问:
Fourth order finite difference
I am studying fourth order central finite difference (CFD) for space discretization of the Black Scholes PDE.
I understood that the standard fourth order  CFD for $N-1$ points is given by
$$\frac{\partial V_{i}}{\partial S}=\frac  {-V_{i+2}+8V_{i+1}-8V_{i-1}+V_{i-2}}{12h} $$
and
$$\frac{\partial ^2 V_{i}}{\partial S^2}=\frac  {-V_{i+2}+16V_{i+1}-30V_{i}+16V_{i-1}-V_{i-2}}{12h^2} $$
both for i between and including $2$ and $N-2$.

However, using the above, the approximation for the first point $V_{1}$ and last point $V_{N-1}$ is not obtained.
The author of the paper I am reading says that a "backward difference method or a unilateral method" is used to obtain  :
$$\frac{\partial V_{1}}{\partial S}=\frac  {-3V_{0}-10V_{1}+18V_{2}-6V_{3}+V_{4}}{12h}$$
and
$$\frac{\partial ^2 V_{1}}{\partial S^2}=\frac  {10V_{0}-15V_{1}-4V_{2}+14V_{3}-6V_{4}+V_{5}}{12h^2} $$
He also gave the derivative for the last point $N-1$ as
$$\frac{\partial V_{N-1}}{\partial S}=\frac  {-3V_{N}-10V_{N-1}+18V_{N-2}-6V_{N-3}+V_{N-4}}{12h}$$
and
$$\frac{\partial ^2 V_{N-1}}{\partial S^2}=\frac  {10V_{N}-15V_{N-1}-4V_{N-2}+14V_{N-3}-6V_{N-4}+V_{N-5}}{12h^2} $$

I did not understand how he used backward difference method to reach the expressions for the derivative at $V_{1}$ or $V_{N-1}$. I tried looking on the web but in vain. If someone could help I'd be grateful.

答:
I will only show the derivation for
$$\frac{\partial V_{N - 1}}{\partial S} \approx \frac{-3V_N - 10V_{N - 1} + 18V_{N - 2} - 6V_{N - 3} + V_{N - 4}}{12h}$$
since a similar procedure can be used to derive all of these formulas.

I will assume the grid is evenly spaced, since that seems to give the results he obtained. WLOG, assume the $V_{N - 1}$ point is at $x = 0$. Then the problem is to approximate $f'(0)$ using function values at $-h, 0, h, 2h, 3h$. Assume the approximation will take the form
$$D_h(f) = \frac{w_1f(-h) + w_2f(0) + w_3f(h) + w_4f(2h) + w_5f(3h)}{h}.$$
From polynomial interpolation theory, we know the approximation can be made exact for $f = 1, x, x^2, x^3, x^4$, and that this uniquely determines the weights (and that the error is $O(h^4)$). Thus we just need to solve the linear system
\begin{align}
&D_h(1) = 0 \\
&D_h(x) = 1 \\
&D_h(x^2) = 0 \\
&D_h(x^3) = 0 \\
&D_h(x^4) = 0. \\
\end{align}
Putting this in a matrix gives (notice the pattern)
$$
\begin{pmatrix}
1 & 1 & 1 & 1 & 1\\
-1 & 0 & 1 & 2 & 3 \\
1 & 0 & 1 & 4 & 9 \\
-1 & 0 & 1 & 8 & 27 \\
1 & 0 & 1 & 16 & 81 \\
\end{pmatrix}
\begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
w_5 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}.
$$
The solution is
$$
\begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
w_5 \\
\end{pmatrix}
=
\begin{pmatrix}
-\frac{1}{4} \\
-\frac{5}{6} \\
\frac{3}{2} \\
-\frac{1}{2} \\
\frac{1}{12} \\
\end{pmatrix},
$$
which matches the formula given.

Title: Fourth order finite difference


问:
Clarifying the definition of a compact subset of a topological space
Let $X$ be a topological space and $A \subset X$. Adams and Franzosa's Introduction to Topology states the definition

$A$ is said to be compact in $X$ or a compact subset of $X$ if $A$ is compact in the subspace topology inherited from $X$.

I want to make sure I understand what this means. Earlier in the text, a compact topological space is defined as follows:

A topological space $X$ is compact if every open cover of $X$ has a finite subcover.

So, how I'm interpreting the definition of $A$ being compact is as follows.

Let $U_i$, $i \in I$ (for some index set $i$) be open sets in $X$, and
$$\mathcal{O} = \left\{A \cap U_i \mid i \in I \right\}\text{.}$$
These sets are open under the subspace topology applied to $A$ inherited by $X$. For notational convenience, let $A_i = A \cap U_i$, $i \in I$. Suppose
$$A \subset \bigcup_{A_i \in \mathcal{O}}A_i\text{.}$$
Then if there exists a finite subcollection $\mathcal{P} \subset \mathcal{O}$ such that
$$A \subset \bigcup_{A_i \in \mathcal{P}}A_i$$
we say that $A$ is a compact subset of $X$.
Is this correct?

答:
Definition. A topological space $X$ is compact if for every collection of open sets (in $X$) $\mathcal U$ such that $X = \bigcup \mathcal U$ there exists a finite subset $\mathcal U'$ of $\mathcal U$ such that $X = \bigcup \mathcal U'$. A subset $A$ of $X$ is said to be compact in $X$ if $A$ is compact in the subspace topology inherited from $X$.

Now we claim that if $X$ is a topological space, a subset $A$ of $X$  is compact if and only if for every collection of open sets (in $X$) $\mathcal U$ such that $A \subseteq \bigcup \mathcal U$ there exists a finite subset $\mathcal U'$ of $\mathcal U$ such that $A \subseteq \bigcup \mathcal U'$.

Proof : First suppose that $A$ is compact, and let $\mathcal U$ be a collection of open sets (in $X$) such that $A \subseteq \bigcup \mathcal U$. Note that the collection $\{A \cap U : U \in \mathcal U\}$ is a collection of open sets (in $A$) whose union is $$\textstyle \bigcup \{A \cap U : U \in \mathcal U\} = A \cap \bigcup \mathcal U = A \quad \textrm{since } A \subseteq \bigcup \mathcal U.$$ Thus, since $A$ is compact, by definition there is a finite subset of $\{A \cap U : U \in \mathcal U\}$ whose union is also $A$, but a finite subset of $\{A \cap U : U \in \mathcal U\}$ is a set of the form $\{A \cap U : U \in \mathcal U'\}$ for some finite subset $\mathcal U'$ of $\mathcal U$. In other words, there is a finite subset $\mathcal U'$ of $\mathcal U$ such that $$\textstyle A = \bigcup \{A \cap U : U \in \mathcal U'\} = A \cap \bigcup \mathcal U',$$ which of course implies that $A \subseteq \bigcup \mathcal U'$. Can you show now the reverse implication?

I hope all of this clarify the things!

ContentLicense: CC BY-SA 4.0


问:
Initial value problem $dy/dx=y^{1/3}$, $y(0)=0$ has one of the following solution
I came across the following problem which says:

The initial value problem $y'=y^{1/3}$, $y(0)=0$ has:

(a) a unique solution,

(b) exactly two solutions,

(c)exactly three solutions,

(d)no solution.

The solution of the problem is given by: $2x=3y^{2/3}$. But I could not come to a conclusion. Clearly, (d) can not be true. But i am not sure about the other options. It will be helpful if someone throws light on it. Thanks in advance.

答:
total $3$ solutions

1.$y=0$ is also a solution.

2.y=0 for $x \le 0$ and $y=(2/3x)^{(3/2)}$ (positive squ root)

3.similarly consider neg squ root.

ContentLicense: CC BY-SA 3.0


问:
Sequence of iteration in Newton Raphson method.
To find  positive square root of $a>0$ by solving equation $x^2-a$ by Newton-Raphson mathod, if $x_n$ denotes $n$th iterate with $x_0>0, x_0\neq\sqrt{a}$, then the sequence $(x_n) $ is

$1.$ strictly increasing.

$2.$ strictly decreasing.

$3.$ constant.

$4.$ not convergent.

According to me if I start from right of  square root of $a$ we get decreasing sequence and if started from left then increasing sequence. How can be some option correct ? For me it seems all options as incorrect. Please comment. Thank you.

答:
The iteration is given by
$$x_{n + 1} = g(x_n)$$
where
$$g(x) = \frac{x}{2} + \frac{a}{2x}.$$
Compute
$$g'(x) = \frac{1}{2}\left(1 - \frac{a}{x^2}\right).$$
Let $c = \sqrt{a}$ and remember that $g(c) = c$ by construction of Newton's method. Then by Taylor's theorem,
\begin{align}
x_{n + 1} - c &= g(x_n) - g(c) \\
&= g'(\xi(c, x_n))(x_n - c)
\end{align}
where $\xi(c, x_n)$ is between $c$ and $x_n$.
If $x_0 > c$, then the sequence clearly decreases and converges to $c$. If $0 < x_0 < c$, then $x_1 > c$ so the sequence decreases from here and converges to $c$.


问:
Limit of mixed partial
Let $u(x,y)$ be continuous for $0\leq x,y$ and twice continuously differentiable on $\mathbb{R}^2_+$ with $u_x$ and $u_{xx}$ continuous on the $x$-axis. Suppose that for $x$ near $x_0>0$, $yu_y(x,y)\rightarrow 0$ as $y\rightarrow 0^+$, does  $yu_{xy}(x,y)\rightarrow 0$ as $y\rightarrow 0^+$? I can show that $\liminf_{y\rightarrow0^+}yu_{xy}(x,y)=0$ but have not been able to rule out pathological behaviour which prevents the limit from existing. I have also not been able to think of an example even with relaxed conditions for which   $yu_{xy}(x,y)\not\rightarrow 0$ as $y\rightarrow 0^+$.

答:
$$u(x,y)=y^{x}$$
$$yu_y(x,y)=xy^{x}\rightarrow0,y\rightarrow0^+$$

$$yu_{xy}(x,y)=y^{x}+xy^{x}\log(y)\not\rightarrow0,y\rightarrow0+ $$ when $$x<1$$

Title: Limit of mixed partial


问:
Prove $\vec a [\vec b\ \vec c\ \vec d] + \vec c[\vec a\ \vec b\ \vec d]=\vec b[\vec a\ \vec c\ \vec d] + \vec d[\vec a \ \vec b \ \vec c]$
Consider the non zero vectors $ \vec a, \vec b,\vec c$  and $\vec d$ such that no three of which are coplanar then prove that
$$\vec a [\vec b\ \vec c\ \vec d] + \vec c[\vec a\ \vec b\ \vec d]=\vec b[\vec a\ \vec c\ \vec d] + \vec d[\vec a \ \vec b \ \vec c]$$
Hence prove that $ \vec a, \vec b,\vec c$  and $\vec d$ represent the position vectors of the vertices of a plane quadrilateral if and only if
$$\frac{[\vec b\ \vec c\ \vec d]+[\vec a\ \vec b\ \vec d]}{[\vec a\ \vec c\ \vec d]+[\vec a \ \vec b \ \vec c]} =1$$

WHere do I begin ?

答:
Let $\vec{a} = x \vec{b}+ y \vec{c}+ z \vec{d}$.

Then 'dotting' with $\vec{c} \times \vec{d}$ on both sides we get

$[\vec{a} \ \ \vec{c} \ \  \vec{d}] = x [\vec{b} \ \ \vec{c} \ \  \vec{d}]$

Thus we get $[\vec{b} \ \ \vec{c} \ \  \vec{d}]\vec{a}=[\vec{a} \ \ \vec{c} \ \  \vec{d}]\vec{b}+ [\vec{a} \ \ \vec{d} \ \  \vec{b}]\vec{c}+[\vec{a} \ \ \vec{b} \ \  \vec{c}]\vec{d} $

which using the properties of the triple product yield upon rearranging that

$[\vec{b} \ \ \vec{c} \ \  \vec{d}]\vec{a}+ [\vec{a} \ \ \vec{b} \ \  \vec{d}]\vec{c}=[\vec{a} \ \ \vec{c} \ \  \vec{d}]\vec{b}+[\vec{a} \ \ \vec{b} \ \  \vec{c}]\vec{d} $


问:
a lemma for Lebesgue outer measure
An interval in $\mathbb{R}$ can be either open, closed, or half closed, $l(I)= b-a$ where $b, a,\ (a<b)$ are exactly the boundaries of $I$.

The following lemma is important in order to prove the countable subadditivity of the Lebesgue outer measure;

Let $(I_m)$ and $(J_n)$ be sequences of intervals such that the $I_m$'s are pairwise disjoint, prove for $k,\ p$ arbitrary in $\mathbb{N}$, that $\bigcup\limits_{m=1}^k I_m\subseteq \bigcup\limits_{n=1}^p J_n$ implies $\sum\limits_{m=1}^kl(I_m)\leq \sum\limits_{n=1}^pl(J_n)$

答:
There is a shorter proof that involves counting. If you were on a finely graded line, the length of a segment would be proportional to the number of gradations in the segment
$$l(I) \approx n(I) \cdot \Delta$$
where $\Delta$ is the distance between the points of the gradation. If we think a little about it, we have the estimates
$$(n(I)-1) \Delta \le l(I) < (n(I)+1) \Delta$$
from where we get
$$\frac{l(I)}{\Delta} -1 < n(I)\le \frac{l(I)}{\Delta}+1$$

Now, if the family of disjoint segments is contained in the union of $J_1$,$\ldots$ $J_p$ then we have

$$\sum n(I_m) \le \sum n(J_n)$$
just be counting . Applying the previous estimates for $n(I)$ we get
$$\sum (\frac{l(I_m)}{\Delta} -1) < \sum(\frac{l(J_n)}{\Delta}+1)$$
that is
$$\sum l(I_m) < \sum l(J_n) + \Delta\cdot (p-k)$$
Now take $\Delta\to 0$ and get the required inequality.

You can give a similar argument in higher dimensions, but one has to be a bit more careful with the estimates for number of lattice points.

One objection to the above argument is that it uses the fact that $\mathbb{R}$ is archimedean. In fact, it the statement is really purely combinatorial, and so, it should be valid for totally ordered abelian groups.

Title: a lemma for Lebesgue outer measure


问:
Over which extensions of $\mathbb Q_2$ is $X^2+Y^2+Z^2$ isotropic?
I was playing around a bit with quadratic forms for a different question here, and among other things had to decide when the form $X^2+Y^2+Z^2$ is isotropic over a certain field $k$, i.e. whether there is $(0,0,0) \neq (x,y,z) \in k \times k \times k$ such that $x^2+y^2+z^2=0$. This is trivially the case if $k$ contains a square root of $-1$, giving a positive answer for $k$ being any $p$-adic field $\mathbb Q_p$ with $p \equiv 1$ mod $4$, and hence all extensions of those too. Playing around a bit with Hensel's Lemma (and, funnily, using that $-1$ is not a square) led me to believe the form is isotropic over all $\mathbb Q_p$ with $p \equiv 3$ mod $4$ as well. As far as $p$-adic fields go, this left the case of extensions of $\mathbb Q_2$.

Here, I think I have an argument that the form has no nontrivial zeroes i.e. is anisotropic over $\mathbb Q_2$ itself. So next I went to its seven distinct quadratic extensions. I found the form is isotropic by coming up with individual solutions over $\mathbb Q_2(\sqrt d)$ for $d=-1, -3, -5, -2, -6, -10$: i.e. all the quadratic extensions except $\mathbb Q_2(\sqrt 2)$. <s>For this one, I believe it is still anisotropic.</s> (Edit: Thanks to the examples in the answer, I no longer believe this.)

Question 1: Are my results so far correct? (Edit: As just said, the form is isotropic over $\mathbb Q_2(\sqrt2) = \mathbb Q_2(\sqrt{-14})$ as well. Otherwise, yes.)

Question 2: Are there easier criteria to decide the question for further extensions of $\mathbb Q_2$? In particular, if the results are correct, for which quadratic extensions of $\mathbb Q_2(\sqrt 2)$ is the form isotropic? Also, is there any high-brow explanation why the field extension $\mathbb Q_2(\sqrt 2)$ is singled out here? (I mean, if the unramified $\mathbb Q_2(\sqrt{-3})$ had behaved differently from the others, that would not have surprised me, but this ...?)

I am aware and know the absolute basics of the theory of Witt groups (or Witt rings), but not enough to be able to apply them here. But I'm happy to learn if somebody can tell me how to apply that machinery here. (I know just enough to guess that the residue characteristic of $2$ here will complicate things.)

Added: There is a theorem of Witt's which I am comfortable with, which says that all quadratic forms in $\ge 5$ variables are isotropic over any $p$-adic field. For a field extension $k \vert \mathbb Q_p$ of degree $n$, it is therefore tempting to express a quadratic form in $d$ variables (here, $d=3$) over $k$ as one in $nd$ variables over $\mathbb Q_p$, however the algebra of the field extension introduces some relations between the coefficients which effectively reduce the number of variables again. Still it seems as if at least in many cases, a "big" enough extension should give us enough variables to just rely on that theorem. I don't have the mindpower to think through this right now, but again would be happy to learn if that works.

Added #2: KCd's answer does not explicitly state it, but from his reformulation of the question in terms of the Hilbert symbol / quaternion algebra $(-1,-1)$ it also follows that the form remains anisotropic over any finite extension $K\vert \mathbb Q_2$ of odd degree, say $n$. Namely, the quaternion algebra $\left(\dfrac{-1,-1}{K}\right) \simeq \left(\dfrac{-1,-1}{\mathbb Q_2}\right)\otimes K$ is a skew field iff the form is anisotropic over $K$; but if $K$ splits $D := \left(\dfrac{-1,-1}{\mathbb Q_2}\right)$, then by definition $D\otimes K \simeq M_{2\times 2}(K)$ whose simple module has $D$-dimension $\frac{n}{2}$ which therefore has to be an integer. [Generalisations of this: A) if $K\vert k$ with $[K:k]=n$ is a splitting field for a $k$-central division algebra $D$ with $\sqrt{\dim_k D}=d$, then $d$ divides $n$; B) a theorem conjectured by Witt (based on our ternary case well-known to him) and proven independently by Artin and Springer (cf. section 1.5 in B. Kahn: Formes Quadratiques Sur Un Corps; also this question and answers) saying that any anisotropic quadratic form remains anisotropic over field extensions of odd degree.]

So the case that remains open is that of extensions of even degree $[K:\mathbb Q_2] \ge 4$. Part of me wants to just mumble something about normal closures, $2$-Sylow groups / Feit-Thompson, solubility and reducing to the quadratic case, but a better part of me says "you know sh*t about these things, ask someone knowledgeable". Anyone?

答:
When $K$ is a field not of characteristic $2$, being able to solve $x^2 + y^2 + z^2 = 0$ nontrivially in $K$ is equivalent to being able to solve $x^2 + y^2 = -1$ in $K$, so I will think of the problem from that point of view.

I have checked explicitly that this can be done when $K$ is a quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$, as you say you did, and it can also be done when $K = \mathbf Q_2(\sqrt{2})$.

I'll list all seven quadratic extensions of $\mathbf Q_2$ in the order you did:
$$
\mathbf Q_2(\sqrt{-1}), \
\mathbf Q_2(\sqrt{-3}), \
\mathbf Q_2(\sqrt{-5}), \
\mathbf Q_2(\sqrt{-2}), \
\mathbf Q_2(\sqrt{-6}), \
\mathbf Q_2(\sqrt{-10}), \
\mathbf Q_2(\sqrt{2}).
$$
In the four fields $\mathbf Q_2(\sqrt{-1})$,
$\mathbf Q_2(\sqrt{-5})$,
$\mathbf Q_2(\sqrt{-2})$, and
$\mathbf Q_2(\sqrt{-10}),$
it is very easy to see a solution to $x^2 + y^2 = -1$:
$$
0^2 + \sqrt{-1}^2, \
2^2 + \sqrt{-5}^2, \
1^2 + \sqrt{-2}^2, \
3^2 + \sqrt{-10}^2.
$$

In $\mathbf Q_2(\sqrt{-3})$ we have a cube root of unity
$\zeta_3 = (-1 + \sqrt{-3})/2$, for which $1 + \zeta_3 + \zeta_3^2 = 0$, so
$$
-1 = \zeta_3 + \zeta_3^2 = \zeta_3^4 + \zeta_3^2.
$$

In $\mathbf Q_2(\sqrt{-6})$, after playing around a bit I got
$$
\left(\frac{2+\sqrt{-6}}{2}\right)^2 +
\left(\frac{2-\sqrt{-6}}{2}\right)^2 = -1.
$$

That leaves us with $\mathbf Q_2(\sqrt{2})$, whose ring of integers is $\mathbf Z_2[\sqrt{2}]$.  I am not going to record here all the numerical work I did (by hand), but just report the final result: we can write
$$
-1 = x^2 + (\sqrt{2} + 2)^2
$$
for some $x \in \mathbf Z_2[\sqrt{2}]$ where $x \equiv 1 \bmod \sqrt{2}^3$. This follows from Hensel's lemma on $\mathbf Q_2(\sqrt{2})$ for the polynomial $f(X) = X^2 + 1 + (\sqrt{2}+2)^2 = X^2 + 7 + 4\sqrt{2}$:
$$
f(1) = 8 + 4\sqrt{2} \Longrightarrow |f(1)|_2 = \frac{1}{4\sqrt{2}}
$$
and
$$
f'(1) = 2 \Longrightarrow |f'(1)|_2^2 = \frac{1}{4}.
$$
Since $|f(1)|_2 < |f'(1)|_2^2$, Hensel's lemma tells us there is a unique solution to $f(x) = 0$ where $x \in \mathbf Z_2[\sqrt{2}]$ and
$|x - 1|_2 < |f'(1)|_2 = 1/2$, or equivalently $x \equiv 1 \bmod (\sqrt{2}^3)$. A more precise form of Hensel's lemma tells us that $|x - 1|_2 = |f(1)/f'(1)|_2 = (1/4\sqrt{2})/(1/2) = 1/2\sqrt{2} = 1/\sqrt{2}^3$, so $x \not\equiv 1 \bmod \sqrt{2}^4$.

We can compute $x$ using the binomial series for the square root:
$$
x = (-7 - 4\sqrt{2})^{1/2} = (1 - 4\sqrt{2} - 8)^{1/2}
$$
so
$$
x = \sum_{n \geq 0} \binom{1/2}{n}(-4\sqrt{2}-8)^n = 1 + \sum_{n \geq 1} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n
$$
Since ${\rm ord}_2\binom{1/2}{n} = -2n + s_2(n)$, where $s_2(n)$ is the sum of the base $2$ digits of $n$, and ${\rm ord}_2((4\sqrt{2}+8)^n) = 5n/2$, the $n$th term of the series
has $2$-adic valuation $5n/2 - 2n + s_2(n) = n/2 + s_2(n)$, so its  $\pi$-adic valuation ($\pi = \sqrt{2}$ over the $2$-adics) is twice that: $n + 2s_2(n)$.  Since $n+2s_2(n) \geq 7$  for all $n \geq 1$ other than $n = 1$, $2$, and $4$,
$$
x \equiv 1  + \sum_{n = 1, 2, 4} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n \bmod \pi^7\mathbf Z_2[\pi].
$$
Working this out modulo $\pi^7$,
$$
x \equiv 1 + \pi^3 + \pi^5 + \pi^6 \bmod \pi^7\mathbf Z_2[\pi].
$$
In "normal" notation, with $\pi = \sqrt{2}$, this approximate root of $f(X)$ is $9+6\sqrt{2} \bmod 8\sqrt{2}\mathbf Z_2[\sqrt{2}]$. As a check,
$$
(9 + 6\sqrt{2})^2 + (7 + 4\sqrt{2}) = 160 + 112\sqrt{2} = 16\sqrt{2}(5\sqrt{2} + 7),
$$
which is a multiple of $\pi^7 = 8\sqrt{2}$.

In what way did your work suggest that there might not be a solution to $x^2 + y^2 = -1$ in $\mathbf Q_2(\sqrt{2})$?

There is a high-brow explanation for why we found a solution in the particular quadratic extension $\mathbf Q_2(\sqrt{2})$. Being able to solve $x^2 + y^2 = -1$  in a local field $F$ (not of characteristic $2$) is the same as saying the Hilbert symbol $(-1,-1)_F$ is $1$ rather than $-1$. Letting $F = \mathbf Q(\sqrt{2})_v$ be a completion of $\mathbf Q(\sqrt{2})$, Hilbert reciprocity says the number of $v$ such that $(-1,-1)_{\mathbf Q_2(\sqrt{2})_v}$ is $-1$ is even.  When $v$ is a place of odd residue field characteristic, $\mathbf Q(\sqrt{2})_v$ is a local field that is an extension of $\mathbf Q_p$ for an odd prime $p$, so that Hilbert symbol is $1$ since we can already solve $x^2 + y^2 = -1$ in $\mathbf Q_p$ (and actually in $\mathbf Z_p$). That leaves us with just three remaining places $v$ on $\mathbf Q(\sqrt{2})$: the two real places (corresponding to the two different embeddings of $\mathbf Q(\sqrt{2})$ into $\mathbf R$) and the unique place over $2$, corresponding to the prime $(\sqrt{2})$ in $\mathbf Z[\sqrt{2}]$.
We obviously can't solve $x^2 + y^2 = -1$ in $\mathbf R$, so
the Hilbert symbol $(-1,-1)_{\mathbf Q(\sqrt{2})_v}$ is $-1$ when $v$ is a real place. That leaves us with just one more place, over $2$, and since the Hilbert symbol is $-1$ at an even number of places it must be $1$ at the place over $2$.

By similar reasoning, if $K$ is an arbitrary real quadratic field, we can describe the non-archimedean places $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:

(1) If $v$ lies over an odd prime then there is a solution in $K_v$ since there already is one in $\mathbf Q_p$ for all odd $p$, and $K_v$ contains some $\mathbf Q_p$.

(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is just one $2$-adic place on $K$. We can solve $x^2 + y^2 = -1$ in $K_v$ by Hilbert reciprocity because we can solve it in every other completion except the two real completions, and two is even but three is not.

(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.

If $K$ is an arbitrary imaginary quadratic field, we can also determine the non-archimedean $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:

(1) If $v$ lies over an odd prime then there is a solution in $K_v$ for the same reason as in the real quadratic case above.

(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is exactly one $2$-adic place on $K$. There is also exactly one archimedean place, with completion $\mathbf C$. We can  solve $x^2 + y^2 = -1$ in $\mathbf C$, so Hilbert reciprocity implies  we can solve $x^2 + y^2 = -1$ in $K_v$.  In other words, we can solve $x^2 + y^2 = -1$ in every completion of $K$, so by Hasse-Minkowski we can solve $x^2 + y^2 = -1$ in $K$.

(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.

The second item of each list for real and imaginary quadratic fields explains why we found solutions in every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ without needing to work in the completion itself (i.e., a solution in $\mathbf Q(i)$, $\mathbf Q(\sqrt{-3})$, and so on): you wrote every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ as a $2$-adic completion of an imaginary quadratic field, and in the imaginary quadratic fields that you chose, you can check that $2$ does not split completely (sometimes $2$ ramifies, sometimes $2$ is inert, but $2$ never splits completely).  In the case of $\mathbf Q_2(\sqrt{2})$, we can write it as $\mathbf Q_2(\sqrt{-14})$, a completion of $\mathbf Q(\sqrt{-14})$ at its unique $2$-adic place.
Then results above say we have to be to solve $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{-14})$, and in fact a solution is $x = (2+3\sqrt{-14})/10$ and $y = (6-\sqrt{-14})/10$.

More generally, if $\mathbf Q_2(\sqrt{2}) = \mathbf Q_2(\sqrt{d})$ for some integer $d$ then ${\rm ord}_2(d) = 1$, so $\mathbf Q(\sqrt{d})$ ramifies at $2$ and thus there is a unique $2$-adic place on $\mathbf Q(\sqrt{d})$, so for $d < 0$ we'd be in the same situation as $\mathbf Q(\sqrt{-14})$: there will be a solution to $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{d})$.


问:
What would be the value of $a$ and $b$ in following rational expression?
If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,

How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?

答:
$$(5 + 2\sqrt{3}) = (7 + \sqrt{3})(a - \sqrt{3b})$$

                  $$=7a-7\sqrt b\sqrt 3 +a\sqrt 3-3\sqrt b$$

compare the coefficients of $\sqrt 3$ of both sides

we get$$7a-3\sqrt b=5$$ and   $$a-7\sqrt b=2$$

solve for a and b

$a=29/46$ and $\sqrt b=-9/46$


问:
Find all the real roots of $ x^2 + x^2/(x+1)^2=3 $

Find all the real roots of $x^2 + \frac{x^2}{(x+1)^2}=3$.

I got a 4th power equation after simplifying but didn't understand what to do afterwards

答:
$$x^2 + \frac{x^2}{(x+1)^2}=3\implies \frac{x^2}{x^2+2x+1}+x^2-3= 0\\ \implies x^2+(x^4+2x^3+x^2)-(3x^2+6x+3)\\
 = x^4 + 2 x^3 - x^2 - 6 x - 3=0\\
\text{ which factors as follows}\\
=(x^2 - x - 1) (x^2 + 3 x + 3) = 0$$

Both of these can be solve by the quadratic equation.

$$(x^2 - x - 1)=0\quad x =\frac{(1 \pm \sqrt{5})}{2}\approx 0.618\lor1.618=\phi$$

The other two roots are complex.


问:
Generalization of standard deviation?
Standard deviation is often taken from a one-dimensional list of numbers. However, how would one evaluate standard deviation of a set of points of $N$-dimensional coordinates?

答: It's defined as $\sqrt{|\Sigma|}$ where $|\Sigma|$ is the determinant of the variance-covariance matrix, a $N\times N$ matrix with the single variances on the main diagonal and covariances elsewhere

Title: Generalization of standard deviation?


Title: Find all the real roots of $ x^2 + x^2/(x+1)^2=3 $

Body:
Find all the real roots of $x^2 + \frac{x^2}{(x+1)^2}=3$.

I got a 4th power equation after simplifying but didn't understand what to do afterwards

Body:
Factorize the equation as follows

\begin{align}
0= & x^2 + \frac{x^2}{(x+1)^2}-3\\=& \frac1{(1+x)^2}\left(x^2(x+1)^2+x^2-3(x+1)^2 \right) \\
= &\frac1{(1+x)^2}\left(x^4 +2(x+1)x^2-3(x+1)^2 \right) \\
= &\frac1{(1+x)^2}(x^2-(x+1))(x^2+3(x+1)) \\
\end{align}
which, from the factor $x^2-x-1$, yields the real solution
$$x=\frac12\pm \frac{\sqrt5}2$$


问:
Deriving Generating function for centered trinomial coefficients
Let $c_n$ denote the $n$-th center trinomial coefficient (OEIS sequence here).

It appears they cannot be generated by a linear recurrence relation, so how should I go about to find the generating function $G(x)$ for the sequence?

$$G(x)=\sum_{n=0}^{∞}c_nx^n=1+x+3x^2+7x^3+19x^4+51x^5+141x^6+...$$

The geometric ratio appears to have a limit close to $$\lim_{n\to ∞}\frac{c_{n+1}}{c_n}=2.95...$$ (these are the successive ratios of the last two listed terms in the OEIS sequence).

Also, what is the interval of convergence (and divergence)? Based on the geometric limit, it seems that $G(1/3)$ will converge.

Edit: The generating function is $$G(x)=\frac{1}{\sqrt{1-2x-3x^2}}$$
Any idea as to how this answer is derived?

答:
$c_n$ is the coefficient of $x^n$ in $(1 + x + x^2)^n$. It follows that its generating function is the diagonal of the rational generating function

$$F(x, y) = \frac{1}{1 - y(1 + x + x^2)} = \sum_{n \ge 0} y^n (1 + x + x^2)^n = \sum f_{n, m} x^n y^m$$

in the sense that $c_n = f_{n, n}$. It's a general fact (which you can find stated, for example, as Theorem 6.3.3 in Stanley's Enumerative Combinatorics, Vol. II) that the diagonal of a bivariate rational generating function is algebraic and can be computed using contour integration, as explained in Stanley, and you can also see my blog post Extracting the diagonal. We can do the computation as follows. Write $C(r) = \sum c_n r^n$. Then for sufficiently small $r$ we have

$$\frac{1}{2 \pi i} \int_{\gamma} \frac{F(rz, rz^{-1})}{z} \, dz = C(r^2)$$

where $\gamma$ is the contour given by the unit circle. In our case the integrand is

$$\frac{F(rz, rz^{-1})}{z} = \frac{1}{z - r - r^2 z - r^3 z^2}$$

which, as a meromorphic function of $z$, has poles given by the zeroes of the denominator. These are the zeroes of a quadratic $r^3 z^2 + (r^2 - 1) z + r$, which are then

$$z_0, z_1 = \frac{(1 - r^2) \pm \sqrt{1 - 2r^2 - 3r^4}}{2r^3}$$

by the quadratic formula. We only need to consider the residue at a pole inside our contour for small $r$, and as $r \to 0$ the $+$ zero goes to infinity so we only need to consider the $-$ zero

$$z_0 = \frac{(1 - r^2) - \sqrt{1 - 2r^2 - 3r^4}}{2r^3}.$$

The residue at this pole is

$$\lim_{z \to z_0} \frac{z - z_0}{-r^3(z - z_0)(z - z_1)} = \frac{1}{-r^3(z_0 - z_1)} = \frac{1}{\sqrt{1 - 2r^2 - 3r^4}}$$

so the residue theorem gives

$$C(r^2) = \frac{1}{\sqrt{1 - 2r^2 - 3r^4}}$$

as desired.

Now some more general facts can be used to deduce asymptotics. The dominant singularity of $C(z) = \frac{1}{\sqrt{1 - 2z - 3z^2}} = \frac{1}{\sqrt{(1 - 3z)(1 + z)}}$ occurs at $z = \frac{1}{3}$. Around this singularity $C(z)$ looks like $\frac{1}{\sqrt{\frac{4}{3}(1 - 3z)}}$ which gives (using e.g. binomial expansion together with Stirling's formula) that the leading order asymptotic of $c_n$ is

$$\boxed{ c_n \sim \sqrt{\frac{3}{4 \pi n}} \, 3^n }.$$

This is in agreement with the comment left by Vaclav Kotesovec on the OEIS page, and in particular implies that the true value of $\lim_{n \to \infty} \frac{c_{n+1}}{c_n}$ is $3$ exactly. For much more on this topic see Chapter VI.1 of Flajolet and Sedgewick's Analytic Combinatorics.


答:
Since these are 3-vectors, any four of them must be linearly-dependent. This linear dependence is preserved by the mapping $(v_x,v_y,v_z)\mapsto (v_x,v_y,v_z,v_x)$, yielding four linearly-dependent 4-vectors. In terms of determinants, this means

$$\begin{vmatrix} a_x & b_x & c_x & d_x \\ a_y & b_y & c_y & d_y \\  a_z & b_z & c_z & d_z \\ a_x & b_x & c_x & d_x\end{vmatrix}=0.$$
Expanding along the last row and employing the [] notation for scalar triple product, we obtain

$$a_x\!\left[\vec{b}\;\vec{c}\;\vec{d}\right]-b_x\!\left[\vec{a}\;\vec{c}\;\vec{d}\right]+c_x\!\left[\vec{a}\;\vec{b}\;\vec{d}\right]-d_x\!\left[\vec{a}\;\vec{b}\;\vec{c}\right]=0$$
which rearranges to
$$a_x\!\left[\vec{b}\;\vec{c}\;\vec{d}\right]+c_x\!\left[\vec{a}\;\vec{b}\;\vec{d}\right]=b_x\!\left[\vec{a}\;\vec{c}\;\vec{d}\right]+d_x\!\left[\vec{a}\;\vec{b}\;\vec{c}\right].$$ Analogous arguments apply to the $y$-,$z$-components, which yields the desired vector equation. (This approach generalizes immediately to $n+1$ vectors in $\mathbb{R}^n$, with $n$-by-$n$ determinants replacing scalar triple products.)

Title: Prove $\vec a [\vec b\ \vec c\ \vec d] + \vec c[\vec a\ \vec b\ \vec d]=\vec b[\vec a\ \vec c\ \vec d] + \vec d[\vec a \ \vec b \ \vec c]$

Body:
Consider the non zero vectors $ \vec a, \vec b,\vec c$  and $\vec d$ such that no three of which are coplanar then prove that
$$\vec a [\vec b\ \vec c\ \vec d] + \vec c[\vec a\ \vec b\ \vec d]=\vec b[\vec a\ \vec c\ \vec d] + \vec d[\vec a \ \vec b \ \vec c]$$
Hence prove that $ \vec a, \vec b,\vec c$  and $\vec d$ represent the position vectors of the vertices of a plane quadrilateral if and only if
$$\frac{[\vec b\ \vec c\ \vec d]+[\vec a\ \vec b\ \vec d]}{[\vec a\ \vec c\ \vec d]+[\vec a \ \vec b \ \vec c]} =1$$

WHere do I begin ?


答:
$x^2+\frac{x^2}{(x+1)^2}=3$ (x≠1)

$\Rightarrow x^2+\frac{x^2}{(x+1)^2}-3=0$

Which simplifies as
$\frac{x^4+2x^3-x^2-6x-3}{(x+1)^2}=0$

$\Rightarrow x^4+2x^3-x^2-6x-3=0$

Factorising, we get

$(x^2-x-1)(x^2+3x+3)=0$

For which we get two equations to solve

$x^2-x-1=0$

$x^2+3x+3=0$

Solving the first eqn. we get $x=\frac{1±\sqrt{5}}{2}$
The second equation has no (real) solution.

$\therefore x=\frac{1±\sqrt{5}}{2}$


问:
Twisted Haar measures on locally compact G-spaces?
Suppose $G$ is a locally compact group, $X$ is a locally compact space equipped with a nice right action $r_g$ of $G$ (I suspect every orbit being compact qualifies as "nice" here, but this probably isn't the greatest generality possible), and $(V,\rho)$ is a finite-dimensional left representation of $G$. Is there any characterization, in terms of some kind of data on the orbit space, of locally-defined $V$-valued measures on $X$ (in the sense of a system of measures on every compact subset of $X$ that behave well under restriction) which respect $G$'s action in the sense that $\left(\left(r_g\right)_*\mu\right)(E)=\rho_g\mu(E)$ for every precompact Borel set $E$? What if we let $V$ be a Banach space or similar? In the case where $X=G$ and $V$ is trivial, these are just Haar measures, but I've been unable to find any discussion of this sort of problem in greater generality.

(Here, all spaces are Hausdorff and all measures are Radon.)

答:
The answer seems a bit silly, which explains why I couldn't find any discussion. Some of what I say may not be strictly correct, but I think it is qualitatively correct. Suppose every orbit is closed in $X$, and suppose $X/G$ is also LCH.

One can construct invariant measures as follows. Take a measure $\mu$ on $X/G$, and take measures $\mu_O$ on every orbit $O$ of $X$ such that for every $f\in C_c(X)$, the function $O\mapsto\int_Of\,d\mu_O$ is in $L^1(\mu)$. Then the Radon measure $\nu$ defined by $\int f\,d\nu=\int_{X/G}\int_Of\,d\mu_O\,d\mu$ is invariant if and only if $\mu$-a.e. $\mu_O$ is invariant.

I think the converse, that every positive invariant measure arises this way, can be argued (probably assuming second-countability) by disintegration and extension theorems, though I'm not up to working through the details at this particular moment. This reduces the untwisted case to studying the homogeneous case, which can be classified in terms of modular functions; see Loomis' book, pages 132-133, or Bourbaki's section on homogeneous spaces.

The twisted case, I think, is a straightforward adaption of these ideas with a little more finagling involving stabilizers; I'll think about this in the morning. You may need to assume the representation is unitary.

If $X/G$ is not Hausdorff, we might have no good invariant measure. For example, if $K$ is a local field, the multiplicative action of $K^*$ on $K$ has a two-point non-Hausdorff quotient space. A standard fact of number theory in this case is if the representation $\rho$ is sufficiently bad, the twisted-invariant measure is not finite near zero, so you cannot integrate normally. One can still define a kind of twisted-invariant integral (this is the local theory in Tate's thesis), but in most modern treatments it lives in a much sketchier space of distributions.


问:
Multivariable limits using $\epsilon-\delta$ definition.
We are supposed to prove that  $\lim_{x,y \to 2,1}(3x+4y)=10$

Therefore, we need to show that $|3x+4y-10|=|3(x-2)+4(y-1)|<\epsilon$, such that for every $\epsilon$, there is some $\delta$,such that $\sqrt{(x-2)^2+(y-1)^2}<\delta.$

Now, my textbook states: "if we take $|x-2|<\delta$ and $|y-1|<\delta$, we have

$|3x+4y-10|=|3(x-2)+4(y-1)|<3|x-2|+4|y-1|<3\delta +4\delta=7\delta<\epsilon$.

So we can chose $\delta <\epsilon/7$, and the proof is finished."

Question:

My only concern is with the statement: if we take $|x-2|<\delta$ and $|y-1|<\delta$. This supposedly accounts for the fact that $\sqrt{(x-2)^2+(y-1)^2}<\delta.$ (i.e, the point $(x,y)$ is inside the circle centered at $(2,1)$ of radius $\delta$ ).

However, $|x-2|<\delta$ and $|y-1|<\delta$ only tells us that the point $(x,y)$ is within the interior of the square circumscribing the required circle. It does not necessarily imply that the point $(x,y)$ is within the circle.

How then, can we justify the fact that $\sqrt{(x-2)^2+(y-1)^2}<\delta.$?

答:
You are confused about the definition of limit. We want $\delta$ such that $\sqrt {(x-2)^{2}+(y-1)^{2}} <\delta$ implies $|3x+4y-10|  <\epsilon$. Choose $\delta$ the way the book suggests. Now start with $(x,y)$ such that $\sqrt {(x-2)^{2}+(y-1)^{2}} <\delta$. Then $|x-2| <\delta$ and $|y-1| <\delta$. Hence, $|3x+4y-10|  <\epsilon$.

[I Have used the fact that $|x-2| \leq  \sqrt {(x-2)^{2}+(y-1)^{2}}$ and  $|y-1| \leq  \sqrt {(x-2)^{2}+(y-1)^{2}}$].


Title: What would be the value of $a$ and $b$ in following rational expression?

Body:
If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,

How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?

Body:
$$\frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - \sqrt{3b}$$
$$5+2\sqrt{3}=(a-\sqrt{3b})(7+\sqrt{3})$$
$(a-\sqrt{3b})(7+\sqrt{3})=7a+a\sqrt{3}-7\sqrt{3b}-3\sqrt{b}$

Now we have two equotation for two unknown $a$ and $b$:

$5 = 7a-3\sqrt{b}$
and
$2\sqrt{3}=(a-\sqrt{b})\sqrt{3} \implies 2=a-\sqrt{b}$.

Now

$\sqrt{b}=\frac{7a-5}{3}=a-2$

$7a-5=3a-6 \implies a = -\frac{1}{4}$

$\sqrt{b} = a - 2 = -\frac{1}{4} - \frac{8}{4}=-\frac{9}{4} \implies b = \frac{81}{16}$

So,
$$a = -\frac{1}{4}, b = \frac{81}{16}$$


问:
A game on a graph
Alice and Bob play a game on a complete graph ${G}$ with $2014$ vertices. They take moves in turn with Alice beginning. At each move Alice directs one undirected edge of $G$. At each move Bob chooses a positive integer number $m,\: 1\leq   m \leq 1000$ and after that directs $m$ undirected edges of $G$. The game ends when all edges are directed. If there is some directed cycle in $G$ Alice wins. Determine whether Alice has a winning strategy.

This problem is from the Turkey JBMO TST 2014. Could someone help? I have not gone anywhere with it. Thanks a lot.

答:
Alice has the winning strategy. We first prove that on each of her turns, if the length of the longest directed chain isn't already $2014$, Alice can always extend it by one. We then use this to give an explicit winning strategy.

To prove our lemma, suppose otherwise. Let $(v_1, v_2, \ldots, v_n)$ be the longest directed chain as of yet, and let $S\neq\varnothing$ be the set of vertices not on the chain. Every single point of $S$ must point to $v_n$ – otherwise, the chain could be trivially extended by Alice. Let $i<n$ be the least integer such that not every single point of $S$ points to $v_i$. Let $a\in S$ such that $a$ does not point to $v_i$. Alice can then connect $v_i$ to $a$, thus creating the new longest chain $(v_1, \ldots, v_i, a, v_{i+1}, \ldots, v_n)$, as we wanted to prove.

Alice thus has the following winning strategy. She builds an ever-increasing chain by the method above. By turn $n=2002$, Alice will have created a chain of length at least $2002$. Since $\binom{2002}{2}>1000\cdot2002$, after Bob's turn, two vertices of the chain will remain unconnected. Alice can then form a cycle with them. $\blacksquare$

Title: A game on a graph


问:
The number of distinct piles of ranked ballots in Ranked-Choice Voting.
This is about Ranked-Choice Voting (RCV) where the ballot has $C$ candidates, and there are $C$ levels of ranking of preference.  Equal ranking of candidates is not allowed and no voter is required to rank any candidates that they don't want to, but any candidate not ranked is tied for lowest preference with all other unranked candidates on the ballot.

We want to know how many operationally distinct ways a ballot can be marked so that all like-marked ballots can go onto a single pile and handled as a block in the Single Transferable Vote operation that occurs in rounds in an RCV election.  Besides the pile of blank ballots in which no candidates were ranked, how do we prove the number of piles is:

$$ \sum\limits_{n=1}^{C-1} \frac{C!}{n!} $$

Then using nothing more advanced than what we get from calculus, what would be a concise proof that:

$$ \sum\limits_{n=1}^{C-1} \frac{C!}{n!} = \big\lfloor (e-1)C! \big\rfloor - 1 $$

for all $C \in \mathbb{Z} \ge 2$ and $e$ is the base of the natural logarithm?

(I'm hoping for something more concise than what I got.  If no one posts, I'll post my not-so-concise proof.)

答:
Okay, I am on board with Dr. Scott's derivation.  It depends on knowing that $e<3$, which is something my calculator says, dunno other than adding a few terms of the Maclauren series and showing that the remaining terms are small enough, I dunno a pure mathematical proof that $e<3$, but we know that it is.

We both have (of course):

$$\begin{align}
(e-1)C! &= \sum\limits_{n=1}^{\infty} \frac{C!}{n!} \\
 &= \sum\limits_{n=1}^{C-1} \frac{C!}{n!} + \frac{C!}{C!} + \sum\limits_{n=C+1}^{\infty} \frac{C!}{n!}\\
 &= \sum\limits_{n=1}^{C-1} \frac{C!}{n!} + \ 1 \ \ \ + \sum\limits_{n=1}^{\infty} \frac{C!}{(C+n)!} \\
\end{align}$$

We know that any integer added to the argument of the floor() function can be taken out and added to what remains:

$$ \big\lfloor n + a \big\rfloor = n + \big\lfloor a \big\rfloor \qquad \forall n\in\mathbb{Z}, a\in\mathbb{R}$$

That means that

$$\begin{align}
 \big\lfloor (e-1)C! \big\rfloor &= \left\lfloor \sum\limits_{n=1}^{C-1} \frac{C!}{n!} + 1 + \sum\limits_{n=1}^{\infty} \frac{C!}{(C+n)!} \right\rfloor \\
&= \sum\limits_{n=1}^{C-1} \frac{C!}{n!} + 1 + \left\lfloor \sum\limits_{n=1}^{\infty} \frac{C!}{(C+n)!} \right\rfloor \\
\end{align}$$

because the first two terms are clearly integers.

The way I dealt with that inequality

$$\sum_{n=1}^{\infty} \frac{C!}{(C+n)!} < 1$$

is

$$\begin{align}
\sum_{n=1}^{\infty} \frac{C!}{(C+n)!} &= \sum_{n=1}^{\infty} \ \prod_{k=1}^{n} \frac{1}{C+k} \\
 &< \sum_{n=1}^{\infty} \ \prod_{k=1}^{n} \frac{1}{C} \\
 &= \sum_{n=1}^{\infty} \ \frac{1}{C^n} \\
 &= \sum_{n=0}^{\infty} \left( \tfrac1C \right)^n - 1 \\
 &=  \frac{1}{1-\tfrac1C}  - 1 \\
 &=  \frac{C}{C-1}  - 1 \\
 &=  \frac{1}{C-1} \qquad \le \ 1 \qquad \forall C \ge 2 \\
\end{align}$$

then that proves

$$ \sum\limits_{n=1}^{C-1} \frac{C!}{n!} = \big\lfloor (e-1)C! \big\rfloor - 1 $$

Number of operationally distinct piles grows proportional to $C!$ which is why we say that Hare RCV is not "precinct summable".  Compare to First-Past-The-Post (FPTP) where the number of subtotals each voting precinct needs to report is $C$.  Essentially most Condorcet methods would require $C(C-1)$ subtotals to report.  Far less than $\big\lfloor (e-1)C! \big\rfloor - 1$ when $C>3$.  I think it's valid to say that both Condorcet and FPTP are precinct summable and Hare RCV is not.


问:
Check whether a given graph is bipartite except for one edge
Given any graph $G=(V,E)$, how to decide if deleting just one of its edges can make it bipartite?

Trivial solution: check for each edge whether deleting it makes the graph bipartite. This is in $O(|E|\cdot(|E|+|V|))$.

Are there more efficient ways?

答:
The edge to remove must belong to all cycles of odd length. If there are disjoint such cycles, then there are no edge satisfying your assertion.

Notice also that several edge choices may be possible. For instance, if $G$ is a ring with odd number of nodes, then removing any edge makes it bipartite.

Let us say that a graph is almost bipartite if it is not bipartite but removing just one of its edges may make it bipartite, and let us assume that $G$ is not bipartite.

My approach would be to first find a cycle of odd length in $G$, which is in $O(|V|+|E|)$. I would remove its edges from $G$ and check in $O(|V|+|E|)$ if the remaining graph is bipartite. If not, then there are disjoint cycles of odd length, and thus $G$ is not almost bipartite.

If the remaining graph is bipartite, then I would compute in $O(|V|+|E|)$ its two parts $\top$ and $\bot$. Then, the vertices of the cycle are in either $\top$, $\bot$, or $V' = V\setminus(\top\cup\bot)$.

EDIT: if the remaining graph is not connected, then there may be several possible bipartitions into $\top$ and $\bot$ above, and the number of bipartitions may be of the order of magnitude of $|V|$ as illustrated by Dániel G. in the comments. Therefore, another approach is needed.

If $V'$ is empty, then we are done: the edge to remove is the one between two $\top$ or two $\bot$ nodes, if its is unique; otherwise, $G$ is not almost bipartite.

If $V'$ is not empty, then either there is only one sequence of consecutive $V'$ nodes in the cycle of odd length, and any of its edges may be removed to make $G$ bipartite; or there are several such sequences and $G$ is not almost bipartite.

Processing the cycle in this way is in $O(|V|)$, which makes the whole process in $O(|V|+|E|)$, if I am correct.


问:
What causes bifurcation?
Suppose let's say we have a recursive function,
$$x_{n+1}=rx_{n}(1-x_{n})$$

From what I understand, the $x_{n+1}$ VS r graph starts to split (bifurcate) after a particular value of r (apparently it's around 3). Does that mean after they split up, we will get two values of the function for the same value of r? Am I getting it right?

How's it possible? (I do understand that it loses its ordered nature and falls into chaos) Could someone illustrate it with an example? Help would be appreciated!  (I'm a high school student, so explaining it in simpler terms would be even more appreciated). Thank you!

答:
That's not quite what this graph represents.
For any given values of $r$ and $x_n$, the formula defines $x_{n+1}$ uniquely .
In other words, given $r$ and an initial value $x_0$, the sequence $(x_n)_{n\in\Bbb N}$ is well and uniquely defined for all its values.

Now, given a sequence, there are things we may ask ourselves about it.

Does it converge, and if so, to what value? How fast?

If it does not converge, does it have accumulation points? Which? How many?

The graph plots on the horizontal axis the parameter $r$, and on the vertical axis the values that are asymptotically visited by iterates of almost all initial conditions $x_0$ (for that parameter $r$).

We see that up until $r=3$, we get a straight limit (which is why we see only one value in the graph), but thereafter behavior changes.
Just after $r=3$, we see that the iterates actually oscillate between two values (which is where we get the bifurcation from).
Up ahead, the iterates then oscillate between $4$ values.
This doubling continues, coming in faster and faster, until behavior changes again.

We get values of $r$ for which the behavior is chaotic, meaning slight changes in the initial conditions $x_0$ yield sequences whose iterates behave in a vastly different ways long term.
Even then, among the chaos there are short 'bursts' of order; in fact, for any value of $P$ there is some $r$ such that the iterates oscillate between $P$ values.

Title: What causes bifurcation?


问:
Why is it true that $\int_0^\infty \sin(qx) dx = 1/q$.
In a Quantum Mechanics class the professor pointed out a calculation from Zetilli's textbook on Quantum Mechanics (page 630).

There he says that:

$$\int_0^\infty \sin(qx) dx = \lim_{\lambda \to 0} \int_0^\infty e^{-\lambda x} \sin(qx) dx$$

Then he writes the $\sin$ as complex exponentials and reaches the result that the integral is $1/q$. WHAT? I was taught at my calculus course that this integral does not converge. For me this result is plain wrong, and I do not understand it at all. I even don't see why the limit can be imposed like that. Any thoughts on this "trick"?

答:
Since
$$
\newcommand{\Im}{\operatorname{Im}}
\int_0^\infty\sin(qx)\,\mathrm{d}x=
\Im\left(\int_0^\infty e^{iqx}\,\mathrm{d}x\right)\tag1
$$
in the spirit of analytic continuation, we can look at
$$
f(q)=\int_0^\infty e^{iqx}\,\mathrm{d}x\tag2
$$
which is well-defined for $\Im(q)\gt0$, and see how $f$ behaves as $\Im(q)\to0$.

When $\Im(q)\gt0$, we get, via contour integration,
$$
\begin{align}
f(q)
&=\frac iq\int_0^{-iq\infty} e^{-x}\,\mathrm{d}x\tag{3a}\\
&=\frac iq\int_0^\infty e^{-x}\,\mathrm{d}x\tag{3b}\\
&=\frac iq\tag{3c}
\end{align}
$$
Explanation:

$\text{(3a)}$: substitute $x\mapsto\frac{ix}q$

$\text{(3b)}$: integration over $\gamma=[0,R]\cup[R,-iqR]\cup[-iqR,0]$ is $0$

$\text{(3c)}$: integrate

Then, by analytic continuation, as $\Im(q)\to0$, $\frac iq$ behaves nicely. Therefore, we say for real $q$:
$$
\int_0^\infty(\cos(qx)+i\sin(qx))\,\mathrm{d}x=\frac iq\tag4
$$
which says not only that $\int_0^\infty\sin(qx)\,\mathrm{d}x=\frac1q$, but also that $\int_0^\infty\cos(qx)\,\mathrm{d}x=0$ (even though neither of these integrals converges in the classical sense).

Let's verify the integral of $\cos(qx)$:
$$
\begin{align}
\int_0^\infty\cos(qx)\,\mathrm{d}x
&=\int_0^\infty\sin(\pi/2+qx)\,\mathrm{d}x\tag{5a}\\[6pt]
&=\int_{\frac\pi{2q}}^\infty\sin(qx)\,\mathrm{d}x\tag{5b}\\
&=\color{#C00}{\int_0^\infty\sin(qx)\,\mathrm{d}x}\color{#090}{-\int_0^{\frac\pi{2q}}\sin(qx)\,\mathrm{d}x}\tag{5c}\\[3pt]
&=\color{#C00}{\frac1q}\color{#090}{+\left.\frac1q\cos(qx)\,\right|_0^{\frac\pi{2q}}}\tag{5d}\\[9pt]
&=0\tag{5e}
\end{align}
$$
Explanation:

$\text{(5a)}$: $\cos(x)=\sin(\pi/2+x)$

$\text{(5b)}$: substitute $x\mapsto x-\frac\pi{2q}$

$\text{(5c)}$: difference of integrals

$\text{(5d)}$: integrate

$\text{(5e)}$: evaluate


问:
Does a sample of a random variable is a sample of a random variable minus a constant?
my question is the following: let $X$ be a real random variable, $b\in\mathbb{R}$ and $X_{1},\cdots , X_{n}$ a sample of $X$ independent and identically distributed. My question is whether or not $|X_{1} - b|,\cdots , |X_{n} - b|$ is a sample of the real random variable $|X - b|$.

Thanks in advance.

答:
Yes. Consider this example:

Let $X \sim \mathsf{Unif}(1,4)$ and $b = 2.$
Then let $Y = |X-2|.$

Simulation in R:

```<code>set.seed(2020)
x = runif(10^5, 1, 4)
y = abs(x - 2)
summary(y)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
0.0000401 0.3695657 0.7426899 0.8283916 1.2438975 1.9999862

hist(y, prob=T, col="skyblue2")
</code>```

What do you suppose is the density function of the
random variable $Y?$

Body:
Yes. Consider this example:

Let $X \sim \mathsf{Unif}(1,4)$ and $b = 2.$
Then let $Y = |X-2|.$

Simulation in R:

set.seed(2020)
x = runif(10^5, 1, 4)
y = abs(x - 2)
summary(y)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
0.0000401 0.3695657 0.7426899 0.8283916 1.2438975 1.9999862

hist(y, prob=T, col="skyblue2")

What do you suppose is the density function of the
random variable $Y?$


答:
To start with, it's worth being really clear on what the equation
$$x_{n+1}=rx_n(1-x_n)$$
actually defines. If you have some fixed value of $r$ and $x_0$, this expression gives some infinite sequence of further values $x_1,\,x_2,\,x_3,\ldots$ via a recurrence. The tricky thing about a bifurcation diagram is that it tries to show how the behavior of this infinite sequence varies with $r$ - which is a lot of information to pack into a single plot. Let's first look at some simpler ways to visualize the behavior of this recurrence. For simplicity, I'll just start at $x_0=0.5$ throughout this post, although this is another parameter you could vary (it turns out that almost all values of $x_0$ lead to similar behavior in the long run - although this is not that easy to prove).

Let me first present several pictures that display this equation from a more familiar angle, where we just start at $x_0=0.5$ and plot the sequence $x_0,x_1,x_2,x_3,\ldots$ for some fixed values of $r$.

At $r=2.9$, before the bifurcation, we see that successive terms in the sequence rather quickly approach some single value (a fixed point of the recurrence).

The long term behavior of this sequence can be described as converging to a single value.

At $r=3$, the terms get closer together, but do so way more slowly - from the first $50$, it is not even apparent that they converge:

If we look at the next $5000$ terms, we can see they continue to get closer together:

And, if we do a bit of mathematics, we can find out that the sequence eventually converges to $2/3$. The sketch of the argument is to observe that if $x_n < 2/3$ some algebra can prove that $x_n < x_{n+2} < 2/3$ and a bit of analytical work can show actual convergence of $x_n$ - and we can even estimate the rate of convergence if we like.

If we keep increasing $r$, strange things start happening. At $r=3.05$, for instance, the terms don't get closer together:

Rather quickly, the terms settle into a pattern of alternation between a high value and a low value - and then don't move any more. The long term behavior here would be described as a two-cycle, where two values keep alternating - and you could find explicitly where that occurs by using the recurrence to solve the quadratic equation $x_{n+2}=x_n$.

Continue increasing more, say to $r=3.5$, and you get a pattern of four values:

And if you sort of start working in smaller and smaller increments of $r$, you find a pattern of eight values, then sixteen and so on - followed by, eventually, cycles of every length (appearing in the same order - and not by coincidence - as described in Sharkovskii's theorem). The last kind of cycle to appear (and of the major events one sees on a bifurcation diagram) is a three-cycle, which appears near $r=3.83$ and gives behavior like:

Where we see three values. There are also values of $r$ that display more complicated behavior that never becomes cyclic - although this poses certain problems in trying to get a computer to accurately plot it. One can get a taste of this sort of thing at $r=3.675$, which gives the first 500 terms as:

There's certainly some patterns - we can see a few shapes repeating irregularly in the middle - but it's not clear that a cycle is emerging (or will emerge). In fact, somewhere near this $r$ value there are some points that are easy-ish to prove do not tend to periodic behavior ever for most starting points $x_0$. This fact is also easy to show at $r=4$, but I promised to start my plots at $x_0=0.5$, and that starting point is not interesting at all at $r=4$.

Now, let me get to your question: a bifurcation diagram tries to display the information contained in these plots over a lot of $r$ all at once. Basically, it reduces each plot to a single vertical line, and plots on that line the values that appear in the long term behavior*. So, for $r\leq 3$, only a single point belongs on the bifurcation diagram, meaning that most points** tend there in the long term. In the next region, we see two points, meaning most points end up alternating between those two points in the long run. So, each vertical slice tells us what the sequence does after a lot of time for a fixed value of $r$ - and, as you can see, this is quite complex. A crucial point here is that the bifurcation diagram isn't trying to show a function - each vertical slice is essentially a distribution*** saying things like "the sequence will spend half of its time on the upper branch and half on the lower branch" or other such statements about behavior more complicated than identifying a single value of interest.

As to why it does that, it helps to define $f_r(x)=rx(1-x)$, which is a function defined to run one step of the iteration. There's a lot to say, but the basic idea is that if you plot this function for $r=2$, you get something like this, where I also plot the line $y=x$:

For this value of $r$, the iteration converges very quickly to $1/2$ - and we can see that, if we started at exactly $1/2$, we would end up exactly back there - which is called a fixed point. More strongly, however, this fixed point is attractive meaning that if we start near 1/2, we end up closer after one iteration - one can see this by the fact that the value of the function changes very slowly near the fixed point - it's just a big flat top, all of it mapping closer and closer to $1/2$ than we started. This leads to very fast convergence of the sequence.

The plot at $r=3$ looks different in substantial ways:

Note that there still is a fixed point - at $2/3$ now - but the behavior of things near $2/3$ is different. The slope of our function $f_3$ near that fixed point is $-1$ - meaning that if we start at $2/3+e$, the next iterate is around $2/3-e$, meaning we didn't get meaningfully closer - although if we look at a quadratic approximation, we can see that we are getting closer, just way more slowly.

At $r=3.1$, the slope near the fixed point is now greater than $1$ in absolute value - meaning if we start to near the fixed point, we actually end up further away after one iteration - this is called a repulsive fixed point.

This is why the behavior changes - it can't be that most points end up at a repulsive fixed point****, so something else must happen. On the plot, we see two branches emerge, so let's consider that.

If we increase $r$ a little more to $3.3$ and plot $f_3(f_3(x))$ as a third line, we get to see a bit more about why now a two-cycle prevails:

While the orange curve ($f_3(x)$) continues to get steeper and steeper near the fixed point - preventing the simple convergent behavior, the green curve ($f_3(f_3(x))$) crosses the $y=x$ line in two other places - and is fairly flat near those locations, meaning that, if we start near one of those places and advance two iterations, we get closer to where those lines cross the $y=x$ line - leading to an attractive two-cycle between the values of those intersections. As we increase $r$ further, what happens is that these new intersections get steeper and steeper until they become repulsive - leading to larger cycles (and eventually to chaotic behavior - but that's really hard to explain in a short post).

(*A common way to create such a plot is to iterate through each $r$, following the following procedure: first calculate $x_{1000}$, then plot the values $x_{1001},\,x_{1002},\ldots,x_{2000}$ on the vertical strip corresponding to that $r$. This basically just waits for the sequence to get close to where it converges, then plots what happens next. Obviously, more terms discarded at the start and plotted at the end give finer results, although at some point you run into the issue that computers do arithmetic with finite precision, which is insufficient to correctly represent the exact mathematical behavior)

(**For $r\leq 3$, this is actually true of every point, but this is stronger than what the bifurcation diagram wishes to communicate)

(***These slices really are distributions - while the really clear bits only involve finite cycles of varying lengths, where it's just as well to talk about some finite set or sequence associated to various values of $r$, if you get a really good quality plot of this, you can see things like, "the iteration behavior chaotically here, but it spends more time near some points than near others" - where there's no simple description of what's happening, yet there's still some structure. This sort of thing explains why, in high-quality plots, one can see various dark curves near things like repulsive cycles, where a value might spend a lot of time slowly getting further away, leading to more time spent near certain points even when those points fail to dominate the long term behavior)

(****It is worth noting that some points do end up at the repulsive fixed point - for instance, the fixed point itself will never leave, even if everything nearby gets pushed further. There is also a countable set of points with the property that, after finitely many iterations, they end up at the fixed point. Similarly, even after the two cycle stops attracting, some countably many points end up there - but this is exceptional behavior if you choose a point at random from $(0,1)$ for $x_0$ this kind of behavior almost never occurs)

Title: What causes bifurcation?

Body:
Suppose let's say we have a recursive function,
$$x_{n+1}=rx_{n}(1-x_{n})$$

From what I understand, the $x_{n+1}$ VS r graph starts to split (bifurcate) after a particular value of r (apparently it's around 3). Does that mean after they split up, we will get two values of the function for the same value of r? Am I getting it right?

How's it possible? (I do understand that it loses its ordered nature and falls into chaos) Could someone illustrate it with an example? Help would be appreciated!  (I'm a high school student, so explaining it in simpler terms would be even more appreciated). Thank you!


问:
Show intersection of non-parallel cylinders is bounded
First let me define what I mean by cylinder. Let $U \subseteq \mathbb{R}^n$ be a one-dimensional subspace of $\mathbb{R}^n$ and $R > 0$ a real number. Then I call the set
$$
C_{R}(U) := \{x \in \mathbb{R}^N\,:\,\|x - y\|_2 < R \;\text{for some y} \in U\}
$$
the cylinder with axis $U$ and radius $R$. Now let $U,V \subseteq \mathbb{R}^n$ be two one-dimensional subspaces of $\mathbb{R}^n$ and $R_1, R_2 > 0$. I want to show that the intersection $C_{R_1}(U) \cap C_{R_2}(V)$ is bounded if $U$ and $V$ are non-parallel subspaces. Intuitively this is clear: When the angle $\alpha$ between $U$ and $V$ is not equal to zero the cylinders intersect in a compact set. I have managed to show this by drawing a picture of the plane spanned by $U$ and $V$ and assuming that $R_1 \geq R_2$. Then the intersection can be seen to be contained in a ball of radius $\sin(\frac{\alpha}{2})\cdot R_1$ centered at the origin.

My question: Can we show the boundedness of the intersection by means of linear algebra alone, i.e. without resorting to drawing a picture? My first idea was to consider the orthogonal decomposition of any vector $x \in \mathbb{R}^n$. For example decomposing $x$ with respect to $U$ we can write $x = u^{||} + u^{\perp}$, i.e. we get a parallel and a orthogonal part. With this we have $x \in C_{R_1}(U) \iff \|u^{\perp}\|_2 < R_1$. We can do the same with respect to the second subspace and obtain $x \in C_{R_2}(V) \iff \|v^{\perp}\|_2 < R_2$. Now it follows that $x = u^{||} + u^{\perp} = v^{||} + v^{\perp}$ is in the intersection of both cylinders iff $\|u^{\perp}\|_2 < R_1 \wedge \|v^{\perp}\|_2 < R_2$. However, the decompositions are of course with respect to different basis and also the angle between $U$ and $V$ doesn't appear anywhere so something is missing in this approach. Can it be salvaged? What is a better approach?

答:
Call $e$ and $f$ unit vectors in the directions of $U$ and $V$. They are linearly independent (otherwise the assertion is false). Choose $\lambda>0$ such that the projection of $\pm\lambda e$ onto $V^\perp$ satisfies
$$
\|P_{V^\perp}(\pm\lambda e)\|=R_1+R_2
$$
which is clearly possible by linearity. The lines play a symmetric role in the problem, so for the same $\lambda$ you have
$$
\|P_{U^\perp}(\pm\lambda f)\|=R_1+R_2
$$
If $x$ is in the intersection of the cylinders, there exist $y_1\in U$ and $y_2\in V$ such that
$$
\|x-y_1\|\le R_1,\quad \|x-y_2\|\le R_2\quad (*)
$$
Then, necessarily $\|y_1\|, \|y_2\|\le\lambda$. Proof: if $\|y_1\|>\lambda$ you have $dist (y_1,V)>R_1+R_2$ and no point $x$ can satisfy (*).

From (*) and the triangle inequality follows
$$
\|x\|-\|y_1\|\le R_1,\quad \|x\|-\|y_2\|\le R_2.
$$
Adding the above conditions,
$$
2\|x\|\le \|y_1\|+\|y_2\|+R_1+R_2\le 2\lambda+R_1+R_2,
$$
proving the assertion.


问:
Is it possible to find non constant, continuous simple function?
Is it possible to find non constant, continuous simple function ? While calculating Lebesgue integral of $f(x) =x^{2}$ on $ \mathbb R $.
We have constructed simple function which is basically function with finite range .
I was wondering will there exist any simple function which is continuous and non constant on any arbitrary space

答:
Proposition. A space $X$ admits a non-constant continuous simple function iff $X$ is not connected.

Proof. $(\Rightarrow)$ Let $f:X\to\Bbb R$ be a non-constant continuous simple function. Pick any $y\in f(X)$. It is easy to check that $f^{-1}(y)$ and $X\setminus f^{-1}(y)$ is a partition of the space $X$ into its disjoint non-empty open subsets.

$(\Leftarrow)$ If $X$ is a union of two its disjoint non-empty open subsets $A$ and $B$ then the characteristic function of the set $A$ is a non-constant continuous simple function on $X$. $\square$


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:
Method$\#1:$

Using Prosthaphaeresis Formulas on $\cos x+\cos5x,$

we get $0=\cos3x(1+2\cos2x)$

$\implies x=(2m+1)\dfrac\pi6$ or $2x=\dfrac{2\pi(3m\pm1)}3\iff x=\dfrac{\pi(6m\pm2)}6$  where $m$ is any integer

Method$\#2:$

Using Prosthaphaeresis Formulas on $\cos3x+\cos5x,$

we get $0=\cos x(1+2\cos4x)$

$\implies x=(2n+1)\dfrac\pi2=\dfrac{(6n+3)\pi}6$ or $4x=\dfrac{2\pi(3n\pm1)}3\iff x=\dfrac{\pi(3n\pm1)}6$  where $n$ is any integer

Method$\#3:$

Like How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

multiply both sides by $\sin\dfrac{3x-x}2=\sin x\ne0,$ and use Werner Formulas

$$0\cdot2\sin x=\sin2x+\sin4x-\sin2x+\sin6x-\sin4x\implies\sin6x=0$$

$$\implies x=\dfrac{r\pi}6\text{ where }r \text{ is any integer, not divisible by }6\text{ as }\sin x\ne0\text{(why?)}$$

So, we need to establish that the following three sets are equivalent :

$$\{r; 6\nmid r\}; \{2m+1,6m+2,6m-2\};\{6n+3, 3n+1, 3n-1\}$$

Take modulo $6,$

$$2m+1\equiv1,3,5; 6m+2\equiv2; 6m-2\equiv4\pmod6$$

$$6n+3\equiv3;3n+1\equiv1,4;3n-1\equiv2,5\pmod6$$


问:
Length Of The Shortest Path Touching 2 Sides And 2 Vertices Of A Triangle.
In $\triangle BAC$, points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. Let $C_1$ be the reflection of $C$ across $\overline{AB}$, and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$. Prove that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$.

This is a problem I contrived from problem 20 on the 2014 AMC12A. It can be solved via a "simple application of the triangle inequality".

The inequalities I set up are:
$BE$ $+$ $ED$ $>$ $BD$,
$CD$ $+$ $ED$ $>$ $CE$,
$AE$ $+$ $ED$ $>$ $AD$,
$AD$ $+$ $ED$ $>$ $AE$,
$AB$ + $AC_2$ > $BC_2$

I then added up the first four to get $BD + CE + 4ED + AE + AD > AE + CE + AD + BD = AB + AC = AB + AC_2 > BC_2$

So $BE+ CD + 4ED + AE + AD > BC_2$, except what I want to prove is that $BE + CD + ED \geq BC_2$

From this point I am stuck. I do not know which inequalities I need to create and combine in order to solve the problem, though I assume there is an inequality I am missing that would allow me to solve it.

答:
This setup is the milkmaid problem, where the sum of distances is minimized by using reflection.

Hint: $BE + ED + DC = BE + ED + DC_1 = BE + ED_2 + D_2 C_2 \leq BC_2$.

Now, what do you think $D_2$ is?


答:
First rewrite $5+2\sqrt{3} = (a-\sqrt{3b})(7+\sqrt{3})$.

Now suppose $b=3n^2$, then we get $5+2\sqrt{3} = (a-3|n|)(7+\sqrt{3})$, which can not be solved for rational $a$ and $n$.

Next, suppose $b=n^2$, then we get $5+2\sqrt{3} = (a-|n|\sqrt{3})(7+\sqrt{3}) = 7a-3|n|+(a-7|n|)\sqrt{3}$ and hence $5=7a-3|n|$ and $2=a-7|n|$. This gives $9=-46|n|$, which has no solutions either for rational $a$ and $n$.

Other $b$ won't work either, since you'll be left with a $\sqrt{b}$. Thus your equation has no rational solutions.


答:
Suppose $X$ is a random point in $\mathbb R^{n\times1}.$ Then $\mu = \operatorname E(X) \in \mathbb R^{n\times1}$ is defined componentwise and then
$$
\Sigma = \operatorname{var}(X) = \operatorname E\big((X-\mu)(X-\mu)^\top\big) \in \mathbb R^{n\times n}
$$
is an $n\times n$ nonnegative-definite symmetric matrix.

It can be shown via the (finite-dimensional version of the) spectral theorem that $\Sigma$ has a nonnegative-definite symmetric square root $\Sigma^{1/2}$ (definitely not defined componentwise).

If $A\in\mathbb R^{k\times n}$ then
$$
\operatorname{var}(X) = A\Sigma A^\top \in\mathbb R^{k\times k}.
$$
From this it follows that if $\operatorname{var}(Y) = I_n$ (the $n\times n$ identity matrix) and $\Sigma$ is any nonnegative-definite symmetric real matrix, then $\Sigma^{1/2} Y$ has variance $\Sigma,$ and therefore every nonnegative-definite symmetric real matrix is a variance.

If $\Sigma$ is invertible then it can be shown that a random vector whose density is
$$
\frac 1 {\sqrt{2\pi}^{\,n}} \frac 1 {\big(\det\Sigma^{1/2}\big)} \exp\left( - \frac 1 2 (x-\mu)^\top \Sigma^{-1} (X-\mu) \right)
$$
has expected value $\mu$ and variance $\Sigma.$

Title: Generalization of standard deviation?

Body: Standard deviation is often taken from a one-dimensional list of numbers. However, how would one evaluate standard deviation of a set of points of $N$-dimensional coordinates?


问:
Why does the cube have the fewest facets among (centrally) symmetric polytopes in $\mathbb{R}^n$?

"A body like the cube, which is bounded by a finite number of flat facets, is called a
polytope. Among symmetric polytopes, the cube has the fewest possible facets, namely $2n$."

I am looking for rigorous proof and also intuition behind why this must be true. I'm thinking of doing something by contradiction, i.e. assuming less than $2n$ facets for a symmetric polytope and then going ahead from there?

I don't have much to add here in terms of how I approached this, because this isn't a problem - it is something I came across while reading some lecture notes! I could not find much on this online, and I'd appreciate any help.

Edit: It seems the author is talking about centrally symmetric polytopes, i.e. $x\in K$ whenever $-x\in K$.

答: By symmetry, any face should have an opposite face. So, your polytope should have at least two opposite faces. But the region bounded between two opposite hyperplanes is not bounded, you still have $n-1$ degrees of freedom inside the "slice" of $\mathbb{R}^n$. You choose another pair of independent (opposite) hyperplanes. These bound your region in an independent direction, but you still have $n-2$ directions along which it is unbounded. You have to continue until you have at least $2n$ faces to get a bounded region. Since a cube has exactly $2n$ faces, it is the polytope with the minimum possible number of them.


答: Can you see that you need to show $BE + ED_1 + D_1C_2 \geq BC_2$?

Title: Length Of The Shortest Path Touching 2 Sides And 2 Vertices Of A Triangle.

Body:
In $\triangle BAC$, points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. Let $C_1$ be the reflection of $C$ across $\overline{AB}$, and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$. Prove that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$.

This is a problem I contrived from problem 20 on the 2014 AMC12A. It can be solved via a "simple application of the triangle inequality".

The inequalities I set up are:
$BE$ $+$ $ED$ $>$ $BD$,
$CD$ $+$ $ED$ $>$ $CE$,
$AE$ $+$ $ED$ $>$ $AD$,
$AD$ $+$ $ED$ $>$ $AE$,
$AB$ + $AC_2$ > $BC_2$

I then added up the first four to get $BD + CE + 4ED + AE + AD > AE + CE + AD + BD = AB + AC = AB + AC_2 > BC_2$

So $BE+ CD + 4ED + AE + AD > BC_2$, except what I want to prove is that $BE + CD + ED \geq BC_2$

From this point I am stuck. I do not know which inequalities I need to create and combine in order to solve the problem, though I assume there is an inequality I am missing that would allow me to solve it.


问:
How does remainder theorem work even though we are technically dividing by zero?
In the polynomial remainder theorem I've noticed one thing that triggers me

$P(x) = Q(x)D(x) + R(x)$

Here we say that the divisor is $D(x) = x-a$ now to show that $P(a) = R(x)$, people simply put $x-a = 0$ which contradicts the fact that we cannot divide by zero.

Another way to justify this is that we just put the value of $x$ as $a$ everywhere.
But this again is the same thing since we are converting the divisor into zero which doesn't make sense.

So is there any way to justify this or some other way to look at it so that it doesn't end up as dividing by zero?

Thank you for seeing my question.

答:
It is called the division algorithm, but no division actually occurs.  Instead we use addition and multiplication to demonstrate what would happen if the division were performed.  So there is no explicit division by zero.

Additionally, the polynomial $x - a$ is not the zero polynomial, so even if there were division by $x-a$, it is not division by zero.

Suppose $a$ is a zero of $P(x)$, so $x-a$ is a factor of $P(x)$.  That is, $P(x) = (x-a)S(x)$ where $S(x)$ has degree one less than $P$.  We have:
\begin{align*}
(x-a)S(x) &= P(x)  \\
    &= Q(x)(x-a) + R(x)  \text{.}
\end{align*}
This is an equality of polynomials, so the value on the left equals the value on the right for each choice of value of $x$.  In particular, we may specialize $x$ to the value $a$.  (There is still no division actually occurring here.  Recall that we define division by "what we would have had to multiply (then add a remainder of bounded degree) to the divisor to obtain the dividend".  We are literally following the definition of division to write down this multiplication and addition.)  After this specialization of $x$, we obtain
\begin{align*}
(a-a)S(a) &= P(a)  \\
    &= Q(a)(a-a) + R(a)  \text{.}
\end{align*}
Of course, $a-a = 0$ and multiplication by zero is very easy, so this simplifies to
\begin{align*}
0 &= P(a)  \\
    &= 0 + R(a)  \text{.}
\end{align*}
We discover that $P$ is zero at one of its zeroes, $a$, and the remainder, $R(a)$, on dividing $P$ by the linear polynomial corresponding to that zero, $x - a$, is zero at $a$ as well.

But this is not as strong as we want.  We want the remainder on dividing by an exact divisor to be the zero polynomial, not just a polynomial that is zero at one place.  That is, we want $R(x) = 0$ for all $x$, not just $R(a) = 0$.  Let's go back.
\begin{align*}
(x-a)S(x) &= P(x)  \\
    &= Q(x)(x-a) + R(x)  \text{.}
\end{align*}
We require that $R(x)$ has degree less than that of $D(x) = x-a$, which has degree $1$.  So $R(x)$ has to have degree at most $0$, so $R(x)$ is constant.  We know the value of $R$ at one point: $R(a) = 0$, so $R$ is the constantly zero function, $R(x) = 0$.  This is what we wanted to happen.


问:
strict ordering a Set
Given:
$(A,<)$ is a strictly ordered set and $b \notin A$.

Define:
a relation $\prec$ in $ B = A \bigcup \{b\} $ as:
$x\prec y $ if and only if $(x,y \in A \text{ and } x<y)$ or $(x \in A \text{ and } y=b)$

Show that $\prec $ is a strict ordering of $B \text{ and } \prec \bigcap A^{2} =<$.

I know that I need to show the relation is asymmetric and transitive.

Here is what I have :

Asymmetric:
let $x \prec y $ then $ x < y \text{ OR } x \in A \text{ and } y = b$

if $x<y$ then $y \not< x$ since $<$ is a strict order.

if $x \in A \text{ and } y=B$ then $y \not\prec x \text{ since } y \notin A \text{ and } x\not= b$

therefore $\prec$ is asymmetric.

Transitive:
I know how to show it is transitive if $x, y, z \in A$ as I would just fall back on the given that $<$ is a strict order.

However I don't know how to show it if $y=b$.

答: Hint: Note that there is no element $z$ such that $b\prec z$.

Title: strict ordering a Set


问:
An algebra with a strictly positive state + condition ? to become a C$^*$-algebra.
Let $(A,*)$ be an involutive $\mathbb{C}$-algebra i.e. a $\mathbb{C}$-algebra with a semi-linear anti-automorphism i.e.
identically

 $(x+y)^*=x^*+y^*$

 $(\alpha.x)^*=\overline{\alpha}.x^*$

 $(xy)^*=y^*x^*$

A strictly positive (or faithful) state $\varphi\in A^*$ is such that for all $x\in A\setminus \{0\}$, $\varphi(x^*x)> 0$.
One can prove easily that $g(x,y):=\varphi(x^*y)$ is a non-degenerate hermitian form. It can be noted that, due to the fact that $g(a.x,y)=g(x,a^*y)$ and with finite dimensionality, one gets that $A$ is semi-simple. So, one can consider $A$ as simple.

My question is the following.

Q) We suppose $A$ to be finite dimensional. Under which additional condition(s) is the Hilbert space $(A,g)$ a C$^*$-algebra ?

Late edit I insist on the fact that I am in search of conditions for
$(A,g)$ to be a C$^*$-algebra, not only $A$. This means that it should be a C$^*$-algebra for the precise norm $||x||=\sqrt{g(x,x)}$.

Late remark The fact that $A$ be a star-algebra of finite dimension, sum of matrix algebras is by no means sufficient to imply that the projectors on the blocks are *-invariant nor $A\simeq \mathbb{C}$ as shows the following counterexample. Take $B=\mathbb{C}^{n\times n}$ (algebra of complex square matrices of dimension $n>0$) and $A=B\oplus B$ with the anti-automorphism
$(X,Y)^\star=(Y^*,X^*)$. Then $(A,\star)$ is easily checked to be a star algebra. It is of finite dimension, sum of matrix algebras but $dim_\mathbb{C}=2n^2\not= 1$. Indeed, the existence of a strictly positive state is crucial as there is none over $A$.

答:
The following statements are equivalent:

$A$ is a C$^*$-algebra with the norm induced by $\varphi$

$\dim A=1$, i.e., $A=\mathbb C$

Proof. if $A=\mathbb C$, then the only state is the identity, which induces the norm.

Conversely, if $A$ is a C$^*$-algebra, its norm satisfies the C$^*$-identity
$$\tag1\|x\|^2=\|x^*x\|.$$ In terms of $\varphi$, this is
$$\tag2\varphi(x^*x)=\varphi((x^*x)^2)^{1/2},\qquad x\in A.$$
So, for each positive $a\in A$,
$$\tag3\varphi(a)^2=\varphi(a^2).$$
As $\varphi$ is a ucp map, the equality $(3)$ says that $a$ is in the multiplicative domain of $\varphi$. Thus the multiplicative domain of $\varphi$ is all of $A$, since a C$^*$-algebra is spanned by its positive elements. So $\varphi$ is a faithful representation of $A$ into (thus onto) $\mathbb C$.


问:
Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $
As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to  $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?

答:
Here a relatively "natural" way to show the inequality. It avoids a time consuming (or rather time wasting) analysis of a quite specific 6th degree polynomial where the amount of learnings and insights to be expected doesn't seem to justify the effort.

To show is equivalently

$$\frac{x^2}y + \frac{y^2}x \geq \sqrt 2$$

\begin{eqnarray*}\frac{x^2}y + \frac{y^2}x
& \stackrel{\frac 1t \text{convex for }t>0}{\geq} & \frac 1{x^2y+y^2x}\\
& = & \frac 1{xy(x+y)}\\
& \stackrel{GM-AM: xy\leq \frac 12}{\geq} & \frac 2{x+y}\\
& \stackrel{C.S.:x+y\leq \sqrt 2}{\geq} & \frac 2{\sqrt 2} = \sqrt 2
\end{eqnarray*}


问:
Question on proof of irrational of $\sqrt{2}$
I am able to follow every step in the standard proof by contradiction that $x^2 = 2$ has no solutions in $\mathbb{Q}$ except for one line (paraphrasing):

Assume there is such an $x$, and write $x = \frac{m}{n}$ where $m$ and $n$ have no common factors.

I know this is trying to say "$m$ and $n$ are relatively prime,'' $\gcd(m,n) = 1$, etc. But $1$ is a common factor even if that is the case. Is it convention to not consider $1$ a "common factor"?

I apologize if this is trivial. I am trying to understand the language used.

答: My guess is that the author meant that $m$ and $n$ have no common prime factors.


问:
Why is my logic incorrect for this probability question?
What is the probability that two people were born on the same day of the week if it is known that one of two was born on a Saturday

So I split it up into two cases: one where person A is born on Saturday and one where person B is born on Saturday. So for the cases where both are born on Saturday, it would be $\frac{1}{7}$ for the first case ie $\frac{1}{7}$ chance that person B is also born on Saturday (fixing that person A is born on Saturday), and also $\frac{1}{7}$ for the second case. So case one or case two $\frac{1}{7}+\frac{1}{7}$ is the probability. Now I know the answer is $\frac{1}{13}$ and I get it correctly when I draw the dot diagram, but I don't see why I get it wrong with the above logic.

答:
You can apply Bayes' theorem given the condition that one of them is born on Saturday.

For explanation, given the information that one of them is born on Saturday, your sample space is now restricted to all the cases where one of them is born on Saturday or both of them are born on Saturday which is $13$.

$6 + 6 + 1 = 13$ possibilities out of $49$.

Now there is only $1$ of the $13$ possibilities where both are born of Saturday. So you get the probability of $\frac{1}{13}$.

Applying Bayes' theorem,

If $P(A)$ is probability of both of them being born on Saturday and $P(B)$ is probability of at least one of them being born on a Saturday.

$P(B) = 2 \times \frac{1}{7} \times \frac{6}{7} + \frac{1}{7} \times \frac{1}{7} = \frac{13}{49} \, \,$.   The first term is one of them having birthday on Saturday and second term is where both of them have birthday on Saturday. Can you see why $\frac{2}{7}$ is wrong even for $P(B)$?

$P(A\cap B) = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49}$

So your answer is $P(A|B) = \displaystyle \frac {\frac{1}{49}}{\frac{13}{49}} = \frac{1}{13}$


问:
What is the expected value of correct placements?
I recently watched a video by Cut on YouTube called "Guess my Zodiac Sign." It's at the following link: https://www.youtube.com/watch?v=bfkwAAi4t2M

So, at the end, two of the people got $4/12$ correct and it got me wondering what the expected value is. They were saying they did well, but I'm not convinced that they did better than average.

I will rephrase this question. Say that I have $12$ distinct cards and $12$ pots; each card corresponds to exactly one pot. Assuming we place the cards randomly in the pots, what is the expected value of correct guesses?

I have attempted this question, but, I must admit, my probability skills are very rusty. There are $12!$ ways one can place the cards in the pots. I started by trying to count the number of ways one could get 0 correct. Going to the first pot, there are $11$ incorrect choices, but then with the second pot it branches. It's possible that we put pot 2's card in pot 1, so there are again $11$ choices. It's also possible that we did not do that, so there are 10 choices. I can see how we would calculate the total number of ways to get 0 correct from there, but it seems it will be very tedious. Is there are a faster/more clever way? I feel like I'm overthinking this.

It would be interesting to see this done in general as well with $n$ pots and $n$ cards, however I would be completely satisfied with the case $n=12$.

Thank you!

答:
What you are looking for is the number of permutations of n with k fixed points, they are also known as partial derangements. You have to find them for k=0 to k=12. For example with k=0 you are going to count the number of derangements. Check this out:
Derivation of the Partial Derangement (Rencontres numbers) formula</p>


问:
Why is it important to have a discrepancy between image and codomain?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.

What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?

I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?

答:
As I see it, the codomain of a function $f$ sets up a context in which the function operates, and tells you what sort of operations you can use to build and inspect the value of $f(x)$. So if I have that $f : \mathbb{R} \to \mathbb{R}$, then I know I can use addition, multiplication, division, order structure, topological structure, etc. when producing the outputs of $f$.

It might help if you view $f : A \to B$ not as giving $A$ and $B$ as properties of the function $f$, but merely as a statement "$f$ is a function from $A$ to $B$", that is to say, if I know $x \in A$ I can deduce that $f(x) \in B$. Then I have that if $B \subset C$ then $f : A \to C$ is also true. In this sense there is no single codomain; what I will call the codomain depends on what statements of the form $x \in A \implies f(x) \in B$ I can prove or need to prove.

Think also of "$f$ is a group homomorphism from $A$ to $B$", where it's really necessary to specify $B$ to say what it even means for $f$ to be a group homomorphism, likewise "continuous map of topological spaces" and various other examples. Functions frequently have nice properties that are defined with reference to some set or structure, and then when we need to discuss that particular structure, that's the codomain we'll use.


答:
A very good spot. Mathematical proof is all about precision in the details, and this is a detail that has been missed by the author.

The way I've known the proof to be written is that $m$ and $n$ are chosen such that the fraction $\frac mn$ is in simplest form (I.e. can't be cancelled down), which essentially conveys that they must have no common factor except $1$.

Title: Question on proof of irrational of $\sqrt{2}$

Body:
I am able to follow every step in the standard proof by contradiction that $x^2 = 2$ has no solutions in $\mathbb{Q}$ except for one line (paraphrasing):

Assume there is such an $x$, and write $x = \frac{m}{n}$ where $m$ and $n$ have no common factors.

I know this is trying to say "$m$ and $n$ are relatively prime,'' $\gcd(m,n) = 1$, etc. But $1$ is a common factor even if that is the case. Is it convention to not consider $1$ a "common factor"?

I apologize if this is trivial. I am trying to understand the language used.


问:
Given an acute triangle in a circle, prove the concyclicity of some points
Given an acute triangle ABC with AB<AC<BC written on a circle c(O, R). The extension of the height AD intersects the circle c(O, R) at E. The circle c1(D, DA) intersects the side AC at the point T, the line AB at the point S, the circle c(O, R) at H and the line OA at Z. Prove that the points S, B, T, C are concyclic on a circle c2 and that the points O, D, E, Z, H and the center of c2 are concyclic.

I attempted to do the question by saying that: If AD intersects the circle c1 at P, then AP is the diameter of c1, hence ASP=ATP=90.

Since BSP=BDP=90 and PDC=PTC=90, then the points B, D, P, S are concyclic, as are the points D, T, C, P. This is where I got stuck. Could you please explain to me how to finish off this question?

答:
Observe that, $\angle BST=\angle AST=\angle APT=90-\angle PAT=90-\angle DAC=\angle C=\angle TCB$.    $\Rightarrow$   Points $B$, $S$, $C$ and $T$ are concyclic.

The perpendicular bisectors of $BC$ and $ST$ meet at $R$;Hence, it is the centre of circle $C_2$. The perpendicular bisectors of $BC$ and $ST$ will pass through point $O$ and $D$ respectively. Let $AZ$ intersect $ST$ at $K$.

Observe that, $\angle SAK+\angle ASK=\angle BAO+\angle AST=90-\angle C+\angle C=90$

$\Rightarrow \angle AKS=90^{\circ}$

$\Rightarrow  AK\parallel DR$

Since $DR\parallel AO$ & $AD\parallel OR$, $AORD$ is a parallelogram.

$\Rightarrow \angle DAO=\angle DRO$

Now, notice that, $\angle DZO=\angle DAO$[ Since $DZ=DA$]$=\angle DRO$; Hence, Points $D$, $O$, $Z$ and $R$ are concyclic.

Also, observe that, $\angle DEO=\angle DAO$[ Since $OE=OA$]$=\angle DRO$.
$\Rightarrow$  Points $D$, $E$, $R$ and $O$ are concyclic.

Since $DH=AD$ and $AO=OH$, $\angle DHO=\angle DAO$

$\Rightarrow   \angle DHO=\angle DAO=\angle DRO$.
$\Rightarrow$  Points $D$, $O$, $H$ and $R$ are concyclic.

Therefore, points $E$, $D$, $O$, $H$, $Z$ and $R$, which is the centre of circle $C_2$, are all concyclic.


问:
Is the set $\Bbb{P}^i \pmod n$ always equal to $(\Bbb{Z}/(n))^{\times}$ or the whole set? Where $\Bbb{P}^i = \{ q_1 \cdots q_i : q_j \in \Bbb{P}\}$.
Is the set $\Bbb{P}^i \pmod n$ always equal to $(\Bbb{Z}/(n))^{\times}$ the whole set $\Bbb{Z}/(n)$?

Where $\Bbb{P}^i = \{ q_1 \cdots q_i : q_j \in \Bbb{P}\} \subset \Bbb{Z}$.

If you like, $\Bbb{P}^i = \Omega^{-1}(i)$ where $\Omega: \Bbb{N} \to \Bbb{N}: n = q_1 \cdots q_i \mapsto i$.

Was intersting in figuring out what $\Bbb{P}^{j} + 1$ (elementwise) does to elements with respect to $i$.

So I figured maybe looking at $\Bbb{Z}/(n)$ and then looking back at the integers in some advanced mathematical fashion (sheaves maybe?).

Anyway, I thought I'd start with what $\Bbb{P}^i$ gets taken to modulo $n$. Since the whole problem $\pmod n$ is then a question of a finite nature.

答:
As stated, the answer is no, because the primes dividing $n$ will appear soon enough that the answers will no longer be elements of $(\Bbb Z/n)^\times$. But if you exclude the primes dividing $n$ from the products $q_1\cdots q_i$, then the answer is yes when $n$ is a power of an odd prime or twice such a power, at least.

D. K. L. Shiu proved (Strings of congruent primes. J. London Math. Soc. (2), 61(2):359–373, 2000), that for any modulus $n$, any element $a\in(\Bbb Z/n)^\times$, and any $r \ge 2$, there are infinitely sets of $r$ consecutive primes all of which are congruent to $a\pmod n$.
If we take $r=\phi(n)$ and $a$ to be a primitive root modulo $n$, then the products $q_1\cdots q_i$ will cycle through all $\phi(n)$ elements of $a\in(\Bbb Z/n)^\times$ during any such set of $\phi(n)$ consecutive primes.

The answer is conjecturally yes (again, once the statement has been suitable refined) for all $n$, but I think it is still an open problem for most moduli $n$.


答:
I wrote some code.  It shows that the answer is negative.

```<code>from sympy import *

while True:
    n = input("n=")
    n = int(n)

    N = input("prime max=")
    N = int(N)
    P = primerange(2, N)
    P = list(P)

    i = input("i=")
    i = int(i)

    Q = P
    for j in range(i-1):
        R = []
        for q in Q:
            for p in P:
                R.append(p*q)
        Q = R

    ZZ_n = set()
    for q in Q:
        ZZ_n.add(q % n)

    print(ZZ_n)

</code>```

It outputs:

```<code>    n=15
    prime max=10000
    i=1
    {1, 2, 3, 4, 5, 7, 8, 11, 13, 14}
    n=
</code>```

Title: Is the set $\Bbb{P}^i \pmod n$ always equal to $(\Bbb{Z}/(n))^{\times}$ or the whole set? Where $\Bbb{P}^i = \{ q_1 \cdots q_i : q_j \in \Bbb{P}\}$.

Body:
Is the set $\Bbb{P}^i \pmod n$ always equal to $(\Bbb{Z}/(n))^{\times}$ the whole set $\Bbb{Z}/(n)$?

Where $\Bbb{P}^i = \{ q_1 \cdots q_i : q_j \in \Bbb{P}\} \subset \Bbb{Z}$.

If you like, $\Bbb{P}^i = \Omega^{-1}(i)$ where $\Omega: \Bbb{N} \to \Bbb{N}: n = q_1 \cdots q_i \mapsto i$.

Was intersting in figuring out what $\Bbb{P}^{j} + 1$ (elementwise) does to elements with respect to $i$.

So I figured maybe looking at $\Bbb{Z}/(n)$ and then looking back at the integers in some advanced mathematical fashion (sheaves maybe?).

Anyway, I thought I'd start with what $\Bbb{P}^i$ gets taken to modulo $n$. Since the whole problem $\pmod n$ is then a question of a finite nature.

Body:
I wrote some code.  It shows that the answer is negative.

from sympy import *

while True:
    n = input("n=")
    n = int(n)

    N = input("prime max=")
    N = int(N)
    P = primerange(2, N)
    P = list(P)

    i = input("i=")
    i = int(i)

    Q = P
    for j in range(i-1):
        R = []
        for q in Q:
            for p in P:
                R.append(p*q)
        Q = R

    ZZ_n = set()
    for q in Q:
        ZZ_n.add(q % n)

    print(ZZ_n)

It outputs:

    n=15
    prime max=10000
    i=1
    {1, 2, 3, 4, 5, 7, 8, 11, 13, 14}
    n=


问:
What causes bifurcation?
Suppose let's say we have a recursive function,
$$x_{n+1}=rx_{n}(1-x_{n})$$

From what I understand, the $x_{n+1}$ VS r graph starts to split (bifurcate) after a particular value of r (apparently it's around 3). Does that mean after they split up, we will get two values of the function for the same value of r? Am I getting it right?

How's it possible? (I do understand that it loses its ordered nature and falls into chaos) Could someone illustrate it with an example? Help would be appreciated!  (I'm a high school student, so explaining it in simpler terms would be even more appreciated). Thank you!

答:
First of all, congratulations for you work and your observations.

Since @Milo Brandt gave very nice explanations, I shall not repeat any of them.

You can "simplify" the quadratic recurrence equation
$$x_{n+1}=r\,x_{n}(1-x_{n})$$ letting
$$x_n=\frac{r-2 z_n}{2 r}\implies z_{n+1}=\frac r 4+\frac 1r z_n^2$$ and you enter in the domain of quadratic maps which, at leat for me, is not the simplest problem.

For example, the simplest
$$c_{n+1}=c_n^2-k$$ with $c_0=0$ used to generate the Mandelbrot Set.

I shall not enter into more details but I suggest you have a look at this paper which, even if too complex for you at the time being, gives very good explanations about the bifurcation diagram. There are a few plots which are very illustrative.

Title: What causes bifurcation?


问:
Compactness of Subset of $\ell^1$.
Let $X$ be a finite-dimensional normed space and let $\ell^1(X)$ denote the Banach space of all sequences in $X$ for which $\sum_{n=1}^{\infty} \|x_n\|_X<\infty$.  For $s\geq 0$, is the set
$
C = \left\{
(x_n)\in \ell^1(X):\, \sum_{n=1}^{\infty} n\|x_n\|\leq s
\right\}
$
compact?

What I've tried:

I tried using the Grothendieck compactness principle to show that $C$ is contained in the closed convex-hull of the sequence $(\frac{s}{n})$ but I'm not sure this is the case..

答:
Let $(x_n^{(i)})$ be  as sequence in $C$. For each $n$ we have $n \|x_n^{(i)}\|^{2} \leq s$. So $(x_n^{(i)})_i$ is bounded. Since $X $ is finite dimensional any bounded sequence in it has a convergent subsequence. By cantor's diagonal procedure we can find $i_1,i_2<...$ such that $\lim_{k \to \infty} x_n^{(i_k)} =x_n$ (say) exists for each $n$.Now $\sum _n n\|x_n^{(i_j)})\|^{2} \leq s$ for each $j$ implies that $\sum n\|x_n\|^{2} \leq s$ by Fatous Lemma.  Hence, every sequence in $C$ has a convergent subsequence.

I leave it to you to check that $\|x^{(i)}-x\|_{\ell^{1}(X)} \to 0$.

[Use the fact that $ \sum\limits_{k=N}^{\infty} \|x_n\| \leq\frac s N$ for all $(x_n) \in C$].

Title: Compactness of Subset of $\ell^1$.


问:
$a^{1/n}$ is uniquely determined
Just wanting to check the reasoning in this proof -- I'm using the notation $(\eta^n)^{-1}$ instead of $\eta^{-n}$ since the latter hasn't yet been introduced in my text.

To be clear I'm not looking for alternative/simpler proofs, just was hoping to confirm the reasoning in this proof, or understand where I've gone wrong.

Theorem. Let $a \geq 0$. For each positive integer $n$ define $a^{1/n}$ to be a number $\beta$ such that $\beta^n = a$, and $\beta \geq 0$. Then such a number $\beta$, if it exists, is uniquely determined.

Proof. Suppose there exist two numbers $\beta\geq0$ and $\alpha\geq 0$ such that $\beta^n = a$, $\alpha^n = a$ for $a \geq 0$.

If $a = 0$, then $\alpha = \beta = 0$.

Otherwise, $a > 0$. Then we have $$\alpha^n = \beta^n  = a$$

Multiplying through by $(\beta^{n})^{-1}$ we obtain

$$
\beta^n (\beta^{n})^{-1} = 1 = \alpha(\beta^{n})^{-1}
$$

Then $(\beta^{n})^{-1} = (\alpha^{n})^{-1}$. Since multiplicative inverses are unique, we have  $(\alpha^{n})^{-1} = (\beta^{n})^{-1}$, and so it must be that $\alpha^n = \beta^n$ and hence $\alpha = \beta$ (which can be shown inductively). $\square$

答:
I don't really know what you are allowed to use but I am suggesting:

Algebraic approach : for $a>0$, if we have $a=\alpha^n = \beta^n$ then $\alpha^n-\beta^n=0$ therefore : $$ (\alpha-\beta)\sum_{i=0}^{n-1}\alpha^i\beta^{n-1-i}=0 $$ and as $\alpha,\beta>0$ (otherwise we would have $a = \alpha^n = \beta^n= 0^n =0$ which is impossible) then all of their positive powers are strictly positive and hence:

$$\sum_{i=0}^{n-1}\alpha^i\beta^{n-1-i}>0$$

Consequently:

$$ \alpha-\beta = 0 \text{ i.e. }\boxed{\alpha = \beta}$$

Title: $a^{1/n}$ is uniquely determined


问:
How do I transform $\frac{1-\exp(-Vx^m)}{1-(1-x^m)^V}-1$ so I can evaluate numerically for $V$ and $m$ between $1$ and $30$, and $0\leq x\leq 1$?
I need to evaluate and plot the following expression:

$$
f(x,V,m)=\frac{1-\exp(-Vx^m)}{1-(1-x^m)^V}-1
$$

$V$ and $m$ are numbers from $1$ to $\sim30$ and $0\leq x\leq1$.

The issue is that for $x$ close to zero, the exponential, and the power term in the denominator take values very close to 1, making it difficult to evaluate $f$ numerically. I've tried MATLAB symbolic toolbox which, just to give one case, gives $f(10^{-3},1,6)=-1$. Mathematica also runs into similar problems. I know this to be incorrect because:

$$
\lim_{x\to 0} f(x,V,m)=0
$$

Is there a manipulation of the equation (the first term is the problem) that I am missing that would make it more amenable to evaluate numerically or simplify it? I want to know the behavior of the function for small values of $x$ but short of plotting I am struggling to conclude anything, because even though both numerator and denominator are monotonic, I cannot say the same about their ratio.

答:
Using Taylor expansion of
$$g(y)= \frac{1-\exp(-Vy)}{(1-(1-y)^V)}-1$$ at order one, you find

$$g(y) = -\frac{1}{2}y + O(y^2)$$ and therefore

$$f(x,V,m)=-\frac{1}{2}x^m + O(x^{2m})$$ around zero. Which provides a practical way to evaluate $f$ for $x$ near zero. Interesting is that this doesn't depend (at order one) on $V$.