问:
Is the golden ratio invariant across non-Euclidean geometries?
I think Pi (the relation between the circumference of a circle and it's diameter) has different values in non-Euclidean geometries. I am interested in what is invariant across all geometries. I wonder if the golden ratio may be invariant across spherical and/or hyperbolic geometries.</p>

答:
Insofar as the Golden Ratio $\phi$ is defined as a number satisfying $\phi^2=\phi+1$ (this is equivalent to the extreme/mean ratio division description), the ratio belongs to any geometry where (loosely speaking) lines are effectively rulers with continuous and uniform real-number labels for points. The existence of $\phi$ in this sense is no more or less remarkable than the existence of $\frac12$. Even $\pi$ as a number exists in such contexts: it lives just past $3$ on the ruler. In a "uniform" geometry (such as the Euclidean, Spherical, and Hyperbolic planes), the rulers don't change from here to there, so the "invariance" of $\phi$ and $\frac12$ and $\pi$ as numbers is baked-in: every segment has a point dividing it in the ratio $1:x$ for any number $x$. As a matter of numerics, there's really nothing to discuss.

Where things get interesting is investigating whether a value like $\phi$ "arises" from the geometry. For instance, one way $\phi$ arises in Euclidean geometry is via the regular pentagram:

Sides $\overline{AB}$ and $\overline{PQ}$ meet at a point $K$ that divides each of those sides into the ratio $1:\phi$. This construction is "invariant" in Euclidean geometry because all regular pentagrams are similar, and corresponding ratios are preserved by proportionality.

Non-Euclidean geometry lacks similarity: for instance, doubling the lengths sides sharpens the angles of a hyperbolic triangle, and dulls those of a spherical one. So, even if we managed to get a pentagrammic "construction" of the Golden Ratio to work once ...

... we couldn't expect more: larger hyperbolic pentagrams have pointier points (contrariwise for spherical pentagrams), so that the ratio determined by how sides cut each other depends upon scale. Dependence is the antithesis of invariance.

That's not to say, however, that there's nothing to discuss. Indeed, I'm about to launch on a (ahem) lengthy digression ...

In non-Euclidean geometries "raw lengths", and/or ratios thereof, aren't a primary concern in establishing relations in triangles and other figures. Instead, lengths tend to be wrapped in (circular or hyperbolic) trig functions.

Consider, for instance, the Laws of Sines for a $\triangle ABC$ in Euclidean, Spherical, and Hyperbolic planes:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \qquad \frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C} \qquad \frac{\sinh a}{\sin A}=\frac{\sinh b}{\sin B}=\frac{\sinh c}{\sin C} \tag1$$
or the circumference of a circle with radius $r$:
$$C_{euc} = 2\pi r  \qquad C_{sph} = 2\pi \sin r \qquad C_{hyp} = 2\pi \sinh r \tag2$$
While it is true that the circumference-to-"raw"-radius ratio is not invariant, we can establish $\pi$ (and/or $\tau:=2\pi$) as an invariant "circle constant" by defining it in terms of the ratio of circumference and an appropriate transmutation of radius:
$$2\pi = \frac{C_{euc}}{r} = \frac{C_{sph}}{\sin r} =\frac{C_{hyp}}{\sinh r} \tag3$$
Likewise for area:
$$\pi = \frac{A_{euc}}{4\left(\tfrac12r\right)^2} = \frac{A_{sph}}{4\sin^2\tfrac12r} = \frac{A_{hyp}}{4\sinh^2\tfrac12r}\tag4$$

Such transmutation is precisely the mechanism that helps us establish the Golden Ratio as a scale-independent —and thus invariant— "pentagram constant" via the above construction, as we can write:

$$\phi = \frac{|KQ|_{euc}}{|PK|_{euc}} = \frac{\sin |KQ|_{sph}}{\sin |PK|_{sph}} = \frac{\sinh|KQ|_{hyp}}{\sinh|PK|_{hyp}} \tag{5}$$
where $|XY|_{geo}$ indicates distance in the corresponding geometry. (I'll omit the subscript when context is clear.)

Importantly, though, this kind of thing doesn't preserve every kind of ratio. In fact, many (most? all?) of the other occurrences of $\phi$ in the Euclidean pentagram don't transfer to non-Euclidean counterparts simply by swapping every length $\ell$ for $\sin \ell$ or $\sinh \ell$.

On the other hand, there are other interesting Euclidean occurrences of the Golden Ratio that do have direct non-Euclidean counterparts. For instance, here is a slight restatement of a delightful result from George Odom:

If a midpoint segment of an equilateral triangle is extended to meet the triangle's circumcircle at $P$ and $Q$, then each midpoint divides the extended segment in the square of the Golden Ratio. In particular, for $K$ in the figure,
$$\frac{|QK|}{|PK|} = \phi^2 \tag{6}$$

(Odom actually says more-simply that, if $K'$ is the other midpoint, then
$$\frac{|KK'|}{|PK|} = \phi \tag{7}$$
a relation equivalent to $(6)$ because $|QK|=|PK|+|KK'|$ and $\phi^2=\phi+1$.)

In hyperbolic geometry (spherical is analogous), the same construction "works", provided we apply the "appropriate transmutation" to the lengths:

$$\frac{\sinh|QK|}{\sinh|PK|} = \phi^2 \tag{8}$$

The best we can do for a counterpart to $(7)$ is
$$\frac{\sinh|QK|-\sinh|PK|}{\sinh|PK|} = \phi \tag{9}$$
Notice that the numerator does not reduce to $\sinh|KK'|$; as I wrote, the transmutation trick doesn't work for all ratios. (This is why I had to restate Odom's result as a construction of $\phi^2$ instead of $\phi$.)

As a final example where the trick does work (and figures into two aspects of a result), there's this (re-discovered?) generalization of Odom's construction (blog link via trigonography.com):

$$\frac{|BK|}{|AK|} = 4\cos^2\theta \quad\iff\quad \frac{|QK|}{|PK|}=\phi^2 \tag{10}$$
which, eg, in the Hyperbolic plane becomes

$$\frac{\sinh|BK|}{\sinh|AK|} = 4\cos^2\theta \quad\iff\quad \frac{\sinh|QK|}{\sinh|PK|}=\phi^2 \tag{11}$$

I'm going to stop typing now. :)


问:
Indeterminate form : $ 0 \cdot \infty$
I have a question about the limit of :
$$\lim_{t \to \infty}t\cdot e^{-t}$$

Why is that equal to $0$ ? Shoudln't $ 0 \cdot \infty$ be an inderminate form ?
It's true that with the graphic we can determine that this limit is equal to  $0$ :

But all I know is that  : $ 0 \cdot \infty$ is supposed to be an inderminate form .

答:
A nice way to show this is:
$$f(t)=te^{-t}=\frac{t}{e^t}$$
and we know that:
$$e^t=\sum_{n=0}^\infty\frac{t^n}{n!}=1+t+\frac{t^2}{2}+\frac{t^3}{6}+...$$
and so:
$$f(t)=\frac{t}{1+t+t^2/2+t^3/3+...}=\frac{1}{1/t+1+t/2+t^2/3+...}$$
now if we look at this we can see that:
$$\lim_{t\to\infty}\frac1t=0$$
and so on the bottom we have something of the form:
$$\frac 1{1+\infty+\infty+...}$$
which will obviously tend to $0$

ContentLicense: CC BY-SA 4.0

Title: Indeterminate form : $ 0 \cdot \infty$


问:
Example of a bounded linear functional on $L^p$ space that is not trivially the dual of an $L^q$ function
I am trying to think of an example of a bounded linear functional on $L^p$ that is not obviously of the form $f \to \int fg $ for some $g \in L^q$. Analogously, maybe an example of a non-trivial operator that is reduced to a matrix in linear algebra would be the derivative operator on polynomials of degree $\leq n$ since the derivative operator isn't an example that came from elementary linear algebra itself.

I want to see an example of how duality adds value.

答:
Take a bounded linear operator $T:L^2\to L^2$ and some $g$ in $L^2$.  Then the functional
$$
f\in L^2 \mapsto \langle T(f),g \rangle \in \mathbb C
$$
is not obviously of the form you mention.  That it turns out to be of that form is a deeply relevant fact of functional analysis underlying  the definition of the adjoint of $T$.


问:
Recurrence Relation General Formula Using Power Functions
Several sources propose finding a general formula for the $n^{\text{th}}$ term of a sequence defined by a recurrence relation by assuming the term is defined by a power function.

E.g. consider the Fibonacci numbers:

$$u_{n+2}=u_{n+1}+u_n$$

$$u_0=u_1=1$$

Let $u_n=kx^n$:

$$x^2=x+1$$

$$x=\frac{1\pm\sqrt{5}}{2}$$

At this point it is usually stated that the double root implies two forms for $u_n$, both of which satisfy the recurrence.

$$u_{n1}=k_1\left(\frac{1+\sqrt{5}}{2}\right)^n$$

$$u_{n2}=k_2\left(\frac{1-\sqrt{5}}{2}\right)^n$$

Therefore the most general solution is a combinaton of the above two:

$$u_n=k_1\left(\frac{1+\sqrt{5}}{2}\right)^n + k_2\left(\frac{1-\sqrt{5}}{2}\right)^n$$

My question is - how can this step be justified? If you check on $u_{n1}$ being a general formula, it does not check out as you get inconsistent values for $k_1$ if $u_0$ and $u_1$ are considered:

$$u_0=k_1=1$$

$$u_1=k_1\left(\frac{1+\sqrt{5}}{2}\right)=1 => k_1 = \frac{2}{1+\sqrt{5}}$$

So this must mean the original assumption of $u_n=kx^n$ is incorrect in the first place?

答:
You’ve misunderstood the purpose of $k_1$ and $k_2$: they serve to pick out one of the infinitely many sequences satisfying the recurrence

$$u_{n+2}=u_{n+1}+u_n\,.\tag{1}$$

Let $\Sigma$ be the vector space of all sequences of real numbers; vector addition is defined termwise, so that

$$\langle x_n:n\in\Bbb N\rangle+\langle y_n:n\in\Bbb N\rangle=\langle x_n+y_n:n\in\Bbb N\rangle\,.$$

The sequences satisfying $(1)$ turn out to be a $2$-dimensional subspace of $\Sigma$. Let $\varphi=\frac12\left(1+\sqrt5\right)$ and $\hat\varphi=\frac12\left(1-\sqrt5\right)$, the two zeroes of $x^2-x-1=0$. Then the sequence $\langle\varphi^n:n\in\Bbb N\rangle$ and $\langle\hat\varphi^n:n\in\Bbb N\rangle$ are linearly independent sequences satisfying $(1)$, so every sequence satisfying $(1)$ is a linear combination of these two sequences. Thus, if $v=\langle v_n:n\in\Bbb N\rangle$ is a particular sequence satisfying $(1)$, there are constants $k_1$ and $k_2$, depending on $v$ such that

$$v_n=k_1\varphi^n+k_2\hat\varphi^n\tag{2}$$

for each $n\in\Bbb N$. One uses the initial values $v_0$ and $v_1$ to find $k_1$ and $k_2$.

The point here is that these values of $k_1$ and $k_2$ apply only to the specific sequence determined by the initial values $v_0$ and $v_1$. If $v_0=1$ and $v_1=\varphi$, then $k_1=1$ and $k_2=0$, and we get the sequence of powers of $\varphi$. If $v_0=1$ and $v_1=1$, substituting $n=0$ and $n=1$ into $(2)$ shows that $k_1+k_2=1$ and $k_1\varphi+k_2\hat\varphi=1$, and we get a very different pair of values for $k_1$ and $k_2$.


问:
Exponential map on 2 dimensional Riemannian manifold restricted on a disk is a diffeomorphism.
Suppose $M$ is a simply-connected and complete Riemannian manifold, with Gaussian Curvature $K\leq \beta$,where $\beta>0$ is a positive real number. Choose a point $O\in M$, using the method of Jacobi field we can know that $|dexp_O(\frac{d}{d\theta}(r,\theta))|\geq \frac{1}{\sqrt{\beta}}sin\sqrt{\beta}r$,thus the exponetial map $exp_O$ is a local diffeomorphism restricted on $B=\{v\in T_OM|
|v|< \frac{\pi}{\sqrt{\beta}}\}$ since exponential map is nondegenerate in radial direction and M is 2-dimensional.
I want to show that $exp_O$ is actually a diffeomorphism from $B$ to $exp_O(B)$(equivalently, $exp_O$ is injective on $B$)，but I can't find a way.

答:
The reason you can't find a way to show this is because it's not true in general. Here's a counterexample. Let $M$ be a dumbbell-shaped surface of revolution in $\mathbb R^3$ obtained by revolving a curve like this around the $z$-axis:

Using bump functions, it's easy to create a profile curve that has arcs of unit circles at the top and bottom, and then transitions smoothly to a straight line in the middle part. The corresponding surface of revolution will have Gaussian curvature equal to $1$ at the bottom and top, equal to $0$ in the cylindrical part, and bounded above by $1$ everywhere. But if $O$ is a point in the cylindrical part, the injectivity radius at $O$ is only $L/2$, where $L$ is the circumference of the cylinder.

In general, lower bounds on the injectivity radius are not easy to come by. As this example shows, just bounding the curvature above isn't enough, even if you assume the manifold is simply connected. (To be sure, if the curvature is bounded above by $0$ and the manifold is complete and simply connected, then the injectivity radius is infinite; but a positive upper bound won't do.)

One useful estimate due to Klingenberg is the following: If $(M,g)$ is a compact Riemannian manifold such that all sectional curvatures are bounded above by $\beta>0$ and all closed geodesics have length at most $L$, then
$$
\operatorname{inj}(M) \ge \min \left( \frac{\pi}{\sqrt{\beta}},\ \frac{L}{2}\right).$$
(See, for example, Petersen's Riemannian Geometry (3rd ed.), Lemma 6.4.7.)


问:
Proving certain equivalent statements related to Borsuk Ulam Theorem
I want to show the following statements are equivalent :

There is no odd function [i.e f(-x) = -f(x) ] from $S^n \to S^{n-1}$

Every odd function from $S^{n-1} \to S^{n-1}$ is not nullhomotopic.

I tried showing that if an odd function is nullhomotopic it can be extended to the cone and by gluing two such cones (upper and lower hemisphere) it can be extended to $S^n$. Can I say this extended map is odd?
And what about the other way?

答:
Both of these theorems can be found in the excellent book "Using the Borsuk-Ulam Theorem".

The key insight, though, is that the projection
$\pi : (x_1, \ldots, x_{n+1}) \mapsto (x_1, \ldots, x_n)$ is a homeomorphism of the upper hemisphere of $S^n$ with the ball $B^n$. You've already made a similar observation (extending a map to the cone, and thus to either hemisphere), so I don't feel too bad showing you how to use this observation to solve the problem.

Recall $f : S^{n-1} \to S^{n-1}$ is nullhomotopic exactly when it can be extended to a map $\tilde{f} : B^n \to S^{n-1}$ agreeing with $f$ on $\partial B^n = S^{n-1}$.

First say you have an odd function $f : S^n \to S^{n-1}$. Define a map
$g : D^n \to S^{n-1}$ by $g = f \circ \pi^{-1}$.

Do you see why $g \restriction S^{n-1}$ is odd? Since $g \restriction S^{n-1}$ can be extended to $g$, then, we see $g \restriction S^{n-1}$ is the desired odd nullhomotopic map.

Now we run the argument in reverse!

Say we have an odd nullhomotopic map $g : S^{n-1} \to S^{n-1}$. Then, by nullhomotopicness, $g$ extends to a map $\tilde{g} : B^n \to S^{n-1}$. Thus
$f = \tilde{g} \circ \pi$ is a map from the upper hemisphere of $S^n$ to $S^{n-1}$ which is odd on the boundary.

Of course, there's only one way to extend this to the lower hemisphere if we want our map to be odd everywhere: defining $f(-x) = -f(x)$ does the job.

Do you see why this map $f$ is well defined and continuous?

I hope this helps ^_^


问:
What is the meaning of 'columns have unit lengths'
What is the meaning of this?

In random projection, the original d-dimensional data is
  projected to a k-dimensional (k << d) subspace through
  the origin, using a random k × d matrix R whose columns
  have unit lengths.

I have already searched around the internet and can't find a straight answer.
Does it mean that each column sum of values is equal to one?

答:
Columns have unit length, typically means, the following:

If $a_i$ is the $i^{th}$ column in a matrix, then $\Vert a_i \Vert_2 = 1$, where $\Vert \cdot \Vert_2$ is the two-norm of a vector.

So for instance, the matrix below has columns of unit length:

$$A = \begin{bmatrix}1/2 & \dfrac1{\sqrt2}\\ -\dfrac12 & 0\\ -\dfrac1{\sqrt2} & \dfrac1{\sqrt2}\end{bmatrix}$$
since
$$\sqrt{\left(\dfrac12\right)^2 + \left(-\dfrac12\right)^2 + \left(-\dfrac1{\sqrt2}\right)^2} = \sqrt{\dfrac14 + \dfrac14 + \dfrac12} = 1$$
$$\sqrt{\left(\dfrac1{\sqrt2}\right)^2 + 0^2 + \left(\dfrac1{\sqrt2}\right)^2} = \sqrt{\dfrac12 + 0 + \dfrac12} = 1$$

ContentLicense: CC BY-SA 3.0


问:
Is the statistical standard deviation of elements of x=(x1,x2,...,xn) a convex function?
For $x\in R^n$, is the statistical standard deviation of its elements, formulated as below, a convex function?

$(\frac{1}{n} \sum_{i=1}^n x_i^2 - (\frac{1}{n}\sum_{i=1}^n x_i)^2)^{\frac{1}{2}}$

答:
Let $\sigma : \Bbb R^n\to \Bbb R$ be the standard deviation. Then, $\sigma$ is convex if and only if for each $x, y\in \Bbb R^n$ we have that $$\sigma(tx+(1-t)y)\leq t\sigma(x)+(1-t)\sigma(y) $$

You just need to verify if this relation holds.


问:
How to prove that $11^{n+2} + 12^{2n+1}$ can be divided by $133$ where $n\in\mathbb N$
I know that I have to get to a point where an summation is multiplied by $133$ but I don't know how to get there.
I tried formating the equation like $11^2\times11^n + 12 \times 12^ {2n}$ but I got stuck there and I don't really know how to prove it with induction. I really need to solve this.

Thank you for your help and sorry if I misspelled anything. Also sorry that you have to edit it, I don't really know how to make superscripts.

答:
Seeing as you don't seem to be familiar with induction, I'll give a quick explanation.

We begin by checking the base case for the statement, ie for the first value we are interested in- in your case, check for when $n=1$. Next, we assume the statement is true for some $n=k$.

Next, we use our assumption to show that if the assumption is true then the statement is also true for $n=k+1$. But we already know the statement is true for $n=1$ (in your case), so it must be true for $n=2,3,4,...$ and so on.

I can provide a proof by induction in this case if you feel you are ready for it?

Proof by induction
I will make this proof quite explicit as this is your first proof by induction.
First, let $f(n)=11^{n+2}+12^{2n+1}$.

Base case

Check for $f(1)$:
$$11^3+12^3=133\times23$$
which is certainly divisible by $133$!

Assumption

Assume true for $n=k$:
$$f(k)=11^{k+2}+12^{2k+1}\text{is divisible by $133$ for all $k\in\mathbb N$}$$

Inductive step

Now to prove that the assumption implies the truth of the statement for $n=k+1$:

$$f(k+1)=11^{k+3}+12^{2(k+1)+1}=11^{k+3}+12^{2k+3}=11(11^{k+2})+144(12^{2k+1})$$
$$=11(11^{k+2}+12^{2k+1})+133(12^{2k+1})=11f(k)+133(12^{2k+1})$$
Now, $133(12^{2k+1})$ is certainly divisible by $133$, but by the assumption, $f(k)$ is also divisible by $133$. Hence we have proved that $f(k+1)$ is divisible by $133$ if $f(k)$ is divisible by $133$. But by the base case, we know it to be true for $n=1$. Hence the result follows for all $n\in\mathbb N$. I'll just formally write the conclusion below.

Conclusion

If $f(n)$ is divisible by $133$ for $n=k$ and this implies $f(k+1)$ is divisible by $133$, as $f(1)$ is divisible by $133$ then $f(n)$ is divisible by $133$ for all $n\in\mathbb N$ by the principle of mathematical induction.


问:
Second order differential equation question.
Consider the equation

$\dfrac{d^2y}{d^2t}+k\cdot\dfrac{y}{m}=0$

a) let $y(t)= \cos (at)$. Under what conditions on $a$ is $y(t)$ a solution?

b) What initial condition in the yv-plane corresponds to this solution?

c) In term of m and n , what is the period of this solution?

d) sketch the solution curve associate to this point.

I have no idea how to start. I really want to understand this so I wonder if you can explain this problem step by step for me please.

答:
You can answer most of this question simply by plugging in $y(t)=\cos(at)$ in and comparing terms:

a)

$$\frac{d^2}{dt^2}[\cos(at)]+\frac{k}{m}\cos(at)=-a^2\cos(at)+\frac{k}{m}\cos(at)=\left(\frac{k}{m}-a^2\right)\cos(at)=0.$$

Since we need the above to hold for all $t$, we require $a^2=\frac{k}{m}$. Thus, either $a=\sqrt{\frac{k}{m}}$ or $a=-\sqrt{\frac{k}{m}}$ (does your problem give some interpretation of what $a$ represents that allows you to rule one of these two options out?).

b) I am assuming that by $v$ you mean $\frac{dy}{dt}=-a\sin(at)$. The initial conditions, $y_0$ and $v_0$, are the values of $y$ and $v$ at the initial time (which I assume is $0$). Hence, $y_0=\cos(0)=1$ and $v_0=-a\sin(0)=0$.

c) I am assuming that you mean $k$ instead of $n$. The period is the inverse of the frequency. From before the frequency is $|a|=\sqrt{\frac{k}{m}}$, hence the period is $\frac{1}{|a|}=\sqrt{\frac{m}{k}}$.

d) Just sketch, $y(t)$ (a cosine function of unit amplitude, frequency $\sqrt{\frac{k}{m}}$ and phase sign$(a)\pi$) and $v(t)$ (a sine function of amplitude $\sqrt{\frac{k}{m}}$, frequency $\sqrt{\frac{k}{m}}$ and phase $\pi$).


问:
Most basic equation: Heart rate is in units of bpm. What are the units of "b"?
This is the most simple question in the world, but it is surprisingly difficult to explain to a student.  The heart rate is measured in bpm, beats per minute.  Minutes are a true unit. Beats are not a true unit. However, we are using beats as if it where a unit.  I'm not sure how to explain this to a student, and I'm confusing myself.   Please help.</p>

答:
@Stork led me to the answer, but he didn't quite get there.  A "beat" is a "counting unit". This is the term I've been looking for!   A "dozen" is also a counting unit.  If you google "counting unit", you will find it is a fairly common term.   So, if a "beat" is a counting unit, then what kind of unit is a "second". A "second" is  a "measuring unit".   This implies that counting and measuring should be treated as distinct activities, which makes sense.  I had never really stopped to realize that counting and measuring are distinct.

I do apologize for dwelling on this, but I think this terminology will clarify matters for students.

ps.  This ancient paper appears to discuss the distinctions.


问:
Why squaring two absolute value functions never gets answers that don't satisfy the domain?

$|x-1| = |3x-1|+1$

Where $f(x)=|x−1|$ then $f(x)=x−1$ in a certain domain, $x≥1$.
And $f(x)=−(x−1)$ in a certain domain, $x≤1$.

And $g(x)=|3x-1|+1$ then $g(x)=3x-1+1$ in a certain domain, $x≥1/3$ and $g(x)= -(3x-1)+1$ in a certain domain, $x≤1/3$.

I know this can be solved by squaring both sides and I'll get no solutions. And drawing them on a graph paper supports that.

$(x-1)^2 = (|3x-1|+1)^2$  →$8x^2-4x+2|3x-1|+1=0$ , $8x^2-4x+1$ can never be negative and so is $2|3x-1|$. So, there's no solution.

But doesn't we mean by squaring both sides that $x-1$ can be either negative or positive.

So it's like we are saying:

$-(x-1)$ can be equal to $-(3x-1)+1$, which has an answer $x = 0.5$.

However, this answer doesn't satisfy the domain of $g(x)$ when its equal to $-(3x-1)+1$, $x≤1/3$ so it's not correct.

Or

$-(x-1)$ can be equal to $3x-1+1$, which also has an answer $x=0.25$.

However this answer doesn't satisfy the domain of $g(x)$ when its equal to $(3x-1)+1$, $x≥1/3$ so isn't correct.

Etc.

But these answers since they are out of the domains of both functions aren't right but how can squaring guess that, why are the answers we get out of squaring always in the domain (if both functions are absolute value, I'm not talking about a case like this when only one function has an absolute value sign: $|x-1|=3x-1+1$) how can squaring know where the domain ends and starts, it seems like magic which surely can't be the case.

I tried to ask my teacher but he couldn't get my point, I hope someone will.

答:
The answer is that squaring does not "know" when $x$ is in the correct domain or not. The method you have used is not a generally correct method; you just got lucky in this case.

Let's try just changing a couple of the numbers to make a problem that looks similar:

$$ |x−1|= \left|x − \frac13\right| + 1. $$

If you draw graphs of $y = |x−1|$ and $y = \left|x − \frac13\right| + 1$
you'll see that the graphs never intersect, so there is no solution.

But if you just square the absolute values on both sides, you get

$$ (x−1)^2 = \left(x − \frac13\right)^2 + 1, $$

which expands to

$$ x^2 - 2x + 1 = x^2 - \frac23 x + \frac19 + 1, $$

which simplifies to

$$ \frac43 x + \frac19 = 0, $$

which has the solution $x = -\frac{1}{12}.$

Here is another example:

$$ |x−1|= \left|\frac43 x − \frac13\right| + 1. $$

This also clearly has no solution (graph it), but if you take
$$ (x−1)^2 = \left(\frac43 x − \frac13\right)^2 + 1, $$

you get a quadratic equation that has two solutions.

What went wrong here is that you did not actually square both sides of the equation. You squared the left side, and you squared at least one part of the right side, but you did not square the entire right side as one thing to be squared.

If you actually square both sides of your original equation,
you get

$$ (|x−1|)^2 = (|3x−1|+1)^2, $$

which simplifies to

$$ 8 x^2 - 4 x + 2 \lvert 3 x - 1\rvert + 1 = 0. $$

It happens that $8 x^2 - 4 x + 1$ by itself is always positive,
so when you add the non-negative number $2 \lvert 3 x - 1\rvert$ you always get another positive number, never zero.

But the squaring function doesn't "know" when the thing it squared was originally the "positive" version of some other function or the "negative" version.
The reason it tells you the correct answer in this case is that it is showing you that there is no solution.

If taking the square of both sides of an equation leads to no solutions,
then you have ruled out all solutions in all four of the following cases:
where the left side is positive and the right is negative,
where the left is negative and the right is positive,
where both sides are positive, and where both sides are negative.
It may actually be that one of those cases is impossible for some other reason; no matter. The cases that actually are possible still have no solution.

In the case where the squared equation has solutions, you have to be more careful.
It is possible that the solutions came from cases that actually cannot occur for some other reason. It is possible that one of the solutions is good and the other is impossible. But you get no solutions at all from the squared equation in the problem you are looking at, so these warnings don't apply.


问:
I am trying to show if $(x-y)(x+y) \leq 0 $ and $ (x+y) \geq 0$ then $(x-y) \leq 0$ real numbers axioms
This is a part of a bigger proof,

I was thinking starting with

$(x-y)(x+y) = (x-y)x + (x-y)y $ [Using distributive law]

Edit:

We also know $x \geq 0$  and $y \geq 0$

答:
You have that

$$x^{2}-y^{2}\leq 0$$

so $$\vert x\vert \leq \vert y\vert$$

If $y$ is negative, then

$$x+y\leq 0$$

so it must be the case that $y>0$. Can you finish from here?


问:
Continuity condition for homogeneous topological spaces
I know that given topological groups $G_1,G_2$ and a homomorphism $f:G_1\to G_2$, $f$ is continuous if it is continuous at a point. If I am not mistaken, this follows using the fact that translation maps of the form $h\mapsto g\cdot h$, 'commute' with the map $f$ and are homeomorphisms. In the sense that
$\require{AMScd}$
$$ \begin{CD}
    \mathbb (x,y) @>\displaystyle{\cdot} >> x\cdot y\\
    @V \displaystyle f\times f V V\ @VV \displaystyle{f} V\\
    \mathbb (f(x),f(y) ) @>>\displaystyle \cdot> f(x\cdot y)
    \end{CD}$$
commutes.
My question is, are there natural generalizations to such results only using this property? i.e., given two topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$ with transitively acting groups $H_X<Homeo(X)$ and $H_Y<Homeo(Y)$, can we say that $f:X\to Y$, which is continuous at a point $x_0\in X$, is a continuous map in general if we have some approriate commutativity of $f$ with the actions?

I think that if one can show somehow that
$$ \begin{CD}
     x @>\displaystyle{f} >> f(x)\\
    @V \displaystyle \phi_x V V\ @AA \displaystyle{\psi_{f(x)}} A\\
    x_0 @>>\displaystyle f> f(x_0)
    \end{CD} $$
commutes in some sense, where $\phi_x\in H_X$ and $\psi_{f(x)}\in H_Y$ satisfy $\phi_x(x)=x_0$ and $\psi_{f(x_0)}\big( f(x_0) \big)=f(x)$, then it will solve the problem. Is there some reasonable conditions on group actions satisfies this?

答:
Continuity at points is preserved under composition:

Given topological spaces $X, Y, Z$ with $x \in X$ and given maps $f : X \to Y, g: Y \to Z$, if $f$ is continuous at $X$ and $g$ is continuous at $f(x)$, then $g\circ f$ is continuous at $x$.

For, letting $y = f(x)$ and $z = g(y)$, if $U$ is a neighborhood of $z$, then $g^{-1}(U)$ is a neighborhood of $y$ by the continuity of $g$ at $y$, and by the continuity of $f$ at $x, f^{-1}(g^{-1}(U)) = (g\circ f)^{-1}(U)$ is a neighborhood of $x$. Since this holds for all neighborhoods of $z=(g\circ f)(x)$, $g \circ f$ is continuous at $x$.

Now for a topological space $X$, define an homogenization $\mathscr H_X$ (all related terminology is my own invention) of $X$ to be a collections of maps $\phi : U \in \tau_X \to X$ such that for every $x, y \in X$, there is a $\phi \in \mathscr H_X$ with $x \in \text{dom}(\phi)$, such that $\phi(x) = y$ and $\phi$ is continuous at $x$. Such a $\phi$ is said to represent $x$ to $y$.

For homogenizations $\mathscr H_X, \mathscr H_Y$ of $X, Y$, call $f:X \to Y$ compatible  with $\mathscr H_X, \mathscr H_Y$ if for all $x, z \in X$, there exist representations $\phi$ of $x$ to $z$ and $\psi$ of $f(z)$ to $f(x)$ such that
$$f = \psi \circ f \circ \phi$$
on some neighborhood of $x$.

Finally, we get the theorem:

Given spaces $X, Y$ with homogenizations $\mathscr H_X, \mathscr H_Y$, if $f : X \to Y$ is compatible with $\mathscr H_X, \mathscr H_Y$, and if $f$ is continuous at some point $z \in X$, then $f$ is continuous on all of $X$.

If $x$ is any point of $X$, then there is a representation $\phi$ of $x$ to $z$, and a representation $\psi$ of $f(z)$ to $f(x)$ such that $f = \psi\circ f\circ\phi$ on some neighborhood of $x$. $\phi$ and $\psi$ are continuous at $x$ and $f(z)$ with respect to their domains, but since those domains are open sets in $X$ and $Y$ respectively, they are also continuous at these points with respect to the topologies of $X$ and $Y$. Hence by the earlier theorem and the continuity of $f$ at $z$, $\psi \circ f \circ \phi$ is continuous at $x$. Since it agrees with $f$ on some neighborhood of $x$, $f$ is continuous at $x$. As $x$ was arbitrary, $f$ is continuous at every point of $X$.

That is the most abstract form of the process that I can think of. Note that I asked far less of $\mathscr H_X$ than you. There is a reason for that. The fewer requirements I can place on $\mathscr H_X$, the bigger it can be, which allows more maps $f$ to be compatible with it, which makes the theorem more widely applicable.

If you want a more restrictive set of maps such as you have described, you just have to show that they none-the-less still satisfy the conditions to be an homogenization.


答:
This is wrong in general.

Take $x=-y=1$ for a counterexample. You have

$(x-y)(x+y) = 0$, $x+y = 0$ but $x-y = 2 \gt 0$.

Title: I am trying to show if $(x-y)(x+y) \leq 0 $ and $ (x+y) \geq 0$ then $(x-y) \leq 0$ real numbers axioms

Body:
This is a part of a bigger proof,

I was thinking starting with

$(x-y)(x+y) = (x-y)x + (x-y)y $ [Using distributive law]

Edit:

We also know $x \geq 0$  and $y \geq 0$


问:
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$
To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?

答:
Bioche's rules say you should set  $\;u=\tan \frac t2,\enspace \mathrm du=\frac12(1+u^2)\,\mathrm dt$, whence the integral of a rational function
$$\int\frac{2(1-u^2)\,\mathrm du}{(1+u^2)(u^2+2u+3)},$$
which is easily calculated using partial fractions decomposition.

Some more details: using the half angle formulæ, we get
\begin{align}
\frac{\cos t}{2 + \sin t + \cos t}\,\mathrm dt&=\frac{\cfrac{1-u^2}{1+u^2}}{2+\cfrac{2u}{1+u^2}+\cfrac{1-u^2}{1+u^2}}\,\frac{2\,\mathrm du}{1+u^2} \\
&=\frac{2(1-u^2)}{(1+u^2)\bigl(2(1+u^2)+2u+1-u^2\bigr)} \\
&=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}. \\
\end{align}


问:
Prove the following statements are equivalent
Prove the following statements are equivalent:

for every $f\geq 0$,$\int_{X}fd\mu<\infty\iff\int_{X}e^{f}d\mu<\infty$

$\mu$ is finite measure and X is not the union of infinitely many sets of positive measure.

I've tried to assume that (2) doesn't holds and I managed to prove that $\int_{X}fd\mu$ doesn't converge but I didn't  managed to prove for the other integral.

any for the second part I had nothing

答:
Let us prove the result.

Let $(X,\Sigma, \mu)$ be a measure space.

$(1 \Rightarrow 2)$. Suppose for every $f\geq 0$,$\int_{X}fd\mu<\infty\iff\int_{X}e^{f}d\mu<\infty$.

Take $f=0$. Clearly, $\int_{X}fd\mu=0<\infty$. So
$$ \mu(X)= \int_X 1 d\mu= \int_X e^0 d\mu= \int_X e^f d\mu < \infty$$

Now, let us prove that $X$ is not the union of infinitely many sets of positive measure.

Suppose that $X$ is the union of infinitely many sets of positive measure. Since $\mu(X)<+\infty$, it follow that for all $\delta >0$ there is $E \in \Sigma$ such that $\mu(E)<\delta$. It follows, from lemma $1$ and lemma $2$ in the Remark 1, that $L^2(X) \subsetneq L^1(X)$. Take a function $f \in L^1(X)$ such that  $f \notin L^2(X)$. But then we have that $|f|\geq 0$, $\int_{X}|f|d\mu<\infty$ and
$$ \int_X e^{|f|} d\mu > \int_X \left (1+|f| + \frac{1}{2}|f|^2 \right) d\mu \geqslant   \frac{1}{2}  \int_X |f|^2 d\mu =+\infty $$
Contradiction to $(1)$.
So $X$ is not the union of infinitely many sets of positive measure.

$(2 \Rightarrow 1)$ In fact, $(2)$ entails a stronger result than $(1)$. Let us assume $(2)$. Given any $f \geq 0$ and given any $n \in \mathbb{N}$ let
$$ A_n=\{x \in X: n \leq f(x) < n+1 \}$$
Clearly $\{A_n\}_n$ is an infinite family of disjoint sets and $X=\bigcup_nA_n$. Since $X$ is not the union of infinitely many sets with positive measure, it is easy to prove that there is $n_0 \in \mathbb{N}$ such that if $n \geq n_0$, $\mu(A_n)=0$. So $f<n_0$ $a.e.$. So $e^f<e^{n_0}$ $a.e.$. So we have $\int_X f d\mu < n_0\mu(X)<\infty$ and $\int_X e^f d\mu < e^{n_0}\mu(X)<\infty$. Then $(1)$ follows immediately.

Remark 1:

Lemma 1. Let $(X,\Sigma, \mu)$ be a measure space. If $\mu(X)<+\infty$ is finite then $L^2(X) \subseteq L^1(X)$.

Proof Since $\mu(X)<+\infty$, we have  $\chi_X \in L^2(X)$. Then for any $f\in  L^2(X)$, applying Hölder inequality, we have
$$ \|f\|_1=  \|f\,  \chi_X  \|_1\leqslant  \|\chi_X\|_2 \|f\|_2 =(\mu(X))^{1/2} \|f\|_2 $$

Lemma 2. Let $(X,\Sigma, \mu)$ be a measure space. If $L^1(X) \subseteq L^2(X)$, then there is $\delta >0$ such that, for any $E\in \Sigma$, if $\mu(E)>0$ then $ \mu(E) > \delta $.

Proof: Let $T: L^1(X) \rightarrow  L^2(X)$ be the inclusion map, that is $T(f)=f$. $T$ is obvious linear. Using the Closed Graph Theorem, we can prove $T$ is continuous (bounded).

In fact, let $\{f_n\}$ be a sequence in $L^1(X)$ which converges to $f$ in the $L^1$ norm, and to $g$ for the $L^2$ norm. We extract a subsequence $\{f_{n_j}\}$ which converges to $f$ $a.e.$; this subsequence still converges to $g$ for the $L^2$ norm; now extract from this subsequence a new subsequence which converges to $g$ almost $a.e.$.
So we have a subsequence which converges $a.e.$ to $f$ and $g$,  hence $f=g$, and, since both $L^1(X)$ and $L^2(X)$ are Banach spaces, by the Closed Graph Theorem we conclude that $T$ is continuous (bounded).

Since $T \neq 0$, we have that $0<\|T\| <+\infty$.

Now, let us prove that, given any $E \in \Sigma$ such that $\mu(E)>0$,
$ \mu(E) \geqslant \frac{1}{\|T\|}$.

If $\mu(E)=+\infty$, then it trivial that $\mu(E) \geqslant \frac{1}{\|T\|}$.

Suppose $0<\mu(E)<+\infty $,
let
$$f_E= \frac{1}{\mu(E)} \chi_E$$
It follows immediately that $\|f_E\|_1=1$ and $\|f_E\|_2=\frac{1}{\mu(E)}$.
So
$$ \frac{1}{\mu(E)}=\|f_E\|_2 =\|T(f_E)\|_2 \leqslant \|T\| \|f_E\|_1 =\|T\| $$
So
$$   \mu(E) \geqslant \frac{1}{\|T\|} $$
Taking $\delta = \frac{1}{2\|T\|}$, we have  $ \mu(E) > \delta $.

Remark 2: In fact, we proved a slightly more general result. We proved that
he following statements are equivalent:

for every $f\geq 0$,$\int_{X}fd\mu<\infty\iff\int_{X}e^{f}d\mu<\infty$

$\mu$ is finite measure and X is not the union of infinitely many sets of positive measure.

for every $f\geq 0$,$\int_{X}fd\mu<\infty$ and $\int_{X}e^{f}d\mu<\infty$


问:
Find all lines on cubic surface
Problem: Find the singular point of the cubic surface

$$ZX^2 - Y^3 + TZ^2 = 0.$$

And also determine all the lines on the surface.

My attempt: I calculated the singular point by setting the partial derivatives to $0$ and I got $P = (0 : a : 0 : b)$ where $a, b$ are not both zero. Did I get this right?

About calculating the lines, I have no idea where to start. Is there perhaps some sort of technique/method/proposition for doing this? Can someone perhaps show me as an answer how to do this for this problem since I have two more of this type to do and then I can practice on those other two.

As a side question: Is there perhaps some criterion of determining the amount of total lines on the surface?

Note: We are in $\mathbb{P}^{3}$ over an algebraically closed field which does not have characteristic $2$.

答:
a) It is quite easy to see that the only singularity of the projective surface $S\subset \mathbb P^3$ given by the equation $ZX^2 - Y^3 + TZ^2 = 0$ is the point $$P=[x:y:z:t]=[0:0:0:1]\in S$$
That type of singularity is called $E_6$ by the specialists.

b) It is also quite clear that the line $L$ given by the two equations $Y=Z=0$ lies on $S$ (i.e. $L\subset S$).

c) What is true but not clear at all however is that $L$ is the only line of $\mathbb P^3$ lying on $S$.

I know no elementary proof of that fact, which follows from the rather complicated classification of  singular cubic surfaces in projective $3$-space.

Reference

Bruce-Wall's article On the Classification of Cubic Surfaces, Lemma 4, page 251  and second table, page 255.

(Beware that their paper is written in a rather old-fashioned  style.)

Title: Find all lines on cubic surface


答:
Mathematica gives:

$$\frac{1}{8} \left(4 \sin ^{-1}(x)+4 \sqrt{2} \tan ^{-1}\left(\sqrt{2}
   (x+1)\right)+\frac{2 i \left(i+2 \sqrt{2}\right) \tan ^{-1}\left(\frac{\left(-16+6 i
   \sqrt{2}\right) x^4-2 i \left(9 \sqrt{1-2 i \sqrt{2}} \sqrt{1-x^2}+10 \left(-3
   i+\sqrt{2}\right)\right) x^3-i \left(36 \sqrt{1-2 i \sqrt{2}} \sqrt{1-x^2}+79
   \sqrt{2}-20 i\right) x^2+3 \left(-9 i \sqrt{1-2 i \sqrt{2}} \sqrt{1-x^2}-20 i
   \sqrt{2}+20\right) x-9 i \sqrt{2}+36}{\left(66 i+32 \sqrt{2}\right) x^4+16 \left(14
   i+3 \sqrt{2}\right) x^3+\left(253 i-14 \sqrt{2}\right) x^2-48 \left(-2
   i+\sqrt{2}\right) x-18 \sqrt{2}+9 i}\right)}{\sqrt{1-2 i \sqrt{2}}}+2 \sqrt{1+2 i
   \sqrt{2}} \tan ^{-1}\left(\frac{2 \left(8+3 i \sqrt{2}\right) x^4+2 \left(-9 i
   \sqrt{1+2 i \sqrt{2}} \sqrt{1-x^2}-10 i \sqrt{2}+30\right) x^3+\left(-36 i \sqrt{1+2 i
   \sqrt{2}} \sqrt{1-x^2}-79 i \sqrt{2}+20\right) x^2-3 i \left(9 \sqrt{1+2 i \sqrt{2}}
   \sqrt{1-x^2}+20 \left(-i+\sqrt{2}\right)\right) x-9 i \sqrt{2}-36}{\left(-66 i+32
   \sqrt{2}\right) x^4+16 \left(-14 i+3 \sqrt{2}\right) x^3-\left(253 i+14
   \sqrt{2}\right) x^2-48 \left(2 i+\sqrt{2}\right) x-9 \left(i+2
   \sqrt{2}\right)}\right)+2 \log \left(2 x^2+4 x+3\right)-\frac{\left(i+2
   \sqrt{2}\right) \log \left(4 \left(2 x^2+4 x+3\right)^2\right)}{\sqrt{1-2 i
   \sqrt{2}}}-\frac{\left(-i+2 \sqrt{2}\right) \log \left(4 \left(2 x^2+4
   x+3\right)^2\right)}{\sqrt{1+2 i \sqrt{2}}}+\frac{\left(i+2 \sqrt{2}\right) \log
   \left(\left(2 x^2+4 x+3\right) \left(-2 \left(-i+\sqrt{2}\right) x^2+2 i
   \left(\sqrt{2-4 i \sqrt{2}} \sqrt{1-x^2}+2\right) x+2 i \sqrt{2-4 i \sqrt{2}}
   \sqrt{1-x^2}+2 \sqrt{2}+3 i\right)\right)}{\sqrt{1-2 i \sqrt{2}}}+\frac{\left(-i+2
   \sqrt{2}\right) \log \left(-\left(2 x^2+4 x+3\right) \left(2 \left(i+\sqrt{2}\right)
   x^2+2 i \left(\sqrt{2+4 i \sqrt{2}} \sqrt{1-x^2}+2\right) x+2 i \sqrt{2+4 i \sqrt{2}}
   \sqrt{1-x^2}-2 \sqrt{2}+3 i\right)\right)}{\sqrt{1+2 i \sqrt{2}}}\right)$$

which strongly suggests no clean simplification.

Title: Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$

Body:
To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?


问:
Fourier Transform - Dirac Delta
How can I compute this Fourier transform?

$$\int \frac{d^3q}{(2\pi)^3}(\vec q \cdot \vec a)(\vec q \cdot \vec b)\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }\,$$

My idea was to write it as

$$-( \vec a \cdot \vec \nabla)(\vec b \cdot \vec \nabla)\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }$$

and use the fact that $\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }=\frac{1}{4\pi r}$, together with $\Delta \frac{1}{r} = -4\,\pi \,\delta(\vec r)$ but is is not exactly the same.

答:
I would go with spherical polar coordinates. Perhaps with $\vec{r}=(r,0,0)$, then the dot products can be evaluated using trigonometry functions of $\phi$ and $\theta$. It would eventually look like a Dirac delta modulated by the dot products among $\vec{a}$, $\vec{b}$ and $\vec{r}$.

Edit:
It is indeed more straightforward with your original idea.

$-(a\cdot\nabla)(b\cdot\nabla) \Delta^{-1} \delta(\vec{r})=-(a\cdot\nabla)(b\cdot\nabla) \frac{1}{4\pi r}=b\cdot\nabla(a\cdot \frac{\vec{r}}{r^3})=\frac{1}{4\pi r^3}[\vec{a}\cdot\vec{b}-3(\vec{a}\cdot\frac{\vec{r}}{r}) (\vec{b}\cdot\frac{\vec{r}}{r})].$

So it's the mutual energy between two dipoles a and b.

I would answer the above comment here (I cannot make comments as I'm new to the site): The Dirac delta is there at $r=0$, it's just that it is also nonzero at other places.

Title: Fourier Transform - Dirac Delta


问:
Number of possible bit sequences of length m with at least n consecutive 1's in them
I have seen similar questions to this but they each seem to be special cases of this general question. Answering this would be beneficial to my research, but I am not a combinatorics expert, and this seemingly simple question eludes me. Is there a simple formula to calculate this? Everything I have seen online has been centered around things like "either 2 consecutive 1's or 0's" or "contains no ..".

If it helps, I know that for $m = 8$ bits and say the sequence is denoted $S(m,n)$
$$
S(m = 8, n = 1) = 255 \\
S(8,2) = 201  \\
S(8,3) = 107  \\
S(8,4) = 48   \\
S(8,5) = 20   \\
S(8,6) = 8   \\
S(8,7) = 3   \\
S(8,8) = 1
$$

Interestingly I'm finding that $S(8,4)=S(9,5)=S(10,6)=S(11,7)=48$
I haven't tested $S(12,8)$ because I don't want my computer to melt but I'm seeing a pattern... However this does not seem to work for $m<8$.

答:
I will not give a formula, but just a recurrence relation. Let T(m, n) be the number of strings of length m with a run of n consecutive 1's.

Consider all the strings of length m-1. Exactly T(m-1, n) of them already contain a string of 'n' consecutive digits. Since we can add a 0 or a 1 we will get double this amount of length m strings strings.

However adding a 1 in the m'th place will give a new good string if the last (n-1) digits are a 1 and the n'th to last digit is a 0 and in addition the digits in place 1, ..., m - n - 1 do not contain a run of n consecutive 1's. i.e. the string looks like this:
$$
\underbrace{xx..xx}_{m - n - 1}0\underbrace{11..11}_{n - 1}
$$
There are 2^{m - n - 1} possibilities for the x-digits, but we should exclude T(m - n - 1, n) of them to avoid the double counting.

Adding it all up we find
$$
T(m, n) = 2\cdot T(m - 1, n) - T(m - n - 1, n) + 2^{(m - n - 1)}
$$

If $m - n - 1 \leq n$, i.e. $m \leq 2n + 1$, the $T(m - n - 1, n)$ term vanishes and you should be able to solve the recurrence relation.


答:
If $x+y>0$ then the result is true, because it is an axiom of the real numbers that inequality is preserved by dividing by a positive quantity.

However, if $x+y=0$, the result is not true in general. Since $x+y=0$, we can choose $x-y$ to be negative and the first two inequalities will remain true. For example, as Professor Vector suggests taking $x=1$, $y=-1$ gives:

\begin{align}
(x-y)(x+y) &= 0 \\
(x+y) &= 0 \\
(x-y) &= 2
\end{align}

In other words,

\begin{align}
(x-y)(x+y) &\le 0 \\
(x+y) &\le 0 \\
(x-y) & > 0
\end{align}

which contradicts the claim.

EDIT: Since we know $x\ge 0$ and $y \ge 0$, we know that if $x+y=0$, then $x=y=0$. That proves the claim.

EDIT 2: Your axioms are phrased in terms of the positive reals $P$. So let's write your statement in terms of $P$. Hopefully it is clear that you want to prove:

If $(x-y)(x+y) \notin P$ and $-(x+y) \notin P$ then $(x-y) \notin P$.

We can prove this by contradiction like so:

Assume $(x-y) \in P$. Then either $(x-y)(x+y) \in P$ or $-(x+y) \in P$.

By P10, either $-(x+y) \in P$ or $(x+y) \in P$ or $x+y = 0$. If $(x+y) \in P$ then by P12 $(x+y)(x-y) \in P$. So we just need to prove $x+y \ne 0$.
Well, either $y=0$ (so $x+y=x-y \in P$), or $y \in P$, so $x-y + y + y \in P$ by P11.


问:
Do the coefficients $a_{i,j}$ of the Gauss-Legendre method form an invertible matrix?
The coefficients of the $s$-step Gauss-Legendre method, which is a special case of a generalized Runge-Kutta method, are given by
\begin{equation*}
a_{i,j}=\int_{0}^{\frac{1}{2}(1+\zeta_{i})}\prod_{k=1,k\neq j}^{s}\frac{t-\frac{1}{2}(1+\zeta_{k})}{\frac{1}{2}(\zeta_{j}-\zeta_{k})}dt\text{,}
\end{equation*}
where $\zeta_{k}$ ($k\in\{1,...,s\}$) are the roots of the Legendre polynomial of degree $s$ on $(-1,1)$.

How do I show that the Matrix $A=(a_{i,j})_{i,j=1}^{s}$ is invertible?

答:
As far as I understand, one can formulate the method as collocation method using a polynomial $p$ of degree $s$ with
$$
p(0)=0, \\
y_n+hp(\tau)\approx y(t_n+h\tau), \\
p'(\tau)\approx y'(t_n+h\tau), ~~\tau\in[0,1].
$$
The step equations are then
$$
k_i=p'(c_i)=f(t_n+c_ih,y_n+hp(c_i)),~~i=1,..,s\\
y_{n+1}=y_n+hp(1).
$$
The relation between the $k_i$ and the coefficients of $p$ is thus linear with a Vandermonde like matrix, which is invertible if the points $c_i$ are all different. This is the case for the Gauss method, as the Legendre polynomial has simple roots. The coefficients $a_{ij}$ are obtained by inverting this relation, giving a likewise invertible matrix.


问:
Is there anything interesting about taking the logarithm of the determinant?
We can view the determinant as a group homomorphism
$$\det : GL_n(F) \to F^\times$$
since $\det(AB)=\det(A)\det(B)$. If the field is the real numbers, we can take the logarithm of the determinant, which by definition maps $\mathbb{R}^\times$ to $\mathbb{R}$ under addition. So then the composition $\log\circ\det$ can be seen as taking the group of $n \times n$ matrices with positive determinant to the group of real numbers under addition.

Is there anything interesting about this map, or is there nothing else to say other than it is just the "logarithm of the dterminant"?

答: Trace of the Lie algebra? Since the determinant is the volume ratio under the linear map, its logarithm is the growth rate indicated by the trace of the corresponding Lie algebra element.</p>


问:
Composition of Linear Approximation
Suppose you have $x,y,z\in\mathbb{R}$ where

$$y:=f(x)$$
$$z:=g(y)$$

for some smooth functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$. Note that there is $h$ (the composition of $g$ and $f$) such that

$$z=h(x)$$

Suppose that I wanted to get a linear approximation to $h$ at $0$.

I could find

$$L_{h}(x ) = h(0)+x h'(0)$$

but I could also find

$$L_{f}(x ) = f(0)+x f'(0)$$

put that into $g$ to get a new function $h_{2}$ where

$$h_{2}(x)=g(L_{f}(x))$$

and then find

$$L_{h_{2}}(x ) = h_{2}(0)+x h_{2}'(0)$$

This way you never approximate the composition function $h$ directly. I don't know any particular reason to do this aside from the fact that it might be easier computationally. Has this ever been looked at? I'm wondering what such an approach would do to the accuracy of the resulting approximation.

Edit: The big picture idea is that I am wondering whether you can apply approximations to compositions of functions in steps. I.e. linearly approximate the inner most function, then go a step out, and so on.

Edit2: Approximating at zero may have been a poor choice. I will add an explicit example of what I mean

$$f(x):=x^{2}$$
$$g(x):=\ln (x)$$

which gets me

$$L_{h}(x)=(x-2)+\ln(4)$$
$$L_{h_{2}}(x)=\frac{1}{3}(x-2)+\ln(12)$$

when approximating at $x=2$.

答:
These two approaches give the same result. You have

$$z=h(x)=g(f(x))$$

If you apply linear approximation directly (standard approach) you get

$$L_h(x)=g(f(0))+g'(f(0))f'(0)x.$$

On the other hand

$$L_f(x)=f(0)+f'(0)x.$$

Assume you define

$$h_2(x)=g(L_f(x))=g(f(0)+f'(0)x)$$

Now

$$L_{h_2}(x)=g(f(0))+g'(f(0))f'(0)x.$$

You can see that $L_{h_2}(x)=L_h(x)$. However, there is a difference if you approximate around some other point than $0$. Typically you would expect performing approximations at interim stages to lower accuracy.

Title: Composition of Linear Approximation


问:
Computing $\iint_{\Omega} (x+y^2) \ dx \ dy$ where $\Omega$ is a parallelogram

Compute $$\iint_{\Omega} (x+y^2) \ dx \ dy$$ where $\Omega$ is a parallelogram bounded by $x+y=0$, $x+y=1$, $2x-y=0$ and $2x-y=3.$

Graphing this in Desmos produces the following plot

I was instructed to use change of variables here, but I'm not familiar with that at all. How can I compute this kind of integral? Any tips would be much appreciated.

答: If you do $x+y=X$ and $2x-y=Y$, then $x=\frac{X+Y}3$ and $y=\frac{2X-Y}3$. Besides, if you define$$g(X,Y)=\left(\frac{X+Y}3,\frac{2X-Y}3\right),$$then $\Omega=g\bigl([0,1]\times[0,3]\bigr)$ and the absolute value of the Jacobian of $g$ is $\frac13$. So,\begin{align}\iint_\Omega x^2+y\,\mathrm dx\,\mathrm dy&=\iint_{g([0,1]\times[0,3])}x^2+y\,\mathrm dx\,\mathrm dy\\&=\iint_{[0,1]\times[0,3]}\frac13\left(\left(\frac{X+Y}3\right)^2+\frac{2X-Y}3\right)\,\mathrm dX\,\mathrm dY\\&=\frac13\int_0^1\int_0^3\left(\frac{X+Y}3\right)^2+\frac{2X-Y}3\,\mathrm dY\,\mathrm dX.\end{align}Can you take it from here?


问:
Determine if the integral converges $\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}$
I want to determine if the following integral converges

$$\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}dx$$

My intuition tells me it diverges. I thought the way to solve that woudln't be by definition but by finding another function that is smaller than that and use integral comparison test. I also know that $$\lim_{x\rightarrow1}\log=0$$ but I don't see how that helps me

答:
It suffices to check that for $x\to 1$, we have
$$\frac{\sin x}{\ln x}\sim \frac{c}{x-1},$$
where $c=\sin(1)$.

Hence, your integral converges iff.
$$\int_1^2\frac{c}{x-1}\mathrm d x=c\int_0^1\frac{1}{x}\mathrm d x$$
converges.

Does it?


答:
Thanks to @Ross Millikan formula, which I searched with Approach Zero, I could find this answer, and using again Approach Zero with that result, this other beautiful answer. Both give the complementary result, so in your case we have:

$$S(m,n) = 2^m-\sum_{q=0}^{\lfloor m/n\rfloor}
{m-nq\choose q} (-1)^q 2^{m-(n+1)q}
+ \sum_{q=0}^{\lfloor m/n\rfloor - 1}
{m-n(q+1)\choose q} (-1)^q 2^{m-n-(n+1)q}$$

See the links for details.

Title: Number of possible bit sequences of length m with at least n consecutive 1's in them

Body:
I have seen similar questions to this but they each seem to be special cases of this general question. Answering this would be beneficial to my research, but I am not a combinatorics expert, and this seemingly simple question eludes me. Is there a simple formula to calculate this? Everything I have seen online has been centered around things like "either 2 consecutive 1's or 0's" or "contains no ..".

If it helps, I know that for $m = 8$ bits and say the sequence is denoted $S(m,n)$
$$
S(m = 8, n = 1) = 255 \\
S(8,2) = 201  \\
S(8,3) = 107  \\
S(8,4) = 48   \\
S(8,5) = 20   \\
S(8,6) = 8   \\
S(8,7) = 3   \\
S(8,8) = 1
$$

Interestingly I'm finding that $S(8,4)=S(9,5)=S(10,6)=S(11,7)=48$
I haven't tested $S(12,8)$ because I don't want my computer to melt but I'm seeing a pattern... However this does not seem to work for $m<8$.


问:
I need to solve $12x-30y+24z=18$ equation as a diophantine equation where $x,y,z$ is whole number
I know i can divide with $6$ so I get $2x-5y+4z=3$, I replaced $2x-5y$ with $u$ so I got $u+4z=3$. I see that $u=3$ and $z=0$ is a solution but I got stuck. How can I continue? Is there a way to solve this using extended euclidean algorithm?

Thank you for your help.

答:
$(-1,-1,0)$ is a solution.

and then we can add multiples $(2,0,-1)$ to any known solution to find other solutions.  Similarly we can add multiples of $(5,2,0)$

$(x,y,z) = (-1+2m + 5n,-1 + 2n, -m)$ should span the space of integer solutions.


答:
In single variable calculus, you used a change of variables called $u$-substitution to make some integrands nicer. In multivariable calculus, you can use a change of variables for that purpose, or for making the region over which you are integrating nicer. Here, the region is not a Type 1 or Type 2 or polar region. But we can change variables to $u$ and $v$ in a clever way to make the region much nicer:$$u=x+y,\quad v=2x-y$$
With this change of variables, the region we integrate over is simply a rectangle: $0\leq u\leq 1, \ 0\leq v\leq 3$.

We have to solve these equations for $x$ and $y$ in terms of $u$ and $v$ so that we can substitute in the integrand as well as find the Jacobian of this change of variables. Here, if we add the equations, notice that $y$ is eliminated, so $x=\frac{1}{3}(u+v)$. Alternatively, if we take twice the first equation and subtract the second equation, $x$ is eliminated, so $y=\frac{1}{3}(2u-v)$.

The Jacobian $\frac{\partial (x,y)}{\partial (u,v)}$ of a change of variables is the determinant of the 2x2 matrix $\begin{pmatrix} \frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}}\\\frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}\end{pmatrix}$.

Finally, the change of variables theorem says that $$\iint_\Omega f(x,y)dA = \int_0^3\int_0^1 f\left(\frac{1}{3}u+\frac{1}{3}v,\frac{2}{3}u-\frac{1}{3}v\right)\cdot \frac{\partial (x,y)}{\partial (u,v)}dudv.$$

Can you finish the calculation?

Title: Computing $\iint_{\Omega} (x+y^2) \ dx \ dy$ where $\Omega$ is a parallelogram

Body:
Compute $$\iint_{\Omega} (x+y^2) \ dx \ dy$$ where $\Omega$ is a parallelogram bounded by $x+y=0$, $x+y=1$, $2x-y=0$ and $2x-y=3.$

Graphing this in Desmos produces the following plot

I was instructed to use change of variables here, but I'm not familiar with that at all. How can I compute this kind of integral? Any tips would be much appreciated.


答:
$f:  x\mapsto \frac{\sin(x)}{\ln(x)} $ is continuous at $ (1,\frac{\pi}{2} ]$ and positive.

in the other hand,

$$\ln(x)=\ln(1+(x-1))$$
thus
$$f(x)\sim \frac{\sin(1)}{x-1} \;\;\;(x\to 1^+)$$

and

$$\int_1^{\frac{\pi}{2}}\frac{dx}{x-1}$$
is divergent. So, your integral is Divergent.

Title: Determine if the integral converges $\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}$

Body:
I want to determine if the following integral converges

$$\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}dx$$

My intuition tells me it diverges. I thought the way to solve that woudln't be by definition but by finding another function that is smaller than that and use integral comparison test. I also know that $$\lim_{x\rightarrow1}\log=0$$ but I don't see how that helps me


问:
Find the least positive integer $x$ such that $x^2 \equiv 30 \pmod{101}$

Find the least positive integer $x$ such that $x^2 \equiv 30 \pmod{101}$

This is the first time I have encountered this kind of problem, I have searched on the Internet but got no results. Is there a formula or an algorithm to solve this kind of problem? I would be very appreciative of your support!

答:
The "brute force" approach would be to try each $x$ from $1$ up to $101$. We can be sure that we won't have to try larger $x$, because if $x \equiv y \pmod{101}$, then $x^2 \equiv y^2 \pmod{101}$.

Here's a less tedious approach that relies on the facts that $101$ is prime and $101 \equiv 5 \pmod{8}$. A proof of the following claim can be found in section 1.5 of this textbook.

Claim: Suppose $p$ is a prime with $p \equiv 5 \pmod{8}$, and suppose $x^2 \equiv a \pmod{p}$. Then $x \equiv \pm a^{(p + 3)/8} \pmod{p}$ or $x \equiv \pm 2a \cdot (4a)^{(p - 5)/8} \pmod{p}$.

In our case, $p = 101$ and $a = 30$. You can quickly compute the values $a^{(p + 3)/8}$ and $2a \cdot (4a)^{(p - 5)/8}$ modulo $101$ using repeated squaring.

If we were working modulo some other prime $p$, the Tonelli-Shanks algorithm would be a good approach. What if we were working modulo a large composite number $n$? It is conjectured that there does not exist an efficient algorithm for solving a given equation of the form $x^2 \equiv a \pmod{n}$. An efficient algorithm for that problem could be used to factor large integers and thereby break important cryptosystems such as RSA.


问:
On approximating the distribution of the distance between two random points on an $n$-sphere
Let $\tau\in(0,2)$ a given threshold value. What is the probability that the Euclidean distance $D$ between two points $\mathbf{x}$ and $\mathbf{y}$ selected uniformly at random on the $n$-sphere $\mathcal{S}_n=\{\mathbf{v}\in\mathbb{R}^n:\|\mathbf{v}\|_2=1\}$, is greater than or equal to $\tau$?

It is well known that the probability density function of the Euclidean distance $d$ between two points on the $n$-sphere
$\mathcal{S}_n=\{\mathbf{v}\in\mathbb{R}^n:\|\mathbf{v}\|_2=1\}$ is

$$f(d)=\frac{\Gamma(n/2) d^{n-2}}{\sqrt{\pi}\,\Gamma((n-1)/2)}\left(1-\left(\frac{d}{2}\right)^2\right)^{\frac{n-3}{2}}~.$$

It is also well known that, as $n$ increases, $f(d)$ approaches the normal distribution $\mathcal{N}\left(\sqrt{2}, \frac{1}{2n}\right)$.

However, even with this knowledge, it is not clear to me how to obtain an analytical expression of $\Pr(d\ge\tau)$ for a finite number of dimensions $n\gg 1$, because it seems hard to evaluate the corresponding integral. Hence, I want to find an approximation of $\Pr(d\ge\tau)$. More precisely I want to obtain a meaningful upper bound of $\Pr(d\ge\tau)$.

答: Given the normal distribution approximation, wouldn't it just be $\rm{erfc}(\frac{\tau-\sqrt{2}}{\sqrt{2/2n}})/2$ if $\tau>\sqrt{2}$, and $1-\rm{erfc}(\frac{\tau-\sqrt{2}}{\sqrt{2/2n}})/2$ otherwise?


问:
Mind too small for quadratic equation problem
I was struggling for a long time with the following question, eventually unable to find it. I'm really frustrated with it because the answer is not difficult per se but to me requires a complete shift in approaching the problem, and I am just unable to make such a leap by myself. So, is it a hard question or am I just stupid?

Here goes: Two pipes together can fill a reservoir in 6 hr 40 min. Find the time each alone will take to fill the reservoir if one of the pipes can fill it in 3 hr less time than the other.

Edit: The solution is in my textbook so not quite asking for that. Once you formulate the quadratic equation it's straightforward but for me this step really takes a leap of the imagination to find it and I wonder if I'm alone in this. Can't really post the answer because then it'll be obvious.

So basically I'm asking if anyone can formulate the equation and tell me if they found it difficult.

Thanks for the willingness to answer already!

答:
Thanks for the comments and answers. The answer my book gives wasn't given, but this is the most straightforward:

Let $x$ = time (hours) required by smaller pipe, $x-3$ = time required by larger pipe. Then $\frac{1}{x}$ = part filled in 1 hr by smaller pipe and $\frac{1}{x-3}$ = part filled in 1 hr by larger pipe.

Since the two pipes together fill $\frac{1}{20/3} = \frac{3}{20}$ of the reservoir in $1$ hr, $\frac{1}{x} + \frac{1}{x-3} = \frac{3}{20}$.

This is the quadratic equation that you can then solve easily to get the answer.

So, the whole idea is that you take think in parts per hour instead of hours per reservoir. Couldn't imagine myself thinking of that.


答:
Notice that $e^x \ge 1+x$ for $x\in(-1, \infty)$. This means that $\log(1+x) \le x$ for $x\in(-1,\infty)$. Hence, $\log(x)\le x-1$ and $1/\log x \ge (x-1)^{-1} $ for $x\in(0, \infty)$. Now, you know that $|\sin x/\log(x)| \ge |\sin(x)|/(x-1)$ on $[1,2]$. The norm of your integral is thus bounded below by
$$\int_1^2\frac{|\sin x|}{x-1}\,dx,$$
which diverges because of the singularity at $x=1$. Conclusion: your integral also diverges.


答: Use the fact that $logx \leq x-1$. So $\frac{sinx}{lnx} \geq \frac{sin1}{x-1}$ for $x \in (1,2]$, and $\intop_{1}^{2}\frac{\sin\left(1\right)}{x-1}$ diverges. So $\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}$ divergent.


答: We have $$\frac{\sin x}{\log x}>\frac1{2\log x}\quad\forall x\in(1,2]$$ and as such, by finding that $$\int_1^2\frac1{2\log x}\,dx=\frac12\left(\text{Li}(2)-\text{Li}(1)\right)$$ is divergent, we can conclude that $$\int_1^2\frac{\sin x}{\log x}\,dx$$ is divergent as well.


问:
Why does the set of submodules of a module that are direct sums of irreducible submodules have a maximal element?
I was reading the proof that every R-module M of an artinian semisimple ring R is the direct sum of all irreducible R-submodules M. In the proof they stated that the set
$$\mathcal{F}:=\{N\subset M: N \text{ is the direct sum of its irreducible submodules}\}$$
has a maximal element. I was trying to see why such a maximal element exists.

My attempt: Let $\{N_i\}_{i\in I}\subset \mathcal{F}$ be a total ordered subset of $\mathcal{F}$ ($I$ is any indexation set). I want to prove that $N=\cup_{i\in I} N_i$ belongs to $\mathcal{F}$. We know that for each $i\in I$ there's an indexation set $J_i$ such that $N_i=\bigoplus_{k_i\in J_i} M_{k_i}$ with $M_{k_i}$ irreducible submodules of $N_i$. Define $$\Lambda :=\cup_{i\in I}\{k_i\in J_i:M_{k_i}=M_{k_{io}} \text{ for some }i_o\in I\setminus\{i\} \text{ and } k_{i_o}\in J_{i_o}\}$$

Obviously $\sum_{k_i\in (\cup_{i\in I} J_i)\setminus \Lambda} M_{k_i}$ is direct sum, but I got stuck for adding only one time such irreducible submodules $M_{i_k}$ associated to $\Lambda$ to asure the direct sum.

答:
It appears that the proof you are reading is incorrect.  There is no straightforward way to show that $\mathcal{F}$ is closed under unions of chains; knowing that each $N_i$ is a direct sum of simple submodules is not particularly helpful for proving that $\bigcup N_i$ is a direct sum of simple submodules.

For a correct proof, you want to use a different poset which forces the direct sum decompositions in such a chain to be "compatible".  Here's one way to formulate it.  Instead of using your poset $\mathcal{F}$, use the poset $$\mathcal{G}=
\left\{(N,S):N\subseteq M\text{ is a submodule, $S$ is a set of simple submodules of $N$, and }N=\bigoplus_{A\in S}A\right\}$$ ordered by saying $(N,S)\leq (N',S')$ if $N\subseteq N'$ and $S\subseteq S'$.

Now suppose you have a chain $\{(N_i,S_i):i\in I\}\subseteq\mathcal{G}$; I claim that $(N,S)\in \mathcal{G}$ where $N=\bigcup N_i$ (or $N=\{0\}$ if the chain is empty) and $S=\bigcup S_i$, and so $(N,S)$ is an upper bound for the chain.  It is clear that $N$ is a submodule of $M$, $S$ is a set of simple submodules of $N$, and $N$ is the sum of the elements of $S$.  It remains to show that this sum is direct.  If the sum were not direct, this would mean there are finitely many elements $A_1,\dots,A_n\in S$ and elements $a_j,b_j\in A_j$ such that $\sum a_j=\sum b_j$ but $a_j\neq b_j$ for some $j$.  But there is some $i\in I$ such that $A_1,\dots,A_n$ are all in $S_i$, and so this would contradict the fact that $N_i$ is the direct sum of the elements of $S_i$.


问:
Specific projective dimension of a module over bound quiver
Suppose $K$ is an algebraically closed field, and $A$ is the algebra presented by the quiver
$$\require{AMScd}
\begin{CD}
1 @>>> 2\\
@V{}VV @V{}VV \\
3 @>>> 4 @>>> 5
\end{CD}
$$
bound by $1\to 2\to 4 = 1\to3\to 4$ and $3\to4\to5 = 0$.

What is the projective dimension of the simple $A$-module $S(1)$?

Here is my work: the projective indecomposables have submodule lattices indicated by the following Alperin diagrams:
$$\begin{array}{c@{}c@{}c@{}c@{}c}
\newcommand{\kem}{\kern-1ex}
\kem&\kem&\kem\kem1\kem&\kem&\kem \\
\kem&\kem\nearrow\kem&\kem&\kem\nwarrow\kem&\kem \\
2\kem&\kem&\kem&\kem&\kem3 \\
\kem&\kem\nwarrow\kem&\kem&\kem\nearrow\kem&\kem \\
\kem&\kem&\kem\kem4\kem&\kem&\kem
\end{array}
\qquad
\begin{array}{c} 2 \\ \uparrow \\ 4 \\ \uparrow \\ 5 \end{array}
\qquad
\begin{array}{c} 3 \\ \uparrow \\ 4 \\ \phantom{\uparrow} \\ \phantom{5}\end{array}
\qquad
\begin{array}{c} 4 \\ \uparrow \\ 5 \\ \phantom{\uparrow} \\ \phantom{5}\end{array}
\qquad
\begin{array}{c} 5 \\ \phantom{\uparrow} \\ \phantom{4} \\ \phantom{\uparrow} \\ \phantom{5}\end{array}
$$

The projective cover of $S(1)$ is $P(1)$ and the kernel is the Heller operator of $S(1)$, namely $$\begin{array}{c@{}c@{}c@{}c@{}c}
\newcommand{\kem}{\kern-1ex}
2\kem&\kem&\kem&\kem&\kem3 \\
&\kem\nwarrow\kem&\kem&\kem\nearrow\kem&\\
&\kem&\kem\kem4\kem&\kem&
\end{array}
$$ with $P(2) \oplus P(3)$ its projective cover.

Now the problem is to find the second Heller operator $\Omega^2(S(1))$. Its composition factors are clear: $S(4)$ and $S(5)$, but how do we know if it is $S(4) \oplus S(5)$ versus $P(4) = \begin{array}{c} 4 \\ \uparrow \\ 5 \end{array}$?

In the former case, the next projective cover is $P(5) = S(5)$ and the resolution terminates with projective dimension 3, but in the latter case the resolution terminates immediately with projective dimension 2.

The former case “uses” the 4 from $P(2)$ leaving $S(5)$ from $P(2)$ and $S(4)$ from $P(3)$. The latter case “uses” the 4 from $P(3)$ leaving the trailing $P(4)$ from $P(3)$. Neither of these is really the kernel, since it is some sort of “diagonal” submodule, but how do we know if the diagonal submodule is split or not?

Does the answer depend on the field?

答:
As far as I know, field should not matter, unless your algebra has "weird relations", and in most case, you are pretty safe as long as characteristic of $k$ is not 2.  I hope somebody can give some supplement answer here...

Back to your specific example, (say if we work over characteristic 0..) the (minimal) projective resolution of $S(1)$ is actually:
$$
0\to P(4)\to P(2)\oplus P(3) \to P(1) \to S(1) \to 0
$$
So the projective dimension is 2.  Eplanation:

(1) Writing out explicitly what the maps are:

first (surjective) map $ae_1 \mapsto a\overline{e_1}$, where $a\in A$, $e_1$ primitive idempotent and $\overline{e_1}=e_1+Rad(P(1))$.

$\Omega(S(1))=ker(ae_1\mapsto a\overline{e_1}) = k-span\{\alpha,\beta,\gamma\alpha\}$, where $\alpha$ is ther arrow from 1 to 2, $\beta$ is the arrow from 1 to 3, and $\gamma$ arrow from 2 to 4.

next map is projective cover of  $\Omega(S(1))$, given by $ae_2 \mapsto a\alpha$ and $ae_3\mapsto a\beta$.

So $\Omega^2(S(1))=k-span\{\delta-\gamma, \epsilon\gamma\}$, where $\delta$ is the arrow $3\to 4$ and $\epsilon$ is $4\to 5$.  You can now construct an isomorphic to $P(4)$ via $ae_4\mapsto a(\delta-\gamma)$.

(2) After enough practice, you should be able to see that everything can be done combinatorially: $\Omega(S(1))$ is the 3-dimensional module with Loewy diagram being the one you have drawn out already.  Well, you can see it comes from "tearing off the radical of $P(1)$ off from $P(1)$".  The same principle applies: $\Omega^2(S(1))$ just come from "tearing off the part in $P(2)\oplus P(3)$ that isn't paste onto $M$ without destroying Loewy structure". Since both 2 and 3 can lie on top of 4, so you need to "tear off half of the 4 from $P(3)$, and half of the 4 from $P(5)$, and glue them back" (c.f. the map written explicitly above).  Sorry this is vague, but this is how people who do computation in these algebras work, for real!

On a side notes, I think "Alperin diagrams" means the diagrams used by Alperin such as those in his book "Local representation theory" (I may be wrong), those are really filtration diagrams, which is different from what you have drawn here (Loewy diagram).  Anyway, if you use "Loewy diagram", most people should know exactly what it is, less ambiguity.

EDIT-Correction:  Thanks to Julian and Jack (the OP)!  Yes, by "weird" I really mean situations like summands in a the relations reducing to zero; or in characteritic 2, minuses become pluses.  When such situation does not occur, and since we are working over algebraically closed field, the homological algebras shouldn't change much.  So this specific example is characteristic independent.


问:
Why are rank-one matrices with nonzero traces diagonalisable?
Given a $n \times n$ matrix $A$ over $\mathbb{C}$, with non-zero trace and rank $1$, why does it always hold that $A$ is diagonalizable? I think it is because of the fact that this matrix has an eigenvalue $0$ with $n-1$ geometric multiplicity, but I don't really get this fact either.

答:
A short answer is that non-zero trace allows us to construct a basis of eigenvectors.

Proof: Since $A$ has rank $1$, there exists a vector $v_1$ such that all columns of $A$ are scalar multiples of $v_1$. In other words, there exists a vector $u_1$ such that

$$
A = v_1u_1^H.
$$

Complete $v_1$ to an orthonormal basis $v_1, v_2, \dots, v_n$ and $u_1$ to an orthonormal basis $u_1, u_2, \dots, u_n$.

Now define $\lambda = u_1^Hv_1$ and note that

$$
\lambda = u_1^Hv_1 = {\rm tr}(A) \ne 0
$$

and so $v_1, u_2, \dots, u_n$ is a basis. Moreover

$$
Av_1 = \lambda v_1 \\
Au_2 = 0 \\
\dots \\
Au_n = 0
$$

so $v_1, u_2, \dots, u_n$ is a basis of eigenvectors of $A$. Conclusion follows from the fact that every matrix with a basis of eigenvectors is diagonalizable.

Remark: Note that if trace is zero then $A$ is nilpotent and the only nilpotent matrix which is diagonalizable is the zero matrix.


问:
What is a non-artificial example of a bifunctor that is contravariant in both arguments?
The Hom functor is a bifunctor that’s covariant in one argument and contravariant in the other.

One can view the construction of group rings as a bifunctor that is covariant in both arguments.

But what is a (hopefully not contrived) example of a bifunctor that is contravariant in both arguments?

答:
Given any covariant bifunctor $A \times B \to C$ and a contravariant functor $C \to D^\text{op}$ we can compose to get a contravariant bifunctor.

One simple example of this might be $(V, W) \mapsto (V \oplus W)^*$ which sends a pair of vector spaces to the dual of their direct sum.

If we do the same construction with the tensor product instead of the direct sum, then we can write this in a slicker way by looking at bilinear maps $V \times W \to k$, which of course gives you $(V \otimes W)^*$. This is a nice observation due to Qiaochu, and is a natural object to study.

I hope this helps ^_^


问:
Under what condition on $a$, the function $\cos ax \cos (ax+\pi)$ will be periodic over $x>0$?
I have this function $\cos ax \cos (ax+\pi)$ for $x>0$, and I want to know under what condition on $a$, it will be periodic?

答:
This function is periodic for every $a$. You can try to find out what is his littlest period.

You can transform the function to $\cos(ax)\cos(ax+\pi)=-\cos^2(ax)$ because $\cos(x+\pi)=-\cos(x)$.

The littelest period of $\cos^2(x)$ is $\pi$.

Hence, the littelest period of the function is $\frac{\pi}{|a|}$ for $a\not =0$.

For $a=0$, the function is constant equals to $-1$, so we can't define a littlest period (the function is periodic for every real).

Indeed, let $T$ the littelest period of our function.

We have : $-\cos^2(ax)=-\cos^2(a(x+T))=-\cos^2(ax+aT)$.

$|aT|$ must be equal to $\pi$.

EDIT :

The op. actually wanted to find for which $a$, $\cos(ax)\cos(a+\pi)x$ is periodic.

We can use this simple argument :

If $\cos(a+\pi)\not = 0$, then $\cos(ax)\cos(a+\pi)x$ is not bounded.

Indeed, $|\cos(a\frac{n\pi}{a})\cos(a+\pi)\frac{n\pi}{a}|\rightarrow +\infty$ if $a\not =0$.

If $a=0$, it's a linear function which is not bounded.

The function can't be periodic with this condition because a periodic function is continuous in a segment $[0,T]$ and then bounded in this segment.
By periodicity, it's bounded everywhere.

If $-\cos(a)=\cos(a+\pi)=0$, the function is constant and then obviously periodic.

So the function is periodic if and only if there exists $k\in \mathbb{Z}$ such as $a=\frac{\pi}{2}+k\pi$.


问:
Unsolved Problems in Elementary Game Theory With Real-World Applications
I am looking for some unsolved problems in elementary game theory (no calculus, etc.) that have real-world applications and are not impractical to solve. It would be great if the games are not well-known.

Thanks for any suggestions.

答:
Coordination games such as

9,9 0,7

7,0 8,8

are unsolvable in a sense (risk dominance vs payoff dominance). But in the real world, they might very well be part of a repeated game or be a widely played game, where that context strongly suggests a conventional coordination point. Real world examples are new electronic product standards, eg, USB standards.


问:
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$
To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?

答:
$$\int\frac{dx}{2+x+\sqrt{1-x^2}}$$
Substitute $x= \sin 2u;\;dx=2\cos 2u$

$$\int \frac{2\cos (2 u)}{2+\sin (2 u)+\cos (2 u)} \, du$$
$$\int \frac{2\cos ^2 u-2\sin ^2 u}{2-\sin ^2 u+\cos ^2 u+2 \sin u \cos u}\,du$$
$$\int \frac{2\cos ^2 u-2\sin ^2 u}{\sin ^2 u+3 \cos ^2 u+2 \sin u \cos u}\,du$$
divide numerator and denominator by $\cos^2 u$
$$\int \frac{2-2\tan ^2 u}{\tan ^2 u +2 \tan u +3}\,du$$
substitute $\tan u = t\to dt=\frac{du}{1+u^2}$
$$\int \frac{2-2t^2}{\left(t^2+1\right) \left(t^2+2 t+3\right)}\,dt$$
using partial fraction
$$\int \left(\frac{1-t}{t^2+1}+\frac{t-1}{t^2+2 t+3}\right)\,dt$$
$$\frac{1}{2} \left(-\log \left(t^2+1\right)+\log \left(t^2+2 t+3\right)+2 \arctan t-2 \sqrt{2} \arctan\left(\frac{t+1}{\sqrt{2}}\right)\right)+C$$
$t=\tan u$ and $x=\sin 2u$ we have $t=\frac{1-\sqrt{1-x^2}}{x}$
$$\frac{1}{2} \log \left(\sqrt{1-x^2}+x+2\right)+\arctan\left(\frac{1-\sqrt{1-x^2}}{x}\right)-\sqrt{2} \arctan\left(\frac{-\sqrt{1-x^2}+x+1}{\sqrt{2} x}\right)+C$$


问:
Intuition for the Lax-Milgram Theorem
I have seen the proof of the Lax-Milgram theorem (and can replicate it with most of the details), I've also applied it for a bi-linear form defined as the weak formulation of a PDE to show the existence of a solution. I still don't get the idea behind it. Is there an equivalent in finite dimensional linear algebra to give me some intuition?

Lax milgram says: $a:H \times H \rightarrow \mathbb{R}$ continuous, coercive, symmetric, then:

$\forall \phi \in H^{\ast}, \exists ! u \in H$ such that $a(u,v) = \phi(v)$ for all $v \in H$.

This $u$ is the only minimiser of $H$ of the problem
$$ \inf_{v \in H}\left\{\frac{1}{2}a(v,v)-\phi(v)\right\}$$

More concretely, my questions are:

What is (1), some kind of dimension reduction?

How did (2) come about? It's so useful in the context of PDEs but its mysterious to me.

What's the finite dimensional intuition I can draw from?

EDIT:

I had some time to think about this.

Take $H:=\mathbb{R}^n$. In this simple finite dimensional context, $a(u,v) := \langle u,v\rangle$ is the natural choice for a continuous, coercive, and symmetric bilinear form.

Then the dual is the space of all row vectors. So Lax-Milgram tells us that $\forall r \in (\mathbb{R}^n)^{\ast}$ there exists a unique $u \in \mathbb{R}^n$ such that $\langle u,v\rangle = r v$ for all $v \in \mathbb{R}$, i.e. the only one is $r=u^T$. It just tells us a different way to write the dot product using elements in the dual.

As for the minimisation problem, Lax-Milgram says that the solution to this minimisation problem would be $$\frac{1}{2} \langle u, u \rangle - \langle u,u\rangle = -\frac{1}{2}\|u\|^2$$ I am at a loss for what this represents, but I now have some intuition for the first result! Hopefully someone can comment on this minimisation.

答:
Regarding the first question:

This $u$ is what we call a weak solution. The most obvious reason why we ask for weak solutions is because sometimes strong solutions don't exist and we still want to make sense of problems in these cases. The other reason that comes to mind is that even when strong solutions exist, it is  still often easier to prove that by showing existence of a weak solution and then a posteriori showing it is a strong solution. This is in part because weak solutions provide a framework for the use of variational methods and so forth.

Regarding the third question:

The finite dimensional analogue of Lax-Milgram is that $Ax=b$ has a unique solution if $A$ is symmetric positive definite. The "weak formulation" of $Ax=b$ is $\langle v,Ax \rangle = \langle v,b \rangle$ for all $v \in \mathbb{R}^n$, and the bilinear form $a$ now is $a(v,x)=\langle v,Ax \rangle$. (I use scare quotes on weak formulation because the two formulations are equivalent.)

Regarding the second question:

Relative to the finite dimensional analogy described in the previous paragraph, you are converting from $Ax=b$ to $x=\arg \min f$ where $f(x)=\frac{1}{2} \langle x,Ax \rangle - \langle x, b\rangle$. The fact that $x=\arg \min f$ implies $Ax=b$ follows from writing out the equation $\nabla f = 0$ and applying the symmetry of $A$. The positive definiteness is what you need to ensure that the resulting critical point is a minimum.

Title: Intuition for the Lax-Milgram Theorem


问:
power series radius of convergence
How can I prove that if the coefficients $\{a_k\}$ of the power series $\sum_{0}^{\infty} \{a_k\}x^k$ form a bounded sequence, then the radius of convergence is at least 1?</p>

答:
Hint:  If the sequence $a_k$ is bounded, then there exists $M$ such that $|a_k|\leq M$ for all $k$.

Then, what can you say about $|\sum_{k=1}^\infty a_k x^k|$ and $\sum_{k=1}^\infty Mx^k$? What is the radius of convergence of the second one?

Title: power series radius of convergence


问:
Can a martingale converge in $L^1$ but not almost surely?
In general, almost sure convergence and convergence in $L^1$ are independent modes of convergence in the sense that neither implies the other. For martingales, at least one implication, namely that almost sure convergence implies convergence in $L^1$ remains false. How about the other one? Does convergence of a martingale in $L^1$ imply almost sure convergence?

答: Yes, for a martingale convergence in $L^1$ implies almost sure convergence.  Suppose $(M_n) \rightarrow M_\infty$ in $L^1$ where $(M_n)$ is a martingale.  Then $\sup_n \mathbb{E}[|M_n|] < \infty$ so by Doob's martingale convergence theorem there exists $N_\infty \in L^1$ such that $(M_n) \rightarrow N_\infty$ almost surely.  But $L^1$ convergence implies a.s. convergence along a subsequence so we can also find a subsequence $(M_{n_k})$ such that $(M_{n_k}) \rightarrow M_\infty$ almost surely, so we must have $N_\infty = M_\infty$ a.s.


问:
Finding the Covariance of a portfolio given the weights of two assets, with their probability and return % for each asset.
I am given the details below:

The following Table reflects data relating to two assets X and Y, which are present in a portfolio in the ratio 60:40

```<code>Probability | Return on X(%) | Return on Y(%)

    0.2            -10            -1.40
    0.4             25            -2.25
    0.3             25            -2.00
    0.1             10            -10.00
</code>```

I had to state what type of portfolio it was and the answer was that it was a nil covariance portfolio.

The calculation to prove this was written as:
Cov = (0.2x-25x1.38)+(0.4x10x0.53)+(0.3x5x0.78)+(0.1x-5x-7.22)=0

The first value in the bracket is the probability, and the third is the return of Y minus the expected return of Y but I think the middle number is incorrect.

Can anyone confirm?

答:
The mean

$E[X] = \Sigma P(Scenario_1) X_i\\
E[Y] = \Sigma P(Scenario_1) Y_i$

$\bar X = (0.2)(-10) + (0.4)(25) + (0.3)(25) + (0.1)(10) = 16.5\\
\bar Y = -2.78$

Each asset has a variance.

$\sigma_X^2 = \Sigma P(Scenario_i)(X_i - \bar X)\\
\sigma_Y^2 = \Sigma P(Scenario_i)(Y_i - \bar Y)$

$\sigma_X^2 = 195.3\\
\sigma_Y^2 = 5.9$

There is a covariance between the two assets.

$\text{Cov}_{X,Y}=\Sigma Scenario_i(X_i - \bar X)(Y_i - \bar Y)$

$\text{Cov}_{X,Y} = 1.17$

And the portfolio with weights $\omega_x, \omega_y$ will have a variance.

$\sigma_P^2=\omega_X^2\sigma_X^2 + \omega_Y^2\sigma_Y^2 + 2\omega_X\omega_Y\text{Cov}_{X,Y} = 71.8$

And $E[P] = 8.8$


问:
Prove that you can make change for any amount greater than 7 cents using only 3 and 5 cent coins.
I am pretty sure this question has something to do with the Least Common Multiple.

I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$.

If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove.

But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.

答:
proof by induction

p(8): 3.(1)+5.(1)=8 holds

Assume p(k) holds => 3.a + 5.b = k

we must prove that p(k+1) holds.

p(k+1): 3x+5y = k+1=>
3x+5y = 3a+5b+1=>
3(x-a) + 5(y-a)

thus suffices to prove that 3z+5ω =1 has an integer solution.
That is true because gcd(3,5)=1 | 1 (1 devdes 1)

Calculating gcd(3,5) gives:
5 = 1*3 + 2

3 = 1*2 + 1

2 = 2*1 + 0

Then applying the Extended Euclidean Algorithm:

1   =   (1 * 3) + (-1 * 2)
=   (-1 * 5) + (2 * 3)=>
z0=2, ω0=-1 =>
3.2+5.(-1) = 1=>
p(k+1) holds as well


问:
Consequences of this powerful result?
Here is a statement about "complete system of projectors" :

Let $E$ a $\mathbb{K}-$vector space and $p_1,...p_n$, $n$ linear maps.
For all $i,j\in\{1,...,n\}$ : $p_i \circ p_j=0$ with $i\neq j$, for all $i\in\{1,...,n\}, \ p_i^2=p_i$ and $p_1+...+p_n=id_E$ iff $E= \bigoplus\limits_{i=1}^{n} \text{Im}(p_i)$.

What are the abstract or concrete applications of this result ?

Maybe it is linked with the kernel's lemma and spectral projectors in reduction's theory.

It is also probably linked with quadratic forms and symmetric endomorphisms. There are Fisher-Cochran's results in probability/statistics.

Maybe we can find similar properties in functional analysis.

And finally I was wondering if there are applications of that statement in finite or infinite groups representation theory ? For instance we have the Reynolds operator.

Thanks in advance !

答:
Certainly the idea that (closed) subspaces of Hilbert spaces have (orthogonal) projectors to them (an immediate corollary of the minimization theorem, a true Hilbert-space analogue of the literally false pseudo-Dirichlet Principle in Banach spaces), is important throughout mathematics.

Yes, representation theory, yes, spectral theory, yes ... everything. In representation theory, e.g., of compact topological groups (so, certainly of finite groups on complex vector spaces), mutually non-isomorphic subrepresentations have orthogonal projectors. This is Schur orthogonality relations (which is a non-trivial fact even for finite groups).

Title: Consequences of this powerful result?


问:
Maximum size of k-uniform set family that satisfies a condition
Let $n \leq 2k$ and $A_1,A_2,...,A_m \subseteq [n]$ be distinct $k$-uniform set where $A_i\cup A_j \neq [n]$ for all $1 \leq i < j \leq m $.

Prove that

$$m \leq (1- \frac{k}{n}) \binom n k$$

and that equality can hold.

I don't know if it is related to Erdős-Ko-Rado theorem. If you have any clue how to come up with this bound please let me know.

答:
It is related to the Erdős-Rado-Ko theorem indeed.

Consider the sets
$$B_i=[n] \setminus A_i \text{ $(1 \leq i \leq m)$}$$
Then we have
$B_i \cap B_j \not= \emptyset$ for any $i\leq j$.

By the Erdős-Rado-Ko
theorem, we deduce

$$
m \leq \binom{n-1}{n-k-1}=\frac{(n-1)!}{(n-k-1)!k!}=
\frac{n-k}{n} \frac{n!}{(n-k)!k!}=(1-\frac{k}{n})\binom{n}{k}
$$

Equality is reached when equality is reached in the Erdős-Rado-Ko
theorem, i.e. when the $B_i$ are all the subsets containing a fixed element
$x$.


问:
Should the Stein-Shakarchi Analysis Volumes be Studied in Order?
I couldn't find this question anywhere else, so just thought I would ask it here. There is a series of 4 analysis books written by Stein and Shakarchi (sometimes called the Princeton Lectures in Analysis), and as I intend to study from them, I was wondering whether it is important or not to learn the material in the order that the books were written:

Fourier Analysis

Complex Analysis

Real Analysis

Functional Analysis

For example, is the Fourier analysis in the first volume a prerequisite to understanding the complex analysis in the second volume? Or can one just proceed straight to complex analysis while having just a basic background in real analysis (like Tao or Rudin)? Thanks.

答:
Most math books contain an introduction that outlines the prerequisites. I have only read Vol III, but that one was generally self-contained. I doubt an introductory complex analysis book would require Fourier analysis.  Quoting from the intro to Vol III:

Despite the substantial connections that exist between the different
volumes, enough overlapping material has been provided so that each of
the first three books requires only minimal prerequisites:
acquaintance with elementary topics in analysis such as limits,
series, differentiable functions, and Riemann integration, together
with some exposure to linear algebra.

In my opinion the chief prerequisite is "mathematical maturity", i.e., familiarity with proving stuff.  They may not be so good as an introduction to proof-writing.


问:
Two representations of a finite group have no irreps in common if and only if their characters are orthogonal.
We want to prove that two representations of a finite group $\Gamma_{1}$ and $\Gamma_{2}$ have no irreducible representation in common if and only if their characters are orthogonal, i.e.,

$\sum_{k=1}^{c} n_k \chi_{1k}\chi_{2k}^{*}=0$,

where $\chi_{1k}$ and $\chi_{2k}$ are the characters of the $k-$th class in $\Gamma_{1}$ and $\Gamma_{2}$ respectively.

We can say that for common irreps the characters are the same and therefore the above relation doesn't equal to zero. But it is trivial. Could you please help?

答:
Let
$$
\Gamma_1 \sim \bigoplus_{\alpha\in\widehat{G}}m_\alpha^{(1)}\Gamma_\alpha, \qquad \Gamma_2 \sim \bigoplus_{\alpha\in\widehat{G}}m_\alpha^{(2)}\Gamma_\alpha
$$
be the decompositions of $\Gamma_1$ and $\Gamma_2$ into irreducible representations. Here $\widehat{G}$ denotes the equivalence classes of irreducible reps of $G$, $\Gamma_\alpha$ denotes a specific representation in the class $\alpha$, and $m_\alpha^{(i)}$ denotes the multiplicity of the $\alpha$ in $\Gamma_i$. Taking the trace of both sides gives
$$
\chi_{\Gamma_i} = \sum_{\alpha\in\widehat{G}}m_\alpha^{(i)}\chi_\alpha
$$
for $i=1,2$. Defining the inner product
$$
\langle a, b \rangle := \frac{1}{|G|}\sum_{g\in G}\overline{a(g)}b(g)
$$
and using orthonormality of the irreducible characters $\chi_\alpha$ gives
$$
\langle \chi_{\Gamma_1}, \chi_{\Gamma_2}\rangle = \sum_{\alpha\in\widehat{G}}m_\alpha^{(1)}m_\alpha^{(2)}.
$$
Since the $m_\alpha^{(i)}$ are non-negative integers, this equals zero iff $m_\alpha^{(1)}m_\alpha^{(2)} = 0$ for all $\alpha\in\widehat{G}$. This is true iff for each $\alpha\in\widehat{G}$, either $m_\alpha^{(1)} = 0$ or $m_\alpha^{(2)} = 0$, i.e. $\Gamma_1$ and $\Gamma_2$ have no irreducible representations in common.


问:
Is this enough information to prove that this figure is a parallelogram?
If one pair of opposite sides in a quadrilateral is parallel but not congruent and the other opposite pair of sides is congruent but not parallel is this enough info to prove that the figure is a parallelogram?</p>

答:
Hint:

Consider the isosceles trapezoid.

I hope this helps ^_^


问:
When do we need ZFC models to be transitive?
I learn set theory mainly through Kunen's classical book: set theory, an introduction to independence proofs. In that book, Almost all models are transitive. I checked some basic lemmas and notions, if the models are not transitive, they may be fail or not absolute.

However, in other books or lectures, authors mainly talk about "models" rather than "transitive models".
Does this mean models need not to be transitive there? Or just omitted "transitive"? Or identify models with their transitive collapsing？I can't guess which one is right.

Thanks for any help.

答:
This is too long for a comment, so:

One exception that should be mentioned is that of small elementary substructures of large, transitive sets.

Fix a regular uncountable cardinal $\kappa$ and let $H_\kappa=\{x:|\text{trcl}(x)|<\kappa\}$. As you probably know if you've been reading Kunen, $H_\kappa$ satisfies ZFC-P, where P is the Power Set Axiom (also, $H_\kappa\models$ ZFC iff $\kappa$ is inaccessible).

Now, take an elementary substructure $M\prec H_\kappa$ with $M=\aleph_0$. These objects are extremely useful in set theory. A certain tension exists between $M$ being small, yet "knowing a lot" of information from the larger structure. This can be exploited to prove many different things. For example, the proof that GCH holds in L uses countable elementary substructures, and a lot facts in infinitary combinatorics (like Fodor's Lemma, the $\Delta$-system Lemma, etc) can be given alternative proofs using countable elementary substructures. In fact, many of the proofs using countable elementary substructures are much slicker, and feel almost magical. See also the proof of Arhangelskii's Theorem for an application to point-set topology.

After that sales pitch, a reasonable question to ask is whether such an $M$ can be transitive? The answer is a resounding no, whenever $\kappa>\aleph_1$. Indeed, for such $\kappa$, $\omega_1\in H_\kappa$. Moreover, one can show that
$$
H_\kappa\models \omega_1\text{ is the least uncountable cardinal}
$$
So that $\omega_1$ is definable without parameters in $H_\kappa$. By elementarity, $\omega_1\in M$. Now, if $M$ were transitive, then $\omega_1\subseteq M$, so $M$ is uncountable, contradiction.

Collapsing these $M$ is also part of the game, and that yields some important information (in the $L$ case, the collapse must be of the form $L_\delta$ for some countable $\delta$), but that "tension" I mentioned before is lost. For example, $\omega_1\in M$ but $\omega_1$ does not belong to the collapse. Thus, whatever ordinal the collapse thinks is the first uncountable cardinal, is not the real $\omega_1$.


问:
A minimal generating set for the order relation of rationals and reals
Does there exist a binary relation $R$ on $\mathbb{Q}$ such that the transitive closure of $R$ is the order relation $<$, and that the transitive closure of any proper subset of $R$ is not the order relation? My second question is the same thing, but with $\mathbb{R}$ instead of $\mathbb{Q}$.

答:
I'll just spell out bof's argument from the comments.

Suppose $(L,<)$ is a dense linear order with more than one element. I claim that there is no minimal binary relation on $L$ whose transitive closure is $<$.

Suppose $R$ is a binary relation on $L$ whose transitive closure is $<$. Since $L$ has more than one element, $R\neq \varnothing$. It suffices to show that for any $(a,b)\in R$, if we let $R' = R\setminus \{(a,b)\}$, then the transitive closure of $R'$ is also equal to $<$, since then $R$ is not minimal.

Since $(a,b)\in R$, we have $a<b$, and since $L$ is dense, there is some $c$ with $a<c$ and $c<b$. Since $<$ is the transitive closure of $R$, there exists $m\geq 0$ and $d_1,\dots,d_m$ with $a R d_1 R d_2\dots R d_m R c$ and there exist $n\geq 0$ and $e_1,\dots,e_n$ with $cR e_1 R e_2 R \dots R e_n R b$. Putting these together, and noting that no instance of $R$ in either chain is $(a,b)$, we have $$a R' d_1 R' d_2\dots R' d_m R' cR' e_1 R' e_2 R' \dots R' e_n R' b$$ so $(a,b)$ is in the transitive closure of $R'$. It follows that the transitive closure of $R'$ is equal to the transitive closure of $R$, which is $<$.


问:
Write ‘There is exactly 1 person…’ without the uniqueness quantifier
During a lecture today the prof. posed the question of how we could write "There is exactly one person whom everybody loves." without using the uniqueness quantifier.

The first part we wrote as a logical expression was "There is one person whom everybody loves.", ignoring the 'exactly one' part of the question initially. From this he wrote

$L(x,y): x$ loves $y$; domain for $x$ and $y$: $\{\text{people}\}$

$\exists x\forall y: L(y,x)$

Which I understand to mean 'There is a person $x$ such that for all $y$, $x$ is loved by $y$' AKA 'There is a person who is loved by everyone'. I get that part.

The part I don't get is how the expression of 'exactly one'.

$\forall z(\forall y(L(y,z))\to x =z)$

which then creates the joint expression

$\exists x\forall y(L(y,x))\land \forall z(\forall y(L(y,z))\to x=z)$

I just can't seem to understand how $\forall z(\forall y(L(y,z))\to x =z)$ means exactly one here.  I suppose you can take $\forall z$ here to mean 'for any given person', which means the $\forall z$ is considering every person in the world.  This would translate the second expression block to something like, "For any given person $z$, if everybody loves $z$ then $z$ is the same person as $x$".

To me though $\forall z$ generally means for every element in the domain which I see as meaning every person in the world simultaneously, as it seems to for $y$.  Is that just plain wrong?  How can I tell when $\forall$ means 'all' and when it means 'for any (one)'?  In the previous English translation the only reason I was able to translate it (if it's even right) is because I already knew what the statement was suppose to mean.

Is it just that $\forall z$ means that this statement could be true for any element, and if so what's the difference between $\forall z$ and $\exists z$?  Someone told me that $\exists z$ would be redundant because the expression says $x=z$ but how do I know that $x$ and $z$ are automatically the same person if $\exists$ is used for both?

Sorry if this is a bit long with too many questions.  I just wanted to try to make the cause of my confusion as clear as possible so you can help me figure this out.

答:
Is it just that $\forall z$ means that this statement could be true for any element,

Yes, exactly.

...and if so what's the difference between $\forall z$ and $\exists z$?

If we only asserted the existence of some particular $z$ such that if $y$ loves $z$, then that particular person $z$ would thus be $x$. But then we are not ruling out that there might be another person, different from $x$, that is also loved by everyone. And we have already asserted the existence of someone (namely, x) who is loved by all. So in that sense, $\exists z$ such that... is reduntant.

We need the universal quantifier for $z$ (to assert a statement $\forall z$) in the second clause to indicate that if there is any $z$ (which means that the claim that follows - as it relates to $z$, is true for every z) $z$ such that $L(y, z)$, then any/every such $z$ must be $x$, since there is exactly one person, namely $x$, who is loved by all $y$.  I.e., for every $z,$ if every y loves z, then z must be x: i.e., that $z$ is not anyone other than $x$. This gives us that $x$ (whose existence we asserted at the start) is therefore the one unique person loved by all.

Now, just one oversight to "clean up" your expression, which you state as:

$$∃x(\forall y L(y,x))\land ∀z(∀y(L(y,z))⟹x=z)$$

But here we have two independent clauses that creates a problem, because in your second clause, you have $x$ appear outside the scope of its quantifier. I.e., it is a free, unquantified varible.

So we want the scope of $\exists x$ to persist over the entire statement, hence the square brackets below.

That is, $$\exists x \big[\forall y(L(y, x))\land \forall z(\forall y(L(y, z)) \rightarrow z = x)\big]$$


问:
Trigonometric power series identity
I've come across the identity:
\begin{equation}
\lim_{z\rightarrow k}\frac{\pi(z-k)}{\sin(\pi z)}=\lim_{z\rightarrow k}\frac{\pi}{\pi\cos(\pi z)}=(-1)^{k}
\end{equation}
for $z\in\mathbb{C}$ and $k\in\mathbb{Z}$. I'm unsure on how the first equality can be derived. How might I use the power series definitions of $\sin(z)$ and $\cos(z)$ to show this?

答: By L'Hôpital, $$\lim_{z\rightarrow k}\frac{\pi (z-k)}{\sin(\pi z)}=\lim_{z\rightarrow k}\frac{\frac{d}{dz}\left (\pi (z-k)\right)}{\frac{d}{dz}\left(\sin(\pi z)\right)}=\lim_{z\rightarrow k}\frac{\pi}{\pi\cos(\pi z)}=(-1)^k$$

Title: Trigonometric power series identity


问:
Does folded cube have a name?
Folding a square to produce a torus has an analogy with "folding" a cube to produce what? Is there a name for the resulting 4D structure?</p>

答:
Yes. Perhaps unfortunately, such an object is also called a torus. If you want to be precise, you might use the word "hypertorus" or say explicitly "$n$-dimensional torus". In your case, you would be interested in the $3$-torus embedded in $4$D.

In general, if we write $S^1$ for the circle, then
$S^1 \times S^1 \times \cdots \times S^1$ is called a torus (by analogy with the classical case $S^1 \times S^1$).

In particular, a square is $[0,1] \times [0,1]$. So when we glue together the boundaries, we're glueing the $[0,1]$s into circles. So we wind up with $S^1 \times S^1$.

What about a cube? It's $[0,1] \times [0,1] \times [0,1]$. Again, we glue opposite sides of the cube together, and we wind up with $S^1 \times S^1 \times S^1$.

I hope this helps ^_^

Title: Does folded cube have a name?


问:
summation on one series and more formal notation
I have a series in my own:

$1+(1+2)+(1+2+3)+...+(1+...+10)$

How I can show these with Sigma notation and Reach to 220?

答:
$$\sum_{m=1}^n(1+2+\cdots+m)=\sum_{m=1}^n\sum_{k=1}^mk$$ Also we can obtain a closed formula as $$\dfrac{n}{6}(n+1)(n+2)$$ for this summation for any $n\in\mathbb{N}.$

Added:

There are multiple ways to establish this general formula. You can get some proof ideas from here, which discusses many different proofs of the fact that $$\color{Green}{1+2+\cdots+n=\dfrac{n}{2}(n+1)}.$$ Since there are many nice internet sources on general techniques to sum (finite) series, I will give you a sort of intuitive explanation using only the formula in green. $$S=1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+n)\tag1$$ First, we can simplify the first equation using the green formula to obtain $$2S=\color{Blue}{2}+\color{Red}{6}+\cdots+\color{Orange}{n(n-1)}+n(n+1)\tag2.$$ Secondly, notice that in the first series one appear $n$ times and two appear $n-1$ times and so on. Therefore we can rearrange the  series as $$S=1.n+\color{Blue}{2.(n-1)}+\color{Red}{3.(n-2)}+\cdots+\color{Orange}{n.1}\tag3$$ Now, lets add second and third equation according to given color codes to get $$3S=n+\color{Blue}{2n}+\color{Red}{3n}+\cdots+\color{Orange}{n^2}+n(n+1)\tag4.$$ Finally factor $n$ from the right hand side of this fourth equation and apply the formula in green once again for $n+1$ terms :)


答: You have $10$ sums, so you want to start with $$\sum_{k=1}^{10}$$  Sum number $k$ is of the form $$1+2+\cdots+k=\sum_{i=1}^ki$$ Altogether, we have $$\sum_{k=1}^{10}\sum_{i=1}^ki$$  As for evaluating the sum, use the formula for the sum of an arithmetic progression for the inner sums.  Then you have $10$ numbers to add up.

Title: summation on one series and more formal notation

Body:
I have a series in my own:

$1+(1+2)+(1+2+3)+...+(1+...+10)$

How I can show these with Sigma notation and Reach to 220?