问:
Continuity of function in $\mathbb{R}^2$
Where should I start if I want to study the continuity of a function in $\mathbb{R}^2$?
Like this one:
$$f(x,y)=
\begin{cases}
      \frac{x^2y}{x^2+\sqrt{y}} & \quad\text{if } y>0,\\
      0 & \quad\text{otherwise.}\\
   \end{cases}
$$
I think $f$ is continuous except at $(0,0)$, so I have to take the limit to see if it's continuous, right? But I'm confused about the piecewise function, which would be $0$ at that point... Could someone help me?

答:
You can notice that for $y \gt 0$

$$\begin{aligned}\vert f(x,y) \vert &= \left\vert \frac{x^2y}{x^2+\sqrt{y}} \right\vert\\
&\le \left\vert \frac{x^2y}{x^2} \right\vert = \vert y \vert
\end{aligned}$$
and that the exact same inequality is also satisfied for $y \le 0$ as in that case $f$ vanishes.

As for any $u \in \mathbb R$ $\lim\limits_{(x,y) \to (u,0)} \vert y \vert = 0$, you can conclude that $f$ is continuous at $(u,0)$ as $f(u,0) = 0$.

Also, $f$ is continuous at $(u,v)$ with $v \neq 0$ using continuity of composition of continuous maps.

Finally, $f$ is continuous everywhere.


问:
Is it possible to multiply out at least by $s^2$ in this root expression? $\sqrt{(a^2-4d)s^2+(4bd-4ac)s+4c^2}$
My goal is to pull out all the $s$ from under the square root.

$$\sqrt{(a^2-4d)s^2+(4bd-4ac)s+4c^2}$$

where $a,b,c,d$ are complex numbers and $s$ is a variable.

I tried solving for $s$, but that only gave me $s_1=\frac{2\left(ac+\sqrt{d\left(db^2-2abc+4c^2\right)}-db\right)}{a^2-4d},\:s_2=\frac{2\left(ac-db-\sqrt{d\left(db^2-2abc+4c^2\right)}\right)}{a^2-4d};\quad \:a^2-4d\ne \:0 $

and i realized that i cannot take the squareroot of $(s-s_1)(s-s_2)$ if $s_1$ and $s_2$ are not equal.

Is it possible to multiply out at least by $s^2$ in this root?

On the same note is there a general way to check whether a parametric expression has a semi-rational form like:

$$(s\cdot \sqrt{\mathscr{\xi}})$$

?

Where $\xi$ is the irreducible part of the expression(preferably made up of only constants).

答:
I think you're asking for some complex $k,l$ such that

$$ks+l=\sqrt{(a^2-4d)s^2+(4bd-4ac)s+4c^2}.$$ Unfortunately, this would make $a,b,c,d$ depend upon $k,l$ and two of their own within the set, or, by excluding $k,l$, have a single dependency. However, we may proceed anyway.

$$k^2s^2+2kls+l^2= (a^2-4d)s^2+4(bd-ac)s+4c^2$$
$$k=\sqrt{a^2-4d}$$ and
$$2\sqrt{a^2-4d}\cdot ls+l^2=4(bd-ac)s+4c^2$$
$$l=\dfrac{2(bd-ac)}{\sqrt{a^2-4d}}$$ leaving the condition
$$(bd-ac)^2c^2=a^2-4d.$$

ContentLicense: CC BY-SA 4.0


问:
How to Solve a Certain Limit Function
I know the answer to this $$\lim_{x\to\infty} \left(\frac{5x-2}{5x+4}\right)^{5x+1}$$ is either 6 or -6, but I can't figure out how to reach it. Any time I try, I get stuck on the $${\infty-\infty}$$ indeterminate form.

答:
Hint

First, apply exponent rules

$$\lim_{x\to \infty}\bigg(e^{(5x+1)\ln(\frac{5x-2}{5x+4})}\bigg)$$

then, would you know how to proceed from here, using the limit chain rule?

Title: How to Solve a Certain Limit Function


问:
AP Calculus multiple choice question
The position of a particle along a line is given by $z(t) = 2t^3 -24t^2 + 90t +7$ for $ t \geq 0$. For what values of $t$ is the speed of the particle increasing?

a) $3 < t < 4$ only

b) $t > 4$ only

c) $t > 5$ only

d) $0 < t < 3$ and $t > 5$

e) $3 < t < 4$ and $t > 5$

For this problem I said d) based on my calculations:

$z'(t) = 6t^2 - 48t + 90 = 6(t-5)(t-3)$. I then did a chart and found out that $z'(0) > 0$ ,$z'(4) < 0$ and $z'(6) > 0$ leading me to conclude d) however the textbook says e). I thought that I did everything right, what is wrong with my conclusion?

答: On a one-dimensional axis representing time, mark the signs of the particle's velocity and acceleration in each of the intervals determined by the zeros of the velocity and acceleration functions. The SPEED (not velocity) is increasing in any interval where the velocity and acceleration have the same sign. That is, the particle is accelerating while moving in what we define to be the positive direction or decelerating in what we define to be the negative direction.</p>

ContentLicense: CC BY-SA 3.0

Title: AP Calculus multiple choice question


问:
What is the logic behind Jacobi iterative method?
The book I follow and on net also, all that I can find is the algorithm to find the solution, but I don't quite understand the physical significance or logic behind the algorithm.

Can someone please help.

答:
You are trying to solve $A \textbf{x} = \textbf{b}$.

The matrix $A$ could be written
$A = D + N$ where $D$ is diagonal.
If the sequence $\textbf{x}^k$ generated by $D\textbf{x}^{k+1}  = b - N\textbf{x}^k$ converges, then you have $D\textbf{x}^{\infty}  = b - N\textbf{x}^{\infty}$ or $A \textbf{x}^{\infty} = \textbf{b}^{\infty}$.

If $A$ is diagonal, then $N=0$ and there is no iteration to do.

The condition for convergence is $|| D^{-1} N || < 1$. Intuitively, not super accurate but helpful, think that in the iteration $\textbf{x}^{k+1} = D^{-1} b- D^{-1} N\textbf{x}^{k}$, the error term $D^{-1} N\textbf{x}^{k}$ gets smaller after each iteration and the equation approaches $\textbf{x} \approx D^{-1} b $.  There is a theorem for this, look up if you want rigor.


问:
Unitary Operator on Hilberspace to show that Fourierbasis is a maximal Orthogonal Set
I have looked at the proof Proving that the Fourier Basis is complete for C(R/$2*\pi$ , C) with $L^2$ norm
but am having trouble understanding the argumentation about the Hilberspace. I think the proof is quite nice and would like to understand it, and also need to refresh my Functional analysis. In particular:

This is probably not the most elegant way, but this method generalizes to showing the unitarity of the Fourier transform. The essential part of this proof is showing that the map L2(R/2πZ)→ℓ2(Z) given by taking the Fourier coefficients from the discrete Fourier transform is an isometry, since the discrete Fourier transform is essentially the orthogonal projection onto the closure of the span of the functions e^inx. Then, by basic Hilbert space theory, it would follow that {e^inx} is a maximal orthogonal set.

The "then by basic Hilber space theory" is not clear to me...

Is this because if you take the Span of $e^{inx}$ it has the same Image under an Unitary Operator  as L2(R/2πZ), and thus {e^inx} must be dense in L2(R/2πZ) - And if yes, whats the name of a theorem showing this/can you direct me to one?

thank you

答:
I actually love Disinegrating's answer, but I do think it may have omitted the Hilbert space theory that OP asked about? For the sake of completeness, maybe it's good if I  describe the underlying theory for future viewers.

Definition: we say that a subset $A$ of an inner product space $H$ is an orthonormal set of vectors if we have $\langle x,y \rangle = 0$ for all distinct $x,y \in A$ as well as $\|x\|^2 = \langle x,x \rangle = 1$ for all $x \in A$.

Definition: in an inner product space $H$, we say that a set $A$ of orthonormal vectors is maximal if, given $x \in H \setminus A$, the set $\{x\} \cup A$ is not an orthonormal set of vectors.

Definition: we say that a countable, maximal, orthonormal set of vectors is an orthonormal basis.

Proposition: if $(e_n)$ is an orthonormal basis for a Hilbert space $H$ (i.e., $(e_n)$ are all distinct and $\{e_n : n\in\mathbb{N}\}$ is an orthornomal basis), then we have $v = \sum_{n\in\mathbb{N}} \langle v,e_n \rangle e_n$ for all $v \in H$.

Theorem: given a sequence $(e_n)$ of mutually orthonormal vectors in a Hilbert space $H$, we have that $(e_n)$ is an orthonormal basis if and only if, for all $v \in H$, $\|v\| = \sum_{n\in\mathbb{N}} |\langle v,e_n \rangle|^2$.

Remark: it is always true that $\|v\| \geq \sum_{n\in\mathbb{N}} |\langle v,e_n \rangle|^2$. The reverse inequality is non-trivial.

In conclusion: in the question cited by OP, the scalars $a_n$ are in fact of the form $\langle f,e_n \rangle$ for some sequence $(e_n)$ of orthonormal vectors. Thus, to be able to conclude that $(e_n)$ are maximal, the problem in the post which OP refers to is to show that $\sum_{|n| \leq N} |a_n|^2 \to \|f\|^2$. as $N\to\infty$.

For a proof of the theorem referenced above, see, perhaps, Real and Complex Analysis by Rudin, or Real Analysis: Measure Theory, Integration, and Hilbert Spaces by Stein and Shakarchi.


答:
Note that

\begin{equation}
\frac{5x-2}{5x+4} = \frac{5x+4-6}{5x+4} = \left( 1 - \frac{6}{5x+4} \right )
\end{equation}

now

\begin{equation}
\left( 1 - \frac{6}{5x+4} \right )^{5x+1} = e^{(5x+1) \log\left(1- \frac{6}{5x+4}\right)}
\end{equation}

and
\begin{equation}
\lim_{x\to \infty} (5x+1) \log\left(1- \frac{6}{5x+4}\right) = - 6
\end{equation}

So by continuity of the exponential function, the original limit is equal to $e^{-6}$.

Title: How to Solve a Certain Limit Function

Body: I know the answer to this $$\lim_{x\to\infty} \left(\frac{5x-2}{5x+4}\right)^{5x+1}$$ is either 6 or -6, but I can't figure out how to reach it. Any time I try, I get stuck on the $${\infty-\infty}$$ indeterminate form.


问:
Rings Isomorphisms
I study ring theory, and I wondered about the next question that I asked myself for
practice:

Let $\mathcal{R}_1, \mathcal{R}_2$ be ring, and let $G_1, G_2$ be their additive groups, and $M_1, M_2$ be their multiplicative monoids.

Suppose there are $f_G: G_1\to G_2$ and $f_M:M_1\to M_2$ isomorphisms ($G_1\cong G_2, M_1\cong M_2$).

Is it true that $\mathcal{R}_1 \cong \mathcal{R}_2?$

My motivation to proving this: $G_1=M_1,G_2=M_2$ as sets, by adding binary operetions they induces the rings $\mathcal{R}_1$ and $\mathcal{R}_2$, but the problem is $f_G$ and $f_M$ not necessarily induce ring homomorphism $f_{\mathcal{R}}$.

Thanks in advance.

答:
The answer is no, but nontrivial in the finite case.

In the infinite case, let $F$ be a field, and let $R=F[x,y]$, $S=F[x]$, the rings of polynomials in one and two (commuting) variables. The additive groups of both are isomorphic to a direct sum of countably many copies of $F$, and the multiplicative structure of nonzero elements is isomorphic to the direct product of multiplicative group of $F$ (the group of units), and a direct sum of $\aleph_0|F|$ copies of the free monoid of rank $1$ (this is given by the irreducible polynomials; that is, the free monoid generated by the irreducible polynomials; since both rings are UFDs). You can then adjoin the $0$ to both with no problems to get the full isomorphism of multiplicative monoids. But $S$ is a PID (every ideal is principal), and $R$ is not, so they cannot be isomorphic as rings.

In the finite case, in this math.overflow question that asked precisely this question, Keith Kearnes gives an explicit example of two finite unital rings with isomorphic additive and multiplicative structures but that are no isomorphic. He mentions that if you allow non-unital rings, then examples exist with an $8$-element ring-without-unity.

Title: Rings Isomorphisms


问:
$e^{\pi i} = -1$ is this a fact or an assumption?
I think it is a fact but can someone explain why is it true intuitively?

I heard a lot of videos on youtube assuming it is the "natural" way of revolving around 0, many other explanations that does not make sense to me.

I only need a direct way of explaining why the imaginary power causes the number to become 1 in its radius (magically).

e.g. 1^i = 1

e.g. 9999^i = 1 (with some rotation)

the general intuition for exponentials is that it gives different numbers, but using an imaginary power always gives a 1 (in the radius)

this can also be proven using the formula: z = r(cos(θ) + i sin(θ)) where θ contains all the information about the number then it gets converted to an angle only without any radius other than 1.

答:
There is a bit of an underlying assumption here, in terms of what $e^z$ even means when $z \in \mathbb{C}$. Everything else, we can prove from that.

In one common systematic development of complex analysis, because we know from real calculus that every real number $x$ satisfies

$$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!} $$

we then define the function $\exp(z)$ similarly for $z \in \mathbb{C}$.

$$ \exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!} $$

Then for a positive real number $b$, we define (assume?)

$$ b^z = \exp(z \ln b) $$

and show that it equals the ordinary meaning of $b^z$ in real numbers when $z = x + 0 i$ is a real number. This allows us to write $\exp(z) = e^z$ and use the $e^z$ notation in place of the $\exp$ function.

So determine the complex value $e^{yi}$, we plug in $yi$ as the argument to the definition of $\exp$:

$$ e^{yi} = \exp(yi) = \sum_{k=0} i^k \frac{y^k}{k!} $$

In this sum, the terms alternate between pure real and pure imaginary. Separating them out,

$$ e^{yi} = \left[\sum_{m=0}^\infty (-1)^m \frac{y^{2m}}{(2m)!} \right] +
i \left[\sum_{n=0}^\infty (-1)^n \frac{y^{2n+1}}{(2n+1)!} \right] $$

Then the Taylor expansions of $\cos y$ and $\sin y$ are plainly seen, leaving Euler's formula

$$ e^{yi} = \cos y + i \sin y $$

In particular, if $y=\pi$, this is Euler's famous identity

$$ e^{i \pi} + 1 = 0 $$


问:
What's the probability of A and B being at most 1 person apart out of 12 randomly positioned persons?
I'm checking if I'm approaching this problem correctly. The problem: there are 12 people seated in a row. Their positions are randomly determined. How likely is it that persons A and B have at most 1 person between them?

Now, it's trivial to just brute force it by visualising how many people there are. There are 11 ways for them to be next to one another and 10 ways to be one apart. For 12 people, there's a total of 12C2 = 66 two person combinations. Ergo P(A) = (11+10)/66 = 21/66 = approx. 31,8%.

My main question is if there's a proper mathematical way to determine the number of combinations in which the two people are either adjacent or one apart, or if I should just use visual aids in situations like this?

答:
It may be simpler (and less "visual") to use the concepts of blocks

For example, if there are $20$ people, and you want the particular two to have $4$ people in between, you have one block of $6$ people and the balance $14$ as blocks of $1$ each.

Here, you only have one block with more than one person, but more complex problems might have more complex situations in which using blocks would prove useful

And then, if Pr is asked for, you can decide whether you want to use combinations or permutations (to take care that the numerator and denominator are of the same type) and proceed.


问:
Suppose $p$ is prime and $m \gt 0$ and $n \gt 0$ and $m \equiv n \pmod{p-1}$. Prove that $a^m \equiv a^n \pmod{p}$ for all $a$.

Suppose $p$ is prime and $m \gt 0$ and $n \gt 0$ and $m \equiv n \pmod{p-1}$. Prove that $a^m \equiv a^n \pmod{p}$ for all $a$.

I know I have to apply Fermat's little theorem to prove this, but I don't know where to start.

答:
Since $m = k(p-1) + n$ for some $k \in \mathbb Z$, $a^m = (a^k)^{p-1}a^n$, so apply Fermat's little theorem. Maybe you should split into two cases: either $p \mid a$ (trivial), or $p \nmid a$.

EDIT: actually, the same question is asked here: Fermat's Little Theorem: $r \equiv s \pmod {p-1} \implies a^r \equiv a^s \pmod p$


问:
Definition of Lie Algebra
I have just start studying Lie Algebra. I want to know the motivation for the conditions on the Lie Bracket in the Definition of Lie Algebra. Please explain me or tell me some references. Thanks. </p>

答:
Two good books are:

Introduction to Lie Algebras by Humphreys.

Introduction to Lie Algebras by Erdmann and Wildon.

The most natural examples of Lie algebras are the matrices and the Lie bracket is the usual commutator $[X,Y] = XY-YX$.

Moreover, the real motivation comes from the study of Lie groups. A Lie algebra of a Lie group is precisely the set of all left invariant vector fields and they have a natural Lie bracket.

There is also some motivation from Physics: see Poisson equation for example.

Title: Definition of Lie Algebra


答:
Continuity at (0,0);

Let $\epsilon >0 $ be given.

Let $\delta=\epsilon, $ and $(x, y) \not =(0, 0);$

$0\le |f(x,y) - 0| \le |\dfrac{x^2|y|}{x^2+\sqrt{|y|}}| \le$

$ |y|\le \sqrt{x^2+y^2} \lt \delta=\epsilon. $

For $y=0$ we have $f(x, 0)=0$ and trivially

$0=|f(x,0)-0| \lt \delta =\epsilon$.

Title: Continuity of function in $\mathbb{R}^2$

Body:
Where should I start if I want to study the continuity of a function in $\mathbb{R}^2$?
Like this one:
$$f(x,y)=
\begin{cases}
      \frac{x^2y}{x^2+\sqrt{y}} & \quad\text{if } y>0,\\
      0 & \quad\text{otherwise.}\\
   \end{cases}
$$
I think $f$ is continuous except at $(0,0)$, so I have to take the limit to see if it's continuous, right? But I'm confused about the piecewise function, which would be $0$ at that point... Could someone help me?


答:
An important (maybe the most important) limit to remember (maybe the only one you really need to remember) is
$$
\lim_{t\to\infty}\left(1+\frac at\right)^t=\operatorname e^a
$$
for any real (actually complex) number $a$.

Try to bring yours to this form starting with the fraction
$$
\frac{5x-2+6-6}{5x+4}=1-\frac{6}{5x+4}
$$
and now pay with the idea of a substitution.  (Don't forget power algebra and factoring "constant" terms out of the limit.)

Title: How to Solve a Certain Limit Function


问:
Finite group with a special property
Let $(G,\cdot)$ be a finite group  with the identity element $e$ such that for each proper subgroup $H$ of $G$ there exists a unique element $g_H\in H$, $g_H\ne e$, such that the following properties hold:

(a) $g_H^2=e$;

(b) for any two distinct proper subgroups $H,K$ of $G$ we have $g_H\ne g_K$.

Then the group $G$ is abelian.

Thus far, I have considered a prime number $p$ dividing the order of $G$. Then, by Cauchy's theorem, there exists an element of $G$ of order $p$ and consider the subgroup $H:=\langle x\rangle=\{e,x,\dots,x^{p-1}\}$. By hypothesis, there exists $g_H=x^r$ (with $r\in \{1,\dots,p-1\}$) and $g_H^2=x^{2r}=e$. Then, we have $p\mid 2r$ and it follows $p=2$. Therefore, the order of $G$ is a power of 2.

答:
Let $G$ be a counterexample. As you have already shown, the order of $G$ is a power of $2$, say $|G|=2^n$. Let $H$ be a maximal subgroup of $G$. From the theory of $p$-groups, we know that $H$ is normal in $G$. Also, by property (a), $H$ contains a unique element $g_H$ of order $2$, and the subgroup $\langle g_H\rangle$ is also normal in $G$.

Let $k\in G\setminus H$, and let $K:=\langle k\rangle$. Since $G$ is not cyclic (because otherwise it would be abelian and would thus not be a counterexample), $K$ is a proper subgroup of $G$ as well and therefore contains a unique element $g_K$ of order $2$.

On the other hand, $K\neq H$, and so by property (b) we have $g_H \neq g_K$.

Since $\langle g_H\rangle$ is normal in $G$, $D:=\langle g_H, g_K\rangle$ is a group of order $4$ containing more than one element of order $2$, and thus by assumption, we have $D=G$. But then $G$ is abelian (since all groups of order $4$ are abelian), a contradiction.

Title: Finite group with a special property


问:
Partial DE Question
I have a quick question.

Question 1). Hence deduce that
$$
u(x,t)=F(x+at)+G(x-at)
$$
satisfies the original PDE for any twice-differentiable functions $F$ and $G$.

c) Hence find the solution of the PDE which satisfies the initial conditions
$$
u(x,0)=e^{-x}\quad\text{and}\quad\dfrac{\partial u}{\partial t}(x,0)=0\quad,
$$

d) If the initial conditions are $~u(x,0)=r(x);\,\,\dfrac{\partial u}{\partial t}(x,0)=s(x)~$ for some functions $r(x)$ and $s(x)$,
show that
$$
F(x)=\tfrac12r(x)+\tfrac1{2a}\Big(\int\limits_0^xs(w)\,\mathrm dw\Big)+k
\quad\text{and}\quad
G(x)=\tfrac12r(x)-\tfrac1{2a}\Big(\int\limits_0^xs(w)\,\mathrm dw\Big)-k
$$
for some arbitrary constant $k$

On part d, there's a bit I'm not too sure of. My workings are:

$$u(x,t) = F(x+at) + G(x-at)$$

$$u(x,0) = F(x) + G(x) = r(x)$$

$$u_t(x,0) = a(F'(x) - G'(x)) = s(x) \Rightarrow F(x) - G(x) =\frac {1}{a} \int s(x) dx + k$$

From there I can use simulataneous equation and solve etc. But in the solution they have a change of variable to change the integral from $s(x)dx$ to $ \int_0^x s(w) dw$. I'm not sure why you can do this. Thanks!

答:
Your formula $F(x) - G(x) =\frac {1}{a} \int s(x) dx + k$ isn't quite right. On the left, $x$ is an actual variable that is plugged into the function. On the right it's the dummy variable of integration. Mixing the two, as in
$$F'(x)=C+\int_0^x F'(x)\,dx$$
is to be avoided. "Integrate with respect to $x$ from $x=0$ to $x=x$?"

The correct notational approach is to proceed from
$$ F'(x) - G'(x) = \frac{1}{a} s(x)$$
to $$F(x)-G(x)=k+\int_0^x(F'(w)-G'(w))\,dw = k+\frac{1}{a} \int_0^xs(w)\,dw$$
and then solve for $F$ and $G$ algebraically.

Title: Partial DE Question


答:
By long division (only one step is needed), we get

$$ \frac{5x-2}{5x+4} \;\; = \;\; 1 \; - \frac{6}{5x+4}, $$

so we get

$$ \left(\frac{5x-2}{5x+4}\right)^{5x+1} \;\; = \;\; \left(1 \; - \frac{6}{5x+4}\right)^{5x+1} \;\; = \;\; \left(1 \; - \frac{6}{5x+4}\right)^{5x}\left(1 \; - \frac{6}{5x+4}\right) $$

Since the last factor (clearly) approaches $1-0=1,$ we'll focus on the first factor (the part with $5x$ as an exponent). With some algebraic work we can reduce this to the standard limit for $e^x.$ Since sums and differences in denominators are a lot harder to work with than sums and differences in exponents, let $u = 5x + 4$ to get a nice denominator. Then $x = \frac{u-4}{5},$ and hence the exponent $5x$ becomes $u-4.$ Also, $x \rightarrow \infty$ if and only if $u \rightarrow \infty,$ so the the operation "$x \rightarrow \infty$" can be replaced with the operation "$u \rightarrow \infty$".

Therefore,

$$ \left(1 \; - \frac{6}{5x+4}\right)^{5x} \;\; = \;\; \left(1 \; - \frac{6}{u}\right)^{u-4} \;\; = \;\;   \left(1 \; - \frac{6}{u}\right)^u \left(1 \; - \frac{6}{u}\right)^{-4}  $$

$$ \longrightarrow  \;\; e^{-6} \cdot 1^{-4} \;\; = \;\; e^{-6}  $$

Finally, we have

$$ \left(\frac{5x-2}{5x+4}\right)^{5x+1} \;\; = \;\; \left(1 \; - \frac{6}{5x+4}\right)^{5x}\left(1 \; - \frac{6}{5x+4}\right) $$

$$ \longrightarrow  \;\; e^{-6} \cdot 1 \;\; = \;\; e^{-6}  $$

Title: How to Solve a Certain Limit Function


问:
If $f \ge 0$ is Riemann integrable on$[a,b]$, with $\int^a_b f =0$ how to show that there must be some $x\in[a,b]$ such that $f(x)=0$?
it is also given that $f(t)\geq 0$ for all $t$ .I'm supposed to first prove that $\exists[c,d]\subseteq[a,b]$ s.t. $f(t)<\varepsilon\forall t\in[c,d],\implies\int^d_c f =0 $. Which I'm struggling to do.

By definition, let $\varepsilon>0$ then $\exists\delta$ s.t.
$$ \left| R(f,P)-\int^b_a f \right| <\varepsilon$$
where $P$ is $\delta$-fine. Note that $R(f,P)$ here denotes the Riemann sum. Suppose $f$ is Riemann integrable, then
$$ \left| R(f,P)-0\right| =|R(f,P)|<\varepsilon$$
$$=\left|\sum^n_{k=1}f(t_k)(x_k-x_{k-1})\right|<\varepsilon$$
then because $x_k>0\forall k$
$$\text{Riemann sum of $f$ from $c,d$ }<\left|\sum^n_{k=1}f(t_k)(x_k-x_{k-1})\right|<\varepsilon$$

but I'm not sure how that implies that $f(t)<\varepsilon$

答:
You can go by contradiction which seems simpler. i.e., assume that there exists $\epsilon'>0$ such that every sub interval $[\alpha,\beta]\subset [a,b]$ has a point $x_{\alpha,\beta}\in [\alpha,\beta]$ satisfying

$$ f(x_{\alpha,\beta})\geq \epsilon' .$$

Then, you obtain that the upper Darboux sum of any partition $P$ will have to satisfy that

$$ U(P,f)\geq \epsilon'\cdot(b-a). $$

You know that

$$ \overline{\int_a^b} f(x)dx= \inf_{P\in \mathcal{P}[a,b]} U(P,f)\geq \epsilon' \cdot (b-a)>0 .$$

If you prefer to work with Riemann sums instead, you can argue that for any partition  $P$, there is a Riemann sum corresponding to $P$ satisfying

$$R(f;P)\geq \epsilon'\cdot (b-a), $$
and obtain a contradiction in a similar manner.


问:
in the derivation of support vector classifier
I was studying the Stephen Boyd's textbook on convex optimization and have a question on the support vector classifier. The book says the following:

In linear discrimination, we seek an affine function $f(x) = a^T x - b$ that classifiers the points, i.e.,

$a^T x_i - b > 0, i = 1,..., N$

$a^T y_i - b < 0, i = 1, ..., M$

Geometrically, we seek a hyperplane that separates the two sets of points. Since the strict inequalities  are homogeneous in a and b, they are feasible if and only if the set of nonstrict linear inequalities

$a^T x_i - b \geq 1, i = 1,..., N$

$a^T y_i - b \leq -1, i = 1, ..., M$

(in the variables a,b) is feasible.

First, what does it mean that the original inequalities are homogeneous in a and b? Second, I have trouble in understanding the if and only if relationship. "If" part is obvious: if a >= 1, a > 0 obviously. However, can anybody explain to me why the "only if" part also holds?

Thanks,

答:
The inequalities are are homogeneous in $a$ and $b$ because you can multiple $a$ and $b$ by any $\alpha >0$ and the inequalities will still hold. In other words, you can always "scale" $a$ and $b$ up or down as you like.

The author is being a little sloppy here and this sloppiness is confusing you. Here's a more formal way of stating what the author wishes to say:

There are $a$ and $b$ such that $a^T x_i - b > 0$ and $a^T y_j - b < 0$ for all $i\in\{1,\ldots,N\}$ and $j\in\{1,\ldots,M\}$ if and only if there are $a$ and $b$ such that $a^T x_i - b \ge 1$ and $a^T y_j - b \le -1$ for all $i\in\{1,\ldots,N\}$ and $j\in\{1,\ldots,M\}$.

In other words, the $a,b$ in the first inequalities are different from the $a,b$ in the second inequalities.

Given that you can scale $a$ and $b$ up by an arbitrarily large positive constant, the statement seems obvious. Please let me know if not and I'll supply a proof.


问:
An honest die is rolled 5 times. What is the probability that you will roll NO 6's?
An honest die is rolled 5 times. What is the probability that you will roll NO 6's?

I now know you can use binomial theorem to get the correct answer but I am failing to see why my original method would not have worked...

I found the total # of ways you could roll zero 6's: $5^5$
(5 ways to roll no 6 on the first roll, 5 ways to roll the second roll, etc.)
and then divided it by the total number of ways to roll a die 5 times: $6^5$
(6 ways to roll the first roll, 6 ways to roll the second roll, etc.).

I got $0.40$.

Why is that incorrect?

答:
The correct answer is already mentioned in comments, however I would have reached it on a different approach.
The odds to score a 6 in one roll are 1/6 therefore the odds to score anything else but a 6 are 5/6.
Since the dice rolling is an event independent of previous or next roll, then the odds to score anything but 6 in 5 consecutive events is ${(\frac{5}{6})}^5$


问:
Principal Value of a divergent integral over $\mathfrak{R}$
I am trying to solve/prove this Principal Value integral
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x=\  -\ i\pi a^2\cdot\cos\left(a\right)\ ,$$
where $a$ takes a finite value.

The way I try to prove this is by noticing there pole is a first order pole at $x=a$, then by calculating the Residue at this value I find that
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x =-i\pi \cdot Res\left(\dfrac{x^2\cdot \sin\left(x\right)}{x-a},a\right)= $$

$$ =-i\pi\cdot\lim_{x\rightarrow a}\left(x^2 \cdot \sin\left(x\right)\right) = -i\pi a^2 \cdot {\sin(a)}\ ,$$
which of course is not correct.

Hence, does anyone knows what I am doing wrong, and if yes could please explain it to me?

NOTE: Just to mention that i end up to this question because a made use of Green's function for quantum many-body system. So if anyone have a clue about this they might be able to help.

Thanks in advance.

*******   EDIT:

I tried to keep it short, but the truth is that I need to prove this
$$-\dfrac{1}{a^3} \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^3\cdot F(x)}{x-a}\dfrac{{\rm{d}}x}{2\pi} = G(a)\ ,$$
where

$$ F(x) = \frac{3}{2}\left[(1-y)\cdot \frac{\sin(x)}{x} + (1-3y)\cdot\left( \frac{cos(x)}{{x^2}} - \frac{sin(x)}{x^3}\right) \right]$$
and
$$ G(a) = \frac{3}{4}\left[-(1-y)\cdot \frac{\cos(a)}{a} + (1-3y)\cdot\left( \frac{sin(a)}{{a^2}} + \frac{cos(a)}{a^3}\right) \right],$$
where $y=const$.

Thanks again for the effort.

答:
According to Maple, the Cauchy principal value integral from $-R$ to $R$ (where $0 \le a < R$) is

$$a^{2} \mathrm{Si}\! \left(R -a \right) \cos \! \left(a \right)+a^{2} \mathrm{Ci}\! \left(R -a \right) \sin \! \left(a \right)+a^{2} \mathrm{Si}\! \left(a +R \right) \cos \! \left(a \right)-a^{2} \sin \! \left(a \right) \mathrm{Ci}\! \left(a +R \right)-2 R \cos \! \left(R \right)+2 \sin \! \left(R \right)
$$

and as $R \to \infty$ this is

$$ -2 R \cos(R) + \pi a^2 \cos(a) + 2 \sin(R) + O(1/R) $$


问:
Number Theory-Find all natural numbers n such that
Find all positive integers n, such that
\begin{equation}
22^n+8^n-11^n-3^n \equiv 0\ (\textrm{mod}\ 209)
\end{equation}
My attempt so far:
\begin{equation}
11^n(2^n-1)\equiv 0\ (\textrm{mod}\ 209)  \Leftrightarrow\ (2^n-1)\equiv 0\ (\textrm{mod}\ 19)
\end{equation}
But from Euler theorem \begin{equation} \Rightarrow\ 2^{18}\equiv 1\ (\textrm{mod}\ 19) \Rightarrow\ 2^{180}\equiv 1\ (\textrm{mod}\ 19).
\end{equation}
In the same fashion we get that \begin{equation}
8^{180}-3^{180}\equiv 0\ (\textrm{mod}\ 209).
\end{equation}
Finally, \begin{equation} \Rightarrow\ 11^{180}(2^{180}-1)+8^{180}-3^{180}\equiv 0\ (\textrm{mod}\ 209).
\end{equation}
But maybe there are even more solutions, so how can I find them?

答:
I'm not sure why you are partitioning the left side, as you are eliminating solutions. All we want is the expression to be equivalent to $0$ mod $11$ and mod $19$.

For mod $11$, we have $$8^n\equiv3^n\bmod{11}\to(-3)^n\equiv3^n\bmod{11}$$So the solutions to this mod equation are when $n$ is even.

For mod $19$, we have $$3^n+8^n-11^n-3^n\equiv0\bmod{19}\to8^n\equiv(-8)^n\bmod{19}$$Which again is true for all even $n$ (and only for even $n$).

So, the system is true for any even $n$.


问:
What's the Fourier transform of these functions?

The Fourier transform of $|x|^{\alpha}$.

This is the Fourier transform of a homogeneous function, and there are several cases of various $\alpha$: when $a\leq -n$, it's not a temperate distribution; when $-n<\alpha<0$,then the Fourier transform is $c_{n}|\xi|^{-n+\alpha}$, where $c_{n}$ is some constant; when $\alpha=2k$,a positive even number, then it's Fourier transformation is $(-\Delta)^{k}\delta_{0}$.

My question is when $\alpha$ is any positive number (not the even case), then what's the Fourier transform of it ?

The Fourier transform of $e^{it|x|}$ ?

(the Fourier transforms I have mentioned here are in the sense of temperate distributions)

答:
${1.} $ The corresponding operator is denoted $(-\Delta)^{\alpha/2}$ and is called fractional Laplacian. Basically the Fourier transform of $|x|^\alpha$ "should be" a homogeneous function $|\xi|^{-n-\alpha}$. Since it is not locally summable it is not a kernel of an integral operator. But it defines an elliptic linear integro-differential operator of order $\alpha$. For $0<\alpha<2$
$$
(-\Delta)^{\alpha/2} u(x) = -c_{n,\alpha} \int_{\mathbb{R}^n}\frac{u(x-y)-2u(x)+u(x+y) }{|y|^{n+\alpha}}dy.
$$

One-dimensional analogue can be this. Function $|x|^{-1}$  is not locally summable and thus doesn't define a regular distribution. But it can be used to define  a distribution as
$$
({\cal P}\frac1{|x|},\varphi)=
\int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|}\,dx+
\int_{|x|> 1}\frac{\varphi(x)}{|x|}\,dx.
$$


问:
Pullback of constant sheaf of abelian groups is isomorphic to another constant sheaf
This is all coming from Example 8 in Lecture 2 of Pramod Achar's lecture notes on perverse sheaves.

Let $f: X \to Y$ be a continuous map of any two arbitrary topological spaces, and let $A$ be any abelian group.
In general, for $\mathcal F$ a sheaf on $Y$, the pull-back $f^{-1}\mathcal F$ is defined to be the sheafification of the presheaf $\text{ps} f^{-1} \mathcal F$, where for $U \subset X$ open, $$ \text{ps} f^{-1} \mathcal F(U) = \{ \text{equivalence classes of pairs }(V,s)\text{ where }V \supset f(U) \text{ and } s \in \mathcal F (V) \},$$
where the equivalence relation is the same as that for stalks.

The constant sheaf $\underline{A}_X$ is defined by $\underline{A}_X(U) = \{\text{locally constant } g: U \to A\}$ for $U \subset X$ open, and a similar definition for $\underline{A}_Y$.
I want to show the following:

There is an isomorphism of sheaves, $f^{-1} \underline{A}_Y \cong \underline{A}_X$.

Since the stalks of a presheaf and its sheafification are isomorphic, I know that to show the above statement it suffices to show that the stalks of $\underline{A}_X$ and $\text{ps} f^{-1} \underline{A}_Y$ are isomorphic, where the stalks are given by $$(\underline{A}_X)_p = \{[U,s] : p \in U, s: U \to A \text{ locally constant}\}$$ and $$(\text{ps} f^{-1} \underline{A}_Y)_p = \{[U, [V,t]] : Y \supset V \supset f(U), U \ni p, t: V \to A \text{ locally constant}\}.$$

To that end, I want to define a morphism of presheaves $F : \text{ps} f^{-1} \underline{A}_Y \to \underline{A}_X$.
For $U \subset X$ open, $$ \text{ps} f^{-1} \underline{A}_Y (U) = \{ [V,s] : V \supset f(U), s: V \to A \text{ locally constant}\}, $$ and $$ \underline{A}_X(U) = \{g: U \to A \text{ locally constant}\}.$$
Define $F([V,s]) = s \circ f$, which induces a homomorphism $F_p$ on the stalks given by $$F_p[U, [V,t]] = [U, F_U[V,t]] = [U, t \circ f].$$
Then all I need to do is check that this is in fact an isomorphism, which does not seem so hard.

My question is: is the outline I've given above all correct for showing the statement? There are a lot of definitions involved at the very beginning of learning about sheaves, and I want to check that my understanding is correct.
Alternate explanations of the statement would also be helpful - I am familiar with the definition of co/limits and how they relate to the pullback sheaf, but am not so clear on proving things e.g. the above with those definitions.

答:
I'll give an alternate proof which also proves some nice and useful facts about constant sheaves.

Lemma 1 : For any space $X$, let $a_X:X\to\{*\}$ be the unique map to the one point set. Then $\underline{A}_X\simeq a_X^{-1}\underline{A}_{\{*\}}$.

Of course, your assertion immediately implies the above lemma. But it is also equivalent. Indeed : let $f:X\to Y$ be any continuous map, then we obviously have $a_Y\circ f=a_X$, so $f^{-1}\underline{A}_Y\simeq f^{-1}a_Y^{-1}\underline{A}_{\{*\}}\simeq (a_Y\circ f)^{-1}\underline{A}_{\{*\}}=a_X^{-1}\underline{A}_{\{*\}}\simeq\underline{A}_X$.

Now, let us prove the lemma 1. By definition $a_X^{-1}\underline{A}_{\{*\}}$ is the sheafification of the presheaf $\operatorname{ps}a_X^{-1}\underline{A}_{\{*\}}$ which is obviously (recall the definition of $\operatorname{ps}a_X^{-1}$) the constant presheaf $U\mapsto A$ with all restriction maps being the identity. To conclude, it suffices to prove :

Lemma 2 : The constant sheaf $\underline{A}_X$ is the sheafification of the constant presheaf $\underline A^p_X : U\mapsto A$.

To prove this, consider the map $\varphi:\underline A^p_X\to \underline A_X$ such that for any $U\subset X$, $\varphi_U:\underline A^p_X(U)\to \underline A_X(U)$ is the map which sends $a\in \underline A^p_X(U)=A$ to the constant function $f:U\to A$ with value $a$. The lemma 2 is proved if we can show that this induces an isomorphism on stalks. But obviously $(\underline A^p_X)_x=A$ and $(\underline{A}_X)_x$ is also isomorphic to $A$ via the map which send a locally constant function $f$ defined in a neighborhood of $x$ to $f(x)$. And clearly, $\varphi_x$ is the identity with the above identification.


答:
This integral is improper both at $a$ and at $\pm \infty$.  We can attempt a principal value using
$$
\lim_{\epsilon \to 0^+}\left(\int_{a+\epsilon}^{a+1} f(x)\;dx
+ \int_{a-1}^{a-\epsilon} f(x)\;dx\right)
+
\lim_{R \to +\infty}\left(\int_{a+1}^R f(x)\;dx
+ \int_{-R}^{a-1} f(x)\;dx\right)
$$
the first limit exists.  The second one diverges by oscillation.
So I conclude this principal value does not exist.

Title: Principal Value of a divergent integral over $\mathfrak{R}$

Body:
I am trying to solve/prove this Principal Value integral
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x=\  -\ i\pi a^2\cdot\cos\left(a\right)\ ,$$
where $a$ takes a finite value.

The way I try to prove this is by noticing there pole is a first order pole at $x=a$, then by calculating the Residue at this value I find that
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x =-i\pi \cdot Res\left(\dfrac{x^2\cdot \sin\left(x\right)}{x-a},a\right)= $$

$$ =-i\pi\cdot\lim_{x\rightarrow a}\left(x^2 \cdot \sin\left(x\right)\right) = -i\pi a^2 \cdot {\sin(a)}\ ,$$
which of course is not correct.

Hence, does anyone knows what I am doing wrong, and if yes could please explain it to me?

NOTE: Just to mention that i end up to this question because a made use of Green's function for quantum many-body system. So if anyone have a clue about this they might be able to help.

Thanks in advance.

*******   EDIT:

I tried to keep it short, but the truth is that I need to prove this
$$-\dfrac{1}{a^3} \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^3\cdot F(x)}{x-a}\dfrac{{\rm{d}}x}{2\pi} = G(a)\ ,$$
where

$$ F(x) = \frac{3}{2}\left[(1-y)\cdot \frac{\sin(x)}{x} + (1-3y)\cdot\left( \frac{cos(x)}{{x^2}} - \frac{sin(x)}{x^3}\right) \right]$$
and
$$ G(a) = \frac{3}{4}\left[-(1-y)\cdot \frac{\cos(a)}{a} + (1-3y)\cdot\left( \frac{sin(a)}{{a^2}} + \frac{cos(a)}{a^3}\right) \right],$$
where $y=const$.

Thanks again for the effort.


问:
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$
Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.

答:
We can rewrite $x^2+y^2+1 \geq xy+x+y \ $ as

$2x^2+2y^2+2 - 2xy - 2x - 2y \geq 0$

or as $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$

which holds for all $x, y \in \mathbb{R}$

Or start from $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$ and expand to show that $x^2+y^2+1 \geq xy+x+y \ $.


问:
For holomorphic function $f(z) = \sum_{n=0}^\infty a_n z^n$, $|a_n|r^n \le 2 \max_{|z|=r} \text{Re}f(z)-2 \text{Re} f(0)$.
Let $f(z) = \sum_{n=0}^\infty a_n z^n \in H(B(0,R))$, for $r\in(0,R)$, let $A(r) = \max_{|z|=r} \text{Re}f(z)$.

Prove that $|a_n|r^n \le 2A(r)-2 \text{Re} f(0)$.

We have $a_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i}\int_{|\xi|=r} \frac{f(\xi)}{\xi^{n+1}}\, d\xi = \frac{1}{2\pi r^n}\int_{0}^{2\pi}f(re^{i\theta})e^{-in\theta}d\theta$, and

$\int_0^{2\pi} \overline{f(re^{i\theta})}e^{-in\theta} d\theta = 0$, so $a_nr^n = \frac{1}{\pi}\int_{0}^{2\pi}\text{Re} f(re^{i\theta}) \ e^{-in\theta} d\theta$.

How can we get that $\text{Re} f(0)$?

答:
Recall that for every $k\in\mathbb{C}, m\in\mathbb{Z}-\{0\}$ we have $$\int_0^{2\pi}ke^{im\vartheta}=0$$
So
$$ a_nr^n=\frac1\pi \int_0^{2\pi}\Re(f)(e^{i\vartheta})e^{-in\vartheta}d\vartheta=\frac1\pi\int_0^{2\pi}\left(\Re(f)(e^{i\vartheta})-\Re(f)(0)\right)e^{-in\vartheta}d\vartheta\\
|a_nr^n|\le 2|A(r)-\Re(f(0))|$$

As the right hand side must be positive by the maximum modulus principle for harmonic functions, the result follows:
$$|a_n|r^n\le 2A(r)-2\Re(f(0))$$


问:
Conjugates of a partition $n$
Is there any easiest way finding of conjugate of a partition $n$ (Except using Ferrers diagram)?

e.g I can find the conjugate of a partition $a=[4,4,1]$ using Ferrers diagram, and I obtain $a^*=[3,2,2,2]$

\begin{equation}
\begin{split}
&****    \ \ \    &***\\
&****   \to   \ &*\,*\\
&*       \ \     &*\,*\\
&             &*\,*
\end{split}
\end{equation}

答:
Given a partition $(\lambda_1,\lambda_2,\dots)$, then
$$
(\lambda^*)_i=\# \{j:\lambda_j\ge i\}
$$
In your example,

There are $3$ entries of $[4,4,1]$ which are $1$ or more,

There are $2$ entries of $[4,4,1]$ which are $2$ or more,

There are $2$ entries of $[4,4,1]$ which are $3$ or more,

There are $2$ entries of $[4,4,1]$ which are $4$ or more,

so $a^*=[3,2,2,2]$.

Title: Conjugates of a partition $n$


答:
I am sure that this has been asked before but there is a truly beautiful proof (folklore, maybe Littlewood?) that goes like this - assume $f$ nonconstant

Using $g(z)=f(z/r)$ we can assume $r=1$ as nothing else changes and then we take $h(z)=f(z)-a_0$ so with $U=A(1), \alpha=\Re f(0), V=U-\alpha$ we have $\Re h \le V, h(0)=0$ and $h$ is analytic on the closed unit disc (hence in a smaller disc - continuity on the boundary would also have been enough by a limiting process).

Then $p(z)=\frac{h(z)}{h(z)-2V}$ self maps the unit disc into itself and $p(0)=0$ so $|p'(0)| \le 1$ by Schwarz. Since $p'(0)=h'(0)/(2V)=f'(0)/(2V)$ we get the inequality $|a_1| \le 2V=2(A(1)-\Re f(0)$ as required.

But now for $n \ge 2$ let $f_n(z)=\frac{f(z)+f(\omega z)+...f(\omega^{n-1}z)}{n}$ where $\omega^n=1$ is a nontrivial root of unity; it is easy to see that $f_n$ still satisfies the inequalities of $f$ and it is a function of $z^n$ so defining $g_n(z)=f_n(z^n)$ still nothing changes but now $g_n=a_0+a_nz+...$ so applying the previous case we get the result for $a_n$ too!

Title: For holomorphic function $f(z) = \sum_{n=0}^\infty a_n z^n$, $|a_n|r^n \le 2 \max_{|z|=r} \text{Re}f(z)-2 \text{Re} f(0)$.

Body:
Let $f(z) = \sum_{n=0}^\infty a_n z^n \in H(B(0,R))$, for $r\in(0,R)$, let $A(r) = \max_{|z|=r} \text{Re}f(z)$.

Prove that $|a_n|r^n \le 2A(r)-2 \text{Re} f(0)$.

We have $a_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i}\int_{|\xi|=r} \frac{f(\xi)}{\xi^{n+1}}\, d\xi = \frac{1}{2\pi r^n}\int_{0}^{2\pi}f(re^{i\theta})e^{-in\theta}d\theta$, and

$\int_0^{2\pi} \overline{f(re^{i\theta})}e^{-in\theta} d\theta = 0$, so $a_nr^n = \frac{1}{\pi}\int_{0}^{2\pi}\text{Re} f(re^{i\theta}) \ e^{-in\theta} d\theta$.

How can we get that $\text{Re} f(0)$?


问:
Axiomatic proof in modal system B
I am trying to prove the statement $\Box(P \rightarrow \Box P) \rightarrow (\Diamond P \rightarrow P)$ within B

I already checked that this statement is valid semantically in B. And by completeness there has to be an axiomatic proof in B using the Axioms K,T,B,mp,nec.

I also have the following derived rules as well

$\phi \rightarrow \Box \Diamond \phi$.

$\phi \rightarrow \Diamond \phi$

$\Box(\phi \rightarrow \psi) \rightarrow (\Diamond \phi \rightarrow \Diamond \phi)$

And of course trivial deductive moves from propositional logic of course and tautologies relating negation of $\Diamond$ and $\Box$ to each other.

I am just wondering if there is a simple proof within B for this system using the above axioms and derived rules, or if I am missing some derived rules that could make the proof easier?

I've tried proving the contrapositive and then simplifying using equivalences of negation and $\Diamond$ and $\Box$, but I couldn't eliminate all of the negations.

答:
Here is the proof of (K $\Diamond$) gives in logic for philosophy by Sider
\begin{align}
1.&~~(\phi\to\psi)\to(\lnot\psi\to\lnot\phi)\tag*{PL(contraposition)}\\
2.&~~\square(\phi\to\psi)\to\square(\lnot\psi\to\lnot\phi)\tag*{1,NEC,K,MP}\\
3.&~~\square(\lnot\psi\to\lnot\phi)\to(\square\lnot\psi\to\square\lnot\phi)\tag*{K}\\
4.&~~\square(\phi\to\psi)\to(\square\lnot\psi\to\square\lnot\phi)\tag*{2,3,PL (syllogism)}\\
5.&~~\square(\phi\to\psi)\to(\lnot\square\lnot\phi\to\lnot\square\lnot\psi)\tag*{4,PL (contraposition)}
\end{align}

which shows $\square(\phi\to\psi)\to(\Diamond\phi\to\Diamond\psi)$, with this result, the statement in the question can be easily proved:

\begin{align}
1.&~~\square(P\to\square P)\to(\Diamond P\to\Diamond\square P)\tag*{K $\Diamond$}\\
2.&~~\Diamond\square P\to P\tag*{B}\\
3.&~~\square(P\to\square P)\to(\Diamond P\to P)\tag*{1,2, PL}
\end{align}

Title: Axiomatic proof in modal system B


答:
In my answer here, I gave a proof of $(\varphi\land \lozenge\psi)\rightarrow \lozenge(\psi\land \lozenge \varphi)$ in system B.

If we take $\varphi = \lnot P$ and $\psi = P$, we get $(\lnot P\land \lozenge P)\rightarrow \lozenge(P\land \lozenge \lnot P)$. The contrapositive of this is $\square(\lnot P \lor \square P)\rightarrow (P\lor \lnot \lozenge P)$. And replacing disjunctions by implications, this is equivalent to $\square(P\rightarrow \square P)\rightarrow (\lozenge P\rightarrow P)$, which is what you wanted to prove.

Note that the linked proof also involves passing to the contrapositive and turning negated implications into conjunctions, and we had to reverse those steps to get the statement you wanted to prove. I'll leave it as an exercise for you to give a more direct proof. (Edit: Actually, it looks like Manx has done just that.)

Title: Axiomatic proof in modal system B

Body:
I am trying to prove the statement $\Box(P \rightarrow \Box P) \rightarrow (\Diamond P \rightarrow P)$ within B

I already checked that this statement is valid semantically in B. And by completeness there has to be an axiomatic proof in B using the Axioms K,T,B,mp,nec.

I also have the following derived rules as well

$\phi \rightarrow \Box \Diamond \phi$.

$\phi \rightarrow \Diamond \phi$

$\Box(\phi \rightarrow \psi) \rightarrow (\Diamond \phi \rightarrow \Diamond \phi)$

And of course trivial deductive moves from propositional logic of course and tautologies relating negation of $\Diamond$ and $\Box$ to each other.

I am just wondering if there is a simple proof within B for this system using the above axioms and derived rules, or if I am missing some derived rules that could make the proof easier?

I've tried proving the contrapositive and then simplifying using equivalences of negation and $\Diamond$ and $\Box$, but I couldn't eliminate all of the negations.


问:
Determine possible coordinates for point $P$ on the terminal arm of angle
a) If angle $\theta\\$ lies in Quadrant II and
$\sin \theta ={3 \over {\sqrt {45} }}$. Determine possible coordinates for point $P$ on the terminal arm of angle $\theta$.

b) Determine the Quadrant in which angle $\theta$ is located if $\cos\theta < 0$ and $\tan \theta > 0$.

How would I do such a problem?

答:
For part a, $\sin\theta=\frac{3}{\sqrt{45}}=\frac{3}{3\sqrt{5}}=\frac{1}{\sqrt{5}}$ and since we know that the sine of an angle is the opposite over the hypotenuse we can imagine a triangle whose hypotenuse is $\sqrt{5}$ and whose opposite side is $1$. The Pythagorean theorem tells us that the adjacent side is then $\sqrt{\sqrt{5}^2-1^2}=\sqrt{5-1}=2$.

Since we are given that the point must be in the second quadrant, this gives a possible point of $(-2,1)$. Note that any positive multiple of this will also work, e.g. $(-4,2)$.

For part b, we use the fact that the cosine function is negative in the second and third quadrants and the tangent is positive in the first and third quadrants. Therefore, $\theta$ must be in the third quadrant for both conditions to be satisfied.


答:
Consider $f(x,y)=x^2+y^2+1-(xy+x+y)$ notice that
$\frac{\partial{f}}{\partial{x}}=2x-y+1$ and $\frac{\partial{f}}{\partial{y}}=2y-x+1$
then $\nabla f=0$ in $A=(-1,-1)$.

The function is convex since $f^{\prime \prime}(x,y)>0 $ hence the function have a minimum value in $1$ and then

$$f(x,y) \geq 1 >0 $$ and hence $$f(x,y)\geq 0$$

Title: Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$

Body:
Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.


问:
Show that $f$ is continuous in $\mathbb{R}$
Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(x+cy)=f(x)+cf(y), \forall \,x,y \in \mathbb{R},\,\forall \,c \in \mathbb{R}$. Show that $f$ is continuous.

My try:

We need to show that $\lim_{x \rightarrow a}f(x)=f(a), \forall a\in\mathbb{R}$.
At first, I tried to show that $f$ is continuous at $0$ and from there I would show for all $a\in\mathbb{R}$. I already managed to prove the second part assuming that $f$ is continuous at $0$, but I can't see how to prove the first part.  I only got that $f(0)=0$. I'm very confused, I appreciate any help!

答: "f is continuous in R" means that, for every a in r, $\lim_{x\to a} f(x)= f(a)$.  Let u= x- a so that becomes $\lim_{u\to 0} f(a+ u)= f(a)$.  If $f(a+ u)= f(a)+ f(u)$  that reduces to $\lim_{u\to 0} f(a)+ f(u)= f(a)+ \lim_{u\to 0} f(u)= f(a)$ so $\lim_{u\to 0} f(u)= 0$.


答:
First Solution

By AM-GM inequality we have
$$
\frac{x^2+y^2}{2} \geq |xy| \geq xy \\
\frac{x^2+1}{2} \geq |x| \geq x \\
\frac{1+y^2}{2} \geq |y| \geq y \\
$$

Add them together.

Second Solution: By Cauchy-Schwarz we have
$$
\left( xy+1\cdot x+y\cdot 1 \right)^2 \leq ( x^2+1^2+y^2)(y^2+x^2+1^2)
$$


问:
Lagrange polynomial interpolation - question about computation from 4 points
I have these sample data points (0,1), (1,4), (2,11), (3,22), and I have to compute the corresponding lagrange polynomial and express it in the form $p(x)= a_nx^n + a_{n-1}x^{n-1}...+a_1x + a_0$

My approach: third order Lagrange polynomial: $ f_3(x)=L_1f(x_1)+L_2f(x_2)+L_3f(x_3)+L_4+f(x_4) $

So to compute $L_1$ I say:

$L_1 = \frac{(x-x2)(x-x3)(x-x4)}{(x1-x2)(x1-x3)(x1-x4)}$

(and for L2, L3, L3, I use same formula with the respective x-values)

This give me a rather "messy" third order polynomial, which I can't see how translate to $p(x)= a_nx^n + a_{n-1}x^{n-1}...+a_1x + a_0$

So, I'm in doubt wether this is this the right approach at all?

答: Substitute the numerical values of $x_i$ and $f(x_i)$ and expand the polynomials in the numerators of $L$'s. Now you have four cubic polynomials, so simply add the coefficients at corresponding powers of $x$.


问:
Find Cross-Tangents to Offset Ellipses
Question:

I am stuying projective geometry and would like to find an analytic solution to my problem. This is a follow-up to my older question.

Given two ellipses with the same major and minor axes, but different origins (written in homogeneous coordinates):

$$
\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1 = 0
$$

and

$$
\frac{(x_2-h)^2}{a^2} + \frac{(y_2-k)^2}{b^2} - 1 = 0
$$

what are the equations of the cross-tangents to the two ellipses? Assume $h, k, a,$ and $b$ are all knowns.

My Attempt:

Following the method from my older question, it makes sense to find the intersection points of the dual curves to each ellipse. Let $f_1$ be the implicit homogeneous function defining our first ellipse (i.e. $f_1(x_1, y_1, z_1) = 0$) and $f_2$ that of our second, respectively. The dual curve of an ellipse is an ellipse (at least, AFAIK, for an ellipse at the origin), which Wikipedia shows nicely.

Hence, our first dual is

$$
a^2 X_1^2 + b^2 Y_1^2 = 1.
$$

I am having trouble finding the second dual (namely, that of the ellipse centered at $(h,k)$). Could someone please help me in deriving this? Is this the simplest way to solve this problem?

Following Wikipedia, about the closest I get to is the following and I don't know how to eliminate $\lambda$:

$$
\frac{a^2}{2\lambda} \left( X_2 + \frac{\lambda h}{a^2} \right)^2 + \frac{b^2}{2\lambda} \left( Y_2 + \frac{\lambda k}{b^2} \right)^2 - \left( \frac{\lambda}{2} \right) \left( \left( \frac{h}{a} \right)^2 + \left( \frac{k}{b} \right)^2 \right)
 + 1 = 0$$

答:
Hint:

Rescale using $au:=x,bv:=y$ and you get two circles,

$$u^2+v^2=1,\\(u-u_c)^2+(v-v_c)^2=1.$$

Rotating around the origin, you can reduce the second to

$$(u-d)^2+v^2=1.$$

The cross-tangents pass through $(\dfrac d2,0)$ and by $(\cos\theta,\sin\theta)$, with

$$\cos\theta\,(\cos\theta-\frac d2)+\sin^2\theta=0.$$

Write the equations of these lines and reverse the transformations.

Title: Find Cross-Tangents to Offset Ellipses


问:
Understanding morphisms of affine algebraic varieties
In class, we defined an affine algebraic variety to be a $k$-ringed space $(V, \mathcal{O}_V)$ where $V$ is an algebraic set in $\bar{k}^n$ defined by a system of polynomial equations over $k$, and the sheaf of regular functions $\mathcal{O}_V$ that assigns an open subset of $V$ to the set of regular functions on that open set.

We then defined a morphism of affine algebraic varieties $(V, \mathcal{O}_V)$, $(W, \mathcal{O}_W)$ to be given by a morphism of the underlying ringed spaces, i.e. a continuous map (which I'm assuming means continuous with respect to the Zariski topology) $\phi: V \to W$, together with a family of $k$-algebra homomorphisms $\phi_U : \mathcal{O}_W(U) \to \mathcal{O}_V(\phi^{-1} (U))$.

This definition is very abstract and I'm having a bit of a hard time wrapping my head around it. However, we were given some equivalent definitions of this, as well as the following correspondence:

For affine algebraic varieties $(V, \mathcal{O}_V)$, $(W, \mathcal{O}_W)$, there is a $1-1$ correspondence between the morphisms of ringed spaces between $(V, \mathcal{O}_V)$, $(W, \mathcal{O}_W)$, and the $k-$algebra homomorphisms $k[W] \to k[V]$, where $k[W], k[V]$ are the coordinate rings of $W, V$ respectively (so $k[W] = k[x_1 , \dots , x_n]/I(W)$).

So, just to make sure I am understanding this correctly. Let's say I am defining a morphism of the affine algebraic varieties of the circle $x^2 + y^2 = 1$ over $\mathbb{C}$ to the hyperbola $xy = 1$. Does the correspondence stated above mean that it is enough to just write a ring homomorphism $\phi : \mathbb{C}[x,y]/(x^2 + y^2 - 1) \to \mathbb{C}[x,y]/(xy-1)$?

答: More or less correct, the thing to note is however, that the functor from affine varieties to rings is contravariant. Meaning that the correspondence is arrow reversing. Intuitively this corresponds to the pullback of functions.</p>


答:
One approach to this problem could be to look at the differences:

$d_1(1)=f(2)-f(1)\\d_1(2)=f(3)-f(2)\\d_1(3)=f(4)-f(3)$

etc. If these differences are constant, then you know you have a linear polynomial, and you can use that constant difference and one of the values (e.g. $f(0)$) to find the equation of the line.

If the differences aren't constant, you can look at the second differences:

$d_2(1)=d_1(2)-d_1(1)\\d_2(2)=d_1(3)-d_1(2)\\d_2(3)=d_1(4)-d_1(3)$

etc.  If these differences are constant, it turns out that the polynomial is quadratic.  (For example, if the polynomial is $f(x)=x^2$, the first differences are 1, 3, 5, 7,..., while the second differences are constantly 2.)

This pattern continues: cubic polynomials have constant third differences, degree four polynomials have constant fourth differences, and so on.

Once you know the degree of your polynomial, you can use various methods to find the coefficients.  You could plug in your points and set up a system of equations; you could compare to the values of $x^d$ where $d$ is the degree you've settled on and see how your function compares; or if you've worked out in general how these successive differences come out, you might have some ideas about how to use the differences to reconstruct the coefficients.

Title: Lagrange polynomial interpolation - question about computation from 4 points

Body:
I have these sample data points (0,1), (1,4), (2,11), (3,22), and I have to compute the corresponding lagrange polynomial and express it in the form $p(x)= a_nx^n + a_{n-1}x^{n-1}...+a_1x + a_0$

My approach: third order Lagrange polynomial: $ f_3(x)=L_1f(x_1)+L_2f(x_2)+L_3f(x_3)+L_4+f(x_4) $

So to compute $L_1$ I say:

$L_1 = \frac{(x-x2)(x-x3)(x-x4)}{(x1-x2)(x1-x3)(x1-x4)}$

(and for L2, L3, L3, I use same formula with the respective x-values)

This give me a rather "messy" third order polynomial, which I can't see how translate to $p(x)= a_nx^n + a_{n-1}x^{n-1}...+a_1x + a_0$

So, I'm in doubt wether this is this the right approach at all?


答:
If you look at the inequality as a quadratic inequality with respect to the variable $x$, then we have

$$x^2+y^2+1 \geq xy+x+y$$

$$\implies x^2+y^2+1 -xy-x-y≥0$$

$$\implies x^2-x(y+1)+(y^2-y+1)≥0$$

$$\implies \left( x -\frac{y+1}{2}\right)^2-\left(\frac{y+1}{2}\right)^2+y^2-y+1≥0$$

$$\implies \left( x -\frac{y+1}{2}\right)^2+ \frac 34 (y - 1)^2≥0.$$

Equality occurs if and only if, when $y=1$ and $x=\dfrac{y+1}{2}=\dfrac 22=1.$


问:
Definition clarification for $k$-colourable graphs
If a graph is $k$-colourable, may we say that it is $k + z$-colourable for any arbitrary positive integer $z$?

Would a graph with $1$ vertex be considered $2$ colourable?

答: The smallest number of colors needed to color the vertices of a graph $G$ is called its chromatic number, $\chi(G) =k$.  And yes, such a graph is $k+z$-colorable, for $z \in \mathbb{N}$.


问:
Can you recommend some books on elliptic function?
I plan to study elliptic function.  Can you recommend  some  books? What is the relationship between elliptic function and elliptic curve？Many thanks in advance!</p>

答:
McKean and Moll have written the nice book Elliptic Curves: Function Theory, Geometry, Arithmetic that cleanly illustrates the connection between elliptic curves and elliptic/modular functions. If you haven't seen the book already, you should.

As for elliptic functions proper, my suggested books tend to be a bit on the old side, so pardon me if I don't know the newer treatments. Anyway, I quite liked Lawden's Elliptic Functions and Applications and Akhiezer's Elements of the Theory of Elliptic Functions. An oldie but goodie is Greenhill's classic, The Applications of Elliptic Functions; the notation is a bit antiquated, but I have yet to see another book that has a wide discussion of the applications of elliptic functions to physical problems.

At one time... every young mathematician was familiar with $\mathrm{sn}\,u$, $\mathrm{cn}\,u$, and $\mathrm{dn}\,u$, and algebraic identities between these functions figured in every examination.

– E.H. Neville

Finally, I would be remiss if I did not mention the venerable Abramowitz and Stegun, and the successor work, the DLMF. The chapters on the Jacobi and Weierstrass elliptic functions give a nice overview of the most useful identities, and also point to other fine work on the subject.


问:
Why do we get $\int_0^{\frac{\pi }{2}} \left\lbrack K^2 {\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)}^n -...\right]$
I have this equivalency:
$$\int_0^{\frac{\pi }{2}} \left(K^2 -12\right){\left(12\;{\mathrm{cos}}^2t+K^2 {\mathrm{sin}}^2 t\right)}^n \;{\mathrm{cos}}^2 t\;\mathrm{dt}=\int_0^{\frac{\pi }{2}} \left\lbrack K^2 {\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)}^n -{\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)}^{n+1}\right\rbrack\mathrm{dt}$$
where $K$and $n$ are constants.

How is the first integral transformed into the second one?

Thanks!

Body:
$$K^2 {\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)}^n -{\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)}^{n+1}\\
=\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)^n[K^2-(K^2\sin^2t+12\cos^2t)]\\
=\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)^n[K^2(1-\sin^2t)-12\cos^2t]\\
=\left(K^2 {\mathrm{sin}}^2 t+12\;{\mathrm{cos}}^2 t\right)^n(K^2-12)\cos^2t$$


问:
How to prove that this function is continuous
I was working on proving that $S^1$\{(1,0)} with the induced topology by $R^2$ ( product topology) is homeomorphic to the interval ]0,2π[..the homeomorphism is the function f(x)=(Cos(x),Sin(x)).. I already proved the bijection but I'm stuck proving the continuity .. Knowing that a function is continuous if the inverse image of every open set is open.
Can anyone help me.

答: The product topology is the smallest topology such that any function at values in the product is continuous if and only if all its coordinates are continuous. Here you only need to check that sin and cosine are continuous.</p>


答:
Cosine and sine are continuous, hence $f$ is continuous.

To show that $f^{-1}$ is continuous, it suffices to show that for each $(x,y)\in C:=S^1\setminus\{(1,0)\}$, there is a neighbourhood of $z$ where $f^{-1}$ is continuous:

In the upper half plane, $f^{-1}$ is given by $(x,y)\mapsto \frac\pi2-\arctan\frac xy$

In the lower half plane, $f^{-1}$ is given by $(x,y)\mapsto \frac{3\pi}2+\arctan\frac xy$

In the left half plane,  $f^{-1}$ is given by $(x,y)\mapsto \pi+\arctan\frac yx$

Title: How to prove that this function is continuous

Body:
I was working on proving that $S^1$\{(1,0)} with the induced topology by $R^2$ ( product topology) is homeomorphic to the interval ]0,2π[..the homeomorphism is the function f(x)=(Cos(x),Sin(x)).. I already proved the bijection but I'm stuck proving the continuity .. Knowing that a function is continuous if the inverse image of every open set is open.
Can anyone help me.


问:
$\mathcal{A}:= \left\{\bigcup_{i \in I} A_i : I \subseteq \mathbb{N} \right\}$ sigma-algebra?
Let $\Omega$ be a set $(A_n)_{n \in \mathbb{N}}$ a sequence of subsets of $\Omega$ such that $A_m \cap A_n = \emptyset$ whenever $m \neq n$ and $\bigcup_{n\in \mathbb{N}}A_n = \Omega$. Consider the family $\mathcal{A}$ of subsets of $\Omega$ defined by:

$\mathcal{A}:= \left\{\bigcup_{i \in I} A_i : I \subseteq \mathbb{N} \right\}$.

To show: $\mathcal{A}$ is a $\sigma$-algebra.

My attempt:

Choose $I = \mathbb{N}$ then $\bigcup_{i \in \mathbb{N}} A_i \in \mathcal{A}$ and by assumption $\bigcup_{i \in \mathbb{N}} A_i = \Omega$

Let $A\in \mathcal{A}$ then $A=\bigcup_{i \in \lambda} A_i $ for $\lambda \subset \mathbb{N}$. Now $A^c = (\bigcup_{i \in \lambda} A_i)^c = \bigcup_{j\in \mathbb{N}\setminus \lambda}A_j $. This is true because  $A_m \cap A_n = \emptyset$ whenever $m \neq n$. But since ($\mathbb{N} \setminus \lambda) \subset \mathbb{N} \implies A^c \in \mathcal{A}$

Let $A_{I_n} \in \mathcal{A}$. Then $A_{I_n} = \bigcup_{i \in I_n}A_i.$ But then $\bigcup_{n \in \mathbb{N}} A_{I_n} = \bigcup_{i \in \mathbb{N}}A_i$ which is by assumption in $\mathcal{A}$.

Is that correct?

答:
Since your proposed sigma algebra consists of the various unions of a specific list of pairwise disjoint sets, it is obvious anyway that if you take a countable union of those it will be in your proposed sigma algebra, merely because that proposed algebra was already defined to be the collection of all subsets of the power set of the basic collection of pairwise disjoint sets.

It still seems there might be a simpler way to say the above; the more I look at it my version seems overly complicated.


问:
Inverse Fourier transform of a function generated from a compactly supported function
Suppose $u$ belongs to a Sobolev space $H^s$ ($s\ge 0$).  Assume $u$ has compact support.  Does $f$ which is defined by $f=\mathcal{F}^{-1}(\hat{u}(\xi^2))$ have also compact support? Or more generally, $f=\mathcal{F}^{-1}(p(\xi)\hat{u}(\xi^2))$ where $p(\xi)$ is polynomial or rational.

答:
For $u\ne 0\in L^2$ supported on $[-r,r]$ then its Fourier transform $$\hat{u}(z)=\int_\Bbb{R} u(t)e^{-i z x}dx$$ is entire and $u(z)=O(e^{r|\Im(z)|})$ as $|z|\to \infty$.

$\hat{u}(z^2)$ is $L^2(\Bbb{R})$ and entire but it isn't $O(e^{R|\Im(z)|})$ for any $R$, whence $\mathcal{F}^{-1}(\hat{u}(z^2))$ is $L^2$ but not compactly supported.


答:
Just to give yet another approach, observe that if we let $x=u+v$ and $y=u-v$, then

$$\begin{align}
x^2+y^2+1\ge xy+x+y
&\iff(u+v)^2+(u-v)^2+1\ge(u+v)(u-v)+(u+v)+(u-v)\\
&\iff2u^2+2v^2+1\ge u^2-v^2+2u\\
&\iff u^2-2u+1+3v^2\ge0\\
&\iff(u-1)^2+3v^2\ge0
\end{align}$$


问:
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$?
I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.

Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.

Can you suggest any ideas how to calculate $S$?

答:
Write down the function
$$ g(z) = \sum_{n\geq1} \frac{z^n}{n}H_n^2, $$
so that $S=g(-1)$ and $g$ can be reduced to
$$ zg'(z) = \sum_{n\geq1} z^n H_n^2 = h(z). $$

Now, using $H_n = H_{n-1} + \frac1n$ ($n\geq2$), we can get a closed form for $h(z)$:
$$h(z) = z + \sum_{n\geq2}\frac{z^n}{n^2} + \sum_{n\geq 2}z^n H_{n-1}^2 + \sum_{n\geq 2} 2\frac{z^n}{n}H_{n-1}. $$
Now, the first and third sums Mathematica can evaluate itself in closed form (the third one evaluates to the function $p(z)$ below, the first one is $\text{Li}_2(z)-z$), and the middle sum is $z h(z)$.

Substituting this into the expression for $g(z)$, we get
$$g(z) = \int \frac{\text{Li}_2(z) + p(z)}{z(1-z)}\,dz, $$
$$p(z) = -\frac{\pi^2}{3} + 2\log^2(1-z)-2\log(1-z)\log(z)+2\text{Li}_2((1-z)^{-1}) - 2\text{Li}_2(z). $$
Mathematica can also evaluate this integral, giving (up to a constant of integration)
\begin{align}
g(z) &= \frac{1}{3} \left(-2 \log(1-z^3+3 \log(1-z)^2 \log(-z)+\log(-1+z)^2 (\log(-1+z)+3 \log(-z) \right. \\
& \hspace{5mm} \left. -3 \log(z))+\pi ^2 (\log(-z)-2 \log(z))+\log(1-z) \left(\pi^2 - 3 \log(-1+z)^2 \right. \right.\\
& \hspace{5mm} \left.\left. +6 (\log(-1+z)-\log(-z)) \log(z)\right)-6 (\log(-1+z)-\log(z)) \left(\text{Li}_{2}\left(\frac{1}{1-z}\right)-\text{Li}_{2}(z)\right) \right.\\
& \hspace{10mm} \left. -3 \log(1-z) \text{Li}_{2}(z)+3 \text{Li}_{3}(z)\right).
\end{align}
The constant of integration is fixed by requiring $g(0)=0$.

Some care needs to be taken, because the function
is multi-valued, when evaluating $g(-1)$. The answer is
$$ \frac{1}{12}(\pi^2\log2-4(\log 2)^3-9\zeta(3)). $$


问:
Find the residue of the following number
Find the residue  :
\begin{equation}
1^{965}+2^{965}+...+2018^{965}
 \equiv x\ (\textrm{mod}\ 2021), x=?
\end{equation}
My attempt so far:
\begin{equation}
\varphi(965)=\varphi(5)\varphi(193)=768.
\end{equation}
But from Euler theorem applied for all numbers coprime with both 5 and 193 we have:\begin{equation} \Rightarrow\ 2^{768}\equiv 1\ (\textrm{mod}\ 2021) \Rightarrow\ 2^{965}\equiv 2^{197}\ (\textrm{mod}\ 2021).
\end{equation}
In the same fashion we get that \begin{equation}
3^{965}\equiv 3^{197}\ (\textrm{mod}\ 2021).
\end{equation}
Unfortunatelly this does not seem to work. Any ideas how to start?

答:
I think with solving mod equivalences, people are afraid of "dirty solutions". But I think just bashing it out works just fine.

Note that $$x^{2p+1}+(-x)^{2p+1}=(x+-x)(x^{2p})=0$$So, the sum is equal to $$1^{965}+2^{965}\bmod{2021}\to1+2^{965}\bmod{2021}$$Note that $2^{11}=2048$, so $2^{11}\equiv3^3\bmod{2021}$

Can you solve from here?


问:
Standard error question. Why can't this be equal to 1?
So I've been working on this question related to standard error and I managed to show that it can't be less than $1$. But how do you show that this has to be greater than $1$ (as in cannot equal $1$)?

$$ \frac{\hat{\text{SE}}_1 + \hat{\text{SE}}_2}{\sqrt{\hat{\text{SE}}_1^2 + \hat{\text{SE}}_2^2}} $$

答: This is merely an algebra question. You may take $a:=\hat{\operatorname{SE}}_1>0$ and $b:=\hat{\operatorname{SE}}_2>0$. Then this expression becomes $$\frac{a+b}{\sqrt{a^2+b^2}}=\sqrt{\frac{(a+b)^2}{a^2+b^2}}=\sqrt{1+\frac{2ab}{a^2+b^2}}>1.$$


问:
Proof identity of differential equation
I would appreciate if somebody could help me with the following problem:

Q: $f''(x)$ continuous in $\mathbb{R}$ show that
$$ \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$

答:
Let $N_{\delta}(c)=(c-\delta,c+\delta)$ be a neighbourhood of $c$ where$f''$ is continuous.Then for any $0<h<\delta$  we have from Taylor's Theorem with $R_2$ , the remainder in Lagrange's form

$f(c+h)=f(c)+hf'(c)+\frac{h^2}{2!}f''(c+\theta h)$  , ($0<\theta<1$  )

$f(c-h)=f(c)-hf'(c)+\frac{h^2}{2!}f''(c-\theta 'h)$ , ($0<\theta '<1$  )

and $\displaystyle\lim_{h\to 0}f''(c+\theta h)=f''(c)$ and $\displaystyle\lim_{h\to 0}f''(c-\theta ' h)=f''(c)$.

Using these results you can easily carry out the calculation I think.

Title: Proof identity of differential equation


问:
Suppose $p$ is prime and $m \gt 0$ and $n \gt 0$ and $m \equiv n \pmod{p-1}$. Prove that $a^m \equiv a^n \pmod{p}$ for all $a$.

Suppose $p$ is prime and $m \gt 0$ and $n \gt 0$ and $m \equiv n \pmod{p-1}$. Prove that $a^m \equiv a^n \pmod{p}$ for all $a$.

I know I have to apply Fermat's little theorem to prove this, but I don't know where to start.

答:
If $a \equiv 0 \pmod{n}$, then automatically $a^m \equiv a^n \pmod{n}$, so let's consider the case when $a \not\equiv 0 \pmod{n}$, i.e., $a$ is not evenly divisible by $p$. Without loss of generality, we can also assume that $m \leq n$.

By using the property that if $a_1 \equiv b_1 \pmod{c}$ and $a_2 \equiv b_2 \pmod{c}$ then $a_1\, a_2 \equiv b_1\, b_2 \pmod{c}$, we can use Fermat's little theorem to multiply any modular arithmetic equation with $a$ on one side and with $a^p$ on the other side.

Since $m \equiv n \pmod{p-1}$, we know that

$$
d = \frac{n-m}{p-1}
$$

is an non-negative integer.

Starting from $a^0 \equiv a^0 \pmod{n}$, we need to create a difference of $n-m$ in the exponent between the LHS and the RHS. Let's multiply the LHS by $a$ and the RHS by $a^p$ for a total of $d$ times (if $d = 0$, do nothing); this should give us the desired difference in exponent. Doing this gives

$$
a^{d} \equiv a^{pd} \pmod{n}\qquad\text{(*)}.
$$

If $d < m$, we need to increase the exponent on both sides. So, for $d \leq m$, if we multiply both sides of this equation by $a^{m - d}$, we get

$$
a^{d}a^{m - d} \equiv a^{pd}a^{m - d} \pmod{n}.
$$

Now, $a^{d}a^{m - d} = a^m$, and $a^{pd}a^{m - d} = a^{p(n-m)/(p-1)}a^{m - (n-m)/(p-1)} = a^{(p-1)(n-m)/(p-1) + m} = a^{n}$, which gives us

$$
a^m \equiv a^n \pmod{n},
$$

as expected. On the other hand, if $d > m$, we would like to divide both sides by $a^{d-m}$ to decrease the exponent. There is a property that if $k\,a \equiv k\,b \pmod{c}$ and $k$ is coprime with $c$, then $a \equiv b \pmod{c}$. Since $a \not\equiv 0 \pmod{p}$, we know that $a$ and $p$ are coprime, so we can decrease the exponent on both sides in (*) by dividing by $a$. Doing this $d-m$ times yields

$$
a^{d}/a^{d-m} \equiv a^{pd}/a^{d-m} \pmod{n},
$$

which after substituting $(n-m)/(p-1)$ for $d$ also gives

$$
a^m \equiv a^n \pmod{n}.
$$


问:
Simplify a sum with a product and multinomial coefficient
This is a follow-up question to my previous post, where I've got a great help! (I created a new one to avoid editing the original one).
Can the following sum be simplified?

$$
 \sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} \left(\prod_{i=1}^{m}k_i\right) \cdot {B \choose k_1, k_2, \ldots k_m} \cdot {W \choose g_1, g_2, \ldots, g_m}
$$
where $B + W = n$, and $\forall_i: g_i = x - k_i$, where $x = \frac{n}{m}$, and $m\mid n$ and $k_i \geq 1$

I tried to follow the steps suggested in my previous post, but I got stuck. For $m=2$ I've got the following
$$
\sum_{\substack{k_1 + k_2 &=& B \\ k_1, k_2 &\geq& 0}} k_1 \cdot k_2 \cdot {B \choose k_1, k_2} \cdot {2N - B \choose N-k_1, N-k_2} \\ = \sum_{\substack{k_1 + k_2 &=& B \\ k_1, k_2 &\geq& 0}} k_2 \cdot {B \choose k_1-1} \cdot {2N - B \choose N-k_1, N-k_2} \\
= \sum_{k_1 = 0}^B \left(B-k_1 \right) \cdot {B \choose k_1-1} \cdot {2N - B \choose N-k_1} \\
$$
From the previous post I know how to handle the binomial coefficients, but I'm not sure what can I do now with the $B-k_1$ factor inside the sum?

答:
We assume positive integers $m,n\geq 2$. Since $m\mid n$ we know that $n=Nm$ is a multiple of $m$ with $N\geq 1$. This way we can write $W=Nm-B$ and $x=N$. We show by induction of $m$ that for all $N\geq 1, m\leq B\leq Nm$ the following is valid:

\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_m=B\\k_1,\ldots,k_m\geq 0}}
\binom{B}{k_1,\ldots,k_m}\binom{mN-B}{N-k_1,\ldots,N-k_m}=B^{\underline{m}}\prod_{q=2}^m\binom{q(N-1)}{N-1}}\tag{1}
\end{align*}

with $B^{\underline{m}}=B(B-1)\cdots(B-m+1)$ the falling factorial.

Base step: $m=2$

We obtain for natural numbers $B,N$ with $2\leq B\leq  2N$:
\begin{align*}
\color{blue}{\sum_{\substack{k_1+k_2=B\\k_1,k_2\geq 0}}}&\color{blue}{k_1k_2\binom{B}{k_1,k_2}\binom{2N-B}{N-k_1,N-k_2}}\\
&=\sum_{k_1=0}^{B}k_1\left(B-k_1\right)\binom{B}{k_1}\binom{2N-B}{N-k_1}\tag{1}\\
&=B\sum_{k_1=1}^{B-1}\left(B-k_1\right)\binom{B-1}{B-k_1}\binom{2N-B}{N-k_1}\tag{2}\\
&=B(B-1)\sum_{k_1=1}^{B-1}\binom{B-2}{k_1-1}\binom{2N-B}{N-k_1}\tag{3}\\
&=B^{\underline{2}}\sum_{k_1=0}^{B-2}\binom{B-2}{k_1}[z^{N-k_1-1}](1+z)^{2N-B}\tag{4}\\
&=B^{\underline{2}}[z^{N-1}](1+z)^{2N-B}(1+z)^{B-2}\tag{5}\\
&\,\,\color{blue}{=B^{\underline{2}}\binom{2(N-1)}{N-1}}\tag{6}
\end{align*}

Comment:

In (1) we write the multinomial as binomial coefficients, substituting $k_2=B-k_1$.

In (2) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{p-q}$ and we skip the indices $k_1=0$ and $k_1=B$ which do not contribute to the sum.

In (3) we use again $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}=\frac{p}{q}\binom{p-1}{p-q}$ as in (2) and (3).

In (4) we shift the index $k_1$ by one to start with $k_1=0$. We use the coefficient of operator $[z^N]$ to denote the coefficient of $z^N$ of a series.

In (5) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and we apply the binomial theorem.

In (6) we select the coefficient of $z^{N-1}$.

Induction hypothesis: $m=M$

We assume the validity of
\begin{align*}
\color{blue}{\sum_{\substack{k_1+\cdots+k_M=B\\k_1,\ldots,k_M\geq 0}}
\binom{B}{k_1,\ldots,k_M}\binom{MN-B}{N-k_1,\ldots,N-k_M}=B^{\underline{M}}\prod_{q=2}^M\binom{q(N-1)}{N-1}}\tag{7}
\end{align*}

Induction step: $m=M+1$

We have to show
\begin{align*}
&\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}
\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,N-k_{M+1}}}\\
&\qquad\qquad\color{blue}{=B^{\underline{{M+1}}}\prod_{q=2}^{M+1}\binom{q(N-1)}{N-1}}\tag{8}
\end{align*}

We obtain for $M\geq 2$:
\begin{align*}
&\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}
\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,N-k_{M+1}}}\\
&\qquad=\sum_{k_{M+1}=1}^{B-1}k_{M+1}\frac{B!}{k_{M+1}!(B-k_{M+1})!}\,\frac{((M+1)N-B)!}{(N-k_{M+1})!(MN-(B-k_{M+1}))!}\\
&\qquad\qquad\cdot\sum_{\substack{k_1+\cdots+k_{M}=B-k_{M+1}\\k_1,\ldots,k_{M}\geq 0}}
\binom{B-k_{M+1}}{k_1,\ldots,k_{M}}\binom{MN-(B-k_{M+1})}{N-k_1,\ldots,N-k_{M}}\tag{9}\\
&=B\sum_{k_{M+1}=1}^{B-1}\binom{B-1}{B-k_{M+1}}\binom{(M+1)N-B}{N-k_{M+1}}\\
&\qquad\qquad\cdot\left(B-k_{M+1}\right)^{\underline{M}}\prod_{q=2}^M\binom{q(N-1)}{q}\tag{10}\\
&=B\sum_{k_{M+1}=1}^{B-M+1}\frac{(B-1)^{\underline{M}}}{(B-k_{M+1})^{\underline{M}}}\binom{B-M-1}{B-k_{M+1}-M+1}\binom{(M+1)N-B}{N-k_{M+1}}\\
&\qquad\qquad\cdot\left(B-k_{M+1}\right)^{\underline{M}}\prod_{q=2}^M\binom{q(N-1)}{q}\\
&=B^{\underline{M+1}}\prod_{q=2}^M\binom{q(N-1)}{N-1}
\sum_{k_{M+1}=0}^{B-M}\binom{B-M-1}{k_{M+1}}\binom{(M+1)N-B}{N-k_{M+1}-1}\\
&=B^{\underline{M+1}}\prod_{q=2}^M\binom{q(N-1)}{N-1}
\sum_{k_{M+1}=0}^{B-M}\binom{B-M-1}{k_{M+1}}[z^{N-k_{M+1}-1}](1+z)^{(M+1)N-B}\\
&=B^{\underline{M+1}}\prod_{q=2}^M\binom{q(N-1)}{N-1}
[z^{N-1}](1+z)^{(M+1)N-B}(1+z)^{B-M-1}\\
&=B^{\underline{M+1}}\prod_{q=2}^M\binom{q(N-1)}{N-1}\binom{(M+1)N-M-1}{N-1}\\
&\,\,\color{blue}{=B^{\underline{M+1}}\prod_{q=2}^{M+1}\binom{q(N-1)}{N-1}}\\
\end{align*}
in accordance with (8) and the claim follows.

Comment:

In (9) we separate $k_{M+1}$ and expand to get binomial coefficients in the next step.

In (10) we apply the induction hypothesis.

Note: We obtain from (1):
\begin{align*}
\color{blue}{B^{\underline{m}}\prod_{q=2}^m\binom{q(N-1)}{N-1}}
&=B^{\underline{m}}\cdot\frac{\prod_{q=2}^m\left(q(N-1)\right)!}{\prod_{q=2}^m(N-1)!\prod_{q=2}^m\left((q-1)(N-1)\right)!}\\
&=B^{\underline{m}}\cdot\frac{\prod_{q=2}^m\left(q(N-1)\right)!}{\left((N-1)!\right)^{m-1}\prod_{q=1}^{m-1}\left(q(N-1)\right)!}\\
&=B^{\underline{m}}\cdot\frac{\left(m(N-1)\right)!}{\left((N-1)!\right)^{m}}\\
&\,\,\color{blue}{=\binom{B}{m}\frac{m!\left(m(N-1)\right)!}{\left((N-1)!\right)^{m}}}
\end{align*}
in accordance with the answer from @GCab.


问:
Efficient algorithm to determine if two lattice elements have an edge between them?
Suppose I have a bounded lattice $L$, a subset $E\subset L$ of elements that are both meet-prime and join-prime, and every $x\in L$ can be represented as a finite expression of $\lor$, $\land$, and elements of $E$. $L$ is guaranteed to not be atomic, complemented, or finite.

If I have $\leq_E:E\times E\rightarrow\mathbb{2}$ to test whether $x\leq y$ for $x,y\in E$, is there an efficient algorithm to extend $\leq_E$ to all of $L$? I have an algorithm that works, but it has a combinatorial expansion that seems like it might have exponential (or worse) complexity in the worst case. (The computational cost of $\leq_E$ varies with $x$ and $y$, but I'm ignoring that for now.)

The current algorithm recursively expands the expressions on both sides to extend $\leq_E$ to $L$. It's straightforward except when dealing with $\land$ on the left and $\lor$ on the right. Only the first matching rule below is used. (I've transposed this from a Haskell function that uses pattern matching.)

Expression Type
Expansion

$\mathbf{x\leq_L y}$ where $x,y\in E$
$x\leq_E y$

$\mathbf{\bigwedge X\leq_L \bigvee Y}$
$\bigvee_{x\in X\cap E}\bigvee_{y\in Y\cap E}\left[x\leq_E y\right]~\lor \\ \bigvee_{x\in X\setminus E}\left[x\leq_L \bigvee Y\right]~\lor \\ \bigvee_{y\in Y\setminus E}\left[\bigwedge X\leq_L y\right]$

$\mathbf{\bigvee X\leq_L y}$ (any $y$)
$\bigwedge_{x\in X}\left[x\leq_L y\right]$

$\mathbf{x\leq_L \bigwedge Y}$ (any $x$)
$\bigwedge_{y\in Y}\left[x\leq_L y\right]$

$\mathbf{\bigwedge X\leq_L y}$ ($y\in E$ by elimination)
$\bigvee_{x\in X}\left[x\leq_L y\right]$

$\mathbf{x\leq_L \bigvee Y}$ ($x\in E$ by elimination)
$\bigvee_{y\in Y}\left[x\leq_L y\right]$

I also preprocess the expressions to cut down on complexity:

Each $\bigvee$ contains only distinct $\bigwedge$ and distinct elements of $E$.

Each $\bigwedge$ contains only distinct $\bigvee$ and distinct elements of $E$.

As far as I know, $\bigwedge X\leq_L \bigvee Y$ is the only case where combinatorial expansion is needed, but I'm not sure if the expansion above is overdoing it. It almost certainly duplicates work in some cases, but I'm not sure if I can transparently cache $\leq_L$, since I'm using a purely functional language.

Overall, this algorithm has been fine in practice, since the expressions are hand-written (they're ephemeral types in a programming language) and are meant to be comprehensible to humans. A hand-written expression with enough nesting to create a combinatorial explosion would be highly unusual, but they might come up in the automatic type inference the type system uses.

答:
After examining the algorithm a bit more, I think it's at worst cubic rather than exponential:

Without the second expansion, the algorithm is quadratic, because at worst it compares every element in the left expression to every element in the right expression.

The second expansion doubles the work done each time it's invoked, provided the left and right have the same number of nested expressions.

With expressions of size $n$, you can have no more than $\lceil\mathrm{log}_2 n\rceil$ levels of nesting in the expression. This means that the comparison count will be multiplied by at most $2^{\lceil\mathrm{log}_2 n\rceil}\in[n,2n)$ in the second expansion.

(Actually, with expression preprocessing, there should be at most $\left\lceil\frac{1}{2}\mathrm{log}_2 n\right\rceil$ levels invoking the second expansion, making it $O\left(n^{2.5}\right)$.)

I did some deterministic empirical tests and it looks like it's between  $\mathbf{O\left(n^{2.5}\right)}$ and $\mathbf{O\left(n^3\right)}$:

Nesting ($k$)
Expression Size ($n=2^k$)
$\leq_E$ Calls

0
1
1

1
2
4

2
4
32

3
8
128

4
16
1536

5
32
6144

6
64
81920

7
128
327680

8
256
4587520

9
512
18350080

I also did some fine-grained sampling over $n\in[1,256]$:

For $\lceil \mathrm{log}_2 n\rceil$ levels of nesting (binary tree with $\left\lceil \frac{1}{2}\mathrm{log}_2 n\right\rceil$ levels invoking the second expansion), the complexity with the best $r^2$ is $O\left(n^{2.589}\right)$.

For no nesting, the number of $\leq_E$ calls is exactly $n^2$, as expected.

So, it looks like the algorithm is already reasonable for checking expressions that are hand-written, since even something with 5 alternating levels of nesting would be a huge pain to both write and interpret. (Keep in mind that preprocessing will only preserve nestings that alternate between $\bigvee$ and $\bigwedge$; everything else will be flattened.)


问:
How find the number of $z$,such that$ |a^2-b^2-b+1|\le 10$
let $f:C\longrightarrow C$ and $f(z)=z^2+zi+1$,find the number of $z$,such $Im(z)>0$
and $|Im(f(z))|\le 10,|Re(f(z))|\le 10$,and $ Im(f(z)) ,Re(f(z)) $are integer numbers.
where $i=\sqrt{-1}$,

$A:399,   B:401,  C:413,   D:431,   E:441$

my idea:let $z=a+bi,b>0$, then
$$f(z)=z^2+iz+1=(a+bi)^2+i(a+bi)+1=(a^2-b^2-b+1)+(2ab+a)i$$

so:
$|a^2-b^2-b+1|\le 10,|2ab+a|\le 10$,and
$a^2-b^2-b+1\in Z,2ab+a\in Z,b>0$
then this problem equalitly find the number pairs$(a,b)$? But I can't find .

Thank you everyone,

答:
From these: $b>0$ and $|2ab+a|\leq 10$ it follows:
$$-10\leq 2ab+a\leq 10 \iff \frac{-10-a}{2}\leq ab\leq\frac{(10-a)}{2}. $$
Then you should take three cases: when $a>0$, $a=0$ or $a<0$. Just for an example of a needed pair let's take $a=0$. Then everything above holds and $|a^2-b^2-b+1|=|-b^2-b+1|$, so $|-b^2-b+1|\leq 10$,  for example $b=1,2$ and pairs $(a,b)=(0,1), (0,2)$. But you can find more pairs by evaluating inequalities.</p>


问:
7 passengers enter the train with 4 carriages numbered 1 to 4.
7 passengers enter the train with 4 carriages numbered 1 to 4. Each passengers chooses
at random a carriage to go into, independently of other passengers.

a) Compute the probability that in one of the carriages there are 4 passengers and that
there is exactly 1 passenger in each of the remaining three carriages.

b) Compute the probability that there is at least one passenger in each carriage.

My rough workings:
a) P(one of the carriages has four passengers and exactly 1 passenger in the remaining 3 carriages) = ((7C4)*4!)/4^7

答:
For $(b)$, please apply Principle of Inclusion-Exclusion which is there in the other answer that user mentioned in comments. In this case as the numbers are small, you can also simply count as below -

As each carriage must have at least one passenger, possibilities are permutations of $\{4 \ 1 \ 1 \ 1\}, \{3 \ 2 \ 1 \ 1\}, \{2 \ 2 \ 2 \ 1\}$

$\displaystyle {7 \choose 4} \cdot 4! + {7 \choose 3}  {4 \choose 2} \cdot 4! + {7 \choose 1}{4 \choose 1} \frac{6!}{2!2!2!} = 8400 \ $ favorable arrangements.

Explanation: First term is what we discussed for $(a)$ in comments. Second term is where we choose $3$ and $2$ passengers out of $7$and then arrange them in carriages. Last term is where we choose $1$ passenger out of $7$ first and place in one of the carriages. Then rest $6$ passengers can be arranged in rest $3$ carriages.

Now divide it by $\displaystyle 4^7$ which is total number of arrangements.


问:
Understanding the Likelihood
I am trying to understand the meaning of the Likelihood function or the value of the Likelihood itself. Let us suppose we are tossing coins and we have a sample data given as

$$\hat{x} = \{H,H,T,T,T,T\}$$

For instance in this case the Likelihood Function can be given as

$$L(\theta|\hat{x}) = C(6,2)\theta^2(1-\theta)^4 = 15\theta^2(1-\theta)^4$$

The function looks like this

Now, my question is what does this graph tells us ? We can say that for $\theta = 0.333$, $L(\theta|\hat{x}) $ is maximum.

In general it seems to me that the Likelihood is way of obtaining  the best parameters that describes distribution the given data. If we change the parameters that describe the distribution, we change the Likelihood.

We know that for fair coin $\theta_{True} = 0.5$. So my question is does the Likelihood means for this data ($\hat{x}$), the best parameter that describes the fairness of the coin is $0.3333$ (i.e when the Likelihood is maximum) ? We know that if the coin is truly fair, for a large sample we will obtain the true $\theta$ parameter (i.e $N \rightarrow \infty, \theta \rightarrow \theta_{True}$).

In general, I am trying to understand the meaning of the likelihood. Does the Likelihood is a way to obtain the best $\theta$ parameter that describes the distribution for the given data. But, depending on the given data the $\theta$ may not be the true $\theta$ (i.e., $\theta \neq \theta_{True}$ for the given data) .

(For a Gaussian $θ = {µ,σ}$, for a Poisson distribution, $θ = λ$ and for a binomial distribution, $θ = p$, the probability of success in one trial.)

答:
Generally speaking, the aim of statistical inference is to construct a model that best illustrates the property of some data. The true underlying model is generally unknown, or we would not have to do such job at all.

In parametric approach, the likelihood function $L(\theta\mid\vec{x})$ illustrates the probability of $\theta$ given data set $\vec{x}=(x_1,\ldots,x_n)$. For instance, in your settings, you assumed each toss $X_i\sim\operatorname{Bernoulli}(\theta)$. Based on this assumption, the likelihood function is constructed as follows: $$L(\theta\mid\vec{x})=\prod_{i=1}^{n}{\mathbb{P}(x_i\mid\theta)}=\theta^{x_1+\cdots+x_n}(1-\theta)^{n-(x_1+\cdots+x_n)}.$$ In your data set, the likelihood function should be $$L(\theta\mid\vec{x})=\theta^2(1-\theta)^4.$$ Then based on this likelihood function, we want to find the value of $\theta$, say $\hat{\theta}$ that maximizes it, which is precisely defined as the MLE of $\theta$. This principle follows from the intuition that the larger the $L(\theta\mid\vec{x})$ is, the most possible such $\theta$ fits the data. By some calculations (see details there), the MLE of $\theta$ is given by $$\hat{\theta}=\bar{x}=\frac{x_1+\cdots+x_n}{n}.$$ In your example, the value is $\hat{\theta}=2/6=1/3$. This means that based on the data set $\{H,H,T,T,T,T\}$ and the Bernoulli assumption, the most plausible estimate for $\theta$ is $1/3$. That is all about this inference procedure.

However, the actual parameter may not be $1/3$. Tossing a fair coin ($\theta=1/2$) can also yield this sequence. However, as you can see, in the likelihood function $L(\theta\mid\vec{x})=\theta^2(1-\theta)^4$, $$L(1/2\mid\vec{x})<L(1/3\mid\vec{x}).$$ We would naturally prefer $1/3$ to $1/2$, because we knew nothing about the true parameter in the inference. Then if the coin is fair, we can only say that the MLE is not accurate and that is all. In other words, the MLE is not always the best choice for us.

Let me demonstrate this again. The true parameter has nothing to do with this inference, so it makes no sense by plugging in the true parameter in the likelihood function and comparing it with other terms. If you know the true parameter, why bother inferring it?

That is why I said you were wrong: "If we change the parameters that describe the distribution, we change the Likelihood." Referring to the setup above, the likelihood function does not change unless

i) data set changes or

ii) the model assumption (i.e., the underlying distribution) is altered.

As for the function value of the likelihood function, we only care about its maximizer.

Title: Understanding the Likelihood


问:
Projection map sends closed sets to closed sets
I am reading Algebraic Geometry, a first course by Joe Harris. In the section on projections, he talks about an application of elimination theory to prove that image of a projection map $\pi: Y \times \mathbb{P}^1 \rightarrow Y$ is a closed set. I do not have strong background in Algebraic Geometry, so I do not understand elimination theory and resultants. Can someone explain what is the author trying to say in the highlighted part and why is it true? Basically, I am interested in understanding its proof.

Also, does the projection map send an open set $Z \subseteq Y \times \mathbb{P}^1$ to an open set $\pi(Z)$? If yes, why?

Thank you very much in advance!

答:
Suppose $k$ is a field and $f,g\in k[x,y]$ are homogeneous polynomials. Claim: the homogeneous resultant of $f,g$ is zero iff $V(f,g)\subset\Bbb P^1_k$ is nonempty.

Proof: The resultant is the determinant of the map $(A,B)\mapsto Af+Bg$ from $k[x,y]_{\deg g}\times k[x,y]_{\deg f} \to k[x,y]_{\deg f+\deg g}$. But the image of this map is exactly $(f,g)\cap k[x,y]_{\deg f+\deg g}$, so $(A,B)\mapsto Af+Bg$ not being surjective is equivalent to $(f,g)$ being contained in homogeneous ideal who's radical does not contain the irrelevant ideal of $k[x,y]$. Such ideals define nonempty subsets of $\Bbb P^1_k$, so by the nullstellensatz we see that $V(f,g)=V(f)\cap V(g)$ is nonempty as a subset of $\Bbb P^1_k$. $\blacksquare$

Now we note that the homogeneous resultant plays nicely with ring homomorphisms, since it's the determinant of a particular matrix: if $\varphi:R\to S$ is a ring homomorphism and $f,g\in R[x,y]$ are homogeneous, then $\varphi(Res(f,g))=Res(\varphi(f),\varphi(g))$. Taking $R=k[Y]$, $S$ to be the residue field of a point $y$, and $\varphi$ the evaluation map, we see that $Res(f,g)$ vanishes at $y$ iff $f,g$ have a common zero in the fiber over $y$. The highlighted result follows.

As for the question about projections of open sets, this has been addressed on MSE before here. The upshot is that for any scheme over a field, the map $X\to\operatorname{Spec} k$ is open, so any projection $X\times Y\to Y$ is open because such a projection is the base change of $X\to\operatorname{Spec} k$ along $Y\to\operatorname{Spec} k$. See Stacks 01TZ for a full proof. A slightly more elementary approach can be found in the comments.


问:
Minimisation problem obeying coupled inequalities
I have a set of 4 two sided inequalities and 3 variables in the form

$ A[0] \le +Z + X \le B[0]$

$ A[1] \le +Z - X \le B[1]$

$ A[2] \le -Z + Y \le B[2]$

$ A[3] \le -Z - Y \le B[3]$

Where all A[j] < B[j], Ideally I would like to find a solution that satisfies these equations while minimising

$S = a |Z| + (1-a) (|X|+|Y|)$

with $0 \le a \le 1$, but I would settle for solutions with just a = 0 and a = 1.  I'm not really sure where to start with this sort of problem, do I derive lots of sets of equations based on the signs elements in A and B?

答:
You can solve this via linear programming by introducing additional variables $X'$, $Y'$, and $Z'$ and constraints:
\begin{align}
X' &\ge X \\
X' &\ge -X \\
Y' &\ge Y \\
Y' &\ge -Y \\
Z' &\ge Z \\
Z' &\ge -Z \\
\end{align}
The new objective is to minimize $aZ'+(1-a)(X'+Y')$.


问:
Proof of the Existence of Integer Solutions for $(f_1)^2 + (f_2)^2 +( f_3)^2... = I^2$

Using fractions($f_n$) where integer $n$ is the number of fractions, prove that the sum of the squares $f_1$ to $f_n$ has no integer($I$) solutions when $n >=  1$ given each fraction has a distinct positive integer numerator and denominator and every fraction is a fully simplified non integer:

$$\left(f_1\right)^2 + \left(f_2\right)^2 + \left(f_3\right)^2 ...  =  I^2$$

I could not find a proof, but using distinct positive integers $a,b,c,d,e,f$ I managed to prove this for the case of
$$\left(\frac{a}{b}\right)^2 + \left(\frac{c}{d}\right) ^2 = \left(\frac{e}{f}\right)^2 $$

by simplifying:

$$\frac{(ad)^2 + (bc)^2}{(bd)^2}  = (\frac{e}{f})^2 $$

Since $(ad)^2 + (bc)^2$  must satisfy a Pythagorean triple, we know that each side length(excluding the hypotenuse) must be the product of two distinct integers, which can be proven through Euclid's formula :

$$ac = (2m)(n)$$
$$bd = (m -n)(m + n)$$

Q.E.D.

Although this still did not prove useful to answering my question, I later found a proof that the sum of the sqaures $f_1$ to $f_2$ has no integer solutions using distinct positive integers $a,b,c,d$ where $(\frac{a}{b})^2 +  (\frac{c}{d} ) ^2 = I^2 $   using the following:

$$\frac {a^2}{b^2} = \frac {I^2d^2 - c^2}{d^2}$$

Both sides are in their simplest form since
$$\gcd(a^2,b^2) = 1 = \gcd(c^2,d^2) = \gcd(I^2d^2 - c^2, d^2)$$

so $a^2 = e^2d^2-c^2, b^2 = d^2$.

Given $b,d > 0$, $b=d$. Therefore no solutions exists.

But this lead me to question what if there were more than $2$ integer fractions? This would mean that
the denominator of some integer fraction could be the product of $2$ or more distinct integers, which suggests that this proof which uses $b = d$ does not nesscacarily need to be true for there to be an integer solution where $n > 2$.

So I made a conjecture that there are no integer solutions for the sum of any number of squared fractions where the numerator and denominator are distinct positive integers and the fractions are fully simplified non-integers.

Edit:

This turned out to be false,

New questions:

Provide a proof

A method to finding integer solutions

(The one that finds the most affective method will be more likely to receive the bounty)

Body:
I did a brute force search in sage to find examples of the form
$$\left(\frac{x}{qr}\right)^2 + \left(\frac{y}{pr}\right)^2 + \left(\frac{z}{pq}\right)^2= n^2$$
with $p, q, r$ primes that are $1$ mod $4$. The condition of this being an integer more or less fixes $x$ and $y$ mod $r^2$ (because $x^2/q^2+y^2/p^2$ must be divisible by $r^2$, and there are only 8 interesting solutions to that). Similarly it fixes $y$ and $z$ mod $p^2$, and $x$ and $z$ mod $q^2$. This gives a system of congruences in $x, y, z$ mod $p^2q^2r^2$. Then brute force search through $p, q, r$ and solutions to the system of congruences by adding multiples of $p^2q^2r^2$ to $x, y, z$, and hope that it gives perfect squares. (I did not but could have separately added multiples of $q^2r^2$ to $x$, of $p^2r^2$ to $y$ and of $p^2q^2$ to $z$. I just translated all of them by $m \cdot p^2q^2r^2$ for simplicity.)

There are plenty of solutions.

The smallest $n$ I saw is $81$:

$(286448/4141)^2 + (1636/205)^2 + (20897/505)^2 = 81^2$

Other solutions (I didn't filter for duplicates)

(221622/481)^2 + (10777/185)^2 + (3186/65)^2 = 467^2

(15246/533)^2 + (28777/205)^2 + (3102/65)^2 = 151^2

(75528/533)^2 + (13248/205)^2 + (2898/65)^2 = 162^2

(11943/793)^2 + (25992/305)^2 + (2848/65)^2 = 97^2

(1188973/1469)^2 + (217148/565)^2 + (2764/65)^2 = 897^2

(332352/1937)^2 + (88104/745)^2 + (1039/65)^2 = 209^2

(1715638/2509)^2 + (819003/965)^2 + (3186/65)^2 = 1091^2

(5821714/2561)^2 + (39784/985)^2 + (2848/65)^2 = 2274^2

(1785867/2977)^2 + (1258284/1145)^2 + (3236/65)^2 = 1253^2

(231632/493)^2 + (941/145)^2 + (6976/85)^2 = 477^2

(151859/1513)^2 + (71094/445)^2 + (2526/85)^2 = 191^2

(498166/1717)^2 + (224522/505)^2 + (907/85)^2 = 531^2

(1723985/2669)^2 + (345746/785)^2 + (3682/85)^2 = 783^2

(231632/493)^2 + (6976/85)^2 + (941/145)^2 = 477^2

(650784/3277)^2 + (204864/565)^2 + (20084/145)^2 = 436^2

(221622/481)^2 + (3186/65)^2 + (10777/185)^2 = 467^2

(80758/4181)^2 + (216998/565)^2 + (8039/185)^2 = 387^2

(3610252/8473)^2 + (1258884/1145)^2 + (12261/185)^2 = 1181^2

(15246/533)^2 + (3102/65)^2 + (28777/205)^2 = 151^2

(75528/533)^2 + (2898/65)^2 + (13248/205)^2 = 162^2

And one where none of the primes is $5$:

(155476/1241)^2 + (164575/949)^2 + (33040/221)^2 = 261^2