问:
Can you solve a linearly independent ODE's homogeneous part without the particular solution.
For example $y''-6y'+9y=xe^{3x},\hspace{5mm} y(0)=0,y'(0)=0 $

I was wondering if it is possible to compute the homogenous solution even before working out the particular solution to this equation.

As an aside, I was wondering if the initial conditions are (0,0) for y' and y then is it not the case that I don't need the particular solution. As, in effect I am only lookin at the homogenous side of the equation.

I think my misunderstanding comes form the fact that in $y_{c}=y_{p}+y_{h}$ i see $y_{p}$ as a translation or rotation of the solution space away from the origin, and therefore if the inital condition are centered at (0,0) then I know the solution must exist in $y_{h}$.

答:
Solving the problem $$y''-6y'+9y=xe^{3x},\hspace{5mm} y(0)=0,y'(0)=0$$

as if it was a homogenous problem and plugging in indeed results in the actual (IVP-solved) general solution to the inhomogenous problem.

General solution:

$$y(x)=\frac{1}{6}e^{3x}x^3+e^{3x}c_1+e^{3x}c_2$$
IVP-General solution:
$$y(x)_{IVP}=\frac{1}{6}e^{3x}x^3$$

A counter-example would be:

$$y''-6y'+9y=e^{3x}+cos(x),\hspace{5mm} y(0)=0,y'(0)=0$$

Where the general solution is:

$$y(x)=\frac{2}{25}\cos(x)-\frac{3}{50}\sin(x)+e^{3x}\left(\frac{1}{2}x^2+\frac{1}{2}c_1x+\frac{1}{2}c_2\right)$$

The IVP-general solution on the other hand is:

$$y(x)_{IVP}=\frac{2}{25}\cos(x)-\frac{3}{50}\sin(x)+e^{3x}\left(\frac{1}{2}x^2+\frac{15}{50}x-\frac{4}{50}\right)$$

In which the constants $c_1$ and $c_2$ are clearly not zero.


问:
What should I learn next for pure math?
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

答:
I think you are correct to lean toward analysis. It is the usual first step to pure math as you describe your inclination. Linear algebra can be done on several levels, but I think is best done in a rigorous context after real analysis.

In real analysis you will become familiar with proofs, begin to accumulate a working, ingrained math vocabulary as well as learn about topology, metric spaces, concepts such as convergence, continuity, etc. There is also usually a component presenting the material of calculus on a rigorous basis. This will give you a feel for the difference between operational mechanics and the pure math behind it.

I would suggest you take a look at this free set of notes of lectures given by Vaughan Jones (Fields Medal winner, equiv. Nobel Prize in math). They are really beautiful, are self-contained, and build nicely from a level that does not  require prior experience.

https://sites.google.com/site/math104sp2011/lecture-notes

ContentLicense: CC BY-SA 3.0

Title: What should I learn next for pure math?


问:
How to understand $\Gamma(\alpha,1)$ distribution when $\alpha$ is large?
It is said for large $\alpha$, $\operatorname{Gamma}(\alpha,1) \approx \operatorname{Normal}(\alpha,\alpha)$. Can anyone tell me why?

We know the mgf of a $\operatorname{Gamma}(\alpha, \beta)$ distribution is given by

$$M_X(t)=\left(\frac{1}{1-\beta t}\right)^\alpha, \quad t<\frac{1}{\beta}.$$

Hence, given $\beta=1$, we have mgf $\left(\frac{1}{1- t}\right)^\alpha$, $t<1$.

Then $$M_X{\left(\frac{t-\alpha}{\sqrt{a}}\right)}=\left(\frac{1}{1- \frac{t-\alpha}{\sqrt{a}}}\right)^\alpha, \quad t<1.$$

How to get the limit when $\alpha\to\infty$ ?

答:
Heuristically speaking, since a $\Gamma(n,1)$ is the sum of $n$ independent exponential variables with mean $1,$ and since the exponential distribution with mean $1$ has variance $1$,  the central limit theorem says that this will be approximately normal for large $n$, with mean $n$ and variance $n.$ More precisely, $ (S_n-n)/\sqrt{n}$ will converge in distribution to a standard normal, where $S_n$ is the sum of $n$ independent exponentials and thus has a $\Gamma(n,1)$ distribution.

Somewhat more formally, the $\Gamma(\alpha,1)$ distribution has pdf $f_X(x) = \frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{-x},$ from which we can compute a characteristic function $\phi_X(t) = E(e^{itX})=(1-it)^{-\alpha}.$ Then, shifting and rescaling, the characteristic function of $Z = (X-\alpha)/\sqrt{\alpha}$ is $$\phi_Z(t) = E(e^{it(X-\alpha)/\sqrt{\alpha}}) = e^{-it\sqrt{\alpha}}(1-it/\sqrt{\alpha})^{-\alpha},$$ which in the limit of $\alpha\to\infty$ is $e^{-\frac{1}{2}t^2},$ exactly the characteristic function of a standard normal.

It's maybe a little bit more burdensome, but you don't need to go through characteristic functions... just transforming the PDF gives $$ f_Z(z) =\frac{\alpha^{\alpha-1/2}}{\Gamma(\alpha)}(1+z/\sqrt\alpha)^{\alpha-1}e^{-\alpha} e^{-z\sqrt \alpha}$$ for $z > -\sqrt\alpha.$ And then with the help of the Stirling formula $\Gamma(\alpha)\sim \sqrt{2\pi}\alpha^{\alpha-1/2} e^{-\alpha}$ you can compute a limit of $\frac{1}{\sqrt{2\pi}}e^{-z^2/2}.$

(Edit in response to request in the comments.) As for how to compute the limit in the MGF case, we want the limit of $e^{-t\sqrt\alpha}(1-t/\sqrt\alpha)^{-\alpha}.$ This is probably easiest seen by taking logs $$ \log(M) = -t\sqrt\alpha -\alpha\log(1-t/\sqrt\alpha).$$ The taylor expansion of $\log(1-x)$ at $x=0$ is $-x-x^2/2+O(x^3),$ so we get $\lim\log(M) = t^2/2,$ so $\lim M = e^{t^2/2}.$

Note in your edit you are not correctly handling the MDF of a shifted and rescaled variable, which is not the same thing as plugging in a shifted and rescaled $t.$ If $Z=(X-\mu)/\sigma,$ $$M_Z(t) = E(e^{tZ}) = E(e^{t(X-\mu)/\sigma}) = E(e^{Xt/\sigma}e^{-\mu t/\sigma})=e^{-\mu t/\sigma}E(e^{X(t/\sigma)})= e^{-\mu t/\sigma} M_X(t/\sigma).$$

ContentLicense: CC BY-SA 4.0


问:
Do Anosov flows exist on two dimensional compact manifolds?
Question:

Let $M$ denote a two-dimensional, smooth, compact Riemannian manifold (without boundary). If $\phi:\mathbb{R}\times M \rightarrow M$ is a smooth flow, then prove that $\phi$ is not an Anosov flow.

Motivation:

I am new to the field of dynamical systems. It seems to be implicit in the literature that Anosov flows do not exist on two dimensional manifolds (see http://www.sciencedirect.com/science/article/pii/S0022039605002767) but I would like to know the proof.

My current understanding:

Assume $\phi$ is Anosov and argue for a contradiction. There is a splitting of the tangent space $T_x M = E_x^s \oplus E_x^u \oplus E_x^c$ at each $x\in M$, where $E_x^c$ is the one-dimensional subspace spanned by the direction of the flow, and the stable and unstable spaces have dimension independent of $x$. Thus, without loss of generality, suppose that $E_x^u = \{0\}$ and $\text{dim}(E_x^s)=1$. The differential of the vectors $v$ in $E_x^s$ satisfy $\|d\phi_tv\| \lesssim e^{-\gamma t} \|v\|$ for all $t\geq 0$ and some $\gamma >0$. I guess that I now need to use the fact that $M$ is compact to obtain a contradiction but I don't see how to proceed. Any help would be much appreciated.

答:
This is by a trivial dimension count.

As I suspect this is missed due to vague definitions, here is a short summary (which is not necessarily canonical for contemporary research, even though it is faithful to the original definition by Anosov, as far as I'm concerned):

If $M$ is a $C^1$ manifold (not necessarily compact) and $\mathfrak{g}$ is a $C^0$ Riemannian metric on $M$, then a $C^1$ flow $\phi_\bullet: \mathbb{R}\to \operatorname{Diff}^1(M)$ (not necessarily volume preserving) with generator $X$ (i.e. a nowhere-zero vector field on $M$ such that $\forall t\in\mathbb{R},\forall p\in M: X(\phi_t(p))=\frac{d}{dt} \phi_t(p)$) is $\mathfrak{g}$-Anosov if there is a continuous $\phi_\bullet$-invariant splitting $TM = S(\phi_\bullet) \oplus \langle X\rangle \oplus U(\phi_\bullet)$, where $\langle X \rangle$ is the rank $1$ subbundle of $TM\to M$ generated by $X$; $S(\phi_\bullet)$ is a nontrivial subbundle consisting of vectors contracting exponentially fast w/r/t/ $\mathfrak{g}$ under the action of $\phi_\bullet$ and $U(\phi_\bullet)$ is a nontrivial subbundle consisting of vector expanding exponentially fast w/r/t/ $\mathfrak{g}$ under the action of $\phi_\bullet$. $\langle X\rangle$ is the center bundle of $\phi_\bullet$, $S(\phi_\bullet)$ is the (strong-)stable bundle of $\phi_\bullet$ and finally $U(\phi_\bullet)$ is the (strong-)unstable bundle of $\phi_\bullet$. By definition both $S(\phi_\bullet)$ and $U(\phi_\bullet)$ have rank at least $1$, meaning that $TM$ has rank at least $3$, i.e. $\dim(M)\geq 3$. Thus the case $U(\phi_\bullet)_x=E^u_x=0$ is discarded by definition.

(When $M$ is compact, the dependency on $\mathfrak{g}$ can be dropped: if $M$ is compact and $\phi_\bullet$ is $\mathfrak{g}_0$-Anosov for some $C^0$ $\mathfrak{g}_0$, then it is $\mathfrak{g}$-Anosov for any $C^0$ $\mathfrak{g}$.)

Anosov originally called Anosov flows (on closed manifolds) (U)-flows (alternatively $(\Upsilon)$-flows). Here is the relevant part from his monograph Geodesic Flows On Closed Riemannian Manifolds With Negative Curvature (p. 6):

Just to tie in with the paper you cited, a $\mathfrak{g}$-Anosov flow $\phi_\bullet$ is codimension 1 if at least one of $S(\phi_\bullet)$, $U(\phi_\bullet)$ is rank $1$.

Finally let me mention that there are generalizations to trivial stable or unstable bundles (partial hyperbolicity, beyond-uniform hyperbolicity, ...) that cover the case you were considering; the Anosov case is very strict and structured compared to these.


问:
Trigonometric Generation Function for $1, 3, 6, 7, 9, 12, 13, 15, 18 \cdots$
I'm looking for a trig function (that is, I want a non-"modular arithmetic" answer) to generate $1, 3, 6, 7, 9, 12, 13, 15, 18...$,
for an example see Function which creates the sequence 1, 2, 3, 1, 2, 3, ...

答:
Starting at $a_0=1$ if we subtract $2n+1$ from $a_n$ you get:

$$(a_n-2n-1)_n=(0,0,1,0,0,1,\cdots)$$

This can be written as:

$$a_n-2n-1 = \frac13\left(1+\cos\frac{2(n+1)\pi}3 +\cos\frac{4(n+1)\pi}3\right)$$


问:
Sum of an arithmetic function over the divisors of an integer is multiplicative
I am trying to prove the following claim:

Let $f$ be an arithmetic function and, for $n \in \mathbb{Z}$ with $n > 0$, let
$$
F(n) = \sum_{d\ \vert\ n,\ d > 0}f(d)
$$
If $F$ is multiplicative, then $f$ is multiplicative.

So far I have the following: given relatively prime integers $a$ and $b$, we can uniquely express each divisor $d$ of $ab$ as $d_1d_2$ where $d_1\ \vert\ a$ and $d_2\ \vert\ b$. This means:
$$
F(ab) = \sum_{d\ \vert\ ab}f(d) = \sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1d_2)
$$
$$
F(a)F(b) = \sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1)f(d_2)
$$
Since $F$ is multiplicative,
$$
\sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1)f(d_2) = \sum_{d_1\ \vert\ a} \sum_{d_2\ \vert\ b} f(d_1d_2)
$$
I am not sure where to proceed from here to show that corresponding terms of the two summations are equal. Am I approaching this the right way?

答: Using the Möbius inversion formula  one gets $f(n)=\sum_{d|n} F(d)\mu(n/d)$ or $f=F*\mu$ where $\mu$ denotes the Möbius function and $*$ the convolution operator. Note that $\mu$ is multiplicative and that the convolution of two  multiplicative functions is multiplicative as well. Thus $f=F*\mu$ is multiplicative.


问:
how to understand mixture distribution
Let $f_1(x)$ and $f_2(x)$ be density functions, and let a be a constant such that $0 \leq a \leq 1$.
Consider the function $f(x) = af_1(x) + (1 − a)f_2(x)$.

suppose $X_i \sim f_i, E(X_i) = \mu_i, Var(X_i) = \sigma^2_i, X$ is a mixture of $X_1, X_2$, find Var(X)

$E(X^2) = \int x^2 f(x)dx$ Why is it okay to express it as $\int x^2(af_1(x)+(1-a)f_2(x))dx = aE(X_1^2)+(1-a)E(X_2^2)$?

Shouldn't it be consist of both $x_1$ and $x_2$?

My understanding is that $E(X^2)$ should be $E((aX_1+(1-a)X_2)^2)=a^2E(X_1^2)+(1-a)^2E(X_2^2)+2a(1-a)E(X_1X_2)$

答:
I think that you are confusing two separate things. When we say that $X$ is a mixture of $X_1$ and $X_2$, with respective probability density functions $f_1$ and $f_2$, we mean that there exists constants $\lambda_1,\lambda_2\ge0$ such that $\lambda_1 + \lambda_2 = 1$ and
$$f(x) = \lambda_1f_1(x) + \lambda_2f_2(x) $$
Where $f$ is the pdf of $X$.

In your question, you have $\lambda_1 := a$ and $\lambda_2 := 1-a$, thus the pdf of $X$ is $f = af_1 + (1 − a)f_2$. To compute $\mathbb E[X^2]$, you can then proceed as usual with the law of the unconscious statistician :
$$\mathbb E[X^2] = \int x^2f(x)\, dx = \int x^2(af_1(x)+(1-a)f_2(x))\,dx $$

If you considered the random variable $\tilde X := aX_1 + (1-a)X_2$ instead, you would not have in general that the pdf of $\tilde X$ is $af_1 + (1-a)f_2$. Consider for instance $X_1$ and $X_2$ to be independent standard normal r.v.s, and $a = 1/2$ :

You have that $f = af_1 + (1-a)f_2 = f_1 = f_2$ so the mixture  has the same distribution as $X_1$ and $X_2$, but $\tilde X = aX_1 + (1-a)X_2 \sim \mathcal N(0,1/2)$ is not a standard normal r.v.

Title: how to understand mixture distribution


问:
If $\omega$ is a faithful state, then the corresponding GNS representation is faithful.
This question is from Pieter Naaijken's "Quantum Spin Systems on Infinite Lattices": Let $\mathcal{U}$ be a C*-algebra and $\omega$ a faithful state on $\mathcal{U}$, meaning that $\omega(A^*A)>0$ if $A\neq 0$. Show that the corresponding GNS representation is faithful (in the sense that the kernel is trivial).

My attempt: Let $A\neq 0$. We wish to show that $\pi_\omega(A)$ is not the zero operator on $\mathcal{H}_\omega$. To that end, let $B$ be any nonzero element of $B(\mathcal{H}_\omega)$. By construction, $\pi_\omega(A)[B] = [AB]$, and we have $\langle [AB],[AB] \rangle = \omega((AB)^*AB) > 0$ by faithfulness of $\omega$; hence $\pi_\omega(A)$ is not the zero operator for any nonzero $A$, which implies that the kernel of $\pi_\omega$ is the trivial kernel $\{0\}$ (i.e. $\pi_\omega$ is injective).

Is it this simple? I am very new to this so I'm doubting myself a little.

答:
Usually, you want to reason the other way around. If $\pi_\omega(A)=0$, then for any $B\in\mathfrak A$ you have that
$$
0=\langle \pi_\omega(A)B,\pi_\omega(A)B\rangle=\omega(B^*A^*AB).
$$
Because $\omega$ is faithful, you get that $AB=0$. As you are free to choose $B$, you can let it be $A^*$, so you get that $AA^*=0$, and so $A=0$ and then $\pi_\omega$ is injective (in case it is not obvious: $\|A\|^2=\|A^*\|^2=\|AA^*\|$).


问:
Analysing Contour Diagrams: Which function represents it?
So I'm trying to analyze the following contour map (visually) and point to the correct function that represents it. I'm not quite sure how to approach this at all, as everything I've tried gives me weird results without really seeing anything clear.

Would appreciate some help, thanks in advance!

答:
The diagram is not symmetrical about the vertical axis, it can't be a contour plot of the two stated function involving $\cos(2x)$ which is an even function.

If $y=-2x$, then we have $\sin(2x)+\sin(y) = \sin(2x)+\sin(-2x)=0$, which is a constant.

If $y=-2x$, we have $\sin(2x)-\sin(y) = \sin(2x)-\sin(-2x) = 2\sin(2x)$ which is not a constant.

Hence the only possible candidate is $\sin(2x)+\sin(y)$.


问:
What are the possible minimal acl-dimensions of strongly minimal models?
The question is as in title. By acl-dimension I understand the cardinality of maximal acl-independent set (well-defined for strongly minimal theories). By minimal I understand that there is no equivalent model of smaller dimension.

It is easy to find examples where it is 0, 1 or $\aleph_0$ (algebraic closure of rationals, rationals as a vector space over themselves, and a countably infinite dimensional vector space over a finite field), and any case can be reduced to 0 or $\aleph_0$, possibly by adding finitely many constant symbols (names for the elements of a basis), and by of course it cannot be more than $\aleph_0$, but I can't seem to think of an example with minimal dimension which is finite, but greater than 1.

I believe it can also be shown that in a strongly minimal model, any infinite, algebraically closed subset is the universe of an elementary substructure, so the question can be restated as the following: what are the possible minimal dimensions of infinite sets in strongly minimal models?

答:
Here is an example of a strongly minimal theory whose prime model has dimension $2$. Let the language $\cal L$ contain just one ternary function symbol $\oplus$. Consider the $\cal L$-theory of $\mathbb Q$, where $\oplus$ is interpreted as $\oplus(x, y, z) = x + y - z$. We also use the infix notation $x \oplus_z y$. Let's look at this theory. In particular it would say

for every $z$ the domain is a non-trivial divisible torsion-free abelian group with identity $z$ under operation $\oplus_z$;

$\forall x, y, z, w, u (\oplus(x, y, z) = u \leftrightarrow \oplus(x, y, w) = \oplus(z, u, w))$

Denote the last theory by $T$. Now $T$ is uncountably categorical. Indeed given any two uncountable models ${\cal M}_1$ and ${\cal M}_2$ of the same cardinality, pick any two elements $w_1 \in M_1$ and $w_2 \in M_2$. Then $M_i$ is a $\mathbb Q$ vector space under $\oplus_{w_i}$. So there is an isomorphism $\alpha$ of structures $\langle M_i, \oplus_{w_1}, w_1\rangle$ and $\langle M_i, \oplus_{w_2}, w_2\rangle$. But then this isomorphism preserves $\oplus$ as the last axiom holds in both structures. In particular $T$ is complete and strongly minimal as an uncountable model of $T$ (which is $\aleph_0$-saturated) is definably interpretable in a strongly minimal set (a $\mathbb Q$ vector space).

Now the prime model of $T$ is $\mathbb Q$. But since both $x \mapsto \lambda \cdot x$ and $x \mapsto x + \lambda$ are automorphisms, we can construct an automorphism that maps any given pair $(x_1, y_1)$ with $x_1 \neq y_1$ to any given pair $(x_2, y_2)$ with $x_2 \neq y_2$. So for every $q \in \mathbb Q$ we have $acl(q) = \{q\}$. Also since $acl(0, 1) = \mathbb Q$ the prime model has dimension $2$.

This is a correction of an exercise in Marker's book by Richard Rast. I think he has taken down the original, so I can't put a link to it.


问:
Proof of derivative of $\frac{\ln x}{x}$ from limits
I have encountered the following proof of derivative from limits:
let $f(x)= \frac{\ln\ x}{x}$

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{\frac{\ln(x+\Delta x)}{x+\Delta x} - \frac{\ln\ x}{x}}{\Delta x} \end{equation}

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}

\begin{equation} f'(x) = lim_{\Delta x \rightarrow 0} \frac{\Delta x - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}

\begin{equation} f'(x) = \frac{1-ln\ x}{x^2}
\end{equation}

From the 2nd to the 3rd line, I don't understand how the term $x\ ln(1+\frac{\Delta x}{x})$ was simplified to $\Delta x$ ?

答:
Hint: $$\ln (1+x) \sim x, x \to 0$$

Addition.

Let's consider $\lim_\limits{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}$ and divide it in 2 parts:

$\lim_\limits{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) }{x\Delta x(x+\Delta x)}=
\lim_\limits{ \Delta x \to 0}\left[\frac{xln(1+\frac{\Delta x}{x})}{\Delta x}\cdot \frac{1}{x(x+\Delta x) }\right]=\\
=\lim_\limits{ \Delta x \to 0} \frac{xln(1+\frac{\Delta x}{x})}{\Delta x} \cdot  \lim_\limits{ \Delta x \to 0} \frac{1}{x(x+\Delta x) }= \frac{1}{x^2}$

$\lim_\limits{\Delta x \to 0} \frac{\Delta x\ ln\ x}{x\Delta x(x+\Delta x)}=\frac{\ln x}{x^2}$

All steps are exact and based on well known theorems.

And separately about limit $\lim_\limits{ \Delta x \to 0} \frac{xln(1+\frac{\Delta x}{x})}{\Delta x} = \lim_\limits{ \Delta x \to 0} \frac{ln(1+\frac{\Delta x}{x})}{\frac{\Delta x}{x}} =\lim\limits_{t\to 0} \frac{\ln (1+t)}{t}=1 $: here is applied a theorem about limit of functions composition i.e. changing variable, which is standard theorem.

Let me say, that it is important to study and remember, that in multiplication it is possible to change sub-expression with its equivalence expression, if all appropriate conditions hold: assume $f \sim \phi$ i.e. $\lim \frac {f}{\phi} = 1$. Then $\lim (f \cdot g)= \lim  \left(\dfrac {f}{\phi} \cdot \phi \cdot g \right) = \lim \frac {f}{\phi} \cdot \lim (\phi \cdot g) = \lim (\phi \cdot g)$ when all written limits exists.


问:
Differentiability of Convex and Monotone Function and $F_{\sigma}$ sets
Consider  $f: I \subseteq \mathbb{R} \to \mathbb{R}$. I know that if $f$ is convex then $f$ is differentiable except in a countable set. Alexandroff theorem states that $f$ is twice
differentiable except at a set of null measure. My first question is: Is the set that $f$ is twice differentiable $F_{\sigma}$?

Similarly, if $f$ is monotone then $f$ is differentiable except at a set of null measure. Is this set $F_{\sigma}$?

答:
Let $D=\{d_j\}_{j \ge 1}$ be a countable dense set in $I=(0,1)$.
Let $f(x)=\sum_{j=1}^\infty 2^{-j} {\mathbf  1}_{[d_j,1)}(x)$ and
$g(x)=\int_0^1 f(t) dt$.
Then $g$ is monotone, and the set of points in $I$ where it is differentiable
is $I \setminus D$. This is not an $F_\sigma$ set, since if it was  a countable union of closed sets, then $I$ would be a countable union of nowhere dense closed sets, contradicting the Baire category theorem. Note that $D$ is also the set of points in $I$ where the convex function $h(x)=\int_0^1 g(t) dt$ is not twice differentiable.


答:
That step is wrong. You can't replace a sub-expression by another unless they are equal. Any appeal to the use of standard limits in this manner is a complete disregard for limit laws. If it's coming from a textbook then shame!!

The correct way to handle this is as follows (using $h$ in place of $\Delta x$ to reduce typing effort)
\begin{align}
f'(x) &=\lim_{h\to 0}\frac{x\log(1+(h/x))-h\log x} {xh(x+h)} \notag\\
&=\lim_{h\to 0}\frac{\log(1+(h/x)) }{h/x} \cdot\frac{1}{x(x+h)}-\frac{\log x} {x(x+h)} \notag\\
&=1\cdot\frac{1}{x\cdot x} - \frac{\log x} {x\cdot x} \notag\\
&=\frac{1-\log x} {x^2}\notag
\end{align}

Title: Proof of derivative of $\frac{\ln x}{x}$ from limits

Body:
I have encountered the following proof of derivative from limits:
let $f(x)= \frac{\ln\ x}{x}$

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{\frac{\ln(x+\Delta x)}{x+\Delta x} - \frac{\ln\ x}{x}}{\Delta x} \end{equation}

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}

\begin{equation} f'(x) = lim_{\Delta x \rightarrow 0} \frac{\Delta x - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}

\begin{equation} f'(x) = \frac{1-ln\ x}{x^2}
\end{equation}

From the 2nd to the 3rd line, I don't understand how the term $x\ ln(1+\frac{\Delta x}{x})$ was simplified to $\Delta x$ ?


问:
Metric for sequences proof
Suppose $X = \{\{x_n\}_{n \in \mathbb{N}} : x_n \in \mathbb{R}$ $\forall n\geq 1\}$. Prove that $d: X \times X \rightarrow \mathbb{R}$ such that $d(\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}}) = \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ is a metric on $X$.

My attempt:

I was given the hint that I should check each of the following: (1) a function $f(x) = \frac{x}{x+1}$ is montone increasing, (2) the triangle inequality applies to $\frac{|x_n - y_n|}{(1 + |x_n - y_n|)}$ and (3) the metric itself satisfies the inequality.

(1) $f$ is monotone increasing:

If $x \geq 0$, $f'(x) = \frac{1}{(1 + x)^2} > 0$.

So $f$ is increasing.

(2) Triangle inequality applies:

We have $|x_n-y_n|≤|x_n-z_n|+|z_n-y_n|$ and so

$\frac {|x_n-y_n|}{1+|x_n-y_n|}≤ \frac {|x_n-z_n|+|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}=\frac {|x_n-z_n|}{1+|x_n-z_n|+|z_n-y_n|} + \frac {|z_n-y_n|}{1+|x_n-z_n|+|z_n-y_n|}≤\frac{|x_n-z_n|}{1+|x_n-z_n|}+\frac {|z_n-y_n|}{1+|z_n-y_n|}$

Therefore $\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|} \geq \frac{|x_n-y_n|}{1 + |x_n-y_n|}$.

(3) Metric satisfies triangle inequality:

We know $\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|} \geq \frac{|x_n-y_n|}{1 + |x_n-y_n|}$. So

$$2^{-n}(\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|}) \geq 2^{-n}\frac{|x_n-y_n|}{1 + |x_n-y_n|}$$

and

$$\sum_{n = 1}^{\infty} 2^{-n}(\frac{|x_n - z_n|}{1 + |x_n - z_n|}+\frac{|z_n-y_n|}{1 + |z_n-y_n|}) \geq \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n-y_n|}{1 + |x_n-y_n|}$$

QED.

First of all, is this correct? If so, is that it? Is that sufficient to show that $d$ is a metric on $X$? Although I understood how to prove the hints, I still am not sure how these hints actually show that $d$ is a metric on $X$, presuming they are sufficient to show this. Any insight is much appreciated.

答:
The definition of a metric $d$ has 3 requirements for all points $x, y \in X$.

$d(x, y)=0 \iff x=y$

$d(x, y)=d(y, x)$

triangle inequality.

You've proved the last requirement successfully, now all you need to do is prove the first two (this is quite straight forward).

Title: Metric for sequences proof


问:
Prove function is strictly decreasing over interval
Prove $f(x) = \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1$ is strictly decreasing over $x \in [-3, -2].$

I remember looking at these type of questions during my undergrad but completely forgot how to do it.

Body:
$f(x)=\frac{x^4}{24}+\frac{x^3}{6}+\frac{x^2}{2}+x+1$

$f'(x)=\frac{x^3}{6}+\frac{x^2}{2}+x+1$

pose $h(x) =f'(x)=\frac{x^3}{6}+\frac{x^2}{2}+x+1
 $

Let's show $ h(x) <0$

$h'(x) =\frac{x^2}{2}+x+1$

we know that : $x\in [-3,-2]$

So
$-3\leq x\leq - 2$$\Rightarrow $

$ 2\leq \frac{x^2} {2}\leq  \frac{9}{2}$

So

$0\leq h'(x)= \frac{x^2} {2}+x+1\leq  \frac{9}{2}-1$

$h(x)$ incremental

$h(-3)=-3$ and $h(-2)=-\frac{4}{3}$

So $\forall x\in [-3,-2]  h(x) <0$

We know $h(x) =f'(x) $

So finally:

$\forall x\in [-3,-2]  f'(x) <0$

$f$ is decreasing in $[-3,-2]$


答:
Hint   Using the power law of logarithm and definition of e, we can change 2nd line to $\to$

$f'(x) = \lim_{\Delta x \to 0} \frac{x\    ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta    x)}$

$=\lim_{\Delta x \to 0}    \frac{\Delta x\ ln(1+\frac{\Delta x}{x})^{\frac{x}{ \Delta    X}}}{x\Delta x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \Delta x\    ln\ x}{x\Delta x(x+\Delta x)}$

$=\frac{\ \lim_{\Delta x \to 0} ln(1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}}}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $

$=\frac{ln(\ \lim_{\Delta x \to 0} (1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}})}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $  limit property of continuous function

$=\frac{\ln e}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $

With these intermediate steps and  definition of e ,  we can go from 2nd line to 3rd line .


问:
Explanation of claim in Dummit and Foote
Dummit and Foote, p. 204

They suppose that $G$ is simple with a subgroup of index $k = p$ or $p+1$ (for a prime $p$), and embed $G$ into $S_k$ by the action on the cosets of the subgroup. Then they say

"Since now Sylow $p$-subgroups of $S_k$ are precisely the groups generated by a $p$-cycle, and distinct Sylow $p$-subgroups intersect in the identity"

Am I correct in assuming that these statements follow because the Sylow $p$-subgroups of $S_k$ must be cyclic of order $p$? They calculate the number of Sylow $p$-subgroups of $S_k$ by counting (number of $p$-cycles)/(number of $p$-cycles in a Sylow $p$-subgroup). They calculate the number of $p$-cycles in a Sylow $p$-subgroup to be $p(p-1)$, which I don't see.

答:
If $k$ satisfies $p\leq k\lt 2p$, then $p$ is the largest power of $p$ that divides $k!$, so the $p$-Sylow subgroups of $S_k$ have order $p$.

If $\sigma\in S_k$ is written as a product of disjoint cycles, then the order of $\sigma$ is the least common multiple of the lengths of these cycles. In particular, an element has order a power of a prime $p$ if and only if it is a product of cycles whose lengths are powers of $p$. An element has order $p$ if and only if it is a product of (at least one) disjoint $p$-cycles.

Since $p\leq k\lt 2p$, the only elements of $S_k$ that are disjoint products of $p$-cycles are single $p$-cycles (not enough points to get another, disjoint, $p$-cycle). So a $p$-Sylow subgroup of such an $S_k$ will be cyclic of order $p$ (since $p|k!$ but $p^2\not|k!$), hence generated by an element of order $p$; and the only elements of order $p$ are the $p$-cycles. Conversely, any $p$-cycle generates a subgroup of order $p$, which will therefore be a $p$-Sylow subgroup of $S_k$. So the $p$-Sylow subgroups of $S_k$ are precisely the subgroups generated by $p$-cycles.

Any two distinct $p$-Sylow subgroups, $P_1$ and $P_2$, must intersect at proper subgroups of $P_1$ and $P_2$; but $P_i$ is of order $p$, so the only proper subgroup is the trivial subgroup, so $P_1\cap P_2 = \{1\}$.

I'm not sure what your final paragraph means, and I don't have Dummit and Foote with me right now, so I'll leave it for now.


问:
expanding convoluted integrand
I have a function on the form $$g(y) = \int_{-\infty}^{\infty}{e^{-v^2}f(y-v)dv}$$ I know that $g(y)$ is linear around $0$, $g(y\approx 0)\approx yG$, and I am interested in finding this gradient $G$.

For this reason I thought that the integrand could be expanded around $y=0$. In general, am I allowed to do that for a convolution? Or should I expand around $y-v=0$?

答:
Yes, you should expand around $y=0$ if you're interested near there! With suitable reasonable restrictions the expansion, which is really just a derivative with respect to $y$ at 0, commutes with the integral (see http://en.m.wikipedia.org/wiki/Differentiation_under_the_integral_sign for example) so everything works out.

$$G= \int e^\cdots \partial_y f(y-v) \text{ at }y=0$$

Incidentally a trick which is sometimes useful is using $\partial_x f(x-y)=-\partial_y f(x-y)$ so that you can integrate by parts,  or similar.

Title: expanding convoluted integrand


问:
Can we Classify the Homomorphic Images of Tori?
Is it known for which topological groups $X$ there exists a positive integer $d$ and a surjective, continuous group homomorphism $([0,1]\text{ mod } 1)^d \to X$?

Certainly, any such $X$ must be compact and path connected. Also, it's set of characters must be countable, but I can't image that these three conditions are sufficient. Is there some known sufficient condition? Are there any further, more or less obvious necessary conditions which are known?

P.S. the set $X^{\wedge}$ of algebraic characters for such a group $X$ is countable because there exists an injection $X^{\wedge} \to (([0,1]\text{ mod } 1)^d)^{\wedge} : \gamma\mapsto\gamma\circ f$ where $f : ([0,1]\text{ mod } 1)^d \to X$ is the mystical continuous, surjective homomorphism.

答:
Well, if $f:T^d\to X$ is a surjective continuous homomorphism, then it induces a topological isomorphism $T^d/\ker f\to X$ where $\ker f$ is a closed subgroup of $T^d$.  In particular, $\ker f$ is a Lie subgroup and so the quotient $X$ is a Lie group.  Being a compact abelian Lie group, it must be a torus itself.

Or alternatively, using Pontryagin duality, the Pontryagin dual of $X$ must be a subgroup of $\mathbb{Z}^d$, the Pontryagin dual of $T^d$.  Every subgroup of a free abelian group is free, so the Pontryagin dual of $X$ is isomorphic to $\mathbb{Z}^e$ for some $e\leq d$ so $X\cong T^e$.

(This is assuming you require topological groups to be Hausdorff.  If you don't, there are a lot of other nasty possibilities--for instance, $X$ could be any divisible abelian group of cardinality at most $2^{\aleph_0}$, with the indiscrete topology.)


问:
Provide a proof or counterexample. If $M$ is an $n \times n$ matrix that is diagonal, then every non-zero value in $M$ is an eigenvalue of $M$.
Provide a proof or counterexample. If $M$ is an $n \times n$ matrix that is diagonal (i.e., only the entries on the main diagonal may be non-zero), then every non-zero value in $M$ is an eigenvalue of $M$.

Checking on MatLab is seems to hold true, but I'm a little stuck on the proof. So far I have:

$Mx = \lambda x$, where x is a non zero vector.

$\lambda x - Mx  = 0$

$\lambda xI - Mx = 0$, where I is the Identity matrix.

$(\lambda I - M)x = 0$

And I don't really know what to do from here.

答: The characteristic polynomial is $p_M(\lambda)=\prod_{i=1}^n(\lambda-m_{ii}))$. The roots of the characteristic polynomial are the eigenvalues. Therefore, the set of diagonal entries is also your set of eigenvalues.


问:
How are all possible poker hands of equal probability?
Please pardon the elementary question, for some reason I'm not grocking why all possible poker hand combinations are equally probable, as all textbooks and websites say. Just intuitively I would think getting 4 of a number is much more improbably than getting 1 of each number, if I were to draw 4 cards. For example, ignoring order, to get 4 of a single number there are only $4 \choose 4$ distinct possibilities, whereas for 1 of each number I would have ${4 \choose 1}^4$ distinct possibilities.

答: Often the books and websites speak a little too loosely. I would say that "getting 4 of a kind" is not a poker hand. It is a set of many different poker hands. A "possible poker hand" is completely specific, e.g. the hand 4 of spades, 4 of hearts, 4 of clubs, 7 of clubs, 8 of diamonds.</p>


答: Yes, that's true, but they mean that any particular hand of 5 cards has the same probability as any other hand of 5 cards. Once you start talking about the probability of a pair or four of a kind, you're talking about the probability of getting one of a number of hands. To put it another way, the probability of drawing a royal flush in spades is exactly the same as the probability of drawing the 2,3 of diamonds, the 6,8 of clubs, and the Jack of hearts.</p>

Title: How are all possible poker hands of equal probability?

Body: Please pardon the elementary question, for some reason I'm not grocking why all possible poker hand combinations are equally probable, as all textbooks and websites say. Just intuitively I would think getting 4 of a number is much more improbably than getting 1 of each number, if I were to draw 4 cards. For example, ignoring order, to get 4 of a single number there are only $4 \choose 4$ distinct possibilities, whereas for 1 of each number I would have ${4 \choose 1}^4$ distinct possibilities.


答:
This is just like flipping a coin. You are just as likely to get exactly
$$HHTHTTH$$
as you are to get
$$HHHHHHH$$
You are intuitively grouping poker hands into their categories, and you are right that for example four of a kind is less likely than high card. Notice for example, that the wikipedia page is careful to distinguish poker hands from "hand-ranking categories".


问:
Whether solutions of a particular matrix equation are only the permutation matrices
Let $V \in R^{d \times d} $ be a real orthogonal matrix. Denote by $V \circ V$ the Hadamard product (elementwise product).

I wish to show that if $V \left(V \circ V  \right)^T V$ is "close" to $V \circ V$, then $V$ is necessarily "close" to $V \circ V \circ V$, meaning $V$ is an orthogonal matrix with entries in $\{-1, 0, 1\}$ (a permutation matrix upto the sign of the entries).

In precise terms, I want to prove that:

$V \left(V \circ V  \right)^T V = V \circ V \implies V_{ij} \in \{-1,0,1\}$.
So, equality implies $V$ is a permutation matrix up to the sign of the entries.

For a given $\epsilon > 0$ to prove there exists an $\hat{\epsilon} >0$ such that:
$||V \left(V \circ V  \right)^T V - V \circ V || < \hat{\epsilon} \implies ||V- V\circ V \circ V|| < \epsilon $. Meaning that approximate equality implies $V$ is close to a permutation matrix (upto the sign). The norm is any standard matrix norm.

答:
I don't think that's true. Let
$$
V=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.
$$
Then $V\circ V=I_2$, and
$$
V(V\circ V)^TV=V^2=I_2=V\circ V.
$$</p>


问:
Prove that R(A) = R(B) iff N$(B^T)$ = N$(A^T)$ where R(A) denotes the range of the matrix A.
This is a bonus question on my assignment and it bothered me quite a while.

I believe the range of a matrix is equivalent to the column space of the matric. That is to prove two matrices have the same row spaces iff their null spaces equal.

However, according to my understanding, the null space of a matrix is related to its column space(I doubted could it be the professor did some typographical error?). Ax means either the linear combination of the column vactors of A or the dot product of its row vectors. I tried to solve the question doing this to transfer between columns and rows(forgive my poor English), but it didn't work as I expected.

Then I run out of ideas

$A^T$ and $B^T$ are row equivalent, how is this indicate they have the same null space?

答:
I've figured out an approach.

when $N(A^T) \subseteq N(B^T)$

let $a_j$ denotes the column vectors of A ; B = [$b_1$...$b_n$]

$\iff$ x$\in$$N(B^T)$ for x$\in$$N(A^T)$

$\iff$ $(B^T,a_j^T)$x = 0 and $B^T$x = 0

$\iff$ $N(B^T) = N(B^T,a_j^T)$

$\iff$ Rank($B^T$) = Rank$(B^T,a_j^T)$

$\iff$ Rank(B) = Rank$(B,a_j)$

$\iff$ $a_j$ = span($b_1$...$b_n$)

$\iff$ R(A) $\subseteq$  R(B)

$\iff$ $N(A^T) = N(B^T)$ gives R(A) = R(B)


问:
Reference for Ergodic Theory
I am looking for a good introductory book on ergodic theory. Can someone recommend  some introductory texts on that? </p>

答:
Arguably this is not a book but Terence Tao's lecture-blogs (the very first of many as I understand) may be worth looking into: https://terrytao.wordpress.com/category/teaching/254a-ergodic-theory/ .

Rudolph's Fundamentals of measurable dynamics. Ergodic theory on Lebesgue spaces is also noteworthy in my humble opinion.

Title: Reference for Ergodic Theory


问:
Looking for a particular advanced high school book; forgot author and title
I am looking for the author and title of a book I used to own. I am not looking for recommendations. I am trying to identify one particular book.

It contains some topics in calculus at the advanced high school or undergraduate level which are not usually covered in the syllabus. A fair bit of it is about polynomials. It is mostly for entertainment and is written in quite a relaxed style.

Some of the topics covered are: the Lambert W function, palindromes, and (I think) the Newton Polytope. I can't remember any others. I have tried searching for these topics but am not getting any hits.

It was a fairly thin hardback, very probably published by Springer between 2000 and 2015. I think it has a single author and the author was an American.

答:
Found it! Uncommon Mathematical Excursions: Polynomia and Related Realms by
Dan Kalman. It was published by AMS and not Springer.

https://bookstore.ams.org/dol-35/


问:
Definition of surjective map $d_K = 1/f_K d : G -> \widehat{\mathbb{Z}}$ in Neukirch 4.4 Abstract Valuation Theory
The setup of the question is the same As the unanswered question here. Here is the relevant part of Neukirch.

Here $G$ is an arbitrary profinite group (but I take it as the Galois group of some extension) and $d : G -> \widehat{\mathbb{Z}}$ is surjective

My question is about the definition of $d_K$ itself. If $f_K$ is finite, then $d(G_K)$ has to be the unique subgroup corresponding to the fixed field of $\mathbb{F}_{p^f}$ and corresponds to those elements of $\widehat{\mathbb{Z}}$ that are $0$ on projecting to $\mathbb{Z}/f\mathbb{Z}$. However the definition of $d_K$, as I understand it, involves computing the $1/f$th element of each $\mathbb{Z}/m\mathbb{Z}$. But then how would this be surjective, as you can't solve $7x = 3$ modulo 7.

Can someone give me the explicit description of this $d_K$ map and explain why it should be surjective?

答:
You can embed $\widehat{\mathbb{Z}}=\prod_p \mathbb{Z}_p$ in $\prod_p \mathbb{Q}_p$, in which every integer $n\ne0$ is invertible, allowing you to divide by $f_K$.

By the definition of $f_K$, we have $d(G_K)=f_k\widehat{\mathbb{Z}}$, so $\frac{1}{f_k}d(G_K)=\widehat{\mathbb{Z}}$.


问:
Im being asked to define a bijection from the set of natural numbers including zero
Im being asked to define a bijection from the set of natural numbers including zero of non negative integers to the set of natural numbers. Then prove that the function is surjective and injective. So far I tried creating a piece wise equation where:

$$f(n) = \begin{cases}2n, &\text{if }n >= 0\\
       -2n-1,& \text{if } n < 0
\end{cases}$$
I'm not sure if thats what they want but I think its correct.

Original question

If possible I would appreciate getting insight on how to do B as well because I genuinely have no clue.

答:
As $n < 0$ never happens (because your domain is $\{0,1,2,3,4,5,.....\}$) that functions is just $f(n) = 2n$ and so $f(n) = odd$ never occurs.

Look,  you just need to map $\{0,1,2,3,4,......\}$ to $\{1,2,3,4,5,.....\}$ that's one to one and onto.  Think of a way to do that.  And don't assume it will be hard.

(It will be easy...... really easy.)

Oh, also the reason your function, which maps $\mathbb Z\to \mathbb N$, works so well, is that by moving all the positives to the even positions, it makes room for all the negatives.

But in this case you don't have an infinite number of negatives.  The only extra element is... $0$.... so you have to make room to fit one element in.

I hope I have given you enough hints.


问:
How to indicate that my proposition implies true/false?
I would like to know the correct notations in logic for the following statements:

"the proposition $\alpha$ is true (or false)"

"the variable $A$ implies true (or false)"

Which of the following sentences are syntactically valid? And why?

Symbol
Proposition case
Variable case

Single arrow ($\rightarrow$)
$\alpha \rightarrow \top$
$A \rightarrow \top$

Double arrow ($\Rightarrow$)
$\alpha \Rightarrow\top$
$A \Rightarrow \top$

Long double arrow ($\implies$)[1]
$\alpha \implies \top$
$A \implies \top$

Single turnstile ($\vdash$)
$\alpha \vdash \top$
$A \vdash \top$

Single turnstile ($\models$)
$\alpha \models \top$
$A \models \top$

[1] Which in LaTeX is properly named "implies".

答:
$\alpha, \beta, \cdots$ are commonly used as syntactic metavariables to stand for well-formed formulas. So using the proposition $\alpha$ is usual in a way that the proposition $A$ would not be. Capital latin letters $A, B, C \cdots$ are normally used as propositional variables, i.e. variables within the propositional logic.

For the proposition $\alpha$ is true, you can use:

$\alpha$ is true.

$\alpha$ holds.

$\models \alpha$

$\alpha$ is a tautology.

To say that $\alpha$ is provably a tautology, meaning that $\alpha$ has a proof in the relevant deduction system, you can say

$\vdash \alpha$

$A$ implies true can be written as a proposition in the following way

$A \to \top$

$A \to \top$, by itself, is a proposition. It is equivalent to $(\lnot A) \lor \top$ and therefore is always true.

$A$ implies false can be written as a proposition in the following way

$A \to \bot$

$A \to \bot$ is also a proposition. It is equivalent to $(\lnot A) \lor \bot$ and is thus equivalent to $\lnot A$.


问:
Time, Speed and Distance : Returning back to pick up problem
$A$,$B$ and $C$ started out in a journey to a place which was $120$ km away. $A$ and $B$ went by car at the speed of $50$ kmph while $C$ travelled by tonga at $10$ kmph. After a certain distance $B$ got off and travelled the rest distance by another tonga at $10$ kmph, while $A$ went back for $C$ and reached the destination at the same time as $B$ arrived. The number of hours required for the trip was ____ $?$

I tried to approach the problem by equating the time taken by $A$ and $B$ as both of them has reached the destination at the same time after starting together. That means they must have taken the same amount of time.

Time taken by $B=\frac{d}{50} + \frac{120-d}{10}$ ; where $d$ is the distance travelled by $A$ and $B$ together on a car together.

Time take by $A=\frac{d}{50}$ + Time taken to return back an pick up $C$ + Time taken to reach the destination after picking up $C$

Time taken to return back an pick up C $\Rightarrow$ this I am not able to figure out as when A will return back to pick up C and until them meet, the C will be moving forward for some distance while A would be coming towards him and here the concept of relative distance will come into the picture. Getting confused here!!!

Please help me out.
Answer Provided : 4.8 hours

答:
I think it's easier to do this if we make times the unknows, rather than distances, but of course, you can do it either way.  Let us say that A and B drive for $t_1$ hours before they stop, and that it takes A $t_2$ hours to go back and pick up C.

It takes B $$t_1+\frac{120-50t_1}{10}=12-4t_1$$ hours to complete the trip.

When A and C meet, C has traveled a distance of $10(t_1+t_2)$ and A's distance from the staring point is $50t_1-50t_2$.  Equating these distances gives $2t_1=3t_2$.

Now the time it take until the car arrives at the destination is $$t_1+t_2+\frac{120-10(t_1+t_2)}{50}$$

Equating the two arrival times, and using $2t_1=3t_2$ should give the right answer.  (I haven't checked this.)


问:
M matrix definition
An $n\times n$ matrix $A$ that can be expressed in the form $A=sI-B$ where $B=(b_{ij})$ with $b_{ij}\geq 0$, $1\leq i,j\leq n$ and $s\geq \rho(B)$, where $\rho(B)$ is the maximum moduli of the eigenvalues of $B$, is called an M matrix.

My doubt is: Can an M matrix be expressed in the above form in multiple ways or is there such a unique expression.

答:
Yes, an M matrix can be expressed in multiple ways. In particular, if $\lambda \ge 0$, then we can write
$$A = (s + \lambda)I - (B + \lambda I). \tag{$\star$}$$
Note that, since the entries of $B$ and $\lambda I$ are non-negative, so too are the entries of $B + \lambda I$. Further, $B + \lambda I$ simply increases each eigenvalue of $B$ by $\lambda$, and so the maximum moduli of $B$ can increase by at most $\lambda$. So, we see that
$$\rho(B + \lambda I) \le \rho(B) + \lambda \le s + \lambda.$$
That is, $(\star)$ is a sum of matrices in the given form, no matter the value of $\lambda \ge 0$, and so the sum is far from unique.

Title: M matrix definition


问:
How do I simplify $\overline{(\bar{X} + Z\bar{Y} + Y\bar{Z})(\bar{X}+Y\bar{Z})}$?
$\overline{(\bar{X} + Z\bar{Y} + Y\bar{Z})(\bar{X}+Y\bar{Z})}$

The farthest I've gotten is DeMorgans:

$X(\bar{Z} + Y)(\bar{Y} + Z) + X(\bar{Y} + Z)$

The answer should be $X\bar{Y} + XZ$, but I'm unable to get anywhere close to it. I tried foiling the binomial but it results in a single X value when the whole expression is evaluated, and that's not correct.

答:
Continuing from where you stopped using $1$ for $True$

\begin{eqnarray*} X(\bar{Z} + Y)(\bar{Y} + Z) + X(\bar{Y} + Z)
& = & X(\bar{Y} + Z)((\bar{Z} + Y) + 1) \\
& \stackrel{x+1=1}{=} & X(\bar{Y} + Z)\\
& = & X\bar{Y} + XZ
\end{eqnarray*}


问:
Evaluate $\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt$ by Laplace transform
$\require{cancel}$

Note: All logarithmic functions in the post include the absolute value function (this doesn't change the convergence of the variables).

For an entrance exam i got the integral

$$\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt$$

The exercise is to calculate it by the Laplace transform.

By the frequency domain integration theorem (only in unilateral), it holds that:

$$\mathcal{L}\left[\frac{f(t)}{t} \right]= \int_s^\infty F(\omega)\; d\omega = \int_s^\infty \mathcal{L}(f(t))\; d\omega $$
$\omega$ being our dummy variable here

Let our $f(t)$ time-domain function be $e^{-\alpha t}\sin(-t)$

$$\mathcal{L}\left[\frac{f(t)}{t}\right]= \mathcal{L}\left[\frac{e^{-\alpha t}\sin(-t)}{t} \right] =\int_s^\infty \mathcal{L}\left[{e^{-\alpha t}\sin(-t)}\right]\; d\omega $$

$$\mathcal{L}\left[e^{-\alpha t}\sin(\beta t)\right]= \dfrac{\beta}{(\omega+\alpha)^2+\beta^2} \hspace{10mm}\alpha,\beta \in \mathbb{R} $$

If $\beta=-1$
$$\mathcal{L}\left[e^{-\alpha t}\sin(-t)\right]= \dfrac{-1}{(\omega+\alpha)^2+(-1)^2} = \dfrac{-1}{(\omega+\alpha)^2+1} $$

$$\int_s^\infty \dfrac{-1}{(\omega+\alpha)^2+1}\; d\omega = -\int_s^\infty \frac{1}{\omega^2 + 2\alpha\omega + \alpha^2 +1}\; d\omega =
- \int_s^\infty \frac{\frac{-i}{2}}{\omega-\alpha-i}\;d\omega \; - \int_s^\infty \frac{\frac{i}{2}}{\omega-\alpha+i}\;d\omega=$$

$$=-\frac{i}{2} \left\{ \int_s^\infty -\frac{1}{\omega-\alpha-i}\;d\omega + \int_s^\infty \frac{1}{\omega-\alpha+i}\;d\omega  \right\}= -\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty       \bigg)= $$

$$\stackrel{\color{red}{\Delta}}{=}-\frac{i}{2} \bigg(  \left[\cancelto{-\infty}{-\ln(\infty)}+\ln(s-\alpha-i)\right] + \left[\cancelto{\infty}{\ln(\infty)}-\ln(s-\alpha+i)\right] \bigg)= \boxed{-\frac{i}{2}\ln\left(\frac{s-\alpha-i}{s-\alpha+i} \right)} $$

Thus the integral

$$\int_0^\infty \frac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt = -\int_0^{\infty}\frac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{t}\; dt$$

is

$$-\left[-\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right) + -\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)\right]= \frac{i}{2}\ln \left(\frac{s-2-i}{s-2+i}\cdot \frac{s-3-i}{s-3+i} \right)=$$

$$= \boxed{\frac{i}{2}\ln\left(\frac{\left(s^2-5s+5\right)+i\left(5-2s\right)}{\left(s^2-5s+5\right)-i\left(5-2s\right)} \right)} $$

The problem here is that i cannot really take the Inverse Laplace Transform of a rational function which is inside a logarithm.

How could i proceed?

If someone is confused by the $\color{red}{\Delta}$ step:
$$\color{red}{\Delta}:=-\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty  \bigg)=-\frac{i}{2} \left\{ \int_s^\infty -\frac{1}{\omega-\alpha-i}\;d\omega + \int_s^\infty \frac{1}{\omega-\alpha+i}\;d\omega  \right\}=$$

$$=-\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty\bigg) = \frac{i}{2}\left(\Bigg[-\ln(\infty-\alpha-i)\underbrace{--}_{+}\ln(s-\alpha-i)\Bigg]+\Bigg[\ln(\infty-\alpha+i)-\ln(s-\alpha+i)\Bigg]\right)=$$

$$=-\frac{i}{2}\left(\color{blue}{\lim_{\sigma\to\infty}\left(-\ln(\sigma-\alpha-i)+\ln(\sigma-\alpha+i)\right)}+\ln(s-\alpha-i)-\ln(s-\alpha+i)\right)=$$

$$=-\frac{i}{2}\left(\color{blue}{\lim_{\sigma\to\infty}\ln\left(\dfrac{\sigma-\alpha-i}{\sigma-\alpha+i}\right)}+\ln\left(\dfrac{s-\alpha-i}{s-\alpha+i}\right)\right)$$

$$=-\frac{i}{2}\left(\color{blue}{\underbrace{\ln\underbrace{\lim_{\sigma\to\infty}\left(\dfrac{\sigma-\alpha-i}{\sigma-\alpha+i}\right)}_{1}}_{0}}+\ln\left(\dfrac{s-\alpha-i}{s-\alpha+i}\right)\right)$$

答:
It turns out that there is a complex logarithmic identity that allows us to turn a logarithm of a complex complex number into an inverse trigonometric function.

$$\color{red}{\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right)} + \color{blue}{\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)}$$

$$\color{red}{\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right)=-\frac{i}{2}\ln\left(\frac{s-2+i}{s-2-i}\right)=\mathrm{arccot}(s-2)=\arctan\left(\frac{1}{s-2}\right)}$$

$$\color{blue}{\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)=-\frac{i}{2}\ln\left(\frac{s-3+i}{s-3-i}\right)=\mathrm{arccot}(s-3)=\arctan\left(\frac{1}{s-3}\right)}$$
Although I'm not really sure how to continue from here now.

From table:

$$\ln\left(\frac{s+a}{s+b}\right)\stackrel{\mathcal{L}^{-1}}{\Rightarrow}\frac{e^{-bt}-e^{-at}}{t}$$


问:
Singularity of a conic combination of rank-$1$ matrices
Given rank-$1$ square matrices $A_1, A_2, \dots, A_n$, determine if there exists $x \in \mathbb R_{>0}^n$ such that

$$ \sum_{i=1}^n x_i A_i = x_1 A_1 + \cdots + x_n A_n $$

is singular, or decide whether none exists (i.e., the positive linear combination is nonsingular for all positive $x$). Is there a well-known method or algorithm to do this?

答:
Given matrices ${\rm A}_1, {\rm A}_2, \dots, {\rm A}_n \in \Bbb R^{d \times d}$, decide the existence of a vector $x \geq {\Bbb 0}_n$ such that

$$\mbox{rank} \left( \sum_{k=1}^n x_k {\rm A}_k \right) <  d$$

Since the nuclear norm is a convex proxy for the rank, we could solve the following convex program

$$\begin{array}{ll} \underset{{\rm x} \in \mathbb R^n}{\text{minimize}} & \left\| \displaystyle\sum_{k=1}^n x_k {\rm A}_k \right\|_* \\ \text{subject to} & x \geq {\Bbb 0}_n\end{array}$$

Let ${\rm x}^{\min}$ denote the minimizer of the aforementioned convex program. If

$$\mbox{rank} \left( \sum_{k=1}^n x_k^{\min} {\rm A}_k \right) <  d$$

we are done. If not, we wasted a few minutes of our lives. Note that we did not require that the given matrices ${\rm A}_1, {\rm A}_2, \dots, {\rm A}_n$ be rank-$1$.


问:
Retract of contractible space is contractible [proof verification]

Retract of contractible space is contractible.

Let $r:X\to A$ where $A\subset X$: subspace and $r$ is retraction, $X$ is contractible.

Statement is so famous anybody can find the proof. But I found that many proofs is much longer than I thought. Here's my proof: $\text{id}_X\simeq C_{x_0}$ and $r\circ i = \text{id}_A$ where $i:A\hookrightarrow X$: inclusion. So, $\text{id}_A =r\circ i  = r\circ\text{id}_X\circ i \simeq r\circ C_{x_0}\circ i=C_{r(x_0)}$ noting the fact that $f\simeq g\Rightarrow f\circ h\simeq g\circ h$ and $k\circ f\simeq k\circ g$ where each composition are well defined. I've never seen such proof in Google so far. They all use explicit homotopy $H:X\times I\to X$ from $\text{id}_X$ to $C_{x_0}$ so I wonder if my proof is wrong or have some serious problem.

答:
Your proof is indeed correct. You rely on the critical notion that if $f \simeq g$ and $h \simeq k$ then $f \circ h \simeq g \circ k$. This observation is critical in another short proof of the fact.

Background:

Consider the "homotopy 1-category", where the objects are topological spaces and the arrows are continuous functions, quotiented by $\simeq$. The fact that this is actually a category is equivalent to the above observation about composition preserving $\simeq$
$\DeclareMathOperator{\id}{id}$

The terminal object of this category is the singleton space 1, because this is the terminal object of the category of topological spaces. In other words, there is only one map from a space $T$ to the singleton space 1, up to homotopy.

It can easily be seen that a space is contractible iff the space is isomorphic to 1 in the homotopy category.

If the space is isomorphic to 1, then consider the homotopy-isomorphism $f : 1 \to A$ and inverse $g : A \to 1$. Then we see that $\id_A \simeq f \circ g$, and $f \circ g$ is a constant function $C_{f(x)}$ where $x$ is the unique element $x \in 1$.

On the other hand, if $\id_A \simeq C_{a_0}$, then define $f : 1 \to A$ by $f(x) = a_0$ and let $g : A \to 1$ be the unique function. Then we see that $C_{a_0} = f \circ g \simeq \id_A$, so $A$ is homotopy-isomorphic to 1.

Thus, a contractible space is exactly a space which is a terminal object in the homotopy category.

End background

In general, if we have a terminal object $1$ and arrows $i : A \to 1$ and $r : 1 \to A$ such that $r \circ i = \id_A$, then $i$ is an isomorphism. This is because we have $i \circ r : 1 \to 1$ and $\id_1 : 1 \to 1$, and hence $i \circ r = \id_1$.

Thus, the retract of a contractible space is contractible.


问:
Is this a group homomorphism?
Problem: Define $\mathbb{Q}/\mathbb{Z}$ by the natural projection $\pi \, : \, \mathbb{Q} \longrightarrow \mathbb{Q}/\mathbb{Z}$, where $\pi(x) = \text{frac}(x) = x - \lfloor x \rfloor$ for all $x \in \mathbb{Q}$.  Show that $\pi$ is a homomorphism.

Question:  Is $\pi$ really a homomorphism by this definition?

$$\pi\left(\frac{5}{3} + \frac{7}{4}\right) = \pi\left(\frac{41}{12}\right) = \frac{41}{12} - \left \lfloor \frac{41}{12}\right \rfloor = \frac{41}{12} - 3 = \frac{5}{12}$$

but

$$\pi\left(\frac{5}{3}\right) + \pi\left(\frac{7}{4}\right) = \left[\frac{5}{3} - \left \lfloor \frac{5}{3}\right \rfloor\right] +  \left[\frac{7}{4} - \left \lfloor \frac{7}{4}\right \rfloor\right] = \left[\frac{5}{3} - 1\right] +  \left[\frac{7}{4} - 1\right] = \frac{2}{3} + \frac{3}{4} = \frac{17}{12}.$$

答:
This is a homomorphism.  Another way of expressing the map is by the rule $x\mapsto x+\Bbb Z$.

$\Bbb Z\triangleleft\Bbb Q$, so the quotient $\Bbb Q/\Bbb Z$ is a group.

$π$ is the canonical projection onto the quotient by a normal subgroup, which is always a homomorphism.

Title: Is this a group homomorphism?


问:
Let $F=\{f: \mathbb{R} \to\mathbb{R}:|f(x)-f(y)|\leq K|(x-y)^\alpha|\}$ for all $x,y\in\mathbb{R}$ and for some $\alpha >0$ and some $K > 0$
Let  $F=\{f: \mathbb{R} \to\mathbb{R}:|f(x)-f(y)|\leq K|(x-y)^\alpha|\}$ for all $x,y\in\mathbb{R}$ and for some $\alpha >0$ and some $K > 0$ .

Which of the following is/are true?

1. Every $ f \in F$ is continuous

2. Every $f\in F$ is uniformly continuous

3. Every differentiable function $ f$  is in $F$

4. Every $f \in F$  is differentiable

The given condition is Lipchitz condition. So 1 and 2 are true. but what can I say about the others

答:
1, 2: $\surd$: For $\epsilon>0$ choose $\delta=\dfrac{\epsilon^{1/{\alpha}}}{2K}>0.$ Then $|x-y|<\delta$$\implies|f(x)-f(y)|$$\le K|x-y|^\alpha\le\epsilon.$ So 2 is true and 2 implies 1.

3: $\times$: Choose $K=1,\alpha=1.$ Then $x^2\notin F:$ $|2^2-1^2|>|2-1|.$

4: $\times$: $|\text{sign }x-\text{sign }y|\le|x-y|$


Title: Is this a group homomorphism?

Body:
Problem: Define $\mathbb{Q}/\mathbb{Z}$ by the natural projection $\pi \, : \, \mathbb{Q} \longrightarrow \mathbb{Q}/\mathbb{Z}$, where $\pi(x) = \text{frac}(x) = x - \lfloor x \rfloor$ for all $x \in \mathbb{Q}$.  Show that $\pi$ is a homomorphism.

Question:  Is $\pi$ really a homomorphism by this definition?

$$\pi\left(\frac{5}{3} + \frac{7}{4}\right) = \pi\left(\frac{41}{12}\right) = \frac{41}{12} - \left \lfloor \frac{41}{12}\right \rfloor = \frac{41}{12} - 3 = \frac{5}{12}$$

but

$$\pi\left(\frac{5}{3}\right) + \pi\left(\frac{7}{4}\right) = \left[\frac{5}{3} - \left \lfloor \frac{5}{3}\right \rfloor\right] +  \left[\frac{7}{4} - \left \lfloor \frac{7}{4}\right \rfloor\right] = \left[\frac{5}{3} - 1\right] +  \left[\frac{7}{4} - 1\right] = \frac{2}{3} + \frac{3}{4} = \frac{17}{12}.$$

Body:
$$\frac{5}{12} = \frac{17}{12}$$

in $\mathbb{Q}/\mathbb{Z}$


问:
How to prove multiplication of natural numbers by induction?
Prove that there exists a unique function $g : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ so that for all $x,y ∈ \mathbb{N}$

a) $g(x,1)=x$

b) $g(x,y')=x+g(x,y)$

So my attempt is(rough sketch):

Define $B$ to be the set of all $x \in \mathbb{N}$ so that one can find a set of function values $\{f(x,i):i \in \mathbb{N}\}$ so that for all $y$ both a) and b) holds( for the fixed $x$)

Base step: let $x=1$(keeping $x$ fixed), for every $y \in \mathbb{N}$, define $g(1,y)=y$. So $g(1,1)=1$ and therefore a) holds.

Now, $g(1,y')=1+g(1,y)=y+1$. So b) also holds.

Now let $x \in B$. To show: $x' \in B$. by definition $g(x',1)= x'=(g(x,1))'$
so a) holds.

Here is my problem, how do I show that $g(x',y')=x'+g(x',y)$

hint: Replace the notation $g(x,y)$ by $xy$ (multiplication of $x$ and $y$)

Thanks in advance.

答:
There are few different ways you can define this type of “multivariate induction”. Since your version of natural numbers does not include $0,$ I will summarize a variation of mathematical induction on two variables accordingly as follows:

$P(1,1)$

$\forall x\quad P(x,1) \Rightarrow P(x+1,1)$

$\forall x,y\quad P(x,y) \Rightarrow P(x,y+1)$

$\forall x,y\quad P(x,y)$

It is readily verify that first two steps of this algorithm holds for your problem. Now for the last step assume that for some arbitrary $x,$ $g(xy)=xy.$ Then by $b)$ we have $$g(x, y+1)=x+g(xy)=x+xy=x(y+1)$$ as expected, and hence the result followed by mathematical induction.


问:
What is the autocovariance of $y_t = \exp{(x_t)}$ where $x_t$ is a stationary normal process.

Let $x_t$ be a stationary normal process with mean $\mu_x$ and
autocovariance function $\gamma(h)$. Define the nonlinear time series
$y_t=\exp{(x_t)}$.

(a) Express the mean function $E(y_t)$ in terms of $\mu_x$ and
$\gamma(0)$. The moment generating function of a normal random
variable $x$ with mean $\mu$ and variance $\sigma^2$ is

$$M_x(\lambda)=E[\exp{(\lambda x)}]=\exp{(\mu \lambda +
> \frac{1}{2}\sigma^2 \lambda^2})$$

(b) Determine the autocovariance function of $y_t$. The sum of the two
normal randomvariables $x_{t+h}+x_t$ is still a normal random
variable.

EDIT:
I figured out the first part, but still struggling with part b.

First we not the MGF for a normal rv $x$ with mean $\mu_x$ and variance $\sigma_x^2$ is
$$E[\exp(xt)] = \exp\left[\mu_x t + \frac{\sigma_x^2 t^2}{2}\right]$$

so that $E[y_t] = E[\exp(x_t)]$ is the same as the normal MGF evaluated at $t=1$, which implies that the mean function of $y_t$ is

$$E[y_t] = E[\exp(x_t)]= \exp\left[\mu_x + \frac{\gamma_x(0)}{2}\right]$$

EDIT 5:  I think I finally see where the dependence of $x_t$ plays in.  How is this?

Note that since $x_t$ and $x_{t+h}$ are identically distributed normal but not independent then the mean and variance of $x_t+x_{t+h}$ are

$$\mu_{x_t+x_{t+h}} = 2\mu_x$$

$$\sigma^2_{x_t+x_{t+h}} = \gamma_x(0) + \gamma_x(0) + 2\rho_x(h)\gamma_x(0) = 2\gamma_x(0)(1+\rho_x(h))$$

Thus the MGF for $x_t+x_{t+h}$ is

$$M_{x_t+x_{t+h}}(\lambda) = E[\exp(\lambda(x_t+x_{t+h}))] = \exp\left[2\mu_x \lambda + \gamma_x(0)(1+\rho_x(h))\lambda^2]\right]$$

By the same argument above using the MGF evaluated at $\lambda = 1$ we have

$$E[y_ty_{t+h}] = E[\exp(x_t+x_{t+h})] = M_{x_t+x_{t+h}}(1) = \exp\left[2\mu_x + \gamma_x(0)(1+\rho_x(h))]\right]$$
And we can calculate the autocovariance:

$$\gamma_y(h) = E[y_ty_{t+h}] - \mu_y^2 = \exp[2\mu_x + \gamma_x(0) + \gamma_x(0)\rho_x(h)] - \exp[2\mu_x +\gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\exp[\gamma_x(0)\rho_x(h)] - \exp[2\mu_x + \gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\left(\exp[\gamma_x(0)\rho_x(h)] - 1\right)$$

答:
Wrong! $E(exp(x_t+x_{t+h}))$ is not $E(x_t)^2$.  You need to take into account the autocorrelation., since they are not independent.

Formula for $A=E(exp(x_t+x_{t+h}))$:  To save writing, simplify notation.  $w=x_t-\mu$, $y=x_{t+h}-\mu$, $\rho=\gamma(h)/\gamma(0)$ = auto-correlation, $\alpha=\sqrt{1-\rho^2}$.

$A=\frac{1}{2\pi \sigma^2 \alpha}\int_{-\infty}^\infty\int_{-\infty}^\infty e^Kdwdy$ where $K=2\mu+\frac{1}{2\sigma^2 \alpha^2}(2\sigma^2 \alpha^2(w+y)-w^2+2\rho wy-y^2)$

The variables can be decoupled by $b=w+y$ and $c=w-y$ so that:

$K=2\mu+\frac{1}{2\sigma^2 \alpha^2}(2\sigma^2 \alpha^2 b-\frac{b^2+c^2}{2}+\rho\frac{b^2-c^2}{2})$

and $dwdy=\frac{1}{2}dbdc$

I'll let you do the calculation using the product of two normal distributions.


问:
Can any(some) generator(s) of adjoint representation of SU(3) (Gell-Mann construction) be reducible under SU(2)?
I was going through another thread on MathStack which was dealing with a specific question from Georgi, Lie Algebras in Particle Physics. I am not specifically looking for an answer/clarification with regards to Particle Physics. Also, the thread came off as somewhat abstract and in the accepted answer I happened to chance across terms which were unexplained. I am putting the question here again. I have answered the first chunk of the question apart from the adjoint part. I haven't seen any resource explain reducibility under a sub-representation.

Show that $T_1$ , $T_2$ and $T_3$ generate an SU(2) subalgebra of
  SU(3). Every representation of SU(3) must also be a representation of
  the subalgebra. However, the irreducible representations of SU(3) are
  not necessarily irreducible under the subalgebra. How does the the
  representation generated by the Gell-Mann matrices transform under
  this subalgebra. That is, reduce, if necessary, the three dimensional
  representation into representations which are irreducible under the
  subalgebra and state which irreducible representations appear in the
  reduction. Then answer the same question for the adjoint
  representation of SU(3).

I am unable to comprehend as to how to address the adjoint part in depth. I have been seeing quite a lot of hand waving arguments as to how the reduction may be done, but I guess I am having difficulty understanding how to show it either through brute force adjoint representation, or by using adjoint generator transformations.

Now, say, I have the three adjoint representation generators of SU(3),constructed by hand, which I did, how do I show that they are a Kronecker product of two irreducible representations of lower dimensions, under SU(2)

答:
Probably of no use to OP at this point but for anyone who might stumble in here like I did, you can "brute force" the problem by putting the adjoint $T_1$, $T_2$, and $T_3$ into block diagonal form with a similarity transformation. Take

$S = \frac{1}{\sqrt{2}}\begin{pmatrix}-i & 0 & -i &  &  &  &  &  \\ 0 & \sqrt{2} & 0& & & & &  \\ -1 & 0 & 1 & & & & &  \\  & & & 0 & -1 & -1 & 0 & \\ & & & -i & 0 & 0& -i & \\ & & & 0 & i & -i & 0 & \\ & & & 1 & 0 & 0& -1 & \\ & & & & & & & \sqrt{2}\end{pmatrix}$

to see how they break down into $SU(2)$ irreducible representations (it's not a product of two by the way, but a direct sum of four). As for the method to derive something like this on your own, I don't have a great answer. I can list the steps I went through though.

Pick one to diagonalize (I chose $T_2$)

Take inventory from its eigenvalues to get an idea of what representations you can make from them

Make a similarity matrix from the normalized eigenvectors

Trial &  error with the phase factors of each vector until a representation from your inventory appears in the block diagonal


问:
How can I prove the following isomorphism?
A profesor of mine said the following: Let $\mathbb{A}_{\mathbb{Q}}$ be the adele group of $\mathbb{Q}$. There exist a isomorphism of topological groups $$\frac{\mathbb{A}_{\mathbb{Q}}}{\mathbb{Q}}\simeq \varprojlim (\frac{\mathbb{R}}{n\mathbb{Z}})$$, where the limit of the right is considered with $\mathbb{N}$ ordered by divisibility. I tried to make a proof using that both spaces are compact and by trying to construct a continuous function. However, I have had no success. Do you know how to prove it or do you have any hint?

答:
Hints:

Show $\displaystyle \varprojlim_n \mathbb{Z}/n\mathbb{Z}\cong \widehat{\mathbb{Z}}$
where $\widehat{\mathbb{Z}}:=\prod_p \mathbb{Z}_p$.

$\displaystyle \varprojlim_n \mathbb{R}/n\mathbb{Z}\cong \mathbb{Z}\setminus \mathbb{R}
\times \varprojlim_n \mathbb{Z}/n\mathbb{Z}$ via noticing $\mathbb{R}/n\mathbb{Z}\cong \mathbb{R}/\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ canonically.

Show $\mathbb{Q}\setminus \mathbb{A}_{\mathbb{Q}}\cong \mathbb{Z}\setminus \mathbb{R}\times \widehat{\mathbb{Z}}$. For this, I think you need strong approximation property of $\mathbb{A}_{\mathbb{Q}}$ to get $\mathbb{A}_{\mathbb{Q}}=\mathbb{Q}+ \mathbb{Q}\times \prod_p \mathbb{Z}_p$, which suggests you how to define this isomorphism.


问:
Proof of derivative of $\frac{\ln x}{x}$ from limits
I have encountered the following proof of derivative from limits:
let $f(x)= \frac{\ln\ x}{x}$

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{\frac{\ln(x+\Delta x)}{x+\Delta x} - \frac{\ln\ x}{x}}{\Delta x} \end{equation}

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}

\begin{equation} f'(x) = lim_{\Delta x \rightarrow 0} \frac{\Delta x - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}

\begin{equation} f'(x) = \frac{1-ln\ x}{x^2}
\end{equation}

From the 2nd to the 3rd line, I don't understand how the term $x\ ln(1+\frac{\Delta x}{x})$ was simplified to $\Delta x$ ?

答:
Using $h$ in place of $\Delta x$

$\begin{align}
f'(x) &=\lim_{h\to 0}\frac{x\log(1+(h/x))-h\log x} {xh(x+h)} \notag\\
\end{align}$

Use series expansion of $\log (1+h/x)$, where $|h/x|\lt 1$,

$\log(1+h/x)=\frac hx-\frac 12(\frac hx)^2+\frac 13(\frac hx)^3-\cdots$

$\begin{align}
f'(x)&=\lim_{h\to 0}\frac{h(1 +\text{higher powers of $h$)}-h\log x}{xh(h+x)}\\
     &= \lim_{h\to 0}\frac{(1+\text{higher powers of $h$})-\log x}{x(h+x)}\\
     &=\frac{1-\log x}{x^2}\end{align}$


问:
If matrix $A_k$ is invertible for all $k$ but $\lim_{k \rightarrow \infty} A_k$ is not invertible, is $\lim_{k \rightarrow \infty} A_k = 0$?
Let $\{A_k\}_{k=1}^{\infty}$ be a sequence of real $n \times n$ matrices. Suppose $A_k$ is invertible for all $k \in \mathbb{N}$, and that $\lim_{k \rightarrow \infty} A_k$ exists but is NOT invertible. Is $\lim_{k \rightarrow \infty} A_k = 0$?

I have a hunch this follows from continuity of the matrix inverse, but can't quite work out a proof. Or maybe it's not true at all!

答: No.  First row of a $2 \times 2~A_k$ is $\left (\frac 1k, 1 \right )$, second row is $\left (0, \frac 1k \right )$.


答:
At the moment A turns around, C will have traveled a distance of $10\frac{d}{50}$, one-fifth the distance as A and B given that it is moving at one-fifth the speed. A and C will meet when they have together covered the distance between them, $4d/5$. We can expect that A will be doing most of the legwork, and that it will cover five times the separating distance as C.

We're interested in both the time ($t_{A\rightarrow C}$) and place ($x_{A\rightarrow C}$) of A and C's reunion. We'll measure time from the moment of the turn-around and distance from the starting point. We can set up a simple system of equations to express the unknowns in terms of d.
$$50t_{A\rightarrow C}+10t_{A\rightarrow C}=\frac{4d}{5}$$
$$x_{A\rightarrow C}=\frac{d}{5}+10t_{A\rightarrow C}$$
The solution is $t_{A\rightarrow C}=2d/150$ and $x_{A\rightarrow C}=d/3$. The total time for A to arrive at the end is
$$T=\frac{d}{50}+\frac{2d}{150}+\frac{120-d/3}{50}=\frac{360+4d}{150}$$
while B gets to the end in
$$T=\frac{d}{50}+\frac{120-d}{10}=\frac{600-4d}{50}.$$
As we are told they arrive at the same instant, then we find

$$4T=960/50$$
$$T=4.8 hrs$$

Title: Time, Speed and Distance : Returning back to pick up problem

Body:
$A$,$B$ and $C$ started out in a journey to a place which was $120$ km away. $A$ and $B$ went by car at the speed of $50$ kmph while $C$ travelled by tonga at $10$ kmph. After a certain distance $B$ got off and travelled the rest distance by another tonga at $10$ kmph, while $A$ went back for $C$ and reached the destination at the same time as $B$ arrived. The number of hours required for the trip was ____ $?$

I tried to approach the problem by equating the time taken by $A$ and $B$ as both of them has reached the destination at the same time after starting together. That means they must have taken the same amount of time.

Time taken by $B=\frac{d}{50} + \frac{120-d}{10}$ ; where $d$ is the distance travelled by $A$ and $B$ together on a car together.

Time take by $A=\frac{d}{50}$ + Time taken to return back an pick up $C$ + Time taken to reach the destination after picking up $C$

Time taken to return back an pick up C $\Rightarrow$ this I am not able to figure out as when A will return back to pick up C and until them meet, the C will be moving forward for some distance while A would be coming towards him and here the concept of relative distance will come into the picture. Getting confused here!!!

Please help me out.
Answer Provided : 4.8 hours


问:
How can we determine the order of zero for $\xi$?
Riemann Xi function, denoted as $\xi$, is given by $\frac{1}2s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$. I know this function is considered as entire, by canceling out the poles at $0,1$ by multiplying $s(s-1)$, but it seems off to me because I cannot literally plug in $0$ and $1$ for $xi$, because I cannot write $\zeta(0)$ since it doesn't exist. Is "cancellation" of poles implied here in this expression?

Also, how can I determine the order of zero for $\xi(z)$ at z=0? This is evaluated as zero at zero, so I think the order is at least 1. But since the pole was cancelled, is it just zero? Thanks for your help.

答:
$$\xi(s)=\frac{1}{2}s(s-1) \, \pi ^{-\frac s2} \,  \Gamma \left(\frac{s}{2}\right)\,\zeta (s)$$ is perfectly defined at $s=0$.

Using series expansions around $s=0$, we have
$$\pi ^{-\frac s2}=1-\frac{1}{2} s \log (\pi )+\frac{1}{8} s^2 \log ^2(\pi )+O\left(s^3\right)$$
$$\Gamma \left(\frac{s}{2}\right)=\frac{2}{s}-\gamma +\frac{1}{24} \left(6 \gamma ^2+\pi ^2\right)
   s+\frac{1}{24} s^2 \left(-\gamma ^3-\frac{\gamma  \pi ^2}{2}+\psi
   ^{(2)}(1)\right)+O\left(s^3\right)$$
$$\zeta(s)=-\frac{1}{2}-\frac{1}{2} s \log (2 \pi )+$$ $$\frac{1}{48} s^2 \left(24 \gamma
   _1+12 \gamma ^2-\pi ^2-12 \log ^2(2)-12 \log ^2(\pi )-24 \log (2) \log (\pi
   )\right)+O\left(s^3\right)$$
Putting all together
$$\color{blue}{\xi(s)=\frac{1}{2}+\frac{1}{4} s (-2-\gamma +\log (4\pi ))}+$$ $$\color{blue}{\frac{1}{32} s^2
   \left(-16 \gamma _1+8 \gamma -6 \gamma ^2+\pi ^2+2 \log (4 \pi ) (-4-2
   \gamma +\log (4 \pi ))\right)+O\left(s^3\right)}$$ This truncated expansion reproduces quite well the $\xi(s)$ function for $0 \leq s \leq 1$.

Just to givr an idea
$$\int_0^1 \Big[\xi(s)- \text{expansion}\Big]^2\, ds=2.81\times 10^{-9}$$


问:
Is it 3-D Catalan numbers?
I am studying Catalan numbers recently but I think that how about 3-D Catalan?
So that I imagine following situation ;

A man travel through the path-way parallel to   $  x, y, z $ axis from O $(0, 0, 0)$ to  $P (n, n, n)$ $ (n \in  N)$

  However he chooses the Points  $(x, y, z)$  satisfy   $ x \leq   y \leq z $  , and the path-way is always the shortest way  $(x, y, z \in  Z)$
  during all his trip.
  What number of path-ways he can choose in his trip?

Abobe situation ,
This is my question.

答: The problem from 2017 Cayley Contest of CEMC of University of Waterloo is about 3D Catalan Number.  The answer is: $\frac{2\cdot(3n)!}{n!\cdot(n+1)!\cdot(n+2)!} = \frac{2\cdot9!}{3!\cdot4!\cdot5!} = 42$

Title: Is it 3-D Catalan numbers?


问:
Why a point $P$ that maximizes $\angle OPQ$ must be a point of tangency of a circle with chord $OQ$?
I've seen the following problem with solution:

Can someone explain:

Why a point $P$ that maximizes $\angle OPQ$ must be a point of tangency of a circle with chord $OQ$?

How do we construct this circle? It seems he takes a circle with center $\left(\frac{1}{2}, k\right)$ but it's not clear to me where this chord $OQ$ must be constructed or where this circle must be constructed so that $OQ$ is a chord.

答:
(1) Let $P_1$ be a point on the tangent line but other than P.

Claim: There is always another point called $P_2$ (in between P and $P_1$) such that $\angle OP_2Q \ge \angle OP_1Q$.

Let the circle passing through $O,Q, P_2$ cut $QP_1$ at X. Then, $\theta = \theta’ \ge \phi$.

Note that $QP_2$ is shorter than $QP_1$ and the shortest that $QP_2$ can have is QP.

(2) The figure shows both P and P’ are points of tangencies but P’ does not meet the requirement and therefore should be rejected.


问:
Dirac delta from polar coordinates to cartesian coordinates
I have:

$$k_x = k \cos\theta\\k_y=k\sin\theta$$

I would like to rewrite in terms of $k_x$ and $k_y$:

$$\exp(in\theta)\,\frac{\delta(k-\alpha)}{k}$$

I start from:

$$\exp(in\theta)=\left(\cos\theta+i\sin\theta\right)^n=\left(k_x+ik_y\right)^n\,k^{-n}$$

but I don't if it helps and if yes how to go on.

答:
Your approach for eliminating $\theta$ seems preferable to the one suggested in jwg's answer. Combining it with jwg's $k=\sqrt{k_x^2+k_y^2}$ yields

$$
\exp(\mathrm in\theta)\frac{\delta(k-\alpha)}k=(k_x+\mathrm ik_y)^n\left(k_x^2+k_y^2\right)^{-(n+1)/2}\delta\left(\sqrt{k_x^2+k_y^2}-\alpha\right)\;.
$$

Depending on what you want to use this for, you may need to use the formula for composition of $\delta$,

$$
\delta(g(x))=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}\;,
$$

where the sum extends over all roots $x_i$ of $g(x)$, which are assumed to be simple.


问:
Relations between factorials
given a $m$ and $n$ both natural numbers such that $1 \leq n \leq m - 1$, how do we relate the factorial $(m - n)!$ in terms of $(n - 1)!$. In other words write $(m-n)!$ in terms of $(n - 1)!$. Im having issues in understanding which terms are leftover.

答:
It usually helps to work an example.  Suppose $m = 8$ and $n = 5$. Then $m - n = 3$.
So you want to express $3!$ in terms of $4!$.  If instead $n = 2$, then $m - n = 6$ and you want to express $6!$ in terms of $1!$.
So either expression could be larger, so it is unclear what you are looking for.
Neither is a factor of the other.
For context, let's imagine you are examining ${{m - 1} \choose {n - 1}} = \frac{(m-1)!}{(n-1)!\cdot (m-n)!}$. Now it becomes clearer that the two are only related in that the sum of the two numbers in the denominator being factorialized is the number in the numerator.
It is very striking to me that the expression always turns out to be an integer, not a fraction.  But I suspect the relation you are looking for does not exist.

Title: Relations between factorials


问:
Euler product over a finite subset of composite numbers
The product $$\prod_{p\text{ prime}}\frac{p^n}{p^n-1}=\zeta(n)$$ is well-known. This works over primes, and only over the entire set of them.

Is any similar expression known for composites? For a finite subset of them?

e.g. $$\prod_{c\text{ composite, c}=c_{i}}^{c=c_{j}}\frac{c^n}{c^n-1}=f(m)$$ for some $f$.

This is generic. If such an expression is not known, is anything known for the case $i=1$ and $n=1$? If not, then for $i=n=1$ and $j=\infty$? If that is also not known, is anything known on $\lim \inf f(m)$ or $\lim \sup f(m)$?

答:
My bit,  with some manipulations one can relate the zeta function $\zeta(m)=f_p(m)$ with the prime zeta function $P$

$f_p(m)
=\prod_p\frac{1}{1-p^{-m}}
=\exp\ln\prod_p\frac{1}{1-p^{-m}}
=\exp\sum_p\ln(\frac{1}{1-p^{-m}})
=\exp\sum_p\sum_k\frac{p^{-mk}}{k}
=\exp\sum_k\frac{1}{k}\sum_p\frac{1}{p^{mk}}
=\exp\sum_k\frac{1}{k}P(mk)
$

Now call $f_n(m)$ the product over all the naturals and let's do the same set of manipulation than before

$f_n(m)
=\prod_n\frac{1}{1-n^{-m}}
=\exp\ln\prod_n\frac{1}{1-n^{-m}}
=\exp\sum_n\ln(\frac{1}{1-n^{-m}})
=\exp\sum_n\sum_k\frac{n^{-mk}}{k}
=\exp\sum_k\frac{1}{k}\sum_n\frac{1}{n^{mk}}
=\exp\sum_k\frac{1}{k}(\zeta(mk)-1)
$

Finally the product over all composites $f_c(m)$ can be obtained as the quotient of the two other expressions since $f_n(m)=\prod_n\frac{1}{1-n^{-m}}=\prod_c\frac{1}{1-c^{-m}}\prod_p\frac{1}{1-p^{-m}}=f_c(m)f_p(m)$ so

$f_c(m)=f_n(m)/f_p(m)=\exp\sum_k\frac{1}{k}(\zeta(mk)-1-P(mk))$

if $i\neq1$ and $j\neq\infty$ instead of $\zeta(mk)$ and $P(mk)$ we should get a partial sum of those.


问:
How to get the weight of the orange sphere?
The problem is as follows:

The figure from below represents two scales, one is not in equilibrium while the other is at equilibrium. Assume both identical objects have the same weight and those with different shapes have different weights and each object weights an integer amount of pounds. With all this information, what would it be the least amount of the orange sphere?.

The choices given are:

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

From looking at the figure what I've attempted to do was as follows:

I'm using the notation:

$\textrm{Circle = C}$

$\textrm{Triangle = T}$

$\textrm{Square = S}$

The first figure shows:

$2c+t>s+2t$

The second figure shows:

$s+c=2t$

Hence it can be inferred that:

$2c>s+t$

$c=2t-s$

$2(2t-s)>s+t$

$4t-2s>s+t$

$3t>3s$

$t>s$

Thus the triangle weighs more than the square.

But from only these it is not possible to know an exact weight for each of these figures. How can I assume which weight goes to which?

If I assume the square weighs $1$ pound, then triangle can be $2$, $3,$4$, etc.

Because the weights of the square and circle must account for two triangles and the weight of all objects is an integer this means that either the square and circle are both odds or both evens.

We don't know which is more heavier than the other so I'm setting that the sphere might be heavier.

Since I set the square to be $1$ pound, then the circle must be $3$, $5$, $7$, etc.

In order to reassure the least possible weight for that sphere, I shall assume that it is ligther in weight than the square.

Thus I rearrange my initial analysis and get to:

The square is $3$ pounds, the circle is $1$ pound, but triangle would be $2$ pounds, this is a contradiction as the triangle weight is more than the square.

$s=2$, $c=2$, $t=2$, also yields a contradiction.

I can't go up for the square as it will remain in contradiction. Thus the only logical choice would be what I originally set. $s=1$, $c=3$, $t=2$.

Therefore the minimum weight that it can have the sphere is three pounds or choice 1.

Is there any way to confirm this?. Is my logic correct?.

答:
You have already found that
$$2c > s+t\quad (1)$$

Now, since you are interested in the weight of $c$, it might be better to search for a relation containing $c$ instead of removing it by using the equation

$$2t = c+s \iff t =\frac{c+s}{2}\quad (2)$$

Plugging $(2)$ into $(1)$ and rearranging gives

$$c >s$$

So, now you are almost done. Since $s$ is natural: $c > s \geq 1\Rightarrow c\geq 2$. But $t=\frac{c+s}{2}$ is also a natural number, hence $\boxed{c\geq 3}$ while choosing $s=1$ also to be minimal.


问:
How to calculate vertical displacement of particle rotating on circular wheel?
I am struggling with the following problem. It is a physics problem, but the step I am stuck at is essentially a trigonometry question.

A particle $P$ rotates on the circumference of a wheel of radius $R$ with the center at height $H$ above the ground. The wheel rotates some angle $\theta$ from its original position, at which point the particle is flung off the wheel with some velocity $v_p$. The problem asks what the maximum height the particle will attain above the ground. To solve for this, I need the initial height of the particle as a function of $\theta$.

To solve for the initial height, I realized that the initial height of the particle was
$H-R$. The height it gains can then be modeled by some $H(\theta)$, where $H(0) = 0$ and $H(90°) = R$. From there, I got stuck. I wasn't sure if at $\theta = 45°$ the particle is at height $\frac{R}{2}$ or not, and I wasn't sure how I could relate the two with a general formula.

Any hints would be greatly appreciated.

答:
Draw the perpendicular from $P$ to the vertical. Then in the right angle triangle formed, you get $$x=R\sin\theta$$ and the distance from the center (below the center) is $$y=R\cos\theta$$
Then the height above ground is $$h=H-y=H-R\cos\theta$$


问:
Apply Lipschitz condition to prove a condition on the coefficients of the exponential Fourier series of f
enter image description here

Question as in the image

note: I have been following Postol's Mathematical Analysis. I used the def to get integral formula for $$c_k=\int_0^{2\pi}f(x)\cdot e^{-ikx}$$, and got another formula of $$c_k=e^{-i\pi}\int_{-\frac{\pi}{k}}^{2\pi-\frac{\pi}{k}}f(t+\frac{\pi}{k})\cdot e^{-ikt}$$ using the change of variable hint. Now, I am supposed to get the average of these two formulas, and somehow apply the above Lipschitz condition to prove. (I got stuck at this step)

答: Hint: $$\int_{-\frac{\pi}{k}}^{2\pi-\frac{\pi}{k}}f(\frac{\pi}{k})\cdot e^{-ikt}=0$$ so $$c_k=e^{-i\pi}\int_{-\frac{\pi}{k}}^{2\pi-\frac{\pi}{k}}[f(t+\frac{\pi} k)-f(\frac {\pi} {k})]\cdot e^{-ikt}.$$  Now apply Lipschitz condition and integrate.


问:
Naively adding a dedicated error value to set-builder notation
Naive set-builder notation includes expressions that don't denote a set (in ZFC), like $\{ x \mathop| \top \}$. I'm wondering if we can do something extremely silly and ad hoc and just say by fiat that such expressions refer to a dedicated error value set $E$. Do we end up with a consistent theory (assuming that ZFC is consistent)?

This question is similar to this one, but I am deliberately making an odd choice for the semantics of set builder notation. Set-builder notation is not part of the syntax of ZFC, but is a bit of notation defined on top of the core theory and justified by the axiom schema of specification, which gives us existence under certain circumstances, and axiom of extensionality, which gives us uniqueness. I'm going to treat it informally as a core bit of syntax in an extended theory. For the purposes of this question, I'm thinking of set builder notation as a core bit of syntax that creates new terms, similar to a function symbol.

Let $E$ be a propertyless constant symbol in the theory of sets. Intuitively, it is an error value, but it is a set, not an urelement or a proper class.

Let $\{ t(\vec{x}) \mathop| \varphi(\vec{x}) \}$ be an expression in set-builder notation where $t(x)$ is a term with $\vec{x}$ as a sequence of free variables and $\varphi(x)$ is a formula with $x$ as a sequence of free variables.

Some expressions in set-builder notation do not name sets in ZFC, for example.

$$ \{ x \mathop| \top \} \;\; \text{is not a set in ZFC} $$

$$ \{ x \mathop| x \not\in x \} \;\; \text{is not a set in ZFC} $$

I'm wondering whether it's possible to make set-builder notation total in a certain sense by saying that any instance of it that would fail to name a set instead refers to $E$.

$$ \{ x \mathop| \top \} = E \;\; \text{because there is no universal set} $$

I've tried to come up with an additional pair of axiom schemas based on the axiom schema of specification that captures this new property. These axiom schemas do not replace the axiom schema of specification; they are additional and possibly redundant axiom schemas.

Every predicate has at most one set equal to its extension.

$$ \forall w_1 w_n \mathop. \exists_{\le 1} B \mathop. \forall x \mathop. (x \in B \leftrightarrow \varphi(x, w_1, w_n)) $$

A term using set-builder notation is equal to $E$ if-but-not-only-if it has no extension as a set. The outermost connective can't be a biconditional because of $\{ x \mathop| x \in E \}$.

$$ \forall w_1 w_n \mathop. (\nexists B \mathop. \forall x \mathop. (x \in B \leftrightarrow \varphi(x, w_1, w_n)) \to \{z \mathop| \varphi(z, w_1, w_n) \} = E ) $$

答:
It seems like you can make the following adjustments to ZF(C) to make this work.

For all the normal axioms of ZF, make sure the quantifiers range over non-E things. EG $\exists x P(x)$ becomes $\exists x (x \neq E \land P(x))$ and $\forall x P(x)$ becomes $\forall x (x \neq E \to P(x))$.

Add the axiom scheme $\forall u_1 \forall u_2 ... \forall u_n [\exists y (\forall x (P(x, u_1, ..., u_n) \iff x \in y)) \iff \{x | P(x, u_1, ..., u_n)\} \neq E]$, where $u_1, u_2, ..., u_n, x, y$ range over non-$E$ things.

Add the axiom scheme $\forall u_1 \forall u_2 ... \forall u_n [\exists y (\forall x (P(x, u_1, ..., u_n) \iff x \in y)) \iff \forall v (v \in \{x | P(x, u_1, ..., u_n)\} \iff P(v, u_1, ..., u_n))]$ where $u_1, u_2, ..., u_n, v, y, x$ range over non-$E$ things.

It seems like a pretty straight-forward relative consistency proof at this point. Start with a model of ZF and add a single element $E$. It should be easy to verify the axioms hold in the new model. However, I wouldn't recommend this.

The most elegant way to do something like what you're trying to do is with classes. The idea is that every piece of notation $\{x | P(x)\}$ defines a class*, but not all classes are sets. One can then say things like "$\{x | \top\}$ is not a set" directly in the theory.

The most well-known class-based theory is NBG set theory. NBG has the nice property that it's a "conservative extension of ZFC". This means the following:

All statements made in ZFC can be translated into statements made in NBG.

A statement can be proved in ZFC if, and only if, its translation can be proved in NGB.

This means that working in NGB is perfectly compatible with working in ZFC.

*as long as all quantifiers in P(x) range only over sets and not over classes


问:
Skew-symmetry $\nabla_{[X,Y]}Z = -\nabla_{[Y,X]}Z$
I'm new to the theory of affine connections and have yet to become comfortable working with. I wanted to confirm to myself that the curvature of an affine connection is skew-symmetric which boils down to showing that $$\nabla_{[X,Y]}Z = -\nabla_{[Y,X]}Z$$

where $X,Y,Z$ are assumed to be smooth vector fields on a manifold $M$ and $[X,Y]$ denotes the Lie bracket.

Can we show this by using the $C^\infty$-linearity of $\nabla$ in the following way?

$$\nabla_{[X,Y]}Z = \nabla_{-[Y,X]}Z = \nabla_{-\operatorname{id}[Y,X]}Z = -\operatorname{id}\nabla_{[Y,X]}Z = -\nabla_{[Y,X]}Z $$

I.e. would this attempt be valid? I hope this does not come across as low effort, I just wanted to clarify wether this is the correct way to handle it.

Thank you very much!

答: Forget connections for one second: $[X,Y]=-[Y,X]$ from the definition of Lie bracket. Then $$\nabla_{[X,Y]}Z = \nabla_{-[Y,X]}Z = - \nabla_{[Y,X]}Z.$$Quoting $C^\infty$-linearity to pull out ${\rm id}$ is meaningless as this is not a real-valued function.


问:
Find $P(4 \leq X^2+Y^2 \leq 9)$
Let $X$ and $Y$ be independent uniformly distributed random variables with $\mathbb{E}X=\mathbb{E}Y=0$ and $\mathbb{D}X=\mathbb{D}Y=4$. The task is to find following probability:
$$P(4\leq (X^2+Y^2) \leq 9)$$
P.S: Here $\mathbb{D}$ stands for variance.

I've tried the following:

Let's find $\varphi_{X^2+Y^2}(t)=(\varphi_X^2(t))^2$
$\varphi_X^2(t)=...=\cfrac{1}{\sqrt{8-it}}$, so $\varphi_{X^2+Y^2}(t)=\cfrac{1}{1-8it}$, and that leads us to: $X^2+Y^2 \sim \text{Exp}(\frac{1}{8})$.

And so on.
However I was told that this solution is not correct and there is more efficient and smart one. May you help me?

答:
Probably $\mathbb{D}X$ is the variance.

The problem is very simple. If you do a drawing, the requested probability is the probability of  an annulus resulting as the difference of the two disks, the greater with radius $3$ and the inner with radius $2$

Knowing mean and variance you can calculate $f_X$ and $f_Y$

Thus the solution is simply the purple area multiplied by the joint density or, what is the same, the purple area divided by the total area of the square.

If I assumed correctly $\mathbb{D}$ as Variance, easy find that

$$f_X(x)=\frac{1}{4\sqrt{3}}\cdot\mathbb{1}_{(-2\sqrt{3};2\sqrt{3})}(x)$$

thus the requested probability is

$$\mathbb{P}[4\leq X^2+Y^2\leq 9]=\frac{1}{16\cdot3}(9-4)\pi\approx 32.72\%$$

Title: Find $P(4 \leq X^2+Y^2 \leq 9)$


问:
Questions about Hatcher 3.2.16
Hatcher's exercise 3.2.16 states

Show that if X and Y are finite CW complexes such that $H^*(X;\mathbb{Z})$ and $H^*(Y;\mathbb{Z})$ contain no elements of order a power of a given prime p, then the same is true for $X \times Y$.

Now the hint says to use the Künneth Formula

$H^*(X;R)\otimes_R H^*(Y;R) \cong H^*(X \times Y ;R)$ when X,Y are CW complexes and $H^k(Y;R)$ is a finitely generated free R module.

for various fields R.

Here's what I need help with (my attempts are also included):

Firstly, I'm looking for a clarification about the question. I know the following logic is flawed but I can't figure out why:
Since X is a finite CW complex, $H^k(X,R) = 0$ for all $k \geq m$ for some number $m$ since there are only finitely many cells. So for $\alpha$ in $H^1(X)$, we have that $\alpha^m \in H^m(X;R) = 0$ (product given by the cup product $\alpha \smile \alpha)$. So pick $n$ such that $p^n > m$, then $\alpha^{p^n} = 0$ and $H^*(X)$ has an element of order $p^n$ always. I'm aware that I'm probably missing something basic.

Assuming the question is well posed and that 1) is not an issue, then my first thought was to use the formula not for a field but with $R = \mathbb{Z}$. If $H^*(X;\mathbb{Z})$ and $H^*(Y;\mathbb{Z})$ have no elements of order $p^n$ for any n, then my first guess is that they are torsion free. If they are indeed torsion free, then by the Universal Coefficient Theorem, we have that $H^*(X;\mathbb{Z}) \cong \mathbb{Z}^r$ and $H^*(Y;\mathbb{Z}) \cong \mathbb{Z}^l$ so $H^*(X\times Y;\mathbb{Z}) \cong H^*(X;\mathbb{Z}) \otimes H^*(Y;\mathbb{Z}) \cong \mathbb{Z}^{rl}$ which is torsion free. But I haven't been able to show why $H^*(X;\mathbb{Z})$ would need to be torsion free. If this is correct, then I'm looking to see why that is the case, and if not, then a counterexample.

Assuming I was wrong and the approach  in 2) was incorrect, I tried thinking about Hatcher's hint more carefully and then decided to look at $H^*(X;\mathbb{Z}/p\mathbb{Z})$ as a $\mathbb{Z}/p\mathbb{Z}$ module. I couldn't get very far but the hope was to extend $H^*(X;\mathbb{Z}) \rightarrow H^*(X;\mathbb{Z}/p\mathbb{Z})$. My guess being that the latter is a free $\mathbb{Z}/p\mathbb{Z}$ module and by the Künneth formula, $H^*(X \times Y;\mathbb{Z}/p\mathbb{Z}) \cong H^*(X;\mathbb{Z}/p\mathbb{Z}) \otimes H^*(Y;\mathbb{Z}/p\mathbb{Z})$, which is an extension of $H^*(X \times Y;\mathbb{Z})$ and was hoping to construct an order $p^n$ element here towards a contradiction. No luck so far, but any advice would be great.

I also found this MSE post which I didn't find too helpful. The comments seem to suggest studying the order of elements or use Lagranges Theorem in the tensor product but wouldn't we expect the tensor to be free over $\mathbb{Z}$ and have elements of infinite order and Lagranges theorem only applies to finite groups?

答:
If $X$ is a finite CW complex, then its rational homology $H^n(X; \mathbb{Q})$ is a rational vector space. By the Universal Coefficient Theorem (Theorem 3.2 in Hatcher, applied with $G=\mathbb{Q}$), its dimension is the rank of $H_n(X)$ as a free abelian group: the finite summands of $H_n(X)$ do not contribute (since there are no maps from a finite group into $\mathbb{Q}$), and nothing from $H_{n-1}(X)$ contributes (since $\mathbb{Q}$ is injective as an abelian group, or you can use the properties of Ext that Hatcher gives just after the theorem to see that the Ext group is zero when $G=\mathbb{Q}$).

If $p$ is a prime, the only way for $H^n(X)$ to have elements of order a power of $p$ is for $H_n(X)$ to also have elements of order a power of $p$, this time by the Universal Coefficient Theorem with $G=\mathbb{Z}$. So if $H^n(X)$ has no elements of order $p^m$ for any $m$, then the same is true for $H_n(X)$. In this case, $H^n(X; \mathbb{F}_p)$ is an $\mathbb{F}_p$-vector space with dimension again equal to the rank of $H_n(X)$ as a free abelian group: any summands of order prime to $p$ in $H_n(X)$ will not contribute to $H^n(X; \mathbb{F}_p)$ by the Universal Coefficient Theorem (with $G=\mathbb{Z}/p$).

Now by the above and the Künneth theorem with various coefficients, the dimension of $H^n(X \times Y; \mathbb{F}_p)$ is equal to
\begin{align}
\dim H^n(X \times Y; \mathbb{F}_p) &=
\sum_{i=0}^n \dim H^i(X; \mathbb{F}_p) \dim H^{n-i}(Y; \mathbb{F}_p) \\
&= \sum_{i=0}^n rk H_i(X) rk H_{n-i}(Y) \\
&= \sum_{i=0}^n \dim H^i(X;\mathbb{Q}) \dim H^{n-i}(Y; \mathbb{Q}) \\
&= \dim H^n(X \times Y; \mathbb{Q}) \\
&= rk H_n(X \times Y).
\end{align}
On the other hand, if $H^*(X \times Y)$ contained an element of order $p^m$ for some $m$, then the same is true for $H_*(X \times Y)$, and so the dimension of $H^n(X \times Y; \mathbb{F}_p)$ will not equal the rank of $H_n(X \times Y)$ for some $n$.

Title: Questions about Hatcher 3.2.16