问:
Lifting a metric on the base space to the principal $U(1)$ bundle
"Let $π:P→M$ be a $U(1)$-principal bundle. We may lift any metric $m$ on $M$ to a $U(1)$-invariant metric $\widehat{m}$ on P such that $π_∗\widehat{m}=m$."

I found a more general form of the above statement in https://mathoverflow.net/questions/372424/lifting-a-metric-on-a-principal-bundle-with-non-compact-fibre. Could anyone tell me what the name of this theorem is? Any reference is welcome.

答:
I don't think this theorem has any particular name (it's a fairly simple construction), although I've occasionally seen the metric referred to as the "Kaluza-Klein metric", since the U(1)-version can be used to construct a Hamiltonian that reproduces the motion of a charged particle in an magnetic field (i.e. the Lorentz force equation - the nonabelian version is called Wong's equation). If you're looking for some specific references to the definition of this metric, you could try "Canonical formulations of a classical particle in a Yang-Mills field and Wong's equations"
 Section 2 or "A Tour of Subriemannian Geometries, Their Geodesics and Applications"
 Chapter 11 (both by Richard Montgomery), or "Riemannian Geometry, Fiber Bundles, Kaluza-Klein Theories and All That..." Chapter 4 by Coquereaux and Jadczyk. I'm sure there are plenty of other references out there.</p>


问:
Limit of Newton's Method on polynomial $Ax^4 + Bx^3+ Cx^2 + Dx + E$?
So if you took the function $f(x) = Ax^4 + Bx^3+ Cx^2 + Dx + E$ and did Newton's Method repeatedly, you would get a sequence that converges to at most $4$ roots. I was wondering what would happen if you took the limit of that sequence, would you get the formula for the roots of a quartic equation? And what if you do it for a polynomial of nth degree. I know it's been proven that there can't be an explicit formula for polynomials of degree higher than $4$ so I'm not sure what would happen if you tried to find it through Newton's Method. Would it not diverge because there are multiple roots and the initial term (the guess) determines which root you find? But shouldn't that just get you a formula for the root in terms of the initial guess? I tried looking at the first few terms for Newton's Method but they became too much to work with so I was wondering if there was a program that could do Newton's Method for functions with unknown coefficients. I think I could do it with Python but it would be a lot of work to do it from scratch so if there's a python library that can support operations with variables, I would then just have to write a program for Newton's Method.

答:
This has been studied:

Doyle, Peter; McMullen, Curt, Solving the quintic by iteration, Acta Math. 163, No. 3-4, 151-180 (1989). ZBL0705.65036.

ContentLicense: CC BY-SA 4.0


问:
A unitary matrix taking a real matrix to another real matrix, is it an orthogonal matrix?
I tried to prove that a real antisymmetric matrix can be taken by an orthogonal tranformation to a form:

where the eigenvalues are $\pm i\lambda_1, \pm i\lambda_2 ...  $

which is a statement I saw on wikipedia in http://en.wikipedia.org/wiki/Antisymmetric_matrix

I also know an antisymmetric matrix can be diagonalized by a unitary transformation, and I found a unitary transformation taking the diagonal matrix to the required form.

So by composing the two transformations (diagonalization, then taking the diagonal matrix to the required form), I'll get a unitary transformation taking the real antisymmetric matrix to another real matrix.

My question is if this transformation must be a real matrix? if so I can deduce that the unitary transformation is in fact an orthogonal transformation.

So is this true?

Is a unitary transformation taking a real matrix to another real matrix necessarily an orthogonal transformation?

EDIT: After receiving in the comment here a counterexample, I'm adding:

Alternatively, if it is not necessarily orthogonal, does there necessarily exist an orthogonal transformation taking the two matrices to each other?

答:
Here is an answer that works for normal matrices, in particular for skew-Germitian matrices.

Every normal matrix is unitarily diagonalizable.  So there is an ONB consisting of eigenvectors.  You can replace this complex-entried ONB by a real-entried ONB that
replaces $2\times 2$ diagonal blocks with conjugated eigenvalues by $2\times 2$ block of the kind you want by using the following observation: If $u\in\mathbb{C}^n$ is an eigenvector corresponding to eigenvalue $\lambda i$, then write $u=x+iy$ where $x, y\in \mathbb{R}^n$.  Since the matrix is real, it is then almost immediate that $x,y$ are perpendicular and of length $1/\sqrt{2}$. Using this, you can then replace the original complex-entried ONB by a real-entries ONB.

ContentLicense: CC BY-SA 3.0


问:
If $A^3B-B$ is invertible then $BA-B$ is invertible.
I assume this is a simple proof but i'm stuck here.

I need to prove that if $A^3B-B$ is invertible then $BA-B$ is invertible.

So $A^3B-B=(A^3-I)B$  and then both $(A^3-I)$ and $B$ are invertible.

So I need to show that $BA-B=B(A-I)$ are both invertible.

I thought maybe I could somehow show that if $(A^3-I)$ is invertible then $(A-I)$ is invertible but I guess there's a simpler way to get there. What am I missing?

Thanks!

Body: Hint $$A^3-I = (A-I)(A^2 + A + I).$$


问:
Cutting a cone with a plane and then tilting the plane
If we cut a cone with a plane parallel to its base we get a circle. Now if we tilt the plane we get various curves like ellipses, parabolas and hyperbolas.
Now my question is that, can we transform our Cartesian coordinate system to another where the effect of tilting the plane is undone. I mean how can I transform the coordinates along with tilting plane to get a circle always.</p>

答: I guess this question comes out by pure curiosity. For what matters yes you can do it, you can imagine to rotate the cone along with the plane so that the plane will be parallel to the base. Bare in mind that if you take a class of linear algebra this question will look silly to you, and going beyond that you will see an actual definition of conics which is not the one you probably encountered in an High school.</p>


问:
Why do we multiply by $-1$ here?
WolframAlpha solves $$\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2}$$ by multiplying by $-1$ as such:
$$\sin\left(-x+\fracπ4\right)=-\frac{1+\sqrt3}{2\sqrt2}$$ then arcsin both sides, etc. but why multiplying by $-1$ and not just directly find the answer? Mainly say this because I've tried without multiplying by $-1$ and haven't managed to got $\frac56 π$ which is the answer (in $0<x≤2π$)

答:
$\sin(x)$ is an odd function, so $\sin(x) = -\sin(-x)$.  This means that

\begin{align*}
\sin\left(x-\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2} &\iff -\sin\left(-x+\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2}\\
&\iff\sin\left(-x+\frac{\pi}{4}\right)=-\frac{1+\sqrt3}{2\sqrt2}.
\end{align*}

In other words, both equations are equivalent.  I'm not sure why WolframAlpha makes this transformation first, it's probably related to whatever algorithm it is using to solve the equation.  For solving on paper it doesn't matter which approach you take.

To solve the equation in its original form, we first note that the equation is in the form $\sin(\theta) = a > 0$, which means that $\theta$ is in the first or second quadrant.  Since the domain of $\arcsin(x)$ is the first and fourth quadrants, we will need to consider the solutions
$$\theta = \arcsin(a)\quad \text{and}\quad \theta = \pi - \arcsin(a).$$
With that in mind, using the first equation, we have

$$
\sin\left(x-\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2} \implies x-\frac{\pi}{4} =\begin{cases}  \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right)\\
\pi - \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right).
\end{cases}
$$
Solving for $x$ gives us
$$x= \begin{cases}\arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right) + \frac{\pi}{4} &= \frac{2\pi}{3}\\\pi - \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right) + \frac{\pi}{4} &= \frac{5\pi}{6}.\end{cases}$$

Title: Why do we multiply by $-1$ here?


问:
Definition of a formula in propositional logic
The set of formulae in propositional logic is characterised by ( at least in Derek Goldreis's book)

Every propositional variable is a formula

If $a$ is a formula then so is $\neg a$

If $a,b$ are formulae then so are $(a \star b)$ for any connective $\star$

Nothing else is a formula and every formula is obtained by a finite number of applications of 1)-3).

From this, does it immediately follow that for any formula $a$ exactly one of the following holds

1') $a$ is some propositional variable

2') $a$ is $\neg b$ for some formula $b$

3') $a$ is $(b \star c)$ for some formulae $b,c$?

I know this is true ( infact the stronger Unique readability theorem holds) but it isn't clear to me how this immediately follows from the definition of a formula I gave above.

I suspect it is clause 4) that makes this work. So perhaps my question really is what exactly does clause 4) mean...

Any help would be appreciated! Thanks!

答:
This relates to a lot of interesting general concepts. Brian Scott outlines the simple answer. I will go a bit deeper into the theory of inductively defining a subset.

For some background, consider a set $A$ and a function $f : P(A) \to P(A)$, where $P(A)$ is the powerset of $A$. Suppose that $f$ is monotonic - that is, if $K \subseteq J \subseteq A$ then $f(K) \subseteq f(J)$.

In this case, $f$ must have a "least fixed point". That is, there must be some $K \in P(A)$ such that $f(K) = K$ and for every $J$, if $f(J) = J$ then $K \subseteq J$.

This least fixed point is constructed by defining $K = \cap \{J \subseteq A | f(J) \subseteq J\}$. For consider the set $f(K)$, and consider an arbitrary set $J \subseteq A$ such that $f(J) \subseteq J$. Then we have $K \subseteq J$, and thus $f(K) \subseteq f(J) \subseteq J$. Therefore, we must have $f(K) \subseteq K$. But since $f(K) \subseteq K$, we also have $f(f(K)) \subseteq f(K)$. And therefore, $K \subseteq f(K)$. Thus, we see that $f(K) = K$, and $K$ is indeed a fixed point.

And clearly, if we have any $J$ such that $J = f(J)$, then we in particular have $f(J) \subseteq J$ and therefore $K \subseteq J$. So $K$ is indeed the "least fixed point" of $f$.

Let's discuss how this applies to the definition of formulas. Consider the alphabet $\Sigma = \{(, ), \neg\} \cup B \cup V$, where $B$ is the set of binary operators and $V$ is the set of variables.

We then define the set $\Sigma^*$ to be the set of all finite sequences with elements in $\Sigma$. We abusively write $V \subseteq \Sigma^*$ to be the set of one-element sequences where the single element is a variable.

Now given a set $U \subseteq \Sigma^*$, define $f(U) = V \cup \{\neg \phi | \phi \in U\} \cup \{(\phi * \psi) | \phi, \psi \in U, * \in B\}$. We see that $f : P(\Sigma^*) \to P(\Sigma^*)$ is an increasing function. Therefore, $f$ has a least fixed point.

That least fixed point is the set of all formulas. Call the least fixed point $K$.

What do we know about $K$? We know that $K = V \cup \{\neg \phi | \phi \in K\} \cup \{(\phi * \psi) | \phi, \psi \in K, * \in B\}$. It follows immediately that every formula can be written as $a$ for $a \in V$, or as $\neg \phi$ for some formula $\phi$, or as $(\phi * \psi)$ for formulas $\phi, \psi$ and binary operator $* \in B$.

We also have the following "induction principle" for $K$. Consider a predicate $P$. Suppose that for all $v \in V$, $P(v)$. Suppose also that if $P(\phi)$, we then have $P(\neg \phi)$. And finally, suppose that if $P(\phi)$ and $P(\psi)$, and if we have $* \in B$, then $P((\phi * \psi))$ holds. Then for every formula $\xi \in K$, $P(\xi)$ holds.

How do we know this? Consider the set $J = \{\xi \in \Sigma^* | P(\xi)\}$. Then $f(J) \subseteq J$. Then $K \subseteq J$.

It turns out that this induction principle, together with the fact that $f(K) \subseteq K$, uniquely characterises $K$. This is the basis of the theory of recursive definitions of a subset and the corresponding principles of induction.

Interestingly, there is another way to define $K$ in this case. It can be shown that $K = \bigcup\limits_{n = 0}^\infty f^n(\emptyset)$. Intuitively, $f^n(\emptyset)$ is the set of all formulas which can be formed using rules 1-3 with a depth of at most $n$. This is the justification for rule 4.


问:
References to Q-curvature
I am working on Q-curvature:

$$
Q^n_g= \frac{1}{2(n-1)} \Delta_g
R_g+\frac{n^3-4n^2+16n-16}{8(n-1)^2(n-2)^2}\:R_g-\frac{2}{(n-2)^2}\:| Ric |^2_g
$$

See, for example, the expression (3) in https://arxiv.org/pdf/0803.4331.pdf. However, the references I encounter only presents the expression of Q-curvature, but they do not indicate how to get it. I would like to know how to get it. Thanks in advance for some collaboration.

答:
The motivation for this formula is that the conformal linearization of the $Q$-curvature is given by the Paneitz operator, a specific formally self-adjoint, conformally covariant operator with leading-order term a power of the Laplacian.  So your question should really be "how do I get the formula for the Paneitz operator".  Branson gave a simple derivation of this as Theorem 1.21 in his article "Differential operators canonically associated to a conformal structure".  For convenience, I will sketch the method.

Let
$$ P = \frac{1}{n-2} \left( \mathrm{Ric} - Jg \right) $$
denote the Schouten tensor, where $J=\mathrm{tr}_g P$ is the trace of the Schouten tensor.  Your formula for $Q$ is equivalently written
$$ Q = -\Delta J - 2\lvert P\rvert^2 + \frac{n}{2}J^2 . $$
The conformal transformation law for the Ricci tensor implies that
$$ \left. \frac{\partial}{\partial t}\right|_{t=0} P^{e^{2tu}g} = -\nabla^2u , $$
where $\nabla^2u$ denotes the Hessian of $u$.
The conformal transformation law for the Laplacian implies that
$$ \left. \frac{\partial}{\partial t}\right|_{t=0} e^{t(2-w)u} \Delta_{e^{2tu}g} (e^{twu}f) = (n+2w-2)\langle\nabla u,\nabla f\rangle + wf\Delta u $$
for any $w\in\mathbb{R}$.
Putting these together (and a similar formula for the conformal variation of the divergence $\delta$) implies that the Paneitz operator
$$ Lf := \Delta^2 f + \delta \left( (4P - (n-2)Jg)(\nabla f) \right) + \frac{n-4}{4}Qf $$
is infinitesimally conformally covariant; i.e.
$$ \left. \frac{\partial}{\partial t}\right|_{t=0} e^{\frac{n+4}{2}tu} L_{e^{2tu}g} \left( e^{-\frac{n-4}{2}tu} f \right) = 0 $$
for all smooth functions $u$ and $f$ and all metrics $g$.  This implies that the Paneitz operator is conformally invariant; i.e.
$$ e^{\frac{n+4}{2}u} L_{e^{2u}g} \left( e^{-\frac{n-4}{2}u}f \right) = L_g(f)  $$
for all smooth functions $u$ and $f$.

Title: References to Q-curvature


问:
Question About Proposition 2.3.12 in Lei Fu Etale Cohomology
I have some trouble with the last step of the proof of the following theorem in Lei Fu's Etale Cohomology.

The statement of the theorem is:

Proposition 2.3.12: Let $S$ be a scheme, $S_0$ a closed subscheme of $S$ with the same underlying topological space. Then the functor $X\to X\times_SS_0$ from the category of etale $S$-schemes to etale $S_0$-schemes is an equivalence of categories.

In the proof he first shows that the functor is fully faithful. Then for any etale $S_0$ scheme $X_0$, he proves that we can find an open covering $\{U_{\alpha_0}\}$ and etale $S$-schemes $U_\alpha$ such that $U_\alpha\times_SS_0=U_{\alpha_0}$. My question is just how can we glue these $U_{\alpha_0}$ together to get the desired $S$ scheme $U$?

答: you have to use fully faithfulness. this is a standard technique when you want to construct an object with some property $P$ over each member of a site you first prove uniqueness (in this case this implied by fully faithfulness) then uniqueness guarantee that you can glue these objects(because there is only one object with property $P$ over the intersection) hence it is enough to construct your object locally.


问:
Is every even number greater than 4 the sum of a prime number with another number that belongs to the set of twin primes?
Thinking about Goldbach conjecture, I have the following question:

Is every even number greater than 4 the sum of a prime number with another number that belongs to the set of twin primes?

For example, as 31 and 17 belong to the set of twin primes,  38=31+7 and 40=17+23.

答: Assume the Twin Prime conjecture is True, It's equivalent to there being infinitely many numbers of form $12m$ that have a Goldbach Partition $(6m-1,6m+1)$. Now consider what happens to the values of 3 times the sum of the values of $m$,  they get 2 Goldbach partitions for $6m$ equidistant for the higher value. Only when the sums of $m$ values are even, can this continue with 3 times the sum of all $m$ used (with duplication possible) over 2 when it uses 4 indices. So the conjecture, relies on the density of the indices $m$ that sum to even values, being dense enough.


问:
Help with arcsin integral
I have the following integral:

$$ I =\int x^2\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) dx$$

Where $a$ and $b$ are non-zero positive integers, and $x<a$. I have started by integration by parts:

$$ I = uv - \int vu'$$

$$u = \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right )$$

$$u' =  \left [ \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) \right ]'$$

We substitute the inside content of the arcsin by $z=\frac{\sqrt{a^2-x^2}}{b}$:

$$u' =  \left [ \sin^{-1}(z) \right ]' = \frac{1}{\sqrt{1-z^2}} \cdot  z'$$

$$z' = - \frac{x}{b} (a^2-x^2)^{- \frac{1}{2}}$$

Hence we obtain:

$$u' =  -\frac{x}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$

On the other hand of the integration by parts we have $v'$:

$$v' = x^2$$

$$v = \frac{1}{3}x^3$$

Hence we end up with:

$$I = \frac{1}{3}x^3\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) + \frac{1}{3}\int \frac{x^4}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$

I cant really figure out how to solve this final integral. Any help would be appreciated

答:
Make the substitution

$$u = \frac{\sqrt{a^2-x^2}}b.$$

Then $x = \sqrt{a^2 - u^2 b^2},$ and so $d x = - \frac{b^2 du}{2 \sqrt{a^2-u^2}},$
and so the integral becomes

$$\frac{b^2}2\int \sqrt{a^2 - u^2 b^2} \arcsin(u) du= \frac{b^2 a}{2} \int \sqrt{1- (\frac{ub}a)^2} \arcsin u du$$ Now integrate by parts.

Title: Help with arcsin integral


问:
Robust numerical solution to linear first order PDE
I have a 2D PDE of the form $$\nabla u \cdot \nabla F(x,y) = G(x,y)$$ with known functions $F$ and $G$ and unknown scalar function $u$, which I would like to solve numerically.  $F$ and $G$ are determined from experimental data, and may have noise in them.  Analytically, this is solved by the method of characteristics, and the obvious way to solve it numerically is to just integrate the characteristic equations.  However, I'm not sure that this is the best approach:
For one thing, it seems like if you integrate $u$ along a characteristic curve, after a long time the value will accumulate errors and diverge from the solution along neighboring curves, leading to a non-smooth solution.  This might be exasperated by the noise in $F$ and $G$.

It seems like there should be a more robust approach (perhaps involving a small artificial diffusive term or something), but I haven't found anything in the usual references (Wikipedia, Numerical Recipes, etc.) on solving this sort of equation.  I'm looking for concise techniques and/or references on robustly solving this sort of PDE.

Also, I'm no expert on this, so perhaps my worries about integrating the characteristics are unfounded, and that's a perfectly good way to solve this.  I welcome someone to tell me that is so, if you know better.

Edit: Given the low uptake on this question, I'm starting a bounty, and I will clarify further what I am looking for:

Please provide references or other verifiable information on how to robustly solve (numerically) a linear, first order PDE.  Preferably the solution should be known to work in the presence of noise on the coefficients.  Conversely, if the method of characteristics is suitable for this purpose, please explain why.

答:
The PDE rewrite as a 2D linear PDE of the form
$$
au_x + b u_y = G
$$
where $a=F_x$, $b=F_y$. A more versatile approach than integration along characteristics could be based on finite difference shemes or on finite volume schemes (see for instance the Lax-Friedrichs scheme). This becomes more obvious if $a\neq 0$, in which case
$$
u_x + c u_y = R
$$
with $c=b/a$ and $R=G/a$.


问:
Convex quadrilateral geometry and circumradius
ABCD is a convex quadrilateral in which AC and BD meet at P. Given PA =1, PB = 2 , PC = 6 and PD = 3 . Let O be the circumcenter of triangle PBC. If OA is perpendicular to AD, find the circumradius of triangle PBC.

I have tried using similarity of the four triangles formed by the two diagonals by utilizing the dimensions given. However I am not able to proceed further.

答:
Since $PC = 2 \, PA \,$ and $\, PB = 2\, PD$ and $\angle\, APD = \angle \, BPC$ the two triangles $\Delta \, PAD$ and $\Delta \, PBC$ are similar, so
$$ \angle \, ADP = \angle \, BCP = \alpha$$
(On the side note, although we do not need this fact, since $\angle\, ADB = \angle \, ADP = \alpha = \angle \, BCP = \angle \, BCA $ the quadrilateral $ABCD$ is inscribed in a circle)

As $O$ is the circumcenter of triangle $\Delta \, PBC$, angle $\angle \, BOP = 2\, \angle \, BCP = 2\, \alpha$. Furthermore, $OB = OP$ so $\Delta\, BOP$ is isosceles and therefore
$$\angle \, OPB = 90 - \angle \, BOP = 90 - \alpha$$
Let line $OP$ intersect line $AD$ at point $A^*$. Then triangle $\Delta \, A^*DP$ has angles
$$\angle\, A^*DP = \angle \, ADP = \alpha \,\, \text{ and } \,\, \angle \, A^*PD = \angle \, OPB = 90 - \alpha$$
Therefore
$$\angle\, DA^*P = 180 - \angle \, A^*DP - \angle \, A^*PD = 180 - \alpha - (90 - \alpha) = 90$$ i.e. the line $$OP \, \text{ is perpendicular to } \, AD$$ However, by assumption, the line $OA$ is also perpendicular to $AD$. Since there can be only one line passing through point $O$ and perpendicular to the line $AD$, the two lines $OP$ and $OA$ coinside, i.e. the points
$$O, \, \, P, \,\, A \,\, \text{ are collinear } \,\,$$ This means that point $O$ lies on the line $AP$. However, point $C$ also lies on the line $AP$, which means that point $$O, \, \, P, \,\, A, \,\, C\,\, \text{ are collinear } \,\,$$
But since $O$ is the circumcenter of triangle $\Delta \, PBC$ and it lies on the triangle's edge $CP$, the edge $CP$ itself is the diameter of the circumcircle of $\Delta\, PBC$ so its radius is
$$OC = OP = OB = \frac{1}{2} \,  CP = \frac{1}{2} \, 6 = 3$$


问:
Joint CDF of 2 Unif(0,1) that are based on 3 other Unif(0,1)?
Let $A,B,C$ be iid Unif(0,1). Let $X,Y$ be random variables:

$X=(A-B)1_{A-B>0}+(1+(A-B))1_{A-B<0}$

$Y=(C-B)1_{C-B>0}+(1+(C-B))1_{C-B<0}$

I am able to show that $X,Y$ are id Unif(0,1). My problem is showing they are iid (i.e. I'm missing the independent).

(I'm not allowed to use measure theory here, but I actually don't see how I would anyway since both $X$ and $Y$ have a '$B$' in the formula.)

Okay so elementary probability stuff only. Let's compute joint cdf and hope it's uniform on unit square. This is

$$P(X \le x, Y \le y) = 1_{x,y > 1} + x1_{0 < x < 1, y > 1} + y1_{0 < y < 1, x > 1} + xy1_{0 < x,y < 1}$$

I believe I get everything except the $xy1_{0 < x,y < 1}$ part.

It seems we have to take cases

Case 1:$ X=A-B, Y=C-B$

Case 2: $X=(A-B)+1, Y=C-B$

Case 3: $X=A-B, Y=(C-B)+1$

Case 4: $X=(A-B)+1, Y=(C-B)+1$

Okay let's try Case 1. (Update: The bounds are wrong, but the questions in re the tags are still valid, I believe.)

$$P(0 < X = A - B \le x, 0 < Y = C - B \le y)$$

What I think is conditional probability but of 2 random variables conditioned on 1. Instead of the usual $$P(z_1 < Z < z_2 | B=b) := \int_{z_1}^{z_2} f_{Z|B=b}(z) dz,$$ with $f_{Z|B=b}(z)=\frac{f_{Z,B}(z,b)}{f_B(b)},$ it looks like we'll have like $$P(z_1 < Z < z_2, u_1 < U < u_2 | B=b) := \int_{u_1}^{u_2} \int_{z_1}^{z_2} f_{(Z,U)|B=b}(z,u) dz du,$$ with $f_{(Z,U)|B=b}(z,u) = $, I think, $\frac{f_{Z,U,B}(z,b,u)}{f_B(b)}$

Note: if any of the definitions ':=' are in fact not definitions, then you'll have to explain conditioning on an event of probability zero to me please.

So here's what I think is next:

$$P(0 < X = A - B \le x, 0 < Y = C - B \le y) = P(B < A \le x+B, B < C \le y+B)$$

$$ = \int_{b=0}^{b=1} P(B < A \le x+B, B < C \le y+B | B=b) f_B(b) db \tag{1?}$$

$$ = \int_{b=0}^{b=1} P(b < A \le x+b, b < C \le y+b | B=b) f_B(b) db \tag{2????}$$

$$ = \int_{b=0}^{b=1} \int_{b}^{x+b} \int_{b}^{y+b} f_{(A,C)|B=b}(a,c) dc da f_B(b) db \tag{3 part 1?}$$

$$ = \int_{b=0}^{b=1} \int_{b}^{x+b} \int_{b}^{y+b} \frac{f_{(A,C,B)}(a,c,b)}{f_B(b)} dc da f_B(b) db \tag{3 part 2?}$$

$$ \text{[details omitted because actually the bounds are wrong]}$$

$$ \text{[details omitted because actually the bounds are wrong]}$$

$$ \text{[details omitted because actually the bounds are wrong]}$$

$$ = xy $$

And then assuming all of the above is correct and all the question mark parts are justified, repeat for the other 3 cases and it looks like we have $xy$ in each. Are these cases supposed to be added up and so I'm missing $\frac14$? Or what?

Case 2: $b-1<A<x+b-1, b<C<y+b$, so again just $xy$

Case 3: $b<A<x+b, b-1<C<y+b-1$, so again just $xy$

Case 4: $b-1<A<x+b-1, b-1<C<y+b-1$, so again just $xy$

About the question marks:

For $(1?)$, I think the rule is like for an event $E$ and continuous random variable $B$, we have $P(E)=\int_{\mathbb R} P(E|B=b) f_B(b) db$. Is this correct?

Oh wait a minute wiki says we can't quite do this. What I understand is that we can't do it for arbitrary $E$, but we can do it when (but not only when I guess) $E=\{Y \in \ \text{some interval or Borel set I guess}\}$, for some continuous random variable $Y$ s.t. the joint pdf $f_{X,Y}$ is well-defined? (I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf.)

For $(2????)$, I think we're doing something like for events $E$, $H$ and $G$ and continuous random variable $B$: we have $P(E|H)=P(E \cap H|H)$, but $P(G|H)$ is defined only for $P(H)>0$. What is being done here when technically $P(H)=0$? I mean of course in the 1st place when we say like '$P(E|B=b)$', this is notational, we're not really conditioning on the $P$-null event $\{B=b\}$. But I still don't get exactly what's being done here.

For $(3 \ \text{parts 1 and 2})$, I'm actually just guessing here, what's the definition of conditional joint cdf of 2 random variables given a 3rd? And please provide a reference.

Wiki just says $F_{(X,Y)|Z=z}(x,y):=P(X \le x, Y \le y|Z=z)$, but it doesn't quite define $P(X \le x, Y \le y|Z=z)$.

For just 1 continuous random variable conditioned on 1 continuous random variable, it's $P(X \le x, |Z=z) := \int_{-\infty}^{x} f_{X|Z=z}(t) dt$, where $f_{X|Z=z}(t) := \frac{f_{(X,Z)}(x,z)}{f_{Z}(z)}$.

For 2 continuous random variables conditioned on 1 continuous random variable, I think it's $P(X \le x, Y \le y|Z=z) := \int_{-\infty}^{x} \int_{-\infty}^{y} f_{(X,Y)|Z=z}(t,u) du dt$, but then...

What's $f_{(X,Y)|Z=z}(t,u)$? (I guess we do the elementary probability way of thinking: define the pdf before the cdf...) According to this site (see problems 1 and 16), it's $f_{(X,Y)|Z=z}(t,u): = \frac{f_{(X,Y,Z)}(t,u,z)}{f_Z(z)}$. So, I guess I'm right about joint cdf/pdf stuff. I'm just hoping for a reference please.

答:
Note that $X = A - B \pmod 1$ and $Y = C - B \pmod 1$.

Now let $\alpha, \beta \in \mathbb R$ be some fixed numbers. Let $A_{\alpha} = A + \alpha$ and $C_{\beta} = C + \beta$.  With these, form $X_{\alpha} = A_{\alpha} - B \pmod 1$ and $Y_{\beta} = C_{\beta} - B \pmod 1$.

Then on the one hand clearly the joint distribution of $(X_{\alpha}, Y_{\beta})$ is equal to that of $(X, Y)$ shifted cyclically $\pmod 1$ over the offset $(\alpha, \beta)$.

On the other hand $A_{\alpha} \pmod 1$, $B$, $C_{\beta} \pmod 1$ are independent and uniform on $[0,1]$.  Therefore $(X_{\alpha}, Y_{\beta})$ and $(X, Y)$ have the same distribution.  Both observations combined show that this joint distribution is invariant under shifts $\pmod 1$ and is therefore uniform.


问:
How to find the maximum value of $f(x,y) = 39 - x^2 - y^2$ on the line $x+6y=37$?
so I'm totally lost how to find this question

find the maximum value of $f(x,y) = 39 - x^2 - y^2$ on the line $x+6y=37$

I know how to find local max and min of multivariable equations, but now I don't know what to do given that line equation.

If someone could help me it would be very much appreciated

Thanks in advance

答:
Lagrange multipliers are a nice way to solve any problem of the form: Find the [maximum/minimum/extrema] of [a function] along [a curve].

In this case, we are trying to find the maximum of the function $f(x,y)=39-x^2-y^2$ over the curve $x+6y=37$.

The first step to solve anything using Lagrange multipliers is to find a new function $g(x,y)$ such that our curve, $x+6y=37$, is the level curve $g(x,y)=0$. The method works for any such function, but an easy way to find a good function is to simply move all the terms to the left and make that our $g$:

$
x+6y=37
\\x+6y-37=0
\\g(x,y)=x+6y-37
$

It is clear that this function satisfies our condition because if we set $g(x,y)=x+6y-37=0$, we can rearrange to get $x+6y=37$.

Next comes the "multiplier" part of Lagrange multipliers. Lagrange tells us that any extrema of $f$ along our curve must have the property $\triangledown f(x,y)=\lambda\triangledown g(x,y)$, so every point satisfying that condition along our curve may be (but is not guaranteed to be) an extrema of $f$.

To find these points, we must find the gradient vectors $\triangledown f(x,y)$ and $\triangledown g(x,y)$:

$
f_x(x,y)=\frac{\partial}{\partial x}\left[29-x^2-y^2\right]=-2x \\
f_y(x,y)=\frac{\partial}{\partial y}\left[29-x^2-y^2\right]=-2y \\
\triangledown f(x,y)=(f_x(x,y),f_y(x,y))=(-2x,-2y)
$

$
g_x(x,y)=\frac{\partial}{\partial x}\left[x+6y-37\right]=1 \\
g_y(x,y)=\frac{\partial}{\partial y}\left[x+6y-37\right]=6 \\
\triangledown g(x,y)=(g_x(x,y),g_y(x,y))=(1,6)
$

Now we can simply solve for points satisfying $\triangledown f(x,y)=\lambda\triangledown g(x,y)$ which are along our curve:

$
\begin{cases}
-2x=\lambda \\
-2y=6\lambda \\
x+6y=37
\end{cases}
$

This gives us $\lambda=-2$, $x=1$, and $y=6$. The value of $\lambda$ doesn't really matter for our purposes, but this tells us that $(x,y)=(1,6)$ is a potential extrema of $f$ along this curve. If our domain curve weren't infinitely long, the ends of our curve would also be potential extrema, but that's not an issue here.

Since we only generated one point from our Lagrange multiplier, either $f(1,6)$ is the maximum of $f$ along this curve and there is no minimum or vice versa. We can easily check by finding that $f(1,6)=2$ and checking the value of $f$ for some other point along the curve. For example, $f(7,5)=-35<f(1,6)$ so $f(1,6)=2$ must be the maximum of $f$ along the curve $x+6y=37$.

P.S. Sorry for the late reply, hopefully this will help someone in the future.


问:
Prob. 57, Chap. 1, in Royden's REAL ANALYSIS: The set of points at which a sequence of continuous functions converges is a ...
Here is Prob. 57, Chap. 1, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

Let $\left\{ f_n \right\}$ be a sequence of continuous real-valued functions defined on $\mathbb{R}$. Show that the set of points $x$ at which the sequence $\left\{ f_n(x) \right\}$ converges to a real number is the intersection of a countable collection of $F_\sigma$-sets.

And, an $F_\sigma$-set is a set that can be expressed as the union of countably many closed sets.

My Attempt:

Let $S$ be the set of all the real numbers $x$ for which the sequence $\left\{ f_n(x) \right\}$ converges to some real number. Let us now define the function $f \colon S \longrightarrow \mathbb{R}$ by the formula
$$
f(x) := \lim_{n \to \infty} f_n(x).
$$

Is the way I've started out likely to reach the desired conclusion? If so, then how to proceed from here?

答:
With problems like this, it is usually more helpful to express the definition of convergence in terms of unions and intersections rather than defining a function on your set. I'll include the full solution below so be warned.

Let $S$ denote the set of points $x$ where $\lim_{n \to \infty} f(x)$ exists. Since a sequence of real numbers is convergent if and only if it is cauchy, we will be using the cauchy condition. As such, $x \in S$ if and only if for all $\epsilon > 0$ there is some $N \in \mathbb{N}$ with $|f_n(x) - f_m(x)| < \epsilon$ for all $n,m \geq N$. We can then restate this as $x \in S$ if and only if for all $M \in \mathbb{N}$ there is some $N$ with $|f_n(x) - f_m(x)| \leq 1/M$ for all $n,m \geq N$. Translating this into unions and intersections we get that $$S = \bigcap_{M \in \mathbb{N}} \bigcup_{N \in \mathbb{N}}\bigcap_{m,n \geq N} |f_n - f_m|^{-1}([0, 1/M]).$$ Since $|f_n - f_m|$ is continuous and $[0,1/M]$ is closed, it follows that $\bigcap_{m,n \geq N} |f_n - f_m|^{-1}([0, 1/M])$ is closed. Hence, $S$ is an intersection of $F_{\sigma}$ sets.


问:
Solving the initial value problem for PDE

Let consider the initial value problem given by $$xu_y-yu_x=u\quad\text{with}\ u(x,0)=h(x).$$ The answer that my teacher put on our exercises is $u=h(\sqrt{x^2+y^2})e^{\arctan(y/x)}$.

How I get the answer? I made some computations here and I find $$u=h\left(\frac{x^2-y^2}{x}\right)e^{y/x}.$$ Probably I'm wrong, but I can't see how to obtain the desired answer. Appreciate any help! :D

Remark: If someome wants to see my computations, I can put it here later.

答:
General solution of pde is
$$u=F(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
Then solution of initial value problem is
$$u=h(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$

In polar coordinates $$x=r\cos\phi,\quad y=r\sin\phi$$ pde is
$$\frac{\partial u(r,\phi)}{\partial\phi}=u(r,\phi)$$
with solution
$$u=F(r)e^\phi$$
If substitute $r=\sqrt{x^2+y^2}$, $\phi=\arctan\frac{y}{x}$ we get general solution.


问:
Expansion of fraction
Is there an expansion for
$$ (x+y)^p \approx x^p + y^p + \cdots $$
for $1<p<2$?

答:
Let $\alpha\in\mathbb{C}$. The series $$(x+y)^{\alpha}
=\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{\alpha - k}y^{k},$$
where
$$\binom{\alpha}{k} = \frac{\alpha (\alpha - 1)(\alpha - 2) \dotsm (\alpha - k + 1)}{k!}.$$
converges for all $x,y\in \mathbb {R}$ with $x>0$ and $\left|{\tfrac {y}{x}}\right|<1$. In the special case $k=0$ we get an empty product, which is defined to be $1$.

Reference: https://en.wikipedia.org/wiki/Binomial_series

Title: Expansion of fraction


问:
Definition of a locally Euclidean space
According to Wikipedia: A topological space $(X,\tau)$ is called locally Euclidean if there is a non-negative integer $n$ such that every point $p$ in $X$ has a neighborhood $N$ which is homeomorphic to $\mathbb{R} ^n.$

But I am not sure what this means to say that $N$ is homeomorphic to $\mathbb{R} ^n$. Does it mean that the subspace topology of $(X,\tau)$ on $N \subseteq X$ is homeomorphic to $\mathbb{R} ^n$ (with Euclidean topology) ?

答:
Yes, that is exactly what it means. In general, we often suppress mention of the subspace topology when referring to a set as a topological space if that set is a subset of exactly one topological space considered so far.

Incidentally, I have a slight stylistic quibble with that definition. In my opinion the following is more natural:

$(X,\tau)$ is locally Euclidean iff for each $x\in X$ there is some $U\in\tau$ with $x\in U$ and some $V\subseteq\mathbb{R}^n$ which is open in the usual topology and $U$ with the subspace topology from $(X,\tau)$ is homeomorphic to $V$ with the subspace topology from $\mathbb{R}^n$.

It's a good exercise to show that this is equivalent to the definition given above, the point being that every open set in $\mathbb{R}^n$ contains an open set homeomorphic to $\mathbb{R}^n$. One advantage of this second definition is that it is frequently more convenient (it's a priori easier to show). But there's a more substantive point as well. Let's say we want a general notion of "$(X,\tau)$ is locally homeomorphic to $(Y,\sigma)$." In general, $(Y,\sigma)$ may not have the same "self-similarity" property of $\mathbb{R}^n$ which makes the two definitions of "locally Euclidean" equivalent. So we get two inequivalent candidates for a general notion of "locally homeomorphic" here. And to my mind, the second is nicer; note for example that it yields an equivalence relation on topological spaces.

Title: Definition of a locally Euclidean space


问:
The meaning of $dx$ in an indefinite integral
This semester I'm taking integral calculus for the first time. We started with the differential (i.e. $dy=f'(x)\,dx$) and right after that with the indefinite integral. Since then, I've been trying to make sense of the $dx$ when it's part of an indefinite integral (i.e. $\int f(x) \, \boldsymbol{dx}$). I know there are already a bazillion answers regarding this question, but all of them refer to the definite integral and the ones that do touch on indefinite integrals only say things like "it's just a syntactical device to tell you the variable to differentiate with respect to or the integration variable" (Ihf, 2012, web). I don't like this answer, especially because I was taught that given two functions $f(x)$, $g(x)$ and the antiderivative of their product $\int f(x)g(x)\,dx$, if I were to assign $f(x)$ to $u$ and $g(x)\,dx$ to $dv$ in order to integrate by parts, then I would have to integrate $dv$ to find $v$. This only makes sense if $dx$ means something by itself and is not just a quirk of the notation, otherwise we would end up with something like the following: $$\int g(x)\,dx\space dx$$

After thinking about it a lot, I believe I finally found a way to make sense of the $dx$ as something that isn't purely and simply a notation device. My reasoning is as follows:

Given a function $f(x)$, let $y=f(x)$. Then $$\frac{dy}{dx}=f'(x)$$

Then from the fundamental theorem of calculus, we know that $$\int\frac{dy}{dx}\,dx=y$$

Let $dy=f'(x)\,dx$. Then the equation $$dy=\frac{dy}{dx}\,dx$$ holds and $$\int\frac{dy}{dx}\,dx=\int dy=y$$

Finally, integrate both sides of the equation $dy=f'(x)\,dx$ in order that $$\int dy=\int f'(x)\,dx\implies y=f(x)$$

From this I conclude that, by integrating a function, what we are really doing is integrating the differential of that function. This makes perfect sense to me, although I'm aware that seemingly logical things aren't necessarily logical. That's why I would appreciate it if someone could tell me whether the above is mathematically correct or pure gibberish.

PS, I had never written a mathematical proof before and I have taken no proofs courses yet, so any suggestions are welcome.

Reference: lhf. (2012, May 9). What does $dx$ mean?. Mathematics Stack Exchange. https://math.stackexchange.com/q/143262

答:
An explanation is given in section 2.9 Fundamental Theorems of the Calculus in Introduction to Calculus and Analysis I by R. Courant and F. John.

It is quite customary to use a notation which is not perfectly clear without comment: we write
\begin{align*}
\color{blue}{F(x) = \int f(x)\,dx}
\end{align*}
when we mean that the function $F(x)$ is of the form
\begin{align*}
F(x) = c + \int_a^xf(u)\,du
\end{align*}
for suitable constants $c$ and $a$, that is, we omit the upper limit $x$, the lower limit $a$ and the additive constant $c$ and use the letter $x$ for the variable of integration.

Strictly speaking, of course, there is a slight inconsistency in using the same letter for the variable of integration and the upper limit $x$ which is the independent variable in $F(x)$. In using
the notation $\int f(x)\,dx$ we must never lose sight of the indeterminacy connected with it, that is, the fact that the symbol always denotes one of the primitive functions of $f$ only. The formula $F(x) = \int f(x)\,dx$ is
just a symbolic way of writing the relation
\begin{align*}
\color{blue}{\frac{d}{dx}F(x)=f(x)}.
\end{align*}


问:
Riemann problem for Burgers' equation with both shock waves and rarefaction waves.
Given the inviscid Burgers' equation with piecewise initial data
$$
u_t + u u_x = 0,\qquad
u(x,0) = \left\lbrace
\begin{aligned}
   &0 && \text{if } x<-1 \\
   &2 && \text{if } {-1}<x<0 \\
   &4   && \text{if } {0}<x<1 \\
   &2 && \text{if } 1\leq x<2 \\
   &0   && \text{if } x\geq 2
\end{aligned}\right.
$$

I want to find the exact solution of this problem represented in a single piecewise function.

Progress and Confusion

I know the Riemann problem has exact solution for rarefaction waves and shock waves respectively.
So for region $(-1,1)$ at $x=-1$ and $x=0$, characteristics separate, I  will obtain the rarefaction waves and the solution will be in same form within different spatial region as following:
$$
u(x,t) = \left\lbrace
\begin{aligned}
   &0   && \text{if } x<0 \\
   &\frac{x}{t} && \text{if } {0}<x<2t \\
   &2   && \text{if } x>2t \\
\end{aligned}\right.
,and\;\;\;
u(x,t) = \left\lbrace
\begin{aligned}
   &2   && \text{if } x<2t \\
   &\frac{x}{t} && \text{if } {2t}<x<4t \\
   &4   && \text{if } x>4t \\
\end{aligned}\right.
$$
and the shock wave solution will occur in region (1,2) and also be in same form within different spatial region as following:
$$
u(x,t) = \left\lbrace
\begin{aligned}
   &4   && \text{if } x<3t \\
   &2   && \text{if } x>3t \\
\end{aligned}\right.
,and\;\;\;
u(x,t) = \left\lbrace
\begin{aligned}
   &2   && \text{if } x<t \\
   &0   && \text{if } x>t \\
\end{aligned}\right.
$$

Those different solution profiles have clearly overleaping in the spatial region, I am not sure what will happen in those overleaping area, and how could I condense them into a single piecewise function to represent the solution for entire region.

For the rarefaction waves case, according to the spatial region profile, I guess the first two rarefaction waves will merge together at first as following:
$$
u(x,t) = \left\lbrace
\begin{aligned}
   &0   && \text{if } x<0 \\
   &\frac{x}{t} && \text{if } {0}<x<4t \\
   &4   && \text{if } x>4t \\
\end{aligned}\right.
$$
But for the two shocks, since the first shock is faster than second, the first shock seems  to swallow up the second shock, and therefore only the first shock waves left:
$$
u(x,t) = \left\lbrace
\begin{aligned}
   &4   && \text{if } x<3t \\
   &0   && \text{if } x>3t
\end{aligned}\right.
$$
Am I got correct in those intermediate state? And also, what will be the case when the rarefaction waves meet with the shock waves?

答:
The problem is somewhat similar to this one for which the basic concepts are explained here. An important aspect is that we need to apply a translation to the Riemann solution. In fact, standard Riemann problems are defined for discontinuities located at $x=0$ only. In other words, for small positive times, the entropy solution reads
$$
u(x,t) =\left\lbrace
\begin{aligned}
&0, && x\leq -1\\
&\tfrac{x+1}{t}, && {-1} \leq x \leq -1+2t\\
&2, && {-1+2t}\leq x\leq 2t\\
&\tfrac{x}{t}, && 2t \leq x \leq 4t\\
&4, && 4t\leq x < 1+3t \\
&2, && 1+3t <x< 2+t \\
&0, && 2+t < x
\end{aligned}\right.
$$
Note that the solution is continuous across rarefactions, and that the initial data must be recovered when setting $t=0$ in the solution's expression. The solution's sub-domains should not overlap at small times, but one notes that there exists a positive time at which the two shock waves interact, and another one at which the rarefaction part interacts with a shock.


问:
Number of generating elements in a finite cyclic group
Let $G$ is finite cyclic group of order  $37$. Then the number of elements in the set  $\{g\in G\mid \langle g\rangle=G\}$  is equal to?

Is it that easy to say that it is 36? Or there is a theory behind it?

答:
In a cyclic group of order $n$, the number of elements which generate the group is given by $\phi(n)$, where $\phi$ denotes the Euler's phi function.

Given any $n$, $\phi(n)$ provides the number of primes to $n$.

Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}$ be the prime factorization of $n$, $p_1,p_2,\dots, p_n$ are distinct primes and $\alpha_1, \alpha_2, \dots, \alpha_n\in \mathbb{N}$; then

$\phi(n)=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}(1-\frac{1}{p_1})(1-\frac{1}{p_2})\dots (1-\frac{1}{p_n})$.

For your example:

$\phi(37)=37(1-\frac{1}{37})=36.$


问:
Normal between two lines
$$l_1:x=(-1,1,2)+t(1,-3,-2)$$
$$l_2:y=(1,1,1)+s(-2,6,-4)$$
Q: Find the equation of plane passing through $l_1$ and $l_2$.

My teacher wrote answer like this:

Since they're parallel, choose vector $u=(1,-3,-2)$ and choose normal vector $n=u$x$(x_0-y_0)=(-3,-5,6)$. Can someone please explain me how did he find $n=(-3,-5,6)$.

No matter how I choose vectors $x_0,y_0$ it gives $n=(-3,3,-6)$. Is there a way to obtain $n=(-3,-5,6)$?

Thanks!

答:
My teacher assigned us this problem in my multivariable calculus class and a classmate and I came to the conclusion that there must be a typo in the problem, because the two lines are skew.
  Using the two lines
$$
_1:=(−1,1,2)+(1,−3,−2)
$$
$$
_2:=(1,1,1)+(−2,6,−4)
$$
We can prove these are skew by first noticing the slope vectors are linearly independent, so the lines cannot be parallel. Next, to prove the lines don't intersect either, we parameterize the two lines to $l_1(t)=(t-1,-3t+1,-2t+2)$ and $l_2(t)=(-2t+1,6t+1,-4t+1)$. Then we can set each $x(t)$, $y(t)$, and $z(t)$ to the other to find where they intersect. This gives 3 $t$ values, $t=2/3,0,-1/2$. Because the 3 $t$ values are not equal, there isn't any one point where the two lines intersect, so the lines are skew.

Alternatively, if we use lines that we assume are the intended lines,
$$
_1:=(−1,1,2)+(1,−3,−2)
$$
$$
_2:=(1,1,1)+(−2,6,4)
$$
These are parallel and we can find a plane that passes through both lines. First we have to find 4 points and subtract the difference to produce 2 vectors later used to find the normal of the plane. The two vectors in between the two lines, when crossed, will produce the vector normal.

To find 4 points, we can first set $t$ and $s$ to 0, and then to 1.
We get $(-1,1,2)$ and $(1,1,1)$ when $t,s=0$ and $(0,-2,0)$ and $(-1,7,5)$ when $t,s=1$.

Subtracting, we get two vectors $(-2,0,1)$ and $(1,-9,-5)$.
Now to find the normal, we cross the two vectors and get $n=(9,-9,18)$, which is parallel to $(−3,3,−6)$.

 Now we use $n=(A,B,C)$ and one of the initial points (we'll use $(1,1,1)$ for simplicity) and plug them into the equation for the plane passing through both lines, $Ax + By + Cz = n \cdot $(1,1,1)$ $

This leaves us with $9x - 9y +18z = 18$.

Title: Normal between two lines


问:
Prob. 48, Chap. 1, in Royden's REAL ANALYSIS: At what points is this function continuous?
Here is Prob. 48, Chap. 1, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

Define the real-valued function $f$ on $\mathbb{R}$ by setting
$$
f(x) = \begin{cases} x & \mbox{ if $x$ is irrational} \\ p \sin \frac{1}{q} & \mbox{ if $x = \frac{p}{q}$ is in lowest terms}. \end{cases}
$$
At what points is $f$ continuous?

I think the sequential criterion for continuity is what we will have to apply here.

The function is not continuous at $x = 0 = \frac{0}{1}$, because $f(0) = 0$, but if we consider the sequence $\left( \frac{1}{2^n} \right)_{n \in \mathbb{N}}$ converging to $0$, then the image sequence
$\left( \sin \frac{1}{2^n} \right)_{n \in \mathbb{N} }$ does not converge to $f(0)$.

Am I right?

How to check if $f$ is or is not continuous at a point $a \neq 0$?

答:
The function is continuous at all irrationals and at $0$, and discontinuous at all nonzero rationals.

In order to prove the last assertion, we can rely on a sequential argument. It is sufficient to produce a sequence that converges to a nonzero rational but whose image does not converge to the image of said rational. To this end, let $\xi = p/q \in \mathbb Q \setminus \{0\}$ be written in minimal terms, and let $y$ be irrational. Define the sequence
$$x_n = \xi + \frac y n; $$
then $x_n$ is irrational for all $n$ and $x_n \xrightarrow{n\to\infty} \ \xi$. However,
$$\lim_{n\to\infty} f(x_n) = \lim_{n\to\infty}  x_n = \lim_{n\to\infty}  \left(\xi + \frac y n\right) = \xi = \frac p q \color{red}\neq p\sin\left(\frac 1 q\right) = f(\xi). $$

Now let $x \notin \mathbb Q$ and $\varepsilon > 0$. Suppose for simplicity that $x>0$ (the argument can be adapted for negative irrationals). We need to make a few observations first:

If $p/q \in \omega$ for some bounded interval $\omega$, then clearly
$$\left|\frac{p}{q}\right| < \sup \omega, \implies |p| < q\sup\omega.$$

For a given positive integer $q_*$, it is always possible to choose a bounded interval $\omega \ni x$ such that, for all rational numbers $p/q \in \omega$ written in minimal terms, one has that $q > q_*$. If it were not so, you could find $q_*$ such that, for all bounded intervals $\omega$ containing $x$, there is a rational $p_\omega/q_\omega \in \omega$ in minimal terms with $q_\omega \leqslant q_*$: for such a rational, by Fact 1,
$$|p_\omega| \leqslant q_*\sup\omega,$$
which sets a bilateral bound to the integer $p_\omega$. Hence there are finitely many possibilities for $p_\omega/q_\omega$, meaning that there exists a neighborhood $\omega'\subset\omega$ of $x$ that does not contain any of them (say one with a diameter strictly smaller than the minimum distance between $x$ and these rationals). Absurd!

Because $|\sin(t)|<|t|$ for all $t\neq 0$, we have by Taylor's theorem that
$$|\sin(t)-t| \leq  t^2. $$

Now consider an integer $q_*$ such that $q_* > 4x/\varepsilon$ (this condition will make sense later) and find $\omega$ as provided by Fact 2. We may shrink this neighborhood to a ball $B_\delta(x)$ with radius $\delta>0$, and if necessary we may force this $\delta$ to be smaller than $\varepsilon/2$ and $x$ (this too will make sense later). This entails that, for all $p/q \in B_\delta(x)$ in minimal terms,
$$|p| < q\sup B_\delta(x) < q (x+\delta) < 2qx ,$$
by Fact 1.

If $\xi = p/q \in B_\delta(x)$ written in minimal terms, by the triangle inequality and Fact 3, seeing as $q_* < q$,
$$\begin{split}
|f(\xi) - f(x)| &\leqslant \left|p\sin\left(\frac 1 q\right) - \frac pq\right| + \left|\frac pq - x\right| \\
&\leqslant |p|\left|\sin\left(\frac 1 q\right) - \frac 1q\right| + \delta < \frac{|p|}{q^2} + \delta \\
&< \frac{2x}{q_*} + \delta < \frac \varepsilon 2 + \frac \varepsilon 2 = \varepsilon.
\end{split}$$
If instead $y \in B_\delta(x)$ is irrational, then naturally
$$|f(y)-f(x)| < |y-x| < \delta < \frac \varepsilon 2 < \varepsilon. $$

Continuity at $0$ is proved in a very similar way.


问:
Does the following always hold $[\hat{A},\hat{B}]=0\Longrightarrow[e^{\hat{A}},\hat{B}]=0$?
I know that if two operators, $A$ and $B$ commute, then also $e^A$ and $e^B$ commute. But would the same condition imply that $A$ commutes with $e^B$?

答:
Yes, $[A,B] = 0$ implies that $[e^A,B] = 0$. How one would prove this depends on one's definition of $e^A$, but with the power series definition the proof is straightforward:
$$
\begin{align}
e^AB &=
\left(\lim_{n \to \infty} \sum_{k=0}^n \frac 1{k!}A^k\right) B =
\lim_{n \to \infty} \sum_{k=0}^n \frac 1{k!}A^k B
\\ &=
\lim_{n \to \infty} \sum_{k=0}^n \frac 1{k!}BA^k = B \left(\lim_{n \to \infty} \sum_{k=0}^n \frac 1{k!}A^k\right) = Be^A.
\end{align}
$$


问:
Integral $\frac{\sqrt{e}}{\sqrt{2\pi}}\int^\infty_{-\infty}{e^{-1/2(x-1)^2}dx}$ gives $\sqrt{e}$. How?
To calculate the expectation of $e^x$ for a standard normal distribution I eventually get, via exponential simplification:

$$\frac{\sqrt{e}}{\sqrt{2\pi}}\int^\infty_{-\infty}{e^{-1/2(x-1)^2}dx}$$

When I plug this into Wolfram Alpha I get $\sqrt e$ as the result. I'd like to know the integration step(s) or other means I could use to obtain this result on my own from the point I stopped. I am assuming that Wolfram Alpha "knew" an analytical solution since it presents $\sqrt e$ as a solution as well as the numerical value of $\sqrt e $. Thanks in advance!

答:
This is because $$\int_{\Bbb R} e^{-x^2}dx=\sqrt \pi$$

Note that your shift $x\mapsto x-1$ doesn't change the value of integral, while $x\mapsto \frac{x}{\sqrt 2}$ multiplies it by $\sqrt 2$, giving the desired result, that is,

$$\int_{\Bbb R} e^{-\frac 1 2(x-1)^2}dx=\sqrt {2\pi}$$


问:
Discontinuities of $u'$ where $u \in W^{1, \infty}([0, T])$
I was having a closer look at the Sobolev space $W^{1, \infty}([0, T])$. I was wondering if $u'$ has a representative that is continuous a.e.. I know that $W^{1, \infty}([0, T]) \hookrightarrow C([0, T])$, but does that tell me anything about the derivatives...

答: Since $v = u'$ can be any bounded measurable function, your question is equivalent to asking if every bounded measurable function $v$ has a representative that is continuous a.e.  The answer is certainly no; one counterexample is $v = 1_C$ where $C$ is a "fat Cantor set" which is nowhere dense but has positive measure.  If $w = v$ a.e. then the set where $w=0$ is still dense, and the set where $w=1$ still has positive measure, so $w$ is discontinuous at every point of that set.


问:
How to find maximum likelihood estimator of this pdf?
A random sample $X_1,X_2\ldots X_n$ is drawn from a population with pdf
$$f(x|\theta)=\frac{1}{2}(1+\theta x),\quad -1<x<1,-1<\theta<1$$
I know how to find MLE, but I feel confused in this pdf. It is obvious that it is an increasing function in terms of $\theta,$ if hold $x$ fixed. But $\theta$ is strictly less than $1.$ So I have to stop.

答:
It is obvious that it is an increasing function in terms of $\theta,$ if hold $x$ fixed.

Wrong, because $x$ can be negative. Let's write the loglikelihood:

$$L(\theta) = \log  \prod_{i=1}^{n} \frac12 (1 + \theta x_i)  \\
= - n  \log 2 + \sum_{i=1}^{n} \log(1 + \theta x_i) \tag1$$

with the domain $[-1 \le \theta \le 1]$.

(Notice that the values $x_i=0$ do not contribute, so we can exclude them).

Then $L(\theta)$ is differentiable inside $(-1,1)$, its critical points are given by $g_1(\theta)=0$ where

$$ g_1(\theta) \triangleq \frac{\partial L(\theta) }{\partial \theta}= \sum_{i=1}^{n} \frac{x_i}{1 + \theta x_i}=0 \tag2$$

Further

$$ g_2(\theta) = \frac{\partial^2 L(\theta) }{\partial \theta^2}= - \sum_{i=1}^{n} \frac{x_i^2}{(1 + \theta x_i)^2}<0 \tag 3$$

Then we can conclude that $g_1(\theta)=0$ has a single solution in the reals. Let's call it $\theta^*$.

Then the MLE is

$$\theta_{ML} = \begin{cases}
\theta^* & \text{if } -1\le \theta^* \le 1\\
-1 & \text{if } \theta^* < -1\\
1 & \text{if } \theta^* > 1\\
\end{cases} \tag 4
$$

To find $\theta^*$ we must resort to numerical methods. A simple procedure that seems to work is Newton–Raphson:

$$ \theta_{k+1} = \theta_{k} + \frac{\sum_{i=1}^{n} \frac{x_i}{1 + \theta_k x_i}}{\sum_{i=1}^{n} \frac{x_i^2}{(1 + \theta_k x_i)^2}} \tag{5}$$

Starting with $\theta_0=0$, this seems to converge quickly.


问:
Automatic additivity of multiplicative maps
There are results which guarantee that a multiplicative bijection between commutative rings is actually additive (that is, it is a ring isomorphism). For example this result of Martindale initiated this kind of research.

I am interested in the case where we consider rings of continuous functions. More specifically,

Let $K$ and $L$ be compact Hausdorff spaces. Suppose that there exists a bijection $p\colon C(K)\to C(L)$ which preserves multiplication. Is $p$ a ring isomorphism?

EDIT: The answer is no.

答:
The answer is no, as already observed, but the existence of such a bijection implies the existence of a ring isomorphism between the respective rings of continuous functions (hence also a homomorphism between the underlying compact spaces). This is a theorem of Milgram.

A.N. Milgram, Multiplicative semigroups of continuous functions, Duke Math. J. 16 (1949), 377–383.


问:
Is the integral over a convex mapping of the solution to the heat equation monotonic?
I have the following problem:

Suppose $u = u(t,x)$ solves the Cauchy-Problem for the (one-dimensional) heat equation: $$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$$ $$u(0,x) = g(x) $$ and satisifes $\int_{\mathbb{R}} u(t,x) \,\text{d}x = 0$ for all $t \geq 0$. Furthermore, define for a convex function $\phi$ satisfying $\phi(0) = 0$ the following:  $$F(t) := \int_{\mathbb{R}} \phi (u(t,x)) \,\text{d}x $$
The goal is to show that $F$ is monotonic (I suppose here that it is decreasing).

What I have tried so far:

The key to this problem seems to be Jensen's inequality, since it relates integrals and convex functions. Therefore, define $T>t$ and apply the inequality:

$$
F(T) - F(t) \geq \phi\left(\int_{\mathbb{R}} u(T,x) \,\text{d}x\right) - F(t) = - F(t)
$$

The minuend being zero by assumtions on $u$ and $\phi$. I don't know how to follow from here. I have also tried representing $u$ with the heat kernel, i.e.:
$$ u(t,x) = \int_{\mathbb{R}} \frac{1}{\sqrt{4\pi t}} \exp{\left\{{\frac{-(x-y)^2}{4t}}\right\}} g(y) \,\text{d}y$$

but I am not able to follow from there either. Any help is appreciated!

答:
only a partial answer

If $\phi$ is twice differentiable, you can use $\lim_{x\to\pm\infty}\partial_xu(t,x)=0$ and $\lim_{x\to\pm\infty}u(t,x)=0$ to show it by partial integration
$$F'(t)=\bigg[\partial_xu(t,x)\phi'(u(t,x))\bigg]_{x=-\infty}^\infty-\int_{\Bbb R}\phi''(u(t,x))\big(\partial_xu(t,x)\big)^2\Bbb dx\leq0.
$$


问:
Limit of Newton's Method on polynomial $Ax^4 + Bx^3+ Cx^2 + Dx + E$?
So if you took the function $f(x) = Ax^4 + Bx^3+ Cx^2 + Dx + E$ and did Newton's Method repeatedly, you would get a sequence that converges to at most $4$ roots. I was wondering what would happen if you took the limit of that sequence, would you get the formula for the roots of a quartic equation? And what if you do it for a polynomial of nth degree. I know it's been proven that there can't be an explicit formula for polynomials of degree higher than $4$ so I'm not sure what would happen if you tried to find it through Newton's Method. Would it not diverge because there are multiple roots and the initial term (the guess) determines which root you find? But shouldn't that just get you a formula for the root in terms of the initial guess? I tried looking at the first few terms for Newton's Method but they became too much to work with so I was wondering if there was a program that could do Newton's Method for functions with unknown coefficients. I think I could do it with Python but it would be a lot of work to do it from scratch so if there's a python library that can support operations with variables, I would then just have to write a program for Newton's Method.

Body:
If the initial point $x_0$ is close enough to a root, the iterations will indeed converge to that root.  However, the formulas for successive iterations (as expressions in the coefficients $A,B,C,D,E$ and $x_0$) just get more and more complicated, and there is no clear way to get from them to a formula for the root.  For a very simple example,
take $f(x) = x^4 - A$; the Newton iteration is the map $$N(x) =  \frac{A+3x^4}{4 x^3}$$
The next couple of iterations give you
$$ \eqalign{N(N(x)) &= \frac{243 x^{16}+580 A x^{12}+162 A^2 x^{8}+36 A^3 x^{4}+3 A^4}{16 x^{3} \left(3 x^{4}+A\right)^{3}}\cr
N(N(N(x))) &= \frac{10460353203 x^{64}+134696910096 A x^{60}+524761849512 A^{2} x^{56}+1030291232688 A^{3} x^{52}+1172817593940 A^{4} x^{48}+847334588688 A^{5} x^{44}+439534043928 A^{6} x^{40}+171389329200 A^{7} x^{36}+51770245554 A^{8} x^{32}+12287794608 A^{9} x^{28}+2311181208 A^{10} x^{24}+346123152 A^{11} x^{20}+41168340 A^{12} x^{16}+3822640 A^{13} x^{12}+262440 A^{14} x^{8}+11664 A^{15} x^{4}+243 A^{16}}{64 x^{3} \left(243 x^{16}+580 A x^{12}+162 A^{2} x^{8}+36 A^{3} x^{4}+3 A^{4}\right)^{3} \left(3 x^{4}+A \right)^{3}}
}$$
and you really don't want to see $N(N(N(N(x))))$.


问:
is there any statistical test for independence in this situation?
I wonder if there is a statistical test which fits my need.

Let $X=(X_1,...,X_n) ,Y=(Y_1,...,Y_n)$ be $n$-dimensional random vectors.

Is there any statistical test to check whether $X_1,...,X_n$ are more mutually independent than $Y_1,...,Y_n$?

答:
As @Robert Israel already stated, this depends on what you mean by "more mutually independent".

In the strict sense of probability theory, random variables are either independent or not. So, the only way $X_1,X_2$ can be more mutually independent than $Y_1,Y_2$ is if the former are independent and the latter aren't.

But if you formally define "independence" to be some quantitative criterion, this will generally get you a test.
Which definition is the right one depends on your problem.

For example, you could be interested in whether the mean-squared correlation in the first group is larger than in the second.
You might use the following statistic:
$$
T=\frac{2}{n(n-1)}\left(\sum_{i=1}^n\sum_{j\neq i}E[(X_i-EX_i)(X_j-EX_j)]^2-\sum_{i=1}^n\sum_{j\neq i}E[(Y_i-EY_i)(Y_j-EY_j)]^2\right)
$$
The empirical version of $T$, say $t$, can be calculated readily.
If $t$ is significantly different from zero, you could say with high probability that one group is more dependent than the other, according to this definition.
Which distribution it follows under $H_0$ depends on your model.

In general, by quantitatively defining what it means to be "more independent", you create a mapping from the random variables $(X_1,\dots,X_n)$ into $\mathbb{R}$.
So you'll get two "independence scores" $T(X_1,\dots,X_n)$ and $T(Y_1,\dots,Y_n)$, one of which will be (weakly) larger.
To test whether this difference is significant, you could then perform a statistical test under distributional assumptions adequate to your situation.


问:
I need to show that this complex series converges in C
I, unfortunately, don't really know where to start. I know I can't use the ratio test. Should I try to split it up into a complex and a real part?

Translation: Show that for all $s\in\Bbb C\;$ such that $\operatorname{Re}s>0\;$ , the series $$h(x)=\sum_{k=0}^\infty\left[(4k+2)^{-s}-(4k+1)^{-s}\right]$$ converges in $\Bbb C$.

Thanks in advance for any suggestions.

答:
Some ideas for you:

First, observe that for any positive real $\;a\;$ , is we write $\;s=x+iy,\,x,y\in\Bbb R\;$, then

$$a^s=e^{(x+iy)\log a}=e^{x\log a}x^{iy\log a}\implies |x^a|=e^{x\log a}=a^x=a^{\text{Re}\,s}$$

Thus, we get that

$$=\left|\sum_{k=0}^\infty\left[(4k+2)^{-s}-(4k+1)^{-s}\right]\right|\le\sum_{k=0}^\infty\left[\frac1{|4k+2|^s}+\frac1{|4k+1|^s}\right]=$$

$$\sum_{k=0}^\infty\left[\frac1{(4k+2)^x}+\frac1{(4k+1)^x}\right]\;,\;\;s=x+iy\;,\;\;x,y,\in\Bbb R\,,\,\,x>0$$

and observe this last series is a positive one...Try to get it from here now.


问:
Math Competition Question on Forming a Triangle from a broken stick
This is a question from a Math-Competition. As the test is multiple choice and the authors only publishe what the correct answer is, but no reason as to why, I would appreciate if someone can tell me why this is correct.

We randomly break a stick in $3$ pieces. What is the probability that you can build a triangle out of those three pieces.

The correct answer: $1/4$

Why is that true? I thought about using the triangle inequality $c<a+b$, but I am not getting anywhere with that...

答:
First, we have to know how we break the stick into three pieces.  There are (at least) two different ways to do it, and I believe they will give different answers.

Assume the sticks has unit length.  We can choose two points independently uniformly at random from $[0,1]$  Assuming the two points are $0<x<y<1$, what is the probability that $x,y-x,1-y$ satisfy the triangle inequality? Look at the three possibilities for the longest, figure out the region where the triangle inequality is satisfied for each, and compute the area, by integration or geometry.  The sum of the three areas is the probability.

The second interpretation is that we break the stick in two, uniformly at random, pick one of the two pieces, uniformly at random, and break it in two, uniformly at random.  Then we ask if the triangle inequality holds.  Note that if we pick up the shorter piece, we have no hope of succeeding.

I would guess that the first interpretation is the one the examiners were looking for, but there's no way to be sure.  It's a badly-phrased question, since it asks for a probability without specifying the probability space.


问:
Show that $N(Q)=N(P)-1$ where $N(P)$ is the number of strictly positive roots
Let $P$ be a real polynomial and $N(P)$ the number of strictly postive roots of $P$

Let $n$ be an integer and suppose $P= \sum_{0}^{n} a_{k}X^{b_{k}}$ where $b_{0}=0$ and $b_{i}$ is an increasing sequence of positive integers.
A certain polynomial $Q$ exists such that $$P'=X^{b_{1}-1}Q$$
Show that $N(Q)=N(P)-1$.

It is clear that $N(Q)=N(P')$ since the roots have to be strictly positive i.e different than $0$, to show that $N(P')=N(P)-1$ I thought of using Rolle's theorem where between two roots of $P$ there exist a root of $P'$ however I need to prove no additional roots come from elsewhere. Any help is greatly appreciated!

Body:
Let $P=X^4-2X^2+2 = (X^2-1)^2+1 \ \ $. We have $\ N(P)=0$.

$P' = XQ \ \ $, with $\ Q=4(X^2-1) \ $. So $N(Q)=1 \ $.

And $N(Q)$ is not equal to $N(P)-1$ ...


问:
Prove square root inequality
Consider: $a,b,c>0$ and $a+b+c=1$, prove $$\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\leq \sqrt{21}. $$

It is supposed to be solved using Cauchy-Schwarz inequality (multiplying by $1$).

答: Hint: Use Cauchy-Schwarz inequality: $(1\cdot\sqrt{4a+1}+1\cdot\sqrt{4b+1}+1\cdot\sqrt{4c+1})^2 \le (1^2+1^2+1^2)(.....).$. Can you finish it ?

Title: Prove square root inequality


问:
Solve two multiplication equations with one multiplication operator
QUESTION

I've been wondering if it's possible to solve two multiplication equations at the same time by only being able to multiply once? It's a very odd question but it's for an algorithm that I'm working on to improve it's time complexity.

THE PROBLEM

2 * 4 = 8

3 * 5 = 15

For example I am multiplying two times to get two separate answers 8 and 15 but what I'm trying to do is multiply ONCE and get the answer for both 2 * 4 and 3 * 5. Though I want the method to work with any number that I'm trying to solve. So long as the solution doesn't require Σ (no extra looping on the equation). What I ideally want is that you can only multiply ONCE and you can add/subtract/divide/mod/etc... beforehand or afterwards.

WHAT I'VE TRIED

I've tried many things that haven't worked. One was reliable method I did was getting the median by manipulating the 4 numbers that I'm going to multiply. Though the median value has the be in between the two solution so I can subtract and add from the median so I can the the answer to both equations. Besides that I've tried other small things that didn't go anywhere because it gave me mixed results with other values.

Hopefully this gives enough context. I'd appreciate the help very much! Thank You!

答:
Expanding on Hagen von Eitzen's hint: $(ax+c)(bx+d)=abx^2+(ad+bc)x+cd$, so if you take $x$ larger than $ad+bc$ (e.g. by ensuring that $x$ has more than twice as many digits as any of the inputs), then $ab$ and $cd$ are the first and last digits of the base-$x$ representation of $(ax+c)(bx+d)$. You could implement this with bit shifting, taking $x$ to be a sufficiently large power of $2$.

This approach won't improve the asymptotic time complexity of a conventional algorithm, though, because multiplying arbitrarily large integers takes $\Omega(n)$ time in the output length and we're throwing away the inner digits of the result. It's also unlikely to be a worthwhile optimization in practice. If all you want is two multiplications, just do them separately.


问:
Is there a model of ZFC such that $\aleph_\omega^{\aleph_0}=\aleph_{\omega+2}$?
By Easton's theorem, it is consistent with ZFC that $2^{\aleph_0}=\aleph_{\omega+1}$ holds, and if we assume $2^{\aleph_0}=\aleph_{\omega+1}$ then
$$\aleph_{\omega+1}\le 2^{\aleph_0}\le \aleph_\omega^{\aleph_0}\le \aleph_{\omega+1}^{\aleph_0} =(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}=\aleph_{\omega+1}$$
So it is consistent with ZFC that $\aleph_\omega^{\aleph_0}=\aleph_{\omega+1}$, and I conjecture that $\aleph_\omega^{\aleph_0}=\aleph_{\omega+2}$ is consistent with ZFC. But how to prove it? Thanks for any help.</p>

答:
Yes. Easily (if one is familiar with forcing). Start with a model of $\sf GCH$. Now force that for every $\alpha\in\omega\cup\{\omega+1\}$:

$$2^{\aleph_\alpha}=\aleph_{\omega+2}$$

Then we have the wanted equality, by the same considerations as in your question.

(In fact, if one starts with $\sf GCH$ then we only need to force $2^{\aleph_0}=\aleph_{\omega+2}$, the rest follows.)


问:
Recursive sequence limit.

$$0<u_1<u_2 \text{ and } u_{n+2} =\sqrt{u_{n+1}. u_n}$$ Show that the limit of this sequence is ${(u_1.u_2^2)}^{1/3}$.

My attempt:
By computing individual terms I see that every term can be written in terms of $u_1$and $u_2$. Also this sequence is monotonic increasing and it also has an upper bound by the AM GM inequality. But it doesnot seem to get me anywhere. Please help.

答:
Hint :
Let $\left(v_n\right)$ be the sequence defined by $v_n=u_{n}u_{n+1}^2$ then
$$
v_{n+1}=u_{n+1}u_{n+2}^2=u_{n+1}\times\left(u_{n+1}u_n\right)=u_nu_{n+1}^2=v_n
$$
As $v_{n+1}=v_n$ for all $n$, the sequence $\left(u_nu_{n+1}^2\right)$ is constant, so for all $n$, $v_n=v_1$ which means
$$
u_nu_{n+1}^2=u_1u_2^2
$$

If you succeed in showing that the sequence $\left(u_n\right)$ converges, you have your limit :
if $u_n \underset{n \rightarrow +\infty}{\rightarrow}\ell$ then you have
$$
\ell\times \ell^2=u_1u_{2}^2 \Rightarrow \ell=\left(u_1u_2^2\right)^{1/3}
$$

Title: Recursive sequence limit.


问:
Use implicit differentiation to find an equation of the tangent line to the curve at the given point

Use implicit differentiation to find an equation of the tangent line to the curve $$x^2+xy+y^2=1$$ at  $(1,1)$.

I am not sure how I should work this out because the given point is not on the curve.

答:
Given $\ x^2 + xy + y^2 = 1$, find tangent line thru (1,1)

$\ Solve \ for \ y $

$ y = \frac{-x \pm \sqrt {4-3x^2}}{2}= \frac{-x}{2}\pm\frac{\sqrt{4-3x^2}}{2}=f(x)$

$\frac{dy}{dx}=\frac{-1}{2}\mp\frac{1.5x}{\sqrt{4-3x^2}}=f'(x)$

Let point (a, f(a)) is tangent point; for the slope, we have the equation

$\ f'(a)=\frac{f(a)-1}{a-1}$

Solve this for a , we obtain

$ a \approx -0.483 \ ; f'(a) \approx -0.10118$

The tangent line y = f '(a)(x - 1) + 1

$y = -0.10118(x - 1) + 1$

By inverse symmetry, also $y = -0.483(x - 1) + 1$

desmos pic


问:
Find the order of $\phi_{(1\,2\,4\,3)}$ in $\text{Inn}(S_5)$

We define $\phi_{(1\,2\,4\,3)}$ as $\phi_{(1\,2\,4\,3)}= (1\,2\,4\,3)x(3\,4\,2\,1)$ where $x\in \text{Inn}(S_5)$ and $\text{Inn}(S_5)=\{\phi_s: s\in S_5\}$. We are interested in finding when $(\phi_{(1\,2\,4\,3)})^n=e$.

In other words when $$(1\,2\,4\,3)x(3\,4\,2\,1)(1\,2\,4\,3)x(3\,4\,2\,1)\cdots (1\,2\,4\,3)x(3\,4\,2\,1)=e.$$ This simplifies to $x^n=e$. I'm not quite sure what to do from this point on. Clearly the identity element of $\text{Inn}(S_5)$ is just $exe=x$. But this doesn't make sense as that would imply that $e^n=e$ and there is no where to go from this.

What am I missing?

答:
You have a surjective homomorphism $\Phi: S_n \to Inn(S_n)$, $\Phi(\sigma)=\phi_{\sigma}$. As pointed out in the comments,
$$\phi_{\sigma}^n=\Phi(\sigma)^n = \Phi(\sigma^n)=\phi_{\sigma^n}.$$

Now, a general property of homomorphism implies that $|\phi_\sigma|$ divides $|\sigma|$ (in your example, $|(1243)|=4$). But in this case, we also know $\ker\Phi=Z(S_n)$. What does that tell you about $\Phi$ and $|\phi_\sigma|$?


问:
Characterisation of longitude and meridian of Knot groups
$V = V(k)$ denotes a tubular neighbourhood of the knot $k$ and $C = S^{3}− V$ is called
the complement of the knot. $H_{j}$ will denote the (singular) homology with coefficients
in $\mathbb{Z}$

Choose the homology class of $k$ as a generator of
$H_1(V)$ and represent it by a simple closed curve l on ∂V which is homologous to 0 in
$H_1(C)$. These conditions determine the homology class of l in $∂V$ ; hence, l is unique
up to isotopy on $∂V$ . A generator of $H_1(C)$ can be represented by a curve m on $∂V$
that is homologous to 0 in $V$ . The curves l and m determine a system of generators
of $H_{1}(∂V ) $= $\mathbb{Z} ⊕ \mathbb{Z}$. By a well-known result, we may assume that m is simple and
intersects l in one point. As m
is homologous to 0 in $V$ it is nullhomotopic in $V$ , bounds a disk, and is a meridian of
the solid torus $V $. These properties determine m up to an isotopy of $∂V$ .
A consequence is that l and k bound an annulus $A ⊂ V $.

With respect to the complement $C$ of a knot the longitude l and the meridian
m have quite different properties: The longitude l is determined up to isotopy and
orientation by $C$ ; this follows from the fact that l is a simple closed curve on $∂C$ which
is not homologous to 0 on $∂C$ but homologous to 0 in $C$. The meridian m is a simple
closed curve on $∂C$ that intersects l in one point; hence, l and m represent generators
of $H_{1}(C) $= $\mathbb{Z}^{2}$. The meridian is not determined by $C$ because simple closed curves on
$∂C$ which are homologous to $m^{±1}l^{r}$, r ∈ $\mathbb{Z}$, have the same properties.

My Question is precisely the last statement, why the meridian isn't determined by $C$?

答:
Yes, the homology groups $H_1(\partial V)$ and $H_1(C)$ along with the map $H_1(\partial V)\to H_1(C)$ only characterize the meridian slope up to some number of longitudes.  There is a general principle called "half lives half dies," which is that for a compact oriented $3$-manifold $M$, the kernel of $H_1(\partial M;\mathbb{Q})\to H_1(M;\mathbb{Q})$ is half-dimensional.  In the case of a torus boundary, like a knot complement, then the kernel is one-dimensional so is generated by a well-defined simple closed curve -- the longitude.  There is ambiguity, however, in the meridian.

The second inclusion $H_1(\partial V) \to H_1(V)$ is what pins down the meridian, again by the generator of the kernel.

The meridian problem is this: given $C$, how do we glue in $V$ to yield $S^3$? The core of $V$ then corresponds to a knot, and the meridian of $V$ is the meridian.  A priori, we also can't be sure that there aren't multiple ways to glue in $V$ that yield different knots, though.

The data of gluing in $V$ ends up, up to homeomorphism, only depending on where a meridian of $V$ is glued into the boundary of $C$ (a procedure called Dehn filling).  This cannot be solved purely homologically, in the sense that for each possible meridian $m^{\pm 1}l^r$, if you glue in $V$ according to this loop you get a space whose homology is the same as that of $S^3$.  That is to say, each of them gives you an integer homology $S^3$, often denoted as a $\mathbb{Z}HS^3$.  For example, you can obtain the Poincaré homology sphere by using $m^{\pm 1}l^{\pm 1}$ (sorry, I forget which!) on a trefoil knot.

That's not to say the problem is not solvable.  Gordon and Luecke proved that there is precisely one boundary slope whose Dehn filling yields $S^3$.  Thus, for knots the meridian is completely determined by $C$.  It is a very delicate argument, and my understanding is that through some careful combinatorial analysis, one shows that if there were two different slopes whose Dehn fillings yield $S^3$, then one could find that $S^3$ had a Lens space connect summand, which violates the fact that $H_1(S^3)=0$.

Gordon, C. McA.; Luecke, J., Knots are determined by their complements, Bull. Am. Math. Soc., New Ser. 20, No. 1, 83-87 (1989). ZBL0672.57009.

For links of more than one component, it turns out that $C$ does not determine the meridians, and there are infinitely many links with the same complement.


问:
Why do we multiply by $-1$ here?
WolframAlpha solves $$\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2}$$ by multiplying by $-1$ as such:
$$\sin\left(-x+\fracπ4\right)=-\frac{1+\sqrt3}{2\sqrt2}$$ then arcsin both sides, etc. but why multiplying by $-1$ and not just directly find the answer? Mainly say this because I've tried without multiplying by $-1$ and haven't managed to got $\frac56 π$ which is the answer (in $0<x≤2π$)

答:
Apply  trig addition identity,we can do this step by step as the following (there is no requirement to multiply -1 for both sides):

\begin{align}\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2} \end{align}
\begin{align}  \sin x\cos\frac{\pi}{4}-\cos x\sin \frac{π}{4}=\frac{1+\sqrt3}{2\sqrt2}  \end{align}\begin{align}(\sin x-\cos x)\frac{\sqrt 2}{2}=\frac{1+\sqrt3}{2\sqrt2}\end{align}  \begin{align}\sin x-\cos x=\frac{1}{2}+\frac{\sqrt3}{2} \end{align}

\begin{align}x = \frac{2\pi}{3},\frac{5\pi}{6}  ( x\in (0,2\pi])  \end{align}

Title: Why do we multiply by $-1$ here?


问:
Is the union of two elementary classes an elementary class?
Suppose $K$ and $K'$ are two classes of structures in the same language which are both elementary classes. Is their union also an elementary class? Note, the classes do not have to be finitely axiomatizable, they just have to be axiomatizable.

答: If $K$ and $K'$ are axiomatized by the sets $\Sigma$ and $\Sigma'$ of sentences, then $K\cup K'$ is axiomatized by $\{\alpha\lor\beta:\alpha\in\Sigma\text{ and }\beta\in\Sigma'\}$.


问:
Not so easy optimization of variables?
What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to $2x^2+5xy+3y^2=2$ and $6x^2+8xy+4y^2=3$. (calculus is not allowed). I tried everything I could but whenever I got for example $or$ $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. However, I also don't see how you could solve it if you leave both variables on one side. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for a much more efficient way than that.

This question came from a math competition from the Math Honor Society, Mu Alpha Theta.

答:
Edit: my suggestion does not work because a solution to the third eqn need not satisfy the original two eqns.

Multiply the first by $8$ to get: $16x^2+40xy+24y^2=16$

Multiply the second by $5$ to get: $30x^2+40xy+20y^2=15$

Subtract the two to get: $14x^2 - 4y^2 = -1$

Thus, you have to maximize $-\frac{1}{14}+\frac{4y^2}{14}+y^2$

Now it should be doable...

Title: Not so easy optimization of variables?


问:
Solve for x and y given $|x-y|, 1-|x-y|, x,y,1-y,1-x \ge m$
Let $m,x,y \in (0,1)$. Suppose $|x-y|, 1-|x-y|, x,y,1-y,1-x \ge m$. Solve for $x$ and $y$ in terms of $m$.

What I tried:

$|x-y| \ge m$ and $1-|x-y| \ge m$ tell me $m \le |X-Y| \le 1-m$

$x,y,1-y,1-x \ge m$ tell me $m \le x \le 1-m$ and $m \le y \le 1-m$

After some long computation I got these. Are these correct?

$x \in (m,1-2m)$, $y \in (x+m,1-m)$, $m \in (0,\frac13)$

$x \in (2m,1-m)$, $y \in (m, x-m)$, $m \in (0,\frac13)$

(Of course you could also write with $y$ as the independent variable, but whatever.)

答:
A geometric interpretation helps:

a. $x, 1-x \ge m$ constitutes a vertical stripe: $m < x < 1-m$

b. $y, 1-y \ge m$ constitutes a horizontal stripe: $m < y < 1-m$

c. $|x-y|, 1-|x-y| \ge m$ constitutes two 45 degree rising  diagonal  stripes:

for $y<x$:  $m < x-y < 1-m$   ;    for $y>x$:  $-m > x-y > -(1-m)$.

From a. and b. it is clear that no solutions exist for $m > 0.5$.

Now the intersection of a. and b. constitute a square, and the intersection of this
square with each of  the diagonal stripes may constitute a triangle.
The "may" results from the question whether the intersection of this
square with each of  the diagonal stripes exists at all. This will be the case if
the upper left corner of the square meets the lower bound of the stripe, i.e.
if  the point $m, 1-m$ is on the line $-m = x-y$. We have that for $-m = m - (1-m)$ or $m = 1/3$.

So if $m\le 1/3$, we can again use the  lower bound of the stripe to compute the bounds for $x$: We have that $x$ can grow up to the point where that line meets the upper bound of the square, i.e. $y = 1-m$, and hence $-m = x-y = x - (1-m)$ gives $x = 1 - 2m$.

So we can summarize:

For  $m > 1/3$, the inequalities allow no solution.

For $1/3 \ge m \ge 0$, the regions allowed by the inequalities are the two triangles (mirrored on the diagonal):

$m\le x \le 1 - 2m$ ; $ m+x \le y \le 1-m$

and $m\le y \le 1 - 2m$ ; $ m+y \le x \le 1-m$ .

So yes this is what you have got.

Here is a picture which illustrates that for $m=0.2$:


问:
How to use implication with universal quantifier?
So im trying to learn some basic logic and I got stuck on how we use implications with the universal quantifier.

My problem is the following:

$\forall x {\in}\varnothing\; P(x)$ is the same as saying

$\forall x\; \big(x\in\varnothing \to P(x)\big)$

However if I've been given a statement of the form $\forall x\in A \; \big(x\in\varnothing \to P(x)\big)$ should I read this as $\forall x\; \big(x\in A \to x \in \varnothing \to P(x)\big)$

or instead $\forall x\; \big(x\in A \wedge x \in \varnothing \to P(x)\big)$.

Thanks in advance .

答: A statement of the form $$\forall x\in A[Q(x)]$$ always means $$\forall x[x\in A\rightarrow Q(x)],$$ by definition (the first statement is just an abbreviation for the second).  So, $$\forall x\in A[x\in\emptyset \rightarrow P(x)]$$ means $$\forall x[x\in A\rightarrow (x\in\emptyset\rightarrow P(x))].$$  This is equivalent to $$\forall x[(x\in A\wedge x\in\emptyset)\rightarrow P(x)]$$ though since $x\in A\rightarrow (x\in\emptyset\rightarrow P(x))$ and $(x\in A\wedge x\in\emptyset)\rightarrow P(x)$ are equivalent.  I'm not sure if this is what you meant with your second statement since you left out the parentheses.


问:
If I have a holomorphic function on the unit disc, do I know anything about the radius of convergence of its series expansion about zero?
I'm looking at a proof that assumes only that

$f : \mathbb{D} \rightarrow \mathbb{D}$ is holomorphic with $f(0) = 0$

The first step in the proof is to "expand $f$ in a power series centered at $0$ and convergent in all of $\mathbb{D}$."

In other words, $f(z) = a_0 + a_1 z + a_2 z^2 + ...$ inside the disc. Is this a valid step? How do I know that the function's series expansion is valid in the whole disc?

答:
Following Tunococ's link, I found the following: "the fact that the radius of convergence is always the distance from the center a to the nearest singularity; if there are no singularities (i.e., if ƒ is an entire function), then the radius of convergence is infinite. Strictly speaking, this is not a corollary of the theorem but rather a by-product of the proof."

Since there are no singularities, then the series is valid for the whole disc!


问:
What type of semi - martingale is local time?
Let $W(t)$ be a Wienner process. Let's define local time as

$$L_t(x) = \lim_{\epsilon \rightarrow0^+}\frac{1}{2 \epsilon}\int_0^t \chi_{\{|W(s) - x| \le\epsilon\}} ds$$

I want to check weather this object is supermartingale, martingale or submartingale.

The hint is to use Tanak's formula:

$$|W(t) - t|= |x|  + \int_0^t\text{sign}(W(s) - x)dW(s)+L_t(x)$$

I tried to do it by definition i.e. to show that $E[L_{t+h} \mid \mathbb{F}_t] = L_t$, but I had no idea how to rewrite this conditional expected value. My second idea was more tricky - to use Ito formula ale see what's the expression in $dt$, however I also end up with nothing and with no idea how the Tanak's formula can be used for simplicity. Could you please give me a hint in which direction should I follow and in which moment Tanak's formula should be used ?

EDIT

I just have several questions about things that I don't fully get.

By facts that $W$ is a local martingale and $\text{sign}(W-x)$ is bounded predictable process you concluded that $H_t$ is local martingale. Could you please tell me to which exactly theorem you are referring to ? I'm afraid that I don't know such.

The second this is that you are rewriting expected value of quadratic variation of $H$:

$$E[(\int_0^t\text{sign}(W_s - x)dWs)^2] = E[\int_0^t \text{sign}(W_s -x)^2dt]$$

This equality is just direct application of Ito Isometry theorem. Now, because $\text{sign}(W_s - x)$ takes value $1$ almost everywhere (it doesn't take it for such $x: W_s = x$) we have that $E[H] = t$. But how this proves that $H_t$ is a martingale ? I was seeking for such theorem but wasn't able to find.

答:
The correct Tanak's formula is given by
$$|W_t - x|= |x|  + \int_0^t\text{sign}(W_s - x)\Bbb dW_s+L_t(x).$$
As $W$ is a local martingale and $\text{sign}(W-x)$ is a bounded predictable process the stochastic integral $H_t:=\int_0^t\text{sign}(W_s - x)\Bbb dW_s$ is a local martingale. By using the generalized Itô-isometrie ($[.]$ denotes the quadratic variation)
$$\forall t\geq0:\Bbb E[H]_t=\Bbb E\int_0^t\text{sign}(W_s - x)^2 \Bbb ds=t<\infty,
$$
$H$ is a martingale. Furthermore $|W - x|$ is a submartingale, as by Jensen's inequality follows
$$\Bbb E\big(|W_t - x|\big|{\cal F_s}\big)\geq \Bbb |E\big(W_t - x\big|{\cal F_s}\big)|=|W_s - x|.
$$ Altogether
$$L_t(x)= |W_t - x| - \int_0^t\text{sign}(W_s - x)\Bbb dW_s-|x|$$
is for each $x$ the difference of a submartingale and a martingale, thus it is a submartingale.


问:
I need to show that this complex series converges in C
I, unfortunately, don't really know where to start. I know I can't use the ratio test. Should I try to split it up into a complex and a real part?

Translation: Show that for all $s\in\Bbb C\;$ such that $\operatorname{Re}s>0\;$ , the series $$h(x)=\sum_{k=0}^\infty\left[(4k+2)^{-s}-(4k+1)^{-s}\right]$$ converges in $\Bbb C$.

Thanks in advance for any suggestions.

答:
One can use that $(4k+2)^{-s}-(4k+1)^{-s}=(-s)\int_{4k+1}^{4k+2}x^{-s-1}dx$ so

$|(4k+2)^{-s}-(4k+1)^{-s}| \le |s|\int_{4k+1}^{4k+2}|x^{-s-1}|dx \le |s|(4k+1)^{-\sigma-1}, \sigma=\Re s$

(since $|x^{-s-1}| \le (4k+1)^{-\sigma-1}, 4k+1 \le x \le 4k+2$)

This means that the series $\sum_{k=0}^\infty|\left[(4k+2)^{-s}-(4k+1)^{-s}\right]|$ is majorized by the absolutely convergent ($-\sigma-1 <-1$) series $\sum_{k \ge 0}|s|(4k+1)^{-\sigma-1}$ so we get absolute convergence for $h$


问:
Prove that two Partial fraction Integration answers are the same
I did the actual Partial Fraction Integration for: $\int \frac{x^2+8}{x^3+8} dx$ and got: $\ln(x+2)+\frac{2}{\sqrt{3}} \arctan(\frac{x-1}{\sqrt{3}})$.

I need to prove that my answer above is equal to: $\ln(x+2)+\frac{2}{3} \sqrt{3} \arctan[\frac{1}{6}(2x-2)\sqrt{3}]$.

答: Note that $\frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{2}{3} \sqrt{3}$. With that and the fact that $\frac{1}{6}(2x-2) = \frac{x-1}{3}$ the two results are basically already the same, just a little bit of variation in the fractions happened.


问:
Translation operator and test function on $\mathbb{R}$
Consider the translation operator $\tau_h, h\in \mathbb{R}$, on the space of test functions $\mathcal{D}(\mathbb{R})$ on $\mathbb{R}$. I would like to prove that for any $\phi \in \mathcal{D}(\mathbb{R})$, we have
$$ \frac{1}{h} \cdot (\phi - \tau_h \phi) \, \to \phi' \quad \text{in $\mathcal{D}(\mathbb{R})$} \quad \text{as $h\to 0$.}$$

I would like to use the characterization of the convergence in $\mathcal{D}(\mathbb{R})$ via compact subsets of $\mathbb{R}$ I'm stucked in how to approach the problem.

Any suggestions? Thanks in advance!

答:
I’m not sure about your conventions, so let’s say $(\tau_h\phi)(x)=\phi(x-h)$, so that the result is correct.

Let $F_h=\frac{\phi-\tau_h\phi}{h}-\phi’$ for $h \neq 0$.

We want to show that in $\mathcal{D}(\mathbb{R})$, $F_h \rightarrow 0$ as $h \rightarrow 0$. This means showing two things:

that there exists a compact subset $K \subset \mathbb{R}$ and $h_0>0$ such that for each $h$ such that $0<|h| < h_0$, $F_h$ has support in $K$.

that for every $n \geq 0$, $\sup_{x \in \mathbb{R}}\,|F_h^{(n)}(x)| \rightarrow 0$ as $h \rightarrow 0$.

For 1), just take $K=L+[-1,1]$ where $L$ is the support of $\phi$ (since $\phi \in \mathcal{D}(\mathbb{R})$) and $h_0=1$.

Let’s see 2). For each $n \geq 0$, for each real number $x$, for each $h \neq 0$, $F_h^{(n)}(x)=\frac{\phi^{(n)}(x)-\phi^{(n)}(x-h)-h\phi^{(n+1)}(x)}{h}=-h^{-1}(\phi^{(n)}(x-h)-\phi^{(n)}(x)-(-h)\phi^{(n+1)}(x))$, so by Taylor, $$F_h^{(n)}(x)=-h^{-1}\int_x^{x-h}{(x-h-t)\phi^{(n+2)}(t)dt}=-h^{-1}\int_0^{-h}{(-t-h)\phi^{(n+2)}(x+t)dt},$$ thus $$F_h^{(n)}(t)=h^{-1}\int_h^0{t\phi^{(n+2)}(x-h+t)dt}=-h\int_0^1{t\phi^{(n+2)}(x-h+ht)dt}.$$

Finally, $|F_h^{(n)}(x)| \leq |h| \|\phi^{(n+2)}\|_{\infty}$, hence the conclusion.


问:
Find the area of an equilateral triangle inscribed in a unit square
If inside the square $ABCD$ there is an equilateral triangle $CMN$ inscribed, as shown in the figure. If the area of the square is 1, find the area of the triangle.

I attempted to say, I state that $x=MD$

then $MC^2=1+x^2$. $AM=1-x$ and I thought that something might come from Pythagoras, but it didn't. $\triangle AMN$ looks isosceles, but I can't prove that it is, although if it is proved to be isosceles, then the question can easily be solved by use of the pythagorean theorem. We also have that $\angle CNB+\angle CMD=150^o$. I got stuck here, could you please explain to me how to solve the question?

答:
How about continuing to compute the lengths of the others segments?

You have seen that if $MD = x$, then $AM = 1-x$ and $MC = \sqrt{1+x^2}$. Now, since $MNC$ is equilateral, it follows that $MN = NC = \sqrt{1+x^2}$.

Applying Pythagoras theorem in the triangle $AMN$, we obtain: $AN = \sqrt{2x}$. It follows that $NB = 1- \sqrt{2x}$.

Now, we know all the lengths of the sides of the triangle $NBC$. Using again the Pythagoras theorem we have:
$$NC^2 = NB^2 + BC^2 \implies 1+x^2 = 1+2x-2\sqrt{2x}+1 \\ \implies 2\sqrt{2x} =-x^2+2x+1 \implies 8x = x^4+4x^2+1-4x^3-2x^2+4x \\ \implies x^4 -4x^3+2x^2-4x+1 = 0 \implies (x^2+1)(x^2-4x+1) = 0$$
The roots of this equation are $i, -i, 2-\sqrt{3}$ and $2+\sqrt{3}$. The only convenient value is $x = 2 - \sqrt{3}$.

Later edit: A simpler way of finding $x$ is to observe that the $\triangle MDC \equiv \triangle DBC$. Hence $MD = BN$, so $x = 1 - \sqrt{2x}$, thus $x^2 - 4x+1 = 0$ and from here you get $x = 2-\sqrt{3}$.


问:
Convergence in distribution to a chi-squared
Let's say I have a sequence of random variables $X_n$ such that $X_n \leadsto N(0,1)$ (convergence in distribution). How can I prove that $X_n^2 \leadsto \chi_{(1)}^2$ without using the continuous mapping theorem?

I'm thinking something like, if $X_n \leadsto N(0,1)$ then $P(X_n \leq x) \rightarrow \Phi(x)$ where $\Phi(x)$ is the CDF of $N(0,1)$, so (suppose $X>0$) $P(X_n^2 \leq x) = P(X_n \leq \sqrt{n}) \rightarrow \Phi(\sqrt{x})$ but now I have the problem to prove that $\Phi(\sqrt{x})$ matches the CDF of the $\chi_{(1)}^2$ (where both CDFs are difficult to write it down in the first place).

Now, if I try with Levy's theorem, $X_n \leadsto N(0,1)$ if and only if $\varphi_{X_n}(t) \rightarrow e^{-t^2/2}$ for all $t$. What I need is $$\varphi_{X_n^2}(t) \rightarrow (1-2it)^{-1/2},$$
but again, I don't know how to conclude this using only the fact that $\varphi_{X_n}(t) \rightarrow e^{-t^2/2}$. Thanks

答:
You were almost there with the CDF, for $x\geq 0$
$$
P(X_n^2 \leq x) = P(-\sqrt{x}\leq X_n \leq \sqrt{x}) \to P(-\sqrt{x}\leq X \leq \sqrt{x}) =2\Phi(\sqrt{x})-1.
$$
Now it is true that the CDF $2\Phi(\sqrt{x})-1$ is little difficult to see but the PDF is easy, take a derivative and get
$\frac{1}{\sqrt{x}}\phi(\sqrt{x})=(2\pi)^{-1/2}x^{-1/2}e^{-x/2}$, when $x>0$, and $0$ otherwise.

If you want to use  Levy's theorem, then
$$
\lim_{n\rightarrow \infty}\mathbb{E}(e^{itX_n^2}) =\mathbb{E}( e^{itX^2}) = (1-2it)^{-1/2}.
$$
The first equality is due to dominated convergence theorem.


问:
Prove that the sum $2^0+2+2^2+\cdots+2^{p-1}(2^p-1)$ is $2^{p-1}(2^p-1)$
Attempt: In an attempt to find the sum of the series given that $2^{p-1}$ is prime,

$$1+2+2^2+\cdots+2^{p-2}+2^{p-1}+(2^p-1)+2(2^p-1)+2^2(2^p-1)+\cdots+2^{p-1}(2^p-1)$$

A teacher grouped elements as follows,

$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}$$

Question: I am confused a little about the last two terms in the summation above as what I think it should be is as follows,

$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$

Edit:
Question: I should get the sum $2^{p−1}(2^p − 1)$, but with the way I wrote it I did not get the sum wanted, but I got $2^p(2^p-1)$ for,

$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$

Edit 2: The only solution to this is that the sum is $2^p(2^p-1)$.

Body:
If I read well you are evaluating the sum
$$ \sum_{k=0}^{p-1}2^k + \sum_{k=0}^{p-1}2^k(2^p-1) = 2^p\sum_{k=0}^{p-1}2^k = 2^p\frac{1-2^p}{1-2} = 2^p(2^p-1)$$


问:
If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $MON$
If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $\triangle MON$

My solution goes as follows, If $D$ is a point on $CB$ such that $MD\perp CB$, we have that the area of $\triangle CMB$ is $\frac{x^2}{4}$. Hence $\frac{MD*x\sqrt{2}}{2}=\frac{x^2}{4}$ we then from this find the height of $\triangle COB$ and $\triangle AMN$, and since we also have the height of $\triangle CAB$ from $CB$, so from this we get the height of $\triangle MON$ and hence we have that its area is $\frac{x^2}{24}$. My solution is very long and tedious, is there a faster and easier approach to this problem?

答:
Another potential solution (which may or may not be faster or easier) involves directly looking at the height of $\triangle MON$ with base $MN$. Let $D$ be the midpoint of $BC$. Then $AD$ passes through $O$ and is also an altitude, because $\triangle ABC$ is isosceles.

Since the centroid splits each median in a $2:1$ ratio, we have that $AO = \frac{2}{3}h$. Now, let $T$ be the intersection of $AO$ and $MN$. Then we have $OT \perp MN$. Furthermore, by similar triangles, we see that $AT = \frac{1}{2}h$. So, $OT = \frac{1}{6}h$.

Since $MN = h = \frac{x\sqrt{2}}{2}$, $$ [MON] = \frac{1}{2}(MN)(OT) = \frac{1}{2}\cdot\frac{x\sqrt{2}}{2}\cdot\frac{1}{6}\cdot\frac{x\sqrt{2}}{2} = \frac{x^2}{24}.$$


答:
Medians of a triangle intersect each other at $2/3$ of their length: $CO=BO=2OM=2ON$.

Hence, if $a$ is the area of $MON$, then the area of $COM$ and $BON$ is $2a$ and the area of $BOC$ is $4a$. Adding those areas together we thus get:
$$
9a={3\over4}\cdot{x^2\over2},
\quad\text{that is:}\quad
a={x^2\over24}.
$$

Title: If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $MON$

Body:
If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $\triangle MON$

My solution goes as follows, If $D$ is a point on $CB$ such that $MD\perp CB$, we have that the area of $\triangle CMB$ is $\frac{x^2}{4}$. Hence $\frac{MD*x\sqrt{2}}{2}=\frac{x^2}{4}$ we then from this find the height of $\triangle COB$ and $\triangle AMN$, and since we also have the height of $\triangle CAB$ from $CB$, so from this we get the height of $\triangle MON$ and hence we have that its area is $\frac{x^2}{24}$. My solution is very long and tedious, is there a faster and easier approach to this problem?


问:
When do formal power series coincide?
Suppose we have two formal power series

$$A(z_1,\ldots ,z_n),B(z_1,\ldots ,z_n)\in\mathbb{C}[[z_1^{\pm 1},\ldots , z_n^{\pm 1}]],$$

where $n>1$.
Hence we can write the given series as

$$A(z_1,\ldots ,z_n)=\sum_{i_1,\ldots ,i_n\in\mathbb{Z}} a_{i_1,\ldots ,i_n} z_1^{i_1}\cdots z_n^{i_n}\quad\text{and}\quad B(z_1,\ldots ,z_n)=\sum_{i_1,\ldots ,i_n\in\mathbb{Z}} b_{i_1,\ldots ,i_n} z_1^{i_1}\cdots z_n^{i_n},$$

where  $a_{i_1,\ldots ,i_n},b_{i_1,\ldots ,i_n}$ are some complex numbers. Suppose there exists a disc

$$
D(x,r)=\left\{z=(z_1,\ldots ,z_n)\in\mathbb{C}^n\,:\,d(x,z)<r\right\}
$$

for some $x\in \mathbb{C}^n$ and positive integer $r$ such that both $A(z_1,\ldots ,z_n)$ and $B(z_1,\ldots ,z_n)$ converge on the disc $D(x,r)$ and such that $A(z)=B(z)$ for all $z\in D(x,r)$.

Do such (or any similar) assumptions imply that all coefficients of the given series coincide, i.e. can we conclude that
$$
a_{i_1,\ldots ,i_n} =b_{i_1,\ldots ,i_n}
\quad\text{
for all integers}\quad i_1,\ldots ,i_n\in\mathbb{Z}?$$

答:
That a $n$-variable Laurent series
$$f(z)=\sum_{\alpha \in \Bbb{Z}^n}a_\alpha z^\alpha$$ converges on some disk $|z-x|<r$ means that for all $z$ in the disk the $a_\alpha z^\alpha$ have exponential decay as $|\alpha|\to \infty$, whence the series converges on the whole annulus $\{ z,\exists \theta\in \Bbb{R}^n,|e^{i\theta}z-x|<r\}$.

If $f=g$ on the disk then by analytic continuation $f=g$ on the annulus.

Conclude from $$a_\alpha = \frac1{(2i\pi)^n}\int_{|z_1|=|x_1|}\ldots \int_{|z_n|=|x_n|}\frac{f(z)}{z_1^{\alpha_1+1}\ldots z_n^{\alpha_n+1}}dz_1\ldots dz_n$$

Title: When do formal power series coincide?


问:
What is the sum of the series $\sum_{n = 1} ^ {\infty} \left( \sqrt[n]3 - \sqrt[n+1]3 \right)$?

What is the sum of the series $$\sum_{n = 1} ^ {\infty} \left( \sqrt[n]3 - \sqrt[n+1]3 \right)\:?$$

I actually know that the answer is $2$, but I don't see how, so I would like to ask for some help with that. Thank you.

答: Hint: Show that $\sum_{n=1}^N (\sqrt[n]{3} - \sqrt[n+1]{3}) = 3 - \sqrt[N+1]{3}$. Then think about how $\sqrt[N+1]{3}$ behaves as $N$ tends to infinity.


问:
complexity analysis with recursive floor function
I have the following function:

$$f(n) = \begin{cases}
k,& n = 1 \\
2f\left(\left\lfloor\dfrac n2\right\rfloor \right)+1, & n > 1
\end{cases}$$
I have to prove that for every $k>0$ we get $f(n)=\Theta(n)$.

So I managed to prove that $f(n)=O(n)$, but I'm having trouble proving the lower bund.
I tried to claim $f(n)\geq0.5kn$ for all $k>0$ using induction.
base case would be $k=1$ so $f(1)=k\geq0.5k$
and then after assuming the induction hypothesis the step will be:
$f(n)=2f(\left \lfloor \frac{n}{2} \right \rfloor)+1\geq 2(0.5k\cdot\lfloor \frac{n}{2}\rfloor)+1\geq k\cdot\lfloor \frac{n}{2}\rfloor+1\geq  k\frac{n}{2}$

but I suppose the last step is illegal since adding 1 to  floor of n/2 won't necessarily be greater than 1.
thanks in advance.!

答:
We want a bound of the form $f(n)\ge An$ and will use $\lfloor x\rfloor>x-1$ to get rid of the floor function. So the induction step looks like:
$$
\begin{align}
f(n)
&=2f\left(\left\lfloor \frac n2\right\rfloor\right)+1\\
&\ge2A\left\lfloor \frac n2\right\rfloor+1\\
&>2A\left(\frac n2-1\right)+1\\
&=An+1-2A\\
&\ge An
\end{align}
$$
where the last step works as long as $A\le\frac12$. For the base case, you need $A\le k$, so take $A=\min(k,\frac12)$.

Alternatively, an exact solution is $f(n)=2^{\lfloor\log_2n\rfloor}(k+1)-1$. This formula can be proved straightforwardly by induction, using that $\left\lfloor\log_2\left\lfloor\frac n2\right\rfloor\right\rfloor+1=\lfloor\log_2n\rfloor$. To see this fact, let $r=\lfloor\log_2n\rfloor$ so that $n\in[2^r,2^{r+1})$, so $\frac n2\in[2^{r-1},2^r)$, so in fact $\left\lfloor\frac n2\right\rfloor\in[2^{r-1},2^r)$, so $\left\lfloor\log_2\left\lfloor\frac n2\right\rfloor\right\rfloor=r-1$.

Then it's clear that $f$ is $\Theta(n)$, since $\frac {n+1}2\le2^{\lfloor\log_2 n\rfloor}\le n$, and each $\le$ is actually an equality for infinitely many $n$. The graph of $f$ consists of horizontal lines through the region bounded by the lines $y=(k+1)\frac{x+1}2-1$ and $y=(k+1)x-1$.