问:
Determining if functions are "Completely Additive"
Let $F([x,x_2,...x_n],...[y,y_2,...y_n])=[z,z_2,....z_n]$ denote a function with $2$ parameters, each is a vector containing $n$ variables. Let $U_k$ be defined by the following recurrence relation:

$U_k = F([x,x_2,...x_n],U_{k-1})$ where the intitial value $U_1=[x,x_2,...x_n]$.

Let $U_{k,k_2} = F(U_k,U_{k_2})$

How can I determine if $F$ is a function such that

$U_{k,k_2} = U_{k+k_2}$

will hold for all integers $k$ and $k_2$? Furthermore, how can I construct one where the number of variables $n$ is a fixed integer?

I'm not sure of the proper terminology for such functions, but for now, I will refer to them as "completely additive" because the addition properties involving $U_{k,k_2}, U_k, U_{k_2}$ will hold.

Simple examples of completely additive functions are addition:

$F(x,y)=x+y$

$U_1 = x$, $U_2 = F(x,x) = 2x$, $U_3= F(x,2x) = 3x,...$

$U_{k}=kx$

$U_{k,k_2}=F(U_k, U_{k_2}) = F(kx, k_2x) = (k + k_2)x$

and multiplication:

$F(x,y)=x\times y$

$U_1 = x$, $U_2 = F(x,x) = x^2$, $U_3= F(x,x^2) = x^3,...$

$U_{k}=x^k$

$U_{k,k_2}=F(U_k, U_{k_2}) = F(x^k, x^{k_2}) = x^{k+k_2}$

Here is another example of a completely additive function involving four variables (this is derived from matrix multiplication):

$F([a,b,c,d],[e,f,g,h])=[ae+bg, af+bh, ce+dg, cf+dh]$

Yet, when the "dot products" are rearranged, in some cases, $F$ is no longer a completely additive function. For example,

$F([a,b,c,d],[e,f,g,h])=[ae+ch, af+bg, bf+dh, cg+de]$

is not completely additive.

I already know matrix multiplication is not commentative, so requiring that $F$ be a communitive function is non-essential for it to be a completely additive function. So then that leaves the question, what properties should $F$ have for it to be completely additive?

答:
Expanding both sides of $U_{k,k_2} = U_{k+k_2}$ gives

$$F(U_k,U_{k_2}) = F(U_1,U_{k+k_2-1})$$

Using $k=2$ gives

$$F(U_2,U_{k_2}) = F(U_1,U_{k_2+1})$$

$$F(F(U_1,U_1),U_{k_2}) = F(U_1,F(U_1,U_{k_2}))$$

Because $U_1$ and $U_{k_2}$ can be anything, this can be seen a special form of the associative property in which the first two variables are equal.

In a similar way to $k=2$, for any $k$, the formula can be expanded on both sides to a mess of $k$ instances of $F$, $k$ instances of $U_1$, and one $U_{k_2}$ at the end. By our modified associative property, these $F$s and $U_1$s can be arranged in any order. Therefore both sides are equal. This means that the modified associative property is both necessary and sufficient to show that $F$ is completely additive.

I am not sure whether the modified version is equivalent to the original associative property.


问:
Is it possible that a clock's three hands divide the clock face into 3 equal parts?
And is there a proof/disproof out of intuition? Here we ignore the structure of the clock and suppose the hands move with constant speed.</p>

答:
Let $\theta$ be the angle of the hour hand past 12 o'clock. Then the angle of the minute hand is $12\theta$ and the angle of the second hand is $720\theta$.

If the face of the clock is to be divided into three equal parts, we need the angle between two hands to be $\frac{2\pi}{3}$ for any two hands. Hence,
$$720\theta - 12\theta \equiv 12\theta - \theta \equiv \theta-720\theta \equiv \pm\frac{2\pi}{3} \pmod {2\pi}$$
or, calculating a little,
$$708\theta \equiv 11\theta \equiv -719\theta \equiv \pm\frac{2\pi}{3}\pmod {2\pi}.$$
It immediately follows that $708\theta-11\theta =697\theta\equiv 0 \pmod {2\pi}$ as well as $11\theta-(-719)\theta = 730\theta \equiv 0\pmod {2\pi}$.
Thus $\gcd(730\theta,697\theta)=\theta\equiv 0\pmod {2\pi}$, which is not a proper solution.

Thus, there is no such configuration of the hands of a clock as required by your problem.

ContentLicense: CC BY-SA 3.0


问:
Confusion on the definition of Tensor product
I know the usual definition of Tensor product (cartesian product of basis of $U$ and $V$, with bilinear property) but I started confusing myself as I was reading this new definition.

The tensor product $u\otimes v$ of two finite-dimensional vector
spaces $u$ and $v$(over the same field) is the dual of the vector
space of all bilinear forms on $u\oplus v$. For each pair of $x$ and
$y$, with $x$ in $u$ and $y$ in $v$, the tensor product $z=x\otimes y$
of $x$ and $y$ is the element of $u\otimes v$ defined by $z(w)=w(x.y)$
for every bilinear form $w$.

What does this function $z$ exactly look like? Does this suggest that the tensor product $z=x\otimes y$ is a function that sends all bilinear forms (which form a vector space) to something that looks like $w(x,y)$, which is a scalar value? How can I understand this definition?

I'd really appreciate any help.

答:
Let $k$ be a field. Recall that if $Z$ is a $k$-vector space, the dual space of $Z$ is the $k$-vector space $Z^*$ of all the linear maps $f : Z \to k$.

Now, if $U$ and $V$ are two $k$-vector spaces, consider the $k$-vector space $\operatorname{Bil}_k(U,V;k)$ of all the bilinear maps$^\color{blue}1$ $b : U \times V \to k$. So, here $U \otimes V$ denotes the dual space of $\operatorname{Bil}_k(U,V;k)$, that is, $$U \otimes V := \operatorname{Bil}_k(U,V;k)^*.$$ Thus, an element $z$ of $U \otimes V$ is a linear map $z : \operatorname{Bil}_k(U,V;k) \to k$. Hence, given $u \in U$ and $v \in V$, the tensor product $u \otimes v \in U \otimes V$ is the linear map $u \otimes v : \operatorname{Bil}_k(U,V;k) \to k$ such that $$\forall b \in \operatorname{Bil}_k(U,V;k) : \quad (u \otimes v)(b) = b(u,v).$$ The linearity of $u \otimes v$ follows from the definition of the operations in $\operatorname{Bil}_k(U,V;k)$: if $b_1,b_2 \in \operatorname{Bil}_k(U,V;k)$ and $\lambda \in k$, then $(\lambda b_1+b_2)(u,v) := \lambda b_1(u,v)+b_2(u,v)$.

Exercise: If $Z$ is a $k$-vector space, and $U,V$ are finite dimensional, prove that for any bilinear map $h : U \times V \to Z$ there exists a unique linear map $\tilde h : U \otimes V \to Z$ sending $u \otimes v$ to $h(u,v)$, for any $u \in U$ and $v \in V$.

$^\color{blue}1$ Which is not the same as $(U \times V)^*$.

ContentLicense: CC BY-SA 4.0


问:
Finiteness and extension of a ring of scalars
Let $A\subset B$ be rings and $P$ a projective left $A$-module. Write $\rho:A\rightarrow B$ for the canonical injection. Suppose $\rho^*(P)=\rho_*(B)\otimes_A P$ is finitely generated. The author says,

Since $\rho^*(P)$ is finitely generated, there exists a finite family
$(y_i)_{i\in I}$ such that $(1\otimes y_i)_{i\in I}$ is a generating
system of $\rho^*(P).$

I am not sure how this follows. The fact that $\rho^*(P)$ is finitely generated implies that there exists an exact sequence
$$B^{\oplus n}\xrightarrow{k}\rho^*(P)\rightarrow0.$$
Writing $(e_i)_{1\leq i\leq n}$ for the canonical basis of $B^{\oplus n}$, we have that $(k(e_i))_{1\leq i\leq n}$ is a generating system for $\rho^*(P)$. On the other hand, we have a canonical injection $\phi:P\rightarrow\rho^*(P)$ sending $x$ to $1\otimes x$. Then $\rho^*(P)$ is generated by $(1\otimes x)_{x\in P}$. So for each $1\leq i\leq n$, we have
$$k(e_i)=\sum_{x\in P}\xi^i_x(1\otimes x)$$
for some $\xi^i\in B^{(P)}$. I don't know how to get $k(e_i)$ in the form $1\otimes y_i$ for some $y_i\in P$. Any suggestions?

答:
The sums in your last equation are finite. That is for each $i$ there is a finite set $P_i \subseteq P$ such that
$$ k(e_i) = \sum_{x \in P_i} \xi_x^i (1 \otimes x) $$
Then $\rho^*(P)$ is generated by $\{1 \otimes x \mid x \in \bigcup_i P_i\}$, a finite set.


问:
Minimum number of trips for a truck with weight limit of $200$ to transport boxes with weights 81, 73, 67, 49, 37, 34, 30, and 26
My 9-year-old son had the following math problem to solve:

A truck can carry $200$kg or less. We have 8 different boxes with given
weights:

$81$kg, $73$kg, $67$kg, $49$kg, $37$kg, $34$kg, $30$kg, and $26$kg

What is the minimum number of trip required to move all boxes from
city A to city B.

I looked at it for 5 minutes and couldn't figure out a meaningful approach other than brute force.

I was in front of my computer and quickly wrote a MATLAB script to bruteforce it.

The solution was 2 trips: one with boxes $81$, $49$, and $67$; and the other with boxes $73$, $37$, $30$, $26$, and $34$.

Is there an obvious way, or at least one more clever than brute force, to solve that kind of problem?

Here, it's not that hard because we only have 8 boxes. But what about a similar problem with a bigger set? Can we only rely on brute force?

Bonus question:

What is this supposed to teach to a 9yo kid ?

答:
Perhaps the main lesson would be:

The greedy approach is not always optimal

What would be the greedy method? Pack the truck with heaviest boxes first until you cannot add any more boxes.
This results in a first tour with 81+73+37=191 and a second tour with 67+49+34+30=180 and then a final tour with the one remaining 26 kg.
The fact that the 26 kg just barely fail to fit into the second tour are perhaps intended to annoy you and make you play around to somehow achieve two tours.

Replacing the 37 of tour with something smaller? There is no way to get four boxes with 81+73 as already 26+30 is too much.

Replace the 73 with something smaller? doing so and continuing greedily, we arrive at 81+67+49=197 and the rest on the second tour - and done!

Given the age, however, I suppose this problem would be more appropriate if the trial-and-error phase could be performed physically (e.g., moving around suitable paper cutouts). But the numbers used in this problem do not seem to allow that easily.


问:
Power analogies to primes
As we all know, a natural number $n$ is prime if and only if there do not exist natural numbers $x, y$ exclusively between $1$ and $n$ such that $xy = n$.

Is there any generally recognized analogy for primes in powers? For example, a natural number $n$ in a set $S$ that is not the $x$th power of $y$ for two natural numbers $x$ and $y$ exclusively between $1$ and $n$? I recognized that the definition is less elegant than that for multiplication, since powers aren't commutative, but I was curious if this set or any similar set has been explored in mathematics.

For clarity, in my example definition, $1, 2, 3, 5, 6, 7 \in S$, but $4 = 2^2, 8 = 2^3, 9 = 3^2 \not\in S$.

答:
A natural number satisfies your criterion if and only if it is not a perfect power.

Consider the function $f\colon\mathbb{R}_{>0}\to\mathbb{R}$ defined by $f(x)=\log_2(x)=\ln(x)/\ln(2)$. It satisfies $f(2)=1$, and
$$f'(x)=\frac{d(\ln(x)/\ln(2))}{dx}=\frac{1}{\ln(2)}\frac{d(\ln(x))}{dx}=\frac{1}{x\ln(2)}.$$
Because $f'(x)<1$ for all $x>\frac{1}{\ln(2)}\approx 1.442$, it will never "catch up" with the function $g(x)=x$, having already started "behind" at $f(2)=1$. Thus, $f(x)<x$ for all $x\geq 2$.

Let $n>1$ be a natural number. Suppose that there exist natural numbers $a$ and $b$ with $n=a^b$, where $b>1$. Because $n>1$, we must have $a>1$, and hence $a\geq 2$. Thus, $b\leq\log_2(n)< n$.

Thus, if a natural number $n>1$ is not equal to $a^b$ for any natural numbers $a$ and $b$ exclusively between $1$ and $n$, then it is not equal to $a^b$ for any natural numbers $a$ and $b$ with $b>1$, period.

Thus, your numbers are precisely those which are not perfect powers.

Title: Power analogies to primes


问:
Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2]
I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.

I tried multiple ways to understand why this assumption was made, but I can't figure it out. My first assumption was that since $m \in S$ it's true that:

$m > 3^{m/3}$

and by consequence:

$m^3 > 3^{m}$

but it still does not prove $3^m < m$. Any help is very appreciated.

答:
$n\le 3^{n/3}$ is true for $n=1$ because $\sqrt[3]{3}>1$

It is true for $n=2$, too, because $\sqrt[3]{3^2}\approx 2.08>2$

Now suppose that $3^{n/3}\ge n$ is true for $n$ and let's prove it for $n+1$

$3^{(n+1)/3}=3^{n/3}\cdot \sqrt[3]{3}\ge \sqrt[3]{3}n>n+1$. Proved.

$\sqrt[3]{3}n>n+1$ because $\sqrt[3]{3}n-n\approx 0.44n>1$ for any $n>2$


问:
Explain why $1 < \frac{c}{p} < c$ where $c$ is a composite number and $p$ is a prime number.
Explain why $1 < \frac{c}{p} < c$ where $c$ is a composite number and $p$ is a prime number.

答:
The issue is, I assume, that $p$ is a prime factor of a composite number $c$.

That means, by definition of being a factor, that $\frac cp$ is an integer.  And because $c$ is composite and $p$ is prime this means $p$ isn't the only factor and $\frac cp \ne 1$.

I assume that is the issue the book is trying to get across.  That $1 < \frac cp < c$ is trivial and obvious.

As $c$ and $p$ are positive and $p$ divides $c$ and $c$ is composite and $p$ is prime we know that $\frac cp$ is a positive integer that is not equal to $1$.  So $1 < \frac cp$.

And as $p > 1$ we know that $\frac cp < \frac c1 = c$.

So that's it.  $1 < \frac cp < c$.

Although the book is talking about integers and factoring; it will always be the case for any type of real numbers that if $0 < w < v$ and $1 < v$ that $1 < \frac wv < w$.

(And if $p > c$ [so $p$ isn't a factor of $c$] the result is not true.)


问:
does the inverse of curl exist for all smooth vector fields?
Let $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a smooth vector field. Does there always exist another smooth vector field $g:\mathbb{R}^3\rightarrow\mathbb{R}^3$ such that $\nabla\times g = f$? Naturally, such a field $g$ is not unique as we can add to it any conservative field and find another solution to $\nabla\times g = f$, however this alone does not imply a solution exists for any $f$. This question is analogous to asking in one dimension if all smooth functions have an anti-derivative.

If a $g$ doesn't exist for all $f$, I would like an example of such an $f$ and a proof that no $g$ satisfies the equation $\nabla\times g = f$, however if there always exists such a $g$ for any $f$, I would like a proof of that. Finally, if the result holds and there exists a weaker condition than smoothness, for example, if continuity is sufficient, I would like to know what this condition is.

答:
Note that
$$ {\rm div} \circ {\rm rot} = 0 $$
Hence, a field $f$ that can be written as $f = {\rm rot}\, g$ has ${\rm div} \, f = 0$. Therefore $f(x) = (x_1, 0, 0)$ cannot be written as ${\rm rot}\, g$ for any $g$.


问:
Equation to find unknown point on arc from radius, starting x/y, and ending y
Basically I want to extend an arc along an axis by a given amount. I can do it easily in CAD, but I'd really like to know if there's an equation I can use to figure it out. It's been about 20 years since I've done this kind of math and it seems like there should be a way, but I haven't found it. I've found similar questions that were trying to do something similar, only with the arc angle, which I do not know. It won't let me embed the image so I had to link it.

In my example, I want to extend the arc along the X axis by 0.100

Reference image

答: The equation for the whole line presumably takes the form $(x-a)^2 + (y-b)^2 = R^2$, so if you just type that into wolfram twice, replacing $x$, $y$ and $R$ with the values you already know, it should give you two solutions for $(a,b)$. Only one of these is the centre of the circle that produced the arc, but it should be clear. Once you have that you have the equation of the line so you can extend the arc by however much you like, Hope this helped!


答:
Not an answer!

Just too long for a comment, and I wanted OP and other parents to know what is the right attitude.

As a teacher/father:

I strongly advice to NOT help your sons/daughters;

NEVER belittle children assignments/homework.

We assign problems/readings to our students not to their families.

Most important:

making unnecessary arguments with teachers harms students and hinders healthy and serene teamwork

Title: Minimum number of trips for a truck with weight limit of $200$ to transport boxes with weights 81, 73, 67, 49, 37, 34, 30, and 26

Body:
My 9-year-old son had the following math problem to solve:

A truck can carry $200$kg or less. We have 8 different boxes with given
weights:

$81$kg, $73$kg, $67$kg, $49$kg, $37$kg, $34$kg, $30$kg, and $26$kg

What is the minimum number of trip required to move all boxes from
city A to city B.

I looked at it for 5 minutes and couldn't figure out a meaningful approach other than brute force.

I was in front of my computer and quickly wrote a MATLAB script to bruteforce it.

The solution was 2 trips: one with boxes $81$, $49$, and $67$; and the other with boxes $73$, $37$, $30$, $26$, and $34$.

Is there an obvious way, or at least one more clever than brute force, to solve that kind of problem?

Here, it's not that hard because we only have 8 boxes. But what about a similar problem with a bigger set? Can we only rely on brute force?

Bonus question:

What is this supposed to teach to a 9yo kid ?


问:
If a polynomial $P$ has only real roots, so does $P'-2xP$
Let $P$ be a polynomial such that all of the roots of $P$ are real. Prove that all the roots of $P'-2xP$ are also real. I know a proof for this fact but it is very computational and messy. Is there an elegant solution?

答:
Let $g(x)=P(x)e^{-x^2}$. Then
$$g'(x)=P'(x)\cdot e^{-x^2}+P(x)\cdot (-2x)e^{-x^2}= (P'(x)-2xP(x))e^{-x^2}. $$
We may consider $g$ to be extended to the infinities per $g(\pm\infty)=0$ because the exponential function outweighs the polynomial factor.

First assume that the roots of $P$ are distinct. Then $g$ has $n+2$ distinct zeroes, namely the $n$ zeroes of $P$ and $\pm\infty$. Between any two consecutive of these, Rolle gives us a zero of $g'$, hence a root of $P'-2xP$. This gives us $n+1$ distinct real roots, and as this corresponds to the degree, the claim follows.

You may contemplate how to make the Rolle argument rigorous when $x=\pm\infty$ is involved, and what to look out for when $P$ has roots of higher multiplicity. For the latter, I suggest to add a tiny "perturbation" to $P$ that splits multiple roots up into separate roots, and let the perturbation tend to $0$.


问:
Is $A$ countable or uncountable?
Let $A=\{0,1\}^\mathbb{N}$, that is the set of all functions from the natural numbers to the set $\{0,1\}$.

Is $A$ countable or uncountable? Prove your answer.

$\textbf{Proof:}$ Let $A = \{0,1\}^\mathbb{N}$. Let $f: \mathbb{N} \rightarrow A$.

Define $\overline{a} = \{\overline{a_1}, \overline{a_2}, \overline{a_3},...\}$ where

$\overline{a_1} = \{a_{11}, a_{12}, a_{13}, ...\}$, where $a_{11} = 0 \, \text{or} \, 1$, $a_{12} = 0 \, \text{or} \, 1$, ...

.

.

.

Pick $\overline{a_n} = \{a_{11}^c, a_{22}^c, a_{33}^c, ... \}$. So if $a_{11} = 1$, then $a_{11}^c = 0$, ...

Then $\nexists n \in \mathbb{N}$ such that $f(n) = \{ a_{11}^c, a_{22}^c, ...\}$ because $\{a_{11}^c, a_{22}^c, ...\} \subset A$, but each element $a_{11}^c, a_{22}^c, ...$ is not mapped by any $n \in \mathbb{N}$ since the compliment does not follow

$f(1) = \overline{a_1} = \{a_{11}, a_{12}, ...\}$

$f(2) = \overline{a_2} = \{a_{21}, a_{22}, ...\}$

Thus, $\nexists$ surjection $f: \mathbb{N} \rightarrow A$.

$\therefore A$ is uncountable.

答: $B$ is equinumerous with the power set of $\Bbb N$ (via characteristic functions).Cantor's diagonal argument (that you sort of reproved) shows it has strictly larger size than $\Bbb N$, hence is uncountable.

Title: Is $A$ countable or uncountable?


问:
Why the differential of a function is a linear transformation?
I try to find some applications of the differential, that weird thing that is a linear transformation, what is the reason of linearity?
And how this linearity works, I mean, there's something fundamental about that linearity? And where we can see the importance of the differential, like in physics or another field of application.</p>

答:
(Initially a comment.)

At a technical level a differential is linear by definition. At a secondary level because linear maps are simple. For example

They scale area by a constant factor, called the Jacobian (even in case of $R^n \to R^m$ is $n\leq m$.

Ultimately (1) is responsible for the change of variables formula, the area formula (think formula for surface area in calculus), and co-area formula.

Linearity of the derivative also gives the notion of tangent space to manifolds. Assuming our manifold is embedded in $R^N$, the tangent space is the image of the derivative of a local parameterization.


答:
$
\newcommand{\N}{\mathbb{N}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\row}[1]{f_{#1}(1) & f_{#1}(2) & f_{#1}(3) & f_{#1}(4) &\cdots & f_{#1}(n) & \cdots \\}
$
At least in principle and spirit, yes, I would say your argument is correct, though it's a bit hard to follow. Here's how I would argue it.

Suppose $A$ is at most countable. Then clearly it is infinite, so let us write $A := \set{f_i}_{i=1}^\infty$ where $f_i : \N \to \set{0,1}$ $\forall i \in \N$. For simplicity, we take the convention that $0 \not \in \N$, but the argument can be modified to accommodate that if we please.

Each $f_i$ may be characterized by its valuation at each $n \in \N$. Form a square table- or matrix-like object as so:

\begin{matrix}
\row 1
\row 2
\row 3
\row 4
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots &\ddots \\
\row n
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots &\ddots \\
\end{matrix}

Note that $f_i(j) \in \set{0,1}$ $\forall i,j \in \N$.

To show $A$ is uncountable, we use a diagonal argument, by defining a function $f : \N \to \set{0,1}$ which is not enumerated by the table above. Define $f(n)$ for each $n \in \N$ as so:

$$f(n) = \begin{cases}
0 & \text{if } f_n(n) = 1 \\
1 & \text{if } f_n(n) = 0
\end{cases}$$

Thus, $f(n) \ne f_n(n)$ $\forall n \in \N$, and therefore $f \ne f_n$ $\forall n \in \N$, but remains as a function $\N \to \set{0,1}$. Thus, $A$ is uncountable.

Note as well that $A$ is bijective to the set of numbers in $[0,1]$ written in binary notation, if we define the mapping

$$f_i \mapsto 0.f_i(1)f_i(2)f_i(3)\cdots$$

i.e. map a function $f_i \in A$ to a binary number in $[0,1]$ defined by where the function maps each $n \in \N$. More explicitly, we could write

$$f_i \mapsto \sum_{n=1}^\infty \frac{f_i(n)}{2^n}$$

(Note that $0$ is included for the case of $f_i(n) = 0$ $\forall n \in \N$, and $1$ is included for the case of $f_i(n) = 1$ $\forall n \in \N$, noting that $0.\overline{1} = 1$ in binary, much as $0.\overline{9} = 1$ in decimal.)

This mapping is surjective, but not injective. Thus intuitively one sees that $|A| \ge |(0,1)| = c > \aleph_0 = |\N|$.

Title: Is $A$ countable or uncountable?

Body:
Let $A=\{0,1\}^\mathbb{N}$, that is the set of all functions from the natural numbers to the set $\{0,1\}$.

Is $A$ countable or uncountable? Prove your answer.

$\textbf{Proof:}$ Let $A = \{0,1\}^\mathbb{N}$. Let $f: \mathbb{N} \rightarrow A$.

Define $\overline{a} = \{\overline{a_1}, \overline{a_2}, \overline{a_3},...\}$ where

$\overline{a_1} = \{a_{11}, a_{12}, a_{13}, ...\}$, where $a_{11} = 0 \, \text{or} \, 1$, $a_{12} = 0 \, \text{or} \, 1$, ...

.

.

.

Pick $\overline{a_n} = \{a_{11}^c, a_{22}^c, a_{33}^c, ... \}$. So if $a_{11} = 1$, then $a_{11}^c = 0$, ...

Then $\nexists n \in \mathbb{N}$ such that $f(n) = \{ a_{11}^c, a_{22}^c, ...\}$ because $\{a_{11}^c, a_{22}^c, ...\} \subset A$, but each element $a_{11}^c, a_{22}^c, ...$ is not mapped by any $n \in \mathbb{N}$ since the compliment does not follow

$f(1) = \overline{a_1} = \{a_{11}, a_{12}, ...\}$

$f(2) = \overline{a_2} = \{a_{21}, a_{22}, ...\}$

Thus, $\nexists$ surjection $f: \mathbb{N} \rightarrow A$.

$\therefore A$ is uncountable.


问:
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$.

I drew the picture but I can't seem to figure out the way to start.. If someone could give me a hint on how to start it would mean a lot..

答:
Given your cylinder is along $y-$axis and radius is $1$, your parametrization should be

$x = \cos t, y, z = \sin t$

$z - 4 \leq y \leq 4 - x$

Also $0 \leq t \leq 2\pi$

Now $|r'_y \times r'_t| = r = 1$ (in this case). You can check that.

Your integral to find surface area is $\iint |r_y \times r_t| \, dy \, dt$


问:
Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

I don't understand the quotient group $G/H$ being cyclic. What does it mean? From what I understand, $G/H$ is a group that includes groups and not elements, right?

I know that the reverse is right (If $G$ is cyclic then $G/N$ is cyclic.)

How do I disprove it?

答:
Of the small finite non-cyclic groups $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ of order $8$ is the smallest which doesn't have a cyclic normal subgroup with a cyclic quotient.

Note that a subgroup of index $2$ is always normal. Do you know any groups which have a cyclic subgroup of index $2$? If they are not themselves cyclic, they will be counterexamples.

You should be able to find both abelian and non-abelian examples of small order, and also some general cases. Also try to find a small example of odd order (the smallest non-abelian example of odd order has order $21$).

This is a test and suggestion to develop a repertoire of examples of groups with/without interesting properties. Working hands-on will teach you a great deal about what is possible and what isn't.


问:
Reliability of a circuit board with components
I have a list of components with their reliability, for example

Resistor [Fixed, Metal Film], Number used 168, Rating 50%, Failure rate 0.005, weighting factor 1.5.

I have calculated the total reliability for all of these components, for the example I think it is 1.26. The next step is M.T.B.F, reliability over 5 days and over 28 days of 8 hour use.

I am currently a little confused as to what I should be doing with these figures I have worked out, do I add them? Multiply them? and then what?

Please help.

答: The reliability seems to be a failure rate, maybe FIT=failures per billion hours (don't ask me where the acronym comes from).  This usually assumes an exponential distribution of failures.  Then MTBF is mean time between failures, which would be just the inverse of the failure rate if the unit were active all the time.  In our business (and I don't know how standard this is) if the unit is powered off the failure rate is 1/10 of the powered on rate.  If that is your model, the chance of non-failure over 5 days the chance of non-failure in 40 powered hours times the chance of non-failure in 80 non-powered hours.  Given your failure rate, can you calculate the chance of failure in 40 hours?</p>


答:
I think you've got the right idea, but the proof has such serious flaws with how it is written that I am unsure. Let's go through the issues and then fix them.

The first line of your proof

Let $A=\{0,1\}^{\mathbb N}$. Let $f:\mathbb N\rightarrow A$.

is good - it's always a good idea to start out with the givens. I would also add in, very clearly, your ultimate goal:

We will show that $f$ is not surjective by constructing an element $g\in A$ that is not in the image of $f$.

This sort of statement is good to focus your proof (and your reader) and to sort of close off your introduction of context and variables. It might save you from the dangers that you run into in your next line:

Define $\overline{a} = \{\overline{a_1}, \overline{a_2}, \overline{a_3},...\}$ where

$\overline{a_1} = \{a_{11}, a_{12}, a_{13}, ...\}$, where $a_{11} = 0 \, \text{or} \, 1$, $a_{12} = 0 \, \text{or} \, 1$, ... \

where you have suddenly stopped talking about $f$ - the only object you introduced - and instead written "define" on something that is patently not a definition given that it just introduces some new variables we've never seen before and said to be equal to an expression involving even more variables that we've never seen before. There's also trouble with misusing the symbols $\{$ and $\}$ to represent a sequence rather than a set.

I think what you mean here is to imagine that each $f(i)$ can be thought of as a sequence of $1$'s and $0$'s - and you want to represent terms of that sequence as $a_{i,j}$ where $j$ is the position in the sequence and $i$ is the input to the function $f$. Note that you could just refer to this value as $f(i)(j)$, which is likely how I would do it, but if you wish to do it this way, I'd write:

We can imagine each element of $A$ to be an infinite sequence of $1$'s and $0$'s - and then imagine a sequence of such elements to be a two dimensional grid of such values. Formally, we define
$$a_{i,j}=f(i)(j)$$
which is an element of $\{0,1\}$.

Note here that our definition is actually a definition - we're taking a given $f$ and writing a new symbol in terms of something that is already known.

Your next line is also not great:

Pick $\overline{a_n} = \{a_{11}^c, a_{22}^c, a_{33}^c, ... \}$.

This creates conflict since you already defined $a_i$ (sort of) - are the values $a_i$ meant to represent the values that $f$ takes or a value that it cannot take? Your proof is not consistent - and this is one reason why I suggest fully writing out your goal at the start. I would rewrite this as, perhaps:

Consider the function $g\in A$ defined by $g(j)=a_{j,j}^c$.

Where we just reuse the notation from our goal. You might also write $g(n)=1-a_{n,n}$ if the $\cdot ^c$ notation is not standard in your context - and you could even define another sequence instead of a function, but I'm not sure why you would. I might avoid sequences entirely and not define $a_{i,j}$ but directly jump to $g(j)=1-f(j)(j)$ - which would shorten the proof a bit (and make it more clear how to prove some generalizations of the theorem).

Then, you would finish your proof somewhat like you have:

We now claim that, for every $n\in\mathbb N$ we have that $f(n)\neq g$.

This is a bit clearer than using quantifiers - and you have a bunch of "fluff" in your proof that doesn't help us move forwards at this point. Better just to state a smaller goal and proceed to prove it. The loss of focus at this point also obscures that you don't really offer up any proof of this fact, beyond repeating definitions and hoping the reader will look at the diagonal.

You ought to, instead, provide this argument clearly using your definitions - as it stands, your proof kind of just defines a bunch of stuff then never uses it, but it's relatively easy to express why the chosen $g$ is not in the image of $f$:

Note that $a_{n,n}=f(n)(n)\neq g(n)=a_{n,n}^c$ by definition of $g$. Therefore, $f(n)\neq g$, completing the proof.

Title: Is $A$ countable or uncountable?


答:
A math-talented high school student can avoid brute force by keying on the comment of Dietrich Burde, and then looking for a first try that involves a group of numbers whose sum is congruent to 0, mod 10.

Within this approach, a plausible first "sub-try" is identifying pairs of numbers whose sum is congruent to 0, mod 10.  Doing this, one gets lucky in observing the two pairs (73,67) and (34,26), since 140 + 60 = 200.

It is unclear whether the problem was designed to allow luck, versus replacing (67,26) with [for example] (68,25).


问:
Is $B$ countable or uncountable?
Let $B=\mathbb{Z}^{\{0,1\}}$, that is the set of all functions from the set $\{0,1\}$ to the integers.

Is $B$ countable or uncountable? Prove your answer.

$\textbf{Proof:}$ Let $B = \mathbb{Z}^{\{0,1\}}$.

$\forall f \in B$, $f \leftrightarrow (a,b) \in \mathbb{Z} \times \mathbb{Z} \implies f(0)=a \in \mathbb{Z}$, $f(1)=b \in \mathbb{Z} \times \mathbb{Z}$

Surjectivity: $\forall (a,b) \in \mathbb{Z} \times \mathbb{Z} \,\, \exists f$ such that $f(0)=a$, $f(1)=b$

Injectivity: If $(a_1,b_1) = (a_2,b_2)$, then $a_1 = a_2$ and $b_1 = b_2$ where $f(0)=a_1$ and $f(1)=b_1$ $\forall (a,b) \in \mathbb{Z} \times \mathbb{Z}$

$\therefore \, \exists$ bijection $f \leftrightarrow \mathbb{Z} \times \mathbb{Z}$.

$\mathbb{Z} \times \mathbb{Z}$ is countable by using the map $f((0,0))=1$, $f((1,0))=1$, $f((1,1))=2$, $f((0,1))=3$, and so on (this can be seen as spiraling the lattice points of $\mathbb{R}^2$) where $f:\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{N}$.

$\therefore B$ is countable. $\blacksquare$

答:
Yes, I think your proof and conclusion are basically correct. Here are some suggested improvements:

You can write the bijection between $B$ and $\mathbb{Z}\times\mathbb{Z}$ in this fashion: Define $\Phi:B\to\mathbb{Z}\times\mathbb{Z}$ by $$\Phi(f)=(f(0),f(1)).$$ It is a good habit to make your notations neat and readable.

For the proof of injectivity, strictly speaking, after you obtained $a_1=a_2$ and $b_1=b_2$, you should say the function $f$ that is mapped to $(a_1,b_1)$ and the function $g$ mapped to $(a_2,b_2)$ coincide. The notation $\forall(a,b)\in\mathbb{Z}\times\mathbb{Z}$ does not make sense at all.

Generally speaking, you did a very nice proof.

Title: Is $B$ countable or uncountable?


问:
Simple question on Mean value theorem
Let $\phi,\psi\in C^1[a,b]$ of bounded variation, and $C$ a curve given by $x=\phi(t),\ y=\psi(t)$. Let $\Gamma=\{t_i\}_{i=0,\cdots,m}$ be a partition of $[a,b]$. By $C$ rectifiable we mean $$ L=\sup_\Gamma S_\Gamma=\sup_\Gamma \sum_{i=1}^m \left( (\phi(t_i)-\phi(t_{i-1}))^{2}+ (\psi(t_i)-\psi(t_{i-1}))^{2}\right)^{1/2}<\infty$$
Here because of bounded variation condition holds, we have $C$ rectifiable. The problem is asking me to prove that $$L=\int_a^b \left( \phi'(t)^{2}+ \psi'(t)^{2}\right)^{1/2}\ dt\ \ \  (*)$$
Since $\phi',\psi'$ are continuous, I know that
$$L=\lim_{|\Gamma|\to 0} S_\Gamma $$
So it seems natural to use, $$\phi(t_i)-\phi(t_{i-1})=\phi'(\xi_i)(t_i-t_{i-1}) \ \mathrm{and}\  \psi(t_i)-\psi(t_{i-1})=\psi'(\rho_i)(t_i-t_{i-1})$$ for some $t_{i-1}\leq \xi_i,\rho_i\leq t_i$ and for all $i=1,\cdots,m$, to form a Riemann sum for $(*)$ that converges to $L$, indeed starting from $S_\Gamma$ and replacing we get
$$S_\Gamma = \sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\rho_i)^{2}\right)^{1/2}(t_i-t_{i-1}) \ \ (**)$$

My question starts here:

Intuitively, if $|\Gamma|\to 0$, is expected that $t_{i-1},\xi_i,\rho_i,t_{i}$ will be close enough, so $(**)$ is very close to a Riemann sum for $(*)$ and we are done. But, is it just like that I mean, can I take $\left( \phi'(\xi_i)^{2}+ \psi'(\rho_i)^{2}\right)^{1/2}$ as if it were $\left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}$? what is the formal justification for this?

答:
Unless by chance it so happens that $\rho_i = \xi_i$ for all $i$, $(\ast\ast)$ will not be a Riemann sum for $(\ast)$. Thus we have a little more work to do. But, since $\phi$ and $\psi$ are assumed to be continuously differentiable, fortunately not much more work.

By the uniform continuity of $\psi'$, $(\ast\ast)$ differs very little from a Riemann sum for $(\ast)$. Using the triangle inequality for the Euclidean norm, we have
$$\bigl\lvert \bigl(\phi'(\xi_i)^2 + \psi'(\rho_i)^2\bigr)^{1/2} - \bigl(\phi'(\xi_i)^2 - \psi'(\xi_i)^2\bigr)^{1/2}\bigr\rvert \leqslant \lvert \psi'(\rho_i) - \psi'(\xi_i)\rvert\,. \tag{1}$$
And by the uniform continuity of $\psi'$, for every $\varepsilon > 0$ there is a $\delta > 0$ such that
$$\lvert \psi'(\rho_i) - \psi'(\xi_i)\rvert \leqslant \varepsilon$$
for all $i$ when $\lvert \Gamma\rvert \leqslant \delta$. Thus, writing
$$\tilde{S}_{\Gamma} = \sum_{i = 1}^{m} \bigl(\phi'(\xi_i)^2 + \psi'(\xi_i)^2\bigr)^{1/2}(t_i - t_{i-1})$$
we have
$$\lvert S_{\Gamma} - \tilde{S}_{\Gamma}\rvert \leqslant \sum_{i = 1}^{m} \lvert \psi'(\rho_i) - \psi'(\xi_i)\rvert (t_i - t_{i-1}) \leqslant \sum_{i = 1}^{m} \varepsilon\cdot (t_i - t_{i-1}) = \varepsilon\cdot (b-a)$$
whenever $\lvert \Gamma\rvert \leqslant \delta$.

Now $\tilde{S}_{\Gamma}$ is a Riemann sum for $(\ast)$, and thus it follows that
$$\Biggl\lvert\, L - \int_{a}^{b} \bigl(\phi'(t)^2 + \psi'(t)^2\bigr)^{1/2}\,dt \Biggr\rvert \leqslant \varepsilon \cdot (b-a)$$
holds for every $\varepsilon > 0$. And that implies
$$L = \int_{a}^{b} \bigl(\phi'(t)^2 + \psi'(t)^2\bigr)^{1/2}\,dt\,.$$

We have used the continuity of only one of the derivatives ($\psi$ here, we could obviously swap the rôles), of the other we only use Riemann integrability (which ensures that indeed $\tilde{S}_{\Gamma}$ tends to the integral as $\lvert \Gamma\rvert \to 0$). The conclusion that the arc-length of the curve is given by the integral holds under still weaker assumptions, but then more work must be put into the proof.

Title: Simple question on Mean value theorem


答:
Note that

$$S_\Gamma = \sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}(t_i-t_{i-1}) + \sum_{i=1}^m (f(\xi_i,\rho_i) - f(\xi_i,\xi_i))(t_i-t_{i-1}),$$

where $f:(s,t) \mapsto  \sqrt{\phi'(s)^2 + \psi'(t)^2}$ is uniformly continuous on $[a,b]\times[a,b]$.

For any $\epsilon > 0$, we have for all sufficiently fine partitions,

$$|f(\xi_i,\rho_i) - f(\xi_i,\xi_i)| < \epsilon,$$

and,

$$|S_\Gamma - I| \leqslant \left|\sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}(t_i-t_{i-1})-I \right|+ \sum_{i=1}^m \epsilon (t_i - t_{i-1}) \\ = \left|\sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}(t_i-t_{i-1})-I \right|+ \epsilon(b-a),$$

where $I = \int_a^b \left( \phi'(t)^{2}+ \psi'(t)^{2}\right)^{1/2}\ dt$.

Title: Simple question on Mean value theorem

Body:
Let $\phi,\psi\in C^1[a,b]$ of bounded variation, and $C$ a curve given by $x=\phi(t),\ y=\psi(t)$. Let $\Gamma=\{t_i\}_{i=0,\cdots,m}$ be a partition of $[a,b]$. By $C$ rectifiable we mean $$ L=\sup_\Gamma S_\Gamma=\sup_\Gamma \sum_{i=1}^m \left( (\phi(t_i)-\phi(t_{i-1}))^{2}+ (\psi(t_i)-\psi(t_{i-1}))^{2}\right)^{1/2}<\infty$$
Here because of bounded variation condition holds, we have $C$ rectifiable. The problem is asking me to prove that $$L=\int_a^b \left( \phi'(t)^{2}+ \psi'(t)^{2}\right)^{1/2}\ dt\ \ \  (*)$$
Since $\phi',\psi'$ are continuous, I know that
$$L=\lim_{|\Gamma|\to 0} S_\Gamma $$
So it seems natural to use, $$\phi(t_i)-\phi(t_{i-1})=\phi'(\xi_i)(t_i-t_{i-1}) \ \mathrm{and}\  \psi(t_i)-\psi(t_{i-1})=\psi'(\rho_i)(t_i-t_{i-1})$$ for some $t_{i-1}\leq \xi_i,\rho_i\leq t_i$ and for all $i=1,\cdots,m$, to form a Riemann sum for $(*)$ that converges to $L$, indeed starting from $S_\Gamma$ and replacing we get
$$S_\Gamma = \sum_{i=1}^m \left( \phi'(\xi_i)^{2}+ \psi'(\rho_i)^{2}\right)^{1/2}(t_i-t_{i-1}) \ \ (**)$$

My question starts here:

Intuitively, if $|\Gamma|\to 0$, is expected that $t_{i-1},\xi_i,\rho_i,t_{i}$ will be close enough, so $(**)$ is very close to a Riemann sum for $(*)$ and we are done. But, is it just like that I mean, can I take $\left( \phi'(\xi_i)^{2}+ \psi'(\rho_i)^{2}\right)^{1/2}$ as if it were $\left( \phi'(\xi_i)^{2}+ \psi'(\xi_i)^{2}\right)^{1/2}$? what is the formal justification for this?


问:
Proportional second derivatives imply proportional first
Let $f:[0,\infty)^n\to[0,\infty),\ f\in\mathcal{C}^{\infty}$, if I assume that the following relation holds:
$${\partial^2 f\over \partial x_1^2}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto{\partial^2 f\over \partial x_2^2}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto\cdots\propto{\partial^2 f\over \partial x_n^2}\bigg\vert_{x_1=x_2=\dots=x_n=t},\ t\in[0,\infty).$$
Does this imply that:
$${\partial f\over \partial x_1}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto{\partial f\over \partial x_2}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto\cdots\propto{\partial f\over \partial x_n}\bigg\vert_{x_1=x_2=\dots=x_n=t},\ t\in[0,\infty)$$
where $\propto$ meaning that the functions are equal up to a constant. Any help is appreciated.

答:
Let $f:[0,+\infty)^2\to [0+\infty):(x,y)\mapsto x^2+x+y^2$.

Then for all $t\ge 0:$ $$\partial_{11}f(t,t)=2=\partial_{22}f(t,t),$$
but $$\partial_1f(t,t) = 2t+1\\
\partial_2f(t,t)=2t.$$


答:
The well ordering principle comes into play in trying to find the first $n$ where $n^3 > 3^n$.

We know $1^3 < 3^1$.

But can there be any $n^3 > 3^n$?

If so, there must be a first $n$ where $n^3 > 3^n$. But if $n$ is the first then it must be that  $(n-1)^3 \le 3^{n-1}$.

Now hopefully we will be able to show $(n-1)^3 \le 3^{n-1}$ while $n^3 > 3^n$ can't ever happen which means we can never have a first case where $k^3 \le 3^k$ is not true, which means $k^3 \le 3^k$ always will be true.

So multiplying both sides by $3$ we have$3(n-1)^3 \le 3^n$. But we also have $3^n < n^3$.  So $3(n-1)^3 \le 3^n < n^3$.

And $3(n^3 - 3n^2 + 3n -1) < n^3$.

Well now simplify that and

$2n^3 - 9n^2 + 9n - 3< 0$ so

$2n^3 +9n < 9n^2 + 3$.

$2n + \frac 9n < 9 + \frac 3{n^2}$.

So $2n < 2n + \frac 9n < 9 + \frac 3{n^2} $.

If $n \ge 2$ then $\frac 3{n^2} < 1$ and so we either have $n =1$ or $2n < 9+\frac 3{n^2}< 9+1 = 10$.  In any event we must have $n < 5$ or in other words, $n=1,2,3,$ or $4$.

So we test that $n=1,2,3,4$ and get $1^3 < 3^1; 2^3 =8 < 9 =3^2; 3^3 = 3^3; 4^3=64 < 81 = 3^4$.  so none of those are the first exception.

So the first exception can not exist.

And if there can't be a first value, there can't be any value.

Title: Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2]

Body:
I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.

I tried multiple ways to understand why this assumption was made, but I can't figure it out. My first assumption was that since $m \in S$ it's true that:

$m > 3^{m/3}$

and by consequence:

$m^3 > 3^{m}$

but it still does not prove $3^m < m$. Any help is very appreciated.


问:
$\sum_{d\mid n}\lambda(d)\sigma(d)=n\lambda(n)\sum_{d^2\mid n}\frac{1}{d^2}$ Solution
Recall that the Liouville function $\lambda$ and $\sigma$ are multiplicative, and the product of multiplicative functions is also multiplicative, thus $\lambda\sigma$ is multiplicative and therefore the function $\sum_{d\mid n}\lambda(d)\sigma(d)$ is multiplicative.

In the same way the function $N$ is multiplicative and therefore the function $\sum_{d\mid n}N(d) = \sum_{d\mid n}N(\frac{n}{d})$, which in particular will remain multiplicative if we only go through some divisors, so the function $\sum_{d^2\mid n} N(\frac{n}{d^2})$ is multiplicative. Therefore $\lambda(n)\sum_{d^2\mid n} N(\frac{n}{d^2}) = \lambda(n)\sum_{d^2\mid n}\frac{n}{d^2} = n\lambda(n)\sum_{d^2\mid n}\frac{1}{d^2}$ is multiplicative because it is the product of two multiplicative functions.

As both sides of the equality is a multiplicative function, it is enough to prove for the case where $n=p^k$ where $p$ is a prime. So let's start from the left side of equality.

Now let's see for cases on $k$.

Case 1: If $k$ is even $\Rightarrow k = 2l$ where $l\in\mathbb{Z}_0 ^ + $.

\begin{align*}
\require{cancel}
    &\sigma(1)-\sigma(p)+\sigma(p^2)-\sigma(p^3)+\ldots+(-1)^{2l-1}\sigma(p^{2l-1})+(-1)^{2l}\sigma(p^{2l})\\
    &= 1-(1+p)+(1+p+p^2)-(1+p+p^2+p^3)+\ldots-(1+p+p^2+p^3+\ldots+p^{2l-1})\\
    &+(1+p+p^2+p^3+\ldots+p^{2l-1}+p^{2l})\\
    &= 1-\cancel{(1+p)}+(\cancel{1+p}+p^2)-\cancel{(1+p+p^2+p^3)}+\ldots-\cancel{(1+p+p^2+p^3+\ldots+p^{2l-1})}\\
    &+(\cancel{1+p+p^2+p^3+\ldots+p^{2l-1}}+p^{2l})\\
    &= 1+p^2+p^4+\ldots+p^{2l-2}+p^{2l}\\
    &=(-1)^{2l}(1+p^2+p^4+\ldots+p^{2l-2}+p^{2l})\\
    &=\lambda(p^{2l})\left(\frac{p^{2l}}{p^{2l}}+\frac{p^{2l}}{p^{2l-2}}+\frac{p^{2l}}{p^{2l-4}}+\ldots+\frac{p^{2l}}{p^{2}}+\frac{p^{2l}}{p^{2(0)}}\right)\\
    &=\lambda(p^{2l})\left(\frac{p^{2l}}{(p^{l})^2}+\frac{p^{2l}}{(p^{l-1})^2}+\frac{p^{2l}}{(p^{l-2})^2}+\ldots+\frac{p^{2l}}{(p)^{2}}+\frac{p^{2l}}{(p^{0})^2}\right)\\
    &=\lambda(p^{2l})\sum_{i=0}^l\frac{p^{2l}}{(p^{i})^2}\\
    &=\lambda(p^{2l})\sum_{d^2\mid p^{2l}}\frac{p^{2l}}{d^2}\\
    &=\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{p^{k}}{d^2}\\
    &=p^k\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{1}{d^2}.
\end{align*}

Observation: $(-1)^{2l}=1$.

Case 2: If $k$ is odd $\Rightarrow k = 2l + 1$ where $l\in\mathbb{Z}_0^+$.

\begin{align*}
\require{cancel}
    &\sigma(1)-\sigma(p)+\sigma(p^2)-\sigma(p^3)+\ldots+(-1)^{2l}\sigma(p^{2l})+(-1)^{2l+1}\sigma(p^{2l+1})\\
    &= 1-(1+p)+(1+p+p^2)-(1+p+p^2+p^3)+\ldots+(1+p+p^2+p^3+\ldots+p^{2l})\\
    &-(1+p+p^2+p^3+\ldots+p^{2l}+p^{2l+1})\\
    &= 1-\cancel{(1+p)}+(\cancel{1+p}+p^2)-\cancel{(1+p+p^2+p^3)}+\ldots+(\cancel{1+p+p^2+p^3+\ldots}+p^{2l})\\
    &-(1+p+p^2+p^3+\ldots+p^{2l}+p^{2l+1})\\
    &= 1+p^2+p^4+\ldots+p^{2l}-(1+p+p^2+p^3+\ldots+p^{2l}+p^{2l+1})\\
    &= \cancel{1}+\cancel{p^2}+\cancel{p^4}+\ldots+\cancel{p^{2l}}-(\cancel{1}+p+\cancel{p^2}+p^3+\ldots+\cancel{p^{2l}}+p^{2l+1})\\
    &= -p-p^3-p^5-\ldots-p^{2l-1}-p^{2l+1}\\
    &= -(p+p^3+p^5+\ldots+p^{2l-1}+p^{2l+1})\\
    &=(-1)^{2l+1}(p+p^3+p^5+\ldots+p^{2l-1}+p^{2l+1})\\
    &=\lambda(p^{2l+1})\left(\frac{p^{2l+1}}{p^{2l}}+\frac{p^{2l+1}}{p^{2l-2}}+\frac{p^{2l+1}}{p^{2l-4}}+\ldots+\frac{p^{2l+1}}{p^{2}}+\frac{p^{2l+1}}{p^{2(0)}}\right)\\
    &=\lambda(p^{2l+1})\left(\frac{p^{2l+1}}{(p^{l})^2}+\frac{p^{2l+1}}{(p^{l-1})^2}+\frac{p^{2l+1}}{(p^{l-2})^2}+\ldots+\frac{p^{2l+1}}{(p)^{2}}+\frac{p^{2l+1}}{(p^{0})^2}\right)\\
    &=\lambda(p^{2l+1})\sum_{i=0}^l\frac{p^{2l+1}}{(p^{i})^2}\\
    &=\lambda(p^{2l+1})\sum_{d^2\mid p^{2l+1}}\frac{p^{2l+1}}{d^2}\\
    &=\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{p^{k}}{d^2}\\
    &=p^k\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{1}{d^2}.
\end{align*}

Observation: $(-1)^{2l+1}=-1$.

In both cases, the same expression is reached, therefore:

\begin{align*}
    \sum_{d\mid p^k}\lambda(d)\sigma(d) = p^k\lambda(p^{k})\sum_{d^2\mid p^{k}}\frac{1}{d^2}.
\end{align*}

From what was said at the beginning, since both sides of the equation were multiplicative, it was enough to test for the power of a prime, so we have tested equality for all $n$.

Note:
I hope that this time Xander Henderson, Professor Vector, Peter and amWhy do not close my publication because apparently they cannot ask for help (I clarify help, not solutions) and I leave my solution in case someone wants to take it as an example of how to address this type of problem, already that if someone asks only for some ideas, they may close the question.

Body:
Probably correct, anyway you got the idea.

I think it is ok to show how I deal with it

$$\sum_{n\ge 1} \sigma(n)n^{-s}=\sum_{n\ge 1} (\sum_{d|n} d)n^{-s} = \prod_p \frac1{(1-p^{-s})(1-p^{1-s})}$$
$$\sum_{n\ge 1} \lambda(n)\sigma(n)n^{-s} = \prod_p \frac1{(1-\lambda(p)p^{-s})(1-\lambda(p) p^{1-s})}= \prod_p \frac1{(1+p^{-s})(1+p^{1-s})}$$
$$\sum_{n\ge 1}(\sum_{d|n} \lambda(d)\sigma(d))n^{-s} =\prod_p \frac1{(1-p^{-s})(1+p^{-s})(1+p^{1-s})}=\prod_p \frac1{(1-p^{-2s})(1+p^{1-s})}$$

$$\sum_{n\ge 1}(\sum_{d^2|n} d^{-2})n^{-s}
= \zeta(s)\sum_{d\ge 1}d^{-2s-2}= \zeta(s)\zeta(2s+2)=\prod_p \frac1{(1-p^{-s})(1-p^{-2s-2})}$$
$$\sum_{n\ge 1}n(\sum_{d^2|n} d^{-2})n^{-s}
= \prod_p \frac1{(1-p^{1-s})(1-p^{-2s})}$$

$$\sum_{n\ge 1}n\lambda(n)(\sum_{d^2|n} d^{-2})n^{-s}
= \prod_p \frac1{(1-\lambda(p)p^{1-s})(1-\lambda(p^2)p^{-2s})}= \prod_p \frac1{(1+p^{1-s})(1-p^{-2s})}$$

Whence $$\sum_{d|n} \lambda(d)\sigma(d)=n\lambda(n)(\sum_{d^2|n} d^{-2})$$


问:
Suggestion for mistake on computation of a pull-back $2$-form on the sphere
Let $f : S^3 \to S^2$ be the function given by
$$
f(x , y , z , t) = (p = (2 (x z + y t) , q = 2 (- x t + y z) , r = - x^2 - y^2 + z^2 + t^2).
$$
Let $\omega = (p \, dq \wedge dr + q \, dr \wedge dp + r \, dp \wedge dq)$ be a $2$-form in $S^2$. How can I show that $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt) \in {\Omega}^2(S^3)$? My attempt is the next:
$$
f^*(\omega) = p f^*(dq) \wedge f^*(dr) + q f^*(dr) \wedge f^*(dp) + r f^*(dp) \wedge f^*(dq).
$$
For instance,
$$
f^*(dq) = \frac{\partial q}{\partial x} \, dx + \frac{\partial q}{\partial y} \, dy + \frac{\partial q}{\partial z} \, dz + \frac{\partial q}{\partial t} \, dt = 2 (- t \, dx + z \, dy + x \, dz + y \, dt).
$$
If we continue with this process, do we reach that $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt)$? I guess no, because
$$
p f^*(dq) \wedge f^*(dr) = 2 (x z + y t) 2 (- t \, dx + z \, dy + x \, dz + y \, dt) \wedge 2 (- x \, dx - y \, dy + z \, dz + t \, dt) =
$$
$$
= \ldots = 8 (x z + y t) (t y + x z) \, dx \wedge dy + \ldots
$$
(the coefficient of $dx \wedge dy$ in the expression of $p f^*(dq) \wedge f^*(dr)$ is $8 (x z + y t) (t y + x z)$. If we take the coefficient of $dx \wedge dy$ from $q f^*(dr) \wedge f^*(dp)$ and $r f^*(dp) \wedge f^*(dq)$, we obtain
$$
8 (- x t + y z) (- x t + y z) \quad \text{and} \quad 4 (- x^2 - y^2 + z^2 + t^2) (z^2 + t^2).
$$
Then using that $x^2 + y^2 + z^2 + t^2 = 1$ in $S^3$, we have that
$$
f^*(\omega) = (8 (x z + y t) (t y + x z) + 8 (- x t + y z) (- x t + y z) + 4 (- x^2 - y^2 + z^2 + t^2) (z^2 + t^2)) \, dx \wedge dy + \ldots = \ldots = 4 (z^2 + t^2) \, dx \wedge dy + \ldots
$$
if I am not wrong, and it is not compatible with $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt)$. Where am I getting mistake? Also I have to keep in mind that $x \, dx + y \, dy + z \, dz + t \, dt = 0$ but it is because $x^2 + y^2 + z^2 + t^2 = 1$.

答:
The result $f^*(\omega) = 4(dx\wedge dy + dz\wedge dt)$ is correct. This is actually a rather well-known fact, since your $f$ is just the Hopf fibration.

The next step in the calculation is to use the relation $x\, dx + y\, dy + z\, dz + t \, dt = 0$, as you've noted. Specifically, with the $dx\wedge dy$ term, you can convert the coefficient $z^2+t^2$ to $1-x^2-y^2$. The $- x^2$ term can be rewritten using this relation as
$$
- x (x\, dx\wedge dy) = x (z\, dz\wedge dy + t\, dt\wedge dy) = - xz\, dy\wedge dz - xt\, dy \wedge dt\ .
$$
Similarly the $-y^2$ term equals $yz \, dx\wedge dz + yt\, dx\wedge dt$.
You can also work out that the coefficient of the $dz\wedge dt$ term is $4(x^2+y^2)$, to which we can apply the same procedure and generate a few more terms.
If you finish computing the coefficients of the other terms (through blood sweat and tears, or alternatively, with Mathematica) in $f^*(\omega)$, you will find that they actually all get cancelled by these extra terms we generated, and what's left will be the desired result.


问:
Prove there exists an open set $U$ with $0 \in U \subset f^{-1}(G)$.
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function with the property that for all sequences $x_n$, if $x_n \rightarrow 0$, then $f(x_n) \rightarrow 1$.

Let $G$ be an open set with $1 \in G$. Prove there exists an open set $U$ with $0 \in U \subset f^{-1}(G)$.

[In this problem we are using the usual topology on $\mathbb{R}$].

$\textbf{Proof:}$ By way of contradiction, assume $G$ is open with $1 \in G$ and $f^{-1}(G)$ is not an open set.

Since $G$ is open and $1 \in G$, then $\exists$ open set of size $\epsilon >0$ containing $1$. We see $(-1,1) \not \subset f^{-1}(G)$ as well as $\left(-\frac{1}{2},\frac{1}{2}\right) \not \subset f^{-1}(G)$. Thus, choose $x_1 \in (-1,1)$, $x_2 \in \left(-\frac{1}{2},\frac{1}{2}\right)$, ..., $x_n \in \left(-\frac{1}{n},\frac{1}{n}\right)$.

By axiom of choice, $x_1$, ... $x_n \not \in f^{-1}(G)$. Hence $x_n \rightarrow 0$.

So, $\forall n \in \mathbb{Z}^+ f(x_n) \not \in G$. Thus, $f(x_n) \not \rightarrow 1$ $\Rightarrow\mspace{-2mu}\Leftarrow$

$\therefore$ By way of contradiction, $f^{-1}(G)$ is open and $0 \in f^{-1}(G)$. $\blacksquare$

答:
Suppose there is no open set such that $0 \in U$ and $U \subseteq f^{-1}[G]$.

This implies that for every open set $U$ that contains $0$, $U \not\subseteq f^{-1}[G]$.

So we can pick for each $n$, $x_n \in (-\frac1n, \frac1n)$ such that $f(x_n) \notin G$ (i.e. $x_n \notin f^{-1}[G]$).

These $x_n \to 0$ but $f(x_n)\not\to 1$, as witnessed by $G$, which does not contain a tail of that sequence.

So this contradiction does how such a $U$ must exist.


问:
Graph $|x| + |y| \leq 1$
I need to graph the inequality in the title but I'm doing something wrong...
so far I have:

$$|y| \leq -|x| + 1 \Leftrightarrow$$

$$\Leftrightarrow |y| \leq -x + 1, \space \forall \space x > 0 \space \vee |y| \leq x + 1, \space \forall \space x < 0 $$

This gives me four different conditions, depending on the value of $x$:

$$y \leq -x + 1, \space \forall \space (x > 0 \space \wedge -x + 1 < 0)$$
$$-y \leq -x + 1, \space \forall \space (x > 0 \space \wedge -x + 1 > 0)$$
$$y \leq x + 1, \space \forall \space (x < 0 \space \wedge x + 1 > 0)$$
$$-y \leq x + 1, \space \forall \space (x < 0 \space \wedge x + 1 < 0)$$

The resulting graph is not the correct one.

答:
Wrong approach; better to forgo all attempts at elegance and apply brute force, by considering the following 4 cases separately:

$x \geq 0, y\geq 0 \implies \left(|x| + |y|\right) = (x + y)$

$x \geq 0, y < 0 \implies \left(|x| + |y|\right) = (x - y)$

$x < 0, y \geq 0 \implies \left(|x| + |y|\right) = (-x + y)$

$x < 0, y < 0 \implies \left(|x| + |y|\right) = (-x - y).$

In $\mathbb{R^2}$, each of these cases nicely fits into one of the four quadrants.  So, in each case, you want the locus of all points that satisfy the inequality and simultaneously are in the corresponding quadrant.

For example, examining Case 3 above, you would be looking for all points $(x,y)$ such that $x < 0 \leq y$ and $(-x + y) \leq 1.$

The first step in solving the Case 3 inequality would be to graph the equation $(-x + y) = 1.$

Then you would consider whether the inequality is satisfied by the region above the line or below the line.  Then, having decided which region is pertinent, since Case 3 involves the second quadrant (when numbering the quadrants counterclockwise starting with the upper right quadrant as number 1), you would look for the intersection of the pertinent region with the second quadrant.

Applying this approach to all 4 cases, you would then have 4 separate regions, whose union represents all possible solutions to the problem.  The fact that this union forms an (elegant) rectangle is totally irrelevant.

The approach is simply to use brute force against each of the 4 cases, and then go wherever this leads.

Addendum

Responding to the OP's comment/question:

Is there a way to present each of the four conditions as an inequality of $y$ in function of $x$?

Yes, but it takes careful consideration of each case separately.  Again, you are forgoing all attempts at elegance.  I emphasize that you are applying brute force to each of the four cases.  Continuing Case 3, where

$(|x| + |y|) = (-x + y)$

and noting that the corresponding quadrant is the second quadrant:

First, you determine the pertinent equation:

$(-x + y) = 1.$

Then, you determine (temporarily ignoring any consideration re quadrants) which half-plane satisfies the inequality $(-x + y) \leq 1.$

Noting that the region below and to the right increases the value of $x$ and decreases the value of $y$, and noting further that as $x$ increases, $(-x)$ decreases, you are supposed to surmise that the entire half-plane below and to the right of the equation $(-x + y) = 1$ will satisfy $(-x + y) \leq 1$.

A moderate alternative approach is to note that $(-x + y) \leq 1 \iff y \leq (1 + x).$  This alternate approach (again) may be used to establish that in Case 3, the half-plane below and to the right is the desired half-plane.

Then, having identified the appropriate half-plane, you then determine the intersection between this half-plane and the second quadrant.

This then gives you a complete solution to the inequality for Case 3 only.  You then rinse and repeat manually for each of the other 3 cases.

Then, you have the additional challenge of identifying the union of the 4 subsets, where each subset corresponds to the solution to the inequality for one of the four cases.

Title: Graph $|x| + |y| \leq 1$


问:
Proving that a Function does not have any extreme points
I need to find a continuous function, such that $$\lim_{x \to \infty}f(x)=L_{1}$$ $$\lim_{x \to -\infty}f(x)=L_{2}$$

And such that $f$ doesn't have any extreme points.

So I found this function here: $f(x)=\frac{x}{|x|+1}$ for every $x \in \mathbb{R}$.

I have proved that the function is continuous and also that  $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$.

Now all I left to do is proving that this function has no extreme points at all.

I want to prove that the function is strictly increasing, and for that reason my theorem holds.

I have tried to choose $x_{1}, x_{2} \in \mathbb{R}$ such that $x_{1}<x_{1}$ and prove that $f(x_{1})<f(x_{2})$ but I find this difficult for me for some reason.

I will appreciate you help to finish my proof!

Thank you so much! :)

答:
Consider the function $f\left(x\right)=\frac{e^{x}}{1+e^{x}}$

$f$ is continuous by arithmetic property of continuous functions. Also We have:

$$\underset{x\rightarrow\infty}{\lim}\frac{e^{x}}{1+e^{x}}=1$$

$$\underset{x\rightarrow-\infty}{\lim}\frac{e^{x}}{1+e^{x}}=0$$

$f$ doesn't have a max because for every $x_1\in\mathbb{R}$ we can take $x_1+1$  such that:

$$\underbrace{f\left(x_1+1\right)}=\frac{e^{x_{1}+1}}{1+e^{x_{1}+1}}=\frac{e\cdot e^{x_{1}}}{1+e\cdot e^{x_{1}}}=\frac{e^{x_{1}}}{\frac{1}{e}+e^{x_{1}}}\underset{\frac{1}{e}<1}{\underbrace >}\frac{e^{x_{1}}}{1+e^{x_{1}}}=\underbrace{f\left(x_{1}\right)}$$

Also, $f$ doesn't have a min because for every $x_2\in\mathbb{R}$ we can take $x_2-1$  such that:

$$\underbrace{f\left(x_2-1\right)}=\frac{e^{x_{2}-1}}{1+e^{x_{2}-1}}=\frac{\frac{e^{x_{2}}}{e}}{1+\frac{e^{x_{2}}}{e}}=\frac{e^{x_{2}}}{e+e^{x_{2}}}\underset{e>1}{\underbrace <}\frac{e^{x_{2}}}{1+e^{x_{2}}}=\underbrace{f\left(x_{2}\right)}$$


问:
$\mathbb{Z}^{\oplus \mathbb{N}} = \mathbb{Z}^{\oplus \mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$.
I am solving the following question from Aluffi chapter 0. I constructed the following counter example.

I know by intuition that the following there exists an isomorphism $\mathbb{Z}^{\oplus \mathbb{N}} = \mathbb{Z}^{\oplus \mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$. I am trying to find explicit isomorphism between these two.

答:
Consider the mapping
\begin{align}
\mathbb Z^{\oplus \mathbb N} \oplus \mathbb Z^{\oplus \mathbb N} & \longrightarrow \mathbb Z^{\oplus \mathbb N} \\ ((a_1,a_2,\dots),(b_1,b_2,\dots)) & \longmapsto (a_1,b_1,a_2,b_2,\dots)
\end{align}


答:
The first and fourth of your conditions are wrong.

For the first one that originates from $|y|\le -x+1$, if RHS $-x+1 < 0$ then no $|y|$ is going to be even equal or less.

$|y|\le -x+1$ should instead imply

$$\begin{align*}
-(-x+1)\le y &\le -x+1\\
x-1 \le y &\le 1-x
\end{align*}$$

with conditions that $x\ge0$ and $-x+1 \ge 0$, i.e. $$0\le x \le 1$$

Title: Graph $|x| + |y| \leq 1$

Body:
I need to graph the inequality in the title but I'm doing something wrong...
so far I have:

$$|y| \leq -|x| + 1 \Leftrightarrow$$

$$\Leftrightarrow |y| \leq -x + 1, \space \forall \space x > 0 \space \vee |y| \leq x + 1, \space \forall \space x < 0 $$

This gives me four different conditions, depending on the value of $x$:

$$y \leq -x + 1, \space \forall \space (x > 0 \space \wedge -x + 1 < 0)$$
$$-y \leq -x + 1, \space \forall \space (x > 0 \space \wedge -x + 1 > 0)$$
$$y \leq x + 1, \space \forall \space (x < 0 \space \wedge x + 1 > 0)$$
$$-y \leq x + 1, \space \forall \space (x < 0 \space \wedge x + 1 < 0)$$

The resulting graph is not the correct one.


问:
Validity of challenge to common Euler product derivation strategy (infinitely long denominator prime factorisations)
Background

The following Euler product for the Riemann zeta function is well known.

$$ \sum_n \frac{1}{n^s} = \prod_p (1-\frac{1}{p^s})^{-1} $$

Here $n$ ranges over all integers, $p$ over a primes, and real $s>1$.

Common Proof Strategy

Many derivations / proofs, found in textbooks and papers, consider the following expression.

$$(1 - \frac{1}{p^s})^{-1} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \ldots$$

The LHS is finite for any given $p$ and the series expansion is valid because $\frac{1}{p} < 1$.

The following takes the product over all primes.

$$\prod_{p_i} (1-\frac{1}{p_i^s})^{-1} = 1 + \frac{1}{p_1^s} + \frac{1}{p_1^{2s}} + \ldots  + \frac{1}{p_1^sp_2^{s}} + \frac{1}{p_1^sp_3^{s}} + \ldots $$

The LHS is a product of finite and non-zero factors.

The RHS has terms of the form $\frac{1}{X}$ where $X$ contains all combinations of the primes, and all combinations of powers of the primes.

It is common to apply the Fundamental Theorem of Arithmetic to see that there is one term X for each integer $n$, and therefore the RHS is the desired $\sum\frac{1}{n^s}$.

Challenge

A challenge (for example here) that has been raised to this very common proof logic is that there are terms $X$ with an infinite number of factors in the denominator, for example:

$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} $$

or another simpler example:

$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} $$

Question

Is the challenge valid?

I am not a trained mathematician, but in my opinion the proof strategy is valid because terms $X$ with denominators with an infinite number of prime factors are equivalent to zero. That is:

$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} = 0$$

and

$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} = 0$$

My assertion is that the proof strategy remains valid because any finite integer $n$ has a single finite non-zero term $X$, and those $X$ with infinitely long denominators can be discarded because they are zero.

答:
The way to prove this rigorously is to show that:

$$\sum_{n\leq N}\frac1{n^s}\leq \prod_{p\leq N}\left(1-1/p^s\right)^{-1}\leq\sum_{n=1}^{\infty}\frac1{n^s}$$ This you can get because the product in the middle is finite, so you can use the argument without worrying about infinitely many primes.

Then use the squeeze theorem as $N\to\infty.$

This technique works only for real $s>1,$ because the inequalities are not true for $s$ complex.

At heart the proof is to let $$A_N=\{n\geq 1\mid n\text{ has no prime factors }>N\}$$

Then you can show that:

$$\prod_{p\leq N}\left(1-1/p^s\right)^{-1} =\sum_{n\in A_N} \frac1{n^s}$$

Then you get the left-hand inequality because $1,2,\dots,N\in A_N$ and the right hand side because $A_N\subseteq \mathbb Z^+.$


问:
Find $D_{A,B}$ satisfying $\int_{-A}^{B}\hat{f}(t)e^{2\pi itx} dt=\int_{\mathbb{R}} f(x-y) D_{A,B}(y) dy$
Find for $A,B > 0$ a function $D_{A,B}:\mathbb{R}\to \mathbb{C}$ so that \begin{align}
\int_{-A}^{B}\hat{f}(t)e^{2\pi itx} dt=\int_{\mathbb{R}} f(x-y) D_{A,B}(y) dy
\end{align}

My attempt:
\begin{align}
\int_{-A}^{B}\hat{f}(t)e^{2\pi itx} dt&=\int_{-A}^{B}\int_{\mathbb{R}}f(x) e^{-2\pi itx}e^{2\pi itx} dt\\
&=\int_{-A}^{B}\int_{\mathbb{R}}f(x) dxdt\\
\end{align}
because of Fubini
\begin{align}
&=\int_{\mathbb{R}}\int_{-A}^{B}f(x) dtdx\\
&=\int_{\mathbb{R}}f(x)(B+A)dx\\
&=\int_{\mathbb{R}}f(x-y)(-B-A)dy
\end{align}

My function $D_{A,B}:\mathbb{R}\to \mathbb{C}$ would be $(-A-B)$. I think I did something wrong because my funtion is independent of $y$.

答:
Your expression for the Fourier transform of $f$ should not re-use the variable $x$, so the cancellation doesn't make sense. Instead, notice that if we let $\chi_{[-A,B]}$ denote the characteristic function on $[-A,B]$, then
$$\int\limits_{-A}^B\hat{f}(t)e^{2\pi itx}\, dt=\int\limits_{\mathbb{R}}\chi_{[-A,B]}(t)\hat{f}(t)e^{2\pi itx}\, dt=\mathcal{F}^{-1}(\chi_{[-A,B]}\mathcal{F} f)(x)=f*\mathcal{F}^{-1}(\chi_{[-A,B]})(x)$$ by the convolution theorem, and
$$\int\limits_{\mathbb{R}} f(x-y)D_{A,B}(y)\, dy= f*D_{A,B}(x).$$ So, $$D_{A,B}(x)=\mathcal{F}^{-1}(\chi_{[-A,B]})(x)=\int\limits_{-A}^B e^{2\pi i tx}\, dt=\frac{e^{2\pi i Bx}-e^{-2\pi i Ax}}{2\pi i x}$$


问:
Computing second order mixed partial derivatives
In this chapter, page 34 of the pdf they give us the equation:
$$ \frac{\partial ^2F(x)}{\partial w_i \partial w_j} = g''(s) \frac{\partial s}{\partial w_i}\frac{\partial s}{\partial w_j} + g'(s) (\frac{\partial ^2F_{l_1q}(x)}{\partial w_i \partial w_j} + ... +  \frac{\partial ^2F_{l_1q}(x)}{\partial w_i \partial w_j})$$
$g(s)$ is the one dimensional function at the output of the network $F(\cdot)$ for input $x$, where $s$ is equal to the sum over $F_{l_kq}(x)$ for $k$ from $1 : m$. Now they claim this is simple differential calculus, but it is slightly lost on me.
Luckily there is an example a few pages further down (page 38 of the pdf).

It's a simple 2 unit network, which is given as:
$F(x, y) = g(w_3x + w_5y + w_4f(w_1x + w_2y))$

Now I am trying to compute $\frac{\partial^2F(x,y)}{\partial w_1 \partial w_2}$ by hand so I can apply what I learned to the problem I actually have. So my approach is:

$$\frac{\partial^2 g}{\partial w_1 \partial w_2} = \frac{\partial}{\partial w_1}\begin{bmatrix} \frac{\partial g}{\partial w_2}\end{bmatrix} = \frac{\partial}{\partial w_1}\begin{bmatrix} \frac{\partial g}{\partial s}\frac{\partial s}{\partial f}\frac{\partial f}{\partial a}\frac{\partial a}{\partial w_2}\end{bmatrix}$$
Where $s$ is the input of $g$, and $a$ is the input of $f$.

Now I try to apply the product rule and get as far as (where the factor in the brackets is being derived):
\begin{align} \frac{\partial}{\partial w_1}\begin{bmatrix} \frac{\partial g}{\partial s} \end{bmatrix}\frac{\partial s}{\partial f}\frac{\partial f}{\partial a}\frac{\partial a}{\partial w_2}
  &= \frac{\partial^2 g}{\partial s^2}\frac{\partial s}{\partial f}\frac{\partial f}{\partial a}\frac{\partial a}{\partial w_1}\frac{\partial s}{\partial f}\frac{\partial f}{\partial a}\frac{\partial a}{\partial w_2}\\
 \frac{\partial g}{\partial s} \frac{\partial}{\partial w_1} \begin{bmatrix}\frac{\partial s}{\partial f}\end{bmatrix}\frac{\partial f}{\partial a}\frac{\partial a}{\partial w_2} &= \frac{\partial g}{\partial s}\frac{\partial^2 s}{\partial f^2}\frac{\partial f}{\partial a}\frac{\partial a}{\partial w_1}  \frac{\partial f}{\partial a}\frac{\partial a}{\partial w_2}\\
&+ \frac{\partial g}{\partial s}\frac{\partial^2 s}{\partial f \partial b}\frac{\partial b}{\partial w_1}  \frac{\partial f}{\partial a}\frac{\partial a}{\partial w_2}\\
&+ .... \end{align}

here $b$ is just some other summand of s, say $w_3x$. Now clearly the third line would $= 0$, but I'm just wondering, is my approach correct? Do I just continue multiplying $\frac{\partial}{\partial w_1}$ with each factor, am I doing that correctly anyway? I.e. would I end with something with a factor $\frac{\partial^2 a}{\partial w_1 \partial w_2}$ ?

Specifically the second line, as I understand it the first line is equal to the first summand of the first equation of this post, then would the second line be part of the second summand?

I hope this is clear enough, I think in typing this out I have understood more of it, but I would be grateful for some confirmation.

答:
Let's keep track of what you are talking about.

You've rewritten $F$ as

$F = g(s)$, where

$s = w_3x + w_5y + w_4f$, and

$f = f(a)$, and

$a = w_1x + w_2y$

where $x, y$ and all the $w_i$ are independent variables.

Now you express
$$\begin{align}\frac{\partial F}{\partial w_1} &= \frac {dg}{ds}\frac{\partial s}{\partial w_1}\\
&=\frac {dg}{ds}\frac{\partial s}{\partial f}\frac{\partial f}{\partial w_1}\\
&=\frac {dg}{ds}\frac{\partial s}{\partial f}\frac{d f}{d a}\frac{\partial a}{\partial w_1}\end{align}$$
Note that in this calculation, you've already made use of the fact that $f$ is the only variable that $s$ depends on that itself may depend on $w_1$.

Now $$\begin{align}\frac{\partial^2 F}{\partial w_2\partial w_1}
&= \frac{\partial }{\partial w_2}\left[\frac{dg}{ds}\frac{\partial s}{\partial f}\frac{df}{da} \frac{\partial a}{\partial w_1}\right]\\
&=\left[\frac{\partial }{\partial w_2}\frac{dg}{ds}\right]\frac{\partial s}{\partial f}\frac{df}{da} \frac{\partial a}{\partial w_1}
 + \frac{dg}{ds}\left[\frac{\partial }{\partial w_2}\frac{\partial s}{\partial f}\right]\frac{df}{da} \frac{\partial a}{\partial w_1}\\
&\quad+ \frac{dg}{ds}\frac{\partial s}{\partial f}\left[\frac{\partial }{\partial w_2}\frac{df}{da}\right]\frac{\partial a}{\partial w_1}
 + \frac{dg}{ds}\frac{\partial s}{\partial f}\frac{df}{da}\left[\frac{\partial^2 a }{\partial w_2\partial w_1}\right]\end{align}$$
Where

$\dfrac{\partial }{\partial w_2}\dfrac{df}{da} = \dfrac{d^2f}{da^2}\dfrac{\partial a}{\partial w_2}$

$\dfrac{\partial }{\partial w_2}\dfrac{\partial s}{\partial f} = \dfrac{\partial^2 s}{\partial f^2} \dfrac{\partial f}{\partial w_2} = \dfrac{\partial^2 s}{\partial f^2}\dfrac{df}{da}\dfrac{\partial a }{\partial w_2}$

$\dfrac{\partial }{\partial w_2}\dfrac{dg}{ds} = \dfrac{d^2g}{ds^2}\dfrac{\partial s}{\partial w_2} = \dfrac{d^2g}{ds^2}\dfrac{\partial s}{\partial f}\dfrac{df}{da}\dfrac{\partial a}{\partial w_2}$

So
$$\begin{align}\frac{\partial^2 F}{\partial w_2\partial w_1} &= \dfrac{d^2g}{ds^2}\dfrac{\partial s}{\partial f}\dfrac{df}{da}\dfrac{\partial a}{\partial w_2}\frac{\partial s}{\partial f}\frac{df}{da} \frac{\partial a}{\partial w_1}\\
&\quad+ \frac{dg}{ds}\dfrac{\partial^2 s}{\partial f^2}\dfrac{df}{da}\dfrac{\partial a }{\partial w_2}\frac{df}{da} \frac{\partial a}{\partial w_1}\\
&\quad+ \frac{dg}{ds}\frac{\partial s}{\partial f}\dfrac{d^2f}{da^2}\dfrac{\partial a}{\partial w_2}\frac{\partial a}{\partial w_1}\\
&\quad+ \frac{dg}{ds}\frac{\partial s}{\partial f}\frac{df}{da}\frac{\partial^2 a }{\partial w_2\partial w_1}\end{align}$$

You worry in your calculation for $\frac{\partial}{\partial w_2}$ about the dependence of other variables in $s$ on $w_2$, but that ship of generality had already sailed when you assumed them independent of $w_1$ earlier. All of the $w_i$ should properly be independent of each other. One might have $x, y$ dependent on them, but none of the calculations appears to assume that, either.

Now for your particular $s$ and $a$,

$\dfrac{\partial s}{\partial f} = w_4$

$\dfrac{\partial^2 s}{\partial f^2} = 0$

$\dfrac{\partial a}{\partial w_1} = x$

$\dfrac{\partial a}{\partial w_2} = y$

$\dfrac{\partial^2 a}{\partial w_2\partial w_1} = 0$

So
$$\begin{align}\frac{\partial^2 F}{\partial w_2\partial w_1} &= g''(s)w_4f'(a)yw_4f'(a) x + 0 + g'(s)w_4f''(a)yx+ 0\\
&=w_4^2xyg''(s)[f'(a)]^2 + w_4xyg'(s)f''(a)
\end{align}$$


问:
$C^\infty$ $f_n$ converge pointwise $\implies$ f is $C^\infty$
Let $f_n : [0, 1] \to \mathbb R$, a sequence of $C^\infty$ functions, and $f : [0, 1] \to \mathbb R$.
If $f_n$ converges pointwise/uniformly to $f$ then $f$ is also $C^\infty$.

I think that if $f_n$ converges pointwise only to $f$ then $f$ is not $C^\infty$ : counter-example : $x^n$ to converges pointwise to $0$ if $x \neq 1$ and $1$ if $x=1$. All $x^n$ are $C^\infty$ but not $f$.

For the uniformly part, I think it is true but I do not know how to deal with all the derivatives and show they are continuous.

答: No, it is not true. Take $f_n(x)=\sqrt{\frac1n+\left(x-\frac12\right)^2}$. Then each $f_n$ is a $C^\infty$ function. But $(f_n)_{n\in\Bbb N}$ converges uniformly to $f\colon[0,1]\longrightarrow\Bbb R$ with $f(x)=\left|x-\frac12\right|$, which is not differentiable.


问:
change of variable for Lp norm of the gradient of a function.
For $v\in H^{1}_0(B_R(0))$ with $B_R(0)$ is a ball in $\mathbb{R}^2,$ we define the following function,
$$\phi(r) = \int_{B_r(0)} |\nabla_x v(x)|^p dx$$
where $r\leq R.$ I want to check if the expression I have derived for the change of variable $x=ry$ is correct.

Using $dx = r^2 dy$ we get,
$$\phi(r) = \int_{B_1(0)} |\nabla_{ry} v(ry)|^p r^2 dy.$$
Now,
$$\nabla_{ry} v(ry) = r^{-1} \nabla_y v(ry)$$
and therefore,
$$\phi(r) = r^{2-p}\int_{B_1(0)} |\nabla_{y} v(ry)|^p dy.$$
Is this correct?

答:
This is correct.

Notice the factor you get $r^{2 - p}$ came in two separate parts: the $r^{2}$ factor and the $r^{-p}$ one.

An easy way to think of the normalization $r^{2}$ is it's what you need to average by to get the right answer (when there are no derivatives around).  For example, if $u : B_{r}(0) \to \mathbb{R}$ and if I define $u_{r} : B_{1}(0) \to \mathbb{R}$ just by rescaling, then the integrals change but the average doesn't.  (Maybe that's not super intuitive, but a nice check is that the average of a constant is always just the constant, no matter the domain.)  In other words,
\begin{equation*}
\frac{1}{|B_{r}(0)|} \int_{B_{r}(0)} u(x) \, dx = \frac{1}{|B_{1}(0)|} \int_{B_{1}(0)} u_{r}(y) \, dy = \frac{1}{|B_{1}(0)|} \int_{B_{1}(0)} u(ry) \, dy.
\end{equation*}
Since $|B_{r}(0)| = r^{2} |B_{1}(0)|$, if we remember this idea about the averages, then we know how to relate the integrals.

The same thing applies when we look at the derivative, except that if we're looking at $\nabla_{y}u_{r}$, there's a factor of $r$ in the mix.  That's because the derivative is "rise over run" (at least in some direction): replacing $u$ by $u_{r}$ changes the "run" by a factor $r$ without changing the rise at all.  Hence $\nabla_{y}u_{r}(y) = r^{-1} \nabla_{x} u(x)$ with $x = ry$.

If you don't like seeing the $r^{-p}$, a solution is to consider the averaged norm in either case.  We get:
\begin{equation*}
\|\nabla_{y} u_{r}\|_{L^{p}_{\text{av}}(B_{1}(0))} = r^{-1} \|\nabla_{x} u\|_{L^{p}_{\text{av}}(B_{r}(0))}.
\end{equation*}
where
\begin{equation*}
\|\nabla_{y} u_{r}\|_{L^{p}_{\text{av}}(B_{1}(0))} := \left( \frac{1}{|B_{1}(0)|} \int_{B_{1}(0)} \|\nabla_{y} u_{r}(y) \|^{p} \, dy \right)^{\frac{1}{p}} \\
\|\nabla_{x} u\|_{L^{p}_{\text{av}}(B_{r}(0))} := \left( \frac{1}{|B_{r}(0)|} \int_{B_{r}(0)} \|\nabla_{x} u(x)\|^{p} \, dx \right)^{\frac{1}{p}}.
\end{equation*}
What's nice about this is it gets us back to the $\nabla_{y} u_{r} = r^{-1} \nabla_{x} u$ picture.


问:
Characterizing open countable subsets of the Cantor set
The Cantor-Bendixson theorem implies that any closed subset of the Cantor set $\mathcal{C}$ can be described as a disjoint union of a set $\mathcal{C}_c$ that is homeomorphic to the original Cantor set, and a countable open set $\mathcal{C}_o$.

The following answer, and the referenced work by Schoenberg & Grunhage therein, implies that all noncompact open subsets of the Cantor set are homeomorphic to the Cantor set minus a point—say, $\mathcal{C}/ \{ 0\}$. But this would mean that $\mathcal{C}_o$ is homeomorphic to $\mathcal{C}/ \{ 0\}$, which would imply the Cantor set minus a point is countable, which seems strange.

Is this true, or am I missing something?

答:
The Cantor-Bendixsson theorem says that $C$ is a union of it's scattered part (open) and its perfect part (closed). The scattered part here is empty (as $C$ has no isolated points, and $C'=\emptyset$, and there is no scattering part) and it's already perfect. The theorem is valid but void here.

If $O \subseteq C$ is an open non-compact subset, it's locally compact (being open in compact) so it has a one-point compactification $\alpha O = O \cup \{\infty\}$. This is compact metric totally disconnected and has no isolated points so is homeomorphic to $C$ by Brouwer's theorem and so $O$ is homeomorphic to $C$ minus (any) point (by homogeneity).


答: Not repeating good analysis from other answers let me show you drawing $|x|+|y| = 1$, which gives

Title: Graph $|x| + |y| \leq 1$


问:
Independence of transformed variables
There are two independent variables X and Y. Y is an input for non deterministic algorithm f, and the output of f(Y) is Z. How to prove that X and Z are independent?</p>

答:
Let $X,Y$ (real-valued) random variables on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and $f: \mathbb{R} \to \mathbb{R}$. Then

$$\begin{align*} \mathbb{P}[X \in B, Z \in C] &= \mathbb{P}[X \in B, Y \in f^{-1}(B)] = \mathbb{P}[X \in B] \cdot \mathbb{P}[Y \in f^{-1}(B)] \\ &= \mathbb{P}[X \in B] \cdot \mathbb{P}[Z \in C] \end{align*}$$

for Borel sets $B,C$. This proves the independence of $Z=f(Y)$ and $X$. (Note that $f^{-1}(B)$ denotes the preimage of $B$ under $f$, there does not necessarily exist a inverse function.)

Title: Independence of transformed variables


问:
inner semidirect and outer semidirect relationship
https://kconrad.math.uconn.edu/blurbs/grouptheory/group12.pdf.

In the text above, the author explains how to find all groups of order 12.

He does so by showing that a group $G$ (of order $12$) is isomorphic to the semi-direct product of the $2$-sylow ($P_2$) and $3$-sylow ($P_3$) subgroups.

Then, he splits to cases and goes through all the possible combinations of semidirect products of $P_2$ ($\Bbb Z_4$ or $\Bbb Z_2 \times \Bbb Z_2$) with $P_3$ ($\Bbb Z_3)$.

Now my problem arises: he also looks for all possible actions $P_3 \to {\rm Aut}(P_2)$ (or $P_2 \to {\rm Aut}(P_3)$).

But why is this necessary?

He proved that $P_2 \cap P_3$ is trivial, $|P_2P_3| = |G|$ and that $P_2$ is normal or $P_3$ is normal. This means that $G$ is isomorphic to the inner semi-direct product, so why there could be more nonisomorphic semidirect products?

An example of what I'm trying to ask:

Let's say we're in the case that $P_2$ is normal and isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, he found that $G$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \rtimes \Bbb Z_3$ or $G$ isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3$.

But, we already know that $G$ is isomorphic to the inner semidirect product, so how is it possible that there are two options but not one?

答:
For any action $\varphi:G\to {\rm Aut}(H)$ we obtain a semidirect product by defining multiplication on elements of $G\times H$ by
$$(g,h)(g',h'):=(gg',\,\varphi_{g'}(h)h')$$
where conjugation $g^{-1}hg$ becomes $\varphi_g(h)$ because
$$(g^{-1},1)(1,h)(g,1)=(1,\varphi_g(h))\,.$$
For distinct actions, we can indeed obtain nonisomorphic semidirect products of the same two groups. For instance, if $\varphi$ is the constant map ${\rm id}_H$, then we obtain the ordinary direct product as a special case.


问:
Does the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ converge for $s>0$ as well as $\Re(s)>\sigma_c>0$?
This question pertains to the Fundamental Theorem of Dirichlet series which is stated on Wikipedia as follows (where  $s=\sigma+i\,t$):

There are now three possibilities regarding the convergence of a Dirichlet series, i.e. it may converge for all, for none or for some values of s. In the latter case, there exist $\sigma_c$ such that the series is convergent for $\sigma>\sigma_c$ and divergent for $\sigma<\sigma_c$.

Assuming the Riemann Hypothesis (RH) I believe the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ defined in formula (1) below is valid for $\Re(s)>\frac{1}{2}$.

(1) $\quad\frac{\zeta(s+1)}{\zeta(s)}=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a_n\ n^{-s}\right),\quad \Re(s)>\frac{1}{2}\quad\text{(assuming RH)}\quad\text{where }a_n=\frac{1}{n}\sum\limits_{d|n} d\ \mu(d)$

I've noticed the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ defined in formula (1) above also seems to converge for $s>0$. Figure (1) below illustrates the reference function $\frac{\zeta(s+1)}{\zeta(s)}$ in blue and formula (1) above evaluated at $N=100$, $N=1000$, and $N=10000$ in orange, green. and red respectively.

Figure (1): Illustration of formula (1) for $\frac{\zeta(s+1)}{\zeta(s)}$ evaluated at $N=100, 1000, \text{and }10000$ (orange, green, and red)

Question: Does the Dirichlet series for $\frac{\zeta(s+1)}{\zeta(s)}$ defined in formula (1) above converge for $s>0$ as well as $\Re(s)>\frac{1}{2}$ (assuming RH), and if so is this inconsistent with the Fundamental Theorem of Dirichlet series?

答:
$$\frac{\zeta(s+1)}{\zeta(s)}= \sum_{n=1}^\infty a_n n^{-s}$$
If it converges at $s_0$ then it converges and is analytic for $\Re(s) >\Re(s_0)$. Whence $\Re(s_0)\ge \sigma_0$ where $\sigma_0=\sup \{ \Re(\rho), \zeta(\rho)=0\}$.

Conversely the main Tauberian theorem plus several growth estimates say that
$$\sum_{k\le K} \mu(k) = O(K^{\sigma_0+\epsilon})$$
So that
$$\sum_{n\le N} a_n = \sum_{d\le N} d^{-1} \sum_{k\le N/d} \mu(k)=
\sum_{d\le N} d^{-1} O((N/d)^{\sigma_0+\epsilon})=O(N^{\sigma_0+\epsilon})$$
And hence $\sum_{n\ge 1}a_nn^{-s}$ converges for $\Re(s)>\sigma_0$.

The RH is $\sigma_0=1/2$. The PNT doesn't imply that $\sigma_0<1$, it can very well be $\sigma_0=1$. The convergence on $\Re(s)=\sigma_0$ is trickier. If some zero has real part $\sigma_0$ then it diverges everywhere on $\Re(s)=\sigma_0$. If no zero has real part $\sigma_0$ then it might depend on the rate of convergence of $\sup \{ \Re(\rho), |\Im(\rho)|\le T,\zeta(\rho)=0\}$.


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:
$$(\mathbb E-1)(\mathbb E-3) y_t = t^2+2t$$

Since the $3^{rd}$ difference of a quadratic polynomial is zero, we have

$$(\mathbb E-1)^3 (t^2+2t) = 0$$

Therefore
$$(\mathbb E-1)^4(\mathbb E-3) y_t = 0$$
is homogeneous. The roots of the characteristic equations are $1$ (with multiplicity $4$) and $3$ (with multiplicity $1$). Now you can solve.

Here $\mathbb E$ is the forward shift operator (i.e., $\mathbb E^i y_t = y_{t+i}, \forall i \in \mathbb N$).


问:
If $\alpha,\beta,\gamma$ are the roots of equation $x^3-7x^2+12x-11=0$, find the last four digits of $\alpha^{21007}+\beta^{21007}+\gamma^{21007}$

If $\alpha,\beta,\gamma$ are roots of the equation $x^3-7x^2+12x-11=0$, then find the last four digits of $\alpha^{21007}+\beta^{21007}+\gamma^{21007}$.

I know that $\alpha+\beta+\gamma=7, \alpha\beta+\beta\gamma+\gamma\alpha=12,\alpha\beta\gamma=11$. From this how I can proceed. please give some idea

答:
COMMENT.-You have to apply Newton's identities giving relations between the polynomials $p_k=x^k+y^k+z^k$ where $x,y,z$ are the roots of your cubic equation (your $\alpha,\beta,\gamma$) and the corresponding elementary simmetric polynomials $e_0=1,e_1=x+y+z=7,e_2=xy+xz+yz=12$ and $e_3=xyz=11$ ($e_n=0$ for all $n\gt3$).

You have the formulae
$$p_1=e_1\\p_2=e_1^2-2e_2\\p_3=e_1^3-3e_1e_2+3e_3\\p_4=e_1^4-4e_1^2e_2+4e_1e_3+2e_2^2-4e_4\\.....................\\.....................\\p_k=(-1)^{k-1}ke_k+\Sigma_{i=1}^{k-1}(-1)^{k-1+i}e_{k-i}p_i$$ I give you the calculation till $p_5$. Surely a certain periodicity must be found between the first few quantities at stake (otherwise it would be impossible to find an answer by this means). My intention is only to help you in your problem but not to find the solution. You have

$$p_1=7\\p_2=25\\p_3=124\\p_4=645\\p_5=3027$$


问:
Why is it that a non-zero determinant indicates linear independence?
I assume this is because the determinant encodes a sort of "test" for linear independence, so that instead of determining if, for example, three vectors are linearly independent with the definition:

$\sum_{i=0}^{\infty}\alpha_{i}v_{i} = 0 \Leftrightarrow \alpha_{1}=\alpha_{2}= ... = 0$

Instead you compute the determinant and get your answer directly. If this is correct, how exactly is it encoded? If not, what's going on?

答: By contrapositive: if three vectors are linearly dependent, their determinant is $0$ since the determinant is a multilinear alternate form.


答: Note that $\sum \alpha_iv_i$ is the image of the vector $(\alpha_1,\ldots,\alpha_n)^T$ under the matrix in question. So a linear dependence is equivalent to a non-zero vector being mapped to the zero vector. The latter exists iff the linear map given by the matirx is not injective, which is equivalent to the matric not beig invertible, which is equivalent to the determinant being $0$ (because an explicit inversion formula exists that has the determinant in the denominator and hence fails if the determinant is $0$).

Title: Why is it that a non-zero determinant indicates linear independence?

Body:
I assume this is because the determinant encodes a sort of "test" for linear independence, so that instead of determining if, for example, three vectors are linearly independent with the definition:

$\sum_{i=0}^{\infty}\alpha_{i}v_{i} = 0 \Leftrightarrow \alpha_{1}=\alpha_{2}= ... = 0$

Instead you compute the determinant and get your answer directly. If this is correct, how exactly is it encoded? If not, what's going on?


问:
Volume of half torus. What's wrong with my solution?
A doughnut has been partially eaten by a meticulous person so that the portion remaining is given by rotating the half-circular region shown above about the y axis. What fraction of the original doughnut remains, assuming that the original doughnut was the volume of revolution of the full circle? Give your answer in exact form in terms of r and R.

My solution: dropbox.com/s/a816w2rgmkq223r/2013-02-12%2015.19.40.jpg

So clearly, volume of a full torus over that of above is $1\over r$

Bu it's wrong? Where did I go wrong?

EDIT: Corrected some mistakes, still getting the wrong answer:

https://www.dropbox.com/s/1sf72pxcm6dk8pp/2013-02-12%2018.45.46.jpg
https://www.dropbox.com/s/wlko0w958plo1pa/2013-02-12%2018.46.22.jpg

EDIT 2: I made a small mistake in the above solution as well. It should be $4 \pi $ and not $2 \pi$. Correcting that, I got the answer:

$$\left(\frac12\right) - \left(8r\over3 \pi R\right)$$

But this is also wrong! What should I do now?

答:
This problem popped back up in the queue, so I had a look. (I'm assuming this was never completely resolved, to judge from OP's final comment on 13 Feb, 16:10.)  In the first page image posted, the construction of the integral on line 5 is largely correct:

$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ . $$

But None has neglected the fact that since the equation for the circle is $ \ (x - R)^2 + y^2 = r^2 \ $ , which does not describe a function, the total "height" of the "vertical slices" runs from $ \ -\sqrt{r^2 - (x - R)^2} \ $ to $ \ +\sqrt{r^2 - (x - R)^2} \ $.  The total volume integral for this "inner semi-torus" should thus be

$$V \ = \ 2 \cdot 2 \pi\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ , $$

after the first substitution. [EDIT: Now it's clear that None's EDIT 2 refers to this.]

The other issue is that the calculation of the first portion of the volume integral goes slightly  astray.  For the substitution $ \ u = r^2 - t^2 \ , \ du = -2t \ dt \ $ , the transformed limits are found correctly as $ \ u = 0 \ $ to $ \ u = r^2 \ $ .  However, the replacement for $ \ t \ dt \ $ should have been $ \ -\frac{1}{2}\ du \ $ , making the result

$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ dt \ \rightarrow \ -\frac{1}{2}\int_0^{r^2} \sqrt{u} \ du \ = \ -\frac{1}{3}r^3 ; $$

the other portion,

$$\int_{-r}^0 \ R \sqrt{r^2 - t^2} \ \ dt \ = \ \frac{\pi}{4} R \ r^2 $$

is determined correctly.  Hence, the volume of this solid of revolution is

$$V \ = \ 4 \pi \cdot (\frac{\pi}{4} R \ r^2 \ - \ \frac{1}{3}r^3 ) \ = \ \pi^2 R \ r^2 \ - \ \frac{4 \pi}{3}r^3 \ .  $$

(Assumingly, since the problem theme concerns consuming donuts, the second term suggests the volume of a spherical "donut hole".)  This result is confirmed by the application of Pappus' (second) centroid theorem: a semi-circle of area $ \ \frac{1}{2} \pi \ r^2 \ $ is revolved around a circle of radius $ \ R - \frac{4}{3 \pi}r \ ^{*} $ , generating a volume of

$$ \ 2 \pi \cdot  (R - \frac{4}{3 \pi}r) \cdot (\frac{1}{2} \pi \ r^2) \ = \ \pi^2 R \ r^2 \ - \ \frac{4 \pi}{3}r^3 \ . $$

$^{*}$ As the centroid of a semi-circle is a familiar result, I shan't derive it here.

To answer the question in the problem statement then, the remaining fraction of the original donut is given by

$$\frac{\pi^2 R \ r^2  - \frac{4 \pi}{3}r^3}{2 \pi^2 R \ r^2 } \ = \ \frac{1}{2}   - (\frac{2}{3 \pi} \cdot \frac{r}{R}) \ ,   $$

so more than half of the donut has been eaten, as discussed in some of the comments.


答:
Simply prove $-1 < \frac {x}{|x| + 1} < 1$ for all $x$.

Thus for any possible $x_0$ we have $-1 < f(x_0) < 1$.

And as $\lim_{x\to \infty} = 1$ we have $x$s where $f(x_0) < f(x) < 1$.

And as $\lim_{x\to -\infty} = -1$ we havve $x$s where $-1 < f(x) < f(x_0)$.

So $f(x_0)$ can not be an extreme value for any $x_0$.

....

If you want to show $f$ is increasing do it in cases

Cases $x < y < 0$, $x < 0 \le y$,  $0\le x < y$.

Note that $\frac {|w|}{|w| + 1} = \frac {|w|+1}{|w|+1} - \frac 1{|w|}=1 -\frac 1{|w|}$

If $0 \le x < y$ then $x=|x|;y=|y|$ and $\frac 1{|y|} < \frac 1{|x|}$ and so $\frac {x}{|x| + 1} =\frac {|x|}{|x|+1} = 1 -\frac 1{|x|} < 1 -\frac 1{|y|} = \frac {y}{|y| + 1}$.

If $x < 0 \le y$ then $\frac {x}{|x|+1} < 0 \le \frac {y}{|x| + 1}$.

And if $x < y < 0$ then $|x| > |y|$ and

$\frac {x}{|x| + 1} = -\frac {|x|}{|x|+1} = -(1-\frac 1{|x|})< -(1-\frac 1{|y|}) = \frac y{|y| + 1}$.

Title: Proving that a Function does not have any extreme points

Body:
I need to find a continuous function, such that $$\lim_{x \to \infty}f(x)=L_{1}$$ $$\lim_{x \to -\infty}f(x)=L_{2}$$

And such that $f$ doesn't have any extreme points.

So I found this function here: $f(x)=\frac{x}{|x|+1}$ for every $x \in \mathbb{R}$.

I have proved that the function is continuous and also that  $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$.

Now all I left to do is proving that this function has no extreme points at all.

I want to prove that the function is strictly increasing, and for that reason my theorem holds.

I have tried to choose $x_{1}, x_{2} \in \mathbb{R}$ such that $x_{1}<x_{1}$ and prove that $f(x_{1})<f(x_{2})$ but I find this difficult for me for some reason.

I will appreciate you help to finish my proof!

Thank you so much! :)


答:
Consider,  for example,  the cyclic group $\langle a\rangle$ of order six. Then

$$\langle a\rangle =\langle a^2\rangle\rtimes \langle a^3\rangle=\{1,a^2,a^4\}\times \{1,a^3\}.$$

But the symmetric group

$$S_3=\langle (123)\rangle\rtimes\langle (12)\rangle=\{{\rm id},(123),(132)\}\rtimes\{{\rm id}, (12)\}.$$

So nonisomorphic groups can be written as semidirect product, whose corresponding factors are isomorphic!

The distinction, as Berci points out, is in conjugation. We have $$(a^2)^{a^3}=a^2$$ but $$(123)^{(12)}=(213)\neq(123).$$

I found this example in Roman's "Fundamentals of Group Theory: An Advanced Approach" early on in the section on semidirect products.

Title: inner semidirect and outer semidirect relationship

Body:
https://kconrad.math.uconn.edu/blurbs/grouptheory/group12.pdf.

In the text above, the author explains how to find all groups of order 12.

He does so by showing that a group $G$ (of order $12$) is isomorphic to the semi-direct product of the $2$-sylow ($P_2$) and $3$-sylow ($P_3$) subgroups.

Then, he splits to cases and goes through all the possible combinations of semidirect products of $P_2$ ($\Bbb Z_4$ or $\Bbb Z_2 \times \Bbb Z_2$) with $P_3$ ($\Bbb Z_3)$.

Now my problem arises: he also looks for all possible actions $P_3 \to {\rm Aut}(P_2)$ (or $P_2 \to {\rm Aut}(P_3)$).

But why is this necessary?

He proved that $P_2 \cap P_3$ is trivial, $|P_2P_3| = |G|$ and that $P_2$ is normal or $P_3$ is normal. This means that $G$ is isomorphic to the inner semi-direct product, so why there could be more nonisomorphic semidirect products?

An example of what I'm trying to ask:

Let's say we're in the case that $P_2$ is normal and isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, he found that $G$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \rtimes \Bbb Z_3$ or $G$ isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3$.

But, we already know that $G$ is isomorphic to the inner semidirect product, so how is it possible that there are two options but not one?


问:
Is the function identically zero?
Let $f(x, y)$ be a continuous, real-valued function on $\mathbb{R}^2$.
Suppose that, for every rectangular region $R$ of area 1, the double integral of $f(x, y)$
over $R$ equals 0. Must $f(x, y)$ be identically 0?</p>

答:
Hint: Consider $g(y)=\int_0^1 f(x,y)dx$.
Since the intgeral of $f$ over $[0,1]\times[y,y+1]$ equals that over $[0,1]\times[y+\epsilon,y+1+\epsilon]$, we conclude $\int_y^{y+\epsilon} g(y)dy=\int_{y+1}^{y+1+\epsilon}g(y) dy$ and in the limit $g(y+1)=g(y)$, that is $g$ is periodic with period $1$.

Title: Is the function identically zero?


问:
Is it possible to prove the quantifier-shift fallacy is not valid using natural deduction?
I am aware that, in general, sentences of the form  $\exists \forall$ logically imply  $\forall \exists$, but not vice versa. An example of the latter case is:

$((\forall x \exists y : loves(x,y)) \nvdash (\exists y \forall x : loves(x,y)))$. (1)

I have found proof by counter example but no direct proof.

The deduction theorem provides a relationship between $\vdash$ and $\implies$.

For example we can prove: If I am loved by everybody then everybody loves someone.

$((\exists y \forall x : loves(x,y)) \vdash  (\forall x \exists y : loves(x,y)))$. (2)

using the corresponding sentence  (3)

$((\exists y \forall x : loves(x,y)) \implies  (\forall x \exists y : loves(x,y)))$. (3)

I do not know how to write a sentence with $\implies$ that corresponds to (1). Is is possible to prove the invalidity expressed in (1) using natural deduction?

答: No. Natural deduction can only be used to show that inferences are valid, not that inferences are invalid. And as you correctly point out, $\forall x \exists y \phi(x,y) \not \vdash \exists y \forall x \phi(x,y)$ (which holds) is not equivalent to $\forall x \exists y \phi(x,y) \vdash \neg \exists y \forall x \phi(x,y)$ (which doesn't hold). To show $\not \vdash$ without a countermodel, you'd have to argue about the non-existence of derivations, which is tricky.


问:
If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?

Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?

Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Suppose if the numbers were very big, I wouldn't have got the answer.
Do you know any ways to find the answer?

答:
If $x,y > 1$ are both odd, then $x+y > 2$ is even, a contradiction.

Then $y = 2$ and by inspection $x=5$ as $x = 3$ results in $x-y = 1$.

Or, we have that $x-2,x,x+2$ are all prime meaning we have three consecutive integers modulo $3$ which is only satisfied by the triple $(3,5,7) \implies y = 5$.


问:
Reference request and proof verification about using a generalized version of a theorem
I'm not sure if what I'm writing is right or wrong, and I feel like I'm walking on water and I'm not feeling well at all. I construct a proof based on modeling and intuition from $L/\mathbb{Q}$, applied to $L/F$. In some sense, my question is: Is the Dedekind-Kummer factorization holds for any extension of number fields? Let $F$ be a number field, and let $L=F(\alpha)$ for some $\alpha \in \mathcal{O}_L$. And let $\mathfrak{p}$ be a prime of $\mathcal{O}_F$ such that $\mathfrak{p} \nmid [\mathcal{O}_L: \mathcal{O}_F[\alpha]]$. Then the factorization of $\mathfrak{p}\mathcal{O}_L$, can be obtained by the factorization of the minimal polynomial of $\alpha$ module $\mathfrak{p}$. Do we have this? If this is true, could you please show a written version of this theorem, in online notes or books or ...?

My original attempt is as follows. Let $F$ be a number field, and let $L=F(\sqrt[3]{u})$ for some $u \in \mathcal{O}_F$. Let $\mathfrak{p}$ be a prime of $\mathcal{O}_F$ such that $\mathfrak{p} \nmid 3u$. Then $\mathfrak{p}$ is unramified in $L=F(\sqrt[3]{u})$.

I made this proof by imitating my observations from the case when the base field is $\mathbb{Q}$. The minimal polynomial of $\sqrt[3]{u}$ is $x^3-u$, which has a discriminant $3u^2$. Since $\mathfrak{p} \nmid 3u$, so we have $\mathfrak{p} \nmid 3u^2$, so $x^3-u$ is separable module $\mathfrak{p}$. Therefore $\mathfrak{p}$ is unramified in $\mathcal{O}_L$.

答: Yes, we have the general version of Kummer-Dedekind's factorisation theorem. You can have a look at Neukirch's Algebraic Number Theory, p.47-48.</p>


问:
Recursive sequence with dependence of $n$
After consideration on monotony of the sequence, I want to find the result of $\lim_{n\to\infty}x_n$ in case it is a recursive sequence like: $x_{n+1}=f(x_n)=\sqrt{\frac{n^2+1}{n^2+3}}x_n$.

In this case the argument for which if $l=\lim_{n\to\infty}x_n$ then $f(l)=l$ does not work. All $l$ are possible . . . what I should do in cases like that?

答:
we would need an initial value $x_0$ if we were to get an exact solution to your problem. But lets just consider arbitrary $x_0$.

Notice that,
$$\begin{align}
x_{n+1} &= \prod_{k=1}^{n}\sqrt{\frac{k^2+1}{k^2+3}}x_0,\\
        &= x_0\sqrt{\prod_{k=1}^{n}\frac{k^2+1}{k^2+3}},\\
&= x_0\sqrt{\prod_{k=1}^{n}\frac{1+\frac{1}{k^2}}{1+\frac{3}{k^2}}}.\\
\end{align}$$

Next due to Euler or Weierstrass we know that $\sin(\pi z) = \pi z\prod_{k=1}^\infty\big(1-\frac{z^2}{k^2}\big)$. Thus by sending $n$ to infinity and setting $z=i$ and $z=i\sqrt3$ would allow us to conclude the result. All that is left to do are the calculations.


问:
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$.

I drew the picture but I can't seem to figure out the way to start.. If someone could give me a hint on how to start it would mean a lot..

答:
If you rotate planes about the y axis to be parallel then the cylinder has a constant height of 8.  The sides are then $2 \pi r h = 16 \pi$.  The planes intersect at a $45^\circ$ angle so the top and bottom are $r1 \cdot r2 \cdot \pi = 1 \sqrt{2} \ pi$.

$$A = (16+2\sqrt{2}) \pi$$