问:
$ef$ theorem for inertial degree and ramification index
$L/K$ is a finite algebraic extension of fields. $O_{P}$ and $O_{\mathcal{P}}$ are DVRs with $P$ and $\mathcal{P}$ maximal ideals respectively, their quotient fields are $K$ and $L$ respectively. Let $e$ be the ramification index of $\mathcal{P} / P$ and $f$ be the inertial degree of $\mathcal{P}/P$.

I am working on showing $ef \leq n=[L: K]$ (without separable condition). Not quite understand several details in the proof.

The proof in the book is: Let $\Pi$ be a generator of $\mathcal{P}$ and choose $\omega_{1},\ldots,\omega_{m}$ such that their modulo $\mathcal{P}$ are linearly independent over $O_{P}/P$. Then show $\omega_{i}\Pi^{j}$ are linearly independent over $K$.

Take a relation
$$\sum_{j=0}^{e-1}\sum_{i=1}^{m}a_{ij}\omega_{i}\Pi^{j}=0.$$
If $a_{ij}$ are not all zero, we can assume the are all in $O_{P}$ and at least one of them is not in $P$ (I understand they could be put in $O_{P}$ since K is the quotient field, but why at least one not in $P$?).

Then consider the elements
$$A_{j}=\sum_{i=1}^{m}a_{ij}\omega_{i}.$$
If some $a_{ij}\not \in P$, then $A_{j}$ is a unit in $O_{\mathcal{P}}$. Otherwise, $A_{j}$ is divisible by $\pi$, the generator of $P$ and so $\text{ord}_{\mathcal{P}}(A_{j}) \geq e$.

Thus, $\text{ord}_{\mathcal{P}}(\sum_{j=0}^{e-1}A_{j}\Pi^{j}) =j_{0}$ for some $j_{0} < e$ (how to get this conclusion? Just becasuce you have some $A_{j}$ with $0$ order and then multiply with $\Pi^{j}$?). Finally this is a contradiction since $\sum_{j=0}^{e-1}A_{j}\Pi^{j}=0$ (why is it a contradiction? Isn't $0$ has $\mathcal{P}$ order less than e?).

Thanks for any kind help you offer!

答:
First question: you have a relation

$$\sum\limits_{i,j} a_{ij} \omega_i \Pi^j =0$$

for some $a_{ij} \in K$, not all zero, and you want to justify the claim that you can multiply this equation by some $c \in K$ so that all of the $ca_{ij}$ are in $\mathcal O_P$ with at least one of these not in $P$.  The only way this would not be possible is if all the nonzero $a_{ij}$ had the same valuation.  That would mean you could choose a $c \in K$ such that all the nonzero terms $ca_{ij}$ were units in $\mathcal O_P$.

Taking the equation $$\sum\limits_{i=1}^m \sum\limits_{j=0}^{e-1} ca_{ij} \omega_i \Pi^j = 0 \tag{1}$$
modulo $\mathcal P$ yields the equation $\sum\limits_{i=1}^{m} ca_{i0} \omega_i \equiv 0 \pmod{\mathcal P}$.  The linear independence of the cosets $\omega_i + \mathcal P$ over the field $\mathcal O_P/P$ implies that all the $ca_{i0}$ are in $P$.  Since the nonzero terms $ca_{i0}$ are units in $\mathcal O_P$, this means that all the $ca_{i0}, 1 \leq i \leq m$ must be $0$.

Equation (1) is then

$$\sum\limits_{i=1}^m \sum\limits_{j=1}^{e-1} ca_{ij} \omega_i \Pi^j = 0.$$

Divide this equation by $\Pi$ and reduce modulo $P$ again.  The same argument as in the previous paragraph implies that all the terms $ca_{i1}, 1 \leq i \leq m$ are all zero.  Iterating this argument shows that all the terms $ca_{ij}$ have to be zero.  This contradicts the assumption that not all the $a_{ij}$ were zero.

Second question:

For each $A_j, 0 \leq j \leq e-1$, we have either $\operatorname{ord}_{\mathcal P}(A_j) =0$ or $\operatorname{ord}_{\mathcal P}(A_j) \geq e$ as they explain.  Since there is at least one $a_{ij}$ which is not in $P$, this means that there is at least one $j$ for which $\operatorname{ord}_{\mathcal P}(A_j) = 0$.

The usual rule

$$\operatorname{ord}_{\mathcal P}(x+y) \geq \operatorname{Min}\{ \operatorname{ord}_{\mathcal P}(x), \operatorname{ord}_{\mathcal P}(y)\}$$

is written for nonzero $x, y \in L$, but it continues to make sense if we make the convention that $\operatorname{ord}_{\mathcal P}(0) = \infty$.  Remember this is an exact equality if $\operatorname{ord}_{\mathcal P}(x) \neq \operatorname{ord}_{\mathcal P}(y)$.  Let $S$ be the set of $0 \leq j \leq e-1$ such that $\operatorname{ord}_{\mathcal P}(A_j) = 0$ (we know it is nonempty), and let $T$ be the set of $0 \leq j \leq e-1$ such that $\operatorname{ord}_{\mathcal P}(A_j) \geq e$.  If we set

$$X_S = \sum\limits_{j \in S} A_j \Pi^j, X_T = \sum\limits_{j \in T} A_j \Pi^j$$
then we have $$\sum\limits_{j=0}^{e-1} A_j \Pi^j = X_S + X_T = 0.$$

Since all the $\operatorname{ord}_{\mathcal P}(A_j)$ for $j \in S$ are $0$, we see that $\operatorname{ord}_{\mathcal P}(A_j \Pi^j) = j$ for all $j \in S$.  Thus all the terms $A_j \Pi^j$ for $j \in S$ have distinct valuations, and so

$$\operatorname{ord}_{\mathcal P}(X_S)= j_0$$
where $j_0$ is the smallest number in $S \subseteq \{0,1, ... , e-1\}$.

Since all the terms $A_j \Pi^j : j \in T$ have order $\geq e$, the element $X_T$ has order $\geq e$.  This implies

$$\operatorname{ord}_{\mathcal P}(X_S + X_T) = j_0 < e.$$

In particular, $\sum\limits_{j=0}^{e-1} A_j \Pi^j$ does not have infinite valuation, so it cannot be zero.


问:
A case when $I \otimes_R M$ and $ IM$ are locally isomorphic at minimal primes?
Let $(R,\mathfrak m)$ be a local Cohen--Macaulay ring of dimension $1$. Let $M$ be a finitely generated $R$-module of depth $1$. Let $I$ be an ideal of $R$ of height $1$.

Then my question is: Is it true that $I_P \otimes M_P \cong (IM)_P$ for every minimal prime ideal $P$ of $R$?

Of-course there is always a natural surjection $I \otimes_R M \to IM$ sending $x \otimes m\to xm$, but I am unable to see if this also becomes injective when localising at minimal primes.

Please help.

答:
As $R$ is Cohen-Macaulay, the set of minimal and associated primes are equal. Also, since $R$ is Cohen-Macaulay, the height of an ideal is equal to the grade.

Let $P$ be a minimal prime and $I$ a height 1 ideal. Since the grade of $I$ is $1$, $I \not\subset P$. Thus, $I_P \cong R_P$. This is sufficient to deduce the isomorphism
$$
I_P \otimes_{R_P} M_P \cong R_P \otimes_{R_P} M_P \cong M_P
$$
and $(IM)_P \cong I_P M_P \cong R_P M_P \cong M_P$.

If you prefer keeping track of the surjective map in your post and familiar with the Tor functor, one can show that $\operatorname{Tor}_1^R (R/I,M)$ becomes zero after localizing at a minimal prime $P$, since $(R/I)_P =0$.

ContentLicense: CC BY-SA 4.0


答:
$L,K$ are discretely valued field, the valuation $v$ is surjective $L\to \Bbb{Z}\cup \infty$ and $K\to e\Bbb{Z}\cup \infty$.

$\Pi$ is chosen such that $v(\Pi)=1$.

$\omega_i$ are taken in $O_\mathcal{P}$ to be a $O_P/P$ basis of $O_\mathcal{P}/\mathcal{P}$, ie. $m=f$.

The $O_P/P$ independence of the $\omega_i$ gives that
$$v(\sum\limits_{i=1}^m a_{ij} \omega_i) = \inf_i v(a_{ij})$$

Thus $v(\sum\limits_{i=1}^m a_{ij} \omega_i \Pi^j) \in j+ e\Bbb{Z}\cup \infty$.

For $v(b)\ne v(c)$, $v(b+c)=\inf( v(b),v(c))$. Thus $$v(\sum_{j=0}^{e-1} \Pi^j\sum\limits_{i=1}^m a_{ij} \omega_i )=\inf_j v(\Pi^j\sum\limits_{i=1}^m a_{ij} \omega_i)=\inf_j (j+\inf_i v(a_{ij}))$$
That it is $\infty$ means that each $v(a_{ij})=\infty$ ie. each $a_{ij}=0$.

Title: $ef$ theorem for inertial degree and ramification index

Body:
$L/K$ is a finite algebraic extension of fields. $O_{P}$ and $O_{\mathcal{P}}$ are DVRs with $P$ and $\mathcal{P}$ maximal ideals respectively, their quotient fields are $K$ and $L$ respectively. Let $e$ be the ramification index of $\mathcal{P} / P$ and $f$ be the inertial degree of $\mathcal{P}/P$.

I am working on showing $ef \leq n=[L: K]$ (without separable condition). Not quite understand several details in the proof.

The proof in the book is: Let $\Pi$ be a generator of $\mathcal{P}$ and choose $\omega_{1},\ldots,\omega_{m}$ such that their modulo $\mathcal{P}$ are linearly independent over $O_{P}/P$. Then show $\omega_{i}\Pi^{j}$ are linearly independent over $K$.

Take a relation
$$\sum_{j=0}^{e-1}\sum_{i=1}^{m}a_{ij}\omega_{i}\Pi^{j}=0.$$
If $a_{ij}$ are not all zero, we can assume the are all in $O_{P}$ and at least one of them is not in $P$ (I understand they could be put in $O_{P}$ since K is the quotient field, but why at least one not in $P$?).

Then consider the elements
$$A_{j}=\sum_{i=1}^{m}a_{ij}\omega_{i}.$$
If some $a_{ij}\not \in P$, then $A_{j}$ is a unit in $O_{\mathcal{P}}$. Otherwise, $A_{j}$ is divisible by $\pi$, the generator of $P$ and so $\text{ord}_{\mathcal{P}}(A_{j}) \geq e$.

Thus, $\text{ord}_{\mathcal{P}}(\sum_{j=0}^{e-1}A_{j}\Pi^{j}) =j_{0}$ for some $j_{0} < e$ (how to get this conclusion? Just becasuce you have some $A_{j}$ with $0$ order and then multiply with $\Pi^{j}$?). Finally this is a contradiction since $\sum_{j=0}^{e-1}A_{j}\Pi^{j}=0$ (why is it a contradiction? Isn't $0$ has $\mathcal{P}$ order less than e?).

Thanks for any kind help you offer!


问:
Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$
I think it's all in the title. $p$ is some random polynomial.
I don't know how to approach this one. I've tried taking the roots of $p$, placing them on the diagonal of a new matrix and reasoning that the linear operator of the new matrix has a common specific value (if it's called that in English) with A, but I have no idea how to prove that that specific value is also a root of $p$.</p>

答: Hint: if $v$ is an eigenvector for eigenvalue $\lambda$, what is $p(A) v$ for polynomials $p$?  </p>

ContentLicense: CC BY-SA 3.0


问:
How to find the perimeter of two trapezoid figures inscribed in a rectangle?
The problem is as follows:

The figure from below represents a rectangle $ABCD$. Assume the area
of the right triangle $BQM$ is $120 \ \text{in}^2$ (in=inches).

Assume it also holds the following equality:

$$\csc \phi - \cot \phi =\frac{1}{5}$$

With all that information find the perimeter of the orange shaded
region.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{1.65 in}\\
2.&\textrm{1.72 in}\\
3.&\textrm{1.88 in}\\
4.&\textrm{1.56 in}\\
\end{array}$

The first part of this problem didn't seem difficult.

All what I did was as follows:

$$\csc \phi - \cot \phi =\frac{1}{5} \Rightarrow \csc \phi = \cot \phi + \frac{1}{5}$$

$$\csc \phi = \left(\cot \phi + \frac{1}{5}\right)^2=\cot^2 \phi+\frac{2}{5}\cot\phi+\frac{1}{25}$$

Then using: $1+\cot^2\phi=\csc^2\phi$

$$1+\cot^2\phi=\cot^2+\frac{2}{5}\cot\phi+\frac{1}{25}$$
$$\cot\phi=\frac{12}{5}$$

But the thing is that with the given information I can find $CN$ and $MA$ and $BQ$.

As the area is: $\frac{a^2\tan\phi}{2}=120$, $BQ=a=24$

But the thing is how to get $BC$? Is it just me or something is strange here?
Can someone help me here, please?

答: Drop the perpendicular from $M$ onto $CD$ and call the foot $F$. Then $\triangle MFN$ and $\triangle BQM$ are similar and $FN+NC=BM=26$; since $NC=\frac{BQ}4=6$, $FN=20$ and hence $BC=MF=48$.


问:
The GCD of $(n^2, 2n +1)$
$\text{GCD}(n^2,2n+1)$

I think I need to use the Euclidean algorithm but $n^2 = (2n+1)(1/2n -1/4)+1/4$ so I have no idea how to get the GCD (the remainder is a fraction).

n is a natural number

Title: The GCD of $(n^2, 2n +1)$

Body:
Note that $2n+1$ is always odd, so $\gcd(n^2,2n+1)=\gcd(4n^2,2n+1)$,

and $4n^2-(2n+1)(2n-1)=1$.


Title: The GCD of $(n^2, 2n +1)$

Body:
$\text{GCD}(n^2,2n+1)$

I think I need to use the Euclidean algorithm but $n^2 = (2n+1)(1/2n -1/4)+1/4$ so I have no idea how to get the GCD (the remainder is a fraction).

n is a natural number

Body: $gc(n^2, 2n+1)$ divides $n(2n+1)-2n^2=n$, it divides $2n+1-2n=1$ so it is $1$


答:
The last nonzero remainder is

$\text{GCD}(x^2,2x+1)=1$

The GCD can be expressed as an integral linear combination of $x^2$ and $2x+1$

$(4)(x^2)+(−2x+1)(2x+1)=1$

Title: The GCD of $(n^2, 2n +1)$


问:
How many 10 letter 'words' can be made with no repeated letters and at least 8 consonants?
Need a little bit of help with this one.

How many 10 letter 'words' can be constructed which have no repeated letters and at least 8 consonants?

where 'word' is defined as any combination of the 26 letters of the english alphabet, and by consonant, (obviously) a letter of the english alphabet that are not either a,i,u,e, or o.

This is where I am at:

We need to choose the positions for
the 8 consonants which is $c(10,8)$

Then we need to choose the 8
consonants which I think is
$p(21,8)$? (not sure about this one)

Choose the remaining 2 letters: We have already used 8 letters so we only have 18 letters to choose from so $p(18, 2)$

What else needs to be calculated? Do we need to choose the position of the remaining 2 letters?

答: The problem with your approach is that words of 10 consonants will get counted many times.  You can pick any two places for the "not necessarily consonants" and finish with the same word.  The approach suggested by user6312 in your previous post will work here, too.</p>


答:
Considering $\sigma(A)$ denotes the set of characteristic(eigen) values of A

As $\lambda$ is an eigen value, $\exists 0\ne v\in V$ such that $Av=\lambda v$

Note that $A^kv=\lambda^k v,\forall k\in N$(easy to prove using Induction)

Let $p(x)=\sum_{i=0}^{n}a_ix^i,a_i\in R$

Then we have ,

$p(A)=\sum_{i=0}^{n}a_iA^i$

So we have ,

$p(A)v=\sum_{i=0}^{n}a_iA^iv=\sum_{i=0}^{n}a_i\lambda^iv=p(\lambda)v$

Given $p(A)=0$

$\Rightarrow 0=p(A)v=p(\lambda)v$

As $v\ne 0\Rightarrow p(\lambda)=0$

Title: Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$

Body:
I think it's all in the title. $p$ is some random polynomial.
I don't know how to approach this one. I've tried taking the roots of $p$, placing them on the diagonal of a new matrix and reasoning that the linear operator of the new matrix has a common specific value (if it's called that in English) with A, but I have no idea how to prove that that specific value is also a root of $p$.</p>


答:
The answer is $\gcd(n^2, 2n+1) =1$.

Let $m = \gcd(n^2,2n+1)$. Suppose $m >1$. Then for each prime $p$ that divides $m$ [now $p$ may be 2], $p$ divides both $n^2$ and $2n+1$. So to show that $\gcd(n^2, 2n+1)=1$ it suffices to show that every prime that divides $n^2$ does not divide $2n+1$. We do this below.

Note that if a prime $p$ divides $n^2$ then it also divides $2n$. Which implies that $p$ does not divide $2n+1$ [make sure you see this]. So every prime $p$ that divides $n^2$, does not divide $2n+1$.

Title: The GCD of $(n^2, 2n +1)$


Title: The GCD of $(n^2, 2n +1)$

Body: $$\gcd(n^2, 2n+1 ) = \gcd(n^2, n^2 + 2n + 1 ) = \left[ \gcd ( n, n+1) \right]^2 = 1^2 = 1.$$


问:
There is exactly one glide reflection and exactly one rotation
Let $g$ be the line with equation $y=1$ and let $A=(0,2)$ and $B=(2,0)$.

Show that there is exactly one glide reflection $\kappa$ with fixed line $g$ that maps $A$ on $B$.

Show that there is exactly one rotation $\delta$ with fixed line $g$ that maps $A$ on $B$.

To show the existence of $\kappa$ we have to show that the distance of $A$ to $g$ and of $B$ to $g$ is the same, or not?

But how can we show the uniqueness?

答:
A glide reflection can be expressed as a composition of a reflection about a line and a translation parallel to that line.

A glide reflection composed of a reflection about a line $L$ followed by translation parallel to $L$ will always map $L$ to $L$, that is, $L$ is a fixed line of the glide reflection.
If the translation is non-trivial (that is, it is a translation through a non-zero distance), then this line $L$ is the only fixed line of the glide reflection.

Therefore when we are told $g$ is a fixed line of glide reflection $\kappa,$
we know that it must be possible to describe $\kappa$ as a reflection about $g$ followed by a translation parallel to $g.$

What happens when we reflect the point $A$ about the line $g$?
Is there more than one possible result?

Find the image of $A$ when $A$ is reflected through $g.$ Call the image $A'$.
What translation (if any) parallel to $g$ takes $A'$ to $B$?

If you work out the composition of $\kappa$ from a reflection about $g$ and a translation parallel to $g,$ you should be able to show that this can be done in only one way, that is, there is only one possible reflection and one possible translation that could compose to $\kappa.$

It is true that it would be impossible to find any such glide reflection $\kappa$ if $A$ and $B$ were at different distances from $g.$ But you don't actually need to prove explicitly that they are. Finding a translation from $A'$ to $B$ parallel to $g$ will prove it implicitly.

For part 2 of the question:

Rotations in the plane are rotations around a single point. It is impossible to rotate a plane about a line in the plane.

In fact, if an isometry of the plane maps $P$ to $P$ and maps $Q$ to $Q$, where $P$ and $Q$ are distinct points, the isometry is either the identity or is a reflection through the line $PQ.$

But my understanding is that an object is "fixed" by an isometry of a plane if the image of the object is the object itself; it is not necessary that every point of the object is a fixed point of the isometry.
This is how the line $g$ is able to be a "fixed line" of the glide reflection in part 1, which has no fixed points at all.

So if we say a rotation has a fixed line then we mean every point on that line is mapped to itself or to another point on the line.

The line $g$ obviously is fixed by the identity transformation, which can also be considered a rotation around some point by a multiple of $2\pi$ radians.

The line $g$ also is fixed if the center of rotation is on $g$ and the angle of rotation is $\pi$ radians. A rotation by $\pi$ radians keeps the center of the rotation fixed and swaps every other point of $g$ with a point of $g$ on the opposite side of the center of rotation.

Suppose the rotation is not by a whole multiple of $\pi$ or is around some point not on the line $g$ (and not the identity transformation). In each case you can show some point on the line $g$ that is mapped to a point not on $g.$

So now you know the angle of rotation and you can limit the location of the center to just a few possible places. In this particular question, however, given that the rotation $\delta$ maps a known point $A$ to a known point $B$, it is sufficient to know the angle of rotation in order to find the center.

If you want to do it with matrices and vectors I suppose you could write a rotation matrix with unknown angle $\theta$ and a translation vector with unknown coordinates, plug in the coordinates of $A$ and $B,$ and solve for the unknowns. The geometric explanation is so compelling, however, that I am not inclined to work through all that algebra.


问:
How to get the angle in a trapezoid formed by two triangles?
The problem is as follows:

Find the value of angle $\angle BCA$ or $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^\circ\\
2.&20^\circ\\
3.&10^\circ\\
4.&40^\circ\\
\end{array}$

The thing here is without any further hints. I'm stuck.

I was only able to spot.

$\triangle ABC$ is isosceles,

$\triangle DBC$ is isosceles

But I can't find any other relationship. What else to do?. How to solve this relying only in euclidean geometry postulates?. Can someone help me here?.

Should congruence be used here or what?.

I'm confused please help me. Please include a drawing in your answer as it would help me to spot what to do or if a construction is needed.

答:
As David K points out, we can easily use the left half-triangle and the right half-triangle to show that that $\angle BAD = 50°$ and $\angle BCD = 70°$. So the left and right triangles are isosceles, $BA = BD$, and $BD = BC$, so all three of $A, D, C$ lie on a circle centered at $B$ (not to scale):
 If we extend segment $BC$ out till it reaches the other side of the circle, then $BE$ is a radius since $B$ is the center (although it doesn't look like it), so we actually have diameter $CE$, and arc $EAC$ is actually a semicircle. From here, basic angle-arc definitions and theorems may be used to compute $x$.

Edit: And as another user points out, once we have $BA = BC$, the top half of the diagram is an isosceles triangle with vertex angle of $120$ and base angles of $x$, $x$, which makes the problem very easy!


问:
Determinant with general case at the diagnal
I tried to calculate the determinant for that matrix:

$\begin{pmatrix}
  0 & 1 & 1 & . &.&.& 1 \\
  1 & 1 & 1 & .  &.&.& 1 \\
  1 & 1 & 2 & . &. & .&1 \\
  .&.&. &. &&&1\\
  .& & & &.&& 1\\
  . & & & && .&1\\
  1 & 1& 1 &.& .& . & n-1
\end{pmatrix}$

and im quit sure for n
\begin{align*}
 n \geq 2
\end{align*}
it will be equal to
\begin{align*}
(n-1)!
\end{align*}

but i'm not sure and can't  understant how to prove that

答:
The determinant of your matrix $A$ is $-(n-2)!$

I apply row operations to $A$ that do not change the determinant.

(1) Subtract $R_1$ from all other rows.

(2) Subtract $R_2$ (new version!) from all later rows.
Ignoring zeros, you only have $\begin{bmatrix}0&1 \\ 1&0 \end{bmatrix}$
in the top left corner and an $(n-2) \times (n-2)$ diagonal matrix
$D$ at the bottom right. Thus, $|A|=(-1)|D|.$


问:
Integration with change of variables (multivariable).
The following are the problems that I have been working on. It involves change in variables with 2,3 variables respectively.

(1)Let $R$ be the trapezoid with vertices at $(0,1),(1,0),(0,2)$ and $(2,0)$. Using the substitutions $u = y-x$ and $v = y +x$, evaluate $$\int\int_R  e^{{y-x}\over{y+x}} dA$$.

(2)Evaluate the following integral $$\int\int\int_D (x^2y + 3xyz)dV$$ where D is the region in 3-space defined by $1 \le x \le 2, 0 \le xy \le 2, 0 \le z \le 1$ using the substitution $u =x , v = xy, w=3z$.

Here are what I tried.

1), I understand the part where the parallel lines  $x+y = 1$ and $x+y=2$ is used to find the limit of integration of $v$, but what am I supposed to do with that of $u$ ?  One side of the trapezoid is vertical and the other side is horizontal and it looks nothing like the equation $v$. I was thinking about trying to doubling the trapezoid so that it becomes a parallelogram, but I am not sure if that even works.

I understand the idea and the fact that $$\int\int_R f(x,y) dA = \int\int_S f(g(u,v),h(u,v))|{ \partial{(x,y)}\over{\partial(u,v)}}|dvdu$$, but this is more of an algebraic thing that I am stuck with.

2), Again, I understand the theory, but I am stuck with the algebraic part. The approach I made was $$\int_0^1 \int_0^2 \int_1^2 (uv+3vw)|{1\over{3u}}|dudvdw$$ but I ended up having to integrate an improper integral with the result being $-\infty$... Can someone help out ?

答:
I am adding some points for the first one so you can find the limits easily. Try to do the rest by yourself. :)

If we consider the 4 points above and mark them on $xy$ plane, we have the area as you see below:

Now, as you probably know, we find $x$ and $y$ respect to $u$ and $v$ as we are indicated. So $$u=y-x,v=x+y,~~~\to y=\frac{u+v}2,~~x=\frac{v-u}2~~~~~~~(I)$$ For finding the limits for $u$ and $v$ in $uv$ plane do as follows and by $I$:  $$(1,0)\longrightarrow u=-1,v=1\\\\\ (2,0)\longrightarrow u=-2,v=2\\\\\ (0,1)\longrightarrow u=1,v=1\\\\\ (0,2)\longrightarrow u=2,v=2\\\\\ $$ This means that you have a new region like:

Obviously $$-1\le v\le +1, ~~~-v\le u\le v$$


问:
Simple cellular homology calculation
Let $X$ be a CW complex and let $X^{n}$ denote the $n$-skeleton of $X$.  Is it true that $H_{k}(X,X^{n+1})=0$ for $k\leq n+1$?  How can I see this using similar calculations that I am familiar with, such as $H_{k}(X^{n},X^{n-1})=0$ if $k\neq n$, and $H_{k}(X^{n})=0$ for $k>n$.  I can show this for $k<n+1$ using the long exact sequence of the pair $(X,X^{n+1})$, but I am stumped on the $k=n+1$ case.

答:
$\newcommand\wt{\widetilde}$
The short answer is that $(X,X^{n+1})$ is a good pair (in the sense of Hatcher), so $H_k(X,X^{n+1})\cong H_k(X/X^{n+1})$ and $X/X^{n+1}$ only has cells in dimension $\ge n+2$, so only has non-trivial homology in dimension $\ge n+2$.

Alternatively, one can look at the LES of the triple $(X^n,X^{n+1},X)$ which includes the exact sequence
$$\widetilde H_k(X^{n+1},X^n)\to\widetilde H_k(X,X^n)\to\widetilde H_k(X,X^{n+1}).$$
We induct (in $n$) downwards from $\dim X$ (in particular, we're assuming $\dim X<\infty$ here). We know the LHS vanishes for $k\le n$ and the RHS vanishes for $k\le n+1$, so the middle term vanishes for $k\le n$.

Finally, maybe the approach that best answers your question is this one. One consequence of cellular homology equalling singular homology is that $\wt H_k(X)\cong\wt H_k(X^n)$ when $n\ge k+1$. Using this, one can compare the long exact sequences of the pairs $(X,X^n)$ and $(X^{n+1},X^n)$ to see that $\wt H_k(X,X^n)\cong\wt H_k(X^{n+1},X^n)$ for $k\le n$, which gives the vanishing result you are after. There is a natural morphism between these long exact sequences
$$\require{AMScd}\begin{CD}
\wt H_k(X^n) @>>> \wt H_k(X^{n+1}) @>>> \wt H_k(X^{n+1}, X^n) @>>> \wt H_{k-1}(X^n) @>>> \wt H_{k-1}(X^{n+1})\\
@VV\sim V @VV\sim V @VVV @VV\sim V @VV\sim V\\
\wt H_k(X^n) @>>> \wt H_k(X) @>>> \wt H_k(X, X^n) @>>> \wt H_{k-1}(X^n) @>>> \wt H_{k-1}(X)
\end{CD}$$
The vertical maps marked $\sim$ are isomorphisms when $k\le n$, so the remaining map is also an isomorphism in this range.

Title: Simple cellular homology calculation


问:
Bounding the tail of a series estimation of $e^x$
Let $a\in[1,\infty)$. I'm trying to bound the (absolute value of the) following sum from above:

$\left(\displaystyle\sum_{k=n+1}^\infty\frac{(2\pi)^k}{k!}\right)(a-a^2)+\left(\displaystyle\sum_{k=n+1}^\infty\frac{(2\pi)^k}{k!}\sum_{j=0}^{k-2}a^j\right)(a^2-a)$

It seems clear that this sum goes to $0$ for any choice of $a$ in the range, but I would like to obtain a good estimate in terms of $n$.

The first part of the sum can be estimated as a tail of a Taylor series for the exponential function, so I'm looking for a way to get a handle on the second part.

Question: What is a reasonable way to estimate the second part of the sum, i.e. $\left(\displaystyle\sum_{k=n+1}^\infty\frac{(2\pi)^k}{k!}\sum_{j=0}^{k-2}a^j\right)(a^2-a)$?

Body:
$$S=\left(\displaystyle\sum_{k=n+1}^\infty\frac{(2\pi)^k}{k!}\sum_{j=0}^{k-2}a^j\right)(a^2-a)$$
$$\sum_{j=0}^{k-2}a^j=\frac{a^k-a}{a^2-a }$$
$$S=\sum_{k=n+1}^\infty\frac{(2\pi)^k}{k!}(a^k-a)=\sum_{k=n+1}^\infty\frac{(2\pi a)^k}{k!}-a\sum_{k=n+1}^\infty\frac{(2\pi)^k}{k!}$$
$$S=e^{2 \pi  a}\frac{ \Gamma (n+1)-\Gamma (n+1,2 a \pi )}{\Gamma (n+1)}- a e^{2 \pi }\frac{ \Gamma (n+1)-\Gamma (n+1,2 \pi )}{\Gamma (n+1)}$$
$$S=e^{2 \pi  a}\Bigg[1-\frac{\Gamma (n+1,2 a \pi )}{\Gamma (n+1)} \Bigg]-ae^{2 \pi  }\Bigg[1-\frac{\Gamma (n+1,2  \pi )}{\Gamma (n+1)} \Bigg]$$


问:
Does every mon-small submodule contain a nonzero supplement submodule of the module.
Definition: Let $M$ be an $R$-module. Then

A submodule $N$ is called small (in $M$) if $N+K=M\implies K=M,\forall K\leq M$.

A submodule $N$ is called a supplement of a submodule $K$ (in $M$) if $N$ is minimal with respect to the property $N+K=M$.

A submodule $N$ is called a supplement submodule if it is a supplement of some submodule $K$.

My Question: Does every mon-small submodule contain a nonzero supplement submodule of the module.

答:
False.

You can take example of $\mathbb{Z}$-module $\mathbb{Z}$. Its every nonzero submodule is nonsmall while it has no nonzero supplement submodule.


答:
I think you meant to say $\triangle ABD$ is isosceles.
Specifically, $AB = BD.$

You also correctly observe that $\triangle BCD$ is isosceles.
Specifically, $BC = BD.$

Putting the equations together, $BA = BC = BD.$
The three points $A, C, D$ are all at the same distance from $B$,
so they all lie on a common circle with center at $B.$

That should be a big enough hint.

Title: How to get the angle in a trapezoid formed by two triangles?

Body:
The problem is as follows:

Find the value of angle $\angle BCA$ or $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^\circ\\
2.&20^\circ\\
3.&10^\circ\\
4.&40^\circ\\
\end{array}$

The thing here is without any further hints. I'm stuck.

I was only able to spot.

$\triangle ABC$ is isosceles,

$\triangle DBC$ is isosceles

But I can't find any other relationship. What else to do?. How to solve this relying only in euclidean geometry postulates?. Can someone help me here?.

Should congruence be used here or what?.

I'm confused please help me. Please include a drawing in your answer as it would help me to spot what to do or if a construction is needed.


问:
Question regarding power series in complex analysis

Fix $a,b,c\in\mathbb{C}$, $c$ not an integer. Show that there is a
unique analytic function $F$ in the unit disk $\mathbb{D}$ such that
$F(0)=1$ and$$z(1-z)F''(z)+(c-(a+b+1)z)F'(z)-abF(z)=0$$ for all
$z\in\mathbb{D}$. [Hint: find a recursion for the power series
coefficients]

I can only think of letting $F=\sum\limits^{\infty}_{0}a_nz^n$ with $a_0=1$. After computing this into the formula, I get a complicated equation, and I'm not sure how to proceed from there to prove existence and uniqueness. Also by letting $F$ be such a power series, I am already assuming that $F$ is analytic on $z=0$.

Another approach is that maybe I need some explicit formula for $F$ and use the analytic continuation. Could you give me any suggestion?

答:
This differential equation has a regular singular point at $0$ with indicial equation $r^2 + (c-1) r = 0$, so at least if $c$ is not an integer, there should be a solution that is a power series.  If you plug in $F(z) = \sum_{n=0}^\infty a_n z^n$ to the differential equation and look at the coefficient of $z^n$ in the result, you should
get a recurrence for $a_{n+1}$ as a function of $a_n$.  The Ratio Test will confirm
that the radius of convergence is $1$.


问:
canonical projection of a subring
Let $R = \mathbb{Z}$ and $R' =\mathbb{Z}_4 = \mathbb{Z}/4\mathbb{Z}$, and $f = \pi$ the canonical projection of $\mathbb{Z}$ on $\mathbb{Z}/4\mathbb{Z}$. And let $A = 6\mathbb{Z}$ and $I=4\mathbb{Z}$.

If I've understood correctly $f$ should be an homomorphism between $R$ and $R'$.

My manual calculates $\pi(A)$ and gets a result different from mine: $\pi(A) = (A + I)/I = {I,2+I}$.

I'm applying the definition of canonical projection so $pi: A \to A/I$. For me is $\pi(A) = 6\mathbb{Z}/4\mathbb{Z}$, I don't understand why is summing up $I$ to $A$.

Can you explain me how to calculate the image $pi(A)$ and if/why it is an ideal of R/I?

答:
If $x \in \pi(6\mathbb Z)$, there is $m \in 6 \mathbb Z$ with $x = \pi(m)$, and since $m \in 6 \mathbb Z$ there is $n \in \mathbb Z$ with $m=6n$. Thus $x = \pi(6n) = 6n+4\mathbb Z = 2n+4\mathbb Z$ (recall that $a+I=b+I \iff a-b \in I$). Now:

if $n$ is even, $n=2k$ for some $k \in \mathbb Z$, and then $x = 2n+4\mathbb Z = 4k+4\mathbb Z = 0+4\mathbb Z = 4\mathbb Z$.

if $n$ is odd, $n=2k+1$ for some $k \in \mathbb Z$, and then $x = 2n+4\mathbb Z = (4k+2)+4\mathbb Z = 2+4\mathbb Z$.

Hence, $x$ is $4\mathbb Z$ or $2+4\mathbb Z$, meaning that $\pi(6\mathbb Z) = \{4\mathbb Z,2+4\mathbb Z\}$.

Title: canonical projection of a subring


问:
Is this true for a system of polynomial equations as well?
For a system of linear equations with $n$ unknowns, if you have as many linearly independent equations as there are unknowns, you can find the only solution that satisfies that system, right? But does this still hold if you have a set of linearly independent polynomial equations in some number $n$ unknowns of some degree $d$ instead of just linear equations?

In other questions the example $x^2 + y^2 = 0$ is given where you only need one equation to solve for both unknowns, but are there cases where you would need more independent equations than unknowns to get to a unique solution?

答:
This is subtle. It is not at all clear a priori what the correct generalization of "linear independence" is. This is because polynomials can nontrivially divide each other; for example, the system in two variables

$$x - y = 0$$
$$x^2 - y^2 = 0$$

has infinitely many solutions $\{ (x, y) : x = y \}$, and the first equation divides the second but they are linearly independent. As a more complicated example, consider the system in three variables

$$x = 0$$
$$y = 0$$
$$xz - yz = 0$$

which still has infinitely many solutions $\{ (x, y, z) : x = y = 0 \}$ ($z$ is arbitrary). This time none of the equations divide each other but the third one is a linear combination of the first two with polynomial coefficients.

From the modern perspective, coming from commutative algebra and/or algebraic geometry, we can understand a system of polynomial equations by understanding the ideal they generate in the ring of polynomials, and one plausible guess as to the appropriate generalization of "linear independence" is most clearly stated in this language: it's that if we have $n$ polynomials $f_1, \dots f_n$ then the ideal $I = (f_1, \dots f_n)$ they generate cannot be generated by fewer than $n$ polynomials. This is related to the notion of a complete intersection except I have only seen complete intersections discussed in projective space. There is also the related notion of a regular sequence which I unfortunately know very little about.

This is still not enough to guarantee a unique solution. Generically one now expects $(\deg f_1) (\deg f_2) \dots (\deg f_n)$ solutions, at least over $\mathbb{C}$. A very simple example is

$$x - y = 0$$
$$x - y^2 = 0$$

which is equivalent to $y = y^2$ and hence which has two solutions $(1, 1)$ and $(0, 0)$. For more complicated examples see the Wikipedia article on Bezout's theorem: there are nice geometric examples coming from intersections of conic sections.

I don't know off the top of my head if "the ideal is not generated by fewer than $n$ polynomials" is enough to guarantee finitely many solutions. This is probably extremely classical and I hope someone else comes along who does know.

Another plausible generalization of "linearly independent" is that the Jacobian of $f_1, \dots f_n$ is generically invertible; equivalently, the differentials $df_1, \dots df_n$ are generically linearly independent. ("Invertible everywhere" is a quite strong condition.) We should be able to conclude something using the inverse function theorem from this but unfortunately my command of the subject is not quite strong enough to tell immediately what.


问:
Calculating the probabilities of different lengths of repetitions of numbers of length 4
I'm trying to calculate the probabilities of different lengths of repetitions of X length number however I know I'm doing it incorrectly since when I add all the probabilities together they don't total to 1

e.g.
Here is my reasoning to calculate the probabilities of the different lengths repetitions for length 4

Probability that there are 0 repeating sequences:
e.g. WXYZ
10/10 * (9/10)^3 = 729

Probability that there is 1 repeating sequence of length 2:
e.g XXYZ or YXXZ or YZXX
10/10 * (9/10)^2 * 1/10 * 3 = 243

Probability that there is 2 repeating sequence of length 2:
e.g XXYY or YYXX
10/10 * 9/10 * (1/10)^2 * 2 = 18

Probability that there is a repeating sequence of length 3:
e.g XXXY or YXXX
10/10 * 9/10 * (1/10)^2 * 2 = 18

Probability that there is a repeating sequence of length 4:
e.g XXXX
10/10 * (1/10)^2 * 3 = 1

When I add the number of outcomes I get 1009, when I should be getting a 1000.
Anyone know what I'm doing wrong?

Thanks in advance!

答: The third entry in the list has a wrong count. One should not multiply by $2$, since shape XXYY is the same as shape YYXX.</p>


答: Angle chasing gives $\angle BAD = 50^{\circ}, \angle BCD = 70^{\circ}$. So $\triangle BAD,\triangle BDC$ is isosceles. Hence we have $BA=BD=BC$. Since $BA=BC$, $\triangle BAC$ is isosceles, so $$ x = \frac{180^{\circ}-(80+40)^{\circ}}{2}=30^{\circ}$$


问:
Simple physical intuition for differential 1-form of a scalar function acting on a tangent vector?
I'm trying to gain some very basic physical intuition as to how the differential 1-form $df$ of a smooth function $f$ acts on a tangent vector $\mathbf{v}$. Say the temperature of a hot plate is given by $f\left(x,y\right)=100-\left(x^{2}+y^{2}\right)$. I'm walking along a curve on the hot plate defined by some parametric equation $\Phi\left(t\right)$. When I get to the point $p\left(3,7\right)$ the tangent vector $\mathbf{v}$ is given by $$\mathbf{v}=\frac{d\Phi}{dt}=\left(\frac{dx}{dt},\frac{dy}{dt}\right)=\left(1,3\right).$$

At that point am I entitled to say that the rate the temperature is changing in the direction of $\mathbf{v}$ is given by$$df\left(\mathbf{v}\right)=\left(-2x\,dx-2y\,dy\right)\left(1,3\right)=-\left(2\times3dx+2\times7dy\right)\left(1,3\right)=-48.$$So if $t$ was time in minutes, and $f\left(x,y\right)$ was in Celsius, that would mean the temperature was changing at the rate of -48 degrees Celsius per minute in the direction of $\mathbf{v}$. Have I got that right?

EDIT 10 Jan 2021

Apologies, I messed up the arithmetic by computing for the wrong point (hopefully now fixed). Also, as pointed out by @Semiclassical the units of temperature are unknown. Having said that, I would like to ask again if my interpretation is correct that at point $p\left(3,7\right)$ the instantaneous rate of change of temperature is equal to -48 units of temperature (whatever they are) per minute in the direction of $\mathbf{v}$?

答:
Based on your calculation that

$\frac{d \Phi}{dt} = (\frac{d \Phi}{dx}, \frac{d \Phi}{dy}) * (\frac{dx}{dt}, \frac{dy}{dt})^T $, then it is interpreted as that the change of temperature at a specific point is equal to the inner product of gradient of $f$ and its mobility at this point.


答:
(diagram not to scale)

Extend line $AF$ so that it meets the circumference at $F$. As $F,C,D,A$ all lie on the circumference of the circle, quadrilateral $FCDA$ is cyclic.

Hence $\angle BFC = 180º - (50º + 70º) = 60º$ and $\angle FBC = 180º - (80º + 40º) = 60º$, hence $\angle BCF = 60º$. Now since $AF$ is also a diameter, $\angle ACF = 90º$, and thus: $\angle ACB = x = 90º - 60º = \boxed{30º}$.

The same can be done with $E$ instead of $F$, but there is one more step as you would need to find $\angle BAC$ before finding $x$.


问:
Show that $\max$ function on $\mathbb R^n$ is convex
I am reading the book Convex Optimization, and I don't understand why a $\max$ function is convex.

The function is defined as:

$$f(x) = \max(x_1, x_2, \dots, x_n)$$

The book offers the proof shown below:

for $0 \leq \theta \leq 1$

$$\begin{aligned} f(\theta x + (1 - \theta)y) &= \max_i \left( \theta x_i + (1 - \theta)y_i \right)\\ & \leq \theta \max_i x_i + (1 - \theta)\max_i y_i\\ &= \theta f(x) + (1 - \theta)f(y) \end{aligned}$$

However, I don't understand why the following inequality holds.

$$\max_i (\theta x_i + (1 - \theta)y_i) \leq \theta \max_i x_i + (1 - \theta)\max_i y_i$$

答:
One line proof:

Since composition of convex functions is convex, we only need to show $\max{(x,y)}$ is convex. But $\max(x,y)=\frac{x+y}{2}+|\frac{x-y}{2}|$ and $|\cdot|$ is obviously convex.


问:
How to mathematically represent an element within a member of a set?
Let's say I have a set $F=\{(h_1,t_1), (h_2,t_2),\ldots, (h_n,t_n)\}$. Then how do I mathematically represent $h_i$, as an object within an element within the set $F$ for $i=1,2,\ldots, n$?

I think $h_i \in F$ is not a correct representation.

答:
It sounds like the question is to find a compact way to say "let $h_i$ be such that there exists a $t_j$ for which $(h_i,t_j) \in F$". If that's the case, then you could call the set $H_F$, so long as you explain this terminology. Or just call it $H$ if there would be no ambiguity.

Or: "let $(h_i, t_i) \in F$", even if you don't care about the $t_i$.


问:
Right-invariance of a volume form on a compact Lie group
The following is a question from the second edition of John M. Lee's Introduction to Riemannian Manifolds.

3-9. Suppose $G$ is a compact Lie group with a left-invariant metric $g$ and a left-invariant orientation. Show that the Riemannian volume form $dV_g$ is bi-invariant. [Hint: Show that $dV_g$ is equal to the Riemannian volume form for a bi-invariant metric.]

Since $G$ is compact, it admits a bi-invariant metric $\tilde g$, and with this and the given orientation, we have a volume form $dV_{\tilde g}$. It's easy to see that $dV_g$ and $dV_{\tilde g}$ are both left-invariant, using the fact that the metrics $g, \tilde g$ and the orientation are left-invariant. Since they are both left-invariant and positive with respect to the given orientation, there is a $c > 0$ such that $dV_g = c dV_{\tilde g}$. Then $dV_g$ is equal to the Riemannian volume form corresponding to the bi-invariant metric $c^{2/n}\tilde g$, as the hint suggests.

I am having trouble showing that $dV_g$ is also right-invariant; here is my work thus far. Since for any $\varphi \in G$ the forms $R_\varphi^*(dV_g)$ and $dV_g$ are left-invariant, there is a function $f \colon G \to \mathbb{R}^\times$ such that $R_\varphi^*(dV_g) = f(\varphi) dV_g$. Evaluating both sides at $e$, one obtains $f(\varphi) = \det(\mathrm{Ad}(\varphi^{-1}))$, a continuous homomorphism. Since $G$ is compact, $f(G)$ is a compact subgroup of $\mathbb{R}^\times$, i.e. $f(G) = \{1\}$ or $f(G) = \{\pm 1\}$.

I do not see how to exclude the second case, i.e. if $R_\varphi$ is orientation-reversing for some $\varphi \in G$. Since $f(e) = 1$, $f$ is identically $1$ on the identity component of $G$; this would finish the problem if $G$ were connected, but unfortunately, it might not be. Since I haven't used the fact that $dV_g$ equals the volume form for a bi-invariant metric yet, I feel it must be used here, but I cannot see how.

Some searching reveals this may be related to the idea of left/right-invariant Haar measures and unimodular Lie groups, but my measure theory knowledge is insufficient to understand that material. A small part of me believes that the problem is incorrect without the connectedness hypothesis (e.g. consider the diagonal matrix $A$ with $-1$ and $1$ in $O(2)$, then $f(A) = -1$?), but the errata for the book reveals nothing. Any hints or suggestions on how to proceed with this problem would be appreciated.

答: The claim is simply false in general. I think, Lee forgot to assume that the group is connected or that the orientation is bi-invariant. As a simple example, as you suggested, consider $G=O(2)$. Then the action of $G$ on itself via conjugation does not preserve any orientation on $G_0=SO(2)$ (since this action contains a reflection). From this, it follows that there is no bi-invariant orientation. From this, it follows that there is no bi-invariant volume form. Of course, there is a bi-invariant measure on $G$, given by a bi-invariant density on $G$.


问:
How can you behold that a 45 degree line's |x-intercept| = |y-intercept|?
I'm not asking for an algebraic explanation...I can do algebra. I know $f(x) = x - c = 0 \iff x = c$. Right now, I have to make my mind process four steps. First, a line with slope $|1|$ must angle 45° with the y-intercept. Second, I must consider the right triangle bound by $y = x$  and $y = 1, 0 \le x \le 1$ and $x = 0, 0 \le y \le 1$. Third, the remaining angle must be $180 - 90 - 45 = 45°$. Thus the triangle is isoceles, and the opposite and adjacent sides have the same length. Fourth, this means $\mid x-intercept\mid = \mid y-intercept \mid $.

I don't know why, but I can't "behold" immediately this. How can I eye-ball all this quickly?

答: Because a 45-45-90 triangle is isosceles.</p>


问:
Examples of Galois Correspondence being inapplicable with non-normal groups and fields
In ``Fields and Galois Theory '' by JM Howie, I cannot think of any example to illustrate strict inclusion for the Galois correspondence in the case of non-normal groups and fields. In particular Howie uses symbols such as $\mathbf{K}, \mathbf{E}, \mathbf{L} $ for fields where often $\mathbf{K} \subset \mathbf{E} \subset \mathbf{L} $ and often uses $\mathbf{H}, \mathbf{G}$ for groups where often $\mathbf{H} \subset \mathbf{G}$. The fixed field  of a subgroup  $\mathbf{H}$  of a Galois group is denoted $\Phi(\mathbf{H})$ and the subgroup of the Galois group which fixes $\mathbf{E}$ is denoted $\Gamma (\mathbf{E})$ I note Howie's theorem 7.11 which shows $\mathbf{E} \subseteq \Phi (\Gamma (\mathbf{E}))$ and $\mathbf{H} \subseteq \Gamma (\Phi (\mathbf{H}))$. and theorem 7.28 which in effect shows that these inclusions are strict for non normal fields and subgroups.
I was then looking at the 8 automorphisms of $D_4$ (Galois group of $x^4-2=0$) as an example. In particular there is a normal and a non-normal subgroup each with one automorphism other than the identity. The normal subgroup's non-identity automorphism doesn't move $i$ but takes $\sqrt[4]{2}$ to its negative and the non normal subgroup's non-identity automorphism takes $i$ to $-i$ as well as taking $\sqrt[4]{2}$ to its negative (This is Howies subgroup $H_7$). I could not use this to find an example to illustrate strict inclusion $\mathbf{E} \subset \Phi (\Gamma (\mathbf{E}))$ and $\mathbf{H} \subset \Gamma (\Phi (\mathbf{H}))$ for non normal fields and subgroups. This makes me suspect that I am missing something important in the whole concept of the Galois correspondence. I would be grateful for an example using this $H_7 \subset D_4$ and the corresponding fields. Thanks David Kault

答:
You're not going to find a counterexample inside a Galois extension, no matter what subgroup or intermediate field is used.  The Galois correspondence holds for all intermediate fields and all subgroups (normal or not normal) in a Galois extension. It does not work when the whole field extension is not Galois.

For example, take $K = \mathbf Q$ and $L = \mathbf Q(\sqrt[4]{2})$.  Then $L/K$ is not a normal extension ($x^4 - 2$ is irreducible over $K$ with degree $4$ and has a root in $L$ without splitting completely over $L$) and the number of $K$-automorphisms of $L$ is $2$ rather than $4$, which is the degree of the extension. Explicitly,
$$
{\rm Aut}(L/K) = \{\sqrt[4]{2} \mapsto \sqrt[4]{2}, \sqrt[4]{2} \mapsto -\sqrt[4]{2}\}.
$$
This group has fixed field $\mathbf Q(\sqrt{2})$, so
$$
\Phi(\Gamma(K)) = \Phi(\Gamma(\mathbf Q)) = \Phi({\rm Aut}(L/K)) = \mathbf Q(\sqrt{2}) \not= K.
$$

You're not going to find a counterexample for the other realtion,  $\Gamma(\Phi(H)) = H$. Artin showed that if $H$ is a finite group of field automorphisms of a field $L$ then $L$ is always a finite Galois extension of $\Phi(H)$ (the subfield of $L$ fixed by $H$) and $\Gamma(\Phi(H)) = {\rm Gal}(L/\Phi(H)) = H$.

So the problem always occurs if you start with a finite extension of fields that is "bad" (not a Galois extension) rather than with a field and a finite group of automorphisms of it.  If $L/K$ is a finite extension of fields and not a Galois extension, the subfield of $L$ fixed by ${\rm Aut}(L/K)$ is always larger than $K$.  The example $\mathbf Q(\sqrt[4]{2})/\mathbf Q$ is a basic one that shows how more elements than the bottom field can be fixed by all the automorphisms of the top field that fix the bottom field. This illustrates what can go wrong if the extension is not a normal extension. What it does not show (and only a proof can really show it) is that the same problem happens every time the field extension is not normal. Also the same problem happens every time the field extension is not separable. That is, a finite extension $L/K$ is Galois if and only if $\Phi(\Gamma(K)) = K$.

For the record, I find the $\Phi$ and $\Gamma$ notation from your textbook very strange. I know $\Phi$ is a reminder of "field" and $\Gamma$ is a reminder of "group", but it is highly nonstandard in my experience. In most discussions of Galois theory, $\Gamma(E)$ is written as ${\rm Gal}(L/E)$ if you know $L/K$ is Galois (or perhaps ${\rm Aut}(L/E)$ if you don't yet know whether $L/K$ is Galois) and $\Phi(H)$ is written as $L^H$.


问:
Calculated $\mathcal{F}\{\operatorname{sinc}(t)\}$ and $\mathcal{F}\{\cos(t)\}$

Let $f: \mathbb{R}_{>0}\to \mathbb{R}$ be a function  defined by $f(t)=\frac{\sin{t}}{t}+\frac{\sin(t)\cos(t)}{t}$. I need to find the Fourier transform of $f(t)$.

My approach: By definition of Fourier's transform, we know that $$\color{blue}{\boxed{\hat{f}(\xi)=\int_{\mathbb{R}}f(t)e^{i\xi t}\operatorname{dx}=\mathfrak{F}\{f(t) \}}}$$
and that the Fourier's transform is linear, it's to say: $\color{blue}{\boxed{\mathfrak{F}(f(t)+g(t))=\mathfrak{F}\{f(t)\}+\mathfrak{F}\{g(t)\}}}$. Then, we have that
\begin{eqnarray*}
\mathfrak{F}\{f(t)\}&=&\mathfrak{F}\left\{ \frac{\sin(t)}{t}+\frac{\sin(t)\cos(t)}{t} \right\}\\
&=&\mathfrak{F}\left\{\frac{\sin(t)}{t} \right\}+\mathfrak{F}\left\{\frac{\sin(t)\cos(t)}{t} \right\}
\end{eqnarray*}
Now, we need calculate  $$\mathfrak{F}\left\{\frac{\sin(t)}{t} \right\} \quad \text{and} \quad \mathfrak{F}\left\{\frac{\sin(t)\cos(t)}{t} \right\}$$
I think, it's enough calculate the first transform, because then we can use after the convolution's theorem to invoke the transformation of $\frac{\sin(t)\cos(t)}{t}$
I know that the convolution's theorem say that: $\color{blue}{\boxed{\mathfrak{F}\{z(t)* \ell(t)\}=\mathfrak{F}\{z(t)\}\cdot\mathfrak{F}\{\ell(t)\}}}$ and taking $z(t)=\frac{\sin(t)}{t}$ and then $\ell(t)=\cos(t)$ we can follow the result.

Now, we can see that $$\mathfrak{F}\left\{\frac{\sin(t)}{t} \right\}=\int_{\mathbb{R}}\frac{\sin(t)}{t}e^{it\xi}\operatorname{dx}:=\int_{\mathbb{R}}\operatorname{sinc}(t)e^{it\xi}\operatorname{dt}$$

but, $\operatorname{sinc}(t)\notin L^{1}(\mathbb{R})$, so I think we can't calculate the Fourier's transform using the definition maybe. So, how can I continue from here? Thanks to @reuns for pointing out that there is a problem that arises when applying the idea of convolution with the Fourier transform of the cosine. Then, how can I solve this problem?

答:
First, we note that we can evaluate the Fourier Transform of $f(t)=\frac{\sin(t)}{t}$ function in a straightforward way by interpreting the Fourier Transform as the Cauchy Principal Value of the Riemann integral.  Proceeding, we write

$$\begin{align}
\mathscr{F}\{f\}(\xi)&=\text{PV}\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{i\xi t}\,dt\\\\
&=2\int_0^\infty \frac{\sin(t)\cos(\xi t)}{t}\,dt\\\\
&=\int_0^\infty \frac{\sin((\xi+1)t)-\sin((\xi-1)t)}{t}\,dt\\\\
&=\frac{\pi}{2}\left(\text{sgn}(\xi+1)-\text{sgn}(\xi-1)\right)\\\\
&=\begin{cases}
\pi,& |\xi|<1\\\\
0,&|\xi|>1\\\\
\pi/2,&|\xi|=1\tag1
\end{cases}
\end{align}$$

Now, let $g(t)=f(t)\cos(t)=\frac{\sin(t)\cos(t)}{t}=\frac{\sin(2t)}{2t}$.  Using $(1)$, the Fourier Transform of $g(t)$ is

$$\begin{align}
\mathscr{F}\{g\}(\xi)&=\text{PV}\int_{-\infty}^\infty \frac{\sin(2t)}{2t}e^{i\xi t}\,dt\\\\
&=\frac12 \text{PV}\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{i\xi t/2}\,dt\\\\
&=\frac{\pi}{4}\left(\text{sgn}(\xi/2+1)-\text{sgn}(\xi/2-1)\right)\\\\
&=\begin{cases}
\pi/2,& |\xi|<2\\\\
0,&|\xi|>2\\\\
\pi/4,&|\xi|=2\tag2
\end{cases}
\end{align}$$

If we wish to use the Convolution Theorem, then we need to appeal to the theory of distributions.  Then, inasmuch as the Fourier Transform of $\cos(t)$ is $\frac12\left(\delta(\xi+1)+\delta(\xi-1)\right)$, where $\delta$ is the Dirac Delta, then we have

$$\begin{align}
\mathscr{F}\{f\cos\}(\xi)&=\frac\pi4\int_{-\infty}^\infty (\delta(\xi-\tau-1)+\delta(\xi-\tau+1))\,\,\left(\text{sgn}(\tau+1)-\text{sgn}(\tau-1)\right)\,d\tau\\\\
&=\frac\pi4 \left(\text{sgn}(\xi+2)-\text{sgn}(\xi-2)\right)\\\\
&=\begin{cases}
\pi/2,& |\xi|<2\\\\
0,&|\xi|>2\\\\
\pi/4,&|\xi|=2
\end{cases}
\end{align}$$

which agrees with the result in $(2)$!


问:
Solution to $ y_{t+2} −4y_{t+1} + 3y_{t} = t^2 + 2t$
I have the following second order difference equation to solve:  $y_{t+2} −4y_{t+1} + 3y_{t} = t^2 + 2t;  y_{0} = 1;  y_{1} = 2$. I have found the solution for the homogeneous part but i have difficulty in solving the non-homogeneous part.
Can anyone advise me on what should i do  in order to find the particular solution? Thanks a lot.

答:
SInce you have a second degree on the rhs, you can assume a third degree for the particular solution.

So, let
$$y_t=a+b t + c t^2+d t^3$$ replace in the lhs and group terms to get
$$(4 d-2 b)-4 c t-6 d t^2=2t+t^2$$ So, comparing the coefficient for each power, $d=-\frac 16$, $c=-\frac 12$, $b=-\frac 13$.

Combining all results
$$y_t=c_1+c_2\, 3^t +a-\frac{t^3}{6}-\frac{t^2}{2}-\frac{t}{3}=c_3+c_2 \,3^t -\frac{t^3}{6}-\frac{t^2}{2}-\frac{t}{3} $$ Use the conditions to get $c_2$ and $c_3$.


问:
How to get the angle in a trapezoid formed by two triangles?
The problem is as follows:

Find the value of angle $\angle BCA$ or $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^\circ\\
2.&20^\circ\\
3.&10^\circ\\
4.&40^\circ\\
\end{array}$

The thing here is without any further hints. I'm stuck.

I was only able to spot.

$\triangle ABC$ is isosceles,

$\triangle DBC$ is isosceles

But I can't find any other relationship. What else to do?. How to solve this relying only in euclidean geometry postulates?. Can someone help me here?.

Should congruence be used here or what?.

I'm confused please help me. Please include a drawing in your answer as it would help me to spot what to do or if a construction is needed.

答: Note that after angle chasing we find that $A,C,D$ lies on a circle with center $B$. Note that $$x=70^{\circ}-\angle ACD =70^{\circ} - \frac{1}{2}\angle{ABD}=30^{\circ}$$ by the inscribed angle theorem.


问:
About Ehrenfeucht's theorem proof
I am re-reading the proof of Ehrenfeucht's theorem on page 90 of Mathematical Logic of Shoenfield. I have the following problem. For the sake of clarity I have highlighted in red the problematic part in the theorem. The question is: How is it possible infer by the lemma 3 that the proof uses special axioms only for $\mathbf r_1$, ... , $\mathbf r_n$ and the special constants in $\mathbf A_1$, ... , $\mathbf A_{k-1}$? (For clarity reasons I have highlighted too in the lemma the part that I consider being referred to by the theorem.) Finally,  for proof ends, it is to be noted that this fact is redundant.
Thanks  in advance for any suggestion.

答: I was wrong, the argument is correct. I give the answer in case someone  were to have my own doubts. First, the part I stressed is essential to the proof for the generality of $\mathbf A$ (i.e., the $\mathbf C'$ must be the same for every $\mathbf A$). Second (I found this irksome), the fact that at most there are special axioms for those special constants, derives by interpreting lemma 3 in such a way that $T_c$ is a conservative extension of $T^*=T \cup \{r_1,...,r_m,$ special axioms of these special constants$\}$ and considering $T'$ starting by $T^*$ by adding the remaining special constants (i.e., we can use the same proof of the lemma adapting it to this case).</p>

Title: About Ehrenfeucht's theorem proof


问:
Question on pointwise convergence of Sequence of functions
In section 8.2 of  textbook "Introduction to real analysis by Bartle" example 8.2.1(c) is as follows:

$f_n : [0,1] →\mathbb{R}$ defined for $n≥2$ by

$$f_n(x) =
  \begin{cases}
     n^2x & \text{for $0≤x≤1/n$} \\[6pt]
    -n^2(x - 2/n) & \text{ for 1/n≤x≤2/n} \\[6pt]
  0 & \text{for 2/n≤x<1}
   \end{cases}$$

then $f_n$ is continuous on $[0,1]$ and hence Riemann integrable______ (upto this i have no problem)

My problem$f_n(x)→0$ for all $x\in\mathbb{R}$ but i unable to see this...

My attempt clearly for $x=0$, $f_n(0)=0$

For $0<x<1/n$, $|f_n(x)-f(x)|=|n^2x|=n^2x$

How, can say this is $<\epsilon$ for sufficiently large $n$?

Further, for other points $x$ how to establish its convergence and how to show integral of $f_n\text{  from $0$ to $1$}=1$  for $n≥2$

Please help and explain this example 8.2.1(c) in detail.

答:
For a fixed $x \in (0,1]$, there exists  a positive integer $N > 2/x$.

Thus, for all $n \geqslant N$ we have $x > 2/n$ and $|f_n(x)-0| = f_n(x) = 0$.


问:
How to solve for unknown variable with summation operator
i had the following equation of which i need to solve for $p_0$. The expression is : $\sum_{m=1}^{M+1}[q_1(m)-q_0(m)]log_2[p_0q_0(m)+(1-p_0)q_1(m)]=\\\sum_{m=1}^{M+1}q_1(m)log_{2}q_1(m)-q_0(m)log_2q_0(m)-[q_1(m)-q_0(m)]log_2e$

答:
The right hand side is just a constant, independent of $p_0$. call it $K$. On the left hand side, the expression can be simplified by letting $a_m = q_1(m) - q_0(m)$ and $b_m = q_1(m)$. And as its our variable, just call $p_0$ by "$x$" instead. The equation becomes
$$\sum_{m=1}^{M+1} a_m\log_2(b_m - a_mx) = K$$

If you raise $2$ to the power of each side, you get
$$\prod_{m=1}^{M+1} (b_m - a_mx)^{a_m} = 2^K$$

If all the $a_m$ happen to be integers $\ge 0$, this would be a polynomial in $x$ of degree $\sum_{m=1}^{M+1} a_m$. Which would mean that there could be as many as $\sum_{m=1}^{M+1} a_m$ different solutions for $x$. Or (if you want real solutions only), there could be no solution at all. Even when the $a_m$ are not subjected to such severe restrictions, these lessons still hold. Depending on the $a_m, b_m$, there could be many solutions or none at all.

Inevitably, when people ask this sort of question, they are wanting some formula $p_0 = ...$ written in terms of functions and operations the questioner is familiar with or can easily learn, and usually because the questioner thinks that this is the way to calculate values. But this hope is more often than not a chimera. One finds such solutions when learning mathematics because the problems you are given to solve have been intentionally limited to the tiny subset that actually have such solutions. But when you stretch your mathematical wings into the broader world, reality comes crashing in. Most problems do not have such "nice" solutions.

But there are alternatives. Root-finding schemes can be used to estimate solutions to any desired accuracy (and another real-life lesson: except when the solution is guaranteed to be integer, or a low denominator rational, you never actually get an exact solution, even when you do have a nice equation for it).

The gold standard for these is Newton's method. Let $$f(x) = \sum_{m=1}^{M+1} a_m\log_2(b_m - a_mx) - K$$
and define
$$f'(x) = \log_2 e\sum_{m=1}^{M+1} \dfrac {a_m^2}{a_mx - b_m}$$
(This is the derivative of $f$. You listed "algebra precalculus", but I've just done the only calculus part, so all you have to do is precalculus.)

In Newton's method, you make some initial guess $x_0$ at a root, then repeatedly calculate
$$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$$
If you are not unlucky, this will quickly approach a constant value that will be an actual solution to your equation. Which solution it will converge to is dependent on your initial guess.


问:
is $f$ constant or dilation
$f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)$ for some $\alpha>0$, $\forall x\in\mathbb{R}$, for any $r\in\mathbb{R}$, could any one tell me which of the following is true?

$1.$ If $f(x)=f(y)$ whenever $||x||=||y||=\beta>0$, then $f(x)=\beta ||x||^{\alpha}$

$2.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)= ||x||^{\alpha}$

$3.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)=c||x||^{\alpha}$ for some constant $c$

$4.$ If $f(x)=f(y)$ whenever $||x||=||y||$, then $f(x)$ must be a constant function.
Thank you.

答:
Let $~f(x)=3~\lVert x\lVert^\alpha~.$

For option $\bf {(1)}$, let $~\beta =2~.$

Clearly, $~f(x)=3\cdot2^\alpha=f(y)~,$ whenever $~\lVert x\lVert=\lVert y\lVert=\beta=2\gt0~.$

But $~f(x)=3~\lVert x\lVert^\alpha\ne \beta~\lVert x\lVert^\alpha~$

$\therefore$ option $(1)$ is not true.

For option $\bf {(2)}$, from the above example, whenever $~\lVert x\lVert=\lVert y\lVert=1~,$ $~f(x)=3=f(y)~,$

But $~f(x)=3~\lVert x\lVert^\alpha\ne \lVert x\lVert^\alpha~$

$\therefore$ option $(2)$ is not true.

For option $\bf {(4)}$, from the above example, $~f(x)=f(y)~,~~~~\forall x,y~~$  for which  $~\lVert x\lVert=\lVert y\lVert~.$

But here $~f~$ is not constant.

$\therefore$ option $(4)$ is not true.

As all other options are incorrect, therefore only option $(3)$ is true.

Title: is $f$ constant or dilation


问:
Convergence of series in p-adic norm
I want to know the convergence of the series

$\sum_{n\geq 0} {{p^{n+1}}\choose{p^n}}$

My idea is to show the convergence of the partial sum $|s_n-s_{n-1}|_p$ but I am stucked in expressing the terms properly.

Is there any hint to this question?

答:
The sum converges in the p-adics if and only if we have $\lim_{n \to \infty} \binom{p^{n+1}}{p^n} = 0$ like you say. Using Legendre's formula we can get an exact answer for what the terms of the series tend to,

$$v_p \left( \binom{p^{n+1}}{p^n}\right) = v_p\left(\frac{p^{n+1}!}{p^n! (p^{n+1}-p^n)!}\right)  = v_p (p^{n+1}!) - v_p(p^n!) - v_p((p^{n+1}-p^n)!)$$

$$=\frac{p^{n+1} - s_p(p^{n+1})}{p-1} - \frac{p^n-s_p(p^n)}{p-1} - \frac{p^{n+1}-p^n-s_p(p^{n+1}-p^n)}{p-1}$$

$$=\frac{- s_p(p^{n+1}) + s_p(p^n)  + s_p(p^{n+1}-p^n)}{p-1}$$

Now since a power of $p$ is simply a single $1$ followed by $0$s the first two terms are easy to evaluate, and since $p^{n+1}-p^n = (p-1)p^n$ this is already written in base $p$, since it is the digit $p-1$ followed by $n$ following $0$s.

$$=\frac{- 1 + 1  + p-1 }{p-1} = 1$$

Because the valuation is the constant $1$, it must be that,

$$\left|\binom{p^{n+1}}{p^n}\right|_p = p^{-1} \not \to 0$$

So it does not converge.

There seemed to be some confusion about how it converges from reading the comments, hopefully I can give some analogies to series of real numbers for how this fails to converge to help clear that up.

Because your series is bounded the entire time, it resembles the way in which the real series $\sum_{n \ge 0} (-1)^n$ fails to converge, because it is always bounded with its partial sums perpetually "meandering around".

Keep in mind we still have sums that "diverge to infinity" here in the p-adics, in the sense that the partial sums get arbitrarily large. For instance the p-adic series $\sum_{n \ge 0} \frac{1}{p^n}$ has its terms $\left|\frac{1}{p^n}\right|_p = p^n \to \infty$ which is how the real series $\sum_{n \ge 0} n$ also fails to converge.

Title: Convergence of series in p-adic norm


问:
canonical projection of a subring
Let $R = \mathbb{Z}$ and $R' =\mathbb{Z}_4 = \mathbb{Z}/4\mathbb{Z}$, and $f = \pi$ the canonical projection of $\mathbb{Z}$ on $\mathbb{Z}/4\mathbb{Z}$. And let $A = 6\mathbb{Z}$ and $I=4\mathbb{Z}$.

If I've understood correctly $f$ should be an homomorphism between $R$ and $R'$.

My manual calculates $\pi(A)$ and gets a result different from mine: $\pi(A) = (A + I)/I = {I,2+I}$.

I'm applying the definition of canonical projection so $pi: A \to A/I$. For me is $\pi(A) = 6\mathbb{Z}/4\mathbb{Z}$, I don't understand why is summing up $I$ to $A$.

Can you explain me how to calculate the image $pi(A)$ and if/why it is an ideal of R/I?

答:
In general, images of ring homomorphims need not be ideals. This applies to ideals as well. For example, consider the ring homomorphism given by the inclusion $\mathbb{Z} \to \mathbb{Q}$. However, preimages of ideals are ideals.

The projection map $\pi: R \to R/I$ is given by $\pi(x) = x + I$. Then, $$\pi(A) = \{\pi(x)\;|\;x \in A\} = \{x + I\;|\;x \in A\}.$$

In the case of $R = \mathbb{Z}$ and $I = 4\mathbb{Z}$ I usually refer to "$x + 4\mathbb{Z}$" as "$x$ mod $4$". We can apply the above to find $$\pi(6\mathbb{Z}) = \{x\;\text{mod}\;4\;|\;x \in 6\mathbb{Z}\}.$$

Thus, if $n \in \pi(6\mathbb{Z})$ then $n = \pi(x)$ for $x = 6 k$ for some $k \in \mathbb{Z}$. Then, $n \equiv 6k \;\text{mod}\;4$. If $k$ is even, then $k = 2l$ for some integer $l$ and $n \equiv 12l \equiv 0 \;\text{mod} \;4$ since $12 \in4 \mathbb{Z}$ and $4\mathbb{Z}$ is an ideal. Similarly, if $k$ is odd, then $n \equiv 2\; \text{mod}\;4$. This accounts for all of the possible integers $k$ so we have $\pi(6\mathbb{Z}) = \{4\mathbb{Z}, 2 + 4\mathbb{Z}\}$. Contrary to the first remark, $\pi(6\mathbb{Z})$ is an ideal of $\mathbb{Z}/4\mathbb{Z}$.

Title: canonical projection of a subring


问:
How can I show that $X_l$ and $X_{f \circ l}$are homotopy equivalent?
Let $(X, \mathcal{T}_X)$ be a topological space.

Let $l$ be a loop so that $p \in X$ is the base point and $f: X \to X$ a continuous function, where $f \simeq id_X$ ($f$ and the identity function is homotopic).

Furthermore, let $X_l = \{x | x = l(t) ( 0 \leq t \leq 1) \}, X_{f \circ l} = \{ x | x = f \circ l(t) (0 \leq t \leq 1) \}$, and their topology defined by reative topology.

Are $X_l$ and $X_{f \circ l}$ homotopy equivalent?

答:
This is not true. Let $X=\Bbb R^2$ and $f:\Bbb R^2\to \Bbb R^2$ be defined as $f(x)=0$ for all $x\in\Bbb R^2$. Define $H:\Bbb R^2\times [0,1]\to\Bbb R^2$ as $H(x,t)=tx$ for all $(x,t)\in\Bbb R^2\times [0,1]$. Then, $H:f\simeq \text{id}_{\Bbb R^2}$.

Next, let $l:[0,1]\ni t\longmapsto e^{2\pi it}\in \Bbb S^1$ be the loop based at $1\in\Bbb S^1$. Note that $X_l=\Bbb S^1$ and $X_{f\circ l}=\{0\}$. But, $\Bbb S^1$ is not homotopically equivalent to $\{0\}$ as $\pi_1(\Bbb S^1)=\Bbb Z$ but $\pi_1(\{0\})=\text{trivial group}$.


问:
How to analyze the elements in a polynomial quotient ring?
I am looking at some old exams, and one thing is to prove that in a quotient ring $R = F[x]/\langle p(x)\rangle$, where $F$ is a field and $\langle p(x)\rangle$ is the ideal generated by a polynomial $p(x) \in F[x]$, that all elements are either units or zero divisors. I have tried looking this up and read several documents on quotient rings, but I haven't seen a mention of this fact.

Here's my attempt at a proof. Suppose $a(x) + \langle p(x)\rangle$ is not a unit. Then this element generates a proper ideal in $R$. This corresponds to a proper ideal in $F[x]$ which contains $a(x)$ and $p(x)$. $F[x]$ is a PID, so this ideal is generated by some other polynomial $f(x)$, which cannot be a unit in $F[x]$. Honestly, I'm not sure where to proceed, and I'm a little tired and scared preparing for this exam.

Any hints would be appreciated, more than a full solution.

Thanks

答:
Say you take an element $\overline{q(x)} \in {F(x)/\langle p(x)\rangle}$. You have two cases:

Case I: $q(x)$ does not share any common factor with $p(x)$ i.e. they are coprime to each other. In this case, the Euclidean algorithm gives you that there are $a(x), b(x)$ such that $a(x) p(x) + b(x) q(x) = 1$. This means that

$b(x) q(x) \equiv 1 ($ mod $p(x))$ which exactly means $\overline{q(x)}$ is a unit.

Case II: $q(x)$ shares a factor $c(x)$ with $p(x)$ i.e. $q(x) = c(x) d(x)$ and $p(x) = c(x) e(x)$ for some $d(x), e(x)$.

In this case, you see that $q(x) e(x) \equiv 0 ($ mod $p(x) )$ which means that $\overline{q(x)}$ is a zero-divisor.


答:
The line segment $y = 1$ for $0 \leq x \leq 1$ seems to me to be completely unnecessary and unhelpful to think about.

For the line $y = x + 4$ the triangle to look at is the green triangle in the figure below. For the line $y = x - 2$ the triangle to look at is the black triangle.

For these two lines, the triangles are isosceles with the equal legs on the $x$ and $y$ axes. Note that $x$ and $y$ intercepts are not equal: they are of equal magnitude (same distance from the origin) but have opposite signs.
For example, for $y = x + 4$ the $x$-intercept is $-4$ but the $y$-intercept is $4$.

For $y = x,$ of course, there is no triangle. The line passes straight through the origin, so it intercepts both axes at the same point, $(x,y) = (0,0)$.

For any other line of the form $y = x - c,$ the triangle is again an isosceles triangle with the apex at the origin like the green or black triangles, but possibly larger or smaller.

Title: How can you behold that a 45 degree line's |x-intercept| = |y-intercept|?

Body:
I'm not asking for an algebraic explanation...I can do algebra. I know $f(x) = x - c = 0 \iff x = c$. Right now, I have to make my mind process four steps. First, a line with slope $|1|$ must angle 45° with the y-intercept. Second, I must consider the right triangle bound by $y = x$  and $y = 1, 0 \le x \le 1$ and $x = 0, 0 \le y \le 1$. Third, the remaining angle must be $180 - 90 - 45 = 45°$. Thus the triangle is isoceles, and the opposite and adjacent sides have the same length. Fourth, this means $\mid x-intercept\mid = \mid y-intercept \mid $.

I don't know why, but I can't "behold" immediately this. How can I eye-ball all this quickly?


问:
A problem about a rank's inequality of a complex matrix
Let $X,Y \in M_{n\times n}(\mathbb{C})$ such that $X^{2}=Y^{2}=I_{n}$. Prove that

$$\operatorname{rank}((X+I_n)(Y-I_n))+\operatorname{rank}((X-I_n)(Y+I_n))=\operatorname{rank}(X-Y)$$

My approach: Using the well-known inequality: $\operatorname{rank}(X)+\operatorname{rank}(Y)\geq \operatorname{rank}(X+Y)$, we can see that $$\operatorname{rank}(X-Y)=\operatorname{rank}(X-XY+XY-Y)\leq \operatorname{rank}(X-XY)+\operatorname{rank}(XY-Y)$$

but, how can I continue from here?

答:
You can show one of the inequalities using your approach, by adding and subtracting $XY$ and $I$ in your expression. For the other inequality to achieve equality is trickier. So I will approach it in a different manner. We will show instead the direct sum decomposition of their images:
$$Im(X-Y) = Im(X+I)(Y-I) \oplus Im(X-I)(Y+I),$$

which implies $rank(X-Y) = rank(X+I)(Y-I) + rank(X-I)(Y+I)$.

To show this direct sum decomposition, we show the sum spaces are equal with trivial intersection.

Part 1: $Im(X-Y) \subset Im(X+I)(Y-I) + Im(X-I)(Y+I).$

Indeed, any vector in $Im(X-Y)$ is of the form $(X-Y)v$ for some $v\in \bf C^n$. Note
$$ (X-Y)v = (XY + Y - X -I)(v/2) + (XY-Y+X-I)(-v/2)\\
=(X+I)(Y-I)(v/2) + (X-I)(Y+I)(-v/2)\\
\in Im(X+I)(Y-I) + Im(X-I)(Y+I).$$

Part 2: $Im(X-Y) \supset Im(X+I)(Y-I) + Im(X-I)(Y+I).$

Indeed, an element in $Im(X+I)(Y-I) + Im(X-I)(Y+I)$ is of the form $(X+I)(Y-I)v + (X-I)(Y+I)w$ for some $v,w\in \bf C^n$. Next, we use the fact that $Y^2 = I$, so
$$(X+I)(Y-I)v + (X-I)(Y+I)w \\
=(XY-X+Y-{\color{blue} I})v +(XY +X -Y -{\color{blue} I})w\\
= (XY-{\color{blue} Y^2} -(X-Y))v + (XY - {\color{blue} Y^2} +(X-Y))w\\
= (X-Y)(Y-I)v + (X-Y)(Y+I)w\\
= (X-Y)[(Y-I)v+(Y+I)w]\\
\in Im(X-Y).$$

Part 3: $Im(X+I)(Y-I) \cap Im(X-I)(Y+I) =(0)$.

Suppose $z$ is in the intersection, so we have $z=(X+I)(Y-I)v = (X-I)(Y+I)w$ for some $v,w$. Then multiply $(X-I)$ from the left to both sides, giving
$$(X-I)(X+I)(Y-I)v = (X-I)(X-I)(Y+I)w \\
\implies (X^2 -I)(Y-I)v = (X-I)^2(Y+I)w\\
\implies 0 = (X-I)^2(Y+I)w
$$ since $X^2 = I$. But
$$(X-I)^2(Y+I)w = (X^2 -2X+I)(Y+I)w \\
 =(2I-2X)(Y+I)w = -2(X-I)(Y+I)w = 0 $$
so $(X-I)(Y+I)w= z=0$. Hence the intersection is trivial.


问:
Question applying Gauss' theorem
Q: Use Gauss' theorem to compute the surface integral
$$
\oint_{S}{\bf r}\cdot{\bf n}\ {\rm d}S
$$
where $S$ is the surface of the unit cube $0 \leq x, y, z \leq 1$ where ${\bf r}$ is the position vector.

Not sure how to apply this, obviously
${\bf r} = \left(x,y,z\right)$ but I'm not sure how to parametrise the surface of the unit cube in this instance $?$. Any help $?$.

答:
By Gauss's theorem
$$\oint_{S}{\bf r}\cdot{\bf n}\ {\rm d}S = \int_V \nabla\cdot{\bf r}\  dV$$

Now, by going to Cartesian coordinates, we get
$$\nabla\cdot{\bf r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3$$

Hence,
$$\oint_{S}{\bf r}\cdot{\bf n}\ {\rm d}S = \int_V 3 \ dV = 3$$

Title: Question applying Gauss' theorem


问:
Prove that if $X$ is a subset Maximally Linearly Independent of $V$, then $X$ is a basis of $V$.
I just want you to check if my proof is correct (or not) please.

Let be $F$ a field and $V$ a vector space over $F$. Prove that if $X$
is a subset Maximally Linearly Independent of $V$, then $X$ is a basis
of $V$.

My attempt:

We suppose that $X$ is not a basis. Then, since $X$ is a subset of $V$, there exist another subset $S$ of $V$ such that, $Z=X \bigcup S$ is linearly independent and generates $V$.

But, that implies $X \subseteq Z$, so $X$ is not Maximally linearly independent. (That's a contradiction).

\begin{align}
\therefore X \text{ is a basis of } V
\end{align}

Is correct my proof? I really appreciate your help!

答:
An alternative way:

Suppose that $X$ is not a basis for $V$, that is, it does not generate the full space. Then there is $v \in V$ which cannot be written as a linear combination of the elements of $X$, meaning that $X \cup \{v\}$ is linearly independent, contradiction.


问:
How can I prove that the $\lim _{n\to \infty }\frac{log_{a}n}{n}=0$ , $a>1$ with the simplest tools of analysis?
I tried solving this problem but I didn't know where to start so I searched it on the internet and the only solution that I found was by using L'Hopital's rule. The thing is we haven't studied that in class, but my professor uses this problem constantly in order to solve other problems and I don't understand why this works. I think I should use the squeeze theorem (since I am only familiar with the limits of sequences and that's the most helpful theorem I know), but I don't know which inequality to use. Could you please help me by giving a hint or some advice?

Thanks in advance.

答:
You know that $\displaystyle \lim_{n\to \infty} n^{1/n}=1$.
$\text{Now}\; \displaystyle \lim_{n\to \infty}{ \log_a n\over n}=\frac{1}{\ln a}\lim_{n\to \infty} {\ln n\over n}={1\over \ln a}\cdot \lim_{n\to \infty} \ln {n^{1/n}}=\frac{1}{\ln a}\cdot \ln \left(\lim_{n\to \infty} n^{1/n}\right)={\ln 1\over \ln a }=0$


问:
Changing bounds of double integral while converting from cartesian to polar form

Evaluate the  integral $\displaystyle \int_{0}^{\sqrt{π}} \int_{y}^{\sqrt{π}} \cos(x^2) \ \mathrm{d}x \, \mathrm{d}y $

I have replaced $x^2$ with $r^2\cos^2(θ)$ but could not change the bounds by using inequalities.

答:
You are integrating over triangle bound by $y = 0, x=y, x = \sqrt {\pi}$ so changing the order your integral becomes

$\displaystyle \int_{0}^{\sqrt{π}} dx \,\int_{0}^{x} \cos(x^2) \, dy$

Now you can use substitution $x^2 = t$ for your second integral.


问:
Problem to understand Hungerford's book
In the Hungerford's book and following the answers of this question:
Help to understand the ring of polynomials terminology in $n$ indeterminates I have troubles to understand the following remark in the page 152:

My question is $\varphi$f is a composition of functions? because if so, it is not well defined because $f\in R[x_1,...,x_n]$ and $s_1,...,s_n\in S$

Thanks a lot

答:
If $\,f\,$ is a polynomial in $\,n\,$ indeterminates with coefficients in $\,R\,$ , we want $\,\phi f\,$ to be an element in $\,S\,$ , and that's why there's written $\,\phi f(s_1,...,s_n)\;,\;\;s_i\in S\,$ . For this to happen, we need  the coefficients of the polynomial $\,f\,$ in $\,R[x_1,...,x_n]\,$ to be mapped to elements in $\,S\,$ , and this is precisely what $\,\phi(a_i)\,$ accomplishes.

So, in short, $\,\phi f\,$ is definitely not composition of function since, as you remark, this wouldn't make any sense in this case.

Title: Problem to understand Hungerford's book


问:
On uniform convergence in normal families of functions
I am just looking at normal families of functions and they have made me somewhat complex to understand, in particular I have two doubts that I consider essential to both distinguish and understand how they are and these are the following:

A) Is there a normal family of holomorphic functions $F$ that has a sequence without subsucessions that converge uniformly in a domain? That is, can you give an example of a normal family of holomorphic functions in the domain $ U $ and a sequence $ \{f_n\} $ contained in $ F $ that does not have a subsucession that converges uniformly in $ U $? Or must they all necessarily have convergent subsucessions in U?

B) Can an example be given of a normal family of holomorphic functions in $U$ domain and a sequence $\{f_n\}$ contained in $F$ that converges uniformly in compacts of $U$ to a point $f$ and such that $f$ is not in $U$?

I hope you can help me please, thank you very much in advance.

For part B), I think this answer works, but I wonder if I am correct.

答:
(A): $U=\{z: |z|<1\}, f_n(z)=z^{n}$. This sequence is normal but no subsequence converges uniformly on $U$.

[This sequence converges uniformly on compact subsets of $U$ to $0$ but not on $U$: Suppose $|z^{n_k}| <\frac 1 2$ for all $z \in U$, for all $k$ sufficiently large. Let $z \to 1$ to get a contradiction].

(B) : The link you have given works.


答:
Going off of the hint given by azif00 and completing the work done by sumanta we have
\begin{align*}
\int_0^{\sqrt{\pi}}\int_{x=y}^{x=\sqrt{\pi}} \cos(x^2)dxdy &= \int_0^\sqrt{\pi}\int_{y=0}^{y=x}\cos(x^2) dydx \\
&= \int_0^\sqrt{\pi}\left[y\cos(x^2)\right]_{y=0}^{y=x}dx \\
&= \int_0^\sqrt{\pi}x\cos(x^2)- (0\cdot\cos(x^2)) \ dx \\
&= \int_0^\pi x\cos(u)\frac{du}{2x}  \tag{substitute $u = x^2$}\\
&= \frac12\int_0^\pi\cos(u)du \\
&= \frac12\Big[\sin(u)\Big]_0^\pi \\
&= \frac12 (0 - 0) \\
&= 0.
\end{align*}
Note that it often helps to draw the region over which we are integrating:

Title: Changing bounds of double integral while converting from cartesian to polar form

Body:
Evaluate the  integral $\displaystyle \int_{0}^{\sqrt{π}} \int_{y}^{\sqrt{π}} \cos(x^2) \ \mathrm{d}x \, \mathrm{d}y $

I have replaced $x^2$ with $r^2\cos^2(θ)$ but could not change the bounds by using inequalities.


问:
How many people do you need to guarantee that two of them have the same initals?
An auditorium has a seating capacity of 800. How many seats must be occupied to guarantee that at least two people seated in the auditorium have the same first and last initials?

I thought $26 \cdot 26$ but the answer key has $26 \cdot 26+1$ Where's the +1 coming from? Is the whole thing about the auditorium a red haring, I don't see how it matters?

答:
The first two intials can be "$aa,ab,ac,ad,\ldots,zw,zx,zy,zz$".
So, total possibilities for different types of first two intials are $26\times 26 = 676$.

And they asked us to have same first two intials for atleast $2$ children, so we can take any of the $676$ possibilities as a common first two intials.
So, number of ways for at least $2$ students who have the same
first two initials are $676+1 = 677$.


问:
Relative prime to $0$
This question is more general, but I am going to use a theorem to motivate it.

Suppose I want to prove that there exists a rational $r$ such that $r^3 + r + 1 = 0$. The first step is to assume that there is such an $r$, so $r = \frac{p}{q}$ where $p,q \in \mathbb{Z}$, $q \neq 0$ where $p,q$ are relatively prime.

Here's my question. If this $r$ were $0$ (it's not, and I can rule out it out, but I'm interested in whether I need to actually rule it out for full rigor), that $r = \frac{0}{q}$. But $0 \cdot 0 = 0$ and $0 \cdot q = 0$, so both $p$ and $q$ have a common factor of $0$.

But $\gcd(p,q) = 1$, still, since $1 > 0$, and it doesn't seem to matter if $q$ is negative.

Based on this, my conclusion is that it doesn't actualy matter if $p = 0$ and I don't need to consider this. Is that right? If I wrote "assume $p$ and $q$ have no common factors," that's already a bit ambiguous because they surely have a common factor of $1$, but the more formal "relatively prime" assumption seems ok.

答:
If we replace "$p,q$ are relatively prime" with "$\frac pq$ is in 'lowest term'" would it change how you think of it?

If $q > 1$ then $\frac 0q = \frac 01$ so $\frac 0q$ is not in lowest terms.

If we use the notation of $\gcd$ and "relative prime" though the arguement is the same.

As $0\cdot q = 0$ we have the $q$ is a divisor of $0$ and so $\gcd(0, q) = q$ and if $q > 1$ then $\gcd(0,q) = q$ and therefore

If $q>1$ then $0$ and $q$ are not relatively prime.

But $\gcd(0,1) = 1$ so

$0$ and $1$ are relatively prime.

And we can just continue.

====

But in your analysis you got confused and made a convolution.

You say:

But 0⋅0=0 and 0⋅q=0, so both p and q have a common factor of 0.

Not quite.  we have $0\cdot q =0$. You do not have $0\cdot something = q$.  So $0$ is NOT a factor of $q$.  So $0$ is not a factor of anything except of itself.

What you do have and should have said is because $0\cdot q = 0$ and $1\cdot q = q$ that it is $q$ (and not $0$)  that is a common factor of $0$ and $q$.

In fact every thing is a factor of $0$ so $\gcd(0,anything) = |anything|$.  (Bear in mind $\gcd(a,b) = \gcd(a,-b) = \gcd(-a, b)=\gcd(-a, -b)$ because if anything divides both $a$ and $b$ it also divides $-a$ and $-b$.)

And $0$ and $q$ are relatively prime means $\gcd(0, q) = 1$.  But $\gcd(0, q) = |q|$ so to have $0$ and $q$ relatively prime we must have $q = \pm 1$.

....

oh, I should point out, as Prasun Biswas correct me, that when we define $\gcd(a,b)$ and the "greatest" common divisor, most texts don't necessarily mean "greatest" in magnitude, but "greatest" in divisibility.  We define $a\preceq b$ to mean that $a$ divides $b$ and that is a partial order (not total, not any two elements compare).  Using this order the "greatest" common divisor is the common divisor that all other common divisor divide into.

For the most part the definition are the same as if $a,b$ are both positive $a\preceq b \implies a \le b$. And if $a,b$ are positive integers the largest common divisor in magnitude and the common divisor greatest in divisibility are the same.

But in this case as everything divides $0$, we always have $q\preceq 0$ and $\max_{\preceq} \mathbb Z = 0$ and $0$ is the greater in divisibility than all integers.  So although all $q$ are common divisors of $0$ and $0$, $\gcd(0,0) = 0$.

Title: Relative prime to $0$


问:
Directed graph, calculation of edges
I have a task "We have a graph G, which is directed and has 10 vertices. It has no parallel edges and has no loops. Also we know that G has 3 components and 5 strongly-connected components.

What is the minimum amount of edges that G can have?

What is the maximum amount of edges that G can have?"

How do I calculate this?

答:
What is the minimum amount of edges that $G$ can have?

Since $G$ has $5$ strongly-connected components, it has a strongly-connected component $H$ consisting of at least $2$ vertices. We can remove an edge $e$ from $H$, keeping its connected. Then $G$ without $e$ still has $3$ connected components. A connected component consisting of $k$ vertices has at least $k-1$ edges, so $G$ has at least $7+1=8$ edges. This value is attained, for instance, when $G$ is a disjoint union of a directed cycle on six vertices and two copies of a graph on two vertices with a single directed edge.

What is the maximum amount of edges that $G$ can have?"

If connected components of $G$ have $n_1$, $n_2$, and $n_3$ vertices then $G$ has at most ${10\choose 2}-(n_1n_2+n_1n_3+n_2n_3)$ edges.

Let us estimate $n_1n_2+n_1n_3+n_2n_3$ from below. Suppose that there exists distinct $i,j$ such that $1<n_i\le n_j$. Say, $1<n_1\le n_2$. Put $n_1’=n_1-1$ and $n_2’=n_2+1$. Then
$$n_1n_2+n_1n_3+n_2n_3- n_1’n_2’+n_1’n_3+n_2’n_3=$$
$$n_1n_2+n_1n_3+n_2n_3- (n_1-1)(n_2+1)+(n_1-1)n_3+(n_2+1)n_3=$$
$$n_1n_2- (n_1-1)(n_2+1)=n_2-n_1+1>0.$$

Thus the minimum of $n_1n_2+n_1n_3+n_2n_3$ is attained when two of $n_i$’s equal $1$, so $G$ has at most ${10\choose 2}-(1\cdot 1+1\cdot 8+1\cdot 8)=28$ edges. This value is attained, for instance, when $G$ is a disjoint union
of two isolated vertices and a tournament on eight vertices, which contains a directed cycle on six vertices and two remaining vertices have only ingoing edges.

Title: Directed graph, calculation of edges


问:
Prove the derivative of the delta distribution is not a 'regular' distribution.
In the terminology of Gel'fand, Generalized Functions Volume 1, a regular distribution, $\Lambda_f$, is one that is obtained from a locally integrable function, $f$, by integration of the function times a test function:
$$\Lambda_f(\phi)=\int_{\mathbb{R}^n}\!f(x)\phi(x)\,dx\qquad(\phi\in C^\infty_c(\mathbb{R}^n)).$$
I know that the Dirac delta distribution, defined by
$$\delta_p(\phi)=\phi(p)\qquad(\phi\in C^\infty_c(\mathbb{R}^n),\,p\in\mathbb{R}^n)$$
is not a regular distribution. For the $n=1$ case, there is a proof in Gel'fand on page 4 of Volume 1, there is this post: How to prove that the Dirac delta is not a function?, and it is not hard to prove using the $C^\infty$-Urysohn Lemma, even for general integer $n$. (Find a mollified sequence of $C^\infty_c(\mathbb{R}^n)$ functions converging pointwise to the characteristic function of $\{p\}$, use Dominated Convergence and the fact that the Lebesgue measure of $\{p\}$ is zero, to get a contradiction.)

I believe that it must be true as well for all derivatives of the delta distribution. Again, for the $n=1$ case, there is this website from Professor Tao: https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/ in which one finds:

Exercise 11 (Derivative of the delta function) Let . Show that the functional  for all test functions  is a distribution which does not arise from either a locally integrable function or a Radon measure. (Note how it is important here that  is smooth (and in particular differentiable, and not merely continuous.) The presence of the minus sign will be explained shortly.

I have been trying to prove that $\delta_p'$ is not regular in $\mathbb{R}^1$; that is, that there does not exist $f\in L^1_{\rm loc}(\mathbb{R})$ such that $\delta_p'=\Lambda_f$, hoping that such a proof would be generalizable to $\mathbb{R}^n$. However, I haven't had any luck even with $n=1$. So my simplified first question, setting $p=0$ and $n=1$, is "why can't $\delta'$ be $\Lambda_f$ for some $f\in L^1_{\rm loc}(\mathbb{R})$?" If that can be answered, then the next question is can it be extended to $\mathbb{R}^n$? Is it true that, given any multi-index $\alpha$, there is no $f\in L^1_{\rm loc}(\mathbb{R}^n)$ such that $D^\alpha\delta_p=\Lambda_f$?

In working on the $n=1$, $p=0$ case, I start with the obvious:
\begin{equation}\tag{1}
  -\phi'(0)=-\delta(\phi')=\delta'(\phi)=\Lambda_f(\phi)=\int_{-\infty}^\infty\!f(x)\phi(x)\,dx
\qquad(\phi\in C^\infty_c(\mathbb{R})),
\end{equation}
and look for a sequence of test functions that generate a contradiction.
The trick of letting $\{\phi_k\}$ be a mollified sequence in $C^\infty_c(\mathbb{R})$ such that $\phi_k$
converges pointwise to $\chi_{\{0\}}$ doesn't seem to help because it appears that $\phi'_k(0)$ will
likely have to approach $0$ and therefore not generate a contradiction. Letting $\{\phi_k\}$ just
be a bump that narrows down to $0$ but whose height remains constant, also doesn't seem to work, since again, the derivatives will be $0$ at the origin. I also hoped I might be able to make use of the fact that
if $\phi_1$ is a fixed element of $C^\infty_c(\mathbb{R})$ such that
$$\int_{-\infty}^\infty\!\phi_1(x)\,dx=1,$$
then any $\phi\in C^\infty_c(\mathbb{R})$ can be expressed as
\begin{equation}\tag{2}
  \phi=\phi_0+\Biggl[\int_{-\infty}^\infty\!\phi(x)\,dx\Biggr]\phi_1,
\end{equation}
where $\phi_0\in C^\infty_c(\mathbb{R})$ and is the derivative of a function in
$C^\infty_c(\mathbb{R})$,
as used by Gel'fand on page 40 of Volume 1; however, I couldn't find a way where that would help.

答:
I want to thank @MaoWao and @copper.hat for their suggestions that lead to this answer. I will rephrase the question to be more general, and then give an answer based on their suggestions.

$\DeclareMathOperator{\supp}{supp}$
Show that, given the nonempty open set $\Omega\subseteq\mathbb{R}^n$, and given various constants, that the distribution
\begin{equation}\tag{3}
  \Lambda=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}c_{i,\alpha}D^\alpha\delta_{p_i}\in\mathscr{D}'(\Omega)
\end{equation}
is not regular. $\alpha$ is a multi-index, and the constants are integers $m>0$, $N\geq 0$, complex $\{c_{i,\alpha}\colon i=1,\dots,m,\,\lvert\alpha\rvert\leq N\}$
(for each $i$, not all $c_{i,\alpha}$ are zero), and $\{p_1,\dots,p_m\}\subseteq\Omega$.

Proof: Let $P=\{p_1,\dots,p_m\}$ and let $\Omega_P=\Omega-P$, an open subset of $\Omega$. Let a tilde over a function variable represent the zero-extension of that function from $\Omega_P$ to $\Omega$. To reach a contradiction, suppose that there is a function $f\in L^1_{\rm loc}(\Omega)$, such that $\Lambda=\Lambda_f$; that is, such that for every $\psi\in\mathscr{D}(\Omega)$,
$$\Lambda(\psi)=\int_\Omega\!f\psi\,dm_\Omega\qquad(\psi\in\mathscr{D}(\Omega))$$
where $m$ is Lebesgue measure on $\mathbb{R}^n$ and $m_\Omega$ is the restriction of $m$ to (the Lebesgue measurable subsets of) $\Omega$. Let $f_P$ be the restriction of $f$ to $\Omega_P$. Then $f_P\in L^1_{\rm loc}(\Omega_P)$, for if $K$ is a compact subset of $\Omega_P$, then it is also a compact subset of $\Omega$ so that
$\chi_Kf\in L^1(\Omega)$, and hence $\chi_Kf_P=(\chi_Kf)|_{\Omega_P}\in L^1(\Omega_P)$.
Let $\phi\in\mathscr{D}(\Omega_P)$. Then $\tilde{\phi}\in\mathscr{D}(\Omega)$ since the support of $\phi$ is contained in $\Omega_P$. We have
\begin{align*}
  \int_{\Omega_P}\!f_P\phi\,dm_{\Omega_P}
  &=\int_\Omega\!(f_P\phi)\tilde{\phantom{a}}\,dm_\Omega\\
  &=\int_\Omega\!\chi_{\Omega_P}f\tilde{\phi}\,dm_\Omega\\
  &=\int_\Omega\!f\tilde{\phi}\,dm_\Omega
  &\qquad&(\text{since $\supp\tilde{\phi}=\supp\phi\subseteq\Omega_P$})\\
  &=\Lambda(\tilde{\phi})\\
  &=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}c_{i,\alpha}D^\alpha\delta_{p_i}(\tilde{\phi})\\
  &=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}(-1)^{\lvert\alpha\rvert}c_{i,\alpha}\delta_{p_i}(D^\alpha\tilde{\phi})\\
  &=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}(-1)^{\lvert\alpha\rvert}c_{i,\alpha}D^\alpha\tilde{\phi}(p_i)\\
  &=0&&(\text{since $p_i\notin\supp\tilde{\phi}$}).
\end{align*}
Therefore, $f_P=0$ almost everywhere on $\Omega_P$. Since $P$ is a finite set, $f=0$ almost everywhere on $\Omega$, and therfore $\Lambda=0$. The inner sum in (3) is a nonzero distribution ($\Lambda_i$) whose support is $\{p_i\}$. If $\omega_i$ is an open neighborhood of
$p_i$ which does not contain $p_j$ for $j\neq i$, then there is a $\phi\in\mathscr{D}(\Omega)$ whose support is contained in $\omega_i$ such that $\Lambda_i(\phi)\neq 0$.
Furthermore, since the support of $\phi$ doesn't intersect the support of
$\Lambda_j$ for $j\neq i$, we have that $\Lambda_j(\phi)=0$ for $j\neq i$, and so
$\Lambda(\phi)=\Lambda_i(\phi)\neq 0$, a contradiction.
$$\tag*{$\blacksquare$}$$


问:
For any $S \subseteq \mathbf{R}$, if $\partial S$ denotes the boundary of $S$, prove that $\partial(\partial S) \subseteq \partial S$.
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I learnt the trick of constructing a $\delta$-neighborhood of a random point $y \in (x - \epsilon, x+ \epsilon)$ to always reside in this $\epsilon$-neighborhood of $x$.

I tried to write a proof for the below assertion, and would like to ask if my proof checks out.

For any $S \subseteq \mathbf{R}$, if $\partial S$ denotes the boundary of $S$, prove that $\partial(\partial S) \subseteq \partial S$.

My Attempt.

Let $x$ be an element of $\partial(\partial S)$.

By definition of the boundary of a set $S$, for every $\epsilon > 0$, the open interval $(x - \epsilon, x + \epsilon)$ contains at least one element of $\partial S$ and at least one element of $(\partial S)^C$.

We are interested to prove that, $\forall \epsilon>0$, the open interval $(x - \epsilon, x + \epsilon)$ contains at least one element of $S$ and at least one element of $S^C$.

Pick an arbitrary element $y \in (x - \epsilon, x + \epsilon)$, such that $y \in \partial S$. Then, for every $\delta > 0$, the open interval $(y - \delta, y + \delta)$ contains atleast one element of $S$ and atleast one element of $S^C$. If we choose,
\begin{align*}
\delta = \delta_0 = \min \left\{\frac{y - (x-\epsilon)}{2},\frac{(x + \epsilon) - y}{2}\right\}
\end{align*}

Then, $(y - \delta_0,y + \delta_0) \subseteq (x - \epsilon, x + \epsilon)$.

So, we found a small $\delta_0$-neighborhood, $(y - \delta_0,y + \delta_0)$ inside of $(x - \epsilon, x + \epsilon)$, which contains at least one element of $S$ and at least one element of $S^C$, for all $\epsilon>0$.

Consequently, we have shown that, $\forall \epsilon > 0$, $(x - \epsilon, x + \epsilon)$ contains at least one element of $S$ and at least one element of $S^C$.

So, $x \in \partial S$.

Hence, $\partial(\partial S) \subseteq \partial S$. This closes the proof.

答: Let $X$ be any topological space. And for any $T\subseteq X$ let $T^c=X\setminus T.$ For any $S\subseteq X$ we have $$(i)\quad\partial S=\overline S \cap \overline {S^c},$$ which is closed, as it is an intersection of closed sets. So we have $$(ii)\quad \overline {\partial S}=\partial S.$$ Therefore, by replacing $S$ with $\partial S$ in $(i),$ and applying $(ii),$ we have $$\partial (\partial S)=\overline {\partial S}\cap \overline {(\partial S)^c}\subseteq$$ $$ \subseteq \overline {\partial S}=$$ $$=\partial S.$$


问:
Showing that there are infinitely many integer solutions to $x^2+y^2=z^2$
Let $x,y,z \in \mathbb{N}$ with $x,y$ relatively prime, $x$ even and $x^{2}+y^{2}=z^{2}$. Show that there are infinitely many $(x,y,z)$ triplets which satisfy these conditions.</p>

答:
I we use Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
we can see that
$$
A^2+B^2=\big((m^2-k^2)+(2mk)^2\\
=(m^4-2m^2k^2+k^4)+4m^2k^2\\
=(m^4+2m^2k^2+k^4)\\
=(m^2+k^2)^2=C^2
$$
One problem with Euclid's formula is that it generates trivial triples
like $f(1,0)=(1,0,1)\quad f(3,0)=(9,0,9)\quad \cdots\qquad$
I developed a formula that generate non-trivial Pythagorean triples for every pair of natural number $(n,k)$ and produces only the subset of triples where $GCD(A,B,C)$ is an odd square -- meaning it includes all primitives where
$GCD(A,B,C)=1^2$.
\begin{equation}
  A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2
\end{equation}

$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$

$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$


问:
Plotting growth functions involving logarithms
How I can plot these functions to compare asymptotic behavior visually?

all of the log here is base $2$.

$1- \log (\log(n^n))$

$2- \log(n!)$

$2- (\log(n))^ {\log(n)}$

We know $1 < 2 < 3$.

答:
I'm assuming you don't mean plotting by hand, in which case you can definitely plot the above functions using online graphing calculators. Try Desmos, Wolfram, or GeoGebra(they're just some of the multiple graphing calculators on the net). I was curious to see what the second function might look like, so you can find the images of the three graphs down below:
$$1-ln(ln(x^x))$$

$$2-ln(x!)$$

$$2-(ln(x))^{ln(x)}$$

All images are from Desmos Graphing Calculator. You can try it out on your own. I'll leave the comparison to you.


问:
Quick Question about Representation of Elements in $\operatorname{Span}(S)$
Let $X$ be a vector space over a field $\mathbb{F}$ and let $S \subset X$ be a nonempty subset of $X$. We know that $\operatorname{span}(S)$ is the set of all linear combination of vectors in $S$.

If $x_0 \in \operatorname{Span}(S)$, then $x_0 = c_1 s_1 + \cdots + c_n s_n$ for some scalars $c_i \in \mathbb{F}$ and vectors $s_i \in S$.

Now suppose $x_1 \in \operatorname{Span}(S)$ and is different from $x_0$. My question is about the representation of $x_1$. In the first scenario, would the same vectors $s_1,\dotsc,s_n$ used in the representation of $x_0$ be used in the linear combination of $x_1$ where only some of the scalars are different? That is, $x_1 = d_1 s_1 + \cdots + d_n s_n$ for some scalars $d_i$?

Or would both the scalars and possibly the vectors in the linear combination of $x_1$ be different, say $x_1 = d_1 s^{\star}_1 + \cdots + d_m s^{\star}_m$?

I'm leaning towards the latter, but I"m not entirely sure. The two texts that I've looked at both begin with a list of vectors $v_1,\dotsc,v_n$, and then talk about the span of that list, but not an arbitrary set.

答:
If you are working in a finite dimensional vector space $X$, or even if $\mathrm{Span}(S)$ is a finite dimensional subspace of $X$, then you can find a finite basis for $\mathrm{Span}(S)$ made up of vectors $s_{1}, \ldots, s_{k} \in S$; then you can write any vector in $\mathrm{Span}(S)$ as a linear combination of these same vectors, that is you can guarantee that
$$x_{0} = d_{1}s_{1} + \cdots + d_{k}s_{k}$$
$$x_{1} = d_{1}^{\prime}s_{1} + \cdots + d_{k}^{\prime}s_{k}$$
(where some of the coefficients may be zero).

If $\mathrm{Span}(S)$ is not finite dimensional, then the normal convention is that the elements of $\mathrm{Span}(S)$ is taken to be the set of linear combinations of elements of $S$ (which have a finite number of terms), in this case you cannot guarantee that each vector in the span will use the same vectors of $S$ in their representation.