问:
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$
Where do I start to solve a equation for x like the one below?

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

After squaring it, it's too complicated; but there's nothing to factor or to expand?

Ideas?

答:
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

$$\implies \sqrt{7x-4}+\sqrt{4x-2}=\sqrt{4x-1}+\sqrt{7x-5} $$

Squaring we get, $$7x-4+4x-2+2\sqrt{(7x-4)(4x-2)}=4x-1+7x-5+2\sqrt{(4x-1)(7x-5)}$$

$$\implies  \sqrt{(7x-4)(4x-2)}= \sqrt{(4x-1)(7x-5)}$$

$$\text{Squaring we get, } (7x-4)(4x-2)=(4x-1)(7x-5)$$

$$\text{ On simplification we get, 8-(16+14)x=5-x(20+7)} $$

$$\implies x=1,\text{  which evidently satisfies the given equation.   }$$

Generalization:

If $$\sqrt{ax+b}-\sqrt{ax+c}=\sqrt{dx+e}-\sqrt{dx+f} \text{ where }b-c=e-f$$

$$\implies \sqrt{ax+b}+\sqrt{dx+f}=\sqrt{dx+e}+\sqrt{ax+c} $$

Squaring we get, $$ax+b+dx+f+2\sqrt{(ax+b)(dx+f)}=dx+e+ax+c+2\sqrt{(dx+e)(ax+c)} $$

$$\implies \sqrt{(ax+b)(dx+f)}=\sqrt{(dx+e)(ax+c)}\text{ as } b-c=e-f\iff b+f=c+e $$

$$\text{Squaring we get, } (ax+b)(dx+f)=(dx+e)(ax+c)$$

$$\implies x(bd+af-cd-ae)=ec-bf $$


问:
What does Euc. VI mean?

You can just tell me what does this Euc. VI mean. I tried searching Euc.VI and also Euc VI postulate. I got nothing. Please tell me what does this mean or how is AB/ab = OA /Oa ? P

答: $\triangle Oab$ and $\triangle OAB$ are both isosceles and they share the same angle $\angle AOB$; Therefore, they are similar. Hence, $\frac{AB}{ab}=\frac{OA}{Oa}$.

ContentLicense: CC BY-SA 4.0

Title: What does Euc. VI mean?


问:
Basic and free variables of a matrix
Suppose I have:

$$  \begin{bmatrix}
 1& 6 &  0 & 3 & 0 & 0\\
 0& 0 &  1 & -4 & 0 & 5\\
 0& 0 &  0 & 0 & 1 & 7
\end{bmatrix} $$

Do the basic variables have to be $x_1, x_3, x_5$? Can they be $x_2, x_4, x_6$ instead and you write your general solution in terms of $x_1, x_3, x_5$?

答:
You have a system (if it's totally reduced with Gauss elimination method) with $6$ variables to determine and three equations, so the vector space generated by the solutions of the system has dimension equal to $6-3=3$.

This means that the system has three free parameters. You usually define the last variable, in your case $x_6$, as a free parameter in $\mathbb K$ in order to obtain all the other variables.

If we look at the system you wrote we have:
$$x_5=-7x_6=-7t$$
$$x_4=s\implies x_3=4x_4-5x_6=4s-5t$$
$$x_2=u\implies x_1=-3x_4-6x_2=-3s-6u$$
This is a quite rapid way to extract the variables but in general you can arbitarily define the free parameters.

Title: Basic and free variables of a matrix


问:
Intuitive explanation of Duhamel's principle
This is with regards to the first section of Wikipedia article Duhamel's principle (revision from July 2012). I want to see if I am understanding this.

Basically the inhomogeneous equation says that heat is being added at a rate of $f(x,t)$, so at each point $x$, $f(x,t)\,dt$ of heat is being added to the system.

And you can think of this as a set of systems $u_t(x,t)-\Delta u(x,t)=0$, $u(x,t_0)=f(x,t_0)$ for all $t_0\in(0,\infty)$, which says the initial heat distribution is $f(x,t)$, and no heat is being added, and if we integrate these solutions over $t_0$ we get the inhomogeneous solution...

It's the third paragraph I am a bit uncomfortable with. Can anyone offer some help?

答: The intuition behind Duhamels principle is, in fact, quite simple. The input function f(x,t) determines,  according to Newtons law, the velocity of the wave u(x.t)at all times.  Instead of regarding f(x.t) as an input function, it might as well be  regarded as a new “initial velocity function” f(x,s) of the wave, restarting repeatedly at all time s less than or equal to t. f(x,s) is, thus, regarded as a  flow of velocities. To find the resulting velocity at any time t, you just add up (integrate) all these (repeting initial velocities) from s= zero to s=t.  Thus, the time from zero to t is divided into infinitesimal small time-slice windows delta s, in each window the new velocity is calculated, and then f(s,t) is integrated from  to zero to t. This may also be regarded as a convolution: https://www.youtube.com/watch?v=acAw5WGtzuk&ab_channel=PhysicsVideosbyEugeneKhutoryansky</p>


问:
Flawed induction proof
I read an example inductive proof as follows:

In the source document, it says this is flawed and has some subtlety that hides the flaw. I am not able to identify the same. What's hidden here?

答: It's not very subtle: the equality does not hold when $n=1$.

Title: Flawed induction proof


问:
I need help with relations
Let $S$ be the power set $P({1,2,...,10})$; that is, $S$ is the set of all subsets of $\{1,2,\dots,10\}$. define the relation $\mathcal R$ on $S$ by:

For all subsets $A,B$ of $\{1,2,\dots,10\}$, $A\mathcal RB$ if and only if $A\cup B$ has exactly $3$ elements.

a) Is $\mathcal R$ reflexive? Symmetric? Transitive?

b) Find and simplify the number of subsets $A$ is contained in $\{1,2,\dots,10\}$ so that $A \mathcal R\{1,2,7\}$.

c) Find and simplify the number of subsets $A$ is contained in $\{1,2,\dots,10\}$ so that $A\mathcal R\varnothing$.

Any help would be appreciated seeing that I don't know where to start

答:
HINT: You can quite easily determine whether $R$ is reflexive, symmetric, or transitive directly from the definition.

Reflexive: If $A\subseteq\{1,\dots,10\}$, is it always true that $A\mathbin{R}A$? In other words, is it always true that $|A\cup A|=3$?

Symmetric: If $A,B\subseteq\{1,\dots,10\}$, and $A\mathbin{R}B$, is it always true that $B\mathbin{R}A$? In other words, if $|A\cup B|=3$, is it always true that $|B\cup A|=3$?

Transitive: If $A,B,C\subseteq\{1,\dots,10\}$, $A\mathbin{R}B$, and $B\mathbin{R}C$, is it always true that $A\mathbin{R}C$? In other words, if $|A\cup B|=3$, and $|B\cup C|=3$, must it be the case that $|A\cup C|=3$? What if $A=C=\varnothing$, for instance?

For the other parts of the question you should begin by trying to understand just how $R$ works. Consider the set $A=\{1,3,7,8\}$; is there any set $B\subseteq\{1,\dots,10\}$ such that $A\mathbin{R}B$? No: no matter what set you choose for $B$, $A\cup B$ is going to have at least $4$ elements, namely, the $4$ elements of $A$. Now what if $A=\{1,3,7\}$? If $B$ contains at least one element that is not already in $A$, $A\cup B$ is going to contain at least $4$ elements. Thus, the only hope of getting $A\mathbin{R}B$ is to have $B\subseteq A$, and that always works: in this case $A\mathbin{R}B$ if and only if $B\subseteq A$. Now what if $A=\{1,3\}$? In order to have $A\mathbin{B}$, $B$ must contain exactly one element that is not in $A$: $B$ must be of the form $X\cup\{b\}$, where $X\subseteq A$, and $b\in\{1,\dots,10\}\setminus A$.

It shouldn’t be too hard to extrapolate from these examples to be able to say for any $A\subseteq\{1,\dots,10\}$ exactly which $B\subseteq\{1,\dots,10\}$ satisfy $A\mathbin{R}B$, and once you’ve done that, it’s not very hard to count those sets $B$.

ContentLicense: CC BY-SA 3.0

Title: I need help with relations


问:
Coordinates of a plane tiling
Consider the plane tiling shown below. All polygons are regular with side $1$. The tiling is centered at the origin.
In terms of complex numbers, the vertices are in $\mathbb Z[\omega,i]$, where $\omega$ is a primitive $6$-th root of unit, and so can be written uniquely as an integer linear combination of $1,\omega,i,\omega i$. However, not all  such linear combinations correspond to vertices of the tiling. In fact, $\mathbb Z[\omega,i]$ is dense in the complex plane.

Do the symmetries of the tiling induce an algebraic structure on the set of vertices?

Can we characterize the set of vertices inside $\mathbb Z[\omega,i]$ ?

Here is some data to get you started:

The central hexagon has vertices $1, \omega, -1+\omega, -1, -\omega, 1-\omega$.

The square vertically adjacent to the central hexagon has vertices $\omega, \omega+i, -1+\omega+i,-1+\omega$.

Multiply these repeatedly by $\omega$ to find the vertices of the other squares adjacent to the central hexagon. Use that $\omega^2=-1+\omega$.

All vertices in the tiling are translations of these central vertices by integer multiples of $-1+2\omega+2i$ and $1+\omega+2i-2\omega i$.

Thus, a not very satisfactory answer to my first question is that the set of vertices is a union of cosets. But I'd like a nice answer to the second question in terms of exactly which integer linear combination of $1,\omega,i,\omega i$ occur.

答:
My answer to the first question remains: the set of vertices is a union of cosets, as illustrated below (the coset representatives are shown in blue).

This is the basis of an integer representation for periodic tilings of the plane by regular polygons, as explained in a paper that has just appeared online. A preprint is available here. Explore a large collection of tilings here.

The vertices in all such tilings can be given integer coordinates in $\mathbb Z[\omega]$, where $\omega$ is the principal 12th root of unity. These coordinates form a  $(2+n)\times4$ integer matrix, where $n$ is the number of cosets in the tiling and the $2$ comes from representing two translation vectors that define a translation cell for the tiling (the red parallelogram in the picture). The cosets are with respect to the discrete subgroup generated by the translation vectors.

The question of which integer matrices are possible appears to be delicate and is related to the enumeration of all possible tilings. Essentially, the area of the parallelogram has to be an integer combination of the area of a square and of the regular triangle.

Title: Coordinates of a plane tiling


问:
Definition of pushforward and pullback for modules
Let $R$ be a commutative ring with unit and $\sigma:R\longrightarrow R$ an endomorphism of the ring $R$. Let $M$ be an $R$-module.

What are the definitions of $\sigma^*M$ and $\sigma_* M$?

How do they behave under tensor product?

Many thanks!

答:
Given a morphism of commutative unital rings $\sigma : R\to S$, if you take the pullback $\sigma^* N$ of an $S$-module $N$ to be $N$ considered as an $R$-module via $\sigma, $ then the pushforward $\sigma_* M$ of an $R$-module $M$ should be $M\otimes_R S.$ Then you get the adjunction
$$\operatorname{Hom}_R(M,\sigma^*N) \cong \operatorname{Hom}_S(\sigma_* M, N).$$

However, this notation is a little at odds with the geometric version: let $f : X\to Y$ be a morphism of schemes, and let $\mathcal{F}$ be a sheaf oof $\mathcal{O}_X$-modules and $\mathcal{G}$ be a sheaf of $\mathcal{O}_Y$-modules. Then we have $$\operatorname{Hom}_Y(\mathcal{G},f_*\mathcal{F})\cong\operatorname{Hom}_X(f^*\mathcal{G},\mathcal{F}).$$

As such, given a morphism of commutative unital rings $\sigma : R\to S,$ I would prefer to define $\sigma_* N$ for an $S$-module $N$ to be $N$ considered as an $R$-module via $\sigma, $ and $\sigma^* M$ for an $R$-module $M$ to be $M\otimes_R S.$ Then we have
$$\operatorname{Hom}_R(M,\sigma_*N) \cong \operatorname{Hom}_S(\sigma^* M, N).$$
If you let $f$ denote the morphism of affine schemes $f : X = \operatorname{Spec}S\to Y = \operatorname{Spec}R$ corresponding to $\sigma,$ and let $\widetilde{M}$ denote the sheaf of $\mathcal{O}_Y$-modules corresponding to an $R$-module $M$ (and similarly for an $S$-module $N$), then we have $\widetilde{\sigma^* M}\cong f^*\widetilde{M}$ and $\widetilde{\sigma_* N}\cong f_*\widetilde{N}.$


问:
Prove that the midpoint of $AH$ lies on the radical axis of $(REC)$ and $(QFB)$.

The incircle of $\triangle ABC$, centered at $I$, touches $AC$ and $AB$ at $E$ and $F$, respectively. Let $H$ be the foot of the altitude from $A$ and let $R = CI \cap AH$ and $Q = BI \cap AH.$ Prove that the midpoint of $AH$ lies on the radical axis of $(REC)$ and $(QFB)$.

The source is Peru TST.

Progress:

Let $X:=$ midpoint of $AH.$ $J,G$ as point of intersection of $(REC)$ and $(QFB).$ Let the perpendicular to $BC$ from $B$ and $C$ intersect $(QFB)$ and $(REC)$ at $K$ and $L.$

Claim: $G$ lies on $BC$

Proof: For this we will define $G':=(QFB)\cap BC.$ And we will show that $G'\in (RFC).$ For this note that since $BQ$ is angle bisector, we get $FQ=QG'.$ ( using incentre-excentre lemma). But $FQ=QD$ by angle bisector property. So $FQ=QG'=QD.$ So $G'$ and $D$ are reflection of each other wrt altitude. Hence $RG'=RD.$ But $RD=RE.$ So $RG'=RE.$ So by converse of incentre -excentre lemma, we get $G=G'$.

Now note that $K-J-L$ are collinear. This is because by cyclicity, $\angle GBK=90=\angle GJK$ and $GCL=90=\angle GJL.$ So $KJL\perp GJ.$

So enough to show $XJ\perp KJL.$ I also thought of introducing $A-$ excentre, because $X-D-I_A$ are collinear( well known).

Any elementary solutions ( not using trig, vectors, complex or any calculative stuff)?

答:
Let's consider point $P$ such that $APDH$ is a parallelogram (actually, a rectangle since $AH\perp BC$). It's clear that $GH=HD$ (as you've mentioned in the question, we have $RG=RD=RE$), so $APHG$ is a parallelogram and $X$ is its center (recall that $X$ is a midpoint of $AH$).

Now, note that $G$ is one of the intersection points of the circles $(REC)$ and $(QFB)$, so it's sufficient to prove that $GX$ is the radical axis of these circles, or to prove that $P$ lies on this radical axis.

Firstly, note that $P$ lies on the circle $AFE$. Indeed, the incenter $I$ lies on $PD$ (since $ID$ and $PD$ are perpendicular to $BC$), so $\angle IPD=90^{\circ}$. Besides that, $\angle IEA=\angle IFA=90^{\circ}$, so $E$, $F$ and $P$ lie on the circle with diameter $AI$

Secondly, the points $P$, $G$ and $J$ are collinear. In irder to prove that, just recall that $J\in(AFE)$ (it's well-known: circles $(AFE)$, $(BFG)$ and $(CGE)$ has common point). Hence,
$$
\angle (PJ, JF)=\angle (PA, AF)=\angle (CB, BA)=\angle (GB, BF)=\angle (GJ, JF).
$$
Therefore, $P$, $G$ and $J$ are collinear and $X$ (the midpoint of $GP$) lies on the radical axis of $(REC)$ and $(QFB)$, as desired.


问:
Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?
In this answer, I give an elementary solution of the Diophantine equation $$y^2=3x^4+3x^2+1.$$

In this post, the related equation $$y^2=3x^4-3x^2+1 \tag{$\star$}$$ is solved (in two different answers!) using elliptic curve theory. Is there an elementary method of proving that $x=1$ is the only positive rational solution to ($\star$)? In particular, I’d love to see a method using descent.

答:
An elementary (and general) method

This is a little-known method which will solve many equations of this type. For this example it will show that the only positive rational solution is $p=1$. We start by letting $p=\frac{y}{x}$, where $x$ and $y$ are coprime integers. Thus we seek positive integer solutions of $$ x^4-3x^2y^2+3y^4=z^2.$$ This is one of  a family of equations which can be dealt with together. (Apologies for using the letters for variables which I am used to when solving these types of equation.)

Theorem

The only positive integer solutions of any equation of either of the forms
$$ Ax^4+6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ ax^4-3x^2y^2+cy^4=z^2,ac=3\tag{2}$$
have $x=y=1$.

Proof

First note that, for either equation, we can suppose that $x,y,z$ are pairwise coprime  since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation can occur.

Also note that precisely one of $A$ and $C$ is divisible by $3$. Without loss of generality we can suppose that $3$ is a factor of $C$ and that $A\in \{\pm 1\}.$ Then $z^2\equiv A\pmod 3$ and so $A=1$. Similarly, we can suppose $a=1$.

An equation of form (1)

$x^4+6x^2y^2-3y^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2+3y^2-z}{2}\right )\left (\frac{x^2+3y^2+z}{2}\right)=3y^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $y$, a contradiction.

Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=3$ and $y=uv$, with $u$ and $v$ coprime. Then $$au^4+cv^4=x^2+3y^2=x^2+3u^2v^2.$$ Therefore $au^4-3u^2v^2+cv^4=x^2$, $ac=3$, an equation of form (2).

It is important to note that the mapping $(x,y,z)\rightarrow (u,v,w)$ is invertible. Only one solution set can map to $(u,v,w)$ by this process.

An equation of  form (2)

Let $u,v,x$ be a pairwise coprime solution of  $ u^4-3u^2v^2+3v^4=x^2$ and let $t$ be the greatest common divisor of $v$ and $2$. Then $(U,V,W)=(\frac{2u}{t},\frac{v}{t},\frac{4x}{t^2})$ is a pairwise coprime solution of $U^4-12U^2V^2+48V^4=W^2$.

This can be rewritten, using completing the square, as $$\left (\frac{U^2-6V^2-W}{2}\right )\left (\frac{U^2-6V^2+W}{2}\right)=-3V^4.$$
The bracketed factors, $L$ and $M$,  are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=-3$ and $V=XY$, with $X$ and $Y$ coprime. Then $$aX^4+6X^2Y^2+cY^4=U^2,ac=-3,$$ an equation of form (1). Again, this mapping of solutions is invertible.

Fermat's infinite descent

We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) leads to another positive integer solution $(X,Y,Z)$, where $$Y=\frac{y}{tuX}.$$
We have to agree with Fermat that "there cannot be a series of numbers [positive integers] smaller than any given [positive] integer we please" and therefore the above process must lead to solutions with $tuX=1$.

It is now straightforward to plug $t=u=X=1$ back into the equations to see that there is only a simple loop consisting of the solutions
$$x^4+6x^2y^2-3y^4=z^2,x=y=1,z=2$$ $$ x^4-3x^2y^2+3y^4=z^2,x=y=z=1.$$


问:
Factorization $a^x-1 | (a^x)^k - 1$
$a^x-1 | (a^x)^k - 1$

Is this expression true, and if yes how can I prove it?

答: Hint: recognize the pattern $\frac{q^n-1}{q-1}$, which is the sum of the first ${n-1}$ terms of a geometric sequence.

Title: Factorization $a^x-1 | (a^x)^k - 1$


问:
Internal Structure of A_7
In $A_7$,

1) Are all subgroups of order 168 are conjugate? ($A_7$ contains a simple group of order 168).

2)Does it contain an abelian group of order 12? What is the largest order of abelian group?

3)What is the Sylow-2 subgroup of $A_7$?

[I am trying to get some other way to prove that there is only one simple group order 168.

In a simple group $G$ of order $168$, the number of Sylow-3 subgroups is $7$, or $28$ . If it is $7$ , then $G$ is contained in $A_7$, and we get an abelian subgroup of order $12$, in $A_7$ . Certainly, $A_7$ does not contain an element of order $12$, hence cyclic subgroup of order $12$. therefore, it may happen that $A_7$ may contain $\mathbb{Z}_2 \times \mathbb{Z}_6$. I want to know whether this is possible? Also, to know about intersection of Sylow-2 subgroups with each other, I want to know their structure.]

答:
Part of answer:

(2) $A_7$ has an abelian subgroup of order $12$, namely $\langle (123)\rangle \times \langle (45)(67), (46)(57)\rangle \cong \mathbb{Z}_3 \times K_4$, $K_4$ is the Klein-$4$ group.

(3) The Sylow-$2$ subgroup of $A_7$ has order $8$ and it should be $D_8$: as $A_7$ has a subgroup $H=\langle (12)(34), (13)(24) \rangle \cong K_4$, it should be contained in some Sylow-$2$ subgroup $K$ of $A_7$. So the Sylow-$2$ subgroup should can not be $\mathbb{Z}_8$ or $\mathbb{Q}_8$, as they do not contain Klein-$4$ group.

If $\sigma \in K$ commutes with every element of this subgroup $H$, then we can see that $\sigma$ should permute the symbols $\{5,6,7\}$, within themselves, and it should permute the symbols $\{1,2,3,4\}$, within themselves; i.e. $\sigma=\sigma_{1} \sigma_{2}$, where $\sigma_1$ permutes $\{1,2,3,4\}$, and $\sigma_2$ permutes $\{5,6,7\}$.

We can see that elements of type $(12)(56)$ (i.e. a transposition on $\{1,2,3,4\}$ followed by a transposition on ${5,6,7}$) can not be in the centralizer of the subgroup $H$; such type of elements commutes with only two elements of $H$- the identity and $(12)(34)$.

So $\sigma=\sigma_1 \sigma_2$, where $\sigma_1 \in H$, and $\sigma_2 = Id$.

Hence the centralizer of the subgroup $H$ in Sylow-$2$ subgroup $K$ should be itself; i.e. Sylow-$2$ subgroup of $A_7$ is non-abelian; hence it must be $D_8$.

Title: Internal Structure of A_7


问:
Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$
Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?

Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.

答:
Such equations, if slick tricks such as lab bhattacharjee's can't apply, are solved with a standard procedure:

\begin{align}
&\sqrt{x+4}-\sqrt{x+1}=1\\[2ex]
&\text{Rearrange}\\
&\sqrt{x+4}=1+\sqrt{x+1}\\[2ex]
&\text{Square}\\
&x+4=1+2\sqrt{x+1}+(x+1)\\[2ex]
&\text{Rearrange}\\
&2=2\sqrt{x+1}\\[2ex]
&\text{Simplify}\\
&1=\sqrt{x+1}\\[2ex]
&\text{Square}\\
&1=x+1\\[2ex]
&x=0
\end{align}

We just need to ckeck that the solution makes the square roots existent, because at each "Square" stage we are dealing with non negative numbers. Of course the conditions are
$$\begin{cases}
x\ge-4\\
x\ge-1
\end{cases}
$$
which boil down to $x\ge-1$, that's satisfied by our solution.

Some care has to be reserved in different situations, when there's no guarantee that at the "Square" staged we have non negative numbers.

Title: Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$


问:
Show that for every $n$ there are integers $a,b$ such that $X_n = a^2 + 2b^2$
We have a sequence $X_0, X_1,...$ which is determined by $X_0=1, X_1=3$ and $X_n = 6X_{n-1} - X_{n-2}$

Show that for every $n$ there are integers $a,b$ such that
$X_n = a^2 + 2b^2$

I thought of changing $a^2 + 2b^2 = (a + \sqrt 2 i)(a - \sqrt 2 i)$, since I thought it might lead to something good, but I'm not getting anything.
Any hints or solutions are appreciated!

答:
Factoring the sum in the complex numbers does not seem to lead us anywhere. Neither does the knowledge of which integers in general can be represented as $a^2+2b^2$. We must try something else.

Well, the roots of the characteristic polynomial are $3\pm2\sqrt2$, which are obviously the squares of $1\pm\sqrt2$, which are related to Pell numbers, hence so must be our $a,b$ - this was my reasoning, about as vague as it gets.

Indeed, for the sequence defined by $P_0=0,\;P_1=1,\;P_n=2P_{n-1}+P_{n-2}$ we have that $P_n+P_{n-1}$ and $P_n$ (which are, coincidentally, numerators and denominators of the best rational approximations to $\sqrt2$) work as $a$ and $b$ for our $X_n$.

$$\begin{array}{l|l|l|l|l|l|l}
n             & 0  & 1& 2&  3&  4&  5& \dots\\ \hline
P_n=b         & 0  & 1& 2& 5 & 12& 29& \dots\\
P_n+P_{n-1}=a & 1  & 1& 3& 7 & 17& 41& \dots\\
X_n=a^2+2b^2  & 1  & 3&17& 99&577&3363& \dots\\ \hline
\end{array}$$

The actual proof never bothered me. It should be easy to do by induction.

So it goes.


问:
〈b, a, b〉 and 〈a, b, a〉: For what non zero values of a and b are these two vectors parallel?
Looking for a hint on how to solve this problem.
I know for the vectors to be parallel, the cross product must equal the zero vector, but I'm unsure on how to use that information to solve for values.

$\langle b, a, b\rangle \times\langle a, b, a\rangle$.

For what non zero values of $a$ and $b$ are these two vectors parallel?

答:
You can use the cross product if you want to use a sledgehammer: the cross product is the formal determinant
$$
\det\begin{bmatrix} e_1 & e_2 & e_3 \\ b & a & b \\ a & b & a \end{bmatrix}
=(a^2-b^2)e_1+(b^2-a^2)e_3
$$
and so the condition is $a^2=b^2$.

You can also use the dot product, because parallelism is equivalent to
$$
|v\bullet w|=|v||w|
$$
that in your case becomes
$$
3|ab|=\sqrt{a^2+2b^2}\sqrt{2a^2+b^2}
$$
that becomes $2a^4+5a^2b^2+2b^4=9a^2b^2$ and so $2(a^2-b^2)=0$.

On the other hand, the problem can be more easily solved by observing that parallelism is the same as requiring
$$
\langle b,a,b\rangle=x\langle a,b,a\rangle
$$
so $b=xa$ and $a=xb$ and easily $x=\pm1$.


问:
Prove that Group Ring RG is not a Division Ring for R commutative and G a finite group
How do I prove that a group ring RG is not a division ring, assuming that R is commutative and G=(g_1,g_2,...g_n) finite?

I understand that I need to show there exists some element of RG that is not a unit. But how do I find such element? Would the element g_1+g_2+g_3+....+g_n work?

答:
Assume that $|G| > 1$ (with identity element $e$) and $R \neq \{0\}$. Consider the map

$$RG \to R, \ \ \ \ \sum_{g \in G} a_g g \mapsto \sum_{g \in G} a_g.$$

It is a ring homomorphism. Its kernel contains all elements of the form $e-g$ for $g \in G$. The kernel is a two-sided ideal in $RG$, and the map is non-trivial (because $G$ and $R$ are non-trivial). Hence $RG$ cannot be a division ring, since a divison algebra only contains the trivial two-sided ideals.


问:
What do you call generating a function out of a graph?
In many physical phenomena, laws an relations to their variables are somehow interpolated (example by statistical data or samples) and then an approximate set of functions are generated to work equivalently within the allowable range. Is there a branch of mathematics that treats this in details?</p>

答:
The branch of mathematic called modeling gives you a idea of the type of curve you may want to use.

Statistics helps you in knowing whether your sample is correct (is the number of experiments enough, what is the experimental errors,... ).

Curve fitting, with optimization is used to adjust the parameters and minimize the difference between your data and you model.

Simulation is when you use your model to mimics reality (how many replicas, what is due to randomness and model inaccuracy...)

This is not a part of mathematics and you will find a lot information there, but be ready to enter in a new universe.


问:
If a set is bounded in $\mathbb{R}^n$, show that each component is bounded
This is the definition of bounded:

A metric space $(X,d)$ is said to be bounded, and $d$ is said to be a bounded metric, if there exists a positive real number $M$ such that $d(x,y)<M$, for all $x,y\in X$.

Let $n$ be a positive integer, $d$ the euclidean metric on $\mathbb{R}^n$, and $X$ a subset of $\mathbb{R}^n$. Prove that $X$ is bounded in $(\mathbb{R}^n,d)$ if and only if there exists a positive real number $M$ such that for all $\langle x_1,x_2,\ldots,x_n\rangle\in X$, $-M<x_i<M$ for all $i=1,2,\ldots,n$.

I am trying to show that the above definition of boundedness is equivalent to showing there is an open ball from the origin large enough that the set is contained in the open ball, but I'm getting stuck doing so. Can someone provide some pointers?

答:
Hint:

Assume that $|x_i|<M$ for all $i$. Then:

Consider the value $x_i^2$. Can you find an upper limit for it?

Use the fact that if $a_1 < b_1$ and $a_2< b_2$, then $a_1 + a_2 < b_1 + b_2$.


答:
Recall that the euclidean metric on $\mathbb{R}^n$ is defined as follows; the distance between $\vec{x}$ and $\vec{y}$ is $$\sqrt{\sum_{i=1}^n(x_i-y_i)^2}.$$

Now suppose that $X$ is bounded, so there is some $M$ such that for every $\vec{x}$ and $\vec{y}$ in $X$ we have $d(\vec{x},\vec{y})<M$. We suppose now that there is some component $i$ which is not bounded. That is, we have some $\vec{x}$ such that $|x_i|$ is not bounded. Can you derive a contradiction?

Title: If a set is bounded in $\mathbb{R}^n$, show that each component is bounded

Body:
This is the definition of bounded:

A metric space $(X,d)$ is said to be bounded, and $d$ is said to be a bounded metric, if there exists a positive real number $M$ such that $d(x,y)<M$, for all $x,y\in X$.

Let $n$ be a positive integer, $d$ the euclidean metric on $\mathbb{R}^n$, and $X$ a subset of $\mathbb{R}^n$. Prove that $X$ is bounded in $(\mathbb{R}^n,d)$ if and only if there exists a positive real number $M$ such that for all $\langle x_1,x_2,\ldots,x_n\rangle\in X$, $-M<x_i<M$ for all $i=1,2,\ldots,n$.

I am trying to show that the above definition of boundedness is equivalent to showing there is an open ball from the origin large enough that the set is contained in the open ball, but I'm getting stuck doing so. Can someone provide some pointers?


问:
Calculating the centralizer of a matrix in a general linear group.
Let $G = GL(3,\mathbb{R})$ be the general linear group over the reals , of order $3$ , and let
$A\in G$ be :

$$
A=\begin{pmatrix}
 -1 & 0 & 0 \\\
 0 & 1  & 0 \\\
 0 &  0 & 2
\end{pmatrix}.
$$

I'm having a hard time calculating $C(A)$ , I've deduced the following :

$$C(A)=\langle A \rangle \cup \{ kI | k \in R , k \neq 0 \}$$

Is there any other elements in $C(A)$? and how do I go about solving such a question?

Note : $\langle A \rangle$ is the cyclic sub-group of $G$, generated by $A$

答:
A matrix $B$ commutes with $A$ iff $AB=BA$ iff $B=ABA^{-1}$; fill the entries of $B$ with unknowns, write out this equation, and then find an equivalent system of relations for the unknowns:

$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}^{-1}=\begin{pmatrix}a & -b & -c/2 \\ -d & e & f/2 \\ -2g & 2h & i\end{pmatrix} $$

$$\implies b,c,d,f,g,h=0,~\text{ no restrictions on }a,e,i.$$

Hence the centralizer is precisely the subgroup of invertible diagonal matrices.


问:
Use the composite Simpson's rule to approximate how far a tricycle traveled from a table with the time and velocity

Use the composite Simpson's rule to find how far the car traveled
between 15:00 and 15:40 from the following table (Hora/s = Hour/s)

I did

$$I(f)\approx \frac{(b-a)}{3N}[f_0+4f_1+2f_2+4f_3+f_4] \\ =
\frac{(b-a)}{3*5}[24.2+4*35+2*41.3+4*42.8+34.8] = \\
(b-a)[30.18667]$$

$$(40)*30.18667 = 1207.45Km$$

That result is ridiculous, obviously, considering this is a tricycle. What went wrong?

答:
If you use hours as the time unit, your data points are:
$$
\left\{(0, 24.2),(\frac 16, 35), (\frac 13 , 41.3),(\frac 12, 42.8), (\frac 23, 34.2) \right\}
$$

The composite Simpson rule for $s = \int_0^{\frac 23} v(t) dt$ is

$$
\frac{1/6}{3}\left(24.2 + 4 \times 35 + 2 \times 41.3 + 4 \times 42.8 + 34.2 \right) = 25.1222 \,\textrm{Km}
$$

if you have any information on the smoothness of $v(t)$, namely about its forth derivative, you can derive an error bound.


问:
Sequential criterion: "... Every sequences $x_n$ in $D$ converging to $c$ ..."
The sequential criterion for the existence of limits states that,

Let $D\subset \mathbb{R}$ and $f:D\rightarrow \mathbb{R}$ be a function. Let $c$ be a limit point of $D$ and $l\in \mathbb{R}$. Then $\lim\limits_{x\rightarrow c} f(x)=l$ if and only if for every sequence $(x_n)$ in $D-\{c\}$ converging to $c$, the sequence $(f(x_n))$ convergs to $l$.

What I want to know is in this part,

$$\textbf{every sequence $(x_n)$ in $D-\{c\}$ converging to $c$}$$

what does it mean for "every" sequence $(x_n)$ to have the same range $\left( D-\{c\}\right)$ and the same limit $(c)$, because in that case wouldn't the sequences become the same?

答:
Maybe it is appropriate to give an example.

Let $f \colon \mathbb R \to \mathbb R$ be given by

$$x \mapsto \begin{cases} 0, & \text{ if } x \geq 0 \\ 1, &\text{ if } x < 0\end{cases}.$$

What is $\lim_{x \to 0} f(x)$? Does it even exist (i.e., does it have a value)?

Well, take the sequence $(x_n)$ given by $x_n = \frac1n$, which converges to $0$, the limit point we are looking at. Then $f(x_n) = 0$ for every $n$, and so $\lim_{n \to \infty} f(x_n) = 0$.

However, if you take the sequence $(y_n)$ given by $y_n = - \frac1n$, then again $(y_n)$ converges to $0$, but $f(y_n) = 1$ for every $n$. Hence $\lim_{n \to \infty} f(y_n) = 1$.

Here you see that $\lim_{n \to \infty} f(z_n)$ depends on the choice of the sequence $(z_n)$, which converges to $0$.

What follows is that $\lim_{x \to 0} f(x)$ does not exist.


问:
$f(X)＝X^p＋mp^2(X-1)(X-2) \cdots( X-p＋2)-p$ has exactly two non-real roots
How to prove $f（X）＝X^p＋mp^2（X-1）（X-2）...（X-p＋2）-p$ has just 2 non-real roots when $m$ is large enough?
I couldn't find even one solution of this polynomial.
But writing graphs with computer,the titled statement seems to be true.Thank you for your help.

答:
The polynomial has exactly two non-real roots for $m$ big enough.

Indeed, using the sign rule of Descartes, we see that, because $f$ has exactly $p-2$ sign changes of the coefficients, $f$ has at most  $p-2$ positive real roots. Since $f(-X)$ has only negative coefficients for $m\ge 0$, there is no negative real root.
Hence $f$ has at most $p-2$ real roots. However, for $m$ big enough
it is clear that $f$ has positive roots nearby $1,2\ldots ,p-2$. So we are done. The polynomial has exactly two non-real roots, so that its Galois group is isomorphic to $S_p$.

Again, here is the polynomial:
$$
f(X)=X^p+mp^2(X-1)(X-2)\cdots (X-p+2)-p
$$


答:
Let $G$ and $R$ be nontrivial.

Let $\alpha = \sum_{g\in G}g$. We claim that $\alpha$ is not a unit in $RG$. To see this, suppose that $x = \sum_{h\in G}r_h h$ is any element of $RG$. We will show that $\alpha x\neq 1$. We have
$$
\alpha x = \sum_{g,h \in G}r_h\cdot gh = \sum_{s\in G}\Big[(\sum_{h\in G}r_h)\cdot s\Big],
$$
where the second equality makes the substitution $s = gh$. We see that each element of $G$ has the same coefficient, namely $\sum_h r_h$, in the expansion of $\alpha x$. However, $1$ has different coefficients for the identity and non-identity elements, so we cannot have $\alpha x = 1$.

Title: Prove that Group Ring RG is not a Division Ring for R commutative and G a finite group

Body:
How do I prove that a group ring RG is not a division ring, assuming that R is commutative and G=(g_1,g_2,...g_n) finite?

I understand that I need to show there exists some element of RG that is not a unit. But how do I find such element? Would the element g_1+g_2+g_3+....+g_n work?


问:
In general, can we prove that admissible quadratic polynomials must yield at least one prime value?
By admissible, I'm referring to an integer-valued quadratic function with no common factors, i.e. one that meets the basic requirements to be able to produce primes. To be more specific, a quadratic polynomial $f(x):=ax^2+bx+c$ with $a>0$ where  $x\in\mathbb Z \implies f(x) \in\mathbb Z$, and where there exist $m,n\in\mathbb Z$ with $\gcd(f(m),f(n))=1$ and $ m \neq n$. In short, a quadratic which has no a priori reason to be prime-free.

First, I know the infinitude of quadratic primes is an open question, so I suspect this will be too, but I'll ask anyway. Do we know of any quadratic functions that should be able to produce primes yet they don't seem to? By this, I mean an admissible quadratic as described above but for which computer searches have been unable to find any prime values.

I'm assuming not, in which case my primary question is: have we reached the point yet where we can prove, in general, that an admissible quadratic function as described above must yield at least one prime value? Or is that still out of reach?

答:
Upon further thought, I realized that the answer must be no, we currently can't prove it.

If we could, a proof of the infinitude of polynomial primes immediately follows, since you can always split one into arbitrarily many parts. For example, the first six values of $x^2+1$ are $\{2,5,10,17,26,37\}$. But by substituting in $3x-2,3x-1,3x$ to $x^2+1$, it can be completely split into

$9x^2-12x+5$ yielding $\{2,17,\ldots\}$

$9x^2-6x+2$ yielding $\{5,26,\ldots\}$

$9x^2+1$ yielding $\{10,37,\ldots\}$

each of which is itself admissible and contains a prime. To ensure admissibility, you either have to split into $k$ pieces where $k$ is not a residue of the original polynomial, or factor out $k$ if it is a residue, but aside from that, I see no reason why this wouldn't work in general, and since I'm sure I'm not the first person to have noticed this, it means that we can't prove that an admissible quadratic must have at least one prime.


问:
Order linear such that extends a partial order
Let $\le$ a partial order on a set $A$. Prove that there exists a linear order $\le ^*$ on $A$ such that $a\le b$ implies $a\le ^*b$.

I considered $\mathcal{F}$ the family of all $\le _0 \subseteq A\times A$ such that $\le\subseteq\le _0$ and $\le _0$ is a partial order on $A$.

Then, using Zorn's lemma I concluded that $\mathcal{F}$ has a maximal element $\le ^*$. I proved that this is a partial order on $A$.

Now, I suppose this is not a linear order, then there exists $a,b\in A$ such that $(a,b)\not\in\le ^*$ and $(b,a)\not\in\le ^*$.

Where is the contradiction here? First, I considered $\le _0^* : = \le ^*\cup \{ (a,b)\}$ and I wanted to prove that this is a partial order, but I can't prove the transitivity.

Any hint? Thanks.

答:
There is the theorem of  Szpilrajn (see E.Harzheim, Ordered Sets, Springer, 2005):

Let $P(\le)$ be a poset. Then there exists a linear order $\le_*$
on $P$ which contains $\le$.

It is proved using the axiom of choice.


问:
Topology Verifying Hausdorff Property Problem
Need some hints on this question because I don't know how to prove it:

On $X = [-1,1]$ we consider the topology $\tau$ that has the family $A = \{{[-1, b): b> 0}\} \cup \{{(a, 1] : a <0}\}$ as subbase.

Does $(X, T)$ verify the Hausdorff property?

Find the limit of ${x_n}=\frac{(-1)^n}{n}$

All help is welcome

答:
Well, every neighbourhood of $-1$ intersects every neighbourhood of $1$ ($0$ is always in both). This answers 1 as one counterexample pair suffices.

It's quite easy to check that $x_n \to -1$ and $x_n \to 1$ by the definitions.
So "the" limit is (on purpose) misleading. Even $x_n \to 0$ is true, I think.


问:
Find $,,$ such that $(−, +, −1)=(4, 2, 3)$.
I am stuck at this question can someone help me?

I've been trying it all morning but I just cant quite get it going
maybe one of the lads can help

答:
We have the equations

$(1) \quad x-y=4$

$(2) \quad x+y=2$

and

$(3) \quad z-1=3.$

Equation $(3)$ gives $z=4.$

$(1)+(2)$ gives you an equation for $x$. Your turn !


答:
You have to solve three equations, being
$$\begin{array}{rcl}x-y&=&4\\x+y&=&2\\z-1&=&3. \end{array}$$
Well, the last one is easy, by adding $1$ to both sides you get $z=4$. To solve $x-y=4$ and $x+y=3$ you rewrite the first equation to $x=4+y$ and then substitute this in the second equation, giving you $$4+y+y=2,$$ which you rewrite to $4+2y=2$, giving you $y=-1$. Then you substitute this in the first equation again to get $x+1=4$ and hence $x=3$.

Title: Find $,,$ such that $(−, +, −1)=(4, 2, 3)$.

Body:
I am stuck at this question can someone help me?

I've been trying it all morning but I just cant quite get it going
maybe one of the lads can help


问:
How to draw samples of two independent iid random variables from real and psudo random number generators?
There are 2 independent and identically distributed random variables $X_1$ and $X_2$. I need 1000 drawings of each one of these variables (in a monte-carlo simulation setting). Given this problem, i am confused if below are equivalent -

(a) draw 1000 samples for $X_1$ and then draw 1000 samples for $X_2$

(b) draw 2000 samples and then assign half to each one of them

Is (a) same as (b) in an ideal scenario when random numbers are truly random?

What happens when, in practice, we use numbers that are psudo random? should I use a different seed in random number generator when following method (a) [i will call this method (c)]. for example, set a random seed -> generate 1000 samples for $X_1$ and then set another seed and then generate another 1000 samples for $X_2$.? How will that compare when using method (a) when i dont change random seed or with method (b) where i generate 2000 numbers and just assign them equally to $X_1$ and $X_2$?

I am adding here work some work that I did using python numpy.
First i generated, random numbers using np.random

```<code>import numpy as np

sample_size = 100000

# method a - drawing samples for x1 and x2
np.random.seed(0)
x1_a = np.random.rand(sample_size)
x2_a = np.random.rand(sample_size)

# method b - drawing 2 * samples and then allocating to x1 and x2
np.random.seed(0)
x_b = np.random.rand(sample_size * 2)
x1_b = x_b[0:sample_size]
x2_b = x_b[sample_size:2 * sample_size]

# method c - drawing samples for x1 and x2 but with different random seeds
np.random.seed(0)
x1_c = np.random.rand(sample_size)
np.random.seed(120)
x2_c = np.random.rand(sample_size)

# now calculating correlations between x_1 and x_2
corr_a = np.corrcoef(x1_a, x2_a)
corr_b = np.corrcoef(x1_b, x2_b)
corr_c = np.corrcoef(x1_c, x2_c)
print(f" correlations are: {corr_a[0][1]}, {corr_b[0][1]}, {corr_c[0][1]}")
</code>```

this returns following

```<code> correlations are: 0.004584252381322772, 0.004584252381322772, -0.0012275661688751346
</code>```

Results make me wonder if :

is coefficient of .005 as witnessed in method a and b good
enough to assume that $X_1$ and $X_2$ are independent with zero
correlation

method c seems to be giving lower correlation. Is method c better in
that case?

答:
If $X_1$ and $X_2$ are genuine random iid variables, a) and b) are equivalent.

But there is no "seed" for a random variable, so you must be dealing with pseudorandom numbers. In this case, with method b) there is some chance that due to unfortunate choices of seeds, the sequences coincide on some drawing, then on all those that follow. In particular, is the two seeds are the same, the sequences will be identical. With a single seed, this is less likely.


答:
To answer the question you need to solve

\begin{align}
x - y = 4, \\
x + y = 2, \\
z - 1 = 3.
\end{align}

The last equation immediately tells you that $z = 4$. Looking at the first equation you can write it as $x = 4 +y$. Plugging this into the second equation you will find
\begin{align}
4 + y + y = 4 + 2y = 2,
\end{align}
or in other words
\begin{align}
2y = -2.
\end{align}
So you know that $y = -1$. For example, looking again at the first equation you get
\begin{align}
x + 1 = 4,
\end{align}
which yields $ x = 3$.

So we found the (unique) solution, namely $x = 3, y = -1$ and $z = 4$.


问:
S_3 is a quotient of the free group F({x,y})
I am self-learning Algebra: Chapter 0 by Paolo Aluffi. He defined a presentation of a group $G$ as follows:

So, according to my understanding, $R$ is the kernel (a normal subgroup) of the surjection $\rho$. and $\mathscr{R}$ must be $R$'s generator in a typical group presentation.

Then the author gave a simple example: the symmetric group $S_3$. We know that $S_3 = \{e, x, y, xy, y^2, xy^2 \}$ with relations $x^2=e$, $y^3=e$, and $yx = xy^2$.

The author said $S_3$ is a quotient of the free group $F(x,y)$ (the free group generated by 2 elements) by the smallest normal subgroup containing $x^2$, $y^3$, and $yx = xy^2$.  Then he claimed that $S_3$ can be presented as $(x,y | x^2,y^3,xyxy)$.

I have 3 questions here:

To me it's not obvious that $S_3 \cong \frac{F(x,y)}{R}$. What is the explicit surjection $\rho: F(x,y) \rightarrow S_3$? What is its kernel $R$?

Why the genrator of $R$( i.e. $\mathscr{R}$) becomes $\{x^2,y^3,xyxy\}$?  Shouldn't it be $\{x^2$, $y^3$, $yx = xy^2\}$ ? For this question, it has been answered here.

Why didn't the author give any proof except the definition above? Maybe it's really obvious? I've never learned group presentation before. And I found many books including Dummit and Foote only touch a little on this subject. Are there any good notes or books that could help me understand this question better?

答:
Hint

The transposition $\tau = (1 \ 2)$ and the $3$-cycle $\sigma =(1 \ 2 \ 3)$ form a generating set of $\mathfrak S_3$. Therefore if you define $\rho$ by $\rho(x) =\tau$ and $\rho(y) = \sigma$, you get the surjection you're looking for.

As $\tau^2 = \sigma^3 = \tau \sigma \tau \sigma=\text{Id}$ (where $\text{Id}$ is the identity of $\mathfrak S_3$), $x^2, y^3$ and $ xyxy$ are in the kernel $R$.

Now $yx = xy^2$ is equivalent to $xyxy=e$. To convince yourself about that, just multiply the initial equality by $x$ on the left and $y$ on the right and use $x^2=y^3=e$.

A Course in the Theory of Groups from Derek Robinson provides a pretty good coverage on free groups. Also, the questions above are becoming clearer is you have some knowledge on the permutations group.


问:
Solve the last part of a diferential equation
I'm trying to solve this problem but I've some doubts about the final steps.

```<code>Solve the initial value problem using matrix functions:
</code>```

$$
\frac {dy}{dx} =
\begin{pmatrix}
1 & a & 1 \\
0 & 1 & a \\
0 & 0 & 1 \\
\end{pmatrix}
y
\text{ ; }
y(0) =
\begin{pmatrix}
 0 \\
 0 \\
-1 \\
\end{pmatrix}
$$

First of all, I'm not sure what do they mean with "matrix functions" but I think I've to do the exponential development.

The way I've tried:

$$
\frac {dy}{dx} = Ay
\text{ }\to\text{ }
\frac {dy}{y} = Adx
\text{ }\to\text{ }
ln(y) = Ax+K
\text{ }\to\text{ }
y(x) = Ce^{Ax}
$$

$$
e^{Ax} = \sum_{n=0}^{\infty} \frac{(Ax)^n}{n!}
$$

I try to get a recurring rule for the powers of the matrix A:

$$
A^2 =
\begin{pmatrix}
1 & 2a & 2+a^2 \\
0 & 1 & 2a \\
0 & 0 & 1 \\
\end{pmatrix}
\text{ }\text{ }
A^3 =
\begin{pmatrix}
1 & 3a & 3+3a^2 \\
0 & 1 & 3a \\
0 & 0 & 1 \\
\end{pmatrix}
\text{ }\text{ }
A^4 =
\begin{pmatrix}
1 & 4a & 4+6a^2 \\
0 & 1 & 4a \\
0 & 0 & 1 \\
\end{pmatrix}
$$

So that, it looks the n-power of A is:

$$
A^n =
\begin{pmatrix}
1 & na & n+\frac{n(n-1)}{2}a^2 \\
0 & 1 & na \\
0 & 0 & 1 \\
\end{pmatrix}
=
I+
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{pmatrix}
na+
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
(n + \frac {n(n-1)}{2}a^2)
$$

I just rename:

$$
\alpha =
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{pmatrix}
\text{ }\text{ }
\beta =
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
$$

So that $A^n$ becomes:

$$
A^n = I + na\alpha + (n + \frac {n(n-1)}{2}a^2)\beta
$$

$$
A^n =
I + na\alpha + n\beta + \frac{n^2}{2}a^2 \beta - \frac{n}{2}a^2 \beta =
I + (a\alpha + \beta -\frac{1}{2}a^2 \beta)n + \frac{a^2 \beta}{2}n^2
$$

Once this is done, I go back to the exponential development:

$$
e^{Ax} =
\sum_{n=0}^{\infty} \frac{(Ax)^n}{n!} =
\sum_{n=0}^{\infty} (I + (a\alpha + \beta -\frac{1}{2}a^2 \beta)n + \frac{a^2 \beta}{2}n^2) \frac{x^n}{n!}
$$

$$
e^{Ax} =
Ie^{x} + (a\alpha + \beta -\frac{1}{2}a^2 \beta) \sum_{n=0}^{\infty} \frac{nx^n}{n!}
+ \frac{a^2 \beta}{2} \sum_{n=0}^{\infty} \frac{n^2x^n}{n!}
$$

From here I've tried to solve both sums:

$$
\sum_{n=0}^{\infty} \frac{nx^n}{n!} =
0 + \sum_{i=1}^{\infty} \frac{nx^n}{n!} =
\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}x =
\sum_{m=0}^{\infty} \frac{x^m}{m!}x = xe^x
$$

$$
\sum_{n=0}^{\infty} \frac{n^2x^n}{n!} =
0 + \sum_{i=1}^{\infty} \frac{n^2x^n}{n!} =
\sum_{n=1}^{\infty} \frac{nx^{n-1}}{(n-1)!}x =
\sum_{m=0}^{\infty} \frac{(m+1)x^m}{m!}x =
\sum_{m=0}^{\infty} \frac{mx^m}{m!}x +
\sum_{m=0}^{\infty} \frac{x^m}{m!}x =
 x^2e^x+xe^x
$$

So that, the solution should be:

$$
e^{Ax} =
Ie^{x} + (a\alpha + \beta -\frac{1}{2}a^2 \beta) xe^x
+ \frac{a^2 \beta}{2} (x^2e^x+xe^x)
$$

$$
e^{Ax} =
Ie^{x} + (a\alpha + \beta) xe^x + \frac{a^2 \beta}{2} x^2e^x
$$

On a matrix representation:

$$
e^{Ax} =
\begin{pmatrix}
1 & ax & x + \frac{a^2}{2}x^2 \\
0 & 1 & ax \\
0 & 0 & 1 \\
\end{pmatrix}
e^x
$$

On the last part, I'm not sure how to get the solution because the constant C should be a row vector instead of a column vector. Is this right? How to get the final solution?

Thanks in advance.

答:
A simple solution:

$$
\frac {dy}{dx} =
    \begin{pmatrix}
    1 & a & 1 \\
    0 & 1 & a \\
    0 & 0 & 1 \\
    \end{pmatrix}
y
\text{ ; }
y(0) =
    \begin{pmatrix}
     0 \\
     0 \\
    -1 \\
    \end{pmatrix}
$$
Let $y=\begin{pmatrix} y_1 \\ y_2 \\y_3 \end{pmatrix}$
Then you have three couples ODEs as
$\frac{dy_1}{dx}=y_1+ay_2+y_3~~~(1), \frac{dy_2}{dx}=y_2+ay_3~~~(2), \frac{dy_3}{dx}=y_3~~~~(3)$
Fist solve (3) to get $y_3=C e^x \implies y_3=-e^{x}$. Put this in (2) and solve the ODE:
$$\frac{dy_2}{dx}=y_2-ae^x \implies y_2=e^{x}\int -ae^x e^{-x}dx+ Be^{x}=-axe^{x}+Be^{x} \implies y_2=-axe^x$$
Putting these solutions in (1), we get
$$\frac{dy_1}{dx}=y_1-a^2 xe^x-e^{x} \implies y_1=e^{x}\int (-a^2xe^x e^{-x}-1) dx+Ae^x$$ $$ \implies y_1= -\frac{a^2x^2}{2} e^{x}-xe^x+Ae^x, y_1(0)=0 \implies A=0.$$ So the solutions of the Matrix ODE are
$$y(x)=\begin{pmatrix} -\frac{a^2x^2}{2}e^x-xe^x \\ -axe^x \\ -e^x \end{pmatrix}$$


问:
What is wrong with my derivation of the general formula of the power series for $\cos^{2}(x)$
I have arduously obtained the series for $\cos^2(x)$ by using long multiplication method (multiplying twice the Taylor series of $\cos(x)$ which is $\sum_{n=0}^{+\infty} \frac{(-1)^nx^{2n}}{(2n)!}$
$$\cos^2(x)=1-x^2+\dfrac{1}{3}x^4-\dfrac{2}{45}x^6+\dfrac{1}{315}x^8+...$$

I wish to manipulate with the sigma notation to reduce the cost of labour, so I square the above general formula for $\cos(x)$:

$$\cos^2(x)=\sum_{n=1}^{+\infty} \left(\frac{(-1)^nx^{2n}}{(2n!)}\right)^2=\sum_{n=1}^{+\infty} \dfrac{(-1)^{2n}x^{4n}}{[(2n)!]^2}$$

But this is plain wrong, the formula that I obtain is not the general formula for the series of $\cos^2(x)$.

So what is wrong with my manipulation with the sigma notation of this series. The power of $x$ and $-1$ is too high, it should be $2n$. Also, is my denominator correct?

How do you obtain the general formulae for this series?

答:
When you square a summation in general, the result is not equal to the sum of the squares of each term of the summation, as you have assumed.

I will demonstrate this with a couple of examples. Firstly, it's clear that
$(a+b)^2=a^2+b^2+2ab$. But according to what you wanted to do, then $(a+b)^2=a^2+b^2$, which is clearly wrong.

Similarly, consider the value of
$$\left(\sum_{r=1}^n 1\right)^2$$
This is clearly $(n)^2=n^2$, but according to what you're saying the value should be equal to
$$\sum_{r=1}^n 1^2=\sum_{r=1}^n 1=n$$
which again is clearly wrong.

Secondly, it's important to realize that $(n!)^2$ is not equal to $n^2!$. It cannot be simplified further I'm afraid; $(n!)^2$ is the best you can do. Again, verify this with some examples, eg $n=4$.

The best strategy to find the power series for your series is to utilise the double angle formulae, as Olivia Oloa has noted.

Firstly, recall that
$$\cos2x=2\cos^2x-1\implies\cos^2x=\frac{1}{2}(1+\cos2x)$$
Now, we know that
$$\cos2x=\sum_{n=0}^\infty (-1)^n\frac{(2x)^{2n}}{(2n)!}$$
Hence,
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\sum_{n=0}^\infty (-1)^n\frac{(2x)^{2n}}{(2n)!}$$

If you have any questions please don't hesitate to ask.


问:
Simplifying the inequality $nx(1-x)^n \leq \epsilon$ to $n > N(x, \epsilon)$?
How can I reduce the following inequality on $x \in [0, 1]$, and $\epsilon > 0$:
$$nx(1-x)^n \leq \epsilon$$
to the this form
$$n > N(x, \epsilon)\quad?$$
We can assume $n = 1, 2, 3, \dots$.

My attempt:

$$\ln\big(nx(1-x)^n\big) \leq \ln \epsilon$$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
\implies \ln n + n\ln(1-x) \leq \ln\frac{\epsilon}{x}$,

and I am stuck. Since $\ln(1-x) < 0$, I think I must divide by it to change the direction of inequality. But I don't know how to simplify/group $n$ and $\ln n$ while decoupling them from $x$ term.

Ultimately, I want to prove that $\forall n \geq N(x, \epsilon), \quad nx(1-x)^n \leq \epsilon$. Perhaps, I do not need to solve for $n$?

答:
As a real, the solution of $$nx(1-x)^n = \epsilon$$ is given in terms of Lambert function
$$n=\frac{1}{\log (1-x)}W\left(\frac{ \log (1-x)}{x}\epsilon \right)$$


问:
questions about excluded values in rational expressions
Two examples, one from my algebra textbook and another from Wolfram that create this question.
I fully realize that zero cannot exist in the denominator, but the book seems to distinguish between removable discontinuities and values of a variable that would cause a zero to exist in the denominator.   My enclosures both show these examples.

Wolfram is as follows:

$(1/x)(x/(x+1)).

The problem shows the x values being canceled and then shows that $x$ cannot $=0$.  But the final answer of $1/(x+1)$ does not exist for $x=-1$, yet this exclusion is not listed.

My text shows the exact same problem and says excluded values of $x$ are $x$ cannot $=0$ not $x$ cannot $=-1$.

Can someone help me understand this dilemma?

答:
The textbook answer to the question is not stating the domain, but stating under what circumstances the simplification is valid.  The step

$$\frac{1}{x}\cdot\frac{x}{x+1} = \frac{1}{x+1}$$

is valid only when $x\neq 0$.  The two expressions are equal otherwise.


问:
What are the implications of Russel's paradox?
In Daniel J. Velleman's How To Prove It there is an exercise which asks us to consider the existence of the set of all sets and its implications:

Suppose U were the collection of all sets. Note that in particular U is
a set, so we would have U ∈ U. This is not yet a contradiction; although
most sets are not elements of themselves, perhaps some sets are elements
of themselves. But it suggests that the sets in the universe U could be
split into two categories: the unusual sets that, like U itself, are elements
of themselves, and the more typical sets that are not. Let R be the set of
sets in the second category. In other words, R = {A ∈ U | A ∈/ A}. This
means that for any set A in the universe U, A will be an element of R iff
A ∈/ A. In other words, we have ∀A ∈ U(A ∈ R ↔ A ∈/ A)

(a) Show that applying this last fact to the set R itself (in other words,
plugging in R for A) leads to a contradiction.

Since $R \in U$, we have that $R \in R ↔ R \notin R$, a contradiction, Russel's famous paradox. We are then asked to reflect on its implications. I used to think that Russel's paradox refuted the existence of a set which contains itself, but it seems to me now that it just refutes the idea that any definable collection is a set, or that one can subdivide any set into any two subsets. In this question I found out that sets that contain themselves can exist even in the context of ZFC.

I  also can't see how this paradox proves that there can't be a set of all sets. I've seen proofs of it based on Cantor's Theorem, but once again the paradox  seems to me not to prove the impossibility of the existence of the set of all sets, but the impossibility of the construction of a certain subset of it.

Concluding, is Russel's paradox just an argument against the idea than any definable collection is a set, or is there more profound implications of it I am not aware of?

答: It does prove that there is no set of sets in ZFC, where we have as an axiom (separation or comprehension) that if $y$ is a set and $\phi(x)$ is some predicate with free variable $x$ (ignoring parameters for now) then $\{x \in y\mid \phi(x)\}$ is also a set. So if $U$ (the set of all sets) existed, so would $\{x \in U\mid x \notin x\}$ and this leads to a contradiction. So $U$ does not exist in a system obeying this axiom (like ZFC, which we do assume in almost all of maths).


Body:
Given $(x-y,x+y,z-1)=(4,2,3)$

$$x-y=4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....1$$
$$x+y=2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....2$$  $$z-1=3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....3$$

so
from $3$ we can see that $z=3+1\;\;$ so $\;\;z=4$

from add $1$ and $2$ you will get
$$x-y+x+y=4+2$$
$$2x=6$$
$$x=3$$

put $x=3$ in equation $1$ you will get $$3-y=4$$
$$-y=4-3$$

$$y=-1$$

so $x=3$, $y=-1$ and $z=4$


答: The claim is false. Consider $p=3$. Then for general $m$ the polynomial has eiher one or three real roots. It never has two real roots (if the multiplicity of the roots is taken into account).

Title: $f(X)＝X^p＋mp^2(X-1)(X-2) \cdots( X-p＋2)-p$ has exactly two non-real roots

Body:
How to prove $f（X）＝X^p＋mp^2（X-1）（X-2）...（X-p＋2）-p$ has just 2 non-real roots when $m$ is large enough?
I couldn't find even one solution of this polynomial.
But writing graphs with computer,the titled statement seems to be true.Thank you for your help.


答:
If $B$ commutes with $A$, it leaves the eigenspaces of $A$ invariant. In this case, these are the one-dimensional subspaces generated by the vectors of the canonical basis $\{e_i\;;\; i=1,2,3\}$. So $Be_i=\lambda_ie_i$ for $i=1,2,3$. That is $B$ is diagonal. The converse is clear.

Conclusion: the commutant of $A$ in $M_3(\mathbb{R})$ is the subalgebra of diagonal matrices. And the centralizer of $A$ in $GL(3,\mathbb{R})$ is therefore the subgroup of invertible diagonal matrices.

Note: this observation generalizes to any diagonal matrix with pairwise distinct diagonal elements. It generalizes also to a any (diagonalizable) matrix with pairwise distinct eigenvalues: the commutant is then the subalgebra of matrices which are simultaneuously diagonalizable with $A$. As soon as the eigenspaces become bigger, the commutant loses its commutativity.

Edit: for more "abstract-algebra", I will add that the commutant of $A$ is, in this case, $\mathbb{R}[A]=\{p(A)\;;\; p\in\mathbb{R}[X]\}$ the algebra of all polynomial evaluations on $A$. It follows from the above after you have observed that, by interpolation, for any $\lambda_1,\lambda_2,\lambda_3$, there exists a polynomial of degree $2$ $p(X)$ such that $p(-1)=\lambda_1, p(1)=\lambda_2, p(2)=\lambda_3$. In particular $\mathbb{R}[A]=\mathbb{R_2[A]}$ is three-dimensional. Ththe latter can be recovered directly from the minimal polynomial.

So the centralizer is $\mathbb{R}[A]^*=\{B=p(A)\;;\; \det B\neq 0\}=\{B=p(A)\;;\; p(-1)p(1)p(2)\neq 0\}$. It is interesting to think about this result in view of the functional calculus and the spectral mapping theorem.

Title: Calculating the centralizer of a matrix in a general linear group.

Body:
Let $G = GL(3,\mathbb{R})$ be the general linear group over the reals , of order $3$ , and let
$A\in G$ be :

$$
A=\begin{pmatrix}
 -1 & 0 & 0 \\\
 0 & 1  & 0 \\\
 0 &  0 & 2
\end{pmatrix}.
$$

I'm having a hard time calculating $C(A)$ , I've deduced the following :

$$C(A)=\langle A \rangle \cup \{ kI | k \in R , k \neq 0 \}$$

Is there any other elements in $C(A)$? and how do I go about solving such a question?

Note : $\langle A \rangle$ is the cyclic sub-group of $G$, generated by $A$


问:
Why do we say this for triangles?
Case 1 ;Let us say there is a triangle $ABC$ and one $DEF$. If $AB = DE$ can we also say that $\angle BCA = \angle EFD $ ? What is the proof and reason for saying this ?

Case 2 : Supose $BC = EF$ along with  $AB=DE$.

答:
It is not true for either case. Let's say $BC = EF = AB=DE$ (Which is a particular case of your conditions). We can choose the angles $\angle ABC$ and $\angle DEF$ at will, so we can make $ABC$ a right triangle at B (meaning $\angle ABC = 90º$) and $\angle DEF$=60º.

If you make the drawing you can clearly see that the two triangles are not equal and that their angles $\angle BCA$ and $\angle EFD$ are diferent. $\angle BCA$ = 45º (since it's a right isosceles triangle) and $\angle EFD$ = 60º (since an isosceles triangle with an angle of $60º$ between two equal sides will be equilateral).

Title: Why do we say this for triangles?


答:
Viewing your last sentence, you can procceed as follows.

Take $\varepsilon>0$ (WLOG, you can assume that $\varepsilon<1$).

for $x=0,1$, it is clear that $nx(1-x)^n=0<\varepsilon$ for all $n$.

For $0<x<1$, since $n(1-x)^n\to 0$ ($n\to\infty$) (recall that $0<1-x<1$) you can find $N=N(x,\varepsilon)$ such that if $n\ge N$, $n(1-x)^n<\varepsilon$.

Now $nx(1-x)^n\le n(1-x)^n<\varepsilon$, since $x\le 1$.

Title: Simplifying the inequality $nx(1-x)^n \leq \epsilon$ to $n > N(x, \epsilon)$?

Body:
How can I reduce the following inequality on $x \in [0, 1]$, and $\epsilon > 0$:
$$nx(1-x)^n \leq \epsilon$$
to the this form
$$n > N(x, \epsilon)\quad?$$
We can assume $n = 1, 2, 3, \dots$.

My attempt:

$$\ln\big(nx(1-x)^n\big) \leq \ln \epsilon$$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
\implies \ln n + n\ln(1-x) \leq \ln\frac{\epsilon}{x}$,

and I am stuck. Since $\ln(1-x) < 0$, I think I must divide by it to change the direction of inequality. But I don't know how to simplify/group $n$ and $\ln n$ while decoupling them from $x$ term.

Ultimately, I want to prove that $\forall n \geq N(x, \epsilon), \quad nx(1-x)^n \leq \epsilon$. Perhaps, I do not need to solve for $n$?


问:
if $A=B \cup C$ and $A,B$ are open and $C$ is closed then $C \subset B$?

if $A=B \cup C$ and $A,B$ are open and $C$ is closed then $C \subset B$?

I thought about this statement for another question that I had.

Suppose $A,B,C$ are subsets of $\mathbb{R}^n $ such that $A=B\cup C$ and $A,B$ are open and aren't closed, $C$ is closed and isn't open, does that necessaraly mean that $C\subset B$?

thanks/

答:
First of all, if you use the usual topology on $\mathbb R^n$, then the space is connected so the only sets that are both open and closed at the same time are $\mathbb R^n$ and $\emptyset$.

For a simple counterexample, consider the intervals $A = (-2, 2), B=(-2, -1) \cup (1, 2)$ and $C = [-1, 1]$.


答:
HINT: You can add either $\langle a,b\rangle$ or $\langle b,a\rangle$, but you have to add more than the single ordered pair. To add $\langle a,b\rangle$, let $L(a)=\{x\in A:x\le^* a\}$ and $R(b)=\{x\in A:b\le^* x\}$, and show that

$$\le^*\cup\;\big(L(a)\times R(b)\big)\;$$

is a partial order.

Title: Order linear such that extends a partial order

Body:
Let $\le$ a partial order on a set $A$. Prove that there exists a linear order $\le ^*$ on $A$ such that $a\le b$ implies $a\le ^*b$.

I considered $\mathcal{F}$ the family of all $\le _0 \subseteq A\times A$ such that $\le\subseteq\le _0$ and $\le _0$ is a partial order on $A$.

Then, using Zorn's lemma I concluded that $\mathcal{F}$ has a maximal element $\le ^*$. I proved that this is a partial order on $A$.

Now, I suppose this is not a linear order, then there exists $a,b\in A$ such that $(a,b)\not\in\le ^*$ and $(b,a)\not\in\le ^*$.

Where is the contradiction here? First, I considered $\le _0^* : = \le ^*\cup \{ (a,b)\}$ and I wanted to prove that this is a partial order, but I can't prove the transitivity.

Any hint? Thanks.


问:
How to find the probability that the variable will be less than the value of the random variable?
There is a random quantity $L$, which is evenly distributed from 0 to $l_{\max}$.

How to find the dependence of the probability that the variable $x$ (positive) will be less than the value of the random variable $L$?

Intuitively, this will be an exponential dependence from zero to $l_{\max}$.
But I cannot prove it to myself. Any thoughts in which direction to think?

答:
The random quantity L has the following distribution functio:

$$
F_{L}(x)=\begin{cases}
0, & x<0\\
\frac{x}{l_{max}}, & 0\leq x<l_{max}\\
1, & x\geq t_{stay}
\end{cases}
$$

The distribution function has the meaning of the probability that the value of a random variable l will be less than an arbitrary number x.

$$
F_{L}(x)=P(l<x).
$$

$l < x$ event, forms a complete group with $l\geq x$ event. Then the probability of this event is found by the formula:

$$
P(x\leq l)=1-P(l<x).
$$

Then the required probability is found by the formula:

$$
p(x)=1-\begin{cases}
0, & x<0\\
\frac{x}{l_{max}}, & 0\leq x<l_{max}\\
1, & x\geq l_{max}
\end{cases}
$$


问:
Relationship between meager subsets of $\mathbb{R}$ and $\omega^\omega$.
I'm reading this book about structure of the real line. One of the things that's proven there is that $\mathfrak{b}\leq non(\mathcal{M}).$ And the proof goes as follows:

For every $f\in\omega^\omega$ let us define $K_f$ as follows:
$$K_f=\lbrace g\in\omega^\omega:\ g\preceq f\rbrace.$$
We observe that
$$K_f=\bigcup_n \bigcap_{k>n} \lbrace g\in\omega^\omega:\ g(k)\leq f(k)\rbrace.$$
So for every $f$ a set $K_f$ is a union of nowhere-dense sets a thus - first category subset of space $\omega^\omega$.

Suppose that $\kappa<\mathfrak{b}$ and that $\lbrace f_\alpha:\ \alpha<\kappa\rbrace\subseteq\omega^\omega$. Then there exists a function $f\in\omega^\omega$ such that
$$(\forall \alpha<\kappa)(f_\alpha \preceq f).$$
But then we have
$$\lbrace f_\alpha:\ \alpha<\kappa\rbrace\subseteq K_f,$$
so $non(\mathcal{M})\geq\mathfrak{b}$.

My main question is why are we proving this for $\omega^\omega$ and not for $\mathbb{R}$? Are meager sets in $\omega^\omega$ somehow connected to meager sets of $\mathbb{R}$?

答: A simple answer: $\omega^\omega$ is just homeomorphic to the set of irrationals in $\Bbb R$. So if we cover it by $\kappa$ many non-meager sets, we can just add, say, one rational to each in the irrationals-version and get a similar cover for the reals and vice versa is similar as well. So the covering numbers for $\omega^\omega$ and $\Bbb R$ are always the same.


问:
Sequence of operators in $L^2(\mathbb{R})$
I am trying to solve the following exercise:

In the space $L^2(\mathbb{R})$ define a sequence of operators $(A_n)$ that are such that $||A_nf||=||f||$ and $\langle A_nf,g\rangle\to0$ for every $f,g\in L_2(\mathbb{R})$.

I have thought of defining $(A_n)f(x)=f(x+n)$ but I don't know how to proceed, can someone give me a hint? Thanks a lot.

答:
I thaught of taking
$$
A_n f=
\begin{cases}
f(x-n),\quad \text{if $x\ge n$}\\
0,\qquad \qquad\text{if $-n<x<n$}\\
f(x+n), \quad \text{if $x\le -n$}
\end{cases}
$$
It is an isometry, so $||A_n f||=||f||$.

$\langle A_n f,g\rangle=\int_{-\infty}^{-n} f(x+n)\overline{g(x)}dx + \int_{n}^{+\infty} f(x-n) \overline{g(x)}dx$.

The two integrals are tails of convergent integrals (since $\langle A_n f,g\rangle<\infty$), so they tend to zero as $n\to +\infty$


问:
Sequence of coprime integers
I am working on this problem:

Let $f(x)=x^2-2x+2$. Define a sequence
$$
\begin{align*}
a_0&=3\\
a_1&=f(a_0)=5\\
a_2&=f(a_1)=17\\
&\vdots\\
a_n&=f(a_{n-1}).
\end{align*}
$$
Prove that, for all $n$, if $p$ is a prime dividing $a_n$, then $p$ does not divide any $a_m$, where $m>n$.

What I've done so far: I know that if $x$ and $y$ are integers then $x-y$ divides $f(x)-f(y)$, since $f$ is a polynomial with integer coefficients. Therefore, if $p$ divides both $a_n$ and $a_m$, then $p$ divides $a_m-a_n$, so $p$ divides $f(a_m)-f(a_n)=a_{m+1}-a_{n+1}$, and so on. However, I am stuck trying to reach a contradiction. Any assistance would be appreciated, thanks!

答:
If $p$ divides $a_n$, then

$$ a_{n+1} = f(a_n) = a_n^2 - 2a_n + 2 \equiv 2 \pmod p $$

What happens next?

Title: Sequence of coprime integers


问:
Help understanding the fundamental system
Given the equation $$y'=Ay \quad \mathrm{for}\ A=\begin{pmatrix} 3 & -1\\1 &  1\end{pmatrix}$$ How does one compute the real fundamental system? So far I have  computed the sole eigenvalue of $\lambda = 2$ with a multiplicity of 2 and the corresponding eigenvector $v_1=\begin{pmatrix} 1\\1\end{pmatrix}$. My question is where to go from here? Since the solution will have the form $t\mapsto e^{\lambda t}(u+tv)$ with $\lambda\in \mathbb{R}; u,v \in \mathbb{R^2}$ where do I get my other vector from and how do I solve for the fundamental system?

答:
Your equations are: $$\frac{dy_1}{dx}=3y_1-y_2~~~(1), \frac{dy_2}{dx}=y_1+y_2~~~(2)$$ the repeated eigenvalue o0f $2$, means
$y_1(x)=(A x+B)e^{2x}$ now put this in (1) to get $y_2(x)=(Ax-A+B)e^{2y}.$
$A,B$ can be found by the initial values of $y_1(x_0)$ and $y_2(x_0)$.
The general solution is:
$$y(x)=\begin{pmatrix} ax+b \\ ax-a+b\end{pmatrix} e^{2x}$$


问:
Prove that $\zeta = \frac {(2+i)}{(2-i)}$ is not a root of unity of any degree, even though $|\zeta |=1$.
Prove that $\zeta = \frac {(2+i)}{(2-i)}$ is not a root of unity of any degree, even though $|\zeta |=1$.

Suppose it is a root of unity of some degree $n\in\mathbb{Z}, n \geq 1$, then (if I'm not wrong) we have $$Im(\zeta ^n) = 0$$ $$ \zeta^n - \zeta ^{n*} = 5^{-n}((3+4i)^n-(3-4i)^n) = 0 , ^*-complexconjugate$$ $$(3+4i)^n=(3-4i)^n$$
If I wanted to prove it's false by induction, would that make a proof? What could be other ways to prove it? I know that $\frac{arg(\zeta)}{\pi} \notin \mathbb{Q}$ (rational numbers), but it seems to be the conclusion of this exercise rather than proof.

答:
We have
$$ \zeta=\frac{2+i}{2-i}=\frac{(2+i)^2}{(2-i)(2+i)}=\frac{3+4i}{5}$$
and so $\zeta$ is a root of the polynomial
$$ X^2-\frac65X+1$$
or equivalently of
$$ 5X^2-6X+5.$$
If $\zeta$ is an $n$th root of unity (with clearly $n>2$), then it must be the case that $5X^2-6X+5\mid X^n-1$. This must hold not only in $\Bbb Q[X]$, but also in $\Bbb Z[X]$, which is absurd.


问:
Finding singular integral for a given pde
The following pde is given and I need to find the complete integral and the singular integral if it exists:

$pqz=p^2(qx+p^2)+q^2(py+q^2)$

Now, I am able to find the complete integral as it is in Clairaut's form. So the complete integral is given by;

$z=ax+by+\frac{a^4+b^4}{ab}$

For finding the singular integral, I need to eliminate the arbitrary constants $a$ and $b$. So, I differentiate the above equation w.r.t. $a$ and w.r.t. $b$ , and I get;

$x+\frac{3a^2}{b}-\frac{b^3}{a^2}=0$

$y+\frac{3b^2}{a}-\frac{a^3}{b^2}=0$

Now, from here I am not able to proceed further and eliminate $a$ and $b$.

Any kind of help or hint will be of great help. Thanking everyone in advance!!

答:
$x+\frac{3a^2}{b}-\frac{b^3}{a^2}=0\quad\implies\quad x=-\frac{3a^2}{b}+\frac{b^3}{a^2}$

$y+\frac{3b^2}{a}-\frac{a^3}{b^2}=0\quad\implies\quad y=-\frac{3b^2}{a}+\frac{a^3}{b^2}$

$z=ax+by+\frac{a^4+b^4}{ab}=a(-\frac{3a^2}{b}+\frac{b^3}{a^2})+b(-\frac{3b^2}{a}+\frac{a^3}{b^2})+\frac{a^4+b^4}{ab}$

After simplification : $\quad z= -\frac{a^4+b^4}{ab}$

The solution is obtained on parametric form :
$$\begin{cases}
x=-\frac{3a^2}{b}+\frac{b^3}{a^2} \\
y=-\frac{3b^2}{a}+\frac{a^3}{b^2} \\
z=-\frac{a^4+b^4}{ab}
\end{cases}$$
There is no need to eliminate the parameters $a,b$ since the solution is already obtained.

Anyways eliminating  $a,b$ would lead to an high degree polynomial equation. Analytical solving for $z(x,y)$ is not possible since the degree is higher than 4. The best way for practical use is numerical calculus starting from the above parametric form of solution.


问:
Is there any connections between boundary points and nowhere dense sets?
Boundary points are defined to be the closure of a set excluding its interior points. A nowhere dense set is when the interior of the closure is empty, so it feels like there should be a theorem liking these two sets. Is that the case?</p>

答:
The boundary of an open set and the boundary of a closed set are both always garantueed to be nowhere dense. If a set is nowhere dense, so is its boundary.

But the boundary of an arbitary set can be large (e.g. $\Bbb Q$ in the reals).


问:
How can a finite integral domain have a non-zero characteristic?
I know that by virtue of being a finite integral domain, $n\cdot 1_{R}=m\cdot 1_{R}$. Hence, $(n-m)\cdot 1_{R}=0$. The smallest positive element in the set of all possible such $(n-m)$'s is the characteristic.

But then again, in an integral domain, if $a\cdot b=0$, then either $a=0$ or $b=0$. Here, neither $n-m=0$ (by definiton, we have a non-zero characteristic), nor $1_{R}=0$ (we're assuming this is not a zero ring). Isn't this a contradiction?

答:
How can a finite integral domain have zero characteristic? It would contain a copy of the integers, which is a contradiction. So a finite integral domain must have non zero characteristic.

When you write $na$, for $n\in\mathbb{Z}$ and $a\in R$ you're not doing a product in $R$, but rather $\underbrace{a+a+\dots+a}_n$ (when $n>0$) or $-((-n)a)$ when $n<0$.

So there's no problem in having $3a=0$ in a domain. This just means that $a+a+a=0$, which happens in $\mathbb{Z}/3\mathbb{Z}$, for instance, which is a field.

I usually consider the unique ring homomorphism $\varepsilon\colon\mathbb{Z}\to R$, defined by $\varepsilon(n)=n1_R$.

In case the ring $R$ is finite, this homomorphism has a non zero kernel, say $k\mathbb{Z}$, so easily $k1_R=0$ and $k$ is the characteristic of $R$.


问:
About $\lim\dfrac{\sigma(n)}{n}$, where $\sigma(n)$ is the sum of divisor of $n$
Denote $p\equiv 1 \mod 3$ is a prime number and $n$ is any number that divisible by $p$. Denote $\sigma(n)$ is the sum of divisor of $n$.
We know that $\dfrac{\sigma (n)}{n}>1$ and
$$\dfrac{\sigma (n)}{n}=\prod\limits_{p_i\mid n}\dfrac{p_i^{\alpha_i+1}-1}{p_i^{\alpha+1}-p_i^\alpha}<\prod_\limits{p_i\mid n\\p_i\neq p}\dfrac{p_i}{p_i-1}.\dfrac{p}{p-1}$$
When $p$ goes to infinitely then $n\rightarrow \infty$, we have $\dfrac{p}{p-1}\rightarrow 1$. But does
$\dfrac{\sigma (n)}{n}\rightarrow 1$ when $p\rightarrow \infty$, since we can't control other $p_i$ and the number of divisor of $n$ ($n$ is arbitrary and $p\mid n$).

答:
The limit of $\dfrac {\sigma(n)}{n}$ does not exist, since for any multiple of $6p$, i.e. $n = 6pk$ we have

$$\frac {\sigma(n)}n > \frac {pk + 2pk + 3pk + 6pk}{6pk} = 2$$

while for $n = p^k$, we have $$\frac {\sigma (n)}n = \frac {p^{k+1} - 1}{p^k(p-1)  } < \frac {p^{k+1} + p^k(p-2)}{p^k(p-1)} = 2$$

Moreover, the ratio for prime powers tends to $1$, so for $n$ an arbitrary multiple of $p$, the limit of the abundancy index does not exist, no matter if $n \to\infty$ or $p \to \infty$.


问:
An affine $K$-algebra factored by a maximal ideal is contained in a finitely generated $K$-domain
I am trying to understand the proof of Proposition 1.2 from Gregor Kemper's "A Course in Commutative Algebra", which says that if $\varphi: A \rightarrow B$ is a homomorphism of the algebras $A,B$ over the field $K$ and if $B$ is an affine $K$-algebra and $\mathfrak{m}$ is a maximal ideal in $B$, then $\varphi^{-1}(\mathfrak{m})$ is a maximal ideal in $A$.

The step I'm having trouble with is when he uses Lemma 1.1,(b), which he proves previously and which states that an if an algebra $A$ over a field $K$ is itself a field and also $A$ is contained in an affine $K$-domain, then $A$ must be algebraic over $K$. This lemma is used for the $K$-algebra $B/\mathfrak{m}$ to conclude that $B/\mathfrak{m}$ is algebraic over $K$. I know that $B/\mathfrak{m}$ is a field because $\mathfrak{m}$ is maximal. But why is $B/\mathfrak{m}$ contained in an affine $K$-domain? Which is this finitely generated $K$-domain or how do we know $B$ is contained in one?

(A similar question was asked here for prime ideals instead of maximals, but got no answer.)

答:
I do not have a copy of your book and cannot comment on "affine $K$-algebras", but the following seems to hold: Let $A,B$ be commutative unital $k$-algebras with $k$ a field, where $B$ is finitely generated over $k$.

Lemma. Let $\phi:A \rightarrow B$ be a map of rings and let $\mathfrak{m}\subseteq B$ be a maximal ideal. It follows $\phi^{-1}(\mathfrak{m})\subseteq A$ is a maximal ideal.

Proof. Assume $k$ is a field and let $\mathfrak{n}:=\phi^{-1}(\mathfrak{m})$. We get an inclusion of rings

$$ k \subseteq \kappa(\mathfrak{n}) \subseteq \kappa(\mathfrak{m})$$

and $\kappa(\mathfrak{n})$ is an integral domain. Since $\kappa(\mathfrak{m})$ is a finite extension of $k$ (this follows from the HNS) the following holds: Let $0\neq x\in \kappa(\mathfrak{n})$ be an element. There is a canonical surjective map

$$ \psi: k[t]\rightarrow k(x)$$

defined by $\psi(t):=x$, with $ker(\psi):=(p(t))$ for a non-zero ideal $p(t)\in k[t]$. Since $k(x)$ is an integral domain it follows $(p(t))$ is a maximal ideal, hence $x$ is a unit in $\kappa(\mathfrak{n})$. Since $x$ was an arbitrary non-zero element, it follows $\kappa(\mathfrak{m})$ is a field and hence $\mathfrak{n}$ is a maximal ideal. QED

I believe a similar proof holds when $k$ is the ring of integers.

There is a class of rings  - Hilbert-Jacobson rings - where the following result holds:

Theorem. Let $A$ be a Hilbert-Jacobson ring and let $\phi: A\rightarrow B$ be a map of commutative unital rings where $B$ is a finitely generated $A$-algebra. It follows $B$ is a Hilbert-Jacobson ring, and for any maximal ideal $I\subseteq B$ it follows $J:=\phi^{-1}(I)\subseteq A$ is a maximal ideal and the canonical map $A/J \rightarrow B/I$ is a finite extension of fields.

A ring $A$ is a Hilbert-Jacobson ring iff every prime ideal in $A$ is the intersection of the maximal ideals containing it. You may find this result proved in Bourbakis "Algebre Commutative" V5.


问:
Implied By/Implies Continuity Using the Definition of Continuity
I have to determine which of some definitions are implied by or imply the continuity of a function in a point.

So I guess I should play with $\varepsilon$ and $\delta$ one time I understood what the definition means (using a graph for example).

Definition $\mathcal A$

$$\forall \varepsilon^* \gt 0 \quad \exists \delta^* \gt 0 : \quad if \quad |x-x_0|\lt \varepsilon^* \Rightarrow |f(x)-f(x_0)| \lt \delta^*$$

I tried to draw some graphs and it seems to me that in this case the definition of continuity implies the definition $\mathcal A$

Then I tried to prove that the continuity implies $\mathcal A$

We have (definition of continuity)

$$\forall \varepsilon \gt 0 \quad \exists \delta \gt 0 : \quad if \quad |x-x_0|\lt \delta \Rightarrow |f(x)-f(x_0)| \lt \varepsilon$$

Then we have that there exists a $\delta$ for the definition of continuity: I pick this $\delta$ and I use it inside $\mathcal A$

So I choose $\varepsilon^* = \delta $, for that $\delta$ I have that $|x-x_0|\lt\delta$, then $|x-x_0|\lt\varepsilon^*$

Before stating that $\delta$ existed I had to pick an $\varepsilon$, so now I can say there is a $\varepsilon$ such that $|f(x)-f(x_0)| \lt \varepsilon$

I choose $\delta^* = \varepsilon$, then I have

$$\forall \varepsilon^* \gt 0 \quad \exists \delta^* \gt 0 : \quad if \quad |x-x_0|\lt \varepsilon^* \Rightarrow |f(x)-f(x_0)| \lt \delta^*$$

The only thing that makes me unsure that this is a correct proof is the fact that I state $\forall \varepsilon^* = \delta$

But not $\forall \varepsilon^* \gt 0$

Does that make sense?

答:
Note that your definition $\cal A$ can be interpreted as following:

Given any $\epsilon^* > 0$, the function $f$ is bounded on $(x_0 - \epsilon^*, x_0 + \epsilon^*)$. If your function is defined on $\Bbb R$, then continuity does imply that, as can be seen by the Extreme Value Theorem.

However, it is not true in general. For example, consider $f : (0, 1) \to \Bbb R$ defined by $$f(x) = \dfrac{1}{x}.$$

Consider $x_0 = \dfrac{1}{2}$ and $\epsilon^* = \dfrac{1}{2}$. You should be able to show that no $\delta^* > 0$ exists.

(Note that $$\sup_{|x - x_0| < \frac12} |f(x) - f(x_0)| = \infty$$

in this case.)


问:
ODE: Initial Value Problem with a Change of Variables
this is a question that's had me sort of stumped. Any points or help would be greatly appreciated, I've never been that great at differential equations

The question at hand

The presence of the alpha seems to throw off any actual working I do such as manipulating for a value of y prime, and I'd rather use an integration factor but the change of variables has been specified for? I also can't seem to find any change of variable questions similar enough to follow the working. Thanks in advance to whoever helps out.

答:
$$y'-2xy=\dfrac {x^3}{y^3}$$
This is Bernoulli's differential equation.
$$y^3y'-2xy^4=x^3$$
Note that $(y^4)'=4y^3y'$. Substitute $v=y^4$ and solve.

You can surely take it from here.


问:
What are the possibilities of a $4 \times 4 \times 4$ array that has two $0$s and two $1$s in each line?
So this is a $4$ by $4$ by $4$ cube consisting of $64$ ($1$ by $1$ by $1$) small cubes, which can have the value of $0$ or $1$.
What we call a line here is strictly straight, and not diagonal. So if we look at one side of the cube we have $4 \times 4$ perpendicular lines from each of the small cubes.

This way we can calculate, that the total lines of the cube are $4 \times 4 \times 3$ (4*4 from the 3 different dimensional sides). And out of the $64$ small cubes $32$ have the value $0$, and $32$ have $1$.
(Just to make sure everyone understand what i am talking about)

The main question is how many different possibilities are there, where you can't rotate one cube to the other. And how can you generate them mathematically?

Other question maybe:
How many are there where it is a different matrix in every aspect (you cant rotate the matrix, swap slabs).

答:
As the other answer already described, this is straightforward to do with a computer program.

The C# code below uses a pre-defined list of the 6 ways to choose 2 out of 4 bits, which are the valid lines in the cube.

It then takes any 3 of those lines to start making a layer, checks that they are a valid combination so far (there are no columns with 3 bits set), calculates the fourth line of the layer (by taking the XOR of the existing lines so that each column has an even number of bits set), and checks that the fourth line is also a valid line. If so, we have a valid layer.

The next stage does the same but with the layers. So it counts all valid triplets of layers that produce a valid fourth layer.

```<code>     ISet<long> lines = new SortedSet<long> { 0b0011, 0b0101, 0b1001, 0b0110, 0b1010, 0b1100 };
     ISet<long> layers = new SortedSet<long>();

     // build layers
     foreach (long l1 in lines)
     foreach (long l2 in lines)
     foreach (long l3 in lines)
        if ((l1 & l2 & l3) == 0)
        {
           long l4 = l1 ^ l2 ^ l3;
           if (lines.Contains(l4))
           {
              long lay = l1 + (l2 << 4) + (l3 << 8) + (l4 << 12);
              layers.Add(lay);
           }
        }

     Console.WriteLine($"{layers.Count} layers");

     long count = 0;
     // build cubes
     foreach (long l1 in layers)
     foreach (long l2 in layers)
     foreach (long l3 in layers)
        if ((l1 & l2 & l3) == 0)
        {
           long l4 = l1 ^ l2 ^ l3;
           if (layers.Contains(l4))
           {
              count ++;
           }
        }

     Console.WriteLine($"{count} cubes");
</code>```

The result is:

90 layers

51678 cubes

Body:
As the other answer already described, this is straightforward to do with a computer program.

The C# code below uses a pre-defined list of the 6 ways to choose 2 out of 4 bits, which are the valid lines in the cube.

It then takes any 3 of those lines to start making a layer, checks that they are a valid combination so far (there are no columns with 3 bits set), calculates the fourth line of the layer (by taking the XOR of the existing lines so that each column has an even number of bits set), and checks that the fourth line is also a valid line. If so, we have a valid layer.

The next stage does the same but with the layers. So it counts all valid triplets of layers that produce a valid fourth layer.

     ISet<long> lines = new SortedSet<long> { 0b0011, 0b0101, 0b1001, 0b0110, 0b1010, 0b1100 };
     ISet<long> layers = new SortedSet<long>();

     // build layers
     foreach (long l1 in lines)
     foreach (long l2 in lines)
     foreach (long l3 in lines)
        if ((l1 & l2 & l3) == 0)
        {
           long l4 = l1 ^ l2 ^ l3;
           if (lines.Contains(l4))
           {
              long lay = l1 + (l2 << 4) + (l3 << 8) + (l4 << 12);
              layers.Add(lay);
           }
        }

     Console.WriteLine($"{layers.Count} layers");

     long count = 0;
     // build cubes
     foreach (long l1 in layers)
     foreach (long l2 in layers)
     foreach (long l3 in layers)
        if ((l1 & l2 & l3) == 0)
        {
           long l4 = l1 ^ l2 ^ l3;
           if (layers.Contains(l4))
           {
              count ++;
           }
        }

     Console.WriteLine($"{count} cubes");

The result is:

90 layers

51678 cubes


问:
Calculating surface integral with sphere
I need to calculate the following surface integral:$\iint x^2dydz+y^2dxdz+z^2dxdy$ on the region $S:(x-a)^2+(y-b)^2+(z-c)^2=R^2$ in the direction of outside normal.
So, first i found that normal has coordinates $(\frac{x-a}{R},\frac{y-b}{R},\frac{z-c}{R})$.(I am not sure should I use $+$ or $-$ outside the bracket.) Then, I am calculating $<(x^2,y^2,z^2),(\frac{x-a}{R},\frac{y-b}{R},\frac{z-c}{R})>$ and get $\frac{x^2(x-a)+y^2(y-b)+z^2(z-c)}{R}$. Using spherical coordinates $x=a+R\cos\phi\sin\theta, y=b+R\sin\phi\sin\theta, z=c+R\cos\theta$ I am proceeding with calculating this integral with bounds $0\le\phi\le2\pi, 0\le\theta\le\pi$. Is this correct?
So, my only doubts is should I use + or -, and is the following at attempt correct?

答:
Alternatively apply divergence theorem.

$\displaystyle \iint_S x^2dydz+y^2dxdz+z^2dxdy = \iiint_V (div \vec{F}) \ dV$

$\vec{F} = (x^2,y^2,z^2)$

$div(\vec{F}) = \nabla \cdot \vec{F} = 2(x+y+z)$

Please use $x = a + \rho \cos \theta \sin\phi, y = b + \rho \sin\theta \sin\phi, z = c + \rho \cos\phi$ for the integral.


答:
Consider what you mean when you write $na=0$ where $n$ is an integer:
$$\underbrace{a+\cdots+a}_{n\text{ times}}=0$$
It very well can be the case that this occurs for a non-zero $n$. This is not a contradiction because the expression $na$ does not mean a product of elements of the ring $R$; it is $\mathbb{Z}$ acting on the ring $R$.

Alternatively, you can think of "$n$" as being  a shorthand for
$$\underbrace{1_R+\cdots+1_R}_{n\text{ times}}$$
If you prefer this approach, you very well can have that $n=0$ in the ring $R$, and therefore it is not a contradiction to have $na=0$.

Title: How can a finite integral domain have a non-zero characteristic?

Body:
I know that by virtue of being a finite integral domain, $n\cdot 1_{R}=m\cdot 1_{R}$. Hence, $(n-m)\cdot 1_{R}=0$. The smallest positive element in the set of all possible such $(n-m)$'s is the characteristic.

But then again, in an integral domain, if $a\cdot b=0$, then either $a=0$ or $b=0$. Here, neither $n-m=0$ (by definiton, we have a non-zero characteristic), nor $1_{R}=0$ (we're assuming this is not a zero ring). Isn't this a contradiction?


问:
Show that $2^n ≤$ $n\choose n/2$ is contradictory
I am writing a proof and I am having issues with this part of the proof. I want to show that $2^n ≤$ $n\choose n/2$ is contradictory.

Body:
Hint:
$$ (1+1)^{2m}=\sum_{k=0}^{2m}{2m\choose k}1^k1^{2m-k}$$


问:
Max a min values of f(xy) on the boundary of a domain
So i have $f(x,y) = x^3 + y^3 - 3xy + 2$

I have a domain $D = \{(x,y) : x,y \geq 0, x^2 + y^2 \leq 4 \}$

I have to determine the maximum and minimum values of $f(x,y)$ on the boundary of the domain and I have no idea where to even start. I've got my partial derivatives and found the stationary/critical points of the function but I can't find the minimum and maximum values.

Thanks in advance

答:
$$
\begin{cases}
z = 2 + x^3 - 3 x y + y^3\\
x^2 + y^2 = 4\\
x\ge 0;\;y\ge 0\\
\end{cases}
$$

Converting to cylindrical coordinates the intersection curve of the two surfaces is, in parametric form
$$\left(2 \cos t,2 \sin t,8 \sin ^3t+8 \cos ^3t-12 \sin t \cos t+2\right),\;t\in[0,\pi/2]$$

that is the curve in blue in the image below.

We can see that $z=8 \sin ^3t+8 \cos ^3t-12 \sin t \cos t+2$ has its absolute maximum $z=10$ at $t=0\to x=2,\;y=0$ or $t=\pi/2\to x=0\;y=2$

It has its absolute minimum when $z'(t)=0$ that is
$$12 \sin ^2t-12 \cos ^2t-24 \sin t \cos ^2t+24 \sin ^2t \cos t=0$$
for $t=\pi/4$ absolute  minimum is $z=4 \sqrt{2}-4$ when $x=\sqrt{2},y=\sqrt{2}$


答: This is clear by a very simple combinatorial argument: Say $S=\{1,2\dots,n\}$. Then $2^n$ is the total number of subsets of $S$, while $C(n,n/2)$ is...

Title: Show that $2^n ≤$ $n\choose n/2$ is contradictory

Body: I am writing a proof and I am having issues with this part of the proof. I want to show that $2^n ≤$ $n\choose n/2$ is contradictory.


问:
Sum of digits of ternary and novenary numbers
When a natural number is represented as a ternary (base 3), the sum of the digits is 2007.  What are the maximal and minimal values of the sum of digits when it is represented as a novenary (base 9)?

I have no clue.

答:
So first we note that to convert from ternary to novenary we simply go from the least significant digit (or "trit") and pair them up and convert each pair into a novenary digit from ternary (I'll leave it to you to see why this is case)

So really we have a bunch of pairs $\in \{0,1,2\}^2$ such that all the numbers have a fixed sum (2007 in this case), and we want to maximise the sum of all the digits when we convert to base 9. So really we want to maximise the base 9 representation of every pair.

So to make the best use of the fixed sum, we need to only put non-zero values in the first element of the pair, as that corresponds to a higher value in base 9.

Interestingly from here (again I'll leave it to you to see why), it's completely irrelevant how we distribute the sum, as the base-9 sum is always the same. So let's just do it the obvious way and have $2007$ pairs all equal to $(1,0)$, so the maximal base-9 sum is $1*3*2007 = 6021$.


问:
Brownian bridge's level reaching probability
Let $X_{t}=\sigma\cdot B_{t}$ ($\sigma\in\mathbb{R}$) be a Brownian
bridge where $B_{t}$ is a standard Brownian bridge with fixed endpoints:
$B_{T_{1}}=a$ and $B_{T_{2}}=b$ on the interval $[T_{1},T_{2}]$,
and $m$ is a real constant.

How can I proove

$$\mathbb{P}\left(\exists t\in\left[T_{1},T_{2}\right]:\sup_{T_{1}\leq t\leq T_{2}}X_{t}>m\right)=\exp\left(-\frac{2\left(m-a\right)\left(m-b\right)}{\left(T_{2}-T_{1}\right)\sigma^{2}}\right)?
$$
I can proove $\mathbb{P}\left(\exists t\in\left[0,1\right]:\sup_{0\leq t\leq1}B_{t}>a+(b-a)\cdot t;\;a,b\in\mathbb{R}\right)=\exp\left(-2ab\right)$,
if $B_{t}$ is a standard Brownian bridge with $0$ startpoint and
endpoint, but I couldn't deal with the asked problem.

答:
To simplify things, I'll set $T_1=0$, $T_2=l$, and $\sigma=1$. Then the law of $\{X(t):0\le t\le l\}$ equals to that of
$$
\left\{a+(b-a)\frac{t}{l}+\frac{l-t}{l}W\!\left(\frac{lt}{l-t}\right):0\le t\le l\right\},
$$
where $\{W(t)\}$ is a standard BM. Thus, for $m\ge a\vee b$,
$$
\mathsf{P}(\exists t\in[0,l]:X(t)> m)=\mathsf{P}\!\left(\sup_{s\ge 0}\left(W(s)-\frac{m-b}{l}\cdot s\right)> (m-a)\right)=e^{-\frac{2(m-b)(m-a)}{l}}.
$$
(See, e.g., Eq. 1.1.4(1) on page 256 here.)


问:
How to derive function of complex variable for a given expression?
Suppose I have a complex-valued expression such as $w(z) = \frac{y}{x}.$ How can I derive the means of representing this expression in terms of $z$ and $\bar{z}$? I'm guessing I might need the Cauchy-Riemann equations if not algebra, but it would be nice to have a more general method for problems like this.

Body: $$W(z)=\frac{y}{x}=-i\frac{z-\bar z}{z+\bar z}$$


答:
When $x=0, 0\leq y \leq 2$, the function becomes $f(0,y)=y^3+2$, attaing its minimum (2) at $(0,0)$ and its maximum (10) at $(0,2)$.

When $y=0, 0\leq x \leq 2$, the function becomes $f(x,0)=x^3+2$, attaing its minimum (2) at $(0,0)$ and its maximum (10) at $(2,0)$.

Regarding the last section of the boundary, we can use Lagrange multipliers (and ignore for now the other restricitons):

The candidate points are the critical points of $L(x,y,\lambda)=x^3+y^3-3xy+2-\lambda(x^2+y^2-4)$. (You can check the assumptions that guarantee this is true). The only relevant critical point in this context will be $(\sqrt{2},\sqrt{2})$, and, at this point, $f=4 \sqrt{2}- 4$.

Putting everything together, the global maximum is 10 and the global minimum is $4\sqrt{2} -4$.

Title: Max a min values of f(xy) on the boundary of a domain

Body:
So i have $f(x,y) = x^3 + y^3 - 3xy + 2$

I have a domain $D = \{(x,y) : x,y \geq 0, x^2 + y^2 \leq 4 \}$

I have to determine the maximum and minimum values of $f(x,y)$ on the boundary of the domain and I have no idea where to even start. I've got my partial derivatives and found the stationary/critical points of the function but I can't find the minimum and maximum values.

Thanks in advance


问:
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$
Where do I start to solve a equation for x like the one below?

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

After squaring it, it's too complicated; but there's nothing to factor or to expand?

Ideas?

答:
Hint: below let $\, a,b,c,d\, =\, \sqrt{7x\!-\!4},\ \sqrt{7x\!-\!5},\ \sqrt{4x\!-\!1},\ \sqrt{4x\!-\!2}$</p>

<p>$$\begin{eqnarray}  & a^2-b^2 &=\,& c^2-d^2 \ne 0 \\
  & a-b  &=\,&  c-d \\
 \Rightarrow &  a+b  &=\,&  c+d\quad \text{via quotient of above two} \\
 \Rightarrow & \qquad 2a &=\,& 2c\qquad\ \text{via sum of above two} \\ \Rightarrow &\qquad a &=\,& c\,\ \Rightarrow\ b=d \\
\Rightarrow &\qquad a^2 &=\,& c^2 \Rightarrow\ b^2\!=d^2 \end{eqnarray}$$  </p>

Body:
Hint: below let $\, a,b,c,d\, =\, \sqrt{7x\!-\!4},\ \sqrt{7x\!-\!5},\ \sqrt{4x\!-\!1},\ \sqrt{4x\!-\!2}$

$$\begin{eqnarray}  & a^2-b^2 &=\,& c^2-d^2 \ne 0 \\
  & a-b  &=\,&  c-d \\
 \Rightarrow &  a+b  &=\,&  c+d\quad \text{via quotient of above two} \\
 \Rightarrow & \qquad 2a &=\,& 2c\qquad\ \text{via sum of above two} \\ \Rightarrow &\qquad a &=\,& c\,\ \Rightarrow\ b=d \\
\Rightarrow &\qquad a^2 &=\,& c^2 \Rightarrow\ b^2\!=d^2 \end{eqnarray}$$


答:
Replace every occurrence of $x$ with $\frac12 (z + \bar{z})$; replace every occurrence of $y$ with $\frac1{2i}(z-\bar{z})$.

In your example, you get
\begin{align}
w(z) &= \frac yx \\
&= \frac{\frac1{2i}(z-\bar{z})}{\frac12 (z + \bar{z})} \\
&= \frac{\frac1{i}(z-\bar{z})}{ (z + \bar{z})} \\
&= -i\frac{(z-\bar{z})}{ (z + \bar{z})} \\
\end{align}

Title: How to derive function of complex variable for a given expression?

Body: Suppose I have a complex-valued expression such as $w(z) = \frac{y}{x}.$ How can I derive the means of representing this expression in terms of $z$ and $\bar{z}$? I'm guessing I might need the Cauchy-Riemann equations if not algebra, but it would be nice to have a more general method for problems like this.


问:
Elements of $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ as Cauchy sequences
There is an isomorphism of abelian groups  $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{R}$, a proof is based on three observations:

$\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ is torsion free.

$\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ is a divisible abelian group.

Hence  $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ is a vector space over $\mathbb{Q}$ and it has the same cardinality as $\mathbb{R}$, hence there is an isomorphism of $\mathbb{Q}$-vector spaces $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{R}$.

This proof is indirect. If we think about elements of $\mathbb{R}$ as classes of Cauchy sequences of elements of $\mathbb{Q}$, can we directly interpret such sequence as an element in $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$?

答:
$$End(\Bbb{Q/Z})=\varprojlim_{n\to \infty}
End(\Bbb{n^{-1}Z/Z})= \varprojlim_{n\to \infty} \Bbb{Z/nZ}$$
(the profinite integers)

$$Hom_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z})=\Bbb{Q\otimes_Z}End(\Bbb{Q/Z})=\Bbb{A_{Q,fin}}$$
(the finite adeles)

It is well-known that the $\Bbb{Q}$-vector space isomorphisms or embeddings $\Bbb{A_{Q,fin}}\to \Bbb{R}$ need the axiom of choice and that none has a concrete expression (no idea in what theory we can prove it, if we can).

$\Bbb{A_{Q,fin}}$ is the completion of $\Bbb{Q}$ for the norm $$\|\pm \prod_j p_j^{e_j}\|=\sup_j p_j^{-e_j}, \qquad e_j\in \Bbb{Z}$$


问:
Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.

Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$

My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx = \int_0^{\pi/2} \frac{\sin^3x}{\cos^2x + \sin^3x}dx.$$
Therefore
$$2I = \int_0^{\pi/2} \left(\frac{\cos^3x}{\sin^2x + \cos^3x} + \frac{\sin^3x}{\cos^2x + \sin^3x}\right) dx.$$

I stucked here.

答: This integral does have a closed form despite being very messy. We have \begin{align}I&=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}\,dx\\&=\frac14\int_{-\pi}^\pi\frac{\cos^3(u/2)}{\sin^2(u/2)+\cos^3(u/2)}\,du\\&=\frac14\oint_{|z|=1}\frac{\frac18(z^{1/2}+z^{-1/2})^3}{-\frac14(z^{1/2}-z^{-1/2})^2+\frac18(z^{1/2}+z^{-1/2})^3}\frac{dz}{iz}\\&=-\frac i4\oint_{|z|=1}\frac{(z+1)^3}{z((z+1)^3-2z^{1/2}(z-1)^2)}\,dz.\end{align} Note that $(z+1)^3-2z^{1/2}(z-1)^2=(w^2+1)-2w(w^2-1)^2=\prod\limits_{i=1}^6(w-z_i)$ where $z_1=\overline{z_2}$, $z_3=\overline{z_4}$ and $z_5=\overline{z_6}$. Hence $$I=-\frac i4\cdot2\pi i\left(\operatorname{Res}(f,0)+\sum_{i=1}^6\operatorname{Res}(f,z_i^2)\right)=\frac\pi2\left(1+\sum_{i=1}^6\operatorname{Res}(f,z_i^2)\right).$$ An interesting property about this polynomial is that it is symmetric about its coefficients \begin{align}(w^2+1)-2w(w^2-1)^2&=w^6-2w^5+3w^4+4w^3+3w^2-2w+1\\&=w^3(w^3-2w^2+3w+4+3w^{-1}-2w^{-2}+w^{-3}).\end{align} This means we can make the substitution $w=v+1/v$ to obtain a cubic equation which is solvable in radicals, as in @Quanto's answer.


问:
Find the limit by L'hospital rule
Finding limit by l'hospital rule

$$
\lim_{x \to 0} \frac{7^{2x}-5^{3x}}{2x - \arctan 3x}
$$

Can anyone to explain me how the L'hospital rule works?

答: $$\frac{7^{2 x}-5^{3 x}}{2 x-\tan ^{-1}(3 x)}\to\frac{2 \times 7^{2 x} \ln 7 -3 \times 5^{3 x} \ln 5}{2-\displaystyle \frac{3}{9 x^2+1}}$$ No more problem. Compute the limit.

Title: Find the limit by L'hospital rule


问:
Suppose that $V = R(T) \oplus W$ and $W$ is $T$-invariant. Show that if $V$ is finite-dimensional, then $W = N(T)$.
Part (a) of the problem has already asked me to prove that $W \subseteq N(T)$.

I did this using the following argument for any arbitray $x \in W$:

$T(x) \in W$ since $W$ is $T$-invariant.

$W$ is a subspace of $V$, so $x \in V$, so $T(x) \in R(T)$.

Thus, $T(x) \in W$ and $T(x) \in R(T) \to T(x) \in W \cap R(T) \to T(x) = 0 \to x ∈ N(T)$.

However, for showing that $W = N(T)$ I can only think of finding a way to prove the other direction of the subset  relation, that is $N(T)\subseteq W$.

The solution says that by observing that both $W$ and $N(T)$ have the same dimension $= \dim(V) - \dim(R(T))$, and combining it with the conclusion of part(a), it can be concluded that $W = R(T)$.

I can not see why this is true.
The book where this question is from explains that if a subspace of a vector space has equal dimension to it, then the subspace must be equal to the vector space, but in this case it is being applied to two separate subspaces of the same vector space?

Can someone please explain why the solution is true or provide another solution?

答:
For anyone else who might be stuck here, here's the solution, thanks to the quick replies.

Let $\beta$ = {$v_1, v_2, v_3, ....., v_k$} be a basis for W.
Observe that $\beta \subseteq$ W $\subseteq$ N(T) and thus, $$\beta \subseteq N(T)$$

Hence, $span(\beta) = W$ is a subspace of N(T).

It now becomes clear that because:

W is a subspace of N(T)

$dim(W) = dim(N(T)) = dim(V) - dim(R(T))$
$$W = N(T)$$


Title: Find the limit by L'hospital rule

Body:
Finding limit by l'hospital rule

$$
\lim_{x \to 0} \frac{7^{2x}-5^{3x}}{2x - \arctan 3x}
$$

Can anyone to explain me how the L'hospital rule works?

Body:
$$L=\lim_{x\to 0}\frac{7^{2x}-5^{3x}}{2x-\tan^{-1}3x}$$
D. up and down separtely w.r.t. $x$
$$L=\lim_{x\to 0} \frac{2.7^{2x} \ln 7 -3.5^{3x}\ln 5}{2-\frac{3}{1+9x^2}}$$
$$L=-\ln(7^2/5^3)=\ln(49/125).$$