问:
Can the nonexistence of a constructive proof be proven when an existential proof exists?
Proofs are usually constructive or existential.  For example, we know there are an infinite number of primes, and therefore the centillionth prime exists.  We don't know what it is, though we can make bounds on its location.

Are there any existential proofs where it has been proven that there is NOT a constructive proof of the same fact?

答:
In the strictest sense it is impossible to prove "there is not a constructive proof of $S$", because there is no formal notion of "constructive proof". There are many different forms of constructivism, and they do not agree on whether various principles are constructively valid (e.g. forms of Markov's principle are accepted by some branches of constructivism and not by others). Moreover, there has been an historical trend where many constructivists avoid formalization altogether.

There are, however, formal systems that model constructive reasoning. It is certainly possible to show that particular theorems cannot be proved in these formal systems. For example, it is often possible to show in a constructive system $T$ that a particular statement $S$ implies the law of the excluded middle, or some other omniscience principle. Because these principles are generally not accepted in constructivism, we interpret this as saying that $S$ is not provable constructively. But really all that we have shown is that $S$ is not provable in $T$; there could well be some other system that someone calls constructive which is able to prove $S$.

For commonly studied systems of constructivism, there are many classical results that are known to be unprovable. user72694 mentioned the principle "every continuous function $[0,1] \to \mathbb{R}$ has a maximum" (when we do not assume the function is uniformly continuous). Others include:

"Every strictly increasing bounded sequence of rational numbers converges"

"Every finite square matrix of complex numbers can be put into Jordan form over $\mathbb{C}$"

"Every ideal in a countable partial order extends to a maximal ideal

Here is an example to illustrate the importance of paying attention to the system. Bishop did prove "Every continuous function $[0,1] \to \mathbb{R}$ has a maximum" in his book on constructive analysis. He was able to do this because, in his system, a continuous function is defined to be uniformly continuous, and that additional information allows for a constructive proof.


问:
Non Pathological examples of non sequential spaces
I'm trying to find simple examples where nets are necessary to describe the space instead of sequences. I know for example that if a space is first countable, then convergence can be described by sequences. However, I think that most examples of spaces which are not first countable are usually pathological, at least the ones I thought of.

Can anyone give me the simple examples, hopefully used in reality and not just given as counter-examples,
such that their properties need nets or filters to be described and can not be described by sequences?

答: Take $\Bbb R^{\Bbb R}$, that is, the space of all functions from $\Bbb R$ into $\Bbb R$. Consider on it the product topology. This space is not first countable. However, a sequence $(f_n)_{n\in\Bbb N}$ of elements of $\Bbb R^{\Bbb R}$ converges to some $f\in\Bbb R^{\Bbb R}$ with respect to this topology if and only if $(f_n)_{n\in\Bbb N}$ converges pointwise to $f$. So, I doubt that you will see this example as a pathological one.

ContentLicense: CC BY-SA 4.0


答: $[0,1]^{\Bbb R}$ is a classic example where nets or filters are needed: it's the set of functions from the reals to $[0,1]$ in the pointwise (product) topology. Such topologies are common in functional analysis. It's compact but not sequentially compact. The Čech-Stone compactification $\beta\Bbb N$ of $\Bbb N$, an important object in many branches of maths, is another example.

Title: Non Pathological examples of non sequential spaces

Body:
I'm trying to find simple examples where nets are necessary to describe the space instead of sequences. I know for example that if a space is first countable, then convergence can be described by sequences. However, I think that most examples of spaces which are not first countable are usually pathological, at least the ones I thought of.

Can anyone give me the simple examples, hopefully used in reality and not just given as counter-examples,
such that their properties need nets or filters to be described and can not be described by sequences?


问:
How can I decide whether this function is integrable?
A function $f$ is defined on $[0,1]$ such that for every

irrational number in $[0,1]$, $f(x)=1$ and

rational number in $[0,1]$, $f(x) = \dfrac{a-2}{a}$, where $a$ is the smallest natural number for which $ax$ is an integer.

Does $\int_0^1 f(x)dx$ exist? If yes, what's the value?

I have a hunch that it is not integrable, but somehow unable to proceed.

答:
Consider $g(x) = 1-f(x)$, so it suffices to solve
$$
\int_{[0,1]}f(x) dx = 1 - \int_{[0,1]} g(x) dx
$$
where $g(x)$ is almost the Thomae's function as comments mentioned, and you can refer to here for a proof, you just need to modify a little.


问:
Give example(s) of elements that inside torsion part of cohomology group after quotient
For a manifold $X$, its (integer-valued) cohomology group $H^k(X,\mathbb Z)=Z^k(X,\mathbb Z)/B^k(X,\mathbb Z)$$=\text{Ker}_{\partial_k^*}/\text{Im}_{\partial^*_{k-1}}~$ is construct from the chain complex:
$$ \cdots\mathop\longrightarrow^{\partial^*_{k-1}} C^k(X,\mathbb Z)\mathop\longrightarrow^{\partial^*_k}C^{k+1}(X,\mathbb Z)\mathop\longrightarrow^{\partial^*_{k+1}}C^{k+2}(X,\mathbb Z)\mathop\longrightarrow^{\partial^*_{k+2}}\cdots $$
where $ C^k(X,\mathbb Z)=\text{Hom}_{\mathbb z}(C_k(X,\mathbb Z),\mathbb Z)~ $ is the group of all $\mathbb Z$-linear maps acts on singular k-chains.

We know that the torsion $ T^k $ part of $ H^k(X,\mathbb Z) $ is equal to $ T_{k-1} $, the torsion part of $H_{k-1}(X,\mathbb Z)$

Also $ f\in Z^k(X,\mathbb Z)\Leftrightarrow f|_{B_k(X,\mathbb Z)}=0 $, $ g\in B^k(X,\mathbb Z)\Leftrightarrow g|_{Z_k(X,\mathbb Z)}=0 $, thus when talking about elements of $H^k(X,\mathbb Z)$, there is no need to consider how an element of $ Z^k(X,\mathbb Z) $ acts on non-closed singular k-chains(this info will be divided, if I'm not wrong here). After this step, I failed to construct torsion inside $H^k(X,\mathbb Z)$.

I wonder what the elements of its torsion part look like in $ Z^k(X,\mathbb Z) $ as maps before quotient, such as the elements of $ H^2(\mathbb{RP}^2,\mathbb Z)\simeq \mathbb Z_2 $. There might need several maps in different equivalent classes of $H^k(X,\mathbb Z)$ to show its torsion.

答:
Again, solved the problem myself...

The problem is: The action info of cochains on non-closed chains is not fully divided in terms of the cohomology group.

For example, if a non-closed singular k-simplex $\sigma$ has a boundary $\partial \sigma = nc^{k-1}$, $n\in\mathbb Z$, $c^{k-1}\in Z_{k-1}(X)$, though a closed cochain $f\in Z^k(X)$ can acting on the simplex whatever it want, a exact cochain $g\in B^k(X)$ is slightly different. There must exist a $k-1$ cochain $h\in C^{k-1}(X)$, such that $g(\sigma^k)=\partial^*h(\sigma)=h(\partial\sigma)$ $=h(nc)=n~h(c)$. If $\partial\sigma$ is decompose properly, $h(c)$ has no constraint. The only problem is n, if $|n|>1$, it can no longer span the ray of all chains $\mathbb Z\sigma^*$ restrict on this simplex $\sigma$. ($\sigma^*(\sigma)=1$), thus the torsion group $\mathbb Z_n=\mathbb Z/n\mathbb Z$ appears.

Consider a more concrete example: $X=\mathbb{RP}^2$. It has the CW structure $\sigma^2\cup\sigma^1\cup\sigma^0$, where $\partial\sigma^2=2\sigma^1$, $\partial\sigma^1=0$, $\partial\sigma^0=0$. Its $Z^2(X)$'s element could simplified as a map $f:\{\sigma^2\}\to\mathbb Z$, because the 3rd chain group $C_3(X)$ is $0$, thus no constraint on this map. But focus on $B^2(X)$, $g(\sigma^2)=\partial^*h(\sigma^2)=2h(\sigma^1)$, it can only maps $\sigma^2$ to a even number. thus $B^2(X)=\text{Hom}_{\mathbb Z}(\sigma^2,2\mathbb Z)\simeq 2\mathbb Z$. We can thus get the cohomology group $H^2(X)=Z^2(X)/B^2(X)=\mathbb Z/2\mathbb Z=\mathbb Z_2$, fits the inference of universal coefficient theorem.


问:
How to resolve this algebra equation?
$$f = X^3 - 12X + 8$$ $a $-  complex number, $a$ is a root for $f$

$b = a^2/2 - 4 $.

Show that $f(b) = 0$

This is one of my theme exercises ... Some explanations will be appreciated ! Thank you all for your time .

ContentLicense: CC BY-SA 3.0

Title: How to resolve this algebra equation?

Body:
As $a$ is root of $f = X^3 - 12X + 8=0, a^3-12a+8=0$

Now, $b=\frac{a^2}2-4\implies a^2=2(b+4)=2b+8$

So, from  $a^3-12a+8=0, a(2b+8)-12a+8=0\implies a=-\frac8{2b+4}=-\frac4{b+2}$

So, $$\left(-\frac4{b+2}\right)^3-12\cdot \left(-\frac4{b+2}\right)+8=0$$

On simplification we  get , $$b^3-12b+8=0$$


问:
Let $f(x)=e^x+x^3+x-3$. This function is invertible. What is the value of $f^{-1}(-2)$?
I am having trouble with this question. I don't have any calculus background.

I think $f^{-1}$ means the inverse of $f(x)$

Does that mean I have to find the inverse of $f(x)$ first? But again, I am trapped when operating the e because there is also $x^3$ and $x$. I cannot think of a way to isolate them.

答: We can just nuzzle around and find some $x$ wifh $f(x)=-2$, then $f^{-1}(-2)=x$. $x=0$ works – and, as it turns out, the function is increasing everywhere so this is the only solution.


问:
Zero divisors of a quotient ring

I want to list zero divisors of the ring $\mathbb{Z}_2[x]/x^3\mathbb{Z}_2[x]$.

I know $\mathbb{Z}_2[x]$ is set of polynomials in one variable whose coefficients from $\mathbb{Z}_2$.

So $x^3\mathbb{Z}_2[x]=\{x^3p(x): p(x)\in \mathbb{Z}_2[x]\}$.

So $\mathbb{Z}_2[x]/x^3\mathbb{Z}_2[x]=\{ax^2+bx+c+ x^3\mathbb{Z}_2[x] : a,b,c\in \mathbb{Z}_2 \}$

Did i define $\mathbb{Z}_2[x]/x^3\mathbb{Z}_2[x]$ right?

$x+x^3\mathbb{Z}_2[x]$ and $x^2+x^3\mathbb{Z}_2[x]$ are the only nonzero zero divisors of $\mathbb{Z}_2[x]/x^3\mathbb{Z}_2[x]$.

Is this correct?

答:
Notation is less heavy if you denote, say $\xi=x+x^3\mathbb{Z}_2[x]$, so every element in $R=\mathbb{Z}_2[x]/x^3\mathbb{Z}_2[x]$ can be written in a unique way as
$$
a+b\xi+c\xi^2
$$
where $a,b,c\in\mathbb{Z}_2$ and $\xi^3=0$.

Certainly $\xi$ and $\xi^2$ are zero divisors, because $\xi\xi^2=0$. But also $(\xi+\xi^2)\xi^2=0$, so you at least forgot to mention one of them.

Let's look when a nonzero $a+b\xi+c\xi^2$ is a zero divisor. It's easy to compute
$$
(a+b\xi+c\xi^2)(u+v\xi+w\xi^2)=au+(bu+av)\xi+(cu+bv+aw)\xi^2
$$
so the product being zero implies $au=0$. Let's first do the case $a=0$, so we get $bu=0$ and $cu+bv=0$. If $b=0$, we need $c\ne0$ and we find $\xi^2$; if $b\ne0$, we need $u=0$ and $v=0$, so we find $\xi$ but also $\xi+\xi^2$.

Suppose $a\ne0$. Then $u=0$ and from $av=0$ we get $v=0$. But then $w=0$: so no luck.

Thus the (nonzero) zero divisors are $\xi,\xi^2,\xi+\xi^2$.

Title: Zero divisors of a quotient ring


问:
$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?
$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?

1.$\{(2n+1){\pi\over2}\}$

2.$\{n\pi\}$

3.$\{{n\pi\over 2}\}$

4.$\{n\pi+2\}$

in all cases $n\in\mathbb{Z}$

well, is there any easy trick to solve this type of probelem in an competitive examination? Do I need to pick $\pi \over 2$ from $1$ and cheked that $f$ is differentiable and same trick from each set? in some what I am lost.

答: Hint: The function $ g(y)=|y|$ isn't differentiable at $ y=0$, and therefore, $ g(\cos x)+g(\sin(2-x))$ isn't differentiable when it's arguments are $0$, meaning, when $ \cos x=0$ or when $ \sin (2-x)=0$</p>


问:
Finding a point given a graph equation and a tangent line
how do I find a point given the graph of the equation (that is $x^2-y^2-z^2=1$) where the tangent plane is parallel to the equation $x+y-z = 0$ ?

what I did was take the gradients of both equation, and equate to the corresponding variables (equate the x value of gradient 1 and gradient 2, etc.). I ended up with a set of points, are they correct tho? Please tell me.

答:
For a point on an implicit surface, the gradient vector and normal vector at that point are parallel. The gradient of $x^2-y^2-z^2=1$ is $(2x,-2y,-2z)$, and we want it to be parallel to the normal of $x+y-z=0$ which is $(1,1,-1)$:
$$(2x,-2y,-2z)=k(1,1,-1)$$
$$2x=k,-2y=k,-2z=-k$$
$$x=-y=z$$
The desired points may then be computed by substituting the last equality into the implicit equation and simplifying, so we get $x^2=-1$ which is impossible. Thus no points of the graph have tangent plane parallel to $x+y-z=0$.


问:
How to simplify this logarithm to find the value of x?
$x^{log_2 x} = 2^4$

Solve for x.

How to do this?

答:
By some easy algebraic computations, we get $x^{\log_2x}=2^{\log_2(x^{\log_2x})}=2^{\log_2x \times \log_2x}=2^{\log_2^2x}$.

$\Rightarrow \log_2^2x = 4$

$\Rightarrow \log_2x = \pm 2$

$\Rightarrow x_1 = 2^2 = 4$ and $x_2 = 2^{-2} = \frac{1}{4}$


问:
how to derive the steps of this intergral for the product of two exponential terms?
For homework, I was looking for closed form solutions of exponential functions from this wiki site.
https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

But I was not sure how to derive the steps for this particular integral,

$\int_{-\infty}^{\infty}e^{-ax^{2}}e^{-\frac{b}{x^{2}}}dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)$

I am fairly new to normal distribution but I assume this has to do with Taylor series..

Any pointers would be greatly appreciated!

答:
As an elementary method, consider
$$
\int_\mathbb{R} e^{-ax^2}e^{-bx^{-2}} \; dx = \int_\mathbb{R} e^{-(\sqrt{a}x - \sqrt{b}/x)^2} e^{-2\sqrt{ab}}\; dx
$$
notice that we changed the sign inside the completed square. A clear observation is that the function is even, so we only need to solve half of it, i.e. find
$$
I/2 = \int_0^\infty e^{-(\sqrt{a}x - \sqrt{b}/x)^2} \; dx
$$
Consider a change of variable
$$
u = \sqrt{\frac{b}{a}}/x, du = -\sqrt{\frac{b}{a}}/x^2 = -\sqrt{\frac{b}{a}}u^{-2}
$$
so
$$
I/2 = \int_0^\infty e^{-(\sqrt{a}x - \sqrt{b}/x)^2} \; dx = \sqrt{\frac{b}{a}}\int_0^\infty e^{-(\sqrt{a}u - \sqrt{b}/u)^2} \frac{1}{u^2} \; du
$$
The negative sign that comes from change of variable cancels out as we changed the order of the integration bounds. As both integral are equal, we can in fact sum them to get
$$
I = \int_0^\infty e^{-(\sqrt{a}x - \sqrt{b}/x)^2} \left(1+ \sqrt{\frac{b}{a}}\frac{1}{x^2}\right) \; dx
$$
Did you see where are we going?
$$
w = \sqrt{a}x - \sqrt{b}/x, dw = (\sqrt{a} + \sqrt{b}/x^2 )dx
$$
this change of variable will result in
$$
I = \int_0^\infty e^{-(\sqrt{a}x - \sqrt{b}/x)^2} \left(1+ \sqrt{\frac{b}{a}}x^2\right) \; dx = \frac{1}{\sqrt{a}} \int_\mathbb{R} e^{-w^2} dw = \sqrt{\frac{\pi}{a}}
$$
:)
$$
\int_\mathbb{R} e^{-ax^2}e^{-bx^{-2}} \; dx = \int_\mathbb{R} e^{-(\sqrt{a}x - \sqrt{b}/x)^2}e^{-2\sqrt{ab}} \; dx = \sqrt{\frac{\pi}{a}} e^{-2\sqrt{ab}}
$$


问:
Unitary and transformation matrix
I have a question that I do not understand how to solve that.

Let $V$ be inner product space.

Let {$e_{1},...,e_{n}$} an orthonormal basis for $V$

Let {$z_{1},...,z_{n}$} an orthonormal basis for $V$

I have to show that the matrix represents

the transformation matrix between {$e_{1},...,e_{n}$} to

{$z_{1},...,z_{n}$} is unitary.

How do I do it ?

I got no clue!!

答: I'd sketch a proof as follows.  Regard the vectors in the orthonormal bases $\left(e_{1},...,e_{n}\right)$ and $\left(z_{1},...,z_{n}\right)$ as column vectors.  Then let $\mathbf{U}_e$ and $\mathbf{U}_z$ be the matrices where the rows are the transposes of the column vectors $e_{1},...,e_{n}$ and $z_{1},...,z_{n}$ respectively.  Then both $\mathbf{U}_e$ and $\mathbf{U}_z$ are unitary, and the matrix which maps between $\left(e_{1},...,e_{n}\right)$ and $\left(z_{1},...,z_{n}\right)$ is going to be given by $\mathbf{U}_e\mathbf{U}_z^{-1}$, which will be unitary because $\mathbf{U}_e$ and $\mathbf{U}_z$ are unitary.</p>

Title: Unitary and transformation matrix


答: A function needs to have  domain and Range (co-domain) defined. But here these are not given so we can take it as $f:R\to R$, it being increasing as $f'(x)=e^x+3x^2+1>0, \forall x \in R$. it is one to one. As $f(-\infty)=-\infty, f(\infty)=+\infty$, it is onto. Hence, it is a bijection so the inverse exists, but it is not obtainable. Noting that $f(0)=-2$ we can claim that $f^{-1}(-2)=0$. $f^{-1}(e-1)=1$ but we cannot find $f^{-1}(2)$ by hand, though it exist and it is numerically obtainable as $f^{-1}(2)=1.04084...$ similarly one can find $f^{-1}(0)=0.68723$. See the plot of $f(x)$ below

Title: Let $f(x)=e^x+x^3+x-3$. This function is invertible. What is the value of $f^{-1}(-2)$?

Body:
I am having trouble with this question. I don't have any calculus background.

I think $f^{-1}$ means the inverse of $f(x)$

Does that mean I have to find the inverse of $f(x)$ first? But again, I am trapped when operating the e because there is also $x^3$ and $x$. I cannot think of a way to isolate them.


问:
Given a bipartite graph $G=(U,V,E)$, what is the bound on $|V|$ given some conditions?
I have a bipartite graph $G=(U,V,E)$ with the following two conditions:

Every node $u\in U$ has a degree of at most a constant $c$, i.e., $1\leq \text{deg}(u)\leq c$;

Every node $v\in V$ has a degree of at most $d$;

No two nodes in $U$ are connected to the same set of nodes in $V$, i.e., for any two distinct $u_1,u_2\in U$, $\text{neighbor}(u_1)\neq\text{neighbor}(u_2)$.

Is it possible to show that the $|V|=\Theta(|U|)$? Note that here $|U|,d\rightarrow\infty$, $d=o(|U|)$, and every $u$ is connected to at least one $v\in V$. Otherwise, is there a way to derive the minimum size of $|V|$ in terms of $|U|$ and $d$?

For the case of $c=1$, it is clear that we have $|V|=|U|$, but things are not so clear to me when $c>1$. The best I could do for a general $c=\Theta(1)$ is $|V|=O(|U|/d)$: the $u$'s contribute at least 1 edge each giving us $|E|\geq|U|$. Dividing $|E|$ by $d$ gives us the bound.

答:
In your situation, there is to every node $u \in U$ a set of nodes $S_u \subseteq V$. Your first condition implies that the cardinality of each $S_u$ is somewhere in between $1$ and $c$. A priori, this means that there are
$$
\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{c}
$$
possible sets $S_u$, where $n = |V|$. When $d \to \infty$, this number can also be attained. From this we infer $|V| = \Omega(|U|^{1/c})$, and this is the best one can do.


问:
ways to walk on a hexagon to return to same point with n steps
So I have an hexagon and I start at point 1 and I can only move right or left, I need to find the formula to reach the same point within n steps,
Assuming I have an odd number of steps allowed, I can't ever return to the point.

But assuming I have even steps, I'm not sure how do I even think about it, I divided by two because I can go from 1>>2 or from 1>>6 so it's easier to just look at it as if i started with 1 step right.

This means that if the sequence is $ a_n $and I took a step right now i have $ a_{n-1}$ steps to come back.
how do I go on from here?

答:
I think I have a solution (although not a closed-form) for the general $N$-gon. Suppose we write out our sequence of moves as a list of $n$ letters, either $L$ or $R$ (for left and right). We want to count the number of possible lists of length $n$ where the number of $L$ exceeds the number of $R$ by a multiple of $N$.

This is done by choosing $k$ of these steps to be $L$ moves, such that the difference between $k$ and $n-k$ is an integer multiple of $N$. Hence, the total number of moves is
$$\sum_{k\in S}\binom{n}{k}$$
where $S=\{0\leq k\leq n:n-2k\in N\mathbb{Z}\}$, which we can rewrite as
$$k=\frac{n+mN}{2}$$
for all integers $|m|\leq n/N$. A closed form for this sum would clearly be desired however.


问:
sum of product of bessel functions with constant difference in index
While doing a calculation on frequency modulation I am encountering the sum
$$
\sum_{m=-\infty}^{+\infty} J_{n+m}(z)J_m(z)
$$
which looks awfully similar to the identity:
$$
J_n(t+r) = \sum_{m=-\infty}^{+\infty} J_k(t) J_{n-k}(r)
$$
Except the former sum is over indices with fixed difference, while the latter is with fixed sum.

For the case of $n=0$, I see my expression has the simplification
$$
J_0^2(z) + 2\sum_{m=1}^{\infty}J_m^2(z)=1
$$
Does a simplification exist for $n\neq 0$?

答:
Ok, after a bit more searching I have found on DLMF:
$$
\mathscr{C}_{\nu}\left(u\pm v\right)=\sum_{k=-\infty}^{\infty}\mathscr{C}_{\nu%
\mp k}\left(u\right)J_{k}\left(v\right)
$$
Where $\mathscr{C}_{\nu}$ is a "cylinder function". The Bessel functions $J_v$ are cylinder functions, so taking the bottom row of the $\pm$ we get:
$$
\sum_{k=-\infty}^{\infty}J_{\nu
+ k}\left(z\right)J_{k}\left(z\right) = J_\nu(z-z) = J_\nu(0)=\delta_{\nu 0}
$$
Which is what I had been finding to be the case numerically.


问:
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain.
$f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$

I know I have to switch the $f(x)$ and the $y$:

$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$

To find the domain:

$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$

$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$

$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$

$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$

The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$

But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.

Thanks

答:
Look at the plot below (first picture). The function can be inverted only in the interval where it is bijective.

Take the derivative
$$f'(x)=\frac{4}{3} \sin \left(\frac{1}{3} (2 x+1)\right) \cos \left(\frac{1}{3} (2 x+1)\right)=\frac{2}{3} \sin \left(\frac{2}{3} (2 x+1)\right)$$
$$f'(x)=0\to  \sin \left(\frac{2}{3} (2 x+1)\right)=0$$
In the interval we are interested in, this happens when
$$\frac{2}{3} (2 x+1)=0;\;\frac{2}{3} (2 x+1)=\pi$$
which gives $$x_1=-\frac12;\;x_2=\frac{3 \pi }{4}-\frac{1}{2}$$
We will invert $f(x)$ in the interval $[x_1,x_2]$.

Set $y=f(x)$ and get
$$\sin ^2\left(\frac{1}{3} (2 x+1)\right)=y\to  \sin \left(\frac{1}{3} (2 x+1)\right)=\sqrt{y}$$
then
$$\frac{1}{3} (2 x+1)=\arcsin\sqrt y\to 2x+1=3\arcsin\sqrt y\to x=\frac{1}{2} \left(3 \arcsin\left(\sqrt{y}\right)-1\right)$$
Only now we swap $x$ and $y$ to get
$$f^{-1}(x)=\frac{1}{2} \left(3 \arcsin\left(\sqrt{x}\right)-1\right)$$
The domain of $f^{-1}(x)$ is the range of $f(x)$ that is $[0,1]$.

Look at the second picture below. The inverse function graph is the symmetric of the function $f(x)$ wrt the line $y=x$.

$$...$$


问:
Question about the stereographic projection. How to find the arc of great circle of the stereographic sphere S.
Let $z$, $w$ be complex numbers and let $P(z)$, $P(w)$ be the corresponding
points on the sphere $S$, associated to $z$, $w$ via the stereographic projection.
Find the length of the arc of the great circle lying on $S$ and joining $P(z)$ to
$P(w)$.

I started to consider the arc of the center should be the greatest length. But I'm so confused.

答:
The Euclidean distance between the projections $P(z)$, $P(w)$ onto the sphere $S$ is the “chordal distance” of $z$ and $w$ and can be computed as
$$
 d(z, w) = \frac{2|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}} \, ,
$$
see for example

prove the chordal distance is: $d(x,y)=\frac{2|z-z'|}{\sqrt{1+|z|^2}\sqrt{1+|z'|^2}}$,

Riemann sphere on Encyclopedia of Mathematics.

The length  $L$ of the arc joining $P(z)$ and $P(w)$ is equal to the angle between the vectors from the origin to $P(z)$ and $P(w)$ respectively, so that
$$
 \sin \frac{L}{2} = \frac 12 \|P(z)-P(w)\| = \frac{|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}}
$$
or
$$
 L = 2 \arcsin \left( \frac{|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}} \right)\, .
$$


问:
problem involving centroid and perpendiculars

in triangle $ABC$, $G$ is the centroid and $L$ is an arbitary line through the centroid.     $K,I,J$ are the foot of the perpendiculars from $A, B, C$

i need to prove that $KA+JC=BI$

how to use the midpoints of the sides and is there any way to use the fact that centroid cut the median in 2:1 ratio

答:
Alternatively, one can prove the statement using the facts you mention. Introduce the midpoint $M$ of $AB$ and its ortogonal projection $N$ onto the line $L$. Since $ABJK$ is a trapezoid $(AK\parallel BJ\perp L)$, it is well known that its midline $MN$ is not only parallel to $AK, BJ$ but also satisfies $2MN=AK+BJ$.

Consider now the similar triangles $\triangle MGN\sim CGI$, and deduce, due to the fact that $CG:GM=2:1$, that $$\frac{CI}{MN}=2\iff CI=2MN=AK+BJ$$


问:
does $d\sim p$ imply $d\approx p$ in the metric space $X$

Let $X\neq \emptyset$ and $M(X)\subseteq [0,\infty)^{X\times X}$ denote the set of all metrics on $X.$ For $d, p\in M(X),$ let

$d\approx p\Leftrightarrow\begin{cases}\text{ there are $m, M > 0$ for which}\\
\text{ $m d(x,y) \leq p(x,y) \leq Md(x,y)$ for all $x,y \in X$ }\end{cases}$

and

$d\sim p\Leftrightarrow \begin{cases}\text{ $\lim_{n\to\infty} x_n = x$ in $(X,d)\Leftrightarrow \lim_{n\to\infty} x_n = x$ in $(X,p)$}\\\text{
 for any sequence $(x_n)_{n=1}^\infty \subseteq X$ and $x\in X$}\end{cases}$

Show that $d\sim p\in M(X)$ iff $(X,d)$ and $(X,p)$ admit the same open sets.

Does $d\sim p$ in $M(X)$ imply that $d\approx p$? Prove or disprove this.

For the first part, I think I need to show that $\{O\subseteq X : \text{ $O$ is open in $(X,d)$}\} =\{O\subseteq X : \text{ $O$ is open in $(X,p)$}\},$ which I think can be done by double inclusion. I know that open sets are defined so that $O$ is open in $(X,d)$ if $\forall x \in O, $ there exists $\delta > 0 $ so that $B(x,\delta) := \{y \in X : d(x,y) < \delta\}\subseteq O$. Let $A$ be open in $(X,d)$. Then $X\backslash A$ is closed in $(X,d)$. Let $(y_n)\subseteq X\backslash A, y_n \to y$ in $(X,p).$ Then since $d\sim p, y_n\to y$ in $(X,d)$ and since $X\backslash A$ is closed in $(X,d), y \in X\backslash A$ so $X\backslash A$ is closed in $(X,p).$ Similarly, one can show that if $A$ is open in $(X,p),$ then $A$ is open in $(X,d).$

However, I'm not sure how to show that if $(X,d)$ and $(X,p)$ admit the same open sets then $d\sim p.$

As for the second part, I think it might be true and that a contradiction might be useful for showing it but I'm not sure about the details.

答:
Your proof is correct. It invokes the fact that a set $C$ is closed in $(X,d)$ if and only if all limits of sequences in $C$ are contained in $C$.

$d \sim p$ means that that the metrics are equivalent in the sense that

$(*) \phantom{x}$ $\lim_{n\to\infty} x_n = x$ in $(X,d)  \Leftrightarrow \lim_{n\to\infty} x_n = x$ in $(X,p)$  for any sequence $(x_n)_{n=1}^\infty \subseteq X$ and $x\in X$.

You certainly define $\lim_{n\to\infty} x_n = x$ in $(X,d)$ by the condition

$(1) \phantom{x}$ For all $\epsilon > 0$ there exists $n_0$ such that $d(x_n,x) < \epsilon$ for all $n \ge n_0$.

Let us show that this condition is equivalent to

$(2) \phantom{x}$ For all open neigborhoods $U$ of $x$ in $(X,d)$ there exists $n_0$ such that $x_n \in U$ for all $n \ge n_0$.

An open neigborhood of $x$ is simply an open set $U$ in $(X,d)$ such that $x \in U$. It is easy to see that all metric balls $B(\xi,\epsilon)$ are open (triangle inequality!). Moreover we clearly have $d(x_n,x) < \epsilon$  if and only $x_n \in B(x, \epsilon)$. Therefore $(2)$ implies $(1)$ because each $B(x,\epsilon)$ is an open neigborhood of $x$. For the reverse implication it suffices to observe that each open neighborhood $U$ of $x$ contains some metric ball $B(x,\epsilon)$.

Using $(2)$ it is clear that if $(X,d)$ and $(X,p)$ have the same open sets, then $(*)$ is satisfied, i.e. $d \sim p$. The converse has already been proved by you.

$d \approx p$ means that the metrics are uniformly equivalent. Clearly uniformly equivalent implies equivalent, but the converse is not true. As an example take $X = \mathbb R$ and $d(x,y) = \lvert x - y \rvert$. Now define $p(x,y) = \min (d(x,y),1)$. it is easy to verify that $p$ is again a metric on $\mathbb R$. Clearly $(*)$ is satisfied, thus $d \sim p$. But $d \approx p$ fails; otherwise there would exist $m > 0$ such $md(x,y) \le p(x,y) \le 1$ for all $x,y$, i.e. $d(x,y) \le 1/m$, which is not true.


问:
Group with exactly two subgroups of index 2

I am looking for a group $G$ such that $G$ has exactly two subgroups of index 2.

I have searched by GAP but I couldn't find it.

答:
There is a very nice result proved by R.R. Crawford and K.D. Wallace, On the number of subgroups of index 2 - an application of Goursat's Theorem for groups, Math. Magazine, 48, no. 3 (1975). Let $I_2(G)=\#\{H \lt G: |G:H|=2\}$.

Theorem Let $n$ be a non-negative integer. Then there exists a group $G$ with $I_2(G)=n$ if and only if $n=2^k-1$ for some non-negative integer $k$.

The theorem can be generalized as follows. Let $p$ be a prime and $N_p(G)=\#\{H \lhd G: |G:H|=p\}$.

Theorem Let $n$ be a non-negative integer. Then there exists a group $G$ with $N_p(G)=n$ if and only if $n=\frac{p^k-1}{p-1}$ for some non-negative integer $k$.


问:
what is $\mathbb{Q}_p^\times/\mathbb{Q}_p^{\times^2}$ called?
I have to do a presentation about $p$-adic numbers and I don't know what to call $\mathbb{Q}_p^\times/\mathbb{Q}_p^{\times^2}$. Can you help me, please?

Thank you!

答: For an arbitrary field $K$, a name used for $K^\times/(K^\times)^2$ is the group of "square classes" (understood not to include $0$).  It is a bit more specific than the more general term "quotient group".  For example, $\mathbf R$ has two square classes while $\mathbf Q_p$ has four square classes when $p \not= 2$ and eight square classes when $p = 2$. The field $\mathbf Q$ has infinitely many square classes.


问:
Multiplying Binomial Terms
$(x+2)(x-1$)

Totally forgot how to do it.
Isn't this the rule:

$(a+b)(c+d)$

$a*c +a*d +b*c +b*d$

The answer I'm getting is $x^2-x+2x-2$ which is wrong.
Please help me out, thanks!

答:
The FOIL "rule" (First, Outer, Inner, Last) simply follows from the distributive laws:

$$k(m+n) = km + kn$$

$$(k+m)n = kn + mn$$

In your case,

$$
\begin{align}
(x+2)(x-1) = x(x-1) + 2(x-1) = x^2 - x + 2x - 2 = x^2 + x - 2,
\end{align}
$$

as you have already noted.

Title: Multiplying Binomial Terms


问:
Image of derivative mapping applied to polynomials of degree at most $n$

Let $P_n$ be the set of polynomials of degree at most $n$ equipped with addition and scalar multiplication. Let $D : P_n \to P_n$ be the derivative mapping. Show that $\forall n \in \mathbb{N}$ $\text{im}D = P_{n-1}$.

Intuitively makes sense since if $p(x) = a_nx^n + \cdots + a_1x +a_0$ then applying the derivative operator to $p(x)$ I have that $\deg(Dp) = n-1$ right? Any hints on how to show this rigorously?

答: You showed that $\operatorname{im} D \subseteq P_{n-1}$. To complete the proof, given a polynomial $q$ of degree at most $n-1$, can you come up with a polynomial $p$ of degree at most $n$ such that $Dp = q$?


问:
Can a generator of the ring of integers of local fields can be chosen so that it is also a uniformizer at the same time?
Let $L/K$ be an extension of local fields. We can find $\alpha$ such that $\mathcal{O}_L=\mathcal{O}_K[\alpha]$. What do we know about this generating element? I think that this $\alpha$ can be selected in such a way that, in addition to the above property, it is also a uniformizing parameter at the same time. But I can not prove it.

答:
If you look at a proof of that $\mathcal O_L = \mathcal O_K[\alpha]$ for some $\alpha$, such as Prop. 3, Chapter III, Section 1 in Lang's "Algebraic Number Theory" then you should be able to see that you can take as $\alpha$ either $\zeta$ or $\zeta + \pi$ where $\zeta$ is an arbitrary generator (modulo the maximal ideal of $\mathcal O_L$) of the residue field extension and $\pi$ is an arbitrary choice of uniformizer.  When $L/K$ is unramified we can take as $\pi$ a prime from $K$ and thus $\mathcal O_L = \mathcal O_K[\zeta + \pi] = \mathcal O_K[\zeta]$. When $L/K$ is totally ramified, we can use $\zeta = 1$ (or $\zeta = 0$), so $\mathcal O_L = \mathcal O_K[\pi]$.

Lang proves the result for an extension of DVRs with a separable residue field extension; no assumption of completeness, so in particular no use of Hensel's lemma.  But the assumption that the DVR downstairs has an integral closure in the extension of its fraction field that is also a DVR is not at all a typical situation in a "global" situation such as the localization of a Dedekind domain, as there can be more than one prime upstairs lying over the prime downstairs.  In the complete setting things are nicer since every valuation ring of a local field is a DVR.


问:
finding the limit $\lim_{x \to 0}[ \lim_{ y \to 0}x\sin\frac{1}{y}]$
limit  $$\lim_{x \to 0}[ \lim_{ y \to 0}f(x,y)]$$

Given :
$$f(x,y) =
\left\{
 \begin{array}{ll}
  x\sin\frac{1}{y}  & \mbox{if } y \ne 0 \\
  0 & \mbox{if } y = 0
 \end{array}
\right.$$
this would mean since: $y\neq0$

we need to find :

$$ \lim_{x\to 0}[\lim_{y\to0}x\sin\frac{1}{y}]$$

My approach was let the limit be L then it's clear that :
$$ \lim_{x\to 0} -x \geq L \geq \lim_{x\to 0} x$$

So the limit must be 0.

But it's given to be not defined.

答:
The problem with the approach you took, is that you suppose that the limit exists by using the letter $L$ to designate it. But for $x \ne 0$ given,

$$\lim_{y\to 0}x\sin\frac{1}{y}$$ doesn't exist as in any neighborhood of $0$ the map $y \mapsto x\sin\frac{1}{y}$ takes both values $x, -x$. Therefore $\lim_{y\to 0}x\sin\frac{1}{y}$ is not defined and consequently $ \lim_{x\to 0}[\lim_{y\to 0}x\sin\frac{1}{y}]$ isn't either.

However:

the map $f(x,y)$ is continuous at $(0,0)$ as $\vert f(x,y) \vert \le
   \vert x \vert$ for all $(x,y) \in \mathbb R^2$.

and is not continuous at any $(x,0)$ when $x \neq 0$.


问:
Proof of -(-v)=v in a vector space
The exercise is: Prove that $-(-v)=v$ for every $v \in V$

Proof

Suppose  $v \in V$ and $V$ is a vector space.

Then $-(-v) \in V$ as result of the scalar multiplication property and

$-(-v)=-(-1\cdot v)=-1 \cdot(-1\cdot v) =(-1 \cdot-1)\cdot v = 1 \cdot v = v $

The desired result $-(-v)=v$ holds.

The solution manual gives the proof:

Proof

I just wanted to make sure that the way I did the proof isn't missing anything?Thanks

答:
Your proof is fine. Although this proof is a long winded version of the proof in the solutions manual, it is a little easier to grasp.

I like your proof, but to be pedantic, one would have to prove the commutativity of scalar multiplication with multiple scalars to a vector for your third equality. This is not presented in the book. So limiting ourselves to a subset of all the axioms and theorems until the point this problem shows up, I believe this is a not so elegant but straightforward proof.

Let $$ -v = w.\tag{1}\label{1}$$
Then we can write $-(-v)=-w$.
Since $-w$ is an additive inverse for $w$ we can write
$$w + (-w) = 0 .\tag{2}\label{2}$$We can also write $$v + (-v) = 0.\tag{3}\label{3}$$ from $\eqref{1}$ and $\eqref{3}$ it is clear that $$v+w=0.\tag{4}\label{4}$$ which means that $$v = -w.$$ Substituting in the value of $w=-v$ we get $$v=-(-v)$$ which concludes this proof.

Title: Proof of -(-v)=v in a vector space


问:
Do the adjoint functor theorems usefully dualise?
The special and general adjoint functor theorems exist to construct left adjoints to particular functors given certain conditions on them. However, I've not been able to find much mention – at least, not in my lecture notes nor in a quick Google search – of using them or something like them to construct right adjoints.

I suppose that if $F \dashv G$ then $G^\mathrm{op} \dashv F^\mathrm{op}$, so I can just apply the adjoint functor theorems to the opposites of the categories I'm interested in. Perhaps something like:

General Adjoint Functor Cotheorem: suppose $\mathcal C$ is locally small and cocomplete. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint if and only if it preserves small colimits and, for each object $A$ of $\mathcal C$, the comma category $G \downarrow A$ has a weakly terminal set.

Special Adjoint Functor Cotheorem: suppose $\mathcal C, \mathcal D$ both locally small, $\mathcal C$ cocomplete, co-well-powered1, and with a separating set. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint iff $F$ preserves small colimits.

Is that right? Is it useful?

1 Not sure if this is standard terminology, but I hope it's clear what I mean: for each object $A$, there is a category of epimorphisms with domain $A$, and a category would be co-well-powered if all these epimorphism categories were equivalent to a partially ordered set (they are already preorders, since if $fg = h$ with $g$ and $h$ epimorphisms, then $f$ is unique)

答:
The dualizations are correct and well-known at least for category theorists. Note that the assumptions in Co-Saft are clearly satisfied if $\mathcal{C}$ is presentable. Thus, left adjoint functors $\mathcal{C} \to \mathcal{D}$ coincide with cocontinuous functors $\mathcal{C} \to \mathcal{D}$. This is a basic and important fact of presentable categories. Here is a nice application: Every presentable monoidal category (i.e. a monoidal category whose underlying category is presentable, and the tensor product should be cocontinuous in each variable) is automatically closed monoidal. In other words, as soon we have a tensor product, we get inner homs for free. In practice, often it is vice versa, but not always.

In my work cocomplete monoidal categories appear where I still have no idea how to describe the inner homs. Nevertheless, they are useful, for example in order to apply Kock's results about monoidal monads. More specifically, if we have a monoidal monad on a cocomplete monoidal category, we can quite easily write down a tensor product on the category of modules, but the proof of associativity is not trivial and actually it seems that we have to impose additional assumptions such as closedness (see     arXiv:1205.0101 for details). You surely have seen this for tensor products of modules over a ring in the usual sense: The proof for $M \otimes_R (N \otimes_R K) \cong(M \otimes_R N) \otimes_R K$ only using the definition of the tensor product as a classifying object for bilinear maps is a bit tricky, because first we have to fix some element of $M$ etc, and this implicitly uses the closed structure.

Here is another application: Let $f : X \to Y$ be a morphism of schemes. Then $f^* : \mathsf{Mod}(Y) \to \mathsf{Mod}(X)$ has a right adjoint $f_*$. The functor restricted to quasi-coherent modules $f^* : \mathsf{Qcoh}(Y) \to \mathsf{Qcoh}(X)$ also has a right adjoint. When $f_*(\mathsf{Qcoh}(X)) \subseteq \mathsf{Qcoh}(Y)$, the right adjoint is $f_*$. This happens, for example, when $f$ is quasi-compact and quasi-separated. But even when this is not the case, $f^*$ has a right adjoint, since it is cocontinuous and $\mathsf{Qcoh}(Y)$ is presentable by a result of Ofer Gabber. On the other hand, it is not so easy to write down the right adjoint and probably it's not really useful. Related to that: The inclusion $\mathsf{Qcoh}(X) \hookrightarrow \mathsf{Mod}(X)$ is cocontinuous, hence has a right adjoint. This means that every sheaf of modules has a "quasi-coherator". For example, if $X$ is affine, then the quasi-coherator of $M \in \mathsf{Mod}(X)$ is $\widetilde{\Gamma(X,\mathcal{O}_X)}$. See here for more about that.

If $f : X \to Y$ is a morphism of schemes, which is affine, then $f_* : \mathsf{Qcoh}(X) \to \mathsf{Qcoh}(Y)$ is cocontinuous (actually the converse is also true by a classical result of Serre, generalized by Rydh), hence has a right adjoint. But it seems that you can only write it down when $f$ is finite. Then the right adjoint is $f^{!}(M):=\underline{\hom}_{\mathcal{O}_Y}(f_* \mathcal{O}_X,M)$. If $f$ is not finite, we have to take the quasi-coherator of $f^!(M)$.

It seems that in each application of Co-Saft, the right adjoint is either "pathological" or we can write it down directly anyway and showing the adjunction directly.


问:
Extracting function from sensor
and sorry if I am on the wrong stackexchange.

I have a problem. I have collected data from a distance sensor and extracted its function with respect to temperature. The data extracted proved that the sensor is completely linear (mx + b).

The problem is that at different distances, the m and b coefficiente of the lines change. Of course b will change, but my problem is with the m coefficient. The slope coefficient changes according to the real distance and as stated above the distance read changes according to temperature in a very linear manner.

How would I go about extracting this behaviour and putting it into a function.

EDIT:

Ok. I was not very clear, so I'll try to add more information.

I noticed that my sensor was behaving very strangely during hot days. This is normal for well ... everything, however this specific distance sensor I'm using is incredibly more sensitive to temperature.

With that problem I tried analyzing my sensor's behaviour over temperature and noticed that the output distance changes with temperature in a line (mx + b). This behaviour repeats itself for several distances. I noticed that m and b coefficients change according to its actual distance (the m coefficient has the Ae^(-Db)+c, where D is the real distance).

So, distance measured changes with temperature, slope (m) coefficient changes with real distance, as well as c.

Since I can't actually know the real distance and my code has to give ma that information, I was trying to regress a function that could give me this information.

However, if I try to regress it according to mx+b, it will work correctly over a small distance range, and I can't use real distance (I think) for the m and c coefficients because I don't know it. How would I go about measuring the distance using only the temperature and given distance by the sensor? Is it possible?

Below it's possible to see what I'm talking about. The x axis is temperature, the y axis is measured distance and the different lines are the lines at 5 different distances (5, 4, 3, 2 and 1 m)

答:
PRELIMINARY COMMENT :

The data is not given on the form of numerical table. A "subsitute" of data was made thanks to scanning the graph provided by the OP. This isn't an accurate way. As a consequence the below numerical results are more or less biaised.

ANSWER TO THE QUESTION :

As far as I can understand the question, it seems that it is a simple problem of fitting a function $y(x,D)$ of two variables. $x$ is the variable on the horizontal axis of the graph. $D$ is the so called "Distance". $y$ is the value of the function on the vertical axis.

A very simple function can be :
$$y(x,D)\simeq (a+bD)x+(p+qD)$$
$a,b,p,q$ are the parameters which can be approximately computed thank to a classical linear regression.

The result is (The fitted function is represented by the black lines)  :

Of course a much better fitting can be obtained with more terms in the polynomial function, for example :

$$y(x,D)\simeq (a+bD+cD^2+dD^3)x+(p+qD+rD^2+sD^3)$$

$a,b,c,d,p,q,r,s$ are the parameters which can be approximately computed thank to a classical linear regression.

The result is (The fitted function is represented by the black lines)  :

Instead of polynomial function one could use other kind of functions $f(D)$ and $g(D)$ :
$$y(x,D)=f(D)x+g(D)$$
For example those functions might include exponentials. But the regression would be non-linear which requires more complicated and difficult methods generally iterative.

Title: Extracting function from sensor


答:
For each polynomial $p(x)$ of $P_n$ which has degree $d\geq 1$, $D(p(x)$ yields a polynomial of degree $d-1$. Moreover, if $p(x)$ is a constant and so has degree $0$, then $D(p(x))=0$.
Thus the mapping $D:P_n\rightarrow P_{n-1}$ is well-defined.

Conversely, if $p(x)=a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ is in $P_{n-1}$, then a $D$-preimage is

$q(x)= \frac{a_{n-1}}{n}x^{n}+\frac{a_{n-2}}{n-1}x^{n-1}\ldots+\frac{a_1}{2}x^2+a_0x + c$

with arbitrary scalar $c$.

Hence,  $D$ is surjective.

Title: Image of derivative mapping applied to polynomials of degree at most $n$

Body:
Let $P_n$ be the set of polynomials of degree at most $n$ equipped with addition and scalar multiplication. Let $D : P_n \to P_n$ be the derivative mapping. Show that $\forall n \in \mathbb{N}$ $\text{im}D = P_{n-1}$.

Intuitively makes sense since if $p(x) = a_nx^n + \cdots + a_1x +a_0$ then applying the derivative operator to $p(x)$ I have that $\deg(Dp) = n-1$ right? Any hints on how to show this rigorously?


问:
Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$
Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}, \forall$ n, k nonnegative integers, $k\le n$.

I know that has something to do with Bernoulli's inequality $(1+\alpha)^x\ge 1+\alpha x, \alpha \ge -1, n\ge1$.

If I reconsider Bernoulli's inequality with $\alpha=\frac{1}{n}$ and $x=k$ it follows that

$(1+\frac{1}{n})^k\ge 1+\frac{k}{n}$, but I don't know how to continue.

I also tried to prove it with induction where I consider
$p(n):(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ to be true and prove that $ p(n+1):(1+\frac{1}{n+1})^k < 1+ \frac{k}{n+1}+\frac{k^2}{(n+1)^2}$ to be also true, but it didn't work.

Thank you!

答:
Fix $n\in\Bbb N$. The claim is trivial in the case $k=1$, so by induction assume the claim is true for some $k<n$. Then we get:
\begin{align*}
\left(1+\frac{1}{n}\right)^{k+1}&<\left(1+\frac{k}{n}+\frac{k^2}{n^2}\right)\left(1+\frac{1}{n}\right)\\
&=1+\frac{k}{n}+\frac{k^2}{n^2}+\frac{1}{n}+\frac{k}{n^2}+\frac{k^2}{n^3}\\
&\overset{k<n}{<}1+\frac{k+1}{n}+\frac{k^2+k}{n^2}+\frac{nk}{n^3}\\
&=1+\frac{k+1}{n}+\frac{k^2+2k}{n^2}\\
&<1+{\frac{k+1}{n}}+\frac{k^2+2k+1}{n^2}\\
&=1+\frac{k+1}{n}+\frac{(k+1)^2}{n^2}
\end{align*}
which proves the claim the case $k+1\leq n$. (Actually we see from this proof that the statement is also correct for $k=n+1$.)


问:
Cokernel within the Category of Groups
This question is in regards to Aluffi's Algebra: Chapter $0$, II.$8.22$

$\textbf{8.22: }$Let $\varphi: G \rightarrow G'$ be a group homomorphism, and let $N$ be the smallest normal subgroup containing im $\varphi$. Prove that $G'/N$ satisfies the universal property of ${\rm coker}~\varphi$ in $\textbf{Grp}$.

My scratch work/thinking:

For $G'/N$ to satisfy the universal property of ${\rm coker}~\varphi$ there must be a group homomorphism
\begin{equation*}
\pi: G' \rightarrow G'/N
\end{equation*}
which is initial with respect to all morphisms $\alpha: G' \rightarrow L$ such that $\alpha \circ \varphi = 0$ . Thus, it follows that im $\varphi \subseteq \ker \alpha$. Now, this is where I get lost.

I think I understand that due to the universal property of quotients with respect to group homomorphisms results in the unique induced group homomorphism $\widetilde{\alpha}: G'/N \rightarrow L$. Does the 'initial' role come from $N$ being the smallest normal subgroup which contains ${\rm im}~\varphi$?

Side question: Precisely what does it mean for every homomrphism $\alpha: G' \rightarrow L$, such that $\alpha \circ \varphi$ is the trivial map, must factor through $G'/N$?

答:
The answer is straightforward if you are familiar with Theorem 7.12 in Aluffi, which is stated as:

Let $H$ be a normal subgroup of group G, and $\pi: G \to G/H$ be defined as $\pi(g) = gH$.  Then for every group homomorphism $\varphi : G \to G'$ such that $H \subseteq \ker\varphi$ there exists a unique group homomorphism $\tilde{\varphi}:G/H \to G'$ so that $\tilde{\varphi} \circ \pi = \varphi$.

The requirement that $H \subseteq \ker\varphi$ makes sure $\varphi$ sends related elements in G to the same element in $G'$, then universal property of quotient is applied to prove the theorem. For this problem, let $\alpha:G' \to L$ be any homomorphism from $G'$ to arbitrary group $L$ such that $\alpha \circ \varphi = 0$. Then $\ker\alpha \supseteq \mathrm{im}\varphi$.

By Theorem 7.12, to show there exists a unique homomorphism $\bar{\alpha}:G'/N \to L $ such that $\bar{\alpha} \circ \pi = \alpha$, we only need to show $N \subseteq \ker\alpha$ . But $\ker\alpha$ is a normal subgroup of $G'$ cantaining $\mathrm{im}\varphi$, and $N$, by definition, is the smallest subgroup of $G'$ containing $\mathrm{im}\varphi$. Thus $N \subseteq \ker\alpha$, and we're done.

Title: Cokernel within the Category of Groups


问:
Question about the terminology "topological invariant" in Engelking's General Topology book
In Engelking's General topology book, (2nd page 34) he states as follows: After some paraphrasing,

Let $\mathcal M$ be a class of continuous mappings. A property $P$ of a topological space is an invariant of class $\mathcal M$ if $P$ is preserved by surjective mappings in the class $\mathcal M$.

On the other hand, Wikipedia article considers topological invariant as a synonym for topological property. However, not all continuous surjective maps are homeomorphisms, so according to Engelking's definition, $P$ being a topological property is a stronger condition.

My question is, are there any standard notation? or terminology on this? Any comments are appreciated.

答:
Invariance is a property of a kind of mapping: a (topological) property can be preserved by $\mathcal{M}$ as the "continuous maps", "open continuous maps" and "closed continuous maps" or "perfect maps" or "quotient maps" and Engelking proves a lot of theorems in his text about properties being an invariant or (more rarely) an inverse invariant of such classes of maps. "Topological property" he uses just as Wikipedia does: an invariant of homeomorphisms, so something a space (or really a class of spaces) can have. If $P$ is a topological property it need not be an invariant of any of the above-mentioned classes of maps, except (by definition) homeomorphisms. All properties of spaces discussed in his book are topological properties and he sometimes explicitly proves it (e.g. after defining a Čech-complete space or a realcompact space, where this is not as immediate..). Then he often devotes times to study different (inverse) invariances under mappings and also closedness of the property when taking constructions like products, subspaces, sums, quotients etc., also discussing some (counter)examples. It's quite systematic that way.

To me the invariance (under maps) idea is a lens or guide to look at results to prove or disprove.


答:
As a hint: exp(x) = 1 + x + x^2/2 + x^3/6 ... The right hand side is 1 + x + x^2, cutting the series short, but not dividing the quadratic term by 2. So for small k not dividing the quadratic term on the right side will make it larger for small k, but for large k adding the missing terms on the left side makes it larger. Try with n = 1 and k = 1, 2, 3, etc.

I’d write down all the terms for fixed n and a given k on the left side, and show that the higher powers are less than the extra k^2/2n^2 on the right hand side, as long as k isn’t too large.

Title: Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$

Body:
Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}, \forall$ n, k nonnegative integers, $k\le n$.

I know that has something to do with Bernoulli's inequality $(1+\alpha)^x\ge 1+\alpha x, \alpha \ge -1, n\ge1$.

If I reconsider Bernoulli's inequality with $\alpha=\frac{1}{n}$ and $x=k$ it follows that

$(1+\frac{1}{n})^k\ge 1+\frac{k}{n}$, but I don't know how to continue.

I also tried to prove it with induction where I consider
$p(n):(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ to be true and prove that $ p(n+1):(1+\frac{1}{n+1})^k < 1+ \frac{k}{n+1}+\frac{k^2}{(n+1)^2}$ to be also true, but it didn't work.

Thank you!


问:
The relationship between regular distribution and tensor product of distribution?
For a regular distribution $T_f$, where $f$ is defined on $\mathbb R^n$,
$$
\langle T_f, \varphi\rangle := \int_{\mathbb R^n}  f(x) \varphi(x) dx.
$$
The tensor product of two distributions $S$ and $T$ is definded by
$$
\langle T \otimes S, \varphi \rangle := \langle T, \langle S, \varphi_\cdot \rangle \rangle,
$$
where
$$
\varphi_\cdot(y) = \varphi(\cdot, y).
$$
If $T_f$ and $T_g$ are two regular distributions, where $f$ is defined on $\mathbb R^n$ and $g$ on $\mathbb R^m$, then
$$
\langle T_f \otimes T_g, \varphi \rangle = \int_{\mathbb R^n \times \mathbb R^m} f(x) g(y) \varphi(x, y) dx dy,
$$
so that the tensor product of distribution is a special case of a regular distribution, ie. $T_f \otimes T_g = T_h$, where $h(x,y)=f(x)g(y)$. Is this correct?

答:
This is correct. By the definition of the tensor product of distributions,

$$
\langle T_f \otimes T_g, \varphi \rangle = \langle T_f, x \mapsto \int_{\mathbb R^m} g(y) \varphi(x, y) dy \rangle = \int_{\mathbb R^m} f(x) \int_{\mathbb R^m} g(y) \varphi(x, y) dy dx,
$$
which by Fubini's theorem equals the integral you describe in your question.


问:
Taylor series expansion or polynomial regression which is the correct term here for finding an analytical estimate for an unknown function?
I have a question regarding the Taylor expansion of an unknown function and I would be tanksful to have your comments on that.

Suppose we want to find an analytical estimate for an unknown function. The available information for this function is; its exact value at $x_0$ $(f_0)$ and first derivative at $x_0$ $(f_0')$, and its exact value at $x=x_1$ $(f_1)$ which is far from $x_0$. This means that we only need to find the second derivative through solving $$f_1 = f_0 + f_0'(x_1-x_0) + \frac12 {f_0}''(x_1-x_0)^2$$ to form the truncated Taylor series for $f$.

My question is that whether the obtained quadratic function is the result of Taylor series expansion or polynomial regression? Is this a proper way of approximating a function considering that f"0 is found by matching the function value at $x_1$ which is far from $x_0$?

答:
This is effectively a Taylor expansion built  around $x=x_0$ but if you want a quadratic form for approximation, write
$$f(x)\sim a+ bx +c x^2$$ and solve the three equations to get
$$c=\frac {f_1-f_0}{(x_1-x_0)^2}-\frac {f'_0}{x_1-x_0}$$
$$b=\frac {f_1-f_0}{x_1-x_0}-c(x_0+x_1)$$
$$a=f_0-bx_0-c x_0^2$$

Let us try for $f(x)=e^x$ with $x_0=2$ and $x_1=4$ which give
$$a=e^2 \left(e^2-4\right)\qquad b=-e^2 \left(e^2-4\right)\qquad c=\frac{1}{4} e^2 \left(e^2-3\right)$$ Computing for $x=3$, we get
$$\frac{1}{4} e^2 \left(5+e^2\right) \sim 22.8859 $$ while $e^3=20.0855$.

For this case, it is not fantastic at all but it could be much better with a function growing slower than the exponential.

Since you have now the formulae for $(a,b,c)$, play with any function of your choice and, please, let me know.


问:
confused about Lemma 5.33 (about Galois Theory) in Rotman's textbook

The confusion occurs at very beginning of the proof (highlighted with red rectangle)。

Why are the roots of $x^p-1\in k[x]$ denoted by $1, \omega, \omega^2, \cdots, \omega^{p-1}$? Why they are all distinct? How to use the condition $p=char(k)$?

答:
Hint: $x^p-1 = (x-1)^p$ if the characteristic is $p>0$.

In the case of characteristic $\ne p$, use the derivative: $(x^p-1)' = px^{p-1}\ne 0$.
It has no roots in common with $x^p-1$ and so the roots are distinct.


问:
Why are some functions called 'forms'?
The question is simple: why are some functions called 'forms'? Modular 'forms', bilinear 'forms', differential 'forms', quadratic 'forms', and so forth. It is not concretely a mathematical question but I suppose the origin of the term must be mathematical, so I ask it here.

Thanks for the comment, it seems the question was already answered here:
What do mathematicians mean when they say "form"?

and here's another page: Earliest use of the word "form" in mathematics?

答:
See links: What do mathematicians mean when they say "form"?

I like this one: Earliest use of the word "form" in mathematics?

Link to wiki of Plato's theory of forms: https://en.wikipedia.org/wiki/Theory_of_forms

Title: Why are some functions called 'forms'?


问:
Parametric simultaneous equations
I stumbled on this one a few days ago and I'm probably missing something obvious...

I basically need to solve those parametric equations for the other coordinate $(x,y)$ other than the point $(t^2,t^3)$:

$$2y = 3tx - t^3$$
$$x = t^2$$
$$y = t^3$$

To have more context, the question started with $x = t^2$, $y = t^3$, asking to find the equation of the tangent at the point $(t^2,t^3)$, then to find the other point where this tangent meets the parametric curve.

It just seems that whenever I try something, I get back to $(t^2,t^3)$ instead of the other point. I've been trying with:

$2(t^3) = 3tx - t^3$, then

$2y = 3t(t^2) - t^3$, then

$2\sqrt{x^3} = 3tx - t^3$ (but this one is intimidating for something which should be much simpler than that)

Is there perhaps a way to exclude $(t^2,t^3)$ to get the other solution? The other point should be $\left(\dfrac{t^2}{4},-\dfrac{t^3}{8}\right)$ by the way.

答:
I think your problem is that the $t$ in the first equation should be fixed, whereas the $t$ in the second and third is actually your parameter.

The tangent line to $(t^2,t^3)$ at the point $s$ is indeed given by $(s^2,s^3)+\lambda(2s,3s^2)$ for $\lambda\in\mathbb{R}$.

This line intersects $(t^2,t^3)$ whenever $t^2 = s^2+2\lambda s$ and $t^3=s^3+3\lambda s^2$. First we can eliminate $\lambda$ to obtain $ 2t^3-3st^2+s^3=0$. Since we already know the solution $t=s$, we may factor it out:
$$2t^3-3st^2+s^3 = (t-s)(2t^2-st-s^2)=0.$$

Now we may solve the resulting quadratic formula by completing the square $$2t^2-st-s^2 = 2\left(t-\frac{s}{4}\right)^2-\frac{9}{8}s^2=0.$$
Thus $t = \frac{s}{4} \pm\frac{3}{4}s$. Thus $t=s$ or $t=-\frac{1}{2}s$.

Hence, the non-trivial intersection point is at $\left(\frac{1}{4}s^2,-\frac{1}{8}s^3\right)$.

Title: Parametric simultaneous equations


问:
Joint Conditional PMF: is this calculation correct?
I am currently working on a review problem which I find a little confusing. The problem goes as follows:
Consider two i.i.d random variables $X$ and $Y$ with values in $\mathbb{N}_0$ such that $\forall n \in \mathbb{N}_0$:
$$\mathbb{P}(X = k | X+Y=n)=\binom{n}{k}2^{-n}$$
What are the possible cummulative distributions for $X$?

My attempt at the first step is to take apart the conditional probability and setting $Z = X+Y$:

\begin{align} \mathbb{P}(X = k | Z=n) &= \frac{\mathbb{P}_{XZ}(k,n) }{\mathbb{P}_Z(n)} \\
&= \frac{\mathbb{P}_{X}(k) \mathbb{P}_{Z}(n) }{\mathbb{P}_Z(n)} \\
&= \mathbb{P}_{X}(k)
 \end{align}
I am not sure if this step is correct, because the conditional probaility is over the sum of two random variables. So my question: is what I have done correct, and if not what would be the correct approach?

答:
The given conditional density is a binomial $B\Big(n;\frac{1}{2}\Big)$

To show that the two marginals are  poisson's law: $X\sim Po(\theta)$  for any $\theta>0$ is not difficult...can you procede by yourself or  you need further help?

Assume that

$X\sim Po(\theta)$

$Y\sim Po(\theta)$

$X\perp\!\!\!\!\!\! \perp Y$

It is evident that $(X+Y)\sim Po(2\theta)$

$$\mathbb{P}(X=k|X+Y=n)=\frac{\mathbb{P}(X=k)\cdot \mathbb{P}(Y=n-k)}{\mathbb{P}(X+Y=n)}=\frac{\frac{e^{-\theta}\theta^k}{k!}\cdot \frac{e^{-\theta}\theta^{n-k}}{(n-k)!}}{\frac{e^{-2\theta}(2\theta)^n}{n!}}$$

$$\mathbb{P}(X=k|X+Y=n)=\frac{n!}{k!(n-k)!}2^{-n}=\binom{n}{k}2^{-n}$$

...as stated

That's all


答:
First of all, if $\omega$ is the $p$-th root of unity, then $\omega^i$ is also a root of $x^p-1$, and this is just because $(\omega^i)^p - 1 = 1 - 1 = 0$. Thus $1, \omega, \omega^2, \cdots, \omega^{p-1}$ are all roots of $x^p - 1$.

They are all distinct because if $\omega^i = \omega^j$, then $\omega^{i-j} = 1$, which means $(i-j) | p$, but $p$ is a prime. So, this can only happen if $i-j = 0$, or $p = 0$, i.e. $p = \text{char}(k)$. This is where you need that condition. So $i = j$, and thus all the above roots are distinct.

Title: confused about Lemma 5.33 (about Galois Theory) in Rotman's textbook

Body:
The confusion occurs at very beginning of the proof (highlighted with red rectangle)。

Why are the roots of $x^p-1\in k[x]$ denoted by $1, \omega, \omega^2, \cdots, \omega^{p-1}$? Why they are all distinct? How to use the condition $p=char(k)$?


问:
Is there a nice definition of a homotopy type?
We often talk about specific mathematical objects, even though the way we formalise them in set theory actually means that we're talking about multiple objects. For example "the" complete graph on $n$-nodes is actually many graphs, one for each set of cardinality $n$. We can solve this problem by taking the proper class of all such graphs, but I find proper classes ontologically unconscionable. In general, in this vein, we could talk about the isomorphism classes in any category. We could also take a representative of each class, but this is "cheating", and probably difficult, if not generally impossible, at least in $\textrm{ZFC}$. I suspect that while keeping a set-theoretic foundation, then we have to make due with this multiplicity. However, graphs and most other objects we talk about are structured sets, and usually they do have the nice property that if they are isomorphic, then only the underlying sets differ, in a sense I make explicit in the appendix. On the other hand, homotopy types are not structured sets. A point and a line are radically different objects but they have the same homotopy type. And, due to my objections, we can't define a homotopy type as the equivalence class of all its spaces, since that's a proper class, and we can't just take a representative. So, is there a nicer way to define homotopy types?

Appendix: We take an underlying set $U$, and build another set out of it. A set can be thought of as a rooted tree with no infinite branches (and an asymmetry condition corresponding to extensionality), so we can think of building a new set as taking such a rooted tree $T$ (without the asymmetry condition), and attaching members of the underlying set to it. Explicitly, we take a surjective (partial) function $f:L\to U$ where $L$ is the set of leaves of $T$. We can then think of the set we've built as that obtained by taking copies of the trees representing the members of $U$ and attaching their roots to the corresponding members of $L$. Of course, this tree needs to satisfy the asymmetry condition to represent a set, although it doesn't really matter. An isomorphism between $f:L\to U$ with $L$ the leaves of $T$ and $f':L'\to U'$ with $L'$ the leaves of $T'$ is a rooted tree isomorphism $i:T\to T'$ and a bijection $b:U\to U'$ such that $bf=f'i$. In all examples I can think of, an isomorphism in a category of structured sets (e.g. graphs, topological spaces, groups) is an isomorphism in this sense, or at least there is a way of representing the objects of the category so that it is one.

答:
For homotopy types we can do the following.

A minimal Kan complex is a Kan complex $K$ with the following property:

Given morphisms $x_0, x_1 : \Delta^n \to K$, if $x_0$ and $x_1$ are homotopic relative to their boundary, then $x_0 = x_1$.

We have the following fact:

Proposition. Minimal Kan complexes are homotopy equivalent if and only if they are isomorphic.

Furthermore:

Proposition. For every Kan complex $K$, there is a simplicial subset $K' \subseteq K$ such that $K'$ is a minimal Kan complex and $K' \hookrightarrow K$ is a strong deformation retract.

Corollary. Every homotopy type can be realised by a minimal Kan complex uniquely up to isomorphism.

I don't think this is nice per se, but it is nicer than using Scott's trick on the class of topological spaces modulo weak homotopy equivalence.

Since it was bothering me for a while, let me remark why this does not contradict Freyd's theorem on the non-concreteness of the homotopy category.
Although it is true that every connected component of a minimal Kan complex has a unique vertex, and therefore that homotopic morphisms into a minimal Kan complex agree on vertices, it is not true that homotopic morphisms are equal.
Nonetheless the definition is just enough to ensure that any endomorphism of a minimal Kan complex homotopic to the identity is an isomorphism, which ensures that any homotopy equivalence of minimal Kan complexes is an isomorphism.

In short, the quotient functor from the category of minimal Kan complexes to the homotopy category is full and conservative but not faithful.


问:
Why is this definite integral divergent?
The integral given is

$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx$$

The steps I have taken to solve it are first applying the double angle formula for the cosine function

$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ (2\cos^2(2x)-1)}dx $$
Then simplifying

$$\int^{\frac{\pi}{3}}_0 \frac{5}{2\cos^2(2x)}dx \iff \frac{5}{2}\int^{\frac{\pi}{3}}_0\sec^2(2x) dx$$

$$ \left[\frac{5}{2}\tan(2x) \right]^{\frac{\pi}{3}}_0 = \frac{-5 \sqrt{2}}{4}$$

However trying to verify this on multiple sources such as integral-calculator.com and wolframalpha have returned no value for the integral and state that this is divergent.

Does this integral really have no finite value over this interval? I'm pretty sure I have not made a method mistake so if there is infact no finite solution to this integral  then what does this value of $\frac{-5\sqrt{2}}{4}$ represent?

答:
The fact that a positive integrand appears to have negative value should ring alarm bells.
The differential $\frac{d}{dx} \tan 2x$ does not exist at $x = \pi/4$ so the use of the fundamental theorem of calculus breaks down.  You could divide the integral into two intervals, $[0,\pi/4- \delta_1)$ and $(\pi/4+\delta_2,\pi/3]$ to see then that the value becomes infinite as $\delta_1, \delta_2 \searrow 0$.

Postscript:  The derivative of $\tan 2x$ is $2 \sec^2 2x$ and   $\tan (2\pi/3) =-\sqrt 3.$


问:
Understanding the modal necessitation rule
I understand that the modal necessitation rule is:

$ \vdash p  \Rightarrow  \vdash \Box p $

That is, if p is a theorem then necessarily p is a theorem.

But I don't quite understand what this means, or how I could use it in a modal proof.  Obviously if I derive $x$ on a line I cannot then move to necessarily $x$.  (Just because it's actually raining does not imply that it's raining in all possible worlds.)

Does the rule mean that only if $p$ is a tautology then I can move to necessarily $p$?  Just trying to figure out exactly what is meant by "theorem" and how i can use the necessitation rule in a proof.  Thanks.

答:
Actually, $p$ has to be a tautology.

For the sake of completeness, let's consider an example about how it can be used in a proof.

Assume the following axioms and inference rules:

A1. All tautologies of propositional calculus

A2. $(\square_i \phi \land \square_i (\phi \Rightarrow \psi) \Rightarrow \square_i \psi \hspace{1cm}i=1,\cdots, n$ $\hspace{1cm}$ [Distribution Axiom]

R1. From $\phi$ and $\phi \Rightarrow \psi$ infer $\psi$ $\hspace{4.5cm}$ [MP]

R2. From $\phi$ infer $\square_i \phi$

Theorem:

$\square (p \land q) \Rightarrow \square p$

Proof:

1. $(p \land q) \Rightarrow p$ $\hspace{13.3cm}$ (A1)

2. $\square ((p \land q) \Rightarrow p)$ $\hspace{12.5cm}$ (1,R2)

3. $(\square ((p \land q) \land \square ((p \land q) \Rightarrow p)) \Rightarrow \square p$ $\hspace{8.9cm}$ (A2)

4. $((\square ((p \land q) \land \square ((p \land q) \Rightarrow p)) \Rightarrow \square p) \Rightarrow (\square ((p \land q) \Rightarrow p)\Rightarrow (\square (p \land q) \Rightarrow \square p))$ $\hspace{1.1cm}$ (A1)

5. $\square ((p \land q) \Rightarrow p)\Rightarrow (\square (p \land q) \Rightarrow \square p)$ $\hspace{8.5cm}$ (3,4,R1)

6. $\square (p \land q) \Rightarrow \square p$ $\hspace{12.2cm}$ (2,5,R1)


问:
$m ∈ M$ such that the image in $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$ is not zero
Let $M$ be a finitely generated $A$-module. For every prime ideal $\mathfrak{p}\in \operatorname{Spec}{A}$, define

$$μ(\mathfrak{p},M)=\dim_{k(\mathfrak{p})}(M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}),$$

where $k(\mathfrak{p}) = A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$

Let $S$ be a finite set of prime ideals of $A$ such that $M_{\mathfrak{p}} \ne 0$ for every $\mathfrak{p} \in S$.

My text reads the following equivalence is clear,

For an element $m \in M$, the image of $m$ in $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$ is not zero for all $\mathfrak{p} \in S$ if and only if $μ({\mathfrak{p} }, M/Am)=μ({\mathfrak{p} }, M)-1$ for all $\mathfrak{p}\in S$.

But why? Do I need to calculate $$μ({\mathfrak{p}}, M/Am)$$ directly? Thank you for your help.

lem1.2 of this page 'in other words' the last paragraph of statement lem1.2 is my question.

https://www2.math.upenn.edu/~chai/602-3_2010-11/notes/forster-swan_notes.pdf

答:
Hints: $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$ is a module over $\kappa(\mathfrak{p})$, i.e. a $\kappa(\mathfrak{p})$-vector space.
The image of an element $m\in M$ in this vector space is non-zero if and only if it spans a one dimensional subspace. The claim follows then by playing around with the isomorphism theorems $(A/B)/(C/B)\cong A/C$ and $(T+S)/T\cong S/(S\cap T)$ and using the fact that localization is exact and thus commutes with quotients.

Edit: (more detailed answer)

Let us introduce some notation:

$\langle m\rangle:=Am$ the $A$-submodule of $M$ generated by $m$.

$\langle \frac{m}{1}\rangle$ the $A_{\mathfrak{p}}$-submodule of $M_{\mathfrak{p}}$ generated by the image $\frac{m}{1}$ of $m$ in $M_{\mathfrak{p}}$.

By definition we have

$$ \mu(\mathfrak{p},M/\langle m\rangle)=\dim_{\kappa(\mathfrak{p})}\left(\frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\right). $$

We want this to be equal to $\dim_{\kappa(\mathfrak{p})}(M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}})-1$.
We are assuming that $m$ has non-zero image in the $\kappa(\mathfrak{p})$-vector space $M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}$, which as we already argued in the hints and comments implies that it spans a linear subpsace of dimension $1$.
This linear subspace is the quotient $\langle \frac{m}{1}\rangle /(\mathfrak{p}M_{\mathfrak{p}}\cap \langle \frac{m}{1}\rangle)$.
So it suffices to show that there is an isomorphism

$$ \frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\cong \frac{M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}}{\langle \frac{m}{1}\rangle /(\mathfrak{p}M_{\mathfrak{p}}\cap \langle \frac{m}{1}\rangle)}. $$

Using the isomorphism theorem $(T+S)/T\cong S/(S\cap T)$ we may rewrite the right-hand side as

$$\frac{M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}}}{(\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle)/\mathfrak{p}M_{\mathfrak{p}}}, $$

and after applying the isomorphism theorem $(A/B)/(C/B)\cong A/C$ we may rewrite this last expression as

$$ \frac{M_{\mathfrak{p}}}{\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle}. $$

There are at least two ways to arrive at this expression from the left-hand side.
The two ways that I have now in mind use as a first step that localization commutes with quotients, one of them going "outward" and the other one going "inward".
Going "inward" and using the isomorphism theorems as before we have

$$ \frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\cong \frac{M_{\mathfrak{p}}/\langle \frac{m}{1}\rangle}{\mathfrak{p}M_{\mathfrak{p}}/(\mathfrak{p}M_{\mathfrak{p}}\cap\langle \frac{m}{1}\rangle)}\cong \frac{M_{\mathfrak{p}}/\langle \frac{m}{1}\rangle}{(\mathfrak{p}M_{\mathfrak{p}}+\langle\frac{m}{1}\rangle)/\langle\frac{m}{1}\rangle}\cong \frac{M_{\mathfrak{p}}}{\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle}. $$

And going "outward" and using the isomorphism theorems as before we have

$$ \frac{(M/\langle m\rangle)_{\mathfrak{p}}}{\mathfrak{p}(M/\langle m\rangle )_{\mathfrak{p}}}\cong \left(\frac{M/\langle m\rangle}{\mathfrak{p}M/(\mathfrak{p}M\cap \langle m\rangle)}\right)_{\mathfrak{p}}\cong \left(\frac{M/\langle m\rangle}{(\mathfrak{p}M+\langle m\rangle)/\langle m\rangle}\right)_{\mathfrak{p}}\cong \left(\frac{M}{\mathfrak{p}M+\langle m\rangle}\right)_{\mathfrak{p}}\cong \frac{M_{\mathfrak{p}}}{\mathfrak{p}M_{\mathfrak{p}}+\langle \frac{m}{1}\rangle}. $$


答:
$$I=\int_0^{\pi/3}\frac{5}{1+\cos(4x)}dx=^{u=4x}\frac54\int_0^{4\pi/3}\frac{du}{1+\cos(u)}$$
now the problem is that when $\cos(u)=-1$ the bottom of this fraction is $0$ so your integrand will tend towards infinity. Whilst an antiderivative does exist for this function, the integral of the function cannot have a point of divergence like this within the domain or as can be clearly seen graphically, it will diverge.

Title: Why is this definite integral divergent?

Body:
The integral given is

$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx$$

The steps I have taken to solve it are first applying the double angle formula for the cosine function

$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ (2\cos^2(2x)-1)}dx $$
Then simplifying

$$\int^{\frac{\pi}{3}}_0 \frac{5}{2\cos^2(2x)}dx \iff \frac{5}{2}\int^{\frac{\pi}{3}}_0\sec^2(2x) dx$$

$$ \left[\frac{5}{2}\tan(2x) \right]^{\frac{\pi}{3}}_0 = \frac{-5 \sqrt{2}}{4}$$

However trying to verify this on multiple sources such as integral-calculator.com and wolframalpha have returned no value for the integral and state that this is divergent.

Does this integral really have no finite value over this interval? I'm pretty sure I have not made a method mistake so if there is infact no finite solution to this integral  then what does this value of $\frac{-5\sqrt{2}}{4}$ represent?


问:
$\sigma=(12345)(67)(89)$, is $\sigma^{2002}$ conjugated with $\sigma^4$?
$\sigma=(12345)(67)(89)$, is $\sigma^{2002}$ conjugated with $\sigma^4$ in $S_9$?

I have $\sigma^{2002}=(13524)(67)(89)$ and $\sigma^4=\sigma^{-1}=(54321)(67)(89)$

Hence, they have same cykeltype and are therefore conjugated, is this correct answer?

Body:
As disjoint permutations commute, we have  $$\sigma^{2002}=(1\,2\,3\,4\,5)^{2002\bmod 5}(6\,7)^{2002\bmod 2}(8\,9)^{2002\bmod 2}=(1\,2\,3\,4\,5)^{2}=(1\,3\,5\,2\,4)$$
whereas
$$\sigma^4=(1\,2\,3\,4\,5)^{4}=(1\,2\,3\,4\,5)^{-1}=(1\,5\,4\,3\,2),$$
and it is easy to check that
$$(2\,4)(3\,5)\sigma^4(2\,4)(3\,5)=\sigma^2.$$


问:
How many equivalence classes are there
$E={1,2,3,4,5,6,7,8}$
Defines the product set E × E the relation R: $(p, q) R (p_0, q_0) $if $ p-p_0$ even and $q-q_0 $divisible by 3

Question :How many equivalence classes are there

My attempt :

$p-p_0$is even $\Rightarrow$$p-p_0=2k$$\Rightarrow$$p=p_0[2]$
So for that exist two equivalent classes 0 and 1

$q-q_0$is divisible by 3 $\Rightarrow$$q-q_0=3k$$\Rightarrow$$q=q_0[3]$
So for that exist  three equivalent classes 0 and 1 and 2

So  finally :

exist 6 equivalent classes :(1, 0),(1,1),(1,2),(0,0)(0,1)(0,2)$

Is that correct?

答:
As JMoravitz commented, $0$ is not an element of your set.

So you seem to change notation between the question and your tentative of answering it.

To keep it tidy, I'll reformulate the question.

We have a set $E = \{ 1, 2, 3, 4, 5, 6, 7, 8\}$ and a binary relation $R$ defined on $E\times E$ by
$$(p,q) R (r,s) \quad\text{ iff }\quad 2|p-r \;\text{ and }\; 3|q-s.$$
Apparently, you already concluded that $R$ is an equivalence relation on $E\times E$, and just want to know what are the equivalence classes.

These are the following $6$:
$$(1,1)/R, (1,2)/R, (1,3)/R, (2,1)/R, (2,2)/R, (2,3)/R.$$
(Here, I'm using the notation in which, for an equivalence relation $\sim$ on a set $X$, we denote by $x/{\sim}$ the equivalence class of $x \in X$; another common notation would be $[x]_{\sim}$, but I'll use the previous one.)

To show that those are exactly the equivalence classes, let us start by describe the class of the element $(1,1)$, that is, to list the elements which are $R$-related with $(1,1)$.

Given the definition, these are the elements whose first coordinate is odd (so that the difference with $1$ is even) and the second coordinate is $1$, $4$ or $7$ (so that subtracting $1$ we get a multiple of $3$).

Hence
$$(1,1)/R =
\{ (1,1), (1,4), (1,7), (3,1), (3,4), (3,7), (5,1), (5,4), (5,7), (7,1), (7,4), (7,7) \}.$$
Analogously you can conclude that
\begin{align}
(1,2)/R
&= \{ (1,2), (1,5), (1,8), (3,2), (3,5), (3,8), (5,2), (5,5), (5,8), (7,2), (7,5), (7,8) \},\\
(1,3)/R
&= \{ (1,3), (1,6), (3,3), (3,6), (5,3), (5,6), (7,3), (7,6) \},\\
(2,1)/R
&= \{ (2,1), (2,4), (2,7), (4,1), (4,4), (4,7), (6,1), (6,4), (6,7), (8,1), (8,4), (8,7) \},\\
(2,2)/R
&= \{ (2,2), (2,5), (2,8), (4,2), (4,5), (4,8), (6,2), (6,5), (6,8), (8,2), (8,5), (8,8) \},\\
(2,3)/R
&= \{ (2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (8,3), (8,6) \}.
\end{align}
Now, just to confirm, you can count that there are $64$ elements listed in those classes (without repetitions), and since that is the same number of elements in $E \times E$, we know that we didn't forget any one.

(These means that you are right, if we increment by one each coordinate of each pair in your list of equivalence classes.
Notice, however, that even so, there would be a problem in your answer:
for example $(1,1)$ is not an equivalence class; what is correct is that $(1,1)/R$ is an equivalence class.)

Title: How many equivalence classes are there


答:
You properly found that
$$\int\frac{5}{1+ \cos(4x)}dx=\frac{5}{4} \tan (2 x)$$ Now, write
$$I=\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx=\int^{\frac{\pi}{4}-\epsilon}_0 \frac{5}{1+ \cos(4x)}dx+\int_{\frac{\pi}{4}+\epsilon}^{\frac \pi3} \frac{5}{1+ \cos(4x)}dx$$ $$I=\frac{5}{4} \cot (2 \epsilon )+\frac{5}{4} \cot (2 \epsilon )-\frac{5 \sqrt{3}}{4}=\frac{5}{2} \cot (2 \epsilon )-\frac{5 \sqrt{3}}{4}$$

Now, what happens when $\epsilon \to 0$ ?


问:
Can I Write It As An Integral?
Is there a way to write $$\Pi _{i=1}^n (i)$$ as an integral?

[ $\because \Pi (x) = \Gamma (x+1) $ and $\Gamma (x+1) $can be written in form of an integral , that's why i am asking, can we write above equation also? If yes, how?]

答:
We have that for $n \in \mathbb{N}$,
$$P_n=\prod_{i=1}^ni=n!=\Gamma(n+1)$$
and since for $x > 0$
$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$$
we also can say that
$$\prod_{i=1}^n i=\int_0^\infty t^ne^{-t}dt$$

Title: Can I Write It As An Integral?


问:
A few questions on an application of the residue theorem
I have got the following closed curve $\gamma$

and I am supposed to calculate
\begin{equation}
\int_\gamma\frac{\cos(z)}{z^3(z^2 +1)}dz
\end{equation}

with the help of the residue theorem.
Now, I've got a few questions

$0$, i and -i are the isolated singularities of our given function. As far as I understand it, the winding number of $0$ is $-2$, the winding number of $-$i is $-1$ and the winding number of i is $0$. Is this correct?

How can I determine the residue in this case? For a start, I tried finding the Laurent expansion for $z_0=0$, but I have no idea how to rearrange the equation further. I only got as far as
\begin{equation}
\frac{\cos(z)}{z^3(z^2 +1)}=\frac{1}{z^2+1}\sum\limits_{n=0}^{\infty}(-1)^n\cdot\frac{z^{2n-3}}{(2n)!}.
\end{equation}
Could someone give me a hint on how to find the Laurent expansion?

Thanks in advance!

答:
You are right.

Let $f(z)=\frac{\cos(z)}{z^2+1}$. Then\begin{align}\operatorname{res}_{z=0}\left(\frac{\cos(z)}{z^3(z^2+1)}\right)&=\operatorname{res}_{z=0}\left(\frac{f(z)}{z^3}\right)\\&=\frac{f''(0)}{2!}\\&=-\frac32.\end{align}


问:
How do you find inflection points
$$f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=x^2e^{-x^2}$$

The second derivative is $f''(x)=e^{-x^2}(4x^4-10x^2+2)$.

So I need the solutions for $f''(x)=0$.
I tried to notate $x^2=t$ but i don't know if it's ok..

Any ideas? Please..

答:
$f''(x) = 0$ iff $(4x^4 - 10x^2 +2) = 0$ (as $e^{-x^2}>0$ always).

To solve the quartic we can indeed set $t=x^2$ and we need to solve

$(4t^2 - 10t + 2) =0$ and we only need to consider $t\ge 0$.

The discriminant is $100 - 4\times 4 \times 2= 68 >  0$, so solutions exist.. Find $t\ge 0$ and thus $x$..

Title: How do you find inflection points


问:
Trigonometry Airplane question. Finding bearing and distance.

A little background(if you don't care for my story, skip straight to the question): I've missed a few lectures from my teacher because I fell ill. Since I have no information to work with other than the bad notes a friend of mine has taken, I don't have much to work with. Now I'd love help with this problem because I've been at it for almost $3$ hours and nothing I try works. I don't need the answer at all, I have the answer, what I need is the process. I need to know how to get that answer.

**Now the question is-**An airplane travels due west for $1.5$ hours at $320$ mph. then it changes course to $S31°W$. Fine the airplane's distance from its point of departure and its bearing, after a  total flight time of $3.5$ hours.

What I know so far, sorry I wasn't as detailed as I should be. I have drawn it out, i know distance between from the starting point and when it changes direction being $480$. And the distance of the direction change to the end point which is $640$. I know the angle where the plane changes direction totals out to $121°$. I could be wrong with this information, but the point is I've taken many different approaches at this problem but have gotten no correct answers. I've taken it "outside of the box" and expanded the 480 line until i can drop it down to the end point and make the spot where those two lines meet a right angle, if that makes sense. I've used "sohcahtoa" and found out the lengths of that small right triangle, added it with the lengths of the large obtuse triangle to put it all together. The missing side is the hypotenuse and I used the Pythagorean theorem to find it, but the answer comes out wrong. I tried so many different values and processes, and nothing. For the missing angle I use the inverse of the tan function and got an answer for that, but it's also wrong. I'd appreciate any input, anything at all.

答:
Cosine Law would be easier, but let's do it your way. Consider the small right triangle (call it $\Delta ABC$), which roughly looks like:

```<code>A----B
|   /
|  /
| /
|/
C
</code>```

Note that $\angle ABC =59^\circ$. Thus, $AB=640\cos(59^\circ)$ and $AC=640\sin(59^\circ)$.

Now consider the large right triangle. Its two legs are $AC$ and $AB+480$. Hence, by the Pythagorean Theorem, we have:
$$x=\sqrt{(AC)^2 + (AB+480)^2}=\sqrt{(640\cos(59^\circ)+480)^2 + (640\sin(59^\circ))^2} \approx 978 \text{ miles}$$
Finally, for the angle bearing:
$$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$

Body:
Cosine Law would be easier, but let's do it your way. Consider the small right triangle (call it $\Delta ABC$), which roughly looks like:

A----B
|   /
|  /
| /
|/
C

Note that $\angle ABC =59^\circ$. Thus, $AB=640\cos(59^\circ)$ and $AC=640\sin(59^\circ)$.

Now consider the large right triangle. Its two legs are $AC$ and $AB+480$. Hence, by the Pythagorean Theorem, we have:
$$x=\sqrt{(AC)^2 + (AB+480)^2}=\sqrt{(640\cos(59^\circ)+480)^2 + (640\sin(59^\circ))^2} \approx 978 \text{ miles}$$
Finally, for the angle bearing:
$$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$


问:
Pole order of $\frac{4}{27}\frac{\left(\lambda^2-\lambda+1\right)^3}{\lambda^2\left(1-\lambda\right)^2}\left(=j(\tau)\right)$
Concerning the relation
$$j=\frac{4}{27}\frac{\left(\lambda^2-\lambda+1\right)^3}{\lambda^2\left(1-\lambda\right)^2},$$
I understand, that the RHS is an element of $\mathbb{C}(j)$, and thus the LHS can be expressed as a polynomial in $\lambda$. I also understand, that if the order of the pole of the RHS at infinity is equal to 1, the highest degree of the polynomial can only be 1.

But no matter how I approach the RHS, the poleorder always seems to be two. For instance: If $\tau \to i\infty$, then $\lambda(\tau)\to 0$, thus we have a $0^2$ in the denominator. Thus the poleorder should be 2.

I suspect, that if I can express $\lambda$ as a 1 periodic Fourie series, that the problem may resolve, but my skills seem to be failing me.

Any help would be greatly appreciated.

答:
The relation defines the modular lambda function, which is a Hauptmodul (generator of the function field) for the congruence subgroup
$$\Gamma(2)=\left\{M\in SL(2,\mathbb Z): M\equiv \mathbb 1\mod 2\right\}.$$ The modular curve $\Gamma(2)\backslash \overline{\mathbb H}$ has three cusps $\tau=i \infty$, $\tau=0$ and $\tau=1$, each of width $2$. This corresponds to the rational function $j(\lambda)$ having second order poles at $\lambda\in\{0,1,\infty\}$.


问:
Prove that the set {¬,∧,∨} is functionally complete.
So far I have been using the given set to prove the functional completeness of other sets, but I don't know how to prove this one. That seems to be the case for similar questions too.Do I need to construct the truth table with every connective and build their equivalent with 3 given connectives?</p>

答: Hint: every Boolean formula can be written in CNF (conjunctive normal form) or DNF (disjunctive normal form), quite constructively.</p>


问:
$l^2$ with the $1$-norm is Banach?
I have to show that $l^2$ is Banach when endowed with the $1$-norm. To this regard, I take a Cauchy sequence $(x_n)_{n \in \mathbb{N}}$. For each component, I have
$$
|x_{n,k}-x_{m,k}|=(|x_{n,k}-x_{m,k}|)^{\frac{1}{2}} \le (\sum\limits_{j=0}^\infty |x_{n,j}-x_{m,j}|)^{\frac{1}{2}}=||x_n-x_m||_2<\infty
$$
So, $x_n \rightarrow x$ for some sequence $x$. In order to prove that $x \in l^2$, I just observe that $x_n$ converges to a unique limit in $l^2$, and this forces $x$ to be such a limit.

I have only a doubt. If $(x_n)_{n \in \mathbb{N}} \in l^2$ is Cauchy with respect to the distance induced by $1$-norm, is it also Cauchy with respect to distance induced by $2$-norm?

Is my argument correct?

答:
The $l^1$ norm is not a norm on $l^2$, because there are elements in $l^2$ that are not in $l^1$, e.g.,
$$
(1,\frac12, \frac13,\dots).
$$
The Cauchy sequence property is not a problem since $\|x\|_{l^2} \le \|x\|_{l^1}$ for all $x\in l^1$:
$$
\sum_{i=1}^N x_i^2 \le \left(\sum_{i=1}^N |x_i| \right)^2 \le \|x\|_{l^1}^2,
$$
which is true for all $N$.

Title: $l^2$ with the $1$-norm is Banach?


问:
Are definitions these of Big-O notation equivalent to the standard?
This definition uses hyper-reals and nonstandard analysis.

Let $k^*(x)$ be the natural extension of $k(x)$. Let $f$ and $g$ be functions.

$f = O(g) := \frac {f^*(H)} {g^*(H)}$ is finite for all positive unlimited $H$.

$f=\Omega(g) := \frac {g^*(H)} {f^*(H)}$ is finite for all positive unlimited $H$.

$f=\Theta(g):=f=O(g) \wedge f=\Omega(g)$

$f=o(g):=\frac {f^*(H)} {g^*(H)}$ is infinitesimal for all positive unlimited $H$.

$f=\omega(g):=\frac {g^*(H)} {f^*(H)}$ is infinitesimal for all positive unlimited $H$.

答:
I am not familiar with the Omega- and Theta- notations, but the big-O and little-o definitions are correct.  There is a bit of an ambiguity in your notation since you didn't specify where $x$ tends to exactly, but I assume $x$ is increasing without bound.

To respond to robjohn's question: as far as comparison between the traditional and the hyperreal approaches, the little-o and big-O example is not particularly enlightening.  A better example is the definition of continuity of a function $f$.  You are probably aware of the fact that a majority of undergraduates never fully master the traditional $\epsilon,\delta$ definition of continuity. It may come as more of a surprise that Cauchy never defined continuity using $\epsilon,\delta$.  In fact, Cauchy defined continuity in terms of the property that an infinitesimal $x$-increment necessarily produces an infinitesimal change in $y=f(x)$.  This is what the hyperreal definition says, as well.


答: The norms are equivalent down to a constant, see more details here, and from there you have what you need, the "Cauchy"-ness of the sequence remains.</p>

Title: $l^2$ with the $1$-norm is Banach?

Body:
I have to show that $l^2$ is Banach when endowed with the $1$-norm. To this regard, I take a Cauchy sequence $(x_n)_{n \in \mathbb{N}}$. For each component, I have
$$
|x_{n,k}-x_{m,k}|=(|x_{n,k}-x_{m,k}|)^{\frac{1}{2}} \le (\sum\limits_{j=0}^\infty |x_{n,j}-x_{m,j}|)^{\frac{1}{2}}=||x_n-x_m||_2<\infty
$$
So, $x_n \rightarrow x$ for some sequence $x$. In order to prove that $x \in l^2$, I just observe that $x_n$ converges to a unique limit in $l^2$, and this forces $x$ to be such a limit.

I have only a doubt. If $(x_n)_{n \in \mathbb{N}} \in l^2$ is Cauchy with respect to the distance induced by $1$-norm, is it also Cauchy with respect to distance induced by $2$-norm?

Is my argument correct?


问:
Finding inflection points of the function $\sqrt[3]{x}$
I have been working on a problem where I have to find critical number and inflection point for the function $\sqrt[3]{x}$

While I have been able to find critical number, I'm not sure how to find the inflection point for the function as for this particular function I cannot assign double derivative to be zero and then solve for x. The double derivative for other points indicates that the inflection point is between -1 and 1, but I'm not able to find any more ideas on how to approach this.

$f''(-1) = 2/9 > 0$

$f''(1) = -2/9 < 0$

答:
The inflection point is $(0,0)$. To see this, note that $f(x) = \sqrt[3]{x}$ is twice differentiable on $\mathbb R^*  =\mathbb R \setminus \{0\}$:

$$f'(x) = \frac{1}{3\sqrt[3]{x^2}}, f''(x) = - \frac{2}{9\sqrt[3]{x^5}} .$$

$f$ is not differentiable at $x = 0$ (but it has the improper derivative $+\infty$ at $0$, see for example Having trouble understanding condition for Rolle's Theorem (russian translation)).

You can now apply 1.3 Theorem 4.4.3 to see that $f$ is concave up on $(-\infty,0]$ and concave down on $[0,\infty)$. According to 1.4 Definition 4.4.4 the point $(0,0)$ is the only inflection point of $f$.

Intuitively this is clear if we look at $f''$: We have $\lim_{x \to 0^+} f''(x) =  -\infty$ and $\lim_{x \to 0^-} f''(x) =  +\infty$, i.e. $f''$ has an "improper change of sign" at $0$.

Update:

You can also easily prove the following theorem:

Let $f : I \to J$ be a bijection between to real intervals. Then $(a,f(a))$ is an inflection point of $f$ if and only $(f(a),a)$ is an inflection point of $f^{-1} : J \to I$.

This true because the graph of $f^{-1}$ is obtained from ther graph of $f$ by a reflection at the diagonal line $y = x$.

In your example $f^{-1}(x) = x^3$ which has exactly one inflection point at $(0,0)$.


答:
The way I addressed this problem was to demonstrate that each of the 16 possible truth functions on 2 variables was derivable using the set of connectives provided. You can cut corners on that by demonstrating equivalence of various combinations of those connectives with other connectives, e.g. showing that $\lnot$ and $\land$ can make $\lor$, and hence exclusive or, equivalence, various implications, and so on.

But first it makes sense to establish exactly what those 16 possible truth functions are, which can be done using a table of some kind, and (if you care to) describing what connective (or combination of connectives) give rise to each one.

When you've done that exhaustively for some subset of connectives, then you can work out how to replace certain connectives with those of a smaller subset, until you have reached the subset you are expected to demonstrate is complete.

Title: Prove that the set {¬,∧,∨} is functionally complete.

Body: So far I have been using the given set to prove the functional completeness of other sets, but I don't know how to prove this one. That seems to be the case for similar questions too.Do I need to construct the truth table with every connective and build their equivalent with 3 given connectives?</p>


问:
Channel Decoding: Gaussian Channel as Time-Varying Binary Symmetric
I am reading Information Theory, Inference and Learning Algorithms by David MacKay. It is available free of charge online: http://www.inference.org.uk/mackay/itila/book.html (official site, not pirated copy).

My question is to solve Exercise 25.1. It is marked as one of the easier ones. I think I'm just missing some fundamental understanding.
Assistance would be enormously appreciated!

I quote the desired section, slightly paraphrased for abbreviation.
It is from Section 25.1 on Page 324 (Page 336 of my pdf).

A codeword $t$ is selected from a code with codewords which form a subset of $\{0,1\}^N$ and transmitted over a noisy channel and $y$ is received.
We have
$$ P(t \mid y) = P(y \mid t) P(t) / P(y) $$
by Bayes's theorem.
Assume that he channel is memoryless.
Thus
$$ P(y \mid t) = \prod_{n=1}^N P(y_n \mid t_n). $$

For example, if the channel is a Gaussian channel with transmissions $\pm x$ and additive noise of standard deviation $\sigma$, then the probability density of the received signal $y_n$ in the two cases $t_n \in \{0,1\}$ are given as follows:
$$
\begin{aligned}
P(y_n \mid t_n = 1) &= (2 \pi \sigma^2)^{-1/2} \exp\bigl( - (y_n - x)^2 / (2 \sigma^2) \bigr);
\\
P(y_n \mid t_n = 0) &= (2 \pi \sigma^2)^{-1/2} \exp\bigl( - (y_n + x)^2 / (2 \sigma^2) \bigr).
\end{aligned}
$$
[Here we are interpreting $t_n = 0$ as "send $-x$" and $t_n = 1$ as "send $+x$" (plus noise).]
From the point of view of decoding, all that matters is the likelihood ratio:
$$
\frac
{P(y_n \mid t_n = 1)}
{P(y_n \mid t_n = 0)}
=
\exp\biggl( \frac{2 x y_n}{\sigma^2} \biggr).
$$

Exercise. Show that from the point of view of decoding a Gaussian channel is equivalent to a time-varying binary symmetric channel with a known noise level $f_n$ which depends on $n$.

For further reference, Section 11.2 (Page 179 of book and 191 of my pdf) discusses inferring the input from a Gaussian channel. It's not so helpful for this, though.
I do not see what is time-varying about this.

PS Multiple different SE networks seemed appropriate for this: maths, cstheory, stats.
I picked this one but if a mod wants to move it, no objection from me.

答:
You are right. It is not time-variant. I just checked MacKay's Example 25.1 and in that configuration, such an assertion is wrong. I just paraphrase what you have already established:

Assume that at instance $n$, the transmitter sends $s_n=+x$ for $t_n=1$, and $s_n=-x$ for $t_n=0$. The receiver gets $y_n=s_n+\nu$, where $\nu \sim \mathcal{N}(0, \sigma^2)$, so, $y_n \sim \mathcal{N}(s_n, \sigma^2)$. The ML decoder is simply:
$$ \exp\left(\frac{2xy_n}{\sigma^2}\right) \overset{+x}{\underset{-x}{\gtrless}} 1 \Longrightarrow y_n \overset{t_n=1}{\underset{t_n=0}{\gtrless}} 0 $$

that is, it decides based on the sign of $y_n$.
Now,
$$\begin{align*}
\Pr\{\text{Error}|t_n=1\} & = \Pr\{y_n<0|t_n=1\}=\Pr\{y_n<0|s_n=+x\} = Q\left(\frac{x}{\sigma}\right) \\
\Pr\{\text{Correct}|t_n=1\} & = \Pr\{y_n>0|t_n=1\}=\Pr\{y_n>0|s_n=+x\} = 1-Q\left(\frac{x}{\sigma}\right) \\
\Pr\{\text{Error}|t_n=0\} & = \Pr\{y_n>0|t_n=0\}=\Pr\{y_n>0|s_n=-x\} = Q\left(\frac{x}{\sigma}\right) \\
\Pr\{\text{Correct}|t_n=0\} & = \Pr\{y_n<0|t_n=0\}=\Pr\{y_n<0|s_n=-x\} = 1-Q\left(\frac{x}{\sigma}\right) \\
\end{align*}$$

which describes a time-invariant binary symmetric channel.


问:
Tensor product of modules and injective map?
$(R,m)$ is a local ring, $M,N$ are $R$-modules.
$M$ is finitely generated and free.
$N$ is generated by only $y$.

Then is $f:M\otimes N\to M, f(x\otimes y)=x$ injective?

答: There is no such map. Consider the case when $M=R$ and $N=R/I$, where $I$ is an ideal. Then $R\otimes_R R/I\cong R/I$ and there's no reason for the existence of any nontrivial homomorphism $R/I\to R$.


问:
Constant curvature and parallel curvature tensor
Let $(M,g)$ be a pseudo-Riemannian manifold. We say that this is a space of constant curvature if the sectional curvature $K(p,S)=\frac{\langle R(X,Y)X,Y\rangle}{X^2Y^2-\langle X,Y\rangle^2}$ is constant for all nondegenerate $2$-directions $S\le T_pM$ for all $p\in M$ ($X,Y\in S$ is any basis for the $2$-direction).

It is well-known and is quite easy to show that an equivalent condition to this is that $$ R_{ijkl}=K\left(g_{ik}g_{jl}-g_{il}g_{jk}\right), $$where $K$ is a constant.

It follows from this that $\nabla R=0$ or in coordinates $R_{ijkl;m}=0$, i.e. the curvature tensor is parallel.

I always had the impression that this is also an equivalent condition, i.e $$ R_{ijkl;m}=0\Longleftrightarrow (M,g)\text{ is of constant curvature}, $$ however I do not know any proof and I skimmed through quite a few books on Riemannian geometry both modern and old-school and I did not see this statement or a proof of it.

So my question is, if one is given a pseudo-Riemannian space with parallel curvature tensor ($R_{ijkl;m}=0$) does it follow that the space is also a space of constant sectional curvature?

If so I'd like to know a reference where this is proven in this direction.

答: The implication does not go the other way. All locally symmetric spaces have parallel curvature and there are locally symmetric spaces of nonconstant sectional curvature, such as the Fubini-Study metric on $\mathbb{CP}^n$ for $n\ge 2$.