问:
Show $\left | \cosh{x}-1 \right |\leq 3 \left | x \right |$ for $\left | x \right |<1/2$
I have to show that $\left | \cosh{x}-1 \right |\leq 3 \left | x \right |$ for $\left | x \right |<1/2$. I can not use derivatives, series of the trigonometric functions etc. I have to use generel inequalities i.e like  $$1+x\le e^x\le\frac{1}{1-x}$$

How would I do this. I am completely lost. I know the definition of hyperbolic cosine but I can not work this one out.

Body:
$\cosh(x)
=\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n)!}
$
so,
if $0 \le x \le a < 1$,

$\begin{array}\\
\cosh(x)-1
&=\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n)!}\\
&=x^2\sum_{n=1}^{\infty} \dfrac{x^{2n-2}}{(2n)!}\\
&=x^2\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n+2)!}\\
&=\dfrac{x^2}{2}\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n+2)!/2}\\
&\le\dfrac{x^2}{2}\sum_{n=0}^{\infty}x^{2n}\\
&\le\dfrac{x^2}{2(1-x^2)}\\
&=x\dfrac{x}{2(1-x^2)}\\
&\le x\dfrac{a}{2(1-a^2)}
\qquad\text{since } \dfrac{x}{1-x^2} \text{ in increasing for }0 < x < 1\\
\end{array}
$

If $a = \dfrac12$,
$\dfrac{a}{2(1-a^2)}
=\dfrac{1/2}{2(1-1/4)}
=\dfrac{1}{4(3/4)}
=\dfrac13
$

so
$|\cosh(x)-1|
\le x/3$
for
$|x| \le \dfrac12$
(actually, according to Wolfy,
for $|x| \lt 0.644$).


问:
Do these sets have the fixed-point property?
I need some help determining if the following sets have the "fixed-point property" (A topological space $X$ has this property if for every continuous function $f: X \to Y,$ there exists an $x_0 \in X$ such that $f(x_0)=x_0).$

$X = (0,1) \times (0,1)$

The continuous function $f:(0,1) \times (0,1) \to Y, \hspace{0.4mm} (x,y) \mapsto (\frac{1}{2}x, \frac{1}{2}y)$ has no fixed point $(x_0,y_0) \in (0,1) \times (0,1)$ such that $f(x_0,y_0) = (x_0,y_0),$ so $X$ does not have the fixed-point property.

$D = \big \{(x,y):(x-1)^2+y^2 < 1 \big \} $

An open disk centered at $(1,0):$ The continuous function $f(x,y) = (\frac{1}{2}x,\frac{1}{2}y)$ has no fixed point $(x_0,y_0) \in D$ such that $f(x_0,y_0) = (x_0,y_0),$ so $D$ does not have the fixed-point property.

$A = \big \{(x,y):1 \leqslant x^2+y^2 \leqslant 2 \big \} $

Define $f:A \to Y$ by $(x,y) \mapsto (x+1,y+1).$ Then $f$ has no fixed points in $A$ and so $A$ does not have the fixed point property.

Questions: Do my functions work here? Do they show that these sets do not have the fixed-point property? (I'm pretty sure all of them don't have it)

答: $f(x,y)=(-x,-y)$ does fix all points with $x=0$ or $y=0$. Consider using polar coordinates instead of Cartesian coordinates to find an example which doesn't fix any points.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Why do we believe the Church-Turing Thesis?
The Church-Turing Thesis, which says that the Turing Machine model is at least as powerful as any computer that can be built in practice, seems to be pretty unquestioningly accepted in my exposure to computer science.  Why?  Do we have any more evidence for its truth than "nobody has been able to think of a more powerful model than Turing's yet?"  If so, why do we accept that argument as truth?  It seems to me that in any other mathematical discipline, a claim like the Church-Turing thesis whose support amounts to "we haven't yet found a counterexample" would not be taken seriously by the community.</p>

答:
A side note: I think it is an error to identify the Church-Turing Thesis with a claim about what machines can do in practice. Some reflections [if I may be forgiven self-quotation.]

(a) It is important to note that there are three levels of concept that are in play here when we talk about computation.

At the pre-theoretic level – and guided by some paradigms of common-or- garden real-world computation – there is a loose cluster of inchoate ideas about what we can compute with paper and pencil, about what a com- puter (in the modern sense) can compute, and about what a mechanism more generally might compute.

Then at what we might call the proto-theoretic level we have, inter alia, one now familiar way of picking out strands from the pre-theoretic cluster while idealizing away from practical limitations of time or amount of paper, giving us the notion of an effectively computable function. So here some theoretic tidying has already taken place, though the concept still remains somewhat vaguely characterized (what makes a step in a step-by-small-step algorithmic procedure ‘small’ enough to be admissible?).

Then at the fully theoretic level we have tightly characterized concepts like the concept of a recursive function and the concept of a Turing-computable total function.

It would be quite implausible to suppose that the inchoate pre-theoretic cluster of ideas at the first level pins down anything very definite. No, the Church–Turing Thesis sensibly understood, in keeping with the intentions of the early founding fathers, is a view about the relations between concepts at the second and third level. The Thesis kicks in after some proto-theoretic work has been done. The claim is that the functions that fall under the proto-theoretic idea of an effectively computable function are just those that fall under the concept of a recursive function and under the concept of a Turing-computable function. NB: the Thesis is a claim about the extension of the concept of an effectively computable function.

(b) There are other strands in the pre-theoretic hodgepodge of ideas about computation than those picked up in the idea of effective computability: in particular, there’s the idea of what a machine can compute and we can do some proto-theoretic tidying of that strand too. But the Church–Turing Thesis is not about this idea. It must not be muddled with the entirely different claim that a physical machine can only compute recursive functions – i.e. the claim that any possible computing mechanism (broadly construed) can compute no more than a Turing machine. For perhaps there could be a physical set-up which somehow or other is not restricted to delivering a result after a finite number of discrete, deterministic steps, and so is enabled to do more than any Turing machine. Or at least, if such a ‘hypercomputer’ is impossible, that certainly can’t be established merely by arguing for the Church–Turing Thesis.

Let’s pause over this important point, and explore it just a little further. It is familiar that the Entscheidungsproblem can’t be solved by a Turing machine. In other words, there is no Turing machine which can be fed (the code for) an arbitrary first-order wff, and which will then decide, in a finite number of steps, whether it is a valid wff or not. Here, however, is a simple specification for a non-Turing hypercomputer that could be used to decide validity.

Imagine a machine that takes as input the (Gödel number for the) wff $\varphi$  which is to be tested for validity. It then starts effectively enumerating (numbers for) the theorems of a suitable axiomatized formal theory of first-order logic. We’ll
suppose our computer flashes a light if and when it enumerates a theorem that matches $\varphi$ . Now, our imagined computer speeds up as it works. It performs one operation in the first second, a second operation in the next half second, a third in the next quarter second, a fourth in the next eighth of a second, and so on. Hence after two seconds it has done an infinite number of tasks, thereby enumerating and checking every theorem to see if it matches $\varphi$ ! So if the computer’s light flashes within two seconds, $\varphi$  is valid; if not, not. In sum, we can use our wildly accelerating machine to decide validity, because it can go through an infinite number of steps in a finite time.

Now, you might very reasonably think that such accelerating machines are a mere philosophers’ fantasy, physically impossible and not to be taken seriously. But actually it isn’t quite as simple as that. For example, we can describe space-time structures consistent with General Relativity which apparently have the following feature. We could send an ‘ordinary’ computer on a trajectory towards a spacetime singularity. According to its own time, it’s a non-accelerating com- puter, plodding evenly along, computing forever and never actually reaching the singularity. But according to us – such are the joys of relativity! – it takes a finite time before it vanishes into the singularity, accelerating as it goes. Suppose we set up our computer to flash us a signal if, as it enumerates the first-order logical theorems, it ever reaches $\varphi$. We’ll then get the signal within a bounded time just in case $\varphi$  is a theorem. So our computer falling towards the singularity can be used to decide validity.
Now, there are quite fascinating complications about whether this fanciful story actually works within General Relativity.  But no matter. The important point is that the issue of whether there could be this sort of Turing-beating physical set-up – where (from our point of view) an infinite number of steps are executed – has nothing to do with the Church–Turing Thesis properly under- stood. For that is a claim about effective computability, about what can be done in a finite number of steps following an algorithm.


问:
Is the negative normalized entropy convex?
The negative normalized entropy is defined as
$$h:\mathbb{R}_{>0}^n \rightarrow \mathbb{R} \ , \ h(x)=\sum_{i=1}^n x_i\log \frac{x_i}{\sum_{j=1}^n x_j} \ .$$
Is this function convex?

Its Hessian is given by
$$H_{ij} = \frac{\partial^2 h}{\partial x_i\partial x_j}(x) = \frac{\delta_{ij}}{x_j} - \frac{1}{\sum_{j=1}^n x_j} \ ,$$
so the question is equivalent to asking if this matrix is positive semi-definite for all $x \succ 0$.

One way to do this would be to show that for any $v \in \mathbb{R}^n$,
$$v^T H v = \sum_{i=1}^{n} \frac{v_i^2}{x_i} - \frac{\left(\sum_{i=1}^{n}v_i\right)^2}{\sum_{j=1}^n x_j} \geq 0 \ ;$$
however, I am not sure if this holds.

With this convexity shown, I would be able to provide a good answer for this open question: Conjugate of negative normalized entropy

答:
It is convex. Since $h$ is homogeneous of degree $1$ on $\mathbb{R}_{>0}^n$ (and $H$ is homgeneous of degree -1) one can wlog assume that $\sum_{i=1}^n x_i = 1$ when proving that $H$ is positive semi-definite.

For such $x$, it boils down to proving that for any $v \in \mathbb{R}^n$,
$$v^T H v = \sum_{i=1}^{n} \frac{v_i^2}{x_i} - \left(\sum_{i=1}^{n}v_i\right)^2 \geq 0 .$$

By letting $v_i =x_i w_i$, this is equivalent to proving that for any $w \in \mathbb{R}^n$,
$$\sum_{i=1}^{n} x_i w_i^2 - \left(\sum_{i=1}^{n}x_i w_i\right)^2 \geq 0 ,$$
which is true by Jensen's inequality.


问:
Any neat way to solve the integral $\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$?
Straight to the point: given the integral

$$\iint_Q \frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$$

where $Q=[-a,a]\times[-b,b]$, can you think of any neat way to solve it?

At a first glance it looked quite innocent. No need to say that I’ve changed my mind.

EDIT:

Integrating first w.r.t. y, using the integral

$$\int \frac{1}{\left(\xi^2+\alpha^2\right)^{3/2}}\,d\xi = \frac{\xi}{\alpha^2\sqrt{\xi^2+\alpha^2}} + \text{constant}, $$

one gets to

$$ 2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx $$

and here I get pretty stuck to be honest, so if someone could give me any hints that would be appreciated. Still, I think there should be some nicer way to approach this from the start.

Since I was asked for the source of this problem: I simply asked myself a basic physics question, and trying to find an answer led to this integral.

Body:
Substitute $x = \sqrt{b^2+z^2}\sinh t$ after integrating over $y$
\begin{align}
&\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy \\
= &2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx \\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\sinh^2 t +z^2}dt\\
 =&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\cosh^2 t -b^2}dt\\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac{d(\tanh t)}{
z^2+b^2\tanh^2 t }\\
 =&\frac 4z \tan^{-1} \left(\frac bz \tanh t\right)\bigg|_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}} }\\
=&\frac 4z \tan^{-1} \left(\frac{ab}{z\sqrt{a^2+b^2+z^2}}\right)
\end{align}


问:
Proving m ≤ α(G)τ in a graph
I was given the following exercise:

Let $G$ be a graph of $m$ edges without simple cycles of length 3. Let $α(G)$ be the number of vertex in a maximum independent set of $G$ and let $τ$ be the number of vertex in any vertex cover of $G$. Prove that $m≤α(G)τ$.

However, I'm stuck trying to find the relationship. I know some properties, but I can't seem to find one that would make sense to use the lack of simple cycles of length 3.

答:
This follows from the following two facts:

If $U$ is a vertex cover, then $m \le \sum_{u \in U} \deg(u)$. (The right-hand side counts each edge at least once.)

If $G$ is triangle-free, then $\deg(v) \le \alpha(G)$ for all vertices $v$. (The neighborhood of any vertex is an independent set.)

The second fact in particular should be something that immediately comes to mind when you think about triangle-free graphs.

Title: Proving m ≤ α(G)τ in a graph


问:
Simple question on exact sequences.
I am just learning about abelian categories and I would like to hear some advice on how to think about the concepts of kernel, cokernel, image etc correctly; I am a bit confused. I am trying to prove the following, probably very easy result, but I am stuck.

Let $0\to A\to B\to C\to 0$ is a short exact sequence of complexes in an abelian category. Show that if two of the three complexes $A$, $B$, $C$ are exact, then so is the third.

Say $A$ and $C$ are exact, and denote the differentials of $B$ by $\beta_{n+1}:B_{n+1}\to B_n$. We need to prove that $\text{im}(\beta_{n+1})=\text{ker}(\beta_n)$. First of all, what does this equality mean? Does it mean we have to prove equivalence as subobjects (one factors through the other and vice versa)? Or do we need to just take the image and the kernel and prove they are equal as arrows (for example by a "pre-compositions with a monic are equal so the arrows are equal" argument)? - but then what is the domain of those arrows and is it the same for both the image and the kernel?

Or am I at a completely wrong path?

Any advice would be highly appreciated, thanks.

答: Since the $\beta_n$ form a chain complex, $\beta_{n+1} \circ \beta_n =0$. This means that $\beta_{n+1}$ will map any arguments to things which $\beta_n$ maps to $0$, i.e im $\beta_{n+1}\subset \text{ ker } \beta_{n}$. If there were something which $\beta_{n}$ mapped to $0$ which $\beta_{n+1}$ did not map anything to but $A$ and $C$ were exact then some part of the image in $B$  from $0\to A \to B$ would fail to end up in the kernel of $B\to C$ and $0\to A \to B \to C \to 0$ would fail to be exact.</p>

Title: Simple question on exact sequences.


答:
I finally figured out the solution. Suppose we have a minimum vertex cover $τ_{min}$, so that
$|E|= \sum_{i=1}^{τ_{min}} E_{V_i} \leq  \sum_{i=1}^{τ_{min}} N(V_i) $

where $E_{V_i}$ is the number of edges incident o $V_i$ in the vertex cover. However, the neighbors of $V_i$ form an independent set, because if there where an edge between two of them, there would be a cycle of length 3. Then:

$|E| \leq \sum_{i=1}^{τ_{min}} N(V_i) \leq \sum_{i=1}^{τ_{min}} α(G) = τ_{min} α(G) \leq τ α(G) $

Title: Proving m ≤ α(G)τ in a graph

Body:
I was given the following exercise:

Let $G$ be a graph of $m$ edges without simple cycles of length 3. Let $α(G)$ be the number of vertex in a maximum independent set of $G$ and let $τ$ be the number of vertex in any vertex cover of $G$. Prove that $m≤α(G)τ$.

However, I'm stuck trying to find the relationship. I know some properties, but I can't seem to find one that would make sense to use the lack of simple cycles of length 3.


答:
First proof We know that $\text{ cosh} \left(x\right)-1 = 2 {\text{ sinh}}^{2} \left(\frac{x}{2}\right)$. Let $t = \left|x\right|/2$, it remains to prove that

\begin{equation}
0  \leqslant  t  \leqslant  \frac{1}{4} \Longrightarrow  {\text{ sinh}}^{2} \left(t\right)  \leqslant  3 t
\end{equation}

For such $t$, we have

\begin{equation}
\text{ sinh} \left(t\right)  \leqslant  \frac{1}{2} \left({e}^{t}-1\right)  \leqslant  \frac{1}{2} \left(\frac{1}{\left(1-t\right)}-1\right)  \leqslant  \frac{t}{2 \left(1-t\right)}
\end{equation}

It remains to prove that $\frac{t^2}{4 \left(1-t\right)^2}  \leqslant  3 t$ for
$t \in  I = \left[0 , 1/4\right]$, but this is equivalent to
$P(t) = 12 t^2 - 25 t + 12\ge 0$, which is true because the roots of $P$ are $3/4$ and $4/3$.

Second proof
It suffices to prove the result for $0\le x\le 1/2$. For such $x$ we have
\begin{equation}
\cosh x - 1 = \frac{e^x + e^{-x}}{2}- 1 \le e^x - 1 \le \frac{1}{1-x} - 1 = \frac{x}{1-x}
\end{equation}
It remains to show that $\frac{x}{1-x}\le 3 x$ but this is equivalent to $1\le 3 (1-x)$, that is to say $x\le \frac{2}{3}$, which is true. QED

Title: Show $\left | \cosh{x}-1 \right |\leq 3 \left | x \right |$ for $\left | x \right |<1/2$

Body:
I have to show that $\left | \cosh{x}-1 \right |\leq 3 \left | x \right |$ for $\left | x \right |<1/2$. I can not use derivatives, series of the trigonometric functions etc. I have to use generel inequalities i.e like  $$1+x\le e^x\le\frac{1}{1-x}$$

How would I do this. I am completely lost. I know the definition of hyperbolic cosine but I can not work this one out.


答:
Let's define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the double integral

$$\mathcal{I}{\left(a,b,c\right)}:=\iint_{\left[-a,a\right]\times\left[-b,b\right]}\mathrm{d}x\,\mathrm{d}y\,\frac{1}{\left(x^{2}+y^{2}+c^{2}\right)^{3/2}}.$$

Suppose $\left(a,b,c\right)\in\mathbb{R}_{>0}^{3}$. As you've already found, the double integral can be boiled down to the following single-variable integral:

$$\begin{align}
\mathcal{I}{\left(a,b,c\right)}
&=\iint_{\left[-a,a\right]\times\left[-b,b\right]}\mathrm{d}x\,\mathrm{d}y\,\frac{1}{\left(x^{2}+y^{2}+c^{2}\right)^{3/2}}\\
&=\int_{-a}^{a}\mathrm{d}x\int_{-b}^{b}\mathrm{d}y\,\frac{1}{\left(x^{2}+y^{2}+c^{2}\right)^{3/2}}\\
&=\int_{-a}^{a}\mathrm{d}x\int_{-b}^{b}\mathrm{d}y\,\frac{d}{dy}\left[\frac{y}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+y^{2}+c^{2}}}\right]\\
&=\int_{-a}^{a}\mathrm{d}x\,\frac{2b}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+b^{2}+c^{2}}}\\
&=2\int_{0}^{a}\mathrm{d}x\,\frac{2b}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+b^{2}+c^{2}}};~~~\small{even\,symmetry}\\
&=4b\int_{0}^{a}\mathrm{d}x\,\frac{1}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+b^{2}+c^{2}}}\\
&=4b\int_{0}^{a}\mathrm{d}x\,\frac{1}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+q^{2}}};~~~\small{\left[q:=\sqrt{b^{2}+c^{2}}>0\right]}.\\
\end{align}$$

Consider the substitution $\frac{x}{\sqrt{x^{2}+q^{2}}}=t$. Then, $x=\frac{qt}{\sqrt{1-t^{2}}}\implies dx=dt\,\frac{q}{\left(1-t^{2}\right)^{3/2}}$ and

$$\begin{align}
\mathcal{I}{\left(a,b,c\right)}
&=4b\int_{0}^{a}\mathrm{d}x\,\frac{1}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+q^{2}}}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{q}{\left(1-t^{2}\right)^{3/2}}\cdot\frac{1}{\left(\frac{qt}{\sqrt{1-t^{2}}}\right)^{2}+c^{2}}\cdot\frac{\sqrt{1-t^{2}}}{q};~~~\small{\left[x=\frac{qt}{\sqrt{1-t^{2}}}\right]}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{1}{\left(1-t^{2}\right)}\cdot\frac{1}{\left(\frac{q^{2}t^{2}}{1-t^{2}}\right)+c^{2}}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{1}{q^{2}t^{2}+c^{2}\left(1-t^{2}\right)}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{1}{c^{2}+\left(q^{2}-c^{2}\right)t^{2}}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}}\mathrm{d}t\,\frac{1}{c^{2}+b^{2}t^{2}}\\
&=\frac{4}{c}\int_{0}^{\frac{ab}{c\sqrt{a^{2}+b^{2}+c^{2}}}}\mathrm{d}u\,\frac{1}{1+u^{2}};~~~\small{\left[t=\frac{cu}{b}\right]}\\
&=\frac{4}{c}\arctan{\left(\frac{ab}{c\sqrt{a^{2}+b^{2}+c^{2}}}\right)}.\blacksquare\\
\end{align}$$

Title: Any neat way to solve the integral $\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$?

Body:
Straight to the point: given the integral

$$\iint_Q \frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$$

where $Q=[-a,a]\times[-b,b]$, can you think of any neat way to solve it?

At a first glance it looked quite innocent. No need to say that I’ve changed my mind.

EDIT:

Integrating first w.r.t. y, using the integral

$$\int \frac{1}{\left(\xi^2+\alpha^2\right)^{3/2}}\,d\xi = \frac{\xi}{\alpha^2\sqrt{\xi^2+\alpha^2}} + \text{constant}, $$

one gets to

$$ 2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx $$

and here I get pretty stuck to be honest, so if someone could give me any hints that would be appreciated. Still, I think there should be some nicer way to approach this from the start.

Since I was asked for the source of this problem: I simply asked myself a basic physics question, and trying to find an answer led to this integral.


答: Each term in $h$ is convex because it is the perspective of the convex function $f(x) = x\log x$ (under an affine transformation of the $y$-variable of the perspective). This insight should make it trivial to derive the conjugate function based on the conjugate of $f$.

Title: Is the negative normalized entropy convex?

Body:
The negative normalized entropy is defined as
$$h:\mathbb{R}_{>0}^n \rightarrow \mathbb{R} \ , \ h(x)=\sum_{i=1}^n x_i\log \frac{x_i}{\sum_{j=1}^n x_j} \ .$$
Is this function convex?

Its Hessian is given by
$$H_{ij} = \frac{\partial^2 h}{\partial x_i\partial x_j}(x) = \frac{\delta_{ij}}{x_j} - \frac{1}{\sum_{j=1}^n x_j} \ ,$$
so the question is equivalent to asking if this matrix is positive semi-definite for all $x \succ 0$.

One way to do this would be to show that for any $v \in \mathbb{R}^n$,
$$v^T H v = \sum_{i=1}^{n} \frac{v_i^2}{x_i} - \frac{\left(\sum_{i=1}^{n}v_i\right)^2}{\sum_{j=1}^n x_j} \geq 0 \ ;$$
however, I am not sure if this holds.

With this convexity shown, I would be able to provide a good answer for this open question: Conjugate of negative normalized entropy


问:
Euclidean Geometry versus Analytic Geometry versus Affine Geometry?
What are the relationships (connections) among:

Euclidean (or Plane) geometry

Analytic geometry

Affine geometry

How do these things relate?

I know that this is a very general question, so I'm looking for a simple overview.

Thanks.

答: Euclidean Geometry is geometry studied by Euclid that is based off of shapes with no values.  Analytical Geometry is geometry on planes with values.  So, the difference is mainly that Euclidean doesn't use specific values</p>


问:
Euler's Rotation Theorem Proof
I want to prove Euler's rotation theorem:

In three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point.

I have just finished studying the matrix proof of Euler's rotation theorem as stated in the wikipedia page here.

The author just takes an arbitrary $3 \times 3$ rotation matrix $R$ and proves that $\lambda = 1$ is eigenvalue for this matrix. I do not understand why this proves that such rigid body displacement is rotation by some axis. For me it only shows rotation matrix properties.

答: Some point is fixed, so take this as the origin. Write the rotation as $x\mapsto Rx$. Since $R$'s eigenvalues are $\pm1$, an arbitrary point is of the form $x_++x_-$ with $Rx_\pm=\pm x_\pm$. Then $Rx=x_+-x_-$, so $x$ is fixed iff $x_-=0$, i.e. iff $x$ is on the axis through $O$ parallel to an eigenvalue-$1$ vector.

Title: Euler's Rotation Theorem Proof


问:
How do I find A if I have the transpose of A?
So I'm thinking about this problem

I know that $(A^T)^T=A$ but when I have a coefficient like in this problem above, what do I do?

答:
Turning my comment hint into into an "answer":

Hint: $(\lambda A)^T = \lambda A^T$.


问:
$\lim_{x\to 3} (x−3)f(x) = 0$ and $\lim_{x\to 3} f(x) = \infty$
I need help with a calculus/limit problem.

I suppose I am looking for a function that:

$f(x)$ such that it has $(x-3)^2$ in the denominator

$f(x)$ such that $f(x)(x-3)$ equals $y=x-3$.

I struggle to understand how the above is possible. How is it possible to have a function that:

$\lim_{x \to 3} f(x) = \infty$

$\lim_{x \to 3} (x-3)f(x) = 0$

Any help or pointers are deeply appreciated.

Body:
Take $$f(x)={1\over \sqrt[3]{(x-3)^2}}$$

then $$\lim _{x\to 3}(x-3)f(x)=   \lim _{x\to 3} {\sqrt[3]{x-3}} = 0$$

and $$\lim _{x\to 3 }f(x) = \infty$$


答:
Hint

You also can try this,
$$[(2A)^T]^T=2A.$$

Title: How do I find A if I have the transpose of A?

Body:
So I'm thinking about this problem

I know that $(A^T)^T=A$ but when I have a coefficient like in this problem above, what do I do?


答:
Focusing on positive $x$ (symmetry takes care of the rest), we need to show that, for $x\in[0,\frac 12]$, we have $\cosh x \leq 3x + 1$.

Assuming that you can use the fact that $\cosh x$ is a convex function, you know that $\cosh x \leq  2 \cosh \frac 12 \cdot x$, when $x \in [0,\frac 12]$. So, we just need to show that $2\cosh \frac 12 \cdot x \leq 3x+1$.


问:
Is it possible to solve $ax + \frac{bx}{y^x} < c $ for $x$?
We have the equation:

$$ax + \frac{bx}{y^x} < c $$

Is it possible to rearrange it to have $x$ on its own on the RHS?

答:
I'm not sure this can be solved unless one of a,b,c is zero.

$$ax + \frac{bx}{y^x} < c, \quad  a=0\qquad
\implies \frac{bx}{y^x} < c\qquad \\
\implies
x < \frac{-W_n\big(-\frac{c \log y}{b}\big)}{\log y}\qquad
b,y,(c \log y)\ne0
\quad n\in\mathbb{ Z}\tag{1}$$

$\\$
$$ax + \frac{bx}{y^x} < c, \qquad b=0\quad
\implies x<\frac{c}{a},\quad a\ne0\tag{2}$$

$\\$
$$ax + \frac{bx}{y^x} < c, \qquad c=0\qquad
\implies a y^x + b < 0\\
\implies x < \frac{\log\frac{-b}{a} + 2 i \pi n}{\log y}
\qquad   a,b,y,\log y\ne0
\quad   n \in\mathbb{ Z} \tag{3}
$$


问:
How to prove inequality including natural log?
Let $0<p<1$ and $0<q<1$ be real numbers and $0 < p + q < 1$.
How can I prove the following inequality is correct or not?

$$q > \frac{\ln(\frac{p}{1-q})}{\ln(\frac{p}{1-q}) + \ln(\frac{q}{1-p})}$$

$\ln$ is natural logarithm.

答:
Since
$$0\lt p\lt 1$$
$$0\lt q\lt 1$$
$$0 \lt p + q \lt 1$$
Then
$$\frac{p}{1-q}\lt 1\Rightarrow\ln\left(\frac{p}{1-q}\right)\lt 0$$
$$\frac{q}{1-p}\lt 1\Rightarrow\ln\left(\frac{q}{1-p}\right)\lt 0$$
So now we have
$$q\gt\frac{\ln\left(\frac{p}{1-q}\right)}{\ln\left(\frac{p}{1-q}\right) + \ln\left(\frac{q}{1-p}\right)}$$
$$\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)\lt\ln\left(\frac{p}{1-q}\right)$$
Here we can use the fact that for $0\lt x\lt 1$, we have
$$\frac{x-1}{x}\lt\ln(x)\lt\frac{(x-1)(x+5)}{2(2x+1)}$$
Which yields
$$\ln\left(\frac{p}{1-q}\right)\lt\frac{(p+q-1)(p-5q+5)}{2(2p-q+1)(1-q)}$$
And
$$\frac{p+q-1}{1-q}\lt\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)$$
If the original inequality were true, one would expect the upper bound to be greater than the lower bound. However
$$\frac{p+q-1}{1-q}\lt\frac{(p+q-1)(p-5q+5)}{2(2p-q+1)(1-q)}$$
$$2(2p-q+1)\gt p-5q+5$$
$$p+q\gt 1$$
Therefore
$$\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)\not\lt\ln\left(\frac{p}{1-q}\right)$$


问:
Show negative exponential of exponential is Lipschitz
Simple analysis question. Let $f(x) = \exp(-\exp(x))$, on $\mathbb R$. Then, how does one show that $f$ is Lipschitz? I tried with the inequality
$$
|\exp(a)-\exp(b)|\leq (\exp(a)+\exp(b))|a-b|,
$$
which is valid for any real numbers $a$, $b$. When I proceed I obtain
\begin{align}
|f(x)-f(y)|&\leq(\exp(-\exp(x))+\exp(-\exp(y)))|\exp(x)-\exp(y)|\\
&\leq 2|\exp(x)-\exp(y)|.
\end{align}
Now the problem is that the exponential is not Lipschitz, so I cannot proceed from here. Another idea would be using $\exp(-a)\leq a^{-1}$ for all $a > 0$, but I do not see how.

答:
In such a case, plotting the curve provides a better acquaintance with the issue:

What do we see on this curve ? That its derivative is negative, bounded and probably at its minimal value (therefore $|f'(x)|$ at its maximal value) for $x=0$.

This has to be confirmed, i.e., made rigorous.

Let us compute :

$$f'(x)=\exp(-\exp(x))(-\exp(x)) \ \ \text{implying}$$  $$f''(x)=\exp(-\exp(x)).\exp(x).(\exp(x)-1)$$

As a consequence:

$$\operatorname{sign}(f''(x))=\operatorname{sign}(\exp(x)-1)$$

which is negative for $x<0$ and positive otherwise ; we can therefore deduce that $f'(x)$ is decreasing until a minimum value $m$  taken in $x=0$, then increasing. This minimum value is

$$m=f'(0)=\exp(-\exp(0))(-\exp(0))=-e^{-1}\approx -0.368$$

Therefore, you can conclude, using the MVT that:

$$\forall a,b \in \mathbb{R}, \ \ \ |f(b)-f(a)| \leq M|b-a|$$

with Lipschitz constant $M=|m|$.


问:
Throwing n-sided fair dice until sum of two throws equals n
I have an $n$-sided fair die and each side has a different number from the set $\{0, 1, 2, ... , n - 1\}$. On each throw I write down number what was thrown. I do this until the sum of two numbers from the list that I wrote down equals $n$. So the question is, what is the probability that this game ends in $k$ throws? It is clear that when $k = 1$ the probability is $0$. I thought of using a Markov chain to solve this problem, but I don't know how to represent states and what is the probability of going from one state to another. Can anybody help me with this?

答:
HINTS

Let $[n-1] = \{k\}_{k=0}^{n-1}$ and one approach is to let $W$ denote the set of all $A \subseteq W$ such that no two elements of $A$ add up to $n$. Then, $S = W \cup \{*\}$ is the state space of the Markov chain, with $*$ denoting the absorbing state of finally generating a set of throws where two members add up to $n$.

Now you can define transition probabilities and solve.

A different approach is to note that if you have thrown $1$, then either you win in the next throw or you cannot throw $n-1$ at any time. So you really have $1,n-1$ and $2,n-2$ etc., and when you throw one number from the pair, you will never throw another one unless you complete the task. So the probability to finish in exactly $k$ steps is the probability of picking $k-1$ of those pairs in the first $k-1$ tries, and then picking one of them again in the last attempt.

You could use a geometric random variable or a Markov chain, or argue from first principles.


问:
Let $f: G_{1} \rightarrow G_{2}$ be a homomorphism. Show that if $H_{1} \leq G_{1}$, then $f\left(H_{1}\right) \leq G_{2}$
If $H_{1} \leqslant G_{1}$ then $f\left(H_{1}\right) \leqslant G_{2}$

I try:

Let $x,y \in H_1$, then $x,y \in G_1$ because $H_{1} \leqslant G_{1}$

$f(x*y)=f(x)\star f(y)$ because $f$ is an homomorphism and how $f(x),f(y)\in G_2 $
then $f(x)\star f(y) \in G_2$

and $f(x*x^{-1})=f(x)\star f(x^{-1})=f(x)\star f(x)^{-1}=e_2 \in G_2$

But I don't know how I can show that the inverse is in $G_2$.

答:
From your notation, I'm assuming that $H_1$ is a subgroup of $G_1$, and $f$ is an homomorphism from $G_1$ and $G_2$.

Now, if $y=f(g)$ for some $g\in H_1$, then as $g^{-1}\in H_1$ and $y^{-1}=(f(g))^{-1}=f(g^{-1})$, one gets that $y^{-1}\in f(H_1)$.

Here we use the fact that if $f:G_1\rightarrow G_2$ is a homomorphism between groups, then $(f(g))^{-1}=f(g^{-1})$. This can be seen from
$$f(\mathbb{1}_{G_1})=\mathbb{1}_{G_2},$$
$$ \mathbb{1}_{G_2}=f(gg^{-1})=f(g)f(g^{-1}),$$
and
$$ \mathbb{1}_{G_2}=f(g^{-1} g)=f(g^{-1})f(g)$$

where $\mathbb{1}_{G}$ is the unit of group $G$.


问:
Why is $\beta=$Probability that we accept the null hypothesis, $H_0$ where $\beta$ is the probability of a type II error in a statistical test?
Is it because it's the likelihood that we get a test statistic under the conditions we set of $H_0$ and $H_1$, the alternative hypothesis that causes us to accept the null hypothesis when it's false?

What is the reasoning behind this?

Thank You

答:
I don't really understand what you are asking..there are two types of errors when testing, they are

rejecting $H_0$ given that $H_0$ is correct (type 1 error)

accepting $H_0$ given that $H_1$ is correct (type 2 error)

The corresponding probabilities are

\begin{align*}
P(\text{reject } H_0| H_0 \text{ correct}) &= \alpha = \text{ level of test (significance)}\\
P(\text{accept } H_0| H_1 \text{ correct}) &= 1-\beta = 1-\text{ power of test}
\end{align*}

So we have
$$
\beta = 1- P(\text{accept } H_0| H_1 \text{ correct})
= P(\text{reject } H_0| H_1 \text{ correct})
$$

When testing, we wish to minimize level, and maximize power, since this would lead to a reduction in both types of errors, but this is in general not possible. So what we do instead is specify an acceptable level $\alpha$, say $\alpha=0.05$, and then search over all possible tests with fixed $\alpha$ and maximum $\beta$.


问:
Parametrization of $y^2=x^3$ to solve a line integral?
Evaluate the following vector line integral given in differential form
$$\int_{\gamma} x^2y\,dx-xy\,dy$$
where $\gamma$ is the curve with equation $y^2=x^3$, from $(1,-1)$ to $(1,1)$. The textbook also says that a parametrization of $\gamma$ is $$\vec{r}(t)=(t^2,t^3),\ -1\le t\le 1$$
My question is: How do I find these parametrizations? I know how to parametrize lines in plane and space, but what about a curve such as $y^2=x^3$? Should I give $x$ any value depending on $t$ and then find $y$? For instance, $$x=t \rightarrow y=t^{3/2}$$ What about the interval in which $t$ varies?

答:
A rough sketch of the curve between start and end points may help.

As you can see, both start and end points $A(1,-1)$ and $B(1,1)$ have $x = 1$ but there are really two parts to the curve, one below x-axis and one above. A parametrization of $(t, t^{3/2})$ will only represent the curve above x-axis and you will need $(t, - t^{3/2})$ to cover the curve below x-axis. This will also mean, you will have to split your line integral in two parts.

i) $(t, - t^{3/2}), 0 \leq t \leq 1$ and your integral will have lower bound of $t=1$ and upper bound of $t = 0$.

ii) $(t, t^{3/2}), 0 \leq t \leq 1$ and your integral will have lower bound of $t=0$ and upper bound of $t = 1$.

That will ensure you line integral is along the curve from $A(1,-1)$ to $B(1,1)$.

But splitting a line integral for the same curve is not ideal and the parametrization $(t^2, t^3)$ addresses this issue because it represents your entire curve from $A$ to $B$ for $-1 \leq t \leq 1$.


问:
The Determinant, Tensors, and Orientation
I am a bit confused about orientation and tensors as exemplified by the determinant.

If we have an inner product or a metric and transform a vector from one coordinate system to another the magnitude of that vector is unchanged as the coordinate representation of the metric also changes. Therefore it is natural to think of a vector as a tensor and it's length doesn't change under orientation reversing transformations.

In contrast, we can consider the determinant as an alternating tensor from the exterior algebra, for instance in $\mathbb{R}^3$, $$det(v_1, v_2, v_3) = e^1\wedge e^2 \wedge e^3(v_1, v_2, v_3)$$

But if we perform a reflection, $A$ then we get,
$$det(Av_1, Av_2, Av_3) = det(A)det(v_1, v_2, v_3) = -det(v_1, v_2, v_3) $$
Does that mean that the 3-covector $e^1\wedge e^2 \wedge e^3$ when contracted with 3 vectors is not invariant, which would seem to violate the idea of contracting tensors to a scalar produces an invariant under transformations.

Or do we instead by transforming our 3-covector and the input vectors get an invariant oriented volume measure?
$$e'^1\wedge e'^2 \wedge e'^3(Av_1, Av_2, Av_3) = e^1\wedge e^2 \wedge e^3(v_1, v_2, v_3)?$$

That would seem to mean that the determinant includes an arbitrary sign convention and that the definition of the determinant is different in left and right handed oriented coordinate systems.

答:
Tu's book, Differential Geometry, answers this question in it's discussion of the Volume Form on a Riemannian Manifold.

If we an $A \in GL(n, \mathbb{R})$ is a transformation between coordinate systems with:
$$ \mathbf{v}' = A\mathbf{v} $$ in the new basis.
Then we similarly have a relation on the dual bases.
$$\mathbf{\omega}' = A^{-1}\mathbf{\omega}$$

So then the volume form transforms as,
$$det(A^{-1})e^1 \wedge e^2 \wedge e^3 = e'^1 \wedge e'^2 \wedge e'^3$$

And then if we transform both the volume form and the vectors we are measuring we get the following.
$$\begin{equation*}
\begin{split}
e^1 \wedge e^2 \wedge e^3(u, v, w) &= det(A^{-1})det(A)e^1 \wedge e^2 \wedge e^3(u, v, w) \\
&= det(A^{-1})e^1 \wedge e^2 \wedge e^3(Au, Av, Aw) \\
&= e'^1 \wedge e'^2 \wedge e'^3(Au, Av, Aw) \\
&= e'^1 \wedge e'^2 \wedge e'^3(u', v', w')
\end{split}
\end{equation*}$$

So the volume form or top form is an invariant volume measure.


答: You want to show  that $f(H_1)\leqslant G_2$. You are confused about the inverse so, I am just solving that part. Note that under homomorphism, identity element $e_{G_1}\in G_1$ maps to identity in $G_2$. This is because $f(e_{G_1})=f(e_{G_1}.e_{G_1})=f(e_{G_1}).f(e_{G_1})\Rightarrow f(e_{G_1})=e_{G_2}$, by cancellation law. Hence, if we take $a\in G_1$, let $b\in G_1$ be the inverse of $a$, then $f(e_{G_1})=f(a.b)=f(a).f(b)=e_{G_2}$. Therefore, $f(b)$ is your inverse.

Title: Let $f: G_{1} \rightarrow G_{2}$ be a homomorphism. Show that if $H_{1} \leq G_{1}$, then $f\left(H_{1}\right) \leq G_{2}$

Body:
If $H_{1} \leqslant G_{1}$ then $f\left(H_{1}\right) \leqslant G_{2}$

I try:

Let $x,y \in H_1$, then $x,y \in G_1$ because $H_{1} \leqslant G_{1}$

$f(x*y)=f(x)\star f(y)$ because $f$ is an homomorphism and how $f(x),f(y)\in G_2 $
then $f(x)\star f(y) \in G_2$

and $f(x*x^{-1})=f(x)\star f(x^{-1})=f(x)\star f(x)^{-1}=e_2 \in G_2$

But I don't know how I can show that the inverse is in $G_2$.


问:
Show that $\langle(1234),(12)\rangle = S_4$
I am trying to show that $\langle(1234),(12)\rangle = S_4$. I can multiply and get all $24$ permutations manually but isn't there a more compact solution?</p>

答:
I hope you are familiar to $S_4$ presentation:
$$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Considering a nice Corollary of Von Dyck's Theorem we can set $a=(1,2),~~~b=(1,2,3,4)$ and see that $(ab)^3=1,a^2=b^4=1$. Also, an small program in GAP's environment could show this fast and easy:</p>

```<code>gap> f:=FreeGroup("a","b");
gap> a:=f.1;; b:=f.2;;
     s:=f/[a^4,b^2,(a*b)^3];;
gap> StructureDescription(s);
                                         "S4"
</code>```

Body:
I hope you are familiar to $S_4$ presentation:
$$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Considering a nice Corollary of Von Dyck's Theorem we can set $a=(1,2),~~~b=(1,2,3,4)$ and see that $(ab)^3=1,a^2=b^4=1$. Also, an small program in GAP's environment could show this fast and easy:

gap> f:=FreeGroup("a","b");
gap> a:=f.1;; b:=f.2;;
     s:=f/[a^4,b^2,(a*b)^3];;
gap> StructureDescription(s);
                                         "S4"


答:
I will use the one-step subgroup test.

Since $H_1\le G_1$, we have $e_1\in H_1$. Since $f$ is a homomorphism, $e_2=f(e_1)\in f( H_1)$, so $f(H_1)\neq\varnothing$.

By definition, $$f(H_1)=\{g\in G_2\mid \exists h\in H_1, g=f(h)\},$$ so we have $f(H_1)\subseteq G_2$.

Let $x,y\in f(H_1)$. Then there exist $a,b\in H_1$ such that $x=f(a)$ and $y=f(b)$. Now

$$\begin{align}
x\star y^{-1}&=f(a)\star (f(b))^{-1}\\
&=f(a)\star f(b^{-1})\\
&=f(a\ast b^{-1}),
\end{align}$$

but $a\ast b^{-1}\in H_1$ since $H_1\le G_1$. Hence $x\star y^{-1}\in f(H_1)$.

Hence $f(H_1)\le G_2$.


问:
Is it true that $\#\{J \cap \mathbb{Z}\} \geq b-a-1$ or $\#\{J \cap \mathbb{Z}\} \geq b-a-2$?
Let $A$ be a bounded subset of $\mathbb{R}$ and $J$ be a bounded interval in $\mathbb{R}$. Define $\lambda(A)=\#\{A \cap \mathbb{Z}\}$. Let $a=\inf J$ and $b=\sup J$. If $[x]$ denotes the integer part of the real number $x$ then:

i) $\lambda(J) \leq \lambda([a,[b]+1[)=[b]-a+1 \leq b-a+1$, if $a \in \mathbb{Z}$;

ii) $\lambda(J) \leq \lambda(][a],[b]+1[)=[b]-[a]\leq b-a+1$, if $a \notin \mathbb{Z}$.

My question: Is it true that $\lambda(J) \geq b-a-1$ or $\lambda(J) \geq b-a-2$?

This is being used to prove that $vol(I) \leq \sum_{j=1}^{n} vol(I_j)$ if $I, I_1, \dots, I_n$ are bounded intervals in $\mathbb{R}^{N}$ and $I \subset \bigcup_{j=1}^{n}I_j$, in some lecture notes on measure theory that I found.

答:
You have that $J \supseteq [\lfloor a \rfloor + 1, \lfloor b \rfloor)$ whenever this makes sense (i.e. whenever $\lfloor a \rfloor + 1 \leq \lfloor b \rfloor$) by the definition of $a, \, b$ and the floor function. It follows that $$\lambda(J) \geq \lambda([\lfloor a \rfloor + 1, \lfloor b \rfloor)) = \lfloor b \rfloor - 1 - \lfloor a \rfloor = \lfloor b \rfloor - \lfloor a \rfloor - 1 \geq b - 1 - a - 1 = b - a - 2.$$ Finally, note that if $\lfloor a \rfloor + 1 > \lfloor b \rfloor,$ then $\lambda(J) \geq b - a - 2$ (since this quantity is negative in this case), so this inequality is always true.

Btw, if you know that your interval $J$ is closed, you have the improved inequality $\lambda(J) \geq b - a - 1.$ I hope this helps. :)


问:
Product of a diagonal matrix with all positive entries and a quadratic matrix
I have a diagonal matrix D with all positive entries on diagonal and a quadratic matrix $A=B^{T}B$, is the product of the two matrices $C=DA$ also positive definite? or does the product has other properties? Thanks!

答:
The product doesn't have to be positive definite. For a simple counterexample take any positive semidefinite $A$ (any positive semidefinite $A$ can be written as $B^T B$). Then take $D$ to be the identity matrix and
$$
C = DA = IA = A
$$
is not positive definite.

Edit
The product need not be positive semidefinite either. For the product to be positive semidefinite one would require $DA$ to be a symmetric matrix. However, this is the case iff $A$ and $D$ commute, i.e. $[A,D]=0$. Which will not be the case for most choices of $A$ and $D$. You will however still have that the eigenvalues of $DA$ are all non-negative.


问:
How does WeBWork determines what is correct and what is not?
Yesterday I was doing a WeBWork and had to check the convergence of $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$; It seemed quite straight forward to me, I answered that it converges but the system told me that my answer was incorrect.

That is not the first time that this happened to me so I searched WeBWork Wiki about how does WeBWork check answers and found this:

Answer Evaluation:

When setting up the answer and solution section, we override the default answer checker in the answer. The replacement is a Perl subroutine that takes as its arguments the correct answer, student answer, and answer hash that is being processed in the answer comparison. Its return value should be $1$ if the student's answer is correct, and $0$ (false) otherwise.

When doing the relational check, you should always put the correct answer on the left and the student answer on the right, e.g., cos($correct) == cos($student) should be used, but cos($student) == cos($correct) should not since the relational operator works from left to right which would mean that the student's answer would be determining the settings (such as domain of function evaluation) instead of the correct answer.

I understand that the answer I enter is cos($student) but I do not know any coding so what's unclear to me is how WeBWork determines what is cos($correct) and what is not. Is it calculated by the computer or is it predetermined by the person who created the question and can a mere typo cause such problems?

答: The "Answer Evaluation" that you're quoting is talking about how WeBWorK evaluates a number or an expression for an answer, neither of which is asked for by this problem.  In your case, the answer was "converges" which means that the question was actually multiple choice (even if it didn't look that way).  This means that either the problem was coded incorrectly or you aren't entering the answer the way that WeBWorK is expecting (for example, you misspell "covnerges").</p>


问:
$-log(x) \in L^p((0,1))$ for $0<p<1$
i proved that $-log(x)\in L^p((0,1))$ for $1<p<\infty$, but how can i prove that it's true for $0<p<1$ too? Any suggestion or hint? Thank you!!

答: Note that $|\log x|\le 1$ on $[1/e,1].$ Hence $|\log x|^p\le 1$ on $[1/e,1]$ for $0<p<\infty.$ So we need worry only about what happens on $(0,1/e],$ where $|\log x|> 1.$ You have shown $|\log x|\in L^2(0,1/e].$ Since $|\log x|^p \le |\log x|^2$ on this interval for $0<p<2,$ we're done.

Title: $-log(x) \in L^p((0,1))$ for $0<p<1$


问:
A New formula For Generating Pythagorean Triples?
(The comments and answers, are no longer relevant to this post other than poestasis's answer, they were caused by misinterpretation)

Using a right triangle with side lengths $(a,b,c)$ where $a , b < c$,
I was thinking about how the area of a Pythagorean triple can be found using the Pythagorean triple right before it and I came across something that worked for a large number of Pythagorean triples,  $12r^2 + a_{r - 1}b_{r-1} = a_rb_r$, a recursive formula in terms of inradius($r$). This seemingly generates a sequence of Pythagorean triples that I could not find used in any other formula. Its important to note that $12r^2$ is  twice the area of Pythagorean triples that stem from side lengths $(3,4,5)$. Using this formula we can find the $1st$ term of sets where the inradius of each Pythagorean triple is $r + r^2k$ ($k$ represents the $k$th term of a sequence) and the relationship between the side lengths are still defined by our recursive formula.

These $1st$ terms are triplets with an even value of $a$ where $r$ increases by $1$:

$(8,15,17),(12,35,37),(16,63,65)...$

Note: We find this using $(8,15,17)$ as we have a recursive formula as well as the knowledge that $r = \frac{a + b - \sqrt{a^2 + b^2}}2$,  which lets us find the side lengths of each Pythagorean triple.

Here is a sample of what they generate:
$$\begin{array}{c|c|c|c}
set_1&15,8,17&33,56,65&51,140,149&69,260,269 \\ \hline
set_2&35,12,37&85,132,157&135,352,377&185,672,697 \\ \hline
set_3 &63,16,65&161,240,289&259,660,709&357,1276,1325& \\ \hline
set_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153  \\ \hline
\end{array}$$

I couldn't seem to find any similar formulas to this one or any method of generating Pythagorean triples that follow this sequence, I am looking for a proof.

答: Connecting Pythagorean triples by inradius is well-known (though kudos to you for finding it on your own!). For example, see this document by Neville Robbins.</p>


问:
Probability of 4 balls in the same jar
I put 10 balls numbered 1 through 10 in a black box. Next to it, I placed two jars, labeled "A" for the first jar and label "B" for the second jar.

I want to distribute the balls in the jars as follows:

1- I take a ball randomly from the box and place it in the "A" jar.

2- I take a ball randomly from the box and put it in the "B" jar.

3- I take a ball randomly from the box and place it in the "A" jar.

...

I will perform these steps until all the balls are distributed in the jars as shown.

Question: What are the chances of balls 1,2,3 and 4 being in the same jar?

答:
Looks like each jar gets $5$ balls.  You want to know the probability that four desired balls end up in jar A or jar B.

Let's look at the probability ($p_\text{A}$) they end up in jar A.  We'll multiply by $2$ at the end.

There are $\pmatrix{10 \\5}$ ways of selecting $5$ balls and putting them in jar A.  How many have $\{1,2,3,4\}$?  Well, there are six other balls, so:

$$p_\text{A} = \frac{6}{\pmatrix{10 \\5}}=\frac{6\cdot5!\cdot5!}{10!}=\frac{ 5 \cdot 4 \cdot 3 \cdot 2}{10\cdot 9 \cdot 8 \cdot 7 }=\frac{1}{42}$$.

The probability that all four balls $\{1,2,3,4\}$ end up in the same jar is thus:

$$p=2p_\text{A} = \frac{1}{21}.$$

Title: Probability of 4 balls in the same jar


问:
Improper Integral via Residue $\int_0^\infty \frac{x^{a-1}\cos(\log_e x)}{e^x+1}dx , 0<a<1$
Evaluate by Residue theory $$\int_0^\infty  \frac{x^{a-1}\cos(\log_e x)}{e^x+1}dx , \quad 0<a<1$$

My try -

$$I= \int_0^\infty  \frac{x^{a-1}\cos(\log_e x)}{e^x+1}dx ,\quad 0<a<1$$
$$I= \int_0^\infty  \frac{x^a\cos(\log_e x)}{x(e^x+1)}dx$$
Take $0<\epsilon<R$. Consider the integration path $C$ formed by the interval $[\epsilon,R]$, the circle $C_R$ of radius $R$ counterclockwise, the interval $[R,\epsilon]$ and the circle $c_\epsilon$ of radius $\epsilon$ clockwise. The function
$$
f(z)=\frac{z^a\cos(\log_e z)}{z(e^z+1)}
$$
is holomorphic in the interior of $C$ (we consider the branch of $z^{a}$ defined on $\mathbb{C}\setminus[0,\infty)$) except for poles at $i\pi (2n+1),n\in \mathbb{Z}$ and branch points of $\log_e z$. Then
$$
\int_C f(z)\,dz=\int_0^Rf(z)\,dz+\int_{C_R} f(z)\,dz+\int_R^0 f(z)\,dz+\int_{c_\epsilon} f(z)\,dz.
$$

答:
Sketch of the residue appproach

Your integral is $\Re w,\,w:=\int_0^\infty g(z)dz$ with $g(z):=\frac{z^{a-1+i}}{e^z+1}$. Replacing $f$ with $g$ in your reasoning, use$$\lim_{\epsilon\to0^+}\int_0^{2\pi}g(\epsilon e^{i\phi})i\epsilon e^{i\phi}d\phi=\lim_{\epsilon\to0^+}\int_0^{2\pi}\frac{\epsilon^{a+i}e^{(-1+ai)\phi}dx}{e^{\epsilon e^{i\phi}}+1}id\phi=0$$to neglect the $c_\epsilon$ contribution; handle $C_R$ similarly. That just leaves$$\int_0^\infty(g(x)-g(xe^{2\pi i}))dx=(1-e^{2\pi (1+(a-1)i)})I.$$Hence$$I=\frac{2\pi i}{1-e^{2\pi (1+(a-1)i)}}\sum_{p\in S}\operatorname{Res}_{z=p}g(z),$$with $S$ the set of poles of $g$, since they each have winding number $1$. I leave you try this using $S=\{\pi i(2n+1)|n\in\Bbb Z\}$.

Alternative approach

@metamorphy suggested using$$\begin{align}\int_0^\infty\frac{z^{s-1}dz}{e^z+1}&=\Gamma(s)\eta(s)\\\implies\int_0^\infty\frac{z^{a-1}\cos\ln zdz}{e^z+1}&=\frac{\Gamma(a-1+i)\eta(a-1+i)+\Gamma(a-1-i)\eta(a-1-i)}{2}.\end{align}$$


问:
What is "X happens whenever Y happens"?
This page says "$X$ happens whenever $Y$ happens" translates to $Y \implies X$. But I feel $Y\implies X$ allows $X$ to be $True$ when $Y$ is $False$, which does not seem to be correct for "$X$ happens whenever $Y$ happens". I feel it should be $X \Longleftrightarrow Y$. Am I correct with this?

答:
I think this might be more of a language problem than a logic problem. The way I understand the sentence, is:

"Everytime $Y$ happens, $X$ happens also."

Looking at my sentence, it may be more clear why this corresponds to $Y \Rightarrow X$, and not $Y \iff X$. Because $X$ can actually be true while $Y$ is false. I don't say anything about $X$ 'acting' on $Y$.

It is probably the usage of the word 'whenever' that causes a problem. In a way this example shows why one must be careful using language when talking about math / logic, since it is sometimes difficult to be absolutely clear what is meant.

I hope this helps.

Title: What is "X happens whenever Y happens"?


问:
Check that a surface over $\mathbb Z$ is regular
Let $X$ be an arithmetic surface over $\mathbb Z$ with smooth generic fibre (which is a curve over $\mathbb Q$). Moreover assume that $X$ is given by one equation $F=0$.

added later: In other words, the generic fibre is expressed as a plane curve over $\mathbb Q$.

Now let $x\in X$ be a closed point. How do I check if $X$ is regular at $X$? Ok, formally I must show that $\mathcal O_{X,x}$ is a regular local ring (of Krull dimension 2). But in more concrete terms what should I do with my $F$? I mean, if $X$ was a variety over a field, I could use the Jacobian criterion, but what about this particular case where the base is $\operatorname{Spec} \mathbb Z$?

Remark: With the term arithmetic surface I mean an integral projective scheme of dimension $2$ which is flat over $\mathbb Z$.

答:
Regularity is a local notion so we can work in affine coordinates $x,y$. The scheme $X$ is automatically regular at every point at which $X/\operatorname{Spec}\mathbb Z$ is smooth by the regularity of $\operatorname{Spec}\mathbb Z$. These are precisely the points satisfying the Jacobian criterion. Now, take a closed point $z$ of $X$ at which $X/\mathbb Z$ is not smooth, you have to show that $X$ is regular at $z$. Since the Krull dimension of $X$ at $z$ is $2$, this is equivalent to saying that the maximal ideal $\mathfrak m$ of $z$ in $\mathcal O_{X,z}$ can be generated by two elements. It is hard to say more without knowing what $F$ you are dealing with. By the answer to this question https://mathoverflow.net/questions/97429/maximal-ideals-of-zx-y you can always write $\mathfrak m=(p,f,g)$ where $p$ is a prime number, $f$ is a polynomial in $x$ irreducible mod $p$, and $g$ is a polynomial in $(x,y)$ irreducible mod $p$; and you could for example try to see if, modulo $F$, one of these three generators is in the ideal spanned by the other two.

As an illustrative example, take $F=xy-n$ for some integer $n\in \mathbb N$. A point $z\in X$ at which $X/\mathbb Z$ is not smooth corresponds to a maximal ideal $\mathfrak m$ of $\mathbb Z[x,y]/(F)$ of the form $(x,y,p)$ where $p$ is a prime dividing $n$. Then, $X$ is regular at $z$ if and only if $\mathfrak m$ is generated by two elements, i.e. if and only if $p^2$ does not divide $n$. In that case, $\mathfrak m$ is spanned by $x$ and $y$.


问:
Finding SCC using DFS algorithm
Where:

SCC - strongly connected components

DFS - depth first search

This is the graph I was trying to "solve" this way is:

So I'm starting from the A vertex and go straight to G, setting proccessing times all the way - A = <1, 0> and G = <7, 0>.

Then I go back, G = <7, 8>, then to H = <6, 9>, I = <5, 10>, E = <4, 11>...is this correct? And what's next? I can go to C = <3, 0>, or to F which was not visited yet...I guess I should do:

F = <12, 0>, D = <13, 0>, D = <13, 14>, F = <12, 15>, E = <4, 16>, C = <3, 17>, B = <2, 18>, A = <1, 19>

Is this correct?

Then I reverse the edges (arrows)...How to get SCC now?

答:
Your question is not so clear. Yes, in the first step of computing strongly connected component you need to do a DFS on the graph. It seems by $$ you denoted $<\mathsf{discovery\_time, finishing\_time}>$ of each vertex in DFS. I think your concept of DFS is not clear. In your graph a valid DFS may proceeded in this order: $A$, $B$, $C$, $E$, $I$, $H$, $G$, $H$, $I$, $E$, $F$, $E$, $C$, $B$, $D$, $B$, $A$. This DFS yields the following $\mathsf{discovery\_time}$ and $\mathsf{finishing\_time}$ pairs: $A=<1, 18>$, $B=<2, 17>$, $C =<3, 14>$, $D = <15, 16>$, $E = <4, 13>$, $F = <11, 12>$, $G = <7, 8>$, $H = <6, 9>$, $I = <5, 10>$.

In the second step you need to reverse the edges of the graph and do another DFS. But in this DFS you need to consider the vertices in decreasing order of $\mathsf{finishing\_time}$s computed during the first DFS. Each DFS tree, in this step, represents a separate SCC---that is all vertices in a particular DFS tree belongs to the same SCC. For example, vertex $A$ has the greatest $\mathsf{finishing\_time}$. If we do a DFS from $A$ with edges reversed, we will get $A, D,$ and $B$ in the DFS tree. And indeed $A, D,$ and $B$ belong to the same SCC. You can find the other SCC yourself.

See Wikipedia or Cormen's Introduction to Algorithms' chapter Elementary Graph Algorithms for detailed discussion both on DFS and finding strongly connected components.

Title: Finding SCC using DFS algorithm

Body:
Your question is not so clear. Yes, in the first step of computing strongly connected component you need to do a DFS on the graph. It seems by $<a,b>$ you denoted $<\mathsf{discovery\_time, finishing\_time}>$ of each vertex in DFS. I think your concept of DFS is not clear. In your graph a valid DFS may proceeded in this order: $A$, $B$, $C$, $E$, $I$, $H$, $G$, $H$, $I$, $E$, $F$, $E$, $C$, $B$, $D$, $B$, $A$. This DFS yields the following $\mathsf{discovery\_time}$ and $\mathsf{finishing\_time}$ pairs: $A=<1, 18>$, $B=<2, 17>$, $C =<3, 14>$, $D = <15, 16>$, $E = <4, 13>$, $F = <11, 12>$, $G = <7, 8>$, $H = <6, 9>$, $I = <5, 10>$.

In the second step you need to reverse the edges of the graph and do another DFS. But in this DFS you need to consider the vertices in decreasing order of $\mathsf{finishing\_time}$s computed during the first DFS. Each DFS tree, in this step, represents a separate SCC---that is all vertices in a particular DFS tree belongs to the same SCC. For example, vertex $A$ has the greatest $\mathsf{finishing\_time}$. If we do a DFS from $A$ with edges reversed, we will get $A, D,$ and $B$ in the DFS tree. And indeed $A, D,$ and $B$ belong to the same SCC. You can find the other SCC yourself.

See Wikipedia or Cormen's Introduction to Algorithms' chapter Elementary Graph Algorithms for detailed discussion both on DFS and finding strongly connected components.


问:
limit proof explanation
Let $\{a_n\}$ be a sequence (for $n \in \mathbb{N}$) such that $\liminf_{n\to\infty} |a_n| = 0$. Show that there is some subsequence $\{a_{n_k}\}$ with $\sum a_{n_k}$ converges.

I found a couple of proofs and was wondering if someone could explain a part of them:

Proof technique 1:

By definition of lim inf, $\liminf_{n\to\infty} |a_n| = 0$ implies that there is an $n_k$ s.t. $|a_{n_k}| < 1$. Arguing by contradiction, suppose $\liminf_{n\to\infty} |a_n| = 0$ and there is no $n$ s.t. $|a_n| < 1$. By definition, $1$ forms a lower bound for $a_n$. For all $n > N$, $1$ then forms a lower bound for $a_n$. Hence, $1 \leq \inf \{a_n: n > N\} \implies \lim_{N\to\infty}\inf \{a_n: n > N\} \geq 1 \implies \liminf_{n\to\infty} |a_n| \geq 1$. Contradiction.

Proof technique 2:

Notice that $\liminf$ is a subsequential limit. So $\liminf_{n\to\infty} |a_n| = 0$ implies that there is an $n_{k2} > n_{k1}$ s.t. $|a_{n_{k2}}| < \frac{1}{2^2}$. Arguing by contradiction, suppose $\liminf_{n\to\infty} |a_n| = 0$ and there is no $n > n_{k1}$ s.t. $|a_n| < \frac{1}{2^2}$. Then for $n > n_{k1}$, $\frac{1}{2^2}$ forms a lower bound for $a_n$. For all $N > n_{k1}$, $\frac{1}{2^2}$ then forms a lower bound for $a_n$, where $n > N$. We have that $\frac{1}{2^2} \leq \inf \{a_n: n>N\} \implies \lim_{N\to\infty}\inf \{a_n: n > N\} \geq \frac{1}{2^2}$. Contradiction.

After either proof above, we have a sequence $\{a_{n_k}\} < \frac{1}{k^2}$. With the comparison test, $\{a_{n_k}\}$ converges.

QED.

Where I'm confused in particular is: Where did those starting conclusions from $\liminf_{n\to\infty} |a_n| = 0$ come from? How do we know that there is an $n_k$ such that $|a_{n_k}| < 1$? How do we know that there is an $n_{k2} > n_{k1}$ such that $|a_{n_{k2}}| < \frac{1}{2^2}$?

Any insight is appreciated.

答: Asserting that a sequence $(x_n)_{n\in\Bbb N}$ of non-negative numbers is such that $\liminf_nx_n=0$ is the same thing as asserting that it has a subsequence whose limit it $0$. So, there is some $n_1\in\Bbb N$ such that $\left|a_{n_1}\right|<1$. And, again since some subsequence of $\bigl(|a_n|\bigr)_{n\in\Bbb N}$ converges to $0$, there is some $n_2$ such that $\left|a_{n_2}\right|<\frac12$ and we may assume, without loss of generality, that $n_2>n_1$. And so on.

Title: limit proof explanation


答:
Your proposition means that if $ Y $ is true, $ X $ will be certainly true.
or in other words, it says that the argument whose premisse is $ Y $ and conclusion is $ X $, is a valid argument.

If $ Y $ is false, we can say nothing about $ X $.

It can be written as
$$Y \implies X$$
it is also what we call a conditionnal.

Title: What is "X happens whenever Y happens"?

Body: This page says "$X$ happens whenever $Y$ happens" translates to $Y \implies X$. But I feel $Y\implies X$ allows $X$ to be $True$ when $Y$ is $False$, which does not seem to be correct for "$X$ happens whenever $Y$ happens". I feel it should be $X \Longleftrightarrow Y$. Am I correct with this?


问:
Field automorphisms send roots into roots
In the post Prove that a field automorphism sends a root into a root, the following problem is discussed:

If $E$ is an extension of $F$, $f(x) \in F[x]$ and $\phi$ is an automorphism of $E$ leaving every element of $F$ fixed, prove that $\phi$ must take a root of $f(x)$ lying in $E$ into a root of $f(x)$ in $E$.

I managed to solve it, using the argument described in the above post, but what I didn't see is where we used that $\phi$ is an auto-morphism. To me, it seems like it should work with any endo-morphism that fixes $F$. Is this true? If not, where were the hypothesis used in the proof of the above fact? And if it is indeed true, why is it?

Thanks in advance and sorry for the silly question!

答: Hint: if you had an endomorphism that wasn't an isomorphism what might its kernel be?</p>


问:
period of a function $\lvert\sin x\rvert+\lvert\cos x\rvert$
I have read that $$y=\lvert\sin x\rvert+ \lvert\cos x\rvert $$ is periodic with fundamental period $\frac{\pi}{2}$.

But Wolfram says it is periodic with period $\pi$.

Please tell what is correct.

Body:
Hint 1: $|\sin(x)|$ and $|\cos(x)|$ have period $\pi$

Hint 2: $(|\sin(x)|+|\cos(x)|)^2=1+|2\sin(x)\cos(x)|=1+|\sin(2x)|$


问:
Find the number of permutations $\sigma \in S_n$ with exactly $c_i$ cycles of length $i$ for each $i$.
Let $n$ be a positive integer, and let $(c_1, \ldots, c_n)$ be a sequence of integers such that $\sum_{i=1}^n ic_i = n$. Show that the number of permutations $\sigma \in S_n$ with exactly $c_i$ cycles of length $i$ for each $i$ is given by $$\frac{n!}{1^{c_1} c_1! 2^{c_2} c_2! \cdots n^{c_n} c_n!}$$

Here, I believe that the approach would be to define a surjection $f: S_n \to X$ and then show that it is surjective, and that $|f^{-1}(\sigma)| = 1^{c_1} c_1! 2^{c_2} c_2! \cdots n^{c_n} c_n!$ for any $\sigma \in X$.

However, I still confused and I don't know how to proceed.

答:
Supposing our sequence $(c_1, c_2, \ldots, c_n)$ is such that
$\sum_{q=1}^n q c_q = n$ with some $c_q$ possibly zero so  that there
are $c_q$ cycles of length $q$ we first have to  select the values from
$n$ that go on every cycle by a multinomial  coefficient

$$\frac{n!}{\prod_{q=1}^n q!^{c_q}}$$

Now a set of $q$ values gives $q!/q$ cycles for a contribution of

$$\prod_{q=1}^n \frac{q!^{c_q}}{q^{c_q}}$$

and the ordering of the $c_q$ cycles of the same length does not matter,
giving

$$\prod_{q=1}^n \frac{1}{c_q!}.$$

Collecting everything we find

$$\frac{n!}{\prod_{q=1}^n q!^{c_q}}
\prod_{q=1}^n \frac{q!^{c_q}}{q^{c_q}}
\prod_{q=1}^n \frac{1}{c_q!}
= \frac{n!}{\prod_{q=1}^n c_q! q^{c_q}}$$

as claimed.


问:
Calculating the normal to a hyper surface
The hyper surface in my lectures is defined as a single function of coordinates which is a constant-

$f(x)= c$

and it’s stated that the normal to the hyper surface is given as

$x_\mu= \nabla_\mu f(x)$

But isn’t the above expression just 0 since $f(x)= constant$ implies

$\nabla_\mu f(x)= \nabla_\mu (constant) =0$

Won’t the above expression be just always 0?

答:
It’s not that $f$ is a constant function, like $f(x, y, z) := 7$. Rather a hypersurface is the set of points where $f$ attains one particular value, like $\{(x,y,z) \in \mathbb R^3: f(x,y,z) = 7\}$. In particular, $f$ is changing in every direction not parallel to the hypersurface so it’s far from a constant.

Edit: Consider $f(x, y, z) = x^2 + y^2 + z^2$. The set $\{(x, y, z) \in \mathbb R^3: f(x, y, z) = 1 \}$ is a sphere with radius 1. The gradient of $f$ is $\nabla f = \langle 2x, 2y, 2z \rangle$, so you see that partial derivatives are nonzero functions. Moreover, for $(a, b, c)$ on the sphere, $(\nabla f)(a, b, c) = \langle 2a, 2b, 2c \rangle$ is a normal vector to the sphere.


答:
The surjection is given by writing the permutation in one line notation, and then breaking into consecutive blocks as follows: $c_1$ blocks of size $1$, $c_2$ block of size $2$, etc. Each of those blocks is then interpreted as a cycle, so the entire thing is a permutation in cycle notation. For example,

\begin{align}
n &= 14\\
(c_1,c_2,c_3,\dots)&=(2,3,2,0,0,0\dots)\\
\pi&=[1, 2, 3, 4, 5, 6, 7, 8, 9 , 10, 11, 12, 13, 14]&\text{one-line notation}\\
f(\pi)&=(1)(2)(3\;4)(5\;6)(7\;8)(9\;10\;11)(12\;13\;14)&\text{cycle notation}
\end{align}

To see why $|f^{-1}(\sigma)|=1^{c_1}c_12^{c_2}c_2!\cdots n^{c_n}c_n!$, note that each of the cycles of length $i$ can be rearranged in $i$ ways, by rotating any of the $i$ elements to the first position, while giving the same cycle. There are $c_i$ cycles of length $i$ for each $i$, contributing the factors of $i^{c_i}$. Then, rearranging the order of the cycles of any given length also results in the same cycle structure. The $c_i$ cycles of any given length can be rearranged in $c_i!$ ways. In the above example, starting with
$$
\pi'=[1,2,5,6,7,8,3,4,10,11,9,14,12,13]
$$
yields the same permutation $f(\pi)$. The only difference is the cycles of length $2$ appear in a different order, and the cycles of length $3$ are written with different elements first.

Title: Find the number of permutations $\sigma \in S_n$ with exactly $c_i$ cycles of length $i$ for each $i$.

Body:
Let $n$ be a positive integer, and let $(c_1, \ldots, c_n)$ be a sequence of integers such that $\sum_{i=1}^n ic_i = n$. Show that the number of permutations $\sigma \in S_n$ with exactly $c_i$ cycles of length $i$ for each $i$ is given by $$\frac{n!}{1^{c_1} c_1! 2^{c_2} c_2! \cdots n^{c_n} c_n!}$$

Here, I believe that the approach would be to define a surjection $f: S_n \to X$ and then show that it is surjective, and that $|f^{-1}(\sigma)| = 1^{c_1} c_1! 2^{c_2} c_2! \cdots n^{c_n} c_n!$ for any $\sigma \in X$.

However, I still confused and I don't know how to proceed.


问:
An example of a "pathological" power-spectral density function?
Suppose that we are given a wide-sense stationary random process $X$ with autocorrelation function $R_X(t)$.  Power spectral density $S_X(f)$ of $X$ is then given by the Fourier transform of $R_X(t)$, i.e. $S_X(f)=\mathcal{F}(R_X(t))$.

I am wondering if there is a valid power spectral density function $S_X(f)$ such that, for a positive integer $n$, the integral over the entire frequency domain of the absolute value of $S_X(f)$ taken to the $n$-th power is not a finite constant.  Formally, is there $S_X(f)$ such that:

$$\int_{-\infty}^{\infty} |S_X(f)|^n df=\infty$$

I know that this is impossible for $n=1$, as $\int_{-\infty}^{\infty} S_X(f) df=R_X(0)=E[X^2]<\infty$, however, I haven't found any result for $n>1$.  Perhaps it's very obvious one way or the other  (though it seems to me that such $S_X(f)$ does not exist, but I can't find a formal proof).  In any case, I would appreciate elucidation.

答:
This is impossible just by definition. </p>

<p>$S_X(f)$, if it exists, is the Radon-Nikodym derivative of the spectral measure $\mu$ with respect to the Lebesgue measure and, being nonnegative, necessarily $L^1$ as you noted. By definition, the Fourier transform of $S_X(f)$ (or $\mu$ in general) is $R_X(t)$. If one insists that $R_X(t)$ lies in the domain of the (inverse) Fourier transform, then Fourier inversion theorem implies that $S_X(f)$ is in fact continuous almost everywhere. A continuous function in $L^1$ lies also in $L^p$ for any $p > 1$.</p>

Body:
This is impossible just by definition.

$S_X(f)$, if it exists, is the Radon-Nikodym derivative of the spectral measure $\mu$ with respect to the Lebesgue measure and, being nonnegative, necessarily $L^1$ as you noted. By definition, the Fourier transform of $S_X(f)$ (or $\mu$ in general) is $R_X(t)$. If one insists that $R_X(t)$ lies in the domain of the (inverse) Fourier transform, then Fourier inversion theorem implies that $S_X(f)$ is in fact continuous almost everywhere. A continuous function in $L^1$ lies also in $L^p$ for any $p > 1$.


问:
How to solve it in a generalised way?
Recently I encountered a sum which is as follows:

Let a number when divided by 9 gives remainder 1 , when divided by 11 gives remainder 2 and when divided by 13 gives remainder 3. so Find ot that number ?

Is there a general method to solve this sum ?
Also if we want to generalize this sum by placing three numbers with 9,11 and 13. How can we do it ? (I think it might be broad question) But How to generalize it ? Hints or hints toward a particular method will be appreciated. Thanks.

My attempt:

I have a quite idea that it might be able to solve using chinese remainder theorem. But I am not able to successfully use it.

答:
Follows immediately by CCRT for congruences in A.P. (Arithmetic Progression), i.e.

$$\begin{align}x \equiv 1\!+\!j\!\!\!\pmod{\color{#c00}{9\!+\!2j}}\! \iff &\, 2x \equiv 2\!+\!\color{#c00}{2j} \equiv 2\!\color{#c00}{-\!9} \equiv -7\!\!\!\pmod{\!1287 \!=\! 9(11)13}\\[.3em]
\iff &\, x \equiv \dfrac{-7}2\equiv\dfrac{1287\!-\!7}2\equiv 640 \end{align}$$

Title: How to solve it in a generalised way?


问:
Find the whole number solutions for W+X+Y+Z = 15 where W, X, Y, Z ≤ 6
I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.

First, the number of non-negative integer solutions for $W+X+Y+Z = 15$ can be calculated using stars and bars:

$$\binom{n+k-1}{n} = \binom{15+4-1}{15}$$

Now, when value of W exceeds 6; i.e., for value of 7 violets the rule of the upper limit for W.
That means the number of violations W can have is among 15-7 = 8 identical items in 4 distinct bins.

Equation for violations becomes W+X+Y+Z = 15-7 = 8

As a result, total number of violations for W = C (n+k-1, k-1) = C (8+4-1, 4-1) = C (11, 3)

⸫ Total number of violations for all 4 bins when 1 bin cross upper limit = C (11, 3) × 4

Since, number of violations 8 is >n/2=15/2=7.5; we need to add in the subtracted repetitions.

When 2 bins cross above upper limit; W+X+Y+Z = 15-7-7 = 1

⸫ Total number of repetitions for all 4 bins when 2 bin cross upper limit = C (1+4-1, 4-1) × 4 = C (4, 3) × 4

When 3 bins cross above upper limit; W+X+Y+Z = 15-7-7-7 = -6 i.e., not possible.

Therefore, the number of integer solutions for $W+X+Y+Z = 15$ where $W, X, Y, Z \leq 6$ is $$\binom{18}{3} – \binom{11}{3} \times 4 + \binom43\times 4$$

答:
The “$\le6”$ constraints are annoying to work with. To fix it,we can substitute $$A=6-W, \quad B=6-X, \quad C=6-Y, \quad D = 6 - Z.$$

Now, the constraints become $A+B+C+D\le 4\times 6 - 15=9$, and $A,B,C,D\ge 0$. Now, you can solve it by standard stars and bars method.

So the answer is $$\binom{9+4-1}{4-1}=220.$$

Whereas the answer I get from your last line is $172$.


答:
Generating the Sets

Given the sets in the question, we can match each triple to one generated by the classical formula: $\left(m^2-n^2,2mn,m^2+n^2\right)$ where $(m,n)=1$, $2\not\mid m-n\gt0$:
$$
\begin{array}{c|c}
\text{set}_0&3,4,5&5,12,13&7,24,25&9,40,41\\\hline
m,n&2,1&3,2&4,3&5,4\\\hline\text{set}_1&15,8,17&33,56,65&51,140,149&69,260,269\\\hline
m,n&4,1&7,4&10,7&13,10\\\hline
\text{set}_2&35,12,37&85,132,157&135,352,377&185,672,697\\\hline
m,n&6,1&11,6&16,11&21,16\\\hline
\text{set}_3 &63,16,65&161,240,289&259,660,709&357,1276,1325\\\hline
m,n&8,1&15,8&22,15&29,22\\\hline
\text{set}_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153\\\hline
m,n&10,1&19,10&28,19&37,28\\\hline
\end{array}
$$
Noticing the pattern in the table above, we get that
$$
\begin{align}
\text{column $j$ of set}_k&=\left(m^2-n^2,2mn,m^2+n^2\right)\\
\text{where }(m,n)&=(1+(2k+1)(j+1),1+(2k+1)j)
\end{align}
$$
Column $0$ is the leftmost column of the table.

To see that $(m,n)=1$, note that $n(j+1)-mj=1$

To see that $2\not\mid m-n\gt0$, note that $m-n=2k+1$

I have added $\text{set}_0$ to the table, following the pattern in the subsequent sets.

These sets do not cover all of the Pythagorean Triples. For example, $(21,20,29)$ is not covered in any of these sets.

Original Answer: Why $\boldsymbol{ab-12r^2=a'b'}$

This answer, and probably many others, shows the following: all relatively prime Pythagorean triples can be written as $\left\{m^2-n^2,2mn,m^2+n^2\right\}$ where $(m,n)=1,\ 2\nmid m-n\gt0$.

The area of a triangle is the inradius times the semi-perimeter. Since a Pythagorean triangle is a right triangle, the area is half the product of the legs. Thus, the inradius is
$$
\begin{align}
\text{inradius}
&=\frac{\text{area}}{\text{semi-perimeter}}\\
&=\frac{mn\left(m^2-n^2\right)}{m^2+mn}\\[9pt]
&=n(m-n)
\end{align}
$$
Suppose that $\{a,b,c\}=\left\{m^2-n^2,2mn,m^2+n^2\right\}$ is a Pythagorean triple. Then
$$
\begin{align}
\overbrace{2mn\left(m^2-n^2\right)}^{\large ab}-12\overbrace{n^2(m-n)^2}^{\large r^2}
&=2m(m+n)n(m-n)-12n(m-n)n(m-n)\\
&=2n(m-n)(m(m+n)-6n(m-n))\\
&=2n(m-n)\left(m^2-5mn+6n^2\right)\\
&=2n(m-n)(m-2n)(m-3n)\\
&=\underbrace{2n(m-2n)\vphantom{\left(n^2\right)}}_{\large e}\underbrace{\left((m-2n)^2-n^2\right)}_{\large d}
\end{align}
$$
where $\{d,e,f\}=\left\{(m-2n)^2-n^2,2n(m-2n),(m-2n)^2+n^2\right\}$ is another Pythagorean triple.

In fact:
$$
\begin{bmatrix}d\\e\\f\end{bmatrix}
=\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}
\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
and, inversely,
$$
\begin{bmatrix}
-1&2&2\\
-2&1&2\\
-2&2&3
\end{bmatrix}
\begin{bmatrix}d\\e\\f\end{bmatrix}
=\begin{bmatrix}a\\b\\c\end{bmatrix}
$$
Note that $d,e,f$ have the same parity as $a,b,c$, respectively.

Because
$$
\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}^T
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&-1
\end{bmatrix}
\begin{bmatrix}
-1&-2&2\\
2&1&-2\\
-2&-2&3
\end{bmatrix}
=\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&-1
\end{bmatrix}
$$
we have $d^2+e^2-f^2=a^2+b^2-c^2$.

Furthermore, as mentioned above,
$$
\begin{align}
\text{inradius}
&=\frac{\text{area}}{\text{semi-perimeter}}\\[6pt]
&=\frac{ab}{a+b+\sqrt{a^2+b^2}}\\
&=\frac{a+b-\sqrt{a^2+b^2}}2
\end{align}
$$

Title: A New formula For Generating Pythagorean Triples?

Body:
(The comments and answers, are no longer relevant to this post other than poestasis's answer, they were caused by misinterpretation)

Using a right triangle with side lengths $(a,b,c)$ where $a , b < c$,
I was thinking about how the area of a Pythagorean triple can be found using the Pythagorean triple right before it and I came across something that worked for a large number of Pythagorean triples,  $12r^2 + a_{r - 1}b_{r-1} = a_rb_r$, a recursive formula in terms of inradius($r$). This seemingly generates a sequence of Pythagorean triples that I could not find used in any other formula. Its important to note that $12r^2$ is  twice the area of Pythagorean triples that stem from side lengths $(3,4,5)$. Using this formula we can find the $1st$ term of sets where the inradius of each Pythagorean triple is $r + r^2k$ ($k$ represents the $k$th term of a sequence) and the relationship between the side lengths are still defined by our recursive formula.

These $1st$ terms are triplets with an even value of $a$ where $r$ increases by $1$:

$(8,15,17),(12,35,37),(16,63,65)...$

Note: We find this using $(8,15,17)$ as we have a recursive formula as well as the knowledge that $r = \frac{a + b - \sqrt{a^2 + b^2}}2$,  which lets us find the side lengths of each Pythagorean triple.

Here is a sample of what they generate:
$$\begin{array}{c|c|c|c}
set_1&15,8,17&33,56,65&51,140,149&69,260,269 \\ \hline
set_2&35,12,37&85,132,157&135,352,377&185,672,697 \\ \hline
set_3 &63,16,65&161,240,289&259,660,709&357,1276,1325& \\ \hline
set_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153  \\ \hline
\end{array}$$

I couldn't seem to find any similar formulas to this one or any method of generating Pythagorean triples that follow this sequence, I am looking for a proof.


问:
Prove the boundedness of a bilinear continuous mapping.
Let $X,Y,Z$ are Banach spaces and $$B:X\times Y\to Z$$ is bilinear and continuous. Prove that there exists $M<\infty$ such that
$$\lVert B(x,y)\rVert \leq M\lVert x\rVert\lVert y\rVert.$$ Is completeness needed here? The bilinearity of $B$ means for a fixed  $x\in X$,$B_x\colon Y\to Z$ is linear and for a fix $y\in Y$, $B^y:X\to Z$ is linear.

Since $X,Y,Z$ are metrizable by norms, I think what we do is to show that $B$ is bounded. Let $E_1,E_2$ are bounded, then $E_1\subset t_1V_1$ and $E_2\subset t_2V_2$ for all large $t_1,t_2$; put $M=\max\{t_1,t_2\}$, so that $$B(E_1\times E_2)\subset B(t_1V_1\times t_2V_2)\subset MB(V_1\times V_2)\subset MW,$$ in which $V_1,V_2$ are balanced neighborhoods of  origins in $X,Y$ respectively and $W$ is a neighborhood of $0$ in $Z$.

Can someone point out if there is some errors in my thought or give a straightforward path?

答:
This post is just a supplement and detailed version of Timon's answer.

I think we may need to change $X_1$ to unit circle. So in this post, $X_1$ will be the unit circle in $X$. And the rest of notations are the same with those in Timon's answer.

Following from Timon's answer, we know that, $\forall x\in X_1,\,y\in Y$, we will have that $|B(x,y)|\le||B_x|| \cdot||y|| \le M \cdot ||y||$, where $M:=\sup_{x\in X_1} ||B_x||<\infty$.

Note that $\forall x\in X_1$, $||x||=1$.

Hence, $\forall x\in X_1,\,y\in Y$, we have that $|B(x,y)|\le M \cdot ||y||\cdot ||x||$

Now for any arbitrary $\alpha\in X,$ we know there exists an $x_0 \in X_1$ and some constant $c$ s.t. $\alpha=c\, x_0.$

Therefore, $$|B(\alpha,y)|=|B(c\,x_0,y)|=|cB(\,x_0,y)|=|c|\cdot|B(\,x_0,y)|\, \le |c|\cdot M \cdot ||y||\cdot ||x_0||$$

$$=M \cdot ||y||\cdot ||c\,x_0||= M \cdot ||y||\cdot ||\alpha||$$

I.e. $|B(\alpha,y)|\le M \cdot ||y||\cdot ||\alpha||,\,\forall \alpha\in X,\,y\in Y$.

$\tag*{$\blacksquare$}$


问:
The functional equation $\delta(n) = \delta(2n) + \kappa(n)$
I am interested in finding a function $\delta(n)$ that solves a functional equation of the form—$$\delta(n) = \delta(2n) + \kappa(n),$$for all integers $n \ge 1$ that are powers of 2, where $\kappa(n)$ is a known function that depends on $n$.  This equation came up when seeking answers to a question of mine about converging polynomials.  For example (Nacu and Peres 2005, proposition 10):

If $\kappa(n) = \dfrac C {\sqrt{2n}}$, then $\delta(n) = \dfrac{(1+\sqrt{2})C}{\sqrt{n}}$, where $C > 0$.

If $\kappa(n) = \dfrac M {4n}$, then $\delta(n) = \dfrac M {2n}$, where $M > 0$.

Unfortunately, I don't know how to solve this functional equation for $\delta(n)$ when $\kappa(n)$ is arbitrary. An example is $\kappa(2^m) = \dfrac M 2 \cdot \dfrac 1 {8\cdot2^m-4}$, which is not a hypergeometric function as rsolve requires. How can I do that?

REFERENCES:

Nacu, Şerban, and Yuval Peres. "Fast simulation of new coins from old", The Annals of Applied Probability 15, no. 1A (2005): 93-115.

答:
As

$$
\delta\left(2^{\log_2 n}\right)=\delta\left(2^{\log_2 (2n)}\right)+\kappa(n)
$$

making $d(\cdot)=\delta\left(2^{(\cdot)}\right)$ and $z=\log_2 n$ we have the recurrence

$$
d(z)=d(z+1)+\kappa\left(2^z\right)
$$

This recurrence has the solution

$$
d(z) = c_1 -\sum_{i=0}^{z-1}\kappa\left(2^i\right)
$$

and going backwards we arrive to

$$
\delta(n) = c_1 - \sum_{i=0}^{\log_2 n -1}\kappa\left(2^i\right)
$$

now choosing $\kappa(n) = \frac{c_0}{\sqrt{2n}}$ we have

$$
\delta(n) = c_1-\frac{\left(1+\sqrt{2}\right)\left(\sqrt{n}-1\right)}{\sqrt{n}}c_0
$$

and for $\kappa(n) = \frac{M}{4n}$ then

$$
\delta(n) = c_1-\frac{M (n-1)}{2 n}
$$

and finally for $\kappa(n) = \frac{M}{2}\frac{1}{8n-4}$ ...


问:
Can probabilities be calculated using cardinality?
The probability that a positive natural number drawn at random is odd, is $\frac{1}{2}$.

Is there a way of calculating this fact using the cardinalities of the sets?  Or is there some measure of cardinality, or of the number of elements within each set, which captures this fact?

If we say the sample space is $\mathbb{N}$ and the successful outcomes are $O=\{2n-1:n\in\mathbb{N}\}$ then we might (obviously incorrectly) infer from the cardinalities of the sets that:

$(\lvert\mathbb{N}\rvert=\lvert O\rvert)\implies P(n$ is odd$)=1$

It would seem to me that the closest we can come is to say the densities of $O$ and $\mathbb{N}\cap O$ in $\mathbb{N}$ are equal to each other and make our deduction by that means.  Is that the normal approach within set theory?

答:
Something slightly deviates, but might still work on your case.

(Here natural numbers refers to positive integer).

Fix the sample space to be no greater than $n$ (denoted $P_n$), it is easy to show that

$\frac{1}{2}\le P_n(odd)\le \frac{n+1}{2n}$.

Interpretation in terms of infimum.

It can be easily shown $\inf\{P_n(odd)\} =\frac{1}{2}$.

However, this does not mean random select a number in $\mathbb{N}$ and the probability of oddity is 0.5.

Interpretation in terms of a limit.

Basic limit property shows $\lim\limits_{n\rightarrow \infty} P_n(odd)=\frac{1}{2}$.

However, the interpretation should be careful:

for all $\epsilon>0$, there exists $N\in \mathbb{N}$ such that, if $n\ge N$, then for the sample space $\Omega: \{i, i\in \mathbb{B}\ and\ i\le n\}$, and random variable $X: \Omega\rightarrow \Omega$ by $X(x)=x$, $|P_n(odd)-\frac{1}{2}|\le \epsilon$.


问:
Solve the equation $\sqrt3\left|\cos x\right|=1+\cot x$
Solve the equation
$$\sqrt3\left|\cos x\right|=1+\cot x$$

My work:

i)
Let $\cos x \ge 0$. Then we have got the equation
$$\sqrt3 \cos x =1+\frac{\cos x}{\sin x}$$
$$\sqrt3 \cos x \sin x =\sin x+\cos x$$
$$\frac {\sqrt3}2\sin 2x=\sqrt2 \sin (x+\frac{\pi}4)$$

What next?

答:
Hint: with $y:=x+\pi/4,\,s:=\sin y$ this is $-\tfrac{\sqrt{3}}{2}\cos(2y)=\sqrt{2}\sin y$, i.e. $\sqrt{3}s^2-\sqrt{2}s-\tfrac{\sqrt{3}}{2}=0$.

Note if we shift $x$ by $\pi$ we change the sign of $\cos x$ while preserving $\cot x,\,|\cos x|$, so the $\cos x\ge0$ solutions easily give the others.


问:
What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

For the case when $x=y$ it is clear, what about the case when $x\ne y$?

答:
The only integer solutions are $x=y$ and also $0^0=1^1$, provided that $0^0$ is defined. This is because $x^x$ is strictly increasing on $(1/e,\infty)$:

On the interval $[0,1]$, there are infinitely many real solutions of $x^x=y^y$ where $x \neq y$. This is because when $x=1/e$, $y=e^{-1/e}$. Then, when $x=1$, $y=1$. Since $x^x$ is continuous on $[1/e,1]$, by the intermediate value theorem, $x^x$ must take on every real value between $e^{-1/e}$ and $1$. Similarly, on the interval $[0,1/e]$, $x^x$ is also continuous, and so must take on every real value between $e^{-1/e}$ and $1$. This means that for every $a \in (e^{-1/e},1]$, there are two possible solutions for $x^x = a$. So there are infinitely many real solutions to $x^x=y^y$, $x \neq y$.

As NoNames has pointed out, there are also infinitely many rational solutions, since the solutions can be parameterised as
$$
x=\left(\frac{t}{t+1}\right)^t,\quad y=\left(\frac{t}{t+1}\right)^{t+1} \, ,
$$
for $t>0$. If $t$ is an integer, then this gives us a valid rational solution.


问:
In set theory what is A* and how does it differ from the power set of A?
I am having a difficult time understanding the use of $A^*$ and how it differs from the usage of a power set.

$A^*$ is defined as "the set of all (finite) lists over A". I understand $\wp(A)$ to be the power set of A defined as "the set of all subsets of A."

How is "all lists over A" different from "all subsets of A"?

Here are both definitions as denoted in my textbook:

$$A^* := \{[a_1,...,a_n]\ |\ n \in \mathbb N;\ a_1,...,a_n \in A\}$$

$$\wp(A) = \{X\ |\ X \subseteq A\}$$

答:
The two have basically nothing in common. $A^*$ is the set of all (finite) words you can write using the elements of $A$ as letters, whereas $P(A) $ is the set of all subsets of $A$.

As a very simple example, if $A = \{a\} $, then $A^* = \{\varepsilon, a, aa, aaa, aaaa, \dots\} $ ($\varepsilon $ is the empty word) whereas $P(A) =\{\emptyset, \{a\} \} $.


问:
Why is the derivative of this Gaussian Function negative before being positive?
From pg. 61 of Principles of Quantum Mechanics, the Guassian $g_Δ$ is defined as

$$
g_Δ(x-x') = \frac{1}{(πΔ^2)^{1/2}} \exp \left[ -\frac{(x-x')^2}{Δ^2} \right]
$$

where the Gaussian is centered at $x'=x$, has width $\Delta$, maximum height $(π Δ^2)^{-1/2}$, and has unit area (independent of $Δ$). In case it matters, we can assume that $Δ ∈ \mathbb{R}$.

Problem: The book goes on to display the derivative of $g_\Delta$ as follows:

It seems to me the derivative graph (b) is wrong, since it is negative when $x' < x$ and positive when $x' > x$. Isn't this exactly backwards, and if not, why?

EDIT for bounty: The crux of my confusion lies in understanding/getting a different viewpoint for why

$$
\frac{dg_\Delta(x-x')}{dx} = -\frac{dg_\Delta(x-x')}{dx'}
$$

Moreover, is this actually a general theorem, or is this using something special about $g_\Delta$ in particular?  That is, if we let $x,' \in \mathbb{R}$ and consider $f(x-x')$ as an arbitrary real function, do we have in general that

$$
\frac{df(x-x')}{dx} = -\frac{df(x-x')}{dx'}?
$$

答:
Rotate your coordinate system (x,x') by 45 degrees. For a function $g_\Delta (x-x')$ even in $x-x'$, you have
$$
\frac{ d g_\Delta (x-x')}{dx} =- \frac{d g_\Delta (x-x')}{d x' },  \tag{1.10.20}
$$
so the ordinate of (b) is an odd function, by convention called $  g_\Delta ' (x-x')$, but really $- d g_\Delta (x-x')/d x' $, so minus the derivative of (a) w.r.t. x'.

To the left of x, this is negative, because $  g_\Delta ' (x-x')>0.$ Its minimum is at $x'=x-\epsilon$, and its maximum is at $x'=x+\epsilon$. So
$$
  g_\Delta ' (x-x') \sim   \delta (x'-\epsilon-x) -\delta(x'-x +\epsilon )
$$
sampling the function f as the author claims, correctly.

Note on your additional edit:

Define $y\equiv x-x'$, the light-cone variable. It then follows that
$$
\frac{d }{dx} f(y) + \frac{d }{dx'} f(y) = \frac{d y }{dx}  \frac{d f(y) }{dy}  + \frac{d y }{dx'}  \frac{d f(y) }{dy}\\
 =   \frac{d f(y) }{dy}  - \frac{d f(y) }{dy}=0.
$$


问:
Let $|X|$ be a random variable in $(\mathbb{R} , \mathbb{B}(\mathbb{R}))$. Is it true that $X$ is also a random variable?
Let $|X|$ be a random variable in $(\mathbb{R} , \mathbb{B}(\mathbb{R}))$. Is it true that $X$ is also a random variable?

I tried to prove it by Borel sigma algebra defnition of an r.v., but that appeared to be a wrng proof. My guess is that an answer is "False", but I have no idea how to prove it

I guess we just need to prove that if f(x) is an r.v. and $\not \exists $ function $ f^{-1}$ and automatically our $X$ wouldn't be an r.v. since $f^{-1}$ isn't a function

答:
Take any non-measurable set $A\subseteq\Omega$ from the sample space, and define:

$X(\omega)=\begin{cases}
1 &: \omega \in A \\
-1 &: \omega \not\in A
\end{cases}$

This is clearly not a random variable, as $\{X=1\}$ is not measurable. But $|X|$ is just the constant function $1$, which is a random variable.

I assumed there is a non-measurable set in $\Omega$. If all the subsets of $\Omega$ are measurable then any function is a random variable.


答: The answer is indeed false. Fix the probability space $(\mathbb{R}, B(\mathbb{R}), \gamma)$ where $\gamma$ is the standard gaussian measure. Let $A$ be a non-measurable subset of $\mathbb{R}$ and let $X = 1_A - 1_{A^c}$. Then $|X| = 1$ is measurable but $X$ is not so that $X$ is not a random variable.

Body:
Let $|X|$ be a random variable in $(\mathbb{R} , \mathbb{B}(\mathbb{R}))$. Is it true that $X$ is also a random variable?

I tried to prove it by Borel sigma algebra defnition of an r.v., but that appeared to be a wrng proof. My guess is that an answer is "False", but I have no idea how to prove it

I guess we just need to prove that if f(x) is an r.v. and $\not \exists $ function $ f^{-1}$ and automatically our $X$ wouldn't be an r.v. since $f^{-1}$ isn't a function


答:
There are several differences:

Finite lists are ordered (and allow duplicates) while sets are not: e.g. $$[1,2,1]\quad\mbox{and}\quad[2,1,1]$$ are different lists, while $$\{1,2\}=\{2,1\}=\{1,2,1\}.$$

Finite lists are finite while elements of $\mathcal{P}(A)$ (= subsets of $A$) need not be: if we take $A=\mathbb{N}$, then the set of even numbers is an element of $\mathcal{P}(A)$ but there is no finite sequence which in any sense "corresponds" to it. Meanwhile, subsets of $A$ are at most as big as $A$, while finite lists can have arbitrarily large finite lengths - this means that if $A$ is finite there will be finite lists of elements of $A$ which are "longer than $A$ is big" (consider e.g. $A=\{1,2\}$ and the finite list $[1,2,1,2,1,2]$).

So $A^*$ and $\mathcal{P}(A)$ are really very different objects.

On a technical note, it's a good idea at this point to give a formal definition of "finite sequence." The definition you've given isn't really rigorous, since the notation "$[...]$" is itself undefined.

The most convenient one winds up being: "a finite list (of elements of some set $X$) is a function $f$ with domain $\{i\in\mathbb{N}:i<n\}$ for some $n\in\mathbb{N}$ (and codomain $X$)." We think of $f(i)$ as the $i$th term of $f$.

This winds up being a very useful definition in that it generalizes broadly - see e.g. here.

Title: In set theory what is A* and how does it differ from the power set of A?

Body:
I am having a difficult time understanding the use of $A^*$ and how it differs from the usage of a power set.

$A^*$ is defined as "the set of all (finite) lists over A". I understand $\wp(A)$ to be the power set of A defined as "the set of all subsets of A."

How is "all lists over A" different from "all subsets of A"?

Here are both definitions as denoted in my textbook:

$$A^* := \{[a_1,...,a_n]\ |\ n \in \mathbb N;\ a_1,...,a_n \in A\}$$

$$\wp(A) = \{X\ |\ X \subseteq A\}$$


问:
Monster double integral
Are there methods to analytically compute this integral, composed of the maximum function, $max(\exp(x_{0})-\exp(z_{0}), 0)$, and exponential functions?

$$\iint_{-\infty}^{\infty}\tilde{A}\max(\exp(x_{0})-\exp(z_{0}), 0)\exp(A_{2}(x_{M}-x_{0})^2)\exp(2A_{2}\alpha_{1}(x_{M}-x_{0}))\exp(-2A_{2}\Omega(x_{M}-x_{0}+\alpha_{1})(z_{M}-z_{0}))\exp(A_{2}\Omega^2(z_{M}-z_{0})^{2})\exp(A_{5}(z_{M}-z_{0}))\exp(A_{6}(z_{M}-z_{0})^{2})dx_{0}dz_{0}$$

$\tilde{A},A_{2},\alpha_{1},\Omega,A_{5},A_{6},x_{M},z_{M}$ are constants.
Specifically, $\tilde{A},A_{2},A_{5},A_{6}$ are negative, while the others take on any real number.

So now let $x_{M}-x_{0}= {x}'$ and $z_{M}-z_{0}= {z}'$.

We transform the integral to:

$$\iint_{\infty}^{-\infty}\tilde{A}\max(\exp(x_{M}-{x}')-\exp(z_{M}-{z}'), 0)\exp(A_{2}({x}')^2)\exp(2A_{2}\alpha_{1}({x}'))\exp(-2A_{2}\Omega({x}')({z}'))\exp(A_{2}\Omega^2({z}')^{2})\exp((A_{5}-2A_{2}\Omega\alpha_{1})({z}'))\exp(A_{6}({z}')^{2})d{x}'d{z}'$$

where the limits have changed because of the substitution.

The exponent can now elegantly be written as:

$$\left ( {x}', {z}' \right )\begin{pmatrix} A_{2} & -A_{2}\Omega \\ -A_{2}\Omega & A_{6}+A_{2}\Omega^2 \end{pmatrix}\left ( {x}', {z}' \right )^{T} +\left ( 2A_{2}\alpha_{1},(A_{5}-2A_{2}\Omega\alpha_{1}) \right )\left ( {x}', {z}' \right )^{T}$$

and by letting $ \hat{A}=\begin{pmatrix} A_{2} & -A_{2}\Omega \\ -A_{2}\Omega & A_{6}+A_{2}\Omega^2 \end{pmatrix}$ and $ J=\left ( 2A_{2}\alpha_{1},(A_{5}-2A_{2}\Omega\alpha_{1})\right )$

The integral becomes
$$\iint_{\infty}^{-\infty}\tilde{A}\max(\exp(x_{M}-{x}')-\exp(z_{M}-{z}'), 0)\exp(\left ( {x}', {z}' \right )\hat{A}\left ( {x}', {z}' \right )^{T}+J\left ( {x}', {z}' \right )^{T})d{x}'d{z}'$$

I have tried doing a simpler integral:

$\iint_{-\infty}^{\infty}\tilde{A}\max(\exp(x_{0})-\exp(z_{0}), 0)\exp(A_{1}(x_{M}-x_{0}-\alpha_{1})^2)\exp(A_{6}(z_{M}-z_{0}-\alpha_{2})^{2})dx_{0}dz_{0}$

but still can not reach a closed-form solution involving error function, dilogarithms, etc ...

答:
HINT.

The first integral can be presented in the form of
$$I=\iint\limits_{-\infty}^{\infty}\tilde{A}\max\left(e^{\large x_0}-e^{\large z_0}, 0\right)e^{\large P_{20}(\vec v_0,x_0,z_0)}\,\text dz_0\,\text dx_0,$$
where
$$\vec v_0 = \{x_M,z_M,A_2,A_5,A_6,\alpha_1,\Omega\},$$
$$P_{20}(\vec v_0, x_0, z_0) = A_{2}\bigg((x_{M}-x_{0})^2+2\alpha_{1}(x_{M}-x_{0})-2\Omega(x_{M}-x_{0})(z_{M}-z_{0}))$$
$$+\Omega^2(z_{M}-z_{0})^{2}\bigg)
+(A_{5}-2\Omega\alpha_1)(z_{M}-z_{0})+A_{6}(z_{M}-z_{0})^{2},$$
or
$$I=\iint\limits_{-\infty}^{\infty}\tilde{A}\max\left(e^{\large x_C+x_M}-e^{\large z_C+z_M
}, 0\right)e^{\large P_{2C}(\vec v_C,x_C,z_C)}\,\text dz_C\,\text dx_C,\tag1$$
where
\begin{cases}
\vec v_C = \{A_2,A_5,A_6,\alpha_1,\Omega\}\\[4pt]
P_{2C}(\vec v_C, x_C, z_C) = A_{2}\big(x_C^2-2\alpha_1x_C-2\Omega x_C z_C+\Omega^2z_C^2\big)\\[4pt]
\qquad\qquad\qquad\;-(A_{5}-2\Omega\alpha_1)z_C+A_{6}z_C^2.\tag2
\end{cases}
Transformation of the coordinates in the form of
$$x_C+iz_C = (x_R+iz_R)e^{\large i\varphi} + x_D +iz_D\tag3$$
should lead from $(2)$ to the expression in the form of
$$P_{2R}(\vec p, x_R, z_R)= p_{xx} x_R^2+p_{zz} z_R^2+p_0.\tag4$$
This allows to get a system of equations for unknowns $\;x_D,z_D,\varphi.\;$

Really, from $(3)$ should

$$\dbinom{x_C}{z_C}=\dbinom{x_R\cos\varphi-z_R\sin\varphi+x_D}
{x_R\sin\varphi+z_R\cos\varphi+z_D},$$

$$P_{2C}(\vec v_C, x_C, z_C) = p_{xx}x_R^2+p_{xz}x_Rz_R+p_{zz}z_R^2+p_x x_R+p_z z_R+p_0,$$
where $p_{xx},p_{xz},p_{zz},p_{x},p_{z},p_{0}$ are functions of $v_C$.

Obtaining $x_D,z_D,\varphi$ from the system $p_{xz}=p_x=p_z=0,$ we should get $(4).$

At the same time, is known closed form of the integral
$$I=\int\limits_{-\infty}^\infty \int\limits_{ax+b}^\infty e^{\large -x^2-z^2} \, \text dx\, \text dz = \dfrac\pi2 \operatorname{erfc}\left(\dfrac b{\sqrt{1+a^2}}\right)\tag5$$
(pointed by DinosaurEgg), and this allows to get the closed form of the given integral.

Let $x=r\cos t,\,z=r \sin t.$

If $\;a>0,\,b>0,\;$ then the angle coordinate corresponds to line $\;z=ax+b,\;$ when $\;t\in(\arctan a,\pi+\arctan a).$

The polar radius can be obtained from the equation $\;a r \cos t +b = r \sin t,\;$
with the result $r=\dfrac b{|\sin t - a\cos t|}=\dfrac b{\sqrt{a^2+1}\cos(t-\arctan a)}.$
Therefore,
$$I=\int\limits_{\arctan a}^{\pi+\arctan a} \int\limits_{\large \frac b{\sqrt{a^2+1}\cos(t-\arctan a)}}^\infty e^{\large -r^2} \, \text dr\, \text dt
=\dfrac12 \int\limits_{0}^{\pi} e^{\left({\large \frac b{\sqrt{a^2+1}\cos t}}\right)^2} \, \text dt,$$
$$I= \dfrac\pi2 \operatorname{erfc}\left(\dfrac b{\sqrt{1+a^2}}\right).$$

About the factor with $\;\max\;$ function:

Equation $(3)$ defines a linear transformation.

Exponential function is monotonic.

$\max\left(e^{\large x_C+x_M}-e^{\large z_C+z_M}, 0\right)=\left(e^{\large x_C+x_M}-e^{\large z_C+z_M}\right)h\left(x_C+x_M-z_C-z_M\right).$

$\;I=I^\star - I^\diamond,\;$ where
$$I^\star=\iint\limits_{-\infty}^{\infty}\tilde{A}h\left(x_C+x_M-z_C-z_M\right)e^{x_C+x_M+\large P_{2C}(\vec v_C,x_C,z_C)}\,\text dz_C\,\text dx_C,$$
$$I^\diamond =\iint\limits_{-\infty}^{\infty}\tilde{A}h\left(x_C+x_M-z_C-z_M\right)e^{z_C+z_M+\large P_{2C}(\vec v_C,x_C,z_C)}\,\text dz_C\,\text dx_C,$$
$\hspace{100mu}h(t)=\begin{cases}
1,\; \text{if} \;t>0\\
0,\; \text{otherwize}
\end{cases}\quad$ is the Heaviside step function.

Therefore, we have $\;P_{2C}^\star =P_{2C}+x_C+x_M,\;P_{2C}^\diamond =P_{2C}+z_C+z_M,\;$ with the same transformation technique.

On the other hand:

Equation $(4)$ can define ellipse, parabola or hyperbola.

Integral $(1)$ diverges in the hyperbolic case.

Integral $(1)$ can be easily calculated in the parabolic case.

Integral $(1)$ can be calculated via $(5)$ in the elliplic case.

Title: Monster double integral