问:
What holds in a deductively complete system?
I read an article which presented some system with axioms and inference rules (I don't know if that type of system have a term for in English).

The article stated that the system is "deductively complete".

I have read in Wikipedia a bit to recall what does it mean (its been a while since I studied
logic).

In Wikipedia in states that $S$ is deductively complete system if
for each formula $\varphi$ either $\varphi$ or $\neg\varphi$ is
a theorem of $S$ .

Do I understand correctly that this system is not neccaseraly consistent
? for example you can add $$p\to\neg p$$ which is a contradiction and
call the new system $S'$. If for a given $\varphi$ you had a proof
for $\varphi$ or $\neg\varphi$ (that did not use the new axiom we
added) then the same proof works in $S'$hence $S'$ is also deductively
complete but is not consistent.

So if I understand correctly, a deductively complete system need not
be sound.

Does it have to be complete ? the name of it suggest so, but I believe
that the answer is no, for example if $S$ is a complete deductively
complete system then either $\varphi$ or $\neg\varphi$ is a theorem
of $S$ . I think there is maybe a way to construct a deductively
complete system $S'$ in which if $\varphi$ is a theorem of $S$ then
$\neg\varphi$ is a theorem of $S'$ and if $\neg\varphi$ is a theorem
of $S$ then $\varphi$ is a theorem of $S'$. Then $S'$ is not complete
(at least in the case where $S$ is consistent).

To sum up: Did I understand correctly that a deductively complete
system need not be consistent, sound or complete ?

答:
The Wikipedia article you linked to describes the different types of completeness fairly well, but there are some interesting connections between them.  First, semantic completeness (everything true is provable) does not imply syntactic completeness (each sentence or its negation is provable), since there are sentences which are contingent (e.g., $A \lor B$), and the propositional calculus (a semantically complete calculus) won't be able to prove either of $A \lor B$ or  $\lnot(A \lor B)$.

To the first question, a deductively complete system need not be consistent.  The trivial counterexample is that an inconsistent system proves everything, and thus proves $\phi$ or $\lnot\phi$ for every $\phi$ (in fact, it proves both!).

To the second question, a deductively complete system need not be (semantically) complete.  We could construct, for instance, an esoteric calculus with bizarre proof rules that guarantee that for every formula $\phi$, either $\phi$ or $\lnot\phi$ is a theorem.  For instance, given the propositional language with variables $A$, $B$, $C$, $\dots$, the connectives $\lnot$, $\land$, $\lor$, and $\to$, where well-formed formulae are defined as usual, take as your system just the following two axiom schemata:

$\phi$ where $\phi$ is a wff in which $\land$ appears an odd number of times

$\lnot\phi$ where $\phi$ is a wff in which $\land$ does not appear an odd number of times

Since every formula $\phi$ either has an odd number of $\land$s, in which case $\phi$ is a theorem, or it doesn't, in which case $\lnot\phi$ is a theorem, the system is deductively complete.  Theorems of this system include
\begin{gather*}
\lnot A \qquad A \land B \qquad \lnot(A \land (B \land C))
\end{gather*}
It's pretty obviously not semantically complete (there are true sentences that it does not prove) and even unsound (there are non-true sentences that it does prove), but it's syntactically consistent (it never proves $\phi$ and $\lnot\phi$).

To sum up, a deductively complete system need not be (syntactically) consistent (never prove $\phi$ and $\lnot\phi$), sound ($\vdash \phi$ implies $\models \phi$), or (semantically) complete ($\models \phi$ implies $\vdash\phi$).

It would be an interesting exercise to determine which combinations of those can be had in a proof system, though.  In the examples above saw deductively complete systems that were:

inconsistent, unsound, and (semantically) complete;

consistent, unsound, and (semantically) incomplete;

Without specifying a particular interpretation for the propositional variables, we cannot have a sound deductively complete system, because the deductively complete system has to make some judgment about the contingent sentences.  Any inconsistent system will automatically be (semantically) complete, so the only possible case that we haven't seen is one which is:

consistent, unsound, and (semantically) complete.

Such a system could be had if a propositional variable assignment were part of the proof system.  For instance, take the standard propositional calculus as a starting point.  Then suppose some particular interpretation $\cal I$ of propositional variables (as an easy one, let every propositional variable be true).  The standard propositional calculus is consistent and semantically complete, so we have those.  To get the unsoundness, let us also add the axiom schema inference rule:

$\phi$, if $\cal I \models \phi$.

This will have the effect of making every otherwise contingent sentence (or its negation) a theorem.  Since the propositional calculus is semantically complete, the only $\phi$ such that neither $\phi$ nor $\lnot\phi$ is a theorem are contingent sentences, but now those have been fixed as well, so the resulting system is consistent, unsound (since, e.g., $\vdash A \land B$ and $\not\models A \land B$), and semantically complete.


问:
Derivative for $\frac{\tan x}x$ in terms of $\sin x$ and $\cos x$
The answer I got is in terms of $\sec x$. $$\left.\left[\dfrac{\sec^2x}{x}-\tan x\right] \right/x^2\text{ ?}$$

I simply used the division formula in differentiation to get it.

Not able to get it in terms of $\sin x$ and $\cos x$.

The answer has to be $\dfrac{x-\sin x\cos x}{x^2\cos^2x}$

答:
You made a mistake in your application of the quotient rule. The derivative is
$$
\frac{x\sec^2x - \tan x}{x^2} = \frac{\frac{x}{\cos^2x}-\frac{\sin x}{\cos x}}{x^2} = \frac{\frac{x}{\cos^2x}-\frac{\sin x\cos x}{\cos^2 x}}{x^2} = \frac{x-\sin x\cos x}{x^2\cos^2 x}.
$$

ContentLicense: CC BY-SA 4.0


问:
Finding minimum value of observation for a given power in hypothesis testing.
Let $X_{1}, \ldots, X_{n}$ be a sample from the distribution $\mathrm{N}(\mu, 1)$ and consider testing $H_{0}: \mu=\mu_{0}$ versus $H_{1}: \mu=\mu_{1}$ where $\mu_{0}<\mu_{1}$ are known numbers.

(a) For a given level $\alpha,$ find the most powerful test.

(b) Calculate the power

(c) For given $\mu_{0}, \mu_{1}, \alpha,$ determine the minimal number of observations needed to have power at least $\beta$ (i.e., to reject $H_{0}$ with probability at least $\beta$ when $H_{1}$ holds).

Using Neyman Pearson lemma I have found the test statistic to be $\bar X$ and reject $H_{0}$ if $\bar X > k$, where $k= z_\alpha / \sqrt{n} + \mu_0$ .

And calculated the power to be Power = P{Reject $H_0$} = P{$Z> z_\alpha - \sqrt{n} (\mu-\mu_0)$}

But for the Part (c) I don't know how to find the minimum number of observation for given condition.

I think I have to solve the equation $\beta = P_{\mu_1}\{\bar X > z_\alpha / \sqrt{n} + \mu_0 \}$ = P{$Z> z_\alpha - \sqrt{n} (\mu_1-\mu_0)$}

But I don't know how to solve this for $n$.

Please help me with this. Thnakyou.

答:
As you found, UMP test is given by Neyman Pearson's Lemma with rejection region

$$\mathbb {P}[\overline{X}_n>k|\mu=\mu_0]=\alpha$$

Now $\overline{X}_n>k$ is your decision rule ($k$ now is fixed) and you can calculate the power ( usually indicated with $\gamma$ because $\beta$ is normally used for type II error)

$$\mathbb {P}[\overline{X}_n>k|\mu=\mu_1]=\gamma$$

Understood this, finally fix $\gamma$ and get $n$

Example

$\mu_0=5$

$\mu_1=6$

$\alpha=5\%$

$n=4$

The critical region is

$$(\overline {X}_4-5)2=1.64\rightarrow \overline {X}_4=5.8224$$

Thus your decision rule is

$$\overline {X}_4\geq 5.8224$$

and you can calculate the power

$$\gamma=\mathbb {P}[\overline {X}_4\geq 5.8224|\mu=6]=1-\Phi(-0.36)\approx 64\%$$

Now suppose you want a fixed power $\gamma \geq 90\%$, simply re-solve the same inequality in $n$

$$\mathbb {P}[\overline {X}_n\geq 5.82|\mu=6]\geq 0.90$$

Getting

$$(5.8224-6)\sqrt{n}\leq-1.2816$$

That is

$$n\geq\Bigg \lceil \Bigg(\frac{1.2816}{0.1776}\Bigg)^2\Bigg\rceil=53$$


答:
You can state it in terms of the CDF $\Phi(z)=P(Z \le z)$; I don't think there is a way to avoid it.

\begin{align}
\beta &\le 1 - \Phi(z_\alpha - \sqrt{n}(\mu_1-\mu_0))
\\
\Phi^{-1}(1-\beta) &\ge z_\alpha - \sqrt{n} (\mu_1 - \mu_0)
\\
\sqrt{n} &\ge \frac{z_\alpha - \Phi^{-1}(1-\beta)}{\mu_1 - \mu_0}
\end{align}

Title: Finding minimum value of observation for a given power in hypothesis testing.

Body:
Let $X_{1}, \ldots, X_{n}$ be a sample from the distribution $\mathrm{N}(\mu, 1)$ and consider testing $H_{0}: \mu=\mu_{0}$ versus $H_{1}: \mu=\mu_{1}$ where $\mu_{0}<\mu_{1}$ are known numbers.

(a) For a given level $\alpha,$ find the most powerful test.

(b) Calculate the power

(c) For given $\mu_{0}, \mu_{1}, \alpha,$ determine the minimal number of observations needed to have power at least $\beta$ (i.e., to reject $H_{0}$ with probability at least $\beta$ when $H_{1}$ holds).

Using Neyman Pearson lemma I have found the test statistic to be $\bar X$ and reject $H_{0}$ if $\bar X > k$, where $k= z_\alpha / \sqrt{n} + \mu_0$ .

And calculated the power to be Power = P{Reject $H_0$} = P{$Z> z_\alpha - \sqrt{n} (\mu-\mu_0)$}

But for the Part (c) I don't know how to find the minimum number of observation for given condition.

I think I have to solve the equation $\beta = P_{\mu_1}\{\bar X > z_\alpha / \sqrt{n} + \mu_0 \}$ = P{$Z> z_\alpha - \sqrt{n} (\mu_1-\mu_0)$}

But I don't know how to solve this for $n$.

Please help me with this. Thnakyou.


问:
How are Taylor and Maclaurin series applied in this situation?
I am a high school student that’s having a hard time figuring out what to carry out in his report. As we know that Newton’s law of cooling shows an exponential curve in other words. This question might be stupid, I haven’t started studying both series but I am really interested on carrying out something new.  Would it be possible to use both or any series into my analysis of Newton’s curve of cooling? If not in your opinion what can I include in the analysis of the curve (even most basic would be fine).

Thanks a lot.

答:
Newton law of cooling write
$$T(t)=T_{\text{env}}+(T_0-T_{\text{env}})\,e^{-\frac t \tau}$$ Since the time constant $\tau$ is positive, you can write
$$e^{-\frac t \tau}=\sum_{n=0}^p \frac{(-1)^n }{n!}\left(\frac{t}{\tau }\right)^n$$

Now, the question is too know how many terms you need to add for a given accuracy. Since this is an alternating series you want to know $p$ such that
$$\frac{1 }{(p+1)!}\left(\frac{t}{\tau }\right)^{p+1} \leq \epsilon$$ that is to say
$$(p+1)! \geq \frac 1  \epsilon \left(\frac{t}{\tau }\right)^{p+1}$$ and there are formulae to approximate very accurately $p$ (if you are concerned by this point, let me know and I shall elaborate).

If you use few terms, it could be better to use for approximation the so called Padé approximants which are sytematically much better than Taylor series. For example, around $x=0$,
$$e^{-x} \sim \frac {2-x}{2+x}$$
$$e^{-x} \sim \frac {12-6x+x^2}{12+6x+x^2}$$
are respectively $O(x^3)$ and $O(x^5)$.


答:
A deductively complete theory can be inconsistent. Indeed, as Pece remarks, a classical inconsistent theory entails every sentence $\varphi$ of the relevant language so (a fortiori) entails at least one of $\varphi$ and $\neg\varphi$ for each sentence $\varphi$. But your argument works too: if $T$ is deductively complete, consider $T'$ which you get by adding a contradiction as a new axiom. $T'$ proves everything which $T$ did, so is complete, but is inconsistent by construction.

A deductively complete theory can be inconsistent, no inconsistent theory is sound, so a deductively complete theory need not be sound.

But a deductively complete theory is, trivially, deductively complete. Of course, if we can show that $\varphi$ is a $T$ theorem iff  $\neg\varphi$ is also a theorem, then we know that either $T$ is incomplete or is inconsistent. Turning that about, if $T$ is deductively complete, then either it is inconsistent or it isn't the case that there is some $\varphi$ such that $\varphi$ is a $T$ theorem iff  $\neg\varphi$ is also a theorem.

A deductively complete theory need not be sound on some given interpretation (if it has false axioms), so its set of theorems and the set of relevant truths are distinct. If completeness is understood in a semantic sense, as a matter of yielding a complete story about the relevant truths, then deductively completeness doesn't imply this kind of semantic completeness.

ContentLicense: CC BY-SA 3.0

Title: What holds in a deductively complete system?

Body:
I read an article which presented some system with axioms and inference rules (I don't know if that type of system have a term for in English).

The article stated that the system is "deductively complete".

I have read in Wikipedia a bit to recall what does it mean (its been a while since I studied
logic).

In Wikipedia in states that $S$ is deductively complete system if
for each formula $\varphi$ either $\varphi$ or $\neg\varphi$ is
a theorem of $S$ .

Do I understand correctly that this system is not neccaseraly consistent
? for example you can add $$p\to\neg p$$ which is a contradiction and
call the new system $S'$. If for a given $\varphi$ you had a proof
for $\varphi$ or $\neg\varphi$ (that did not use the new axiom we
added) then the same proof works in $S'$hence $S'$ is also deductively
complete but is not consistent.

So if I understand correctly, a deductively complete system need not
be sound.

Does it have to be complete ? the name of it suggest so, but I believe
that the answer is no, for example if $S$ is a complete deductively
complete system then either $\varphi$ or $\neg\varphi$ is a theorem
of $S$ . I think there is maybe a way to construct a deductively
complete system $S'$ in which if $\varphi$ is a theorem of $S$ then
$\neg\varphi$ is a theorem of $S'$ and if $\neg\varphi$ is a theorem
of $S$ then $\varphi$ is a theorem of $S'$. Then $S'$ is not complete
(at least in the case where $S$ is consistent).

To sum up: Did I understand correctly that a deductively complete
system need not be consistent, sound or complete ?


问:
$||f||=\max\limits_{0 \leq x \leq 1}|x^2f(x)|< \infty$ is not complete?
$f$ is continuous function on $[0,1]$ and $||f||=\max\limits_{0 \leq x \leq 1}|x^2f(x)|< \infty$

Indeed, this defines a norm.

Now, I need to show that this norm is not complete.

Can anyone help?

答:
Here is a slightly more general,  albeit  abstract  approach.

Theorem.  If $\varphi $ is a continuous function on $[0,1]$ such that the set
$$
  Z= \{x\in [0,1]: \varphi (x) = 0\}
  $$
has empty interior, define
$$
  \|f\|_\varphi  = \sup_{0\leq x\leq 1}|\varphi (x)f(x)|,
  $$
for   $f\in C([0,1])$.  Then $\|\cdot\|_\varphi $ is a norm.  It is complete iff $Z$ is empty.

Proof.
Consider the linear transformation
$$
  T: f\in  C([0, 1]) \mapsto  \varphi f \in  C([0, 1]).
  $$
Notice that  $T$ is an injective map because if $T(f)=0$, then $f$ must vanish on $[0, 1]\setminus Z$, but
since this set is dense, one must have that $f=0$.

Observing taht
$$
  \|f\|_\varphi  =   \|T(f)\|, \quad \forall f\in C([0, 1]),
  \tag 1
  $$
it is clear that   $\|\cdot \|_\varphi $ is a norm.

Regarding the last sentence
we will just prove the "only if part" since the converse is much easier.

It  follows from (1) that
$\big (C([0, 1]),  \|\cdot \|_\varphi \big )$ is isometrically isomorphic to the range of $T$.  Assuming that $\|\cdot \|_\varphi $ is
complete, we deduce that the range of $T$ is also complete, and hence closed in $C([0, 1])$.

By the Open Mapping Theorem, we
then conclude that there exists a constant $K>0$ such that
$$
  \|T(f)\|\geq K\|f\|, \quad \forall f\in C([0, 1]).
  $$
Assuming by contradiction that $Z$ is nonempty, choose a point $x_0$ in $Z$, as well as  a neighborhood $U$
of $x_0$, such that $|\varphi (x)|<K/2$, for $x$ in $U$. Choosing  a continuous function $f$ supported on $U$, with $\|f\|=1$, we then have that
$$
  K/2\geq \|f\varphi \|= \|T(f)\|\geq K\|f\| = K,
  $$
a contradiction.  QED

Needless to say,  if $\varphi (x)=x^2$,  if follows from the above that $\|\cdot\|_\varphi $ is not complete.


问:
How to find all integer $x$ such that $x^{2}+5145x+2332\equiv 0\pmod{ 9797}$
How to find all integer $x$ such that $x^{2}+5145x+2332\equiv 0\pmod{ 9797}$. I think $9797$ can factor, then $9797=97\cdot11$. So I can convert to $$x^{2}+5145x+2332\equiv 0\pmod{97}$$
$$x^{2}+5145x+2332\equiv 0\pmod{11}$$
How do to next? Thanks.

答:
The coefficients may be factored as follows:
$$x^2+(15×7^3)x+(44×53)\equiv0\bmod(97×101)$$
Modulo $97$, the equation becomes
$$x^2+4x+4\equiv(x+2)^2\equiv0\bmod97$$
Modulo $101$,
$$x^2-6x+9\equiv(x-3)^2\equiv0\bmod101$$
(We have used the smallest absolute value possible for calculations to simplify the work.) Thus $x\equiv-2\bmod97\equiv3\bmod101$, or by the Chinese remainder theorem $x\equiv2326\bmod9797$.


问:
Defining the exponential map without explicit affine connection
Given a smooth manifold $M$, a smooth vector field $V$, and some $x_0 \in M$, we can define the exponential map through the map $\phi$, where $\phi$ is defined on some open interval of $\mathbb{R}$, $\phi(t_0) = x_0$, and $x = \phi(t)$ satisfies the differential equation
$$\frac{dx}{dt} = V(x),$$
and we let $\phi(t) = \exp(t\cdot V, x_0)$, so $\exp(v, x) = \phi(1)$.

This is taken from Theorem 3.1 of J-M Souriau's Structure of Dynamical Systems. At this point of the book, he hasn't introduced any concept of an affine connection, only assuming a Hausdorff manifold, but here we have somehow defined (and shown the existence of a unique) exponential map. Is there some natural affine connection he has assumed, or can we find the connection corresponding to this given definition of an exponential map? I saw this related question: Exponential maps depends on Riemannian metric?, but I'd like to pinpoint where exactly the metric or associated connection would show up.

I know that the exponential map can be defined in terms of geodesics, and I guess the ODE $dx/dt = V(x)$ is the geodesic equation, which can be solved componentwise with respect to some a suitable coordinate basis on the tangent bundle (e.g., Prop 6.1.2 here). But what metric or connection does this geodesic equation correspond to?

答:
He should not have called this map exponential. The most common name for it is the "time 1 flow of the vector field $V$."

Here is how this relates to the exponential map in differential geometry. What you need for exponential map is an affine connection $\nabla$ on the tangent bundle $TM$, not a Riemannian metric. (Although, a Riemannian metric defines the Levi-Civita connection that you can then use.)
The connection $\nabla$ defines a certain vector field $W$ on $TM$. The time one flow $F$ of $W$ is closely related to the exponential map of $M$: Given a point $p\in M$ and an tangent vector $v\in T_pM$, consider the vector $W(v)\in T_vTM$. Then $F(W(v))=(p',v')$, where $p'\in M$ and $v'\in T_{p'}M$. Lastly,   $\exp_p(v)=p'$.

My favorite reference for this is "Riemannian Geometry" by do Carmo.


问:
Understanding $P_i$ primary component.
From Abstract Algebra by Dummit and Foote, after proving

It says the following.

In class, we defined $N_i$ to be the $p_i$-primary component of $M$. But I'm having trouble understanding what grouping all the cyclic factors corresponding to the same prime $p_i$.

My guess is that in the decomposition, we get primes $p_1, \ldots p_t$. And out of those, we pick out distinct $p_1, \ldots p_n$. And do we say $N_i = R/(p_i^{\alpha_s}) \oplus \ldots \oplus R/(p_i^{\alpha_k})$? And is there no more simplification possible?

Thanks.

答:
You are correct. You let $N_i$ be the direct sum of those components corresponding to the prime $p_i$ (where the $p_i$ are now distinct primes).  There is no more simplification possible.

One thing to be careful of: The $N_i$ are uniquely determined by the group $M$, but the components within an $N_i$ are not uniquely determined.  For example, there are many ways to decompose $\mathbb{Z} / p\mathbb{Z} \oplus \mathbb{Z} / p\mathbb{Z}$ as a direct sum of that form.

Title: Understanding $P_i$ primary component.


问:
Decompose $\frac{1}{a \sin x +b}$ into sum of trigonometric functions
I am wondering if it is possible to write

$$f(x)=\frac{1}{a \sin x +b}$$

as a sum of functions that don't have trigonometric functions in the denominator? Or just pure trigonometric functions, without a constant added. So maybe something like this:

$$\frac{1}{a \sin x +b}=A\sin^3 x \cos^2x + B\sin x \cos ^4 x + C\frac{1}{\cos x}+ D\frac{\cos^2 x }{\sin x}+\dots $$

The reason I'm asking is that I need to integrate products of $f(x)$ with powers $\cos^n x$, and it would be really neat to get rid of the constant in the denominator, as integrals of the form $\int\frac{\cos^n x}{a \sin x +b}dx$ are extremely hard to solve, whereas integral tables contain solutions to $\int \sin^p x \cos^q x dx$ and similar integrals.

One potential starting point could be

$$\frac{1}{a \sin x +b}=\frac{a \sin x -b}{(a \sin x +b)(a \sin x -b)}=\frac{a \sin x -b}{(a^2 \sin^2 x -b^2)}=...$$
with the goal to eventually make the replacement $\sin^2 x +\cos^2 x =1$. Can such a method be formalized? Or is the decomposition I have in mind provable not possible? Thanks for your help!

答:
Consider
$$f(x)=\frac{1}{\sin (x)+k}$$
If $k\ll 1$, you could use  expansions and write
$$f(x)=\sum_{n=0}^\infty (-1)^n \csc^{n+1} (x)\,k^n$$
If $k\gg 1$, you could use  expansions and write
$$f(x)=\sum_{n=0}^\infty (-1)^n \sin^{n}(x) \,k^{-(n+1)} $$


问:
Use the Handshake Lemma to determine the number of edges in GK_n
In chess, a knight's move consists of two spaces either vertically or horizontally, followed by one space in the perpendicular direction. In this way, every knight's move results in an L shaped displacement from the original position. To model all of the legal knight's moves on an n x n chessboard, we can create a graph where each vertex represents a square on the board, and two vertices are adjacent if an only if a knight can legally move between the corresponding squares.
Use the handshake lemma to determine the number of edges in GK_n

Is GK_n always, sometimes or never Eulerian

Does GK_n always, sometimes or never contain an Euler trail

By use of the Handshake Lemma edges are twice the amount of degree sum so if you had a graph GK_4 with 16 vertices, it would have degree sum 48 and 24 edges. I'm not sure how to use this information to determine the number of edges in GK_n. The abstractness is what confuses me.

答:
If I was going to do this problem (find the number of edges in the graph), I'd look at the small cases, and find a formula for the number of vertices of each degree.  (There will be some small cases which need to be handled separately.)

So, let's take a look at the $5 \times 5$, $6 \times 6$, and $7 \times 7$ graphs:

(I've ascribed the vertices with their degrees.)

E.g. for $n \geq 3$, there will always be $4$ vertices of degree $2$ (the four corner vertices).  I would find such formulae for the number of vertices of degrees $3,4,6,8$, and argue that degrees $1,5,7$ are impossible.

To make it formal, we'd need to ensure that every square was accounted for.  Depending on how much formality is required, an argument might also be presented as to why each square has a given degree.

Once we have a formula for the number of vertices of each degree, we can use the Handshaking Lemma to find the number of edges.


问:
Imagine graph that has 10 vertices and 38 edges. Prove that there exist $K_4$ induced subgraph.

Imagine graph that has 10 vertices and 38 edges. Prove that there exist $K_4$ induced subgraph.

Such question already was asked here but I wanted to clarify some things.

One of people who answered claims that "There are only 7 edges missing from $K_{10}$" (what is obviously true). However is next statement is so: Each missing edge can prevent up to 28 different 4-vertex sets from inducing a $K_4$. If this is actually correct then we are done with proof. But I don't see why this is the case. Can you, please, explain this statement?

I was trying to prove it a bit differently: since we have 38 edges we have total sum of degrees equal to $38 \cdot 2 = 76$. Let us find pair of vertices with total degree equals 16 (such one definitely exists otherwise our total degree is not greater than 75 $\Rightarrow$ contradiction). There are basically two situations that we should consider. When one of vertices of this pair has degree 8 and another has degree 8 either and when one has degree 7 and another has degree 9. The second case is trivial. However, I am not managing to finish the first one.

Any help on that? (Please, do not use Turan's theorem if you want to show some alternative proof)

答: The cliques prevented by removing the edge $ab$ are those whose four vertices are $a,b$ and a choice of two from the remaining eight vertices, so they number $\binom82=28$.


答: Let $u$ and $v$ be any two vertices of $K_{10}$. There are $\binom82=28$ remaining pairs of vertices, so $K_{10}$ has $28$ distinct $4$-vertex sets containing the vertices $u$ and $v$. If you remove the edge $\{u,v\}$ from $K_{10}$, you’ve killed off these $28$ possible $K_4$ subgraphs.

Body:
Imagine graph that has 10 vertices and 38 edges. Prove that there exist $K_4$ induced subgraph.

Such question already was asked here but I wanted to clarify some things.

One of people who answered claims that "There are only 7 edges missing from $K_{10}$" (what is obviously true). However is next statement is so: Each missing edge can prevent up to 28 different 4-vertex sets from inducing a $K_4$. If this is actually correct then we are done with proof. But I don't see why this is the case. Can you, please, explain this statement?

I was trying to prove it a bit differently: since we have 38 edges we have total sum of degrees equal to $38 \cdot 2 = 76$. Let us find pair of vertices with total degree equals 16 (such one definitely exists otherwise our total degree is not greater than 75 $\Rightarrow$ contradiction). There are basically two situations that we should consider. When one of vertices of this pair has degree 8 and another has degree 8 either and when one has degree 7 and another has degree 9. The second case is trivial. However, I am not managing to finish the first one.

Any help on that? (Please, do not use Turan's theorem if you want to show some alternative proof)


问:
Parseval's identity
How to prove the Parseval's identity , I know the formal way but how to justify the interchange between the integral and the sum in a rigorously way , in addition what extra condition does the function have to satisfy , is the continuity enough in this case ?</p>

答: Parseval's formula holds for every $L^2$-function, even if it's not continuous. You should be able to find the proof in any decent (sufficiently sophisticated) textbook on Fourier analysis.</p>

Title: Parseval's identity


问:
Why does the residue method not work straight out of the box here?
I'm trying to evaluate the integral $$I = \int_0^{\infty} \frac{\cos(x)-1}{x^2}\,\mathrm{d}x $$ The way I've done this is by rewriting $\frac{\cos(x)-1}{x^2}$ as $\Re\left[\frac{e^{iz}-1}{z^2}\right]$, and then using the residue method to get $$I = \Re\left[\frac{i\pi}{2}\cdot\mathrm{Res}\left(\frac{e^{iz}-1}{z^2}; 0\right)\right]$$ It's clear from the series expansion of $e^{x}$ that the residue evaluates to $i$, which gives us an answer of $-\frac{\pi}{2}$. My question is, why was it necessary to change $\frac{\cos(x)-1}{x^2}$ to a complex-valued function? I see that if I attempt to jump straight to the residue theorem before converting the function, I end up with a residue of $0$, which is clear from the series expansion of $\cos(x)$, which gives $$\frac{\cos(x)-1}{x^2} = -\frac{1}{2}+\frac{x^2}{24}+\cdots$$ But, still, why doesn't the residue theorem immediately apply in this example?</p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{\epsilon > 0}$, I'll "close" the integration around a quarter circle in the first quadrant of the complex plane:

Its radius $\ds{\to \infty}$

An indent $\ds{\pars{~\mbox{arc of radius}\ \epsilon~}}$ around the origen the coordinates is included.

The above mentioned contour doesn't include any pole.

Namely,
\begin{align}
&\bbox[5px,#ffd]{\int_{\epsilon}^{\infty}
{\expo{\ic x} - 1 \over x^{2}}\,\dd x}\ =\
\overbrace{\left.-\
\lim_{R \to \infty}\,\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}{\expo{\ic z} - 1 \over z^{2}}\,\dd z\,
\right\vert_{\ z\ =\ R\exp\pars{\ic\theta}}}^{\ds{0}}
\\[2mm] &\
-\int_{\infty}^{\epsilon}{\expo{\ic\pars{\ic y}} - 1 \over \pars{\ic y}^{2}}\,\ic\,\dd y -
\int_{\pi/2}^{0}{\expo{\ic\epsilon\expo{\ic\theta}} - 1 \over
\epsilon^{2}\expo{2\ic\theta}}\epsilon\expo{\ic\theta}\ic
\,\dd\theta
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&
-\ic\int_{\epsilon}^{\infty}{\expo{-y} - 1 \over  y^{2}}\,\dd y -
\int_{0}^{\pi/2}\,\dd\theta
\\[5mm] = &\
-\ic\int_{\epsilon}^{\infty}{\expo{-y} - 1 \over  y^{2}}\,\dd y -
{\pi \over 2}
\end{align}
Then,
\begin{align}
& \mbox{}
\\ &\ \lim_{\epsilon \to 0^{+}}\,\,\,\Re
\int_{\epsilon}^{\infty}{\expo{\ic x} - 1 \over x^{2}}\,\dd x =
\bbox[5px,#ffd]{\int_{0}^{\infty}{\cos\pars{x} - 1 \over x^{2}}
\,\dd x} =
\bbx{-\,{\pi \over 2}} \\ &
\end{align}


问:
Evaluating $\cos (i)$
This is a question out of my own curiousity.

We know that by Euler $$e^{ix}=\cos x+i\sin x$$ $$e^{-ix}=\cos x-i\sin x$$ .Adding $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ if $x=i$:$$\cos (i)=\frac{e+\frac{1}{e}}{2} \tag!$$.Now this result is surprising.Indeed by  taking cos of complex number i was expecting something more strange on RHS,but here we get a real number.Is such an operation even defined or is it that euler's formula is applicable only when $x$ is real This result is confusing me a bit as i never expected a real number on RHS.What would be an intuitive explanation on this result

Thank you!(i apologise if this is  a stupid question)

答:
Subtracting you should get $$2i\sin(x)=e^{ix}-e^{-ix}$$

Replace $x$ with $i$

Title: Evaluating $\cos (i)$


答:
Euler's formula is valid even for complex $x$. You have shown that if $x=ia$ is purely imaginary,
$$\cos x=\cosh-ix=\cosh a$$
where $\cosh$ is the hyperbolic cosine. This is the link between trigonometric and hyperbolic functions.

Title: Evaluating $\cos (i)$

Body:
This is a question out of my own curiousity.

We know that by Euler $$e^{ix}=\cos x+i\sin x$$ $$e^{-ix}=\cos x-i\sin x$$ .Adding $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ if $x=i$:$$\cos (i)=\frac{e+\frac{1}{e}}{2} \tag!$$.Now this result is surprising.Indeed by  taking cos of complex number i was expecting something more strange on RHS,but here we get a real number.Is such an operation even defined or is it that euler's formula is applicable only when $x$ is real This result is confusing me a bit as i never expected a real number on RHS.What would be an intuitive explanation on this result

Thank you!(i apologise if this is  a stupid question)


问:
Improving convergence of the Newton-Raphson method
How can the Newton-Raphson method (that is, the multivariate generalization of Newton's method, used in the solution of nonlinear systems) be improved so as to attain better convergence? As-is, in most cases a fairly good initial value is required to ensure convergence.

Note that I'm aware other method exist. Here, I'm simply interested in alterations/modifications to Newton's method.

答:
If you want faster than Newton, the next step is Halley method.

It is amazing since, on yesterday, I found a very recent paper on this topic (which is also among my concerns).

For Householder, another paper.


问:
Find all elements in ideal of $\Bbb Q$ generated by $J$ as follows.
Let $\Bbb Q$ be a field of all rational numbersa and its ideal $J = \langle \Bbb Z \cup \{\frac{1}{2}\} \rangle$.

What are the all elements in J?
Is it like this?

$J = \{z + \frac{q}{2} + \frac{n}{2} \mid z,n \in \Bbb Z, q \in \Bbb Q \}$.

答:
A field $K$ never has nontrivial ideals. Indeed as an ideal $I$ is closed under multiplication with field elements and any nonzero element in a field is invertible we find
$0 \neq k \in I \Rightarrow 1=k^{-1} k \in I \Rightarrow j = jk^{-1}k \in I {\forall j \in K} \Rightarrow K=I$

This shows that any ideal in a field $K$ is either $0$ or $K$.


问:
How to understand the Galois *-action on a Dynkin diagram
Let $L/k$ be a (Galois) quadratic field extension, and let $\sigma \in \operatorname{Gal}(L/k)$ be the nontrivial automorphism. Let $h$ be a Hermitian form on $L^{4}$, and let $G = \operatorname{SU}_{4}$ be the special unitary group associated to $h$. Concretely, if $H$ is the matrix of $h$ in some fixed basis, then the $k$-points of $G$ are
$$G(k) = \{ X \in \operatorname{SL}_{4}(L) : X^* H X = H \}$$
After extension to $L$, $G$ becomes isomorphic to $\operatorname{SL}_{4}$.
$$G_L \cong \operatorname{SL}_{4}$$
In particular,
$$G(L) \cong \operatorname{SL}_{4}(L)$$
Let $T \subset G$ be the diagonal subgroup, which is a maximal torus. $T$ is not split over $k$, but is split over $L$. On $L$-points, $T(L)$ corresponds to the usual diagonal subgroup in $\operatorname{SL}_{4}(L)$. For $i=1, 2, 3, 4$, let
$$\eta_i:T(L) \to L^\times$$
be the character which picks off the $i$th diagonal entry. The (absolute) roots for $G$ are
$$
\Phi = \{ \eta_i - \eta_j : i \neq j \}
$$
which is a root system of type $A_3$. A set of simple roots is
$$
\Pi = \{ \eta_1 - \eta_2, \eta_2 - \eta_3, \eta_3 - \eta_4 \}
$$
and the associated Dynkin diagram $A_3$ has the root $\eta_2 - \eta_3$ corresponding to the middle vertex.

I have been unable to understand any sources which describe the $*$-action of the Galois group $\operatorname{Gal}(L/k)$ on the Dynkin diagram. Possibly this should instead focus on the absolute Galois group $\operatorname{Gal}(k^{\operatorname{sep}}/k$), I am not sure.

There is not that much that can possibly happen here, since the automorphism group $\operatorname{Aut}(D)$ of the Dynkin diagram is just $\mathbb{Z}/2\mathbb{Z}$, with the nontrivial element given by reflection across the middle node. So either $\sigma$ acts as this nontrivial automorphism of $D$, or it acts trivially on $D$. However, I can't find any definition concrete enough to tell me which it should be in this particular case.

One definition I've been told is that the $*$-action here should be given by taking a character $\eta$ and then
$$
\sigma \cdot \eta = \sigma \circ \eta \circ \sigma^{-1}
$$
If this definition is correct, then the action is trivial, since conjugation commutes with picking off matrix entries. Is this definition correct? If not, how else to think about this?

答:
Choose $y \in L$ with $\sigma(y)=-y$.
Let's concretely look at the two examples case 1: $H = \pmatrix{0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0}$ and case 2: $H = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$.

Part 1: Seeing the standard Galois action $\alpha \mapsto \sigma \cdot \alpha$ on roots

Method 1: With the Lie algebra

Case 1: The Lie algebra of $G(k)$ is

$$\mathfrak g = \{ X \in M_4(L): Tr(X)=0 \text{ and } XH + H X^* =0\} = \{X \in M_4(L): Tr(X) = 0 \text{ and } x_{ij} = -\sigma(x_{5-j,5-i})\}$$

 $$= \{ \pmatrix{a+by&c+dy&e+fy&gy\\h+iy&j-by&ky&-e+fy\\l+my&ny&-j-by&-c+dy\\oy&-l+my&-h+iy&-a+by} : a, ..., o \in k \}.$$

Surely its scalar extension $\mathfrak g_K$ is (isomorphic to) the split Lie algebra $\mathfrak{sl}_4(L) = \{X \in M_4(L): Tr(X)=0\}$, but let us look closely how Galois operates on that; namely, it does not just act on the entries of the matrix (when written as above), but rather as

$$\sigma_{\mathfrak g_L} := \sigma \otimes id_{\mathfrak g} \qquad \qquad (1)$$

on $\mathfrak g_L:= L \otimes_k \mathfrak g$. That is, e.g. the matrix
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0} \in \mathfrak g_L$$ "correctly viewed" as an element of the tensor product $L \otimes_k \mathfrak g$ is

$$=\frac12 \otimes \underbrace{\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0}}_{\in \mathfrak g} + \frac{1}{2y} \otimes \underbrace{\pmatrix{0&y&0&0\\0&0&0&0\\0&0&0&y\\0&0&0&0}}_{\in \mathfrak g}$$

so that $$\sigma(\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0})=
\frac12 \otimes \pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} + \frac{1}{2\color{red}{\cdot(-y)}} \otimes \pmatrix{0&y&0&0\\0&0&0&0\\0&0&0&y\\0&0&0&0} = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} $$

from which one gets $\sigma (\pmatrix{0&x&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}) = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&-\sigma(x)\\0&0&0&0}$ for all $x \in L$. Likewise, for example,

$$ \sigma(\underbrace{\pmatrix{1&0&0&0\\0&-1&0&0\\0&0&0&0\\0&0&0&0}}_{=:H_{\alpha_1}})= \underbrace{\pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&-1}}_{=:H_{\alpha_3}})$$

whereas $\pmatrix{0&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&0}) \in \mathfrak g$ and hence is actually fixed by $\sigma$.

Now in this case, indeed the diagonal matrices in $\mathfrak g$ form a maximal toral subalgebra (a.k.a. Cartan subalgebra) which contain a maximal split toral subalgebra (namely, the two-dimensional $\mathfrak s := \{ \pmatrix{a&0&0&0\\0&j&0&0\\0&0&-j&0\\0&0&0&-a} : a,j \in k \}$), so we can use the diagonal as above to see how Galois acts on the roots. Turns out that $\sigma$ maps the root space $\mathfrak g_{L, \alpha_1}$ onto the root space $\mathfrak g_{L, \alpha_3}$ (I call $\alpha_i = \eta_i - \eta_{i-1}$ the standard roots of the diagonal in $\mathfrak{sl}_4(L)$); and, accordingly, interchanges the coroots $H_{\alpha_1}$ and $H_{\alpha_3}$; in short, now there are many ways to define the action of Galois on the roots via $\sigma \cdot \alpha_1 := \alpha_3$ and $\sigma \cdot \alpha_2 =\alpha_2$. The actual technical formula to do it is

$$\sigma \cdot \alpha := \sigma \circ \alpha \circ \sigma_{\mathfrak g_L}^{-1}$$

where the first $\sigma$ is just the normal Galois on $L$, but the second (inner) $\sigma_{\mathfrak g_L}$ is defined via $(1)$ and in particular is not just Galois action on entries, so that this normal Galois action $\alpha \mapsto \sigma \cdot \alpha$ is not trivial.

Case 2: Now the Lie algebra of $G(k)$ is

$$\mathfrak g = \{ X \in M_4(L): Tr(X)=0 \text{ and } X = - X^* \} = \{X \in M_4(L): Tr(X) = 0 \text{ and } x_{ij} = -\sigma(x_{j,i})\}$$

 $$= \{ \pmatrix{ay&b+cy&d+ey&f+gy\\-b+cy&hy&i+jy&k+ly\\-d+ey&-i+jy&my&n+oy\\-f+gy&-k+ly&-n+oy&(-a-h-m)y} : a, ..., o \in k \}.$$

Again its scalar extension $\mathfrak g_K$ is (isomorphic to) the split Lie algebra $\mathfrak{sl}_4(L)$, but this time Galois operates on that via

$$\sigma_{\mathfrak g_L} : X \mapsto -X^* (= -\sigma(X)^{tr}).$$

This seems to immediately give us an action $\sigma \cdot \alpha_i = -\alpha_i$ ($i=1,2,3$) on the roots, however we have to be very careful here. The problem is that although again the diagonal matrices in $\mathfrak g$ define a maximal toral subalgebra (CSA) $\mathfrak t$, and certainly $\mathfrak t_L$ is a split CSA in $\mathfrak g_L$, this $\mathfrak t$ does not necessarily contain a maximal split toral subalgebra of $\mathfrak g$ (e.g. if $k$ is a $p$-adic field, one can show there are non-zero ad-diagonalisable elements in $\mathfrak g$, but obviously none is in $\mathfrak t$): and we need that to set up the proper Galois action on the roots.

But if we assume that $k=\mathbb R$, $y=i$ (imaginary unit), then one can show that $\mathfrak g$ has no non-zero split toral subalgebra at all, and we are good to go, and indeed $\sigma$ operates on the entire root system as $-id$.

Method 2: With tori and algebraic geometry

(References: Galois action on character group and Galois action on the character group of a torus, but I find these a bit misleading because they focus on cases where the Galois action is just the one on points / matrix entries. Actually Definition of the unitary group via a cocycle on $\operatorname{GL}_n$ shows better what's going on here.)

Case 1: Our group has a torus with $T(L) = \{ \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} : x_i \in L, x_1 x_2 x_3 x_4 =1\}$ which is not split in $G(k)$, rather we get

$$T(k) = \{ \pmatrix{a&0&0&0\\0&b&0&0\\0&0&\sigma(b)^{-1}&0\\0&0&0&\sigma(a)^{-1}} : a,b \in L, ab \in k\}.$$

Accordingly, the Galois action on $T(L)$ is again not given by $\sigma$ acting on entries, but rather via

$$\sigma_{T(L)} : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \pmatrix{\sigma(x_4)^{-1}&0&0&0\\0&\sigma(x_3)^{-1}&0&0\\0&0&\sigma(x_2)^{-1}&0\\0&0&0&\sigma(x_1)^{-1}}.$$

Again we define the $\sigma$-action on a character $\chi : T(L) \rightarrow L^\times$ of the torus via

$$\sigma \cdot \chi := \sigma \circ \chi \circ \sigma_{T(L)}^{-1}$$

and again we can see that rather than being trivial, e.g. the character $\chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_1}{x_2}$ (i.e. the root $\alpha_1$ written multiplicatively) gets sent to

$$ \sigma \cdot \chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_3}{x_4}$$

i.e. the root $\alpha_3$ written multiplicatively, etc.

Case 2: (But again assuming that the maximal $k$-split torus of $G$ is trivial, e.g in the case $k=\mathbb R$, but not for $p$-adic fields $k$.) Again we can choose $T(L) = \{ \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} : x_i \in L, x_1 x_2 x_3 x_4 =1\}$ but this time in $G(k)$ we get

$$T(k) = \{ \pmatrix{a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&d} : a,b,c,d \in L, abcd =1, a^{-1} = \sigma(a), b^{-1}=\sigma(b), c^{-1}=\sigma(c)\}.$$

Accordingly, the Galois action on $T(L)$ is again not given by $\sigma$ acting on entries, but rather via

$$\sigma_{T(L)} : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \pmatrix{\sigma(x_1)^{-1}&0&0&0\\0&\sigma(x_2)^{-1}&0&0\\0&0&\sigma(x_3)^{-1}&0\\0&0&0&\sigma(x_4)^{-1}}.$$

Again we define the $\sigma$-action on a character $\chi : T(L) \rightarrow L^\times$ of the torus via

$$\sigma \cdot \chi := \sigma \circ \chi \circ \sigma_{T(L)}^{-1}$$

and again we can see that rather than being trivial, e.g. the character $\chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_1}{x_2}$ (i.e. the root $\alpha_1$ written multiplicatively) gets sent to

$$ \sigma \cdot \chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_2}{x_1}$$

i.e. the root $\color{red}{-}\alpha_1$ written multiplicatively, etc.

Part 2: The $*$-action

(References: 6.2 in Borel-Tits, 2.3 in Tits' Classification article in the Boulder Proceedings, ...)

Now this can be done in various ways, using maximal parabolic subgroups and whatnot, but here is a way of doing it just on the root level:

Say we are given a root system $R$ in an Euclidean space $V$, and an action of a finite Galois group $\Gamma$ via automorphisms of $R$.

Set $V_0 := \{v \in V: \sum_{\sigma \in \Gamma} \sigma \cdot v =0 \}$. (To make things work, here it is crucial that in the above one had chosen a maximal Torus $T$ which contains a maximal $k$-split torus $S$: Because then and only then will $V/V_0$ be well-behaved enough and realise the $k$-relative roots as quotient ("folding") of the original root system $R$.)

A $\Gamma$-linear order on $R$ (or $V$) is a choice of a positive half-space in $V$ such that for all $v \in V \setminus V_0$, we have $v$ positive $\implies \sigma \cdot v$ positive for all $\sigma \in \Gamma$. One can show that such $\Gamma$-linear orders always exist.

A $\Gamma$-basis for our root system $R$ is a set of simple roots among the ones positive for a $\Gamma$-linear order. I.e. a basis consisting of roots which are positive for a $\Gamma$-linear order.

Examples: In case 1, there are eight possible $\Gamma$-bases: $\{\alpha_1, \alpha_2, \alpha_3\}$ and its negative, $\{\alpha_1, -\alpha_1-\alpha_2-\alpha_3, \alpha_3\}$ and its negative, $\{\alpha_1+\alpha_2, -\alpha_2, \alpha_2+\alpha_3\}$ and its negative, and a fourth one I leave to you. Whereas e.g. $\{\alpha_2, \alpha_3, -\alpha_1-\alpha_2-\alpha_3\}$ is a basis, but not a $\Gamma$-basis of $R$.

In case 2, every basis of the root system is a $\Gamma$-basis (because $V=V_0$, the extra condition for positive systems to be $\Gamma$-linear is empty).

Now the $*$-action is defined as follows: For any basis $\Delta$ of the root system, $\sigma \cdot \Delta$ is again a basis. As is well-known, the Weyl group operates simply transitively on the set of bases of the root system; further, let's take a $\Gamma$-basis for $\Delta$. Then there is a Weyl group element $w_{\sigma}$ (actually, a unique one) such that $w_{\sigma}(\sigma \cdot \Delta) = \Delta$. Then we define the $*$-action as

$$\sigma * \alpha = w_{\sigma} (\sigma \cdot \alpha).$$

So to say, the $*$-action is the normal Galois action but "bent back" from a $\Gamma$-basis into itself via the Weyl group.

Now in our examples, in case 1, the normal action already stabilises each $\Gamma$-basis, and so the $*$-action is just the normal action. In case 2, the normal action actually sends each basis to its own negative, so as $w_\sigma$ we have to take the longest element of the Weyl group, which again gives the non-trivial automorphism of the Dynkin diagram.

In both cases, the $*$-action of the non-trivial element of the Galois group is given by $\alpha_1 \mapsto \alpha_3, \alpha_2 \mapsto \alpha_2$.

Now you will notice that a lot of this seems like overkill. Indeed, in both cases, it seems as if one does not need $\Gamma$-bases at all! Indeed, one can readily tell whether the "normal" Galois action acts through the Weyl group or not, and if not, then it necessarily gives the non-trivial diagram automorphism.

I am sure though that in more complicated examples (different root systems but more importantly, more intricate Galois groups!) this will make a difference. And then it is a crucial fact remarked in the above sources that the $*$-action, unlike the normal action, does not depend on many choices, in particular of the involved tori.


问:
Differentiate $\frac13\tan^3x-\tan x+x$
$\frac13\tan^3x-\tan x+x$

I solved it and got $\frac13\cdot3\tan x-\sec^2x+1$ by using chain rule.

I got $\tan x+\tan^3x-\tan^2x$.

The answer is $\tan^4x$. I am not able to get this.

答:
You forgot to multiply the first term by $\sec^2x$…
$$\frac d{dx}=\tan^2x(1+\tan^2x)-(1+\tan^2x)+1=\tan^4x$$

Title: Differentiate $\frac13\tan^3x-\tan x+x$


答:
Others answered about how $cos(i)$ can be calculated using Euler's formula. But I will elaborate from a different perspective.

We know that cosine function can be defined geometrically for certain real numbers and using further geometric arguments can be extended to entire real numbers.

But how to extend it to complex plane?

One of the natural ways to do is by defining
$$
cos z = \frac{e^{iz} + e^{-iz}}{2}
$$

How this can be justified? Remember that for real $z$ our $cos z$ will match with cosine function for reals.
Similarly $sin z$ can be defined.  Also we see that almost all trignometric identities will hold for $cos z$ and $sin z$ as if our complex domain doesn't effect them. But there is an notable exception here, $sinz $ and $cosz$ are not bounded in complex plane.

Further if you know analysis , you will find that power series expansion for sine and cosine functions in complex case is similar to that of real case.

Title: Evaluating $\cos (i)$


问:
Weighted directed graph clustering
I had a really huge sets of molecules and it's I'd like to compare according to various factors. So I created a similarity measure which was very informative and fitted for comparing different types of molecules. After I introduced comparing between two elements I'd like to go further and try to cluster them, however I've got a problem, my similarity measure is not reflexive, i.e. $dist(A,B)!=dist(B,A)$ (though it satisfies triangle inequality). And I have no problem about it since it represents the problem of my interest perfectly. I generated the distance matrix ( or dissimilarity, my measure is bounded $[0,1]$ so it could be converted to similarity as well ).

I thought that clustering with asymmetric distances isn't common thing, so there is not much algorithms implemented (only ones that I know are OPTICS and DBSCAN, but as I tried they are no good), so I decided maybe I should change this problem to the problem of clustering strongly connected directed graph with weights. Maybe someone is aware of such issue and knows some usable implementation of it?

I found Tarjan's article, so maybe someone is familiar with the approach described there and knows possible implementation.

This question is also methodological. Maybe I should check another clustering algorithms, however I don't want to change my measure formula.

答: You can check the Knowledge Discovery Toolbox (KDT) here (a link to a related paper: A Flexible Open-Source Toolbox for Scalable Complex Graph Analysis). Also, take a look at this article: Preconditioners Based on Strong Subgraphs.</p>

Title: Weighted directed graph clustering


答:
This is too long for a comment. So I will write it as an answer.

Lets assume the definition of $\exp$ function via power series. Then it is well defined on the complex plane as well and Euler's formula gives us its real and imaginary parts. However Euler's formula still valid for any complex number and makes bridges between ordinary trigonometric functions, hyperbolic trigonometric functions, and complex logarithmic function.

The formula $\cos(i)=\dfrac{e^2+1}{2e}=\csc(2\tan^{-1}(e))$ is nothing more surprise than $i^2=-1$ or $e^{2i\pi}=1$ if you think of this for sometime. On the other hand, given any real number $r$ you can solve $\cos(z)=r$ over complex numbers. See here for an example.

Since $\exp, \sin, \cos$ all are well defined in any Banach algebra by means of power series, you can get similar results in much more generally. For example in quaternions. See here for more details on this.

Title: Evaluating $\cos (i)$


问:
A question about congruence in a cyclotomic coset
Let $q\neq 2$ be a prime number. Let $\gcd(q^m-1,e)=1$ for all $1\leq e \leq d-1$.

Proposition: If $1\leq i<j \leq d-1$, then $j \not\equiv q^si \pmod{q^m-1}$ for any integer $s\geq 0$.

Proof: Otherwise, we would have $j-i \equiv (q^s-1)i \pmod{q^m-1}$ which results in
$j-i \equiv 0 \pmod{q-1}$ which is a contradiction to the condition $\gcd(q^m-1,j-i)=1$.

Two questions:

How the result $j-i \equiv 0 \pmod{q-1}$ is obtained?

In which part of proof the condition $q\neq 2$ is used?

Thanks in advances

Source:  Proposition 8.1.12 in Page 164 in Coding Theory A First Course.

答:
Good questions:

Since $(q-1) \mid (q^m-1)$, the congruence $j-i \equiv (q^s-1)i \pmod{q^m-1}$ implies the congruence $j-i \equiv (q^s-1)i \pmod{q-1}$. However, $(q-1)\mid(q^s-1)$ as well, and so this latter congruence is equivalent to $j-i \equiv 0i \equiv 0 \pmod{q-1}$.

The conguence $j-i \equiv 0 \pmod{q-1}$ proves that $(q-1)\mid(j-i)$; since $(q-1)\mid(q^m-1)$ as well, we conclude that $(q-1)\mid\gcd(q^m-1,j-i)$, and therefore that $\gcd(q^m-1,j-i) \ge q-1$. However, this only contradicts $\gcd(q^m-1,j-i)=1$ when $q-1>1$, which is where the assumption $q\ne2$ is used.


答: Recall the power series for cosine:  $$\cos z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}$$  It turns out that this series converges for all complex numbers $z$, and can be taken as the definition of the complex cosine.  Since it only involves even powers of $2$, substituting $z=i$ will give you a real number.

Title: Evaluating $\cos (i)$


问:
Do there exist numbers of the form $n^2+1$ with arbitrarily many prime factors?
This is actually two questions:

1.Do there exist numbers of the form $n^2+1$ with arbitrarily many unique prime factors?

2.Do there exist numbers of the form $n^2+1$ with arbitrarily many prime factors?

I think an approach using gaussian primes might help,
$$n^2+1=(n+i)(n-i)$$
let $n=c^2$
$$(c^2 + i)(c^2-i)...$$
but from here I haven't managed to make anything productive out of this approach.

Note: the primes in the question are normal natural number primes.

答: Yes and yes.  In fact, given any finite set of primes $p_i \equiv 1 \pmod 4$, there exists an integer $n$ such that $n^2 + 1$ is divisible by all of the $p_i$.  Proof:  each of the congruences $n_i^2 \equiv -1 \pmod{p_i}$ individually has two solutions modulo $p_i$; choose one of the two for each $i$, and use the Chinese Remainder Theorem to find an $n$ congruent to each $n_i$ modulo $p_i$.


问:
Perfect Square Number, prime
I want to show that $p^{p+1}+(p+1)^p$ is not perfect square number. ($p$ is prime number)

This is what I do

Assume $p^{p+1}+(p+1)^p=k^2$.

$(p+1)^p=(k+p^{\frac {p+1} 2})(k-p^{\frac {p+1} 2})$.

Let $2ab=p+1, gcd(a,b)=1$.

$k+p^{\frac {p+1} 2}=2^{p-1}a^{p}, k-p^{\frac {p+1} 2}=2b^p$.

Therefore $p^{\frac {p+1} 2}=|2^{p-2}a^p-b^p|$.

By Fermat's little theorem, $a\equiv 2b \pmod{p}$, and $2ab=p+1$, so $a=2b$. In this case I can show contradiction.

but, if $a=1$ I don't know how show contradiction.

答:
Your proof is quite faulty (mathematically speaking, you omitted 2-3 cases i think), so I will reweite it entirely.

$$p^{p+1}+(p+1)^p=k^2\Leftrightarrow (p+1)^p=\big(k-p^{\frac{p+1}{2}}\big)\big(k+p^{\frac{p+1}{2}}\big)$$

Notice that $\text{gcd}(p+1;p)=1$ and $\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)|2\cdotp^{\frac{p+1}{2}}$, thus leading to $$\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)|2$$

$(p+1)^p$ is even so at least one of $k-p^{\frac{p+1}{2}}$ and $k+p^{\frac{p+1}{2}}$ is even, thus $$\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)=2$$

Note that case $p=2$ does not work, so from now on we will only work with $p\geq 3$

We can now safely assume that we have $2$ cases:

$1)$ $k-p^{\frac{p+1}{2}}=2\cdot a^p$ and $k+p^{\frac{p+1}{2}}=2^{p-1}\cdot b^p$ with $2ab=p+1$

$2)$ $k-p^{\frac{p+1}{2}}=2^{p-1}\cdot a^p$ and $k+p^{\frac{p+1}{2}}=2\cdot b^p$ with $2ab=p+1$

(notice that $a$ abd $b$ are not divisible by $p$)

Case $1)$: Using Fermat's little theorem, we get $k\equiv2a\pmod{p}$ and $k\equiv b\pmod{p}$ so $2a\equiv b\pmod{p}$ but $2ab\equiv 1\pmod{p}$ so either $b=1$ and $a=\frac{p+1}{2}$ or $b=p-1$ and $a=\frac{p+1}{2(p-1)}$.

Clearly when $a=\frac{p+1}{2(p-1)}$ we get a contradiction (unless $p=3$, which is a special case, we can easily see that $4^3+3^4$ is not a perfect square)

So $b=1$ and $a=\frac{p+1}{2}$ But $k=2^{p-1}b^p-p^{\frac{p+1}{2}}$ so if $b=1$ then $2^{p-1}> p^{\frac{p+1}{2}}$ which is clearly false for all primes $p$.

So we only get contradictions in this case.

Case $2):$ Again, using the same method we get $2b\equiv a\pmod{p}$ so with the same reasoning $a=1$ and $b=\frac{p+1}{2}$

We know $k=2\cdot a^p+p^{\frac{p+1}{2}}=2^{p-1}\cdot b^p-p^{\frac{p+1}{2}}$ so we get $$2+p^{\frac{p+1}{2}}=2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p-p^{\frac{p+1}{2}}$$ which can we rewritten as $$2(1+p^{\frac{p+1}{2}})=2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p$$

We assumed $p\geq 3$ so $p\equiv 1$ or $3\pmod{4}$

If $p\equiv 1\pmod{4}$, $1+p^{\frac{p+1}{2}}\equiv 2\pmod{4}$ so $$v_2(2(1+p^{\frac{p+1}{2}})=2$$ but $$v_2(2(1+p^{\frac{p+1}{2}})=v_2(2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p)\geq p-1$$ giving us $$3\geq p$$ so contradiction; we assumed $p\equiv 1\pmod{4}$

If $p\equiv 3\pmod{4}$ then $\frac{p+1}{2}$ is even, so $p^\frac{p+1}{2}\equiv 1\pmod{4}$ (any odd perfect square is $1\pmod{4}$) so we have again that $1+p^{\frac{p+1}{2}}\equiv 2\pmod{4}$ so $$v_2(2(1+p^{\frac{p+1}{2}})=2$$ giving us the same result, $3\geq p$, which would lead to $p=3$, which we established is a contradiction.

Thus, there is no prime $p$ for which $p^{p+1}+(p+1)^p$ is a perfect square.

Title: Perfect Square Number, prime


问:
Harmonic conjugate of $u(r,\theta)=r^3\cos(3\theta)$

Verify that the following function $u$ are harmonic, and in each case give a conjugate harmonic function $v$ (i.e $v$ such that $u+iv$ is analytic)

$$u(r,\theta)=r^3\cos(3\theta)$$

I need to find the harmonic conjugate of the equation but I can't seem to prove it is harmonic. Does anybody got any idea? Thank you so much.

答:
By de Moivre's theorem
$$r^3\cos3\theta=\operatorname{Re}(r^3(\cos3\theta+i\sin3\theta))=\operatorname{Re}(r^3(\cos\theta+i\sin\theta)^3)=\operatorname{Re}((re^{i\theta})^3)=\operatorname{Re}(z^3)$$
where $z$ is an arbitrary complex number. Thus $r^3\cos3\theta$ is harmonic and its harmonic conjugate is $r^3\sin3\theta$.

The surface obtained as $z=r^3\cos3\theta$ is called the monkey saddle.


问:
group cohomology and singular cohomology for $K(G,1)$
For $K(G,1)$ space, I know that there is an isomorphism between group cohomology and singular cohomology. Is there an example for which this fact is useful (for example, it is hard to calculate group cohomology but easy to calculate singular cohomology).

答:
I mean... Off the top of your head do you see what the cohomology of $\mathbb{Z}$ should be? Now ask yourself if you can see immediately what the cohomology of $S^1$ is. For me, at least, one of these questions is much easier. More broadly, what's the cohomology of a free group $F_k$? What about a wedge of $k$ circles?

These examples are slightly silly because they're one dimensional, but the idea is still there:

$K(G,1)$ spaces (when we can get a handle on what they look like) let us leverage our geometric intuition to solve group cohomological problems. I would argue that $K(\mathbb{Z}/2, 1) = \mathbb{R}P^\infty$ is a similar example (which is less silly, seeing as it's infinite dimensional). It admits a famous cellular decomposition, from which is cohomology follows immediately (especially over $\mathbb{Z}/2$ coefficients!).

I find that much easier to conceptualize and compute than working with a projective resolution of $\mathbb{Z} / 2$ directly.

I'll agree with you, though. Perhaps it's because I'm not an algebraic topologist, but I don't often find examples "in the wild" where $K(G,1)$ is a much simpler thing to consider than $G$ itself. When it happens, though (as in the cases above), it's really convenient.

Group cohomology is, in general, a rather complicated thing. After all, it encodes the solution to the extension problem in $H^2$! So it might be asking a bit much for $K(G,1)$ to be substantially simpler for a random group $G$.

I hope this helps ^_^


问:
How can I show that: $\text{Tr}(A^*A)=\sum_{\lambda\in\sigma(A^*A)}\lambda$?
Let $A\in \mathbb{K}^{n\times n}$, how can we show that:
$$ ||A||^2_F=\text{Tr}(A^*A)=\sum_{\lambda\in\sigma(A^*A)}\lambda$$

Where $A^*=\bar A^T.$

Thanks in advance for any help!

答:
The trace of a matrix is the sum of the eigenvalues, up to multiplicity. As $A^*A$ is Hermitian, it is diagonalisable, hence there is an invertible matrix $P$ and diagonal matrix $D$, whose diagonal entries are the eigenvalues of $A^*A$, up to multiplicity, such that
$$A^*A = P^{-1} D P.$$
Using the fact that $\operatorname{Tr}(CD) = \operatorname{Tr}(DC)$, we get
$$\operatorname{Tr}(A^*A) = \operatorname{Tr}(P^{-1} D P) = \operatorname{Tr}(DPP^{-1}) = \operatorname{Tr}(D),$$
which is the sum of the diagonal entries of $D$, i.e. the sum of the eigenvalues of $A^*A$ up to multiplicity.

Here are two things worth noting:

The notation $\sum_{\lambda \in \sigma(A^* A)} \lambda$ could be misleading, in that it doesn't make it clear that eigenvalues must be summed up to multiplicity. For example, if $A = I_{2 \times 2}$, then $A^*A = I_{2 \times 2}$, which has one eigenvalue of $1$, up to a multiplicity of $2$, i.e. $\sigma(A^* A) = \{1\}$. One could reasonably interpret the sum as simply $1$, when the trace of $A^*A$ is actually $1 + 1 = 2$.

The fact that $\operatorname{Tr}(A)$ is the sum of the eigenvalues up to multiplicity is true for any matrix, even non-diagonalisable matrices. We may not be able to get a diagonal matrix $D$, but we can get an upper-triangular matrix $D$ (e.g. the Jordan Normal Form).


问:
Determine the primitive $F(x)$ of $f(x) = xe^{-\lvert x \rvert}$ which satisfies $\lim_{x \to +\infty} F(x) = -5$
I have tried splitting the primitive into two cases, based on the definition of the absolute value function:

When $x \geq 0$ then $\int_{0}^{+\infty} xe^{-x} dx $. Using integration by parts I got: $-xe^{-x} - e^{-x}$.

When $x < 0$ then $\int_{-\infty}^0 -xe^{+x} dx $. Using integration by parts I got: $-xe^{x} + e^{+x}$.

After applying the limits and using L'Hôpital rules:

$\lim_{a \to +\infty} \left[ -xe^{-x} - e^{-x}  \right]_0^{a} = +1$

$\lim_{b \to -\infty} \left[ -xe^{x} + e^{+x}  \right]_{b}^{0} = +1$

Summing both results I got $+2$.

Is the problem correct? if yes, is it possible to modify the intervals to get $-5$?

答:
Note that when you calculate $F$ you don't do definite integrals. So, similar to what you wrote in your question, $$F(x)=\begin{cases}−xe^{−x}−e^{−x}+C_1, x\ge 0\\ −xe^x+e^x+C_2, x<0\end{cases}$$
Here $C_1$ and $C_2$ are some constants. From your condition at $x\to\infty$, you get $C_1=-5$. To get $C_2$, you must put the condition that $F$ is continuous at $0$:
$$-0e^0-e^0-5=-0e^0+e^0+C_2$$
Therefore $C_2=-7$


问:
Some questions about eigenvalues, eigenvectors, and diagonalization
I've been studying for my linear algebra final and was going through the review sheet the professor gave us. Most of the content was easy to understand, but I couldn't get my head around a few concepts. I've listed the questions below:

Provide an example of a 2x2 matrix with one linearly independent eigenvector

Provide an example of           matrices which have the same eigenvalues but are not similar.

Provide an example of a matrix with eigenvalues 3 and 4, occurring at multiplicities 5 and 6, respectively.

Suppose that A and B are n × n matrices, and that A is similar to B.
Show that A and B have the same characteristic polynomial.

I think I've sort of figured out 4, so I'll show my working:

If $A = PBP^{-1}$, then to show that A and B have the same characteristic polynomials we must find $det (A-\lambda I)$ and prove that it is equal to $det (B-\lambda I)$

$\to$ $det (A - \lambda I) $ = $det (PBP^{-1} - \lambda I) $

= $det (P^{-1}(B- \lambda I)P)$

= $(det P^{-1})(det B- \lambda I)(det P)$

= $ (det B- \lambda I)$

Since the determinants of A and B are equal, we know that their characteristic polynomials will be the same.

I hope this is a coherent explanation; I'm a freshman in college so my exposure to rigorous, inductive mathematics isn't very high.

答:
The standard example is $\begin{pmatrix}1&1\\0&1\end{pmatrix}$.

Looking at the matrix above. can you think of another matrix with the same eigenvalues that is not similar to it?

Use a diagonal matrix with enough $3$'s and $4$'s on the diagonal.


问:
What do I know when the curl of a vector field equals 0?
I'm currently doing a multiple-choice exercise which includes the following question:

Let $F: \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}^2$ be defined by $F(x,y) = \left( - \frac{y}{x^2+y^2}, \frac{x}{x^2+y^2} \right)$. Which of the following statements are true?

There exists a closed path $\gamma$ in $\mathbb{R}^2 \setminus \{0\}$ such that
  $$\int_\gamma F \cdot \mathrm ds < 0.$$

There exists a $\epsilon \in (0,1)$ such that for all closed $C^1$-paths $\gamma$ with Im $\gamma \subseteq B_\epsilon((1,0))$
  $$\int_\gamma F \cdot \mathrm ds = 0. $$

F is a gradient field.

Now up to now I thought that whenever the curl of a vector field equals 0, firstly the vector field is a gradient field and secondly the integral around every closed paths equals 0. So this would make the second and the third statement to be correct whilst the first statement obviously would be wrong.

However, I don't feel like my reasoning is correct. It would be somewhat too easy, so there has to be some mistake in my train of thought, some point I've been missing.

Where did I make a mistake or is it really that easy?

Thanks for any answer in advance.

答:
I'm making my comments into an answer. As you said in a comment, there is the theorem:

If $\Omega \subset \mathbb{R^{2}}$ is an open and star-shaped domain then a $C^1$-vector field $K: \Omega \to \mathbb{R}^2$ is conservative if and only if $\operatorname{rot} K = 0$.

But as $\Omega = \mathbb{R}^2 \smallsetminus \{0\}$ is not star-shaped with respect to any point [proof: if $x$ is any point of $\Omega$ the line connecting $x$ and $-x$ crosses the origin, thus $\Omega$ is not star-shaped], this theorem is not applicable and its conclusion can turn out to be false, as we will see in a moment.

Now let's look at the vector field $F(x,y) = \left( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}\right)$. If $(x_{0},y_{0})$ lies on the unit circle, i.e., $x_{0}^2 + y_{0}^2 = 1$, then $F(x_{0},y_{0}) = (-y_{0},x_{0})$. Note that $F(x_{0},y_{0})$ has length one and is perpendicular to the vector $(x_{0},y_{0})$ and points in the counterclockwise direction (seen from the origin). So geometrically, if we move around the origin clockwise, the line integral should evaluate negatively (since the vector field points in the opposite direction of our movement). Formally, if we parameterize the unit circle by $\gamma(t) = (\cos{t}, -\sin{t})$ with $t \in [0,2\pi]$ (signs are chosen so that we walk clockwise) the line integral of $F$ along $\gamma$ evaluates to

\begin{align*}
\int_{\gamma} F\cdot ds & = \int_{0}^{2\pi} F(\gamma(t)) \cdot \dot{\gamma}(t)\,dt = \int_{0}^{2\pi} \begin{pmatrix} \sin{t} \\ - \cos{t} \end{pmatrix} \cdot \begin{pmatrix} -\sin{t} \\ \cos{t} \end{pmatrix}\,dt \\ & = \int_{0}^{2\pi} \left(-\sin^2{t} - \cos^{2}\right)\,dt = -2\pi
\end{align*}
since $\sin^2 + \cos^2 = 1$.

This computation solves the question almost completely: We have found a path such that the integral evaluates negatively and thus the first bullet point is correct. The third bullet point is wrong since our path is closed and a gradient vector field is conservative, hence evaluates to zero on all closed paths.

Finally, note that for every $\varepsilon \in (0,1)$ the ball $B_{\varepsilon}(0,1)$ is entirely contained in $\Omega$ and it is star-shaped (with respect to $(0,1)$). So the theorem is applicable to the restriction $K$ of $F$ to $B_{\varepsilon}(0,1)$. We conclude that the vector field $K$ is conservative on $B_{\varepsilon}(0,1)$ and hence for every closed $C^1$-path $\gamma$ entirely contained in $B_{\varepsilon}(0,1)$ the integral $\int_{\gamma} F \cdot ds = 0$. So the second bullet-point is correct.

To summarize:

true

true

false


问:
Help proving Calderón reproducing formula (simple version)
Let $\phi$ be a real compactly supported smooth function on $\mathbb R$ with total integral zero. Define $\phi_t=\frac{1}{t} \phi(\frac{x}{t})$. I also suspect that they must be even, but the notes I am working from are imprecise.

I want to show that $$f(x) \left(\int_0^\infty |\hat\phi(t)|^2 \frac{dt}{t}\right)=\int_0^\infty (\phi_t\ast\phi_t\ast f)(x) \frac{dt}{t}.$$

This can be thought of as a reproducing formula for $f$ because the integral on the left side is a constant (with respect to $x$).

I am attempting to prove this but coming up short. I began by taking the Fourier transform of the right hand side and recalling that it takes convolutions into pointwise multiplication. We get

$$f(\xi) \int_0^\infty \hat\phi_t(\xi)^2 \frac{dt}{t}.$$

We can rewrite the term being integrated because the Fourier transform of ${\frac{1}{t}\phi(\frac{x}{t})}$ is $\hat\phi(t\xi)$.

$$f(\xi) \int_0^\infty \hat\phi(t\xi)^2 \frac{dt}{t}.$$

I do not see how to continue. In particular, I do not see how to make use of the fact that the total integral of $\phi$ is 0.

EDIT: A proof has been given in the comments, but I still don't know why the integral of $\phi$ must be zero! This condition appears in other statements of this formula I have found, not just this simplified one, but it does not seem necessary. I still welcome a definitive answer on this matter.

答:
The first thing you should determine is in what sense do you want this integral to hold.  Do you want it pointwise:

For all $x\in\mathbb R$,

$$f(x)=\lim_{R\rightarrow\infty}\int_{1/R}^R\phi_t*\phi_t*f(x)\frac{dt}{t}.$$

Or do you want it in $L^p$:

$$\lim_{R\rightarrow\infty}\left|\left|\int_{1/R}^R\phi_t*\phi_t*f\frac{dt}{t}-f\right|\right|_{L^p}=0.$$

Or do you want it in the topology of $\mathcal S(\mathbb R)$, the Schwartz class:

For all multi-indices $\alpha,\beta\in\mathbb N_0^n$,

$$\lim_{R\rightarrow\infty}\rho_{\alpha,\beta}\left(\int_{1/R}^R\phi_t*\phi_t*f\frac{dt}{t}-f\right)=0.$$

Each of these things are true for appropriate functions $f$.  It is not hard to show that the formula holds in $L^2$ when $f\in L^2$.  Plus, if it holds in $L^2$, then it holds pointwise a.e.  First fix an $L^2$ function $f$.  Then you must show

$$\lim_{R\rightarrow\infty}\left|\left|\int_{1/R}^R\phi_t*\phi_t*f\frac{dt}{t}-f\right|\right|_{L^2}=0.$$

Since $f\in L^2$ and $\phi_t$ is "nice," it follows that

$$f_R(x):=\int_{1/R}^R\phi_t*\phi_t*f(x)\frac{dt}{t}$$

is in $L^2$ for all $R>0$ (a priori with $L^2$ norm depending on $R$).  Then by Plancherel's theorem, it is enough to show $\hat f_R\rightarrow \hat f$ in $L^2$ as $R\rightarrow\infty$, i.e. it is enough to show the following goes to zero as $R\rightarrow \infty$

\begin{align*}
\int_{\mathbb R}\left|\hat f_R(\xi)-\hat f(\xi)\right|^2d\xi&=\int_{\mathbb R}|\hat f(\xi)|^2\left|\int_{1/R}^R\hat\phi(t\xi)^2\frac{dt}{t}-1\right|^2d\xi.
\end{align*}

One way to do this is to apply dominated convergence in the following way:  First uniformly bound

\begin{align*}
\left|\int_{1/R}^R\hat\phi(t\xi)^2\frac{dt}{t}-1\right|&\leq1+\int_0^{|\xi|^{-1}}\hat\phi(t\xi)^2\frac{dt}{t}+\int_{|\xi|^{-1}}^\infty\hat\phi(t\xi)^2\frac{dt}{t}\\
&\leq1+||\nabla\hat\phi||_{L^\infty}^2\int_0^{|\xi|^{-1}}(t\xi)^2\frac{dt}{t}+||\nabla\phi||_{L^\infty}^2\int_{|\xi|^{-1}}^\infty(t\xi)^{-2}\frac{dt}{t}\\
&=1+\frac{1}{2}||\nabla\hat\phi||_{L^\infty}^2+\frac{1}{2}||\nabla\phi||_{L^\infty}^2.
\end{align*}
I'm not completely sure that the $||\nabla\phi||_{L^\infty}$ term is correct here, but it should be some constant depending on the smoothness of $\phi$.  Then

$$|\hat f_R(\xi)-\hat f(\xi)|^2\leq|\hat f(\xi)|^2\left(1+\frac{1}{2}||\nabla\hat\phi||_{L^\infty}^2+\frac{1}{2}||\nabla\phi||_{L^\infty}^2\right)^2,$$

which is an $L^1$ function independent of $R$ since $f\in L^2$.  So by dominated convergence,

$$\lim_{R\rightarrow\infty}||\hat f_R-\hat f||_{L^2}^2=\int_\mathbb R|\hat f(\xi)|^2\lim_{R\rightarrow\infty}\left|\int_{1/R}^R\hat\phi(t\xi)^2\frac{dt}{t}-1\right|^2d\xi=0.$$

This proves that the formula holds in $L^2$, and hence it follows that it holds pointwise.  To prove the $L^p$ convergence is a little more difficult since we cannot make use of Plancherel's theorem, and to prove convergence in the topology of Schartz functions is even more difficult since you must work with Schwartz semi-norms.

A word of caution:  In order for this argument to work, you must know that $f\in L^2 $ to start.  If $f$ is not in $L^2$ and you prove that $f_R$ (as defined above) is Cauchy in $L^2$, then indeed it does converge to some function $\tilde f\in L^2$, but it is possible that $f\neq\tilde f$.  In particular, these two function may differ by a polynomial.  This is essentially because $\hat\phi(0)=0$ and the distributional Fourier transform of polynomials are supported at the origin.


问:
Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps.
I know this question has been posted before, there is a solution in my text-book for this question and also this is posted on so many websites with full solution but I still don't understand. So can someone please help me?

BASIS STEP: We can form postage of 12, 13, 14, and 15 cents using three 4-cent stamps, two
4-cent stamps and one 5-cent stamp, one 4-cent stamp and two 5-cent stamps, and three 5-cent
stamps, respectively. This shows that P (12), P (13), P (14), and P (15) are true. This completes
the basis step.

INDUCTIVE STEP: The inductive hypothesis is the statement that P (j )is true for 12 ≤ j ≤ k,
where k is an integer with k ≥ 15. To complete the inductive step, we assume that we can form
postage of j cents, where 12 ≤ j ≤ k. We need to show that under the assumption that P (k + 1)
is true, we can also form postage of k + 1 cents. Using the inductive hypothesis, we can assume
that P (k − 3) is true because k − 3 ≥ 12, that is, we can form postage of k − 3 cents using
just 4-cent and 5-cent stamps. To form postage of k + 1 cents, we need only add another 4-cent
stamp to the stamps we used to form postage of k − 3 cents. That is, we have shown that if the
inductive hypothesis is true, then P (k + 1) is also true. This completes the inductive step.

I am stuck at this part:

how did they get k> 15 ?
and i don't get how did they get P(k-3)

like can someone please explain this to me?

答:
The main question here is, why $4$ cases in the basic step. @Angelo Randina's answer explains this. I will, here, try to explain the problem that will arise if we don't do the 4 base cases.

Basic Step: $P(12) = 4\times 3$ or four stamps of 3 cents

Inductive Step: Now, since we have only done the basic step for $P(12)$, we will have to let $12\le k$ instead of $15\le k$ as it was done with the $4$ base cases. Now, since $k\ge 12$, so $k-3\ge9$. It means that $k-3$ may fall below $12$. This is the problem because now we cannot argue that $P(k-3)$ is true all the time because it is not true when $k-3<12$.

Since, we cannot argue that $P(k-3)$ is true, we cannot say that $P(k+1) = P(k-3) + 4$. So we have only two options 1) Either, we do $4$ base cases and let $k\ge15$ and get to $P(k+1) = P(k-3) + 4$ or 2) we do $5$ base cases and let $k\ge 16$ and get to $P(k+1) = P(k-4)+5$

Here, we can see how many base cases have to be solved is derived solely by the value of the postage stamp. If the minimum value postage stamp is $4$ cents, then we need at least $4$ base cases.

Another Example: To see this, consider another example, every amount of postage greater than $12$ cents can be paid just by $3$ cents and $7$ cents stamps. Here, the minimum value of a stamp is $3$ cents so we need $3$ base cases and we will let $k\ge 12+3-1=14$ so that we can get to $P(k+1) = P(k-2) + 3$. So, lets do the basic and inductive steps

Basic Step:
$P(12) = 4\times 3$, i.e using four $3$ cents stamps

$P(13) = 2\times 3 + 1\times 7$, i.e using two $3$ cents stamps and one $7$ cents stamp

$P(14) = 2 times 7$, i.e using two $7$ cents stapms

Inductive Step:
Assume $12\le j\le k, P(j)$ and prove $P(k+1)$. We can get to $P(k+1)$ by going back two steps adding a $3$ cents stamp, i.e $P(k+1) = P(k-2)+3$ because we are sure that $k-2\ge 12$ and we have assumed $\forall j>12 P(j)$


答:
I will elaborate what Ravi said for you.

$n_i^2\equiv -1\pmod{p_i}$ has a solution if $-1$ is a quadratic residue. A known formula is $$\bigg(\frac{-1}{p}\bigg)=(-1)^\frac{p-1}{2}$$ thus, if and only if $p\equiv 1\pmod{4}$ $-1$ is a quadratic residue.

(search up quadratic residues and laws for more information).

Now simply consider $n$ according to the Chinese Remainder Theorem.

Body:
This is actually two questions:

1.Do there exist numbers of the form $n^2+1$ with arbitrarily many unique prime factors?

2.Do there exist numbers of the form $n^2+1$ with arbitrarily many prime factors?

I think an approach using gaussian primes might help,
$$n^2+1=(n+i)(n-i)$$
let $n=c^2$
$$(c^2 + i)(c^2-i)...$$
but from here I haven't managed to make anything productive out of this approach.

Note: the primes in the question are normal natural number primes.


问:
Division of power series
Let $f(x) =\sum_{n=0}^{\infty} a_n x^n$ and $g(x) =\sum_{n=0}^{\infty} b_n x^n$ in an open interval $(-r, r)$, and $g \neq 0 $ in $(-r, r)$. How can I prove that $h = f/g$ can be expressed as a power series $\sum_{n=0}^{\infty} c_n x^n$ that converges at least in $(-r,r)$?
I know how to 'calculate' the power series expansion of $h$, for example I can use the determinant method in https://en.wikipedia.org/wiki/Power_series#Multiplication_and_division. The question is, how can I prove that the series expansion 'does exist'(that is, converges)?

答:
If there is a purely real analytic approach to finding the radius of convergence of the quotient of two power series, it would be interesting to see.  However, the only approach I know relies on Cauchy's formula, so it requires a bit of complex analysis. The fact that the radius of convergence is equal to the distance from the center of the expansion to the nearest singularity in $\mathbb{C}$, indicates some dependence on complex analysis.

Here is a proof that the power series for an analytic function has a radius of convergence equal to the distance from the center of the expansion to the closest singularity.

Let $R$ be the radius of convergence of
$$
f(z)=\sum_{k=0}^\infty a_n(z-z_0)^n\tag1
$$
where $f$ is analytic inside, and continuous on, $B(r,z_0)=\{z:|z-z_0|\le r\}$. Set $\gamma=z_0+re^{i\theta}$, then
$$
\begin{align}
\frac1R
&=\limsup_{n\to\infty}|a_n|^{1/n}\tag2\\
&=\limsup_{n\to\infty}\left|\frac{f^{(n)}(z_0)}{n!}\right|^{1/n}\tag3\\
&=\limsup_{n\to\infty}\left|\frac1{2\pi i}\oint_\gamma\frac{f(z)}{(z-z_0)^{n+1}}\,\mathrm{d}z\right|^{1/n}\tag4\\[3pt]
&\le\frac1r\tag5
\end{align}
$$
Explanation:

$(2)$: the Cauchy–Hadamard theorem (real analysis)

$(3)$: Taylor's Theorem (real analysis)

$(4)$: Cauchy's Integral Formula (complex analysis)

$(5)$: $|z-z_0|=r$ on $\gamma$, the length of $\gamma$ is $2\pi r$, and $|f|$ is bounded

That is, $R\ge r$, where $f$ is analytic inside, and continuous on, $B(r,z_0)$.

Thus, if the series for $f(z)$ and $g(z)$ converge inside $B(R,z_0)$ and and $g(z)\ne0$ on $B(R,z_0)$, then both $f(z)$ and $1/g(z)$ are analytic inside $B(R,z_0)$, and neither have poles inside $B(R,z_0)$. Therefore, the series for $f/g(z)$ converges inside $B(R,z_0)$.

Title: Division of power series


问:
How many types of continuity are there?
Currently I'm familiar with continuity on the reals and uniform continuity, which releases the dependency of $\delta$ on $a$, all from a Calculus perspective.

What other types of continuities are there?

答:
There are many types of continuity and most of them are subsumed under topology where we give the domain and codomain a certain topology and a fixed general notion of continuity always applies.
Here the topologies are usually given by open sets, or equivalently by nets (which generalise sequences).

One form of continuity that does not fall into this category is uniform continuity and which requires uniform spaces and which has three standard definitions - one of them is by entourages.

Another kind of continuity that also does not fall into this category, is completeness which requires Cauchy sequences. This is axiomatised by Cauchy spaces.

Moreover, there are more general notions of topology; for example, proximity and pseudo-topological spaces. The former generalises uniform spaces, as it is equivalent to allowing more than one uniformity in a space (upto a class worth) and the latter generalises topological spaces in its definition by nets, here it uses ultranets or ultrafilters.

Title: How many types of continuity are there?


问:
What would be the immediate implications of a formula for prime numbers?
What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer, however, I would like to know some of these implications - like the breaking of many security systems based on the primes. Moreover, there are examples of practical Math's implication, not just theoretical, of a possible prime number formula discovery? There is for example in Physics, Chemistry, Geography or Astronomy any field which would be very improved with a so great and dreamed Eureka?

答:
There are in fact many 'formula's which always generate prime numbers.  Among the simplest ones listed by Wikipedia are:

There is some real number $A$ such that $\left\lfloor A^{3^n}\right\rfloor$ is always prime.

There is some real number $\alpha$ such that $\left\lfloor 2^{2^{\cdots^{\alpha}}} \right\rfloor$ is always prime.

The formula $2 + [2(n!) \bmod (n+1)]$ always gives a prime, where here '$\bmod$' (nonstandardly) denotes the remainder.

The third in particular generates every prime.

These formulas are mathematically valid, fun to prove, and highly interesting.  However, they are actually completely useless, and not just because we don't know the values of $A$ and $\alpha$.  So (as some have suggested in the comments) a better question is how fast one can generate primes.

The ability (or inability) to generate or check for primes in a certain amount of time is fundamentally important to cryptographic systems such as RSA. However, the "practical" applications of prime numbers (to fields like physics, chemistry, etc.) are, as far as I understand, very few -- cryptography is the major application.


问:
Prove that there is no a such that $a^{-1}(1,2,3)a = (1,3)(5,7,8)$
Herstein 2.10.8 (b)

Prove that there is no a such that $a^{-1}(1,2,3)a = (1,3)(5,7,8)$

My initial thought was that order of RHS and LHS permuation should be same.

For RHS, the cycles are disjoint, so order = lcm(2,3) = 6

And for LHS, I got to know that it has order 3.  And thus there is no such 'a' possible.

What I don't understand is how did we come to conclusion that $a^{-1}(1,2,3)a$ has order 3 ? Clearly (1,2,3) has order 3, but after taking product with some other permutation, will the order remain same, even if the cycles are not disjoint ?

答: $a^{-1}pa$ must yield an element in the same conjugacy class as $p$, and conjugate elements have the same order.


答: In any group $(x^{-1}gx)^{n}=x^{-1}g^n x$ and so $g^n=e$ if and only if $x^{-1} g^n x=e$ if and only if $(x^{-1}gx)^{n}=e$.

Body:
Herstein 2.10.8 (b)

Prove that there is no a such that $a^{-1}(1,2,3)a = (1,3)(5,7,8)$

My initial thought was that order of RHS and LHS permuation should be same.

For RHS, the cycles are disjoint, so order = lcm(2,3) = 6

And for LHS, I got to know that it has order 3.  And thus there is no such 'a' possible.

What I don't understand is how did we come to conclusion that $a^{-1}(1,2,3)a$ has order 3 ? Clearly (1,2,3) has order 3, but after taking product with some other permutation, will the order remain same, even if the cycles are not disjoint ?


问:
fundamental theorem of calculus involving multivariable calculus
The fundamental theorem of calculus states that $$\dfrac{d}{dx}\int_{a}^{f(x)}g(t)dt=g(f(x))f'(x)$$ with the appropriate prerequisites ($g$ is continuous and real-valued, etc).

1)By some counterexamples (for example $\dfrac{d}{dx}\int_{a}^{x}(t+x)dt$) we can easily know that the statement $\dfrac{d}{dx}\int_{a}^{f(x)}g(t,x)dt=g(f(x),x)f'(x)$ is false. Why, though? Or is this the wrong question to ask? It was never correct in the first place and the question should never have been raised?

2)Suppose $u=q(x)$. Then if $$F(x)=\int_{a}^{q(x)}g(t,x)dt$$, let $$P(u,x)=\int_{a}^{u}g(t,x)dt$$

Then, $$\dfrac{\partial P}{\partial u}=g(u,x)$$

I can't really get the logic behind this-- is this something to do with differentiating under the integral sign? But that wouldn't work, would it? $u$ is not a constant-it's a function, isn't it?

Appreciate the help!

答:
We have that $$P(u,x) = \int_a^u g(t,x)\,{\rm d}t \implies \frac{\partial P}{\partial u}(u,x) = g(u,x) \quad\mbox{and}\quad \frac{\partial P}{\partial x}(u,x) = \int_a^u \frac{\partial g}{\partial x}(t,x)\,{\rm d}t.$$The first derivative is just the fundamental theorem of Calculus without anything else behind the scenes, because $x$ is frozen! The second equality is indeed differentiation under the integral sign, and holds provided $g$ is smooth enough.

If you want to compute the derivative of $$F(x) = \int_a^{q(x)}g(t,x)\,{\rm d}x,$$a chain rule is required. Because $F(x) = P(q(x),x)$, so $$F'(x) = q'(x)\frac{\partial P}{\partial u}(q(x),x) + 1\frac{\partial P}{\partial x}(q(x),x).$$This means that $$F'(x) = q'(x)g(q(x),x) + \int_a^{q(x)} \frac{\partial g}{\partial x}(t,x)\,{\rm d}t.$$Keeping track of all evaluation points helps you understand better what is going on.


问:
$\sum_{\zeta^p=1}(\zeta-1)^n$
Given $n\geq0$ let
$$
z_n=\sum_{\zeta^p=1}(\zeta-1)^n
$$
where $p$ is an odd prime number (summation extended to all $p$-th roots of 1). It is clear that:

$z_n\in\Bbb Z$ (it's a Galois invariant sum of integral elements in the extension $\Bbb Q(\zeta)$).

$p^m\mid z_n$, where $p^m$ is the maximal power of $p$ in $\frak p^n$ where $\frak p$ is the unique prime in $\Bbb Z[\zeta]$ over $p$ (see Jyrki Lahtonen's comments below), i.e. $m=\lceil\frac n{p-1}\rceil$).

It is an obvious computation that
$$z_n=\sum_{\zeta^p=1}\sum_{k=0}^n\binom nk\zeta^k(-1)^{n-k}=\sum_{k=0}^n\binom nk(-1)^{n-k}\sum_{\zeta^p=1}\zeta^k
$$
and since $\sum_{\zeta^p=1}\zeta^k=0$ if $(p,k)=1$ and $=p$ if $p\mid k$ we eventually get
$$
z_n=(-1)^np\sum_{\ell=0}^{n/p}(-1)^\ell\binom n{\ell p}
$$
(clearly the summation stops at the greatest integer smaller than $n/p$ if $n/p\notin\Bbb Z$). Now my questions are:

can we get a more compact expression for $z_n$ and/or some characterizing description?

The divisibility mentioned above translates implies that $lim_{n\to}|z_n|_p=0$ ($p$-adic absolute value). Can we be more precise about the speed of convergence to $0$?

答:
Many of the results can be easily seen through using a Mahler series expansion as a generating function. Specifically, we have the continuous function $\chi_{p^t\mathbb{Z}_p} :\mathbb{Z}_p \to \{0,1\}$, which is the locally constant indicator function,

$$\chi_{p^t\mathbb{Z}_p}(x)  = \begin{cases} 1 & x \in p^t\mathbb{Z}_p \\  0 & x \not\in p^t\mathbb{Z}_p\end{cases}$$

Clearly, its Mahler series has coefficients $a_n = \nabla^n \chi_{p^t\mathbb{Z}_p}(0)$ in $\mathbb{Z}$,

$$\chi_{p^t\mathbb{Z}_p}(x)=\sum_{n\ge 0} a_n \binom{x}{n}$$

However, notice that we have the alternative representation in $\mathbb{Q}_p(\zeta_{p^t})$,

$$\chi_{p^t\mathbb{Z}_p}(x) = \frac{1}{p^t}\sum_{k=0}^{p^t-1} \zeta_{p^t}^{kx}$$

We can now evaluate its finite differences at $x=0$,

$$\nabla^n \chi_{p^t\mathbb{Z}_p}(0) = \frac{1}{p^t}\sum_{k=0}^{p^t-1}(\zeta_{p^t}^k-1)^n  = \frac{z_n}{p^t}$$

At this point, the original question is concerned with the $t=1$ case, but I'll keep the argument general. Because Mahler series representations are unique, the integer (in $\mathbb{Z}$) $a_n = \frac{z_n}{p^t}$ already shows us that $z_n$ is an integer divisible by $p^t$. Furthermore, Mahler series have coefficients that satisfy $\lim_{n \to \infty} |a_n|_p = 0$, so $\lim_{n \to \infty} |z_n|_p\ = 0$.

We can also be more precise on the speed of convergence since $\chi_{p\mathbb{Z}_p}$ is strictly differentiable. It implies that $\lim_{n \to \infty} n|a_n|_p = 0 $ so we have,

$$\lim_{n \to \infty} n|z_n|_p = 0 $$

While lying in bed this morning I realized we also get the same formula from the question as well just by the formula for the Mahler coefficients, the characteristic function simply screens out anything that is not a multiple of $p^t$ for us.

$$a_n = \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} \chi_{p^t\mathbb{Z}_p}(k)= \sum_{k=0}^{ \lfloor n/p^t \rfloor} \binom{n}{p^tk} (-1)^{n-p^tk} $$

$$z_n = (-1)^n p^t \sum_{k=0}^{ \lfloor n/p^t \rfloor} \binom{n}{p^tk} (-1)^{pk} $$
The $(-1)^{pk}$ is left to account for the case when $p=2$.

Title: $\sum_{\zeta^p=1}(\zeta-1)^n$


问:
How to prove that
Given that  fulfills the following condition: there is a real number 0<<1 so for all-natural ≥2:

|$a_{n+1}−_|<c |(_^2−_{−1}^2)|$ where  $|a_n|<2$

To prove : ($a_n$) is convergent

Now I was able to prove that ($a_n$) is convergent for the case where sum of consecutive terms of the sequence is more than or equal to one
How should I go about solving the case where their sum will be less than one ??

答: This is false. Let $a_n=0$ for $n$ even and $a_n=1.5$ for $n$ odd. Let $\frac  2 3 <c<1$. Then $|a_{n+1}-a_n|=1.5 <c(1.5)^{2} <|a_n^{2}-a_{n-1}^{2}|$ for all $n$ but $(a_n)$ is not convergent.

Title: How to prove that


问:
Integrate $\frac{1}{ \sqrt{T^2 - \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} - \frac{b^2}{4 (T-\tau )}\right)$
I want to compute the integral
$$
\int_t^T \int_{-\infty}^\infty \frac{1}{ \sqrt{\tau - t} (T-\tau)} \exp\left(-\frac{(z-x)^2}{2(\tau - t)} -\frac{(z-v)^2 + (z-w)^2}{2(T-\tau)} \right) d z d \tau.
$$
First integrating with respect to $z$ I get
$$
\int_t^T \frac{\sqrt{2\pi}}{ \sqrt{T - \tau}\sqrt{-2 t + \tau + T}}\exp\left(-\frac{(v + w - 2 x)^2}{4 (-2 t + \tau + T)} - \frac{(v-w)^2}{4 (T-\tau )}\right) d \tau.
$$
So to go further I think I should be able to integrate
$$
\int_0^T \frac{1}{ \sqrt{T^2 - \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} - \frac{b^2}{4 (T-\tau )}\right) d \tau
$$

答:
$$I=\underbrace{\int_0^{T}\frac{d\tau}{\sqrt{T^2-\tau^2}}\exp\left(-\frac{a^2}{4(T+\tau)}-\frac{b^2}{4(T-\tau)}\right)}\\{\tau\rightarrow \frac{T(1-y^2)}{1+y^2}}$$

$$=\exp\left(-\frac{a^2+b^2}{8T}\right)\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\frac{a^2y^2+\frac{b^2}{y^2}}{8T}\right)$$

Let's parametrize the integral in order to build two ODE: $$ f(\alpha,\beta)=\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\alpha^2y^2-\frac{\beta^2}{y^2}\right)$$

$$g(\alpha,\beta)=f(\alpha ,\beta)-\frac{1}{2\alpha}\frac{\partial f}{\partial\alpha}=\int_{0}^{1}\exp\left(-\alpha^2y^2-\frac{\beta^2}{y^2}\right)dy$$

Now, let's use $g(\alpha,\beta)$ to solve the remaining integral: $$\alpha g(\alpha,\beta)+\frac{1}{2}\frac{\partial g}{\partial\beta}=\exp\left(2\alpha\beta\right)\int_{0}^{1}\left(\alpha-\frac{\beta}{y^2}\right)\exp\left[-\left(\alpha y+\frac{\beta}{y}\right)^2 \right]dy=-\frac{\sqrt{\pi}}{2}\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$

$$\frac{\partial g}{\partial\beta}+2\alpha g(\alpha,\beta)=-\sqrt{\pi}\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$

$$\mu_1(\beta)=\exp\left(\int 2\alpha d\beta\right)=\exp\left(2\alpha\beta\right)$$

$$\exp\left(2\alpha\beta\right)\frac{\partial g}{\partial\beta}+2\alpha\exp\left(2\alpha\beta\right) g(\alpha,\beta)=-\sqrt{\pi}\exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$

$$\left(\exp\left(2\alpha\beta\right)g(\alpha,\beta)\right)'=-\sqrt{\pi}\int \exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)d\beta=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\operatorname{erf}\left(\alpha-\beta\right)\right)$$

$$g\left(\alpha,\beta\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)\right)+c$$

$$g\left(\alpha,0\right)=\int_{0}^{1}\exp\left[-\left(\alpha y\right)^2 \right]dy=\frac{\sqrt{\pi}}{2\alpha}\ \operatorname{erf}\left(\alpha\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\ \operatorname{erfc}\left(\alpha\right)-\ \operatorname{erf}\left(\alpha\right)\right)+c\\ c=\frac{\sqrt{\pi}}{4\alpha}$$

$$\boxed{g\left(\alpha,\beta\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)}$$

Using this result in the first ODE: $$f(\alpha ,\beta)-\frac{1}{2\alpha}\frac{\partial f}{\partial\alpha}=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)$$

$$\frac{\partial f}{\partial\alpha}-2\alpha f(\alpha ,\beta)=\frac{\sqrt{\pi}}{2}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)$$

$$\mu_2(\alpha)=\exp\left(-\int 2\alpha d\alpha\right)=\exp\left(-\alpha^2\right)$$

$$\left(\exp\left(-\alpha^2\right)f(\alpha,\beta)\right)'=\frac{\sqrt{\pi}}{2}\int\left(\exp\left(-\alpha^2+2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-\alpha^2-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-\exp\left(-\alpha^2\right)\right)d\alpha$$

$$=-\frac{\operatorname{erf}\left(\alpha\right)\sqrt{\pi}}{2}+\frac{e\sqrt{\pi}}{2}\int\left(\underbrace{\exp\left(-\left(\alpha-\beta\right)^2\right)\ \operatorname{erfc}\left(\alpha+\beta\right)}_{IBP}-\exp\left(-\left(\alpha+\beta\right)^2\right)\ \operatorname{erf}\left(\alpha-\beta\right)\right)d\alpha$$

$$=-\frac{\operatorname{erf}\left(\alpha\right)\sqrt{\pi}}{2}+\frac{e\sqrt{\pi}}{2}\left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\int\left(\underbrace{\operatorname{erf}\left(\beta-\alpha\right)+\operatorname{erf}\left(\beta+\alpha\right)}_{0}\right)\exp\left(-(\alpha+\beta)^2\right)d\alpha\right)$$

$$f(\alpha ,\beta)=-\frac{\sqrt{\pi}\operatorname{erf}\left(\alpha\right)e^{\alpha^2}}{2}-\frac{\pi e^{a^2+1}}{4}\left(\operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)\right)+k$$

$$f(0,0)=\int_0^1\frac{dy}{1+y^2}=\frac{\pi}{4}=k$$

$$\boxed{f(\alpha,\beta)=-\frac{\sqrt{\pi}e^{\alpha^2}}{2}\operatorname{erf}\left(\alpha\right)-\frac{\pi e^{\alpha^2+1}}{4}\left(\operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)\right)+\frac{\pi}{4}}$$

Therefore, the original integral is equal to: $$I=\exp\left(-\frac{a^2+b^2}{8T}\right)f\left(\frac{a}{2\sqrt{2T}},\frac{b}{2\sqrt{2T}}\right)=$$

$$ \bbox[5px,border:2px solid red] { \frac{\pi}{4}\exp\left(-\frac{a^2+b^2}{8T}\right)-\frac{\sqrt{\pi}}{2} \exp\left(-\frac{b^2}{8T}\right)\ \operatorname{erf}\left(\frac{a}{2\sqrt{2T}}\right)-\frac{\pi}{2} \exp\left(-\frac{b^2}{8T}+1\right)\ \operatorname{erf}\left(\frac{b-a}{2\sqrt{2T}}\right)\ \operatorname{erfc}\left(\frac{b+a}{2\sqrt{2T}}\right) } $$


问:
Variance and Expectation of a biased coin toss with different distributions
I would love to verify if my thought process is correct regarding $E(X)$ and $Var(X)$:

Problem: A biased coin is toss with $P(\{H\}) = p$ and $P(\{T\}) = 1-p$. $X$ is a random variable.

If $H$ then $X \sim U(a,b)$ with expectation and variance $E(X_H)$ and $Var(X_H)$.

If $T$ then $X \sim Exp(\lambda)$ with expectation and variance $E(X_T)$ and $Var(X_T)$.

The expectation and variance of X is then:

\begin{equation}
  \begin{aligned}
    E(X) =& \quad p*E(X_H) + (1-p)*E(X_T)  \\
    \text{and}  \\
    Var(X)=& \quad E(X^2) - E(X)^2  \\
          =& \quad p*E(X_H)^2 + (1-p)*E(X_T)^2 - E(X)^2 \\
  \end{aligned}
\end{equation}

答:
Your formulae are correct. Here is a proof.

If we note $Y$ the result of the coin draw. Heads is denoted by $Y=1$, tails $Y=0$, and let $U \sim U(a,b)$,$Z \sim \exp(\lambda)$. Thus formally :

$$X=\mathbf 1_{Y=1}U+\mathbf 1_{Y=0}Z$$

Thus :

$$E(X)=E(\mathbf 1_{Y=1}U+\mathbf 1_{Y=0}Z)$$

$\mathbf 1_{Y=1}$ and $\mathbf 1_{Y=0}$ are independent of $U$ and $Z$, the outcome of the draw will chose if you look at $U$ or $Z$ but the value of these r.v. are not impacted. Thus by linearity :

$$E(X)=E(\mathbf 1_{Y=1})E(U)+E(\mathbf 1_{Y=0})E(Z)$$

Since $E(\mathbf 1_{Y=1}) =P(Y=1)=p$, we arrive at your formula. For the variance we can extend this reasoning.


问:
Finding stationary Distribution
I need to know how to find the stationary distribution for this matrix:
$$
Q=
\begin{bmatrix}
-2 & 2  & 0 & 0 \\
1  & -2 & 1 & 0 \\
0 & 1 & -2 &1\\
0  & 0  & 2 & -2 \\
\end{bmatrix}$$
in three different ways.

I only know one, which is to replace the last column with $1$'s and then take the inverse. The last row will be the stationary distribution.

Thank you

Update: Sorry, I forgot a row. Thanks for the comments!

答:
Since you are asking for different proofs, here is a proof by exhibition: the unique stationary distribution $\pi$ is
$$
\pi=\left(\tfrac16,\tfrac13,\tfrac13,\tfrac16\right).
$$
And a proof by uniformization: assume that the state space is $S=\{1,2,3,4\}$, duplicate state $2$ into states $2$ and $6$, and duplicate state $3$ into states $3$ and $5$. Thus, the transition rates from state $1$ are now $+1$ to state $2$ and $+1$ to state $6$. Likewise, the transition rates from state $4$ are now $+1$ to state $3$ and $+1$ to state $5$. The transition rates from states $2$ and $3$ are unchanged and the transition rates from states $5$ and $6$ are those from states $3$ and $2$ respectively.

Phew! Almost done... The resulting Markov chain performs a standard random walk on the state space $\bar S=\{1,2,3,4,5,6\}$, best viewed as the discrete circle $\mathbb Z/6\mathbb Z$. Its  stationary distribution $\bar\pi$ is obviously uniform on $\bar S$ and $\pi$ is the projection of $\bar\pi$ on $S$. States $1$ and $4$ correspond to one state in $\bar S$ while states $2$ and $3$ correspond to two, hence the ratios $1:2:2:1$ in $\pi$.

And finally a proof by definition (probably the most direct in this context): solve $\pi Q=0$.

Title: Finding stationary Distribution


问:
Projection map: change of coordinates
Let $X\subset\mathbb{P}^n$ be a projective variety,and let $P\in\mathbb{P}^n$ be a point not in $X$.

By a change of coordinates we can assume that $P=[0:\cdots:1].$

Question. what change of coordinates are we talking about? How can you write it explicitly?

Let $H\cong\mathbb{P}^{n-1}\subset\mathbb{P}^n$ be a linear subspace of codimension $1$ tha does not contain $P$; again by a change of coordinates we can assume that $H=\{x_n=0\}$.

Question. what change of coordinates are we talking about? How can you write it explicitly?

Thanks!

答:
I think the  way to think about this is to realize that there's nothing special about any point in $\mathbb P^n$. The same goes for any (n-1)-plane in $\mathbb P^n$.

To expand upon @Mummy the turkey's comment: every point in $\mathbb P^n$ is represented by a point $\overline P$ in $\mathbb A^{n+1}$. There is always a linear transformation sending $\overline P$ to the point $(1, 0, \ldots, 0)$. An explicit example is found as follows: find a basis for $\mathbb A^{n+1}$ as a vector space, where $\overline P$ is the first vector. Make the matrix $A=[\overline P, \ldots]$. Then the matrix $A^{-1}$ takes $\overline P$ to $(1, 0, \ldots, 0)$.

All this  carries over to $\mathbb P ^n$ since matrix multiplication is linear.

Title: Projection map: change of coordinates