问:
Hermiticity/self-adjointness of the Laplacian operator
I'm trying to understand the concept of hermicity/self-adjointness but even after reading through some threads here, I just don't get it (first problem being that I don't understand the difference between the two, if there is any). But anyway, let's look e. g. at the one-dimensional eigenvalue problem of the Laplace operator $$\Delta f(x)=\frac{d^2}{dx^2} f(x)=cf(x)$$ where $c$ is a constant and the domain $\Omega$ for $x$ is e. g. $[x_0;x_1]$. From what I've read, the Laplace operator is self-adjoint/hermitian for boundary constraints $f(x_0)=f(x_1)=0$ (though I don't know why that is). Could someone expain why? (And maybe also showcase a set of BCs for which the Laplacian operator is not self-adjoint/hermitian?)

答:
If $f,g$ are twice absolutely continuous on $[x_0,x_1]$ with $f,g,f'',g''\in L^2[x_0,x_1]$, then
$$
       \langle f'',g\rangle-\langle f,g''\rangle=\int_{x_0}^{x_1}f''\overline{g}-f\overline{g''}dx \\
   = \int_{x_0}^{x_1}\frac{d}{dx}(f'\overline{g}-f\overline{g'})dx \\
   = \left.\det\left(\begin{array}{cc}\overline{g(x)} & f(x) \\ \overline{g'(x)} & f'(x)\end{array}\right)\right|_{x_0}^{x_1}.
$$
$\Delta$ is symmetric if the determinants given above at $x_1,x_2$ are $0$ for all $f,g$ in the domain. This typically involves imposing endpoint conditions on $f,g$ such as
$$
                    \cos\alpha f(x_0)+\sin\alpha f'(x_0) = 0, \\
                    \cos\beta f(x_1)+\sin\beta f'(x_1) = 0.
$$
For $\alpha = 0 = \beta$, this amounts to imposing $f(x_0)=f(x_1)=0$. If $f,g$ satisfy these conditions, then $\langle f'',g\rangle = \langle f,g''\rangle$ for all $f,g$ satisfying $0$ endpoint conditions at $x=x_0,x_1$. But, as you can see, many other conditions give rise to a symmetric operator $\frac{d^2}{dx^2}$.

Symmetry does not imply self-adjointness of $Lf=f''$. However, symmetry is necessary. It turns out that $L_{\alpha,\beta}$ given by $Lf=f''$ with $f$ being twice absolutely continuous with $f''\in L^2$ is self-adjoint. There are periodic conditions that can imposed that will also result in self-adjoint operators. Examples of non-selfadjoint versions of $L$ include conditions
$$
               f(x_0)=f'(x_0)=f(x_1)=f'(x_1)=0.
$$
Then $L^*$ is the unrestricted operator $L$ (i.e., one with no endpoint conditions.) In particular $\langle Lf,g\rangle = \langle f,L^*g\rangle$ for all $f$ satisfying $f(x_0)=f'(x_0)=f(x_1)=f'(x_1)=0$ and $g$ unrestricted.

If $A : \mathcal{D}(A)\rightarrow\mathcal{H}$ is a linear operator on a dense subspace of a Hilbert space $\mathcal{H}$, then one way to define the adjoint $A^*$ is that the domain $\mathcal{D}(A^*)$ consists of all $y\in\mathcal{H}$ such that
$$
      |\langle Ax,y\rangle | \le M \|x\|,\;\;\; x\in\mathcal{H}.
$$
Then, $x\mapsto \langle Ax,y\rangle$ extends by continuity to a continuous linear function on $\mathcal{H}$ and, so, there exists a unique $z\in\mathcal{H}$ such that $\langle Ax,y\rangle = \langle x,z\rangle$ for all $x\in\mathcal{H}$. Define the adjoint operator $A^*$ on $y$ by $A^*y=z$. Then,
$$
              \langle Ax,y\rangle = \langle x,A^*y\rangle,\;\; x\in\mathcal{D}(A),\;y\in\mathcal{D}(A^*).
$$
$0$ is always in the domain of the adjoint with $A^*0=0$. In order to prove more about the domain of $A^*$, you can write the adjoint relation as an inner product condition in $\mathcal{H}\times\mathcal{H}$:
$$
         \left\langle \left[\begin{array}{c}Ax \\ x\end{array}\right],\left[\begin{array}{c}y\\-A^*y\end{array}\right]\right\rangle = 0
$$
Using this observation, it can be shown that, if $A$ is densely defined and closed (which means its graph is closed), then $A^*$ is closed and densely-defined. This is done by looking at orthogonal complements in $\mathcal{H}\times\mathcal{H}$.


问:
Is this system of geometric points solvable?
Apologies for the vague title, but I am not sure how otherwise to explain my question with a title. I have a set of 3 known points (x, y), and a set of 3 unknown points and a few known relationships between the points.

The positions of the red points are known.

The context is that a Quadratic Bezier curve drawn from s' to e' with control point a' will pass through a at its midpoint period (I've solved the math for this part already).

s' is at the intersection of the line between a' and s and of the circle with known radius r around s.

e' is at the intersection of the line between a' and e and of the circle with known radius r around e.

a' can be solved from the three unknowns: $a_x = (2a'_x + s'_x + e'_x) / 4$. With the same for the y coordinate.

Putting that all together, I need a' to solve for s' and e', but I need those two terms to solve for a'. I've tried to put together a system of equations.

$s'_x = s_x + r * cos(arctan(\frac{a'_y - s_y}{a'_x - s_x}))$

$s'_y = s_y + r * sin(arctan(\frac{a'_y - s_y}{a'_x - s_x}))$

$e'_x = e_x + r * cos(arctan(\frac{a'_y - e_y}{a'_x - e_x}))$

$e'_y = e_y + r * sin(arctan(\frac{a'_y - e_y}{a'_x - e_x}))$

$a'_x = (2*a_x + s'_x + e'_x) / 4$

$a'_y = (2*a_x + s'_x + e'_x) / 4$

So I have a system of 6 equations with 6 unknowns that I am not able to solve. The trig gets way too crazy for me, and I am sure that I am missing many simplification steps or tricks somewhere.

In some situations, I know a' instead of a, and in those cases, I can solve for a. So I believe there should be at least (and probably only) one solution to the problem above that I can't solve. Also, there is a situation with doing a sign flip somewhere after the arctan that I've figured out, but not sure how to represent in math notation.

Any help is appreciated. Thanks in advance.

Supplementary information:

For even more context, in case anyone ever runs into this. I am
designing a graph (nodes and edges graph) visualization interface,
where I would like to allow the user to bend the edges manually when
the layout algorithm has not produced a desirable result (as is
common). The offsets s' and e' are  to account for the node radii
such that the terminal point of the edge is always directed to the
center of the node (which, especially with arrow-heads, looks the
best). a is fixed to the mouse position, but the svg engine (and underlying Bezier function of course) needs
to be specified the a' parameter. My current implementation was to
start off with a straight line where $a = a'$, and at each iteration
of edge-bending, to approximate a' by assuming the s' and e' of
the previous iteration. Basically, $s'_i = s'_{i-1}$ for the purposes of calculating $a'_i = (2 * a + s'_{i-1} + e'_{i-1}) / 4$. This performs pretty well in most cases, but
there are certain regions of a where the approximation becomes
wildly off and causes bouncing-back and forth effect until a moves
out of the region.

The region of terrible-approximation is something like this:
, and I would be OK with falling back to a
straight-line in this admittedly uninteresting region, but I also
can't seem to detect the boundaries of the region either.

答:
Treat all points as complex numbers. There is then really only one variable $a'$, for the curve endpoints may be derived from it as follows.

$s'$ lies on the line between $s$ and $a'$. The distances $ss'$ and $sa'$ are known (in terms of $a'$), so writing $u=\frac r{|s-a'|}$ as the ratio of those two distances, we have $s'=(1-u)s+ua'$. Similarly, if $v=\frac r{|e-a'|}$ then $e'=(1-v)e+va'$.

Then the single complex equation to solve is $s'+2a'+e'=4a$. Since the number of variables and equations is the same whether you split this into real and imaginary parts or not, the system is solvable… but in general you have to solve a sextic for each of real and imaginary parts, so you are better off solving numerically. An mpmath program doing this is below.

```<code>#!/usr/bin/env python3
from mpmath import *
mp.dps = 50

def solve(s0, a0, e0, r):
    def f(p): # p = (x, y)
        u = r / abs(p - s0)
        s1 = (1-u)*s0 + u*p
        v = r / abs(p - e0)
        e1 = (1-v)*e0 + v*p
        return s1 + 2*p + e1 - 4*a0
    # get a reasonable starting point
    u = r / abs(a0 - s0)
    s1 = (1-u)*s0 + u*a0
    v = r / abs(a0 - e0)
    e1 = (1-v)*e0 + v*a0
    p0 = (4*a0 - s1 - e1) / 2
    return findroot(f, p0, method="muller")

print(solve(-4+1j, 1+5j, 3-2j, 1))
</code>```

ContentLicense: CC BY-SA 4.0

Body:
Treat all points as complex numbers. There is then really only one variable $a'$, for the curve endpoints may be derived from it as follows.

$s'$ lies on the line between $s$ and $a'$. The distances $ss'$ and $sa'$ are known (in terms of $a'$), so writing $u=\frac r{|s-a'|}$ as the ratio of those two distances, we have $s'=(1-u)s+ua'$. Similarly, if $v=\frac r{|e-a'|}$ then $e'=(1-v)e+va'$.

Then the single complex equation to solve is $s'+2a'+e'=4a$. Since the number of variables and equations is the same whether you split this into real and imaginary parts or not, the system is solvable… but in general you have to solve a sextic for each of real and imaginary parts, so you are better off solving numerically. An mpmath program doing this is below.

#!/usr/bin/env python3
from mpmath import *
mp.dps = 50

def solve(s0, a0, e0, r):
    def f(p): # p = (x, y)
        u = r / abs(p - s0)
        s1 = (1-u)*s0 + u*p
        v = r / abs(p - e0)
        e1 = (1-v)*e0 + v*p
        return s1 + 2*p + e1 - 4*a0
    # get a reasonable starting point
    u = r / abs(a0 - s0)
    s1 = (1-u)*s0 + u*a0
    v = r / abs(a0 - e0)
    e1 = (1-v)*e0 + v*a0
    p0 = (4*a0 - s1 - e1) / 2
    return findroot(f, p0, method="muller")

print(solve(-4+1j, 1+5j, 3-2j, 1))


问:
On the calculation of curve integral
Let $\vec{n}$ be the out of unit normal vector on the curve $\varGamma$,and the $\varGamma:x^2+y^2=R^2$.define$$r=\sqrt{x^2+y^2}.$$Calculate$$\oint_{\varGamma}\dfrac{\partial\ln r}{\partial{\vec{n}}}\mathrm{d}s.$$
I made the following attempts,Known that $\vec{n}=(\dfrac{x}{R},\dfrac{y}{R})$.So$$\dfrac{\partial\ln r}{\partial{\vec{n}}}=\dfrac{\partial\ln r}{\partial{x}}\cdot\dfrac{x}{R}+\dfrac{\partial\ln r}{\partial{y}}\cdot\dfrac{y}{R}$$
But I don't understand why
$$\dfrac{\partial\ln r}{\partial\vec{n}}=\dfrac{d\ln r}{d r}$$
Can you explain it to me, thank you.

答:
Note that $R$ belongs to $\Gamma$.

At first the function $\frac{\partial \ln r}{\partial \vec n}$ is considered independently of $\Gamma$.

Now, interpreting $\frac{\partial \ln r}{\partial \vec n}$ as the derivative of $\ln r$ in the direction of $\vec n = \left( \frac x{\color{blue}{r}}\:\: \frac y{\color{blue}{r}} \right)$ you get

\begin{eqnarray*} \frac{\partial \ln r}{\partial \vec n}
& = & \frac 12 \left(\frac{\partial \ln (x^2+y^2)}{\partial x}\cdot\frac xr + \frac{\partial \ln (x^2+y^2)}{\partial y}\cdot\frac yr  \right) \\
& = & \frac 12 \left(\frac {2x}{r^2}\cdot\frac xr + \frac {2y}{r^2}\cdot\frac yr\right) \\
& = & \frac 1r\\
& = & \frac{d\ln r}{dr}
\end{eqnarray*}

Title: On the calculation of curve integral


问:
What does $p+q=k$ mean in the index of summation?
I need help solving something I don't understand. OK so the problem is this:

$$H^k(X,C)=\bigoplus_{p+q=k} H^{p,q}(X),$$

What does the $\;p+q=k\;$ mean?

Thank you anybody that helps! :)

答:
There is something to be said for writing
$$
\bigoplus_{p,\,q\, :\, p+q=k} H^{p,q}(X).
$$
For example, in case $k=3$, this would mean
$$
H^{0,3}(X) \oplus H^{1,2}(X)\oplus H^{2,1}(X)\oplus H^{3,0}(X)
$$
(or if only positive indices were indicated by the context, then discard the first and last terms, etc.).

Sometimes people write
$$
\sum_{i<j} a_{ij}
$$
when $j$ is fixed by the context and it means $i$ runs through the list of all values less than $j$.  But sometimes they mean the pair $i,j$ runs through the list of all values in which $i<j$, and neither is fixed.  It is sometimes convenient to make the notation explicit about the difference, writing
$$
\sum_{i\,:\,i<j} a_{ij}
$$
in one case, and
$$
\sum_{i\,j\,:\,i<j} a_{ij}
$$
in the other.

ContentLicense: CC BY-SA 3.0


问:
Unbiased estimator of variance from one data point
Let $X_1,...,X_n\sim\mathcal{N}(0,\sigma^2)$. I know that the sample variance
$$\hat{\sigma}^2=\frac{1}{n-1}\sum_{i=0}^{n}X_{i}^2$$
of this data is the UMVUE for $\sigma^2$.

But what if we have only a single data point? $\hat{\sigma}^2$ is not defined because of the $n-1 $ factor in the denominator. I'm wondering whether there exists some other unbiased estimator for $\sigma^2$ in this case.

It might seem like a weird question, but I think I've constructed such an estimator for the standard deviation $\sigma$. We know that $E(|X_1|)=\sigma\sqrt{2/\pi}$, so rearranging gives $E(|X_1|\sqrt{\pi/2})=\sigma$.

答:
Dividing by $n-1$ instead of $n$ is called Bessel's correction. You should only be using Bessel's correction when you are also subtracting by the sample mean. If you already know the population mean and the sample mean is unnecessary then you shouldn't be using the correction (see the Wikipedia page).

In the case where you didn't know the population mean you'd need to subtract the sample mean which in the one data point case would be that single data point. You would then end up with $0/0$ which is undefined and that makes sense intuitively since one point can't tell you anything about the variability of the population.


问:
If a random variable converge in distribution then scalar multiplication also converges in distribution.
Let c be a constant and suppose that $X_n\xrightarrow{d}X$, prove that

i)  $cX_n\xrightarrow{d}cX$

ii) $X_n+c\xrightarrow{d}X+c$

I've tried to use that:

$lim_{n\rightarrow\infty} F_{cX_n}(x) = lim_{n\rightarrow\infty} P(cX_n\leq x) = lim_{n\rightarrow\infty} P\left(X_n\leq \dfrac{x}{c}\right) = lim_{n\rightarrow\infty} F_{X_n}\left(\dfrac{x}{c}\right) $

But that's not exactly what I was looking for.

答:
Remember that $X_n \xrightarrow{d} X$ it's by definition:
$\lim\limits_{n \rightarrow \infty} F_{X_n} (x) = F_X(x)$

Hence:

a)
\begin{equation*}
\begin{split}
\lim\limits_{n \rightarrow \infty} F_{cX_n} (x) & = \lim\limits_{n \rightarrow \infty} \mathbb{P} (cX_n \leq x)  \\
 & = \lim\limits_{n \rightarrow \infty} \mathbb{P} \left( X_n \leq \dfrac{x}{c} \right) \\
 & = \lim\limits_{n \rightarrow \infty} F_{X_n} \left(\dfrac{x}{c}\right) \\
 & =  F_{X} \left(\dfrac{x}{c}\right) \\
 & =  \mathbb{P} \left( X \leq \dfrac{x}{c} \right) \\
 & =  \mathbb{P} \left( cX \leq x \right) \\
 & =  F_{cX} \left({x}\right) \\
\end{split}
\end{equation*}

\begin{equation*}
\therefore \lim\limits_{n \rightarrow \infty} F_{cX_n} (x) =  F_{cX} \left({x}\right) \end{equation*}

\begin{equation*}
\therefore cX_n \xrightarrow{d} cX
\end{equation*}

Analogously:

b)
\begin{equation*}
\begin{split}
\lim\limits_{n \rightarrow \infty} F_{X_{n}+c} (x) & = \lim\limits_{n \rightarrow \infty} \mathbb{P} (X_n + c \leq x)  \\
 & = \lim\limits_{n \rightarrow \infty} \mathbb{P} \left( X_n \leq x-c \right) \\
 & = \lim\limits_{n \rightarrow \infty} F_{X_n} \left( x-c \right) \\
 & =  F_{X} \left( x-c \right) \\
 & =  \mathbb{P} \left( X \leq x-c \right) \\
 & =  \mathbb{P} \left( X+c \leq x \right) \\
  & =  F_{X+c} \left( x \right) \\
\end{split}
\end{equation*}

\begin{equation*}
\therefore \lim\limits_{n \rightarrow \infty} F_{X_{n}+c} (x) =  F_{X+c} \left({x}\right)
\end{equation*}

\begin{equation*}
\therefore X_n+c \xrightarrow{d} X+c
\end{equation*}


问:
Invertibility of $I-A^s$ for each $s \geq 1$ implying nilpotence
If $R$ is a unital ring and $x \in R$ a nilpotent element, then $1-x$ is a unit, and so is $1-x^s$ for any $s \geq 1 $ since powers of $x$ are again nilpotent.

My question concerns the converse of this statement in the case in which $R = M_n(\mathbb{C})$, that is:

Question 1: if $A \in M_n(\mathbb{C})$ is a square matrix with complex entries and $I-A^s$ is invertible for all $s \geq 1$, what conditions can we impose on $A$ to guarantee that it is nilpotent?

Some condition has to be in place, as the general statement is already false in dimension $1$. In that case, the matrices $1-a^n$ will be always invertible whenever $a$ is not a root of unity, and the only nilpotent element is zero. More generally, any diagonal matrix $\mathrm{diag}(a_1, \ldots, a_n)$ of non-zero elements that are not roots of unity will do the trick.

Here's a partial result,

If $\{\det(I-A^n)\}_n \subset \mathbb{R} \setminus \{0\}$ converges, then either the spectrum of $A$ is contained in $B_1(0)$. In particular,
if $A$ has integer coefficients then it is nilpotent.

A proof follows: let $c_n := \det(I-A^s)$ and $\lambda_1, \ldots, \lambda_n \in \mathbb{C}$ be the eigenvalues of $A$, counted with multiplicity. Then

$$
(1-\lambda_1^s) \cdots (1-\lambda_n^s) = c_s
$$

is a convergent subsequence. Noting $d_s = \prod_{i \ : \ |\lambda_i| \neq 1}(1-\lambda_i^s)$, we then have

$$
\prod_{i \ : \ |\lambda_i| = 1}(1-\lambda_i^s) = c_s/d_s \to L.
$$
with either $L = 0$ or $L = \lim_s c_s$, depending on whether there are eigenvalues with absolute value greater than $1$. If has some eigenvalue outside the unit disc, the right hand side converges and moreover the left hand side has some non-trival factor.

Observe that since $c_n$ is never zero, no $\lambda_i$ can be a root of unity; and so we are left with showing that if $k \geq 1$ and $\alpha_1, \ldots, \alpha_k \in [0,1)$ are irrational then

$$
a_s := \prod_{j = 1}^k(1-e^{i2\pi\alpha_j s})
$$
does not converge to $0$ nor $\lim_s c_s$.

By Dirichlet's approximation theorem, for each $k \geq 1$ we can find a natural number $s_k$ and an integer $p_k$ for which $|s_k\alpha_j - p_k| < 1/k$, and thus there is a subsequence $(s_k)_k$ for which $e^{i2\pi\alpha_j s_k} \to 1$ for all $j$, and $a_{s_k}\to 0$. But then taking $s'_k = s_k+1$ yields $e^{i2\pi\alpha_j s_k'} \to e^{i2\pi\alpha_j s}$ and $a_{s_k'} \not \to 0$. Hence $(a_s)$ never converges.

Are there results in this direction whose proof is purely algebraic? In particular, can we recover the former partial results without doing analysis?

It seems to me like there is some relation between the spectrum of $A$ and the fact that $I-A^s$ being invertible implies the invertibility of $\sum_{i = 0}^s A^i$, the limit of the latter - were to exist- being the inverse of $I-A$.

Is there a reference for this statement, at least in the case in which $\det(I-A^s) = \pm 1$ for each $s$ and $A$ has integer coefficients?

I suspect this specific case can be proved in a purely algebraic fashion but so far I haven't been able to come up with a proof. Here's what I've done so far:

Suppose that $f = \chi_A = X^m g, g = (X-\lambda_1) \cdots (X-\lambda_k)$ with $X \not \mid g$. Note $p^{(m)}$ the polynomial whose roots are the $m$-th powers of $p$'s roots. Then our hypothesis is precisely that $f^{(m)}(1)$ is always a unit in $\mathbb{Z}$. Thus, the same holds for $g$, write $g = p_1 \cdots p_r$
where $p_j \in \mathbb{Z}[X]$ are irreducible. Since no eigenvalue $\lambda_i$ is a root of unity, once again by Kronecker's  theorem  each $p_j$ has a - possibly complex - root of absolute value greater than $1$. Moreover the proof notes that in general if $p \in \mathbb{Z}[X]$ then $p^{(m)} \in \mathbb{Z}[X]$, so in any case, we have
$$
\pm 1 = g^{(m)}(1) = p_1^{(m)}(1) \cdots p_r^{(m)}(1)
$$
and so $p_j^{(m)}(1) = \pm 1$. This allows us to reduce the original problem to:

Attempted Subproblem 1. Let $f = \prod_{i=1}^k (X-\alpha_i) \in \mathbb{Z}[X]$ be an irreducible monic polynomial. If $f^{(m)}(1) = \pm 1$ for each $m \geq 1$, where $f^{(m)} = \prod_{i=1}^k (X-\alpha_i^m)$, then $f = X$.

Since $\det(I-A^s) = \pm 1$ for all $s$, dividing by $\det(I-A)$ gives
$\det(\sum_{i=0}^s A^i) = \pm  1$ for all $s$. This yields a pretty large sum of all monomials $x_{i_1}^{j_1} \ldots x_{i_r}^{j_r}$ where $\{i_1, \ldots, i_r\} \subset \{1,\ldots, k\}$ and $j_t \in \{0,\ldots, s\}$, being equal to $\pm 1$. Maybe there's some underlying interpretation in terms of symmetric polynomials.

答:
You can definitely say that if $I-A\cdot P(A)$ is invertible for every $P$ polynomial in $A$, then $P$ is nilpotent. Or: if $I - Q(A)$ is invertible for every polynomial $Q$ with $Q(0) = 0$, then $A$ is nilpotent. Since otherwise, let $\lambda\ne 0$ an eigenvalue of $A$, and $Q$ polynomial, $Q(0) = 0$, $Q(\lambda) = 1$. We get
$I- Q(A)$ has an eigenvalue $1- Q(\lambda) = 1-1=0$.


问:
How many elements does the intersection of $n$ circles contain?
For the real problem that I'm solving is

For given three circles there are seven elements, then how many elements that 6 circles have?

For the first approche I just draw the picture and count all the elements and get the answer of $31$. But then I wonder what would the $n$ circles contain? So I draw the circle from $2$ circles to $10$ circles and count them all again. I see the pattern of $n(n-1)+1$ elements in $n$ circles, if I'm not wrong. But still, I wonder if one could find it using Algebra. Could someone help please

答:
You can do this with Euler's formula, $V-E+F=2$, since the arrangement of circles forms a planar graph, where the vertices are intersection points, and the edges are arcs of the circle connecting intersection points. The following analysis assumes that every pair of circles intersects at exactly two points, and no three intersect at a point.

The number of vertices, $V$, is $\binom{n}2\cdot 2$, since there are two intersection points for each pair of circles.

The number of edges, $E$, is $n\cdot 2(n-1)$, since each of the $n$ circles is cut into $2(n-1)$ arcs by the other $n-1$ circles.

Therefore,
$$
F=E-F+2=n\cdot 2(n-1)-2\cdot \binom{n}2+2=n(n-1)+2
$$
Finally, you need to subtract one, since the faces counted by Euler's formula includes the unbounded external face.


问:
Number of solutions to $x^k \equiv h \pmod {q^n}$
Could someone please give me a hint/solution to
the question, say $q$ is a prime and $(q,h)=1$, then
$$
x^k \equiv h \pmod {q^n}
$$
has at most $k$ solutions $1 \leq x < q^n$?

Thanks!

答:
This is false as stated. Let $q=2$, $h=1$, $k=2$, $n=3$. Then
$$x^2\equiv 1\bmod 2^3$$
has four solutions $1\leq x<2^3$, namely $x=1,3,5,7$.</p>


问:
Can one say that $\int_{r_{a}}^{r_{b}} \mathrm{d}\vec{r}$ is the infinite sum of infinitesimal magnitudes, $\mathrm{d}r$, with direction, $\hat{r}$?
Say, for example, that you had a situation where you needed to some up an infinite number of infinitesimally small portions of some vector, $\vec{r}$. This vector, $\vec{r}$, can be then defined as

\begin{equation}
\vec{r} = \int_{r_{a}}^{r_{b}} \mathrm{d}\vec{r}
\end{equation}

which states that the infinite summation of these infinitesimally small vector contributions, $\mathrm{d}r$, will total to the vector, $\vec{r}$.

We can say that this vector, $\vec{r}$, can be split into its components of some direction, $\hat{r}$, and magnitude, $r$, given by the following relation

\begin{equation}
\hat{r} = \dfrac{\vec{r}}{r}
\end{equation}

Using this, we could say that the vector, $\vec{r}$, can be expressed by the following
\begin{equation}
\vec{r} = r\hat{r}
\end{equation}

From the integral which was stated above, we know that, $\vec{r}$, can be expressed as the infinite sum of the infinitesimally small vector contributions, $\mathrm{d}\vec{r}$, so perhaps we can express this differential as

\begin{equation}
\mathrm{d}\vec{r} = \hat{r}\mathrm{d}r
\end{equation}

If this is true, then I could rewrite the previous integral as

\begin{equation}
\vec{r} = \int_{r_{a}}^{r_{b}} \hat{r}\mathrm{d}r
\end{equation}

This makes more sense to me geometrically. I can picture this as going along some function interval which describes the direction of the vector at any point, and providing it with a magnitude at that direction. Summing up all these magnitudes pointing in some, $\hat{r}$, direction should, in turn, provide me with the total vector, $\vec{r}$.

答: Yes, you can. In physics documentation, it is very common to see vectors as a magnitude with an associated direction $r\hat r$ since having constant "track" over the direction of where the calculation is taking place is useful at times.


问:
What is the connection of a sigmoid with probability?
In high school probabilities for an event are the chance of an event to happen. The easiest is a coin with 1/2 for each event.

I've been reading about Machine Learning and sigmoid functions, and it seems sigmoid has something to do with probability.

I've noticed the derivative is a Bell-like curve, and sure enough the integral of bell curves is normally a sigmoid.

But apart from being between 0 and 1 I see no real connection to a probability.

Any hints or useful places to get started?

答:
The sigmoid function $\sigma(u) = \frac{e^u}{1 + e^u}$ is useful in machine learning because it converts a real number into a probability (that is, just a number between $0$ and $1$). If you try to come up with the simplest possible differentiable function that does this job for us, I don't think you can find anything much simpler or more elegant than the sigmoid function. One of the elegant properties of the sigmoid function is that $\sigma'(u) = \sigma(u)(1 - \sigma(u))$.

When doing binary classification (with two classes called class $0$ and class $1$), we want to find a prediction function $f$ that takes a feature vector $x$ as input and returns as output a probability that an example described by this feature vector belongs to class $1$. We use the sigmoid function to force the output of $f$ to be between $0$ and $1$ so that it can be interpreted as a probability.

By the way, another elegant property of the sigmoid function is that it "plays well" with the cross-entropy loss function $\ell(p,q) = -p\log(q) - (1 - p) \log(1 - q)$, which is used in machine learning to measure how well a probability $q$ agrees with another probability $p$. Notice that if $0 \leq p \leq 1$ then the function $L(u) = \ell(p, \sigma(u))$ satisfies
\begin{align}
L(u) &= \ell(p, \sigma(u)) \\
&= -p \log\left(\frac{e^u}{1 + e^u} \right) - (1 - p) \log\left( 1 - \frac{e^u}{1 + e^u} \right) \\
&= -p \log\left(\frac{e^u}{1 + e^u} \right) + (1-p) \log\left(1 + e^u \right) \\
&= -pu + \log(1 + e^u) .
\end{align}
Look how much that simplified! Beautiful.
It follows that
\begin{align}
L'(u) &= \sigma(u) - p.
\end{align}
That's a beautiful and simple result. When using gradient descent to train a logistic regression model, the iteration is very simple and intuitive thanks to the simplicity of this result.


问:
$R$ is a ring with unity such that for all $x,y \in R: xy^2=xy$, then show that $R$ is commutative ring.
Assume $R$ is a ring with unity such that for all $x,y \in R: xy^2=xy$, then show that $R$ is commutative ring.

I was able to show that if $(xy)^2=(yx)^2$ then the ring is a commutative ring, or the case where $(xy)^n=x^ny^n$.

Or the case where  $(S; ·)$ is a semigroup which satisfies the identities $x^3 = x$ and
$x^2y = yx^2$ for all $x, y ∈ S$, then $S$ is commutative.

But I don't know how to show that for all $x,y \in R: xy^2=xy$, then show that $R$ is commutative ring,moreover the ring is a ring with unity which is different from the other questions.

答:
By setting $x = 1$ and $y=-1$, one has $x = -x$ and $y^2 = y$ for all $x, y \in R$.

Let $a,b \in R$. We want to show $ab = ba$. Since $(a+b)^2 = (a+b)$, one as $ab + ba = 0$. Equivalently, $ab = -ba$. Since $x = -x$ for all $x \in R$, $-ba = ba$.


问:
Regarding a well-known tensor product formula
As we know, for a commutative ring $R$, an ideal $I$ of $R$ and an $R$-module $A$ we have:

$R/I \otimes_R A = A/IA$

My question is that does this result still hold in case $R$ is not a commutative ring?

Thank so much in advance.

答:
Recall that for a right $R$-module $A$ and a left $R$-module $B$ the tensor product $A \otimes_R B$ is, by definition, an abelian group which represents the functor $\mathsf{Ab} \to \mathsf{Set}$ which maps an abelian group $C$ to the set of balanced maps $f : A \times B \to C$, i.e. maps which are additive in each variable and satisfy $f(ar,b)=f(a,rb)$.

The isomorphism $R/I \otimes_R A \cong A/IA$ holds for every ring $R$ and every ideal $I \subseteq R$: A balanced map $f : R/I \times A \to B$ gives a homomorphism $f([1],-) : A \to B$ which vanishes on $IA$. Conversely, if $g : A \to B$ is a homomorphism which vanishes on $IA$, then one checks that $f([r],a)=r \cdot g(a)$ is well-defined and balanced. Thus, there is a natural bijection between balanced maps on $R/I \times A$ and homomorphisms on $A/IA$. In other words, $A/IA$ is a model for $R/I \otimes_R A$.


问:
Why is an indefinite matrix tough to solve for direct and iterative solvers?
One can sometimes read the rather vague comment that indefinite matrices are "tough to solve" by iterative and direct solvers.

An example would be a saddle point system that arises from the numerical solution of the Stokes equations. For this problem, a discretization can lead to a symmetric linear system and one could apply the conjugated gradient method. In that case, the error can be bounded by
\begin{align}
  \left\| \mathbf{e}_k \right\|_\mathbf{A}
  &\leq 2 \left( \frac{ \sqrt{\kappa(\mathbf{A})}-1 }{ \sqrt{\kappa(\mathbf{A})}+1 } \right)^k \ \left\|  \mathbf{e}_0 \right\|_\mathbf{A}
  \,,
\end{align}
where $ \mathbf{e}_k := \mathbf{x}_k - \mathbf{x}_* $ is the error and $\kappa(\mathbf{A})$ is the condition number. For symmetric matrices $\kappa(A) = \frac{\left|\lambda_\text{max}(A)\right|}{\left|\lambda_\text{min}(A)\right|}$.

I understand that $\kappa(\textbf{A})$ increases as e.g. $|\lambda_\text{min}(A)|$ decreases (while $|\lambda_\text{max}(A)|$ remains constant), this makes the system "tougher to solve" in the sense that more iterations are required. But:

How is the indefinite property relevant for iterative solvers (with respect to the example)?

Why is the indefinite property relevant when applying direct solvers?

答:
The conjugate gradient method works only for symmetric positive definite matrices, so there's one reason why indefinite problems are harder to solve by iterative methods. The issue is that conjugate gradient, in a fundamental way, uses the fact that the matrix $A$ defines an inner product $(x,y) \mapsto y^\top A x$. The conjugate gradient method essentially computes a sequence of search directions which are "$A$-orthogonal" to each other. Once definiteness breaks down, you're toast because then a vector can be "orthogonal to itself" and satisfy $x^\top A x = 0$. If you study the CG method, you see that this $A$-orthogonality idea isn't just a minor point of the analysis and is really essential to why the method works. With an indefinite problem, you lose access to this $A$-inner product and you have to do some uglier things to compensate.

Another factor which is a little harder to put into a neat box is the fact that a lot of the most commonly solved symmetric positive definite matrices possess really good preconditioners. The discretized Poisson equation, for example, has incredibly effective multigrid and graph-theoretic preconditioners. Again, with indefiniteness you lose this nice structure and these problems have empirically proven to be much harder to precondition.

As for direct solvers, there's not nearly as big a difference, at least for dense matrices. The nice thing about positive definite matrices is the need to do pivoting for numerical stability is significantly obviated by the positive definiteness. One structural fact that might help illustrate why this is the case is that the largest entry in any row or column of a positive definite matrix is on the diagonal, which is definitely not true in general for indefinite problems. This means you need to do pivoting for symmetric indefinite matrices for stability reasons, which might negatively some structural properties you might want to maintain (e.g. sparsity, low-rank structure, etc.)


答:
You have $\mathrm d\vec r = \hat r\,\mathrm dr + r\,\mathrm d\hat r$, as you would expect from the product rule.  It doesn't make sense for the limits of integration of $r$ to depend on $r$, but we can say that $\vec r(b) - \vec r(a) = \int_a^b \hat r\,\mathrm dr + \int_a^b r\,\mathrm d\hat r$, because the chain rule shows that, since $\hat r = \frac1 r\vec r$, we have $\mathrm d\hat r = \frac1 r\mathrm d\vec r - \frac1 r\hat r\,\mathrm dr$.

You already noted that the formula $\mathrm d\vec r = \hat r\,\mathrm dr$ works in a special case; you said “all portions of $\mathrm dr$ are pointing in the direction of $\vec r$”, which doesn't directly make sense to me ($\mathrm dr$ is scalar, so doesn't have ‘portions’ in an obvious sense), but perhaps you meant “if $\mathrm d\vec r$ is always parallel to $\vec r$”.  We describe this by saying that all the acceleration of $\vec r$ is tangential.  In this case $\hat r$ is constant, so $r\,\mathrm d\hat r$ is $0$, and we recover your formula.

At the opposite extreme, if $\vec r$ has constant length $r$ (i.e., it moves around a circle centred at the origin), then $\mathrm dr$ is $0$.  We describe this situation by saying that the acceleration of $\vec r$ is normal, and we have the formula $\vec r(b) - \vec r(a) = \int_a^b r\,\mathrm d\vec r$.

In general, the acceleration of $\vec r$ has some tangential component and some normal component.  You have basically discovered the 2D part of the Frenet–Serret frame!  Congratulations!

Title: Can one say that $\int_{r_{a}}^{r_{b}} \mathrm{d}\vec{r}$ is the infinite sum of infinitesimal magnitudes, $\mathrm{d}r$, with direction, $\hat{r}$?

Body:
Say, for example, that you had a situation where you needed to some up an infinite number of infinitesimally small portions of some vector, $\vec{r}$. This vector, $\vec{r}$, can be then defined as

\begin{equation}
\vec{r} = \int_{r_{a}}^{r_{b}} \mathrm{d}\vec{r}
\end{equation}

which states that the infinite summation of these infinitesimally small vector contributions, $\mathrm{d}r$, will total to the vector, $\vec{r}$.

We can say that this vector, $\vec{r}$, can be split into its components of some direction, $\hat{r}$, and magnitude, $r$, given by the following relation

\begin{equation}
\hat{r} = \dfrac{\vec{r}}{r}
\end{equation}

Using this, we could say that the vector, $\vec{r}$, can be expressed by the following
\begin{equation}
\vec{r} = r\hat{r}
\end{equation}

From the integral which was stated above, we know that, $\vec{r}$, can be expressed as the infinite sum of the infinitesimally small vector contributions, $\mathrm{d}\vec{r}$, so perhaps we can express this differential as

\begin{equation}
\mathrm{d}\vec{r} = \hat{r}\mathrm{d}r
\end{equation}

If this is true, then I could rewrite the previous integral as

\begin{equation}
\vec{r} = \int_{r_{a}}^{r_{b}} \hat{r}\mathrm{d}r
\end{equation}

This makes more sense to me geometrically. I can picture this as going along some function interval which describes the direction of the vector at any point, and providing it with a magnitude at that direction. Summing up all these magnitudes pointing in some, $\hat{r}$, direction should, in turn, provide me with the total vector, $\vec{r}$.


问:
How to find angle in between two right triangles when one side is twice of other?
The problem is as follows:

$\begin{array}{ll}
1.&10^\circ\\
2.&12^\circ\\
3.&15^\circ\\
4.&16^\circ\\
\end{array}$

I'm not sure which sort of construction can be used here to solve this problem?.

I've attempted to draw a perpendicular line from $B$ to segment $AC$. But this did not yielded good results.

Then I've attempted tracing a perpendicular segment to $BC$ intersecting $AC$, from looking on this possibility. I got still stuck.

I don't know how to use the given angles, they suspiciously add up to $3\alpha$.

I still don't know how to use $QC=2HC$. Can someone give me some ideas on what to do?. Should congruence be used here?.

I'm assuming that the intended approach is try to spot triangles such as:

$3-4-5$ or $1-\sqrt{3}-2$ or something along those special right triangles.

I do hope someone could help me on how to solve this problem relying on euclidean geometry can this be done?.

答:
Construct isosceles triangle $CQD$ and connect $PD$. Note that
$$\angle HPD = \angle HPB + \angle BPD =(90^\circ +\angle HCP) + (90^\circ -\angle BDP)=180^\circ
$$
Thus, $H$, $P$ and $D$ are colinear and
$$\cos\angle HCD = \cos5\alpha = \frac{HC}{DC} = \frac12$$
which yields
$$\alpha = 12^\circ$$


问:
convergence of a sub-sequence in norm infinity
The question is, true or false: if fn(x) is bounded in (C[0,1],|| ||infinity) for all n, then there exist a sub-sequence of fn(x), fnk(x) such that fnk(x) converges uniformly in (C[0,1],|| ||infinity).

my answer was false, because {sin(nx)} is a sequence of continuous functions in [0,1] but the set does not have a sub-sequence that converges uniformly in norm infinity.

i want to know if i'm thinking correctly.

答: Your idea is fine but you have to prove that $\sin (nx)$ does not converge uniformly. Actually, it does not even converge pointwise. To see this take $x=\frac {\pi} 4$ which lies in $[0,1]$. When $n$ is a multiple of $4$ we see that $\sin (nx) \to 0$ whereas when $n$ is an odd number multiplied by $2$ it dose not tend to $0$.


问:
Prove that if $\varphi$ is injective then $G \cong \varphi(G)$
Let $G$ and $H$ be groups and let $\varphi : G \to H$ be  a homomorphism  and  the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .
Prove that if $\varphi$ is injective then  $G   \cong \varphi(G)$

My attempt:

If $\varphi$ is injective, then for every $a \in \varphi(G)$,there exist $b \in G$ such that $\varphi(b)=a$.

After that I'm not able to proceed any further.

答: Define $f:G\to \varphi(G)$ such that $f:g\mapsto \varphi(g)$. Clearly, $f$ is surjective. We are given that $\varphi$ is injective, hence $f$ is injective. So, $f$ is a bijection. Since $\varphi$ is a homomorphism, so is $f$. Thus, $G\cong\varphi(G)$.


问:
The formula of quadratic non-linear recurrence?
I'm currently working on the recurrence equation $a_{n+1}=a_n^2-a_n+1, a_1=3$, and I wonder if there is a formula for this. It would be better if anyone can give a primary solution.

答:
First of all, Welcome to the site !

In fact, what you have is a Sylvester sequence with $a_0=2$.

If you use $a_1=3$, the very first terms are
$$\{3,7,43,1807,3263443\}$$ Have look in $OEIS$ and this is sequence $A000058$ (have a look here).

If you have a look at the formula section, you will notice a formula
$$a_n \sim \left\lfloor c^{2^{n+1}}+\frac{1}{2}\right\rfloor \qquad \text{with}\qquad c=1.2640847353053011130795995...$$ $c$ is Vardi's constant (have a look here for its decimal representation).

A good rational approximation is $c=\frac{136530}{108007}$ which is error by $5.66\times 10^{-11}$.


问:
$\pi$ & $\phi$ (Golden ratio), Pentagon inscribed in unit circle
Everyone is aware that square inscribed in unit circle and infinite product giving rise to $\pi$.
One of the simplest way to represent $\pi$ with the help of nested radical as follows
$$\pi = \lim_{n\to \infty} 2^n\cdot \sqrt{2-{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\tag1$$

where number of 2 inside the nested radicals is n. Viete's formula for $pi$

Approach to get similar result with inscribed initial polygon as PENTAGON will be as follows..

According to Chord in Trigonometry

A perpendicular line from the centre of a circle bisects the chord. Or otherwise can be stated as, 2. The line segment through the centre bisecting a chord is perpendicular to the chord.

The length of the chord inside the unit circle for Pentagon will be $2\sin36^\circ$. Therefore circumference of Regular pentagon will be $5\times2\sin36^\circ$.

For the sake of simplicity let us apply rules which are basic

$$2\sin\frac{\pi}{5} = \sqrt{2 - 2\cos\frac{2\pi}{5}}$$

Now doubling the sides the length of the each chord inside the unit circle will be $2\sin(\frac{\pi}{2\cdot5})$ and therefore circumference of decagon is $5\cdot2^2\sin18^\circ$. where $2\sin18^\circ = \sqrt{2-2\cos\frac{\pi}{5}}$

Therefore circumference can be represented as $C_{10} = 5\cdot2^1\sqrt{2-2\cos\frac{\pi}{5}} = 5\cdot2^1\sqrt{2-\phi}$

(Here we get $2\cos\frac{\pi}{5} = \phi $, the Golden ratio)

As $\pi = \frac{C}{D}$

Iterating the sequence we get the circumference of infinitely increasing polygons as follows and simplifying for $\pi$

Assuming doubling from Pentagon as initial polygon

For Pentagon, calculating $\pi$ as $\frac{C}{D}$

Circumference of Pentagon is $5\times 2\sin\frac{\pi}{5} = 5\cdot\sqrt{2-2\cos\frac{2\pi}{5}}$ (by Half angle cosine formula). As the diameter is 2 in unit circle. Value of $\pi_{n}$ where n represents the number of sides of the polygon of increasing order, i.e. doubling everytime from Pentagon

From Regular Pentagon
$\pi_{5}$ = $\frac{1}{2}\cdot 5\cdot\sqrt{2-2\cos\frac{2\pi}{5}} = \frac{5}{2}\cdot\sqrt{3-\phi} = 2.938926...$

From decagon
$\pi_{5\cdot2^1} = 5\cdot2\sin\frac{\pi}{5\cdot2} = 5\cdot\sqrt{2-2\cos\frac{\pi}{5}} = 5\cdot\sqrt{2-\phi} = 3.0901699...$

From icosagon (20 sided polygon)
$\pi_{5\cdot2^2} = 5\cdot2^2\sin\frac{\pi}{5\cdot2^2} = 5\cdot2^1\cdot\sqrt{2-2\cos\frac{\pi}{5\cdot2^1}} = 5\cdot2^1\cdot\sqrt{2-\sqrt{2+2\cos\frac{\pi}{5}}} = 5\cdot2^1\sqrt{2-\sqrt{2+\phi}} = 3.128689...$

From 40gon

$\pi_{5\cdot2^3} = 5\cdot2^2\sqrt{2-\sqrt{2+\sqrt{2+\phi}}} = 3.138363...$

From 80gon

$\pi_{5\cdot2^4} = 5\cdot2^3\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\phi}}}} = 3.140785...$

And so on

The value of $\pi$ for $5\cdot2^{10}$gon is $\pi_{5\cdot2^{10}}=3.14159245...$

$$\pi = \lim_{n \to \infty} 5\cdot2^n\sqrt{2-\sqrt{2+\sqrt{2+...\text{(n terms)}+\phi}}} \tag2$$ by applying limits $2^n$ becomes infinitely large and nested radical becomes infinitely small. The resulting value of $\pi$ is increasingly accurate and converges little faster compared to $\pi$ from incribed polygon as square.

The question is

$Equation1 \over Equation2$ $= 5$ explained as follows

$\frac{\pi}{\frac{\pi}{5}} = 5 = \frac{\lim_{n\to\infty}2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{n times}}}}}{\lim_{n\to\infty}2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{n times}+\phi}}}}$ (The $2^n$ gets cancelled out)

How to explain the fact that $\sqrt{2-\sqrt{2+\sqrt{2+...+\phi}}}$ is an infinitely small number different from $\sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$ another infinitely small number which is 5 times larger than the the first one? (By the geometric method we are deriving the value of $\pi$ from right side proceeding towards left to get the value of nested square root. (Computing also does the same). It seems that we should change the conventional thinking of representing infinite nested radicals in this respect. I need comments on this)

Infinite expansion happens between both ends. Can we call these as INFINITELY EXPANDING BALLOON NESTED RADICALS? (Bear with me for poor image quality)

When trigonometry is more concerned with unit circle practically, is $\epsilon = \frac{\pi}{\infty}$(all operations as in real number systems, here also it is applicable) different from traditional $\epsilon = \frac{1}{\infty}$? (Even though it looks trivial) (Of course there are different ways to get the infinitesimals in surreal/hyperreal number systems)

By deriving $\pi$ from polygons other than square, equilateral triangle and pentagon, probably it may be possible to get infinite number of infinitesimals by the way of representing as nested square roots. Please comment or clarify on these.

答:
This has nothing to do with nested radicals.

Let the sequence $a_n$ be $1,1/2,1/3,1/4,1/5,1/6,1/7,\dots$.

Let the sequence $b_n$ be $1/5,1/10,1/15,1/20,1/25,1/30,1/35,\dots$.

Then $a_n/b_n$ is the sequence $5,5,5,5,5,5,5\dots$.

So $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=0$, but $\lim_{n\to\infty}(a_n/b_n)=5$.

What's the problem?


问:
Air fan problem - fluid dynamic
For preparing an exam me and my classmates are unable to solve this question. Can anybody help us. Thanks.

At a conference my colleagues had a room without air conditioning. At
the reception they could get a fan to cool down the warm room.

(a) Assume that the open part of the window is a vertical opening not
much wider than the fan, but much higher in height. Where would it be
better, for my colleagues to put the fan, at the bottom of the
opening, in the middle or at the top? Why? What physical limitation do
you have to make on the blowing speed of the fan to get to physically
relevant results?

(b) The flow rate of the inflow caused by the fan is Q and the inflow
air has a temperature Ti , and the volume of the chamber is Vk. Assume
(as opposed to question a) that the outflow from the window at any
time is the average temperature of the chamber T(t) (and thus there is
a very efficient mixing of the air). Set up a differential equation
for T(t) and calculate how long it takes for the room to cool from a
high initial temperature $T_0$ to $T = (T_0 + T_i)/2$\

(c) The cooling effect of a fan is caused by the layer of air around
the skin (heated by thermal conduction) being replaced by cool air of
the passing air flow of the fan. We can write the influence of the
thermal conduction with the help of the P´eclet number, which takes
the ratio of the advection transport of thermal energy and the
conductive transport. Calculate this dimensionless number (analog of
the Reynolds number) using the equation for temperature transport:
$\frac{dT}{dt}=k \nabla^2T$ where κ is the thermal conduction coefficient

答:
Welcome to MSE. Please note that at least the first part of your question may be more suitable for a physics forum. Please also note that in here we expect you to show your efforts towards solving your own problems. Otherwise it may seem that you are having others do your homework for you! That said, I will be happy to give you a boost with part (b). You can use the hints in this comment and try to solve the problem yourselves. And if you still find it difficult, you can share your work with us, and hopefully some of us will help you again. Alright, let's do this:

We are assuming air flow to be incompressible in this example. Therefore, the flow rates of inlet and exit are equal ($Q$). In the rest of this post we assume that $Q$ is volume flow rate (and not mass flow rate). In an infinitesimally short duration, $dt$ , volume of in-coming air is $Qdt$ . This air, with temperature $T_i$ mixes with the air in the rest of the room (with volume $V-Qdt$) at temperature $T(t)$ . As a result, the temperature of the room changes by $dT$. The temperature of the mixture is a weighted average of the temperatures of the two volumes of air, with weights being proportional to their respective volumes.
$$T(t)+dT = \frac{Qdt}{V}T_i + \frac{V-Qdt}{V}T(t) $$
Rearranging the terms, we will have:
$$\frac{dT}{dt} + \frac{Q}{V}T(t) = \frac{Q}{V}T_i \qquad(1)$$
This is a first-order linear ordinary differential equation (ODE) that is not homogeneous. From the text of the problem I assume you are familiar with ODEs, so I will suffice to a hint for solving equation (1). You must find the general solution of (1) and then apply the initial condition $T(0) = T_0$ . The general solution is the sum of the solution of the homogeneous equation (in which the right hand side is zero) and a particular solution of the non-homogenous equation. The solution of the homogeneous equation $y'+ry=0$ is $y_h=e^{-rx}$. To find a particular solution to the non-homogeneous equation $y'+ry=A$ , try $y_p=y_h+B$ and use it in the equation to find $B$. The general solution will be $y_g=cy_h+y_p$ , wherein $c$ is a constant that should satisfy the initial condition.

There, you now have the equation and you know how to solve it. Good luck!

Title: Air fan problem - fluid dynamic


问:
Property of $f(x, y)= z$ for all permutations of $x, y, z$
What property would you say a function $f(x, y)= z$ has if it is true for all permutations of values $x, y, z$?

答:
Let $f\colon A\to A$ be a function. Assume that for all $x,y,z$ with $f(x,y)=z$, we also have $f(y,x)=z$, $f(x,z)=f(z,x)=y$ and $f(y,z)=f(z,y)=x$.

Equivalently, $f$ is symmetric, i.e., $f(x,y)=f(y,x)$ for all $x,y\in A$, and additionally $f(x,f(x,y))=y$ for all $x,y\in A$, i.e., for every fixed $a\in A$, $f(a,\cdot)$ is involutory.


问:
Convergent subsequences of $\{ n \alpha \bmod 1 \}$
It's well known that if $\alpha \not\in \mathbb{Q}$, then the sequence $\{n \alpha \bmod 1\}_{n \geq 0}$ is dense in the torus $\mathbb{T} = \mathbb{R}/\mathbb{Z}$.  Does every convergent subsequence necessarily have zero density?  To be a bit more precise, if the sequence $\{n_j \alpha \bmod 1\}_{j \geq 1}$ converges, does the set $\{n_j : j \geq 1\}$ have density zero in $\mathbb{N}$?

答: Suppose $n_j\alpha\bmod1$ converges to $\beta$ and $\{n_j\}$ has positive lower density, i.e., there exists a $d>0$ such that $\{n_j\}\cap [1,N]$ has at least $dN$ elements if $N\gg0$. Fix $M\in \Bbb N$. Then there must exist infinitely many intervals $[k+1,k+M]$ such that $\{n_j\}\cap [k+1,k+M]$ has at least $dM$ elements. So if we pick our $M$ such that $dM>2$, there are infinitely many pairs of numbers $n,m\in\{n_j\}$ with $n<m<n+M$. By pigeon-hole, there exists a $1\le \Delta<M$ such that there are infinitely many $n$ with $n,n+\Delta\in\{n_j\}$. For large enough such $n$, the difference $(n+\Delta)\alpha\bmod 1-n\alpha\bmod 1=\Delta\alpha\bmod 1\pm1$ must become arbitrarily close to $0$. We conclude that $\Delta\alpha\equiv 0\pmod 1$, i.e., $\alpha\in\Bbb Q$.


问:
If $X_n \to X$ in Hausdorff distance then does $X_n \cap E \to X \cap E$ in Hausdorff distance?
Provided all of the sets mentioned are compact subsets of $\mathbb{R}^n$.

I know that $d_{H}(X_n \cap E,X \cap E)$ need not be less than or equal to  $d_{H}(X_n,X)$ but that alone does not seem to be enough to disprove the claim (if it is false).

Body: Take $X = [0,1]$, $X_n = [1/n, 1]$, and $E = \{0,1\}$.  Then $d_H(X_n, X) = 1/n$ but $X\cap E = \{0,1\}$ and $X_n \cap E = \{1\}$ for all $n$, so $d_H(X_n \cap E, X \cap E) = 1$ for all $n$.


问:
number of permutations in which all number changes

What is the number of permutations in which all number change
position?

For example, suppose a ordered set (12345678...n)

What is the number of permutations this (below) is not allowed:

Let P be, let's say, the permutation operator, and a_{i} the element
at the i position

Apply now P on the set, if $ai_{after, p} = ai_{before, p}$ for some
i, so it is not allowed.

I am a little stuck because, i think we can not separate in little cases as, for example, "n! - (number o situations in which 1 remains 1)..." Since if we try that, we will need to consider the cases in which, let's say, 2 go for 1 or for another, and so on...

答: This is called a "derangement."</p>

Body: This is called a "derangement."


问:
Spectrum of weighted right shift operator on $\ell^{\infty}(\mathbb N)$ - solution verification
I'm interested in finding the spectrum of the following weighted right shift operator on $\ell^{\infty}(\mathbb N)$:
$$T(a_1,a_2,a_3,...)=\left(0,\frac{a_1}{1},\frac{a_2}{2},\frac{a_3}{3},...\right)$$
If first noticed that $T-\lambda I$ is always injective:
$$(T-\lambda I)(a_1,a_2,a_3,...)=\left(-\lambda a_1,\frac{a_1}{1}-\lambda a_2,\frac{a_2}{2}-\lambda a_3,\frac{a_3}{3}-\lambda a_4,...\right)$$
And this vector is equal to $0$ if and only if $0=a_1=a_2=a_3=\dotsb$

To find whether it is surjective, I ask myself if there is a solution to the problem:
$$ \left(-\lambda a_1,\frac{a_1}{1}-\lambda a_2,\frac{a_2}{2}-\lambda a_3,\frac{a_3}{3}-\lambda a_4,...\right)=(b_1,b_2,b_3,b_4,...)$$
Clearly, for $\lambda=0$ there isn't always a solution (so $0$ is definitely in the spectrum). For $\lambda \neq 0$, I can try to solve this equation by induction and get the solution via the following recursion relation:
$$a_1=-\frac{b_1}{\lambda},a_2=-\frac{b_2-a_1}{\lambda},...,a_n=-\frac{1}{\lambda}\left(b_n-\frac{a_{n-1}}{n-1}\right),...$$
Since we've found a solution for any $(b_n)$, the operator is presumably surjective. But the problem is, the solution can possibly be unbounded, and then $T$ cannot really be surjective (since the solution does not belong to our space).

Since our formula for $a_n$ above has a leading factor which behaves like $\frac{1}{\lambda^n}$ (at least it seems so), then for $|\lambda|<1$ the solution is unbounded and for $|\lambda|>1$ it is bounded. So the spectrum includes the open unit disk (since then $T$ is not surjective), and not the exterior of the closed unit disk. From compactness of the spectrum, it must be exactly the closed unit disk.

Does my solution seem correct? I'm not entirely sure about my work. If I'm not mistaken, the spectrum of the usual right shift operator is the closed unit disk as well, and intuitively I'd expect the coefficients to affect the spectrum somehow. But I guess it doesn't have to happen necessarily.

Thanks a lot in advance.

答:
So thanks to the advice by Ruy, I saw a problem with my computation - the solution is actually bounded for $|\lambda|<1$ as well, so the spectrum does not contain the unit disk.

A simpler way to find the spectrum is noticing that the spectral radius is $0$. It's not too hard to see that $||T^n||=\frac{1}{n!}$, which means that the spectral radius is $lim_{n\rightarrow \infty}(||T^n||)^{1/n}=0$, which means that the spectrum must be $0$ (since it's non-empty and bounded by $0$).


问:
Prove $ |\frac{\text{Cov}\left(X,Y\right)}{\sqrt{\text{Var}[X]\text{Var[Y]}}}|=1\iff\exists a,b\in\mathbb{R}:(\mathbb{P}[Y=aX+b]=1) $

Let $ \rho\left(X,Y\right)=\dfrac{\text{Cov}\left(X,Y\right)}{\sqrt{\text{Var}[X]\text{Var[Y]}}}$. Prove that $$ |\rho\left(X,Y\right)|=1\iff\exists a,b\in\mathbb{R}:(\mathbb{P}[Y=aX+b]=1). $$

I prove the $ \Leftarrow $ direction. But I'm struggeling to prove the other direction.

Here's what I've tried:

Assume that $ |\rho\left(X,Y\right)|=1 $

Then

$ \left(\text{Cov}\left(X,Y\right)\right)^{2}=\text{Var}[X]\text{Var[Y]} $.

Now :

$ \require{cancel}
\mathbb{E}^{2}[XY]+2\mathbb{E}[XY]\mathbb{E}[Y]\mathbb{E}[X]+\cancel{\left(\mathbb{\mathbb{E}}[X]\mathbb{E}[Y]\right)^{2}}=\mathbb{E}[X^{2}]\mathbb{E}[Y^{2}]-\mathbb{E}[X^{2}]\left(\mathbb{E}[Y]\right)^{2}-\left(\mathbb{E}[X]\right)^{2}\mathbb{E}[Y^{2}]+\cancel{\left(\mathbb{E}[X]\mathbb{E}[Y]\right)^{2}}
$

So

$ \mathbb{E}^{2}[XY]+2\mathbb{E}[XY]\mathbb{E}[Y]\mathbb{E}[X]=\mathbb{E}[Y^{2}]\text{Var}[X]-\mathbb{E}[X^{2}]\left(\mathbb{E}[Y]\right)^{2} $

$ \mathbb{E}[Y^{2}]\text{Var}[X]=\mathbb{E}[\mathbb{E}^{2}[XY]+2\mathbb{E}[XY]\mathbb{E}[Y]X+\left(\mathbb{E}[Y]\right)^{2}X^{2}] $

$ \mathbb{E}[Y^{2}]=\mathbb{E}[\frac{\left(\mathbb{E}[XY]+\mathbb{E}[Y]X\right)^{2}}{\text{Var}[X]}] $

Define: $ b:=\frac{\mathbb{E}[XY]}{\sqrt{\text{Var}[X]}},\thinspace\thinspace\thinspace a:=\frac{\mathbb{E}[Y]}{\sqrt{\text{Var}[X]}} $

$ \mathbb{E}[Y^{2}]=\mathbb{E}[\left(aX+b\right)^{2}] $

$ \mathbb{E}\left(Y^{2}-\left(aX+b\right)^{2}\right)=0 $

I feel close because if the random variable in the expectation would have been non-negative I guess I could proceed from there, but it is not necessarily true.

Im looking for the most direct way to prove it, preferable without any tricks that I couldnt think of really fast (this is a question from an exam), is there a way to solve it using what I started?

Thanks in advance.

答:
In the special case where $X$ and $Y$ each have variance $1$, the fact that their correlation is $1$ would imply $\operatorname{Cov}(X,Y)=1$, so that
$$
\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)-2\operatorname{Cov}(X,Y)=1+1-2=0,
$$
which implies $X-Y$ is constant.

In the general case, try to modify $X$ and $Y$ so they each have variance $1$. Now apply the previous (special) case.


问:
Cluster point of the sequence $a_n = (-1)^n$
I am trying to understand an example given of the cluster points of the sequence $a_n=(-1)^n$.It is given that -1 and 1 are the only cluster points of this sequence.But i am unable to understand mathematically why they(-1 &1) are the cluster point.The definition of the cluster point that i know of is as follows:
A real no. $p$ is said to be a cluster point of a sequence <$a_n$> if every interval around $p$ contains infinitely many points of <$a_n$>

答: For example, for all $\epsilon>0,$ $(1-\epsilon,1+\epsilon)$ contains all $a_n$ for $n$ even, which is infinitely many terms. But take $p=1-\epsilon,$ then the interval $(p-\epsilon/3,p+\epsilon/3)$ contains no points of $\{a_n\}$ so is not a cluster point.


问:
Spanning trees implies isomorphism?
I'm having trouble solving this question:

Prove or disprove the following statement: Given a graph G, if T and U are spanning trees
of G, then T and U are isomorphic.

I know that two graphs are isomorphic if they have an "edge-preserving bijection." I also know that a spanning tree of a graph G is a connected graph that can be defined as a maximal set of edges of G that contains no cycle.

How would I go about proving or disproving?

答:
Actually, graph which possesses this property would be quite a rare thing. Still, there are some connected graphs which are not trees but any two spanning trees in them are isomorphic. An easy example---any cyclic graph. Another one---take a triangular graph $ABC$ (complete graph with three vertices) and an arbitrary tree $T$ with root $r$. Then "glue" $T$ to each of the vertices of $ABC$ by identifying $r$ and $A$, $r$ and $B$, $r$ and $C$ respectively, so you get graph $G$ which has rotational symmetry (it maps into itself when "rotated" by $120$ degrees). Obviously, $G$ has three spanning trees---namely, $G \backslash AB$, $G \backslash BC$, and $G \backslash CA$, and they are all isomorphic to each other. This can be generalized to taking a cyclic graph with $n$ vertices and gluing $n$ copies of $T$ to its vertices.

Here is another question, though---does there exist a graph (not a multigraph, mind you, but a regular graph, so double edges and loops are not allowed) with more than one simple cycle which possesses this property? I haven't really thought too long about it---but it could have a nice elementary solution...

Title: Spanning trees implies isomorphism?


问:
Upper bound on trace of solution to discrete lyapunov or discrete matrix ricatti equation
Take a discrete Lyapunov equation
$$
P = R + \beta A' P A
$$
for $\beta \in (0,1)$, $R$ and $A$ symmetric.

Quetsion: Can I put an upper bound on $tr(P)$ given properties of the primitives?  In particular, I have the $tr(R)$ and bounds on the maximum eigenvalue of $A$.

I can probably deal with something that doesn't generate the tightest bound exploiting the $A$ as long as the $tr(R)$ is involved.

Alternatively:  I am happy to use similar results for the solution to the Discrete matrix ricatti equation that comes out of optimal LQ control if that is available.

答:
By using a cyclic permutation it can be shown that the matrix part of the trace of the second term of the right hand side can also be written as

$$
\text{tr}(A'\,P\,A) = \text{tr}(A\,A'\,P). \tag{1}
$$

When assuming that $\sqrt{\beta} A$ is Hurwitz (this would be the case for all $\beta$ if $A$ is Hurwitz) and $R$ positive definite, then the solution for $P$ should also be positive definite. In that case one can use the Cauchy-Schwarz inequality, such that $(1)$ satisfies

$$
\text{tr}(A'\,P\,A) \leq \text{tr}(A\,A')\,\text{tr}(P). \tag{2}
$$

The trace of a matrix is equal to the sum of its eigenvalues, thus an upper bound for $\text{tr}(A\,A')$ would be $n$ times the maximum eigenvalue of $A\,A'$. If $A$ is symmetric then the maximum eigenvalue of $A\,A'$ would simply be the square of the the maximum eigenvalue of $A$, so

$$
\text{tr}(A\,A') \leq n\,\lambda_\text{max}\!(A)^2, \tag{3}
$$

with $A\in\mathbb{R}^{n\times n}$.

Taking the trace of the Riccati equation and substituting in $(2)$ and $(3)$ yields

$$
\text{tr}(P) \leq \text{tr}(R) + \beta\,n\,\lambda_\text{max}\!(A)^2\,\text{tr}(P). \tag{4}
$$

Solving $(4)$ for $\text{tr}(P)$ thus yields

$$
\text{tr}(P) \leq \frac{\text{tr}(R)}{1 - \beta\,n\,\lambda_\text{max}\!(A)^2}, \tag{5}
$$

assuming that $\beta\,n\,\lambda_\text{max}\!(A)^2 < 1$.

Or a less tight bound could be obtained by only substituting in $(2)$ which yields

$$
\text{tr}(P) \leq \frac{\text{tr}(R)}{1 - \beta\,\text{tr}(A\,A')}. \tag{6}
$$


问:
how to compute a series whose terms are a rational function times an exponential function?
How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}

I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n = \left(\frac{11}{14n-7}-\frac{1}{n+3}\right)\left(-\frac{1}{3}\right)^n\text{.}
\end{equation}
But I don't know what to do next.

Body:
$$\frac{n+10}{2n^2+5n-3}=\frac{3}{2 n-1}-\frac{1}{n+3}$$

Consider the first sum
$$S_1=\sum_{n=1}^\infty\frac{x^n}{2 n-1}=y\sum_{n=1}^\infty\frac{y^{2n-1}}{2 n-1}=y\tanh ^{-1}(y)=\sqrt{x} \tanh ^{-1}\left(\sqrt{x}\right)$$ With $x=-\frac 13$ we then have
$$S_1=-\frac{\pi }{6 \sqrt{3}}$$

For the second sum
$$S_2=\sum_{n=1}^\infty\frac{x^n}{ n+3}=\frac 1 {x^3}\sum_{n=1}^\infty\frac{x^{n+3}}{ n+3}=-\frac 1 {x^3}\left(\log(1-x)+x+\frac {x^2}2 +\frac {x^3}3\right)$$
With $x=-\frac 13$ we then have
$$S_2=\frac{47}{6}-27\log \left(\frac{4}{3}\right)$$

For the total
$$\frac{47}{6}-\frac{\pi }{2 \sqrt{3}}-27\log \left(\frac{4}{3}\right)$$


问:
Show that $u$ is increasing if $u''=a(t)u+b(t)u'$
All functions are defined in an open interval $I$, and $a,b$ are continuous. The problem asked me to show that $u$ is monotonically increasing in $\{t\geq t_0:t\in I\}$, provided that $u''=a(t)u+b(t)u'$, and $u(t_0)=0$, $u'(t_0)=1$, $a(t)>0$ for all $t$.

I have so far found that:
$$u''(t_0)= a(t_0)u(t_0)+b(t_0)u'(t_0)=b(t_0) $$ Also I wrote the problem as $$u'(t)=1+\int_{t_0}^t a(s)u(s)+b(s)u'(s)ds $$ I tried exploring the methods $u(t)>u(s)$ if $t>s$, and even $u'(t)>0$ for $t\geq t_0$, but goes nowhere. I'm don't see more directions to take not sure how to go from here, any idea is appreciated
I have added the original problem came from some chapters I found out somewhere, but without author, problem and chapter are added.

答: Let $U$ be the open interior of $\{\,t>t_0\mid u'(t)\le 0\,\}$. Assume $U\ne\emptyset$  and let $\tau=\inf U$. As $u'(t_0)>0$ and $u'$ is continuous, we have $\tau>t_0$ and clearly $u'(\tau)=0$. As $u'$ is non-negative on $[t_0,\tau)$ and positive near $t_0$, we conclude that $u$ is positive on $(t_0,\tau]$. Therefore $u''(\tau)=a(\tau)u(\tau)>0$. That means that $u'$ changes sign from negative to positive at $\tau$, contradiction the definition of $\tau$. Hence $U$ is empty and so $u$ is strictly increasing.


Title: how to compute a series whose terms are a rational function times an exponential function?

Body:
How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}

I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n = \left(\frac{11}{14n-7}-\frac{1}{n+3}\right)\left(-\frac{1}{3}\right)^n\text{.}
\end{equation}
But I don't know what to do next.

Body:
$$S=\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n$$
$$S=\sum_{n=1}^{\infty} \left(\frac{3}{2n-1}-\frac{1}{n+3}\right)(-3^{-1})^{n}=S_1-S_2$$
$$S_2=-27\sum_{n=1}^{\infty} \frac{(-3^{-1})^{n+3}}{n+3}$$
Use $\ln(1+z)=\sum_{k=1}^{\infty} (-1)^{k-1} \frac{z^k}{k}$
$$=-27\sum_{m=4}^{\infty}\frac{(-1)^m(3^{-1})^m}{m}=\frac{-47}{6}+27\ln(4/3)$$
Next use $\tan^{-1} x=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{z^{2k-1}}{2k-1}$
$$S_1=\sqrt{3}\sum_{n=1}^{\infty} (-1)^n \frac{(1/\sqrt{3})^{2n-1}}{2n-1}=-\sqrt{3} \tan^{-1}(1/\sqrt{3})=\frac{-\pi}{2 \sqrt{3}}$$
Finally $$S=\frac{-\pi}{2 \sqrt{3}}+\frac{47}{6}-27\ln(4/3)$$


问:
Finding a basis of an image of a linear transformation
Let f be a linear transformation such that $$f: \mathbb{R}^3 \rightarrow \mathbb{R}^4, V = \operatorname{span}(\vec v_1,\vec v_2)\subset\mathbb{R}^3.$$ My question is if I have the matrix $A$ of $f$ and I check the value of $f|_V(\vec v_1)$ and $f|_V(\vec v_2)$ by calculating $A\vec v_1, A\vec v_2 $, are these two vectors $A\vec v_1, A\vec v_2 $ (if they are linearly independent) a basis of an image of $f|_V$? And if yes, then does that imply that $f|_V$ is a monomorphism since $\dim(\operatorname{im}(f|_V)) = \dim(V) = 2$ ? Thanks in advance.

答: Yes, and actually if $\vec v_1,\vec v_2$ are a basis for $V$ then whether $f(\vec v_1),f(\vec v_2)$ are linearly independent or not, they span $f(V)$, so they form a basis unless they are linearly dependent, in which case if either is nonzero it forms a basis on its own. You're entirely right about the monomorphism as well, because if we express $f(V)$ as the span of $f(\vec v_1),f(\vec v_2)$ and $V$ as the span of $\vec v_1,\vec v_2$ there is an obvious bijection between the two spaces except in the case of linear dependence as described above.


问:
Countability of $\mathbb{Z} \times \mathbb{Z}$ using isomorphism with $\Bbb Z[i]$ - how to make this rigorous?

Visualise the set $\mathbb{Z}\times \mathbb{Z}$  as points on the complex plane,  and take origin as the center of a circle of radius $r$. With $r$ going from $0$ to $\infty$ - as you encounter points on the circle, write them down in counterclockwise order. This way you have a bijective mapping, thus $\mathbb{Z}\times \mathbb{Z}$ is countable.

This proof about the countability of $\mathbb{Z}\times \mathbb{Z}$ doesn't look rigorous enough to me. It was proposed by one of my colleagues. The proposal is to write down elements of $Z[i]$ as we encounter them in some sort of a spiral order.

I need help making this more rigorous.

Is there an explicit bijection for this? It'd be great if someone could help me construct the same.

Perhaps we must show that every such circle has only a countable number of such points and that only countably many such circles, i.e. values of $r$ matter. Thanks!

答:
I came up with this - does it look fine enough?

Consider $z = x+iy \in \mathbb{Z}[i]$, with $x^2 + y^2 = r^2$.  As $r^2$ goes from zero to $\infty$, we cover the entire complex plane. If $x,y\in\mathbb{Z}$ then $r^2 \in \mathbb{Z}$. Define $P = \{ r^2: r\in \mathbb{R}^{\geq 0}, r^2 \in\mathbb{Z}\}$. Since $\mathbb{Z}$ is countable,  so is $P$. Now, for some fixed $r_1$such that $r_1^2\in P$, think about $x^2 + y^2 = r_1^2$. This gives $|x| \le r_1, |y| \le r_1$. Since $r_1$ is finite, the set of values $(x,y) $ such that  $x^2 + y^2 = r_1^2$ holds, is finite. Knowing that a countable union of countable sets is countable, we see that the set $\mathbb{Z}[i]$ is countable. As a result, $\mathbb{Z} \times \mathbb{Z}$ is also countable.


问:
Multiplying a ring(or an ideal) by an element.
I was looking at a proof of the statement: Given a ring $R$ and an ideal $M$, $R/M$ is a field$\iff$M is maximal.

The backwards direction (assume $M$ maximal) proof involved a statement where we take an $r\in R$ such that $r\notin M$ and take $M + r*R$ which is an ideal as $M$ and $R$ are ideals, which implies that this construction must be $R$ itself.

I had two questions about this statement. First of all, I realized very quickly that for a given ring $r*R\neq R$. Is this simply because $R$ is not a group under $*$, i.e., if $R$ were a field, would $r*R=R$, or is there something else going on? If it's because of the group assumption, what would the reasoning behind a proof be (I assume something to do with existence and uniqueness of inverses but not 100% sure)?

The second was assuming that $r*R$ is an ideal. By definition of an ideal I know $r*R\in R$, but by the above question there's no guarantee that $r*R=R$. How then do we know that $r*R$ is an ideal?

答: It's become from definition directly, $r*R \ \& \ R*r$ are right and left ideals generated by $r$, respectively.


问:
Let $S = \{1,2,3,4\}. X, Y \in \mathcal{P}(S)$ and R be a relation $R(X,Y): |X \cap Y| = 1$. Is the relation R transitive?
I'm reading up on Relations and Functions with the book Discrete Mathematics by G. Chartrand, P. Zhang and one of the exercises there proposes that the relation R over $\mathcal{P}(S)$ where $R(X,Y): |X \cap Y| = 1$ is symmetric only — which I can make sense with since my mental model with this is that $(a, b) \in R$ never occurs because $|X \cap Y| = 1$ is only true when $|X|$ or $|Y|$ is 1, therefore only $(a,a) \in R$, thus the implication for symmetry is vacuously true.

My concern is that with my thinking of how the relation R is symmetric, then this can also mean that it's transitive as well, since $(a,b) \in R \land (b,c) \in R$ never occurs and therefore is vacuously true also.

Is my thinking correct? How is relation R symmetric only and why can't it be transitive as well?

Edit
Okay. I had $(a,a) \in R(X \cap Y, X \cap Y)$ in mind for whatever reason when I was thinking $(a, a) \in R$.

答:
I have absolutely no idea why you think $|X\cap Y|$ only if $|X|$ or $|Y|$ is $1$.

Simple case $X=\{1,2\}$ and $Y=\{2,3\}$ then $X\cap Y = \{2\}|$ and $|X\cap Y| = 1$ and $R(X,Y)$.

Further more if $Z = \{3,4\}$ then $R(Y,Z)$ but $X\cap Z =\emptyset$ so $\not R(X,Z)$ even though $R(X,Y)$ and $R(Y,Z)$.  So not transitive.

....

Any way  $X \cap X = X$ so $R(X,X)$ only if $|X| = 1$.  So if $|X| \ne 1$ we do not have $R(X,X)$ so it is not reflexive.

$X\cap Y=Y\cap X$ so $|X\cap Y| = |Y\cap X|$ so $R(X,Y) \iff R(Y,X)$ so symmetry holds.

Transitivity:  There no way to determine $X\cap Z$ from $X\cap Y$ and $X\cap Z$.  It is certainly possible for $X$ and $Y$ to have only one thing in common and $Y$ and $Z$ to have one thing in common but $X$ and $Z$ to have nothing or many things in common.

Let $X = $ the prime natural numbers.  Let $Y = $ the even natural numbers.  And let $Z=$ then numbers between $11$ through $13$ inclusively.

$X \cap Y= \{2\}$.  $Y\cap Z=\{12\}$ and $X\cap Z = \{11,13\}$.  Not transitive.


问:
How to factor a four term polynomial without grouping?
$$2x^3 + 9x^2 +7x -6$$

This equation doesn't factor by grouping, and other than that I have no idea how to solve this problem. Will someone please help?

答:
The way to factor a four-term polynomial like this is to apply Rational Root Theorem along with synthetic division or substitution to determine whether a rational root works for the polynomial or not.

Here is how Rational Root Theorem works.  The number of rational roots is determined by the quotient of the factors of the last term and the factors of the first term.  That is:

$$\{\pm \text{factors of last term} / \text{factors of first term}\} = \{\pm 1, 2, 3, 6 / 1, 2\}$$

So the possible rational roots are:

$$\pm 1, \pm2, \pm3, \pm6, \pm1/2, \pm3/2$$

Select one of the rational roots and apply synthetic division.  The number of times you need to test a root for the polynomial depends on whether you get $0$ for evaluating the polynomial or factored polynomial at some rational root or not.  This is done computationally.

For example, suppose that $x = -3$ is one of the rational roots.  Evaluate the polynomial at $x = -3$, so we have $0$.  This means that the given polynomial has the factor $(x + 3)$, so we can apply synthetic division to obtain the factored polynomial, which is $2x^2 + 3x - 2$.

You can either work out via Rational Root Theorem or by simply brain work of factorizing that polynomial.  Try this one with Rational Root Theorem and check that factored polynomial has two rational roots.


答:
We know $aR$ is an ideal because for $r,r^\prime\in R,x\in aR$ we have $ar+ar^\prime=a(r+r^\prime)\in aR$ and $(ar)b=a(rb)\in aR$ (the proof for left ideals is similar).

We don't necessarily expect $r\cdot R$ to be equal to $R$, in fact, if it is then the ideal is trivial. It is precisely when $r\cdot R\neq R$ that things begin to become interesting. Fields, because every (nonzero) element is invertible, do not have any interesting ideals.

Title: Multiplying a ring(or an ideal) by an element.

Body:
I was looking at a proof of the statement: Given a ring $R$ and an ideal $M$, $R/M$ is a field$\iff$M is maximal.

The backwards direction (assume $M$ maximal) proof involved a statement where we take an $r\in R$ such that $r\notin M$ and take $M + r*R$ which is an ideal as $M$ and $R$ are ideals, which implies that this construction must be $R$ itself.

I had two questions about this statement. First of all, I realized very quickly that for a given ring $r*R\neq R$. Is this simply because $R$ is not a group under $*$, i.e., if $R$ were a field, would $r*R=R$, or is there something else going on? If it's because of the group assumption, what would the reasoning behind a proof be (I assume something to do with existence and uniqueness of inverses but not 100% sure)?

The second was assuming that $r*R$ is an ideal. By definition of an ideal I know $r*R\in R$, but by the above question there's no guarantee that $r*R=R$. How then do we know that $r*R$ is an ideal?


问:
Confusion in definition of real valued function?
A function that has range as set of real numbers

OR

A function which has either ${R}$ or one of its subsets as its range.

So , in 2nd definition

Does the 2nd statement mean that the functions range is either a real number or a subset of real numbers ?

Could you give an example for 2nd definition.Especially in subset case

答: A real function is a function whose codomain is $\Bbb R$. So, its range is necessarily a subset of $\Bbb R$, but it doesn't have to be the whole $\Bbb R$.


答:
As $\mathbb R$ is a subset of itself and is a set of real numbers both those definitions are the same.

If $f(x) = \sin x$ has $[-1,1]\subset \mathbb R$ as it's range.

$g(x) = x^2$ has $[0,\infty)\subset \mathbb R$ as its range

and $h(x) =x^3$ has $\mathbb R\subset \mathbb R$ as its range.

And $j(x) = 7$ has $\{7\} \subset \mathbb R$ as its range.

All are real functions.

But as Jose Carlos Santos points out the actual range doesn't matter.  So long as it is specified that the function's codomain (a set of which all the values of the function are mapped to) is the real numbers you don't actually have to specify the exact range.

In all my above examples it's enough to say they all map to the real numbers.

Title: Confusion in definition of real valued function?

Body:
A function that has range as set of real numbers

OR

A function which has either ${R}$ or one of its subsets as its range.

So , in 2nd definition

Does the 2nd statement mean that the functions range is either a real number or a subset of real numbers ?

Could you give an example for 2nd definition.Especially in subset case


问:
Volume of a solid region enclosed by elliptic cylinder using triple integral
Find the volume of a solid limited laterally by an elliptical cylinder $$\frac{(x-2)^2}{4}+\frac{(y-1)^2}{9}=1$$ and by the planes $z+x=5$ and $z+x=6$.

I tried doing by cylindrical coordinates and Cartesian, but I could not get to the answer, which is $6\pi$. Can anyone give me some ideas on how to proceed?

答:
Please use change of variable $x = 2 + 2r \cos \theta, y = 1 + 3r \sin\theta$

Your ellipse transforms to a circle of radius $1$ and so we have $0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi$.

and the limits of $z$ will be given by planes

$3 - 2 r \cos \theta \leq z \leq 4 - 2r \cos \theta$.

Jacobian for the change of variables,

$|J| = 6r$

Can you take it from here?


问:
In any tree, what is the maximum distance between a vertex of high degree and a vertex of low degree?
In any undirected tree $T$, what is the maximum distance from any vertex $v$ with $\text{deg}(v) \geq 3$ to the closest (in a shortest path sense) vertex $y$ with $\text{deg}(y) \leq 2$? That is, $y$ can be leaf.

It seems to me that this distance can be at most $\dfrac{\text{diam}(T)}{2}$, and furthermore that the maximum distance will be attained from a graph center. Is this true? There's probably simple argument for it somewhere.

答:
In your second question (shortest path to vertex of degree $\le 2$), the bound $\operatorname{diam}(G) / 2$ holds, simply by noticing that the ends of the longest path in a tree are leaves, and the "worst" that a graph can do is have the vertex of degree $3$ or higher right in the center.  But in fact, this holds for the shortest path from any vertex of to vertex of degree $1$.

Can we do better?  No, in fact.  Take a look at a tournament bracket, for example (just delete the leaves on one side to create a unique center).  Viewed as a tree, all the vertices are of degree $3$ except for the leaves, which are all at least distance $d/2$ from the center of this modified graph.


问:
$M = (M_t)_{t\geq 0}$ obtained by Itô's formula is a martingale
Here I defined a non-negative stochastic process. Now, taking $F(t,x) = tx^2$, I would like to find that the continuous local martingale ``part'' obtained by Itô's formula is indeed a Martingale. To do so, I obtained, by applying Itô's formula, and
$$
d X_t = 3dt + 2\sqrt{X_t}d B_t, \quad\quad d\langle X,X\rangle_t = 4X_t\langle B,B\rangle_t,
$$
the following formula:
$$
F(t,X_t) = \int_0^2 [X_s^2 + 10s X_s ] ds + 4\int_0^t sX_s \sqrt{X_s}d B_s.
$$

Hence, I defined $M = (M_t)_{t\geq 0}$ by
$$
\int_0^t sX_s \sqrt{X_s}d B_s.
$$
To see that it is a martingale, it is sufficient to see that $E\sqrt{\langle M,M\rangle_s}< \infty$, but, by (concave function version) Jensen's inequality, $E\sqrt{\langle M,M\rangle_s}\leq \sqrt{E\langle M,M\rangle_s}$, and then it is enough to see that $E\langle M,M\rangle_s<\infty$. Hence,
$$
E\langle M,M\rangle_s = E\int_0^t s^2 X_s^3 ds.
$$
Since $X_s\geq 0$, we can apply Fubini's theorem, and hence
$$
 E\int_0^t s^2 X_s^3 d s = \int_0^t s^2 EX_s^3 ds.
$$
My question is: How can I asssure that $EX_s^3 <\infty$? Is there any other way of doing it? Thank you in advance and I hope you find it interesting.

答:
There are general statements on the existence of moments of solutions to SDEs. If you are lucky, you have one of those results at your disposal to deduce that $\sup_{s \leq t} \mathbb{E}(|X_s|^3)<\infty$ for all $t>0$.  If this is not the case, then you need to do some further calculations.

In the following, I will consider $\mathbb{E}(X_t^4)$ (rather than $\mathbb{E}(|X_t|^3)$) since this saves us the trouble to deal with the modulus. By Itô's formula, we have

$$X_t^4 = 8 \int_0^t X_s^3 \sqrt{X_s} \, dB_s +30  \int_0^t X_s^3 \, ds.$$

Now we want to take the expectation, but since we do not yet know whether the expectation is finite, we first need to some stopping. To this end, define

$$\tau_r := \inf\{t \geq 0; |X_t| \geq r\},$$

note that $\tau_r \uparrow \infty$ as $r \to \infty$ and $|X_{t \wedge \tau_r}| \leq r$. Since

$$t \mapsto \int_0^{t \wedge \tau_r} X_s \sqrt{X_s} \, dB_s$$

is a martingale (because of the stopping!), and, hence, has expectatio zero, we get

$$\mathbb{E}(X_{t \wedge \tau_r}^4) = 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} X_s^3 \, ds \right).$$

Using the elementary estimate

$$x^3 \leq |x|^3 \leq 1+x^4, \qquad x \in \mathbb{R},$$

for $x=X_s$, we find that

$$\mathbb{E}(X_{t \wedge \tau_r}^4) \leq 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} (1+X_s^4) \, ds \right) \leq 30 \int_0^t \mathbb{E}(1+X_{s \wedge \tau_r}^4) \, ds.$$

If we define $u(t) := \mathbb{E}(1+X_{t \wedge \tau_r}^4)$ for fixed $r>0$, then

$$u(t) \leq 1+ 30 \int_0^t u(s) \, ds$$

and so Gronwall's lemma yields

$$u(t) \leq c e^{Mt}$$

for suitable constants $c,M>0$, which do not depend on $r>0$. Consequently, we have shown that

$$\mathbb{E}(1+X_{t \wedge \tau_r}^4) \leq c e^{Mt}, \qquad t \geq 0.$$

By Fatou's lemma, this implies

$$\mathbb{E}(1+X_{t}^4) \leq c e^{Mt}, \qquad t \geq 0.$$

In particular,

$$\sup_{t \leq T} \mathbb{E}(X_t^4) < \infty, \qquad T>0,$$

which also yields

$$\sup_{t \leq T} \mathbb{E}(X_t^3) < \infty, \qquad T>0.$$


问:
Prove $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$
Show that $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$ for $x>0$. My idea is:

$(1+\frac{1}{x})^x(1+\frac{1}{x})>(1+\frac{1}{x})^x\rightarrow e$, as $x\rightarrow \infty $. So from the permanence theorem of the sign I conclude that $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$.

to do: check my idea

答:
Similar to @egreg's answer, consider that you look for the minimum value of
$$f(x)=(x+1)\log\Bigl(1+\dfrac{1}{x}\Bigr) $$ We have
$$f'(x)=\log \left(1+\frac{1}{x}\right)-\frac{1}{x}\qquad \text{and} \qquad f''(x)=\frac{1}{x^2(x+1)}$$ The first derivative is always negative and the second derivative is positive.

Expanding $f(x)$ as a series
$$f(x)=1+\frac{1}{2 x}-\frac{1}{6 x^2}+\frac{1}{12
   x^3}+O\left(\frac{1}{x^4}\right)$$
$$e^{f(x)}=e\left(1+\frac{1}{2 x}-\frac{1}{24 x^2}+\frac{1}{48
   x^3}+O\left(\frac{1}{x^4}\right)\right)$$


问:
If $B \subseteq A$ and $f:A \xrightarrow{1-1} B$, prove that $\bigcap_{n \lt \omega}f^n[A]=\bigcap_{n\lt \omega}f^n[B]$
If after confirming the validity of my proof, someone could comment on the possibility of creating a direct proof...that would be awesome. At any rate:

Prove that $\bigcap_{n \lt \omega}f^n[A]=\bigcap_{n\lt \omega}f^n[B]$ using the following information:

i. $B \subseteq A$

ii. $f:A \xrightarrow{1-1} B$

iii. $f^n$ is simply saying iterate $f$ $n$ times (and $f^0$ is the identity function $I_A$)

iv. $A \supseteq B \supseteq f^1(A) \supseteq f^1 (B) \supseteq f^2(A) \supseteq f^2(B) \supseteq...$

The proof to iv. can be found here (as well as a more formal definition of $f^n$): If $B \subseteq A$ and $f:A\overset{1-1}\longrightarrow B$ then $A \supseteq B \supseteq f^1(A) \supseteq f^1 (B) \supseteq f^2(A) \supseteq...$

Importantly, iv. amounts to proving the following two induction claims, which will be used in this proof:

Statement 1. $\forall n \in \omega\big( f^{n+1}[A] \subseteq f^n[B]\big)$

Statement 2. $\forall n \in \omega\big( f^n[B] \subseteq f^n[A]\big)$

Proceed by contradiction: assume that $\bigcap_{n \lt \omega}f^n[A]\neq \bigcap_{n\lt \omega}f^n[B]$ and derive a contradiction.

I have broken this into two cases (based on what is meant by "not equal"):

$ \Big [\bigcap_{n \lt \omega}f^n[A]\neq \bigcap_{n\lt \omega}f^n[B] \Big] \rightarrow \Big [\bigcap_{n \lt \omega}f^n[A]\nsubseteq \bigcap_{n\lt \omega}f^n[B] \ \lor \ \bigcap_{n\lt \omega}f^n[B] \nsubseteq \bigcap_{n \lt \omega}f^n[A] \Big]$

Case One: $\bigcap_{n \lt \omega}f^n[A]\nsubseteq \bigcap_{n\lt \omega}f^n[B]$

Case Two: $\bigcap_{n\lt \omega}f^n[B] \nsubseteq \bigcap_{n \lt \omega}f^n[A]$

Case One:

$\Big [\bigcap_{n \lt \omega}f^n[A]\nsubseteq \bigcap_{n\lt \omega}f^n[B] \Big ] \rightarrow \exists x \Big( x \in \bigcap_{n \lt \omega}f^n[A] \land x \notin \bigcap_{n\lt \omega}f^n[B] \Big ) $ Call one such instance $x'$.

$x' \in \bigcap_{n \lt \omega}f^n[A] \rightarrow \forall n \in \omega \Big (x' \in f^n[A] \Big)$ $\dagger$

$x' \notin \bigcap_{n\lt \omega}f^n[B] \rightarrow \exists n \in \omega \Big (x' \notin f^n[B] \Big )$ Call one such instance $n'$.

So we have $x' \notin f^{n'}[B]$

However, we proved earlier through induction (Statement 1) that $\forall n \in \omega\big( f^{n+1}[A] \subseteq f^n[B]\big)$.

Consider when $n=n'$: we then have $f^{n'+1}[A] \subseteq f^{n'}[B]$

By our earlier universal claim $\dagger$, we know that $x' \in f^{n'+1}[A]$

But $\Big[x' \in f^{n'+1}[A] \land f^{n'+1}[A] \subseteq f^{n'}[B] \Big ] \rightarrow x' \in f^{n'}[B]$.

A contradiction

Case Two works similarly to Case One, except we will use Statement 2 instead of Statement 1. With both claims leading to contradiction, we must conclude that $\bigcap_{n \lt \omega}f^n[A]=\bigcap_{n\lt \omega}f^n[B]$.

I will include the proof for completeness.

Case Two:

$\Big [\bigcap_{n \lt \omega}f^n[B]\nsubseteq \bigcap_{n\lt \omega}f^n[A] \Big ] \rightarrow \exists x \Big( x \in \bigcap_{n \lt \omega}f^n[B] \land x \notin \bigcap_{n\lt \omega}f^n[A] \Big ) $ Call one such instance $x'$.

$x' \in \bigcap_{n \lt \omega}f^n[B] \rightarrow \forall n \in \omega \Big (x' \in f^n[B] \Big)$ $\dagger\dagger$

$x' \notin \bigcap_{n\lt \omega}f^n[A] \rightarrow \exists n \in \omega \Big (x' \notin f^n[A] \Big )$ Call one such instance $n'$.

So we have $x' \notin f^{n'}[A]$

However, we proved earlier through induction (Statement 2) that $\forall n \in \omega\big( f^n[B] \subseteq f^n[A]\big)$

Consider when $n=n'$: we then have $f^{n'}(B) \subseteq f^{n'}(A)$

By our earlier universal claim $\dagger\dagger$, we know that $x' \in f^{n'}[B]$

But $\Big[x' \in f^{n'}[B] \land f^{n'}[B] \subseteq f^{n'}[A] \Big ] \rightarrow x' \in f^{n'}[A]$.

A contradiction $\square$

答:
By straightforward induction,
$$ f^{n+1}[A]\subseteq f^n[B]\subseteq f^n[A]$$
hence
$$ \bigcap_{n<\omega}f^n[A]\subseteq\bigcap_{0\ne n<\omega}f^n[A]=\bigcap_{n<\omega}f^{n+1}[A]
\subseteq \bigcap_{n<\omega}f^{n}[B]\subseteq\bigcap_{n<\omega}f^{n}[A].$$


问:
Not able to understand this graph for rational function .

I am not getting why there are 3 lines drawn in that way .

答: As the final sentence above "Example 2" says, those are the asymptotes of the function, lines which the graph approaches as $x$ approaches $\pm\infty$, or as $x\to\pm2$ (in which case $y$ approaches $\pm\infty$).


问:
Weak $L^{p}$ spaces are quasi-normed?
Let $(X,\mathcal{B}, \mu)$ be a measure space.  Then for $0< p < \infty$ by definition

$L^{p,\infty}(X,\mathcal{B}, \mu)$ is the class of all measureable functions $f$ such that

\begin{eqnarray*}
\|f\|_{p,\infty} &:=& \text{inf}\{c > 0: d_{f}(\alpha)\leq \frac{c^{p}}{\alpha^{p}}\text{ for all }\alpha > 0\}\\
 &=& \text{sup}\{\gamma d_{f}(\gamma)^{\frac{1}{p}}:\gamma > 0\}
\end{eqnarray*}
where
$$d_{f}(\alpha) = \mu(\{x\in X:|f(x)| > \alpha\})$$

I'm trying to verify that $\|\cdot\|_{p,\infty}$ is a quasi-norm on $L^{p,\infty}$.

The non-trivial thing to check is that for all $f,g\in L^{p,\infty}(X,\mathcal{B}, \mu)$, we have $\|f + g\|_{p,\infty} \leq c_{p}(\|f\|_{p,\infty} + \|g\|_{p,\infty})$, where $c_{p} = \text{max}\{2,2^{\frac{1}{p}}\}$.

For $1\leq p < \infty$, I was able to show that $\|f + g\|_{p,\infty} \leq 2(\|f\|_{p,\infty} + \|g\|_{p,\infty})$ using the supremum definition.

For $0 < p < 1$, I need to show that $\|f + g\|_{p,\infty} \leq 2^{\frac{1}{p}}(\|f\|_{p,\infty} + \|g\|_{p,\infty})$, but I'm stuck.

The idea is supposed to be to use the following property of $d_{f}$:

$$d_{f + g}(\alpha + \beta)\leq d_{f}(\alpha) + d_{g}(\beta)$$

which implies in particular that

$$d_{f + g}(\alpha)\leq d_{f}\left(\frac{\alpha}{2}\right) + d_{g}\left(\frac{\alpha}{2}\right)$$

Can anyone help me finish the proof?  Thanks in advance!

答:
Fix $\gamma>0$ and $0<p<1$. Since $t^{1/p}$ is an increasing function for $t>0$, then using the last inequality in your question we get
$$
\left(\frac{d_{f+g}(\gamma)}{2}\right)^{1/p}
\leq\left(\frac{d_{f}(\gamma/2)+d_{g}(\gamma/2)}{2}\right)^{1/p}
$$
Since $t^{1/p}$ is a convex function we get
$$\left(\frac{d_{f}(\gamma/2)+d_{g}(\gamma/2)}{2}\right)^{1/p}
\leq\frac{d_{f}(\gamma/2)^{1/p}+d_{g}(\gamma/2)^{1/p}}{2}
$$
Therefore
$$
d_{f+g}(\gamma)^{1/p}\leq 2^{1/p}\left(\frac{1}{2}d_f\left(\frac{\gamma}{2}\right)^{1/p}+\frac{1}{2}d_g\left(\frac{\gamma}{2}\right)^{1/p}\right)
$$
Multiplying by $\gamma$ we obtain
$$
\gamma d_{f+g}(\gamma)^{1/p}
\leq 2^{1/p}\left(\frac{\gamma}{2}d_f\left(\frac{\gamma}{2}\right)^{1/p}+\frac{\gamma}{2}d_g\left(\frac{\gamma}{2}\right)^{1/p}\right)
\leq 2^{1/p}(\Vert f\Vert_{p,\infty}+\Vert g\Vert_{p,\infty})
$$
Tking supremum over $\gamma>0$ in the last inequality we get
$$
\Vert f+g\Vert_{p,\infty}\leq 2^{1/p}(\Vert f\Vert_{p,\infty}+\Vert g\Vert_{p,\infty})
$$</p>

Title: Weak $L^{p}$ spaces are quasi-normed?


问:
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively.
So here is the Question :-

Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .

What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.

Any hints or explanations for this problem will be greatly appreciated !!

Body:
$f(x)$ leaves remainder $2x-5$ when divided by $x^2-1.$
Then,
$$f(1)=2(1)-5=-3$$
$$f(-1)=2(-1)-5=-7$$

$f(x)$ leaves remainder $-3x+4$ when divided by $x^2-4.$
So we have,
$$f(x)=(x^2-4)\,q(x)-3x+4$$
Since $f(x)$ is of degree $3$, the degree of $q(x)$ should be $1.$

Let $q(x)=ax+b.$ Then,
$$f(x)=(x^2-4)(ax+b)-3x+4$$
$$f(x)=ax^3+bx^2-(4a+3)x-(4b-4)$$
Let $x=1$ then,
$$f(1)=a(1)^3+b(1)^2-(4a+3)(1)-(4b-4)=-3$$
$$a+b=\dfrac{4}{3}\quad\quad\dots(1)$$
Let $x=-1$ then,
$$f(-1)=a(-1)^3+b(-1)^2-(4a+3)(-1)-(4b-4)=-7$$
$$a-b=-\dfrac{14}{3}\quad\quad\dots(2)$$
Solving (1) and (2) we get,
$$a=-\dfrac{5}{3}$$
$$b=3$$
Therefore,
$$f(x)=-\dfrac{5}{3}x^3+3x^2+\dfrac{11}{3}x-8$$
Now,
$$\underline{\underline{f(-3)=53}}$$


问:
Solve the inequality $\cos 2\theta -3\sin \theta-2\geq 0$ for $0^{\circ}\le \theta \le 360^{\circ}$.
This question is in A-levels chapter 3 trigonometry. I have no teacher as I am doing self studying. This sub chapter I learned about all the double angle formulae but I don't know how it is related and how to solve this question. If someone could help me that would be great. I got $-2\sin^2\theta-3\sin\theta-1\geqslant 0$ so far don't know if it is correct. then got −1⩽sin⩽−1/2 how do you solve this. Please help me I have been stuck on his for 2 hours.

答: If you're using a calculator do arcsin(-1) and arcsin(-1/2). Sin has a period of 360 degrees so if the calculator outputs something outside of $0\le\theta\le360$ you might need to add a multiple of 360.


问:
Expected value of pulling blue marble from sack on nth pull
I am doing a refresher for a work course and I'm a little stuck the question asks the following

You have a sack with 7 marbles inside 4 Blue and 3 Green

You draw all the marbles (without replacing them) and check their color.

If you let X represent the value of n if the first blue marble appears on the nth draw.

Obviously P(b) is 4/7 and P(g) = 3/7

My assumption was that:
4/7*(1+2+3+4+5+6+7) = (4/7)*1 + (4/7)*2 + (4/7)*3 ... (4/7)*7 = 16

But that seems entirely nonsensical value for X to take given that you can only pull 7 marbles from the sack.

I assume that is because the probability is obviously not static at 4/7.

If I look at the probability of the draws being green before being blue I end up with:
X=1 = 3/7
X=2 = 2/6
X=3 = 1/5

[(3/7)*1+ (2/6)*2 + (1/5)*3 ] = 1.67

But after this I'm a little stuck

Where am I going wrong?

答:
A hint

$X=1$ has probability $4/7$. Then $X=2$ has probability $3/7\times 4/6$ and so on.

Can you see what to do now?


答:
If $x=2$ or $x = -2$ you get $\frac {x^2-3x-2}{x^2 - 4}$ will have $x^2 - 4 = (\pm 2)^2 -4 = 0$ and the whole thing is undefined and can not be plotted.  For values where $x$ is close to $0$ then $x^2 - 4$ will be very small in magnitude and so $\frac 1{x^2- 4}$ will be very large in magnitude and the function will stretch to very large positive or negative values and as $x\to 0$ the $|\frac {x^2-3x-2}{x^2 - 4}| \to \infty$.  The vertical lines $x =2$ and $x=-2$ are the vertical asymptotes and as $x$ is very close to $\pm 2$ the graph will get very close to those vertical lines.

$y = 1$ is the horizontal asymptote.  As $x$ gets very very large then the the proportional difference between the numerator and the denominate will get insignificant and the value of $\frac {x^2 - 3x-2}{x^2 -4}$ will get closer and closer to $1$. (for example if $x=10$ you have $\frac {x^2-3x-2}{x^2-4} =\frac {100-30 -2}{100 -4} = \frac {68}{96}$ but and $x =100$ you have $\frac {x^2-3x -2}{x^2 - 4}=\frac {10000-300 -2}{10000 - 2}=\frac {9698}{9992}$.)

More formally $\frac {x^2 -3x -2}{x^2 - 4} = \frac {x^2 -4 -3x +2}{x^2 -4}=1 -\frac {3x-2}{x^2-4}$ and as $x \to \pm \infty$ we have $\frac {3x-2}{x^2 -4} \to 0$.

so the graph gets close to $y=1$ for large positive and negative values of $x$.

This is the nature of assymptotes.  It's pretty important you understand them.  Keep studying.

Title: Not able to understand this graph for rational function .

Body: I am not getting why there are 3 lines drawn in that way .


问:
Spanning trees implies isomorphism?
I'm having trouble solving this question:

Prove or disprove the following statement: Given a graph G, if T and U are spanning trees
of G, then T and U are isomorphic.

I know that two graphs are isomorphic if they have an "edge-preserving bijection." I also know that a spanning tree of a graph G is a connected graph that can be defined as a maximal set of edges of G that contains no cycle.

How would I go about proving or disproving?

答: Find two nonisomorphic trees with the same number of vertices. Draw them on the same vertex set. Their union is a graph with two nonisomorphic spanning trees.</p>

Title: Spanning trees implies isomorphism?


问:
Prove/disprove $(\int_0^{2 \pi} \!\!\cos f(x) \, d x)^2+(\int_0^{2 \pi}\!\!\! \sqrt{(f'(x))^2+\sin ^2 f(x)} \, dx)^2\ge 4\pi^2$

Let $f(x)$ be a differentiable function on $[0,2\pi]$ s.t. $0\leq f(x)\leq 2\pi$ and $f(0)=f(2\pi)$. Prove or disprove that
$$
\left(\int_0^{2 \pi} \cos f(x) \,d x\right)^2+\left(\int_0^{2 \pi} \sqrt{(f'(x))^2+\sin^2 f(x)} \, d x\right)^2 \geq(2 \pi)^2
$$

It seems that when $f$ is an arbitrary constant, the left side equals $(2\pi)^2$ and seems to be the minimum. But how can I show that there's no other $f$ that makes the left side equal (or be less than) $(2\pi)^2$?

A geometric interpretation of the inequality has been found: Consider a closed curve on a sphere: $C=\{(\cos x\cdot\sin f(x),\,\sin x\cdot\sin f(x),\,\cos f(x))\mid x\in[0,2\pi)\}$, we have its perimeter $\displaystyle L=\int_0^{2\pi}\sqrt{(f'(x))^2+\sin^2 f(x)}\,dx$ and its area $\displaystyle S=2\pi-\int_0^{2\pi}\cos f(x)\,dx$. From spherical isoperimetric inequality $L^2\ge S(4\pi-S)$, we have $(2\pi-S)^2+L^2\ge(2\pi)^2$, and the equality holds iff $C$ is any circle on the sphere. In this way we get the original inequality in the sense of geometry.

Now the question is, how to prove the inequality with only pure analysis methods?

答:
<This is not an answer!! This is just some of my thoughts, and also I am not good at English. If there are any things to fix, please be my guest.>

Prediction: For two functions $A(x)$ and $B(x)$, $$(\int_{\alpha}^{\beta} A(x) dx)^2+(\int_{\alpha}^{\beta} B(x) dx)^2 \geq (\int_{\alpha}^{\beta} \sqrt{A(x)^2+B(x)^2} dx)^2 $$
holds.

If this is true(which I am not able to prove...), the problem can be solved easily.

Let $A(x)=\cos f(x)$ and $B(x)=\sqrt{{f'(x)}^2+\sin^2 f(x)}$, and $\alpha=0, \beta=2\pi$.

Then, by the Prediction, $$(\int_{0}^{2\pi} \cos f(x) dx)^2+(\int_{0}^{2\pi} \sqrt{(f'(x))^2+\sin^2 f(x)} dx)^2\geq (\int_{0}^{2\pi} \sqrt{(f'(x))^2+1}  dx)^2 $$
We know that the right hand side is the form of the length of a curve $f(x)$.

As $y=f(x)$ satisfies $f(0)=f(2\pi)$, the shortest length of the curve $y=f(x)$ in the interval $[0, 2\pi]$ would be just a simple line connecting $(0, f(0))$ and $(2\pi, f(2\pi)=f(0))$, so the length will be just the x-coordinate difference, which is $2\pi$. Therefore, the RHS will have the minimum of $(2\pi)^2=4\pi^2$, and the problem is solved.

However, I couldn't think of a sharp way to prove this Prediction, or even it is true at all. I managed to think of the integrals as sequences, and tried solving this.
$$(\sum_{k=0}^{n} a_k)^2+(\sum_{k=0}^{n} b_k)^2 \geq (\sum_{k=0}^{n} \sqrt{(a_k)^2+(b_k)^2})^2$$
I think this equation can be solved by mathematical induction, but I am not sure.


问:
Prove that the splitting field of $x^p-1$ for p an odd prime contains a unique intermediate field of degree 2
Let $p$ be an odd prime. Let $F$ be the splitting field of $x^p-1$ over $\mathbb{Q}$. Prove that there is a unique field $K$ bewtween $\mathbb{Q}$ and $F$ which is of degree 2 over $\mathbb{Q}$.

I know that $F = \mathbb{Q}(\xi_p)$ and  $[F : \mathbb{Q}] = p-1$, which is divisible by 2. My intuition tells me that this field $K$ corresponds to the subgroup of order 2 of $D_n$, which would be generated by complex conjugation but I don't know how to go abour showing it.

I appreciate any help.

答: As you noted, the splitting field is $\mathbb{Q}(\zeta_p)$. The Galois group of $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ is $(\mathbb{Z}/p\mathbb{Z})^{\times}$, which is cyclic. Therefore it has a unique subgroup of index $2$, and thus by Galois correspondence, there is a unique sub-extension of degree $2$.


问:
What does "well defined" mean?
On page 153 of Linear Algebra Done Right the second edition, it says:

Define a linear map $S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$ by:

7.43:   $S_1 (\sqrt{T^* T}v)=Tv$

First we must check that $S_1$ is well defined. To do this, suppose $v_1, v_2 \in V$ are such that $\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$. For the definition given by 7.43 to make sense, we must show that $Tv_1=T v_2$.

It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?

Thanks

答:
What exactly a function $f:X\to Y$ is? Typically we define it as a subset of $X\times Y$ such that for any $x\in X$ there is precisely one $y\in Y$ such that $(x,y)\in f$.

And so "well defined function" means: "a subset of $X\times Y$ we just defined is actually a function" which boils down to showing that (a) for any $x\in X$ there is $y\in Y$ such that $(x,y)\in f$ and (b) if $(x,y)\in f$ and $(x',y)\in f$ for some $x,x'\in X$ then $x=x'$.

Or equivalently for any $x\in X$ the set $\{y\in Y\ |\ (x,y)\in f\}$ has exactly one element.

A common example is when we deal with equivalence relationships. For example consider integers $\mathbb{Z}$ with the following relationship: $x\sim y$ if and only if $2$ divides $x-y$. Now consider the quotient set $X=\mathbb{Z}/\sim$ and

$$f:X\to\mathbb{Z}$$
$$f([x]_\sim)=x$$
$$g:X\to\mathbb{Z}$$
$$g([x]_\sim)=x\text{ mod }2$$

Our first $f$ is not well defined. Because $[0]_\sim=[2]_\sim$ but $f([0]_\sim)=0$ while $f([2]_\sim)=2$ are different values for the same argument.

But our $g$ is well defined. That's because $[x]_\sim=[y]_\sim$ if and only if $2$ divides $x-y$. Which is if and only if $(x-y)\text{ mod }2=0$ and this is if and only if $x\text{ mod }2=y\text{ mod }2$.

Title: What does "well defined" mean?