问:
Basis of $SL_q(2)$
While trying to show that $SL_q(2)$ is noncocommutative, I needed to prove the following fact:

Show that the set $\{a^ib^jc^k\}_{i,j,k\geq 0}\cup\{b^ic^jd^k\}_{i,j\geq 0,k>0}$ is a basis of $SL_q(2)$.

I was stuck with proving the above fact. (Incidentally, the above is exercise IV.9.7 in Kassel's book on Quantum Groups.)

I have tried to search for a similar proof in the book, and found that Lemma IV.4.5 is rather similar ($\{a^ib^jc^kd^l\}_{i,j,k,l\geq 0}$ is a basis of $M_q(2)$). However, the proof involves Ore extensions, which I do not really understand.

Sincere thanks for any help or hint to all.

Edit: For those who have Kassel's book, $a,b,c,d$ are four variables described in Chapter IV.3 (pg. 77).

For those without Kassel's book, I have attached the following paraphrase of Kassel's book from my report.

The Algebra $M_q(2)$

For any algebra $A$ we denote by $M_2(A)$ the algebra of $2\times 2$ matrices with entries in $A$. As a set, $M_2(A)$ is in bijection with the set $A^4$ of 4-tuples of $A$. We have a natural bijection $$Hom_{Alg}(M(2),A)\cong M_2(A)$$ for any commutative algebra $A$, where $M(2)$ is defined as the polynomial algebra $k[a,b,c,d]$. This bijection maps an algebra morphism $f:M(2)\to A$ to the matrix $$\begin{pmatrix} f(a) & f(b) \\ f(c) & f(d) \end{pmatrix}.$$

From now on, we assume that $q^2 \not= -1$. We define a $q$-analogue of the algebra $M(2)$. In addition to variables $x,y$ subject to the quantum plane relation $yx=qxy$, consider four variables $a,b,c,d$ commuting with $x$ and $y$. Define $x',y',x''$, and $y''$ using the following matrix relations

$\begin{pmatrix} x' \\ y'\end{pmatrix}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} \text{and} \begin{pmatrix}x'' \\ y'' \end{pmatrix}=\begin{pmatrix}a & c \\ b& d \end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$.

答:
The widely used tool for proving such statements is the so-called Bergman's diamond lemma. It is explained in the book of Klimyk–Schmüdgen in §4.1.5 and they give exactly the example of $SL_q(2)$, so you can check it there. Let me briefly explain the idea also here.

I am going to prove that the basis of $SL_q(2)$ is the set
$$\{b^ia^jd^kc^l\mid j=0\text{ or }k=0\}=\{b^ia^jc^l\}\cup\{b^id^kc^l\}.$$
You can then use the commutation relations to reorder those monomials as you wish.

So, given this, I will order the generators as follows $b<a<d<c$. Now I will rearange my commutation relations in such a way that on the left hand side I have always a the lexicographically largest monomial and on the right-hand side the rest:
$$\eqalign{ab&=q\,ba\cr ca&=q^{-1}ac\cr db&=q^{-1}bd\cr cd&=q^{-1}dc\cr cb&=bc\cr ad&=q\,bc+1\cr da&=q^{-1}bc+1 }$$

Now you can see that if I start with any monomial, then using those relations, I can reduce this monomial to some ordered form $b^ia^jd^kc^l$ (or maybe some linear combination of those). In addition, I can also see that the occurence of $ad$ can be further reduced, so actually I can manage that $j=0$ or $k=0$. This proves that the given set of monomials is a generating set.

Now the diamond lemma assures that under certain conditions this set is also linearly independent. And the condition is that all ambiguities are resolvable. Check Klimyk–Schmüdgen for a precise definition. But the idea is that if I have two options how to resolve some monomial, then, in the end, both must lead to the same result. For instance, having the monomial $ada$, I can go two ways: either first applying the rule for $ad$ or first applying the rule for $da$. Both are however equivalent, so the ambiguity is resolved:
$$(ad)a=(qbc+1)a=qb(ca)+a=bac+a$$
$$a(da)=a(q^{-1}bc+1)=q^{-1}(ab)c+a=bac+a$$

By the way. In order to prove that such an algebra is indeed non-commutative, you do not necessarily need to find its basis (which may sometimes be hard or impossible; for instance, if you deal with free quantum groups). It is enough to find its representation and show that it is non-commutative. Basically, you just need to find a quadruple of matrices that satisfy these relations and they mutually do not commute (at least two of them). For example, the following matrix is such a "point" in $SL_q(2)$, which proves its non-commutativity (see also Some concrete examples of $M_q(2)$ points):

$$\pmatrix{
\pmatrix{q^{-1/2}&0\cr0&q^{1/2}}&\pmatrix{0&1\cr0&0}\cr
\pmatrix{0&1\cr0&0}&\pmatrix{q^{1/2}&0\cr0&q^{-1/2}}
}.$$

Title: Basis of $SL_q(2)$


问:
Restricted probability measure definition and property
Let $(\Omega, \mathcal{F}, \mu)$ be a probability measure space.

Let $\mathcal{F}'\subset \mathcal{F}$ be a sub-sigma-algebra.

According to WikiProof, the restricted measure $\mu|_{\mathcal{F}'}$on $\mathcal{F}'$ is defined as

$$
\forall E \in \mathcal{F}', \quad \mu(E) = \mu|_{\mathcal{F}'}(E).
$$

The proof shows that $\mu|_{\mathcal{F}'}$ is a measure on $(\Omega, \mathcal{F}')$.

Now I would like to define an extended restricted measure such that $\tilde{\mu}|_{\mathcal{F}'}(A)=0$ for any $A\in\mathcal{F}$ and $A\notin \mathcal{F}'$.

Questions are

Is $\tilde{\mu}|_{\mathcal{F}'}(A)$ a measure on $(\Omega, \mathcal{F})$? I found the pairwise disjoint union axiom difficult to prove.

If not, how to extend $\tilde{\mu}|_{\mathcal{F}'}(A)$ so that it is a measure on $(\Omega, \mathcal{F})$?

答:
$\Omega$ belongs to every sigma algebra on $\Omega$. It belongs to $\mathcal F'$ in particular and  $\mu|_{\mathcal F'}(\Omega)=\mu (\Omega)=1$. So $\mu|_{\mathcal F'}$ is a probaility measure.

$\mu|_{\mathcal F'}$ is defined only for sets in $\mathcal F'$, so $\mu|_{\mathcal F'}(A)$ does not even make sense when $A \notin \mathcal F'$.
Just defining it as $0$ for sets not in $\mathcal F'$ will not yield a measure.

An obvious extension of the restricted measure to $\mathcal F$ is $\mu$ itself.

For the edited version of 1) the answer is the restriction of $\mu$ is not defined on $\mathcal F$, so the question of its countbale additivity does not arise. Its countable aditivity on $\mathcal F'$ is obvious.

ContentLicense: CC BY-SA 4.0


问:
What is the equivalent of a change of basis in geometric algebra?
A change of basis of a matrix is

$$
M'=PMP^{-1}
$$

But a vector is

$$
v'=Bv
$$

A multivector $u$ can be represented as a matrix, but it is also a vector.

So...

$$
u'=Vu
$$

or

$$
u'=VuV^{-1}
$$

Is $V$ an arbitrary multivector (provided it has an inverse - thus a general linear group multi-vector...)?

答:
I'll assume we are talking about orthonormal basies.  Then a change of basis is just a rotation.   Rotate each basis vector and you have the new basis.

The computer system I have can't deal with that.  The basis in unchangeable.  It seems to me that what is needed instead is to transform every element in your space as though it were in the new basis, even though as far as the computer is concerned the basis hasn't changed.  That means you apply the inverse of that rotation to every element.


问:
Basel Problem approximation error bounded by $\mathcal O(1/x)$?
In this answer it is stated that
$$
\sum_{n\geq1}\frac{1}{n^2}=\sum_{n\leq x}\frac1{n^2}+\mathcal O(1/x).
$$
Is this statement true as $x\to\infty$?

What I've done: If $x$ is fixed, then I think the answer is almost trivial, because we may set $C=\pi^2x/6$, so
$$
\sum_{n=x}^\infty\frac1{n^2}\leq\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}=\frac{C}{x},
$$
therefore
$$
\sum_{n\geq1}\frac1{n^2}=\sum_{n\leq x}\frac{1}{n^2}+\sum_{n=x}^\infty\frac{1}{n^2}\leq\sum_{n\leq x}\frac{1}{n^2}+C/x=\sum_{n\leq x}\frac{1}{n^2}+\mathcal O(1/x).
$$
But is there a constant independent of $x$ that makes this true?

答:
Euler-Maclaurin Sum Formula

What is being referred to here is that
$$
\sum_{k=n+1}^\infty\frac1{k^2}=\frac1n-\frac1{2n^2}+\frac1{6n^3}+O\!\left(\frac1{n^5}\right)
$$
This can be gotten using the Euler-Maclaurin Sum Formula. Thus, the constant is $1$.

Integral Estimate for the Tail of the Series

We also have the basic integral estimate for the tail of a monotonically decreasing series. This is based on the fact that
$$
\frac1{k^2}\le\int_{k-1}^k\frac1{x^2}\,\mathrm{d}x
$$
Summing this inequality yields
$$
\begin{align}
\sum_{k=n+1}^\infty\frac1{k^2}
&\le\int_n^\infty\frac1{x^2}\,\mathrm{d}x\\
&=\frac1n
\end{align}
$$

Telescoping Series

Using the simple inequality
$$
\frac1{k^2}\le\frac1{k-1}-\frac1k
$$
we can sum, using the telescoping series, to get
$$
\sum_{k=n+1}^\infty\frac1{k^2}\le\frac1n
$$

Example
$$
\sum_{k=1}^{1000}\frac1{k^2}=1.6439345667
$$
whereas
$$
\frac{\pi^2}6=1.6449340668
$$
The error is $0.0009995001\lt\frac1{1000}$.


问:
How many positive multiples of 2013, and have exactly 2013 divisors?
The full question:

How many positive integers are multiples of 2013 and have exactly 2013
divisors (including 1 and the number itself)?

I thought of 2013!, but I think it wouldn't work out well, since it would have more than 2013 divisors. I don't have any other way to go through this problem, so it would be greatly appreciated if anyone can answer this with an explanation...

答:
$2013 = 3 \times 11 \times 61$.

Now, the number of divisors of $n=\prod p_i^{\alpha_i}$ is $\prod(\alpha_i+1)$.

Clearly, $\prod(\alpha+1)=2013 \implies n=p_1^2 p_2^{10} p_3^{60}$ for some $p_1,p_2,p_3$. ALso, since $2013|n$, $\{p_1,p_2,p_3\}=\{3,11,61\}$.

So, the number of such integers is $3!=6$.


问:
Just want to make it clear for the Defintion of Splitting Field
If $E = \mathbb{Q}(\alpha_1, \alpha_2, \alpha_3)$ the splitting field of $f(x)$ over $\mathbb{Q}$.

Then, does that have to be the case which $\alpha_1$, $\alpha_2$ and $\alpha_3$ are distinct roots of $f(x)$ ?

And also, does that have to be the case which $\alpha_1$, $\alpha_2$ and $\alpha_3$ are the only roots of $f(x)$ ?

答:
No, it only means that $f(x)$ splits into linear factor in $E$ and in no proper subfield  of $E.$

For example, $\mathbb Q(\sqrt 2,\sqrt 3)$ is the splitting field of $f(x)=x^4-10x^2+1.$ Neither $\sqrt 2,\sqrt 3$ are roots of $f(x).$ (The roots are $\pm\sqrt 2\pm \sqrt 3.$)


问:
Collatz Conjecture: Does it follow that since there is no $1$-cycle, if $ab > 1$, then $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be an integer?
For me, one of the most interesting results of the Collatz Conjecture is that if an $m$-cycle exists, then $m > 68$.  So, there are no non-trivial $1$-cycles, $2$-cycles, ..., up to $68$-cycles.  The details behind this argument can be found in this paper by Simons & De Weger (2003).

Let:

$\nu_2(x)$ be the 2-adic valuation of $x$.

$u = 3^kt - 1$

$b = \nu_2(u)$

$a = \nu_2(\frac{u}{2^b}+1)$

If a $1$-cycle exists, then the maximum value of the cycle is equal to $3^kt - 1$ where $k,t$ are positive integer and the lowest value of the cycle is equal to $2^kt - 1$ characterized by the following:

$$u = \left(\frac{3}{2}\right)^a\left(\frac{u}{2^b} + 1\right)-1$$

So that:

$$\left(\frac{2}{3}\right)^a(u+1) = \frac{u}{2^b} + 1$$

And:

$$\frac{u(2^{a+b} - 3^a)}{3^a2^b} = \frac{3^a - 2^a}{3^a}$$

Which reduces to:

$$\frac{u}{2^b} = \frac{3^a - 2^a}{2^{a+b} - 3^a}$$

It seems to me that since no non-trivial $1$-cycle exists, it follows that if $a,b$ are positive integers where $ab > 1$, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not a positive integer.

Here's the argument:

(1) Assume that there exists positive integers $a,b$ with $ab > 1$ such that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is an integer.

(2)  Let $x = \frac{3^a - 2^a}{2^{a+b} - 3^a}$

(3)  Let $u = 2^bx$

(4)  Let $k = a$, $t = \left(\frac{1}{2^a}\right)\left(\frac{u}{2^b}+1\right)$ which are both integers.

(5)  So it follows that $u = 3^kt - 1$

(6)  But then we have a contradiction since no $1$-cycle exists.  Therefore, we can reject our assumption at step(1).

Is this argument valid?  Does the result from Simons & De Weger (2003) prove that with the given assumptions, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not an integer?

Edit:  I added the clarification that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be a positive integer.

@rukhin makes a great point.  For positive integers $a=2$ and $b=1$, $\dfrac{3^a - 2^a}{2^{a+b} - 3^a} = -5$

This is correlated with the negative $1$-cycle with $k=1$ and $t=\left(\frac{1}{4}\right)\left(\frac{-10}{2} + 1\right) = -1$

The standard collatz sequence defined as:

$T(x) = \begin{cases}
\frac{1}{2}(3x + 1), && \text{if }x\text{ is odd}\\
\frac{1}{2}x, && \text{if }x\text{ is even}\\
\end{cases}$

$T_1(x) = T(x)$

$T_{c+1}(x) = T(T_c(x))$

So that:

$$-1, -2, -1, \dots$$

答:
Say you start with $n_0=2^kt-1$, apply $a$ times the Collatz function $f(n)=\frac{3n+1}{2}$, than $b$ times the Collatz function $f(n)=\frac{n}{2}$, you can easily see that you end up with $$n=\frac{3^a}{2^{a+b}}n_0+\frac{3^a-2^a}{2^{a+b}}$$
or
$$n\cdot{2^{a+b}}-3^a\cdot n_0=3^a-2^a$$
and in the case of a cycle $n=n_0$
$$n=\frac{3^a-2^a}{2^{a+b}-3^a}$$
Note: $a=k$, the value $a = \nu_2(\frac{u}{2^b}+1)$ you gave is for the next iteration

Now in case of a cycle, it has been shown that (you can find it by noticing that $\frac{1}{2}<\frac{3^a}{2^{a+b}}<1$):
$$a+b=\lceil a\cdot \log_2(3)\rceil $$
So indeed, if there is such an integer $n=\frac{3^a-2^a}{2^{a+b}-3^a}$, it must be part of a 1-cycle, and you can conclude there is none (outside the trivial one), but it would only be valid within the above constraint on $a+b$.

Also notice that the question has already been raised.....by you
Is this argument valid for showing given integers $n>1$ and $2^m > 3^n$, $(2^m - 3^n) \nmid 3^n - 2^n$

Also note that a m-cycle does not mean m successive 1-cycles, so there is no such constraint on $a+b$ within the m-cycle (but in this case you don't have $n=n_0$ or the equation you mentioned)


问:
How many times to differentiate to obtain differential equation?
Consider the following question:

Find the differential equation of the family of curves, $x^2=4b(y+b)$ ($b$ is a real arbitrary constant).

Now, if we differentiate the given equation twice, we get $xy''=y'$. However, if we only differentiate once, we get $x(y')^2=x+2yy'$

Why is this so?

答:
Always remember: A differential equation of the $n^\text{th}$ order cannot have more than $n$ arbitrary constants in its solution.

Here in your question, the solution curve of the differential equation $~x^2=4b(y+b)~$ contains only one arbitrary constant $b$ and hence the order of the differential equation is exactly one. So you need not differentiate the function for more than one time.

Now given curve is $$x^2=4b(y+b)\implies y+b=\dfrac{x^2}{4b}\tag1$$
Differentiate with respect to $x$ we have $~y'=\dfrac{x}{2b}\implies b=\dfrac{x}{2y'}~$

Putting the value of $b$ in $(1)$ we have the required differential equation as, $$x^2=2\dfrac{x}{y'}\left(y+\dfrac{x}{2y'}\right)\implies x\left(y'\right)^2=x+2yy'\implies x\left(y'\right)^2-2yy'=x~.$$


问:
Does taking restriction and left derived functors commute?
Let $R$ be a commutative Noetherian ring. Let Mod$(R)$ denote the category of $R$-modules, and mod$(R)$ denote the category of finitely generated $R$-modules; notice that both of these categories are abelian and has enough projective objects.

For a functor $F: \text{Mod}(R) \to \text{Mod}(R)$, if $F(M)\in$mod$(R)$ for every $M\in \text{mod}(R)$, then let $F|_\text{mod}$ denote the restricted functor $F|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$ where for every $M,N \in \text{mod}(R)$,
$F|_\text{mod}(M):=F(M)$ and $F|_\text{mod}(f):= F(f) $ for every $f\in \text{Hom}(M,N)$.

Now let $F: \text{Mod}(R) \to \text{Mod}(R)$ be an $R$-linear right exact functor satisfying $F(\text{mod}(R))\subseteq \text{mod}(R) $. So $F|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$ is also right exact and $R$-linear.

We can consider the left derived functors (https://en.m.wikipedia.org/wiki/Derived_functor) $L_i F: \text{Mod}(R) \to \text{Mod}(R)$ and $L_i (F|_\text{mod}): \text{mod}(R) \to \text{mod}(R)$ . Since

$F(\text{mod}(R))\subseteq \text{mod}(R) $ holds, so

$L_iF(\text{mod}(R))\subseteq \text{mod}(R) $ holds for every
$i\ge 0$, so for every $i\ge 0$, I can consider the restriction
$(L_iF)|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$.

My question is: Do we have an isomorphism of functors $L_i (F|_\text{mod}) \cong (L_iF)|_\text{mod}$ for all $i\ge 0$ ?

Since different choices of projective resolutions ultimately give Isomorphic derived functors, so I can clearly see that $L_i (F|_\text{mod}) (M) \cong (L_iF)|_\text{mod}(M)$ for all $M \in \text{mod}(R)$ and for all $i\ge 0$.  However, I am not sure how to show if the corresponding functors are Isomorphic.

Please help.

答:
The key property for defining left derived functors is that the homology modules don't depend on the particular projective resolution.

If you have a module $M$ and two projective resolutions thereof, say
\begin{align}
\dots\to P_n\to P_{n-1}\to\dots\to P_1&\to P_0\to M\to 0 \\[1ex]
\dots\to P_n'\to P_{n-1}'\to\dots\to P_1'&\to P_0'\to M\to 0
\end{align}
then upon applying $F$, the complexes you obtain have the same homology.

Since a finitely generated module has a projective resolution consisting of finitely projective modules, due to $R$ being Noetherian, the left derived functors of a functor that restricts to $\mathrm{mod}(R)$ are the same when computed over $\mathrm{Mod}(R)$ or when computed for the restriction.


问:
Solving a system of differential equations with the eigenvector but no eigenvalues?
Solving a system of differential equations by substitution

I originally asked for help with this question which ended up just being me missing a simple identity. I've moved on and am stuck again, so hopefully you all can be as helpful as you were previously.

Define the vectors $\vec y$ = [$y_1$, $y_2$], $\vec y$' = $\left[ \frac{dy_1}{dt}, \frac{dy_2}{dt}\right]$ and the coefficient matrix A = $\begin{bmatrix} a & -b\\ b & a\end{bmatrix}$.

Then, $\vec y$' = Ay defines a system of linear homogeneous differential equations. Note that A has complex eigenvalues.

Hint: To solve the system of equations, write $z = y_1 + iy_2$ and calculate $z'$. For some $u$ defined in terms of $a$ and $b$ you should find that $z'=uz$, which is easily solved. Now, write the solution in terms of the original variables $y_1$ and $y_2$. You will need to use the identity $e^{itb}=\cos tb+i\sin tb.

Solve the system of equations

Locate all equilibrium points, and provide a stability analysis

Given your answers to 1. and 2., sketch the phase plane for this system in $\mathbb{R}^2$.

I've since found that since $z' = uz = (a + ib)(y_1 + iy_2)$, we can say that $z = e^{(a+ib)t} = e^{at}e^{ibt} = e^{at}\cos(bt) + ie^{at}\sin(bt)$

From this we can go back to our original vector $\vec{y}$ and say:
$\vec{y} =
\begin{bmatrix}
e^{at}\cos(bt)\\
e^{at}\sin(bt)
\end{bmatrix}$

Of course we now want to solve the system, so I observe that for some system of differential equations we have that:

$\vec{x}(t) = \vec{\eta}e^{\lambda t}$

As such I feel that we have:

$\vec{y} =
\begin{bmatrix}
\cos(bt)\\
\sin(bt)
\end{bmatrix}e^{at}$

I'm concerned that here I am making a mistake or am getting myself confused. How do I find the eigenvalues for this system?

If I pursue this thread, I end up with the typical steps for finding eigenvalues and eigenvectors, where:

$\det(A-\lambda I) = 0$

But above I have that $a$ is $\lambda$, so I end up solving the Wronskian such that:

$\begin{vmatrix}
a-\lambda & -b\\
b & a - \lambda
\end{vmatrix} =
\begin{vmatrix}
a-a & -b\\
b & a - a
\end{vmatrix}  =
\begin{vmatrix}
0 & -b\\
b & 0
\end{vmatrix} = b^2 = 0$

Now I'm spinning around and around and the anxiety has me frozen.
Any advice at this point would again be greatly appreciated!

EDIT: Okay - so in the case where $u = (a+ib)$ and its conjugate $\bar{u}$ are my eigenvalues, I have that:

$\begin{bmatrix}
a - (a-ib) & -b\\
b & a-(a-ib)
\end{bmatrix}
\begin{bmatrix}
\eta_1\\
\eta_2
\end{bmatrix} = \begin{bmatrix}
0\\0\end{bmatrix}$

This gives:

$ib\eta_1 - b\eta_2 = 0$
and
$b\eta_1+ib\eta_2 = 0$

which gives that $\vec{\eta^{(1)}} = \begin{bmatrix}
-i\\1\end{bmatrix}$, and similar for $\vec{\eta^{(2)}}$

答:
The eigenvalue decomposition serves to decompose the system into scalar components, real scalar or complex scalar. The eigenvalues are then the coefficients for these scalar equations.

Here from the start you have a transformation into a complex scalar equation. Its complex conjugate is likewise a scalar transformation of the system. The scalar coefficients $a\pm ib$ are then the eigenvalues.


Title: Groebner Basis in Macaulay2

Body:
This is a follow-up question to this question (calling on @Jan-Magnus Okland for help). Plugging the following into Macaulay2:

```<code>R = QQ[a,b,c,d,e,f]
S = R[l]
T = S[p,q,r,X,Y,Z,MonomialOrder=>Lex]
I=ideal(X-l*(2*a*p+b*q+d*r),Y-l*(b*p+2*q*c+e*r),Z-l*(d*p+e*q+2*f*r),X*p+Y*q+Z*r)
</code>```

and then

```<code>J = groebnerBasis I
</code>```

Edit: I made a mistake initially that yielded a 1 x 17 Groebner basis. I have corrected this and the correct 1 x 23 basis is shown below.

I get the following output:

```<code>o5 = | (e2-4cf)X2+(-2de+4bf)XY+(4cd-2be)XZ+(d2-4af)Y2+(-2bd+4ae)YZ+(b2-4ac)Z2
     --------------------------------------------------------------------------
     (2cd2-2bde+2ae2+2b2f-8acf)lr+(-2cd+be)X+(bd-2ae)Y+(-b2+4ac)Z
     --------------------------------------------------------------------------
     (e2-4cf)lrX+(-de+2bf)lrY+(2cd-be)lrZ-eXY+2cXZ+dY2-bYZ
     --------------------------------------------------------------------------
     (de-2bf)lrX+(-d2+4af)lrY+(bd-2ae)lrZ-eX2+dXY+bXZ-2aYZ
     --------------------------------------------------------------------------
     (2cd-be)lrX+(-bd+2ae)lrY+(b2-4ac)lrZ-2cX2+2bXY-2aY2
     --------------------------------------------------------------------------
     (2cd-be)lq+(de-2bf)lr-dY+bZ (bd-2ae)lq+(d2-4af)lr-dX+2aZ
     --------------------------------------------------------------------------
     (b2-4ac)lq+(bd-2ae)lr-bX+2aY
     --------------------------------------------------------------------------
     (2cd2-2bde+2ae2)qY2+(2b2e-8ace)qYZ+(-2b2c+8ac2)qZ2+(de2-4cdf)rXY+(-be2+
     --------------------------------------------------------------------------
     4bcf)rXZ+(-d2e+4aef)rY2+(4cd2-2ae2-8acf)rYZ+(-4bcd+b2e+4ace)rZ2
     --------------------------------------------------------------------------
     (2cd-be)qX+(-bd+2ae)qY+(b2-4ac)qZ+(de-2bf)rX+(-d2+4af)rY+(bd-2ae)rZ
     --------------------------------------------------------------------------
     elqX-dlqY+2flrX-dlrZ-XZ 2clqX-blqY+elrX-blrZ-XY blqX-2alqY+dlrX-2alrZ-X2
     --------------------------------------------------------------------------
     eqXY-2cqXZ-dqY2+bqYZ+2frXY-erXZ-drYZ+brZ2
     --------------------------------------------------------------------------
     eqX2-dqXY-bqXZ+2aqYZ+2frX2-2drXZ+2arZ2
     --------------------------------------------------------------------------
     2cqX2-2bqXY+2aqY2+erX2-drXY-brXZ+2arYZ dlp+elq+2flr-Z blp+2clq+elr-Y
     --------------------------------------------------------------------------
     2alp+blq+dlr-X 2apZ-eqX+dqY+bqZ-2frX+2drZ dpY-bpZ+eqY-2cqZ+2frY-erZ
     --------------------------------------------------------------------------
     2apY-2cqX+2bqY-erX+drY+brZ pX+qY+rZ |

             1       23
o5 : Matrix T  <--- T

</code>```

I am trying to calculate the dual curve, but $\lambda$ (i.e. l in the Macaulay2 equations above) still appears in the basis. I know from my question that the final result should be

$$
(-\frac14e^2+cf)x^2+(\frac12de-bf)xy+(-\frac14d^2+af)y^2+(-cd+\frac12be)x+(\frac12bd-ae)y-\frac14b^2+ac=0.
$$

My question has 2 parts:

How do I go from the Groebner basis to the final answer? Does this equivalent system of equations need to be solved? (i.e. if the original system was $F = 0$, do I need to solve $G = 0$ and how would I do so?)

Once I have my 2 dual curves (from their respective Groebner bases), how do I solve the two simultaneously? Can I do it in Macaulay2, or do I have to do it in another tool?

Body:
I don't see what goes wrong with the above, what version of M2 did you use?

I get with the same commands (and toString oo)

matrix {{(e^2-4*c*f)*X^2+(-2*d*e+4*b*f)*X*Y+(4*c*d-2*b*e)*X*Z+(d^2-4*a*f)*Y^2+(-2*b*d+4*a*e)*Y*Z+(b^2-4*a*c)*Z^2,
(2*c*d^2-2*b*d*e+2*a*e^2+2*b^2*f-8*a*c*f)*l*r+(-2*c*d+b*e)*X+(b*d-2*a*e)*Y+(-b^2+4*a*c)*Z,
(e^2-4*c*f)*l*r*X+(-d*e+2*b*f)*l*r*Y+(2*c*d-b*e)*l*r*Z-e*X*Y+2*c*X*Z+d*Y^2-b*Y*Z,
(d*e-2*b*f)*l*r*X+(-d^2+4*a*f)*l*r*Y+(b*d-2*a*e)*l*r*Z-e*X^2+d*X*Y+b*X*Z-2*a*Y*Z,
(2*c*d-b*e)*l*r*X+(-b*d+2*a*e)*l*r*Y+(b^2-4*a*c)*l*r*Z-2*c*X^2+2*b*X*Y-2*a*Y^2,
(2*c*d-b*e)*l*q+(d*e-2*b*f)*l*r-d*Y+b*Z,
(b*d-2*a*e)*l*q+(d^2-4*a*f)*l*r-d*X+2*a*Z,
(b^2-4*a*c)*l*q+(b*d-2*a*e)*l*r-b*X+2*a*Y,
(2*c*d^2-2*b*d*e+2*a*e^2)*q*Y^2+(2*b^2*e-8*a*c*e)*q*Y*Z+(-2*b^2*c+8*a*c^2)*q*Z^2+(d*e^2-4*c*d*f)*r*X*Y+(-b*e^2+4*b*c*f)*r*X*Z+(-d^2*e+4*a*e*f)*r*Y^2+(4*c*d^2-2*a*e^2-8*a*c*f)*r*Y*Z+(-4*b*c*d+b^2*e+4*a*c*e)*r*Z^2,
(2*c*d-b*e)*q*X+(-b*d+2*a*e)*q*Y+(b^2-4*a*c)*q*Z+(d*e-2*b*f)*r*X+(-d^2+4*a*f)*r*Y+(b*d-2*a*e)*r*Z,
e*l*q*X-d*l*q*Y+2*f*l*r*X-d*l*r*Z-X*Z,
2*c*l*q*X-b*l*q*Y+e*l*r*X-b*l*r*Z-X*Y,
b*l*q*X-2*a*l*q*Y+d*l*r*X-2*a*l*r*Z-X^2,
e*q*X*Y-2*c*q*X*Z-d*q*Y^2+b*q*Y*Z+2*f*r*X*Y-e*r*X*Z-d*r*Y*Z+b*r*Z^2,
e*q*X^2-d*q*X*Y-b*q*X*Z+2*a*q*Y*Z+2*f*r*X^2-2*d*r*X*Z+2*a*r*Z^2,
2*c*q*X^2-2*b*q*X*Y+2*a*q*Y^2+e*r*X^2-d*r*X*Y-b*r*X*Z+2*a*r*Y*Z,
d*l*p+e*l*q+2*f*l*r-Z,
b*l*p+2*c*l*q+e*l*r-Y,
2*a*l*p+b*l*q+d*l*r-X,
2*a*p*Z-e*q*X+d*q*Y+b*q*Z-2*f*r*X+2*d*r*Z,
d*p*Y-b*p*Z+e*q*Y-2*c*q*Z+2*f*r*Y-e*r*Z,
2*a*p*Y-2*c*q*X+2*b*q*Y-e*r*X+d*r*Y+b*r*Z,
p*X+q*Y+r*Z}}

Here you see that $p,q,r$ (and $\lambda$) are eliminated from the first element which is what you want up to a scalar.

When you have two duals (I show what I did for the original problem) you can get the groebner basis for the ideal of the intersection:

R=QQ[h,k,a,b]
S=R[x,y,MonomialOrder=>Lex]
I=ideal(k^2*y^2-b^2*y^2+2*h*k*x*y+2*k*y+h^2*x^2-a^2*x^2+2*h*x+1,(-b^2*y^2)-a^2*x^2+1)
gens gb I -- matrix {{(k^4*a^4+2*h^2*k^2*a^2*b^2+h^4*b^4)*y^4+(4*k^3*a^4+4*h^2*k*a^2*b^2)*y^3+(-2*h^2*k^2*a^2+4*k^2*a^4-2*h^4*b^2+4*h^2*a^2*b^2)*y^2-4*h^2*k*a^2*y+h^4-4*h^2*a^2,
 (2*h^3*k^2*a^2-2*h*k^2*a^4-2*h^3*a^2*b^2)*x+(-k^5*a^4-2*h^2*k^3*a^2*b^2-h^4*k*b^4)*y^3+(-3*k^4*a^4-2*h^2*k^2*a^2*b^2+h^4*b^4)*y^2+(3*h^2*k^3*a^2-2*k^3*a^4+h^4*k*b^2-2*h^2*k*a^2*b^2)*y+3*h^2*k^2*a^2-h^4*b^2,
 2*h*k*a^2*x*y+2*h*a^2*x+(k^2*a^2-h^2*b^2)*y^2+2*k*a^2*y+h^2, 2*h^3*a^2*b^2*x*y+(2*h^3*k*a^2-2*h*k*a^4)*x+(-k^4*a^4-2*h^2*k^2*a^2*b^2-h^4*b^4)*y^3+(-3*k^3*a^4-h^2*k*a^2*b^2)*y^2+(3*h^2*k^2*a^2-2*k^2*a^4+h^4*b^2)*y+3*h^2*k*a^2,
 (k^2*a^2-h^2*b^2)*x*y^2+2*k*a^2*x*y+h^2*x-2*h*k*b^2*y^3-2*h*b^2*y^2+2*h*k*y+2*h, 2*h^3*b^2*x*y^2+(-2*h^3+2*h*a^2)*x+(k^3*a^2+3*h^2*k*b^2)*y^3+(3*k^2*a^2+3*h^2*b^2)*y^2+(-3*h^2*k+2*k*a^2)*y-3*h^2, a^2*x^2+b^2*y^2-1,
 h^2*x^2+2*h*k*x*y+2*h*x+k^2*y^2+2*k*y}}

But then I switch to maxima CAS to get the intersection points:

solve([(k^4*a^4+2*h^2*k^2*a^2*b^2+h^4*b^4)*y^4+(4*k^3*a^4+4*h^2*k*a^2*b^2)*y^3+(-2*h^2*k^2*a^2+4*k^2*a^4-2*h^4*b^2+4*h^2*a^2*b^2)*y^2-4*h^2*k*a^2*y+h^4-4*h^2*a^2,(2*h^3*k^2*a^2-2*h*k^2*a^4-2*h^3*a^2*b^2)*x+(-k^5*a^4-2*h^2*k^3*a^2*b^2-h^4*k*b^4)*y^3+(-3*k^4*a^4-2*h^2*k^2*a^2*b^2+h^4*b^4)*y^2+(3*h^2*k^3*a^2-2*k^3*a^4+h^4*k*b^2-2*h^2*k*a^2*b^2)*y+3*h^2*k^2*a^2-h^4*b^2],[x,y]);


问:
Intuition for polynomial long division
So I am trying to learn some basic algebra and go over some precalc material and I am terribly confused on why or how polynomial long division works.

I have looked at many sources online and they seem to all suggest that division of numbers is identical to division of polynomials since a number in its decimal expansion can be viewed as a sort of polynomial e.g $a_n10^n+\cdots+a_2100+10a_1+a_0$

What i don't understand is why we only use the term with the highest degree in the denominator as the divisor for the whole polynomial as shown here  Polynomial Long Division whereas with normal division we divide the whole divisor into the dividend for example when calculating $88÷32$ we look at how many times $32$ goes into $88$ and not how many times $30$ does whilst with polynomial division only the highest degree term is considered which corresponds to 30 in this example ignoring 2.

So how are they the same? Some intuition behind why polynomial division works would be greatly appreciated.

Thanks in advance.

答:
If you really want to understand the mathematics, do not do the kind of long division that you were taught. Instead, you must learn to write down true equalities. For example:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

  $\lfrac{4x^3-7x^2-6}{2x^2+5}$

  $ = \lfrac{\color{blue}{2x}(2x^2+5)-10x-7x^2-6}{2x^2+5}$

  $ = 2x + \lfrac{-7x^2-10x-6}{2x^2+5}$

  $ = 2x + \lfrac{\color{blue}{-3.5}(2x^2+5)+17.5-10x-6}{2x^2+5}$

  $ = 2x-3.5 + \lfrac{17.5-10x-6}{2x^2+5}$

  $ = 2x-3.5 + \lfrac{-10x+11.5}{2x^2+5}$.

Where did the blue terms come from? First you need to recognize that what I wrote is true. Next the reason for choosing those blue terms was to eventually get an equal expression where the fractional part is 'simpler'. You should try other terms in place of the blue terms and see why they do not work, and then you will understand why those blue terms work.

Also, there is absolutely nothing forcing you to 'simplify' the given polynomial fraction in that way. The following is also true:

  $\lfrac{4x^3-7x^2-6}{2x^2+5}$

  $= \lfrac{-6-7x^2+4x^3}{5+2x^2}$

  $= \lfrac{\color{blue}{-1.2}(5+2x^2)+2.4x^2-7x^2+4x^3}{5+2x^2}$

  $= -1.2 + \lfrac{2.4x^2-7x^2+4x^3}{5+2x^2}$

  $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$.

Why might you want to do this? Well, if you want to know the behaviour of $\lfrac{4x^3-7x^2-6}{2x^2+5}$ as $x→0$, then you want to extract out the 'big' terms, and leave the 'residual' fraction with a 'small' numerator, so that it is clear what the asymptotic behaviour is. For instance as $x→0$ we have:

  $\lfrac{4x^3-7x^2-6}{2x^2+5}$

  $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$

  $= -1.2 + 0.2·\lfrac{-4.6x^2+4x^3}{1+0.4x^2}$

  $∈ -1.2 + 0.2·(-4.6x^2+4x^3)(1-0.4x^2+O(x^4))$

  $⊆ -1.2 + 0.2·(-4.6x^2+4x^3+O(x^4))$

  $= -1.2-0.92x^2+0.8x^3+O(x^4)$.

Title: Intuition for polynomial long division


问:
ODE solving mistake in homogeneous and particular solution method
When I was solving the ODE $x^{'}+2x=e^t$, $x(0)=5$, if using convolution method I got below derivation:

$x=x(0)e^{-2t}+\int_0^t {e^{-2(t-\tau)} e^{\tau} d\tau}=5e^{-2t} +\frac{1}{3}e^t-\frac{1}{3} \\$

Above seems to be correct since I.C. is satistifed. But when I practice the method of homogeneous with particular solution, I came up with wrong solution:

$x_h(t)=x(0)e^{-2t}=5e^{-2t}\\$

$x_p(t)=\frac{1}{3}e^t\\$

$x(t)=x_h(t)+x_p(t))=5e^{-2t}+\frac{1}{3}e^t\\$

The particular solution $x_p(t)$ was by assuming $x_p(t)=Le^{t}$, and plug it into DE so that:

$x_p^{'}(t)+2x_p(t)=e^t\\$

$Le^t+2Le^t=e^t\\$

$L=\frac{1}{3}\\$

$x_p(t)=\frac{1}{3}e^t$

Hopefully my mistake can be pointed out.

答:
Both your approaches are slightly wrong.

In the first you have to apply the factor $e^{-2t}$ to the full result of the integration, thus $e^{-2t}\frac{e^{3t}-1}3=\frac13(e^t-e^{-2t})$.

In the second approach you do not evaluate the initial condition on the homogeneous solution. The constant is left as unknown. Only at the end in the general inhomogeneous solution is the constant determined from the initial condition.


问:
In a PID if the rank (finite) of a free module and its submodule are the same, then are they necessary equal?
let $R$ be a PID and $M$ a free (unital) R-module of finite rank $a$. If $N$ is a free $R$-submodule of $M$ with $\operatorname{rank}(N) = a$, then is $N = M$?

I think it doesn't holds if rank is not finite for eg if $F=\Bbb Z$ and $M=\langle 2\rangle$ then $\operatorname{rank}F= \operatorname{rank} M$ but $F$ is not equal to $M$.

Any help will be appreciated.
Thanks in advance.

答: Given a commutative ring with unit $R$ and $x\in R$ which is not a zero divisor, the $R$-modules $R$ and $xR$ are isomorphic and free of rank 1. If $x\notin R^*$ as well, then $xR\subsetneq R$.


问:
I'm trying to prove $ M \subset N \Rightarrow M \cup(N \setminus M) = N $
I'm trying to prove:
$$
M \subset N \Rightarrow M \cup(N \setminus M) = N
$$
My trial:

$
\{x|x\in M \cup(N \setminus M)\} \Leftrightarrow \{x|x\notin \overline{M \cup(N \setminus M)}\}
$

using de Morgan
$ \Leftrightarrow \{x|x\notin \overline{M} \cap \overline{(N \setminus M)}\}$

$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \notin \overline{(N \setminus M)}\}$

$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in (N \setminus M)\}$

$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in N \wedge x\notin M\}$

aand we got a problem here. $x$ cannot be both $x\notin \overline{M}$ and  $x\notin M$.
My steps seems logical to me. Can someone please spot my mistake. Thanks a lot!

答:
Your error: $x\notin A\cap B$ is the negation of $x\in A\cap B$. $$x\in A\cap B \text{ iff }x\in A \land x\in B$$
So,$$x\notin A\cap B \text{ iff }x\notin A \lor x\notin B$$
Please notice how the $\land$ becomes $\lor$ on negation! In general, $\lnot(p\land q) = \lnot p\lor \lnot q$.

Alternative proof strategy: Whenever I want to show $A = B$, where $A,B$ are two sets, I aim to establish $A\subset B$ and $B\subset A$. Works on most occasions!

Suppose $M\subset N$. Then, for every $x\in M$, $x\in N$. Note that
$$N\setminus M = \{x\in N: x\notin M\}$$
Suppose $x\in M \cup N\setminus M$. Then, $x\in M$ or $x\in N\setminus M$. If $x\in M$, then $x\in N$ by $M\subset N$. If $x\in N\setminus M$, then $x\in N$ by definition of $N\setminus M$. So, $M\cup N\setminus M \subset N$.

Next, suppose $x\in N$. Then, either $x\in M$ or $x\notin M$. If $x\in M$, $x\in M\cup N\setminus M$ by definition of the union. If $x\notin M$, then clearly $x\in N\setminus M$, since we assumed $x\in N$, to begin with (refer to the definition of $N\setminus M$ again). As a result, $x\in M\cup N\setminus M$ by definition of the union, once again. So, we have $N\subset M\cup N\setminus M$.

$M\cup N\setminus M \subset N$ and $N\subset M\cup N\setminus M$ together, yield $M\cup N\setminus M = N$. So we have shown $$M\subset N\implies M\cup N\setminus M = N$$ Done!

Lastly, the following image says more than words ever can:


答:
Your mistake :

$x \notin A \cap B \implies (x \notin A) \lor (x \notin B) $ not what you wrote. Think of it like this, $x$ not being in one of those keeps it out of the intersection.

As for the proof,
$$M \cup (N/M) = M \cup (N \cap \overline M) = (M \cup N) \cap (M \cup \overline M) = M \cup N = N$$

Title: I'm trying to prove $ M \subset N \Rightarrow M \cup(N \setminus M) = N $

Body:
I'm trying to prove:
$$
M \subset N \Rightarrow M \cup(N \setminus M) = N
$$
My trial:

$
\{x|x\in M \cup(N \setminus M)\} \Leftrightarrow \{x|x\notin \overline{M \cup(N \setminus M)}\}
$

using de Morgan
$ \Leftrightarrow \{x|x\notin \overline{M} \cap \overline{(N \setminus M)}\}$

$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \notin \overline{(N \setminus M)}\}$

$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in (N \setminus M)\}$

$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in N \wedge x\notin M\}$

aand we got a problem here. $x$ cannot be both $x\notin \overline{M}$ and  $x\notin M$.
My steps seems logical to me. Can someone please spot my mistake. Thanks a lot!


问:
Determine convergence of $\sum_{n=1}^{\infty}a_n$
Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$.

Does $\sum_{n=1}^{\infty}a_n$ converge or diverge when compared to $\sum_{n=1}^{\infty}b_n$ if $p < 1$?

My answer to this is that it diverges, since $b_n$ diverges if $p<1$, and by a comparison test $a_n$ does as well.

Is this correct? Thank you.

答:
Comparing to $\sum_nb_n$ with $p<1$ is not useful. What you will find is that $a_n/b_n\to 0,$ but this is not sufficient to decide whether $\sum_na_n$ converges.

Comparing to $\sum_nb_n$ with $p>1$ is not useful either. What you will find is that $a_n/b_n\to \infty,$ but this is  not sufficient to decide whether $\sum_na_n$ converges.

For brevity let $s_n=\frac {\ln 2}{n}.$ When $n\in \Bbb N$ we have $1>s_n>0$ so $$2^{1/n}=\exp (s_n)=1+s_n/1!+s_n^2/2!+... >1+s_n>0.$$ So when $n\in \Bbb N$ we have $2^{-1/n}<1/(1+s_n)$ so $$1-2^{-1/n}>1-1/(1+s_n)=s_n/(1+s_n)>s_n/2=\frac {\ln 2}{2n}.$$ Compare this to $b_n$ with $p=1.$


答:
My advice is that you should prove each inclusion separately (in one you will have to use the hypothesis, but in the other you don’t need to).

Let $M$ and $N$ be sets. Suppose that $M \subseteq N$. We want to prove that $M \cup (N \setminus M) = N$. By definition of equality of sets, we must prove that

\begin{align*}
x \in M \cup (N \setminus M) \iff x \in X.
\end{align*}

Let’s prove each of the implications. Let $x \in M \cup (N \setminus M)$. Hence

\begin{align*}
x \in M \cup (N \setminus M) & \implies x \in M \vee x \in N \setminus M & (\text{definitions of union}) \\
& \implies x \in N \vee x \in N \setminus M & (\text{by hypothesis, }M \subseteq N) \\
& \implies x \in N \vee (x \in N \wedge x \notin M) & (\text{definition set difference}) \\
& \implies x \in N \vee x \in N & (\text{definition of conjunction})\\
& \implies x \in N & (\text{idempotent property of } \vee)
\end{align*}

which allows us to conclude that $M \cup (N \setminus M) \subseteq N$. Now, let $y \in N$. Then

\begin{align*}
y \in N & \implies y \in N \setminus M & (\text{definition of set difference}) \\
& \implies y \in M \vee y \in N \setminus M & (\text{addition}) \\
& \implies y \in M \cup (N \setminus M) & (\text{definition of set union})
\end{align*}

from which we conclude that $N \subseteq M \cup (N \setminus M)$.

Hence, we have that $N \subseteq M \cup (N \setminus M) \subseteq N$. Therefore, $M \cup (N \setminus M) = N$. $\square$

Any doubt on these, just let me know.


问:
Prove that series $\sum_{n\in M(P)} \frac{1}{n}$ converges and find its sum
Let $P = \{p_1, p_2, \ldots, p_k\}$ be a finite set of prime numbers, and $M(P)$ be a set of natural numbers, whose prime divisors are in $P$. How can I prove that $$\sum_{n\in M(P)} \frac{1}{n}$$ converges and how can I find its sum? I would highly appreciate any help on this subject.

答:
Hint: convince yourself that

\begin{align}
\sum_{n \in M(P)} \frac1n &= \prod_{j=1}^k \left(\sum_{n=0}^\infty \frac1{p_j^n}\right) \\&= \left(1 + \frac1{p_1} + \frac1{p_1^2} + \dots\right)\left(1 + \frac1{p_2} + \frac1{p_2^2} + \dots\right) \dots \left(1 + \frac1{p_k} + \frac1{p_k^2} + \dots\right)
\end{align}

i.e. in the expansion of the product the reciprocal of each $n\in M(P)$ appears once and only once. (You may need to use the Fundamental Theorem of Arithmetic) Then use the sum of geometric series formula.

EDIT: since $M(P)$ may not contain all numbers with prime divisors only in $P$, the sum cannot be obtained without further information, but the upper bound still stands.


问:
Example of a $\kappa$-long sequence of disjoint club subsets of regular cardinal $\kappa$
I'm self-studying set theory and got stuck on this exercise:

Let $\kappa$ be a regular cardinal. Give an example of a sequence $\langle C_\alpha\mid\alpha<\kappa\rangle$ such that $C_\alpha$ is club in $\kappa$ for every $\alpha<\kappa$, but
  $$\bigcap\{C_\alpha\mid\alpha<\kappa\}=\emptyset.$$

To make some progress at all, I've tried specific cases. For $\kappa=\omega$, I define $f:\omega\to\omega$ as taking $n$ to the $n$'th prime, and then define

$$C_n:=\{f(n)k\mid k<\omega\},$$

which is clearly unbounded ($|C_n|=\aleph_0$) and closed (every $n\cap C_n$ will be finite, so $\sup(n\cap C_n)=\max(n\cap C_n)\in C_n$). Furthermore due to the $f(n)$'s being primes, we have

$$\bigcap\{C_n\mid n<\omega\}=\emptyset.$$

I'm trying to do the case with $\kappa=2^{\aleph_0}$ now, where I'm trying to exploit that $|\mathbb{R}|=2^{\aleph_0}$ and $|[0,1]|=2^{\aleph_0}$, so if we set $C_n:=[n,n+1]$ then the $C_n$'s are clubs (unbounded due to cardinality and closed due to closed intervals preserve limits) and disjoint. But I can only construct countably many such $C_n$'s, so it doesn't satisfy the example of needing $2^{\aleph_0}$ many disjoint clubs.

If you guys would be so kind as to hint me in the right direction, whether it be about the $2^{\aleph_0}$ case, the general case, or about how I might generalize my methods from my special cases to the general, I would appreciate it a great deal.

Thanks.

答:
First of all, when talking about clubs we generally require that $\kappa$ has uncountable cofinality. There is a problem with clubs for cardinals whose cofinality is $\omega$, because one unbounded sequence without limit points is not really a club (but still closed and unbounded). It's true this question is true for the countable case, but its solution doesn't necessarily generalize. It's better to start with the uncountable case, and deduce from it the countable case (and seeing how you solved the countable case, you only need to solve the uncountable case anyway).

Secondly, you can't use $\Bbb R$. It's not a well-order. The idea is to use the order topology of the cardinal, not of an ordered set of the same cardinality. If you note, the real numbers is an ordered set whose cofinality is $\omega$, as the natural numbers is a cofinal sequence in that order. Furthermore, $[0,1]$ is not unbounded in $\Bbb R$ as it is very much bounded by $0$ and by $1$.

Thirdly, $2^{\aleph_0}$ can be singular. It is consistent, for example that $2^{\aleph_0}=\aleph_{\omega_1}$.

Fourthly, clubs cannot be disjoint. The whole idea is that clubs represent "almost everything", and when you intersection two "almost everything", you should get "almost everything" again. The exercise asks for you to show that it is possible to have a sequence of $\kappa$ clubs, that the intersection of all of them is empty.

Finally, to hint you for the solution, think about end-segments of $\kappa$. (with its order as an ordinal!)

ContentLicense: CC BY-SA 3.0


问:
$f:R\to \text{Im}\:f,\:\pi : R\to R/\ker f,\;f=g\circ \pi$. Check if $g:\:R/\ker f\to \text{Im}\:f$ is an isomorphism
$\DeclareMathOperator{\Im}{Im}$
I can't really think of a way to prove it, I know that we have to prove that

$$g\bigl([x_1]\bigr)=g\bigl([x_2]\bigr)\:\Rightarrow [x_1]=[x_2]$$

And

$$\forall y\in \Im f\enspace\exists x \in  R/\ker f,\;y=g(xt)$$

But I don't really know how to start.

答:
definition of $g$

we define $g$ as
\[ g \colon R / \ker f \ni [x_1] \mapsto f(x_1) \in \operatorname{Im}f . \]
We have to show that this definition doesn't depend on representation, i.e. if $[x_1] = [x_2]$, then $g([x_1]) = g([x_2])$. In other words, if $x_1 - x_2 \in \ker f$, show that $f(x_1) = f(x_2)$, which follows from definition of a kernel. This is what is called well-definedness, and often matters when there is quotient.

surjectivity

Let us show surjectivity:

for all $y \in \operatorname{Im}f$, is there $[x] \in R / \ker f$ such that $g([x]) = y$ ?

Since $y \in \ker f$, there is $x' \in R$ such that $f(x') = y$. Does $\pi(x') = [x']$ suit for $[x]$ mentioned above ?

injectivity

Now, injectivity:

Let $[x_1], [x_2] \in R / \ker f$ and $g([x_1]) = g([x_2])$. To show that $g$ is injective, it is sufficient to show that $[x_1] = [x_2]$.

Since $g([x_1]) = g([x_2]) = f(x_1) = f(x_2)$, $f(x_1) - f(x_2) = f(x_1 - x_2) = 0$, which means $x_1 - x_2 \in \ker f$. Hence $[x_1] = [x_2] \in R/ \ker f$.

Of course, it is ok to show that $\ker g = \{0\}$. This is a good exercise.


问:
A reference book to study Thom-Pontryagin Theory from basic
I am a beginner in Algebraic Topology and Differential Topology. Please suggest me a reference book to study Thom-Pontryagin Theory from basic and which book is easy to understand for a beginner. Thank you.</p>

答: The standard reference for this would be Robert Stong's book "Notes on Cobordism Theory". The first chapter sets up the theory of cobordism categories, and then in the second chapter, the Pontryagin-Thom Theorem is proved in the generality of cobordism classes of manifolds with normal/tangential structure. I think this book is not too difficult to understand. Nonetheless, the first chapter requires some basic category theory.</p>


答:
If $M(P)$ is the set of all natural numbers whose prime divisors in $P$, we can use a similar approach to the Euler product formula for the Riemann Zeta function.

Let $S = \sum_{n \in M(p)} \frac {1}{n}$.  Then $\frac{1}{p_1} S$ is the sum over the numbers whose prime factora are in $P$ and the $p_1$ exponent in each of these numbers is at least 1. So $\left( 1-\frac{1}{p_1} \right) S $ removes the terms from $S$ that are divisible by $p_1$. Repeating this sieving process we get

$$ \left( 1 - \frac{1}{p_1} \right) \left( 1- \frac{1}{p_3} \right) \cdots \left( 1-\frac{1}{p_k} \right) S = 1.$$

You can finish from here.

Title: Prove that series $\sum_{n\in M(P)} \frac{1}{n}$ converges and find its sum

Body: Let $P = \{p_1, p_2, \ldots, p_k\}$ be a finite set of prime numbers, and $M(P)$ be a set of natural numbers, whose prime divisors are in $P$. How can I prove that $$\sum_{n\in M(P)} \frac{1}{n}$$ converges and how can I find its sum? I would highly appreciate any help on this subject.


答:
So clearly if $x\in N\setminus M$ then $x\in N$ and if $x\in M$ then $x\in N$ from the hypothesis $M\subseteq N$ so that

$$M\cup N\setminus M\subseteq N$$

and thus we have only to show the other inclusion. So clearly if $x\in N$ then either $x\in M\cap N=M$ or $x\notin M\cap N=M$ (any event is true or false!) and thus $x\in M$ or $x\in N\setminus M$ so that $x\in M\cup N\setminus M$  and thus the statement follows immediately.


问:
Find a maximal ideal in $\mathbb Z[x]$ that properly contains the ideal $(x-1)$
I'm trying to find a maximal ideal in ${\mathbb Z}[x]$ that properly contains the ideal $(x-1)$.

I know the relevant definitions, and that "a proper ideal $M$ in ${\mathbb Z}[x]$ is maximal iff ${\mathbb Z}[x]/M$ is a field."

I think the maximal ideal I require will not be principal, but I can't find it.

Any help would be appreciated.

Thanks.

答: Hint: the primes containing $\,(x-1)\subset \Bbb Z[x]\,$ are in $1$-$1$ correspondence with the primes in $\,\Bbb Z[x]/(x-1)\cong \Bbb Z,\,$ by a basic property of quotient rings.</p>


问:
Definition of equality amongst elements of vector space
Suppose we are given an arbitrary vector space $V(\mathbb{F})$. We define by a set of axioms what are the properties that a given vector space has to in the form of binary operations of vector addition and scalar multiplication for it to be considered as a vector space.

However, we do not strictly define what is equality in the context of vectors.

Different vector spaces have different definitions of equality. For eg. the vector space $\mathbb{R}^n(\mathbb R)$ considering elements as n-tuples we call two vectors equal iff they are component-wise equal. Similarly, we define equality on the vector space $P_n(\mathbb R)$ i.e. the set of polynomials with degree $\le n$, as equality of coefficients. While for $M_{m\times n}(\mathbb R)$ it is defined as element-wise equality.

So I wanted to ask that is it not necessary to define equality of vectors beforehand and bind it as a set of axioms just like scalar multiplication or vector addition as the definition of equality is unique to a vector space based on the structure it defines. In a more fundamental sense, what do I mean when I say that $u+v=v+u$ when I have not even defined what equality to that specific vector space means.

答: The easy variant is to take equality as identity and as integral parto of logic (First Order Logic with Equality). In effect $a=b$ is true whenever $\phi(a)\leftrightarrow\phi(b)$ for every logical predicate $\phi$.


问:
Find the density function of X given the joint density distribution X and Y
Given the joint density function f of X and Y, find the density of X:
$$
f(x, y) =
     \begin{cases}
       xe^{-x + y} &\quad x>0, y>0 \\
       0, &\quad \text{otherwise}
     \end{cases}
$$

My approach to computing the marginal is the following:
$$
f_X = \int_{-\infty}^{\infty} f(x, y) dy =
    \begin{cases}
        \int_0^{\infty} f(x, y) dy  = xe^{-x}\int_0^{\infty} e^y dy = \infty \textbf{???!!!}, &\quad x > 0 \\
        0, &\quad \text{otherwise}
    \end{cases}
$$

My problem is that the pdf goes to $\infty$ if x>0. Am I correctly computing the pdf and the integral, or am I missing something?

答:
I suspect the joint density should have been $$
f(x, y) =
     \begin{cases}
       xe^{-(x + y)} &\quad x>0, y>0 \\
       0, &\quad \text{otherwise}
     \end{cases}
$$
since this integrates to $1$, i.e. the author probably forgot the parentheses.

Your approach to computing the marginal density w.r.t. $X$ is fine.


答:
First show that if $A\subset C$ and $B\subset C$, then $(A\cup B) \subset C$.

Then we have $M\subset N$ and $(N\setminus M)\subset N$.

To prove $N\subset M\cup (N\setminus M)$, note that $N=(N\cap M) \cup (N\setminus M)\subset M\cup (N\setminus M)$.

Alternatively, we have $M\cup (N\setminus M)=N\cup (M\setminus N)$, and as $M\subset N$, then $M\setminus N=\emptyset$.


问:
Does there exist any $C^{\infty}$-function with the following property?
Does there exists a $C^{\infty}$ function $f : \Bbb R \longrightarrow \Bbb R$ such that $f(x) = 1,$ for all $x \in \Bbb R$ with $|x| \lt 1$ and $f(x) = 0,$ for all $x \in \Bbb R$ with $|x| \gt 5\ $?

If the answer is "yes" can anybody give me some hint as to how to construct such a function? Any help in this regard will be warmly appreciated.

Thanks for your time.

答:
With a $C^\infty$ helper function $h\colon\Bbb R\to \Bbb R$ such that $h(x)=0$ for $x\le 0$ and $h(x)>0$ for $x>0$, you can easily construct such and similar examples.
For example,
$$g(x):= \frac{h(x+1)}{h(5-x)+h(x+1)}$$
(note that the denominator is always $>0$!) has the property that $g(x)=0$ for $x\ge5$ and $g(x)=1$ for $x\le 1$. Then
$$ f(x):=g(x)g(-x)$$
has your desired property.

Remains to find a suitable $h$. Here, $e^{-\frac1x}$ is your friend.


问:
explaining the derivative of $x^x$
You set the following exercise to your calculus class:

Q1. Differentiate $y(x) = x^x$.

A student submits the following solution:

Let $g(a)=a^x$ and $f(x)=x$.  Then $y(x) = g(f(x))$, so by the chain rule, $y'(x) = f'(x) g'(f(x)) = 1 \cdot x \cdot (x^{x-1}) = x^x$.

How would you explain to the student why their solution is incorrect?

To be clear, I know why this is wrong but am interested in good ways to explain it to undergraduate or high school students.

In this question someone has problems differentiating $x^x$, but they didn't take the approach of my hypothetical student.

答:
I think the easiest way is to show why $g'(a) = x * a^{x-1}$ is incorrect

Its common for students to think that this can be done because of the similar differentiation rule.

for

$y = x^n$

$y' = n * x^{n-1}$

The key is to understand that the function g in NOT in the form of $y = x^n$

See here for the neatest three line proof of the power rule I've ever seen

Title: explaining the derivative of $x^x$


问:
Given this limit involving T can I imply that T is injective?
Recently, trying to solve a problem, I got stuck in the following fact: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$ such that $$\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=0.$$ Is it true that $T \equiv 0$? I think I was able to prove it by supposing there is some $w \in \mathbb{R}^m \setminus \{0\}$ such that $T \cdot w \neq 0$. Then the sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ given by $v_k=\frac{1}{k} \cdot w$ is such that $\lim\limits_{k \to \infty} v_k=0$ and$$0=\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=\lim\limits_{k \to \infty}T \cdot \frac{v_k}{|v_k|}=\lim\limits_{k \to \infty} T \cdot \frac{w}{|w|}=T \cdot \frac{w}{|w|}\neq 0.$$ Hence $T \cdot w =0 $ for all $w \in \mathbb{R}^m$ and $T$ must be the zero transformation.

In weakening the hypothesis, I tried to find an answer to the following problem: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$, a sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ such that $\lim\limits_{k \to \infty}v_k=0$ and $$\lim\limits_{k \to \infty} T \cdot \frac{v_k}{|v_k|}=0.$$ Is it true that $Ker(T) \neq 0$?
First I tried to find a counterexample, but I couldn't. And since It is easy to prove the converse (i.e. if $Ker(T) \neq 0$ then there is sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ satisfying the above conditions) I thought this might be true but have no idea how to prove it. Any help with a proof or counterexample is appreciated.

答:
Well, that's a high mathematics!  But ...can we make it simpler, maybe?

The expression $\frac v{|v|}$ is a unit vector, for any $v$. All possible values of the fraction make a unit sphere centered at $0$.

And $T$ transforms every point of the sphere to $0$. (*)

Pairs of antipodes among them.

If a linear transformation transforms some $q$ and $-q$ to $0$, then it transforms the whole line to $0$. And due to (*) it does so to all lines, hence it is zero everywhere.

Am I right?


答:
This is a really good question! So, what you're talking about is an example of a first-order theory with equality. We define the axioms for equality beforehand (I'll state them in a minute), and then add a set of proper symbols $\mathcal P$ and proper axioms. Other examples of first-order theories with equality include that of groups, abelian groups, lattices, posets, etc.

Axioms for Equality:

$\forall x (x=x)$

$\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (p^nx_1...x_n \leftrightarrow p^ny_1...y_n))$ for each n-ary predicate symbol $p^n$ in the first order language.

$\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (f^nx_1...x_n = f^ny_1...y_n))$ for each n-ary function symbol $f^n$ in the first order language.

Now, we add to the above axioms a set of proper axioms and proper symbols to describe what is known as a vector space:

Proper symbols, functions, relations:

There is an element $0\in V$, called zero.

A function symbol $\alpha: V\times V\to V$, written as $\alpha(v,w) = v+w$.

Another function symbol $\mu: \mathbb F\times V\to V$, written as $\mu(\lambda,v) = \lambda\cdot v$.

You may be wondering why I'm calling this a first-order theory. The following proper axioms will make that evident:

Vector Space (Proper) Axioms:

$\forall x\forall y (x+y = y+x)$

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

$\forall x(0+x = x)$

$\forall x\exists y (x+y = 0)$

$\forall x(1\cdot x = x)$

$\forall a\forall b \forall x((ab)\cdot x = a\cdot (b\cdot x))$

$\forall a\forall x\forall y (a\cdot(x+y) = a\cdot x + a\cdot y)$

$\forall a\forall b\forall x ((a+b)\cdot x = a\cdot x + b\cdot x)$

Whenever I say $\forall x, \forall y, \exists y$ I really mean $\forall x\in V, \forall y\in V, \exists y\in V$ respectively. Whenever I say $\forall a,\forall b$ I really mean $\forall a\in\mathbb F,\forall b\in\mathbb F$, respectively.

Clearly, all the axioms of vector spaces are expressible in terms of the first-order language (note the quantifiers, $\forall,\exists$) endowed with the special symbol of equality (and its axioms). Hence, this is a first-order theory with equality. I hope this helps. Please let me know if you've any questions!

To summarize: We don't worry about defining $=$ among the vector space axioms, because the intention is to complement the vector space axioms with the equality axioms - and their amalgamation is the first-order theory (with equality) of vector spaces.

Title: Definition of equality amongst elements of vector space

Body:
Suppose we are given an arbitrary vector space $V(\mathbb{F})$. We define by a set of axioms what are the properties that a given vector space has to in the form of binary operations of vector addition and scalar multiplication for it to be considered as a vector space.

However, we do not strictly define what is equality in the context of vectors.

Different vector spaces have different definitions of equality. For eg. the vector space $\mathbb{R}^n(\mathbb R)$ considering elements as n-tuples we call two vectors equal iff they are component-wise equal. Similarly, we define equality on the vector space $P_n(\mathbb R)$ i.e. the set of polynomials with degree $\le n$, as equality of coefficients. While for $M_{m\times n}(\mathbb R)$ it is defined as element-wise equality.

So I wanted to ask that is it not necessary to define equality of vectors beforehand and bind it as a set of axioms just like scalar multiplication or vector addition as the definition of equality is unique to a vector space based on the structure it defines. In a more fundamental sense, what do I mean when I say that $u+v=v+u$ when I have not even defined what equality to that specific vector space means.


问:
Compute the homology group of $S^{1}\times D^{2}$
I'm having troubles computing the homology groups of $X=S^{1}\times D^{2}$ where $D^{2}$ is the $2$-disk. Actually I'm not sure what is the best idea here: if we want to apply cellular homology then we (should) have: as $0$-cells $e^{1}_{0}, e^{2}_{0}$, as $1$-cells: $e^{1}_{1}, e^{2}_{1}$ and just one $2$-cell $e_{2}$, but I'm not sure how to proceede now. On the other hand, I tried to find an excisive couple of $X$: $S^{1}$ can be taken as the union of two $1$-cells, say $a_{1}$ and $a_{2}$, in such a way that $A_{1}=a_{1}\times D^{2}$ and $A_{2}=a_{2}\times D^{2}$ is an excisive couple, and both are $A_{1}$ and $A_{2}$ are acyclic. My problem now is, what about the homology of $A_{1}\cap A_{2}$?

答:
Thanks to @Pedro for the suggestion. The question is not actually hard but it took me off guard, so here is the idea:

since $D^{2}$ is convex, then it is contractible to a single point, so that we can write:
$$D^{2}\simeq\{x_{0}\}$$
Thus:
$$S^{1}\times D^{2}\simeq S^{1}\times\{x_{0}\}\simeq S^{1}$$
which turns into an isomorphism in homology, hence the work is done.


问:
How to check if $x^TCx\geq0$?
I have the next 3x3 block matrix C, where each block is a square matrix.

$$
C = \begin{bmatrix}
0 & A & B \\
0 & A+K_1 & B \\
0 & A & B+K_2
\end{bmatrix},
$$
where $K_i$ is also a squared diagonal matrix (I have freedom choosing their elements). I want to be sure that $x^TCx \geq 0$ choosing the appropriated $K_i$.

EDIT

The answer is no

答:
Unless $A=B=0$, this is an impossible mission.

Let $x^\top =(u^\top ,v^\top ,w^\top )$, where the partitioning of $x$ conforms to the partitioning of $C$. If $x^\top Cx\ge0$ for all $x$, then in particular, this is true for $w=0$, i.e. $x^\top Cx=u^\top Av + v^\top (A+K_1)v\ge0$. However, if $A\ne0$, then for every $v\notin\ker(A)$, we can always find a vector $u$ (e.g. $u=-kAv$ for a large positive $k$) such that $u^\top Av$ and in turn $x^\top Cx$ are as negative as possible.

Put it another way, $x^\top Cx\ge0$ for all $x$ if and only if $P=C+C^\top $ is positive semidefinite. For a positive semidefinite matrix $P$, if it has a zero on its diagonal, the row and column containing this zero diagonal entry must also be zero, otherwise $P$ will contain a $2\times2$ submatrix that has negative determinant. Now the $(1,1)$-th subblock of $C+C^\top $ is zero, but the $(1,2)$-th and $(1,3)$-th subblocks are $A$ and $B$. So, $C+C^\top $ cannot be positive semidefinite when $A\ne0$ or $B\ne0$.

Title: How to check if $x^TCx\geq0$?


问:
l'Hôpital and substitution give different limits
Find $$\lim_{x \to 0}\frac{\sin x-x+\frac{x^3}6}{x^3}$$

L'Hospital 3 times:

$$\lim_{x \to 0}\frac{\sin x-x+\frac{x^3}6}{x^3} = \lim_{x \to 0 } \frac{\cos x - 1 + \frac{x^2}2}{3x^2}=\lim_{x \to 0}\frac{-\sin x+x}{6x}=\lim_{x \to 0}\frac{-\cos x +1}{6}=0$$

L'Hospital once and Substitution:
$$\lim_{x \to 0}\frac{\sin x-x+\frac{x^3}6}{x^3} = \lim_{x \to 0 } \frac{\cos x - 1 + \frac{x^2}2}{3x^2} = \lim_{x \to 0}\frac{x^2}{6x^2}=\frac16$$

答:
The second one is wrong. How did you drop $\frac {\cos x-1} {3x^{2}}$? The limit of this part is $-\frac  16$ which cancels with the limit of the other term.

It is also not clear as to what you mean by 'substitution'.


问:
I'm having some trouble in the below propositional logic question.
How to write the statement for the following?

If it does not rain, then if we go swimming then we have to take the bus.

I have been confused with the above "then" repetition...

答:
Let

Does not rain $\equiv A$

Go swimming $\equiv B$

Take the bus $\equiv C$

the n your statement becomes $A → [B →  C]$. For better understanding, you can say: Assume that it doesn't rain, then if we go swimming then we take the bus.


问:
How many compositions of $n \in N$ are there where each part is greater than $1$?
Can someone help me with this?

Let $n \in N$. How many compositions of $n$ are there where each part is greater than $1$? (Number of parts are not restricted)

答:
We know that number of compositions of $n$ in $k$ positive parts is $$\binom{k-1}{n-1}$$ and to each composition of $n$ in $k$ parts correspondes a composition of $n+k$ into $k$ parts greater than $1$ if each part we increase by 1. Conversely: to each composition of n into k parts greater than two each part we decrease by 1 we get a composition of $n-k$ into k parts greater than 1 so number of compositions of $n$ into $k$ parts greater than $1$  is $$\binom{k-1}{n-k-1}$$ and the total number is
$$\sum_{k\geq1}\binom{k-1}{n-k-1}=\sum_{k\geq1}\binom{k-1}{n-2-(k-1)}=F_{n-2}$$

$F_n$ is Fibonacci number


答:
Let $p =$ “It is raining.”, $q =$ “We will swim.” and $r =$ “We are taking the bus.”. The proposition that we have is

If it does not rain, then if we go swimming then we have to take the bus.

As pointed at the answer given by Jotabeta, we have that this proposition has the form

\begin{align*}
\neg p \implies (q \implies r)
\end{align*}

You have an hypothesis that is $p$ (If it does not rain), but then at the conclusion you other implication, i.e., the conclusion itself is another implication, namely $q \implies r$. Then, you just put all together.

Also, you can check by yourself, it would be the same if we have written $(\neg p \wedge q) \implies r$. You can use a truth table to immediately conclude that

\begin{align*}
(\neg p \Rightarrow (q \Rightarrow r)) \iff ((\neg p \wedge q) \Rightarrow r)
\end{align*}

Title: I'm having some trouble in the below propositional logic question.

Body:
How to write the statement for the following?

If it does not rain, then if we go swimming then we have to take the bus.

I have been confused with the above "then" repetition...


问:
Dimension of solution set for $Ax=b$. But the solution set isn't a vector space so why talk about it's dimensions?
The solution set of a system of $n$ linear equations written as a matrix equation $AX=b$,$b\neq0$ cannot have zero vector in it. This means that the solution set cannot be a vector space and talking about it's dimensions is absurd.

But why do then people talk about the dimension of the solution set as in here :https://math.stackexchange.com/questions/1080941/determining-the-dimension-of-the-set-of-solutions-of-the-system-ax-b-without-s#:~:text=solution%20of%20the%20homogeneous%20equation,is%2C%20of%20course%2C%200.

答: That is true. In fact, if $b \neq 0$, then the set of solutions of the system $Ax = b$ is not a vector space (because $0$ will not be an element as you well pointed out). Although not being a vector space, that set will constitute an affine space. And the dimension of this affine space, will be the dimension of the vector space generated by the set of the solutions of the homogenous system associated $Ax = 0$.


问:
In how many ways can a team be formed when some players can play multiple positions?
I want to form a lineup where we have 1 goalkeeper, 2 defenders and 3 forwards. We have 3 goalkeepers, 7 defenders and 10 forwards available, as well as 4 players who can play as defender or forward. How many possible lineups are there?

My approach was to separately consider the cases where $n$ of the 4 players (lets call them hybrid players) who can play as defender or forward are first chosen as defenders or forwards. If we first pick the 2 defenders, we can pick 0, 1 or 2 of the hybrid players as defenders, and then either 4, 3 or 2 of these players will be left available for forward. I tried calculating each case separately, and summing the combinations for each case together. However, if I calculate it such that we first choose the forwards, and we can choose 3, 2, 1 or 0 of the hybrid players to be forwards, I get a different result. I assume there is a smarter way to do this, but I can't seem to figure it out.

答:
Assuming you consider the positions within each position group irrelevant, you would still want to consider, for a given selection of hybrids, which position group each of the hybrids play in. Further, given that the GK selection is independent of the rest of the selections, that is simply a factor of 3 on the final answer, so I'll ignore it from here. The solution is then broken up based on how many hybrids are selected, where the hybrids play, and any remaining selections. I will include a sample term with reasoning, and the remaining terms are left as an exercise.
Consider the case where 3 hybrids are selected , two of them playing as forwards, one as a defender. Then the term is:
$${4 \choose 3} \cdot {3 \choose 1} \cdot {10 \choose 1} \cdot {7 \choose 1}$$
With each factor corresponding respectively to combinations of:

hybrid selections

hybrid positions

forward selections

defender selections

By iterating over the number of selected hybrids and the possible position combinations they play, I reach 16972 combinations before considering the GK.


问:
Show that every ring with three elements is commutative

Show that every ring with three elements is commutative

I'm unsure how can I proceed with the problem? I found some answers online on $x^3=x$ being commutative, but not sure if this is equivalent to the property I want to show?

答:
In order to even express non-commutativity, you need a lot of elements:
You need $a$ and $b$ with $ab\ne ba$. Then of course we must have $a\ne b$, and neither $a$ nor be may be $=0$ (because $0x=x0=0$ for all $x$). Thus our three elements must be $0,a,b$. Moreover, the additive group must be cyclic of order $3$, so we must have $b=-a$. But then $a\cdot(-a)=(-a)\cdot a$.


答:
If $A$ is a $m \times n$ matrix and if $b \in \mathbb R^m,$ then the soltion set $L$ of $Ax=b$ has the form

$$ L=x_0 +U,$$

where $U$ is a subspace of $ \mathbb R^n$ and $x_0$ is a special solution of $Ax=b.$ Then

$$ \dim L := \dim U.$$

Note that $U$ is the solution space of $Ax=0.$

Title: Dimension of solution set for $Ax=b$. But the solution set isn't a vector space so why talk about it's dimensions?

Body:
The solution set of a system of $n$ linear equations written as a matrix equation $AX=b$,$b\neq0$ cannot have zero vector in it. This means that the solution set cannot be a vector space and talking about it's dimensions is absurd.

But why do then people talk about the dimension of the solution set as in here :https://math.stackexchange.com/questions/1080941/determining-the-dimension-of-the-set-of-solutions-of-the-system-ax-b-without-s#:~:text=solution%20of%20the%20homogeneous%20equation,is%2C%20of%20course%2C%200.


问:
Bounds of a trigonometric expression
Find the bounds of the function $\frac{|\sec x|}{\sqrt{2}} (\sqrt{2-4\cos 2x} + 2\sqrt{2}\sqrt{2+3\cos2x})$

I first tried putting conditions for the square roots to exist to obtain bounds on $\cos 2x$. Then wrote $\sec {x}$ in terms of $\cos 2x$. But the resulting expression seems very complicated

答:
Consider the positive side because of the symmetry and make $x=\cos ^{-1}(t)$. The expression becomes
$$f(t)=\frac{2 }{t}\left(\sqrt{3-4 t^2}+\sqrt{6 t^2-1}\right)$$ The first derivative
$$f'(t)=\frac{2 }{t^2}\left(\frac{1}{\sqrt{6 t^2-1}}-\frac{3}{\sqrt{3-4 t^2}}\right)$$
Let $y=t^2$ and we need to solve for $y$
$$\sqrt{3-4 y}=3\sqrt{6 y-1} \implies y=\frac 6 {29}\implies t_*=\sqrt{\frac 6 {29}}$$ So, the maximum is
$$f(t_*)=4 \sqrt{\frac{14}{3}}$$ and this is the upper bound.

Now, it is your turn.

Title: Bounds of a trigonometric expression


问:
How to Solve for t and C in S'(t)=-420t^2/5
Annual sales of a certain cell phone has been declining at a rate of S′(t) = −420t2/5 phones per year, where S(t) is the number of cell phones sold in t years. The company is planning on withdrawing this model when annual sales reach 60,000 cell phones. If current annual sales of these phones are 100,000, find S(t). How long will the company continue to manufacture this phone?

I started by integrating S'(t) = -300t^7/5 + C

But this is where im stuck. Since im not given any indication of the starting t when s(t) = 100000 nor s(t) = 60000, how would I go about solving this equation.

Should I start by finding dS/dt? but yet again I don't have any indication for that.
any pointers would be appreciated. thank you.

答:
You got that $S(t)=-300 t^{7/5}+C.$ Since you know the current $S$ you can assume that now is when $t=0.$ That is: $S(t)=100000-300 t^{7/5}.$

How long will the company continue to manufacture this phone? We have to solve $S(t)=60000.$ In other words: $$100000-300 t^{7/5}=60000.$$


问:
Finding a basis with Change of Basis

Find the coordinate vector for v relative to the basis S = {v1, v2, v3} for $R^3$.
  $$v = (2,-1,3);$$
  $$v1 = (1,0,0); v2 = (2,2,0); v3 = (3,3,3);$$

So I did and I got the coordinate vector space as (v)s = (3, -2, 1). I do not believe that this answer is wrong. So now I am at the next question.

Consider the coordinate vector:
  $$ [w]s =
\left[ \begin{array}{c}
6 \\
-1 \\
4
\end{array} \right]$$
  Find w if S is the basis of the first question.

I have no idea how to do this as my textbook is not accurate. Can someone please help me.

答:
Your answer to the first part is correct.

The second question is essentially the reverse process. If you know what the coordinate vector for w is relative to the basis S, can you "unpack" that coordinate vector to recover the original vector w?

Title: Finding a basis with Change of Basis


答:
The set of all solutions for $Ax = b$ is only a vector space iff  $b=0$, as you said, and if that is the case you will have your normal definition of dimension.

But if $b \neq 0$, the set of all solutions will be an Affine space.
An affine space is basically a point plus a vector space.

For example, let $V$ be the set of solutions for $Ax = 0$, and let $k$ be just one solution for $Ax = b$. Then, the set of ALL solutions of $Ax = B$ will be: $$\{k\} + V$$

And because this is a point plus a vector space this is an Affine space. By definition, the dimension of an Affine space $\{k\} + V$ is just the dimension of the vector space $V$ and that is why even if the space of all solutions for $Ax = B$ isn't a vector space, you can still define its dimension.


问:
What's the torsion subgroup of $y^2 = x^3+2015^2x+2015^3$?

$Q$. (a) Let $E$ be the elliptic curve given by $y^2 = x^3+2015^2x+2015^3$. Prove that $E(\mathbb Q)_{\text{tors}}$ is the trivial group. (b) Now let $E$ be the elliptic curve $y^2 = x^3+x+2$. Determine $E(\mathbb Q)_{\text{tors}}$

What's a good method for part (a)? In the attempt below, at some point I've to show that $E(\mathbb F_7)_{\text{tors}} \cong \mathbb Z/ 11 \mathbb Z$. During this do I've to check that every point that I find is torsion? Or is there some fact that allows me to know the points on my list are indeed torsion without any checking? In addition I can't finish part (b) after showing $E(\mathbb F_5)  \cong \mathbb Z/ 4 \mathbb Z \cong E(\mathbb F_3).$

[EDIT: I've realised that part (b) follows from Nagell-Lutz -- in fact $E(\mathbb Q)_{\text{tors}} \cong  \mathbb Z/ 4.$ However, I'm wondering if it can be done with reduction? Does the fact that $E(\mathbb F_5)  \cong \mathbb Z/ 4 \mathbb Z \cong E(\mathbb F_3)$ imply that $E(\mathbb Q)_{\text{tors}} \cong  \mathbb Z/ 4?$]

(a) The discriminat $\Delta = -5^6 \cdot 13^6 \cdot 31^7$. This means $E$ has good reduction for all primes $P \ne 2,5,13,31.$  In particular, for ${E}(\mathbb F_3):$ $y^2= x^3+x+2$ we have $E(\mathbb F_3)_{\text{tors}} \cong \mathbb Z/ 4 \mathbb Z, $ as the torsion points in this case are $4(1,1) = 4(1, 2) = 2(2,0) =  \mathcal{O}$.

Since the next good prime is $7$ we can reduce ${E}(\mathbb F_7):$ $y^2= x^3+x+6$, which has points:

$\left\{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)\right\} $

Then I find the order of every point on this list and conclude that:

and $\left\{(1 , 1), (1 , 6), (2 , 3), (2 , 4), (3 , 1), (3 , 6), (4 , 2 ), (4 , 5), (6 , 2), (6 , 5)\right\} \cup \left\{\mathcal O\right\} \cong \mathbb Z/ 11 \mathbb Z.$

Therefore $E(\mathbb F_7)_{\text{tors}} \cong \mathbb Z/ 11 \mathbb Z. $ Then the order of $E(\mathbb Q)_{\text{tors}}$ divides $\gcd(4,11) = 1$ therefore $E(\mathbb Q)_{\text{tors}}$ is the trivial group.

(b) If ${E}/ \mathbb Q:$ $y^2= x^3+x+2$  then $E(\mathbb F_3)  \cong \mathbb Z/ 4 \mathbb Z$ as above, but also similarly $E(\mathbb F_5)  \cong \mathbb Z/ 4 \mathbb Z$. I can't finish this but I'm guessing it somehow follows it from part (a). If so, how? (See the edit in paragraph 2 please).

答:
Yes, your answer to part (a) is correct. It's probably good to be slightly more careful in the following way: Let $p$ be an odd prime at which $E$ has good reduction. We know that $E(\mathbb{Q})[m]$ injects into $E(\mathbb{F}_p)$ for each $m$ coprime to $p$. In particular we have that $\#E(\mathbb{Q})_{tors}$ divides $ p^r \# E(\mathbb{F}_p)$ for some $r$. Then applying this with $p = 3, 7$ gives the result.

For part (b) you have showed (via the same observation above) that $\#E(\mathbb{Q})_{tors}$ divides $4$. It remains to find a point of order $4$ to show this is the entrie thing. Notice that $P = (-1, 0)$ is an obvious (unique) point of order $2$. Now, when is $2Q$ a $2$-torsion point? Precisely when the tangent to $E$ at $Q$ meets $E$ at the $x$-axis.

All you need to do at this point is to solve the arising equation to find that $(1,2)$ is a point of order $4$.

Aside: A (generalisation of a) similar approach to what I mention for classifying $4$-torsion points at the end is used in Silverman VIII Ex8.13 to provide a cute way of constructing $X_1(N)$ for $N \geq 4$.


问:
How to derivate the arc length of manifold curve？
I would like to find the arc length of a manifold $a\left( \theta  \right)$. The manifold can be regarded as a curve embedded in an N-dimensional complex space.

The manifold is given as
$a\left( \theta  \right) = {e^{-jr\pi \sin \theta }} = \left[ {\begin{array}{*{20}{c}}
{{e^{-j\pi \sin \theta }}}\\
{{e^{-j2\pi \sin \theta }}}\\
 \vdots \\
{{e^{-jN\pi \sin \theta }}}
\end{array}} \right]$ , and $r = {\left[ {1,2, \cdots ,N} \right]^T}$.

The related reference defines the arc length s as
$s = \int_0^\theta  {\left\| {\frac{{da\left( \theta  \right)}}{{d\theta }}} \right\|d\theta } $, and the result is given as $s\left( \theta  \right) = \pi \left\| r \right\|\left( {1 - \cos \theta } \right)$.

The problem is that I know little about differential geometry, and I don't know how to derivate the above formulation. I need help with the derivation.

答:
This is a differential geometry question, but you don't need differential geometry to solve it. Just apply the definition
$$ \left\|{da(\theta) \over d\theta}\right\| = \sqrt{ {da(\theta) \over d\theta} \cdot \overline{da(\theta) \over d\theta}} = \sqrt{\sum_{i=1}^N {da_i(\theta) \over d\theta}\overline{da_i(\theta) \over d\theta}}$$
and you'll get a simple expression for the integrand that you can then integrate.


问:
Ambiguous Question in Mendelson's Intro to Topology
I'm reading Bert Mendelson's Intro to Topology and there's a lemma that's kinda worded ambiguously.

So the lemma goes as follows

Lemma Given a subset $A$ of a topological space and a point $x \not \in \bar A,$ then $x \not \in F$ for some closed set $F$ containing $A$.

My question is: Is Mendelson saying that this lemma is true for all sets $F$ such that A $\subset$ F?

OR

Is he saying that this lemma says that there exists a set $F$ satisfying the above properties?

答:
He clearly says some so it's certainly not meant to hold for all such subsets.

The proof is obvious of course: take $F=\overline{A}$ and you're done. This can only mean that in this text he hasn't shown yet that $\overline{A}$ is closed, maybe it's a step towards that fact?


问:
What happens outside radius of convergence
A real power series $\sum_{n=0}^\infty a_n z^n$ has radius of convergence $R$. I am able to prove that for any real number $r>R$, the sequence $|a_n|r^n$ must be unbounded. Must it also tend to $\infty$? Please give me some hints, thanks.</p>

答:
Let $a_n=\frac{1}{n!}$ for $n$ even, and let $a_n=1$ for $n$ odd. </p>

<p>Then the radius of convergence of the series $\sum_{n=0}^\infty a_nz^n$ is $1$, but for fixed $r\gt 1$ it is not the case that $|a_n|r^n\to\infty$. </p>

Body:
Let $a_n=\frac{1}{n!}$ for $n$ even, and let $a_n=1$ for $n$ odd.

Then the radius of convergence of the series $\sum_{n=0}^\infty a_nz^n$ is $1$, but for fixed $r\gt 1$ it is not the case that $|a_n|r^n\to\infty$.


问:
Conditional probability with Bernoulli and Normal
Let $X$ be a normal variable with mean $0$ and variance $1$. Let $Y$ be Bernoulli with $p = \frac{1}{2}$, and suppose that $X$ and $Y$ are independent. Let $Z = (Y + 1)X$. Find the conditional probability that $Y = 1$ given that $Z = 3$.

I know I need to find $P(Y = 1 | Z = 3) = \frac{P(Y = 1 \cap Z = 3)}{P(Z = 3)}$.

Attempt:

For the numerator, $P(Y = 1 \cap Z = 3) = P(Y = 1 \cap (Y + 1)X = 3) = P(2X = 3) = P(X = \frac{3}{2})$. Since $X$ is standard normal, $P(X = \frac{3}{2}) \approx 0.130$

As for the denominator, $P(Z = 3) = $?

I can't figure this out.

Any assistance is much appreciated.

答: $$P(Z=3)=P(X=3,Y=0)+P(X=\frac  3 2, Y=1)$$ $$=P(X=3,P(Y=0)+P(X=\frac  3 2)P(Y=1)=0+0=0$$ since $X$ is a continuous random variable.  [$P(X=a)=0$ for every real number $a$].


问:
ODE solving mistake in homogeneous and particular solution method
When I was solving the ODE $x^{'}+2x=e^t$, $x(0)=5$, if using convolution method I got below derivation:

$x=x(0)e^{-2t}+\int_0^t {e^{-2(t-\tau)} e^{\tau} d\tau}=5e^{-2t} +\frac{1}{3}e^t-\frac{1}{3} \\$

Above seems to be correct since I.C. is satistifed. But when I practice the method of homogeneous with particular solution, I came up with wrong solution:

$x_h(t)=x(0)e^{-2t}=5e^{-2t}\\$

$x_p(t)=\frac{1}{3}e^t\\$

$x(t)=x_h(t)+x_p(t))=5e^{-2t}+\frac{1}{3}e^t\\$

The particular solution $x_p(t)$ was by assuming $x_p(t)=Le^{t}$, and plug it into DE so that:

$x_p^{'}(t)+2x_p(t)=e^t\\$

$Le^t+2Le^t=e^t\\$

$L=\frac{1}{3}\\$

$x_p(t)=\frac{1}{3}e^t$

Hopefully my mistake can be pointed out.

Body:
$$x'+2x=e^{t}$$
Is a linear ODE whose integrating factor is $I=e^{2t}$, then
$$x(t)=e^{-2t} \int e^{2t} e^{t} dt+C e^{-2t}$$
$$ \implies x(t)=\frac{e^{t}}{3}+C e^{-2t}.$$
$x(0)=5$ gives $C=14/3$. Finally
$$x(t)=\frac{e^{t}}{3}+\frac{14}{3} e^{-2t}.$$


问:
Definition of the Coefficients of the connection and the covariant derivative
I am watching this general relativity course on youtube and they define the coefficients of the connection on the first line of this timestamp of the video:

https://youtu.be/X5TU0eNz5cU?t=1665

Notice that the $\Upsilon$ symbols represent elements of an arbitrary basis (not necessarily coordinate/holonomic).

My questions are two:

1-First in this course they get that the covariant derivative of a vector field in an arbitrary basis has a term contracting the first lower index of the coefficient of the connection with he vector field. This is opposite to what I see in most other places where they contract the second one, but it is true that in most other sources they work with a torsion free connection and in a coordinate basis for which the symbol is symmetric. Is the expression they obtain the correct one?

2-How does one get from the second step of the first line (right after the definition symbol) to the next step? I am not sure I can prove it or how to see it is consistent. I was wondering if anybody could help me out here. I can follow everything else in the course but this pesky bit :).

Thank you very much in advance!

Edit: In case the link doesn't work, I am referring to timestamp 27:45

答:
You're correct that the more usual conventional of the $\Gamma$'s would be to take $\nabla_{\Upsilon_b}\Upsilon_c = \Gamma^a_{\ bc}\Upsilon_a$. However it's just a convention, so as long as the lecturer is consistent, it shouldn't be a problem.

To go from the first step to the second: presumably $\Upsilon^a$ denotes the dual basis to $\Upsilon_a$, i.e. the set of 1-forms satisfying $\langle \Upsilon^a, \Upsilon_b\rangle = \delta^a_b$. Then since the vector fields $\Upsilon_a$ form a basis at each point, $\nabla_{\Upsilon_b}\Upsilon_c$ must be expressible as a linear combination of the $\Upsilon_a$, i.e.
$$
\nabla_{\Upsilon_c}\Upsilon_b = A^a_{\ cb}\Upsilon_a
$$
for some functions $A^a_{\ bc}$. Then apply the dual form $\Upsilon^d$ to both sides to yield
$$
\langle \Upsilon^d,\nabla_{\Upsilon_c}\Upsilon_b\rangle = A^a_{\ cb}\langle\Upsilon^d,\Upsilon_a\rangle\\
\implies \Gamma^d_{\ bc} = A^a_{\ bc}\delta^d_a = A^d_{\ cb}
$$
giving the expression.


问:
Show that probability of a biased coin returning heads more than half the time is less than a certain number
Let X be the number of successes of a biased coin returning head $\frac{1}{4}$ of the time within $k$ tosses. Show that $P(X\geq\frac{k}{2})\leq\frac{1}{2^{k/5}}$.

So I know that $X$ is $B(k,\frac{1}{4})$. And I was thinking about re-expressing $P(X\geq\frac{k}{2})$ as $\sum_{r=\lceil k/2\rceil}^k \binom{k}{r}(\frac{1}{4})^r(\frac{3}{4})^{k-r} =\sum_{r=\lceil k/2\rceil}^k \binom{k}{r}(\frac{3^{k-r}}{4^k})$  but I don't really know what to do afterwards.

I was told I can use $\binom{2r}{r}\leq\frac{4^r}{2}$ but the only way I can think of is to let $\binom{k}{r}\leq\binom{k}{\frac{k}{2}}\leq2^{k-1}$ for all $r$ and after that I get a very high upper bound much larger than required.

Would appreciate any help towards solving the question. Thank you.

Edit: Made a typo should have been $P(X\geq\frac{k}{2})$ instead of $P(X>\frac{k}{2}).$

Edit 2: Fixed more typos

答:
That is quite a tight upper bound: for example $\frac{1}{2^{k/4}}$ would not work for large $k$ such as $k= 68$.

A possible approach with some gaps for you to fill:

Show the probability is higher for $k=2m$ than for $k=2m-1$, so can concentrate on even $k$

For even $k$, show the probability is less than $2 P\left(X = \frac k2\right)$

Use your $\binom{k}{k/2}\leq \frac{4^{k/2}}{2}$ to say $2 P\left(X = \frac k2\right) = 2{k \choose k/2} \frac{3^{k/2}}{4^k} \le 2\frac{4^{k/2}}{2} \frac{3^{k/2}}{4^k} =\left(\sqrt{\frac{3}{4}}\right)^k$

$\sqrt{\frac{3}{4}}\approx 0.866 < 0.871\approx \frac{1}{2^{1/5}}$


答:
$\{Y=1\cap (Y+1)X=3\}=\{Y=1\cap X=3/2\}$ and $\{Y=0\cap (Y+1)X=3\}=\{Y=0\cap X=3\}$.

However these are null events, since $X$ is continuous.  So the expression needs to use the appropriate probability density functions.

$$\mathsf P(Y=1\mid Z=3) =\dfrac{f_{X}(3/2)\,\mathsf P(Y=1)}{f_{X}(3/2)\,\mathsf P(Y=1)+f_{X}(3)\,\mathsf P(Y=0)}$$

Title: Conditional probability with Bernoulli and Normal

Body:
Let $X$ be a normal variable with mean $0$ and variance $1$. Let $Y$ be Bernoulli with $p = \frac{1}{2}$, and suppose that $X$ and $Y$ are independent. Let $Z = (Y + 1)X$. Find the conditional probability that $Y = 1$ given that $Z = 3$.

I know I need to find $P(Y = 1 | Z = 3) = \frac{P(Y = 1 \cap Z = 3)}{P(Z = 3)}$.

Attempt:

For the numerator, $P(Y = 1 \cap Z = 3) = P(Y = 1 \cap (Y + 1)X = 3) = P(2X = 3) = P(X = \frac{3}{2})$. Since $X$ is standard normal, $P(X = \frac{3}{2}) \approx 0.130$

As for the denominator, $P(Z = 3) = $?

I can't figure this out.

Any assistance is much appreciated.


问:
A covering map is a homeomorphism
Let $p : E \rightarrow X$ be a covering map. Suppose $E$ is path connected and that $\pi_1(X, b) = \{1\}$. Prove that $p$ is a homeomorphism.

I have no idea how to approach this problem.

答:
As $p$ is already open, continuous and onto, we only need that $p$ is injective.

So suppose $p$ is not and we have $e_1 \neq e_2$ in $E$ with $p(e_1)=p(e_2)$. Let $x_0 = p(e_1)=p(e_2)$. Now appply 54.4 in Munkres (2nd ed p. 345) where it says that $\pi_1(X,x_0)$ has a surjection onto $p^{-1}[\{x_0\}]$ and note that we have an outright contradiction, as the former set has size $1$ by assumption and the fibre on the right has at least size $2$, containing $\{e_1,e_2\}$. So $p$ is indeed 1-1.

Title: A covering map is a homeomorphism


问:
Lower bound on $\log \left( \sum_{i=1}^tx_i \right)$

I am wondering if there is a known lower bound on $\log \left( \sum_{i=1}^tx_i \right)$ in terms of some function of $\log(x_i), i=1,\cdots,t$?

My attempt:

Let $t=2$, then:

\begin{align*}
\log(x_1+x_2) &= \log x_1\left(1+\frac{x_2}{x_1} \right) \\
&=\log x_1 + \log \left(1+\frac{x_2}{x_1} \right) \\
&\geq \log x_1 + \left( \frac{\frac{x_2}{x_1}}{1+\frac{x_2}{x_1}} \right) \\
&=\log x_1 + \frac{x_2}{x_1+x_2}
\end{align*}

So I am wondering if I can get a similar or a better lower bound for a general $t$ i.e. $t>2$?

答: $\log \, (\sum\limits_{i=1}^{t}x_i)=\log n+\log \, (\frac 1 n\sum\limits_{i=1}^{t}x_i)) \geq \log n+  (\frac 1 n \sum\limits_{i=1}^{t} \log x_i))$ by Jensen's inequality.


问:
solving differential equations : $ay' + by^2 + cy = u$
solving differential equations: $ay' + by^2 + cy = u$ with $a,b,c$ are positive constants and $u$ is arbitrary constant.

I can solve this equation:

$ay' + by^2 + cy = 0$

$\Rightarrow -a\frac{dy}{y(by+c)}=dx $

$\Rightarrow \frac{a}{c}(\frac{bdy}{by+c}-\frac{dy}{y})=dx$

$\Rightarrow ln(\frac{by+c}{y})=\frac{c}{a}x$

$\Rightarrow y = \frac{c}{e^{\frac{c}{a}x}-b}$

But with an arbitrary constant at the righthand side, I have no idea.

答:
With a constant in the RHS, the equation is still separable and you solve it the same way !

$$\frac{a\,dy}{by^2+cy-u}=-dx.$$

After integration,

$$\frac{2a}{\sqrt{c^2+4 b u}}\arctan\frac{c + 2 b y}{\sqrt{c^2 + 4 b u}}=c-x$$ from which you can draw $y$.


问:
Every algebraic extension of a perfect field is separable and perfect
I am trying to prove this statement in the characteristic $p>0$ case.

Every algebraic extension of a perfect field is separable and perfect.

This is stated as a corollary of Proposition V.6.11 in Lang.

Let $k$ be a field and let $K$ be a normal extension of $k$. Let $G$ be the group of automorphisms of $K$ that fix $k$. Let $K^G$ be the fixed field of $G$ acting on $K$, and $K_0$ the maximal separable subextension of $K$. Then $K^G$ is purely inseparable over $k$, $K$ is separable over $K^G$, $K_0\cap K^G =k$  and $K_0K^G=K$.

Lang doesn't write many details in the proof except that every finite extension is contained in a normal extension.

This is what I have so far. Suppose, $k=k^p$. Let $E$ be an algebraic extension of $k$. Let $\alpha \in E$. Now, $\alpha$ is contained in the field, say $K$ generated by the roots of the minimal polynomial of $\alpha$ over $k$ (and is hence normal as it's the splitting field of the minimal polynomial of $\alpha$). At this point, I don't know how to proceed. I am tempted to use the Frobenius map, but it does not fix every element of $k$ although it does stabilize it in this perfect case. Any hints will be appreciated. Thanks.

答:
It is enough to prove that every finite extension of a perfect field is perfect and separable. Thus let $K$ be a perfect field and $L/K$ a finite extension and consider $\Phi:L\to L,x\mapsto x^p$.

$\mathbf{L}$ is perfect. We endow $L$ with two $K$-vector space structures: the standard one (using the product of $L$) and the following $K$-vector space structure:
$$\forall c\in K,\forall x\in L,~~ c.x \overset{\text{def}}= c^p\times x$$
Let's write $L'$ for $L$ with this $K$-vector space structure. Every linearly independent family in $L$ remains linearly independent when viewed as a family in $L'$, and $K$ being perfect, every spanning family of $L$ remains so when viewed as a family in $L'$. In other words: every $K$-basis of $L$ is a $K$-basis of $L'$, so that
$$\dim_K L=\dim_KL'.$$
Now $\Phi$ is a one to one $K$-linear mapping $L\to L'$. Since both $L$ and $L'$ have the same finite dimension over $K$, $\Phi$ is an isomorphism, and so $L^p=\mathrm{Im}(\Phi)=L$ and $L$ is perfect.

$\mathbf{L/K}$ is separable. Let $x\in L$ have minimal polynomial $\mu$. Let $d\geq 0$ be maximal with the property that there exists some polynomial $Q=\sum_{i=0}^{n}a_i X^i\in K[X]$ with $\mu=Q(X^{p^d})$. Note that $Q$ is separable over $K$. Let us prove that $d=0$ (so that $\mu=Q$ is separable). Since $K$ is perfect we can find $\alpha_0,\dots,\alpha_n\in K$ with $\alpha_i^{p^d}=a_i$ for $i=0,\dots,n$, thus
$$
\mu=R^{p^d}
$$
where $R=\sum_{i=0}^n\alpha_iX^{i}\in K[X]$. Since $\mu$ is irreducible over $K$, $p^d=1$ i.e. $d=0$ and $\mu=R=Q$ is separable.


问:
Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$
$\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

I guess it is convergent, so I apply comparsion test for this.

$\frac{log^2}{n^2} < \frac{n^2}{n^2} = 1$

So it is bounded by 1 and hence it is convergent.

Can I do in this way? It kind of make sense, but I have never deal with constant in this case

Usually we do bounding using another term depend on $n$?

答:
Since $\log n\le n-1$ for all $n$, we get $\frac{1}{4}\log n \le \sqrt[4]{n}-1\le \sqrt[4]{n}$ for all $n$. So
$$\frac{\log^2 n}{n^2}\le \frac{16\sqrt{n}}{n^2}=\frac{16}{n\sqrt{n}}$$
and we know that $\sum \frac{16}{n\sqrt{n}}$ converges. So by comparison test, $\sum \frac{\log^2 n}{n^2}$ converges.</p>


答:
This is an equation of the Riccati type. One can set $y=\frac{az'}{bz}$, then
$$
u=ay'+by^2+cy=\frac{a^2z''}{bz}+\frac{caz'}{bz}\\~\\
\iff
a^2z''+acz'-ubz=0
$$
This now is a second order linear DE with constant coefficients, so standard solution methods apply. The characteristic roots are solutions of the quadratic polynomial
$$
0=4a^2\lambda^2+4acλ-4ub=(2aλ-c)^2-(c^2+4ub)
$$

Title: solving differential equations : $ay' + by^2 + cy = u$

Body:
solving differential equations: $ay' + by^2 + cy = u$ with $a,b,c$ are positive constants and $u$ is arbitrary constant.

I can solve this equation:

$ay' + by^2 + cy = 0$

$\Rightarrow -a\frac{dy}{y(by+c)}=dx $

$\Rightarrow \frac{a}{c}(\frac{bdy}{by+c}-\frac{dy}{y})=dx$

$\Rightarrow ln(\frac{by+c}{y})=\frac{c}{a}x$

$\Rightarrow y = \frac{c}{e^{\frac{c}{a}x}-b}$

But with an arbitrary constant at the righthand side, I have no idea.


问:
Number of multiset permutations where no element occurs twice in a row
Karlheinz Stockhausen's Klavierstück XI consists of 19 fragments. The performer may begin with any fragment, and continue to any other, until a fragment has been reached for the third time, when the performance ends.

I need to calculate how many ways it can be played.

Some observations:

minimum length of performance is 5: let $a$, $b$, $c$ be any fragments, where $b\neq a, c \neq a$ and order would be $a$ $b$ $a$ $c$ $a$. There is $19 \cdot 18  \cdot 18 = 6156$ possible ways to play it in most short form.

maximum length is 39: every fragment is played twice then any is played for the third time, and the performance ends. If fragment can be played twice in a row, then number of variants would be equal to $$\frac{19! \cdot 18}{2^{19}}$$ (count of multiset permutations of $\{1, 1, 2, 2, ... 19, 19\}$  multiplied  to 18 final fragments, because it can not be same as previous).

Also when length is non-maximal, some elements can occur only once and therefore number of possible final elements is lesser than 19. I suppose it is possible to split that calculation in two parts:

For each $n$ from $1$ to $19$:

calculate number of multisets, where exactly $n$ elements occurs twice

calculate number of permutations with restrictions from that multiset (that part cause troubles)

UPD: thanks to user, correct formula for length 19 orders is $\frac{38! \cdot 18}{2^{19}}$, not $\frac{19! \cdot 18}{2^{19}}$

答:
First note:  if the order of fragments could be arbitrary the correct answer would be:
$$
18\times\frac {38!}{2^{19}},
$$
where 18 counts the number of ways to choose the last fragment (it cannot be the same as the last of the previous 38).

If no fragment can be played in a row, we should exclude such paired events. This can be done using inclusion exclusion principle and the final answer is:
$$
18 \times \sum_{j=0}^{19}(-1)^j\binom{19}j\frac {(38-j)!}{2^{19-j}}.
$$


问:
Projection of triangle in 3D to 2D, and reverse
In a software I am developing, I use a triangle defined in 3D by three points $A$, $B$, and $C$. In order to display this triangle on the screen, I use a simple projection as follow (this is basically a projection with a bit of perspective):

$$
x^{\prime}_i = -Sx_i\frac{z_i+N}{F-N}
$$

$$
y^{\prime}_i = -Sy_i\frac{z_i+N}{F-N}
$$

$x_i$, $y_i$ and $z_i$ are the coordinates of any points inside the triangle $ABC$. $x^{\prime}_i$ and $y^{\prime}_i$ are the corresponding coordinates on the screen after projection. $S$, $N$ and $F$ are three constants. I know the coordinates of $A$, $B$ and $C$.

Now, I have a projected point $P^{\prime}$ with coordinates $(x^{\prime}_P, y^{\prime}_P)$. I would like to find $P$, the unprojected version of $P^{\prime}$ with coordinates $(x_P, y_P, z_P)$. I know that $P$ is inside the triangle $ABC$, and I know the coordinates of $P^{\prime}$. What I don't know, is the value of $x_P$, $y_P$ and $z_P$. Is there a way to find them?

What I have tried so far, is to use the dot product of $\vec{AP}$ and $\vec{n}$, the normal of the triangle. Because they are perpendicular, the dot product is equal to zero:
$$
\vec{AP}\cdot\vec{n} = 0
$$
which gives me this after decomposition:
$$
n_{x}\left(\frac{x^{\prime}_{P}(F-N)}{S(-z_{P}-N)}-x_{A}\right) + n_{y}\left(\frac{y^{\prime}_{P}(F-N)}{S(-z_{P}-N)}-y_{A}\right) + n_{z}(z_{P}-z_{A})=0
$$
Then I obtain a quadratic equation: $az^{2}_{P} + bz_P + c$ with:
$$
a=-Sn_{z}
$$
$$
b=S(n_{x}x_{A}+n_{y}y_{A}+n_{z}z_{A}-n_{z}N)
$$
$$
c=n_{x}(SNx_{A}+x^{\prime}_{P}(F-N)) + n_{y}(SNy_{A}+y^{\prime}_{P}(F-N)) + SNn_{z}z_{A}
$$

I am a bit stuck here, because most of the time, I have two solutions. Considering the nature of the problem, I was not expecting two solutions, and I am not sure why I get a quadratic equation. Also, I was expecting to get the same result whether I use $A$, $B$ or $C$ as a reference point for the dot product, but I don't. I suppose I made a mistake somewhere, but I just can't find a good way to get the coordinates of $P$ knowing $P^{\prime}$, $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$, $C^{\prime}$.

答:
As kindly suggested by @Claude in comments, I checked the OpenGL documentation. They use barycentric coordinates to find the value I am looking for.

With the projected triangle $A^{\prime}B^{\prime}C^{\prime}$, and the point $P^{\prime}$ located inside the triangle, we want to find the coordinate $z_P$ that corresponds to the $z$ coordinate of the point $P^{\prime}$ in the unprojected space.

First, we compute the barycentric coordinates in the projected space (page 478 of the documentation):
$$
\alpha = \frac{\mathrm{area}(PBC)}{\mathrm{area}(ABC)},\space\space\space \beta = \frac{\mathrm{area}(PAC)}{\mathrm{area}(ABC)},\space\space\space \gamma = \frac{\mathrm{area}(PBA)}{\mathrm{area}(ABC)}\tag{1}
$$

Second, the documentation gives the following equation that interpolates any parameter $f$ in the triangle while taking into account the perspective. In the manual, the equation has ref $(14.9)$ page 479.
$$
f_P = \frac{\alpha f_A\mathbin{/}w_A+\beta f_B\mathbin{/}w_B+\gamma f_C\mathbin{/}w_C}{\alpha\mathbin{/}w_A+\beta\mathbin{/}w_B+\gamma\mathbin{/}w_C} \tag{2}
$$

In my case, $f_A$, $f_B$, $f_C$ and $f_P$ are the $z$ coordinates of the points $A$, $B$, $C$ and $P$ respectively. The values $w_A$, $w_B$ and $w_C$ are the clip coordinates. To get these coordinates, I had to change multiple things in the projection method. Using the information from here, I reconstructed a projection matrix like this:

$$
M_{proj} =
\begin{pmatrix}
\frac{N}{S} & 0 & 0 & 0 \\
0 & \frac{N}{S} & 0 & 0 \\
0 & 0 & -\frac{F + N}{F - N} & -2\frac{FN}{F-N} \\
0 & 0 & -1 & 0
\end{pmatrix}
\tag{3}$$

with $F$ the far plan, $N$ the near plan, $S$ the tangent of the field of view (in my case 45º but many people use 60º). The projected coordinates is obtained by multiplying the matrix and the unprojected coordinates of $A$, $B$ and $C$ and using $1$ for the 4th element of the matrix:
$$
A^{\prime}=\begin{pmatrix} x^{\prime}_A \\ y^{\prime}_A \\ z^{\prime}_A \\ w_A \end{pmatrix}=M_{proj}
\begin{pmatrix} x_A \\ y_A \\ z_A \\ 1 \end{pmatrix}
\tag{4}$$

Now we can compute $z_P$ using the equation $(2)$, the barycentric weights found in $(1)$ and the clip coordinates found in $(4)$.


问:
posterior predictive probability of heads on the third coin toss
There are three types of coins which are indistinguishable apart from their probability of landing heads when tossed.

Type A coins are fair, with probability .5 of heads

Type B coins have probability .6 of heads

Type C coins have probability .9 of heads

You have a drawer containing 4 coins: 2 of type A, 1 of type B, and 1 of type C. You reach into the drawer and pick a coin at random.  Assuming the first two tosses were heads, what is the posterior predictive probability of heads on the third toss?

(the answer is 0.71522 but I can't figure out how)

答:
Since you have provided the answer, I suppose it is an exercise for self-learning, so I think it's fine to help a bit. In order to solve this, we need to compute the following conditional probability: given two heads, the probability of the coin being A, B, or C given the first two tosses are heads (we denote this as $P(A|HH), P(B|HH), P(C|HH)$ respectively). This can be computed using the definition of conditional probability, which yields (for $B$ and $C$ analogically)
$$P(A|HH) = \frac{P(A \cap HH)}{P(HH)}$$
where P(HH) can be computed using Law of total probability, which gives $P(HH) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B) + P(H|C) \cdot P(C)$. Probability of head on the third toss depends on what coin do we have in hand, which is $0.5\cdot P(A|HH) + 0.6 \cdot P(B|HH) + 0.9 \cdot P(C|HH)$.