问:
Does the following identity hold?
I encountered a claim in the proof a a claim in a paper that seemed to imply/use the following:

Suppose we have two subgroups $K$ and $H$ of a larger group $G$.  Let $T$ be a translation of $K$, which is defined as a subset of $G$ such that every $g \in G$ can be uniquely written as $t k$, for some $t \in T$ and $k \in K$. For example, if $G = \mathbb{Z} / 12 \mathbb{Z}$ and $K = \{0, 2, 4, \cdots, 10\}$, then $T = \{0, 1\}$ is a possible translation of $K$.

Let us also for simplicity define the notation $t K = \{tk ~|~ k \in K\}$.

The claim is then the following: for any translation of $T$ and any subgroup $H \leq G$ which is neither a subset nor a superset of $K$, we have the following:
$$\sum_{t \in T} |tK \cap H|^2 = |H| \cdot |K \cap H|$$

Is this claim true? It does seem to work out for example with $G, T$ as given above and $H = \{0, 3, 6, 9\}$, but I can't seem to find a proof of this fact or even an intuition as to why this should be true. If it is not true, then are there any other conditions we can impose (maybe I have missed them in reading the paper) so that it is true?

EDIT: The paper in question can be found on ArXiv at this link. The question stems from a step in the proof of lemma $4$.

答:
This is correct. In order to prove it, we need a lemma first:

I claim that for any $t\in G$ either $tK\cap H$ is empty or $|tK\cap H|=|K\cap H|$. To prove this, let $|K\cap H|=n$ and then let $k_1,k_2,\dots,k_n$ be the distinct members of $K\cap H$. And assume $tK\cap H$ is nonempty, so there is some $y=tk\in tK\cap H$. Then the elements $yk_1,yk_2,\dots yk_n$ must all be distinct. But $y,k_i\in H$, so $yk_i\in H$, and $yk_i=(tk)k_i=t(kk_i)\in tK$. So we have identified $n$ distinct elements of $tK\cap H$, which means that if we let $m=|tK\cap H|$ then $m\geq n$.

Now, let $tk_1,tk_2,\dots,tk_m$ be the distinct elements of $tk\cap H$. But if $tk_1\in H$ then $(tk_1)^{-1}=k_1^{-1}t^{-1}\in H$, and so the elements $(tk_1)^{-1}tk_1,(tk_1)^{-1}tk_2,\dots,(tk_1)^{-1}tk_m$ must all be distinct elements of $H$. But we can simplify those elements down to $k_1^{-1}k_1,k_1^{-1}k_2,\dots,k_1^{-1}k_m$, which clearly must be elements of $K$. So we have identified $m$ distinct elements of $K\cap H$, meaning that $n\geq m$, and therefore $n=m$.

Now, we prove the main result. We know that the cosets $tK,\ t\in T$ are a partition of $G$, and therefore the sets $tK\cap H$ are a partition of $H$. Let $T'=\{t\in T|tK\cap H\neq \varnothing\}$. Then if $|K\cap H|=n$, we have for every $t\in T'$ that $|tK\cap H|=n$. We also know that $|H|=mn$ for some integer $m$, and therefore $|T'|=m$. Finally, we know that restricting $t$ to $T'$ doesn't change the sum $\sum |tK\cap H|^2$, since if $t\not\in T'$ then $|tK\cap H|=0$. Putting it all together:

\begin{align*}
\sum_{t\in T} |tK\cap H|^2&=\sum_{t\in T'} |tK\cap H|^2\\
&=\sum_{t\in\{t_1,t_2,\dots,t_m\}}n^2\\
&=mn^2\\
&=n\cdot(mn)\\
&=|K\cap H|\cdot |H|
\end{align*}

Title: Does the following identity hold?


问:
Inverse of Homomorphism
I am a little confused when working with the inverse of a homomorphism, this specifically came up when proving the following claim:

Let $\phi$ be some homomorphism from $G_1\rightarrow G_2$:
$$H_2 \subset G_2\textrm{ is a normal subgroup } \Rightarrow \phi^{-1}(H_2)\subset G_1 \textrm{ is a normal subgroup}$$

The proof for this begins by stating that
$$e_1\in\phi^{-1}(H_2)\textrm{ since } \phi(e_1)=e_2\in H_2$$
But since $\phi$ is a homorphism, it is not nessesarily surjective or injective. So there could be many terms in $g_1,g_2,...\in G_1$ such that $\phi(g_1)=\phi(g_2)=...=e_1$. So does $\phi^{-1}(e_1)$ return all of these values? Can the inverse function of a Homomorphism return multiple values per input? In general, when dealing with inverse functions, should I think of the inverse as a function that will return every coresponding value?

答: The expression $\phi^{-1}(H_2)$ means the set $\{g\in H_1\mid\phi(g)\in H_2\}$, for any function $\phi$, even if it has no inverse.

ContentLicense: CC BY-SA 4.0

Title: Inverse of Homomorphism


问:
Solve the recurrence relation given by: $A_{n+2} + 4A_n = 0; A_0 = A_1 = 1$
I am having trouble coming up with the general solution. It seems to be given by: x2 + 4, but if this is the case I don't see how the problem can be solved with real numbers. Thanks for any advice or guidance in advance.

Body:
$$A_{n+2}=-4A_n$$
Let $A_n=t^n$, then $t=\pm 2 i \implies A_n =C(2i)^n+D (-2i)^n=2^n [ C  e^{in\pi/2}+D e^{-i n\pi/2}]$
$$A_0=0 \implies A_0=C+D=1 ~ \text{and}~A_1=C-D=-i/2 \implies C=(1/2-i/4), D=(1/2+i/4)$$
So $$A_n=2^n[(1/2-i/4)e^{in\pi/2}+(1/2+i/4)e^{i\pi n/2}]=2^{n} [\cos(n\pi/2)+\frac{1}{2}\sin(n\pi/2)]$$


问:
Difference between $\implies$ and $\;\therefore\;\;$?
I've seen both symbols used to mean "therefore" or logical implication. It seems like $\therefore$ is more frequently used when reaching the conclusion of an argument, while $\implies$ is for intermediate claims that imply each other. Is there any agreed upon way of using these symbols, or are they more or less interchangeable?</p>

答:
$$A \implies B \implies C \tag{1}$$ has a different meaning and reading from $$A;\:\:\therefore B;\:\:\therefore C \tag{2} \\\text{($A$ is true; therefore $C$ is true)}.$$ Line $(2)$ asserts premise A, that $A \implies B$ and that $B \implies C,$ and concludes that $C$ is true.

Line $(1),$ on the other hand, neither specifies that sentences $A$ and $B$ are actually true, nor asserts that sentence $C$ is true. It in fact has three possible meanings, no pair of which is equivalent (convince yourself using the truth assignments $(0,0,0)$ and $(1,0,0)$): $$A \implies (B \implies C).\tag{1a}$$
$$(A \implies B) \implies C\tag{1b}$$
$$(A \implies B)\: \text{ and } \:(B \implies C)\\\text{(A being true implies that C is true)}\tag{1c}$$ The third is the most common interpretation, and can be read as “in case $A$ is true, then $C$ too is true”.

If the difference between lines $(2)$ and $(1\text c)$ seems trivial or just a matter of contextually inferring the stronger meaning as necessary, then observe that  $$x>-4;\:\:\therefore\, x>4$$ means that $x\in(4,\infty),$ whereas $$x>-4\implies x>4$$ means that $x\in(-\infty,-4]\cup(4,\infty).$

Another example: \begin{align}\lvert2x\rvert=x-1&\implies x=-1 \;\text{ or }\; \frac13\end{align} is true (replace the LHS with $\pm2x$) while $$\lvert2x\rvert=x-1; \;\text{therefore, }\; x=-1 \;\text{ or }\; \frac13$$ is false (explanation here and here).

And two illustrations that overloading the $\text‘{\implies}\text’$ symbol imposes a burden of disambiguation on the reader, breaking their flow of reading and skimming.

All in all, I think it is bad practice to treat $\text‘{\implies}\text’$ and $\text‘\therefore\text’$ as interchangeable.

Here are several articles expounding their distinction:

implication versus conclusion

“implies” versus
“therefore”

guide to writing mathematics (p. 17)

abuse of the implication
symbol

ContentLicense: CC BY-SA 3.0


问:
$C_c^0(\Omega)$ is not Banach!?! Also density requires completeness?
Today I was very surprised to learn that $C_c^0(\Omega)$ is not a Banach space with the supremum norm. Why is that, when $C^\infty_c(\Omega)$ is?

Also (from here, bottom of page 18), I learn that
even though we can approximate any function in $C_c^0(\Omega)$ by a $C_c^\infty(\Omega)$ function as closely as we want, there is no dense inclusion since the space is not Banach. Is it really necessary for "density" that being complete is required?

If completeness is required for density, then the statement I read somewhere

$C_c^\infty(0,T;V)$ ($V$ is Hilbert) is dense in $W(0,T)$ (some space)

does not make sense either. Or is it different because $[0,T]$ is compact?

答:
It should not be a surprise that $C^0_c(\Omega)$ is not a Banach space. The sequence of functions with compact supports can easily converge to a function which is nonzero on all of $\Omega$. For example, the function $f(x)= \mathrm{dist}\,(x,\partial \Omega) $ does not belong to $C_c^0(\Omega)$, but $f_n(x)=\max(0,f(x)-1/n)$ do, and $f_n\to f$ uniformly.

Concerning density: on the top of page 12 the author defines the concept of a dense subset only for Banach spaces. Therefore, a subset of a $C^0_c(\Omega)$ cannot be dense, simply because the notion of being dense is not defined in this space. This is not a position that I would take, but the author of a book  gets to decide what to define and how.


问:
Is an ellipse determined by three points and tangent lines at two of them?
I have three points, $A$, $B$, and $C$, on an ellipse; two tangent lines at $A$ and $B$ are also known. Are these enough to determine the whole ellipse?

答:
Answered in comments by OP:

[See] Paris Pamfilos, A Gallery of Conics by Five Elements, Forum
Geometricorum, Volume 14 (2014) 295–348. My problem was completely
solved


问:
Is is true that $\mathbb{Q}(x^2)(x) = \mathbb{Q}(x^2)[x] = \mathbb{Q}(x)$?
I am working on a proof and want to show this equality. It appears to be true, but I'm not sure of how to show it rigorously. I know the first equality hold since $x \in \mathbb{Q}(x)$ is algebraic over $\mathbb{Q}(x^2)$.

答:
We obviously have $\Bbb Q(x^2)(x) = \Bbb Q(x)$, as $x^2$ belongs to $\Bbb Q(x)$.

Thus it suffices to show that $\Bbb Q(x^2, x) = \Bbb Q(x^2)[x]$.

But this is simly because the element $x$ is algebraic (of degree $2$) over $\Bbb Q(x^2)$, as it is root of the polynomial $T^2 - x^2 \in \Bbb Q(x^2)[T]$.


问:
Manifold structure on a set of matrices
Let $Ω$ denote the $((n+1)×(n+1))$ matrix given by
\begin{bmatrix}
-1&0\\0&I_n
\end{bmatrix}
where $I_n$ is the $n×n$ identity matrix.

Let $X=\{A∈M_{n+1}(\mathbb{R}):A^tΩA=Ω\}$. How can I show that $X$ is a manifold? What will its dimension be? I am thinking of using the determinant map. It is easy to see that the determinant of each element of $X$ is $1$ or $-1$. What next? Any help will be appreciated.

答:
$X$ is a group, so a neighbourhood of the identity matrix $I$ is mapped to a neighbourhood of any $A \in X$ by the map
$B \mapsto AB$.  So it suffices to prove that a neighbourhood of $I$ is a manifold.  This can be done using the implicit function theorem.

Title: Manifold structure on a set of matrices


问:
Distance between point and a line - problems simplifying the minimised distance equation
Someone asked how to prove the distance between a point $(x_1,y_1)$ and a line $Ax + By + C = 0$ is:$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$ The currently accepted answer shows that a point on the line can be expressed as $(x, {-Ax-C\over B})$, then explains an approach using calculus to find the value of $x$ when the square distance is minimised, and then plug it into the distance formula: $$\tag{1}D=\sqrt{ (x_1-x)^2 + {\Bigl(y_1-  { \textstyle{-Ax-C\over B}}\Bigr)^2 }}$$ and simplify.

I understand the approach. My problem is with the simplification part. I found $x$ to be minimised when $x = \dfrac{B^2x_1 - A(By_1 + C)}{A^2 + B^2}$. How would you plug this into $(1)$ and simplify effectively?

My first approach was to plug $x$ into the two places it appears in the $(1)$ but expanding $(x_1 - \frac{B^2x_1 - A(By_1 + C)}{A^2 + B^2})^2$ (let alone the rest of it) made me give up as it just seemed to get even more complex. I tried some substitutions:

but this route didn't seem to be getting me closer to the formula to be proven.

I understand there are other far simpler ways to arrive at the formula (see other answers to the linked question), but it irked me that my algebraic manipulation skills are too weak to follow through this one. What I'm wondering is how an expert would approach this. How would you break the problem down and proceed?

I was deliberating over whether to post this question because part of me is telling me I'm lazy for not just trying to plough on through it.

答:
Since you wanted to see how this is done algebraically, we'll need to look at a particular property of quadratic functions.  As I've said in a comment, if $r > 0$, then $ f(x) = rx^2 + sx + t \ $ represents an "upward-opening" parabola.  By completing the square, we can write this as

$$f(x) \ = \ r \cdot (x^2 \ + \ \frac{s}{r} \cdot x \ + \ \frac{s^2}{4r^2}) \ + \ (t  \ - \ r \cdot \frac{s^2}{4r^2}) \ = \ r \cdot ( \ x  +  \frac{s}{2r} \ )^2 \ + \ ( \ t  -  \frac{s^2}{4r} \  ) \ . $$

The vertex of this parabola lies at  $ \ x = -\frac{s}{2r} \ $ , with the y-coordinate of the vertex being the minimum value of the function $ \ f(-\frac{s}{2r}) \ = \ t  -  \frac{s^2}{4r} .$  We will exploit this to reduce the amount of writing we will need to do.

We will label the closest point on the line to $ \ (x_1 , y_1) \ $ as $ \ (X,Y) \ $ . The distance-squared function is

$$D^2 \ = \ ( X - x_1 )^2 \ + \ (Y - y_1)^2 \ = \ ( X - x_1 )^2 \ + \ ( \ [-\frac{AX + C}{B} \ ] - y_1)^2 \    $$

$$= \ ( X - x_1 )^2 \ + \ (  \frac{A}{B} \cdot X  \ + \ \frac{C}{B}  + \  y_1)^2 \ = \ ( X - x_1 )^2 \ + \ (  \frac{A}{B} \cdot X  \ + \ K \ )^2 \ ,  $$

where we will make a temporary substitution, $ \ K = \frac{C+By_1}{B} \ $ , to further save some writing. (This is often a good idea when working out complicated algebraic expressions, as it reduces the risk of embarrassing, time-wasting errors.)  We will see that we can hold out for some time before needing to replace $K$ .

From here, we have

$$D^2 \ = \ ( \ X^2 \ - \  2x_1X \ + \ x_1^2 \ ) \ + \ ( \ \frac{A^2}{B^2} \cdot X^2  \ + \ 2\frac{A}{B} K \cdot X \ + \ K^2 \ ) \   $$

$$= \   ( \ 1 + \frac{A^2}{B^2} \ ) \cdot X^2 \ + \  2 ( \ \frac{A}{B} K - x_1 \ ) \cdot X \ + \ (  \ K^2 \ + x_1^2 \ ) ,   $$

which now has the "general form" for a quadratic function.  Since we already know how to convert this to "standard form" by completing the square, we can simply use the results we discussed back at the start.  The vertex of the parabola described by this equation lies at

$$X \ = \ -\frac{s}{2r} \ = \ -\frac{2 ( \ \frac{A}{B} K - x_1 \ )}{2( \ 1 + \frac{A^2}{B^2} \ )} \cdot \frac{B^2}{B^2} \ = \ \frac{ B^2 x_1 \ - \ AB K}{A^2 \ + \ B^2 \ } $$

$$= \ \frac{ B^2 x_1 \ - \ AB \cdot (\frac{C+By_1}{B})}{A^2 \ + \ B^2 \ } \ = \ = \ \frac{ B^2 x_1 \ - \ A \cdot (C \ + \ By_1)}{A^2 \ + \ B^2 \ } \ . $$

So here is the x-coordinate of the point on the line that is closest to $(x_1,y_1)$, confirming the stated result without the use of calculus.

We do not bother substituting this into either the quadratic function or the equation for the line, since we know from the standard form of the quadratic function that its minimal value is

$$D^2_{min} \ = \ t  -  \frac{s^2}{4r} \ = \ ( \ K^2 \ + \ x_1^2 \ ) \ - \ \frac{[ \ 2 ( \ \frac{A}{B} K - x_1 \ ) \ ]^2}{4( \ 1 + \frac{A^2}{B^2} \ )} \cdot \frac{B^2}{B^2}  $$

$$= \ ( \ K^2 \ + \ x_1^2 \ ) \ - \ \frac{ ( \ A K \ - \ Bx_1 \ )^2}{ A^2 + B^2} \ = \  \frac{( \ K^2 \ + \ x_1^2 \ )(A^2 + B^2)  \ - \ ( \ A K - Bx_1 \ )^2}{ A^2 + B^2} $$

$$= \  \frac{( \ A^2K^2 \ + \ B^2K^2 \ + \ A^2x_1^2 \ + \ B^2x_1^2 \ ) \ - \ ( \ A^2K^2 \ - \ 2ABKx_1 \ + \ B^2x_1^2 \ )}{ A^2 + B^2} $$

$$= \  \frac{ A^2x_1^2 \ + \ 2ABKx_1 \ + \ B^2K^2 \ }{ A^2 + B^2} \ = \  \frac{ ( \ Ax_1 \  + \ BK \ )^2 }{ A^2 + B^2}$$

$$= \  \frac{ ( \ Ax_1 \  + \ B \cdot [\frac{C+By_1}{B}] \ )^2 }{ A^2 + B^2} \ = \frac{ ( \ Ax_1 \  + \ By_1 \ + C  \ )^2 }{ A^2 + B^2} \ .$$

Since $ \ \sqrt{x^2} = |x| \ , $ we at last have the relation for the minimum distance,

$$D_{min} \ = \ \frac{ | \ Ax_1 \  + \ By_1 \ + C  \ | }{ \sqrt{A^2 + B^2}} \ .$$

It took me a few goes to find a reasonably clean way to develop this by algebra alone.  The straight answer to your question about how "an expert" would set this up and proceed is that they wouldn't, but instead would use a more efficient method (such as the ones I linked to in my other answer).  That may also explain why one doesn't have much luck finding an approach like this by looking around the internet.  In any event, I hope this is what you were inquiring about.


答:
A conic by five distinct points can be constructed as follows: consider the lines $d_{12}=0,d_{34}=0,d_{13}=0$ and $d_{24}=0$ and form the pencil of conics

$$(1-\lambda)d_{12}d_{34}+\lambda d_{13}d_{24}=0.$$ these are all the conics through $1,2,3,4$. Then plug the coordinates of $5$ in the equation and determine $\lambda$ to get the conic by $1,2,3,4,5$.

Now if $1$ and $2$ get closer and closer, $d_{12}$ tends to a tangent, let $t_1$, and the pencil becomes

$$(1-\lambda)t_{1}d_{34}+\lambda d_{13}d_{14}=0.$$

If $4$ also migrates to $3$, defining the tangent $t_3$,

$$(1-\lambda)t_{1}t_3+\lambda d_{13}^2=0.$$

Title: Is an ellipse determined by three points and tangent lines at two of them?

Body: I have three points, $A$, $B$, and $C$, on an ellipse; two tangent lines at $A$ and $B$ are also known. Are these enough to determine the whole ellipse?


问:
Is this mapping a contraction?
I need to show whether $f:X \rightarrow X$ is a contraction mapping where $$f(x,y)=(-y,x^3), X = \{(x,y) \in \mathbb R^2 : x+y\geq 1 \}$$

I know we have to show that $d(f(x),f(y)) \leq c.d(x,y)$ for some $0 \leq c < 1$ but this is my attempt at it: $$d((-y_1,x_1^3),(-y_2,x_2^3)) = \sqrt{(-y_1+y_2)^2+(x_1^3-x_2^3)^2}=\sqrt{(y_1-y_2)^2+[(x_1-x_2)(x_1^2+x_1x_2+x_2^2)]^2} \leq d((x_1,y_1),(x_2,y_2)).(x_1^2+x_1x_2+x_2^2)$$

which implies that $f$ is a contraction mapping if and only if $0 \leq x_1^2+x_1x_2+x_2^2 < 1$. So then would this mean that $f$ isn't actually a contraction since the above is not true $\forall (x,y) \in X$. Thank you in advance for the help.

答:
As said in my comment $f$ is not a well defined (So it is not even a function). And a contraction is a function F defined from a metric space $(M,d)$ to itself with a constant $0 \leq c <1$ such that $$d(F(x),F(y)) \leq cd(x,y)$$ for all $x,y \in M$. So if it is not a function there is no point in trying to show it is a contraction. Take a look at these links if you want more information about what it means to be well defined. What are well-defined functions?
"Well defined" function - What does it mean?

Title: Is this mapping a contraction?


问:
Solving a differential equation in distribution theory
Let $$a_n\frac{d^ny(t)}{dt^n} + a_{n-1}\frac{d^{n-1}y(t)}{dt^{n-1}} + \dots +a_0y(t) = \delta(t)$$What's the general solution for $y(t)$? I don't know how to interpret and solve this equation in the sense of distribution theory. Certainly $\delta(t)$ is not a function, so it can't be evaluated at point. I'm looking for a rigorous solution which gives general solution. Here is a specific example which is solved by md2perpe but doesn't seem can be generalized to the general case.

答:
If you take the solution $y_0$ with initial values $0=y_0(0)=y_0'(0)=...=y_0^{n-2}(0)=0$ and $y_0^{n-1}(0)=1$, then you can confirm that $y(x)=y_0(x)u(x)$, with the unit ramp function $u$, inserted into the differential operator will give a right side of
$$
(n-1)!a_n\delta(t).
$$
Now divide $y$ by the unwanted factors.


问:
Curves admitting Belyi maps are defined over $\overline{\mathbb{Q}}$.
Belyi's theorem states that a complex algebraic curve $X$ admits a model over $\overline{\mathbb{Q}}$ if and only if it admits a map to $\mathbb{P}^1$ which is ramified over at most three points.

In fact, Belyi proves that if the curve X admits such a model, then there is a map to $\mathbb{P}^1$. The converse, which was previously known, follows from the existence of the étale fundamental group.

I have to admit that I don't see why the converse is true just from knowing basic facts about the étale fundamental group, and the references I've found all approach it obliquely, with a summary more or less like what I said in the two paragraphs above.

Would someone mind writing the argument out carefully?

答:
Let $U$ be a variety over $\overline{\mathbb{Q}}$, where variety means finite type separated integral scheme over $\overline{\mathbb{Q}}$. Then the category of finite 'etale covers of $U$ is equivalent to the category of finite 'etale covers of $U_{\mathbb{C}}$. This is stated in Szamuely's book "Galois groups and fundamental groups" (but not proven). In any case, it implies the "converse of Belyi's theorem" by taking $U = \mathbb{P}^1\setminus \{0,1,\infty\}$.

The way you prove this equivalence of categories is essentially by using the "rigidity" of finite etale covers of bounded degree. Much like Richard D. James explains in the comments, the basic idea is to look at the set of isomorphism classes of $\mathrm{Aut}(\mathbb{C})$-conjugates of a given  finite etale morphism $V\to U_{\mathbb{C}}$. Any such conjugate $V^{\sigma}\to U^{\sigma}_{\mathbb{C}} = U_{\mathbb{C}}$ (with $\sigma \in\mathrm{Aut}(\mathbb{C})$) is finite etale of the same degree as $V\to U_{\mathbb{C}}$. By the finite generation of $\pi_1(U_{\mathbb{C}})$ (proven in SGA7) this  implies that the set of conjugates is finite. From this is follows (not immediately, though) that $V\to U_{\mathbb{C}}$ can be defined over $\overline{\mathbb{Q}}$.

It's the same proof as Richard D. James explained, except that I invoke a more general finite generation result for arbitrary varieties.


问:
Uniqueness of $G$-invariant complement of a $G$-invariant subspace?
Let $V$ be a finite dimensional vector space over a field $k$ of characteristic $0$. Let $G$ be a finite group and $\rho : G\to GL(V)$ be a group homomorphism (so $V$ is a linear representation of $G$). Let $W$ be a $1$-dimensional $G$-invariant subspace of $V$. Then by Maschke's theorem, we can write $V=W\oplus U$ for some $G$-invariant subspace $U$ of $V$. My question is, is the existence of $U$ unique? i.e. if $U_1, U_2$ are $G$-invariant subspaces with $V=W\oplus U_1=W \oplus U_2$, then is $U_1=U_2$ ? I can show that $U_1,U_2$ are  $G$-equivariant isomorphic, but I am not sure if they have to be exactly equal or not.

I am especially interested in the case where $G=S_n$ is the permutation group, $V=k^n$ and $\rho$ is the permutation representation i.e. $\rho(g)$ is the invertible matrix whose $ij$-th entry is $1$ when $j=g(i)$ and $0$ otherwise. And $W$ is the subspace spanned by the vector $(1,...,1)$ (all entries are $1$).

答:
Consider for example a 2-dimensional vector space with a trivial action of $G$.  Then any pair of distinct 1-dimensional subspaces gives a decomposition into irreducible $G$ subspaces, so the complement is far from unique.  This same behavior happens whenever you have multiple copies of the same irreducible representation inside $V$.

However for each irreducible representation $W$ of $G$ its isotypic component in a representation $V$ is the subspace $V_W$ spanned by all copies of $W$ in $V$. $V$ always decomposes as a direct sum of its isotypic components $V_W$ for different irreducible representations $W$, and moreover this decomposition is unique. In fact this isotypic decomposition is the common eigenspace decomposition for the action of the center of $\mathbb{C}[G]$ on $V$, and the character table tells you the eigenvalues.

In the case you are interested in for $S_n$, since $V$ decomposes as a direct sum of two non-isomorphic irreducible subspaces then indeed this decomposition is unique as each of the two factors is its own isotypic component.


问:
Let $X_1,X_2,...,X_{100}$ be a random sample from a distribution with pdf $f(x) = \cfrac{4}{3} - x^2, 0 \leq x \leq 1$
Use the central limit theorem to find an approximate probability of $P(0.4 < \bar X < 0.5)$

So I've found the mean ($\mu$) to be $\cfrac{5}{12}$

and I've found the Variance ($\sigma^2$) to be $\cfrac{17}{240}$

So standard deviation ($\sigma) $ is $ .2661$ (approximately)
So my thought process was
$P(0.4 < \bar X < 0.5)$ = $P(\cfrac{0.4-\cfrac{5}{12}}{.2661/10} < \bar X < \cfrac{0.5-\cfrac{5}{12}}{.2661/10})$  = $P(-.63 < Z < 3.13) = .7348$

But my book says that the answer should be $.9342$

What on earth am I doing wrong here? Any help would be greatly appreciated.

答:
I agree with your calculations.

As a further check, here is a simulation in R:

```<code>set.seed(2020)
cases <- 10^7
samplesize <- 100
densitytest <- function(x,t){ return(t*4/3 < 4/3 - x^2) }
dat <- runif(cases)
testdat <- runif(cases)
okdat <- dat[densitytest(dat,testdat)]
okdat <- okdat[1:(samplesize * floor(length(okdat)/samplesize))]
mean(okdat)
# 0.4166633
sd(okdat)
# 0.2661291
matdat <- matrix(okdat, ncol=samplesize)
meandat <- rowMeans(matdat)
mean(meandat)
# 0.4166633
sd(meandat)
# 0.02664449
mean(0.4 < meandat & meandat < 0.5)
# 0.7317877
</code>```

which is close to (though slightly lower than) your normal approximation

Body:
I agree with your calculations.

As a further check, here is a simulation in R:

set.seed(2020)
cases <- 10^7
samplesize <- 100
densitytest <- function(x,t){ return(t*4/3 < 4/3 - x^2) }
dat <- runif(cases)
testdat <- runif(cases)
okdat <- dat[densitytest(dat,testdat)]
okdat <- okdat[1:(samplesize * floor(length(okdat)/samplesize))]
mean(okdat)
# 0.4166633
sd(okdat)
# 0.2661291
matdat <- matrix(okdat, ncol=samplesize)
meandat <- rowMeans(matdat)
mean(meandat)
# 0.4166633
sd(meandat)
# 0.02664449
mean(0.4 < meandat & meandat < 0.5)
# 0.7317877

which is close to (though slightly lower than) your normal approximation


问:
Looking for some basic examples of finite subcovers of various compact topological spaces
I'm not sure exactly what to put here, because my question isn't about a particular space. I'm not really sure if I fully understand the concept of compactness, and I was hoping someone might know an example that would help me with this.

Here's my basic understanding so I can verify I have the right idea. Many (or all?) topological spaces have an infinite number of open coverings we could consider. We say our topological space is compact if every single one of these open coverings has a finite refinement. In that sense, for a given open covering we require that each open set in that covering has a finite refinement. I'm still unsure what that last part means, because the definition as I understand it seems extremely lax. For a given open set, how is the open set itself not a "finite subcover"?

答:
$\{(1/n,1-1/n)\mid n\in\mathbb N\}$ is an open cover of $(0,1)$ --

a collection of open sets whose union contains $(0,1)$.

However, there is no finite subcover; that is, you cannot select finitely many open sets

in this particular collection that are an open cover of $(0,1)$.  Therefore, $(0,1)$ is not compact.

On the other hand, $[0,1]$ is compact, so any open cover of $[0,1]$ has a finite subcover.

For example, the open cover $\{[0,1/n)\mid n\in\mathbb N\}\cup\{(1/n,1]\mid n\in \mathbb  N\}$ of $[0,1]$

has $\{[0,1),(0,1]\}$ as a finite subcover.


问:
Concept and decomposition of semisimple rings
I started studying the idea of semi-simple and I wanted to double-check why this composition makes sense but before that

Def: A ring with $1_R$ is said to be semi-simple if the free left $R$-module underlying $R$ is a (direct) sum of simple $R$- modules.

The decomposition of a semi-simple ring is given as follows

$$A=\bigoplus_{i=1}^n\bigoplus_{j=1}^{k_i} L_{i,j}$$ where each $L_i$ is a simple left ideal and $A$ is any ring.

My Uncertainties:

To start of, I wanted to make sure my understanding on a semi-simple ring is correct. Would I be correct in thinking that a semi-simple ring $A$ is basically a semi-simple module where the ring $A$ is viewed as a $A$-module? Then adopting the ideas from modules, this is basically saying any semi-simple ring $A$ can be decomposed into simple $A$-submodules, which in this context, are just simple left ideals and so the composition makes sense?

A Question:

Why does the definition that I saw online stressed about the left $R$-module being free? Would the existence of the identity element automatically guarantees the module to be free because we can multiply everything by $1_R$ to get everything back?

Is my understanding correct? Please let me know and many thanks in advance!

答:
Would I be correct in thinking that a semi-simple ring  is basically a semi-simple module where the ring  is viewed as a -module?

To say that $_AA$ is a semisimple module, or that $A_A$ is a semsimple module, is equivalent to saying that $R$ is a semisimple (Artinian) ring, yes.

Why does the definition that I saw online stressed about the left -module being free?

Possibly it is just a redundancy to make sure the reader knows they are talking about the regular module $_RR$.  The freeness is immaterial to the definition.

Would the existence of the identity element automatically guarantees the module to be free because we can multiply everything by $1_$ to get everything back?

It is extremely uncommon (and unproductive, at the beginning) to consider the theory of semisimple rings and modules without identity.  My guess is your book wants you to assume identity.


问:
Cauchy's integral formula computation
I want to compute
$$
\int_C \frac{|z|e^{z}}{z^{2}} \,dz,
$$
where $C$ is the circumference of the circle of radius $r$ around the origin.

I don't know how to proceed with this exercise because of the lack analyticity of the function in the numerator near $C$. What can I do?

Thanks in advance.

答:
On the path of integration, we have $|z|=r$, therefore
$$\int_C \frac{|z|e^z}{z^2}=r\int_C \frac{e^z}{z^2}.$$

Using Cauchy's Integral Formula for derivatives we obtain
$$r\int_C \frac{e^z}{z^2}=2\pi ir.$$

Title: Cauchy's integral formula computation


答: We can prove by induction $A_{2n}=A_{2n+1}=(-4)^n$, i.e. $A_n=(-4)^{\left\lfloor\tfrac{n}{2}\right\rfloor}$.

Title: Solve the recurrence relation given by: $A_{n+2} + 4A_n = 0; A_0 = A_1 = 1$

Body: I am having trouble coming up with the general solution. It seems to be given by: x2 + 4, but if this is the case I don't see how the problem can be solved with real numbers. Thanks for any advice or guidance in advance.


问:
Finding $\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+ \sin(x))} \right)$
I am trying to evaluate the following limit:
$$L=\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))} \right)$$

Begin by rewriting the limit as:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\sin(x)-3\sinh(x)+2x}{x^2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{1}$$
Applying L'Hospital's Rule to the numerator only:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\cos(x)-3\cosh(x)+2}{2x} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{2}$$
The numerator is still in an indeterminate form, applying L'Hopital to the numerator again:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{-\sin(x)-3\sinh(x)}{2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{3}$$

Rewriting as a single limit:
$$L=-\frac{1}{2}\lim_{x \to 0}\frac{\sin(x)+3\sinh(x)}{\tanh(2x)+\sin(x)} \tag{4}$$
And applying  L'Hospital's Rule...
$$L=-\frac{1}{2}\lim_{x \to 0}\left(\frac{\cos(x)+3\cosh(x)}{2\operatorname{sech}^2(2x)+\cos(x)} \right)=-\frac{2}{3} \tag{5}$$
But according to Wolfram Alpha, $L=-\frac{2}{9}$

So something must be wrong in my calculation (I guess it's the limit of a product bit)?

答:
We can't separate the limit in this way and then apply l'Hospital's rule only to a single part.

In this case we can proceed as follows

$$\frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))}= \frac{x}{\tanh(2x)+\sin(x)} \frac{\sin(x)-3\sinh(x)+2x}{x^3}$$

and use standard limit for this one

$$ \frac{x}{\tanh(2x)+\sin(x)} =  \frac{1}{2\frac{\tanh(2x)}{2x}+\frac{\sin(x)}{x}} $$

and then apply l'Hospital's rule for the second part to obtain the result according to the product rule

$$\lim_{x\to x_0} f(x)g(x)=\lim_{x\to x_0} f(x)\cdot \lim_{x\to x_0} g(x)$$


问:
Find the probability of the event that we run out of white balls first
I have $a$ white balls, $b$ black balls, $c$ red balls in a basket.
We take balls one by one and do not return.
I have to find the probability of the event that we run out of white balls first.
If I define:

$A_1$ - The event that we run out of white balls first.

$B_2$ - The event that we run out of black balls second.

$B_3$ - The event that we run out of black balls third.

$C_2$ - The event that we run out of red balls second.

$C_3$ - The event that we run out of red balls third.

I get that:

$P(A_1)=P(B_2 \cap C_3) + P(C_2\cap B_3)$

$P(B_2 \cap C_3) = P(B_2|C_3) * P(C_3)$

$P(C_2\cap B_3) = P(C_2|B_3) * P(B_3)$

I also found out that:

$P(C_3)=\frac{c}{a+b+c}$

$P(B_3)=\frac{b}{a+b+c}$

I need help finding $P(B_2|C_3)$ and $ P(C_2|B_3$

Thanks in advance!

答:
I think you're on the right track.  Your calculations for $\Pr(B_3)$ and $\Pr(C_3)$ are correct.  I assume you did it something like this: line the balls up in the order they wee drawn.  What is the probability that the last ball in the sequence is black (or red)?

The same idea will work to compute $\Pr(B_2)$ and $\Pr(C_2)$.  If we run out of black balls second, then either,

The last ball in the sequence is red and the last ball among the white and black balls is black, or

The last ball in the sequence is white and the last ball among the red and black balls is black.

Can you finish it now?


答:
$$\sinh x=x+\frac{x^3}{6}+O\left(x^4\right)$$
$$\tanh(2x)=2 x-\frac{8 x^3}{3}+O\left(x^4\right)$$
$$\sin x=x-\frac{x^3}{6}+O\left(x^4\right)$$
The limit can be rewritten as
$$\frac{x-\frac{x^3}{6}-3 \left(\frac{x^3}{6}+x\right)+2 x}{x^2 \left(2x-\frac{8 x^3}{3}+x-\frac{x^3}{6}\right)}\to -\frac{2}{9}\text{ as }x\to 0$$

Title: Finding $\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+ \sin(x))} \right)$

Body:
I am trying to evaluate the following limit:
$$L=\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))} \right)$$

Begin by rewriting the limit as:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\sin(x)-3\sinh(x)+2x}{x^2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{1}$$
Applying L'Hospital's Rule to the numerator only:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\cos(x)-3\cosh(x)+2}{2x} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{2}$$
The numerator is still in an indeterminate form, applying L'Hopital to the numerator again:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{-\sin(x)-3\sinh(x)}{2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{3}$$

Rewriting as a single limit:
$$L=-\frac{1}{2}\lim_{x \to 0}\frac{\sin(x)+3\sinh(x)}{\tanh(2x)+\sin(x)} \tag{4}$$
And applying  L'Hospital's Rule...
$$L=-\frac{1}{2}\lim_{x \to 0}\left(\frac{\cos(x)+3\cosh(x)}{2\operatorname{sech}^2(2x)+\cos(x)} \right)=-\frac{2}{3} \tag{5}$$
But according to Wolfram Alpha, $L=-\frac{2}{9}$

So something must be wrong in my calculation (I guess it's the limit of a product bit)?


问:
Solve the recurrence relation given by: $A_{n+2} + 4A_n = 0; A_0 = A_1 = 1$
I am having trouble coming up with the general solution. It seems to be given by: x2 + 4, but if this is the case I don't see how the problem can be solved with real numbers. Thanks for any advice or guidance in advance.

答:
Following your approach I think you got it backwards. Notice that the polynomial is $4x^2+x+1,$ you have to take care of the initial conditions so you get that if $A=\sum _{n=0}^{\infty}A_nx^n,$ then
$$A=1+x-4x^2A,$$
so $$A=\frac{1+x}{1+4x^2}=(1+x)\cdot\sum _{n=0}^{\infty}(-4)^nx^{2n}=\cdot\sum _{n=0}^{\infty}(-4)^nx^{2n}+\cdot\sum _{n=0}^{\infty}(-4^n)x^{2n+1}$$
so gives you that $A_{2n}=A_{2n+1}=(-4)^n.$


问:
Why is the following statement involving predicates false?

I dont really see why the answer should be false. The first part of the statement is that 'for all $x$ all $y$..'. Then if $x$ is not equal to $y$, either $P(x,y)$ is true or $P(y,x)$ is true. This seems to be true to me. If $x$ is $1$, the statement is true for $y = 1,~2,~3$ and the same for $x=2,~3$. So where am I going wrong here? Thanks!

答: You are correct.  The formula evaluates to true.</p>


问:
$x^2$ with removable discontinuities has bounded variation
Define
$$f(x) = \begin{cases}0 & \text{ if } x=1/n \text{ for some } n\in\mathbb{N}
\\x^2 & \text{ else}\end{cases}$$
on $[0, 1].$ I want to show that this has bounded variation. It is straightforward to show that $x^2$ has bounded variation on $[0,1]$, but the discontinuities at $1/n$ cause problems. I was thinking of splitting up the variation over the interval $[0,1]$ into the variation over each subinterval $[1/n, 1/(n+1)]$ as follows:
$$V_{[0,1]}(f) = \sum_{n=1}^\infty V_{[1/n, 1/(n+1)]}(f) = 2\sum_{n=1}^\infty \frac{1}{n^2} <\infty.$$
However, I'm stuck in proving the first equality. In fact, I'm not even completely sure that it is true. Any help would be appreciated.

答:
According to wiki:

The total variation of a real-valued (or more generally complex-valued) function $f$, defined on an interval $[0, 1] \subset \mathbb {R}$ is the quantity
$$ V_{[0,1]} (f) = \sup_P \sum_{i=0}^{n_P - 1} |f(x_{i+1}) - f(x_i)|,$$
where the supremum runs over the set of all partitions $P$ of the given interval.

Every partition of $[0,1]$ is a subset of the union of some partitions of your intervals, namely if $P = \{0 = x_0, x_1, \dots, x_n = 1\}$ then $x_1 \ge \tfrac{1}{m}$ for some $m$, and you can add points $\tfrac{1}{m}, \tfrac{1}{m-1}, \dots, \tfrac{1}{2}$ to $P$, and then take $x_i$ into the corresponding interval, showing that $$V(P) \le \sum_{n = 1}^{m_P} V_{[1/n,1/(n+1)]}(f).$$

Taking $\sup_P$ on the left corresponds to taking $\sup_{m_P}$ on the right, which is the same as changing the finite sum to the series because the total variation is nonnegative. Therefore, you get $$V_{[0,1]}f(V) \le \sum_{n = 1}^\infty V_{[1/n,1/(n+1)]}(f).$$

The reversed inequality follows from considering a sequence of partitions $(P_n)$ with $m_P \to \infty$, one possible example being $$P_n = \left\{ 0, \tfrac{1}{n}, (\tfrac{1}{n}+\tfrac{1}{n-1})/2, \tfrac{1}{n-1}, (\tfrac{1}{n-1}+\tfrac{1}{n-2})/2, \dots, 1 \right\}.$$

It appears to me that a similar approach can be used to prove the general statement $$V_{[a,b)} = \sum_{i = 1}^\infty V_{[a_i, b_i)}(f), \quad \bigsqcup_{i = 1}^\infty [a_i, b_i) = [a, b),$$ but I don't know whether this result has a name and whether or not it holds in other measure spaces.


问:
How many people at a party if everyone knows 22 other people and any two who don't know each other have exactly 6 mutual friends?

You are in a strange party where everyone knows exactly 22 other person and any two who don’t know each other have exactly 6 mutual friends. How many are there in the party?

Source 40- Strange Party

My research:

I know that the adjacency matrix $M$ is symmetrical and if $M[i,j] = 0$ then i-th and j-th rows have 6 times 1 at the same index position. I am thinking about encoding problem into SAT formulae and trying to brute force SAT solver with different number of literals (elements of upper triangle matrix of adjacency matrix).

I have read some articles about Moore graphs and it seems that the upper limit for the vertices is $22^2 + 1 = 485$.

Do you have any hints to the solution?

答:
A partial answer: the possible numbers of vertices are $23, 40, 100$, and some unknown subset of $\{51,52, \dots, 99\}$.

For a vertex to have degree $22$, we need at least $23$ vertices. This is possible: take the clique $K_{23}$. The second condition is satisfied trivially.

If we want the second condition to be satisfied nontrivially, then there must be non-adjacent vertices $v$ and $w$. This requires $6$ more vertices in $N(v) \cap N(w)$, $16$ more in $N(v) \setminus N(w)$, and $16$ more in $N(w) \setminus N(v)$, for at least $40$ total.

$40$ vertices is possible: consider the graph with vertex set $\{v_1, \dots, v_{20}, w_1, \dots, w_{20}\}$, with a complete graph on the first $20$ and the last $20$ vertices. Additionally, add edges $v_i w_i$, $v_i w_{i+1}$, and $v_i w_{i-1}$ (with arithmetic modulo $20$). Two non-adjacent vertices $v_i$ and $w_j$ have $\{w_{i-1}, w_i, w_{i+1}, v_{j-1}, v_j, v_{j+1}\}$ as common neighbors.

If there are more than $40$ vertices, then in the lower bound above there must be at least one vertex $u$ not adjacent to $v$ or $w$. Let's say that $|N(u) \cap N(v) \cap N(w)| = k$. Then there are $6-k$ vertices in each pairwise intersection of $N(u), N(v), N(w)$ leaving $22 - k - 2(6-k) = k+10$ vertices adjacent to only one of $u,v,w$. That's a total of at least $51$ vertices.

I don't know if this is achievable, but there's an upper bound. On the high end, the Higman–Sims graph has $100$ vertices, and is an example because it is strongly regular. Every vertex has degree $22$, every two non-adjacent vertices have $6$ common neighbors, and for a bonus every two adjacent vertices have no common neighbors. As the other answer argues, this is the largest possible number of vertices.


问:
Shear inverse transformation
Find the inverse of the shear transformation, and describe it geometically.

Matrix transformation T $\begin{bmatrix}x\\y\end{bmatrix}$=$\begin{bmatrix}x+ay\\y\end{bmatrix}$, induced by matrix A = $\begin{bmatrix}1&a\\0&1\end{bmatrix}$.

答:
It is clear that inverting:

$$\begin{cases}x'&=&x+ay\\ y'&=&y\end{cases} \ \iff \ \begin{cases}x&=&x'-ay'\\ y&=&y'\end{cases}$$

is transcripted into matrix notations:

$$\begin{bmatrix}x'\\y'\end{bmatrix}=\underbrace{\begin{bmatrix}1&a\\0&1\end{bmatrix}}_S\begin{bmatrix}x\\y\end{bmatrix} \ \iff \ \begin{bmatrix}x\\y\end{bmatrix}=\underbrace{\begin{bmatrix}1&-a\\0&1\end{bmatrix}}_{S^{-1}}\begin{bmatrix}x'\\y'\end{bmatrix}$$

$S^{-1}$ is thus obtained in a natural way...

More generally, it is interesting to see that the set of transvections with the general multiplication rule:

$$\underbrace{\begin{bmatrix}1&u\\0&1\end{bmatrix}}_{T_u}\underbrace{\begin{bmatrix}1&v\\0&1\end{bmatrix}}_{T_v}=\underbrace{\begin{bmatrix}1&u+v\\0&1\end{bmatrix}}_{T_{u+v}}$$

constitute a group isomorphic to $(\mathbb{R},+)$.

Fig. 1: The action of operators $T_u$ on a unit rectangle (colored in cyan). The neutral element corresponds to this unit rectangle. The multiplication law is illustrated here on $T_1 T_{2.5}=T_{3.5}$. Please note as well the visualization of the action of $T_{-1}$, inverse of $T_{1}$.

Title: Shear inverse transformation


问:
Calculating Lie bracket
Let $f \in C^{\infty}(\mathbb{R}^2)$. For which $a_1$ and $a_2$ does this hold $[a_1(x,y) \frac{\partial}{\partial x}+a_2(x,y) \frac{\partial}{\partial y}, f \frac{\partial}{\partial x}]$=0.

The problem for me is what to doe with the f in the lie Bracket. I found online that $$[X, fY ] = X(f)Y + f[X, Y ].$$
Here $X(f) := f_{*}(X)$. But I don't really how to use it?

答: For all $g\in C^\infty(\Bbb R^2)$,$$\begin{align}0&=[a_1\partial_x+a_2\partial_y,\,f\partial_x]g\\&=a_1(fg_x)_x+a_2(fg_x)_y-f(a_1g_x+a_2g_y)_x\\&=a_1f_xg_x+a_1fg_{xx}+a_2f_yg_x+a_2fg_{xy}-fa_{1x}g_x-fa_1g_{xx}-fa_{2x}g_y-fa_2g_{yx}\\&=(a_1f_x+a_2f_y-fa_{1x})g_x-fa_{2x}g_y,\end{align}$$so $a_1f_x+a_2f_y=a_{1x}f$ and $a_{2x}f=0$. If this works for all $f\in C^\infty(\Bbb R^2)$, $a_1=a_2=0$. I'll leave you to consider which $a_1,\,a_2$ work for a specific $f$.

Title: Calculating Lie bracket


问:
Some inequality with complex variables and a concavity of a complex function.
I am doing some project. I have to calculate the estimates of an operator. But I was stuck on a part. I need to show the following form of inequality to derive a conclusion what I want to show.

$|\sqrt{z_{1}} - \sqrt{z_{2}}| < C|\sqrt{z_{1} - z_{2}}|$ for a constant $C \in \mathbb{R}$ and $z_{1}, z_{2} \in \mathbb{C}$

Does anyone give me some comments related to this inequality? $C$ might depend on the values of $z_{1}$ and $z_{2}$.

Here, $z_{1}$ and $z_{2}$ are values of complex functions and so I am not sure how I can deal with this property. I know, for the real-valued, it is trivial because $f(x) = \sqrt{x}$ is a concave function. But, how can I define a concavity of a complex function?

I scanned Functions of One Complex Variables 1 by J. B. Conway, but I couldn't find any related theorems to this case.

Thanks.

答:
As you probably know, there is no continuous branch of $\sqrt{z}$ on the complex plane. No matter which branch we take, it will involve a cut somewhere in the plane, for example along the negative real axis. If $z_1$ and $z_2$ lie near each other on different sides of the cut, then $|\sqrt{z_1-z_2}|=\sqrt{|z_1-z_2|}$ will be small while $|\sqrt{z_1}-\sqrt{z_2}|$ will be approximately $2\sqrt{|z_1|}$. So the stated inequality cannot work for a fixed branch independent of $z_1,z_2$.

However, we can approach the problem differently: pick a branch of $\sqrt{z}$ at $z_1$ and extend it to $z_2$ in a reasonable way. To this end, let $\zeta=z_2/z_1$ and define $\sqrt{\zeta}$ to be the branch with $\sqrt{1}=1$ in $\mathbb C\setminus (-\infty,0]$. (Annoyingly, $\zeta$ could be negative; let's take $\sqrt{-1}=i$ in this case.) Then it's reasonable to set  $\sqrt{z_2}=\sqrt{z_1}\sqrt{\zeta}$

Now, the question reduces to showing  that
$$ \left |1-\sqrt{\zeta}\right|\le C\sqrt{|1-\zeta|} \tag1$$
Since you allow for $C$ here, I will not sweat too much over sharpness, although $C=1$ should work.

When $|\zeta|\le 1/2$, inequality (1) holds with $C=\sqrt{2}+1$ because we have at most $1+\sqrt{1/2}$ on the left and at least $\sqrt{1/2}$ on the right.

When $|\zeta|\ge 1/2$, the function
$$f(\zeta)=(1-\sqrt{\zeta})^2=1+\zeta-2\sqrt{\zeta}$$
satisfies $|f'(\zeta)|\le 1+1/\sqrt{|\zeta|}\le 1+\sqrt{2}$. Every point of the domain $\{|\zeta|\ge 1/2\}$ can be connected to $1$ by a curve of length at most $\pi|\zeta-1|$ (rough estimate). Integrating $f'$ along this curve, we obtain

$$|f(\zeta) |\le (1+\sqrt{2})\pi |\zeta-1|$$
and therefore (1) holds with $C=\sqrt{(1+\sqrt{2})\pi}$.


问:
Cross Multiplying and Simplifying
In a YouTube video by Michael Penn, at the 3:56 mark, he gets the equation: $$ \frac{4}{2} = \frac{2}{2-x}\Rightarrow 8-4x=4$$ Could someone explain to me how he does this and the process behind it? Thanks for your help! (Im only freshman year of high school so sorry if this question seems basic)

答:
The domain is $x\neq2$ and we obtain:
$$\frac{4}{2}\cdot2(2-x)=\frac{2}{2-x}\cdot2(2-x),$$ which gives
$$4(2-x)=4$$ or
$$8-4x=4.$$
Here $2(2-x)$ it's a common denominator of the left side and the right side of the equation.

Title: Cross Multiplying and Simplifying


问:
What is the rank of $M- \lambda U U^*$
Let $M$ be an $n×n$ Hermitian matrix of rank $k,k \neq n$.If $\lambda \neq 0$ is an eigenvalue of M with corresponding unit column vector $U$. Then prove that,
$rank(M- \lambda U U^*)=k-1$.

Now I have ,
$MU=\lambda U$, so
$M-(\lambda U) U^*=M-MUU^*=M(I-UU^*).
$
I can get only this far, after this I have tried my best to get that rank but nothing significant I have achieved.THANK YOU.

答:
Since $M$ is Hermitian, it admits an orthonormal eigenbasis $B = \{v_1, v_2, \cdots, v_n\}$. Since $M$ is of rank $k$, we know that $n - k$ of these eigenvectors have eigenvalue $0$. Let us suppose without loss of generality that $U = \vec{v}_1$. Then for any $\vec{v} \in B$ with $M \vec{v} = \vec{0}$, we have $(M - \lambda U U^{*})(\vec{v}) = M\vec{v} - \lambda \langle \vec{v}_1, \vec{v} \rangle \vec{v}_1$. But since by construction of the basis $B$ $\vec{v} \perp \vec{v}_1$, this expression is just the zero vector. Furthermore, it should be clear that $M \vec{v}_1 = \vec{0}$ as well, so we have found $n - k + 1$ linearly independent vectors in the null space of $M$.

Now we can also check that for any $\vec{v} \in B$ with $M \vec{v} \neq \vec{0}$ and $\vec{v} \neq \vec{v}_1$, we have
$$(M - \lambda \vec{v} \vec{v}^*)\vec{v} = M \vec{v} - \lambda \langle \vec{v}_1, \vec{v} \rangle \vec{v}_1 = \mu \vec{v}$$
where $\mu$ is the eigenvalue associated with $\vec{v}$. There are $k - 1$ such vectors, so indeed we have verified there are $k - 1$ linearly independent vectors in the column space of $M$.

These two facts are enough to show that $\text{rank}(M) = k - 1$. $\square$

Remark: In fact, the spectral decomposition of $M$ can also be written as
$$M = \sum_{k = 1}^n \lambda_k \vec{v}_k \vec{v}_k^*$$
where $\vec{v}_k$ is an eigenvector of $M$ with eigenvalue $\lambda_k$. Note that each term of the summation is a rank one matrix. Often times, we approximate these Hermitian matrices by truncating this sum to a fewer number of terms (usually the terms with the largest eigenvalues). The rank of this approximation is then exactly the number of terms which have non-zero eigenvalues.

In your problem, we are effectively removing the term of the summation with $\lambda_k = \lambda$ and $\vec{v}_k = U$. Since the eigenvalue corresponding to this term is nonzero, we would be reducing the rank by one, exactly as claimed.

Title: What is the rank of $M- \lambda U U^*$


问:
Power series with non-negative i.i.d. random coefficients
I have the power series

$$\sum_n e^{X_n}x^n$$

where $X_n$ are non-negative, independent, identically distributed.
I'm supposed to find the a.s. radius of convergence in the cases (1) $E[X_n]<\infty$ and (2) $E[X_n] = \infty$.

Case (1): By Cauchy-Hadamard, we know that $r^{-1} = \limsup_n e^\frac{X_n}{n} = \exp\left(\limsup_n\frac{X_n}{n}\right)$. If we assume
$$\limsup_n \frac{X_n}{n}\ne 0 \text{ a.s.}\implies X_n \text{ not bounded a.s.}$$
which is a contradiction as the expected value is finite and $X_n \ge 0$, so we have:
$$\limsup_n \frac{X_n}{n} = 0 \text{ a.s.}\implies r^{-1} = r = 1 \text{ a.s.}$$
Is this argument correct as it stands? I feel I may have missed something, as I never used independence.

Case (2): I don't know how exactly to handle this case, as the same approach as (1) has some problems, as the radius of convergence should be $0$ as far as I can tell. Again: can we use independence somehow?

答:
You used independence implicitly when you deduced that if $Y:=\limsup_{n\to\infty}X_n/n>0$ a.s. then $\operatorname{E}[X_1]=\infty $. For this assertion independence is required, just note that $\{X_n/n>0\text{ i.o. }\}=\{X_n>0\text{ i.o. }\}$, so setting $X_n:[0,1]\to \mathbb{R},
\, x\mapsto 1$ for all $n$, where the probability measure is the Lebesgue measure, then we have that $\operatorname{P} [X_n/n>0]=1$ for all $n$, so $\operatorname{P} [X_n/n>0\text{ i.o. }]=1$, however $\operatorname{E}[X_n]=1<\infty $.

But if the $X_n$ are independent then $Y$ is a tail function, so $Y$ is constant a.s. Thus asserting that $\operatorname{P} [Y>0]=1$, when independence is in play, is asserting that $Y=k$ for some $k\in(0,\infty ]$, and from here we can deduce that $\operatorname{E}[X_n]=\infty $ using the first Borel-Cantelli lemma and a formula for the expectation when the r.v. are non-negative (see below).

A way to show the result is noticing that as the $X_n$ are non-negative, then $\mathrm{E}[X_1]=\int_{[0,\infty )}\Pr [X_1\geqslant t]\mathop{}\!d t$, and so for any chosen $\epsilon >0$ we have that
$$
\begin{align*}
\epsilon \Pr [X_1\geqslant (n+1) \epsilon ]\leqslant \int_{n \epsilon }^{(n+1) \epsilon }\Pr [X_1\geqslant t] \mathop{}\!dt \leqslant \epsilon \Pr [X_1\geqslant n \epsilon ]\\
 \therefore\quad \epsilon\sum_{n\geqslant 0}\Pr [X_1\geqslant (n+1)\epsilon ]\leqslant \mathrm{E}[X_1]\leqslant \epsilon \sum_{n\geqslant 0}\Pr [X_1\geqslant n \epsilon ]
\end{align*}\tag1
$$
and
$$
\Pr [e^{X_k}\geqslant c^{-k} \text{ i.o. }]=\Pr [X_k\geqslant k\log(c^{-1})\text{ i.o. }]\tag2
$$
So we can use (1) and (2) and the Borel-Cantelli lemmas to reach the result, that is: if $\operatorname{E}[X_1]<\infty $ then the radius of convergence is one, and if the expectation diverges to infinity then it is zero.


问:
If $f$ is bounded then convolution $f*g$ is continuous
Define convolution $f$ of two functions $f_1, f_2: \mathbb R \rightarrow \mathbb R$:
$$f(x) =  \int_{-\infty}^{\infty} f_1(t) f_2(x-t)dt$$

Show that if $f_1$ is bounded, then $f$ is continuous.

I'm totally out of ideas to be honest. I only suppose that the previous problem in my assignment might have to do something with this (I'm allowed to link to this problem if needed):

Let $f_1, f_2 \in L_1(\mathbb R)$. Show that $f_1(t)f_2(x-t)$ is measurable for all $t\in\mathbb R$ and $f(x) \in L_1(\mathbb R)$

答:
Use the fact that for every $f\in L_1(\mathbb{R})$ and $t\in \mathbb{R}$, if $f_t(x)=f(x-t)$ then
$$\tag{1}||f-f_t||_1=\int_{-\infty}^{\infty}|f(x)-f(x-t)|dx\to 0$$
as $t\to 0$. (1) can be proved by using the fact that continuous functions with compact support are dense in $L_1(\mathbb{R})$. For these functions its easier to prove (1).

Now fix $\delta\in \mathbb{R}$ and $x$ and write
\begin{align}
|f(x)-f(x-\delta)|=&\biggl|\int_{-\infty}^{\infty}f_1(t)[f_2(x-t)-f_2(x-\delta-t)]dt\biggr|\\
&\leq M\int_{-\infty}^{\infty}|f_2(x-t)-f_2(x-\delta-t)|dt
\end{align}

where $M=\sup\{|f_1(t)|:t\in \mathbb{R}\}$. Now use (1) and conclude that $f$ is indeed continuous.


答:
If $X = X^i \partial_i$ and $Y = Y^j\partial_j$ in Einstein summation convention, then
\begin{align}
[X,Y] = (X^j\partial_jY^i - Y^j\partial_jX^i)\partial_i
\end{align}
Here, $X^1 = a_1$, $X^2=a_2$, $Y^1 = f$ and $Y^2=0$. Thus
\begin{align}
[X,Y] = (a_1\partial_xf + a_2\partial_yf - f\partial_x a_1)\partial_x + (-f\partial_xa_2)\partial_y
\end{align}
and this vector field is identically $0$ if and only if its components are zero, that is
\begin{align}
 a_1\partial_xf + a_2\partial_yf &= f\partial_xa_1& f\partial_ya_2&=0
\end{align}

Title: Calculating Lie bracket

Body:
Let $f \in C^{\infty}(\mathbb{R}^2)$. For which $a_1$ and $a_2$ does this hold $[a_1(x,y) \frac{\partial}{\partial x}+a_2(x,y) \frac{\partial}{\partial y}, f \frac{\partial}{\partial x}]$=0.

The problem for me is what to doe with the f in the lie Bracket. I found online that $$[X, fY ] = X(f)Y + f[X, Y ].$$
Here $X(f) := f_{*}(X)$. But I don't really how to use it?


问:
How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$?
How can we compute the following limit:

$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$

Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it.

答:
We will estimate
$$f(n) = \sum_{k=1}^n \log \left(1 + \dfrac{k}{n^2}\right)$$
Recall that $\log(1+x) = x + \mathcal{O}(x^2)$. Hence, we get that
$$f(n) = \sum_{k=1}^n \left(\dfrac{k}{n^2} + \mathcal{O}\left(\dfrac{k^2}{n^4}\right)\right) = \dfrac{n+1}{2n} + \mathcal{O}(1/n) \tag{$\star$}$$
where we made use of the fact that $\displaystyle \sum_{k=1}^n k = \dfrac{n(n+1)}2$ and $\displaystyle \sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}6$.

Now letting $n \to \infty$ in $(\star)$, we get that
$$\lim_{n \to \infty} f(n) = \dfrac12$$
The product you are interested in is $e^{f(n)}$ and since $e^x$ is continuous, we have
$$\lim_{n \to \infty} e^{f(n)} = e^{\lim_{n \to \infty} f(n)} = e^{1/2} = \sqrt{e}$$


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
I am not sure if this is the "geometric" interpretation you hope, but here is a way to see why the strong "symetry" of the expression implies that it must be $0$.

Let's denote $\phi : (\mathbb{R}^3)^4 \rightarrow \mathbb{R}$ the application defined for all $A,B,C,D \in \mathbb{R}^3$ by
$$\phi(A,B,C,D) = \overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}$$

You can see that $\phi$ is a $4-$linear form on $\mathbb{R}^3$. Moreover, you have easily
$$\phi(B,A,C,D) = \overrightarrow{AC} \cdot \overrightarrow{BD}\ +\ \overrightarrow{CB} \cdot \overrightarrow{AD}\ +\ \overrightarrow{BA} \cdot \overrightarrow{CD} = -\phi(A,B,C,D)$$

and this generalizes by saying that for every permutation $\sigma$ of the set $(A,B,C,D)$, one has $$\phi(\sigma(A),\sigma(B),\sigma(C),\sigma(D)) = \varepsilon(\sigma) \phi(A,B,C,D)$$

So $\phi$ is a $4-$linear antisymetric form on $\mathbb{R}^3$. And because $4 > 3$, the only antisymetric form on $\mathbb{R}^3$ is the null form, so $\phi \equiv 0$.


问:
Interpreting derivative questions in context

I have these two questions and I am able to approximate $S''(10)$ like this:

$\frac{S'(12)-S'(8)}{12-8}=-0.05$

I am just unsure how to interpret the meaning of $S''(x)$ in context. I know that the second derivative tells you whether your graph is concave up or concave down, which tells you whether or not the the first derivative (slope of the tangent line) is increasing or decreasing.

So would an adequate explanation be: $S''(x)$ shows the change in rate of snow deposition with respect to time?

For the second problem, I have the same question. I have $D'(t) = \frac{92}{40}e^{-t/40}$

and $D'(10)\approx 1.79$ cm/hr

Is the correct interpretation: $D'(10)$ is the rate that the snow is depositing in centimeters per hour at $t=10$ hours.

答:
In this case S"(x) represents the rate of the rate of snow piling up. For example, when we say a car is accelerating, it means that the rate at which its going forward is positive.

Note that the units for rate of rate would be units/time^2, since the units would be units/time/time.

For your second question, you are correct when you claimed that D'(10) represents the rate at which the snow is depositing in centimeters per hour.


问:
Let $k$ be a field then $kG$ k-algebra
In my course of commutative algebra, the professor told us the follow:

Let $k$ be a field and let $G$ be a group. then $kG$ is a $k$-algebra.

I don't understand this  totally. I know that a $k$-algebra is a ring with structure of module, in this sense, what is the action for $kG$?

Is $k \times kG \to kG$? or
is $k \times G \to G$ ?

My question arise because i don't understand the structure of $\mathbb{C}\mathbb{Z}_2$ as a $\mathbb{C}$-algebra. I need to know very good this structure because I want to show that $\mathbb{C}\mathbb{Z}_2$ is isomorphic to $\mathbb{C}^2$

答: A module over a field is just a vector space. Here, the scalar multiplication is simply $\lambda (\sum_{g\in G} a_gg)=\sum_{g\in G} (\lambda a_g)g$.

Title: Let $k$ be a field then $kG$ k-algebra


问:
Find the length of the loop of the given parameterized curve: $x=12t-4t^3$ and $y=12t^2$
Find the length of the loop of the given parameterized curve: $x=12t-4t^3$, $y=12t^2$

I'd like to get some views on different ways of solving this problem. Is there an easy way to know how to sketch this curve? Using a computer I saw what the graph looks like and that indeed there is a loop, so I could find the values of $t$ that parameterize that loop, but I'd like to be able to do this without computer aide.

Would eliminating the parameter be a good idea here? I couldn't find a nice way to do that.

Thank you very much for your time.

答:
If there is a "loop," it means the curve self-intersects.  So we have $s$ and $t$, $s \neq t$, with
$$12 s - 4 s^3 = 12 t - 4t^3$$
$$12s^2 = 12t^2$$

Looking at the second equation, we have $s^2 = t^2$, i.e. $s = \pm t$.  If $s \neq t$, it follows that $s = -t$.  Plugging this in, we have

$$12(-t) - 4(-t)^3  =12t - 4t^3$$
$$-12 t + 4t^3 = 12 t - 4t^3$$
$$8 t^3 - 24 t = 0$$
$$8t(t^2 - 3) = 0$$

The solutions to this equation are $t = -\sqrt{3}$, $t = 0$, and $t = \sqrt{3}$.  Interpreting these in the context of the problem, it's no surprise that when $t = 0$ we get the same result as when $t = -0$, but now we know that there is also a loop between $t = - \sqrt{3}$ and $ = \sqrt{3}$.

Of course this analysis had some parts that are special to this particular curve.  In general, if the equations were more complicated we might not be able to get a straightforward solution like this, because (for instance) we might not be able to eliminate anything, and/or we might have to solve a very high-degree polynomial.  The general problem itself is just inherently difficult, and there is a whole area of math (computational algebraic geometry) devoted to its generalizations.  So your best hope for a problem like this is to look at the exact equations you are dealing with and hope that you can find a trick that works for them in particular.


问:
Problem books in ODE
I'm studying Ordinary differential equations right now in the level of Hartman's book.

I've never seen problem books in ODE in this level even if you consider it without solutions.

I would like to know if one of you know any problem book in ODE in this level?

Thanks a lot

答:
I do not know of another book like this one.

I think you can do several things, like finding books with the main themes, for example:

Example 1: About Exponential Dichotomy of Linear Autonomous Differential Equations, Mircea Reghis

Example 2: Perturbation Methods for Differential Equations, Bhimsen Shivamoggi

Example 3: Countable Systems of Differential Equations by A. M. Samoilenko, Yu. V. Teplinskii

Another Suggestion: Review the extensive list of references in his book and see other books on the main themes that he used, including the referenced papers

Title: Problem books in ODE


问:
Vague convergence VS Weak convergence of probability measure.
Something looks strange to me (I put a screen shot of my lecture, because I want to be sure to not miss any information). So I have the following text :

And then I have an an exemple that if $X_n$ is uniformly on $[-n,n]$, then the sequence of measure $(\mu_n)$ defined by $$\mu_n(A)=\mathbb P\{X_n\in A\},\quad A\in \mathscr B(\mathbb R),$$
(i.e. $A$ a Borel set) converges vaguely but not weakly.

Question If you see Remark 3.25, it is say that weak convergence and vague convergence for probability measure is equivalent, but $(\mu_n)$ is a probability sequence that is vaguely convergent, but not weakly convergent. Did I misunderstood something or is my lecture has contradiction and is wrong ?

答:
$(\mu_n)$ converges vaguely to the measure $0$, which is obviously not a probability measure. So, there is no contradiction. What you have is :

Let $(\mu_n)$ being a sequence of probability measure and $\mu$ being a probability measure. Then, $(\mu_n)$ converges weakly to $\mu$ if and only if $(\mu_n)$ converges vaguely to $\mu$.

As your counter-example shows, if $\mu$ is not a probability measure, the converse is not true. Nevertheless, the implication is always true.


问:
Prove $v(t)$ is a Markov Process, find its' transition density and its' generator.
If $v(t)$ is the velocity of the spherical particle is $v(t) = v(0)e^{-\alpha t} + \frac{\sigma}{m} \int_0^t{e^{-\alpha(t-s)}}dw(s)$ with mass $m$, radius $r$, $\alpha = \frac{\sigma \pi r \eta}{m}$ and $\sigma = \sqrt{2m\alpha KT}$. $w(s)$ is the Wiener process. The process $V(t)$ is supposed to be the Ornstein-Ulenbek process. But, I don't see how? The definition does not match the definition of the process in Wikipedia. And I am not sure how to go about proving that it is a Markov Process.

答: There is a mistake in your formula. The upper bound of integration in the stochastic integral has to be $t$ not $1$. Then, differantiating in $t$ you can easily check that you $v_{t}$ satisfies the OU-SDE. Apart, you do also find that formula on wikipedia.


问:
Compute the spectrum of an explicit integral operator
Let $H = L^2[0,1]$. Define an operator $K \in B(H)$ by $Kf(x) := x^2 \cdot \int_{0}^{1} y f(y) \; \text{d} y$. Show that $K$ is compact and compute its spectrum.

I already showed that $ ||K || \leq \sqrt{\frac{1}{15}}$ and as K is an bounded integraloperator  with continuous kernel $k(x,y) = x^2y$ it is also compact since it can be approximated by finite rank operators. To compute the spectrum we now want to know  for which $\lambda \in \mathbb{C}$ the operator $ ( \lambda Id - K ) $ is not invertible.
So I tried to solve the equation $ \lambda f(x) - x^2 \int_0^1 y f(y) \; \text{d} y  = g(x) $ for $f(x)$ but I didn't get anything useful on the paper and I dont know if thats the right way to do it.

How can I continue from here? Thank you very much !

答:
As observed by Ruy in the comment, $Kf \in \mbox{Span}(x^2)$ for all $f \in L^2([0,1])$ and thus the unique nonzero eigenvalue $\lambda=\frac 1 4$.

In fact $K x^2=x^2 \int_0^1y^3dy=x^2 \frac 1 4$. This also prove that $K$ is compact.


问:
Marking the prime points on a circle
If you travel around a circle and mark all the points on the circle where the distance you travelled is a prime number, where you would go through many rotations*, do you end up marking the entire circumference of the circle as the number of points you pick goes to infinity?

Or do you leave  "holes" in the circle by virtue of only marking the prime points of the circumference?

*I understand that you would have to define the circumference of your circle. Define it as a finite number, prime or not. How does the circumference you pick affect the answer/whether the circumference is prime or not?

Please let me know how I can explain this question further or whether this question does not make sense.

答:
No, you'll never fill the circle. This is because:

The set of prime is countable.

But, the set of point on the circle is uncountable, i.e its cardinarlity is continuum.

It's like you can never fill the real line with rationals, because the set of rationals is countable, whereas the set of point in the real line is not.

You can have a quick look here.

Title: Marking the prime points on a circle


问:
Check series for convergence or divergence
$$\sum_{n=1}^{\infty} \frac{2+(-3)^n}{4^n} $$

I had an idea to decompose this series into the sum of two series
$$ \sum_{n=1}^{\infty} \frac{2+(-3)^n}{4^n}  = \sum_{n=1}^{\infty} \frac{2}{4^n} + \sum_{n=1}^{\infty} \frac{(-3)^n}{4^n} $$
and already consider the convergence of each of the series.
But as far as I understand, this is not quite true, because we are changing the summation order. And it is not clear what the convergence of an ordinary series and an alternating series will tell us about convergence.

The main problem arises from the fact that

$$ \left | \frac{2+(-3)^n}{4^n} \right | \neq \frac{2+3^n}{4^n} $$
that's why I can't use the Leibniz attribute and work with it any further.
Maybe I don't understand something and then everything is just solved.

答:
Your way is fine since both series converge indeed

$$ \sum_{n=1}^{\infty} \frac{2}{4^n} + \sum_{n=1}^{\infty} \frac{(-3)^n}{4^n}= 2\sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n + \sum_{n=1}^{\infty} \left(-\frac 3 4\right)^n$$

which are two geometric series.


问:
Why are these two numbers relatively prime
I am studying Linear Congruential Generators (LCG). In the proof of the main theorem of these generators:

Theorem: The sequence defined by $X_{n+1}=(aX_n+c) \;mod\;m$, has full period $m$, provided that:

$(i)$ $c$ is relatively prime to $m$

$(ii)$ $a\equiv 1 \pmod{p}$ if $p$ is a prime factor of $m$.

$(iii)$ $a\equiv 1 \pmod {4} $, if 4 is a factor of $m$

Thet sequence can be written as
$$
\frac{(a^n-1)(x_0(a-1)+c)}{a-1}\equiv 0\pmod{m}
$$
To reduce that expression to the following:
$$
\frac{(a^n-1)}{a-1}\equiv 0\pmod{m}
$$
The author says that that can be done because due to "the conditions of the theorem (I think $i$ and $ii$), $x_0(a-1)+c$ is relatively prime to $m$.

Why that's true? And why with that can the congruence be simplified?

答:
Suppose otherwise. Let $n=x_0(a-1)+c$ and $p$ prime such that $p|m$ y $p|n$. By $ii$, we have that $a\equiv 1\,(\,mod\,p)$, that is:
$$
a = 1+kp, \;k\in\mathbb{N} \Rightarrow x_0(1+kp-1)+c\;(\;mod\;p) = c \Rightarrow n\equiv c\,(\,mod\,p)
$$
But we have assumed that $p|n$, therefore $n\equiv 0\,(\,mod\,p)$, but $p\nmid c$ (by $i$), so we have a contradiction.


问:
$x^2$ with removable discontinuities has bounded variation
Define
$$f(x) = \begin{cases}0 & \text{ if } x=1/n \text{ for some } n\in\mathbb{N}
\\x^2 & \text{ else}\end{cases}$$
on $[0, 1].$ I want to show that this has bounded variation. It is straightforward to show that $x^2$ has bounded variation on $[0,1]$, but the discontinuities at $1/n$ cause problems. I was thinking of splitting up the variation over the interval $[0,1]$ into the variation over each subinterval $[1/n, 1/(n+1)]$ as follows:
$$V_{[0,1]}(f) = \sum_{n=1}^\infty V_{[1/n, 1/(n+1)]}(f) = 2\sum_{n=1}^\infty \frac{1}{n^2} <\infty.$$
However, I'm stuck in proving the first equality. In fact, I'm not even completely sure that it is true. Any help would be appreciated.

答:
The total variation of a differentiable function $f(x)$ on the interval $[a.b]$ is
$$V_b^a(f)=\int_a^b \left| f'(x)\right|  \, dx$$
The function $f(x)=x^2$ on intervals $\left(\frac{1}{n+1},\frac{1}{n}\right)$ has variation
$$\int_{\frac{1}{n+1}}^{\frac{1}{n}} 2 x \, dx=\frac{1}{n^2}-\frac{1}{(n+1)^2}$$
The sum of these variations is
$$ \sum _{n=1}^{\infty }\left( \frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=1$$
Look here for further details.

This result fits with intuition, actually. The function goes from $0$ to $1$ and the total variation is $1$


问:
(clarify this answer?) Prove that for all subsets of the reals A, there exists a Borel set $A \subset B$ such that $\lambda(B) = \lambda^*(A) $
This (or a similar) question has been asked already on stackexchange. I am asking for a variation on the accepted answer.

Prove for each $A\subset \mathbb{R}$ there exists Borel set whose measure is same as outer measure of $A$

how do we relate this to the question above, where in we must consider $\lambda(B)$ instead of $\lambda^*(B)$?

答:
If you analize the definition of $\lambda^*(A)$, you get $\lambda^*(A)=\inf \{\sum_n \lambda(E_n) \mbox{ where } (E_n)_{n \in amthbb{N}} \mbox{ is a countable covering of } A   \}$ and thus for all $n \in \mathbb{N}$ there exists $(E_k^n)_k$ such that $\sum_k \lambda (E_k^n) \leq \lambda^*(A)+\frac 1 n$.
If you take $E=\bigcap_n\bigcup_k E_k^n$ you have that $E$ is what you are searching for. In fact $\lambda^*(A) \leq \lambda(E) \leq \lambda^*(A)+\frac 1 n$ for all $n \in \mathbb{N}$.

Note that this holds in general for every $(X,\mathcal{A},\mu)$ measure space the "borel" set is replaced by $E \in \mathcal{A}$.


问:
Understanding Property of PSD Matrix
For some reason I can't wrap my mind around several concepts.

I'm reading the proofs:

The column space of a positive semi-definite matrix A is contained in the column space of A+B?

Rank of the sum of two positive semi-definite matrices

I can't seem to understand one line. Suppose $X \succeq 0$ is positive semidefinite. Let $v \neq 0$ and suppose $v^{T}Xv=0$. The second and first link say then that $v$ is then in the null space of $X$, meaning $Xv=0$

Why must this be the case?

Thanks.

答:
Any positive semidefinite matrix $X$ has a positive semidefinite square root $Y$, that is, $Y^2=X$.

 $X$ is symmetric, hence diagonalizable in an orthonormal basis, with nonnegative eigenvalues. Take the diagonal matrix with the square roots of the eigenvalues, in the same basis.

Consequently, $0=v^TXv=v^TY^2v=v^TY^TYv=\|Yv\|^2$ implies $Yv=0$ whence $Xv=Y(Yv)=0$ as well.

Title: Understanding Property of PSD Matrix


问:
Use a calculator to approximate $\int_0^{7\pi} \sqrt{5-4\cos(t)}dt$ to four decimal places
Use a calculator to approximate $\int_0^{7\pi} \sqrt{5-4\cos(t)}dt$ to four decimal places

Can somebody explain what this means? Do they want me to use a series expansion or something?? What are some ways this question could be interpreted and could I have a bit help getting through it??

Maybe I'm suppose to use Simpsons rule? I feel like if that were so they would have explicitly said so..

Thank you for your time.

答:
I think you are just supposed to use any approximation method that yields that particular accuracy, for instance Simpon's composite method. It may be beneficial if you take into consideration that
$$
\int_0^{7 \pi} \sqrt{5 - 4\cos t}dt = 7 \int_0^{\pi} \sqrt{5 - 4\cos t}dt.
$$

If you choose Simpson's method with a certain step $h$, you know that the error satisfies
$$
|E(h)|=\frac{\pi h^4 |f''''(\xi)|}{180}\leq \frac{7 \pi h^4}{80},
$$
and you can choose a value for $h$ in such a way that the answer has four correct decimal digits. This will however implicate using a few hundred points... I hope the calculator is a figure of speech and you can use a computer!


答:
As for eliminating the parameter, $x^2$ has only even powers of $t$, so its terms are equivalent to powers of $y$: $$x^2=12y-\frac{2}{3}y^2+\frac{1}{108}y^3$$

Perhaps (or perhaps not) at this point you would find it easier to sketch $$x=\sqrt{12y-\frac{2}{3}y^2+\frac{1}{108}y^3}$$ plotting $x$ as a function of $y$. And then plot its reflection as well.

Title: Find the length of the loop of the given parameterized curve: $x=12t-4t^3$ and $y=12t^2$

Body:
Find the length of the loop of the given parameterized curve: $x=12t-4t^3$, $y=12t^2$

I'd like to get some views on different ways of solving this problem. Is there an easy way to know how to sketch this curve? Using a computer I saw what the graph looks like and that indeed there is a loop, so I could find the values of $t$ that parameterize that loop, but I'd like to be able to do this without computer aide.

Would eliminating the parameter be a good idea here? I couldn't find a nice way to do that.

Thank you very much for your time.


问:
Sign of lagrange multiplier with inequality constraints
While I have read on several places that the sign of lagrange multiplier $\lambda$ is not that important I'm reading now on Patten recognition and machine learning by Bishop the following:

If we wish to minimize (instead of maximize) rather than maximize the function $f(\text{x})$ subject to an inequality constraint $g(\textbf{x})$ then we minimize the lagrangian function $L(\textbf{x},\lambda) = f(\textbf{x}) - \lambda g(\textbf{x})$ with respect to $\bf{x}$ with $\lambda \ge 0$.
It actually specifies that in this situation the sign of lagrange multiplier is crucial.

Is this sign distinction important when I consider inequality constraints, while this is not important when considering equality constraints like $g(\textbf{x}) = 0$?

Can anyone explain me this a little bit better? Thank you

答:
The sign comes from the following reasoning:

With equality constraints $g(x) = 0$, for a point $x$ to be optimal, any perturbation to $x$ that changes $f$ must also violate constraints $g$ become (no matter if $g$ becomes positive or negative, important thing is that it is no longer zero), hence the gradient of $f$ must be parallel to that of $g$. It follows that $\nabla f(x) = \lambda \nabla g(x)$, for some (potentially negative) $\lambda$.

With inequality constraints $g(x) \ge 0$:

when minimizing, for a point $x$ on the boundary $g(x) = 0$ to be optimal, the gradient $\nabla f$ must point in the same direction of the gradient of $g$; otherwise, following the antigradient of $f$ along the boundary would decrease $f$. It follows that $\nabla f(x) = \lambda \nabla g(x)$ for some positive $\lambda$, and subtracting you get $f(x) - \lambda g(x)$.

when maximizing, for a point $x$ on the boundary $g(x) = 0$ to be optimal, the gradient $\nabla f$ must point in the opposite direction of the gradient of $g$; otherwise, following the gradient of $f$ along the boundary would increase $f$. It follows that $\nabla f(x) = -\lambda \nabla g(x)$ for some positive $\lambda$, and subtracting you get $f(x) + \lambda g(x)$.

Bishop has several illustrations about this, but I don't remember the exact page. Feel free to edit if you do.

Update (example): take $f(x, y) = x$ and $g(x, y) = 1 - x^2 - y^2$. If you want to minimize $f$ on the unit disk $g(x, y) \ge 0$ but take $$L(x, y; \lambda) = f(x, y) + \lambda g(x, y) = x + \lambda (1 - x^2 - y^2)$$ and then take the derivatives then you will get $$\frac{\partial L}{\partial x} = 1 - 2 \lambda x = 0, \frac{\partial L}{\partial y} = -2 \lambda y = 0, \frac{\partial L}{\partial \lambda} = g(x, y) = 1 - x^2 - y^2 = 0.$$ It follows that $y = 0$, $x = -1$ or $x = 1$, and $\lambda = -\frac{1}{2}$ or $\lambda = \frac{1}{2}$. You would then discarded $\lambda = -\frac{1}{2}$ and (optimal solution) $x = -1$ because the corresponding $\lambda$ is negative, and choose $x = 1$ which is the worst possible value (it maximizes $f$ instead of minimizing).

So yeah, the sign is important because you want $\lambda \ge 0$. Afaik, in the equality case it is not demanded, so it does not matter which sign do you use.


问:
Prove that a subgroup which contains half of all elements is a normal subgroup.
I have been going through some material on elementary group theory (Group Theory for Physicists, Christoph Ludeling) and in one place it states:

A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.

Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$.
How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.

答:
Not as slick but perhaps more direct (as given in Rotman):

It suffices to prove that if $h \in H$, then the conjugate
   $ghg^{-1}\in H$ for every $g \in G$.  If $g \in H$, then $ghg^{-1}\in
 H$, because $H$ is a subgroup.  If $g \notin H$, then $g = ah_0$,
   where $h_0\in H$ (for $G = H \cup aH$).  If $ghg^{-1} \in H$, we are
   done.  Otherwise, $ghg^{-1} = ah_1$ for some $h_1\in H$.  But $ah_1 =
 ghg^{-1} = ah_0hh_0^{-1}a^{-1}$.  Cancel $a$ to obtain $h_1 =
 h_0hh_0^{-1}a^{-1}$, contradicting $a \notin H$.

This was the proof I came up with before reading the slick proof in the text I was reading, seeing it appear in Rotman later made me feel less silly.


问:
Simple algebra issue with regards to finding the correct $Z$ value when using the Central Limit Theorem
Suppose that the number of traffic accidents, $N$, in a given period of time is distributed as a Poisson random variable with $E(N) = 100$. Use the normal approximation to the Poisson to find $\Delta$ such that $P(100-\Delta < N < 100 + \Delta) \approx 0.9$.

I understand the procedure on how to figure out this problem, but for some reason the way I went about figuring it out clashes with the final solution:

Attempt

Given that we have to approximate it by the normal we will be standardizing using the theory of the Central Limit Theorem. After the algebra I arrive at:

$$P(\frac{-\Delta}{10} < Z < \frac{\Delta}{10}) \approx 0.9$$

So I proceeded to treat things as follows:

$$P(\frac{-\Delta}{10} < Z < \frac{\Delta}{10})\\
\Rightarrow\  P(Z< \frac{\Delta}{10}) - P(\frac{-\Delta}{10} < Z) \\
\Rightarrow \ (1 - P(Z \leq \frac{-\Delta}{10})) \\
\Rightarrow \ 2P(Z < \frac{\Delta}{10}) - 1\ \text{(used the symmetry of the normal distribution)} \\ \Rightarrow\ 2P(Z < \frac{\Delta}{10}) - 1 \approx 0.9 \\ \Rightarrow P(Z < \frac{\Delta}{10}) = \frac{1.9}{2} = 0.2375$$

So taking the compliment of this value $(1 - 0.2375 = 0.7625)$ and using the tables of standard normal $Z$ values I arrive at an expression of:

$$0.72 = \frac{\Delta}{10} \\ 7.2 = \Delta$$

But this isn't what the solution arrived at. The solution did the following:

Which algebraically is the same thing, they instead took $\frac{-\Delta}{10}$ and worked with that while I worked with the positive side. So in theory the same result should be achieved. So what gives with what I did?

答:
I discovered the mistake I made with regards to interpreting the question. It had to do with my interpretation of the symmetry of the normal distribution.

I was thinking that:

$$P(Z < \frac{\Delta}{10}) = P(Z < \frac{-\Delta}{10})$$

By the symmetry of the normal distribution. The only thing that is similar here is how they are written syntactically. This statement is not true. On the other hand

$$P(Z > \frac{\Delta}{10}) = P(Z < \frac{-\Delta}{10})$$

This uses the symmetry of the normal distribution. A quick look at a picture of a normal distribution will confirm this heuristically.

And as a result:

$$P(Z < \frac{\Delta}{10}) - P(Z < \frac{-\Delta}{10})  \\ \Rightarrow\ 1 - P(Z > \frac{\Delta}{10}) - P(Z < \frac{-\Delta}{10}) \\ \text{(now using the symmetry of the normal distribution)}\\ \Rightarrow 1 - P(Z < \frac{-\Delta}{10}) - P(Z < \frac{-\Delta}{10}) \\ \Rightarrow \ 1 - 2P(Z < \frac{-\Delta}{10})$$

And the rest of the algebra will follow easily.


问:
A problem of optimization
Determine the maximum of $f(x_1,x_2,x_3,\dots,x_n)=\displaystyle\prod_{i \ne j}|x_i-x_{j}|$ for $0\le x_1<x_2<x_3< \dots<x_n<1 $.

It seems that you can always translate $x_i$'s such that $x_n \to 1$.
At the same time one always maximizes the value of $f$ when stretching on the right $x_i$'s, ie when $x_1=0$
So we just need to consider $n-2$ values.
Using dumb calculations (derivatives...), we get that:

for $n=2$, $\max=1$

for $n=3$, $\max=\frac 1 4$

for $n=4$, $\max=\frac{\sqrt{5}}{125}$...

Is there a way to find it in general ?

答:
Using Arithmetic Mean as an upper bound, using a simpler case

Using $x_i$'s with
$$0\le x_1<x_2<x_3< \dots<x_n\leq1$$
let's look instead at the max value of a simpler product
$$g(x_1,x_2,x_3,\dots,x_n) = \prod (x_{i+1} - x_i)$$
We can see that $g$ is the product of the lengths of the line segments formed between all the $x_i$'s:
$$\prod (x_{i+1} - x_i) = (x_n - x_{n-1})(x_{n-1} - x_{n-2})\dots(x_2 - x_1)$$
Furthermore, because $x_i < x_{i+1}$, we know each term is positive.

To simplify things, let's call each of these terms $a_i$ where

$$a_i = x_{i+1} - x_i$$

We want to know the max value of
$$a_1a_2\dots a_{n-1}$$

For positive numbers we know that the Geometric Mean (GM) is less than or equal to the Arithmetic Mean (AM).

Looking at the line we can see that, as long as we keep $x_1 = 0$, $x_n = 1$, and obey the general other constraints, the sum of all the segment lengths will remain maxed out at $1$, regardless of our placement of the interior $x_i$'s.

$$AM = \frac{\sum a_i}{n-1} = \frac {1}{n-1}$$

Moving around the $x_i$'s (aka increasing and decreasing the interior segment lengths) will change the value of the product of all the $a_i$'s, and consequently will increase or decrease the value of the GM.

GM will reach its maximum value when it's equal to the AM.

This occurs when

$$GM = \sqrt[n-1]{\prod a_i} = \frac {1}{n-1}$$
$$\prod a_i = \frac {1}{(n-1)^{n-1}}$$

$$g_{\max} = \frac {1}{(n-1)^{n-1}}$$

You can see that this happens when the $x_i$'s are evenly spaced and the intervals all have length $\frac {1}{n-1}$.

More complicated product

I don't know if this approach might be helpful with the nastier product $\prod_{i \ne j}|x_i-x_{j}|$, but it might be worth looking into. If the sum of the lengths of all the intervals in this product is similarly constant regardless of moving around interior $x_i$'s, you might be able to do the same trick.

I'm sorry if this isn't helpful!

Title: A problem of optimization