问:
finding expected value and variance using conditional expected value and variance
Let $X \sim G(p)$ and $Y|X \sim B(j,p)$, I need to express $E[Y]$ and $Var(Y)$ using $p$

first I tried expressing $E[Y]$:

$$E[Y|X=j]=\sum_{i=0}^{j}i\cdot\binom{j}{i}p^i(1-p)^{j-i}=\sum_{i=0}^{j}i\frac{j!}{i!(j-i)!}p^i(1-p)^{j-i}=jp\sum_{i=0}^j\frac{(j-1)!}{(i-1)!(j-i)!}p^{i-1}(1-p)^{j-i}=jp\sum_{i=0}^{j}\binom{j-1}{i-1}p^{i-1}(1-p)^{j-i}=jp(p+1-p)^{j-1}=jp$$
hence
$$E[Y]=E[E[Y|X]]=\sum_{j=0}E[Y|X=j]p_x(j)=\sum_{j=0}jp(1-p)^{j-1}p=p^2\sum_{j=0}j(1-p)^{j-1}$$
But I don't know how to reach a solution for this last series, and get an answer for $E[Y]$

I then tried expressing $Var(Y)$

Let $Y_i$ be an indicator, therefore $Y=\sum Y_i$ so
$$Var(Y|X)=Var(\sum_{i=1}^{j}Y_i|X=j)=\sum_{i=1}^j Var(Y_i)=jp(1-p)$$
I already know that $E[Y|X]=jp$ so all I need is to calculate
$$Var(Y)=E[Var(Y|X)]+Var(E[Y|X])=E[jp(1-p)]+Var(jp)$$
but I have no idea how I can calculate this either

In both cases I followed the same steps that were in the examples I got, so I fail to understand what did I do wrong, this is an introduction exercise so it shouldn't be too complicated, yet I am at a complete loss

答:
First question

But I don't know how to reach a solution for this last series

$$\sum_{j=0} j(1-p)^{j-1}$$

set $(1-p)=q$ and observe that

$$jq^{j-1}=\frac{d}{dq}q^j$$

...the sum of the derivative is the derivative of the sum...the resulting sum is easy because it is a geometric series....and the solution easy follows

Second question

Your text is the following

$X\sim G(p)$ and $Y|X=j\sim Bin (j,p)$

Thus using the variance decomposition you get, as you found

$$\mathbb{V}[Y]=\mathbb{V}[jp]+\mathbb{E}[jpq]=p^2\cdot\frac{q}{p^2}+pq\frac{1}{p}=2(1-p)$$

EDIT: I think it is easy for you to rewrite the text in the following way

$X\sim G(p)$

$Y|X\sim Bin(x,p)$

Thus it is easy to realize that, i.e.

$$E(x pq)=pqE[X]=pq\frac{1}{p}=q$$


问:
Problem calculating the integral $\int_{0}^{\infty}x^ne^{-xt}dx$
I want to calculate the Laplace transform of the function $ x ^ n $ but I have a problem with the limit that I get from applying integration by parts:

\begin{align*}
\int_{0}^{\infty}x^ne^{-xt}dx&=\int_{0}^{\infty}x^n\left(-\cfrac{1}{t}e^{-xt}\right)'dx\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}-\int_{0}^{\infty}-\cfrac{1}{t}e^{-xt}(nx^{n-1})dx\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}\int_{0}^{\infty}x^{n-1}e^{-xt}dx\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}y_{n-1}(t)\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}y_{n-1}(t)\\
&=\lim_{x\to\infty}-\cfrac{x^n}{t}e^{-xt}-\left[-\cfrac{0^n}{t}e^{-(0)t}\right]+\cfrac{n}{t}y_{n-1}(t)\\
&=\lim_{x\to\infty}-\cfrac{x^n}{t}e^{-xt}+\cfrac{n}{t}y_{n-1}(t)\\
\end{align*}

But the problem is that I don't know how to take this limit, my intuition tells me that it is $ 0 $ because lowering the term $ e ^ {- xt} $ presents an indeterminacy of the type $ \cfrac{\infty}{\infty} $ so if we apply l'Hôpital's rule of thumb enough times to eliminate the term "$ x $" from above the fraction and since the term below is never affected by the derivatives other than by the term $ t ^ m $ that multiplies it, then the limit would be 0 but I don't know how to prove it properly.

Although I also tried that it should be possible to demonstrate this limit by induction, but in that case I do not know how to make such a demonstration, especially I do not know how to use the fact that it is true for $ k $ and with it demonstrate that it is true for $ k + 1 $. I appreciate any help.

答: To prove $\lim_{x\to\infty}x^ne^{-xt}=0$ for all $n>0,\,t>0$, it suffices viz. $y:=x/n\implies x^ne^{-xt}=(nye^{-yt})^n$ to check the case $n=1$, and then it suffices viz. $z=yt\implies ye^{-yt}=t^{-1}ze^{-z}$ to check the case $n=t=1$, i.e. $\lim_{z\to\infty}\tfrac{z}{e^z}=0$. You can do this with e.g. L'Hôpital's rule.

ContentLicense: CC BY-SA 4.0


问:
Exponentiation of infinite cardinals: rules involving $\max\{\kappa_1, \kappa_2\}$?
Th question is "simple". Let $\kappa_1, \kappa_2$ be two infinite cardinals. We have the simple rules for addition and multiplication:
$$
\kappa_1+ \kappa_2=\kappa_1 \kappa_2=\max\{\kappa_1, \kappa_2\}.
$$
Now do we have a similar rule for
$$
\kappa_1^{\kappa_2}\space?
$$
Could we share some reference for this?

答:
No. We don't even have a rule for $2^\kappa.$ This is related to the independence of the continuum hypothesis: the value of $2^\kappa$ for regular $\kappa$ is (almost) completely independent of ZFC. For example, $2^{\aleph_0}$ (the cardinality of the reals) can consistently be any cardinal with uncountable cofinality.

See Jech Set Theory chapter 5 for information about what we do know, for instance, about how to reduce certain cardinal exponentiations to others. (e.g. look at theorem 5.20.) Under GCH, where we assume $2^\kappa=\kappa^+$ for all $\kappa,$ this leads to a general formula for $\kappa^\lambda$ (theorem 5.15). Under the SCH, $\kappa^\lambda$ is fully determined by the continuum function $\kappa\mapsto 2^\kappa$ on regular cardinals (theorem 5.22), but the continuum function on the regular cardinals can still consistently be almost anything. In ZFC alone, there are some nontrivial results on $\kappa^{\operatorname{cf}\kappa}$ and $2^\kappa$ for singular cardinals, though these take the form of bounds under assumptions on what the continuum function looks like for regular $\kappa,$ so this isn't anything approaching a "formula". These results are much harder, though, see for instance Jech chapter 24 or the book by Holz, Steffens, and Weltz.


问:
The gaussian integers modulo a pair of complex numbers

Taking a pair of Gaussian integers $u,v$ satisfying

There exist $a,b \in \mathbb Z$

$v^2 = av+bu$

$u^2 = av+bu$

$uv = av+bu$

(Diffrent $a,b$)

define the equivalence relation $\sim$ by $z\sim w \iff \exists a,b \in \mathbb Z, z-w=au+bv$
Let $\mathbb Z[i]/\sim$ be the gaussian integers modulo $u,v$

For example, if $u=2i,v=2$ we get a four element ring with $0,1,i,1+i$ with the multiplication table:
$$\begin{matrix} \cdot && 0 && 1 && i && 1+i \\ 0 && 0 && 0 && 0 && 0 \\ 1 && 0 && 1 && i && 1+i\\ i&& 0 && i && 1 && 1+i \\ 1+i && 0 && 1+i && 1+i && 0\end{matrix}$$
Are these objects studied?

Do any of you have a reference to an article about them?

if not, I have some questions:

is there a way to use $u,v$ to find properties of the ring? for example, in the example there was a nilpotent element, $1+i$, how could I have predicted it using $2,2i$?

Do there exist pairs for which this is a field? if so, is there a criterion for these?

Is there a criterion for when this is the zero(trivial) ring?

Can this be expressed as some "multiplication" of normal cyclic rings? For example, the ring I mentioned is isomorphic to $\mathbb Z_4$($0 \mapsto 0, 1 \mapsto 1, 2 \mapsto i+1, 3\mapsto i$), I've been trying to do it on $2-2i,2+2i$ and have not managed it.

答:
Another way to state your assumption about $u$ and $v$ is that the additive subgroup generated by $u$ and $v$ is closed under multiplication. Thus, this additive subgroup is close to being an ideal, but in general it need not be!

As a result, multiplication on $\mathbb{Z}[i]/\sim$ is not always well-defined. For example, let $u = 2, v = 3$. Then $u^2 = 4 = 2u + 0v$, $v^2 = 9 = 0u + 3v$, and $uv = 6 = 3u + 0v$. Note that $1 - 0 = (-1)u + 1v$, so $[1] = [0]$ in $\mathbb{Z}[i]/\sim$ (where $[\bullet]$ means "the equivalence class of $\bullet$"). If multiplication were well-defined, we would get
$$[m] = [m][1] = [m][0] = [0]$$
for all $m \in \mathbb{Z}[i]$, but this is not the case, for example $[i] \neq [0]$.

In fact, $([x],[y]) \mapsto [xy]$ is a well-defined ring multiplication operation on $\mathbb{Z}[i]/\sim$ if and only if the additive subgroup generated by $u$ and $v$ is an ideal. This turns out to be equivalent to the following:

There exist integers $a$ and $b$ such that $iu = au + bv$

There exist integers $a$ and $b$ such that $iv = au + bv$

These two conditions actually imply the three conditions you gave! Note that your example of $u = 2i$, $v = 2$ does satisfy these conditions, which is why you got an actual ring out in the end.

For the rest of this answer, I'll suppose we've imposed the two conditions above instead of the three conditions you gave, so that these quotients always make sense as rings. Let $I = \{au + bv : a,b \in \mathbb{Z}\}$, which we now know is an ideal of $\mathbb{Z}[i]$, and let $R = \mathbb{Z}[i]/I = \mathbb{Z}[i]/\sim$. Let me try to answer your questions:

Are these objects studied?

Yes! It turns out that these objects are just arbitrary quotient rings of $\mathbb{Z}[i]$. Quotient rings are a basic construction in ring theory, and $\mathbb{Z}[i]$ is very well-studied.

Why are these things arbitrary quotients of $\mathbb{Z}[i]$? This is the same as saying that every ideal of $\mathbb{Z}[i]$ is generated, as an additive group, by two elements. This is true because every ideal of $\mathbb{Z}[i]$ is principal ($\mathbb{Z}[i]$ is a Euclidean domain), and the principal ideal $(x)$ is generated by $x$ and $ix$ as an additive group.

Do any of you have a reference to an article about them?

No, this would be like finding an article about cyclic groups -- these objects are too basic to be the focus of an article. To learn more about them, I'd recommend reading some introductory abstract algebra books (ring theory in particular). To dive a little deeper, you could read some introductory algebraic number theory, where $\mathbb{Z}[i]$ is an important classical example.

Is there a way to use $u,v$ to find properties of the ring? for example, in the example there was a nilpotent element, $1+i$, how could I have predicted it using $2,2i$?

Yes, there is a way to use $u$ and $v$ to find properties of the ring. Since $u$ and $v$ generate $I$ as an additive subgroup, they also generate $I$ as an ideal. Now $I$ is generated (as an ideal) by $x = \gcd(u,v)$. Note: the choice of $\gcd$ is non-unique, but any $\gcd$ will generate $I$, and you can compute a $\gcd$ using the Euclidean algorithm. In your specific example, we can take $x = 2 = \gcd(2,2i)$.

Now $R = \mathbb{Z}[i]/I = \mathbb{Z}[i]/(x)$ will have nilpotent elements if and only if $(x)$ is not a radical ideal (this is just by definition of "radical"). $(x)$ is radical if and only if $x$ is square-free (standard argument about PID's). An element of a ring is said to be square-free if $y^2 \mid x \implies y$ is a unit, and in $\mathbb{Z}[i]$ the units are $1, -1, i, -i$. In your specific example, $x = 2$ is not square-free, because $2 = i(1-i)^2$, so $R$ has a nilpotent element. In this case, we can choose the nilpotent element to be $[1-i]$ (which is equal to $[1+i]$), because $(1-i)^2 = (-i)2 \in I$. In general, to check whether or not $R$ has nilpotent elements, we do the following:

Compute the prime factorizations $u$ and $v$ in $\mathbb{Z}[i]$ (there is an algorithm for doing this).

Use these prime factorizations to compute $x = \gcd(u,v)$ and its prime factorization.

If $x$ has any repeated prime factor, so $x = p^2 y$ for some gaussian prime $p$, then $[py]$ will be nilpotent in $R$ (in particular, its square will be zero). Otherwise, $R$ will not have any nilpotent elements.

Do there exist pairs for which this is a field? if so, is there a criterion for these?

Yes, $R = \mathbb{Z}[i]/I$ will be a field if and only if $I = (x)$ is a maximal ideal, if and only if $x$ is prime. This is actually particularly easy to check: $x = a + bi$ is prime if and only if one of the following holds:

$a = 0$, $\lvert b \rvert$ is prime, and $\lvert b \rvert \equiv 3 \pmod{4}$

$b = 0$, $\lvert a \rvert$ is prime, and $\lvert a \rvert \equiv 3 \pmod{4}$

$a \neq 0$, $b \neq 0$, and $a^2 + b^2$ is prime

Is there a criterion for when this is the zero(trivial) ring?

Yes. $R = \mathbb{Z}[i]/I$ is the zero ring if and only if $I = \mathbb{Z}[i]$, if and only if $1 \in I = (x)$, if and only if $x$ is a unit, if and only if $u$ and $v$ are coprime. Again, this is computable by just finding the factorizations of $u$ and $v$.

Can this be expressed as some "multiplication" of normal cyclic rings?

This is a little tricker, and the answer is no! See here for a general discussion of how to compute the isomorphism type of the quotient. Important examples are:

If $x = a + bi$ with $a$ and $b$ coprime, then $R \cong \mathbb{Z}_{a^2 + b^2}$

If $x = 3$, then $R \cong \mathbb{Z}_3[X]/(X^2+1)$ is a field (as discussed earlier), and we have $\lvert R \rvert = 9$ since this field is $2$-dimensional over $\mathbb{Z}_3$. Thus, $R \cong \mathbb{F}_9$ is not a product of quotients of $\mathbb{Z}$.

All of this is (more or less) elementary abstract algebra, so I'd recommend learning more abstract algebra if you want to understand these things more deeply.


问:
Question on equivalence of definitions of quasi-affine varieties
I'm hoping to introduce some basic concepts of classical algebraic geometry to some students in a seminar soon. To avoid the usual ambiguity and confusion around the terms in this picture, I was hoping to present things in a more consistent, if atypical, fashion. Since I have a tendency to overlook some subtleties, I'd like to know if there are any issues with the following outline:

Fix a field $k=\overline{k}$ and define $\mathbb{A}^n$

Define the Zariski topology on $\mathbb{A}^n$ and use it to distinguish closed and irreducible closed subsets.

Define a morphism $f:X \to Y$ between subsets $X \subseteq \mathbb{A}^n$ and $Y \subseteq \mathbb{A}^m$ to be a set-theoretic map $f =(f_1/g_1, \ldots, f_m/g_m)$, where $f_i,g_i \in k[x_1,\ldots,x_n]$ and $\forall x \in X, g_i \neq 0$ (a regular map). Isomorphisms are biregular bijections.

Give the following extrinsic definitions: $X \subseteq \mathbb{A}^n$ is a quasi-affine variety if it's isomorphic to a closed subset of some $\mathbb{A}^m$, and an affine variety if it's isomorphic to an irreducible closed subset of some $\mathbb{A}^m$.

I think this captures the picture with some consistency, but I'm not sure if the definition of quasi-affine above is equivalent to the usual definition as locally closed subset. If $X$ is locally closed then I believe it has a closed image in some $\mathbb{A}^m$ (if it's an open subset of a closed set $X \subset \mathbb{A}^n$, then  it can be written as a finite intersection of closed sets in $\mathbb{A}^{n+1}$). I'm not sure about the converse, but the idea that quasi-affine means we have a solution of a polynomial system minus some other solution set should mean it's a union of solutions somewhere.

答:
$X \subseteq \mathbb{A}^n$ is a quasi-affine variety if it's isomorphic to a closed subset of some $\mathbb{A}^m$

This is not equivalent to the usual definition of quasi-affine; for example, $\mathbb{A}^2 \setminus \{ 0 \}$ is quasi-affine in the usual sense but not in this sense. (Personally I would call this "affine.")


问:
Only solution to Diophantine equation with primes?
Find all solutions to:

$$p_1 ^2 - p_2^2 = p_3^2 + 1$$

where $p_i$ is a prime number (positive of-course)
The only solution that seems to exist is $(3,2,2)$ for $(p_1, p_2, p_3)$ accordingly. But I have no clue how to prove that no other solutions exist..
The only thing that jumps to my mind is assuming that another solution exists and somehow get to a contradiction...

I would highly appreciate your help, thank you!

答: Taken mod $8$, the squares of primes leave remainders $4$ (for $p=2$) and $1$ (for all odd primes). The only combination of those two remainders that satisfy the congruence $a-b\equiv c+1$ mod $8$ is $1-4\equiv4+1$ mod $8$, which means we need $p_2=p_3=2$, which in turns implies $p_1=3$.


问:
How to find the generating function of a recurrent series: $b_{0} = 4$, $b_{n}=3b_{n-1}-2^{n-1}$
The series is defined as such:
$$
b_{0}=4
$$$$
b_{n}=3b_{n-1}-2^{n-1}\text{ for }n\geq1
$$

How do I go about finding its generating function?
I started like this:
$$
f(x)=\sum_{n=0}^{\infty} b_{n} \cdot x^n =4+\sum_{n=1}^{\infty}(3b_{n-1}-2^{n-1})\cdot x^n
$$
And now I am lost. How should I proceed?

答:
You are almost there! Keep in your mind that you should always look for two things when manipulating generating functions:

Expressions like $\sum b_k x^k$, with exponent and index equal; beside the first terms that could not be there, this is your generating functions;

Known generating functions; for example, keep in your pocket the identity

$$\sum t^j = \frac{1}{1-t}$$

And in general, if the coefficient of your series do not appear, you should be able to algebraically solve this expression independently from the recurrence formula you are transforming.

With this in mind, let's start from your last passage:
$$f(x)=\4+\sum_{n=1}^{\infty}(3b_{n-1}-2^{n-1})\cdot x^n = 4+ \sum_{n=1}^{\infty} 3b_{n-1} x^n - \sum_{n=1}^{\infty} 2^{n-1} x^n $$

The left part seem to be related to $\sum b_k x^k$; indeed, factoring a $3x$ you get
$$ \sum_{n=1}^{\infty} 3b_{n-1} x^n = 3x \sum_{n=1}^{\infty} b_{n-1} x^{n-1} = 3x \sum_{n=0} b_n x^n = 3x f(x) $$
Note the shift in the indices in the last passage, so that we recover the expression for $f(x) $.

The right part does not contain $b_k$; it should be a known expression. It is a slight variant of the identity I presented you above:
$$ \sum_{n=1}^{\infty} 2^{n-1} x^n = x \sum_{n=1}^{\infty} 2^{n-1} x^{n-1} = x \sum_{n=1}^{\infty} (2x) ^{n-1} = x\sum_{n=0}^{\infty} (2x) ^n = \frac{x}{1-2x} $$

Putting all together we have
$$ f(x) = 4+3x f(x) - \frac{x}{1-2x} = 3x f(x) + \frac{4-9x}{1-2x} $$

Solving for $f(x) $ we have

$$ f(x) =\frac{4-9x}{ (1-2x) (1-3x) }  $$


问:
Prove Lindeberg condition holds
If $X_1, ..., X_n$ is an iid sequence where $P(X_i = \sqrt{i}) = 1/2$ and $P(X_i = -\sqrt{i}) = 1/2$ for $i = 1,...,n$, how do I verify the Lindeberg condition?

I know that $E(X_i) = 0$ and $Var(X_i) = i$, and hence $S_n = Var(\sum{X_i})  = \frac{n(n+1)}{2}$. But I am getting stuck trying to prove that, as $n \rightarrow \infty$:

$$\frac{1}{S_n^2}\sum_{i = 1}^n E\left(X_i^2I\{|X_i| > \epsilon S_n\}\right) \rightarrow 0$$

答:
A possible hint:

You have $S_n^2=\sum\limits_{k=1}^n\operatorname{Var}( X_k)=\frac{n(n+1)}{2}$.

Now $|X_k|=\sqrt k\le \sqrt n$ almost surely for every $k=1,\ldots,n$.

Observe that $\frac{\sqrt n}{S_n}\to 0$ as $n\to \infty$. So given $\varepsilon>0$, you have $|X_k|<\varepsilon S_n$ almost surely for every $k=1,\ldots,n$ where $n$ is large. In other words, $P\left[|X_k|>\varepsilon S_n\right]= 0$ for all $k=1,\ldots,n$ and for large enough $n$.

Title: Prove Lindeberg condition holds


问:
How to get the value of a function dependant on $\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$?
The problem is as follows:

Find the value of:

$$R=\frac{\sec^2B-\cot A}{4+\csc^2 A}$$

Where $A$ is given by:

$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot
> \left(\frac{A}{2}\right)}$

Also assume this follows the relationship seen in the figure from
below:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{1}{2}\\
2.&1\\
3.&\frac{1}{4}\\
4.&\frac{1}{3}\\
\end{array}$

My main issue with this problem is the half angle which is presented in the right side of this equation:

$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$

I was able to spot that the figure might intend to say that $A$ and $B$ are complementary thus:

$A+B=90^{\circ}$

Therefore this makes the previous expression into:

$\tan A = \frac{2+\tan \left(\frac{A}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$

This helps a bit but again the half angle would produce a square root. I tried to follow such path. But it didn't ended very well.

After squaring and using other identities I ended up having a 8th degree equation. And certainly this wasn't intended.

So what's the trick here?. So far I'm still stuck and all sorts of manipulations don't give me results. Can someone guide me in the right path?. I'm also confused on how to better use the figure. Those other letters don't seem to help much.

It would really help me alot a method explained step by step so I can understand.

答:
We will use the identity: $$\cot (A/2) = \csc A + \cot A$$

We will equate

$$(4+\cot(A/2))\tan A = 4\tan A + \csc A \tan A + \cot A \tan A = 4\tan A + \sec A + 1$$

with

$$2+\cot(B/2) = 2+\csc B + \cot B = 2 + \sec A + \tan A$$

as given. (Here we used $A+B = 90^\circ.$) Hence $1 = 3\tan A \implies \tan A = 1/3$.

Using the diagram, we have $b = 3a$ and $c = \sqrt{10} a$. Using this information we can easily determine $\sec B$, $\cot A$ and $\csc A$; thus:

$$\frac {\sec^2 B - \cot A}{4+\csc^2 A} = \frac {\sqrt{10}^2 - 3}{4+\sqrt{10}^2}=\frac12$$


问:
IVP - Differential equation
Working on this. Have to find a solution. Very new to DE:

$$\begin{cases}
 2yx'(y)-1=x^2(y) \\
 x(y=1)=1 \\
 \end{cases}
$$

Writing this as:

\begin{align}
2y \frac{dx}{dy}-1&=x^2y\\
\frac{1}{x^2}2y\,dx-1&=y\,dy\\
\frac{1}{x^2}\,dx &= 1+y\,dy\\
\int \frac{1}{x^2}\,dx &= \int 1+y\,dy\\
-\frac{1}{x}+C &= \frac{y^2}{2}+y+C
\end{align}

Is this correct? Feel like I'm just doing algebra, but don't really understand if it's correct. Would love feedback.

答:
Careful, when you have $$2y \frac{dx}{dy}-1=x^2y$$ and divide both sides by $x^2$ you should get $$\frac{2y}{x^2}\frac{dx}{dy} - \frac{1}{x^2} = y$$ which is not the same as $$\frac{2y}{x^2}\frac{dx}{dy} - 1 = y.$$

Also, are you certain you've written the problem correctly?

Title: IVP - Differential equation


问:
Find the equation of a plane containing line $L$ and equidistant from two points
I want to find the equation for a plane that is equidistant from the points $P_0 = (1, 2, 3)$ and $P_1=(3, 2, 1)$ and contains the line $\ell=(1, 1, 1) + t(0, 1, -1).$

I know that the midpoint of the two points $P_0$ and $P_1$ must be on the plane, which is $MP=(2, 2, 2).$ However, I notice this point is on the line $\ell$. I need another vector in the plane to preform the cross product and find a normal to the plane, however I can't think of a way to find such vector. I know that the length of the orthogonal complements of the two points on the normal must be the same for all points on the plane, but setting up such an equation doesn't seem like the right way to go about it. Any help would be appreciated.

答:
There are actually two planes satisfying the given condition.

I use the following notations:

$n=\begin{pmatrix}n_x \\ n_y\\ n_z\end{pmatrix}$ a normal vector of a plane satisfying the given condition

$d=\begin{pmatrix}0 \\ 1\\ -1\end{pmatrix}$ the direction vector of line $l$

$P_l =\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$ the position vector of point $P_l(1,1,1)$ on line $l$

Then we must have

$$n\cdot d = 0 \Rightarrow n_y = n_z$$

Furthermore, because of the equal distances of points $P_0$ and $P_1$ to the plane, we have

$$\left|(P_0-P_l)\cdot n\right| = \left|(P_1-P_l)\cdot n\right|$$
$$\Rightarrow \left|n_y+2n_z\right| = \left|2n_x+n_y\right|$$
or since $n_y = n_z$
$$3|n_y| = |2n_x+n_y|$$

Solving this, you get (up to a non-zero scalar) the normal vectors

$$n =\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} \text{ or } n =\begin{pmatrix}-2 \\ 1\\ 1\end{pmatrix}$$

Hence, the two planes are

$$x+y+z = 3 \text{ or } -2x+y+z = 0$$


问:
Proof of identity in the sense of distributions
I am trying to prove this:

$\bullet$ Prove that $\Delta(\varrho_\epsilon \star u) = \varrho_\epsilon \star f $ in the sense of distributions, if $\Delta u = f$ in the sense of distributions, $ u \in L^1_{loc}(\mathbb{R}^n)$.

Can anybody help me?
Thank you in advance and I take advantage of the situation to wish you a happy new year :D

答:
Suppose that $\Delta u=f$ in the sense of distributions. One can easily verify that $$P (v*u)=(Pv)*u=v*(Pu)$$ in the sense of distributions for $v\in \mathcal{S},u\in\mathcal{S}'$, and any constant-coefficient differential operator $P$. If we take $P=\Delta$, $v=\rho_\epsilon,$ and $u$ to be itself, then it follows directly that

$$ \Delta (\rho_\epsilon*u)=\rho_\epsilon*(\Delta u)=\rho_\epsilon*f$$ in the sense of distributions.


问:
If $x-y$ were written as $-y+x$, then would its conjugate be $-y-x$?
The conjugate of $x-y$ is $x+y$. But, if I wrote it as $-y +x$, would the conjugate be $-y-x$?

Furthermore, is the conjugate of $x + y +z$

$(x+y)+z \to (x+y) - z $

$(x+z)+y \to (x+z) - y $

$(z+y)+z \to (z+y) - x $

Is there a way to formalize the idea of a conjugate?

答:
The term "conjugate" is meant to be used with numbers, where in some context there is a natural other number that they pair with. Usually, it's pairs of numbers that are roots of the same quadratic polynomial with coefficients in a more restrictive field.

So when $a,b$ are real, $a+bi$ and $a-bi$ are conjugates because they are both roots of $x^2-2ax+(a^2+b^2)$, a quadratic polynomial with real coefficients. It's not really about changing a plus sign to a minus sign.

Similarly when $a,b$ are rational, $a+b\sqrt{2}$ and $a-b\sqrt{2}$ are conjugates because they are both roots of $x^2-2ax+(a^2-2b^2)$, a quadratic with rational coefficients. Again, it's not really about changing a plus sign to a minus sign.

So if you rearrange to $bi+a$, it still has the same conjugate: $a-bi$ (or if you want to write it as $-bi+a$, fine.) Because $bi+a$ and $a-bi$ are the two roots of the same polynomial.

And with abstract $x$ and $y$ without more context, there is not a good meaning for the conjugate of $x+y$.


答:
Continuation of my comment and an alternative solution, from FLT
$$3 \mid p^2 -1$$ for $p \ne 3$.
Then check
$$ p_1^2 - p_2^2=p_3^2+1 \tag{1}$$
$$\iff p_1^2 -1 - p_2^2+1=p_3^2-1+2$$
This will lead to $3 \mid 2$, if none of $p_1,p_2$ and $p_3$ is $3$. Thus, at least one of $p_1,p_2$ or $p_3$ must be $3$.

if $p_1=3$ then $9-p_2^2=p_3^2+1$ yields $p_2=2$, since any other value will make $9-p_2^2\leq 0$, whether the RHS of $(1)$ is always positive.

for this case, either $p_2=3$ or $p_3=3$ doesn't matter, we can swap them. So we have $$(p_1-p_2)(p_1+p_2)=p_1^2-p_2^2=10=2\cdot5$$ just 2 case to analyse $p_1-p_2=1,p_1+p_2=10$ and $p_1-p_2=2,p_1+p_2=5$.

Title: Only solution to Diophantine equation with primes?

Body:
Find all solutions to:

$$p_1 ^2 - p_2^2 = p_3^2 + 1$$

where $p_i$ is a prime number (positive of-course)
The only solution that seems to exist is $(3,2,2)$ for $(p_1, p_2, p_3)$ accordingly. But I have no clue how to prove that no other solutions exist..
The only thing that jumps to my mind is assuming that another solution exists and somehow get to a contradiction...

I would highly appreciate your help, thank you!


问:
Examples of computational tools/theories for graph theorists?
It's been several months since I learned about Razborov's theory behind flag algebras. I realized that I enjoyed learning the theory in all of its abstraction, but equally enjoyed the part where I got to write code and produce some concrete results and numbers.

What other theories are out there that go hand-in-hand with computation and writing code (preferably in the area of graph theory)?

I am not looking for things like common algorithms on graphs (BFS, graph traversal, min/max cut, etc.), and I am already aware of plenty of mathematical software packages that support writing code for abstract mathematics (Sage, Magma, Mathematica, etc.).

Instead, I am interested in active research areas where a mathematical theory may be realized in the form of code or software which produces results in graph theory. For example, computational software for flag algebras has already yielded results in Ramsey theory, hypergraph Turan densities, permutations, etc.

答:
I tried to assemble a list for the areas in which I have worked or am currently working in. This is by far not a complete list and it is very subjective to my research areas. There is of course a lot more.

I hope this list covers some ideas your are seeking for.

Extremal Graph Theory

The first one, you named it yourself, is the nice concept called Flag Algebras by Razborov. It is specifically designed for solving Turán-type problems in extremal combinatorics.

Checkout the tool flagmatic. You most likely already know it. Flagmatic is used in extremal graph theoretical research and yielded to new results. Checkout the following paper (starting with a nice interim report by Razborov himself. He is specifically mentioning flagmatic.)

Flag Algebras: an Interim Report by Alexander A. Razborov,

On applications of Razborov's flag algebra calculus to extremal 3-graph theory by Falgas-Ravry and Vaughan,

On the Caccetta-Häggkvist Conjecture with a Forbidden Transitive Tournament by Grzesik in the Electronic Journal of Combinatorics,

Minimizing the number of 5-cycles in graphs with given edge-density by Bennett et. al,

There are many many more...

Notice, and you might also already know, that the Cauchy-Schwarz calculus is not complete (which might or might not be very surprising).

Checkout this important result

Undecidability of Linear Inequalities in Graph Homomorphism Densities by Hatami and Norin in Journal of the American Mathematical Society.

Graph Generation

The topic of graph generation is strongly connected to the topic of graph recognition. However, I have decided to split these topics into two.

You said that you are "[...] interested in active research areas where a mathematical theory may be realized in the form of code or software which produces results in graph theory".

Now, this is very important. Graph generation is a research area by its own, hence this is not a direct answer to your question, but the generation of graphs, and most importantly the capability of generating specific classes of graphs, is one of the most useful things in providing new results in structural graph theory.

There are several papers around, but the most known are the ones whose results are implemented in the tool plantri.

You might check out the follwing:

Generation of simple quadrangulations of the sphere by Brinkmann et. al,

Fast generation of planar graphs by Brinkmann and McKay,

Generation of Cubic Graphs and Snarks with Large Girth by Brinkmann and Goedgebeur.

There are many many more.

Graph Recognition

The graph recognition problem deals with the problem of telling whether a given graph belongs to a certain graph class. This is by far not an easy task. For example, recognizing $1$-planar graphs is proven to be $\mathcal{NP}$-hard. Again, it is of course an research area on its own, but there are several examples on how developing a recognition algorithm lead to new insights in graph theory. To name one: Chudnovsky et. al implemented an algorithm for detecting odd-holes in a given graph, which could be very useful for solving open problems in chromatic graph theory.

Here are some more references on this exciting topic:

Outer 1-Planar Graphs by Auer et. al,

Recognizing Optimal 1-Planar Graphs in Linear Time by Brandenburg,

Detecting a Long Odd Hole by Chudnovsky et. al.

Graph Isomorphism

Graph generation is strongly related to the Graph Isomorphism Problem, which is an really exciting research topic for several reasons.

It is one of the most researched problems in theoretical computer science with great theoretical interest, since its complexity still remains unkown. In fact, the problem is even discussed as a candidate for modern quantum computing. It is believed that the GI is neither $\mathcal{NP}$-complete nor in $\mathcal{P}$, which would make it a candidate for a $\mathcal{NP}$-intermediate, according to Ladner’s theorem under the assumption that $\mathcal{NP}  \neq \mathcal{P}$.

However, it is very important for graph generation, if not the most important, because we want to generate graphs of a certain graph class without generating isomorphic copies. We need isomorphism rejection of some sort.

This topic is very well explored, and it is very challenging to do research here. The most famous algorithm is given by McKay and Piperno. It is based on canonical labeling and automorphism group computation. It is implemented in the tool Nauty and Traces.

Check out the following paper:

Practical graph isomorphism, II by McKay and Piperno.

Automated Conjecture-Making

Now, this is a really exciting topic and has led to numerous results in mathematics and especially in graph theory.

In addition to definitions and axiomatic systems, the process of discovering structural relationships or patterns in some kind of objects plays a fundamental role in mathematical research. A new conceivable relation is then formalized in a statement, which is initially open to formal proofs. These unproven statements are called conjectures.

Even though conjectures have many shapes, they often describe a relation between different invariants of the object of interest, whereby an objects invariant is a property which remains unchanged under certain classes of transformations. Often, the conjectures are of the form $I \leq J$, $I \leq J+K$ and $I+J \leq K+L $ with $I,J,K$ and $L$ being either an invariant of the mathematical object or a constant value.

Due to the increasing complexity in modern mathematical research, the establishment of relations between these invariants in the above forms is becoming increasingly important. Therefore, in addition to pure scientific interest, researchers try to automate the process of conjecture-making using methods of computer science. Then, we also speak of computerized mathematical discovery.

There is a famous tool by Larson and van Cleemput called "conjecturing", which generates conjectures (of the form given above) for a set of mathematical objects. There are many new results in matrix theory, number theory and graph theory, which used automated conjecture-making.

In order to give some references:

A Survey of Research in Automated Mathematical Conjecture-Making by Larson,

Automated conjecturing I: Fatjlowicz's Dalmatian heuristic revisited by Larson and Van Cleemput,

A survey of automated conjectures in spectral graph theory by Aouchiche and Hansen (they present some more tools for automated conjecture-making, namely "GRAPH", "Graffiti", "Ingrid", "newGRAPH", and "AutoGraphiX".

There are many more.

To give a last nice example of how automated conjecture-making discovered new bounds for the domination number of Benzenoids in chemical graph theory, you should check out the following paper:

Automated Conjecturing VI: Domination Number of Benzenoids by Hutchinson et. al.

If I remember correctly, this is a publication of a PhD-Thesis which goes into more detail.


问:
Let $G$ be a non-abelian group, $|G|=12$. If $G$ has a normal $2$-Sylow, then $G$ is isomorphic to $A_4$.
Let $G$ be a non-abelian group, $|G|=12$. If $G$ has a normal $2$-Sylow, then $G$ is isomorphic to $A_4$.

What I've done so far: I know that $A_4$ has $8$ $3$-cycles, $3$ elements of order $2$, and the unit. Thanks to the Sylow's theorems I've also found out that $n_3=4$ (if it was $n_3=1$, then $G\simeq D\times T$, where $D \in {\rm Syl}_2(G)$, $T \in{\rm Syl}_3(G)$, and it'd be abelian), hence $G$ has $8$ elements of order $3$. I'd like to conclude now that $D \in{\rm Syl}_2(G)$ is not cyclic, meaning that $D$ has $3$ elements of order $2$ and $G$ has basically the same structure as $A_4$.

Any hints?

答:
You don't need to try to determine the structure of the Sylow $2$-subgroup.

You've used the fact that $G$ has a normal Sylow $2$-subgroup to conclude that it must have four Sylow $3$-subgroups. Great. Now let $G$ act by conjugation on the set of Sylow $3$-subgroups; this induces a morphism $\varphi\colon G\to S_4$. Because there are four Sylow $3$-subgroups, each of them is equal to its own normalizer (because the index of the normalizer needs to be equal to the size of the orbit). Use that to show the kernel of $\varphi$ is trivial and hence that it is an isomorphism onto $A_4$ (as that is the only subgroup of index $2$ in $S_4$).


答:
Alternative approach

Using the diagram and half-angle formulas:

$$\cot(A/2) = \frac{1 + \cos(A)}{\sin(A)}
~~\text{and}~~ \cot(B/2) = \frac{1 + \cos(B)}{\sin(B)} = \frac{1 + \sin(A)}{\cos(A)} \implies $$

$$\frac{\sin(A)}{\cos(A)} = \frac{2 + \frac{1 + \sin(A)}{\cos(A)}}{4 + \frac{1 + \cos(A)}{\sin(A)}} =
\frac{2\cos(A) + 1 + \sin(A)}{4\sin(A) + 1 + \cos(A)} \times
\frac{\sin(A)}{\cos(A)} \implies$$

$$2\cos(A) + 1 + \sin(A) = 4\sin(A) + 1 + \cos(A) \implies$$

$$\cos(A) = 3\sin(A)\implies$$

$\displaystyle \cot(A) = 3$

$\displaystyle 10\sin^2(A) = 1.$

Therefore,

$$R = \frac{\frac{1}{\cos^2(B)} - 3}{4 + \frac{1}{\sin^2(A)}}
= \frac{\frac{1}{\sin^2(A)} - 3}{4 + \frac{1}{\sin^2(A)}}
= \frac{10-3}{4 + 10} = \frac{1}{2}.$$

Title: How to get the value of a function dependant on $\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$?

Body:
The problem is as follows:

Find the value of:

$$R=\frac{\sec^2B-\cot A}{4+\csc^2 A}$$

Where $A$ is given by:

$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot
> \left(\frac{A}{2}\right)}$

Also assume this follows the relationship seen in the figure from
below:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{1}{2}\\
2.&1\\
3.&\frac{1}{4}\\
4.&\frac{1}{3}\\
\end{array}$

My main issue with this problem is the half angle which is presented in the right side of this equation:

$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$

I was able to spot that the figure might intend to say that $A$ and $B$ are complementary thus:

$A+B=90^{\circ}$

Therefore this makes the previous expression into:

$\tan A = \frac{2+\tan \left(\frac{A}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$

This helps a bit but again the half angle would produce a square root. I tried to follow such path. But it didn't ended very well.

After squaring and using other identities I ended up having a 8th degree equation. And certainly this wasn't intended.

So what's the trick here?. So far I'm still stuck and all sorts of manipulations don't give me results. Can someone guide me in the right path?. I'm also confused on how to better use the figure. Those other letters don't seem to help much.

It would really help me alot a method explained step by step so I can understand.


问:
torsion sheaf/ quasi-coherent sheaf on projective scheme
Assume $S$ is a graded ring generated by finite elements of $S_1$ as $S_0$ algebra. I came across a claim that $\tilde{M}$ is not determined by $M$, with a counterexample $M =\underset{n\geq 0}{\sum} M_n $ and $N=\underset{n\geq n_0}{\sum} M_n$, the goal is to prove that $M_{(f)} = N_{(f)}$ for any $f$ homogeneous in $S_+$. For $f$ of degree 1, it is easy. But for general $f$ is it still true? I can't construct the isomorphism.

For degree 1, it is just $\psi:M_{(f)}\rightarrow N_{(f)}$ sending $\frac{m}{f^k}$ to $\frac{m}{f^{k-n_0}}$ with the obvious converse by multiplying $f^{n_0}$. Many thanks!

It turns out that the confusion comes from the fact that the grading on $N$ is the old grading, and we view the lower degree part as $0.$

答:
As pointed out by Shivering Soldier in the comments, the key fact here is $$\frac{m}{f^n}=\frac{f^km}{f^{k+n}}.$$
Consider the obvious injection $N\to M$. Since localization is exact, $N_f\to M_f$ is also an injection, and as $N\to M$ preserves the grading, we have that the image of $N_{(f)}\subset N_f$ lands in $M_{(f)}$, so we have an injective map $N_{(f)}\to M_{(f)}$. To see that it's surjective, let $\frac{m}{f^n}\in M_{(f)}$: then $\frac{m}{f^n}=\frac{f^km}{f^{k+n}}$ for any $k$, and so fixing $k>n_0$, we see that $f^km\in M_{(n+k)\deg f} = N_{(n+k)\deg f}$ and so $\frac{m}{f^n}$ is in the image of $N_{(f)}\to M_{(f)}$. So this map is surjective and therefore an isomorphism.

Note that no assumptions on $f$ besides being homogeneous of positive degree are necessary.


问:
$\int^{\pi/2}_{0}\log|\sin x| \,dx = \int^{\pi/2}_{0}\log|\cos x| \,dx $
Prove that :

$$\int^{\pi/2}_0 \log|\sin x| \,dx = \int^{\pi/2}_0 \log|\cos x| \,dx $$

I tried to cut the integral into a sum of parts and changing variable but it didn't work out right, i dont know how to solve this kind of problems in any other way, any hint will be much appreciated!

答: Set $t = \pi/2-x$ and make use of the fact that $\displaystyle \int_a^b f(y) dy = - \int_b^a f(y) dy$ to conclude what you want.</p>

ContentLicense: CC BY-SA 3.0


问:
Use of rank theorem
Let $$|rank(P)-rank(Q)|^2=(rank(P)-rank(Q))^2=rank(P)^2-2(rank(P))(rank(Q))+rank(Q)^2$$
$$|rank(P)-rank(Q)|^2\le |rank(P)|^2+2|rank(P)||rank(Q)|+|rank(Q)|^2$$
$$|rank(P)-rank(Q)|^2\le |rank(P)|+|rank(Q)|$$
Since, $rank(P)\ge 0$ and $rank(Q)\ge 0$
$$|rank(P)-rank(Q)|^2\le rank(P)+rank(Q)$$
I'm stuck here and I'm sure if my proof thus far is correct since we know that
$$rank(P+Q)\le rank(P)+rank(Q)$$

答:
I would not say that your proof is on the right track. I would suggest that you instead use the fact (or try to prove yourself) that
$$
\operatorname{rank}(P + Q) \leq \operatorname{rank}(P) + \operatorname{rank}(Q).
$$
Now, if we set $P = A + B$ and $Q = -B$ and note that $\operatorname{rank}(B) = \operatorname{rank}(-B)$, we find that
$$
\operatorname{rank}((A + B) - B) \leq \operatorname{rank}(A + B) + \operatorname{rank}(-B) \implies\\
\operatorname{rank}(A) \leq \operatorname{rank}(A + B) + \operatorname{rank}(B) \implies\\
\operatorname{rank}(A) - \operatorname{rank}(B) \leq \operatorname{rank}(A + B).
$$
Now, if we do the same thing but reverse the roles of $A$ and $B$, we end up with
$$
\operatorname{rank}(B) - \operatorname{rank}(B) \leq \operatorname{rank}(A + B).
$$
Putting these two results together gives us the desired inequality
$$
|\operatorname{rank}(A) - \operatorname{rank}(B)| \leq \operatorname{rank}(A + B).
$$

Title: Use of rank theorem


问:
What is the canonical isogeny?
Given a (split connected reductive) root datum $\Psi$, there is another associated root datum $\Psi'$ and a morphism of root data $\Psi \to \Psi'$ called the canonical isogeny. My question is, is the canonical isogeny ever not an isomorphism?

I ask this question because I believe the answer is yes—simply from the name—but I don't see how it's possible. Specifically, it seems to me that $X \cong X/X_0 \oplus X/\tilde{Q}$, implying $\Psi \cong \Psi'$. What am I missing?

In the first chapter of the Corvallis proceedings (by Springer), a root datum is defined as a quadruple $\Psi = (X, \Phi, X^{\vee}, \Phi^{\vee})$, satisfying some axioms. Most relevant here are that $X$ and $X^{\vee}$ are dual free abelian groups of finite rank and $\Phi \subset X$ is a finite subset in bijection with $\Phi^{\vee} \subset X^{\vee}$. Given a root datum $\Psi$, we define $Q = \mathbb{Z} \Phi \subset X$ and $X_0 = (\Phi^{\vee})^{\perp} \subset X$. More generally, for any subgroup of a free abelian group $L \subset M$ we have $\tilde{L}$ is the maximal subgroup $L \subset \tilde{L} \subset M$ such that $\tilde{L}/L$ has finite order. Springer notes that $L = \tilde{L}$ if and only if $L$ is a direct summand of $M$, but note also that every element $\ell \in \tilde{L}$ has a multiple $m \ell \in L$, as well as $\tilde{\tilde{L}} = \tilde{L}$.

We have (Lemma 1.2) that $X_0 \cap Q = 0$, and $X_0 + Q$ is of finite index in $X$. It is easy to check that $X_0 = \tilde{X}_0$, and so both $X_0$ and $\tilde{Q}$ are direct summands of $X$, they  span $X$, and they must have trivial intersection. Therefore we have both that $X \cong X_0 \oplus \tilde{Q}$ and $X \cong X/X_0 \oplus X/\tilde{Q}$. (Note: I believe this paragraph is correct as written, but it is not as strong as it appears. I was trying to claim that $X = X_0 \oplus \tilde{Q}$, which is not necessarily true—only that $X \supset X_0 \oplus \tilde{Q}$.)

How is it not the case that the canonical isogeny $f: X \to X/X_0 \oplus X/\tilde{Q}$ induces an isomorphism of root data? Is it some slightly more subtle possible difference in $\Phi$ "$\subset$" $X/X_0$ vs $\Phi \subset \tilde{Q}$ (and dual)? Or is the canonical isogeny actually necessarily an isomorphism of root data?

答:
I think I figured this out with the simplest example imaginable, $GL_2$. We have $X = \mathbb{Z}^2$, with basis $\{\lambda_1, \lambda_2\}$, where $\lambda_i \begin{pmatrix} t_1 & 0 \\ 0 & t_2 \end{pmatrix} = t_i$, and $\Phi = \{\pm(\lambda_1 - \lambda_2)\}$. Then $X_0 = \mathbb{Z}(\lambda_1 + \lambda_2)$ and $\tilde{Q} = Q = \mathbb{Z}(\lambda_1 - \lambda_2)$. Each is a direct summand and the two subgroups do have trivial intersection, but they are not complementary. Most immediately, $X_0 \oplus \tilde{Q} \subsetneq X$.

Basically I was thinking of lattices too much like vector spaces. In particular, the word "span" that I used in the question does not apply—despite being linearly independent and full rank, the subgroups $\tilde{Q}$ and $X_0$ do not necessarily generate $X$.

Title: What is the canonical isogeny?


问:
Is it necessarily true that$P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$
I've been trying my luck at the seemingly trivial and intuitive statement that $P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$ but I have struggled to prove it.

My attempts:

$$P(A\lvert C)= \frac{P(A\cap C)}{P(C)}\geq\frac{P(A\cap C)}{P(C)}\times \frac{P(C)}{P(B)}=\frac{P(A\cap C)}{P(C)}\times\frac{P(C\cap B)}{P(B)}=\frac{P(A\cap C)}{P(B)}\times\frac{P(C\cap B)}{P(C)}\\ $$

$$=\frac{P(A\cap C)}{P(B)}\times P(B\lvert C)$$

I do not know where I am going with this... any hints?

答:
If $P(A\mid B)=1$ then $P(AB) = P(B)$. But also $P(B)=P(AB)+P(A^cB)$, so this means $P(A^cB)=0$. Now, since $C\subset B$, we have $P(C)=P(BC)$, which we can write as
$$
P(C)=P(BC)=P(ABC) + P(A^cBC).$$
Now argue that the first term on the RHS equals $P(AC)$, while the second term is zero.


问:
Finding the coordinates of a vector in a specific basis
In an exercise I'm asked the following:

Consider $\mathbb R^2$ with an inner product whos metric matrix relative to the standard basis is:
$$\left( \begin{matrix} 4 & 1  \\ 1 & 1  \end{matrix} \right)$$
and the basis of $\mathbb R^2$, $\mathcal {B} = ((1/2,0) , (1/\sqrt{12},-4/\sqrt{12}))$.

What is the second coordinate of the vector $(1,\sqrt{12})$ in the basis $\mathcal B$

The thing that's bothering me in this question is: Why do we need to use the metric matrix to find out the coordinates of the vector? Wouldn't just a simple change of basis work, or is there something that I'm missing? Thanks

答:
You are right, a simple basis transformation would do it.

However, you can alternatively aim for a specific $i$th coordinate in an orthonormal basis $e_j$ by simply evaluating $\langle x,e_i\rangle$.

So now if you have verified that the given basis is orthonormal for the given inner product (which it is), then
$$\pmatrix{1&\sqrt{12}}\pmatrix{4&1\\1&1}\pmatrix{1/{\sqrt{12}} \\ -4/\sqrt{12}}$$
will give you the second coordinate in $\mathcal B$.


问:
Find the Lyapunov function for the nonlinear second order system.
Consider the system
$$
\begin{aligned}
\dot{x}_1 &= -x_1+x_2,\\
\dot{x}_2 &= -x_2^3.
\end{aligned}
$$
The origin is obviously globally asymptotically stable. Indeed, $x_2\to 0$ and the dynamics of $x_1$ is linear and thus converges to zero for the converging input $x_2$.

However, I got stuck trying to find a Lyapunov function for this system. Could you please help?

Body:
Best I could find is using SOSTOOLS (see also the manual, especially chapter 4.2), but its not a "nice" solution:

$$
V(x) = v(x)^TP v(x)
$$

$$
v(x) = \begin{pmatrix}
x_1 \\
x_2 \\
x_1^2 \\
x_1 x_2 \\
x_2^2
\end{pmatrix}
$$

$$
P = \begin{pmatrix}
0.2270 & -0.2270 & 0.0000 & 0.0000 & -0.0000 \\
-0.2270 & 0.9522 & -0.0000 & -0.0000 &0.0000 \\
0.0000 & -0.0000 & 0.3215 & -0.0000 & -0.0278 \\
0.0000 & -0.0000 & -0.0000 & 0.5520 & 0.0293 \\
-0.0000 & 0.0000 & -0.0278 & 0.0293 & 0.3014
\end{pmatrix}
$$

At least $P > 0$ and $\dot{V}(x) \leq 0$. So it should be a Lyapunov function, but unfortunatly not enough for asymptotic stability...

EDIT: If you include not the full monomials up to fourth degree but only

$$
m(x) = \begin{pmatrix}
x_1^2 \\
x_2^2 \\
x_1 x_2 \\
x_2^4
\end{pmatrix}
$$

for the optimization then SOSTOOLS computes

$$
\begin{align}
V(x) &= (0.51201 x_1 - 0.21772 x_2)^2 + (0.21772 x_1 - 0.90981 x_2)^2 + 0.41052 x_2^4 \\
&= 0.30955 x_1^2 - 0.61911 x_1 x_2 + 0.41052 x_2^4 + 0.87515 x_2^2
\end{align}
$$

and

$$
\begin{align}
\dot{V}(x) = &-(-1.268 x_2^3 + 0.00000073 x_1 x_2^2 - 0.1309082 x_2 + 0.130908 x_1)^2 \\
&-(0.1309082 x_2^3 + 0.0000034 x_1 x_2^2 + 0.5486238 x_2 - 0.5486187 x_1)^2 \\
&-(0.130908 x_2^3 + 0.0000036 x_1 x_2^2 + 0.5486187 x_2 - 0.5486244 x_1)^2 \\
&- (1.063507 x_2^2 + 0.00006736922 x_1 x_2)^2 \\
&- (0.00000073 x_2^3 - 0.0000019 x_1 x_2^2 - 0.0000034 x_2 + 0.0000036 x_1)^2 \\
&- (0.00006736922 x_2^2 + 0.002748198 x_1 x_2)^2
\end{align}
$$

If you scale $V$ this is just:

$$
\frac{V(x)}{0.30955} =  x_1^2 - 2 x_1 x_2 + 1.326 x_2^4 + 2.827 x_2^2
$$

Your solution is:

$$
V(x) = x_1^2 - 2 x_1 x_2 + \frac{1}{2} x_2^4 + x_2^2
$$

So it finds a somewhat similar solution but the $\dot{V}$ is of course quite horrendous.


问:
For a filter on set, ultrafilters are exactly prime filters.
I'm following through a book, and I'm understanding the proof of the proposition that ultrafilters are exactly prime filters.

The proof that ultrafilters are prime goes as the following.

Suppose $F$ is ultrafilter on a set $X$ that is not prime. Then there are $A, B \subseteq X$ with $A \cup B \in F$ but $A, B \notin F$. We have proved before in the text that a filter $U$ on set $X$ is an ultrafilter if and only if for every $A \subseteq X$, $A \notin U$ if and only if there is $B \in U$ with $A \cap B = \emptyset$.

So we may find $A', B' \in F$ with $A \cap A' = \emptyset = B \cap B'$. The proof says that this implies $(A \cup B) \cap (A' \cup B') = \emptyset$. I'm having trouble with this line as $A \cap A' = \emptyset = B \cap B'$ does not imply $(A \cup B) \cap (A' \cup B') = \emptyset$. For example, take $A = \{0,1\}$ $A' = \{2,3\}$, $B = \{2,4\}$, $B' = \{1,5\}$.

So with what condition do we know that $(A \cup B) \cap (A' \cup B') = \emptyset$?

After this line, the proof is easy as $(A \cup B) \cap (A' \cup B') = \emptyset$ if and only if $A \cup B \notin F$.

Thanks!

答:
$$ (A\cup B)\cap (A'\cup B')
=(A\cap A')\cup (A\cap B')\cup (B\cap A')\cup (B\cap B')
=(A\cap B')\cup (B\cap A')$$


答:
Ok. I have now tried this:

\begin{align}
2y \frac{dx}{dy}-1&=x^2y\\
\frac{1}{x^2}2y\,dx-\frac{1}{x^2}&=y\,dy\\
-x^{-2}+x^{-2}dx= \frac{y}{1}dy
\end{align}

Does this make sense..?

Title: IVP - Differential equation

Body:
Working on this. Have to find a solution. Very new to DE:

$$\begin{cases}
 2yx'(y)-1=x^2(y) \\
 x(y=1)=1 \\
 \end{cases}
$$

Writing this as:

\begin{align}
2y \frac{dx}{dy}-1&=x^2y\\
\frac{1}{x^2}2y\,dx-1&=y\,dy\\
\frac{1}{x^2}\,dx &= 1+y\,dy\\
\int \frac{1}{x^2}\,dx &= \int 1+y\,dy\\
-\frac{1}{x}+C &= \frac{y^2}{2}+y+C
\end{align}

Is this correct? Feel like I'm just doing algebra, but don't really understand if it's correct. Would love feedback.


答:
I think the confusion arises because you don't have a definition of conjugate. Unfortunately, a "cool enough" definition of conjugate is quite abstract. I think it's still a good idea to try to get a feeling of the argument behind conjugation without understanding all the details.

If you know some field theory, the field-theoretic definition of "conjugate" goes as follows: given a Galois field extension $K/F$ and an element $\alpha \in K/F$, you can look at the Galois group $\mathrm{Gal}(F(\alpha)/F)$, and for any $\sigma \in \mathrm{Gal}(F(\alpha)/F)$, the operation $\alpha \mapsto \sigma(\alpha)$ will be called a "conjugation map" and the element $\sigma(\alpha)$ is called a conjugate of $\alpha$. When $[F(\alpha):F] = 2$ (or in other words, when the extension is quadratic and Galois), the group $\mathrm{Gal}(F(\alpha)/F)$ has order $2$, and its non-trivial element $\sigma$ sends $\alpha$ to "its (non-trivial) conjugate" $\sigma(\alpha)$.

If this made no sense to you, just know that in the most famous quadratic extensions that you might have heard of, such as $\mathbb C / \mathbb R$ or $\mathbb Q(\sqrt 2)/\mathbb Q$, the conjugation operation does what you think it does, namely
$$
a+bi \mapsto a-bi, \qquad a+b \sqrt 2 \mapsto a-b\sqrt 2.
$$
What it does behind the scenes is that it looks at the roots of the minimal polynomial of the element and "swaps the roots". So for instance, the polynomial with real coefficients of minimal degree that has $a+bi$ as a root (where $a,b \in \mathbb R$) is
$$
(x-(a+bi))(x-(a-bi)) = x^2 - 2ax + (a^2+b^2) = 0
$$
and the conjugation operation maps one root of that polynomial to the other one. The same idea happens for $\mathbb Q(\sqrt 2)/\mathbb Q$; the polynomial with rational coefficients of minimal degree that has $a+b\sqrt 2$ as a root (where $a,b \in \mathbb Q$) is
$$
(x-(a+b\sqrt 2))(x-(a-b\sqrt 2)) = x^2 - 2ax + (a^2-b^2) = 0
$$
and again, the conjugation operation maps one root to the other.

The first moment calculus students usually encounter the word "conjugate" is when trying to work with difference of square roots, something like
$$
\lim_{x \to \infty} \frac{\sqrt{2x+1} - \sqrt{2x-1}}2
= \lim_{x \to \infty} \frac{(\sqrt{2x+1} - \sqrt{2x-1})(\sqrt{2x+1} + \sqrt{2x-1})}{2(\sqrt{2x+1} + \sqrt{2x-1})} \\
= \lim_{x \to \infty} \frac{(2x+1)-(2x-1)}{2(\sqrt{2x+1} + \sqrt{2x-1})}
\underset{x \to \infty}{\longrightarrow} 0.
$$
This is a general trick, and I would argue that the use of the word "conjugate" is interpreted by most people as "the technique I use to clear the square roots", but it has nothing to do with the change of sign $(x+y) \mapsto (x-y)$ and more to do with the identity $x^2-y^2 = (x-y)(x+y)$ where you want to consider "the other factor" and make it appear in your problem to square both $x$ and $y$, which are usually wrapped in a square root in some way ($\sqrt 2$, $i = \sqrt{-1}$, etc.). The other factor is called "the conjugate factor".

Formally speaking though, this is always at least the image of some element considered a conjugate of something in some field extension. Let $F$ be a field, and consider the field extension $F(\sqrt b)/F(b)$ where $b$ can be anything (an element of $F$, an algebraic element over $F$, transcendental, indeterminate, whatever). If $\sqrt b$ is not in $F(b)$ (or in other words, if you can't compute the square root within $F(b)$), this extension is Galois of degree $2$ with minimal polynomial $m_{\sqrt b}(t) = t^2 - b$. Over $F(\sqrt b)$, this polynomial splits as $t^2-b = (t-\sqrt b)(t+\sqrt b)$, so that $-\sqrt b$ is conjugate to $\sqrt b$. Any element of $F(\sqrt b)$ can be uniquely written as $u+v \sqrt b$ where $u,v \in F(b)$. The conjugation operation is what you expect it to be, namely $u+v \sqrt b \mapsto u - v \sqrt b$. This is a field automorphism of $F(\sqrt b)$ which fixes $F$, i.e. an element of $F(\sqrt b)$ is equal to its own conjugate if and only if $v=0$, i.e. if and only if the element belongs to the fixed field $F(b)$ of the automorphism. In the field extension $F(t,\sqrt b)/F(t,b)$, elements can be uniquely written as $u(t) + v(t) \sqrt b$ where $u(t), v(t)$ are ratios of polynomials in $t$. The conjugation operation still exists here (for the same reasons) and it maps $(t+\sqrt b)$ to $(t-\sqrt b)$ and vice-versa.

All this to say that even in the most abstract settings, the equation $(x^2-y^2) = (x-y)(x+y)$ is still of crucial importance, and the two factors $x-y$ and $x+y$ are called conjugate of each other. If you should remember one thing of that long discussion, it is that conjugation has more to do with $x^2-y^2$ than it has to do with changing the sign of $x+y$ to $x-y$.

Évariste Galois himself studied all the possible ways one could look at a general multivariate polynomial (not just $x^2-y^2$) and see if there were any restrictions on how to permute the factors while respecting the algebraic structure that the elements $x$ and $y$ could have. Turns out there are, because if there were no restrictions, the number of permutations would always be $n!$ (such as $2! = 2$ in the case of $x^2-y^2 = (x-y)(x+y) = (x+y)(x-y)$). Some Galois extensions are very different from this, and it's not easy to picture them without diving into the theory! When the factorization of the polynomial admits enough "good" permutations to catch all the possible permutations of the factors, the field generated by using the factors is called a Galois extension, and Galois theory (the theory of Galois field extensions) is probably one of the most influential field of abstract algebra to this day.

Hope that helps,

Title: If $x-y$ were written as $-y+x$, then would its conjugate be $-y-x$?

Body:
The conjugate of $x-y$ is $x+y$. But, if I wrote it as $-y +x$, would the conjugate be $-y-x$?

Furthermore, is the conjugate of $x + y +z$

$(x+y)+z \to (x+y) - z $

$(x+z)+y \to (x+z) - y $

$(z+y)+z \to (z+y) - x $

Is there a way to formalize the idea of a conjugate?


答: $P(A\mid B)=1$ means that $B\setminus A$ is a set of (probability) measure zero. Then so is $C\setminus A$.

Title: Is it necessarily true that$P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$

Body:
I've been trying my luck at the seemingly trivial and intuitive statement that $P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$ but I have struggled to prove it.

My attempts:

$$P(A\lvert C)= \frac{P(A\cap C)}{P(C)}\geq\frac{P(A\cap C)}{P(C)}\times \frac{P(C)}{P(B)}=\frac{P(A\cap C)}{P(C)}\times\frac{P(C\cap B)}{P(B)}=\frac{P(A\cap C)}{P(B)}\times\frac{P(C\cap B)}{P(C)}\\ $$

$$=\frac{P(A\cap C)}{P(B)}\times P(B\lvert C)$$

I do not know where I am going with this... any hints?


问:
Residue fields of gaussian primes
I just started with algebraic number theory and need some help: In the ring of Gaussian integers a prime $p$ with $p=1\bmod 4$ splits into a product of two irreducible elements $(a+bi)(a-bi)$, with $a^2+b^2=p$, if $p=3\bmod 4$ then $p$ stays prime in $\mathbb{Z}[i]$ and if $p=2$ then $p$ ramifies as $(1+i)^2$. I wonder how I can see that the different residue fields that you get from the prime ideals generated by the irreducible elements just described have $p$, $p^2$ and $2$ elements respectively. I.e. the norm of the generator gives the size of the field. Why?</p>

答:
Besides the wonderful answer of KCD, we can also avail of the theorem of Dedekind. Also see IV.XXXI.

Now we take $f(x)=x^2+1$.

If $p\equiv 1\pmod4$, then there is an integer $a$ such that $a^2\equiv -1\pmod p$. So, according to the theorem, $(p)=(p,i-a)(p,i+a)$. Now $\mathbb Z[i]/(p,i-a)\cong (\mathbb Z[i]/(i-a))/(p)\cong \mathbb Z/(p)$, so the residue field in question has $p$ elements.

If $p\equiv -1\pmod 4$, then $x^2+1$ is irreducible over $Z/(p)$, so $(p)\mathbb Z[i]=(p,i^2+1)=(p)$, and hence the residue field $\mathbb Z[i]/(p)\cong \mathbb Z[x]/(p,x^2+1)\cong (\mathbb Z/p\mathbb Z)[x]/(x^2+1)$ is an extension of $\mathbb Z/p\mathbb Z$ of degree $2$, namely, it has $p^2$ elements.

If $p=2$, then $x^2+1\equiv (x-1)^2\pmod p$, so $(2)=(2,i-1)^2$, and the residue field $\mathbb Z[i]/(2,i-1)\cong \mathbb Z/(2)$ has $2$ elements.

Inform me, if the notations are too ambiguous, or if some errors take place here.

Thanks for any attention.

Title: Residue fields of gaussian primes


问:
Cauchy-Riemann conditions implied by harmonicity and orthogonal gradients?
If I have two functions $\eta:\mathbb{R}^2\to\mathbb{R}$ and $\xi:\mathbb{R}^2\to\mathbb{R}$ for which the following relations hold:

$$
\begin{align}
\left( \frac{\partial\xi}{\partial x} \right)^2 + \left( \frac{\partial\xi}{\partial y} \right)^2 &= \left( \frac{\partial\eta}{\partial x} \right)^2 + \left( \frac{\partial\eta}{\partial y}\right)^2 \neq 0\,,\\
\frac{\partial^2\xi}{\partial x^2} + \frac{\partial^2\xi}{\partial y^2} &= \frac{\partial^2\eta}{\partial x^2} + \frac{\partial^2\eta}{\partial y^2} = 0\,,\\
\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial x} + \frac{\partial\xi}{\partial y}\frac{\partial\eta}{\partial y} &= 0\,.
\end{align}$$

Do these conditions imply that these two functions satisfy Cauchy-Riemann conditions?

$$
\begin{align}
\frac{\partial\xi}{\partial x} &= \frac{\partial\eta}{\partial y}\,,\\
\frac{\partial\xi}{\partial y} &= -\frac{\partial\eta}{\partial x}\,.
\end{align}
$$

If so, how to show that? If not, is there some simple condition that would need to be added?

答:
A proof for the statement that it $u,w$ are real harmonic and satisfy the two conditions above:

$u_x^2+u_y^2=w_x^2+w_y^2$ (and not identically zero) and $u_xw_x+u_yw_y=0$ then if $v$ is a harmonic conjugate of $u$ we have $w=v+C$ or $w=-v+C$ and conversely those functions work, so one needs an extra condition on $w$ to ensure it is a harmonic conjugate and not a negative of such, for the C-R to hold wr $u,w$.

First, we note that by C-R if $w= \pm v+C$ the identities work.

Then taking $v$ a conjugate of $u$ and substituting the C-R equations in the orthogonality relation, we get $w_xv_y=v_xw_y$ while also $v_x^2+v_y^2=w_x^2+w_y^2$

But now assume at a point $w_y(z) \ne 0$ then it follows that $v_y(z) \ne 0$ (otherwise we must have $v_x(z)=0$ but then $w_x^2(z)+w_y^2(z)=0$ contradicting $w_y(z) \ne 0$) so we get at $z$:

$w_x^2/w_y^2=v_x^2/v_y^2$ or adding $1$ to each side $(w_x^2+w_y^2)/w_y^2=(v_x^2+v_y^2)/v_y^2$ and since the numerators are equal and non zero, we get $w_y^2(z)=v_y^2(z)$ so $w_y = \pm v_y$

If $w_y(z)=0$ then automatically $v_y(z)=0$ (since if it would not, then $w_x(z)=0$ and then the sum of squares is zero etc) so the equality holds everywhere.

By the same reasoning $w_x= \pm v_x$ everywhere and of course, if we are at a non-zero value of both $v_x, v_y$ the signs must coincide which means that $w_y=v_y$ or $w_y=-v_y$ on any connected component of the complement of $v_x=0$ (set being same with $w_x=0$ by the above, while the common zeroes of $v_x,v_y$ are discrete of course as they are zeroes of a non-constant analytic function $(u+iv)'$); but now $w_y+v_y$ is a real harmonic function and same with $w_y-v_y$ so if either is zero on any open set, it is zero in the plane, hence we must have the same sign everywhere and then the same holds for $w_x,v_x$ and we are done as we get that either $w=v+C$ or $w=-v+C$


答:
It appears to be a typo. $A',B'\in\mathscr{F}$, so $A'\cap B'\in\mathscr{F}$, and clearly

$$(A\cup B)\cap(A'\cap B')=\varnothing\,,$$

implying that $A\cup B\notin\mathscr{F}$.

Title: For a filter on set, ultrafilters are exactly prime filters.

Body:
I'm following through a book, and I'm understanding the proof of the proposition that ultrafilters are exactly prime filters.

The proof that ultrafilters are prime goes as the following.

Suppose $F$ is ultrafilter on a set $X$ that is not prime. Then there are $A, B \subseteq X$ with $A \cup B \in F$ but $A, B \notin F$. We have proved before in the text that a filter $U$ on set $X$ is an ultrafilter if and only if for every $A \subseteq X$, $A \notin U$ if and only if there is $B \in U$ with $A \cap B = \emptyset$.

So we may find $A', B' \in F$ with $A \cap A' = \emptyset = B \cap B'$. The proof says that this implies $(A \cup B) \cap (A' \cup B') = \emptyset$. I'm having trouble with this line as $A \cap A' = \emptyset = B \cap B'$ does not imply $(A \cup B) \cap (A' \cup B') = \emptyset$. For example, take $A = \{0,1\}$ $A' = \{2,3\}$, $B = \{2,4\}$, $B' = \{1,5\}$.

So with what condition do we know that $(A \cup B) \cap (A' \cup B') = \emptyset$?

After this line, the proof is easy as $(A \cup B) \cap (A' \cup B') = \emptyset$ if and only if $A \cup B \notin F$.

Thanks!


问:
Prove subset equivalence
Given $A,B$ and $C$ are subspace to a vector space $T$ over $\mathbb{R}$

If $A\subset T$ then show

$A\cap T=A$

答:
Why is $ (X \cap Z) + (Y \cap Z)$ equal to $(X + Y) \cap Z$? Did you try it before? Because otherwise you would have to prove it.

$\Rightarrow )$ We know that $X = X \cap Z$ because $X \subset Z$ so
\begin{equation}
\begin{aligned}
& X= X\cap Z \subset X \ \ \ \mbox{and} \ \ \ Y \cap Z \subset Y \ \ \Rightarrow \ \ X+(Y \cap Z) \subset X+Y \\
& X= X\cap Z \subset Z \ \ \ \mbox{and} \ \ \ Y \cap Z \subset Z \ \ \Rightarrow \ \ X+(Y \cap Z) \subset Z
\end{aligned}
\end{equation}
then $X + (Y \cap Z) \subset (X+Y)\cap Z$.

$\Leftarrow )$ Let $a \in (X+Y)\cap Z$ then $a \in Z$ and $a\in X+Y$ so exist $x\in X, y\in Y$ such that $a = x+y$. Remember that $X \subset Z$ then $x \in Z$. Since $Z$ is a vector subspace it follows
\begin{equation}
y = x+y - x = a - x \in Z
\end{equation}
Hence $y \in Y \cap Z$, i. e., $a = x+y \in X+ (Y\cap Z)$.

Title: Prove subset equivalence


问:
The measure of the image of the exponential of real matrices
It is known that the matrix exponential over the real matrices $\exp : M_n(\mathbb{R}) \to GL_n(\mathbb{R})$ is not surjective and that its image $S
$ is the subset of all invertible matrices that are the square of a real matrix.

My question is about the "size" of that set within $GL_n(\mathbb{R})$ equipped with the Lebesgue measure of $\mathbb{R}^{n^2}$. Do we know if $S$ has full measure ? Or is it a null set ?

答: Note that the image includes all real matrices $n\times n$ matrices that have $n$ distinct positive eigenvalues.  The set of all such matrices is open and nonempty, and therefore has positive Lebesgue measure.  Indeed, since this set is closed under multiplication by positive scalars, the Lebesgue measure must be infinite.


答: Hint:  Use the fact $$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$  You can show this by letting $u = a-x$.

Title: $\int^{\pi/2}_{0}\log|\sin x| \,dx = \int^{\pi/2}_{0}\log|\cos x| \,dx $

Body:
Prove that :

$$\int^{\pi/2}_0 \log|\sin x| \,dx = \int^{\pi/2}_0 \log|\cos x| \,dx $$

I tried to cut the integral into a sum of parts and changing variable but it didn't work out right, i dont know how to solve this kind of problems in any other way, any hint will be much appreciated!


答:
Remember that when we say $P(A) = 1$, it means that $A$ happens "almost surely" (or in other words, as far as the probability is concerned, $A$ always happens). So $P(A|B)$ means that whenever $B$ happens, $A$ happens. Saying that $C \subseteq B$ means that if $C$ happens, $B$ surely happens (not even almost surely, this time, surely!). So if $C$ happens, $B$ surely happens, and if $B$ happens, $A$ happens almost surely...  so it's a bit like $C \implies B \implies A$ but with some probability jargon around it.

More formally, $1 = P(A|B) = \frac{P(A \cap B)}{P(B)}$, so
$$
P(A \cap B) = P(B) = P(A \cap B) + P(A^c \cap B)
$$
The second equality follows from the measurability of $B$, which is required to consider probabilities. It follows that $P(A^c \cap B) = 0$. Since $C \subseteq B$, we also have
$$
0 \le P(A^c \cap C) \le P(A^c \cap B) = 0 \implies P(A^c \cap C) = 0.
$$
Running our above equalities in reverse, we get
$$
P(C) = P(A \cap C) + P(A^c \cap C) = P(A \cap C) \implies P(A|C) = 1.
$$
Hope that helps,


问:
Only solution to Diophantine equation with primes?
Find all solutions to:

$$p_1 ^2 - p_2^2 = p_3^2 + 1$$

where $p_i$ is a prime number (positive of-course)
The only solution that seems to exist is $(3,2,2)$ for $(p_1, p_2, p_3)$ accordingly. But I have no clue how to prove that no other solutions exist..
The only thing that jumps to my mind is assuming that another solution exists and somehow get to a contradiction...

I would highly appreciate your help, thank you!

答: Let $(p_1,p_2,p_3)\ne(3,2,2)$ If $p_1=2$ then we have $p_2^2+p_3^2=3$ which is impossible by the known theorem on the representation of a natural integer as a sum of two squares. Thus $p_1$ must be odd but in this case we have $p_2^2+p_3^2\equiv0\pmod4$ which is impossible since $p_2$ or $p_3$ must be distinct of $2$.


问:
differentiation with respect to a quaternion
My question is as follows: if I have a unit quaternion(used to represent the rotation) $q=[q_{0}\,\textbf{q}]$ and a pure quaternion $v=[0\,\textbf{v}]$,where $\textbf{v}$ is a 3x1 vector.* is the quaternion multiplication.

what is $\frac{\partial (q*v*q^{-1})}{\partial v}$? could you show me which one of following options is right?

I have searched some materials about quaternion derivatives, and one of them(https://web.cs.iastate.edu/~cs577/handouts/quaternion.pdf) show an result like:$\frac{\partial (q*v*q^{-1})}{\partial v}=(q_{0}^2-||\textbf{q}||^2)I_{3}+2\textbf{q}\textbf{q}^T-2q_{0}\textbf{q}\times$

Another potential answer is:$\frac{\partial (q*v*q^{-1})}{\partial v}=q*e_{i}*q^{-1}$, where $e_{i}$ are basis of inertial frame.

what is $\frac{\partial (q*v*q^{-1})}{\partial q}$?

what is $\frac{\partial (q*p^{-1})}{\partial q}$? p is also a unit quaternion.

答: I'll change your notation slightly, denoting multiplication as$$[a,\,u][b,\,v]=[ab-u\cdot v,\,av+bu+u\times v]$$so$$\begin{align}qvq^{-1}&=\frac{[q_0,\,q][0,\,v][q_0,\,-q]}{q_0^2+q^2}\\&=\frac{[-q\cdot v,\,q_0v-q\times v][q_0,\,-q]}{q_0^2+q^2}\\&=\frac{[0,\,2(q\cdot v)q+(q_0^2-q^2)v+2q_0v\times q]}{q_0^2+q^2},\end{align}$$with $q^2=q\cdot q$ in terms of the $3$-vector $q$. Differentiation with respect to $v_a,\,a\in\{1,\,2,\,3\}$ gives a pure quaternion whose $b$th imaginary entry, $b\in\{1,\,2,\,3\}$, is$$\frac{2q_aq_b+(q_0^2-q^2+2q_0\times q)\delta_{ab}}{q_0^2+q^2}.$$You can treat parts 2, 3 similarly.


问:
Composition of Lebesgue measurable function $f$, with a continuous function $g$ having a certain property, is Lebesgue measurable

Suppose that $f$ is Lebesgue measurable and $g$ is real valued, continuous, and has the property that for any null set $N$, $g^{-1} (N)$ is measurable. Then $f \circ g$ is also Lebesgue measurable.

That seems entirely too strong? Any idea why this is true?

答:
Any Lebesgue measurable set can be written as the symmetric difference $B \vartriangle N$ of a Borel set $B$ and a null set $N$.  Since $g$ is continuous, $g^{-1}(B)$ is then also Borel, and, by assumption, $g^{-1}(N)$ is measurable, so $g^{-1}(B \vartriangle N)=g^{-1}(B) \vartriangle g^{-1}(N)$ is also Lebesgue measurable.  This means that applying $g^{-1}$ to a Lebesgue measurable set gives another Lebesgue measurable set.  Also, since $f$ is Lebesgue measurable, applying $f^{-1}$ to an open set gives a Lebesgue measurable set.  So, if $O$ is open,
$$(f\circ g)^{-1}(O)=g^{-1}(f^{-1}(O))$$
must be Lebesgue measurable, meaning that $f\circ g$ is Lebesgue measurable.</p>


问:
Why are fixed points of a discrete time system backward-invariant?
Consider a discrete time system given by $x(t+1) = F(x(t))$.

A backward invariant set $S$ is one such that if the solution $x(t)\in S$ at time $T$, for all $t<T$, $x(t) \in S$.

I see why a fixed point is forward invariant, but why is it backward invariant? Is there proof that it cannot be reached from another point?

Thanks in advance!

答:
This is false. For example, take $F(x) : \mathbb{R} \to \mathbb{R}$ to be the function $\text{max}(x - 1, 0)$; in other words, $F(x)$ subtracts $1$ until we get a number less than or equal to $0$ and then just outputs $0$. The unique fixed point of this function is $\{ 0 \}$, and there are trajectories such as

$$x(t) = \begin{cases} -t & \text{ if } t \le 0 \\ 0 & \text{ otherwise} \end{cases}$$

showing that $0$ is not backwards invariant (here I'm assuming $t \in \mathbb{Z}$).

It is, of course, true if $F$ is assumed to be invertible, since then $x(t) = F^{-1}(x(t+1))$, and $F(x) = x$ implies $F^{-1}(x) = x$, so we can run time evolution backwards uniquely in this case.


问:
Can PMF of transformed discrete RV sum to less than $1?$
I'm trying to understand an exercise about transformation of random variables. Because the result is giving a value lower than 1 (at least in my head), and I'm not sure that's possible, given that random variables should always total to 1. Here's the example:

Let $X$ be a discrete R-valued random variable with p.m.f. $f_X,$ where $\mathrm{supp}(f_X) = \{-2,-1,0,1,2\},$ and $f_X(-2) = 3/20,$ $f_X(-1) = 7/20,$ $f_X(0) = 5/20,$ $f_X(1) = 4/20,$ $f_X(2) = 1/20.$

Let $g: R \rightarrow R$ be defined by:
$g(x) := |x|,$ for $ -3/2 <= x <= 3/2,$ and $x - 1,$ otherwise. Let $Y = g(X)$

The exercise asks to give all values of $f_Y(y)$

So my thought process here is that when you apply $g(x),$ you get:

$g_X(-2) = -2 -1 = -3,$ which is not in $f_X,$ therefore equal to $0.$

$g_X(-1) = 1 = f_Y(1) = 4/20 + 7/20 + 1/20$ (since we have to sum all values that will result in 1, which is  $12/20.$

Likewise, $g_X(1) = 1 = f_Y(1) = 12/20.$

Also, $g_X(2) = 2 - 1 = f_Y(1) = 12/20.$

Then, $g_X(0) = 5/20.$

The problem is that $12/20 + 5/20 = 17/20,$ therefore it could not be a random variable.

What am I missing here? Or which step am I doing wrong?

This is my first post to math stackexchange, so suggestions on how to better format my questions are appreciated! I hope it's clear enough to understand tho.

答:
For verifying your work:  A simulation in R of 100,000 realizations of each random variable gives good approximate
probabilities and provides the opportunity for a
graphic display of the distributions of $X$ and $Y$.

```<code>set.seed(120)
x = sample(-2:2, 10^5, rep=T, p=c(3,7,5,4,1)/20)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 -2.000  -1.000  -1.000  -0.352   0.000   2.000
table(x)/10^5
x
     -2      -1       0       1       2
0.14947 0.35154 0.25046 0.19857 0.04996
</code>```

The table shows approximate probabilities
$p_X(x) = (3/20,7/20,5/20,4/20,1/20)$ for $x=(-2,-1,0,1,2),$
respectively.

```<code>table(x)/10^5

     -2      -1       0       1       2
0.14947 0.35154 0.25046 0.19857 0.04996
</code>```

Transformation:

```<code>y = abs(x)
y[x==-2] = -3
y[x==2] = 1

summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-3.0000  0.0000  1.0000  0.1517  1.0000  1.0000

table(y)/10^5
y
     -3       0       1
0.14947 0.25046 0.60007 # aprx 3/20, 5/20, 12/20
</code>```

The table shows approximate values of probabilities $(3/20, 5/20, 12/20).$

In the histograms below, histogram bars show simulated
proportions of values of $X$ (beige) and $Y$ (blue). Colored dots show
exact probabilities, and illustrate that $X$-values
$-1, 1,$ and $2$ are transformed to $Y$-value $1.$

R code for figure:

```<code>par(mfrow=c(1,2))
 hist(x, prob=T, br=-2.5:2.5, ylim=c(0,.6), col="wheat")
  farb.x = c("red", "blue", "green3", "blue", "blue")
  points(-2:2, c(3,7,5,4,1)/20, pch=19, col=farb.x)
 hist(y, prob=T, br=-3.5:1.5, col="skyblue2")
  farb.y = c("red", "black", "black", "green3", "blue")
  points(-3:1, c(3,0,0,5,12)/20, pch=19, col=farb.y)
par(mfrow=c(1,1))
</code>```

Note: This exercise can be used to illustrate
that while $E(Y)$ can be found by definition as

$$E(Y) = \sum_{\mathrm{all\,}y}  yp_Y(y)
= (-3)\frac{3}{20} + (0)\frac{5}{12}+(1)\frac{12}{20} = \frac{3}{20}.$$

$E(Y)$ can also be found by the 'law of the unconscious statistician' as follows:

$$E(Y) = E(g(X)) =
\sum_{\mathrm{all\;}x}g(x)p_X(x)\\
= g(-2)\frac{3}{20}+g(-1)\frac{7}{20}+g(0)\frac{5}{20}+g(1)\frac{4}{20}+g(2)\frac{1}{20}\\
= \frac{-9}{20} + \frac{7}{20}+\frac{4}{20}+\frac{1}{20} = \frac{3}{20}.$$

Thus, it is not necessary to know the distribution of $Y$ in order to find $E(Y),$ if $Y = g(X)$ and the distribution of $X$ is known.

Body:
For verifying your work:  A simulation in R of 100,000 realizations of each random variable gives good approximate
probabilities and provides the opportunity for a
graphic display of the distributions of $X$ and $Y$.

set.seed(120)
x = sample(-2:2, 10^5, rep=T, p=c(3,7,5,4,1)/20)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 -2.000  -1.000  -1.000  -0.352   0.000   2.000
table(x)/10^5
x
     -2      -1       0       1       2
0.14947 0.35154 0.25046 0.19857 0.04996

The table shows approximate probabilities
$p_X(x) = (3/20,7/20,5/20,4/20,1/20)$ for $x=(-2,-1,0,1,2),$
respectively.

table(x)/10^5

     -2      -1       0       1       2
0.14947 0.35154 0.25046 0.19857 0.04996

Transformation:

y = abs(x)
y[x==-2] = -3
y[x==2] = 1

summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-3.0000  0.0000  1.0000  0.1517  1.0000  1.0000

table(y)/10^5
y
     -3       0       1
0.14947 0.25046 0.60007 # aprx 3/20, 5/20, 12/20

The table shows approximate values of probabilities $(3/20, 5/20, 12/20).$

In the histograms below, histogram bars show simulated
proportions of values of $X$ (beige) and $Y$ (blue). Colored dots show
exact probabilities, and illustrate that $X$-values
$-1, 1,$ and $2$ are transformed to $Y$-value $1.$

R code for figure:

par(mfrow=c(1,2))
 hist(x, prob=T, br=-2.5:2.5, ylim=c(0,.6), col="wheat")
  farb.x = c("red", "blue", "green3", "blue", "blue")
  points(-2:2, c(3,7,5,4,1)/20, pch=19, col=farb.x)
 hist(y, prob=T, br=-3.5:1.5, col="skyblue2")
  farb.y = c("red", "black", "black", "green3", "blue")
  points(-3:1, c(3,0,0,5,12)/20, pch=19, col=farb.y)
par(mfrow=c(1,1))

Note: This exercise can be used to illustrate
that while $E(Y)$ can be found by definition as

$$E(Y) = \sum_{\mathrm{all\,}y}  yp_Y(y)
= (-3)\frac{3}{20} + (0)\frac{5}{12}+(1)\frac{12}{20} = \frac{3}{20}.$$

$E(Y)$ can also be found by the 'law of the unconscious statistician' as follows:

$$E(Y) = E(g(X)) =
\sum_{\mathrm{all\;}x}g(x)p_X(x)\\
= g(-2)\frac{3}{20}+g(-1)\frac{7}{20}+g(0)\frac{5}{20}+g(1)\frac{4}{20}+g(2)\frac{1}{20}\\
= \frac{-9}{20} + \frac{7}{20}+\frac{4}{20}+\frac{1}{20} = \frac{3}{20}.$$

Thus, it is not necessary to know the distribution of $Y$ in order to find $E(Y),$ if $Y = g(X)$ and the distribution of $X$ is known.


问:
The symmetry of conjugacy classes of a subgroup of a finite group
I have been solving some exercise in group theory and came up with two observations, the first and second below. Then I tried to generalize the fact from that exercise and got an idea of other two observations (the third and fourth below), but was unable to prove or disprove them. Any further observations about them is welcome, and a proof or disproof would resolve my problem.

Let $G$ be a finite group, let $H\leq G$ be its subgroup, and let's define the notation $[x]_A$ to be the conjugacy class of element $x\in A$ of group $A\in \{G,H\}$.

It holds that $(\forall x\in H)\Big(\frac{|[x]_G|}{|[x]_G\cap H|}\text{ divides }[G:H]\Big)$.

It holds that $(\forall x\in H)[x]_H\subseteq [x]_G$.

Does it hold that $(\forall x,y\in H)\big(\text{some of the numbers }\frac{|[x]_G|}{|[x]_G\cap H|}\text{ and }\frac{|[y]_G|}{|[y]_G\cap H|}\text{ divides the other one}\big)$?

Does it hold that $(\forall x\in H)(\forall a,b \in[x]_G\cap H)|[a]_H|=|[b]_H|$?

The exercise that I was solving was the next one:

Let $n\in\mathbb{N}$ be a natural number. And let $H\leq S_n$, $[S_n:H]<n$ be a subgroup of symmetric group $S_n$ of order less than $n$. If $n=5$, prove that $H$ is either $A_n$ or $S_n$.

To prove that, I used the first statement from above. Then I tried to generalize this exercise and replace the condition $n=5$ with $n\geq 5$ and came up with the third and fourth statements as possibly useful lemmas.

答:
3 is false. Take $G=S_6$, and let $H$ be the stabilizer of $6$, which is isomorphic to $S_5$ and has index $6$ in $G$. Let $x=(1234)$; the number of elements in the conjugacy class of $x$ in $G$ is the number of $4$-cycles in $G$, which is $90$. Of those, the ones that are in $H$ are precisely the $4$-cycles that fix $6$, which is the same number as the number of $4$-cycles in $S_5$, namely $30$ of them. So your first quotient is
$$\frac{|[x_G]|}{|[x_G]\cap H|} = \frac{90}{30} = 3.$$
Now let $y=(123)$. Its conjugacy class in $G$ consists of all $3$-cycles, and there are $40$ of them; the ones in $H$ are the ones that fix $6$, and there are as many of them as there are $3$-cycles in $S_5$, namely $20$ of them. So the second quotient is
$$\frac{|[y_G]|}{|[y_G]\cap H|} = \frac{40}{20} = 2.$$
Of course, neither of $2$ and $3$ divide the other.

(I think the answer for 4 is "yes", but I'm still checking) As has been shown, the answer for 4 is "no" in general.

But it is true if $H$ is normal, because in that situation the $G$-conjugacy classes of elements of $H$ split into equal-sized conjugacy classes in $H$.

To see this, assume that $a,b\in H$ are conjugate in $G$, so that there exists $g\in G$ with $gag^{-1}=b$. The number of elements in $[a]_H$ is the index of its centralizer in $H$,
$$C_H(a) = \{h\in H\mid ha=ah\},$$
and the number of elements in $[b]_H$ is likewise the index in $H$ of
$$C_H(b)=\{h\in H\mid hb=bh\}.$$
Now consider the action of $g$ on $H$ by conjugation: I claim that $gC_H(a)g^{-1} = C_H(b)$. Indeed, if $h\in C_H(a)$, then $ghg^{-1}\in C_H(b)$:
$$(ghg^{-1})b = ghg^{-1}gag^{-1} = ghag^{-1} = gahg^{-1} = gag^{-1}(ghg^{-1}) = b(ghg^{-1}).$$
Thus, $gC_H(a)g^{-1}\leq C_H(b)$. Symmetrically, $g^{-1}C_H(b)g\leq C_H(a)$, which gives the converse inclusion. Thus, $[H:C_H(a)]=[H:C_H(b)]$, and so $|[a]_H|=|[b]_H|$, as desired.

(More generally, if $\varphi$ is an automorphism of $G$, then $\varphi(C_G(a)) = C_G(\varphi(a))$, and here $\varphi$ is the automorphism of $H$ induced by conjugation by $g$.)


问:
Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$

Problem 1: Let $a, b, c \ge 0$ with $c = \min(a, b, c)$ and $ab + bc + ca = 2$. Prove that
$\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$.

Background: This problem was proposed by csav10@AoPS
(https://artofproblemsolving.com/community/q1h2321937p18543869)

which is a variant of the following:

Problem 2: Let $a, b, c \ge 0$ with $ab + bc + ca = 2$. Prove that $\sqrt{a+ab} + \sqrt{b+bc} + \sqrt{c+ca} \ge 3$.

(Proposed by KaiRain@AoPS. See https://artofproblemsolving.com/community/c6h1905412.

Also see: https://artofproblemsolving.com/community/q1h2308247p18537783.)

For Problem 2, yleo@AoPS and csav10@AoPS gave nice solutions, respectively.
Also, I solved Problem 2 by the Buffalo Way (BW) which is a computer solution.

For Problem 1, I have not yet solved it by the Buffalo Way (BW).

By the way, Equality case: $(a, b, c) = (2, 1, 0)$.

Any comments and solutions are welcome and appreciated.

Add the Buffalo Way solution (outline) for Problem 2

Fact 1: If $b \ge a \ge c \ge 0$, then
$$\sqrt{a + ab} + \sqrt{b + bc} + \sqrt{c + ca} \ge \sqrt{b + ba} + \sqrt{a + ac} + \sqrt{c + cb}.$$

From Fact 1, we assume that $a \ge b \ge c$.

After homogenization, we need to prove that
$$\sqrt{\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}} \ge 3.$$
Denote
$$X = \frac{1}{4}\cdot\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}},
Y = \frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}},
Z = \frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}.$$
We need to prove that $2\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.

By using $\sqrt{u} \ge \frac{2u}{1+u}$ for $u\ge 0$, it suffices to prove that
$$\frac{4X}{1+X} + \frac{2Y}{1+Y} + \frac{2Z}{1+Z} \ge 3.$$
After some manipulation, it suffices to prove $A\sqrt{\frac{ab+bc+ca}{2}} + B \ge 0$ where $A, B$ are some polynomials and $A \ge 0$.

It suffices to prove that $A^2\cdot \frac{ab+bc+ca}{2} \ge B^2$.

BW works.

答:
Full answer

This is an answer with a different method than my other (partial) answer, therefore a new one.

Outline: this answer establishes a polynomial inequality and then shows that the polynomial will have no zeroes which gives the proof. To do so, we apply a method where all coefficients of this polynomial are inspected, which makes the proof somewhat technical.

Note that for $a\ge b$, we observe for the exchange of $a\leftrightarrow b$ that  $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3 \le \sqrt{b + ab} + \sqrt{a} + \sqrt{c} - 3$, hence we need to consider the case $a\ge b$ only. Since $c$ is required to be the smallest variable, we consider $a\ge b \ge c$.

As expressed in the problem statement, a single minimum of the target function is expected at $b=1$ and $c=0$, with equality.

To establish a polynomial, first replace all variables by their squares. Then we need to show that
$$
f = a \sqrt{1 + b^2} + b + c - 3 \ge 0
$$
with the condition $ g = a^2b^2 + b^2 c^2 + c^2 a^2 - 2 = 0$. The condition for $f$ can be written
$$
a^2 (1 + b^2) - (b + c - 3)^2 \ge 0
$$
(Note $b + c - 3 <0$ in the case of concern, $a\ge b\ge c$, so this is indeed equivalent to $f \ge 0$.)
From $g$, replacing  $a^2 = \frac{2 - b^2c^2}{b^2 + c^2}$ then gives that we have to prove
$$
F = (2 - b^2c^2) (1 + b^2) - (b^2 + c^2)(b + c - 3)^2 \ge 0
$$
which is now a bivariate polynomial with conditions on the range of $b$ and $c$ as follows (for qualitative illustration see also the first figure in my other answer):

We turn to the condition $c = \min(a, b, c)$.
Due to the condition $g$, we have that $c \le b$ for $b \in [0 \quad \sqrt[4]{\frac23}]$. If $b$ gets larger  than $\sqrt[4]{\frac23}$, we have to ensure $a \ge b$ which entails the limit $a=b$ in $g$, giving $c \le \sqrt{\frac{2 - b^4}{2 b^2}}$ for  $b \in [ \sqrt[4]{\frac23}  \quad \sqrt[4]{2} ]$. The upper limit for $b$ is given when $a=b$ and $a^2 b^2 = 2$, which forces $c=0$ in $g$. We  will come back to these two ranges for $b$ later.

We can now replace $c = x b$ with $x \in (0 \quad 1]$. Doing so we have to show

$$
F = (2 - b^4 x^2) (1 + b^2) - b^2(1 + x^2)(b + b x - 3)^2 \ge 0
$$
Regard $F$ as a polynomial in $x$ with parameter $b$. For some permitted values $(x, b)$ we have $F > 0$. For proving the claim, we establish now that for $x \in (0 \quad 1]$, this polynomial has no roots, hence $F>0$ everywhere in the permitted range of $b$ and $c$. To do so, we use Budan's theorem. The theorem says that if the number of sign changes of the coefficients in $F(x)$ equals the number of sign changes of the coefficients in $F(x+1)$, then the polynomial $F(x)$ has no roots in  $(0 \quad 1]$.

We have
$$
F(x)  = [- b^4 + 6b^3 - 7b^2 + 2] + [6 - 2b] b^3 x + [- b^4 - 3b^2 + 6b - 9] b^2 x^2 +  [6 - 2b] b^3 x^3 +
[-1]b^4 x^4
$$
Regardless of the value of $b \in [0  \quad \sqrt[4]{2} ]$, we have the following signs, symbolically denoted by
$$
F(x)  = ["+"] + ["+"]  x + ["-"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
which makes three sign changes.

For $b \in [0 \quad 1]$, we have
$$
F(x+1)  = [- b^6 - 9 b^4 + 24 b^3 - 16 b^2 + 2] + [- 2 b^4 - 18 b^2 + 36 b - 18] b^2 x + [- b^4 - 15 b^2 + 24 b - 9] b^2 x^2 +  6 [1 - b] b^3 x^3 +
[-1]b^4 x^4
$$
Again we have the following signs, symbolically denoted by
$$
F(x+1)  = ["+"] + ["-"]  x + ["-|+"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
where $["-|+"]$ indicates that, depending on the value of $b \in [0 \quad 1]$, this coefficient has different signs. No matter what this sign, however, there are three sign changes in the coefficients of $F(x+1)$.

For $b \in [1 \quad \sqrt[4]{2} ]$, we have   $c \le \sqrt{\frac{2 - b^4}{2 b^2}}$. To make matters easier, we increase the range of $c$ by the wider limit $c \le 3 - 2 b$. We use this limit to introduce $c = x( 3 - 2 b)$ with  $x \in [0 \quad 1]$.
Then we have

$$
F (x) = (2 - x^2 b^2(3 - 2 b)^2) (1 + b^2) - (b^2 + x^2(3 - 2 b)^2)(b + x(3 - 2 b) - 3)^2
$$
This gives
$$
F(x)  = [- b^4 + 6 b^3 - 7 b^2 + 2] +[4 b^2 - 18 b + 18] b^2 x + [- 4b^6 + 12 b^5 - 21 b^4 + 60 b^3 - 135 b^2 + 162 b - 81] x^2 +  [16 b^4 - 120 b^3 + 324 b^2 - 378 b + 162] x^3 +
[- 16 b^4 + 96 b^3 - 216 b^2 + 216 b - 81] x^4
$$
We have the following signs, symbolically denoted by
$$
F(x)  = ["+"] + ["+"]  x + ["-"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
for all $b \in [1 \quad \sqrt[4]{2} ]$, which makes three sign changes in the coefficients of $F(x)$.

For $x+1$, this gives
$$
F(x+1)  = [- 4 b^6 + 12 b^5 - 18 b^4 + 24 b^3 - 16 b^2 + 2] +[ - 8 b^5 + 24 b^4 - 54 b^3 + 126 b^2 - 144 b + 54] b x + [- 4 b^6 + 12 b^5 - 69 b^4 + 276 b^3 - 459 b^2 + 324 b - 81] x^2 +  [- 48 b^4 + 264 b^3 - 540 b^2 + 486 b - 162] x^3 +
[- 16 b^4 + 96 b^3 - 216 b^2 + 216 b - 81] x^4
$$
Again we have the following signs, symbolically denoted by
$$
F(x+1)  = ["+"] + ["-"]  x + ["-"] x^2 +  ["+"] x^3 +
["-"] x^4
$$
for all $b \in [1 \quad \sqrt[4]{2} ]$, which makes again three sign changes in the coefficients of $F(x+1)$.

This establishes that for all $b$, the polynomial $F$ has no roots, which completes the proof. $\qquad  \Box$

** Remark: For the two ranges of $b$, another proof of the inequality  $F\ge 0$ with the help of Buffalo Way has been given by River Li here **


问:
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$

I did:

I take $S=\{x\in \mathbb{R} \mid x^3+x<3\}$

So, this set is bounded and by completeness of the real numbers exists $a=\sup(S)$

Then I want to prove that $a^3+a=3$

And then in order to get a contradiction I suppose that $a^3+a>3$

and how $a $ is the supremum then $\forall\epsilon\in\mathbb{R} $ exists $b \in S$ such that
$a-\epsilon<b$ then $(a-\epsilon)^3+b<b^3+b<3$

But I'm stuck from here. I'm not able to use derivatives, or continuity so I can't use IVT

答:
Hint 1:

Apply the intermediate value theorem (which does not require taking derivatives) to the continuous function $f(x)=x^3+x-3$ on some closed interval.

Hint 2:

For the function defined above, $f(-1)<0$ and $f(2)>0$.

I don't understand what you mean by "I'm not able to use derivates or continuity". If you wish, you can unwrap the proof of the intermediate value theorem and apply the steps to the function $f$ above.

Alternatively, you may use the fundamental theorem of algebra and the fact that all complex roots come up as conjugate pairs. (Well, you only have the real-analysis tag...)


问:
Prove $\sin x +\frac{1}{(\sin x)^{\sin x}}<2, \quad x \in(0,\pi/2)$
No idea how to prove that. What should I try?
$$\sin x +\frac{1}{(\sin x)^{\sin x}}<2, \quad x \in(0,\pi/2)$$</p>

答:
let $$\sin{x}=t$$

use$ AM-GM$,we hve
$$\dfrac{1}{t^t}=\left(\dfrac{1}{t}\right)^t\cdot 1^{1-t}\le t\cdot\dfrac{1}{t}+(1-t)\cdot 1=2-t$$


问:
Stationary points of a bivariate quadratic function along the standard unit circle
Where along the unit circle $\mathbf{x}(\theta)=\begin{bmatrix}cos\theta\\sin\theta\end{bmatrix}$ does the quadratic function $f(\mathbf{x})=(\mathbf{x}-\mathbf{\mu})^TP(\mathbf{x}-\mathbf{\mu})$ attain its stationary points?

My attempt

$$\frac{\partial f}{\partial \mathbf{x}}=(\mathbf{x}-\mathbf{\mu})^T(P+P^T)$$

$$\frac{\partial\mathbf{x}}{\partial\theta}=\begin{bmatrix}-sin\theta\\cos\theta\end{bmatrix}$$

$$\frac{df}{d\theta}=(\begin{bmatrix}cos\theta\\sin\theta\end{bmatrix}-\mathbf{\mu})^T(P+P^T)\begin{bmatrix}-sin\theta\\cos\theta\end{bmatrix}=0$$

How do I proceed to solve for $\theta$ interms of $\mu$ and $P$? Do I have to break up $P$ into its entries or can I get an explicit matrix/vector formula? (We can assume that $P$ is symmetric if it helps somehow.)

答:
Let us start by considering an example to build up intuition. Take $P$ to be the identity matrix, so that $f(x) = \|x-\mu\|^2$. It follows that $\nabla f = 2(x-\mu)$, so that for $\frac{df}{d\theta} = 0$ to have a solution, we have to have $\nabla f\left( \frac{dx}{d\theta} \right) = 0$. Since $x(\theta)$ and $\frac{dx}{d\theta}$ are always orthogonal, the stationary points are those for which $\frac{dx}{d\theta}$ and $\mu$ are orthogonal. Since a line and a circle intersect in two points, there are two solutions. If $\nu$ is a vector spanning the line orthogonal to $\mu \neq 0$, then the two solutions are $\frac{\pm \nu}{\|nu\|}$ (this also includes the case where $\mu$ is on the unit circle; in this case, $\mu$ is a stationary point because $(x-\mu) = 0$, so that $\nabla f = 0$ is orthogonal to $\frac{dx}{d\theta}$). If $\mu = 0$, all points are stationary and $f(x(\theta))$ is a constant function of $\theta$.

It's not so hard to picture this example: picture a paraboloid centered at $\mu$ and intersect it with a cylinder centered around the $z$-axis of radius $1$. The two surfaces will intersect, and the curve of intersection's $z$-coordinate will have minimal $z$-value closest to $\mu$ and maximal $z$-value farthest from $\mu$, given that $\mu$ is outside the unit circle. (At least that's the picture I have in my head.)

We can assume without loss of generality that $P$ is symmetric, because if we write $f(P,x,\mu) = (x-\mu)^{\top} P (x-\mu)$, then $f(P,x,\mu)$ is a real number, therefore
$$
f(P,x,\mu) = f(P,x,\mu)^{\top} = f(P^{\top},x,\mu),
$$
and by linearity we deduce
$$
\frac {f(P, x,\mu) + f(P^{\top}, x, \mu)}2
= \frac 12 \left((x-\mu)^{\top} P (x-\mu) + (x-\mu)^{\top} P^{\top} (x-\mu)
\right) \\
= \frac 12 (x-\mu)^{\top} \left( P(x-\mu) + P^{\top} (x-\mu) \right) \\
= \frac 12 (x-\mu)^{\top} \left( P + P^{\top} \right) (x-\mu) \\
= (x-\mu)^{\top} \left( \frac {P+P^{\top}}2 \right) (x-\mu) \\
= f((P+P^{\top})/2, x,\mu)
$$
Since the matrix $\frac{P+P^{\top}}2$ is symmetric, the operation $P \mapsto \frac{P + P^{\top}}2$ has no effect on the problem and replaces $P$ by a symmetric matrix.

We assume without loss of generality that $P$ is a symmetric matrix. By the spectral theorem, $P$ is orthogonally diagonalizable, i.e. we can write $P = Q^{\top} D Q$ where $Q$ is an orthogonal matrix. It follows that we can re-write
$$
f(P,x,\mu)
= f(Q^{\top}DQ, x, \mu) \\
= (Q(x-\mu))^{\top} D Q(x-\mu) \\
= (Qx, Q\mu)^{\top} D (Qx - Q\mu) \\
= f(D, Qx, Q\mu)
$$
so that up to a change of basis, we can assume without loss of generality that $D$ is diagonal, because the orthogonal matrix just "turns the problem around"; if $x$ is a stationary point of $x \mapsto f(D,Qx,Q\mu)$, then $Qx$ is a stationary point of $y \mapsto f(D,y,\mu)$. Note that the orthogonality of $Q$ is important; orthogonal matrices map the unit circle to itself.

Write $D = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}$ and $x = (x_1,x_2)$ and $\mu = (\mu_1,\mu_2)$. We have $\nabla f = (\lambda_1(x_1-\mu_1), \lambda_2(x_2-\mu_2))$, and since $\frac{dx}{d\theta}$ is parallel to $(-x_2, x_1)$, we see that the stationary points will satisfy
$$
\lambda_1(x_1-\mu_1)x_2 - \lambda_2(x_2-\mu_2)x_1 = 0 = (\lambda_1-\lambda_2) x_1x_2 - \lambda_1\mu_1 x_2 + \lambda_2\mu_2 x_1.
$$
We have three cases, up to a permutation matrix (note that a permutation matrix maps the unit circle to itself because it is orthogonal). Note that the unit circle is compact and our function is continuous, so there is a minimum and a maximum, which means at least $2$ stationary points for sure.

I considered splitting into five cases

$\lambda_1, \lambda_2 > 0$.

$\lambda_1 > 0 > \lambda_2$.

$\lambda_1, \lambda_2 < 0$.

$\lambda_1 > 0, \lambda_2 = 0$

$\lambda_1 = \lambda_2 = 0$

but I couldn't get more specific than the polynomial equation I wrote above. In the last two cases (which you could call trivial), we can already say a little bit: set $\lambda_2 = 0$ and assume $\lambda_1 \neq 0$ (there is nothing to say when $\lambda_1 = 0$), so that $f(D,x,\mu) = \lambda_1 (x_1-\mu_1)^2$ and the equation for stationary points becomes
$$
0 = \lambda_1 x_1x_2 - \lambda_1\mu_1 x_2 = \lambda_1(x_1 - \mu_1) x_2
$$
so that either $x_2 = 0$ (and $x_1 = \pm 1$) or $x_1 = \mu_1$, which is only possible if $\mu_1 = \pm 1$, and in this case $x_1 = \mu$. In both cases, the solutions are $(\pm 1, 0)$. So there are two solutions in this case. The extrema $(\pm 1, 0)$ take the values $\lambda_1 (\pm 1 - \mu_1)^2$, so one can tell the maxima apart from the minima by comparing those two values.

In the first three cases, all I could say is that this is an equation of the form
$$
a x_1 x_2 + b x_1 + c x_2 = 0
$$
which we can interpret as a linear equation in $x_1$ (or in $x_2$), and solving the system of polynomial equations
$$
a x_1 x_2 + b x_1 + c x_2 = 0 \\
x_1^2 + x_2^2 = 1
$$
is an intersection of two quadrics, which has at most four solutions. More than that I could not say.

Hope that helps,


问:
Union Intersection Test
I am trying to understand the union intersection test shown in my class. Originally we were attempting to test for $X_i\sim  N_p(\mu, \Sigma)$ that $H_0:\mu=u_o$ the professor proposed an equivalent test
$H_o: \cap_{a\in\mathbb{R}^p}\{a^t\mu=a^t\mu_0\}$ vs the alternative of just finding one $a$ such that this does not hold. She mentioned that we must be mindful of accounting for multiple comparison tests and want to control for error. That  is we want to make sure that $P(\cup_{a\in\mathbb{R}^p}|\frac{\sqrt{n}a^{T}(\bar{X}-\mu)}{\sqrt{a^{T}Sa}}|>c)\le \alpha$ which are the  rejection regions for the t-tests resulting from the projection using $a$. Then she stated something that made no sense to me she said that this is equivalent to $P\bigg(\underset{a\in\mathbb{R}^p}{\text{sup}}|\frac{\sqrt{n}a^{T}(\bar{X}-\mu)}{\sqrt{a^{T}Sa}}|>c\bigg)$. I can't see how these probabilities are equivalent. The only sense I can make of this is that the equivalency follows from instead of testing multiple tests we are going to find one such $a$ that we can reject therefore we can reject the null. That is we are only going to perform one test using the sup as our projection and we can control type 1 error. Is this correct?

答:
Let $f(\omega,a):=\left|\sqrt{n}a^{\top}(\bar{X}_n(\omega)-\mu)/\sqrt{a^{\top}S_n(\omega)a}\right|$, and let
\begin{align}
A&:=\bigcup_{a\in \mathbb{R}^p}\{\omega:f(\omega,a)>c\} \quad\text{and} \\
B&:=\left\{\omega:\sup_{a\in\mathbb{R}^p}f(\omega,a)>c\right\}.
\end{align}
If $\omega\in A$, then $\exists a'$ s.t. $f(\omega,a')>c$. This means that $\sup_{a\in\mathbb{R}^p}f(\omega,a)\ge f(\omega,a')>c$, i.e., $\omega\in B$. This proves the first inclusion, $A\subseteq B$. On the other hand, when $\omega\in B$, $\exists a''$ s.t. $f(\omega,a'')>c$ so that $\omega\in A$ (note that we couldn't conclude that if we used the weak inequality instead). This proves the second inclusion, $B\subseteq A$. Combining both inclusions, we get $A=B$.

Title: Union Intersection Test