triangular numbers gives us the number of such combinations: +(1, 3, 6, 10, 15, 21, 28, 36)=120.  Similarly, it should logically follow that for x digit numbers in base z, where x < z, or x=z, there exist +[T$_1$, ..., T$_ (z-(x+1))$] such combinations, where T$_n$ indicates the nth triangular number.

I guess the underlying idea I've used here lies in following the natural number sequence across the rows, and down the columns for the digits e and f also.


问:
How do interpret the notation $\operatorname{Max}(\mathopen|x\mathclose|,\mathopen|y\mathclose|)$
I just started calc 3 and there are problems with the notation $\operatorname{Max}(a,b)$. I haven't seen it in other courses, for example, calc 2, yet there is nothing in my book about it so I suspect it's supposed to be prerequisite knowledge.

An example question is, sketch the following set in $\mathbb{R}^{2}$: $\{(x,y); \operatorname{Max}(\mathopen|x\mathclose|,\mathopen|y\mathclose|) \leq1\}$.

What do I do/how should I think?

Body:
$\max (A,B) = $ the larger value of $A$ or $B = \begin{cases}A&&if\ A> B\\B&&if\ B > A\\A&which\ is\ = B&if\ A=B\end{cases}$

So $\{(x,y)|\max(|x|,|y|) \le 1\}=$

$\{(x,y): |y| \le |x|$ and $|x| \le 1$ or $|x| \le |y|$ and $|y|\le 1\}=$

$\{(x,y): |y|\le |x|\le 1$ or $|x|\le |y|\le 1\}=$

$\{(x,y): 0\le |y|\le |x|\le 1$ or $0\le |x|\le |y|\le 1\}=$

$\{(x,y): 0\le |y|\le 1$ and $0\le |x|\le 1\} =$

$\{(x,y): -1\le y \le 1$ and $-1\le x \le 1\}=$

the solid rectange from the corner $(-1,-1)$ to the corner $(1,1)$.


问:
Trace of $U'U$ with $U$ upper-triangular
I have found the following formula for $U$ $(m\times m)$ upper-triangular matrix.
\begin{equation}\operatorname{tr}[U'U]=\sum_{i\leq j}^m U_{ij}^2\end{equation}
where $\operatorname{tr}[\cdot]$ is the trace operator, $'$ denotes the transposition, $U_{ij}$ the $ij$-th entry of $U$. I can't show this formula, I need an help.

答:
For a general $m\times m$matrix $A$ the trace of $A^TA$ is simply the sum of squares of all elements of $A$:

$$\operatorname{Tr}( A^TA) =\sum_{i=1}^m (A^TA)_{ii} = \sum_{i=1}^m \sum_{j=1}^m (A^T)_{ij}A_{ji} = \sum_{i=1}^m \sum_{j=1}^m A_{ji}A_{ji} = \sum_{i,j=1}^m A_{ij}^2.$$

Now, if your matrix $U$ is upper-triangular, this means that $U_{ij} = 0$ whenever $i > j$ so the sum reduces to
$$\operatorname{Tr}( U^TU) = \sum_{i,j=1}^m U_{ij}^2 = \sum_{i \le j}^m U_{ij}^2.$$


问:
Uniform continuity choosing delta problem
I am trying to show that $x^2$ is uniformly continuous on the set union of intervals [n,n+$n^{-2}$] for all positive integer n. I have also checked the previous posts. But my question is that for a given epsilon we need to choose $\delta$ smaller than $\frac{\epsilon}{2(n+n^{-2})}$. This will ensure uniform continuity.  Here delta depends on interval where two elements are. for n=5 delta changes or for n=6 also another delta. In uniform continuity for every epsilon there exists a delta which satisfies for all points in delta distance. Here there are several deltas if I am not mistaken. Can you clarify this issue? I am bit confused thanks a lot.

答:
Suppose that you have a function $f\colon[-1,1]\longrightarrow\Bbb R$ and consider the problem: is the restriction of $f$ to $[-1,0]$ continuous? Asserting that it is means that, for any $\varepsilon>0$, there is some $\delta_1>0$ such that$$(\forall x,y\in[-1,0]):|x-y|<\delta_1\implies\bigl|f(x)-f(y)\bigr|<\varepsilon.$$Now, consider the problem: is the restriction of $f$ to $[0,1]$ continuous? Asserting that it is means that, for any $\varepsilon>0$, there is some $\delta_2>0$ such that$$(\forall x,y\in[0,1]):|x-y|<\delta_2\implies\bigl|f(x)-f(y)\bigr|<\varepsilon.$$So, you have two $\delta$'s, $\delta_1$ and $\delta_2$, which may be distinct. There is nothing wrong with that. You wanted to establis that there was one $\delta$ for each interval, and, again, it may will happen that they are not the same.

So, in your situation, on any interval $\left[n,n+\frac1{n^2}\right]$ you shall have a certain number $\delta_n$. And there's nothing wrong if there is no $\delta$ that works for all thos intervals. There is no such $\delta$ if the intervals are of the for $[n,n+1]$, but the restriction of $x^2$ to any such interval is still uniformly continuous.


问:
Boundedness of a linear functional
I am trying to prove the following result.

Let $X$ be a normed vector space over $\mathbb K$. If $f: X \rightarrow \mathbb{K}$ is a linear functional such that $(x_n) \rightarrow 0$ implies in the sequence $(f(x_n))$ being bounded, then the functional $f$ is bounded.

What I have done: Let $(x_n)$ be a sequence such that $(x_n) \rightarrow 0$. Since $x_n$ converges to $0$, there must be some $n_0 \in \mathbb{N}$ such that $||x_n|| \leq ||x_{n_0}||, \forall n \in \mathbb{N}$. Also, there must be a constant $M > 0$ such that $||f(x_n)|| \leq M , \forall n \in \mathbb{N}$.

If $||x_{n_0}|| \geq M$ then the result follows, but I am stuck in the case $||x_{n_0}|| < M$. Any help would be appreciated.

答:
Assume that $f$ is unbounded. You can then pick a sequence $(x_n)_n$ in $X$ such that $\|x_n\| = 1$ but $|f(x_n)| \ge n^2$ for all $n\in\Bbb{N}$.

Consider $y_n  := \frac{x_n}n$. Then $y_n \to 0$ but $|f(y_n)| \ge n$ which contradicts your assumption.

Title: Boundedness of a linear functional


问:
About uniqueness in the Caratheodory's Theorem
Caratheodory's Theorem asserts that if $\mathcal{A}$ is an algebra of subsets of $X$ and $\mu_0:\mathcal{A}\to \mathbb{R}\cup\{\infty\}$ is $\sigma-$additive on $\mathcal{A}$, then there is a measure $\mu$ defined on $\sigma(\mathcal{A})$ (the sigma-algebra generated by $\mathcal{A}$). Furthermore, this extension is unique if $\mu_0$ is $\sigma-$finite.

Let me remind you that the idea of the proof is to define an outer measure $\mu_*$ and prove that restricted to the caratheodory measurable sets is a measure. Of course one also have to prove that this class of sets is a $\sigma-$algebra. The outer measure is defined for all subsets of $X$:

$$\mu_*(E) = \inf \left \{\sum_i \mu_0(A_i) \ \Bigl|\Bigr. \ E\subset \bigcup A_i, A_i \in \mathcal{A} \right \}$$

I know two proofs of the uniqueness. One of these use the so called monotone class and the fact that the smallest monotone class that contains $\mathcal{A}$ is $\sigma(\mathcal{A})$. My question is with the second proof. The same is as follows:

Let $\nu$ is another measure on $\sigma(\mathcal{A})$ which coincides with $\mu_0$ on $\mathcal{A}$. Put $F \in \sigma(\mathcal{A})$ of finite measure. If $F\subset \bigcup A_i$ with $A_i \in \mathcal{A}$ hence

$$\nu(F) \leq \sum_i\nu(A_i) = \sum_i\mu(A_i) $$

taking infimum $\nu(F)\leq\mu(F)$ (remember $\mu$ comes from the outer measure). Now consider $F\subset \bigcup_i A_i$ such that $\mu(\bigcup A_i)\leq \mu(F) + \epsilon$. Let me call $E:= \bigcup A_i$. As $F$ has finite measure then this condition implieas $\mu(E\backslash F) \leq \epsilon$. Furthermore, $\mu(E) = \nu(E)$ because $\mu$ y $\nu$ agree on $\bigcup_{i=1}^n A_i \ \forall n$ and this is an increasing sequence of sets. Now:

$\mu(F) \leq \mu(E) = \nu(E) = \nu(E\backslash F) + \nu(F) \leq \mu(E\backslash F) + \nu(F) \leq \epsilon + \nu(F)
$

since $\epsilon$ is arbitrary the proof is done for all $F \in \sigma(\mathcal{A})$ of finite measure. Using this and the condition that $\mu_0$ is $\sigma-$finite the result is extended to all $F \in \sigma(\mathcal{A})$.

What i can't see is where we are using the condition $F \in \sigma(\mathcal{A})$. I mean, we have $\mathcal{M}$ the $\sigma-$algebra of caratheodory measurable sets which is larger and maybe one can obtain uniqueness here. If a examine the previous proof I can't see why we can't prove the result for this $\sigma-$algebra. Indeed, we have the existence of the measure since, as I comment before, $\mu_*$ is a measure on $\mathcal{M}$. To prove the uniquess I repite the same proof, word by word. Of course in this case $\nu$ is a measure on $\mathcal{M}$.

What is wrong?

答:
Your proof is correct. If $\mu_0$ is $\sigma$-finite, your proof shows that, for any sigma-algebra $\mathcal{S}$, such that $\mathcal{A}\subseteq \mathcal{S} \subseteq \mathcal{M}$, there a unique extension of $\mu_0$ to $\mathcal{S}$.

Answering to your comment: your proof does not work for sigma-algebras larger than $\mathcal{M}$, because it uses the fact that $\mu_*$ when restricted to the sigma-algebra is a measure (and $\mathcal{M}$ is the largest sigma-algebra containing $\mathcal{A}$ such that  $\mu_*$ , restricted to it, is a measure -- see proof below). In fact, there are examples of sigma-finite $\mu_0$ having more than one extension to a sigma-algebra larger than $\mathcal{M}$. Of course such extensions don't come from the outer measure. Let us see one such example.

Example:  Let $X =\{a,b\}$, where $a \neq b$. Let $\mathcal{A}=\{\emptyset, X\}$. Clearly,  $\mathcal{A}$ is an algebra (in fact a sigma-algebra).  Let us define $\mu_0(\emptyset)=0$ and $\mu_0(X)=1$. Clearly $\mu_0$ $\sigma-$additive on $\mathcal{A}$.

Let $\mu_*$ be the outer measure induced by $\mu_0$. Then, we have $\mu_*(\emptyset)=0$ and, for all $E \subseteq X$, if $E \neq \emptyset$, $\mu_*(E)=1$.

It is easy to check that $E$ is $\mu_*$-measurable if and only if $E=\emptyset$ or $E=X$.
That means, $\mathcal{M}= \{\emptyset, X\}= \mathcal{A}$.

Now, let $\mathcal{S} = \{\emptyset, \{a\}, \{b\}, X\}=2^X$. Clearly, $\mathcal{S}$ is a
sigma-algebra.

Now, for each $r \in [0,1]$, consider $\mu_r$ defined on $\mathcal{S}$ by
$\mu_r(\emptyset)=0$, $\mu_r(\{a\})=r$,  $\mu_r(\{b\})=1-r$ and  $\mu_r(X)=1$.

For each $r \in [0,1]$, $\mu_r$ is different extension of $\mu_0$ to $\mathcal{S}$.

Now let us prove that

$\mathcal{M}$ is the largest sigma-algebra containing $\mathcal{A}$ and such that  $\mu_*$ , restricted to it, is a measure.

Proof: Let $\mathcal{S}$ be a sigma-algebra of subsets of $X$ such that $\mathcal{A} \subseteq \mathcal{S}$ and such that $\mu_*$ , restricted to it, is a measure.

Given $E\in \mathcal{S}$, we want to prove that, for all $A \subseteq X$,
$$\mu_*(A) = \mu_*(A\cap E) + \mu_*(A\cap E^c)$$
If $\mu_*(A)=\infty$, this an immediate consequence of the sub-additivity of the outer measure. Now suppose that  $\mu_*(A)<\infty$. Then there is $B\in \sigma(\mathcal{A})$ such that $A \subseteq B$ and $\mu_*(B)=\mu_*(A)$ (such $B$ is called a measurable cover of $A$).

Since $\mathcal{A} \subseteq \mathcal{S}$ and $\mathcal{S}$ is a sigma-algebra, we have that  $\sigma(\mathcal{A}) \subseteq \mathcal{S}$. So $B\in \mathcal{S}$. Since $E \in \mathcal{S}$ and  $\mu_*$ , restricted to $\mathcal{S}$, is a measure, we have, using the monotonicity of the outer measure:
$$\mu_*(A\cap E) + \mu_*(A\cap E^c) \leqslant \mu_*(B\cap E) + \mu_*(B\cap E^c)=\mu_*(B)= \mu_*(A)$$
and, by the sub-additivity of the outer measure, we have
$$ \mu_*(A) \leqslant \mu_*(A\cap E) + \mu_*(A\cap E^c)$$
So
$$ \mu_*(A) = \mu_*(A\cap E) + \mu_*(A\cap E^c)$$

Thus, we have proved that given any  $E\in \mathcal{S}$, then $E$ is $\mu_*$-measurable, that is, $E\in \mathcal{M}$. So $\mathcal{S} \subseteq \mathcal{M}$.


答:
HINT.-A way to study the curve is to consider it as a union of two graphs of functions (similar to how you would see two functions $y=\pm\sqrt{r^2-x^2}$ for a circle) so you get two explicits functions:
$$y=-x+\sqrt{b^2-x^2}\\y=-x-\sqrt{b^2-x^2}$$ whose domains are $[-b,b]$.

This way  you have an ellipse for each value of $b$ whose axis are (approximately) in the lines
$$y=0.625x\\y-1.6x$$

For example, for $b=2$ you have (approximately) the semiaxis $1.2358$ and $3.2358$.

Title: Finding the locus of $b^2-2x^2=2xy+y^2$

Body:
I was asked to determine the locus of the equation
$$b^2-2x^2=2xy+y^2$$

This is my work:

Add $x^2$ to both sides:
$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
 b^2-x^2 &=\left(x+y\right)^2
\end{align}$$

I see that this is similar to the equation of a circle. How can I find the locus of this expression?


问:
How to prove an upper bound inequality for the $l_1$ norm of $Ax$
I am stuck trying to show the following inequality, which seems obvious but I am not sure how to prove formally:

$||Ax||_1 \le \sum^{n}_{j=1} (\sum^{m}_{i=1} |a_{ij}|)|x_j|$

for any $A \in \mathbb R^{m × n} $ and $x \in \mathbb R^n$

I am only supposed to use the $l_1$ vector norm definition in the proof.

Thanks.

答:
Recall the definition of matrix-vector multiplication and $\ell^1$-norm:
\begin{align}
\|Ax\|_1 &= \left\|\left(\sum_{j=1}^n a_{1j}x_j, \ldots \sum_{j=1}^n a_{mj}x_j\right)\right\|_1\\
&= \sum_{i=1}^m \left|\sum_{j=1}^na_{ij}x_j\right|\\
&\le \sum_{i=1}^m \sum_{j=1}^n|a_{ij}||x_j|\\
&\le \sum_{j=1}^n \left(\sum_{i=1}^m|a_{ij}|\right)|x_j|.
\end{align}


问:
Why $\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}{y}$ holds?
Let $(M_s)$ a martingale s.t. $\mathbb E[M_t^2]<\infty $. Regarding the proof of $$\mathbb E[\sup_{0\leq s\leq t}M_s]\leq 4\mathbb E[M_t^2],$$

the person who answer the question in this post mention that $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}{y},$$
and I don't understand why this should be true. The best I know is
$$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|)}{y},$$
which is Doob inequality. But I don't get how we can get to
$$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}{y}.$$
Any idea ?

答:
I guess that your martingale is at least right continuous, otherwise your statement is not true. Under this hypothesis, let $\tau=\inf\{s>0\mid |M_s|\geq y\}\wedge t$. The process $(|M_t|)_{t\geq 0}$ is a submartingale, i.e. by Optional Stopping theorem $$\mathbb E|M_\tau|\leq \mathbb E|M_t|\tag{1}.$$ Notice that \begin{align*}
|M_\tau|&\geq y\\
&=y\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s|\geq y\}}+y\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s| <y\}}\\
&\geq y\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s|\geq y\}}+|M_t|\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s| \}},
\end{align*}

and thus, $$\mathbb E|M_\tau|\geq y\mathbb P\left\{\sup_{0\leq s\leq t}|M_s|\geq y\right\}+\mathbb E\Big[|M_t|\boldsymbol 1_{\{\sup_{0\leq s\leq t}|M_s| <y\}}\Big]\tag{2}.$$

Combine $(1)$ and $(2)$ gives the wished result.


问:
Conditions for a bipartite graph
Is a 8 vertices graph with degree $3$ and $12$ edges a bipartite graph? For instance, $24$ is the sum of degree and it is twice the number of edges and theorem 1 it says any graph is bipartite if the sum of the degree of all vertices is equal to twice the number of edges. The problem is that when I give the partition some of the vertices are adjacent to each other and even if they are in the same set. Like $V1 = \{a,c,f,h\}$ and $V2 = \{b,d,e,g\}$ and $a$ is adjacent to $h$ or $e$ is adjacent to $d$. Does it matter if they are adjacent in the same set?

答:
The theorem I think you are talking about doesn't have anything to do with whether a graph is bipartite or not - its the handshaking lemma which just observes that the sum of the vertex degrees must count all the edges twice.

Graphs where every vertex has the same degree are called regular. If all the vertices have degree $k$ they are $k$-regular and in particular $3$-regular graphs are called cubic.

Typically, knowing the degree sequence of a graph (which gives the degree of every vertex) is not enough to establish whether or not it is bipartite - that usually requires knowing  how the vertices are connected, since - to specifically answer a question there - the defining quality of a bipartite graph is that there are no connections between nodes in the same part.

Here's a couple of graphs that fit your description - one is bipartite, one is not. In the bipartite graph on the left, the red nodes are only ever connected to blue nodes and vice-versa. I've highlighted a cycle of length $5$ in the non-bipartite graph, since you never find odd-length cycles in bipartite graphs - it would be impossible to make the colouring described for the other graph.

Title: Conditions for a bipartite graph


问:
Any practical difference between the metrics $d_1=\sup\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ and $d_2=\max\{\left|x_j-y_j\right|:j=1,2,...k\}$?
I've been asked to do a proof showing that $d_1\left({x,y}\right)$ is a metric on $\mathbb{R}^k$, but is there any difference between this and $d_2\left({x,y}\right)$, for which I've already done a proof? </p>

答: No, because the sets are finite.</p>

Body: No, because the sets are finite.


问:
Show that $T$ contains at most two points
Take $n \geq 2$ and $f:S^{n} \to \mathbb{R}$ a continuous map.Call $T$ the set of points $t\in f(S^{n})$ such that the fibre $f^{-1}(t)$ has finite cardinality.
Show that $T$ contains at most two points.

I try understand the problem, but I can´t make a example of the proposition I try functions like projections and $x+y+z$, but I don´t get catch the statement of the proposition.

Any example that helps to me to see the proposition or in other form help me to catch the idea behind of the proof

答:
Here's an example: view $S^2 \subset \mathbb{R}^{3}$ in the standard way, and consider the height map $(x, y, z) \mapsto z$. The fibers are parallels that are the same height above the $xy$-plane. The fibers $f^{-1}(1)$ and $f^{-1}(-1)$ each consist of one point, namely the poles $\{(0, 0, 1)\}$ and $\{(0, 0, -1)\}$, whereas every other fiber has an infinite number of points (such as $f^{-1}(0)$, which is the equator or unit circle in the $xy$-plane).

So in this case, there are exactly two values in $\mathbb R$ with finite cardinality preimages.


答: HINT: $S^n$ is compact and connected, so $f[S^n]$ must be a closed interval $[a,b]$. If $|T|\ge 3$, there must be some $t\in T\cap(a,b)$. Consider the sets $f^{-1}\big[[a,c)\big]$ and $f^{-1}\big[(c,b]\big]$.

Body:
Take $n \geq 2$ and $f:S^{n} \to \mathbb{R}$ a continuous map.Call $T$ the set of points $t\in f(S^{n})$ such that the fibre $f^{-1}(t)$ has finite cardinality.
Show that $T$ contains at most two points.

I try understand the problem, but I can´t make a example of the proposition I try functions like projections and $x+y+z$, but I don´t get catch the statement of the proposition.

Any example that helps to me to see the proposition or in other form help me to catch the idea behind of the proof


问:
Intersection curve parametrization
How can we find parametrization of the intersection curve between a pseudo-sphere and sphere ?

Elimination of variables appears not readily amenable/elegant. Thanks for the same.

答:
Three circles having center on $z$ axis at about $C_1(0,0,-0.95936);\;C_2(0,0,0);\;C_3(0,0,-0.95936)$ and radius  about $r_1=r_3=0.28786$
and $r_2=1$

This is the best I could do, since are involved transcendent function an exact solution is impossible to get.

Hope it helps

Title: Intersection curve parametrization


答: No, $\sup$ and $\max$ are the same for finite sets. In fact, they are the same for infinite sets as well whenever both exist, however an infinite set of real numbers may fail to have a maximum even if it's bounded above (consider $\{1-\frac1n:n\in\mathbb N\}$).</p>

Title: Any practical difference between the metrics $d_1=\sup\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ and $d_2=\max\{\left|x_j-y_j\right|:j=1,2,...k\}$?

Body: I've been asked to do a proof showing that $d_1\left({x,y}\right)$ is a metric on $\mathbb{R}^k$, but is there any difference between this and $d_2\left({x,y}\right)$, for which I've already done a proof? </p>


问:
Hölder inequality and interpolation
I was seeking the other day for intuition about the Hölder inequality and more precisely the geometric intuition behind the relation $\frac{1}{p} + \frac{1}{q} = 1$.

I found this interpretation of the Hölder inequality.

It says that we are trying to find an estimate of :

$$ \int f^a g^b$$ with the knowledge of $\int f^p$ and $\int g^q$.

For $(a, b) = (1,1)$, we need the relationship $\frac{1}{p} + \frac{1}{q} = 1$ so that the points : $(p, 0), (1,1), (0, q)$ lie on the same line. We can know use interpolation to have the Hölder inequality which give an estimate at the point $(1, 1)$.

But I feel like an argument is missing here. Why the estimate is using the geometric mean of the  function $f$ and $g$ ?  Why we don't have something like (using arithmetic mean) :
$$
\int \mid fg \mid \leq \frac{1}{p} \int f^p + \frac{1}{q}\int g^q \;\;?
$$
Moreover the article is saying we can generalize this approach to approximate $\int f^ag^b$ but in this case we need three points such that $(a,b)$ is in the convex hull of these three points. Why for $(1, 1)$ two points are needed while for general $(a, b)$ we need three points ?

答:
Focusing more on intuition, and landing on the geometry how integrals scale, let us check out a simple case.

Consider $\int_{0}^{r} 1 dx = r$. Applying Holder's inequality we would get on the opposite side $\left( \int_{0}^{r} 1 dx \right)^{\frac{1}{p}} \left( \int_{0}^{r} 1 dx \right)^{\frac{1}{q}}$. So, what can we say to relate $r$ to $r^{\frac{1}{p} + \frac{1}{q}}$?

Case 1: ($\frac{1}{p} + \frac{1}{q} = 1$)

Evidently if $\frac{1}{p} +\frac{1}{q} =1$, then we have the equality that $r = r^{\frac{1}{p} + \frac{1}{q}}$

Case 2: ($\frac{1}{p} + \frac{1}{q} < 1$)

If $\frac{1}{p} + \frac{1}{q} < 1$ then for $r < 1$ we have $r < r^{\frac{1}{p} +\frac{1}{q}}$ and for $r > 1$ we have $r > r^{\frac{1}{p} + \frac{1}{q}}$.

Case 3: ($\frac{1}{p} + \frac{1}{q} > 1$)

In this case, similar to Case 2, you can quickly rule at an inequality that holds for all values of $r$.

Conclusion:
If an inequality like Holder's is true, the only possible option is when $\frac{1}{p} + \frac{1}{q}=1$.

Title: Hölder inequality and interpolation


答:
Let's first take one of you're attempted examples and work through it to see what's happening: take the 2-sphere $S^2 \subseteq \mathbb{R}$ and consider the projection onto the $z$-co-ordinate. The image of this map is $[-1,1]$.

Now at $-1$ and $1$, the only points that are sent to each of those by this projection map are the poles: $(0,0,-1)$ and $(0,0,1)$ respectively. So these two points both have a finite preimage.

But for any $z$ in between $-1$ and $1$, looking at the preimage of these from the projection map means taking an entire slice through the sphere at that height. You get an entire circle, all at the same height, so all sent to the same $z$ when you project out. All of these $z$s have an infinite set of points sent to them.

Thus for this projection map, exactly two points have finite preimage: $-1$ and $1$.

How are we going to prove that this works for any sphere with $n \geq 2$, and any continuous map?

Start off by noting that the sphere is closed and bounded subset (thus compact) and also connected. Therefore the image has to be a closed bounded interval $[a,b]$.

Can you show that $a$ and $b$ are the only two numbers that can have finite preimage? That is, knowing that there are points on the sphere sent to $a$ and $b$, can you show that for all the numbers between $a$ and $b$ there are infinitely many points on the sphere that are sent to that number?

Title: Show that $T$ contains at most two points


问:
Time function of Laplace tranform $F(s) = \frac{1}{s^2-1}$
I have the following question:
What are the time functions corresponding to the Laplace transforms below? What values will the time functions approach as time goes to infinity?

$$F(s) = \frac{1}{s^2-1}$$

I assumed it was a standard Laplace transformed and wrote $\sinh(t)$, which is unbounded when $\lim_{t\to \infty} \sinh(t) = \infty$

But the expected answer is $f(t) = −0.5e^{−t} + 0.5e^t$, why is that?

答: See Wikipedia for the definition of $\sinh(t)$ :$$\sinh(t) = −0.5e^{−t} + 0.5e^t$$


问:
equivalence of purely transcendental extension
Sorry for my bad English.

Let $K\subset L$ be field extension.

Now, are the following equivalent?

(1)L/K is purely transcendental. Namely, there is subset $S\subset L$ such that $S$ is algebraically independent over $K$ and $L=K(S)$.

(2)forall $x\in L\backslash K$, $x$ is not algebraic over $K$.

Please tell me relationship (1) and (2).

答:
Assume that $L/K$ is purely transcendental with $L=K(S)$ for some algebraically independent set $S$. Let $x\in L$ be algebraic over $K$, i.e. $$x^n+b_1x^{n-1}+\dots+b_n=0$$ for some $b_i\in K$. Now if any element of $S$ would appear non-trivially in $x$ then after clearing denominators in the equation above we would get a non-trivial polynomial equation involving elements of $S$ which is impossible as they were assumed to be algebraically independent over $K$. Hence $x\in K$. This proves (1)$\implies$ (2).

The other implication is not true, let for example $K$ be an algebraically closed field. Then any field extension $L/K$ satisfies (2), but there are field extensions which are not purely transcendental (which is in general actually not that trivial to show, see here for an example).


答:
Well the image of $f$ is going to be a connected compact subset of $\mathbb{R}$, so it will be a closed interval $[a,b]$ (or 1 point if $f$ is constant).

So it would be natural to think that $a$ and $b$ are the only possible values of $t\in[a,b]$ such that $f^{-1}(t)$ is finite. To prove this, let $p,q$ be points of $\mathbf{S}^n$ with $f(p)=a,f(q)=b$.

There are clearly infinite different paths $g_k:[0,1]\to\mathbf{S}^n$ in the sphere that go from $p$ to $q$. We can pick them so that they only meet at $p$ and $q$ and don´t share any other points (take for example circunference arcs that go from $p$ to $q$ in $\mathbf{S}^n$). Moreover, by the intermediate value theorem, $f\circ g_k:[0,1]\to[a,b]$ has $f\circ g_k(0)=a$ and $f\circ g_k(1)=b$, so $f\circ g_k$ must take the value $t$ at some point if $t\in(a,b)$ (thus, $f$ takes the value $t$ at some point of the path). That is, for given $t\in(a,b)$, you have at least one point $x$ in each path from $p$ to $q$ such that $f(x)=t$, so there must be infinitely different such points, showing that $f^{-1}(t)$ is infinite.


问:
Is't true that for a linear transformation $T:\mathbb R^n\to\mathbb R^n,~T$ is positive definite $\iff\langle Tx,x\rangle>0~\forall~x\ne 0$
Is't true that for a linear transformation $T:\mathbb R^n\to\mathbb R^n,$

$T$ is positive definite $\iff\langle Tx,x\rangle>0~\forall~x\ne 0$

$T$ is negetive definite $\iff\langle Tx,x\rangle<0~\forall~x\ne 0$

答:
For a square matrix, $T$, $T$ is positive/negative definite if $\langle Tx,x\rangle$ is greater/less than $0$ for all $x\ne0$. $T$ is positive/negative semi-definite if $\langle Tx,x\rangle$ is greater/less than or equal to $0$ for all $x\ne0$.

Wikipedia has a lot more about positive definite matrices.


答:
It's just the definition of the Hyperbolic Sine:

$$\sinh(x) = \dfrac{e^x - e^{-x}}{2}$$

Whence your result.

As you can see

$$\lim_{t\to +\infty} \sinh(t) = \lim_{t\to +\infty} \dfrac{e^t - e^{-t}}{2} \to +\infty$$

Title: Time function of Laplace tranform $F(s) = \frac{1}{s^2-1}$

Body:
I have the following question:
What are the time functions corresponding to the Laplace transforms below? What values will the time functions approach as time goes to infinity?

$$F(s) = \frac{1}{s^2-1}$$

I assumed it was a standard Laplace transformed and wrote $\sinh(t)$, which is unbounded when $\lim_{t\to \infty} \sinh(t) = \infty$

But the expected answer is $f(t) = −0.5e^{−t} + 0.5e^t$, why is that?


问:
Prove that any finite poset is isomorphic to a subset of $\mathbb {N}$ ordered by divisibility rule
Prove that any finite poset is isomorphic to a subset of $\mathbb {N}$ ordered by divisibility rule.

I have no idea how to prove this so any hints would be much appriciated.

答: Given any finite poset $\{x_1,x_2,...,x_n\}$ and any set of $n$ distinct primes $\{p_1,p_2,...,p_n\}$ (e.g. the first $n$ primes), the poset can be embedded in $(\mathbb{N}, \vert)$ via the map that sends $x_i$ to the product of the primes $p_j$ over those indices $j$ for which $x_j \le x_i$.


问:
$\mathbb{C}(V)$ is a finite module over $\mathbb{C}(V)^G$?
Is it true that $\mathbb{C}(V)$ is a finite module over $\mathbb{C}(V)^G$ for any finite subgroup $G \subset GL(V)$ and, moreover, $\dim_{\mathbb{C}(V)^G} \mathbb{C}(V) = |G|$? It possibly follows from the well-known theorems of Hilbert and Noether in invariant theory, but doesn't seem obvious to me. Maybe I missed something?

答:
Yes. There is the following result:

Prop Let $V$ be a finite dimensional representation of a finite group $G$ over a field $k$. Then $k(V)$ is galois over $k(V)^G$ with galois group $G$.

The proof is just that $G$ acts as automorphisms on $k(V)$, so we may apply all the standard field theory.

It is immediate from this that your result follows. This result is in Benson's Polynomial Invariants of Finite Groups Prop 1.1.1 (i.e., the very first thing anyone proves in the subject :) )


问:
How to Compute the Rotation Matrix?
Dear Math Stackexchange community, how can I compute $R$ rotation matrix in this equation ($P=Rp+t$)?

$p$ consists of 3 points of $p_1,p_2,p_3$. And $p_1,p_2,p_3$ each has 3 coordinates of $x,y,z$. Therefore p is 3x3 matrix.

$p=\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}$

t is 3x1 vector.  $t=\begin{bmatrix} t_x \\ t_y \\ t_z \end{bmatrix}$

P is a 3x3 matrix.
$P=\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix}$

R is a 3x3 matrix.
$R=\begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$

Therefore, all in all:
$P=Rp+t$

$\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix} = \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}.\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}+\begin{bmatrix} t_x \\ t_y \\ t_z \end{bmatrix}$

I tried to take the inverse and calculate R. However; when I put numeric values in it, that method doesn't work with numeric values.

Do you have any suggestion?

edit1:
I did the following after DavidK, and Cryo's suggestion:

$P=Rp+t$

$(P-t)=R.p$

$(P-t).p^{-1}=R$

However there is a problem now: last raw of $p$ is always fully zero for my case. And I can't take the inverse of $p$.

答:
Thanks to @David K. I was suspecting that Matlab was doing something convenient, but I don't use it any more, so wanted to avoid guessing.

So you have

$\mathbf{Q}=\left[\mathbf{q}_1\,\mathbf{q_2}\,\mathbf{q_3}\right]=\mathbf{P}-\mathbf{t}=\mathbf{R}.\boldsymbol{\mathcal{P}}=\mathbf{R}.\left[\mathbf{p}_1\,\mathbf{p}_2\,\mathbf{p}_3\right]$

Where $\mathbf{Q}$ is some $3\times 3$ real-valued matrix, which we can also view as 3 column vectors $\mathbf{q}_{1,2,3}$. $\mathbf{R}$ is a real-valued rotation matrix, which means it is orthogonal ($\mathbf{R}^T\mathbf{R}=\mathbf{Id}=identity$) and has determinant of $det\left[\mathbf{R}\right]=+1$. Finally, matrix $\boldsymbol{\mathcal{P}}$ is made up from three 3d real-valued column vectors $\mathbf{p}_{1,2,3}$. The aim is to determine the rotation matrix by choosing the vectors $\mathbf{p}_{1,2,3}$.

If you can pick three linearly independent vectors $\mathbf{p}_{1,2,3}$ you can have a unique solution for $\mathbf{R}$ given $\mathbf{Q}$. You seem to say that all z-components all your three vectors are zero. Then the three vectors are not linearly independent.

Why can you not have three linearly independent vectors? Are there any limitations?

PS: This is not so much an answer as restatement of the problem, but I am still struggling to understand the question

Title: How to Compute the Rotation Matrix?


问:
Density argument for a densely bounded operator
Let $H$ be a Hilbert space and Let $A$ be a bounded linear operator on a dense domain $D$, i.e. There exists a positive constant $C$ such that
$$||Ax||_H\leq C||x||_H, x \in D.$$
My question is: What assumtpions should $A$ satsisfies so that the boundedness can be extended on $H$?.
For me it is straightforward but I have some doubts about that.
Thank you in advance.

答:
No assumptions are needed on $A$ whatsoever  - there exists one unique extension of $A$ to a bounded operator on (say $\tilde{A}$) on $H$ with norm $||\tilde{A}|| \leq C$.

Let $x\in H$ and $x_i \in D$ be such that $x = \lim x_i$.
One checks that the sequence $A x_i$ is Cauchy  since $||A x_i - A x_j|| \leq C ||x_i - x_j|| \to 0$ if $i, j$ are large enough (recall that any convergent sequence is Cauchy). Hence the sequence $A x_i$ converges and we set $\tilde{A}(x) := \lim A x_i$.


问:
Number of binary square matrices whose every contiguous 2 by 2 submatrix has exactly 2 ones (zeroes)

What is the number of binary $n\times n$ matrices whose every contiguous  $2\times 2$ submatrix has exactly two ones (zeroes)?

That is, if $a_{ij}\in\{0,1\}$ for $i,j\in\{1,2,\dots,n\}$ are matrix entries, then $$a_{i,j} + a_{i,j+1} + a_{i+1,j} + a_{i+1,j+1}= 2$$

for $i \in \{1,\dots,n-1\}$ and $j \in \{1,\dots,n-1\}$, where $n\ge 2$.

After writing out some cases for small $n$, it appears to me that the answer might be $2^{n+1}-2$.

Is there an elegant way to prove this?

For example, for $n=2$ we have $6$ matrices

$$
\begin{bmatrix}
1 &0 \\
1 &0 \\
\end{bmatrix},
\begin{bmatrix}
1 &1 \\
0 &0 \\
\end{bmatrix},
\begin{bmatrix}
1 &0 \\
0 &1 \\
\end{bmatrix},
\begin{bmatrix}
0 &1 \\
1 &0 \\
\end{bmatrix},
\begin{bmatrix}
0 &0 \\
1 &1 \\
\end{bmatrix},
\begin{bmatrix}
0 &1 \\
0 &1 \\
\end{bmatrix}.
$$

For example, for $n=3$ we have $14$ matrices:

$4$ rotations of
$\begin{bmatrix}
1 &0 &1 \\
1 &0 &1 \\
0 &1 &0 \\
\end{bmatrix}$, $2$ rotations of
$\begin{bmatrix}
1 &0 &1 \\
1 &0 &1 \\
1 &0 &1 \\
\end{bmatrix}$, $1$ matrix
$\begin{bmatrix}
1 &0 &1 \\
0 &1 &0 \\
1 &0 &1 \\
\end{bmatrix}$.

One additional $(J-M)$ matrix for every of $7$ matrices $M$ from previous case.

$J$ is a matrix filled with $1$'s at every entry.

答:
We will casework on the numbers in the first row of the $n*n$ table in two cases:

The first row is $1,0,1,0,\dotsb$ or $0,1,0,1,\dotsb$

There are two consecutive elements in the first set.
First of all, it is clear that in the first case the answer is exactly $2^n$:

Name every row without consecutive elements as simple rows. Now if the first case happens, each row can be a simple row and it is all of the cases without consecutive elements in the first row so, the first case is of $2^n$ cases.

On the other in the second case we will prove that the rest of the table can be constructed uniquely:

The two unit squares under the two consecutive elements of the first row are chosen uniquely and based on these two, the rest of the second row is constructed uniquely. Similarly, the second row has at least two consecutive elements (the two unit squares under the two consecutive elements in the first row are the same) so the third row is constructed uniquely and so on. So in the second case, the rest of the table is constructed uniquely. So the number of cases of the tables of the second kind is equal to the number of binary sets except the two simple sets which is $2^n-2$ so the answer is equal to $2^n+2^n-2=2^{n+1}-2$


答:
The first row can be any of the $2^n$ members of $\{0,1\}^n$.
Given the $k$'th row $(x_1, \ldots, x_n)$, the $k+1$'th row can always be
$(1-x_1, \ldots, 1-x_n)$.  The only other possibility is that the $k+1$'th row is
$(x_1,\ldots, x_n)$, but only if the $k$'th row is alternating $(0,1,0,1,\ldots)$ or $(1,0,1,0,\ldots)$. Now use induction on the number of rows...

Title: Number of binary square matrices whose every contiguous 2 by 2 submatrix has exactly 2 ones (zeroes)

Body:
What is the number of binary $n\times n$ matrices whose every contiguous  $2\times 2$ submatrix has exactly two ones (zeroes)?

That is, if $a_{ij}\in\{0,1\}$ for $i,j\in\{1,2,\dots,n\}$ are matrix entries, then $$a_{i,j} + a_{i,j+1} + a_{i+1,j} + a_{i+1,j+1}= 2$$

for $i \in \{1,\dots,n-1\}$ and $j \in \{1,\dots,n-1\}$, where $n\ge 2$.

After writing out some cases for small $n$, it appears to me that the answer might be $2^{n+1}-2$.

Is there an elegant way to prove this?

For example, for $n=2$ we have $6$ matrices

$$
\begin{bmatrix}
1 &0 \\
1 &0 \\
\end{bmatrix},
\begin{bmatrix}
1 &1 \\
0 &0 \\
\end{bmatrix},
\begin{bmatrix}
1 &0 \\
0 &1 \\
\end{bmatrix},
\begin{bmatrix}
0 &1 \\
1 &0 \\
\end{bmatrix},
\begin{bmatrix}
0 &0 \\
1 &1 \\
\end{bmatrix},
\begin{bmatrix}
0 &1 \\
0 &1 \\
\end{bmatrix}.
$$

For example, for $n=3$ we have $14$ matrices:

$4$ rotations of
$\begin{bmatrix}
1 &0 &1 \\
1 &0 &1 \\
0 &1 &0 \\
\end{bmatrix}$, $2$ rotations of
$\begin{bmatrix}
1 &0 &1 \\
1 &0 &1 \\
1 &0 &1 \\
\end{bmatrix}$, $1$ matrix
$\begin{bmatrix}
1 &0 &1 \\
0 &1 &0 \\
1 &0 &1 \\
\end{bmatrix}$.

One additional $(J-M)$ matrix for every of $7$ matrices $M$ from previous case.

$J$ is a matrix filled with $1$'s at every entry.


问:
Find value of $\lim_{n \to \infty} \sum_{k=1}^{101} \frac{(n+k)^{100}}{n^{99}}-101n$
$$ \lim_{n\to\infty} \left(\frac{(n+1)^{100} + (n+2)^{100} + ... + (n+101)^{100}}{n^{99}} - 101n\right) $$

I have tried multiplying n in numerator and denominator in $$ \left(\frac{(n+1)^{100} + (n+2)^{100} + ... + (n+101)^{100}}{n^{99}}\right) $$ so that both numerator and denominator raised to $100$

Another thing I have thought about is multiplying $n^{99}$ with $101n$ and then subtracting $n^{100}$ from each, that is, $$ \left(\frac{(n+1)^{100} - n^{100} + (n+2)^{100} - n^{100} + ... + (n+101)^{100} - n^{100}}{n^{99}}\right) $$ Will that lead to the solution?

Body:
$$(n+1)^{100} + (n+2)^{100} + \ldots + (n+101)^{100}\sim 101n^{100}+\sum_{k=1}^{101}100k \;n^{99};\;n\to\infty$$
$$\lim_{n\to\infty} \left(\frac{101n^{100}+\sum_{k=1}^{101}100k \;n^{99}}{n^{99}}-101n\right)=\sum_{k=1}^{101}100k =515100$$


问:
How to determine quartile points in the right way?
One way to get quartile points is get median(Q2), then the upper and lower sets and then find their median to get Q1 and Q3.

But, I have found this formula which says Q1 = (1/4)(n+1)th term where n is no. of terms in the set.

Similarly for Q2 = (2/4)(n+1)th term

and Q3 = (3/4)(n+1)th term

Let's say there is a set x = 5,10,15,20,25,30,35,40,45,50

So I know the Median is (25+30)/2 = 27.5 = Q2

and to find Q1 we can take median of lower set, i.e 5,10,15,20,25, which is 15 = Q1.

Similarly to find Q3 we can take median of upper set, i.e 30,35,40,45,50, which is 40 = Q3

But by using the formula, we get Q1 = 1/4(11) = 2.75th term. = 13.75

and Q2 = 2/4(11) = 5.5th term. i.e = 27.5

and Q3 = 3/4(11) = 8.25th term. i.e = 41.5

which gives me different inter quartile range. so which method should I use in the exam?

here is an example of the formula if you haven't heard of it before.

http://facstaff.cbu.edu/wschrein/media/M201%20Notes/M201L23.pdf

答:
Comment: As you have discovered, various sources have slightly different formulas for finding quantiles (including quartiles). Here are answers from R for your data.

```<code>x = c(5,10,15,20,25,30,35,40,45,50)
sort(x)
[1]  5 10 15 20 25  30 35 40 45 50
quantile(x)
   0%   25%   50%   75%  100%
 5.00 16.25 27.50 38.75 50.00
</code>```

According to some definitions any number between 25 and 30 (inclusive)
would be a possible median. And 15 and 40 would be
values for lower and upper quartiles, respectively.

The difficulty, of course, is that it is impossible to divide ten sorted observations into
four groups of equal size. (And if there are any ties,
it can get messier.)

R documentation discusses briefly about eight other
slightly different methods of finding quantiles.
For small samples these methods can show substantial
differences. The results shown above are for the default
method, called type 7.

Here are results from three other 'types' in R:

```<code>quantile(x, type=1)
  0%  25%  50%  75% 100%
   5   15   25   40   50
quantile(x, type=2)
  0%  25%  50%  75% 100%
 5.0 15.0 27.5 40.0 50.0
quantile(x, type=4)
  0%  25%  50%  75% 100%
 5.0 12.5 25.0 37.5 50.0
</code>```

While you're a student in a particular class, use
the method specified there. In practice, quantiles
are most often used for large datasets, and
differences among definitions are mostly unimportant.

```<code>y = rnorm(1000, 100, 15)
quantile(y, type=1)
       0%       25%       50%       75%      100%
 54.82376  89.56435  99.39657 109.90225 148.54398
quantile(y, type=2)
       0%       25%       50%       75%      100%
 54.82376  89.58146  99.44074 109.92691 148.54398
quantile(y, type=7)  # R default
        0%       25%       50%       75%      100%
  54.82376  89.59001  99.44074 109.91458 148.54398
</code>```

Body:
Comment: As you have discovered, various sources have slightly different formulas for finding quantiles (including quartiles). Here are answers from R for your data.

x = c(5,10,15,20,25,30,35,40,45,50)
sort(x)
[1]  5 10 15 20 25  30 35 40 45 50
quantile(x)
   0%   25%   50%   75%  100%
 5.00 16.25 27.50 38.75 50.00

According to some definitions any number between 25 and 30 (inclusive)
would be a possible median. And 15 and 40 would be
values for lower and upper quartiles, respectively.

The difficulty, of course, is that it is impossible to divide ten sorted observations into
four groups of equal size. (And if there are any ties,
it can get messier.)

R documentation discusses briefly about eight other
slightly different methods of finding quantiles.
For small samples these methods can show substantial
differences. The results shown above are for the default
method, called type 7.

Here are results from three other 'types' in R:

quantile(x, type=1)
  0%  25%  50%  75% 100%
   5   15   25   40   50
quantile(x, type=2)
  0%  25%  50%  75% 100%
 5.0 15.0 27.5 40.0 50.0
quantile(x, type=4)
  0%  25%  50%  75% 100%
 5.0 12.5 25.0 37.5 50.0

While you're a student in a particular class, use
the method specified there. In practice, quantiles
are most often used for large datasets, and
differences among definitions are mostly unimportant.

y = rnorm(1000, 100, 15)
quantile(y, type=1)
       0%       25%       50%       75%      100%
 54.82376  89.56435  99.39657 109.90225 148.54398
quantile(y, type=2)
       0%       25%       50%       75%      100%
 54.82376  89.58146  99.44074 109.92691 148.54398
quantile(y, type=7)  # R default
        0%       25%       50%       75%      100%
  54.82376  89.59001  99.44074 109.91458 148.54398


答:
Note that in every binomial of the numerator, only $n^{100}$ and $n^{99}$ terms will be significant in the limit $n \to  \infty$. So its worth computing only these.

$$(n+k)^{100}=n^{100}+100\cdot k \cdot n^{99} + \cdots$$

Therefore on summing,
$$ \lim_{n\to \infty}\dfrac{101n^{100}+100\cdot\tfrac{1}{2}\cdot101\cdot102\cdot n^{99}+o(n^{98})}{n^{99}}-101n=515100$$

Title: Find value of $\lim_{n \to \infty} \sum_{k=1}^{101} \frac{(n+k)^{100}}{n^{99}}-101n$

Body:
$$ \lim_{n\to\infty} \left(\frac{(n+1)^{100} + (n+2)^{100} + ... + (n+101)^{100}}{n^{99}} - 101n\right) $$

I have tried multiplying n in numerator and denominator in $$ \left(\frac{(n+1)^{100} + (n+2)^{100} + ... + (n+101)^{100}}{n^{99}}\right) $$ so that both numerator and denominator raised to $100$

Another thing I have thought about is multiplying $n^{99}$ with $101n$ and then subtracting $n^{100}$ from each, that is, $$ \left(\frac{(n+1)^{100} - n^{100} + (n+2)^{100} - n^{100} + ... + (n+101)^{100} - n^{100}}{n^{99}}\right) $$ Will that lead to the solution?


问:
Distribution of the sum of random variables (are those dependent or independent?)
The weight of the citizens in a city is distributed normally with expected value of 70kg and Standard deviation of 6kg. How is the weight of 9 people is distributed?

Attempt:

Let X be the weight of a person in that city. So $X\sim\mathrm{N}\left(70,36\right)$.

Now Let Y=9X, so as Y is the sum of normally distributed variables, it is also distributed normally. then:

$E\left[Y\right]=E\left[9X\right]=9E\left[X\right]=9\cdot70=630$

$Var\left(Y\right)=Var\left(9X\right)=9^{2}Var\left(X\right)=9^{2}\cdot36=2916$

So we conclude that $Y\sim\mathrm{N}\left(630,2916\right)$

On the other side:

My teacher told me that $Var\left(Y\right)=9\cdot36=324$ as we can conclude it from the Central Limit Theorem. I don't understand how is the CLT related here, as it is a theorem about a limit and also its premise is that the variables are independent no?

He added me the following theorem:

Let $X_1,\ldots,X_n$ be a sample out of a distribution with expected value $\mu$ and variance $\sigma^2$. For large $n$ we have that the sum $\sum_{i=1}^n X_i$ is closely distributed as $N(n\mu,n\sigma^2)$

Please help me figure it out

Thanks

答:
If this is a random sample from the population you should probably assume independence and so your teacher's method

Your solution would be correct if all nine had the same weight

If there was partial dependence (say an individual and their siblings cousins) but not total dependence, then the answer might be somewhere in between


问:
Can a signed arithmetic progression reach all integers?
$F:=\big\{f\,\big|\,\text{function }f: \mathbf N\rightarrow \{1,2\}\big\}$. $S:=\big\{\sum_{i=1}^k (-1)^{f(i)}i\,\big|k\in\mathbf N, f\in F\big\}$. Is $S=\mathbf Z$?

I had thought of using dynamic programming to solve this problem. However, the fact that the number of summands $k$ is unbounded seems to make this unfeasible.

答: By induction, $S_k = \{\sum_{i=1}^k (-1)^{f(i)} i \mid f \in F\}$ consists of all integers $x$ with $|x| \le k(k+1)/2$ and $x \equiv k(k+1)/2 \mod 2$.


问:
What does it mean when the rate of convergence is not two when using the trapezoidal rule?
For $f(x) = \sqrt{x}$ on the interval $[0,1]$ and $N = 16,32,64,128$, the ratio of my errors is not $4$, so the rate of convergence is not 2. Instead I get ~$2.8$ which means the rate of convergence is ~1.485. What causes this to happen?

答: $f$ doesn't have a bounded second derivative (or a bounded first derivative for that matter), so the errors close to $x=0$ decay rather slowly.


问:
Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$
We have a parabola and a circle with the following equations and their graph placed at the end of my question.

Parabola: $y^2 = 4x -4$

Circle: $(x-2)^2 + y^2 = 9$

My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got

$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$

$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?

What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?

update

When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:

$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$

$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$

$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.

2nd update

I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:

$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} +  1)^2=9 \implies
(\frac{y^2}{4} +  1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$

答:
When you are searching for intersection of two figures the intersection dot must satisfy parabola equation which indicates that

$$ y^2 = 4x - 4$$

Which means that

$$ 4x - 4 \geq 0$$
$$ x \geq 1 $$

The last inequality will exclude $x = -3$ solution.


Body:
Let
$a_n=\sum_{k=1}^{101}(n+k)^{100}-101n^{100}$, then by Stolz-Cesaro theorem,
\begin{align}
\lim_{n\to\infty} \left(\frac{\sum_{k=1}^{101}(n+k)^{100}}{n^{99}} - 101n\right)&=\lim_{n\to\infty}\frac{a_n}{n^{99}}
=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)^{99}-n^{99}}\\
&=\lim_{n\to\infty}\frac{(n+102)^{100}-(n+1)^{100}-101((n+1)^{100}- n^{100})}{(n+1)^{99}-n^{99}}\\
&=
\lim_{n\to\infty}\frac{\frac{102\cdot 101\cdot 100\cdot 99}{2}n^{98}+o(n^{98})}{99n^{98}+o(n^{98})}\\
&=\frac{102\cdot 101\cdot 100}{2}=515100.
\end{align}


问:
Finding the locus of $b^2-2x^2=2xy+y^2$
I was asked to determine the locus of the equation
$$b^2-2x^2=2xy+y^2$$

This is my work:

Add $x^2$ to both sides:
$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
 b^2-x^2 &=\left(x+y\right)^2
\end{align}$$

I see that this is similar to the equation of a circle. How can I find the locus of this expression?

答:
The equation is clearly of a conic section. However,since the coefficient of $xy$ is non-zero, the conic section is tilted by some angle $\theta$. The value of $\theta$ can be determined as:
$$\theta = \dfrac{1}2 \arctan (\dfrac{C}{A-B})$$
, where $A,\ C $ and $B$ are the coefficients of $x^2$, $y^2$ and $xy$ respectively in the equation of the conic section. Substitution yields $\theta = 0.5535\dots$ (in radians).
Now rotate the conic section by $-\theta$ to get its standard equation with vertical or horizontal orientation.
$$  x  =  x^\prime \cos \theta  - y^\prime \sin \theta $$
$$y  =  x^\prime \sin \theta + y^\prime \cos \theta$$

After substitution and (ugly) computation, you will get the following equation:

$$2.62 (x')^2 + 0.38(y')^2 = b^2$$
This means the original equation is an equation of an ellipse with major axis of length $a^\prime= b\sqrt{1/0.38}$ tilted by $.5535\dots$ radians (about $31$ degrees) and minor axis of lengh $b^\prime = b\sqrt{1/2.62}$

But,of course, there are easier tests you can check with what conic section formula it is.

Title: Finding the locus of $b^2-2x^2=2xy+y^2$


问:
Prove that you can choose three of spotlights.
The plane is illuminated by several spotlights, each of which illuminates a half-plane. Prove that you can choose three of these spotlights, which also illuminate the entire plane?</p>

答:
Rephrasing the other answer to make it more understandable (to me). The proof is by induction on $n$, the base case $n=3$ being obvious.

Remove one of the spotlights, which covers a half-plane $H$. If the remaining $n-1$ spotlights cover the plane, then you can select $3$ of them which cover the plane by the inductive hypothesis.

If not, then there will be an uncovered region which is a convex polygon. Let $P$ be a vertex of this polygon whose distance to $H$ is minimal, and let $e_1$ and $e_2$ be the edges adjacent to $P$. There is a spotlight whose boundary contains $e_1$, and points away from the uncovered region, therefore pointing towards the complement of $H$. Same goes for $e_2$. Therefore, some thought should convince you that $H$, together with the spotlights for $e_1$ and $e_2$, will cover the whole space.


问:
Assertion about Compact sets : True or false with Justification
Problem 3.3.5 from Understanding Analysis by Stephen Abott poses the following question. I'd like someone to verify if my justification is sound and rigorous.

Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.

(a) The arbitrary intersection of compact sets is compact.

(b) The arbitrary union of compact sets is compact.

(c) Let $A$ be arbitrary and $K$ be compact. Then, the intersection $A \cap K$ is compact.

(d) If $F_1 \supseteq F_2 \supseteq F_3 \supseteq F_4 \supseteq \ldots$ is a nested sequence of nonempty closed sets, then the intersection $\bigcap_{n=1}^{\infty}F_n \ne \emptyset$.

My Attempt.
(a) The countably infinite intersection of closed sets is closed. If each of these sets are bounded, the intersection is also bounded. Hence, $\bigcap_{n=1}^{\infty}K_i$ is compact.

(b) The finite union of closed sets is closed. The countably infinite union of closed sets need not be closed (since the infinite intersection of open sets is not always open, for example $\bigcap_{n=1}^{\infty} \left(0,\frac{1}{n}\right) = \emptyset$, which is closed). As a result, the finite union of compact sets is compact.

(c) $A \cap K \subseteq K$. $A \cap K$ is a closed and bounded subset of $K$. $A \cap K$ is compact.

(d) This proposition is false. Let $F_n = [n,\infty)$. Then,
\begin{align*}
 \bigcap_{n=1}^{\infty} F_n = \emptyset
\end{align*}

答:
(a) It is correct, by I think that it's more natural to say that the intersection is a closed subset of $K_1$ and that therefore it is compact.

(b) You did not really answer the question. An arbitrary union of compacts doesn't have to be compact. For instance, $\Bbb R$, which is not compact, is equal to $\bigcup_{x\in\Bbb R}\{x\}$.

(c) This is false: take $A=(0,2)$ and $K=[1,3]$.

(d) Your answer is fine.


问:
Correct cumulative distribution function notation
If I have a random variable, $v_{A}$, distributed by CDF $G$ on $[v_1,v_2]$, then is it correct to refer to the CDF as $G(v_{A}),$ or should I instead use a generic $v$, and so $G(v)?$ Thank you.

答:
Random variables are actually functions (yes, really confusing!).

A more precise notation would be something like this:

$$G_{\nu_A}(x), \;x \in [\nu_1,\nu_2]$$

This emphasizes that G is an ordinary function.


问:
Simple random Sample Without Replacement (SRSWOR)
Note that SRSWOR refers to Let U be a population of size N. From U, we first select a Simple Random Sample Without Replacement (SRSWOR), S_1 , of size n_1 . Then, from S_1 , we select a SRSWOR, S_2 , of size n_2 . Show that S_2 is a SRSWOR of size n_2 selected from U.

I wasn't familiar with the notion of SRSWOR. But during my research I found that the probability of selecting S_1 for example is P(S_1)= 1/[N!/n_1!(N-n_1)!] I also thought that the answer deals with the fact that S_2 depends actually on S_1? I explored the path of conditional probability and proof for inclusion and subset.

答: This is a two-phase Simple Random Sampling Without Replacement (SRSWOR). the SRSWOR, $S_2$ of size $n_2$ has been chosen from $S_1$ which is a SRSWOR of size $n_1$ drawn from U. We can see the Sample $S_2$ as a set which is a subset of $S_1$, and also $S_1$ a subset of U. Then $S_2$ is also a subset of U. So the second phase of the sampling which requires to draw $S_2$ of size $n_2$ from $S_1$ is equivalent to a one phase sampling which requires to draw $S_2$ of size $n_2$ from the original set U. Therefore we can then conclude that $S_2$ is a SRSWOR of size $n_2$ selected from U.