问:
Determine if a transformation is self-adjoint with respect to a non-orthonormal basis
I have matrix that represent $T:V \to V$ (linear map over $\mathbb{R}$) according to basis $E$.

$E$ is not an orthonormal set.

how can I know if this $T$ is self-adjoint.

I know that if $E$ was orthonormal basis we would take the transpose matrix.

but what about here ?

the matrix is :
$$\begin{bmatrix}
1 & 2 \\
2 & 1 \\
\end{bmatrix}$$

答:
Without an orthonormal matrix I'm afraid you'll have to do it the long way: we're looking for a matrix $\,T^*\,$ s.t. (I'm assuming the usual, euclidean inner product):

$$\overline x:=\binom xy\,,\,\,\overline y=\binom ab\implies\langle T\, \overline x\,,\,\overline y\rangle=\langle \overline x\,,\,T^*\,\overline y\rangle\;\iff $$

$$\left\langle\,\begin{pmatrix}1&2\\2&1\end{pmatrix}\binom xy\,,\,\binom ab\,\right\rangle =\left\langle\,\binom xy\,,\,\begin{pmatrix}p&q\\r&s\end{pmatrix}\binom ab\,\right\rangle\iff$$

$$\left\langle\,\binom{x+2y}{2x+y}\,,\,\binom ab\,\right\rangle=\left\langle\,\binom xy\,,\,\binom{pa+qb}{ra+sb}\,\right\rangle\iff$$

$$ax+2ay+2bx+by=pax+qbx+ary+bsy\iff$$

$$(1-p)ax+(1-s)by+(2-r)ay+(2-q)bx=0$$

As the above has to be true for any two vectors $\,\overline x\,,\,\overline y\,$ , you can now choose particular cases and deduce the possible values of $\,p,q,r,s\,$ (which, of course, you then need to check in the general case), for example: with $\,x=a=1\,,\,y=b=0\,,\,$ :

$$(1-p)=0\implies p =1\;\;,\;\;etc.\ldots$$


问:
Visualizing complex roots of a quadratic equation
Is there some way to visualize or plot Complex roots of a quadratic equation ( of real coefficients):

$$ y= x^2 + v x + c =0 $$

whose product is constant? .. in a complex plane?

Shown are cases together with real/ coincident /double roots.

Here $v$ is variable and the constant product $c= 4. $

```<code>Plot[{x^2 + 4, x^2 - x + 4, x^2 - 2 x + 4, x^2 - 3 x + 4,
  x^2 - 4 x + 4, x^2 - 5 x + 4}, {x, -3, 6}, GridLines -> Automatic]
</code>```

答:
For $c>0$:

$$z=\frac{-v\pm\sqrt{v^2-4c}}2\text{ or }\frac{-v\pm i\sqrt{4c-v^2}}2.$$

In a $(v,x,y)$ representation, an hyperbola in the $(v,x)$ plane and an ellipse in the $(v,y)$ plane. In the $(x,y)$ plane, a circle of radius $2\sqrt c$ and the portions of the axis $x$ outside of it.

ContentLicense: CC BY-SA 4.0


问:
{x} = [x] - how many numbers?
{$x$} is like the fractional part: x - floor(x)

$[x]$ is the whole part or whatever it's called: floor(x)

答:
$$x - [x] = [x]$$
$$x = 2[x]$$

Since $[x]$ is an integer, then $x$ is also an integer. But then $x=[x]$, so you have
$$x = 2x$$

to which $x=0$ is the only solution.

Title: {x} = [x] - how many numbers?


问:
Show that matrix of $T$ has at least dim range $T$ nonzero entries.
Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V, W)$. Show that with respect to each choice of bases of $V$ and $W$, the matrix of $T$ has at least dim range $T$ nonzero entries.

I have no idea on how to solve this.

答: Suppose we have a basis {$v_{1},...,v_{n}$} of $V$ and {$w_{1},...,w_{m}$} of $W$, and so $M(T)$ is an $m$ by $n$ matrix. Let $p$ denote the number of zero-columns in the matrix. Thus, we know the dimension of the null space is at least $p$. And so by the fundamental theorem of linear maps, $n = \dim \operatorname{ran}(T)+\dim \operatorname{null}(T) \geq \dim \operatorname{ran}(T)+p$. Therefore, $n-p \geq \dim \operatorname{ran}(T)$, and we are done.


答:
You're right.

$[x]$ is an integer and $\{x\}$ is in the interval $[0,1)$. The only integer in that interval is $0$.

Thus, $\{x\} = [x]$ implies $\{x\}=0$ and so $x=0$.

Title: {x} = [x] - how many numbers?

Body:
{$x$} is like the fractional part: x - floor(x)

$[x]$ is the whole part or whatever it's called: floor(x)


问:
Whether an infinite series can be tested by integral test
I am asked whether the following infinite series can be proved to be convergent by integral test.

$$\sum_{n=1}^\infty n e^{6 n}$$

so I integrate it

$$\int_1^{\infty}\ n e^{6n}\, dn$$

and find it diverges so I concluded that the above series also diverges by the integral test. However, the answer is that the integral test cannot be used to test this infinite series. What is wrong in my deduction?

答:
Look carefully at the statement of the integral test. I suppose that $n \mapsto n e^{6n}$ should be decreasing, which is clearly false.

By the way, since $\lim_{n \to \infty} n e^{6n} = +\infty$, how can the series converge?

ContentLicense: CC BY-SA 3.0


答: Call the $x$-coordinate of the vertex $a$ and the roots $a\pm bi$.  From the point $(a,0)$ which is the perpendicular projection of the vertex onto the $x$-axis, draw a tangent to the parabola and drop a perpendicular from the tangent point back to the $x$-axis.  The distance between the two projection points you have constructed is $b$.

Title: Visualizing complex roots of a quadratic equation

Body:
Is there some way to visualize or plot Complex roots of a quadratic equation ( of real coefficients):

$$ y= x^2 + v x + c =0 $$

whose product is constant? .. in a complex plane?

Shown are cases together with real/ coincident /double roots.

Here $v$ is variable and the constant product $c= 4. $

```<code>Plot[{x^2 + 4, x^2 - x + 4, x^2 - 2 x + 4, x^2 - 3 x + 4,
  x^2 - 4 x + 4, x^2 - 5 x + 4}, {x, -3, 6}, GridLines -> Automatic]
</code>```


问:
Convergence of $\int_{0}^{\infty} x^{a-\frac{3}{2}}e^{-\frac{1}{2x}} \,dx$
I am trying to determine the values of $a$ for which the following integral converges:

$$\int_{0}^{\infty} x^{a-\frac{3}{2}}e^{-\frac{1}{2x}} \,dx.$$

My lecturer has stated that this converges for $a<\frac{1}{2}$ but I want to understand why. I did not get anywhere by substituting this value of $a$ into the integral.

Body:
Use $$\int_{0}^{\infty}z^c e^{-bz} dz=\frac{\Gamma(c+1)}{b^{c+1}}$$
Let $x=1/z \implies dx=-dz/z^2,$ then the given integral is
$$I=\int_{0}^{\infty} x^{a-1/2} e^{-1/(2x)}dx=\int_{0}^{\infty} z^{-a-3/2} e^{-z/2} dz= 2^{-a-1/2} \Gamma(-a-1/2).$$
As $\Gamma(s)$ is finite if $s \ne 0,-1,-2,-3,...$. So $I$ converges for $a \ne n-1/2$, if $n=1,2,3,...$


问:
Determine to point of maximum and minimum of a function
I am learning calculus by watching some videos that I have downloaded some time ago. Some of the videos are missing.I cannot find the part where it teaches point of maximum and minimum.   Along with the videos I have some questions. One of the questions is this

determine the local maximum and minimum points of the function
$$
g(x) = \frac{1}{3}x^3 - 3x^2 + 8x + 5
$$
Can you say how to solve one so I can solve the others.

答:
Hint: Find derivative and equate it to zero:

$g'(x)=x^2-6x+8=0$

Which gives $x_1=4$ and $x_2=2$

Now put these values in $g(x)$:

$g(4)=..?$

$g(2)=..?$


问:
Span of $(a x + b y )^n $ is exactly the space of $n$-dimensional homogeneous polynomials in $x$ and $y$
Actually, I need to prove that the Delta operator

$$ \Delta = \frac{\partial^2}{\partial x^2 } + \frac{\partial^2}{\partial y^2 } + \frac{\partial^2}{\partial z^2 } $$

maps the space $P_k$ ($k$-dimensional homogeneous polynomials in $x,y,z$) surjectively to $P_{k-2}$.

We have

$$ \Delta [ (ax + b y + c z )^k] = k(k-1)(a^2 + b^2 + c^2 )(ax + b y + c z )^{k-2} . $$

Now the proof would be done if $(ax + b y + c z )^{k-2}$ spans $P_{k-2}$.

It seems very reasonable. But I cannot figure out a proof. Could anyone help me?

答:
We will show that $$\{(x+ky)^n : 1 \le k \le n+1\}$$
spans the $n$-dimensional homogeneous polynomials in $x$ and $y$. Note that polynomials $$v_j := {n \choose j}x^{n-j}y^j, \quad 0 \le j \le n$$
form a basis for our space. We have
$$(x+ky)^n = \sum_{j=0}^n {n \choose j} k^j x^{n-j}y^j = \sum_{j=0}^n k^jv_j, \quad 1 \le k \le n+1$$

so these polynomials written as rows in the basis $\{v_j\}_{0\le j \le n}$ have coordinates
$$\begin{bmatrix}
1 & 1^2 & 1^3 & \cdots & 1^{n}\\
1 & 2^2 & 2^3 & \cdots & 2^{n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & (n+1)^2 & (n+1)^3 & \cdots & (n+1)^n
\end{bmatrix}$$
This is a Vandermonde matrix and hence invertible so we conclude that $\{(x+ky)^n : 1 \le k \le n+1\}$ is linearly independent. Since there are precisely $n+1$ vectors, which is the dimension of our space, we conclude that it is a basis. In particular, it spans the space.


答:
Hint:

Is the sequence $\,\{ne^{6n}\}_{n\in\Bbb N}\,$ monotone decreasing?

Title: Whether an infinite series can be tested by integral test

Body:
I am asked whether the following infinite series can be proved to be convergent by integral test.

$$\sum_{n=1}^\infty n e^{6 n}$$

so I integrate it

$$\int_1^{\infty}\ n e^{6n}\, dn$$

and find it diverges so I concluded that the above series also diverges by the integral test. However, the answer is that the integral test cannot be used to test this infinite series. What is wrong in my deduction?


问:
To prove that $(H,\cdot)$ is a subgroup of the group $(G,\cdot)$ it is sufficient to show that $H$ is closed with multiplication?
I have proved that the set of invertible triangle matrices form a group (elements in $\mathbb{R}$). I now want to show that the subset where all diagonal-entries are $1$ make a subgroup. And I was wondering whether I also have to prove that the inverse-elements of elements in this subset are also in the subset or if it is sufficient to show that the subset is closed under multiplication.

Is there a theorem that says if a subset is closed than it is also a subgroup? I.e. that the inverses are also in this subset?

答: You have to prove that the inverses are in there. For a simple example, note that both $\Bbb N$ and $\Bbb N\setminus\{0\}$ are additively closed subsets of $(\Bbb Z,+)$.


问:
How do you prove $ \sum_{n=1}^{\infty}{a_n} = \sum_{n=1}^{\infty}{a_{2n}} + \sum_{n=1}^{\infty}{a_{2n-1}} $
How do you prove that by adding two series, both convergent, one
$$ \sum_{n=1}^{\infty}{a_{2n}} $$

and the other

$$ \sum_{n=1}^{\infty}{a_{2n-1}} $$

You get

$$\sum_{n=1}^{\infty}{a_n}$$

Body:
$A_{2n}:=\sum_{k=1}^{n}a_{2k};$

$B_{2n-1}:=\sum_{k=1}^{n}a_{2k-1};$

$c_k:=a_{2k-1}+a_{2k},$ $k=1,2,...n;$

$C_n:=\sum_{k=1}^{n}c_k= B_{2n-1}+A_{2n};$

$\lim_{n}C_n=\lim_{n}(B_{2n-1}+A_{2n})=$

$\lim_{n}B_{2n-1}+\lim_{n}A_{2n}$, since both limits exist.


问:
Is Big-O closed under composition?
to show: if $f,g\in O(h)$ then $f(g)\in
O(h)$

Proof: let $h= n^2$ then $f(g)=(n^2)^2= n^4$ now its easy to see that $n^4\notin O(n^2)$ which implies the proposition is false this completes the proof

Is this correct?

答: You’ve got the right idea, but the notation is a little confused. Using your counterexample, let $g(x) = f(x) = x^2$. Trivially, $f, g \in O(h(x))$ where $h(x) = x^2$. Naturally, $f \circ g (x) = x^4$, but clearly $x^4 \notin O(x^2)$ (in fact the opposite is true), so this provides a sufficient counterexample.

Title: Is Big-O closed under composition?


问:
Circles $C_1,C_2$ have radii $3$ and $7$ respectively. Point $Q$ is chosen on $C_2$ such that $PQ$ is tangent at $Q$. Find $PQ$.

Circles $C_1,C_2$ have radii $3$ and $7$ respectively. The circles intersect at distinct $A$ and $B$. A point $P$ outside $C_2$ lies on the line determined by $A$ and $B$ at a distance $5$ from center $C_1$. Point $Q$ is chosen on $C_2$ such that $PQ$ is tangent at $Q$. Find $PQ$.

What I Tried: Here is a picture in Geogebra :-

I have no good idea on how to start this. I cannot, for example, find the lengths of $C_1C$ and $CC_2$ from pure deduction, ($C_1$ and $C_2$ are centres of the circles) and even if Pythagorean Theorem may look like they can work, I cannot understand how, because I don't have many required lengths. I also tried similarity but in fact I could not find any similar triangles here.

Can anyone help me? Thank You.

Body:
$$
|PQ|^2=|PA||PB|= (|PC_1|+r_1)(|PC_1|-r_1)
$$


问:
Expected values of balls at position equal to ball number
This is the question that I'm trying to solve

Balls are drawn one after the other uniformly at random without replacement from a set of eight balls numbered 1,2,...,8 until all balls drawn what is expected number of balls whose value matched their ordinality (i.e their position in the order in which balls were drawn)

I came up with this solution

Let P(1) be the probability of drawing 1st ball at 1st draw, clearly
$P(1) = \frac{1}{8}$

Similarly $P(2) = \frac{7}{8}\times\frac{1}{8}=\frac{7}{8^2}$

& $P(3) =
> \frac{7}{8}\times\frac{6}{8}\times\frac{1}{8}=\frac{7\times6}{8^2}$

It can be seen that, the probability of drawing ith ball at ith draw
is $P(i) = \frac{^7P_{i-1}}{8^i}$

The expected number of balls whose value match their position =
$\sum^8_{i=1}i\times P(i)$

$ \sum^8_{i=1}i\times\frac{^7P_{i-1}}{8^i}$,

Which when calculated comes out to be 1.

Is this solution correct? Instead of calculating by hand, i wrote this python script,

```<code>def fact(n):
    f = 1
    for i in range(1, n+1):
        f *= i
    return f

def P(n, r):
    return fact(n) / fact(n - r)

def foo(n):
    sum = 0
    for i in range(1, n+1):
        sum += i * P(n-1, i - 1) * (n**(n-i))
    return sum/n**n

for i in range(1, 15):
    print(i, foo(i)) # always prints 1.0
</code>```

To my surprise it gave the answer 1 despite the number of balls n, which is why I think there must be some mistake or is $\sum_{i=1}^{n}i\times\frac{^{n-1}P_{i-1}}{n^i} = 1$ some identity I'm unaware of.
Even if my solution is correct it seems over complicated and hard to compute for such a simple problem.

答:
Use the linearity of expectation.

Expectation is linear even when the random variables are not independent.

Let $X$ be an indicator random variable such that $X_i$ assumes a value of $1$ if number $i$ is drawn on the $i^{th}$ draw, and $0$ otherwise.

Then $\mathbb{E}(X_i) = P(X_i) = \frac 1 8$

and $\mathbb{E}(X) = \Sigma\mathbb{E}(X_i) = 8\cdot \frac 1 8  = 1$

This also explains why you got the answer of $1$ for $n$ numbers, too.


问:
Proving Inf with epsilon
Let X denote a nonempty bounded subset of $\mathbb{R}$. Prove that for each $\epsilon$ > 0,
there exists x in X such that inf X $\leq$ x < inf X + $\epsilon$

答:
If $ \epsilon >0$, then $ \inf X + \epsilon $ is not a lower bound of $X$, hence there is some $x \in X$ such that $ x < \inf X + ϵ.$

Furthermore it is clear that $\inf X ≤ x.$

Title: Proving Inf with epsilon


问:
Rank and Dimension of Kernel for non-square matrix
suppose a 7x5 matrix has B and Bx=0 has a general solution of x=$\lambda$v where v is some non-zeo vector and an element of $\mathbb{R}^5$. What's the dimension of the kernel for this matrix B and its rank?

I'm not quite sure about this but the way I interpret this is by looking at the homogenous solutions given by $\lambda$v. Does it only have one linearly independent vector as a solution which is v? So the dimension of the kernel is 1? if so, there are still infinite solutions right? So how would I get to know what the rank of this matrix is, cause I'm all confused about this and would really appreciate some guidance.

答:
Let $T: \mathbb R^5 \to \mathbb R^7$ be defined by $T(x)=Bx.$

Then $ker(B)=ker(T)= span(v)$, hence $\dim ker(B)=1.$

From the nullity - rank - theorem we derive

$$5= \dim ker(T)+ rank(T)=dim ker(B)+ rank(B)=1+rank(B).$$

Thus

$$rank(B)=4.$$


问:
Showing the equivalence of two statements regarding the limits of sequences
I need to show the equivalence of the following two statements

$(i)$ For any sequence $(a_n)_n$ with $a_n \in D$ and $a_n > a$ for all $n \in N$ such that
$\lim\limits_{n\to\infty} a_n = a$ we have $\lim\limits_{n\to\infty} f(a_n) = L.$

$(ii)$ For any sequence $(a_n)_n$ with $a_n \in D$ and $a_n \ge a_{n+1}>a$ for all $n \in N$ such that
$\lim\limits_{n\to\infty} a_n = a$ we have $\lim\limits_{n\to\infty} f(a_n) = L.$

My attempt:

Going from $(i)$ to $(ii)$ is obvious.

To show that $(ii)=>(i)$, take a sequence $(a_n)_n$ with $a_n \in D$
and $a_n > a$ for all $n \in N$ and $\lim\limits_{n\to\infty} a_n =
> a$. We want to show that $\lim\limits_{n\to\infty} f(a_n) = L$.

$(a_n)_n$ has a non-increasing subsequence $(a_{k_n})_n$. Then, we
have $a_{k_n} \in D$ and $a_{k_n} \ge a_{k_{n+1}}>a$ for all $n \in N$
and $\lim\limits_{n\to\infty} a_{k_n} = a$. Then, we can apply $(ii)$
to this sequence and get $\lim\limits_{n\to\infty} f(a_{k_n}) = L$.

I am stuck here and I don't know how to go back to the limit of the original sequence. Am I on the right path? Can I get some hints?

答:
Take a sequence $(a_n)_n$ with $a_n \in D$ and $a_n > a$ for all $n \in N$ and $\lim_{n\to\infty} a_n =  a$.

Assume that $\lim_{n\to\infty} f(a_n) = L$ does not hold. Then there is a $\epsilon > 0$ and a subsequence $(a_{n_k})$ of $(a_n)$ such that $|f(a_{n_k}) - L | \ge \epsilon$ for all $k$.

Now apply (ii) to a decreasing subsequence of $(a_{n_k})$ to get a contradiction.

This is an example of the following principle, applied to $x_n = f(a_n)$:

Let $(x_n)$ be a sequence in a topological space $X$ and $L \in X$. If every subsequence of $(x_n)$ has itself a subsequence converging to $L$, then the complete sequence $(x_n)$ converges to $L$.


问:
Is this statement true: $x^2<0 \ \land \ x^2 \in \mathbb{R} \ \Rightarrow \ x \in \mathbb{C} \setminus \mathbb{R} $?
So basically, is there a set containing elements not in the complex numbers that squared is a negative real number?

$ \{ x \ \vert \ x^2 \in \mathbb{R} \land x^2<0 \} \stackrel{?}{=} \emptyset $

答:
For all $x \in \mathbb R$ we have that $x^2 \ge 0.$ Hence

$$\{ x \ \vert \ x^2 \in \mathbb{R} \land x^2<0 \} = \emptyset.$$


问:
Alternative definition of real valued random variable using Cartesian product
I know that $X: \Omega \rightarrow \mathbb{R}^n$ is a real valued random variable if and only if for any Borel set $B \in \mathcal{B}(\mathbb{R}^n)$ the set $X^{-1}(B) \in \mathcal{F}$.

However, I would like to prove the alternative definition using using Cartesian product, that is:

$X$ is a real valued random variable, if and only if for any numbers $\lambda_1,\ldots,\lambda_n \in \mathbb{R}$ it's true that $$X^{-1}((-\infty,\lambda_1] \times \ldots \times (-\infty,\lambda_n]) \in \mathcal{F}.$$

I realize that proving ''$\Rightarrow$'' is easy: the set $(-\infty,\lambda_1] \times \ldots \times (-\infty,\lambda_n]$ is a Cartesian product and therefore a subset of a family of Borel sets. However, I fail to see how to prove ''$\Leftarrow$''. I' ve seen some proofs of this in the literature for a one-dimensional case and while they're neat and short, I have an inkling I'm missing some deeper underlying concepts in those proofs.

答:
Show that
$$
\mathcal{A}=\{B\in\mathcal{B}(\mathbb{R}^n)\mid X^{-1}(B)\in\mathcal{F}\}.
$$
is a $\sigma$-field. Then since
$$
\{(-\infty,\lambda_1]\times\cdots\times(-\infty,\lambda_n]\mid \lambda_1,\ldots,\lambda_n\in\mathbb{R}\}\subseteq\mathcal{A}
$$
we automatically get that
$$
\sigma(\{(-\infty,\lambda_1]\times\cdots\times(-\infty,\lambda_n]\mid \lambda_1,\ldots,\lambda_n\in\mathbb{R}\})\subseteq\mathcal{A}.
$$
Finally, use that the left-hand side is $\mathcal{B}(\mathbb{R}^2)$ to conclude.


问:
Does statements about multigraphs imply statements about simple graphs
Simple question, which is already given in the heading. "If we know a question hold for multigraphs does it mean that this theorem hold for simple graphs as well"?</p>

答: Is a simple graph a multigraph?  Yes, so any true statement about multigraphs is true of a simple graph.</p>


问:
Find the Laurent Series centered at $z=0$ for all the analitical region of $\frac{\sinh(z) - z}{z^4}$
Find the Laurent Series centered at $z=0$ for all the analitical region of

$\frac{\sinh(z) - z}{z^4}$

答:
Step 1: consider the Taylor- series for $sinh.$

Step 2: subtract $z$.

Step 3: now divide by by $z^4.$

Whoops, the  Laurent Series centered at $z=0$  is completed.

This series converges on $ \mathbb C \setminus \{0\}.$


问:
Some converse of Jensen inequality for some extreme random variable
This problem arises in my research in a more general form. It concerns information theory and statistical bounds using information theory. I tagged this question with Jensen inequality since the statement can be seen as a converse for Jensen inequality for the following extreme random variable $X$.

Suppose that $X$ is a random variable on $[0,1]$ that averages to $1/2$ such that
\begin{align*}
X=\begin{cases}
0&\text{with probability } 1/2-\varepsilon (1-p)\\
p&\text{with probability } \varepsilon\\
1&\text{with probability } 1/2-\varepsilon p\\
\end{cases}
\end{align*}
for some $p\in(0,1)$ and $\varepsilon \leq \frac{1}{2\max(p,1-p)}$. Let for any $q\in (0,1)$, $h_{q}(x)=\min\left( \frac{x}{q}, \frac{1-x}{1-q} \right)$.

Show (or disprove) that for any random variable $Y$ on $[0,1]$ that averages to $1/2$, if we have $\mathbb E[h_p(X)]\leq \mathbb E[h_p(Y)]$ then for any $q\in (0,1)$, $\mathbb E[h_{q}(X)]\leq \mathbb E[h_{q}(Y)]$.

Due to the construction of $X$, for any $q$, $\mathbb E[h_{q}(X)]=\varepsilon \cdot h_{q}(p)$ and so $\mathbb E[h_p(X)]\leq \mathbb E[h_p(Y)]$ implies that $\varepsilon\leq \frac{\mathbb E[h_{p}(Y)]}{h_p(p)}=\mathbb E[h_{p}(Y)]$, now this means that
\begin{align*}
\mathbb E[h_{q}(X)]&=\varepsilon \cdot h_{q}(p)\\
&\leq \mathbb E[h_{p}(Y)]\cdot h_{q}(p)
\end{align*}
Now observe that $h_{q}$ is linear on $[0,q]$ and on $[q,1]$ so we can partition $[0,1]$ in three intervals, suppose without loss of generality that $p < q$, then denote the event $A=\{ Y\in [0,p]\}$, $B=\{ Y\in (p,q) \}$ and $C=\{ Y\in[q,1] \}$
\begin{align*}
\mathbb E[h_p(Y)] &= \mathbb P[A] \frac{\mathbb E[Y|A]}{p}+\mathbb P[B]\frac{1-\mathbb E[Y|B]}{1-p}+\mathbb P[C]\frac{1-\mathbb E[Y|C]}{1-p}
\end{align*}
and
\begin{align*}
\mathbb E[h_{q}(Y)] &= \mathbb P[A] \frac{\mathbb E[Y|A]}{q}+\mathbb P[B]\frac{\mathbb E[Y|B]}{q}+\mathbb P[C]\frac{1-\mathbb E[Y|C]}{1-q}
\end{align*}
We can now compare those three expressions one by one when we multiply the first one by $h_{q}(p)=\frac{p}{q}\leq \frac{1-p}{1-q}$.
\begin{align*}
\mathbb P[A] \frac{\mathbb E[Y|A]}{p}\cdot h_{q}(p) &= \mathbb P[A] \frac{\mathbb E[Y|A]}{q}\\
\mathbb P[C]\frac{1-\mathbb E[Y|C]}{1-p}\cdot h_{q}(p)&\leq \mathbb P[C]\frac{1-\mathbb E[Y|C]}{1-q}
\end{align*}
for the middle one, set $a=\mathbb E[Y|B]$, then $p\leq a \leq q$ so that
\begin{align*}
p\leq a &\Leftrightarrow (1-a)p\leq a (1-p)\\
&\Leftrightarrow \frac{1-a}{1-p} \frac{p}{q} \leq \frac{a}{q}
\end{align*}
This finishes the proof that $\mathbb E[h_{p}(Y)]\cdot h_{q}(p)\leq \mathbb E[h_{q}(Y)]$ and so finishes the proof of the statement.

It would be nice to have some feedback on that proof and since I want to generalize to more dimensions I would appreciate to have a much simpler proof if someone have an idea.

答:
I just have found a much more satisfying proof of the fact that $\mathbb E[h_p(Y)]\cdot h_q(p)\leq \mathbb E[h_q(Y)]$ :

\begin{align*}
h_p(x)\cdot h_q(p) &= \min\left(\frac{x}{p},\frac{1-x}{1-p}\right)\cdot \min\left(\frac{p}{q},\frac{1-p}{1-q}\right)\\
&= \min\left( \frac{x}{p}\cdot\frac{p}{q},\frac{1-x}{1-p}\cdot\frac{p}{q},\frac{x}{p}\cdot\frac{1-p}{1-q},\frac{1-x}{1-p}\cdot\frac{1-p}{1-q}\right)\\
&\leq \min\left( \frac{x}{p}\cdot\frac{p}{q},\frac{1-x}{1-p}\cdot\frac{1-p}{1-q}\right)\\
&=\min\left(\frac{x}{q},\frac{1-x}{1-q}\right)\\
&=h_q(x)
\end{align*}


问:
Write the expression in coordinates of the projection on W parallel to U
Consider the following subspaces of $\in R^{3}$.

$U: x_{1}+2x_{2}+3x_{3}=0$

$W= span$ {$(1, -1, 1)$}

Show that $\in R^{3}= W \bigoplus U$ and write the expression in coordinates of the projection on $W // U$

After deducing the span of the subspace $U$ I've verified the linear independency by

$a(1,-1,1) + b(-2,1,0) + c(-3, 0, 1)= (0,0,0)$ $\leftrightarrow a=b=c=0$

Then, I concluded that $W \cap U=0$

So, consequently for the dimension theorem we have that $ R^{3}= W \bigoplus U$

Now, I'm having problems with the second part of the exercise.

I would be very thankful if someone can explain me geometrically what does it mean a projection on $W$ parallel to $U$

答:
Geometrically we start out from a point $v\in\Bbb R^3$ and move it parallelly with $U$ to arrive to a point $P(v)$ in $W$.

Put in other words, $P(v)$ is the intersection of $W$ and the affine subspace (plane in this case) $U+v$, which is a unique point if $\Bbb R^3=U\oplus W$.

Algebraically we can simply define $P(v)=P(u+w)=w$ where $u\in U,\,w\in W$ are the unique elements satisfying $v=u+w$.

Now either you directly calculate $P(e_i)$ for the standard basis vectors $e_1,e_2,e_3$, or write up the matrix of $P$ in the basis you found and apply basis transformation.


问:
Rank of elliptic curves used in cryptography
I was playing around with an implementation of the NIST P-256 elliptic curve and I noticed that the field order and the order of the base point in the implementation are "close" :

Field order:
$115792089210356248762697446949407573530086143415290314195533631308867097853951$

Group order:
$115792089210356248762697446949407573529996955224135760342422259061068512044369$

I see this as a consequence of Hasse theorem, and as an evidence that the base point is generating the whole group of the curve, which is then of rank 1.

Is it just a coincidence or is it something desirable for cryptographic purpose that the group of the underlying curve should have a rank equal to 1 ?

答:
In Cryptography we choose a curve $E$ over a finite field $K$ then the $E(K)$ is the groups of the points and $|E(K)|$ is the order of $E(K)$, i.e. cardinality, or the number of the rational points.

Then we choose a base point $G$, which is usually chosen as nothing-in-my-slave number to prevent suspicious curve parameters. It is clear that the multiples of $[k]G$ ($\langle G\rangle$) form a subgroup of $E(K)$. Let call order of this subgroup $r = r = \left|\langle G\rangle\right|$ than by the Lagrange theorem
$$h = \frac{|E(K)|}{r}.$$ We call this $h$ as the cofactor

There is a small subgroup attack (Lim–Lee active small-subgroup attacks ).  The attacker chooses $P$ and the user reveals $[k]P$. The obvious method for the attacker is choosing $P$ in a small group where the ECDLP is easy.  If the curve $|E(K)|$ is prime then there are no non-trivial subgroups for the attacker. If near the prime like 8 in the Curve25519 the prevention is easy (see below).

We also want the $r$ to be prime to resist the Pohlig–Hellman algorithm.

Does cofactor affect security?

It is clear that when the cofactor is 1 then $\langle G\rangle$ is the whole curve and any non-zero point can be a generator. Lim–Lee, and Pohlig-Hellman will fail.

What if the cofactor >1? There is only some special need for checking that the incoming point is not on the small subgroup, rather than the $\langle G\rangle$. The designers of the curves take this into consideration, for example for Curve25519 has cofactor 8. When used with  Diffie-Hellman key exchange then while selecting the private key, the three least significant bit set to zero to guarantee that the private key in the $\langle G\rangle$

So $cofacor =1$ is easy to use. However, having larger some other advantages like the Montgomery Curves

has the side-channel resistant Montgomery Ladder,

every Montgomery Curve has a point of order 2 i.e. the cofactor is always >1.

Note that: the curve P-256 is suspicious since the coefficients generated by hashing the unexplained seed c49d3608 86e70493 6a6678e1 139d26b7 819f7e90


问:
Finding the orthonormal basis of $\Bbb R^3$ using the Gram-Schmidt algorithm
I'm relatively new to inner product and I was asked to obtain an orthonormal basis of $\Bbb R^3$ using

$$
    \begin{pmatrix}
    1 \\
    1 \\
    1 \\
    \end{pmatrix}   ,  \begin{pmatrix}
    1 \\
    0 \\
    1 \\
    \end{pmatrix} , \begin{pmatrix}
    0 \\
    1 \\
    2 \\
    \end{pmatrix}
$$

by applying the Gram Schmidt algorithm which I have found to be

$$
    (\frac{1}{\sqrt 3})\begin{pmatrix}
    1 \\
    1 \\
    1 \\
    \end{pmatrix}   ,  (\frac{\sqrt 6}{2})\begin{pmatrix}
    1/3 \\
   -2/3 \\
    1/3 \\
    \end{pmatrix} , (\frac{1}{\sqrt 2})\begin{pmatrix}
    -1 \\
    0 \\
    1 \\
    \end{pmatrix}
$$

In addition to that my inner product was defined to be $\langle u,w\rangle = u_1w_1+ 2u_2w_2+ 3u_3w_3 $.

I'm not sure what I'm supposed to do with the defined innerproduct as such. What do i have to do to solve this question or have i gotten the concept all wrong?

答: Note that the inner product is an abstract concept. Check out the Wikipedia article. You can use the GS process with any inner product, just use your inner product definition to calculate $\langle\cdot,\cdot\rangle$ where it appears!


答: Which inner product did you use for your orthogonal projections during Gram-Schmidt ? Make sur you stay consistent with using the same inner product all the time. Recall : the orthogonal projection of $u$ onto $v$ is given by $\text{proj}_vu=\frac{\langle u,v\rangle}{\langle v,v \rangle}v$, and the norm of a vector $w$ as defined is this context is equal to $||w||=\sqrt{\langle w,w \rangle}$

Title: Finding the orthonormal basis of $\Bbb R^3$ using the Gram-Schmidt algorithm

Body:
I'm relatively new to inner product and I was asked to obtain an orthonormal basis of $\Bbb R^3$ using

$$
    \begin{pmatrix}
    1 \\
    1 \\
    1 \\
    \end{pmatrix}   ,  \begin{pmatrix}
    1 \\
    0 \\
    1 \\
    \end{pmatrix} , \begin{pmatrix}
    0 \\
    1 \\
    2 \\
    \end{pmatrix}
$$

by applying the Gram Schmidt algorithm which I have found to be

$$
    (\frac{1}{\sqrt 3})\begin{pmatrix}
    1 \\
    1 \\
    1 \\
    \end{pmatrix}   ,  (\frac{\sqrt 6}{2})\begin{pmatrix}
    1/3 \\
   -2/3 \\
    1/3 \\
    \end{pmatrix} , (\frac{1}{\sqrt 2})\begin{pmatrix}
    -1 \\
    0 \\
    1 \\
    \end{pmatrix}
$$

In addition to that my inner product was defined to be $\langle u,w\rangle = u_1w_1+ 2u_2w_2+ 3u_3w_3 $.

I'm not sure what I'm supposed to do with the defined innerproduct as such. What do i have to do to solve this question or have i gotten the concept all wrong?


问:
When does a partial isometry have unitary extension?
$u$ extends $v$ means that $u^*v=v^*v$.

If $v^2=0$ or $v^*v=vv^*$ it certainly does. If $v^*v<vv^*=1$ it certainly does not.

Does finite Von Neumann algebra sound like to have the property that every partial isometry has a unitary extension?

答: Yes. In a finite von Neumann algebra, if $v^*v=p\sim q$ then $1-p\sim 1-q$. If $w$ is a partial isometry for the second equivalence, then $v+w$ is a unitary.


答:
We can actually make a much stronger conclusion.

Let $z = re^{i\theta} \in \mathbb{C}$. We know $z^{2} = r^{2} e^{i 2 \theta} \in \mathbb{R}$ so $2\theta = k \pi$ so $\theta = k \pi/2$. We also know $z^2 < 0$ so $r > 0$ and $\theta = \pi/2$ or $\theta = 3\pi/2$. That is, we know $z$ is of the form $\pm b i$ where $b$ is a positive real number.

Title: Is this statement true: $x^2<0 \ \land \ x^2 \in \mathbb{R} \ \Rightarrow \ x \in \mathbb{C} \setminus \mathbb{R} $?

Body:
So basically, is there a set containing elements not in the complex numbers that squared is a negative real number?

$ \{ x \ \vert \ x^2 \in \mathbb{R} \land x^2<0 \} \stackrel{?}{=} \emptyset $


问:
Why is the origin the only zero of this class of functions in the unit disk?
Let f be analytic in $\mathbb{D}, f(0) = 0, f'(0) = 1$ and assume that $Re(\frac{zf'(z)}{f(z)}) > 0$. Then f has a simple zero at the origin and no other zeros in the unit disk. Why?

答: $\frac{f'}{f}$ has a pole of order one at every zero of $f$, so $g(z):=z\frac{f'}{f}$ has a simple pole at every zero of $f$ different from $0$. Near a zero ($\neq 0$) of $f$, $g$ can be written as $\frac c{z-z_0}+\psi(z)$, where $\psi$ is holomorphic in a neighbourhood of $z_0$. It is easy to see that for every value of $c\in\mathbb{C}-\{0\}$, $g(z_0-c\varepsilon)=-\frac{1}{\varepsilon}+\psi(z_0-c\varepsilon)$. Taking $\varepsilon>0$ small enough, we get that $\Re(g(z_0-c\varepsilon))<0$, contradicting the requirement. Thus $f$ cannot have any zero on $\mathbb{D}-\{0\}$.


问:
Dependence of Axioms of Equivalence Relation?
This question is problem 11(a) in chapter 1 in 'Topics in Algebra' by I.N. Herstein.

These are the properties of equivalence relation given in this book.

Prop 1 $a \sim a$

Prop 2 $a \sim b$ implies $b \sim a$.

Prop 3 $(a \sim b$ and $b \sim c)$ imply $a \sim c$.

Statement Property 2 of an equivalence relation states that if $a \sim b$ then $b \sim a$. By property 3, we have the transitivity i.e. $a \sim b$, and $b \sim c$ then $a \sim c$. What is wrong with following proof that property 2 and 3 imply property 1? Let $a \sim b$; then $b \sim a$, whence by property 3 (using $a = c$), $a \sim a$.

I think I can prove this to be wrong. Without proving equivalence relation first, one can not use $a = c$. Right? After all, equality is equivalent to 'equivalence relation' and 'axiom of substitution' are satisfied. If this is right, then I have trouble with the next part of this problem.

Part 2 Can you suggest an alternative of property 1 which will insure us that prop 2 and prop 3 do imply 1?

Can one give such a formulation without using the idea of '=' or otherwise?

EDIT : Italics are my comments. Rest is as it appeared in the book.

Notion of Equality

I have read in Terry 'Analysis 1' in Appendix A.7 published by Hindustan Book Agency that there are four axioms of 'equality'. First 3 are same as equivalence relation where $\sim $ in replaced by $ = $. The fourth one is known as axiom of substitution. Given any two objects $x$ and $y$ of some type, if $ x = y $, then $f(x) = f(y) $ for all functions or operations $f$.

答:
For the first part: The "proof" assumes that for $a$ there is a $b$ such that $a\sim b$. This of course is not necessarily given. The empty relation i.e. for no $a,b$ we have $a\sim b$ is transitive, symmetric but not reflexive. (This example seems to many students not very explanatory although it is the simplest example of this situation. See the example of Arturo Magidin if that helps.)

For the second part: I think the book might ask for something like this.

Prop 1': For any $a$, for which there is any $b$ such that $a\sim b$, we also have $a\sim a$.

This might seem to be an weird property, but we could also take "suffices Prop 2 and Prop 3", which wouldn't make much more sense.

Or as Arturo Magidin suggested

Prop 1'': For any $a$, there is a $b$ such that $a\sim b$

together with Prop 2 and Prop 3 implies Prop 1.


问:
What is $\pi_{31}(S^2)$?
What is $\pi_{31}(S^2)$ - high homotopy group of the 2-sphere ?

This question has a physics motivation:

There are relations between (2nd and 3rd) Hopf fibrations and (2- and 3-) qbits (quantum bits) entanglement, see  this reference: https://arxiv.org/pdf/0904.4925v1.pdf

Maybe  there are relations between classification of qbits entanglements and sphere homotopy groups ?

We are interested in the classification of 4-qbits entanglements.

答:
In general, finding the isomorphism class of higher homotopy groups of the spheres is an extremely hard problem. Luckily however, there seem to be some techniques available to the low dimensional spheres which aren't available to the higher dimensional spheres. In particular, there appears to be an isomorphism between $\pi_n(S^2)$ and the co-kernel of a homomorphism from the Brunnian braid group in $n$ strings on the disk to the equivalent Brunnian braid group on the sphere. The homomorphism is induced by the canonical geometric embedding of the disk in the sphere.

That is, let $f\colon D\rightarrow S^2$ be the canonical embedding of the disk in to the sphere and let $C(n,D)$ and $C(n,S^2)$ be the configuration space of $n$ points in the the disk $D$ and sphere $S^2$ respectively. It's a well-known theorem that $\pi_1(C(n,M))$ is isomorphic to the pure braid group on $n$ strings in the manifold $M$ and so $f$ induces a homomorphism $f_*\colon\pi_1(C(n,D))\rightarrow\pi_1(C(n,S^2))$ on braid groups.

Let $BB_n\leq\pi_1(C(n,D))$ be the subgroup of Brunnian braids in the disk and similarly let $BB_N^{S^2}\leq \pi_1(C(n,S^2))$ be the subgroup of Brunnian braids in the sphere. A Brunnian braid is a pure braid such that the removal of any one string leaves the remaining braid isotopic to the identity on $n-1$ strings. We may restrict $f_*$ to the Brunnian braids and so get a homomorphism $f_*'\colon BB_n\rightarrow BB_n^{S^n}$. For $n$ large enough (greater than 3 i think), there is an isomorphism $$\pi_n(S^2)\cong \mbox{coker}f_*'=BB_n^{S^2}/\mbox{Im}f_*'.$$

This theorem is attributed to Berrick, Cohen, Wong, Wu in their paper Configurations, braids, and homotopy groups.

What makes this theorem so useful is that it reduces the hard problem of finding homotopy groups of the sphere to what is essentially computational group theory. This is because we have pretty good knowledge of both the disk and sphere braid groups thanks to work done in the 60s stemming from the analysis of the Fadell-Neuwirth fibrations between configuration spaces of $n$ points in a manifold. I've not done any computational group theory before so I don't know if identifying the isomorphism class of the above mentioned cokernel is computationally efficient, but it at least gives us a known algorithm for calculating it (it's possible that identifying the Brunnian subgroup of the respective braid groups might be difficult).

The following is just to keep a comprehensive collection of information we have, and I take no credit for finding these references.

For those that haven't seen the cross-posted question on MathOverflow Mark Grant gives a reference to a paper which appears to give computations of $\pi_i(S^3)$ for all $i\leq 64$ which are of course isomorphic to the $\pi_i(S^2)$ for suitably large $i$ by the isomorphism induced by the hopf fibration:

Curtis, Edward B.,Mahowald, Mark, The unstable Adams spectral sequence for $S^3$, Algebraic topology (Evanston, IL, 1988), 125–162, Contemp. Math., 96, Amer. Math. Soc., Providence, RI, 1989.

In the comments, the OP also gives a reference to a paper in which the $2$ primary part of $\pi_{31}(S^3)$ is given as isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$:

Nobuyuki Oda, On the 2.Components of the Unstable Homotopy Groups of Spheres I, Proc. Japan Acad. Ser. A Math. Sci. Volume 53, Number 7 (1977), 215-218. (link)

Title: What is $\pi_{31}(S^2)$?


答: You are supposed to use the inner product that you were given. With respect to that inner product, $\|(1,1,1)\|=\sqrt6$ and therefore the first vector that you get when you apply Gramm-Schmidt is $e_1=\frac1{\sqrt6}(1,1,1)$. Then, let$$a_2=(1,0,1)-\langle(1,0,1),e_1\rangle e_1=\left(\frac13,-\frac23,\frac13\right).$$It norm is $\sqrt{\frac43}$ and so you take $e_2=\frac1{2\sqrt3}(1,-2,1)$. Finally, you take$$a_3=(0,1,2)-\langle(0,1,2),e_1\rangle e_1+\langle(0,1,2),e_2\rangle e_2=\left(-\frac32,0,\frac12\right).$$Its norm is $\sqrt3$ and so you take $e_3=\frac1{\sqrt3}\left(-\frac32,0,\frac12\right)$.


问:
Convergence of $\int_{0}^{\infty} x^{a-\frac{3}{2}}e^{-\frac{1}{2x}} \,dx$
I am trying to determine the values of $a$ for which the following integral converges:

$$\int_{0}^{\infty} x^{a-\frac{3}{2}}e^{-\frac{1}{2x}} \,dx.$$

My lecturer has stated that this converges for $a<\frac{1}{2}$ but I want to understand why. I did not get anywhere by substituting this value of $a$ into the integral.

答:
Since the integral is improper for $x\to 0$ and $x\to \infty$ you may first split it into two parts:

$x\geq 1$:

Here you can estimate $1 \leq e^{\frac 1{2x}}\leq e$ and it follows

$$\int_1^{\infty}x^{a-\frac{3}{2}} \,dx \leq \int_1^{\infty}\frac{x^{a-\frac{3}{2}}}{e^{\frac{1}{2x}}} \,dx \leq \frac 1e\int_1^{\infty}x^{a-\frac{3}{2}} \,dx$$
But $\int_1^{\infty}x^{a-\frac{3}{2}} \,dx$ converges if and only if $a-\frac 32 < -1 \Leftrightarrow \boxed{a<\frac 12}$.

$0<x\leq 1$:

Note that using the series expansion of the exponential function you have $e^{\frac 1{2x}} \geq \frac{\left(\frac 1{2x}\right)^n}{n!} = \frac 1{2^nn!x^n}$ for each $n \in \mathbb{N}$. Hence, choosing $n$ large enough, you have

$$0\leq \int_{0}^{1}
\frac{x^{a-\frac{3}{2}}}{e^{\frac{1}{2x}}}  \,dx \leq 2^nn!\int_{0}^{1}x^{a-\frac{3}{2}+n}  \,dx <\infty$$


问:
Is there an explicit formula for the inverse of $\cot\left(\frac{x}{2}\right)\sqrt{1-\cos(x)}$?
I apologize if this is trivial but I am stuck.

Given the bijective function $f:(0,2\pi) \to (-2,2)$ with
$$
f(x)=\cot\left(\frac{x}{2}\right)\sqrt{1-\cos(x)}
$$
where $\cot$ is the cotangent, how can I find an inverse $g:(-2,2)\to (0,2\pi)$? Is there an explicit formula?

答: Hint: Use this fact that $$\frac{1-\cos(x)}{2}=\sin^2(x/2)$$ and that if $x\in(0,2\pi), |\sin(x/2)|=\sin(x/2)$</p>


问:
What does $e^{i \pi \mathbf x} \in \mathbb C^2$ mean?
I saw this in an book (coding theory) without explanation and I need help figuring out the meaning.

Consider $\mathbf x\in \{0,1\}^2$ and
$$
\mathbf s = e^{i \pi \mathbf x} \in \mathbb C^2 \tag 1
$$

I know $\{0,1\}^2$ and $\mathbb C^2$ are the cartesian products, i.e.
\begin{align}
\{0,1\}^2&=\{0,1\}\times \{0,1\}\\
&=\{(0,0),(0,1),(1,0),(1,1)\} \tag 2
\\
\mathbb C^2 &=\mathbb C \times \mathbb C = \{(z,y): z,y\in \mathbb C\} \tag 3
\end{align}
So $\mathbf x= (x_1,x_2)$ is a vector which take a value in the set $\{0,1\}^2$, for instance, $\mathbf x = (1,1)$.

So far so good, but what is $e^{i \pi \mathbf x} \in \mathbb C^2$?
Any examples?

Say $\mathbf x = (1,1)$, what does $e^{i \pi (1,1)} \in \mathbb C^2$ mean?

Any clarification would be appreciated.

Body:
$e^{0} = 1$ and $e^{i\pi} = -1$ implying for example $e^{i \pi (1,1)} = e^{ (i \pi,i \pi)} = (-1,-1)$

so if $\mathbf x \in \{(0,0),(0,1),(1,0),(1,1)\}=\{0,1\}^2$

then $e^{i \pi \mathbf x} \in \{(1,1),(1,-1),(-1,1),(-1,-1)\} = \{1,-1\}^2 \subset \mathbb C^2$ and $e^{i \pi \mathbf x} \in \mathbb C^2$


问:
meaning of II symbol in a math equation
I have a math equation that has II in it I cannot understand what the II function is and what it does. I really appreciate it if you can explain what it is.

答:
It is a Capital i, as "indicator function"

you can also find $\mathbb{1}$ and indicates that the function is 1 for the values in the braces and zero elsewhere

Example,

Suppose you have the following function

$$ y =
\begin{cases}
x,  & \text{if $0<x<1$ } \\
2-x, & \text{if $1\leq x\leq 2$}\\
0, & \text{elsewhere}
\end{cases}$$

you can compact the notation and write

$$y=\{1-|1-x|\}\cdot\mathbb{1}_{[0;2]}(x)$$

Title: meaning of II symbol in a math equation


问:
Local connectedness is preserved under retractions
I want to show that if $X$ is a locally connected topological space, $A\subseteq X$ is a subspace and $f:X \rightarrow A$ is continuous such that $f|_{A} = Id_{A}$, then $A$ must be locally connected as well.

My progress so far:

Take $U\subseteq A$ $A-$open and $x\in U$. Since $f^{-1} (U)$ is $X-$open and $x\in f^{-1} (U)$, there exists a connected, $X-$open subset $V\subseteq X$ such that $x\in V\subseteq f^{-1}(U)$.

Now we have $x\in V\cap A \subseteq f^{-1}(U)\cap A = U$

My guess is that $V\cap A$ should be connected, but I am unsure if this is correct.

Any help would be appreciated!

答:
It is easier to instead show that components of open sets are open, which is equivalent to local connectedness. Let $r\colon X\to A$ be the retraction and suppose $N$ is an open set in $A$ and $C$ is a component of $N$.

Let $p\in N$ and $U = r^{-1}(N)$. By continuity, $U$ is open and by local connectedness, there is a connected open $V$ such that $p\in V\subseteq U$. Again by continuity, $r(V)$ is a connected set containing $p$, hence $r(V)\subseteq C$.

Now, $p\in V\cap A\subseteq r(V)$ and $V\cap A$ is open in $A$, therefore, $p\in V\cap A\subseteq C$, hence $C$ is an open set.

Therefore, the connected components of $N$ are open, hence $A$ is locally connected.


问:
Use of the Reciprocal Fibonacci constant?
The Reciprocal Fibonacci constant ($\psi$) is defined as

$$\psi=\sum_{k=1}^{\infty} \frac{1}{F_k}$$

where $F_{k}$ is the $k^{th}$ Fibonacci number.

The irrationality of $\psi$ has been proven.  Does the Reciprocal Fibonacci constant have a use in mathematics or is is notable simply because it is the value of an interesting sum?

答: As far as I know, the constant $\psi$ doesn't arise naturally other than in just studying it from the outset. However, a related constant does make its appearance in a surprising way. That is, $$\beta:=1+\sum_{k=1}^\infty \dfrac{1}{\mathrm F_{2k}},$$or rather its reciprocal $1/\beta\approx0.3944196702$, arises as the solution to a problem in the analysis of bounded real sequences, which is the title question of my paper that addresses it: "How slowly can a bounded sequence cluster?". It is fully answered there and in the follow-up paper "A maximally separated sequence". The two papers are published in Functiones et Approximatio, volumes 46.2 (2012) and 48.1 (2013), which you can read if you have access to a library that subscribes to Project Euclid (the abstracts are freely available on-line). When I set out on the research, I had no inkling that a number with this sort of structure would come up. If this interests you, I can provide more details.


问:
Limit theorem for integral of quotient of sums
I want to show the following limit theorem:

Let $f:[0,1]\rightarrow\mathbb{R}$ and $g:[0,1]\rightarrow(0,\infty)$ be integrable functions such that $\lvert f\rvert\le Cg$ for some $C>0$. Then
$$\lim_{n\rightarrow\infty}\int_{[0,1]^n}\frac{\sum_{i=1}^nf(x_i)}{\sum_{i=1}^ng(x_i)}dx_1,...,dx_n = \frac{\int_0^1 f(x)dx}{\int_0^1g(x)dx}$$

This is an exercise in my advanced probability course, even though it's a purely analytical result, and I'm stuck on the following step:
It seems clear that the result should be derived somehow using the strong law of large numbers.

(EDIT) This is my most promising attempt so far: for independent, uniformly distributed $X_i$ and using the dominated convergence to theorem to pull the limit in we have
$$\begin{aligned}
\lim_{n\rightarrow\infty}\int_{[0,1]^n}\frac{\sum_i^nf(x_i)}{\sum_i^ng(x_i)}dx_1,...,dx_n
&= \lim_{n\rightarrow\infty}E\left(\frac{\sum_i^nf(X_i)}{\sum_i^ng(X_i)}\right)\\
&=\lim_{n\rightarrow\infty}\sum_j^nE\left(\frac{f(X_j)}{\sum_i^ng(X_i)}\right) \\
&=\lim_{n\rightarrow\infty}E\left(\frac{nf(X_1)}{\sum_i^ng(X_i)}\right)  \\
&=E\left(\lim_{n\rightarrow\infty}\frac{f(X_1)}{n^{-1}\sum_i^ng(X_i)}\right)\\
&=E\left(\frac{f(X_1)}{E(g(X_1))}\right)\\
&=\frac{E(f(X_1))}{E(g(X_1))} = \frac{\int_0^1f(x)dx}{\int_0^1g(x)dx}
\end{aligned}$$

Is the "transformation" from the n-dimensional integral to the expected value allowed as I wrote it? Did I make any errors?

答: The solution from my post is in fact complete.</p>


问:
Show $\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$
I am tasked with the following:

Show the following inequality for $\lvert b \rvert < \frac{\lvert a \rvert}{2}$ where $a,b \in \Bbb R$:
$$\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$$

I'm a bit at a loss as to where to even start with this one. I'm mostly looking for a little nudge.

答:
$|a-b|>\left||a|-|b|\right| \ge |a|-|b| >|a|-|a|/2=|a|/2$.
Take the inverse of both sides, and you get what you want.


问:
Finding expected value of operating units
Given $N$ operating units, every unit consists $k$ elements.

If one element is missing from one operating unit, it stops working.
The units operate independently.

When $m$ elements are removed from the system, what will be the expected value of the number of operating units?

So, we have to choose $m$ element out of $k$ groups.

If we choose so that $\,m=k\,$, the result can be minimum $\,0\,$ or maximum $\,N-1\,$ operating unit.

In case $\,m<k\,$, it is minimum $\,1\,$ or maximum $\,N-1\,$ operating units possible.

If $\,m>k\,$ it can be minimum $\,0\,$ or maximum $\,N-2\,$ operating units.

How should I continue the solution?

答:
This isn't stated well.  I'm assuming that each element has a probability $p$ that it is operating.  If the elements fail independently, then the probability that a given $k$-unit is working is $p^k$, and by linearity of expectation, the expected number of working units is $Np^k$.

To see that the last statement is true, use indicator functions.  For $i=1,2,\dots,N$, let $X_i$ be a random variable that is $1$ if unit number $i$ is working and $0$ if is not, so that the number of working units is $\sum_{i=1}^NX_i$.  By linearity of expectation,, $$E(\sum_{i=1}^NX_i=\sum_{i=1}^NE(X_i)=Np^k$$


问:
a.s. convergence in $\ell_2$ space
Assume $(X_{n})_{n\geq 1}$ is a sequence of random variables defined on the same probability space and taking values at $\ell_{2}$ (space of square-summable sequences). Next, assume that for some $a\in \ell_{2}$
$$
\sqrt{n}(X_{n} - a) \xrightarrow{a.s.} Z,
$$
where $Z$ is a random variable.

Is this true that
$$
X_{n} \xrightarrow{a.s.} a?
$$

答:
Defining
$$
  Y_n= \sqrt n(X_n-a) - Z,
  $$
we have
by hypothesis that $Y_n\xrightarrow{a.s.} 0$, so
$$
X_n=\frac{Z + Y_n}{\sqrt n}+a \xrightarrow{n\to0} a.
$$

Title: a.s. convergence in $\ell_2$ space


问:
Subgroup of $S_{n}$ that is not in $A_{n}$
I'm trying to prove this following statement:

Let $G\leq S_{n}$ for $n\geq 5$.

I need to show that if $G\not\subseteq A_{n}$ then $GA_{n}=S_{n}$.

I was thinking to show it implies $G\cap A_{n}=e$, and then using direct product, but I don't know if it is necessarily correct. If it doesn't, can I get a hint?

Thank you.

答: As $G \not \subseteq A_n $, there exists an odd permutation $\sigma \in G $. We can see that $\sigma A_n \subseteq GA_n$ must then consist of only odd permutations, and since it has $\frac{n!}{2}$ elements, it consists of all odd permutations. Moreover, $e \in G$, and thus, $eA_n \subseteq GA_n$ consists of all even permutations. Hence, $GA_n=S_n$.


问:
Is $\{(x, y) \mid y \in \text{Range}(\phi_x)\}$ decidable?
Is the following language decidable? A decidable language must be recursive, right? How should I show that the following is or is not recursive?

$$\{(x, y) \mid y \in \text{Range}(\phi_x)\}$$

答:
Let $A$ denote the set above.

As usual, you can show that $A$ is not computable (recursive, decidable - all the same) if you can reduce it to a set that you know is not computable.

$K$ is a r.e. not computable set.

An equivalent characterization of r.e. sets, is that $B$ is r.e. if and only there is a total computable function $f$ such that the range of $f$ is $B$. This is in fact why r.e. sets are call recursively enumerable since $f(0)$, $f(1)$, $f(2)$, ... is a computable listing of $B$.

So since $K$ is r.e., there is a computable function $f$ such that the range of $f$ is $K$. Since $f$ is computable, it has an index $e$. That is, $f = \Phi_e$.

Then one has that $x \in K$ if and only if $x$ is in the range of $\Phi_x$ if and only if $(e,x) \in A$. This gives a many to one reduction of an incomputable set $K$ to $A$. So $A$ can not be computable.


问:
Determine to point of maximum and minimum of a function
I am learning calculus by watching some videos that I have downloaded some time ago. Some of the videos are missing.I cannot find the part where it teaches point of maximum and minimum.   Along with the videos I have some questions. One of the questions is this

determine the local maximum and minimum points of the function
$$
g(x) = \frac{1}{3}x^3 - 3x^2 + 8x + 5
$$
Can you say how to solve one so I can solve the others.

答:
I'm giving the theory of maximum and minimum first and then you should be able to relate.
If you sketch a graph between an independent variable x and a function of that variable f(x) = y (say), and that can be any arbitrary curve. If you carefully investigate the curve, you will find that, at the point on the curve, where it is maximum or minimum, ie. the height of that point above the x-axis is either maximum or minimum, the curve is kind of flat there.
More specifically, if you look on the tangent at every point on the curve, you will see that at "such" points, the tangent to the curve is parallel to the x-axis.
Conversely, you can say that the point on the curve is maximum or minimum, if the "tangent to the curve at that point" is parallel to the x-axis.
To check that mathematically, you could search for the "slope of the tangent on every point on the curve", and slope of tangent to the curve at any point (x, y) on the curve is dy/dx.
For tangents parallel to x-axis, it's slope is 0, so dy/dx is 0, and remember y=f(x).
Procedure:
Evaluate the first derivative of y=f(x), equate that to 0, to find the x for which f'(x) is 0, and then find the corresponding y for that x, and you are done.</p>


答:
In general you can without comment extend any function $f: A \to B$ to a function of the same name $f: A^n \to B^n$ by defining $f(a_1, a_2, ..., a_n) = (f(a_1), f(a_2), ... f(a_n)).$

In your particular case, $A = \left\{0, 1\right\}$, $B = \mathbb{C}$, $f(a) = e^{i \pi a}$, and $n = 2$.

So $e^{i \pi (x_1, x_2)} = (e^{i \pi x_1}, e^{i \pi x_2})$.

A simpler way to write exactly the same transformation is $f(x) = (-1)^x$. Basically the idea is to send all the $0$ bits to $1$ and send all the $1$ bits to $-1$.

Title: What does $e^{i \pi \mathbf x} \in \mathbb C^2$ mean?

Body:
I saw this in an book (coding theory) without explanation and I need help figuring out the meaning.

Consider $\mathbf x\in \{0,1\}^2$ and
$$
\mathbf s = e^{i \pi \mathbf x} \in \mathbb C^2 \tag 1
$$

I know $\{0,1\}^2$ and $\mathbb C^2$ are the cartesian products, i.e.
\begin{align}
\{0,1\}^2&=\{0,1\}\times \{0,1\}\\
&=\{(0,0),(0,1),(1,0),(1,1)\} \tag 2
\\
\mathbb C^2 &=\mathbb C \times \mathbb C = \{(z,y): z,y\in \mathbb C\} \tag 3
\end{align}
So $\mathbf x= (x_1,x_2)$ is a vector which take a value in the set $\{0,1\}^2$, for instance, $\mathbf x = (1,1)$.

So far so good, but what is $e^{i \pi \mathbf x} \in \mathbb C^2$?
Any examples?

Say $\mathbf x = (1,1)$, what does $e^{i \pi (1,1)} \in \mathbb C^2$ mean?

Any clarification would be appreciated.


问:
Is function $ f(x)= e^{-(x+x^{2})} $ uniformly continuous on $R$
Is function $ f(x)= e^{-( x+ x^{2}) } $ uniformly continuous on $ R $?

As we know for any function $f$ which is continuous on $R$ and $ \lim_{ x\to  \infty} f(x)  $ ,  $ \lim_{x \to - \infty } f(x) $ exist finitely, then $ f$ is uniformly continuous.

I am confused about the limit of $f$ at $ - \infty $.

Please help.

答:
$g(x)=-(x^2+x)=-(x+\frac{1}{2})^2+\frac{1}{4}$ $\Rightarrow$ $\lim_{x \to -\infty}g(x)=-\infty$

$\lim_{x \to -\infty}g(x)=-\infty$   $\Rightarrow$  $\lim_{x \to -\infty}e^{g(x)}=0$


答: Subtract $|a|$ from both sides and use $|x| = |x+y-y| < |x+y|+|y|$, so that  $|x|-|y| < |x+y|$

Title: Show $\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$

Body:
I am tasked with the following:

Show the following inequality for $\lvert b \rvert < \frac{\lvert a \rvert}{2}$ where $a,b \in \Bbb R$:
$$\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$$

I'm a bit at a loss as to where to even start with this one. I'm mostly looking for a little nudge.


问:
Are the sequences $\frac{ih}{2}((1-ih)^{k}-(1+ih)^{k})$ and $\frac{1}{(1+h^{2})^{k}}\frac{ih}{2}((1-ih)^{k}-(1+ih)^{k})$ bounded?
I want to understand the explicit and the implicit Euler method better. Assume I have the initial value problem $y''+y=0$, $y(0)=0$, $y'(0)=1$, which is of course solved by $y=\sin(x)$, and I convert this into a system of first-degree linear equations
\begin{equation*}
\left(\begin{matrix}y'\\z'\end{matrix}\right)=\left(\begin{matrix}0 & 1\\-1 & 0\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)\text{.}
\end{equation*}
When I apply the explicit and the implicit Euler methods to the differential equation $y''+y=0$, with stepwidth $h>0$, I get approximation points that satisfy:
\begin{eqnarray*}
y_{k} & = & \frac{ih}{2}((1-ih)^{k}-(1+ih)^{k}) & \text{for the explicit Euler method}\\
\tilde{y}_{k} & = & \frac{ih}{2(1+h^{2})^{k}}((1-ih)^{k}-(1+ih)^{k}) & \text{for the implicit Euler method}
\end{eqnarray*}
I am interested in seeing how fast these sequences grow with $k$. My guess is that the first sequence is unbounded, whereas the second converges to zero. At least that is what I get for , but I have no idea how to prove this. Any suggestions?

答:
Let $\theta=\arctan h, r=\sqrt{1+h^2}$ and then
$$ 1+ih=r(\cos\theta+i\sin\theta), 1-ih=\sqrt{1+h^2}(\cos\theta-i\sin\theta). $$
So
\begin{eqnarray*}
y_{k} & = & \frac{ih}{2}((1-ih)^{k}-(1+ih)^{k}) \\
&=& hr^k\sin(k\theta).
\end{eqnarray*}
Since $h>0$, you have $r>1$ and $\theta\in(0,\frac\pi2)$ and hence $\{y_k\}$ diverges. Similarly
\begin{eqnarray*}
\tilde{y}_{k} & = & \frac{ih}{2(1+h^{2})^{k}}((1-ih)^{k}-(1+ih)^{k}) \\
&=&hr^{-k}\sin(k\theta).
\end{eqnarray*}
So $\{\tilde{y}_k\}$ converges to $0$ since $r>1$.


问:
Prove that there is a quadratic irreducible in $\mathbb Z_p[x]$ for $p$ prime.
I am kind of stuck at this problem. I know how to find irreducible polynomials in fields like $\mathbb Z_1,\mathbb Z_2$ or where the specific field is given.

I can't figure out a way to prove this. I am thinking to use theorems like:
If an irreducible polynomial $p(x)$ belongs to a field $F$, then $F[x]/\langle p(x)\rangle$ is a field.

Or should I use arguments like if $\mathbb Z_p$ is $\mathbb Z_7$, then polynomials like $x^2 + 3$ are irreducible. So, there exists a quadratic irreducible in $\mathbb Z_p[x]$?

Or is there any other method that I should follow? Any suggestions will be helpful.

答:
A quadratic polynomial is irreducible if and only if it has no roots.

This easily implies that $x^2+x+1$ is irreducible if $p=2$.

If $p\neq 2$, then for any nonzero $a\in \mathbf Z_p$, you have $a\neq -a$ and $a^2=(-a)^2$. Thus, the function $x\mapsto x^2$ is non-injective, so (because $\mathbf Z_p$ is finite), it is non-surjective. It follows that for some $a$, the polynomial $x^2-a$ has no root.

Note that this proof works for any finite field of characteristic $p>2$.


问:
What is the answer of $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$
I found a limit question in my textbook but i could not obtained the result given by answer key.

Let me introduce the question :  $ \lim_{n\to \infty}\frac{{(-1)}^{(n-1)}\cdot n}{(n+1)!}$

I tried to separate it into two part such that $ \lim_{n\to \infty}{(-1)}^{(n-1)}\cdot\lim_{n\to \infty}\frac{n}{(n+1)!}$

As you see, this partition gave me "$\text{oscillating (undetermined)}$".$0$

However, the answer is zero. What am I missing? Thank you for your helps ...

答:
You managed to rewrite your sequence as $x_n=y_n\cdot z_n$ where $y_n$ is bounded and $z_n\to0$. (In your case, $y_n=(-1)^n$ and $z_n=n/(n+1)!$.

In general, the following is true: If $y_n$ is bounded and $\lim\limits_{n\to\infty} z_n=0$ then $$\lim\limits_{n\to\infty} y_nz_n=0.$$

Proof. Let $M$ be such that $|y_n|\le M$ for each $n$. If we are given $\newcommand{\ve}{\varepsilon}\ve>0$, then there is $n_0$ such that for $n\ge n_0$ we have
$$|z_n| < \frac\ve{M}.$$
Then for every $n\ge n_0$ we also get
$$|y_nz_n| < M\cdot \frac\ve{M} = \ve.$$
This shows that the sequence $(y_nz_n)$ converges to zero. $\square$


答:
Observe that, since $A_n$ is normal, $GA_n$ is necessarily a subgroup of $S_n.$

Also $|G|\ge2$ and $|G\cap A_n|=1$ gives us $$|GA_n|=\dfrac{|G||A_n|}{|G\cap A_n|}=?$$

In fact, this tells you more about the order of such groups $G.$

Title: Subgroup of $S_{n}$ that is not in $A_{n}$

Body:
I'm trying to prove this following statement:

Let $G\leq S_{n}$ for $n\geq 5$.

I need to show that if $G\not\subseteq A_{n}$ then $GA_{n}=S_{n}$.

I was thinking to show it implies $G\cap A_{n}=e$, and then using direct product, but I don't know if it is necessarily correct. If it doesn't, can I get a hint?

Thank you.


问:
Show that the tangents at P & Q meet on the curve...
A straight line is drawn parallel to the conjugate axis of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to meet it and the conjugate hyperbola in the points P & Q. Show that the tangents at P & Q meet on the curve $\frac{y^4}{b^4}\left(\frac{y^2}{b^2}-\frac{x^2}{a^2}\right)=4\frac{x^2}{a^2}$.

So here is my approach:

taking the point P as

$P\left(a.sec\theta, b.tan\theta \right)$,

since the line was parallel to the conjugate axis, i.e., the y axis, the x coordinate of Q will also be $(a.sec\theta)$ and by putting this in the equation of the conjugate hyperbola, I was able to get the y-coordinate of Q and finally, the point was:

$Q\left(a.sec\theta, b\sqrt{1+sec^2\theta}\right)$

So finally I got the respective tangents by putting the values of P and Q in the general tangent equations:

Tangent at P: $\frac{x.sec\theta}{a}-\frac{y.tan\theta}{b}=1$

Tangent at Q:  $\frac{x.sec\theta}{a}-\frac{y.\sqrt{1+sec^2\theta}}{b}=-1$

This is where I got stuck as solving these equations yields a weird result. Can someone please let me know how I can proceed or if this approach is wrong?

答:
Hint:

Let $P(a\tan u,b\sec u);Q(a\tan v,b\sec v)$

The gradient of $PQ$ is  $$\dfrac{b(\sec u-\sec v)}{a(\tan v-\tan u)}=\dfrac{b(\cos v-\cos u)}{a\sin(u-v)}=\dfrac{2b\sin\dfrac{u+v}2}{2a\cos\dfrac{u+v}2}$$

As $PQ|| $ the conjugate axis, $\cos\dfrac{u+v}2=0\implies v=(2n+1)\pi-u$ where $n$ is any integer

$\implies Q(-a\tan u,-b\sec u)$

Now form the equation of the tangents  at $P,Q$ and solve for $x,y$


问:
Solving a congruence equation
Let $\pi$ be a primitive $p$-th root of unity, $p$ some prime and
$$ x= \sqrt{ \left(\frac{ \pi^k -1}{\pi-1}\right) \overline{\left(\frac{ \pi^k -1}{\pi-1}\right)} } \in \mathbb{Z}[\pi]$$
for some natural exponent $k<p$. Is it really true that $x \equiv k \operatorname {mod} (\pi-1)$ ?

Thanks for helping

edit: to use a different notation, let $\pi:=\vartheta$

Title: Solving a congruence equation

Body:
Some ideas\hints :

$$\left(\frac{ \pi^k -1}{\pi-1}\right) \overline{\left(\frac{ \pi^k -1}{\pi-1}\right)}=\left|\,\frac{\pi^k-1}{\pi-1}\,\right|^2\implies x=\left|\,\frac{\pi^k-1}{\pi-1}\,\right|=|\pi^{k-1}+\pi^{k-2}+\ldots+\pi+1|$$

But since

$$1+\pi+\pi^2+\ldots+\pi^{p-1}=0\implies x=-|\pi^k+\pi^{k+1}+\ldots+\pi^{p-1}|$$

Finally :

$$\pi^k+\ldots+\pi^{p-1}=(\pi^k-1)+(\pi^{k+1}-1)+\ldots+(\pi^{p-1}-1)-(p-k-1)$$

and for $\,k\le j\le p-1\,$ :

$$\pi^j-1=(\pi-1)(\pi^{j-1}+\ldots+\pi+1)$$


问:
If $q < 0$, prove that the power mean $M_q$ is lesser than the geometric mean $G$
Let $p, q \in \mathbb{Z}$. Assuming the lemma that when $p > 0$ then $M_p > G$, I need to prove that if $q < 0$ then $M_q < G$.

I'm trying to prove it this way:

Let $q = -k$ and $k > 0$. Then $(M_k)^k > G^k$ then $(M_{-q})^{-q} > G^{-q}$ then $G^q > (M_{-q})^q$ then $G > M_{-q}$, but now that negative subindex is bothering me and I can't find a way to start with a hypothesis that allows me to conclude that $G > M_{q}$.

I found a solution here, in the end of the article, but I don't know how is this implication correct:

$G^{-q} < (M_{-q})^{-q} \implies G^q > M_{q}^q$

Shouldn't $G^{-q} < (M_{-q})^{-q}$ just imply $G^q > M_{-q}^q$? How can I prove this?

PD: For context, this is the problem 20 of Apostol's Calculus, volume I first chapter, so I would like to keep the proof as basic as possible. Thanks!

答:
Let $p=-q>0$, and apply your lemma to $\frac{1}{x_i}$:
$$\left(\frac{\sum\left(\frac{1}{x_i}\right)^{p}}{n}\right)^{\frac 1p} \geqslant \left(\Pi \frac{1}{x_i}\right)^{1/n} \implies \left(\frac{\sum\left(\frac{1}{x_i}\right)^{-q}}{n}\right)^{\frac {-1}{q}} \geqslant \left(\Pi \frac{1}{x_i}\right)^{1/n}\\
\implies \left(\Pi x_i\right)^{1/n} \geqslant \left(\frac{\sum x_i^q}{n}\right)^{\frac {1}{q}}
$$


答: We have $[S_n:A_n]=2$, so $GA_n$ can only be $A_n$ or $S_n$, since there are no other subgroups in between $A_n$ and $S_n$. Since $G\not\subseteq A_n$, we cannot have $GA_n=A_n$, so it must be $S_n$.


问:
How to correct the azimuth obtained from coordinates of two points.
Assuming we have points $A (-0.4664, 0.7104)$ and $B (-1.8112, 2.8032)$.
To calculate their azimuth the following equation may be used: $$\text{Azimuth}=\tan^{-1}\frac{X_B-X_A}{Y_B-Y_A}=\tan^{-1}\frac{-1.3448}{2.0928}=-32.72415526582776^\circ$$
However, based on the sign of $X_B-X_A<0$ and $Y_B-Y_A>0$ this angle must be found in quadrant 4 and thus should range between $270$ and $360$ degrees. (The azimuth is measured clockwise from the north or positive y-axis). To correct this angle, we can add/subtract $180$ or $360$ to the calculated azimuth. In this case, adding 360 will solve the issue since $360-32.73272415526582776=327.2672$ degrees.

My question is the following: If for the same example, the calculated azimuth was found to be $32$ degrees. How do we correct this sort of angle? Since adding/subtracting $180$ and $360$, in this case, will not help.

答:
I guess you look for the function arctan2(y,x):

```<code>from numpy import arctan2
azimuth = - arctan2(-1, 0) + math.pi/2
if azimuth < 0:
    azimuth = azimuth + 2 * math.pi

</code>```

will give you $\pi$ (x=0, y=-1) and it returns the correct angles from 0 to 360 degrees.

Body:
I guess you look for the function arctan2(y,x):

from numpy import arctan2
azimuth = - arctan2(-1, 0) + math.pi/2
if azimuth < 0:
    azimuth = azimuth + 2 * math.pi

will give you $\pi$ (x=0, y=-1) and it returns the correct angles from 0 to 360 degrees.


问:
Generalisation of IMO 1990/P3:For which $b $ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

It was from my LTE/ Zsigmondy handout.

By taking examples, it looks like for $b= 2^k-1 , 2$ it's not true .

Here's my progress:

I got $b=4,5,6,8,9$ works ( $2,3,7$ doesn't )

$n$ is odd : If not then $4\mid b^n-1$ , but $b^n \equiv 0,1 \mod 4$

If $b+1$ is a power of $2$ and $n^2\mid b^n+1$, then $n=1$ : Let $p$ smallest prime dividing $n$ ( note that it can't be $2$) . Then $n^2\mid b^{2n}-1 \implies p\mid b^{2n}-1 \implies p\mid b^{\gcd(2n,p-1)}-1 \implies p\mid b^2-1\implies p|b-1 \text{(since $b+1$ is power of $2$)}\implies p\mid b^n+1 \implies p|2.$ Contradiction.

Any hints? Thanks in advance!

答:
We can construct b, we want:

$b^n+1\equiv 0 \mod (n^2)$

If $b=t.n^2-1$ and $n =2k+1$ is odd, then we have:

$(t.n^2-1)^n+1=M(n^2)-1+1=M(n^2)\equiv 0 \ mod (n^2)$

Where $M(n)$ is a multiple of $n^2$.

The condition is that n must be odd.


答:
The hypothesis $n \geqslant 5$ is not necessary. For $n \geqslant 2$ we have that $(\Sigma_n\colon\mathrm{A}_n)=2$ and we appeal to the following general:

Lemma. Let $G$ be an arbitrary group and $H \leqslant G$ be a subgroup such that the index $(G \colon H)$ be a prime natural number. Then $H$ is a maximal subgroup of $G$ (i.e. a maximal element of the set of proper subgroups of $G$ ordered by inclusion).

We gather in particular that the alternate group $\mathrm{A}_n$ is maximal in $\Sigma_n$. If $\Gamma \leqslant \Sigma_n$ we have on the one hand that $\Gamma\mathrm{A}_n \leqslant \Sigma_n$ by normality of $\mathrm{A}_n$. By virtue of the maximality of $\mathrm{A}_n$, this relation entails $\Gamma\mathrm{A}_n=\Sigma_n$ or $\Gamma\mathrm{A}_n=\mathrm{A}_n$, the latter being equivalent to $\Gamma \leqslant \mathrm{A}_n$.

For $n=0, 1$ the statement is trivially true, as $\Sigma_0$ and $\Sigma_1$ are both trivial groups. The claim you make generalises to finitary groups of permutations $\Sigma_0(A)$ on arbitrary sets $A$.


问:
parametric equation of the circle in polar form with center different from origin
To find the parametric equation of the circle in polar form of radius $1$ with center $(-1,-1)$ where we start at the point $(-1,0)$ at $\theta = 0$ and travel counterclockwise up to $\theta = 2 \pi$.

I have found the following:

$$x = -1 + \sin(\theta), y = -1 + \cos(\theta).$$ Is the parametrization correct?

答:
Sure, indeed
$$
x^2+y^2=(-1+\sin(\theta))^2+(-1+\cos(\theta))^2=3-2\sin(\theta)-2\cos(\theta)
$$
and knowing that $\sin(\theta)=x+1$ and $\cos(\theta)=y+1$, you get
$$
x^2+y^2=-2x-2y-1
$$
i.e.,
$$
x^2+2x+1-1+y^2+2y+1-1=-1
$$
or
$$
(x+1)^2+(y+1)^2=1
$$
which is the cartesian equation of the circumference you want to parameterize.

In general, applying the same argument, you can prove that a parameterization of the circumference with center $(x_C,y_C)$ and radius $R$ is
$$
\begin{cases}
x=x_C+R\sin(\theta)\\
y=y_C+R\cos(\theta)\\
\end{cases}
$$
with $\theta\in [0,2\pi)$ (or any other interval of the same amplitude).
Note that also
$$
\begin{cases}
x=x_C+R\cos(\theta)\\
y=y_C+R\sin(\theta)\\
\end{cases}
$$
is a parameterization of the same cirumference, but with a different starting point.


问:
Is $ds^2=\frac{dxdy}{xy(x-1)(y-1)}$ a valid semi-Riemannian metric?

Is $ds^2=\frac{dxdy}{xy(x-1)(y-1)}$ a valid semi-Riemannian metric?

I got it from adding together four semi-Riemannian metrics:

$ds^2=\frac{dxdy}{xy}+\frac{dxdy}{x-xy}+\frac{dxdy}{y-yx}+\frac{dxdy}{(x-1)(y-1)}=\frac{dxdy}{xy(x-1)(y-1)}$

I have checked that these four metrics are indeed valid semi-Riemannian metrics describing Minkowski space in a unique coordinate system. For example, the first metric in the sum, can be verified to be Minkowski upon inspection. One way to see this is to do a change of coordinates:  $u=\frac{dx}{x}$ and $v=\frac{dy}{y}.$

I have verified that the metric $ds^2$ is undefined at $x,y=0,1.$

答:
Your tensor $g$ at a point $(x_0,y_0)$ with $x_0,y_0 \neq 0,1$ is given by
$$g = \frac{1}{x_0y_0(x_0-1)(y_0-1)}\mathrm{d}x\mathrm{d}y$$

On each connected component of $\mathbb{R}^{2} \setminus\{(x,y) ~|~ x=0 \text{ or } x=1 \text{ or } y=0 \text{ or } y=1\}$, $g$ is conformal to the tensor $\bar{g} = \pm\mathrm{d}x\mathrm{d}y$, which is symmetric and non-degenerate, of type $(1,1)$. The $\pm$ is because the function $f(x_0,y_0) = \frac{1}{x_0y_0(x_0-1)(y_0-1)}$ is not of constant sign, but it is on every connected component.

Of course the natural coordinates are not globally adapted to the tensor everywhere because of the change of the sign. But the signature of $\pm \mathrm{dx}\mathrm{d}y$ is unvariant and equal to $(1,1)$ so the result follows.

I would add that it seems weird to me to add "the same tensor written in different coordinates system but in the same coordinate system". I mean it has no geometric sense. For example, take the tensor $g=\mathrm{d}x \mathrm{d}y$ on $\mathbb{R}^2$ with natural coordinates. If one takes the $(-x,y)$ coordinates, then $g$ is then written $-\mathrm{d}x \mathrm{d}y$. But the tensor $h = \mathrm{d}x\mathrm{d}y + (-\mathrm{d}x\mathrm{d}y)$ has nothing to do with $"g+g"$ even if $g$ can be written as one or the other in some particular coordinates system.


问:
What is the norm of the gradient of $f$ in normal coordinate?
Let $M$ be a Riemannian manifold and $f$ a smooth function on $M$. The Bochner formula proved in Schoen-Yau's book "lectures in Differential Geometry": (prop. 2.2)
$$
\Delta |\nabla f|^2(p)=2\sum _{ij} |f_{ij}|^2 +2 R_{ij}f_i f_j +2 \sum f_i(\Delta f)_i
$$
The proof assumes that ${x_i}$ are normal coordinates around $p$. and $f_i$ is the co variant differential of $f$ with respect to $\partial/\partial x_i$. $R_{ij}$ is the Ricci tensor.

Then the proof proceed as follows: Since $|\nabla f|^2=\sum f_i^2$. Hence at $p$ one has
$$
\Delta |\nabla f|^2 =\sum_j(\sum f_i^2)_{jj}=.....
$$

My question is actually the first claim $$|\nabla f|^2=\sum f_i^2$$
It seems for my that in general $$\nabla f=f_sg^{si}\frac{\partial}{\partial x_i}$$
In normal coordinate we can only get the expression $|\nabla f|^2=\sum f_i^2$ only holds at the point $p$, not other points. But clearly we need to take derivative, the value at the point $p$ is not enough.

It's most likely I miss some point here, but I can't figure out why. Anyone can help?

答:
If I am not mistaken, the point that achille hui was trying to make is that you can commute tracing with respect to the metric $g$ and covariant derivatives. In other words, you can take the terms $g^{ij}$ out of the covariant derivatives.

In general, you have $\left|\nabla f\right|^2=g^{ij}f_if_j$. Then,

$$\Delta\left(\left|\nabla f\right|^2\right)=tr_g\nabla^2\left(\left|\nabla f\right|^2\right)=g^{kl}\nabla^2_{k,l}(g^{ij}f_if_j)=g^{kl}g^{ij}\nabla^2_{k,l}(f_if_j).$$

So if you are working in geodesic normal coordinates centered at $p$, the RHS corresponds to the term $\sum\limits_j\left(\sum\limits_if_i^2\right)_{jj}$ in your question.


问:
Compute the flow of a Killing field
I know that the vector field $$X = a_1\partial_1 + a_2\partial_2$$ where $a_1,a_2 : \mathbb{R}^2 \rightarrow \mathbb{R}$ are smooth,  is a Killing field on $\mathbb{R}^2$ with the Euclidean metric $dx_1^2 + dx_2^2$.

I have to show that the flow of X is $$\phi_t^X(x) = A_tx+c_t$$ where $A_t$ is a rotation and $c_t$ is a constant vector in $\mathbb{R}^2$.
Can someone help me ?

答:
In your other question, I answered you that a vector field $X =a_1\partial_x + a_2 \partial_y$ is Killing in the euclidean plane if and only if
$$
\partial_x a_1 = \partial_ya_2 = 0,~ \partial_ya_1 = -\partial_x a_2
$$
By some (easy) calculations, one then shows that $X$ is Killing if and only if there exist a constant $\lambda$ and a constant vector field $V$ such that
$$
X = \lambda(y\partial_x - x\partial_y) + V
$$
It is then easy to compute the flow of $X$. Let $\gamma : t \mapsto \gamma(t) = (x(t),y(t))$ be a smooth path. Then $\gamma$ is an integral curve of $X$ if and only if

$$\gamma'(t) = X_{\gamma(t)}$$

This leads to a linear differential equation that is easy to solve. The equation is
$$
\begin{pmatrix}x(t) \\ y(t) \end{pmatrix}' = \begin{pmatrix}0 & -\lambda\\
\lambda & 0 \end{pmatrix}\begin{pmatrix}x(t) \\ y(t) \end{pmatrix} + V
$$

The solutions of this differential equation are
$$
\begin{pmatrix}
x(t) \\ y(t)
\end{pmatrix}= \begin{pmatrix}
\cos\lambda t & -\sin \lambda t \\
\sin \lambda t & \cos \lambda t
\end{pmatrix} \begin{pmatrix}
x_0 \\ y_0
\end{pmatrix} + tV
$$

Thus, $\phi_t^X (x_0,y_0) = A_t(x_0,y_0) + V_t$ where $A_t = \begin{pmatrix}
\cos \lambda t & -\sin \lambda t \\
\sin \lambda t & \cos \lambda t
\end{pmatrix} $ and $V_t = tV$.
This shows that the flow of $X$ acts by isometries. More generally, one can show that in a general riemannian manifold, the flow of a Killing vector field acts by riemannian isometries. The converse is more easy to show: if the flow $\phi_t^X$ acts by isometries, then $(\phi_t^X)^*g = g$ and $L_Xg = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}((\phi^X_t)^*g) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}g=0$, and hence $X$ is Killing.

Comment: During the calculations, I was wondering why would it be mandatory for the rotations to have constant speed $\lambda$ and for $V$ to be constant. I mean, why would $\phi^X_t$ not have the general form
$$
\phi_t(x_0,y_0) = \begin{pmatrix}
\cos \lambda(t)& \sin \lambda(t) \\
\sin\lambda(t) & \cos \lambda(t)\end{pmatrix}(x_0,y_0) + V(t)
$$
and why $\lambda$ and $V$ are forced to be linear in $t$? Indeed this general forms are isometries. But the reason is the following: for $\phi_t$ like above to be a flow, it is mandatory to have $\phi_0 = \mathrm{id}$ and $\phi_t\circ \phi_s = \phi_{t+s}$. Thus, $\lambda$ and $V$ must satisfy
\begin{align}
\lambda(0) &= 0& \lambda(t+s)&=\lambda(t) + \lambda(s) \\
V(0) &= 0& V(t+s) &= V(t)+V(s)
\end{align}
and then they have to be linear.

Title: Compute the flow of a Killing field


问:
Prove by definition that $H_a$ is a Group
I've got this question about groups, I know all definitions of a group that it has to be (Closure, Associativity,   Identity and Inverse), but I have no clue how to start this question. I'll be happy if someone could help.

Let $〈G,*〉$ be a group and $a\in G$, let $H_a=\{a*g*a^{-1} \mid g\in G\}$.

Prove by definition that $H_a$ is a group.

答:
Clearly $H_a\subset G$. The question is whether it is a subgroup: does it contain the identity element $e$ of $G$, and is it closed under $*$ and under inverses? (You get associativity for free, since you're using the same operation $*$ as in $G$.)

So you want to show that:

$e$ is in $H_a$.

If $h_0$ and $h_1$ are in $H_a$, then $h_0 * h_1$ is also in $H_a$.

If $h$ is in $H_a$, then $h^{-1}$ is also in $H_a$.

For example, for 3: if $h \in H_a$, that means there's a $g \in G$ such that $h = a g a^{-1}$. Can you use this $g$ to show that $h^{-1}$ is in $H_a$?


问:
Using Stone-Weierstrass theorem prove that $f\equiv 0$ in $[0,1]$ under certain conditions.

Let $f$ be a continuous function over $[0,1]$ and $\displaystyle \int_0^1 f(x)\,dx=0$. If for any positive integer $n$, $\displaystyle \int_0^1 x^{12+3n}f(x)\,dx=0$ then using Stone-Weiersterass Theorem prove that $f\equiv 0$ in $[0,1]$.

Since $f$ is contnuous on $[0,1]$ so by Stone-Weierstrass Theorem there exists a sequence of polynomials $\{p_n(x)\}_n$ which converges uniformly to $f$ on $[0,1]$.  That is  , $p_n\to f$ , implies  $p_n(x)f(x)\to f^2(x)$ as $n\to \infty$, which implies that $\displaystyle \int_0^1p_n(x)f(x)\,dx \to \int_0^1 f^2(x)\,dx$. But from the given condition I can't say that $\displaystyle \int_0^1 f^2(x)\,dx =0$. So how to proceed to solve this problem ? Any hint. please.

答:
The span of the monomials $\{1\} \cup \{x^{3n+12} : n\in\Bbb{N}\}$ is dense in $C[0,1]$ because
$$\sum_{n=1}^\infty \frac1{3n+12} = \frac13 \sum_{n=1}^\infty \frac1{n+4} = \frac13 \sum_{n=5}^\infty \frac1n = +\infty$$

so the Müntz–Szász theorem gives the conclusion.

Now approximate $f$ by polynomials from $\operatorname{span}(\{1\} \cup \{x^{3n+12} : n\in\Bbb{N}\})$.


问:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?

I know that $289 = 17 \times 17$

so would it be $2^8\equiv 256\bmod17 =1$

and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this

Body:
$2^8\equiv1\bmod17$, so

$2^{128}+2^{120}+2^{112}+\cdots+2^{16}+2^{8}+1\equiv1+1+1+\cdots+1+1+1=17\equiv0\bmod17,$

so $2^{136}-1=(2^{128}+2^{120}+2^{112}+\cdots+2^{16}+2^{8}+1)(2^8-1)\equiv0\bmod289$,

but $2^8-1=255\not\equiv0\bmod289$,

and $2^{68}-1\not\equiv0\bmod289$ because $2^{68}-1\equiv2^4-1=15\not\equiv0\bmod17$,

so, by the order test (linked in Bill Dubuque's answer), the order of $2$ mod $289$ is $136$.


问:
Show that $x(t) = e^{- \gamma t} \cos(\omega t) x(0)$ is the solution a diffusion equation.
I am working through Kurt Jacob's Stochastic Processes for Physicists and came across a question I am unsure about on linear diffusions.

The equation for the damped harmonic oscillator is

\begin{align} m \frac{\mathrm{d}^2 x}{\mathrm{d}^2 t}  + 2 \gamma m
 \frac{\mathrm{d} x}{\mathrm{d} t} + kx = 0 . \end{align}

Assume that $\gamma < \sqrt{k/m}$, in which the resonator is said to
be "under-damped". Show that

$$ x(t) = e^{- \gamma t} \cos(\omega t) x(0)$$

is the solution to the equation above, and find the expression for the
frequency $\omega$ in terms of $m,k$ and $\gamma$.

(pg 24 Question 1)

答:
An explanation "keeping in mind" for the students:

The equation for your damped oscillator is:
$$m \frac{\mathrm{d}^2 x}{\mathrm{d}^2 t}  + 2 \gamma m \frac{\mathrm{d} x}{\mathrm{d} t} + kx = 0$$
To show that $$ x(t) = e^{- \gamma t} \cos(\omega t) x(0)$$  is the solution to the above equation, recall the fact that $x(t)$  can only be a solution to the equation if it satisfies the differential equation. I am not solving the differential equation as it has already been solved previously. If you want to solve this equation there are many ways of doing it, from complex exponentials to Laplace transform and what not. Usually sometimes we also assume functions of some specific form to be the solution and carry on the analysis to find it.

Calculate the derivatives :
$$ x(t) =  x(0)\ e^{- \gamma t} \cos(\omega t)$$
$$ x'(t) = x(0)[- \gamma  e^{- \gamma t} \cos(\omega t) - \omega e^{- \gamma t} \sin(\omega t)]$$
$$ x''(t) = x(0)[ \gamma^2  e^{- \gamma t} \cos(\omega t) +\gamma \omega e^{- \gamma t} \sin(\omega t) - \omega^2e^{- \gamma t} \cos (\omega t) + \gamma \omega e^{- \gamma t} \sin(\omega t)]$$

Now, put $x(t)$ in the differential equation to get:
$$m x(0)[ \gamma^2  e^{- \gamma t} \cos(\omega t) +\gamma \omega e^{- \gamma t} \sin(\omega t) - \omega^2e^{- \gamma t} \cos (\omega t) + \gamma \omega e^{- \gamma t} \sin(\omega t)] + 2 x(0)[- \gamma  e^{- \gamma t} \cos(\omega t) - \omega e^{- \gamma t} \sin(\omega t)] + kx(0)\ e^{- \gamma t} \cos(\omega t) = 0$$

When you solve it further, you will get this equation:
$$[m(\gamma^2 - \omega^2) - 2m \gamma^2 + k]e^{-\gamma \omega t}\cos{\omega t} = 0$$
or
$$[k - m(\gamma^2 + \omega^2)]e^{-\gamma \omega t}\cos{\omega t} = 0$$
This is the condition that should be satisfied for $x(t)$ to be a solution of the equation. What does this mean? This means that this particular complex exponential should be 0.

The factor $\cos{\omega t}$ becomes 0 when $\omega t = (2n+1) \frac{\pi}{2}$

The factor $e^{-\gamma \omega t} = 0$ when $\gamma \omega t$ becomes infinitely large.

But the first factor becomes 0 when,
$$k - m(\gamma^2 + \omega^2)=0$$
or
$$\omega^2 = \frac{k}{m} - \gamma^2$$
giving you the result for $\omega$ as
$$\omega = \sqrt{\frac{k}{m} - \gamma^2}$$

In the question it has been given that
$$\gamma^2 \lt \frac{k}{m}$$ which makes your system underdamped and if you see the equation for $\omega$ you will see it will come out as positive. Now, there are lots of things in this, first there is a relation between these oscillations and phasors. Then there are ways to solve this equation involving many techniques. One of the technique has been shared as the answer. You can also have a pole plot analysis for looking at the stability. But from the student's point of view I think this is enough.
PS: You can solve this equation using complex exponentials, and then make the imaginary part as $0$ in the previous answer.
Hope this helps...


问:
What are the rules for changing indexes and powers in a series?
My TA told me that when I want to change indexes in a series, I can make a substitution just like when I'm doing integral by substitution. But I think I have encountered some cases where this goes wrong. So can anybody tell me what substitutions are allowed if I want to rewrite a series?

As an example, if I have a geometric series:
$$
\sum_{n=2}^{\infty} z^{n-2}=\sum_{k=0}^{\infty} z^{k}
$$
I can change the index and power by substituting k=n-2 and change the limits correspondingly. This substitution is valid.

But If you instead consider a power series of the form:
$$
\sum_{n=0}^{\infty} c_{n} x^{2 n+1} \neq \sum_{k=1}^{\infty} c_{\frac{k-1}{2}} x^{k}
$$
I believe it's generally untrue, that I can just substitute $k=2n+1$, so $n=(k-1)/2$.

A specific case where I think this is wrong is this series, which I tried to rewrite:
$$\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}= \sum_{k=0}^{\infty}(-1)^{k / 2} \frac{x^{k}}{k !}=\sum_{k=0}^{\infty} \frac{i^{k} x^{k}}{k !}$$
With k=2n.
But I'm pretty certain that the equal sign is wrong (since the first is the cosine series, and the second is some complex series), and thus what my TA told me must be wrong?

The reason why I wanted to learn how to rewrite the series, is that I often have formulas, that only work for specific expressions. For example the sum of a geometric series, or how to find the radius of convergence:

Theorem: Let $\sum_{n=0}^{\infty} a_{n} x^{n}$ Be a power series. The limit
$$
R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|
$$ is the radius of convergence for the series.

答:
Consider the power series$$\sum_{n=0}^\infty2^nx^{3n}.\tag1$$I suppose that this is one of those power series to which you would like to apply the formula$$\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\tag2$$in order to compute the radius of convergence. But you can't. That is, yes, you can write $(1)$ as a power series of the type$$\sum_{n=0}^\infty a_nx^n.$$The problem is that then$$a_n=\begin{cases}2^{n/3}&\text{ if }3\mid n\\0&\text{ otherwise.}\end{cases}$$and those $0$'s will not let you use formula $(2)$.

But you can solve it in another way: compute$$\left|\lim_{n\to\infty}\frac{2^{n+1}x^{3(n+1)}}{2^nx^n}\right|=\left|\lim_{n\to\infty}{2x^3}\right|,$$which will be $\infty$ if $|x|>\frac1{\sqrt[3]2}$ and $0$ if $|x|<\frac1{\sqrt[3]2}$. Therefore, the radius of convergence is $\frac1{\sqrt[3]2}$.


问:
Minkowski Theory
In Jurgen Neurich, under the section of Minkowski Theory, the following thing is mentioned,

Let $K/\mathbb{Q}$ be an algebraic extension of degree $n$ over $\mathbb{Q}$ and let $\tau$ denote complex embedding of $K$ in  $\mathbb{C}$. Let $K_{\mathbb{C}}=\prod_{\tau}\mathbb
{C}$, where $\tau$ varies over $n$ complex embeddings and let galois group $G(\mathbb{C}/\mathbb{R})$ be generated by complex conjugation, $F:z\mapsto \bar{z}$. Now, $F$ acts on $K_{\mathbb{C}}\mapsto K_{\mathbb{C}}$, by conjugating each factor and also by conjugating $\tau$ according to Neurich.

The definition of $K_{\mathbb{C}}$ is little confusing. The way I understood it is, $K_{\mathbb{C}}$ is the direct product of the embedded fields of $K$ in $\mathbb{C}$.

Can anyone explain why $F$ acts on $K_{\mathbb{C}}$, by conjugating each factors as well as $\tau$? In other words, why $F(z)=\bar{z}_{\bar{\tau}}$, where $z\in K_{\mathbb{C}}$. Again I am not quite sure about the notation, so I may have not explained it properly.

答:
Your definition of action is correct. One might say that there is some abuse of notation but everything is clear once we understand what is going on. I explain it a bit here.

Think of $K_{\mathbb C} $ as direct product of $\mathbb C $s with complex embeddings $\tau $s as labels. It may be a good idea to write it as $$K_{\mathbb C} = \prod _{\tau} \mathbb C_{\tau} $$

When $F$ acts, it acts on the elements as well as the labels. It shuffles the elements in the direct product (and also acts on the elements themselves, by conjugation), via conjugation.

So for example suppose we only have a pair of complex embeddings, say for $\mathbb Q [i] / \mathbb Q $, then $K_{\mathbb C} = \mathbb C \times \mathbb C $. Let $\tau $ and $\bar \tau $ be the natural embedding and conjugate of that.

Here an element of $K_{\mathbb C}$ looks like $(\__{\tau}, \__{\bar \tau}) $. Say we have $(i,0) $, then $F$ acts on this as $F(i,o)=(0,-i) $.

Title: Minkowski Theory


问:
Is this language decidable?
Is this language decidable?

$$\{x\mid \text{$x$ is the code of a Turing machine that always halts on $y$
    in less than $y^3$ steps}\}$$

I think it is, because it halts in a finite number of states. Am I right? Is there a more classic way of saying so?

答:
No, it is not computable. First, the intuition is that you can not know that $\Phi_x$ halts on all $y$ in less than $y^3$ steps without actually checking it for all of the infinitely many $y$.

Let $A$ denote the set above.

Now let $\bar{K}$ denote the complement of $K$. Given $x$, define a Turing machine $T_x$ as follows: On input $y$, run $\Phi_x(x)$ for $y$ steps. If $\Phi_x(x)$ has not halted, then $T_x(y)$ will halt in $y$ steps. If $\Phi_x(x)$ has halted by $y$ steps, then $T_x(y)$ will keep running without ever halting.

It is clear that there is a computable $f$ that takes $x$ to the index of the Turing $T_x$ constructed above.

If $x \in \bar{K}$, then $f(x)$ is the index of a Turing program that halts on all input in $y < y^3$ steps. So $f(x) \in A$.

If $x \notin \bar{K}$, then $x \in K$. So $\Phi_x(x)$ will halt at some stage $k$. Then $T_x$ is a Turing that does not even halt on any $y > s$. So $f(x) \notin A$.

This is a many to one reduction of $\bar{K}$ to $A$. Since $\bar{K}$ is not computable, neither is $A$.

Title: Is this language decidable?


问:
Find a matrix with given row and column sums
Please forgive my intrusion. I've been working for days on this problem and it's vexing me. It doesn't seem to have a solution and I could really use some help. I have a "math square" (not a magic square) that looks like this when filled in

$$
\begin{array}{ccccc|c}
9 & 4 & 8 & 4 & 7 & 32 \\
7 & 9 & 15 & 7 & 5 & 43 \\
3 & 2 & 9 & 10 & 9 & 33 \\
5 & 3 & 5 & 6 & 4 & 23 \\
\hline
24 & 18 & 37 & 27 & 25 & 131 \\
\end{array}
$$

However, the puzzle when empty looks like this.

$$
\begin{array}{ccccc|c}
x & x & x & x & x & 32 \\
x & x & x & x & x & 43 \\
x & x & x & x & x & 33 \\
x & x & x & x & x & 23 \\
\hline
24 & 18 & 37 & 27 & 25 & 131 \\
\end{array}
$$

I need an equation of some sort to fill in the unknowns ($X$'s) and recreate the missing values.

There appears to be some symmetry with the puzzle as the columns and row all add up to be the same which in this case is $131$. If you separate them by every other row to perhaps break it down to make it easier to solve you get. You could also do this with the columns but for simplicity I haven't written it here.

$$
\begin{array}{ccccc|c}
9 & 4 & 8 & 4 & 7 & 32 \\
3 & 2 & 9 & 10 & 9 & 33 \\
\hline
12 & 6 & 17 & 14 & 16 & 65 \\
\end{array}
\dots
\begin{array}{ccccc|c}
7 & 9 & 15 & 7 & 5 & 43 \\
5 & 3 & 5 & 6 & 4 & 23\\
\hline
12 & 12 & 20 & 13 & 9 & 86   \\
\end{array}
$$

using these givens in the puzzle is allowed, but not the $X$'s

If it is indeed unsolvable I would like to know but if it can what missing component can be add to achieve that goal?

Your help is very much appreciated and thank you!

Regards,

Tony

p.s. sorry about the formatting.

答:
Here is a theorem I found referenced by Alexander Barvinok in his "Matrices with prescribed row and column sums" where he in turn references Section 6.2 of "Combinatorial Matrix Theory" by Brualdi and Ryser:

Let $R = (r_1, r_2, \dots, r_m)$ and $S = (s_1, s_2, \dots, s_n)$ be non-negative integral vectors. There exists an $m$ by $n$ non-negative integral matrix with row sum vector $R$ and column sum vector $S$ if and only if $\sum_{i = 1}^{m}r_i = \sum_{j = 1}^n s_j$

Proof: if there exists such a matrix, then the equality of the sums holds. Conversely, suppose that the equality holds. Now let's inductively construct the matrix $A = [a_{ij}]$ with row sum vector $R$ and column sum vector $S$:

If $m = 1$, then we let $A = [s_1, s_2, \dots, s_n]$

If $n = 1$, then we let $A = [r_1, r_2, \dots, r_n]$

Now we assume $n > 1, m > 1$ and proceed by induction on $m + n$ (this is the tricky bit):

Let $a_{11} = min(r_1, s_1)$, then there are two options:

Suppose $a_{11} = r_1$, then let $a_{12} = \dots = a_{1n} = 0$ and the new vectors become: $R' = (r_2, r_3, \dots, r_m)$ and $S' = (s_1 - r_1, s_2, \dots, s_n)$ For these vectors we can check that the equality holds:

$$ r_2 + r_3 + \dots + r_m = (s_1 - r_1) + s_2 + \dots + s_n$$

So, there exists a matrix $A'$ such that it has non-negative elements and has row sum vector $R'$ and column sum vector $S'$. If we add the row $[r_1, 0, 0, \dots, 0]$ "on top" of that matrix $A'$, we've constructed the desired matrix $A$.

Suppose $a_{11} = s_1$, then we have a symmetrical case. We just construct the matrix $A$ by adding a column on the left of $A'$ this time.


问:
What are these groups known as, and are their orders multiplicative functions?
This question has arisen in my work, and I imagine it is fairly standard to those more algebraic than myself.

Suppose for simplicity that $n$ is odd and squarefree, let $\mathbb{Z}_n^{\times}$ be the unit group of $\mathbb{Z}/n\mathbb{Z}$ and let $H\leq \mathbb Z_n^{\times}$ be a subgroup.

Associated with each divisor $d|n$, one can form a new subgroup, say $H_d$, defined by

$$H_d=\{h\in H: h\equiv 1 \pmod d\}.$$

In other words, the elements of $H_d$ are precisely those that are trivial on $\mathbb{Z}_d^{\times}$.

Firstly, I would like to ask: do these subgroups have a name/standard interpretation in algebraic terms?

Secondly, having tried a few cases (for example when $H$ is a Boolean subgroup) it seems that the $|H_d|$ might be multiplicative functions, that is

$$|H_{de}|=|H_d||H_e|,$$

so I would like to ask if this a known result?

答:
Note that for every positive integer $n$, every positive divisor $d$ of $n$, and every subgroup $H\subset(\Bbb{Z}/n\Bbb{Z})^{\times}$, we have $H_d=H\cap\ker\varphi_d=\ker\varphi_d|_H$, where $\varphi_d$ is the natural group homomorphism
$$\varphi_d:\ (\Bbb{Z}/n\Bbb{Z})^{\times}\ \longrightarrow\ (\Bbb{Z}/d\Bbb{Z})^{\times}.$$
I don't know of any standard name for such a subgroup.

It is easy to see that the identity $|H_{de}|=|H_d||H_e|$ cannot hold in general, because $H_{de}\subset H_d$.


问:
Proving that exponential is growing faster than polynomial
I want to prove this limit:$\lim_{x \to +\infty}(|P(x)|-e^x)=-\infty$.
I can't use Hopital's rule, but only the fact that $a_{n}=(1+\frac{x}{n})^n$ and $b_{n}=(1+\frac{x}{n})^{n+1}$ converge to $e^x$.

答:
Use the definition of a limit. $\lim_{x \to \infty} f(x) = -\infty$ means that given any arbitrarily large $N$, you can come up with an $M$ such that $x > M \to f(x) < -N$.

Let $d$ be the degree of $P(x)$. So $P(x) = \sum_{i = 0}^d a_i x^i$. Then $|P(x)| \leq \sum_{i = 0}^d |a_i| |x|^i$ and if $|x| > 1$ then $|P(x)| \leq |x|^d \sum_{i = 0}^d |a_i|$. Define $A = \sum_{i = 0}^d |a_i|$ as the sum of the absolute values of the coefficients of $P(x)$. We have $|x| \gt 1 \implies |P(x)| \geq A |x|^d$.

Given $N$ we want to find an $M$ that is big enough so that $x \gt M \implies A x^d - e^x \lt -N$ which is to say $e^x \gt Ax^d + N$.

At this point we need to use the definition of $e^x$, in particular the fact that $e^x \gt (1 + x/n)^n$. Take $n = d + 1$ and we have $x \gt 1 \implies e^x \gt (1 + x/(d + 1))^{d + 1} \gt x^{d + 1}/(d + 1)^{d + 1}$. (We could have taken $n$ much higher but that's all we need.)

So now we want to show that we can make $x^{d + 1}/(d + 1)^{d + 1} \gt Ax^d + N$. Multiply both sides by $(d + 1)^{d + 1}/x^d$ to get $x \gt (d + 1)^{d + 1}(A + N/x)$. Since we are assuming $x \gt 1$ it suffices to take $M = (d + 1)^{d + 1}(A + N)$.

Putting it together:

Given any $N$ let $M = \max \left\{ 1, (d + 1)^{d + 1}(A + N) \right\}$ where $d$ is the degree of $P(x)$ and $A$ is the sum of the absolute values of the coefficients in $P(x)$. Then $x \gt M \implies P(x) - e^x \lt A x^d - (1 + x/(d + 1))^{d + 1} \lt A x^d - x^{d + 1}/(d + 1)^{d + 1} = (A - x/(d + 1)^{d + 1}) x^d \lt (A - (A + N)(d + 1)^{d + 1}/(d + 1)^{d + 1})x^d = -N x^d \lt -N$

QED


答:
All mathematical libraries define an azimuth $A$ which is 0 along $+x$
and increasing counter-clock-wise such that it is 90 deg at $+y$.
If you're defining your own azimuth convention with $A'=0$ along $+y$
and $A'=90$ deg along $+x$, the relation between the two azimuths
is A'= -A+90 deg. So you need to invert the angle obtained
with the usual programming languages and add 90 or subtract 270 degrees afterwards.

Title: How to correct the azimuth obtained from coordinates of two points.

Body:
Assuming we have points $A (-0.4664, 0.7104)$ and $B (-1.8112, 2.8032)$.
To calculate their azimuth the following equation may be used: $$\text{Azimuth}=\tan^{-1}\frac{X_B-X_A}{Y_B-Y_A}=\tan^{-1}\frac{-1.3448}{2.0928}=-32.72415526582776^\circ$$
However, based on the sign of $X_B-X_A<0$ and $Y_B-Y_A>0$ this angle must be found in quadrant 4 and thus should range between $270$ and $360$ degrees. (The azimuth is measured clockwise from the north or positive y-axis). To correct this angle, we can add/subtract $180$ or $360$ to the calculated azimuth. In this case, adding 360 will solve the issue since $360-32.73272415526582776=327.2672$ degrees.

My question is the following: If for the same example, the calculated azimuth was found to be $32$ degrees. How do we correct this sort of angle? Since adding/subtracting $180$ and $360$, in this case, will not help.


问:
Given $y'(t)=\frac{t}{4}-\frac{1}{y^2+1},y(0)=1$ and let $x(t)=1-\frac{t}{2}$ approx $|x(t)-y(t)|$
Given $y'(t)=\frac{t}{4}-\frac{1}{y^2+1},y(0)=1$ and let $x(t)=1-\frac{t}{2}$

Approximate $|x(t)-y(t)|,|t|<1/2$

I need to do this with $\epsilon$ solution which is corollary of Gronwall thm but I don't know how to use it.

答:
If I understand you right, you apply the decomposition
$$
(y-x)'(t)=[f(t,y(t))-f(t,x(t))]+[f(t,x(t))-x'(t)]
$$
and use it to get the inequality
$$
|(y-x)'(t)|\le L|y(t)-x(t)|+|f(t,x(t))-x'(t)|
$$
which can then be treated via the Grönwall lemma. One can easily conclude $L=1$, using $2|y|\le 1+y^2$, and with some effort $L=0.65$. For the second term,
$$
f(t,x(t))-x'(t)=\frac{t}4-\frac1{1+1-t+\frac{t^2}4}+\frac12=\frac{(1+\frac t2)(1-\frac t2+\frac{t^2}8)-1}{2-t+\frac{t^2}4}
\\
=\frac{-\frac{t^2}8+\frac{t^3}{16}}{2-t+\frac{t^2}4}=-\frac{t^2}{16}\frac{2-t}{2-t+\frac{t^2}4}
$$
so that
$$
|f(t,x(t))-x'(t)|\le \frac{t^2}{16}
$$
for $2-t>0$. Thus by applying Grönwall you get
\begin{align}
e^{-Lt}|y(t)-x(t)|&\le\frac1{16}\int_0^ts^2e^{-Ls}\,ds
\\
&=\frac{6}{16L^4}-e^{-L t} \frac{L^3 t^3 + 3 L^2 t^2 + 6 L t + 6}{16L^4}
\\
|y(t)-x(t)|&\le\frac{3}{8L^4}\left(e^{Lt}-1-Lt-\frac12(Lt)^2- \frac16(Lt)^3\right)
=\frac3{32}t^4+\frac3{160}Lt^5+...
\end{align}


问:
Show that $w_3=\frac{-1+i\sqrt3}2$ without $\cos$ and $\sin$
I struggle a lot to show that $w_3 = e^{i2\pi/3}$ is equal to $\frac{-1+i\sqrt{3}}{2}$  without using cos and sin, knowing only that $f :(\mathbb R, +) \to (\mathbb C(1), \times)$ that from an angle $\theta$ gives the complex number $e^{i\theta}$ is a morphism so $\operatorname{ker} f = 2\pi \mathbb Z$, also knowing $\cos(\theta) = \Re(e^{i\theta})$, $\sin(\theta) = \Im(e^{i\theta})$, knowing $e^{i\pi}=-1$ all summation and duplication rules of $\cos$ and $\sin$. The indication tells to verify that $w_3$ is a solution of $X^{2} + X + 1 = 0$ then compute the complex solutions with $\Delta$ and find that since $w_3$ is a solution so it must be one of them, without knowing what $\cos(\pi/3)$ is. Tried all possible methods but still need to find that value of $\cos(\pi/3)$ but the whole goal of the exercice is to do it without knowing it. I also wrote $f(\pi) = f(2\pi/3 + \pi/3) = f(2\pi/3)\cdot f(\pi/3)$ but leads me nowhere... I would like your help to show that $w_3^2 + w_3 + 1 = 0$.

答:
show that $w = e^{i2\pi/3}$ is equal to $\dfrac{-1+i\sqrt3}{2}$  without using cos and sin

Well, $w^3=1$ and $w\ne1$, so $\dfrac{w^3-1}{w-1}=w^2+w+1=0.$

Now, by the quadratic formula, $w=\dfrac{-1\pm\sqrt{-3}}2.$

(Note that $w^2=\dfrac{-1\mp\sqrt{-3}}2$, and you can verify that $w^3=w^2w=1$.)