问:
Inverting a diagonal (almost) linear relationship in a numerically stable way
I have a relationship of the form

$$y=Dx$$

where $D$ is a diagonal, square matrix with  non-negative entries. Some of the diagonal elements of $D$ may be zero. However, since $D$ is calculated by a computer, I don't know which ones are "really" zero, and which ones are just very small. That is, in practice none of them are exactly zero, just -1.34534634e-16 or something, but I believe that in the "true" matrix, were I to calculate it by hand, that entry would be a zero. Since I know from the theory that the matrix has non-negative entries, I can assume that tiny negative values are "really" zero, but it's less clear with a small positive value like 1.0234234e-05 or something.

I need to calculate $\tilde x$, where:

If $D_{i,i}>0$, then $\tilde x_i=x_i=D_{i,i}^{-1} y_i$.

If $D_{i,i}=0$, then $\tilde x_i=a_i$, where $a$ is some otherwise unrelated, constant vector.

How can I do this? Notice that if $D_{i,i}$ is very small, the two rules will give wildly different results depending on whether I assume it's actually zero or not.

For context: I have a linear ODE $u'=Du$, and I want to compute $u$ from $u'$. The vector $a$ mentioned above is in fact $u(0)$: when an entry of $D$ is zero I know that the corresponding entry in $u(t)$ doesn't change over time, so I can obtain it from $u(0)$. The reason I want to compute $u$ from $u'$ is because I have another ODE $v'=Au$, so by rewriting this as something like $v'=AD^{-1}u'$ I can integrate it to compute $v(t)$. $D$ is the diagonalization of a symmetric semi-postive-definite real matrix, as computed by some linear algebra package.

答:
Hint:

As to discriminate between actual zero and non-zero values of D, a practical way allowed on modern computers is to calculate D at two different levels of digits.  At increasing the digits, the actual zeros shall move towards 0, while small values shall remain relatively stable.

In reply to your comment, in my old Mupad CAS I can fix the number of floating point digits for the whole session (for whichever calculation, by putting DIGITS:=20 or else).

Now, if the matrix you are diagonalizing is input precisely (in terms of rationals or irrationals that can be computed at the desired accuracy) and if it is not ill-conditioned,
then if you get a D value of , say, $1 \cdot 10^{-16}$ with 20 digits,  changing to e.g. 25 digits you should get approx. $1 \cdot 10^{-21}$ if that is an actual zero.

If it is an actual $1 \cdot 10^{-21}$ it should remain approximately so.

If instead the starting matrix is input as rounded floating point numbers, the above considerations are not applicable.

And if the matrix is ill conditioned, then none of the D values are reliable.


问:
Does Gödel's second incompleteness theorem provide a specific sentence for his first theorem?
Ebbinghaus' Mathematical Logic  on p184

7.8 Gödel's First Incompleteness Theorem.  Let $\Phi$ be  consistent and
R-decidable  and  suppose  $\Phi$  allows  representations.  Then  there  is  an  $S_{ar}$-
sentence $\phi$  such that neither $\Phi \vdash \phi$  nor  $\Phi \vdash \neg \phi$.

p185

7.10 Gödel's Second Incompleteness Theorem.  Let $\Phi$  be consistent  and R-decidable  with $\Phi \supset \Phi_{PA}$.  Then  $$ \text{not} \quad \Phi  \vdash Consis_\Phi.$$

On the relation between the two theorems, p184 says

A refinement of the above argumentations (in Godel's first Incompleteness
Theorem) leads to results concerning the
consistency of mathematics. In particular, Godel's Second Incompleteness
Theorem, which we shall now derive, shows that the consistency of a suffi-
ciently rich system cannot be proved using only the means available within
the system.

and https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem says

This theorem (Godel's second incompleteness theorem) is stronger than the first incompleteness theorem because the statement constructed in the first incompleteness theorem does not directly express the consistency of the system. The proof of the second incompleteness theorem is obtained by formalizing the proof of the first incompleteness theorem within the system F itself.

I have trouble understanding the relation between the two theorems in the above sources.

Does the first theorem roughly say that if a set of sentences can represent all the computable functions, the deductive theory derived from the set can't have both proof consistency and proof completeness: either one of them, or neither?

Does the second theorem  say almost the same as the first theorem, except that while the first theorem does not give a specific sentence which can be neither proved nor disproved from the set, the second theorem provides $Consis_\Phi$ as such a sentence?

Does the second theorem imply the first?

答:
Does the first theorem roughly say that if a set of sentences can represent all the computable functions, the deductive theory derived from the set can't have both proof consistency and proof completeness: either one of them, or neither?

Sure, that's one way to think about it (though you left out the crucial assumption that the set of sentences should also be decidable).

Does the second theorem  say almost the same as the first theorem, except that while the first theorem does not give a specific sentence which can be neither proved nor disproved from the set, the second theorem provides $Consis_\Phi$ as such a sentence?

Not quite.  The second theorem provides $Consis_\Phi$ as a sentence that cannot be proved from $\Phi$.  However, it's possible that $Consis_\Phi$ could be disproved from $\Phi$.  Intuitively, this means that "$\Phi$ proves its own inconsistency".  You might think that this means $\Phi$ must be inconsistent (contrary to the assumption that it was consistent), but that's not necessarily the case--the sentences that $\Phi$ proves don't necessarily have to be true (when interpreted in the natural numbers).  For instance, if you start with a set of sentences $\Phi$ and then let $\Phi'=\Phi\cup\{\neg Consis_\Phi\}$, then $\Phi'$ proves $\neg Consis_\Phi$ and thus also $\neg Consis_{\Phi'}$.  But $\Phi'$ is still consistent, exactly because $\Phi$ did not prove $Consis_\Phi$, so adding its negation to the theory does not lead to a contradiction.

Does the second theorem imply the first?

No, as explained above.  The second theorem implies a weaker version of the first where you add an additional assumption that implies $\Phi$ does not prove its own inconsistency.  For instance, you could assume that every sentence in $\Phi$ is actually true when interpreted in the natural numbers, and so $\Phi$ cannot prove any false statement about the natural numbers.  In particular, since $\neg Consis_\Phi$ is false (i.e., $\Phi$ is not actually inconsistent), this means $\Phi\not\vdash \neg Consis_\Phi$.

ContentLicense: CC BY-SA 4.0


问:
Composition is an immersion?
Considering a map $w$ from $ \mathbb{R}^3$ to $ \mathbb{R}^6$ defined as follows:

$$ w(x,y,z)= (x,y,z, \sqrt{2}xy,\sqrt{2}yz,\sqrt{2}zx )$$

in order to show that it defines an immersion of $S^2$ into $ \mathbb{R}^6$ I first proved that it defines an immersion from $ \mathbb{R}^3/{0}$ to $ \mathbb{R}^6$ by, first, computing the Jacobian matrix an proving that is has rank $3$ for all the points in $ \mathbb{R}^3/{0}$.

$$ J = \left(\begin{array}{cccc}
2x & 0& 0\\
 0 & 2y & 0\\
 0 & 0 & 2z  \\
\sqrt{2}y  & \sqrt{2}x & 0 \\
0 & \sqrt{2}z & \sqrt{2}y \\
\sqrt{2}z & 0& \sqrt{2}x
\end{array}\right) $$

It is also easy to verify that it is injective. After that, I ddefined a map that goes from $S^2$ to $ \mathbb{R}^6$, call it $ \psi$, that looks like:

$$ \psi(x,y,z) =(r^2\cos^2{\theta}\sin^2{\phi},r^2\sin^2{\theta}\sin^2{\phi},r^2\cos^2{\phi}, \sqrt{2}r^2\cos{\theta}\sin{\theta}\sin^2{\phi},\sqrt{2}r^2\sin{\theta}\cos{\phi}\sin{\phi},\sqrt{2}r^2\cos{\theta}\cos{\phi}\sin{\phi}) $$

Where I basically substituded the usual definition for spherical coordinates of $x$ . $y$ and $z$ in $w(x,y,z)$.I did the Jacobian matrix of this and since $(0,0,0)$ is not included then the rank of this Jacobian is also always $3$, therefore its an immersion. Is this attempt correct?

答:
It depends on what you mean by "since $(0, 0, 0)$ is not included". What you would like to check is that that Jacobian is injective. (And then the product of Jacobian matrices is also injective.)

But note that this can be done much simpler: First, the map $w : \mathbb R^3 \to \mathbb R^6$ is an immersion because already the projection onto the first three coordinates gives an immersion, it is just the identity map. Perhaps you saw this, and this is what you meant by "it is easy to verify".
Second, because $S^2 \subset \mathbb R^3$ is a submanifold, the inclusion map $S^2 \to \mathbb R^3$ is an immersion. (Its differential is just the inclusion map on tangent spaces.)
Therefore, the composition $S^2 \to \mathbb R^3 \to \mathbb R^6$ is also an immersion.

One way to prove that $S^2 \subset \mathbb R^3$ is a submanifold, is to define charts (spherical coordinates) and check that these define embeddings (are injective with injective differential). This is almost the same as what you have done. But it is not necessary to do that, if we are allowed to use that $S^2$ is a submanifold.

Title: Composition is an immersion?


问:
Is the derivative of the Riemann integral continuous wherever it exists?
Let $f$ be Riemann integrable on $[a,b]$ and let $F(x)$ be its integral on $[a,x]$. (Note the minimal assumption of integrability, not continuity.)

Must $F’$ be continuous wherever it’s defined? (Nota bene: here I mean continuous in the general sense of continuous on its domain. On this definition, it’s possible for a function to be continuous at a point without there being a deleted interval of the point on which the function is defined.)

Context

The claim is plausible on these heuristic grounds: $F$ ignores trouble at discontinuities (which happen on a set of measure zero), so even if $f$ fails to be continuous somewhere, $F’$ should be continuous there (if it exists). In other words, $F’$ should be “at least as nice” as $f$ (again, given that it exists).

As an example, take an $f$ continuous except at $c$, where it has a removable discontinuity. Let $g$ be the everywhere continuous function that removes the discontinuity. Since $f=g$ almost everywhere, their integrals $F$ and $G$ are everywhere the same. By the FTC, $G’=g$; combining with the previous sentence we have $F’=g$. Thus $F’$ will be continuous at $c$ even though $f$ isn’t: it plugs the hole.

This is just intuition. Any proof will need to deal with two cases: one at points where $f$ is continuous and the other at points where $f$ is discontinuous.

In the former case we know by the FTC that $F’$ exists at $c$ (and moreover $F’(c)=f(c)$). Can we prove that it’s continuous there?

In the latter case, we could break into subcases depending on the type of discontinuity. If $f$ has a removable discontinuity we can reduce to the previous case. If $f$ has a jump discontinuity, I think we can show $F’$ doesn’t exist. And if $f$ has an essential (oscillatory) discontinuity, I also think we can show $F’$ doesn’t exist. (I haven’t produced a rigorous argument but can’t find counterexamples. I’m stuck with $f(x)=\sin(1/x)$ on $[-1,0)\cup(0,1]$ and $0$ at the origin. Will $F’$ exist at the origin?)

If all that’s right, the proof depends essentially on whether the following implication holds: $f$ continuous at $c$ implies $F’$ continuous at $c$.

If we strengthen the hypothesis by assuming $f$ is not just continuous but differentiable at $c$, it does indeed follow that $F’$ is continuous at $c$. The question is whether we can relax the hypothesis to bare continuity of $f$ at $c$.

EDIT

Paramanand Singh points out in comments that $F'$ can indeed exist and be discontinuous where $f$ has an oscillatory discontinuity. (The right example, as I guessed, was $f(x)=\sin(1/x)$ on $[-1,0)\cup(0,1]$ and $0$ at the origin, but it turns out that $F'$ exists here, contrary to my hunch.)

So the answer to the question in the title is no. But this question is still open: if $f$ is continuous somewhere, must $F'$ be continuous there?

答:
Yes, if $f$ is continuous at $c,$ then $F'$ is continuous at $c$ within the domain of $F'.$ (Somewhat humorously, the proof of this is easier than my previous proof in the case where $f'(c)$ exists. See the link the OP provides.)

Proof: Suppose $x_1> x_2 > \cdots \to c$ and $F'(x_n)$ exists for each $n.$ WLOG, $[x_n,x_n+1/n]\subset (c,b]$ for all $n.$

Fix $n$ for the moment. Then

$$F'(x_n)-f(c) = \lim_{h\to 0^+}\frac{1}{h}\int_{x_n}^{x_n+h} (f(x)-f(c))\,dx.$$

We can assume $0<h<1/n$ above. Thus

$$\tag 1| F'(x_n)-f(c)| \le \sup_{[x_n,x_n+1/n]}|f-f(c)|.$$

Now unfreeze $n.$ Because $f$ is continuous at $c,$ the right side of $(1)$ $\to 0$ as $n\to \infty.$ Hence so does the left side. Since $F'(c)=f(c),$ we have shown the desired continuity of $F'$ at $c.$


问:
$A$ is $p\times n$, $B$ is $q\times n$, what can we say about Systems $1$ and $2$ of equations?

Problem: Let $A$ be a $p\times n$ matrix, and $B$ be a $q\times n$ matrix.
 System 1: $Ax < 0, Bx= 0$ for some $x\in\mathbb R^n$.
 System 2: $A^Tu + B^Ty = 0$ for some non-zero $(u,y)$ with $u\ge 0$.

Is it true that if System $1$ has no solution, System $2$ has a solution? Prove or disprove.

Further, show that if $B$ has full rank, exactly one of the systems has a solution. Is this necessarily true if $B$ is not of full rank? Prove or disprove.

Let's say that system $1$ has no solution. Then, $(Ax)_i\ge 0$ for some $1\le i\le p$, for all $x\in\mathbb R^n$. I don't know if System $2$ has a solution or not, so let's suppose it doesn't (if it does, we should get a contradiction). So, for all $u \ne 0, y\ne 0, u\ge 0$ we have $A^Tu + B^Ty \neq 0$. What's next? I'm stuck here.

It would be nice if you could help me solve it or at least point me in the right direction!

Clarification: For $x\in\Bbb R^n$, $x>0$ is interpreted element wise, i.e. $x_i > 0$ for all $1\le i \le n$. Similarly for other symbols, i.e. $\le, \ge, <, \not <, \not >, =$ etc.

答:
Let $e$ be the all-one vector.

Consider the optimization problem $(P)$: $$\max 0$$

$$Ax \le - e $$$$ Bx=0$$

The corresponding dual $(D)$ is

$$\min - e^Tu$$

$$A^Tu+B^Ty=0$$

$$u \ge 0$$

If $(P)$ is infeasible, then $(D)$ is either infeasible or unbounded. Since $(0,0)$ is a solution to $(D)$, hence $(D)$ must be unbounded. At least one component of $u$ can be chosen to be positive.

That is we have proven that if System $1$ has no solution, then System $2$ has a solution.

If $B$ has full rank and  suppose System $1$ has a solution. Then the optimal value of $(P)$ is $0$. Hence by strong duality, the optimal value of $(D)$ is $0$ as well. That is we must have $u=0$ and we end up wiht $B^Ty=0$. Since $B$ is of full rank, the unique solution of $B^Ty=0$ is $y=0$. That is to say, the feasible point of $(D)$ is the zero vector. System $2$ is not feasible.

This is not true if $B$ is not of full rank. For example, if $B=0$ and let $A=1$, then System $1$ is feasible since $x$ can take any negative number. System $2$ remains feasible since we can let $u=0$ and $y=1$.


问:
Spliting Field over $\mathbb{F}_3$
How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?</p>

答:
I'll try to help you with one, you do the other one and try to build an isomorphism between the corresponding fields (knowing it exists must help...).

Since $\,f(x)=x^3-x+1\in\Bbb F_3[x]\,$ is irreducible, we get that $\,\Bbb F_{3^3}:=\Bbb F_3[x]/\langle f(x)\rangle\,$ is a field with $\,3^{\deg f}=27\,$ elements. Let $\,w\in \overline{\Bbb F_3}\,$ be a root of $\,f\,$ , so :

$$(**)\;\;\;\;w^3=w-1=w+2\pmod 3\implies$$

$$\implies \Bbb F_{27}=\{0,1,2,w, 2w,...\}=\{q(w)\;;\;q(x)\in\Bbb F_3[x]\;,\;\deg q\le 2\} $$

with the sum and product operations done modulo three and with the relation (**) taken into account.

Thus, for example, we have that

$$(2w+1)(w^2+w+1)=2w^3+\color{red}{2w^2}+\color{blue}{2w}+\color{red}{w^2}+\color{blue}w+1=2(w-1)+1=2w-1=2w+2$$

or also

$$w(aw^2+bw+c)=1\;,\;\;a,b,c\in\Bbb F_3\;\implies aw-a+bw^2+cw=1\implies$$

$$\implies a = 2\,,\,c=1\,,\,b=0\implies\;\text{in the field}\;\;\Bbb F_{27}\;,\;\;w^{-1}=2w^2+1$$

Well, try to do the same with the other polynomial and then try to come up with some isomorphism between the respective fields.

Further hint: each one of the above fields has an element (a primitive element) s.t. every non-zero element in the field is the power of that elements, which is just a lengthy way to say that the multiplicative groups of the non-zero elements of each field is a cyclic group of order $\;3^3-1=26\;$ .

If you understand how the product is done from the examples above then that'll help you to find those primitive elements in each field...

ContentLicense: CC BY-SA 3.0

Title: Spliting Field over $\mathbb{F}_3$


问:
How to simplify following the boolean expression $A+\bar{A} B \bar{C}$
Equation to minimize using Boolean Algebra Laws:
$A+\bar{A} B \bar{C}$

I have tried doing this but i am unsure of the answer:
$$
\begin{array}{l}
\text { Let } K=B \bar{C} \\
A+\bar{A} K=A+K=A+B \bar{C}
\end{array}
$$

答:
You can always add a lesser expression as in

$$
    A = A + AX
$$
Therefore

\begin{align}
  A + \bar{A}B\bar{C} & = (A + AB\bar{C}) + \bar{A}B\bar{C} \\
                      & = A + (AB\bar{C} + \bar{A}B\bar{C}) \\
                      & = A + (A+\bar{A})B\bar{C} \\
                      & = A + B\bar{C}
\end{align}


问:
Divisor sums of multiplicative functions
Suppose that $f$ is a multiplicative function, i.e., $f(ab)=f(a)f(b)$ whenever $\gcd(a,b)=1$. Are there any techniques for estimating
$$\sum\limits_{d \mid n}f(d)$$
in terms of $f(n)$?

I am interested in general lower and upper bounds, aside from the obvious ones
$$\tau(n) \leq \sum\limits_{d \mid n}f(d) \leq \tau(n)f(n).$$
The ideal would be to find a function $g$ such that
$$\lim\limits_{n \to \infty}\frac{\sum_{d \mid n}f(d)}{g(f(n))}=1.$$

A particular case of interest to me is when $f$ is the divisor function $\tau$, i.e., the number of distinct divisors of an integer.

Many thanks.

答:
This does not answer your question but it did not fit in a comment. There is a large literature on maximal orders of multiplicative functions, for example, Wigert proved that there exists $c>0$ such that $d(n) \leq n^{c/\log \log n}$ for all large enough $n$, and he also found the best possible value for $c$. The literature gives bounds using calculus functions, not multiplicative functions. The function $F$ becomes very large especially when $n$ has many small divisors so any inequality that you want to prove should be first checked in the case that $n$ is the product of the first primes. To be precise let $n_x:=\prod_{p\leq x }p$ so that $F(n_x)=3^{\pi(x)}$. Then by the prime number theorem $$\log F(n_x) \sim (\log 3) \frac{x }{\log x },$$ as $x\to \infty $.
Note also that by the prime number theorem again we have $\log n_x=\theta(x) \sim x$, hence,  $$\log F(n_x) \sim (\log 3) \frac{\log n_x }{\log \log n_x }.$$ This means that for any $b<3$ there are infinitely many $n$ for which $$ F(n) >  n^{ \frac{\log b  }{\log \log n  }}.$$ This is in a way best possible. By the inequality $F(n)\leq \tau^2(n)$ and Wigert's result we also get that there exists $c>0$ such that for all $n $ the bound $$ F(n) \leq   n^{ \frac{c }{\log \log n  }}$$ holds.It would be nice to determine the best value for $c$. It would have to be at least $\log 3 $ from the previous argument and I would be surprised if it was strictly larger than $\log 3 $.

EDIT: this paper of Shiu shows that $c$ is indeed $\log 3$. Namely, it shows that if we have any $c>\log 3 $ then $ F(n) \leq   n^{ \frac{c }{\log \log n  }}$ holds for all large enough $n$. This can be seen by taking $\theta=1/2$ in his Theorem 1 and noting that his $M$ is achieved for $a=1$, thus, $\log M=\log 3 $.


问:
Prove that if $G$ is a finite simple group containing a subgroup $H$ of order $27$, then $|G : H| \ge 9.$

Prove that if $G$ is a finite simple group containing a subgroup $H$ of order $27$, then $|G : H| \ge 9.$

I have so far been solving this by looking at $|G:N|=1, 2, 3...$ etc., all the way up to $8$ and showing that each case is not possible. This is a method that I believe will work in solving the problem if I continue using it (I have shown it does not work for $1, 2$, and $3$).

However, I was wondering if there was a more efficient way to go about a problem like this.

答: If $|G:H|<9$, then since  $G$ is a simple group, it is isomorphic to a subgroup of the symmetric group $S_8$. But the group $S_8$ has a Sylow $3$-subgroup of  order $9.$ We have a contradiction.


问:
Creating models from $P'=kP^a$
So we were given values of $a=1,2,\frac{1}{5}$ and
$P'=kP^a$. I got $P=Ce^{kt}$ for $a=1$.
How do you put $a=\frac{1}{5}$ with it being a model that makes sense? Are there integral rules I don’t know that keep it from getting overly complicated?

答:
The general solution of the differential equation:

$$u'=Pu^{n+1}+Qu$$ is

$$u=\dfrac{e^{\int Qdx}}{\sqrt[n]{C-n\int P e^{n\int Qdx}dx}}$$ for some constant $C$. This is derived in the following manner:
$$z'+Az=B$$ has the general solution of $$z=e^{-\int A dx}(C+\int Be^{\int A dx}dx)$$ By putting $u^q=z$, we obtain
$$qu^{q-1}u'+Au^q=B$$
$$qu'+Au=Bu^{1-q}$$ or
$$qu' = Bu^{1-q}-Au.$$ Substitute $q=-n$, $A = nQ, B = -nP$
$$-nu' = -nPu^{n+1}-nQu$$
$$\boxed{u'=Pu^{n+1}+Qu}$$ Therefore,
$$u^{-n} = e^{-n\int Qdx}\left[C-n\int Pe^{n\int Qdx}dx\right]$$ or
$$\boxed{\boxed{u=\dfrac{e^{\int Qdx}}{\sqrt[n]{C-n\int P e^{n\int Qdx}dx}}}}$$ assuming $n\neq 0$. Putting $Q=0,$ we obtain
$$u=\dfrac{1}{\sqrt[n]{C-n\int P dx}}$$ which, for a constant $P=k$, is
$$u= \dfrac{1}{\sqrt[n]{C-nkx}},$$ and taking the limit as $n\to 0$ corresponds to the exponential solution.

Title: Creating models from $P'=kP^a$


问:
Evaluation of sum of power series $\sum \frac{n}{n^2-1}x^n$
Evaluation of sum of power series $\sum \frac{n}{n^2-1}x^n$ ?

I know $\sum_{n\geq0} nx^n = x/(1-x)^2$, how do I Include the $\frac{1}{n^2-1}$ ?

答:
You have to exclude $n=1$ since you don't want $0$ in the denominator.

For $|x| < 1$,

$$ \sum_{n=2}^\infty \frac{n}{n^2-1} x^n = - \frac{x^2+1}{2x} \ln(1-x) - \frac{1}{2} - \frac{x}{4} $$

You get this by first writing $n/(n^2-1)$ in partial fractions:
$$ \frac{n}{n^2-1} = \frac{1}{2(n+1)} + \frac{1}{2(n-1)}$$

EDIT: Then
$$ \eqalign{\sum_{n=2}^\infty \frac{x^n}{2(n+1)} &= \frac{1}{2x} \sum_{n=2}^\infty \frac{x^{n+1}}{n+1}\cr
&= \frac{1}{2x} \sum_{n=3}^\infty \frac{x^n}{n}\cr
&= \frac{1}{2x} \left(-\ln(1-x) - x - \frac{x^2}{2} \right)}$$
while
$$ \eqalign{\sum_{n=2}^\infty \frac{x^n}{2(n-1)} &= \frac{x}{2} \sum_{n=2}^\infty \frac{x^{n-1}}{n-1}\cr
&= \frac{x}{2} \sum_{n=1}^\infty \frac{x^n}{n}\cr
&= -\frac{x}{2} \ln(1-x) }$$


问:
Convergence of modified geometric series
I have the following problem:

Let $\{a_n\}$ be a sequence of real numbers such that
$\sum_{n=0}^\infty a_n = 0$.

Prove that $\sum_{n=0}^\infty a_n x^n$ converges if $|x| < 1$.

Prove that $\lim_{x \to 1^-} \left( \sum_{n=0}^\infty a_n x^n \right) = 0$.

Hint: $\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty s_n x^n (1 - x)$ where $s_n = a_0 + a_1 + \cdots + a_n$.

My reasoning for 1 is as follows: by the ratio test on $\sum a_n$, we must have $\lim \frac{a_{n+1}}{a_n} \leq 1$. Then we apply the ratio test to $\sum a_n x^n$ to find that
$$ \lim \frac{a_{n+1} x^{n+1}}{a_n x^n} \leq \lim \frac{x_{n+1}}{x_n} = x < 1. $$

For 2. I'm not sure what to do exactly. The equality in the hint is clear to me, but that's about it.

答:
Because $a_n$ can vanish, the rati test doesn't work in this case.

The condition : $\sum _{n=0}^{\infty} a_n$ converges to $0$ means that the radius of convergence of  $\sum _{n=0}^{\infty} a_nx^n$ is $\geq 1$ which means that $\sum _{n=0}^{\infty} a_nx^n$ is perfectly defined when $|x|<1$.

For 2 i suggest you want to use the fact that $\sum _{k=0}^{n} a_k \rightarrow 0$ as $n \rightarrow \infty$.

Let $\epsilon > 0$ and   $N_0$ be an integer such that for all $N\geq N_0$ : $$|\sum _{n=0}^{N} a_n |\leq\epsilon$$

We have $$\sum _{n=0}^{\infty} s_nx^n(1-x) = \sum _{n=0}^{N_0} s_nx^n(1-x)+\sum _{n=N_0+1}^{\infty} s_nx^n(1-x)$$

and we have :
$$|\sum _{n=N_0+1}^{\infty} s_nx^n(1-x) |\leq \sum _{n=N_0+1}^{\infty} |s_nx^n(1-x) | \leq \epsilon( \sum _{n=N_0+1}^{\infty} x^n(1-x) ) \leq \epsilon $$

Therefore :

$$|\sum _{n=0}^{\infty} s_nx^n(1-x)|\leq \epsilon+\sum _{n=N_0+1}^{\infty} s_nx^n(1-x)  $$

Now if we let $x$ be sufficiently close to $1$, the term  ( $\sum _{n=N_0+1}^{\infty} s_nx^n(1-x)  $)   will also be $\leq \epsilon $.

Therefore

Given that $x$ is sufficiently close to 1 the following estimate holds : $$|\sum _{n=0}^{\infty} s_nx^n(1-x)| \leq 2\epsilon $$

This is exactly :

$$lim _{x\rightarrow 1^- } \sum s_n x^n(1-x) = 0 $$


答:
You have check that $f,g$ are irreducible over $\mathbb{F}_{3}$.

You know that in this case $\mathbb{F}_{3}[x]/\langle f\rangle$ is a
field extension of degree $\deg(f)=3$ over $\mathbb{F}_{3}$. The
same applies for $g$.

Moreover, you know that if $x$ is a root of $f$ in the above extension
then $\{1,x,x^{\deg(f)-1}\}$ is a basis for this extension.

The same applies for $g$: if $x'$ is a root of $f$ in the above
extension then $\{1,x',(x')^{\deg(f)-1}\}$is a basis for $\mathbb{F}_{3}[x]/\langle g\rangle$.

Hint: Try to find an isomorphism that maps $1\mapsto 1$ and $x \mapsto x'$

Title: Spliting Field over $\mathbb{F}_3$

Body: How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?</p>


问:
Non-isomorphic localized rings
I want to know why $\mathbb{R}[x,y]_{x^2+y^2}\not\cong\mathbb{R}[x,y]_{xy}$?

(I mean $\mathbb{R}[x,y,(x^2+y^2)^{-1}]\not\cong \Bbb{R}[x,y,(xy)^{-1}]$)

I'm reading about weil restriction. Let $G_{m,k}$ be the multiplicative group, where $k$ is field and $\mathbb{S}:= R_{\mathbb{C}/\mathbb{R}}G_{m,\mathbb{C}}$ to be the Weil restriction of $G_{m,\mathbb{C}}$. Now, I'm trying to prove that $\mathbb{S}\not\cong G_{m,\mathbb{R}}\times G_{m,\mathbb{R}}$(as affine algebraic varieties over $\mathbb{R}$).

My attempt: I was thinking of finding some maximal ideal $m$ such that $m/m^2$ has different $\mathbb{R}-$dimension in the above two rings.

答:
If there were an isomorphism of rings $\mathbb{R}[x,y]_{x^2 + y^2} \cong \mathbb{R}[x,y]_{xy}$ then there would be an induced isomorphism of their unit groups.

Since $\mathbb{R}[x,y]$ is a UFD and $x^2 + y^2$ is irreducible, the units of $\mathbb{R}[x,y]_{x^2 + y^2}$ are precisely the elements of the form $r (x^2 + y^2)^m$ where $0 \not= r \in \mathbb{R}$ and $m \in \mathbb{Z}$.  Hence the group of units of $\mathbb{R}[x,y]_{x^2 + y^2}$ is $\mathbb{R}^{\times} \times \mathbb{Z}$.  (The first factor is the non-zero reals with multiplicative group structure, the second factor is the integers with additive group structure).

Similarly one calculates that the group of units of $\mathbb{R}[x,y]_{xy}$ is $\mathbb{R}^{\times} \times \mathbb{Z} \times \mathbb{Z}$.

The problem thus reduces to checking that there is no group isomorphism $\mathbb{R}^{\times} \times \mathbb{Z} \cong \mathbb{R}^{\times} \times \mathbb{Z} \times \mathbb{Z}$.  In fact there are no such surjections.
To check this you could proceed as follows: suppose that such a surjection exists and let $(q,a), (r, b)$ be the elements mapping to $(1, 1, 0), (1, 0, 1)$.
This implies that $(q^b/r^a, 0)$ maps to $(1, b, a)$.  Since $n$th roots exist in $\mathbb{R}$, conclude that $a,b$ are divisible by arbitrarily large integers, hence $a = b = 0$.  But then $(q, 0)$ maps to $(1,1,0)$, which by the same reasoning is absurd.

Title: Non-isomorphic localized rings


问:
What are the names of numbers in the binary system?
The names we use are very much related to the radix we use

$0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9$

zero - one - two - three - four - five - six - seven - eight -nine

We repeat the names

$21$ twenty one, $22$ twenty two .. and so on.

This is not suitable for binary system

If we used the same nomenclature

$1 - \text{one}$

$10 - \text{two}?$

$11 - \text{three}?!$

This is very hard to read and write

Maybe if we used something else it would be very convenient something like:

$1 - \text{John}$

$10 - \text{Watson}$

$11 - \text{Watson John}$

$100 - \text{Kevin}$

$101 - \text{Kevin John}$

$110 - \text{Kevin Watson}$

$111 - \text{Kevin Watson John}$

and so on ..

Is there any official nomenclature? or should I make it up myself?

答:
Here's a compromise that will never catch one.

1: one

10: two

11: twoty-one

100: four

101: forty-one

110: forty-two

111: forty-twoty-one

1000: eight

....
1111: eighty-forty-twoty-one

eventually we'll have to stop calling $10^k$ but then base ten equivalence.  "sixteen" which means "ten and six" is no longer a single unitary word so that's a good place to stop

1,0000: ant (huh?  be patient it will become clear)

1,0000: instead of giving this name it is two-ant.(because it is ... two ant.  This  is equivalent to base tens word for $10,000$ as "ten thousand".  Just as at this stage only have new names for $10^{3k}$ in base ten: thousand, million, billion, trillion.... Here we will only have names for $10^{4k}$.  And for convenience we well use commas for blocks of $4$, not $3$.)

1111,1111: eighty-forty-twoty-one ant eighty-forty-twoty-one.

1,0000,0000: bear

1,0000,0000,0000: cat

.....

$1,0000^{26}$: zebra.

So we run out of names when we don't have a name for $1,0000^{2\times 26} = 2^{208}\approx 4.114\times 10^{62}$.  But if that becomes an an issue it can be "zebra-zebra" and we can hope will expressing the names of binary numbers with over 208 digits that our lack of pause at the first name of a number will indicate we actually mean a magintude of zebra times that number. (Equivalent to "seven million 5 thousand million and twenty seven"$=7,005,000,000,027$ and the first "million" is a "million million"; not a single million).  Hopefully it will become easy with practice.

Not bad, eh?

I'll just sit here waiting for it to catch on.


答:
One more possible way
$$A+\overline{A}B\overline{C} =A+B\overline{A+C}=A+ \overline{\overline{B}+A+C}=\\
=\overline{\overline{A}(\overline{B}+A+C)} = \overline{\overline{A}(\overline{B}+C)}=A+B\overline{C}$$

Title: How to simplify following the boolean expression $A+\bar{A} B \bar{C}$

Body:
Equation to minimize using Boolean Algebra Laws:
$A+\bar{A} B \bar{C}$

I have tried doing this but i am unsure of the answer:
$$
\begin{array}{l}
\text { Let } K=B \bar{C} \\
A+\bar{A} K=A+K=A+B \bar{C}
\end{array}
$$


问:
Does $\sum^{\infty}_1 k \tan(1/k^3)$ converge?
I'm trying to determine if $$\sum^{\infty}_1 k \tan(1/k^3)$$ converges.

Since $1/k^3 \to 0$ as $k \to \infty$, I'm thinking that all the terms will be positive, so I'm thinking the alternating series test is out of the question. I'm trying to think of an appropriate series to compare it to, but I am not sure what to compare a tan-series to! Do I compare it to another tan-series or something else? Thanks!

答: Hint: Asymptotically $$\tan\left( \frac{1}{k^3}\right)\sim \frac{1}{k^3}$$


问:
Infinite product of Zetafunctions
It is well-known (I learned about this first in a video by Papa Flammy) that $\sum_{n=2}^\infty (\zeta(n)-1) = 1$. This result on its own is quite remarkable, but it also implies convergence of the infinite product
$$P = \prod_{n=2}^\infty \zeta(n).$$
I somehow got invested in this. Upon trying to work out the exact value of $P$, I found that
$$P = \sum_{n=1}^\infty \frac{c_2(n)}n,$$
where $c_2$ is the multiplicative continuation of $c_2(p^e) := b_2(e),$ where $b_2(e)$ is the amount of partitions of $e$ not containing $1$, i.e.
$$b_2(n) = \# \{(a_2, a_3, a_4, \dots ) \in \mathbb N_{\geq 0}^* | 2a_2 + 3a_3 + \dots = n\}.$$

Furthermore, the function
$$Q(s) = \prod_{n=1}^\infty \zeta(ns),$$
converges compactly on $\Re s > 1,$ and one has
$Q(s) = \sum_{n=1}^\infty c_1(n) n^{-s}.$ Here again $c_1(p^e) = b_1(e),$ the amount of partitions of $e$ (this time including 1). Also, $Q$ has Residue
$$\text{Res}_{s=1} (Q(s)) = P.$$

Doing some numerical analysis with python I found that $P \approx 2.29$, but i would love to know if there is some exact form. Does anyone know about such a result?

答:
This is practically just a rewriting, and I also doubt there is a closed form.

We have $$\prod_{n\ge 2} \zeta(n) = \prod_p \frac{p-1}{p\,\phi(1/p)}$$

where $\phi(q)$ is the Euler function (the one used in modular forms, not the totient function).

Title: Infinite product of Zetafunctions


问:
$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$
Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdot(a+b+c)(ab+bc+ca)(a+bc)(b+ca)(c+ab)}$$
$$P \ge3\sqrt[3]{\frac{8}{3}\cdot(a+b+c)(a+bc)(b+ca)(c+ab)} $$
I can't prove $(a+bc)(b+ca)(c+ab)\ge 8$
Could you help me ?

答:
Hints :

We have :

$$P=(a+b)(b+bc) + (b+c)(c+ca) + (c+a)(a+ab) \ge 12$$

Using :$$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$$

Now we use Cauchy-Schwarz inequality to get :

$$P\geq \sum_{cyc}^{}(a+\sqrt{abc})^{2}\geq 12$$

Now expand and use the substitution $3u=a+b+c$,$3v^2=ab+bc+ca$ and $w^3=abc$

Conclude using uvw's method .


问:
How to solve $x'=A(t)x+b(t)$
How to solve $x'=A(t)x+b(t)$ where $$A(t)=\left(\begin{array}{cc} 2t & 0 \\ 1 & t \end{array}\right) $$ and $$b(t)=\left(\begin{array}{c} e^{t^2}  \\ -4e^{t^2} \end{array}\right)  $$ I tried puting $$x = (y ,z)^t $$ we get the system $$y' = 2ty+2tz+e^{t^2} \\ z'=y+tz-4e^{t^2}$$ but I don't think that can do something with this, I mean if  I do $$y'=2t(z'-tz+4e^{t^2})+2tz+e^{t^2}$$ does not seems to help any idea?

答:
The first differential equation is easy to integrate:
$$y'=2ty+e^{t^2}$$
$$y'e^{-t^2}-2te^{-t^2}y=1$$
$$(ye^{-t^2})'=1$$
Integrate both sides. Then solve the second DE:
$$z'=y+tz-4e^{t^2}$$

Title: How to solve $x'=A(t)x+b(t)$


问:
Is '$M$ is a model of cardinality $\kappa$' a type definable property?
I think that for any given theory $T$, the statement in the question should be true.

My attempt of proof:

Let us fix a theory $T$ and a cardinal $\kappa$, we introduce variables $\{ z_\alpha : \alpha<\kappa \}$. We would like to construct a type $p$ in this variables such that $(a_\alpha)_{\alpha<\kappa}$ satisfy $p$ if and only if $\{ a_\alpha : \alpha<\kappa \}$ is the universe of a model of $T$.

I first looked at sentences with only one quantifier. Let $\varphi(x)$ be q-f. If $T$ implies $\forall x \varphi(x)$, then we add to $p$ the formula $\varphi(z_\alpha)$ for every $\alpha$. But if we are in the case $\exists x \varphi(x)$ I don't really know how to proceed. I would like to add something equivalent to $\bigvee_{\alpha<\kappa} \varphi(z_\alpha)$.

答:
What you ask for, literally, is impossible, as in the example given by Noah in the comments --- in a type $p(\bar x)$ in infinitely many variables, you cannot assert in a first order way that any one of the variables in $\bar x$ represents the constant $c$.

What is possible is finding a type $p((x_\alpha)_{\alpha<\lambda})$ over $\emptyset$ such that the elementary submodels (of cardinality $\leq \lambda$ or $=\lambda$) are exactly the sets enumerated by realisations of $p$ (usually non-uniquely and non-injectively!).

Note that I will assume $\lambda\geq \lvert T\vert$. I suspect this should not be hard to extend to small models of large theories, but there will likely be some technicalities involved that I would rather avoid.

For a given set $M$ to be an elementary submodel, it is sufficient (and necessary) to know that it satisfies the Tarski-Vaught test, i.e. that for any formula $\varphi(x,\bar y)$ and any tuple $\bar b\in M$ such that $\models\exists x\varphi(x,\bar b)$, there is an $a\in M$ such that $\models \varphi(a,\bar b)$.

You are given a tuple $I = (x_\alpha)_{\alpha<\lambda}$ of variables, where $\lambda$  is an infinite cardinal $\geq \lvert T\rvert$.

To ensure that the Tarski-Vaught test is satisfied, you want to have, for each formula $\varphi(x,\bar y)$ and each tuple $\bar y\subseteq I$, there is some variable $z=z_{\bar y, \varphi}\in X$, such that your type has the formula $\tilde\varphi(\bar y, z_{\bar y,\varphi})=(\exists x \varphi(x,\bar y))\rightarrow \varphi(z_{\bar y,\varphi}, \bar y)$.

Now, in order to ensure that this is satisfiable, you need to be careful with your choice of $z_{\bar y,\varphi}$. The easiest way to do this is to first enumerate all pairs $(\bar y, \varphi)$ as a $\lambda$-sequence $(\bar y_\alpha,\varphi_\alpha)_{\alpha<\lambda}$ (this is why I wanted to assume that $\lambda\geq \lvert T\rvert$ --- otherwise, there would be too many formulas). To ensure that there is never any clash, you should choose $z_\alpha= z_{\bar y_\alpha,\varphi_\alpha}$ in such a way that $z_\alpha\neq z_\beta$ for $\beta<\alpha$ and $z_\alpha\notin \bar y_\beta$ for any $\beta\leq \alpha$. This is standard book-keeping. Given this, you can recursively find a realisation of this type (in any model of $T$, in fact), and it is not hard to see that it enumerates a submodel (by the Tarski-Vaught test).

Now, every realisation of this type will enumerate an elementary submodel. However, it is not obvious (to me) that every elementary submodel of cardinality $\leq \lambda$ has an enumeration satisfying this type. In order to have that, you need to ensure that there are $\lambda$-many variables which are not $z_\alpha$ for any $\alpha$. This is, again, straightforward book-keeping. Then given any elementary submodel, you can enumerate the model by the "un-assigned" variables, and then extend the enumeration to all variables in accordance with the Tarski-Vaught test.

(In order to ensure that you do not get any models of smaller cardinality than $\lambda$, you should additionally inject a requirement that $\lambda$ many variables refer to different elements, but that is also quite straightforward.)


答:
Another possible hint:

$$\lim_{n\to\infty} n^{2}\times a_n=1<\infty.$$ So, according to Quotient test, it converges.

Title: Does $\sum^{\infty}_1 k \tan(1/k^3)$ converge?

Body:
I'm trying to determine if $$\sum^{\infty}_1 k \tan(1/k^3)$$ converges.

Since $1/k^3 \to 0$ as $k \to \infty$, I'm thinking that all the terms will be positive, so I'm thinking the alternating series test is out of the question. I'm trying to think of an appropriate series to compare it to, but I am not sure what to compare a tan-series to! Do I compare it to another tan-series or something else? Thanks!


问:
Prove that $\lim_{r \to 0} \frac{1}{r^2} \int_{|z| =r} f(z)dz = 2\pi i\frac{\partial f}{\partial \overline z}(0) $.
Given a Frechet differentiable function $f:U\to \mathbb{C}$, where $U$ is a open disk centered at 0 with some radius. Prove that $$\lim_{r \to 0} \frac{1}{r^2} \int_{|z| =r} f(z)dz = 2\pi i\frac{\partial f}{\partial \overline z}(0) $$

I know that the integral inside can be expressed in a parameterized form $z = re^{i\theta}$, which makes the integral $\int_{0}^{2\pi} f(re^{i\theta})rie^{i\theta}d\theta$. After that I have no idea  how to use the Frechet differentiability. Any help will be appreciated.

答:
This expression can be used to define a more general derivative (see aereolar derivative).

Recall the definition of Wirtinger derivatives. Let $\mathrm{U}$ be an open subset of $\mathbb{C}$ and $f:\mathrm{U}\to\mathbb{C}$ be Fréchet differantiable. The Wirtinger derivatives with respect to $z=x + i y$ are:
$$ \frac{\partial f}{\partial z} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + \frac{1}{i}\frac{\partial f}{\partial y} \right)\, ,$$
$$ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} - \frac{1}{i}\frac{\partial f}{\partial y} \right)\, .$$
For an Holomorphic function holds $\frac{\partial f}{\partial \bar{z}} =0$ (as well as $\frac{\partial f}{\partial z} = f'$) and the integral $\int_{\gamma} f(z)dz$ is zero by Cauchy's theorem. The problem is trivial in this situation.

In a general case, write $f=u+iv$. The Wirtinger derivative is
$$ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right)+\frac{i}{2} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right)\, .$$

Let $\Gamma$ be an open circle centered at the origin and with radius $r$ and $R$ be the disk that is bounded by $\Gamma$. The integral of $f$ in $\Gamma$ can be re-written using Stokes theorem
\begin{align}
\int_{\Gamma} f(z)dz &= \int_{\Gamma} (u dx - v dy ) \; + \; i \int_{\Gamma} (v dx + u dy ) \\[6pt]
&= \int_{R} \left[  - \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) + i \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right)
\right] dx dy\\[6pt]
&= 2 i   \int_{R}  \frac{\partial f}{\partial \bar{z}} dx dy \; .
\end{align}
Since $\frac{\partial f}{\partial \bar{z}}$ is continuous at $z=0$, as $r \to 0$ the right-hand-side of the last equation goes to $2 i  \cdot \frac{\partial f}{\partial \bar{z}}(0) \cdot \pi r^2$. Thus, dividing by $2\pi i r^2$ and taking the limit we obtain:
$$  \frac{\partial f}{\partial \bar{z}}(0) = \frac{1}{2 \pi i} \lim_{r \to 0} \frac{1}{r^2} \int_{\vert z \vert = r} f(z)dz  \; .$$

References:

Henrici, Peter [1986], Applied and Computational Complex Analysis Volume 3 - Section 15.10.


答:
HINT:

Using $\sin(x)\le x$ for $x>0$, we note that for $0<x<1$, $$\tan(x)\le x/\cos(1)$$

What happens when you set $x=1/k^3$?


问:
Technique to expand polynomials and equate coefficients
This question concerns my attempts to re-implement parts of this paper on developing wavelet filters. Though I don't think it is necessary to look at that paper, for full context, I'm trying to compute the contents of table II (using equation 9 and the description under the heading A Spline Variant with Less Dissimilar Lengths).

To summarize the technique, we have a trigonometric polynomial in $e^{-i\xi}$ (with real coefficients) given by the product of two factors, $H\left(\xi\right)$ and $\hat{H}\left(\xi\right)$.

These factors are defined as follows:
$$
H\left(\xi\right) = 2^{-\frac{1}{2}}\left.\sum_n h_ne^{-in\xi}\right.\qquad \hat{H}\left(\xi\right) = 2^{-\frac{1}{2}}\left.\sum_n \hat{h}_ne^{-in\xi}\right.
$$
in which the coefficients are symmetric about $n=0$, that is $h_n = h_{-n}$ and $\hat{h}_n = \hat{h}_{-n}$.

We wish to derive those factors so they have as close a number of non-zero coefficients as possible (which I know to be 9 and 7 for $h_n$ and $\hat{h}_n$ respectively for the case described below).

The polynomial we have is of a simple symmetric form (derived from equation 9 in the above reference) and is as follows (I might well have made a mistake in the preceding algebra, but I'm sure the form is correct):

Defining $B = e^{-i\xi}$
$$
H\left(\xi\right)\hat{H}\left(\xi\right) = \frac{B^{-7}}{2^8}\left[-\frac{5}{16}\left(B^{14} + 1\right) + \frac{49}{16}\left(B^{12} + B^{2}\right) - \frac{245}{16}\left(B^{10} + B^{4}\right) + \frac{1225}{16}\left(B^{8} + B^{6}\right) + 128B^{7}\right]
$$

Given this, I can define $H\left(\xi\right)$ and $\hat{H}\left(\xi\right)$ as follows:
$$
H\left(\xi\right) = B^{-4}2^{-\frac{1}{2}}\left[h_0B^4 + \sum_{n=1}^{4} h_n\left(B^{4+n} + B^{4-n}\right)\right]\\
\hat{H}\left(\xi\right) = B^{-3}2^{-\frac{1}{2}}\left[\hat{h}_0B^3 + \sum_{n=1}^{3} \hat{h}_n\left(B^{3+n} + B^{3-n}\right)\right]
$$
in which I wish to find $h_n$ and $\hat{h}_n$.

Now, I can always multiply and expand out the above and attempt to equate the coefficients by hand, but this seems long, tedious and error prone.

Is there some technique I can use to automate this task? I initially thought a matrix technique, but it's not obvious to me how. It might be really trivial, so feel free to point it out as such!

Edit: I'm pretty sure I need an additional constraint to solve this uniquely. In this case it's:
$$
\sum_n (-1)^n h_n = 0 \\
\sum_n (-1)^n \hat{h}_n = 0 \\
$$
My apologies, I've only just come to this conclusion.

答:
It's not a full solution, but with a helpful pointer from a good friend, the following seems to be part way there...

It's easier to see the symmetry when the equations are written in the following form:
$$
H\left(\xi\right)\hat{H}\left(\xi\right) = \frac{1}{2^8}\left[-\frac{5}{16}\left(B^{7} + B{-7}\right) + \frac{49}{16}\left(B^{5} + B^{-5}\right) - \frac{245}{16}\left(B^{3} + B^{-3}\right) + \frac{1225}{16}\left(B^{1} + B^{-1}\right) + 128\right]
$$
and
$$
H\left(\xi\right) = 2^{-\frac{1}{2}}\left[h_0 + \sum_{n=1}^{4} h_n\left(B^{n} + B^{-n}\right)\right]\\
\hat{H}\left(\xi\right) = 2^{-\frac{1}{2}}\left[\hat{h}_0 + \sum_{m=1}^{3} \hat{h}_m\left(B^{m} + B^{-m}\right)\right]
$$
If we define
$$
c\left(p\right) = B^{p} - B^{-p}
$$
we can see that
$$
c\left(p\right)c\left(q\right) = c\left(p + q\right) + c\left(p - q\right)
$$
We can now write the product of $H\left(\xi\right)$ and $\hat{H}\left(\xi\right)$ neatly as
$$
H\left(\xi\right)\hat{H}\left(\xi\right) = \frac{1}{2}\left[h_0 + \left.\sum_{n=1}^{4}h_nc\left(n\right)\right.\right]\left[\hat{h}_0 + \left.\sum_{m=1}^{3}\hat{h}_mc\left(m\right)\right.\right]
$$
and expand that into a neat sum of products of $h_n$ and $\hat{h}_m$:
$$
H\left(\xi\right)\hat{H}\left(\xi\right) = \frac{1}{2}\left[h_0\left.\sum_{m=1}^{3}\hat{h}_mc\left(m\right)\right. + \hat{h}_0\left.\sum_{n=1}^{4}h_nc\left(n\right)\right. + \left.\sum_{n=1}^{4}\left.\sum_{m=1}^{3}h_n\hat{h}_m\left(c\left(n + m\right) + c\left(n - m\right)\right)\right.\right. + h_0\hat{h}_0\right]
$$
So, we're now in a position to see how the coefficients combine (remembering that $c\left(0\right) = 2$), equating them to the known coefficients of $H\left(\xi\right)\hat{H}\left(\xi\right)$:
\begin{align}
2\hat{h}_{1}h_{1} + 2\hat{h}_{2}h_{2} + 2\hat{h}_{3}h_{3} + \hat{h}_0h_0 &= 0.5\\
h_0\hat{h}_1 + \hat{h}_0h_1 + \hat{h}_{1}h_{2} + \hat{h}_{2}h_{1} + \hat{h}_{2}h_{3} + \hat{h}_{3}h_{2} + \hat{h}_{3}h_{4} &= \frac{1225}{2^{12}}\\
h_0\hat{h}_2 + \hat{h}_0h_2 + \hat{h}_{1}h_{1} + \hat{h}_{1}h_{3} + \hat{h}_{2}h_{4} + \hat{h}_{3}h_{1} &= 0\\
h_0\hat{h}_3 + \hat{h}_0h_3 + \hat{h}_{1}h_{2} + \hat{h}_{1}h_{4} + \hat{h}_{2}h_{1} &= \frac{-245}{2^{12}}\\
\hat{h}_0h_4 + \hat{h}_{1}h_{3} + \hat{h}_{2}h_{2} + \hat{h}_{3}h_{1} &= 0\\
\hat{h}_{1}h_{4} + \hat{h}_{2}h_{3} + \hat{h}_{3}h_{2} &= \frac{49}{2^{12}}\\
\hat{h}_{2}h_{4} + \hat{h}_{3}h_{3} &= 0\\
\hat{h}_{3}h_{4} &= \frac{-5}{2^{12}}
\end{align}
I think with the additional constraints:
$$
\sum_n (-1)^n h_n = 0 \\
\sum_n (-1)^n \hat{h}_n = 0 \\
$$
that should be solvable...


问:
Given $v \in V$ and $dim(V) = 1$, show that exists a subspace $U$ of $V$ where the orthogonal projection of $v$ is equal to $\alpha$
I've been asked to prove or disprove the following:

Given $v \in V$ and $dim(V) = 1$, show that exists a subspace $U$ of $V$ such that the length of the orthogonal projection of $v$ onto that subspace is equal to $\alpha$ and that $0 \leq \alpha \leq \|v\|$.

How does one approach this question?

答:
That question is weird, since it is a property of the orthogonal projection that the norm of a point's image is no larger than the norm of said point, regardless of the subspace on which it is projected. So it's not only possible to find a subspace $U$ that satisfy your conditions, but every possible subspace works, regardless of $V$'s dimension. But let's assume you are not supposed to know that.

Then, let's check what happens for the simplest value of $U$ you can think of. Namely, $\{0\}$ looks simple enough. The projection of $v$ on $U$ belongs to $U$, therefore it is $0$. Therefore, $\alpha=0$ satisfy the $0 \leq \alpha \leq ||v||$.


答:
All we need is $$f(z)= f(0)+x\partial_x f(0)+y\partial_y f(0)+o(|z|)$$
$$ = f(0)+x(\partial_z f(0)+\partial_\overline{z} f(0))+iy(\partial_z f(0)-\partial_\overline{z} f(0))+o(|z|)$$
$$=f(0)+z\, \partial_z f(0)+\overline{z}\, \partial_{\overline{z}} f(0)+o(|z|)$$
Then $$r^{-2}\int_{|z|=r}\overline{z}\ \partial_{\overline{z}} f(0)dz=2i\pi \partial_{\overline{z}} f(0)$$
$$r^{-2}\int_{|z|=r} f(0)dz= 0,\qquad r^{-2}\int_{|z|=r} z \partial_z f(0)dz= 0,\qquad r^{-2}\int_{|z|=r} o(|z|)dz= o(1)$$

Title: Prove that $\lim_{r \to 0} \frac{1}{r^2} \int_{|z| =r} f(z)dz = 2\pi i\frac{\partial f}{\partial \overline z}(0) $.

Body:
Given a Frechet differentiable function $f:U\to \mathbb{C}$, where $U$ is a open disk centered at 0 with some radius. Prove that $$\lim_{r \to 0} \frac{1}{r^2} \int_{|z| =r} f(z)dz = 2\pi i\frac{\partial f}{\partial \overline z}(0) $$

I know that the integral inside can be expressed in a parameterized form $z = re^{i\theta}$, which makes the integral $\int_{0}^{2\pi} f(re^{i\theta})rie^{i\theta}d\theta$. After that I have no idea  how to use the Frechet differentiability. Any help will be appreciated.


问:
1D Green's function: from interval to infinite line
Let's consider two problems for diffusion equation.

The first one:
$$
u_t = a^2u_{xx},\qquad 0<x<l,\quad 0<t\leq T
$$
$$
u(x,0) = \phi(x), \qquad 0 \leq x \leq l
$$
\begin{equation}
u(0,t)=0,\quad u(l,t)=0, \quad 0 \leq t \leq T
\end{equation}
and the second one:
$$
u_t = a^2u_{xx},\qquad -\infty <x<+\infty,\quad t>0
$$
$$
u(x,0) = \phi(x), \qquad -\infty < x < +\infty
$$
For both these cases there are well known expressions for Green's function:
$$
G_1(x,\xi,t) = \frac{2}{l}\sum_{n=1}^{\infty} \exp\left({-\left(\frac{\pi n}{l}\right)^2}a^2t\right)\sin\frac{\pi nx}{l}\sin\frac{\pi n\xi}{l}
$$
$$
G_2(x,\xi,t) = \frac{1}{\sqrt{4\pi a^2 t}}\exp{\left(-\cfrac{(x-\xi)^2}{4a^2t}\right)}
$$
Thus we obtain solutions for first (finite) problem:
\begin{equation}
u(x,t)=\int\limits_0^lG_1(x,\xi,t)\phi(\xi)\,d\xi
\end{equation}
and for the second (infinite):
\begin{equation}
u(x,t)=\int\limits_{-\infty}^{+\infty}G_2(x,\xi,t)\phi(\xi)\,d\xi
\end{equation}

Is there any way to obtain Green's function for infinite case from Green's function for range $[0,l]$ (for example by stating  $\quad l\rightarrow+\infty$, but I have not achieved any success with this idea)?
Or may be second solution from first?

答:
Yes, $G_2$ can be obtained as a limit of $G_1$ but I will do a slightly easier limit first.

Write
$${G}_1(x,\xi,t)=\frac{2}{l}\sum_{q} \exp\left(-q^2a^2t\right)\sin q x\sin q\xi,$$
where $q=\frac{\pi n}{l}$, $n=1,2,\ldots$ Now since $q_{j+1}-q_j=\frac{\pi}{l}$, for $l\rightarrow\infty$ we have
$$\frac{\pi}{l}\sum_{q}\rightarrow \int_0^{\infty}dq,$$
and therefore
\begin{align}\lim_{l\rightarrow\infty}{G}_1(x,\xi,t)&=\frac{2}{\pi}\int_0^{\infty}\exp\left(-q^2a^2t\right)\sin q x\sin q\xi\;dq=\\
&=\frac{1}{\sqrt{4\pi a^2t}}\left[\exp{\left(-\cfrac{(x-\xi)^2}{4a^2t}\right)}-
\exp{\left(-\cfrac{(x+\xi)^2}{4a^2t}\right)}\right].
\end{align}
As you may guess, this is the Green function for the heat equation on the half-line. Now the modifications for the whole line will be as follows:

Instead of $G_1(x,\xi,t)$ one should start with translated Green function $$\tilde{G}_1(x,\xi,t)=G_1(x-l/2,\xi-l/2,t).$$ It corresponds to the solution of the heat equation on $[-l/2,l/2]$.

Similarly transform the sum over $n$ into an integral over $q$ (you will need to consider odd and even values of $n$ separately). Evaluating two resulting gaussian integrals, you will recover $G_2(x,\xi,t)$.


问:
How to calculate $(G/H)/(K/H)$?
I have a group $G=\mathbb{Z}_{36}$ with normal subgroups $K$ and $H$, where $H$ is also a normal subgroup of $K$. What I'm really supposed to find is a bijection between $G/K$ and $(G/H)/(K/H)$ by using the third isomorphism theorem. I have already calculated $G/K$, $G/H$ and $K/H$, but I do not really know how to calculate $(G/H)/(K/H)$.

Say that one of the elements in $G/H$ is $\{0,18\}$ and one of the elements in $K/H$ is $\{0,18\}$. Will one of the elements in $(G/H)/(K/H)$ be as follows?
$$\{0,18\}+\{0,18\} = \{0+0,18+18\}=\{0,36\}=\{0\}$$

Another way I'm thinking it's possible to caluclate is
$$\{0,18\}+\{0,18\} = \{0+0,0+18,18+0,18+18\}=\{0,18,18,36\}=\{0,18\}$$

Is any of these a correct way of calculating an element in $(G/H)/(K/H)$?

答:
Though we really don't want to work with the explicit elements-as-cosets (it's too unwieldly), perhaps you want to really get a feel for them? Note that the Third Isomorphism Theorem gives you, explicitly (in conjunction with the first) the bijection you want. Trying to find one from scratch is an exercise in ignoring what you've already managed to prove.

So, first let's discuss how you want to use the Third Isomorphism Theorem to get this done. I'm using additive notation, since you are working with an additive group.

The elements of $G/H$ look like sets of the form $g+H = \{g+h \mid h\in H\}$. The set $g+H$ is the same as the set $g'+H$ if and only if $g-g'\in H$. The elements of $K/H$ are the sets where $g\in K$; that is, of the form $k+H$, with $k\in K$.

We have a map from $G/H$ to $G/K$, obtained by sending the set $g+H$ to the set $g+K$. Note that this is well defined because if $g+H=g'+H$, then $g-g'\in H\subseteq K$, so $g-g'\in K$, and therefore $g+K=g'+K$.

This map is a  group homomorphism. The kernel consists of precisely the sets $g+H$ for which $g+K=0+K$. That is, the sets $g+H$ with $g\in K$. That is, $K/H$.

The First Isomorphism Theorem then tells you that you have a bijection
$$\frac{G/H}{K/H} \to \frac{G}{K},$$
given by sending the element $(g+H) + (K/H)$ to the element $g+K$. That's your bijection.

Now, explicitly what are you doing, is that you have $G=\mathbb{Z}/36\mathbb{Z}$ (note: this is already a quotient! but you probably don't think of it as a quotient, but rather as consisting of number $0$ through $35$ with addition modulo $36$); I will denote the elements of $G$ by $\overline{k}$, with $0\leq k\lt 36$. And your $H$ is clearly $\{\overline{0},\overline{18}\}$. You don't tell us who is $K$, but let's say that it is $K=\{\overline{0},\overline{6},\overline{12},\overline{18},\overline{24},\overline{30}\}$.

The elements of $G/H$ are:
$$\begin{array}{rcccl}
\overline{0}+H &=& \{\overline{0},\overline{18}\} &=&\overline{18}+H,\\
\overline{1}+H &=& \{\overline{1},\overline{19}\} &=&\overline{19}+H,\\
\overline{2}+H &=& \{\overline{2},\overline{20}\} &=& \overline{20}+H,\\
&\vdots & &\vdots &\\
\overline{17}+H &=& \{\overline{17},\overline{35}\} &=&\overline{35}+H.
\end{array}$$
And we have
$$K/H = \{ \overline{0}+H,\overline{6}+H,\overline{12}+H\}.$$
(Because $\overline{18}+H=\overline{0}+H$, $\overline{24}+H=\overline{6}+H$, and $\overline{30}+H = \overline{12}+H$.)

Since the elements of $(G/H)/(K/H)$ are of the form $(\overline{a}+H)+(K/H)$, we have:
$$\begin{array}{rcccccl}
(\overline{0}+H)+(K/H) &=& \{ \overline{0}+H,\overline{6}+H,\overline{12}+H\} &=& (\overline{6}+H)+(K/H) &=& (\overline{12}+H)+(K/H)\\
(\overline{1}+H) + (K/H) &=& \{ \overline{1}+H, \overline{7}+H, \overline{13}+H\} &=& (\overline{7}+H) + (K/H) &=& (\overline{13}+H)+(K/H)\\
(\overline{2}+H) + (K/H) &=& \{ \overline{2}+H, \overline{8}+H, \overline{14}+H\} &=& (\overline{8}+H) + (K/H) &=& (\overline{14}+H)+(K/H)\\
(\overline{3}+H) + (K/H) &=& \{ \overline{3}+H, \overline{9}+H, \overline{15}+H\} &=& (\overline{9}+H) + (K/H) &=& (\overline{15}+H)+(K/H)\\
(\overline{4}+H) + (K/H) &=& \{ \overline{4}+H, \overline{10}+H, \overline{16}+H\} &=& (\overline{10}+H) + (K/H) &=& (\overline{16}+H)+(K/H)\\
(\overline{5}+H) + (K/H) &=& \{ \overline{5}+H, \overline{11}+H, \overline{17}+H\} &=& (\overline{11}+H) + (K/H) &=& (\overline{17}+H)+(K/H)
\end{array}$$

Now, if we really wanted to expand that, we would need to express each of the elements in the coset as sets. So for example,
$$\begin{align*}
(\overline{1}+H)+(K/H) &= \{ \overline{1}+H, \overline{7}+H, \overline{13}+H\}\\
&= \Bigl\{ \{\overline{1},\overline{19}\} , \{\overline{7},\overline{25}\}, \{\overline{13},\overline{19}\} \Bigr\},
\end{align*}$$
a set of sets. (And if we were looking at $G$ as a quotient, it would be a set of sets of sets of integers).

Which is pretty much why you don't want an explicit description. You want to work with the cosets symbolically, not directly.

Title: How to calculate $(G/H)/(K/H)$?


问:
A ping-pong ball lying inside a wine glass
Show that a ping-pong ball with radius $r$, lying inside a wine glass described by the function $x^2$, has its center at $r^2+\frac{1}{4}$ units above the bottom of the glass.

Here is a visualization of the problem

My best attempt is trying to find the derivative of the circle and the function to find some relationship at the point where they meet. The problem looked very simple at first, but I can't figure it out now.

答:
There is a solution using the concept of envelope. Indeed, parabola $P$ with equation $y=x^2$ can be considered as the envelope of all possible ping-pong balls with generic equation:

$$x^2+(y-f(r))^2=r^2$$

where we have to check that $f(r)=r^2+\frac14$.

The classical method for the determination of envelopes is obtained by working with a system of 2 equations, the initial one (1) and the equation obtained by differentiating it with respect to the parameter, here $r$, which is :

$$-2(y-f(r))f'(r)=2r\tag{2}$$

Extracting

$$y-f(r)=-\ \dfrac{r}{f'(r)}\tag{3}$$

and plugging this expression into (1), we get

$$x^2=r^2-\left(\dfrac{r}{f'(r)}\right)^2\tag{4}$$

Replacing now $y$ in the LHS of (3) by $x^2$ in (4), we get:

$$r^2-\left(\dfrac{r}{f'(r)}\right)^2-f(r)=-\ \dfrac{r}{f'(r)}$$

which is equivalent to:

$$r^2f'(r)^2-r^2-f(r)f'(r)^2=-rf'(r)$$

We will not solve this differential equation.

It is sufficient to check that $f(r):=r^2+\tfrac14$ is a solution.


问:
Two ants walk on the plane starting from the same point. Find all different ways they can travel, if they end their journey at the same point.

Two ants start at the same point in the plane. Each minute they choose
whether to walk due north, east, south or west. They each walk $1$
meter in the first minute. In each subsequent minute the distance they
walk is multiplied by a rational number $q>0$. They meet after a whole
number of minutes, but have not taken exactly the same route within
that time. Determine all possible values of $q$.

The solution goes as follows:

Consider the ants' positions $a_k$ and $b_k$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $a_{k+1}-a_k=a_kq^k$ and $b_{k+1}-b_k=b_kq^k$ with $a_k,b_k\in \{1,-1,i,-i\}$.

If $a_n=b_n$ for some $n>0$, then:

$\sum\limits_{k=0}^{n-1}(a_k-b_k)q^k=0$, where $a_k-b_k\in\{0, \pm1 \pm i, \pm2, \pm2i\}$.

Note that the coefficients $a_k-b_k$ is always divisible by $1+i$ in Gaussian integers:

Hence $c_0+c_1q+...+c_{n-1}q^{n-1}=0$. Therefore if $q=\frac{a}{b}$ we have $a|c_0$ and $b|c_{n-1}$ in Gaussian integers, which is only possible if $a=b=1$.

Could you please explain a solution without the use of complex numbers and Gaussian integers?

答:
Let's simplify the problem to the 1-D case first:

Two ants start at the same point in the plane. Each minute they choose whether to walk due east or west. They each walk $1$ meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exactly the same route within that time. Determine all possible values of $q$.

The approach that we take is similar to the given solution, where we apply the Rational Root Theorem for the appropriate field:

Let $a_k, b_k \in \{ 1, -1 \}$ denote whether the ant walked east or west for the kth step.

Then, we are given that $ \sum_{k\geq 0}  a_k q^k = \sum b_k q^k$,  and $ a_k \neq b_k$ for some $k$.

We have $ (a_k - b_k) q^k = 0$, and $ a_k - b_k \in \{ -2, 0, 2 \}$.

Since all of the coefficients are integers, we apply the rational root theorem to conclude that roots $q > 0 $ could only be $ \frac{1}{2}, 1, $ or $2$.

If $ q = 1$, we have a solution, eg $(a_1, a_2 ) = (-1, 1), (b_1, b_2 ) = (1, -1)$.

If $ q = 1/2$, then by considering the smallest $K$ in which $ a_K \neq b_K$, show that these ants can never meet at any point in time, not just whole minutes. (Basically, $2 \times \frac{1}{2^K} > \sum_{k> K } 2 \frac{1}{2^k}$ because we have finitely many steps.)

If $ q = 2$, then by considering the largest $K$ in which $a_K \neq b_K$, likewise show that these ants can never meet.

This solution also works for the 2-D case:

Let $a_k, b_k \in \{ 1, -1 , 0\}$ denote whether the ant walked east or west or north/south for the kth step. (In particular, we don't differentiate between the ants walking north or south.)

By focusing on the x-coordinate, we have $ (a_k - b_k) q^k = 0$, and $ a_k - b_k \in \{ -2, -1, 0, 1, 2 \}$.

The rest of the solution is identical to the 1-D case.

Note:

In the 1-D case, we could have divided the polynomial by 2 to conclude that $q > 0 $ could only be $1$. If so, the solution to the 2-D case needs to be modified with that last bit.

This approach extends to the n-D case, using the same argument as the 2-D case.


问:
Quadratic forms and polarization
Maybe it is a trivial question but I cannot see it. I an working on Riemannian manifolds and I have a bilinear form $Q$ on a conformally flat manifold that for orthogonal vector fields $X,Y$ it verifies $Q(X,Y)=0$.
In the paper I am reading, it proves that this implies that $Q=\lambda g_0$, where $\lambda : M \rightarrow \mathbb{R}$ and $g_0$ a flat metric. In the proof, it says

This is a generality on quadratic forms and follows easily by "polarization".

Why is this?

答:
By linear algebra, there is a linear map $P: T_x\to P_x$ so that
$$
\langle Y, P(X)\rangle =Q(X, Y).
$$
In fact, under an orthonormal basis of $T_x$, $P$ has the same matrix as $Q$.

Therefore, $\langle Y, P(X)\rangle=0$ for all $Y\perp X$. In another word, $P(X)\perp Y$ for all $Y\in X^\perp$, where $X^\perp\subset T_x$ is the (codimension one) subspace of all vectors perpendicular to $X$. So the vector $P(X)$ is perpendicular to the space $X^\perp$; such vecor must be a multiple of $X$, i.e. $P(x)=\lambda_x X$.

Thus every vector $X$ is a real eigenvector of $P$. Then all the eigenvalues must be the same: when $X, Y$ are independent, from $P(X+Y)=P(X)+P(Y)=\lambda_1 X+\lambda_2Y$ one the one hand, and $P(X+Y)=\lambda_3(X+Y)$ on the other hand we see $\lambda_1=\lambda_2=\lambda_3$. So $P(X)=\lambda X$ for all $X$, thus
$$Q(X, Y)= \langle Y, P(X)\rangle=\lambda \langle X, Y\rangle.
$$

Title: Quadratic forms and polarization


问:
Can you go backwards from a Morley triangle?
Definition 1. If T is a triangle, let E(T) be the triangle formed by the intersections of the adjacent trisectors of the (interior) angles of T. Synonymously, E(T) will be called the Morley triangle for T.

Remark. By the theorem known as “Morley’s Miracle”, E(T) is always equilateral.

Motivation: I think it is a fairly natural and appropriate question to ask in its own right, but in fact for me it is a stepping stone (provided that the answer is in the affirmative) of another question that is the one that I REALLY want to ask. I don’t know whether it is appropriate for a user to “cascade” questions in a single post, but I’ll go ahead and do so. If it is not appropriate, please let me know. In that case the question that follows can be ignored as an official question for now, and regarded as simply part of the background/motivation for the present question.

While adding entries to the “what surprised you” community wiki, I noticed that one of the entries was on covering an equilateral triangle with squares. Here is the link:

Which one result in mathematics has surprised you the most?

The heads-up on this fascinating result (“One surprise for me – What is the optimal way to cover an equilateral triangle with two squares? It wasn’t solved correctly until 2009. https://erich-friedman.github.io/packing/squcotri/ ”) is due to Silas Pike.

I immediately wondered whether there was a way to relate this to Morley’s Miracle. So, here is what I came up with:

Definition 2. For each positive integer n, let f(n) be the maximal positive number s such that an equilateral triangle of side of length s can be covered by n unit squares.

Definition 3. For each positive integer n, let g(n) be the area of any triangle T such that E(T) has side of length f(n). (Notice that for g to be well-defined requires an affirmative answer to the official question of this post.)

Question: What is the behavior of g? (i.e., “What does its graph look like?”) Is it logarithmic, or exponential, or what? THAT’S what I really wanted to ask.

Regards,

Mike Jones

22.May.2011 (Beijing time)

答:
I will take the question in the title:

Can you go backwards from a Morley triangle?

The simple answer is no. Start with any asymmetric triangle, and find its Morley triangle. There is an equilateral triangle which also gives the same Morley triangle (just put a triangle with angles $4\pi/9$, $4\pi/9$, and $\pi/9$ on each edge of the Morley triangle to find its vertices), so the Morley triangle does not define the shape of the original triangle.

An example of an equilateral triangle and a half-square with the same Morley triangle:


问:
Realizing a vector bundle as a summand of a trivial bundle
A very interesting statement, in my opinion, that is proved in the famous Serre-Swan paper is the following lemma

Lemma: Let $X$ be a compact topological space and $\xi \rightarrow X$ a vector bundle . Then there exists a trivial vector bundle $\eta$, i.e. $\eta=X\times \mathbb{R}^n$ for some n, and a vector bundle epimorphism  $\eta \rightarrow \xi$  over the identity map.

It is an immediate corrolary that any vector bundle over a compact topological space can be completed to a trivial vector bundle, in the sense that it is a direct summand of a trivial vector bundle over X.

It seems to me that the compactness assumption is critical in the proof of this Lemma, however, I was wondering if the statement is also true for vector bundles over arbirtary smooth manifolds.

答:
This is true. In fact, any vector bundle over something homotopy equivalent to a finite dimensional CW complex (for example, over a non compact manifold) is a summand of a trivial bundle.

To see this, note that you just need to prove it for tautological bundle of a Grassmannian, since every vector bundle over such a space is pulled back from the tautological bundle over a Grassmannian. The tautological $k$-plane bundle $E_k \to \mathrm{Gr}_{k} \mathbb{R^n}$ is a summand of the trivial $n$-plane bundle over $Gr_{k} \mathbb{R}^n$, which finishes the proof.

By the way, its not true that vector bundles over any space have a complement, consider the universal line bundle over $\mathbb{RP}^{\infty}$. Since $H^*(\mathbb{RP}^{\infty}, \mathbb{F}_2 )$ is a polynomial ring with generator the first Stieffel-Whitney class of the universal bundle, this bundle doesn't have a complement.


问:
Field cut out by a character
Let $p$ be an odd prime and let $\Delta:=\textrm{Gal}(\mathbb Q(\zeta_p)/\mathbb Q)$, where $\zeta_p$ is a primitive $p$-th root of unity. Consider the character:
$
\chi: \textrm{Gal}(\overline{\mathbb Q}/\mathbb Q) \rightarrow \Delta \rightarrow \mathbb F_p^{\times},
$
where the first map is the projection and the second map is the canonical isomorphism. I am interested in the character $\chi^{1-k}$, where $k \geq 2$ even. The kernel of $\chi^{1-k}$ is of the form $\textrm{Gal}(\overline{\mathbb Q}/E)$, where $E$ is a finite extension of $\mathbb Q$. Why is $E$ contained in $\mathbb Q(\zeta_p)$ and moreover why is the degree of $E$ over $\mathbb Q$ equal to $\frac{p-1}{ \textrm{gcd}(p-1,k-1)}$?

答:
At the end of the day, you have a character $\chi$ from a group $G$ to $\mathbb F_p^\times$ of order $p-1$. If $g\in \ker\chi$, then $\chi(g) = \chi(g)^{1-k} = 1$, so $g\in \ker\chi$. We see immediately that
$$\mathrm{Gal}(\overline{ \mathbb Q}/E) = \ker\chi^{1-k} \supset \ker\chi= \mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q(\zeta_p))$$
from which it follows that $E\subset \mathbb Q(\zeta_p)$.

The degree of $E$ over $\mathbb Q$ is equal to the size of the image of $\chi^{1-k}$ which is exactly the order of $\chi^{1-k}$, i.e. $\frac{p-1}{\mathrm{gcd}(p-1, k-1)}$.

Title: Field cut out by a character


问:
Slowpoke and Doubles probability -- does it converge as the numbers of laps increases?
I posted this on puzzling stack exchange three months ago and it was immediately closed as off-topic, for being a "fairly straightforward probability calculation".

I have the solution to the problem, which I do not give here in case you might want to solve it yourself. My two questions for this forum are: 1) is this indeed an interesting problem to solve and not "fairly straightforward", and 2) how can I prove that the answer for the probability of N laps converges over time (i.e. as N increases)? (This second part I have been unable to solve.)

=== Here is the original post: ===

I created this dice and paper game for my kids years ago, and at the same time gave myself an interesting puzzle.

This game is a horse race between two horses: Slowpoke and Doubles.

With a sheet of lined writing paper, draw vertical lines to make six columns down the paper. The first five columns belong to Slowpoke, and the sixth column belongs to Doubles.

Each row across the paper represents one lap around a racetrack. Before the race starts, decide on the number of laps that will be run. You can then have fun predicting which horse will win the race.

To begin the race, roll a pair of dice. If you roll a double -- (1,1), (2,2), (3,3), (4,4), (5,5), or (6,6) -- then draw an X in Doubles' column in the first lap. Doubles has just completed the first lap!

If you roll anything other than doubles, then draw an X in Slowpoke's first column in the first lap. Slowpoke has made it one fifth of the way around the track.

Keep rolling the dice. Each time a doubles comes up, Doubles completes another lap.

Each time anything else comes up, Slowpoke advances across the current lap, left to write. Draw an X in the next free column. It will take five non-doubles rolls for Slowpoke to complete the lap.

In this way, roll after roll, one or the other horse advances with another X, until finally, one of the two horses completes their last lap and wins!

My puzzle for you is:

Is one of the two horses favored to win the race?

If so, what is the probability that a given horse wins a race of N laps?

Enjoy!

=== UPDATE:

@user2661923 I found solving the probability for Doubles is easier than for Slowpoke. It follows a simple pattern. (I'm using spoiler blocks in case others find it interesting to solve on their own.)

For one lap:

 $$ \frac {6^4 + (6^3\cdot 5) + (6^2\cdot 5^2) + (6\cdot 5^3) + 5^4} {6^5} $$

For two laps:

 $$ \frac {6^9 + (6^8\cdot 5) + (6^7\cdot 5^2) + (6^6\cdot 5^3) + (6^5\cdot 5^4) + (6^4\cdot 5^5) + (6^3\cdot 5^6) + (6^2\cdot 5^7) + (6\cdot 5^8) + 5^9} {6^{11}} $$

For n laps:

 $$ \sum_{k=0}^{5n - 1} \frac{5^k\cdot 6^{(5n - 1) - k}} {6^{6n - 1}} $$

Which simplifies to:

 $$ \sum_{k=0}^{5n - 1} \frac{5^k} {6^{k + n}} $$

This last equation I would like to prove continually approaches, but never reaches, 1/2.

答:
Let $Y_n$ be the number of nondoubles in a sequence of dice-rolls ending upon the $n$th double. Then $Y_n$ is a $\text{NegativeBinomial$(n,p)$}$ random variable with $p={1\over 6},$ having the following probabilty mass function (with $q=1-p$):
$$P(Y_n=k) = \binom{k+n-1}{n-1} p^nq^k\quad(k=0,1,2,...).$$

Letting $P_n$ be the probability that Doubles wins an $n$-lap game, we then have:
$$\begin{align}
P_n&=P(\text{fewer than $5n$ nondoubles occur before the $n$th double occurs})\\
&=P(Y_n< 5n)\\
&=p^n\sum_{k=0}^{5n-1}\binom{k+n-1}{n-1} q^k\tag{1}\\
&=\sum_{k=0}^{5n-1}\binom{k+n-1}{n-1} {5^k\over 6^{k+n}}
\end{align}$$

(The OP's formula for $P_n$ has omitted the binomial coefficient.)

Example ($n=2$):
Denote each occurrence of a double (resp. nondouble) by $1$ (resp. $0$). Then $P_2$ is the sum of the probabilities of just those sequences that end on the $2$nd occurrence of $1$ and have fewer than $2\times 5=10$ occurrences of $0$, as follows:

```<code>sequences                                    probabilities

11,                                          pp,
011, 101,                                    qpp, pqp,
0011, 0101, 1001,                            qqpp, qpqp, pqqp,
00011, 00101, 01001, 10001                   qqqpp, qqpqp, qpqqp, pqqqp,
000011, 000101, 001001, 010001, 100001       qqqqpp, qqqpqp, qqpqqp, qpqqqp, qqqqpp
...                                          ...
00000000011, 00000000101, ..., 10000000001   qqqqqqqqqpp, qqqqqqqqpqp, ..., pqqqqqqqqqp
                                             ------------------------------------------
                                             p^2 (1 + 2q + 3q^2 + 4q^3 + ... + 10q^9)
                                           = 12909191/22674816
                                           = 0.569318445627078...
</code>```

To obtain the limit of $P_n$ as $n\to\infty,$ we use the fact that a $\text{NegativeBinomial$(n,p)$}$ random variable is distributed as the sum of $n$ i.i.d. $\text{Geometric}(p)$ random variables; thus,
$$\begin{align}
P_n&=P(Y_n< 5n)\\
&=P(G_1+...+G_n< 5n)\\
&=P(X_1+...+X_n<0)
\end{align}$$

where the $G_i$ are i.i.d. $\text{Geometric}(p)$ random variables, and we have defined $X_i=G_i-5.$

Now the $X_i$ are i.i.d. with $E(X_i)=(1-p)/p-5=0,\ V(X_i)=(1-p)/p^2=30,\ $ so the Central Limit Theorem implies convergence in distribution to a standard normal random variable $Z$ as $n\to\infty:$
$${(X_1+...+X_n) - E(X_1+...+X_n)\over \sqrt{V(X_1+...+X_n)}}={X_1+...+X_n\over \sqrt{30n}}\xrightarrow{d} Z.$$
Hence, $$\begin{align}
\lim_{n\to\infty}P_n&=\lim_{n\to\infty}P(X_1+...+X_n< 0)\\
&=\lim_{n\to\infty}P\left({X_1+...+X_n\over \sqrt{30n}}<0\right)\\
&=P(Z<0)\\
&={1\over 2}.
\end{align}$$

Here are some values of $P_n$ (computed from (1) and rounded to $5$ digits) showing the convergence:

```<code>n       P_n

1       0.59812
10      0.53079
100     0.50971
1000    0.50307
10000   0.50097
</code>```

TBD (Strict monotonic convergence of $P_n\searrow{1\over 2}$?)

It appears from numerical computations -- but remains to be proved -- that the convergence to ${1\over 2}$ is strictly monotonic, such that for all $n$, $P_{n}>P_{n+1}>{1\over 2}.$ As described earlier, this is equivalent to saying that for all $n$,
$$P(X_1+...+X_{n}<0)> P(X_1+...+X_{n+1}<0) $$
where the $X_i=G_i-5$ are the i.i.d. "shifted $\text{Geometric}({1\over 6})$" random variables with $E(X_i)=0$ and $V(X_i)=30.$ Note that
$$P(X_i<0)=P(G_i<5)=1-\left({5\over 6}\right)^5=0.59812...$$
so each of the independent $X_i$ is more likely to be negative than nonnegative -- yet, the more of them that are added together, the less likely it appears to be that the sum is negative! (If independent quantities are each most likely negative, wouldn't one expect that the more of these that are added together, the more likely it would be that the sum is negative?)

Generalization to races with arbitrary $p$

This more-general game is played with a pair of dice for which the probability of a double is any fixed arbitrary $p$ in the interval $(0,1)$, and there are now $m$ columns and $n$ rows, where $n$ is the number of laps to be run and $m$ is any fixed integer greater than $1$. Now Slowpoke has the first $m-1$ columns and Doubles has only the rightmost column to "X out", as before; i.e., Doubles wins the race if $n$ doubles occur before $(m-1)n$ nondoubles occur, otherwise Slowpoke wins.

Now $$\begin{align}P_n
&= P(\text{Doubles wins an $n$-lap race}) \\
&=P(Y_n<(m-1)n)\\
&=P(G_1+...+G_n<(m-1)n)\\
&=P(X_1+...+X_n<0)\\
\end{align}$$
where $X_i=G_i-(m-1)$ and the $G_i$ are i.i.d. $\text{Geometric}(p),$ for which we find $E(X_i)=1/p-m$ and $V(X_i)=(1-p)/p^2.$

By the CLT we have
$${(X_1+...+X_n) - E(X_1+...+X_n)\over \sqrt{V(X_1+...+X_n)}}={(X_1+...+X_n)-n(1/p-m)\over \sqrt{n(1-p)/p^2}}\xrightarrow{d} \text{Normal($0,1)$}$$
Hence,
$$\begin{align}\lim_{n\to\infty}P_n
&=\lim_{n\to\infty}P(X_1+...+X_n< 0)\\
&=\lim_{n\to\infty}P\left({(X_1+...+X_n)-n(1/p-m)\over \sqrt{n(1-p)/p^2}}<{-n(1/p-m)\over \sqrt{n(1-p)/p^2}}\right)\\
&=\lim_{n\to\infty}P\left(\text{Normal($0,1)$}<{-n(1/p-m)\over \sqrt{n(1-p)/p^2}}\right)\\
&=\lim_{n\to\infty}P\left(\text{Normal($0,1)$}<\sqrt{n}{m-1/p\over \sqrt{1-p}}\right)\\
\end{align}$$

and therefore:
$$\lim_{n\to\infty}P_n=\begin{cases}
0&\text{if $m<1/p$}\\
1/2&\text{if $m=1/p$}\\
1&\text{if $m>1/p$}.
\end{cases}$$

Although this proves convergence, it says nothing about monotonicity; however, computations do suggest that the convergence is strictly monotonic if $m=1/p$, but that it need not be so otherwise.

Example (effects of biased dice):

Here's a plot showing the three behaviors of $P_n\ (1\le n\le 1000)$ for $n$-lap races according to the OP's original rules (i.e. $m=6$) with $$ P(\text{double})=\begin{cases}
1/6+0.01 & \color{green}{\text{ (green)}} \\
1/6 & \color{blue}{\text{ (blue)}} \\
1/6-0.01 & \color{red}{\text{ (red)}} \end{cases}$$

As illustrated, convergence is to $1/2$ only when $p=1/6,$ otherwise convergence is to $1$ or $0$. The "zoomed view" on the right shows the initial non-monotonic behavior when $p=1/6+0.01$ (green).

Body:
Let $Y_n$ be the number of nondoubles in a sequence of dice-rolls ending upon the $n$th double. Then $Y_n$ is a $\text{NegativeBinomial$(n,p)$}$ random variable with $p={1\over 6},$ having the following probabilty mass function (with $q=1-p$):
$$P(Y_n=k) = \binom{k+n-1}{n-1} p^nq^k\quad(k=0,1,2,...).$$

Letting $P_n$ be the probability that Doubles wins an $n$-lap game, we then have:
$$\begin{align}
P_n&=P(\text{fewer than $5n$ nondoubles occur before the $n$th double occurs})\\
&=P(Y_n< 5n)\\
&=p^n\sum_{k=0}^{5n-1}\binom{k+n-1}{n-1} q^k\tag{1}\\
&=\sum_{k=0}^{5n-1}\binom{k+n-1}{n-1} {5^k\over 6^{k+n}}
\end{align}$$

(The OP's formula for $P_n$ has omitted the binomial coefficient.)

Example ($n=2$):
Denote each occurrence of a double (resp. nondouble) by $1$ (resp. $0$). Then $P_2$ is the sum of the probabilities of just those sequences that end on the $2$nd occurrence of $1$ and have fewer than $2\times 5=10$ occurrences of $0$, as follows:

sequences                                    probabilities

11,                                          pp,
011, 101,                                    qpp, pqp,
0011, 0101, 1001,                            qqpp, qpqp, pqqp,
00011, 00101, 01001, 10001                   qqqpp, qqpqp, qpqqp, pqqqp,
000011, 000101, 001001, 010001, 100001       qqqqpp, qqqpqp, qqpqqp, qpqqqp, qqqqpp
...                                          ...
00000000011, 00000000101, ..., 10000000001   qqqqqqqqqpp, qqqqqqqqpqp, ..., pqqqqqqqqqp
                                             ------------------------------------------
                                             p^2 (1 + 2q + 3q^2 + 4q^3 + ... + 10q^9)
                                           = 12909191/22674816
                                           = 0.569318445627078...

To obtain the limit of $P_n$ as $n\to\infty,$ we use the fact that a $\text{NegativeBinomial$(n,p)$}$ random variable is distributed as the sum of $n$ i.i.d. $\text{Geometric}(p)$ random variables; thus,
$$\begin{align}
P_n&=P(Y_n< 5n)\\
&=P(G_1+...+G_n< 5n)\\
&=P(X_1+...+X_n<0)
\end{align}$$

where the $G_i$ are i.i.d. $\text{Geometric}(p)$ random variables, and we have defined $X_i=G_i-5.$

Now the $X_i$ are i.i.d. with $E(X_i)=(1-p)/p-5=0,\ V(X_i)=(1-p)/p^2=30,\ $ so the Central Limit Theorem implies convergence in distribution to a standard normal random variable $Z$ as $n\to\infty:$
$${(X_1+...+X_n) - E(X_1+...+X_n)\over \sqrt{V(X_1+...+X_n)}}={X_1+...+X_n\over \sqrt{30n}}\xrightarrow{d} Z.$$
Hence, $$\begin{align}
\lim_{n\to\infty}P_n&=\lim_{n\to\infty}P(X_1+...+X_n< 0)\\
&=\lim_{n\to\infty}P\left({X_1+...+X_n\over \sqrt{30n}}<0\right)\\
&=P(Z<0)\\
&={1\over 2}.
\end{align}$$

Here are some values of $P_n$ (computed from (1) and rounded to $5$ digits) showing the convergence:

n       P_n

1       0.59812
10      0.53079
100     0.50971
1000    0.50307
10000   0.50097

TBD (Strict monotonic convergence of $P_n\searrow{1\over 2}$?)

It appears from numerical computations -- but remains to be proved -- that the convergence to ${1\over 2}$ is strictly monotonic, such that for all $n$, $P_{n}>P_{n+1}>{1\over 2}.$ As described earlier, this is equivalent to saying that for all $n$,
$$P(X_1+...+X_{n}<0)> P(X_1+...+X_{n+1}<0) $$
where the $X_i=G_i-5$ are the i.i.d. "shifted $\text{Geometric}({1\over 6})$" random variables with $E(X_i)=0$ and $V(X_i)=30.$ Note that
$$P(X_i<0)=P(G_i<5)=1-\left({5\over 6}\right)^5=0.59812...$$
so each of the independent $X_i$ is more likely to be negative than nonnegative -- yet, the more of them that are added together, the less likely it appears to be that the sum is negative! (If independent quantities are each most likely negative, wouldn't one expect that the more of these that are added together, the more likely it would be that the sum is negative?)

Generalization to races with arbitrary $p$

This more-general game is played with a pair of dice for which the probability of a double is any fixed arbitrary $p$ in the interval $(0,1)$, and there are now $m$ columns and $n$ rows, where $n$ is the number of laps to be run and $m$ is any fixed integer greater than $1$. Now Slowpoke has the first $m-1$ columns and Doubles has only the rightmost column to "X out", as before; i.e., Doubles wins the race if $n$ doubles occur before $(m-1)n$ nondoubles occur, otherwise Slowpoke wins.

Now $$\begin{align}P_n
&= P(\text{Doubles wins an $n$-lap race}) \\
&=P(Y_n<(m-1)n)\\
&=P(G_1+...+G_n<(m-1)n)\\
&=P(X_1+...+X_n<0)\\
\end{align}$$
where $X_i=G_i-(m-1)$ and the $G_i$ are i.i.d. $\text{Geometric}(p),$ for which we find $E(X_i)=1/p-m$ and $V(X_i)=(1-p)/p^2.$

By the CLT we have
$${(X_1+...+X_n) - E(X_1+...+X_n)\over \sqrt{V(X_1+...+X_n)}}={(X_1+...+X_n)-n(1/p-m)\over \sqrt{n(1-p)/p^2}}\xrightarrow{d} \text{Normal($0,1)$}$$
Hence,
$$\begin{align}\lim_{n\to\infty}P_n
&=\lim_{n\to\infty}P(X_1+...+X_n< 0)\\
&=\lim_{n\to\infty}P\left({(X_1+...+X_n)-n(1/p-m)\over \sqrt{n(1-p)/p^2}}<{-n(1/p-m)\over \sqrt{n(1-p)/p^2}}\right)\\
&=\lim_{n\to\infty}P\left(\text{Normal($0,1)$}<{-n(1/p-m)\over \sqrt{n(1-p)/p^2}}\right)\\
&=\lim_{n\to\infty}P\left(\text{Normal($0,1)$}<\sqrt{n}{m-1/p\over \sqrt{1-p}}\right)\\
\end{align}$$

and therefore:
$$\lim_{n\to\infty}P_n=\begin{cases}
0&\text{if $m<1/p$}\\
1/2&\text{if $m=1/p$}\\
1&\text{if $m>1/p$}.
\end{cases}$$

Although this proves convergence, it says nothing about monotonicity; however, computations do suggest that the convergence is strictly monotonic if $m=1/p$, but that it need not be so otherwise.

Example (effects of biased dice):

Here's a plot showing the three behaviors of $P_n\ (1\le n\le 1000)$ for $n$-lap races according to the OP's original rules (i.e. $m=6$) with $$ P(\text{double})=\begin{cases}
1/6+0.01 & \color{green}{\text{ (green)}} \\
1/6 & \color{blue}{\text{ (blue)}} \\
1/6-0.01 & \color{red}{\text{ (red)}} \end{cases}$$

As illustrated, convergence is to $1/2$ only when $p=1/6,$ otherwise convergence is to $1$ or $0$. The "zoomed view" on the right shows the initial non-monotonic behavior when $p=1/6+0.01$ (green).


问:
Linearizing equations of motion
I have been looking at the operations of a quadrotor drone. I am reading the maths behind it and in one section it mentions:

"These equations of motion are linearized with respect to an equilibrium point" . I have some questions regarding this.

1) What exactly does it mean to linearize an equation of motion ? how exactly is this usually done?

2) If say I linearize the equations of motion for a drone or car or whatever, why is this useful, what information can it tell me?

Thank you

Here is a sort of idea of it, I cant post them all since it is a 70 page document (Don't think I can attach documents) but it mostly relates to the translational and altitude dynamics.

答:
A time-independent equation of motion can be written as
$$
{d^2x\over dt^2}=F(x),
$$
where $x$ is a vector representing the various variables involved. If function $F$ is sufficiently regular it can be expanded about an equilibrium point $x_0$ as
$$
F(x)=F(x_o)+F'(x_o)\,(x-x_0) + \dots
$$
But $F(x_0)=0$ by definition of equilibrium point, hence we can approximate the equation of motion with its linearised version:
$$
{d^2x\over dt^2}=F'(x_o)\,(x-x_0).
$$
This is useful because the linearised equation is much simpler to solve and it will give a good approximation if $\|x-x_0\|$ is small enough.

Title: Linearizing equations of motion


问:
Spliting Field over $\mathbb{F}_3$
How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?</p>

答:
First show that $K=\mathbb{F}_3[x]/\langle f(x)\rangle$ is a splitting field of $f(x)$. Answers by others will help you there.

Hint (for the isomorphism): You can easily verify that
$$
f(-x)=-g(x).
$$
It follows that the coset of $-x$ in the field $\mathbb{F}_3[x]/\langle f(x)\rangle$ is a zero of $g(x)$.

The rest is routine, I hope.

Title: Spliting Field over $\mathbb{F}_3$


问:
Collection of co-finite subsets has finite intersection property

Let $W$ be an infinite set. $X\subseteq W$ is co-finite if $W\setminus X$ is finite. Prove that the collection of co-finite subsets of $W$ has the finite intersection property (FIP).

My work:

Let $$S: = \{X\subseteq W: W\setminus X \text{ is finite}\} $$
To show that $S$ has FIP, we want to show that for all $X_1,X_2,\ldots,X_n$ where $n\in\mathbb N$, we have $\bigcap_{i=1}^n X_i \neq \varnothing$. Let's suppose the contrary, i.e. for some $n\in\Bbb N$, suppose there are elements $X_1,\ldots, X_n$ in $S$ such that $\bigcap_i X_i = \varnothing$. So, $\bigcup_i X_i^c = \bigcup_i W\setminus X_i = W$. $W$ is infinite, so $W\setminus X_i$ is infinite for at least one $1\le i\le n$ (if not, we have a contradiction). This is a contradiction, because all $W\setminus X_i$ are finite since $X_i\in S$ for $1\le i\le n$.

Does this look good? Are there any other ways of approaching this problem?

答: $$\left(\bigcap_{i=1}^n X_i\right)^\complement = \bigcup_{i=1}^n X_i^\complement$$ by de Morgan. The finite union of finite sets is finite, so $\mathcal{S}$ is closed under finite intersections and $\emptyset \notin \mathcal{S}$ because $W$ is infinite. No need for a proof by contradiction.


问:
Is Calkin algebra $B(H)/\mathcal K(H)$ isomorphic to some subalgebra of $B(H)$?
$B(H)$ is the operator algebra on separable Hilbert space. $\mathcal K(H)$ is the compact operator algebra.

I have no idea on this. Could someone please give me a hint?

答:
From the Wikipedia page on the Calkin Algebra:

"As a C*-algebra, the Calkin algebra is not isomorphic to an algebra of operators on a separable Hilbert space"


问:
How to find the middle point of an arc slice?
I need to find centre of different arc in the Pie Chart. I have following details -

Similar question is here -
How to find the middle point of an arc slice?

here is the reference image from above post -

My Scenario. When I move to different slice I need to calculate centre of that slice. This is the piechart centre coordinates -

```<code>cx = 330.5;
cy = 140;
</code>```

First (Green) slice details -

```<code>{
    "cx": 330.5,
    "cy": 140,
    "innerRadius": 0,
    "middleRadius": 54,
    "outerRadius": 108,
    "maxRadius": 352.3850876526985,
    "startAngle": 0,
    "endAngle": 17.358490566037737,
    "midAngle": 8.679245283018869,
    "paddingAngle": 0,
    "percent": 0.04821802935010482,
}
</code>```

Second (Orange) slice details -

```<code>{
    "cx": 330.5,
    "cy": 140,
    "innerRadius": 0,
    "middleRadius": 54,
    "outerRadius": 108,
    "maxRadius": 352.3850876526985,
    "percent": 0.36547868623340324,
    "startAngle": 17.358490566037737,
    "midAngle": 83.14465408805033,
    "endAngle": 148.93081761006292,
    "paddingAngle": 0,
    "percent": 0.36547868623340324,
}
</code>```

Third (Gray) slice details -

```<code>{
    "cx": 330.5,
    "cy": 140,
    "innerRadius": 0,
    "middleRadius": 54,
    "outerRadius": 108,
    "maxRadius": 352.3850876526985,
    "startAngle": 148.93081761006292,
    "midAngle": 254.46540880503147,
    "endAngle": 360,
    "paddingAngle": 0,
    "percent": 0.5863032844164919,
}
</code>```

I tried following formula but it did not work and not getting the expected result -

```<code>x = cx + outerRadius * Math.cos(midAngle),
y = cy + outerRadius * Math.sin(midAngle),
</code>```

Update

After changing the formula to

```<code> x = cx + outerRadius * Math.cos(midAngle * 3.14159265358979323846 / 180.0);
 y = cy + outerRadius * Math.sin(midAngle * 3.14159265358979323846 / 180.0);
</code>```

I am getting following output -
Completed -

```<code>{x: 437.2632503544796, y: 156.29749593485772}
</code>```

In progress -

```<code>{x: 343.3912133107331, y: 247.2278723997505}
</code>```

Not started -

```<code>{x: 301.5754291795817, y: 35.9453547271694}
</code>```

You can see in all above three images tooltip position is not correct.

答:
What seems to be the issue

To create the above image, I simply grabbed the descriptions at their calculated positions and superimposed them on top of the original pie chart. Then, I shaded the areas where it seems that the back-end thinks the positions are. As you can see, both description boxes are positioned at the midpoint of the (mirrored) shaded regions.

How to interpret the image

It seems that the calculated angle increases in the clockwise direction, while the shown slices somehow still increase in the anticlockwise direction. In other words, when you tell it to show something at midAngle = 8.679, it thinks it should position it at an angle of -8.679 or alternatively 351.321 degrees when using the conventions of the unit circle.

What to change

To get the angle right in your program, simply change every occurrence of midAngle to -midAngle. In the comments I said changing your code to 360 - midAngle could also work, but simply making it negative will result in the same thing.

Once you've got the angle, all you need to do is tweak how far along the radius (at said angle) the description box must appear. Currently you're using the full value of outerRadius to calculate the values of x and y, but as NickD says in his answer, you could use 0.5 * outerRadius to make the boxes appear halfway the pie chart's radius.

In other words, this should work:

```<code>x = cx + 0.5 * outerRadius * Math.cos(-midAngle * 3.14159265358979323846 / 180.0);
y = cy + 0.5 * outerRadius * Math.sin(-midAngle * 3.14159265358979323846 / 180.0);
</code>```

Although using Math.cos(midAngle * ...) would work just as well as Math.cos(-midAngle * ...), because $cos(\theta)=cos(-\theta)$ for all values of $\theta$.

Body:
What seems to be the issue

To create the above image, I simply grabbed the descriptions at their calculated positions and superimposed them on top of the original pie chart. Then, I shaded the areas where it seems that the back-end thinks the positions are. As you can see, both description boxes are positioned at the midpoint of the (mirrored) shaded regions.

How to interpret the image

It seems that the calculated angle increases in the clockwise direction, while the shown slices somehow still increase in the anticlockwise direction. In other words, when you tell it to show something at midAngle = 8.679, it thinks it should position it at an angle of -8.679 or alternatively 351.321 degrees when using the conventions of the unit circle.

What to change

To get the angle right in your program, simply change every occurrence of midAngle to -midAngle. In the comments I said changing your code to 360 - midAngle could also work, but simply making it negative will result in the same thing.

Once you've got the angle, all you need to do is tweak how far along the radius (at said angle) the description box must appear. Currently you're using the full value of outerRadius to calculate the values of x and y, but as NickD says in his answer, you could use 0.5 * outerRadius to make the boxes appear halfway the pie chart's radius.

In other words, this should work:

x = cx + 0.5 * outerRadius * Math.cos(-midAngle * 3.14159265358979323846 / 180.0);
y = cy + 0.5 * outerRadius * Math.sin(-midAngle * 3.14159265358979323846 / 180.0);

Although using Math.cos(midAngle * ...) would work just as well as Math.cos(-midAngle * ...), because $cos(\theta)=cos(-\theta)$ for all values of $\theta$.


问:
Rearranging equation to solve for $C$ in an integral

I tried solving for y, and I get the resulting equation:

However, once plugging in $(0, 6)$ you get a power of $e$ that is equal to a negative number. How would you solve this? Is there a different way to find $y$ where $e^c$ does not equate to a negative number?

答: You have $$\ln|y-7| = \frac{x^2}{2} - 8x +C$$which implies $$|y-7| = e^{\frac{x^2}{2} - 8x +C} \implies y = e^{\frac{x^2}{2} - 8x +C} + 7 \ \  \text{or} \ \ y = -e^{\frac{x^2}{2} - 8x +C}+7$$If you want the real solution and $y(0) = 6$ then $$6 = -e^{C} + 7 \implies e^C = 1 \implies C = 0$$


问:
What is the Lie group infinitesimal generator?
Our prof gave us the definition of the Lie group infinitesimal generator and it's kth-coordinate, he also explained why the kth-coordinate is called "the kth-coordinate", but nonetheless he didn't want to explain the Lie group infinitesimal generator because "we will understand it later as the course will continue".

I'm fine with that explanation but nevertheless I'd really like to at least know why we are defining the Lie group infinitesimal generator in the first place, and where it is used. Maybe nothing too deep as we just started working with Lie groups. But I want to grasp the intuition behind the Lie group infinitesimal generator

答:
A Lie group is a group that is also a manifold. Formally the generators are a basis for the tangent space of the identity element in the group.

Informally you can think of the generators as corresponding to small group transformations which are "near" the identity. Every element in the connected component of the group can be "generated" by repeatedly applying these transformations.

As an example consider the positive real numbers with multiplications as the group operation. The identity element of this group is the real number $1$. The tangent space at $1$ is just the real line, and the tangent vector $\sigma=+1$ can be taken as a basis for this 1D space.

Now consider the "infinitesimal transformation" which is "in the neighborhood of the identity element" of our group,

$g = 1+\epsilon \sigma,$

where $\epsilon$ is some "small" real number (can be positive or negative).

For a finite epsilon we can approximately represent most members of our group from repeated multiplication of $g$.

For instance suppose $\epsilon=0.001$ and we want to represent the group element $2$.

Then our approximate $g$ would be $1.001$ and we could represent $2$ by $g^{693}=1.999013$. Now of course we don't need to be so approximate. Using calculus we can get an exact representation.

Let $g_\theta$ be a member of the group labelled by a real number $\theta$. Then we can express $g_\theta$ in terms of our generator $g$ by the following limit.

$$g_\theta = \lim_{N\rightarrow \infty} (1+\theta \cdot \sigma /N)^N = e^{\theta \cdot \sigma}$$

Here I have replaced $\epsilon$ with $\theta/N$ and taken the limit as the exponent $N$ approaches infinity.

Now we have a way of representing our entire group in terms of this function $exp(\theta \cdot \sigma)$ where $\theta$ labels the different elements of the group.

The essential idea of Lie groups is to represent the connected component of your group-manifold in terms of the exponential map applied to a linear combination of your generators. The relations between the generators themselves can reveal a lot about the structure of the group.


问:
Exterior multiplication with parallel form commutes with Laplacian
Let $(M,g)$ be a Riemannian manifold, let $\nabla$ be the Levi-Civita connection, and let $\Delta=dd^*+d^*d$ be the Laplacian. Suppose $\omega \in \Omega^*(M)$ is a parallel differential form (i.e. $\nabla \omega=0$), is it true that $\Delta (\alpha \wedge \omega)=\Delta(\alpha) \wedge \omega$?

I have seen this fact being used a few times in literature, but I cannot find or work out a full proof myself. The sources I have consulted simply say 'use Weitzenbock's formula', but I cannot see how the formula can prove this. Any help is appreciated!

答:
A rather simple proof based on graded Lie superalgebra and by computing the commutator of  linear operators $L_{\omega}: \eta\mapsto \eta\wedge \omega$ and $\Delta$ (i.e. $[L_\omega,\Delta]$) then using Jacobi identity of graded Lie superalgebra, can be found in

Verbitsky, Misha, Manifolds with parallel differential forms and Kähler identities for $G_{2}$-manifolds, J. Geom. Phys. 61, No. 6, 1001-1016 (2011). ZBL1214.58002.

corollary 2.9 page 6. According to this paper, the wanted result is due to S. S. Chern.


问:
Use $\varepsilon$-$\delta$ to prove $\lim_{x\to 1} \frac{1}{x^2} = 1$

Use $\varepsilon$-$\delta$  to prove that $\lim_{x\to1}$ $\frac{1}{x^2} = 1.$

I set up the proof as follows: Let $\varepsilon>0$ s.t. $0<|\frac{1}{x^2} -1|<\delta.$ which simplifies to $0<|\frac{1-x^2}{x^2}|<\delta$. However I'm not sure where to take this next.

答: Suppose that $\left|x-1\right|<\frac12$. Then $x>\frac12$ and therefore $x^2>\frac14$. On the other hand,$$|1-x^2|=|x-1||x+1|\leqslant\frac32|x-1|.$$Therefore$$\left|\frac{1-x^2}{x^2}\right|<6|x-1|.$$So, take $\delta=\min\left\{\frac12,\frac\varepsilon6\right\}$.


问:
Cosine of angle between diagonals of a cube
I am having trouble in this question:

The cosine of angle between two diagonals of a cube is:

a) $\frac{1}{2}$

b)$\frac{1}{3}$

c)$\frac{2\sqrt{2}}{3}$

d)None of the above

Upon solution I found out that the only possible answers are $\frac{1}{2}$, $\frac{2\sqrt{2}}{3}$, and 0 so option B should be the answer but the sheet tells the answer to be option A. Please can anyone help me with this?

PS. Please do let me know if you guys want to see my solution or if you guys think the answer mentioned in the sheet is wrong, also I have to solve this question by assuming position vectors of the vertices.

答:
Two diagonal vectors are $\hat A$= <1,1,1>,  $\hat B$=<-1,1,1>

cos $\theta$ = $\frac{\hat A.\hat B}{|\hat A|.|\hat B|}$= $\frac{1}{\sqrt(3).\sqrt(3)}$=$\frac{1}{3}$