问:
Is every finite connected set equipped with a binary relation of the form Aut(X)\X?
Is every finite connected set equipped with a binary relation of the form Aut(X)\X?

Say that a related set is a couple $(X,R_X)$ where $X$ is a set and $R_X\subset X^2$ a binary relation on $X$. Abusing the notation we often write $X$ instead of $(X,R_X)$ and $R$ instead of $R_X$. We also abbreviate the condition $(x,y)\in R$ by $xRy$. If $Y$ is another related set we say that a map $f:X\to Y$ is a morphism if $xRx'$ implies $f(x)Rf(x')$ for all $x,x'\in X$. We define the notions of isomorphism and automorphism in the usual way. We say that $X$ is connected if the equivalence relation generated by $R$ has exactly one class.

Let $X$ be a finite connected related set, $G$ its automorphism group and $X_*$ the set of all $G$-orbits in $X$. For $A,B\in X_*$ decree that $ARB$ holds in $X_*$ if and only if $aRb$ holds in $X$ for some $a\in A,b\in B$. Equipped with this relation $X_*$ is again a finite connected related set, whence the question

Is every finite connected related set $X$ isomorphic to $Y_*$ for some finite connected related set $Y$?

[Without the connectedness assumption, the related set with two elements and no relation would be a counterexample.]

Clearly, if $X$ is a poset, so is $X_*$.

This answer shows that for every connected poset $X$ there is a connected poset $Y$ such that $Y_*\simeq X$.

Proposition. Every finite connected related set $X$ of size at most 3 is isomorphic to $Y_*$ for some finite connected related set $Y$.

Let us introduce some notation for the statement of Lemma 1 below.

Assume that $m,n\in\mathbb Z_{\ge0}$ with $m\ge1$; that $a_1$, ... $a_m$, $b_1$, ..., $b_m$, $c_1$, ..., $c_n$, $b'_1$, ..., $b'_m$, $c'_1$, ..., $c'_n$ are distinct elements; that the set $X$ defined by
$$
X=\{a_1,\ldots,a_m,b_1,\ldots,b_m,c_1,\ldots,c_n\};
$$
is equipped with a binary relation; and that the $\operatorname{Aut}(X)$-orbits in $X$ are the $\{a_i,b_i\}$ and the $\{c_j\}$. Form the related set
$$
X'=\{a_1,\ldots,a_m,b'_1,\ldots,b'_m,c'_1,\ldots,c'_n\}
$$
such that the obvious bijection $X\to X'$ is an isomorphism. Equip the set $Y:=X\cup X'$ with the unique binary relation such that

(a) the binary relations of $X$ and $X'$ are induced by that of $Y$,

(b) if $x\in X\setminus X'$ and $x'\in X'\setminus X$ then $x$ and $x'$ are unrelated in $Y$.

The proof of Lemma 1 and Lemma 2 below a tedious but easy exercise using the list of isomorphism classes of related sets of size 3 given on this page, which is part of the entry number of nonisomorphic unlabeled binary relations on n labeled nodes of The On-Line Encyclopedia of Integer Sequences.

Lemma 1. In the above setting, assume that $X$ is connected and that $2m+n\le3$. Then $Y$ is a finite connected related set satisfying $Y_*\simeq X$.

Lemma 2. Let $X=\{a_1,a_2,a_3\}$ be a related set of size three and $G$ its automorphism group. Assume that $G$ acts transitively on $X$. Let $Y=\{a_{11},a_{21},a_{22},a_{31},a_{32},a_{33}\}$ be a set of size $6$. Define the binary relation $R$ on $Y$ by
$$
a_{ij}Ra_{pq}\iff
\begin{cases}
a_iRa_p&\text{if }i\ne p\\
j=q\text{ and }a_iRa_i&\text{if }i=p.
\end{cases}
$$
Then $Y_*$ is isomorphic to $X$.

I haven't been able to decide if Lemma 2 holds without the assumption that $2m+n\le3$.

The proposition follows from the lemmas.

答:
This is a partial answer, too long for a comment.

If $X$ is a poset we write $X^-$ for the subset of minimal elements, and $X^+$ for the subset of maximal elements. An element of $X^-\cap X^+$ is called an isolated point.

Proposition. If $X$ is a nonempty finite poset without isolated points, then there is a finite poset $Y$ without isolated points such that $Y_*\simeq X$.

To prove the proposition we will define a poset $Y$ and show that it has the required properties.

Denote by $|S|$ the cardinality of any set $S$ and put $[n]:=\{1,2,\ldots,n\}$ for any nonnegative integer $n$. Let $X$ be a nonempty finite poset without isolated points. Set
$$
m:=|X^-|,\quad n:=|X|,\quad X^-=\{x_1,\ldots,x_m\}.
$$
Let $x_{m+1},x_{m+2},\ldots,x_n$ be an enumeration of the elements of $X\setminus X^-$ such that $x_i<x_j$ implies $i<j$.

We will first define integers $c_1<\cdots<c_n<r$ and then define $Y$ using these integers.

For $i\in[m]$ we set $c_i:=i$.

$\bullet$ Definition of $c_{m+1}<c_{m+2}<\cdots<c_n$: Let $\Pi$ be the set of all non-constant polynomial $P(t)\in\mathbb Q[t]$ with positive leading coefficient; here $t$ is an indeterminate. Set for $m<i\le n$
$$
P_i(t):=\sum_{x_j<x_i}\binom tj,
$$
where the sum runs over the $j\in[m]$ such that $x_j<x_i$. Since $P_i\in\Pi$ for all $i$, there are integers $c_{m+1},\ldots,c_n\in m\mathbb Z$ such that $m<c_{m+1}<\cdots<c_n$ and $P_i(c_i)\ne P_j(c_j)$ whenever $m<i<j\le n$.

$\bullet$ Definition of $r$: For $i\in[m]$ denote by $\Pi_i$ the set of all the polynomials in $\Pi$ whose degree is congruent to $-i$ modulo $m$, and set
$$
Q_i(t):=\sum_{x_j>x_i}\binom{t-i}{c_j-i},
$$
where the sum runs over the $j\in[n]$ such that $x_j>x_i$. Since each $Q_i$ is in $\Pi_i$, the $Q_i$ are pairwise distinct. Hence there is an integer $r>c_n$ such that the $Q_i(r)$ are pairwise distinct.

$\bullet$ Definition of $Y$: For $i\in[n]$ set
$$
Y_i:=\{S\subset[r]\ |\ |S|=c_i\},\quad Y:=\bigcup Y_i.
$$
For $S\in Y_i$, $T\in Y_j$ set
$$
S<T\iff x_i<x_j\text{ and }S\subset T.
$$
Then $\le$ is a partial order on $Y$. Note that for $S\in Y_i$ we have: $S$ is minimal in $Y$ if and only if $x_i$ is minimal in $X$ and $S$ is maximal in $Y$ if and only if $x_i$ is maximal in $X$. In particular $Y$ is a nonempty finite poset without isolated points.

$\bullet$ Proof of the isomorphism $Y_*\simeq X$: Write $S\sim T$ for $S,T\in Y$ to indicate that there is an automorphism of $Y$ which maps $S$ to $T$.

There is a strictly increasing surjection $f:Y\to X$ mapping $S\in Y_i$ to $x_i$. We claim that $f$ induces an isomorphism $Y_*\to X$. To prove this it suffices to show that for $S\in Y_i$ and $T\in Y_j$ we have $S\sim T$ if and only if $i=j$. If $i=j$, then, since $S$ and $T$ are two cardinality $c_i$ subsets of $[r]$, there is a permutation $\sigma$ of $[r]$ which moves $S$ to $T$, and $\sigma$ induces an automorphism of $Y$ which maps $S$ to $T$. It only remains to prove that $S\sim T$ implies $i=j$.

Recall that we have $S\in Y_i$, $T\in Y_j$, $S\sim T$ and we claim $i=j$.

Case 1: $S$ is minimal. Then $T$ is also minimal and we get $i,j\in[m]$. The number of elements of $Y$ which are greater than $S$ (respectively greater than $T$) is $Q_i(r)$ (respectively $Q_j(r)$). But the assumption $S\sim T$ implies $Q_i(r)=Q_j(r)$, and thus $i=j$.

Case 2: $S$ is not minimal. Then $T$ is also not minimal and we get $m<i,j\le n$. The number of minimal elements of $Y$ which are less than $S$ (respectively less than $T$) is $P_i(c_i)$ (respectively $P_j(c_j)$). But the assumption $S\sim T$ implies $P_i(c_i)=P_j(c_j)$, and thus $i=j$. This completes the proof of the proposition.


问:
Solving $\cos(40+\theta) = 3\sin(50+\theta)$
Question
Solve the following equation for $0≤\theta≤180$
$$
\cos(40+\theta) = 3\sin(50+\theta)
$$
Hint: $\cos(40) = \sin(50)$

Solution

$$
\cos(40+\theta) = 3\sin(50+\theta)
$$
$$
\cos(40)\cos\theta -\sin(40)\sin\theta = 3(\sin(50)\cos\theta +\sin\theta \cos(50))
$$
$$
\sin(50)\cos\theta-\cos(50)\sin\theta = 3(\sin(50)\cos\theta+\sin\theta \cos(50))
$$
$$
2\sin(50)\cos\theta+4\sin\theta \cos(50) = 0
$$
What do I do next?

答:
Continuing your work:
$$\sin50^\circ\cos\theta + 2\sin\theta \cos50^\circ = 0 \\ \tan\theta = -\frac{\tan50^\circ}{2} \\ \theta = -\arctan\left(\frac{\tan50^\circ}{2} \right)$$

ContentLicense: CC BY-SA 4.0


问:
Poisson Distribution - sum of RVs
Question:

$X$ balls are thrown to $n$ bins (each ball has an equal chance to get to each bin). Let $X_1,\dots, X_n$ be the amount of balls in each cell.
a. Show that if $X \sim \text{Poisson}(\lambda)$ then $\displaystyle X_i \sim \text{Poisson} \left(\frac \lambda n\right)$ and that the $X_i$'s are independent.

My thought:
I know how to prove it the other way around, but not from the sum to the parts of it...

Here's a chance to reverse the steps (I'm not sure how I can justify the last move, and how I should figure indepndence:
$P(X=k)=e^{-\lambda} \frac {\lambda^k}{k!}=
e^{-(\frac \lambda n)n} \frac {((\frac \lambda n)n)^k}{k!}=
\frac {e^{-(\frac \lambda n)n}}{k!} \sum_{m=0}^k \frac {k!}{m!(k-m!)}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=
 {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac 1{m!(k-m)!}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=$
$ {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac {(\frac \lambda n)^m}{m!}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$ \sum_{m=0}^k {e^{-(\frac \lambda n)}} \frac {(\frac \lambda n)^m}{m!} {e^{-(\frac \lambda n(n-1))}}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$\sum_{m=0}^k P(X_1=m)P(\sum_{i=2}^nX_i=k-m)$

答:
You need to try and reverse the steps in your previous answer.

The important point is that each ball is goes independently and uniformly (equally as likely to go in any of the buckets)

So for a sequence $X_1,\dots, X_n$, with $\sum_{i=1}^n X_n = X$ you can write down the probability that I get $X$ from a Poisson distribution  and the probability that when I throw $X$ balls randomly into $n$ buckets I get the sequence $X_1,\dots X_n$.

So you can write down the probability that you get the sequence $X_1,\dots,X_n$, but you need to manipulate it so it looks like a product of Poisson probabilities.

You've done all the equations, you just need to write them down in the reverse order and keep your head clear to make sure each step is justified.

ContentLicense: CC BY-SA 3.0

Title: Poisson Distribution - sum of RVs


问:
The portion of the tangent to any conic intercepted between the point of contact and the directrix subtends a right angle at the focus of the conic?
Is there a general proof for this?

The portion of the tangent to any conic intercepted between the point of contact and the directrix subtends a right angle at the focus of the conic?

For parabola, it can be solved this way: the tangent at $P=(at^2, 2at)$ is
$ty = x + at^2$. This meets the directrix at the point $Q=(-a,a(t^2 - 1)/t)$. Now, $\text{Slope of }SQ\times\text{Slope of }SP = -1$. Hence, $∠PSQ = 90°$. A similar way of writing the parametric coordinates and equating the product of slopes to $-1$ yields the desired result for other conics.

But is there a geometric proof for any general conic?

答:
The theorem we want to prove can be restated as follows:

If the tangent to a conic at any point $P$ intersects the
directrix at point $T$, and F is the corresponding focus, then
$\angle PFT$  is a right angle.

Let $P'$ be a point on the conic near to $P$ (see figure below) and suppose line $PP'$ to intersect the directrix at $T$. Let then $M$, $M'$ be the orthogonal projections of $P$, $P'$ on the directrix, and $Q$ a point on line $PF$ produced.

As triangles $TPM$, $TP'M'$ are similar we have:
$$
TP:TP'=MP:MP'=FP:FP'
$$

and this implies, by the angle bisector theorem applied to triangle $FPP'$, that $TF$ bisects the exterior angle $\angle P'FQ$.

In the limit $P'\to P$, line $PT$ tends to the tangent at $P$ and $\angle P'FQ$ tends to $\angle PFQ$, which is two right angles. Therefore $\angle PFT$, which is the limit of $\angle P'FT$, is one half of $\angle PFQ$, that is a right angle, as it was to be proved.


问:
When can i get away with approximating the expected value of a ratio as the ratio of expected values
I'm actually an engineering student so I'm not too good with probability and was hoping someone may be able to help with the following:

So I have a ratio of discrete random variables. I want to be able to know when I can get away with approximating its expected value as a ratio of expected values. For example, for the continuous case I came across the following formulation from a Taylor series expansion

$\text{E}[x/y]=\text{E}[x]/ \text{E}[y]- \text{Cov}[x ,y]/ \text{E}[y]^2 + \text{E}[x] \text{Var}[y]/ \text{E}[y]^3$

Does this apply for discrete random variables? if not, what is its analogous expression?

thank you!
hadi

答:
That last term doesn't make sense to me.  Let $E[x] = \mu_x$ and $E[y] = \mu_y$.
From the Taylor series of $x/t$ around $t=\mu_y$, $x/y = x/\mu_y - x(y -\mu_y)/\mu_y^2 + x (y - \mu_y)^2/\mu_y^3 + \ldots $ resulting in
$E[x/y] = \mu_x/\mu_y - {\rm Cov}[x,y]/\mu_y^2 + E[x (y - \mu_y)^2]/\mu_y^3 + \ldots$, but $E[x (y - \mu_y)^2] \ne \mu_x {\rm Var}[y]$ unless $x$ and $(y - \mu_y)^2$ are uncorrelated.

In any case, it has nothing to do with discrete vs. continuous: it is just as valid for discrete random variables as it is for continuous ones.  The main point is that to have a good approximation (apart from that last problematic term) you'll want
$y - E[y]$ to be (with high probability) small compared to $|E[y]|$.  You especially need to avoid $y$ near 0, which could make $E[x/y]$ be undefined.


问:
Are there other uncommon calculation methods?
enter image description here

$\lim_{h\to0} \frac{（(x+h)^h-1）}{h}$

$\lim _{h\to0} \frac{(1+h/x)^{(x/h)h}}{h}$

What method can I use to get the result? I don't understand what method is used here.

I'm a little curious, is there any way.Honestly, I’m not sure because there is no other process here.

Thanks for your replies.I think I am missing something, this is a $\frac{0}{0}$,Can I not apply the L'Hospital's rule.I think how he used other methods to calculate this limit.

$t=(x+h)^{h}-1；h=\log _{(x+h)}{(t+1)}=\frac{\ln {(t+1)}}{\ln {(x+h)}};t\to0$

$\lim_{h\to0} \frac{（(x+h)^h-1）}{h}=\lim _{t\to0}{h\to0} \frac{t}{\log _{(x+h)}{(t+1)}}=\ln{(x+h)}\frac{t}{ln{(t+1)}}$

$\lim_{t\to0}(1+t)^{1/t}=e$

$\lim _{t\to0} \frac{t}{\ln(t+1)}=\frac{1}{\frac{\ln(t+1)}{t}}=\frac{1}{\ln((1+t)^{1/t})}=1$

答: Hint: most of the time, $a^b$ is defined as $e^{b\log(a)}$. So while it is possible to compute the derivative of $x^x$ using differentiation from first principles, it is also possible to compute it directly using the chain rule.


问:
Extremal distribution of random variable that averages to a given value
Denote by $\mathcal M$ the set of probability measures over $[0,1]$ that average to $p$. This is a convex set and I am trying to characterize its extreme points. For any $x$, $y$ such that $0\leq x< p< y\leq 1$ define
\begin{align*}
\lambda_x^y = \frac{p-y}{x-y}\cdot\delta_x+\frac{x-p}{x-y}\cdot\delta_y
\end{align*}
where for any measurable $A\subseteq [0,1]$ and $a\in[0,1]$, $\delta_a(A)=\mathbf 1(a\in A)$. Since $\delta_x$ (resp. $\delta_y$) averages to $x$ (resp. $y$) we get that $\lambda_x^y$ averages to $\frac{p-y}{x-y}\cdot x+\frac{x-p}{x-y}\cdot y=p$ and so is in $\mathcal M$.

I am very tempted to say that the extreme points of $\mathcal M$ is the set $\{ \lambda_x^y : 0\leq x< p < y \leq 1 \}\cup \{ \delta_p \}$ without being able to prove it. If I am not mistaking I am supposed to prove that (ignoring the $\delta_x$ component)

If for some $x$ and $y$ and $a\in[0,1]$, $\lambda_x^y=a \mu+(1-a) \nu$ then $\mu=\nu=\lambda_x^y$.

For any $\mu\in\mathcal M$ there is a probability measure $\kappa$ over $[0,p]\times[p,1]$ such that $\mu(A)=\int \lambda_x^y(A) d\kappa(x,y)+\mu(\{ p \})\delta_p(A)$ for all measurable $A\subseteq [0,1]$.

For the first point, it is quite clear by positivity of the measures $\mu$ and $\nu$ that their support $\{ x,y \}$, then there is only one probability measure with such a support that averages to $p$ and it is $\mu_x^y$.

For the second statement however it is less clear to me, Maybe one way to proceed is computing the integral
\begin{align*}
\int \lambda_x^y(A) d\kappa(x,y)&=\int \left( \frac{p-y}{x-y}\cdot\delta_x(A)+\frac{x-p}{x-y}\cdot\delta_y(A) \right) d\kappa(x,y)\\
&=\int_{\left(A\cap [0,p]\right)\times [p,1]}\frac{p-y}{x-y} d\kappa(x,y)+\int_{[0,p]\times\left(A\cap [p,1]\right)}\frac{x-p}{x-y}d\kappa(x,y)
\end{align*}
It now feel easier to separate cases $A\subseteq [0,p]$ and $A\subseteq [p,1]$ to set one of the two integral to $0$ but I still don't know how to proceed.

Actually thinking about how I could use it in my research it would be much more powerful to not enforce $x\leq p \leq y$ but just that $p$ is in the convex hull of $\{ x,y \}$ and then we get repetition $\lambda_x^y=\lambda_y^x$ but the of extreme points is the same. This makes $\kappa$ a probability measure over $[0,1]^2$ where we have to be careful assigning $0$ probability to sets $A\times B\subseteq [0,1]^2$ where the convex hull of $A\cup B$ does not contain $p$.

Here is the last thing I tried :
Denote $a_x^y=\frac{p-y}{x-y}$, for $\mu \in\mathcal M$ and $A$ measurable
\begin{align*}
\mu(A)&=\int_{[0,p)} \delta_x(A) d\mu(x)+\int_{(p,1]} \delta_y(A) d\mu(y)+\mu(\{ p \})\delta_p(A)\\
&=\int_{[0,p)\times (p,1]} \left(\frac{a_x^y}{a_x^y}\delta_x(A) + \frac{1-a_x^y}{1-a_x^y}\delta_y(A) \right)d\mu\otimes\mu(x,y)+\mu(\{ p \})\delta_p(A)\\
\end{align*}
It feels like we can define $\kappa$ as a function of $\mu\otimes \mu$ and $a_x^y$, I cannot finish the argument though.

答:
We can adapt the results in the paper "Extreme Points of Certain Sets of Probability Measures, with Applications" by A.F. Karr (https://www.jstor.org/stable/3689412) mentionned by @JohnDawkins to the following more general result.

Suppose that $X$ is a closed compact convex subset of a locally convex space $E$.

Definition. The subset $F\subseteq X$ is called a face if for all $x\in F$ such that there is $y,z\in X$ and $\lambda\in(0,1)$ with $x=\lambda y+(1-\lambda)z$, then $y,z\in F$.

One can show that faces are closed compact convex subsets of $X$ and that $x$ is extreme in $X$ if and only if $\{ x \}$ is a face. Indeed if $x$ is extreme and $\{ x \}$ is a face is really the same statement. We are now ready to characterize the extreme points of a linear subspace of $X$. Let $\mathcal F(A)$ be the set of faces of $A$.

Lemma. Let $f$ be an affine functional from $X$ to some other vector topological space $Y$. Define $X_f=\{ x\in X:f(x)=0\}$, then $\mathcal F(X_f)=\{ F\cap X_f:F\in\mathcal F(X) \}$.

Corrolary. $x$ is an extreme point of $X_f$ if and only if there exist a face $F\in\mathcal F(X)$ such that $F\cap X_f=\{x\}$.

Suppose that $F\in\mathcal F(X)$, then for $x\in F\cap X_f$, $y,z\in X_f$ and $\lambda\in(0,1)$ such that $x=\lambda y+(1-\lambda)z$, we know that $y,z\in F$ which means that $x,z\in F\cap X_f$ and so $F\cap X_f$ is a face.

Suppose that $F\in\mathcal F(X_f)$ and let $G=\{ y\in X : \exists x\in F,z\in X,\lambda\in(0,1):x=\lambda y+(1-\lambda)z \}$, then we can show that $F=G\cap X_f$ and $G\in \mathcal F(X)$ which finishes the proof. To show the first statement, it is clear that any $y\in F$ is also in $G\cap X_f$ by taking $x=z=y$ and any $\lambda$. Now take $y\in G\cap X_f$, then there is $x\in F$, $z\in X$ and $\lambda\in(0,1)$ such that $x=\lambda y+(1-\lambda)z$, and since $\lambda\neq 1$, $z\in X_f$. But since $y$ and $z$ are in $X_f$, then they are in $F$ and in particular $y\in F$.

In order to show that $G\in\mathcal F(X)$, we take any $x\in G$ and any $y,z\in X$, $\lambda\in (0,1)$ such that $x=\lambda y+(1-\lambda)z$. Since $x\in G$ there is $x'\in F$, $z'\in X$ and $\lambda'$ such that \begin{align*}
x'&=\lambda'x+(1-\lambda')z'\\
&=\lambda'\lambda y+\lambda'(1-\lambda)z+(1-\lambda')z'\\
&=\alpha y+(1-\alpha)\tilde z
\end{align*}
with $\alpha=\lambda' \lambda\in(0,1)$ and $\tilde z =\frac{\lambda'(1-\lambda )z+(1-\lambda')z'}{1-\alpha}\in X$, this implies that $y\in G$ and similarly we can write $x'=\beta z+(1-\beta)\tilde y$ with $\beta=\lambda'(1-\lambda)\in(0,1)$ and $\tilde y=\frac{\lambda'\lambda y+(1-\lambda')z'}{1-\beta}\in X$ and so $z\in G$, so $G\in\mathcal F(X)$. $\square$

In my case we can have $X$ to be the set of probability measures over $[0,1]$ and if $f:\mu\rightarrow \int xd\mu(x)-p$ then $\mathcal M=X_f$. One can show that for any Borel measurable $A\subseteq [0,1]$, $F_A=\{ \int_A \delta_x d\mu(x):\mu(A)=1 \}$ is a face (it is all measures with support inside $A$) and all faces are of this form. Now we can also show that $F_A\cap X_f=\{ x\}$ if and only if $1\leq|A|\leq 2$ and $p$ is in the convex hull of $A$. This indeed gives the set that I mentioned in the question.


问:
Would it ever be acceptable to omit the constant of integration? Particularly in coding.
In my coding (python if that makes a difference) lecture we have been given a recursive definition for a sequence a(n), and then given f(x) = a(n) + a(n)^2 + a(n)^3 and asked to integrate it. In the examples the lecturer gave throughout the lecture they didn't always add a constant of integration, and when asked if we should include one he gave a vague answer as to make us think. He said we add a constant in mathematics when trying to describe the family of functions that when differentiated would give a specific set of functions.

I'm in two minds. I'm thinking maybe we shouldn't include a constant because we aren't integrating again so the constant wouldn't change anything and differentiating nothing and differentiating gives the same result. But also surely we should include a constant, or at least shouldn't be penalised for including one because that's the rules of mathematics

*edit, i forgot to say the integral is an indefinite integral

答:
The difference is between definite and indefinite integrals. Consider
$$\int f(x)\,dx =F(x)+C$$ and you must put the $C$.
Now
$$\int_a^b f(x)\,dx =\left( F(b)+C\right)-\left( F(a)+C\right)=F(b)-F(a)$$

Is this clear ?


答:
A reason why it is important to add a constant when we compute primitives: consider
$$I:=\int\frac{1}{t\ln(t)}\mathrm{d}t.$$
Using the (rough version of) integration by parts formula $\int u'v=uv-\int uv'$ with $u'(t)=\frac{1}{t}$ and $v(t)=\frac{1}{\ln(t)}$, we get:
$$I=1-\int\frac{\ln(t)}{-\ln^2(t)}\frac{1}{t}\mathrm{d}t=1+I.$$
We "conclude" that $0=1$, which is true modulo an additive constant.

So we can do our computations without writing the constant, but we have to keep in mind that the result has to be understood modulo an additive constant.

Title: Would it ever be acceptable to omit the constant of integration? Particularly in coding.

Body:
In my coding (python if that makes a difference) lecture we have been given a recursive definition for a sequence a(n), and then given f(x) = a(n) + a(n)^2 + a(n)^3 and asked to integrate it. In the examples the lecturer gave throughout the lecture they didn't always add a constant of integration, and when asked if we should include one he gave a vague answer as to make us think. He said we add a constant in mathematics when trying to describe the family of functions that when differentiated would give a specific set of functions.

I'm in two minds. I'm thinking maybe we shouldn't include a constant because we aren't integrating again so the constant wouldn't change anything and differentiating nothing and differentiating gives the same result. But also surely we should include a constant, or at least shouldn't be penalised for including one because that's the rules of mathematics

*edit, i forgot to say the integral is an indefinite integral


问:
A problem about zero derivatives in the weak sense implying zero almost everywhere in $W_{0}^{1,1}$
In my PDE course, I encountered a problem stated below:

Exercise. Suppose that $\Omega \in \mathbb{R}^{n}$ is an open and bounded domain. Prove that if $u \in W_{0}^{1,1}(\Omega)$ and $\nabla u \equiv 0$ in $\Omega,$ then $u=0$ a.e. in $\Omega$.
Here $W_{0}^{1,1}(\Omega)$ is defined to be the closure of smooth functions of compact support in $W^{1,1}(\Omega)$ in $W^{1,1}(\Omega)$ norm.

I know that zero weak derivatives imply constant almost everywhere in the domain, but I don't know how to see $u=0$ a.e. in $\Omega$ using the assumption that  $u \in W_{0}^{1,1}(\Omega)$.

Is there a convincing proof?

答:
Let's make an example in the one dimensional case $I=(0,1)$. You already know that $u$ has to be constant, we will see that $u=0$ is the only constant possible. Assume not, then there exists a sequence $\varphi_n \in C_c^\infty(I)$ that converges to $u$ in $W^{1,1}$. Without loss of generality we assume $u=1$. The convergence of $\varphi$ to $1$ in $W^{1,1}$ means that $$\lim_{n\rightarrow \infty} ||\varphi_n-1||_{L^1} + ||\varphi_n' -0||_{L^1}=0.$$ In particular, by the positivity of the norm we see that $$\lim_{n\rightarrow \infty} ||\varphi_n'||_{L^1}=0.$$ But since $\varphi_n$ is compactly supported this is not possible. Let's take the following sequence as an illustrating example
$$\varphi_n(x) = \begin{cases} nx, &x\in (0,1/n), \\ 1, &x \in (1/n,1).\end{cases}$$
This function is neither smooth nor compactly supported but it is good enough for the purpose of demonstration (otherwise use tools like mollification to get a compactly supported smooth function). We see that
$$||\varphi_n'||_{L^1} = \int_0^{1/n} n \, \mathrm{d}x = 1.$$ This contradicts the observation made above. We conclude that $u=1$ is not possible. Similarly, we deduce the same for the other constants except $0$.

Summary: Constant functions as for example $u=1$ are not possible in your problem. The problem is that if you want to approximate $u=1$ by a compactly supported smooth function, then the function will be $1$ at most points but since it is compactly supported, it has to be zero somewhere. To get from $1$ to $0$ the function has to change direction, hence the function has a positive gradient. It is not possible to neglect this term. Indeed, if you try try to make the region where you go from $0$ to $1$ smaller, then, as a compensation the gradient gets bigger.

Primarily, this answer should give you an intuition why $u=0$ is the only possible answer. See gerw's comment about Poincaré's inequality for a more sophisticated answer.


答:
$f(x)=x^x$

take logarithm

$$\log f(x)=x \log x$$

differentiate both sides

$$\frac{f'(x)}{f(x)}= \log x + 1$$

$$f'(x)=f(x)(\log x+1)$$
$$f'(x)=x^x(\log x +1)$$

Title: Are there other uncommon calculation methods?

Body:
enter image description here

$\lim_{h\to0} \frac{（(x+h)^h-1）}{h}$

$\lim _{h\to0} \frac{(1+h/x)^{(x/h)h}}{h}$

What method can I use to get the result? I don't understand what method is used here.

I'm a little curious, is there any way.Honestly, I’m not sure because there is no other process here.

Thanks for your replies.I think I am missing something, this is a $\frac{0}{0}$,Can I not apply the L'Hospital's rule.I think how he used other methods to calculate this limit.

$t=(x+h)^{h}-1；h=\log _{(x+h)}{(t+1)}=\frac{\ln {(t+1)}}{\ln {(x+h)}};t\to0$

$\lim_{h\to0} \frac{（(x+h)^h-1）}{h}=\lim _{t\to0}{h\to0} \frac{t}{\log _{(x+h)}{(t+1)}}=\ln{(x+h)}\frac{t}{ln{(t+1)}}$

$\lim_{t\to0}(1+t)^{1/t}=e$

$\lim _{t\to0} \frac{t}{\ln(t+1)}=\frac{1}{\frac{\ln(t+1)}{t}}=\frac{1}{\ln((1+t)^{1/t})}=1$


问:
Boundedness of the integral
Let $f$ be a real-valued function that is Riemann-integrable in every closed, bounded subinterval of $(0, \infty)$. Suppose that the improper integral $\int_0^{\infty}f(x)dx$ exists. Then how can I prove that there exists a real number $ M > 0$ such that for all $ 0\leq a < b$, $|\int_a^{b}f(x)dx| < M$?

答:
The function $F(t)=\int_0^t f(x)\,dx$ is continuous by the hypothesis that $f$ is Riemann-integrable in every closed, bounded subinterval of $(0,\infty)$. Indeed, consider some $t_0>0$ and fix $\delta>0$ such that $$I_{t_0,\delta}=[t_0-\delta,t_0+\delta]\subseteq (0,\infty)$$
Since, $f$ is Riemann-integrable on $I_{t_0,\delta}$ by definition it should be bounded also in that interval. Hence, there is some $M_1>0$ (depending on $t_0,\delta$) such that $|f(x)|\leq M_1$ for every $x\in I_{t_0,\delta}$. Then, if $|t-t_0|<\delta$ we have
\begin{align}
|F(t)-F(t_0)|&=\biggl| \int_t^{t_0} f(x)\,dx \biggr|\\
&\leq \int_{t}^{t_0}|f(x)|\,dx\\
&\leq M_1|t_0-t|
\end{align}
Hence, $\lim_{t\to t_0}|F(t)-F(t_0)|=0$ which shows that $F$ is continuous at $t_0$. Since, $F$ is continuous on $[0,\infty)$ and $\lim_{t\to \infty}F(t)$ exists you can can conclude that there exists some $M>0$ such that
$$\biggl|\int_{\alpha}^\beta f(x)\,dx\biggr|\leq M$$
for every $\alpha<\beta$ (check this step).

Title: Boundedness of the integral


问:
Sum of probabilities of all events less than 1?
I solved this question from the Bertsekas probability book:

On a given day, your golf score takes values uniformly from the range
101 to 110 (i.e. the probability of having a score of 101, 102, . . . 110 are
all 0.1), independent of other days. Determined to improve your score,
you decide to play on three different days and declare as your score the
minimum X of the scores X1, X2, and X3 on the different days. Calculate
the PMF and the expectation of X.

I have calculated the PMF of X as follows:

$P(X=x) = \frac{(111-x)^3-(110-x)^3}{1000}$

Now that I compute the exact value of probabilities for each $x=101,...,110$, the sum of all of these probabilities doesn't add up to 1.

Am I supposed to divide each probability by the sum of them?

Hypothetically, if the sum of the probabilities were 0.75 and P(X=101)=0.3, should I divide 0.3 by 0.7 to reach the correct probability of P(X=101)? Because this way, the sum of all probabilities will add up to 1.

答:
I don't see a problem with your PMF for $x = 101, \ldots, 110$, maybe you just had a mistake when adding those PMF for all $x$? Anyway here's my rework.

For each of $x = 101, \ldots, 110$, first consider the probability that $X$ is at least $x$:

$$\begin{align*}
P(X\ge x) &= P(\min(X_1, X_2, X_3)\ge x)\\
&= P(\{X_1\ge x\}\cap\{X_2\ge x\} \cap \{ X_3\ge x\})\\
&= \prod_{i=1}^3P(X_i\ge x)\\
&= \prod_{i=1}^3 \frac{111-x}{10}\\
&= \frac{(111-x)^3}{1000}
\end{align*}$$

And also consider the probability that $X$ is strictly more than $x$:

$$\begin{align*}
P(X>x) &= P(X\ge x+1)\\
&= \begin{cases}\dfrac{(111-(x+1))^3}{1000}&x = 101,102,\ldots, 109\\
0&x= 110\end{cases}\\
&= \frac{(110-x)^3}{1000}
\end{align*}$$

So the PMF for $x = 101, \ldots, 110$, as you already found out, is

$$\begin{align*}
P(X=x) &= P(X\ge x) - P(X>x)\\
&= \frac{(111-x)^3 - (110-x)^3}{1000}
\end{align*}$$

As a check, the probability that $X$ is between $101$ and $110$ inclusive is

$$\begin{align*}
P(101\le X \le 110) &= \sum_{x=101}^{110}P(X=x)\\
&= \sum_{x=101}^{110} \frac{(111-x)^3 - (110-x)^3}{1000}\\
&= \frac{(10^3-9^3) + (9^3 - 8^3) + \cdots + (1^3 - 0^3)}{1000}\\
&= \frac{10^3 - 0^3}{1000}\\
&= 1
\end{align*}$$


答:
There are many ways we can interpret the statement
$$
\int f(x) \, dx = F(x)+C \, .
$$
In some treatments of calculus, all this means is that $\frac{d}{dx}\left(F(x)+C\right)=f(x)$, in which case the constant of integration is not strictly necessary. But often, $\int f(x) \, dx$ refers to a set of antiderivatives (i.e. a set of functions whose derivative is $f(x)$). This is what your lecturer was probably referring to. If antiderivatives are treated more formally, then writing
$$
\int f(x) \, dx = F(x)
$$
is just incorrect, even if people understand what you mean.

Although the constant of integration is often just an annoyance, there are cases when not writing it can cause problems (see Nicolas' answer). However, sometimes, you might only require one antiderivative (e.g. when trying to compute (definite) integrals). Although it is correct to write
$$
\int_{a}^{b} f(x) \, dx = (F(b)+C)-(F(a)+C)=F(b)-F(a) \, ,
$$
it is not necessary to write in a constant of integration, since it will be cancelled anyway. Since you only require one antiderivative to compute the integral, why not choose the simplest one?


问:
Prove that if $z$ in $\mathbb{C}$ is a solution of $z^5= 1$ then the conjugate of z is also a solution
I have to show that $(z^*)^5=1$, if $z^5=1$, but how do I do that?

I tried to show it by inserting $x-yi$ into the equation, so $(x-yi)^5=1$.

After using the binomial formula for a power of $5$, I thought I might be able to see similarities between the imaginary and real parts but I just got a complete mess.

I'd be happy if someone could help me.

答:
We can easily verify(by the definition of complex addition and multiplication) that
$$
\overline{w+z}=\bar{w}+\bar{z} \text { and } \overline{w z}=\bar{w} \bar{z}
$$

Here's a stronger version of your claim.

Theorem(Polynomials with real coefficients have roots in pairs). Suppose $p\in \mathcal P(\mathcal C)$ is a polynomial with real coefficients. If $\lambda \in \mathbb C$ is a root of $p$, then so is $\bar \lambda $.

Proof. Let
$$
p(z)=a_{0}+a_{1} z+\cdots+a_{m} z^{m}
$$
Assume $\lambda$ is a root of the polynomial, then
$$
a_{0}+a_{1} \lambda+\cdots+a_{m} \lambda^{m}=0
$$
Take the complex conjugate of both sides of this equation, obtaining
$$
a_{0}+a_{1} \bar{\lambda}+\cdots+a_{m} \bar{\lambda}^{m}=0
$$
so $\bar \lambda$ is also a root of $p$.

The polynomial mentioned in your case is $z^5=1$, thus for one of its roots $\lambda$, $\bar \lambda$ is also a root of $z^5=0$.


问:
Finding $y$ value of canonical ellipse.
I have an ellipse:
$$
 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
This may be a simple question, but my mind plays tricks on me at the moment;

Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?

Hope someone can help me - thanks in advance :)!

答:
You can rearrange $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ to $$y^2 = b^2(\frac{x^2}{a^2} - 1)$$

With knowledge of $a,b$ and $x$ you can evaluate $y^2$ and $$y_{1,2} = \pm \sqrt{y^2} = \pm \sqrt{b^2(\frac{x^2}{a^2} - 1)}$$

Title: Finding $y$ value of canonical ellipse.


问:
Simplification of multiplying fractions
Simplify $(1+\frac{1}{20\times22})(2+\frac{2}{21\times23})(2+\frac{2}{22\times24})(2+\frac{2}{23\times25})(13+\frac{13}{24\times26})$

I noticed the denominator of the five fractions are all in the form of:$$\frac{...}{(n-1)(n+1)}$$
Which can be simplified to:$$\frac{...}{n^2-1}$$

I futher attempted to move the integral part into the fraction, but it just got more complicated. How should I start from there?

Title: Simplification of multiplying fractions

Body:
$$m+\frac{m}{(n-1)(n+1)}=m\times(1+\frac{1}{(n-1)(n+1)})=m\times (\frac{n^2-1+1}{(n-1)(n+1)})=m\times\frac{n^2}{(n-1)(n+1)}$$

$$(1+\frac{1}{20\times22})(2+\frac{2}{21\times23})(2+\frac{2}{22\times24})(2+\frac{2}{23\times25})(13+\frac{13}{24\times26})=\frac{21^2}{20\times 22}\times 2\times \frac{22^2}{21\times 23}\times 2\times\frac{23^2}{22\times 24}\times 2\times \frac{24^2}{23\times 25}\times 13 \times \frac{25^2}{24\times 26}=105$$


答: Using the polar form we can write the fifth roots of unity as $z = e^{2k\pi i /5}$ with $k$ an integer and $-2 \leq k \leq 2$ which has conjugate $z^* = e^{-2k\pi i /5}$. Then we calculate that $$(z^*)^5 = (e^{-2k\pi i /5})^5 = e^{-2k \pi i} = (e^{2k \pi i })^{-1} = 1^{-1}=1$$ and we are done. Alternatively we can repeatedly use that $x^*y^*=(xy)^*$ by expanding $(z^*)^5$.

Title: Prove that if $z$ in $\mathbb{C}$ is a solution of $z^5= 1$ then the conjugate of z is also a solution

Body:
I have to show that $(z^*)^5=1$, if $z^5=1$, but how do I do that?

I tried to show it by inserting $x-yi$ into the equation, so $(x-yi)^5=1$.

After using the binomial formula for a power of $5$, I thought I might be able to see similarities between the imaginary and real parts but I just got a complete mess.

I'd be happy if someone could help me.


答:
What you can use is the general fact $$\overline{zw} = \overline z\cdot \overline w.$$ If you want to prove it, just let $z = a+ib$, $w = x + iy$ and calculate both sides.
By induction, it then follows $$\overline {z^n} = \overline z^n$$ for any $n\in\mathbb N.$ You can use it to prove the statement you need.

More generally, if you combine the above with the fact $$\overline{z+w} = \overline z + \overline w,$$ it follows that if $p$ is a polynomial with real coefficients, then $$\overline{p(z)} = p(\overline{z}),$$ which proves that $z$ is a root of $p$ if and only if $\overline{z}$ is. The key is that the coefficients of $p$ have to be real, so they are not affected by complex conjugation.


问:
Predicate calculus question
I have got the next structure: $M =<${1,2,3,4,5}$,S^M>$ such that: $S^M$ = {$<x,y>\mid y-x=1$}. Need to find the truth value of the next sentence: $$\exists x \forall y \exists z[\lnot [S(x,z)\lor x=z\ \lor S(y,z)\ \lor y=z]]$$

How do you suggest to do that?

Thanks for the help!

答:
First of all note that $\phi = \exists x \forall y \exists z(\lnot (S(x,z)\lor x=z\ \lor S(y,z)\ \lor y=z))$ is a sentence, hence its truth set can be either

$M^3$ if $M$ satisfies $\phi$; or

the empty set if $M$ doesn't satisfy $\phi$.

In the first case the truth value is $1$, $0$ otherwise. Is it true that $\exists x \forall y \exists z(\lnot (S(x,z)\lor x=z\ \lor S(y,z)\ \lor y=z))$? Well, I think so. Fix $x=1$. If you choose $y=1,2,3$ I pick $z=3,4,5$; if you choose $y=4,5$ I pick $z=3,3$ and $\lnot (S(x,z)\lor x=z\ \lor S(y,z)\ \lor y=z))$ is satisfied.

Title: Predicate calculus question


问:
Riccati equation $y'+\frac{y^2}{x}=1$
I have a Riccati equation:
$$y'+\frac{y^2}{x}=1$$
How can I solve it? Where do I begin?

答:
[First] :  Rewrite the ODE in the form of :
$$
y'(x)=p(x)+q(x)y(x)+r(x)(y(x))^2\implies y'=1-\frac{y^2}{x}
$$
Where $p(x)=1$, $q(x)=0$, and $r(x)=-\frac{1}{x}$

[Second] : Let $y(x)=-\frac{u'(x)}{r(x)u(x)}$, we obtain :
$$
r(x)u''(x)-\left(r'(x)+q(x)r(x)\right)u'(x)+p(x)(r(x))^2u(x)=0\implies-\frac{1}{x}u''+\frac{1}{x^{2}}u'+\frac{1}{x^{2}}u=0
$$
That is assuming $x\neq0$, we get :
$$
xu''-u'-u=0
$$
You can now proceed from here.

Title: Riccati equation $y'+\frac{y^2}{x}=1$


问:
Linearity of derivative
Let $U \subset \mathbb{R}^{m}$ be an open set and $f: U \to \mathbb{R}^{n}$. We say that $f$ is differentiable at $x \in U$ if there exists a linear map $T: \mathbb{R}^{m}\to \mathbb{R}^{n}$ such that:
$$f(x+h)=f(x)+T(h)+r(h)$$
where $\lim_{h\to 0}r(h)/|h| = 0$. The above equality holds provided $x+h \in U$. From now on, $T$ will be denoted by $f'(x)$.

Now, my advanced calculus book states the following: if $f$ is differentiable at $x \in U$ then, for every $h \in \mathbb{R}^{m}$ we have:
$$f'(x)(h) = \frac{f'(x)(th)}{t} = \frac{f(x+th)-f(x)}{t} \pm \frac{r(th)}{|th|}|h| $$
for every $t \neq 0$ real. Thus:
\begin{eqnarray}
f'(x)(h) = \lim_{t\to 0}\frac{f(x+th)-f(x)}{t}. \tag{1}\label{1}
\end{eqnarray}

Question: $f'(x)$ is supposed to be linear in $h$ and, at a first glance, (\ref{1}) does'nt seem linear in $h$ to me. How can I prove $f'(x)$ is indeed linear in $h$ using (\ref{1})?

答:
You can't. You state in your question that differentiability requires the existence of a linear $T$ such that
$$ f(x+h) = f(x) + Th + o(h), h \to 0 $$
If that holds (that is if $f'(x) = T$ is linear), then $f'(x)$ is given by (1). The other direction is not true, you can't prove from the existence of the limits in (1) that $f'(x)$ is linear.

Title: Linearity of derivative


问:
Prove if $\nabla f(x_0) = 0$ and $\nabla^2 f(x_0)$ is positive definite, then x$_0$ is a point of local minimum
Let $f: \mathbb R ^n\to\mathbb R$ be a differentiable function. </p>

<p>If $f$ is twice differentiable, and there exists a point $x_0\in\mathbb R^n$ such that $\nabla f(x_0) = 0$ and $\nabla^2f(x_0)$ is positive definite, then $x_0$ is a point of local minimum. </p>

答:
In elementary calculus for a function to have a local minimum the following condition must hold around $x_0$
$$f(x_0+\epsilon )\gt f(x_0)\Rightarrow f(x_0+\epsilon )- f(x_0)\gt 0$$
By Taylor's approximation we have
$$f(x_0+\epsilon)\approx f(x_0)+\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2+O(\epsilon^3)$$
We can substitute the approximation of $f(x_0+\epsilon)$ to the definition of minimum
$$f(x_0+\epsilon )- f(x_0)=f(x_0)+\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2-f(x_0)\gt 0$$
$$\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2\gt 0$$
We can generalize it to n variables such as
$$\nabla f(x_0)\vec \epsilon+ \vec \epsilon^T \frac 12\nabla^2 f(x_0)\vec \epsilon\gt 0$$
Since it is given that $\nabla f(x_0)=0$ the condition reduces to
$$\vec \epsilon^T \nabla^2 f(x_0)\vec \epsilon\gt 0$$
Since the term $\vec \epsilon^T (.)\vec \epsilon\gt 0$ is always positive definite for every $\epsilon$ the condition for local minimum reduces to
$$\nabla^2 f(x_0)\gt 0$$</p>


问:
If $(a,b)=1$ and $a|c$ and $b|c$ then $ab|c$
I'm studying unique factorization domains and I was able to prove that in a UFD if $gcd(a,b)=1$ and $a|bc$ then $a|c$ with the help of the experts here Prove general Euclid's Lemma in a UFD using prime factorization

Now since I'm taking a number theory class too, I've learned this lemma: if $gcd(a,b)=1$ and $a|c$ and $b|c$ then $ab|c$.

My question:
Does this lemma hold in a UFD too??
If not can you explain why?
It sounds natural to work but I don't know how to prove it....

答:
Yes, this holds in any UFD as well. Write down

$$a=p_1\cdots p_n$$
$$b=q_1\cdots q_m$$
$$c=z_1\cdots z_t$$

where each element on the right side is prime/irreducible.

Then $a|c$ implies that for each $p_i$ there is $z_j$ such that $p_i\equiv z_j$ (i.e. equal up to invertible). And so we have an injective function $A:\{1,\ldots,n\}\to\{1,\ldots, t\}$ with the property that $p_i\equiv z_{A(i)}$.

The same works for $b|c$ relationship, i.e. for any $q_i$ there is $z_j$ such that $q_i\equiv z_j$, giving us an injective function $B:\{1,\ldots, m\}\to\{1,\ldots,t\}$ with the property $q_i\equiv z_{B(i)}$.

Now $gcd(a,b)=1$ implies that $im(A)\cap im(B)=\emptyset$. Indeed, otherwise, if $A(i)=B(j)$ then $p_i\equiv z_{A(i)}\equiv z_{B(j)}\equiv q_{j}$ would be a common divisor of both $a$ and $b$. Therefore we have an injective function

$$A\sqcup B:\{1,\ldots,n,n+1\ldots,n+m\}\to\{1,\ldots, t\}$$
$$A\sqcup B(k)=\begin{cases}
A(k) &\text{if }k\leq n \\
B(k-n) &\text{otherwise}
\end{cases}$$

Which gives us that $z_i\equiv p_{A\sqcup B(i)}$ when $i\leq n$ and $z_{i-n}\equiv q_{A\sqcup B(i-n)}$ when $i> n$. This means that the decomposition of $a$ and the decomposition of $b$ can be found (disjointly) in the decomposition of $c$. Meaning $ab|c$.


问:
Unboundedness in linear programming
Let $K = \{ x :  Ax \le b, x \ge 0 \}$, $b \ge 0$. If there are solutions $x \ge 0$ different from zero of $A x \le 0$, show that $K$ is unbounded.(A is mxn, x is 1xn b is 1xm matrix.)

答:
Let $x_0 \ge 0$ be a non-zero solution of $Ax \le 0$. Then, for each $t \ge 0$, we have $tx_0 \ge 0$ and
$$ A(tx_0) = tAx_0 \le 0 \le b $$
hence $tx_0 \in K$. As
$$ \| tx_0 \| = t\|x_0\| \to \infty, \quad t \to \infty $$
$K$ is unbounded.

Title: Unboundedness in linear programming


问:
Prove $\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$
I am trying to prove that the following sum equals to zero:
$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$$
This actually surprised me a-lot. Usually when these kinds of infinite sums equal to zero, it's because of some sort of symmetry, or something like that. But I can't see it here. The $n=0$ term isn't zero, and we don't have $n\mapsto -n$ symmetry. The only thing I thought of is that maybe the Taylor series of $\arctan(x)$ might help, but the sum there starts from $n=1$ (or $n=0$ if you look at $\arctan(x)/x^2$), and needs to be integrated. It was kind of complicated (maybe it's not and I missed something).

P.S. - The original thing I need to compute is the following integral $I$, which I proved that is connected to the mentioned sum through the Residue Theorem:

$$I=\int_{0}^{2\pi}e^{i\theta}\sec{(e^{-i\theta})}\text{d}\theta=\frac 8\pi \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(2n+1)^2}=0$$

I added this because maybe the integral might be useful somehow (maybe easier to compute). Again I searched for symmetry (the integrand is clearly $2\pi$ symmetric, so I changed the integration interval to be $[-\pi,\pi]$, and then hoped that the integrand would be odd or something. But it's not, if I'm not mistaken).

Thanks

Body:
Let us check some terms:

For $n=-1$ and $n=0$ we get the terms of the series:
$$
\frac{(-1)^{-1}}{(-2+1)^2}=-\frac 1{1^2}
\text{ and }
\frac{(-1)^{0}}{(0+1)^2}=+\frac 1{1^2}\ .
$$

For $n=-2$ and $n=1$ we get the terms of the series:
$$
\frac{(-1)^{-2}}{(-4+1)^2}=+\frac 1{3^2}
\text{ and }
\frac{(-1)^{1}}{(2+1)^2}=-\frac 1{3^2}\ .
$$

For $n=-3$ and $n=2$ we get the terms of the series:
$$
\frac{(-1)^{-3}}{(-6+1)^2}=-\frac 1{5^2}
\text{ and }
\frac{(-1)^{2}}{(4+1)^2}=+\frac 1{5^2}\ .
$$

For $n=-4$ and $n=3$ we get the terms of the series:
$$
\frac{(-1)^{-4}}{(-8+1)^2}=+\frac 1{7^2}
\text{ and }
\frac{(-1)^{3}}{(6+1)^2}=-\frac 1{7^2}\ .
$$
The cancelling scheme should be clear now.


答: Alternatively, $I=-i\oint_{|z|=1}z^{-2}\sec zdz=0$ because $\sec z$ has no poles in the contour, and the $z^{-2}$ factor doesn't contribute either. (The $-$ sign comes from the contour being clockwise.)

Title: Prove $\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$

Body:
I am trying to prove that the following sum equals to zero:
$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$$
This actually surprised me a-lot. Usually when these kinds of infinite sums equal to zero, it's because of some sort of symmetry, or something like that. But I can't see it here. The $n=0$ term isn't zero, and we don't have $n\mapsto -n$ symmetry. The only thing I thought of is that maybe the Taylor series of $\arctan(x)$ might help, but the sum there starts from $n=1$ (or $n=0$ if you look at $\arctan(x)/x^2$), and needs to be integrated. It was kind of complicated (maybe it's not and I missed something).

P.S. - The original thing I need to compute is the following integral $I$, which I proved that is connected to the mentioned sum through the Residue Theorem:

$$I=\int_{0}^{2\pi}e^{i\theta}\sec{(e^{-i\theta})}\text{d}\theta=\frac 8\pi \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(2n+1)^2}=0$$

I added this because maybe the integral might be useful somehow (maybe easier to compute). Again I searched for symmetry (the integrand is clearly $2\pi$ symmetric, so I changed the integration interval to be $[-\pi,\pi]$, and then hoped that the integrand would be odd or something. But it's not, if I'm not mistaken).

Thanks


问:
Is $f(x,y) = x + y$ surjective?
I would like to know if the multivariable function $f(x,y) = x+y$ is surjective. I think I've figured out it is not injective as $g(2,3)$ and $g(5,0)$ both map to $5$, but I am not sure how to check for surjection on a multivariable function.

Thanks in advance.

答: Hint: $(\forall x\in\Bbb R):f(x,0)=x$.

Title: Is $f(x,y) = x + y$ surjective?

Body: Hint: $(\forall x\in\Bbb R):f(x,0)=x$.


问:
longest increasing subsequence with at most k deletion
Given a number array of length n and a number k, what is the longest non-decreasing contiguous subarray if we allow at most k deletions in our original array?

example:
n, k = 5, 2
array = 5 2 1 3 4

answer = 2 3 4
with deleting 1 from index 2.

答:
Suppose the initial array is named $R$.
The problem can be solved by storing $k+1$ lists $L_i, 0\leq i \leq k$. Element $L_i[t]$ denotes the length of the longest non-decreasing subarray ending with (and using) element $t$ of $R$, with at most $i$ deletions. These lists can be filled in a dynamic-programming fashion. $L_i[0]=1$, for all $i$. Suppose all lists $L_i$ are filled up to index $t$. Then we can determine $L_i[t+1]$ for all $i$ by simple reasoning.

For example: if $R[t+1]<R[t]$ then $L_0[t+1]=1$ because we cannot append element $t+1$ to the subarray ending at element $t$. Otherwise, $L_0[t+1]=L_0[t]+1$.

For $L_1[t+1]$ we need to take the maximum of two possibilities: either we skip element $R[t]$ and end up with a subarray of length $L_0[t-1]+1$ (if $R[t-1]\leq R[t+1]$), or we allow an earlier skip and use $L_1[t]+1$ (if $R[t]\leq R[t+1]$).

Filling in $L_i[t+1]$ for $i>1$ is completely analogous, and requires to take the maximum over the lengths reached by the $i+1$ possible skips. The longest subarray with at most $k$ skips is then equal to the maximal value found in the list $L_k$.


答:
Hint

You can use the definition. Take two real numbers $x_0$ and $c$. Now, pick $y_0=c-x_0$.

Title: Is $f(x,y) = x + y$ surjective?

Body:
I would like to know if the multivariable function $f(x,y) = x+y$ is surjective. I think I've figured out it is not injective as $g(2,3)$ and $g(5,0)$ both map to $5$, but I am not sure how to check for surjection on a multivariable function.

Thanks in advance.


答:
You can plot the summands:

So evidence suggests there are opposing terms, it's just that they're symmetric about $n=-1/2$ not $n=0$. Symbolically,
$$\sum_{n=-\infty}^{-1}\frac{(-1)^n}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}.$$
Now just change the index of the first sum,
$$-\sum_{n=-\infty}^0\frac{(-1)^n}{(2n-1)^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}.$$
Hopefully you can see why this equals zero now.


问:
Solve integral $\int^1_0 \frac{1-x^2}{{(1+x^2)}\sqrt{1+x^4}}dx$ using subsituition $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$
Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with different substitutions

The problem is when using the given substitution I end up with integral

$$\int^{\pi/4}_0 \frac{\sin{\theta}}{2x}d\theta $$

Solving for $x$ : $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$

According to WolframAlpha: $$x = \frac{|\csc{\theta}||1 \pm \sqrt{\cos{2\theta}}|}{\sqrt{2}} $$

I do know the sign for x is $+ve$ so I'll take $+ve$ solutions after opening the modulus, I do not know which sign to prefer in  $ \pm \sqrt{\cos{2\theta}} $  term

Eitherway, I solved for both cases :

$$  \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 + \sqrt{\cos{2\theta}}} d\theta -(I) $$
$$  \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 - \sqrt{\cos{2\theta}}} d\theta -(II) $$

Which leads to :
$$  \frac{1}{\sqrt{2}}\int^{\pi/4}_0 1 \pm \sqrt{\cos{2\theta}} d\theta = \frac{\pi}{4\sqrt{2}}
\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}} $$

Here I cheated and used WolframAlpha as I was exhausted.

Both the cases lead to a similar answer with difference of a function unknown to me: $\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}}$

The correct answer is :

$$\frac{\pi}{4\sqrt{2}}$$

My main questions are:

What am I doing wrong?

How did author arrive at this ingenious  substitution $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ , this doesn't appear to me trial n' error kind of substitution, I have spent hours doing algebraic transformation on this and no matter how you procced and plug the variable your integrand will only have one of these terms only $2x$, $1-x^2$, $1+x^2$ and $ \sqrt{1+x^4}$ and $2x$ being the simplest.

Related question: How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$

答:
Let's begin with the substitution suggested in the question. We have $$\cos t=\frac{\sqrt{1+x^4}} {1+x^2}$$ ($\theta$ is replaced by $t$ to reduce burden of typing mathjax). Differentiating we get
\begin{align}
-\sin t \, dt&=\dfrac {(1+x^2)\cdot \dfrac{4x^3} {2\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\
&=\dfrac {(1+x^2)\cdot \dfrac{2x^3} {\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\
&= 2x\cdot\dfrac {(1+x^2)\cdot \dfrac{x^2} {\sqrt{1+x^4}} - \sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\
&= 2x\dfrac { \dfrac{x^2(1+x^2)-(1+x^4)} {\sqrt{1+x^4}}} {(1+x^2)^2} \, dx\notag \\
&= 2x\dfrac { \dfrac{x^2-1} {\sqrt{1+x^4}}} {(1+x^2)^2} \, dx\notag \\
\implies \sin t \, dt&=\frac{2x(1-x^2)}{(1+x^2)^2\sqrt{1+x^4}}\, dx\notag\\
\end{align}
You need to double check your calculations as there is no cancellation in above to reduce the $(1+x^2)^2$ in denominator to $1+x^2$.

Next we note that $$\sin t =\frac{\sqrt {2}x}{1+x^2}$$ and therefore $$dt=\sqrt{2}\cdot\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx$$ or $$\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx=\frac{dt}{\sqrt{2}}$$ Integrating the above (noting that interval $[0,1]$ for $x$ maps to $[0,\pi/4]$ for $t$) we get $$I=\int_0^1\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx=\frac{1}{\sqrt {2}}\int_0^{\pi/4}\,dt=\frac{\pi}{4\sqrt{2}}$$


问:
$ \int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t$ is Convergent
Define $$I= \int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2} \mathrm{d}t.$$ Balazard, Saias and Yor showed that the Riemann Hypothesis is equivalent to the statement that $I=0$.

I want to prove that I is convergent.

My try:-

$$\zeta(s)= s\int_1^\infty \frac{1-x+[x]}{x^{s+1}}dx +\frac{1}{s-1}, \Re(s)>0$$

So $$\mid \zeta(1/2+it)|\leq 2\sqrt{1/4+t^2}+\frac{1}{\sqrt{1/4+t^2}}$$

$$\mid \zeta(1/2+it)|\leq \frac{2(1/4+t^2)+1}{\sqrt{1/4+t^2}}$$

$$log \mid \zeta(1/2+it)\mid \leq log(\frac{3/2+t^2}{\sqrt{1/4+t^2}})$$

$$\int_{-\infty}^{\infty}\frac{log \mid \zeta(1/2+it)\mid}{1/4+t^2}dt \leq \int_{-\infty}^{\infty} \frac{log(\frac{3/2+t^2}{\sqrt{1/4+t^2}})}{1/4+t^2}dt$$
Since $$\int_{-\infty}^{\infty}\frac{ log (1/4+t^2)}{1/4+t^2}=0$$
$$\int_{-\infty}^{\infty}\frac{log \mid \zeta(1/2+it)\mid}{1/4+t^2}dt \leq \int_{-\infty}^{\infty} \frac{log({3/2+t^2})}{1/4+t^2}dt$$

The last integral converges  see this link.

答:
Note first that the OP proof above actually shows that $I$ cannot be $\infty$ but it doesn't show that $I$ converges or it is not $-\infty$ since $I$ is an oscillating integral with $\log |\zeta|=-\infty$ infinitely many times on the critical line of course.

From general results about Hardy spaces and some changes of variable, it actually follows immediately that $I$ is finite but much more is known, namely that

$0 \le I \le 0.146$

(and much tighter bounds can be easily found by better numerics) as from the results of Balazard, Saias and Yor we know that:

$I=2\pi \sum_{\Re \rho >1/2}\log \frac{|\rho|}{|1-\rho|}=\pi \sum_{\Re \rho >1/2}\log \frac{|\rho|^2}{|1-\rho|^2}$

But now if $\rho=\sigma+it, \frac{|\rho|^2}{|1-\rho|^2}=\frac{\sigma^2+t^2}{(1-\sigma)^2+t^2}=1+\frac{\sigma^2-(1-\sigma)^2}{(1-\sigma)^2+t^2}$, so we have

$ 1 < \frac{|\rho|^2}{|1-\rho|^2} < 1+\frac{1}{(1-\sigma)^2+t^2}$ whenever $\sigma >1/2$ hence $0 < \log \frac{|\rho|^2}{|1-\rho|^2} < \frac{1}{(1-\sigma)^2+t^2}=\frac{1}{|1-\rho|^2}$

But now it is well known (see Broughan, Equivalents if The Riemann Hypothesis, vol 1, page 35) that: $\sum_{\rho} \frac{1}{|\rho|^2} \le  .046191441$, where the sum is taken on all the non-trivial zeroes of zeta.

Since either $I=0$ if RH is true, or $0 <I < \pi \sum_{\Re \rho <1/2}\frac{1}{|1-\rho|^2}< \pi \times .046191441=0.145...$ if RH is false and the sum is non-void, we immediately see the claimed result; using that all the zeroes up to some huge $t$ are known to be on the critical line so not in the sum above, we can probably reduce the estimate of $I$ to a ridiculously low nonnegative number, showing the tightness of the equivalence.


问:
existence 3-regualar sub graph in 6- regular graph.
I want prove that:

6-regular graph with even vertices have 3-regular sub-graph.

if fact there is a 6-regular graph whit n vertexes that n is even. we want create a 3-regular sub-graph for it.

答: The disjoint union of two $K_7$'s satisfies the premises of the question – $6$-regular and even number of vertices – but has no spanning $3$-regular subgraph. This is easy to see by the handshaking lemma applied to each connected component.


问:
Residue at infinity and quotient of complex polynomials
A textbook problem: Say I have $f(z) = P(z)/Q(z)$, where $P(z) =a_0+a_1z+...+a_nz^n$ and $Q(z)=b_0+b_1z+...+b_mz^m$ are both polynomials such that $m\geq n+2$, and that $a_n\neq 0$, $b_m\neq0$. I want to show that $$\int_C f(z) dz=0$$
if all zeros of $Q$ are inside a simple closed contour $C$.

A solution is provided this way: We of course firstly compute

$$\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}=\frac{a_0z^{m-2}+a_1z^{m-3}+...+a_nz^{m-n-2}}{b_0z^m+b_1z^{m-1}+...+b_m}$$

The solution then says that "As $m-n-2\geq0$ so that the numerator is a polynomial, and since $b_m\neq 0$, this $\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$ is represented by a series of the form $d_0+d_1z+d_2z^2+...$." From here the result is clear by the theorem using residue at infinity, since this $d_i$ series has $z^{-1}$ coefficient $0$.

I have two questions:

How and why did it say that $\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$ is represented by a series of that form?

I try concluding in another way instead: From the above computation, $\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$ is clearly analytic at $0$ because $b_m\neq 0$ and the numerator is indeed a polynomial as above. Thus, the residue of $\frac{1}{z^2}\frac{P(1/z)}{Q(1/z)}$ at $0$, which is just the integral of it over any simple closed contour small enough only enclosing $0$ but not the zeros of the denominator (since they must be finitely many), is zero by Cauchy-Goursat. Is this argument valid? It may or may not be longer once I know the answer to question 1 above.

Thank you so much for the guidance!

答:
Some ideas:

The function $\;\cfrac1{z^2}\cfrac{P\left(\cfrac1z\right)}{Q\left(\cfrac1z\right)}\;$ is analytic at $\;z=0\;$ and thus it has a Taylor series centered at $\;z=0\;$ , as said there.

Google "Residue at infinity", there you'll be able to see that if $\;f\;$ is a meromorphic function on and inside a closed, simple curve $\;C\;$ with positive orientation in the complex plane, and its singularities inside $\;C\;$ are all simple poles, then

$$\frac1{2\pi i}\oint_C f(z)\,dz=\text{Res}_{z=0}\,\left(\frac1{z^2}f\left(\frac1z\right)\right)$$

The above is very handy and saves lots of work when trying to calculate the residues at many finite, simple poles to evaluate an integral. Instead of that, only one resiude!


答:
Note that

$$\sum_{n=-\infty}^\infty{(-1)^n\over(2n+1)^2}=\sum_{m\equiv1\text{ mod }4}{1\over m^2}-\sum_{m\equiv3\text{ mod }4}{1\over m^2}$$

and $m\equiv1$ mod $4$ if and only if $-m\equiv3$ mod $4$, so, since $m^2=(-m)^2$, the two sums on the right hand side are equal, hence cancel each other out.


问:
Determinant of Adjoint matrix
Suppose $A$ is an $n \times n$, $n > 0$ matrix with complex entries, and $A^*$ is the adjoint matrix of $A$. Show that $\det(A^*) = \overline{\det(A)}$.

My attempt:

By definition,

$$\det(A) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$

Similarly for $A^*$, by complete expansion,

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{\sigma_1 1} \cdots A_{\sigma_n n}}$$

Consider the product $A_{\sigma_1 1} \cdots A_{\sigma_n n}$. If $j = \sigma_i$, then $A_{\sigma_i i} = A_{j \sigma^{-1}_{j}}$. Since $\sigma$ is a bijection, both $i, j$ range over $1, ..., n$. So,

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma^{-1}) \overline {A_{1 \sigma^{-1}_1} \cdots A_{n \sigma^{-1}_n}}$$

However, $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\sigma^{-1})$. So:

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma^{-1}_1} \cdots A_{n \sigma^{-1}_n}}$$

As before, $\sigma^{-1}$ ranges over all degree n permutations just as $\sigma$ does. Thus,

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma_1} \cdots A_{n \sigma_n}}$$

$$\det(A^*) = \overline{\sum_{\sigma} (\operatorname{sgn} \sigma) A_{1 \sigma_1} \cdots A_{n \sigma_n}} = \overline{\det(A)}$$

QED.

Is this proof correct? I'm not sure because I derived it from the proof involving real entries. Any assistance is much appreciated.

答:
Why put everything together? It is easier to use some known results.

For $A\in \mathbb F^{n\times n}$, we have
$$\det A=\det A^T,$$
where $F$ is any field and $A^T$ is the transpose of $A$.

For $A\in \mathbb C^{n\times n}$, we have
$$\det \overline{A}=\overline{\det A}.$$
The reason is exactly as you wrote.
$$\begin{aligned}
\det \overline{A}&=\sum_{\sigma} \prod_{i} \overline{a_{i\sigma(i)}}\\
&=\overline{\sum_{\sigma} \prod_{i} a_{i\sigma(i)}}\\
&=\overline{\det A}.
\end{aligned}$$

The adjoint matrix you mean, is just $A^*=\overline{A^T}$.
It follows that
$$\det \overline{A^T}=\overline{\det A^T}=\overline{\det A}.$$

It is not necessary to proof $\det{A^T}=\det{A}$ again.

Title: Determinant of Adjoint matrix


问:
Prove that if $z$ in $\mathbb{C}$ is a solution of $z^5= 1$ then the conjugate of z is also a solution
I have to show that $(z^*)^5=1$, if $z^5=1$, but how do I do that?

I tried to show it by inserting $x-yi$ into the equation, so $(x-yi)^5=1$.

After using the binomial formula for a power of $5$, I thought I might be able to see similarities between the imaginary and real parts but I just got a complete mess.

I'd be happy if someone could help me.

答:
For a general polynomial with real coefficients

$a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$

If $z$ is a soltuion for this equation then $z*$ will also be a sloution for the equation.

Proof:

Let $z_1 = a_1+ib_1$ and $z_2 =a_2+ib_2$ be two arbitary complex numbers

Therefore $z_1* = a_1 - ib_1$ and $z_2* = a_2 - ib_2$

$z_1*+ z_2* = (a_1+b_1)-i(b_1+b_2) = (z_1+z_2)*$

$z_1*.z_2* = (a_1a_2-b_1b_2)-i(a_1b_2+a_2b_1)=(z_1.z_2)*$

Thus we have established the fact that

$z_1*+ z_2*=(z_1+z_2)*$ and
$z_1*.z_2* =(z_1.z_2)*$

If $z$ is the solution of $a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$,

$a_0z^0+a_1z^1+a_2z^2+...a_{n-1}z^{n-1}+a_nz^n = 0$

$a_0{z*}^0+a_1{z*}^1+a_2{z*}^2+...a_{n-1}{z*}^{n-1}+a_n{z*}^n$

$=(a_0z^0+a_1z^1+a_2z^2+...a_{n-1}z^{n-1}+a_nz^n)*$

$=0$*

$=0$

Thus,

For a general polynomial with real coefficients

$a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$

If $z$ is a soltuion for this equation then $z*$ will also be a sloution for the equation.

Now, talking specifically about $z^5=1$.

$(z*)(z*)(z*)(z*)(z*)$

$=(z.z.z.z.z)*$

$=(z^5)*$

$=1$

A very important tool here is ${z*}^k = {z^k}*
$

Body:
For a general polynomial with real coefficients

$a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$

If $z$ is a soltuion for this equation then $z*$ will also be a sloution for the equation.

Proof:

Let $z_1 = a_1+ib_1$ and $z_2 =a_2+ib_2$ be two arbitary complex numbers

Therefore $z_1* = a_1 - ib_1$ and $z_2* = a_2 - ib_2$

$z_1*+ z_2* = (a_1+b_1)-i(b_1+b_2) = (z_1+z_2)*$

$z_1*.z_2* = (a_1a_2-b_1b_2)-i(a_1b_2+a_2b_1)=(z_1.z_2)*$

Thus we have established the fact that

$z_1*+ z_2*=(z_1+z_2)*$ and
$z_1*.z_2* =(z_1.z_2)*$

If $z$ is the solution of $a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$,

$a_0z^0+a_1z^1+a_2z^2+...a_{n-1}z^{n-1}+a_nz^n = 0$

$a_0{z*}^0+a_1{z*}^1+a_2{z*}^2+...a_{n-1}{z*}^{n-1}+a_n{z*}^n$

$=(a_0z^0+a_1z^1+a_2z^2+...a_{n-1}z^{n-1}+a_nz^n)*$

$=0$*

$=0$

Thus,

For a general polynomial with real coefficients

$a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$

If $z$ is a soltuion for this equation then $z*$ will also be a sloution for the equation.

Now, talking specifically about $z^5=1$.

$(z*)(z*)(z*)(z*)(z*)$

$=(z.z.z.z.z)*$

$=(z^5)*$

$=1$

A very important tool here is ${z*}^k = {z^k}*<br>$


问:
what does it mean for a subset graph to be contructed from the power set of the vertices of a graph
Definition: The subset graph of a set $S$ is constructed as follows. The vertex set is the powerset $2^S$
and two vertices $X,Y \in 2^S$ are adjacent if they have nontrivial intersection i.e. $X\cap Y$ is non-empty.

I am not sure what this even means. Conceptually, how can you construct subgraph vertices from the power set, unless this definition is meant to introduce a new idea?

答:
The intuition is as follows. Using some set $S$, we create a graph, showing the subsets of this set $S$. For every possible subset of $S$ we create a vertex in this graph, and we connect two vertices only if they have at least one element in common.

Example for the set $S = \{1,2\}$:

```<code>(1,2) ----- (1)
  |
  |
 (2)        ()
</code>```

Body:
The intuition is as follows. Using some set $S$, we create a graph, showing the subsets of this set $S$. For every possible subset of $S$ we create a vertex in this graph, and we connect two vertices only if they have at least one element in common.

Example for the set $S = \{1,2\}$:

(1,2) ----- (1)
  |
  |
 (2)        ()


问:
Self-adjoint operator and inner products
Suppose $V$ is a nonzero finite dimensional vector space over $F = R$. If $T: V \rightarrow V$ is linear, and $B = \{v_1, ..., v_n\}$ is an ordered eigenbasis of $V$ with respect to $T$, show that there exists some inner product on $V$ where $T$ is self-adjoint.

My attempt:

By the orthogonality theorem, given $B = \{v_1, ..., v_n\}$, there is an orthonormal basis $\{u_1, ..., u_n\}$ of $V$.

By the spectral theorem, $V$ has an orthonormal basis with respect to $T$ if and only if $T$ is self-adjoint.

So, $[T]_B = [T*]_B \implies T = T*$.

From here, I'm not sure how to proceed. I can't think of how to prove that there is an inner product where $T$ is self-adjoint. Any tips or assistance is much appreciated!

答:
There is a unique inner product $\langle -,-\rangle$ on $V$ satisfying
$$
\langle v_i, v_j \rangle = \delta_{ij} = \begin{cases} 1 & \text{if $i=j$},\\ 0 & \text{otherwise.} \end{cases}
$$
With respect to this inner product, your basis $B$ is an orthonormal basis.

Since $B$ is also an eigenbasis of $T$, the spectral theorem yields that $T$ is self-adjoint with respect to $\langle - , -\rangle$.

Note that the basis $B=\{v_1,\dots,v_n\}$ of $V$ yields an isomorphism of vector spaces $\Phi\colon V\to \mathbb R^n$ such that $\Phi(v_i)=e_i$ is the $i$-th standard basis vector of $\mathbb R^n$ for all $i$. The inner product given above is now obtained by transferring the standard dot product from $\mathbb R^n$ along this isomorphism:
$$
\langle v, v' \rangle := \Phi(v)\cdot\Phi(v').
$$
We have
$$
\langle v_i, v_j\rangle = \Phi(v_i)\cdot\Phi(v_j) = e_i \cdot e_j = \delta_{ij}
$$
so this does indeed give the desired product.

Since the dot product is an inner product on $\mathbb R^n$, this product is an inner product on $V$, rendering $\Phi$ an isometry.


答:
Here's an example for $S=\{1,2,3\}$. There are $8=2^3$ vertices, labelled with all possible subsets of $\{1,2,3\}$. If two subsets have an element in common, there is an edge between the vertices with those labels. In particular this means vertex $\{\}$ is adjacent to nothing, but $\{1,2,3\}$ is adjacent to everything except $\{\}$.

(I would normally write $\varnothing$ rather than $\{\}$, but the software I made the image in couldn't cope with that!)

Title: what does it mean for a subset graph to be contructed from the power set of the vertices of a graph

Body:
Definition: The subset graph of a set $S$ is constructed as follows. The vertex set is the powerset $2^S$
and two vertices $X,Y \in 2^S$ are adjacent if they have nontrivial intersection i.e. $X\cap Y$ is non-empty.

I am not sure what this even means. Conceptually, how can you construct subgraph vertices from the power set, unless this definition is meant to introduce a new idea?


问:
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$
I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$

My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.

Edited: I found some variants of the original problem.

Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.

Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.

答:
By using Schur's inequality and the identity
\begin{align}
&(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca)\\
 =\ & a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b),
\end{align}
we have $(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca) \ge 0$ which results in
$$\frac{(a+b+c)^3 + 9abc }{4(a+b+c)} \ge ab + bc + ca. \tag{1}$$

We need to prove that $(a+b+c)^2 - 2(ab + bc + ca) + 6 \ge 3(a+b+c)$.
By using (1), it suffices to prove that
$$(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \ge 3(a+b+c)$$
that is (using $abc = 1$)
$$\frac{(a+b+c - 3)[(a+b+c)^2 - 3(a+b+c) + 3]}{2(a+b+c)} \ge 0$$
which is true (using $a+b + c \ge 3\sqrt[3]{abc} = 3$).

We are done.

Remark: Actually, it is the pqr method.


问:
Understanding some Galois theoretic identification of a quotient over some 3-adic extension
Let $K = \mathbb{Q}_3$. Let $L=K(\alpha)$ given by $\min_K(\alpha) = x^4-3x^2+18$. This extension is cyclic of degree $4$ with ramification index $2$. In particular, the intertial degree is also $2$. See also this page.

Let $L'/L$ be the unique unramified extension of degree $2$. Since $L'/L$ is generated by a primitive $5$-th root of unity $\zeta_5$, it is $L'= K(\alpha,\zeta_5)$.

If I am not wrong, the Galois group of $L'/K$ is
$$
\operatorname{Gal}(L'/K) \simeq C_4 \times C_2 = \langle \sigma' \rangle \times \langle \sigma \rangle.
$$

Suppose that $\sigma: \alpha \mapsto -\alpha, \, \zeta_5 \mapsto \zeta_5$ (which has order $2$) and $\sigma': \alpha \mapsto \alpha, \, \zeta_5 \mapsto \zeta_5^3$ (which has order $4$).

According to the Fundamental Theorem of Galois Theory, it is $Gal(L/K) = Gal(L'/K)/Gal(L'/L)$. A generator of $Gal(L'/L)$ is given by $\alpha \mapsto \alpha, \, \zeta_5 \mapsto \zeta_5^9 = \zeta_5^{-1}$.

If I look at the explicit elements of the quotient, they must be $\{ Id, \bar{\sigma}, \bar{\sigma'}, \bar{\sigma} \bar{\sigma'} \}$ (it is $(\sigma')^2 \in Gal(L'/L)$) But none of these elements seem to be a generator of the quotient. (if my computations are correct, all these elements have order $2$ at most). Since $L/K$ is indeed cyclic, this cannot be possible.

I assume I made a mistake with identifying my elements. Could you please say where my line of thought is wrong?

答:
You can show that $$L = \Bbb{Q}_3( \zeta_8, \zeta_{16} 3^{1/2}) \qquad \implies \qquad L' =   \Bbb{Q}_3( \zeta_{80}, 3^{1/2})$$
Then it is clear that
$$Gal(L'/\Bbb{Q}_3)=Gal(L'/\Bbb{Q}_3(3^{1/2}))\times Gal(L'/\Bbb{Q}_3(\zeta_{80}))= C_4\times C_2$$
and $Gal(L'/L) = \langle (2,1)\rangle$.

Any element of $Gal(L'/\Bbb{Q}_3)$ whose reduction is the Frobenius will have order $4$ both in $Gal(L'/\Bbb{Q}_3)$ and $Gal(L/\Bbb{Q}_3)$


问:
meaning of the notation $\mathbb{R}^{\omega}$
I have been self studying Topology from Wayne patty and I don't know meaning of this notation which came up in the theorem  2.50 on Page 92.

Statement of the theoerm: Let $\bar{d}$ be th estandard bounded metric on $\mathbb{R}$ and define D(x,y)=$ lub [\bar {d}( x_i ,y_i) /i ] $for all x,y $\in \mathbb{R}^{\omega}$ .

What is meaning of the notation $\mathbb{R}^{\omega}$ ?

答:
In general $$X^{\omega}=X\times X \times \cdots $$
is an infinite countable product of $X$, it's to say the set of all countable sequences over $X$.


答:
In your proof, when you apply the change of variables $\sigma^{-1}\mapsto\sigma$ in the sum, the coefficient $\operatorname{sgn}(\sigma)$ should change to $\operatorname{sgn}(\sigma^{-1})$. In fact, you don't need to artificially use $\operatorname{sgn}(\sigma^{-1})=\operatorname{sgn}(\sigma)$ before doing the change of variables.

A different proof would be: By definition $A^*:=(\overline A)^T$, where $\overline A$ denotes the componentwise conjugation. Then $\det(A^*)=\det((\overline A)^T)=\det(\overline A)$, and then, by linearity of conjugation,
$$\det(\overline A) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma_1} \cdots A_{n\sigma_n}} = \overline {\sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}}=\overline{\det A}$$

Title: Determinant of Adjoint matrix

Body:
Suppose $A$ is an $n \times n$, $n > 0$ matrix with complex entries, and $A^*$ is the adjoint matrix of $A$. Show that $\det(A^*) = \overline{\det(A)}$.

My attempt:

By definition,

$$\det(A) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$

Similarly for $A^*$, by complete expansion,

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{\sigma_1 1} \cdots A_{\sigma_n n}}$$

Consider the product $A_{\sigma_1 1} \cdots A_{\sigma_n n}$. If $j = \sigma_i$, then $A_{\sigma_i i} = A_{j \sigma^{-1}_{j}}$. Since $\sigma$ is a bijection, both $i, j$ range over $1, ..., n$. So,

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma^{-1}) \overline {A_{1 \sigma^{-1}_1} \cdots A_{n \sigma^{-1}_n}}$$

However, $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\sigma^{-1})$. So:

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma^{-1}_1} \cdots A_{n \sigma^{-1}_n}}$$

As before, $\sigma^{-1}$ ranges over all degree n permutations just as $\sigma$ does. Thus,

$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma_1} \cdots A_{n \sigma_n}}$$

$$\det(A^*) = \overline{\sum_{\sigma} (\operatorname{sgn} \sigma) A_{1 \sigma_1} \cdots A_{n \sigma_n}} = \overline{\det(A)}$$

QED.

Is this proof correct? I'm not sure because I derived it from the proof involving real entries. Any assistance is much appreciated.


问:
Isomorphism between (even, odd) and (positive, negative)
This is a problem in Conceptual Mathematics: A First Introduction to Categories

Find an isomorphism
$$
(\{{\rm odd}, {\rm even}\}, +) \stackrel{f}{\to} (\{{\rm positive}, {\rm negative} \}, \times)
$$
HINT: There are only two invertible maps of sets from $\{{\rm odd}, {\rm even}\}$ to $\{{\rm positive}, {\rm negative} \}$. One of them 'respects the combining rules', but the other doesn't

I've a little stumped with this question. This is one of the maps I've come up with

$$
f(x): \begin{cases}
-k & {\rm if} & x = 2k - 1, ~~k=1,2, ...\\
k & {\rm if} & x = 2k, ~~ k = 0, 1, 2,
\end{cases}
$$

This will map odd to positive numbers, and even to negative numbers, and is invertible. The issue is that it does not preserve the combining rules, e.g. if $x = 2m - 1$ is odd, and $y = 2n$ even

$$
f(x + y) = f(2(m + n) - 1) = -(m + n) \tag{1}
$$

but on the other hand

$$
f(x) \times f(y) = f(2m + 1) \times f(2n) = (-m) \times (n) = -mn \tag{2}
$$

Clearly

$$
f(x + y) \not= f(x)\times f(y)
$$

Any other suggestion?

答:
Let g be the map of sets from {odd,even} to {pos, neg} that sends odd to pos and even to neg.

g is an isomorphism for obvious reason (existence of an inverse)

Now, suppose you want to "add structure" to g and make it an isomorphism between $(\{{\rm odd}, {\rm even}\}, +)$ and $(\{{\rm pos}, {\rm neg} \}, \times)
$. Let's prove that the following equation doesn't not hold:

$g(odd + even) \stackrel{?}{=}  g(odd) * g(even)$

$g(odd + even)    = g(odd)$  (odd + even = odd, combining rule for elements in {odd,even})

```<code>             = pos             g(odd) = pos by construction of g
</code>```

g(odd) * g(even) = pos * neg       by construction of g

```<code>             = neg             (pos * neg = neg, combining rule in ({pos, neg}, *))
</code>```

g(odd+even) is different from g(odd) * g(even) that's why the combining rule is not followed by g.

Body:
Let g be the map of sets from {odd,even} to {pos, neg} that sends odd to pos and even to neg.

g is an isomorphism for obvious reason (existence of an inverse)

Now, suppose you want to "add structure" to g and make it an isomorphism between $(\{{\rm odd}, {\rm even}\}, +)$ and $(\{{\rm pos}, {\rm neg} \}, \times)
$. Let's prove that the following equation doesn't not hold:

$g(odd + even) \stackrel{?}{=}  g(odd) * g(even)$

$g(odd + even)    = g(odd)$  (odd + even = odd, combining rule for elements in {odd,even})

             = pos             g(odd) = pos by construction of g

g(odd) * g(even) = pos * neg       by construction of g

             = neg             (pos * neg = neg, combining rule in ({pos, neg}, *))

g(odd+even) is different from g(odd) * g(even) that's why the combining rule is not followed by g.


问:
Is $\mathbb Q[[x]]$ compact with the $(x)$-adic topology?
I'd like to understand what the $(x)$-adic topology looks like here. I know that $\mathbb Q[[x]]$ is the completion of $\mathbb Q[x]$.

I feel like this should not be compact. To show its not compact we want to find some open cover which cannot be refined to a finite sub cover. The open sents in this topology are $0+(x^n)$ and translations. It seems a good candidate for an open cover would be include something like $0+(x^p)$ with $p$ running through the primes but this is not an cover.

答: I believe the answer is negative. The $(x)$-adic topology is induced by a norm, so compactness is equivalent to sequentual compactness. Taking an enumeration of $\mathbb{Q}$, $<q_i>$ gives us a sequence in which every 2 elements are at a fixed distance from each other, so it can not have a convergent subsequence.


问:
Range of a function
Find the range of

$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$ where $a, b, c$ are distinct real numbers such $a\neq b\neq c\neq a$.

答:
Observe that $g(x) = f(x) - 1$ is a polynomial of degree $2$.

But $g(a) = g(b) = g(c) = 0$. Thus a quadratic has three distinct roots. This would imply that $g(x)$ is identically $0$. Therefore $f(x)=1$. And the range, therefore, is the singleton $\{1\}$.

Title: Range of a function


问:
Sin in Cross Product Formula
I know that $ |A\times B| = ||A|| ||B|| \sin(\theta)$,
wherein the case of 3D Vectors and I wanted to calculate $\sin(\theta)$ from the previous formula I will get a [1x3] vector as a result, my question is, can the 3 values in the resulted sin vector be treated as 3 separate angles with the primary axes, say $X, Y, Z$?

答: Your formula is not correct. It should be$$\|A\times B\|=\|A\|\|B\|\sin(\theta)$$and therefore, unless $A=(0,0,0)$ or $B=(0,0,0)$, you can compute $\sin\theta$ by doing$$\sin(\theta)=\frac{\|A\times B\|}{\|A\|\|B\|}.$$

Title: Sin in Cross Product Formula


问:
Show that this quasi-geodesic ray is not Gromov hyperbolic?
Consider the spiral (t, log(1+t)) (given in polar coordinates); it inherits the Euclidean metric from the plane. I have to show that this spiral (a quasi-geodesic) is not Gromov hyperbolic. In other words, it isn't  hyperbolic for any  > 0. I've been stuck on this problem for a while now. My professor suggested that I should use the 4-point condition, but I haven't gotten anywhere with that either. Any ideas?</p>

答:
Hints:

Step 1. Show that your spiral is Hausdorff-close to the spiral $S$ given in polar coordinates by the parameteric equation
$$
(r, \log(r)), r\ge 1.
$$

Step 2. Verify that the spiral $S$ is not 0-hyperbolic.

Step 3. Verify that the spiral $S$ maps into itself under the sequence of dilations
$$
f_n: (r, \theta)\mapsto (e^{2\pi n} r, \theta), n\in {\mathbb N},
$$
i.e. $f_n(S)\subset S$ for each $n$.

Step 4. Combine steps 2 and 3 to conclude that $S$ is not $\delta$-hyperbolic for any $\delta<\infty$.

Step 5. Make the same conclusion for the original spiral using Step 1 and your knowledge (I assume you know this staff) that the 4-point hyperbolicity is preserved under $(1,C)$-quasi-isometries.


答:
It is equal to $1$.

Actually this polynomial is the the approximate polynomial you get in  Lagrangian Interpolation for a curve at three data points a,b,c taking value $1$ and so $f(x)$ is identically equal to $1$ as this is $2nd$ order interpolation and it takes $1$ at all three points.

Title: Range of a function

Body:
Find the range of

$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$ where $a, b, c$ are distinct real numbers such $a\neq b\neq c\neq a$.


问:
Prove that $1,t-1,t^2-t,\cdots,t^{n-1}-t^{n-2}$ a basis for any polynomial $P_{n-1}$

Prove that $1,t-1,t^2-t,\cdots,t^{n-1}-t^{n-2}$ a basis for any polynomial $P_{n-1}$ represented by $a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}$

$$
p(t)=a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}=
$$

Would be helpful I someone can give me a hint on how to rearrange the terms to show that the given set is a basis ?

答:
The easiest way to show that some set is a basis for the set of polynomials, is to show that every element of the usual basis for the set of polynomials can be written as a sum of elements of your new basis.

In this case, it is therefore sufficient to show that $1, t, t^2, ..., t^{n-1}$ can be written as sums of elements of your new basis.

As $1 = 1, t = (t-1) + 1, t^2 = (t^2-t) + (t-1) + 1, ...$, this condition is satisfied, and therefore your new set is a basis for the set of polynomials.


问:
Solving $\cos(40+\theta) = 3\sin(50+\theta)$
Question
Solve the following equation for $0≤\theta≤180$
$$
\cos(40+\theta) = 3\sin(50+\theta)
$$
Hint: $\cos(40) = \sin(50)$

Solution

$$
\cos(40+\theta) = 3\sin(50+\theta)
$$
$$
\cos(40)\cos\theta -\sin(40)\sin\theta = 3(\sin(50)\cos\theta +\sin\theta \cos(50))
$$
$$
\sin(50)\cos\theta-\cos(50)\sin\theta = 3(\sin(50)\cos\theta+\sin\theta \cos(50))
$$
$$
2\sin(50)\cos\theta+4\sin\theta \cos(50) = 0
$$
What do I do next?

答:
Observe that $\cos(\theta)=0$ cannot give a solution. Rearranging from your last line as suggested by comments, we get $\tan(-50)=2\tan(\theta)$. Recall that $\tan(-150)=\dfrac1{\sqrt 3}$ and the trigonometric identity $\tan(3\phi)
=\dfrac{3\tan(\phi)-\tan^3(\phi)}{1-3\tan^2(\phi)}$. So $\dfrac{6\tan(\theta)-8\tan^3(\theta)}{1-12\tan^2(\theta)}=\dfrac1{\sqrt 3}\implies \dfrac1{\sqrt 3}-6\tan(\theta)-4\sqrt3\tan^2(\theta)+8\tan^3(\theta)=0$. We transform it into a monic depressed cubic in $\tan(\theta)$ as follows:

$\dfrac1{8\sqrt 3}-\dfrac34\tan(\theta)-\dfrac{\sqrt3}2\tan^2(\theta)+\tan^3(\theta)=0$. Let $x+\dfrac{\sqrt3}6=\tan(\theta),$ then the equation may be rewritten as $f(x)=x^3-x-\dfrac{\sqrt3}9=0$.

The discriminant of $f(x)$ is then $\delta^2=-4(-1)^3-27(-\dfrac{\sqrt3}9)^2=3$, and by Cardano formula, $\chi=\sqrt[3]{\dfrac{\sqrt3}{18}+\sqrt{\dfrac{-3}{108}}}+\sqrt[3]{\dfrac{\sqrt3}{18}-\sqrt{\dfrac{-3}{108}}}=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}$ is a root of $f$.

Observe that $f(\dfrac{\sqrt3}3)\gt 0,f(1)\lt 0, f(2)\gt 0, $ so by intermediate value theorem, $f$ has at least two real roots. However since $f(x)\in \Bbb R[x]$, if $\alpha$ is a non real root of $f$, then so is its complex conjugate $\bar{\alpha}\ne\alpha$. Therefore $f$ has three real roots, and from this we see that $\chi=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}\in \Bbb R$.

Write $p=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}, q=\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}, $ then $\chi=p+q\implies \chi^3=(p+q)^3=p^3+q^3+3pq(p+q)$. Now, $p^3+q^3=\dfrac {\sqrt 3}9, pq=\sqrt[3]{(\dfrac{\sqrt 3}{18})^2+\dfrac 1{36}}$. Suppose that $p+q<0$, then $|p+q|^3\lt 3pq|p+q|\implies (p+q)^2=p^2+2pq+q^2\lt 3pq \implies p^2+q^2\lt pq$, but by A.M.-G.M.inequality we have $p^2+q^2\gt 2|pq|=2pq$, which gives $0\lt 2pq\lt p^2+q^2\lt pq$, a contradiction.

Hence $\chi=p+q\gt 0$ and $\tan(\theta)=\chi+\dfrac{\sqrt3}6=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}+\dfrac{\sqrt3}6\gt 0$. Since $\arctan(\phi)\gt 0\iff \phi \gt 0, \theta=\arctan(\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}+\dfrac{\sqrt3}6) \gt 0$, and such $\theta$ is a desired solution.


答:
Write their trivial linear combination: $$a_11 +a_2(t-1)+a_3(t^2-t)+\cdots +a_n(t^{n-1}-t^{n-2})=0$$

so $$(a_1-a_2)1 + (a_2-a_3)t+(a_3-a_4)t^2+\cdots+(a_{n-1}-a_n)t^{n-2}+a_nt^{n-1}=0 $$

is valid for all $t$ which means that this polynomial is actually zero polynomial, so $$a_1-a_2=a_2-a_3=...= a_{n-1}-a_{n}=a_n=0$$

so all $a_i$ are zero and thus a conclusion.

Body:
Prove that $1,t-1,t^2-t,\cdots,t^{n-1}-t^{n-2}$ a basis for any polynomial $P_{n-1}$ represented by $a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}$

$$
p(t)=a_0+a_1t+a_2t^2+\cdots+a_{n-1}t^{n-1}=
$$

Would be helpful I someone can give me a hint on how to rearrange the terms to show that the given set is a basis ?


问:
Axiom of infinity exceptions?
It is my understanding that the axiom of infinity, or axioms that depend on the axiom of infinity, are the only axioms that are not representable in finite First-Order Predicate Logic.  Is this correct?  If so, how can the axiom of infinity be justified?  Furthermore, it is my understanding that the axiom of infinity requires infinite induction, which requires the axiom of infinity.  How is this circular definition resolved?  Is there a way for the axiom of infinity to be defined so that it does not require infinite induction?</p>

答:
The axiom of infinity can be represented perfectly well in first-order logic, for example by asserting the existence of an inductive set (a nonempty set closed under the successor relation, like $\omega$). This does not require "infinite induction": it's just an ordinary statement in the language of set theory stating that something satisfying a certain condition exists. There is no circularity in the definition or the axiom.

The question of justifying the axiom is a different matter. One approach would be to say simply that lots of mathematics we want to do requires this axiom, so we assume it. Most mathematicians are happy enough with this pragmatic position, although it leaves unanswered the question of whether the axiom of infinity, like the other axioms of set theory, is just an assumption under which we carry out mathematical reasoning or whether we believe that there really are an infinity of sets.

Title: Axiom of infinity exceptions?


问:
Show a complex function is local Lipschitz
Let $F(z)=z\left\lvert z\right\rvert^a$ for $a\geqslant0$. Show that it is local Lipischitz on $\mathbb{C}$ and that
$$\left\lvert F(z)-F(z')\right\rvert\leqslant C_a\left\lvert z-z'\right\rvert\left(\left\lvert z\right\rvert^a+\left\lvert z'\right\rvert^a\right)$$

My work
It is of class $\mathcal{C}^1$ on $\mathbb{C}\setminus\left\{0\right\}$ so it is local Lipschitz, however I cannot prove the inequality.

答:
Let $a>0$ as otherwise, the required inequality is trivial and also assume $zz' \ne 0$ for the same reason; applying the mean value theorem to $x^a$ on we get that if $0<x \le y$ then $|x^a-y^a|=y^a-x^a=ac^{a-1}|x-y| \le a\max \{x^{a-1}, y^{a-1}\}|x-y|$ since $c \in [x,y]$

So $|z|z|^a-z'|z'|^a|\le|z-z'||z|^a + |z'(|z|^a-|z'|^a)| \le |z-z'||z|^a+a||z|-|z'|||z'|\max \{|z|^{a-1}, |z'|^{a-1}\}\le |z-z'|(|z|^a+a|z'|\max \{|z|^{a-1}, |z'|^{a-1}\}$

Repeating same with $z,z'$ switched we get that:

$|z|z|^a-z'|z'|^a|\le |z-z'|(|z'|^a+a|z|\max \{|z|^{a-1}, |z'|^{a-1}\}$

But now clearly either $|z'|\max \{|z|^{a-1}, |z'|^{a-1}\}=|z'|^a$ or $|z|\max \{|z|^{a-1}, |z'|^{a-1}\}=|z|^a$, while $|z'|^a, |z|^a \le |z'|^a +|z|^a$  so in any event we get:

$|z|z|^a-z'|z'|^a|\le |z-z'|(a+1)(|z'|^a +|z|^a)$ so we can take $C_a=a+1$


答:
Note that $S=\{1, t, t^2, \dots, t^{n-1}\}$ is a bais of $P_{n-1}$.

Now consider the linear map $f\colon P_{n-1}\to P_{n-1}$ given by
$$
f(1) = 1,\ f(t) = t-1,\ f(t^2) = t^2-t,\ \dots,\ f(t^{n-1}) = t^{n-1} - t^{n-2}.
$$

The matrix of $f$ with respect to $S$ is obtained as
$$
\begin{pmatrix}
1 & -1 & 0 & \cdots & 0 \\
0 & 1 & -1 & \ddots & \vdots\\
0 & 0 & 1 & \ddots & 0 \\
\vdots & \vdots & \ddots  & \ddots & -1 \\
0  & 0 & \cdots & 0 & 1
\end{pmatrix}.
$$
Since this matrix is invertible as an upper triangular matrix with non-zero diagonal entries, the linear map $f$ is an automorphism.

Hence, $f(S)$ is a basis as well.

Title: Prove that $1,t-1,t^2-t,\cdots,t^{n-1}-t^{n-2}$ a basis for any polynomial $P_{n-1}$


问:
Finding $E\left(\frac{X_1 + X_2}{2}\mid X_1+\cdots+X_n=y_1\right)$ where $X_i$'s are i.i.d Binomial$(1,\theta)$
Let $X_1,X_2,\ldots,X_n\,, n > 2$  be a random sample from  the binomial distribution $b(1, \theta)$.

I have shown $$Y_1 = X_1 + X_2 + \cdots + X_n$$ is a complete sufficient statistic for $\theta$ and $$Y_2 = \frac{X_1 + X_2}{2}$$ is unbiased estimator of $\theta$.

The question ask also to find $$E(Y_2\mid Y_1 = y_1)$$

So I am thinking to find the conditional distribution then find the expectation.

\begin{align}
P(Y_2=y_2\mid Y_1=y_1)&=\frac{P(Y_2=y_2,Y_1=y_1)}{P(Y_1=y_1)}
\\\\&=\frac{P(Y_2=y_2)P(Y_1-Y_2=y_1-y_2)}{P(Y_1=y_1)}
\end{align}

But I do not know the distribution of $Y_2 = (X_1 + X_2)/2$.

The question from Introduction to Mathematical Statistics Hogg 7ed page 401 problem 11.

答:
You do not need the biniomal knowledge.  In general, if the $X_i$ are i.i.d. with any distribution with finite expectation, then clearly $E\left[X_1+\cdots+X_n \mid X_1+\cdots+X_n=y\right] = y$, so by symmetry and exchangeability

$$E\left[X_i \mid X_1+\cdots+X_n=y\right] =  \frac1ny$$

and thus with $n \ge 2$ $$E\left[\tfrac{X_1+X_2}{2} \mid X_1+\cdots+X_n=y\right] =  \frac1ny$$ too


答:
Let $t = \sqrt{ab}$ and
$$f(a,b,c) = a^2+b^2+c^2 + 6 - 3(a+b+c).$$
Suppose $c = \min\{a,\,b,\,c\},$ then $c \leqslant 1,$ so
$$\sqrt{a}+\sqrt{b} \geqslant 2\sqrt{t} \geqslant 2.$$
We have
$$f(a,b,c) - f(t,t,c) = a^2+b^2-2t^2 - 3(a+b-2t) = \left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\right]\left(\sqrt{a} - \sqrt{b}\right)^2 \geqslant 0.$$
Therefore $f(a,b,c) \geqslant f(t,t,c),$ and
$$f(t,t,c) = f\left(t,t,\frac{1}{t^2}\right) = \frac{(2t^4-2t^3+2t+1)(t-1)^2}{t^4} \geqslant 0.$$
The proof is completed.

Title: $a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$

Body:
I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$

My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.

Edited: I found some variants of the original problem.

Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.

Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.


问:
Recurrence $a_{n}=a_{\lfloor 2n/3\rfloor}+a_{\lfloor n/3\rfloor}$
I am considering the sequence
$$a_n=a_{\lfloor 2n/3\rfloor}+a_{\lfloor n/3\rfloor}$$
with $a_0=1$, and I would like to calculate the limit
$$\lim_{n\to\infty} \frac{a_n}{n}$$
I have seen this famous question and its answer, but since the recurrence in this question has only two terms on the RHS instead of three, I was wondering if there is a more elementary solution that does not use specialized knowledge like renewal theory.

I have not made much progress; all I have managed to prove so far is that the sequence contains runs of arbitrarily long length, and this is probably not relevant to the desired limit.

答:
The sequence $a_n/n$ is convergent.

We adapt the idea given in this paper A master theorem for discrete divide and conquer recurrences by Michael Drmota, Wojciech  Szpankowski.

The general theorem and proof are given in the paper, so I will focus on this specific sequence in this posting.

Two main theorems are

Theorem 1 (Landau)

Let $F(s)=\sum a_n n^{-s}$ be a Dirichlet series with a finite abscissa of convergence $\sigma_c$. If $a_n\geq 0$ for all $n$ then the point $\sigma_c$ is a singularity of the function $F(s)$.

Note that if $a_n\geq 0$ for all $n$ then $\sigma_c=\sigma_a$.

Theorem 2 (Wiener-Ikehara)

Let $F(s)=\sum a_n n^{-s}$ with $a_n\geq 0$ for all $n$. Suppose $F(s)$ converges for $\sigma>1$ and $F(s)-c/(s-1)$ extends to a continuous function in the closed half-plane $\sigma \geq 1$. Then
$$
\sum_{n\leq x}a_n=cx+o(x).
$$

Both theorems are in "Multiplicative Number Theory I. Classical Theory, by Montgomery & Vaughan". They are Theorem 1.7 and Corollary 8.8 in the book.

Solution to the problem

Let $b_n=a_n-a_{n-1}$, we have $b_1=1$ and
$$
b_n= {\bf 1}_{3|n} b_{2n/3} + {\bf 1}_{n\equiv 2(3)} b_{(2n-1)/3} + {\bf 1}_{3|n} b_{n/3} \text{ if } n>1. \ (1)
$$

By induction, we have $0\leq b_n\leq n$ for all $n$. Thus, the Dirichlet series $F(s)=\sum b_n n^{-s}$ has a finite abscissa of absolute convergence $\sigma_a$.

By (1), we have for $\sigma>\sigma_a$,
$$
\sum b_n n^{-s} = 1+\sum_{3|n} b_{2n/3} n^{-s} + \sum_k b_{2k-1} (3k-1)^{-s} + \sum_{3|n} b_{n/3} n^{-s}.
$$
Then
$$
\begin{align}
(1-3^{-s})& \sum b_n n^{-s} = 1 + \sum_k b_{2k} (3k)^{-s} + \sum_k b_{2k-1} (3k-1)^{-s}\\
&=1+(3/2)^{-s} \left( \sum_k b_{2k} (2k)^{-s} + \sum_k b_{2k-1} (2k-\frac23)^{-s} \right).
\end{align}
$$
The last sum can be written as
$$
\sum_k b_{2k-1}(2k-1)^{-s} + \sum_k b_{2k-1}\left((2k-\frac23)^{-s}-(2k-1)^{-s}\right).
$$
Then we have
$$
\left(1-3^{-s}-(\frac32)^{-s}\right) \sum b_n n^{-s} $$
$$= 1+ (\frac32)^{-s} \sum_k b_{2k-1} \left((2k-\frac23)^{-s}-(2k-1)^{-s}\right).  \ \ (2)
$$
By ML-inequality,
$$\left|(2k-\frac23)^{-s}-(2k-1)^{-s}\right|$$
$$=\left|\int_{2k-1}^{2k-\frac23} (-s)x^{-s-1} dx\right|=O(|s|k^{-\sigma-1})$$
with an absolute implied O-constant.

Thus, the RHS of (2) defines an analytic function on $\sigma>\sigma_a-1$. On the LHS of (2), $1-3^{-s}-(3/2)^{-s}$ has a unique real zero at $s=1$ which is simple, and $\sum b_n n^{-s}$ is analytic on $\sigma>\sigma_a$ with singularity at $\sigma_a$ by Landau's theorem. Hence, the singularity at $\sigma_a$ must be removable. This implies $\sigma_a = 1$. Also, $1-3^{-s}-(3/2)^{-s}$ does not have complex zero $1+it$ with $t\neq 0$.

Thus, we obtain the expression
$$
F(s)=\frac{1+ (\frac32)^{-s} \sum_k b_{2k-1} \left((2k-\frac23)^{-s}-(2k-1)^{-s}\right)}{1-3^{-s}-(\frac32)^{-s}}. \ \ (3)
$$
Moreover, $F(s)-c/(s-1)$ extends to a continuous function on $\sigma\geq 1$ where $c$ is
$$
c=\frac{3+2\sum_k b_{2k-1} \left( (2k-\frac23)^{-1}-(2k-1)^{-1}\right)}{\log\frac{27}4}.
$$

By Wiener-Ikehara's theorem, we obtain
$$
\sum_{n\leq x} b_n \sim cx.
$$
For the original problem, we have
$$
\lim_{n\rightarrow\infty}\frac{a_n}n=\frac{3+2\sum_k b_{2k-1} \left( (2k-\frac23)^{-1}-(2k-1)^{-1}\right)}{\log\frac{27}4}.
$$

Remark.
The convergence may be very slow and difficult to observe. However, the convergence is mainly due to the following
$$\frac{\log(2/3)}{\log(1/3)} \notin \mathbb{Q}.$$


问:
Convergence in distribution to a chi-squared random variable with 1 degree of freedom
Question: If $(X_i, i \geq 1)$ is a sequence of independent random variables satisfying $P(X_i = 1)=P(X_i = -1)$ then $\frac{{({\sum_{i=1}^{n}{X_i}}})^2}{n}$ converges in distribution to a random variable having the chi-sqaured distribution with 1 degree of freedom.

Is this true?

So far I have: $P(X_i = 1)=P(X_i = -1)= \frac{1}{2}$

So, $M_{X_i}(t)=\frac{e^t +e^{-t}}{2}$.

Then letting $S_n= ({\sum_{i=1}^{n}{X_i}})^2$, we have that $M_{S_n}(t)= \prod_i^n E(e^{t{X_i}^2}) = e^t \dots e^t = e^{nt}$

Now let ${S_n}^*= \frac{({\sum_{i=1}^{n}{X_i}})^2}{n}$ and so $M_{{S_n}^*}(t)= e^{{n}{\frac{t}{n}}}= e^t$

Is what I have so far correct? If not, what is the correct way of working this out and how do I conclude if it does converge in distribution to a chi-squared random variable with 1 degree of freedom?

答:
The answer is positive.

It will be more convenient to use the next approach.

It follows from the Central limit theorem that
$$\xi_n = \frac{\sum_{i=1}^n X_i - n EX_1}{\sqrt{n DX_1}} = \frac{\sum_{i=1}^n X_i}{\sqrt{n}} \to \xi \sim N(0,1)$$
in distribution.  Thus $\frac{(\sum_{i=1}^n X_i)^2}{\sqrt{n}}  = \xi_n^2 \to \xi^2 \sim \chi_1^2$, where we imply convergence in distribution.


问:
Geometric Interpretation of the Permanent
The Permanent of a square matrix M = ($m_{i,j}$) is defined as follows:

$Perm(M) = \sum_{\sigma\in S_n}\prod_{i=1}^{n} m_{i,\sigma(i)}$

The Permanent is quite similar to the Determinant of a square matrix, defined as follows:

$Det(M) = \sum_{\sigma\in S_n}sign(\sigma)\prod_{i=1}^{n} m_{i,\sigma(i)}$

The Determinant has an intuitive geometric interpretation. Is anything similar known about the Permanent? If not, why does the signed sum of the permutations lend itself to a geometric interpretation and the unsigned sum does not?

答:
In the article by Terrance Tao linked in Keeler's answer, the first commenter, "nobody", points out that

As for a geometric interpretation of the permanent, we can think of the determinant of a vector space $\mathbb{R}^n$ (which is just a vector bundle over a point) as a section of the top (n-th) exterior power of that vector bundle; equally we can regard the permanent as a section of the n-th symmetric tensor bundle – and given a Young diagram, we can play the same sort of game for any immanant...

I wouldn't say this is a particularly intuitive explanation (I'm not sure I even really understand it), but I believe it adds something.


问:
Help understanding a proof (Linear Transformation)

On a V vectorial space

we have that

$T(x+y) = Tx - Ty$

$T(\alpha x) = \alpha Tx$ , for $\alpha \geq 0$

Prove that T is linear.

I tried some stuff by myself and then I found an answer online, but I can't completely understand it.

It goes like this:

Suppose $\alpha < 0 $. We have $T[(\alpha-\alpha)x]= T(0x)=0Tx=0$

That part I get.

But then it goes,

so that $0=T( \alpha x- \alpha x) = T( \alpha x) + T(- \alpha x) = T(\alpha x) - \alpha T(  x) $

Thus, $T(\alpha x)= \alpha Tx$ if $\alpha < 0 $

The part I don't quite get is, how you get this equality:

$T( \alpha x- \alpha x) = T( \alpha x) + T(- \alpha x)  $

答:
For any vector $v$, let $u=\frac v2$.  Then $$Tv=T(u+u)=Tu-Tu=0$$  by the first property.
Therefore, $T$ must be the zero mapping, which is trivially linear.


问:
Cayley-Hamilton theorem and $\prod_{i=1}^nA-\lambda_iI$ ($\lambda_i$ is the $i$th eigenvalue)
How does Cayley Hamilton theorem link to $\prod_{i=1}^nA-\lambda_iI$? All definitions of the theorem I can find say that $p(A)=0$ (the characteristic polynomial of $A$ is $0$). Thanks!

答:
Over an algebraically closed field like $K=\mathbb C$, the characteristic polynomial of a square matrix $A\in K^{n\times n}$ factors as
$$
p_A(x) = \prod_{i=1}^n (x-\lambda_i) = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n),
$$
where the $\lambda_i$ are the eigenvalues of $A$ (counting multiplicity).
Hence, $p_A(A)=0$ is equivalent to
$$
\prod_{i=1}^n (A-\lambda_i I) = (A-\lambda_1 I)(A-\lambda_2 I)\cdots(A-\lambda_n I) = 0 \in K^{n\times n}.
$$


问:
Differential equations and proof
I have solved this. But I don't know how to came up with the (1\2) beside m. I have tried I but I was unable to do that. Please help me in this.</p>

答:
The trick is to just to realize that $\int x \dot{x}dt = \frac{x^2}{2} + C$ where $C$ is an integration constant (try integrating by parts with $u(t) = x(t)$ and $v(t) = \dot{x} dt$ if you have troubles with it). The same is true if we change $x \mapsto \dot{x}$ and $\dot{x} \mapsto \ddot{x}$. So when you multiply by $\frac{dx}{dt}$ and the integrate, you are just creating two integrals of the same functional form:
$$
\begin{aligned}
&\int \dot{x} \ddot{x} dt + \omega^2\int x \dot{x} dt = 0 \\
&\iff \frac{1}{2} \left( \frac{dx}{dt} \right) ^2 + C_1 + \omega^2 \left( \frac{1}{2}x^2 + C_2 \right) = 0
\end{aligned}
$$

You get to the result just by using the fact that $\omega^2 = \frac{k}{m}$, multiplying by $m$ and summing all the corresponding constants to $E$.

I hope it helps :)

Title: Differential equations and proof