问:
Bourbaki: Quantification Expansion
I'm reading Theory of Sets where existential quantification is defined:

a)
$$
\exists x R = (\tau_x(R)|x)R
$$

and I have questions regarding the expansion for an example relation:

b)
$$
\exists y \exists x \in x y
$$

using the definition to expand.  Link bars are placed over and between symbols $\tau_{\alpha} \rightarrow \square_{\alpha}$.  $\alpha$ is only used in this post to identify the initial and final point of the link bar.  If the link bars are added on top of the symbols, $\alpha$ can be removed.

c)
$$
\exists y (\in (\tau_x \in \square_x y) y)
$$
Use definition (a) for $\exists y$:
$$
y_{sub} = \tau_y \in (\tau_x \in \square_x \square_y) \square_y
$$
and the resulting expression:

d)
$$
(\in (\tau_{x_1} \in \square_{x_1} (\tau_{y_1} \in (\tau_{x_2} \in \square_{x_2} \square_{y_1}) \square_{y_1})) (\tau_{y_2} \in (\tau_{x_3} \in \square_{x_3} \square_{y_2}) \square_{y_2}))
$$

$\tau_x(R)$ represents a fully qualified object such that $R$ holds.  If there is no such object, then $\tau_x(R)$ represents an arbitrary object.  In the latter case, such an arbitrary object would cause $R$ to be false.

Question
Does the final form (d) look correct?  I have doubts because the outer most choice operators $\tau_{x_1}$ and $\tau_{y_2}$ may result in different values which may not hold for the expression $\in x y$.  For instance, the sub-expression starting with $\tau_{y_1}$ and $\tau_{y_2}$ is the same, but the link bars are self contained to each sub-expression, meaning $y$ may hold for multiple values, which may not correspond to $x$.

I think either form (d) is incorrect or Bourbaki intended that duplicate $\tau_x$ linked expressions all choose the same value.  In the later case, it seems one would have to search the expression for duplicates somehow.

Thanks

UPDATE
The form (d) above is correct and if there are duplicate $\tau$ expressions the choice operator must choose the same for $\square$.

答:
Hyper long comment

IMO, Bourbaki's cumbersome notation (squares and links) is only needed in order to define the usual concepts of scope and binding of a quantifier.

The $\tau$ operator is exactly like the set-builder one: $\{ x \mid \varphi(x) \}$; when applied to a formula with $x$ free produces a term, i.e. the name of an object.

In the same way, we have that [page 20]:

if $B$ is an assertion and $x$ a letter, then $\tau_x(B)$ is an object. [When used with] an assertion $B$ that expresses a property of an object $x$, it represents a distinguished object which has that property, if such an object exists.

Bourbaki uses it to define the quantifiers, and specifically: $\exists x A := (\tau_x (A) \mid x)A$, which we can write as: $A[\tau_x(A) /x]$.

What happens with a double quantification $\exists y \exists x R(x,y)$?

We have to keep track of the formula to which we apply the $\tau$ operator; thus, with the first choice, the operator is not relative to formula $R(x,y)$ but to formula $S(y)=\exists x R(x,y)$, where only $y$ is free.

In order to keep track of the formula $A$ to which the $\tau$ operator applies, I'll use $\tau_x^A$.

Thus, the first step will produce: $\exists x R(x,\tau_y^S)$, and the second step will produce:

$R(\tau_x^R,\tau_y^S)$.

In conclusion, IMO we have to consider not different variables $x_1,x_2$, but different formulas.

Now, going back to your example, we have that $R$ is $\in$ and thus: $\in(\tau_x^R,\tau_y^S)$, but what $\tau_x^R$ and $\tau_y^S$ look like?

$\tau_y^S$ will be $\tau_y(\exists x \in(x,y))$, and thus: $(\tau_y \exists x \in (x, \square_y)$. Does it work? I think so, because the formula $\exists x \in (x, y)$ has only $y$ free, and thus we can correctly apply the $\tau$ operator (that will bind it).

Up to now, we have "reduced" the original formula to $\in (\tau_x^R, (\tau_y \exists x \in (x, \square_y))$.

The second step will apply $\tau$ operator to $\exists x R(x,\tau_y^S)$, i.e. to $\exists x \in (x,(\tau_y \exists x \in (x, \square_y))$.

Formula $R$ is $\in (x,(\tau_y \exists x \in (x, \square_y)))$ and again it has only $x$ free: the second and third occurrences of $x$ are bound and thus we cannot use them in the substitution process (we can rewrite them with $z$ and nothing will change).

Thus, again, we have the term $\tau_x[\in (\square_x,(\tau_y \exists x \in (x, \square_y)))]$ that we have to replace in place of $x$ into $R$ to get:

$\in (\tau_x[\in (\square_x,(\tau_y \exists x \in (x, \square_y)))],(\tau_y \exists x \in (x, \square_y)))$.

Now, we have to try to read it...

According to the semantics of the $\tau$ operator, the term $\tau_y \exists x \in (x, \square_y)$ represents a distinguished objcet that has property $\exists x \in (x, y)$, if any. But this amounts to saying that it represents a non-empty set.

Assuming a "naive" set universe with only two sets: $\emptyset$ and $\mathcal P(\emptyset)$, the formula $\exists x \in (x, y)$ is satisfied by $\mathcal P(\emptyset)$, and thus we may choose it as value for $\tau_y \exists x \in (x, \square_y)$.

The second step is obvious: $\tau_x[\in (\square_x,(\tau_y \exists x \in (x, \square_y)))]$ represents a distinguished object that satisfy formula $\in (x, \mathcal P(\emptyset))$ and thus we may choose $\emptyset$ as value for it.

In conclusion, in our universe formula $\exists y \exists x (x \in y)$ is true because: $\emptyset \in \mathcal P (\emptyset)$.

Title: Bourbaki: Quantification Expansion


问:
How can I reduce a $2$nd order ODE as $1$ st order ODE
I'm trying to learn differential equations, I have some troubles with the following $2$nd order ode :

Let

$$\ddot{x}=-7\dot x +\omega(t)^2x(t) \:, \:\: x(0)=7,\dot x(0)=0$$

Where $\omega$ is a differentiable and periodic function on $\mathbb R$.

So my two questions are : How can I express this ODE as a  $1$ st order ODE ?  And how can I tell if the solution is global ?

Any help would be a lot appreciated. Best wishes.

EDIT: As marked in the comment there might be an error in the formula..

How would the same question be answered for

$$\ddot{x}=-\dot x -\omega(t)^2x(t) \:, \:\: x(0)=1,\dot x(0)=0?$$

答:
Hint

Set $y=\dot x$. Then $$\begin{pmatrix}\dot x\\ \dot y\end{pmatrix}=\begin{pmatrix}y\\ -7y+\omega (t)x\end{pmatrix}=\begin{pmatrix}0&1\\\omega (t)^2&-7\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}.$$

ContentLicense: CC BY-SA 4.0


问:
Hatcher A.1 , difficulty in understanding a certain step
In hatcher (algebraic topology) proposition A.1 page 520 in the appendix section, I don't see why we have $\phi_{\alpha}^{-1}(S)$ necessarily closed in $\partial D_{\alpha}^n$, once we have this I can see why $$\Phi_{\alpha}^{-1}(S)$$ contains at most one more point and is thus closed, but I struggle to understand the first part.

Here is the proposition for reference:

答: The map given by the attaching map $\partial D_{\alpha}^n \rightarrow X^{n-1}$ composed with the inclusion $X^{n-1} \hookrightarrow X$ is equal to the map $\Phi_{\alpha}$ restricted to $\partial D_{\alpha}^n$. Therefore this maps preimage on the set $S$ is necessarily closed, as we assumed that $S \cap X^{n-1}$ is closed inductively.


问:
Uniform $f$ is uniformly continuous
$f$ is a map from $f:\mathbb{R} \rightarrow \mathbb{R}$. Let $f$ be continous and assume $f$ is both uniformly contiuous on $(-\infty,0]$ and $[0,\infty)$ . Show that $f$ is uniformly continuous on all of $\mathbb{R}$.

I thought that the function $f$ is actually uniformly continuous over all the real numbers since it is in both the intervals $(-\infty,0]$ and $[0,\infty)$. I guess the "problem" arises on $f(0)$.

I have been trying to prove this for hours upon hours now but I just do not get it even though I get the definitions and that you should assign 3 $\delta$‘s to cover every case (I think) and then a $\delta$ to be the minimum of one of thoes. I can not show how to pick the $\delta$’s as this is hard for me.

答: Given $\varepsilon>0$, take $\delta>0$ such that $|x-y|\implies\bigl|f(x)-f(y)\bigr|<\frac\varepsilon2$ both on $[0,\infty)$ and on $(-\infty,0]$. Then if $x,y\in\Bbb R$ are such that $|x-y|<\delta$ and if $x\leqslant0\leqslant y$, then$$\bigl|f(x)-f(y)\bigr|\leqslant\bigl|f(x)-f(0)\bigr|+\bigl|f(0)-f(y)\bigr|<\varepsilon$$and you have a similar argument if $x\geqslant0\geqslant y$.

Title: Uniform $f$ is uniformly continuous


问:
Commutation of symmetric and skew-symmetric part of orthogonal matrix
Can we claim that for an orthogonal matrix $A$ (satysfying $AA^T=I$),
its symmetric and skew-symetric parts are always commuting?

Symmetric part is calculated as $S=\frac{1}{2}(A+A^T)$ and skew symmetric part as $K=\frac{1}{2}(A-A^T)$.

It's easy to show the claim for dimension $2$.

Also for dimension $3$ we can see commutation for orthogonal matrices with determinant equal to 1.

For a rotation matrix $R$ it's evident from the Rodrigues formula,

when $R=\color{blue}{I+(1-\cos\theta) V^2 }+ \color{green}{\sin\theta V }$ for some skew-symmetric matrix $V $.

Here commutation  follows because   symmetric part   and skew-symmetric part are expressed as polynomials of the same matrix $V $.

Is it possible to find similar explanation for higher dimensions of orthogonal matrices?

答:
One can verify it using a direct computation:

$$ SK = \frac{1}{2} \left( A + A^T \right) \cdot \frac{1}{2} \left(A - A^T \right) = \frac{1}{4} \left( A^2 - AA^T + A^TA - \left( A^T \right)^2 \right) = \frac{1}{4} \left( A^2 - \left( A^T \right)^2 \right)$$
while
$$ KS = \frac{1}{2} \left( A - A^T \right) \cdot \frac{1}{2} \left( A + A^T \right) = \frac{1}{4} \left( A^2 + AA^T - A^TA - \left( A^T \right)^2 \right) = \frac{1}{4} \left( A^2 - \left( A^T \right)^2 \right)$$

where we used the fact that $AA^T = A^TA$ (as they both are equal to $I_n$). More generally, the symmetric and and skew-symmetric part of a matrix commute iff the matrix is normal and an orthogonal matrix is normal.


答:
Let $\epsilon >0$. There exist $\delta_1$ abnd $\delta_2>0$ such that

$|f(x)-f(y)| <\epsilon/2$ if $x, y \leq 0$ and $|x-y| \leq \delta_1$

$|f(x)-f(y)| <\epsilon/2$ if $x, y \geq 0$ and $|x-y| \leq \delta_2$

Let $\delta=\min\{\delta_1,\delta_2\}$.
Suppose $|x-y| <\delta$. We claim that  $|f(x)-f(y)| <\epsilon$. The only case where you face difficulty is when one of $x,y$ is $\leq 0$ and the other is $\geq 0$. Supose $x \leq 0$ and $y \geq 0$. Then $|f(x)-f(y)| \leq |f(x)-f(0)|+|f(0)-f(y)|<\epsilon /2+\epsilon /2=\epsilon$.

Title: Uniform $f$ is uniformly continuous

Body:
$f$ is a map from $f:\mathbb{R} \rightarrow \mathbb{R}$. Let $f$ be continous and assume $f$ is both uniformly contiuous on $(-\infty,0]$ and $[0,\infty)$ . Show that $f$ is uniformly continuous on all of $\mathbb{R}$.

I thought that the function $f$ is actually uniformly continuous over all the real numbers since it is in both the intervals $(-\infty,0]$ and $[0,\infty)$. I guess the "problem" arises on $f(0)$.

I have been trying to prove this for hours upon hours now but I just do not get it even though I get the definitions and that you should assign 3 $\delta$‘s to cover every case (I think) and then a $\delta$ to be the minimum of one of thoes. I can not show how to pick the $\delta$’s as this is hard for me.


答:
The solution of any linear DE with leading coefficient $1$ exists where or as long as the coefficients are continuous. Here that only depends on $ω$, if that function is continuous on $\Bbb R$, then any solution can be extended to the whole of $\Bbb R$, growing in a rule-of-thumb like $$\exp\left(\int_0^r(\sqrt{3.5^2+ω(s)^2}-3.5)ds\right)$$
where the integrand is $ω(s)^2/7$ for small $ω$ and about $(ω(s)-3.5)$ for large $ω$.

This estimate follows the WKB approach, try $x=e^U$ to find $u'+u^2+7u-ω^2=0$, $u=U'$, assume the changes in $ω$ and thus in $u$ are slow enough that $u'$ can be neglected (that is the reason for it being hand-wavy or a rule-of-thumb), and solve the quadratic equation for $u$, then integrate. There will be a second solution due to the quadratic equation that is strictly falling and thus not relevant for the growth of the solution.

If, as in the second equation $\ddot x+\dot x+ω^2x=0$, the sign of the last term was correct for an oscillation, one would get oscillating solutions of the approximate shape
$$
\frac{e^{-0.5·t}}{\cos\phi_0}·\cos\left(\phi_0+\int_0^t\sqrt{ω(s)^2-0.5^2}ds\right)
$$
if the frequency has a lower bound $ω(t)>0.5$. Here $\phi_0$ was introduced to be chosen so that the initial derivative becomes zero. (This is somewhat superfluous as the formula is an approximation anyway and will quickly deviate, probably first in phase, the amplitude shape remaining valid for a longer time.)

Title: How can I reduce a $2$nd order ODE as $1$ st order ODE

Body:
I'm trying to learn differential equations, I have some troubles with the following $2$nd order ode :

Let

$$\ddot{x}=-7\dot x +\omega(t)^2x(t) \:, \:\: x(0)=7,\dot x(0)=0$$

Where $\omega$ is a differentiable and periodic function on $\mathbb R$.

So my two questions are : How can I express this ODE as a  $1$ st order ODE ?  And how can I tell if the solution is global ?

Any help would be a lot appreciated. Best wishes.

EDIT: As marked in the comment there might be an error in the formula..

How would the same question be answered for

$$\ddot{x}=-\dot x -\omega(t)^2x(t) \:, \:\: x(0)=1,\dot x(0)=0?$$


问:
Math problem using cups..
I have the following question:

Mike has 58 white cups and 198 green cups. He wants to place his cups in stacks by color so there are the same number in each stack and same color.

What is the greatest number of cups he can place in each stack?

How would I answer this guys?

答: Edit: It seems that you want a given stack to consist of only one color of cup, and for each stack to have the same number of cups. (This is a slightly more interesting problem, now.) We have to divide all the white cups up evenly and make the stack size as big as possible, so the stack size needs to be as large a divisor of $58$ as we can manage. Likewise, the stack size needs to be as large a divisor of $198$ as we can manage. Thus, the stack size needs to be the greatest common divisor of $58$ and $198$--that is, $2$.

ContentLicense: CC BY-SA 3.0

Title: Math problem using cups..


Title: Commutation of symmetric and skew-symmetric part of orthogonal matrix

Body:
Can we claim that for an orthogonal matrix $A$ (satysfying $AA^T=I$),
its symmetric and skew-symetric parts are always commuting?

Symmetric part is calculated as $S=\frac{1}{2}(A+A^T)$ and skew symmetric part as $K=\frac{1}{2}(A-A^T)$.

It's easy to show the claim for dimension $2$.

Also for dimension $3$ we can see commutation for orthogonal matrices with determinant equal to 1.

For a rotation matrix $R$ it's evident from the Rodrigues formula,

when $R=\color{blue}{I+(1-\cos\theta) V^2 }+ \color{green}{\sin\theta V }$ for some skew-symmetric matrix $V $.

Here commutation  follows because   symmetric part   and skew-symmetric part are expressed as polynomials of the same matrix $V $.

Is it possible to find similar explanation for higher dimensions of orthogonal matrices?

Body:
We have $AA^T=A^TA=I.$ Hence

$$(A+A^T)(A-A^T)=A^2-AA^T+A^TA-(A^T)^2=A^2-I+I+(A^T)^2=A^2-(A^T)^2.$$

A similar computation gives

$$(A-A^T)(A+A^T)=A^2-(A^T)^2.$$


问:
If each of 'n' integers is chosen at random from the set {$1,2,...,f$}, then find the probability that there sum is '$s$'.
If each of 'n' integers is chosen at random from the set {$1,2,...,f$}, then find the probability that there sum is '$s$', ( $n$ $\leq$ $s$ $\leq$ $nf$ )

Here, there is another particular case of this problem, If 'n' unbiased dice are cast, we have to show the probability that the sum is '$s$' will be,

$6^{-n}$ $\sum_{i=0}^{s-n/6}(-1)^i\binom ni\binom{s-6i-1}{n-1}$, i.e I think here the set is {$1,2,3,4,5,6$} to choose the values of $n$ .

答:
The claim seems trivial to me.

We have $n$ integers $x1, x_2, \dots, x_n$, each of them is in the set $\{1,2,\dots,f\}$, so
$$ 1 ≤ x_i ≤ f \qquad \forall \, i = 1, \dots, n .$$

Then we can easily bound the sum s:
$$ s = \sum_{i=1}^n x_i ≥ \sum_{i=1}^n 1 = n$$
and
$$ s = \sum_{i=1}^n x_i \leq \sum_{i=1}^n f = nf .$$

This means
$$ n ≤ s ≤ nf .$$

Notice that I didn't use any info about the randomness here, it is a general result which holds regardless of how the integers are chosen.


问:
Explicit formula for codifferential
I want to show that $$\delta\eta \overset{!}{=} \nabla^*\eta = -\sum\limits_{i=1}^{n} E_i \lrcorner \nabla_{E_i}\eta$$ where $\delta:\Omega^{k+1} \to \Omega^k$ is the adjoint of the exterior derivative $d:\Omega^k \to \Omega^{k+1}$ and $\nabla^*:\mathcal{T}^0_{k+1} \to \mathcal{T}^0_{k}$ is the adjoint of the covariant derivative $\nabla:\mathcal{T}^0_k \to \mathcal{T}^0_{k+1}$ of $(0,k)$-tensor fields, both w.r.t to the averaged inner product $(\cdot,\cdot) = \int_M g(\cdot,\cdot)$ (assume $M$ is compact for simplicity).

I think usually one proves this using the Hodge star operator (see e.g. [Poor], Differential Geometric Strucutres, p.152--154).

What I tried to do is this:

I know that
$$
 d{\omega}(X_0,\dots,X_k) = \sum\limits_{j=0}^k (-1)^j(\nabla\omega)(X_j,X_0,\dots,\widehat{X_j},\dots,X_k).
$$

Note that $\nabla\omega$ is not skew-symmetric, so we can not simplify this any further.

Let $\tau_{i,j} : \mathcal{T}^0_k \to \mathcal{T}^0_k$ denote the transposition of the $i$th and $j$th entry, i.e.,
$$
 (\tau_{i,j} \circ T)(X_1,\dots,X_i,\dots,X_j,\dots,X_k) := T(X_1,\dots,X_j,\dots,X_i,\dots,X_k)
 $$
if $T\in \mathcal{T}^0_k$ and $X_1,\dots,X_k \in \mathfrak{X}(M)$. The adjoint of $\tau_{i,j}$ with respect to the  inner product $g$ on $\mathcal{T}^0_k$ is $\tau_{i,j}$ itself. Hence the same holds for the averaged inner product $(\cdot,\cdot)$ on $\mathcal{T}^0_k$.  With this notation the above formula for $d$ becomes
$$
  d{\omega} = \sum\limits_{j=0}^k (-1)^j (\tau_{0,j} \circ \nabla\omega)
 $$
and it follows that for all $\omega \in \Omega^k$ and $\eta \in \Omega^{k+1}$
\begin{align*}
  (d{\omega},\eta) &= \sum\limits_{j=0}^k (-1)^j (\tau_{0,j} \circ \nabla\omega, \eta) \\
  &= \sum\limits_{j=0}^k (-1)^j (\nabla\omega, \tau_{0,j} \circ \eta) \\
  &= \sum\limits_{j=0}^k (-1)^j (\omega, \nabla^*(\tau_{0,j} \circ \eta)) \\
  &= (\omega,\nabla^*\eta) + \sum\limits_{j}^{k} (-1)^j (\omega, \nabla^*(\underbrace{\tau_{0,j} \circ \eta}_{=-\eta})) \\
  &= \left(\omega,\left(1+\sum\limits_{j}^{k} (-1)^{j+1}\right)\nabla^*\eta \right),
 \end{align*}
so $\delta\eta = 2\nabla^*\eta$ if $k$ is odd and $\delta\eta = \nabla^*\eta$ is $k$ is even.

This is of course the wrong result and it also feels really weird. After reading this I realized that the problem probably is that $\tau_{i,j}$ is not globally defined (and obviously not independent of the frame (not even of its order)).

EDIT: I don't think that this is the problem. $\tau_{i,j}$ is globally defined. Then I wonder even more: Where is this going wrong?

But my hope is that this could be fixed somehow. The usual proof with the Hodge star or, I think, another possibility is to define the 1-form $\alpha(X) = g(X \lrcorner \eta, \omega)$ and then go on from
$$
  \text{div } \alpha^\sharp = g(\underbrace{\sum\limits_{i=1}^{n} E_i \lrcorner\nabla_{E_i}\eta}_{=-\nabla^*\eta}, \omega) + \sum\limits_{i=1}^{n}g(E_i \lrcorner \eta, \nabla_{E_i} \omega),
 $$
both involve "hardcore" combinatorics. (In the latter case I need to show that $\sum\limits_{i=1}^{n}g(E_i \lrcorner \eta, \nabla_{E_i} \omega)$ is $g(d\omega,\eta)$ which feels like an index war.)

I guess the combinatorics must be somewhere and the above approach with the $\tau_{i,j}$ might just be "too simple".

Still I'd like to know if this approach can be realized somehow and if not what's the real reason for this to go wrong.

Also, please correct me if anything else is wrong.

答:
I think I solved it. First of all, the formula for $d$ with the $\tau_{0,j}$'s was of course wrong. It should be
$$
  d{\omega} = \sum\limits_{j=0}^k (-1)^j (\tau_j \circ \nabla\omega).
 $$
where $\tau_j := \tau_{0,1} \circ \dots \circ \tau_{j-1,j}$.
The other problem was the (hidden) interchange of inner products. Let me elaborate:

For $k$-forms $\omega,\eta \in \Omega^k$ we defined the inner product as
$$
  g^\Omega(\omega,\eta) = \sum\limits_{1 \le i_1 < \dots < i_k \le n} \omega_{ i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k}
 $$
if we write $\omega$ and $\eta$ locally in an orthonormal frame. On the hand, if we would interpret $\omega$ and $\eta$ as skew-symmetric tensor fields, we would have
\begin{align*}
  g^\mathcal{T}(\omega,\eta) &= \sum\limits_{1 \le i_1,\dots,i_k \le n} \omega_{i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k} \\
  &= \sum\limits_{1 \le i_1\neq\dots\neq i_k \le n} \omega_{i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k} \\
  &= k! \sum\limits_{1 \le i_1<\dots<i_k \le n} \omega_{i_1,\dots,i_k} \cdot \eta_{i_1,\dots,i_k} \\
 \end{align*}
by skew-symmetricity of $\omega$ and $\eta$. This pushes through to the averaged inner product, i.e., we have
$$
  (\omega,\eta)^\mathcal{T} = k!(\omega,\eta)^\Omega.
 $$
Then everything works out:

Let $\omega \in \Omega^k$ and $\eta \in \Omega^{k+1}$. The claim now follows from the definition of $\delta$ and
\begin{align*}
  (d{\omega},\eta)^\Omega &= \tfrac{1}{(k+1)!}\sum\limits_{j=0}^k (-1)^j (\tau_j \circ \nabla\omega, \eta)^\mathcal{T} \\
  &= \tfrac{1}{(k+1)!} \sum\limits_{j=0}^k (-1)^j (\nabla\omega,  \underbrace{\tau_{j-1,j}\circ\dots\circ\tau_{0,1} \circ \eta}_{(-1)^j\eta})^\mathcal{T} \\
  &= \tfrac{k+1}{(k+1)!} (\nabla\omega, \eta)^\mathcal{T} \\
  &= \tfrac{1}{k!} (\omega, \nabla^*\eta)^\mathcal{T} = (\omega,\nabla^*\eta)^\Omega.
 \end{align*}

Title: Explicit formula for codifferential


问:
Stopped martingale
Given a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathbb{P}-$Brownian Motion $B$ starting from 0, consider $\tau$ the first hitting time at 1.

As $\tau<\infty$ $\mathbb{P}$ a.s., the optional stopping theorem tells us that the stopped process $B^\tau$ is a martingale starting from zero and ending at 1.

However,
$1=\mathbb{E}[B_\tau]\neq \mathbb{E}[B_{\tau\wedge 0}]=0$, indicating it should not be a martingale. How can we reconcile this? Where the argument goes wrong?

答: Yes, the process $(B_{t \wedge \tau})_{t \geq 0}$ is a martingale, but you are considering the random variable $B_{\tau}$, which you cannot find among the family $B_{t \wedge \tau}$, since $\tau$ is not bounded.

Title: Stopped martingale


答: The martingale property is only concerned with $(B_{\tau \wedge t}: 0 \leq t <\infty)$. You are looking at the limiting random variable $B_{\tau}=\lim_{t \to \infty} B_{\tau \wedge t}$ and there is no reason why its expectation has to be $0$.

Title: Stopped martingale

Body:
Given a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathbb{P}-$Brownian Motion $B$ starting from 0, consider $\tau$ the first hitting time at 1.

As $\tau<\infty$ $\mathbb{P}$ a.s., the optional stopping theorem tells us that the stopped process $B^\tau$ is a martingale starting from zero and ending at 1.

However,
$1=\mathbb{E}[B_\tau]\neq \mathbb{E}[B_{\tau\wedge 0}]=0$, indicating it should not be a martingale. How can we reconcile this? Where the argument goes wrong?


问:
Show $\cosh{[0,\infty)}=[1,\infty)$ and $\sinh{\mathbb{R}}=\mathbb{R}$
Show $\cosh{[0,\infty)}=[1,\infty)$ and $\sinh{\mathbb{R}}=\mathbb{R}$.

I have shown that both functions are strictly increasing (hyperbolic cosine strictly decreading on the negatives but that should’t matter here?). Also, it is shown that the functions are continuous. From thid, how can I show that they are indeed having the domain and thus range. I sm sure that I have to use the intermediate value theorem but do not know how to formally show this.

答:
First of all, both are continuous.

$\cosh$ is strictly increasing and s.t. $\cosh(0)=1$ and $\lim_{x\to \infty }\cosh(x)=+\infty $. Therefore, Intermediate Value Theorem (IVT) allows you to conclude.

Same with $\sinh$, it's strictly increasing and s.t. $\lim_{x\to -\infty }\sinh(x)=-\infty $ and $\lim_{x\to \infty }\sinh(x)=+\infty $. So, IVT allows you to conclude.


答:
Take $y\in[0,\infty)$.

You have $\cosh(0)\leqslant y$ and, since $\lim_{x\to\infty}\cosh(x)=\infty$, there is some $x_0\in[0,\infty)$ such that $\cosh(x_0)>y$. So, by the intermediate value theorem, there is some $x\in[0,x_0]$ such that $\cosh(x_1)=y$. The case of $\sinh$ is similar.

Body:
Show $\cosh{[0,\infty)}=[1,\infty)$ and $\sinh{\mathbb{R}}=\mathbb{R}$.

I have shown that both functions are strictly increasing (hyperbolic cosine strictly decreading on the negatives but that should’t matter here?). Also, it is shown that the functions are continuous. From thid, how can I show that they are indeed having the domain and thus range. I sm sure that I have to use the intermediate value theorem but do not know how to formally show this.


问:
Function of two Brownian Motion
I am new to Brownian motion, and I have problem with verifying a Brownian motion.
Let $B(t)$ be the standard Brownian motion. Is true that $W(t)=B(5t)-B(t)$ is also a Brownian motion (not necessarily standard). Thank you for any comment/hint.

My attempt: We know $W(0)=0$ holds. I tried to check the independent increments property: for any $s<t<r$, $W(t)-W(s)=B(5t)-B(t) - (B(5s)-B(s)) = B(5t)-B(5s)- (B(t)-B(s))$, and also $W(r)-W(t)=B(5r)-B(r)-(B(5t)-B(t))$. I am not sure how to proceed to prove/disprove the independence of $W(t)-W(s)$ and $W(r)-W(t)$.

答:
What is a standard Brownian Motion? One of the properties is that Var(B(t)) = t.

This process $W(t) = B(5t)-B(t)$ does not satisfy this property, so it cannot be a standard Brownian Motion.

However, a general Brownian Motion has stationary increments, but the process $W(t)$ is not even stationary:
$$\begin{aligned}
W(t+h)-W(t) &= B(5t+5h)-B(t+h)-(B(5t)-B(t)) = \\
            &= [B(5t+5h)-B(5t)] - [B(t+h)-B(t)] .
\end{aligned} $$

For $h$ small enough, $5t > t+h$, then the random variables $B(5t+5h)-B(5t)$ and $B(t+h)-B(t)$ are independent since $B$ has independent increments. Furthermore, by definition of $B$,
$$ B(5t+5h)-B(5t) \sim \mathcal{N}(0,5h) $$
$$ B(t+h)-B(t) \sim \mathcal{N}(0,h) .$$

If we sum (resp. subtract) two independent gaussian random variables, we obtain another gaussian random variable with mean equals to the sum (resp. the difference) of the means and variance equals to the sum of the variances. So
$$ W(t+h)-W(t) = [B(5t+5h)-B(5t)] - [B(t+h)-B(t)] \sim \mathcal{N}(0,6h) ,$$
while
$$ W(h) = B(5h) - B(h) \sim \mathcal{N}(0,4h). $$

Title: Function of two Brownian Motion


问:
On the number of ways to draw kissing circles
So I was watching Numberphile with Neil Sloane of OEIS fame on the number of ways to make circles intersect. During the intro, he explicitly forbid kissing (touching, tangent) circles. This made me think about the number of ways to draw circles that must all kiss, where intersections are not allowed. Kind of the dual problem, if you will. So in a nutshell:

Circles in the affine plane can be of any finite size

Circles must kiss at least one other (no separate circles, all must be connected)

Solutions that can be transformed into each other through mirroring, rotation, stretching, moving circles without making new tangent points, are considered the same

For $n=2$ circles there are two ways: touching on the outside, and touching on the inside of the larger circle.

For $n=3$ I believe the answer is 9. Try it, should be easy.

For $n=4$ I started drawing and came up with, so far, and with help from you guys, 38, 39, 42, 43, 44, 45, 47, 44 ways. The colored dots represent the number of kissing circles at the tangent points.

Questions:

Did I miss any? I'm sure I did.

Do I have inadvertent duplicates?

Does an integer sequence exist, or even a formula mapping $n = 2, 3, 4, ...$ to $2, 9, 44, ...$?

答: Thanks to @DanUznanski here is the complete solution with 58 ways, using his sibling relation to group diagrams.


问:
Is an empty set clopen or neither?
I thought the set $\mathbb{R}$ is clopen, so its complement, $\emptyset$, is neither open nor closed. But from my research, it seems $\emptyset$ is also clopen. Can someone explain why?

And if a set has only one element, it's closed, right?

答:
It is part of the axioms of a topology that the empty set is open and that the whole space is open. Since a closed set is per definition a set whose complement is open and since the complement of the empty set and the whole space is the respective other, this immediatly yields that both the emtpy set and the whole space are clopen.

Regarding your second question, it is not necessarily true that every one element set is closed. Consider for example $\mathbb{R}$ with the topology $\mathcal{O}=\{\emptyset, \mathbb{R} \}$. Nevertheless the condition is quite important and is called $T_1$.

Title: Is an empty set clopen or neither?


答: Since a set $A$ is closed if and only if $\Bbb R\setminus A$ is open, it follows from the fact that $\Bbb R$ is clopen that $\emptyset$  is clopen too.

Title: Is an empty set clopen or neither?

Body:
I thought the set $\mathbb{R}$ is clopen, so its complement, $\emptyset$, is neither open nor closed. But from my research, it seems $\emptyset$ is also clopen. Can someone explain why?

And if a set has only one element, it's closed, right?


答:
$\emptyset$ and $\Bbb R$ are open by definition of a topology, and as $\emptyset$ is the complement of $\Bbb R$ it is also closed, and $\Bbb R$ is the complement of $\emptyset$ so it is also closed. So both sets are clopen in any topology on $\Bbb R$.

In a metric topology, like the usual topology, indeed all sets of the form $\{x\}$ are closed.

If a topology on $\Bbb R$ has the property that the only clopen (closed-and-open) subsets are $\emptyset$ and $\Bbb R$, we call this topology "connected". The usual topology on $\Bbb R$ has this property.

Title: Is an empty set clopen or neither?


问:
Show that $\det(AC) \ge 0$ for real matrices with $(A+iB)^{-1} = C+iD$.
Let $H=A+Bi$ be a complex $n \times n$ invertible matrix where $A,B$ are real matrices, with inverse $H^{-1}=C+Di$ for $C,D$ real.

Prove $\det(AC)\geq 0$.

My attempt so far, $$I=HH^{-1}=(A+Bi)(C+Di)=AC-BD+(AD+BC)i$$
So $$AC-BD+(AD+BC)i=I$$
Solving by $AC$ and taking the determinant in both sides, I end up with $$
\det(AC)=\det(I-(AD+BC)I+BD)$$
Since  $AC-BD+(AD+BC)i$ is the identity then it is a real matrix so it must be true that $(AD+BC)=O$,
So finally I get $$\det(AC)=\det(I+BD).$$

But this is where I am stuck, is $\det(I+BD)$ non-negative?

Or did I take a completely wrong approach?

答:
We can assume that $A$ is non-singular. Then
$$
AD + BC = 0 \implies D = -A^{-1}BC
$$
and therefore
$$
I = AC-BD = AC+AA^{-1}BA^{-1}BC = A(I+X^2)C
$$
with $X= A^{-1}B$. We conclude that
$$
 1 = \det(AC) \cdot \det(1+X^2) \, .
$$
The second factor cannot be zero. The second factor is also not negative, see for example

is $\det(A^2 + I)$ always non negative? or

Polynomial with no real roots implies that $\det(P(A))\ge 0$.

It follows that $\det(AC) \ge 0$.

Remark: In addition to $\det(AC) \ge 0$ we also have $\det(-BD)\ge 0$. This can be obtained from the above result in various ways:

Apply the above result to $I = (A+iB)(C+iD) = (-B+iA)(D-iC)$.

Or write the matrix identities as
$$
 AC + (-B)D= I \, , \, AD = (-B)C \, .
$$
and use a symmetry argument.

Or use that
$$
 \det(AC) \cdot \det(BD) = \det(AD) \cdot \det(BC) = -\det(AD)^2 \le 0 \, .
$$

With respect to your approach: It is correct that $\det(AC)=\det(I+BD)$, but that alone cannot give the desired conclusion. The other identity $AD+BC=0$ must be used as well.


问:
looking for help with the proof of the sum of 3 consecutive natural numbers
prove that the sum of the cubes of any 3 consecutive natural numbers is a multiple of 3.

I know that I would start this by the process of induction,

however I am not sure how to go about this. I would appreciate any help or guidance.

答:
Hint

Although induction approach plausible it is not necessary.

For any integer $x$, $x^3 \equiv x \pmod{3}.$

Further, any integer $x$ will have a residue $\pmod{3}$ that is some element in $\{0,1,2\}$.

Therefore, all that you have to do, is check each of the 3 residues for $x$ separately.

So, the direct (non-inductive) approach only involves 3 separate cases.


答:
You could do it in a simpler way by avoiding induction too.
Just take three consecutive numbers as $k, k+1, k+2$, cube each of them using the binomial theorem, and get the resultant sum. You will get a factor of 3 in the resultant summation, and you can proceed it from here.

Title: looking for help with the proof of the sum of 3 consecutive natural numbers

Body:
prove that the sum of the cubes of any 3 consecutive natural numbers is a multiple of 3.

I know that I would start this by the process of induction,

however I am not sure how to go about this. I would appreciate any help or guidance.


答:
Let $n-1, n, n+1$ be three consecutive natural numbers. Then

$$
(n-1)^3+n^3+(n+1)^3=n^3-3n^2+3n-1+n^3+n^3+3n^2+3n+1=3n^3+6n
$$

which is clearly a multiple of $3$


问:
Please if someone help me to find the winning Probability
you are playing a chess match against a skilled opponent .for a given game ,there is a 75% chance the game will end in a draw.of the game that do not end in a draw,there is a 2/3 chance you will win,and a 1/3 chance you will lose.the first player to win 2 games in a row is declared the winner of the chess match.what is the probability you will win the match?</p>

答:
Firstly let us find out your winning probability.

$\frac{1}{4}$ of games end in a draw in which you win $\frac{2}{3}$ of them hence your winning probability is $\frac{1}{4}\cdot\frac{2}{3} = \frac{1}{6}$

Similarly the opponents probability of winning is $\frac{1}{12}$

We can break down  your probability of winning overall, i.e having 2 consecutive wins before your opponent on the three states the game can be in depending on the last result.

Let $\{ D,L,W\}$ be the states that indicate respectively the last game was a draw, a loss or a win.

Firstly our transition probabilities are as follows:

$D\to D \frac{3}{4}, D\to W\frac{1}{6}, D\to L \frac{1}{12}$

$L\to D\frac{3}{4}, L\to W \frac{1}{6}$

$W \to D \frac{3}{4}, W \to L \frac{1}{12}$

Hence:

$\mathbb{P}_D[\text{You win the match}] := \mathbb{P}[\text{You win the match} | \text{Most recent game was draw}] = \frac{3}{4}\mathbb{P}_D[\text{You win the match}] +\frac{1}{6}\mathbb{P}_W[\text{You win the match}] + \frac{1}{12}\mathbb{P}_L[\text{You win the match}]$

For shorthand let the event of you winning the match be $S$.

We have:

$\mathbb{P}_D[S] = \frac{3}{4}\mathbb{P}_D[S] +\frac{1}{6}\mathbb{P}_W[S] + \frac{1}{12}\mathbb{P}_L[S]$

$\mathbb{P}_W[S] = \frac{3}{4}\mathbb{P}_D[S] +\frac{1}{6}\cdot 1 + \frac{1}{12}\mathbb{P}_L[S]$

$\mathbb{P}_L[S] = \frac{3}{4}\mathbb{P}_D[S] +\frac{1}{6}\mathbb{P}_W[S] + \frac{1}{12}\cdot 0 $

Solving these equations yields: $\mathbb{P}_D[S] = \frac{26}{33} $

Notice that the most recent game being a draw is identical to the state when the game starts so your probability of winning is $\frac{26}{33}$


问:
A property of Infinite sets
I am trying to prove the following:

If $X$ is an infinite set, then for every $f:X \rightarrow X,$ there exists a nonempty subset of $X$, which is denoted as $A$, such that $A \ne X$, and $f(A) \subset A$.

I am thinking about using the property that every infinite set is equivalent to a proper subset, but fail to make a breakthrough.

Thank you for your help!

答:
HINT(1): use the following theorem. EDIT(1): Based on the comments, it may be worth mentioning that the Axiom of Choice is not needed to prove the theorem.

Theorem. Let $X$ be a set. Fix $f:X\to X$. Fix $a\in X$. There exists a unique function $f^{(\cdot)}(a):\mathbb{N}\to X$ such that $$f^{0}(a)=a\qquad\text{and}\qquad f^{n+1}(a)=f(f^{n}(a))$$ for every $n\in\mathbb{N}$.

HINT(2): If there exists $n'\ne 0$ such that $f^{n'}(a)=a$, then $\{f^{n}(a):n\in\mathbb{N}\}$ is a finite set.

HINT(3): If for each $n'\ne 0$, $f^{n'}(a)\ne a$, then $\{f^{n}(a):n\in\mathbb{N}\}\setminus\{a\}\subsetneq X$.

Title: A property of Infinite sets


问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
To make an even more ridiculous example, consider the function sequence
$$
\phi_n(x)=\begin{cases}
0&  \text{for }~n|x|<1\\
n/2& \text{for }~1\le n|x|<2\\
0&  \text{for }~2\le n|x|.
\end{cases}
$$
This has the required properties for an "approximation of unity" (delta being the unit element for convolution) of being

non-negative,

having integral one, and

pointwise converging to zero for $x\ne 0$,

so that for any continuous $f$
$$
\int_{-\infty}^\infty f(x)\phi_n(x)dx=\int_1^2\frac{f(s/n)+f(-s/n)}2ds~~\xrightarrow{n\to\infty}~~f(0).
$$
But in the pointwise limit $\phi_n(x)\to 0$ everywhere, also at $x=0$, there is no infinite value.


问:
Show that $a^2 - 2021b^2 = 13$ has no integer solutions

Show that $a^2 - 2021b^2 = 13$ has no integer solutions.

I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$
A square is $0$ or $1 \bmod 4$, so we get $a^2\equiv 1\bmod 4$ and $b^2 \equiv 0 \bmod 4$. Thus, $2|b$ and $a^2 = 4k+1$. I do not see a contradiction yet. What else should I try?

Also tried to use the fact that squares have units digit only $0,1,4,5,6,9$, so if $a,b$ exist that satisfy the equation, we must have $a^2$ ending in $\{0,1,4,5,6,9\} \cap \{3,4,7,8,9,2\} = \{4,9\}$. Moreover, we must have $b^2$ ending in $\{0,1,4,5,6,9\} \cap \{7, 8, 1,2,3,6\} = \{1,6\}$.

If $b^2$ ends in $1$, then $a^2$ ends in $4$ - but we just saw that $a^2 = 4k+1$, so this is not a possibility.

If $b^2$ ends in $6$, then $a^2$ ends in $9$. How do I contradict this? Thank you!

答:
A slightly different alternative that does not require separate consideration of zero residues:

Because any odd number $n$ is a product of only odd primes, quadratic reciprocity implies $(p|n)=(n|p)$ for any $4m+1$ prime $p$.  Thus $(13|2021)=(2021|13)=(6|13)=-1$, and then we can't have $a^2\equiv13\bmod 2021$.


问:
Proof that if $f(x)=o(|g(x)|)$ for $x\to a$, then also $f(x)=O(|g(x)|)$ for $x\to a$.
Let $f,g$ be continuous functions on $\mathbb{R}$. Proof that if $f(x)=o(|g(x)|)$ for $x\to a$, then also $f(x)=O(|g(x)|)$ for $x\to a$.
I know that if $f(x)=o(|g(x)|)$ for $x\to a$, that
\begin{equation*}
\lim_{x\to a} \frac{f(x)}{g(x)}=0.
\end{equation*}
I get intuitively that this means that there exists a $C>0$ such that $|f(x)|\leq C|g(x)|$ (and thus $f(x)=O(|g(x)|)$). But I wouldn't know how to write a formal proof.
Any help would be great.

答:
From

$$\begin{equation*}
\lim_{x\to a} \frac{f(x)}{g(x)}=0
\end{equation*}$$

we see that $f/g$ is bounded in a neighborhood $U$ of $a$. Hence, there is $C>0$ such that

$$|\frac{f(x)}{g(x)}| \le C$$

for all $x \in U.$ This gives $f(x)=O(|g(x)|)$ for $x \to a.$


问:
What do you call a linear operator whose image is bounded?
I was just curious, is there any specific name for linear operators $T: X \rightarrow Y$ where $X$ and $Y$ are normed linear spaces, whose image is bounded? (In the sense that there exists $M > 0$ such that $||Tx||_Y \leq M$ for every $x \in X$.)

This was how I initially thought how bounded operators must be defined as, until I saw the actual definition.

答: If such an operator exists it is trivial. Indeed, suppose $v\in T(X)$ is non-zero. Then $v=Tu$ for some $u\in X$. Furthermore $\alpha v=T\alpha u$ for all $\alpha\in \mathbb R$. Thus $\alpha v\in T(X)$ for all $\alpha$, but $\|\alpha v\|\to \infty$ as $\alpha\to \infty$.


答:
Let $x \in X$ and suppose that $Tx \ne 0.$ We can assume that $||Tx||_Y=1.$  For $n \in \mathbb N$ we have

$$ n=||T(nx)||_Y \le M.$$

Since this holds for all $n$ we get a contradiction. Thus $Tx=0$. Since $x \in X$ was arbitrary, $T=0.$

Title: What do you call a linear operator whose image is bounded?

Body:
I was just curious, is there any specific name for linear operators $T: X \rightarrow Y$ where $X$ and $Y$ are normed linear spaces, whose image is bounded? (In the sense that there exists $M > 0$ such that $||Tx||_Y \leq M$ for every $x \in X$.)

This was how I initially thought how bounded operators must be defined as, until I saw the actual definition.


问:
An odd determinant and a parameter

In a matrix of size $n$, there are exactly $m$ odd elements, the rest of them being even. Find all the values of $0\le m \le n^2$ such that the determinant of the matrix is odd.

My first idea was to begin from the formula $\det A = \sum_{\sigma \in S(n)} \mathrm{sgn}(\sigma)a_{1\sigma(1)}\cdots a_{n\sigma(n)}$, but I didn't know how to grasp the solution from here. Can you help me?

答:
You have explained that what you want to find are those $m$ for which every $n\times n$ integer matrix with exactly $m$ odd entries has odd determinant.

For $n>2$ there are no such $m$.

To see this first note that $m\geqslant 2n$ is impossible: the matrix with $1$'s in the first two rows and $m-2n$ $1$'s anywhere else has determinant zero.

But $m<2n$ is also impossible: consider a matrix with $m$ $1$'s placed anywhere in the first two rows, and all other entries $0$, and note that this has determinant $0$ (since there are more than $2$ rows).

As in my comment I think "Find for each $n$ the possible values of $m$ such that there is a matrix with exactly $m$ odd entries whose determinant is odd" is a lot more interesting.

Title: An odd determinant and a parameter


问:
Roots of parabola
I have a parabola with the equation $y = x^2 + 6x + 7$ and I am trying to calculate the $x$-intercept points.

Here is my working so far...

let $y = 0$,

$x^2 + 6x + 7 = 0$

$x(x + 6) = -7$

After this I have no idea where to go next. Any help is appreciated.

答:
You should probably find the roots by using this formula. For a quadratic equation $ax^{2}+bx+c$ we have the root's as: $$x = \frac{-b \pm \sqrt{D}}{2a}$$

where $D$ is the discriminant given by $D=b^{2}-4ac$. So in your case note that $a=1$, $b=6$ and $c=7$. Hope you can find your way from here. So $D=6^{2}-4\times 7 \times 1=8$. So you have the value of the roots as: $$ x= \frac{-6 \pm \sqrt{8}}{2 \times 1} = \frac{-6 \pm 2\sqrt{2}}{2} = -3 \pm \sqrt{2}$$

For more information regarding Quadratic Equations Please see:

http://en.wikipedia.org/wiki/Quadratic_equation

Here is the Image of how your parabola will look like:

Title: Roots of parabola


问:
Inverse of Eigen value
What is the physical meaning of inverse square root of the eigen value? Is it possible to use it as stretch factor to decorrelate the data.</p>

答: Given a matrix $A$, one calculates the eigenvalues of $A^TA$; the square roots of eigenvalues are the singular values of $A$, $\sigma_1\ge \dots \ge \sigma_n$. They give the stretch factors of $A$ in various directions. The image of the unit ball under $A$ is an ellipsoid with semi-axes $\sigma_j$. Applying the reciprocal factors $\sigma_j^{-1}$, one can bring the shape back to the unit ball. From some angle this can be seen as removing correlations in the data. The relevant term is the Principal Component Analysis.</p>

Title: Inverse of Eigen value


答:
Given that: \begin{equation*}
\lim_{x\to a} \frac{f(x)}{g(x)}=0
\end{equation*}

By definition of limit we have that for $\epsilon=\frac12, \exists \delta\gt 0: 0\lt|x-a|\lt \delta\implies |\frac{f(x)}{g(x)}|\lt \frac 12 $

So $|f(x)|\lt \frac 12 |g(x)|\implies |f(x)|\le \frac 12 |g(x)|\implies f(x)=O(g(x))$

Body:
Let $f,g$ be continuous functions on $\mathbb{R}$. Proof that if $f(x)=o(|g(x)|)$ for $x\to a$, then also $f(x)=O(|g(x)|)$ for $x\to a$.
I know that if $f(x)=o(|g(x)|)$ for $x\to a$, that
\begin{equation*}
\lim_{x\to a} \frac{f(x)}{g(x)}=0.
\end{equation*}
I get intuitively that this means that there exists a $C>0$ such that $|f(x)|\leq C|g(x)|$ (and thus $f(x)=O(|g(x)|)$). But I wouldn't know how to write a formal proof.
Any help would be great.


问:
Non singular degenerate conics in the projective plane
I was trying to count degenerate conics in the projective plane $\mathbb{P}^2(\mathbb{F}_q)$ and I discovered what ooks like a "paradox" (I know it's not but I can't solve it).

Given a point (let's say (0:0:1)) in $\mathbb{P}^2(\mathbb{F}_q)$ the space of singular conics passing throught it is nothing but the spaace of solutions of $ax_0+bx_0x_1+cx_1^2$ that has cardinality $q^2+q+1$.

On the other hand I thought that degenerate conics were of the form $L_1*L_2$ (the two lines can also be the same), there are q+1 lines through any point and so there should be $(q+1)^2=q^2+2q+1$ degenerate conics.

Why are the two numbers different? Are there q degenerate conics that are not singular?

EDIT: The first formula, that is indeed correct, comes from the fact that a point is said to be "smooth", and a point s that is not smooth is singular, if the conic equation vanishes only to order 1 at s, or equivalently, if at least one of the partial derivatives
of it is nonzero at s.

Let's now consider the generic conic equation $ax_0^2+bx_1^2+cx_2^2+dx_0x_1+ex_0x_2+fx_1x_2=0$, if we evaluate it at (0:0:1) we get $c=0$ and so the equation becomes $ax_0^2+bx_1^2+dx_0x_1+ex_0x_2+fx_1x_2=0$, we can now compute the partial derivative and find they are $P_0=2ax_0+dx_1+ex_2$, $P_1=2bx_1+dx_0+gx_2$, $P_2=ex_0+fx_1$, for the point to be singular we need all of these to be 0 if valuated at (0:0:1), now $P_0(0:0:1)=e\Rightarrow e=0$,  $P_1(0:0:1)=g\Rightarrow g=0$,  $P_2(0:0:1)=0$ and so the right equation we need to look for solutions to is indeed $ax_0^2+dx_0x_1+ex_1^2$.

My question is why doesn't the second way work too

答:
To count the conics in $\mathrm{PG}(2,q)$, I will consider the set of homogeneous degree 2 polynomials
$$c_{0}X_{0}^{2} + c_{1}X_{1}^{2} + c_{2}X_{2}^{2}
+ c_{3}X_{0}X_{1} + c_{4}X_{0}X_{2} + c_{5}X_{1}X_{2}$$
there are $q^{6}-1$ such nonzero polynomials, and since we are associating a conic with the zero locus of such a polynomial, we get that there are $\frac{q^{6}-1}{q-1}$ conics in the plane (in the sense of their algebraic equation; it is possible that some of these equations define the same point set of $\mathrm{PG}(2,q)$).

If we want to count the number of degenerate conics, it is worth noticing that these come in three forms (I borrowed the example from Stanley Payne's extensive Topics In Finite Geometry, which sadly seems to no longer be available online):

a line (counted twice); e.g. $X_{0}^{2}$

a pair of distinct lines (contrary to what you state in the question body, order does not matter); e.g. $X_{0}X_{1}$

a single point (this is a possibility since, unlike $\mathbb{C}$, $\mathbb{F}_{q}$ is not algebraically closed; this case corresponds to a pair of conjugate lines in $\mathrm{PG}(2,q^{2})$); e.g. $X_{0}^{2}+aX_{0}X_{1}+bX_{1}^{2}$ with $a$ and $b$ chosen so this polynomial is irreducible over $\mathbb{F}_{q}$.

There are $q^{2}+q+1$ degenerate conics of the first form, and ${q^{2}+q+1 \choose 2}$ of the second form.

The number of conics of the third form is given by $\frac{1}{2}(q^{2}+q+1)(q^{2}-q)$; I'm being a little hand-wavy here; there are of course only $q^{2}+q+1$ point sets of $\mathrm{PG}(2,q)$ of the third type but this counts the number of these degenerate conics considered over $\mathrm{PG}(2,q^{2})$. This count comes from noticing that each of the $q^{2}+q+1$ points of $\mathrm{PG}(2,q) \subset \mathrm{PG}(2,q^{2})$ is incident with $q^{2}+1$ lines, $q+1$ of which intersect $\mathrm{PG}(2,q)$ in $q$ other points, and the remaining $q^{2}-q$ containing no other points of $\mathrm{PG}(2,q)$; we cut this number in half because we are counting the conjugate pairs of lines. Of course, if you are defining a conic as a point set of $\mathrm{PG}(2,q)$ then there are only $q^{2}+q+1$ degenerate conics of this type.


答:
In $\mathbb{R^n}$ a set is open if and only if it contains none of its boundary points.

Further, in $\mathbb{R^n}$ a set is closed if and only if it contains all of its boundary points.

Therefore, in $\mathbb{R^n}$, the only sets that can be simultaneously open and closed, are sets that have no boundary points.

The only such sets are $\mathbb{R^n}$ itself, and $\emptyset$.

Title: Is an empty set clopen or neither?


问:
Linear multistep methods have even order of convergence
A linear multistep method
$$
\sum_{j=0}^n\alpha_jx_{n-j}=h\sum_{i=0}^n\beta_if(x_{n-j},t_{n-j})
$$
is called symmetric, if
$$
\alpha_{n-j}=-\alpha_{j}, \quad \beta_{n-j}=\beta_{j}, \quad j=0,...,n
$$
Now how can one show that the order of convergence for symmetric linear multistep methods is always an even number?

答: Formulate the order condition for the symmetric test functions $x(t)=(t-nh/2)^m$ to see that the conditions for $m$ even are trivially satisfied due to symmetry.


问:
Proving a point inside a triangle is the centroid using ratios.
Let P be a point inside triangle ABC. Let the ray AP intersect with BC at A’, BP intersect with AC at B’, and CP intersect with AB at C’.

Prove that if AP/PA’=BP/PB’=CP/PC’ then P is the centroid of ABC.

答:
We use barycentric coordinates. Let $P=(x,y,z)$. Then
$$ \frac{AP}{PA'}=\frac{y+z}{x}$$
and cyclic permutations so
$$\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}.$$
Multiplying through by $xyz$ gives
$$y^2z+yz^2=xz^2+x^2z=x^2y+xy^2$$
The first equation is equivalent to
$$z(y-x)(x+y+z)=0$$
and since $P$ is inside the triangle we have $x,y,z>0$ so, necessarily, $x=y$. Similarly, we can get $y=z$ from another pair of equations, hence $x=y=z$ and $P$ is the centroid.


问:
Meaning of = in antisymmetric relation
We have a set of students.

$$
\{\text{Bob}, \text{John}, \text{Tom} \}
$$

Scores of a exam were:

$$
\begin{align}
\text{Bob} & : 5 \\
\text{John} & : 10 \\
\text{Tom} & : 10 \\
\end{align}
$$

In the real life, we can sort by scores and say students are ordered.

$$\text{Bob} \le \text{John}$$
$$\text{Bob} \le \text{Tom}$$
$$\text{John} \le \text{Tom}$$
$$\text{Tom} \le \text{John}$$

But order relations are said to be antisymmetric.

$$
xRy \land yRx \Rightarrow x = y
$$

In the example,

$$
\text{John} \le \text{Tom }\land \text{Tom} \le \text{John} \Rightarrow \text{John} = \text{Tom}
$$

So, John and Tom must be equal, if the example is an ordered set. But it is only the scores that matches. If you compare the elements, they are different and John $\ne$ Tom.

Is it allowed to compare scores and say $\text{John} = \text{Tom}$? Can't we say that the example is an ordered set?

Adding some comment to make my question clear:

As nachosemu said, I am doing

$$
x R y \Leftrightarrow score(x) \leq score(y), \,\,\, x,y\in \{\text{Bob}, \text{John}, \text{Tom} \}
$$

For this example, should I read antisymmetric relation as comparing elements:

$$
xRy \land yRx \Rightarrow x = y
$$

or as $=$ in $\lt, =, \gt$:

$$
xRy \land yRx \Rightarrow score(x) = score(y)
$$

答:
In this case,

The (partial order) relation R is defined:

$$
x R y \Leftrightarrow score(x) \leq score(y), \,\,\, x,y\in \{Bob, John, Tom\}
$$

After this, you are comparing this relation with another relation R':

$$
x R' y \Leftrightarrow x = y, \,\,\, x,y\in \{Bob, John, Tom\}
$$

which is not the same one. Here, R' is an equivalence relation.

Title: Meaning of = in antisymmetric relation


答:
Another solution using areas: The condition is equivalent to
$$\frac{AA'}{PA'}=\frac{BB'}{PB'}=\frac{CC'}{PC'}$$
And hence this gives the area ratios
$$\frac{[BCP]}{[BCA]}=\frac{[CAP]}{[CAB]}=\frac{[ABP]}{[ABC]}.$$
On the other hand, however, $[BCP]+[CAP]+[ABP]=[ABC]$ and so each of these ratios must be $1/3$. In turn, each of the ratios of lengths above equals $3$ and hence $P$ divides each of $AA', BB'$ and $CC'$ in the ratio $2:1$. Thus, $P$ is the centroid.

[To see why $P$ must now be the centroid, you can again use areas. Since $AP/A'P=2$, we have $[BAP]/[A'BP]=2$. On the other hand, we already established $[BAP]/[ABC]=1/3$ and so $[A'BP]=1/6[ABC]$. Similarly, $[CA'P]=1/6[ABC]$ and thus $[A'BP]=[CA'P]$, which means $CA'=A'B$ and $A'$ is the midpoint of $BC$. Similarly, $B'$ and $C'$ are the midpoints of the other two sides.]

Title: Proving a point inside a triangle is the centroid using ratios.

Body:
Let P be a point inside triangle ABC. Let the ray AP intersect with BC at A’, BP intersect with AC at B’, and CP intersect with AB at C’.

Prove that if AP/PA’=BP/PB’=CP/PC’ then P is the centroid of ABC.


答:
Starting with the bottom line: You can say either $\mathop{score}(\mathrm{John}) = \mathop{score}(\mathrm{Tom})$, or $\mathrm{John} \sim \mathrm{Tom}$ (not equal, cause they are different persons, but equivalent). Here $\sim$ is simply defined as meaning "having the same score".

You can also define an ordering directly on the students by $x\lesssim y\iff score(x)\le score(y)$ (this is basically what you're doing in the question). If you do this, what you get is in fact a total preorder on the set of students. A total preorder is the same as a total order, except it's not required to be anti-symmetric. But we can say $x\lesssim y\wedge y\lesssim x \iff x\sim y$. This is closely related to a total order (which is anti-symmetric): If you group the students into equivalence classes corresponding to points, you get a total order of two sets
$$
\{\mathrm{Bob}\} <  \{\mathrm{John, Tom}\}
$$
As such, total preorders are total orders of equivalence classes. See also the top picture on the linked wiki-page.

Title: Meaning of = in antisymmetric relation

Body:
We have a set of students.

$$
\{\text{Bob}, \text{John}, \text{Tom} \}
$$

Scores of a exam were:

$$
\begin{align}
\text{Bob} & : 5 \\
\text{John} & : 10 \\
\text{Tom} & : 10 \\
\end{align}
$$

In the real life, we can sort by scores and say students are ordered.

$$\text{Bob} \le \text{John}$$
$$\text{Bob} \le \text{Tom}$$
$$\text{John} \le \text{Tom}$$
$$\text{Tom} \le \text{John}$$

But order relations are said to be antisymmetric.

$$
xRy \land yRx \Rightarrow x = y
$$

In the example,

$$
\text{John} \le \text{Tom }\land \text{Tom} \le \text{John} \Rightarrow \text{John} = \text{Tom}
$$

So, John and Tom must be equal, if the example is an ordered set. But it is only the scores that matches. If you compare the elements, they are different and John $\ne$ Tom.

Is it allowed to compare scores and say $\text{John} = \text{Tom}$? Can't we say that the example is an ordered set?

Adding some comment to make my question clear:

As nachosemu said, I am doing

$$
x R y \Leftrightarrow score(x) \leq score(y), \,\,\, x,y\in \{\text{Bob}, \text{John}, \text{Tom} \}
$$

For this example, should I read antisymmetric relation as comparing elements:

$$
xRy \land yRx \Rightarrow x = y
$$

or as $=$ in $\lt, =, \gt$:

$$
xRy \land yRx \Rightarrow score(x) = score(y)
$$


问:
How does matrix multiplication work when you have two parenthesis of two calculations?
How do I calculate the following example:
$$(\lambda A-I)(\lambda A-I)$$

where A is any $n\,x\,n$ matrix of the same size as $I$? Do I multiply as regular algebra or what?

Is is:
$$(\lambda A)^2-2\lambda A + I$$

Why, why not?

Edit:
Or for the example
$$(A-2BC)(A-2BC)$$

where $A, B, C$ are three $3\,x\,3$ matrices.

答:
Matrix multiplication is distributive but since order matters you have to be careful not to commute any of the elements. Lets look at the example $(A-B)(C-D)$ since it's more general and assume all the matrices are square. We can do this two ways. First we can calculate $$(A-B)(C-D)=(A-B)C -(A-B)D=\\AC-BC-(AD-BD)=AC-BC-AD+BD$$

Or we could have distributed the other way with $$(A-B)(C-D)=A(C-D)-B(C-D)=\\AC-AD-(BC-BD)=AC-AD-BC+BD$$

And we can see by inspection that they are the same regardless of our choice in left or right distributivity.


问:
How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$
How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$

Are there $9$ additions, because of the nine numbers that are added together.
Or can you also say that there are $8$ additions, because there are only $8$ '$+$' signs.

答: Addition is a binary operator that uses two operands to execute.so if there are $n$ number of elements then number of addition will be  $$\;n-1$$.In other answers people uses "addition of $0$",because $0$ is addition identity,it will give $n$ addition but it is only in case of certain operator like $+,-$ but there are other binary operator also like $\times ,/$ with these operator we can not use $0$ there.For these operator we have multiplicative identity $1$.But regardless we can say there are minimum $$n-1$$ operations.</p>


问:
Problem with system of equations (a;b;c)
I wonder how to solve this system of equations:

\begin{cases}a = \left(b-c\right)^{2}\\b = \left(a-c\right)^{2}\\c = \left(a-b\right)^{2}\end{cases}

答: We have $$a - b = (b-c)^2 - (a-c)^2 = (b -c -a+c)(b-c+a - c) = (b-a)(b+a - 2c)$$ Assuming $a \not = b$, $$1 = -(b+a - 2c) = -b-a+2c \implies c = \frac{1+a+b}{2}$$Plugging in the last equation, $$\frac{1+a+b}{2} = (a-b)^2$$ Note that $a,b\ge 0$. Solving this quadratic equation when $a \ge -\frac{9}{16}$, $$b = \frac{4a - \sqrt{16a + 9} + 1}{4}, b = \frac{4a + \sqrt{16a + 9} + 1}{4}$$ Now use these solutions and the first equation to find $a = 0,b=1$ or $a = 1,b = 0$. If $a = b$ then $c = 0$ and $$a = b^2 \\ b = a^2 $$ which implies $a = b = 1$ or $a = b = 0$.


问:
PDF of the sum distribution of a Laplace and a Gamma distributions
Let $X,Y$ be independent randomly distributed continuous variables with the following PDF

$f_X(x) = \frac{\lambda}{2}e^{-\lambda\mid x \mid}$ for $x \in \mathbb{R}$

$f_Y(y) = \lambda^2ye^{-\lambda y}$

What's the PDF of the sum distribution $Z = X + Y$?

My working:

I've found out that $X \sim Laplace(0,\frac{1}{\lambda})$ and $Y \sim Gamma(2,\frac{1}{\lambda})$.

I'm willing to apply the following formula (which I think is the correct way to go):

$f_Z(z) = \int_\mathbb{R} f_X(x)f_Y(z-x) dx$.

So I calculate $f_X(x)f_Y(z-x)$ (using MATLAB) and I get

$f_X(x)f_Y(z-x) = -\frac{1}{2}(\lambda^3e^{-\lambda\mid x \mid}e^{\lambda(x-z)}(x - z)$.

Now, when I integrate with respect to x, I dont get any sensible results, and I assume that's because I'm not getting the integration limits correctly.

How are the appropiate integration limits found?

答:
long calculations to me (I am not familiar with long calculations) but easy exercise:

Set $Z=X+Y$ and $U=Y$. After verifying that Jacobian is 1, the joint density of $(U,Z)$ is

$$f_{UZ}(u,z)=\frac{\lambda^3}{2}ue^{-\lambda u}e^{-\lambda|z-u|}$$

Now to get $f_Z(z)$ you have to integrate in $du$. To do that observe that:

IF $Z<0$ the expression inside the module is always $<0$ thus

$$f_{UZ}(u,z)=\frac{\lambda^3}{2}e^{\lambda z}\cdot ue^{-2\lambda u}$$

and

$$f_Z(z)=\frac{\lambda^3}{2}e^{\lambda z}\int_0^{\infty} ue^{-2\lambda u}du=\frac{\lambda}{8}e^{\lambda z}$$

IF $Z>0$ you have to integrate accordingly splitting the module...

After some easy but tedious calculations the result is the following:

$$  f_Z(z) =
\begin{cases}
\frac{\lambda}{8}e^{\lambda z},  & \text{if $z<0$ } \\
\frac{\lambda^3}{4}e^{-\lambda z}\left[z^2+\frac{2\lambda z+1}{2\lambda^2}\right], & \text{if $z\geq 0$ }
\end{cases}$$

Just as an example, setting $\lambda=1$ this is the resulting density


答:
Hint for a possible approach:

Substitute  $a=(b-c)^2$ into $b=(a-c)^2$

Expand this into an quartic equation involving $b$ and $c$

That can be factorised into two quadratics involving $b$ and $c$

So you can get four possible expressions for $b$ in terms of $c$

You can then use each of these in $c=(a-b)^2$ to get corresponding values of $c$ (messy but simplifies nicely), and thus working backwards of $b$ and $a$

so you end with a set of solutions for $(a,b,c)$

Title: Problem with system of equations (a;b;c)

Body:
I wonder how to solve this system of equations:

\begin{cases}a = \left(b-c\right)^{2}\\b = \left(a-c\right)^{2}\\c = \left(a-b\right)^{2}\end{cases}


问:
Let $f, g$ be polynomials of degree n such that $\int_0^1 x^kf(x)dx =\int_0^1x^k g(x)dx$ holds for each $k = 0, 1, . . . , n.$ Show that $f = g.$

Let $f, g$ be polynomials of degree n such that $\int_0^1 x^kf(x)dx =\int_0^1x^k
g(x)dx$ holds for each $k = 0, 1, . . . , n.$ Show that $f = g.$

I considered the function $ h(x)=f(x)-g(x)$ and got $\int_0^1 x^kh(x)dx=0, k=0,1,...,n$. How do I proceed further?

答:
Denote $h(x)= \sum_{k=0}^n a_k x^k.$

By linearity of the integral, you can write that :

$$\int_{0}^1 h(x) h(x) dx = \sum_{k=0}^n a_k \underbrace{\int_0^1 h(x) x^k dx}_{=0} = 0$$

Then you have $\int_0^1 |h(x)|^2 =0$, which implies that $h(x)=0$ for all $x$ in $[0,1]$, which should be enough for you to conclude.


问:
Calculate directional derivative at $(0, 0)$
I need to calculate the partial derivative for the function $f$, that is $f(x, y) = \frac{x^2y}{x^2+y^2}$ when $(x, y) \neq (0,0)$ and $0$ when $(x, y) = (0, 0)$.

I can calculate the partial derivative for $(x, y) \neq (0, 0)$. However, how do I calculate the partial derivative at $(0, 0)$ in the direction of the vector $v = <1, 1>$?

答:
We have

$$ \frac{f(t,t)-f(0,0)}{t}=1$$

for all $t \ne 0.$ Hence

$$\frac{\partial f}{\partial v} (0,0)=1.$$


问:
Find two finite and two infinite subgroups of $GL_4(\mathbb C)$
As the question says I have to find two subgroups of $GL_4(\mathbb C)$ which are finite and two which are infinite. For each of them I have to also give some properties(abelian, normal, order). This is my first algebra course so I do have a lot of problems grasping the concepts.

I thought about:

Identity Matrix

$GL_4(\mathbb C)$ itself

this are both the trivial subgroups where the first might be finite and the second infinite.

Group of orthogonal matrices $4 \times 4$

$SL(4,\mathbb C)$ all $4 \times4$ matrices wih determinant 1.

The problem is that i think now I have 3 infinite and 1 finite.

The second problem is regarding the properties that I have to write.
Abelian: I think 1) is, 2) isn´t, I don´t know about 3) and 4).
Order and if normal or not, I am completely clueless.

Thanks for any help

答:
Finite subgroups of $GL_4(\Bbb C)$ and more generally of $GL_n(K)$ have been studied a lot, see for
example the following posts:

Finite-order elements of $\text{GL}_4(\mathbb{Q})$

Finite Subgroups of $GL(n,\mathbb{C})$

Consider for example subgroups for $n=2$ generated by
$$
\begin{pmatrix}\cos 2\pi/m &-\sin2\pi/m \\ \sin2\pi/m & \cos 2\pi/m\end{pmatrix}
$$
and embed into $GL_4(\Bbb C)$.

As for proper non-trivial infinite subgroups, we have $SL_n(\Bbb C)$, and also the infinite cyclic subgroup $C_{\infty}$, which is abelian. The group  $SL_n(\Bbb C)$ is semisimple, as an algebraic group:

The Radical of $SL(n,k)$

It is a normal subgroup of $GL_n(\Bbb C)$, because it is the kernel of the determinant homomorphism
$$
\det\colon GL_n(\Bbb C)\rightarrow \Bbb C^{\times}.
$$

The orthogonal group is not normal in $GL_n$, see here:

Is $O(n)$ normal in $GL(n)$?


答:
Remember how matrix multiplication is defined:

Given a matrices $\mathbf A$ and $\mathbf B$ of size $m\times n$ and $n \times p$, we define the product $\mathbf C = \mathbf {AB}$ as the matrix of size $m \times p$, whose entries, $c_{ij}$ are given by the dot product between the $i$th row of $\mathbf A$ and the $j$th column of $\mathbf B$.

That is

$$
c_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}.
$$

Using this definition, we can satisfy ourselves that matrix multiplication does distribute over addition. Given matrices $\mathbf{L}$, $\mathbf{M}$ and $\mathbf{N}$, of suitable sizes, the product $\mathbf{L}(\mathbf M + \mathbf{N})$ will have coefficients given by

$$ \sum_k l_{ik}(m_{kj} + n_{kj}) = \sum_k l_{ik}m_{kj} + \sum_k l_{ik}n_{kj}$$

The two terms on the right are the coefficients of $\mathbf {LM}$ and $\mathbf{LN}$ respectively, and we conclude that $\mathbf{L}(\mathbf M + \mathbf{N}) = \mathbf{L}\mathbf M + \mathbf{L}\mathbf{N}$

Title: How does matrix multiplication work when you have two parenthesis of two calculations?

Body:
How do I calculate the following example:
$$(\lambda A-I)(\lambda A-I)$$

where A is any $n\,x\,n$ matrix of the same size as $I$? Do I multiply as regular algebra or what?

Is is:
$$(\lambda A)^2-2\lambda A + I$$

Why, why not?

Edit:
Or for the example
$$(A-2BC)(A-2BC)$$

where $A, B, C$ are three $3\,x\,3$ matrices.


问:
Question on invertible matrices $(E-A)$
Let $A$ be a $n^{th}$ order square and skew-symmetric matrix, if $(E-A)$ is an invertible matrix show that
$(E+A)(E-A)^{-1}$ is an invertible matrix (where $E$ is an identity matrix)

This is a part of a question I have proved that $||(E-A)u|| \geq ||u||$ for any column vector $u \in \mathbb{R^n}$

If we can show that $(E+A)$ is invertible then the proof will be trivial. Any ideas would be appreciated.

答:
I am writing an answer to my own question please let me know if I am wrong

1.
\begin{align*}
        ||(E-A)u|| &= \sqrt{((E-A)u)^T(E-A)u}\\
        &= \sqrt{u^T (E-A)^T (E-A)u}\\
        &= \sqrt{u^T(E^T- A^T)(E-A)u}\\
        &= \sqrt{u^T (E^TE - E^TA -A^TE + A^TA)u} \qquad \qquad \text{since}\,\, A^T = -A\\
        &= \sqrt{u^T (E - EA + EA + A^TA)u}\\
        &= \sqrt{u^T (E + A^TA)u}\\
        &= \sqrt{u^TE u+ u^TA^TAu}\\
        &= \sqrt{u^Tu + (Au)^TAu}\\
        &= \sqrt{||u||^2 + ||Au||^2}\\
         ||(E-A)u|| &\geq ||u||
    \end{align*}

If $E-A$ is invertible $(E-A)^T$ is invertible as well (it is a property of inverse of a matrix)\begin{align*}
        (E-A)^T &= (E^T-A^T)\\
        &= (E-A^T) \qquad\qquad \text{since $A$ is a skew-symmetric matrix}\\
        &= (E+A)
    \end{align*}

Now we can say $(E+A)$ is an invertible matrix. Since both $(E-A)$ and $(E+A)$ are invertible

\begin{align*}
        (E+A)(E-A)^{-1}\big[(E+A)(E-A)^{-1}\big]^{-1} &= (E+A)(E-A)^{-1} (E-A)(E+A)^{-1} \\
        &= (E+A)E(E+A)^{-1}\\
        &= E
    \end{align*}

Title: Question on invertible matrices $(E-A)$


问:
Prove $1+2^{1}+2^{2}+\dots+2^{\left(d-1\right)}=\frac{1\left(2^{d}-1\right)}{2-1}=2^{d}-1\ \lt \ n$

$1+2^{1}+2^{2}+\dots+2^{\left(d-1\right)} < n$ implies
$d\le\log_{2}\left(n\right).$

$1+2^{1}+2^{2}+\dots+2^{\left(d-1\right)} ≥ n$ implies
$d>\log_{2}\left(n\right)-1.$

Show that$\quad d = ⌊\log_{2}\left(n\right)⌋.$

Hi, I'm not sure about my solution, specially the part where I substitute "$\ge$" with "$\gt$" and the last proof with the floor.. I feel like I did the wrong direction there and was wondering if it's valid, and if not, how should I approach it?

A.

$1+2^{1}+2^{2}+\dots+2^{\left(d-1\right)}=\frac{1\left(2^{d}-1\right)}{2-1}=2^{d}-1\ \lt \ n$

As $(2^{d}-1),n\in N$, we have

$2^{d}-1\lt n\ \ \ \ \implies\ \ \ \ 2^{d}\le n$

$\log_{2}\left(2^{d}\right)\ \le\log_{2}\left(n\right)$

$d\le\log_{2}\left(n\right)$

B.

$1+2^{1}+2^{2}+\dots+2^{\left(d-1\right)}=\frac{1\left(2^{d+1}\right)}{2-1}=2^{d+1}-1\ \ge\ n$

As $(2^{d+1}-1),n\in N$, we have

$2^{d+1}-1\ge n\ \ \ \ \implies \ \ \ \ 2^{d+1}\gt n$

$\log_{2}\left(2^{d+1}\right)\ \gt \log_{2}\left(n\right)$

$d+1\gt \log_{2}\left(n\right)$

$d\gt \log_{2}\left(n\right)-1$

C.

Let $x\in R$.

$⌊x⌋ =$ max$\{n\in Z | n\le x\}$

Therefore $⌊x⌋\le x$.

Assume toward contradiction $x-1\ge ⌊x⌋$, then $x\ge ⌊x⌋+1\ge ⌊x⌋$

But by the closure of $Z$, $⌊x⌋+1\in Z$ in contradiction to the maximality of $⌊x⌋$.

Therefore $x-1\lt ⌊x⌋\lt x$. Substituting $x = d = \log_{2}\left(n\right)$ we get

$\log_{2}\left(n\right)-1\lt ⌊\log_{2}\left(n\right)⌋\lt \log_{2}\left(n\right)$, as required.

答:
Since the sum is a geometric series we have that $$1+2+2^2+\ldots+2^{d-1}=2^d-1.$$
If we assume that $$2^d-1<n \Longrightarrow 2^d<n+1 \Longleftrightarrow 2^d\leq n \quad \text{(just add $1$ in both sides)}.$$
This last equivalence is a consequence of $n$ belonging in $\mathbb{N}$ and not in $\mathbb{R}-\mathbb{N}$. Finally, taking by applying logarithms... $$d<\log_2(n+1)\quad \text{and} \quad d\leq \log_2n.$$
These are two different ways of saying the same thing. The same can be applied to the '$\geq$' part of the statement, and by adding both you can write $d$ as $$d=\lfloor \log_2n \rfloor$$ in a very compact way.


问:
Proof that polynomials are dense in $H^p$ spaces
In some papers and books I read they used that the polynomials are dense in $H^p$ for $1\le p<\infty$, but I could not find a proof for this stament. I tried to proof it myself but failed. My idea was to use the theorem of Stone Weierstrass, but I did not come any futher.
So my question is, how can I proof it?

Edit:
For $1\le p<\infty$ and $H(\mathbb{D})$ the set of all holomorphic functions on $\mathbb{D}=\{z\in \mathbb{C}: \lvert z\rvert<1\}$ we define
$$
 H^p=H^p(\mathbb{D})=\{f\in H(\mathbb{D}):\lVert f\rVert_{H^p}<\infty\}
 $$
with
$$
 \lVert f\rVert_{H^p}=\sup \limits_{0\le r<1}\left(\frac{1}{2\pi} \int \limits_0^{2\pi}\lvert f(re^{i\theta})\rvert^pd\theta\right)^{\frac{1}p}.
 $$

答:
Let $f_r(z):=f(rz)$. Recall that, for $f\in H^p$ with $p\ge 1$, there exists the radial limit $f^*\in L^p(\partial D)$ and $||f^*||_{L^p}=||f||_{H^p}$. It is easy to see that $\lim_{r\to 1}||f_r^*- f^*||=\lim_{r\to 1}||f(re^{i\vartheta})-f^*(e^{i\vartheta})||=0$, by definition of radial limit. Thus $\lim_{r\to 1} ||f_r-f||_{H^p}=\lim_{r\to 1} ||f_r^*-f^*||_{L^p}=0$.

Now let $0<r_n<1$ be an increasing sequence such that $r_n\to 1$. By elementary results, there exists a polynomial $p_n:\sup_{D}|f_{r_n}-p_n|<\frac1{2^n}$. Since $||f_{r_n}-p_n||_{H^p}\le \sup_D|f_{r_n}-p_n|<\frac{1}{2^n}$. Since by the previous result we have $||f-f_{r_n}||_{H^p}\to 0$, the triangular inequality implies
$$\lim_{n\to \infty}||f-p_n||_{H_p}\le\lim_{n\to \infty} ||f-f_{r_n}||_{H^p}+||f_{r_n}-p_n||=0$$


问:
Fixpoint iteration of $ x + \ln(x) = 0 $
I want to solve the equation  $ x + \ln(x) = 0 $ which has a root $ x^* \approx 0.5$.

I am given three iterations:

$$ \quad  x_{n+1} = - \ln(x_n)\\  \quad x_{n+1} = e^{-x_n} \\ x_{n+1} =\frac{x_n + e^{-x_n}}{2}
 $$

What method do I use to find equation 3?

答:
I think it is a typo and should be $$x_{n+1} = \frac{x_n +e^{-x_n}}2$$

In this case the third iteration is a smoothed version of the second one.

We remember $1/2$ of the last estimate and add $1/2$ of a new estimate.


问:
Continuity of $f(x)=[x]+ \sqrt{x-[x]}$
Consider the function $f(x)=[x]+ \sqrt{x-[x]}, \, x\in \Bbb R$ ; where "$[ \space ]$" denotes the greatest integer function. It is obvious that if $b$ is an integer, then </p>

<p>$$\lim_{x\to b-} f(x)=\lim_{x\to b-}[x] + \sqrt { \lim_{x\to b-}x - \lim_{x\to b-}[x] } =b-1+\sqrt {b-(b-1)}=b-1+1=b$$ </p>

<p>$$\lim_{x\to b+} f(x)=\lim_{x\to b+}[x] + \sqrt { \lim_{x\to b+}x - \lim_{x\to b+}[x] } =b+\sqrt {b-b}=b,$$ and </p>

<p>$$f(b)=[b]+ \sqrt{b-[b]}=b+\sqrt {b-b}=b,$$ hence $f(x)$ is continuous for all integer values of $x$.  Is $f(x)$ continuous for all non-integer values of $x$ also, i.e. is it true that $f(x)$ is continuous for all $ x \in \Bbb R$ ? </p>

答: Since continuity is a local property, for any $x_0$ not an integer, $f(x)$ is continuous at $x_0$ if and only if $g(x)=[x_0]+\sqrt{x-[x_0]}$ is continuous at $x_0$.  This is because $g(x)=f(x)$ in a neighborhood of $x_0$.  Since $g(x)$ is continuous at $x_0$, so is $f(x).$</p>

Title: Continuity of $f(x)=[x]+ \sqrt{x-[x]}$


问:
A special type of linear regression
Usually linear regression involves two variables $(x,y)$, i.e. an independent variable $x$ and a dependent variable $y$, and they are related by the following expression
\begin{equation}
y = a_0 + a_1 x,
\end{equation}
where $a_0$ and $a_1$ are parameters that define the linear model. In linear regression we have one equation of this form for each couple of observed variables $(x_i,y_i)$, thus we have a linear system and its solution gives us $a_0$ and $a_1$.

Let's consider that we have two set of independent-dependent variables, namely $(x,y)$ and $(w,z)$. The first two variables $(x,y)$ are related by the previous equation, while the second two variables $(w,z)$ are related by the following
\begin{equation}
z = b_0 + b_1 w,
\end{equation}
where $b_0$ and $b_1$ are parameters that define the linear relation between $z$ and $w$. Also in this case a set of observation $(w_j,z_j)$ leads to a linear system and its solution gives us $b_0$ and $b_1$.

In general, if $a_0$, $a_1$, $b_0$ and $b_1$ are independent, then we can solve the two linear systems separately. But now, let's suppose that $a_0$ and $b_0$ are independent, while $a_1=b_1$. In this case, the two linear systems should be solved simultaneously.

I've solved this problem just definying one linear system of equation involving both the two sets of equations, but I would like to know if this problem has a specific name and how to correctly approach it. In particular, I want to know how to assessing the fit quality (for example, with an equivalent of the $R^2$).

Thank you!

答:
If your model is $$y=a_0+c_1x,\\z=b_0+c_1w$$ you can minimize

$$\sum(a_0+c_1x-y)^2+\sum(b_0+c_1w-z)^2,$$

giving the equations

$$\sum a_0+c_1x-y=0,\\\sum b_0+c_1w-z=0,\\\sum x(a_0+c_1x-y)+\sum w(b_0+c_1w-z)=0.$$

Now solve this $3\times3$ system for $a_0,b_0,c_1$.

The fit quality is still given by the ratio of the explained variance over the total variance.

Title: A special type of linear regression


问:
If Banach space is countable union of closed sets, Can I conclude space should be one of the closed set?
Let $X$ be Banach space. $F_n$ be closed subsets of X. If $X = \cup F_n$, Can I conclude $X = F_{n_0}$ for some $n_0$? If right, suggest a hint.

答:
Proof in case the $F_n$ are closed linear subspaces.  Assume $\bigcup_{n=1}^\infty F_n = X$.  Since $X$ is a complete metric space, by the Baire category theorem $X$ is not the union of a countably many closed sets with empty interior.  Therefore, there is $n_0$ so that $F_{n_0}$ has nonempty interior.

There is $x_0 \in F_{n_0}$ and $r_0 > 0$ so that $F_{n_0} \supseteq B_{r_0}(x_0) := \{x : \|x-x_0\| < r_0\}$.  Since $F_{n_0}$ is closed under substraction, $F_{n_0} \supseteq B_{r_0}(x_0) - x_0 = B_{r_0}(0)$.  Then, since $F_{n_0}$ is closed under scalar multiples, $F_{n_0} \supseteq X$.


问:
Ordninary Differential equation ( global solution )
I want to check if the equation: $$x''(t)=-3x'(t) + (w(t))^{2}x(t)$$$$x(0)=3\ \&\ x'(0)=0$$ has a global solution knowing that w(t) is periodic and differentiable.

I started First by writing it into two first order differential equations as follows $$x'(t) = y(t)$$ $$y'(t)= (w(t))^{2}x(t) -3y(t)$$$$x(0)=3\ \&\ y(0)=0$$

Now i was trying to calculate a first integral by trying to find a solution for $$\frac{dy}{dx}=\frac{w^{2}x-3y}{y}$$ So that I can either bound the first integral or show that the solution cannot be contained in any compact set as t tends to infinity. However I cannot find the first integral as I couldn't solve the differential equation above. Is my approach correct but hard to solve and is there another way to find a global solution ?

答:
Hint

Set $y=\dot x$. Then $$\begin{pmatrix}\dot x\\ \dot y\end{pmatrix}=\begin{pmatrix}y\\ -3y+\omega (t)^2x\end{pmatrix}=\begin{pmatrix}0&1\\\omega (t)^2&-3\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}.$$

Using Picard-Lindelof Theorem allows you to conclude.


问:
Solve linear system of ODEs using Laplace transform
I need to solve the following initial value problem via Laplace transform
\begin{align*}
\dot{\mathbf{x}} = \begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix} \mathbf{x} + \begin{pmatrix} \sin t \\ \tan t \end{pmatrix} , \: \mathbf{x}(0) = \begin{pmatrix} -1 \\ 0 \end{pmatrix}
\end{align*}
but I don't know how to do that since there's no Laplace transform for the tangent function.</p>

答:
There is Laplace transform for tangent function, but it's complicated and using digamma function.
Laplace transform</p>


问:
Intersection of cyclotomic fields over an arbitary number field
Let $K$ be an arbitrary number field and $m,n$ are two co-prime integers. Suppose we know that, $[K(\zeta_m):K]=\phi(m), [K(\zeta_n):K]=\phi(n).$ Then is it true that $[K(\zeta_{mn}):K]=\phi(mn)?.$ I can see a positive answer under certain conditions, like, if the discriminant to $K$ is co-prime to $mn,$ or if $K$ contains no proper abelian extension of $\mathbb{Q}.$ But I am not able to see if it is a general phenomenon.

答:
At least the following phenomenon may ruin your day.

It may well happen that $K$ contains a non-trivial subfield of $\Bbb{Q}(\zeta_{mn})$ but does not contain non-trivial subfields of either $\Bbb{Q}(\zeta_m)$ or $\Bbb{Q}(\zeta_n)$. For example, if $m=3$, $n=5$. Then $K=\Bbb{Q}(\sqrt{-15})$ is a subfield of $\Bbb{Q}(\zeta_{15})$ because $\sqrt5\in\Bbb{Q}(\zeta_5)$ and $\sqrt{-3}\in\Bbb{Q}(\zeta_3)$. Here
$$
[K(\zeta_3):K]=2=\phi(3),\quad [K(\zeta_5):K]=4=\phi(5),
$$
but
$$[K(\zeta_{15}):K]=4=\phi(15)/2$$
as $K(\zeta_{15})=K(\sqrt{-3},\zeta_5)$ and both $K(\sqrt{-3})/K$ as well as
$K(\zeta_5)/K(\sqrt{-3})$ are quadratic.

Your discriminant condition would, of course, catch this example.


问:
Definite integration includes Exponential and Logarithmic functions
Does anyone have an idea of having solution for the following integration?

$$I=\int_0^\infty x^a e^{-b x} \log(1+cx)dx$$

where $a$ is a positive integer and $b,c>0$.

I found similar integral in eq. (2.6.23.4) in book " Integrals and Series: elementary functions. Vol.1", which is only valid for non-integer $a$.

Further, if I give same integral for known $a$ values such as $a=1,2,3,..$, then MATHEMATICA gives answers one by one. But it did not give for general $a$.

答:
Start with $x=\frac t c$ and $k=\frac b c$ to make
$$I=\int_0^\infty x^a e^{-b x} \log(1+cx)\,dx=\frac 1{c^{a+1}}\int_0^\infty t^a\, e^{-k t} \,\log(1+t)\,dt$$ Even specifying that $a$ is integer and $k>0$, you will recieve for
$$J=\int_0^\infty t^a\, e^{-k t} \,\log(1+t)\,dt$$
$$ k \left(-k^2\right)^a\, J=k (-k)^a \Gamma (a) \, _2F_2(1,1;2,1-a;k)-$$ $$\pi  k^a \csc (\pi  a) \Gamma (a+1,-k)+$$ $$a
   \Gamma (a) \left(\pi  k^a \csc (\pi  a)-(-k)^a \log (k)+(-k)^a \psi
  (a+1)\right)$$ which present problems of indeterminate forms when you try $a=1,2,\cdots$.

I must confess that I prefer the individual cases which show some interesting patterns.

$$J_a=\frac 1 {k^{a+1}} \Big[P_{a-1}+Q_a\,e^k\, \text{Ei}(-k) \Big]$$ where $P_n$ and $Q_n$ are simple polynomials of degree $n$ in $k$.


问:
Expand $(\frac{1}{4} + \frac{3}{4})^n$
Construct a flow diagram to output the terms of the binomial expansion of $(\frac{1}{4} + \frac{3}{4})^n$, where n is positive integer. Test your procedure using a calculator and the value n = 3.

I don't want to do a flow chart but I do want to do the calculation.

This problem has now been solved but I leave the question here, with the edited answer.

My reasoning is: $(\frac{1}{4} + \frac{3}{4})^3 = [\frac{1}{4}(1 + 3)]^3$

I have expanded to get

$(\frac{1}{4})^3(1 + 3)^3 = \frac{1}{4}(1 + 3.3 + \frac{(3)(2)(3)^2}{2!} + \frac{(3)(2)(1)(3)^3}{3!})$

The answers are:

$0.015625, 0.140625, 0.421875, 0.421875$

答:
This is actually a comment, but I post it here because it's too long to a comment space.

Your error is not following the instructions given. The task says: 'Test your procedure using a calculator.'

Once again: 'Test your procedure....', not 'replace it'.

If you did it as described, you would most probably have noticed discrepancy between results and found which one was wrong. You did only half of the task and as a result you have no material for comparison, which prevents you from spotting your error.

Title: Expand $(\frac{1}{4} + \frac{3}{4})^n$


问:
Riemann Zeta Function at $z=1+i \cdot t$
First, we define the Riemann Zeta functon as a infinite sum:

$$\zeta(z)=\sum_{n=1}^{\infty}\frac{1}{n^z}, \ Re(z)>1$$

This is an analytic function in this domain. We can not extend this function with this summation because we have

$$\left|\frac{1}{n^z}\right|\leq \left|\frac{1}{n^{Re(z)}}\right|$$

and then by comparision the entire sum converges when $Re(z)>1$.

We also know that we can extend $\zeta$ via holomorphic continuation to a meromorphic function with a simple pole in $z=1$.

My question is, what about the values of the extended $\zeta$ function in $1+it$, where $t\in\mathbb{R}$? I mean, if you take the limit of the summation by the right way, you have this

$$\lim_{z\rightarrow (1+it)^+}\sum_{n=1}^{\infty}\frac{1}{n^z}$$

and this converges to $\zeta(1+it)$ because is an analytic continuiation, hence continuous, right? But the sum goes to infinity, or am I wrong?

Thank you very much!

答:
Intuitive reasoning: Given a complex number $z=1+\varepsilon+it$ for $0<\varepsilon$ and $t\neq0$, the partial sums
$$
\sum_{n=1}^k\frac1{n^z}
$$
form a spiral-like pattern, spiralling inwards to $\zeta(z)$. There is, a priori, no reason why the limit of the partial sums should go to infinity as $\varepsilon\to0$. And in fact, it turns out that doesn't happen. The spiral just spirals in slower and slower, and the center it spirals towards doesn't fly off to infinity. At $1+it$ the partial sums all lie along a circle (this circle is larger the smaller $|t|$ is)


答:
If $x \notin \mathbb{Z}$, then for some $\epsilon >0$, $B(x,\epsilon) \cap \mathbb{Z} = \emptyset$. Then $\lfloor y \rfloor $ is constant for $y \in B(x,\epsilon)$. Let $k = \lfloor y \rfloor $. Note that if $y \in B(x,\epsilon)$, we have $y > k$.

For $y \in B(x,\epsilon)$, we have $f(y) = k + \sqrt{y-k}$, which is continuous (in fact smooth, since $y > k$).

$f$ is not differentiable at the integers.

Title: Continuity of $f(x)=[x]+ \sqrt{x-[x]}$

Body:
Consider the function $f(x)=[x]+ \sqrt{x-[x]}, \, x\in \Bbb R$ ; where "$[ \space ]$" denotes the greatest integer function. It is obvious that if $b$ is an integer, then </p>

<p>$$\lim_{x\to b-} f(x)=\lim_{x\to b-}[x] + \sqrt { \lim_{x\to b-}x - \lim_{x\to b-}[x] } =b-1+\sqrt {b-(b-1)}=b-1+1=b$$ </p>

<p>$$\lim_{x\to b+} f(x)=\lim_{x\to b+}[x] + \sqrt { \lim_{x\to b+}x - \lim_{x\to b+}[x] } =b+\sqrt {b-b}=b,$$ and </p>

<p>$$f(b)=[b]+ \sqrt{b-[b]}=b+\sqrt {b-b}=b,$$ hence $f(x)$ is continuous for all integer values of $x$.  Is $f(x)$ continuous for all non-integer values of $x$ also, i.e. is it true that $f(x)$ is continuous for all $ x \in \Bbb R$ ? </p>


问:
How to show $\sum\limits_{j=1}^\infty(\sum\limits_{i=1}^\infty v_i a_{ij})^2$ is convergent?
Assuming
\begin{align}
&\sum\limits_{i=1}^\infty v_i^2 <\infty  \\
&\sum\limits_{j=1}^\infty  a_{ij}^2 <\infty    ~~~~~~\forall i\ge 1  \\
&\sum\limits_{i=1}^\infty  a_{ij}^2 <\infty    ~~~~~~\forall j\ge 1
\end{align}
then how to show
$$
\sum\limits_{j=1}^\infty(\sum\limits_{i=1}^\infty v_i a_{ij})^2
$$
is convergent?    I can get the convergence of
$$
\sum\limits_{i=1}^\infty v_i a_{ij}~~~~~~\forall j\ge 1
$$

But, I still  suspect it is not right, although I   fail to find counter-example after spending an afternoon.  Therefore, I asked here.  Thanks for any help.

答: This is not true. Let $a_{ij}=j$ if $i =j$ and $0$ is $i \neq j$. Let $v_j=\frac 1  j$ . Then $\sum_i v_ia_{ij}=1$ and $\sum_j (\sum_i v_ia_{ij})^{2}=\infty$.


问:
Height of a tetrahedron
How do I calculate the height of a regular tetrahedron having side length $1$ ?

Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table?

答:
The centroid, a vertex, and the midpoint of the base form a right triangle on the base of the tetrahedron. Let the distance from a vertex to the centroid be $d$, and the side length be $2$. Observing that the side lengths in this $30º-60º-90º$ triangle are $1 - \sqrt{3} - 2$, $d^2 = 1^2 + \left(\frac{\sqrt3}{3}\right)^2 = \frac{4}{3}$ by Pythagoras: the centroid is one-third the distance from the base to the vertex.

Then using the other right-angled triangle, the height $h^2 = 2^2 - d^2$, where the edges are all $2$, so $h = \sqrt{4 - 4/3} = \sqrt{8/3}$. Now the height of the original tetrahedron with side length $1$ is half of $\sqrt{8/3}$ by similarity, which is $\frac{2\sqrt{2/3}}{2} = \sqrt{2/3}$.

Title: Height of a tetrahedron


问:
Can I write the integral of a function in terms of its level sets?
I have a function like this $f:\mathbb{R}^2\to\mathbb{R}_+$

$$f(x, y) = e^{-\frac{x^2+y^2}{2}}$$

Its level sets $f(x, y) = c$ are simply circles centered at the origin
$$
x^2 + y^2 = \log\left(\frac{1}{c^2}\right) \qquad \forall\, c\in\mathbb{R}_+
$$

Here's my question:

Usually we would take the integral of this function as a double integral where one dimension of integration is $x$ and one is $y$
$$
\int_{\mathbb{R}^2} f(x, y) dxdy = \int_{\mathbb{R}} \int_{\mathbb{R}} f(x, y) dx dy
$$
Can we rewrite this integral as an integral over its level sets? For instance something like this
$$
\int_{\mathbb{R}^2} f(x, y) dxdy = \int_{\mathbb{R}_+} \int_{x^2 + y^2 = \log\left(\frac{1}{c^2}\right)} f(x, y) dt dc
$$

I am not sure how to write it and I am pretty sure the "inner" integral should be modified so that we are integrating in a "curvilinear" way along the circle.

答:
In a way, it's something called a Lebesgue integral. Assuming that $f(x,y) > 0$ we have

$$ \int_{\mathbb{R}^2} f(x,y) dx dy = \int_0^\infty \mu\big(\{(x,y): f(x,y) > t\}\big) dt $$

where $\mu(A)$ is generally called the Lebesgue measure of a set, and in the case of 'nice' sets it's just the area; in this case, the area of the region where $f(x,y) > t$.

With the function $f(x,y) = e^{-\frac{x^2+y^2}{2}}$ we have:
\begin{align}\{(x,y): f(x,y) > t\} &= \{(x,y): e^{-\frac{x^2+y^2}{2}} > t\} = \{(x,y): x^2+y^2 <  - 2 \ln t\} =\\
&= \left\{\begin{array}{ll} \varnothing & \text{for }t\ge 1 \\ \{(x,y): x^2+y^2 <  2 |\ln t|\} & \text{for }0<t< 1\end{array} \right. \end{align}
so
$$ \mu\big(\{(x,y): f(x,y) > t\}\big) = \left\{\begin{array}{ll} 0 & \text{for }t\ge 1 \\ 2\pi| \ln t| & \text{for }0<t< 1\end{array} \right.$$
and you get

$$ \int_{\mathbb{R}^2} f(x,y) dx dy = \int_0^1 2\pi| \ln t| dt = 2\pi(t-t\ln t)\big|_0^1 =2\pi$$

The Lebesgue integral can be defined on other sets, not only $\mathbb R^n$, as long as you can measure the size of sets somehow.

ANOTHER APPROACH:

Another way you can try to calculate this integral is to parametrize $\mathbb R^2$ with new coordinates $x= x(c,t)$, $y=y(c,t)$ such that
$$ f(x(c,t),y(c,t)) = c$$
Then we have
$$ \int_{\mathbb{R}^2} f(x,y) dx dy = \iint f(x(c,t),y(c,t)) |J(c,t)| dc dt = \iint c |J(c,t)| dc dt$$
where $J(c,t)$ is the jacobian:
$$ J(c,t) = \det\left[\begin{array}{cc} \frac{\partial x}{\partial c}& \frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial c}& \frac{\partial y}{\partial t}\end{array}\right]$$

In this case, we could use coordinates $(c,t)$ such that $c\in(0,1]$, $t\in[0,2\pi]$ and
$$ x= \sqrt{-2\ln c} \cos t$$
$$ y= \sqrt{-2\ln c} \sin t$$
Then
$$ |J(c,t)| = \frac{1}{c} $$
so
$$ \int_{\mathbb{R}^2} e^{-\frac{x^2+y^2}{2}} dx dy = \iint_{(c,t)\in(0,1]\times[0,2\pi]} c \frac{1}{c} dc dt = \int_0^{2\pi} \Big(\int_0^1 dc\Big) dt = 2\pi $$
In this case, the integral turns out to be very simple, but in general it's difficult to find parametrization $x= x(c,t)$, $y=y(c,t)$. This case was special because the level sets were concentric circles, so just a slight modification of well-known radial coordinates did the job.


问:
Find all cosets of $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ in $\operatorname{GL}_2(\mathbb{R})$
I am trying to tackle this question:

Find all left and right cosets of
$$ H = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} $$
in $G = \operatorname{GL}_2(\mathbb{R})$, where $a, c \in \mathbb{R}^*$ and $b \in \mathbb{R}$.

Supposedly, the solution is to take
$$ g_1 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G $$
and
$$ h_1 = \begin{pmatrix} 1 & -\frac{b}{a} \\ 0 & 1 \end{pmatrix} \in H $$
to get
$$ g_2 = g_1 h_1 = \begin{pmatrix} a & 0 \\ c & \frac{ad-bc}{a} \end{pmatrix}, $$
and then take
$$ h_2 = \begin{pmatrix} \frac{1}{a} & 0 \\ 0 & \frac{a}{ad-bc} \end{pmatrix} \in H $$
to yield
$$ g_3 = g_2 h_2 = \begin{pmatrix} 1 & 0 \\ \frac{c}{a} & 1 \end{pmatrix} . $$
Apparently $g_3$ for all $c \in \mathbb{R}$ is a coset. It's also claimed
$$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$
is the only other coset.

Two questions:

How are you supposed to go about choosing matrices like $g_1$, $h_1$, and $h_2$? It almost seems like they were pulled out of thin air. How would you approach a question like this (still in $\operatorname{GL}_2(\mathbb{R})$) that has a lot of cosets, e.g. 4 or more, without being bogged down by the sheer number of possibilities to compute?

How do we know $g_3$ is a complete coset and can't be "broken down" further?

答:
I would approach this question using group actions. The group $GL_2(\Bbb{R})$ acts on the set $X=\Bbb{P}^1(\Bbb{R})$ of 1-dimensional subspaces of the plane $\Bbb{R}^2$. That is, lines through the origin. Basic linear algebra says that this action is transitive. In other words, if $L_1$ and $L_2$ are two such lines, there exists a matrix $g\in GL_2(\Bbb{R})$ such that $g(L_1)=L_2$.

What's more, the subgroup is the stabilizer of the special line
$$
L_0=\left\{\binom x0\mid x\in\Bbb{R}\right\}.
$$
That is, $g(L_0)=L_0$ if and only if $g\in H$.

Basic properties of group actions then imply that the cosets of $H$ are in a bijective correspondence with the lines from $\Bbb{P}^1(\Bbb{R})$. If $g(L_0)=L'$ then
$$
gH=\left\{g'\in GL_2(\Bbb{R})\mid g'(L_0)=L'\right\}.
$$
So it suffices to enumerate the lines in $\Bbb{P}^1(\Bbb{R})$ and find matching coset representatives.

Well, from school years we know that a line through the origin is uniquely determined by its slope, and then we have the vertical line $L_{\infty}=\{\binom 0y\mid y\in\Bbb{R}\}$
of infinite slope. The line with slope $k$ is
$L_k=\{\binom x{kx}\mid x\in\Bbb{R}\}.$ To get all the cosets of $H$ you need to find a matrix $g_k$ such that $g_k(L_0)=L_k$, one for each $k$. And then one matrix $g_\infty$
such that $g(L_0)=L_\infty$.