问:
Limit of $\lim_{x\to\infty}\frac{\int_0^xt-\lfloor t\rfloor dt}{x}$
As shown in the title, what is the limit $\displaystyle \lim_{x\to\infty}\frac{\int_0^xt-\lfloor t\rfloor \mathrm{d}t}{x}$?

Both numerator and donominator goes to infinity as $x\to\infty$ but the numerator is not always differentiable on $[0,\infty)$, so I think we can't use L'Hopital's Rule?  I think the answer should be that the limit DNE but how do we prove that?

答: Write $n:=\lfloor x\rfloor,\,r:=\{x\}$ so $\int_0^x(t-\lfloor t\rfloor)dt=\tfrac12n+\int_0^rydy=\tfrac12(n+r^2)$, so the limit is squeezed between those of $\frac{n}{2(n+r)}$ and $\frac{n+r}{2(n+r)}=\frac12$, and is $\frac12$.


问:
Sierpinski Triangle as Finite Union of Dendrites
Can the Sierpinski Triangle be written as a finite union of dendrites?  If so, can it also be verified what the minimal number is (assuming you can't do it with just two)?

This is a small piece from a more sophisticated thread, but the particular example can possibly be solved by 'geometric cleverness' so I wanted to make it a separate question.  The main thread is:

Finite Unions of Dendrites

答:
No, it can't be written as a finite union of dendrites.  Actually, it can't be written as a countable union of dendrites.  Since every neighborhood in the Sierpinski Triangle $S$ contains a simple closed curve, a dendrite has empty interior in $S$, so by the Baire Theorem you can't cover $S$ with countably many dendrites.  This is also true of dendroids, $\lambda$-dendroids etc. by the same argument; any class of continua which are acyclic.

This Baire trick was too clever for me, a friend hit me up with this one.

ContentLicense: CC BY-SA 4.0


问:
Why there's a different answer for $\int x\ln{x^2}dx$ in wolfram alpha?
I'm attempting to solve $\int x\ln{x^2}dx$

Using integration by parts I was able to do the following steps.

$$2\int x \ln x dx=2\left(\frac{x^2}{2}\ln x -\int\frac{x^2}{2}dx\right)=x^2\ln x - \frac{x^3}{3}+c$$

But when I verified it with wolfram alpha, I'm getting a different answer.

i.e. $\frac{1}{2}x^2(\log x^2 -1)+c$

Can anyone please explain me why there's a difference? Thank you.

wolfram alpha link: https://www.wolframalpha.com/input/?i=integrate+xln%28x%5E2%29

答:
There is an error; it should be
$$\int x\ln x^2 dx \overset{t=x^2}= \frac12\int \ln t dt= \frac12(t\ln t-t)+C$$


问:
Find the equation of the hyperbola, if it passes through $(1;0)$ and has the asymptotes $x=0$ and $y=1$.
I know I have to rotate it 45 degrees I just don't know how. I don't even know how this rotation thing works. I would really appreciate any help and if you can help me understand the rotation thing I promise to pray for you and your family every night. Thanks in advance!

P.S.Please do not bully me if I'm asking stupid stuff.I am a sensitive person.

答:
The equation of the asymptotes of $xy=c^2$ are $xy=0$

So, the equation of the hyperbola will be $$x(y-1)=d$$

As it passes through $(1,0);$
$$d=1\cdot(0-1)$$


问:
What is the correct term for sets that dont contain other sets?
As the title asks, what is the correct term for a set that don't contain another set?

I have called them "flat" but what is the correct term?

答:
I've not heard of any term for this in particular. But the term urelements is used for elements that are not sets, so you could call your set a set of urelements.

One warning though: As you can see by following that link, the standard ZF axioms of set theory are formulated in such a way that urelements do not exist, i.e. everything is a set, and therefore (as said in the comments) the only set whose elements are not sets is the empty set. The reason for this formulation is explained in that link: "the urelements were not needed because they can easily be modeled in a set theory without urelements."

One thing to keep in mind: in analysis or geometry, it is not of particular importance whether numbers are or are not sets. Nonetheless, when analysis and geometry are modelled using the standard ZF axioms of set theory, every number is a set, every geometric point is a set, and so on.

You might be interested, for example, in the Von Neumann definition of the natural numbers, which is formulated in set theory. And perhaps you already know how one can, within set theory, use the natural numbers to form the integers, the integers to form the rationals, and the rationals to form the reals. And, of course, one can use the reals to form the points of the Euclidean plane and so on.


问:
Find the value of $\theta$ in parallelogram $ABCD$.
Question:

If $ABCD$ is a parallelogram then find the angle $\theta$ in degrees.

What I tried:

I assumed $BD$ a straight line as I don't think this question can be solved if $BD$ is not a straight line.(I will be happy if someone solves this without this assumption)

Construction: Drew $EF$ perpendicular to ($CD$ and $AB$) and $GH$ to ($BC$ and $AD$).

$$\angle DAO=\angle DCO=y$$ Alternate interior angles:$$\angle BDA=\theta\\\angle ABD=40^\circ$$

By Angle Sum Property:
$$\angle DOG=90^\circ-\theta\\\angle GOA=90^\circ-y\\\angle AOF=70^\circ\\\angle FOB=50^\circ\\\angle BOH=90^\circ-\theta\\\angle HOC=70^\circ\\\angle COE=90^\circ-y\\\angle EOD=50^\circ$$
Adding all these and equating to $360^\circ$:
$$240^\circ+360^\circ-2\theta-2y=360^\circ$$
$$\theta+y=120^\circ$$

After this, I drew parallel lines to ($AD$ and $BC$) and ($AB$ and $CD$) through $O$ but didn't get the value of $\theta$.

How to solve this question?
Can this be solved?

Thanks!

答: The assumption is not necessary. Construct $P$ on the parallel to $AB$ by $O$ such that $PAD\cong OBC$. Then $\angle PDA = 20°$, but $\angle POA = \angle OAB = 20 °$, so $AODP$ is cyclic. But then $\angle CBO = \angle DAP = \angle DOP = \angle CDO = 40°$.


问:
Good reference texts for introduction to partial differential equation?
As the title, are there any good reference texts for introduction to partial differential equation?</p>

答: I would recommend "Partial Differential Equations with Fourier Series and Boundary Value Problems" written by Asmar. It is a text for beginners in the subject, but eventually it gets very far, to a thorough treatment of Green's functions. It does also include a lot of material on Bessels functions.   </p>

ContentLicense: CC BY-SA 3.0


答:
First, observe that $f(t)=t-\lfloor t\rfloor$ is periodic and bounded:
\begin{align}
f(t) &= f(t+1)\\
\lvert f(t)\rvert &\le 1\ \forall t
\end{align}

Therefore, the primitive of $f$ can be expressed as
$$F(x) = ax + B(x)$$
where $a$ is the average value of $f$ over a period, and $B$ is a periodic and bounded function.

$$\therefore\ \lim_{x\to\infty}\frac{\int_0^x f(t)\mathrm{d}t}{x} = \lim_{x\to\infty}\frac{ax+B(x)}{x}$$

Since $B$ is bounded, it is clear that this limit is just equal to $a$, which is given by
$$a = \int_0^1 f(t)\mathrm{d}t = \tfrac{1}{2}$$

Title: Limit of $\lim_{x\to\infty}\frac{\int_0^xt-\lfloor t\rfloor dt}{x}$

Body:
As shown in the title, what is the limit $\displaystyle \lim_{x\to\infty}\frac{\int_0^xt-\lfloor t\rfloor \mathrm{d}t}{x}$?

Both numerator and donominator goes to infinity as $x\to\infty$ but the numerator is not always differentiable on $[0,\infty)$, so I think we can't use L'Hopital's Rule?  I think the answer should be that the limit DNE but how do we prove that?


问:
Does the law of excluded middle imply the axiom of choice?
I know that the Axiom of Choice implies Excluded Middle, but I haven't been able to find any discussion of the other direction.

Does the law of excluded middle imply the axiom of choice?

Here's a stab at a proof, but I'm not sure if it's correct:

AC says that there is always a function f that can take a set $X$ with $\geq 1$ elements and chooses for each $x \in X$ whether $x \in A$ or $x \in B$ (exclusive).

For any $x \in X$, let P be the proposition that $x \in A \land \neg (x \in B)$ and let Q be the proposition that $x \in B \land \neg (x \in A)$.

By LEM, $P \vee \neg P$ and $Q \vee \neg Q$. So there are the following cases:

$P \land Q$ . Therefore $x \in A \land \neg (x \in A)$. Contradiction.

$P \land \neg Q$. Therefore $f(x) \in A \land \neg (f(x) \in B)$.

$\neg P \land Q$. Thefore $f(x) \in B \land \neg (f(x) \in A)$.

$\neg P \land \neg Q$. Therefore $x \in B \land \neg (x \in B)$. Contradiction.

答:
Your meaning of "AC" is clearly wrong.

Of course, we usually accept LEM as a postulate, but it is known that AC is independent of ZF. So, the answer is "no".


问:
Show that if $f$ is analytic on $B(z_0,r)\setminus\{z_0\}$, and $ \lim_{z\to z_0} f(z)(z-z_0)=A$ , then, $\oint_{|z-z_0|=r} f(z) dz=2\pi iA$
Show that if $f$ is analytic on $B(z_0,r)\setminus\{z_0\}$ , and $$ \lim_{z\to z_0} f(z)(z-z_0)=A$$
then $$\oint_{|z-z_0|=r} f(z) dz=2\pi iA$$ for any $r$ : $0<r<R$

I am not sure how to start, first of all, it looks like the solution requires Cauchy's Formula, but  since we don't know that $f$ is analytic in $z_0$, doesn't that make the domain not simply connected, and thus doesn't allow the use of the theorem?

答: Let$$g(z)=\begin{cases}(z-z_0)f(z)&\text{ if }z\ne z_0\\A&\text{ if }z=z_0.\end{cases}$$Then $g$ is continuous at $z_0$ and therefore, by Riemann's extension theorem, it is holomorphic. So, Cauchy's integral formula telles us that$$\frac1{2\pi i}\oint_{|z-z_0|=r}\frac{g(z)}{z-z_0}\,\mathrm dz=g(z_0)=A.$$So,$$\frac1{2\pi i}\oint_{|z-z_0|=r}f(z)\,\mathrm dz=A.$$


问:
Equidistant points by modular arithmetic
If $\frac{a}{b}$ is an irreducible rational number, how does one show that the collection of points $\{ \frac{na}{b}\ \mbox{mod}\, 1\}$, $1\leqslant n \leqslant b$ are equidistant along the real line with distance $\frac{1}{b}$?

I’ve tried numerous examples to see how it works but I’m sure how to give a rigorous argument.

答:
The key fact here is the notion of invertibility modulo $b$.
Firstly, note that you will hit some $c/b$, for $c<b$ iff there exist $n, k$ such that
$$ \frac{na}{b} + k = \frac{c}{b} $$

Which is verified iff
$$ na +bk = c$$

Here modular arithmetics comes into play. Since $k$ is arbitrary, this is verified if exist $n$ such that $na \equiv c \pmod{b}$.
Since $a$ is coprime with $b$, there exist a number $a' $ such that $a a' \equiv 1 \pmod b$, called its inverse modulo b.

Now let's show that we can hit any $c$ we want. By taking $n = ca'$ you will get

$$ na \equiv (ca') a \equiv c (a'a) \equiv c \pmod{b}$$

Yey!


答:
The axiom of choice states that given a family of non-empty sets, there is a function mapping each of those sets to an element of the set. The idea is to be able to move from $\forall a\in A\exists x(x\in a)$ to $\exists f\forall a\in A(f(a)\in a)$.

What you're talking about is existential instantiation, i.e. if $X$ is not empty, then we can "name" an element of it. There is also no $A$ and $B$ involved (and in your statement there is no information about them either). People sometimes mistake EI to be an instance of $\sf AC$, but that is not the case.

And of course, we know that LEM does not imply $\sf AC$, since we know that $\sf ZF$ is consistent with $\lnot\sf AC$ while LEM holds.

Body:
I know that the Axiom of Choice implies Excluded Middle, but I haven't been able to find any discussion of the other direction.

Does the law of excluded middle imply the axiom of choice?

Here's a stab at a proof, but I'm not sure if it's correct:

AC says that there is always a function f that can take a set $X$ with $\geq 1$ elements and chooses for each $x \in X$ whether $x \in A$ or $x \in B$ (exclusive).

For any $x \in X$, let P be the proposition that $x \in A \land \neg (x \in B)$ and let Q be the proposition that $x \in B \land \neg (x \in A)$.

By LEM, $P \vee \neg P$ and $Q \vee \neg Q$. So there are the following cases:

$P \land Q$ . Therefore $x \in A \land \neg (x \in A)$. Contradiction.

$P \land \neg Q$. Therefore $f(x) \in A \land \neg (f(x) \in B)$.

$\neg P \land Q$. Thefore $f(x) \in B \land \neg (f(x) \in A)$.

$\neg P \land \neg Q$. Therefore $x \in B \land \neg (x \in B)$. Contradiction.


问:
Given $X,U$ random variables, find $F_Y$ f $Y=U+X$ (attemp added)
Let $X,U$ be independant random variables s.t $X\sim\mathsf{Geo}\left(\frac12\right)$ and $U\sim\mathsf{U}\left[-1,0\right]$. Define $Y=X+U$, find $F_Y$ of $Y$

$$P_x(k)=(\frac{1}{2})^k$$ $$F_Y(t)=
\mathbb{P}(X+U \leq t)=\sum_{k=1}^n \mathbb{P}(X+U \leq t \mid X=k) \mathbb{P}(X=k)=\sum_{k=1}^n \mathbb{P}(X=k) \mathbb{P}(U \leq t-k)
$$

now subtitude $P(X=k)=(\frac{1}{2})^k$ $$\mathbb{P}(X+U \leq t)=\sum_{k=1}^n (\frac{1}{2})^k \mathbb{P}(U \leq t-k)$$ now I am stuck how to write the expression for $\mathbb{P}(U \leq t-k)$

If some one can explain it details it will be great

答:
As you have correctly obtained, we have, $\displaystyle\Pr(X + U \leq t) = \sum_{k = 1}^{\infty} 2^{-k} \Pr(U \leq t - k)$.

Let us evaluate this expression for different values of $t$. Note that for $t \leq 0$, $t - k \leq -1$ and hence RHS is zero. Thus, $\Pr(X + U \leq t) = 0$ for $t \leq 0$. For the case $t > 0$, we consider two cases. Firstly, let us consider the case when $t \in \mathbb{N}$. Note that under this scenario $\Pr(U \leq t - k) = 1$ for all $k \leq t$ and $\Pr(U \leq t - k) = 0$ for $k \geq t + 1$. Hence, for $t \in \mathbb{N}$, we have,
$$\Pr(X + U \leq t) = \sum_{k = 1}^t 2^{-k} = 1 - 2^{-t}.$$
We can similarly solve for the case when $t > 0$ and $t \notin \mathbb{N}$. This, time note that $\Pr(U \leq t - k) = 1$ for all $k \leq \lfloor t \rfloor$ and $\Pr(U \leq t - k) = 0$ for $k \geq \lfloor t \rfloor + 2$. For the case when $k = \lfloor t \rfloor + 1$, we have $\Pr(U \leq t - k) = \Pr(U \leq t - \lfloor t \rfloor - 1 ) = t - \lfloor t \rfloor$. Thus, for $t \notin \mathbb{N}$, we have,
$$\Pr(X + U \leq t) = \sum_{k = 1}^{\lfloor t \rfloor} 2^{-k} + 2^{-\lfloor t \rfloor - 1}(t - \lfloor t \rfloor)  = 1 - 2^{-\lfloor t \rfloor} + 2^{-\lfloor t \rfloor - 1}(t - \lfloor t \rfloor).$$

We can combine all the cases and write
\begin{align*}
F_Y(t) = \Pr(X + U \leq t) = \begin{cases} 0, & \text{ if } t \leq 0, \\ 1 - 2^{-\lfloor t \rfloor} + 2^{-\lfloor t \rfloor - 1}(t - \lfloor t \rfloor), & \text{ if } t > 0. \end{cases}
\end{align*}


问:
Checking exercise on uniform convergence
I have solved the following exercise and I would like someone to check and tell me if my solution is right. Thank you.

Let $f_n(x)=\frac{nx}{1+nx^2}$

(a) Find the pointwise limit of $(f_n)$ for all $x\in (0,\infty)$

(b) Is the convergence uniform on $(0,\infty)$?

(c) Is the convergence uniform on $(0,1)$?

(d) Is the convergence uniform on $(1,\infty)$?

My solution (NOTE: corrected parts (b) and (c) thanks to the answers below):

(a) $f_n(x)=\frac{nx}{1+nx^2}=\frac{1}{\frac{1}{nx}+x}\xrightarrow[]{n\to\infty} \frac{1}{x} $

In what follows we use the fact that if $f_n \to f$ uniformly on $(0,\infty)$ or $(0,1)$ then $|f_n(a_n)-f(a_n)|\xrightarrow{n\to\infty}0$ (where $(a_n)$ is a sequence of points belonging to these spaces.)

(b) Consider the sequence $(a_n)=\frac{1}{n}$; then $|f_n(a_n)-\frac{1}{a_n}|=|\frac{n}{n+1}-n|=|-\frac{n^2}{n+1}|=\frac{n^2}{n+1}\xrightarrow{n\to\infty}\infty$ thus the convergence is not uniform here.

(c) The convergence is not uniform by the same reasoning as in part (b)

(d) $|f_n(x)-\frac{1}{x}|=\frac{1}{x(1+nx^2)}\leq\frac{1}{1\cdot (1+1\cdot n)}=\frac{1}{1+n}\leq\frac{1}{n}$. Thus, given $\varepsilon > 0$ we can choose $N >\frac{1}{\varepsilon}$ (independent of $x$) and it follows that $n\geq N$ implies $|f_n(x)-\frac{1}{x}|\leq\varepsilon$ for all $x\in (0,1)$ so here we have uniform convergence.

答:
It is almost correct. Here is a problem in parts (b) and (c), which is this: you used the inequality$$\left|f_n(x)-\frac1x\right|\leqslant\frac1{nx^3}$$and then you stated that, since $\frac1{nx^3}$ is not arbitrarily small uniformly, then the convergence is not uniform. That is not correct. For instance, $\frac xn$ converges uniformly to $0$ on $[0,1]$. But you also have $|f_n(x)-0|\leqslant1$ on $[0,1]$, and the constant function $1$ is not arbitrarily small uniformly.

In order to do (b) and (c), just use the fact that$$(\forall n\in\Bbb N):f_n\left(\frac1n\right)=\frac n{n+1}\to_{n\to\infty}1.$$


问:
Reference request: Classification of 2-dimensional quadratic forms
My professor recently gave me an outline of a correspondence of the following sort: Let $K$ be a field with $\operatorname{char}{K} \neq 2$. Then, there is a correspondence
$$ \{ (V,Q) : 2\text{-dim non-degenerate quadratic module} \} \cong \{(L/K, \mathrm{disc}) : L/K \text{ quadratic étale algebra, }\mathrm{disc} \in K^{\times}/N_{L/K}L^{\times} \}$$ up to isomorphism. (There may be some mistakes here but those would be entirely due to me.)

I think that this is an extremely cool result! I would also expect it to be quite fundamental. However, when I tried googling this, I surprisingly found almost no references. My professor also didn't have a reference of the top of his head.

Therefore, I want to ask for references regarding this result.

答: This might be classified as shameless advertising but this statement can be found in a talk of mine and the proof is sketched on the last two slides.</p>


问:
What bizarrities lurk within the decimal expansion of an irrational number?
i.e. can we write

$\pi = 3.14159\dots X\dots$

where $X$ consists of (say) $10^{100}$ consecutive zeroes?

[Originally asked on reddit without response :-( ]

答: To expand on Dan Brumleve's comment; it is widely believed, but not proved, that, given any finite string of digits, that finite string appears somewhere in the decimal for $\pi$, in fact, occurs infinitely often, in fact, occurs about once every $10^n$ digits, where $n$ is the length of the string. The same is believed to hold true for $e$, $\sqrt2$, in fact, for pretty much any number known to be irrational and not constructed specifically to falsify the belief, but, again, nothing has been proved. For all we know, all the digits of $\pi$ from some point on are sixes and sevens. </p>


问:
Showing characteristic and minimum polynomial coincide for this particular matrix
Given a matrix such as  $$\pmatrix{0 & 0 & 2 \\ 1 & 0 & -1 \\0 & 1 & 1   \\ },$$ whose characteristic polynomial is $-X^3+X^2-X+2.$
 $$$$How it could be deduced that it equals minus the minimum polynomial ?$$$$Thanks in advance. </p>

答:
It is easier to see that this is the minumum polynomial than that it is the characteristic polynomial. In fact this is a companion matrix, of which the final column always gives minus the non-leading coefficients of the minimum polynomial, which is also the characteristic polynomial, see this question
or this answer.

Explicitly, if $e_i$ are the canonical basis vectors, then $n-1$ repeated multiplications by any companion matrix send $e_1\mapsto e_2\mapsto e_3\mapsto\cdots e_{n-1}\mapsto e_n$; one sees that no polynomial of degree $d$ with $0\leq d<n$ applied to $e_1$ gives $0$, and the minimal polynomial must be of degree at least $n$ (in fact exactly $n$ by Cayley-Hamilton, but I'm avoiding use of that much more complicated theorem). If the final column has successive entries $c_0,\ldots c_{n-1}$, this means that $$A\cdot e_n=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n=c_0A^0\cdot e_1+c_1A^1\cdot e_1+\cdots+c_{n-1}A^{n-1}\cdot e_1=(c_0I+c_1A+\cdots+c_{n-1}A^{n-1})\cdot e_1,$$ so the indicated polynomial $P$ of $A$ annihilates $e_1$. Since it commutes with $A$, it also annihilates $A^i\cdot e_1=e_{i+1}$ for $i=1,\ldots,n-1$, in other words it annihilates as basis and therefore the whole space.


问:
Reference Request: Witt Vectors
I'm interested in a nice (and possibly gentle) introduction to Witt vectors. It doesn't need to go into too much detail, for now, I'd just like to roughly know what they are about. So essentially my wish is something along the following lines: a definition/construction along with some motivation and possibly some suggestive examples/applications. (Though I would fancy references that do not satisfy all of this points.)

The Wikipedia article seems decent but I'm interested to know where members from MSE first learned about Witt vectors and what you would recommend.

(I was surprised by the way that there was no reference request about Witt vectors on MSE or MO. Usually, those forums would be the first place for me to check when I am looking for good book suggestions.)

答:
For completeness, another reference that I know is Bosch's Algebra, Chapter 4.10. (It is written in German and a classic for general/introductory algebra for German students.)

I haven't read the Witt vectors section yet, but I have invested a lot of time in other parts of the book and these were usually splendid, so I assume the Witt vectors section is also good.

Title: Reference Request: Witt Vectors


问:
Proof from Holt's Linear Algebra and its Application Textbook
I was having trouble figuring out the proof for the following question:

Let $\{u_1, u_2,...,u_m\}$ be a set of vectors in $\mathbb{R}^n$. If $m < n$, then this set does not span $\mathbb{R}^n$.

I have tried various examples and created augmented matrices for the examples and I've noticed that the last row is always a row of zeros. This indicates that the matrix does not have pivot points in every row which means that the set does not span $\mathbb{R}^n$. The only issue I have is that I don't know how to show this rigorously. I started learning Lin Alg yesterday so a little help would be appreciated!

答: In my version of the book you are using, Theorem 2.8 shows that $\langle u_1,\ldots,u_m\rangle = \Bbb{R}^n$ iff a row echelon form of the matrix whose columns are the $u_i$ has a pivot in each row. A matrix with $m$ columns can have at most $m$ pivots, so if $m < n$ then there can't possibly be $n$ pivots.


问:
Do these equations represent the distance of a point from the origin on both axes?
Triangle and the equations

Does these equations somehow represent the distance between the leftmost point (assumed as origin) and the point (x,y), with the distance on the x axis being the first equation and the distance on the y axis being the second equation? If so, how; and if not, what do they represent?

答:
That is a strange picture.

The formulas give the new coordinates of the point you get when you rotate the point $(x,y)$ through angle $\theta$. about the origin.

Try some examples to confirm that.


问:
Prove that $x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right]$ is measurable for a Brownian motion $\mathrm B$

Given a (one-dimensional) Brownian motion $\mathrm{B}$ and any $A \in \mathcal B\left(\mathbb R^{[0,1)}\right)$, how may one prove that the map, from $\mathbb R$ to $[0,1]$, given by
$$
x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right]
$$
is measurable?

The notation
$$
\boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x) \in A \right]
$$
means that we consider the whole path; or in other words, if the function
$$
t \mapsto \mathrm{B}(t + x)
$$
is in $A$.

Most grateful for any advice provided!

答:
As suggested by John Dawkins, we may start with show that the map is measurable when $A$ is of the form
$$
\bigcap_{k=1}^n \pi_{t_k}^{-1}(G_k),
$$
where $\pi_t$ denotes the projection from $\mathbb R^{[0,1)}$ to $\mathbb R$ given by
$$
\pi_t(f) = f(t),
$$
and each $G_k$ is an open set in $\mathbb R$.

When $A$ is of the specified form, the map is measurable:

We say that the map
$$
x \mapsto \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right]
$$
is semicontinuous if for any $t$, the set
$$
\left\{ x: \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right] > t  \right\}
$$
is open. And we note that a semicontinuous map is Borel-measurable as any Borel-set in $\mathbb R$ may be written as a countable union of sets of the form $(t, \infty)$ and the preimage of a countable union is the countable union of the (individual) preimages.

To prove semicontinuity we need to show that if
$$
x_0 \in \left\{ x: \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right] > t  \right\}
$$
then there exists an open interval $(x_1, x_2)$ about $x_0$. Assume that $A$ is of the form
$$
A = \bigcap_{k=1}^n\left\{B(t_k)\in G_k\right\}, \quad 0<t_1<\cdots <t_n\le 1, \, G_k \text{ open}.
$$
Suppose that
$$
\omega \in \left \{ \mathrm{B}(\boldsymbol\cdot + x_0 ) \in A \right \} = \bigcap_{k=1}^n  \left \{ \mathrm{B}(t_k + x_0) \in G_k \right \}
$$
and that $\omega$ is in the set of probability one such that the paths of $\mathrm{B}$ are continuous. Then, since each $G_k$ is open, by letting $x_1 < x_0 < x_2$ be sufficiently close to $x_0$ we get that for $x \in (x_1, x_2)$
$$
\omega \in  \bigcap_{k=1}^n  \left \{ \mathrm{B}(t_k + x) \in G_k \right \} =
 \left \{ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right \}.
$$
Hence, for $x \in (x_1, x_2)$
$$
 \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right]= \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x_0) \in A\right] > t .
$$

Define the class $\mathcal A$ as the collection of set $A \subset \mathbb R^{[0,1)}$ such that,
$$
x \mapsto \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right]
$$
is measurable.

As shown above, the collection of sets
$$
\left\{\bigcap_{k=1}^n \pi_{t_k}^{-1}(G_k): \; n \in \mathbb N, t_1, \dots t_n \in [0,1), G_k \subset \mathbb R \text{ open}, \right\}
$$
is contained in $\mathcal A$. We also know that this collection of sets forms a $\pi$-system for the Borel-sigma algebra on $\mathbb R^{[0,1)}$. We may use Dynkin's $\pi$-$\lambda$ theorem, by showing that the $\mathcal A$ also is a $\lambda$-system and then conclude that it thus must contain the sigma-algebra generated by the $\pi$-system $\mathcal A$ - i.e. the Borel-sigma algebra.

Let us show that $\mathcal A$ is a $\lambda$-system:

clearly $\mathbb R^{[0,1)}$ is in $\mathcal A$ as we may take any $t \in [0,1)$ and consider
$$
\mathbb R^{[0,1)} = \pi_t^{-1}(\mathbb R ).
$$

suppose that $A, B \in \mathcal A$, and $A\subsetneq B$ From the \textit{excision property} of probability measures
$$
\boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \setminus A \right ] = \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \right ] - \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in  A \right ],
$$
hence
$$
x \mapsto \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \setminus A \right ] = \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \right ] - \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in  A \right ]
$$
is a measurable mapping as the difference of two measurable mappings is a measurable mapping.

Suppose that $A_1, A_2, \dots \in \mathcal A$ is a sequence of pairwise disjoint sets. Since measures are countable additive,
$$
\boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in \bigcup_{n=1}^\infty A_n \right ] = \sum_{n=1}^\infty \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in A_n \right ]
$$
and hence
$$
x \mapsto \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in \bigcup_{n=1}^\infty A_n \right ] = \sum_{n=1}^\infty \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in A_n \right ]
$$
is the limit of a monotone sequence of measurable functions and hence measurable.

We conclude that $\mathcal A$ is a $\lambda$-system and hence contains the Borel sigma algebra.


问:
Probability that you picked at least $2$ correct out of $8$, whereby the $8$ are again a sample out of $11$.
The question goes like this: You write a test and have $11$ topics in total. In the test, the teacher picks $8$ out of the $11$ topics. For each number of topics you study, what are the probabilities that you studied for at least two topics asked for in the test.
I'm trying to wrap my head around this, but am stuck since a while now.

答:
First, one thing to note is that the question indicates is without replacement, so when a teacher picks a topic she won't choose it again, this is usually a good hint for you to think about the hypergeometric distribution as that is the only probability distribution with no replacement.

In general, the distribution takes the following form:

$\frac{{K \choose k} {N-K \choose n-k}}{N \choose n} $

Where:

$N$ is the population size,

$K$ is the number of success states in the population,

$n$ is the number of draws (i.e. quantity drawn in each trial),

$k$ is the number of observed successes

Now getting to your question:

Since it is worded as "what are the probabilities" I am assuming they want each probability separate, even though that is unusual in such exercises, we have the following, the probability that you studied exactly two correct topics is (note that I am assuming that :

When you study $0$ and $1$ topic then the probability of having $2$ correct is $0$, I guess that is pretty trivial.

Otherwise, the probability of getting $2$ topics correct when you studied for $k$ topics is:

$\frac{{8 \choose 2} {3 \choose k-2}}{11 \choose 8} $

Otherwise, the probability of getting $3$ topics correct when you studied for $k$ topics is, obviously this is $0$ when $k<3$ but otherwise it is:

$\frac{{8 \choose 3} {3 \choose k-3}}{11 \choose 8} $

Can you continue?


问:
Direct proof for the irrationality of $\sqrt 2$.

Prove that $\sqrt 2$ is irrational using direct proof.

I have seen TONS indirect proofs (e.g. proof by contradiction) for it, and people say that it's difficult to proof this directly. So is this impossible?

Thank you.

答:
It has an infinite continued fraction.

$$ \!\ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}. $$


问:
Algorithms : What is the difference between complexity of $T(n)=T(\log n)+cn$ and $T(n)=T(\log n)$ + Θ(1)
The solution to $T(n)=T(\log n) +  \Theta(1)$ is $O(\log n)$. Does adding $cn$ to this make any difference?

Similarly, is the time complexity of $F(n)=F(\sqrt{n}) + \Theta(1)$ different from $F(n)=F(\sqrt{n}) +cn$ , and if yes what is the difference?

Thanks in advance.

答:
The solution of $T(n) = T(\log(n)) + \Theta(1)$ is $T(n)=\log^*(n)$.
You are essentially adding a $\Theta(1)$ each time you apply $\log(\cdot)$ to $n$, which you can do at most $\log^*(n)$ times. See iterated logarithm.

$$
T(n) = T(\log(n)) + \Theta(1) = T(\log\log(n)) + \Theta(1) + \Theta(1) = \underbrace{\Theta(1) + \ldots + \Theta(1) + \Theta(1).}_{\text{$\log^*(n)$ times}}
$$

For $F(n)=F(\sqrt{n}) + \Theta(1)$ we have $F(n) = \log_2(n)$, since, as before, we add $\Theta(1)$ for each iteration of $\sqrt{\cdot}$ on $n$, which are in total $\log_2(n)$.

$$
F(n) = F(\sqrt{n}) + \Theta(1) = F(\sqrt[4]{n}) + \Theta(1) + \Theta(1) = \underbrace{\Theta(1) + \ldots + \Theta(1) + \Theta(1).}_{\text{$\log_2(n)$ times}}
$$

On the other hand, if you set $cn$ in stead of $\Theta(1)$ the first term overpowers everything and you obtain

$$
T(n) = T(\log(n)) + cn = T(\log\log(n)) + c\log(n) + cn = \sum_{i=0}^{i=\log^*{n}}c\: \underbrace{\log\log\cdots\log(n)}_{\text{$i$ times}} = \Theta(n),
$$

$$
F(n) = F(\sqrt{n}) + cn = F(\sqrt[4]{n}) + c\sqrt{n} + cn = \sum_{i=0}^{i=\log_2(n)} c\:n^{1/2^i} = \Theta(n).
$$


问:
Derivatives of sin wave function with respect to amplitude, frequency, and horizontal/vertical shifts?
I am trying to find the derivatives of a sine wave function $$f(x,A,F,H,V) = A\sin(F(x-H))+V$$ with respect to each variable, where $A$ is amplitude, $F$ is frequency, $H$ is horizontal shift, and $V$ is vertical shift. So far I believe I have correctly found the derivatives with respect to amplitude and vertical shift.

$\frac{\partial f}{\partial A} = \sin(F(x-H))$

$\frac{\partial f}{\partial V} = 1$

However I am having trouble finding the derivative with respect to frequency and horizontal shift. Kindly let me know how I should go about this and let me know if I have made any errors in derivatives with respect to amplitude and vertical shift.

答:
I'll show you how to do the partial derivative with respect to $F$ and leave $H$ as an exercise.

$$\begin{align*} \frac\partial{\partial F} \big(A\sin(F(x-H))+V\big) &=
A\frac\partial{\partial F} \sin(F(x-H)) + 0 \quad \text{(linearity)}\\
&= A\cos(F(x-H))\frac\partial{\partial F}(F(x-H)) \quad \text{(chain rule)}\\
&=A\cos(F(x-H))(x-H).
\end{align*}$$


问:
Question on sums of products of divisors
Let us consider some positive integer $n$ with divisors (other than $1$) $d_1, d_2, ..., d_n$. Lately, it came to me the following question:

Does it exist some positive integer $n$ such that a sum of $k$ products of its divisors greater than $1$ equal $n$ itself, when $k$ is not a divisor of $n$ itself?

My "gut feeling" is that is not possible, but I have not been able to ellaborate any proof of it. If all products added are equal, it is clear that its sum is not equal to $n$ unless $k$ divides $n$, but I am having trouble when considering distinct products of divisors.

It can be noticed easily that if $n$ is some prime number, a perfect power, or a semiprime, then a sum as the defined is not possible, as either there are no possible products of divisors, or all the products of divisors are equal, or all the possible sums of products have a number of terms $k$ which is a divisor of $n$.

Any hint or a sketch of a proof would be really welcomed. Thanks!

答:
If I understand your question correctly, the answer is yes.

For $n=54=2\times 3^3$, we can have
$$54=2\times 3+2\times 6+2\times 9+3\times 6$$
where $2,3,6$ and $9$ are its divisors larger than $1$.

There are $k=4$ terms in the sum, and $4$ does not divide $54$.

Added :

There are infinitely many such $n$.

Proof :

For $n=2\cdot 3^p$ where $p$ is a prime number larger than $3$, we can have
$$2\cdot 3^p=2\times (2\times 3^1)+2\times (2\times 3^2)+\cdots +2\times (2\times 3^{p-1})+2\times 3$$

There are $k=p$ terms in the sum, and $p$ does not divide $n=2\cdot 3^p$.


问:
concatenation of rotations in spherical coordinates
Any rotation in $R{^3}$ can be expressed by a point $(\alpha,\beta)$ on the sphere in spherical coordinates. Is there any way to compute the concatenation of two rotation in spherical coordinates without translating back and forth to cartesian?

答:
The set of points of the sphere $S^2$ is 2-dimensional; the set $SO(3)$ of all rotations of 3-space fixing the origin is 3-dimensional. The map from $S^1 \times S^1$ to $S^2$ given by polar coordinates is continuous.

And there's no (bi)continuous bijection between a 2-manifold and a 3-manifold.

So all this goes to show that you first statement is false, so the second is sort of unanswerable.

Sorry to be disappointing.


问:
Minimizing maximum is equivalent to minimizing sum over strictly convex set?
Let $f: \mathbb{R}^n \to \mathbb{R}$ be a strictly concave function. Consider the following quantity $\mathbf{a} \in \mathbb{R^n}$

\begin{align}
\DeclareMathOperator{\argmin}{argmin}
\mathbf{a} = \argmin\limits_{\mathbf{x} : f(\mathbf{x}) \geq 1} \max_{i= 1, \dots, n} x_i
\end{align}

and the following quantity $\mathbf{b} \in \mathbb{R}^n$:

\begin{align}
\DeclareMathOperator{\argmin}{argmin}
\mathbf{b} = \argmin\limits_{\mathbf{x} : f(\mathbf{x}) \geq 1} \sum_{i=1}^n x_i
\end{align}

Is it true that $\mathbf{a} = \mathbf{b}$? I tried to come up with a few counter examples but I could not find any so I feel like $\mathbf{a} = \mathbf{b}$. Any hints?

Body: Take $f(x) = 2-x^Tx$ on $\mathbb{R}^2$, so $\{x \in \mathbb{R}^2 : f(x) \geq 1\} = \{ x \in \mathbb{R}^2 : x^Tx \leq 1 \}$. You get $a=\{(1,0),(0,1)\}$ and $b=\{(1/\sqrt{2}, 1/\sqrt{2})\}$.


问:
Seeing that application in the PCA $A[f]$ is really a partial function
According to Jaap van Oosten's "A General Form of Relative Recursion," given a PCA $A$ and a partial function $f:A\rightharpoonup A$, one can define a new PCA $A[f]$ whose elements are those of $A$, but with a new application operation $a\cdot^f b$ defined as follows:

Let $[u_0,\ldots,u_{n-1}]$ denote the code for the finite sequence of elements $u_0,\ldots,u_{n-1}\in A$.

An $f$-dialogue is a sequence $[u_0,\ldots,u_{n-1}]$ such that for any $i<n$ there's a $v_i\in A$ with $a\cdot [b,u_0,\ldots,u_{i-1}]=p\bot v_i$ and $f(v_i)=u_i$

Finally, $a\cdot^f b$ is defined and equal to $c$ if there is an $f$-dialgue between $a,b$ with $a\cdot [b,u_0,\ldots,u_{n-1}]=p\top c$.

(Here $p$ is the pairing combinator and $\bot,\top$ are the coded booleans, e.g. $ki$ and $k$.)

Van Oosten's paper contains a proof that the resulting structure has the $S$ and $K$ combinators, but it doesn't, that I can see, contain a proof that this new operation is actually functional; that is, that there is at most one such $c$ in the above definition. The doubt that springs to mind is I can see no reason to think there wouldn't be an $f$-dialogue $[w_0,\ldots,w_k]$ such that $a\cdot[b,w_0,\ldots,w_k]=p\top d$ with $d\neq c$, possibly even with the $w_j$ just being the $u_i$ in a different order.

I'm guessing that my concern, and not Jaap, is incorrect. So how do I see that there is at most one $c\in A[f]$ with $a\cdot^f b=c$? Have I accidentally misconstrued some part of the definition?

答:
The answer is that $f$-dialogues of length $n$ between a pair of elements, if they exist, are unique.

Let $[u_0,\ldots,u_{n}]$ and $[w_0,\ldots,w_n]$ be $f$-dialogues between $a$ and $b$. Say the two sequences agree up to the $(i-1)$-th coordinate (where . This means that $$a\cdot [b,u_0,\ldots,u_{i-1}]=a\cdot[b,w_0,\ldots,w_{i-1}]=p\bot v_i$$ for some $v_i\in\operatorname{dom}(f)\subseteq A$ by the defintion of $f$-dialogues. But then by definition the $i$-th coordinate of either dialogue must be $f(v_i)$, so they also agree on the $i$-th coordinate; and by an inductive argument, every coordinate must be the same.


问:
Is this way to prove that union of powersets is not equal to powerset of union correct?
I am trying to disprove this statement:

P(C) ∪ P (D) = P(C ∪ D)

Union of powersets is equal to powerset of union

and here is my proof:

```<code>Let C = {2}, D = {3}

P(C) = {{}, {2}}

P(D) = {{}, {3}}

P(C) ∪ P (D) = {{}, {2}, {3}}

P(C ∪ D) =  {{}, {2}, {3}, {2, 3}}

P(C) ∪ P (D) ≠ P(C ∪ D)
</code>```

Is there another way to proof this? Thank you

答:
if $P(C) \cup P(D)=P(C \cup D)$ then $P(C) \cup P(D) \subseteq P(C \cup D)$ and $P(C \cup D) \subseteq P(C) \cup P(D)$

So , you must also know that if $(X \subseteq Z )\land (Y \subseteq Z) $ then $(X \cup Y) \subseteq Z$

So , using thtis property , we can show that $P(C) \cup P(D) \subseteq P(C \cup D)$

but $P(C \cup D) \subseteq P(C) \cup P(D)$   is $NOT $ always true because we know that each set is subset of itself ,i.e,   $P(C \cup D)$ includes the subset constituted by the combination of whole elements but  $P(C) \cup P(D)$ cannot contain the subset which is constituted by all elements.


问:
$ 3^2 + 4^2 = 5^2 $. Are there more of these equations?
I found this equation and it actually makes sense once you figure it out:
$$ 3^2 + 4^2 = 5^2 $$
$$ 9 + 16 = 25 $$
$$ 25 = 25 $$

Are there any more of these kinds of equations?

Here's a list I made of all I can find:

$ 3^2 + 4^2 = 5^2 $

$ 12^2 + 5^2 = 13^2 $

答:
Although this isn't quite what you asked, it's interesting that
$$
3^2+4^2=5^2
$$
is the only 'consecutive' Pythagorean triple. In other words, there are no other positive integer solutions to
$$
(n-1)^2+n^2=(n+1)^2 \, .
$$
This can be proven in the following way:
\begin{align}
(n^2-2n+1)+n^2&=(n^2+2n+1) \\
2n^2-2n+1&=n^2+2n+1 \\
n^2-2n+1&=2n+1 \\
n^2-4n+1&=1 \\
n^2-4n&=0 \\
n(n-4)&=0 \\
n&=0 \text{ or }n=4 \, .
\end{align}
While $(-1)^2+0^2=1^2$ does satisfy the equation, it is usually ignored because we are trying to find triples that can form the lengths of the sides of a triangle. Thus, we are left with $3^2+4^2=5^2$.


答: It may be a terminological issue, by rotations I do not mean all rotation positions of the sphere, I mean rotations around an axis.</p>

Title: concatenation of rotations in spherical coordinates

Body: Any rotation in $R{^3}$ can be expressed by a point $(\alpha,\beta)$ on the sphere in spherical coordinates. Is there any way to compute the concatenation of two rotation in spherical coordinates without translating back and forth to cartesian?


问:
Are the two sets equal?
Is the following statement True for any sets A and B?
$$(A\setminus B)^\complement= A^\complement\setminus B^\complement$$

Attempt:

Let $x\in (A\setminus B)^\complement$. Then, $x\notin A\setminus B$. So, $x\notin A$ and $x\in B$.

Let $x\in A^\complement\setminus B^\complement$. Then, $x\in A^\complement$ and $x\notin B^\complement$. So, $x\notin A$ and $x \in B$.

So, this is a true statement. But the Venn diagram I drew says different. So maybe the attempt was wrong.

Note: $A^\complement$ is complement of $A$ and $A\setminus B$ is the set difference between $A$ and $B$.

答:
A counter example:

Let $A, B \subset X$. and let $A=B;$

LHS: $(A$ \ $ A) ^c=(\emptyset) ^c=X;$

RHS:: $A^c $ \  $A^c=\emptyset;$

A contradition

Title: Are the two sets equal?


问:
Fourier transform wrt spatial variable of a distribution
Let $T\in\mathcal{S}'(\mathbb{R}^{n+1})$ be a tempered distribution, i think that:
$$ \mathbb{R}^{n+1}=\{(x,y):x\in\mathbb{R}^n,y\in\mathbb{R}\}.$$
How i can define the Fourier transform of $T$ wrt $x$? There is way to make this definition rigours? Any help is appreciated:

答:
You can define the Fourier transform for only one variable as following:

If $T$ is a tempered distribution, $\mathbb{F}_yT$ is the only distribution such that for any $\psi \in \mathcal{S}(\mathbb{R}^{n+1})$ we have:

$$\langle\mathbb{F}_yT,\psi\rangle = \langle T,\mathbb{F}_y\psi\rangle$$

Where on the right hand side $\mathbb{F}_y\psi$ is the usual fourier tranform in only $y$ of $\psi$.


问:
Minimal polynomial and Fredholm operator.
Let $X$ be a Banach space and let $m_{T}$ be the minimal polynomial of $T \in\mathcal{L}(X)$.
We have $$\sigma_{e}(T)=\{\lambda \in \mathbb{C}, \lambda-T \not \in \Phi(X)\}.$$
Where, $\Phi(X)$ is the set of Fredholm operators on $X$.
Do we have this inclusion :
$$\sigma_{e}(T)\subset \{\lambda \in \mathbb{C}: m_{T}(\lambda)=0 \}.$$

答:
The answer is affirmative, but please see the comments below.  To prove it let us first show the following result.

Lemma. Let $A$ be a unital complex algebra and let $a\in A$.  Suppose that $p$ is a polynomial in   $\mathbb C[X]$ such
that $p(a)=0$.  Then the spectrum of $A$, namely
$$
  \sigma (a) = \{\lambda \in \mathbb C: a-\lambda  \text{ is not invertible}\},
  $$
is contained in the set of  roots of $p$.

Proof.  Given $\lambda $ in $\sigma (a)$, suppose by contradiction that $p(\lambda )\neq 0$.  Observig that the polynomial $p(\lambda )-p(z)$ vanishes on $\lambda $, we may
write
$$
  p(\lambda )-p(z) = (z-\lambda )q(z),
  $$
for some polynomial $q$.  Substituting $z$ for $a$ we get
$$
  p(\lambda ) = (a-\lambda )q(a),
  $$
and hence also
$$
  1 =  (a-\lambda )\frac{q(a)}{p(\lambda )},
  $$
which implies that $a-\lambda $ is invertible,  contradicting the fact that $\lambda $ is in $\sigma (a)$.  QED

The next  question is how to define the minimal polynomial of an operator $T$ acting on an infinite dimensional Banach space.  Mimicking the finite matrix definition I'd say the most natural one is to let $m_T$ be the generator of the ideal formed by all polynomials $p$  in $\mathbb C[X]$ such that $p(T)=0$.

Of course the main trouble is that it is impossible to define the characteristic polynomial in infinite dimensions and hence there is no chance
of even stating the Cayley-Hamilton Theorem.  In other words, more often than not, only the zero  polynomial will vanish
on $T$, in which case $m_T=0$.  If so,  the question of whether or not
$$
  \sigma_{e}(T)\subseteq \{\lambda \in \mathbb{C}: m_{T}(\lambda)=0 \}
  $$
has an uninteresting, trivial affirmative answer.

In the rare instances that some nonzero polynomial vanishes on $T$, meaning that $m_T\neq 0$, then $m_T$ also vanishes on
the image of $T$ under the quotient map
$$
  \pi:\mathcal L(H) \to \mathcal L(H)/  \mathcal K(H),
  $$
where $\mathcal K(H)$ is the ideal of compact operators.

By the lemma we then have that
$$
  \sigma(\pi (T))\subseteq \{\lambda \in \mathbb{C}: m_{T}(\lambda)=0 \}.
  $$
Combining this with the fact that   $\sigma_e(T) =\sigma(\pi (T))$, we have arrived at an affirmative answer.


问:
Evaluating sums and integrals using Taylor's Theorem
Taylor's theorem states that

$$f(x)-\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt $$

We can use this to evaluate integrals. For example, consider $f(x)=\frac{b!x^{b+n+1}}{(b+n+1)!}$. This has $f^{(k)}(0)=0$ for $k\leq n$ and $f^{(n+1)}(x)=x^b$. Hence, by Taylor's theorem,

$$f(1)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!} = \int_0^1 \frac{f^{(n+1)} (t)}{n!} (1 - t)^n \, dt $$
$$\implies \frac{b!}{(b+n+1)!}=\frac1{n!}\int_0^1t^b(1-t)^n\,dt$$
$$\implies\beta(b+1,n+1)=\int_0^1t^b(1-t)^n\,dt=\frac{b!\, n!}{(b+n+1)!}$$

So we have determined an explicit expression for the Beta function very efficiently using Taylor's theorem.

Taylor's theorem could also be used to evaluate partial sums using knowledge of the infinite sum; for example, considering $f(x)=\sum\limits_{k=0}^\infty kx^k=\frac x{(1-x)^2}=\frac1{(1-x)^2}-\frac1{1-x}$ for $|x|<1$, and noting that $f^{(n+1)}(x)=(n+2)!(1-x)^{-(n+3)}-(n+1)!(1-x)^{-(n+2)}$, we can evaluate its partial sum as

$$\sum_{k=0}^nkx^k=\frac x{(1-x)^2}-(n+2)(n+1)\int_0^x(1-t)^{-(n+3)} (x - t)^n \, dt$$ $$-(n+1)\int_0^x(1-t)^{-(n+2)} (x - t)^n \, dt ={\frac {x ( 1-{x}^{n} ) }{ \left( 1-x \right) ^{2}}}-{
\frac {n{x}^{n+1}}{1-x}}
$$
after evaluating the integrals and simplifying. In fact, this holds for all $x$ except $x=1$, and if a limit towards $1$ is taken, we correctly obtain $\frac{n(n+1)}{2}$.
What other sort of sums and integrals can be evaluated using Taylor's Theorem? Is this an effective method for summing or integrating? Are there certain limitations we should be aware of with this method?

答:
For what concerns the first question, it would be a nice exercise trying to evaluate the following integral
$$\int_{0}^{1}x^x\textrm{d}x.$$
For what concerns the second question, there are different necessary and sufficient conditions actually i.e. If you are integrating over an interval, the uniform convergence of Taylor series is sufficient to expand the function and interchange integral with series.


问:
Formal Definition of Summation over a Set
Formal definition of summation over a sequence is very clear. Suppose we have a sequence $\left\{a_{i}\right\}_{0}^{\infty}$. Then the summation of this sequence can be defined as a new sequence $\left\{ b_i \right\}_0^\infty$ in a recursive way:

$$b_0 = a_0,\ b_{i+1} = b_{i} + a_{i+1},\ i \geq 0.$$

Then we may write
$$\sum_{i=0}^n a_i = b_n,$$

and

$$\sum_{i=0}^\infty a_i = \lim_{i\to\infty} b_i.$$

However, sometimes we see summation over a set. For example, in defining the determinant, we use the following notation:

$$\det(A) = \sum_{\sigma\in S_n} \left(\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma_i}\right),$$

where $S_n$ is the set of all permutations of the sequence $\left\{1,2,\dots,n\right\}$. How are summation symbols over sets like $\sum_{\sigma\in S_n}$ formally defined?

答:
Assume that $\lvert S \rvert = n$. Then there is a bijection between the set $\left\{1,2,\dots,n\right\}$ and $S$. That is, there exists a mapping $f: \left\{1,2,\dots,n\right\} \mapsto S$ and the inverse $f^{-1}: S \mapsto \left\{1,2,\dots,n\right\}$ exists. Then we may write
$$\sum_{s \in S}h_{s} = \sum_{i=1}^{n}h_{f_{i}},$$ where $h: S \mapsto \mathbb{R}$ is the sequence we'd like to sum. This definition incorporates the definition of determinants, as a bijection between permutations of the sequence $\left\{1,2,\dots,n\right\}$ and a subset of natural numbers always exists.


问:
How do I find a plane that is tangent to two given spheres and passes through a given point?
My problem is the following:
Given two spheres: $$(x-6)^2+(y-1)^2+(z+1)^2=1$$ and $$x^2+(y-5)^2+(z-4)^2=9$$ find a plane that is tangent to both of those spheres and passes through the point $$(5;2;0)$$
I tried plugging all the information in formulas but then I get a scary-looking system of equation which I don't know how to solve.
My problem book says that the correct answer is $$x+2y+2z-9=0$$ or $$y-2=0$$
Whatever I did I couldn't get the answer.

I would really appreciate any help.
Thanks in advance!

答:
Here is a solution that, IMHO, is in the spirit of the question.

Working on the expanded version of the spheres' equations, the generic equations of planes tangent to these spheres are resp.:

$$\begin{cases}xx_1+yy_1+zz_1-6(x+x_1)-(y+y_1)+(z+z_1)+37&=&0\\xx_2+yy_2+zz_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
 -5(y+y_2)-4(z+z_2)+32&=&0\end{cases}\tag{1}$$

where $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are the contact points on the resp. spheres. Therefore, we have already the two equations

$$\begin{cases}x_1^2+y_1^2+z_1^2-12x_1-2y_1+2z_1+37&=&0\\x_2^2+y_2^2+z_2^2-10y_2-8z_2+32&=&0\end{cases}\tag{2}$$
for unknowns
$x_1,y_1,z_1,x_2,y_2,z_2$.

Let us express that the two equations (1) describe the same plane by writing the proportionality of coefficients of $x,y,z$ and constant terms in (1):

$$\dfrac{x_1-6}{x_2}=\dfrac{y_1-1}{y_2-5}=\dfrac{z_1+1}{z_2-4}=\dfrac{-6x_1-y_1+z_1+37}{-5y_2-4z_2+32}\tag{3}$$

yielding 3 more equations.

We have at present 5 equations for 6 unknowns. It is enough to obtain the general equation of a plane tangent to both $S_1$ and $S_2$ (remember that the equation of a plane is defined up to a multiplicative constant).

But we have the supplementary fact that $(5,2,0)$ verifies the two equations, giving two more equations!

$$\begin{cases}5x_1+2y_1-6(5+x_1)-(2+y_1)+z_1+37&=&0\\5x_2+2y_2-5(2+y_2)-4z_2+32&=&0\end{cases}\tag{4}$$

We are in the unusual situation of 7 (compatible, we wish so) equations for 6 unknowns...

I leave you the computations... that can be easily handled using a CAS (Computer Algebra System, which, honestly, is the future of this kind of tedious final computations).


问:
Intuitive interpretation of the Laplacian Operator
Just as the gradient is "the direction of steepest ascent", and the divergence is "amount of stuff created at a point", is there a nice interpretation of the Laplacian Operator (a.k.a. divergence of gradient)?</p>

答:
Although this thread is quite old, I think a quick picture would add up to the excellent answers here. An explanation is also below. I've tried to be as non-technical as I could get, since this would also add up intuition to the other more technical answers. Anyway, let's see how come that the Laplacian is a local averaging operator.

So, let's imagine the points $A$, $B$, $C$, $D$ and $O$ are some points on a metal sheet and we have an associated temperature at each (call it $a$, $b$, $c$, $d$, $o$).

Cool.

Now we want to learn something about how this temperature is distributed on the sheet and specifically on these five points.
Specifically, how fast the temperature changes from $A$ to $O$? Well, this should be the total change in temperature from $A$ to $O$ (call it $o-a$) divided by the distance from $A$ to $O$ (call it $h$, and the distance from $A$ to $C$ - $2h$). Thus, as we are going from $A$ to $O$, the temperature changes with a rate of $(o-a)/h$ per unit distance. Similarly, we can calculate $(b-o)/h$, $(c-o)/h$ and $(o-d)/h$. For brevity, let these be $a', b', c'$ and $d'$.

What these tell us is whether the temperature is increasing in a given direction (and how fast). For example, if $a'>0$ then the temperature at $O$ is higher than the temperature at $A$, and it is increasing as we go from $A$ to $O$ at a rate of $(o-a)/h$. Obviously, as $h$ tends to zero, this becomes the first derivative.

...but we can do more.
What if we wanted to know how the change of temperature changes as we move on the metal sheet? In other words, the "second derivatives"? (Formally, we are still "discrete", so these are not exactly derivatives, but let's call them so for convenience.)

Now let's find this for the temperature between $A$ and $C$ by taking the average change of $a'$ and $c'$: $$\frac{(a'+b')}{2h} = \frac{1}{2}\frac{\frac{c-o}{h}-\frac{o-a}{h}}{h} = \frac{1}{2}\frac{a-2o+c}{h^2}$$
We can calculate the second derivative between $B$ and $D$ analogously.

This doesn't seem too big of a step toward the Laplacian, but let's continue by adding what happens in the $y$ direction. Which is, let's add the second derivative in the y-direction to the second derivative in the x-direction, giving: $$\frac{1}{2}\frac{a-2o+c}{h^2}+\frac{1}{2}\frac{b-2o+d}{h^2}=\frac{1}{2h^2}(a+b+c+d-4o)$$
The last thing is the sum of the two second partial derivatives, which is - the Laplacian. Let's see what happens when it is equal to zero... $$\frac{1}{2h^2}(a+b+c+d-4o)=0$$discarding the $1/2h^2$ term: $$a+b+c+d-4o=0$$
...which gives:$$o=\frac{a+b+c+d}{4}$$
Wait... but this means that saying that the Laplacian being equal to zero is the same as saying that the temperature at $O$ is equal to the average of the temperatures at $A,B,C$ and $D$! Exactly as we wanted.


问:
Pullback in the top dimension
I am solving exercises from Loring Tu.

Show that if $L : V \rightarrow V$ is a linear operator on a vector space V of dimension n, then the pullback $L^{\wedge} : A_n(V) \rightarrow A_n(V)$ is multiplication by determinant of L.

Attempt:

This is a linear operator between the same spaces of the dimension 1. It follows that it must be multiplication by a constant. I don't understand why is it the determinant ?

答:
Choose a basis $v_1,\dots,v_n$ for $V$, and consider the linear map $\phi : \mathbb R^n \to V$ given by $\phi(a_1,\dots,a_n) = a_1v_1 + \cdots + a_nv_n$. Now, for $\omega \in A_n(V)$ define $\eta_\omega \in A_n(\mathbb R^n)$ by
$$\eta_\omega(x_1,\dots,x_n) = \omega(\phi(x_1),\dots,\phi(x_n)).$$
Since $\det \in A_n(\mathbb R^n)$ form a basis for $A_n(\mathbb R^n)$, there exists $c \in \mathbb R$ such that $\eta_\omega = c \det$. Evaluating in the standard basis for $\mathbb R^n$ we see that $c = \omega(v_1,\dots,v_n)$, and so we have
\begin{align} L^\vee(\omega)(v_1,\dots,v_n)
&= \omega(Lv_1,\dots,Lv_n) \\
&= \eta_\omega(\phi^{-1}(Lv_1),\dots,\phi^{-1}(Lv_n)) \\
&= \det(\phi^{-1}(Lv_1),\dots,\phi^{-1}(Lv_n)) \omega(v_1,\dots,v_n) \\
&= (\det L) \omega(v_1,\dots,v_n)
\end{align}
meaning that $L^\vee(\omega) = (\det L) \omega$.

Title: Pullback in the top dimension


问:
Difference equation - counting problem
I need to to define difference equation for following problem and solve that equation using generating function.

Border of length 10cm is made of small bricks (10cm long) and large bricks (20cm long). Small bricks come in 2 colours and large bricks in 3 colours. How many different borders of length 10n cm can be made.

I'm having problems defining difference equation for this. I've computed:

$u_{1} = 2$

$u_{2} = 5$

答:
$u_n$ should be the number of ways to make a border of length $n$.  We have $u_0=1$ as a starting value because there is one way to make a border of length $0$.  $u_1=2$ is not correct because there is no way to make a border of length $1$.  $u_2=2$ because there are $2$ ways to make a border of length $2$-you start with a border of length $0$ and have $2$ choices of a brick of length $2$.  In general, the way to make a border of length $n$ is to start with a border of length $n-2$ and add a length $2$ block (which you can do in $2$ ways) or start with a border of length $n-5$ and add a length $5$ block (which you can do in $3$ ways).  The result is
$$u_n=2u_{n-2}+3u_{n-5}$$

Title: Difference equation - counting problem


问:
Compactness of $ Lu = -\partial_x^2 u + u $
I am trying to apply the spectral theorem to the operator:
$$Lu = -\partial_x^2u + u $$
in $L^2([0,1])$ with domain $D(L)=\{u\in H^2([0,1]) s.t. \partial_xu(0)=\partial_xu(1)=0\}$.

I need to prove that $L$ is self-adjoint and compact. For the first point, denote $\langle \cdot,\cdot\rangle$ the standard inner product in $L^2$ and let $u,v\in D(L)$:
$$
\langle Lu,v\rangle = \int_0^1 (-\partial_x^2u)v + uv = \int_0^1 u(-\partial_x^2v) +uv = \langle u,Lv\rangle.
$$

Now for the compactness, I am no certain about my argument. Here is the definition of compact operator I have:

A bounded linear operator $K:X\rightarrow Y$ ($X,Y$ Banach spaces) is
called  compact if for any bounded sequence $(u_k)_k$ in $X$, there
exists a subsequence such that $\bigl(Ku_{\phi(k)}\bigr)_k$ converges
in $Y$.

In my case, $X=D(L)$ with the norm $\lVert u\rVert_{H^2}$, $Y=L^2([0,1])$. Thanks to the Reilich-Kondrachov theorem, the embedding $H^2([0,1])\subset L^2([0,1])$ is compact.

Then I use the fact that $L$ is continuous, i.e. $\forall u\in D(L)$:
$$
\lVert Lu\rVert_{L^2}^2=\int_0^1\lvert -\partial_x^2 u + u \rvert^2 \leq \lVert u \rVert^2_{H^2}
$$
to show that $\forall (u_k)_k$ bounded in $D(L)$, $\exists (u_{\phi(k)})_k$ converging in $L^2$ (compact embedding) and by continuity $Lu_{\phi(n)}$ converges in $L^2$.

Is my proof correct? In particular, I am always confused by which norm I should use at each step. Thank you.

答:
Yes your proof is correct, let's just walk real quick through the steps and the norm you are using:

Remember to show the operator $L$ is compact you want to show that if $\{u_k\}_{k}$ is bounded in the domain (here $X$ which has the $H^2$ norm) then you want to show that within $\{Lu_k\}_k$ has a convergent subsequence, convergent in its own space which is $Y$ with the $L^2$ norm. Since as you said by the Reilich-Kondrachov theorem, the embedding $H^{2}([0,1])⊂L^{2}([0,1])$ is compact, then the boundedness of $\{u_k\}_k$ in the $H^2$ norm is sufficient to show that $\{Lu_k\}_k$ has a convergent subsequence in the $L^2$ norm, thus your operator $L$ is compact.


问:
Mean value theorem for curves
Suppose we have a smooth non constant curve

$$
f(t) = (x(t),y(t))
$$

from $\mathbb{R}\to\mathbb{R}^2$ with $t\in(0,1)$. Suppose we have the limits at $0$ and $1$ of the curve is the same. That is

$$
\lim_{t\to 0} f(t) = \lim_{t\to 1} f(t) = \textbf{a}.
$$

Then is true that there exists a $t_0\in(0,1)$ such that $f’(t_0) = (c,c)$ for some $c>0$?

More generally, is it true that the tangent vector faces every direction at least once? This seems geometrically ok, but I couldn’t find a rigorous proof of this statement.

Thank you in advance.

答:
If $f([0,1])$ is an immersed closed curve (suitably defined --- you need $f'(1) = f'(0)$ --- or, in other words, $f$ should descend to a $C^{1}$  map of $S^{1}$), then the answer is yes.  Otherwise, it's false.

Suppose that $\Gamma = f([0,1])$ is an immersed closed curve.  Precisely, assume that $f(0) = f(1)$, $f'(0) = f'(1)$, and $f'$ is non-vanishing in $[0,1]$.  Up to redefining $f$, we can then assume that $f(t + 1) = f(t)$ and $f'(t + 1) = f'(t)$ and $f$ remains at least $C^{1}$.

Henceforth, given $e \in S^{1}$, denote by $e^{\perp} \in S^{1}$ the vector with $e \cdot e^{\perp} = 0$ and $(e,e^{\perp})$ right-hand oriented.

Pick $e \in S^{1}$.  Define $\ell_{e} : \mathbb{R}^{2} \to \mathbb{R}$ by $\ell_{e}(x) = \langle x,e \rangle$.  Since $\Gamma$ is compact, $\ell_{e}$ achieves its maximum at some point $x_{M} = f(t_{M})$ and its minimum at $x_{m} = f(t_{m})$.  Thus, an advanced calculus exercise shows that $D \ell_{e}(x_{0})$ is parallel to the "normal vector"* $f'(t_{m})^{\perp}$ and $f'(t_{M})^{\perp}$ to $\Gamma$ at $x_{m}$ and $x_{M}$.  This tells us that $f'(t_{M})^{\perp} = \pm e$ and $f'(t_{m})^{-} = \mp e$ since $x_{M}$ is a maximum and $x_{m}$, a minimum.

Since $e \mapsto e^{\perp}$ is a surjective map onto $S^{1}$, it follows that the tangent vector always points in each direction at least once.  (Note we proved that the normal vector always points in each direction at least once.)

A counter-example if $\Gamma$ is not immersed: take $f$ so that it traces out a square.  (This can be done smoothly by asking that $f$ stops and takes a break at each corner point.)  The normal vector (where defined) only takes on four values.  (If it's the coordinate square, then these four are $(1,0)$, $(-1,0)$, $(0,1)$, and $(0,-1)$ so you won't see $(1/\sqrt{2},1/\sqrt{2})$.)  The problem is $f'$ vanishes.

"Normal vector" in quotes since $\Gamma$ is immersed.  For example, if $\Gamma$ is a figure-eight, there is no well-defined normal vector at the intersection point between the two loops.  However, $f'(t)$ and $f'(t)^{\perp}$ are always well-defined and the non-vanishing condition on $f'$ ensures that, locally in $t$, we only see one of the possibilities.

Title: Mean value theorem for curves


问:
Limit of a integral with arctan function
Let the sequence $I_n$ , $n\ge 2$ and $$I_n=\int^n_0 x^n\arctan(x^\frac{1}{n})\,dx$$.
Calculate:
$\displaystyle \lim\limits_{n\to\infty}\frac{I_n}{n^n}$
I tried to divide the integral in two different ones , or to apply the Mean Value Theorem for Integrals but I didn't get to any point.

答:
Hint: Let $x=y^n.$ The integral becomes

$$\int_0^{n^{1/n}} y^{n^2}(\arctan y) ny^{n-1}\,dy$$ $$ \le (\arctan n^{1/n})n\int_0^{n^{1/n}} y^{n^2+n-1}\,dy.$$

For an estimate from below, integrate over $[1,n^{1/n}].$


问:
Equidistant points by modular arithmetic
If $\frac{a}{b}$ is an irreducible rational number, how does one show that the collection of points $\{ \frac{na}{b}\ \mbox{mod}\, 1\}$, $1\leqslant n \leqslant b$ are equidistant along the real line with distance $\frac{1}{b}$?

I’ve tried numerous examples to see how it works but I’m sure how to give a rigorous argument.

答:
Hint: scaling the fractions by $\,b\,$ yields all the multiples of $\,d := \gcd(a,b)\ $ [by Bezout], i.e.

$$b \underbrace{\!\!\left(\frac{a}b \Bbb Z + \Bbb Z\right)\!\!\!}_{\textstyle\left\{\frac{an}b\!\bmod 1\right\}}\ \, =\, a\Bbb Z + b\Bbb Z\, =\, \gcd(a,b)\, \mathbb{Z} =: d\:\!\Bbb Z\qquad$$

Thus $\,\left\{\dfrac{na}b\bmod 1\right\} =\left\{\dfrac{nd}b\bmod 1\right\} = \left\{ \dfrac{d}b,\ \dfrac{2d}b,\,\ldots,\, \dfrac{b}b\right\} =\, $ multiples of $\,\dfrac{d}b\bmod 1$


问:
Help me prove $K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$ where $K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$
Hi this problem is posed in Sean M. Stewart's How to Integrate It book pg.143

$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$$ where $ 0 \leq k < 1$

Based on the substitution:

$$ \tan{u} = \frac{\sin{2\theta}}{k+k\cos{2\theta}} =\frac{\tan{\theta}}{k}  $$

Show that:

$$K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$

$K(k) $ is some function defined in term of definite-integral called complete elliptic integral which is not easily computable, so attacking $K(k)$ gave no fruitful result.

Well then proceeding with substitution ( It is easy to lose track as it  involved a lot of trigonometric manipulations )

I reduced the integral into this form :
$$K(k) = k\int^{\pi/2}_0 \frac{\sec^2{\theta}d\theta}{\sqrt{k^2 + \tan^2{\theta}}\sqrt{k^2 + \tan^2{\theta} - k^2\tan^2{\theta}}}$$

Using substitution : $\tan{\theta} = p$

$$K(k) = k\int^{\infty}_0 \frac{dp}{\sqrt{k^2 +p^2}\sqrt{k^2 +p^2 - (kp)^2}}$$

To my naive eyes this looked manageable but WolframAlpha says otherwise.

Any hint or ideas on how to attack this problem?

I thought  the trig manipulations to find a more suitable form but in the I end up transform $\sin {u}$ term into $\tan{u}$ to use the given substitution

EDIT:  It has been suggested in commens that there is typo in the substituion(given in book) , it should be :

$$ \tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}} $$

答:
I think that you are on the right track. But you have to be aware that the completed elliptic integral of this type can be defined by any of three integrals:

$$
K(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2\theta}}\ d\theta\\=\int_0^1\frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\ dt\\
=\int_0^\infty \frac{1}{\sqrt{(1+t^2)[(1+(1-k^2)t^2)}]}\ dt
$$

This is from K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Springer . Chapter 61 covers the Complete Elliptic Integrals. Can you take it from here?


问:
Let $F(t)=\int_{0}^{\infty}\frac{\arctan(tx)-\arctan(x)}{x}dx$ Prove $F$ is $C^1(0,\infty)$ and find $F'(t)$

Let $F(t)=\int_{0}^{\infty}\frac{\arctan(tx)-\arctan(x)}{x}dx$

Prove $F$ is $C^1(0,\infty)$ and find $F'(t)$

I've thought that maybe I should prove that $f(t,x)=\frac{\arctan(tx)-\arctan(x)}{x}$ is riemann integrable and therefore it's lebesgue integrable on $(0,\infty)$ and then use differentiation under integral (in the real analysis sense) and to show $\frac{\partial f}{\partial t}$ is bounded.

答:
Fix $t>0$, for $\Delta t$ such that $|\Delta t|<\frac{t}{2}$, there is
$\theta\in(0,1)$ such that
$$ \arctan((t+\Delta t)x)-\arctan(tx)=\frac{1}{1+(t+\theta\Delta t)^2x^2}\Delta t x. $$
So
\begin{eqnarray}
&&\frac{F(t+\Delta t)-F(t)}{\Delta t}\\
&=&\int_{0}^{\infty}\bigg[\frac{\arctan((t+\Delta t)x)-\arctan(tx)}{x}dx\\
&=&\int_{0}^{\infty}\frac{1}{1+(t+\theta\Delta t)^2x^2}dx.\\
\end{eqnarray}
Since
$$ |t+\theta\Delta t|\ge t-\theta |\Delta t|>\frac{t}{2} $$
we have
$$ \bigg|\frac{1}{1+(t+\theta\Delta t)^2x^2}\bigg|<\frac4{4+t^2x^2}. $$
Since
$$ \int_{0}^{\infty}\frac4{4+t^2x^2}dx $$
converges, by the DCT,
\begin{eqnarray}
\lim_{\Delta t\to0}\frac{F(t+\Delta t)-F(t)}{\Delta t}=\int_{0}^{\infty}\lim_{\Delta t\to0}\frac{1}{1+(t+\theta\Delta t)^2x^2}dx=\int_{0}^{\infty}\frac{1}{1+t^2x^2}dx=\frac{\pi}{2t}\equiv f(t).
\end{eqnarray}
which impies $F(t)$ is differentiable and $F'(t)=f(t)$. Since $f(t)$ is continuous for $t>0$, we conclude $F(t)\in C^1(0,\infty)$.


问:
Solving Equation Using Algebraic Method
How to solve these equations using an algebraic method?
I need to show my working, don't you do something in reverse, like 7 multiplies by something. I haven't done it in class.

$$\dfrac{5(3y-4)}{2y}=7$$

答:
Multiply each side of the equation by $2y$ to get:

$$\dfrac{5(3y-4)}{2y}=7 \iff 5(3y - 4) = 7\cdot 2y = 14 y$$

Now, distribute, and then gather "like terms", and simplify:

$$
\begin{align} 5(3y - 4)  = 14 y & \iff 15y - 20 = 14y \\ \\
& \iff 15y - 14y = 20 \\ \\
& \iff y = 20.
\end{align}
$$

Title: Solving Equation Using Algebraic Method


答:
Use the substitution $t = \left(\frac{x}{n}\right)^{n+1}$

$$ \frac{I_n}{n^n} = \frac{n}{n+1}\int_0^1 \arctan\left(n^{\frac{1}{n}}t^{\frac{1}{n^2+n}}\right)\:dt \to \int_0^1 \arctan(1)\:dt = \frac{\pi}{4}$$

Title: Limit of a integral with arctan function

Body:
Let the sequence $I_n$ , $n\ge 2$ and $$I_n=\int^n_0 x^n\arctan(x^\frac{1}{n})\,dx$$.
Calculate:
$\displaystyle \lim\limits_{n\to\infty}\frac{I_n}{n^n}$
I tried to divide the integral in two different ones , or to apply the Mean Value Theorem for Integrals but I didn't get to any point.


问:
Alternative proof for "linear map from product of subspaces to a sum of subspaces is injective if and only if the sum is a direct sum"
For the theorem 3.77 in Axler's linear algebra done right:

Suppose that $U_1, U_2, \ldots U_m$ are subspaces of $V$. Define a
linear map $\Gamma: U_1 \times U_2 \times U_3 \ldots U_m \to U_1 + U_2
> \ldots + U_m$ by $\Gamma(u_1, u_2, \ldots u_m) = u_1 + u_2 + \ldots
> u_m$ Then $U_1 + U_2 + \ldots + U_m$ is a direct sum if and only if
$\Gamma$ is injective.

Is this a valid proof?

Proof that injective $\Gamma$ implies direct sum of subspaces

$\Gamma$ is also surjective

For each $u_1 + u_2 + \ldots u_m \in U_1 + U_2 \ldots U_m \ \exists (u_1, u_2, \ldots, u_m) \in U_1 \times U_2 \times U_3 \ldots U_m$. So $\Gamma$ is surjective.

$\Gamma$ is invertible

$\Gamma$ is both surjective and injective, so it is invertible.

Isomorphism of $U_1 \times U_2 \times U_3 \ldots U_m$ and $U_1 + U_2 \ldots + U_m$

Because $\Gamma$ is invertible, this is equivalent to saying these two spaces are isomorphic. But then, two finite dimensional spaces are invertible if and only if their dimensions match. The product of subspaces has a dimension $\sum_{i} \text{dim } U_i$. This is only possible for the sum if and only if $\text{dim } U_1 \cap U_2 \ldots \cap U_m = 0 \implies U_1 \cap U_2 \ldots \cap U_m = \{ 0 \}$. This implies that the sum of these subspaces is a direct sum. (this is the part I feel unsure about. Since Axler mentions a proof only for two subspaces, I'm not entirely sure if this is true. But I can see how it can be generalized by using $U_1 + U_2, U_3$ as the two subspaces and then imply the same for $m$ subspaces.)

Proof that the sum of subspaces is a direct sum implies injective $\Gamma$

I would start again with the dimensional implication. Since the dimension of the intersection of subspaces is zero, this means dimension of the sum of subspaces is $m$. This is equal to the dimension of the product of subspaces. This implies an isomorphism exists. Since $\Gamma$ is surjective and invertible, this implies it must also be injective.

答:
No, the condition $U_1 \cap \cdots \cap U_m = \{0\}$ does not imply that $U_1 + \cdots + U_m$ is direct, and, as you said, this only works when $m=2$. For example, in $\mathbb R^2$ take $U_1 = \{(x,0) : x \in \mathbb R\}$, $U_2 = \{(0,y) : y \in \mathbb R\}$ and $U_3 = \{(x,y) \in \mathbb R^2 : x=y\}$. Clearly $U_1 \cap U_2 \cap U_3 = \{0\}$ but $\mathbb R^2 = U_1 + U_2 + U_3$ is not direct since there are no a unique way of writting $(1,1)$ as a sum $u_1+u_2+u_3$ with $u_i \in U_i$ for $i=1,2,3$: $$(1,1) = (1,0)+(0,1)+(0,0) = (0,0)+(0,0)+(1,1).$$
I suggest you use the proposition 1.44:

Suppose $U_1,\dots,U_m$ are subspaces of $V$. Then $U_1 + \cdots + U_m$ is direct if and only if the only way to write $0$ as a sum $u_1+\cdots+u_m$, where each $u_j$ is in $U_j$, is by taking each $u_j$ equal to $0$.

and the proposition 3.16:

Let $T \in \mathcal L(V,W)$. Then $T$ is injective if and only if $\operatorname{null} T = \{0\}$.

Observe that $\operatorname{null} T = \{0\}$ is equivalent to saying that for each $v \in V$ the following holds: if $Tv=0$ then $v=0$.


答:
$C_1(6,1,-1);\;r_1=1$

$C_2(0,5,4);\;r_2=3$

A generic plane $\pi$ has equation
$$\pi:ax+by+cz+d=0$$
The distance from the plane to the center of the spheres must be equal to the respective radius. Furthermore the plane must pass through $P(5,2,0)$

$$\begin{cases}
\frac{|6a+b-c+d|}{\sqrt{a^2+b^2+c^2}}=1\\
\frac{|5b+4c+d|}{\sqrt{a^2+b^2+c^2}}=3\\
5a+2b+d=0&\\
\end{cases}$$

$$\begin{cases}
(6 a+b-c+d)^2=\frac{1}{9} (5 b+4 c+d)^2\\
(6 a+b-c+d)^2=a^2+b^2+c^2\\
d=-5a-2b\\
\end{cases}$$

Two solutions $a=0,b=-d/2,c=0;\;a=-d/9,b=-2 d/9,c=-2 d/9$

Two planes

$-d/2 y + d = 0\to y-2=0$

$-d/9 x -2d/9 y -2 d/9 z + d=0 \to x+2 y+2 z-9=0$

Title: How do I find a plane that is tangent to two given spheres and passes through a given point?

Body:
My problem is the following:
Given two spheres: $$(x-6)^2+(y-1)^2+(z+1)^2=1$$ and $$x^2+(y-5)^2+(z-4)^2=9$$ find a plane that is tangent to both of those spheres and passes through the point $$(5;2;0)$$
I tried plugging all the information in formulas but then I get a scary-looking system of equation which I don't know how to solve.
My problem book says that the correct answer is $$x+2y+2z-9=0$$ or $$y-2=0$$
Whatever I did I couldn't get the answer.

I would really appreciate any help.
Thanks in advance!


问:
Is there any known lower bound for the number of polynomia functions over a finite field with fixed number of variables?
Say we have a finite field $K$ of size $q$, is there an easy lower bound for the number of distinct polynomial functions in $K[x_1,x_2,\ldots,x_n]$? What about if we specifically fix the degree $d$, so that we only count those polynomial functions of degree $d$ on $n$ variables?

答:
The number of polynomials of at least degree $d$ is the number of homogeneous polynomials of degree $d$ with $n$ variables.

There are ${n+d\choose n}$ monomials of degree $d$, so the number of polynomials at most degree $d$ with $n$ variables is $$q^{n+d\choose n}$$


问:
Every irreducible polynomial $f\in\mathbb{F}_p[x]$ splits completely in $\mathbb{F}_{p^n}$
My question is why does every irreducible polynomial $f\in\mathbb{F}_p[x]$ of degree $n$ split completely in $\mathbb{F}_{p^n}$?

I know that since $f$ has degree $n$, we have $\mathbb{F}_p[x]/(f)\cong \mathbb{F}_{p^n}$, and I also know that $\mathbb{F}_{p^n}$ is the splitting field for the polynomial $x^{p^n}-x$, but I am having trouble connecting the dots.

答:
Let $\xi$ be a root of $f$ in $\mathbb F_{p^n}$. $\{1, \xi, \dots, \xi^{n-1}\}$ is a linearly independent family over $\mathbb F_p$. Therefore $f$ is the minimal polynomial $\mu_{\mathbb F_p, \xi}$ of $\xi$ over $\mathbb F_p$.

And as $\xi$ is a root of $q(x)=x^{p^n}-x$, $f$ divides $q$ which splits in $\mathbb F_{p^n}$. Consider the long division of $q$ by $f$ to be convinced by this statement.


问:
Showing some subset of a topological space is disconnected
It may be trivial but I am going to show that a set $S$ that has the following property is disconnected:

Consider a topological space $(X, \tau)$. $S$ is a subset of $X$ and $A$, $B$ are disjoint open subsets of $X$.
$$
S \subseteq A \cup B\\
S \cap A \neq \varnothing \land S \cap B \neq \varnothing
$$
Then, $S$ is a disconnected set.

My proof: $S = S \cap (A\cup B) = (S\cap A) \cup (S\cap B)$. $S$ is open in the subspace $S$ of $X$. Hence, $S \cap A$ and $S \cap B$ are open in $S$ and because $A$, $B$ are disjoint, so are they. $S$ is the union of two disjoint open subsets of $S$. Therefore, $S$ is disconnected.

Is this correct?

答: It is perfectly fine. Another way to prove it (less elegant though) is to assume by contradiction it is connected, then since it is in $A \cup B$ and $A \cap B = \emptyset$ then it must be in only one of them, contradicting the fact that $S \cap A \neq \emptyset$ and $S \cap B \neq \emptyset$.


问:
Problem involving the computation of the following integral
I was solving the past exam papers and stuck on the following problem:

Compute the integral $\displaystyle \oint_{C_1(0)} {e^{1/z}\over z} dz$,where $C_1(0)$ is the circle of radius $1$ around $z=0.$

Here,$z=0$ is a pole of order $1$ and so  Res$(f,0)=\lim_{z \to 0 } z f(z)=\lim_{z \to 0}e^{1/z}=?$ ,where $f(z)={e^{1/z}\over z}$.

Can someone point me in the right direction with some explanation?

答:
$z=0$ is an essential singularity because of the factor $\mathrm e^{1/z}$. You can compute the integral by transforming to $w=1/z$, which yields

$$
\oint_{C_1(0)}\mathrm e^{1/z}\frac{\mathrm dz}z=\oint_{C_1(0)}\mathrm e^w\frac{\mathrm dw}w=2\pi\mathrm i\;.$$


问:
how to write this linear transformation as a matrix?
I was trying to write this linear transformation as a matrix,
the exercise is the following:
let "f: $R^3 --> R^4$" be a linear transformation and Kerf = (x, y , z) in $R^3$: $x-2y+z = 0$  is its null space. Determine the canonical matrix.

I know how to solve this type of exercises (generally), as follow:

I use the basis of the first vector space (R3 in this case),

I plug x, y, z, in the equations of the linear transformation,

then, I'm ready to use each component of the vectors (of R4, in this case) to create the matrix. (the canonical matrix has basis of the entire space, for example, 1 0 0, 0 1 0, 0 0 1).

But, when I'm about to solve these exercises, I simply cannot solve them. I don't know why, because I know how to do them theoretically.

Can you help me to solve this exercise?

答: I'll give you a hint or perhaps a topic you will want to understand better. What is the vector corresponding to your transformation (x-2y+z) and why?</p>


答:
This answer is motivated by what Peter wrote in his answer. I realize now that when I was thinking about this question, I was thinking of something along the lines of intermediate value theorem, but didn't know how to apply it for a curve in $\mathbb{R}^2$.

Suppose that $f'(t)$ be a continuous non zero function of $t$. Choose a unit vector $e_1$ and let $e_2$ be perpendicular to $e_1$. Then it follows that $t\mapsto \langle e_2, f'(t)\rangle$ is a continuous function of $t$. Now if $\langle e_2, f'(t)\rangle|_{t=0} < 0$ and $\langle e_2, f'(t)\rangle|_{t=1} > 0$, we can apply the intermediate value theorem for the function $t\mapsto \langle e_2, f'(t)\rangle$ and get the required result.

Title: Mean value theorem for curves

Body:
Suppose we have a smooth non constant curve

$$
f(t) = (x(t),y(t))
$$

from $\mathbb{R}\to\mathbb{R}^2$ with $t\in(0,1)$. Suppose we have the limits at $0$ and $1$ of the curve is the same. That is

$$
\lim_{t\to 0} f(t) = \lim_{t\to 1} f(t) = \textbf{a}.
$$

Then is true that there exists a $t_0\in(0,1)$ such that $f’(t_0) = (c,c)$ for some $c>0$?

More generally, is it true that the tangent vector faces every direction at least once? This seems geometrically ok, but I couldn’t find a rigorous proof of this statement.

Thank you in advance.


问:
What's the sign of $\det\left(\sqrt{i^2+j^2}\right)_{1\le i,j\le n}$?
Suppose $A=(a_{ij})$ is a $n×n$ matrix by $a_{ij}=\sqrt{i^2+j^2}$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it？

答:
The beautiful idea of @fedja: (also used by @Swanson:) is to show that
$$\sum_{i,j} \sqrt{\lambda_i+\lambda_j}\, x_i x_j$$
is negative definite on the subspace $\sum x_i = 0$.  Here is another way to
check this. We have the integral formula

$$\sqrt{\lambda} = \frac{1}{2 \sqrt{\pi}} \int_{0}^{\infty}\frac{1 - e^{-\lambda s}}{s^{3/2}} ds$$
so we can write
$$\sum \sqrt{\lambda_i+\lambda_j}\, x_i x_j = \frac{1}{2 \sqrt{\pi}}\int_{0}^{\infty} \frac{\sum_{i,j} (1- e^{-(\lambda_i+\lambda_j) s})x_i x_j} { s^{3/2}} d\,s $$
But now observe that since $\sum x_i = 0$ we also have $\sum_{i,j} x_i x_j=0$, so the numerator in the integral equals
$$-\sum _{i,j}  e^{-(\lambda_i+\lambda_j) s}x_i x_j= - (\sum_i e^{-\lambda_i s} x_i)^2 \le 0$$
so the integral is $\le 0$. But since the functions $(e^{-\lambda s})_{\lambda}$ are a linearly independent system, the integral is is fact $<0$ if not all the $x_i$ are $0$. Therefore, the quadratic form is negative definite on $\sum_i x_i = 0$

Now, using Cauchy interlacing theorem we conclude that the matrix $(\sqrt{\lambda_i+\lambda_j})$ has one positive eigenvalue and the rest negative. ( or: there is no $2$ subspace on which this form is $\ge 0$, since any such subspaces intersects $\sum x_i=0$ non-trivially).

This can be generalized, using other Bernstein functions, for instance $\lambda\mapsto \lambda^{\alpha}$  ($0<\alpha < 1$), since we have the integral representation
$$\lambda^{\alpha} = \frac{\alpha}{\Gamma(1-\alpha)} \int_{0}^{\infty}\frac{1- e^{-\lambda s}}{s^{\alpha+1}} d s$$

One can take a look at the book of Schilling et al -- Bernstein functions theory and applications

Note: from numerical testing, it seems that matrices of form $((\lambda_i +\mu_j)^{\alpha})$  ($0<\alpha<1$, $\lambda_i$, $\mu_i$ positive and ordered in the same way), have one eigenvalue positive and the rest negative. It would imply that the matrix $(-(\lambda_i +\mu_j)^{\alpha})$ is totally negative.

$\bf{Added:}$ Sketch of proof for the integral representation:

For $\beta> 0$ we have
$$\Gamma(\beta) = \int_{0}^{\infty} e^{-s} s^{\beta}\frac{d s}{s} = \int_{0}^{\infty}e^{-\lambda s} (\lambda s)^{\beta} \frac{d s}{s}= \lambda^{\beta} \int_{0}^{\infty}e^{-\lambda s} s^{\beta} \frac{d s}{s}$$
and so
$$\lambda^{-\beta} = \frac{1}{\Gamma(\beta)} \int_0^{\infty} e^{-\lambda s} s^{\beta-1} d s$$
Now integrate wr to $\lambda$ from $0$ to $\lambda$ and get the formula for $1-\beta = \alpha>0$.

$\bf{Added:}$  Let's show that for $\alpha$ not an integer, $(\lambda_i)$, $(\mu_i)$ two $n$-uples of distinct positive numbers, the determinant $\det((\lambda_i + \mu_j)^{\alpha})_{ij}$ is $\ne 0$. Now, the determinant is $0$ is equivalent to its columns being linearly dependent, that is
there exist $a_i$ not all zero such that the function
$$\sum a_j (x+\mu_j)^{\alpha}$$
is zero at the points $x= \lambda_i$.
Let us show by induction on $n$ that such a function  $\sum_{j=1}^n a_j (x + \mu_j)^{\alpha}$ cannot have $n$ positive zeroes.

For $n=1$ it's clear. Assume the statement is true for $n-1$ and let's prove it for $n$.

Write $f(x) =\sum a_j (x + \mu_j)^{\alpha}$. We may assume (a shift in argument $x$) that the smallest $\mu_1 =0$.  Now the positive zeroes of $f(x)$ are the positive zeroes of
$$\frac{f(x)}{x^{\alpha}} = a_1 + \sum_{j>1} a_j (1 + \frac{\mu_j}{x})^{\alpha}=a_1 + \sum_{j>1} a_j \mu_j^{\alpha}\, (\frac{1}{\mu_j} + \frac{1}{x})^{\alpha} $$
So now we get another function $g(x) = a_1 + \sum b_j( x+ \xi_j)^{\alpha}$ that has $n$ positive roots, so its derivative will have $n-1$ positive roots. But the derivative is a sum of $n-1$ terms (involving the exponent $\alpha-1$). Now apply the induction hypothesis.

Note: with a bit more care, we  show that the determinant $\det ((\lambda_i + \mu_j)^{\alpha})$ is $\ne 0$ whenever $(\lambda_i)$, $(\mu_i)$ are $n$-uples of distinct positive numbers and $\alpha\not = 0, 1, \ldots, n-2$. Its sign will depend only on $n$, $\alpha$  (perhaps the integral part of $\alpha$ ) ( and the ordering of the the $\lambda_i$, $\mu_i$).

An interesting problem is to determine this sign. We understand the case $0< \alpha<1$ (and  $\alpha < 0$).  Perhaps $\alpha \in (1,2)$ next?  The determinant $\det((\lambda_i + \lambda_j)^{\alpha})$ should be negative if $n\ge 2$.


问:
Show map is not a isomorphism
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)

Show that $f:U \rightarrow K^2$ is not a isomorphism

My try

I have only been told what a isomorphism is in a general sense; that it has to be "the same" i.e some structure is still the same. For vectors I guess it could be addition, multiplication and the null vector? I am thinking $f$ must be bijective as well. Other than that I have not been taught much more than "it is the same" – which really does not help me in the slightest.

答:
Yes, what you have been thinking of, is correct.
Here, the map $f$ is between the vector space $V$ and the field $K^2$, hence, we are asked to show that $f$ is an isomorphism of vector spaces.
(I leave it to you to check that $V$ is indeed a vector space)

Now, an isomorphism between vector spaces is nothing but an injective and a subjective map, i.e. a bijection between them. Here, it is enough to show that it is not either one of them.

Hint: Show that $f$ is not injective. Can you use the relation between $ x_1,x_2,$ and $x_3$ in the definition of $V$, to write the image of $f$ as something that gives you this?

Title: Show map is not a isomorphism


答: Note that $(1,1,0),(-2,0,1)\in V$, that $f(1,1,0)=(1,1)$, and that $f(-2,0,1)=(-3,3)$. Therefore, $f(-3,-3,0)=f(-2,0,1)\bigl(=(-3,3)\bigr)$ and so $f$ is not injective. It follows that $f$ is not an isomorphism.

Title: Show map is not a isomorphism

Body:
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)

Show that $f:U \rightarrow K^2$ is not a isomorphism

My try

I have only been told what a isomorphism is in a general sense; that it has to be "the same" i.e some structure is still the same. For vectors I guess it could be addition, multiplication and the null vector? I am thinking $f$ must be bijective as well. Other than that I have not been taught much more than "it is the same" – which really does not help me in the slightest.


问:
What is the algebraic structure of all the quadtrees under these operations?
I was implementing a quadtree in a programming language, and I realized I could define operations such as negation $\bar{x}$, reunion $a\cup b$ and intersection $a \cap b$ on these objects.

These operations have all the propreties of a boolean algebra, but the number of elements is not 2, in fact the set of all elements is not finite.

I think this is equivalent to this situation: you have shapes on the plane with an inside, an ouside and a border. I can take the negation of a shape (the inside became the outside), the intersection of 2 shapes and the reunion of 2 shapes.

To what algebric structure does the set of all possibles shapes under this 3 laws correspond to ?

答:
There is an algebraic structure also named "Boolean Algebra" which can be defined as a complemented distributive bounded lattice (See the definition here). The "Boolean Algebra" as engineers, programmers, etc. call it (the one with 2 elements) is a special case (indeed a very important one, see here) of this "Boolean Algebra" algebraic structure.

Note: In the definition linked above, there are two special elements of $A$ called $0$ and $1$. It is important, then, to emphasize in the definition that $A$ may contain more than just $0$ and $1$! These 2 special elements are just specified in the definition because they are used the axiom of complements $a\wedge\neg a=0$ and $a\vee\neg a=1$. In the shape analogy, $0$ and $1$ correspond to the "empty" shape, and the "whole plane" shape, respectively.


答: Your notion is actually correct. Observe that for all $x \in V, x_1 - x_2 + 2 x_3 = 0$ which implies  $x_1 - x_3 = x_2- 3 x_3 $ . Thus, this map is not surjective as well.

Title: Show map is not a isomorphism


问:
Proof of a property of conditional expectation
I'm new here and this is my first question.

I'm struggling with a proof for my statistics course.

I need to show the following proposition:

$E[Y^2|\bar{X}=\bar{x}]\geq[E(Y|\bar{X}=\bar{x})]^2$

I tried to show it using Jensen's inequality but my teacher told me that is not correct.

Thank you :)

答:
Hint:

This can be shown using $\textrm{Var}(Y \mid \bar X = \bar x) \ge 0$

So if you want more detail, expand and rearrange  $\mathbb E[(Y - \mathbb E[Y \mid \bar X = \bar x])^2 \mid \bar X = \bar x] \ge 0$, which is true since  it is the conditional expectation of a square term


问:
Reduce the radical:
From: Lumbreras Editors

So I proceeded:

$
 \left(\sqrt{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}+\sqrt[4]{6 }\right)^4=\left(\sqrt{\sqrt{\frac{(1 +\sqrt{3})^2}{2}}+\sqrt{\frac{(1 -\sqrt{3})^2}{2}}}+\sqrt[4]{6}\right)^4\\
  =\left(\sqrt{\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{1-\sqrt{3}}{\sqrt{2}}}+\sqrt[4]{6}\right)^4
    \left(\sqrt{\frac{2}{\sqrt{2}}} +\sqrt[4]{6}\right)^4 =\left(\sqrt[4]{2}+\sqrt[4]{2}\sqrt[4]{3}\right)^4\\
  =2\left(1+\sqrt[4]{3}\right)^4
$

What else can be done?

Title: Reduce the radical:

Body:
Let

$$x=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}$$

$$x^2=2+\sqrt{3}+2-\sqrt{3}+2\sqrt{2+\sqrt{3}}\times \sqrt{2-\sqrt{3}}=6$$

$$x=\sqrt6$$

$$
 (\sqrt{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}+\sqrt[4]{6 })^4=(2\sqrt[4]{6 })^4=96$$


问:
Understanding the proof of ${\rm ord}(G)=p^n \Rightarrow Z(G) \neq \{e\}$.
Let $p$ be a prime number and $G$ be a group. I have some troubles to understand the proof of ${\rm ord}(G)=p^n \Rightarrow Z(G) \neq \{e\}$. The proof is given as follow:

Observe the Group $G$ acting on itself by conjugation, then we have:

$$x \in Z(G) \iff Z_G(\{x\})=G \iff (G:Z_G(\{x\}))=1$$

Until here it is okay, but now it goes on with:
So $$x \neq Z(G) \Rightarrow p\, \mid \frac{{\rm ord}(G)}{{\rm ord}(Z_G(\{x\}))}.$$

Why we get this inclusion? The last is because of the lagrange theorem but why we take $x \neq Z(G)$?

Many thanks for some help.

答: By the class formula, $|G|=|Z(G)|+\sum_i [G:Z_G(x_i)]$. If $x\in G\setminus Z(G)$, its centralizer $Z_G(x)$ is a subgroup of $G$, so it has order $p^k$. So $[G:Z_G(x_i)]=p^{n-k}$, and thus if $|Z(G)|=1, |G|\equiv 1\pmod{p}$, a contradiction. Hence $Z(G)$ is a nontrivial p-subgroup of $G$.


问:
What does this problem mean? And why is C the correct answer and not B?
So I tried to solve this problem in the 1955 AHSME:

In order to pass $B$ going $40$ mph on a two-lane highway, $A$, going $50$ mph, must gain $30$ feet. Meantime, $C, 210$ feet from $A$, is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, then, in order to pass safely, $A$ must increase his speed by:

$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$

I do not understand some parts of the problem clearly, but I did try to solve the problem. So I came up with answer B. But the correct answer is C. After some research, C only makes sense if the cars are mathematical points, but if they have a volume, then C shouldn't be correct.

So, why is C the correct answer? And why isn't B the correct answer?

答:
Let $x$ be speed of $A$ at which it can pass safely. Then we can show that
$$\frac{210}{50+x} = \frac{30}{x-40}=t,$$ where $t$ is time of passing.
Solving this equation gives us $x = 55$, and if A was going $50$ mph, then answer is $55-50=5 \to C$


问:
Sierpinski Triangle as Finite Union of Dendrites
Can the Sierpinski Triangle be written as a finite union of dendrites?  If so, can it also be verified what the minimal number is (assuming you can't do it with just two)?

This is a small piece from a more sophisticated thread, but the particular example can possibly be solved by 'geometric cleverness' so I wanted to make it a separate question.  The main thread is:

Finite Unions of Dendrites

答:
I have probably horribly misunderstood the question but here goes:

if you consider a triangle fractal (https://i.stack.imgur.com/shOK1.gif) with depth 1 then there are 3 dendrites, depth two there are 6 dendrites, depth three there are 15 dendrites.

so the Sierpinski triangle is this triangle fractal with depth n=inf. The formula for the number of dendrites for a triangle fractal with depth n is:
$$ 3+3^1+3^2+3^3+...+3^{n-1}=-3(1-3^n)/2 $$

the limit as n->inf for the nth term diverges to infinity, so the Sierpinski triangle can't be written from a finite amount of dendrites.


答:
Use Frullani's Integral: https://en.wikipedia.org/wiki/Frullani_integral
$$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x}dx=(f(\infty)-f(0))\ln\frac{a}{b}$$
$$F(t)=\int_{0}^{\infty} \frac{\tan^{-1}(tx)-\tan^{-1}{x}}{x}dx=(\pi/2-0)\ln t=\frac{\pi}{2}\ln t$$
$$F'(t)=\frac{\pi}{2t}$$

Body:
Let $F(t)=\int_{0}^{\infty}\frac{\arctan(tx)-\arctan(x)}{x}dx$

Prove $F$ is $C^1(0,\infty)$ and find $F'(t)$

I've thought that maybe I should prove that $f(t,x)=\frac{\arctan(tx)-\arctan(x)}{x}$ is riemann integrable and therefore it's lebesgue integrable on $(0,\infty)$ and then use differentiation under integral (in the real analysis sense) and to show $\frac{\partial f}{\partial t}$ is bounded.


问:
How to determine whether a two-variable function is injective, surjective or bijective?
```<code>1.f(x,y) = x^2 +1
2.f(x,y) = x + y +2
from Z x Z to Z
</code>```

I have never learned how to determine the type of two-variable functions before, and they're quite confusing for me.
At first, I intended to pick tow random values to prove that the first function is not injective, but it has a second variable y, and I am not sure if same technique would work. Please explain to me, thank you

答:
The general way to prove injectivity or surjectivity is as follows (this is specialized to your case where $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and all variables are real):

To prove a function is injective: Assume that $f(x_1,y_1)=f(x_2,y_2)$ and prove that $x_1=x_2$ and $y_1=y_2$.

To prove that a function is not injective: Find an example of $(x_1,y_1)\not=(x_2,y_2)$ where $f(x_1,y_1)=f(x_2,y_2)$.

To prove a function is surjective: Pick any $z$ and construct a formula for $x$ and $y$ so that $f(x,y)=z$.

To prove a function is not surjective: Find an example of a $z$ and show that it is impossible for $f(x,y)=z$ for any $(x,y)$.


问:
Find the general solution of $x^3y^{\prime \prime \prime} + y = 0$
I tried to use the Frobenius technique, but I am unable to find the solution by infinite series.</p>

答: If $a_k:=x^k[y]$ then $x^k[x^3y^{(3)}+y]=a_kp(k),\,p(k):=k^3-3k^2+2k+1$. Each of the three roots of $p$ provides a linearly independent solution $y=x^k$. These span the $3$-dimensional vector space of solutions.


答:
Found your mistake!

$\sqrt{\frac{(1 -\sqrt{3})^2}{2}}\ne \sqrt{\frac{1-\sqrt{3}}{\sqrt 2}}$

Because the quantity under the square root in the RHS is negative

If you simplify like this

$\sqrt{\frac{(1 -\sqrt{3})^2}{2}}= \frac{\sqrt{3}-1}{\sqrt 2}$

everything works

Title: Reduce the radical:

Body:
From: Lumbreras Editors

So I proceeded:

$
 \left(\sqrt{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}+\sqrt[4]{6 }\right)^4=\left(\sqrt{\sqrt{\frac{(1 +\sqrt{3})^2}{2}}+\sqrt{\frac{(1 -\sqrt{3})^2}{2}}}+\sqrt[4]{6}\right)^4\\
  =\left(\sqrt{\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{1-\sqrt{3}}{\sqrt{2}}}+\sqrt[4]{6}\right)^4
    \left(\sqrt{\frac{2}{\sqrt{2}}} +\sqrt[4]{6}\right)^4 =\left(\sqrt[4]{2}+\sqrt[4]{2}\sqrt[4]{3}\right)^4\\
  =2\left(1+\sqrt[4]{3}\right)^4
$

What else can be done?


答: By construction, $\mathbb{F}_p[x]/(f)$ is an extension of $\mathbb{F}_p$ containing a root $a$ of $f$. Since $\mathbb{F}_p[x]/(f)\cong\mathbb{F}_{p^n}$, and every element of the latter field is a root of $x^{p^n}-x$, it follows that $f$ and $x^{p^n}-x$ both have $a$ as a root. But since $f$ is irreducible, we must have that $f|x^{p^n}-x$ and so all the roots of $x$ are roots of $x^{p^n}-x$, which in turn implies that all the roots of $f$ are in $\mathbb{F}_{p^n}$ and so $f$ splits there.

Title: Every irreducible polynomial $f\in\mathbb{F}_p[x]$ splits completely in $\mathbb{F}_{p^n}$

Body:
My question is why does every irreducible polynomial $f\in\mathbb{F}_p[x]$ of degree $n$ split completely in $\mathbb{F}_{p^n}$?

I know that since $f$ has degree $n$, we have $\mathbb{F}_p[x]/(f)\cong \mathbb{F}_{p^n}$, and I also know that $\mathbb{F}_{p^n}$ is the splitting field for the polynomial $x^{p^n}-x$, but I am having trouble connecting the dots.


答: The function $zf(z)=e^{1/z}$ also has a singularity at $0$; The laurent series of $f(z)$ is $\sum_{k=0}^\infty \frac{1}{k!}\frac{1}{z^{k+1}}=\frac{1}{z}+\sum_{k=1}^\infty \frac{1}{k!}\frac{1}{z^{k+1}}$. The residue is therefore $1$.</p>

Title: Problem involving the computation of the following integral

Body:
I was solving the past exam papers and stuck on the following problem:

Compute the integral $\displaystyle \oint_{C_1(0)} {e^{1/z}\over z} dz$,where $C_1(0)$ is the circle of radius $1$ around $z=0.$

Here,$z=0$ is a pole of order $1$ and so  Res$(f,0)=\lim_{z \to 0 } z f(z)=\lim_{z \to 0}e^{1/z}=?$ ,where $f(z)={e^{1/z}\over z}$.

Can someone point me in the right direction with some explanation?


问:
Probability to get 5 successes in at most 7 trials
The night before the party, you plan to rehearse the song you are to sing and would like to sing it perfectly 5 times before sleeping. Since you are a great singer, the probability of you singing it perfectly on each attempt is 5/6. Suppose your attempts are mutually independent, what is the probability that you'll rehearse the song at most 7 times?

Here's my approach:
\begin{align*} \text{Probability to get 5 successes in 5 tries} = 4C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^0\dfrac{5}{6} &\approx 0.402\\ \text{6 tries} = 5C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^1\dfrac{5}{6} &\approx 0.334\\ \text{7 tries} = 6C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^2\dfrac{5}{6} &\approx 0.167\\\\ \text{Probability to get 5 successes in at most 7 tries} &= 0.903 \end{align*}

Was my approach correct? May I also know what type of Probability Distribution is suitable to model this problem?

答:
If the problem is asking for the $5^{th}$ success in the $7^{th}$ trial then your approach is incorrect, continue reading for the correct solution. If the question is like the body "$5^{th}$ success in $7$ trials then what you did is correct in terms of choosing the binomial distribution, not the arithmetic, I think the arithmetic mistakes you did show some misunderstanding of the binomial distribution. You did the following:

\begin{align*}
        7C5\left(\dfrac{5}{6}\right)^5\left(\dfrac{1}{6}\right)^2 &\approx 0.2344\\
      +\ 7C6\left(\dfrac{5}{6}\right)^6\left(\dfrac{1}{6}\right)^1 &\approx 0.3907\\
      +\ 7C7\left(\dfrac{5}{6}\right)^7\left(\dfrac{1}{6}\right)^0 &\approx 0.2791\\
      &= 0.9042
    \end{align*}

However, you have two things wrong: your combinations and powers.

Your first equation seems correct to me (in case you meant $5$ successes in $7$ trials), for the second equation what you did gives the probability of $6$ heads in $7$ trials, but in your question you should calculate the probability of $5$ heads in $5$ trials + the probability of $5$ heads in $6$ trials+ the probability of $5$ heads in $5$ trials, so the second combination should be:

$\binom{6}{5}\cdot (\frac{5}{6})^5 \cdot \frac{1}{6}$

We used this combination as you have $5$ successes that can happen anywhere during $6$ trials.

Note that you should always have $(\frac{5}{6})^5$ since you always have $5$ successes, however, you should have $(\frac{1}{6})^{n-5}$ where $n$ is the total number of trials. Similarly, your third equation should be the one of $5$ success in $5$ trials which is:

$\binom{5}{5}\cdot (\frac{5}{6})^{5} \cdot (\frac{1}{6})^0=(\frac{5}{6})^5$

I will leave the numerical answers to you.

Moving to the second interpretation of the question, when you want to find the probability that the $k^{th}$ success happens in the $n^{th}$ trial you use the negative binomial distribution.

For example. the probability that the $5^{th}$success happens in the $6^{th}$ trial is:

$\binom{5}{4} \cdot (\frac{5}{6})^5 \cdot \frac{1}{6}$

We use the following combination as the first four successes can happen anywhere in the first five trials but the last success is for sure in the fifth trial so we don't include it in the combination as it has a fixed position at the end, can you continue the rest of the cases?