问:
Find general solution in radians of $\cos x + \cos 3x + \cos 5x = 0$
I study maths as a hobby. I have this problem:

Find in radians the general solution of:

$$\cos x + \cos 3x + \cos 5x = 0$$

I have said:

$\cos 3x + \cos 5x = 2\cos\left( \frac{1}{2}(3x + 5x)\right)\cos\left( \frac{1}{2} (3x - 5x)\right) = 2\cos 4x\cos x$

$\cos x + 2\cos 4x \cdot \cos (x) = 0$

$(1 + 2\cos 4x)\cos x = 0$

$\cos x = 0$ or  $\cos 4x = - \frac{1}{2}$

$x = \frac{\pi}{2}$ or $x = \frac{\pi}{3} $

My book gives the answers as:

$(2n + 1)\frac{\pi}{6}, n\pi \pm \frac{\pi}{3}$

So we agree on the second answer but is the book wrong when it says $(2n + 1)\frac{\pi}{6}$?

I get the answer as $x = n\pi + \frac{\pi}{2}$, which can be written as $(2n +1)\frac{\pi}{2}$

答:
$$\cos 4x =-1/2 \iff \cos4x +\cos \frac{\pi}{3} =0
\iff 2\cos\left(2x +\frac{\pi}{6} \right)\cos\left(2x-\frac{\pi}{6} \right)=0 \\
\iff \cos\left(2x +\frac{\pi}{6} \right) = 0 \text{ or } \cos\left(2x -\frac{\pi}{6} \right) = 0 \\
\iff 2x +\frac{\pi}{6} = \frac{\pi}{2}+n\pi \text{ or } 2x -\frac{\pi}{6} = \frac{\pi}{2}+n\pi \\
\iff 2x = \frac{\pi}{3}+n\pi \text{ or } 2x = \frac{2\pi}{3}+n\pi\\
\iff x=\frac{\pi}{6} + n\frac{\pi}{2} \text{ or }x= \frac{\pi}{3}+n\frac{\pi}{2} \\
\iff x= (3n+1) \frac{\pi}{6} \text{ or } x= (3n+2) \frac{\pi}{6}$$ And as you said, for $\cos x=0$ you need to take $$x=(2n+1) \frac{\pi}{2} = (6n+3)\frac{\pi}{6}$$

This is, in fact, equivalent to the book’s answer. To see this, observe that the union of $3n+1, 3n+2, ,6n+3$ is all the integers $m$ such that $m\equiv 1,2,3,4,5 \pmod 6$, which is further equivalent to the union of the odd numbers and numbers of the form $6n+2$ and $6n-2$. So, the solution set can be written as $$x=(2n+1)\frac{\pi}{6} \ \text{or} \ (6n\pm 2)\frac{\pi}{6} = (2n+1)\frac{\pi}{6} \ \text{or} \ n\pi \pm \frac{\pi}{3}$$


问:
Showing that a function has a certain absolute minimum.
Suppose we have the function $$f(x) = \frac{x}{p} + \frac{b}{q} - x^{\frac{1}{p}}b^{\frac{1}{q}}$$ where $x,b \geq 0 \land p,q > 1 \land \frac{1}{p}+\frac{1}{q} = 1$

I am trying to show that $b$ is the absolute minimum of $f$.

I proceeded as follows:

$$\frac{df(x)}{dx} = \frac{1}{p} - \frac{x^{\frac{1}{p}-1}}{p} b^{\frac{1}{q}} = \frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px}$$

Now I will look for critical points by searching for the zeros of this function.

$$\frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px} = 0 \iff x - x^{\frac{1}{p}} b^{\frac{1}{q}} = 0 \iff x = x^{\frac{1}{p}} b^{\frac{1}{q}}$$.

Now I can see that $b$ is a critical point.

How ever when I continue my calculations to check whether there are any other critical points
$$x = x^{\frac{1}{p}} b^{\frac{1}{q}} \implies x^p = b^{\frac{p}{q}}x \implies x^{p-1} = b^{\frac{p}{q}} \implies x = b^{\frac{p}{(p-1)q}}$$

But this could not be equal to $b$, where did I go wrong?

答: Note that $$\dfrac{p}{q(p-1)}=1$$ since $\dfrac1p + \dfrac1q = 1$. This is because, we have $\dfrac1q = 1 - \dfrac1p = \dfrac{p-1}p \implies \dfrac{p}{q(p-1)} = 1$.</p>

ContentLicense: CC BY-SA 3.0


答:
Breaking up the multiple angles.
$$\cos3x=4\cos^3x-3\cos x$$
$$\cos5x=16\cos^5x-20\cos^3x+5\cos x$$
Plugging everything together:
$$16\cos^5x-16\cos^3x+3\cos x=0$$
Dividing by $\cos x$, and $\cos x$ can't be zero. Defining and plugging in we get:
$$y=\cos^2x$$
$$16y^2-16y+3=0$$
Can you take it from here?

ContentLicense: CC BY-SA 4.0

Title: Find general solution in radians of $\cos x + \cos 3x + \cos 5x = 0$

Body:
I study maths as a hobby. I have this problem:

Find in radians the general solution of:

$$\cos x + \cos 3x + \cos 5x = 0$$

I have said:

$\cos 3x + \cos 5x = 2\cos\left( \frac{1}{2}(3x + 5x)\right)\cos\left( \frac{1}{2} (3x - 5x)\right) = 2\cos 4x\cos x$

$\cos x + 2\cos 4x \cdot \cos (x) = 0$

$(1 + 2\cos 4x)\cos x = 0$

$\cos x = 0$ or  $\cos 4x = - \frac{1}{2}$

$x = \frac{\pi}{2}$ or $x = \frac{\pi}{3} $

My book gives the answers as:

$(2n + 1)\frac{\pi}{6}, n\pi \pm \frac{\pi}{3}$

So we agree on the second answer but is the book wrong when it says $(2n + 1)\frac{\pi}{6}$?

I get the answer as $x = n\pi + \frac{\pi}{2}$, which can be written as $(2n +1)\frac{\pi}{2}$


问:
Proving $\bigcap_n I_n \neq \emptyset$ for monotonically smaller closed intervals $(I_n)_n$
Theorem: Let $(I_n)_n$ be monotonically smaller closed intervals. i.e. $I_{n+1}\subseteq I_n$
Then $$\bigcap_n I_n \neq \emptyset$$
I easily grasped this geometrically, But I doubt of my proof. This is my proof:

Since $I_n$'s are real closed intervals, $I_n \neq \emptyset$. Then we can take $x \in I_n$. Since $I_n\subseteq I_{n-1}$, $x \in I_{n-1}$.
And generally from that,
$$x \in I_{n-2} \Rightarrow x \in I_{n-3} \Rightarrow x \in I_{n-4} \Rightarrow...\Rightarrow x \in I_2 \Rightarrow x \in I_1$$

Thus $x \in I_{n}$ is also in other closed intervals $I_i$s where $i < n-1$. Thus,
$$\bigcap_{i \leq n} I_i \neq \emptyset$$ Then, if we just take $\lim_{n\to \infty}$ equation will still hold and consequently,
$$\bigcap_{n} I_n \neq \emptyset$$

To me, it looks right, but I could'nt be sure. I would appreciate your short-time look. Thanks from now.

答:
For the case $I_n=[0,1/n]$ the number $1/100$
belongs to $I_1,I_2,\ldots,I_{100}$ but not to $I_{101}$

Hint: Use Bolzano-Weierstrass.


问:
Prove that $\int_0^\infty\frac1{x^x}\, dx<2$

Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$

 Note: This inequality is rather tight. The integral approximates to $1.9955$.

Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.

Expressing the left-hand side as a Frullani integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ means that $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ that satisfies the functional equation. Is there such a function?

 (For context, user371838's post below proves sophomore's dream which I also asked about originally.)

答:
I don't have a solution, but somehow interesting observation:
$$\int_0^\infty x^{-x}dx=\int_0^1 x^{-x}dx+\int_1^2 x^{-x}dx+\int_2^3 x^{-x}dx+\dots=$$
$$\int_0^1 x^{-x}dx+\sum_{n=1}^\infty\int_0^1(x+n)^{-(x+n)}dx.$$
The second term $\int_0^1\sum_{n=1}^\infty(x+n)^{-(x+n)}dx$ looks like "double sophomore dream"; in $n$ as summation ($\sum_{n=1}^\infty n^{-n}$) and in $x$ as integral ($\int_0^1 x^{-x}dx$). But I don't know what to do with this next.

It would be nice to be able to find a tight bound, e.g., in terms of "sophomore dream" value. Note that $\left(\int_0^1x^{-x}dx\right)^{\int_0^1x^{-x}dx}=1.29128599^{-1.29128599}\approx 0.71885$.


问:
How to I have to understand the Cut in an graph in this case?
How is $\delta$ defined in (#)?

Let $G = (V, E)$ be a connected undirected graph and $c : E \rightarrow \mathbb{R}$ be a cost function.  Let $T = (V, F)$ be a minimal spanning tree.
(#) Let $f = \{x, y\} \in F$,
and let $A$ be the set of all nodes reachable from $x$ in $(V, F \setminus {f})$.
Then for every edge $e \in \delta(A)$ it holds that $c(e) \leq c(f)$.

Normally the cut $\delta$ is defined like $\delta(A) := \{ e = \{v, w\} \in E
\mid v \in A \text{ and } w \in V \setminus A \}$.
But then $c(e) \geq c(f)$ is not always true, because you could compare an edge in $T$ with an edge in $T$ and there could be a case like $c(e)$ < $c(f)$.

答:
Your definition of $\delta(A)$ is correct, and the statement is indeed false (although not for the reason you seem to say: $\delta(A)$ contains no edge of $T$ other than $f$).

The missing condition is that $T$ is a minimum cost spanning tree. In that case the statement is true: if there were an edge $e \in \delta(A)$ with $c(e) < c(f)$, then $T - f + e$ would be another spanning tree (this has to be proved but it is not difficult) with smaller total cost than $T$, which is a contradiction.


问:
What is the sum of this infinite series? Which one is it, Taylors? Binomial?
I am trying to figure which formula to use for this one.
$$\displaystyle\sum\limits_{x=-1}^{-\infty} -x(1-y)p(1-p)^{-x}+\sum\limits_{x=1}^\infty xyp(1-p)^x$$
where $0<y<1$, and $0<p<1$.

My work:
$$=\displaystyle\sum\limits_{x=1}^{\infty} -x(1-y)p(1-p)^{x}+\sum\limits_{x=1}^\infty xyp(1-p)^x$$
next step,
$$=\displaystyle\sum\limits_{x=1}^{\infty} p(1-p)^{x}(2xy-x)$$
This doesn't look like anything.  Does anyone recognize this?

答:
We have, provided $|a|<1$, that $$\sum_{n=0}^\infty na^n=\frac{a}{(a-1)^2}$$

In your problem, we may rewrite $$\sum_{x=1}^\infty p(1-p)^x(2xy-x)=(2y-1)p\sum_{x=1}^\infty x(1-p)^x$$

Applying that first identity, we find the sum to be $$\frac{(2y-1)p(1-p)}{p^2}=\frac{(2y-1)(1-p)}{p}$$

Edit: Deriving the formula, per request.

We start with the geometric series $$\sum_{n=0}^\infty a^n=\frac{1}{1-a}$$
We take $\frac{d}{da}$ of both sides to get $$\sum_{n=0}^\infty na^{n-1}=\frac{1}{(1-a)^2}$$
We multiply both sides by $a$ to get $$\sum_{n=0}^\infty na^{n}=\frac{a}{(1-a)^2}$$


问:
The integral $\int_{U}u^{4}$ is well-defined for $u\in H_{0}^{2}(U)$ and $U\subset\mathbb{R}^{n}$ open, bounded, $n\leq3$.
I'm trying to prove that the integral $\int_{U}u^{4}$ is well-defined for $u\in H_{0}^{2}(U)$ and $U\subset\mathbb{R}^{n}$ open, bounded, $n\leq3$. In other words, I need to prove that $H_{0}^{2}(U)$ ($=W_{0}^{k,p}(D)$ with $k=p=2$) can be embedded in $L^{4}(U)$. Note that there are no smoothness assumptions on the boundary.

I tried to distinguish the cases $n=1,2,3$. For $n=1$ I think we can use Morrey's inequality (which requires $1=n<p=2$) to prove that $u$ is Hölder continuous (up to a set of measure zero). Thus $$H_{0}^{2}(U)\hookrightarrow C^{0}(\overline{U})\hookrightarrow L^{\infty}(U)\hookrightarrow L^{4}(U).$$

However, I cannot come up with the right arguments for the cases $n=2,3$. The only seemingly useful embedding-theorem I can find requires $n>kp=4$, which we do not have.

答:
If $n=3$, then the Sobolev conjugate of $2$ is equal to $6$, and since $\nabla^2 u\in L^2(U)$, we have $\nabla u\in L^6(U)$. Hence, Morrey's theorem implies that $u\in L^{\infty}(U)$.

If $n=2$, then $\nabla u$ belongs to every $L^p(U)$ space, for any $p\geq 1$. In particular $\nabla u\in L^3(U)$, and Morrey's theorem shows that $u\in L^{\infty}(U)$.


问:
Number theory problem with sets of elements - $x = (1 + 3 + \dots + m)p + (m + 2) + (m + 4) + \dots + n$

Given $p \in \mathbb{N}$, $p > 1$ and the set:
$$A_p = \{x \ | \ (\exists) \ m ,n \in 2\mathbb{N} + 1, x = (1 + 3 + \dots + m)p + (m + 2) + (m + 4) + \dots + n\}$$
Prove that, if $x, y, \in A_p$, then $xy \in A_p$.

Source: Romanian magazine in mathematics.

Attempt: Note that $1 + 3 + \dots + k = z^2$, if $k \in 2\mathbb{N} + 1$. So, the set $A_p$ may be redefined as:
$$A_p = \{x \ | \ x = p\alpha^2 + \beta^2 - \alpha^2, \alpha, \beta \in \mathbb{N}^* \}$$

Therefore, $x = (p - 1)\alpha^2 + \beta^2$ and $y = (p - 1)\gamma^2 + \delta^2$. So, $xy$ should be written as $(p - 1)\varepsilon^2 + \varphi^2$, but multiplying the $x$ and $y$ equations, the general form doesn't look good.

I'm also thinking at induction, as $p$ natural. We may start with $p = 2$ (it may be simpler) and than a general $p \to p + 1$ case.

Body: Use $(a^2+n b^2)(c^2+n d^2) = (ac+nbd)^2 + n(ad-bc)^2$


问:
$G=D_8$ and $H=\langle r^2 \rangle $ why is $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$
In my homework I have:

$G=D_8$ (Dieder) and $H=\langle r^2 \rangle$

It it said is that $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

Why is that? Why not $\mathbb{Z}/4\mathbb{Z}$?

答:
Since $Z(D_8)$ is generated by $r^2$ we have $G/H=D_8/Z(D_8)$. If it would be cyclic then $D_8$ would be abelian, a contradiction. Hence $D_8/Z(D_8)$ must be isomorphic to $C_2\times C_2$.

Reference: If $G/Z(G)$ is cyclic, then $G$ is abelian


问:
Sum of subsets in linear algebra
I am working with subspaces and using Linear Algebra done right as literature. Axler defines the sum of subsets as:

Suppose $U_1,...,U_m$ are subsets of V. The sum of $U_1,...,U_m$, denoted $U_1+\cdots+U_m$, is in the set of all possible sums of elements of $U_1,...,U_m$. More precisely,
  $$U_1+\cdots+U_m=\{u_1+\cdots+u_m:u_1\in U_1,\dots,u_m\in U_m\}$$

Now suppose that $U = \{(x,x,y,y)\in\mathbb{F}^4:x,y\in\mathbb{F}\}$ and $W = \{(x,x,x,y)\in\mathbb{F}^4:x,y\in\mathbb{F}\}$. Then:

$$U+W=\{(x,x,y,z)\in\mathbb{F}^4:x,y,z\in\mathbb{F}\}$$

When I try to verify the assertion I get:

$$U+W = \{(x+x,x+x,x+y,y+y):x,y\in\mathbb{F}\}$$

I just apply the definition of the sum of subsets. What am I missing?
Any help greatly appreciated.

答:
I also had trouble understanding this one, but I think it all comes down to how you name the elements of the set.

As in this answer, and this comment, we can use different letters for one of the sets. Let's suppose that $U = \{(x,x,y,y)\in\mathbb{F}^4:x,y\in\mathbb{F}\}$ and $W = \{(v,v,v,w)\in\mathbb{F}^4:v,w\in\mathbb{F}\}$.

Now, we get the same as your result:

$$U+W = \{(x+v,x+v,y+v,y+w):x,y,v,w\in\mathbb{F}\}$$

Then, by setting $x=x+v$, $y=y+v$, $z=y+w$, we have

$$U+W = \{(x, x, y, z):x,y,z\in\mathbb{F}\}$$

Which is what the book directly states.

Title: Sum of subsets in linear algebra


问:
Can every square matrix be expressed as a unique product of a symmetric and an orthogonal matrix?

Background

My undergrad education in linear algebra ended with eigenvalues, symmetric matrices, and orthogonal matrices. Now I'm trying to gain an intuitive understanding of singular values, and reached a new perspective/understanding of square matrices described below. However in case I have some misconceptions, or in case I am misguiding myself, I would like to test it.

Questions

Can every square matrix be expressed (uniquely?) as a product:

$$A=SR$$

where

$S$ is a symmetric matrix (and therefore has orthogonal eigenspaces, like an inertia tensor or variance matrix, i.e. can be diagonalized by an orthogonal matrix)

$R$ is an orthogonal matrix (and therefore is a distance-preserving rotation/reflection)

If yes, is this decomposition unique?

I like this perspective because:

A symmetric matrix and an orthogonal matrix both exert actions associated with simple geometric intuitions.

It makes evident what singular values are. (They are just the eigenvalues of $S$.)

The geometric effect of the transpose of a matrix becomes more apparent: $A^T=R^{-1}S$. As does its relationship to the inverse $A^{-1}=R^{-1}S^{-1}$, as inverting a symmetric matrix corresponds to simple geometric intuitions (inverting each eigenvalue).

Any guidance/comment on a shortcoming, caveate, or misconception you see in my understanding would be appreciated, or any pointer to related concepts you think could be useful to be aware of, before I internalize this perspective.

答:
Sure, use the SVD decomposition: if $A=U\Sigma V^\top$ for $U,V$ orthogonal and $\Sigma$ diagonal, then we can rearrange this to get $U\Sigma U^\top \cdot U V^\top$, while $U \Sigma U^\top$ is symmetric whose spectral decomposition is given, and $U V^\top$ is unitary as a product of such.

In general the geometric intuition of what you said is already evident from $U \Sigma V^\top$ without the need to do this trick, $A \vec{x}$ is just:

Projecting $\vec{x}$ onto the columns of $V$ to get $\vec{\pi}=V^\top \vec{x}$

We then stretch each entry $i$ of the result $\vec{\pi}$ by $\sigma_i$

Use the resulting vector entries as coordinates, to make a linear combination of the columns of $U$


答:
Two methods of attack (that haven't been mentionned) for such issues with angles in arithmetical progression:

a) It's more efficient to group extreme expressions, here $\cos(x)+\cos(5x)=2 \cos(3x)cos(2x)$, then factor $\cos(3x)$, which gives you instantly the answer.

b) or (longer and advisable only for more complicated cases) use complex numbers

$$\Re(e^{ix}+e^{3ix}+e^{i5x})=\Re(e^{ix}(1+e^{2ix}+e^{i4x}))$$

and recognize the sum of an arithmetic progression...


问:
Zeckendorf with Negative Fibonacci Numbers
Zeckendorf : Every positive integer N can be expressed uniquely as a sum of distinct non-consecutive Fibonacci numbers

I was wondering if this theorem can be applied with the extended Fibonacci numbers, and especially I am looking for a way to find the Zeckendorf-like representation of $ N $, with Fibonacci numbers $ F_n $ having only negative indexes. (First for $ N \in \mathbb{N} $ then maybe $ N \in \mathbb{Z} $)

Mathworld states that this theorem only applies on positive numbers but do not says if Fibonacci numbers can be negative.
Online Zeckendorf Representation pages show positive indexes only and mostly use Binet's formula.

I've read that Binet's formula can be applied to generalized Fibonacci numbers (Proof), but it does not prove that Zeckendorf theorem can be used as well.

EDIT : I missed a paragraph in this Wikipedia page called Representation with Negafibonacci numbers that states this Zeckendorf-like representation exists and is unique. I probably can use NegaFibonacci coding method to find the indexes.

答:
The negative fibbonacci numbers run as F(-n)=F(n) for 2|n and F(-n)=-F(n) for n odd.

So we get for fibonacci numbers

```<code> -21  13  -8   5  -3   2  -1   1
                          -1
                  -3           1
                  -3
                  -3      -1
          -8           2       1
          -8           2
          -8       t0 12
  -21          5       2       1
  -21     -8       -3     -1
</code>```

So yes, all of the natural numbers can be expressed as a sum of non-consecutive fibonacci numbers of negative index.

It ought be a straight-forward conversion from the positive to negative schema.

```<code>  eg  -21            5     2      1
       21           -5    -2     -1
           13   8   -5    -2    -1
           13           3 -2     -1
           13
</code>```

It's possible to go both ways, and there is a carry rule to prevent adjacent columns being filled with more than one counter.

Body:
The negative fibbonacci numbers run as F(-n)=F(n) for 2|n and F(-n)=-F(n) for n odd.

So we get for fibonacci numbers

 -21  13  -8   5  -3   2  -1   1
                          -1
                  -3           1
                  -3
                  -3      -1
          -8           2       1
          -8           2
          -8       t0 12
  -21          5       2       1
  -21     -8       -3     -1

So yes, all of the natural numbers can be expressed as a sum of non-consecutive fibonacci numbers of negative index.

It ought be a straight-forward conversion from the positive to negative schema.

  eg  -21            5     2      1
       21           -5    -2     -1
           13   8   -5    -2    -1
           13           3 -2     -1
           13

It's possible to go both ways, and there is a carry rule to prevent adjacent columns being filled with more than one counter.


问:
Combinatorics problem with students choosing elective courses. (Prove by assuming the opposite).
There are 2009 different elective courses in the university. Each one of the courses has exactly 45 students in it. Every 2 courses has exactly 1 student in common. Prove that there is a student enrolled in all courses.

I think it should be proven by assuming the opposite.
Maybe there is a connection that 2009/45 is less than 45 but I'm not sure about that.

答:
We will see the courses as a sets of students, where $x \in Y$ means that $x$ is enrolled in course $Y$ and $x = Y \cap Y'$ means that $x$ is the common student of $Y$ and $Y'$ (we omit curly braces around $x$).

We can prove the following lemma.

Lemma: There is a student enrolled in all courses, or each student is enrolled in at most 45 courses.

Proof: For a contradiction, let there be a student $a$ who is enrolled to more than $45$ courses, but not to all courses. The courses of $a$ will be denoted by $C_1, \dots C_k$, where $k > 45$. Let $C'$ be a course to which $a$ is not enrolled.

We have $\forall i \not = j < k: C_i \cap C_j = a$. Therefore, the student which a $C_i$ shares with $C'$ must be different for each $i$ (otherwise, the $|C_i \cap C_{i'}| > 1$). That is impossible by the pigeonhole principle ($C'$ have only $45$ students, while the range of $i$ is greater), which concludes the proof of the lemma.

$$\tag*{$\blacksquare$}$$

Thus, we may assume that each student is enrolled in at most $45$ courses, otherwise, we have the desired student.

Now, we can count the pairs $(\text{x}, \text{Y})$, such that the student $x$ is enrolled to course $Y$. Each course has exactly $45$ students, which gives $45 × 2009 = 90405$ pairs. However, each student can take up to $45$ courses by our assumption, which gives as most $45 × 45 = 2045$ pairs, which is a contradiction.

Therefore, the assumption regarding the maximal number of courses for a student cannot hold, which implies (by the lemma) that there is a student who took all courses.

$$\tag*{$\blacksquare$}$$


问:
Understanding probability of independent events
I am trying to learn probability, and having a hard time to be honest. What I try to do is to break a problem in terms of event space. For eg, event space for rolling a die is {1,2,3,4,5,6}. For rolling two dice it would be 36 elements with pairs from 1,1 to 6,6. Then I try to calculate probability of a given event using simple basic formula of occurrences/total events.

On topic of probability of independent events, suppose I roll 2 dice and want to calculate probability of 2 on first die and 5 on second die. Book I am reading uses "And" rule for independent events:

```<code>P(2 and 5) = P(2) * P(5)
</code>```

I am not sure of this approach. When we say event "2 in first roll, 5 in second roll", we basically are talking of an event "2 rolls of die". And so our event space consists of 36 elements, each representing a pair as discussed above. But in above formula, we are multiplying probabilities of events which come from  a different event space(containing 6 elements). So basically we are trying to solve problem belonging to some event space using probabilities from different event space.

Am I wrong in my thinking here?

答:
Essential is that probability spaces are chosen. This in order to model real-life situations in a natural and convenient way.

If we want to apply probability theory on the throw of a single die then it is quite natural and convenient to go for sample space $\Omega=\{1,2,3,4,5,6\}$. In that case the real event that the die shows a $5$ corresponds with the theoretical event $\{5\}\subseteq\Omega$.

But do not think we are stick to that choice.

If we are interested in a model for the rolling of two dice then it is (again) natural and convenient to go for sample space $\Omega=\{(i,j)\mid i,j\in\{1,2,3,4,5,6\}\}$ and this with $P(\{(i,j)\})=\frac1{36}$ for every singleton event.

In that case the real event that the first die shows a $5$ does corresponds  with the theoretical event $\{(5,j)\mid j\in\{1,2,3,4,5,6\}\}=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\subseteq\Omega$.

Realize that we could have chosen this space also to manage the first situation.

Only there is no reason to do that, and it is handsome to keep the sample space as simple as possible.


问:
Expressing a permutation as a product of transpositions
Express $(123) \in S_3$ as the product of four transpositions $$(123) = \tau_1\tau_2\tau_3\tau_4$$ such that $\tau_i \neq \tau_{i+1}$ for $i = 1,2,3$.

I am confused as to how I should approach this question. In the examples from lectures we have gone over examples where there are 2 variables but not only one. Any help would be appreciated.

Body: $(1 2 3) = (2 3) (3 1) (2 3) (3 1)$


问:
Density in negative Sobolev spaces
Consider $\Omega = (0,2 \pi)^d$ and the negative Sobolev space $H^{-s}(\Omega)$, defined as the dual of $H^s_0(\Omega)$ for the $L^2$ inner product. Due to the simple shape of $\Omega$, we can see $H^{-s}(\Omega)$ as a Banach/Hilbert space with the inner product:
$$
\langle f, g \rangle_{H^{-s}} := \sum_{n \in \mathbb{Z}^d}  f_n g_n (1 + | n |^2)^{-s},
$$
where $(f_n)$ and $(g_n)$ are the Fourier coefficients of $f$ and $g$.

Question: Is $C^\infty_0(\Omega)$ dense in $H^{-s}(\Omega)$?

答:
The answer is affirmative. In a previous version of this post I claimed the opposite, but I was wrong. (The previous version is below. Following my usual policy, I am NOT deleting it).

Claim. The space $C^\infty(\mathbb T)$ of smooth and $2\pi$ periodic functions is dense in $H^{-s}(\mathbb T)$ for all $s\in \mathbb R$.

Proof. We will prove that $C^\infty(\mathbb T)^\bot$ is reduced to $\{0\}$. So, let $g\in H^{-s}(\mathbb T)$ be such that
$$
\sum_{-\infty}^\infty \frac{\hat{g}(n)\overline{\hat\phi(n)}}{(1+|n|^2)^s}=0, \qquad \forall \phi \in C^\infty(\mathbb T).$$
In particular, letting $\phi(x)=e^{imx}$, so that $\hat\phi(n)=0$ for $n\ne m$ and $\hat\phi(m)=1$, we see that
$$\frac{\hat{g}(m)}{(1+|m|^2)^s}=0, $$
which implies $\hat{g}(m)=0$. Since $m$ is arbitrary, we conclude that $g=0$, as claimed. $\Box$

Remark. This is exactly the proof suggested by the OP dgontier in comments below, and also by ktoi in comments to the main question.

Remark. I find this result a bit surprising. If $f\in H^{-s}(\mathbb T)$, its Fourier coefficients $\hat{f}(n)$ are allowed to grow as $n\to \infty$. If $s>1/2$, the space $H^{-s}(\mathbb T)$ contains singular distributions such as the Dirac comb $\delta$. On the other hand, if $\phi\in C^\infty(\mathbb T)$, then its Fourier coefficients decay faster than any polynomial. So, I find it  surprising that $C^\infty$ functions suffice to approximate arbitrary $H^{-s}$ elements.

PREVIOUS VERSION OF THIS ANSWER.

WARNING. There is a mistake, marked below. Now I have to go, but I am not deleting this answer, maybe someone will point out a solution, or it will be useful in some other way.

The answer is negative. To ease typing, let me assume $d=1$. Let $f_n\in C^\infty_0(0, 2\pi)$ and suppose that there is a $f\in H^{-s}$ such that $\lVert f_n-f\rVert_{H^{-s}}\to 0$. This means that
$$
\sum_{k=-\infty}^\infty \frac{\lvert \hat{f}_n(k)-\hat{f}(k)\rvert^2}{1+k^{2s}}\to 0, \qquad n\to \infty.$$
In particular, for each $k_0\in\mathbb Z$,
$$\tag{*}
\lim_{n\to \infty} \frac{\lvert \hat{f}_n(k_0)-\hat{f}(k_0)\rvert^2}{1+k_0^{2s}} =0.$$
Now, integration by parts shows that, since $f_n$ is smooth, $\hat{f}_n(k)$ decays as $\lvert k \rvert\to 0$ (faster than any power of $k$, but we won't need this quantification). For sufficiently big $n$, we have from (*) that
$$
\frac{\lvert \hat{f}_n(k_0)-\hat{f}(k_0)\rvert^2}{1+k_0^{2s}} \le 1.$$

(WARNING: this seems to be wrong). So, taking the limit as $k_0\to \infty$ shows that  $$ \lim_{k\to
 \infty} \hat{f}(k)=0.$$

However, not all $f\in H^{-s}$ satisfy this. For example, if $s>1/2$ then the Dirac comb is in $H^{-s}$, and $\hat{\delta}(k)=1$ for all $k$.

Title: Density in negative Sobolev spaces


问:
How is Stromberg Wavelet S orthogonal to Simple tents?
$\newcommand{\inprod}[2]{\left\langle{#1}\,|\,{#2}\right\rangle}$
$\newcommand{\eqd}{\triangleq}$
$\newcommand{\setn}[1]{{\left\{{#1}\right\}}}$
$\newcommand{\setu}{\cup}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brbl}[1]{\left\{#1\right.}$
How do they come about the difference equation when demonstrating a Strömberg wavelet $S(t)$ is orthogonal to a simple tent function
$$\lambda_{\sigma}(t) \eqd \brbl{\begin{array}{ll}
      1             & \text{if $t=\sigma$}
    \\0             & \text{if $t\in A_0$ and $t\neq\sigma$}
    \\\text{linear} & \text{otherwise}
\end{array}}$$
in the computation of the Strömberg Wavelet? Here,
$$\begin{array}{lcll}
  A_0 &\eqd& \setn{\cdots,-\frac{3}{2},-1,-\frac{1}{2},0,1,2,3,\cdots}&\text{and moreover}\\
  A_1 &\eqd& A_0 \setu \setn{\frac{1}{2}}
\end{array}$$
In particular, how can we arrive at this set of equations?
$$0 = \inprod{S}{\lambda_{\sigma }} =
  \brbl{\begin{array}{l ll}
    S\brp{\sigma -1}+ 4S\brp{\sigma}+S\brp{\sigma +1}
    &   \text{for $\sigma\in\setn{2,3,4,\ldots}$}
    &   \text{(1.10)}
    \\
    S\brp{\sigma-\frac{1}{2}}+ 4S\brp{\sigma}+S\brp{\sigma+\frac{1}{2}}
    &   \text{for $\sigma\in\setn{-\frac{1}{2},-1,-\frac{3}{2},-2,\ldots}$}
    &   \text{(1.11)}
    \\
    2S\brp{-\frac{1}{2}} + 9S\brp{0} + 6S\brp{\frac{1}{2}} + S\brp{1}
    &   \text{for $\sigma=0$}
    &  \text{(1.12)}
    \\
    S\brp{0} + 6S\brp{\frac{1}{2}} + 13S\brp{1} + 4S\brp{2}
    &   \text{for $\sigma=1$}
    &  \text{(1.13)}
\end{array}}$$

You can check the equations (1.10)–(1.13) here.

答:
$\newcommand{\inprod}[2]{\left\langle{#1}\,|\,{#2}\right\rangle}$
$\newcommand{\eqd}{\triangleq}$
$\newcommand{\setn}[1]{{\left\{{#1}\right\}}}$
$\newcommand{\setu}{\cup}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\brbl}[1]{\left\{#1\right.}$
$\newcommand{\intcc} [2]  {{\left[#1:#2\right]}}$
$\newcommand{\dx}{{\;dx}}$
I don't have a huge background in Strömberg Wavelets$\ldots$ but assuming the equation for (1.10) is correct, it should be possible to prove it by writing down the equations for $\Lambda_\sigma$ and $S(\sigma)$ and integrate their product over $\sigma-1\leq t\leq\sigma+1$. In particular, note that for $\sigma\in\setn{2,3,4,\ldots}$
$$\Lambda_{\sigma}(x) = \brbl{\begin{array}{ll}
      1-\sigma+x   & \text{for $\sigma-1\leq x\leq\sigma$}
    \\1+\sigma-x   & \text{for $\sigma\leq x\leq\sigma+1$}
    \\0            & \text{otherwise}
\end{array}}$$
and
$$S(x) = \brbl{\begin{array}{ll}
      \frac{S(\sigma)-S(\sigma-1)}{1}\brp{x-\sigma}+S(\sigma) & \text{for $\sigma-1\leq x\leq\sigma$}
    \\\frac{S(\sigma+1)-S(\sigma)}{1}\brp{x-\sigma}+S(\sigma) & \text{for $\sigma\leq x\leq\sigma+1$}
\end{array}}$$
Then
$$\begin{align*}
  \inprod{S(x)}{\lambda_\sigma(x)}
    &= \int_{x\in\mathbb{R}}S(x)\lambda_\sigma(x)dx
    && \text{by definition of $\inprod{f(x)}{g(x)}$}
  \\&= \int_{x=\sigma-1}^{x=\sigma+1}S(x)\lambda_\sigma(x)dx
    && \text{because support of $\lambda_\sigma$ is $\intcc{\sigma-1}{\sigma+1}$}
  \\&= \int_{x=\sigma-1}^{x=\sigma}S(x)\lambda_\sigma(x)dx
     + \int_{x=\sigma}^{x=\sigma+1}S(x)\lambda_\sigma(x)dx
\end{align*}$$

Actually, for the case $\sigma\in\setn{2,3,4,\ldots}$, I get a similar result as the linked reference, but different by a factor of $\frac{1}{6}$:

(lemma 1):
$$    \begin{align}
        &\int_{x=\sigma-1}^{x=\sigma}   \brs{ (A-B)(x-\sigma)+A           } (1-\sigma+x)    \dx
    \\&= \int_{x=\sigma-1}^{x=\sigma}   \brs{ A(1-\sigma+x) + B(\sigma-x) } (1-\sigma+x)    \dx
    \\&= \int_{x=\sigma-1}^{x=\sigma}  A\brs{ x^2 + 2x(1-\sigma)        + (1-\sigma)^2    } + B\brs{-x^2 +  x(\sigma+\sigma-1) + \sigma(1-\sigma)} \dx
    \\&= \brs{ A\brs{ \frac{1}{3}x^3 +             x^2(1-\sigma ) + x(1-\sigma)^2    } + B\brs{-\frac{1}{3}x^3 + \frac{1}{2} x^2(2\sigma-1) + x\sigma(1-\sigma)}}_{x=\sigma-1}^{x=\sigma}
    \\&= A\brs{ \frac{1}{3}\brp{\sigma^3-\brp{\sigma^3-3\sigma^2+3\sigma-1}} +            \brp{\sigma^2-\brp{\sigma^2-2\sigma+1}}(1-\sigma ) + (1-\sigma)^2\brp{\sigma-(\sigma-1)} }
\\&+ B\brs{-\frac{1}{3}\brp{\sigma^3-\brp{\sigma^3-3\sigma^2+3\sigma-1}} + \frac{1}{2}\brp{\sigma^2-\brp{\sigma^2-2\sigma+1}}(2\sigma-1) + \sigma(1-\sigma)}
\\&= A\brs{\brp{\sigma^2 - \sigma + \frac{1}{3} } +(2\sigma-1)(1-\sigma)+(-\sigma+1)^2}
\\&+B\brs{-\brp{\sigma^2 - \sigma + \frac{1}{3}} + \frac{1}{2}(2\sigma-1)(2\sigma-1) + \sigma(-\sigma+1)}
\\&=A\brs{\sigma^2( 1-2+1)+\sigma(-1+2+1-2)+\brp{ \frac{1}{3}-1+1}}
\\&+B\brs{\sigma^2(-1+2-1) + \sigma( 1-1-1+1) + \brp{-\frac{1}{3}+\frac{1}{2}}}
\\&= \boxed{\frac{1}{3}A + \frac{1}{6}B}
\end{align}$$

(lemma 2):
$$\begin{align}
  &\int_{x=\sigma}^{x=\sigma+1}  \brs{ (C-A)(x-\sigma)+C      } (1+\sigma-x)  \dx
\\&= \int_{x=\sigma}^{x=\sigma+1}  \brs{ C(1-\sigma+x) + A(\sigma-x) } (1+\sigma-x)  \dx
\\&= \int_{x=\sigma}^{x=\sigma+1} C\brs{-x^2 + x(\sigma+\sigma+1) + \sigma(-\sigma-1)  }
\\&+ A\brs{ x^2 + x(-\sigma-1-1-\sigma) + (1+\sigma)^2} \dx
\\&= \brs{ C\brs{-\frac{1}{3}x^3 + \frac{1}{2} x^2(2\sigma+1) - x\sigma(\sigma+1)} +A\brs{ \frac{1}{3}x^3 -       x^2( \sigma+1) + x(\sigma+1)^2  }}_{x=\sigma}^{x=\sigma+1}
\\&= C\brs{-\frac{1}{3}\brp{\brp{\sigma^3+3\sigma^2+3\sigma+1}-\sigma^3} + \frac{1}{2}\brp{\brp{\sigma^2+2\sigma+1}-\sigma^2}(2\sigma+1) - \sigma(\sigma+1) }
\\&+ A\brs{ \frac{1}{3}\brp{\brp{\sigma^3+3\sigma^2+3\sigma+1}-\sigma^3} -      \brp{\brp{\sigma^2+2\sigma+1}-\sigma^2}( \sigma+1) + (\sigma+1)^2   }
\\&= C\brs{-\sigma^2 + \sigma - \frac{1}{3} + \frac{1}{2}(2\sigma+1)(2\sigma+1) - \sigma^2 -\sigma }
\\&+ A\brs{ \sigma^2 + \sigma + \frac{1}{3} - (2\sigma+1)( \sigma+1) + \sigma^2 + 2\sigma + 1   }
\\&= C\brs{\sigma^2(-1+2-1) + \sigma(-1+1+1-1) + \brp{-\frac{1}{3}+\frac{1}{2}}} \\&+ A\brs{\sigma^2( 1-2+1) + \sigma( 1-2-1+2) + \brp{ \frac{1}{3}-1+1}}
\\&= \boxed{\frac{1}{6}C + \frac{1}{3}A}
\end{align}$$

(definitions): Let $A\eqd S(\sigma)$, $B\eqd S(\sigma-1)$, and $C\eqd S(\sigma+1)$.

Then$\ldots$
$$\begin{align}
  \inprod{S(x)}{\Lambda_\sigma(x)}
    &= \int_{x\in\mathbb{R}}S(x)\Lambda_\sigma(x)\dx
    && \text{by definition of $\inprod{f(x)}{g(x)}$}
  \\&= \int_{x=\sigma-1}^{x=\sigma+1}S(x)\Lambda_\sigma(x)\dx
    && \text{because support of $\Lambda_\sigma$ is $\intcc{\sigma-1}{\sigma+1}$}
  \\&= \int_{x=\sigma-1}^{x=\sigma}S(x)\Lambda_\sigma(x)\dx + \int_{x=\sigma}^{x=\sigma+1}S(x)\Lambda_\sigma(x)\dx
  \\&\eqd \int_{x=\sigma-1}^{x=\sigma  } \brs{ (A-B)(x-\sigma)+A } (1-\sigma+x) \dx
   \\&+ \int_{x=\sigma  }^{x=\sigma+1} \brs{ (C-A)(x-\sigma)+C } (1+\sigma-x) \dx
    && \text{by (definitions)}
  \\&= \brs{\frac{1}{3}A + \frac{1}{6}B} + \brs{\frac{1}{6}C + \frac{1}{3}A}
    && \text{by (lemma 1) and (lemma 2)}
  \\&= \frac{1}{6}\brs{B + 4A + C}
  \\&\eqd \boxed{\frac{1}{6}\brs{S(\sigma-1) + 4S(\sigma) + S(\sigma+1)}}
    && \text{by (definitions)}
\end{align}$$


问:
What is the mathematical foundation of Control Theory?
There is a question which I'm wondering again and again in recent months. I have taken courses on elementary differential equations, signals and systems, linear control systems, general theory of circuits and networks but I still do not know how can a simple linear integral transformation (Laplace transform) be so much helpful in analyzing and designing linear systems. For example, I think I somewhat know what "time domain" is, but how about the "frequency domain"?

Consider a block diagram in the frequency domain. Conceptually, what do the blocks represent? What is traveling(?) on the connections between blocks? First, I have thought that instead of taking care of the statement "when does some phenomenon happens" we consider "how many times does a phenomenon oscillate", but how can this approach be more helpful? Is it the best (optimal) way of treating such problems?

I think control systems and even systems have a more general meaning in mathematics. I even think that the very problem of "controlling a system" does have multiple "paradigms" — A.I., neural networks, state space approach, fuzzy logic, robust control, adaptive control, etc.

What I'm looking for is a book (preferred) or paper discussing the very beginning notions of system, control system, its history, and more importantly, what is the field or language in mathematics to study systems rigorously? Dynamical systems? Game theory? Differential equations? Linear algebra? A.I?

Any help is appreciated. Thanks.

P.S.: I've already studied these books:

Alan Oppenheim's Signals and Systems (I think it is the best in the field)

Ogata's Modern Control Engineering

Desore's Basic Circuits Theory

Charles A - Kuh and Ernest S's A rather brief book on the history of control systems ($H_\infty$ methods and so on, but I didn't get much; at least, this book induced some information on me that nobody actually knows what "feedback" is, why is it so helpful and why we should take samples of the output! It wasn't a precise book, I admit)

26 May 2013:

As I said, I am looking for a book or article discussing the origins of Control Theory and Systems. Seven days are passed since I started the bounty, yet I didn't find an answer. I'll get to the bottom of this issue as soon as I get some free time. Thanks.

答:
Arbib and Manes once made an attempt to apply the Automata Theory to Control Systems, see:

Manes, E.G.(ed.)
Category theory applied to computation and control. Proc. $1$st Internat. Symp., San Francisco, 1974. Lecture Notes in Comp. Sci., 25.

I am not an expert in the Control Systems and can not judge how successful this attempt is, but from the point of view of the Automata Theory, it looks pretty well.


问:
Dirichlet's Integral to find Volume with gamma function values
Evaluate volume $\iiint_D dx dy dz$ over a domain $D$ where $D$ is the region bounded by $x\ge 0$, $y\ge 0$ and $z\ge 0$ and ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$.

My approach:

I used the RMS$\ge$AM inequality (since $x,y,z\ge 0$ and $a,b,c>0$ without loss of generality) to arrive at $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\le\sqrt{3}$
and then did a substitution $X=\frac{x}{a}$, $Y=\frac{y}{b}$ and $Z=\frac{z}{c}$
And I got answer to be
$\frac{abc(\cdot\Gamma(1)\Gamma(1)\Gamma(1)\cdot(\sqrt{3})^3)}{\Gamma(4)}$ ie $abc\sqrt{3}/2$

My friend's approach:

Direct substitution of $X=\frac{x^2}{a^2}$, $Y=\frac{y^2}{b^2}$ and $Z=\frac{z^2}{c^2}$ and his answer was
$(\frac{abc}{8})\frac{(\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}))}{\Gamma(\frac{5}{2})}$ ie $\frac{abc\cdot\pi}{6}$

Professor says mine is wrong and friend's method is right but I am not satisfied without the reason.

答:
Did you just replace the ellipsoid $(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2\le 1$ with the octahedron $|\frac{x}{a}|+|\frac{y}{b}|+|\frac{z}{c}|\le\sqrt{3}$ ?

No wonder you are getting a different result for its volume. The RMS$\ge$AM
inequality guarantees that the ellipsoid is inscribed in the octahedron, i.e. fully contained in it - but of course, they are not the same bodies.


问:
Show that a convex polygon is contained within the largest circle determined by three consecutive vertices

Given a convex $n$-gon. The circumcircle is constructed for every triple of consecutive vertices of the polygon. We get the $n$ circles. Select the circle with the largest radius. Prove that the circle contains the polygon.

My work so far:

$n=3 -$ triangle - obviously.

$n=4 -$

If $\angle B = \max \left\{A,B,C,D \right\}$ then $ABCD \in \omega_{ABC}$

$n\ge 5$. I need help here.

答:
The crux of the argument is contained in the following proposition.

Proposition 1 Let $p$ and $q$ be points in the plane, $H$ be one half-space separated by the line through $p$ and $q$, and $D$ be a disk with $p$ and $q$ on its boundary and centered in $H$. If point $r\in H$ forms a circumcircle with $p$ and $q$ that has a smaller radius than that of $D$, then $r\in D$.

See the configuration in this proposition in the figure below. Any circle through $p$ and $q$ has a center on the line orthogonal to the line through $p$ and $q$ and passing halfway between those points. For that circle to have a radius smaller than $D$ and be centered in $H$, the center must be on dotted segment between $c$ (the center of $D$) and the midpoint between $p$ and $q$. And we can see that the portion of any such circle that is in $H$ (the dotted circle in the figure) is contained in $D$.

Next, we establish that ``adding vertices'' to a polygon only increases the size of the largest circumradius formed by consecutive vertices.

Proposition 2 Let $P$ be a polygon with vertices in clockwise order $v_0$, ..., $v_{n-1}$ and let $P'$ be a polygon with vertices $v_0$, ..., $v_{n-1}$, $v'$ (where $v'$ is not necessarily listed in clockwise order). The largest circumradius of three consecutive vertices of $P'$ is at least as large as that for $P$.

Assume the largest circumradius for $P$ is formed by vertices $v_{i-1}$, $v_i$, $v_{i+1}$. If $v'$ does not occur between vertices $v_{i-1}$ and $v_{i+1}$, then those vertices remain consecutive and thus the largest circumradius associated with $P'$ is at least as large as $P$. So we focus on the case that  $v'$ causes $v_{i-1}$, $v_i$ and $v_{i+1}$ to no longer be consecutive. Without loss of generality, suppose $v'$ lies between $v_i$ and $v_{i+1}$. Now there are two cases each providing a larger circumradius of a new set of consecutive vertices:

If $v'$ lies inside the circumcircle of $v_{i-1}$, $v_i$, $v_{i+1}$, then the circumradius of $v_i$, $v'$, $v_{i+1}$ is larger.

If $v'$ lies outside the circumcircle of $v_{i-1}$, $v_i$, $v_{i+1}$, then the circumradius of $v_{i-1}$, $v_i$, $v'$ is larger.

Now we can proceed to the main result. Let $P$ be a convex polygon with vertices $v_0$, $v_1$, ..., $v_n$ ordered in clockwise order. Without loss of generality, assume that $v_0$, $v_1$, $v_2$ form the largest circumcircle of any three consecutive vertices of the polygon.

The proof follows by induction, evaluating the sequence of polygons $P_i$ with vertices $v_0$, ..., $v_i$. The base case holds trivially for a single triangle $P_2$. The inductive step involves adding a single new vertex $v_i$ to the polygon and checking that vertex lies in the circumcircle of $v_0$, $v_1$, $v_2$. Proposition 2 ensures that $v_0$, $v_1$, $v_2$ provides the largest circumradius for each polygon, i.e., the largest circumradius for $P_i$ cannot be realized by a different set of consecutive vertices.

We complete the result by applying the Proposition 1 with $p=v_0$, $q=v_1$, $r=v_i$, and $D$ being the circumcircle of $v_0$, $v_1$, $v_2$. Convexity of the polygon ensures the assumptions of the proposition. Specifically, $v_2$ and $r$ must lie in the same half-space and the center of $D$ lies in the same half-space as $v_2$.


问:
A form of the Cauchy functional equation

Find all the functions $f :\mathbb R \to \mathbb R$ that satisfy the
conditions:

$$f(x+y)=f(x)+f(y), \enspace \forall x,y \in \mathbb R;$$

$$\exists \lim_{x\to \infty}f(x).$$

This problem is important for the community because on the forum I only saw the case where $f$ is continuous, but not the case when we only know that $\displaystyle\exists \lim_{x\to \infty}f(x).$

I found a solution to this problem in a book of analysis I own, which hints the following:

It is easy to see that $f(q)=q\cdot f(1), \enspace \forall q\in \mathbb Q$. Let $f(1)=a$.
If $a>0$, then consider the sequence $a_n=n.$ It follows that $$\lim_{x\to \infty}f(x)=\lim_{n\to \infty}f(a_n)=\lim_{n\to \infty}an=\infty.$$

The next step in the book is that this implies $f$ is increasing. I couldn't understand this line properly. Why is $f$ increasing? Of course, it is easy to check that $f(q)\le f(r), \forall q<r \in \mathbb Q$, but why is this also true for reals?

Please help me understand this! Thank you so much!

It is also clear that if $f$ is increasing then $f(x)=ax$ for all $x\in \mathbb R$ (suppose there exists an $x\in \mathbb R\setminus \mathbb Q$ such that $f(x)<x$. Then by density there is a $q\in \mathbb Q$ such that $f(x)<aq=f(q)<ax$, and this clearly gives the contradiction that $x< q <x$).

答:
So we have that lim $f = +\infty $.

Let's $x<y$ be real numbers. Then $f(y)-f(x) = f(y-x)$, so $n(f(y)-f(x)) = f(n(y-x))$.

Since $f(n(y-x)) \rightarrow +\infty$ when $n\rightarrow +\infty$, it is positive for $n$ big enough.

So $n(f(y)-f(x))$ is positive for $n$ big enough. But the sign of $n(f(y)-f(x))$ does not depend on $n$.

So $n(f(y)-f(x))$ is always positive, and especially $f(y) > f(x)$.


问:
$p\Bbb Z + x\Bbb Z[x]$ is a prime/maximal ideal
Let $\Bbb Z$ : ring of integers, $p$ : a prime number.

I know that $p\Bbb Z + x\Bbb Z[x]$ is an ideal. But how can I check if it is a prime/maximal ideal?

答:
Hint:
Let $I$ be the ideal in question. What is
$\mathbb{Z}[X]/ I$? Is it an integral domain? Is it a field?


问:
Proof of the following norm inequality
Let $x,y ∈ \mathbb{R}^n$ be vectors, let $A ∈ \mathbb{R}^{n×n}$ be a regular matrix. We denote $\lambda_{min}$ as the smallest eigenvalue of matrix $A^TA$ and $\lambda_{max}$ as the largest eigenvalue of $A^TA$. Then $
\frac{\| Ay\|^2}{\| Ax\|^2}≤\frac{\lambda_{max}}{\lambda_{min}}\frac{\| y\|^2}{\| x\|^2}$.

I have a problem with this proof since I can't seem to be able to connect eigenvalues of $A^TA$ with norms. Can someone point me in the right direction?

答: Hints: $\left\lVert Ax\right\rVert^2=x^{T}A^{T}Ax$ and $A^TA$ is symmetric, hence orthogonally diagonalizable.

Title: Proof of the following norm inequality


问:
Is there any general criteria to find $e$'th roots modulo a primer number $p$?
It is known that if $p$ is an odd prime and $\gcd(e, p-1) = 1$, then one can easily find an $e$'th root of every $a \in \mathbb{Z}_p$ by computing $d \equiv e^{-1} \pmod{p-1}$: $$a^{de} = a^{k(p-1)+1} = (a^{p-1})^k a = a. $$

Also, if $e = 2$ (meaning that $\gcd(2, p-1) \neq 1$) we know how to efficiently compute square roots modulo a prime $p$.

But note that the first sentence is just an implication: $$\gcd(e, p-1) = 1 \implies \exists x \in \mathbb{Z}_p,~ x^e \equiv a \pmod{p}$$ What about the other implication? Do we know something? Do we have a general "if and only if" criteria for the existence of an $e$'th root, even if it is proven using a constructive proof?

答:
For any $n\in\mathbb{Z}$ the map
$$
\mathbb{Z}_p^\times\longrightarrow\mathbb{Z}_p^\times\qquad
x\mapsto x^n
$$
is a homomorphism which is surjective if and only if ${\rm gcd}(n,p-1)=1$.
This is a straightforward consequence of the fact that the multiplicative group $\mathbb{Z}_p^\times$ is cyclic.

So, given any $n>0$ there are $\frac{p-1}{{\rm gcd}(n,p-1)}$ elements in $\mathbb{Z}_p^\times$ that have a $n$-th root.

The problem is that it is hard to get a practical characterization of this set since we don't easily know a generator of $\mathbb{Z}_p^\times$, e.g. see Artin's Conjecture

Similar considerations hold for the multiplicative group of any finite field.


问:
How to calculate $\int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta$
How to calculate:

$$ \int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta $$

答: Use the fact that $\cos^{2}\theta = 1-\sin^{2}\theta$ and the fact that integral of $\cos\theta$ is $\sin\theta$. Also $\sqrt{1-\sin^{2}\theta} = |\cos{\theta}|$. And note that $\cos\theta$ is positive in the first and the fourth quadrant.</p>


问:
Prove that all graphs $G=(V,E)$ with $|V|=4$ have non-trivial automorphism group
My understanding of this so far is that for $G$ to have a non-trivial automorphism group, ${\rm Aut}(G)$ must be greater than $1$ and therefore there exists a non-trivial symmetry of the vertices of $G$, i.e. not the identity. Hence, $G$ is symmetric (not asymmetric). So to prove all graphs on $4$ vertices have non-trivial automorphism group I can prove that there are no asymmetric graphs on $4$ vertices.

I believe my reasoning is correct but I don't know how to proceed from here. If anyone could help I would really appreciate it.

答:
This is,of course, a matter of a finite check that you can leave to a computer. However, for a "manual" proof, we need a way to classify the (numerous) cases.

First notice that a graph with $4$ nodes cannot have all the degrees of those nodes different. Namely, there is no graph with precisely the degrees $0,1,2,3$, because the node with degree $0$ is not linked to any nodes, while the node with degree $3$ is linked to all nodes, including the former - a contradiction.

Thus, we can distinguish the cases based on the degree of two nodes with the same degree:

If there are two nodes of degree $0$, a non-trivial automorphism will just swap those two nodes.

If there are two nodes (say $A$ and $B$) of degree $1$, then: either they are linked to the same node (say $C$) and a non-trivial automorphism just swaps $A$ with $B$, or they are linked to different nodes (say $C$ and $D$, respectively), and a non-trivial automorphism maps $A\mapsto B, B\mapsto A, C\mapsto D, D\mapsto C$.

If two nodes (say $A$ and $B$) are of degree $2$ or $3$, then consider the complement graph of the original graph. In the complement graph, $A$ and $B$ are of degree $1$ or $0$, respectively. Thus (as per the previous two points), there is a nontrivial automorphism of the complement graph. The same map is then a nontrivial automorphism of the original graph.


问:
Can a sum of an elementary and non-elementary definite integral be elementary or evaluated without special functions?
Hi I'm asking this in context of this:

Suppose we have non-elementary integral $\chi(x) = \int K(x)dx$ and we want to find

$$\int^b_aK(x)dx$$

Now lets say there is some function $F(x)$  which has elementary integral $\int F(x)$

such that $K(x) + F(x) $ has elementary integral so we can do this :

$$\int^b_aK(x)dx = \int^b_a(K(x)+0)dx$$

$$ = \int^b_a(K(x)+ F(x) - F(x))dx$$

$$=\int^b_aK(x)+ F(x)dx + \int^b_a-F(x)dx$$

Maybe $\int^b_a K(x) + F(x)dx$ is not elementary but lets say I dilute this condition to there exist some definite integral property transformation or substitution which allows us to arrive at numerical value for e.g just like $\int \ln(\sin(x))dx$ which is non-elementary according to WolframAlpha but it gives into some computable value for appropriate limits like

$\int^{\pi/2}_0 \ln(\sin(x))dx$

So my end goal goal is converting $\int^b_a K(x) + F(x)dx$ into some function like that atleast.

答:
$$\int_2^3 \Gamma'(x)dx=1$$

Does this fit your question?


答:
I do not think you need the increasing property and can argue more directly:
As before, $f(qx)=qf(x)$ for $q\in \Bbb Q$ and $x \in\Bbb R$, and let $a=f(1)$
Let $y\notin\Bbb Q$. Assume $f(y)\ne ay$, say $\epsilon:=\left|\frac{f(y)}y-a\right|>0$. By density of $\Bbb Q$ in $\Bbb R$, we find $u,v\in \Bbb Q$ with $y-\epsilon<u<y<v<y+\epsilon$.
Then $|f(y)-ay|>|a|\epsilon$ whereas $|f(u)-ay|$ and $|f(v)-ay|$ are both $<|a|\epsilon$. So
$f(u)$, $f(v)$ are either both $>f(y)$ or both $<f(y)$. At any rate, we find two positive numbers $x_1=y-u$ and $x_2=v-y$ such that exactly one of $f(x_1)$, $f(x_2)$ is positive and the other negative. Then one of $\lim_{n\to\infty}f(nx_1)$, $\lim_{n\to\infty}f(nx_2)$ is $+\infty$ and the other is $-\infty$. This contradicts the existence of $\lim_{x\to\infty}f(x)$.

Title: A form of the Cauchy functional equation

Body:
Find all the functions $f :\mathbb R \to \mathbb R$ that satisfy the
conditions:

$$f(x+y)=f(x)+f(y), \enspace \forall x,y \in \mathbb R;$$

$$\exists \lim_{x\to \infty}f(x).$$

This problem is important for the community because on the forum I only saw the case where $f$ is continuous, but not the case when we only know that $\displaystyle\exists \lim_{x\to \infty}f(x).$

I found a solution to this problem in a book of analysis I own, which hints the following:

It is easy to see that $f(q)=q\cdot f(1), \enspace \forall q\in \mathbb Q$. Let $f(1)=a$.
If $a>0$, then consider the sequence $a_n=n.$ It follows that $$\lim_{x\to \infty}f(x)=\lim_{n\to \infty}f(a_n)=\lim_{n\to \infty}an=\infty.$$

The next step in the book is that this implies $f$ is increasing. I couldn't understand this line properly. Why is $f$ increasing? Of course, it is easy to check that $f(q)\le f(r), \forall q<r \in \mathbb Q$, but why is this also true for reals?

Please help me understand this! Thank you so much!

It is also clear that if $f$ is increasing then $f(x)=ax$ for all $x\in \mathbb R$ (suppose there exists an $x\in \mathbb R\setminus \mathbb Q$ such that $f(x)<x$. Then by density there is a $q\in \mathbb Q$ such that $f(x)<aq=f(q)<ax$, and this clearly gives the contradiction that $x< q <x$).


问:
Showing that a function has a certain absolute minimum.
Suppose we have the function $$f(x) = \frac{x}{p} + \frac{b}{q} - x^{\frac{1}{p}}b^{\frac{1}{q}}$$ where $x,b \geq 0 \land p,q > 1 \land \frac{1}{p}+\frac{1}{q} = 1$

I am trying to show that $b$ is the absolute minimum of $f$.

I proceeded as follows:

$$\frac{df(x)}{dx} = \frac{1}{p} - \frac{x^{\frac{1}{p}-1}}{p} b^{\frac{1}{q}} = \frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px}$$

Now I will look for critical points by searching for the zeros of this function.

$$\frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px} = 0 \iff x - x^{\frac{1}{p}} b^{\frac{1}{q}} = 0 \iff x = x^{\frac{1}{p}} b^{\frac{1}{q}}$$.

Now I can see that $b$ is a critical point.

How ever when I continue my calculations to check whether there are any other critical points
$$x = x^{\frac{1}{p}} b^{\frac{1}{q}} \implies x^p = b^{\frac{p}{q}}x \implies x^{p-1} = b^{\frac{p}{q}} \implies x = b^{\frac{p}{(p-1)q}}$$

But this could not be equal to $b$, where did I go wrong?

答:
This is an instance of the Weighted Arithmetic Mean-Geometric Mean Inequality. It uses the concavity of $\log(x)$ and can be generalized to
$$
\prod_{i=1}^nx_i^{\alpha_i}\le\sum_{i=1}^n\alpha_ix_i\tag{1}
$$
where $\alpha_i\ge0$ and $\sum\limits_{i=1}^n\alpha_i=1$.

$(1)$ follows from Jensen's Inequality because $\log(x)$ is concave. Just exponentiate the following instance of Jensen
$$
\sum_{i=1}^n\alpha_i\log(x_i)\le\log\left(\sum_{i=1}^n\alpha_ix_i\right)\tag{2}
$$


问:
If X can be topologically embedded in a metric space Y how would one define the metric on X?
I have an idea of how one might go about doing it, but it's never explicitly mentioned in the books or anywhere on the net for that manner (at least not where I looked which includes this site), so I just wanted to make sure. So suppose we have a homeomorphism $f$ of $X$ into $Y$ which is equipped with a metric $d$; how would one define the metric on $X$? And show that this metric induces the same original topology on $X$.

答:
Let $X$ be a topological space and $i: X\to Y$ a topological embedding into a metric space $(Y, d)$.

The most important thing to note is that $S$ is open in $X$ if and only if $f(S)$ is open in $Y$. Equivalently, $S$ is a neighborhood around $x\in X$ if and only if $f(S)$ is a neighborhood around $f(x)\in Y$.
Therefore, intuitively speaking, having an embedding means that we can effectively treat $X$ as a subspace of $Y$.

So to define a metric $\delta$ on $X$, we can transport our points to $Y$ and measure the distance there: $$
  \delta(x, x') := d(f(x), f(x')).
$$ Proving that this indeed is a metric is essentially just transporting the metric axioms on $Y$ through the injection $f$.

To show that the topology induced by $(X, \delta)$ is the same as the original one, we can just verify that the notion of „neighborhood“ is the same (as it is purely definable by the notion of open sets, hence only depends on the topology): note that the following are equivalent for $x\in X, S\subseteq X$:

$S$ is a neighborhood around $x$

$f(S)$ is a neighborhood around $f(x)$

$f(S)$ contains an $\epsilon$-ball around $f(x)$ with regards to $d$

$S$ contains an $\epsilon$-ball around $x$ with regards to $\delta$

Edit: Your title said „topological embedding“, and your post just spoke of „homeomorphic“. Therefore I chose to show the more general principle.


问:
Write in matrix form
$D_{βν} =∑^{n}_{µ=1}∑^{n}_{a=1}A_{µν}B_{αµ}C_{αβ}$

Since the first index of $B$ and $C$ is the same, I believe what we have here is:

$$\sum^{n}_{a=1}B_{αµ}C_{αβ} = B(C^{T})$$

So that the equation given means:

$D = ABC^{T}$

I am not sure if this is right, after the conclusion that $C$ is a transpose matrix, the index become confusing to me.

答:
If $D = ABC^T$ then:

$$D_{i,j} = (ABC^T)_{i,j} = \displaystyle \sum_{k=1}^n(AB)_{i,k} (C^T)_{k,j} =\displaystyle \sum_{k=1}^n \sum_{\ell=1}^n A_{i,\ell}B_{\ell,k} C_{j,k} $$

So this is not right. The problem is that in fact:

$$ \sum_{\alpha=1}^nB_{\alpha\mu} C_{\alpha \beta} = \sum_{\alpha=1}^n(B^T)_{\mu \alpha} C_{\alpha \beta} = (B^TC)_{\mu \beta} $$

Title: Write in matrix form


问:
Every symmetric convex body in $\mathbb{R}^n$ is the closed unit ball of a norm on $\mathbb{R}^n$ - why?
I have trouble understanding the following proof:

Theorem: Every symmetric convex body in $\mathbb{R}^n$ is the closed unit ball of a norm on $\mathbb{R}^n$

Proof: Given a symmetric convex body $K$ in $\mathbb{R}^n$ we define its (so-called) Minkowski functional $$p_K(x) = \inf\{t>0, x\in tK\}$$
It is clear that $\{p_K(x)\leq 1\} = K$, so it remains to check that $p_K$ is a norm (exercise).

In the definition of $p_K(x)$, there is $``x\in tK"$ - what's $tK$? The convex body scaled in all directions by a factor of $t\in\mathbb{R}$? Also what does $p_K(x)$ really mean intuitively?

Why is $\{p_K(x)\leq 1\} = K$?

To see that $p_K$ is a norm:
(i) Consider $p_K(x) = 0$. This means $\inf\{t>0, x\in tK\} = 0$. I need to show that $x=0$. This is a little weird, innit? Perhaps the definition of $p_K(x)$ should have been $p_K(x) = \inf\{t\geq0, x\in tK\}$?

(ii) To show $p(ax) = |a|p(x), a\in\mathbb{R}$: we must show $\inf\{t>0, ax\in tK\} = |a|\inf\{t>0, x\in tK\}$. Why does this hold?

(iii) Lastly we also need to show the triangle-inequality, i.e. $p(x) + p(y) \ge p(x+y), x,y\in\mathbb{R}^n$.

I'm not necessarily interested in a detailed solution (though that would help), some hints or pointers to get me started would suffice too!

答:
In the definition of $p_K(x)$, $tK = \{tk ; k\in K\}$.

Why is $\{p_K(x)\leq 1\} = K$?

First, if $x\in K$ then $x\in 1K$ so $\inf \{t>0 ; x\in tK\} \leqslant 1$, so $p_K(x) \leqslant 1$.

If $x\in \{p_K(x)\leq 1\}$, then there exists $t\leqslant 1$ such that $x\in tK$. So there exists $k\in K$ such that $x = tk = tk + (1-t)0$ and since $K$ is convex, we have that $x\in K$ ($0$ is in $K$ since $K$ is symmetric).

To see that $p_K$ is a norm:
(i) $p_K(x) = 0$ means $\inf\{t>0, x\in tK\} = 0$, which means there exists a sequence $(t_n)$ of positive numbers such that $t_n \rightarrow 0$ and for all $n$, $x\in t_n K$. So for $n$, there exists $k_n \in K$ such that $x = t_n k_n$. As $K$ is compact (finite-dimension), we can extract $k_{\varphi (n)}$ which converges to $k$. Since $t_{\varphi (n)} \rightarrow 0$, we have $x = \lim t_n k = 0$.

(ii) To show $p(ax) = |a|p(x), a\in\mathbb{R}$: as you said, we must show $\inf\{t>0, ax\in tK\} = |a|\inf\{t>0, x\in tK\}$. It's true because $x\in tK$ also means $ax \in (at)K$.

(iii) For the triangle-inequality, remark that if $x = t_x k_x$ and $y= t_y k_y$, then after some work, we get $x+y = (t_x + t_y)k$ for some $k$. So $p_k (x+y) \leqslant t_x + t_y$. Then since this is true for all $t_x$ and $t_y$ such that $x\in t_x K$ and $y\in t_y K$, we can take the inf of the equation and we get the result.


问:
What is a hypersingular integral kernel?
While reading literature about boundary element and finite element methods I have repeatedly seen that some integral kernels are singular and others are hypersingular. Could you explain what is the difference between singular and hypersingular and how to tell where a certain kernel belongs?

"Weakly singular" also seems unclear to me, what is that one?

答:
Consider an integral of the sort
$$
\int_{\mathbb R^n}K(x,y)f(y)\,dy
$$
for some smooth function $f$ and a kernel $K$. To allow the integral to converge at infinity we'll assume that $f$ has compact support. Integrals of that sort appears in the integral equations to which, in particular, boundary value problems for PDE are reduced (not always in $\mathbb R^n$, but on surfaces too).

If $|K(x,y)|\sim |x-y|^{-\alpha}$ when $y\to x$, $\alpha>0$, the kernel is called singular, because it tends to infinity as $y\to x$. If $\alpha<n$ the kernel (and the integral) is called weakly singular, since the integrand is absolutely summable. For $\alpha>n$ a kernel is called hypersingular, because in this case the estimate on $|K|$ is not enough to guaranty convergence. Such integrals are often taken in the sense of the principal value. It can be done under some additional assumptions on $K$ such as
$$
\int_{|x-y|=\varepsilon}K(x,y)\,ds_y=0, \quad \varepsilon>0.
$$
For example, the values of the simple-layer potential for the Laplace equation (n=3)
$$
\frac{1}{4\pi}\int_S\frac{f(y)}{|x-y|}ds_y, \quad x\in S,
$$
is a weakly singular integral on a smooth bounded surface $S$ and the double layer potential
$$
\frac{1}{4\pi}\int_Sf(y)\frac{\partial }{\partial n_y}\frac1{|x-y|}ds_y, \quad x\in S, \qquad (*)
$$
is singular, but not weakly singular, because  $\left|\frac{\partial }{\partial n_y}\frac1{|x-y|}\right|\sim|x-y|^{-2}$,
$S$ is a 2-dimensional surface and
$$
\frac{1}{4\pi}\int_S\left|\frac{\partial }{\partial n_y}\frac1{|x-y|}\right|ds_y=+\infty
$$
for some points $x\in S$.
Nevertheless for smooth $f$ the integral $(*)$ converges in the sense of the principal value.


问:
Solving Matrix Riccati equation?
I was solving the Riccati equation $XAX=B$ where $A$ and $B$ are positive matrices, with $A$ being strictly positive.

Solving this, I did

$U'A^{1/2}XA^{1/2}UU'A^{1/2}XA^{1/2}U = U'A^{1/2}BA^{1/2}U$

$U'A^{1/2}XA^{1/2}U = (U'A^{1/2}BA^{1/2}U)^{1/2}$

$X = A^{-1/2}U(U'A^{1/2}BA^{1/2}U)^{1/2}U'A^{-1/2}$

where $U$ is a unitary matrix.

It's clear that $X = A^{-1/2}(A^{1/2}BA^{1/2})^{1/2}A^{-1/2}$ is a solution but why is it positive and furthermore the unique positive solution?

答:
In general, if $A$ is positive and $Y$ arbitrary, then $Y^\ast A Y$ is positive: It is clearly symmetric (or Hermitian, if you're working over $\mathbb{C}$) and $\langle Y^\ast A Y\xi,\xi\rangle=\langle A(Y\xi),Y\xi\rangle\geq 0$. Also the square root of a positive matrix is positive by definition. This proves that the matrix $X$ from your question is positive.

Uniqueness actually follows from the computation in your question. Just note that for unitary $V$ and positive $Y$ you have $(V^\ast Y^{1/2} V)^2=V^\ast YV$, which implies $(V^\ast Y V)^{1/2}=V^\ast Y^{1/2}V$ (remember that $V^\ast Y^{1/2}V$ is positive). Thus the solution you got is $X=A^{-1/2}(A^{1/2}B A^{1/2})^{1/2}A^{-1/2}$.

Title: Solving Matrix Riccati equation?


问:
All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$

Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.

I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$

I think that a geometric solution might be possible, please help me to proceed.

Thanks.

答:
I went ahead and drew an example in Geogebra. Let $\alpha = \frac{\sqrt{2}-1}{2}$. It was clearly chosen specially, since it looks like almost no choices of $q$ (labelled $E$ in the plot) allow any solutions.

With that in mind, I'll consider a necessary condition for there to be any solutions $p$ for a given value of $q$.

Condition: The line joining $(0,q)$ to $(1,\alpha)$ must not cross the lower circle.

When does that happen? Well, experimenting suggests it never happens, and the chord is tangent when $q=1+\alpha$. So maybe we try to prove that the line $AB$ is tangent to the circle $DC$.

The line $AB$ has equation $y=1+\alpha - x$. Does this intersect the circle $DC$? That is equivalent to asking for solutions of $1+\alpha-x = \sqrt{1-x^2} - \alpha$, or:

\begin{align}
1-x^2 &= (1+2\alpha-x)^2 \\
&= (\sqrt{2}-x)^2 \\
&= 2-2\sqrt{2}x + x^2 \\
2x^2 - 2\sqrt{2} x + 1 &= 0 \\
(\sqrt{2}x - 1)^2 &= 0.
\end{align}

So we have a point of tangency. Clearly, therefore, if $q < 1+\alpha$ then there is no value of $p$ such that the line is contained in the figure $ABCD$. This means the only solution is $(-1, 1+\alpha)$.


答:
Using the hint provided by leoli1, you can prove the following result:
\begin{equation}
\left\lVert Ax\right\rVert \leq \lambda_{max}\left\lVert x\right\rVert
\end{equation}
You can try setting the coordonates of x$\in \mathbb{R}^n$ in an orthonormal basis of eigenvectors of $A^TA$ and see what happends when computing $x^TA^TAx$.

The same idea apply for the inequality with $\lambda_{min}$.

Hope it helps!

Title: Proof of the following norm inequality

Body:
Let $x,y ∈ \mathbb{R}^n$ be vectors, let $A ∈ \mathbb{R}^{n×n}$ be a regular matrix. We denote $\lambda_{min}$ as the smallest eigenvalue of matrix $A^TA$ and $\lambda_{max}$ as the largest eigenvalue of $A^TA$. Then $
\frac{\| Ay\|^2}{\| Ax\|^2}≤\frac{\lambda_{max}}{\lambda_{min}}\frac{\| y\|^2}{\| x\|^2}$.

I have a problem with this proof since I can't seem to be able to connect eigenvalues of $A^TA$ with norms. Can someone point me in the right direction?


问:
What is the sum of this infinite series? Which one is it, Taylors? Binomial?
I am trying to figure which formula to use for this one.
$$\displaystyle\sum\limits_{x=-1}^{-\infty} -x(1-y)p(1-p)^{-x}+\sum\limits_{x=1}^\infty xyp(1-p)^x$$
where $0<y<1$, and $0<p<1$.

My work:
$$=\displaystyle\sum\limits_{x=1}^{\infty} -x(1-y)p(1-p)^{x}+\sum\limits_{x=1}^\infty xyp(1-p)^x$$
next step,
$$=\displaystyle\sum\limits_{x=1}^{\infty} p(1-p)^{x}(2xy-x)$$
This doesn't look like anything.  Does anyone recognize this?

答:
It works like @vadium123 mentioned, I just wanna give you the idea how one derives that result.

The first idea is that
\[ \sum_{n=0}^\infty n \cdot x^n  \]
looks like the goemetric series (at least a bit). In fact if we take the derivative the geometric series in respect to $x$ we get
\[ \sum_{n=1}^\infty n\cdot x^{n-1} \]
So when we take
\[ x \cdot \sum_{n=1}^\infty n \cdot x^{n-1} = \sum_{n=1}^\infty n\cdot x^n\]
Furthermore we know that
\[ \sum_{n=0}^\infty x^n = \frac{1}{1-x} \]
When we take the derivative on both sides we have
\[ \frac{1}{(1-x)^2}\]
As we need to multiply with $x$ our final sum will be
\[ \frac{x}{(1-x)^2}\]


问:
Error Propagation And Multiplication by Zero
So I'm trying to create a python module to be used for Error Propagation just for sake of learning.

let:
$$A=a \pm \delta a$$
$$B=b \pm \delta b$$
As explained here Error Propagation of a function as shown
$$y=f(A, B) = A \times B$$
is
$$\delta y= \sqrt{ (\frac{\delta a}{A})^2 + (\frac{\delta b}{B})^2}$$

I understand this part. But what if one of the values is 0.
So let temperature measurements be:
$$T_1=1 \pm 0.5$$
$$T_2=0 \pm 0.7$$

When I multiply these two values I'll get division by zero naturally.

How to solve this problem?

答:
When one of the values is zero, you can't use relative errors. Then use

$$AB\,\delta y=\sqrt{B^2(\delta a)^2+A^2(\delta b)^2}.$$


问:
Is $\log(z-\alpha)$ well defined?
Could you help me with the following problem please:

Show that if $U\subset{\mathbb{C}}$ is simply connected domain and $\alpha \in \mathbb{C}\setminus U$, then log$(z-\alpha)$ is well defined for all $z\in U$

I have no idea how to solve this problem, it arose at the time of proving the Riemann mapping theorem, but it does not occur to me how to do the proof, also, I have tried to see it taking the case in which U is an open ball, but not I can understand well what the exercise says, since if, for example, I took the unit ball centered on the origin, there would still be values in which the logarithm would not be well defined. I hope you can help me with this, thank you.

My try:

We fix a point $z_0\in U$ and let's define:
$$
\log(z-\alpha)=\displaystyle\int_{\gamma(z)}\dfrac{1}{w-\alpha}dw+\log(z_0-\alpha)
$$
with $\gamma(z)$ a path in $U$ that joins $z$ with $z_0$.

And it would be necessary to see that in this case the integral does not depend on the path, I think that would suffice, but I have not managed to prove it, it occurs to me that it is because it is a simply connected domain and applying Cauchy's theorem

答:
I think you are on the right track. I'd proceed by doing a further step remembering the following facts:

Cauchy's integral theorem: if $f: U\to \Bbb C$ is a holomorphic function on the domain $U$, then
$$
\oint\limits_\gamma f(w)\mathrm{d}w=0
$$
for each simple closed curve contained in $U$.

If $\alpha \in \Bbb C\setminus U$ then
$$
f(w)=\frac{1}{w-\alpha}\quad \forall w\in U
$$
is holomorphic on the whole $U$. To see this note that, by putting $w\equiv x+iy$, we have
$$
\begin{split}
\frac{\partial}{\partial x} f(x,y) &= \frac{\partial}{\partial x} \frac{1}{x+iy-\alpha} \\
& = -\frac{1}{(x+iy-\alpha)^2}\\
& = -i\frac{\partial}{\partial y} \frac{1}{x+iy-\alpha}= -i\frac{\partial}{\partial y} f(x,y)
\end{split}
$$
thus $f$ satisfies the Cauchy-Riemann equations on the whole $U$ since it is continuous with its partial derivatives for any $w\in U$ due to the hypothesis $\alpha \in \Bbb C\setminus U$.

Now the definition of $\log(z-\alpha)$ as given above is seen to be path independent since, for any two continuous paths $\gamma_1[z],\gamma_2[z]: [0,1]\to U$ joining $z_0$ to $z$ in $U$
$$
\oint\limits_{\gamma_1[z]}\frac{1}{w-\alpha}\mathrm{d}w = \oint\limits_{\gamma_2[z]}\frac{1}{w-\alpha}\mathrm{d}w \iff \oint\limits_{\gamma_1\cup [-\gamma_2]}\frac{1}{w-\alpha}\mathrm{d}w=0
$$
provided having defined the closed curve $\gamma_1\cup [-\gamma_2]:[0,1]\to \Bbb C$ as
$$
\gamma_1\cup [-\gamma_2](t)=
\begin{cases}
\gamma_1[z](2t) & t\in \big[0,\frac{1}{2}\big[\\
\gamma_2[z](2- 2t) & t\in \big[\frac{1}{2},1\big]
\end{cases}
$$
Thus $\log(z-\alpha)$ is univocally defined.

Final notes

As @Haus said in the comments, the choice of $z_0\in U$ is exactly the choice of the branch of the logarithm we are willing to use: once you chose $z_0$, the additive constant becomes $\log(z_0-\alpha)$ and is fully determined for the whole simply connected domain $U$. If we consider $\alpha=0$ and $U=\Bbb C\setminus {0}$ (which is not simply connected) we see that it is not so: depending on how we encircle $0$ by the path $\gamma[z]$ (clockwise or counterclockwise, starting from an arbitrary $z_0\neq 0$), we have that
$$
\log z =\log |z| + i\arg z
$$
where the imaginary part is only known modulo $2n\pi i$ for all $n\in\Bbb N$ and thus not uniquely determined.

The curve $\gamma_1\cup[-\gamma_2]$ has been defined in perhaps a cumbersome way in order to show how it is closed. Intuitively speaking, when the parameter $t$ rises from $0$ to $\frac{1}{2}$, the mapped point "moves" along $\gamma_1$ from $z_0$ to the (chosen and fixed) $z$: when it further rises beyond $\frac{1}{2}$ to $1$, the mapped point "moves" from $z$ to $z_0$ "reversing its motion". This later behavior is the reason for the highly nonstandard and perhaps unfortunate notation $[-\gamma_2]$ I've chosen for this part of the composite path: I wanted to find something that would remind the reader of the addition of simplexes in euclidean spaces, in order to make the concept intuitively clear, but I've obviously failed.

Strictly speaking, the reasoning above requires the two paths $\gamma_1[z]$ and $\gamma_2[z]$ to be non intersecting: however, the integral calculated on them can be seen to be equal to the one calculated on an auxiliary path joining $z$ to $z_0$ and not interesting any of them, thus the independence of the definition of the complex logarithm holds without any restriction on the used path.

Title: Is $\log(z-\alpha)$ well defined?


问:
A set of Real numbers that contains all elements of a closed Interval except an open interval within it. x in {[0,4]- (1,2)}
I don't know if this question has been asked before.
Suppose we have a set of Real Numbers such that

$ A =[0,4] \\ B = (1,2) $

$ X = \{x \in \{A - B\} \} $

or in general it contains all the elements of a closed interval except the elements of an open interval thats a proper subset of the closed interval.

Is X now a closed set?

By the definition a Real Number Set is closed if it contains all of its its limit points.

and s is a limit point of set S if

$ \forall \epsilon \gt 0 , \exists \  a\ne s  $ such that
$ a \in \{ V_{\epsilon(s)} \cap S \} $

Now I can intuitively think of that any elements in (1,2) cannot be a limit point of set X. And neither can be any elements of Real numbers outside of the closed set [0,4].

Did it come out as a valid conclusion?

答:
If you mean by {$x \in {A - B}$} the set of all subtractions with $a\in A, b\in B$ so $a-b \in X$, then you have an open set, because the least number you can get is $0 - (2)$ but $-2$ is not in the set and the biggest number you can get is $4-1$ but as $1$ is also not in the set, the final set (or interval) is $X= (-2,3)$.

But if you mean $X=A \setminus B$ then you have $X=[0,1] \cup [2,4]$ as 1 and 2 has to be in the set, but all the other numbers between aren't.


问:
How can I prove $2^n+10n^2 \le 12 \cdot 2^n, n \ge 1$ is true by induction?
I'm having trouble with this problem.
$2^n+10n^2 \le 12 \cdot 2^n, n \ge 1$

Here is my process.

n = 1, it is true <-- basis step

Assume that arbitrary k is true. $2^k+10k^2 \le 12 \cdot 2^k$.
And I make  $2^{k+1}+20(k)^2 \le 12 \cdot 2^{k+1}$ by multiplying 2.

But if k become k+1,the equation should be $2^{k+1}+10(k+1)^2 \le 12 \cdot 2^{k+1}$..
I can't translate $20k^2$ into $10(k+1)^2$. How can I prove it by induction?
Or is this equation false?

答:
Base case : $2+10 \leqslant 24$ and $4 + 10 \cdot4 \leqslant 12 \cdot 2^2$ and $8+10 \cdot 9 \leq 12 \cdot 8$ so it's ok for $n\in \{1,2,3\}$.

Inductive step :

Let's suppose it's true for some $n\geqslant 3$.

Then $2^{n+1} + 10 (n+1)^2 = 2.2^n + 10 n^2 + 10(2n+1)$

$\leqslant 2 (12.2^n - 10 n^2) + 10 n^2 + 10(2n+1)$

$\leqslant 12.2^{n+1} - 10 n^2 + 10(2n+1)$

and this last expression is $\leqslant 12.2^{n+1}$ since $10 n^2 \geqslant 10 (2n+1)$ for $n\geqslant 3$ (you can easily prove that by induction).


答:
Since there is already an answer on your questions, I will try to explain the intuition behind the minkowski functional.

You can think  of the minkowski functional $ ρ_Α$  (geometrically) as the fraction $ \frac{d(x,0)}{d_x'}$ where $d(x,0)$ is the distance from x to $0$ and $d_x'$ is the distance from $0$, to the "farthest" point of $A$, in the direction of $x$.
See here for some visual examples.

In the first example: $A \subset \mathbb R^n$ is a ball of radius $R$ centered at $0$. It turns out that  $ ρ_A(x) = \frac{||x||}{R}$. Notice that     $ d(x,0)=||x||$ and $ d_x' =R$.

In the second example: $A$ is a half plane in $\mathbb R^2$. The distance from $0$ to the farthest point of $A$ in the direction of $x$ is actually infinity (since $A$ is unbounded in that direction) and $ρ_A(x) =0 = \frac{(OM)}{\infty} $. On the other hand, the  distance from $0$ to the farthest point in $A$ in the direction of $y$ is $N'$. So, $ρ_A(y) = \frac{(ON)}{(ON')}= 1/μ $.

Title: Every symmetric convex body in $\mathbb{R}^n$ is the closed unit ball of a norm on $\mathbb{R}^n$ - why?

Body:
I have trouble understanding the following proof:

Theorem: Every symmetric convex body in $\mathbb{R}^n$ is the closed unit ball of a norm on $\mathbb{R}^n$

Proof: Given a symmetric convex body $K$ in $\mathbb{R}^n$ we define its (so-called) Minkowski functional $$p_K(x) = \inf\{t>0, x\in tK\}$$
It is clear that $\{p_K(x)\leq 1\} = K$, so it remains to check that $p_K$ is a norm (exercise).

In the definition of $p_K(x)$, there is $``x\in tK"$ - what's $tK$? The convex body scaled in all directions by a factor of $t\in\mathbb{R}$? Also what does $p_K(x)$ really mean intuitively?

Why is $\{p_K(x)\leq 1\} = K$?

To see that $p_K$ is a norm:
(i) Consider $p_K(x) = 0$. This means $\inf\{t>0, x\in tK\} = 0$. I need to show that $x=0$. This is a little weird, innit? Perhaps the definition of $p_K(x)$ should have been $p_K(x) = \inf\{t\geq0, x\in tK\}$?

(ii) To show $p(ax) = |a|p(x), a\in\mathbb{R}$: we must show $\inf\{t>0, ax\in tK\} = |a|\inf\{t>0, x\in tK\}$. Why does this hold?

(iii) Lastly we also need to show the triangle-inequality, i.e. $p(x) + p(y) \ge p(x+y), x,y\in\mathbb{R}^n$.

I'm not necessarily interested in a detailed solution (though that would help), some hints or pointers to get me started would suffice too!


答:
$p_K(x)$ is measuring, roughly speaking, the smallest dilation $t$ for which $x$ sits on the boundary of the convex body. It might help to consider the case where $K=\overline{B(0,1)}$, the closed Euclidean unit ball. We look for the greatest lower bound of $t$ for which $x\in tK$--but this $t$ is precisely the $t$ for which $x$ lies on the sphere of radius $t$. But that radius tells us the distance of $x$ to the origin, so it's in this way scalings of $K$ can give us a notion of norm.

If $x\in 1\cdot K$, then $p_K(x)\leq 1$ since $p_K(x)$ is the infimum of $t$ for which $x\in tK$. Thus $K\subset\{x: p_K(x)\leq 1\}$. The reverse inclusion is a little trickier. Suppose $0<p_K(x)\leq 1$. There is a minimizing sequence $t_n\to p_K(x)$ such that $x\in t_nK$ for each $n$. But then $x/t_n\in K$, and since $K$ is closed, we conclude $x/p_K(x)\in K$. Thus $x\in p_K(x)K$, which is a subset of $K$ since $p_K(x)\leq 1$ and $K$ is convex. If $p_K(x)=0$, then it is not hard to see $x=0$ (if one assumed $x\neq 0$ and $x\in (1/n)K$ for all positive integers $n$, then we get $nx\in K$ for all positive integers $n$--this will contradict boundedness of $K$).

(i) I already mentioned this case. You'll want to use boundedness of $K$.

(ii) Just show that, for $a\neq 0$, $\{t>0: ax\in tK\}=\{t>0: x\in (t/|a|)K\} = \{|a|s>0: x\in sK\}$. The infimum of the last set is $|a|p_K(x)$, so the result will follow.

(iii) If $x\in t_1K$ and $y\in t_2K$, then show that $(x+y)\in (t_1+t_2)K$. It follows $p_K(x+y)\leq t_1+t_2$. Now take infimums over $t_1,t_2$.


Title: How can I prove $2^n+10n^2 \le 12 \cdot 2^n, n \ge 1$ is true by induction?

Body:
I'm having trouble with this problem.
$2^n+10n^2 \le 12 \cdot 2^n, n \ge 1$

Here is my process.

n = 1, it is true <-- basis step

Assume that arbitrary k is true. $2^k+10k^2 \le 12 \cdot 2^k$.
And I make  $2^{k+1}+20(k)^2 \le 12 \cdot 2^{k+1}$ by multiplying 2.

But if k become k+1,the equation should be $2^{k+1}+10(k+1)^2 \le 12 \cdot 2^{k+1}$..
I can't translate $20k^2$ into $10(k+1)^2$. How can I prove it by induction?
Or is this equation false?

Body:
$ k^2 > 2k+1 \forall k>2$

$2k^2 > (k+1)^2$
$\Rightarrow 10(k+1)^2 < 20k^2$
$2^{k+1} + 10(k + 1)^2 < 2^{k+1} + 20k^2$


问:
Questions about set cardinalities (infinite/finite).
I have a couple questions that I undertstood (maybe) intuitively and tried to prove them somehow, I would like if someone can give feedback and tell me if I'm doing it the right way.

True or false:

1)$ |\{ \frac {1} {2^n} \mid n\in \mathbb{N} \}| = |\mathbb{N}|$
.
Intuitively this seems true, and I thought of proving it by introducing the function $f: \mathbb{N} \longrightarrow \{ \frac {1} {2^n} \mid n\in \mathbb{N} \}$, $f(n)=\frac {1} {2^n}$.

2) $K= \{A\in P(\mathbb{N}) \mid A \space is \space finite \} $, Assumption: $K$ is finite.

Intuitively I can see it's infinite, since the power set of natural numbers have infinity elements, but I really don't know how to prove it.

3) $L= \{A\in P(\mathbb{N}) \mid A^c \space is \space finite \}$,
Assumption $L$ is finite.

I think this is true, since for all $A\in P(\mathbb{N})$, $A^c$ is infinite, so theres no elements in $L$.

I would really appreciate it if someone can approve my work, and tell me how to formalize the proofs, and point out my mistakes. Thanks in advance

答:
Just give an infinite number of elements of $K$, for example all $\{k\}$ are in  for $k\in \mathbb{N}$.

If $A^c$ is the complementary of $A$, then I think the proposition is false, since the $\{n; n\geqslant k\}$ are in $L$ for all $k$.


问:
The convergence of this integral $\int_{1}^{\infty}\frac{1}{x+x^{2}\sin^{2}(\pi x)}~dx$
The elementary question is below :

Q : Is this integral convergent $?$
\begin{align}
\int_{1}^{\infty}\frac{1}{x+x^{2}\sin^{2}(\pi x)}~dx
\end{align}

But it seems to be not easy to deduce that this integral converges or not. I think the key point is that sin function is in the denominator. Are there any suggestions of this problem ? I will be appreciated.

答:
The convergence of

$$I = \int_{1}^{\infty}\frac{1}{x+x^{2}\sin^{2}(\pi x)}\ dx$$ is equivalent to the convergence of the series $\sum_{n=1}^\infty I_n$ where

$$I_n = \int_{n-1/2}^{n+1/2}\frac{1}{x+x^{2}\sin^{2}(\pi x)}\ dx.$$

Following inequalities / equalities hold for $n \in \mathbb N$:

$$\begin{aligned}
0 \le I_n &=  \int_{-1/2}^{1/2}\frac{1}{(y+n)+(y+n)^{2}\sin^{2}(\pi x)}\ dy\\
&\le \frac{1}{n-1/2}\int_{-1/2}^{1/2}\frac{1}{1+(n-1/2)\sin^{2}(\pi y)}\ dy\\
&=\frac{2}{n-1/2}\int_{0}^{1/2}\frac{1}{1+(n-1/2)\sin^{2}(\pi y)}\ dy\\
&\le \frac{2}{n-1/2}\int_{0}^{1/2}\frac{1}{1+4(n-1/2)z^2}\ dz\\
&\le \frac{2}{n-1/2}\int_{0}^\infty\frac{1}{1+4(n-1/2)z^2}\ dz\\
&= \frac{\pi}{2(n-1/2)^{3/2}}
\end{aligned}$$

as:

$x \mapsto \frac{1}{1+(n-1/2)\sin^{2}(\pi x)}$ is an even function.

$\sin (\pi x) \ge 2x$ on the interval $[0, 1/2]$. This is a convexity inequality.

$\int_0^\infty \frac{dt}{1+b^2t^2} = \frac{\pi}{2b}$ for $b \gt 0$.

Therefore $I$ converges as $\sum I_n$ converges.


问:
Absolutely integrable implies almost surely finite
If $X$ is absolutely integrable, i.e., $\mathbb {E}\vert X \vert<\infty$, I want to show that $X$ is almost surely finite, i.e., $\mathbb {P}(\vert X\vert<\infty)=1$. For now, I have the following:
\begin{align*}
\mathbb {E}\vert X\vert=\int_{0}^{\infty}\mathbb {P}(\vert X\vert\geq u)\, du
\end{align*}

I wonder how to argue based on this.

答: Suppose that $P(|X| = \infty) = c > 0$. Thus for all $u \ge 0$ we have  $P(|X| > u) \ge P(|X| = \infty) = c$. Thus $\infty > E|X| = \int_{0}^{\infty} P(|X|\ge u)du \ge \int_{0}^{\infty} c du =\infty$. It's contradiction.


问:
How do I evaluate $\iiint_{R} (xy+z^2) \ dV\,$?
The question states:

Sketch the region $$R = \left \{ (x,y,z) : 0 \leq z \leq 1- |x| -|y| \right \}$$

and evaluate the following integral $$\iiint_{R} (xy+z^2) \ dV$$

So my thoughts are that the sketch is straightforward just a pyramid with a square base. But then I get stuck with the limits of the integral. All I know is that we can take the positive segments and double and add on to find R since it's symmetrical but if someone could explain the limits involved.

The answer has been given to be: $$\frac{1}{15}$$

Thank you!

答: You are right about the shape of $R$. Since$$|x|+|y|\leqslant1\iff|y|\leqslant1-|x|\iff|x|-1\leqslant y\leqslant1-|x|,$$your integral is$$\int_{-1}^1\int_{|x|-1}^{1-|x|}\int_0^{1-|x|-|y|}(xy+z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$$


Body:
We can use

$10n^2 - 20n - 10 = 10 [(n-1)^2 - 2] \gt 0$ for $n \gt 2$

i.e $20n^2 \gt  10n^2 + 20n + 10 \implies 10n^2 \gt  \frac{1}{2} (10n^2 + 20n + 10)$

We have $2^n+10n^2 \le 12 \times 2^n$

and we need to show $2^{n+1}+10(n+1)^2 \le 12 \times 2^{n+1}$

$2^{n+1}+10(n+1)^2 = 2 [2^n + \frac{1}{2}(10n^2 + 20n + 10)] \lt 2(2^n + 10n^2)$

i.e $2^{n+1}+10(n+1)^2 \leq 12 \times 2^{n+1}$


问:
How did the approximation $\left(1+\frac{h}{R}\right)^{-2} \approx 1-\frac{2h}{R}$ come about?
In physics there is a formula calculate the variation in the acceleration due to gravity at a height, it is given by:

$$g=g_0\left(1+\frac{h}{R}\right)^{-2}$$

But it is approximated as $$g=g_0\left(1-\frac{2h}{R}\right)$$

Some how the terms in the brackets are almost equal could anyone show why ?

答:
We want to show that for small $|x|$,
$$(1+x)^{-2}\approx 1-2x$$
Equivalently,
$$(1+x)^2(1-2x)\approx 1$$
Now,
$$\begin{align}
&(1+x)^2(1-2x)\\
&=(1+2x+x^2)(1-2x)\\
&=1+2x+x^2-2x-4x^2-2x^3\\
&=1-3x^2-2x^3\\
\end{align}$$

When $x$ is sufficiently small, we can ignore the $x^2$ and $x^3$ terms.

Also,
$$\frac1{(1+x)^2}
= \frac{1-2x}{1-3x^2-2x^3}$$
Once again, for small $x$ we can ignore the higher powers of $x$, so the denominator of the RHS is very close to $1$.

If we look at the graphs of these functions, we can see that the line $y=1-2x$ is tangent to the curve $y=(1+x)^{-2}$ at $(0, 1)$.

So the slope of the curve at $(0, 1)$ is $-2$. When you learn calculus, you will learn how to find the slope of a tangent at a point on a curve. But here's a "sneak preview". Let's find the slope $m$ of the line through $(0, 1)$ and $(x, 1/(1+x)^2)$ for some small $x$.
$$m=\frac{1-1/(1+x)^2}{x}$$
$$m=\frac{-2-x}{(1+x)^2}$$
As $x$ gets close to $0$, $m$ gets close to $-2$.

FWIW, we can find the series

$$\frac1{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+\dots$$
which is valid for $|x|<1$, using polynomial long division.

$$\begin{align}
&\underline{1-2x+3x^2-\dots}\\
1+2x+x^2)&1\\
&\underline{1+2x+x^2}\\
&-2x-x^2\\
&\underline{-2x-4x^2-2x^3}\\
&3x^2+2x^3\\
&\underline{3x^2+6x^3+3x^4}\\
&-4x^3-3x^4\\
\end{align}$$
Etc.

And we can verify it by multiplication:

$$\begin{array}{r|rrrrrrr}
1 & 1 & -2x & +3x^2 & -4x^3 & +5x^4 & -6x^5 & \dots\\
2x &  & 2x & -4x^2 & +6x^3 & -8x^4 & +10x^5 & \dots\\
x^2 &  &  & x^2 & -2x^3 & +3x^4 & -4x^5 & \dots\\ \end{array}$$


问:
Why are there factorials in Taylor series?
I looked up this question online and some people mentioned that they are used to cancel out the constants from differentiating $f(x)$. I don't see how this would work when $f(x) = \sin(x)$ or $f(x) = e^x$.

Am I misunderstanding something?

答:
Let's say you want to write $f(x)$ as a power series $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$. You want to find the coefficients $a_i$.

To do this, set $x=0$, then $a_0=f(0)$.

So $f(x) = f(0)+a_1 x+a_2x^2+\cdots$

To find $a_1$ first differentiate to get $f'(x) = 0+a_1 +2 a_2 x + \cdots$. Now set $x=0$ to get $a_1 = f'(x)$.

To find $a_2$ you need to differentiate  twice and set $x=0$. And so on.

In general, in the $n$-th step you'll have $f^{(n)}(0) = n!a_n$.
The $n!$ comes from differentiate $n$ times $x^n$. From here, $a_n =f^{(n)}(0)/n!$.


问:
Formula to Find All Possible Variations
I've been trying to solve a coding problem for several days now and I'm fairly stuck and I'll blame this on my inferior knowledge of math.  I've been trying to do this by iterating over each number, finding the differences, and listing each iteration; however, this is proving to be incredibly complicated as well as non-performant.

Given a monotonic array of numbers, count the number of variations that can exist between these numbers where the distance can be 1, 2, or 3.

So if I have a list of numbers:

```<code>   1,  2,  3,  4,  7,  8,  9, 10, 11,
  14, 17, 18, 19, 20, 23, 24, 25, 28,
  31, 32, 33, 34, 35, 38, 39, 42, 45,
  46, 47, 48, 49
</code>```

If I were to list out each number that can have a variation and the distance between it and the next number I would get a list like:

```<code>// num => paths: [...distance to next number]
  1 => [ 1, 2, 3 ],
  2 => [ 1, 2 ],
  7 => [ 1, 2, 3 ],
  8 => [ 1, 2, 3 ],
  9 => [ 1, 2 ],
  17 => [ 1, 2, 3 ],
  18 => [ 1, 2 ],
  23 => [ 1, 2 ],
  31 => [ 1, 2, 3 ],
  32 => [ 1, 2, 3 ],
  33 => [ 1, 2 ],
  45 => [ 1, 2, 3 ],
  46 => [ 1, 2, 3 ],
  47 => [ 1, 2 ]
</code>```

I'm fairly certain there is a way that I can use the list above to grab the different variations that would exist.  An example of a particular variation would be:
1,4,7,10,11,14,17,19,20,23,25,28,31,34,35,38,39,42,45,48,49

A couple of other takeaways:

The order of the numbers must remain in a monotonic pattern.

I know that this particular list has a possible of number of variations of 19208.

The distance between one number and the next cannot be greater than 3

答:
It seems like the question has an extra condition that the first number must be less than or equal to $3$ and the last number must be the largest number in the set, based on the two test cases given with results $19208$ and $8$,

Here's how you find the number:

```<code>#include <iostream>

using namespace std;

int num = 0;
void traverse(int* list, int size, int i) {
    if(i >= size) {
        return;
    }

    if(i == size - 1) {
        num ++;
        return;
    }
    if(i <= size - 2 && list[i+1] - list[i] <= 3) {
        traverse(list,size,i+1);
    }
    if(i <= size - 3 && list[i+2] - list[i] <= 3) {
        traverse(list,size,i+2);
    }
    if(i <= size - 4 && list[i+3] - list[i] <= 3) {
        traverse(list,size,i+3);
    }
}

int main()
{
    int arr[]= {1,  2,  3,  4,  7,  8,  9, 10, 11,
  14, 17, 18, 19, 20, 23, 24, 25, 28,
  31, 32, 33, 34, 35, 38, 39, 42, 45,
  46, 47, 48, 49};
    int size = 31;

    for(int i = 0; i < size; i ++) {
        if(arr[i] > 3) break;
        traverse(arr,size,i);
    }
    cout << num;
    return 0;
}
</code>```

You can paste it here and test (modify the arr and size parts to change the array): https://www.onlinegdb.com/online_c++_compiler

DP-optimized way of the above algorithm (In order to deal with the "first number must be less than or equal to three" constraint you will need to add a zero at the beginning of the array and increase size by 1):

```<code>#include <iostream>

using namespace std;

int main()
{
    int arr[]= {0,  1,  2,  3,  4,  7,  8,  9, 10, 11,
  14, 17, 18, 19, 20, 23, 24, 25, 28,
  31, 32, 33, 34, 35, 38, 39, 42, 45,
  46, 47, 48, 49};
    int size = 32;

    int* nums = new int[size];
    for (int i = 0; i < size; i ++) {
        nums[i] = 0;
   }
    nums[0] = 1;
    for (int i = 0; i < size; i ++) {
        if(i >= 3 && arr[i] - arr[i - 3] <= 3) {
            nums[i] += nums[i - 3];
        }
        if(i >= 2 && arr[i] - arr[i - 2] <= 3) {
            nums[i] += nums[i - 2];
        }
        if(i >= 1 && arr[i] - arr[i - 1] <= 3) {
            nums[i] += nums[i - 1];
        }
    }

    cout << nums[size - 1];
    return 0;
}
</code>```

Title: Formula to Find All Possible Variations

Body:
It seems like the question has an extra condition that the first number must be less than or equal to $3$ and the last number must be the largest number in the set, based on the two test cases given with results $19208$ and $8$,

Here's how you find the number:

#include <iostream>

using namespace std;

int num = 0;
void traverse(int* list, int size, int i) {
    if(i >= size) {
        return;
    }

    if(i == size - 1) {
        num ++;
        return;
    }
    if(i <= size - 2 && list[i+1] - list[i] <= 3) {
        traverse(list,size,i+1);
    }
    if(i <= size - 3 && list[i+2] - list[i] <= 3) {
        traverse(list,size,i+2);
    }
    if(i <= size - 4 && list[i+3] - list[i] <= 3) {
        traverse(list,size,i+3);
    }
}

int main()
{
    int arr[]= {1,  2,  3,  4,  7,  8,  9, 10, 11,
  14, 17, 18, 19, 20, 23, 24, 25, 28,
  31, 32, 33, 34, 35, 38, 39, 42, 45,
  46, 47, 48, 49};
    int size = 31;

    for(int i = 0; i < size; i ++) {
        if(arr[i] > 3) break;
        traverse(arr,size,i);
    }
    cout << num;
    return 0;
}

You can paste it here and test (modify the arr and size parts to change the array): https://www.onlinegdb.com/online_c++_compiler

DP-optimized way of the above algorithm (In order to deal with the "first number must be less than or equal to three" constraint you will need to add a zero at the beginning of the array and increase size by 1):

#include <iostream>

using namespace std;

int main()
{
    int arr[]= {0,  1,  2,  3,  4,  7,  8,  9, 10, 11,
  14, 17, 18, 19, 20, 23, 24, 25, 28,
  31, 32, 33, 34, 35, 38, 39, 42, 45,
  46, 47, 48, 49};
    int size = 32;

    int* nums = new int[size];
    for (int i = 0; i < size; i ++) {
        nums[i] = 0;
   }
    nums[0] = 1;
    for (int i = 0; i < size; i ++) {
        if(i >= 3 && arr[i] - arr[i - 3] <= 3) {
            nums[i] += nums[i - 3];
        }
        if(i >= 2 && arr[i] - arr[i - 2] <= 3) {
            nums[i] += nums[i - 2];
        }
        if(i >= 1 && arr[i] - arr[i - 1] <= 3) {
            nums[i] += nums[i - 1];
        }
    }

    cout << nums[size - 1];
    return 0;
}


答:
Taylor identifies a smooth function to a polynomial by equating the derivatives at zero.

E.g.

$$\sin(x)=p_0+p_1x+p_2x^2+p_3x^3+\cdots$$

giving

$$\sin(0)=\left.p_0+p_1x+p_2x^2+p_3x^3+\cdots\right|_{x=0}=p_0$$
$$\sin'(0)=\left.p_1+2\,p_2x+3\,p_3x^2+\cdots\right|_{x=0}=p_1$$
$$\sin''(0)=\left.2\,p_2+3\cdot2\,p_3x+\cdots\right|_{x=0}=2\,p_2$$
$$\sin'''(0)=\left.3\cdot2\,p_3+\cdots\right|_{x=0}=3!\,p_3$$
$$\cdots$$

Title: Why are there factorials in Taylor series?

Body:
I looked up this question online and some people mentioned that they are used to cancel out the constants from differentiating $f(x)$. I don't see how this would work when $f(x) = \sin(x)$ or $f(x) = e^x$.

Am I misunderstanding something?


问:
Find a function $f$ such that $\lim_{x\to{}0}{f(x^2)}$ exists, but $ \lim_{x\to{}0}{f(x)}$ does not.
Context:

I'm brushing up on some Analysis and currently going the exercises in M. Spivak's Calculus book, specifically chapter 5 on limits. Everything was going fine until I came across this question. I've been thinking about it for some time with no luck.

Question:
"Give an example where $\lim_{x\to{}0}{f(x^2)}$ exists, but  $\lim_{x\to{}0}{f(x)}$ does not."

My attempts:

A previous question showed that $\lim_{x\to{}0}{f(x^3)}=\lim_{x\to{}0}{f(x)}$, which I believe works because we can find the third root of any real number (which was useful in the epsilon - delta proof for it). Which makes me believe the above fails because we can't square root negative reals. This lead me to play around with functions involving $\sqrt{x}$ and utilising its 'undefinedness' on the negatives.

I started with $f(x)=\sqrt{x-1}$ which clearly has an undefined limit at $0$. But this is of course no different (considering the limit at $0$ that is) to $f(x^2)$.

Any hints? I feel as though I'm overlooking something so simple.

答: $$\begin{align}f\colon \Bbb R\setminus\{0\}&\to \Bbb R\\x&\mapsto \frac x{|x|}\end{align}$$


问:
Subsequences of convergent series
We know that all subsequences $x_{\phi(n)}$ of a convergent sequences $x_n$ have the same limit :
$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} x_{\phi(n)}$$
Now we consider $x_n=\frac{n^3}{n!}$, we have that $S_n = \sum_{k=0}^n x_k = \sum_{k=0}^n\frac{k^3}{k!}$.

Suppose we already know that $\sum_{n=0}^\infty \frac{n^3}{n!}= \lim_{n \to \infty} S_n=5e$.

My question is : as the series converges (absolutely), i.e. $S_n$ has a limit when $n \to \infty$, why does the subsequence $S_{2n} = \sum_{k=0}^n\frac{(2k)^3}{(2k)!}$ seem to have a different limit that is not $5e$ ? I thought we should have
$\lim_{n \to \infty} S_n = \lim_{n \to \infty} S_{2n}$ but I also see that $S_{2n}$ is missing half of the terms from the original sum so its limit "must" be lower. So how does subsequences work with series ?
I am sure there is something very stupid that I don't get.

Title: Subsequences of convergent series

Body:
You made a mistake:
$$S_{2n} = \sum_{k=0}^{2n} \frac{k^3}{k!}$$ and not:

$$S_{2n} = \sum_{k=0}^{n} \frac{(2k)^3}{(2k)!}$$


Title: Subsequences of convergent series

Body:
We know that all subsequences $x_{\phi(n)}$ of a convergent sequences $x_n$ have the same limit :
$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} x_{\phi(n)}$$
Now we consider $x_n=\frac{n^3}{n!}$, we have that $S_n = \sum_{k=0}^n x_k = \sum_{k=0}^n\frac{k^3}{k!}$.

Suppose we already know that $\sum_{n=0}^\infty \frac{n^3}{n!}= \lim_{n \to \infty} S_n=5e$.

My question is : as the series converges (absolutely), i.e. $S_n$ has a limit when $n \to \infty$, why does the subsequence $S_{2n} = \sum_{k=0}^n\frac{(2k)^3}{(2k)!}$ seem to have a different limit that is not $5e$ ? I thought we should have
$\lim_{n \to \infty} S_n = \lim_{n \to \infty} S_{2n}$ but I also see that $S_{2n}$ is missing half of the terms from the original sum so its limit "must" be lower. So how does subsequences work with series ?
I am sure there is something very stupid that I don't get.

Body:
Because
$$S_{2n} = \sum_{k=1}^n \frac{(2k)^3}{(2k)!}$$
is false. The correct $S_{2n}$ is
$$S_{2n} = \sum_{k=1}^{2n} \frac{k^3}{k!}$$


答: You are wrong when you assert that$$S_{2n}=\sum_{k=0}^n\frac{(2k)^3}{(2k)!}.$$In fact,$$S_{2n}=\sum_{k=0}^{2n}\frac{k^3}{k!},$$which is a different thing. And there is no reason for you to assume that$$\lim_{n\to\infty}\sum_{k=0}^n\frac{(2k)^3}{(2k)!}=\lim_{n\to\infty}\sum_{k=0}^{2n}\frac{k^3}{k!}.$$

Title: Subsequences of convergent series


问:
Fixed points and iterates of an invertible function
Suppose that $g : [0,1] \rightarrow [0,1]$ is a continuous and strictly increasing function such that $g(0)=0$ and $g(1)=1$. Under these hypotheses $g(x)$ has an inverse function $g^{-1} :[0,1] \to [0,1]$ such that $g^{-1}(g(x)) =x$ and $g(g^{-1}(x)) =x$ for all $x \in[0,1]$. How do I show that the fixed point of $g(x)$ is also a fixed point of $g^{-1}(x)$. Also would $g^{-1}(x)$ have the same iterates of $g(x)$?</p>

答:
The inverse function $g^{-1}$ "undoes" the action of $g$. A fixed point of $g$ is a point to which $g$ doesn't do anything. How do you undo not-doing-anything? By not doing anything! So fixed points of $g$ must be fixed points of $g^{-1}$.

In general, if $x_0$ is sent to $x_1$ by $g$ then $g^{-1}$ sends $x_1$ back to $x_0$.

Another way to think about this: in this particular case, you can draw the graph of $g$ as a line from one corner of a square to the other that is always going upwards. Then the graph of $g^{-1}$ is what you get when you reflect this square in the diagonal line $y = x$ (basically because if $(x,y)$ is a point on the graph of $g$ then $(y,x)$ is a point on the graph of $g^{-1}$). The points that stay the same are exactly the points that are already on the diagonal, which is exactly the fixed points.


问:
If $a,b,c \in \mathbb{Z}$ and $a^2+b^2=c^2$ show that $abc \equiv 0 \pmod{60}$.

If $a,b,c \in \mathbb{Z}$ and $a^2+b^2=c^2$ show that $abc \equiv 0 \pmod{60}$.

I once read, on a number theory textbook - forget the title, in one of the problems list that all Pythagorean triplets when multiplied are divisible by 60

答:
In $\mathbb{Z}/4$ the square are $0$ and $1$, we cannot have $a=b=1$ mod $4$, in this case $c^2=2$ mod $4$ contradiction.

In $\mathbb{Z}/3$ the square are $0$ and $1$, we cannot have $a=b=1$ mod $3$, since in this case $c^2=2$ mod $3$.

In $\mathbb{Z}/5$ the square are $0,1,4$, we cannot have $a=b=1$ mod $5$ and $a=b=4$ mod $5$, since in this case $c^2=2$ mod $5$, we deduce that $a=1, b=4$ mod $5$, $b=1, a=4$ mod $5$, this implies that $c^2=0$ mod $5$ and $c=0$ mod $5$,

This implies $3,4$ and $5$ are divsors of $abc$ which implies that $abc$ is divisible by $60$.

$a=0$ mod $5$ or $b=0$ mod $5$.


问:
Suppose $log_{5}(11)=b$ Use change-of-base formula & properties of logs to rewrite each expression in terms of a and b. Show the steps for solving.
Screen shot of the exercise I'm working on (exercise 27):

I am able to arrive at an equation of the form $\frac{1}{b}=...$

I'd like to check if I'm on the right track and also how to get the expression for a whole $b$, not it's reciprocal like I currently have:

My working:
$$\log_{11}(5)=\frac{\log_5(5)}{\log_5(11)}$$

Since $\log_5(11) = b$:
$$\frac{\log_5(5)}{b}=\log_{11}(5)$$
$$\frac{1}{b}=\log_{11}(5)$$

Am I on the right track? How can I express my left hand side as a whole $b$ rather than $\frac{1}{b}$?

答: That's completely correct. The questions asks you to write your answer in terms of $a$ and $b,$ so $1/b$ is completely valid. No need to worry that your answer isn't in the form $kb$ (or similar). In fact, there's no simpler way to write $1/b$ unless I'm missing something.