问:
Finding log likelihood
Consider $X_1,X_2,…,X_n\sim N(\mu,\sigma^2)$.

The density of a single $X_i\sim N(\mu,\sigma^2)$ is
$$f_{X_i}(x_i;\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{−(x_i−\mu)^2}{2σ^2}\right).$$

How do I find the log-likelihood $\ln(L(\mu,\sigma^2|x_1,…,x_n))$?

答:
$$\begin{align}
L(\mu, \sigma^2)&=\prod_{i=1}^n f_{X_i}(x_i)\\
&=\prod_{i=1}^n\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{−(x_i−\mu)^2}{2σ^2}\right)\\
&=\frac{1}{(\sqrt{2\pi\sigma^2})^n}\exp\left(\sum_{i=1}^n\frac{−(x_i−\mu)^2}{2σ^2}\right)\\
\end{align}
$$

So we have
$$\begin{align}
\ln L(\mu, \sigma^2)&=\underbrace{-n\ln(\sqrt{2\pi})}_{\text{constant}} - n\ln(\sqrt{\sigma^2}) + \sum_{i=1}^n\frac{−(x_i−\mu)^2}{2σ^2}\\
\end{align}
$$

Title: Finding log likelihood


问:
Suppose $2y+3x=3y-4x$, and $x$ and $y$ are not both zero. Prove that $y$ does not equal $0$.
I was wondering if my explanation makes sense. I solved the equations each if one or the other was $0$ like the following:

$2y+3(0) = 3y-4(0) \implies 0=y$, meaning both $y$ and $x$ equal zero to make the statement correct.

$2(0)+3x = 3(0)-4x \implies 0=x$, meaning both $x$ and $y$ equal zero to make the statement correct.

Therefore, $y\neq 0$ because if one of them were to be zero, the other one has to be zero for the statement to be true.

答: You only needed that last equation to prove it. You did a proof by contradiction where you assumed the opposite of the conjecture and found a contradiction. In this case, it asks for you to prove $y\neq 0$, so for a proof by contradiction you will assume that $y=0$. Your second equation already does that and finds that $x=0$, but you know that both $x$ and $y$ cannot be zero. You have found a contradiction and are done. Your first equation is unnecessary for this problem because you have already solved it.

ContentLicense: CC BY-SA 4.0


问:
Are $SO(3)$ and $SU(2)$ abstractly isomorphic?
The groups $SO(3)$ and $SU(2)$ are usually "the" classic example of nonisomorphic Lie groups whose corresponding Lie algebras are isomorphic. While I understand the isomorphism of their algebras, it was not obvious at all to me that we must have $SU(2) \not\cong SO(3)$, I think one can conclude that by looking at the topological structure, for example the fact that their fundamental groups are nonisomorphic should suffice (I think).

What if we forget about the topology and consider them just as abstract groups? I've been trying to find some obstruction but all the facts I know about them, for example $SO(3) \cong SU(2)/Z(SU(2))$, don't seem to be enough! Am I missing something obvious?

答:
The standard way to distinguish these two is that $\pi_1(\operatorname{SU}(2))=0$ while $\pi_1(\operatorname{SO}(3))=\Bbb{Z}/2$. Actually, more is true. You can realize $\operatorname{SU}(2)$ as the unit quaternions (hence diffeomorphic to $S^3$), and $\operatorname{SO}(3)=\operatorname{SU}(2)/Z(\operatorname{SU}(2))$ realizes $\operatorname{SU}(2)\to\operatorname{SO}(3)$ as a $2-$fold covering map. From what we know about the topology of $S^3$ we get indeed that $\operatorname{SU}(2)$ is simply connected and hence that $\pi_1(\operatorname{SO}(3))\cong \Bbb{Z}/2.$

Clearly, I misread part of the question. Sorry – we can analyze the centres of these groups to find that $Z(\operatorname{SO}(3))=\{I_3\}$, while $Z(\operatorname{SU}(2))=\{\pm I_2\}$. This is a good exercise to do for yourself. The fact about $\operatorname{SU}(2)$ follows from thinking about the identification as quaternions, while the fact for $\operatorname{SO}(3)$ can be understood by thinking about $\operatorname{SO}(3)$ acting on $S^2$. You want to show that the centres are of the form $\{\text{Scalar matrices}\}\cap G$.

I think there is also a nice proof of this using Schur's Lemma.


问:
Generating functions and the Riemann Zeta Function
The generating function for the terms of the harmonic series:

$\frac{1}{n}$

is $-\ln(1 - x)$.

Does an ordinary generating function exist for the terms of the zeta function $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$
for any $ s > 1$?

That is, does there exist any function $f(x)$ that can be expressed in terms of elementary functions such that $f(x) = \sum_{n=1}^\infty \frac{1}{n^s}x^n$ for some $s > 1$?

I'm assuming that such a function in fact does not exist.  Can this be proven to be the case?

答:
This is an answer to the initial question (asking for a generating function of $\zeta(s)$) :

$−\ln(1−x)$ is the generating function of $\ \frac 1n$.

$-\frac{\ln(1−x)}{1-x}$ is the generating function of the harmonic number $\ H_n=\sum_{k=1}^n\frac 1k$

The generating function of the generalized harmonic number $\ H_{n,s}:=\sum_{k=1}^n\frac 1{k^s}\ $ is given by :
$$\frac{\operatorname{Li}_s(x)}{1-x}$$
with $\operatorname{Li}_s$ the polylogarithm.

Should you simply want $\ \displaystyle\sum_{n=1}^\infty \frac{x^n}{n^s}=\operatorname{Li}_s(x)\ $ then dot dot's answer is right of course !

Note that a generating function for $\zeta(n)$ is known as the digamma function :
$$\psi(1+x)=-\gamma-\sum_{n=1}^\infty \zeta(n+1)\;(-x)^n$$
while the reflection formula allows to get the even values of $\zeta$ directly as :
$$\pi\;x\;\cot(\pi\;x)=-2\sum_{n=0}^\infty \zeta(2n)\;x^{2n}$$

A generating function for the polylogarithm was obtained too :
$$z\,\Phi(z, 1, 1-x)=\sum_{n=0}^\infty\;\operatorname{Li}_{n+1}(z)\;x^n$$
using the Lerch zeta function $\displaystyle\;\Phi(z, s, \alpha) := \sum_{k=0}^\infty \frac { z^k} {(k+\alpha)^s}$.

ContentLicense: CC BY-SA 3.0


问:
Why are all prime numbers are congruent to 1 or 5 (mod6)?
I understand that any natural number congruent to 2 or 4 (mod6) is even and that 3 is the only prime congruent to 3(mod6). Thus, any prime, p > 3, must be congruent to either 1 or 5 (mod 6).

I just don't understand the reasoning behind why any natural number congruent to 2 or 4 (mod6) is even and that any natural number congruent to 3(mod6) is a multiple of 3.

Could someone explain why this is true?

答:
All you have to do is eliminate the other cases, and use the definitions of modular congruence, divisibility, and primality. Recall:

$$x \equiv y \pmod z \iff z \mid x-y \iff \exists k \in \Bbb Z^+ \text{ such that } x-y = kz$$

(Or at least, we use the positive integers in this context usually.)

So, if $p \equiv 0 \pmod 6$, then $p = 6k$ for an integer $k$. That means $p$ isn't prime.

If $p \equiv 3 \pmod 6$, then $p = 6k+3$ for an integer $k$. But then $p=3(k+1)$, and thus is not prime (assuming $p > 3$).

If $p\equiv 2,4 \pmod 6$ then you have that $p=6k+2 = 2(3k+1)$ or $p = 6k+4 = 2(3k+2)$, respectively, for integers $k$, and thus no primality (assuming $p > 3$).

Thus, the only remaining cases are $1$ and $5$.


问:
Maximum distance of Bezier curve
I have a Bezier curve given by four points: a start point ($P_0$), an end point ($P_3$), and two control points ($P_1$ and $P_2$). The points lie in a certain way. $P_1$ and $P_3$ lie on the x axis. $P_1$ lies on a vertical line with $P_0$ and $P_2$ lies on a vertical line with $P_3$. $P_1$ and $P_2$ also lie on the same horizontal line. Let's say the following

\begin{aligned}
P_0 &= (0,0), & P_1 &= (x_1,y_1), & P_2 &= (x_2,y_1), & P_4 &= (x_2,0)
\end{aligned}

How do I get the maximum distance between the baseline (line from $P_0$ and $P_3$) and the Bezier curve? In other words, how do I get the top of the bezier curve?

答:
If I understand correctly, your points form a rectangle aligned with the coordinate axes. I think your formula for $P_1$ has a typo — it should be $P_1 = (0,y_1)$.

In that case, the height of the Bézier curve is $3/4$ the height of the rectangle.

Why? Well, by symmetry, the highest point must occur where $t = 1/2$. If you plug $t=1/2$ into the equation of the Bézier curve, you’ll get the coordinates of the highest point. The $y$ coordinate of this point is the value you want.

Title: Maximum distance of Bezier curve


问:
How to prove that $f(x) = x^2_i$ is differentiable?
Let $f:\mathbb{R}^m \longrightarrow\mathbb{R}$ denoted by $f(x) = x^n_i$, for $i \in \{1,2,\dots,m\}$ and $n \in \mathbb{N}$, both fixed. So, let $x, h \in \mathbb{R}^m$, where $h \not = 0$. Then we have
\begin{align}
f(x+h) - f(x) &= (x_i - h_i)^n - x^n_i\\
&= \sum_{j = 0}^n{\binom{n}{j}x^{n-j}_ih^j_i} - x^n_i\\
&= \sum_{j = 1}^n{\binom{n}{j}x^{n-j}_ih^j_i}\\
&= nx^{n-1}h_i + \sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}.
\end{align}
So, basically, we have to prove that
$$\lim_{h \to 0}\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|} = 0.$$
I tried to manipulate the following expression
$$\left|\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|}\right|$$
but no success. Is there any hint or strategy to help me?

答:
$$f(x)=x_i^n.$$

You want to show that

$$\frac{\partial f}{\partial x_i} = \frac{\partial x_i^n}{\partial x_i} = n x_i^{n-1}$$
and
$$\frac{\partial f}{\partial x_j} = \frac{\partial x_i^2}{\partial x_j} = 0, \quad (i\ne j).$$

The first partial derivative is computed just like an ordinary derivative.  The other expression should also be clear.


问:
Solve $\frac{dy}{dx}=\frac{y+y^2+x^2}{x}$
Suppose I didn't know to begin with that $y=x\tan x$ satisfies the differential equation below; how I would I go about solving this differential equation?

$$\frac{dy}{dx}=\frac{y+y^2+x^2}{x}$$
It seems very unlikely that I can use some clever manipulation to convert this into a differential equation that can be solved by separating the variables, and I also don't think that I'd be able to use the integrating factor method here.

The only thing that comes to mind as possibly useful is to use some substitution, but I cannot yet see any substitution that is useful.

Thank you for your help.

答:
$$\frac{dy}{dx}=\frac{y+y^2+x^2}{x}$$
$$xy'={y+y^2+x^2}$$
$$xy'-y=y^2+x^2$$
$$\left (\dfrac yx \right)'=1+\dfrac {y^2}{x^2}$$
$$d\left (\dfrac yx \right)=\left(1+\dfrac {y^2}{x^2} \right) dx$$
This is separable.


问:
Counting subsets of fixed size which differ by more than one element.
The board game Dominion is randomized at the start of play to allow players to purchase from 10 different piles, each pile chosen at random from 26 different choices.  This number (26) has increased greatly over the years as more and more expansions have been added, and it now sits around 360.

The number of distinct 'games' is easy to count: ${26\choose 10}$.  But games which differ by one card do not, in my experience with the early expansions, differ greatly in play style.  Is there a reasonable estimate to the number of games which differ by two cards?  To be more precise, is there a reasonable estimate to the number of games one could play which never (pairwise) overlap by 9 cards?

答:
The question as stated is not well posed. There are many possible sets which fit the description "set of games where each pair differ by two or more cards," you can cannot ask "the" size of such a set. It is reasonable to ask what the largest possible such set of games is, so I will assume you mean that.

This is likely an open problem. In the literature, a collection of subsets where every pair differ in a prescribed number of elements is known as a constant weight binary code. According to https://www.win.tue.nl/~aeb/codes/Andw.html, there exists $222775$ subsets of $\{1,2,\dots,26\}$ of size $10$ where each pair have at most  $8$ elements in common (see the $A(n,4,w)$ table, $n=26,w=10$). It is unknown if you can push this number any higher. They cite [Brouwer et. al.] for the construction of this set of sets. Note that the data goes no where near close to $n=360$, so I have no answer on how many such games you can have for the full Dominion set.

A. E. Brouwer, J. B. Shearer, N. J. A. Sloane & W. D. Smith,
A new table of constant weight codes,
IEEE Trans. Inform. Theory 36 (1990) 1334-1380. (http://www2.research.att.com/~njas/codes/Andw/)


问:
Symbol for "if and only if": $\implies$ or $\iff$?
I was wondering about the iff sign in maths.
I've never learned about it in school & see it a lot online. Usually the sign looks like this: $\implies$, but in math.stackexchange I always see this: $\iff$.

Does that sign means iff?

Am I using iff right here:

$$8x + 2x - 44 = 220 + 4 - x \implies 11x = 268 \implies x = 24.36$$

答:
To complete the answer to include what you mentioned in the title:

"${}\Longleftarrow{}$" means "if",

"$\implies$" means "only if", and

"$\iff$" means "if and only if", which is sometimes written "iff".

A bare "if" not followed by "and only if" is actually rather rarely used. Instead of  "$A$ only if $B$" it is common to say "$A$ implies $B$".


问:
Can I interpret $\models$ as $\implies$ while proving logical expressions using different techniques?
In the book by Artificial Intelligence by Norvig and Russel, I came across following problem:

Prove if correct: $(A ∧ B) \models (A ⇔ B).$

I quickly interpreted $\models$ as $\implies$ and tried to prove it using the truth table:

It seems that at least while interpreting $\models$ as $\implies$, the statement is true. Then I gave second thought and did some more reading to come across this thread. Now I know that the two are not same. But, it turns out, when we don't interpret $\models$ as $\implies$, the statement is still true (my course TA uploaded answers without proof).

So I am wondering:

Q1. How exactly the given statement is correct (given that $\models$ and $\implies$ are not same)?

Q2. Is my method to interpret both same and then forming truth table, a correct method for such  problems? If not then how I should solve it?

Q3. If answer to Q2 is no, then will above method of interpreting $\models$ as $\implies$ and forming truth table always given correct answer? If not, when it will fail to give correct answer?

Q4. I also tried to solve the same using resolution:

$$\neg (A\Longleftrightarrow B)\equiv \neg((A\wedge B)\vee(\neg A\wedge\neg B))\equiv\neg(A\wedge B)\wedge \neg(\neg A\wedge \neg B)\equiv (\neg A\vee\neg B)\wedge (A \vee B)$$
So my clauses will be:

$A$ (from $A\wedge B$)

$B$ (from $A\wedge B$)

$(\neg A\vee\neg B)$ from $(\neg A\vee\neg B)\wedge (A \vee B)$

$(A \vee B)$ from $(\neg A\vee\neg B)\wedge (A \vee B)$

So I was able to derive empty clause, so my assumption $\neg (A\Longleftrightarrow B)$ was incorrect. So $(A ∧ B) \implies (A ⇔ B)$. Will application of resolution technique for $\models$ be exactly same?

Update

[This is my updated understanding based on Graham's answer]

(a) After reading Graham's answer, I felt that the truth table above is proving the "tautology" $(A ∧ B) \implies (A ⇔ B)$, but not $(A ∧ B) \models (A ⇔ B)$.

(b) To prove $(A ∧ B) \models (A ⇔ B)$, I need another truth table, something like this:

(c) Also, I guess the resolution technique is used to prove tautology $(A ∧ B) \implies (A ⇔ B)$ and not $(A ∧ B) \models (A ⇔ B)$. However, I feel we can use the (first) truth table and resolution proving $\implies$ to also prove $\models$, because of the following fact:

$\varphi\vDash \psi$ iff $M(\varphi)\subseteq M(\psi)$: that is, iff every truth assignment that makes $\varphi$ true also make $\psi$ true. This is the case iff $\vDash \varphi\Rightarrow\psi$, i.e., if the formula $\varphi\Rightarrow\psi$ is true in all truth assignments (is a tautology). - source

Can someone please confirm if my understanding in the above points is correct?

答:
$\mathcal A \vDash \varphi$ says: "statement $\varphi$ is a semantic consequence of the list of statements $\mathcal A$ (often called the premises, or assumptions)" and a semantic consequence must be true when all the premises are interpreted as true.  [Note: this list may be of zero, one, or several statements.]

In propositional logic, you can use a truth table to establish such a semantic consequence: and $(A\land B)\vDash(A\leftrightarrow B)$ since indeed "every assignment that values $A\land B$ as true will also value $A\leftrightarrow B$ as true," will be shown by a table of every assignment of $A,B$ and the corresponding evaluations of the statements.

$$\boxed{\begin{array}{c4|c}A&B & A\land B& A\leftrightarrow B\\\hline\top&\top &\top &\top&\star\\\top&\bot&\bot &\bot\\\bot&\top&\bot&\bot\\\bot&\bot&\bot&\top\\\end{array}}\\\text{the consequent is true in the only row where the premise is true}$$

[Note: the trend this century is to use single bar arrows, $\to,\leftrightarrow$, for logical connectives, and double edge for metalogic inferences, though not all authors do so.]

$(A\land B)\to(A\leftrightarrow B)$ is a statement, and moreover one which is a tautology.   That means that it is a logical consequence of no premises at all.  Thus:$$\vDash (A\land B)\to(A\leftrightarrow B)$$

This can also be demonstrated by a truth table, where we show that "every assignment of $A,B$ will value the whole statement as true."

$$\boxed{\begin{array}{c4|c}A&B & (A\land B)\to(A\leftrightarrow B)\\\hline\top&\top &\top\\\top&\bot&\top\\\bot&\top&\top\\\bot&\bot&\top\\\end{array}}\\\text{the consequent is true in every row}$$

That is not exactly the same test, but it is clearly the case that where one works the other will too.

It should be noted, truth tables work as semantic proofs only in classic propositional logic.  They are less applicable in higher order logics, modal logics, among others.


问:
If $\int_a^\infty f(x)dx$ exists (converges), then $\int_a^\infty \frac{f(x)}{x}dx$ exists
This is a true or false statement:

If $\displaystyle \int_a^\infty f(x)dx$ exists (converges), then $\displaystyle\int_a^\infty \frac{f(x)}{x}dx$ converges for $a>0$ (and $a\in\mathbb{R^+}$).

In first place I thought this was false, but tried a numerous amount of functions but none of them works as a counterexample. It also confuses me that $a>0$, since I'm not sure if an inegral where I take $a=10^{-50}$ (for instance) it should diverge if it's a "huge" number or not.

It might be true, but I can't prove it. Any hint?

PD: I don't know if working with bounded functions in the interval $[a,1]$ may also work.

答:
This is true if  the improper integral $\displaystyle\int_a^\infty f(x) \, dx$ is conditionally convergent, even if $f \not\in L^1([a,\infty))$ and $f$ is neither nonnegative nor nonpositive.

Note that  $x \mapsto \frac{1}{x}$ is decreasing.  By the second mean value theorem for integrals, for any $c_2 > c_1 > a$  there exists $\xi \in (c_1,c_2)$ such that

$$\int_{c_1}^{c_2}\frac{f(x)}{x} \, dx = \frac{1}{c_1} \int_{c_1}^\xi f(x) \, dx,$$

and

$$\left|\int_{c_1}^{c_2}\frac{f(x)}{x} \, dx\right| = \frac{1}{c_1} \left|\int_{c_1}^\xi f(x) \, dx\right|\leqslant \frac{1}{a} \left|\int_{c_1}^\xi f(x) \, dx\right| $$

Since the improper integral of $f$ converges, it follows by the Cauchy criterion that for any $\epsilon > 0$ there exists $C(\epsilon)> a$ such that for all $\xi > c_1 > C(\epsilon)$,
$$ \left|\int_{c_1}^\xi f(x) \, dx\right|< a\epsilon$$

Thus, for all $c_2> c_1 > C(\epsilon)$, we have

$$\left|\int_{c_1}^{c_2}\frac{f(x)}{x} \, dx\right| < \epsilon,$$

and $\displaystyle\int_a^\infty \frac{f(x)}{x} \, dx$ is convergent.


问:
Uniform convergence of the inverse (solution verification)
Consider a continuous, increasing function $f : \mathbb R \rightarrow [0,1]$ that is onto $(0,1)$. Let $f_n$ be a sequence of non-decreasing functions converging uniformly to $f$. Show that $f_n^{-1}$ converges uniformly to $f^{-1}$ over the open interval $(0, 1)$ where we define $g^{-1}(\alpha) := \inf \{y | g(y) \geq g(\alpha)\}$.

What I've tried

We know that $f^{-1}$ is continuous (Inverse of a continuous increasing function is continuous (solution verification)).

We need to show that: $\sup_{\alpha \in (0,1)}|f_n^{-1}(\alpha) - f^{-1}(\alpha)| \rightarrow 0$ as $n\rightarrow \infty$.

Notice that:
$$|f_n^{-1}(\alpha) - f^{-1}(\alpha)| = |f_n^{-1}(f(x)) - f^{-1}(f(x))|$$

for some $x$ such that $f(x) = \alpha$ (since $f$ is onto $(0, 1)$). Now we have:

\begin{equation}\tag{1}
= |f_n^{-1}(f(x)) - x|
\end{equation}

Applying the definition of $g^{-1}$:

$$f_n^{-1}(f(x)) = \inf\{ y | f_n(y) \geq f(x)\} = \inf\{ y | f(y) + o(1) \geq f(x)\}$$

where the second equality comes from the uniform convergence of $f_n$ to $f$. Finally, since $f^{-1}$ is continuous: $y \geq f^{-1}(f(x)  + o(1)) \rightarrow x$ so the RHS above converges to $x$. So we have that expression (1) goes to $0$ showing the desired convergence.

I could used some feedback on my reasoning.

答:
Having proved the pointwise convergence you are in fact done, because of the following

Claim: Let $(h_n)$ be a sequence of non-decreasing functions, each mapping $(0,1)$ into $\Bbb R$ and converging pointwise to a continuous strictly increasing function $h$ mapping $(0,1)$ onto $\Bbb R$. Then the convergence is  uniform on compact subsets of $(0,1)$.

Fix $0<a<b<1$ and consider $h$ on the compact interval $[a,b]$.   For a large integer $m$  partition $[h(a),h(b)]$ into $m$ equal-length subintervals, with endpoints $h(a)=y_0<y_1<\cdots<y_m=h(b)$. Define  $x_k:=h^{-1}(y_k)$. Suppose $u\in[x_k,x_{k+1}]$ for some $k\in\{0,1,2,\ldots,m-1\}$. Then
$$
h_n(x_k)-h(x_{k+1})\le h_n(u)-h(u)\le h_n(x_{k+1})-h(x_k),
$$
and so
$$
\sup_{a\le u\le b}|h_n(u)-h(u)|\le\left[\max_{0\le k\le m-1}|h(x_k)-h_n(x_{k+1})|\vee \max_{0\le k\le m-1}|h_n(x_k)-h(x_{k+1})|\right].
$$
Sending $n\to\infty$ we obtain
$$
\limsup_n\sup_{a\le u\le b}|h_n(u)-h(u)|\le\max_{1\le k\le m}|h(x_k)-h(x_{k+1})|=1/m.
$$
As $m$ can be made as large as we please, $\lim_n\sup_{u\in[a,b]}|h_n(u)-h(u)|=0$.


问:
Question about a definition $S_2 = \{x \in N : -x\in N\}$ from the book Mathematical Proofs
I'm wondering about the following from the book Mathematical Proofs

$$S_2 = \{x \in N : -x\in N\} = \emptyset$$

I'm not really sure what this is even saying I'm reading it as follows the set S2 has elements {X that are members of the natural numbers such that negative x are in the natural numbers}.
This statement seems nonsensical to me. What is this actually saying?

Another example from the book that seems absurd to me is the following.

$$\{x \in Z: |x| = -x\}$$

How can this ever be the case, that $|x| = -x$.

Thank you in advance for any help,

答: This is just an example of an empty set, because, as you said, there is no natural number $x$ that, $-x$ is a natural number. Thus, there is no element $S_2$. Another way to see that: assumed that $0 \not \in \mathbb{N}$, suppose that there is a $x \in S_2$. So $x \in \mathbb{N}$ and $-x \in \mathbb{N}$. Then, $x + (-x) \in \mathbb{N}$, and, $0$ would be an element of $\mathbb{N}$. That's a contradiction. So, there is no elements in $S_2$.


问:
Probability of Maximum of Subset of Multinomial Random Variable
Suppose that I have a random variable $X \sim \text{Multinomial}(N, M, \mathbf{p})$, where $\mathbf{p} = \frac{1}{M} \mathbf{1}$, $N$ is the number of trials, and $M$ is the number of bins.

I want to find out:

$$ P\left(\text{max}\left\{ x_{2}, x_{3}, \ldots, x_{M} \right\} \leq \gamma \right)  $$

In other words, I want to find out the probability that the maximum of all but one of the $x_{i}$ are less than a threshold.

My immediate thought is that because $x_{i} \sim \text{Binomial}\left( N, p \right)$, that the answer would simply be:

$$ P\left(\text{max}\left\{ x_{2}, x_{3}, \ldots, x_{M} \right\} \leq \gamma \right) = \prod_{i=2}^{M} P(x_{i} \leq \gamma) =  P(x_{i} \leq \gamma)^{M-1} $$

Therefore, if I wanted the maximum of all but $k$ of the $x_{i}$, I would have:

$$ P\left(\text{max}\left\{ x_{k+1}, x_{k+2}, \ldots, x_{M} \right\} \leq \gamma \right) = \prod_{i=k+1}^{M} P(x_{i} \leq \gamma) =  P(x_{i} \leq \gamma)^{M-k} $$

However, this doesn't seem to be the case when I compare against a numerical experiment. The figure shows $P(\text{max}\left\{ x_{2}, x_{3} \right\} \leq \gamma)$ when $X \sim \text{Multinomial}\left(N=1000, M=3, p = [\frac{1}{3}, \frac{1}{3}, \frac{1}{3}]^{T} \right)$. The red curve represents my thoughts above and the blue curve represents the hard data.

So, what am I missing?

答:
What you're missing is that the $x_i$ are not independent binomials. The sum $x_1 + x_2 + \dots + x_n$ is fixed to $N$, so if one of them is small, that makes the rest likely to be larger.

Consider the $\gamma=0$ case of your example, chosen because it's the easiest to compute. We have $\max\{x_2, x_3\} \le \gamma$ precisely when $X = (N,0,0)$, which happens with probability $(\frac13)^N$. But the individual probability that $x_i \le \gamma$ is $(\frac23)^N$, so your method gives $(\frac23)^{2N} = (\frac49)^N$ as the answer instead.

I don't expect a closed form to exist for your expression, because even $\Pr[x_i \le \gamma]$ has no closed form.


问:
finding $2$-cycle in logistic difference equation
The logistic difference equation is given by $x_{n+1}=Cx_n(1-x_n)$. I'm supposed to show that when $C>3$, the logistic difference equation has a $2$-cycle.

Letting $f(x)=Cx(1-x)$, if $\{x_0,x_1\}$ is our $2$-cycle we have that $x_{1}=f(x_0)$. Thus, if this is a $2$-cycle, we want to find values of $C$ so that $f^2(x_0)=x_0.$

So what we want is some value of $C$ so that $$f(f(x_0))=Cf(x_0)(1-f(x_0))=C(Cx_0(1-x_0))(1-(Cx_0(1-x_0))) = x_0.$$

My thinking is that, after some manipulation, I should be able to show that this equality can only hold when $C>3$, but actually performing the manipulation which shows that is proving to be a bit of a challenge.

Any thoughts would be greatly appreciated.

Thanks in advance!

答:
Iterating we have:

$$x_{n} = C^{2}x_{n-2}(1-x_{n-2})(1-Cx_{n-2}(1-x_{n-2}))$$

$$x_{n} = C^{2}x_{n-2}(1-x_{n-2})(1-Cx_{n-2} + Cx^{2}_{n-2})$$

Solving for equilibria we have a zero equilibrium and the roots of the polynomial:

$$C^{2} - 1 - C^{2}x - C^{3}x + C^{3}x^{2} + C^{3}x^{2} - C^{3}x^{3} $$

You can factor the known positive equilibrium $\frac{C-1}{C}$ using polynomial long division and then use the quadratic formula to get the periodic solution of period 2.

$$P_{1} = \frac{C + 1 - \sqrt{C^{2} - 2C - 3}}{2C}$$

$$P_{2} = \frac{C + 1 + \sqrt{C^{2} - 2C - 3}}{2C}$$

Which will exist whenever $C^{2} - 2C - 3 > 0$. Assuming $C > 0$ this will occur whenever $C > 3$.


问:
Solving inequality by induction associated with a recurrence
Consider the sequence an defined by $a_1=1$, $a_2=3$ and

$$a_n=a_{n-1}+a_{n-2} \text{ for } n>2.$$

Prove that $$a_n<
\left(\frac{7}{4}\right)^n\text{ for every positive integer }n.$$

My attempt: By induction, we assume that $a_k<(\frac{7}{4})^k$

Now, we need to prove to $n=k+1$. I did: $a_k<(\frac{7}{4})^k$, and $\frac{7}{4}.a_k<(\frac{7}{4})^{k+1}$.

I assumed that $a_{k+1}\le a_k\cdot\frac{7}{4}$

I replaced the minor cases and it always worked, but I am not able to prove this inequality by induction

答:
Because the recurrence depends on $2$ previous steps, you need to assume $2$ steps in the inductive hypothesis.

Inductive hypothesis: $a_k \le (7/4)^k$ and $a_{k+1} \le (7/4)^{k+1}.$

Inductive step: you need to prove that $a_{k+2}\le(7/4)^{k+2}$. Can you do this step using the recurrence $a_{k+2}=a_{k+1}+a_{k}$?

 $$a_{k+2}=a_{k+1}+a_{k}\le(7/4)^{k+1}+(7/4)^k=(7/4)^k(1+7/4)=(7/4)^k(44/16)<(7/4)^k(49/16)=(7/4)^{k+2}$$

Note: you’ll also need to prove $2$ base cases.

 $a_1 = 1 < 7/4, \quad a_2 = 3 = 48/16 < 49/16$

If you’re interested, these are almost the Lucas numbers and the proof works essentially because $7/4 > \phi$ the golden ratio.


问:
Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$
Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.

and please don't use any $\sum_{cyc}$ because I get confused with it.

答:
We can rewrite the LHS :

$$LHS = \frac1{1+ \frac{bc}{a^2}} + \frac1{1+\frac{ac}{b^2}} + \frac1{1+\frac{ab}{c^2}}$$

Let $u = \frac{b}{a}$, $v = \frac{a}{c}$, $w = \frac{c}{b}$. Then :

$$LHS = \frac{v}{u+v} + \frac{u}{w+u} + \frac{w}{w+v}$$

We suppose that $u \geq v$ and $u \geq w$ (other cases are similar).

If $w > v$ :
$$LHS < \frac{v}{u+v} + \frac{u}{v+u} + \frac{w}{w} = 2$$

If $w \leq v$ :

$$LHS < \frac{v}{v+v} + \frac{u}{u} + \frac{w}{w+w} = 2$$

Taking $u = n^2$, $w = n$, $v = 1$ :

$$LHS = \frac1{1+n}+\frac{n^2}{n+n^2}+\frac{n}{n+1} \to 2 \quad [n \to \infty]$$

Thus $2$ can't be improved.


答:
You're on the right track, the way to go really is to show that
$$\lim_{h\to0}\left|\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|}\right|=0.$$
To this end, you just have to use the triangle inequality and the fact that $\vert h_i\vert\leq\Vert h\Vert$ for components $h_i$ of $h$. This will give you
$$\left|\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|}\right|
\leq\frac{1}{\Vert h\Vert}\sum_{j=2}^n\binom{n}{j}\vert x_i\vert^{n-j}\vert h_i\vert^j
\leq\frac{1}{\Vert h\Vert}\sum_{j=2}^n\binom{n}{j}\vert x_i\vert^{n-j}\Vert h\Vert^j,$$

where every term in the sum in the last upper bound is at least quadratic in $\Vert h\Vert$, so after cancelling $\frac{1}{\Vert h\Vert}$, every term is at least linear in $\Vert h\Vert$. You can use this to prove that the limit is in fact $0$.

Title: How to prove that $f(x) = x^2_i$ is differentiable?

Body:
Let $f:\mathbb{R}^m \longrightarrow\mathbb{R}$ denoted by $f(x) = x^n_i$, for $i \in \{1,2,\dots,m\}$ and $n \in \mathbb{N}$, both fixed. So, let $x, h \in \mathbb{R}^m$, where $h \not = 0$. Then we have
\begin{align}
f(x+h) - f(x) &= (x_i - h_i)^n - x^n_i\\
&= \sum_{j = 0}^n{\binom{n}{j}x^{n-j}_ih^j_i} - x^n_i\\
&= \sum_{j = 1}^n{\binom{n}{j}x^{n-j}_ih^j_i}\\
&= nx^{n-1}h_i + \sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}.
\end{align}
So, basically, we have to prove that
$$\lim_{h \to 0}\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|} = 0.$$
I tried to manipulate the following expression
$$\left|\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|}\right|$$
but no success. Is there any hint or strategy to help me?


问:
How to find the basis of an intersection of vector spaces, along with vector spaces relative to one another.

The Question

I would appreciate a  nudge in the right direction. I have two vector spaces $V_1$ and $V_2$, given spanning sets for both respectively. I am trying to find a basis for $V_1 \cap V_2$ and a basis for $V_1$ relative to $V_1\cap V_2$.

Consider a vector space $V=K^4$ and its subspaces

$$V_1 = span \left\{ \left(
  \begin{matrix}
    3 \\ 3 \\ 2 \\ 2
  \end{matrix}
  \right), \left(
   \begin{matrix}
    2 \\ 1 \\ 2 \\ 2
  \end{matrix}
  \right),\left(
   \begin{matrix}
    0 \\ 1 \\ 2 \\ 0
  \end{matrix}
  \right),\left(
   \begin{matrix}
    2 \\ 3 \\ 2 \\ 1
  \end{matrix}
  \right) \right\},V_2 = span \left\{ \left(
  \begin{matrix}
    2 \\ 3 \\ 1 \\ 1
  \end{matrix}
  \right), \left(
   \begin{matrix}
    2 \\ 1 \\ 1 \\ 2
  \end{matrix}
  \right),\left(
   \begin{matrix}
    2 \\ 2 \\ 0 \\ 1
  \end{matrix}
  \right)\right\}$$
Find the bases of $V_1$ and $V_2, V_1 \cap V_2$ and $V_1$ relative to $V_1 \cap V_2$

My Understanding

I calculated the basis of the intersection to be the column vectors $(0,-2,0,1)^T$ and $(2,2,0,1)^T$, I did this by constructing the matrix $(Base(V_1)|-Base(V_2))$ and finding a basis for the kernel, of the form =( ). I then proceeded to calculate the column vectors by taking $Base(V_1)\cdot$ for each basis vector of the kernel. However, I can not see how the column vectors $(0,-2,0,1)^T$,$(2,2,0,1)^T$ can form a basis of $V_1$ as they do not have a third entry.

答: I'm not sure about a switch of basis because the basis $\{(0,-2,0,1)^T,(2,2,0,1)^T\}$ of $V_1 \cap V_2$ doesn't have three vectors in it, like the basis $\{(3,3,2,2)^T,(2,1,2,2)^T,(0,1,2,0)^T\}$ of $V_1$ does. But you could always find a third linearly independent vector $\vec{v_3}$ in $V_1$ such that $\{(0,-2,0,1)^T,(2,2,0,1)^T,\vec{v_3}\}$ is a basis for $V_1$, then you could write every vector in $V_1$ as a linear combination of the basis vectors of $V_1 \cap V_2$ and $\vec{v_3}$.


答:
To make your life easier, you can do the following:

In order to find the $2$-cycle, you can directly consider the following equation in $x$ [Note the denominator; you don't want the $1$-cycles]

$$\frac{f^2(x) - x}{f(x) - x} = C^2x^2 - C(C+1)x + (C+1) = 0$$

The roots will be real and unequal if the discriminant $$C^2(C+1)^2 - 4C^2(C + 1) > 0 \implies C^2(C+1)(C-3) > 0 \implies C > 3$$ since $C > 0$

Note

The intuition is as follows: if you calculate the second-order fixed points directly, then the first-order fixed points are also calculated. You will get a $4$th degree equation in $x$, ignore the two linear factors corresponding to the two first-order fixed points and consider the quadratic (i.e., the discriminant of the quadratic) in $x$. Here, we avoid these details and directly consider the quadratic. For more details, please see equations $15$-$23$ here

Title: finding $2$-cycle in logistic difference equation

Body:
The logistic difference equation is given by $x_{n+1}=Cx_n(1-x_n)$. I'm supposed to show that when $C>3$, the logistic difference equation has a $2$-cycle.

Letting $f(x)=Cx(1-x)$, if $\{x_0,x_1\}$ is our $2$-cycle we have that $x_{1}=f(x_0)$. Thus, if this is a $2$-cycle, we want to find values of $C$ so that $f^2(x_0)=x_0.$

So what we want is some value of $C$ so that $$f(f(x_0))=Cf(x_0)(1-f(x_0))=C(Cx_0(1-x_0))(1-(Cx_0(1-x_0))) = x_0.$$

My thinking is that, after some manipulation, I should be able to show that this equality can only hold when $C>3$, but actually performing the manipulation which shows that is proving to be a bit of a challenge.

Any thoughts would be greatly appreciated.

Thanks in advance!


问:
The solution to the differential equation doesn't work? Help?
I have the differential equation:

$$\frac{dN_b}{dt}=\lambda_aNa(0)exp(-\lambda_at)-\lambda_bN_b(1)$$

I am given that the solution to this is:

$$N_b(t)=\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[exp(-\lambda_at)-exp(-\lambda_bt)](2)$$

I am told that differentiating (2) against time, and then substituting into (1) will return the same answer. However, no matter what I do I cannot figure out how this works. I do not understand what I am doing wrong. I know the derivative of (2) is:

$$\frac{dN_b}{dt}=\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[-\lambda_aexp(-\lambda_at)+\lambda_bexp(-\lambda_bt)](3)$$

Substituting (3) into (1) and (2) into (1) provides:

$$\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[-\lambda_aexp(-\lambda_at)+\lambda_bexp(-\lambda_bt)]=\lambda_aNa(0)exp(-\lambda_at)-\lambda_b\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[exp(-\lambda_at)-exp(-\lambda_bt)]$$

No matter how much I simplify, I never get out $$0=0$$

This is a well-known and well-proven differential equation relating to radioactive decay. Equation 2 has been shown to be a solution to Equation 1, but my professor wants us to solve it through differential form instead of solving the ordinary differential equation. Can someone please help me? Thank you so much.

答:
Starting point:
$$ \dot{N}_b = \lambda_a N_{a,0} e^{-\lambda_a t} - \lambda_b N_b $$
$$ N_b = \frac{\lambda_a N_{a,0}}{\lambda_b-\lambda_a}\Big( e^{- \lambda_a t} - e^{-\lambda_b t}\Big) $$
Take a derivative of the second equation:
$$\dot{N}_b = \frac{\lambda_a N_{a,0}}{\lambda_b-\lambda_a}\Big(-\lambda_a e^{-\lambda_a t} + \lambda_b e^{-\lambda_b t}\Big).$$
Plug the latter two equations into the first equation:
$$ \frac{\lambda_a N_{a,0}}{\lambda_b-\lambda_a} \Big( -\lambda_a e^{-\lambda_a t} + \lambda_b e^{-\lambda_b t}\Big) = \lambda_a N_{a,0}e^{-\lambda_a t} - \frac{\lambda_a \lambda_b N_{a,0}}{\lambda_b-\lambda_a}\Big( e^{- \lambda_a t} - e^{-\lambda_b t}\Big) .$$
Cancel common factors:
$$ \frac{1}{\lambda_b-\lambda_a} \Big( -\lambda_a e^{-\lambda_a t} + \lambda_b e^{-\lambda_b t}\Big) = e^{-\lambda_a t} - \frac{\lambda_b}{\lambda_b-\lambda_a}\Big( e^{- \lambda_a t} - e^{-\lambda_b t}\Big) .$$
Collect terms involving linearly independent factors
$$ \Big(\frac{-\lambda_a}{\lambda_b-\lambda_a} -1 +  \frac{\lambda_b}{\lambda_b-\lambda_a}\Big)e^{-\lambda_a t} = \Big(\frac{\lambda_b}{\lambda_b-\lambda_a}-\frac{\lambda_b}{\lambda_b-\lambda_a}\Big)e^{-\lambda_b t}$$
Simplify:
$$ 0 = 0$$


答:
$S_2$ is the set of all natural numbers $x$ where $-x$ is also an natural number.

There are no such natural numbers at all[*].  So the set $S_2$ does not have any elements at all.

So $S_2 = \emptyset$.  $\emptyset$ is the set that has no elements at all.

It's not nonsensical to consider natural numbers have negatives that are natural numbers, any more than it is nonsensical to consider a rational square root of $2$.  The fact that there aren't any doesn't make it nonsense.  It just makes them non-existent.

How can $|x| = -x$?

Why shouldn't it be.

$|x|$ is always $0$ or positive.  So we just have to find an $x$ where $-x$ is $0$ or positive.

If $x = 0$ then $-x = -0 = 0$.

And if $x < 0$ then $-x > 0$.

SO what is $|0|$?  It is $0$.  ANd $|0| = 0 = -0$.  So that's one example.

And if $x < 0$ then $|x|=....$ well it equals the number that has the same magnitude as $x$ but in the positive rather than negative direction. In other words if $x < 0$ then $|x| = -x$.

In fact.... how was $|x|$ ever defined.  Most books define it as $|x| =\begin{cases}x& x\ge 0\\-x&  x< 0\end{cases}$.

So $\{x\in \mathbb Z: |x|= -x\} =\{0,-1,-2,-3,-4,......\}$.

But even if $|x| = -x$ was impossible that wouldn't be a problem.

We can have sets with no elements.  It's the empty set.

$\{$ even primes that are larger than $2\} =\emptyset$

$\{$ composite numbers that are prime $\} = \emptyset$

$\{$ square circles$\} = \emptyset$.

Nothing wrong with that.

Title: Question about a definition $S_2 = \{x \in N : -x\in N\}$ from the book Mathematical Proofs

Body:
I'm wondering about the following from the book Mathematical Proofs

$$S_2 = \{x \in N : -x\in N\} = \emptyset$$

I'm not really sure what this is even saying I'm reading it as follows the set S2 has elements {X that are members of the natural numbers such that negative x are in the natural numbers}.
This statement seems nonsensical to me. What is this actually saying?

Another example from the book that seems absurd to me is the following.

$$\{x \in Z: |x| = -x\}$$

How can this ever be the case, that $|x| = -x$.

Thank you in advance for any help,


问:
Dot product terminology
How would you grammatically correctly describe the computation of the dot product of two n-dimensional vectors.

Is this correct:

We multiply the corresponding elements of the two vectors. Then we sum the products together.

答: Corresponding to what? To be more precise, I would say multiply the components that have the same index and them take the sum.</p>

Title: Dot product terminology


问:
Does diffeomorphism change sectional curvature too much?
Pertaining to sectional curvature:

$f: M \to M$ is a $C^1$-diffeomorphic map on Riemannian manifold $M$, $K_p(u,v)$ is the sectional curvvature at $p \in M$ and $u,v \in T_pM$ independent. Want to see if the following hold:

$$K_{f(p)}(Df(u), Df(v)) \le C_p \cdot K_p(u,v),$$

where the constant in $ C_p$ only depends on $p$?

答:
No: $K_p(u,v)$ could be $0$ while $K_{f(p)}(f_*u,f_*v)$ is positive (for instance, $M$ could be a surface that is flat in some parts and has positive curvature in other parts).  More generally, there is nothing you can say relating $K_p(u,v)$ and $K_{f(p)}(f_*u,f_*v)$ that you cannot say relating sectional curvature functions at any points of any manifolds; that is, $f$ is totally irrelevant to your question.  To be precise, given two $n$-manifolds $M_0$ and $M_1$, points $p\in M_0$ and $q\in M_1$, and a linear isomorphism $g:T_pM_0\to T_qM_1$, you can take small neighborhoods $U$ of $p$ and $V$ of $q$ and a diffeomorphism $f:U\cap V\to U\cap V$ which swaps $U$ and $V$, maps $p$ to $q$, and maps $T_pM_0\to T_qM_1$ via $g$.  You then have $K_p(u,v)=K_q(g(u),g(v))$ for all $u$ and $v$.

(On the other hand, if you assume that $K_p$ is never $0$, then trivially such a $C_p$ exists since $K_{f(p)}$ is bounded and $K_p$ is bounded away from $0$ and always has the same sign, since for instance it suffices to consider the case where $u$ and $v$ are orthonormal and the space of unordered pairs of orthonormal tangent vectors $u$ and $v$ is compact and connected.  Or, to use a bit more machinery, you can think of $K_p$ and $K_{f(p)}$ as continuous functions on a Grassmannian of $2$-planes, which is compact and connected.)


问:
Finding $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles
I had an exercise in my text book:

Find $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles.

The answer that they gave was as follows:

$$\lim_{h\to 0} \frac{6-6}{h} = 0$$

However surely that answer would be undefined if we let $h$ tend towards $0$, as for finding all other limits like that we substitute $0$ in the place of $h$?

答:
When evaluating the limit:
$$\lim_{h\to 0} \frac{6-6}{h}$$
notice that (provided that $h \ne 0$), we have:
$$\frac{6-6}{h}=\frac{0}{h}=0$$
Thus, since it is possible to evaluate $\lim_{x\to a}f(x)$ even if $f(a)$ is undefined, we have:
$$\lim_{h\to 0} \frac{6-6}{h} = \lim_{h\to 0} 0 = 0$$</p>


问:
Basis of a vector space and subspace
It is given that basis B = { v1 ,...,vn} for a vector space V and U a subspace of V, then there C be a subset of B that is a basis for U.
Would this be true? Struggling to wrap my mind around this. A proof/counterexample would help immensely.

答:
No. Given a basis B $= \{ v_1 ,...,v_n\}$ of a vector space V, and a subspace U of V, then there is not necessarily a subset C of B such that C is a basis for U.

leslie townes gives a standard example in the comments: There is no subset of B = $\{(0,1),(1,0)\}$ which is a basis of U = $\{(x,x): x \in \mathbb{R}^2 \},\ $ yet U is a subspace of V = span(B) $= \mathbb{R^2}.$

What is true that for any subspace U of V,  there exists a basis B of the vector space V such that there is a subset of B which is a basis of U. Notice in this statement, you are not given the basis (of V), you can choose the basis - and you have a wide choice - to be one that conveniently has a subset that is a basis of U.

Title: Basis of a vector space and subspace


问:
Two aeroplanes $1$ and $2$ bomb a target in succession. Probability of $1$ and $2$ hitting correctly are $0.3$ and $0.2$.

Two aeroplanes $1$ and $2$ bomb a target in succession. Probability of $1$ and $2$ hitting correctly are $0.3$ and $0.2$. The second plane will bomb only if first misses the target. Find the probability that target is bombed by the second plane

I want to know why the complementary method cannot be used here i.e. $$P(B) = 1-0.3=0.7$$

Which gives the probability that the target is never hit by the first plane.

Answer is 0.318

Note: I know the correct way to do this, I just want to know why this is wrong.

答: There's a second way to reach the correct answer that avoids needing to sum a series.  The key insight is to note that if both planes miss on their first pass (which happens with probability $0.56$), you're right back where you started.  Thus, $P(A)=0.3+0.56 P(A)$, so $P(A)=\frac{0.3}{0.44}$ and $P(B)=1-P(A)=\frac{0.14}{0.44}=\frac{7}{22}$.


问:
Finding the derivative of this function with natural log and sums
Find the derivative of this function with respect to alpha

$n \ln(\alpha + 1) + \alpha \sum \ln x$

The natural log and sum is confusing me.

答:
By the chain rule, the derivative of $n\log (a+1)=\frac n{a+1}*\frac{d}{da}(a+1)=\frac{n}{a+1}*1$

$$\frac{n}{a+1}*1+\sum\log x=0\\
\frac{n}{a+1}=-\sum\log x\\
-\frac{n}{\sum \log x}=a+1\\
a_{MLE}=-\left(1+\frac{n}{\sum \log x}\right)$$


Title: Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$

Body:
Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.

and please don't use any $\sum_{cyc}$ because I get confused with it.

Body:
Suppose $a \geqslant b \geqslant c,$ we have
$$\frac{a^2}{a^2+bc} < 1,$$
and
$$\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} \leqslant \frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+b^2} = 1.$$
Therefore
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} < 2 .$$


问:
E[E[A|B]|C], law of total expectation with two conditional expectations
I have a conditional expectation inside of a conditional expectation.

Let A, B, C be real-valued random variables.

I am trying to simplify E[E[A|B]|C]. I am not sure much can be done in general; however, in this case, I know that C$\in$B.

I also know from TLOTE: E[E[A|B]]= E[A].

Intuition tells me that since C is a subspace of B, then I should wind up with E[A|C], but I am short on ideas to show it mathematically.

Thank you.

答:
One may show that if $\mathbb{E}[\lvert X \rvert] < \infty$, then for $\mathcal{G}_1\subset\mathcal{G}_2\subset\mathcal{F}$ it holds that
$$
\mathbb{E}\!\left[\left. \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right] \, \right| \mathcal{G}_1\right] = \mathbb{E}\!\left[X \, | \, \mathcal{G}_1\right].
$$
To proof this, note e.g. that for $G \in \mathcal{G}_1$,
$$
\int_G \mathbb{E}\!\left[\left. \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right] \, \right| \mathcal{G}_1\right] \mathrm{d}\mathbb{P}
=
\int_G \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right]\mathrm{d}\mathbb{P}
=
\int_G X \, \mathrm{d}\mathbb{P},
$$
since $\mathbb{E}\!\left[\left. \mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right] \, \right| \mathcal{G}_1\right]$ is the conditional expectation of $\mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right]$ given $\mathcal{G}_1$, since $\mathbb{E}\!\left[X \, | \, \mathcal{G}_2\right]$ is the conditional expectation of $X$ given $\mathcal{G}_2$, and since $\mathcal{G}_1 \subset \mathcal{G}_2$ implies $G \in \mathcal{G}_2$.


问:
Simplifying the sequence $(n-1)x_1+\cdots+(n-1)x_{n-1}$

Background

I am currently working through Concrete Mathematics by Donald Knuth (1988) and in Chapter 1, Exercise 9a, there is a part I do not understand:

$$(n-1)x_1+\cdots+(n-1)x_{n-1}$$

seems to simplify to just $n$ according to one of the answers for this exercise. How? The actual exercise is:

$$P(n): x_1\dots x_n\le \bigg(\frac{n(x_1+\cdots+x_{n-1})}{n}\bigg)^n,\quad \text{if}\ x_1,\dots,x_n \ge 0.$$

By setting $x_n = \frac{(x_1 +\dots + x_{n-1})}{(n − 1)}$ , prove that $P(n)$ implies $P(n − 1)$ whenever $n \gt 1.$

My Attempt

\begin{align}
&\bigg(\frac{(n-1)x_1+\cdots+(n-1)x_{n-1}+x_1+\cdots+x_{n-1}}{n(n-1}\bigg)^n \\
&=\bigg(\frac{n(x_1+\cdots+x_{n-1})}{n(n-1)}\bigg)^n\\
&=\bigg(\frac{x_1+\cdots+x_{n-1}}{n-1}\bigg)^n\\
\end{align}
Am I on the right track?

答:
You should follow the steps more carefully. The step you’re confused about is elaborated upon below:

$$\begin{align}
&(n-1)x_1 + \ldots + (n-1)x_{n-1} + x_1+\ldots + x_{n-1} \\
&=(n-1)(x_1+\ldots+x_{n-1})+1(x_1+\ldots + x_{n-1})\\
&= ((n-1)+1) (x_1+\ldots + x_{n-1})\\
&=n(x_1+\ldots + x_{n-1})
\end{align}$$

Exercises on  factorisation, simplification,  algebraic manipulation  can  help you see these things in one step.

And no, $(n-1)x_1 + \ldots (n-1)x_{n-1}$ doesn’t simplify to $n$. Rules of arithmetic don’t break simply because of presence of “$\ldots$”.


问:
Symbol for "if and only if": $\implies$ or $\iff$?
I was wondering about the iff sign in maths.
I've never learned about it in school & see it a lot online. Usually the sign looks like this: $\implies$, but in math.stackexchange I always see this: $\iff$.

Does that sign means iff?

Am I using iff right here:

$$8x + 2x - 44 = 220 + 4 - x \implies 11x = 268 \implies x = 24.36$$

答:
As others have said, $A$ if and only if $B$ means that $A$ implies $B$ and $B$ implies $A$.  That is, if $A$ is true then $B$ is true, and if $B$ is true then $A$ is true.

To make sense of why we say 'if and only if': If we have that $A$ if and only if $B$, then we have $A$ IF $B$ since $B$ implies $A$, i.e. IF $B$ is true, then $A$ is true.

We also have $A$ ONLY IF $B$, since we have $A \implies B$, and using basic logic (not sure if you know this - you would see it in an introductory abstract math course) we see that the negation of this is $\neg B \implies \neg A$, or in other words if $B$ is NOT true, then $A$ is not true.  Therefore, $A$ is only true if $B$ is true, or $A$ only if $B$.

So combining $B \implies A$ and $A \implies B$ (logically equivalent to $\neg B \implies \neg A$), we arrive at $A \iff B$, i.e. $A$ is true if and only if $B$ is true.


问:
How to prove that $f(x) = x^2_i$ is differentiable?
Let $f:\mathbb{R}^m \longrightarrow\mathbb{R}$ denoted by $f(x) = x^n_i$, for $i \in \{1,2,\dots,m\}$ and $n \in \mathbb{N}$, both fixed. So, let $x, h \in \mathbb{R}^m$, where $h \not = 0$. Then we have
\begin{align}
f(x+h) - f(x) &= (x_i - h_i)^n - x^n_i\\
&= \sum_{j = 0}^n{\binom{n}{j}x^{n-j}_ih^j_i} - x^n_i\\
&= \sum_{j = 1}^n{\binom{n}{j}x^{n-j}_ih^j_i}\\
&= nx^{n-1}h_i + \sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}.
\end{align}
So, basically, we have to prove that
$$\lim_{h \to 0}\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|} = 0.$$
I tried to manipulate the following expression
$$\left|\frac{\sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}}{\|h\|}\right|$$
but no success. Is there any hint or strategy to help me?

答:
You have shown that $$f(x+h)-f(x)=nx^{n-1}h_i + \sum_{j = 2}^n{\binom{n}{j}x^{n-j}_ih^j_i}=nx^{n-1} h_i + \frac{n(n-1)}{2} x^{n-2} h_i^2 + \text{higher order terms}.$$

So, now take $T=\pmatrix{0 & 0 & \cdots & \underbrace{n x^{n-1}}_{i^{\text{th}} term} & \cdots & 0}$ and $r(h)=  \frac{n(n-1)}{2} x^{n-2}  h_i^2+\text{h.o.t.}$

In the comment to another answer we showed that $\frac{r(h)}{||h||}\to 0.$


问:
integration table not working out for IBP
Given $F'(x)=f(x)$ and $G'(x)=F(x)$, evaluate $\displaystyle \int_a^b xf(x)\ dx$, using the information on the following table.

\begin{array}{|c|c|c|c|}
    \hline
     x & f(x) & F(x) & G(x) \\
     \hline
     a & 2 & 4 & 10 \\
     \hline
     b & -1 & 5 & -11 \\
     \hline
\end{array}

Here is where I went with it.

\begin{align*}
                            \int_a^b & xf(x)\ dx\\
                            u&=x\\
                            du&=x \,dx\\
                            dv&=f(x)\, dx\\
                            v&=F(x)\\
                            &=xF(x)\bigg\rvert_a^b - \int_a^b xF(x) \,dx\\
                            w&=x\\
                            dw&=x \,dx\\
                            dp&=F(x)\, dx\\
                            p&=G(x)\\
                            \int_a^b xf(x)\, dx &=xF(x)\bigg\rvert_a^b - \left(xG(x)\bigg\rvert_a^b-\int_a^b xG(x) \,dx\right)\\
                            \int_2^{-1} xf(x)\, dx &=xF(x)\bigg\rvert_4^5 - \left(xG(x)\bigg\rvert_{10}^{-11}-\int_{10}^{-11} xG(x) \,dx\right)\\
                        \end{align*}

But the equality is not working out at all. Did I do the IBP incorrectly? Currently looking to see if I can do anything with it before plugging the bounds in.

答:
If you let $u=x$, then $du=1\,dx$, not $du=x\,dx$.

Hint.

$$
\int xF'(x)\,dx=xF(x)-\int F(x)\,dx=xF(x)-G(x)+C
$$

In general,
$$
\int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx
$$


问:
Question about the notation $\{x \in \mathbb{Z}: p(x)\}$
A book I'm currently working through has the following exercise.

$1.5$ Write each of the following sets in the form $\{x \in \mathbb{Z} : p(x)\}$, where $p(x)$ is a property concerning $x$.

$$(a)\ A = \{-1, -2, -3,...\}\\
(b)\ B = \{-3, -2, ..., 3\}\\
(c)\ C = \{-2, -1, 1, 2\}
$$

I'm not exactly sure what the $p(x)$ is asserting here, do they want me to come up with a transformation for each $x$ by means of some function?

The only way I can even conceive of answering these would be something like the following

$(a)\ A = \{x\in \mathbb{Z}: x< 0\}$
But this isn't some transformation $p(x)$. I don't know a function that is not piecewise who will take some $x$ and transform it to $-x$ but leaves $x$ positive if it's positive?

$(b)\ B = \{x\in \mathbb{Z} : |x|\leq3\}$

$(c)\ C = \{x\in \mathbb{Z} : 0 \lt|x|\leq2\}$

Again neither of my solutions $(c)$ or $(b)$ are of the form $p(x)$ I'm just giving some hard rules on what x has to be here. If anyone has any ideas on how I can approach these differently I would appreciate it.
Regrettably, my book comes with exactly zero solutions.

答:
Your work looks fine; you can interpret $p(x)$ as a statement imposing a restriction on $x$. It is not a transformation.

It's fine even to use words:
$$A = \{x \in \Bbb Z : x < 0\}\\
A' = \{x \in \Bbb Z : x \textrm{ is negative}\}$$

The sets $A$ and $A'$ are exactly the same; it's just that one is more compact to write. (You won't often see "$x$ is negative" written out like that; a more common example would be something like "$x$ is a prime number".)


问:
Showing $a^2=b^2$ in a group with 8 elements isomorphic to Quaternion group
I am working in a group $G$ with 8 elements such that $a, b\in G$ and $ba=a^3b$. Moreover, I am given that both $b$ and $a$ have order 4 and $\langle a \rangle \neq \langle b \rangle$. I understand this group is isomorphic to Quaternion group, but this is not given and I am trying to prove this. I want to show $a^2=b^2$, but I am stuck.
Here is what I have done so far:

$$(ab)^2=abab=a(ba)b=aa^3bb=a^4b^2=b^2$$

Also from the equality $ba=a^3b$ I can derive many other equalities, but they do not seem to give me anything useful. For instance

$$aba=b \text{ or } a^3ba^3=b$$

Another observation I have made is that both $b^2$ and $a^2$ have order 2. But I am not sure if that is enough to conclude $a^2=b^2$. Please guide me in the right direction.

答:
Similar to @GerryMyerson's comment, use the fact that there are only two nonabelian groups of order $8$.

For instance, the quaternions only have one element of order $2$.  The dihedral group has $5$.


问:
Recurrence relation with Z-Transform
I'm revising the Z-Transform. I am looking at the book which gives an example of how to solve the recurrence relation

$$x_{k+2} - 3x_{k+1} +2x_k = 1$$

where $x_0 = 0$ and $x_1 = 1$. The book uses the left-shift theorem which states that if $Z[x_k] = F(z)$ then $Z[x_{k+m}] = z^m F(z) - (z^m x_0 + z^{m-1} x_1 + z^{m-2} x_2 + \dots + z x_{m-1})$. Using this, we can find the solution.

However, I remember 10 years ago when I did a lot of signal processing, I used to invert the recurrence relation so that I can use the right-shift theorem which is simpler. So I would convert the above recurrence relation to

$$x_k - 3 x_{k-1} + 2 x_{k-2} = 1$$

and then use the right-shift theorem which states: if $Z[x_k] = F(z)$ then $Z[x_{k-m}] = z^{-m}F(z)$. Using this technique, I get a different result to the book. I must have made a mistake because I also don't get the opportunity to input the initial conditions using the right shift theorem. What have I done wrong? I don't need the recurrence relationship solved - the book solves it. I just don't know if what I remember is true or if I'm mistaken.

答:
In this case the result of the question in using the right-shift form is correct in concept. In this case the initial terms are not required.

The process becomes more clear when solving this recurrence relation:
$$ x_{k} - 3 \, x_{k-1} + 2 \, x_{k-2} = 1 $$
by the following method.
\begin{align}
x_{k} - 3 \, x_{k-1} + 2 \, x_{k-2} &= 1 \\
F - \frac{3}{t} \, F + \frac{2}{t^2} \, F &= \frac{t}{t-1} \\
(2 - 3 t + t^2) \, F &= \frac{t^3}{t-1} \\
F &= \frac{t^3}{(t-1)^2 \, (t-2)} \\
&= \frac{1}{\left(1 = \frac{1}{t}\right)^2 \, \left(1 - \frac{2}{t}\right)} \\
&= - \frac{1}{(\left(1 - \frac{1}{t}\right)^2} - \frac{2}{1- \frac{1}{t}} + \frac{4}{1- \frac{2}{t}} \\
&= \sum_{n=0}^{\infty} ( 2^{n+2} - n - 3 ) \, t^{-n},
\end{align}
where
$$ F(t) = \sum_{n=0}^{\infty} x_{n} \, t^{-n}. $$

The solution of the difference equation is then
$$ x_{k} = 2^{k+2} - k - 3. $$
This result may be a shifted form of result.

When considering the original form
$$ y_{n+2} - 3 \, y_{n+1} + 2 \, y_{n} = 1$$
the general solution is
$$ y_{n} = c_{0} \, 2^n - n - 1 + c_{1} $$
which maintains the same form as the result found for $x_{k}$.

Title: Recurrence relation with Z-Transform


问:
Category theory for sensorimotor learning?
Robotics, machine learning, inference, control, decision theory, system identification. There are many different views on how the information would flow from the environment into a robot, and the other way around. One example, Ralf Der is working on "homeokinesis". A robot is driven towards these regions in state space that cause large perceptual variations. To be able to do so, it necessarily must understand enough of the environment and hence won't just exhibit chaotic behavior.

Is there a branch of category theory that considers the fundamentals of a "self-learning" system? A description of a system which performs its own "system identification". It would be great if there is a formal approach to it, because many of the scientist working in this field (Oudeyer, Pfeiffer, Wolpert, Metta, Friston to name just a few) tend to explain their concepts in a narrative manner. It would be great to have a thorough conceptual review using monads, etc.

Challenges (from my laymen's perspective in increasing order of difficulty for a categorical approach):

Describe Pearl's calculus of interventions.

Describe principle of optimality using category theory (http://en.wikipedia.org/wiki/Principle_of_optimality) or the related notion of optimal substructure (http://en.wikipedia.org/wiki/Optimal_substructure).

Describe the curse of dimensionality (http://en.wikipedia.org/wiki/Curse_of_dimensionality).

Describe "embodied cognition". For example "Extending Dynamical Systems Theory to Model Embodied Cognition" (Hotton, Yoshima) is a tad more formal, but reaches in no way the rigorous accuracy of the definition of concepts as in category theory.

Pointers to people who only tried(!) to apply category theory to these challenging problems will be also appreciated.

答:
Regarding "Describe Pearl's calculus of interventions" and other related ideas, you might find this thesis interesting:

B. Fong, Causal Theories : A Categorical Perspective on Bayesian Networks, 2013.

Depending upon the level of speculation you are interested in considering, you might also like to look at this book:

A. C. Ehresmann and J. P. Vanbremeersch, Memory Evolutive Systems; Hierarchy, Emergence, Cognition, Volume 4 (Studies in Multidisciplinarity). Elsevier Science, 2007, p. 402.


问:
How to check the matrix is Hermitian?
I have the following matrix

$$A = \pmatrix{1 & 0 & 1\cr 0 & 2 & 0\cr 1 & 0 & 1\cr}$$

and

$$adj(\bf{A})=\pmatrix{2 & 0 & -2\cr 0 & 0 & 0\cr -2 & 0 & 2\cr}$$

MATLAB says the matrix $A$ is Hermitian. However the conjugate transpose is not equal the matrix $A$.
How to check the matrix is Hermitian ?

答:
If you mean $$A = \pmatrix{1 & 0 & 1\cr 0 & 2 & 0\cr 1 & 0 & 1\cr}$$
that is Hermitian: its conjugate transpose (which in this case is the transpose, because the matrix is real) is itself.
As for $\pmatrix{2 & 0 & -2\cr 0 & 0 & 0\cr -2 & 0 & 2\cr}$, that is the "classical adjoint" or adjugate.  It has nothing to do with the Hermitian adjoint or conjugate transpose.  The terminology is unfortunately confusing.

The Matlab command

```<code>adjoint(A)
</code>```

gives you the classical adjoint of $A$, while

```<code>A'
</code>```

gives you the Hermitian adjoint.

Title: How to check the matrix is Hermitian?

Body:
If you mean $$A = \pmatrix{1 & 0 & 1\cr 0 & 2 & 0\cr 1 & 0 & 1\cr}$$
that is Hermitian: its conjugate transpose (which in this case is the transpose, because the matrix is real) is itself.
As for $\pmatrix{2 & 0 & -2\cr 0 & 0 & 0\cr -2 & 0 & 2\cr}$, that is the "classical adjoint" or adjugate.  It has nothing to do with the Hermitian adjoint or conjugate transpose.  The terminology is unfortunately confusing.

The Matlab command

adjoint(A)

gives you the classical adjoint of $A$, while

A'

gives you the Hermitian adjoint.


问:
Uniform distribution on the unit sphere rotated by a random orthogonal matrix
Question 1. Let $u\in \mathbb{R}^n $ be a random vector uniformly distributed on $\mathbb{S}^{n-1}$, and $T\in \mathbb{R}^{n\times n}$ be a random orthogonal matrix. If $u$ and $T$ are independent, is $Tu$ uniformly distributed on $S^{n-1}$ and statistically independent of $T$?

I get this question when thinking about the following one.

Question 2. Let $u, v\in\mathbb{R}^n$ be two independent random vectors, and $u$ be uniformly distributed on $\mathbb{S}^{n-1}$. Consider the inner product $u^\top v$. Take a constant vector $v_0\in \mathbb{S}^{n-1}$. It seems that

$u^\top v$ has the same distribution as $u^\top v_0$;

$u^\top v$ and $v$ are statistically  independent.

Is this true?

Towards a positive answer to Question 2, we let $T$ be the Housholder matrix such that $Tv = v_0$. Note that $T$ is independent of $u$. Then
$$
u^\top v = u^\top (T^\top v_0) = (Tu)^\top v_0.
$$
If the answer to Question 1 is yes, then $Tu$ and $u$ are identically distributed, and $Tu$ is independent of $v$, and hence the answer to Question 2 is yes. (BTW, are Question 1 and Question 2 equivalent?)

Any comments or criticism will be appreciated. Thank you.

(A different but related question: $X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, $Y$ is $\mathcal{G}$-measurable, then $f(X,Y)$ is independent of $\mathcal{G}$?)

答:
The answer to question #1 is yes.  Here's the proof.

Let $\mu$ be the uniform distribution on $S^{n-1}$, and let $\nu$ be the distribution for $T$ on $O(n)$.  Given any measurable set $A\subseteq S^{n-1}$, let
$$
A' = \{(u,T)\in S^{n-1}\times O(n) \mid T(u)\in A\}
$$
and let $I_{A'}$ be the indicator function for $A'$.  Then
$$
P(v\in A) = (\mu\times \nu)(A') = \int_{S^{n-1}\times O(n)} I_{A'} d(\mu\times \nu) = \int_{O(n)} \int_{S^{n-1}} I_{A'}(u,T)\,\mu(du)\,\nu(dT)
$$
But for any $T\in O(n)$, we have
$$
\int_{S^{n-1}} I_{A'}(u,T)\,\mu(du) = \mu(T^{-1}(A)) = \mu(A)
$$
where the last equality follows from the fact that $\mu$ is invariant under the action of $O(n)$.  We conclude that
$$
P(v\in A) = \int_{O(n)} \mu(A)\,\nu(dT) = \mu(A).
$$
This holds for every measurable set $A\subseteq S^{n-1}$, so the distribution for $v$ is also $\mu$.

For independence of $v$ and $T$, let $A$ be any measurable subset of $S^{n-1}$, let $A'$ be as above, let $B$ be any measurable subset of $O(n)$, and let $B'=S^{n-1}\times B$.  Then
\begin{align*}
P(v\in A\text{ and }T\in B) &= \int_{S^{n-1}\times O(n)} I_{A'} I_{B'}\, d(\mu\times \nu) \\[6pt]
&= \int_{O(n)}\int_{S^{n-1}} I_{A'}(u,T)\, I_{B'}(u,T) \,\mu(du)\,\nu(dT) \\[6pt]
&= \int_{O(n)}\int_{S^{n-1}} I_{A'}(u,T)\, I_{B}(T) \,\mu(du)\,\nu(dT) \\[6pt]
&= \int_{O(n)} I_B(T) \int_{S^{n-1}} I_{A'}(u,T) \,\mu(du)\,\nu(dT) \\[6pt]
&= \int_{O(n)} I_B(T) \,\mu(A)\,\nu(dT) \\[6pt]
&= \mu(A)\, \nu(B) \\[6pt]
&= P(v\in A)\,P(T\in B).
\end{align*}


问:
How can I compute a monad from this adjunction?
What are the steps to computing a Monad from an adjunction?

There has to be the following steps:

Compute the endofunctor on the base category given the adjunction

Compute the product natural transformation

Compute the unit natural transformation

I would like to see each of these steps and I am interested in the following adjunction.  There is a functor from the category, Gpd, of Groupoids and Cat, of Categories.  It has an adjoint called the core which goes in the opposite direction.  Can I see this computation done with this adjunction?

答:
So, this is not going to be a particularly enlightening example, but we can definitely walk through it together. In the interest of showing something a bit more complex as well, I've also included a worked version of another pair of adjoints.

So you're totally right. Every groupoid is a category, so there is an inclusion functor

$$\iota : \mathsf{Grpoid} \to \mathsf{Cat}$$

Moreover, by ignoring any non-isos, we can look at "the biggest groupoid living inside $\mathsf{Cat}$". This gives us a functor

$$\text{core} : \mathsf{Cat} \to \mathsf{Grpoid}$$

The first question to ask yourself after "is this an adjunction?" is "which one goes on which side?". Since a functor sends isos to isos, the image of any groupoid under a functor must be a groupoid itself. So a functor $\iota G \to C$ must factor through the core of $C$ (which is the largest possible groupoid in $C$). Succinctly:

$$\iota \dashv \text{core}$$

So now we want to look at the monad associated to this adjunction. I always remember this in terms of the free-forgetful adjunction of an algebraic structure. For groups, say, we have

$$(F : \mathsf{Set} \to \mathsf{Grp}) \dashv (U : \mathsf{Grp} \to \mathsf{Set})$$

This should give us a monad on $\mathsf{Set}$, so our monad must be $UF : \mathsf{Set} \to \mathsf{Set}$.

By analogy, our monad is

$$\text{core} \circ \iota : \mathsf{Grpoid} \to \mathsf{Grpoid}$$

And now we see why this is not a particularly enlightening example: If we look at the largest groupoid contained in $G$.... Well, $G$ is already a groupoid. So we're actually looking at the identity monad $I$!1

Of course, we can still work out the things that you've asked for.

The endofunctor $\text{core} \circ \iota$ sends $G \mapsto G$ and sends a groupoid hom to itself.

The product is a natural transformation from $\text{core} \circ \iota \circ \text{core} \circ \iota \Rightarrow \text{core} \circ \iota$. But this is a natural transformation $I^2 \Rightarrow I$ from the identity monad to itself. In particular, each component of this natural transformation is the identity.

The unit is also somewhat silly. We need a map from $I \Rightarrow \text{core} \circ \iota$. But this is just a map from $I \Rightarrow I$, and again taking the identity arrows on each component works.

As a slightly more instructive example, here is something that I thought about a little while ago:

Given a graph $\Gamma = (V,E)$, we can form the Right Angled Artin Group

$$A\Gamma = \langle V~|~[x,y] \iff xEy \rangle$$

This is the free group on the vertices, where we force two vertices to commute when they're connected by an edge.

Conversely, given a group $G$, we can form is Commutation Graph $CG$ whose vertices are group elements, and we connect two group elements exactly when they commute.

You can convince yourself that these form a pair of adjoint functors $A \dashv C$ between $\mathsf{ReflGph}$ and $\mathsf{Grp}$. Here $\mathsf{ReflGph}$ is the category of graphs (sets equipped with a reflexive, symmetric binary relation). We really need reflexivity here (do you see why?) but that's fine. Graph homomorphisms are the natural thing: Functions on the vertices which preserve (but don't necessary reflect) the edge relation.

Now I claim $A \dashv C$ is a pair of adjoint functors. So then the monad will be $CA : \mathsf{ReflGph} \to \mathsf{ReflGph}$. It sends a graph $\Gamma$ to a new graph $CA\Gamma$ with

a vertex for every element of $A\Gamma$. So we roughly have a vertex for each word in the vertices of $\Gamma$, but we have to declare certain words equivalent
(like $xy$ and $yx$ for adjacent $x,y \in \Gamma$).

an edge connecting any two commuting elements of $A \Gamma$. So, for instance, we have a countable clique connecting all of the $x^n$ for $n \in \mathbb{Z}$. Moreover, we have an edge from the empty word to every other vertex.

Ok. So this is a monad on $\mathsf{ReflGph}$. What are the operations?

The product should send $CACA \Rightarrow CA$. So we need to know how to map the vertices of $CACA\Gamma$ to vertices of $CA\Gamma$. But vertices of $CACA\Gamma$ are words in the vertices of $CA\Gamma$. That is, words of words of vertices of $\Gamma$. So concatenation gives us a map from $CACA\Gamma \to CA\Gamma$. This may sound complicated at first, but after some meditation it really is simple: Given a word of words, say $[[x,y],[z],[w,x^{-1},y^{-1}]]$ we concatenate these to get the word $[x,y,z,w,x^{-1},y^{-1}]$.

The unit, though, is simple even on first hearing! $\Gamma$ sits naturally as a subgraph of $CA\Gamma$ by sending each vertex to the word of length $1$ containing that vertex.

1: As an aside, this situation happens often enough to have a name.
When the inclusion functor $\mathcal{C} \hookrightarrow \mathcal{D}$ admits a right adjoint we call $\mathcal{C}$ a coreflective subcategory of $\mathcal{D}$.

I hope this helps ^_^


问:
Compare two normal distribution that are unrelated
I have two normal distributions, about the incomes of 2 stores (assuming they are unrelated).

For each of the 2 stores I have mean and standard deviation over a 40 days income samples.

Now, I'd like to compute the probability of Store A to perform better than Store B on a new day.

I was thinking to use

$$ (A-B) \sim \mathscr{N} (\mu_A - \mu_B, \sigma_A^2 + \sigma_A^2) , $$

And then do something like probability of B performing better than A is as $P(B>A) = P(A-B<0)$, and we use the normal distribution, to evaluate $A-B < 0$.

But I want to evaluate if my reasoning is correct.

Thanks!

答:
Several comments:

I agree with @heropup (+1) that you need data for differences $D$ of Store B sales minus Store A sales on the same days. But then you don't have independent data unless the 40 days
are quite far apart. (A time-series approach might look for weekly, monthly, or
seasonal effects.

If you you have independent normal differences d then you could use $\bar d$ and $S_d$ to estimate $\mu_d, \sigma_d.$ It's not quite clear what you mean by the probability
that $B > A$ or $D > 0.$ Are you asking for daily predictions or are you willing
to deal with historical trends and just hope they 'predict' the future. You might want
to look at distinctions between confidence intervals (for the difference on a typical randomly past day) or prediction intervals for a new difference (with it's own
hypothetical variability, in view of what is known from the past).

If you just want $P(D > 0),$ for a randomly chosen day, then that's easy to compute for normal $D$ with with
$\hat\mu_d = 0.2, \sigma_d = 2.12,$ from a random normal sample. (In R.)

```<code>1 - pnorm(0, 0.2, 2.12)
[1] 0.5375803
</code>```

Body:
Several comments:

I agree with @heropup (+1) that you need data for differences $D$ of Store B sales minus Store A sales on the same days. But then you don't have independent data unless the 40 days
are quite far apart. (A time-series approach might look for weekly, monthly, or
seasonal effects.

If you you have independent normal differences d then you could use $\bar d$ and $S_d$ to estimate $\mu_d, \sigma_d.$ It's not quite clear what you mean by the probability
that $B > A$ or $D > 0.$ Are you asking for daily predictions or are you willing
to deal with historical trends and just hope they 'predict' the future. You might want
to look at distinctions between confidence intervals (for the difference on a typical randomly past day) or prediction intervals for a new difference (with it's own
hypothetical variability, in view of what is known from the past).

If you just want $P(D > 0),$ for a randomly chosen day, then that's easy to compute for normal $D$ with with
$\hat\mu_d = 0.2, \sigma_d = 2.12,$ from a random normal sample. (In R.)

1 - pnorm(0, 0.2, 2.12)
[1] 0.5375803


问:
How to check if $x^TMx > 0$? for some square block matrix $M$
I want to check if the next expression holds for some $K_1$ and $K_2$, being positive diagonal matrices.

$x^T M x > 0$

where $M = \begin{bmatrix}A^TA-A^TB & A^TB \\ A^TA-A^TB & A^TB + K_1B^TB+K_2 \end{bmatrix}$

This matrix is not symmetric, and the four blocks are squared. Note that if $K_1 = K_2 = 0$, then the two "row or column blocks" are equal.

So far, I have not success checking if $M+M^T$ is positive definite, probably I am missing something.

$M+M^T = \begin{bmatrix}2A^TA-A^TB - B^TA& AA^T-B^TA+A^TB \\
                        A^TA-A^TB+B^TA & A^TB+B^TA+2K_1B^TB+2K_2\end{bmatrix}$

Any ideas?

答:
Let $A = I$, $B = 2I$, $x = \begin{bmatrix} x_1 \\ 0 \end{bmatrix}$, where $I$ is the identity matrix and $x \ne 0$ is partitioned conformally with $M$ (i.e., it has the same length as is the dimension of $A$).

Then
$$\begin{align*}
x^T M x &= \begin{bmatrix} x_1^T & 0 \end{bmatrix} \begin{bmatrix}
I - 2I & 2I \\
I - 2I & 2I + K_1 (2I) (2I) + K_2
\end{bmatrix} \begin{bmatrix} x_1 \\ 0 \end{bmatrix} \\
&= \begin{bmatrix} x_1^T & 0 \end{bmatrix} \begin{bmatrix}
-I & 2I \\
-I & 2I + 4K_1 + K_2
\end{bmatrix} \begin{bmatrix} x_1 \\ 0 \end{bmatrix} \\
&= \begin{bmatrix} -x_1^T & 2x^T \end{bmatrix} \begin{bmatrix} x_1 \\ 0 \end{bmatrix} = -x^T x < 0.
\end{align*}$$

So, for some $A$, $B$ and $x$ the above is not true, regardless of $K_1$ and $K_2$.


问:
What is the large lowercase $n$ symbol in math?
What is $\prod$ and how can I use it?
I have seen it used under numerous occasions and on the Desmos graphing calculator keyboard-menu. I could not get it to work, and wanted some pointers.

答: The Greek "S" is $\Sigma$ is typically used for sum. Analogously, The greek capital "P" is $\Pi$ and is typically used for product. We can use it to index some collection of numbers that we wish to multiply together. Say for example I wish to define $n! = n \cdot (n-1)\cdot (n-2)...\cdot 2\cdot 1$. I could write this as $\Pi_{k=1}^n k$ which says that we take every integer value $k$ between $1$ and $n$ and multiply them together.


答:
As pointed out by multiple users above, this is the uppercase version of $\pi$ in the greek alphabet, and it usually means "product".

In Desmos, you use it like this, similar to the summation sign ($\Sigma$):

$$ \prod_{i=a}^b {c} $$

where $a$ is the lower bound (bound included), $b$ is the upper bound (bound included), and $c$ is the component to take the product of.

An example:

$$ \prod_{i=1}^{10} {(i+2)}
= (1+2) \cdot(2+2) \cdot(3+2) \cdot(4+2) \cdot(5+2) \cdot(6+2) \cdot(7+2) \cdot(8+2) \cdot(9+2) \cdot(10+2)$$
$$ = 239500800 $$

Hope this helps.

Title: What is the large lowercase $n$ symbol in math?

Body:
What is $\prod$ and how can I use it?
I have seen it used under numerous occasions and on the Desmos graphing calculator keyboard-menu. I could not get it to work, and wanted some pointers.


问:
Maximum and minimum function on area
Find maximum and minimum value of function $f(x,y) = 3x+14y$ on $ \left\{ (x,y): 3x^4 + xy + y^4 =6\right\} $.

I will grateful for hints and yours help.

答:
What could be said about this "skewed superellipse"  $ \ 3x^4 \ + \ xy \ + \ y^4 \ = \ 6 \ $ is that since the curve has symmetry about the origin [if a point $ \ (x,y) \ $ lies on the curve, so does $ \ (-x,-y) \ $ ], we should expect that extrema are located at corresponding points in opposing quadrants.  user64494  finds, by implicit differentiation of the curve equation,  the slope of a tangent line to be  $ \ \frac {12x^3+y}{4y^3+x} \ $ , and from the linear function to be extremized, the "useful" equation

$$\frac {12x^3 \ + \ y} {4y^3 \ + \ x} \ \ = \ \ \frac{3}{14} \ \ . $$

As the OP, Thomas, indicated that this was an exam problem, I wondered whether some indication was given that the student was to locate a rational point on this curve in answering the question, since the time would be severely (hopelessly?) limited to find real zeroes of the related polynomial without a computational aid.  If it were given that the coordinates of a rational point have the same denominator, so that $ \ x \ = \frac{a}{c} \ \ , \ \ y \ = \frac{b}{c} \ \ , $ we would be able to write

$$\frac {12 \left( \frac{a}{c} \right)^3 \ + \ \frac{b}{c} } {4\left( \frac{b}{c} \right)^3 \ + \frac{a}{c} } \ \ = \ \ \frac {12 a^3 \ + \ bc^2 } {4 b^3 \ + \ ac^2 } \ \ = \ \ \frac{3}{14} \ \ . $$

We might then begin a search with $ \ c  \ = \ 2 \ $ (much as we do in "hunting" rational zeroes of a polynomial), giving us

$$ \  \frac {12 a^3 \ + \ 4b  } {4 b^3 \ + \ 4a } \ \ = \ \ \frac{3}{14} \ \ . $$

Since we need $ \ a \ $ and $ \ b \ $ to be integers, we may require that $ \ 6 a^3 \ + \ 2b \ \ = \ \ 3k \ $ and $ \ 2 b^3 \ + \ 2a \ \ = \ \ 14k \ \ , $ with $ \ k \ $ being a positive integer.  It takes only a little effort to find $ \ 6  · 1^3 \ + \ 2·3 \ \ = \ \ 12 \ \ = \ \ 3·4 \ $ ; we then note that  $ \ 14 · 4 \ \ = \ \ 56 \ \ = \ \ 2 ·3^3 \ + \ 2·1 \ \  \ \ . $

This gives us the desired ratio $ \ \frac{12}{56} \ \ = \ \ \frac{3}{14} \ \ . $  As all of the exponents in the slope expression are odd, the signs of both $ \ a \ $ and $ \ b \ $ can both be changed without altering the sign of the ratio.  This then yields two rational points $ \ \left( \ \frac{1}{2} \ , \ \frac{3}{2} \ \right) $ and $ \ \left( \ -\frac{1}{2} \ , \ -\frac{3}{2} \ \right) \ \ , $  as expected from the symmetry of the curve.  The maximum and minimum of $ \ f(x,y) \  = \ 3x \ + \ 14y \ $ occur, respectively, for those two points as $ \ \frac{45}{2} \ $ and $ \ -\frac{45}{2} \ $ .  A representative graph is shown below.  A calculation of this sort may have been possible, with the clues mentioned, in the available time.

$$ $$

I can't help feeling that this is one of those exam problems, if it is correctly reproduced here, that the problem-poser had not worked out in advance to estimate how much work would be necessary.  (I've had to reject many prospective "interesting" exam problems after seeing the amount of solution-writing that would be called for...)  An optimization problem for an exam would be rather more tractable if the curve were a "rotated ellipse", say,   $ \ 3x^2 \ + \ xy \ + \ y^2 \ = \ C \ $ .

Title: Maximum and minimum function on area


问:
Proximal Point of the absolute value function
Absolute value function is defined as follows:
\begin{equation*}
   f(x) = |x| =
\left\{
 \begin{array}{ll}
  x  & \mbox{if } x \geq 0 \\
  -x & \mbox{if } x < 0
 \end{array}
\right.
\end{equation*}

Prox of $|x|$ is

\begin{equation*}
   \text{prox}_{|\cdot|}^{\gamma}(x) = \text{argmin}_{y \in \mathcal{H}}
\left\{
 \begin{array}{ll}
  y + \frac{1}{2\gamma} ||y-x||^2 & \mbox{if } y \geq 0 \\
  -y + \frac{1}{2\gamma} ||y-x||^2& \mbox{if } y < 0
 \end{array}
\right.
\end{equation*}

Since the objective function of the minization problem is convex, the first order condition of optimality of a convex function is applied to get the minimizing point $y$.

Derivate w.r.t $y$ and using first order condition, $1+ \frac{1}{\gamma} (y -x) = 0 $ implies that the minimizer (proximal point) is $y = x - \gamma$ when $x - \gamma \ge 0$

Similary, $ -1+ \frac{1}{\gamma} (y -x) = 0 $ implies that the proximal point is $y = x + \gamma$ when $x < -\gamma$

\begin{equation*}
   \text{prox}_{|\cdot|}^{\gamma}(x) =
\left\{
 \begin{array}{ll}
  x - \gamma & \mbox{if } x \geq \gamma \\
  x + \gamma & \mbox{if } x \leq -\gamma \\
  ? & \mbox{if } |x| < \gamma \\
 \end{array}
\right.
\end{equation*}

Question: When $|x| < \gamma $ then what is the proximal point of the absolute value function?.

答: Let us denote by $g^\gamma(y)$ the function that you are interested in minimizing. If $\vert x \vert < \gamma,$ then the derivative of $g^\gamma$ is always non-zero. However, we know that a proximal point must exist since $g^\gamma$ is convex, so it follows that the minimizing value of $g^\gamma$ must be exactly the point of non-differentiability of $g^\gamma,$ i.e. $\text{prox}^\gamma_{\vert \cdot \vert}(x) = 0$ when $\vert x \vert < \gamma.$ I hope this helps. :)


问:
Recursive Function for Pyramid-Scheme
consider a group which has 1 user.
each month, every user can bring another user to join the group.
the user that has been joined for 3 months, should leave the group.
calculate the total membership count, after 30 months.

```<code>MONTH    0    1    2    3    4    5    6    7
TOTAL    1    2    4    8    14   26   48   88
</code>```

If I'm right with those numbers, from month 1, the recursive function would be f(n) = f(n-1) + f(n-2) + f(n-3)

Is it correct?
if yes then how to calculate the new memberships and leavings counts?

UPDATE

I found a solution..., correct?

```<code>n > 3 | G(n) = G(n-1) + G(n-2) + G(n-3) //total count, remove after month 3

n > 3 | L(n) = L(n-1) + L(n-2) + L(n-3) //sum of people left the group per month

n > 3 | N(n) = T(n-1) //new people per month

n > 3 | T(n) = G(n) - L(n) //total count, remove on month, third

MONTH    1    2    3    4    5    6    7
------------------------------------------
G        2    4    8    14   26   48   88
N        1    2    4    7    13   24   44
L        0    0    1    1    2    4    7
T        2    4    7    13   24   44   81
</code>```

答:
I think you're whit numbers but the recursive algorithm is:
($F^0_n:$new members , $F^1_n:$1 month members , $F^2_n:$2 month members , $F^3_n:$3 month members)

$F^0_n=F^0_{n-1}+F^1_{n-1}+F^2_{n-1}$

$F^1_n=F^0_{n-1}$

$F^2_n=F^1_{n-1}$

$F^3_n=F^2_{n-1}$

$f_n=F^0_{n}+F^1_{n}+F^2_{n}+F^3_{n}$

This equations can reduce to

$A_n= 2A_n-2B_n$

$B_n=A_{n-4}-B_{n-4}$

    ($A_n=f_n$)

    ($B_n=F^3_n$)

Which is easier to solve.

Title: Recursive Function for Pyramid-Scheme


问:
"The p-power bounding" right multiplication in a finite group G under some special conditions
[Roughly speaking, the following question considers a special setting in which we want to prove a property in the form of $ord(g \sigma)\ |\ p^k$.]

The Problem in Detail:

Let $G$ be a finite group, and $H \leq G$ any subgroup with order $|H|=p^k$ (where $p$ is a prime, and $k \in \mathbb{N}$. Note that $H$ may not be a Sylow $p$-subgroup). Let $g \notin H$ be an element in $G$ s.t. $ord(g)\ |\ p^k $; and $\sigma\in G$ be s.t. for every integer $n$ we have $g^n \sigma g^{-n} \in H$. We claim that $ord(g \sigma)\ |\ p^k$.

(Edit: The previous claim is $ord(g) = ord(g \sigma)$, which is wrong due to the counterexample provided by @jpvee.)

My Attempts:

I can virtually do nothing other than expanding $(g \sigma)^l$, which does not contribute much.

I am quite confused on how to even start out the first step properly.

Can anyone please give me some hints on how I should proceed?

Thank you very much for your help!
:)

答:
I have found a reference for proving this claim.
My claim is in fact a weaker version of the first lemma given in the article "On A Theorem of Frobenius" by Richard Brauer in 1969.

Since the proof is clearly presented in the original article (which can be easily accessed). I would not copy it down here :)


问:
How to define the differential on a manifold/How to proove a function is differentiable over a manifold?
How to define the differential on a manifold ?
How to prove a function is differentiable over a manifold ?

I've seen some definitions but can't get to a proper understanding of them.

To what i've understand , we could prove a function is differentiable over a manifold if it's a local restriction of a differentiable function over a vectorial space for example , but as i don't have any proposition or theorem to support on , that's complicated.

What would be a good way to define it ?
Or , source for a good definition ,

答:
Given two smooth manifolds $(M,\mathcal{A}_M)$ and  $(N,\mathcal{A}_N)$ of dimensions $m$ and $n$ respectively, a function $f:M\to N$ is defined to be of class $C^r$ ($r \geq 0$ any integer, usually we consider $C^{\infty}$), if for every chart $(U,\alpha)\in\mathcal{A}_M$ and $(V,\beta)\in\mathcal{A}_N$ with $f(U)\subset V$, we have that the composite map
\begin{align}
\beta\circ f\circ \alpha^{-1}:\alpha[U]\subset \Bbb{R}^m\to \Bbb{R}^n
\end{align}

is of class $C^r$ in the usual sense that one learns in multivariable calculus.

If this is your first time studying smooth manifolds, this can all be very abstract, so I highly recommend you watch this lecture series by Frederic Schuller on General Relativity. Particularly relevant for your question are lectures 4 and 5 (though if you have time I'd suggest you watch atleast lectures $1$-$6$). Following this, you should of course supplement it with any standard book on the subject (for example the first few chapters of John Lee's smooth manifolds book).


问:
Prove that the normalizing constant of the continous categorical distribution is positive

Problem statement

Given a vector $\mathbf{x} \in \mathbb{R}^{K}$, where $x_{i} \neq x_{j}, \forall i \neq j$. Prove that:

\begin{equation}
\sum_{k=1}^{K} \frac{\exp(x_{k})}{\prod_{i \neq k} (x_{k} - x_{i})} \ge 0.
\end{equation}

If it is wrong, I would appreciate if a counter example is provided.

Background

I am reading the paper "The continuous categorical: a novel simplex-valued exponential family". In the paper, the authors proposed a continuous version for the categorical distribution. They then provided the normalizing constant. To calculate the log-likelihood, we need the log of that normalizing constant. However, I do not know how to prove that it is always positive, so that we can take a log. Here, I reformulate the expression of the normalizing constant for convenience. If one read the paper, we can analogize $\mathbf{x}$ as $\pmb{\eta}$. Note that although $\eta_{K} = 0$, we could simply consider it as any vector, and just shift $\eta_{i} - \eta_{K}$ to obtain $\pmb{\eta}$ as in the paper. The non-shifting version does not  change the positivity nature of the expression of interest.

In addition, I tried the implementation provided by the authors, but it was unclear to me since the authors calculated the denominator $\propto (x_{k} - x_{i})$, not $(x_{i} - x_{k})$ as specified in the paper. Of course, it would lead to an additional factor $(-1)^{K - 1}$ in the denominator. This factor would then neutralizes the $(-1)^{K + 1}$. However, the implementation still have something with $(-1)^{K-1}$ (Note that the implementation denotes a variable $K$ as the value $(K - 1)$ in the paper.)

I did try the implementation by passing a vector as input to the provided function. However, it does not always return a value, but sometimes, not-a-number (NaN).

答:
Good question! The simple answer is that the expression you wrote corresponds to the integral of a positive function, therefore it must be positive. To be more explicit, Gordon-Rodriguez et al. showed (in Section A of the supplement for the original paper):
$$
\int_{\Delta^{K-1}}
 \exp\left( \sum_{k=1}^{K} \eta_k z_k \right) d\mu(\mathbf z)
=
(-1)^{K-1} \sum_{k=1}^K
\frac
 { { \exp\left(\eta_k\right)} }
 { \prod_{i \ne k } \left(\eta_i - \eta_k\right)},
$$
where $\mu$ is the Lebesgue measure, $(\eta_1, \dots, \eta_{K})$ are real numbers such that $\eta_i \ne \eta_j$ whenever $i \ne j$, and the integral is taken over the simplex:
$$
\Delta^{K-1} = \left\{ \mathbf z \in \mathbb R^{K} : z_k \ge 0, \sum_{k=1}^{K} z_k = 1 \right\}.
$$
Now, as you correctly pointed out, the $(-1)^{K-1}$ factor corresponds to the changes in sign on the denominator, and so our two expressions are equivalent. Noting that the integrand $\exp\left( \sum_{k=1}^{K} \eta_k z_k \right)$ is always positive, we can conclude that the RHS is also positive, as required.

Let me add a couple of remarks:

The proof that our expression can indeed be written as such an integral is algebraically involved, but it is helpful to assume WLOG that $\eta_K = 0$ (as you already pointed out), and note that our integral is over a $K-1$ dimensional set and can be written as follows:
$$
\int_{\Delta^{K-1}}
 \exp\left( \sum_{k=1}^{K} \eta_k z_k \right) d\mu(\mathbf z)
= \int_0^1 \int_0^{1-z_1} \cdots \int_0^{1-z_1-\cdots-z_{K-2}} \exp\left(\sum_{k=1}^{K-1} \eta_k z_k\right)
dz_{K-1} \cdots dz_2 dz_1.
$$
You can then proceed by induction (details in the supplement).

As for your question about NaN values, you can take a look at section 3.4 from the original paper, as well as section 5 from this other paper. To summarize - computing the normalizing constant is numerically unstable, a problem which becomes worse as $K$ increases (and as a result, these papers only considered examples with $K \le 10$). In terms of the implementation, note that the product in this line is taken over axis=1, which does in fact index rows (not columns), since the 'zeroth' dimension indexes the minibatch samples.


答:
Let $A=\langle a \rangle$ and $B=\langle b\rangle$. Note that because $ba=a^3b$, we cannot have $A=B$. Thus, we know that $A\cap B$, which is a proper subgroup of both $A$ and $B$, has either one or two elements.

Thus, we have
$$8 \geq |AB| = \frac{|A||B|}{|A\cap B|} = \frac{(4)(4)}{|A\cap B|}.$$
Therefore, we must have $|AB|=8$, and $|A\cap B|=2$. But $A\cap B$ is a subgroup of both $A$ and $B$, which being cyclic have each only one subgroup of order $2$. Thus, $\langle a^2\rangle = \langle b^2\rangle$, and we cannot have $a^2=e$ or $b^2=e$, so $a^2=b^2$ must hold.

Title: Showing $a^2=b^2$ in a group with 8 elements isomorphic to Quaternion group

Body:
I am working in a group $G$ with 8 elements such that $a, b\in G$ and $ba=a^3b$. Moreover, I am given that both $b$ and $a$ have order 4 and $\langle a \rangle \neq \langle b \rangle$. I understand this group is isomorphic to Quaternion group, but this is not given and I am trying to prove this. I want to show $a^2=b^2$, but I am stuck.
Here is what I have done so far:

$$(ab)^2=abab=a(ba)b=aa^3bb=a^4b^2=b^2$$

Also from the equality $ba=a^3b$ I can derive many other equalities, but they do not seem to give me anything useful. For instance

$$aba=b \text{ or } a^3ba^3=b$$

Another observation I have made is that both $b^2$ and $a^2$ have order 2. But I am not sure if that is enough to conclude $a^2=b^2$. Please guide me in the right direction.


问:
Correspondence between continuous maps and convergent sequences
I am trying to find a topological space $X$ such that there is a bijection between $C(X,Y)$ and $\{convergent$ $sequences$ $in$ $Y\}$ for any Hausdorff space $Y$. The natural idea is to choose $X=\mathbb N$, but then I cannot find a suitable topology. I tried the co-finite topology since it feels like convergent sequence, but with it, any continuous map from $X \to Y$ would be a sequence fluctuating in $Y$ instead of a convergent sequence. Any help would be appreciated. Looking for a explicit construction.

答:
The definition of a convergent sequence is the following. $(x_n)_n$ converges to $x$ iff for each open $U$ around $x$ there exists an $N\in\mathbb{N}$ such that for all $n>N: x_n\in U$. This is the same as saying that the preimage of $U$ is cofinite. But to talk about neighborhoods of the limit $x$ it would be convenient if the limit was also a part of the map in $X\to Y$.

This motivates the bijection between convergent sequences in $Y$ and $C(\mathbb{N}\cup \{\infty\}, Y)$ where the topology on $\mathbb{N} \cup \{\infty\}$ is generated by $\{\{n\}| n\in \mathbb{N}\} \cup \{[n, \infty)\cup\{\infty\}| n\in\mathbb{N}\}$. The important part is that each point in $\mathbb{N}$ is isolated and that each neighborhood of $\infty$ is cofinite.

The bijection sends each convergent sequence to the function $$f:X\to Y: \begin{cases}n\mapsto x_n\\\infty\mapsto \lim_{n\to\infty}x_n\end{cases}$$
The continuity of the function $f$ follows from the initial observation that the preimage of any open $U$ containing the limit is cofinite.*

This bijection is injective, because limits are unique in Hausdorff spaces. And it is also surjective since every continuous function $X\to Y$ gives rise to a convergent sequence by considering the image of $\mathbb{N}\subset X$.

*EDIT:
To explain the continuity of $f$ in more detail, the argument is as follows.
For $f$ to be continuous we need the preimage of any open to be open. So consider any open $V\subset Y$. We split the argument into two cases. Either $V$ contains the limit or $V$ doesn't contain the limit.

Case 1:
$V$ contains the limit, this implies the preimage of $V$ is a cofinite set and the point $\infty$. By definition this set is open.

Case 2:
$V$ does not contain the limit, in this case the preimage of $V$ is a subset of $\mathbb{N}$ because $X$ has the discrete topology on $\mathbb{N}$ all subsets of $\mathbb{N}$ are open, therefore the preimage of $V$ is open.

This proves that the preimage of any open in $Y$ is open in $X$.


问:
Determine whether the given points are limit points of the given set
Let $P = \{(x,y) \in \Bbb R^2 : 0 \le x^2 < 1, 2 < y \le 4 \}$ and $S = \{(x,y) \in \Bbb R^2 : 1\le x \le 3, y = 2 \}$.

Are $(0,3)$ and $(0,4)$ be the limit points of $P$?

Are all points of $S \cup P$ be the limit points of $S \cup P$?

Attempt:
For 1, I tried on $(0,3)$ first like this.

Let $r > 0 $. To check: $(0,3)$ is a limit point of $P$. In other words, $B((0,3),r) \cap P \setminus \{(0,3) \} \ne \emptyset$.

Let $(x,y) \in B((0,3),r) \setminus \{(0,3)\}$. Then, we have
\begin{align*}
x^2 + (y-3)^2 < r^2 \\
x^2 < r^2 \ \wedge (y-3)^2 < r^2 \\
-r < x < r \ \wedge 3-r < y < 3+r
\end{align*}
Consider the interval $(3,3+r)$. We take a smallest $y$ in $(3,3+r)$, say $y = 3 + \frac{r}{3}$.
Then, we have
\begin{align*}
x^2 + ((3 + \frac{r}{3}) - 3)^2 = \frac{r^2}{9} < r^2
\end{align*}
which means that $(0,3+\frac{r}{3}) \in B((0,3),r) \setminus \{(0,3) \}$. On the other hand, it's clear that $(0,3+\frac{r}{3}) \in P$. Hence, we have $B((0,3),r) \cap P \setminus \{(0,3)\} = \{(0,3+\frac{r}{3}) \} \ne \emptyset$. Therefore, $(0,3)$ is a limit point of $P$.

Is it true? If yes, then the analog could be used on $(0,4)$.

For 2, I'm not sure that the answer is true since there exists point that would be outside of $S \cup P$. Here's what I tried.

I'm considering the point $(3,2) \in S \cup P$ and take $r = 1 > 0$.

To show: $B((3,2), 1) \cap (S \cup P) \setminus \{(3,2) \} = \emptyset$.
Let $(x,y) \in B((3,2),1) \setminus \{(3,2)\}$. Then, we have
\begin{align*}
(x-3)^2 + (y-2)^2 < 1 \\
-1 < x-3 < 1 \ \wedge -1 < y-2 < 1 \\
2 < x < 4 \ \wedge 1 < y < 3.
\end{align*}
But, the point $(3.5, 2) \notin S \cup P$. Hence, we have
$B((3,2),1) \cap (S \cup P) \setminus \{(3,2)\} = \emptyset$.

Am I true?

答:
The fact that $\lambda$ is a limit point of a set $S$ is also equivalent to the fact there exists a sequence $(x_n)_{n \in \mathbb{N}} \subseteq S \setminus \{\lambda\}$ such that $x_n \to \lambda.$ This is usually easier to check than the exact definition of $S'$ (they are in fact equivalent). For example, you might have noticed that $((\frac1n, 3 + \frac1n))_{n \in \mathbb{N}} \subseteq P \setminus \{(0, 3)\}$ and the limit of this sequence is of course equal to $(0, 3),$ so this is a limit point indeed. Analogously one shows that $(0, 4) \in P'.$

Your reasoning of the second part is not entirely sound. $(3, 2) \in (S \cup P)'$ because $((3 - \frac1n, 2))_{n \in \mathbb{N}} \subseteq (S \cup P) \setminus\{(3, 2)\}.$ It is a general fact (and also not hard to prove) that the closure of a set is the union of its limit and isolated points (disjoint union, of course). Note that $S$ is closed and has no isolated points. Therefore, any point of $S$ is a limit point of $S,$ so it is also a limit point of $S \cup P.$ The closure of $P$ is $\overline{P} = [-1, 1] \times [2, 4]$ and it clearly has no isolated points, so $P \subseteq \overline{P} = P',$ hence any point in $P$ is a limit point of $P,$ so also a limit point of $S \cup P.$ I hope this helps. :)


问:
Taylor expansion of $(1+x)^α$ to binomial series – why does the remainder term converge?
For $α ∈ ℝ$ the function $g_α \colon B_1(0) → ℝ, x ↦ (1+x)^α$ is $C^∞$ and $g_α^{(n)}(x) = n! \tbinom{α}{n}(1+x)^{α-n}$, where $\tbinom{α}{n} = \frac{α(α-1)\cdots(α-n+1)}{n!}$ is the generalized binomial coefficient.
I want to show that $g_α$ has a Taylor expansion $g_α (x) = \sum_{k=0}^∞ \tbinom{α}{k}x^k$. I can easily show that the series really converges for $|x| < 1$, but can I use this fact to show that the remainder term
$$R_n(x) = \int_0^x \frac{(x-t)^n}{n!} (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1} dt$$
or in Lagrange form
$$\quad R_n(x) = \frac{x^{n+1}}{(n+1)!} (n+1)! \binom{α}{n+1}(1+ξ)^{α-n-1} = x^{n+1} \binom{α}{n+1}(1+ξ)^{α-n-1}$$
converges to zero for $|x| < 1$?

How else can I show that the remainder term converges to zero for $|x| < 1$?

Progress: Using the Lagrange form, for $x ∈ (-1,1)$ and $n ∈ ℕ$ there is an $h=h(n,x) ∈ (0,1)$ such that:

$$|R_n(x)| = \Big| x^{n+1} \binom{α}{n+1}(1+hx)^{α-n-1} \Big| = \Big| \binom{α}{n+1}\big(\frac{x}{1+hx}\big)^{n+1}(1+ξ)^{α} \Big| $$
If $x > 0$, then $\big|\frac{x}{1+hx}\big| < {\big|\frac{x}{1+x}\big|}< 1$ and so for $q := \big|\frac{x}{1+x}\big|$, since $\sum_{k=0}^∞ \tbinom{α}{n} q^n$ converges:
$$|R_n(x)| ≤ \Big| \binom{α}{n+1} · q^{n+1} ·(1+ξ)^{α} \Big| \overset{n → ∞}{\longrightarrow} 0$$
The problem is still open for $x < 0$.
Mhenni suggested to use Stirling approximation, but I want to do this more elementary and I think it is possible.

答:
Following the technique in the problem. Let $ g(n,\alpha)= \frac{(n+1)!}{n!} \tbinom{α}{n+1}$, for not to have a mess

$$ \Big|R_n(x)\Big| \leq g(n,\alpha)\int_0^x {(x-t)^n} (1+t)^{α-n-1} dt$$

$$ \leq g(n,\alpha)x^n\int_0^x (1+t)^{α-n-1} dt $$

$$ < g(n,\alpha)\int_0^1 (1+t)^{α-n-1} dt,\quad \mathrm{since}\quad |x|<1,\, \int_{0}^{x}\leq \int_{0}^{1}. $$

$$\implies \Big|R_n(x)\Big| < g(n,\alpha)\frac{2^{\alpha-n}-1}{\alpha-n}$$

$$  \Big|R_n(x)\Big| = \frac{\alpha!(2^{\alpha-n}-1)}{(\alpha-n)!n!}. $$

Now, you can see that the remainder goes to $0$ as $n\to \infty$ with some restrictions on $\alpha$. Try to use Stirling approximation $n! \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}$ and see what you get.

Note:

1) we used the inequality $ x-t \leq x $ which follows from

$$ 0\leq t \leq x \implies -x \leq -t \leq 0 \implies 0<x-t < x. $$

2)

$$ \tbinom{α}{n}(2^{α-n} - 1)= \frac{\alpha!(2^{\alpha-n}-1)}{(\alpha-n)!n!}. $$


问:
Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$
Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.

and please don't use any $\sum_{cyc}$ because I get confused with it.

答: let $p=\frac{a^2}{a^2+bc},q=\frac{b^2}{b^2+ac},r=\frac{c^2}{c^2+ab}$ then easy to see $p,q,r\in (0,1)$ Now notice that $$\frac{(1-p)(1-q)(1-r)}{pqr}=1$$ so $$p+q+r=1+pq+qr+rp-2pqr$$  It remains to prove $$pq+qr+rp-2pqr\le 1$$ $$\iff \underbrace{p(1-q)(r-1)}_{\le 0}+\underbrace{q(1-r)(p-1)}_{\le 0}+\underbrace{(1-p)(q-1)}_{\le 0}\le 0$$ which is obvious as each term is $\le 0$ as shown


问:
For $\gcd(p,q) = 1$ and $\gcd(r,s) = 1$ show $\gcd(ps,rq) = 1$.

For $\gcd(p.q)=1$ and $\gcd(r,s)=1$ show $\gcd(ps,rq)=1$.

I tried via contradiction. Let $\gcd(ps,rq) \neq 1$, so let $d > 1$ a common divisor of $ps$ and $rq$, i.e. $ps = dx$ and $rq = dy$ for some integers $x,y$. So $ p = \frac{dx}{s} $ and $q = \frac{dy}{r}$. It follows $\gcd(p,q) = \gcd(\frac{dx}{s}, \frac{dy}{r}) \geq d > 1$. That should do it, but I don't think the last but one inequality holds. If $s \nmid x$ or $r \nmid y$ this won't work...
Might be the wrong way, but I would be very thankful, if someone could help me!

答: This statement is simply untrue. As, Arturo notes, if you have $p=2,q=3,r=2,s=3$, we have that $gcd(2,3) =1$ for both number sets, but $gcd(2*3,3*2) = 6$, thus proving your statement wrong. How about adding the extra condition that all $p,q,r,s$ must be distinct? However, we can have $p=4,q=7,r=1,s=21$ and have $gcd(4,7) =1, gcd(1,21)=1$, but $gcd(4*21,1*7) = gcd(84,7) = 7.$ Thus, this statement is untrue.