问:
Representing matrix exponential in terms of matrix powers via Cayley-Hamilton

Using the Cayley-Hamilton theorem, show that for a $2 \times 2$ matrix $A$, $$e^A = c_1 A + c_0 I$$ where $c_1$ and $c_2$ are constants.

This problem came to me absolutely from nowhere and I have no clue how to solve it. Basically the theorem tells that every square matrix satisfies its own characteristic equation. By this theorem, we can find the exponent of a square matrix and its inverse. But does it really give a clue how to find anything like $e^A$? Help me to solve this

答:
There are two parts to the question:

Lemma 1 For any polynomial $P,$ $P(A) = c_1(P) A + c_2(P)I.$

Proof sketch The right hand side is equal to the remainder when $P(x)$ is divided by $\chi(x),$ where $\chi$ is the characteristic polynomial of $A.$

Lemma 2 If $P_n$ is the $n$-th Taylor polynomial of $\exp,$ the coefficients  $c_{1, 2}(P_n)$ converge as $n$ goes to infinity.

You can go from here.


问:
Find ∂x/∂u if I know ∂u/∂x?
I have the following transformation

$\it{u=x^2-y^2}$

$\it{v=2xy}$

I want to calculate $\frac{∂x}{∂u}$ $(0,2)$ and I know that the transformation is bijective around the point $(1,1)$. The answer is $\frac{1}{4}$, but I am unsure how to tackle this problem. Help would be appreciated.

答:
Write $J$ for the Jacobian matrix
$$
J = \pmatrix{
\partial{u}/\partial{x} & \partial{u}/\partial{y} \\
\partial{v}/{\partial{x}} & \partial{v}/\partial{y}
}.
$$
You want to find the upper left entry of the inverse matrix to $J$.

ContentLicense: CC BY-SA 4.0

Title: Find ∂x/∂u if I know ∂u/∂x?


问:
Tiny doubt with square root
I have a small doubt. Is $\sqrt{x^2}= |x|$ true? For, put $x=-1\implies x^2=1$. Now, $\sqrt 1=\pm 1$, whereas $|1|=+1$ only. $x\in\mathbb{R}$

P.S. This doubt came @Daniel's answer from this post. To quote-

The problem appears when you take square roots, since it is not still true that $\sqrt{(-1)^2}=-1$, in fact, $\sqrt{x^2}= |x|$ so in this case $\sqrt{(-1)^2}=|-1|=1$, which is not equal to $x-2$ when $x=1$.

答: The definition of the square root symbol is to indicate the positive square root only, which is why we write $\pm \sqrt{x}$ when we want to indicate both answers. So $\sqrt{1} = 1$ only.

Title: Tiny doubt with square root


答:
Compute the Jacobian of the map $(x,y) \mapsto (u(x,y),v(x,y))$ at $(x,y)=(1,1)$.

$\frac{\partial x}{\partial u}$ is the entry $1,1$ of the inverse of that Jacobian.

The Jacobian is

$$
J = \begin{pmatrix}
\partial{u}/\partial{x} & \partial{u}/\partial{y} \\
\partial{v}/{\partial{x}} & \partial{v}/\partial{y}
\end{pmatrix} = \begin{pmatrix}
2& -2 \\
2 & 2
\end{pmatrix}
$$

and its inverse
$$\begin{pmatrix}
1/4& 1/4 \\
-1/4& 1/4
\end{pmatrix}
$$

Leading  to the desired result.

Title: Find ∂x/∂u if I know ∂u/∂x?

Body:
I have the following transformation

$\it{u=x^2-y^2}$

$\it{v=2xy}$

I want to calculate $\frac{∂x}{∂u}$ $(0,2)$ and I know that the transformation is bijective around the point $(1,1)$. The answer is $\frac{1}{4}$, but I am unsure how to tackle this problem. Help would be appreciated.


问:
If $|G|=n>1$, then $|{\rm Aut}(G)|\le\prod_{i=0}^k(n-2^i)$ for $k=[\log_2(n-1)]$.
This is Exercise 1.5.16 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

The Details:

The exercise comes with a hint:

Use Exercise 1.3.4.

On page 16, ibid., we have

Exercise 1.3.4: Let $d(G)$ be the smallest number of elements necessary to generate a finite group $G$. Prove that $\lvert G\rvert\ge 2^{d(G)}$.

Proof (sketch): Each element of a minimal generating set $\{g_1,\dots, g_{d(G)}\}$ of $G$ maps to two elements in $G$: itself and its inverse (assuming they are distinct), so

$$\begin{align}
\lvert G\rvert &\ge \left\lvert \{{\rm itself}, {\rm inverse}\}^{\{g_1,\dots, g_{d(G)}\}}\right\rvert\\
&=2^{d(G)}.
\end{align}$$ "$\square$"

(I'm aware that this sketch sweeps a lot under the rug. However, I think it illustrates the general idea.)

The Question:

On page 31, ibid., we have

Exercise 1.5.16: If  $G$ has order $n>1$, then $$\lvert {\rm Aut}(G)\rvert\le\prod_{i=0}^k(n-2^i)$$ for $k=[\log_2(n-1)]$.

Thoughts:

Each automorphism of $G$ is determined by what it maps each of $G$'s generating elements $g_\lambda$ (as above) to. This is a key concept, I'm sure.

My first guess was that the $2^i$ in the product corresponds to $2^{d(H)}$ for $H\le G$ generated by a subset of the $g_\lambda$. I don't understand where $k$ comes from if this is the case though.

My postgraduate research degree specialises in combinatorial group theory, the study of groups as given by their presentations, so I feel as if I ought to be able to do this exercise without too much help, although I don't work with automorphism groups often.

I'm interested in understanding this exercise, in particular, because I think it's rather beautiful, especially with the value of $k$.

Perhaps I might have to give Exercise 1.3.4 some more thought in order to do the exercise at hand.

Please help :)

答:
Hint:
You are right that this exercise plays a key role. If $g_1,\ldots,g_k$ are generators of the group $G$, and $f$ is a automorphism of $G$, then $f(g_1),\ldots,f(g_k)$ are also generators of $G$. Now consider. The element $f(g_1)$ we can choose by $n-1$ ways, the element $f(g_2)$ we can choose by $n-2^1$ ways. And so on.


答: I don't quite understand your reasoning after the question, therefore I'll state my point without addressing it. The standard notation for the real variable (which is the one used in user Daniel's comment) is that, given $\alpha\in\Bbb R$, $\sqrt\alpha$ is the non-negative real number $u$ such that $u^2=\alpha$, if such a number exist. Therefore the domain of $\sqrt\bullet$ is $[0,\infty)$ and its range is $[0,\infty)$ as well. It follows that $\sqrt{x^2}=x$ if $x\in[0,\infty)$ and $\sqrt{x^2}=-x$ if $x\in(-\infty,0]$.

Title: Tiny doubt with square root

Body:
I have a small doubt. Is $\sqrt{x^2}= |x|$ true? For, put $x=-1\implies x^2=1$. Now, $\sqrt 1=\pm 1$, whereas $|1|=+1$ only. $x\in\mathbb{R}$

P.S. This doubt came @Daniel's answer from this post. To quote-

The problem appears when you take square roots, since it is not still true that $\sqrt{(-1)^2}=-1$, in fact, $\sqrt{x^2}= |x|$ so in this case $\sqrt{(-1)^2}=|-1|=1$, which is not equal to $x-2$ when $x=1$.


问:
Proving Lyapunov unstability of a $C^3$ discrete dynamical system
In proving that if $f$ is a $C^3$ function from $\mathbb{R}$ to $\mathbb{R}$ and $f(p) = p$ and $f^{'}(p) = 1$ and $f^{''}(p) = 0$ and $f^{'''}(p) > 0$ then $p$ is a not Lyapunov stable fixed point of $f$, I have arrived at $(p,p+\delta) \subseteq W^u(p)$ for some $\delta > 0$. However, I can't prove the unstabillity. I know I should use the definition and have tackled with the problem but I can't solve it. Can anybody help? Thanks.

答:
So your map has a Taylor expansion of the form
$$
f(x)=p+1(x-p)+0(x-p)^2+\frac16f'''(y)(x-p)^3=x+\frac16f'''(y)(x-p)^3
$$
for some $y$ close to $p$. Since $f$ is of class $C^3$ and $f''(p)>0$, also $f''(y)>0$ for any $y$ sufficiently close to $p$. This implies that for $x>p$ sufficiently close to $p$ we have $f(x)>x>p$. This gives you the instability.

You should avoid writing $W^u(p)$ since there is no hyperbolicity in this case.


问:
Converting parametric equations to xy equation
I have these parametric equations:
$$
x=1+t \\
y=1-t
$$

How do I make this to the xy equation?

David

ContentLicense: CC BY-SA 3.0

Body: $\text{AD}$$\text{D}$ $\text{EM}$.


问:
Intuition for multiplying vectors by complex scalars
There is a very pleasing interpretation of multiplying a vector by a real scalar. I can ''stretch or shrink'' and also ''flip over the origin (in the case of negative scalars)''.

Is there some analogous way to think of scalar multiplication when the vector field is over $\mathbb{C}$ instead? Intuitively I feel like its the same idea, just with more freedom over where scalars can point the vector (in the same way that complex number multiplication rotates and stretches other complex numbers) but I can't seem to write down anything concrete.

答: In the real case scalar multiplication takes a vector $v$ to another vector $w$ which lies in the linear subspace spanned by the vector $v$, which is a line, thus the multiplication stretches or shrinks the vector but does not change the line it spans. Similarly in the complex case, scalar multiplication takes a vector $v$ to another vector $w$ which lies in the complex span of $v$, this is now a 2-dimensional real vector space, therefore your intuition is correct, multiplying by a complex scalar can stretch, shrink or rotate the vector (where rotations are only possible in the plane spanned by the inital vector). You can think of it this way: given any vector $v\in\mathbb{C}^n$ and a scalar $a+ib=\lambda\in\mathbb{C}$ then multiplication by $\lambda$ changes $v$ to the vector $v = av+ibv$. Thus we can see that we can stretch or shrink $v$, but also add a shrunken or stretched version of $iv$ which gives the rotation.


问:
How to prove $a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta$ has $2n$ different zeros, $\theta \in (0,2\pi)$.
$0 < a_0 < a_1 < \cdots < a_n$.

Prove that $a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta$ has $2n$ different zeros, $\theta \in (0,2\pi)$.

[Hint: First prove that $P_n(z)=a_o+a_1z+a_2z^2+\cdots+a_nz^n$ has $n$ zeros in unit ball $B(0,1)$.]

This is an assignment I copied from my textbook. It's in the section "The Argument Principle & Rouche Theorem".

Though I followed this hint, I still can't see how this would imply the desired conclusion. Help needed.

答:
The problem you have provided does not appear to be a simple application of Rouche's theorem or equivalent. I would be curious what the author's thoughts were for a proof. In particular, I believe the problem is equivalent to the following non-trivial theorem.

Theorem. If $0<a_0<a_1<\cdots<a_n$ and $p(z)=a_0+\cdots+a_nz^n$, then $$q(z):=z^{-n}p(z^{-1})+z^np(z)$$ has only simple zeros located on the unit circle.

The equivalency can be easily verified. Indeed, if we define $$c(z):=a_0+a_1\cos(z)+a_2\cos(2z)+⋯+a_n\cos(nz)\ \ \ ,$$ then the statement "$c(z)$ has $2n$ simple zeros in the interval $(0,2\pi)$" is equivalent to the statement that "$q(z)$ has $2n$ simple zeros on the unit circle". This is because $\cos(z)=(e^{iz}+e^{-iz})/2$, hence $$c(z)=e^{-inz}/2*q(e^{iz})\ \ \ .$$ Thus $c(z)$ has a zero in $(0,2\pi)$ if and only if $q(z)$ has a zero on the unit circle.

I would gladly welcome a simple proof of the above theorem with an application of Rouche's theorem or similar, but this appears to not be the case. In fact, if you generalize $p(z)$ to polynomials with zeros in the unit disc and throw in a rotation coefficient of $|a|=1$, the above theorem is precisely the complete characterization of all polynomials with zeros on the unit circle, something that was only established in 1995. See the first answer here To prove this complex polynomial has all zeros on unit circle, Chen (J. of Math. Anal. and Appl. vol 190, 714-724 (1995)).


问:
Understanding mathematics imprecisely
For a long time, it has been a complete mystery to me how any of my peers understood any math at all with anything short of filling in every detail, being careful about every set theoretic detail down to the axioms.  That's a slight exaggeration, but I certainly did much worse in courses where I attempted to replicate this myself by not reading every proof to save time.

It's only recently, and only within the field of probability theory, that I've developed the ability to do this myself.  I am currently following Grimmett's book on Percolation theory.  There are far too few details for someone of my level to fill it in completely, but I am getting more than nothing out of it.

Question 1: I would like to learn how to get even more out of such "incomplete" studying.

What tends to happen even now, and more before, is that as soon as I don't understand something, I lose focus and everything just flies over my head.  I imagine this is partly psychological, since from a logical perspective if I have to accept proposition $P$ to derive $Q$, I could just think of myself as having proven merely that $P$ implies $Q$, and then no "acceptances" are being made.

Most of the time, professors simply stare blankly at me, wondering how I could persist like this, and all they say is to stop.  But it's not that simple, because it appears my intuition is also primarily symbolic.  Sure, I think of some geometric pictures when they're called for, but most of my problem-solving creativity comes from pattern matching methods and tricks with situations.

Question 2: How does one distill out the important ideas of a mathematical reading, such as a proof or paper?

Grimmett's book is very helpful in this regard.  He will always tell me what's important, and as long as I'm willing to believe him, then I don't have to do anything.  But what if I need things that are different than he emphasizes?  I always worry that by not understanding everything, I will eventually reach some point in my life when I need to use some fact/method I glossed over and forgot, and that it could be framed in such a way that I would not even be aware of what is missing.  That way I wouldn't even be able to do a huge review to rescue the fact from the depths of my ignorance.  My current way of thinking about this subproblem of question 2 is that mathematicians always take this risk by not studying everything.  So it's a risk-minimization game with time as the constraint.  If so, how do I make smart choices with regard to this game?

Question 3: With my recent ability to learn imprecisely in probability, I've started to see many connections, even with outside fields.  Many of them are probably fictitious.  Many of the questions that I think are highly-motivated might actually be not really worth answering.  How does one decide what questions are interesting?  As a graduate student who has barely popped out above what the traditional classroom has to offer, I am very lost in this regard.

Question 4: The revelations that enabled me to understand math imprecisely came all at once.  A similar comment about the abruptness of my coming upon the ability to proof-check without significant error could be made 2 years ago.  Most of my peers seem to learn rather continuously, but the evolutions of my way of thinking seem to come all at once.  Is there anything bad or good about this?  If so, how do I minimize the bad and maximize the good?

As always, answers to subsets are appreciated.

答:
I've observed how my closest colleague did it back in the day of my Phd and postdocs and I took over his technique. What he did was to sit down, read the paper superficially and then try to work out simple stuff he understood on his own. Then he would try to build his own version of what he got from the paper, often without fully understanding what had been going on. But he just had a general idea of the gist of the paper and tried to rebuild the idea in his own words, math, etc...

I remember I then proceeded to do the same later when studying some ecological model that we were trying to pour into mathematical formulas. I felt that the work that had been done was not very rigorous or even incomplete. So I rebuilt the model for myself superficially imitating others at first but gradually abandoning their approach for my own. And this without ever fully learning the necessary techniques of Markov processes, stochastic equations, etc... I feel that by doing this work, my understanding of the material is much deeper than it would have been if I had read a book about it or followed a standard course.

What also helped was the countless conversations and presentations I had to do about my work that forced me to put my thoughts into words understandable to others. They might not have gotten much out of it, but it has been very beneficial to myself for sure.

Title: Understanding mathematics imprecisely


问:
How to make an hyperellipsoid touch a hyperplane
Let's say we have an hyperellipsoid $\mathcal{E}$ defined by the symmetric, positive definite matrix $P$, i.e.,

$$\mathcal{E} = \{ x \in \mathbb{R}^n \mid x^\top P x = c \}$$

where $c>0$. I would like to find the biggest $c$ possible to make $\mathcal{E}$ touch the hyperplane $\{x \in \mathbb{R}^n \mid a^\top x = b\}$. According to a German book I read, this problem can be solved using Lagrange multipliers. Stated as an optimization problem, this should read
$$
\begin{align}
&\max_{x \in \mathbb{R}^n} x^\top P x = c \\
&\mathrm{s.t.} \: a^\top x = b
\end{align}
$$
The corresponding Lagrangian is $\mathcal{L} = x^\top P x - \lambda (a^\top x-b)$ with $\lambda$ being the Lagrange multiplier. The necessary condition for optimality leads to the linear system
$$
\underbrace{\begin{bmatrix}
-a & 2P \\
0 & a^\top
\end{bmatrix}}_{ =A}
\begin{pmatrix}
\lambda \\
x
\end{pmatrix}
=
\begin{pmatrix}
0 \\
b
\end{pmatrix}
$$
and I have no idea how to invert $A$. Am I even on the right track? Any hint/help would be greatly appreciated.

Remark: The aforementioned book states that the solution is $c = {b^2 \over a^\top P^{-1} a }$ if that helps.

答:
No need to maximize. We know that at tangency

$$
2 P x = \mu a\Rightarrow 2x^{T}P x = \mu x^T a\Rightarrow 2c=\mu b\Rightarrow \mu = 2\frac cb
$$

substituting the last result into $2 P x = \mu a$ we get $x = \frac cb P^{-1}a$ and finally substituting this result into $a^T x = b$ we obtain the solution

$$
c=\frac{b^2}{a^T P^{-1}a}
$$


问:
The minimum of $f(x)=1+x+\cdots+x^{2n}$
For the function $$
f(x)=\sum_{k=0}^{2n}{x^k}
$$
I think this function has no zeros, and if ${f(x_0)}$ is the minimum point, then ${x_0\in \left[-1,0 \right] }$.

For $1+x+x^2+x^3+x^4$ and $1+x+x^2$, I can get the minimum by calculating $f'(x)$ and then using the root fomula. But this method doesn't work for $n\geqslant 3$ since there is no root fomula for equations higher than the fifth degree.

I have no idea for the situation where  $n\geqslant 3$, so any advice could be helpful

答:
Yes, it has no zeros, because $f(1)=2n+1\ne0$ and because, if $x\ne1$, $f(x)=\frac{x^{2n+1}-1}{x-1}$, which is never equal to $0$.

However, it is not true that the minimum is attained at some of $[0,1]$. For instance, of $n=2$, the the minimum is attained at about $-0.606$.

Title: The minimum of $f(x)=1+x+\cdots+x^{2n}$


问:
How do you divide a complex number with an exponent term?
Ok, so basically I have this:

$$ \frac{3+4i}{5e^{-3i}} $$

So basically, I converted the numerator into polar form and then converted it to exponent form using Euler's formula, but I can have two possible solutions.

I can have $5e^{0.972i}$ (radian) and $5e^{53.13i}$ (angle).

So my real question is, the exponent, does it have to be in radian or angle?

I ended up with $e^{3.927i}$.

答:
A complex number $z=x+iy$ can be written as $z=re^{i\theta}$, where $r=|z|=\sqrt{x^2+y^2}$ is the absolute value of $z$ and $\theta=\arg{z}=\operatorname{atan2}(y,x)$ is the angle between the $x$-axis and $z$ measured counterclockwise and in radians. In this case, we have $r=5$ and $\theta=\arctan\frac{4}{3}$ (since $x>0$, see atan2 for more information), so $3+i4=5e^{i\arctan\frac{4}{3}}$ and

$$\frac{3+i4}{5e^{-i3}}=\frac{5e^{i\arctan\frac{4}{3}}}{5e^{-i3}}=e^{i(\arctan\frac{4}{3}+3)}\approx e^{i3.927}.$$


问:
Matrix Multiplication of $(AB) (CD)$
For Matrix Multiplication of $(AB) (CD)$ with dimension of $A_{30\times 10}, B_{10\times 40}, C_{40\times 50}, D_{50\times 30}$, what is the number of multiplications we need for $(AB) (CD)$?

For $(AB)$, we need $30\times10\times40=12000$ multiplications.

For $(CD)$, we need $40\times50\times30=60000$ multiplications.

Question: the solution stops and said that the number of multiplications of $(AB) (CD)$ would be 72000, but I guess it should be

For $(AB)$, we need $30\times10\times40=12000$ multiplications.

For $(CD)$, we need $40\times50\times30=60000$ multiplications.

For $(AB)(CD)$, we need $30\times40\times30=36000$ multiplications.

And that the number of multiplications of $(AB) (CD)$ would be $12000+60000+36000=108000$, is not that right?

答:
Well, that depends.

Since matrix multiplication is associative, you can get different results.

In your example, there are few different ways of doing the multuplication:

$((A\times B)\times(C\times D))$:

$A\times B$ costs $30\cdot 10\cdot 40=12,000$ multiplications and results in a matrix $E_{30\times 40}$

$C\times D$ costs $40\cdot 50\cdot 30=60,000$ multiplications and results in a matrix $F_{40\times 30}$

$E\times F$ costs $30\cdot 40\cdot 30=36,000$ multiplications and results in a matrix $G_{30\times 30}$

so the total cost is $108,000$ multiplications

$(((A\times B)\times C)\times D)$:

$A\times B$ costs $30\cdot 10\cdot 40=12,000$ multiplications and results in a matrix $E_{30\times 40}$

$E\times C$ costs $30\cdot 40\cdot 50=60,000$ multiplications and results in a matrix $F_{30\times 50}$

$F\times D$ costs $30\cdot 50\cdot 30=45,000$ multiplications and results in a matrix $F_{30\times 30}$

so the total cost is $117,000$ multiplications

$(A\times (B\times (C\times D)))$:

$C\times D$ costs $40\cdot 50\cdot 30=60,000$ multiplications and results in a matrix $E_{40\times 30}$

$B\times E$ costs $10\cdot 40\cdot 30=12,000$ multiplications and results in a matrix $F_{10\times 30}$

$A\times F$ costs $30\cdot 10\cdot 30=9,000$ multiplications and results in a matrix $F_{30\times 30}$

so the total cost is $81,000$ multiplications

There are also $((A \times (B \times C)) \times D)$ and $(A \times ((B \times C) \times D)$

Title: Matrix Multiplication of $(AB) (CD)$


问:
A binary operation $f$ that is associative and satisfies $f(f(x,y),x)\approx x$
"Let $f$ a binary function symbol, and $$E=\{f(x,f(y,z)) \approx f(f(x,y),z), f(f(x,y),x) \approx x\}$$ a set of identities. Show that $f(x,x) (\leftrightarrow_E)^* x$ and $f(f(x,y),z) (\leftrightarrow_E)^* f(x,z)$."
I am using the definitions in the Baader Book of Rewriting Systems, where this $(\leftrightarrow_E)^*$ relationship is closed under substitutions and compatible with $\sum$-operations. I managed to do the first part, doing some manipulations, but I'm not able to do the second part.

答:
First, some psychological simplification. I'll use juxtaposition notation in place of $f$, so "$f(t_1,t_2)$" will be written "$t_1t_2$." This makes it a bit easier to write out the relevant steps. Moreover, given the associativity of $f$ (your first axiom), we can include or omit parentheses as we please.

With this in mind, you want to get from "$xyz$" to "$xz$" using associativity and the equation $uvu=u$. Well, by omitting parentheses we've essentially "maximally used" associativity already, so the only thing left to do is find some clever way to apply that remaining equation. There are no terms of the form $uvu$ involved, so we have to pick some existing term $u$ and replace it with $uvu$ for some term $v$ of our choice.

A big part of our goal is to get rid of $y$, and the only way we can get rid of a symbol is by using the $uvu\leadsto u$-direction of the axiom. Since $y$ is flanked on the right by $z$, this tells us that we'd be really happy if we could get a $z$ to show up somewhere to the left of $y$. The only thing to the left of $y$ right now is $x$, so this suggests the following choice of $u$ and $v$ for applying the "lengthening" rule $u\leadsto uvu$:

 Set $u=x$ and $v=z$. This will give as our new string $$xzxyz.$$

Now this has a substring of the form $z[stuff]z$, so we can - by applying the "shortening" rule $uvu\leadsto u$, with $u=z$ and $v=[stuff]$ - collapse that to just $z$. And at that point:

 We're done: throwing in parentheses for clarity, we have $$xzxyz=x(z(xy)z)=xz.$$

Hopefully the explanation above makes this seem not ad hoc; I think the "goal-directed" language I used is often helpful, but that's just my experience. (Of course I haven't actually written the above in the literal form of an $\leftrightarrow_E^*$-derivation, but the translation into that format is straightforward.)


问:
Radius of Convergence of Complex Series (Real parts)
I want to show that if the radius of convergence of $\sum_{n=0}^\infty a_nz^n$ is R

Then the radius of convergence for $\sum_{n=0}^\infty (\operatorname{Re}(a_n))z^n$ is greater than or equal to R

I know that $|\operatorname{Re}(a_n)|\le |a_n|$ but I'm not confident in being able to find the radius of convergence so don't know how to use this information to prove this?

Any suggestions of how to start a proof like this would be brilliant

答: For each $n\in\Bbb N$, $|\operatorname{Re}(a_n)|\leqslant|a_n|$, and therefore $\limsup_n\sqrt[n]{|\operatorname{Re}(a_n)|}\leqslant\limsup_n\sqrt[n]{|a_n|}$. But the radius of convergence of the power series $\sum_{n=0}^\infty a_nz^n$ and $\sum_{n=0}^\infty\operatorname{Re}(a_n)z^n$ are equal to$$\frac1{\limsup_n\sqrt[n]{|a_n|.}}\quad\text{and to}\quad\frac1{\limsup_n\sqrt[n]{|\operatorname{Re}(a_n)|}}$$respectively.


问:
Three simple questions on measure theory
A couple of days ago, I posted this question, which remains unanswered. The comments that I have received make me think that this is so because it is ill-defined. Hence, I post this other question with the hope of understanding the basics of measure theory so I can go back to the other question and edit it accordingly.

Let $(\mathbb{R}^n,\mathcal{A})$ be any measurable space. Then,

$X\in\mathcal{A}\Leftrightarrow X$ measurable.

Let $(\mathbb{R}^n,\mathcal{L})$ be a measurable space, where $\mathcal{L}$ is the Lebesge $\sigma$-algebra. Then,

There exist two measures $\mu:\mathcal{L}\to\mathbb{R}$ and $\mu':\mathcal{L}\to\mathbb{R}$ satisfying $\mu\neq\mu'$.

Let $(\mathbb{R}^n,\mathcal{L},\lambda)$ be a measure space where $\lambda$ is the Lebesgue measure. Then,

$X\subset\mathbb{R}^n$ infinite and bounded $\Rightarrow \lambda(X)\in(0,\infty)$.

Are (1), (2) and (3) true statements? I think that (1) is true by definition of $\sigma$-algebra. I think that (2) is also true because as shown in the accepted answer to this question, one can define a measure that is not the Lebesgue measure on $(\mathbb{R}^n,\mathcal{L})$. For (3), I do not have an intuition and I'm not sure whether it's true.

Thank you all!

EDIT! As noted in the comments, (3) is false. Hence, let me replace (3) for (3'). Let $(\mathbb{R}^n,\mathcal{L},\lambda)$ be a measure space where $\lambda$ is the Lebesgue measure. Then,

3'. $X\subset\mathbb{R}^n$ infinite, uncountable and bounded $\Rightarrow \lambda(X)\in(0,\infty)$.

答:
In a measurable space $(Y,\mathcal A)$, the term $\mathcal A$-measurable sets indicates the elements of $\mathcal A$, and obviously the term is shortened to measurable sets when there isn't any ambiguity.

There are always at least two measures on a non-empty measurable space $(Y,\mathcal A)$: the zero measure and the restriction to $\mathcal A$ of the counting measure, namely $$\mu(X)=\begin{cases}\operatorname{card} X&\text{if }X\text{ is finite}\\ \infty&\text{if }X\text{ is an infinite set}\end{cases}$$

Some bounded sets aren't Lebesgue-measurable, therefore discussing their Lebesgue measure is somewhat idle talk (it's still possible to talk about their Lebesgue outer measure, but I digress). For Lebesgue-measurable subsets, $A\subseteq B$ implies $\lambda(A)\le\lambda(B)$ and therefore, since $\lambda(a,b)=b-a$, bounded Lebesgue-measurable subsets have finite Lebesgue measure. Obviously, countable additivity makes it so that countably infinite subsets are Lebesgue-null sets but, more to the point, Cantor's set is a compact and uncountable Lebesgue-null set.


Title: Converting parametric equations to xy equation

Body:
I have these parametric equations:
$$
x=1+t \\
y=1-t
$$

How do I make this to the xy equation?

David

Body:
$x = 1 + t
\\y = 1 - t
\\x = -(1- t) + 2
\\x = 2 - y  $


问:
Find Cross-Tangents to Offset Ellipses
Question:

I am stuying projective geometry and would like to find an analytic solution to my problem. This is a follow-up to my older question.

Given two ellipses with the same major and minor axes, but different origins (written in homogeneous coordinates):

$$
\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1 = 0
$$

and

$$
\frac{(x_2-h)^2}{a^2} + \frac{(y_2-k)^2}{b^2} - 1 = 0
$$

what are the equations of the cross-tangents to the two ellipses? Assume $h, k, a,$ and $b$ are all knowns.

My Attempt:

Following the method from my older question, it makes sense to find the intersection points of the dual curves to each ellipse. Let $f_1$ be the implicit homogeneous function defining our first ellipse (i.e. $f_1(x_1, y_1, z_1) = 0$) and $f_2$ that of our second, respectively. The dual curve of an ellipse is an ellipse (at least, AFAIK, for an ellipse at the origin), which Wikipedia shows nicely.

Hence, our first dual is

$$
a^2 X_1^2 + b^2 Y_1^2 = 1.
$$

I am having trouble finding the second dual (namely, that of the ellipse centered at $(h,k)$). Could someone please help me in deriving this? Is this the simplest way to solve this problem?

Following Wikipedia, about the closest I get to is the following and I don't know how to eliminate $\lambda$:

$$
\frac{a^2}{2\lambda} \left( X_2 + \frac{\lambda h}{a^2} \right)^2 + \frac{b^2}{2\lambda} \left( Y_2 + \frac{\lambda k}{b^2} \right)^2 - \left( \frac{\lambda}{2} \right) \left( \left( \frac{h}{a} \right)^2 + \left( \frac{k}{b} \right)^2 \right)
 + 1 = 0$$

答:
I wanted to expand on the answer I accepted for my own sake and the sake of those reading this in the future. My post here is also instructive for this problem.

If the equation of a line in line coordintes is

$$
Xx+Yy+1=0 \tag{1}
$$

and the tangent to our curve at point $(x_0)$ (identically $(p,q,r)$ on Wikipedia) $<f_x \left. \right|_p, f_y \left. \right|_q, 0>$, denoted $(x')$, is coincident with the line if its inner product with the line coordinates is $0$ per Coxeter

$$
Xx'+Yy'=0
$$

where $(x')=(x',y',0)$, then

$$
Xf_x \left. \right|_p + Yf_y \left. \right|_q = 0. \tag{2}
$$

Solving $(1)$ and $(2)$ simultaneously yields

$$
X = \frac{-f_y \left. \right|_q}{xf_y \left. \right|_q-yf_x \left. \right|_p}
$$
$$
Y = \frac{f_x \left. \right|_q}{xf_y \left. \right|_q-yf_x \left. \right|_p}
$$

Title: Find Cross-Tangents to Offset Ellipses


问:
Evaluate the Jacobian if $u = xy, v = y-x$
The following equations are given:
$u=xy$ and $v = y - x$.
To evaluate the Jacobian, you have to express $x$ in terms of $u$ and $v$, and $y$ in terms of $u$ and $v$.
However, I don't see how you would exactly do this.

I tried using the ABC formula as follows:
$$u = xy$$
$$x = \frac{u}{y}$$
$$x = \frac{u}{v+x}$$
$$x(v+x) = u$$
$$vx + x^2 = u$$
$$x^2 + vx - u = 0$$
$$x = \frac{-v \pm \sqrt{v^2+4u}}{2}$$
$$y = v + x = v - \frac{v \pm \sqrt{v^2+4u}}{2}$$

But then you have two equations for x and y? So can someone tell me if my method is incorrect and if yes, how you would do this with the correct method to evaluate the Jacobian?

答:
Much easier: if we denote by $\;J\;$ the Jacobian you need, then with inverse matrices we're done:

$$J:=J\frac{(x,y)}{(u,v)}=\left[J\frac{(u,v)}{(x,y)}\right]^{-1}=\begin{vmatrix}y&x\\-1&1\end{vmatrix}^{-1}=\frac1{x+y}$$


问:
Trying to find if this trigonometry problem is wrong or not
I know That we have a triangle ABC with $$A=90°  $$ $$ b+c=18 $$ $$ \cot(B)=4\sin^2(C) $$ $$ 0<B,C<90 $$

But , upon calculating i end up a = 18, b = 18, c = 0 which should not be possible

Body:
$$\frac{c}{b}=\cot(B)=4\sin^2(C)=\frac{4c^2}{a^2}$$

$$a^2=4bc=b^2+c^2$$

Now use $c=18-b$ and get:

$$b^2-18b+54=0$$

$$b=9\pm3\sqrt{3}$$

Therefore, $a=6\sqrt{6}$; $b,c=9\pm3\sqrt{3}$.


答:
You get two possible solutions because if $(x,y)$ is one solution for given $(u,v)$, then $(-y,-x)$ is another solution. That means that $(u,v)$ aren't good coordinates forr the whole plane, because they do not differentiate point $(x,y)$ and $(-y,-x)$. You can only use them on some subsets of a plane, for example with $x+y>0$.

If you have condition $x+y>0$ then from
$$ (x+y)^2 = (y-x)^2 +4xy = v^2+4u$$
you can conclude that
$$ x+y = + \sqrt{v^2+4u}$$
so $$ x= \frac{-v+ \sqrt{v^2+4u}}{2}$$
$$ y= \frac{v+ \sqrt{v^2+4u}}{2}$$
If you have $x+y<0$ then you use the other solutions:
$$ x+y = - \sqrt{v^2+4u}$$
$$ x= \frac{-v- \sqrt{v^2+4u}}{2}$$
$$ y= \frac{v- \sqrt{v^2+4u}}{2}$$
From this you can calculate the Jacobian.

Another way to calculate Jacobian is to calculate the inverse of the Jacobian first:
$$ J^{-1} = \frac{\partial (u,v)}{\partial (x,y)} = \left|\begin{array}{cc}y & x \\ -1 & 1\end{array}\right| = x+y$$
which means that
$$ J = \frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{x+y} = \pm\frac{1}{\sqrt{v^2+4u}}$$
where you need to choose the sign depending on whether $x+y>0$ or $x+y<0$.

Title: Evaluate the Jacobian if $u = xy, v = y-x$

Body:
The following equations are given:
$u=xy$ and $v = y - x$.
To evaluate the Jacobian, you have to express $x$ in terms of $u$ and $v$, and $y$ in terms of $u$ and $v$.
However, I don't see how you would exactly do this.

I tried using the ABC formula as follows:
$$u = xy$$
$$x = \frac{u}{y}$$
$$x = \frac{u}{v+x}$$
$$x(v+x) = u$$
$$vx + x^2 = u$$
$$x^2 + vx - u = 0$$
$$x = \frac{-v \pm \sqrt{v^2+4u}}{2}$$
$$y = v + x = v - \frac{v \pm \sqrt{v^2+4u}}{2}$$

But then you have two equations for x and y? So can someone tell me if my method is incorrect and if yes, how you would do this with the correct method to evaluate the Jacobian?


问:
Proving an expression for $T^n.$

Let $\mathcal H = L^2[0,1].$ Define an operator $T : \mathcal H \longrightarrow \mathcal H$ by $Tf(x) = \displaystyle {\int_{0}^{x} f(y)\ dy.}$ Show that for all $n \geq 1$ $$T^n f(x) = \int_{0}^{x} \frac {(x-t)^{n-1}}{(n-1)!} f(t)\ dt.$$

Clearly this result holds for $n=1.$ Let the result hold for $n.$ Then $$T^{n+1} f(x) = T^n(Tf(x)) = \int_{0}^{x} \frac {(x-t)^{n-1}}{(n-1)!} \left (\int_{0}^{t} f(y)\ dy \right )\ dt.$$ Another way of writing it down is $$T^{n+1} f(x) = T(T^nf(x)) = \int_{0}^{x} \left (\int_{0}^{y} \frac {(y-t)^{n-1}} {(n-1)!} f(t)\ dt \right )\ dy.$$ From here how do I conclude that $$T^{n+1} f(x) = \int_{0}^{x} \frac {(x-t)^n} {n!} f(t)\ dt\ ?$$

Any help in this regard will be warmly appreciated. Thanks for your time.

答:
Change the order of integration for the induction step: \begin{eqnarray*}
(T^{n+1}f)(x) & = & \int_0^x(T^nf)(y)\mathrm dy
\\ & = & \int_0^x\int_0^y\frac{(y-t)^{n-1}}{(n-1)!}f(t)\mathrm dt \mathrm dy \\
 & = & \int_0^x \int_t^x\frac{(y-t)^{n-1}}{(n-1)!}f(t)\mathrm dy \mathrm dt
\end{eqnarray*} Can you finish?

Title: Proving an expression for $T^n.$


问:
Is there an unlimited number of sequences of consecutive numbers with $\mu(n)=0$ of any length?
EDIT

I have received interesting comments to my post. Especially the comment of @Martin Hopf showed that this problem is a "classic" and by no means new.

Here is Eric Weissten's article on "squareful (nice term!) numbers" https://mathworld.wolfram.com/Squareful.html from which you can find the topic exposed.

This article https://arxiv.org/pdf/1210.3829.pdf is an exhaustive study of the problem.

Original post

Consider the subset of the natural numbers $n\ge 1$ which have at least one square prime factor. These can be formally defined by $\mu (n) = 0$, where $\mu$ is the Möbius function.

The first 40 of these are

$$s_{0}=\{4,8,9,12,16,18,20,24,25,27,28,32,36,40,44,45,48,49,50,52,54,56,60,63,64,68,72,75,76,80,81,84,88,90,92,96,98,99,100,104\}$$

Closer inspection of $s_{0}$ shows that there are sequences of consecutive such numbers. We call them compact sequences. If sorted by the length $m$ of the compact sequence we find in $s_{0}$ the following

$$m=2: \{8,9\}, \{24,25\}, \{27,28\},\{44,45\},\{48,49\},\{49,50\},\{63,64\},\{75,76\},\{80,81\},\{98,99\},\{99,100\}$$
$$m=3:  \{48,49,50\}, \{98,99,100\}$$

We see that for a given length $m$ there is more than one sequence, and that for the given length 40 of $s_{0}$ there are no sequences with length $m\gt3$.

Prolonging the list $s_0$ we find also longer compact sequences, and, again, for a given length there is more than one sequence.

Quoting only the first member of the first appearance of the corresponding sequence
I found numerically in the format $\{m,\text{first term of first appearance}\}$

$$c=\left(
\begin{array}{cc}
 2 & 8 \\
 3 & 48 \\
 4 & 242 \\
 5 & 844 \\
 6 & 22'020 \\
 7 & 217'070 \\
 8 & 1'092'747 \\
 9 & 8'870'024 \\
10 & 221'167'422\\
\end{array}
\right)$$

Remark: this sequence is contained in OEIS (https://oeis.org/A045882). Thanks to @Martin Hopf for pointing this out in a comment.

Conjecture

(1)  there is a compact sequences for any given length and

(2)  there are infinitely many compact sequences for any given length

Unfortunately, I was not able to prove or disprove the conjecture. Can you do better?

Additional question

(3) can you devise a formula for the first appearance of the compact series for given $m$?

答:
There is an elementary way to show that for all $m$ there are infinitely many integers such that $$|\mu(n)|=|\mu(n+m+1)|=1,\qquad  \mu(n+1)=\ldots=\mu(n+m)=0$$

Let $N_m=\prod_{j\le m} p_{j+m}^2$. With the CRT take $$a_m\in [1,N_m], \qquad a_m\equiv -j \bmod p_{j+m}^2,\qquad j \in 1\ldots m$$ so that $\mu(a_m+j+lN_m)=0$ for $j\in 1\ldots m$.

Look at the Dirichlet series $$\begin{eqnarray}F_m(s)&\\ =&\sum_{l\ge 0} |\mu(a_m+lN_m)| (a_m+lN_m)^{-s}+|\mu(a_m+m+1+lN_m)| (a_m+m+1+lN_m)^{-s}\end{eqnarray}$$
Note that $a_m,a_m+m+1$ are coprime with $N_m$. By the orthogonality of Dirichlet characters we have $$F_m(s)= \sum_{\chi \bmod N_m}\frac{\overline{\chi(a_m)}+\overline{\chi(a_m+m+1)}}{\varphi(N_m)}\sum_{n\ge 1}\chi(n)|\mu(n)| n^{-s} $$
For $\chi$ a non trivial Dirichlet character, $\sum_{n\ge 1}\chi(n)|\mu(n)| n^{-s}$ is holomorphic at $s=1$.

Whence the asymptotic as $s\to 1$ is determined by the trivial character $1_{\gcd(n,N_m)=1}$:

$$\begin{eqnarray}F_m(s)&\sim& \frac{2}{\varphi(N_m)}\sum_{n\ge 1,\gcd(n,N_m)=1} |\mu(n)| n^{-s}\\&=&\frac{2}{\varphi(N_m)} \frac1{\prod_{p\ |\ N_m} (1+p^{-s})} \frac{\zeta(s)}{\zeta(2s)}\\&\sim& \frac{2}{N_m} \frac1{\prod_{p\ | \ N_m} (1-p^2)} \frac{1}{\zeta(2)}\zeta(s)\end{eqnarray}$$

Since $\frac1{\prod_{p\ | \ N_m} (1-p^2)}\frac{1}{\zeta(2)}>1/2$ this implies that for infinitely many $l$ $$|\mu(a_m+lN_m)|=|\mu(a_m+m+1+lN_m)|=1$$
so that $a_m+lN_m,\ldots, a_m+m+1+lN_m$ is our chain of length $m$.


问:
Equality validation in vector calculus
I have getting myself into the area of fluid mechanics.

I would like to know if the following relationship holds.

$\left (\vec{\mathbf{a}} \cdot \nabla  \right ) \vec{\mathbf{b}} = \vec{\mathbf{a}} \cdot \nabla \vec{\mathbf{b}}$

The left side of the equation is also known as the convective operator (https://mathworld.wolfram.com/ConvectiveOperator.html).

Reading:

(vector $\vec{\mathbf{a}}$ dot product with the $\nabla$ operator) times vector $\vec{\mathbf{b}}$ is equal to the dot product of vector $\vec{\mathbf{a}}$ and the gradient of vector $\vec{\mathbf{b}}$.

Sorry if this a very basic question!

答:
It's not the same. Let's look at the $\hat x$ component for example in Cartesian coordinates. The left hand side (from the link that you have) is $$a_x\frac{\partial b_x}{\partial x}+a_y\frac{\partial b_x}{\partial y}+a_z\frac{\partial b_x}{\partial_z}$$
For the right hand side, $$\nabla\vec b=\frac{\partial b_x}{\partial x}+\frac{\partial b_y}{\partial y}+\frac{\partial b_z}{\partial_z}$$
This is a scalar. So when we multiply with vector $\vec a$, the $\hat x$ component will be $$a_x\frac{\partial b_x}{\partial x}+a_x\frac{\partial b_y}{\partial y}+a_x\frac{\partial b_z}{\partial_z}$$
Notice the difference: in the right hand side you have each term of the sum multiplied with a different component of $\vec a$, but each derivative acts on $b_x$. On the right hand side each term contains the same $a_x$, but the derivatives act on different components of $\vec b$.

Title: Equality validation in vector calculus


问:
Question on the proof of a subspace of Polish space is Polish, iff it's a $G_\delta$ set.
Suppose, $X$ is a Polish space, $Y$ is a Polish subspace of $X$. $\{U_n\}_{n \in \Bbb N}$ is a basis of open sets of $X$.

Let $A = \{ x \in \overline {Y} : \forall \epsilon \exists {n}(x \in U_n \land \operatorname{diam}{(Y \cap U_n)} < \epsilon) \}$

$\overline {Y}$ is the closure of $Y$. $\operatorname{diam}{(Y \cap U_n)}$ is the diameter of $Y \cap U_n$.

Why $A$ is different from $\overline {Y}$?

Added: The proof is from an online note (page 11)about descriptive set theory. Since it's posted free online by the author, I take the liberty to paste a screenshot of it for convenience.

答: Let's look at an example.  Let $X = \{1, 1/2, 1/3, \dots, 0\}$ with the Euclidean topology, and take $Y = X \setminus \{0\}$.  Then the discrete metric $d(x,y) = \delta_{xy}$ on $Y$ is complete and compatible with the subspace topology.  Now $0 \in \bar{Y}$.  However, if $U_n$ is any open neighborhood of 0 (in the topology of $X$), then it contains infinitely many points of $Y$; in particular, at least two points, so $\operatorname{diam}(U_n \cap Y) = 1$ (where the diameter is computed with respect to $d$).  Therefore $0 \notin A$, and we see explicitly that $\bar{Y} \ne A$.</p>


问:
Probability of Sum being $4$ with unfair dice
You roll two six-sided dice.  Die $1$ is fair.  Die $2$ is unfair such that the probability of rolling an odd number is $2/3$ and the probability of rolling an even number is $1/3$, though the probability rolling of each odd number is the same, and the probability of rolling each even number is the same.  What is the probability of the sum of both dice adding up to $4$?

My answer:

I wrote out the combinations to sum $4$ between both dies:
(F1, U3), (F2, U2), (F3, U1) [F = Fair die, U = Unfair die].

My thinking is we need to find the OR probability of these pairs:
P((F1 and U3) or (F2 and U2) or (F3 and U1)):
$$(\frac16 \times \frac23) + (\frac16 \times \frac12) + (\frac16 \times \frac23) = 0.30555555555.$$

Is this correct?

答:
There is a mistake. For the unfair die, the probability of an odd number is $\frac{2}{3}$ and of an even number is $\frac{1}{3}$. It also says that the probability of each odd number is same and of each even number is same.

So for the unfair die, probability of each odd number is $\frac{2}{9}$ and of each even number is $\frac{1}{9}$.

So the desired probability should be $ = \displaystyle \small 2\cdot \frac{1}{6} \cdot \frac{2}{9} + \frac{1}{6} \cdot \frac{1}{9} = \frac{5}{54}$


答:
No.

With the fair die, the chance of $1,2,3$ is $(1/6), (1/6), (1/6)$, respectively.

With the unfair die, the chance of $1,2,3$ is $(2/9), (1/9), (2/9)$, respectively.

Roll of 4 by 1-3 or 2-2 or 3-1.  Chances are

$(2/54) + (1/54) + (2/54) = (5/54).$

Title: Probability of Sum being $4$ with unfair dice

Body:
You roll two six-sided dice.  Die $1$ is fair.  Die $2$ is unfair such that the probability of rolling an odd number is $2/3$ and the probability of rolling an even number is $1/3$, though the probability rolling of each odd number is the same, and the probability of rolling each even number is the same.  What is the probability of the sum of both dice adding up to $4$?

My answer:

I wrote out the combinations to sum $4$ between both dies:
(F1, U3), (F2, U2), (F3, U1) [F = Fair die, U = Unfair die].

My thinking is we need to find the OR probability of these pairs:
P((F1 and U3) or (F2 and U2) or (F3 and U1)):
$$(\frac16 \times \frac23) + (\frac16 \times \frac12) + (\frac16 \times \frac23) = 0.30555555555.$$

Is this correct?


问:
$f^{p_1}$ belongs to $L^p(E)$ and $g = \chi_{E}$ belongs to $L^q(E)$?
I was reading a Corollary in Royden, which states:

In the proof,

The book mentioned by observation that $f^{p_1}$ belongs to $L^p(E)$ and $g =  \chi_{E}$ belongs to $L^q(E)$. I am not sure how this fact can be observed?

答:
Notice that:
$$||f^{p_1}||^{p}_p = \int\limits_{E} \, |f^{p_1}|^p \, \mathrm{d} \mu = \int\limits_{E} \, |f|^{\frac{p_1 p_2}{p_1}} \, \mathrm{d} \mu = \int\limits_{E} \, |f|^{p_2} \, \mathrm{d} \mu = ||f||^{p}_{p_2} < \infty \Rightarrow f^{p_1} \in L^p(E)$$
Additionally:
$$||\chi_{E}||^{q}_q = \int\limits_{E} \, |\chi_{E}|^q \, \mathrm{d} \mu = \int\limits_{E} \, \chi_{E} \, \mathrm{d} \mu = \mu(E) < \infty \Rightarrow \chi_E \in L^q(E)$$


问:
Chain Rule of Logarithm?
The students are taught the well known  change base rule of logarithm:

\begin{align}\log_a b = \frac{\log_c b}{\log_c a}\end{align} Most text books proves it by invoking  $(a^x)^y=a^{xy}$ to show:
\begin{align}\log_c a\times\log_a b=\log_c b\end{align}

Question:

Why don't we call it  Chain Rule of Logarithm (at least as a second name)?
Is it because the way we use it is almost alway in the change of base format or something else?
\begin{align}
\log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z\end{align}

It is easy to remember and the students  can have fun to continue chaining  it (similar to the change rule of derivative).

答:
A side note:

The argument of the current factor must be equal to the base of following factor. I´ve colored the corresponding parameters.

$$\begin{align}\log_{\color{blue}{a}} \color{red}{b}\times\log_{\color{red}{b}} \color{orange}{c}\times\log_{\color{orange}{c}} \color{green}{d}\times\cdots\times\log_y \color{yellowgreen}{z}=\log_{\color{blue}{a}} \color{yellowgreen}z\end{align}$$

I hope you see the difference to your term. The chain rule is about derivatives and the concept is very different from the rule you´ve posted.

It is more related to the overall growth rate $r$, if you have n consecutive growth rates ($r_i$), with $r_i=\frac{y_{i+1}}{y_i}-1$.

$$1+r=\frac{y_{1}}{y_0}\cdot \frac{y_{2}}{y_1}\cdot \ldots \cdot \frac{y_{n-1}}{y_{n-2}} \cdot \frac{y_{n}}{y_{n-1}}=\frac{y_{n}}{y_{0}}$$

The rule you´ve posted is related to the concept of the  geometric mean.

Title: Chain Rule of Logarithm?


问:
Intuition for multiplying vectors by complex scalars
There is a very pleasing interpretation of multiplying a vector by a real scalar. I can ''stretch or shrink'' and also ''flip over the origin (in the case of negative scalars)''.

Is there some analogous way to think of scalar multiplication when the vector field is over $\mathbb{C}$ instead? Intuitively I feel like its the same idea, just with more freedom over where scalars can point the vector (in the same way that complex number multiplication rotates and stretches other complex numbers) but I can't seem to write down anything concrete.

答:
Assume that you have
$$w=\lambda\,z\ ,\tag{1}$$
where everything is complex. In addition we imagine that $\lambda\in{\mathbb C}$ is constant, preferably $\ne0$, whereas $z$ is our independent variable, and $w$ changes when $z$ changes. Of course you can compute all of this algebraically through $\lambda=\mu+i\nu$,$\>z=x+iy$, $\>w=u+iv$.

But there is also a geometric interpretation of the transformation $z\mapsto w:=\lambda\,z$. Of course you have heard of such things when you first met complex numbers. You can draw a chosen point $z$ and its image point $w$ in the same Argand plane. The scaling factor $\lambda$ can be written in the form $\lambda=|\lambda|\,e^{i\phi}$ for a certain $\phi\in[0,2\pi[\>$. The equation $(1)$ then means that you obtain $w$ (resp., the vector $\vec{0w}$) from $z$ by scaling the vector $\vec{0z}$ by the $\geq0$ factor $|\lambda|$ and rotating this intermediate result by $\phi$ counterclockwise around $0$.

Of course I have just given another version of the rule "under complex multiplication the absolute values multiply, and the arguments add".


问:
Average of a function over random uniform sample.
I'm trying to understand some implications concerning the Weak Laws of Large Numbers, and I'm stuck in this problem.

If $f:[0,1]\to\mathbb{R}$ is a measurable function with $\int_0^1\mid f(x)\mid dx<\infty$ and $U_1,...,U_n,...$ is a sequence of independent uniformly distributed random variables, consider the following average:

\begin{equation}
I_n = \frac{f(U_1) + ... + f(U_n)}{n}
\end{equation}

Then, $I_n\to \int_0^1f(x)dx$ in probability.

I was trying to apply some kind of Weak Law of Large Numbers, since we can compute using the uniform distribution that:

\begin{equation}
E[f(U_j)] = \int_0^1f(x)dx
\end{equation}

And hence:

\begin{equation}
E[I_n] = I
\end{equation}

However, I was not able to get some bound on the variance $var[f(U_j)]$ in order to apply the Weak Law of Large Numbers.

So instead I also tried to prove $L^1$ convergence, but got stuck as well: assume $I=0$ wlog, and then I got the following chain of inequalities:

\begin{equation}
E[\mid I_n \mid] \leq \frac{1}{n}\sum_{j=1}^{n}E[\mid f(U_j)\mid] = \int_0^1 \mid f(x) \mid dx < \infty
\end{equation}

Which is far for enough to prove $L^1$ convergence. When trying to estimate the actual probability convergence, I had no luck either, since my only idea was to use Chebyshev's Inequality, which would be useful only if I knew about some $L^p$ convergence.

What would be a better approach here?

答: Hint: recall that simple functions are dense in $L^1,$ so it is enough to prove the result for the indicator function of an interval $[a, b].$


问:
How does the average energy from equation (41.15) of Feynman's lectures approach $kT$ as omega goes to zero?
Feynman said the following equation should approach $kT$ as $\omega$ goes to zero, or $T$ goes to infinity.

$$\langle E \rangle = \frac{\hbar \omega}{e^{\hbar \omega/kT}-1}$$

Does anybody know how to prove this?

答:
Use the Taylor series for the exponential function,

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots.$$

In addition to giving you the $\omega\to 0$ limit, it will give you the low-frequency corrections.

OwnerDisplayName: Physics Student

OwnerDisplayName: G. Smith


答:
Get use of L'Hopital's rule where ' stands for the derivative

$$\lim_{\omega \to 0} \langle E\rangle= \lim_{\omega \to 0}\frac{(\hbar \omega)'}{(e^{\frac{\hbar \omega}{kT}}-1)'}  = \lim_{\omega \to 0} kT e^{-{\frac{\hbar \omega}{kT}}}=kT$$

Title: How does the average energy from equation (41.15) of Feynman's lectures approach $kT$ as omega goes to zero?

Body:
Feynman said the following equation should approach $kT$ as $\omega$ goes to zero, or $T$ goes to infinity.

$$\langle E \rangle = \frac{\hbar \omega}{e^{\hbar \omega/kT}-1}$$

Does anybody know how to prove this?

OwnerDisplayName: Monopole


答:
The expression for $\langle E \rangle$ is not defined at $\omega = 0$ as it would be ${0 \over 0}$. So we need to take the limit $\omega \to 0$ i.e.

$$\lim_{\omega\to 0} \langle E \rangle = ?$$

To solve this, very simply, use the fact that the Taylor expansion of the exponential function is

$$e^x\approx 1+x+{1\over 2}x^2 + ...$$ which in your case is
$$e^{\hbar\omega \over k_B T}\approx 1 + {\hbar\omega \over k_B T}+{1\over 2}\left({\hbar\omega \over k_B T}\right)^2+...$$

so
$$\langle E \rangle \approx {\hbar\omega \over 1 + {\hbar\omega \over k_B T}+{1\over 2}\left({\hbar\omega \over k_B T}\right)^2+... -1}$$
which becomes

$$\langle E \rangle  \approx  {\hbar\omega \over {\hbar\omega \over k_B T}+{1\over 2}\left({\hbar\omega \over k_B T}\right)^2} = {\hbar\omega\over\hbar\omega} {k_BT \over 1+{1\over 2}{\hbar\omega \over k_B T}} $$
and finally

$$\langle E \rangle \approx {k_BT \over 1+{1\over 2}{\hbar\omega \over k_B T}} $$
This expression is valid if $\omega\approx 0$ and describes the behavior of $\langle E  \rangle$ at low frequencies.

From this, just take the limit $\omega\to 0$, and it should be clear that
$$\lim_{\omega\to 0} \langle E \rangle = \lim_{\omega\to 0} \langle E \rangle {k_BT \over 1+{1\over 2}{\hbar\omega \over k_B T}} \to k_B T $$

OwnerDisplayName: JalfredP


问:
Why $\frac{\sqrt{x}-1}{\sqrt{x}+1} \leq \exp{\left(-\frac{C}{\sqrt{x}}\right)}$?
I am reading that

$$\frac{\sqrt{x}-1}{\sqrt{x}+1} \leq \exp{\left(-\frac{C}{\sqrt{x}}\right)}.\tag{1}$$

Doing the graph, we can see that $\exp{\left(-\frac{C}{\sqrt{x}}\right)}$ upper bounds $\frac{\sqrt{x}-1}{\sqrt{x}+1}$ when $C$ is small enough. Could you please someone provide some analytic explanation on how we can get $(1)$ and what is the maximum value of $C$ to achieve the upper bound?

答: Observe that $\dfrac{\sqrt{x}-1}{\sqrt{x}+1} = \dfrac{1-\frac{1}{\sqrt{x}}}{1+\frac{1}{\sqrt{x}}}$. Thus if you set  $u = \dfrac{1}{\sqrt{x}}$, then you want to prove: $e^{-Cu} \ge \dfrac{1-u}{1+u}$ for small $C > 0$. Consider: $f(u) = e^{-Cu} + \dfrac{u-1}{u+1}$ on $(0,\infty)$. Taking $1$st derivative of $f$: $f'(u) = -Ce^{-Cu} +\dfrac{1}{(u+1)^2} $. Your aim is to show: $f'(u) > 0$ on $(0,1)$ because if $u \ge 1$ then $f(u) > 0$ and the inequality you are trying to prove holds. But now $f'(u) > 0 \iff \dfrac{1}{\sqrt{C}}\cdot e^{\frac{Cu}{2}} \ge u+1$. Replace $C$ by $2C$ in this new inquality to get: $\dfrac{1}{\sqrt{2C}}\cdot e^{Cu} \ge u+1$. Now using the inequality: $e^{Cu} \ge 1+ Cu + \dfrac{C^2u^2}{2} \ge \sqrt{2C}u+\sqrt{2C}$. Consider $g(u) = 1+Cu+\dfrac{C^2u^2}{2}-\sqrt{2C}u - \sqrt{2C}$ on $(0,1)$. Then $g'(u) = C+C^2-\sqrt{2C}> 0 \iff C(1+C)^2 > 2$. Thus if you choose $C^{*} = 2C$ small $> 0$ and $C$ satisfy this condition then you have the inequality you want to show.


问:
Let $T,S: C[0,1]\rightarrow C[0,1]$, such that $S(x)(z)=z\int_{0}^{1}x(t)dt$, $T(x)(z)=s\cdot x(s)$. Calculate $||S||, ||T||, ||ST||$ e $||TS||$.
I know that $||T||=||S||=||TS||=1$, because I was able to calculate this. But, for $||ST||$, we had $||ST||\leq ||S|| ||T||=1$ and  $$||(S\circ T)(x)||_{\infty}=\sup_{z\in [0,1]}|z|\left|\int_{0}^1tx(t)dt\right|=\left|\int_{0}^1tx(t)dt\right|.$$

How $ST\neq TS$ trying to find a function $x_0\in C[0,1]$ such that $||ST(x_0)||_{\infty} \geq 1$. However, I'm starting to give up on that idea.

答:
Let $\|x\|_{\infty}=1$. Then
$$\|(S\circ T)(x)\|_{\infty}=\left|\int_{0}^1tx(t)dt\right| \leq \int_{0}^1\left|tx(t)\right|dt = \int_{0}^1 t \left|x(t)\right|dt \leq \|x\|_{\infty}\cdot \int_0^1 t dt = \int_0^1 t dt = \frac{1}{2} $$

Thus $\| S\circ T \|_{op} \leq \frac{1}{2}$.

On the other hand, if $x_0\equiv 1$, then $\|x_0\|_{\infty}=1$ and $\|(S\circ T)(x_0)\|_{\infty}=\frac{1}{2}$

Hence, $\| S\circ T \|_{op}=\frac{1}{2}$


问:
Solving a Polynomial Equation by Factoring.
$n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.

答:
Hint: remember the binomial theorem $(a+b)^3 = a^3 + 3 a^2 b + 3 b^2 a + b^3.$

Now guess what $a$ and $b$ might be here.


问:
Help with Cartesian product subsets
I want to prove that if  $A \subseteq C\,$ and $\,B \subseteq D,\,$ then $\,A \times B \subseteq C \times D.$

I know that $A \subseteq C \iff a \in A \rightarrow a \in C$ and that $B\subseteq D\iff b \in B \rightarrow  b \in D$  I also know that $A \times B = \{(a, b)\mid a\in A, b\in B\}$ and that $C\times D = \{(c, d) \mid c \in C, d \in D\}$.

So keep that in mind, how do I connect the dots?

答:
Here is a slightly more formal way to connect the dots: starting with the most complex expression, let's try to expand the definitions and simplify:
\begin{align}
& A \times B \subseteq C \times D \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& \langle \forall p : p \in A \times B : p \in C \times D \rangle \\
\equiv & \;\;\;\;\;\text{"basic property of $\times$, twice: split $p$ into its components $x$ and $y$"} \\
& \langle \forall x,y : x \in A \land y \in B : x \in C \land y \in D \rangle \\
\equiv & \;\;\;\;\;\text{"logic: split range -- to try and separate $x$ and $y$"} \\
& \langle \forall x,y : x \in A \land y \in B : x \in C \rangle \;\land\; \langle \forall x,y : x \in A \land y \in B : y \in D \rangle \\
\equiv & \;\;\;\;\;\text{"logic: in LHS, bring $y$ to the only place where it is used; same for $x$ in RHS"} \\
& \langle \forall x : x \in A \land \langle \exists y :: y \in B \rangle : x \in C \rangle \;\land\; \langle \forall y : \langle \exists x :: x \in A \rangle \land y \in B : y \in D \rangle \\
\equiv & \;\;\;\;\;\text{"logic: in LHS, extract $x$-less part from $\forall x$; same for $y$ in RHS"} \\
& (\langle \exists y :: y \in B \rangle \Rightarrow \langle \forall x : x \in A : x \in C \rangle) \;\land\; (\langle \exists x :: x \in A \rangle \Rightarrow \langle \forall y : y \in B : y \in D \rangle) \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$, twice; definition of $\emptyset$, twice; write $P \Rightarrow Q$ as $\lnot P \lor Q$, twice"} \\
& (B = \emptyset \lor A \subseteq C) \;\land\; (A = \emptyset \lor B \subseteq D) \\
\equiv & \;\;\;\;\;\text{"add $\lor A = \emptyset$ to LHS since $A = \emptyset$ implies $A \subseteq C$; same for RHS; simplify"} \\
& A = \emptyset \lor B = \emptyset \lor (A \subseteq C \land B \subseteq D) \\
\end{align}
This immediately implies the original statement.

Yes, this is longer than the other answers.  However, note how at most points there is really only one thing we could do to make progress: this kind of proof is easy to (re)construct.  And also note that we have not lost any information, so we proved something more general than was originally asked, viz., an equivalence which correctly accounts for the handling of the special case $A \times B = \emptyset$.

Title: Help with Cartesian product subsets


Title: Solving a Polynomial Equation by Factoring.

Body:
$n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.

Body:
Hint: \begin{align} n^3+12n^2+48n+64&= (n^3+64)+(12n^2+48n)\\  &= (n^3+4^3)+12n(n+4)\\
&=... \end{align}


答:
Only the case $x > 1$ is interesting, so we can substitute $y = 1/\sqrt x$. Then the inequality becomes
$$
 \frac{1-y}{1+y} \le e^{-Cy}
$$
or
$$
 \ln \left( \frac{1-y}{1+y} \right) \le -Cy
$$
for $0 < y < 1$.  Using the Taylor series for the logarithm we get
$$ \tag{*}
\ln \left( \frac{1-y}{1+y} \right) = -2(y + \frac{y^3}{3} + \frac{y^5}{5} + \ldots) \le -2y.
$$
It follows that the desired estimate holds with $\boxed{C=2}$. That is the best possible constant because
$$
 \lim_{y \to 0} \frac 1y \cdot \ln \left( \frac{1-y}{1+y} \right) = -2 \, .
$$

Remark: An alternative method to obtain $(*)$ is
$$
 \ln \left( \frac{1-y}{1+y} \right) = - \int_0^y \frac{2}{1-t^2} \,dt
\le - \int_0^y 2 \, dt =  -2y \, .
$$

Title: Why $\frac{\sqrt{x}-1}{\sqrt{x}+1} \leq \exp{\left(-\frac{C}{\sqrt{x}}\right)}$?

Body:
I am reading that

$$\frac{\sqrt{x}-1}{\sqrt{x}+1} \leq \exp{\left(-\frac{C}{\sqrt{x}}\right)}.\tag{1}$$

Doing the graph, we can see that $\exp{\left(-\frac{C}{\sqrt{x}}\right)}$ upper bounds $\frac{\sqrt{x}-1}{\sqrt{x}+1}$ when $C$ is small enough. Could you please someone provide some analytic explanation on how we can get $(1)$ and what is the maximum value of $C$ to achieve the upper bound?


答:
Here it is an approach based on the Binomial Theorem:

\begin{align*}
n^{3} + 12n^{2} + 48n + 64 = {3\choose 0}n^{3}4^{0} + {3\choose 1}n^{2}4^{1} + {3\choose 2}n4^{2} + {3\choose 3}n^{0}4^{3} = (n+4)^{3}
\end{align*}


问:
How to show that ET is finite in Gambler's ruin?
I am looking at the simplest application of optional stopping, the Gambler's ruin problem. In particular, in order to apply OST, we need to check that the martingale $X_n$ has bounded increments (this is easy to check for Gambler's ruin) and also $ET<\infty$. But I am not sure how to prove this. I found the this lecture slides online where on page 15 it says we don't need to check $ET<\infty$ but it suffices to know that $EX_{n\wedge T}<\infty$. Well, this makes sense but now we are applying the OST to a different martingale $X_{n\wedge T}$ rather than the original $X_n$. I wonder if there is a direct way to prove that $ET<\infty$?

答:
You have not defined $T$, and this matters.  Nor have you defined your Gambler's Ruin problem, but slides 4 and 14 suggest starting from $K$ and at each round you win $1$ with probability $\frac12$ and otherwise you lose $1$.

If you had defined $T$ as the first time you are ahead, as in slides 20 and 21 of your line, or if you had defined it as the first time after the start that you hit $K$, i.e. breaking even, then in either of those cases you would have $\mathbb E[T]=\infty$, even though in both cases you have $\mathbb P(T <\infty)=1$.

But slides 5 and 14 suggests that $T$ is the first time you hit $N$ or $0$ starting from $K$ with $0 < K < N$.  Try this argument:

You will hit at least one of $N$ or $0$ before you get $N$ consecutive wins, since in that case you must either start at $0$ or below, or end above $N$

The probability of $N$ consecutive wins starting at a particular point is $\frac1{2^N}$

The expected time until you get $N$ consecutive wins is then less than or equal to $N 2^N$, because you can bound this by: making $N$ attempts, seeing if you won all of them, and if not trying again

So for this $T$ you have $\mathbb E[T] \lt N2^N \lt \infty$

In fact $E[T]$ is much less than this and is in fact $K(N-K)$: for example if $N=5$ and $K=2$ then $E[T]=6$ rather than $160$.


问:
Show that $(1,2, ...,n)$ has at most $\varphi(n)/2$ of $\varphi(n)$-course numbers.
Given is the number $n$ which is divisible by the square of a prime number $p$.
Let a positive integer $a$ be euler-course when $n\mid (a^n -a)/a$.
Show that $(1,2, ...,n)$ has at most $\varphi(n)/2$ of  $\varphi(n)$-course numbers.

I was trying to solve this using primitive roots, by looking at the primitive root of $p^2$, but I couldn't quite do it.

Any hints or solutions are greatly appreciated.

答:
Clearly, $n | (a^n-a)/a=a^{n-1}-a \implies ord(a) | n-1$.

Also, $ord(a) | \lambda(n)$ where $\lambda(n)$ is the Carmichael function.

So, $ord(a) | \gcd(\lambda(n),n-1)=d$ (say).

Now, since $n$ is not square-free, $\lambda(n) \not | n-1$, as proven here.

Hence, $d$ is a proper divisor of $\lambda(n)$, so $d < \lambda(n)$.

Now, $a$ is a root of $x^d-1$, and the roots of $x^d-1$ form a subgroup of $\mathbb Z_n ^ \times$. The subgroup is proper, since $d < \lambda(n)$, S (say). So, $|S| \leq \lambda(n)/2 \leq \phi(n)/2$ i.e. the maximum possible number of such $a$'s is $\phi(n)/2$.


Body:
If $$P(n)=(n-a)^3$$
then $ a $ is a root of

$$P''(n)=0$$
with

$$P'(n)=3n^2+24n+48$$
and

$$P''(n)=6n+24$$
thus $$a=-4 \;;\;P(n)=(n+4)^3$$


问:
Meaning of certain cohomology classes and morphisms in Prop 3.3 of Deligne-Illusie's paper on mod $p^2$ liftings and decompos. of the de Rham complex
I am still struggling with Deligne and Illusie's paper (https://eudml.org/doc/143480). They say on page 261, in the course of the proof of theorem 3.3:

The class $e(K)$ (which is associated to the degree-$1$ morphism in $H^{0}(K)\longrightarrow K\longrightarrow H^{1}(K)[-1]\longrightarrow^{+1}$, where $K$ is a decomposable complex with $K^{i}=0$ for $i\notin\left\{0,1\right\}$ over a ringed site $(T,\mathcal{O})$) is also the class of the composite morphism
\begin{equation*}
H^{0}(T,\check{N}\otimes N)\longleftarrow^{q}_{\cong}H^{1}(T,\check{N}\otimes K)\longrightarrow^{-\operatorname{Id}_{N}\otimes\operatorname{pr}}H^{2}(T,\check{N}\otimes M)
\end{equation*}
where $M=H^{0}(K)$, $N=H^{1}(K)$, and $\check{N}$ denotes the dual of $N$.

Thus
\begin{equation*}
c=(h,-g,f)\in\check{C}^{1}(U_{\bullet},\check{N}\otimes K)=\check{C}^{2}(U_{\bullet},\check{N}\otimes M)\oplus\check{C}^{1}(U_{\bullet},\check{N}\otimes K^{0})\oplus\check{C}^{0}(U_{\bullet},\check{N}\otimes K^{1})
\end{equation*}
is a $1$-cocycle of image $\operatorname{Id}_{N}$ by $q$ and, and $-h$ through $-\operatorname{Id}_{N}\otimes\operatorname{pr}$. Therefore $e(K)=-\operatorname{cl sc}(K)$.  ($U$ is a hypercovering of $T$, and $\operatorname{cl sc}$ is the class associated to the gerbe $\operatorname{sc}(K)$ of decompositions of $K$.)

Questions:
(1.) How do the morphisms $q$ and $-\operatorname{Id}_{N}\otimes\operatorname{pr}$ come about and why is $q$ a quasi-isomorphism?

(2.) How does this $c$ come about and what does it have to with everything else?

(3.) Why does that imply that $e(K)=\operatorname{cl sc}(K)$?

I promise this will be the last question concerning their paper. This paper is actually quite beautiful and easy to read, but I am having problems with chapter 3, probably because I am only superficially acquainted with sheaf cohomology on sites. Thanks in advance anyway.

答:
Let's follow the argument. first you want the inverse image of $id_N\in H^0(N\otimes\check N)=Hom(N,N)$ with the map $q$. the map $E\to N$ by definition factors through $K^1$, and $K^1\otimes\check N \to N\otimes \check N$ is surjective and first cohomology is always computed by Čech cohomology. so as usual you have to first find a cover $U$ and a lift of $id_N$ to an element of $C^0(K^1\otimes\check N)=Hom_U(N,K^1)$ which the call $f$,in now consider$$C^1(E\otimes \check N)=C^2(M\otimes\check N)\oplus C^1(K^0\otimes\check N)\oplus C^0(K^1\otimes\check N) $$(this is the usual way you compute the Čech complex of a complex,you have to first write Čech complex of each term which gives you a double complex and then you take the total complex of that double complex).

now you want a cocycle  here which has $f$ as the third component, if you write the boundary map you understand that the cocycle should be of the form $c=(h,g,f)$ where $$dg=d_0f+d_1f,h=d_0g-d_1g+d_2g.$$

the upshot is that $$e(K)=-id\otimes pr_0(q^{-1}(Id_N))=id\otimes pr_0(c)=-h.$$ now you explicitly computed $e(K)$. you can also explicitly compute the class in $H^2$ associated to a gerb(in our case sc(K)) and see that this is also $h$. in the first paragraph after proposition they give a reference for this fact:(12,IV 3.5) hence we win.

note: there is a typo in the diagram in the paper it should be $H^1(T,E\otimes \check N)$ instead of $H^1(T,K\otimes \check N)$.


问:
Evaluating limit of sequence of integrals
I want to find the limit of

$$  \lim_{n \rightarrow \infty} \int_{a}^{b} (1+\frac{x}{n})^{n} e^{-x}dx  $$ where $ b > a >0$

Since, $\lim_{n \rightarrow 0} (1+\frac{x}{n})^{n}= e^{x}$, the sequence of functions do not converge uniformly so I cannot exchange the limit and the integral. If I could do that, then $e^{x}$ and $e^{-x}$ would cancel out and the final result would be $b-a$. I know that by using the Lebesgue Dominated Convergence Theorem I can show that the answer would still be $b-a$. But, is there any way to show that without using that theorem?

答:
First a comment

I don’t understand your argument Since, $\lim\limits_{n \rightarrow \infty} (1+\frac{x}{n})^{n}= e^{x}$, the sequence of functions do not converge uniformly. For example, the constant sequence of maps $a_n(x)=e^x$ is obviously uniformly converging to the map $x \mapsto e^x$ on $\mathbb R$.

About the question itself

You can prove that for $0 \lt a \lt b$, $\left(1+\frac{x}{n}\right)^n$ converges uniformly to $e^x$ on $[a,b]$. Hence you can switch the limit and the integral to conclude by using a classical Riemann integration theorem.


问:
A theorem on hermitian forms
In Lang's Linear Algebra the following theorem was mentioned:

I am not sure if I truly understand the result. The author here is referring to the Representation theorem, so I wrote the following proof:
For any functional $L$ on $V$ there exists a unique $v$ in $V$ such that $L(w)=\langle w,v\rangle$. In particular if we define for a fixed $w$ in $V$ the functional $L_w$ by $L_w(v)=\langle Av,w\rangle$ then there is a unique $w'$ in $V$ such that $L_w(v)=\langle v,w'\rangle$ for each $v$ in $V$.

The failure of the association mentioned in the remark to be an isomorphism is due to the fact that the same vector could be associated to different functionals on $V$ which makes the association not even a linear map on $V$. Is that right? And is my proof fine? Any help is appreciated!

答:
No. The failure of the association to be an isomorphism is due to the fact that the form is hermitian which in particular means that it is linear in one argument but anti-linear in the other, i.e.

$$
\langle \alpha v, w\rangle = \alpha\langle v, w\rangle \\
\langle v, \alpha w\rangle = \overline{\alpha} \langle v, w\rangle.
$$

Consequently, upon fixing $w$ the associated $L_w = \langle ., w\rangle$ is linear, but the mapping $w \mapsto L_w$ is antilinear. To see the former, write

$$
L_w(\alpha v) = \langle \alpha v, w\rangle = \alpha\langle v, w\rangle = \alpha L_w(v)
$$

and to see the latter write

$$
L_{\alpha w} = \langle ., \alpha w\rangle = \overline{\alpha}\langle ., w\rangle = \overline{\alpha} L_w.
$$

Note that the mapping $w \mapsto L_w$ is a bijection and hence an anti-isomorphism, see Riesz representation theorem.

Title: A theorem on hermitian forms


答:
You can simplify expression to see what happens.

For example, you see that $12,48,64 $ have a common divisor: $4$

If you accept $4=n$, then you get

$$m^3+3m^2n+3mn^2+n^3=(m+n)^3.$$


问:
Ring homomorphisms which map a unit to a unit map unity to unity?
this is the third part of a question I've been working on from Hungerford's Algebra. It is exercise 15 in the first section of Chapter III.

$(c)$ If $f\colon R\to S$ is a homomorphism of rings with identity and $u$ is a unit in $R$ such that $f(u)$ is a unit in $S$, then $f(1_R)=1_S$ and $f(u^{-1})=f(u)^{-1}$.

I see how $f(u^{-1})=f(u)^{-1}$ follows from $f(1_R)=1_S$, for if that is so, then
$$f(1_R)=f(uu^{-1})=f(u^{-1}u)=1_S\implies f(u)f(u^{-1})=f(u^{-1})f(u)=1_S.
$$
It seems easy, but I can't manage to show $f(1_R)=1_S$. I was hoping that someone could perhaps give me a hint on how to proceed on showing the first part? Thank you.

答: You can cancel $f(u)$ in the equation $f(u)=f(1_R)f(u)$.</p>


问:
For any solutions $\phi_1(t)$ and $\phi_2(t)$ of $y''+py'+qy=f(t)$ we have $\lim_{t\to\infty}[\phi_1(t)-\phi_2(t)]=0$.
How can I prove this statement?

If $p$ and $q$ are positive constants and $f:(0,+\infty)\rightarrow \mathbb{R}$ is a continuous function, show that for any solutions $\phi_1(t)$ and $\phi_2(t)$ of $y''+py'+qy=f(t)$ we have
$$\lim_{t\to\infty}[\phi_1(t)-\phi_2(t)]=0.$$

答:
Hint.

As

$$
\cases{
\phi_1''+p\phi_1'+q\phi_1 = f\\
\phi_2''+p\phi_2'+q\phi_2 = f
}\Rightarrow (\phi''_2-\phi''_1)+p(\phi'_2-\phi'_1)+q(\phi_2-\phi_1)=0
$$

now calling $\psi =\phi_2-\phi_1$ the ODE $\psi''+p\psi'+q\psi=0$ has as differential operator $s^2+ps+q = (s+p/2+\sqrt{p^2-4q}/2) (s+p/2-\sqrt{p^2-4q}/2)$ so the real component of both roots is negative and the solution is exponentially decreasing with time.


问:
Prove that an infinte ring with identity must have an infinite number of non-invertible elements
The problem is to prove that an infinte ring $K$ with identity can not contain a finite number $n \ge 1$ of non-invertible elements.

The hint advices to use a proof by contradiction. Assume that $N = \{a_1, \ldots, a_n\}$ is the set of all non-invertible elements of $K$ without $0$. A map $p_x: a_i \to xa_i$ is a bijection $N \to N$, $\forall x \in K \setminus(N \cup{0})$. And the kernel of the map $p: x \to p_x$ is infinite.

I wanted to show that in this case the map $p$ is an isomorphism from $K \setminus(N \cup{0})$ to a subset $H \subset S_n$ of all permutations on $N$ and therefore $K \setminus(N \cup{0})$ must contain a finite number of elements, but the kernel of $p$ which is a susbset of $K \setminus(N \cup{0})$ has an infinite number of elements by contradition. I could make a mistake, but I see that $p$
is indeed the isomorphism. I'm not able to prove that its kernel has an infinite number of elements. By its kernel I assume $\ker p = \{ x \in K \setminus(N \cup{0})| p_x = e \}$, where $e$ is the identity permutation on $N$, but the author could think of something different. I can only think of $p_1$ where $1$ is an identity element in $K$ .

答:
Assume that $K$ is an infinite ring whose set of nonzero nonunits is finite. Let $N= \{a_1,\ldots,a_n\}$ be the set of nonzero nonunits. Let $K^{\times} = K\setminus (N\cup\{0\})$ be the group of units of $K$. As noted in the problem, the infinite group $K^{\times}$ acts by left multiplication on the finite set $N$, yielding a permutation representation $p\colon K^{\times}\to S_N\colon x\mapsto p_x$. This representation must have infinite kernel.
If $g_1, g_2, \ldots \in \ker(p)$ are infinitely many distinct elements of this kernel, then
$$A = \{g_2-g_1, g_3-g_1, g_4-g_1\ldots\}$$
is an infinite set of distinct nonzero
elements which annihilate $N$ from the left.
(That is, $(g_i-g_1)a_j = g_ia_j-g_1a_j = a_j-a_j = 0$ for all $i, j$.) But the only nonzero elements of $K$ that can annihilate nonzero elements of $K$ lie in $N$, so we are led to the conclusion that the infinite set $A$ is a subset of the finite set $N$.


问:
Solution for $y''+4y'+8y=x+e^x$.
How can I solve this ODE?

$$y''+4y'+8y=x+e^x$$

After I find the roots of the characteristic and create the solution for the homogeneous, how can I find the solution for this ODE?

答:
hint

Find a particular polynomial solution $ y_1$ of the equation

$$y''+4y'+8y=x$$

and a particular solution $ y_2=z_2e^x$ of
$$y''+4y'+8y=e^x$$

and sum them to get a particular solution $ y_p $.

The general solution is $$ y_g=y_h+y_p$$
$y_h $ is your homogeneous solution.

You should find that
$$y_p=\frac{2x-1}{16}+\frac{e^x}{13}$$

Title: Solution for $y''+4y'+8y=x+e^x$.


问:
For $n,k \in \mathbb{N}$, prove that $\left \lceil \frac{n}{k} \right \rceil=\left \lfloor \frac{n-1}{k} \right \rfloor +1 $.
For $n,k \in \mathbb{N}$, prove that $\left \lceil \frac{n}{k} \right \rceil=\left \lfloor \frac{n-1}{k}  \right \rfloor+1 $.

答:
The number $a = \lfloor \frac{n-1}{k} \rfloor$ is the unique nonnegative integer satisfying $n-1 = ak + r$ with $0 \leq r < k$.
This condition on $r$ implies the fraction $\frac{r+1}{k}$ is in the interval $[1/k, 1]$. Therefore,
$$
\left\lceil \frac{n}{k} \right\rceil
= \left\lceil \frac{ak+r+1}{k} \right\rceil
= a + \left\lceil \frac{r+1}{k} \right\rceil
= a + 1 .
$$


问:
For which $\alpha$ this sum converges? $\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$
Given:

$$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$$
I am asked: For what values of $\alpha$ does this summation converge?

So I said, $f(n) = \frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}$. $f(n)$ is obviously monotonically decreasing function. Then, by using the integral test, this summation converges if and only if $I = \int_{3}^{\infty} {\frac{dn}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$ converges. (has a value)

But I am finding this very hard to go on with. Any direction will be appreciated!

答:
By the Cauchy Condensation Test:
$$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}} \text{ converges} \iff $$
$$\sum_{n=3}^{\infty} {\frac{2^n}{2^n \cdot \ln(2^n) \cdot \ln(\ln(2^n))^{\alpha}}} =
\sum_{n=3}^{\infty} {\frac{1}{n \ln(2) \cdot \ln(n\ln(2))^{\alpha}}} \text{ converges} \iff $$
$$\sum_{n=3}^{\infty} {\frac{1}{(n\ln(2)+\ln(\ln(2)))^{\alpha}}} \text{ converges} \iff \alpha>1$$</p>


答:
If $k\mid n$, then this result is immediate.

Suppose $k\nmid n$, then $\lceil \frac{n}{k}\rceil = 1 + \lfloor \frac{n}{k} \rfloor$.

Lastly, can you reason why $\lfloor \frac{n}{k} \rfloor = \lfloor \frac{n-1}{k} \rfloor$.

Title: For $n,k \in \mathbb{N}$, prove that $\left \lceil \frac{n}{k} \right \rceil=\left \lfloor \frac{n-1}{k} \right \rfloor +1 $.

Body: For $n,k \in \mathbb{N}$, prove that $\left \lceil \frac{n}{k} \right \rceil=\left \lfloor \frac{n-1}{k}  \right \rfloor+1 $.


问:
Prove that curve is a straight line iff curvature vanishes.
For my class in classical differential geometry I have to solve the following problem:

Let $\alpha$ be a regular curve in $\mathbb{R}^2$. Proof that:

a) $\quad$ $\alpha$ is a straight line if and only if the curvature vanishes.

b) $\quad$ If $\dot{\alpha}$ and $\ddot{\alpha}$ are linearly dependent, then $\alpha$ is a straight line.

Conisder a). At first this problem seemed easy to me. Because if the curvature is defined via

$$\ddot{\alpha}(t) = \kappa(t) \cdot n(t) \qquad \qquad (1)$$

where $n(t)$ is the unit vector perpendicular to the tangent vector, then it is easy to see that $\kappa = 0$ implies $\ddot{\alpha} = 0$ and therefore $\alpha$ is a straight line. However, I realsied, that the above equation is derived by using the fact

$$\big< \dot{\alpha} , \dot{\alpha} \big> = 1 \quad \Longrightarrow \quad \big< \dot{\alpha}, \ddot{\alpha} \big> = 0$$

in the case of $\alpha$ being parameterised by arc length. Since my given $\alpha$ is not parameterised by arc length, I cannot assume equation (1) to hold and therefore cannot assume my $\ddot{\alpha}$ to be perpendicular to the tangent vector. My conclusion would be wrong then. Any ideas on how to solve this problem for general curves $\alpha$?

b) I have no idea on how to show this. Any tips here?

答:
Bearing in mind that any regular curve can be parametrized by the arc length then

a) if $\alpha$ is a straight line, i.e., $\alpha(t)=\text{p}+\text{v}t$ and so  $k(t)=||\alpha''(t)||=||0||=0$.

If $k(t)=0$ then $\alpha''(t)=0$. If you calculate the antiderivative twice you obtain $\alpha'(t)=\text{v}$ and finally $\alpha(t)=\text{p}+\text{v}t$ for some $\text{p,v}\in\mathbb{R^2}$.

b) As you saw $\langle\alpha',\alpha''\rangle=0$ then $\alpha''$ is null vector or $\alpha'$ is perpendicular to $\alpha''$, but as they are linear dependent then $\alpha''=0$ and by a) it can only be a straight line.


问:
Roots of a certain sixth order polynomial
I am looking for the roots (or basically any information regarding them) of the sixth order polynomial
$$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$
for positive, real constants $a,b$.
Since $p(0)=-b^2<0$ and $\lim_{x\to\pm\infty}p(x)=+\infty$ we have at least two real roots $x_0,x_1$. Moreover, plotting the function for several parameters suggests, that the remaining four roots are non-real (i.e. two pairs of complex conjugate solutions).

Unfortunately this is as far as I got. I am especially interested in the complex solutions and their behavior for large parameters $a,b$.
Any idea on how to tackle this problem is very welcome!

答:
By Descartes rule of signs, the polynomial $$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$ has exactly one positive root. It is so because there is exactly one change of sign in $p(x).$

Similarly, there is exactly one change of sign in $p(-x).$ Our polynomial has one negative root.

Some bounds of real roots are found below.

Consider a monic polynomial $$q(x):=x^6+\frac{a+1}{a}x^4+2\frac{b}{a}x^3-\frac{b^2}{a}.\tag 1$$ Clearly, $q(x)$ and $p(x)$ share the roots.

I assume that $a+1<2b$ and $b>2,$ from where $2\frac{b}{a}<\frac{b^2}{a}.$

By Lagrange theorem 1.3., an upper bound for the positive real root is
$\left(\frac{b^2}{a}\right)^{1/6}.$
Replacing $x$ by $-x$ in $(1)$ we get a lower bound for negative real root
$-\left(2\frac{b}{a}\right)^{1/3}-\left(\frac{b^2}{a}\right)^{1/6}.$
Putting together, real roots $r_i$ satisfy
$$-\left(2\frac{b}{a}\right)^{1/3}-\left(\frac{b^2}{a}\right)^{1/6}<r_i<\left(\frac{b^2}{a}\right)^{1/6}.$$

The remaining four roots are in conjugate pairs, $\;(\alpha\pm i\beta)\;$ and $\;(\gamma\pm i\delta).$ From the coefficients of $(1)$ we deduce that the sum of real parts of all six roots is $0,$ and the product of the roots is $-b^2\over a.$ The latest can be formulated as "the product of modules of the roots equals $b^2 \over a$".

"Lagrange over $\mathbb{C}$" (Theorem 1.5 in the above mentionned article) gives an upper bound for the absolute values of the roots $w_i,$ real or complex:
$$|w_i|<\left(2\frac{b}{a}\right)^{1/3}+\left(\frac{b^2}{a}\right)^{1/6}.$$


问:
Let $m,n,b \in \mathbb{N} $ with $b > 1$ and $m \neq n$.
Let $m,n,b \in \mathbb{N} $ with $b > 1$ and $m \neq n$.

If $b^{m}-1 $ and $b^{n}-1 $ have the same prime divisors, prove that $b+1$ is a power of 2.

I know that $b^{m}-1 = (b-1)(b^{m-1}+b^{m-2}+\cdots+ 1 )$ and same for $b^{n}-1$

Or maybe, I should consider $b^{m}-1 -b^{n}+1 = b^{m} - b^{n}$.

I really have no idea on how to tackle this problem.

答:
Please note that this solution is due to Kamil Duszenko.

Consider first the case $n = 1$. So suppose first that $a - 1$ and $a^m - 1$ have the same prime divisors. If a prime p divides m, then $a - 1$ divides $a^p - 1$, and $a^p - 1$ divides $a^m - 1$, so a - 1 and $a^p - 1$ have the same prime divisors. But $a^p - 1 = (a - 1) L$, where $L = a^{p-1} + a^{p-2 }+ ... + a + 1$. So if a prime q divides L, then it must divide $a^p - 1$ and hence $a - 1$. But $L = (a - 1) (a^{p-2} + 2 a^{p-3} + ... + p-1) + p$, so any prime q dividing L and $a - 1$ must also divide p. So L must be a power of p, m must be a power of p, and p must divide $a - 1$. In other words, $a = 1 \, mod \, p$. Hence $(a^{p-2}+ 2 a^{p-3 }+ ... + p-1) = 1 + 2 + ... + p-1 = p(p-1)/2 \, mod \, p$. But if $p > 2 \, , p(p-1)/2 = 0 \, mod \, p$, so $L = p^2 + p \, mod \, p^2$. But L is a power of p, so $L = p$. Contradiction, since L is obviously > p (or p > 2). Hence p must be 2. Hence $L = a + 1$, and $ a + 1$ is a power of 2.

Now take the general case. Suppose d is the greatest common divisor of m and n. Put $m = dM \, ,  n = dN$ and $ a = b^d$, so that $a^M - 1$ and $a^N - 1$ have the same prime divisors, and M, N are relatively prime. Take positive integers h, k such that $hM - kN = 1$, then $(a^{hM}- 1) - (a^{kN} - 1) = a^{hM} - a^{kN} = a^{kN}(a - 1)$. So if s > 1 divides $a^M - 1$ and $a^N - $1, then it also divides $a^{hM} - 1$ and $a^{kN} - 1$ and hence also $a^{kN}(a - 1)$. It cannot divide $a^{kN}$, so it must divide $a - 1$. So the greatest common divisor of $a^M - 1$ and $a^N - 1$ must divide $a - 1$. But $a - 1$ divides both, so the greatest common divisor is $a - 1$. So $a- 1$ must also have the same prime divisors as $a^M - 1$. It follows from the special case that $a + 1 = b^d + 1$ is a power of 2. Since $b > 1, b^d + 1 > 2$, so $ b^d + 1$ is certainly divisible by 4. If d is even, then $odd^d = 1 \, mod \, 4$ and $even^d = 0 \, mod \, 4$, so $b^d + 1 = 2 $ or $1 \, mod \, 4$. Hence d must be odd. Hence $b + 1$ divides $b^d + 1$, so $b + 1$ is also a power of 2.


答: Hint for $(a)$ : WLOG Assume $\alpha(0)=0$ (Curvature is invariant under isometries) and try looking at $\gamma(s)=(\int_0^scos(\theta(u))du, \int_0^ssin(\theta(u))du)$.

Title: Prove that curve is a straight line iff curvature vanishes.

Body:
For my class in classical differential geometry I have to solve the following problem:

Let $\alpha$ be a regular curve in $\mathbb{R}^2$. Proof that:

a) $\quad$ $\alpha$ is a straight line if and only if the curvature vanishes.

b) $\quad$ If $\dot{\alpha}$ and $\ddot{\alpha}$ are linearly dependent, then $\alpha$ is a straight line.

Conisder a). At first this problem seemed easy to me. Because if the curvature is defined via

$$\ddot{\alpha}(t) = \kappa(t) \cdot n(t) \qquad \qquad (1)$$

where $n(t)$ is the unit vector perpendicular to the tangent vector, then it is easy to see that $\kappa = 0$ implies $\ddot{\alpha} = 0$ and therefore $\alpha$ is a straight line. However, I realsied, that the above equation is derived by using the fact

$$\big< \dot{\alpha} , \dot{\alpha} \big> = 1 \quad \Longrightarrow \quad \big< \dot{\alpha}, \ddot{\alpha} \big> = 0$$

in the case of $\alpha$ being parameterised by arc length. Since my given $\alpha$ is not parameterised by arc length, I cannot assume equation (1) to hold and therefore cannot assume my $\ddot{\alpha}$ to be perpendicular to the tangent vector. My conclusion would be wrong then. Any ideas on how to solve this problem for general curves $\alpha$?

b) I have no idea on how to show this. Any tips here?


问:
Vectorial $L^p$ spaces
Consider the space

$$Ł^p(\Omega):=(L^p(\Omega))^N=L^p(\Omega)×L^p(\Omega)×...×L^p(\Omega), \, N \ge 1,\, \Omega \subset \mathbb{R^n},$$

as the vectorial $L^p$ space associated to the scalar one.

My questions are:

What is the norm $||\cdot||_{Ł^p}$ and can we relate it to $||\cdot||_{L^p}$?

How to apply the Hölder Inequality on

$$\int_{\Omega}{u(x)\cdot v(x)dx},$$

where

$u \in Ł^q$, $v \in Ł^p$,
$\frac{1}{p}+\frac{1}{q}=1$, $u=(u_1,...,u_N)^t$,
$v=(v_1,....,v_N)^t$,
$u_i \in L^q(\Omega)$, $v_i \in L^p(\Omega)$, and $u(x)\cdot v(x)$ is the scalar product between?

答:
A nice norm is
$$
\lVert u \rVert_{L^p(\Omega)^N} := \sqrt[p]{\sum_{i = 1}^N \lVert u_i \Vert_{L^p(\Omega)}^p}.
$$
If $\lVert \cdot \rVert_{\mathbb{R}^N}$ is any norm on $\mathbb{R}^N$, then it is easy to see that  $\lVert u \rVert_{L^p(\Omega)^N}$ is equivalent to
$$
\lVert u \rVert := \lVert (\lVert u_1 \rVert_{L^p(\Omega)}, ...,  \lVert u_N \rVert_{L^p(\Omega)})^\top \rVert_{\mathbb{R}^N}.
$$
In the case of $\lVert \cdot \rVert_{L^p(\Omega)^N}$ we had set $\lVert \cdot \rVert_{\mathbb{R}^N} := \lVert \cdot \rVert_p$.

The Hölder inequality also holds in the following sense: We first use the Hölder-inequality of $\mathbb{R}^N$ inside the integral and then use the Hölder-inequality for Lebesgue-integrals:
$$
\int_\Omega u \cdot v ~\mathrm{d}x \leq \int_\Omega \lVert u \Vert_p \lVert v \Vert_q ~\mathrm{d}x \leq \sqrt[p]{\int_\Omega \lVert u \rVert_p^p ~\mathrm{d}x} \sqrt[q]{\int_\Omega \lVert v \rVert_q^q ~\mathrm{d}x} =
$$
$$
\sqrt[p]{\sum_{i = 1}^N \int_\Omega \lvert u_i \rvert^p~\mathrm{d}x}\sqrt[q]{\sum_{i = 1}^N \int_\Omega \lvert v_i \rvert^q~\mathrm{d}x} = \sqrt[p]{\sum_{i = 1}^N \lVert u_i \rVert_{L^p(\Omega)}^p} \sqrt[q]{\sum_{i = 1}^N \lVert v_i \rVert_{L^q(\Omega)}^q} =
\lVert u \rVert_{L^p(\Omega)^N} \lVert v \rVert_{L^q(\Omega)^N}
$$

Title: Vectorial $L^p$ spaces


问:
Prove or refute - FO logic and structures

Let $\sigma$ be a signature with a constant symbol $c$ and let $\Phi \in FO[\sigma]$ and $\psi(x) \in FO[\sigma]$. Prove or refute the following statements:

$(i)$ If $\Phi \models \forall x \psi(x)$, then $\Phi \models \psi(c)$.

$(ii)$ If $\Phi \models \exists x \psi(x)$, then $\Phi \models \psi(c)$.

$(iii)$ If $\Phi \models x \neq x$ , then $\Phi \models x = x$.

$(iv)$ $\exists x \psi(x) \models \exists x (\psi(x) \land x \neq c)$.

My attempt:

$(i)$ is correct, since we're quantifying over all the values of the free variable $x$ in the structure's universe and the constant $c$ has to come from the same universe, so at some point $x$ has to be equal to $c$.

$(ii)$ is incorrect, because we don't know anything about $c$. (Question: Since we don't know what our universe is, how can we give a counterexample here?)

$(iii)$ is correct, since $x \neq x$ is always false, the statement $\Phi \models x \neq x$ is also false, making the implication $\Phi \models x \neq x \rightarrow \Phi \models x = x$ true.

$(iv)$ is incorrect, because $c$ can be the only $x$ for which $\psi(x)$ is true.

I'm very sceptical of my answers, what do you think?

答:
I’m convinced by what you did.

Regarding point (ii). To find a counterexample, we just have to find a model in which $\Phi$ and $\exists x \psi(x)$ hold but $\psi(c)$ doesn’t.

For this take $M=\{1,2\}$, $\Phi \equiv c=c$, $\psi(x) \equiv x=1 \wedge \neg x=2$ and interpret $c $ as $2$.


问:
Communicating mathematics
As a TA, my ratings have been fairly mediocre.  There's nothing to suggest I'm a problem, but I have room to improve.

What's more disturbing to me is that I even struggle to communicate math with people who know a lot more than I do.  It's no surprise that they struggle to communicate with me.  They know more, probably forgot what it's like to know less, and generally do not give the details on everything, sometimes even neglecting to mention that a particular detail needs to be filled in.  But when I communicate, I try to leave out as few details as possible and clearly enunciate when I am leaving them out, and to minimize the number of mistakes I make.  I am extremely nitpicky about this.

So there must exist other obstacles to communication other than just misrepresentations.  I suspect that what I'm lacking is a pedagogical element, which would also explain my interaction with students.  How can I go about finding out what is lacking, and then fixing it?  What are some common teaching mistakes made by people with the attitude that mathematics should be precise and airtight?  It's not that I'm such an adherent of my own philosophy that I refuse to "dumb things down."  That used to be the problem, but it appears that when I attempt to convey vague ideas, people do not understand me.

答:
Most people, when presented with a claim, need a certain amount of time and/or contextual information to understand not why that claim holds, but what it means and why it matters.

Simple example: if you were to present to me a theorem about simple groups, having defined them, I would be staring at you wondering why I should care, because I have no concept of what makes simple groups important.

Title: Communicating mathematics


问:
There exists a nonzero element $x \in W$ such that $x \perp V$ if $\dim(W) > \dim(V)$
Consider an $n$-dimensional Euclidean space $U$. Let $V, W \subseteq U$ be two subspaces with dimensions $\dim(V) = k$ and $\dim(W) = k+1$ for some $k < n$.

How can we prove that there exists a nonzero element $x \in W$ such that $x \perp V$, i.e., the inner product $x^T y = 0$ for any $y \in V$?

If $V$ is a subspace of $W$, the above is obvious. But what if it is not? Since $W$ has a higher dimension, we know there must exist a nonzero element $x \in W$ that does not belong to $V$. But I failed to go any further...

答:
We have $\dim (V^{\perp} + W) = \dim V^{\perp}+\dim W - \dim (V^{\perp}\cap W) $.

Then $\dim V^{\perp}=n-k$, $\dim W=k+1$, and $\dim (V^{\perp}+W) \leq n$.

Thus, $\dim (V^{\perp} \cap W) \geq 1$.


问:
5 and 7 point regular 3d shape
Assume a situation where there are n balls tied to a point individually using strings of equal lengths.

Each ball repels the other ball.

Now,

For n=2, the balls would form a line - 180˚,

n=3, the balls would form a triangle - 120˚,

n=4, the balls would form a tetrahedron - 108˚30'28", and so on.

n=6 - octahedron - 90˚

n=8 - cube - 54.74˚

My question is, what would the shape be for n=5 and n=7?

Is there a name for that shape? What would the angle formed be?

答:
Here's a paper that answers that question for $n = 5$, I believe (at least if you assume that all the strings have the same length):

https://www.math.brown.edu/reschwar/Papers/electron.pdf

The shape is apparently called "the triangular
bi-pyramid" and consists (after a suitable rotation) of the north and south poles, and three points lying on an equilateral triangle on the equator.

My guess is that there's no known-and-proved answer for $n = 7$, and given the difficulty of the $n = 5$ case (and Schwartz's remark about a "sense of diminishing returns"), I wouldn't guess that there's anyone seriously working on it, but I could easily be wrong.

Title: 5 and 7 point regular 3d shape


答:
Write

$$V+W = V_1 \oplus (V \cap W ) \oplus W_1$$

where $V = V_1 \oplus (V \cap W) $ and $V_1 \perp (V \cap W)$ and similar equalities for $W_1$. With finally also $V_1 \perp W_1$. For this, we can take for $V_1$ the orthogonal complement of $V \cap W $ in $V$. And for $W_1$ the orthogonal complement of $V_1 \oplus (V \cap W)$ in $V +W$.

Based on the given dimensions for $V,W$, we have $\dim W_1 \ge 1$ and any non zero element $x \in W_1$ satisfies the desired result.

Title: There exists a nonzero element $x \in W$ such that $x \perp V$ if $\dim(W) > \dim(V)$

Body:
Consider an $n$-dimensional Euclidean space $U$. Let $V, W \subseteq U$ be two subspaces with dimensions $\dim(V) = k$ and $\dim(W) = k+1$ for some $k < n$.

How can we prove that there exists a nonzero element $x \in W$ such that $x \perp V$, i.e., the inner product $x^T y = 0$ for any $y \in V$?

If $V$ is a subspace of $W$, the above is obvious. But what if it is not? Since $W$ has a higher dimension, we know there must exist a nonzero element $x \in W$ that does not belong to $V$. But I failed to go any further...


问:
Is the identity map a retraction?
I'm sort of wrestling with the definition of retraction in my head. The definition I was given was the following:

A retraction of a space $X$ onto a subspace $Z \subseteq X$ is a continuous map $r$ s.t. $r(z) = z \in Z, \forall z \in Z$.

Now to me this just seems like the definition of an identity map, so essentially the map $r$ is an identity on $Z$ but doesn't really care about what happen to points outside of $Z$.
So my question is the identity map a retraction? If it is then wouldn't any subspace of a topological space be a retract of the identity map? Can someone clarify or confirm whether my reasoning is correct or not?

答: The retraction must be onto $\color{red}Z$, it can't leave the points outside of $Z$ unchanged, they must go into $Z$.

Title: Is the identity map a retraction?


问:
Derivative of Zonal Polynomial w.r.t. Matrix Argument
Let $\mathbf{T}$ be a real symmetric positive definite matrix. And let $\mathbf{Z}_\kappa\left(\mathbf{T}\right)$ denote the zonal polynomial as defined here.

Does
$$
\begin{equation}
\frac{\partial \mathbf{Z}_\kappa\left(\mathbf{T}\right)}{\partial \text{vech}(\mathbf{T})}
\end{equation}
$$
exist in closed form?

I couldn't find any results in the academic literature.

答:
Your question asks for the derivative of $Z_\kappa(\mathbf{T})$ with respect to {\it every} entry $t_{i,j}$ of the $m \times m$ positive definite (symmetric) matrix $\mathbf{T}$.  This is generally an intractable problem even for small $m$ and partitions $\kappa$ of low weight.  For instance, suppose $m=3$ and $\kappa = (2,2,1)$, the use of {\sl Mathematica} to calculate $\partial Z_\kappa(\mathbf{T})/\partial t_{1,1}$, the derivative with respect to $t_{1,1}$, leads to results that are impenetrable, i.e., do not seem to have any structure.

So this raises the issue of whether the problem that motivates your question really require all such derivatives.  Perhaps you may wish to elaborate on the motivating problem.

Noting that the zonal polynomials $Z_\kappa(\mathbf{T})$ are othogonally invariant functions of $\mathbf{T}$, and therefore depend only on the eigenvalues of $\mathbf{T}$, perhaps you only need the derivatives of $Z_\kappa(\mathbf{T})$ with respect to the eigenvalues of $\mathbf{T}$.  If this is the case then see the book, Aspects of Multivariate Statistical Theory'' (Wiley, New York, 1982), or the paper, Applications of invariant differential operators to multivariate distribution theory,'' SIAM J. Appl. Math., 45 (1985), 280--288.

Or perhaps you only need to calculate $\phi(\partial/\partial\mathbf{T}) Z_\kappa(\mathbf{T})$, where $\phi(\cdot)$ is an orthogonally invariant function of $\mathbf{T}$.  In that case, a generating-function approach to evaluating such derivatives is given in two papers, Differential operators associated with zonal polynomials. I,'' Ann. Inst. Statist. Math. 34 (1982), 111-–117, and Differential operators associated with zonal polynomials. II'' Ann. Inst. Statist. Math. 34 (1982), 119--121.


答: You've correctly deduced that the identity map $i:X \rightarrow X $ is a retraction. Note that if $Z$ is a proper subset of $X$ then the function $f: X \rightarrow Z$ cannot be the identity as it would fail to be a function.

Title: Is the identity map a retraction?

Body:
I'm sort of wrestling with the definition of retraction in my head. The definition I was given was the following:

A retraction of a space $X$ onto a subspace $Z \subseteq X$ is a continuous map $r$ s.t. $r(z) = z \in Z, \forall z \in Z$.

Now to me this just seems like the definition of an identity map, so essentially the map $r$ is an identity on $Z$ but doesn't really care about what happen to points outside of $Z$.
So my question is the identity map a retraction? If it is then wouldn't any subspace of a topological space be a retract of the identity map? Can someone clarify or confirm whether my reasoning is correct or not?


问:
Centered-Difference Scheme to approximate BVP
I have the following boundary value problem:

$\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}=0, \; 1<x<2, \;0<y<1;$

$u(x,0)=2\ln x, \; u(x,1)=\ln(x^2+1), \; 1 \leq x \leq 2;$

$u(1,y)=\ln(y^2+1), \; u(2,y)=\ln(y^2+4), \; 0 \leq y \leq 1$

and I need to use the scheme of centered-differences to approximate the second partial derivatives and find the difference scheme for the elliptic equation. Once I have this, I must verify that it's second-order accurate with respect to $\Delta x$ and $\Delta y$.

My ideas:

I know that the following is the centered-difference equation for the first derivative (please correct me if I'm wrong!):

$f'(x) \approx \frac{f(x+h)− f(x−h)}{2h}.$

So I think this would be the formula for the second equation (again, please correct me if I'm wrong):

$f''(x) \approx \frac{f(x+h)− 2f(x)+f(x−h)}{h^2}.$

So how would I apply this to the above BVP? I suppose I'm most confused about when to use the boundary conditions, how to actually derive the equations based off this formula, and how those equations relate to the elliptic equation.

Hopefully verifying the second-order accuracy won't be too bad. I'm guessing that will come in the form of some error term.

Any help would be appreciated!

Thank you in advance.

答:
As you noted the central difference scheme can be written as
$$\frac{\partial^2 u}{\partial x^2}\bigg|_{m,n}=\frac{u_{m+1,n}-2u_{m,n}+u_{m-1,n}}{\Delta x^2}$$
$$\frac{\partial^2 u}{\partial y^2}\bigg|_{m,n}=\frac{u_{m,n+1}-2u_{m,n}+u_{m,n-1}}{\Delta y^2}$$
The finite difference approximation becomes
$$\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}\approx \frac{u_{m+1,n}-2u_{m,n}+u_{m-1,n}}{\Delta x^2}+\frac{u_{m,n+1}-2u_{m,n}+u_{m,n-1}}{\Delta y^2} =0$$
Further if $\Delta x=\Delta y$ the equation becomes
$$\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}\approx u_{m+1,n}-2u_{m,n}+u_{m-1,n}+u_{m,n+1}-2u_{m,n}+u_{m,n-1} =0$$
$$\Rightarrow u_{m,n}=\frac{u_{m+1,n}+u_{m-1,n}+u_{m,n+1}+u_{m,n-1}}4$$

Consider a simple case with $h=\Delta x=\Delta y=\frac 13$. For boundary conditions
$$u_{1,1}=\frac{u_{2,1}+u_{0,1}+u_{1,2}+u_{1,0}}4=\frac{u_{2,1}+u_{0,1}+u_{1,2}+u_{1,0}}4$$
where
$$u_{0,1}=u(x=1,y=1/3)=\ln(\frac 19+1)$$
$$u_{1,0}=u(x=4/3,y=0)=2\ln(\frac 43)$$


答:
You can try the following approach:
Having a nonhomogeneous equation you must determine the homogeneous part and a particular solution:

We write as follows $y=y_h+y_p,$ where $y_h$ is the solution of the homogeneous equation and $y_p$ is a particular solution.

As you said, using the characteristic polynomial we get the roots: $-2+2i$ and $-2-2i.$

So the solution of the homogeneous equation will be: $y_h(x)=c_1e^{-2x}cos(2x)+c_2e^{-2x}sin(2x).$

Now it is up to us to finish the particular solution. Because your equation has an expression like that $x+e^x$ in the right side, you can try to find a solution in the form of: a first degree polynomial plus an exponential function. Something like this: $y_p(x)=Ax+B+Ce^x.$
Put this particular solution in your equation, force it to check your equation and find the coefficients.
After you have found the coefficients replace them in the expression for $y_p$.

After all this, combine the two solutions $y_h$ and $y_p$ and so we have the final solution of the equation.
If you need more explanations you can write a comment.

Title: Solution for $y''+4y'+8y=x+e^x$.

Body:
How can I solve this ODE?

$$y''+4y'+8y=x+e^x$$

After I find the roots of the characteristic and create the solution for the homogeneous, how can I find the solution for this ODE?