问:
Chances of getting exactly one heads in two coin flips?
I know the chances of getting exactly one "heads" in two coin flips is 50%, as there are only four possible outcomes: HH, HT, TH, TT; and only two of those four meet the restraint I've placed. To figure this out, I think you can use a binomial distribution: 2C1 times 1/2 times 1/2 gives the probability of 1 in 2 that I'm looking for.

But why doesn't the formula $$P(A \text{ or }B) = P(A) + P(B) - P(A \text{ and } B)$$ (where A and B are heads on the first and second attempts, respectively) work? I get 75% (1/2 + 1/2 -1/4) which would work if I wanted the restraint "at least one 'heads,'" or simply what are the chances of getting one "heads."

Moreover, if you were to find the chances of getting "heads," wouldn't you take 1 minus the probability of getting two "tails" (25%) to get the 75% you wanted? So when would you use P(A or B)?

答:
None of these "solutions" is correct because when a coin is flipped 2 times, there ARE NOT  4 possible outcomes. There are only 3 possible outcomes: 2 heads, 2 tails, or 1 head and 1 tail.
The answer is 5 out of 12, with the probability of getting exactly 2 heads (or exactly 2 tails) 7 out of 24.
A statistician (long since deceased) calculated this for me, but unfortunately did not share his calculations with me.
The answer can be verified by flipping a coin a large number of times (100 times is sufficient, but the more flips the more precise) and recording the results. Next record the results of consecutive "pairs" of results i.e. HH, HT, TH,TT which will confirm that 1 head and 1 tail occurs 41.6% of the time.</p>


问:
How to tell whether two functions are linearly independent by inspection of their graphs?
In $\mathbb{R}^2$, it's known that linearly independent vectors means that they are not parallel.  But with functions, how do the graphs of linearly independent functions look like?

For instance, how can I know by inspection that $p_1(x) = 1-x^2$ and $p_2(x)= 2-x^2$ are linearly independent?

答:
Vector spaces are abstract entities in mathematics, $\mathbb{R}^{2}$ and even a little $\mathbb{R}^{3}$ are intuitive and rich in geometry, however vector spaces like $P_{n}(\mathbb{R}), M_{m\times n}(\mathbb{R})$,etc, are difficult to analyze geometrically.

For this, it is best to obtain the coordinates of the vectors and then analyze the geometry of the vectors.

Note that
$$[1-x^{2}]_{B}=\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$$
and $$[2-x^{2}]_{B}=\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} $$
considering vector space $P_{2}(\mathbb{R})$, $S=\{1-x^{2},2-x^{2}\}$ and a basis for $\mathbb{R}^{3}$ given by $B=\{(1,0,0),(0,1,0),(0,0,1)\}$ and since that $P_{2}(\mathbb{R})$ is isomorphic to $\mathbb{R}^{3}$, so you can get a geometric idea of the linear independence dependence of this set of vectors you mention.

ContentLicense: CC BY-SA 4.0


问:
Las Vergnas theorem

I am supposed to prove that the graph satisfies the Las Vergnas
theorem. Here is the graph:

Las Vergnas theorem is the following:

$G$ is a graph and $V(G)=(v_1,...,v_n)$. There is no such $i,j$ such that $i<j, i+j \geq n, v_iv_j \notin E(G),
deg(v_i) \leq i, deg(v_j) \leq j-1, deg(v_i) + deg(v_j) \leq n-1$. Then $G$ is hamiltonian

But if I label the vertices like this:

I found such $i$, $j$: $i=2$ and $j=4$, so the graph does not satisfy Las Vergnas theorem.

Is there something, which I misunderstood or where is the mistake.

Thanks

答:
Let me quote the actual theorem (as stated in Berge's Graphs and Hypergraphs, on p. 208 of the 1973 edition):

Let $G=(X,E)$ be a simple graph of order $n \ge 3$. Let the vertices $x_i$ of $G$ be indexed arbitrarily, and let $q$ be an integer, where $0 \le q \le n-1$. If $$\left.\begin{array}{l}1 \le i < j \le n \\ i+j \ge n-q \\ d_G(x_i) \le i+q \\ d_G(x_j) \le j+q-1 \\ [x_i,x_j] \notin E\end{array}\right\} \implies d_G(x_i) + d_G(x_j) \ge n+q$$ then, for each $F \subset E$ with $|F|=q$ such that the connected components of $(X,F)$ are elementary chains, there exists a hamiltonian cycle that contains $F$.

Here, we want $q=0$, so the conditions are the same as in the question; the point that I am making is that the hypotheses in the question are satisfied if there is any ordering of the vertices that works.

In the case of your graph, if we swap the order of vertices $v_1$ and $v_2$ in your labeling, there will no longer be any pair $(i,j)$ violating the conditions, because the only two vertices in the right positions with the right degrees will be adjacent. So the graph does satisfy the hypotheses of the theorem.

As a side note: it is incorrect to say "the graph satisfies the theorem" or "the graph does not satisfy the theorem". All graphs satisfy the theorem: it's a theorem! Some graphs don't satisfy the hypotheses of the theorem, but that's okay. The only counterexample would be a graph that did satisfy the hypotheses of the theorem, but wasn't hamiltonian: such a graph would not satisfy the theorem. Such graphs do not exist.

Title: Las Vergnas theorem


问:
Lebesgue integral equals Riemann Integral on $\mathbb{R}$
I know that if a function $f:[a,b]\rightarrow \mathbb{R}$ is Riemann Integrable  then it is Lebesgue Integrable there with respect to the lebesgue measure and the integrals coincide.
Does this theorem also apply to improper Riemann integrability on unbounded intervals of the real line e.g. over $\mathbb{R}=(-\infty,\infty)$ itself?

答:
No. remember that $f$ is Lebesgue integrable if and only of $|f|$ is Lebesgue integrable. But this isn't true for Riemann (improper) integrability.

So theere may exists functions that are Riemann (improper) integrable but such that its absolute value is not integrable. The classical example is $\dfrac{\sin x}{x}$ over $(0,+\infty)$.

It can be shown that $\int_0^{+\infty}\dfrac{\sin x}{x}$ converges but $\int_0^{+\infty}\dfrac{|\sin x|}{x}$ diverges. It is known as Dirichlet Integral).

One can see this function as the equivalent to the alternating Harmonic series.


问:
Thermodynamics for math majors
I'm about to wrap a course in partial differential equations. We've discussed the heat/wave equations and introductory Fourier Analysis.

I'd like to do some reading into the field of thermodynamics. Would it be best to start with an introductory text given to most lower-division under graduates or should I begin with something with a little more meat on the bones? By meat on the bones, I mean a bit more rigor and explanation for the fundamental theorems and applications. If this is not clear enough please ask for clarification, I'll be glad to provide it.

答:
In my opinion Thermodynamics is physics, chemistry and engineering.
In thermodynamics  mathematical operations are used that are unwarranted and would not be acceptable anywhere else. (eg applying a Legendre transform, without solving the differential equation involved.) I would start with 'Cengel and Boles' not with Callen.
As Gibbs noted:  "A mathematician may say anything, but a physicist must be partially sane."</p>

ContentLicense: CC BY-SA 3.0

Title: Thermodynamics for math majors


问:
Struggling to find solution: question 2
$$xy'-2y=\frac{x^6}{(y+x^2)}$$

Can someone please help with this one?

It looks like a bernoulli equation, I have tried to multiply both sides by denominator and gather them in parantheses to leave the $y'$ alone but to no avail.

Title: Struggling to find solution: question 2

Body:
$$xy'-2y=\frac{x^6}{(y+x^2)}$$
$$x^2y'-2xy=\frac{x^5}{(1+y/x^2)}$$
$$\left (\dfrac {y}{x^2}\right )'=\frac{x}{(1+y/x^2)}$$
It's separable:
$$\left ( \dfrac y{x^2}+1 \right)d\left (\dfrac {y}{x^2}+1\right )={x}dx$$
Integrate.
$$\boxed {\left ( \dfrac y{x^2}+1 \right)^2-x^2=C}$$


问:
Solving $a(a + 1) + b(b + 1) = 12, (a+ 1)(b+1) = 4$, is there any trick?
What are all (possibly complex) solutions to the following two equations?

$a(a + 1) + b(b + 1) = 12$

$(a+ 1)(b+1) = 4$

Clearly one can substitute the second equation into the first then solve the quartic. But I am wondering if there is a shorter/sneakier way to solve this.

答:
Let $u = a+1, v=b+1$. The equations become

$$(u-1)u+(v-1)v = 12, uv = 4$$

Thus $uv+(u-1)u+uv+(v-1)v = u(v+u-1) + v(u+v-1) = (u+v)(u+v-1) = 20$.

Let $x = u+v$ and we have a quadratic, which yields $u+v = 5$ or $-4$.

Solve the rest using this info and $uv=4$, which is just another quadratic.

EDIT: "Solving the quartic" doesn't seem too bad either, by substitution:

$$(u-1)u+\frac4u\left(\frac4u-1\right)=12$$

$$u^2+\frac{16}{u^2}-u-\frac4u=12$$

Note that $\left(u+\frac4u\right)^2=u^2+8+\frac{16}{u^2}$. Hence we can rewrite the above into:

$$\left(u+\frac4u\right)^2 - \left(u+\frac4u\right)=20$$

which, of course, is a quadratic equation with $u+\frac4u = 5$ or $-4$, and thus we arrive at essentially the same conclusion as the above (slightly unnatural) method.


问:
Spectral theorem, inner products and adjoints/transposes: what's the relation?
I only consider operators on finite-dimensional vector spaces.

I'll write down my question and explain my motivation afterwards:

The spectral theorem relates diagonalizability, which is a purely algebraic notion, with adjoints, which requires inner products. Why is that so? Is there a non-complex/real version of the spectral theorem?

Consider the following version of the Spectral Theorem:

Spectral Theorem: Every self-adjoint operator on a finite-dimensional (complex) inner product space admits a basis consisting of eigenvectors ("eigenbasis")

I am teaching Linear Algebra for the first time, so I am revisiting lots of concepts I studied several years ago, and just got to the Spectral Theorem. So this is a little funny, since "diagonalizability"/existence of an eigenbasis does not relate directly to inner products in any obvious manner.

Yes, I understand the point that self-adjointness of an operator $A$ takes in the proof: Every $A$-invariant subspace has an $A$-invariant complement. A similar statement holds for normal operators and eigenspaces.

But the point seems to be that it just so happens that a condition related to inner product implies in a condition related to invariant subspaces, and it is not clear to me why that is so from a non-purely formal perspective.

So that got me thinking: How would one be able to get a spectral theorem for non-complex/real spaces? The more I think the more I convince myself the condition needs to deal with some sort of transpose. But the problem is that transposes do not make much sense in general, and this is also another point: Inner products seem to just-so-happen allow us to define adjoints/transposes of operators.

The point seems to be that, in general, transposes "live" on dual spaces, whereas Riesz representation allows us to go back to the level of the original spaces.

So I would imagine the abstract version of the Spectral Theorem (if there is one) would need a condition relating an operator $T\colon V\to V$ and its formal transpose $T'\colon V'\to V'$, perhaps passing to the bidual if necessary (as this is naturally isomorphic to the ground space), but I can't find any possible formula involving both $T$ and $T'$.

答:
There are a few ways one could go with this question, but I will focus on the following.

Is there a satisfying explanation how the "geometric" property of a transformation (being self-adjoint or normal) ensures the "algebraic" property that each invariant subspace has an invariant complement?

First, let's consider normality and the complex version of the spectral theorem.
Let $\langle \cdot, \cdot \rangle$ denote our inner product. The statement that $A:V \to V$ is normal means that $A^*A = AA^*$. As it turns out, this is equivalent to the statement that for all $x \in V$, we have
$$
\|Ax\|^2 = \langle Ax, Ax \rangle = \langle A^*x, A^*x \rangle = \|A^*x\|^2.
$$
If you haven't seen this characterization before, you might find coming up with a proof to be an engaging exercise. This property ensures that $(A - \lambda I)$ and $(A - \lambda I)^*$ have the same kernel. After all, it is easy to see that $(A - \lambda I)$ is normal, and it follows that $\|(A - \lambda I)x\| = 0 \iff \|(A - \lambda I)^*x\| = 0$.

With that established, we can see the algebraic property as follows. We note that every non-zero $A$-invariant subspace contains an eigenspace associated with some eigenvalue $\lambda$; that is, the (subspaces of) eigenspaces are the minimal invariant subspaces. We see that each eigenspace has an invariant complement since
$$
\ker(A - \lambda I) = \ker([A - \lambda I]^*) = \operatorname{im}(A - \lambda I)^\perp.
$$
Note that $\operatorname{im}(T)^0 = \ker(T^*)$ usually relates invariant subspaces of a transformation to the invariant subspaces of its dual. However, the normality condition (defined relative to an inner product) means that we now have a relation between two invariant subspaces of $T$.

Because eigenspaces have invariant complements, it follows that all invariant subspace have invariant complements. First of all, note that all subspaces of an eigenspace are invariant, which means that subspaces of eigenspaces will also have invariant complements (since we can combine the complement of the subspace within the eigenspace with the complement to the eigenspace as a whole).
From there, note if $U,W$ are subspaces with respective complements $P,Q$, then $U + W$ will have complement $P \cap Q$.

With the complex version of the spectral theorem addressed, the condition of being self-adjoint can be characterized as "normal with real eigenvalues", which lets us arrive at the same result for the same invariant subspace properties in the real version of the spectral theorem.


问:
How long does it take for an object to slide down an inclined plane?
Suppose you have an object that is at height h and slides down a length l. How long does it take to slide this distance?

From my understanding, on an inclined plane, $s=\frac{1}{2}sin(\theta)\alpha t^2$, Hence, $l=\frac{1}{2}sin(\theta)\alpha t^2$ where $\theta$ is the angle of the plane and $\alpha$ is the downward acceleration.

The answer provided is $\frac{l}{4\sqrt{h}}$ But I do not know how to get to this expression.

答:
Taking a coordinate system with the $x$-axis parallel to the slope, you get according to Newton's second law:

$$mx^{\prime \prime}(t) = m \alpha \sin \theta$$ where $\alpha$ is the gravity.

And therefore $x(t) = \frac{1}{2} \alpha \sin \theta t^2$ if the object starts at time $t=0$ at the origin of the coordinate system with no velocity. You have $\frac{h}{l} = \sin \theta$.

We're looking for $T$ such that $x(T) = l$ which means

$$l=  \frac{1}{2} \alpha \frac{h}{l} T^2$$ and finally

$$T = l \sqrt{\frac{2}{\alpha h}}$$

...and I don't understand where the expression you were given is coming from. By the way, if I understood correctly the question, if $\theta = \pi/2$ you get $l=h$ and the expression I got is coherent with a free falling motion.


问:
Counting cycles in undirected graphs
I'm trying to write a program that given adjacency matrix of an undirected graph, computes the number of cycles of different sizes using matrix operations and linear algebra. I read here that

If $A$ is the adjacency matrix of the directed or undirected graph $G$,
then the matrix $A^n$ has an
interesting interpretation: the element $(i, j)$ gives the number of
(directed or undirected) walks of length $n$ from vertex $i$ to vertex $j$.
If $n$ is the smallest nonnegative integer, such that for some $i$, $j$, the
element $(i, j)$ of $A^n$ is positive, then n is the distance between
vertex $i$ and vertex $j$.

The text concludes that

This implies, for example, that the number of triangles in an undirected
graph $G$ is exactly the trace of $A^3$ divided by $6$.

I was wondering if this can be extended to cycles of larger size.

答:
Triangles are special in this regard.

The only possible closed walk of length $3$ is a 3-cycle, so the only closed walks from a vertex to itself counted by $A^3$ are the desired cycles.

However for longer closed walks, not all of these are cycles. For all $k>3$, there exists at least one up-to-isomorphism closed walk that is not a cycle. (Exodd's example shows you how to find ones of even length).

Title: Counting cycles in undirected graphs


问:
Evaluation of hard expression with repeated exponentiation and modulo
Assume we want to calculate

$$
N = z^{1998000^{100^{10}}} \pmod{10^m}
$$

where $z, m$ are known ($z$ can reach up to $10$ digits and let's assume that $m$ is around $10$).

According to similar questions, it seems that the usual method is to find a $k$ such that:

$$
z^k \equiv 1 \pmod{10^m} \tag{1}
$$

Then proceed with finding:

$$
19988000^{100^{10}} \pmod{k}
$$

For solving $(1)$, the only thing that comes to mind is Euler's generalization of Fermat's little theorem, but that carries the assumption that $\text{gcd}(z,10^m) = 1$, which is not generally known.

Can you suggest a way to solve $(1)$ and how would you proceed with finding $N$ from there?

答:
First suppose $\gcd(z,10^m)=1$. Then we can take $k=4\cdot 10^{m-1}$, and $19988000^{100^{10}} \pmod{k}$ is clearly $0$, so $N=1$.

Now suppose, say, that $z$ is even and not divisible by $5$. Then $N\equiv 0\pmod {2^m}$; and to compute $N\bmod 5^m$, we can take $k=4\cdot 5^{m-1}$, so $N\equiv 1\pmod{5^m}$. Now you can use the Chinese Remainder Theorem to find $N\pmod{10^m}$.

The case when $z$ is an odd multiple of $5$ is similar.


问:
Does smoothness depend on parameterization?
Consider an algebraic curve:

$$ \vec{r(t)} = t \vec{i}  + t^2 \vec{j}$$

It has a tangent vector:

$$ \frac{d \vec{r}}{dt} = \vec{i} + 2 t \vec{j} \tag{1}$$

Now, consider parameterizing the first equation by sub $ t \to t^3$, then:

$$ \vec{r(t^3)} = t^3 \vec{i} + t^6 \vec{j}$$

Now the derivative of this is also (w/ chain rule considerations):

$$ \frac{dr}{dt}|_{t^3} = \vec{i} + 2t \vec{j} \tag{2}$$

This leads me to a weird conclusion, if I put t=2 then the tangent vector corresponding to that equation in (1) is $ \vec{i} + 2t \vec{j}$. However in the second parameterization, for t=2, he points on the curve is where $t=8$,* at that point it is also having the same tangent vector!! How can different points on the curve have the same tangent vector?**

Secondly, consider calling:

$$ \vec{q(t)} =  \vec{r(t^3) } = t^3 \vec{i} + t^6 \vec{j}$$

Then, q is not smooth! because the $ \frac{dq}{dt}|_0 = 0$, how is it that when I reparamterizaed the curve using another function , it became not smooth? Aren't they both representing the same geometric object?

答:
The behavior of a parametric curve depends not only on the set of points that the parametric curve passes through (the "geometric curve"), but also the chosen parameterization of that curve. Properties of the geometric curve can be defined in terms of the set of points only. But if a property requires the parameter to define, it is a property of the parametrization, not of the geometric curve.

It is possible to define geometric properties in terms of parameters. But such properties will not change if you change the parametrization. If a change of parametrization also changes the property, then that property is not geometric. It is a property of the parametrization.

The tangent vector to a curve is defined in terms of the parameter:
$$\left.\dfrac{d\vec r}{dt}\right|_{t_0} := \lim_{t\to t_0} \dfrac{\vec r(t) - \vec r(t_0)}{t-t_0}$$

And a change of parametrization changes the tangent vector. If $t = t(u)$ and $\vec r(t(u)) = \vec q(u)$, then

$$\left.\dfrac{d\vec q}{du}\right|_{u_0} = \left.\dfrac{d\vec r}{dt}\right|_{t(u_0)}\left.\dfrac{dt}{du}\right|_{u_0}$$

Thus if $\frac{dt}{du} \ne 1$ (which is easily accomplished), you get a different vector for one parametrization than for the other.

What you can define without respect to the parametrization is the tangent line. Note in the equation above that while the magnitude of the tangent vector changes, the direction does not - except for possibly flipping to point in the anti-parallel direction. But both parallel and anti-parallel directions point along the same line. Thus the tangent line at the point is unchanged by the reparametrization. (Yes, you can reparametrize to make a $0$ tangent vector, or to make it undifferentiable at the point, but those just fail to properly define a tangent line, they don't change it.) The tangent vector is a parametric property, but the tangent line is a geometric property.

"Smoothness" is the existence of the tangent line, not the tangent vector, so it is also a geometric property.

Also, when you change the parametrization, you are changing which points are assigned to which values of the parameter. So to compare the values of two different parametrizations when the parameter is $2$ is nonsensical - at least as far as geometrical properties are concerned. These are at two diffent points on the geometric curve. They might have some properties in common, but that would only be by coincidence.

Finally, you are mistaken in thinking your $\vec r(t)$ and $\vec r(t^3)$ parametric curves have the same tangent vector at the two points. Though you claimed to have accounted for the chain rule, you did not do so correctly.


问:
Find the limit of a recursive sequence with parameter
Let the following recursive sequence:
$$a(0)=\alpha\in\mathbb R;\ a(n+1)=\max\{1/2,a(n)^2\}$$
find the limit as n goes to infinity.

Work: (wondering what's wrong on my answer. Thank you very much)

答:
I think there are three cases,

$|\alpha| > 1$ then $\lim_{n \to \infty} a(n) = +\infty$

$|\alpha| = 1$ then $\lim_{n \to \infty} a(n) = 1$

$|\alpha| < 1$ then $\lim_{n \to \infty} a(n) = 1/2$


问:
Counting the Number of Solutions Satisfying the Inequality, $x_1 + x_2 +\dots+ x_7 \leq 86$
I am trying to solve find the number of solutions to the inequality

$x_1 + x_2 + \dots + x_7 \leq 86$

Where $x_i$'s are $\textbf{odd}$ positive integers, and the order matters.

If we ignore the constraint that $x_i$'s have to be odd, we can introduce another variable $x_8$ such that we have an equality: $x_1 + x_2 + \dots + x_7 + x_8 = 86$. Now a solution to this equation corresponds to a way of choosing $86$ items from a a set with $8$ elements such that there are $x_1$ items of type one, $x_2$ items of type two, etc. Thus, the number of solutions is equivalent to the number of $86-$combinations with repetition allowed from a set with $8$ elements, which is given by:

$\binom{86+8-1}{86} = \binom{93}{86} = \binom{93}{7}$

However, I am confused about how I can now find the number of solutions if $x_i$'s are odd. I would appreciate any help on this!

答:
They all have to be odd, so put 1 into each of the 7 variables.

That leaves $86-7 = 79$

Every variable needs to remain odd so you can only change them by amounts of 2. So you discard the extra 1. That leaves 78. $78/2 = 39$ groups of 2 to distribute amount 7 variables.

Now do the same trick of adding an 8th variable for the $\le$:

${{8+39-1} \choose 39} = {46 \choose 39} = {46 \choose 7}$


问:
Permutations with fixed point conditions
I am trying to find the number of permutations $P(10,10)$ and we have the condition that $\sigma(2i) = 2i$ for every $i\in [5]$. I though about inclusion exclusion or that we essentially hold 5 points fixed and permute the other 5 points so we would have $5!$ ways. Can anybody tell me if I am going down the right path?

答: Yes, you are right. If you fix the $5$ even, you just have to permute the $5$ odd in $5!$ ways. Inclusion-exclusion would be useful if you have at least one even fixed.


问:
When can we say that $e(LK/L)\leq e(K/\mathbb{Q}_p)$?
Fix a prime $p$ and let $L$ and $K$ be two finite extensions of $\mathbb{Q}_p$ with ramification degrees $e_L$ and $e_K$, respectively. Let $e'$ and $e''$ be the ramification degrees of $LK/L$ and $LK/K$, respectively. Is it true that $e'\leq e_K$, or are there some additional constraints we can add to $L$ and $K$ which make this true? It is trivially true if $L=\mathbb{Q}_p$ (then $e'=e_K$), or if $K$ is unramified over $\mathbb{Q}_p$ (then $e_K=e'=1$), but I'm not exactly sure what to do if $L$ is strictly larger than $\mathbb{Q}_p$ and $K$ is ramified.

Clearly we have $e_Ke''=e_Le'$ by multiplicativity of ramification indices in towers, so the ratios $e_K/e'$ and $e_L/e''$ are equal. Intuitively, it seems like $e'\leq e_K$ should be true, in that by adjoining elements of $K$ to $L$ we should not end up with "any more" ramification than we began with by adjoining elements of $K$ to $\mathbb{Q}_p$. In other words, it would be nice to say that the ramification occuring in $LK/L$ is no worse than the ramification occuring in $K/\mathbb{Q}_p$. I can't seem to make this precise though, and am not sure whether it's true. Help or counterexamples would be much appreciated. I imagine the case where $K/\mathbb{Q}_p$ is tamely ramified might be the next easiest case, since then $LK/L$ is also tamely ramified, but I haven't had much success with that so far either.

答:
$$e(K/F) = \frac{v(\pi_F)}{v(\pi_K)}$$

It is also $$[K:F(\zeta_{q_K-1})]$$ where $q_K$ is the cardinality of $O_K/(\pi_K)$.

Clearly $$[K:F(\zeta_{q_K-1})]\ge [LK:LF(\zeta_{q_{LK}-1})]$$


问:
Finding the Posterior Distribution
I am trying to solve this problem: A new drug has p chance of working, where p is a random variable with density $f_p(a)\sim Beta(2,3)$. They test the drug on five animals, three of whom show that the drug works. What is the posterior distribution on $p$ given this information?

I found the beta density as: $2p(1−p)2⋅\mathbb(p \in [0,1]$ But I am not sure where to begin?

Can I just add $(2,3)+(3,2)$ to get $Beta(5,5)$?

答:
The prior is

$$\pi(\theta)\propto \theta(1-\theta)^2$$

The likelihood is

$$p(\mathbf{x}|\theta)\propto \theta^3(1-\theta)^2$$

Multiplying you get

$$\pi(\theta|\mathbf{x})\propto \theta^4(1-\theta)^4$$

Thus

$$\pi(\theta|\mathbf{x})\sim  Beta(5;5)$$

Title: Finding the Posterior Distribution


问:
Let $z$ be a complex number satisfying $|z|^4-16|z|^2-3(z^2+{\overline{z}}^2)+9=0$ ..

Let $z$ be a complex number satisfying $$|z|^4-16|z|^2-3(z^2+{\bar {z}}^2)+9=0$$ and let $\min(|z|)=a,b=\max(|z|)$ then find $ab$.

We notice that $$\Im(z^2+{\overline{z}}^2)=0$$
$$\Im({(z+\overline{z})}^2-2z\bar z)=0$$ but this is obvious as both $z+\bar z,z \bar z$ are real.
Hence i was unable to find any additional condition

Also substituition $z=x+iy$ and taking $x^2=l,y^2=m$ we have the mess $$l^2+m^2-22l+10m+2lm+9=0$$I recognise this as a quadratic and setting $D\ge 0$  may give the max and min value of $l+m$.

But surely there must be something easier?!

答:
First let $t = z^2$, so that we can rewrite that as
$$|t|^2-16|t| - 3(t+\overline t) + 9 = 0 \iff (|t| - 8)^2 = 3(t + \overline t) + 55$$

Now it may be easier to note that $|t|$ can find extrema when $\Im(t)=0$, so that we may consider only the case $t \in \mathbb R$.  So we have for extreme $|t| = k$, the quadratic:
$$k^2 -22k +9 = 0 \implies |t|_\min \cdot |t|_\max = 9$$
Hence $a\cdot b = \sqrt{9}=3$


问:
Write a statement that will be true if A = $\Bbb Z$ and false if A = $\Bbb N$
Write a statement that will be true if A = $\Bbb Z$ and false if A = $\Bbb N$, using only quantifiers, logical connectives, variables, parentheses, equals sign and addition symbol. Any variable is assumed to be a member of A.

Here is what I have so far:

$\forall x\exists y \exists z((x+y=x) \land \exists z(x+z=y))$

However it seems that this is only valid if A contains only even numbers.

答:
If your definition of $\mathbb{N}$ does not contain $0$, then the formula could be
$$\exists x\,\, \forall y\,\,x+y=y.$$


问:
Construct an algorithm which checks whether it is true that  [] =  (, ) for all  ∈ .
Given a digraph $ = (, )$ with integer weights on the edges, vertex $ ∈ $, and an array $$ of $||$ size. Also, $[] \geqslant  (, )$ (shortest path between  and  in given graph) holds for all $ ∈ $.
The digraph is represented by adjacency lists with weights. Construct an algorithm which checks whether it is true that $ [] =  (, )$ for all $ ∈ $.
Algorithm running time must be linear, that is, $ (|| + ||)$.

First of all, I thought of BFS, but here we have a weighted graph. Then there was an idea to use the shortest paths algorithm for DAGs, but there is no way I can prove that given graph is acyclic. Bellman-Ford algorithm can calculate an array of shortest paths, but it performs for  $ (||)$. To be honest, I really don't know how without calculating shortest distancies solve this problem in linear time.

答:
HINT:

Claim: Let $a$ be an array on $V$ where $a[v] \ge \delta(V,s)$ for all $v \in V(G)$. Then $a[v] = \delta(v,s)$ iff both (a) $a[s]=0$ and (b) $a[v] \le a[w]+\ell(wv)$ for each $w \in V(G)$ and each $v$ such that $wv$ is an arc in $G$, and where $\ell(wv)$ is defined to be the length of the arc $wv$.

Can you see why this is? If not, here is another hint:

 Let $v$ be any vertex in $V(G)$, and let $P$ be any path from $s$ to $v$ in $G$, including a shortest such path. Then if $a$ satisfies (a) and (b), then $a[v] \le a[s] + \ell(P)$ $=$ $\ell(P)$, where $\ell(P)$ is defined to be the length of $P$ in $G$. So then the inequality $a[v] \le \ell(P)$ must follow for each such $v$ and each such $P$. But then we are also given $a[v] \ge \delta(v,s)$ for each such $v$. So what must one conclude?


问:
Group proofs with $g^N=e$.
Let $G$ be a finite group.

I want to prove, that:
$$\forall g \in G, \exists N \in\Bbb N>0, g^N = e $$
as well as
$$\exists N \in\Bbb N>0, \forall g \in G, g^N = e.$$

My approach:
If $g=e$ than it is trivial:
$$g^N = e^N = e$$

But it is hard to unterstand the rest, in this case $g\neq e$.

答:
First, note that if $G$ is a finite group then for all $g \in G$ there is $n_g \in \mathbb{N}$ such that $g^{n_g}=e$. In fact, suppose not: then at some point, since $G$ is finite, you must have $g^i = g^j$ for some $i < j$. But then $g^{j-i}=e$ (multiply $i$ times left and right by $g^{-1}$).

So to each $g$ you can associate a natural number $n_g$. Now take $N=\prod_{g \in G} n_g$. For any $h \in G$, what is $h^N$? Well:

$$h^N = h^{n_h \cdot \prod_{g \in G, h \neq g} n_g} = (h^{n_h})^{\prod_{g \in G, h \neq g} n_g} = e^{\prod_{g \in G, h \neq g} n_g} = e$$

This is far from the optimal $N$ for which this holds, but it avoids using Lagrange's theorem or any other result really.

Title: Group proofs with $g^N=e$.


答:
First part : Let $n$ be the cardinality of $G$. Consider the $n+1$ elements $e,g,g^2, ..., g^n$. You have $n+1$ elements from a set of $n$ elements, therefore two of them are equal. So there exists $k > l \in \lbrace 0, ..., n \rbrace$ such that $g^k=g^l$. Now let $N_g = k-l > 0$ : for this $N_g$, you have
$$g^{N_g}=e$$

Second part : Apply Lagrange's theorem to the subgroup generated by $g$, to show that each $N_g$ must divide the order $n$ of $G$, and therefore that $n$ works for all the $g$'s in $G$.

Title: Group proofs with $g^N=e$.

Body:
Let $G$ be a finite group.

I want to prove, that:
$$\forall g \in G, \exists N \in\Bbb N>0, g^N = e $$
as well as
$$\exists N \in\Bbb N>0, \forall g \in G, g^N = e.$$

My approach:
If $g=e$ than it is trivial:
$$g^N = e^N = e$$

But it is hard to unterstand the rest, in this case $g\neq e$.


问:
How can I prepare effectively for this test?
I live in Australia and I will be doing my final high school mathematics test in 5 months from today. The test is about the following topics:

Differentiation and Integration and their applications (Maxima/minima problems, volumes, motion [simple harmonic, projectile, circular and resisted motion], conical pendulum and banked tracks)

Complex numbers

Polynomials

Conic sections

Series and sequences, mathematical induction and analysis

Inequalities (similar to ones in IMO)

Circle geometry

Probability, combinatorics and binomial probability.

Here are some examples of past exam papers:

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2012exams/pdf_doc/2012-hsc-exam-maths-ext2.pdf

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2011exams/pdf_doc/2011-hsc-exam-mathematics-ext2.pdf

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-mathematics-extension-2.pdf

The test starts of very easy then gradually gets harder. I'm confident that I can get most of the questions correct but timing is a problem as it is difficult to get the whole test done in the time allowed.

Many people, including my teacher, told me that the best way to prepare is to do as many of these papers as possible in timed conditions. Do you agree with that? The problem is that doing the same thing with slightly different questions over and over is very boring :(

答:
Yes, it is the best way to get the highest mark on the test and yes it is very boring. At the very high percentiles, the ranks are decided by who did the most past papers, who saw the most familiar questions and who has practiced their speed - not necessarily by who has the most solid understanding of all the mathematical principles involved. With enough practice you should aim to get to the point where you don't even need to think about how to do the problems until Q8, everything before that is just writing down what comes to your mind as fast as you can.

Not saying is this the best way to learn or enjoy mathematics, but it is the best way to rank the highest on the tests. Unfortunately sometimes we all have to jump through some hoops.


问:
Proving $ \sum_{l=1}^{2n} (-1)^{l+1}\frac{1}{l}=\sum_{k=1}^{n} \frac{1}{n+k}$ by induction
How to prove this by induction?

$$ \sum_{l=1}^{2n} (-1)^{l+1}\frac{1}{l}=\sum_{k=1}^{n} \frac{1}{n+k} \text{ for } n \in \mathbb N$$

I got the beginning $A(1)$ is true, but I'm getting stuck somewhere in the middle of the proof for $A(n+1)$.

This is how far I got:

答:
You are almost there; you just need to see that

\begin{align}\sum_{k=1}^{n+1}\frac1{n+k+1}=\sum_{k=2}^{n+2}\frac1{n+k} &= \frac1{n+n+2}+\frac 1{n+n+1}+\sum_{k=1}^n\frac1{n+k} - \frac1{n+1}
\\&=\frac1{2n+2}+\frac 1{2n+1}- \frac2{2n+2} + \sum_{k=1}^n\frac1{n+k}
\\&=\frac 1{2n+1}- \frac1{2n+2}+ \sum_{k=1}^n\frac1{n+k}\end{align}

and you can finish the induction.


问:
Confusion regarding the logarithm of identity function on simply connected domain
Define  $D=\mathbb C-\{x+i(\sin x)|x\in (-\infty,0]\}$.

Clearly $D $ is simply connected domain, define
$f:D\to \mathbb C$ by
$$f(z)=z$$

We know that for if we have a non-vanishing holomorphic function on simply connected domain $ D $ then we have a holomorphic logarithm of that function (i.e. a holomorphic function $g$ on $D $ such that $e^g=f$ on $D$) .

Now I am confused what will my $g$ look in this case and how will it differ from the principal logarithm.

I understand $g'(z)=\frac{1}{z}$ but that is not leading me anywhere.

Any help or reference shall be appreciated. Thank you.

答:
A comment that got too long - not sure about $D$ as it seems you wanted $x\in (-\infty,0]$ - anyway assuming that and assuming you take the branch that gives the real logarithm on the positive axis, it is easy to see that for example in the right half-plane $f$ is the same as the principal value (identity theorem), while in the 2nd quadrant it is same above the curve $x+i(\sin x)$ and shifted by $-2\pi$ below (identity theorem above, continuity plus identity theorem applied for third quadrant below), while in the third quadrant it is same below the curve $x+i(\sin x)$ and shifted by $2\pi$ above (the same reasoning as for the second quadrant with above/below switched)

In particular on the negative axis it is $\pi$ in $(-(2k+1)\pi, -2k\pi)$ so $(-\pi,0), (-3\pi, -2\pi)$, etc and then by continuity when we go a little above/below the axis, and $-\pi$ in $(-(2k+2)\pi, -(2k+1)\pi)$, so $(-2\pi, -\pi), (-4\pi, -3\pi)$ and by continuity when we go a little  above/below there etc

Looking at the maximum values of $\sin x /x$ on $k\pi, (k+1)\pi$ one also sees that $\arg f \in (-\pi - \arctan \frac{2}{3\pi}, \pi + \arctan \frac{2}{\pi})$ so it is still a bounded function


问:
Please help me understand this linear algebra proof.
Please help me understand the proof below.

Here are my questions:

1.) what does r$_1$ represent?

2.) is $A_1$ just matrix $A$ with the first row and first column deleted?

3.) what does $\tilde{r_1}$ represent?

4.) why does r$_1$$Y$=[r$_{12}$$\tilde{r_1}$]

I'm sorry if this is a difficult question to answer. I can't tell if I am confused by some notational stuff or if I am not understanding some concepts. Please help me arrive at an understanding of this proof.

答:
Regarding 1,2: in the equations
$$
\begin{align}
AX &= \cdots = \pmatrix{\lambda \mathbf x_1 & A \mathbf x_2 & \cdots & A \mathbf x_n}
\\ & = \pmatrix{\mathbf x_1 & \mathbf x_2 & \cdots & \mathbf  x_n} \pmatrix{\lambda_1 & \mathbf r_1 \\ 0 & A_1} = \cdots,
\end{align}
$$
the point being made is that there exists some matrix $M$ for which $AX = [\mathbf x_1 \  \ \cdots \ \  \mathbf x_n] M$, and this $M$ must have the first column $[\lambda_1,0,\dots,0].$ An efficient way to convey this is to avoid introducing the matrix $M$ and instead use the block-partitioned matrix
$$
M = \pmatrix{\lambda_1 & \mathbf r_1 \\ 0 & A_1}.
$$
So, to answer your question, $\mathbf r_1$ is a $1 \times (n-1)$ row-vector and $A_1$ is an $(n - 1) \times (n-1)$ matrix. Both of these are submatrices of the (unique) matrix $M$ satisfying $AX = XM$.

Regarding 3,4, we have the same thing going on. The author would like to break up the $1 \times (n-1)$ matrix $\mathbf r_1 Y$ into two submatrices. $r_{12}$ is the first entry of $\mathbf r_1 Y$, and $\tilde {\mathbf r}_1$ is the $1 \times (n-2)$ matrix built from the remaining entries of $\mathbf r_1 Y$.


问:
continuous functions above and below a characteristic function
If $\chi_{[a,b)}:[0,1]\rightarrow \{0,1\}$ is the characteristic function of a subinterval $[a,b)\subseteq [0,1)$ then how can one show that for arbitrary $\varepsilon >0$ that there exist functions $f,g \in C[0,1]$ such that $f(x) \leqslant \chi_{[a,b)}(x) \leqslant g(x) $ on $[0,1]$ and $\int_{0}^{1}\left[g(x)-f(x)\right] \, dx \leqslant \varepsilon$?

This makes sense intuitively to me and I can create $f,g$ that satisfy the latter properly but I don't know how to ensure $f,g$ also satisfy the former property (continuous). Is it to do with the Darboux Criterion of integrability?

答:
Suppose $0 < a < b < 1$.  Take $\delta \leqslant \epsilon/2$ such that $a - \delta \geqslant 0$.  Define continuous piecewise linear functions,

$$g(x) = \begin{cases}0,& 0 \leqslant x \leqslant a - \delta\\ 1+\delta^{-1}(x - a ), &a- \delta < x \leqslant a\\1,& a < x \leqslant b\\1 - \delta^{-1}(x - b),&b < x \leqslant b+\delta\\0,& b+\delta < x \leqslant 1 \end{cases} \\f(x) = \begin{cases}0,& 0 \leqslant x \leqslant a \\ \delta^{-1}(x - a ), &a < x \leqslant a+\delta\\1,& a+\delta < x \leqslant b-\delta\\\delta^{-1}(b - x),&b-\delta < x \leqslant b\\0,& b < x \leqslant 1 \end{cases} $$

Then $f(x) \leqslant \chi_{[a,b)}(x) \leqslant g(x)$ for all $x \in [a,b]$.  We have $g(x) = f(x)$ except on the intervals $(a - \delta,a+ \delta)$ and $(b- \delta, b+\delta)$.  The region between the graphs over these two intervals are parallelograms with height $1$ and base $\delta$  and, thus,

$$\int_0^1 |g(x) - f(x)| \, dx = 2 \cdot \delta \cdot 1 = 2\delta \leqslant \epsilon.$$

The case where $a = 0$is handled in a similar way.

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问:
Finding $g'(2)$ given $f(x)$ and $f(g(x))$

Let $f(x) = e^x + x + 1$ and $f(g(x)) = x$. Find the value of $g'(2)$

Here is my attempt:

Because $f(g(x)) = x$, that means I need to find what value of x will make $g(x)$ satisfy $f(x) = x$.

Using inverse differentiation I can find this:

$$f^{-1}(x) = e^y + y + 1$$

which means:
$$g(x) = e^x + x + 1$$

so the derivative of g(x) is:
$$g'(x) = e^x + 1$$

which means:

$$g'(2) = e^2 + 1$$

Which step did I go wrong?

Body:
First, note that $f(g(x)) = x$ means $g(x) = f^{-1}(x)$. (We can skip $g(f(x)) = x$ because $f(x)$ is a"nice" function.) Then, $g'(x) = \frac{1}{f'(f^{-1}(x))}$, so $g'(2) = \frac{1}{f'(f^{-1}(2)}$. To find $f^{-1}(2)$, note that $f(0) = 2$ and $f(x)$ is one to one, so $f^{-1}(2) = 0$. Thus:

$$g'(2) = \frac{1}{f'(0)} = \frac{1}{e^{0} + 1} = \boxed{\frac{1}{2}}$$


问:
Derivative of a function $f({\bf X})$ w.r.t. matrix ${\bf X}$ when ${\bf X}^\prime {\bf X}$ is a diagonal matrix
I wish to compute the derivative and the Hessian of the function ${\bf f}(X)$ where

$$ {\bf f}({\bf X}) = {\bf X} \, {\bf a}, $$

${\bf X}$ is an $(m \times n)$ matrix and ${\bf a}$ is a vector of constants of size $n$. From "Matrix Differential Calculus" by Magnus and Neudecker, it is relatively straightforward to obtain the derivative and the Hessian of ${\bf f}({\bf X})$ w.r.t. ${\bf X}$. For example, the first differential is

\begin{align}
\partial {\bf f}({\bf X}) &= (\partial {\bf X}) {\bf a} = {\rm vec} (\partial {\bf X}) {\bf a} = ({\bf a}^\prime \otimes {\bf I}_m) \,\partial {\rm vec}  {\bf X} \\
\end{align}

In my case however the matrix ${\bf X}$ has a special structure. Specifically, I know that that ${\bf X}^\prime {\bf X}$ is a diagonal matrix, say, ${\bf X}^\prime {\bf X} = {\rm diag}(d_1, \ldots, d_n)$, which is not necessarily equal to the identity matrix. In other words, I know that the columns of ${\bf X}$ are mutually orthogonal, but each column vector can be of any length. How do I compute the derivative and the Hessian of $f({\bf X})$ w.r.t. ${\bf X}$ to take this structure into account?

Any references on how to proceed would be greatly appreciated.

I could not find a similar question posted before. The closest I found was Hessian of $f(X)$ when $X$ is a symmetric matrix, but I was not able to apply it to my problem.

答:
$\def\p{\partial}$
Use an orthogonal matrix $Q$ and a vector $y$ to construct $X$
$$\eqalign{
Y &= {\rm Diag}(y) &\;=\; \sqrt{X^TX} \\
X &= QY  &\implies X^TX = YQ^TQY = Y^2 \\
Q &= X(X^TX)^{-1/2} \\
}$$
Also define the diagonal matrix $\;A = {\rm Diag}(a)$.

Write the function in terms of these new variables
and calculate its gradient.
$$\eqalign{
f &= Xa = QYa = QAy\\
\p f &= QA\,\p y \\
\frac{\p f}{\p y} &= QA \\
}$$
This can be interpreted as a gradient wrt $X$ by using
$$\eqalign{
\p{\rm vec}(X) &= (I\otimes Q)\,{\rm vec}\Big({\rm Diag}(\p y)\Big) \\
}$$


问:
Prove that $f(x)=\frac{1}{x}$ is uniformly continuous on $(\frac{1}{2},\infty)$

Prove that $f(x)=\frac{1}{x}$ is uniformly continuous on $(\frac{1}{2},\infty)$

Attempt:

We must show that for any $\epsilon>0$ we can find a $\delta > 0$, such that if $|x-a|<\delta$ and $|\frac{1}{x}-\frac{1}{a}|<\epsilon$ for all $a \in (\frac{1}{2},\infty)$.

$\quad$Scratchwork for $\delta$:.

$\begin{align}
\quad \quad |\frac{1}{x}-\frac{1}{a}|=|\frac{a-x}{xa}|&=\fbox{$\frac{1}{|xa|}$}|x-a|\\&=4|x-a|<\epsilon \\ &\Rightarrow|x-a|<\frac{\epsilon}{4}
\end{align}$

$\quad \quad $ **In order to bound $\frac{1}{|xa|}$ (or $\frac{1}{xa}$; since $a,x>0$), I used the fact that $x,a>\frac{1}{2}$.

$\quad \quad $Which would result in $\frac{1}{|xa|}=\frac{1}{xa}=\frac{1}{\frac{1}{2}*\frac{1}{2}}=4$.

$\quad$Choose $\delta=\frac{\epsilon}{4}\\$

$\quad$Proof:.
Let $\epsilon>0$ be given and let $\delta=\frac{\epsilon}{4}$. If $|x-a|<\delta$ we have:

$\begin{align}
\quad \quad |\frac{1}{x}-\frac{1}{a}|&=\frac{1}{|xa|}|x-a|\\&\leq4|x-a|\\ &<4\delta=4\frac{\epsilon}{4}=\epsilon.
\end{align}$

$\rule{18cm}{.03cm}$

Is it enough to use the fact that that $x,a>\frac{1}{2}$, in order to place bounds on the function in order to reach a $\delta$? Any guidance or advice to strengthen the integrity of this proof would be greatly appreciated.

答:
Actually, here is a much faster way to prove this result. Indeed, $f$ is $\mathcal{C}^1$ and one has, over $\left(\frac{1}{2},+\infty \right)$, $$|f'(x)|=\left| \frac{1}{x^2}\right| \leq 4$$

so $f$ is $4-$lipschitz, so $f$ is uniformly continuous over $\left(\frac{1}{2},+\infty \right)$.


答:
Hint:

Since $\;g=f^{-1}\;$ (why? Why is $\;f\;$ even invertible?) then by the theorem of the derivative of the inverse funcion

$$g'(x)=\frac1{f'(f^{-1}(2))}$$
so...

Title: Finding $g'(2)$ given $f(x)$ and $f(g(x))$

Body:
Let $f(x) = e^x + x + 1$ and $f(g(x)) = x$. Find the value of $g'(2)$

Here is my attempt:

Because $f(g(x)) = x$, that means I need to find what value of x will make $g(x)$ satisfy $f(x) = x$.

Using inverse differentiation I can find this:

$$f^{-1}(x) = e^y + y + 1$$

which means:
$$g(x) = e^x + x + 1$$

so the derivative of g(x) is:
$$g'(x) = e^x + 1$$

which means:

$$g'(2) = e^2 + 1$$

Which step did I go wrong?


问:
Integral over $[-1,0 ]$ with parametrized exponent
The question is when the following integral converges or diverges:

$\int_{-1}^{0} x^{-p}$ for $p > 0$.

I am stuck at the case where $0 < p < 1$, where I end up with

$$\frac{1}{1-p} - \frac{(-1)^{1-p}}{1-p}$$

I am not sure how to split this funciton. should it be from 0 < p < 1 or p <= 1. I cant tell what to do at the value '1'.

答:
The problem with non-integer powers of negative numbers is that you'll need to use complex numbers.

Also, in the complex numbers, $(-1)^z$ for a given value of $z$ is a multi-valued function.

You could always choose one particular value and, for example, represent $(-1)^p$ as $e^{i \pi p}$.


问:
Shading A Venn Diagram Using A Specific Equation
The expression is: $A\triangle(B\cap C')$. The $\triangle$ refers to $-$ and the $\cap$ refers to an intersection, whilst the $\;{}'$ refers to the prime of $C$.

There are no numbers or items included in this problem. I have drawn a Venn diagram with a background (for the universal set) and circles $A$, $B$ and $C$ which equally overlap. The problem is, I am unsure how to shade it. I also must provide the logic used to solve the equation. Thanks a bunch!

答:
It might help to do this in steps, from the inside out, try drawing tree diagrams:

in the first, simply shade C', that is all the area out side of $C$.

next, you'd like to draw $B\cap C'$ i.e. the intersection of $B$ with the shaded area of the previews diagram, more simply it would be the area inside $B$ that was shaded in the first diagram.

finally, in the third diagram, you need to shade the area inside A that was not shaded in the second diagram.

put into words the expression means, everything that is in A with the exception of($\Delta$) things that are in B and($\cap$) not($′$) in $C$.

At first this might be confusing, and you might need to draw the steps, but with practice you will learn to do it in your head, Venn diagrams are a very powerful tool for intuitively understanding logic.


问:
How do we show the vector $v = (0,1,1)$ belongs (or not) to the space spanned by $w = (1,1,-1)$ and $z = (1,0,2)$?
I have found the range to be these to vectors [1,1,-1] and [1,0,2]

How do I show/find out if this vector [0,1,1] belong to that range?

答:
HINT

Consider the following linear combination
\begin{align*}
(0,1,1) = \alpha(1,1,-1) + \beta(1,0,2) = (\alpha + \beta,\alpha,-\alpha + 2\beta)
\end{align*}

Does the corresponding system of equations have a solution?


问:
Meaning of notation $\frac{1}{N}\mathbb{Z}$
I am reading a set of notes (on modular curves: here), which seems to suggest on page 5 that  $\frac{1}{11}\mathbb{Z}$ is a cyclic group of order 11. Am I missing something? Doesn't this notation just mean the integers multiplied by $11$, which certainly doesn't have order 11?

答: It is of course the rational numbers whose denominator is $11$.  Its image in the complex torus $\Bbb{C/(Z+\tau Z)}$ is a cyclic subgroup with $11$ elements.


问:
Proving that a $p$-Sylow subgroup is unique
Let $G$ be a group and $|G|=84$. How can I prove that the $2$-Sylow subgroup is unique?

I know that the $7$-Sylow subgroup in unique, but I don't know how I can use it. So far, I know that $|G|=84=2^2\cdot3\cdot7$, $n \equiv 1 \pmod{2}$ and $n\mid 21$.

答: Counterexample: $G=C_7 \times D_6$, where $D_6$ is the dihedral group of order $12$. The latter group (and hence the group $G$) has $3$ Sylow $2$-subgroups.


问:
Proving or disproving basic facts about sequences in Real Analysis
I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.

\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):

(a) sequences $(x_n)$ and $(y_n)$, which both diverge, but whose sum $(x_n + y_n)$ converges;

(b) sequences $(x_n)$ and $(y_n)$, where $(x_n)$ converges, $(y_n)$ diverges, and $(x_n + y_n)$ converges;

(c) a convergent sequence $(b_n)$ with $b_n \ne 0$ for all $n$ such that $1/b_n$ diverges;

(d) an unbounded sequence $(a_n)$ and a convergent sequence $(b_n)$ with $(a_n - b_n)$ bounded

(e) two sequences $(a_n)$ and $(b_n)$, where $(a_n b_n)$ and $a_n$ converge but $(b_n)$ does not.

Solution.

(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.

Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.

(b) This request is impossible. If the $(x_n + y_n)$ is to be convergent, it implies we are able to make the distance $\vert{(x_n + y_n) - (x + y)}\vert$ as small as like. However, we cannot make $y_n$ to lie eventually in a set $(y - \epsilon, y + \epsilon)$. Hence, the sum cannot be convergent.

(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.

(d) This request is impossible. The key here is to show that, assuming here $(a_n)$ is bounded leads to the contradiction, leading to their difference also being bounded.

If $(a_n)$ is a bounded sequence, there exists a large number $M > 0$, such that $\vert{a_n}\vert < M$ for all $n \in \mathbf{N}$. If $(b_n)$ is a bounded sequence, there exists a large number $N > 0$, such that $\vert{a_n}\vert < N$ for all $n \in \mathbf{N}$.

Thus,

\begin{align*}
    \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\
    &\le \vert{a_n}\vert + \vert{-b_n}\vert\\
    &< M + N
\end{align*}

(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not.

答:
Thank you for a detailed, well-asked question!

For (a), (c), and (e): your examples are all correct. I would say that you asserted several facts that haven't been justified (which would make the proof incomplete, if you were writing for someone else's judgment/understanding). For example, the asserted limits of your examples in part (a) would need to be justified, as would the convergent/divergent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that case they're pretty obvious).

While the idea is reasonable, your proof for part (b) isn't rigorous. One detail: assuming that $(x_n+y_n)$ is convergent means that it converges to some number $z$, not to $x+y$ (indeed you didn't define either $x$ or $y$). Then you asserted, without proof, that $y_n$ can't be made to lie inside a small set. Your ideas are heading in the right direction, but you would need to use the precise definitions of convergence/divergence in exploiting the assumptions and in setting out what needs to be proved.

Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $(x_n)$ converges and $(y_n)$ diverges, then $(x_n+y_n)$ diverges". As it turns out, that statement is logically equivalent to "if $(x_n)$ converges and $(x_n+y_n)$ converges, then $(y_n)$ converges"—which you might well find easier to prove! (By logically equivalent, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.)

Your proof for (d) seems to prove the following statement: "if $(a_n)$ is bounded and $(b_n)$ is bounded, then $(a_n+b_n)$ is bounded". That is a true fact, but is that what you want to prove here?


问:
Exercise 4.2.5 [HDP Book Vershynin]: Packing the balls into K
Suppose $T$ is a normed space. Prove that packing number $P(K,d,\epsilon)$ of $K \subset T$  is the largest number of closed disjoint balls with centers in $K$ and radii $\epsilon/2$. Show by example that the previous statement may be false for a general metric space $T$.

Def:(Packing numbers). A subset $\cal{N}$ of a metric space $(T,d)$ is $\epsilon$-separated if $d(x,y)>\epsilon$ for all distinct points $x,y\in \cal{N}$. The largest possible cardinality of an $\epsilon$-separated subset of a given set $K\subset T$ is called the packing number of $K$ and is denoted by $P(K,d,\epsilon)$

Proof: Suppose $P(K,d,\epsilon)= n$, so there are $n$ points in $K$ which are $\epsilon$-separated. Each of these $n$ points can be thought as a center of a $\epsilon/2$ radius ball. Then there are $n$ closed disjoint balls of radius $\epsilon/2$, that have their center in K. It remains to show that $n$ is largest number.
Assume that there are $n+1$ such $\epsilon/2$ radius closed disjoint balls that have centers in $K$, which implies that $K$ has $n+1$ $\epsilon$-separated points, which is contradictory to the fact that $K$ has a packing number $n$. Hence there can be at-most $n$ such balls.

Doubt: The above proof holds true for any metric space, which is not correct according to the question. Please help.

答:
Popescu Claudiu gave a great counter-example in general metric space.

Now I discussed why your original proof can only work in normed space: notice that the difference between normed space and metric space is that normed space must be a linear space, which means any $a,b \in T$ must have, say, $(a+b)/2 \in T$. But metric space does not need to, just as above counter-example shows.

Thus, for above example, if it is also a normed space then we must have $(a+b)/2 \in T$ and so the Ball $B(a,\epsilon/2)$ and $B(b, \epsilon/2)$ are not disjoint.


问:
Number of vertices in tree
I'm trying to prove the following problem:

$v_i$ is number of vertices of degree $i$ in tree $T$. Prove that
$v_1=v_3+2v_4+...+(\Delta -2)v_\Delta +2$; $\Delta$ is the max degree.

But am I not sure how. Could you please help me? My idea is to use the consequence of Eulers formula but not sure how to begin.

答:
Hint: $v_3+2v_4+3v_5+\dots+(\Delta-2)v_\Delta+2\\=v_3+2v_4+3v_5+\dots+(\Delta-2)v_\Delta+\color{red}{2(v_1+v_2+\dots+v_\Delta)}-\color{blue}{2(v_1+v_2+\dots+v_\Delta)}+2\\=(\color{red}{v_1+2v_2}+\color{red}3v_3+\color{red}4v_4+\dots+\color{red}\Delta v_\Delta)+\color{red}{v_1}-\color{blue}{2V}+2\\=2E+\color{red}{v_1}-\color{blue}{2V}+2\\=2(E-\color{blue}V+1)+\color{red}{v_1}$

where $E,V$ is the total number of edges and vertices in the tree respectively.

Title: Number of vertices in tree


问:
Minimum value of $\sqrt{x^4 + 3x^2 - 6x + 10} + \sqrt{x^4 - 5x^2 + 9}$ without using calculus?
Hi mathematics stack exchange, what is the minimum value of $\sqrt{x^4 + 3x^2 - 6x + 10} + \sqrt{x^4 - 5x^2 + 9}$? I know how to solve this problem using calculus, you take a derivative, but I am wondering if there is an elementary method to find the minimum using precalculus methods.

答:
Let $y = x^2$, notice

$$\begin{align}
x^2 + (y-3)^2 & = x^2 + (x^2-3)^2 = x^4 - 5x^2+9\\
(x-3)^2+(y+1)^2 & =  (x-3)^2 + (x^2 + 1)^2 = x^4 - 3x^2 -6x + 10\end{align}$$

The problem at hand can be rephrased as:

Given $A = (0,3)$, $B = (3,-1)$ and $P = (x,y)$ be a point on the parabola $y = x^2$. What is the minimum value of $AP + PB$?

If one make a plot of the parabola $y = x^2$, one will notice the parabola intersect with the line segment $AB$, this means the minimum value of $AP + PB$ is $$AB = \sqrt{(0-3)^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = 5$$


答:
You can prove it by induction on the number of vertices in the tree. Suppose that it’s true for all trees with fewer than $n$ vertices, and $T$ is a tree with $n$ vertices. Let $u$ be a vertex of $T$ of degree $\Delta$, and let $u_1,\ldots,u_\Delta$ be the vertices adjacent to $u$. For $k=1\ldots,\Delta$ let $T_k$ be the subtree of $T$ induced $u$, $u_k$, and all of the vertices $w$ of $T$ such that the path from $w$ to $u$ goes through $u_k$.

Each of these trees has fewer than $n$ vertices, so the result holds for each of them. Collectively they have the same number of vertices of degrees $2$ through $\Delta-1$ as $T$, one fewer vertex of degree $\Delta$, and $\Delta$ more vertices of degree $1$, so

$$v_1+\Delta=\sum_{k=3}^\Delta(k-2)v_k-(\Delta-2)+2\Delta\,.\tag{1}$$

The numbers $v_k$ are the vertex counts for $T$. The $\Delta$ on the left is for the new vertices of degree $1$. The $\Delta-2$ on the right accounts for the loss of a vertex of degree $\Delta$. And the $2\Delta$ is the extra $+2$ terms in the $\Delta$ equations for the trees $T_1,\ldots,T_\Delta$.

Simplifying $(1)$ yields the desired result for $T$:

$$v_1=\sum_{k=3}^\Delta(k-2)v_k+2\,.$$

Title: Number of vertices in tree

Body:
I'm trying to prove the following problem:

$v_i$ is number of vertices of degree $i$ in tree $T$. Prove that
$v_1=v_3+2v_4+...+(\Delta -2)v_\Delta +2$; $\Delta$ is the max degree.

But am I not sure how. Could you please help me? My idea is to use the consequence of Eulers formula but not sure how to begin.


问:
Which matrix belongs to the stabiliser of a vector?
We consider the action of $GL_2(\mathbb{R})$ on the plane $\mathbb{R}^2$ by linear transformation. Which of these matrices belong to the stabiliser of the vector $\begin{pmatrix} 1\\3\end{pmatrix}$?

A: $\begin{pmatrix} -1 & 1  \\ 4 & -1 \end{pmatrix}$

B: $\begin{pmatrix} 3 & -1  \\ 0 & 2 \end{pmatrix}$

C: $\begin{pmatrix} 1 & 0  \\ -3 & 2 \end{pmatrix}$

D: $\begin{pmatrix} 1 & 0  \\ 0 & 3 \end{pmatrix}$

What I have tried so far:

Given a group action of a group $G$ on a set $X$. For $x \in X$, the stabiliser of $x$, denoted $\text{Stab}(x)$ is such that $g \cdot x=x $. Hence I think it is D Since $\begin{pmatrix} 1 & 0  \\ 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1\\3\end{pmatrix}= \begin{pmatrix} 1\\3\end{pmatrix}$

Is this method correct?

答:
I think you miscalculated the value of
$$\begin{pmatrix} 1 & 0  \\ 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1\\3\end{pmatrix}$$
But if $g \in G_x$, then $g \cdot x = x$. So your method is correct but there is an error in your calculation.


问:
Hölder continuity of gradient
i would to like know what fact in the pde theory is used in the following article , pg 15

"Since $Dw$ is Hölder continuous, we can see this equation as a linear ellip-
tic equation with $C^{\alpha}$ coefficients. The standard Calderon-Zygmund theory
provides local $C^{2,\alpha}$ regularity on $w$ (bootstrapping the argument gives even
C∞ regularity on w)."

https://arxiv.org/abs/1603.06391

Thnaks

答:
The result the paper references is commonly referred to as Schauder estimates, which are a collection of results which (roughly) assert that if $u$ is a solution to a linear elliptic PDE of the form $Lu=f$ whose coefficients in $C^{\alpha},$ then $f \in C^{\alpha}$ implies that $u \in C^{2,\alpha}.$

You can find the details in most texts on elliptic PDE, such as Chapter 6 of Gilbarg and Trudinger.

Title: Hölder continuity of gradient


问:
Confusion over semantics behind ∃-elimination
$\exists-$ states that: if $\Sigma, A(u) \vdash B$ where u occurs nowhere else, then $\Sigma, \exists x A(x) \vdash B$.

Why does this translate to $\exists$ and not $\forall$? Intuitively, $A(u)$ where $u$ occurs nowhere else should mean that $u$ can be any arbitrary thing, so shouldn't $\Sigma, A(u) \vdash B$ be valid for any conceivable $u$ (a.k.a, all $x$) meaning $\Sigma, \exists xA(x) \vdash B$?

This is giving me trouble mostly because of the below theorem I've already learned:

Why is it that in this case $A(u)$ can translate in to a "for all" while in $\exists-$ it only gives us "there exists"? In both cases $u$ is arbitrary, so it should refer to anything$-$"for all"$-$in my  head.

答:
The term $u$ is arbitrary yes, but what isn't arbitrary is the fact that we have a derivation $\Sigma, A(u) \vdash B$.

What does it mean to have $\Sigma, A(u) \vdash B$? It means that, assuming $\Sigma$ and $A(u)$ (for some particular $u$), we can prove $B$. Moreover, since $u$ doesn't appear anywhere else, it can't be a "proper" part of the proof: all that is necessary to prove $B$ is that $A(x)$ is true for some x. The $\exists- $ says precisely this, that we don't care about the specific $u$ that satisfies $A(x)$, only that there is something which satisfies it.

On the other hand, suppose we can prove $A(u)$ from $\Sigma$ without assuming $u$. This means that the proof of $A(u)$ can't really "depend" on $u$, so we should be able to prove $A(x)$ for any $x$. And this is what the $\forall +$ rule says.


问:
Consider $x^4 \pmod {pq}$, with $p = q = 3 \pmod4$.
Consider $x^4 \pmod {pq}$, with $p = q = 3 \pmod 4$.

Would someone explain to me why exactly one of the four square roots of $x^4 \pmod {pq}$ is also a square?

This result was given without proof and I do not understand.

What I have tried:
Edit: This boils down to proving that some $\exists$ $a: a^2 \cong x^4 \pmod {pq}$, $a$ is a quadratic residue $\pmod {pq}$

Since $p = q = 3 \pmod 4$, I know that the value of the Legendre symbol ($\frac{-1}{pq})$ is $-1$.

So, if the four candidates are $x,-x,y,-y$, one of $x,-x$ and one of $y,-y$ is not a quadratic residue, due to the multiplicity of the Legendre Symbol.

Without loss of generality, let's assume $x$ and $y$ are left as candidates. Without loss of generality, how can we prove $y$ is not a square/quadratic residue $\pmod {pq}$?

答:
I work under the assumption that $x^4$ has to be replaced by $x^2$ in the formulation of the problem, otherwise $x^2$ is the required square root of $x^4$.

There are four distinct square roots of $1$ in $\Bbb{Z}_{n}$, where $n = p q$, say $1, -1, a, -a$. (Addendum Recall that we (can choose to) have $a \equiv 1 \pmod p$ and $a \equiv -1 \pmod{q}$.)

The four square roots of $x^2$ modulo $n$ are $x, -x, a x, -a x$. Now with the choice of $a$ as in the Addendum above, we have that the four $\left(\dfrac{p-1}{2}\right)$-th and $\left(\dfrac{q-1}{2}\right)$-th powers of these elements are
$$
x^{(p-1)/2}, - x^{(p-1)/2}, x^{(p-1)/2}, - x^{(p-1)/2} \quad\text{modulo $p$}
$$
and
$$
x^{(q-1)/2}, - x^{(q-1)/2}, - x^{(q-1)/2}, x^{(q-1)/2}  \quad\text{modulo $q$}.
$$
This is because $\dfrac{p-1}{2}$ and $\dfrac{q-1}{2}$ are odd, so if $\epsilon \in \{ 1, -1, a, -a \}$, then $\epsilon \equiv \pm 1 \pmod{p}$ (see the Addendum above), so that
$$
\epsilon^{{(p-1)}/{2}} \equiv \epsilon \pmod{p},
$$
and similarly for $q$.

Now recall that
$$
x^{(p-1)/2} = \begin{cases}
1 & \text{if $x$ is a square modulo $p$,}\\
-1 & \text{if $x$ is not a square modulo $p$,}
\end{cases}
$$
and similarly for $q$.

Checking all four possibilities for the signs, we see that for exactly one of the elements $b$ among $x, -x, a x, -a x$ we have we have that both its $\left(\dfrac{p-1}{2}\right)$-th and $\left(\dfrac{q-1}{2}\right)$-th powers are $1$, so $b$ is a square modulo $p$ and $q$, so it is a square modulo $n = p q$.


问:
Representation of cyclotomic polynomial over finite field with prime characteristic
If I have a finite field $F_p$ with characteristic $p$ prime (and size $p$), the cyclotomic polynomial over $F_p$ is defined as $$\Phi_r(X) = \prod\limits_{i=1}^k (X - \eta_i)$$ where $\eta_i$ is a primitive $r$th root of unity (maybe not an element of $F_p$ but certainly an element from the cyclotomic field). Is it true that $\Phi(X)$ can always be represented over $F_p[X]$? I mean, are the coefficients of $F_p[X]$ always elements of $F_p$?

I know there are lots of question regarding factorization of cyclotomic polynomials over finite fields, but I couldn't find the answer to this question. I don't really care if it's irreducible or not, just if it can be represented in the field.

Thanks!

答:
Your definition of $\Phi_r(X)$ a priori makes sense over some finite field $\mathbb{F}_{p^k}$ which is a splitting field of the polynomial $x^r - 1$ over $\mathbb{F}_p$, so its coefficients a priori live in $\mathbb{F}_{p^k}$. To check whether or not they live in $\mathbb{F}_p$ it's necessary and sufficient to check whether or not they're fixed by the Frobenius map $x \mapsto x^p$, whose fixed points on any field extension of $\mathbb{F}_p$ are exactly $\mathbb{F}_p$ (this is an easy special case of Galois theory). So we need to check whether the primitive $r^{th}$ roots of unity are permuted by $x \mapsto x^p$.

If $\gcd(r, p) = 1$ this is clear because $p$ is invertible $\bmod r$, so multiplication by $p$ is an isomorphism $(\mathbb{Z}/r\mathbb{Z})^{\times} \to (\mathbb{Z}/r\mathbb{Z})^{\times}$ and sends generators to generators.

If $p \mid r$ then there aren't any primitive $r^{th}$ roots of unity, because there aren't any primitive $p^{th}$ roots of unity.

This argument suggests but does not prove that the polynomial you've defined is actually the reduction $\bmod p$ of the cyclotomic polynomial $\Phi_r(X)$ in the usual sense (defined over $\mathbb{C}$ a priori, whose coefficients live in $\mathbb{Z}$). This is also true if $\gcd(r, p) = 1$, but if $p \mid r$ then $\Phi_r(X) \bmod p$ has repeated roots. For example

$$\Phi_p(X) = \frac{X^p - 1}{X - 1} \equiv (X - 1)^{p-1} \bmod p.$$


问:
Is the space of continuous functions which are uniformly continuous on bounded subsets complete wrt the sup-norm?
Let $H$ be a separable $\mathbb R$-Hilbert space and $C^2_b(H)$ denote the space of all $f\in C^2(H)$ such that $f$, ${\rm D}f$ and ${\rm D}^2f$ are bounded. We know that $C^2_b(H)$ equipped with $$\left\|f\right\|_{C^2_b(H)}:=\left\|f\right\|_\infty+\left\|{\rm D}f\right\|_\infty+\left\|{\rm D}^2f\right\|_\infty$$ is a $\mathbb R$-Banach space.

Now consider the subset $U^2_b(H)$ of all $f\in C^2_b(H)$ such that $f$, ${\rm D}f$ and ${\rm D}^2f$ are uniformly continuous on bounded subsets of $H$. Can we show that $U^2_b(H)$ is a closed subset of $C^2_b(H)$?

If I'm not missing, the claim is trivial whenever $H$ has the nearest-point property (e.g. if $H=\mathbb R^d$, $d\in\mathbb N$, or, more generally, if $H$ has the Heine-Borel property), since then every continuous function on $H$ into any other metric space is uniformly continuous on every bounded subset of $H$.

答:
I think all you need here is the standard argument that a uniform limit of uniformly continuous functions is uniformly continuous.

Suppose that $f=\lim_nf_n$, with each $f_n$ uniformly continuous on bounded subsets. Let $\varepsilon>0$ and $Y\subset H$ be bounded. Then there exists $n$ with $\|f-f_n\|_{C_b^2(H)}<\varepsilon/3$. As $f_n$, $Df_n$, $D^2f_n$ are uniformly continuous on $Y$, there exists $\delta>0$ such that $\|D^kf_n(x)-D^kf_n(y)\|<\varepsilon/3$, $k=0,1,2$, whenever $x,y\\in Y$ and $\|x-y\|<\delta$.

Then, if $x,y\in Y$ with $\|x-y\|<\delta$, for $k=0,1,2$ we have
\begin{align}
\|D^kf(x)-D^kf(y)\|
&\leq\|D^kf(x)-D^kf_n(x)\|+\|D^kf_n(x)-D^kf_n(y)\|+\|D^kf_n(y)-D^kf(y)\|\\[0.3cm]
&\leq \|f-f_n\|_{C_b^2(H)}+\frac\varepsilon3+\|f_n-f\|_{C_b^2(H)}\\[0.3cm]
&\leq\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.
\end{align}
So $f$, $Df$, $D^2f$  are uniformly continuous.


问:
A convergence test for series
I need to prove the following result: let $\left( a_{n}\right) \subset
\left( 0,\infty \right) $ such that
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\quad
(X)
\end{equation*}

where $x_{n}\geq x>1$. Prove that
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is convergent. In addition, if
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\end{equation*}
where $x_{n}\leq x<1$, then
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is divergent.

For the first asertion it is clear. More precisely, the condition (X) can be
rewritten as follows
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n=\left(
x_{n}-1\right) a_{n}.
\end{equation*}
Since $x_{n}\geq x>1$, then $x_{n}-1\geq x-1>0$
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n\geq \left(
x-1\right) a_{n},
\end{equation*}
and, using the inequality
\begin{equation*}
\left( n-1\right) \ln \left( n-1\right) >\left( n-1\right) \ln n-1,
\end{equation*}
we obtain that
\begin{equation*}
a_{n}\left( n-1\right) \ln \left( n-1\right) -a_{n+1}n\ln n\geq \left(
x-1\right) a_{n}>0.
\end{equation*}
Hence $u_{n}=a_{n}\left( n-1\right) \ln \left( n-1\right) ,n\geq 2$, is
strictly decreasing, so it converges. Whence the telescoping sum associated
with $\left( u_{n}\right) $, that is
\begin{equation*}
\sum_{n\geq 1}\left( u_{n}-u_{n+1}\right) ,
\end{equation*}
is convergent. But
\begin{equation*}
0<\left( x-1\right) a_{n}\leq u_{n}-u_{n+1},
\end{equation*}
for all $n\geq 2$. By comparison test, it follows that the series
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is convergent.

Now the problem is the second assertion. Since
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n=\left(
x_{n}-1\right) a_{n}.
\end{equation*}
and since $x_{n}\leq x<1$, then $x_{n}-1\leq x-1<0$,
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n\leq \left(
x-1\right) a_{n}<0,
\end{equation*}

If one can guarantee that
\begin{equation*}
a_{n}\left( n-1\right) \ln \left( n-1\right) -a_{n+1}n\ln n\leq \left(
x-1\right) a_{n}<0,
\quad(XX)
\end{equation*}
for all $n\geq 2$, then it follows that $\left( u_{n}\right) $ is
increasing. So
\begin{equation*}
u_{n}=a_{n}\left( n-1\right) \ln \left( n-1\right) >\alpha ,
\end{equation*}
for all $n\geq 2$ and some $\alpha \in \left( 0,\infty \right) $. Hence, for
all $n\geq 2$, we have
\begin{equation*}
\frac{\alpha }{\left( n-1\right) \ln \left( n-1\right) }<a_{n}
\end{equation*}
and, since $\sum \frac{1}{\left( n-1\right) \ln \left( n-1\right) }$ is
divergent, it follows that the series $\sum a_{n}$ is divergent. How can I
prove that (XX), if it is correct of course. If not, how can we tackle the
proof of the second assertion?

答:
Here is a method which can be applied to both cases : let's do the first one, the other one will be similar. We have $a_n > 0$ such that
$$\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n} - \frac{x_n}{n \ln (n)}$$

with $x_n \geq x > 1$ for a real number $x$. Let $y=(1+x)/2$. Consider $$u_n = \frac{1}{n \ln^{y}(n)}$$

One has
$$\frac{u_{n+1}}{u_n} = \frac{n \ln^y(n)}{(n+1)\ln^y(n+1)} = \left(\left(1+ \frac{1}{n} \right)\left( \frac{\ln(n+1)}{\ln(n)}\right)^y \right)^{-1}$$ $$ =\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{\ln(1+\frac{1}{n})}{\ln(n)}\right)^y \right)^{-1}=\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{1}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right)\right)^y \right)^{-1}$$
$$=\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right)\right) \right)^{-1} =\left(1+ \frac{1}{n} +\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right) \right)^{-1} $$ $$=1- \frac{1}{n} -\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right) $$

Now, because $$x_n - y \geq  \frac{x-1}{2} > 0$$
you see that from a certain rank, you will have
$$\frac{u_{n+1}}{u_n} \geq \frac{a_{n+1}}{a_n}$$

Because the series $\sum u_n$ is convergent, you deduce that $\sum a_n$ is also convergent by comparison.

Title: A convergence test for series


问:
Open Maps in Topology
I can not seem to figure out why one mapping is not an Open Map, whereas the one is an Open Map. Below will be the following two examples.

In both examples, the the intervals inherit the Subspace Topology from $\mathbb{R}_{std.}$ and $\mathbb{S}^1$ is the unit circle with Subspace Topology from $\mathbb{R}_{std.}^2$. Also, the function in both examples is defined as $f(t):=(\cos{2\pi t},\sin{2\pi t})$.

Example 1: The mapping $f:[0,1)_{std.}\to\mathbb{S}^1$ is not an Open Map.

Example 2: The mapping $f:[0,\frac{1}{2})_{std.}\to f([0,\frac{1}{2}))\subseteq\mathbb{S}^1$ is an Open Map.

I believe that Example 1 is not an Open Map, since, for $1>\epsilon > 0$ the set $U:=[0,+\epsilon)$ is open in $[0,1)_{std.}$, however, $(1,0) \in f(U)$ and there is no open set around $(1,0)$ contained in $f(U)$. Though, I am not sure if this is correct, and I am not sure why Example 2 becomes an Open Map.

答:
You’re right about the first map. Let $S$ be the semicircle consisting of the points $\langle 1,\theta\rangle$ (in polar coordinates) with $\theta\in[0,\pi)$. The second map is open as a function from $\left[0,\frac12\right)$ to $S$, because sets of the form $\{\langle 1,\theta\rangle:\theta\in[0,\alpha)\}$ are open in $S$ for $0<\alpha\le\pi$.

This second map is just a homeomorphism from a half-open interval to a half-open interval that has been bent around into a semicircle. The first map fails to be open because the half-open interval $[0,1)$ was bent so far around that the open end now converged to the closed end, so to speak.

Title: Open Maps in Topology


问:
Prove that a family of subsets of a probability space is a $\sigma$-algebra
Let $(\Omega,\mathcal{F}, P)$ be a probability space. We define $\mathcal{N}=\{S\subseteq \Omega\ |\ \exists B\in \mathcal{F}: \ P(B)=0, \ S\subseteq B\}$.

I want to prove that the following family is a $\sigma$-algebra:
$$\hat{\mathcal{F}}=\{A\subseteq \Omega\ |\ \exists S \in \mathcal{F}:\ A\triangle S\in \mathcal{N}\}$$
(where $\triangle$ denotes the symmetric difference between two sets: $A\triangle B=(A\setminus B)\ \cup\ (B\setminus A))$

Clearly $\hat{\mathcal{F}}\neq\emptyset$ and, for every $A \in \hat{\mathcal{F}},\ A^c=\Omega\setminus A$ is also an element of $\hat{\mathcal{F}}$. The part of the proof that I'm having trouble with is the fact that, for every countable family $\{A_n\}_{n \in \mathbb{N}}\subseteq \hat{\mathcal{F}}$, the union $\bigcup_{n \in \mathbb{N}}A_n$ is in $\hat{\mathcal{F}}$. Probably I'm missing something obvious, but I'm stuck.

答: HINT: remember that if $P(B_k)=0$ for each $k\in \mathbb N\setminus\{0\}$ then $P(\bigcup_{k\geqslant 1}B_k)\leqslant \sum_{k\geqslant 1}P(B_k)=0$. Now observe that if $A_k\in\hat{\mathcal{F}}$ then there exists some $C_k\in \mathcal{F}$ such that $B_k:=A_k\,\triangle\, C_k\in \mathcal{N}$.


问:
Let $n = p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$. Show that $d(n) = (a_1+1)(a_2+1)\dots(a_k+1)$

Let $n = p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$. Show that $d(n) = (a_1+1)(a_2+1)\dots(a_k+1).$

How should I go about this? I tried to first show that the divisor function is multiplicative and use that result to prove the statement, but it seems that the multiplicativity depends on the result I'm actually trying to show? Any hints would be appreciated.

答: Any factor of $n$ is of the form $d=p_1^{\beta_1}p_2^{\beta_2}...p_k^{\beta_k}$ where  $0\le\beta_i\le a_i$. It is clear that there are $a_i+1$ choices (including $0$) for each $\beta_i$ giving us a total of $(a_1+1)(a_2+1)...(a_k+1)$ factors.


答: If $d $is a divisor of $n, $we can take any one of $0,...,a_{i} $for each $p_{i} $. Since these are all powers of primes, by uniqueness of prime factorizations each way to choose is distinct.

Title: Let $n = p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$. Show that $d(n) = (a_1+1)(a_2+1)\dots(a_k+1)$

Body:
Let $n = p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$. Show that $d(n) = (a_1+1)(a_2+1)\dots(a_k+1).$

How should I go about this? I tried to first show that the divisor function is multiplicative and use that result to prove the statement, but it seems that the multiplicativity depends on the result I'm actually trying to show? Any hints would be appreciated.


Body:
Note that if $d\ \vert \ n$:

$$d = p_{1}^{b_{1}}p_{2}^{b_{2}}...p_{k}^{b_{k}}\ \ \text{where}\  0\leq b_{i}\leq a_{i}$$

Then, there are $a_{i} + 1$ (from $0$ to $a_{i}$) possible values for each $b_{i}$. Thus, $\boxed{d(n) = (a_{1} + 1)(a_{2} + 1)...(a_{k} + 1)}$


问:
Using cross ratios to show how an old projection technique used by artists was wrong

These questions are taken from Stillwell's The Four Pillars of Geometry. For $Q 5.7.3$, I am confused about what the question is asking. Is at asking me to:

1) Compute the cross ratio's between four points $P,$ $Q,$ $R,$ $S$, where the distance $PQ=1$, $QR=e$, $RS=e^2$

OR

2) Is it asking me to compute first the cross ratio between four point $P$,$Q$,$R$,$S$ each separated by distance $1$, then compute the cross ratio for another four points $A,B,C,D$ each separated by distance $e$, and then compute the cross ratio between another 4 points $W,X,Y,Z$ each separated by distance $e^2$.

I need clarification of what is being asked of me, but I would also really appreciate some sort of dumbed down intuition (I am a complete beginner at Projective Geometry) about the technique mentioned in the question would be incorrect. Because as it stands, I really don't see how either one of 1) or 2) would show that the technique is incorrect? In fact, what does it even mean for the technique to be incorrect? What does it do incorrectly?

In response to rschwieb's answer:

What I understood from your answer:

The points on the tiles in the real world are $P=0, Q=1, R=2, S=3$. When making his drawing, the artist thinks that right way to draw this would be to have each width a constant fraction $e$ of the one before it. So on this drawing board he puts the point points $P'=0, Q'=1, R'=1+ e, S'=1+ e + e^2$. If we calculate the cross ratio of $P',Q', R', S'$, we know that it should be $\frac{4}{3}$ because the original points $P,Q,R,S$ were equally spaced. But when we calculate $\frac{(1+e)(e^2+e)}{(e)(e^2+e+1)} =\frac{4}{3}$, we get $e=0,1$, which contradicts our original assumption that $e$ was a fraction. So we conclude the technique is wrong, and doesn't correctly project the original points $P, Q, R, S$ in terms of perspective.

答:
Your first interpretation is fine. It's saying that along the line receding into the distance, the first gap has length $1$, the second $e$ and the third $e^2$.  The artist is succeeding in showing the gaps are shrinking, but it does not reflect reality.

The cross-ratio is a quantity for the positions of the four points, one which is invariant under projective transformations.  We know that a projective transformation can transform the line into one which has the three points equally spaced because we're supposed to assume a rectangular or square tiling.

We also know that equally spaced points always give a cross ratio of $\frac 43$, so if you compute the cross ratio of $P,Q,R,S$ as you have given, you should find it's $\frac{(1+e)^2}{1+e+e^2}$. If you equate this to $\frac43$ and solve, you should find the only solution is $e=1$, which of course isn't under consideration (we're using $e< 1$.)

You might follow up by computing, given the first gap $1$, the second gap $e$ what that third gap is in terms of $e$ for the cross-ratio to work out to $\frac43$.

In desmos, I plotted the artist's prediction (blue) and the cross-ratio's prediction (red) for $x$ in the interval $[0,1]$

They're close, but not the same!

For fun I picked the value $e=0.3$ and generated a few plots that demonstrate what the perspective looks like, when drawn that way. Keep in mind that they plot the same sequence of colored lines, just at different $y$ values.

Here's the naïve artist's formula:

And here's what projective geometry says:

With a little more work one should be able to predict the $y$ coordinate of the horizon. From what it looks like here, the artist's prediction looks like it might converge way too fast, below the true horizon.


问:
Prove that the tensor product of non algebraic extensions is not a field

Suppose that $E,F $ are two extensions of $K$ and  $E\otimes _K F$ is a field. Prove that $E$ or $F$ are algebraic over $K$.

Is there any hint to prove this?
Thanks.

答:
Here's a more hands-on proof. We'll prove the contrapositive: if $E$ and $F$ are both transcendental over $K$ we'll show that $E \otimes_K F$ is not a field by exhibiting a nontrivial ideal. Let $e_i, i \in I$ and $f_j, j \in J$ be transcendence bases of $E$ and $F$ respectively over $K$, so that $E$ is an algebraic extension of $K(\{ e_i \})$ and $F$ is an algebraic extension of $K(\{f_j\})$.

Assume WLOG that $|I| \le |J|$ and pick an embedding $g : I \hookrightarrow J$. This induces an embedding of both $E$ and $F$ into the algebraic closure $\overline{K(\{f_j\})}$, which induces a map

$$E \otimes_K F \to \overline{K(\{f_j\})}.$$

The kernel of this map is nontrivial because it contains, for example, elements of the form $e_i \otimes 1 - 1 \otimes f_{g(i)}$ (and these elements are nonzero because, for example, instead of using the embedding $g$ above we can write down a map from $E \otimes_K F$ to $\overline{K(\{e_i\} \cup \{ f_j\})}$, and the image of $e_i \otimes 1 - 1 \otimes f_{g(i)}$ here is $e_i - f_{g(i)}$ where $e_i$ and $f_{g(i)}$ are algebraically independent). So $E \otimes_K F$ can't be a field.


问:
Calculate an integral in a measurable space
Let $(X,\mathcal{M})$ a measurable set with measure  $\mu$. Let $f$ be an integrable non negative function, such that $K:=\int_{E}f \mathrm d\mu<\infty$, where $E\in(X,\mathcal M)$. Let $\alpha>0$. Calculate the following integral

$\displaystyle \lim_{n\rightarrow\infty}\int_{E} n \ln\left(1+\left(\frac{f}{n}\right)^{\alpha}\right)\mathrm{d}\mu$

I prove that for $\alpha=1$ the previous integral is $K$, and $\forall \alpha>1$ is zero. It follows by the identity $(1+x^{\alpha})\leq\alpha x$, $\forall x\geq0$ $\forall \alpha\geq1$. And the Dominated Convergence Theorem. But I don´t know how to take the case $0<\alpha<1$.
    Thanks in advance!

Suppose that $\alpha>1$. We take the sequece
 $\{f_{n}\}_{n\in\mathbb N}$ of measurable functions, $f_{n}:=n\ln(1+(f/n)^{\alpha})$.  Moreover by the identity, for $\alpha>1$,
 we have the following $|f_{n}|=\left|n\log\left(
 1+\left(\frac{f}{n}\right)^{\alpha}\right) \right|\leq n \alpha\frac{f}{n}=\alpha f$,
  $\alpha f$ is measurable. We need only to prove that$\{f_{n}\}_{n\in\mathbb N}$ converges pointwise to a function.

In this case $f\equiv 0$
   $\forall x\in Dom(f)$, $\displaystyle\lim_{n\rightarrow \infty}n\log\left(\left(\frac{f}{n}\right)^{\alpha}+1\right)=
 \lim_{t\rightarrow 0}\frac{\log((ft)^{\alpha}+1)}{t}$.
  Making the change of variables $t=\frac{1}{n}$. And by L'Hopital, it arises
  $\lim_{t\rightarrow 0}\alpha f^{\alpha}
 \frac{t^{\alpha-1}}{(ft)^{\alpha}+1}=0$. ($\alpha-1>0$). Finally, using the Dominated Convergence Theorem, it leads to $
 \displaystyle\lim_{n\rightarrow \infty} \int f_{n}  \mathrm d \mu=\int 0 \mathrm d\mu =0$

答:
By a familiar property of $e$ and the exponential function, we have:
$$
\lim_{n\to\infty} n^\alpha \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) = f^\alpha
$$

Hence:
$$
\lim_{n\to\infty} n \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) = f^\alpha \lim_{n\to\infty} n^{1-\alpha}
$$

When $0 < \alpha < 1$ and $f \ne 0$, the the limit diverges to $\infty$.

Apply Fatou's lemma:
$$
\int \liminf_{n\to\infty} n \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) \,d\mu \le \liminf_{n\to\infty} \int n \ln\left(1 + \left(\frac{f}{n}\right)^\alpha\right) \,d\mu
$$

Assuming $f$ is not zero a.e., the limit on the LHS diverges to $\infty$ for a set of positive measure. It follows that the limit on the RHS is $\infty$.


问:
nonexistence of a function satisfying $f(x)^{f(y)}=y^x$
prove that there is no function $f:\mathbb{N}\mapsto \mathbb{N}$
such that $\forall x, y \in \mathbb{N}\ (f(x))^{f(y)}=y^x$

My attempt: If such function exists, so for $x,y,z \in \mathbb{N}$ $(f(x))^{f(y+z)}=(y+z)^x$ but in the other hand $(f(x+y))^{f(z)}=z^{(x+y)}=z^xz^y=(f(x))^{f(z)}(f(y))^{f(z)}$

and I need a Hint to accomplish this.

答: Denote $a=f(1)$ then $a^a = 1^1 = 1$. Therefore $a=f(1)=1$ and $1 = 1^{f(2)} = 2^1=2$, a contradiction.


问:
Find Fourier sine series
I have to find a Fourier sine series of $f(x)=x^2$ for $0<x\leq2$. And show that converges to $0$ at $x=\pm 2$.

(UPDATE) My attempt was :

extended the function to $0<x\leq-2$ s.t. period is 2.

Then, $b_n=\frac{2}{2}\int_{-2}^{0}x^2sin\frac{\pi nx}{2}dx$

Solving this I got: $$-\frac{8}{\pi n}cos(\pi n)+\frac{16}{(\pi n)^3}cos(\pi n)-\frac{16}{(\pi n )^3}$$
$$=-\frac{8}{\pi n}(-1)^n+\frac{16}{(\pi n)^3}((-1)^n-1)$$

is this correct? how do I show this converges to zero?

Thanks!

答:
If you define $f(x)$ to be odd over $-2<x\le 2$, and periodic, with period 4, then the Fourier sine series will converge to $f(x)$ at all points where $f$ is continous, and will converge to $\lim_{x\downarrow x_0} f(x) + \lim_{x\uparrow x_0}f(x)$ at points $x_0$ where $f(x)$ is discontinuous.

At $x_0=2$, the limit from the left is $2^2=4$ and the limit from the right is $-2^2=-4$ so the series converges to $4-4=0.$  Similarly at $x_0=-2.$

Title: Find Fourier sine series


问:
How to show that if two Platonic solids have the same number of edges, vertices, and faces, then they are similar in $\mathbb{R}^{3}$?
Note: It appears that some of the terms here do not have standardized definitions, so some sources may give conflicting info.

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

We define a Platonic solid as a simple, regular polyhedron homeomorphic to $S^{2}$.

We define a simple polyhedron to be a polyhedron that does not self-intersect itself.

We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$.
This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

答: Another way to interpret the $5$ Platonic solids is that they are the only configurations of at least $3$ regular polygons around each vertex satisfying that the total sum of angles at that vertex is less than $180^{\circ}$ Also note that each Platonic solid is uniquely determined by the number of faces around each vertex and the number of sides of each face because all Platonic solids are vertex-transitive and rotationally symmetric about each vertex. Then, note that (a) the number of sides to each face is determined by $n_{f} = \frac{2e}{f}$, and (b) the number of faces around each vertex is determined by $n_{v} = \frac{2e}{v}$. Thus, each Platonic solid is uniquely determined by the choice of $(v, e, f)$, or any two of them given Euler's relation. $\blacksquare$


问:
How can I calculate the length of intersection on a 2d line?
The following diagram explains my question:

$\text{Line}A = [2,7]$

$\text{Line}B = [5,9]$

$X =$ The length of the intersection between LineA and LineB

Thanks

Update: I ended up using this algorithm
https://en.wikipedia.org/wiki/Marzullo%27s_algorithm

Body:
It is $$\ell ([A_1,A_2],[B_1,B_2])=\begin{cases}
\min \{A_2,B_2\}-\max \{B_1,A_1\} & \text{If }\min \{A_2,B_2\}\geq\max \{B_1,A_1\} \\
0 & \text{Otherwise}
\end{cases}$$
In your example $7-5=2.$


问:
Expected Value from Joint Density Distribution
Let X and Y be continuous random variables with joint density function f(x,y) and marginal density function fx and fy respectively, that are nonzero only on the interval (0,1). Which one of the following statements is always true**?

(A)
$$E[X^{2}Y^{3}]=( \int_{0}^{1}x^{2}  \, dx ) ( \int_{0}^{1}y^{3}  \, dy ) $$
(B)
$$E[X^{2}]= \int_{0}^{1}x^{2}f(x,y)  \, dx $$
(C)
$$E[X^{2}Y^{3}]=( \int_{0}^{1}x^{2}f(x,y)  \, dx ) ( \int_{0}^{1}y^{3}f(x,y)  \, dy ) $$
(D)
$$E[X^{2}]= \int_{0}^{1}x^{2}f(x)  \, dx $$
(E)
$$E[Y^{3}]= \int_{0}^{1}y^{3}f(x)  \, dx $$

Q1. The question said fx and fy are nonzero only on the interval (0,1), so 0<=x<=1,0<=y<=1, but the question didn't say f(x,y)=1, So, we can't say it's a joint uniform distribution right?

Q2. With the information provided, how do we know whether x,y are independent?

Q3. If f(x,y)=1, then (A) is correct?

Q4. If x,y are independent, is (C) correct? Can we use f(x,y) instead of f(x)?

答:
I'm not sure why you concern with whether $f(x,y)=1$ or anything about independence and uniform distribution. The question is about what is always true. So for each of the 5 choices, just check whether the LHS always equals RHS, not just in special cases.

Hint: In general,
$$
E[g(X,Y)] = \int_0^1\int_0^1g(x,y)f(x,y)dydx=\int_0^1\int_0^1g(x,y)f(x,y)dxdy. \tag{$*$}
$$
For example, (E) is incorrect because:
$$
\text{LHS}=E[Y^3]=\int_0^1\int_0^1y^3f(x,y)dxdy=\int_0^1y^3\left[\int_0^1f(x,y)dx\right] dy=\int_0^1y^3f_Y(y)dy
$$
which is not the RHS of (E). You can also use ($*$) to verify / refute the smaller questions that you have.

Hint 2: the answer is

(D).


问:
How to show that this languange is not turing-recognizable?
Let $S=\{<M> | M  \textrm{ is a } TM \textrm{ and } L(M) = \{<M>\}\}$ how can I show that $S$ isn't turing-reconizable? Besides that, what $L(M) = {<M>}$ means?

答:
L(M) = < M> means that the only string the M accepts is its own description < M>  , hopefully you can see now that S is the language of TMs that accept only their own descriptions

Proof  :

We provide a reduction from ATM to S ,  ATM ≤m S to show that s is not Turing-recognizable.

Assume we have a TM S the recognises S , we construct a TM A to recognise ATM

A on input < M,w> : "

Construct  a TM T , T on input x : "

If x ≠ < T> , reject

If x = < T> , simulate M on w , if M accepts w accept , else reject "

Now run S on  < T>  , accept if S accepts , reject otherwise "

If M accepts w , then T accepts < T> , hence S accepts < T> , and thus A accepts < M,w>

If M does not accept w , then T rejects < T> , hence S does not accept < T> , and A does not accept  < M,w>

Clearly A recognises ATM , which is a contradiction , thus S is not Turing recognisable

You may want to try to showing S' is also non-Turing-recognisable , the proof will be very similar


问:
Linear programming - Objective function is a multiple of one of the constraints
I wondered if someone could explain to me the intuition of what it would mean if the objective function in a Linear Programme is a multiple of one of the constraints?

I am thinking it means that the lines are parallel with the objective function being at a different contour. Does this imply anything about the uniqueness of an optimal solution? Would appreciate some intuition

答:
It means indeed that the optimal solution is not unique, if it exists. The corresponding constraint must be active. In a linear program with two variables it can be shown graphically. An example. Le´t suppose you have the following program:

$\textrm{max} \  Z=500x_{1}+300x_{2}$

s.t.

$15x_{1}+5x_{2}\leq 300$

$10x_{1}+6x_{2}\leq 240$

$8x_{1}+12x_{2}\leq 450$

$x_{1},x_{2}\geq 0$

We can solve the objective funktion for $x_2$: $ \, \, \, x_2=\frac{Z}{300}-\frac53x_1$. This function has a slope of $\color{red}{-\frac53}$. And we can solve the second constraint (equality) for $x_2$ as well:

$10x_{1}+6x_{2}=240\Rightarrow x_2=40\color{red}{-\frac53}x_2$

For more detailed information with a graph to this example see here.


问:
Archimedean ordered field and non-expressible in first order logic
In Chang and Keisler's model theory, they define that an ordered field $(F,+_{F},\cdot,0_{F},1_{F},\le)$ is Archimedean iff for any two positive elements $a$, $b$ in $F$ there is an $n$ such that $b\le na$.

(Here the notation $na$ is an abbreviation for the term $a+_{F}\ldots+_{F}a$, $n$ times.)

And they say that this definition is not expressible in first order logic.
My question is why it is impossible to express the definition in first order logic. In $\mathrm{ZF}$, we can define recursively a function $f$ such that $\mathrm{dom}(f)=\omega$, $f(0)=a$ and $f(n+1)=f(n)+_{F}a$. So the definition can be formalized in $\mathrm{ZF}$. I don't know what I'm missing..

答:
As Alex Kruckman says, the intended meaning is that this definition can't be expressed in the first-order language of ordered fields. The problem is that the first-order language of ordered fields does not obviously allow you to quantify over all positive integers $n$; you can talk about

$$na = \underbrace{a + \dots + a}_{n \text{ times}}$$

for any fixed $n$, but the first-order language of ordered fields $F$ only allows you to quantify over $F$, and doesn't obviously supply you with a way of talking about $\mathbb{N} \subset F$.

I say "obviously" because a priori there could be a clever trick that does it, but in fact there can't be such a trick. This is because the class of models of a first-order theory must be closed under ultraproducts, but it's possible for an ultraproduct of archimedean fields to be non-archimedean. For example the ultrapower $\mathbb{R}^{\infty}/U$ of countably many copies of $\mathbb{R}$ (the nonstandard reals) is non-archimedean: we can take $a = (1, 1, 1, \dots)$ and $b = (1, 2, 3, \dots)$ and then there is no positive integer $n$ such that $b \le na$ (in other words, $b$ is infinitely large compared to $a$).

Ignorable additional comment: You can extend the first-order theory of ordered fields to a two-sorted theory consisting of one sort for the natural numbers $\mathbb{N}$ and another sort for the field $F$, together with, say, an embedding $\mathbb{N} \hookrightarrow F$; this would allow you to talk about the operation $(n, a) \mapsto na$ where $n \in \mathbb{N}$ and $a \in F$, and to quantify over $\mathbb{N}$. Now the problem is that it's impossible to write down axioms guaranteeing that the first sort is exactly $\mathbb{N}$ and not some larger nonstandard model of the natural numbers; for example, whatever axioms we write down, we can take an ultrapower of the pair $(\mathbb{N}, \mathbb{R})$ and we'll have infinitely large nonstandard natural numbers, so the ultrapower $\mathbb{R}^{\infty}/U$ will still be "archimedean" but with respect to the nonstandard natural numbers $\mathbb{N}^{\infty}/U$, not the standard ones. So this doesn't work either.


答: OK so we can imagine two solids, each with the same triple of $(V,E,F)$ each has the same number of faces. And you calculate the number of edges that each face has as $2E/F$ (each edge can be in two faces). So they have the same number of faces and each face has the same number of edges. How many edges meet at each vertex? This is $2E/V$ (each edge hits two vertices). So these two polyhedra have the same number of faces, each with the same number of sides and the same number of edges meeting at each vertex so they must be the same polyhedron. We could build each to the precise formula of faces, number of edges per face and number of edges per vertex and we have no room to make choices. The recipe precisely defines the polyhedron.

Title: How to show that if two Platonic solids have the same number of edges, vertices, and faces, then they are similar in $\mathbb{R}^{3}$?

Body:
Note: It appears that some of the terms here do not have standardized definitions, so some sources may give conflicting info.

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

We define a Platonic solid as a simple, regular polyhedron homeomorphic to $S^{2}$.

We define a simple polyhedron to be a polyhedron that does not self-intersect itself.

We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$.
This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.