问:
Prove the existence of conditional expectations in $\mathcal{L}^2$
Let $X \in \mathcal{L}^2(\Omega,\mathcal{A},P)$ be a square integrable real random variable. Then for any sub-$\sigma$-algebra $ \mathcal{C}$ of $\mathcal{A}$ and any $ \mathcal{C}$ measurable $Y \in \mathcal{L}^2(\Omega,\mathcal{A},P)$ we have

$$E[(X-X_0)^2]\leq E[(X-Y)^2]$$

with equality holding if and only if $Y=X_0$ a.s.. Here $X_0$ is a version of the conditional expectation $E[X|\mathcal{C}]$.

The exercise asks to use this result to show the existence of $E[X|\mathcal{C}]$ for any $X \in \mathcal{L}^2(\Omega,\mathcal{A},P)$ using the theory of Hilbert spaces.

My argument is

Pass over to the quotient space $L_2:=\mathcal{L}^2(\Omega,\mathcal{A},P) / \mathcal{N}(\Omega,\mathcal{C},P_{\mathcal{C}}) $ where

$$\mathcal{N}(\Omega,\mathcal{C},P_{\mathcal{C}}):=\{X\in \mathcal{L}^2(\Omega,\mathcal{C},P_{\mathcal{C}}):X=0 \text{  }P_{\mathcal{C}}\text{-a.s.} \} \subset \mathcal{L}^2(\Omega,\mathcal{A},P) $$
and $P_{\mathcal{C}}$ denotes the restriction of $P$ to $\mathcal{C}$. This is a Hilbert space with respect to the norm induced by the following inner product:

$$\langle \tilde{X},\tilde{Y} \rangle \mapsto \int XYdP$$

where $X$,$Y \in \mathcal{L}^2(\Omega,\mathcal{A},P)$ are arbitrary representatives of the classes $\tilde{X}$,$\tilde{Y} \in L_2$ respectively.

Let $L^{\mathcal{C}}_2:=\mathcal{L}^2(\Omega,\mathcal{C},P_{\mathcal{C}}) / \mathcal{N}(\Omega,\mathcal{C},P_{\mathcal{C}})$. Since $\mathcal{L}^2(\Omega,\mathcal{C},P_{\mathcal{C}}) \subset \mathcal{L}^2(\Omega,\mathcal{A},P)$, every equivalent class of  $L^{\mathcal{C}}_2$ is an equivalent class of $L^2$, and so $L^{\mathcal{C}}_2$ is subspace of $L^2$.

Now, $L^{\mathcal{C}}_2$ is a Hilbert space in its own right. It is complete, and in particular closed. Hence given $X\in \mathcal{L}^2(\Omega,\mathcal{A},P)$ we can apply the projection theorem to an arbitrary $\tilde{X} \in L^2$ and obtain a unique vector $\tilde{X}_0 \in L^{\mathcal{C}}_2$ such that

$$ || \tilde{X} - \tilde{X}_0 || \leq || \tilde{X} - \tilde{Y} || $$

for all $\tilde{Y} \in L^{\mathcal{C}}_2$, with equality holding if and only if $\tilde{Y}=\tilde{X}_0$.

Translating back into expectations and using the result above we see that each $X_0 \in \tilde{X}_0$ is a version of $E[X|\mathcal{C}]$. Conversely if $Z_0$ is a version $E[X|\mathcal{C}]$, then we see that we must have $\tilde{Z}_0=\tilde{X}_0$.

Hence $\tilde{X}_0$ is precisely the class of all conditional expectation versions of $X$.

Am I missing something? My definition of $L^2$ is not the usual one, since I use $\mathcal{N}(\Omega,\mathcal{C},P_{\mathcal{C}})$ instead of $\mathcal{N}(\Omega,\mathcal{A},P)$. I feel like this is necessary.

答:
As noted by Ian, my definition of $L^2$ is problematic because the inner product $\langle \tilde{X},\tilde{Y} \rangle \mapsto \int XYdP$ is not positive definite. This is because if $X=0$ $P$-a.s. and $X$ is not $\mathcal{C}$ measurable, then $\tilde{X}\neq\tilde{0}$ but $\langle \tilde{X},\tilde{X} \rangle =0$.

We can alternatively proceed thus:

Define $$L_2:=\mathcal{L}^2(\Omega,\mathcal{A},P) / \mathcal{N}(\Omega,\mathcal{A},P)$$

$$L^{\mathcal{C}}_2:=\mathcal{L}^2(\Omega,\mathcal{C},P_{\mathcal{C}}) / \mathcal{N}(\Omega,\mathcal{C},P_{\mathcal{C}})$$

with the same notation as before. Then both $L_2$ and $L^{\mathcal{C}}_2$ are Hilbert spaces with the usual inner product.

Since $\mathcal{L}^2(\Omega,\mathcal{C},P_{\mathcal{C}}) \subset \mathcal{L}^2(\Omega,\mathcal{A},P)$ and $\mathcal{N}(\Omega,\mathcal{C},P_{\mathcal{C}}) \subset  \mathcal{N}(\Omega,\mathcal{A},P)$, we see that each equivalent class of $L^{\mathcal{C}}_2$ is included in some equivalent class of $L^2$. Moreover, if $X,Y\in \mathcal{L}^2(\Omega,\mathcal{C},P_{\mathcal{C}})$ belong to the same equivalent class in $L^2$, then $X=Y$ $P$-a.s. (hence also $P_{\mathcal{C}}$-a.s.), and so $X$ and $Y$ are from the same equivalent class in $L^{\mathcal{C}}_2$.

The previous argument shows that there exist an linear injection $I:L^2_{\mathcal{C}}\to L^2$. Because only null sets are involved, we see that

$$||\tilde{X}||_{L^{\mathcal{C}}_2}=||I(\tilde{X})||_{L_2} \hspace{1cm} \text{and} \hspace{1cm} \langle \tilde{X},\tilde{Y} \rangle_{L^{\mathcal{C}}_2}= \langle I(\tilde{X}),I(\tilde{Y}) \rangle_{L^{}_2}$$

for all $\tilde{X},\tilde{Y} \in L^{\mathcal{C}}_2$. Hence from the completeness of $L^2_{\mathcal{C}}$ follows the completeness of $I(L^{\mathcal{C}}_2)$, i.e. $I(L^{\mathcal{C}}_2)$ is a Hilbert subspace of $L^2$.

Now the argument can proceed as before with the conclusions unchanged. For example the projection theorem now reads

For every $\tilde{X} \in L^2$ and there is a unique vector $\tilde{X}_0 \in I(L^{\mathcal{C}}_2)$ such that

$$ || \tilde{X} - \tilde{X}_0 ||_{L^2} =\min_{\tilde{Y} \in I(L^{\mathcal{C}}_2)} || \tilde{X} - \tilde{Y} ||_{L^2} $$

which, on account of the injectivity of $I$, is translated as

For every $\tilde{X} \in L^2$ and there is a unique vector $\tilde{X}_0 \in L^{\mathcal{C}}_2$ such that

$$ || \tilde{X} - \tilde{X}_0 ||_{L^2} =\min_{\tilde{Y} \in L^{\mathcal{C}}_2} || \tilde{X} - \tilde{Y} ||_{L^2} $$


问:
Is the countable union of measure-zero sets zero?
Let $E=\cup_{i=1}^\infty A_i$ where the measure of $A_i$ is zero. How can I conclude that $E$ has measure zero?</p>

答:
If your measure $\mu$ is  sub sigma additive you can surely say that
\[ \mu(E)\leq \sum_{i=1}^\infty \mu (A_i)=0\]
If it's not the statement doesn't need to be true.</p>

ContentLicense: CC BY-SA 3.0


问:
Putting Pins in a Grid - UK Senior Maths Challenge

In an $8 \times 2$ grid, you can place $5$ blue pins, $2$ red pins and $1$ yellow pin. The only rule is that the two red pins cannot share an edge. How many different arrangements of the pins can be made?

Looking at the arrangements top to bottom are considered different arrangements, as are using left and right columns.

I think you can have 36 different arrangements for the red pins and 8 for the yellow pin, and since the blue pins are all the same, does this mean that the answer would be 288? I feel like my reasoning has gone wrong somewhere though, so if you could offer any advice, thanks.

答:
If one red pin goes to one of four corners, the other red can go to $16-2-1=13$ squares. This is $4\cdot13/2 = 26$ ways for the pair.

If a red pin goes to any of $12$ non-corner squares, the other red can go to $16-3-1=12$ squares. This is $12\cdot 12/2 = 72$ ways for the pair.

After $26+72=98$ ways of arranging two red pins, five blue can be arranged in any of $14$ empty squares, followed by one yellow in one of remaining $9$ empty squares in following ways$$ 98 \cdot \binom{14}{5} \cdot 9$$

ContentLicense: CC BY-SA 4.0


问:
Integration by parts is not working.
I an trying to solve $$\int_{-R}^{R}\tan^{-1}(\sin^2(x))dx$$ with Integration by Parts, by integrating 1 and differentiating our integrand, yielding $$2R\tan^{-1}(\sin^2(x))-2\int_{-R}^{R}\frac{x\sin(x)\cos(x)}{\sin^4(x)+1}$$ and when I plug these into Desmos they yield different results. This probably has something to do with my assumption of the principal value of $\tan^{-1}(\sin^2(x))$, but I don't know how to fix it.

答: It turns out the error was with Desmos. My plan was to divide both sides by $\frac{1}{2R}$ to make this an average value and take $R\to\infty$ which allows me to use Jordan's lemma of the integral. The issue is, the integral is kind of messy so for large values of $R$ the value will not be equal, but in reality the integration by parts is correct and there is no error in the integration.

Title: Integration by parts is not working.


问:
Estimating $|\mu_1 - \mu_2|$ in a mixture of two Gaussians with known variance.
Consider $n$ samples denoted with $X^n = X_1, \ldots, X_n$ from mixture of two Gaussians with equal variance and equal mixture proportions:

$X^n \sim 1/2 \mathcal{N}(\mu_1, 1) + 1/2 \mathcal{N}(\mu_2, 1)$.

Now the goal is to estimate the difference in means $|\mu_1 - \mu_2|$ from this $n$ samples. One idea is to rank them from lowest to highest, and take the difference between the 1/4th highest and lowest sample point. But how accurate is this estimate, and is there a better way?

答:
The first two moments of this mixture are:
$$\mathbb{E}[X] = \frac{\mu_1+\mu_2}{2}$$
$$\mathbb{E}[X^2] = \frac{\mu_1^2+\mu_2^2}{2}+1$$
Substituting the true expectation by their respective means:
$$S_1=\frac{1}{n}\sum_{k=1}^nX_k$$
$$S_2=\frac{1}{n}\sum_{k=1}^nX_k^2$$
We can transform the system of equations:
$$
\begin{cases}
S_1 = \frac{\mu_1+\mu_2}{2} \\
S_2 = \frac{\mu_1^2+\mu_2^2}{2}+1
\end{cases}
$$
To get:
$$
\begin{cases}
\mu_1+\mu_2 = 2S_1 \\
\mu_1^2+\mu_2^2 = 2S_2 - 1
\end{cases}
$$
Now, knowing that $(x+y)^2+(x-y)^2=2(x^2+y^2)$ or equivalently $(x-y)^2=2(x^2+y^2) - (x+y)^2$ we can obtain the value of:
$$(\mu_1-\mu_2)^2 = 4(S_2-S_1^2-1)$$
And so:
$$|\mu_1-\mu_2| = 2\sqrt{S_2-S_1^2-1}$$
This estimator is at least consistent. Moreover, both $S_1$ and $S_2$ will be asymptotically normal. They are definitely not independent though and I don't even know about their covariance (though I suspect that their asymptotic distribution is jointly normal so the covariance is the only factor deciding their dependence). It is therefore not easy (at least not for me) to compute the bounds for the error of this estimator other than the MC simulation.


问:
Factorization of $f(x) = x^n - 1$ into irreducible polynomials $\mathbb{R}[x]$
Knowing that $e^{2\pi i \frac{k}{n}} = \cos(2 \pi k/n) + i \sin(2 \pi k/n) = \alpha $ is complex roots of $f(x) = x^n - 1$ and therefore the complex factorization is

\begin{align}
    f(X) &= \displaystyle\prod^{n - 1}_{k = 0} (x - \alpha_k)
\end{align}

How do I factorize into irreducible polynomials in $\mathbb{R}$ for odd and even n's? I know that for $(x-1) \mid f(x)$ for all n and if n is even $(x+1)\mid f(x)$ as well. I'm also aware that $(x - \alpha)(x-\bar{\alpha})$ is an irreducible polynomial in $\mathbb{R}$ that also divides $f(x)$. So the remaining factors most take the form $x^2 - 2x\cos(2 \pi k/n) + 1$. But I fail to see which k's to use in the factorization?

答:
If I understand your question correctly, if $\exp[2\pi ik/n]$ has a non-zero imaginary component, re $\sin[2\pi k/n]$, then you have to combine it with its complex conjugate.  Otherwise, you can leave it alone.

Addendum
Responding to the OP's comment:

Let $c_1 \equiv \cos(2\pi/5), ~c_2 \equiv \cos(4\pi/5).$

Let $s_1 \equiv \sin(2\pi/5), ~s_2 \equiv \sin(4\pi/5).$

Then $(z^5 - 1)$ factors into

$(z-1) ~\times $

$(z - [c_1 + i(s_1)]) ~\times~ (z - [c_1 - i(s_1)]) ~\times $

$(z - [c_2 + i(s_2)]) ~\times (z - [c_2 - i(s_2)]).$

This equals

$(z-1) ~\times $

$[z^2 - (2c_1)z + ([c_1]^2 + [s_1]^2)] ~\times~ $

$[z^2 - (2c_2)z + ([c_2]^2 + [s_2]^2)].$

This equals

$(z-1) ~\times $

$[z^2 - (2c_1)z + (1)] ~\times~ $

$[z^2 - (2c_2)z + (1)].$


问:
Problem on implicit functions
I'm asked in Ted Shifrin's Textbook Multivariable Mathematics to :

Suppose $h: \mathbb{R}^2 \to \mathbb{R}$ is $\mathcal{C^1}$ and $\frac{\partial h}{\partial x_2} \neq 0$. Show that the equation $h \begin{pmatrix}y/x \\z/x \end{pmatrix}=0$ defines $z$  $\\$ (locally) implicitly as a $\mathcal{C^1}$ function $z= \phi \begin{pmatrix}x \\y \end{pmatrix} $,

This is easy enough: I fix a $x \neq 0$, use the first assumption and then apply the implicit function Theorem.

However I'm also asked to show that $$x \ \frac{\partial \phi}{\partial x} + y \ \frac{\partial \phi}{\partial y} = \phi \begin{pmatrix}x \\y \end{pmatrix}$$

Which has me stuck. I thought about trying to show that the equation of the tangent plane at $\begin{pmatrix}x \\y \end{pmatrix}$ is zero at the point $\pmb 0$ but I can't go further than that.
$$\\$$

Edit after Ted Shifrin asked me to.

Since $h$ does change as we wiggle $x_2$ we can fix $x \neq 0$ to obtain a continuous one to one onto function $\mathbb{R} \to \mathbb{R}$, namely: $z/x$, and so we know that if we wiggle $z$ then $h \begin{pmatrix}y/x \\z/x \end{pmatrix}$ changes too (the partial derivative isn't zero) using the previous fact. This, together with the assumptions in the exercise about $h$, let us deduce that there is a $\mathcal{C^1}$ function $\phi \begin{pmatrix}x \\y \end{pmatrix} = z$ .

The equation of the tangent plane of the graph of $\phi$ at $\pmb x$ is:
$$g\begin{pmatrix}v \\ w \end{pmatrix}= \phi\begin{pmatrix}x \\ y \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}v-x\\ w-y \end{pmatrix}$$
So, evaluating $g$ at $\pmb 0$ we obtain:

$$g\begin{pmatrix}0 \\ 0 \end{pmatrix}= \phi\begin{pmatrix}x \\ y \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}-x\\ -y \end{pmatrix}$$

$$g\begin{pmatrix}0 \\ 0 \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} = \phi\begin{pmatrix}x \\ y \end{pmatrix}$$

But I can't show how $g\begin{pmatrix}0 \\ 0 \end{pmatrix}=0$

答:
Because $\dfrac{\partial h}{\partial x_2}\ne 0$, the Implicit Function Theorem tells us that locally we can express the level curve $h(x_1,x_2)=0$ as a graph $x_2 = F(x_1)$ for some smooth function $F$. Substituting $x_1=y/x$ and $x_2=z/x$, we get $\frac zx = F\big(\frac yx)$, so we have
$$z=\phi(x,y)= xF(\frac yx),$$
with $F$ smooth and hence $\phi$ smooth.
Now we just differentiate this equation (using product and chain rule), obtaining
$$\frac{\partial\phi}{\partial x} = F(\tfrac yx) -\frac yx F'(\tfrac yx) \quad\text{and}\quad \frac{\partial\phi}{\partial y} = F'(\tfrac yx).$$
Now the rest is easy.

An alternative solution is to define $g(x,y,z)=(\frac yx,\frac zx)$ and set $H(x,y,z) = (h\circ g)(x,y,z)$. Now apply the chain rule to $H$ and use the implicit function theorem to show that since $\partial H/\partial z\ne 0$, we can locally express the level surface $H(x,y,z)=0$ as $z=\phi(x,y)$ for a smooth function $\phi$. This approach is good practice with the techniques.

Title: Problem on implicit functions


问:
Is every matrix norm compatible with a vector norm?
in $\mathbb{C}^n$, I know every vector norm induces a matrix norm, and an induced matrix norm is compatible to its dedicated vector norm. So for every vector norm there exists a matrix norm, where the matrix norm is compatible with the vector norm. But is the other implication true too?

Is there for every matrix norm a vector norm it is compatible with?

答:
Yes. This is the result of theorem 5.7.13 on p.373 of Horn and Johnson's Matrix Analysis (2/e).

A (submultiplicative) matrix norm $\|\cdot\|_M$ is said to be compatible with a vector norm $\|\cdot\|_v$ if $\|Ax\|_v\le\|A\|_M\|x\|_v$ for every matrix $A\in M_n(\mathbb C)$ and every vector $v\in\mathbb C^n$. Now, given any matrix norm $\|\cdot\|_M$, we pick a fixed but arbitrary nonzero vector $y$ and define a vector norm $\|\cdot\|_v$ by $\|x\|_v=\|xy^\ast\|_M$. Then $\|\cdot\|_M$ is compatible with $\|\cdot\|_v$ because
$$
\|Ax\|_v=\|Axy^\ast\|_M\le\|A\|_M\|xy^\ast\|_M=\|A\|_M\|x\|_v.
$$


问:
Formula for perimeter of a figure formed by cutting 4 inscribed congruent figures.
The area of a cross formed by cutting 4 congruent squares from a large square is $20$.  If the cross is a regular polygon, its perimeter equals:

a.)$24$

b.)$48$

c.)$64$

d.)$12\sqrt{5}$

e.)$16\sqrt{5}$

答:
A picture will tell you everything. The "regular polygon" part is intended to tell you that all sides of the cross are of equal length.

The picture will tell you that the cross is made up of $5$ squares. The total area is $20$, so each of the $5$ squares has area $4$, and therefore side $2$. Now you know enough to find the perimeter.

There is no need for formulas, just add up. But you might notice something nice. The cross was made by cutting out $2\times 2$ squares from the corners of a $6\times 6$ square. How is the perimeter of the cross related to the perimeter of the original square?


问:
Weak operator topology and matrix representation
For simplicity let $H$ be the separable Hilbert space and $(e_i)_i$ be a fixed orthonormal basis. Weak operator topology on $B(H)$ is usually defined using seminorms $|(Tu,v)|$ where $u,v$ are vectors in $H$. Let us denote $\mathcal{T}$ the topology induced by the seminorms $|(Te_i,e_j)|$. I believe this is also the topology induced by all $|(Tu,v)|$ where $u,v$ are finite linear combination of $e_i$'s. What is the relation between $\mathcal{T}$ and WOT? Do they agree on norm bounded sets? I guess yes because

$||(Tu,v)|-|(Tu_0,v_0)||\leq|(Tu,v)-(Tu_0,v_0)|=|(Tu-Tu_0,v)+(Tu_0,v-v_0)|$

$\leq||T||\cdot||u-u_0||\cdot||v||+||T||\cdot||u_0||\cdot||v-v_0||$

So for any fixed $u,v$ and $\epsilon>0$, if $u_0,v_0\in \text{span}\{e_i\mid i\in\mathbb{N}\}$ are close enough to $u,v$ and there is a uniform bound on $||T||$, then $|(Tu_0,v_0)|\leq \epsilon$ implies $|(Tu,v)|\leq 2\epsilon$.

Now since we fix a basis $(e_i)_i$ we can regard operators as infinite matrices and $(Te_i,e_j)$ is the $j,i$-th entry. Therefore if the above calculation is correct then on bounded sets WOT can be viewed as “entrywise convergence topology”. This looks more comfortable and intuitive for me, for example it becomes even clearer that unit ball is compact because we can view it as a subspace of $[-1,1]^{\mathbb{N}^2}$ (I found that I don't know how to argue it is a closed subspace without using the fact that it is same as WOT). Similarly strong operator topology may be viewed as “columnwise convergence topology”.

答:
The answer is no.  Although $\mathcal T$ coincides with the WOT on bounded sets, as observed by the OP, these topologies
do not coincide over the whole of
$B(H)$.

One very nice point of view from which to understand this question is the theory of duality.  Suppose $B$ is a complex
vector space (I am using the letter $B$ here because the application will be for $B=B(H)$).   Suppose also we are given
a set   $\Phi $  of linear functionals on $B$.

One may then define a topology $\mathcal T$ on $B$ using the seminorms
$$
  T\in B\mapsto |\varphi (T)|\in \mathbb R,
  $$
for every $\varphi $ in $\Phi $.

It is evident that every $\varphi $ in $\Phi $ is $\mathcal T$-continuous and hence also every linear combination of functionals in
$\Phi $.

The key fact is that nothing else is continuous!  In other words, the topological dual of $(B,\mathcal T)$ is precisely
the linear span of $\Phi $ within the algebraic dual of $B$.

Taking $B=B(H)$ and $\Phi $ the set of linear functionals of the form
$$
  T\in B(H)\mapsto \langle Te_i, e_j\rangle ,
  \tag{1}
  $$
we may ask whether or not a linear  functional of the form
$$
  T\in B(H)\mapsto \langle Tu, v\rangle ,
  \tag {2}
  $$
is $\mathcal T$-continuous,  for given vectors $u$ and $v$ in $H$.

The answer will be negative if $u$ and $v$ are
infinite linear combinations of basis elements since one may easily prove that  (2) is not a finite linear combination
of functionals as in (1).


问:
Are the two connected components of orthogonal flag bundles isomorphic?
Let $X$ be a scheme and $V\to X$ be a vector bundle over $X$. Fix a line bundle $L$ on $X$. We equip $V$ with a symmetric non-degenerate bilinear form $\omega: V \times V \to L$ and say that a subbundle $W \subset V$ is isotropic if $\omega$ restricts to zero on $W$. Let $\dim V = 2n$ and
$$OG_X(n,V)= \{W \subset V \ \vert \ W \textrm{ is an isotropic subbundle of rank } n \}.$$
If $X=spec(k)$ for a field $k$ of characteristic $\neq 2$ it is known that $OG_k(nV)$ splits into two connected components, both being isomorphic to $OG_k(n-1,2n-1)$.

$\textbf{my question}$: Does this hold in general? I could not find such a statement for flag bundles. Pragacz discusses this bundle in his "FORMULAS FOR LAGRANGIAN AND ORTHOGONAL DEGENERACY LOCI" and mentions two families, but not wether they are isomorphic.

答:
The Stein factorization for the map $OG_X(n,V) \to X$ takes the form
$$
OG_X(n,V) \to \tilde{X} \to X,
$$
where the second arrow is an 'etale double covering, which in general is not trivial. So, in general there is only one component (even over a field, if the field is not algebraically closed and the determinant of the quadratic form is not a square).


问:
Prove that$\lim_\limits{n\to \infty}\frac{n}{|\sin(n^2+5)|}=\infty$
I need to prove the following limit:
$$\lim_\limits{n\to \infty}\frac{n}{|\sin(n^2+5)|}=\infty$$

I don't know how to do it, because $\frac{1}{|\sin(n^2+5)|}$ is unbounded, even though $\sin(n^2+5)\ne0$ for every natural number $n\ge1$.

Body:
$$
0 < |\sin(n^2+5)| < 1 ,
\\
\frac{1}{|\sin(n^2+5)|} > 1 ,
\\
\frac{n}{|\sin(n^2+5)|} > n .
$$


问:
Proof of closed set
I had today a discussion with my professor about A being closed or not

let $ A = \left \{ {(x,y)\in \mathbb{R}^2: x>0 , y\geq 1/x} \right \} $

I claim A is not closed since $ (x,y) = (\frac{1}{n+1}, n+1)\in A  \xrightarrow[n\rightarrow + \infty]{} (0, n+1)\notin A $, therefore A is not closed.
Could you enlighten me how A is closed, our professor showed us some proof that I don't accept, because it supposes A closed in first place and then does the whole proof around that and claiming that our hypothesis is right.

答:
As $n \to \infty$, the point $(\frac1{n+1}, n+1)$ does not converge to any point in $\mathbb R^2$, since its $y$-coordinate goes off to $+\infty$. In particular, it cannot converge to $(0,n+1)$ because $n+1$ does not exist as a variable outside the limit you're taking!

We can describe the set $\{(x,y) : x>0, y \ge \frac1x\}$ equally well as the set $\{(x,y) : x \ge 0, xy \ge 1\}$ and since these inequalities are preserved by taking limits, the set is closed.

Title: Proof of closed set


问:
Proving that the coefficents of the inverse formal power series takes the form $b_{n} = \frac{-1}{a_{0}} (\sum_{i=1}^n a_{i} b_{n-i}). $
Proving that the coefficents of the inverse formal power series takes the form $b_{n} =  \frac{-1}{a_{0}} (\sum_{i=1}^n a_{i} b_{n-i}). $ \

I found this proof online:

But I guess it is incorrect because it did not use the order of the power series $ a = \sum_{n}a_{n}x^n,$ am I correct? If so, below is my trial for the solution, If not please tell me how the argument used is the same as order argument.

My trial:

I got $a = a_{0}( 1 + xb(x))$ then the inverse $\frac{1}{1+xb(x)} = \sum (-xb(x))^k$ but then what happen to the first non-zero term I do not know, I got a hint that it starts further and furthur away but I do not know why and how that will help me in my proof? could any one help me in clarifying my confusions please?

答:
Looking here we have the formula to multiply two series

$$f(x)=\sum _{n=0}^{\infty } a_n x^n;\;g(x)=\sum _{m=0}^{\infty } b_m x^n$$
$f$ and $g$ are inverse if $(fg)(x)=I(x)$, identity
$$f(x)g(x)=\sum _{n=0}^{\infty } a_n x^n\sum _{m=0}^{\infty } b_m x^m=\sum _{k=0}^{\infty } c_k x^k$$
where coefficients $c_k$ are computed as follows
$$c_k=\sum _{j=0}^{k } a_jb_{k-j}$$
In our problem we have the first coefficent $c_0=1$ and all $c_k=0$ for $k>0$.

Thus $a_0b_0=1\to b_0=\frac{1}{a_0}$

$$c_1=a_0b_1+a_1b_0= 0\to b_1=-\frac{1}{a_0}\left(\frac{a_1}{a_0}\right)$$
Coefficient $n$-th is zero, so we have, if we know the previous $(n-1)$ coefficients
$$a_0b_n+a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0=0$$
$$b_n=-\frac{1}{a_0}\left(a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0\right)$$
and finally
$$b_n=-\frac{1}{a_0} \sum_i^n a_ib_{n-i}$$

Hope it is useful


问:
Determine all prime numbers p such that p^2+2 is also a prime number
Determine all prime numbers p such that p^2+2 is also a prime number.

I have only found 3, and I think that's the only number but I have trouble writing proof for this. But I know that if p is a prime number other than 3 then p^2+2 will be divisible with 3.

答: Suppose $p$ is prime and different from $3$. Write $p^2+2$ as $$p^2+2=(p^2-1)+3=(p-1)(p+1)+3.$$ Since $3\nmid p$, $3$ has to divide either $p-1$ or $p+1$. We conclude that $3$ divides $(p-1)(p+1)+3=p^2+2$, and since $p^2+2\ne 3$, it follows that $p^2+2$ is not prime.


答:
$\displaystyle \forall x_0>0, B_{x_0}=(x_0,y) \text{ with } y\ge \frac{1}{x_0}$ is closed.

To show that $B_{x_0}$ is closed, we can express, $B_{x_0}^C$, the complement of $B_{x_0}$ as a finite union of open sets.

$\displaystyle B^C_{x_0} $ is the union of three half planes:

$\displaystyle B^C_{x_0} =  \{ (x,y)\in\text{Reals}|x<x_0\} \bigcup
\left\{(x,y)\in\text{Reals}|x>x_0\right\} \bigcup
\left\{(x,y)\in\text{Reals}|y<1/x_0\right\} .$

Thus, $\displaystyle A=\bigcup_{x_0 \in \{x>0\}} B_{x_0}$ is closed.

Title: Proof of closed set

Body:
I had today a discussion with my professor about A being closed or not

let $ A = \left \{ {(x,y)\in \mathbb{R}^2: x>0 , y\geq 1/x} \right \} $

I claim A is not closed since $ (x,y) = (\frac{1}{n+1}, n+1)\in A  \xrightarrow[n\rightarrow + \infty]{} (0, n+1)\notin A $, therefore A is not closed.
Could you enlighten me how A is closed, our professor showed us some proof that I don't accept, because it supposes A closed in first place and then does the whole proof around that and claiming that our hypothesis is right.


问:
Let $f : A \to B$ and $g : B \to C$ be functions, Suppose that $g \circ f$ is bijective and $g$ is bijective. Prove $f$ is bijective.
My work so far is:

for 1-1:
for all $x,y \in A$, if $f(x)=f(y)$, then
since $g \circ f$ is bijective, $g(f(x))=g(f(y))$.
Since we know $g$ is bijective, then
$f(x)=f(y)$

is there a way I can get this down to $x=y$? I'm not sure I'm doing this the right way but this was all I could think of to at least start the problem. I'm not sure how to start solving onto; should I use substitution from knowing that $g \circ f$ and $g$ are surjective to try to get $f(a)=b$ ?

答:
$$g\circ f\text{ is bijective }\iff f\text{ is injective }\&\text{ g is surjective}\tag 1$$

Since $g\circ f$ is bijective, $(\forall c\in C)(\exists !\space a\in A)$ s. t. $c=g(f(a))$.

Since $g$ is bijective, $(\forall c\in C)(\exists!\space b\in B)$  s. t. $c=g(b)$.

As $c=g(f(a))=g(b)$ and $b$ is unique, then $b=f(a)$.

Look at $(1)$.

Hence the statement.


问:
How to differentiate step responses of 1st-order and 2nd-order systems?
How can one determine if a system is of 1st or 2nd order by merely looking at the step response of the system or device?

One can easily spot an underdamped response straightaway by the overshoots, but the step response of first-order and second-order critically damped and overdamped systems look very similar.

Is there a key to differentiating these responses from a graph?

Figures

First order step response

Second order step response

答: If the curve has an inflection point, then the system is second order. Otherwise, I am afraid that there is little you can do.</p>


问:
Creating a converging series of functions
I have to create a series of functions ($\sum_{k=0}^{\infty}f_{k}$) such that it converges uniformly in $[1,2], [3,4]$ and diverges in $(2,3)$. I tried creating two different power series and taking their sum but I can't promise convergence that way (if one converges in [1,2] and the second in [3,4] it doesn't promise me convergence). So I'm stuck and I don't have an idea how to approach this question. Every hint would help.

答:
What about defining for all $k \in \mathbb{N}$
$$f_k(x)=1 \text{ if } x \in (2,3) \quad \quad \text{and} \quad \quad f_k(x)=0 \text{ otherwise} \quad ?$$


问:
Evaluate $\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$
Here is a some what challenging log sine integral.

$$I=\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$$

The upper limit of integration is $\frac{\pi}{4}$ instead of the usual $\frac{\pi}{2}$.

I tried the famous identity $\displaystyle\ln(\sin(x))=-\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}-\ln(2)$

But, the log is squared and I do not think Parseval is applicable here. But, I may be wrong.  By Parseval, I mean $\displaystyle\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(kx)\cos(nx)}{kn}$.

I have tried other means.  I made the sub $t=\sin(x)$. This led to $\displaystyle\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\sin^{-1}(t)\ln^{2}(t)}{\sqrt{1-t^{2}}}dt$

Now, I thought perhaps the famous sum $\displaystyle\frac{\sin^{-1}(t)}{\sqrt{1-t^{2}}}=\sum_{k=1}^{\infty}\frac{(2t)^{2k-1}}{k\binom{2k}{k}}$ could be used.

So, I integrated and got the form

$$I=\frac{\ln^{2}(2)}{16}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{2}(2k)!}+\frac{\ln(2)}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{3}(2k)!}+\frac{1}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{4}(2k)!}$$

These sums actually evaluate to the correct integral result, so it would appear they are equivalent to said integral. But evaluating them may be even more daunting. Though the first one from the left evaluates to $\frac{{\pi}^{2}}{8}$ and the center one evaluates to

$$\frac{\ln(2)}{8}\left(\pi G-\frac{35}{16}\zeta(3)+\frac{{\pi}^{2}}{8}\ln(2)\right)\;.$$

This center one looks to be related to the integral:

$\displaystyle8\ln(2)\int_{0}^{\frac{\pi}{4}}\ln(\sin(x))dx=\frac{35}{16}\ln(2)\zeta(3)-\pi\ln(2)G-\frac{{\pi}^{2}\ln^{2}(2)}{4}$, which I got by expanding the square of the log sine and integrating.

This just leaves the sum to the right.  It is tougher and I can find no closed form.

Alas, If I could evaluate it then I think the integral would be solved.

I also thought...how about a closed form of some sort for

$$\int_{0}^{\frac{\pi}{4}}x\sin^{a}(x)dx$$

that could be differentiated w.r.t 'a' and applied?. I do not know of a closed form for
this, though.

I got an encouraging looking solution by expanding, making the sub $t=x/2$, then solving for the integral on the far right (which is the integral in question):

$\displaystyle  \int_{0}^{\frac{\pi}{2}}x\ln^{2}(2\sin(x/2))dx$
$=\displaystyle \ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx+8\ln(2)\int_{0}^{\frac{\pi}{4}}x\ln(\sin(x))dx+\color{red}4\int_{0}^{\frac{\pi}{4}}x\ln^{2}(\sin(x))dx$

But, it did not check numerically. (Error has now been fixed. - Editor) Numerically, the integral in question should evaluate to $I = 0.263605....$

Then again, I may just have a computation error in all of that.

So, does anyone know of a method for evaluating this log-sine integral?.

EDIT:  Here is what Mathematica gave for the solution.  It may be too involved to be worth the effort.

$\displaystyle I=\frac{37{\pi}^{4}}{92160}-\frac{\pi}{4}\sum_{k=1}^{\infty}\frac{\sin(\pi k/4)}{2^{k/2}k^{3}}$ $\displaystyle +\frac{\pi}{8}\ln(2)G+\frac{25}{768}{\pi}^{2}\ln^{2}(2)+\frac{\ln^{4}(2)}{384}+\sum_{k=1}^{\infty}\frac{\cos(\pi k/4)}{2^{k/2}k^{4}}-\frac{35}{128}\ln(2)\zeta(3)$.

Sometimes with these math engines the solution can be simplified down. But, maybe not in this case. Those complex polylogs may pose a problem.

答:
Thanks for the input, Fred. I managed to evaluate these types of integrals by using

$\displaystyle \ln(1-e^{ix})=\ln(2\sin(x/2))-i/2(\pi-x)$.  Mathematica often gives

hypergeometric solutions,  which I have no interest in. These log-sin integrals can be done,

but they are extremely onerous. I managed to work out the above integral except the upper limit was $\frac{\pi}{2}$.

It was extremely tedious, but does have a closed form in terms of

polylogs, zeta, and so on. As does the $\frac{\pi}{4}$ case.

For instance, evaluating $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln^{2}(2\sin(x/2))dx$ gives

$\displaystyle \frac{5}{96}\ln^{4}(2)-\frac{{\pi}^{2}\ln^{2}(2)}{48}+\frac{5}{4}Li_{4}(1/2)+\frac{19{\pi}^{4}}{1152}+\pi\mathfrak{I} \left[Li_{3}\left(1/2+i/2\right)\right]+\frac{35}{32}\zeta(3)\ln(2)-\frac{\pi\ln(2)}{2}G$.

This required a lot of patience and work on my part to arrive at this. The $\frac{\pi}{4}$ case is even worse to evaluate.

But, regarding the math engine result: take the indefinite integral

$\displaystyle \int\frac{1-x}{(1+x)\sqrt{x^{3}+x^{2}+x}}dx$.

This has a rather simple closed form of

$\displaystyle-\sin^{-1}\left(\frac{x^{2}+1}{(x+1)^{2}}\right)-\frac{\pi}{2}$,

but all I could get Mathematica and Maple to return was a long, complicated EllipticF mess.


问:
prove : The tangent to the curve ${x}^3+{y}^3=3axy$ at $(\frac{3a}{2},\frac{3a}{2})$ makes obtuse angle with the positive direction of $x$ axis?
I have differentiated the equation from both side and put the value of coordinates given. Is there any better way? Can anyone help me to solve the problem?</p>

答:
Hint:

Point $(\frac {3a}2, \frac{3a}2)$ and also point (0, 0) i.e origin are exactly the points of intersection of line $y=x$ with curve $x^3+y^3=3axy$. if you plot the curve and line you  see that the tangent line at that point is perpendicular on line $y=x$, that is the gradient of tangent lines is $m=-1$ which means its angle with positive direction of x axis is obtuse.The equation of tangent line is:

$y-\frac 32=-1(x-\frac 32)$

Or: $y=-x +3$


问:
How to show that $2+\sqrt{5}i\not | 3$?
How can I prove that there is no $x\in \mathbb{Z}[\sqrt{-5}]$ such that $x\cdot (2+\sqrt{5}i)= 3$ ?

答:
Write $x=a+ib\sqrt{5}$. You have

$$x.(2+\sqrt{5}i)=(a+ib\sqrt{5})(2+\sqrt{5}i)=(2a-5b)+i\sqrt{5}(a+2b)$$

This is equal to $3$ iff $a+2b=0$ and $2a-5b=3$. This is easy to see that there are no solution in $\mathbb{Z}$.

Title: How to show that $2+\sqrt{5}i\not | 3$?


问:
Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$?
I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem.
But in several occasions I couldn't follow the answers because of this step.
In both of these excerpts (taken from answered questions on this site, the links are below) two different square roots are combined like this.
\begin{align}
\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}\tag{Now it's an inequality}
\end{align}

Why is this possible even though the square roots are not identical?

Here the excerpts:

\begin{align}
= & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\
> & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\
\end{align}
taken from here

or

\begin{align}
\sqrt{k+1}-\sqrt{k}&=\frac{1}{\sqrt{k+1}+\sqrt{k}}\\
&\le\frac{1}{2\sqrt{k}}
\end{align}

taken from here

I hope someone can explain this to me or point me in the right direction

答: No, the linked examples use $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, which is equivalent to the trivial $\sqrt{n+1}>\sqrt{n}$.


问:
Condition that shows when there are elements in $[G,G]$ which are not commutator.(Looking for a proof)
It's known that for a given group $G$ and the derived subset $[G,G]$ it's not true that every element of $[G,G]$ is a commutator. I've seen a condition that shows when there are elements in $[G,G]$ which are not commutator.

For any group $G$ with
$$\left|\left[G:Z\left(G\right)\right]\right|^{2}<\left|\left[G,G\right]\right|$$
there are elements in $[G,G]$ that are not commutator.

I have not seen any proof of this, it would be really nice if someone proof that or give me a link about the problem.

答:
Since $[gk_1,gk_2] = [g,h]$ for all $k_1,k_2 \in Z(G)$, there are at most $[G:Z(G)]^2$ distinct commutators.

So if $[G:Z(G)]^2 < |[G,G]|$ then $|[G,G]|$ is larger than the number of commutators, so there must be elements of $[G,G]$ that are not commutators.


问:
How to find the coordinates with the given data
In the following figure, Is there a way to calculate the coordinates of $X$ if we know the coordinates of $A$, $B$ and $C$ and the distances $AX$, $AB$ and $AC$?

答:
In this case, $AB$ and $AC$ are already determined by the coordinates of $A,B$, and $C$ so the information is redundant. In fact $X$ can lie anywhere on the circle with center $A$ and radius $AX$, so we cannot determine the location of $X$ with the information given.

If we were given the distances $AX, BX,$ and $CX$ instead, it would be possible to calculate the location of $X$ be determining the intersection point of the circles with centers $A, B,$ and $C$ and radii $AX, BX,$ and $CX$. This is called trilateration, and versions of it are used in the GPS system and in seismology.


Title: How to show that $2+\sqrt{5}i\not | 3$?

Body: How can I prove that there is no $x\in \mathbb{Z}[\sqrt{-5}]$ such that $x\cdot (2+\sqrt{5}i)= 3$ ?

Body: Hint: If $(a+b\sqrt{5}i)(2+\sqrt{5}i)=3$, then $a^2+5b^2=1$ and so $a=\pm1$ and $b=0$.


答:
We have that

$$\sqrt{n} + \sqrt{n + 1} >\sqrt{n}+\sqrt{n}= 2\sqrt{n}$$

since $\sqrt{n + 1}>\sqrt{n}$ or also dividing by $\sqrt n$

$$\sqrt{n} + \sqrt{n + 1}> 2\sqrt{n} \iff 1+\sqrt{1+\frac1n}>2$$

which is true since $\sqrt{1+\frac1n}>1$.

Title: Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$?

Body:
I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem.
But in several occasions I couldn't follow the answers because of this step.
In both of these excerpts (taken from answered questions on this site, the links are below) two different square roots are combined like this.
\begin{align}
\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}\tag{Now it's an inequality}
\end{align}

Why is this possible even though the square roots are not identical?

Here the excerpts:

\begin{align}
= & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\
> & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\
\end{align}
taken from here

or

\begin{align}
\sqrt{k+1}-\sqrt{k}&=\frac{1}{\sqrt{k+1}+\sqrt{k}}\\
&\le\frac{1}{2\sqrt{k}}
\end{align}

taken from here

I hope someone can explain this to me or point me in the right direction


问:
Does every element of an integral domain have an inverse?
I am reading a first course in algebra 7th edition written by John B. Fraleigh. I have seen the following two definitions:

1) A field is a commutative ring in which every nonzero element has
  multiplicative inverse.

2) An integral domain is a commutative ring with unity 1 and
  containing no zero divisors.

Then i saw a picture in the book that shows fields as subsets of integral domains like in the following picture:

My question is, how do we understand from these two definitions that fields are subsets of integral domains? In the definition of integral domain, i do not see anything saying that every element in the ring should have an inverse, it just says that there must be a multiplicative identity. Am i missing something or is there something missing in the definitions?

Thank you

答:
As already noticed, every field is an integral domain but the converse doesn't hold (take for example $\mathbb{Z}$). In order to show that a field $F$ is an integral domain, suppose $a,b\in F$ are such that $ab=0$ and assume $a$ is not zero. Then since every non zero element in a field is invertible, one has
$$ab=0\implies a^{–1}ab=0\implies b=0. $$ Similarly if $b\neq 0$ one gets that a must be zero and hence $F$ is an integral domain. </p>


问:
Cumulative distribution function of $\sqrt{X^2+4Y^2}$ without $\chi$ distribution
I want to find the cumulative distribution function of the random variable $Z=\sqrt{X^2+4Y^2}$, where $X\sim N(0,4)$, $Y\sim N(0,1)$ are independent. I know it is equal to $\chi_{[0,\infty)}(t)\left(1-\exp\left(-\frac{t^2}{8}\right)\right)$, but I don't know how to show it.

My research has led me to the Chi-square distribution, however, this has never come up in class, and also the CDF for this distribution looks complicated enough on its own, without even taking the square root.

Noting that $$P(\sqrt{X^2+4Y^2}\le t) = 1-P(\sqrt{X^2+4Y^2} > t)$$
it would be sufficient to show
$$P(\sqrt{X^2+4Y^2} > t) = \exp\left(-\frac{t^2}{8}\right)$$
but I don't know how that could be done. How could I show this relation?

答: As @dimebucker noted, we need to assume independence here. Let $W:=X/2$. The claim is equivalent to $R:=T/2=\sqrt{W^2+Y^2}$ having PDF $r\exp(-r^2/2)$ for $r\ge0$. This is easily proven with a switch to polar coordinates:$$\frac{1}{\sqrt{2\pi}}\exp\frac{-r^2}{2}\frac{1}{\sqrt{2\pi}}\exp\frac{-y^2}{2}dwdy=r\exp\frac{-r^2}{2}dr\frac{d\theta}{2\pi}.$$Since this is separable in $R,\,\Theta$, the PDF of $R$ is $r\exp\frac{-r^2}{2}$ for $r\ge0$, as required.


答:
I think you forgot to mention that $X$ and $Y$ are independent. Then note that $X = 2 Z$ where $Z \sim N(0,1)$, and so $X^2 = (2Z)^2 = 4 Z^2 \sim 4 \chi^2_1$, further, $4Y^2 \sim 4\chi^2_1$. We are therefore looking at the sum of two independent $4\chi^2_1$ random variables, which has distribution $4 \chi^2_2$. Then, $\sqrt{4 \chi^2_2} \sim \sqrt{2} \chi_2$, which is the distribution of the quantity of interest.

To show the result you had in the end we can use the above argument to note that
\begin{align*}
P(\sqrt{X^2 + 4Y^2} \le t)
&=P( \chi^2_2 \le t^2/4)\\
&= \frac{1}{\Gamma(1)} \gamma(1, \frac{t^2}{8})\\
&= \int_0^{t^2/8} y^{1-1} e^{-y} dy\\
&= \int_0^{t^2/8} e^{-y} dy\\
&= 1-e^{-t^2/8}
\end{align*}
where $\gamma(a,b) = \int_0^a y^{a-1} e^{-y}$ is the lower incomplete gamma function, and I have used the fact that $P(\chi^2_k \le x) = \frac{1}{\Gamma(k/2)} \gamma(k/2, x/2$ (basically using the notation on the wiki page)

Title: Cumulative distribution function of $\sqrt{X^2+4Y^2}$ without $\chi$ distribution

Body:
I want to find the cumulative distribution function of the random variable $Z=\sqrt{X^2+4Y^2}$, where $X\sim N(0,4)$, $Y\sim N(0,1)$ are independent. I know it is equal to $\chi_{[0,\infty)}(t)\left(1-\exp\left(-\frac{t^2}{8}\right)\right)$, but I don't know how to show it.

My research has led me to the Chi-square distribution, however, this has never come up in class, and also the CDF for this distribution looks complicated enough on its own, without even taking the square root.

Noting that $$P(\sqrt{X^2+4Y^2}\le t) = 1-P(\sqrt{X^2+4Y^2} > t)$$
it would be sufficient to show
$$P(\sqrt{X^2+4Y^2} > t) = \exp\left(-\frac{t^2}{8}\right)$$
but I don't know how that could be done. How could I show this relation?


问:
For what matrices $B,C$ is there an $A$ such that $AB=C$?
Let $B,C$ be arbitrary matrices of dimensions $n_2\times n_3$ and $n_1\times n_3$, respectively. What are the conditions for the existence of some $A$ of dimensions $n_1\times n_2$ such that $AB=C$? And if we can prove the existence of such $A$, can we write an explicit expression for it?

In the case $n_3=1$ this amounts to asking, given vectors $x\in\mathbb R^{n_2}$ and $y\in\mathbb R^{n_1}$, whether there is $A$ such that $Ax=y$. This is simple to answer: any $A$ such that $x\in\mathrm{supp}(A)$ and $y\in\mathrm{range}(A)$ will do. One such matrix is $A=yx^*$ (here, $x^*$ denotes the dual of $x$ and $yx^*$ is the outer product).
I'm still not sure if there is a good way to write a general form for the full set of solutions for $A$.

What about the general case? Clearly we must have $\mathrm{ker}(B)\subseteq\mathrm{ker}(C)$ and thus $\mathrm{supp}(C)\subseteq\mathrm{supp}(B)$, and also $\mathrm{range}(A)=\mathrm{range}(C)$. This implies that $\min(n_1,n_3)\le \min(n_2,n_3)$, but this is not enough to answer the question.

I don't think whether the matrices are real or complex matters here, but if it does, I would be interested in the solution in both cases.

答:
Note that for the case that $A$ is a single row we have
$$
(a_{11}\ a_{12}\ \dots\ a_{1,n_2})
\begin{pmatrix}
—\ B_1\ —\\
—\ B_2\ —\\
\vdots \\
—\ B_{n_2}\ —
\end{pmatrix}
= a_{11}B_1+a_{12}B_2+\cdots+a_{1,n_2}B_{n_2},
$$
where $B_i$ denotes the $i$th row of $B$. Hence, for $n_1=1$ the product $AB$ is a linear combination of the rows of $B$ and $AB=C$ has a solution (in $A$) if and only if the row $C$ is a linear combination of the rows of $B$.

For $n_1>1$, the equation $AB=C$ splits up into $n_1$ equations, one for each row.

Therefore, in general, the equation $AB=C$ has a solution in $A$ if and only if each row of $C$ is a linear combination of the rows of $B$. This might be expressed as
$$
\operatorname{rowspace}(C) \subseteq \operatorname{rowspace}(B),
$$
or by taking orthogonal components as
$$
\ker(C) \supseteq \ker(B).
$$
Or as a characterization using ranks:
$$
\operatorname{rank}(B) = \operatorname{rank}\begin{pmatrix} B\\C\end{pmatrix}.
$$


问:
Joint distribution of order statistics, need some more details?
I was reading about how to compute the joint distribution of an order statistics but I don't really follow the beginning. How do we arrive to the formula?

$f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k}f_x(x)$

(it is not very detailed and I am bad in probabilities).

答:
This is strictly meant to give you some intuition behind the following equation:
$f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k}f_X(x).$

Let's break it down piece-by-piece. $f_{X_{(k)}}(x) $ is the PDF of the $k$th order statistic, i.e., if you draw $n$ independent identically distributed random variables from some distribution, arrange them in ascending order by their size, we are looking at how the $k$th one is distributed ($k$th smallest). Informally, it is the "probability" that the $k$th order statistic is $x$. Let's come back to $\frac{n!}{(k-1)!(n-k)!}$ in a bit. Next we have $[F_X(x)]^{k-1}$, which is the CDF of the original distribution raised to the $(k-1)$th power---the idea here is that in order for the $k$th order statistic to be $x$, then we need $k-1$ of the random variables to be less than or equal to $x$. The way we measure this probability is with the CDF and because our random variables are independent, we can just multiply these probabilities. The next term, $[1-F_X(x)]^{n-k}$, is similar; it gives the probability that $n-k$ of the random variables are greater than $x$. At the end, we have $f_X(x)$ which is just the "probability" that one of the random variables is $x$. Returning to $\frac{n!}{(k-1)!(n-k)!}$, this is combinatorial bookkeeping. The $n!$ gives the number of ways that the $n$ random variables might be ordered, and the denominator says we aren't concerned about the actual order of the first $k-1$ or the last $n-k$ random variables.


问:
Flux of a vector field in a closed parabolic surface
Considering S to be a closed surface consisting of a paraboloid $z=x^2+y^2$, with ($0≤z≤1$), and capped by the disc $x^2+y^2 ≤1$ on the plane $z=1$.

If the vector field is $\vec F (x,y,z) = z\hat j − y\hat k$, is it correct to say that the flux in the direction that points out across the surface S is zero?

答:
Yes, since $\operatorname{div}\vec F=0$, so the result follows from the divergence theorem.

Just to convince ourselves further, we can explicitly compute the integral over the surface. Let $S_1$ and $S_2$ denote the "faces" of the surface $S=S_1\cup S_2$, where

$$\begin{align}
S_1&=\left\{\vec s_1(u,v)=(u\cos v,u\sin v,u^2)\mid0\le u\le1,0\le v\le2\pi\right\}\\[1ex]
S_2&=\left\{\vec s_2(u,v)=(u\cos v,u\sin v,1)\mid0\le u\le1,0\le v\le2\pi\right\}
\end{align}$$

Take the normal vectors to be

$$\begin{align}
\vec n_1&=\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath-u\,\vec k\\[1ex]
\vec n_2&=\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}=u\,\vec k
\end{align}$$

Then the flux of $\vec F$ across $S$ is

$$\begin{align}
\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S&=\int_0^{2\pi}\int_0^1\vec F(u\cos v,u\sin v,u^2)\cdot(\vec n_1+\vec n_2)\,\mathrm du\,\mathrm dv\\[1ex]
&=\int_0^{2\pi}\int_0^12u^4\sin v\,\mathrm du\,\mathrm dv\\[1ex]
&=0
\end{align}$$

since the integral with respect to $v$ vanishes.


问:
Differentiate logarithmically the expressions for $\sin{\theta}$ and $\cos{\theta}$ in factors and deduce the sums to infinity.
(i)$$\frac{1}{ \theta^2 - \pi^2} + \frac{1}{ \theta^2- 2^2\pi^2} + \frac{1}{ \theta-  3^2\pi^2 }... $$
(ii)$$\frac{1}{1^2 + x^2} + \frac{1}{2^2 +x^2} + \frac{1}{3^2 + x^2}... $$
These are the problems given in "Differential Calculus for Beginners"" by Joseph Edwards. I am beginner and have no idea how to approach such problems. The phrase 'differentiate logarithmically' is confusing me.

答:
The phrase "the expressions for $\sin\theta$ and $\cos\theta$ in factors" seems to refer to the infinite product representations of those functions.

For (i):

\begin{align*}
\sin\theta &=
\theta \; \prod_{n=1}^{\infty}\left(1 - \frac{\theta^2}{n^2 \pi^2}\right)\\
\frac{d}{d\theta}\ln(\sin\theta) &=
\frac{d}{d\theta}\left[\ln\theta \;+\; \sum_{n=1}^{\infty}\ln\left(1-\frac{\theta^2}{n^2\pi^2}\right)\right]\\
&=
\frac{1}{\theta} \;+\; 2\theta \sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2}
\end{align*}
From there you can solve for the sum in question, to yield:
$$
\sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2} \;=\;
\frac{1}{2\theta}\left(\cot\theta \;-\; \frac{1}{\theta}\right)
$$

For (ii), the simplest thing to do is to use the above result with $\theta\rightarrow i \pi x$.


问:
$ \sum_{a \mid n} \Bigl( f(a) \sum_{b \mid \frac{n}{a}} g(b) \Bigr) = \sum_{a \mid n} \Bigl( g(a) \sum_{b \mid \frac{n}{a}} f(b) \Bigr) $
Is the following equation true?

$$
\sum_{a \mid n} \Bigl( f(a) \sum_{b \mid \frac{n}{a}} g(b) \Bigr) =
\sum_{a \mid n} \Bigl( g(a) \sum_{b \mid \frac{n}{a}} f(b) \Bigr)
$$

If this equation is true, how can I prove it?

答: Hint: both sides are equal to $\displaystyle\sum_{\substack{a,b,c\\abc=n}}f(a)g(b)$.


问:
Series at $\infty$ of $\ln\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)$
I'm currently looking for the series expansion of $\ln(1 + \frac{1}{n} + \frac{1}{n^2})$ when n goes to $\infty$.

I've got this:

$$\frac{1}{n} + \frac{1}{n^2} + o\left(\frac{1}{n} + \frac{1}{n^2}\right)$$

but that's very ugly and wolframalpha gives me:

$$\frac{1}{n} + \frac{1}{2n^2}- \frac{2}{3n^3}+ \frac{1}{4n^4}+O\left(\frac{1}{n^5}\right)$$

How can i get the same result? I wanna get rid of that ugly little-o.

Thanks!

答:
Recall that as $x \to 0$

$$\log(1+x)=x-\frac12x^2+\frac13 x^3-\frac14x^4+O(x^5)$$

that is by $x=\frac{1}{n} + \frac{1}{n^2}\to 0$

$$\ln\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)=$$
$$=\left(\frac{1}{n} + \frac{1}{n^2}\right)-\frac12 \left(\frac{1}{n} + \frac{1}{n^2}\right)^2+\frac13 \left(\frac{1}{n} + \frac{1}{n^2}\right)^3-\frac14 \left(\frac{1}{n} + \frac{1}{n^2}\right)^4+O\left(\frac{1}{n^5}\right)$$

since

$$O\left(\left(\frac{1}{n} + \frac{1}{n^2}\right)^5\right)=O\left(\frac{1}{n^5}\right)$$

then expand keeping only the terms less than the fifth order.


问:
Given marginal of $U$ and $V$, how can we know if there is joint such that $\mathbb E[U|V]=V$
Suppose we are given two random variable $U$ and $V$ on the interval $[0,1]$ with respective probability measure $p_U$ and $p_V$.

When is there a coupling measure $p_{U,V}$, that respect the marginals $p_U$ and $p_V$, such that $\mathbb E[U|V]=V$.

I have tried few approach, like expressing this as a linear program and looking at the dual, but the following seemed to be the most promising.

Suppose we have a measurable convex set $S\subseteq [0,1]$, if we are given $V$, then what is the maximal amount of probability we can give to $U\in S$. Observe that the conditional expectation constraint gives
\begin{align*}
V &= \mathbb E[U|V]\\
&=\mathbb E[1_S(U)|V]\cdot \inf_{a\in S}a+\mathbb E[(U-\inf_{a\in S} a) 1_S(U)|V]+\mathbb E[U1_{S^c}(U)|V]\\
&\geq \mathbb E[1_S(U)|V]\cdot \inf_{a\in S}a
\end{align*}
So that we have $\mathbb E[1_S(U)|V]\leq\sup_a\frac{V}{a}$, similarly, we can also get
\begin{align*}
1-V&= 1-\mathbb E[U|V]\\
&\geq \mathbb E[1_S(U)|V]\cdot \inf_{a\in S}(1-a)
\end{align*}
So that $\mathbb E[1_S(U)|V]\leq\sup_a\frac{1-V}{1-a}$, and finally we also know that $\mathbb E[1_S(U)|V]\leq 1$.

If we average we get that
\begin{align*}
P_U(S) &= \mathbb E[\mathbb E[1_S(U)|V]]\\
&\leq \mathbb E\left[ \min\left(\sup_{a\in S}\frac{V}{a},\sup_{a\in S}\frac{1-V}{1-a},1\right) \right]
\end{align*}
Now if we have $\mathbb E[U|V]=V$, this needs to be satisfied for all $S$, but it feels like if we add the constraint that $\mathbb E[U]=\mathbb E[V]$, then we get that this is an if and only if. In order to prove this I don't think I have another choice than giving the conditional $p_{U|V}$ but I am not sure how to do that. It would also be nice to have a modification of that constraint that will also enforce that $\mathbb E[U]=\mathbb E[V]$ (because I don't think it is the case right now).

In the following I will try to motivate the problem and explain why it would be useful to solve. In short it would improve a lot my (and hopefully other's) understanding of Markov chains.

Let us be given a random variable $X$ with support $\mathcal X$. For $x\in\mathcal X$, define $\delta_x$ as the probability measure that gives $1$ to any set containing $x$ and $0$ otherwise. We can see $p_{X|X}=\delta_X$ as a random measure. for any random variable $U$, let $p_{X|U}=\mathbb E[p_{X|X}|U]$ where this conditional expectation is defined as $\forall A$, $p_{X|U}(A)=\mathbb E[p_{X|X}(A)|U]$.

We can prove a lot of things of those object and we can convince ourselves that this is indeed a conditional distribution since conditioned on $U=u$, we get that $p_{X|U}$ is the measure $p_{X|U=u}$. We can also show that this is a minimally sufficient statistic i.e. $X-U-p_{X|U}$ and $X-p_{X|U}-U$ are both Markov chains plus for any other sufficient statistic with $X-U-V$ and $X-V-U$, we have that $p_{X|U}=p_{X|V}$ almost surely (hence minimality).

Now of course $p_{X|U}$ is a random variable on the simplex of $\mathcal X$ with some distribution $\mu$, similarly we can have $p_{X|V}$ with distribution $\nu$ and I think it is important to wonder about when there can be a Markov chain $X-p_{X|U}-p_{X|V}$ by looking only at the distribution $\mu$ and $\nu$. And this is answered by

Is there a joint measure $\lambda$ with marginals $\mu$ and $\nu$ such that $\mathbb E[p_{X|U}|V]=p_{X|V}$.

And hence my question (which corresponds to the $X$ binary case), I think if I can solve the binary cases, there is chances the solution generalizes.

答:
This is the subject of chapter 15 of Phelps's 1966 Lectures on Choquet's Theorem.  The answer is,  this happens exactly when the distributions of $U$ and $V$ are such that $E[f(U)]\ge E[f(V)]$ for all continuous convex functions $f$. There is a whole series of results, by   Hardy, Littlewood, Polya, Blackwell, Stein, Sherman, Cartier, etc; what you want is contained in Blackwell's 1953 "Equivalent comparisons of experiments",  Ann. Math. Statistics 24 (1953).

The buzz word is that the distribution of $U$ is a dilation of that of $V$, written $\mu_U\succ\mu_V$; the intuition is that the distribution of $U$ is more "spread out" than that of $V$.  It is, in a sense, a kind of converse to Jensen's inequality.

To check the condition it suffices to restrict yourself to $f$ of form $f(x)=\max(0,ax+b).$

As the discussion in the comments show, the result (the "Blackwell-Sherman-Stein theorem") is not easy to prove, and I don't know of a simple way to understand (or explain) it.  You might find these references helpful: The result  is Theorem 2 in V. Strassen's 1965 paper The existence of probability measures with given
marginals.  The 1996 survey paper  by Le Cam might also be useful.


问:
prove that $(1 + x)^\frac{1}{b}$ is a formal power series
How to prove that $(1 + x)^\frac{1}{b}$ (where $b$ is an integer) can be expressed as a formal power series without using Binomial theorem?</p>

答:
We are looking for a power series $f=1+a_1x+a_2x^2+\cdots$ such that $f^b=1+x$. Writing out $f^b$, you get $1+ba_1x+g_2(a_1,a_2)x^2+g_3(a_1,a_2,a_3)x^3+\cdots$, where $g_i(a_1,a_2,\ldots,a_i)$ is a linear function of the coefficients. Then for $f^b=1+x$, we need to solve a system of equations
$$ba_1=1$$
$$g_2(a_1,a_2)=0$$
$$g_3(a_1,a_2,a_3)=0$$
$$\vdots$$
This gives us an infinite system of equations which can be solved iteratively, starting with $a_1=1/b$.</p>


问:
Details of Hecke Summation for weight 2 Eisenstein Series
In reference to https://math.stackexchange.com/a/51467/356443

I don't understand the details of the Hecke summation trick in order to define an Eisenstein Series of weight $2$.

For convergence issues, we can't just define
$$ E(z) = \sum_{c, d}'' \dfrac{1}{(cz+d)^2}. \qquad (z \in \mathbb{H}) $$
The trick is to consider
$$ E(z,s) = \sum_{c, d}'' \dfrac{1}{(cz+d)^2 |cz+d|^{2s}}, \qquad (z\in\mathbb{H}; \Re(s) > 0) $$
and try to find an analytic continuation for $s$.
This justifies setting $s=0$ to get a holomorphic function that should satisfy the right transformation rule.

It is noted that it is not always possible to continue $E(z,s)$ analytically.
In the cases that such an analytic continuation does exist, what is the strategy for the computation to construct such a continuation?

答:
In response to your question in the last paragraph, there are at least two strategies in the literature to obtain the analytic continuation of $E(z,s)$ (which, by the way, really should read $E_2(z,s)$ to remind oneself of the exponent of $2$ in the series, the weight) to $s=0$. One is by writing $E_2(z,s)$ as the Mellin transform of a suitable theta function - this is analogous to Riemann's famous proof of the analytic continuation (and functional equation!) of the Riemann zeta function, and the key technical tool is the Poisson summation formula. As pointed out in the comments, this is explained, e.g., in Diamond--Shurman, 4.10, and at least in the weight zero case, is Exercise 1.6.2 in Bump's "Automorphic forms and representations".

The second approach would be to compute the Fourier expansion of $E_2(z,s)$ and to verify the analytic continuation as well as the functional equation on the level of Fourier coefficients. Again, the Poisson summation formula is key to this. This is explained in great detail, but again in the weight zero case only, in Bump, 1.6, and in Zagier's Introduction to modular forms, 3.A, and probably in many other places as well.

However, if, as it would seem, you really are only interested in understanding what the value at $s=0$ of $E_2(z,s)$ is, you neither need to know Mellin transforms, nor Poisson summation, indeed not even analytic continuation at all. Instead, in that case Hecke's trick boils down to a skilful (but elementary!) computation of $\lim_{s\to 0^+}E_2(z,s)$ in terms of the (non-modular but holomorphic) $G_2(z)$. See Zagier's Elliptic modular forms and their applications, 2.3.


问:
Bijection From $S = \{ -2, -1 \} \cup [0, 1]$ To $T = (0, 1)$
A bijection is a function that is both one-to-one and onto.

I need to find such a function that maps $\{ -2, -1 \} \cup [0, 1] \rightarrow (0, 1)$ or equivalently, $ (0, 1) \rightarrow \{-2, -1 \} \cup [0, 1]$.

This problem is very confusing because if I exclude any function output for $x \in [0, 1]$ (output occupied by $\{ -1, -2 \}$) I have to find another function output in $(0, 1)$ that associates with $x$, even though all numbers in $(0, 1)$ have already been covered.

Help needed.

答:
We can compose bijections. First let's get a bijection from $[0,1]\to(0,1)$.

To do it, we can consider the following function (see this answer): $f(x) = \left\{
 \begin{array}{1 1}
  \frac{1}{2}  & \mbox{if } x = 0\\
  \frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}\\
                x & \mbox{otherwise}
 \end{array}
\right.$

So we have the bijection $f:[0,1]\to(0,1)$. Now consider $g:[0,1]\cup\{-1,-2\}\to[0,1]$ defined by $g(x) = \left\{
 \begin{array}{1 1}
  f(x)  & \mbox{if } x \in[0,1]\\
  0 & \mbox{if } x = -1\\
                1 & \mbox{if } x=-2
 \end{array}
\right.$

Then $h=f\circ g:[0,1]\cup\{-1,-2\}\to(0,1)$ is a bijection.


问:
How to solve Quadratic Equations with an Unknown C and other variables.
For instance $3x^2 - 11x + r$, I understand the value of $r$ is $6$ through trial and error but trial and error is extremely inefficient and time consuming thus not useful in exam situations, How would I solve this?

And also : If $x^2 +px + q$ is a factor of $x^3 + ax^2 + b$,

prove that:

$b = q(a-p)$

答:
A:  $3x^2-11x +r=0$

The  codition is that discreminet $\Delta\geqslant 0$, so we must have:

$\Delta= 11^2-12 r\geqslant0$

Or: $r\leqslant 10$

Now $121$ is odd and $12r$ is even and we have to find odd perfect squares less than $121$ which are $81, 49, 25$ and $9$. Only $49$ and $25$ give integers for $r$:

$r= 121-49=72=12r$ so $r=6$

$r=121-25=96=12r$ so $r=8$

B: $x^2+px+q|x^3+ax+b$

By direct division you find that thr remainder is $(a-p)x^2-qx+b$ which must be equal to zero, so we have:

$x[(a-p)x-q]=-b$

Therefore b must divide both $q$ and $(a-p)$ because $(a-p)$ and $q$ have no common divisors, hence we must have $b|q(a-p)$


问:
prove : The tangent to the curve ${x}^3+{y}^3=3axy$ at $(\frac{3a}{2},\frac{3a}{2})$ makes obtuse angle with the positive direction of $x$ axis?
I have differentiated the equation from both side and put the value of coordinates given. Is there any better way? Can anyone help me to solve the problem?</p>

答:
I trust that you tried. Only the last step did not connect somewhere.

Differentiate $ x^3+y^3-3 a x y =0$ implicitly and solve for the slope

$$ y'=y'(x)=\frac{ay-x^2}{y^2-ax}$$

The given point has

$$x=y$$

Plug in and the slope reduces to $-1$, so the the tangent makes CCW  $ 3 \pi/4$ obtuse angle to x-axis.


答:
For your second question, long division gives us:
$$
\require{enclose}
\begin{array}{rll}
    x + k \phantom{0000000000i} &&  \\[-3pt]
   x^2+px+q \enclose{longdiv}{x^3+ax^2+0x+b}\kern-.2ex \\[-3pt]
      \underline{x^3+px^2+qx\phantom{00ii}} &&  \\[-3pt]
      \phantom{0} kx^2-qx+b \phantom{0000i}&& \hbox{(set $k=a-p$)} &&  \\[-3pt]
      \underline{kx^2+kpx+kq}\phantom{0ii} && \\[-3pt]
      -(q+kp)x+b-kq &&  \\[-3pt]
      \  \\[-3pt]
      \phantom{00}
  \end{array}
$$
So we expect $-(q+kp)x+b-kq$ to be $0$. Therefore we've established that $$(x^2+px+q)(x+a-p)=x^3+ax^2+b.$$ In this arrangement, $b$ is clearly the result of multiplication of the constants, i.e. $b = qa - qp = q(a-p)$.

Title: How to solve Quadratic Equations with an Unknown C and other variables.

Body:
For instance $3x^2 - 11x + r$, I understand the value of $r$ is $6$ through trial and error but trial and error is extremely inefficient and time consuming thus not useful in exam situations, How would I solve this?

And also : If $x^2 +px + q$ is a factor of $x^3 + ax^2 + b$,

prove that:

$b = q(a-p)$


问:
Suggestions for books on combinatorics
I am very weak in combinatorics and I have not got my understanding clear to think a problem in combinatorics and come up with a logic/analysis.I want to develop both the interest and the understanding so that I can proceed further and solve questions.Are there books on combinatorics that are both interesting and has good problems(problems of all difficulties)</p>

答:
I second awkward’s recommendation of Bóna.

Kenneth P. Bogart’s Combinatorics Through Guided Discovery is freely available online and is a copious source of problems, many of which have hints at the back of the book.

Title: Suggestions for books on combinatorics


问:
Determine the $n$-th power of matrix
Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1  \\ }$$ May you help me with the answer since the book states :  $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n &  3\cdot4^n & 2+4^n  \\ }$$ while mine misses the ${1\over 6}$. So who is right?

$$$$Thanks in advance for any help you are able to provide.

答: Check with $n=0$ and $n=1$. Both are wrong.</p>

Title: Determine the $n$-th power of matrix


问:
Calculating expectation of a function of a continuous random variable?
Assume we have a continuous random variable $X$, with a probability density function

$$ f(x) = 1 - \frac{x}{2},\; 0 < x < 2  $$

Let's say I want to calculate $E(2X)$. I will use two different methods, and get different results, so I obviously understand something wrong.

METHOD 1:

Using the formula for the expectation of a function of a random variable, I get:

$$ E(2X) = \int_{-\infty}^{+\infty} 2xf(x)dx = \int_0^2 (2x-2x^2)dx = \frac{4}{3}  $$

I understand this formula intuitively, as it's pretty much exactly the same as in the discrete case, if we replace the integral with a finite sum and the probability density function with the probability mass function. Only the values change, and not the distribution, so we only apply the function 2x to the values, and not the distribution $f(x)$.

METHOD 2:

Let's say I declare a new continuous random variable, $Y = 2X$. I can calculate the probability density function of $Y$:

$$ g(y) = |\frac{dx}{dy}|f(x) = \frac{1}{2}f(\frac{y}{2}) = \frac{1}{2}(1-\frac{y}{4})$$

for $0<y<4$.

Then I can calculate:

$$ E(2X) = E(Y) = \int_{-\infty}^{+\infty} y\;g(y)dx = \int_0^4 \frac{1}{2}(1-\frac{y}{4})dx = \frac{2}{3}  $$

I don't understand what I did wrong here.

But I also don't understand the formula for $g(y)$ intuitively - shouldn't $g(y)$ and $f(x)$ be the same functions if they're injective, similar to the discrete case?

Anyway, if someone can point out what I did wrong, I would be very grateful.

答:
You forgot to multiply by $y$ for the second computation.

$g(y)$ and $f(x)$ are different functions, but they are related by $g(y)=\frac{1}{2}f\big(\frac{y}{2}\big)$ and are both decreasing (also see here). For instance, $h_{1}(x)=x$ and $h_{2}(x)=2x$ are injective, but $h_{1}(1)=1$ and $h_{2}(1)=2$.

It is a good idea to check if $g(y)=\frac{1}{2}(1-\frac{y}{4})$ for $0<y<4$ is a valid pdf. Indeed we have $g(y)\geq 0$ for $0<y<4$ and $\int_{-\infty}^{\infty}g(y)dy=\frac{1}{2}\int_{0}^{4}(1-\frac{y}{4})dy=1,$ so $g(y)$ is a valid pdf.

Next we have $$ E(2X) = E(Y) = \int_{-\infty}^{+\infty} y\;g(y)dx = \int_0^4 \frac{\color{red}{y}}{2}(1-\frac{y}{4})dx = \frac{4}{3}  $$


问:
Tensor Calculus: Divergence of the inner product of two vectors.
In my undergraduate course I had to prove this:$ \nabla \cdot (u \cdot v) = u \cdot \nabla(v) + v \cdot \nabla(u)$

But I believe that statement is wrong, I think it should be the following:
$\nabla \cdot (u \cdot v)= \partial^{j} (u^i v^{i}) = u^{i} \partial^{j} v^{i} + v^{i} \partial^{j} u^{i} = \nabla v \cdot u + \nabla u \cdot v
$

Here the "dot product" does not commute since the gradient of a vector is a matrix and the dot product of a vector with a matrix is non commutative like this:

$\nabla v \cdot u = (\partial ^{j} v^{i} e^{j} \otimes e^{i}) \cdot (u^{k} e^{k}) = \partial ^{j} v^{i} u^{i} = u^{i} \partial ^{j} v^{i}$

$u \cdot \nabla v = (u^{k} e^{k}) \cdot  (\partial ^{j} v^{i} e^{j} \otimes e^{i}) = u^{j} \partial ^{j} v^{i}$

$\therefore \nabla v \cdot u \neq u \cdot \nabla v$

My questions is, I'm doing something wrong for reaching this result:

$\nabla \cdot (u \cdot v)= \nabla v \cdot u + \nabla u \cdot v
$

Instead of this:

$ \nabla \cdot (u \cdot v) = u \cdot \nabla(v) + v \cdot \nabla(u)$

答:
I think I found out whats going on. First of all, we can all agree that $\nabla\cdot(u\cdot v)=\nabla(u\cdot v)$, so we are computing the gradient of a dot product. Now, it all depends on how you define the gradient of a vector: if you define it as $\nabla v = \partial^i v^j e^i\otimes e^j$ or as $\nabla v = \partial^j v^i e^i\otimes e^j$. I have found references that differ between these two. Note that one is just the transpose of the other. I prefer the first one which is the one that you are using actually. With the first notation you obtain the second identity ($\nabla \cdot (u \cdot v)=\nabla v \cdot u + \nabla u \cdot v
$) just as you showed.

However, if we look into wikipedia here we see that for two vector fields $A,B$ we get
$$
\nabla(A\cdot B) = A\cdot \nabla B + B\cdot \nabla A
$$
where its important not to confuse $A\cdot \nabla B$ with $(A\cdot \nabla) B$ as I did in the comments. Note that this equation coincides with your first identity (the problematic one), and not the second one. This can be explained by how the gradient of a vector is defined in that wikipedia page here in which the gradient is basically defined as $\nabla v = \partial^j v^i e^i\otimes e^j$ ($\nabla A = \left[\frac{\partial A_i}{\partial x^j}\right]_{ij}$ in the wikipedia notation) different from what you used as the definition of gradient in your procedure.

Under this setting:
$$
u\cdot \nabla v = (u^ke^k)\cdot(\partial^jv^ie^i\otimes e^j) = (u^i\partial^jv^i)e^j
$$
with the term $u^i\partial^jv^i$ as what you obtained with $\nabla v\cdot u$ with the other definition of $\nabla v$. (remember that $e^k\cdot (e^i\otimes e^j) = \delta^{ki}e^j$ with $\delta^{ki}$ the Kronecker delta, so you missed the $e^j$ in your procedure).

This is the most probable explanation of whats going on: your teacher may have defined the gradient as the transpose of what you understand as the gradient. Sounds like a reasonable confusion to me but let me know your thoughts.


问:
A property about Riemann integral
Let $f:[a,b] \subset \mathbb{R} \longrightarrow \mathbb{R}$ a integrable function. I know that if $|f|$ is integrable, then
$$\Bigg| \int_{a}^{b} f(x)\; dx\Bigg| \leq  \int_{a}^{b} |f(x)|\; dx. \tag{1}$$

Question. The property $(1)$ implies
$$\Bigg| \int_{a}^{b} f(x)\; dx\Bigg|^2 \leq  \int_{a}^{b} |f(x)|^2\; dx? \tag{2}$$

答:
Property (2) is not true. For example, take $f(x) = 1$ on the interval $[0,2]$. We have

$$ \left| \int_0^2 f(x) dx \right|^2 = 4$$
while
$$ \int_0^2 |f(x)|^2 dx = 2.$$

Title: A property about Riemann integral


问:
Is there a relationship between associators and commutators?
Let $A$ a algebra over a field $K$. We define the associator of elements $a,b,c \in A$ as
$$(a,b,c) = (ab)c - a(bc).$$
We define the commutator too by
$$[a,b] = ab - ba.$$

I'm looking for some relationship between them, more precisely to try to solve a Jordan algebra problem that I already know has a solution for Lie algebras.

I researched some time and I didn't find a reference that deals with this subject and more, I don't even know if there is such a relationship.

I decided to ask here, because sometimes someone has already worked or seen something in that sense.

Thank you very much in advance.

答:
In any algebra over $K$ we have the following relationship:

\begin{align*}
[[a,b],c]+[[b,c],a]+[[c,a],b] & =
(a,b,c)+(b,c,a)+(c,a,b)-(b,a,c)-(a,c,b)-(c,b,a),
\end{align*}

This is of course in the context of the Jacobi identity for Lie algebras. But there are several more such identities, also in the context of Jordan algebras (with $[[x,y]]=xy+yx$).


问:
Can an arc be greater than 360 degrees or less than 0 degrees.
Can an arc be greater than 360 degrees or less than 0 degrees? Is the (angular) arc length a signed scalar?

I'm creating a Graphics and basic geometry software library. I believe an arc can be defined by 5 scalars, the x and y components of the start point, the x and y components of the centre point and the signed value of the rotation. However I'm concerned that this data definition could lead to unnecessary rounding errors on the end point of the arc, the precise value of which is generally of greater concern than the precise value of the arc's centre point. So I'm leaning towards defining arcs by the 6 scalar definition of the start point, the end point, and the mid point of the circumference of the arc. This data definition has the added advantage that you don't have to worry about reflections, but would exclude arcs of greater than 360 degrees. Is this mathematically sound?

A successful software library will get used in ways unimagined by the library authors. So this where I place a strong priority or weighting on following mathematical definitions, because those more rigorous,robust and tried and tested definitions can hopefully pre-empt problems further down the road.

答:
So I think the issue here is the difference between an angular position and an angular displacement. The suggested What is the exact and precise definition of an ANGLE? triggered me to towards What is the difference between a point and a vector?. Prior to this I had been using the same class to represent 2 dimensional points and 2 dimensional vectors with the same class. I'm now separating them. In the same way that 2 o'clock is not the same thing as 2 hours duration. While you can kind get away with it with points, you can't with angles.

An angular position of -10 degrees is equal to an angular position of 350 degrees. But an angular rotation of -10 degrees is not equal to an angular rotation of 350 degrees.

So I think the answer is that I need separate classes for angles and rotations.


答: Hint: try Jordan normal form of your matrix.</p>

Title: Determine the $n$-th power of matrix

Body:
Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1  \\ }$$ May you help me with the answer since the book states :  $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n &  3\cdot4^n & 2+4^n  \\ }$$ while mine misses the ${1\over 6}$. So who is right?

$$$$Thanks in advance for any help you are able to provide.


问:
How to prove that the spectral norm is unitarily invariant?
How to prove that $\| U A \|_2 = \| A U \|_2 = \| A \|_2$ for any unitary matrix $U$?

It can be proved that

$$\| UA\|_2 = \sqrt{\lambda_{\max}({(UA)}^*(UA))} = \sqrt{\lambda_{\max}(A^*U^*UA)} = \sqrt{\lambda_{\max}(A^*A)} = \| A\|_2\\$$

but how to prove $\| A U \|_2 = \| A \|_2$?

Body: Hint: Prove $\|M\|_2 = \|M^*\|_2$ <s>for square matrices</s>. Then $\|AU\| = \|U^* A^*\| = \|A^*\| = \|A\|$.


问:
infinite sums and inequalities
We are in $(L^2[0,1],\|.\|_{L^2})$ with some basis $\{\phi_k\}$. For $f\in L^2$ define $f_k:=\langle f,\phi_k\rangle$. Furthermore assume some smoothness condition on $f$:
$$
\sum_{k=1}^\infty k^{2s}f_k^2<1\,\,\,\text{for some s > 0 }.
$$
From this it follows $$\sum_{M=k}^\infty f_k^2\leq M^{-2s}$$ for any $M\in\mathbf{N}$.

Now define M such that
$$
\|f\|^2_{L^2}=\sum_{k=1}^\infty f_k^2= 2 M^{-2s}
$$

I need to proof for some $\gamma>0$ that
$$
\sum_{k=1}^\infty k^{-\gamma}f_k^2 \lesssim M^{-2s-\gamma}
$$
Thank you for any suggestions.

答:
What i tried is to define N as
$$
\sum_{k=1}^Nk^{-\gamma} f_k^2=\sum_{k=N+1}^\infty k^{-\gamma}f_k^2
$$
Then the left hand side $\sum_{k=1}^\infty k^{-\gamma}f_k^2$ is smaller than $2N^{^-2s-\gamma}$ so it remains to show that N<M but i have no clue how to proof this.

Title: infinite sums and inequalities


问:
Weaker/Stronger Topologies and Compact/Hausdorff Spaces
In my topology lecture notes, I have written:

"By considering the identity map between different spaces with the same underlying set, it follows that for a compact, Hausdorff space:

$\bullet$ any weaker topology is compact, but not Hausdorff

$\bullet$ any stronger topology is Hausdorff, but not compact "

However, I'm struggling to see why this is. Can anyone shed some light on this?

答:
Hint. $X$ being a set, a topology $\tau$ is weaker than a topology $\sigma$ on $X$ if and only if the application
$$ (X, \sigma) \to (X,\tau), x \mapsto x $$
is continuous.


问:
Factorization of $f(x) = x^n - 1$ into irreducible polynomials $\mathbb{R}[x]$
Knowing that $e^{2\pi i \frac{k}{n}} = \cos(2 \pi k/n) + i \sin(2 \pi k/n) = \alpha $ is complex roots of $f(x) = x^n - 1$ and therefore the complex factorization is

\begin{align}
    f(X) &= \displaystyle\prod^{n - 1}_{k = 0} (x - \alpha_k)
\end{align}

How do I factorize into irreducible polynomials in $\mathbb{R}$ for odd and even n's? I know that for $(x-1) \mid f(x)$ for all n and if n is even $(x+1)\mid f(x)$ as well. I'm also aware that $(x - \alpha)(x-\bar{\alpha})$ is an irreducible polynomial in $\mathbb{R}$ that also divides $f(x)$. So the remaining factors most take the form $x^2 - 2x\cos(2 \pi k/n) + 1$. But I fail to see which k's to use in the factorization?

答:
Very simply, you have two cases:

If $n$ is even: $n=2m$, the complex roots are pairwise conjugate, except for $k=0$ and $k=m$, so the factorisation is
$$x^{2m}-1=(x-1)(x+1)\prod_{k=1}^{m-1}\Bigl(x^2-2\cos\frac{2k\pi}n x+1\Bigr).$$

If $n$ is odd: $n=2m+1$, they're pairwise conjugate, except for $k=0$, so we obtain
$$x^{2m+1}-1=(x-1)\prod_{k=1}^{m}\Bigl(x^2-2\cos\frac{2k\pi}n x+1\Bigr).$$


问:
Approximating $E[g(Y)]$ with a Taylor series is easy, but $E[g(YX)]$ seems much tougher (potentially impossible)?
Let $X$ and $Y$ be continuous random variables with $Y \sim \mathcal{N}(\mu_Y,\sigma_Y^2)$.

I want to approximate $E[g(YX)]$ where $g(x) = e^{-x^2}$.

Specifically, my plan is to use a Taylor expansion of $Y$ at $\mu_Y$, which I expect should be accurate when the variance $\sigma_Y^2$ is small.

If we had $E[g(Y)]$ instead of $E[g(YX)]$ we could use the Taylor expansion directly to get
\begin{align}
E[g(Y)] &\approx g(\mu_Y) + \frac{\sigma_Y^2}{2}g''(\mu_Y).
\end{align}

But the $X$ in $E[g(YX)]$ complicates matters. I made an attempt at it and obtained the following expression:
\begin{align}
E[g(YX)] & \approx E[g(\mu_Y X)] + \frac{\sigma_Y^2}{2}E[g''(\mu_Y X)].
\end{align}
This seems to match up well intuitively with what we have for the simpler expecation $E[g(Y)]$ above. But I'm not certain whether its correct. Here's what I did (note that I'm still just getting to grips with conditional probability):
\begin{align}
E[g(YX)] & = E[E[g(YX)|X]] \\
& = \int E[g(YX)|X=x] p_X(x) dx \\
\end{align}
Since the expectation is conditioned on $x$, I treat the $X$ in $g(XY)$ like a constant and Taylor expand with respect to $Y$. That is, I define $h(y;x) = e^{- y^2 x^2}$, so that $h(y;x) = g(yx)$, and then write:
\begin{align}
g(Yx) = h(Y;x) &\approx h(\mu_Y;x) + \frac{h''(\mu_Y;x)}{2} (Y-\mu_Y)^2.
\end{align}
Taking the conditional expectation gives
\begin{align}
E[g(YX)|X=x] = E[h(Y;X)|X=x] &\approx E\bigg[h(\mu_Y;X) + \frac{h''(\mu_Y;X)}{2} (Y-\mu_Y)^2\bigg|X=x\bigg] \\
& = h(\mu_Y;x) + \frac{h''(\mu_Y;x)}{2} \sigma_Y^2.
\end{align}

Then I substitute this expansion into the integral from earlier to get
\begin{align}
E[g(YX)] & \approx \int \bigg(h(\mu_Y;x) + \frac{h''(\mu_Y;x)}{2} \sigma_Y^2\bigg) p_X(x) dx \\
& = E[h(\mu_Y;X)] + \frac{\sigma_Y^2}{2}E\bigg[h''(\mu_Y;X)\bigg] \\
& = E[g(\mu_Y X)] + \frac{\sigma_Y^2}{2}E[g''(\mu_Y X)].
\end{align}

I'm not sure whether what I've done is valid, and if so, if I have fully justified what I have done.

If this is not valid, is it possible to correct it such that we can obtain an approximation $E[g(YX)]$ by way of taking a Taylor expansion of $Y$..or is what I am trying to fundamentally impossible?

答:
The second-order Taylor expansion of $g(xy)$ around $y=\mu_Y$ is
$$
g(xy)=g(x\mu_Y)+g'(x\mu_Y)x(y-\mu_Y)+\frac{1}{2}g''(x\mu_Y)x^2(y-\mu_Y)^2+R.
$$
Thus, since $X$ and $Y$ are independent,
$$
\mathsf{E}g(XY)=\mathsf{E}g(X\mu_Y)+\frac{\sigma_Y^2}{2}\mathsf{E}[g''(X\mu_Y)X^2]+\mathsf{E}R.
$$


答:
This comes from the fact that the eigenvalues of $AB$ and $BA$ are the same:

Proof:

If $\lambda\neq0$ is an eigenvalue of $AB$ then there is some nonzero eigenvector $v$ such that $ABv=\lambda v$. Now consider $(BA)Bv=B(ABv)=\lambda Bv$. So $w=Bv$ is a nonzero eigenvector (since $ABv=A(Bv)=\lambda v$) with $\lambda$ as eigenvalue. Of course you repeat the argument for $BA$.

If $\lambda=0$ is an eigenvalue of $AB$ then $\det(AB)=0$, so $\det(BA)=\det(B)\det(A)=\det(AB)=0$ and $\lambda=0$ is an eigenvalue of $BA$. Again, this can be repeated for $BA$.

We conclude they have the same eigenvalues. $\Box$

Knowing this, can you use it to prove what you want?

Title: How to prove that the spectral norm is unitarily invariant?

Body:
How to prove that $\| U A \|_2 = \| A U \|_2 = \| A \|_2$ for any unitary matrix $U$?

It can be proved that

$$\| UA\|_2 = \sqrt{\lambda_{\max}({(UA)}^*(UA))} = \sqrt{\lambda_{\max}(A^*U^*UA)} = \sqrt{\lambda_{\max}(A^*A)} = \| A\|_2\\$$

but how to prove $\| A U \|_2 = \| A \|_2$?


问:
Show that $C(\mathbb R, \mathbb R)$ is not connected
$\bar{d}: \mathbb R × \mathbb R → \mathbb R$ denotes the standard bounded metric on $\mathbb R$ defined
by $\bar{d}(x, y) := \min\{|y − x|, 1\}$.

Given a topological space $X$, let $C(X, \mathbb R)$ denote the set of all continuous functions $f : X → \mathbb R$ topologized with the topology of the uniform metric
$ρ(f, g) :=\sup\{\bar{d}(f(x),g(x)) :x \in X\}$ for $f, g \in C(X, \mathbb R)$

Show that $C(\mathbb R, \mathbb R)$ is not connected.

Hint: let $p \in C(X,\mathbb R)$ be bounded and let $q \in C(X, \mathbb R)$ be unbounded. Are $p, q$ in the same connect led component?

My attempt:

$p$ is bounded, then there exists $M \in \mathbb R$ such that $|p(x)| \leq M$.

$q$ is unbounded, then $|q(x)| < inf$.

$ρ(p, q) :=\sup\{\bar{d}(p(x),q(x)) :x \in X\} = \sup\{\min\{|q(x)-p(x)|,1\}:x \in X\}$ for $p, q \in C(X, \mathbb R)$

Now $|q(x) - p(x)| \leq |q(x)| + |p(x)| < inf + M = inf$

So, $\min\{|q(x) - p(x)|, 1\} = 1$, and $\sup\{\min\{1\}\}= 1$. Does this tell me that $p, q$ are in different components, and so $C(\mathbb R, \mathbb R)$ is not connected?

答:
Let $P,Q$ be the subspaces respectively of bounded and unbounded continuous functions on $\mathbb{R}$.

$P$ and $Q$ are disjoint and $P\cup Q=\mathcal{C}(\mathbb{R},\mathbb{R})$: a function cannot be both bounded and unbounded, but it must be one of the two

Every $p\in P$ satisfies $\overline{d}(p,Q)=1$, because necessarily for all $q\in Q$, $\sup (|p-q|)=+\infty$, and thus $\overline{d}(p,q)=1$. This implies that $P$ is open, since every ball of radius less than $1$ is totally contained in $P$. The same result holds for $Q$ by a similar argument

Thus, the space is not connected, since it is the disjoint union of open sets.

Note that one can prove (as hinted by TheSilverDoe in the comments to the OP, and proved here), that $P$ is path connected.


问:
Is every geodesic-preserving map of the sphere an isometry?
Let $\mathbb{S}^n$ be the $n$-dimensional unit sphere, equipped with the standard round Riemannian metric.

Let $f:\mathbb{S}^n \to \mathbb{S}^n$ be a diffeomorphism and suppose that for every (parametrized) geodesic $\gamma$, $f \circ \gamma$ is also a (parametrized) geodesic.

Must $f$ be an isometry? (that is the restriction of an orthogonal matrix on $\text{O}(n+1)$).

An equivalent condition on $f$ is that $\nabla df=0$ where $\nabla=\nabla^{(T\mathbb{S}^n)^*} $ $ \otimes \nabla^{f^*T\mathbb{S}^n}$ is the relevant tensor product connection.

答:
Yes, such an $f$ must be an isometry.

Let $p\in S^n$, $v\in T_p S^n$ with $v\neq 0$.  By polarization, it is enough to show that  that $\|d_p f(v)\|= \|v\|$.

So, let's establish this equality.  To do so, first note that $\gamma:\mathbb{R}\rightarrow S^n$ defined by $\gamma(t) = \exp_p(tv)$ is injective on $\left[0,\frac{2\pi}{\|v\|}\right)$.

Because $f$ is a diffeomorphism, $f\circ\gamma$ is injective on the same interval.  But since $f$ maps geodesics to geodesics, we know $f\circ \gamma = \exp_{f(p)}(td_p f(v))$.  This implies that $\|d_p f(v)\| \leq \|v\|.$

Further, because $\gamma$ fails to be injective on $\left[0, \frac{2\pi}{\|v\|}\right]$, the same must be true of $f\circ \gamma$.  This implies that $\|d_pf(v)\| \geq \|v\|$.  The two inequalities together now give the result.


问:
Why are "~B" and "$\{x : x\notin B\}$" not the same thing? Trying to understand basic set theory.
Why are they different: "~B" vs. $\{x:x \notin B\}$?  They both mean elements not in B. As long as that's satisfied, it should work, right?

How we use them in class:

$A\setminus B$ = {$x: x \in A \text{ and } x\notin B $}

$x \in A\setminus B$  iff $(x \in A) \land (x \notin B)$

Example:

Let B = {2, 4, 6, 8}.

the integers 1, 3, 5, 7, 9

the real numbers greater than 25

the function $f(x)=x^2+3$

the circle of radius 1 centered at the origin of the plane
the Empire State Building
my uncle Wilbur

Solution explains they are different:

$\{x:x \notin B\}$ contains all of the above.

1, 3, 5, 7 should be included in "~B", and perhaps real numbers greater than 25. But, most likely not the others. And certainly, knowing that my uncle Wilbur is not a member of the set B would contribute little to any discussion of B.

Question:
What's wrong with including the others? They are clearly not in B. I'm so confused right now.

答:
When taking the complement of some set we are (at least implicitly) working with some sort of universal set in the background. In your example, the universal set $X$ could be the set of all integers $\mathbb{Z}$, or the set of natural numbers $\mathbb{N}$, or the set $\{2,4,6,8\}\cup C$, where $C$ is the set of all cows. You can't meaningfully take the complement of a set unless you know what its background set is.

(In some sense, Russell's paradox is an extreme example of how things can break when you haven't specified your background set.)


问:
How can we describe all the maximal normal subgroups of $G$?
if $G$ is cyclic of order $n>1$ describe all the maximal normal subgroups of $G$. (We say that $H\trianglelefteq G$ is maximal normal if it doesn't exists $K\trianglelefteq G$ such that $H \nsubseteq  K \nsubseteq  G$)

My Proof:
I tried that if $G=\langle a \rangle$ with $|a|=n=P_1^{\alpha_1}P_2^{\alpha_2}...P_p^{\alpha_p}$ and we let be $\langle b_m \rangle \leq \langle a \rangle$ such that $b_m=a^m$ and we observe that $g\langle b_m \rangle=\langle b_m \rangle g$ for all $g\in\langle a \rangle$ then we have that $\langle b_m \rangle \trianglelefteq\langle a \rangle$ and if $x\mid y$ then $\langle b_y \rangle\subseteq\langle b_x \rangle$. Therefore we have that all of the maximal normal subgroups are of the form $\langle b_{p_i} \rangle$ with $i=1, 2, ..., p$

($p$ because we are talking of the primes without power)

But I'm not sure if this is correct

答:
You've got it! Thinking about the divisibility order on $n$ is the exact right thing to consider. You have also noticed that $\mathbb{Z}/n$ is abelian, so every subgroup is normal. This means we're trying to find all the maximal subgroups, since they are automatically normal. That's a great observation.

Then the $\langle a^p \rangle$ for each fixed prime $p | n$ are the maximal normal subgroups! Another way to see that they're maximal is to notice $\frac{|G|}{| \langle a^p \rangle |} = p$ (do you see why?). Do you see why this observation also proves that $\langle a^p \rangle$ is maximal?

Again, your proof is perfect! I just don't like leaving answers that say "you did it", so I'm giving another approach too.

I hope this helps ^_^


答:
As the comments indicate, the expression ~$B$ implicitly assumes a universe $U$ of possible elements some of which may be in the subset $B$.

Both the first two possible answers require a very improbable universe $U$. The third even more so. The last two are not even statements about mathematical objects, so make no sense at all.

Title: Why are "~B" and "$\{x : x\notin B\}$" not the same thing? Trying to understand basic set theory.

Body:
Why are they different: "~B" vs. $\{x:x \notin B\}$?  They both mean elements not in B. As long as that's satisfied, it should work, right?

How we use them in class:

$A\setminus B$ = {$x: x \in A \text{ and } x\notin B $}

$x \in A\setminus B$  iff $(x \in A) \land (x \notin B)$

Example:

Let B = {2, 4, 6, 8}.

the integers 1, 3, 5, 7, 9

the real numbers greater than 25

the function $f(x)=x^2+3$

the circle of radius 1 centered at the origin of the plane
the Empire State Building
my uncle Wilbur

Solution explains they are different:

$\{x:x \notin B\}$ contains all of the above.

1, 3, 5, 7 should be included in "~B", and perhaps real numbers greater than 25. But, most likely not the others. And certainly, knowing that my uncle Wilbur is not a member of the set B would contribute little to any discussion of B.

Question:
What's wrong with including the others? They are clearly not in B. I'm so confused right now.


问:
Show that sets are pairwise disjoint
Let $A, B \subset X$. Show that the sets $A$ \ $B$, $B$ \ $A$, $A \cap B \subset X$ are pairwise disjoint and that their union is $A \cup B$.

I know that $A$ \ $B$ is $A \cap B^{c}$ and I can write  $B$ \ $A$ as $B \cap A^{c}$, but how can I show that they are pairwise disjoint?

答:
I don't know any shortcut, sorry.

By definition, $A\setminus B=\{x\in X\mid x\in A \wedge x\notin B\}$ and $B\setminus A=\{x\in X\mid x\in
B \wedge x\notin A\}$. As far as the disjointnesses concerns, note that:

$x\in (A\setminus B)\cap (B\setminus A)$; then, $(x\in A \wedge x\notin B)\wedge (x\in B \wedge
x\notin A)$, which is equivalent to $(x\in A \wedge x\notin A)\wedge (x\in B \wedge x\notin B)$: false;
then, $(A\setminus B)\cap (B\setminus A)=\emptyset$.

$x\in (A\setminus B)\cap (A\cap B)$; then, $(x\in A \wedge x\notin B)\wedge (x\in A \wedge x\in B)$,
which is equivalent to $(x\in A) \wedge (x\in B\wedge x\notin B)$: false; then, $(A\setminus B)\cap (A\cap
B)=\emptyset$.

$x\in (B\setminus A)\cap (A\cap B)$; then, $(x\in B \wedge x\notin A)\wedge (x\in A \wedge x\in B)$,
which is equivalent to $(x\in B) \wedge (x\in A\wedge x\notin A)$: false; then, $(B\setminus A)\cap (A\cap
B)=\emptyset$.

Moreover:

\begin{alignat}{1}
(A\setminus B)\cup (B\setminus A) &= \{x\in X\mid (x\in A \wedge x\notin B) \vee (x\in B \wedge x\notin
A) \} \\
&= \{x\in X\mid [(x\in A \wedge x\notin B) \vee (x\in B)] \wedge [(x\in A \wedge x\notin B) \vee (x\notin A)]
\} \\
&= \{x\in X\mid [(x\in B\vee x\in A)\wedge (x\in B\vee x\notin B)] \wedge [(x\notin A\vee x\in A)\wedge
(x\notin A\vee x\notin B)] \} \\
&= \{x\in X\mid (x\in B\vee x\in A) \wedge (x\notin A\vee x\notin B) \} \\
&= \{x\in X\mid (x\in B\vee x\in A) \wedge \neg(x\in A\wedge x\in B) \} \\
&= (A\cup B)\cap (X\setminus (A\cap B)) \\
\end{alignat}

whence:

\begin{alignat}{1}
(A\setminus B)\cup (B\setminus A) \cup (A\cap B) &= [(A\cup B)\cap (X\setminus (A\cap B))]\cup (A\cap B) \\
&= [(A\cap B)\cup (A\cup B)]\cap [(A\cap B)\cup (X\setminus (A\cap B))] \\
&= (A\cup B)\cap X \\
&= A\cup B
\end{alignat}

Title: Show that sets are pairwise disjoint


问:
Solving a determinant equation for the properties of a matrix $C$
Suppose I have two square $N \times N$ matrices which a satisfy
$$
\frac{\det( R + C )^2}{\det(R - (C + C^{\ast})/2) \det( R + (C + C^{\ast})/2 )} \ = \ 1
$$
where $R$ is a real matrix, and $C$ has entries which are complex-valued in general (the operation $\ast$ is just complex-conjugation here).

What can one say about the matrix $C$ from this? I can see that $C=0$ is a solution to this equation, but do there exist other solutions for $C$?

答: Whenever $R$ is invertible and skewsymmetric, then $C=-2R$ is a solution


问:
Supremum of product of matrices with norm less than one
We consider 2 invertible matrices $A_{1}$ and $A_{2}$, and a submultiplicative norm $\|.\|$ such that $\|A_{1}\|\le 1$ and $\|A_{2}\|\le 1$. My goal is to find:
$$
\sup\limits_{i_{1}\in \mathbb N^{*},i_{2}\in \mathbb N^{*},i_{3}\in \mathbb N}\|A_{1}^{i_1}A_{2}^{i_2}A_{1}^{i_3}\|
$$

My intuition says that the supremum is achieved when $i_{1}=1, i_{2}=1,i_{3}=0$. I tried to prove this by induction but no luck. Does any one can help?

答:
Using the submultiplicativity of the norm, if $p,q,r$ are three integers such that $p,q\ge1$ and $r\ge0$, then
\begin{aligned}
\|A_1^pA_2^qA_1^r\|
&\le\|A_1^pA_2^q\|\|A_1\|^r\\
&\le\|A_1^pA_2^q\|\\
&\le\|A_1^pA_2\|\|A_2\|^{q-1}\\
&\le\|A_1^pA_2\|\\
&\le\|A_1\|^{p-1}\|A_1A_2\|\\
&\le\|A_1A_2\|.
\end{aligned}


问:
Trouble finding the level curves of $f(x,y)=\frac{2x}{x^2+y^2}$
I am trying to find the level curves of $f(x,y) = \frac{2x}{x^2+y^2}$ for $f(x,y) = {-1,0,1}$.

I can not find examples in Stewart Calculus for rational equations.

Any help would be greatly appreciated.

答:
We have that

$$\frac{2x}{x^2+y^2}=k \implies x^2+y^2-\frac2kx=0 \implies \left(x-\frac1k\right)^2+y^2=\frac1{k^2}$$

which is a circle centered at $\left(\frac1k, 0\right)$ with radius $R=\frac1{|k|}$.


问:
Area under a curve with two functions given
Find the area of the region bounded by the curve and line in the diagram

I've found the upper function to be ($-x + 6$) and the lower function is ($x^2+4$)

I did $-x + 6 = x^2 + 4 $ to get the upper and lower. a = 1 b = -2

$$
\int_{-2}^1 (-x + 6) - (x^2 +4) dx
$$

Would like to know if this is correct.

Graph / Diagram given by the question

答:
Yes you are correct.

In general the area between two curves, say $f(x)$ and $g(x)$, on an interval $[a,b]$ is given by $$\int_{a}^{b}|f(x)-g(x)|dx$$

Since $f(x)=-x+6\geq x^2+4=g(x)$ for $x\in[-2,1]$, then $|f(x)-g(x)|=f(x)-g(x).$

Thus we have $$
\int_{-2}^1 |(-x + 6) - (x^2 +4)| dx
=\int_{-2}^{1}(-x+6)-(x^2+4)dx=\frac{9}{2}$$


问:
Prove that if subsequences converge to L, then the original sequence converges to L
Here's the question:

We know that all subsequences of an original sequence $(x_n)$ converge to $L$. For example, $(x_{2n})$, $3n$, ... $mn$ for $m \geq 2$ converge to $L$.

How do I prove that $(x_n)$ converges to $L$?

Here's my attempt:

Assume that $(x_n)$ diverges, i.e. $\exists \epsilon > 0$ such that $\forall N \in \mathbb{N}$, $\exists n \geq N$ such that $|x_n - L| \geq \epsilon$.

Then (tldr) I show that $mn > n \geq N$ which implies $x_{mn}$ diverges which is a contradiction.

My problem is that $x_n$ might also oscillate, but not diverge, and I'm not sure how to go about showing that - my prof mentioned that it might be difficult to show that it doesn't oscillate (at my skill level).

Is there a better way to go about this problem?

答:
In fact if there are two convergent subsequences $(x_{n_k}),(x_{n_l})$ whose union is the total sequence and converging to the same limit $L$, then $(x_n)$ converges to $L$.

Proof:

Given $\epsilon>0$ there are $n_{k_0},n_{l_0}>0$ such that $|x_{n_k}-L|<\epsilon$ and $|x_{n_l}-L|<\epsilon$ for every $n_k\ge n_{k_0}$ and $n_l\ge n_{l_0}$ respectively.

Then for every $n\ge n_0=\max\{n_{k_0},n_{l_0}\}$ we have that $|x_n-L|<\epsilon$, since either $x_n\in(x_{n_k})$ or $x_n\in(x_{n_l})$. Therefore $x_n$ converges to $L$.$\Box$

Of course, the result is true if you have any finite number of subsequences all converging to the same limit, and whose union is the total sequence. This is often used to prove the convergence and calculate the limit of alternating sequences. For example, to see the sequence of Fibonacci numbers' ratios is convergent one can consider the disjoint subsequences $(x_{2n})$ and $(x_{2n+1})$.


问:
partial differential equation $y^2\frac{\partial ^2 u}{\partial x\partial y} + \frac{\partial ^2 u}{\partial y^2}$
Helle everybody, i have a problem with a partial differential equation, I tried to work but it's not true
$$\begin{array}{l}
 y^2\frac{\partial ^2 u}{\partial x \, \partial y} + \frac{\partial ^2 u}{\partial y^2} - \frac{2}{y}\frac{\partial u}{\partial y} = 0 \\[6pt]
 u(x,1) = 1 - x \\[6pt]
 \left. \frac{\partial u}{\partial y} \right|_{y = 1} = 3
 \end{array}$$</p>

答:
I have done this as; howerver i am embarrass when use the condition:
$$\begin{array}{l}
 a = 1;b = y^2/2;c = 0 =  > b^2 - ac = \frac{y^2}{4} =  > \frac{dx}{dy} = \frac{b \pm \sqrt{ b^2 - ac }}{a} = \frac{y^2/2 \pm y^2/2}{1} = \left[ \begin{array}{l}
 0 \\
 {y^2} \\
 \end{array} \right. \\
 \left\{ \begin{array}{l}
 \frac{{dx}}{{dy}} = 0 =  > dx = 0 =  > x = C =  >  \\
 \frac{{dx}}{{dy}} = {y^2} =  > dx = {y^2}dy =  > C = x - \frac{{{y^3}}}{3} =  > {\rm{  }} \\
 \end{array} \right.{\rm{ =  > }} \\
 \frac{\partial ^2 u}{\partial y^2} = \frac{\partial ^2 u}{\partial \xi ^2}{\left( {\frac{{\partial \xi }}{{\partial y}}} \right)^2} + \frac{{{\partial ^2}u}}{{\partial {\eta ^2}}}{\left( {\frac{{\partial \eta }}{{\partial y}}} \right)^2} + 2\frac{{{\partial ^2}u}}{{\partial \xi \partial \eta }}\left( {\frac{{\partial \xi }}{{\partial y}}} \right)\left( {\frac{{\partial \eta }}{{\partial y}}} \right) + \frac{{\partial u}}{{\partial \xi }}\left( {\frac{{{\partial ^2}\xi }}{{\partial {y^2}}}} \right) + \frac{{\partial u}}{{\partial \eta }}\left( {\frac{{{\partial ^2}\eta }}{{\partial {y^2}}}} \right) \\
  =  > (1) \\
 \frac{{{\partial ^2}u}}{{\partial x\partial y}} = \frac{{{\partial ^2}u}}{{\partial {\xi ^2}}}\left( {\frac{{\partial \xi }}{{\partial x}}\frac{{\partial \xi }}{{\partial y}}} \right) + \frac{{{\partial ^2}u}}{{\partial {\eta ^2}}}\left( {\frac{{\partial \eta }}{{\partial x}}\frac{{\partial \eta }}{{\partial y}}} \right) + \frac{{{\partial ^2}u}}{{\partial \xi \partial \eta }}\left( {\frac{{\partial \xi }}{{\partial x}}\frac{{\partial \eta }}{{\partial y}} + \frac{{\partial \xi }}{{\partial y}}\frac{{\partial \eta }}{{\partial x}}} \right) + \frac{{\partial u}}{{\partial \xi }}\left( {\frac{{{\partial ^2}\xi }}{{\partial x\partial y}}} \right) + \frac{{\partial u}}{{\partial \eta }}\left( {\frac{{{\partial ^2}\eta }}{{\partial x\partial y}}} \right) \\
  =  > (2) \\
 \frac{{\partial u}}{{\partial y}} = \frac{{\partial u}}{{\partial \xi }}.\frac{{\partial \xi }}{{\partial y}} + \frac{{\partial u}}{{\partial \eta }}\frac{{\partial \eta }}{{\partial y}} =  > (3) \\
 (1)(2)(3) =  > (*): - {y^4}\frac{{{\partial ^2}u}}{{\partial {\eta ^2}}} - {y^4}\frac{{{\partial ^2}u}}{{\partial \xi \partial \eta }} + {y^4}\frac{{{\partial ^2}u}}{{\partial {\eta ^2}}} + 0 + 0 - 2y\frac{{\partial u}}{{\partial \eta }} - \frac{2}{y}\left[ {{y^2}\frac{{\partial u}}{{\partial \eta }}} \right] = 0 \\
  =  >  - {y^4}\frac{{{\partial ^2}u}}{{\partial \xi \partial \eta }} - 4y\frac{{\partial u}}{{\partial \eta }} = 0 =  > {\rm{    }}(**) \\
 \left\{ \begin{array}{l}
 \eta  = x - \frac{{{y^3}}}{3} \\
 \xi  = x \\
 \end{array} \right. =  > \left\{ \begin{array}{l}
 {y^3} = 3(\xi  - \eta ) \\
 x = \xi  \\
 \end{array} \right. =  > (**) =  >
 \end{array}$$

$$\begin{array}{l}
 Let:\frac{{\partial u}}{{\partial \eta }} = V(\xi ,\eta ) =  > \frac{{{\partial ^2}u}}{{\partial \xi \partial \eta }} = \frac{{\partial V}}{{\partial \xi }} =  > 3(\xi  - \eta )\frac{{\partial V}}{{\partial \xi }} + 4V = 0 =  >  \\
 \frac{{\partial V}}{V} =  - \frac{4}{3}\frac{{\partial \xi }}{{(\xi  - \eta )}} =  > \ln V =  - \frac{4}{3}\ln(\xi  - \eta ) - \underbrace {\ln F(\eta )}_{C{\rm{ons t    C}}} =  >  {V(\xi ,\eta ) = F(\eta ){{(\xi  - \eta )}^{ - 4/3}}}\\
 From:\frac{{\partial u}}{{\partial \eta }} = V(\xi ,\eta ) =  > \frac{{\partial u}}{{\partial \eta }} = F(\eta ){(\xi  - \eta )^{ - 4/3}} =  > \partial u = F(\eta ){(\xi  - \eta )^{ - 4/3}}\partial \eta  \\
  =  > {u(\xi ,\eta ) = \underbrace {\int {F(\eta ){{(\xi  - \eta )}^{ - 4/3}}\partial \eta } }_{G(\eta )} + \underbrace {\Phi (\xi )}_{C{\rm{ons}}t{\rm{ C}}}} =>{u(\xi ,\eta ) = G(\eta ) + \Phi (\xi )}
{\rm{   =  > }} \\
 {\rm{ =  >   }}u(x,y) = G(x - \frac{{{y^3}}}{3}) + \Phi (x){\rm{     (***)}} \\
 {\rm{ Now we apply condition of u into (***)}} \\
 \left\{ \begin{array}{l}
 u(x,1) = 1 - x =  > G(x - \frac{1}{3}) + \Phi (x){\rm{  }} \\
 {\left. {\frac{{\partial u}}{{\partial y}}} \right|_{y = 1}} = 3 =  >  - {y^2}G'(x - \frac{{{y^3}}}{3}){|_{y = 1}} + 0 = 3 \\
 \end{array} \right. =  > \left\{ \begin{array}{l}
 G\left( {x - \frac{1}{3}} \right) + \Phi (x){\rm{   = 1 - x}} \\
  - G'\left( {x - \frac{1}{3}} \right) = 3 \\
 \end{array} \right. =  > \left\{ \begin{array}{l}
 G\left( {x - \frac{1}{3}} \right) + \Phi (x){\rm{   = 1 - x}} \\
 G\left( {x - \frac{1}{3}} \right) =  - 3y + C \\
 \end{array} \right. \\
  \\
 \end{array}$$
$${ =  > \left\{ \begin{array}{l}
 \Phi (x) = 1 - x + 3y - C \\
 G\left( {x - \frac{1}{3}} \right) =  - 3y + C \\
 \end{array} \right. =  > \left\{ \begin{array}{l}
 \Phi (x) = 1 - x + 3y - C \\
 G\left( {x - \frac{{{y^3}}}{3}} \right) = ?????????????????????? \\
 \end{array} \right.}$$