问:
When is a polynomial map over the complex plane proper?
I am trying to prove the following for a polynomial map $p$:

$p:\mathbb{C}\to\mathbb{C}$ is proper $\iff$ $p$ is non-constant

How would I go about doing this, with definitions alone? With the first assumption we can get that if $p$ were constant then the mapping doesn't necessarily send compact sets to compact sets, but I'm not quite sure on how to go about proving the other direction, that if $p$ is non-constant then the map is proper.

Any help is appreciated, thank you.

答: You want preimages of compact sets to be compact. A compact set (in the complex plane) is closed and bounded. Can a preimage of a bounded set be unbounded? (hint: what happens to polynomials at infinity?) Similarly for closed (hint: take subsequences).


问:
How wide can a unit-length planar curve be?
The width of a bounded set in the plane is the minimum distance between two parallel lines bounding the set.

Suppose that we have some curve $C: [0,1]\to\mathbb{R}^2$ of unit length. How large can its width be? Equivalently, if we would like to bend a wire into a shape of unit width, how much wire do we need? (I'll use the latter framing in this post because it makes the numbers a little nicer.)

If one tries forming the wire into a circle or semicircle, $\pi$ units of length are needed. All but one of the sides of a unit square only takes $3$ units. Using $270$ degrees of a circular arc one can do better, with length $\frac{3\pi(2-\sqrt{2})}{2}=2.7604\ldots$.

After playing around with different natural options for a bit, the winner appears to be $\frac{4}{\sqrt{3}}\approx2.3094$, given by forming the wire into a $60$-degree angle so that its convex hull is the equilateral triangle of width 1.

However, it turns out that this is not optimal! Consider the following construction: Fix $0.5<x<\frac{\sqrt{3}}{2}$. Start with the triangle formed by the points $(0,1), (-x,0), (x,0)$. Then, add the circular arcs of radius $1$ from each of the latter two points as they pass through the opposite side of the triangle. Take the convex hull of these three points and two circular arcs, and omit the line segment of length $2x$ at the bottom; the resulting set is our wire.

Above is a diagram of this construction: the original triangle is marked in blue, the circular arcs (of which only part make it to the convex hull) are marked in red, the straight-line tangents to said arcs are in black, and the portion of the convex hull not on the wire is marked with a dashed line.

It is fairly easy to see that this construction has width $1$, and that the resulting length is

$$2\left(x+\sqrt{4x^2-1}+\arctan(1/x)-\arctan(x)-\arccos(1/2x)\right)$$

As it turns out, this function is minimized around $x=0.521795$, with a wire length of around $2.2783$. (In fact, the derivative of the above function is a rational function, so we can be more precise: $x$ is the unique positive real root of the polynomial $3x^6 + 9x^4 + x^2 - 1$.)

Is this construction optimal? I do not believe there are any obvious local improvements one can make, but it's possible that a radically different wire shape could improve on this configuration.

Obviously, there is a lower bound of $1$ on the length of any solution, and by considering the width in two orthogonal directions one can increase this to $\sqrt{2}$. It's tempting to say that the length must be at least $2$, because there must be a point at distance $1$ from the midpoint of the wire, but this only follows if one makes the assumption that the wire lies on the boundary of its convex hull, which I cannot see a good way to justify.

I would be curious to see any improved lower bounds, better solutions than the one given above, or pointers to discussion of this question in the literature.

答: A simple lower bound is $\sqrt 5>2.236$. Just choose the $x$-axis along the line joining the endpoints and $y$-axis perpendicular to it. Then, if your wire is $(x(t),y(t)), t\in I$, we must have $\int_I|y'(t)|dt\ge 2$ (you make a swing of width $1$ in the $y$-coordinate but you have to return to the start) and $\int_I|x'(t)|\ge 1$, so $L=\int_I\sqrt{x'(t)^2+y'(t)^2}\ge\sqrt{2^2+1^2}=\sqrt 5$.

ContentLicense: CC BY-SA 4.0


问:
Find the density function of $X+Y$ when $X,Y \sim_{i.i.d} U[0,1]$
I'm using the transformation of R.V and calculating marginal density approach and it confuses me how there's two cases.

In my working I let $U=X+Y, V=Y$

$f_{U,V} = f_{X,Y}|J|=f_{X,Y}=1$

from $0<x<1,0<y<1$ we deduct that $vu-1,0<v< 1$
by calculating marginal density for any joint density function we calculate $\int_{Y}f_{X,Y}dy$ and here it would be $f_U=\int_{V}1dv = \int_0^u1dv+\int_{u-1}^11dv=u+2-u$.

from the graph it kinda make sense that $f_U$ is not a constant, but I can't figure out from the calculation where to introduce the two cases and why.

答:
Think geometrically.  Let $(X,Y)$ be a random coordinate in the unit square $[0,1] \times [0,1]$.  Then if $U = X+Y \le u$, this means $(X,Y)$ is a point in the subset of the unit square that is "below" the line $X+Y \le u$.  For $0 < u \le 1$, this is a triangle with vertices $(0,0), (u,0), (0,u)$.  If $1 < u \le 2$, then this is a pentagon with vertices $(0,0), (1,0), (1,u-1), (u-1,1), (0,1)$.  Its complement is the triangle $(1,u-1), (1,1), (u-1,1)$.  What is the area of this region as a function of $u$?

Now with this insight, can you work with your method in a way that allows you to arrive at the same conclusion?


问:
How do I prove that the sine function on the domain $[-1/2, 1/2]$ is injective?
How do I prove that the sine function on the domain $[-1/2, 1/2]$ is injective?

I completely understand the concept but I'm having trouble writing a proof for this. Thanks in advance.

答:
The defining series for $\sin u $ and $\cos u$ are alternating for small $u>0$. It follows that
$$\eqalign{\sin u&\geq u-{u^3\over 6}=u\left(1-{u^2\over6}\right)\geq{23 u\over24}\qquad\bigl(0<u\leq{1\over2}\bigr),\cr
\cos u&\geq1-{u^2\over2}\geq {3\over8}\qquad\bigl(0<|u|\leq{1\over2}\bigr)\ .\cr}$$
Assume now that $-{1\over2}\leq x<y\leq{1\over2}$. Then $0<{y-x\over2}\leq{1\over2}$ and $\left|{x+y\over2}\right|\leq{1\over2}$. Therefore
$$\sin y-\sin x=2\sin{y-x\over2}\cos{x+y\over2}\geq 2\ {23\over24}{y-x\over2}\ \cdot{3\over8}>0\ .$$</p>

ContentLicense: CC BY-SA 3.0


问:
Showing all boolean functions can be expressed by only conjunctions or negations
Let $f:$ {T,F}$^n \rightarrow$ {T,F}, i.e a function of n boolean variables.

Show that each $f$ can be expressed as a formula of only conjunctions and negations, and give an upper bound for the number of conjunctions aswell as a lower bound.

For example n=2 has 2^4=16 different functions and I can express them all using only  conjunctions and negations. But, for the general case I'm really lost on how to proceed.

答:
Welcome to MSE!

Hint:

You mention in the comments that you're proceeding by brute force for the $16$ functions when $n=2$... What's preventing you from doing this in the general case?

Given $n$ variables you have a (big) truth table with $2^n$ rows.

Can you find a way to implement any particular row? That is, can you find a combination of $\land$s and $\lnot$s which gets a fixed row correct?

Can you then modify each row function to be $0$ if you plug in a different row of the truth table, but still give the correct answer on your row of interest?

Once you have these, can you show that "or-ing" all of the rows together will give you the function you want?

You might take inspiration from the Lagrange Interpolation Formula. In fact, the problem you're describing is exactly showing that every boolean function is a polynomial over $\mathbb{Z}/2$!

I hope this helps ^_^


问:
Help needed with minimum cost flow problem
I am trying to solve a very simple minimum cost flow problem using optimization software. However, the software keeps on complaining about the problem being infeasible, which means I have most certainly not expressed the model/constraints correctly. The network is illustrated below:

Supply nodes: 0 and 1 (75 units for each node)

Demand nodes: 4 (50 units),5 (60 units), 6 (40 units)

Costs in red along the arcs

Node 3 has a max capacity of 50 units

I tried to solve the problem by defining the following equations:

Minimize the objective function:
$5 x_{02}+8x_{03}+7x_{12}+4x_{13}+x_{24}+5x_{25}+8x_{26}+3x_{34}+4x_{35}+4x_{36}$

I have also defined the following constraints:

$x_{02}+x_{03} = 75$

$x_{12}+x_{13} = 75$

$x_{24}+x_{34} = 50$

$x_{25}+x_{35} = 60$

$x_{26}+x_{36} = 40$

At the transshipment nodes 2 and 3:

$x_{02}+x_{12}-x_{24}-x_{25}-x_{26} = 0$ (at node 2)

$x_{03}+x_{13}-x_{34}-x_{35}-x_{36} = 0$ (at node 3)

Maximum capacity constraints at node 3:

$x_{03} \leq 50$

$x_{13} \leq 50$

$x_{34} \leq 50$

$x_{35} \leq 50$

$x_{36} \leq 50$

Am I missing any other constraints? Are any of the above constraints expressed in a wrong way?

Thanks for your help!

答:
Node capacity of 50 means the sum of incoming flows (which is equal to the sum of outgoing flows) at that node is at most 50.  This is a tighter constraint than the five $\le$ constraints that you specified.

An optimal solution, with objective value 1250, is
\begin{matrix}
i &j &x_{i,j} \\
\hline
0 &2 &75 \\
0 &3 &0 \\
1 &2 &25 \\
1 &3 &50 \\
2 &4 &50 \\
2 &5 &50 \\
2 &6 &0 \\
3 &4 &0 \\
3 &5 &10 \\
3 &6 &40 \\
\end{matrix}


问:
Given an arbitrary quadratic polynomial $ax^2 + bx + c$, how do I find the path the vertex of the parabola travels upon when b is varied?
https://www.desmos.com/calculator/ysuklqqwfe

I've made a desmos to illustrate the problem. If you press play on b you'll see what I mean.

I'm pretty sure the arc is parabolic because the y coordinate varies quadratically with respect to x. But I'm not sure how to go about proving it, or to find the equation of the path.

Any insights?

答: The vertex of the parabola is$$\left(-\frac b{2a},-\frac{b^2}{4a}+c\right).$$Therefore, the set of all vertices is $\left\{(x,c-ax^2)\,\middle|\,x\in\Bbb R\right\}$, which is indeed a parabola.


问:
rank function on Spec (help with definition)
one definition of the line bundle over a ring is: a finitely generated projective A-module such that the rank function Spec A → N (positive integers) is constant with value 1. We call A itself the trivial line bundle.

so here i think that spec is equipped with zariski topology and N with the discrete one. Does this mean that in general the rank function is not continuous? Does one know about a basic example of non constant rank function to illustrate the peculiarities implied by the definition above?

Many thanks

答:
I am only going to talk about the case that $A$ is Noetherian.

Here is part of Theorem A3.2 of Eisenbud's "commutative algebra with a view toward Algebraic geometry": a finitely generated module $M$ over a noetherian ring $A$ is projective if and only if there exists a finite set of elements $x_1,\ldots, x_r$ in $A$ that generates the unit ideal of $R$ such that $M[x_i^{-1}]$ is free over $A[x_i^{-1}]$ for each $i$.

It follows that the rank is a locally constant function. So if $\mathrm{Spec}(A)$ is connected, then the rank has to be constant.

I'll leave it to you to construct an example of a projective module with nonconstant rank.


问:
Existence of a smooth homotopy
Suppose I have a contractible neighborhood $B$ inside a manifold $M$. Then I would like to justify that there exists a family of smooth maps $F_t:M\rightarrow M$ such that $F_0=id$ and $F_1(B)=\{p\}$, where $p\in M$.

We know that $i:B\rightarrow M$ is homotopic to a constant map but other than this I am bit lost on how to get the homotopy,since I am not sure I can use the homotopy extension property , and then how to make it smooth.

Any help is appreciated. Thanks in advance.

答:
As requested, I’m turning my hint into a fully-fledged solution.

First, as Connor Malin’s comment points out, the statement is false in general, because some manifolds can have a dense contractible open subset. A general example would be any compact manifold which has a finite cellular decomposition with a single top-dimensional cell (the interior of which is open, dense and contractible) – real and complex projective spaces, spheres, tori are in this case.

A natural restriction (not the only one) is if we require $B$ to be small enough. If $B$ is small enough, we can assume that there is a smooth, open, injective map $\pi: 3B^n \rightarrow M$ which is a diffeomorphism into its image $B’$ and such that $\pi(B^n) \supset B$.

Let $\psi:[0,3] \rightarrow [0,1]$ as a smooth function such that $\psi([0,1])=\{0\},\psi([2,3])=\{1\}$. Let, for $0 \leq t \leq 1$, $g_t: x \in 3B^n \longmapsto (1-t+t\psi(\|x\|))x \in 3B^n$. Then if $\|x\| \geq 2$, $f_t(x)=x$; $f_0(x)=x$ for any $x$; and $f_1(x)=0$ if $\|x\|=1$.

Define, for $q \in M$, $f_t(q)=q$ if $q \notin B’$, and $f_t(q)=\pi(g_t(\pi^{-1}(q)))$ if $q \in B’$. Then $f_t$ is a smooth sequene of smooth maps $M \rightarrow M$; $f_0=id$ and $f_1(B)=\{\pi(0)\}$.

Title: Existence of a smooth homotopy


问:
MLE of $\frac{1}{\theta}$ Uniform distribution.
Let $X_1,X_2...X_n$ be a random sample from $U(\theta-5,\theta+5)$ where $\theta \in(0,\infty) $ is unknown. Let $T=\max(X_1,X_2...,X_n)$ and $U=\min(X_1,X_2...,X_n)$. Then which of the following statements is TRUE?

$(A) \dfrac{2}{T+U}$ is M LE of $\dfrac{1}{\theta}$

$(B) MLE$ of $\dfrac{1}{\theta}$ doesnt exist

$(C)U+8$ is MLE of $\theta$

$(D)\dfrac{T+U}{2}$ is unique MLE of $\theta$

$\theta-5<x_{(1)} < x_{(2)}....x_{(n)}< \theta+5$

$\theta<x_{(1)}+5$ and $\theta>x_{(n)}-5$

$x_{(n)}-5<\theta<x_{(1)}+5$

Any statistic satisfying above condition is MLE of $\theta$ so there is no unique MLE here option $(D)$ rules out

$x_{(1)}+8<x_{(1)}+5$

Option $(C)$

$8<5$ which is false so option $C$ is false.

I am not sure how to prove $(A)$ and $(B)$ but in the exam, I have ticked $(B)$ so am I right?   If I am wrong can anyone prove how?

答:
This question really boils down to whether you're allowed to divide by 0 or not. Suppose that you have a sample $X_1,...,X_n$ with $U=\min\{X_1,...,X_n\}=-1$ and $T=\max\{X_1,...,X_n\}=1$. Then we can agree that any $\theta\in[-4,4]$ is an MLE of $\theta$ because the indicator variables in the likelihood is 1 (as opposed to 0). That is, the inequality

$$T-5\le\theta\le U+5$$

holds. Then by the invariance property $\left(\frac 1 \theta\right)_{MLE}=\frac 1 {\theta_{MLE}}$. That is, $\left(\frac 1 \theta\right)_{MLE}\in(-\infty,-\frac 1 4]\cup[\frac 1 4,\infty)$. In particular, $\frac{T+U}{2}=0$ is an MLE of $\theta$, but $\frac{2}{T+U}$ is not an MLE of $\frac 1 \theta$ because it is undefined. However $\left(\frac 1 \theta\right)_{MLE}$ do exist, as previously mentioned being in the interval of anything smaller than -1/4 or greater than 1/4.

It is helpful to determine the difference between maximum likelihood estimators and maximum likelihood estimates. Note that $\frac{T+U}{2}$ is a MLE of $\theta$ that works for all samples - in fact it is the only estimator that works for all samples. By the invariance property of maximum likelihood estimators, we can conclude that $\frac{2}{T+U}$ is a maximum likelihood estimator of $\frac 1 \theta$, even though as in the example in the previous paragraph, its maximum likelihood estimates that it produces may be undefined when U=-T, whilst other MLEstimates exist for 1/theta.

Conclusion
$\frac 2{T+U}$ is a Maximum Likelihood Estimator of $\frac 1 \theta$. Since it does not exist when the denominator is 0, it is also true that it is a MLEstimator of $\frac 1 \theta$ when this happens.

Eliminate:

(B) It is too specific. A MLE of 1/theta only doesn't exist when U=-5 and T=5. 2/(U+T) doesn't work for when U=-T and $U\ne-5$, but MLEs of 1/theta always exist in these cases.


问:
Let T be a tree such that every leaf is adjacent to a vertex of degree at least 3. Show that there are two leaves with a common neighbor.
I came up with a proof for this but I think it might be a little hand-wavy and I can't figure out why. Some advice on how to make this proof clearer will be really appreciated!

Problem: Let $T$ be a tree such that every leaf is adjacent to a vertex of degree at least $3$. Show that
there are two leaves with a common neighbor.

Proof: Suppose by contradiction that there are no two leaves with a common neighbor in T. Then let there be $k$ leaves in T and because each leaf is adjacent to a vertex of degree at least $3$, then there are k vertices in T that has degree at least $3$. For vertices other than the leaves and the leaves' neighbor, there are $V(T)-2k$ of those vertices and because they are not leaves they have degree of at least 2. So the minimum sum of degree of vertices in $T$ is $$\sum_{v\in T}d(v)=k+3k+2\cdot (V(T)-2k)=2V(T)$$ however by degree sum formula there is exactly $\sum_{v\in T}d(v)=2V(T)-2$, a contradiction.

答:
Lemma: Any tree has more leaf nodes than degree-3 and higher nodes.

Proof: Let a graph $G$ be a tree on $n$ vertices with $\ell$ leaf nodes and $h$ vertices of degree $3$ or more. Then $G$ has $n{-}1$ edges and by handshaking we know:

$2(n-1) = \sum d(v_i) \;  \geq \; \ell + 2(n-\ell-h) + 3h =  2n-\ell+h,\quad$ so $\ell-2\geq h$ as required.

Now since in your question every leaf in $T$ is attached to a degree-$3+$ node, by the pigeonhole principle some leaves must be attached to the same such node.


问:
find the value of k at the point P
Whilst using this figure:
enter image description here

It's a sketch showing the part of the curve C with equation:

$y = (2x-1)^{\frac{3}{4}}$, $x \ge\frac{1}{2}$

The curve C cuts the line y = 8 at the point with coordinates (k, 8), where k is a constant. Find the value of k.

I'm wondering If I approached this correctly?

my workings:

$8 = (2x-1)^{\frac{3}{4}}$

$4096 = (2x-1)^3$

$\sqrt[3]{4096}=2x-1$

$16 = 2x-1$

$\frac{17}{2}=x$ such that, ($\frac{17}{2}$, 8)?

Title: find the value of k at the point P

Body:
hint

$$8=(2k-1)^\frac 34=2^3$$

$$\implies (2k-1)^\frac 14=2$$
$$\implies 2k-1=2^4=16$$


问:
Vanishing of $\pi_n(X\times Y, X\vee Y)$ for $n\leq p+q+1$
Let $X,Y$ be (based) CW-complexes (assume $X,Y$ are nice so that there is no problem with $X\times Y$). If $X$ is $p$-connected and $Y$ is $q$-connected, how to show that $(X\times Y, X\vee Y)$ is $(p+q+1)$-connected?

I think I need to use the following result (Theorem 4.23 of Hatcher):
Let $X = A\cup B, C = A\cap B$. If $(A,C)$ is $m$-connected and $(B,C)$ is $n$-connected, then $\pi_i(A,C)\to \pi_i(X,B)$ is an isomorphism for $i<m+n$.

But I failed to decompose $X\times Y$ into two subsets that are useful to solve this problem. I know $\pi_{n+1}(X\times Y, X\vee Y)$ is a direct summand of $\pi_n(X\vee Y)$, but this is not very useful. Maybe I am missing something very obvious.

Any help is appreciated.

答: If $X$ is $p$-connected, you should choose a CW model for it with no cells in dimensions between 1 and $p$, and similarly for $Y$. Also choose models with a single $0$-cell (that's probably guaranteed already unless $p=0$ or $q=0$). The first interesting cells $e \times e'$ in $X \times Y$ (by which I mean the first cells that are not in $X \vee Y$) will have dimension at least $p+q+2$, since $\dim e \geq p+1$ and $\dim e' \geq q+1$.


问:
All idempotents are central then $KG$ has no nilpotent
Given $G$ a finite group and $K$ a field of characteristic zero such that all idempotents in $KG$ are central, is it true that $KG$ has no nilpotent element or equivalently $KG$ has only division ring in its Wedderburn decomposition?

答: Yes. If there is a factor $M_r(D)$ with $r\geq2$ in the decomposition, then any elementary matrix $E_{ii}$ is a non central idempotent.


问:
How to express the probability $P(X \text{ mod } 2 = 0)$ that an integer-valued $X$ is even using characteristic function?
Suppose $X$ is an integer-valued random variable. My question is, how can I express the probability $P(X \text{ mod } 2 = 0)$ that $X$ is even knowing the characteristic function $\phi_X(t)$? Can I do it without using a general inversion formula by summing up $P(X=2k)$ over $k \in \mathbb{Z}$? Maybe, I could use some sort of simpler argument and end up with some nice expression?

I almost don't have any idea. I tried to use $\phi_X(2t) = \sum\limits_{k \in \mathbb{Z}} e^{2ikt}P(X=k)$, but it just doesn't work. Do I miss something trivial here?

Actually, this is only a part of the problem that I'm trying to solve. Another part is the following:

Let $\{X_n\}_{n=1}^{+\infty}$ is a sequence of i.i.d random variables, and $P(X_1 \text{ mod } 2 = 0) = p \in (0, 1)$. Let $Y_n = X_1 + X_2 + ... + X_n$. Prove that $P(Y_n \text{ mod } 2 = 0) = 1/2$.

And I want to solve this problem using characteristic functions. I suspect, for this I would need to find $\phi_{Y_n}(t) = (\phi_{Z_1}(t))^n$ with $Z_i = X_i \text{ mod } 2$.

答:
You're making this way too complicated. $Y_n$ is even if and only if an even number of $\{X_1, \dots, X_n\}$ are odd. So

$$P(Y_n \text{ is even}) = \sum_{k \text{ even}} \binom{n}{k} (1 - p)^kp^{n-k}.$$

So one strategy is to compare this sum to the expansion of $(p + q)^n$ where $q = 1 - p$ (so of course $p + q = 0$).

For instance, one can show that

$$\sum_{k \text{ even}} \binom{n}{k} = \sum_{k \text{ odd}} \binom{n}{k},$$

by looking at the binomial expansion of $(1 - 1)^n$.


问:
Show that $\mathcal{F}_{X}$ is the smallest $\sigma-$field containing the inverse images $X^{-1}(B)$ of all Borel sets $B$
I am currently working on the following problem for my measure theory course and I wanted some feedback on my work:

Show that $\mathcal{F}_{X}$ is the smallest $\sigma-$field containing the inverse images $X^{-1}(B)$ of all Borel sets $B$ (note that $\mathcal{F}_X :=\{S \subset \mathcal{F}:S = X^{-1}(B),\text{for some} \ B \in \mathcal{B} \}$ where $X$ is a random variable and $(\Omega,\mathcal{F},P)$ is a probability space).

My (short) work is as follows:

Take any $\sigma-$field $\mathcal{L} := \{S \subset \mathcal{F}:S = X^{-1}(B), B \in \mathcal{B} \}$. Then we can see that $\mathcal{F}_X \subset \mathcal{L}$ (since $\mathcal{F}_X$ is defined for some Borel sets $B$), hence $\mathcal{F}_X$ is the smallest $\sigma-$field containing the inverse images $X^{-1}(B)$ of all Borel sets $B$.

My question regarding my work: my professor said everything looks good but I need to justify that $\mathcal{L}$ is in fact a sigma-field (which I thought to be given). Is there any easier way to show this without going through the requirements for a set to constitute a sigma-field? Regardless, any tips or feedback of any kind is welcome.

答:
It's quite clear that $$\mathcal{F}_X:=\{X^{-1}[B]\mid B \in \mathcal{B}\}$$

is a $\sigma$-field, because $\mathcal{B}$ is one and we have identities like

$$X^{-1}[\bigcup_n B_n]=\bigcup_n X^{-1}[B_n]; X^{-1}[B^\complement] = \Omega\setminus X^{-1}[B]; X^{-1}[\bigcap_n B_n]=\bigcap_n X^{-1}[B_n]$$

This is a trivial verification. And it's trivially the minimal one that contains all inverse images under $X$ of all Borel sets. If $X$ is measurable, it's a sub-$\sigma$-field of $\mathcal{F}$.


问:
Existence of periodic solution in an ODE system
Let the ODE system,

\begin{cases}x'(t)=x^3+x^5 \\ y'(t)=y+y^7 \end{cases}

defined in $\mathbb{R}^2$.

I need to say if this system supports periodic solution with period $T>0$.

This is a question of Analysis in $\mathbb{R}^n$ not ODE, that is, I need to demonstrate this using concepts presented in Analysis in $\mathbb{R}^n$, my textbook is Analysis on Manifolds by Munkres. The hint is,

A periodic solution of the system above is a closed curve $\gamma(t) = (x(t), y (t))$ that satisfies the system.

I believe I have to use Stokes, but I am not able to proceed with the solution.

答:
If there is a solution that is periodic in $(x(t),y(t))$, then also functions $x(t)$ and $y(t)$ have to be periodic separately, as real functions.

Further, the system is decoupled, you have two scalar, non-coupled autonomous ODE of first order. Scalar autonomous first-order ODE do not have periodic solutions, their solutions are constant or (strictly) monotonous (assuming the right side function is locally Lipschitz or differentiable).


问:
Help with this probability density function

(a) Verify that$ f $is a PDF. That is,  check that $f(x)$ is always non-negative and that the area under $f(x)$ is $1$. My understanding is that if $a < x < b $then it is $1/b-a$ so the area would have to be non negative, the graph would look somewhat like this

$a$ being $20$ and $b$ being $40$. $1/20 = .05$ which is non negative but I am not sure how to prove it is $1$?

(b) Find the $CDF$ of $X$, $F(y)$, for all $y$.
Finding the $PDF$ first,

| x  | P(x)|
| ---| --- |
| 20 | .1  |
| 21 | .1  |
| 22 | .1  |
And so on

So $CDF$ would be
| x       | F(x)|
| ---     | --- |
| x<20    | .1  |
| 20<=x<21| .2  |
| 21<=x<22| .3  |
and so on

(c) Find $P(30<X<50)$.

Since $(20<=X<40)$ is $1/20$ would $P(30<X<50)$ just be $1/30$?

(d)Find expected value of X

$.1*20+.1*21....+.1*40$ (I have not calculated it yet but I believe this is the correct form)

It also asks to find standard deviation and variance, but as long as I am correct thus far I do not need help figuring it out. Thank you in advance!

答:
All of your calculations are for if f is discrete, but here f is continuous.

(a) To show that f(x) is a pdf, you have to verify that the area under it is 1. You can do it like this: $\int_{20}^{40}\frac{1}{20}dx=\frac 1 {20}[x]_{20}^{40}=\frac 1 {20}(40-20)=1$ Since the area under the pdf integrates to 1, it is a valid pdf. Or you can observe that the pdf is a rectangle that has height y=1/20 for 20<x<40 and is flat y=0 when x is not in that range. So the pdf is given by the area of the rectangle, 20*1/20=1. Your picture is right, now you need to multiply b-a by 1/(b-a).

(b) The cdf F(y) is the probability that X is less than y, aka P(X<y). For a uniform distribution, it is given by

$$F(y)=\begin{cases}0&y<a\\
\frac{y-a}{b-a}&a\le y<b\\1&y\ge b\end{cases}$$

Here we have

$$F(y)=\begin{cases}0&y<20\\
\frac{y-20}{40-20}&20\le y<40\\1&y\ge40\end{cases}$$

This is a flat line at 0 when y<20, a line with constant slope $\frac 1{20}$ when 20<y<40, and then a flat line at 1 when y>40.

(c) $P(30<X<50)=P(30<X<40)+\underbrace{P(40<X<50)}_{=0}=(40-30)*1/20=\frac 1 2$

where the second term is 0 because X has 0 density above 40.

(d) The expected value or mean of a continuous uniform distribution is $\frac{a+b}{2}=\frac{20+40}{2}=30$. It is the tick that is between the two ends.


问:
Solving $y'' = -y \cdot (y')^2$ analytical
I was trying to solve the following differential equation:

$$y'' = -y \cdot (y')^2$$

But I always have some trouble solving nonlinear ODE. I used wolfram alpha to check the solution but it used complex number and the $\text{erf}$ function and I'm not very used to work with it. Is that the only way of solving this analytical? If so, how Can I solve this equation?

答:
This equation is separable,
$$
\frac{y''}{y'}=-yy'
$$
The resulting first-order equation might not as easy to integrate. And will possibly contain the inverse of the error function in its solution.


问:
Proofs: If-then vs modus ponens
I am just beginning to study proofs. What is the difference between modus ponens and implies/if-then when relating to proofs?

I figure that we are using modus ponens to advance the proof. If we were using $P \implies Q$, couldn't I start with a false P in order to falsely "prove" Q? In the limited number of proofs I have seen it seems like we just use $P \implies Q$. I am thinking it is either shorthand for modus ponens if we assume what we start out with is true?

How should I understand $\implies$ in proofs?

答:
Modus ponens, specifically is an inference rule. An inference rule is says that a conclusion is true if its premises are true. In other words, the inference rules of a system tell you what manipulations you are allowed to perform that don't introduce erroneous conclusions.

Here's modus ponens symbolically.

$$ \frac{A \to B \;\; \text{and} \;\; A}{B} $$

$A \to B$ is a premise. $A$ is a premise as well. $\text{and}$ exists outside the formal language and just tells us that we have both premises. $B$ is the conclusion. The line also exists outside the formal language and tells us how the premises and the conclusion are related.

Modus ponens allows you to take a conditional that you already know and its premise and infer its conclusion. This rule is fundamental to how conditionals are used.

Implies, which I've written $\to$, is how you represent the meaning of if A, then B. It is equivalent to the following:

$$ A \to B \;\; \text{is equivalent to} \;\; (\lnot A) \lor B $$

Another way of looking at it is that $A \to B$ is false if and only if $A$ is true and $B$ is false.

Title: Proofs: If-then vs modus ponens


问:
Find the density function of $X+Y$ when $X,Y \sim_{i.i.d} U[0,1]$
I'm using the transformation of R.V and calculating marginal density approach and it confuses me how there's two cases.

In my working I let $U=X+Y, V=Y$

$f_{U,V} = f_{X,Y}|J|=f_{X,Y}=1$

from $0<x<1,0<y<1$ we deduct that $vu-1,0<v< 1$
by calculating marginal density for any joint density function we calculate $\int_{Y}f_{X,Y}dy$ and here it would be $f_U=\int_{V}1dv = \int_0^u1dv+\int_{u-1}^11dv=u+2-u$.

from the graph it kinda make sense that $f_U$ is not a constant, but I can't figure out from the calculation where to introduce the two cases and why.

答:
$f_{\small X,Y}(x,y)$ does not equal 1 except when the values are in the support.

Always include the support in a density function.  This is usually done by using an indicator function. That is a piecewise function equal to one when the indicated condition holds, but zero when it does not: $$\begin{align}f_{\small X,Y}(x,y)&=\mathbf 1_{0<x<1,0<y<1}\\[1ex]&=\begin{cases} 1 & : & 0<x<1, 0<y<1 \\ 0 & : & \textsf{otherwise}\end{cases}\end{align}$$

So, indeed you have found the support for $\langle U,V\rangle$ is: $$\{\langle u, v\rangle: v<u<1+v, 0<v<1\}$$

Which can also be written as:

$${\{\langle u,v\rangle: 0<u<2,\max\{0,u-1\}<v<\min\{1,u\}\}\\[2ex]\text{or}\\[2ex]\{\langle u,v\rangle: 0<u<1, 0<v<u\}\cup\{\langle u,v\rangle:1\leqslant u<2, u-1<v<1\}}$$

So we have that :$$\begin{align}f_{\small U,V}(u,v) & = f_{\small X,Y}(u-v, v)\\[1ex]&=\mathbf 1_{0<v<1, v<u<1+v}\\[1ex] &= \mathbf 1_{0<u<1, 0<v<u}+\mathbf 1_{1\leqslant u<2, u-1<v<1}\end{align}$$

Thus $f_{\small V}(v)$ is found by using the former:$$\begin{align}f_{\small V}(v) &=\int_\Bbb R f_{\small X,Y}(u-v,v) \mathrm d u\\& = \int_v^{1+v} \mathbf 1_{0<v<1}\,\mathrm d u\\&= \mathbf 1_{0<v<1}\end{align}$$

Likewise $f_{\small U}(u)$ is found by using the later: $$\begin{align}f_{\small U}(u) &=\int_{0<v<u}\mathbf 1_{0<u<1} \,\mathrm d v+\int_{1-u<v<1} \mathbf 1_{1\leqslant u<2}\,\mathrm d v\\[1ex]&=u\,\mathbf 1_{0<u<1}+(2-u)\,\mathbf 1_{1\leqslant u<2}\\[1ex]&=\begin{cases} u&:& 0<u<1\\ 2-u&:& 1\leqslant u<2\end{cases}\end{align}$$


问:
Limiting Function for uniform convergence
$\lim_{n\to\infty} \frac{x+n}{n}$  I assumed that this limit equaled 1, correct? I am trying to determine a limiting function.

答:
As Kavi points out the limit is $1$. We can write $\frac{x+n}{n} = \frac{x }{n}+ \frac{n}{n} = \frac{x }{n}+1 = 1+ \frac{x }{n}$. The second term tends to 1.
So the functions $f_n:\mathbb R \to \mathbb R; f_n(x) =\frac{x+n}{n} $ converge to $f(x)=1$ pointwise.

However the convergence is not uniform since uniform convergence means $\max_x |f_n(x)-f(x)|  \to 0$ but in this case $\max_x |f_n(x)-f(x)| = \max_x |x|/n = \infty$.


答:
You can consider the curve $\phi = (x, y)$ and take the squared distance to the origin $x^2 + y^2$. The derivative of this is
$$2(x\dot{x} + y\dot{y}) = 2(x^4 + x^6 + y^2 + y^8)\ge 0.$$
If this curve is periodic, then it is actually a closed curve and so a compact subset of the plane so there must be a minimum of this distance, corresponding to a point closest to the origin.
Since it is a sum of squares this means $x = y = 0$ at that point, so your curve goes through the origin, but clearly the constant curve $\phi\equiv 0$ satisfies the equation and it can be the only one doing that.

If you wanted to conclude this from Stokes Theorem you need to integrate the 1-form
$$(x^3 + x^5)dx + (y + y^7)dy$$ along the line given by a potential closed curve, i.e. around $\phi$ and if $D$ is the interior, assuming $T$ is the minimum period so that you avoid self intersections, you conclude
$$\int_C (x^3 + x^5)dx + (y + y^7)dy = \int\int_D \dfrac{d(y + y^7)}{dx} - \dfrac{d(x^3 + x^5)}{dy} dxdy = 0.$$
But, by definition of evaluating a line integral, this is the same as
$$\int_C (x^3 + x^5)dx + (y + y^7)dy = \int \phi'(t)\cdot\phi'(t) dt,$$
where we have used that $\phi$ satisfies the differential equation to get the first $\phi'$. The integrand is nonnegative and since it integrates to $0$, it is equal to $0$ everywhere in the interval it is defined. That is, $\phi \equiv 0$.

Title: Existence of periodic solution in an ODE system

Body:
Let the ODE system,

\begin{cases}x'(t)=x^3+x^5 \\ y'(t)=y+y^7 \end{cases}

defined in $\mathbb{R}^2$.

I need to say if this system supports periodic solution with period $T>0$.

This is a question of Analysis in $\mathbb{R}^n$ not ODE, that is, I need to demonstrate this using concepts presented in Analysis in $\mathbb{R}^n$, my textbook is Analysis on Manifolds by Munkres. The hint is,

A periodic solution of the system above is a closed curve $\gamma(t) = (x(t), y (t))$ that satisfies the system.

I believe I have to use Stokes, but I am not able to proceed with the solution.


答: Hint: $x^3 + x^5 > 0$ when $x > 0$, $< 0$ when $x < 0$.  Similarly for $y + y^7$.


答:
Using Wolfram Alpha's solution, set $v(y) = y'$. Then $$y'' = \frac{dv}{dx} = \frac{dv}{dy}\frac{dy}{dx} = v\frac{dv}{dy}$$ so the equation becomes $$vv' = -yv^2$$

Which means either $v = 0$ or $$v' = -yv \to \frac{1}{v}dv=-ydy$$

Which has solution $\ln(v) = -\frac{1}{2}y^2+c \to $ $$v = Ce^{-y^2/2}$$

Then the equation is $$y' = Ce^{-y^2/2} \to e^{y^2/2}dy=Cdx$$

which has the solution of $$\sqrt{\frac{\pi}{2}}\text{erfi}\left( \frac{y}{\sqrt{2}} \right) = c_1x + c_2$$

Or equivalently, $$y = \sqrt{2}\text{erfi}^{-1}\left(c_1x + c_2\right) = -i\sqrt{2}\text{erf}^{-1}\left(c_1x + c_2\right)$$

Title: Solving $y'' = -y \cdot (y')^2$ analytical

Body:
I was trying to solve the following differential equation:

$$y'' = -y \cdot (y')^2$$

But I always have some trouble solving nonlinear ODE. I used wolfram alpha to check the solution but it used complex number and the $\text{erf}$ function and I'm not very used to work with it. Is that the only way of solving this analytical? If so, how Can I solve this equation?


问:
Finding vectors within Underdetermined Linear Systems
Consider the underdetermined linear system

```<code>x + 2y − z + 2w = 6

5x + 10y − 3z + 16w = 34

3x + 6y + 2z + 21w = 28
</code>```

Find three vectors l, m and n in R^4
such that the solutions to the system have the form λl+µm+n
where λ, µ ∈ R are free variables.

I have reduced the matrix down to

```<code>1 2 0 5 8

0 0 1 3 2

0 0 0 0 0
</code>```

and have determined that

x = 8-2y-5w

y=y

z = 2-3w

w=w

are the free variables here x and z? and where do I go from here to conclude?

答:
are the free variables here x and z? and where do I go from here to conclude?

Not quite, the free variables are $y$ and $w$ here, notice that once you choose $y$ and $w$ all of $x,y,w,z$ are determined. You are very close to concluding, just one small thing left to do...

If you don't mind, to make this a little clearer, (at least to myself), I will say $y = \alpha$ and $w = \beta$ for free variables $\alpha$ and $\beta$.

Then the solutions take the form
$$\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 8 - 2\alpha - 5\beta \\ \alpha \\ 2 - 3\beta \\ \beta \end{pmatrix} = \begin{pmatrix} 8 \\ 0 \\ 2 \\ 0 \end{pmatrix} + \alpha\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \beta\begin{pmatrix} -5 \\ 0 \\ -3 \\ 1 \end{pmatrix}$$


问:
Solving linear equations with coefficients under the natural logarithm function
A friend of mine sent me his homework in economics since I put myself forward for solving it. One of the exercises was formulated as follows:
$$  x = x(r_1,r_2) = 8 * r_1^\alpha *r_2^\beta$$
Knowing that: $ x(14400,32768) = 61440 $ and $ x(5184,7776) = 20736$ find the values of parameters $\alpha,\beta$.

My approach to the problem consists in applying the natural logarithm to both equations and then solving them as linear ones for the parameters. Since the numbers were cumbersome to work with i denoted them with capital letters so it came up to:
$$ln(E) = ln(8)+\alpha ln(A)+\beta ln(B)$$
$$ln(F) = ln(8)+\alpha ln(C)+\beta ln(D)$$
I tried to solve it by multiplying the first equation by  $log_BD$  but it turned out to be a real headache. After some calculations I came up with:
$$\alpha = \frac{ln(\frac{F}{8(\frac{E}{8})^{log_BD}})}{ln(\frac{C}{A^{log_BD}})}$$
The exercise gave the opportunity of checking the results that were $\alpha =0.5, \beta=0.4$, therefore invalidating my answer as a longshot. What am I doing wrong?

答: Never mind it turns out that the ratio is indeed rational and 0.5 . How can I delete the question?</p>


问:
Determine velocity & vector of particle

A particle is moving with constant velocity along a straight line
through space. It starts at $(8, 5, 10)$ when the time $t = 0$. It is
at $(2, 9, 5)$ when $t = 1$. The measurement units are meters, and the
time is in measured in seconds.

I need to determine the velocity vector of the particle and the velocity. How do I go about figuring this out? I found this topic which I think puts me on the right path but I'm not completely sure.

What do I do to find the solution here?

答:
Velocity in three dimensions is just the same as velocity in 1 dimention, just everything becomes a vector, $\vec{v}= \frac{d}{dt}\vec{s} =\frac{\vec{s_2} -\vec{s_1}}{\Delta t}$

That should be easy enough to solve. And if you are interested in the magnitude of velocity, well then same as any other vector: $|\vec{v}|=\sqrt{\vec{v} \cdot \vec{v}}$

Title: Determine velocity & vector of particle


问:
Solving the next differential equation
How can I solve the next differential equation?
$$\cos(x+y)dx=x\sin(x+y)dx+x\sin(x+y)dy$$
I dont know what kind of equation it is.
It's not homogeneous, separable differential equation or linear. Any suggestions?</p>

答:
use $u=x+y$, $y=u-x, dy =du-dx $ to get $\cos u dx =x \sin u dx +x \sin u (du-dx)$ which simplifies to $\cos u dx=x\sin u du$.

So $dx/x= \sin u du /\cos u$, or $\ln x= -\ln \cos u +C$, or $x= K/\cos u$, or $x\cos(x+y)=K$, or $x+y=\arccos (K/x)$, or $y=-x+\arccos(K/x)$.

Title: Solving the next differential equation


问:
Do holomorphic mappings always preserve/induce Kähler forms?
I was thinking in several related questions for which I haven't been able to find a proof nor a formal statement (Basically when do holomorphic mappings induce Kähler forms and how does Kählerness behave under holomorphic mappings).

So, let's suppose $X$ and $Y$ are complex manifolds. Now, I will denote by $\pi\colon X\to Y$ and $f\colon Y\to X $ some random holomorphic mappings. My questions are

If $(X, \omega_X)$ is Kähler, is it true that $(Y, f^*\omega_X)$ is always Kähler? (it should be true if $f$ is an immersion, but I want to know if that holds as well for other conditions, e.g., properness, submersions, etc.)

Does the existence of a holomorphic mapping $f\colon Y\to (X, \omega_X)$ impose some restrictions on the Kählerness of $Y$ or the fibres $Y_x:= f^{-1}(x)$? (in case the latter happen to be complex manifolds as well). For example, can we find another Kähler form on $Y$ other than $f^*\omega_X$  (whenever $f^*\omega_X$ is Kähler of course)? In general, does the existence or the properties of the mapping $f$ itself impose restrictions on the the Kählerness of $Y$ and/or the fibres $Y_x$?.

If $\pi\colon (X, \omega_X)\to Y$ is holomorphic, when can we say that $Y$ and/or the fibres $X_y:=\pi^{-1}(y)$ are Kähler as well? (and how do we construct their Kähler forms if it can be done?) [this one is particularly relavant when $\pi$ is a quotient map or a covering map]
Edit: I found this exercise (aptly labelled hard) in some online notes a few days ago. As for the weird counterxample mentioned below (a deformation of Kähler manifolds that is not a Kähler), one can trace it back to Hironaka (An Example of a Non-Kahlerian Complex-Analytic Deformation of Kahlerian Complex Structures)

A slightly related question: is it true that the pullback of positive forms is always a positive form, i.e., if $\omega$ is a positive $(p, p)$-form on $X$, does it hold as well for $f^*\omega$ on $Y$? Edit: This seems to be true at least for vector spaces, as you can see in page 129 of Demailly's  Complex analytic and differential geometry.

I will appreciate if you can give me some hints/proofs/counterexamples or point me out to some solid references with fully elaborated proofs/examples, not just some yes/no kind of answers.
Regards!

答:
$1.$ This is not true. For example, if $\dim Y > \dim X$, then $f^*\omega_X$ is degenerate (in directions in the kernel of $f_*$). Even if $\dim Y = \dim X$, this issue can still occur. For example, let $Y$ be the blowup of $X$ at a point $x$ (as Ted Shifrin alluded to in the comments). If $\dim X \geq 2$, then $(f^*\omega_X)_y$ is degenerate for all $y$ with $f(y) = x$. To see this, note that for a tangent vector $v \in T_yY$ along the exceptional divisor, we have $f_*(v) = 0$, so $(f^*\omega_X)_y(v, w) = (\omega_X)_x(f_*(v), f_*(w)) = (\omega_X)_x(0, f_*(w)) = 0$ for all $w \in T_yY$.

In general, a two-form $\alpha$ is the Kähler form of a Kähler metric if and only if it is a real positive $(1, 1)$-form, see this question. As $f$ is holomorphic, $f^*\omega_X$ is a real $(1, 1)$-form, so it is the Kähler form of a Kähler metric if and only if it is positive. In the examples above, $f^*\omega_X$ is degenerate and hence not positive. If $f : Y \to X$ is a holomorphic immersion, then $f^*\omega_X$ is positive and therefore gives rise to a Kähler metric on $Y$.

$2.$ The existence of a holomorphic map $f : Y \to X$ with $X$ Kähler has no bearing on the existence of Kähler metrics on $Y$ or the fibers of $f$. One reason for this is that we always have constant maps $X \to Y$. Even if $f : Y \to X$ is non-constant, it could be the case that neither $Y$ nor the fibers of $f$ admit Kähler metrics. For example, let $Y = Y_0\times X$ where $Y_0$ is a Hopf surface and $X$ is a compact Kähler manifold, and consider the holomorphic map $f : Y \to X$ given by projection onto the second factor. The complex manifold $Y$ does not admit a Kähler metric, and neither do the fibers of the map as they are Hopf surfaces.

$3.$ The existence of a holomorphic map $\pi : X \to Y$ with $X$ Kähler does not imply that $Y$ is Kähler. For example, take $X$ and $Y$ as in point $2$ and let $\pi(x) = (y_0, x)$ where $y_0 \in Y_0$ is some chosen point. On the other hand, if $\pi^{-1}(y)$ is a complex submanifold of $X$ which is the case if $y$ is a regular value of $\pi$ for example, then the restriction of the Kähler metric on $X$ to $\pi^{-1}(y)$ is a Kähler metric. More generally, the restriction of a Kähler metric to a complex submanifold is a Kähler metric, see Proposition $3.1.10$ of Huybrechts' Complex Geometry: An Introduction for example.

If $\pi : X \to Y$ is a finite covering map and $X$ admits a Kähler metric, then so does $Y$. This is true because there is a finite covering $\rho : X' \to X$ such that the induced finite covering $\pi' : X' \to Y$ is a normal covering. Then we can average the Kähler metric $\rho^*\omega_X$ on $X'$ over the deck transformations of $\pi'$ to obtain a Kähler form which descends to $Y$.

$4.$ This is false as the counterexamples in point $1$ demonstrate (for the case of $\omega = \omega_X$ a positive $(1, 1)$-form).


问:
Question - Covariance between T=X+Y and U=X-Y
Two days ago a friend of mine questioned me about these simple statistics problem:

Q: Take X and Y as two independet random variables with $X \sim N(0,1)$ and $Y \sim N(0,2)$. Calculate de covariance between $T=X+Y$ and $U=X-Y$.

Well, one can clearly conclude:

$$Cov(T,U)= Cov(X+Y,X-Y)= Cov(X,X) - Cov(X,Y) + Cov(Y,X) - Cov(Y,Y)=Var(X)-Var(Y) $$

But, $Var(x)=1$ and $Var(Y)=2$, therefore:

$$Cov(T,U)=-1$$

Nevertheless, we know from the Central Limit Theorem that $T \sim N(0,3)$ and $U\sim N(0,3)$. So I wonder how two variables with same distribution have a negative covariance.

Thank you very much.

答:
There's norhing impossible about it. It just means that if you measure $T$ and $U$ seperately, each of them has a normal distribution, but when you comapre the results, when $T$ is positive, then $U$ is more likely to be negative than positive, and when $T$ is negative then $U$ is more likely to be positve than negative.

Consider a simpler case, where $T \sim N(0,1)$, $U=-T$. Then also $U \sim N(0,1)$ and
$$ Cov(T,U) = -Cov(T,T) = -Var(T) =-1$$
You also have the same normal distribution and negative corelation.

Or you can consider $ X,Y\sim N(0,1)$, with $Cov(X,Y)=0$ and $T=2X\sim N(0,4)$, $U=-X+\sqrt{3}Y \sim N(0,4)$. Then you also have
$$Cov(T,U) = -2Cov(X,X)+2\sqrt{3} Cov(X,Y) = -2 $$

Remember, having the same distribution doesn't mean that two variables are equal.


问:
Proving the interval $(0, 1)$ and $(1, 3)$ have the same cardinality.

Prop: Show that the intervals $(0, 1)$ and $(1, 3)$ have the same cardinality.

What I have tried: I showed that the intervals $[2, 4]$ and $[0, 5]$ have equal cardinality by creating a function $F(x) = \frac52x - 5$. I did this by getting the length of both intervals, however, I don't understand why this works. In my notes, it shows that the length of $[2, 4]$ is $2$ but if we are including both endpoints shouldn't it be $3$? Same goes for the interval $[0, 5]$ we should have a length of $6$, not $5$, but when using $5/2$ and shifting everything by $-5$ it seems to work just fine.

Trying to solve this current prop:
Because it is an open interval we cant use $0$ and $1$ from what I understand so I don't quite understand what to use as the length of this interval.

I have tried reading other questions similar to this one however I am still stumped.

Thanks for the support!

答:
It is a basic fact that $(0,1)$ has the same cardinality with any interval $(a,b)$, for any $-\infty<a<b<\infty$.

The reason is that we can build a bijection from $(0,1)$ to $(a,b)$. Why does the bijection matter? Because it is our definition: if there exists a mapping $\varphi:A\rightarrow B$, $\varphi$ is a bijection (i.e.,$\varphi$ is 1-1 and onto), then we say $A$ and $B$ have the same cardinality. Notation: $|A|=|B|$.

As long as we find such a bijection, then our proof is finished. We do find a bijection, that is $\varphi(x)=a+(b-a)x$.

Now we check $\varphi$ is a bijection.
First, suppose $\varphi(x_1)=\varphi(x_2)$, then we have
$$a+(b-a)x_1=a+(b-a)x_2$$ that is
$$(b-a)x_1=(b-a)x_2$$
thus $$x_1=x_2$$
So $\varphi$ is one-to-one.

Second, to say $\varphi$ is onto, $\forall y\in(a,b), a<y<b$, there must exist $x\in(0,1),s.t. \varphi(x)=y$. That is,
$$a+(b-a)x=y$$
Let $x=\frac{y-a}{b-a}$. You can verify that $x\in(0,1)$ and $\varphi(x)=y$. So $\varphi$ is onto.

Together, we prove that $\varphi$ is bijection. Then according to definition, our proof is finished.


问:
Definition of a primitive polynomial
I understand there are already some questions (A, B) on primitive polynomials.  But none of these clears my confusion.

In page 84 of Handbook of Applied Cryptography, primitive polynomial has been defined as,

Now, if I try to understand the definition by dissecting the parts,

This is an irreducible polynomial, that means, it cannot be factored into the product of two or more non-trivial polynomials.

The polynomial $f(x) \in \mathbb{Z}_p[x]$. So, the polynomial belongs to the  polynomial ring $\mathbb{Z}_p [x]$, where, $\mathbb{Z}_p [x]$  is the ring formed by the set of all polynomials in the indeterminate $x$ having coefficients from $\mathbb{Z}_p$. Here $\mathbb{Z}_p$, will be the integers modulo $p$,  set of (equivalence classes of) integers
$\{0, 1, 2, . . . , p − 1\}$.

$x$ is a generator of $\mathbb{F}^*_{p^m}$: I am coming to this part regarding $x$ later on. $\mathbb{F}^*_{p^m}$,  is the multiplicative group of $\mathbb{F}_{p^m}$ such that $ \{a \in \mathbb{F}_{p^m} | \gcd(a, p) = 1\}$.

$\mathbb{F}_{p^m} = \mathbb{Z}_p[x]/(f(x))$, denotes the set of (equivalence classes of) polynomials in $\mathbb{Z}_p[x]$
of degree less than $n = \deg f (x)$. Addition and multiplication are performed modulo $f (x)$.

Now, coming back to the point of $x$, I began to realize that I must have some
serious flaw in my understanding above. So far, I have seen that generators have always been numbers.
Here the generator is $x$, which is an indeterminate.

Could you please point out the where I have gone off the track?

Perhaps the best way to salvage me will be to simply rewrite my points 1-4. Adding an example will make things perfect.

答:
What is meant is that the equivalence class of $x$ in $F = \Bbb{Z}_{p}(x)/(f(x))$ is a generator of the multiplicative group $F^{\star}$.

For instance if $p = 2$ and $f(x) = x^{2} + x + 1$, we have in $F = \Bbb{Z}_{2}(x)/(x^2+x+1)$
$$
x^{0} = 1, x^{1} = x, x^{2} \equiv x + 1, x^{3} = x^{2} x \equiv (x+1) x = x^2 + x \equiv 1,
$$
where congruences are modulo $x^{2} + x + 1$. So (the class of) $x$ is a generator of $F^{\star}$.

Title: Definition of a primitive polynomial


问:
How to prove that a function produces only integer results
Given the sequence $a_n = 2^n-1,$ prove that $$\frac{a_na_{n-1}...a_{n-k+1}}{a_ka_{k-1}...a_1},$$is always an integer for $0\leq k \leq n$.

In other words, show that the product of $k$ consecutive terms of the sequence is an integer multiple of the product of the first $k$ consecutive terms.

What if the sequence is $a_n = n(n+1)$?

答: This is a special form of the $q$-binomial coefficient (special, because $q=2$ in the question), and these are integer, as described in gory detail here: https://mathworld.wolfram.com/q-BinomialCoefficient.html


问:
If $(1,1)$ is an eigenvector of $A=\begin{pmatrix}2 &5\\3&k\end{pmatrix}$,then one of the eigenvalues of $A$ is :-
If $(1,1)$ is an eigenvector of $A=\begin{pmatrix}2 &5\\3&k\end{pmatrix}$,then one of the eigenvalues of $A$ is :-

$0,-1,1,2$

Can I get some hints please.

答:
$(1,1)$ is an eigen vector.let $\lambda$ be corresponding eigen value.then

$\begin{pmatrix}2 &5\\3&k\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\lambda\begin{pmatrix}1\\1\end{pmatrix}$ solving you can get the values of $k$ and $\lambda$ and then it is easy.


问:
Given that $z$ is a complex number, find the least integer greater than $z^{2000} + (\frac{1}{z})^{2000}$ if $z + \frac1z = 2\cos3$.
Problem: Given that $z$ is a complex number, find the least integer greater than $z^{2000} + (\frac{1}{z})^{2000}$ if $z + \frac1z = 2\cos3$.

I found $z$ but then just found the answer to be infinity. Can someone help?

Body:
$z + \frac 1z = 2\cos 3$  multiply both sides by $z$

$z^2 + 1 = 2\cos 3z$

$z^2 - 2\cos 3z + 1=0$

$x = \frac {2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}2=\cos 3\pm \sqrt{\cos^2 3-1}$

$\cos 3\pm i\sqrt{1-\cos^2 3}=$

$\cos 3\pm i\sin 3=$

$e^{\pm 3i}$

So $\frac 1z = e^{\mp 3i}$

And $z + (\frac 1z)^{2000}=$

$e^{\pm 3i} + e^{\mp 6000i} =$

$\cos 3 \pm i\sin 3 + \cos 6000 \mp i\sin 6000=$

$(\cos 3+\cos 6000) \pm (\sin 3 - \sin 6000) i$.


问:
Any Gaussian prime dividing a product of gaussian primes is equal to one of those gaussian primes
I was asked to show that if any Gaussian prime $p$ divides the product $abc$ of Gaussian primes, then $p$ is equal to one of those primes, or one of them multiplied by a unit.

I know that a Gaussian prime are primes with a norm that is a real prime of the form $4n+3$. It seems to me that the norm of an integer $a+bi$ is $a^2+b^2$.  How can I use this information to help me solve the given problem? Will finding the norms help?

答:
DISCLAIMER I am unsure what have been told you can assume about Gaussian integers (for example that they form a Euclidean Ring) but this is a reasonably basic proof.

A Gaussian prime is one of $\pm q,\pm qi\text { or } u+vi$ where $q$ is an integer prime of the form $4k+3$ and $u^2+v^2$ is either $2$ or an integer prime $r$ of the form $4k+1$. Therefore the norm of a Gaussian prime is of the form either $2,q^2$ or $r$.

Now suppose the Gaussian prime $p$ is a factor if the product $abc$ of three Gaussian primes. The $N(p)$ is a factor of $N(abc)=N(a)N(b)N(c)$.

If $p$ is one of $\pm q,\pm qi$ then one of $a,b,c$ must have norm $q^2$ and therefore be  of the form $\pm q,\pm qi$. Then this prime is a multiple of $p$ by an element of $\{\pm1,\pm i\}$ and all these elements are units.

If $p=u+vi$ where  $u^2+v^2$ is an integer prime $r$ of the form $4k+1$ (and similarly if $u^2+v^2=2$), then one of $a,b,c$ is $x+yi$ where $x^2+y^2=r$. Then $$(u-vi)p=r=(x+yi)(x-yi)$$ and so the prime $p$ is a divisor of either $x+yi$ or $x-yi$. Without loss of generality let it be a divisor of $x+yi$. Then $\frac {x+yi}{p}=l+mi$ is a Gaussian integer with norm $1$.

Therefore $l^2+m^2=1$ and so $l+mi\in \{\pm1,\pm i\}$ is a unit.


问:
Succinct Proof: All Pentagons Are Star Shaped

Question: What is a succinct proof that all pentagons are star shaped?

In case the term star shaped (or star convex) is unfamiliar or forgotten:

Definition Reminder: A subset $X$ of $\mathbb{R}^n$ is star shaped if there exists an $x \in X$ such that the line segment from $x$ to any point in $X$ is contained in $X$.

This topic has arisen in the past in my class discussions around interior angle sums; specifically, for star shaped polygons in $\mathbb{R}^2$ we can find their sum of interior angles as follows:

By assumption, there is an interior point $x$ that can be connected to each of the $n$ vertices. Drawing in these line segments, we construct $n$ triangles; summing across all of their interior angles gives a total of $180n^\circ$, but this over-counts the angle sum for the polygon by the $360^\circ$ around $x$. Therefore, the sum of interior angles is $(180n - 360)^\circ = 180(n-2)^\circ$.

This formula for the sum of interior angles holds more generally (often shown by triangulating polygons) but the proof strategy above already fails for some (obviously concave) hexagons.

For example:

The above depicted polygon is not star shaped.

Moreover, it is a fact that any polygon with five or fewer sides is star shaped. And so I re-paste:

Question: What is a succinct proof that all pentagons are star shaped?

Edit: Since it has come up as a counterexample (of sorts) for each of the first two responses, here is an example of a concave pentagon that may be worth examining in thinking through a proof.

答:
This is several years late, but I think I have a nicer proof. I thought I would post it.

Let's believe we can triangulate the 5-gon into three triangles. Note that these three triangles all must have a vertex $v$ in common: To see this, remove one of the triangles $T$, s.t. we have a 4-gon left. Then whichever way we triangulate the 4-gon every edge will contain a vertex from both of its triangles. In particular, the joint edge of $T$ and the 4-gon will have such a vertex $v$.

Then $v$ sees all three triangles and hence the entire 5-gon.


问:
Prove that the tangent space of a hyperplane is itself
I know this might sound really stupid: I was trying to show that the tangent space of a hyperplane is itself.

I started by parametrising the hyperplane locally at $x$ with a diffeomorphism $\phi : U \rightarrow X$, where $U\subset R^n$ and $X\subset R^m$ is the hyperplane we're considering, such that $\phi(0)=x$.

Now the tangent space $T_x(X)$ is given by the image of the derivative map $d\phi_0: R^n \rightarrow R^m$. If we can prove that $\phi$ is a linear map, then we are done, because the derivative map of a linear map is itself. But it seems that I can't prove that $\phi$ is linear.

Also I realised there is a loophole in my proof. Since we're proving a global result, not just at $x$, parametrising $X$ locally won't work...

Any ideas?

答: You can choose $\phi$ to be a global and linear map. Since the hyperplane is a linear subspace of $\mathbb{R}^m$, you may pick a basis for your hyperplane. That defines a linear isomorphism from $\mathbb{R}^n$ to the hyperplane taking the standard basis of $\mathbb{R}^n$ to your chosen basis, where $n$ is the dimension of the hyperplane. Since a linear isomorphism is automatically a diffeomorphism, $\phi$ indeed serves as a chart.</p>


问:
Deriving origin centred circle's equation using symmetry
I want to derive that $x^2 +y^2 -r^2 = 0$ must be the implicit equation for the shape of a circle by usage of symmetries. My idea is inspired from how the roots of a polynomial completely determine the polynomial upto scaling, so to begin I consider the implicit equation of circle as : $$C(x,y)=0$$

Now, suppose we take a point $(x,y)$ on the circle then this means that new points generated by rotating original point by any angle $\theta$ would solve the implicit equation

$$ \begin{bmatrix} x \cr y \end{bmatrix} \xrightarrow[]{\text{rotation by  } \theta }\begin{bmatrix} x \cos \theta -y \sin \theta \cr y \cos \theta + x \sin \theta\end{bmatrix}$$

Hence, $$ C(x,y) =0\implies C(x \cos \theta - y \sin \theta, y \cos \theta + x \sin \theta)=0$$

Now.. I'm not sure how to get the functional representation...

An alternate approach:

I came up with another way to do with actually works but it first involves assuming that it is some general conic:

$$C(x,y) = Ax^2 + Bxy +Cy^2 + Dy +Gx + H=0$$

Now, I considered transformations which still keeps me on the circle, for example reflection of a point $(x,y)$ about the line $y=x$ is a solution , hence $(y,x)$ is a solution to curve:

$$C(y,x) = Ay^2 + Byx + Cx^2 + Dx + Gy + H = 0$$

Similarly we can observe that if $C(x,y)$ is a solution to the equation, then $C(-x,-y)$ is also a solution (from the figure):

$$C(-x,-y) = Ax^2 +Bxy+Cy^2 -Dy -Gx + H=0$$

And also, we can see that reflecting our point about $x$ and $y$ axis keeps on circle so : $$ C(-x,y)=Ax^2 -Bxy+Cy^2+Dy-Gx+H=0$$ $$ C(x,-y)=Ax^2 -Bxy + Cy^2 - Dy +Gx+H=0$$

Now, composing any of that with a reflection along $y=x$ line ( I choose $C(x,-y)$:

$$C(-y,x) = Ay^2 - Byx + Cx^2 + Dx - Gy + H = 0$$

And now we have six equations and six unknowns, writing in matrix form:

$$ \begin{bmatrix} A&B&C&D&G&H \cr C&B&A&G&D&H \cr A&B&C&-D&-G&H \cr  A&-B&C&D&-G&H \cr A&-B&C&-D&G&H \cr  C&-B&A&-G&D&H\cr\end{bmatrix} \begin{bmatrix} x^2 \cr xy \cr y^2 \cr y \cr x \cr 1 \end{bmatrix}= 0$$

Solving this system we will get the conditions for second degree curve to be a circle.

My question: I am looking if it was possible to derive the equation in first way directly without having such an algebra mess.

答: You cannot get an explicit form for $C$ using your first argument, since any relation of the form $f(x^2+y^2)=c$ Also defines a circle for nice $f$. You are requiring that a certain level set of $C$ be rotationally invariant but this is satisfied by any radial function and by non radial ones having a circular $0$-level set


问:
Steps to find $\lim_\limits{n\to +\infty} \frac{n^n+3n!}{n^n+(3n)!} = 0$
I don't understand why $\lim_\limits{n\to +\infty} \frac{n^n+3n!}{n^n+(3n)!} = 0$

It's taken from Basisboek wiskunde (8.22d) from Jan van de Craats.

The solution given is:

$0$ (because $(3n)! = n! \times (n+1)(n+2)...3n>n! \times n^n$, etc.)

Actually, I don't understand this answer at all, I'd need some more steps to get it, could someone help me to make it clearer ?

答:
We have

$(3n)! = n! \times (n+1)(n+2)...(3n) > n! \times \underbrace{(n)(n)...(n)}_{\text{2n times}} = n! \times n^{2n} \geq n! \times n^n$.

The key step above is replacing $n+1$ , $n+2$ , $n+3$ etc with $n$ each time.

Thus we have:

$$\frac{n^n+3n!}{n^n+(3n)!} < \frac{n^n+3n!}{n^n + (n! \times n^n)}$$

Can you finish it up from here?


问:
For what values of n, will $1 + 2^n + 3^n + 4^n$ be divisible by $5$ where $0 \leq n \leq 100$
There's this problem that I am trying to find the solution for:

For what values of n, will $1 + 2^n + 3^n + 4^n$ be divisible by $5$ where $0 \leq n \leq 100$

So that doesn't hold for $n=0$, but then I am stuck as to what approach to take. It has to do with the last digit being $0$ or $5$, or not! I can't seem to factorize it either, $4^n = 2^{2n}$ then what? I want to know some hints and approaches to take, the final answer doesn't matter to me.

Body:
$1 + 2^n + 3^n + 4^n \equiv 1 + 2^n + (-2)^n + (-1)^n\pmod 5$.

If $n$ is odd then $1+2^n +(-2)^n + (-1)^n \equiv 0 \pmod 5$ and $5$ will divide $1+2^n + 3^n + 4^n$ if $n$ is odd.

If $n=2k$ is even then $1+ 2^n + (-2)^n + (-1)^n\equiv 2\cdot 1 +2\cdot 2^{n}\equiv 1 + 2\cdot 4^{k}\equiv 2 + 2(-1)^k \equiv \begin{cases} 4&k\ even\\0&k\ odd\end{cases}$. So $5  \mid 1 + 2^n + 3^n + 4^n$ if $n=2k$ and $k$ is odd.

So the solution is $5$ divides $n$ if $n$ is odd or if $n$ is $2$ times an odd number.

.......

Alternatively if you don't know modular arithmetic.

let $n = 4k + i$ where $i= 0,1,2,3$

Then $1^n + 2^n + 3^n + 4^n =$

$1 + 2^{4k+i} + 3^{4k+i} + 4^{4k +i} =$

$1 + 16^k*2^i + 81^k*3^i + 256^k*4^i$.

Now $16 = 15 + 1$ so $16^k = (15 + 1)^k = 5M + 1$ for some $M$.

$81^k = (80+1)^k = 5N +1$ for some $N$

$256^k = 5W + 1$ for some $W$ so

$1 + 2^n + 3^n + 4^n = 1 + (5M+1)2^i + (5N+1)3^i + (5W+1)4^i =$

$5(2^iM + 3^iN + 4^iW) + (1 + 2^i + 3^i + 4^i)$.

That is divisible by $5$ if and only if $1+2^i +3^i +4^i$ is.

$1 + 2^0 + 3^0 + 4^0 = 4$ is not.

$1 + 2^1 + 3^1 +4^1 = 10$ is

$1 + 2^2 + 3^2 + 4^2 = 1+4 + 9 + 16= 30$ is.

$1 + 2^3 + 3^3 +4^3 = 1 + 8 + 27+ 64 = 100$ is.

So if $n$ is not divisible by $4$ then $1 + 2^n +3^n + 4^n$ is divisible by $5$.

(Which is the exact same thing as $n$ is odd ($n = 4k+1$ or $n = 4k + 3$) or $n$ is $2$ times an odd number ($n = 4k + 2$)).


问:
When rolling a die, are we equally as likely of getting 1 twice as getting exactly a single 1 and a single 2?
The die is unbiased and has $6$ sides. I want to know if the probability of getting $1$ twice (which is $(\frac{1}{6})^2$) is the same as getting a $1$ and a $2$ (order does not matter).

答:
On the first roll, there is a $\frac{1}{6}$ chance that a one is rolled. On the second roll, there is  also a $\frac{1}{6}$ that a one is rolled. Multiplying the probabilities, there is a $\frac{1}{36}$ chance that we roll two ones.

Now let us consider the other possibility. On the first roll, there is a $\frac{1}{3}$ chance that either a one or a two is rolled. No matter what happens, there is a $\frac{1}{6}$ chance we roll the number that was not rolled before (e.g. if one was rolled first, we rolled a two next). Multiplying the probabilities, there is a $\frac{1}{18}$ chance that we roll two ones.

Therefore, it is more likely we roll a two and a one than two ones.


问:
Is that true If I say posterior = prior multiply likelihood?
I see a formula that we should also have marginal distribution as denominator. But some tutorial said posterior = prior multiply likelihood?</p>

答:
It is not equal to, but it is proportional to:

$$
\text{posterior} = \frac{\text{likelihood} \times \text{prior}}{\text{marginal}} \propto \text{likelihood} \times \text{prior}.
$$
The reason is that we are interesting in the parameter, say $\theta$, and this parameter only appears in the likelihood and prior, but not the marginal, which is purely a function of the data $x$. In other words, the marginal is just a normalisation quantity that allows the posterior to be a proper probability.


问:
Proving $f(x) := \sin(ax) - \sin(bx)$ has absolute value greater or equal to 1 for some $x>0$
I am trying to prove that for any choice of reals $a$ and $b$  such that
$a \neq b $
the function $f(x) := \sin(ax) - \sin(bx)$ has $|f(x)| \geq 1$ for some $x>0$

(This is not an exercise question but something I am trying to prove for myself. Plots with Wolfram Alpha seem to indicate it might be true, and so counterexamples are very much welcome)

Here is my attempt.
For this I wrote $f$ of 3 variables $a,b,x$. We find that value of $x_0$
for which $f$ attains a maximum and minimum and then show that $|f(x_0)|$
greater than 1.

$g(a,b,x)=f(x)$ and then zeroing the gradient. (pending the 2nd derivative test)

$f_a = x \cos(ax)$

$f_b = -x \cos(bx)$

$f_x = a\cos(ax) - b\cos(bx)$

Since the point of minima/maxima are conjectured $x>0$ , this means $\cos(ax)$ and $\cos(bx)$ are both zero
and the third equation is redundant.

If $a$ and $b$ are rational then obviously we can find such an $x$.
But when one of them is irrational, I don't know how I would prove this.

Another attempt was using the $\sin(u)-\sin(v)$ product formula, but that
led nowhere too.

Any pointers?

答:
If $a=b$, then $f(x)=0$. If $a=0$ or $b=0$ then we have just a single sine function and hence $|f(x)|\le 1$ for all $x$. Therefore these cases must be excluded.

If $a=-b$, we have $f(x)=2\sin (ax)$, hence $f(\frac{\pi}{2a})=2$ and we are done. We may therefore assume that
$$\tag1a,b,a+b,a-b\text{ are nonzero.}$$
Note that $$\tag2\sin (ax)-\sin (bx)=2\cos{\left(\frac{a+b}2x\right)}\sin\left(\frac{a-b}2x\right).$$
The cosine is $\pm1$ at $x\in \frac{2k\pi}{a+b}\mathbb Z$. Thus it is sufficient to find $k\in\mathbb Z$ with
$$\tag3 \left|\sin\frac{(a-b)k\pi}{a+b}\right|>\frac12.$$
For which $c\in\mathbb R$ can we conclude that there exists $k\in\mathbb Z$ with $|\sin( ck\pi)|>\frac12$? Since $|\sin(ck\pi)|=|\sin((c+1)k\pi)|$, we may restrict to the case $c\in[0,1)$.
And since $|\sin((1-c)k\pi)|=|\sin(ck\pi)|$, we can restrict further to $c\in[0,\frac12]$. If $\frac16<c\le \frac12$, we succeed by choosing $k=1$.
If $0<c\le\frac16$, let $k=\lceil\frac1{6c}\rceil$, i.e. we have $\frac16<kc\le\frac16+c\le \frac13$ and thus obtain $\sin(ck\pi)>\frac12$.
Allowing arbitrary $c\in\mathbb R$ again, we therefore conclude that there exists $k\in\mathbb Z$ with $|\sin(ck\pi)|>\frac12$ unless $c$ is not an integer.
Back to our original problem, we can thus show the existence of $x$ with $|f(x)|>1$ unless $c:=\frac{a-b}{a+b}$ is an integer.
Note that $c\notin\{-1,0,1\}$ because of $(1)$.
Therefore $\left|\frac{\pi}{2c}\right|\le\frac\pi 4 3$ and
$$f\left(\frac{\pi}{a-b}\right)=2\cos \frac{\pi}{2c}\sin\frac\pi2\ge2\cos\frac\pi4=\sqrt2>1. $$


问:
What is the advantage of Runge Kutta (or any other) method over just solving the o.d.e?
An example in task had a model with "runge kutta RK4 approximation applied" so I naturally went to Google what Runge Kutta method is. I found how to use it and why it's better over euler method but I didn't find why/what it is used for.

From what I read, Runge-Kutta is used to approximate ordinary differential equation solutions.
And here's my question: Why use Runge-Kutta, if we can just solve the differential equation using integral and get infinitely more precise solution?
Question is not limited to Runge Kutta, it is just the one I'm currently working with.

Perhaps my understanding of RK or o.d.e is totally wrong.

答:
Most real world applications of differential equations are not easily analytically solvable. When I took numerical methods an example problem we had was simulating a giant asteroid called Apophis to determine the likelihood of a collision with Earth using other planets and the moon as well. This ends up being a system of differential equations. It isn't super hard but it's also not something most people want to do.

Real world problems are much more complex than that. Applications in engineering typically have you simulating different forces on buildings or products by generating complex meshes to evaluate them. The reality is you could never do any of that by hand. If you watched the movie "Hidden Figures" she ends up using a simple Runge-Kutte method. More complex Runge-Kutte methods that you use in practice are more complex and require more evaluations.

Runge-Kutte is also a relatively simple way that doesn't apply to all differential equations. There are much more complex and time consuming methods.


问:
What are the prime and maximal ideals of the cartesian product of commutative rings?

Let $R$ and $S$ be two commutative rings. Describe the prime and maximal ideals of $R\times S$ in terms of the prime and maximal ideals of $R$ and $S$.

I am asked to solve this question. Under the influence of this question and the hint given by Michael Burr I'm now trying to prove that all maximal and prime ideals are of the form $P \times S$ or $R\times P$ or $M\times S$ or $R\times M$ for $P,M$ a prime and maximal ideal respectively. I was wondering if my ideas are correct, or if there were any holes in my proof ? ._.

I am able to deduce that all ideals of $R\times S$ must be of the form $I\times J$ for $I,J$ ideals of $R$ and $S$ respectively, from here. (If $I$ is an ideal of $R$ and $J$ is an ideal of $S$ then $I\times J$ is an ideal of $R\times S$, and if $I$ is an ideal of $R\times S$ then there are ideals of $R$ and $S$ such that $I=I_R \times I_S$).

From Corollary 14 of chapter 7 of Dummit and Foote:

Assume $R$ is commutative. Every maximal ideal of $R$ is a prime ideal.

As mentioned in the comments if $P_1,P_2$ are proper prime ideals, then $P_1 \times P_2$ cannot be a prime ideal because given $r \not\in P_1$ and $s \not \in P_2$ then $(r,0)\times(0,s) = (0,0) \in P_1\times P_2$ but $(r,0),(0,s) \not\in P_1 \times P_2$. Since maximal ideals are prime ideals, given $M_1, M_2$ maximal/prime proper ideals, then $M_1 \times M_2$ is not prime, and thus not maximal. A two for one...

If the above is correct, both ideals cannot be proper, thus only one ideal must be the whole ring $R$ or $S$, $R\times S$ would not be proper and prime and maximal ideals are proper ideals. (All prime and maximal ideals must be of the form $P \times S$ or $R\times P$ or $M\times S$ or $R\times M$).

Following through to the next step in the hint I try taking quotients (to show that they are actually maximal or prime ideals):

$P \times S$ or $R\times P$ must be prime ideals for $P$ a prime ideal:

${R\times S}/{R\times P} = \{(r,s)+R\times P | r\in R \text{ and } s\in S\}$

If $s \in P$ then $(r,s) + R\times P = 0 + R\times P$

suppose $(r,s),(r',s')\not\in R\times P$ and $[(r,s)+R\times P]\times [(r',s')+ R\times P] = (r\cdot r', s\cdot s') + R\times P = 0 + R\times P$

then $s,s' \not\in P$ and $s\cdot s' \in P$, but $P$ is prime, contradiction. ${R\times S}/{R\times P}$ has no zero divisors. I guess that makes ${R\times S}/{R\times P}$ into an integral domain if it is assumed that $R,S$ have identity $1\not= 0$... thus $R\times P$ is a prime ideal, similarly $P\times S$ is prime.

I'm not so sure about a two for one in this case,

Is there a way to show that $R\times S / M\times S$ or $R\times S / R\times M$ is a field?

I think the latter would be maximal by a direct proof?

Suppose there is an ideal $M\times S \subset I \subset R$ then $I$ must be of the form $J\times S$ for some ideal $J$ of $R$ where $M \subset J \subset R$. Contradiction because $M$ is maximal. Thus $M\times S$ is maximal in $R\times S$ and similarly $R\times M$ is maximal in $R\times S$. Does this work?

答:
In general, if $I \subset R$ and $J \subset S$ are left ideals of rings $R$ and $S$, then the product of the quotient maps

$$
p_I \times p_J \colon R \times S \to R/I \times S/J
$$

has in its kernel the precisely the elements $(x,y)$ such that $x \in I, y \in J$. By the first isomorphism theorem, we get

$$
\frac{R \times S}{I \times J} \simeq R/I \times S/J.\tag{1}
$$

Now, a product $A \times B$ of two non-zero rings $A$ and $B$ aways has zero divisors: taking $a \in A$ and $b \in B$ both non-zero, we get $(0,0) = (a,0)(0,b)$.

Hence if $I \times J$ is prime (resp. maximal), one of the factors in the right hand side of $(1)$ must be trivial, and the other must be a domain (resp. field). This translates to saying that one of the ideals must be the whole ring and the other prime (resp. maximal).


问:
Cardinality != Density?
I was in a discussion where I argued that the density of two sets of the same cardinality could be different in respect to the infinite range of non-negative integers. Does cardinality imply that any set of ${\aleph_0}$ has equal density to any other set of the same ${\aleph_0}$?

Also, does cardinality imply equivalence in set size? Or is a formal ranking of different tiers which cannot be compared?

Ok, got that so far. Expanding question now:

Densities can be different but the count is the same. Infinite == Infinite iff cardinality is equivalent. Why doesn't a different density imply infinite set != infinite set even if cardinality is equivalent?

答:
Let $\mathbb{N}$ refer to the positive integers.  Usually we define the density of a subset $A\subset \mathbb{N}$ with respect to the integers to be $$\lim_{N\rightarrow\infty}\frac{|\{m\in A:m\leq N\}|}{N}.$$  For example, the set of even numbers $$\{2,4,6,\dots\}$$ has density $\frac{1}{2}$.  From this, it is not hard to see that for any $c\in[0,1]$ you can find a set with density $c$.

Notice that if $c>0$ this automatically implies the set has cardinality $\aleph_0$, but it is also possible to have a set of density $0$ with cardinality $\aleph_0$.  For example, the set of powers of $2$, or the set of prime numbers.

Hope that helps,

Title: Cardinality != Density?


问:
If $\dim({\rm Im}(T))$ and $\dim({\rm Ker}(T))$ are finite $\Longrightarrow$ $\dim(V)$ is finite

Let be $F$ a field, $V$ and $W$ are vector spaces over $F$. Prove that
if it exist a linear tranformation $T:V\rightarrow W$ with
$\dim({\rm Im}(T))$ and $\dim({\rm Ker}(T))$ both finite, then $\dim(V)$ is finite.

What we know is that:

There exists a finite set of vectors $\left \{ v_1,...,v_n \right \}$ such that generates ${\rm Ker}(T)$

There exists a finite set of vectors $\left \{ w_1,...,w_m \right \}$ such that for each one exists $u_i \in V$, such that $T(w_i)=u_i$

Since there exists a finite quantity of $u_i \in V $ such that $T(w_i)=u_i$, and $\dim({\rm Ker}(T))<\infty$, then $\dim(V)$ must be finite. But I'm not sure about how can I conclude, can you help me please?

答:
Hint: proceed as in the usual proof that $\dim V = \dim \ker T + \dim \text{im} \ T$, that is, consider a basis $B_0$ of $\ker T$ and extend it to a basis $B$ of $V$. Now, show that the image of $B \setminus B_0$ is a basis for $\text{im} \  T$.

 These remarks and the hypothesis tells us that both $B_0$ and $B\setminus B_0$ must be finite, hence the basis $B$ of $V$ is finite. This completes the proof.


问:
Factor a rank one operator on a Hilbert C*-module.
Let $A \subset \mathcal{L}(\mathcal{H}_0)$ be a concrete $C^*$-algebra and $X$ be a right Hilbert A-module. For each $x,y \in X$ we have rank one operators
\begin{align}
\theta_{x,y}: X & \to X \\
 z & \mapsto x\langle y, z\rangle_A
\end{align}
The set $\mathcal{K}_A(X):=\overline{\mathrm{span}\{\theta_{x,y} : x,y \in X\}}$ is well known to be a $C^*$ algebra.

Question: Can I always find a representation of $\mathcal{K}_A(X)$ on some Hilbert space $\mathcal{H}_1$ such that for any $x \in X$ the map $\theta_{x,x}$, regarded as an element of $ \mathcal{L}(\mathcal{H}_1)$, "factors trhough $\mathcal{H}_0$"? To be precise, can I always find an element $b_x \in \mathcal{L}(\mathcal{H}_1, \mathcal{H}_0)$ such that $\theta_{x,x}=b_x^* b_x$?

I can't seem to find this in any of my usual references for Hilbert modules. It seems to be true for all the examples I've tried. For instance, for a usual Hilbert space regarded as a right Hilbert $\mathbb{C}$-module, we have $b_x=\langle x|$ so that $b_x^*$ is multiplication by $x$.

However, I can't quite prove it in general nor find a counterexample. I don't mind assuming that $X$ is full, but I am not sure this is needed. Also, it might be that this only works for a carefully chosen $\mathcal{H}_1$, but this is fine as I care for a "concrete" result rather than a necessarily abstract one.

Any reference or help to prove/disprove the claim will be much appreciated.

答:
Given $\mathcal{H}_0$ define $\mathcal{H}_1 := X \otimes_{A} \mathcal{H}_0$, which is a Hilbert space carrying a representation of $\mathcal{K}(X)$ given by $T(x \otimes h) = (Tx) \otimes h$. Define $b_x$ to be the "creation" operator $b_x \in \mathcal{L}(\mathcal{H}_0,\mathcal{H}_1)$ satisfying $b_x(h) = x \otimes h$ and then $b_x^*$ is the corresponding "annihilation" operator  $b_x^*(y \otimes h)  = \pi((x \mid y)_A) h$, where $\pi$ is the representation of $A$ on $\mathcal{H}_0$. Then
$$
b_x b_x^* (y \otimes h) = x \otimes \pi((x \mid y)_A)h = x(x \mid y)_A \otimes h = \Theta_{x,x}(y) \otimes h.
$$
Of course there's some checking to be done to make sure everything is well-defined.

I should also mention that the representation of $\mathcal{K}(X)$ on $\mathcal{H}_1$ is faithful since $\pi$ is faithful.


问:
Circle/ Ellipse geometry problem from Ramanujan
Does someone have detail of the problem and his solution? I saw it in youtube. A rough hand sketch remade seeing YouTube video SR geometry problem  (17 min 51 sec). I have no access to the  TIFR book cited.

Thanks in advance.

The original image is here :

答:
Background

This can be found in Mathematics Magazine, Vol. 51, No. 3 (May, 1978), pg 161:

A similar sketch can be found in the Manuscript Book 1 of Srinivasa Ramanujan,  pg. 54:

The Problem

This can be found in the Journal of the Indian Mathematical Society, V, 1913, pg 24:

Let $PQR$ be a circle with center $O$, of which a diameter is $PR$.
Bisect $PO$ at $H$ and let $T$ be the point of trisection of $OR$
nearer $R$. Draw $TQ$ perpendicular to $PR$ and place the chord $RS =TQ$.

Join $PS$, and draw $OM$ and $TN$ parallel to $RS$. Place a chord $PK= PM$, and draw the tangent $PL = MN$. Join $RL, RK$ and $KL$. Cut off $RC = RH$. Draw $CD$ parallel to $KL$, meeting $RL$ at $D.$

Then the square on $RD$ will be equal to the circle $PQR$
approximately. For

$$RS^2=\frac{5}{36}d^2,$$

where $d$ is the diameter of the circle. Therefore

$$PS^2=\frac{31}{36}d^2.$$

But $PL$ and $PK$ are equal to $MN$ and $PN$ respectively. Therefore

$$PK^2=\frac{31}{144}d^2,\quad \text{and} \quad PL^2=\frac{31}{324}d^2.$$

Squaring the circle

Hence

$$RK^2=PR^2-PK^2=\frac{113}{114}d^2,$$

and

$$RL^2=PR^2+PL^2=\frac{355}{324}d^2.$$

But

$$\frac{RK}{RL}=\frac{RC}{RD}=\frac{3}{2}\sqrt{\frac{113}{355}},$$

and

$$RC=\frac{3}{4}d.$$

Therefore

$$RD=\frac{d}{2}\sqrt{\frac{355}{113}}=r\sqrt{\pi},\quad \text{very nearly}$$

Perhaps this will help you in your studies


问:
Need help solving this log equation $\log_{5}(x-4) = \log_{10}(x+5)$
I need help solving for $x$:

$$\log_{5}(x-4) = \log_{10}(x+5)$$

答:
Just to help close this question, let me demonstrate the difficulty in solving this exactly. "$\lg$" denotes the base $10$ logarithm.

By change of base, we can transform the LHS. The equation becomes:

$\frac{\lg(x-4)}{\lg 5} = \lg (x+5)$

$\lg(x-4) = \lg(x+5)^{\lg 5}$

Take inverse logs of both sides:

$x-4 = (x+5)^{\lg 5}$

And that's where you have to throw in the towel. That is a transcendental equation, no way to simplify that algebraically with elementary functions. Expressing $\lg 5 = 1 - \lg 2$ won't help.

The only time when you can solve your form of equation exactly is when you have the same expression in $x$ on either side. So, for instance, if you had $x+5$ on both sides, you could let $y=x+5$ and write:

$\lg y = \lg 5 \cdot \lg y$

Giving us the immediate solution by inspection: $\lg y = 0 \implies y = 1 \implies x = - 4$. No other solution exists.


答:
Try to convert $\log_5 K = \log_{2\cdot 5=10} M$ how does that work.

$\log_5 A = k$ means $5^k = A$ so $10^k = \frac {A}{2^k}$ so $\log_{10}\frac {A}{2^{\log_5 A}} = \log_5 A$

So $\log_{10}\frac {x-4}{2^{x-4}} = \log_{10} (x+5)$

So $\frac {x-4}{2^{x-4}} = x+5$.

$x-4 = (x+5)2^x\frac 1{16}$

$16\frac {x-4}{x+5} = 2^x$

$1- \frac 9{x+5} = 2^{x-4}$  and ... I quit.

You can google how to solve these with lambert functions but.... I quit.

Hmm... actually $1-2^{x-4} = \frac 9{x+5}$ gives us a good test to guess by trial and error..... but I still quit.

Title: Need help solving this log equation $\log_{5}(x-4) = \log_{10}(x+5)$

Body:
I need help solving for $x$:

$$\log_{5}(x-4) = \log_{10}(x+5)$$


答: By the first isomorphism theorem, $V/\ker T\cong\rm{im}\,T$.  Thus $\dim V=\dim(\rm{im}\,T)+\dim(\ker T)\lt\infty$.

Title: If $\dim({\rm Im}(T))$ and $\dim({\rm Ker}(T))$ are finite $\Longrightarrow$ $\dim(V)$ is finite

Body:
Let be $F$ a field, $V$ and $W$ are vector spaces over $F$. Prove that
if it exist a linear tranformation $T:V\rightarrow W$ with
$\dim({\rm Im}(T))$ and $\dim({\rm Ker}(T))$ both finite, then $\dim(V)$ is finite.

What we know is that:

There exists a finite set of vectors $\left \{ v_1,...,v_n \right \}$ such that generates ${\rm Ker}(T)$

There exists a finite set of vectors $\left \{ w_1,...,w_m \right \}$ such that for each one exists $u_i \in V$, such that $T(w_i)=u_i$

Since there exists a finite quantity of $u_i \in V $ such that $T(w_i)=u_i$, and $\dim({\rm Ker}(T))<\infty$, then $\dim(V)$ must be finite. But I'm not sure about how can I conclude, can you help me please?


问:
$\text{Inn}(G)$ cannot be nontrivial cyclic group

Let $G$ be any group, and let $Z$ be its center.

(a) Show that $G/Z\cong \text{Inn}(G)$.

(b) Conclude that $\text{Inn}(G)$ cannot be a nontrivial cyclic group.

I've already gotten part (a) by considering the mapping $\pi:G\rightarrow\text{Inn}(G)$ such that $\pi(g)$ is the automorphism that takes $x$ to $g^{-1}xg$ for all $x\in G$. The mapping $\pi$ is clearly a surjective homomorphism with kernel $Z$, and part (a) follows from the isomorphism theorem.

For part (b), I must prove that $G/Z$ cannot be a nontrivial cyclic group. If it were, the group would equal $\{Z,Zg,Zg^2,\ldots,Zg^{n-1}\}$ for some $g\in G$. Also, $G/Z$ would be an abelian group, and it follows that the commutator subgroup $G'$ belongs to $Z$. I don't see how to derive a contradiction from there.

Body:
Hints:

$$\forall x,y\in G\;\exists z_1,z_2\in Z(G)\;,\;n_1,n_2\in\Bbb N\;\;s.t.\;\; x=g^{n_1}z_1\;,\;y=g^{n_2}z_2\implies$$

$$xy=g^{n_1}z_1g^{n_2}z_2=g^{n_1}g^{n_2}z_1z_2=g^{n_2}g^{n_1}z_2z_1=\ldots$$

The above thus proves $\,G\,$ is abelian, but then $\,Z(G)=G\,$ , so$\;\ldots\;$


问:
Kind of equality on the sphere in Holder inequality
Lets suppose p > 1 and q as $\frac{1}{p} + \frac{1}{q} = 1$

$g$ is function of $L^q(\Omega, A, \mu) $

We suppose that exists $M$ positive number as : $M = sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p = 1 \right \}$

$\epsilon(A)$ is the set of staircase functions

The goal is to show that : $\left \| g \right \|_q = M $

With Holder inequality, we show easily that $\left \| g \right \|_q \geq  M $

In other sens, I'm searching for range of function $(g_n)$ in $\epsilon_+(A)$ to show that : $ \int \left | g \right |^q d\mu \leq M^q $ (then we can conclude easily)

Any idea ?

答:
Let $p > 1$ and $q$ as $\frac{1}{p} + \frac{1}{q} = 1$

Let $g$ be a function of $L^q(\Omega, A, \mu) $

Suppose that exists $M$ positive number as :
$$M = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p = 1 \right \}$$

$\epsilon(A)$ is the set of simple functions. It is clear that $M \geqslant 0$.

Lemma:
$$M = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p \leqslant 1 \right \}$$

Proof:
Let $M' = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p \leqslant 1 \right \}$.  It follows immediately that $M' \geqslant M$.

If $M' > M$ then there is $f_1 \in \epsilon(A) ;  \left \| f_1 \right \|_p \leqslant 1$ such that $\int f_1g\: d\mu > M \geqslant 0$. So $0< \|f_1\|_p \leqslant  1$. Let $f_2=\frac{f_1}{\|f_1\|_p}$. It is immediate that $\|f_2\|_p=1$ and
$$\int f_2g\: d\mu = \frac{1}{\|f_1\|_p}\int f_1g\: d\mu \geqslant \int f_1g\: d\mu > M $$
Contradiction, since $M = \sup\left \{ \int fg d\mu  ;  f \in \epsilon(A) ;  \left \| f \right \|_p = 1 \right \}$. So $M'= M$. $\square$

Now we want to prove that : $ \| g \|_q = M $

From Holder inequality, we show easily that $\| g \|_q \geqslant  M $.

So we need to prove that $\left \| g \right \|_q \leqslant  M$.

Now, if $\|g\|_q=0$, then $M=0$ and it is trivial that $\|g\|_q=M$. Let us suppose  $\|g\|_q>0$

Now, we have two cases: $q>1$ and $q=1$.

Case $q>1$.
Let
$$h = sgn(g) \frac{|g|^{q-1}}{\|g\|_q^{q/p} }$$
where $sgn(g) = \chi_{\{x: g(x)>0\}} - \chi_{\{x: g(x)<0\}} $.

Since $q=(q-1)p$, it follows that $h\in L^p(\Omega, A, \mu) $ and $\|h\|_p=1$.
Moreover, since $q-(q/p)=1$, we have
$$\int h g \:d\mu = \frac{1}{\|g\|_q^{q/p}} \int |g|^q d\mu =\|g\|_q^{q-(q/p)} = \|g\|_q$$

Now, let $\{f_n\}_n$ be a sequence of simple function converging such that $\{f_n^+\}_n$ converges monotonically from below to $h^+$  and $\{f_n^-\}_n$ converges monotonically from below to $h^-$.

It follows that $\|f_n\|_p \leqslant \|h\|_p =1$  and
$$\lim_n \int f_n g d\mu = \int h g \:d\mu = \|g\|_q$$
But, for all $n$, $\lim_n \int f_n g d\mu \leqslant M$. So $\|g\|_q \leqslant M$. So $\|g\|_q = M$.

Case $q=1$.
Let
$$h = sgn(g) $$
where $sgn(g) = \chi_{\{x: g(x)>0\}} - \chi_{\{x: g(x)<0\}} $.

It follows that $h\in L^\infty(\Omega, A, \mu) $ and $\|h\|_\infty=1$.
Moreover, we have
$$\int h g \:d\mu = \int |g| d\mu =\|g\|_1$$

Now, let $\{f_n\}_n$ be a sequence of simple function converging such that $\{f_n^+\}_n$ converges monotonically from below to $h^+$  and $\{f_n^-\}_n$ converges monotonically from below to $h^-$.

It follows that $\|f_n\|_1 \leqslant \|h\|_1 =1$  and
$$\lim_n \int f_n g d\mu = \int h g \:d\mu = \|g\|_1$$
But, for all $n$, $\lim_n \int f_n g d\mu \leqslant M$. So $\|g\|_1 \leqslant M$. So $\|g\|_1 = M$.


问:
Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$

$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?

If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.

答:
Well, this is definitely a brute-force method but here's what I can think of:

Draw your triangle with P inside, along with the lines PA, PB, PC. There are 9 angles in total. There are also 4 unknown lengths, $x,y,z$, and the side-length of the equilateral triangle, $L$. In total, this is 13 unknowns.

Let's just call the six outer angles d,e,f,g,h,i and the 3 inner angles (around P) j,k,l.

It's not hard to come up with 13 equations: you know the angles making up the corners of the equilateral triangle add up to 60, i.e. $d+e = 60, f+g = 60, h+i = 60$, and the three on the inside add up to 360, i.e. $j+k+l = 360$. Also, the three "subtriangles" each total angle $180$. So that's 7 equations.

You can also use the sin and cosine rules of triangles incorporating the information about side lengths. This is an additional 3+3 = 6 equations.

In the end, there are 13 unknowns and 13 equations (which do not depend on one another). These can be solved by substitution/elimination to solve for L.


问:
When can I say $\int\limits_{-T}^{T} \sum_n\sum_m f(m.n.t)dt=\sum_n\sum_m \int ^T_{-T}f(m,n,t)dt$?
When can I say $\int\limits_{-T}^{T} \sum_n\sum_m f(m.n.t)dt=\sum_n\sum_m \int ^T_{-T}f(m,n,t)dt$?

All I know is that for each $t\in[-T,T]$, $\sum_n\sum_m f(m,n,t)$ is absolutely convergent, which I think is farily strong condition. But now I can't find any easy argument. Could you give me any tips?

答:
I like to remember this as a special case of the Fubini/Tonelli theorems, where the measures are counting measure on $\mathbb{N}$ and Lebesgue measure on $\mathbb{R}$ (or $[0,\infty)$ as you've written it here).  In particular, Tonelli's theorem says if $f_n(x) \ge 0$ for all $n,x$, then $$\sum \int f_n(x) \,dx = \int \sum f_n(x) \,dx$$ without any further conditions needed.  (You can also prove this with the monotone convergence theorem.)

Then Fubini's theorem says that for general $f_n$, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$ (by Tonelli the two conditions are equivalent), then $\int \sum f_n = \sum \int f_n$.  (You can also prove this with the dominated convergence theorem.)

Source: When can a sum and integral be interchanged?


问:
Is there a topology $T$ on the set of complex numbers such that the class of $T$-continuous functions and the class of analytic functions coincide?
Is there a topology $T$ on the set of complex numbers such that the class of $T$-continuous functions and the class of analytic functions coincide?

答:
For infinitely differentiable functions on $\mathbb{R}$, there is no such topology $T$:

If there is, as rohit has noted, due to the bicontinuity of $ax+b$, translations and dilations of open sets will be open. If $U\in T$ such $U$ is bounded, then we can construct the usual topology on $\mathbb{R}$ from $U$ (see rohit's comment)

Now, let $V \in T$, such that $V\neq \mathbb{R}$ (wlog, assume $0\notin V$). Then, take any $f \in C_c^{\infty}(\mathbb{R})$.  Now, $f^{-1} (V)$ is bounded.

From this, it follows that any such topology $T$ is finer than the usual topology. But, as Berci said, in the usual topology, pasting doesn't work in general for differentiable functions. (Take $x$ on $[0,\infty)$ and $-x$ on $(-\infty,0]$)

Edit: As NielsDiepeveen pointed out, my earlier answer (about the complex case) was wrong.


问:
Prove that for any integer $n>0$ which is a perfect square, $n+4$ is not a perfect square.
I have been told that this proof is incorrect, however, I'm having a hard time seeing the issue.

Prove that for any integer $n > 0$ which is a perfect square, $n+4$ is not a perfect square:

Proof:

$n > 0$

Since $n$ is a perfect square, $a^2=n$

The subsequent perfect square can be written as ${b^2=\left(a+1\right)}^2$

The difference between consecutive squares is then ${|b}^2-a^2|=|2a+1|$

$\forall a\in\mathbb{Z}(|2a+1|\neq 4)$

$\square$

Edit: Thanks everyone, appreciate it :)

答: Let $n = a^2 \implies n+4 = a^2+4 = b^2 \implies 4 = (b-a)(b+a) \implies b-a = 1, b+a = 4 \implies 2b = 5$, contradiction since $2b$ is even and $5$ is odd. Note that the case $b-a = 2, b+a = 2$ is ruled out because it leads to $b+a = b-a \implies a = 0$, but $a > 0$ since $n > 0$.


问:
$\int_0^{2 \pi} \sinh^{-1} \frac{h}{\sqrt{s^2 + R^2 -2sR \cos(\phi)}} \, d\phi $
The definite integral is the following one:

$$I_0 = \int_0^{2 \pi} \sinh^{-1} \frac{h}{\sqrt{s^2 + R^2 -2sR \cos(\phi)}} \, d \phi $$

Where $h, s, R$ are positive real quantities and $\sinh^{-1}$ is the arcsinh function. Basically, I have a cylinder of height $h$ and radius $R$ centered at the $xy$ plane origin. It goes up along the $z$ axis. I want to calculate the magnetic potential at a radial distance $s$ from the $z$ axis. And $\phi$ is the argument in respect to the $x$ axis. In other words, I am using the usual cylindrical coordinates, $s, \phi, z$.

This integral appeared when I was calculating the magnetic vector potential $A$. I tried to make substitutions with inverse functions ($\phi = \sin(x)$ and so on...) and also tried to substitute the whole argument of arcsinh so as to achieve an easier one. I also tried to use the approximation $\ln(2x) = \sinh^{-1}(x)$. But nothing worked out well. Also tried to expand in a power series the integrand itself and just the cosine function...

One of the results I found was:

$$I_0 = 4h^2 \int_{\sinh^{-1}(\frac{h}{s-R})}^{\sinh^{-1}(\frac{h}{s+R})} \frac{y \coth(y)}{\sqrt{-(\sinh^4(y)(s^2-R^2)^2-2(s^2+R^2)h^2 \sinh^2(y)
 +h^4  )}}dy$$

It appears to be a binomial expansion, but it isn't. It then becomes:

$$I_0 = 4h^2 \int_{\sinh^{-1}(\frac{h}{s-R})}^{\sinh^{-1}(\frac{h}{s+R})} \frac{y \coth(y)}{\sqrt{ (\sinh^2(y) 2sR)^2  - (\sinh^2(y)(s^2-R^2)-h^2)^2  }}dy$$

Any suggestions would be very much appreciated! Thanks!

MAJOR EDIT

It seems very unlikely that this integral above is easily solvable. That made me go back and realize the following:

$A(s, 0, 0) = \underbrace{\frac{u_0 nI h R}{4 \pi}}_{C} \displaystyle\int_0^{2 \pi} \displaystyle\int_{-1}^1 \frac{ \bar{\phi}}{\sqrt{s^2-2sR \cos(\phi) +R^2+h^2z^2}} dz d \phi$,

where $\bar{\phi}$ is the unit vector (cylindrical coordinates). Notice that, if we solve the integration in respect to $z$ we will achieve the nasty integral above ($I_0$). However, let's approach differently this time. Let's invert the order of integration and perform the following substitution: $\phi = \arccos(x)$. Then it becomes:

$A(s, 0, 0) = 2C \displaystyle\int_{-1}^1 \displaystyle\int_{-1}^1 \frac{\bar{\phi}}{ \sqrt{1-x^2} \cdot \sqrt{(s^2+R^2+h^2z^2)-2sRx}} dxdz$

Let's take both radicands to be non-negative and we then have:

$A(s, 0, 0) = 2C \displaystyle\int_{-1}^1 \displaystyle\int_{-1}^1 \frac{\bar{\phi}}{ \sqrt{ 2sRx^3 - (s^2+R^2+h^2z^2)x^2 - 2sRx + (s^2+R^2+h^2z^2) } } dxdz$

Which seems to be an elliptic function (if my quick search was correct). If these steps are correct, how should I proceed? Maybe residue theorem? Although I could recognize, I don't have the knowledge to solve it properly.

So far, I have found the following websites:

http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf ;
https://functions.wolfram.com/EllipticIntegrals/EllipticE/introductions/CompleteEllipticIntegrals/ShowAll.html ;
https://en.wikipedia.org/wiki/Elliptic_integral

Although they mentioned a similar integral to $A(s, 0, 0)$ and the need of elliptic integrals, they do not explain much.

答:
You said numerical methods is an option so I'm only posting this because of that. This just uses the generic quad method which I think is Gauss-Legendre integration.

```<code>from scipy.integrate import quad
from numpy import sqrt,arcsinh, pi,cos

def integrand(x, h, s,R):

    return arcsinh(h/sqrt(s**2 +R**2 -2*s*R*cos(x)))
### parameters
h = 1
s = 1
R = 10

I = quad(integrand, 0, 2*pi,args=(h,s,R))
</code>```

the results are given as a tuple. For instance this $(0.6288320765849464, 3.145647904875375e-09)$ means it approximates it as 0.6288 whatever within $3.25e-9$. You need to be careful about the constants because it doesn't evaluate well in some instances. You could add in something to catch negative parameters if you want

Body:
You said numerical methods is an option so I'm only posting this because of that. This just uses the generic quad method which I think is Gauss-Legendre integration.

from scipy.integrate import quad
from numpy import sqrt,arcsinh, pi,cos

def integrand(x, h, s,R):

    return arcsinh(h/sqrt(s**2 +R**2 -2*s*R*cos(x)))
### parameters
h = 1
s = 1
R = 10

I = quad(integrand, 0, 2*pi,args=(h,s,R))

the results are given as a tuple. For instance this $(0.6288320765849464, 3.145647904875375e-09)$ means it approximates it as 0.6288 whatever within $3.25e-9$. You need to be careful about the constants because it doesn't evaluate well in some instances. You could add in something to catch negative parameters if you want


问:
Failure of Unique Factorization into Primes

Show that in the domain of integers in the form $a+b\sqrt{-17}$ unique factorization into primes fails.

It was given that $169=13\bullet 13=(4+3\sqrt{-17})(4-3\sqrt{-17})$. This is an example that falls into the required domain.

In order for unique factorization into primes to fail, it needs to be shown that $(4+3\sqrt{-17})(4-3\sqrt{-17})$ is not prime - meaning it is irreducible.

If it is irreducible, then it cannot be factored into two other integers in that domain.

The norm of $(4+3\sqrt{-17})$ is $25$. What can I do now?

答:
Actually the norm of $4 + 3\sqrt{-17}$ is $169 = 13^2$. But anyway, if it had a nontrivial factorization, say $4 + 3\sqrt{-17} = \alpha\beta$, then by multiplicativity of norms you would have:

$$N(\alpha)N(\beta) = 13^2$$

Which means that $N(\alpha) = \pm 13$ (since otherwise $\alpha$ or $\beta$ would have unit norm and would have to be a unit). Remember that $\alpha$ must be of the form $\alpha = x + y\sqrt{-17}$ for some integers $x, y$.

But there are no integers $x, y$ such that $x^2 + 17y^2 = \pm 13$.


答:
What's wrong is no-one said it was the next perfect square.

So if $n = a^2$ and $n+4 = b^2$ we have no reason to assume $b = a+1$.

But we can assume $b \ge a+1$ and get a similar problem.

$n=a^2$ and $n+4 = b^2 \ge (a+1)^2 = a^2 + 2a + 1$

So $4 \ge 2a+1$ so $a \le 1.5$.  Well nothing wrong with that.  So $a = 1, 0$ and $a=0\implies n =a^2 = 0$ that's out (but it is a case where $n =0^2$ and $n+4 = 2^2$) and $a=1\implies n=1^2 = 1$ and $n+4 = 5$ not a perfect square.

So that can fix the proof.

Or we can do:

You have $n=a^2$ and let's suppose $n+4 = b^2$ then

$b^2 - a^2 = 4$ and

$(b-a)(b+a) =4$ and... now we are talking.  Assume $a,b$ are positive and $b > a$ we get $(b-a)(b+a) = (1,4), (2,2)$.

The first means $2b = 5; 2a = 3$ and $a = 1.5$ and $b = 2.5$ (and indeed $1.5^2 = 2.25$ and $2.5^2 = 6.25 = 2.25 + 4$.  But $a, b$ are not integers.

The second means $a=0$ and $b =2$. So $n = 0^2 = 0 \not > 0$. So that is out.

Title: Prove that for any integer $n>0$ which is a perfect square, $n+4$ is not a perfect square.

Body:
I have been told that this proof is incorrect, however, I'm having a hard time seeing the issue.

Prove that for any integer $n > 0$ which is a perfect square, $n+4$ is not a perfect square:

Proof:

$n > 0$

Since $n$ is a perfect square, $a^2=n$

The subsequent perfect square can be written as ${b^2=\left(a+1\right)}^2$

The difference between consecutive squares is then ${|b}^2-a^2|=|2a+1|$

$\forall a\in\mathbb{Z}(|2a+1|\neq 4)$

$\square$

Edit: Thanks everyone, appreciate it :)


问:
Proof of Cauchy Criterion

How did they get $|x_{n} - x_{n_{k}}| = \dfrac{\epsilon}{2}$ in the proof?

答: You know that $|x_n - x_m| < \epsilon/2$ for $n,m \geq N$. In the proof, $n$ and $n_K$ are greater than or equal to $N$. They don't mention that $n \geq N$, but that is what they have used.</p>

Title: Proof of Cauchy Criterion