问:
Invertible modules over a noetherian ring

Let $A$ be a commutative noetherian ring and let $M, N$ be two
finitely generated $A$-modules such that $M\otimes_A N\cong A$

I would like to know which are all such $A$-modules.

It seems to me that $M$ (and so $N$) has to be free and then since $M\otimes_A N\cong A$, it follows that they are both of rank $1$, i.e. isomorphic to $A$ itself.

My understanding of the problem

Using the noetherian hypothesis on $A$ we know that $M, N$ are also finitely presented and so we have two exact sequences \begin{equation*}A^{\oplus^{m_1}}\to A^{\oplus^{m_2}}\to M\to 0 \\ A^{\oplus^{n_1}}\to A^{\oplus^{n_2}}\to N\to 0 \end{equation*} We could try to tensor the first sequence by $N$ and the second one by $M$ to obtain \begin{equation*}N^{\oplus^{m_1}}\to N^{\oplus^{m_2}}\to A\to 0 \\ M^{\oplus^{n_1}}\to M^{\oplus^{n_2}}\to A\to 0 \end{equation*}but maybe this is a dead end. I don't have any further idea. Any hint/solution is appreciated.

答:
Perhaps I found another way to prove it. First of all we can assume that $A$ is a local ring and let $m$ be its unique maximal ideal.

The isomorphism $M\otimes_A N\cong A$ induced an isomorphism $M/mM\otimes N/mN\cong \mathcal{K}$ where $\mathcal{K}:=A/m$ is the residue field and both factor in the latter tensor product are $\mathcal{K}$-vector spaces. In particular they are of dimension $1$ and by Nakayama's lemma $M$ (and $N$) is cyclic, i.e. is generated by one element $x$. So we can write $M=(x):=\{ax \ | a\in A\}$. In particular is easy to prove that $M\cong A/Ann_A(x)$ where $Ann_A(x):=\{a\in A \ | ax=0\}$ is the annihilator of $x$ in $A$. Suppose now that there exists an element $a\in A$ such that $ax=0$ in $A$ ( so $a\in Ann_A(x)$). Consider the map $f:A\to A$ given by $f(b):=ab$, then it follows that \begin{equation*}1_M\otimes f:A\otimes_A M\cong M\to A\otimes_A M\cong M \end{equation*}is the zero map since $M\cong (x)$ and $ax=0$. From this follows that $f$ is the zero map and since $1\in A$ then $a=0$. We conclude that $M\cong A$ as desired.

If anyone can check my argument I appreciate.


问:
Optimisation problem in a browser game
I am playing a game, and I have an optimisation problem I would like to solve. I can send soldiers to plunder resources. There are two ways I can do it: In the first way, they gather less, but they return to my base faster. In the second they gather more, but it takes them longer.

The numbers look like this:

I can use both ways simultaneously. So if I decide to send 30 soldiers using first method I can still send another 20 using the second method. I would like to know what is the most efficient way to send the troops. Even though Method 2 is clearly a better than Method 1, I don't think that using the Method 2 only is the best way forward

答:
There might be any number of situations where you would choose to use the first method or send fewer than $50$ soldiers: maybe you need the soldiers to defend your base; maybe they're sleeping; maybe you absolutely need the resources right now; ...

But without further context or constraints, you should obviously send $50$ soldiers using the second method.

ContentLicense: CC BY-SA 4.0

Title: Optimisation problem in a browser game


问:
Is it possible that a space which isn’t an affine variety is the union of countable affine varieties?
This question is distinct from other questions on the topic about unions as I am specifying it is a union of countable varieties. Is there any simple (or more abstract) example of this? Or is it not possible that the union of countable affine varieties is not an affine variety? My gut feeling is that the union should be affine but would like to be sure of it</p>

答: Take $\Bbb Z \subset \Bbb A^1_\Bbb C$. It is a union of singletons, and each singleton is an affine variety because $\{n\} = V(x-n)$. However, $\Bbb Z$ is not an affine variety. (Why?)


问:
Euler product over a finite subset of composite numbers
The product $$\prod_{p\text{ prime}}\frac{p^n}{p^n-1}=\zeta(n)$$ is well-known. This works over primes, and only over the entire set of them.

Is any similar expression known for composites? For a finite subset of them?

e.g. $$\prod_{c\text{ composite, c}=c_{i}}^{c=c_{j}}\frac{c^n}{c^n-1}=f(m)$$ for some $f$.

This is generic. If such an expression is not known, is anything known for the case $i=1$ and $n=1$? If not, then for $i=n=1$ and $j=\infty$? If that is also not known, is anything known on $\lim \inf f(m)$ or $\lim \sup f(m)$?

答:
Surely we can say something in the case of the set of all composite numbers and for an integer exponent. From the following representation of Gamma function $$\Gamma\left(z\right)=\lim_{n\rightarrow+\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ it is possible to prove that $$\prod_{n\geq0}\frac{\left(n+a_{1}\right)\cdots\left(n+a_{k}\right)}{\left(n+b_{1}\right)\cdots\left(n+b_{k}\right)}=\frac{\Gamma\left(b_{1}\right)\cdots\Gamma\left(b_{k}\right)}{\Gamma\left(a_{1}\right)\cdots\Gamma\left(a_{k}\right)},\,a_{1}+\dots+a_{k}=b_{1}+\dots+b_{k}$$ for $a_{i},b_{j}\in\mathbb{C}\setminus\mathbb{Z}^{-},$ (see Whittaker and Watson, A course of modern analysis, $4$th edition, Cambridge University Press, Cambridge, $1927$, section $12.13$) then if $k>1$ is a positive integer we have $$\prod_{n\geq2}\frac{n^{k}}{n^{k}-1}=\prod_{n\geq0}\frac{\left(n+2\right)^{k}}{\left(n+2-\omega_{0}^{k}\right)\cdots\left(n+2-\omega_{k-1}^{k}\right)}=\frac{\Gamma\left(2-\omega_{0}^{k}\right)\cdots\Gamma\left(2-\omega_{k-1}^{k}\right)}{\Gamma\left(2\right)^{k}}$$ where $\omega_{j}^{k}=\exp\left(\frac{2\pi ij}{k}\right)$. Then, since $\Gamma(2)=\Gamma(1)=1$, $$\prod_{c\text{ composite}}\frac{c^{k}}{c^{k}-1}=\color{red}{\frac{\Gamma\left(2-\omega_{1}^{k}\right)\cdots\Gamma\left(2-\omega_{k-1}^{k}\right)}{\zeta\left(k\right)}}.$$
ADDENDUM: I think it is interesting to see how the answer of Daniel D. is linked to mine; from the power series $$\sum_{n\geq2}\frac{\left(-1\right)^{n}x^{n}\zeta\left(n\right)}{n}=\gamma x+\log\left(\Gamma\left(x+1\right)\right),\,-1<x\leq1$$ we have $$\sum_{n\geq2}\frac{\left(-1\right)^{n}x^{n}\left(\zeta\left(n\right)-1\right)}{n}=\left(\gamma-1\right)x+\log\left(\Gamma\left(x+1\right)\right)+\log\left(\left(x+1\right)\right)$$ which can be extended to $x=-1$ and where $\gamma$ is the Euler-Mascheroni constant. Now recalling that if $f(x)=\sum_{n\geq0}a_{n}x^{n}$ then we have the multisection formula $$\sum_{n\geq0}a_{qn+p}x^{qn+p}=\frac{1}{q}\sum_{j=0}^{q-1}\omega_{q}^{-jp}f\left(\omega_{q}^{j}x\right)$$ where $\omega_{q}^{j}$ is a primitive $q$-root of the unity, we get, assuming that $f(x):=\left(\gamma-1\right)x+\log\left(\Gamma\left(x+1\right)\right)+\log\left(\left(x+1\right)\right)$ and $m>1$, that $$\sum_{n\geq1}\frac{\zeta\left(mn\right)-1}{n}=\lim_{x\rightarrow-1^{+}}\sum_{j=0}^{m-1}f\left(\omega_{m}^{j}x\right)=\color{red}{\sum_{j=1}^{m-1}\log\left(\Gamma\left(2-\omega_{m}^{j}\right)\right)}.$$ Notice that we used the relations $\sum_{j=0}^{q-1}\omega_{q}^{j}=0$ and $$\lim_{x\rightarrow-1^{+}}\left[\log\left(\Gamma\left(x+1\right)\right)+\log\left(\left(x+1\right)\right)\right]$$ $$=\lim_{x\rightarrow-1^{+}}\left[\log\left(\Gamma\left(x\omega_{q}^{0}+1\right)\right)+\log\left(\left(x\omega_{q}^{0}+1\right)\right)\right]=0.$$


问:
The signum function expressed using a single formula
The signum function is defined as a piecewise function:  $\operatorname{sgn} (x)=\left\{ \begin{matrix}
   1,x>0  \\
   0,x=0  \\
   -1,x<0  \\
\end{matrix} \right.$.

Naturally, we could simplify this: $\operatorname{sgn}  (x)=\left\{ \begin{matrix}
   \frac{\left| x \right|}{x},x\ne 0  \\
   0,x=0  \\
\end{matrix} \right.$.

I’m just wondering is it possible to define this function using a single formula? I understand that the absolute value $\left| x \right|$ is a piecewise function itself but for the sake of this question let’s assume we are allowed to use modulus to define the signum function.

Edit:

I completely agree with the answers given below; however, my original question was about the possibility to define the signum function using only elementary functions and the absolute value function (which itself is not an elementary function) in such a way that would require exactly one formula (i.e., avoiding a piecewise function).
For example, the following function $g\left( x \right)=\left\{ \begin{matrix}
   x,x>0  \\
   0,x\le 0  \\
\end{matrix} \right.$ can be defined in such a manner: $g\left( x \right)=\frac{x+\left| x \right|}{2}$.

答: Using only the arithmetic operations, the identity function and the absolute value, this does not seem possible, because the $\text{sgn}$ function is discontinuous. As far as I see, a discontinuity can only appear with a division by zero, which results in an undefined value in at least one point.


问:
Is there an image to the point at infinity through this map?
I encountered a conformal mapping on the complex plane:$$z\rightarrow e^{i\pi z}$$
and I am not sure about where it does send the point at infinity. If I could say something along the lines: $$\text{Im}(\infty) = \infty$$
Then it would map it to the origin but there is still a voice in my head saying that this equality is non-sense. And from the usual definition of the imaginary part:$$\text{Im}(z)=\frac{z+\bar{z}}{2i}$$ it makes even less sense.
I'd appreciate some enlightenment.</p>

答: If we allow $z$ to approach the point at infinity along the positive imaginary axis, we find that $e^{i\pi z}$ tends to $0$. Approaching along the negative imaginary axis, we find that $e^{i\pi z}$ gets big without bound. It turns out that we can make $e^{i\pi z}$ tend toward anything we like, just by allowing $z$ to approach the point at infinity along an appropriate path, and along some paths (such as the positive real axis) it won't tend toward anything at all. Thus, there's no nice way to extend the map to the point at infinity.

ContentLicense: CC BY-SA 3.0


问:
Question about the proof for "The rational line Q does not have the completeness property"
Referring to the attached png file, I have questions about the inequality:

$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$

How is this inequality derived?  How did the n get into the inequality?  Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$, and later in the proof, $2q^2/(p^2 - 2q^2)$ come from?   There were no derivations for these 3 expressions.  Could someone please provide the derivations?

答: You have $\frac{p^2}{2q^2-p^2}$ fixed. Since $\lim_{n \rightarrow \infty} \frac{n^2}{2n+1}=\infty$, using definition of limit at $\infty$ the inequality involving $n$ is obtained.


问:
Notation for n-qubits.
I have been asked to, given a n-qubit state and measuring  in the computational basis on the first qubit, to calculate the probability of obtaining the outcome 1 and writing the state of the system after the measurement.

The problem is I do not understand the notation of it. Could someone help? This is the system:

$|ψ⟩ = \sum_{x∈\{0,1\}^n} a_x|x⟩ ∈ C^{2^n}$

I know how to do it for 1-2-3...qubits, but do not know how to unroll this formula.

Thanks in advance!

答:
We may represent the state as follows:
$$ |\psi\rangle = \sum_{b_1,b_2,\dots,b_n\in\{0,1\}} \alpha_{b_1b_2\cdots b_n}|b_1b_2\dots b_n\rangle $$
where $\alpha_{b_1b_2\dots b_n}\in\mathbb{C}$ are the coefficients in the computation basis. Assuming this state is normalized, we have
$$
\langle \psi|\psi\rangle = \sum_{b_1b_2,\dots,b_n\in\{0,1\}} |\alpha_{b_1b_2\cdots b_n}|^2 = 1.
$$

If we were to measure the first qubit and obtain the outcome that the first qubit is in the $|1\rangle$ state, this collapses the state $|\psi\rangle $ into superpositions of states whose first bit is $|1\rangle$ (i.e., $b_1=1$). That is, the resulting (unnormalized) state is
$$
|\psi'\rangle = \sum_{b_2,\dots,b_n\in\{0,1\}} \alpha_{1b_2\cdots b_n}|1b_2\dotsb_n\rangle.
$$
The probability of having obtained this result is the norm $\langle \psi'|\psi'\rangle = \sum_{b_2,\dots,b_n\in\{0,1\}} |\alpha_{1b_2\cdots b_n}|^2$.

Does this help?

Title: Notation for n-qubits.


问:
Let $\eta$ be $4 X 1$ vector and $C$ a $4 X 4$ symmetric matrix. Is it possible to simplify the following expression? (expression too big for title!)
Let:
$\eta = \begin{bmatrix} \eta_1 \\ \eta_2 \\ \eta_3 \\ \eta_4 \end{bmatrix}$ and its derivative in respect to time: $\dot{\eta} = \frac{d\eta}{dt} = \begin{bmatrix} \dot{\eta_1} \\ \dot{\eta_2} \\ \dot{\eta_3} \\ \dot{\eta_4} \end{bmatrix}$.

Let $C$ be a $4 \times 4$ symmetric matrix whose elements are all functions of $\eta_1$, $\eta_2$, $\eta_3$ and $\eta_4$.

I have the following system of 4 equations (one for each element):

$\begin{cases}
a_1 = \dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_1}\right) \dot{\eta} \\
a_2 = \dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_2}\right) \dot{\eta} \\
a_3 = \dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_3}\right) \dot{\eta} \\
a_4 = \dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_4}\right) \dot{\eta} \\
\end{cases}$

I need to rewrite this system in Matrix form, so I defined $A = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \\ a_4 \end{bmatrix} $ and rewrote the equation as:

$A = \begin{bmatrix}
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_1}\right) \dot{\eta} \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_2}\right) \dot{\eta} \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_3}\right) \dot{\eta} \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_4}\right) \dot{\eta} \\
\end{bmatrix}$

And managed to show it is equivalent to:

$A = \begin{bmatrix}
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_1}\right) \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_2}\right) \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_3}\right) \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_4}\right) \\
\end{bmatrix} \dot{\eta}$

It is possible to rewrite the expression $\begin{bmatrix}
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_1}\right) \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_2}\right) \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_3}\right) \\
\dot{\eta}^T \left(\cfrac{\partial C}{\partial \eta_4}\right) \\
\end{bmatrix}$ in a more simplified/compact way, defining maybe some kind of tensor operation or matrix derivative operator?

Thanks a lot in advance!

答:
$\def\v{\operatorname{vec}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$Define
the scalar
$$\eqalign{
\phi &= \dot\eta^T C\dot\eta \\
 &= (\dot\eta\otimes\dot\eta)^T \v(C) \\
 &= v^T c \\&= c^Tv \\
}$$
where $\otimes$ denotes the Kronecker product and the vec operation stacks the columns of $C$ to create one long column vector $c$.

The vector in question is the gradient of this scalar, i.e.
$$\eqalign{
A &= \p{\phi}{\eta} &= \left(\p{c}{\eta}\right)^Tv \\
A_k &= \p{\phi}{\eta_k}
 &= \left(\p{c}{\eta_k}\right)^Tv
 \,\;=\; \dot\eta^T\left(\p{C}{\eta_k}\right)\dot\eta \\\\
}$$
NB:   This assumes that $\{\eta,\dot\eta\}$ are independent variables, such as position and velocity.


问:
Can all points in the plane be represented like this?
Solving a task regaring affine geometry, I've come across a problem:

Is it  true that, for every point $(x,y)\in \mathbb{R}^2$, there  exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that

$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$

I don't have any idea how to prove this or find a counterexample. All hints appreciated!

答:
Here is an attempt at a visual interpretation.

The equation is equivalent to

$$ \frac{ x - \sin \alpha} { \cos \alpha} = t = \frac{ y - \cos \alpha} { \sin \alpha},$$

$$ \frac{ y - \cos \alpha } { x - \sin \alpha} = \frac{ \sin \alpha} { \cos \alpha}$$

What this means, is that the slope of the point $(x,y)$ to the point $A=(\sin \alpha, \cos \alpha)$ on the unit circle, is $\tan \alpha$.


问:
Drawing the grid without lifting the pen
The below shape consist of $24$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw?
$1)25\quad\quad\quad\quad\quad\quad2)26\quad\quad\quad\quad\quad\quad3)27\quad\quad\quad\quad\quad\quad4)28\quad\quad\quad\quad\quad\quad5)29\quad\quad\quad\quad\quad\quad$

I think we can do it somehow by segments with the length $28$ (by passing twice the four middle segments on each side of outer big square).

I'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?

答: Notice that if you draw it without lifting the pen, except for the starting and ending point you must draw the same number of lines going into each point as going out. This means that you must draw an even number of segments meeting every point except possibly two. How many extra segments does this imply you need at a minimum? Can you achieve this?</p>


问:
if I have 100 numbers, and 2 lists each with a certain number of random numbers from the set of 100. How can I tell what the overlap should be.
To clarify, say I have 100 numbers.

group 1 has a random 30 out of those 100 numbers.

group 2 has a random 50 out of those SAME 100 numbers.

Is there a way to tell how many numbers should overlap?

(The 30 and 50 are just examples. Hopefully, there's a formula where I'd be able to swap out those numbers for others).

答:
The expected overlap under the assumption of independent random selection can be computed from the simple proportions selected by each group.

In your example, Group 1 consists of a random 30% of the cohort, so it's therefore expected to contain 30% of Group 2 (since we assume Group 2 is independently randomly selected from the cohort at large) - the expected overlap is 30% of 50% of 100, or 15 samples.

You can also compute this the other way, reasoning that Group 2 is a random 50% of the cohort, and therefore is expected to contain 50% of Group 1. The expected overlap is 50% of 30% of the samples, yielding the same answer of 15 samples.

This number represents the expected, most likely size of the overlap, but it's actually unlikely to find exactly 15 samples overlapping due to random variation. Seeing 14 or 16 samples overlapping would also be reasonably likely. One could perform a hypothesis test to determine the size of overlap that would be deemed too unlikely to occur by chance. If you see that both groups overlap perfectly or not at all, for example, it's highly unlikely that the null hypothesis that the groups were selected randomly and independently from one another is true. To perform such a test, you'd use the hypergeometric test, which requires four data points - the size of Group 1, the size of Group 2, the size of the overlap, and the size of total cohort from which the groups are selected.


问:
How to solve the differential equation for the motion equation of a body in a gravitational field from one fixed source
I want to develop the motion equation of a body in a classic gravitational field ($F=\frac{Gm_1m_2}{r^2}$).

Starting by creating the lagrangian of a body under gravitational force, in polar coordinates. The speed in direction $\hat{r}$ is $\dot{r}$ and the speed in direction $\hat{\theta}$ is $r\dot{\theta}$. So the kinetic energy of the body is $K=\frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)$ and the potential energy is $U=-\frac{GMm}{r}$.

$M$ is the mass of the source generating the gravitational field (a star), and $m$ is the mass of the body (a planet).

Creating the lagrangian we get:

$$\mathcal{L}=K-U=\frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)+\frac{GMm}{r}$$

Writing down the Euler–Lagrange equation ($\frac{{\partial}\mathcal{L}}{{\partial}q}=\frac{d}{dt}\left(\frac{{\partial}\mathcal{L}}{{\partial}\dot{q}}\right)$)we get
$$mr^2\ddot{\theta}=0$$
$$\dot{\theta}=\frac{p_{\theta}}{mr^2}$$
$p_\theta$ is the angular momentum which is conserved
$$\theta=\frac{p_{\theta}}{mr^2}t+\theta_0$$

$$m\ddot{r}=mr\dot{\theta}^2-\frac{GMm}{r^2}$$

$$\ddot{r}=\frac{p_{\theta}^2}{m^2}r^{-3}-GMr^{-2}$$

How do I solve the differential equation?

答:
Some of the  computations in the middle seem wrongly to rely on $r$ being a constant. In the Euler-Lagrange equation for the angle, one gets
$$
\frac{d}{dt}(mr^2\dot\theta)=0\implies mr^2\dot\theta=p_\theta=const.
$$
The next step is likewise only possible for circular orbits, constant radius. Else the angle is not linear in time.

Then on the correctly performed Euler-Lagrange equation for the radius transform to
$$
2r\dot r = \frac{p_\theta\dot r}{mr^2}\left(\frac{p_\theta}{mr}-\frac{GMm}{p_\theta}\right)
$$
which integrates nicely to
$$
\dot r^2+\left(\frac{p_\theta}{mr}-\frac{GMm}{p_\theta}\right)^2=C^2\tag{*}
$$
Now set
$$
\dot r=C\sin\phi,~~~\left(\frac{p_\theta}{mr}-\frac{GMm}{p_\theta}\right)=C\cos\phi\tag{**}
$$
so that
$$
C\dot\phi\sin\phi=\frac{p_\theta\dot r}{mr^2}=\frac{p_\theta C\sin\phi}{mr^2}
\\
\implies mr^2\dot\phi=p_\theta=mr^2\dot\theta\\\implies \phi=\theta+\psi
$$
with a constant phase angle $\psi$. The second equation in (**) can be written as
$$
r=\frac{R}{1+E\cos\phi}
$$
with $R$, $E$ constants for the reference radius and eccentricity.

The remaining dynamic is then contained in
$$
\frac{mR^2\dot\phi}{(1+E\cos\phi)^2}=p_\theta
$$


问:
$\int_0^1f(x)dx= \lim_{c\rightarrow0}\int_c^1f(x)dx$
I am reading Rudin's "Priciples of Mathematical Analysis", I am trying to solve exercise 6.7 which says:

Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathcal R$ on $[c, 1]$ for every $c > 0$.

Define:
$$\int_0^1f(x)dx= \lim_{c\rightarrow0}\int_c^1f(x)dx$$
if this limit exists (and is finite).

(a) If $f \in \mathcal R$ on $[0, 1]$, show that this definition of the integral agrees with the old one.

My solution is:

Since $f \in \mathcal R$, we know that $f$ is bounded, let $|f| \leq M$.

Let $c < \epsilon / M$.

$$\left| \int_0^1f(x)dx - \int_c^1f(x)dx \right| = \left| \int_0^cf(x)dx \right|$$
$$\leq M.c = \epsilon$$

And we are done.

Is there is a flaw in this solution?

答:
Expanding on the comment, to complete this proof, you also require $f$ to be Riemann Integrable on $[0,1]$. To observe this, we will use the Riemann's Criteria of Integrability.

Let $\epsilon >0$. Choose $c = \frac{\epsilon}{2(M-m)}$, where $M$ and $m$ are the maximum and minimum values of the function $f$ on the interval $[0,1]$.

Note that since $f \in \mathcal R[c,1]$, for this $\epsilon$, $\exists \ P=\{c,x_1,x_2....1\}$ such that
$$U(P,f) - L(P,f) < \frac{\epsilon}{2}$$

Now, consider the partition $P_c = \{0,c,x_1,x_2,...1\}$ of $[0,1]$ (i.e. the partition $P$ with the additional point $0$) Corresponding to this partition we have:

$$U(P_c,f) - L(P_c,f) = \sum_{i=1}^{n} (M_i-m_i)\Delta x_i$$

Suppose the maximum and minimum values of $f$ on $[0,c]$ are $M_1$ and $m_1$ respectively. We can therefore split the summation as:

$$U(P_c,f) - L(P_c,f) = (M_1-m_1)c + \sum_{i=2}^{n} (M_i-m_i)\Delta x_i$$

Note that $M_1-m_1 \leq M-m$. Also, the sum $\sum_{i=2}^{n} (M_i-m_i)\Delta x_i$ is nothing but $U(P,f) - L(P,f)$. Thus,

$$U(P_c,f) - L(P_c,f) < (M-m)c + U(P,f) - L(P,f) < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon$$
Thus, $f \in \mathcal R  \blacksquare$ and you can proceed as you did in your solution.


问:
Do falling factorials form a Schauder basis for formal power series in some topology?
We usually talk about $F[[x]]$, the set of formal power series with coefficients in $F$, as a topological ring.  But we can also view it as a topological vector space over $F$ where $F$ is endowed with the discrete topology.  And viewed in this way, $\{x^n:n\in\mathbb{N}\}$ is a Schauder basis for $F[[x]]$.

Now in contrast, $\{(x)_n:n\in\mathbb{N}\}$, where $(x)_n$ denotes the falling factorial, is not a Schauder basis for $F[[x]]$.  That’s because if $\Sigma_na_n(x)_n$ never converges in the standard topology on $F[[x]]$ if infinitely many of the $a_n$’s are nonzero.  But my question is, does there exist some alternate topology on $F[[x]]$ which makes $\{(x)_n:n\in\mathbb{N}\}$ a Schauder basis for $F[[x]]$ as a topological vector space over $F$ endowed with the discrete topology?

答:
I don't see the point of the question.

With the axiom of choice there is some $F$-vector subspace such that $$F[[x]]=W\oplus F[x]$$

For a sequence $c_n\in F$ we get the unique decomposition $$\sum_{n\ge 0}c_n x^n= w(c)+\sum_{k=0}^{d(c)} a_k(c) x^k$$

When $c_n$ is zero for $n> N$ we have $w(c)=0$ thus $d(c)=N$,$a_k(c)=c_k$.

Sending $c_n$ to $$w(c)+\sum_{k=0}^{d(c)} a_k(c) (x)_k$$
gives an alternative topology on $F[[x]]$, the topology inherited from that of $F^\Bbb{N}$.

By definition of the alternative topology $$\lim_{N\to \infty} \sum_{n=0}^N c_n (x)_n =w(c)+\sum_{k=0}^{d(c)} a_k(c) (x)_k$$


问:
Percentiles of sums of random variables
Heads up, I'm asking this question after having read Percentile of sum of random variables and other questions. Please don't close this as a duplicate. It's not the same question.

Suppose I have two positive random variables $X_1$ and $X_2$, and the following percentiles or upper bounds on them. The joint probability assumes independence.

\begin{aligned}
P(X_1<35) &= 0.95 \\
P(X_2<40) &= 0.95 \\
P(X_1<35, X_2<40) &= 0.95^2
\end{aligned}

$X_1$ is concentrated beneath $35$, and $X_2$ beneath $40$. They're both under their respective limits with probability $0.0925$. That means that their sum can not reach $\ge70$ with $p=0.0925$, right? How can I prove this? I have a feeling I'm wrong. I know I'm skipping steps, yet don't know the steps.

$$P(X_1+X_2<35+40) = 0.95^2$$

答:
No. You can't. Because it is not true.

Because it is possible to get less than 75 in other ways also. For example, 10+65 = 75.

$$P(X_1+X_2<35+40) = 0.95^2 $$
$$\implies P(X_1+X_2<35+40 \land X_1<35 \land X_2<40) $$$$+ P(X_1+X_2<35+40 \land X_1<35 \land X_2\ge40) $$$$+ P(X_1+X_2<35+40 \land X_1\ge35 \land X_2<40) $$$$+ P(X_1+X_2<35+40 \land X_1\ge35 \land X_2\ge40) $$$$= 0.95^2$$

In this, the fourth term is obviously 0.

In the fist term, $X_1<35 \land X_2<40 \implies X_1+X_2<35+40$. So the first term becomes $P(X_1<35 \land X_2<40)$ which is nothing but $ 0.95^2$.

Substituting the first and fourth term, we get
$$ 0.95^2 + P(X_1+X_2<35+40 \land X_1<35 \land X_2\ge40) $$$$+ P(X_1+X_2<35+40 \land X_1\ge35 \land X_2<40) +0 $$$$ = 0.95^2 $$

Which means
$$ P(X_1+X_2<35+40 \land X_1<35 \land X_2\ge40) + P(X_1+X_2<35+40 \land X_1\ge35 \land X_2<40) =0 $$

Which obviously not true.
What could be true is
$$P(X_1+X_2<35+40) > 0.95^2 $$

Title: Percentiles of sums of random variables


问:
Function from [0,1] to rational number
Suppose that $f: [0,1]\to \mathbb{Q}$ is a function such that: for any sequence $x_1,x_2,\ldots,\in [0,1]$ such that $f(x_1)=f(x_2)=\ldots$ and $\lim x_n=x\in [0,1]$ then $f(x)=f(x_1)$. Prove that there exists a non-empty open interval $I$ of $[0,1]$ such that $f$ is constant on that interval.

So far, by the property of $f$, I have proved that for any $r\in \mathbb{Q}$ then the preimage $f^{-1}(r)$ is compact in $[0,1]$. But I don't know how to proceed from there. Any suggestion?

答: If you look at this powerful theorem of Sierpinsky, and consider the countable, pairwise disjoint, closed cover of $[0,1]$ given by $\{f^{-1}(q)\mid q\in\mathbb{Q}\}$, you can conclude that $f$ is a constant function.

Title: Function from [0,1] to rational number


问:
Value of a zero-sum matrix games, if one matrix has smaller entries than another
Suppose that we have two zero-sum games represented by matrices $A = (a_{ij})$ and $B = (b_{ij})$. I am curious to know that if we assume $a_{ij} \leq b_{ij}$ for all $i,j$, does this imply that the value of nash equilibria in game $A$ has a smaller value than equilibria in game $B$?

This seems intuitive to me but I have not been able to find any information, nor prove or disprove the statement.

答:
For any arbitrarily fixed $x\geq 0$ and $y\geq 0$, we can say that $x^TAy\leq x^TBy$.

$\Rightarrow$ $\min_y x^TAy\leq \min_y x^TBy $ (as the above inequality is true for all $y$)

$\Rightarrow$ $\max_x \min_y x^TAy\leq \max_x\min_y x^TBy$. Thus the nash equilibrium of the first game is always smaller than that of the last game. It seems this is true because of specific properties that $x$ and $y$ carry, namely $1^Tx=1, 1^Ty=1, x,y\geq 0$. Hope this helps.


问:
$\mathbb N$ a Banach space?
Is $\mathbb N$ a Banach space with the norm $|x-y|$ from $\mathbb R$? I think is Banach space because there is no convergent sequence that is not constant after some $N$. Then all limit points are in the space. But I am not sure. </p>

答: It surely is a complete metric space, your proof is correct. Another justification of this would be that it is a closed subspace of the Banach space $(\Bbb R,|\cdot|)$. But it has no linear structure, so it's not a Banach space.</p>

Title: $\mathbb N$ a Banach space?


问:
Probability math question, what is the chance that player 1 will win more matches than player 2?
player 1 and player 2 play a series of 4 games. for each game the probability of player 1 winning is 0.5 and the probability of player 2 winning is 0.5. What are the chances that player 1 will win more of the games than player 2???</p>

答:
First, we calculate the probability they have an EQUAL number of wins. First of all, there is a $\frac1{16}$ chance for each possible win pattern because you have a $\frac12$ chance multiplied by itself $4$ times.

Now, count the number of ways to have an equal number of wins. We can choose $2$ out of the $4$ wins to be Player $1$s win, and $ {4  \choose 2 }=6$. So now we multiply $6*\frac{1}{16}$ to get $\frac38$. Since the probability they have an equal number of wins is $\frac38$, then the probaiility of having an UNEQUAL number of wins is $\frac58$.

If they have an unequal number of wins, both of them have the same chance of winning more than the other because they are equally skilled, so the probability that one player wins, specifically Player $1$, is $\frac58 /2 = \frac{5}{16}$.


问:
Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$

Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$

Attempt:

First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...

Now I try to find the sum and I can't get to the solution $\frac{5}{2}$.

Let's try:

$$\sum_{k=2}^{\infty} \frac{5}{2^k}=5\sum_{k=2}^{\infty} \frac{1}{2^k}=5\sum_{k=2}^{\infty} 2^{-^k}=5\cdot2^{-2}\sum_{k=2}^{\infty}2^{-k}$$

I don't know how to work with the $2^{-k}$ in here, hope for your help^^

Body: $$\sum_{k=2}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} - (1 +\frac{1}{2}) = \frac{1}{1 - \frac{1}{2}} -\frac{3}{2} = \frac{1}{2}$$So the answer is $5\times\frac{1}{2} = \frac{5}{2}$


问:
Inequality induction proof with two variables
For all $ ∈ $ and all $>0$ it holds that $1+≤(1+)^$

I've done the base-case already

Induction steps.

答:
If $1+ks\leq(1+k)^{s}$ for $s\in\mathbb{N}$ then we need to prove that $1+k(s+1)\leq(1+k)^{s+1}$.

$(1+k)^{s+1}=(1+k)(1+k)^{s}=(1+k)^{s}+k(1+k)^{s}\geq(1+ks)+k=1+k(s+1)$, where the inequality above holds because of the induction step and because $1+k>1$.


Title: Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$

Body:
Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$

Attempt:

First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...

Now I try to find the sum and I can't get to the solution $\frac{5}{2}$.

Let's try:

$$\sum_{k=2}^{\infty} \frac{5}{2^k}=5\sum_{k=2}^{\infty} \frac{1}{2^k}=5\sum_{k=2}^{\infty} 2^{-^k}=5\cdot2^{-2}\sum_{k=2}^{\infty}2^{-k}$$

I don't know how to work with the $2^{-k}$ in here, hope for your help^^

Body: Actually,$$\sum_{k=2}^\infty2^{-k}=2^{-2}\sum_{k=0}^\infty2^{-k}=2^{-2}\times2=\frac12.$$


答: No, a Banach space has to be a vector space over $\mathbb R$.  This is not true for $\mathbb N$ because in general $n\in \mathbb N$ does not mean that $\lambda n\in \mathbb N$ for every $\lambda \in\mathbb R$.</p>

Title: $\mathbb N$ a Banach space?

Body: Is $\mathbb N$ a Banach space with the norm $|x-y|$ from $\mathbb R$? I think is Banach space because there is no convergent sequence that is not constant after some $N$. Then all limit points are in the space. But I am not sure. </p>


问:
Graph puzzles: Constructing graphs from tiles
There is a vast literature on the reconstruction conjecture which says that two graphs with the same deck $D$ are isomorphic. The deck is the multi-set of vertex-deleted subgraphs of a graph (which are called cards).

I'd like to take the other perspective and start with the multi-set $T$ of so-called tiles of a graph which are the vertex-induced subgraphs of the graph containing a distinguished vertex, all of its neighbours, and all the edges between them. As for $D$, $T$ contains as many elements as there are nodes in the graph: one tile for each.

Different from $D$, the tile set $T$ of a graph can be the tile set of many other non-isomorphic graphs. Needless to say that for a given multi-set $T$ of possible tiles there may be no graph at all which it is the tile set of.

In case there are many graphs with the same tile set $T$, these may share or may share not a lot of graph properties like being connected or having a small diameter. The only property theses graphs do share for sure is the degree distribution which is immediately given by the tile set.

Some tile sets may define a unique graph. I assume this is the case for this one, but I may be wrong:

Puzzle 1

Can you find a solution to this puzzle? Solving the puzzle means to put together tiles by identifying  nodes or edges while the red dots must not match black ones. Note, that the Euclidean lengths of edges in the final graph doesn't have to match the lengths of edges in the tiles when embedding them in some Euclidean space: you can stretch them however you want. (By the way, this is not necessary in these examples.)

Try out this one:

Puzzle 2

My questions are:

If $T$ is guaranteed to be the tile set of a graph which you don't know (like in the examples above): by which algorithm would you construct any graph which $T$ is the tile set of?
While this algorithm is probably some kind of a search algorithm, I added the [searching] tag.

How to prove that for any given set $T$ of tiles

there is no graph with tile set $T$

there are exactly $n$ graphs with tile set $T$ (for small $n$)

there are really many graphs with tile set $T$ (e.g. $\mathcal{O}(|T|^\alpha)$)

there is a graph with tile set $T$, but only with a small fraction $\epsilon$ of tiles left.

In case there are really many graphs with the same tile set $T$, these may be considered as a family of random graphs (= all graphs with tile set $T$). What may be said about this family of random graphs? Note, that in case you have an algorithm that generates a solution (question 1) you will have to randomize it to generate other random graphs with tile set $T$.

Since these questions seem so natural, I would imagine that they have been asked before. In this (probable) case, I would be happy to get a reference where I will find them answered. And a search term I can scholar-google for.

For those who like to puzzle and find out from which (non-random) graphs some given tile sets originated, here are two more puzzles. (Spoiler alert: Solutions are given in an answer below.)

Puzzle 3

Puzzle 4

答:
As a partial answer, here is how we can solve the sampling problem in cases like your first tile set: triangle-free tile sets. When lots of triangles get involved, the problem becomes much harder.

A random graph with the degree sequence $$(2, 2, \underbrace{3, 3, \dots, 3}_{20}, \underbrace{4,4,\dots,4}_{25})$$ can be generated by sampling from the configuration model. Generally, this will produce multigraphs with loops and parallel edges, but we're unlikely to get many of these, and repeated sampling will typically give us a simple graph after only a few trials.

In a similar way, repeated sampling will typically give us a graph with no $3$-cycles after only a few trials: short cycles are unlikely in sparse graphs with a given degree sequence. If a graph with no $3$-cycles has the degree sequence above, its tile set should be exactly the one you gave.

In practice, the Mathematica code

```<code>RandomGraph[DegreeGraphDistribution[{2, 2}~Join~Table[3, 20]~Join~Table[4, 25]]]
</code>```

produces a graph with no $3$-cycles pretty efficiently (over $100$ experiments, the average number of trials required was $19.94$). By the way, all $100$ graphs I generated were different, so the solution is far from unique.

Body:
As a partial answer, here is how we can solve the sampling problem in cases like your first tile set: triangle-free tile sets. When lots of triangles get involved, the problem becomes much harder.

A random graph with the degree sequence $$(2, 2, \underbrace{3, 3, \dots, 3}_{20}, \underbrace{4,4,\dots,4}_{25})$$ can be generated by sampling from the configuration model. Generally, this will produce multigraphs with loops and parallel edges, but we're unlikely to get many of these, and repeated sampling will typically give us a simple graph after only a few trials.

In a similar way, repeated sampling will typically give us a graph with no $3$-cycles after only a few trials: short cycles are unlikely in sparse graphs with a given degree sequence. If a graph with no $3$-cycles has the degree sequence above, its tile set should be exactly the one you gave.

In practice, the Mathematica code

RandomGraph[DegreeGraphDistribution[{2, 2}~Join~Table[3, 20]~Join~Table[4, 25]]]

produces a graph with no $3$-cycles pretty efficiently (over $100$ experiments, the average number of trials required was $19.94$). By the way, all $100$ graphs I generated were different, so the solution is far from unique.


答: Every node must have the same number of edges entering and leaving, except for the end points. There are 8 nodes on the graph with an odd degree, but we can make two of them the endpoints, so we have 6 nodes that need an "extra" edge. Finally, we can note that the odd-degree nodes come in adjacent pairs, so the "extra" edge leaving one node can be the "extra" edge entering another, so we only need half as many "extra" edges as we have non-endpoint, odd-degree nodes.

Title: Drawing the grid without lifting the pen

Body:
The below shape consist of $24$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw?
$1)25\quad\quad\quad\quad\quad\quad2)26\quad\quad\quad\quad\quad\quad3)27\quad\quad\quad\quad\quad\quad4)28\quad\quad\quad\quad\quad\quad5)29\quad\quad\quad\quad\quad\quad$

I think we can do it somehow by segments with the length $28$ (by passing twice the four middle segments on each side of outer big square).

I'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?


问:
How should I glue together the sides of the complex plane when adding a "single point" at infinity?
If we construct the Riemann sphere by taking a single point at infinity and adding it to the complex plane, then if I start with the point
$$z = 1 + i$$
in the finite part of the complex plane, and then push that point off to the top right and to infinity on a straight line at a 45 degree angle, then which side of the finite part of the complex plane does that point come back from?  Is it the bottom left at a 45 degree angle?  I thought that was topologically the projective plane.  Is there some kind of very subtle distinction here related to the metric, or do we need to use 8 arrows instead of 4 in the fundamental rectangle?  Even so, I still can't see which side a point pushed off to the top right would come back from if its not the bottom left(except maybe it comes back from the top right going the other way, but I find that very confusing, although conceptually possible).

In order to glue together the sides of the complex plane into a projective plane, you glue the top to the bottom like a mobius strip, and the left and the right like a mobius strip.

If the Riemann sphere is topologically distinct from a projective plane, then how can this distinction be represented in terms of gluing together the sides of a fundamental rectangle as in the picture above?

Thanks!

(NOTE:  I recently asked this question Is the Riemann sphere really just the entire complex plane where you add a "single point" at infinity? and feel like the confusion I had was resolved by a comment, and now I have a different question.)

答:
Topologically, the Riemann sphere is the one point compactification of the complex plane, which is not usually constructed via a gluing procedure. It is defined as $\mathbb C\cup \{\infty\}$ where the open sets are either ordinary open subsets of $\mathbb C$ or sets containing $\infty$ whose complement is compact in $\mathbb C$ (that is, $\infty$ and all of $\mathbb C$ except a closed, bounded portion). This means that the Riemann sphere is topologically equivalent to the standard $2$-sphere $S^2$, not the projective plane.

Using gluing to construct the Riemann sphere is not natural because a gluing procedure involves identifying edges of polygons, but the complex plane does not have edges--it is not compact. There are many ways you could add edges to the complex plane and then construct a sphere via gluing, but none of them extend the geometry of the complex plane naturally because instead of adding just one point, you have to add a whole system of edges.

For example: find a homeomorphism from the complex plane to the interior of square, and treat the edges of that square as the "edges" of the complex plane. Then you could label the edges $a$, $a^{-1}$, $b$, $b^{-1}$. Gluing them together would result in a sphere, but it would not extend the geometry of the complex plane in the way you describe (i.e., pushing a point to the top right, and having it reappear on the bottom left).


问:
identity proof for partitions of natural numbers

Definition:

A tuple $\lambda = (\lambda_1, \cdots, \lambda_k)$ of Natural Numbers is called a numeric partition of n if $1 \leq \lambda_1 \leq \cdots \leq

\lambda_k$ and $\lambda_1 + \cdots + \lambda_k = n$ and is written as $\lambda \vdash n$. A numeric partition $\lambda = (\lambda_1, \cdots, \lambda_k)

\vdash n$ can be as well written as $\lambda = [m_1, \cdots, m_n]$ with $m_i = \# \{j | \lambda_j = i\}$ and $\sum_{k=1}^{n} m_k*k = n$.

Exercise:

$g_k(\lambda) := \# \{i | m_i(\lambda) \geq k\}$.

Example:

$\lambda = [5,3,3,3,2,2,0,\cdots,0] \Rightarrow m_3(\lambda) = 3, g_2(\lambda) = 6$

Prove for fix $n,k \geq 1$
$$\sum\limits_{\lambda \vdash n} m_k(\lambda) = \sum\limits_{\lambda \vdash n} g_k(\lambda)$$

First I tried to write down all numeric partitions of $n = 3$

$$ \{\lambda | \lambda \vdash 3 \} = \{(1,1,1),(1,2),(3)\}$$

$$
\begin{array}{l|cccccc}
\lambda & m_1 & m_2 & m_3 & g_1 & g_2 & g_3 \\\hline
(1,1,1) & 3 & 0 & 0 & 1 & 1 & 1 \\
(1,2) & 1 & 1 & 0 & 2 & 0 & 0 \\
(3) & 0 & 0 & 1 & 1 & 0 & 0 \\\hline
\sum & 4 & 1 & 1 & 4 & 1 & 1
\end{array}
$$

答:
Suppose we could show that the number of partitions with $m_l \geq k$ is equal to the number of partitions with $m_k \geq l$. It then follows that
$$ \sum_{p \vdash n} m_k(p) = \sum_{l \geq 1} \sum_{p \vdash n}[m_k(p) \geq l] = \sum_{l \geq 1} \sum_{p \vdash n}[m_l(p) \geq k] = \sum_{p \vdash n} g_k(p). $$
In order to prove the claim, it is enough to restrict ourselves to partitions whose only parts are $k,l$. Indeed, an arbitrary partition of $n$ decomposes uniquely as a partition of $n_1+n_2$, where $n_1$ is a partition using $k,l$, and $n_2$ is a partition not using $k,l$. Furthermore, we can assume that $k,l$ are relatively prime (otherwise, cancel the GCD).

We show that the number of $k,l$-partitions with $m_l < k$ is equal to the number of $k,l$-partitions with $m_k < l$. Notice that since $k,l$ are relatively prime, there's at most one $k,l$-partition with $m_l < k$ and at most one $k,l$-partition with $m_k < l$. We show that if there is a $k,l$-partition with $m_l < k$ then there is one with $m_k < l$.

Suppose there's a $k,l$-partition with $m_k < l$. If for that partition also $m_l < k$, we're done. Otherwise, repeatedly replace $k$ parts of size $l$ with $l$ parts of size $k$ until you reach a partition with $m_l < k$.

Here is an alternative proof for the claim about $k,l$-partitions, for coprime $k,l$.

Let
$$\alpha = (k^{-1}n) \bmod{l}, \quad \beta = (l^{-1}n) \bmod{k}.$$
Here $k^{-1}$ is taken with respect to $l$, and vice versa.

Define the following operation ("conjugation") on pairs $(a,b)$ that solve $ak+bl=n$:
$$(a,b)' = \left(\frac{l}{k}(b - \beta) + \alpha, \frac{k}{l}(a - \alpha) + \beta\right).$$
One can check the following properties:

Conjugation takes a solution of $ak+bl=n$ to another solution.

Conjugation is an involution (it is its own inverse).

If there's an integral solution $(a,b)$ with $0 \leq a < l$ then its conjugate $(a',b')$ has $0 \leq b < k$, and vice versa.

If $kl|n$ then $(a,b)' = (b,a)$.

Here are some examples, with $k = 5$ and $l = 3$. One can check that $k^{-1} \bmod{l} = 2$ and $l^{-1} \bmod{k} = 2$.

If $n = 30$ then $(a,b)'=(b,a)$. We have $(0,10)' = (6,0)$ and $(3,5)' = (3,5)$.

If $n = 40$ then $\alpha = 80 \bmod{3} = 2$ and $\beta = 80 \bmod{2} = 0$. Thus $(a,b)' = (3/5b + 2, 5/3(a - 2))$. We have $(8,0)' = (2,10)$ and $(5,5)' = (5,5)$.

If $n = 7$ then $\alpha = 14 \bmod{3} = 2$ and $\beta = 14 \bmod{2} = 0$. This time there are no minimal solutions. Instead, we have the pair $(2,-1)' = (7/5,0)$.

Here's a proof using generating series. The advantage of this proof is that it requires no thought.

The usual generating series for partitions is
$$P = \frac{1}{(1-x)(1-x^2)(1-x^3)\cdots}.$$
The coefficient of $x^n$ is the number of partitions of $n$.

Suppose we want to sum over $m_k$. So we need to replace the factor $1/(1-x^k)$ with
$$ x^k + 2x^{2k} + 3x^{3k} + \cdots = \frac{x^k}{(1-x^k)^2}. $$
So the generating series for $\sum m_k$ is
$$ M = \frac{x^k}{1-x^k} P. $$

Now let's sum over $g_k$. How much do $1$-parts contribute to the sum? We only count them if there are at least $k$ of them. So we need to replace $1/(1-x)$ with
$$ x^k + x^{k+1} + \cdots = \frac{x^k}{1-x}. $$
So $1$-parts contribute $x^k P$. Similarly, $2$-parts contribute $x^{2k} P$, and so on. In total, the generating series for $\sum g_k$ is
$$ G = (x^k + x^{2k} + x^{3k} + \cdots) P = \frac{x^k}{1-x^k} P = M. $$


问:
A circle is added to the equally spaced $5*5$ grid of vertices, alongside. Find the largest number of dots that the circle can pass through :-

A circle is added to an equally spaced $5*5$ grid of vertices, alongside . Find the largest number of dots that the circle can pass through.

What I Tried: I experimented with the problem by trying to make circles in Geogebra.

The most number of points I was able to join was $8$.

However, there were $4$ options given to me and they were :-

$i)$ $4$

$ii)$ $6$

$iii)$ $18$

$iv)$ $10$

So $8$ is not the answer, and I am thinking it is $10$.

I tried in many ways to get $10$ points intersected, but I could not do that even in Geogebra.

Can anyone help me give the correct answer? Also it would be better if someone explains how to solve problems like these, without Geogebra and on normal paper.

Note that circles can intersect from outside the $5*5$ grid too, and I tried in that way as well.

答:
Let me first restate the problem: what is the maximum size of a concyclic subset of $\{1,2,3,4,5\}^2 \subseteq \mathbb{R}^2$?

Let's first consider some easy bounds on the problem. Since there is a circle through any three noncollinear points, the answer is at least 3. For an upper bound, notice that the point set is contained in five lines, and a line intersects a circle in at most two points. So the answer is at most 10.

The example you found is natural to come up with, since the point set has 8-fold symmetry. There are other similar examples, like a circle of radius $\sqrt{5}$ centered at the center of the point set. So the answer is at least 8.

This question may seem difficult because there are infinitely many circles, but since the point set is finite, there are actually a tractable number of circles that contain many points. I claim that 10 is not the answer. Suppose there was a circle going through ten points -- it must go through exactly two in each row and two in each column. The number of choices for these points in the first row is $\binom{5}{2} = 10$, and given these points, the number of choices for another point in the same row as the lower point is $4$. So you only need to check 40 circles. One could come up with longer arguments excluding more circles, but 40 is not too big.

In fact, you can use the exact same argument to show that 9 is not the answer. So 8 is the correct answer.


问:
Show that $\lambda(E_1 \cap ... \cap E_n) > 0$
We use the Lebesgue measure $\lambda$ on $\mathbb{R}$. Let $E_t \subset (0,1)$, $t=1, ..., n$ be measurable subsets such that $\lambda(E_1)+...+\lambda(E_n) > n-1$.

How do I show that $\lambda(E_1 \cap ... \cap E_n) > 0$?

My idea is to let  $S_t=(0,1)$ \ $E_t$.  Then we also have  an inequality  for $\lambda(S_t)$. But I don't know how to express $\bigcap_t E_t$ with $S_t$.

答:
Hint If $A, B \subseteq (0,1)$ then
$$1 \geq \lambda(A \cup B)=\lambda(A)+ \lambda(B)-\lambda(A \cap B)$$
and hence
$$
\lambda(A \cap B) \geq \lambda(A)+ \lambda(B) -1
$$

Prove now by induction that if $A_1,.., A_k \subseteq (0,1)$ then
$$
\lambda (A_1 \cap A_2 \cap ... \cap A_k)\geq \lambda (A_1) +\lambda( A_2)+... +\lambda(A_k)-k+1
$$


问:
Drawing the grid without lifting the pen
The below shape consist of $24$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw?
$1)25\quad\quad\quad\quad\quad\quad2)26\quad\quad\quad\quad\quad\quad3)27\quad\quad\quad\quad\quad\quad4)28\quad\quad\quad\quad\quad\quad5)29\quad\quad\quad\quad\quad\quad$

I think we can do it somehow by segments with the length $28$ (by passing twice the four middle segments on each side of outer big square).

I'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?

答:
The figure has 4 nodes with 2 lines connected to node. 8 nodes with 3 lines connected, and 4 nodes with 4 lines connected.

If you want to complete the circuit without lifting your pen, you can start at a node with an odd number of connections, and you can end at a node with an odd number of connections, for all the rest you must have an even number of connections.  That is, if you get to the node on one path you leave the node on a different path.

This means that you must double up on some of the lines where it appears that there is a odd number of lines to make it an even number.  What is the minimum number of lines you need to double up such that there are at most two nodes with an odd number of connections?


答:
You can combine the answers from S.H.W and José Carlos Santos to figure out a solution if you don't know the sum $\sum^\infty_{k=0} 2^{-k}$.

$$
\begin{align}
\sum^\infty_{k=2} 2^{-k} &= \sum^\infty_{k=0}2^{-k} - 1 - \frac{1}{2} \tag{S.H.W} \\
&=2^{-2}\sum^\infty_{k=0}2^{-k} \tag{José Carlos Santos}
\end{align}
$$

If we write $L=\sum^\infty_{k=0}$ then $L-\frac{3}{2} = \frac{L}{4}$. This has as only solution $L=2$. Therefore:
$$\sum^\infty_{k=2} = \frac{L}{4} = \frac{1}{2}$$


问:
If $u\otimes v_1 = v_2\otimes u$; $v_1,v_2,u \in V-\{0\}$ then $v_1 = v_2$?
On the beginning I though that it was false. But, making several trials for counter-example I have failed in find one. Anyone has any suggestions to prove or disprove?</p>

答:
Another solution:

It suffices to show that $u, v_1$ are linearly dependent. In fact, if for some $\lambda \neq 0$ we have  $v_1 \otimes \lambda v_1 = \lambda v_1 \otimes v_2 $, then  $v_1\otimes(\lambda v_1 - \lambda v_2 )= 0 $ implies $v_2 = v_1.$

Thus, we are left to show that  $u, v_1$ cannot be linearly independent. Let $\Gamma: V\otimes V \rightarrow \mbox{Hom}(V^*,V)$  be the canonical map satisfying $\Gamma(w_1\otimes w_2)(f) = f(w_1)w_2$ for every $f\in V^*$ and $w_1,w_2\in V$.

Take any $f\in V^*$ such that $f(u) =1 , f(v_1) =0$. Such a $f$ exists, since $u$ and $v_1$ are suppposed to be linearly independent. Thus,
$$   v_2 = f(u)v_2= \Gamma(u\otimes v_2)(f) = \Gamma (v_1\otimes u ) = f(v_1)u = 0$$
is a contradiction.


问:
Generating Function for Recursive Sequences (generalization to the binomial series)
Let $S_k$ be a recursive sequence defined by

$S_k = (x+1)S_{k-1}+S_{k-2}$ with $k>1$

with initial conditions:

$S_0 = 2, S_1=x+1$.

Then the first few terms are

$2, x + 1, x^2 + 2x + 3, x^3 + 3x^2 + 6x + 4, x^4 + 4x^3 + 10x^2 + 12x + 7,\\ x^5 + 5x^4 + 15x^3 + 25x^2 + 25x + 11, x^6 + 6x^5 + 21x^4 + 44x^3 + 60x^2 + 48x + 18$

For simplicity, the first term $S_0$ will be set to $1$ instead of $2$. We can arrange the coefficients of all the terms of $S_k$ in a triangle like format (Similar to Pascals Triangle).

$[1]$

$[1, 1]$

$[1, 2, 3]$

$[1, 3, 6, 4]$

$[1, 4, 10, 12, 7]$

$[1, 5, 15, 25, 25, 11]$

$[1, 6, 21, 44, 60, 48, 18]$

$...$

Let $P_{t,n}$ denote the $t$-th coefficient of $S_k$ (starts at $t=0$). For example, $P_{3,3} = 4$.

If $t$ is fixed, then $P_{t,n}$ is a polynomial in $n$.

$P_{0,n} = 1$, $P_{1,n} = n$, $P_{2,n} = \frac{n^2 + n}{2}, P_{3,n} = \frac{n^3 + 3n^2 - 10n}{6}$

and so on.

If $t > 1, n > 2$, then $P_{t,n} = P_{t,n-1}+P_{t-1,n-1}+P_{t-2,n-2}$. This is similar to Pascal's Identity for binomial coefficients.

What is the generating function for the power series

$$G(x,n) = \displaystyle\sum_{t=0}^{\infty}P_{t,n}x^t = 1 + nx + \frac{n^2 + n}{2}x^2 + \frac{n^3 + 3n^2 - 10n}{6}x^3 \\+ \frac{n^4 + 6n^3 - 37n^2 + 30n}{24}x^4 + \frac{n^5 + 10n^4 - 85n^3 + 50n^2 + 264n}{120}x^5 + O(x^6)$$

Is there a general formula for generating functions involving other polynomial recursive functions?

Edit:

$$\displaystyle\sum_{t=0}^{\infty}P_{t,n}x^t = (\sum_{j=0}^{\lceil{t/2}\rceil} \frac{n}{n-j}\binom{n-j}{j}\binom{n-2j}{t-2j})x^t $$

答:
The solutions to $X^2 -(x+1)X-1=0$ are $X_1= \frac{x+1+\sqrt{(x+1)^2+4}}{2}$ and $X_2= \frac{x+1-\sqrt{(x+1)^2+4}}{2}$. Then, with the given initial values, we have

$$S_k(x)= X_1^k+X_2^k.$$
Since by definition $$S_n(x):=\sum_{t\ge 0}x^{n-t}P_{t,n},$$
we have $$S_n(x^{-1})=\sum_{t\ge 0}x^{t-n}P_{t,n},$$
that is $$\sum_{t\ge 0}P_{t,n}x^{t}=x^nS_n(x^{-1}),$$
that is
$$\sum_{t\ge 0}P_{t,n}x^{t}=\big(\frac{1+x+\sqrt{5x^2+2x+1}}{2}\big)^{n}+\big(\frac{1+x-\sqrt{5x^2+2x+1}}{2}\big)^{n},$$
if I did not make a mistake.


问:
Show that if $x_{t+l}=Ax_t$, then $ \rho(l)=1$ if $A>0$, and $\rho(l)=−1$ if $A<0$.
Given a single stationary series $x_t$ with zero mean and autocovariance function $\gamma(h)$ and autocorrelation function $\rho(h)$

Show that if $x_{t+l}=Ax_t$, then $ \rho(l)=1$ if $A>0$, and $\rho(l)=−1$ if $A<0$.

What I have done is

$$ \text{By definition we know that}\quad \rho(l) = \frac{\gamma(l)}{\gamma(0)}$$

So for $x_{t+l}=Ax_t$ we have

$$\gamma(l) = cov(x_t, x_{t+l}) = cov(x_t, Ax_t) = A\gamma(0)$$

and it follows that

$$\rho(l) = A\frac{\gamma(0)}{\gamma(0)} = A$$

which is not what I want. Where did I go wrong?

EDIT: Can I say that because of stationarity that

$$\frac{x_{t+l}}{x_t} = \pm 1$$

EDIT 2:  How about this?

Because of stationarity

$$Var(x_t) = Var(x_{t+l}) = Var(Ax_t) = A^2Var(x_t)$$

which implies that $A^2 = 1$, and thus

$$\rho(l) = \sqrt{1} = \pm 1$$

答:
$$ \text{By definition we know that}\quad \rho(l) = \frac{\gamma(l)}{\gamma(0)}$$

So for $x_{t+l}=Ax_t$ we have

$$\gamma(l) = cov(x_t, x_{t+l}) = cov(x_t, Ax_t) = A\gamma(0)$$

and it follows that

$$\rho(l) = A\frac{\gamma(0)}{\gamma(0)} = A$$

Because of stationarity

$$Var(x_t) = Var(x_{t+l}) = Var(Ax_t) = A^2Var(x_t)$$

which implies that $A^2 = 1$, and thus

$$\rho(l) = \sqrt{1} = \pm 1$$


问:
Equilibrium Distribution of Reducible Markov Chain
Let's say a Markov chain has the state-space $S = \{A,B,C,D\}$ Transition Matrix as follows:

$$
\begin{bmatrix}
 0     & 0     & 0 & 1 \\
 1     & 0     & 0  & 0  \\
 0     & 0    & 0  & 1 \\
 0     & 1     & 0  & 0
\end{bmatrix}
$$

This is NOT irreducible as far as I am aware due to state C being transient.

To find an equilibrium distribution for the chain, would I just treat it the same way as if it were irreducible in which case I get $\pi = (1/3)(1,1,0,1)$. Is this unique? I'm not sure.
Thanks in advance!

答: This Markov chain is periodic, so no stationary distribution exists. If we start in state A we see the pattern ADBADBADB... so we know every third time step we will be in either A, D or B. And if we start the chain in state C then the pattern is CADBADBADB...</p>


问:
Is all algebraic system is monoid?
Is all Algebraic system is monoid?
I cross-checked the properties of both monoid and algebraic systems. Here is what I found:

Properties of Algebraic system:
1.closure property
2.Associativity
3.commutativity
4.existence of identity
5.Existence of inverse

Property of monoid
1.Associativity
2.existence of identity element.

Considering above mentioned properties I think all Algebraic systems are monoid.
But the problem is,
Questions are asked "Given an algebraic system... Check whether it's a monoid? or show that it's a monoid. Please clear my doubt!
Edit
To be clear about algebraic system and algebraic structures ,here is what I found:

An algebraic system is a system consisting of a nonempty set A and one or more n-ary operations on the set A. It is denoted by ⟨A,f1,f2,...⟩.
An algebraic structure is an algebraic system, ⟨A,f1,f2,...,R1,R2,...⟩, wherein addition to operations fi, the relations Ri are defined on A. This leads to a structure on the elements of A.
Note: My doubt is about Algebraic system (not algebraic structure) and monoids!

答:
No.

Consider a magma $(M,\ast)$, which is simply a set $M$ with a binary operation $\ast:M\times M\to M$. It needn't be associative.

Title: Is all algebraic system is monoid?


答:
You could think of the complex plane as the interior of the closed disk in the plane. If we then squeeze the boundary circle to a point (the point at infinity ) we get the Riemann sphere. So we do have a circle at infinity (all directions) but we identify all directions to one "infinity" point.

We could put a compatible metric on the disk defined by $d(z,w)  = |f(z)-f(w)|$ where $z,w$ lie in the interior and $f(z)=\frac{z}{1-|z|} \in \Bbb C$ and $d(z, \infty) = \frac{1}{1-|z|}$. Then a sequence that converges to the boundary in the disk becomes a sequence converging to $\infty$ as it ought.

Title: How should I glue together the sides of the complex plane when adding a "single point" at infinity?

Body:
If we construct the Riemann sphere by taking a single point at infinity and adding it to the complex plane, then if I start with the point
$$z = 1 + i$$
in the finite part of the complex plane, and then push that point off to the top right and to infinity on a straight line at a 45 degree angle, then which side of the finite part of the complex plane does that point come back from?  Is it the bottom left at a 45 degree angle?  I thought that was topologically the projective plane.  Is there some kind of very subtle distinction here related to the metric, or do we need to use 8 arrows instead of 4 in the fundamental rectangle?  Even so, I still can't see which side a point pushed off to the top right would come back from if its not the bottom left(except maybe it comes back from the top right going the other way, but I find that very confusing, although conceptually possible).

In order to glue together the sides of the complex plane into a projective plane, you glue the top to the bottom like a mobius strip, and the left and the right like a mobius strip.

If the Riemann sphere is topologically distinct from a projective plane, then how can this distinction be represented in terms of gluing together the sides of a fundamental rectangle as in the picture above?

Thanks!

(NOTE:  I recently asked this question Is the Riemann sphere really just the entire complex plane where you add a "single point" at infinity? and feel like the confusion I had was resolved by a comment, and now I have a different question.)


问:
Proof of Gauss Seidel converges for any initial vector if the Coefficient matrix is diagonally dominant.
Friends

I came through the proof of Gauss-Seidel method i understood except the points marked in blue, and in the last line how the inequality is $<1$, it seems obvious but still it looks complex.

Source - Numerical methods for Scientific computation by S.K.Jain, proof of Theorem 3.7.

EDIT : Already figured out the last inequality but still trying to understand the WLOG part, any intuition on why we can assume $\|x\|_{\infty} = 1$?

Any help !

答:
I think that @MartinSleziak was already crystal clear but... The proof starts by taking an eigenvalue $\lambda$ and one of its eigenvectors. If $x$ is an eigenvector associated with $\lambda$, the same happens with any scalar multiple $k x, k \ne 0$. Since we get to choose $x$, we decide to choose it in such a way the largest component (in absolute value) is equal to 1.

Imagine that  $y = (1, 2, 3)$ is an eigenvector, then we can take $x = (\frac 13, \frac 23, 1)$ which is also an eigenvector and satisfies $\|x\|_{\infty} = 1$.


Title: Inequality induction proof with two variables

Body:
For all $ ∈ $ and all $>0$ it holds that $1+≤(1+)^$

I've done the base-case already

Induction steps.

Body:
Hint

Induction step:

$$(1+k)^{s+1}=(1+k)(1+k)^{s}\ge(1+k)(1+sk)=[1+k(s+1)]+sk^2$$

But, $s\ge 1$ and $k>0$, so $sk^2>0$.


问:
Basic Property of Finite Presentation of Modules
I'm struggling to prove some rudimentary facts about finite presentations of a module $A$ over a ring $\Lambda$, which is defined in Hilton/Stammbach's A Course in Homological Algebra as a short exact sequence $0 \to N \to P \to A \to 0$ with $N$ finitely-generated and $P$ finitely-generated projective.

I'd like to show that if $A$ has a finite presentation, then for all short exact sequences $0 \to R \to S \to A \to 0$ with $S$ finitely-generated, $R$ is also finitely-generated. Here's my progress so far: If we denote the surjective maps in the two exact sequences above by $\epsilon : P \to A$ and $\epsilon' : S \to A$, then since $P$ is projective we have a map $\alpha : P \to S$ such that $\epsilon = \epsilon'\alpha$. We can identify $N$ with $\operatorname{Ker}{\epsilon}$ and $R$ with $\operatorname{Ker}{\epsilon'}$. We then have $\alpha(\operatorname{Ker}{\epsilon}) = \alpha(P) \cap \operatorname{Ker}{\epsilon'}$, and here I get stuck. We know that $\alpha(\operatorname{Ker}{\epsilon})$ and $\alpha(P)$ are finitely-generated, but how do we use this to show that $\operatorname{Ker}{\epsilon'}$ is finitely-generated? A hint would be much appreciated.

答:
The simplest way is to make a pullback diagram
$$\require{AMScd}
\begin{CD}
{} @. {} @. 0 @. 0 \\
@. @. @VVV @VVV \\
{} @. {} @. N @= N \\
@. @. @VVV @VVV \\
0 @>>> R @>>> Q @>>> P @>>> 0 \\
@. @| @VVV @VVV \\
0 @>>> R @>>> S @>>> A @>>> 0 \\
@. @. @VVV @VVV \\
{} @. {} @. 0 @. 0 \\
\end{CD}
$$
The rows and columns are exact; the arrows marked with equality are easily seen to be isomorphisms when $Q$ is the pullback of the maps $P\to A$ and $S\to A$.

Since $N$ and $S$ are finitely generated by assumption, $Q$ is finitely generated as well. Since $P$ is projective, the sequence $0\to R\to Q\to P\to 0$ splits and therefore $R$ is finitely generated.

Another way is to use Schanuel's lemma. In order to do this, take a finitely generated projective module $S'$ and an epimorphism $S'\to S$; let $K$ be the kernel of the composition $S'\to A$. Then, by Schanuel's lemma, $K\oplus P\cong N\oplus S'$, which implies $K$ is finitely generated because $N\oplus S'$ is.

The homomorphism theorem provides a surjective homomorphism $K\to R$.


问:
What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?
What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?

Sorry for the basic question. I am a little confused when solving such problems. They are very easy, I know, but still... Which is the easiest algorithm and how would you solve it? We are supposed to use the rule that $$\sqrt{a^2b}=a\sqrt{b},a\ge0, b\ge0,$$ right? What would you do to solve the problem? I was thinking about $$8=4\times2\\32=8\times4\\(-3)^2=3^2$$ but are we supposed to continue this to $$8=4\times2=2\times2\times2=2^3\\32=8\times4=2^3\times2^2=2^5?$$ How can we still use $\sqrt{a^2b}=a\sqrt{b}$ then? Thank you in advance!

Body: $\sqrt{8\cdot32\cdot(-3)^2} = \sqrt{8\cdot32\cdot9} = 3 \cdot \sqrt{8\cdot32} = 3 \cdot \sqrt{16\cdot16} = 3 \cdot 16 = 48$


Body:
What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?

Sorry for the basic question. I am a little confused when solving such problems. They are very easy, I know, but still... Which is the easiest algorithm and how would you solve it? We are supposed to use the rule that $$\sqrt{a^2b}=a\sqrt{b},a\ge0, b\ge0,$$ right? What would you do to solve the problem? I was thinking about $$8=4\times2\\32=8\times4\\(-3)^2=3^2$$ but are we supposed to continue this to $$8=4\times2=2\times2\times2=2^3\\32=8\times4=2^3\times2^2=2^5?$$ How can we still use $\sqrt{a^2b}=a\sqrt{b}$ then? Thank you in advance!

Body:
It looks like you have the right idea... $8 = 2^3, 32 = 2^5, 8\cdot 32 = 2^8, (-3)^2 = 3^2...$

$\sqrt {2^8\cdot 3^2} = (2^8\cdot 3^2)^\frac 12 = 2^4\cdot 3 = 48$


答:
Since $(-3)^2=3^2$, the expression simplifies to
$$
\sqrt{8 \cdot 32 \cdot3^2} \, ,
$$
meaning that you can set $a=3$ and $b=8\cdot32=256$.

Title: What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?


问:
Existence of an element of order $p$
Let $G$ be an abelian finite non trivial group and $p$ a prime number which devises o(G). We denote $H$ the set of non trivial subgroups of $G$.

Could you please help me to show that if there exists $M\in H$ such that $p$ devises $o(M)$  then there exists $g\in G$ of order $p$.

Thanks in advance

答: This is true by Cauchy's Theorem: If $p$ divides the order of $G$ then $G$ contains an element of order $p$. You don't need the hypothesis that $M$ exists, and I'm not sure that assumption allows for a simpler proof anyways.

Title: Existence of an element of order $p$


问:
Number of N step Random walks on 2D infinite grid or lattice
What is the number of N step random walks starting from a point (x0,y0) to a point (x1,y1) assuming each direction (right,left,up,down) has equal probability. I know the expression for the one dimension case is:

$$ \binom{N}{\frac{N+(y1-y0)}{2}} $$

is there a similar expression for two dimensions?
I saw this post (Probability Distribution of a 2D lattice random walk) but the answer given is clearly wrong.

答:
The linked answer has a formula, but I wanted to provide a different perspective. I am generalizing the method used by Brian M. Scott in this answer.

Suppose the vertices your walk visits are $(x_0,y_0),(x_1,y_1),\dots,(x_N,y_N)$. (I am using different notation for the final vertex). Note that

The sums $x_0+y_0, x_1+y_1,\dots,x_N+y_N$ are a one-dimensional random walk from $x_0+y_0$ to $x_N+y_N$. This is because the values change by $\pm1$ each time.

The differences $x_0-y_0,x_1-y_1,\dots,x_N-y_N$ are also a $1$D random walk from $x_0-y_0$ to $x_N-y_N$.

Any pair of $1$D random walks can be generated in this fashion. You just need to check that the four possible choices of north, south, east and west for the $2$D walk generate all possible pairs of $\pm1$ for the two $1$D walks described above.

It follows that the number of $2$D random walks is the product of the number of ways to complete these two $1$D random walks, which is
$$
\binom{N}{\frac12({N+x_0+y_0-x_N-y_N})}\times \binom{N}{\frac12({N+x_0-y_0-x_N+y_N})}.
$$


问:
Question about the proof for "The rational line Q does not have the completeness property"
Referring to the attached png file, I have questions about the inequality:

$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$

How is this inequality derived?  How did the n get into the inequality?  Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$, and later in the proof, $2q^2/(p^2 - 2q^2)$ come from?   There were no derivations for these 3 expressions.  Could someone please provide the derivations?

答:
The development of a mathematical proof often has an "exploratory" phase in which various formulas (or other facts) are discovered, and then those formulas are used in the formal proof.

What often happens when the proof gets written up (in a textbook, or in a research journal) is that the "exploratory" phase is ignored and only the formal proof is written up, and the formulas appear to pop out of nowhere. The justification for this is that it doesn't matter where those formulas came from (see the comment of @fleablood for a guess as to their origin), only that they actually work.

This can be unsatisfying to a student who is trying to understand a proof. "Where did that formula come from?" is a common question. And if the proof writer does not say where it came from, we are left to guess; it's impossible to know for sure.

In this case, though, there's maybe just enough information to guess where the formula came from. Let me try to explain what could have happened when writing the paragraph of the proof which starts with the sentence "Suppose first that $x^2 < 2$."

In that paragraph, one is assuming that $x = \frac{p}{q}$ is a positive rational number (so $p,q \in \mathcal N$), and that $x^2 < 2$, and using those assumptions one wishes to prove that $x$ is not an upper bound for the set $A$.

This will follow if one can prove the existence of $n \in \mathcal N$ such that
$$y = \left( \frac{n+1}{n} \right) \frac{p}{q} \in A
$$
because that quantity $y$ is clearly larger than $x=\frac{p}{q}$.

To say that $y \in A$ is equivalent to the inequality
$$(*) \qquad \left( \frac{n+1}{n} \right)^2  \, \frac{p^2}{q^2} < 2
$$
At this point, the author pretty clearly manipulated the inequality $(*)$ with the purpose of isolating all the $n$'s on one side and all the $p$'s and $q$'s on the other side, rewriting the inequality in the form
$$\frac{n^2}{2n+1} > \frac{p^2}{2q^2-p^2}
$$
And then the author, who is very good with sequences, probably thought:

The left hand side of this inequality diverges to $+\infty$ as $n \to +\infty$. But $p$ and $q$ are fixed. The definition of divergence to $+\infty$ therefore guarantees the existence of $n \in \mathcal N$ which makes this inequality true. So that's how I'll start my proof, and then I'll use that to derive the inequality $(*)$, by working backwards through my exploratory manipulations.

or something to that effect.

And later, when cleaning up their desktop, the author tossed all that lovely exploratory derivation in the garbage pan, leaving us to make our wild guesses as to the origin of that final formula.

Something quite similar is going on in the later paragraph which starts "It follows that we must have $x^2 \ge 2$." I'll leave that for you to think about, and if you like to try to reconstruct the possible exploratory phases which led to the proof in that paragraph.


问:
Stirling numbers of first order
Show that the number of permutations of $n$ numbers, for $n ≥ 2$, with two cycles is at most $(n−1)!\log n$.

I know so far that from my notes that $S(n,1) = (n-1)!$ and that $S(n,2) = (n-1)!H_{n-1}$ also according to them. I assume you can make $H_{n-1}$ equal $\log n$. And also understand that $n ≥ 2$. Any help would be appreciated.

Title: Stirling numbers of first order

Body:
If $n\geq 2$, then
$$
H_{n - 1}  = 1 + \sum\limits_{k = 2}^{n - 1} {\frac{1}{k}}  \le 1 + \sum\limits_{k = 2}^{n - 1} {\int_{k - 1}^k {\frac{{dt}}{t}} }  = 1 + \int_1^{n - 1} {\frac{{dt}}{t}}  = 1 + \log (n - 1).
$$


问:
Finding the value of C in probability density function
$$f(x) =  \begin{cases} cx  & : 0\leq x\leq 1/2 \\ c(1-x) & : 1/2 \leq x \leq 1 \\ 0 &: \textsf{otherwise} \end{cases}$$

Is $\int_0^{1/2}cx\,dx + \int_{1/2}^1c(1-x)\,dx = 1$ the correct approach to finding the value of $c$?

Body:
Your approach is correct.
$$\int_0^{1/2}cx\,dx + \int_{1/2}^1c(1-x)\,dx = 1$$
So you get,
$$\left [  \frac{cx^{2}}{2}\right ]_{0}^{\frac{1}{2}}+\left [  cx-\frac{cx^{2}}{2}\right ]_{\frac{1}{2}}^{1}=1$$
$$(\frac{c}{8}-0)+(c-\frac{c}{2}-\frac{c}{2}+\frac{c}{8})=1$$
$$\frac{c}{4}=1$$
$$c=4$$


问:
Finding equilibrium point from a differential equation
I have the differential equation of a system as given below, $$ m\frac{d^2H}{dt^2} = mg-k\frac{I^2}{H^2} $$

How do I find the equilibrium points of this?

答:
The equilibrium points of the ordinary differential equation

$m \dfrac{d^2H}{dt^2} = mg -  k \dfrac{I^2}{H^2} \tag 1$

are characterized by the conditions

$ \dfrac{dH}{dt} = 0, \tag 2$

$ \dfrac{d^2H}{dt^2} = 0;\tag 3$

substituting (3) into (1) we find that the equilibria occur where

$ mg = k \dfrac{I^2}{H^2}, \tag 4$

that is, where

$H^2 = \dfrac{kI^2}{mg}; \tag 5$

thus, at equilibria

$H = \pm \sqrt{\dfrac{kI^2}{mg}}. \tag 6$

Note Added in Edit, Friday 10 September 2021 8:32 PM PST:  Tbe reader might well be curious as to why, when the equilibria of (1) are characterized by conditions (2), (3), we only need find the values of $H$ corresponding to (3).  The answer may be found in the structure of the phase plane for the differential equation (1).  A second-order equation such as (1) may be thought of as a case of the general second-order system

$\ddot x = f(x, \dot x), \tag 7$

which in turn may be thought as a first-order system in two variables by taking

$y = \dot x, \tag 8$

so that

$\dot y =\ddot x, \tag 9$

and hence (7) takes the form

$\dot x = y, \tag{10}$

$\dot y = f(x, y), \tag{11}$

and the equilibria of this system ((10)-(11)) occur at those points $(x, y)$ where

$\dot x = \dot y = 0; \tag{12}$

(10) and (12) taken in concert yield

$y = \dot x = 0, \tag{13}$

and now (12) and (13) allow (11) to be written

$f(x, 0) = 0, \tag{14}$

which defines and constrains the relevant values of $x$, the ones corresponding to critical points of the system of ODEs (11)-(12).  We note that in accord with (4), all these critical points lie on the $x$-axis.

The developments expressed in this note are directly applicable to equation (1) above, which may indeed be written in the form

$\dfrac{d^2H}{dt^2} = g -  k \dfrac{I^2}{mH^2} \tag{15}$

End of Note.


问:
Things around $(\forall x)(\Psi \Rightarrow \Phi) \Leftrightarrow ((\exists z)\Psi) \Rightarrow \Phi)$
Let $M,N$ be sets, $Q_M$ a set that depends on $M$ and $P(\cdot,\cdot)$ a property with two parameters (is it ok, to call these "parameters" or would "variables" have been better ?). Consider the following three statements:

i) $(\forall x\in M)(\forall y\in Q_M)[P(x,y)\Rightarrow x \in N]$

ii) $(\forall x\in M)[((\exists y\in Q_M)[P(x,y)])\Rightarrow x \in N]$

iii) $N=\{ x\in M \mid (\exists y\in Q_M)[P(x,y)]\}$

I would very much like to know the following:

I have been told that i) and ii) are equivalent, since they are an instance of a general scheme
$$
(\forall x)(\Psi \Rightarrow \Phi) \Leftrightarrow  ((\exists z)\Psi) \Rightarrow \Phi),
$$
which holds if $z$ is not mentioned in $\Phi$, as Carl Mummert I think explained a long time ago  - I can't find the entry right now.

What are the "values" the $x,z,\Psi$ and $\Phi$ have to take, such that this scheme shows the equivalence of i) and ii) ?

What kind of setting do you need in which you can prove this general scheme ? Is it only predicate logic ? Or do you need ZFC to formalize this logic so you can you use set-theory to prove it ?

How to you prove it (or a corrected version of it, if the above is flawed; it seems to me that quantifiers are missing) ?

Is ii) equivalent to iii) ? How do you rigorously prove that ? Which ZFC axioms are needed for that proof ?

答:
Hints:

1]  Assuming that $x$ is not free in $\varphi$, here's a chain of equivalences:

(a) $\forall x(\psi(x) \to \varphi)$

(b) $\forall x(\neg\psi(x) \lor \varphi)$

(c) $\forall x\neg\psi(x) \lor \varphi$

(d) $\neg\forall x\neg\psi(x) \to \varphi$

(e) $\exists x\psi(x) \to \varphi$

The only step that might give you pause is the equivalence of (b) and (c): it might help to compare $(\neg\psi(a) \lor \varphi) \land (\neg\psi(b) \lor \varphi)$ and $(\neg\psi(a) \land \neg\psi(b)) \lor \varphi$. Note that equivalence of (a) and (e) is just a matter of first-order logic, and nothing to do with ZFC.

2] These are also equivalents:

(a')  $(\forall x \in A)(\psi(x) \to \varphi)$

(b')  $(\exists x \in A)\psi(x) \to \varphi$

by just the same kind of reasoning, and again just by appeal to first-order logic. You just need to unpack the abbreviations for restricted quantifiers.

3] To see your (i) implies (ii), instantiate (i) -- use a parameter to name an arbitrary object in $M$ -- apply an (a'/b')-type equivalence, and generalize to get (ii). You can then do the reverse equivalence.


问:
Proof that a number and its multiplicative inverse have the same sign
I am trying to prove that if a number $a > 0$, then its multiplicative inverse $a^{-1}$ is also $>0$.

What I have done thus far is this:

Using the trichotomy property for the real numbers, I consider the following cases, where $\mathbb{P}$ is the set of all positive real numbers:

$a^{-1} = \frac{1}{a} = 0$.  I'm trying to show this for ALL a, but when $a=1$, this possibility violates the Nontriviality Property, so $a^{-1} \neq 0$.

$a^{-1} \in \mathbb{P}$. In this case, there is nothing to show.

$-a^{-1} \in \mathbb{P}$. Or, in other words, $a^{-1}$ is negative. In this case, I assume that I would need to show that this implies that $a < 0$ (or, equivalently, $-a \in \mathbb{P}$, but I am not sure how to prove this part. I want to use the Positivity Axiom of $\mathbb{R}$ that states that "if $a$, $b > 0$, then both $ab>0$ and $a+b>0$". I was thinking of perhaps using the contrapositive, that "if $ab<0$, then either $a<0$ or $b<0$", but I am not sure how to express it in a mathematically rigourous manner.

If anyone could please assist me on this, I would be eternally grateful. Thank you!

答:
Another way to prove it would be by making use of the Density Theorem of rational numbers,
that is, if $a>0\Rightarrow \exists k\in\mathbb{Q}$ such that $0<k<a$, then applying arithmetic operations to the inequality we obtain the desired result $0<\frac{1}{a}<\frac{1}{k}$.


问:
quaternions as right or left $\mathbb C$-vector space.
I got confused when I read the question titled "Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$" here on this site.

I know that the OP did not specify the tensor product over which field in the main question (but he specified it in the comments of one answer), so my question, just to clarify, if we want to prove that $$\mathbb{C} \otimes_{\mathbb R} \mathbb H \cong_{\mathbb R} M_2(C),$$

which is an isomorphism of $R$-algebras for the field of real numbers.

1- Do we have to take into account $\mathbb H$ as a left $\mathbb C$-vector space or $\mathbb H$ as a left $\mathbb C$-vector space as far as I understand there is no distinction between left and right vector spaces so why here there is a distinction?

2- Do we have to prove first the isomorphism  $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ and then prove the required isomorphism  $\mathbb{C} \otimes_{\mathbb R} \mathbb H \cong_{\mathbb R} M_2(C)$ as mentioned in the answer of @reuns?

Here is the answer of @reuns:

$\Bbb{H=C+Cj}$ is a 2-dimensional complex vector space so $End_\Bbb{C}(\Bbb{H})\cong M_2(\Bbb{C})$ as complex algebras.

Then we are willing to check that $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ as complex algebras, where $\Bbb{H\otimes_R C }$ comes with its own complex algebra structure.

For this, send $1\otimes a+i\otimes b+j\otimes c+ij\otimes d$ to $a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$

3-How can we show $End_\Bbb{C}(\Bbb{H})\cong M_2(\Bbb{C})$?

Any clarification will be appreciated!

EDIT:

Also, I was looking at that question: Why is $\mathbb{H}\otimes\mathbb{H}\cong\text{End}_\mathbb{R}\mathbb{H}$?

Why is $\mathbb{H}\otimes\mathbb{H}\cong\text{End}_\mathbb{R}\mathbb{H}$?

答:
It is unclear what is unclear to you.

I am saying that $\Bbb{H}$ is just a 2-dimensional (left) vector space so we can replace $End_\Bbb{C}(\Bbb{H})$ by $End_\Bbb{C}(\Bbb{C}^2)$.

It means that I am interpreting $End_\Bbb{C}(\Bbb{H})$  as $$End_{left\ \Bbb{C}\ vector\ space}(\Bbb{H})$$

Then $1\otimes a+i\otimes b+j\otimes c+ij\otimes d\mapsto a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$ gives the isomorphism $\Bbb{H\otimes_R C }\to End_\Bbb{C}(\Bbb{C}^2)$.

The $i$ of $i\otimes b$, the $i$ hidden in $a=A+iB$, and the $i$ in  $\pmatrix{i&0\\0&-i}$ are not the same, that's part of the game to understand what we identity the latter two.

The distinction between left and right vector space is because $\Bbb{C}$ is a subring of $\Bbb{H}$ but which doesn't commute with $\Bbb{H}$.


问:
Trigonometric Manipulation with inequality:$x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.$

Find all angles $\theta,$ $0 \le \theta \le 2 \pi,$ with the following
property: For all real numbers $x,$ $0 \le x \le 1,$

$$x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.$$

I am not exactly sure how to solve this problem. A friend gave me the suggestion to use the substitution $\sin^2(y)=x,\cos^2(y)=1-x$. Then, $\sin^4(y)\cos(\theta)-\sin^2(y)\cos^2(y)+\cos^4(y)\sin(\theta) > 0$. I tried to exploit symmetry by dividing by $\sin^2(y)\cos^2(y)$ to get $\tan^2(y)\cos(\theta)+\cot^2(y)\sin(\theta)>1$. I am not sure how to continue from here. Anything I tried from here was fruitless. What should I do to solve this problem?

答:
Hint

$$f(x)=x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0\Leftrightarrow$$
$$f(x)=x^2(\sin \theta+\cos \theta+1)-x(1+2\sin \theta)+\sin \theta>0.$$

Let's look at the discriminant,
$$\Delta=(1+2\sin \theta)^2-4(\sin \theta+\cos \theta+1)\sin \theta$$
$$\Delta=1-2\sin (2\theta)\to -1\le \Delta\le 3.$$

First case of solutions,

$\Delta<0$ and $\sin \theta+\cos \theta+1 >0.$

Second case,

$\Delta\ge 0$: $x_1\le x_2$ are the roots of $f(x)$,

a) If $\sin \theta+\cos \theta+1 >0$ then $x_2\le0$ or $x_1\ge 1$;

b) If $\sin \theta+\cos \theta+1 <0$ then $[0,1]\subset[x_1,x_2]$.


Title: $R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$

Body:
So the question is:

$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.

I don't really know what to do from here. If I substitute

$$R_n = 3(2^n)-4(5^n)$$

into

$$Rn = 7R_{n-1}-10R_{n-2}$$

I end up getting

$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$

Dont know what to do...

EDIT: Thanks to Zev, what I did was:

$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$

Body:
We need to eliminate $2^n,3^n$

$$R_n=3\cdot 2^n-4\cdot 5^n \ \ \ \ (1)$$

$$(1)\implies 3\cdot2^n-5\cdot 5^n-R_n=0\ \ \ \ (2)$$

$$(1)\implies R_{n+1}=3\cdot 2^{n+1}-4\cdot 5^{n+1}=6\cdot2^n-20\cdot 5^n$$
$$\implies 6\cdot 2^n-20\cdot 5^n-R_{n+1}=0 \ \ \ \ (3)$$

Solving $(2),(3)$ for $2^n,3^n$ we get $$2^n=\frac{5R_n-R_{n+1}}9\text{ and } 3^n=\frac{2R_n-R_{n+1}}{12}$$

$$(1)\implies R_{n+2}=3\cdot 2^{n+2}-4\cdot 5^{n+2}=12\cdot2^n-100\cdot 5^n\ \ \ \ (4)$$

Put the values of $2^n,3^n$ in $(4)$


问:
Algebra question involving quadratic
I dont know how to approach this problem. I tried considering letting w be a common root then I substituted into both equations and equate them and I get into (a-b)(w-1)=0 but what does that exactly means in the context of the problem and are these all the solutions?

答:
You're on the right track. If $w$ is the common root, then
$$
w^2+aw+b=0,\qquad w^2+bw+a=0
$$
and subtracting the two equations yields
$$
(a-b)w+b-a=0
$$
hence
$$
(a-b)(w-1)=0
$$
What does this mean? Well, it can be $a=b$ and so the original equations are the same, so they have common roots.

Otherwise the common root is $1$, so we need $1^2+a\cdot1+b=0$ and the condition becomes $b=-a-1$.

Hence the only cases when the equations have common roots are $b=a$ or $b=-a-1$.

If you need the roots to be real, the first case requires  $a^2-4b\ge0$, that is $a^2-4a\ge0$, and therefore $a\le0$ or $a\ge4$.

In the second case we don't need to compute the discriminant, because the equations do have at least a real root, so both roots are real (assuming $a,b$ are real).

Title: Algebra question involving quadratic


问:
Derivative by $a$ of the solution to $x’ = x^2 - x\cos t$, $x(0) = a$
Question:
Find the derivative with respect to $a$ of the solution to the following IVP:
\begin{align}
x’ &= x^2 - x\cos t \\
x(0) &= a
\end{align}
at $a = 0$.

What I tried:

I'm not sure how to solve the quation explicitly (I am not sure I need to).

Yet, if I try to solve, I don't successeed.

Separation of variables is a problem because I have
$x\cos t$.

Linear of course not, Bernoulli doesn't work, substitioution I don't see how.

I would like a hint please.

Thank you.

答:
$$x’ = x^2 - x\cos t $$
Note that $x=0$ is a solution. This differential equation is Bernoulli's differential equation
$$\dfrac {x’}{x^2} = 1 - \dfrac {\cos t }x$$
$$-\left(\dfrac {1}{x}\right)' = 1 - \dfrac {\cos t }x$$
$$u' - u {\cos t }=-1$$
Where $u=\dfrac 1 x$. You should be able to solve this linear first order DE.


问:
Fractional Infinite Power Tower
I have been having fun with infinite power towers and I came across this problem:

Let $a_2=\frac12$, and for $n>2$, $a_n=a_{n-1}^{\frac{1}{n}}$. Find $\lim_{n\to\infty}a_n$.

So I have absolutely no idea how to deal with the fractional exponents. I tried the standard power tower method where you assign it to a variable but I can't get anywhere with that. Does anyone have any ideas?

Title: Fractional Infinite Power Tower

Body:
You have look at the first terms of the recurrence sequence to be able to understand the explicit expression of $a_n$:

$a_3=\frac{1}{2}^{\frac{1}{3}}=\frac{1}{4}^{\frac{1}{6}}$

$a_4=({\frac{1}{2}^{\frac{1}{3}}})^{\frac{1}{4}}=\frac{1}{4}^{\frac{1}{24}}$

$a_5=\{[({{\frac{1}{2}^{\frac{1}{3}}})]^{\frac{1}{4}}}\}^{\frac{1}{5}}=\frac{1}{4}^{\frac{1}{120}}$

.
.
.

$a_n=\frac{1}{4}^{\frac{1}{n!}}$.

So $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{4}^{\frac{1}{n!}}=1$ (since $\lim_{n\to\infty}\frac{1}{n!}=0$).


问:
How to find $y'$ in terms of $x$ and $y$ for the curve $5x^{2} - xy - 4y^{2} = 0$
Find $y'$ in terms of $x$ and $y$ for $5x^{2} - xy - 4y^{2} = 0$.

Quite confused about how I'd go about this, the solution will be helpful (I'm good at working backwards) but an explanation will be appreciated if anyone has the time.

答:
use the implicit function theorem
$5x^2-xy-4y^2=0\Rightarrow 10x-y-x\frac{dy}{dx}-8y\frac{dy}{dx}=0$ and solve $\frac{dy}{dx}$, that is to say:
\begin{equation}
y'=\frac{dy}{dx}=\frac{10x-y}{x+8y}
\end{equation}


Title: How to find $y'$ in terms of $x$ and $y$ for the curve $5x^{2} - xy - 4y^{2} = 0$

Body:
Find $y'$ in terms of $x$ and $y$ for $5x^{2} - xy - 4y^{2} = 0$.

Quite confused about how I'd go about this, the solution will be helpful (I'm good at working backwards) but an explanation will be appreciated if anyone has the time.

Body:
$5x^{2} - xy - 4y^{2} = 0$
Diff w.r.t x on both sides

$5(2x) - (xy' + y(1)) - 4(2y)y' = 0 $

$10x - xy' -y -8yy' = 0$

Rearranging

$y' = \frac{10x - y}{x + 8y}$


答:
I will provide some explanation from the point when the author makes an assumption that $x^2<2$.

The key here is to observe that any upper bound $x$ of $A$ must satisfy $x^2\geq 2$ (a good guess because all the $x$ with $x^2<2$ are in $A$). To show that one assumes $x^2<2$ and derives a contradiction.

The way to get a contradiction here is to find some $y\in A$ such that $y>x$ (a contradiction is obtained because $x$ is an upper bound for $A$ and no member of $A$ can exceed $A$).

Now the author takes a suitable form for $y$ as $y=(n+1)x/n$ for some positive integer $n$. This automatically ensures that $y>x$, but more importantly $y$ is near $x$ (in terms of limits $\lim_{n\to\infty} y=x$). Another form could have been $y=x+(1/n)$ and a proof may be given based on this.

To obtain the contradiction we not only need $y>x$ but also $y\in A$ ie $y^2<2$. This means that we need $$\frac{(n+1)^2}{n^2}<\frac{2}{x^2}$$ or $$\frac{(n+1)^2} {n^2} -1<\frac {2} {x^2} -1$$ or $$\frac{2n+1}{n^2}<\frac{2q^2-p^2}{p^2}$$ or $$\frac{n^2}{2n+1}>\frac {p^2}{2q^2-p^2}$$
Also note that this is not an inequality which is logical consequence of our assumptions, but this is a target/goal which needs to be achieved if we want our contradiction. If we can't find an $n$ which can satisfy this inequality then we have no hope.

Since we have $$\frac {n^2}{2n+1}>\frac{4n^2-1}{4(2n+1)}=\frac{2n-1}{4}>\frac{n}{2}-1$$ we can see that the desired goal is achieved if we take $$n>2\left(\frac{p^2}{2q^2-p^2}+1\right)=\frac{4q^2}{2q^2-p^2}$$
The inequality in question arose from the special form of $y$ as $y=(n+1)x/n$  which ensures $y>x$ and an additional requirement that $y^2<2$.

Also the entire argument above does not need $p, q$. It could have worked by just using the symbol $x$. This is typical of many textbook authors who uses too much symbols instead of natural language.

Using this desired value of $n$ we obtain the contradiction and hence we must have $x^2\geq 2$.

The next part of the proof mentions that since $x$ is rational we can't have $x^2=2$ and therefore we must have $x^2>2$. Here we again obtain a contradiction by finding a $y<x$ such that $y\notin A$. Can you figure out why finding such a $y$ would lead to a contradiction? Also you should be able to guess that the chosen form of $y$ should now be $y=nx/(n+1)$ because it automatically ensures $y<x$ (another option would be have $y=x-1/n$).

I personally do not like the way the proof has been presented by the author. It is best if the proof and its ideas are developed by author taking the reader also into confidence. No point trying to play smart with your readers.

Title: Question about the proof for "The rational line Q does not have the completeness property"

Body:
Referring to the attached png file, I have questions about the inequality:

$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$

How is this inequality derived?  How did the n get into the inequality?  Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$, and later in the proof, $2q^2/(p^2 - 2q^2)$ come from?   There were no derivations for these 3 expressions.  Could someone please provide the derivations?


问:
When dealing with a group, is it possible that we have $\forall _{a\in G}\:\:a^{-1}=fixed$ (and no "individual" invertible elements?)
For example if we have $\phi \left(a,b\right)=\left(1+a\right)\left(1+b\right)$ the neutral element is equal to $e = 0$, and if we look for an element $a^{-1}$ for which $a\cdot a^{-1}=e$ I get that $\forall _{a\in G}\:\:a^{-1}=-1$

Is G a group then? If the invertible element is "fixed" for all a?

答:
First of all, $0$ is not the identity of the operation $\phi$ you wrote down. $\phi(a, 0) = (1+a)(1+0) = 1+a$, but this is not $a$ (unless $1=0$ in whatever ring you're defining $\phi$ on).

To get to your general question, no this is not possible except in the trivial group. In a group $G$, every element has a precisely one inverse. That is, if $ab = ac = e$ then $b = c$. This allows us to have a well defined operation $G \longrightarrow G$ via $a \mapsto a^{-1}$. This map is always a bijection, so every element has precisely one inverse that is uniquely determined by that element. In your claimed example, the map $a \mapsto a^{-1}$ has only $-1$ in the image, but for this to be the case that would mean that $|G| = 1$. That would be the case $1=0$ I mentioned above, but ignore that if you're unfamiliar with ring theory. If you meant $\phi$ to be an operation on something like $\mathbb R$ then this is just false.

You can prove that $a \mapsto a^{-1}$ is a bijection by proving that it is its own inverse, i.e. that $(a^{-1})^{-1} = a$.


答:
Then for any $a,b \in G$ we have $e = ac = bc$ where $c$ is the fixed inverse. As $c$ is also invertible,

$$\begin{align}
ac = bc &\implies a=b\\
 &\implies |G| = 1 \\
&\implies G = \{e\}.
\end{align}$$

Title: When dealing with a group, is it possible that we have $\forall _{a\in G}\:\:a^{-1}=fixed$ (and no "individual" invertible elements?)

Body:
For example if we have $\phi \left(a,b\right)=\left(1+a\right)\left(1+b\right)$ the neutral element is equal to $e = 0$, and if we look for an element $a^{-1}$ for which $a\cdot a^{-1}=e$ I get that $\forall _{a\in G}\:\:a^{-1}=-1$

Is G a group then? If the invertible element is "fixed" for all a?


问:
Achilles and the tortoise problem
You know the story - Achilles and the tortoise are having a race. The tortoise is given a head start of 100m. After some time, Achilles will arrive at where the tortoise was at, but the tortoise will have moved further. Question: find how long it will take for Achilles to catch up to the tortoise in seconds, and how far Achilles will travel when he reaches the tortoise in meters. Achilles runs at $10m/s$ and the tortoise $0.2m/s$. Use an infinite series.

Intuition

Obviously I would have to use infinite series to solve. As I see it, I should create two series: one for Achilles and one for the tortoise. For the turtle, it would roughly be $100 + \sum_{n=1}^{\infty}a_n$. The 100 is the 100 meter head start, and $a_n$ would be the series that represents the movement of the tortoise. For Achilles it would be similar.

Goal

The goal is to find $n$, which would depict the time in seconds, where Achilles will catch up to the tortoise. To do this, I would have to equate Achilles' series and the tortoise's series, evaluate each, and solve for n. To find the total distance traveled by Achilles, it would be substituting $n$ into the equation and evaluating the series.

Answers

So, any answers for $n$? I'm mainly confused as to how to formulate the infinite series, and how to solve for each series (which would involve seeing if it converges and such).

答:
There is no reason that you have to use an infinite series.  Achilles gains on the tortoise at a speed of $10-0.2 = 9.8 \frac {m}{s}$ and catches the tortoise in $\frac {100}{9.8} s$

If you really want to set this up as a series.

In 10 seconds the Achilles is at the tortoise's starting point.  And the tortoise has moved 2 meters away.  The new distance is 2 percent of the original distance.  In $(10)(0.02)$ seconds Achilles will be an this new point, The tortoise will have move again creating a new gap $2\%$ of the previous gap.

$10\sum_\limits{i=0}^{\infty} 0.02^i$

Title: Achilles and the tortoise problem


答:
Keeping in mind that $y$ is just a shorthand for $y(x)$, we can write the equation as
$$
5x^2-xy(x)-4y(x)^2=0 \, ,
$$
and so
\begin{align}
\frac{d}{dx}\left(5x^2-xy(x)-4y(x)^2\right)&=\frac{d}{dx}(0) \\
10x-y(x)-xy'(x)-8y(x)y'(x) &= 0 \, ,
\end{align}
using the chain and product rules. Rearranging this equation should leave you with the answer.


问:
When there exists a lifting $\rho e^{i\theta}$?
Assume $f : \mathbb{T}^N \to \mathbb{C}$ is a complex  function defined on a torus. I wonder when can we write $f = \rho e^{i\theta}$ (this is called a lifting for $f$) for some real functions $\rho $ and $\theta$. There is some conditions of $f$ that assures there is a lifting?. For instance, if the function does not vanish ($f (x) \neq 0$ for all $x\in\mathbb{T}^N$), then is there a lifting?.

Thanks in advance. Any comment is welcome.

答:
It is easy if you use the machinery of covering projections.

It is well-known that $\exp : \mathbb C \to \mathbb C \setminus \{0\}$ is a covering projection; see for example How do I prove $e^z$ is a covering map using this fact? .

Then each map $f : X \to \mathbb C \setminus \{0\}$ defined on a simply connected $X$ has a lift. See for example Hatcher's book on Algebraic Topology. Clearly $\mathbb R^n$ is contractible, thus simply connected.

Update:

The question was modified: Instead of $X = \mathbb R^n$ we consider $X = \mathbb T^n$. Then in general there is no lift. Take for example $n = 1$. Then $\mathbb T^n = S^1 \subset \mathbb C \setminus \{0\}$. Now let $f$ be the inclusion map. It does not have a lift.

See also A path from $S^1$ to $\mathbb{C} \setminus \{0\}$ can be expressed in "polar" form if and only if, the winding number of the normalized path is zero .


问:
Existence of an element of order $p$
Let $G$ be an abelian finite non trivial group and $p$ a prime number which devises o(G). We denote $H$ the set of non trivial subgroups of $G$.

Could you please help me to show that if there exists $M\in H$ such that $p$ devises $o(M)$  then there exists $g\in G$ of order $p$.

Thanks in advance

答:
We know that there exist $q \in \mathbb{Z}$ such that $o(M) = pq$, we also know that $o(M), p > 0$ then $q >0$. Since H is non-trivial, there exists $m \in M \setminus\{e\}$ such that $o(m) \neq q$. If $q =1$ take  $g=m$ and if $q>1$ take $g = m^q$
\begin{equation*}
g^p = m^{pq} = m^{o(M)} = e.
\end{equation*}
This implies that $o(g)$ divides $p$ but $p$ is prime then $o(g) = 1$ or $o(g) = p$ but if $o(g) = 1$ then $m^q = e$ which is not possible due to the choice of $q$. Therefore $o(g) = p$.

Title: Existence of an element of order $p$


答:
Here is a trick... differentiate with respect to x, treating y as a constant.  Put it all in brackets and write "dx" at the end.  Then do the same thing over again except you will be differentiating with respect to y and and treating x as a constant.

$\frac {\partial}{\partial x}(5x^2 - xy - 4y^2)\ dx + \frac {\partial}{\partial y}(5x^2 - xy - 4y^2)\ dy = 0\\
(10x - y) \ dx + (-x - 8y)\ dy = 0$

Now solve for $\frac {dy}{dx}$

$\frac {dy}{dx} = \frac {10 - y}{x+8y}$