问:
Prove that $S$ has the same property $P_k$ of $majority$ for all positive integers $k$.

Let $n$ be a positive integer and let $S \subseteq \{0, 1\}^n$ be a set of binary strings of length $n$. Given an odd number $x_1, \dots, x_{2k + 1} \in S$ of binary strings (not necessarily distinct), their $\textit{majority}$ is defined as the binary string $y \in \{0, 1\}^n$ for which the $i^{\text{th}}$ bit of $y$ is the most common bit among the $i^{\text{th}}$ bits of $x_1, \dots,x_{2k + 1}$. (For example, if $n = 4$ the majority of 0000, 0000, 1101, 1100, 0101 is 0100.)

Suppose that for some positive integer $k$, $S$ has the property $P_k$ that the majority of any $2k + 1$ binary strings in $S$ (possibly with repetition) is also in $S$. Prove that $S$ has the same property $P_k$ for all positive integers $k$.

This is my progress..

We will prove it by induction .

For base case , take $n=2$, which can be verified by checking .

Suppose it's true for $n=l$ i.e for any $S \subseteq \{0, 1\}^l$ be a set of binary strings of length $l$ ,when satisfying the property $p_k$ that for some positive integer $k$ ,then $S$ will have   property $p_k$  for all $k$.

Now we will show that for any  $S \subseteq \{0, 1\}^{l+1}$ will be a set of binary strings of length $l+1$  , which satisfies the property that for some positive integer $k$ , $S$ has property $p_k$ , we will show it's true for all $k$

Now Consider a new set $S'$ which formed by deleting the last digit of the strings in $S$
and also consider another new set $S''$ which is formed by deleting the first digit of the strings in $S$

then note that both $S'$ and $S''$ are $ \subseteq \{0, 1\}^l$ will be a set of binary strings of length $l$.

Now since we were given that $S$ satisfies the property that for some positive integer $k$  ,so   $S'$ and $S''$ will also satisfy for property $p_k$ for some $k$ and since $S'$ and $S''$ are $ \subseteq \{0, 1\}^l$ are a set of binary strings of length $l$ ,by induction hypothesis $S'$ and $S''$ satisfies property of $p_k$ for all $k$.

After this I couldn't have nice progress. Thanks in advance.

答:
So well, the idea of removing bits from the sequences and forming sets like $S',S''$ sounds cool! However, what you showed that $S',S''$ follow $P_k$ for all $k$s can actually be generalized a bit. Like we would have the same result even if we remove $i$th bit from every binary sequence of $S$. So lets define
$$S_i:=\{(a_1a_2\ldots a_{i-1}a_{i+1}\ldots a_{l+1})_2 | (a_1a_2\ldots a_{l+1})_2\in S\}$$
So basically $S'\equiv S_{l+1}$ and $S''\equiv S_{1}$. Now, its easy to see that from the induction hypothesis, $S_i$ satisfies $P_n$ for all $n$. Now, we need to show that $S$ satisfies $P_n$ for all $n$ as well. So FTSOC, assume that $S$ doesn't follow $P_m$ for some $m$. Thus, there exists a sequence of $2m+1$ binary numbers $B_i$ of length $l+1$ for which
$$(B_1,B_2,B_3,\ldots, B_{2m+1})\in S^{2m+1}\text{ for which }\mathcal{M}(B_1,B_2,\ldots,B_{2m+1})\notin S$$ where $\mathcal M$ denotes the majority sequence. However, we know that
$$\mathcal{M}(B_1(i),B_2(i),\ldots, B_{2m+1}(i))\in S_{i}$$where $B_j(i)$ is the corresponding binary number of $B_j$ of $S$ in $S_i$ (i.e. removing the number at $i$th bit in $B_j$). Let the $i$th bit in $B_j$ be $z_j(i)$ and let,
$$A_i:=|\underbrace{\mathcal M(z_1(i),z_2(i),\ldots , z_{2m+1}(i))}_{\text{we call this number $\omega(i)$}}-1|$$Also, let,
$$\mathcal{M}(B_1(i),B_2(i),\ldots, B_{2m+1}(i))=(b_1b_2\ldots b_{i-1}b_{i+1}\ldots b_{l+1})_2$$
and hence,
$$(b_1b_2\ldots b_{i-1}A_ib_{i+1}\ldots b_{l+1})_2\in S,~\forall i\in\{1,2,\ldots, l+1\}$$
because $\underbrace{(b_1b_2\ldots b_{i-1}\omega(i)b_{i+1}\ldots b_{l+1})_2}_{\text{note that this is nothing but $\mathcal M(B_1, B_2,\ldots, B_{2m+1})$}}\notin S$.
Now, by we know that $S$ follows $P_k$ for some $k$. Let
$$X_i:=(b_1b_2\ldots b_{i-1}A_ib_{i+1}\ldots b_{l+1})_2$$
and thus, as $(X_1,X_2\ldots, X_{l+1})\in S^{l+1}$ and $l\geq 3$, applying the property $P_k$ on binary numbers $X_1,X_2,X_1,X_2,\ldots , X_1, X_2, X_3$ (here number of $X_1$'s and $X_2$'s used are $k$ each and one $X_3$), we get
$$\mathcal{M}(X_1,X_2,X_1,X_2,\ldots , X_1, X_2, X_3)\in S$$however, its easy to note that this majority is nothing but $\mathcal M(B_1,B_2,\ldots ,B_{2m+1})$. Thus, $\mathcal M(B_1,B_2,\ldots ,B_{2m+1})\in S$ which is a contradiction. Thus, $S$ follows $P_k$ for all $k$. This completes the proof.$$\tag*{$\blacksquare$}$$


问:
Binomial series problem $\lim_{n\to\infty}\left[\sum_{i,j}{2i\choose i}{2j\choose j}\right]^{1/n}=4$
Prove that

$$\lim_{n\to\infty}\Bigg[\sum_{i,j}{2i\choose i}{2j\choose j}\Bigg]^{1/n}=4$$
such that, $i+j=n$
and $0\le i,j \le n$.

My idea was to first sum the series when $i<j$ and when $j<i$. But I am unable to proceed. Please enlighten me.

答:
$$\begin{aligned}\lim_{n\to \infty}\left[\sum_{i+j=n\\ 0\le i,j\le n}{2i\choose i}{2j\choose j}\right]^{1/n}&=\left[\left(x^{i} \text{ in }(1+x)^{2i}\right)\left(x^{j} \text{ in }(1+x)^{2j}\right)\right]^{1/n}\\&= \left(x^{i+j} \text{ in }(1+x)^{2(i+j)}\right)^{1/n}\\ &= \left(x^{n}\text{ in } (1+x)^{2n}\right)^{1/n}=\lim_{n\to \infty}\left[{2n\choose n}\right]^{1/n}\end{aligned}$$

Now it is a trivial task to show that $\left[{2n\choose n}\right]^{1/n}\to 4$ as $n\to \infty$.

Edit:
I must add that the trivial task of the computation of the limit of the binomial coefficient, can be done by a simple application of Stirling's Approximation.

ContentLicense: CC BY-SA 4.0


问:
Solving $\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$ without series expansion or L'Hopital's rule
I want to find the following limit without using series expansion or L'Hopital's rule. I tried replacing $x$ with $2x$.

$$\begin{aligned}L_{1}&=\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)\end{aligned}$$

$$\begin{aligned}L_{2}&=\lim_{x\to 0}\left(\frac{4x+2x(\cos^2x-\sin^2x)-6\sin x\cos x}{32\sin x\cos x}\right)\end{aligned}$$

How to proceed. Any hints are appreciated. Thanks.

答:
HINT

To simplify the derivation we can proceed starting to prove the followings

$$L_1: \frac{\cos x-1}{x^2}\to \frac12 \implies L_2: \frac{x\cos x-x+\frac12x^3}{x^5}\to \frac1{24}$$

$$L_3:\frac{x-\sin x}{x^3}\to \frac16 \implies L_4:\frac{3(x-\sin x)-\frac12 x^3}{x^5}\to -\frac1{40}$$

and then use that

$$\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{x}{\sin x}\left(\frac{x\cos x-x+\frac12x^3}{x^5}+\frac{3(x-\sin x)-\frac12 x^3}{x^5}\right)$$

Refer to the related

Are all limits solvable without L'Hôpital Rule or Series Expansion


问:
Asymptotic expansion of an integral with exponent times another function
According to this book page number 192

The asymptotic expansion of

$$\int_{0}^{\infty}[1-t \log t+(1-C-2 \log 2) t] e^{-2 t \log \left(\mu H / T_{1}\right)} d t$$

is

$$\frac{1}{2 \log \frac{\mu H}{T_{1}}}-\frac{\log 2}{\left(2 \log \frac{\mu H}{T_{1}}\right)^{2}}+\frac{\log \log \frac{\mu H}{T_{1}}}{\left(2 \log \frac{\mu H}{T_{1}}\right)^{2}}+\cdots$$

How is this computed?

Here is my try

The integrand can be written as

$$e^{{-2 t \log \left(\mu H / T_{1}\right)}+\log\left([1-t \log t+(1-C-2 \log 2) t] \right)}$$

Now, most of the contribution comes from the extremum of the function
which is given by

$$0=\frac{-C-\log (t)-2 \log (2)}{t (-C+1-2 \log (2))-t \log (t)+1}-2 \log \left(\frac{H \mu }{\text{T1}}\right)$$

Now, I do not know how to invert this relation to find an expression for $t_{ext}$.

Is there any other method to calculate the asymptotes other than the steepest decent method?

答:
Assuming that $\Re(\lambda )>0$, we have
$$\int_0^\infty (a t-t \log (t)+1)\,e^{-\lambda  t}=\frac{\lambda +\log (\lambda )+(a+\gamma -1)}{\lambda ^2}$$ which is exact and gives your expression.


问:
Show that the harmonic conjugates are collinear - Menelaus and Ceva's Theorems with homogeneous coordinates
Suppose we have a triangle $\triangle ABC$ such that $D$ is an arbitrary point on $BC$, $E$ is an arbitrary point on $AB$ and $F$ is an arbitrary point on $AC$. Let $G$ be the harmonic conjugate of $E$ with respect to $AB$ (the cross ratio $(A,B,E,G)=-1$), $I$ the harmonic conjugate of $D$ with respect to $BC$ (the cross ratio $(B,C,D,I)=-1$) and $H$ the harmonic conjugate of $F$ with respect to $AC$ (the cross ratio $(A,C,F,H)=-1$). Show that $G,I,H$ are collinear if and only if $AD$, $CE$ and $BF$ have a point in common.

The picture would look like this (I amended the pic as pointed out in the comments).

My idea is to convert $A,B,C$ to homogeneous coordinates, for example, $A=(1:0:0), B=(0:0:1)$ and $C=(0:1:0)$, but I am not sure where to go from there. I read that these are equivalent statements from Menelaus and Ceva's theorems but I cannot find anything on the internet with homogeneous coordinates, so any help is appreciated.

答:
First have a look here showing that this property, called "trilinear polarity" is explained using trilinear coordinates (cousin of barycentric coordinates) with the very nice correspondence:

$$\text{Common point's trilinear coord.} \ (p,q,r) \ \leftrightarrow \ \text{line GIH trilinear equ. } \ \tfrac{x}{p}+\tfrac{y}{q}+\tfrac{z}{r}=0$$


问:
How to estimate the oscillation range of this differential equation
Consider the system of differential equation
$$\left\{\begin{aligned}\frac{\mathrm{d}x}{\mathrm{d}t} &= y\\\frac{\mathrm{d}y}{\mathrm{d}t} &= x - x^3 - y^3\end{aligned}\right.$$
with initial value $x(0) = y(0) = 1$.

Numerical simulation shows that $(x, y)$ oscillates around $(1, 0)$, with no sign of convergence even at $t = 1000$. I would like to estimate the range of this oscillation.

My attempt: Let $t_0$ be the smallest positive $t$ at which $y(t) = 0$. Then when $0 \leq t \leq t_0$ we have $y > 0$, hence $x$ is increasing, so $x \geq 1$. Then $x - x^3 < 0$, so $y$ is decreasing, $y \leq 1$.

Note that
$$\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\mathrm{d}y}{\mathrm{d}t} = x + y - x^3 - y^3 = (x + y)(1 - x^2 - xy - y^2)$$
Since $x \geq 1$, we have $1 - x^2 \leq 0$, so $x + y$ is decreasing, so $x + y \leq 2$. We have $\mathrm{d}y/\mathrm{d}t \geq x - x^3 - (2-x)^3 =
 -6x^2+13x-8$.

Let $t_1$ be the smallest positive $t$ such that $x(t_1) = 7/6$. When $0 \leq t \leq t_1$, we have $\mathrm{d}y/\mathrm{d}t \geq -1$, so $y \geq 1-t, x \geq -t^2/2 + t + 1$. Hence $t_1 \leq 1 - \sqrt{6}/3$.

On the other hand $t_0 \geq 1 - \sqrt{6}/3$, because whenever $t \leq t_0$ and $t \leq 1/2$, we have $x \leq 1 + t \leq 3/2$, so $\mathrm{d}y/\mathrm{d}t \geq -2$, hence $y \geq 1 - 2t$. If $t_0 < 1/2$, then $y(t_0) \geq 1 - 2t > 0$, contradicting $y(t_0) = 0$. Hence $t_1 \leq 1 - \sqrt{6}/3 < 1/2 \leq t_0$.

I'm not sure how to proceed. Numerical experiment seems to show that $\mathrm{d}y/\mathrm{d}t \leq -1$ when $t_1 \leq t \leq t_0$, but I don't know how to prove that. If that's the case, then since $y(t_1) \leq 2 - x(t_1) = 5/6$, we have $t_0 \leq t_1 + 5/6 \leq 11/6 - \sqrt{6}/3$. Since $y(t_1 + t) \leq 5/6 - t$, we have $x(t_1 + t) \leq 1 + t_1 + 5/6t - 1/2t^2$. Plug in $t_1 \leq 1 - \sqrt{6}/3$ and $t \leq 5/6$ to get $x(t_0) \leq x_0$ where $x_0 \approx 1.531$. That's pretty close to what I saw in the numerical experiment.

How to estimate the next $x$ value where $\mathrm{d}x/\mathrm{d}t = y = 0$? The long-term behavior of this equation seems to be that $x$ oscillates between these two extreme values. Also, can we say anything about the behavior when $t$ goes to negative? Mathematica thinks this is a stiff system when $t \leq -0.55$

答:
Mechanical interpretation

As per the comment, you can see that as a second order equation of the form
$$
x''+(x'^2)x'+(x^3-x)=0,
$$
which can be interpreted/visualized as a mechanical system with a conservative force $F(x)=-V'(x)$ where $V=\frac14(x^2-1)^2$ and a friction/dissipation term that has a coefficient $c(x,x')=x'^2\ge 0$. This means that the system is continuously losing energy as long as it moves, crossing the level curves of
$$
E(x,x')=\frac12x'^2+V(x)
$$
towards ever lower energies to end up in one of the minima $x=\pm 1$ of the potential energy.
$$
\frac{d}{dt}E(x,x')=x'(x''+V'(x))=-x'^4.
$$
However, the closer to the minimum the solution gets, the slower it gets and thus the smaller the energy loss term. Thus it may look as if the solution becomes periodic. In the other time direction the dissipation term now "collects" energy from the environment, making the solution behave more and more lively. Now the 4th power of $x'$ gives a positive feed-back which in general leads to a divergence to infinity in finite time. Then also all derivatives take large values, so the system may appear as "stiff", but more correctly it would be "run-away".

A first approach to the energy decay

To be a little more precise, close to the equilibrium one can try $x'^2\approx 2E$ for the qualitatively close equation
$$
E'=-4E^2\implies E=\frac1{c+4t}.
$$
This is likely quantitatively deviating from numerical results, what should stay true is that $E^{-1}$ is approximately linear.

A second approach using averaging over a period

As the distance to the equilibrium decays slowly, one can average over periods. The linearization close to $x=1$ gives $u''+2u=0$ for $x=1+u$. This has solutions $x=1+r\sin(\sqrt2 t+\phi)$ and $x'=\sqrt2r\cos(\sqrt2 t+\phi)$ so that $E=r^2+O(r^3)$. Then the average of
$$
x'^4=4r^4\cos^4(\sqrt2 t+\phi)=\frac{r^4}2(\cos(4\sqrt2 t+4\phi)+4\cos(2\sqrt2 t+2\phi)+3)
$$
over the period $\frac{2\pi}{\sqrt2}$ is the constant term, $\frac32r^4$. This implies that $E^{-1}$ or $r^{-2}$ grows like $c+\frac32t$, which is slower than the first guess above.

The plot of $E^{-1}-\frac32t$ about confirms this behavior.


问:
Is a complex-valued function on an interval constant iff its derivative is zero?
I know that this is true for real-valued functions. However, the proof is based on the mean value theorem and the latter is not valid for complex-valued functions (here is a counter-example).

That being said, the following generalization was proven in one of my lectures:

Let $V$ and $W$ be normed vector spaces over the same field $F$. Let $U\subset V$ be open and connected, then a map $f\colon U\to W$ is constant iff $Df(x)=0$ for all $x\in U$, where $Df(x)\in L(V,W)$ is the Fréchet-derivative.

In the title of my question, I consider the case $V=\mathbb{R}$ and $W=\mathbb{C}$. The problem is that the Fréchet derivative is not defined when $V$ is real vector space and $W$ is a complex vector space, since the set $L(V,W)$ is only defined when $V$ and $W$ are vector spaces over the same field.

Is there another definition of the derivative in this situation and does it allow a similar theorem?

答:
I found an answer for the case $V=\mathbb R$ and $W=\mathbb C$. Yes, there is a similar theorem.

Lemma. $f\colon U\subset\mathbb R\to\mathbb C$ is differentiable if and only if $u:=\text{Re}f$ and $v:=\text{Im}f$ are differentiable and $f'=(u+iv)'=u'+iv'$.

Corollary. $f'=0\Leftrightarrow u'=0$ and $v'=0\Leftrightarrow u$ and $v$ are constant $\Leftrightarrow f$ is constant.

Proof:

To proof the Lemma, we prove the following two Lemmas:

Lemma 1. If $u,v\,\colon U\to\mathbb R$ are differentiable, $f=u+iv$ is differentiable and $f'=u'+iv'$.

Proof.
\begin{equation}
\left|z-\frac{f(a+\delta)-f(a)}{\delta}\right|\leq\left|x-\frac{u(a+\delta)-u(a)}{\delta}\right|+\left|y-\frac{v(x+\delta)-v(x)}{\delta}\right|
\end{equation}
Lemma 2. If $f$ is differentiable, $u:=\text{Re}f$ and $v:=\text{Im}f$ are differentiable and $f'=u'+iv'$.

Proof.
\begin{equation}
\left|x-\frac{u(a+\delta)-u(a)}{\delta}\right|=\left|\text{Re}\left(z-\frac{f(a+\delta)-f(a)}{\delta}\right)\right|\leq\left|z-\frac{f(a+\delta)-f(a)}{\delta}\right|
\end{equation}
\begin{equation}
\left|y-\frac{v(a+\delta)-v(a)}{\delta}\right|=\left|\text{Im}\left(z-\frac{f(a+\delta)-f(a)}{\delta}\right)\right|\leq\left|z-\frac{f(a+\delta)-f(a)}{\delta}\right|
\end{equation}


问:
Convergence of $\max$ and $\min$ sequences of first $n$ terms of a convergent sequence
Let $(x_n)$ be a sequence of real numbers converging to $x$.

Define sequences $y_n:=\max\{x_1,x_2,\ldots,x_n\}$ and $z_n:=\min\{x_1,x_2,\ldots,x_n\}$

Now do the sequences $(y_n),(z_n)$ converge? If $(x_n)$ is monotonic , then both  these sequences are convergent. What can we say in general case? Please help me with this. Thanks!

答:
$(y_n)$ is monotically increasing. It is also bounded: There exists $N$ such that $\lvert x_n - x \rvert < 1$ for $n \ge N$. Thus $x_n < x + 1$ for $n \ge N$ and thefore $y_n \le \max(y_N,x+1)$ for all $n$. Thus $(y_n)$ converges to some $y \in \mathbb R$. Similarly $(z_n)$ converges to some $z \in \mathbb R$.

However, $z < y$ unless $(x_n)$ is constant: If $(x_n)$ is not constant, then $x_N \ne x_M$ for suitable $N, M$. Let $R = \max(N,M)$. Then $z_R < y_R$ and thus
$$z \le z_R < y_R \le y .$$


问:
Painting a cube using N different colors?

In how many different ways can a cube be painted by using N different colors of
  paint?

Note that this question is not same to Painting the faces of a cube with distinct colours as the colours here may not be distinct (i.e. used many times). For Painting the faces of a cube with distinct colours, we must have $N>=6$, and the result will be:

$$\binom{n}{6}(5)(3!)$$

Can anyone help how to tackle such problem?

答:
I'm not sure if the answer is correct, but here's some of my idea:
if we are painting a cube with 6 distinct colours.
fixing one of the faces, the face on the opposite side has 5 colour choices
using the circular arrangement for the rest of the 4 sides we have 3! ways.
In total, we have 5*6=30 ways.

Now if we have n colours, after fixing the first face the opposite face has (n-1) colour choices. There are now (n-2) colours left to choose and for the four faces left we might choose 4 colours from the (n-2) colours and do the circular arrangements again.

In total, we have (n-1)*[(n-2)C4]*3! ways.

ContentLicense: CC BY-SA 3.0


问:
Quadrilateral with given angles

We are looking for angles x and y.

I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,

plus the relationship $x+y = 68$.

All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!

I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!

Does anyone have any ideas?

Thank you!

答:
my geometric solution:
http://www.davdata.nl/math/geopuzzle28.html

arc calculation

arc calculation(2)

Title: Quadrilateral with given angles


问:
Laurent Expansions about $z_0=0$
I'm really struggling to understand the concept of Laurent Expanaions.. I have gone throu lectures notes and a couple of examples, but don't seem to be able to compute fully understand..

Any help with the following example would be greatly appreciated...

$f(z)= \dfrac{1}{z(z-1)(z-2)} $

On

(i) $0< |z| < 1$

(ii) $ 1< |z|< 2$

(iii) $ 2<|z|$

Workings

I have worked out the partial fractions for this equation as follows....

$\dfrac{1}{z(z-1)(z-2)} = \dfrac{A}{z} + \dfrac{B}{z-1} + \dfrac{C}{z-2}$

$\dfrac{1}{z(z-1)(z-2)}= \dfrac{1}{z} -\dfrac{1}{z-1}+ \dfrac{1}{2(z-2)}$

But now I'm lost as to what to do..

答:
Hint:

Expand separately $\dfrac{1}{z-1}$ and $\dfrac{1}{z-2}$ using the geometric series and taking into account the domain of convergence.

Title: Laurent Expansions about $z_0=0$


问:
$20\{x\} = x + [x] + [x + 0.5]$

Let $A = \{x \, | \; 20\{x\} = x + [x] + [x + 0.5] \, \}$, if S = sum of elements of $A$, find $[S]$ where $[x]$ integer part of $x$ and $\{x\} = x - [x]$.

Source: Stage 1 of India MO

I was actually trying to get better at solving these types of questions involving [.] and {x} so i found it confusing.

My attempt:

I took two cases, one for $\{x\}<0.5$ and another for $\{x\}>0.5$ and on writing $[x]$ as $x-\{x\}$, I came up with $19x = 22[x]$ for one of the cases but I am not sure what to do next.

If you could also tell me general methods to tackle such problems, I would be thankful. Thank you!

答:
Rewriting the constraint
$$19 \{x\} = [x] + [x] + [x + 0.5]$$
we see, that left part is in range $[0; 19]$, and right part is integer. Let that integer be $k$.

Now we can through all $k=0, 1, ..., 19$, and find $x$ for each of them (of course, not for all $k$ such $x$ exists). It's straightforward to express $\{x\}$ and $[x]$ in terms of $k$:

$$
\{x\} = \frac{k}{19} \\
k = [x] + [x] + [x + 0.5] = 2 [x] + [ [x] + \{x\} + 0.5] = \\
= 3 [x] + [\{x\} + 0.5] = 3 [x] + \left[ \frac{k}{19} + \frac12 \right] \\
[x] = \frac13 \left( k - \left[ \frac{k}{19} + \frac12 \right]\right)
$$

Now with some diligence (which I lack) we can find all $x$ in $A$ and get the answer.

Title: $20\{x\} = x + [x] + [x + 0.5]$


问:
Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's Rule.
How to solve $\lim\limits_{x\to 0} \frac{x - \sin(x)}{x^2}$ Without L'Hospital's Rule?
you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.</p>

答: 1.Divide both nominator and denominator by $(1-\cos x)$ and apply product rule for limits $$\lim_{x\to 0}\Biggl(\frac{\frac{x-\sin x}{1-\cos x}}{\frac{x^2}{1-\cos x}}\Biggl)\space=\space\lim_{x\to 0}\Biggl(\frac{x-\sin x}{1-\cos x}\Biggl)\space \lim_{x\to 0}\Biggl(\frac{1-\cos x}{x^2}\Biggl)$$ 2. Notice the standard limit in this expression and use it in the next step $$\lim_{x\to 0}\frac{1-\cos x}{x^2}\space=\space\frac12$$ 3.$$\frac12\lim_{x\to 0}\Biggl(\frac{x-\sin x}{1-\cos x}\Biggl)$$ 4. Notice that denominator is derivative of nominator. We can modify the equation a little bit by $$f(x)\space=\space x-\sin x$$ $$f'(x)\space=\space 1-\cos x$$ 5. If we substitute $x_0$ for $0$ and accept $\frac{f(x)}{f'(x)}$ is continuous at $x_0$ , we get $$\frac12\lim_{x\to x_0}\frac{f(x)}{f'(x)}\space=\space \frac{f(x_0)}{f'(x_0)}\space=\space (x-x_0)\space+\space f(x_0)$$  7. Finally substitute $0$ for $x_0$ $$\lim_{x\to 0}(x-0)\space+\space\lim_{x\to 0}f(0)\space=\space 0\space+\space f(0)\space=\space 0\space+\space 0 \space-\space \sin 0 \space=\space 0$$


问:
generator of first homology group of cylinder
Let $X = S^1 \times I$. I know that $H_1(X) = \mathbb{Z}$, which implies that $H_1(X)$ has 1 generator. It is intuitive to see that the generator of $H_1(X)$ is the circle that wraps around the cylinder. However, when I look at the definition of $C_1(X)$, it is the free module over the set of all continuous functions $\sigma: \triangle^1 \rightarrow X$. There are infinitely many $\sigma$,i.e,circle that wraps around $X$. So somehow, passing from $C_1(X)$ into $H_1(X)$, all of these circles become 1 equivalent class. Can anyone show me how this is true from the definition of chain complexes and homology.

答:
Observe that the cylinder $X= S^1\times [0,1]$ is homotopy equivalent to $S^1\times\{0\} = S^1$, therefore $$H_1(S^1\times[0,1]) = H_1(S^1) = \mathbb{Z}$$

Now if you want to justify $H_1(S^1\times[0,1]) = \mathbb{Z}$ purely from the properties of the singular chain complex $(C_\bullet(X), \partial_\bullet)$, it's much more complicated, that's because computing singular homology groups directly from the chain complex is complicated.

Fortunately, we can equip the cylinder with a $\Delta$-complex structure. The simplicial homology $H_*^\Delta(X)$ which we get from the simplicial chain complex $(C_\bullet^\Delta(X), \partial_\bullet^\Delta)$ computes up to isomorphism singular homology $H_*(X)$.
That means, simplicial homology and singular homology are isomorphic homology theories
$$H_*^\Delta(X) \cong H_*(X).$$

Using the $\Delta$-complex structure of the cylinder allows you to directly compute the homology groups $H_n(X)$ from the respective chain groups $C_n(X)$.

The $\delta$-complex structure of the cylinder is pictured below

$\hskip2in$

Your chain groups are therefore $$C_0(X) = \langle v,w\rangle,\quad C_1(X) = \langle a,b,c,d\rangle,\quad C_2(X) = \langle A,B\rangle$$

the boundary operator $\partial_n$ is defined analougously as for the singular chain complex, therefore you can evaluate $\partial_n$ on the given generators in each degree and compute your homology groups via $$H_1(X) = \ker \partial_1/{\operatorname{im} \partial_{2}}.$$


问:
Questions about foundations of mathematics
It seems to me that in trying to create a foundation of mathematics, mathematicians are trying to create a formal system that models the language used in what I view as common-sense, real mathematics. Whatever the number $1$ is, I doubt it's $\{\{\}\}$, no? The set containing the empty set is merely a representation of $1$ in this formalism. So then is a foundation of mathematics merely a formalism used to model 'real' mathematics? If not, what is it?

I thought I was in on a little-known secret when I was (perhaps not intentionally) lead to believe that $1$ is the same as $\{\{\}\}$, but now I see how preposterous a claim that is, especially considering that the empty set is probably but a linguistic construct.

If the answer to my first question is 'yes', then presumably the reason for the foundation is to check it for self-consistency, for if the foundation is self-consistent, the thought process probably goes, then 'real' mathematics is self-consistent, yes? How can one know that the foundation (always) parallels 'real' mathematics? How do we know that the two discourses are isomorphic? If we can't know then why bother with a foundation in the first place? If we can know then presumably it's due to some metalanguage, but then... is the metalanguage self-consistent? How do we know? Isn't this just kicking the can down the road?

答:
There are various approaches to a foundation of mathematics. Russell and Whitehead in Principia Mathematica defined $1$, for example, as the set of all singletons, not just $\{\{\}\}$.

Gödel showed there are limits to what we can prove with formal systems, e.g. there are true statements of arithmetic that are unprovable.

Research on  a foundation for mathematics using axiomatic systems, has led to some major results in logic and set theory. From a more practical perspective, first-order logic and the Simple Theory of Types are applied in program verification and hardware verification, e.g. Hoare Logic, HOL and Isabelle.


问:
Verification of proof involving the lcm of consecutive numbers.
$\newcommand{\lcm}{\operatorname{lcm}}$
Just a quick one.

I have to prove that the lcm of two consecutive numbers is its product.

Using the identity $\gcd(a,b) \cdot \lcm(a,b) = a \cdot b$, you can find $\lcm(a,b) = \frac{a \cdot b}{\gcd(a,b)}$

I can prove separately that the gcd of consecutive numbers is 1 (coprime) -- which leaves $a \cdot b$ and thus proving the statement.

I'm pretty sure this is complete but for the sake of clarity is there something I'm missing?

答:
Nothing. That is correct.

On the other hand, when two coprime numbers $a$ and $b$ divide a number $c$, then $ab\mid c$ too. So, $ab$ divides every common multiple of $a$ and $b$ and this also proves that $\operatorname{lcm}(a,b)=ab$, without using the fact that $\gcd(a,b)\operatorname{lcm}(a,b)=ab$.


问:
why Rudin said that let $\Delta$ be closed disc?
I have some confusion in RCA book

my confusion is given below and marked in red box

My attempt :  I think Rudin is wrong . Here  it  is given that  $S$ is closed   we know that complement of closed set is open  so $S^c$ is an  open  I think  it should  be let $\Delta$  be open disc because any open ball  is  a countable union of closed ball

My doubt : why Rudin said  that let $\Delta$  be closed  disc ?

答:
It doesn't really matter. First of all, it's true you could use an open disc and you would get $|f-\alpha|<r$, but then you would need to use some integral inequality to complete the proof anyway.

It think, but I'm not entirely sure, that at that point in the book you know that a positive function on a set with positive measure has positive Lebesgue integral. This is: if $g>0$ on a measurable set $A$ such that $\mu(A)>0$, then $\int_A g\,\text{d}\mu>0$. If you know this then you can use $|f-\alpha|<r|\Rightarrow r-|f-\alpha|>0\Rightarrow\int_E (r-|f-\alpha|)\,\text{d}\mu>0\Rightarrow\int_E |f-\alpha|\,\text{d}\mu<\int_E r\,\text{d}\mu$

and get $\left|A_E(f)-\alpha\right|<r$, which would be a contradiction.

Now, on the other hand, I am entirely sure the at that point in the book you know that $g\le h$ on $A$ implies $\int_A g\,\text{d}\mu\le\int_A h\,\text{d}\mu$. In a way, this property is "more fundamental" than the previous one, and that's why I think Rudin wants to use a closed disc, since that gives you $|f-\alpha|\le r$. But, why can he? Because any open set can also be generated as a countable union of closed discs.

Imagine you have an open set $A$ as a countable union of open discs $U_i$, each centred at $\alpha_i$ with radius $r_i$. So $A=\bigcup_i U_i$. Now take one $U_i$ and consider the collection of closed discs $V_{j_i}=\overline B(\alpha_i,r_i-\frac{1}{j})$. Each $V_{j_i}$ is a closed disc, but the countable union $\bigcup\limits_{j=1}^\infty V_{j_i}=\bigcup\limits_{j=1}^\infty \overline B(\alpha_i,r_i-\frac{1}{j})=B(\alpha_i,r_i)=U_i$. So you have each $U_i$ as a countable union of closed discs. Then $A=\bigcup_i U_i=\bigcup_i\big(\bigcup\limits_{j=1}^\infty V_{j_i}\big)$, which is the countable union of countable unions, hence countable. Therefore we have $A$ as a countable union of closed discs.


问:
How to decompose the intecept of a linear regression?
Problem

I'm given a linear regression model. I need to to obtain the decomposition of intercept coefficient in the form of

$$ \hat{\beta_1} = \beta_1 + \sum c_i u_i$$

where $c_i = 1/n - a_i\bar{X}$ and $a_i = \frac{(X_i - \bar{X})}{\sum_{i=1}^n (X_i - \bar{X})^2}$

Can anyone give some ideas?

答:
Note that I will be using $\beta_1$ for intercept and $\beta_2$ for slope since it is more appropriate to do so.

Using the common formula we can write: $\hat\beta_1 = \bar{Y} - \hat\beta_2\bar{X}$.

Now we can substitute $\bar{Y} = \beta_1 + \beta_2\bar{X} + \bar{u}\:\:$ (since we are given a linear regression)

And $\hat{\beta_2} = \beta_2 + \sum c_i u_i \:\:$ (which is given by the task):

$\hat\beta_1 = \beta_1 + \beta_2\bar{X} + \bar{u} - \bar{X}\beta_2 - \bar{X}\sum a_iu_i = \beta_1 + \bar{u} - \bar{X}\sum a_iu_i = \beta_1 + \frac{1}{n}\sum u_i - \sum \bar{X}a_iu_i = \beta_1 + \sum(\frac{u_i}{n} - \bar{X}a_iu_i) = \beta_1 + \sum(u_i(\frac{1}{n} - \bar{X}a_i)) = \beta_1 + \sum c_i u_i.$

The last step was the substitution of $c_i = 1/n - a_i\bar{X}$, which was given in the task.


问:
Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$
Find the group A such that  $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?

Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$

but how i can from here find all x such that $x \in \mathbb{Z}$ ??

i did use the hint for $y = 2x+1$ so that  $x =\frac{y-1}{2}$

made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$

$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$

I can not understand with the help of the clues, can some 1 some formal proof ?

Body:
$2x+1|x^3-3x+2\iff 2x+1|8(x^3-3x+2)$

$\iff 2x+1|8(x^3-3x+2)-(2x+1)^3=-12x^2-30x+15$

$\iff 2x+1|3(2x+1)^2-12x^2-30x+15=-18x+18$

$\iff 2x+1|9(2x+1)-18x+18=27$

$\iff x\in \{-14,-5,-2,-1,0,1,4,13\}.$


问:
Finding PDF and CDF and probability distribution for the transformation / change of RV
Consider a continuous RV $X$ with CDF $F_X(x) = e^{-(e^{-x})}$,

$\forall x \in R$ (note that $X$ takes values over the entire real line).

(a) Find the PDF of $X$

(b) Calculate the probability $P(|X| > 1)$.

(c) Define the transformation $Y = g(X) = e^{-X}$. Find the CDF of $Y$ and identity its distribution, i.e. give the name and parameter(s) of the distribution.

$\mathbf{My~Attempts:}$

(a) Let $f_X(x)$ be the PDF of $X$.

Since, by the $\mathbf{Properties~of~PDFs}$,

$~\hspace{11mm}$ we know that $F_X(x) = \int_{-\infty}^{x} f_X(u) du \implies f_X(x) = \frac{d}{dx} F_X(x) = F_X'(x)$.

So, I find that the PDF of $X$ is $f_X(x) = \begin{cases} e^{-x - e^{-x}}, & \text{if $x \geq 0$} \\ 0, & \text{if $x < 0$} \end{cases}$

(b) $P(|X| > 1) = P(X < -1 or X > 1) = 1 - P(-1 < X < 1) = 1 - (F_X(1) - F_X(-1)) \approx 0.3738$.

(c) I will use the PDF method which $f_Y(y) = \frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}$.

Since, $g^{-1}(y) = -log(y)$ and $g'(y) = -e^{-x}$.

So, $g'(g^{-1}(y)) = -e^{-(-log(y))} = -e^{log(y)} = -y$.

Also, $f_X(g^{-1}(y)) = e^{-(-log(y)) - e^{-(-log(y))}} = e^{log(y) - y} = y e^{-y}$.

So, $f_Y(y) = \frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|} = \frac{y e^{-y}}{|-y|} = e^{-y}$.

Since, I know that the Exponential Distribution $Z \sim Exponential(\lambda)$ has PDF of $f(z) = \begin{cases} \lambda e^{-\lambda z}, & \text{if $z \geq 0$} \\ 0, & \text{if $z < 0$} \end{cases}$,

$~\hspace{100mm}$ for some $\lambda > 0$.

So, I observe that $\lambda = 1$ has PDF of $f(y) = \begin{cases} e^{-y}, & \text{if $y \geq 0$} \\ 0, & \text{if $y < 0$} \end{cases}$.

Therefore, I can say that $Y \sim Exponential(1)$. Also, the CDF would be $F_Y(y) = \int_{-\infty}^{y} f_{Y} (v) dv = 1 - e^{-y}$ because from the plot of $f(x) = e^{-x}$, I notice that $x = 0$ makes $f(x) = 1$ which is the maximum value of $f(x)$.

$\mathbf{Questions:}$

$\mathbf{Is~that~my~attempt~of~part~(a)~to~(d)~all~correct~?}$

$\mathbf{And~are~there~anything~I~can~improve~or~add~?}$

Also, I notice a problem with (c), since I want to use the CDF method to get CDF of Y which $F_Y(y) = P(Y \leq y) = P(X \leq g^{-1}(y)) = F_X(g^{-1}(y)) = F_X(-log(y)) = e^{-(e^{-(-log(y))})} = e^{-y}$.  Which is not $1 - e^{-y}$. I am quite confuss on where I am doing wrong ?

答:
$(a)$ is not correct. The CDF is non-constant over $\Bbb R$ so the PDF is  not $0$ for $x<0$. It is $e^{-x-e^{-x}}$ over all of $\Bbb R$.

$(b)$ is okay.

Your answer to $(c)$ is okay because you have made an error here which offsets the error you made in part $(a)$. Note that in $(a)$, you took $X\ge0$ so $Y=e^{-X}\le1$. Thus the "PDF" you should have got is $e^{-y},0<y\le1$, which is not a PDF at all since it doesn't integrate to give $1$. The error you made here is that you didn't find the range of $Y$ according to the range of $X$ you wrote in $(a)$ and automatically assumed $Y\ge0$.

With the correct bounds on $X$ i.e. $-\infty<X<\infty$, $Y$ would vary from $0\to\infty$, which corresponds to the exponential dsitribution with parameter $1$.

So the PDF is$$f(y)=\begin{cases}e^{-y},&y>0\\0,&y\le0\end{cases}$$The CDF is$$F(y)=\int_{-\infty}^yf(u)du=\begin{cases}0,&y\le0\\\int_\color{red}0^yf(u)du=1-e^{-y},&y>0\end{cases}$$

The error you have made in the CDF approach is that $Y\le y$ does not translate to $X\le g^{-1}(y)$ for $g$ which is not strictly increasing. In our case, $Y\le y\iff e^{-X}\le y\iff -X\le\ln y\iff X\color{red}\ge-\ln y=g^{-1}(y)$.

So $P(Y\le y)=\begin{cases}0,&y\le0\\P(X\ge-\ln y),&y>0\end{cases}\\=\begin{cases}0,&y\le0\\1-P(X\le-\ln y),&y>0\end{cases}\\=\begin{cases}0,&y\le0\\1-F_X(-\ln y),&y>0\end{cases}$

which gives the same answer.


问:
If $G$ is a connected simple graph and mean of degrees is $2$. How many cycle in $G$?
Proof:

Let G is connected simple graph with n vertex and $\sum \dfrac{d(v)}{n}=2$.Since $\sum{d(v)} =2ε(G)$ then $n=ε(G)$.

I want to prove that $G$ has cycle by contradiction.Let $G$  has no cycle, I get $G$ is tree with $n$ vertex then $G$ has $n-1$ edges but $G$ has $n$ edges  that mean $G$ has cycle
and I have no idea to proof how many cycle in $G$.

答: You have correctly figured out that $G$ has $n$ vertices and $n$ edges. If $G$ is connected, then it has a spanning tree $T$. Since $T$ has exactly $n-1$ edges, there is exactly one edge $e$ in $E(G)\setminus E(T)$. Since there are no cycles in $T$, any cycle in $G$ must contain the edge $e$. In fact there is just one such cycle; it is formed by the edge $e$ and the unique path in $T$ which connects the endpoints of $e$.


问:
Show that ($\lnot$q $\implies$ p) $\implies$ (p $\implies$ $\lnot$q) $\equiv$ ($\lnot$p $\lor$ $\lnot$q) using the conditional-disjunction equivalence
Conditional-disjunction equivalence:

(p $\implies$ q) $\equiv$ ($\lnot$p $\lor$ q)

To show:

($\lnot$q $\implies$ p) $\implies$ (p $\implies$ $\lnot$q) $\equiv$ ($\lnot$p $\lor$ $\lnot$q)

Attempt:

$\lnot$($\lnot$q $\implies$ p) $\lor$ (p $\implies$ $\lnot$q) by conditional-disjuction equivalence

$\equiv$ $\lnot$($\lnot$($\lnot$q) $\lor$ p) $\lor$ ($\lnot$p $\lor$ $\lnot$q) by conditional-disjunction equivalence

$\equiv$ $\lnot$(q $\lor$ p) $\lor$ ($\lnot$p $\lor$ $\lnot$q) by double negation law

$\equiv$ ($\lnot$q $\land$ $\lnot$p) $\lor$ ($\lnot$p $\lor$ $\lnot$q) by DeMorgan law

I am stuck at the last step.

答:
You are nearly there.

$(\neg q\wedge\neg p)\vee(\neg p\vee\neg q)\equiv[(\neg q\wedge\neg p)\vee\neg p]\vee\neg q$ by the associativity of $\vee$.

$\color{red}{[(\neg q\wedge\neg p)\vee\neg p]}\vee\neg q\equiv\neg p\vee \neg q$ by absorption on the red term.


答:
Let $x=n+\dfrac m{20}$, with $n$ integer and $0\le m<20$.

The equation reads

$$m = n+\frac m{20} + n + n$$ ($m<10$) or $$m = n+\frac m{20} + n + n+1$$ otherwise.

So

$$0\le\frac{19}{60}m=n<\frac{19}3$$ or
$$\frac{17}6\le \frac{19m-20}{60}=n<6.$$

From the possible values of $n$, you draw $m$ and $x$.

Title: $20\{x\} = x + [x] + [x + 0.5]$

Body:
Let $A = \{x \, | \; 20\{x\} = x + [x] + [x + 0.5] \, \}$, if S = sum of elements of $A$, find $[S]$ where $[x]$ integer part of $x$ and $\{x\} = x - [x]$.

Source: Stage 1 of India MO

I was actually trying to get better at solving these types of questions involving [.] and {x} so i found it confusing.

My attempt:

I took two cases, one for $\{x\}<0.5$ and another for $\{x\}>0.5$ and on writing $[x]$ as $x-\{x\}$, I came up with $19x = 22[x]$ for one of the cases but I am not sure what to do next.

If you could also tell me general methods to tackle such problems, I would be thankful. Thank you!


答:
Let call $n=\lfloor x\rfloor\in\mathbb Z$ and $r=\{x\}\in[0,1)$ to avoid writing all these brackets.

$x\in A\iff 20r=n+r+n+\lfloor n+r+0.5\rfloor=3n+r+\lfloor r+0.5\rfloor$

for $r<0.5$ we have $19r=3n\implies n=\frac {19r}3<\frac{19}6\implies n\le 3$

for $r\ge0.5$ we have $19r=3n+1\implies \frac{19-2}6\le n=\frac{19r-1}3<\frac{19-1}3\implies 3\le n\le 5$

Since there are so few $n$ possible, let calculate the corresponding $x$:

For small $r$ first $n=0,1,2,3$ then $r=0,\frac 3{19},\frac 6{19},\frac 9{19}$

For big $r$ now $n=3,4,5$ then $r=\frac {10}{19},\frac {13}{19},\frac {16}{19}$

$S=(0+1+2+3+3+4+5)+\dfrac 1{19}(0+3+6+9+10+13+16)=18+\dfrac{57}{19}=21$

I let you find $[S]$.

Title: $20\{x\} = x + [x] + [x + 0.5]$


问:
Find the distance of the solution of a system of differential equations with the $0$ solution
I am given the system of ODE
$$x'(t)=x(x^2 +y^2)-x$$
$$y'(t)=y(x^2+y^2)-y$$
for $t \geq 0$

I want to find $$d(t)=\| v(t)-v_0 (t) \|$$ without solving the system.

where $v(t)=(x(t),y(t))$ is a solution to the system and $v_0=(0,0)$ is the zero solution
($\| \cdot \|$ is the euclidean norm).

My attempt, by multiplying the first equation by $x$ and the second by $y$, adding them together and switching to polar coordinates I ended up with
$$r'=r^4-r^2$$
Now, I Know that $$d(t)=r$$ so if I solve $$r'=r^4-r^2$$ I will get the solution, but the problem says that I should find $d(t)$ without solving the system and this is where I am stuck.

How can I find $d(t)$?

答:
$$\frac{dx}{dt}=x(x^2 +y^2-1)$$
$$\frac{dy}{dt}=y(x^2+y^2-1)$$
Trivial solution : $x(t)=y(t)=0.$

Solving in the case $x\neq 0$ and $y\neq 0$ :
$$\frac{dy}{dx}=\frac{y(x^2 +y^2-1)}{x(x^2 +y^2-1)}$$
$$\frac{dy}{dx}=\frac{y}{x}$$
$$y=c_1x$$
$$\frac{dx}{dt}=x(x^2 +(c_1x)^2-1)=x^3+(c_1^2-1)x$$
This ODE is separable. The solution is :
$$x(t)=\pm\frac{1}{\sqrt{c_1^2+1+c_2e^{2t}}}$$
$$y(t)=\pm\frac{c_1}{\sqrt{c_1^2+1+c_2e^{2t}}}$$
This general solution includes the trivial solution $x=y=0$ corresponding to the particular case $c_2=\infty$.


问:
it's possible to invert summation/ series limits?
If the summation  just sum every term i was thinking that for instance 1+2+3+4 = 4+3+2+1

so why this $$\sum\limits_{i=1}^{n} (3i)\       = \sum\limits_{i=n}^{1} (3i) $$ is not true ?

And how i can invert the limits of the summation without changing the result ?

答: As a matter of convention, indices rise. If you want to reverse the order of the terms, you need to write $\sum_{i=1}^na_i=\sum_{i=1}^na_{n+1-i}$.


问:
The diameter of the set of projections in a C*-algebra is at most 1
Let $p,q$ be projections in a $C^*$-algebra $A$. I am trying to show that $\|p-q\|\leq1$, but I can't.

If the projections $p,q$ commute, then this is easy: we set $C=C^*(1,p,q)$ and this is an abelian $C^*$-algebra. By the Gelfand representation, we have that $\sigma(x+y)\subset\sigma(x)+\sigma(y)$ in an abelian C*-algebra, thus $\sigma_A(p-q)=\sigma_C(p-q)\subset\sigma_C(p)-\sigma_C(q)\subset\{-1,0,1\}$ and therefore $\|p-q\|\leq1$. But what about the general case?

答:
For projections $p,q \in A$ we have $0 \le p,q \le 1$ so

$$-1 \le -q \le p-q \le p \le 1$$

which implies $\sigma(p-q) \subseteq [-1,1]$ and therefore $\|p-q\| \le 1$.


问:
Analyticity implying Constant
Question: $f(z)$ is analytic in $C$ and $Im(f(z))\leq 0$. I want to show that $f(z)$ is a constant.

Approach: I know that if $f$ is analytic on a closed curve then the line integral along that curve of $f$ is $0$. But I how do I use the fact that the $f$ is analytic on the lower half plane? And how exactly does it imply that $f$ is constant?

Any help would be greatly appreciated.

Thankyou

答:
Hint: Is $\exp(-i f(z))$ bounded?

Let $f(z) = u(z) + i v(z)$ where $u$ and $v$ are real functions. We have:

$$
\exp(-if(z)) = \exp(-iu(z))\exp(v(z))
$$

Hence:

$$
\left|\exp(-if(z))\right| = \exp(v(z))
$$

Since $v(z) \le 0$. It follows that $\exp(-if(z))$ is bounded. By Liouville's theorem, it's constant. Thus $f$ is constant.

Title: Analyticity implying Constant


答:
To be consistent with the summation identity $\sum\limits_a^{b-1}+\sum\limits_b^c=\sum\limits_a^c$ when $a<b<c$

You need to set $\displaystyle \sum\limits_{i=M}^m x_i=-\sum\limits_{i=m+1}^{M-1} x_i$ for the case $m<M\ $ (and $0$ if $|M-m|\le 1$).

Have a look at this paper for a detailed approach.

As for integrals when bounds are reversed, a negative sign appears before the sum.

Title: it's possible to invert summation/ series limits?

Body:
If the summation  just sum every term i was thinking that for instance 1+2+3+4 = 4+3+2+1

so why this $$\sum\limits_{i=1}^{n} (3i)\       = \sum\limits_{i=n}^{1} (3i) $$ is not true ?

And how i can invert the limits of the summation without changing the result ?


问:
Prove the convergence of $\{a_n\}$ with $0 \le a_{n+1}\le \frac{1}{2}(a_{n}+a_{n-1})$.
Note that $a_{n+1}\le \frac{1}{2}(a_{n}+a_{n-1})\le \max(a_n,a_{n-1})$. Hence, if we define $b_n:=\max( a_{n},a_{n+1})$, then $b_{n+1}=\max(a_{n+1},a_{n+2})\le\max(a_{n+1},a_{n})=b_n, $ which implies $\{b_n\}$ is decreasing. Moreover, $ b_n\ge a_{n+1}\ge0,$ which implies $\{b_n\}$ is down-bounded. As per the the monotone bounded convergence theorem, $\{b_n\}$ is convergent.

But how to prove $\{a_n\}$ has the same limit as $\{b_n\}$?

答:
You already showed that $b_n=\max( a_{n},a_{n+1})$ is convergent. Now
$$
 b_{n+1} = \max(a_{n+1}, a_{n+2}) \le \max(a_{n+1}, \frac{a_n+a_{n+1}}{2}) \\
= \frac{a_{n+1}}{2} + \max(\frac{a_{n+1}}{2}, \frac{a_{n}}{2}) = \frac 12 (a_{n+1} + b_n)
$$
so that
$$
 2  b_{n+1} - b_n \le a_{n+1} \, .
$$
We also have $ b_n=\max( a_{n},a_{n+1}) \ge a_{n+1}$. So
$$
 2  b_{n+1} - b_n \le a_{n+1} \le b_n
$$
which shows that $(a_n)$ is convergent as well, with $\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n$.


答:
Your error is that your left side is incomplete. It is
$$
rr'=xx'+yy'=r^4-r^2
$$
This now is a Bernoulli equation for $r$ or $u=r^2$, or at the basis of both, a linear equation in $v=r^{-2}-1$.
$$
r(t)^{-2}-1=(r_0^{-2}-1)e^{2t}
\implies
r(t)=\frac{r_0}{\sqrt{r_0^2+(1-r_0^2)e^{2t}}}
=\frac{r_0e^{-t}}{\sqrt{1-r_0^2(1-e^{-2t})}}
$$

Title: Find the distance of the solution of a system of differential equations with the $0$ solution

Body:
I am given the system of ODE
$$x'(t)=x(x^2 +y^2)-x$$
$$y'(t)=y(x^2+y^2)-y$$
for $t \geq 0$

I want to find $$d(t)=\| v(t)-v_0 (t) \|$$ without solving the system.

where $v(t)=(x(t),y(t))$ is a solution to the system and $v_0=(0,0)$ is the zero solution
($\| \cdot \|$ is the euclidean norm).

My attempt, by multiplying the first equation by $x$ and the second by $y$, adding them together and switching to polar coordinates I ended up with
$$r'=r^4-r^2$$
Now, I Know that $$d(t)=r$$ so if I solve $$r'=r^4-r^2$$ I will get the solution, but the problem says that I should find $d(t)$ without solving the system and this is where I am stuck.

How can I find $d(t)$?


问:
Distribution of $n$ balls to 10 cells; Inclusion-exclusion problem
So I got another ( :[ ) problem I got stuck with. So before I get going with that, I would like to know if you know any places where I can learn the principles of these subjects (compositions, inclusion-exclusion, etc') online?

So the question: How many ways are there to distribute $n$ balls to 10 cells, so the first and second cells will have at least 4 balls, the third, fourth and fifth have at most 3 balls.

So for the first step I said I put 4 balls in each of the first and second cells, so now we have $n-8$ balls, and we got $|U|={n-8+9 \choose 9}={n+1 \choose 9}$ ways to distribute the rest of the balls to the 10 cells. I got all messed up with the rest :\

Thanks in advance for any help!

答:
You’re off to a good start; now you have to account for the upper bound on the numbers of balls in the third, fourth, and fifth cells. This requires an inclusion-exclusion argument.

First count the distributions that exceed the limit on cell $3$, i.e., that have at least $4$ balls in cell $3$. Put $4$ balls in cell $3$ (and of course $4$ in each of the first two cells as well), so that you’re distributing $n-12$ balls; you can do this in $\binom{n-3}9$ ways. There are also $\binom{n-3}9$ distributions that violate the upper limit on cell $4$ and another $\binom{n-3}9$ that violate the upper limit on cell $5$, so the next approximation to the desired number is

$$\binom{n+1}9-3\binom{n-3}9\;.\tag{1}$$

Unfortunately, every distribution that violates two of the upper limits is subtracted twice in $(1)$ and therefore counted $-1$ times. You want to count such distributions $0$ times, so you need to add those back in. How many are there that violate the upper limit on cells $3$ and $4$? Place $4$ balls in each of cells $1,2,3$, and $4$, and distribute the remaining $n-16$ arbitrarily; you can do this in $\binom{n-7}9$ ways. And there are $\binom32$ pairs of cells with upper bounds, so you need to add this quantity back in $\binom32$ times:

$$\binom{n+1}9-3\binom{n-3}9+\binom32\binom{n-7}9\;.$$

One more correction is needed, to fix up the count for the distributions that violate all three of the upper bounds; can you make that one yourself?


问:
Greatest coefficient of $x$ in expansion of $(1+2x)^8$

If the sum of the coefficients in the expansion of $(1+2x)^n$ is 6561, then what is  the value of the greatest coefficient in the expansion?

Solution

Standard Result:If $n$ is even positive integer then the greatest coefficient in the Binomial Expansion of $(x+a)^n$ is ${n\choose \frac{n}{2}}$

Sum of the Coefficients in the expansion of $(1+2x)^n=(1+2\times1)^n=3^n$.But it is given that Sum of the Coefficients in the expansion of $(1+2x)^n$ is 6561.

So,$3^n=6561=2^{11}\implies n=8$

Using the Standard result,we get Greatest Coefficient in the expansion of $(1+2x)^8=2^8(x+\frac{1}{2})$ is $2^8 \times {8\choose \frac{8}{2}}$=$2^8 \times {8\choose 4}=2^8 \times \frac{8\times 7\times 6 \times 5}{4\times3\times2\times1}=17920$

But the Correct answer is $1792$.

Please point out my mistake.

Thank you

答:
The coefficient of $x$ in $(1+2x)^8$ is $^8C_r 2^r$ -- not simply $^8C_r$. So it's maximum won't occur at usual maximum of binomial coefficients alone (which occurs at $r=n/2$).

One instead has to maximize the general term as
$$ \dfrac{\text{coefficient of } T_{r+1}}{\text{coefficient of } T_r} > 1  $$

and solve for $r$.

I got this way that $^8C_6 2^6$ is the largest coefficient. Indeed $$ ^8C_6 2^6 = 1792$$


答:
as others pointed out it is common (not convention) that indexes start from low values to high in that order, but when it is obvious it is undertood

$\sum\limits_{i=1}^{n} (3i)\       = \sum\limits_{i=n}^{1} (3i) $ is true, not sure why you think $3+6+9+\cdots +3n \neq 3n+\cdots+9+6+3$?

Just remember it is a shorthand, it is meaningless without defining what it means, you coud even define it with additional properties $\sum\limits_{k \text{ is prime} \le10000} k $  means $2+3+5+7+11+\cdots$ only primes less than 10000.


问:
A geometry question...
In the given figure, $ABCD$ is a square of side $3$cm. If $BEMN$ is another
square of side $5$cm & $BCE$ is a triangle right angled at $C$. Then the
length of $CN$ will be:-

I plotted this on GeoGebra:

I can't figure out how to use the smaller square for the question. I did try finding the area of triangle $CBN$. For that I tried subtracting the area of $CEMN$, but to no avail. I want to approach this problem mathematically rather than graphically.

I highly suspect that there is a way to involve the smaller square somehow. Or, if I somehow prove that $\triangle BCE\sim  \triangle CBN$, that could solve it.

答:
Use coordinate geometry.

Let $B$ be the origin. Recall that for two lines to be perpendicular, $m_1 m_2 = -1$. Now let the line perpendicular to the line passing through $(0, 0)$ and $(4, -3)$ pass through $(a, b)$. We have that $m_1 = -\frac{3}{4}$ and $m_2 = \frac{b}{a}$, so $m_1 m_2 = -\frac{3}{4} \frac{b}{a} = -1$, and  $\frac{b}{a} = \frac{-1}{-3/4} = \frac{4}{3}$ (note that there are an infinite number of points satisfying this condition).

$b = 4, a = 3$ or $(3, 4)$ satisfies this condition, and it has the same distance to $B$ as $(4, -3)$, as the absolute $x$ and $y$ distances are the same, thus Pythagoras's theorem is not affected.

Now move the origin $(0, 0)$ back to $D$. Therefore $N$ is $3$ units right and $4$ units up from $B$, which is at $(3 + 3, 3 + 4) = (6, 7)$. Now just find the distance between $C = (3, 0)$ and $N = (6, 7)$.

Title: A geometry question...


问:
Prove that $b_n$ converges if $a_n$ converges and $a_n+b_n$ converges
I’ve been given this assignment and I’m having a really hard time figuring out how to answer it.
I’ve tried proving it in the same way as proving that if $a_n$ converges and $b_n$ converges then $a_n+b_n$ converges, but I got stuck halfway through.

I’d really appreciate any help. Thanks in advance!

答:
Define $L := \lim_{n \to \infty} a_n$, $L' := \lim_{n \to \infty} (a_n + b_n)$.

Suppose that $\varepsilon > 0$. Then there exists $k \in \mathbb{N}$ such that for all $n \geq k,$ $|a_n - L| < \varepsilon/2$. There also exists a natural number $k'$ such that for all $n \geq k'$, $|a_n + b_n - L'| < \varepsilon/2$. Can you continue the argument from here?

Define $K := \text{max}(k, k')$, then for all $n \geq K$ we have that

\begin{align}|b_n - (L' - L)| &= |(a_n + b_n - L') - (a_n - L)| \\ &\leq |a_n + b_n - L'| + |a_n - L| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align} This shows that $b_n \to L' - L$ as $n \to \infty$.


问:
Show a formula is equivalent in a theory to a universal formula iff it is preserved under passing to submodels of models of the theory
I'm trying to solve the following exercise in my lecture notes.

Let $T$ be a theory, $\phi(x)$ a formula. Show that the following
conditions are equivalent:

(1) $\phi$ is $T$-equivalent to a universal formula, i.e. there exists a universal
formula $\phi'$ such that $T \models (∀x)(\phi ↔ \phi')$.

(2) $\phi$ is preserved under passing to submodels of models of $T$, i.e. whenever
$A$, $B$ are models of $T$ with $A ≤ B$, $a ∈ A$, and $B \models \phi(a)$, we have
$A \models \phi(a)$.

I think I've made some good progress, but I'm unsure how to prove one step in the proof I've got.

Firstly, it's clear that if we can prove this for sentences $φ$ then the result will follow for arbitrary formulas, since the more general result will follow if we change the formula to a sentence by replacing the free variables with constants.

The (1) $\implies$ (2) is simple and follows easily by working with definitions.

This leaves me with the (2) $\implies$ (1) direction.

I've seen a reasonably similar proof to my attempt here in a related result, which I think I can adapt and make work here.

I let $\Sigma$ be the set of all universal sentences $\psi$ such that $T,\phi \models \psi$. If $T \cup \Sigma \models \phi$, then there's a finite subset $\Delta \subset \Sigma$ such that $T \cup \Delta \models \phi$. Clearly the conjunction of all $\psi \in \Delta$ will be equivalent to a universal formula, and this universal formula will be the one I'm looking for.

The only step I'm unsure about is proving that $T \cup \Sigma \models \phi$ (which I think is true). Presumably to do so I'd want to show $T \cup \Sigma \cup \lbrace\neg\phi\rbrace $ is unsatisfiable, but I can't see how I'd do that.

I'd appreciate any help anyone could offer - either, if my thinking is right, in showing that $T \cup \Sigma \models \phi$ or, if I'm mistaken and need to try another strategy, pointing me in the right direction.

答:
You are very close indeed. You are right that the only gap is that $T \cup \Sigma \models \phi$, so I will fill that gap and then the rest of the argument goes through as you have written.

Let $M \models T \cup \Sigma$, we will show that $M \models \phi$. Consider $T' = T \cup \{\phi\} \cup \operatorname{Th_{qf}}(M)$, where $\operatorname{Th_{qf}}(M)$ is the set of all quantifier-free formulas in the language where we added constants for $M$. Suppose for a contradiction that $T'$ is inconsistent. Then there are $\psi_1, \ldots, \psi_n \in \operatorname{Th_{qf}}(M)$ such that $T \cup \{\phi\} \models \neg \psi_1 \vee \ldots \vee \neg \psi_n$. Since the constants from $M$ do not appear in $T \cup \{\phi\}$ we get that $\neg \psi_1 \vee \ldots \vee \neg \psi_n$ is equivalent to some universal sentence $\theta$ modulo $T \cup \{\phi\}$. So we have $\theta \in \Sigma$, but then $M \models \neg \psi_1 \vee \ldots \vee \neg \psi_n$, a contradiction. We conclude that $T'$ is consistent, so there is a model $N \models T'$. Then in particular $N \models T$ and $N \models \phi$, while also $M \leq N$. Now we can use (2) to conclude that $M \models \phi$, as required.


问:
Vectors: Finding third point in a right angle triangle
Given A(1,4) and B (3,-5) use the dot product to find point C so that triangle ABC is a right angle triangle.</p>

答:
Find the vector connecting between the two:

$$\vec{c}=<1-3,4+5>=<-2,9>$$

You can use the dot product between $\vec{c}$ and a unit vector $\hat{b}=<-1,0>$:
$$\cos{\theta}=\frac{\vec{c}\cdot\hat{b}}{||\vec{c}||\space||\hat{b}||}$$

The length of the opposing two angles are $||\vec{c}||\sin{\theta}$ and $||\vec{c}||\cos{\theta}$.


问:
Expansion of the Frobenius norm
this might be very elementary question. I was confused by looking at some different sources when expanding the Frobenius norm into trace. Would these two expressions below always be the same? Or only under certain conditions?

\begin{aligned}
\left\|X-Y\right\|_{F}^{2}&=\operatorname{tr}\left(\left(X-Y\right)\left(X-Y\right)^{\top}\right)
\end{aligned}

\begin{aligned}
\left\|X-Y\right\|_{F}^{2}&=\operatorname{tr}\left(\left(X-Y\right)^{\top}\left(X-Y\right)\right)
\end{aligned}

Thanks

答: For any two $n\times n$ matrices $A$ and $B$, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ and, in particular,$$\operatorname{tr}\left(AA^\top\right)=\operatorname{tr}\left(A^\top A\right).$$

Title: Expansion of the Frobenius norm


问:
How to maximize $\sum x_i\times x_j$ as $1\leq i,j\leq n$ with $i\neq j$ subject to $\sum x_i=1$?
I want to maximize $$\sum_{i,j \in [n], i \ne j} x_i \times x_j$$ where $\forall i ~~$  $0\le x_i \le 1$ and $x_1 + x_2 + x_3 + \ldots +x_n = 1$

I want to prove that the sum will be maximized when $\forall i ~~$ $x_i = \frac{1}{n}$. I don't know whether this statement is true or not.

Note:-
It would be more helpful if you can prove that using shifting of weights from one $x_i$ and $x_j$. Even the proof is not following that method it is fine.

答:
Your summation is $x_1(\sum_{j\ne1}x_j)+x_2(\sum_{j\ne2}x_j)+...+x_n(\sum_{j\ne n}x_j)$ which is $x_1(1-x_1)+...+x_n(1-x_n)=(x_1+...+x_n)-(x_1^2+...+x^2_n)=1-(x_1^2+...+x^2_n)$. To maximize this, we need to minimize $\sum_{i\in[n]}x_i^2$.

We know $\sum_{i=1}^nx_i^2\ge \frac{(\sum_{i=1}^nx_i)^2}n=1/n$.

And indeed this minimum is observed when $ x_1=x_2=...=x_n=1/n$. The maximum value is $1-1/n$.


答:
Extend $CB$ and drop a perpendicular from $N$ to this extension. Let the intersection be $H$.

We see that $\triangle BCE \cong \triangle NHB$. Indeed, $\angle NHB = \angle BCE = 90^\circ$ and $NB = BE$.

We also have $\angle NBH = 180^\circ - \angle NBE - \angle CBE = 180^\circ - \angle BCE - \angle CBE = \angle BEC$. Hence congruence follows by AAS.

Now $BC = NH = 3cm$ and $NB = BE = 5cm$. By Pythagoreas theorem, $HB = CE = 4cm$, and $NC^2 = HC^2+NH^2 = (3+4)^2+3^2$.

Title: A geometry question...

Body:
In the given figure, $ABCD$ is a square of side $3$cm. If $BEMN$ is another
square of side $5$cm & $BCE$ is a triangle right angled at $C$. Then the
length of $CN$ will be:-

I plotted this on GeoGebra:

I can't figure out how to use the smaller square for the question. I did try finding the area of triangle $CBN$. For that I tried subtracting the area of $CEMN$, but to no avail. I want to approach this problem mathematically rather than graphically.

I highly suspect that there is a way to involve the smaller square somehow. Or, if I somehow prove that $\triangle BCE\sim  \triangle CBN$, that could solve it.


问:
Every smooth vector field generates a smooth flow
How Every smooth vector field generates a smooth flow?

Definitions

Smooth Vector Field: A vector field on a smooth manifold $M^n$ is a function $\xi$ on
$M^n$, such that $\xi(p)\in T_p(M)\ \forall p\in M^n$ and which is smooth in the following sense: Given local coordinates $x_1,...,x_n$ near $p\in M$, we can write
$$\xi(p)=\sum\limits_{i=1}^{n} a_i(p)\frac{\partial}{\partial x_i}$$
and smoothness of $\xi$ means that the $a_i$ are smooth functions.

Smooth Flow: A smooth flow on a smooth manifold $M^n$ is a smooth map
$\theta: \mathbb{R} \times M\to M$ such that:
(i) $\theta(0,x)=x\ \forall x\in M$ and
(ii) $\theta(s+t,x)=\theta(s,\theta(t,x))\ \forall x\in M$ and $s,t\in \mathbb{R}$.

In many places I saw they are constructing a system of ODE's and saying the solution of the system is the required smooth flow. But I didnot understand how it is actually happening?

答:
Fix $p \in M$. Consider the linear first order differential equation :
\begin{align}
\forall t \in I, ~ \gamma'(t) = X(\gamma(t)),~ \gamma(0) = p
\end{align}
In charts, this is an ordinary differential equation in $\mathbb{R}^n$ and it has a unique maximal solution, say $\gamma_p : I_p \to M$, with $I_p$ open.

Now, let $U = \bigcup_{p \in M} I_p\times \{p\}$. This is an open subset of $\mathbb{R}\times M$ containing $\{0\}\times M$ (this is the difficult statement of the theorem, I will not provide a proof here). Define on $U$:
\begin{align}
\varphi^X : U &\to M \\
(t,p) & \mapsto \gamma_p(t)
\end{align}
One can show that it is a flow and that it generates the vector field $X$.

First, it is a flow because of Cauchy-Lipschitz theorem on uniqueness of solution of differential equation. For $t,s$ such that $\varphi^X(t+s,p)$ and $\varphi^X(t,\varphi^X(s,p))$ is defined, one can show, differentiating with the $t$ variable, that they both are solutions of the same first order ODE with initial data $\varphi^X(s,p)$ at $t=0$, and consequently they are equal.

Moreover, by the very definition of $\varphi^X$:
\begin{align}
\left.\frac{\mathrm{d}}{\mathrm{d}t}\varphi^X(t,p)\right|_{t=0} = \left.\frac{\mathrm{d}}{\mathrm{d}t} \gamma_p(t)\right|_{t=0}=\gamma_p'(0) = X(\gamma_p(0))=X(p)
\end{align}
and $\varphi^X$ generates $X$.


答: Just put $(C-A).(B-A)=0$, that should solve it. Note that there are infinitely many such triangles.</p>

Title: Vectors: Finding third point in a right angle triangle

Body: Given A(1,4) and B (3,-5) use the dot product to find point C so that triangle ABC is a right angle triangle.</p>


答:
$BC = 3, BN = 5, \sin (\angle CBE) = \frac{4}{5}$

Using Cosine law, $CN^2 = BN^2 + BC^2 - 2 BN \times BC \times \cos (\angle CBN) = 5^2 + 3^2 + 2 \times 5 \times 3 \times \sin (\angle CBE)$

$= 34 + 24 = 58 \implies CN = \sqrt{58}$

[$\cos (\angle CBN) = \cos (90^0 + \angle CBE) = - \sin (\angle CBE$)]

Title: A geometry question...


问:
Why is $\sum\limits_{n=1}^{\infty}e^{-(n/10)^2}$ almost equal to $5\sqrt\pi-\frac12$ (agreeing up to $427$ digits)?
The following I saw as an exercise in a text on modular forms. I am lacking the understanding for why the following is true, but it is nevertheless astonishing:

Why is the numerical value of

$$x:=\sum_{n=1}^{\infty} \exp(-(n/10)^2)$$

so ridiculously close to the value of

$$y:=5\sqrt\pi-\frac12$$

while (and this is the truly surprising aspect) not being equal to it?

And by "ridiculously close" I mean that $x$ and $y$ agree up to the 427-th digit after the decimal point:

\begin{align}
x=8.
&3622692545275801364908374167057259139877472806\\
&1193564106903894926455642295516090687475328369\\
&2723327081134118121412853331180764328622113012\\
&6254685480139353423101884932655256142496258651\\
&4475413114466047689633981400087319507675739860\\
&2583500950926170092927234872474563201569608877\\
&6295310820270966625045319920380686673873757671\\
&6833994894682925918204397725582580869380029533\\
&6967158956664049274231240924510273274260978066\\
&257808237337\color{red}{62}
\end{align}

They first disagree in the red digits.

Question: Can someone explain this in "simple" words?
It does not have to be elementary, but I want to understand what machinery has to interact to arrive at this.
And how would one even come up with such an example, or similar ones?

答: Define $$\psi(q)\equiv-\frac12+\frac12\vartheta_3(0,e^{-\pi q})=\sum_{n\ge1}e^{-\pi n^2q}$$ where $\vartheta_3(z,q)$ is Jacobi's third theta function. An application of the Poisson summation formula leads to the identity $$\vartheta_3(0,e^{-\pi/q})=\sqrt q\,\vartheta_3\left(0,e^{-\pi q}\right)$$ whose proof is outlined in Jacobi (1828). Substituting $q:=(\pi k^2)^{-1}$ yields $$\sum_{n\ge1}e^{-n^2/k^2}=-\frac12+\frac12\vartheta_3(0,e^{-1/k^2})=-\frac12+\frac12\cdot\sqrt{\pi k^2}\vartheta_3(0,e^{-\pi^2k^2})$$ so that $$\sum_{n\ge1}e^{-n^2/k^2}=-\frac12+\frac k2\sqrt\pi(1+2\psi(\pi k^2))=-\frac12+\frac k2\sqrt\pi+k\sqrt\pi\sum_{n\ge1}e^{-(\pi kn)^2}.$$ When $k=10$ the error has an order of magnitude of $$\log_{10}\left(10\sqrt\pi e^{-100\pi^2}\right)=1+\log_{10}\sqrt\pi-100\pi^2\log_{10}e=\bf-427$$ which corresponds to your observation.


问:
A regular domain is a Krull Domain.
I don't know how to prove that a regular domain is a Krull domain.

If we say that $\mathcal{P}=\{p\in \operatorname{Spec}(A)\mid \operatorname{ht}(p)=1 \}$ a Krull domain is a domain s.t.:

$A_p$ is a DVR for any $p\in\mathcal{P}$;

$A=\bigcap_{p\in\mathcal{P}}A_p\subseteq \operatorname{Frac}(A)=k$;

$\forall\ a\in A$ the cardinality of $\{p\in\mathcal{P}\mid a\in p\}\}$ is finite.

The first and the third condition are okay for regular rings. In a intuitive way also the second is okay but I have a doubt:

I take $\frac{a}{s}\in k\ s.t.\ \frac{a}{s}\in A_q\cap A_p$, this means that $\exists\ \frac{a_1}{s_1}\in A_p$, $\frac{a_2}{s_2}\in A_q$ with $\frac{a_1}{s_1}=\frac{a_2}{s_2}=\frac{a}{s}\in k$. How can I conclude that there exists $a_3\in A$ $s_3\in A\setminus(p\cup q)$ $s.t.\ \frac{a}{s}=\frac{a_3}{s_3}$ ?

Because I don't know if $s_1\in A\setminus p$ or $s_2\in A\setminus q$.

thank you.

答:
I find a solution:

$A_p\subseteq\{\frac{a}{b}\in k\mid \frac{b}{a} A\cap A\not\subseteq p\}\cup\{0\}$. One cn check that it is an equality, but the inequality is immediate, because if $\frac{a}{b}\in A_p\setminus\{0\}$ we have that $\frac{b}{a}a=b\in A\setminus p$.

So $A\subseteq\bigcap_{p\in \mathcal{P}} A_p \subseteq
\bigcap_{p\in \mathcal{P}}\{\frac{a}{b}\in k\mid \frac{b}{a} A\cap A\not\subseteq p\}\cup\{0\}
= \{\frac{a}{b}\in k\mid b A\subseteq\frac{b}{a} A\cap A\not\subseteq p\ \forall p\in\mathcal{P}\}\cup\{0\}
\subseteq\{\frac{a}{b}\in k\mid b A \not\in p\ \forall p\in\mathcal{P}\}\cup\{0\}\subseteq A$

Where the last inequality holds because if $b\in A\setminus\bigcup_{p\in \mathcal{P}}p$ the ideal $b A$ could be not contained in a prime ideal, otherwise (Krull's Hauptidealsatz) there is a $p\in\mathcal{P}$ s.t. $b\in p$ that is absurd, so $bA=A$ and $b$ is invertible in A.

Title: A regular domain is a Krull Domain.


问:
Martingale whose distribution is independent of n
Claim: A one-point stationary sequence $(X_n)_{n\in\mathbb{N}}$ of square integrable r.v's is a martingale w.r.t the natural filtration iff it is surely constant in time.

I am convinced that this is a false statement, and therefore I'd like to provide a counterexample. As a simple counterexample I wanted to take a sequence of $i.i.d. ~ Bernoulli(p)$.

Then $E[X_{n+1} | \mathcal{F}_n$] = $E[X_{n+1}] = E[X_n] = p$, independent of $n$, and therefore one-point stationary. On the other hand, $P(X_{n+1} = X_{n}) \neq 1$.

As I'm not very familiar yet with martingales, I am wondering if the sequence of Bernoulli i.i.d. r.v's is indeed a martingale, and if it suffices as a counterexample to disprove the claim.

EDIT: $E[X_{n+1} | \mathcal{F}_n] = E[X_n] = p \neq X_n$. Therefore, this sequence is not a martingale. What is an appropriate counterexample?

答: $E(X_{n+1}|\mathcal F_n) =X_n$. So $X_n^{2}=(E(X_{n+1}|\mathcal F_n))^{2} \leq E(X^{2}_{n+1}|\mathcal F_n)$. Since both sides have the same expecatation (by stationarity) we get $X_n^{2}= E(X^{2}_{n+1}|\mathcal F_n)$. Now $E((X_{n+1}-X_n)^{2} |\mathcal F_n)=E(X^{2}_{n+1} |\mathcal F_n)-2X_nE(X_{n+1} |\mathcal F_n)+X_n^{2}=0$ and taking expectation we get $X_{n+1}=X_n$ a.s.


答:
Hint: Use generating functions. We can work directly with the problem as is, or put $4$ balls into each of the first two cells. Then we have the problem of distributing $m=n-8$ balls into $10$ cells, with at most $3$ in each of the last three. We will take that approach.

The generating function for this problem is
$$(1+x+x^2+x^3+x^4+\cdots)^{7} (1+x+x^2+x^3)^3.$$
To make it easier to find the coefficient of $x^m$,  use the formula for the sum of a finite geometric progression to simplify this to
$$(1-x^4)^3 (1-x)^{-10}.$$
Then expand $(1-x^4)^3$ and write down the general binomial theorem expansion of $(1-x)^{-10}$. Out of this we can find an expression for the coefficient of $x^m$.

Title: Distribution of $n$ balls to 10 cells; Inclusion-exclusion problem

Body:
So I got another ( :[ ) problem I got stuck with. So before I get going with that, I would like to know if you know any places where I can learn the principles of these subjects (compositions, inclusion-exclusion, etc') online?

So the question: How many ways are there to distribute $n$ balls to 10 cells, so the first and second cells will have at least 4 balls, the third, fourth and fifth have at most 3 balls.

So for the first step I said I put 4 balls in each of the first and second cells, so now we have $n-8$ balls, and we got $|U|={n-8+9 \choose 9}={n+1 \choose 9}$ ways to distribute the rest of the balls to the 10 cells. I got all messed up with the rest :\

Thanks in advance for any help!


答:
If they are one-point stationary, there are a lot of things we can do with them to show there invariance.

In this case, you have further that they're square-integrable, which makes thing even easier.

Just observe that:

$E[X_{n+1}^2]= E[X_n^2]+E[ (X_n-X_{n+1})^2]$

Done

Title: Martingale whose distribution is independent of n

Body:
Claim: A one-point stationary sequence $(X_n)_{n\in\mathbb{N}}$ of square integrable r.v's is a martingale w.r.t the natural filtration iff it is surely constant in time.

I am convinced that this is a false statement, and therefore I'd like to provide a counterexample. As a simple counterexample I wanted to take a sequence of $i.i.d. ~ Bernoulli(p)$.

Then $E[X_{n+1} | \mathcal{F}_n$] = $E[X_{n+1}] = E[X_n] = p$, independent of $n$, and therefore one-point stationary. On the other hand, $P(X_{n+1} = X_{n}) \neq 1$.

As I'm not very familiar yet with martingales, I am wondering if the sequence of Bernoulli i.i.d. r.v's is indeed a martingale, and if it suffices as a counterexample to disprove the claim.

EDIT: $E[X_{n+1} | \mathcal{F}_n] = E[X_n] = p \neq X_n$. Therefore, this sequence is not a martingale. What is an appropriate counterexample?


问:
Can some please help me?
Why I am getting nan in python.

```<code>from sympy import *
x,f,g=symbols('x f g')

def f(x):
    return (1+x**2)**2
def g(x):
    return (1-f(x))/x
expand(g(x))
g(0)

</code>```

I know the $x$ is in division but after manual simplification I found $$g(x)=−^3−2$$ and $$g(0)=0$$
I could not understand why python result give me $$g(0) = NaN$$ How can I get g(0)=0 in python.

答:
Possible workaround:</p>
```<code>from sympy import *
x,f,g,t=symbols('x f g t')

def f(x):
    return (1+x**2)**2
def g(x):
    return expand((1-f(t))/t).subs(t,x)
print(g(0))
</code>```

Title: Can some please help me?

Body:
Possible workaround:

from sympy import *
x,f,g,t=symbols('x f g t')

def f(x):
    return (1+x**2)**2
def g(x):
    return expand((1-f(t))/t).subs(t,x)
print(g(0))


答: Rotation around $B$ for $-90^{\circ}$ takes $C$ to $A$ and $N$ to $E$, so it takes $CN$ to $AE$. So $$CN = AE = \sqrt{3^2+7^2}=\sqrt{58}$$

Title: A geometry question...


问:
2 -form invariant under the flow of a vector field
I have this exercise.

Let $(x_0, x_1, x_2, x_3)$ be coordinates in $\mathbb{R}^4$ and let
$$ w = dx_0 \wedge dx_1 + dx_2 \wedge dx_3 $$.
Let  $u: \mathbb{R}^4 \rightarrow \mathbb{R}$ be a smooth function and $X$ be the following vector field:
$$
X = -\partial_1(u)\partial_0 +\partial_0(u)\partial_1 -\partial_3(u)\partial_3 +\partial_2(u)\partial_3
$$

Show that if $\phi_t^X$ defines the flow of X and $\phi_1^X$ is defined, then $(\phi_1^X)^* w = w.$

[Hint: Consider $\frac{d}{dt}(\phi_t^X)^* w$]

答:
Hint
Define $\gamma : \mathbb{R} \to \mathbb{R}^4$ to be the solution of $\gamma' = X(\gamma)$ and $\gamma(0) = p \in \mathbb{R}^4$. The flow you want to compute is $\varphi_t^X(p) = \gamma(t)$.

Write $\gamma = (x_0(t),x_1(t),x_2(t),x_4(t))$ in coordinates and find the differential equations verified by the coordinates. For example, for the first and second coordinate:
\begin{align}
{x_0}'(t) &= -\partial_1u(x_0(t),x_1(t),x_2(t),x_3(t))\\ {x_1}'(t) &= \partial_0 u(x_0(t),x_1(t),x_2(t),x_3(t))
\end{align}

Remark it appears constant are not integral curves of the vector field if $\partial u$ is not zero.

Edit To answer the (new) question there is no need to compute the flow, but just to use its definition. By the very definition of the Lie derivative of a differential form in the direction of a vector field, one has
\begin{align}
\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi^*_t\omega = \mathcal{L}_X\omega
\end{align}
By the Cartan's (magic) formula,
\begin{align}
\mathcal{L}_X\omega = \mathrm{d}\left(\omega(X,\cdot)\right) + \mathrm{d}\omega\left(X,\cdot\right)
\end{align}
As $\mathrm{d}\omega = 0$ (easy computation), it is
\begin{align}
\mathcal{L}_X\omega = \mathrm{d}\left(\omega(X,\cdot)\right)
\end{align}
But
\begin{align}
\omega(X,\cdot) &= -\partial_0u\mathrm{d}x^0 -\partial_1u\mathrm{d}x^1-\partial_2u\mathrm{d}x^2-\partial_3u\mathrm{d}x^3 \\
&= -\mathrm{d}u
\end{align}
and thus
\begin{align}
\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi^*_t\omega = \mathrm{d}\left(-\mathrm{d}u\right) = -\mathrm{d}^2u = 0
\end{align}
and consequently, $\varphi_t^*\omega$ is constant. As $\varphi_0 = \mathrm{id}$, it is equal to $\omega$.


问:
$\phi(x+1)=\phi(x)$ and $\phi(\frac{x}{2})\phi(\frac{x+1}{2}) =d\phi(x)$ then $\phi$ is constant
For $0 < x < \infty$, let $\phi(x)$ be positive and continuously twice differentiable satisfying $\phi(x+1)=\phi(x)$ and $\phi(\frac{x}{2})\phi(\frac{x+1}{2}) =d\phi(x)$ with $d$ being a constant.

I have to show that $\phi$ is constant, but I don't know how to start the proof here. Any hints?

答:
Iterating the second condition one you get
\begin{align*}
\phi\left(\frac x2\right)\phi\left(\frac{x+1}2\right)
&\phi\left(\frac{x+2}2\right)\phi\left(\frac{x+3}2\right)
\cdots\phi\left(\frac{x+2n}2\right)\phi\left(\frac{x+2n+1}2\right)\\
&=d^n\phi(x)\phi(x-2)\cdots\phi(x-2n)\\
&=d^n\left(\phi(x)\right)^n
\end{align*}
where the last equality follows from the first condition on $\phi$.

Can you conclude from here?


问:
How to represent the floor function using mathematical notation?
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

答: $\left \lfloor x \right \rfloor = a \in \mathbb{Z}, a \le x \ni a\ge b \;\forall \;b \in \mathbb{Z}, b\le x$</p>


问:
Equivalence class involving lcm
I'm struggling to find a beginning for finding an equivalence class involving lcm. Given a fixed $n \in \mathbb{N}$ and relation $R = \{(a,b) \in \mathbb{Z}^2 \: | \: lcm(a,n) = lcm(b,n) \}$, find the equivalence class of $p$ where $p$ is prime.

I don't have much experience dealing with lcm's in regards to equivalence relations. I understand the ideas between them separately but have a bit of trouble figuring how they relate in this case. How does $p$ relate to $R$?

答:
If p|n, then lcm(p,n) = n and
[p] = { q : q|n }.

If not p|n, then lcm(p,n) = pn and
[p] = { pq : q|n }.

Title: Equivalence class involving lcm


答:
If you take the derivative wrt $x_i$ you get, for each $i$

$$ \sum_{j=1}^n x_j-x_i-\lambda=0\implies x_i=1-\lambda$$

where $\lambda$ is the Lagrange multiplier. This means that all the $x_i$ need to take the same value. Adding the $n$ FOC's results in

$$ 1=n(1-\lambda)\implies \lambda=\frac{n-1}{n},$$

and, therefore, $$ x_i=\frac{1}{n}$$

Title: How to maximize $\sum x_i\times x_j$ as $1\leq i,j\leq n$ with $i\neq j$ subject to $\sum x_i=1$?

Body:
I want to maximize $$\sum_{i,j \in [n], i \ne j} x_i \times x_j$$ where $\forall i ~~$  $0\le x_i \le 1$ and $x_1 + x_2 + x_3 + \ldots +x_n = 1$

I want to prove that the sum will be maximized when $\forall i ~~$ $x_i = \frac{1}{n}$. I don't know whether this statement is true or not.

Note:-
It would be more helpful if you can prove that using shifting of weights from one $x_i$ and $x_j$. Even the proof is not following that method it is fine.


问:
Is Petersen Graph without an edge planar
Let G be the Petersen graph. Is G-e planar? If no, explain why. If G-e is planar, then draw a plane graph isomorphic to it.

So we can remove 3 types of edges. 1) Connecting 2 vertices on the outside (eg. 0-1). 2) Connecting an inside and an outside vertice (eg. 4-9). 3) Connecting 2 inside vertices (eg. 5-7). It's also clear that we need the edges of type 3 to not cross over each other.

Removing 1) adds no benefit. Because even if we remove the edge, we'd have to cross another outer edge to connect 2 inner vertices. For eg. removing 0-1 so we can connect 6-9. This would not yield results since we'd still have to cross either 0-4 or 0-5 (or any corresponding pair).

Removing 2) is also useless. We'd still have other internal edges (type 2) to cross. For example if we remove 0-5, we can loop 6-9 around 5 to connect them. But we can't do the same for 6-8 which is seperated by 2 regions.

Quick inspection shows same result for 3)
This is my working theory. That G-e is nonplanar because removing any one edge still leaves other vertices seperated by 2 regions. Is there a better way to articulate this. Am I just wrong?

答:
It is still not planar.

First, we delete three edges as follows, and label the vertices.

Then, we reduce the vertices of the same color: 1,6; 5,7; 3,4 and obtain a $K_{3,3}$


问:
Exercise 4.(a) on p.30 in "Analysis on Manifolds" by James R. Munkres. Is my solution ok? Please tell me a simpler or smarter solution.
I am reading "Analysis on Manifolds" by James R. Munkres.

There is the following exercise in this book:

Exercise 4.(a) on p.30

Show that if $Q$ is a rectangle, then $Q$ equals the closure of $\mathrm{Int}\, Q$.

The definition of a rectangle in this book is the following:

$Q = \{(x_1, \dots, x_n)\in\mathbb{R}^n\mid a_i \leq x_i \leq b_i \text{ for all }i\}$.

My solution for Exercise 4.(a) on p.30 is the following:

$Q = \bigcap_{i=1}^{n}{(\{(x_1, \dots, x_n)\in\mathbb{R}^n\mid a_i \leq x_i\}\cap\{(x_1, \dots, x_n)\in\mathbb{R}^n\mid x_i \leq b_i\})}$.

In general, $\mathrm{Int}\, (S_1\cap\cdots\cap S_n)=\mathrm{Int}\, S_1\cap\cdots\cap\mathrm{Int}\, S_n$ holds.

So, $\mathrm{Int}\, Q = \bigcap_{i=1}^{n}{(\{(x_1, \dots, x_n)\in\mathbb{R}^n\mid a_i < x_i\}\cap\{(x_1, \dots, x_n)\in\mathbb{R}^n\mid x_i < b_i\})}$.

In general, $\overline{S_1\cap\cdots\cap S_n}\subset\overline{S_1}\cap\cdots\cap\overline{S_n}$ holds.

So, $\overline{\mathrm{Int}\, Q} \subset \bigcap_{i=1}^{n}{(\{(x_1, \dots, x_n)\in\mathbb{R}^n\mid a_i \leq x_i\}\cap\{(x_1, \dots, x_n)\in\mathbb{R}^n\mid x_i \leq b_i\})}=Q$.

Conversely, let $(x_1, \dots, x_n) \in Q$.

If $a_i < x_i$ and $x_i < b_i$ for all $i\in\{1,\dots,n\}$, then $(x_1,\dots,x_n) \in\mathrm{Int}\, Q\subset\overline{\mathrm{Int}\, Q}$.

If $a_i \leq x_i$ and $x_i \leq b_i$ for all $i\in\{1,\dots,n\}$ and ($a_i = x_i$ or $x_i = b_i$ for some $i\in\{1,\dots,n\}$),

then $a_i < x_i < b_i$ or $a_i = x_i < b_i$ or $a_i < x_i = b_i$ holds for each $i\in\{1,\dots,n\}$.

If $a_i < x_i < b_i$, then $((a_i, b_i)\cap(x_i-\frac{x_i-a_i}{2}, x_i+\frac{b_i-x_i}{2}))\setminus\{x_i\}\neq\emptyset$.

If $a_i = x_i < b_i$, then $((a_i, b_i)\cap(x_i-\frac{b_i-x_i}{2}, x_i+\frac{b_i-x_i}{2}))\setminus\{x_i\}\neq\emptyset$.

If $a_i < x_i = b_i$, then $((a_i, b_i)\cap(x_i-\frac{x_i-a_i}{2}, x_i+\frac{x_i-a_i}{2}))\setminus\{x_i\}\neq\emptyset$.

So, $(x_1,\dots,x_n) \in\overline{\mathrm{Int}\, Q}$.

So, $Q=\overline{\mathrm{Int}\, Q}$.

Is my solution ok?

I think my solution is not smart.

At first glance, I wondered this exercise was very easy.

But now I think this exercise is not very easy.

I cannot judge if this exercise is very easy or not.

Is this exercise very easy or not?

Please tell me a simpler or smarter solution for this exercise?

答:
Stinking Bishop, thank you very much for your comment.

Let $x := (x_1, \dots, x_n)$.

If there exists $i \in \{1,\dots,n\}$ such that $x_i\leq a_i$ or $b_i\leq x_i$, then it is obvious that $x$ is not an interior point of $Q$.

Conversely, if $a_i < x_i < b_i$ for all $i\in\{1,\dots,n\}$, then it is obvious that $x$ is an interior point of $Q$.

So, $\mathrm{Int}\, Q = (a_1,b_1)\times\dots\times(a_n,b_n)$.

If there exists $i \in \{1,\dots,n\}$ such that $x_i< a_i$ or $b_i< x_i$, then it is obvious that $x$ is not a limit point of $\mathrm{Int}\, Q$.

Conversely, if $a_i \leq x_i \leq b_i$ for all $i\in\{1,\dots,n\}$, then it is obvious that $x$ is a limit point of $\mathrm{Int}\, Q$.

So, $Q=\overline{\mathrm{Int}\, Q}$.


问:
How to solve the following periodic ODE
Let $a\in \mathbb{R}$ and  $y$ is a $\pi-$periodic function

how we can solve the following ODE

$$y(x)\sin(x)-y^{'}(x)\cos(x)=a\sin(x)$$

what I have try:

I take $Y(x)=y(x)\cos(x)$ this gives $Y^{'}(x)=-a \sin(x)$ so $Y(x)=a \cos(x)+C$
this implies $y(x)=a+\frac{C}{\cos(x)}$ I am little confused about this solution (if I put $y(\frac{\pi}{2})=0$ this solution will be note defined )

答:
Suppose $Y(x)=(-cos(x)y(x))$, so it holds that $$Y'(x)=sin(x)y(x)-cos(x)y'(x)$$
That means $$ Y'(x)=asin(x)$$
And hence $$Y(x)=-acos(x)+ cst$$
Therefore $$y(x)=a-\frac{cst}{cos(x)}$$ of course whenever $x\neq \pi/2+k \pi$ where $k \in \mathbb{Z}$.

Title: How to solve the following periodic ODE


问:
prove that the unit circle $x^2+y^2=1$ is a closed set
I need to prove that the unit circle $x^2+y^2=1$ is a closed set in $\mathbb{R}^2$ is closed using convergent sequence method.</p>

答: Consider the set $S=\{(x,y)\in\mathbb{R}^2\;|\; x^2+y^2=1\}$. Consider a sequence $(x_n,y_n)$ in S, s.t. it converges in $\mathbb{R}^2$.  We have $(x_n,y_n)\rightarrow (x,y)\in\mathbb{R}^2$ iff $x_n\rightarrow x\in\mathbb{R}$ and $y_n\rightarrow y\in\mathbb{R}$. This means that $x_n+y_n\rightarrow x+y$. Since $(x_n,y_n)\in S$ for all $n\in\mathbb{N}$ we have $x_n^2+y_n^2=1$ for all $n\in\mathbb{N}$. Therefore $x_n^2+y_n^2$ is a constant sequence (given equality to 1). Since $x_n^2+y_n^2\rightarrow x^2+y^2$ we get $x^2+y^2=1$. Therefore $(x,y)\in S$. This completes our proof. </p>


答:
This question is somehow interesting, as for $\phi$ is not a positive function, you can always find a solution of form :

$\phi(x)= ce^{ i2\pi x}$

To solve it in case, I'll assume $d=1$ and $\phi$ is also continuous at $0$( Note that this function is 1-periodic)

Define $h$ as $h(x)=ln(\phi(x))$

So we have easily that:

$h(x)= h(\frac{x}{2})+h( \frac{x+1}{2})$

Which leads to, by induction,:
$ h(x) = \left( \sum_{k=0}^{2^n-1} h \left( \frac{x}{2^n}+\frac{k}{2^n}\right)\right)$

So,

$  h'(x)=\frac{1}{2^n}\left( \sum_{k=0}^{2^n-1} h' \left( \frac{x}{2^n}+\frac{k}{2^n}\right)\right) \longrightarrow h(1)-h(0)$ when $n \rightarrow \infty$ (Because h is continuous)

So $h'(x)=0$
QED.

Title: $\phi(x+1)=\phi(x)$ and $\phi(\frac{x}{2})\phi(\frac{x+1}{2}) =d\phi(x)$ then $\phi$ is constant

Body:
For $0 < x < \infty$, let $\phi(x)$ be positive and continuously twice differentiable satisfying $\phi(x+1)=\phi(x)$ and $\phi(\frac{x}{2})\phi(\frac{x+1}{2}) =d\phi(x)$ with $d$ being a constant.

I have to show that $\phi$ is constant, but I don't know how to start the proof here. Any hints?


问:
$X$ is complete and $T$ is an onto open map, then $Y$ is complete
I am trying to prove sort of a converse to open mapping theorem. If $X, Y$ are normed linear spaces where $X$ is complete, and $T \in B(X, Y)$ is open, onto then I have to show Y is complete. I came across this answer here: https://math.stackexchange.com/a/1446979/698050
and my doubt is the following:

I have proved the proposition that is mentioned, it says
$∀y ∈ Y$  there is an $x \in T^{−1}(y)$ such that
$∥x∥_X ≤ C ∥y∥_Y$.

But now to prove pullback of a Cauchy sequence in $Y$ is Cauchy in $X$, say I have $x_n$ and $x_m$ in $X$ having their norms bounded by corresponding $C||y_n||$ and $C||y_m||$. This does not imply $||x_n - x_m|| \le C||y_n - y_m||$ as we can have some other $x \in X$ satisfying the proposition for $(y_n - y_m)$. How do I conclude the sequence in $X$ is Cauchy then?

答:
You need to do a little bit more. Since $(y_n)$ is a Cauchy sequence in $Y$. Then there is a subsequence $y_{k_n}$ of $(y_n)$ such that
$$\tag{1} ||y_{k_{n+1}}-y_{k_n}||<\frac{1}{2^n}$$

for all $n$. Now, take $x_n\in X$ such that $Tx_n=y_{k_{n+1}}-y_{k_n}$ and $||x_n||\leq C||y_{k_{n+1}}-y_{k_n}||$. Then by $(1)$ it follows that $\sum_{n=1}^{\infty}||x_n||<\infty$. By the completeness of $X$ there is some $x\in X$ such that $s_n\to x$ where $s_n$ are the partial sums of $\sum_{n=1}^{\infty}x_n$. Then for all n
$$Ts_n=y_{k_{n+1}}-y_{k_1}$$
and by the continuity of $T$, $Ts_n\to Tx$. It follows that $y_{k_{n+1}}\to Tx+y_{k_1}$, which means that $(y_n)$ has a convergent subsequence, hence $(y_n)$ converges. So, $Y$ is complete.


答: Hint for a different approach: Define $S:X|Ker (T) \to Y$ by $S(x+Ker(T))=Tx$. Verify that $S$ is also an open map. It is also an injective continuous linear map, so it is an isomorphism. Since $X|M$ is complete for any close subspace $M$ of $X$ it follows that $X|Ker (T)$ is complete. Since an isomorphic image of any complete space is complete we are done.

Title: $X$ is complete and $T$ is an onto open map, then $Y$ is complete

Body:
I am trying to prove sort of a converse to open mapping theorem. If $X, Y$ are normed linear spaces where $X$ is complete, and $T \in B(X, Y)$ is open, onto then I have to show Y is complete. I came across this answer here: https://math.stackexchange.com/a/1446979/698050
and my doubt is the following:

I have proved the proposition that is mentioned, it says
$∀y ∈ Y$  there is an $x \in T^{−1}(y)$ such that
$∥x∥_X ≤ C ∥y∥_Y$.

But now to prove pullback of a Cauchy sequence in $Y$ is Cauchy in $X$, say I have $x_n$ and $x_m$ in $X$ having their norms bounded by corresponding $C||y_n||$ and $C||y_m||$. This does not imply $||x_n - x_m|| \le C||y_n - y_m||$ as we can have some other $x \in X$ satisfying the proposition for $(y_n - y_m)$. How do I conclude the sequence in $X$ is Cauchy then?


问:
Techniques for backwards-Leibniz theorem of product of derivative?
If I have a function like $$u(t)f'g + w(t) fg'$$ and I know that for general $f,g$: $$(fg)' = f'g + fg'$$

Is there some general way for me to modify the expression so that I can end up with derivative of a product?

Own work:
$$\frac{\partial }{\partial t}\left\{f(a(t)) \cdot g(b(t))\right\} = a'(t)f'(a(t))\cdot g(b(t)) +f(a(t))\cdot b'(t)g'(b(t))$$

So if I can get $b',a'$ to match $u,w$ then it shall work.

A special case when this works should be "time-flipping" $a(t) = -t$

$$\sin(t) y(t) + \cos(t) y'(t) =/y(-t) = u(t)/= (u(-t)(-\cos(t)))'$$

But this is just a toy example to see that it works for a simple function. I am curious if it always will be possible or at least for some reasonably large family of functions.

答:
One way to obtain such a product will be to multiply and divide through by an integrating factor $v:=v(t)$ to obtain
$$ugf’+wg’f=\frac{1}{v}(uvgf’+vwg’f)=\frac{1}{v}(uvfg)’\,,$$ where, for example, $(uvg)’=vwg’$.
It follows that $v’ug+v(ug)’=vwg’$, or equivalently $$\frac{v’}{v}=\frac{wg’-(ug)’}{ug}= \frac{wg’}{ug}-\frac{g’}{g}-\frac{u’}{u}\,.$$ Upon integrating, the sought-after $v$ becomes $$v=\frac{1}{ug}\exp\left(\int\frac{w}{u}d(\ln(g))\right)\,.$$ If this is a less tractable integral, then consider $uvf’=(vwf)’$ and solve for $v$ instead as $$v=\frac{1}{wf}\exp\left(\int\frac{u}{w}d(\ln(f))\right)\,.$$

Added Later
In case you’d rather prefer derivative of a product but without having $v$ as quotient then you’re looking at a solution of the form $(vfg)’=uf’g+wfg’$, which can/may be solved for $v$ by integrating, though this may be less tractable than the previous approach.


问:
Find equation of a plane, that contains a point, is perpendicular to plane and parallel to a line.
I have to solve this problem from Analytic Geometry, but I'm pretty much stuck.

Find equation of plane π, that contains point A, is perpendicular to plane ρ, and is parallel to line p.

Point:

```<code>A = [1, 1, 1]
</code>```

Plane:

```<code>ρ:  x - 2y + z - 3 = 0
</code>```

Line:

```<code>p:  x + y - 2z - 2 = 0
   2x - y + 3z - 1 = 0
</code>```

We should get 3 equations in total, to get the plane π.

We get the first equation from point A:

```<code>A = [1, 1, 1] = [x, y, z] -> xa + yb + zc + d= 0
a + b + c + d= 0
</code>```

The second one is the normal vector of plane ρ:

```<code>n = (1, -2, 1) -> xa + yb + zc = 0
a - 2b + c = 0
</code>```

And the third from the vector line p (this is where I get stuck):

```<code> x + y - 2z - 2 = 0
2x - y + 3z - 1 = 0
Some mind blowing calculation...
</code>```

答:
$$p:\begin{cases}
 x + y - 2z - 2 = 0\\
   2x - y + 3z - 1 = 0\\
\end{cases}
$$
Let $x=t$ we get
$$p:\begin{cases}
 y - 2z  = -t+2\\
 - y + 3z= 1-2t\\
\end{cases}
$$
Add the equations to have $z=3-3t$

Plug in the first equation and solve $y = 8-7t$

Line $p$ has parametric equation $[t, 8 - 7 t, 3 - 3 t]$

So the direction vector is $\vec{u}=[1,-7,-3]$

The parallel plane $\pi$ has normal vector $\vec{n}=[a,b,c]$. $\pi$ is parallel to the line $p$ if $\vec{n}$ is
perpendicular to $\vec{u}$
that is $\vec{n}\cdot \vec{u}=0\to a-7b-3c=0$

$\pi$ must be perpendicular to $\rho$ which has normal vector $\vec{v}=[1,-2,1]$, thus $\vec{n}\cdot \vec{v}=0\to a-2b+c=0$
$$
\begin{cases}
a-7b-3c=0\\
a-2b+c=0\\
\end{cases}\to \begin{cases}
7b+3c=a\\
2b-c=a\\
\end{cases}
$$
Solution is $\left[a,\frac{4 a}{13}, -\frac{5a}{13}\right]$

Plane $\pi$ has equation

$a(x-1)+b(y-1)+c(z-1)=0$

Plug the values of $b,c$

$$a(x-1)+\frac{4 a}{13}(y-1)-\frac{5a}{13}(z-1)=0$$
cancel $a$ and get
$$\pi:13 x+4 y-5 z=12$$


答:
If you know how the cross product works, you only need to calculate two cross products.

The line $p$ is given as the intersection of two planes: you could convert this into a vector (or parametric) equation but you're only interested in its direction. The cross product of the normal vectors of both planes gives you exactly that:
$$\left(1,1,-2\right) \times \left(2,-1,3\right) = \left(1,-7,-3\right)$$
Since $\pi$ has to be parallel to $p$, a normal vector of $\pi$ should be perpendicular to this vector. On the other hand, since $\pi$ has to be perpendicular to $\rho$, their normal vectors should be perpendicular too. Hence, a normal vector of $\pi$ should be perpendicular to $\left(1,-2,1\right)$ and $\left(1,-7,-3\right)$, so it is given by their cross product:
$$\left(1,-2,1\right) \times \left(1,-7,-3\right) = \left(13,4,-5\right)$$
Now you have a point $A=(1,1,1)$ and a normal vector $\left(13,4,-5\right)$ so the plane is given by:
$$13(x-1)+4(y-1)-5(z-1)=0 \iff 13 x + 4 y - 5 z - 12 = 0$$

Title: Find equation of a plane, that contains a point, is perpendicular to plane and parallel to a line.

Body:
I have to solve this problem from Analytic Geometry, but I'm pretty much stuck.

Find equation of plane π, that contains point A, is perpendicular to plane ρ, and is parallel to line p.

Point:

```<code>A = [1, 1, 1]
</code>```

Plane:

```<code>ρ:  x - 2y + z - 3 = 0
</code>```

Line:

```<code>p:  x + y - 2z - 2 = 0
   2x - y + 3z - 1 = 0
</code>```

We should get 3 equations in total, to get the plane π.

We get the first equation from point A:

```<code>A = [1, 1, 1] = [x, y, z] -> xa + yb + zc + d= 0
a + b + c + d= 0
</code>```

The second one is the normal vector of plane ρ:

```<code>n = (1, -2, 1) -> xa + yb + zc = 0
a - 2b + c = 0
</code>```

And the third from the vector line p (this is where I get stuck):

```<code> x + y - 2z - 2 = 0
2x - y + 3z - 1 = 0
Some mind blowing calculation...
</code>```


问:
How to prove that there exists a $z_0 \in U_{1} [0]$ such $ \prod_{k=1}^{n} |z_0 - a_k | \geq 1 $ for $a_1, \dots , a_n \in U_{1} [0] $?
Let $a_1 , \dots , a_n $ be points in the unit circle/ball in $\mathbb{C}$ around $(0,0)$ (also known as $U_{1} [0]$), which do not necessarily differ from one another. How to prove that there exists some $z_0 \in U_{1} [0] $ such that $$ \prod_{k=1}^{n} |z_0 - a_k| \geq 1 ? $$
Here's what I tried thus far: First, pick a $z_0$ such that $|z_0 - a_j| \geq 1 $ for some $1 \leq j \leq n$. Then pick another point $a_k$ on the unit cirle. By the triangle inequality, we know that $|a_j - a_k| + |z_0 - a_k | \geq |z_0 - a_j| \geq 1 $, so $|z_0 - a_k | \geq 1 - |a_j - a_k| $. Therefore, we find that $$ \prod_{k=1}^{n} |z_0 - a_k | \geq |z_0 - a_j| \cdot \prod_{k=1 , k \neq j }^{n} (1 - |a_j - a_k| ) \geq  \prod_{k=1 , k \neq j }^{n} (1 - |a_j - a_k| ) .$$

I don't know how to proceed from here though. Any suggestions?

答:
Let $f(z)=\Pi(z-z_k)=z^n+\dots+f_0$. Let $\xi_j$ be the roots $\xi^{n+1}=1$. Since
$$
\sum_{j=0}^n f(\xi_j)=(n+1)f_0,
$$
you can find some $\xi_j$ such that $|f(\xi_j)|\ge 1$.</p>


问:
How to solve this uncomplicated equation (inspired by election arithmetic / election "math")?
[{(Disclaimer:  If it's more appropiate in a different section, please don't yell, just move it.}])

Problem:

If a candidate Smith trails a candidate Jones by 20K votes, and there's 50K votes still outstanding, what is the percentage needed of the still outstanding 50K votes to overcome Jones?

If:

"x" < 1

S = number of votes for Smith

J = number of votes for Jones

then:

Yes, intuitively I knów the answer (x = 0,7 ), but how do I get there mathematically?
Something like:

x * S - (1-x) * J = 20

x * S + (1-x) * J = 50

I don't how to solve 2 equations with 3 unknowns, but I also don't know if that is a good mathematical representation of the problem.

答: You don't need to know $S,J$. You are given $J-S=20$ (measuring in thousands). Suppose Smith gets additional $x$ votes, then Jones gets the remaining $50-x$ votes. We need Smith to overcome Jones, so the only relation needed is$$(S+x)>(J+50-x)\\2x>J-S+50=70$$giving $x>35$. So Smith must get more than $35k$ of the remaining votes to beat Jones.


问:
Limit Find the value of C given function
Limit Find the value of $c$ given function

Find $c$, such that the function

$$f(x) = \begin{cases}
 & {1 - \sqrt{x} \over x - 1},&0\le x < 1\\
 & c,& x= 1
\end{cases}$$

is continuous for all $x \in [0, 1 ]$.

I try to solve that question on interval $x = 0$ i get answer $-1$, with $f(0)$, and get undefined on limit so this is discontinous.

but when the $x$ is $1$ how can i solve? and How i get value of $c$?

i try the following approach

on left hand limit we get $- 1 / 2 $ but on right hand limit we have nothing!. to me this function seems to  not complete. this was the assignment i told our teacher to upload solution of this, but he didnt do that till yet.

Sorry, if it is nonsense question; i am software engineer student and on 2nd sem we're studying calculus subject.

答: As $x \to 1^-$, $\frac{1 - \sqrt x}{x-1} \to -\frac{1}{2}$. Therefore, for $x$ to be continuous at $x=1$, $c = -\frac{1}{2}$.


问:
IOQM/INMO- TRIANGLE GEOMETRY

In triangle $ABC$ the altitude from $A$ to $BC$ meets $BC$ at $D$ and the altitude from $B$ to $AC$ meets $AD$ at $H.$ It is also given that $AD=4, BD=3, CD=2.$ Find $HD.$

This is an IOQM-INMO question and it is complete, though if you search on net it is also given that $AB/BD=AH/HD$ but I want to know if it can be solved without using this.

答:
After four weeks of cooldown time, I hope giving this answer will not raise any issue. So, as promised, here are three solutions, and you will find no square roots (except the 3-4-5 Pythagorean Triple).

I did not mark $AD = 4$ since the placement would be awkward.

Method 1: The shortest solution: use the Angle Bisector Theorem.

Using $AC \perp BE$, $AB = BC$ and the common side $BE=BE$, $\triangle ABE \cong \triangle CBE$ by RHS. This gives $\angle ABE = \angle CBE$. Applying the theorem on $\triangle ABD$, we have $AB/BD = AH=HD$, which was the "not given" ratio.

Method 2: The simplest solution: use similar triangles.

$\angle ADB = \angle ADC = 90^\circ$. We also have $\angle DBH = 90^\circ - \angle BHD = 90^\circ - \angle AHE = \angle DAC$. By AAA similarity we have $\triangle DBH \sim \triangle DAC$. By comparing corresponding sides:

$$\frac {BD}{DH} = \frac {AD}{DC}\implies \frac 3{DH} = \frac 42$$

so it is immediate that $DH = 1.5$.

Method 3: The most general solution: use Menelaus' Theorem.

We have shown above that $\triangle ABE \cong \triangle CBE$, and thus $AE = EC$.

Applying Menelaus' Theorem on $\triangle ADC$ and the transversal $BHE$, we have

$$\frac {AE}{EC} \times \frac {CB}{BD} \times \frac{DH}{HA} = 1$$

This gives

$$\frac {DH}{HA} = \frac{BD}{CB} \times \frac {EC}{AE} = \frac3{3+2}\times \frac11 = \frac35$$

which is the ratio you seek.

Title: IOQM/INMO- TRIANGLE GEOMETRY


问:
Length of Triangle BCD

Hey, well I'm doing some higher level revision and I'm stuck...

In the diagram triangle BCD is mathematically similar to triangle ACE.

So what is the length of BD? How do you work it out?

答: The ratio DB/EA must equal the ratio DC/EC. Take it from there.</p>

Title: Length of Triangle BCD


问:
counterexample of $\rho(AB) \leq \rho(A)\rho(B)$
Suppose $A$ and $B$ are normal complex $n \times n$ matrices. Then we know that $\rho(AB) \leq \rho(A)\rho(B)$ where $\rho(.)$ is the spectral radius of a matrix.
Is there any counterexample if $A$ or $B$ is not normal?

答:
Consider $A = \begin{bmatrix} 0 &1 \\ 0 &0 \end{bmatrix}$ and $B = A^T$.
$\rho(A) = 0 = \rho(B)$ but $AB = \begin{bmatrix} 1 &0 \\ 0 &0 \end{bmatrix}$ then $\rho(AB)=1$.


问:
Compute approximately the probability that more than 50 of the observations of the random sample are less than 3.
Let $X_1,X_2,...,X_n$ be a random sample of size $n = 72$ from a distribution
with probability density function

$f(x) =\begin{cases} 1/x^2,& 1 < x < infinity\\0,&\text{otherwise}\end{cases}$

Compute approximately the probability that more than $50$ of the ob-
servations of the random sample are less than $3$.

答:
Step I :

Define $Z_i = I_{\{X_i < 3\}}$ and find the probability that $Z_i = 1$ which is equal to $\int_{1}^{3}{f(x)dx}$.

Step II :

Apply Central Limit Theorem on $Z_i$'s to approximate $P[\sum_{i=1}^{72}{Z_i} > 50]$. If you do not know how to do that, here is a reference for the same (page 6) https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter9.pdf


问:
Why does the following probability identity hold (probability of at least one sucess)
Assume we have a coin, which has probability $p$ for showing heads. Why does it hold that
$$P(at ~least ~one ~head) = P(first ~flip ~is ~head) + P(second~flip~is~head)\cdot P(first~flip~is~tail) + P(third~flip~is~head)\cdot P(first~and~second~flip~is~tail)?$$
I know it would be simpler to write it as $1-P(no~tails)$, but I was wondering why this works? Do we need to assume independence of the single flips here?

Thanks in advance

答:
Flipping 3 times a coin the sample space is the following

$$\Omega=\{TTT;HTT;THT;TTH;HHT;HTH;THH;HHH\}$$

The probability to get "at least one H" is

$$1-\mathbb{P}(TTT)=\mathbb{P}(HTT)+\mathbb{P}(THT)+\mathbb{P}(TTH)+\mathbb{P}(HHT)+\mathbb{P}(HTH)+\mathbb{P}(THH)+\mathbb{P}(HHH)$$

EDIT: your equation holds because it is only a different partition of the sample space

$$\underbrace{\mathbb{P}(HTT)+\mathbb{P}(HTH)+\mathbb{P}(HHT)+\mathbb{P}(HHH)}_{\text{=first flip H}}+\underbrace{\mathbb{P}(THT)+\mathbb{P}(THH)}_{\text{=First T, second H}}+\underbrace{\mathbb{P}(TTH)}_{\text{FIrst and Second T, third H}}$$

but this is wrong

I know it would be simpler to write it as $1−P(no tails)$

The correct one is

$$\mathbb{P}[\text{at least one H}]=1-\mathbb{P}[\text{no Heads}]$$