问:
Understand representations of c*-algebras from a categorical point of view
In my lecture on von Neumann algebras we have defined a representation of a c*-algebra $\mathcal{A}$ as a *-homomorphism $\pi$ into $\mathcal{B}(\mathcal{H})$ for some Hilbert space $\mathcal{H}$. Then we have defined subrepresentations as the restriction of $\pi$ to invariant subspaces, i.e.: If $\mathcal{K}$ is a closed subspace of $\mathcal{H}$ such that for every $a\in\mathcal{A}$ we have $\pi(a)(\mathcal{K}) \subseteq \mathcal{K}$, then $$\pi|_\mathcal{K}: \mathcal{A}\to\mathcal{B}(\mathcal{K}),  a\mapsto \pi(a)|_\mathcal{K}$$ is a subrepresentation of $\pi$ on $\mathcal{K}$.

My question is: How can we understand this from a categorical point of view. First I thought that just a morphism is a representation of one object of a category as another. However in this generality it wouldn't be obvious to extend the term of a subrepresentation to this larger context.

I did some research and figured out that the aim of a representation is indeed to represent a structure as linear maps on a vector space, in this case as a subalgebra of $\mathcal{B}(\mathcal{H})$, in the case of groups as a subgroup of $\mathrm{Aut}(X)$ for some vector space $X$.

In the case of groups I have found two ways to obtain representations: First via a group action $\varphi: G \times X \to X$, then $\pi(g)=\varphi(g,\cdot)$ is a representation. Secondly if one considers a group as a category with one object $*$, then $G=\mathrm{Mor}(*)$ and a functor from this one object category into $\mathrm{Vect}$ yields a representation.

However, both constructions didn't enlighten me in such a way that I would see how the c*-algebra case fits into this construction or how one could express the idea of representations in a larger class of categories uniformly.

Disclaimer: I have no deep knowledge in category theory. I am totally convinced that thinking in categories is essential to really understand mathematics, otherwise I wouldn't rise such questions, nevertheless it would be nice to get an answer on a level that I am able to understand ;)

Kind regards,
Sebastian

答:
Not sure if this is what you are looking for, but if $A$ is a $C^*$-algebra, then we can define the category $\textbf{rep}(A)$ of the representations of $A$. More precisely:

The objects of $\textbf{rep}(A)$ are pairs $(H,\pi)$ where $H$ is a Hilbert space and $\pi:A\to\mathcal{B}(H)$ is a $*$-homomorphism.

The morphisms between two representations are the bounded operators that intertwine the images of the representations, i.e.
$$\text{Hom}_{\textbf{rep}(A)}((H,\pi), (K,\rho)):=\{u\in\mathcal{B}(H,K): u\circ\pi(a)=\rho(a)\circ u \text{ for all }a\in A\} $$

One can easily verify that $\text{rep}(A)$ is a pre-abelian category, i.e. we have kernel representations and co-kernel representations. More precisely, if $(H,\pi)\xrightarrow{u}(K,\rho)$ is a morphism in $\textbf{rep}(A)$, then the pair $(\ker(u), \pi(\cdot)\vert_{\ker(u)})$ is a representation of $A$ and the inclusion operator $i:\ker(u)\to H$ is a morphism in $\textbf{rep}(A)$. This is the kernel of $u$, i.e.
$$\ker\big((H,\pi)\xrightarrow{u}(K,\rho)\big)=(\ker(u),\pi(\cdot)\vert_{\ker(u)})\xrightarrow{i}(H,\pi).$$
Likewise, one shows that we have cokernels: the pair $(u(H)^\bot,\rho(\cdot)\vert_{u(H)^\bot})$ is a representation of $A$ and the projection operator $p:K\to u(H)^\bot$ is a morphism in $\textbf{rep}(A)$, so
$$\text{cokernel}\big((H,\pi)\xrightarrow{u}(K,\rho)\big)=(K,\rho)\xrightarrow{p}(u(H)^\bot,\rho(\cdot)\vert_{u(H)^\bot}).$$
The category is not really abelian and the explanation is the same as why the category $\textbf{Hilb}$ is not abelian, but c'est la vie!

We can say a few more things: irreducible representations are identified with the objects $(H,\pi)$ that are simple, in the sense that the Hom-sets $\text{Hom}_{\textbf{rep}(A)}(-, (H,\pi))$ do not contain any non-trivial monomorphisms, i.e. injective intertwining operators.

When we say that two representations are "unitarily equivalent" we simply mean that they are isomorphic in the category $\textbf{rep}(A)$.

There are probably a few more things that we can say, for example about what the role of cyclic representations is, about the universal representation and so on, but I think the above cover the main idea.

EDIT: As noted in the comments by QuantumSpace, the fact that isomorphism in this category is exactly unitary equivalence is not obvious. I include a proof below, the idea of which was communicated to me by QuantumSpace (thank you).

Let $(H,\pi), (K,\rho)\in\textbf{rep}(A)$ with $(H,\pi)\cong(K,\rho)$, so there exists a bijective $s\in B(H,K)$ (i.e. an invertible operator) such that $s\pi(a)=\rho(a)s$ for all $a\in A$.

We then do polar decomposition for $s$, say $s=u|s|$ where $|s|:=(s^*s)^{1/2}\in B(H)$ and $u\in B(H,K)$ is a partial isometry. Now by invertibility of $s$ we have that $|s|$ is also invertible in $B(H)$ and thus $u=s|s|^{-1}$. One can easily show that $u$ is a unitary: it is obviously surjective and
$$\langle u\xi,u\eta\rangle_K=\langle s^*s(s^*s)^{-1/2}\xi,(s^*s)^{-1/2}\eta\rangle_H=\langle|s|\xi,|s|^{-1}\eta\rangle_H=\langle\xi,|s||s|^{-1}\eta\rangle_H=\langle\xi,\eta\rangle_H,$$
where we used that $|s|$ is self-adjoint.

Now observe that $s^*s\pi(a)=s^*\rho(a)s=(\rho(a^*)s)^*s=(s\pi(a^*))^*s=\pi(a)s^*s$, so $s^*s$ commutes with $\pi(A)$, and hence so does $|s|$ and thus so does $|s|^{-1}$. Therefore
$$u\pi(a)=s|s|^{-1}\pi(a)=s\pi(a)|s|^{-1}=\rho(a)s|s|^{-1}=\rho(a)u$$
and thus $u$ is a unitary intertwining operator, i.e. $(H,\pi)$ and $(K,\rho)$ are unitarily equivalent.


问:
Prove converse Thales theorem, proportional sides imply parallel lines
I'm going through John Stillwell's Four Pillar's of Geometry and trying to follow the book's structure when doing the exercises.  Generally, a 'pillar' is divided into two chapters; the first chapter states useful results and motivation while the second chapter brings the machinery to do the proofs.

Thales' Theorem: Suppose that the line EF is parallel to BC. Then AE/EB=AF/FC.

Question

Prove Converse Thales' Theorem: Suppose that EG is not parallel to BC. Then AE/EB not equal to AG/GC.  Equivalently, AE/EB=AG/GC is sufficient for EG parallel BC.

The issue is that I don't think I'm 'allowed' to use much.  The ideas of similar triangles, SAS, etc. have not been developed.  I know that without having the book in front of you it's hard to judge what can be used but essentially I'm looking for the dirt simplest proof.  From what I can see the only known facts are Thales theorem and that was only stated.

If clarification is needed I'll be glad to answer in the comments.

答:
Proof of the converse of "Thales Theorem":

First, let the line $d$ intersect the sides $AB$ and $AC$ of $\triangle ABC$ at distinct points E and G, respectively, such that $\displaystyle \frac{AE}{EB} = \displaystyle \frac{AG}{GC}.$  We now need to prove that $EG \parallel BC$.

Assume $EG \nparallel BC$.  Then there must be another line intersecting point $E$ of side $AB$ as well as some point, say $F$, of side $AC$ that is parallel to $BC$.  So, let $EF \parallel BC$.

By Thales Theorem, since $EF \parallel BC$, it follows that:
$$\frac {AE}{EB} = \frac{AF}{FC}\quad\quad (1)$$
But we are given
$$\frac{AE}{EB} = \frac{AG}{GC}\quad\quad (2)$$
Hence, from (1) and (2), it must follow that $$\frac{AF}{FC} = \frac{AG}{GC}\quad\quad (3)$$
Adding "1" to both sides of equation (3) gives us: $$ \frac {AF}{FC} + \frac{FC}{FC} = \frac {AG}{GC} + \frac{GC}{GC}$$ which simplifies to $\displaystyle \frac{AF+FC}{FC} = \frac{AG+GC}{GC} \implies \frac{AC}{FC} = \frac{AC}{GC} \implies FC = GC$

But $FC = GC$ is only possible when points $F$ and $G$ coincide with one another, i.e. if $EF$ is the line $d = EG$ itself.

But $EF\parallel BC$, and hence it cannot be the case that $EG = EF \nparallel BC$.

Therefore, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side: i.e. the converse of Thales Theorem has been established.

ContentLicense: CC BY-SA 3.0


问:
Classification of nonzero prime ideals of $\mathbb{Z}[i]$
I know the classification of Gaussian primes: let $u$ be a unit of $\mathbb{Z}[i]$. Then the following are all Gaussian primes:

1) $u(1+i)$

2) $u(a+ib)$ where $a^2+b^2=p$ for some prime number p congruent to $1$ modulo $4$

3) $uq$ where $q$ is a prime number congruent to $3$ modulo $4$.

If i knew that $\mathbb{Z}[i]$ is a Principal ideal domain, i could say: a nonzero ideal $P$ of $\mathbb{Z}[i]$ is prime iff it is principal prime, say $P=p\mathbb{Z}[i]$ iff $p$ is a Gaussian prime, and conclude that every nonzero prime ideal $P$ of $\mathbb{Z}[i]$ belongs to one of the following families:

1) $P=(1+i)\mathbb{Z}[i]$

2) $P=(a+ib)\mathbb{Z}[i]$, where $a,b\in\mathbb{Z}$ and $a^2+b^2=p$ is an odd prime congruent to $1$ modulo $4$

3) $P=q\mathbb{Z}[i]$ where $q$ is an odd prime congruent to $3$ modulo $4$.

Now, suppose that for some reason i don't know that $\mathbb{Z}[i]$ is a PID. Can i still prove the classification above? I can't say an ideal is prime iff its generator is prime, so, how could i proceed in this case?

答:
To prove this result, you must prove along the way that $\mathbb{Z}[i]$ is a PID, to guarantee that you have not missed any non-principal ideals.  One cheap way to get around your restriction is to implicitly use the implication
$$\text{Euclidean domain} \Longrightarrow \text{PID}.$$
In the case at hand, we obtain a Euclidean algorithm on $\mathbb{Z}[i]$ as follows.  Given non-zero $x,y \in \mathbb{Z}[i]$, let $z \in \mathbb{Z}[i]$ denote the lattice point nearest $x/y$.  Geometrically, it's clear that
$$\vert z - x/y \vert \leq 1/\sqrt{2},$$ in which the metric above is the standard metric on the plane.  Then $x-yz \in \mathbb{Z}[i]$ satisfies $\vert z-xy \vert < \vert y \vert$, so that $\vert \cdot \vert$ defines a Euclidean function on $\mathbb{Z}[i]$.  As in the case with the integers, a Euclidean algorithm can be used to find GCD (up to units).

Now, suppose that $\mathfrak{p}=(a,b) \subset \mathbb{Z}[i]$ is a prime ideal.  If $c$ denotes any GCD of $a$ and $b$, we have $c=(a,b)$.  Induction (and the fact that integer rings are Noetherian) implies that all ideals are principal.  For your problem, it would then suffice to classify all the principal ideals in $\mathbb{Z}[i]$, which is standard practice (look at norms).

Note: while this approach seems contrary to the question asked, I nevertheless believe that the best way to classify the prime ideals in $\mathbb{Z}[i]$ is to first prove that such ideals must be principal.  Other techniques would simply reprove this result, without the benefit of machinery.

Edit: This edit includes a proof of the fact that $\mathbb{Z}[i]$ is Noetherian.  Let $$\mathfrak{p}_1 \subsetneq \mathfrak{p}_2 \subsetneq \ldots $$
be an infinite chain of ideals in $\mathbb{Z}[i]$.  We note that $a_k:=\#\mathbb{Z}[i]/\mathfrak{p}_k$ is finite for all $k$, because both $\mathbb{Z}[i]$ and $\mathfrak{p}_k$ are abelian groups of rank $2$.  The sequence of positive integers $\{a_k\}$ is decreasing and does not stabilize, a contradiction.  Thus no such chains exist.


问:
Submodules of a $\mathbb{C}[x]$-module
Let $V=\mathbb{C}^{2}$ and let $$T=
  \left[ {\begin{array}{cc}
   1 & 1 \\
   0 & 1 \\
  \end{array} } \right]:V \rightarrow V.$$
Consider the $\mathbb{C}[x]$-module structure on $V$ where complex numbers act by scalar multiplication and $x$ act by $T$.

(a) Find the matrix of the action of $2x^3-x+3 \in \mathbb{C}[x]$ on $V$.

(b) Does the same assignment define a $\mathbb{C}[x]/(2x^2+x+1)$-module structure on $V$?

(C) Describe all submodules of $V$.

I know that for $\mathbb{C}$-vector space $V$ and linear map $T:V \rightarrow V$ we can turn $V$ in to an $\mathbb{C}[x]$-module by defining scalar multiplication by (($\sum a_{i}x^{i})v)=\sum a_{i}T^{i}(v)$ where $T^{0}(v)=v$ and $T^{i+1}(v)=T^{i}(T(v))$ and I can solve part (a) but about part (b) and (c) I don't have any ideas and I don't know the connection of part (a), (b) and (c).

答:
Out of sheer laziness and for cleaner reading, define
$$
A := \frac{\mathbb{C}[x]}{(2x^3 -x + 3)}.
$$
Note that in $A$, the polynomial $2x^3 - x + 3 \equiv 0$. It can be checked (!!!) that a module $M$ over the ring $A$ is equivalent to a pair $(V,T)$, where $V$ is a complex vector space and $T:V \to V$ is a matrix for which $2T^3 - T + 3 = 0$. The reason why this happens is essentially from the fact that $A$ is a quotient ring of $\mathbb{C}[x]$. Because modules over $\mathbb{C}[x]$ are equivalent to pairs $(V,T^{\prime})$ where $T^{\prime}$ is a matrix (no restrictions), adding the relation that defines $A$ means that you have to enforce that same relation on the matrix $T^{\prime}$ you choose when giving the pair $(V,T^{\prime})$.

With this observation, your problem (b) reduces to the following: Is it true that
$$
2T^3 - T + 3\cdot I = 0?
$$
Note that this connects to (a) because your problem (a) asks you to compute exactly this qunatity.

Finally, for your problem (c) you need to ask what it means to be a submodule. Generally, a submodule $M \hookrightarrow N$, after translating through this ''vector space and matrix'' trick becomes a function between pairs $f:(V,S) \to (W,T)$ where $T:W \to W$ and $S:V \to V$ are matrices and $f:V \to W$ is an injective linear function such that
$$
f(S(v)) = T(f(v))
$$
for all $v \in V$. This is perhaps seen via the more elegant but more abstract commuting diagram:
$$
\begin{array}
 VV & \xrightarrow{T} & V \\
 f\downarrow & & \downarrow f \\
 W & \xrightarrow[S]{} & W \\
 {}
\end{array}
$$
Let's see how this relates to your problem. For this you have
$$
(W,S) = \left(\mathbb{C}^2, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\right).
$$
What you need to do is now find all vector spaces $V$ with injective linear functions $f:V \to \mathbb{C}^2$ such that the given matrix $S$ that acts on $V$ satisfies the following property for all $v \in V$:
$$
T(f(v)) = f(S(v)).
$$
Expanding this out asks you to find all $(V,S)$ for which $f:V \to \mathbb{C}^2$ is an injective linear function and
$$
f(S(v)) = T(f(v)) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}f(v)= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} f(v)_1 \\ f(v)_2 \end{pmatrix} = \begin{pmatrix} f(v)_1 + f(v)_2 \\ f(v)_2 \end{pmatrix}$$
whenever $v \in V$.

ContentLicense: CC BY-SA 4.0

Title: Submodules of a $\mathbb{C}[x]$-module


问:
Prove that two distinct points are contained by at most one line?
Definition. A line is the set of points $(x,y)\in\mathbb{R}^2$ that satisfy the equation $$ax+by+c=0,$$ where at least one of $a$ or $b$ is non-zero.

How to prove that given two distinct points $(p,q)$ and $(r,s)$, there is at most one line that contains both points? (Edit: Ideally without much more knowledge than high school level algebra and arithmetic, so ideally without linear algebra.)

Unfortunately my proof (posted below) is long and inelegant, so I'm hoping for a short and elegant one.

答:
Suppose that $Α(x_1,y_1)$ και $Β(x_2,y_2)$ belong to
\begin{align*}
\varepsilon_1&:a_1x+b_1y+c_1=0\\
\varepsilon_2&:a_2x+b_2y+c_2=0.
\end{align*}
Without loss of generality, let $y_1\neq y_2$.
Multiplying the first equality with $a_2$, the second equality with $a_1$ and subtracting the first equality from the second equality, we get the equation
\begin{equation}
(a_1b_2-a_2b_1)y+(a_1c_2-a_2c_1)=0.
\end{equation}
which has two different solutions $y_1$ και $y_2$. Since the last equation has a degree equal to one, this can only happen when
\begin{align*}
a_1b_2-a_2b_1&=0\\
a_1c_2-a_2c_1&=0
\end{align*}
or equivalently
\begin{equation*}
\begin{vmatrix}
a_1 & b_1  \\
a_2 & b_2
\end{vmatrix}
=
\begin{vmatrix}
a_1 & c_1  \\
a_2 & c_2
\end{vmatrix}
=0
\end{equation*}
This means that there exists $k \in \mathbb{R}$ such that
\begin{align*}
a_2&=ka_1\\
b_2&=kb_1\\
c_2&=kc_1\\
\end{align*}
Hence
\begin{align*}
\varepsilon_1&:a_1x+b_1y+c_1=0\\
\varepsilon_2&:ka_1x+kb_1y+kc_1=0.
\end{align*}
and so $\varepsilon_1$ and $\varepsilon_2$ are actually the same lines.


问:
Can a single chart on a real number, ($\mathbb{R}, f:x\mapsto x^{1/3}$), be allowed to be considered as an atlas (differential structure)?
I might misunderstand something. But it seems that if we choose an atlas ($\mathbb{R}, f:x\mapsto x^{1/3}$), there is nothing to check for compatibility, because there is only one in there, hence even though $f$ is not smooth, it still qualifies as an atlas. Also, $f$ is homeomorphism between a real number to a real number. Am I thinking right?

答: There is a priori no notion of smoothness on the domain. While this is not smooth with respect to the standard manifold structure on $\mathbb R$, it doesn't matter for the purpose of defining an atlas. You're right that we don't have a compatibility condition to check in this case (perhaps besides the trivial transition function from the given chart to itself), but we still need to make sure that the chart is a homeomorphism onto its image (which is not difficult to see in this case).


问:
solving diophantine equations using congruences
So, a question in Elementary Number Theory by David M. Burton:

```<code>    Problems 4.4
    2(a) Using congruences, solve the diophantine equation 4x + 51y = 9

</code>```

Hint: $4x \equiv 9(mod \ 51) $ gives $ x = 15 +51t$, whereas $ \ 51y \equiv 9 (mod \ 4)$ gives $y = 3+4s $. Find the relation between $s$ and $t$

From this hint, I substitute $ x = 15 +51t$ and $y = 3+4s$ in the given equation $4x + 51y = 9$ to get the relation $s = -t-1$ and using which I can get the solutions $ x = 15 + 51t, \ y =-1-4t$

But we know that the solutions of $ax \equiv b (mod \ n)$ is equivatent to the solutions of the diophantine equation $ax -ny = b$,

so, solutions of $4x \equiv 9(mod \ 51) $ is equivalent to solutions of the equation $4x-51y=9$

whereas, solutions of $51y \equiv 9(mod \ 4) $ is equivalent to solutions of the equation $-4x+51y=9$

Now, my question is:

as, the above equations $4x-51y=9$ and $-4x+51y=9$ are different, then what is the point of substituting $ x = 15 +51t$ (solution of $4x-51y=9$) and $y = 3+4s $ (solution of $-4x+51y=9$) in the given equation $4x+51y=9$ to find a relation between $t$ and $s$ ? Isn't it absurd as $ x = 15 +51t$ and $y = 3+4s $ are solutions of two different equations? Can anyone please explain this?

PS: I did it like this $4x \equiv 9(mod \ 51) $ gives $ x = 15 +51t$ and by substituting it in $4x+51y=9$ we get $y=-1-4t$

答:
Here's the argument outlined in equational logic (vs. less precise English language). No use is made of $\,4x-51y = 9.\,$ The elimination steps use $E_z\!:\,z = c,\, az = b\iff z=c,\, ac = b,\,$ familiar as an elementary row operation in Gaussian elimination (triangularization) in linear algebra.

$$\begin{align}
 4x+51y\, =\, 9\, \ \ \ \ \ \ \ \ \ \ \ \ \ & \\[.2em]
\iff\ 4x+51y\, =\, 9\, \ \ \ \ \ \ \ \ \&\ \ &x = 15+51t\ \ \&\ \ y = 3+4s\\[.2em]
\iff  204(t+s) = -204\ \ \&\ \ &x = 15+51t\ \ \&\ \ y = 3+4s,\ \ \ \ {\rm by}\ \ E_x,\,E_y\\[.2em]
\iff\ \ \ \ \ \ \ \ \   t+s = -1\ \ \ \ \ \ \&\ \ &x = 15+51t\ \ \&\ \ y = 3+4s,\ \ \ \ {\rm by\ cancel}\ 204\\[.2em]
\iff  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phantom{t+s = -1\ \&}\ &x = 15+51t\ \ \&\ \ y = -1-4t,\ \ {\rm by}\ \ E_s \
\end{align}$$

What you did - solve one congruence first, then substitute - is generally simpler and the more common way to proceed, i.e. in equational language as above

$$\begin{align}
 4x+51y\, =\, 9\, \ \ \ \ \ \ \ \ \ \ \  & \\[.2em]
\iff\ 4x+51y\, =\, 9\, \ \ \ \ \ \ \&\ \ &x = 15+51t\\[.2em]
\iff    51y = -51\!-\!204t\ \ \&\ \   &x = 15+51t,\ \ \ {\rm by}\ \ E_x\\[.2em]
\iff\ \ \  \ \:\!   y = -1-4t\ \ \ \ \ \ \&\ \ &x = 15+51t,\ \ \ {\rm by\ cancel}\ 51
\end{align}\qquad$$

Remark  $ $ If instead of the above (bidirectional) equivalences we use only unidirectional arrows $(\Rightarrow)$ then we obtain only necessary conditions on the solutions. Then to verify sufficiency we have to check that the candidate  solutions actually work (are not extraneous roots). See here for more on the insufficiency of unidirectional inferences.


问:
Proving $\lim\limits_{x \to 0} \frac{1 - \sqrt{1 - x^2}}{x^2} = \frac{1}{2}$ from the definition
I am trying to prove that
$$\lim\limits_{x \to 0} \frac{1 - \sqrt{1 - x^2}}{x^2} = \frac{1}{2}$$
Given an $\epsilon$, I need to find $\delta$ so that $|x| < \delta \implies \left \lvert \frac{1 - \sqrt{1 - x^2}}{x^2} - \frac{1}{2} \right \rvert < \epsilon$. I can't figure out how to work backwards to $\delta$. I have:
\begin{align*}
 \left \lvert \frac{1 - \sqrt{1 - x^2}}{x^2} - \frac{1}{2} \right \rvert & = \left \lvert \frac{x^2}{x^2(1 + \sqrt{1 - x^2})} - \frac{1}{2} \right \rvert \\
& = \left \lvert \frac{2 - 1(1 + \sqrt{1 - x^2})}{2(1 + \sqrt{1 - x^2})} \right \rvert \\
& = \frac{1}{2} \left \lvert \frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}}  \right \rvert \\
& = \frac{1}{2} \cdot \frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}} \\
& < \frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}}
\end{align*}
The result is true provided that $\frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}} \leq \epsilon$. The only quantity I can control is $|x|$, but it is nowhere to be found in this expression. I can write it in by noticing that $x^2 = |x|^2$, but that only "artificially" works in $\delta$ without a lear end in sight.

Should that be my next step, or did I miss a more simple estimate?

Body:
HINT

Based on the assumption that $0 < |x| < \delta \leq 1$, we deduce that
\begin{align*}
0 < |x| < \delta & \Rightarrow 0 < x^{2} < \delta^{2}\\\\
& \Rightarrow 1 - \delta^{2} < 1 - x^{2} < 1\\\\
& \Rightarrow \sqrt{1 - \delta^{2}} < \sqrt{1-x^{2}} < 1\\\\
& \Rightarrow 1 + \sqrt{1 - \delta^{2}} < 1 + \sqrt{1 - x^{2}} < 2\\\\
& \Rightarrow \frac{1}{2} < \frac{1}{1+\sqrt{1-x^{2}}} < \frac{1}{1 + \sqrt{1-\delta^{2}}}\\\\
& \Rightarrow 0 < \frac{1}{1+\sqrt{1-x^{2}}} - \frac{1}{2} < \frac{1}{1 + \sqrt{1-\delta^{2}}} - \frac{1}{2}
\end{align*}

Consequently, we conclude that
\begin{align*}
\left|\frac{1-\sqrt{1-x^{2}}}{x^{2}} - \frac{1}{2}\right| & = \left|\frac{1 - (1-x^{2})}{x^{2}(1 + \sqrt{1-x^{2}})} - \frac{1}{2}\right|\\\\
& = \frac{1}{1+\sqrt{1-x^{2}}} - \frac{1}{2}\\\\
& < \frac{1}{1+\sqrt{1-\delta^{2}}} - \frac{1}{2} = \varepsilon
\end{align*}

Can you take it from here?


问:
Operations on subsets of Rings
I'm reading Algebraic number theory and Fermat's last theorem, and I just reached the text below (pp 13)

If $X$ and $Y$ are subsets of a ring $R$ we write $X+Y$ for the set of all elements $x+y$ ($x \in X$, $y \in Y$), and $XY$ for the set of all finite sums $\sum x_i y_i$($x_i \in X$, $y_i \in Y$).

I think I understand the $X+Y$. If $X=\{a,b,c\}$ and $Y=\{w,z\}$, then $X+Y=\{a+w, a+z, b+w, b+z, c+w, c+z\}$. The $XY$ bit doesn't make much sense to me (particularly because $x$ ad $y$ have the same index $i$). Can someone please tell me what would be the set $XY$ in this case?

Thanks in advance.

答:
In your example case, the way I read that summation, the product $XY$ would contain
$$
aw,az,bw,bz,cw,cz
$$
as well as any element you can reach by adding together finitely many of these six, in any combination, with repeats allowed. Like, for instance,
$$
0\\
aw+bz\\
aw + az + bw + bz + cw + cz\\
aw+aw+aw+cz
$$

Title: Operations on subsets of Rings


答:
Let $u = \sqrt{1 - x^2}$. You want to see when ${1 - u \over 1 + u} < \epsilon$. This is equivalent to
$$1 - u < \epsilon + \epsilon u$$
Equivalently,
$$1 - \epsilon < u(\epsilon + 1)$$
This is the same as
$$u > {1 - \epsilon \over 1 + \epsilon}$$
Squaring, this becomes
$$1 - x^2 > \bigg({1 - \epsilon \over 1 + \epsilon}\bigg)^2$$
Solving for $|x|$, we get
$$|x| < \sqrt{1 - \bigg({1 - \epsilon \over 1 + \epsilon}\bigg)^2}$$
These steps are reversible, so the expression on the right above can be used as your $\delta$.

Title: Proving $\lim\limits_{x \to 0} \frac{1 - \sqrt{1 - x^2}}{x^2} = \frac{1}{2}$ from the definition

Body:
I am trying to prove that
$$\lim\limits_{x \to 0} \frac{1 - \sqrt{1 - x^2}}{x^2} = \frac{1}{2}$$
Given an $\epsilon$, I need to find $\delta$ so that $|x| < \delta \implies \left \lvert \frac{1 - \sqrt{1 - x^2}}{x^2} - \frac{1}{2} \right \rvert < \epsilon$. I can't figure out how to work backwards to $\delta$. I have:
\begin{align*}
 \left \lvert \frac{1 - \sqrt{1 - x^2}}{x^2} - \frac{1}{2} \right \rvert & = \left \lvert \frac{x^2}{x^2(1 + \sqrt{1 - x^2})} - \frac{1}{2} \right \rvert \\
& = \left \lvert \frac{2 - 1(1 + \sqrt{1 - x^2})}{2(1 + \sqrt{1 - x^2})} \right \rvert \\
& = \frac{1}{2} \left \lvert \frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}}  \right \rvert \\
& = \frac{1}{2} \cdot \frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}} \\
& < \frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}}
\end{align*}
The result is true provided that $\frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}} \leq \epsilon$. The only quantity I can control is $|x|$, but it is nowhere to be found in this expression. I can write it in by noticing that $x^2 = |x|^2$, but that only "artificially" works in $\delta$ without a lear end in sight.

Should that be my next step, or did I miss a more simple estimate?


问:
For any conditionally convergent series $\sum _{n=1}^\infty a_n,\ \exists\ k\geq 2\ $ such that the subseries $\sum _{n=1}^\infty a_{nk}$ converges.
A subseries of the series $\displaystyle\sum _{n=1}^\infty a_n$ is defined to be a series of the form $\displaystyle\sum _{k=1}^\infty a_{n_k}$, for $n_k \subseteq \Bbb N$.

Prove or disprove: For any conditionally convergent series $\displaystyle\sum _{n=1}^\infty a_n,\ \exists\ k\geq 2\ $ such that the subseries $\displaystyle\sum _{n=1}^\infty a_{kn}$ converges.

Certainly, every conditionally convergent series has some subseries that converge (decreasing alternating terms), and some subseries that diverge (the subseries of all of the positive terms; also the subseries of all of the negative terms).

$$$$

My original "answer" which I now doubt is a counter-example to the proposition:

Hint: Find a subset $\ A \subset \mathbb{N}\ $ with the property that, for each $\ n \in \mathbb{N},\ $ the set $ \{kn\}_{k \in \mathbb{N} }$  has finitely many elements in common with $A.$

Answer: The prime numbers is the standard example of such a subset of $\mathbb{N}$ with the property in the hint. This point is to make the 2nd, 3rd, 5th, 7th, ... members of our sequence have one sign, and every other member of our sequence have another sign. For example, $$ a_1 =  \frac{1}{1},\ a_2 =  \frac{-1}{4},\ a_3 =  \frac{-1}{4},\ a_4 =  \frac{1}{3},\ a_5 =  \frac{-1}{4},\ a_6 =  \frac{1}{5},\ a_7 =  \frac{-1}{6},\ a_8 = a_9 = a_{10} =  \frac{1}{3} \times \frac{1}{7},\ a_{11} = \frac{-1}{8},\ ... $$ This is simply the alternating harmonic series split up in a way that (I thought) answers the question.

That was my answer, but now I'm not sure the series above is a valid counter-example. I constructed it so that every subseries of the form $\displaystyle\sum _{n=1}^\infty a_{kn}$ contains finitely many negative numbers. And then I'm not sure what my logic was. It was either that "every subseries of a monotonic divergent series diverges", but now I realise that this is not what the link says, and is also not true. Or my reasoning was that, for each $k \in \mathbb{N},\ $ every subseries $\displaystyle\sum _{n=1}^\infty \frac{1}{kn}$ of the harmonic series, diverges, which is true, but the subseries of my series defined above do not match these subseries of the harmonic series.

So the problem remains open...

答:
(This is a simplified version of my previous solution when I used harmonic series to fill in "gaps")
The statement is not true.

To see this fix a increasing sequence of primes $p_{l}$ such that $lim_{l\to\infty} (p_{l+1} - p_{l}) = \infty$ and define $a_{n}$ such that
$a_{n} = \frac{1}{l}$ if $n=p_{l}$.

To fill in the gaps in the definition of $a_{n}$ where n is between $p_{l}$ and $p_{l+1}$ set $a_{n} = - \frac{1}{l*(p_{l+1} - p_{l})}$
In words, $a_{n}$ is positive only for $n=p_{l}$ and its value for such $n$ is $\frac{1}{l}$; then it is followed by small, negative, and identical entries which add up to $\frac{1}{l}$ and then again $a_{n}$ for $n= p_{l+1}$ which is $\frac{1}{l+1}$ followed by negative entries adding up to $\frac{1}{l+1}$.
We note that the sum of terms in "negative block" is $\frac{1}{l}$ hence sum of a sub-block which is represented by an arithmetic progression with difference $k$ which fall into such block is $\frac{1}{lk}$.

Lastly, if we take a sub series of the form $a_{nk}$ then indexes are never primes so all terms are negative. Further, this sub series, on each sub block adds up to $-\frac{1}{kl}$ and when summed up by l is divergent.

Addendum, please note that the original series $\sum a_{n}$ is convergent.
This follows easy by looking at the difference between partial sums (i.e. checking that partial sums forms Cauchy sequence).


问:
On the definition of absolute maximum
In my life, I have been considering the definition of absolute maximum as "the highest point of a graph", if it does not diverge to infinity, of course.

However, I attended a lesson in which it was taught that "absolute maximum is the highest of the relative maxima".

In other words, they're defining global maximum as the highest local maximum.

This can be true for functions defined in the whole real domain. Nevertheless, if we constrain the domain to a certain closed interval $[a,b]$, it can be that $f(a)$ or $f(b)$ is actually higher than a possible local maximum located inside the interval.  For example, $f(x)=x^3-2x;\qquad x\in[-1,2]$.

According to the first definition, $(2, f(2))$ is an absolute maximum, but according to the second one, it would not be. On the contrary, the absolute maximum would be located on the local maxium, as it is the only one in the graph.

The questions are: which  one of the two definitions is true? Are those two definitions coexisting, forming two streams? Which books can support each theory? Thanks in advance.

答:
The absolute maximum is a value $M$ in the image such that $f(x) \leq M$ for all $x$ in the domain. So, the first definition is accurate here and reflects how the term is used generally.

It depends on what you define as a local maximum here. If local maxima can only exist within the interior of the interval, then yes, in general the maximum of the local maximum is not the global maximum. However, if a local maximum is defined to occur at a point $x^*$ where $f(x^*) \geq f(x)$ for all $x$ in a neighborhood of $x^*$, then it follows that the global maximum is a local maximum, and must be the greatest among these.

Title: On the definition of absolute maximum


问:
Looking for a proper proof that two sets with sub sets are equivalent
Basically I would just like to know how to prove the following equation

$$ A \subseteq B \cup C \iff A \cap \overline B \subseteq C $$

I understand that I have to prove that the left-hand side should equal the right-hand side, and I can perform all the basic logical equivalencies, like Association ad Distribution, but I am unsure how to express the subset ($ \subseteq $) in the proof. I'm tempted to treat the subset sign as intersection ($ \cap $), but since my proof doesn't work... I'm pretty sure this is the wrong approach.

答:
I completely agree with Peter Smith's advice to "sweat the definitions".  Here is the way in which I would do this: start at the most complex side, expand the definitions, and see whether you can get to the other side.

Note that below, $x$ ranges over the 'universe' that is implicit in this question.

\begin{align}
& A \cap \overline B \subseteq C \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& \langle \forall x :: x \in A \cap \overline B \;\Rightarrow\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\cap$"} \\
& \langle \forall x :: x \in A \land x \in \overline B \;\Rightarrow\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\overline{\phantom\square}$"} \\
& \langle \forall x :: x \in A \land x \not\in B \;\Rightarrow\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"expand $\Rightarrow$, using De Morgan on its left hand side"} \\
(*) \; \phantom\equiv & \langle \forall x :: x \not\in A \;\lor\; x \in B \;\lor\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"reintroduce $\Rightarrow$ at the first $\lor$ -- suggested by the shape of our goal"} \\
& \langle \forall x :: x \in A \;\Rightarrow\; x \in B \lor x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\cup$"} \\
& \langle \forall x :: x \in A \;\Rightarrow\; x \in B \cup C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& A \subseteq B \cup C \\
\end{align}

This completes the proof.

So by expanding the definitions, and also writing $\Rightarrow$ which often is difficult to calculate with, we arrived at $(*)$.  Then it was not difficult to see how we could get to the goal.  Alternatively, one could also start at both ends and see that both end up at $(*)$.


问:
Equinumerosity of strings and natural numbers
so I'm trying to show that there exists a bijection from the infinite set of finite strings composed of elements from $\{a, b\}$ to $\mathbb{N}$. So I was thinking about showing that $\mathbb{N}$ and the infinite set of finite binary sequences are equinumerous (each have a unique representation), and then constructing a bijection between $\{a, b\}$  and $\{0, 1\}$. Then by the transitive property of equinumerosity, I would be done with my goal.

However, when you're writing something in binary, like 00001 (which would be equivalent to aaaab), and so I think this proof would exclude some cases? How else can I go about this?

答:
If your alphabet is $\{0, 1\}$, a simple way to describe the bijection with $\mathbb{N} = \{1, 2, \dotsc\}$ is to stick an 1 in front of the string, and read the result as a binary number. To go the other way around, write the number in binary and chop off the first digit.

The general case for an arbitrary alphabet is called bijective numeration


问:
Question about an example of $(W^{\perp})^{\perp} \neq W$.
Let $\mathbb{R}[X]$ be a vector space endowed with an inner product $\langle p,q \rangle = \int_{0}^{1}p(x)q(x)dx$ and let $W = \{ p(x) \in \mathbb{R}[X]: p(0) = 0 \}$. Show that $W^{\perp} = \{0\}$.

It could be very simple (and intuitive), but I don't know how to show that if $\int_{0}^{1}p(x)q(x)dx =0$ for all $p(x) \in W$ then $q(x) = 0$. I was thinking about write generical expressions for p and q but when we make $p \cdot q$ it is not usefull.

Can you give me an easy way?? I mean, some hint?

答: Take $p(x)=xq(x)\in W$ and note that $xq^2(x)$ is continuous and non-negative in $[0,1]$.


问:
The set $C= \{x \in X: \text{dist} (x,K) ≤\frac 12 \}$ is not compact

Give an example of a complete metric space $(X,d)$ and its compact subset $K$, that the set $C= \{x \in X: \text{dist} (x,K) ≤\frac 12 \}$ is not compact.

From Heine-Borel theorem: completeness + totally boundation = compactness. The set $C$ is closed so also complete. Therefore, lest C be compact, then C cannot be tatally bounded. However I have no idea which $(X,d)$ and $C$ to choose to fulfill this.

答:
Let me write some comments in order to address the difficulty of finding a "nice" example and, in contrast to this, give you a really neat one.

Consider the function $f:X\to \mathbb{R}$ given by $f(x)=\text{dist}(x,K)$. It is easy to see that $f$ is a Lipschitz continuous map and that our set $C$ is simply $f^{-1}\left[(-\infty, \frac{1}{2}]\right]$. It is immediate from this, as you pointed out, that $C$ is a closed subset of $X$. Another easy but maybe not immediate fact is that $C$ is bounded.

To see this, let's use that $K$ is a bounded subset of $X$ to fix $z\in X$ and $r>0$ in such a way that $K\subseteq B_{X}(z,r)$. Now, I would like to prove that $C\subseteq B_{X}(z,r+1)$. For any $x\in C$, there exists a $y\in K$ such that $d(x,y)\leq 1$ (if this wasn't the case, then we can prove that $1\leq f(x)$, contradicting that $x\in C$). Then, the triangle inequality gives us

$$d(x,z) \leq d(x,y) + d(y,z) < 1+r.$$

This proves that $C$ is a bounded subset of $X$.

Now, if $X$ was a finite-dimensional Banach space then, since the Heine-Borel Theorem holds for every finite-dimensional normed vector space, $C$ would automatically be a compact subset of $X$, and we don't want this. So, if we try to find an example of a set $C$ which is not compact in a Banach space $X$, we need $X$ to be infinite-dimensional. This rules out our favorite spaces, such as $\mathbb{R}^{n}$ and $\mathbb{C}^{n}$.

Consider now the space of all bounded sequences of real numbers $\ell_\infty$. It is well known that $\ell_\infty$ is an infinite-dimensional Banach space. Let $\overline{0}$ denote the element of $\ell_\infty$ with zeroes in every entry, and put $K=\{\overline{0}\}$. Observe that, since $K$ is finite, $K$ is a compact subset of $\ell_\infty$. Lastly, notice that $$C=\overline{B}_{\ell_\infty}(\overline{0},\frac{1}{2})=\left\{x\in \ell_\infty : d_{\ell_\infty}(x,\overline{0})\leq \frac{1}{2}\right\}.$$

From this we deduce that $C$ is homeomorphic to the closed unit ball in $\ell_\infty$. Since $\ell_\infty$ has infinite dimension, the closed unit ball is not compact (this is a nice corollary of Riesz's Lemma; check out the section of "Some consequences" in https://en.wikipedia.org/wiki/Riesz%27s_lemma) and so, $C$ is not compact.

Hope this helps.


问:
Stationary distribution for Amount of Cash in an ATM
I'm a little stuck on how to solve this problem. Suppose I have an ATM that starts with zero dollars. Every morning, a truck will deposit a random amount of dollar bills into the ATM so people can withdraw. The amount of cash deposited is Poisson with mean parameter $\lambda$. The total amount withdrew for the day is binomially distributed with parameter $p$. That is, every dollar in the ATM has a probability $p$ of being withdrawn. Cash that is not withdrawn at the end of the day is left in the ATM. Find the stationary distribution of the amount of cash in the ATM.

I have already shown that the amount of cash in the ATM is a Markov Chain. What I tried was defining the amount of cash in the ATM on day $n$ as a sum of a Binomial and Poisson variables, since the amount of cash on day $n$ is equal to the number of dollar bills not withdrawn from the previous day plus the new cash deposited in the morning. So I had

$$X_n = Binom(X_{n-1}, 1-p) + Poisson(\lambda)$$
$$ X_{n} = Y_{n} + Z_{n} $$
$$ e^{X_{n}} = e^{Y_{n} + Z_{n}} $$
$$ e^{X_{n}} = e^{Y_{n}}e^{Z_{n}} $$
$$ G_{X_n}(s)= G_{Y_{n}}(s)G_{Z_{n}}(s) $$
$$ G_{X_n}(s)= G_{Y_{n}}(s) G_{Z_{n}}(s) $$
$$ G_{X_n}(s)= ((1-p)s + p)^{X_{n-1}} \cdot e^{\lambda (s-1)} $$
$$ G_{X_n}(s)= G_{X_{n-1}}((1-p)s + p) \cdot e^{\lambda (s-1)} $$

Where $G_{X_n}(s) = E[s^{X_n}]$, the probability generating function. I plugged in the recurrence relation to get

\begin{align*}
    G_{X_n}(s) &= G_{X_{n-1}}((1-p)s + p) \cdot e^{\lambda (s-1)} \\
               &= G_{X_{n-2}}((1-p)[(1-p)s + p)] + p) \cdot e^{\lambda ([(1-p)s + p)]-1)} \cdot e^{\lambda (s-1)} \\
               &= G_{X_{n-2}}((1-p)^2 s + (1-p)p + p) \cdot e^{\lambda ((1-p)s + p - 1)} \cdot e^{\lambda (s-1)} \\
               &= G_{X_{n-3}}((1-p)^3 s + (1-p)^2 p + (1-p)p + p) \cdot e^{\lambda ((1-p)^2 s + (1-p)p + p - 1)} \cdot e^{\lambda ((1-p)s + p - 1)} \cdot e^{\lambda (s-1)} \\
               &= G_{X_{n-3}}((1-p)^3 s + (1-p)^2 p + (1-p)p + p) \cdot e^{\lambda ((1-p)^2 s + (1-p)p + p - 1)} \cdot e^{\lambda ((1-p)s + p - 1)} \cdot e^{\lambda (s-1)} \\
               &= G_{X_{n-3}}((1-p)^3 s + (1-p)^2 p + (1-p)p + p) \cdot e^{\lambda ((1-p)^2 s + (1-p)p + p - 1)} \cdot e^{\lambda ((1-p)s + p - 1)} \cdot e^{\lambda (s-1)} \\
               &\vdots \\
               &= \left[p \sum_{i}^{k} (1-p)^i + (1-p)^k\right]^{X_{N-K}} \cdot \prod e^{\lambda ((1-p)^k s + p\sum_{i=0}^{k}(1-p)^i - 1)}  \\
\end{align*}

If we take the limit as $K \xrightarrow{} \infty$, then we are left with

$$ G_{X_n}(s) = \prod e^{\lambda ((1-p)^k s + p\sum_{i=0}^{k}(1-p)^i - 1)} $$

My friend told me he got $G_{X_n} = e^{\lambda/p(1-s)}$, which is crazy but works out when you graph it via simulation. Can anybody point out how to continue from here?

答:
Given a random variable $X$ let $M_{X}(t)=\mathbb{E}\big(e^{tX}\big)$ and $G_{X}(t)=\mathbb{E}\big(t^{X}\big)$. Evidently we have $G_{X}(t)=M_{X}\big(\ln(t)\big)$.

You have $Y_{n}|X_{n-1}\sim \text{Binomial}(X_{n-1},1-p)$ so with this and the total law of expectation $$M_{Y_n}(t)=\mathbb{E}(e^{tY_n})=\mathbb{E}\Big(\mathbb{E}(e^{tY_n}|X_{n-1})\Big)=\mathbb{E}\Big(\big((1-p)e^t+p\big)^{X_{n-1}}\Big)$$ In other words, $$M_{Y_n}(t)=G_{X_{n-1}}\big((1-p)e^t+p\big)=M_{X_{n-1}}\Big(\ln\big((1-p)e^t+p\big)\Big)$$ Combining this result with this we get $$M_{X_n}(t)=M_{Y_n}(t)M_{Z_n}(t)=M_{X_{n-1}}\Big(\ln\big((1-p)e^t+p\big)\Big)\cdot e^{\lambda(e^t-1)}$$ It's relatively easy to show that $$M_{X_{n-k}}\Big(\ln\big((1-p)^ke^t+1-(1-p)^k\big)\Big)=M_{X_{n-k-1}}\Big(\ln\big((1-p)^{k+1}e^t+1-(1-p)^{k+1}\big)\Big)e^{\lambda (1-p)^k (e^t-1)}$$ Combining the previous two lines yields $$M_{X_n}(t)=M_{X_0}\Big(\ln\big((1-p)^n e^t +1-(1-p)^n\big)\Big)e^{\frac{\lambda(e^t-1)(1-(1-p)^n)}{p}}$$ If $n$ is large then $$M_{X_n}(t)\approx M_{X_0}(0)e^{\frac{\lambda(e^t-1)}{p}}=e^{\frac{\lambda(e^t-1)}{p}}$$ We now see the limiting distribution of $X_n$ is $\text{Poisson}\Big(\frac{\lambda}{p}\Big)$. Note $M_{X_0}(0)=1$ since $X_0 \sim \text{Poisson}(\lambda)$


问:
Convergence of a quadratic form in probability
Suppose $x$ is a random vector in $R^n$ with positive definite covariance matrix, $R$. I'm interested in the convergence in probability of the quadratic form, $\dfrac{1}{n}x^TR^{-1}x$. If $x$ were Gaussian, this would be simple enough since were can write, $x^TR^{-1}x = \sum\limits_{i=1}^ny_i^2$ where $y_i \sim N(0,1)$ i.i.d. Then, $\dfrac{1}{n}x^TR^{-1}x \xrightarrow{P} 1$ by the law of large numbers. How about for a non-Gaussian $x$? The only thing that I can calculate is,

$$ E\left[\dfrac{1}{n}x^TR^{-1}x\right] = 1, \;\;\; \forall n$$

But I don't think this is enough to ensure that $\frac{1}{n}x^TR^{-1}x \xrightarrow{P} 1$ (I don't think I can apply Chebyshev?). Are there other conditions on $x$ that I can impose here?

EDIT: I guess a more general way of asking the question would be, let $\{X_n\}$ be a non-negative sequence of random variables such that,
$$ \lim\limits_{n\rightarrow\infty}E[X_n] = 1 \; (\text{or} \;E[X_n] = 1 \;\; \forall n)$$
What conditions ensure that, $X_n \xrightarrow{P} 1$?

答:
Suppose the covariance matrix of $X$ is $R\in \mathbb{R}^{n\times n}$ and the expectation of $X$ is $\mu \in \mathbb{R}^{n}$

Because $R$ is a symetric definite positive matrix, then its square root matrix $R^\frac{1}{2}$ defined as $R^\frac{1}{2} R^\frac{1}{2} =R $ is also a symetric definite positive matrix. The matrix $R^\frac{1}{2}$ is also invertible.

Denote $Y = R^{-\frac{1}{2}}X$, then $X = R^\frac{1}{2}Y$. We prove that the covariance matrix of $Y$ is the identity matrix $I$. Indeed, we have
\begin{align}
Cov(Y) &= E((R^{-\frac{1}{2}}X)(R^{-\frac{1}{2}}X)^T) - E(R^{-\frac{1}{2}}X)(E(R^{-\frac{1}{2}}X))^T \\
&= E(R^{-\frac{1}{2}}XX^TR^{-\frac{1}{2}}) - R^{-\frac{1}{2}}E(X)E(X)^TR^{-\frac{1}{2}} \\
&= R^{-\frac{1}{2}}E(XX^T)R^{-\frac{1}{2}} - R^{-\frac{1}{2}}E(X)E(X)^TR^{-\frac{1}{2}} \\
&= R^{-\frac{1}{2}} Cov(X) R^{-\frac{1}{2}} \\
&= R^{-\frac{1}{2}}R R^{-\frac{1}{2}} \\
&= I \\
\end{align}
In particular, we have $V(Y_i) = E(Y_i^2)-E(Y_i)^2 =1\tag{1}$ with $Y_i$ is the i-th element of  the random vector $Y$.

Because $Y = R^{-\frac{1}{2}}X$, hence $E(Y)= R^{-\frac{1}{2}}\mu \tag{2}$

From (1) and (2), we deduce that
$$\sum_{i=1}^n E(Y_i^2) = n- (\mathbf{1}^T  R^{-\frac{1}{2}}\mu)$$
with $\mathbf{1} \in \mathbb{R}^{n}$ is the all-ones vector. For information, the term $\mathbf{1}^T  R^{-\frac{1}{2}}\mu$ is the sum of all elements of the $R^{-\frac{1}{2}}\mu$ (the expectation vector of $Y$).

We have now
\begin{align}
\frac{1}{n}X^TR^{-1}X &= \frac{1}{n}(R^\frac{1}{2}Y)^TR^{-1}(R^\frac{1}{2}Y) \\
&= \frac{1}{n} Y^T R^\frac{1}{2}R^{-1}R^\frac{1}{2}Y \\
&= \frac{1}{n} Y^T Y \\
&= \frac{1}{n} \sum_{i=1}^n Y_i^2 \\
\end{align}

By denoting $S_n = \frac{1}{n} \sum_{i=1}^n Y_i^2 $ the problem becomes: what is the condition which ensures that $S_n$ converges in probability given that $E(S_n)= 1 - \frac{1}{n}(\mathbf{1}^T  R^{-\frac{1}{2}}\mu)$.

For sufficient conditions, for example, if we have 2 following conditions:

the term $\frac{1}{n}(\mathbf{1}^T  R^{-\frac{1}{2}}\mu)$ converges (to $a  \in \mathbb{R}$ for example)

$E(|S_n - (1-a)|) \xrightarrow{n\rightarrow +\infty} 0$ (convergence in $L_1$)

Then $S_n$ converges in probability to $1-a$, as the convergence in $L_1$ implies the convergence in probability.


问:
Football Pool - odds of buying (2) squares where the numbers received are "boxed"
There are 100 squares in a standard football pool..  with the heading on each side 0-9.  So in a standard 10 x 10 square football pool, I just bought (2) boxes and the numbers drawn for the (2) boxes were:

1st square:   7,4

2nd square:    4,7

So..  I have the numbers 4 and 7 "boxed" no matter which team ends up winning.   What are the odds of that?   Buying (2) football pool squares and getting the numbers drawn "boxed" like that for both teams ?    No matter which team is on top at the end, if the last digits of the final score are either 7,4 or 4,7 then we win.   Curious about the statistics because this has never happened before.

Another way to ask the same question would be:

What are the odds, when buying (2) squares in a standard football pool, of having the numbers drawn "reversed"? or "boxed"?     i.e.

1st square:  7, 4

2nd square:  4, 7

答:
Assuming no box is allocated more than once:

You want your first box $x,y$ to have distinct values $y\not =x$.  The probability of that happening is $\dfrac{90}{100}=\dfrac{9}{10}$

Given that, you want your second box to be the reverse of your first box, i.e.  $y,x$.  The conditional probability of that happening is $\dfrac{1}{99}$ as it is impossible to get your first box again and the others are equally likely

So the overall probability of getting this pattern is the product of these,  which is $\dfrac{1}{110}$


问:
Exponential function of a complex variable
For real $z$, It is true that $f(z) = \exp(z) > 0$. However, if $z$ is now complex, is this statement still true?

Thanks.

答:
$$e^{\pi i}  =-1, \text{ so no, not positive } \forall z.$$

Actually, $f(z)=\exp(z)$ is not real for many $z$.


问:
Express the product of a binomial distribution for i = 1 to n as a sum
I am told that
$$f(y|p) = \prod_{i=1}^n (1-p)^yp $$
$$=p^n(1-p)^{\sum_{i=1}^n y_i}$$
What are the steps to get there?

Body:
Steps:

$$\begin{split}f(y|p)&=\prod_{i=1}^n(1-p)^{y_i}p\\
&=\underbrace{p*p*...*p}_{\text {n}}*(1-p)^{y_1}*(1-p)^{y_2}*...*(1-p)^{y_n}\\
&=p^n(1-p)^{y_1+y_2+...+y_n}\\
&=p^n(1-p)^{\sum_{i=1}^ny_i}\end{split}$$


问:
Solve $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$
I want to find the solutions of  $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$.

It is not hard to see $x=0$ is a root of the equation. but I don't know how to solve this equation in general. I can see terms of the equation looks very similar to binomial expansion of $[(x+1)+(x-1)]^{63}$ except the coefficient of each term is $1$ rather than $63\choose k $ (for $k=0,1,\cdots,63$ ). is it possible to use binomial theorem to solve the equation? (or other approaches)

答: Note that $x=-1$ is not a solution. This is a GP with common ratio $r=(x-1)/(x+1)\ne1$ and the sum is$$(x+1)^{63}\left[\frac{r^{64}-1}{r-1}\right]=0\implies r=\pm1$$Since $r\ne1$ it must equal $-1$ giving only real solution as $x=0$.


答:
You can solve it by using the formula of GP

Here $a = (x+1)^{63}$ and $r = \dfrac{x-1}{x+1}$

Title: Solve $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$

Body:
I want to find the solutions of  $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$.

It is not hard to see $x=0$ is a root of the equation. but I don't know how to solve this equation in general. I can see terms of the equation looks very similar to binomial expansion of $[(x+1)+(x-1)]^{63}$ except the coefficient of each term is $1$ rather than $63\choose k $ (for $k=0,1,\cdots,63$ ). is it possible to use binomial theorem to solve the equation? (or other approaches)


问:
Ellipse formula
In the video https://youtu.be/eI4an8aSsgw?t=16354 the professor says that the following equation
$$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a$$
simplifies to
$$
(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)
$$
I don't get it. I need someone who can explain it step by step

Title: Ellipse formula

Body:
Let's call $\begin{cases}U=\sqrt{(x+c)^2+y^2}\\V=\sqrt{(x-c)^2+y^2}\end{cases}\quad$ and we start from $U+V=2a$

Squaring we get $\quad U^2+2UV+V^2=4a^2$

$U^2+V^2=(x+c)^2+y^2+(x-c)^2+y^2=2(x^2+y^2+c^2)$

Now we'll divide by $2$ and put the roots on one side and square again:

$$U^2V^2=(2a^2-\tfrac 12(U^2+V^2))^2$$
$\begin{align}U^2V^2
&=((x+c)^2+y^2)((x-c)^2+y^2)\\
&=((x+c)^2+(x-c)^2)y^2+(x+c)^2(x-c)^2+y^4\\
&=(2x^2+2c^2)y^2+(x^2-c^2)^2+y^4\\
\require{cancel}&=\cancel{2x^2y^2}+\cancel{2c^2y^2}+\cancel{x^4}-2c^2x^2+\cancel{c^4}+\cancel{y^4}
\end{align}$

It is equal to:

$(2a^2-x^2-y^2-c^2)^2
\require{cancel}=4a^4+\cancel{x^4}+\cancel{y^4}+\cancel{c^4}-4a^2x^2-4a^2y^2-4a^2c^2+\cancel{2x^2y^2}+2x^2c^2+\cancel{2y^2c^2}$

Gathering remaining terms:

$-2x^2c^2=4a^4-4a^2x^2-4a^2y^2-4a^2c^2+2x^2c^2\iff 4\Big(a^4-a^2x^2-a^2y^2-a^2c^2+x^2c^2\Big)=0$

$$(-a^2+c^2)x^2-a^2y^2+a^2(a^2-c^2)=0$$

Which is the desired expression.


Body:
$$\frac {a^n - b^n}{a-b} = a^{n-1} + a^{n-2} b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1}$$

Hence:

$$(x+1)^{63} + (x+1)^{62}(x-1) + \dots+ (x-1)^{63} = \frac {(x+1)^{64}-(x-1)^{64}}{(x+1)-(x-1)} = \frac12((x+1)^{64} - (x-1)^{64})$$

If the above expression is zero, we must have $(x+1)^{64} = (x-1)^{64}$. Hence $\dfrac {x+1}{x-1}$ must be one of the 64-th roots of unity (or if you are only considering real roots, $\pm 1$).


问:
Prove that a homeomorphism of finite order have a fixed point.

Let $\phi: \mathbb{R} \to \mathbb{R}$ a homeomorphism of finite order, it is $\phi^p(x)=x$ for $p>0$.Prove that $\phi$ has a fixed point.

Attempt

Let $g: \mathbb{R} \to \mathbb{R}^p$ given by
$$g(x)=(\phi(x)-x,\phi^2(x)-\phi(x), \cdots, \phi^p(x)-\phi^{p-1}(x))$$

$(i)$ Notice that $g$ is continuous since is linear combination of continuous functions.

$(ii)$ WLOG assume that $0< \phi(0)$ in otherwise ($\phi(0)<0$) and notice that if  $g$ don't have a fixed point in $[0,\phi(0)]$ then for each $i\in 0,1,\cdots,p-1$ we will have $\phi^{i+1}(x)-\phi(x)>0$
(in otherwise in $[ \phi(0),0]$we should have  $\phi^{i+1}(x)-\phi(x)<0)$

But by the continuity from $(i)$ it implies that $\phi$ maintain the sign in each enter of $g$.

$(iii)$ Finally notice  that if $g$ maintain the sign in all the entries then it would implies that $\phi$ is not periodic which is not true, then $g$ don't maintain the sign in all their entries and therefore by $(ii)$ we get that $\phi$ have a fixed point.

I have two questions about my proof.

In $(ii)$ is valid the arguments to show that if $\phi$ have a fixed point then each of the entries maintain the same sign (Wich depends that if $0< \phi(0)$)?.

The part $(iii)$ I felt that is the more frail part of the proof.
There are other more strong argue to show that is impossible that the signs of all the entries of $g$ maintain constant ?

答:
In (ii), I don't see why $g$ not  having a fixed point would imply that each component is strictly positive, it could be that one of them is strictly positive and another is strictly negative.
In (iii), it seems like you are not arguing the point.

Here's a neat proof of the claim:
Assume that $\phi$ does not have a fixed point, then  WLOG we can assume that $ \phi(x)-x > 0$ for all $x \in \mathbb{R}.$(By the mean value theroem)  This implies that

$$ \phi(\phi^{k-1}(x)) -\phi^{k-1}(x) >0$$ for  $1 \le k\le p.$
But now  $$0= \phi^p(x) -x= \sum_{k=1}^p (\phi^{i}(x) -\phi^{i-1}(x)) >0,$$ a contradiction.


问:
find the maximum of the value $\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\frac{a_{i}}{i}\right)^2$
let $a_{i}\ge 0$ and such $$a_{1}+a_{2}+\cdots+a_{n}=1$$
find the maximum of the value
$$\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\dfrac{a_{i}}{i}\right)^2$$
I  try to use
From Pólya-Szegö’s inequality, we have for $0 < m_1 \leqslant u_k \leqslant M_1$ and $0 < m_2 \leqslant v_k \leqslant M_2$,
$$\left(\sum u_k^2 \right) \left( \sum v_k^2 \right) \leqslant \frac14 \left( \sqrt{\frac{M_1 M_2}{m_1m_2}} + \sqrt{\frac{m_1 m_2}{M_1 M_2}} \right)^2 \left( \sum u_k v_k\right)^2$$

But I can't it.Thanks

答:
Alternative solution

Problem: Let $a_i\ge 0, \forall i$ with $a_1 + a_2 + \cdots + a_n = 1$. Prove that the maximum of
$$\left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2$$
is $\frac{4(n+1)^3}{27n^2}$.

We may use the approach in my answer to this equation:
An upper bound of product of two inner products

Consider the optimization problem
$$\max_{a_i\ge 0, \forall i; \ \sum_{i=1}^n a_i = 1} \left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2.$$
Let $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$ be a global maximizer.

We claim that if $1 < k < n$, then $a_k^\ast = 0$.
Indeed, if $a_k^\ast > 0$, let
$$a_1' = a_1^\ast + (1 - y) a_k^\ast, \quad a_k' = 0, \quad a_n' = a_n^\ast + y a_k^\ast$$
where $\frac{k-1}{n-1} < y < \frac{n}{k}\cdot \frac{k-1}{n-1}$. We have
$$a_1' + a_k' + a_n' = a_1^\ast + a_k^\ast + a_n^\ast,$$
and
\begin{align}
&a_1' + ka_k' + na_n' - (a_1^\ast + ka_k^\ast + na_n^\ast)\\
=\ & (n-1)\left(y - \frac{k-1}{n-1}\right) a_k^\ast\\
>\ & 0,
\end{align}
and
\begin{align}
&a_1' + \frac{a_k'}{k} + \frac{a_n'}{n} - \left(a_1^\ast + \frac{a_k^\ast}{k}
+ \frac{a_n^\ast}{n}\right)\\
=\ & \frac{n-1}{n}\left(\frac{n}{k}\cdot \frac{k-1}{n-1} - y\right)a_k^\ast\\
>\ & 0.
\end{align}
However, this contradicts the optimality of $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$.

Now, since $a_2^\ast = a_3^\ast = \cdots = a_{n-1}^\ast = 0$, we have
\begin{align}
&\left(\sum_{i=1}^n i a_i^\ast\right)\left(\sum_{i=1}^n \frac{a_i^\ast}{i}\right)^2\\
=\ & (a_1^\ast + n a_n^\ast)\left(a_1^\ast + \frac{a_n^\ast}{n}\right)^2\\
=\ & (1 - a_n^\ast + na_n^\ast)\left(1 - a_n^\ast + \frac{a_n^\ast}{n}\right)^2\\
=\ & \frac{1}{2n^2}[2 + 2(n-1)a_n^\ast][n - (n-1)a_n^\ast]^2\\
\le\ & \frac{1}{2n^2} \left(\frac{2 + 2(n-1)a_n^\ast + n - (n-1)a_n^\ast
+ n - (n-1)a_n^\ast}{3}\right)^3\\
=\ & \frac{4(n+1)^3}{27n^2}
\end{align}
with equality if $a_n^\ast = \frac{n-2}{3(n-1)}$, where we have used AM-GM inequality.

Thus, the desired maximum is $\frac{4(n+1)^3}{27n^2}$
at $a_1 = \frac{2n-1}{3(n-1)}, a_2 = a_3 = \cdots = a_{n-1} = 0, a_n = \frac{n-2}{3(n-1)}$.


问:
A simple method to calculate minimal area enclosed between a tangent to $f(x)$ and coordinate axes
Given a function $y=f(x)$, take a tangent to the function at some point. We are to find the minimum area between this line and the coordinate axes. An example question is the coordinate axes have minimum area.

I faced two different algorithms to find the solution. The first one is straight-forward:

Pick a point on the function: $(x_1, y_1) = (x_1, f(x_1))$.

Find the derivative of the function at $x_1$ to calculate the slope of the line. ($m=f'(x_1)$)

Derive the tangent-line formula. $y-y_1=m(x-x_1)$

Find the formulations of intersections with coordinate axes $(0, y_0)$, $(x_0, 0)$.

Calculate the formulation of the area of triangle as $A = x_0 y_0/2$.

$A$ shall be a function of $x_1$, minimize that to calculate minimum area.

Second algorithm is very short compared to that:

Take the function $g(x) = 2 x f(x)$.

Minimize that function to calculate the result.

I cannot figure out how the second algorithm works, or when it works. I checked both against the following family of functions, and both algorithms give the same result:

$f(x) = ax+b$

$f(x) = k/x$

$f(x) = 3b^2 - a^2 x^2$

$f(x) = \frac{b}{a} \sqrt{a^2-x^2}$

Question is: Can we prove/disprove the second algorithm? If disproved, under what conditions does the second algorithm work?

答:
We know that it cannot be correct to minimize the product $2x f(x)$
in order to minimize the area between the coordinate axes and the curve
$y = f(x)$ in the first quadrant, because the product $2x f(x)$ is zero when
$x=0$ and when $f(x) = 0$ and is positive everywhere else in the first quadrant.
Hence it has no minimum in the first quadrant except at those two points,
and we can easily show by example that the tangent enclosing the minimum area
is not always tangent to $y = f(x)$ where it intercepts a coordinate axis.

But let's see what method is correct.

Consider the curve $xy = k$ in the first quadrant.
The value at $x = a$ is $\frac ka$ and the slope is $-\frac k{a^2}.$
The tangent line therefore is $y - \frac ka = -\frac k{a^2}(x - a),$  which has $y$-intercept $\frac{2k}a$ and $x$-intercept $2a.$
The area bounded by this line and the two axes therefore is
$\frac12 \left(\frac{2k}a\right)(2a) = 2k,$
independent of the choice of tangent point $\left(a,\frac ka\right).$

Now consider a function $f$ on the interval $[0,x_0]$,
whose graph $y=f(x)$ connects the points $(0,y_0)$ and $(x_0,0)$,
and which is positive and has a decreasing derivative on the interval $(0,x_0).$
At any $x$ between $0$ and $x_0,$ if we take a line tangent to the graph of $f$
at $(x,f(x)),$ the graph of $f$ is below that tangent line at every other point.

The graph of $f$ between $(0,y_0)$ and $(x_0,0)$
intersects curves of the form $xy = k$ for many values of $k$;
let $m$ be the maximum such value of $k$.
Then the graph of $f$ will meet the graph of $xy = m$ at exactly one point,
$(x_m,f(x_m)),$
where the two graphs will be tangent to each other.
The area between the axes and the line tangent to the two graphs at that point
has area $2m.$

Now consider any other value of $x$ in the interval $[0,x_0]$
and take a line tangent to the graph of $f$ at $(x,f(x)).$
This tangent line passes above the point $(x_m,f(x_m)),$
so it also passes above the graph of $xy = m$ at that point.
This line is tangent to a graph of $xy = k$ for some $k > m$,
so the area between that line and the axes is greater than $2m.$

So the area under the tangent to the graph of $f$ is minimized at the point where the graph of $f$ is tangent to the graph of $xy = m$.
Moreover, $m$ is the maximum value of $xy$ at any point on the curve $y = f(x)$.

So the way to minimize the area between the axes and the tangent to the curve
$y = f(x)$ is to maximize (not minimize!) the product $xy = x f(x)$ for
$x$ between $0$ and $x_0.$

Naturally this also maximizes $2x f(x),$ and that maximum value also happens to be the minimum area between the axes and any tangent to the graph of $f.$

The reason it seems OK to try to minimize the product $2x f(x)$
is that the "obvious" way to minimize $2x f(x)$ in the first quadrant (if you forget how the minimum actually occurs) is to find the value of $x$ for which
$\frac{\mathrm d}{\mathrm dx} (2x f(x)) = 0,$
and this happens actually to be the value of $x$ that maximizes $2x f(x).$


问:
How to solve this simultaneous equation of $3$ variables.
I've stuck in this equation system. No clue how to start ?

$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$

Find the value of $x,y,z$ is in the form of $a,b$ and $c$.
I want to know steps of solution.

答:
This is a slightly interesting question. Let me try and provide some motivation for the ugly values that Americo calculated.

Let me assume that $a\neq b, b \neq c, c \neq a $. We'd deal with this issue separately.

Note that the first 2 equations give the equation on a plane. Hence, their intersection is a line, which we can calculate. The first 2 equations are equivalent to
$$ \begin{array} {llll}
&&(x-a) &+ &(y-b) &+ &(z-c) &= 0\\
&a&(x-a) &+b&(y-b) & +c&(z-c) & = 0 \\ \end{array}$$

Since

$$ \begin{pmatrix} 1\\1\\1\\ \end{pmatrix} \times  \begin{pmatrix} a\\b\\c\\ \end{pmatrix} = \begin{pmatrix} c-b\\a-c\\b-a\\ \end{pmatrix}, $$

we know that the line is given by $\frac{x-a}{c-b} = \frac{ y-b}{a-c} = \frac{z-c}{b-a} = k $. (This is valid because the denominator is never 0.) This is equivalent to $x=a + k(c-b), y = b+k(a-c) , z = c+k(b-a)$.

Next, we are interested in when this line intersects the ellipsoid $ax^2 + by^2 + cz^2 = a^3 + b^3 + z^3 $, which happens in at most 2 places. (Of course, one place is $k=0$, which leads to the solution $x=a, y=b, z=c$.) Using the equation of the line above, we get that

$$\begin{array}{l l l l}
ax^2 &= a^3 & + 2a^2k(c-b) &+k^2 a(c-b)^2 \\
by^2 &= b^3 & + 2b^2k(a-c) & + k^2 b(a-c)^2   \\
cz^2 &= c^3 & + 2c^2k(b-a) & + k^2 c(b-a)^2  \\
\end{array} $$

Summing up these three equations, and using the last equation given, we get that

$$-2k(a-b)(b-c)(c-a) + k^2 (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc) = 0. $$

One solution is $k=0$ as mentioned, and the other is

$$ k = \frac{ 2 (a-b)(b-c)(c-a) } { a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc} .$$

With this value of $k$, we get the answer that Americo calculated above.

Now, what happens when $a=b$ or $b=c$ or $c=a$? Interestingly, this forces there to be only 1 solution, namely $k=0$. You can also get this result by substituting $a=b$ into the value of $k$ and showing that the numerator is 0. The geometric interpretation is that the line becomes tangential to the ellipsoid.

Similarly, if $a=b=c$, then there is only 1 solution.


问:
prerequisites for calculus
What are the prerequisites for learning calculus?
I have experience in proof writing, trigonometric functions,permutations and combinations,naive set theory, mathematical induction,
and the abstract notion of functions. Should I start calculus or should I complete some other prerequisites?I want to learn calculus more by theorems and proofs than computation. Can I learn
calculus theorems without much experience in computational calculus? Is it mandatory to do computational calculus first?How many hours should i devote to it?Can I self study calculus?</p>

答: Calculus has many levels. The prerequisite is pre-calculus. Here is a free online pre-calculus textbook. If you are above this level, then you are ready for Calculus I.</p>

Title: prerequisites for calculus


问:
What is the Z-transformation of following discrete time signal of this?
I am learning how $Z$-Transforms work, but I have no encountered a situation in which the bound does not account for any signal.

Take for example the following discrete time signal:
\begin{cases}
x(-2)=4\\
x(-1)=0\\
x(0)=2\\
x(1)=4\\
x(2)=6\\
x(3)=8\\
x(4)=-2\\
\end{cases}
I need to find the $Z$-transform for $x(n)=0$ for $n$ from $-\infty$ to $-3$ and also for $n$ from $5$ to $\infty$.

My answer for this would be $0$ for both cases.

Is this correct?

答:
For finding the $Z$ transform $\sum_{n=-\infty}^{\infty}x(n)z^{-n}$ you need the value of a sequence $x(n)~\forall n\in\Bbb Z$. So those "two cases" are actually the description of the same sequence.

The $Z$-transform is given by $x(-2)z^2+x(-1)z^1+\cdot\cdot\cdot+x(4)z^{-4}=4z^2+2+4/z+6/z^2+8/z^3-2/z^4$ since the remaining $x(i)$'s are all zero.


问:
Integral and differential calculus, after a good (and mathematically correct) explanation of the two.
My son is beginning his final year of secondary school this year and one of the major components is calculus. He's always been better at visual methods and I'm struggling to remember how I learnt this stuff some forty years ago.

My (vague) understanding is that differential calculus has to do with the slope of an equation at a point and this is achieved by choosing two coordinates (say at x=4 and x=5 when you're wanting the slope at the former) and working out the slope of the line between them. This gives an approximation of the slope which is perfect for a linear equation but not so good for any other equation.

Then, as we slide the second point closer to 4 (say using 4 and 4.5), the approximation tends to come closer to the actual slope (wildly curvy lines like those with terms up to x99 may not do this immediately but they'll get there eventually).

Hence we're looking for the limit of dy/dx as dx -> 0.

Similarly, integral calculus is area under the line between two points, down to the x-axis. The idea there is to form a shape under the line (triangle on top of rectangle, which I'm reliably informed in the comments by the good folk here that it's a right trapezoid).

The area of that shape is an approximation since its top is a single straight line which may not match the actual shape of the equation. Just like in the differential calculus described above, that top line of the right trapezoid will approach the actual equation as the two x co-ordinates get closer.

So, in this case, we still use limits but the idea is to gradually increase the number of right trapezoids in the region. Hence the first step would be to add the areas of the two between x = 4 to 4.5 and x = 4.5 to 5. This would make the two top lines a better approximation of the actual equation line, hence make the resultant area calculation closer (again, restricted at first by any wild gyrations in the actual equation).

Then the sum of the area of the four right trapezoids bounded by x = 4 to 4.25, x = 4.25 to 4.5, x = 4.5 to 4.75, and x = 4.75 to 5. Hence this is the sum of all areas as the number of right trapezoids approaches infinity (or their width approaches zero, if you'd prefer).

Is that a correct understanding of the two types of calculus, in terms of visualisation? Can anyone point out any obvious (or not so obvious) flaws in my reasoning?

As I said, it's been a while since I've done this stuff. I was quite good with all sorts of math concepts even through Uni but lack of practice has dimmed the memories somewhat. He's doing Comp Sci which is handy given my background, but I'd like to be able to help out with other subjects as well, just not that :-)

答:
You are correct. Just some further explanation:

Regarding the derivative part, here is a more precise notation: we usually say the slope is $\frac{\Delta{y}}{\Delta{x}}$ as $\Delta{x} \to 0$. And, we usually use the notation $\frac{dy}{dx}$ to show the exact derivative.

Regarding the integral, your explanation and approach is again correct. "Riemann sum" is also another kind of approximation for calculating the area under a curve and I think it might be easier due to the fact that it uses rectangular to the best of my knowledge. And I think rectangular is an easier shape.

But, overall, you are good to go.


问:
Probability of getting two consecutive tails on the $n^{th}$ and $(n-1)^{th}$ toss for a biased coin.
The experiment goes like this.

The coin is tossed repeatedly and independently until two successive tails are obtained. And the probability of getting a head is $p$.
Let $q_n$ be the probability that we stop at $n^{th}$ coin toss.

What is the recursion for $q_n$?

答:
Let $q=1-p$. Clearly $q_1=0$ and $q_2=q^2$. Now consider $n>2$.  If the first toss yields heads, you are left with $n-1$ tosses. If the first toss yields tails, examine the second toss. This must yield heads in order for the first pair of consecutive tails to occur at time $n$. Thus for $n>2$,
$$q_n=p\cdot q_{n-1}+q\cdot p \cdot q_{n-2} \,.$$


问:
If $X \in L_{1}(\Omega,\mathcal{F},P)$ and $B \in \mathcal{F}$, how can I show that $X \mathbf{1}_{B} \in L_{1}(\Omega,\mathcal{F},P)$?
If $X \in L_{1}(\Omega,\mathcal{F},P)$ and $B \in \mathcal{F}$ then how can I show that: $X \mathbf{1}_{B} \in L_{1}(\Omega,\mathcal{F},P).$

I am thinking of :

$||X\mathbf{1}_B||_{1}=(\int |X\mathbf{1}_B|^1 dP)^1)$

Where
$$\mathbf{1}_{B} = \begin{cases}
1 &\text{if B $\in \Omega$} \\
0 &\text{otherwise}
\end{cases}
$$

答:
Let $Y=X\boldsymbol{1}_{B}$.
We need to show that $Y$ is (i) measurable and (ii) integrable.

Remark.
One way to establish (i) is to show that indicator functions on members of $\mathcal{F}$ are measurable and so too are products of measurable functions.
Below, I opt for a direct approach instead.

Proof of (i).
Let $A\subset\mathbb{R}$ be a Borel set.
If $A$ does not contain zero, then $Y^{-1}(A)=X^{-1}(A)$.
Otherwise, let $A^{\prime}=A\setminus\{0\}$ and note that
$$
Y^{-1}(A)=Y^{-1}(\{0\}\cup A^{\prime})=Y^{-1}(\{0\})\cup Y^{-1}(A^{\prime})=B^{c} \cup X^{-1}(\{0\}) \cup X^{-1}(A^\prime).
$$
In both cases, the final expression is a union of sets in $\mathcal{F}$, and hence the desired result follows by the closure of $\mathcal{F}$.

Proof of (ii).
Since $X$ is integrable,
$$
\mathbb{E}\left[\left|X\boldsymbol{1}_{B}\right|\right]=\mathbb{E}\left[\left|X\right|\boldsymbol{1}_{B}\right]\leq\mathbb{E}\left[\left|X\right|\right]<\infty.
$$


问:
Struggling with proof of obvious statement regarding an inequality
Given that $n_1(n_1-1)\leq n_2(n_2-1)$ for $n_1, n_2 \in \{1, ..., n\}$ is true. How can I prove that $n_1 \leq n_2$ must also be true?

I thought about saying if $n_1 \leq n_2$ than $n_1^2 \leq n_2^2$ and therefore $n_1^2-n_1 \leq n_2^2 - n_2$, but here I'm already using the solution.

Body:
Because $n_1+n_2-1 \ge1$ we have
$$0 \ge n_1(n_1-1)-n_2(n_2-1)=n_1^2-n_2^2 - (n_1-n_2)\\=(n_1-n_2)(n_1+n_2-1) \ge n_1 - n_2$$
Done. No need to complete square.


问:
Finding a general expression for a recursively defined sequence.
I have a sequence of numbers that I would like to find a general expression for. The sequence is defined as follows.
$$C_m=\frac{1}{2m-3}\left[C_{m-1}+\frac{m^2-9m+16}{9\cdot 2^{m+4}(m-4)!}\right], m\geq 6$$
With the the seed being $C_5=-\frac{1}{5760}$.

The strategy would be to calculate several terms to try to find any patterns and then use induction to try to prove some general expression for $C_m$.
I have calculated - by hand - the first 7 terms with no promising pattern in sight.

$$C_5=-\frac{1}{2^7\cdot3^2\cdot5}$$
$$C_6=-\frac{13}{2^{10}\cdot3^4\cdot5}$$
$$C_7=-\frac{1}{2^{11}\cdot3^4\cdot5}$$
$$C_8=\frac{1}{2^{12}\cdot3^4\cdot5}$$
$$C_9=\frac{1}{2^{10}\cdot3^5\cdot5^2}$$
$$C_{10}=\frac{19}{2^{17}\cdot3^5\cdot5^2}$$
$$C_{11}=\frac{29}{2^{18}\cdot3^5\cdot5^2\cdot7}$$

It it also tempting to use the recursive definition to try to get a sum representation for $C_m$, but this just gets ugly really quick.

I should also explain why I have reason to think that a general expression exists. The sequence is one sequence of a family of sequences that are in some sense connected to each other. All the previous sequences have had a provable general expression so thats why I suspect this would be possible for this sequence as well.

Does anyone see any patterns in the given terms or have any other strategies that could work?

答:
One way is to use Mathematica. Upon giving Mathematica the code

```<code>RSolve[{a[n] ==
   1/(2 n - 3) (a[n - 1] + (n^2 - 9 n + 16)/(9*2^(n + 4)*(n - 4)!)),
  a[5] == -1/5760}, a[n], n]
</code>```

It outputs (in plain mathematical language)
$$a_n=\frac{2^{-4-n}\left(5n^3-60n^2+199n-192\right)\Gamma\left(\frac{2n-1}{2}\right)}{405\sqrt{\pi}(n-3)!\prod_{k=1}^{n-1}\left(k-\frac{1}{2}\right)}$$

Body:
One way is to use Mathematica. Upon giving Mathematica the code

RSolve[{a[n] ==
   1/(2 n - 3) (a[n - 1] + (n^2 - 9 n + 16)/(9*2^(n + 4)*(n - 4)!)),
  a[5] == -1/5760}, a[n], n]

It outputs (in plain mathematical language)
$$a_n=\frac{2^{-4-n}\left(5n^3-60n^2+199n-192\right)\Gamma\left(\frac{2n-1}{2}\right)}{405\sqrt{\pi}(n-3)!\prod_{k=1}^{n-1}\left(k-\frac{1}{2}\right)}$$


问:
For every natural number n, prove that the vector $x = (n, n + 1, n (n + 1)) ∈ \Bbb R^3$ has as a norm a natural number, that is, $\|x\| \in \Bbb N$.

For every natural number $n$, prove that the vector $x = (n, n + 1, n (n + 1))$ ∈ $\mathbb{R}^3$ has as a norm a natural number, that is, $\|x\| \in \mathbb{N}$.

I made the internal product of the vector $x$ which gave $n^4 + 3n^3 + n^2$. Now I can't prove by induction. Does anyone have an idea?

答: Here's one way you can go about it.  You can assume that the expression inside the radial is a perfect square.  Then, $n^4+2n^3+3n^2+2n+1 = (n^2+an+b)(n^2+an+b)$.  Then, solve for $a$ and $b$.  If they are natural numbers, then that completes the proof.


问:
Positive function for complex variable
I was curious whether it makes sense to talk about positive functions of a complex variable. I.e, does it make sense to talk about $f(z) > 0$, if $z$ is complex, say? As an example, what if $f(z) = \cos z + i \sin z$?

答: Only if $f(z)$ is a real number. If it's a complex number, then no. See discussion here.

Title: Positive function for complex variable


问:
every integers from 1 to 121 can be written as 5 powers of 3
We have a two-pan balance and have 5 integer weights with which it is possible to weight
exactly all the weights integers from 1 to 121 Kg.The weights can be placed all on a plate but you can also put some in a dish and others with the goods to be weighted.

It's asked to find the 5 weights that give us this possibility. It also asks you to prove that the group of five weights is the only one that solves the problem.

Easily i found the 5 weights: 1 , 3 , 9 , 27 , 81

but i can' demonstrate that this group of five weights is the only one that solves the problem.

Can you help me ?

Thanks in advance !

答:
Your five weights add to 121. If there were another set of weights, they would have to sum to at least 121. You can obtain a contradiction. If the lower four weights are the same, you need 81 to hit 121.

Show inductively that with one weight you can get to 1, with two to 4, with three to 13, with four to 40, with five to 121.

It works as follows - there are three possibilities for each weight - in either pan or in neither. There is one possibility for the case where no weights are in either pan, which weighs zero. The remaining cases come in pairs (swap the weights between pans, they weigh the same weight). These are all distinct (the ternary expansion of a number tells you how to allocate the weights) so the maximum number for $n$ weights is $\cfrac {3^n-1}2$.

In response to comment below - now work as follows. The above shows that the set $\{1, 3, 9, 27, 81\}$  works, and that it is efficient (each weight can be weighed in just one way). Suppose we have another set $a \le b \le c \le d \le e$ which does the same. We can immediately replace $\le$ by $\lt$ throughout, because if $a=b$ our set is not efficient, and we can't weigh $121$ distinct weights.

We naturally need $a+b+c+d+e=121$ since this will be the maximum we can weigh. In order to weigh $120$ we need $a=1$. We can't have $b=2$ else we can weigh $1=a=b-a$ (this is inefficient - note, this is where it is useful to have the inductive step in place) but to weigh $118$ we then need $b=3$. This enables us to weigh up to 4. If $c<9$ then $c-a-b \le 4$, and we can weigh the same using just $a$ and $b$ - so we are not efficient (using the inductive step again). If $c>9$ we can't weigh $121-9=112$.


Title: For every natural number n, prove that the vector $x = (n, n + 1, n (n + 1)) ∈ \Bbb R^3$ has as a norm a natural number, that is, $\|x\| \in \Bbb N$.

Body:
For every natural number $n$, prove that the vector $x = (n, n + 1, n (n + 1))$ ∈ $\mathbb{R}^3$ has as a norm a natural number, that is, $\|x\| \in \mathbb{N}$.

I made the internal product of the vector $x$ which gave $n^4 + 3n^3 + n^2$. Now I can't prove by induction. Does anyone have an idea?

Body:
\begin{align*}
\sqrt{n^2+(n+1)^2+n^2(n+1)^2}&=\sqrt{(n^2+1)(n+1)^2+n^2}\\
&=\sqrt{(n^2+1+2n)(n+1)^2+n^2-2n(n+1)^2}\\
&=\sqrt{(n+1)^4-2n(n+1)^2+n^2}\\
&=\sqrt{[(n+1)^2-n]^2}\\
&=(n+1)^2-n\\
&=n^2+n+1.
\end{align*}
This is obviously a natural number for $n\in\mathbb{N}$.


问:
How to prove $ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{(-1)^{n+1}n} \,= \frac{\pi}{2e}$
How can I prove that

$$ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \,= \frac{\pi}{2e}$$

The result is given here (result 48).

The source: Prudnikov et al. 1986, p. 757 is given, however I have been unable to find the book online.

Some of my attempts include:

Multiplying known infinite products for $\frac{\pi}{2}$  and
$\frac{1}{e}$

$$ \frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdots} $$
$$ \frac{1}{e} =  \frac{1}{2} \left(\frac{3}{4}\right)^{\large{\frac{1}{4}}}\left(\frac{5\cdot7}{6\cdot8}\right)^{\large{\frac{1}{4}}} \cdots$$

as well as others given in https://arxiv.org/pdf/1005.2712.pdf

Trying to find the partial products from ${n=1} $ to $k $, Mathematica Gives:

$$\scriptsize{ \frac{\exp\left(2\left(-\zeta(-1,k+\frac{1}{2})^{(1,0)}+\zeta(-1,k+\frac{3}{2})^{(1,0)}+\zeta(-1,k+1)^{(1,0)}-\zeta(-1,k+2)^{(1,0)}\right)\right)\pi\, \Gamma(k+2)^2}{2\,\Gamma(k+\frac{3}{2})^2}} $$

However, I have been unsuccessful producing partial products on my own.

Taking the $\ln$ of the product to try to find the partial sums;

Trying to transform the following integral into infinite product.

$$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} = \frac{\pi}{2e} $$

From what I have seen in some papers, the partial products are found, then the Stirling's Approximation is used to find the limit.

Question: How can I prove the value of the given Infinite Product? Proofs or hints are both welcome.

Thank you kindly for your help and time.

答:
The stated product does not converge and therefor the answer is incorrect.

It's only valid if the final value is even.

$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \neq  \frac{\pi}{2e}$$

Most likely you ment:

$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} e^{2(-1)^n} =  \frac{\pi}{2e}$$

For the proof of even value's or the product above:
Consider:
$$\prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} e^{2(-1)^n} =  $$

$$\exp\bigg(\sum_{n=1}^{\infty}(-1)^{n+1} \big(n \ln(1+2/n)-2\big)\bigg)$$
We know the factorial (e.g.  stirlings formula, but people don't seem to write it this way, derived by writing it as sums, alternatively you can use divergent products, see: Divergent products.)

$$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(\sqrt{2 n\pi}\big) \exp\bigg(-\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{j}}\bigg) $$

$$\exp\bigg(\sum_{n=1}^{\infty}(-1)^{n+1} 2\ln\bigg(\frac{(1+2/n)^{n/2})}{e}\bigg)=$$

$$\exp\bigg(2\sum_{n=1}^{\infty} \ln\bigg(\frac{(1+1/n)^{n+1/2-1/2})}{e}\bigg)-\ln\bigg(\frac{(1+1/(n-1/2))^{n-1/2})}{e}\bigg)=$$

$$\exp\bigg(2\sum_{n=1}^{\infty} \ln\bigg(\frac{(1+1/n)^{n+1/2})}{e}\bigg)-\ln\bigg(\frac{(1+1/(n-1/2))^{n})}{e}\bigg)+\frac{-1}{2}\ln\bigg((1+1/n))\bigg)+\frac{-1}{2}\ln\bigg(1+1/(n-1/2)\bigg)=$$

$$\exp\bigg(2\sum_{n=1}^{\infty} \bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n+1)^j}-\frac{\zeta(-j)}{j(n)^j}\bigg)-\bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n+1/2)^j}-\frac{\zeta(-j)}{j(n-1/2)^j}\bigg)+\frac{-1}{2}\ln\bigg((1+1/n))\bigg)+\frac{-1}{2}\ln\bigg(1+1/(n-1/2)\bigg)=$$

if $\Delta_n^+=f(n)-f(n+1)$ and if f(n)= 0 as n goes to infty.

$$\exp\bigg(2\sum_{n=1}^{\infty} \sum_{j=1}^\infty \Delta_n^+\frac{\zeta(-j)}{j(n)^j}-\Delta_n^+\sum_{j=1}^\infty \frac{\zeta(-j)}{j(n-1/2)^j}- \Delta_n^+\frac{1}{2}\ln\big(\frac{n-1/2}{n}\big)\bigg)=$$

$$\exp 2 \bigg(\sum_{j=1}^\infty \frac{\zeta(-j)}{j}-\sum_{j=1}^\infty \frac{\zeta(-j)}{j(1/2)^j}+\frac{\ln(2)}{2}\bigg)=$$
$$\exp 2 \bigg(\ln\big(\frac{\sqrt{2\pi}}{e}\big)-\ln(\frac{\sqrt{e}}{2\sqrt{2}})+\frac{\ln(2)}{2}\bigg)=$$

$$\frac{2\pi}{e^2}\frac{e}{8} 2=\frac{\pi}{2e}$$


问:
Let $(2x^2 +3x+4)^{10} = \sum _{r=0}^{20} a_rx^r$. Then find $\frac{a_7}{a_{13}}$
I tried using the multinomial theorem where the n’th term is

$$\frac{10!}{r_1!r_2!r_3!}\times 2^{r_1}3^{r_2} 4^{r_3} x^{2r_1 + r_2}$$
($r$ is an integer)

Where $$r_1+r_2+r_3=10$$ and
For first part ie. $a_7$
$$2r_1+r_2 =7$$

But I don’t have enough equations to solve for all the $r$. Where am I going wrong?

答:
You are not going wrong. There are different ways to create $x^{7}$ term. For instance, you may choose $x^2$ once and choose $x$ five times (look at Case(II) below).

So, the goal is to find all possible ways to satisfy your equation. In other words:

$$Case(I): {r_{1} = 0}, \space {r_{2}=7} $$
$$Case(II): {r_{1} = 1}, \space {r_{2}=5} $$
$$Case(III): {r_{1} = 2}, \space {r_{2}=3} $$
$$Case(IV):  {r_{1} = 3}, \space {r_{2}=1} $$

Now, for each case, let's calculate the coefficient behind $x^7$:

$$Case(I): {10 \choose 0}{10 \choose 7}{3 \choose 3} \times {2^{0}} \times {3^{7}} \times {4^{3}} $$

Do the same things for other cases and then sum them together.


问:
Infinitely many pairs $(a,b)$ where if r is a root it's inverse is also a root.
Prove that there exist infinitely many pairs $(a,b)$ of integers such that $x^{2020}=ax+b$ has among its solutions two distinct real numbers whose product is 1.

I tried to construct $a$ and $b$ by taking two real roots whose product is one and the rest of the roots as complex roots of unity but still wasn't working for me.

Can this be extended for any $n$ instead of $2020$? Maybe once I have the proof then it will be clear.

答:
Notice that two roots of product $p=1$ and sum $s$ verify the equation $x^2-sx+p=0$

In particular $x^2=sx-1$

$x^3=x^2x=(sx-1)x=sx^2-x=s(sx-1)-x=(s^2-1)x-s$

You can prove by induction that $x^n=f_n(s)x+g_n(s)$ with $\begin{cases}f_{n+1}(s)=sf_n(s)+g_n(s)\\g_{n+1}(s)=-f_n(s)\end{cases}$

For for these two roots $x,y$ you get $\begin{cases}x^{2020}=f_{2020}(s)x+g_{2020}(s)=ax+b\\y^{2020}=f_{2020}(s)y+g_{2020}(s)=ay+b\end{cases}$

By difference and since $x-y\neq 0$ as the problem specifies explicitely two distinct real numbers

we get: $\begin{cases}a=f_{2020}(s)\\b=g_{2020}(s)\end{cases}$

As long as $s$ is an integer, it follows that $f_n(s)$ and $g_n(s)$ are also integers.

Therefore you can generate $(a,b)$ integers and $(x,y)$ real roots of product $1$ that answer the problem by starting from $x^2-sx+1=0$ and varying $s\in\mathbb Z$.

We still need to prove injectivity to get that they are in infinite quantity.

But notice that $f_n(s)=\dfrac{x^n-y^n}{x-y}$ and that $xy=1\implies |x|>1\text{ and }|y|<1\quad$

(or vice-versa, WLOG let assume $|x|>|y|$)

It ensues that $|y|^{2020}\ll 1$ and that $a=\operatorname{round}(\frac{x^{2020}}{x-y})\ge\operatorname{round}(\frac 12x^{2019})$

And since $x=\frac{s+\sqrt{s^2-4}}{2}\nearrow$ when $s\nearrow$ we have also $a\nearrow$, and strict monotonicity gives injectivity.

Edit:

Note that for $s$ large, $x\sim s$ and $a\sim s^{n-1}$, so you are pretty much ensured of injectivity for any exponent instead of $2020$ as long as your $s$ is large enough, which is not an issue since $s$ can be taken arbitrarily in $\mathbb Z$.


问:
Algebraic expression in its most simplified form
I am trying to simplify the algebraic expression:

$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$

I am having trouble though. My current thoughts are:

$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$

$$=\bigg(\dfrac{x(x-3)}{1(x-3)}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)}{1(x-3)}+\dfrac{2+6x}{(x-3)}\bigg)$$

$$=\bigg(\dfrac{x(x-3)+(-4)}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)+2+6x}{(x-3)}\bigg)$$

$$=\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x}$$

$$=\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} $$

$$\boxed{=\dfrac{-4(x-3)}{2(1+3x)} }$$ Which does not appear is not the answer. Am I close? Where exactly did I go wrong? I have tried this question multiple times.

Edit: Figured it out!

$\dfrac{x(x-3) - 4}{x(x - 3) + 2(1 + 3x)}\implies\dfrac{x^2-3x-4}{x^2+3x+2}\implies \dfrac{(x-4)(x+1)}{(x+2)(x+1)}$

$(x+1)$'s cancel leaving us with: $\boxed{\dfrac{x-4}{x+2}}$

答:
You did not distribute the term $(x - 3)$ in the denominator when you wrote:

$$\begin{align} & =\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x} \\ \\ & =\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} \end{align}$$

What would be correct is the following denominator:
$$\begin{align} & \quad\color{blue}{(x-3)}[x(x-3)+2+6x] \\ \\ & = \color{blue}{(x-3)}x(x-3)+\color{blue}{(x-3)}(2+6x)\end{align} $$

But note

$$\dfrac{x(x-3)+(-4)}{\color{blue}{\bf (x-3)}}\times \dfrac{\color{blue}{\bf(x-3)}}{x(x-3)+2+6x}$$

The highlighted terms cancel, leaving you:

$$\begin{align} & =\dfrac{x(x-3)+(-4)}{1}\times \dfrac{1}{x(x-3)+2+6x} \\ \\
& = \frac{x(x-3) - 4}{x(x - 3) + 2 + 6x} \\ \\
& = \dfrac{x^2-3x-4}{x^2+3x+2} \tag{$\diamondsuit$}
\end{align} $$

Now, all both the numerator and denominator of $\diamondsuit$ factor very nicely, and in fact, share a common factor, and hence, can be further simplified.

Recall: $$\frac{[b + c]a}{a[d+e]} = \frac{a[b+c]}{a[d+e]} = \frac{b+c}{d+e}$$

Or: $$\frac{a[b+c]}{a[d+e]} = \frac{ab+ac}{ad+ae}= \frac{b+c}{d+e}$$


问:
Error in official paper about bat algorithm?
I need help from a person that knows the so-called bat algorithm well. I have a problem with the third equation listed in the publication of Xin-She Yang from 2010. There it says that the velocity is updated adding:

(current_position - best_position) * random _number.

Why is it (current_position - best_position) and not (best_position - current_position)? I can't think of any case where the proposed variant is consistently better than the intuitively more appealing second variant. (best_position - current_position) can be interpreted as making a step with random step size into the direction of the currently best solution. The variant from the paper is the exact opposite, i.e. making steps into the direction that lead into the exact opposite direction than the one to the best.

In another paper that I found, they did switch both variables, i.e. they are using the second more intuitive variant. Did they understand it wrong, or did the author make an mistake?

Can anyone explain this to me, please? Thanks soo much in Advance!

答:
Note

First of all, I am not an expert on this area, but, as an engineering student, I've faced the same problem as you have, and also have spent hours thinking about it.

Answer

I agree with you and think that Yang made a mistake in his original article. In my opinion, the correct form would be:
$$
\mathbf{v}^t_i = \mathbf{v}^{t-1}_i + (\mathbf{x}_∗ - \mathbf{x}^{t-1}_i)f_i \tag{1}\label{modif}
$$
instead of the original:
$$
\mathbf{v}^t_i = \mathbf{v}^{t-1}_i + (\mathbf{x}^t_i - \mathbf{x}_∗)f_i \tag{2}\label{orig}
$$
because the behavior of bats would be to move away from the best solution using $\eqref{orig}$ and to approach using $\eqref{modif}$

Why

Using $\mathbf{x}^t_i$ is wrong, for sure. Because it would lead to an impossible situation of interdependence (since you calculate $\mathbf{x}^t_i$ from $\mathbf{v}^t_i$). Yang even corrects this error on page 8 of his book:

Yang, Xin-She (ed.), Nature-inspired algorithms and applied optimization, Studies in Computational Intelligence 744. Cham: Springer (ISBN 978-3-319-67668-5/hbk; 978-3-319-67669-2/ebook). xi, 330 p. (2018). ZBL1375.68026.

Now, about the inversion of positions, let's say: $\quad \mathbf{x}_∗ = \mathbf{x}_{best} \quad $ and $ \quad \mathbf{x}^{t-1}_i = \mathbf{x}_{current}$

As you have already said Yang uses: $\quad \mathbf{v}^t_i = \mathbf{v}^{t-1}_i + (\mathbf{x}_{current} - \mathbf{x}_{best})f_i$

and in his article it is said: $\quad  f_i = f_{min} + (f_{max} - f_{min})\beta$

However, if you pay attention on page 146 of other Yang's book (in which he implements the code):

Yang, Xin-She, Nature-inspired optimization algorithms, Amsterdam: Elsevier (ISBN 978-0-12-416743-8/hbk; 978-0-12-416745-2/ebook). xii, 263 p. (2014). ZBL1291.90005.

It is said:

$\texttt{Q(i)=Qmin+(Qmin-Qmax)*rand;}$

$\texttt{v(i,:)=v(i,:)+(Sol(i,:)-best)*Q(i);}$

and he also uses $\texttt{Qmin=0}$.

See? $\texttt{Qmin}$ and $\texttt{Qmax}$ are inverted with respect to the definition!! So he is making $(f_{min} - f_{max})$ instead of making $(f_{max} - f_{min})$ as written in the article.

As $f_{min}=0$, that inversion will invert the sign of $f_i$, and then
$$\mathbf{v}^t_i = \mathbf{v}^{t-1}_i + (\mathbf{x}_{current} - \mathbf{x}_{best})f_i$$
will become
$$\mathbf{v}^t_i = \mathbf{v}^{t-1}_i + (\mathbf{x}_{current} - \mathbf{x}_{best})(-f_i)$$
or, rearranging:
$$\mathbf{v}^t_i = \mathbf{v}^{t-1}_i + (\mathbf{x}_{best} - \mathbf{x}_{current})f_i$$

I think he fixed a mistake with another one, but... Who am I to say that?

Note: You can find the mentioned code (a very similar one) posted by Yang (I suppose) at MathWorks File Exchange

Conclusion

I must confess that I have no idea if this mistake actually helps the BA's search, once it makes the bats to spread out from the best solution, but I agree with you that to converge is much more intuitive.

P.S. 1

I am almost sure that there are other mistakes in this paper like doing $rand > r_i$ instead of $rand < r_i$ at the local search part, but I don’t feel too confident to give you a certainty yet.

P.S. 2

I have used Bat Algorithm once with the "corrected" equation $\eqref{modif}$ once and it went well. I haven't tested the original version $\eqref{orig}$ though.


答:
The polynomial $x^{2020} - ax - b$ has a pair of roots multiplying to $1$ if and only if it is divisible by a quadratic of the form $x^2 - px + 1$. To enforce these roots be real and distinct, we further insist on discriminant of this quadratic being positive, i.e. $p^2 - 4 > 0$.

So, we are assuming that $x$ is a real number satisfying $x^2 = px - 1$ where $p^2 > 4$ and we wish to have $x^{2020} = ax + b$. Let us do some investigation as to the powers of $x$. The fact that $x^2$ can be expressed as $px - 1$ allows any power of $x$ to be expressed in terms of $x$ and $1$. Let's say that we have sequences $c_n, d_n$ such that
$$x^n = c_n x + d_n$$
Then
$$x^{n+1} = x(c_n x + d_n) = c_n x^2 + d_n x = c_n (px - 1) + d_n x = (pc_n + d_n)x - c_n.$$
If we restrict our attention to integers $p \ge 3$, then $x^n$ can be expressed as an integer combination of $x$ and $1$. We will get $x^{2020} = c_{2020}x + d_{2020}$, where $c_{2020}, d_{2020} \in \Bbb{Z}$. Plus, $x^2 - px + 1$ will divide the polynomial, and hence there will be two distinct real roots whose product is $1$. So, simply by choosing $p$ to be one of the infinitely many integers greater than or equal to $3$, we obtain corresponding $a = c_{2020}$ and $b = d_{2020}$ so that the roots of $x^2 - px + 1$ are roots of $x^{2020} - ax - b$.

There's a slight potential kink here, which is that different values of $p$ may not produce different values of $a$ and $b$. However, if distinct $p_1, \ldots, p_n \ge 3$ produce the same $a, b$ then that says both $x^2 - p_ix + 1$ divides the polynomial $x^{2020} - ax - b$ for all $i = 1, \ldots, n$. Since each $p_i$ is distinct, and produces a quadratic with completely different positive roots, the quadratics $x^2 - p_ix + 1$ and $x^2 - p_j x + 1$ share no common factors, and so we can conclude
$$\prod_{i=1}^n (x^2 - p_ix + 1) \mid x^{2020} - ax - b.$$
But obviously this poses a problem if $n > 1010$. Therefore, while it is possible for distinct $p \ge 3$ to produce the same $a, b$, it still must produce infinitely many distinct $a, b$ pairs.

Title: Infinitely many pairs $(a,b)$ where if r is a root it's inverse is also a root.

Body:
Prove that there exist infinitely many pairs $(a,b)$ of integers such that $x^{2020}=ax+b$ has among its solutions two distinct real numbers whose product is 1.

I tried to construct $a$ and $b$ by taking two real roots whose product is one and the rest of the roots as complex roots of unity but still wasn't working for me.

Can this be extended for any $n$ instead of $2020$? Maybe once I have the proof then it will be clear.


Title: Let $(2x^2 +3x+4)^{10} = \sum _{r=0}^{20} a_rx^r$. Then find $\frac{a_7}{a_{13}}$

Body:
I tried using the multinomial theorem where the n’th term is

$$\frac{10!}{r_1!r_2!r_3!}\times 2^{r_1}3^{r_2} 4^{r_3} x^{2r_1 + r_2}$$
($r$ is an integer)

Where $$r_1+r_2+r_3=10$$ and
For first part ie. $a_7$
$$2r_1+r_2 =7$$

But I don’t have enough equations to solve for all the $r$. Where am I going wrong?

Body:
Method without finding coefficient

given$$\color{blue}{(2x^2 +3x+4)^{10}} = \sum _{r=0}^{20} a_rx^r$$ replace $x\to \frac{2}{x}$  $$2^{10} \frac{{(2x^2+3x+4)}^{10}}{x^{20}}=\sum_{r=0}^{20} \frac{2^ra_r}{x^r}$$  $$\iff 2^{10}\color{blue}{{(2x^2+3x+4)}^{10}}=\sum_{r=0}^{20} 2^ra_r x^{20-r}$$ Now as this is an identity the coefficient of $x^7$ in the LHS should be same as in RHS $$2^{10}a_7=2^{13}a_{13}\implies \frac{a_7}{a_{13}}=?$$


问:
Find the Maximum Trigonometric polynomial coefficient $A_{k}$

Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.

I don't know if this question has been studied

If  $n=2$ it is easy to solve it.

答:
My second answer:

Edit 2021/03/11

According to NTstrucker@AoPS's result (https://artofproblemsolving.com/community/c6h2477208), I give the following conjecture.

Conjecture 2: $\max A_1 = 2\cos \frac{\pi}{n+2}$.

When $n = 2$, $\max A_1 = \sqrt{2} = 2 \cos \frac{\pi}{4}$.

When $n = 3$, $\max A_1 = \frac{\sqrt{5} + 1}{2} = 2\cos \frac{\pi}{5}$.

When $n = 4$, $\max A_1 = \sqrt{3} = 2\cos \frac{\pi}{6}$.

When $n = 5$, $\max A_1 = 2\cos \frac{\pi}{7}$.

When $n = 6$, $\max A_1 \approx 1.8478$ (I only get numerical result) should be $2\cos \frac{\pi}{8}
\approx 1.847759065$.

When $n = 7$, $\max A_1 \approx 1.8794$ (I only get numerical result) should be $2\cos \frac{\pi}{9}
\approx 1.879385242$.

When $n = 8$, $\max A_1 \approx 1.9021$ (I only get numerical result) should be $2\cos \frac{\pi}{10}
\approx 1.902113033$.

When $n = 9$, $\max A_1 \approx 1.9190$ (I only get numerical result) should be $2\cos \frac{\pi}{11}
\approx 1.918985947$.

$\phantom{2}$

Let us prove that $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.

We need Theorem 1. The proof is given at the end.

Theorem 1: Let $n\ge 2$ be a given integer. Let $A_i, i=1, 2, \cdots, n$ be given real numbers such
that $A_1\cos x + A_2 \cos 2x + \cdots + A_n \cos n x \le 1, \ \forall x\in \mathbb{R}$. Then
$A_m \le 1$ for $m = \lfloor \frac{n}{2}\rfloor + 1,  \cdots, n$.

By Theorem 1, we have immediately $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$
(simply let $A_m = 1$ and $A_k = 0$ for all $k\ne m$
and the condition $A_m \cos m x \le 1, \forall x\in \mathbb{R}$ is clearly satisfied.).

$\phantom{2}$

Proof of Theorem 1: First, we have Fact 1. The proof of Fact 1 is easy and thus omitted.

Fact 1: Let $M\ge 2$ be an integer. Let $K$ be a positive integer. Then
$$\frac{1}{M} + \frac{\cos K\pi}{M}\cdot \frac{1 + (-1)^M}{2}
+ \frac{2}{M}\sum_{j=1}^{\lfloor \frac{M+1}{2}\rfloor - 1} \cos \frac{2jK\pi}{M} = \left\{\begin{array}{cc}
                                                              0 & M \nmid K \\[5pt]
                                                              1 & M \mid K.
                                                            \end{array}
\right.$$

Second, we have Fact 2.

Fact 2: Let $n\ge 2$ be a given integer. Let $m$ be an integer with $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.
Let $A_i, i= 1, 2, \cdots, n$ be given real numbers.
Let $f(x) = \sum_{k=1}^n A_k\cos (kx)$. Then
$$\frac{f(0)}{m} + \frac{f(\pi)}{m}\cdot \frac{1 + (-1)^m}{2}
+ \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} f\left(\frac{2j\pi}{m}\right) = A_m. $$
(Proof of Fact 2: It suffices to prove that
$$\frac{1}{m} + \frac{\cos k\pi}{m}\cdot \frac{1 + (-1)^m}{2}
+ \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} \cos \frac{2j k \pi}{m} = \left\{\begin{array}{cc}
                                                              0 & k \ne m \\[5pt]
                                                              1 & k = m.
                                                            \end{array}
\right. $$
This is true by Fact 1.)

By Fact 2, Theorem 1 is proved. We are done.


问:
Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.
Let $F$ be a field and let $f(x) \in F[x]$. Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the
quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.

$\textbf{My Attempt:}$

prove "$\implies$":

Assume $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$.

Then, $(f(x) + \langle x^2 − x \rangle) \cdot (f(x) + \langle x^2 − x \rangle) = f(x)f(x) + \langle x^2 − x \rangle = 1 + \langle x^2 − x \rangle$.

Then, $f(x)f(x) = 1$ $\implies$ $f(x) \neq 0$.

Then, neither $f(0)$ nor $f(1)$ can be equal to $0$. So does $f(0)f(1)$. So, $f(0)f(1) \neq 0$.

prove "$\Longleftarrow$":

Assume $f(0)f(1) \neq 0$. Then, $f(0) \neq 0$ and $f(1) \neq 0$.

Which I am getting stucked on how to related $f(0) \neq 0$ and $f(1) \neq 0$ to proving that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring.

Also, I don't know if the prove of "$\implies$" is correct as well.

答:
We first unravel the definition.  Let us write $[f(x)]$ for the image of $f(x)$ in the quotient ring.  Then $[f(x)]$ is a unit if and only if there exists a $g(x)$ such that $[f(x)][g(x)]=[f(x)g(x)]=[1]$.  So $[f(x)]$ is a unit if and only if there exist $g(x),h(x)$ such that $f(x)g(x)=1+h(x)(x)(x-1)$.

In one direction, evaluating this expression at $x=0$ and $x=1$ shows that $f(1)g(1)=1$ and $f(0)g(0)=1$, so $f$ is non-zero at both $0$ and $1$.  In the other direct, assume that $f(0)=a, f(1)=b$, $a,b\neq 0$.  Then let $g(x)$ be any polynomial such that $g(0)=1/a, g(1)=1/b$.  Then $f(x)g(x)-1$ evaluates to $0$ at both $0$ and $1$, so it is divisible by bother $x$ an $(x-1)$.  Since those are relatively prime polynomials, $f(x)g(x)-1$ is divisibly be $x^2-x$.  Thus, $[f(x)][g(x)]=[1]$, so $f$ is a unit.