Title: Expansion of $(x+y)^n+(x+z)^n+(y+z)^n-x^n-y^n-z^n$ in terms of elementary symmetric polynomials

Body:
Consider the symmetric polynomial in $3$ variables
$$
f_n(x,y,z)=(x+y)^n + (x+z)^n+(y+z)^n - x^n-y^n-z^n
$$
where $n\geq 0$ is an integer. I'm inquiring if there is a closed formula for the coefficients of $f_n$ in the basis of elementary symmetric polynomials? In other words, writing
$$
f_n(x,y,z)=\sum_{\nu_1+2\nu_2+3\nu_3=n} c(\nu_1, \nu_2,\nu_3)e_1^{\nu_1}e_2^{\nu_2}e_3^{\nu_3}
$$
with $e_1=x+y+z$, $e_2=xy+xz+yz$ and $e_3=xyz$, is there a formula for $c(\nu_1, \nu_2, \nu_3)$?

Addendum: As per requested, here are the $f_n$'s for $n\leq 10$
$$
\begin{aligned}
f_0 &= 0\\
f_1&=e_1\\
f_2&=e_1^2\\
f_3&=e_1^3-6e_3\\
f_4&=e_1^4-12e_{1}e_3\\
f_5&=e_1^5-20e_1^2e_3+10e_2e_3\\
f_6&=e_1^6-30e_1^3e_3+30e_1e_2e_3\\
f_7&=e_1^7 - 42 e_1^4 e_3 + 63 e_1^2 e_2 e_3 - 14 e_2^2 e_3 + 7 e_1 e_3^2\\
f_8&=e_1^8 - 56 e_1^5 e_3 + 112 e_1^3 e_2 e_3 - 56 e_1 e_2^2 e_3 + 28 e_1^2 e_3^2\\
f_9&=e_1^9 - 72 e_1^6 e_3 + 180 e_1^4 e_2 e_3 - 144 e_1^2 e_2^2 e_3 + 18 e_2^3 e_3 +
 72 e_1^3 e_3^2 - 18 e_1 e_2 e_3^2 - 6 e_3^3\\
f_{10}&=e_1^{10} - 90 e_1^7 e_3 + 270 e_1^5 e_2 e_3 - 300 e_1^3 e_2^2 e_3 +
 90 e_1 e_2^3 e_3 + 150 e_1^4 e_3^2 - 90 e_1^2 e_2 e_3^2 - 30 e_1 e_3^3
\end{aligned}
$$

Body:
In addition to @Somos' answer, is the following useful?
$p\left( w \right) = \left( {w - x} \right)\left( {w - y} \right)\left( {w - z} \right) = {w^3} - {e_1}{w^2} + {e_2}w - {e_3}$

${g_n}\left( {x,y,z} \right) = {x^n} + {y^n} + {z^n} = \sum\limits_{{\nu _1} + 2{\nu _2} + 3{\nu _3} = n} {{a_n}\left( {{\nu _1},{\nu _2},{\nu _3}} \right)} \;e_1^{{\nu _1}}e_2^{{\nu _2}}e_3^{{\nu _3}}$

${h_n}\left( {x,y,z} \right) = {\left( {y + z} \right)^n} + {\left( {z + x} \right)^n} + {\left( {x + y} \right)^n} = \sum\limits_{{\nu _1} + 2{\nu _2} + 3{\nu _3} = n} {{b_n}\left( {{\nu _1},{\nu _2},{\nu _3}} \right)} \;e_1^{{\nu _1}}e_2^{{\nu _2}}e_3^{{\nu _3}}$

$\sum\limits_{n = 0}^\infty  {{g_n}{t^n}}  = \frac{1}{t}\frac{{d\log p}}{{dw}}\left( {\frac{1}{t}} \right)$

$\sum\limits_{n = 0}^\infty  {{h_n}{t^n}}  =  - \frac{1}{t}\frac{{d\log p}}{{dw}}\left( {{e_1} - \frac{1}{t}} \right)$

So,

${c_n} = {b_n} - {a_n}$

Then

$${a_n} = {\left( { - 1} \right)^{{\nu _1} + {\nu _3}}}\left( {3\frac{{\left( {{\nu _1} + {\nu _2} + {\nu _3}} \right)!}}{{{\nu _1}!{\nu _2}!{\nu _3}!}} + 2\frac{{\left( {{\nu _1} + {\nu _2} + {\nu _3} - 1} \right)!}}{{\left( {{\nu _1} - 1} \right)!{\nu _2}!{\nu _3}!}} + \frac{{\left( {{\nu _1} + {\nu _2} + {\nu _3} - 1} \right)!}}{{{\nu _1}!\left( {{\nu _2} - 1} \right)!{\nu _3}!}}} \right)$$
$$\begin{array}{c}
{b_n} = {\left( { - 1} \right)^{{\nu _2} + {\nu _3}}}{2^{{\nu _1} - 2}}\left( {4\sum\limits_{r = 0}^{\left( {{\nu _1} + {\nu _3}} \right) \wedge \left\lfloor {\frac{{{\nu _1}}}{2}} \right\rfloor } {{{\left( { - 1} \right)}^r}\left( {2{\nu _2} + 3{\nu _3} + 3r} \right)\frac{{\left( {{\nu _2} + {\nu _3} + r - 1} \right)!\left( {{\nu _1} + {\nu _3} - r} \right)!}}{{\left( {{\nu _3} + r} \right)!\left( {{\nu _1} - 2r} \right)!{\nu _2}!{\nu _3}!r!}}} } \right.\\
 - 8\sum\limits_{r = 0}^{\left( {{\nu _1} + {\nu _3} - 1} \right) \wedge \left\lfloor {\frac{{{\nu _1} - 1}}{2}} \right\rfloor } {{{\left( { - 1} \right)}^r}\frac{{\left( {{\nu _2} + {\nu _3} + r} \right)!\left( {{\nu _1} + {\nu _3} - r - 1} \right)!}}{{\left( {{\nu _3} + r} \right)!\left( {{\nu _1} - 2r - 1} \right)!{\nu _2}!{\nu _3}!r!}}} \\
\left. { + \sum\limits_{r = 0}^{\left( {{\nu _1} + {\nu _3} - 2} \right) \wedge \left( {\left\lfloor {\frac{{{\nu _1}}}{2}} \right\rfloor  - 1} \right)} {{{\left( { - 1} \right)}^r}\frac{{\left( {{\nu _2} + {\nu _3} + r} \right)!\left( {{\nu _1} + {\nu _3} - r - 2} \right)!}}{{\left( {{\nu _3} + r} \right)!\left( {{\nu _1} - 2r - 2} \right)!{\nu _2}!{\nu _3}!r!}}} } \right)
\end{array}$$


问:
Why does this integral equals to given output?
I was reading about Gaussian distribution function:

and I saw this formula:

I don't understand why that integral is equal to $$\frac{\phi (t)}{t}$$ It seems like it must be equal to infinity. And how does $$\phi (t)$$ function works there?

my question is to explain how last step in that formula works. why does that last integral is equal to $\frac{\phi (t)}{t}$

答:
Your question is how we can evaluate the integral
$$\int\limits_{t}^{\infty}x\left(\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{x^2}{2} \right\}\right)\,\mathrm{d}x = \frac{1}{\sqrt{2\pi}}\int\limits_{t}^{\infty}x\exp\left\{-\frac{x^2}{2} \right\}\,\mathrm{d}x.$$
Substitute $u=-x^2/2$. The differential is going to be $\mathrm{d}x=-x^{-1}\mathrm{d}u$. This will yield
$$-\frac{1}{{\sqrt{2\pi}}}\int\limits_{-t^2/2}^{\infty}e^u\,\mathrm{d}u = \frac{1}{{\sqrt{2\pi}}}\exp\left\{-\frac{t^2}{2} \right\}.$$
This is our desired $\phi(t)$.

ContentLicense: CC BY-SA 4.0


问:
Faithfully flat map of schemes
Let $f:X \to Y$ be a faithfully flat  morphism of schemes. If necessary, one can put moderate mild assumptions on these as they are locally noetherian and the map is locally of finite type.

It occurred to me the necessity of a statement as follows: there exists  AFFINE open coverings $Y_i$ of $Y$ and $X_i$ of $X$ such that $f(X_i) \subseteq Y_i$ and $f|_{X_i}: X_i \to Y_i$ is faithfully flat. To obtain that, one should prove in particular that the maps $ f|_{X_i}$ are surjective. I wasn't able to ind such a statement in the literature neither to prove this. Is this even true?

答:
The general answer is "no". Depending on what you are trying to do exactly, you may want to require $f$ to be affine, or quasi-compact. If $f$ is affine, then for any affine open $Y'\subset Y$, the preimage $X'$ of $Y'$ in $X$ is an affine open subscheme of $X$ and the canonical map $X'\to Y'$ is faithfully flat. If $f$ is quasi-compact, then $X'$ can be covered by finitely many affine open subschemes of $X$, and the map from their disjoint union to $Y'$ is a faithfully flat morphism of affine schemes.

For a counterexample to the general statement, consider $Y=\operatorname{Spec} \mathbb Z$. For each closed point $y$ of $Y$, let $X_y$ be the localization of $Y$ at $y$. Call $X$ the disjoint union of the $X_y$ and $f$ the natural map $X \to Y$. Then, $f$ is faithfully flat, but there exist no nonempty affine open subschemes $Y'\subset Y$ and $X'\subset X$ such that $f(X')=Y'$ set-theoretically.

To see this, suppose given such $X',Y'$. Since $Y'$ contains infinitely many closed points of $Y$, we know that $X'$ must contain infinitely many of the $X_y$, so $X'$ is not quasi-compact and in particular not affine: we have a contradiction.

Title: Faithfully flat map of schemes


问:
Is a function useful that has zeros on the prime integers?
So I want a function that is zero on the reals only on the prime integers and which doesn't depend on knowing the primes. I construct:

$$f(x) = e^{-x^2} - \sum\limits_{n=2}^\infty e^{-n^2} \frac{ \sin(\pi x)^2 }{ n^2\sin(\pi x/n)^2}$$

Which has zeros on the real line only on the positive and negative prime integers.

( $c_n(x)=\frac{\sin(\pi x)^2}{n^2\sin(\pi x/n)^2}$ has a well defined Taylor series and can be defined everywhere. It is $1$ when $x$ is a multiple of $n$ and 0 otherwise. The exponentials are just to make the whole thing converge.)

So my question is, is a function like this useful? As in, would it tell us anything about the primes?

Edit: if it is analytic, as well as zeros on the real axis it will have many complex zeros.

I also note that it is quite "easy" to convert this into a series of the form:
$f(x)=\sum\limits_{k=0}^\infty a_{k} x^{k}$  where the coefficients $a_{2k} =\frac{(-1)^k}{k!} - \sum\limits_{n=2}^\infty c^{(2k)}_n(0)e^{-n^2}$. It begins $f(x)=0.981-0.97722x^2+...$ although you would need a lot of terms when $x$ is big!!! But we could say $f^{-1}(0)\subset$primes.

Edit 2: I think the function actually also has zeros at points very close to the primes and only the zeros where $f'(x)<0$ are primes. (Basically every second zero.) This works equally well replacing the exponentials with $1/x^2$.

答:
Yes: In case all zeroes have same order, in doubt consider f/f', one could derive a function with simple poles and residual 1 in primes; then a circular integral would deliver the number of enclosed primes.
Concerning convergence rate, just an idea: May be there is a representation with elliptic functions, then fast Jacobi elliptic functions could serve.</p>


问:
How to prove (using integration by parts)
Could you help to prove this:

$$Y(\lambda) = \int_{a}^{b} g(x) e^{i \lambda   f(x)} \,dx$$

we should use integration by parts to get
$$Y(\lambda) = \frac{1}{i \lambda }e^{i \lambda   f(x)} \frac{g(x)}{f'(x)}\bigg\vert _{a}^{b}-\frac{1}{i \lambda }\int_{a}^{b} dx \ e^{i \lambda   f(x)} \frac{d}{dx} \bigg[\frac{g(x)}{f'(x)}\bigg] $$

Thanks in advance

答:
Starting with $$\int_{a}^{b} g(x) e^{i \lambda   f(x)} \,dx$$ we set
\begin{align}
u &= g(x), &&du = g'(x)\,dx\\
dv &= e^{i \lambda   f(x)}\,dx, &&\phantom{d}v = \frac{1}{f'(x)i\lambda}e^{i \lambda   f(x)}.
\end{align}

This then gives us
\begin{align}
\int_{a}^{b} g(x) e^{i \lambda   f(x)} \,dx &= uv\bigg|_{a}^{b} - \int_{a}^{b}v\,du\,dx\\
&=\frac{1}{i\lambda}e^{i \lambda   f(x)}\frac{g(x)}{f'(x)}\bigg|_{a}^{b} - \frac{1}{i\lambda}\int_{a}^{b}e^{i \lambda   f(x)}\frac{g'(x)}{f'(x)}\,dx.
\end{align}

Note that this is close, but not the same as what you have, so I suspect there is either a typo in what you entered or in your original source.


问:
Questions on Fraenkel models
Halbeisen on page 172 contains a section entitled "The Second Fraenkel Model". The original paper by Fraenkel containing this model can be found here. I have several questions regarding this model and $\mathsf{FM}$-models in general.

If this 1922 paper by Fraenkel is not the first describing $\mathsf{FM}$-models, can someone please point me to his very fist paper in which he invents the technique? (given that Halbeisen calls it the second model I suppose there must be an earlier paper containing his first model although looking at this list of publications it appears to be his first.)

In this 1922 paper he writes that for this model construction we can keep the axioms of Zermelo I,II,IV-VII as they are but in Halbeisen on page 168 it is stated that we need to use modified versions of Empty Set and Extensionality. Hence my question 2.a) is: Is this a mistake in Fraenkel's paper? (Unlikely, I would think) and my question 2.b) is: Empty Set is a consequence of all the other axioms hence wouldn't it be enough to just use a modified version of Extensionality?

Last but not least the question that is most important to me: What exactly did Fraenkel's 1922 paper prove? The aim of course was to show that $\mathsf{AC}$ is independent of $\mathsf{ZF}$ by constructing a model of $\mathsf{ZF}$ in which $\mathsf{AC}$ fails. But the model he constructed is a model of $\mathsf{ZFA}$ and not of $\mathsf{ZF}$ and it seems to be the case that those transfer methods used to embed these Fraenkel models into symmetric models weren't known for another 30 years or so. Hence: what is the relation of $\mathsf{ZFA}$ and $\mathsf{ZF}$? (without using transfer methods)

答:
Historically the list of axioms known as $\sf ZF$ wasn't written until mid-1920's when von Neumann wrote his Ph.D. dissertation, where he proved the relative consistency of regularity, and defined the ordinals as we know them today, and so on.

Fraenkel's original proofs contained mistakes. I don't know the exact mistake, but I do know that Mostowski found a mistake; Fraenkel rewrote his arguments; Mostowski found further mistakes, and ended up writing the arguments using supports which is very similar to what we know today, Specker was the one putting the finishing touch on the technique. Do note that not once the constructions were done in $\sf ZFC-Reg$, where sets of the form $x=\{x\}$ were used for atoms. (This is why in quite a few places this method is called Fraenkel-Mostowski or even Fraenkel-Mostowski-Specker.)

Lastly, $\sf ZFA$ contains atoms. That's huge. Even more so when we only "disorder" the impure sets. Note that no matter how hard you'll try, you can never form a permutation model in which the real numbers cannot be well-ordered. Because the power of the integers can be well-ordered in permutation models of $\sf ZFA$. But Fraenkel did "prove" (or at least lay foundations for the proof) that the axiom of choice is not provable from the other axioms when we allow atoms, and that was a big step. It shows that this axiom is not a consequence of Zermelo's set theory (sans choice itself, of course) if we allow atoms.

The historical overview, including references, by the way, can be found in both Jech's "The Axiom of Choice" and in Halbeisen's book, in the same chapter (the notes section at the end).

ContentLicense: CC BY-SA 3.0

Title: Questions on Fraenkel models


问:
When does $x^3 +Ax + B + Cy + y^3$ factorise into three distinct linear factors?
I am attempting to understand equations of the form $z^2 = x^3 + Ax + B + Cy + y^3$, which is a generalisation of an elliptic curve $y^2 = x^3 + Ax + B$. I know that to make sure an elliptic curve has nice properties, we require that $f(x) = x^3 + Ax + B$ has three distinct roots. This leads to $4A^3 + 27B^2 \neq 0$. I wondered if the generalised equation has a similar inequality in A, B and C which forces the surface to have particularly "nice" properties, such as a group law.

I guessed that if $g(x,y) = x^3 + Ax + B + Cy + y^3$ could be factored into distinct linear factors then it would be sufficiently "nice". Trying to find the required restriction on the coefficients, I first set $g(x,y) = (a_1 x + b_1 y + c_1)(a_2 x + b_2 y + c_2)(a_3 x + b_3 y + c_3)$ and expanded the RHS. This results in the following system of equations,
$$a_1 a_2 a_3 = 1$$
$$b_1 b_2 b_3 = 1$$
$$a_3 b_2 c_1 + a_2 b_3 c_1 + a_3 b_1 c_2 + a_1 b_3 c_2 + a_2 b_1 c_3 + a_1 b_2 c_3 = 0$$
$$a_2 a_3 b_1 + a_1 a_3 b_2 + a_1 a_2 b_3 = 0$$
$$a_3 b_1 b_2 + a_2 b_1 b_3 + a_1 b_2 b_3 = 0$$
$$a_2 a_3 c_1 + a_1 a_3 c_2 + a_1 a_2 c_3 = 0$$
$$b_2 b_3 c_1 + b_1 b_3 c_2 + b_1 b_2 c_3 = 0$$
$$A = a_3 c_1 c_2 + a_2 c_1 c_3 + a_1 c_2 c_3$$
$$B = c_1 c_2 c_3$$
$$C = b_3 c_1 c_2 + b_2 c_1 c_3 + b_1 c_2 c_3$$
But I don't see how I can use these to deduce a restriction on the coefficients. Hopefully someone can find a simpler approach to this problem!

By "nice" I mean that every line meeting the surface must meet it in exactly three points (counting multiplicities and involving points at infinity if necessary). My ultimate goal is to find the weakest, necessary condition for such a surface to have this property.

答:
Turns out the reference for part reduction was a book by Schinzel, he of the Hypothesis.

there is a nice answer to this sort of thing. Throw in a variable $z$ to homogenize,
$$ f = x^3 + A x z^2 + B z^3 + C y z^2 + y^3  $$

Next, find the second partial derivatives of $f$ by $x,y,z$ and write out the Hessian matrix. The entries in $H$ are linear in $x,y,z,$ as they begin degree three.

Finally, calculate the determinant of $H.$  This is a homogeneous degree three. The form $f$ factors into linear terms over $\mathbb C$ if and only if the determinant is a constant multiple of $f.$  The reference is by Brookfield, I believe I saved part of it, let me look.

Gary Brookfield (2016) Factoring Forms, The American Mathematical
Monthly, 123:4, 347-362


答:
The problem is $\int_a^b g(x)e^{i\lambda f(x)}dx$. When you see e raised to the power of something, it's usually going to be set to the dv term in the integration by parts. Let's use

$$\begin{matrix}u=g(x)&dv=e^{i\lambda f(x)}dx\\du=g'(x)dx&v=\frac{1}{i\lambda f'(x)}e^{i\lambda f(x)}.\end{matrix}$$

By the integration by parts formula, we have:

$$\begin{split}uv|_{a}^b-\int_a^b v du&=\frac{1}{i\lambda}\frac{g(x)}{f'(x)}e^{i\lambda f(x)}|_{a}^b-\frac{1}{i\lambda}\int_a^b \frac{\frac{d}{dx}g(x)}{f'(x)}e^{i\lambda f(x)}dx.\end{split}$$

Title: How to prove (using integration by parts)

Body:
Could you help to prove this:

$$Y(\lambda) = \int_{a}^{b} g(x) e^{i \lambda   f(x)} \,dx$$

we should use integration by parts to get
$$Y(\lambda) = \frac{1}{i \lambda }e^{i \lambda   f(x)} \frac{g(x)}{f'(x)}\bigg\vert _{a}^{b}-\frac{1}{i \lambda }\int_{a}^{b} dx \ e^{i \lambda   f(x)} \frac{d}{dx} \bigg[\frac{g(x)}{f'(x)}\bigg] $$

Thanks in advance


问:
quadratic over Z7 no solutions?
I was given the question
Solve this formula for X1 and X2 over Z7 (mod7)

$$
3x^2+x+1 = 0
$$

So I went on and converted it to the quadratic formula over Z7 like and solved it like so:
$$
\frac{6\pm \sqrt{1+3*3*1}}{6} =
\frac{-1\pm \sqrt{1-4*3*1}}{6} =
\frac{-1\pm \sqrt{3}}{6} =
$$

And... now what?
Can this even be solved? is it correct to say that there is no solution as $$\sqrt{3}$$ does not exist in Z7?
Am I missing something?

答: You are correct that there is no solution, but not for the reason you're giving here. The quadratic equation only holds if we're working in $\mathbb R$ or $\mathbb C$ or something similar (like a subring of one of those). In the case of $\mathbb Z / 7\mathbb Z$, there are only seven elements to consider, so you can simply try plugging each of them into $3x^2 + x + 1$ and verify that none of them is a zero.

Title: quadratic over Z7 no solutions?


问:
Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled

Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).

I am looking for a quick/efficient way to resolve this problem. I thought of fixing "$aa$" in the beginning then draw a tree of the next possibilities but this tree will end up to be a whole forest. Can you help me ?

答:
A word of length $n$ is either a word of length $n - 1$ and a $b$ or a $c$, or a word of length $n - 2$ and $aa$. Call the number of words of length $n$ $x_n$, you see $x_0 = 1$, $x_1 = 2$, and:

$\begin{align*}
   x_{n + 2}
     &= 2 x_n + x_{n - 2}
\end{align*}$

Solving this using generating functions is routine. Define:

$\begin{align*}
    g(z)
      &= \sum_{n \ge 0} z_n z^n
\end{align*}$

Multiply your recurrence by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:

$\begin{align*}
   \sum_{n \ge 0} x_{n + 2} z^n
     &= 2 \sum_{n \ge 0} x_{n + 1} z^n + \sum_{n \ge 0} x_n z^n \\
   \frac{g(z) - x_0 - x_1 z}{z^2}
     &= 2 \frac{g(z) - x_0}{z} + g(z)
\end{align*}$

Solve for $g(z)$ using the initial values, express as partial fractions:

$\begin{align*}
   g(z)
     &= \frac{1}{1 - 2 z - z^2} \\
     &= \frac{1}{(1 - (1 - \sqrt{2}) z) (1 - (1 + \sqrt{z}) z)} \\
     &= \frac{\sqrt{2} - 1}{2^{3/2} (1 + (\sqrt{2} - 1) z)}
          + \frac{\sqrt{2} + 1}{2^{3/2} (1 - (\sqrt{2} + 1) z)}
\end{align*}$

We need the coefficient of $z^n$, as this is just geometric series:

$\begin{align*}
  [z^n] g(z)
    &= (-1)^n \frac{\sqrt{2} - 1}{2^{3/2}} (\sqrt{2} - 1)^n
         + \frac{\sqrt{2} + 1}{2^{3/2}} (\sqrt{2} + 1)^n
\end{align*}$

Noting that $\rvert \sqrt{2} - 1 \lvert < 1$, and furthermore the respective coefficient is also less than $1$, you have finally that:

$\begin{align*}
   x_n
     &= \operatorname{round}
          \left( \frac{\sqrt{2} + 1}{2^{3/2}} (\sqrt{2} + 1)^n \right)
\end{align*}$

(No, this isn't the best way to compute $x_n$ for large $n$, but it shows how the value grows).


问:
A property for some finite groups (especially ${\rm PSL}(2,13)$)
Let $G$ be a finite group of order $n$ and consider the following property:

(P) for every factorization
 $n=ab$ there exist subsets $A$ and $B$ such that $|A|=a$, $|B|=b$ and $G=AB$.

($AB=\{ab:a\in A, b\in B\}$)

Note. If $G$ has the property that for every divisor $d$ of $n$  there exists a subgroup of $G$ with order $d$ or $n/d$, then we can show that $G$ enjoys (P).

Therefore, (P) is true for all finite abelian groups, and also one can check that the groups $S_n$, $A_n$ and ${\rm PSL}(2,n)$, where $n\leq 8$, have the property.

Now, is (P) true for ${\rm PSL}(2,13)$?

(we think this is a good candidate for a probable counterexample.)

Thanks in advance.

答:
I don't know whether this can be generalized in any way, but for the $G=PSL_2(13)$ case, you can get sets $A$ and $B$ of sizes $21$ and $52$ with the property $G=AB$ by simply multiplying Sylow-$3$ and Sylow-$7$ subgroups for $A$ and Sylow-$2$ and Sylow-$13$ subgroups for $B$, as GAP shows:

```<code>gap> G:=PSL(2,13);;
gap> A := Concatenation(List(SylowSubgroup(G, 7), s -> AsList(RightCoset(SylowSubgroup(G, 3), s))));;
gap> B := Concatenation(List(SylowSubgroup(G, 13), s -> AsList(RightCoset(SylowSubgroup(G, 2), s))));;
gap> Size(G);
1092
gap> Size(A);
21
gap> Size(B);
52
gap> Size(Concatenation(List(A, a -> List(B, b -> a*b))));
1092
</code>```

Body:
I don't know whether this can be generalized in any way, but for the $G=PSL_2(13)$ case, you can get sets $A$ and $B$ of sizes $21$ and $52$ with the property $G=AB$ by simply multiplying Sylow-$3$ and Sylow-$7$ subgroups for $A$ and Sylow-$2$ and Sylow-$13$ subgroups for $B$, as GAP shows:

gap> G:=PSL(2,13);;
gap> A := Concatenation(List(SylowSubgroup(G, 7), s -> AsList(RightCoset(SylowSubgroup(G, 3), s))));;
gap> B := Concatenation(List(SylowSubgroup(G, 13), s -> AsList(RightCoset(SylowSubgroup(G, 2), s))));;
gap> Size(G);
1092
gap> Size(A);
21
gap> Size(B);
52
gap> Size(Concatenation(List(A, a -> List(B, b -> a*b))));
1092


答:
Since we are dealing with models of $\sf{ZFA}$, we have to keep those atoms in mind.  In particular, atoms have no elements, and so they have the same elements as the empty set.  But they are different from the empty set.  And from each other.  Therefore a modified form of Extensionality is needed.  Usually $\sf{ZFA}$ includes a predicate to the effect of $A(x)$ holds iff $x$ is an atom.  So Extensionality should be rephrased

$( \forall x ) ( \forall y ) ( ( \neg A(x) \wedge \neg A(y) ) \rightarrow ( ( \forall u ) ( u \in x \leftrightarrow u \in y ) \rightarrow x = y ) )$.

To put it into words: any two non-atoms (i.e., sets) which have the same elements are equal.

The Empty Set Axiom is similarly altered to ensure a set without elements:

$( \exists x ) ( \neg A(x) \wedge ( \forall u ) ( u \notin x ) )$.

As to the question of its necessity? This is a bit tricky.

Note that unchanged, the Axiom Schema of Separation applied to some set $X$ and the formula $\varphi (u) \equiv u \neq u$ will yield some object $Y$ with the property that $u \notin Y$ for all $u$.  However by unchanged this means that there is no specification that the object $Y$ is itself a set. In particular, the object $Y$ could be an atom.

Note also the following construction:  Given any atom $a$, consider, by the Axiom of Power Set (unchanged) $$\mathcal{P} ( a ) = \{ X : ( \forall u ) ( u \in X \rightarrow u \in a ) \}.$$ Note that $\mathcal{P} ( a )$ contains all atoms, and no nonempty sets.  Then applying an instance of (a slightly expanded) Axiom Schema of Separation, we get the set $$Y = \{ X \in \mathcal{P} ( a ) : \neg A ( X ) \}.$$ Then either $Y$ is empty, or contains the empty set.  In particular, the empty set must exist.  But this construction only works if you allow the predicate $A$ in instance of the Axiom Schema of Separation, which hasn't been explicitly stated by Halbeisen.

After some extra consideration, it appears to me that the way to get the existence of the empty set in $\sf{ZFA}$ making the least number of changes to $\sf{ZF}$ might be to include the (modified) Axiom of Empty Set.  Of course, one could formulate the axioms of $\sf{ZFA}$ by going through all the axiom of $\sf{ZF}$ and ensure that when objects are constructed they are sets, and then the usual method of showing that the empty set exists would go through.

One should note that in Jech's The Axiom of Choice, $\sf{ZFA}$ is described as the theory obtained by making the following alterations to $\sf{ZF}$:

include a unary predicate symbol $A$, and a constant symbol $0$.

Axiom of Empty Set. $\neg ( \exists x ) ( x \in 0 )$.

Axiom of Atoms. $( \forall z ) ( A(z) \leftrightarrow ( z \neq 0 \wedge \neg ( \exists x ) ( x \in z ) ) )$.

Axiom of Extensionality. $( \forall \text{ set } X ) ( \forall \text{ set } Y ) ( ( \forall u ) ( u \in X \leftrightarrow u \in Y ) \leftrightarrow X = Y )$.

Axiom of Regularity. $( \forall \text{ nonempty } S ) ( \exists x \in S ) ( x \cap S = 0 )$.

where

$( \forall \text{ set } X ) \ldots$ abbreviates $( \forall X ) ( \neg A ( X ) \rightarrow \ldots )$; and

$( \forall \text{ nonempty } X ) \ldots$ abbreviates $( \forall X ) ( ( \exists x ) ( x \in X ) \rightarrow \ldots )$.

What did Fraenkel's paper do?  You're right that it was a weaker result, proving that $\sf{AC}$ is not provable from $\sf{ZFA}$, but it at least hinted at the possibility of Cohen's later result.  And one should see a similarity between the method of permutation models, and symmetric submodels of generic extensions.  Note, also, that any model of $\sf{ZF}$ is a model of $\sf{ZFA}$, with an empty collection of atoms.

Title: Questions on Fraenkel models

Body:
Halbeisen on page 172 contains a section entitled "The Second Fraenkel Model". The original paper by Fraenkel containing this model can be found here. I have several questions regarding this model and $\mathsf{FM}$-models in general.

If this 1922 paper by Fraenkel is not the first describing $\mathsf{FM}$-models, can someone please point me to his very fist paper in which he invents the technique? (given that Halbeisen calls it the second model I suppose there must be an earlier paper containing his first model although looking at this list of publications it appears to be his first.)

In this 1922 paper he writes that for this model construction we can keep the axioms of Zermelo I,II,IV-VII as they are but in Halbeisen on page 168 it is stated that we need to use modified versions of Empty Set and Extensionality. Hence my question 2.a) is: Is this a mistake in Fraenkel's paper? (Unlikely, I would think) and my question 2.b) is: Empty Set is a consequence of all the other axioms hence wouldn't it be enough to just use a modified version of Extensionality?

Last but not least the question that is most important to me: What exactly did Fraenkel's 1922 paper prove? The aim of course was to show that $\mathsf{AC}$ is independent of $\mathsf{ZF}$ by constructing a model of $\mathsf{ZF}$ in which $\mathsf{AC}$ fails. But the model he constructed is a model of $\mathsf{ZFA}$ and not of $\mathsf{ZF}$ and it seems to be the case that those transfer methods used to embed these Fraenkel models into symmetric models weren't known for another 30 years or so. Hence: what is the relation of $\mathsf{ZFA}$ and $\mathsf{ZF}$? (without using transfer methods)


问:
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.

Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator

答:
Note that $\cos(\frac{\pi}5)=\frac{1+\sqrt5}4=\frac{\phi}2$ where $\phi$ is Golden ratio . also note that $\phi^2=\phi+1$, we use this in the following calculations

$$\cos(\frac{\pi}5)-\cos(\frac{2\pi}5)=\frac{\phi}2-(2(\frac{\phi}2)^2-1)=\frac{\phi-(\phi+1)+2}{2}=\frac12$$
$$\cos(\frac{\pi}5)\;\cos(\frac{2\pi}5)=\frac{\phi}2\times(\frac{\phi^2}2-1)=\frac{\phi}2\times\frac{\phi-1}2=\frac{\phi^2-\phi}4=\frac14$$


问:
How to solve quadratic inequalities with two boundaries
I was solving a math question on an online test, and the question was asking what is possible values of $x$ if the perimeter of some rectangle tile is between $20$ cm and $54$ cm. I wondered how we would find possible values of $x$ if the area was between $20$ and $54$ cm.

I thought we could use sign analysis for each inequality, but I understood that won't work because we will be having different inequalities in each part and that doesn't make sense.

Inequality looks like this: $20<2x^2+12x<54$

How could we solve quadratic inequalities with two boundaries like this?

答:
The values of $x$ that verify the inequality, will verify the following inequalities at the same time:
$$
\begin{array} 020<2x^2+12x, & 2x^2+12x<54\end{array}
$$
So all we have to do is work with both and we will intersect the solution spaces, getting the solution of the original inequality. Let's start with the first one:
$$
20<2x^2+12x \implies 0<(x+3+\sqrt{19}) ·(x+3-\sqrt{19}) \implies x>-3+\sqrt{19} \  \lor \ x<-3-\sqrt{19}
$$
The second one will have the following result:
$$
2x^2+12x<54 \implies x^2+6x-27<0 \implies -9<x<3
$$
The intersection is all of the $x\in \mathbb{R}\ $ such that:
$$
-9<x<-3-\sqrt{19} \ \lor -3+\sqrt{19}<x<3
$$


问:
Integral beginner proof
Question:

Let $f:[0,\pi]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^{\pi} f(t)\sin(t)dt =0,$$ then the equation $f(x)=0$ admits a solution in $[0,\pi].$

What I've thought of:

I'd suppose that $f(x) > 0$ and try to get a contradiction, but I'm not really sure how to go about it.

I'm still new to integrals and would appreciate if someone could help me. Thank you.

答:
If $f(x)>0$ for all $x\in [0, \pi]$, then we have $f(x)\sin(x)>0$ for all $x\in (0, \pi)$, and therefore we have
$$
\int_0^\pi f(x)\sin(x)dx > 0
$$
Similarly for $f(x)<0$ for all $x$.

So by contraposition, there must be some $x_1$ such that $f(x_1)\leq0$ and some $x_2$ such that $f(x_2)\geq0$. If any of these two are equalities, we are done. If neither is an equality, the intermediate value theorem says there is some $x_3$ between them with $f(x_3)=0$.

Title: Integral beginner proof


答:
You can solve it by splitting it up into two inequalities, and solving them each separately. If you start from $20<2x^2+12x<54$, that means you need both of the following to be true:

$2x^2+12x-20 > 0$

$2x^2+12x-54 < 0$

The first quadratic inequality is true for $x<-3-\sqrt{19}$ or for $x > -3+\sqrt{19}$. The second quadratic inequality is true for $-9<x<3$. You can get to these solutions by graphing the quadratic or using the quadratic formula, then checking signs, as you mention in your question.

Combining everything, we need:

$$\left( x<-3-\sqrt{19} \ \cup \ x > -3+\sqrt{19} \right) \ \cap \ (-9<x<3) , $$

where $\cup$ means "or" and $\cap$ means "and."

Then, $-3-\sqrt{19}\approx -7.4$, and $-3+\sqrt{19} \approx 1.4$. So, we end up at the following set of values:

$$ -9<x<-3-\sqrt{19} \ \cup \ -3+\sqrt{19}<x<3. $$

If that last step doesn't make sense and it's hard to visualize, try plotting both of the inequalities on a number line, and wherever they're both true, that's the solution.

Title: How to solve quadratic inequalities with two boundaries

Body:
I was solving a math question on an online test, and the question was asking what is possible values of $x$ if the perimeter of some rectangle tile is between $20$ cm and $54$ cm. I wondered how we would find possible values of $x$ if the area was between $20$ and $54$ cm.

I thought we could use sign analysis for each inequality, but I understood that won't work because we will be having different inequalities in each part and that doesn't make sense.

Inequality looks like this: $20<2x^2+12x<54$

How could we solve quadratic inequalities with two boundaries like this?


问:
The integral over a subset is smaller?
In a previous question I had $A \subset \bigcup_{k=1}^\infty R_k$ where $R_k$ in $\Bbb{R}^n$ are rectangles I then proceeded to use the following inequality $\left|\int_A f\right| \le \left|\int_{\bigcup_{k=1}^\infty R_k} f \right|$ which I am not really certain of. Does anyone know how to prove it? If its wrong what similar inequality should I use to prove the result here.</p>

答:
For $A \subset B$, $$
  \left|\int_A f\right| \leq \left|\int_B f \right|
$$
is true if either $f(x) \geq 0$ for all $x \in B$, or $f(x) \leq 0$ for all $x \in B$.

For a counter-example if this isn't the case, let $f(x) = x$, $A = [0,2]$, $B=[-2,2]$. Then $$
   \left|\int_A f\right| = 2 > 0 = \left|\int_B f \right| \text{.}
$$

In particular, you thus always have that $$
  \int_A |f| \leq \int_B |f| \
$$
if $A \subset B$.

Title: The integral over a subset is smaller?


答:
Hint:

You just have to solve the system of quadratic inequalities:
\begin{cases}
2x^2+12x-20 >0,\\[0.5ex] 2x^2+12x-54<0.
\end{cases}
The first quadratic polynomial has  real roots  and its leading coefficient is positive,  hence the first inequality is satisfied outside of the interval of the roots.

The second quadratic polynomial  does have real roots, hence it has negative values in the interval of the roots.


问:
The Variance Rule of $V(cX) = c^2V(X)$
As a statistics major, I learned a while back that the $V(cX) = c^2 V(X)$, where $V(X)$ is the variance of $X$. However, I never quite understood why that was the case. Consider this: We have an event $X$ such that each event is independent of any previous events. Take for instance, a roll of copper wire. We know that the mean length is $150 \text{m}$ and the standard deviation is $4\text{m}$. If we want to find the variance of $5$ rolls of copper wire, we could write this as $V(5X) = 25V(X) = 400\text{m}^2$. However, an alternative way to write the problem is doing $V(X + X + X + X + X) = V(X) + \ ... \ + V(X)$ - as they are all independent. Thus, we can simplify to $5V(X)$ and get $80\text{m}^2$. This second way seems like it should be the same as the first way, but they produce different answers. Can anyone explain this? I was helping a friend out with a similar problem and the second way actually produces the correct answer, so can anyone explain why this is the case?

答:
If the $5$ rolls of copper wire are chosen independently of each other, then the variance of the sum of their lengths is just $5$ times the population variance.

But multiplying the random variable by $5$ means you're not looking at $5$ independently chosen rolls of wire, but rather at $5$ times the length of that one roll that appeared first. You don't have independence here: it's as if you had five identical copies of that one roll.

Title: The Variance Rule of $V(cX) = c^2V(X)$


答:
You are on the right track. In $(0,\pi)$

$$f(x)>0\implies f(x)\sin(x)>0\implies\int_0^\pi f(x)\sin(x)\,dx>0,$$ a contradiction.

Title: Integral beginner proof

Body:
Question:

Let $f:[0,\pi]\rightarrow\mathbb{R}$  be a continuous function. Show that, if $$\int\limits_0^{\pi} f(t)\sin(t)dt =0,$$ then the equation $f(x)=0$ admits a solution in $[0,\pi].$

What I've thought of:

I'd suppose that $f(x) > 0$ and try to get a contradiction, but I'm not really sure how to go about it.

I'm still new to integrals and would appreciate if someone could help me. Thank you.


问:
Calculating this series integral $\int (\frac{x^{2n-2}}{(2n-2)!})\cdot \frac{x^{2n-1}}{(2n-1)!}$
I've calculated this integral and would like the communities feedback and support towards my solution:

$$\int \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\cdot \left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\cdots\frac{x^{2n-1}}{(2n-1)!}\right)\,\mathrm{d}x$$

My attempt:

I performed substitution where $u = \left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\cdot\frac{x^{2n-1}}{(2n-1)!}\right)$.
Then taking the derivative of both sides $\mathrm{d}u = \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\mathrm{d}x$.
This yields $$\int u\space \mathrm{d}u = \frac{u^2}{2}+C.$$

Then substituting back in the variable in $u$:

$$\frac{1}{2}\left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\frac{x^{2n-1}}{(2n-1)!}\right)^2$$

Although I'm not entirely sure how to distrbute the $2$ properly. Do I multiply the single $x$ onto $\frac{x^3}{2!}$, then distribute the power, and multiply the $2$ on the numerator using this fraction rule like this

$$\left(\frac{2(x^6)}{3!} \cdot \frac{2(x^7)}{5!}\frac{2(x^{2n+1)}}{(2n-1)!}\right) ?$$

答:
Please keep in mind: the derivative of a product is not the product of the derivatives!

In this case you should simply multiply both factors obtaining
$$\int\prod_{k=0}^{2n-1}\frac {x^k}{k!}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{\sum_{k=0}^{2n-1}k}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{n (2n-1)}dx=
\frac{x^{n (2n-1)+1}}{(n (2n-1)+1)\prod_{k=0}^{2n-1} k!}.
$$


问:
How do I find the limit of $\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))}$?
How do I find the limit of

$$\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))}$$

I know it is equal to $\frac{1}{4}$, but how did we get that? Without using L'Hopital's rule.

I tried canceling $x$ with $\sin(2x)$ and $\ln(x+1)$, but still got the wrong  result...

答:
here's my attempt to solve your problem without using L'Hospital rule :

Reminder :

$$\lim_{x\to0}\frac{\sin(ax)}{x}=a$$

From this post here you can see that :
$$\lim_{x\to 0} \frac{\ln(x+1)-x}{x^2}=-\frac{1}{2}$$
We'll need these properties, so let's start :

Let our limit be $\Lambda$
\begin{align}
\Lambda&=\lim_{x\to 0}\frac{x-\ln(x+1)}{x\sin(2x)}\\
&=\lim_{x\to 0} \frac{1}{\sin(2x)}\times \frac{x-\ln(x+1)}{x}\\
&=\lim_{x\to 0}\frac{x}{\sin(2x)}\times \frac{x-\ln(x+1)}{x^2}\\
&=\lim_{x\to 0}\frac{1}{\frac{\sin(2x)}{x}}\times \frac{x-\ln(x+1)}{x^2}\\
&=\frac{1}{2}\lim_{x\to 0} -\frac{\ln(x+1)-x}{x^2}\\
&=\frac{1}{2}\times \left((-1)\times \left(-\frac{1}{2}\right)\right)\\
&=\frac{1}{4}
\end{align}
Hence your limit
$$\Lambda=\lim_{x\to 0}\frac{x-\ln(x+1)}{x\sin(2x)}=\frac{1}{4}$$


问:
Equivalent sets for naturals
Prove that the sets $\{0, 1\}^\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$ are equivalent.

My friend gave me the following proof:

Proof.

It suffices to show that $\{0, 1\}^\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$ have the same cardinality so we need to find a function. In particular, we need

$$\{f:\mathbb{N} \rightarrow \{0, 1\}\} \longleftrightarrow \{f:\mathbb{N} \rightarrow \mathbb{N}\}$$

Consider $f: \mathbb{N} \rightarrow \mathbb{N}$.

Define $\phi(f) = (1 ... 1 0 1 ... 1 0 ...) = ((1 ... 1) 0 (1 ... 1) 0 ...)$, where the $i$th grouping of $1$s can be treated as $f(i)$.

We have $\phi:\{f:\mathbb{N} \rightarrow \mathbb{N}\} \rightarrow \{0, 1\}^{\mathbb{N} \times \mathbb{N}}$, with $\{0, 1\}^{\mathbb{N} \times \mathbb{N}}$ having the same cardinality as $\{0, 1\}^\mathbb{N}$.

Then we also have $\phi(f) = 1_{\{(n, f(n)) | n \in \mathbb{N}\}}$.

End.

He barely explained this and did not explain any further. I have been trying to make sense of it but cannot figure out what is happening or how to formalize it. Can somebody write it down as a clear, formal proof and/or explain it? Any assistance is appreciated.

答:
I wonder if your friend is a CS major or likes computability theory, because I feel the idea resembles how you would pass multiple integer parameters to a Turing machine.

First, use Schröder–Bernstein. Changing the codomain from $\{0,1\}$ to $\mathbb N$ gives an injection from $\{0,1\}^{\mathbb N}$ to $\def\N{\mathbb N}\N^{\N}$, so it suffices to find another injection the other way around, i.e. from $\def\N{\mathbb N}\N^{\N}$ to $\{0,1\}^{\mathbb N}$.

Now this is how you would do it: given $f:\N\rightarrow\N$, think of the output as an infinitely (on one side) long tape, where any entry has either 0 or 1 written on it. How would you code an integer sequence $\{f(n)\}_n$ onto it? Well, start by writing $f(0)+1$ 1's, then write a 0 as a separator, and then write $f(1)+1$ 1's, etc. It is clear that this gives an injection (it "codes" an infinite tuple).

EDIT: More explicit expressions are given here. For changing the codomain, you could view it as post-composing the function by the inclusion map $i:\{0,1\}\rightarrow\N$. Since $i$ is injective (i.e. a monomorphism), given any two maps $f,g, i\circ f=i\circ g$ implies $f=g$. Hence $f\mapsto i\circ f$ gives an injection.

For the converse side, given $f:\N\rightarrow\N$, define $g_f:\N\rightarrow\{0,1\}$ (associated with $f$) given by $g_f(x)=0$ when $x=2n+1+\sum_{k=0}^nf(k)$ for some $n\in\N$ and $g_f(x)=0$ otherwise. Check that $g_f$ agrees with the above description, and $f\mapsto g_f$ is an injection.

Title: Equivalent sets for naturals


问:
Showing unique solution to an Initial Value Problem
I am trying to show that the equation
$$dy/dt = \cos^2(y)$$
has a unique solution when $$0 \le t \le1$$
and
$$y(0) = \pi/4$$ I don't believe I can just solve this by integrating both sides like I can with simple IVPs, yet it seems from some searching on the web that the "Lipschitz condition" can be used to demonstrate that this equation has a unique solution. I'm finding it very difficult to understand what the Lipschitz condition means, and would be very grateful both for an explanation of what it means conceptually as well as how it can be used to solve this problem.

答:
Don't fret when you see a $y$ inside parentheses in a differential equation!

The equation

$$\dfrac{dy}{dt}=\cos^2(y)$$

looks nonlinear but is actually 'trivially' integrable.

As long as a ODE is of the form

$$y'(x)=A(y(x))\cdot B(x)$$

its integrable (regardless of whether your variables are $y(x)$, $x{t}$ or anything else). Or more generally

$$y^{(n)}(x)=A(y(x))\cdot B(x)$$

where $y^{(n)}$ is the $n$-th derivative of $y$.

Solution by separation of variables

$$\dfrac{dy}{dx}=\cos^2(y)$$
$$\dfrac{dy}{cos^2(y)}=dx$$
$$\int \dfrac{dy}{cos^2(y)}=\int dx$$

Now you can use $u$-substitution to get the antiderivative of $\frac{1}{cos^2(x)}$ or remember what is the derivative of $\tan(x)$.

$$\tan(y)=x+C \hspace{10mm} C\in \mathbb{R}$$

Explicitly:

$$y=\arctan(x+C)$$

To find the IVP you need to plug in $0$ and $\pi/4$, solve for $C$ and put that $C$ value back to the solution.

Let's prove the solution by the Lipschitz continuity

This condition is usually introduced when learning about solution of systems of differential equations.
The theorem states that if there is a positive real number $K$ and $K\cdot |x_1-x_2|$ is greater or equal to $|f(x_1)-f(x_2)|$, then $f(x)$ is Lipschitz continous on set $X$ for any $x_1,x_1\in X$.

Let's dissect this:

Imagine you have a function and you draw a secant of it (a line that crosses the function twice) you take. The absolute value of the absolute value of its steepness is essentially the number $K$:

$$\dfrac{|f(x_1)-f(x_2)|}{|x_1-x_2|}\leq K$$

If $x_2\to x_1$ then this is just the absolute value of its derivative at $x_1$:

$$\lim\limits_{x_2 \to x_1}\dfrac{|f(x_1)-f(x_2)|}{|x_1-x_2|}=\left|\dfrac{d\hspace{1mm}f(x_1)}{dx}\right|$$

I'm sure you can imagine a number that has a greater value than the steepness of a function's derivative. The reason you see a cone at the Wikipedia page of it is because we took the absolute value of the differences of $x_1$ and $x_2$, thus always having a steeper $K$.

$$\dfrac{d\hspace{1mm}f(x_1)}{dx}\leq K$$

If there was no greater or equal $K$ the function wouldn't really be well-defined.

If $x_1\neq x_2$ we have more problems. Imagine the function

$$f(x)=x^2$$

We know that its steepness continously grows, thus we cannot find an upper bound where the function won't cross the cone. This function is globally not Lipschitz continous, it's not Lipschitz continous on its entire domain.

Another example would be

$$f(x)=\sqrt{x}$$
on the region $[0,1]$ as we get closer to $0$ the steepness of the function grows to $\infty$ thus we'll always have an intersection.

Let our function be

$$y(t)=\tan(t)$$

on the domain $[0,1]$

From the graph of $tan(t)$ it's clearly visible that the function is Lipschitz continous (locally) on the domain.

If you want prove it:

$$\dfrac{|tan(t_1)-tan(t_2)|}{|t_1-t_2|}\leq K$$

$$\exists K\in\mathbb{R}^+ \hspace{2mm}\forall t_1, t_2 \in \mathbb{R}$$

Ironically the famous universal solution of linear DE, the exponential function is not Lipschitz continous. It is analytic though.

Additional fun exercise: Try solving

$$\frac{dy}{dt}=cos^2(t)$$

Compare it to your original problem. Is there a connection between the two.


问:
Few topological operations under which the classes of CDH-spaces
It's known that $\mathbb{R}$ is a CDH space. Does anyone have any idea why $\omega^{\omega}$ is a CDH space? Then, Does anyone have any suggestion why $\mathbb{R}\times\omega^{\omega}$ is not a CDH space?

Also, the Cantor set $K$ in $[0,1]$ is CDH, and any compact metrizable space is continuous image of $K$. Then why this implies that continuous images of CDH-spaces need not be CDH?

Finally, note that $\mathbb{Q}$ is not CDH, so the property of being CDH is not hereditary.

Thanks

答:
Theorem $5.2$ of this PDF says that every topologically complete separable metric space that is strongly locally homogeneous is CDH. A space $X$ is strongly locally homogeneous if it has a base $\mathscr{B}$ such that if $x,y\in B\in\mathscr{B}$, there is an autohomeomorphism $h$ of $X$ such that $h(x)=y$, and $h(z)=z$ for all $z\in X\setminus B$. Since $\omega^\omega$ is strongly locally homogeneous, it is also CDH.

To see why $\Bbb R\times\omega^\omega$ is not CDH, note that $\Bbb R$ is connected, $\omega^\omega$ is zero-dimensional, and autohomeomorphisms permute the connected components. The components of $\Bbb R\times\omega^\omega$ are the sets $\Bbb R\times\{x\}$ for $x\in\omega^\omega$. If $D$ is a dense subset of $\Bbb R\times\omega^\omega$ that has at most one point in each component, and $E$ is one that has more than one point in some component, there is no autohomeomorphism of $\Bbb R\times\omega^\omega$ that sends $D$ to $E$.

Finally, $[0,1]$ is the continuous image of $K$ but is not CDH.


问:
Solution of SDE $dX(t)=a(t)dt+b(t)dW(t)$ is gaussian?
A stochastic process $X(t)$ by definition is gaussian iff all its finite-dimensional joint probability density functions are multivariate gaussian. Namely iff given the times $(t_1,t_2,...,t_n)$ , the random variables $(X(t_1),X(t_2),...,X(t_n))$ are jointly gaussian (n is arbitrary).

My intent is understanding why the solution of the following simple stochastic differential equation is a gaussian process:

$$dX(t)=a(t)dt+b(t)dW(t)$$

where W(t) is a Wiener process with diffusion coefficient D and the initial condition is $X(t_0)=X_0$, where $X_0$ is a random variable that is independent of W(t). a(t) and b(t) are deterministic "regular"
functions.

The explicit solution is:

$$X(t)=X_0+ \int_{t_0}^{t}a(s)ds+\int_{t_0}^{t}b(s)dW(s)$$

If the integral $\int_{t_0}^{t}b(t)dW(t)$ were a gaussian process (I'm not sure about this) and the initial condition were a gaussian random variable, than at fixed t, the random variable $X(t)$ would be the sum of 3 independent gaussian random variables ( $\int_{t_0}^{t}a(t)dt$ is in fact degenerate) and would be therefore gaussian. But what about the multiple-times joint densities of $(X(t_1),X(t_2),...,X(t_n))$?

答:
You can get rid of the drift applying the Girsanov's theorem (see for instance Oksendal's introductory book on SDEs, or "Introduction to stochastic integration" by H.H. Kuo), so you end up with $X(t)$ being a Wiener integral (plus the initial condition, but since it's independent and Gaussian I will ignore it hereafter ).

As you may know the Wiener integrals are gaussian random variables, and by the linearity of the integral any linear combination of Wiener integrals will be in turn a Wiener integral and hence gaussian. This allows to conclude that variables $X(t_1),...X(t_n)$ are jointly gaussian.


问:
Which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses?
How would I find which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses? I assume I would find them by finding which elements in $\mathbb{Z}_{26}$ have a $\gcd(a, m) \neq 1$. Is this the correct approach? I have gotten that 2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, and 24 don't have multiplicative inverses in $\mathbb{Z}_{26}$.

答:
hint

$$26=2\times 13$$

$$\phi(26)=26(1-\frac 12)(1-\frac{1}{13})=12$$

So, there are $ 26-12=14 $ elements with no inverse.

$\phi(n) $ is the number of inversible elements in $ \Bbb Z_n$.


问:
Is $K=\{f_\lambda(x)=e^{\lambda x}\mid \lambda \in [0,a], x\in [0,b]\}$ equcontinuous?
I am trying prove that the following set is equcontinuous.

$K=\{f_\lambda(x)=e^{\lambda x}\mid \lambda \in [0,a], x\in [0,b]\}$

I read two prove of equcontinuty in two theorem and twice of theme have the hypothesis of compactness. So I have no idea prove this without compactness. Please help me with Hints or similar.

答:
Another one:  if you want to avoid saying "compact".  The derivative of $e^{\lambda x}$ is $\lambda e^{\lambda x}$.  The value of $e^{\lambda x}-e^{\lambda y}$ is that derivative at some point $\xi$ of $[0,b]$ times $(x-y)$.  So
$$
\left|e^{\lambda x} - e^{\lambda y}\right| = \lambda e^{\lambda \xi}|x-y|
\le a e^{ab} |x-y| .
$$
Can you see how to finish?</p>


答:
That is exactly the right approach.

If $\gcd(a,m) =1$ then there will exist integers  $x,y$ so that $ax + ym = 1$ and $ax = 1 + ym$ so $ax \equiv 1$ so $x$ is the inverse.

If $a$ has a multiplicative inverse $w$ then $aw \equiv 1 \pmod m$ so there is an integer $v$ so that $aw = 1 + mv$.  So $aw -mv = 1$ but that's only possible if $\gcd(a,m) =1$.

So $a$ has an inverse $\pmod m$ if and only if $\gcd(a,m)=1$.

So you answer (if you include $0$) is correct.

Title: Which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses?

Body: How would I find which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses? I assume I would find them by finding which elements in $\mathbb{Z}_{26}$ have a $\gcd(a, m) \neq 1$. Is this the correct approach? I have gotten that 2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, and 24 don't have multiplicative inverses in $\mathbb{Z}_{26}$.


问:
Where does $\sqrt{n}\left(\frac{n}{S_n} - \frac{1}{\mu}\right)$ approach by distribution?
Let $\{X_n\}_{n=1}^{+\infty}$ be a sequence of i.i.d. random variables with expectation $\mathbb{E}{X_1} = \mu$ and variance $\text{Var}({X_1}) = \sigma^2$. Now consider a sequence $S_n = X_1 + X_2 + \dots + X_n$. Where does $\sqrt{n}\left(\frac{n}{S_n} - \frac{1}{\mu}\right)$ approach by distribution?

I think I should use the central limit theorem (CLT) here, but I'm not sure how can I use the fact that $\dfrac{S_n - n\mu}{\sigma\sqrt{n}} \to Z \sim \mathcal{N}(0, 1)$ by distribution.

Any ideas or help on how to approach this problem would be greatly appreciated.

答:
From CLT, you have

$$\sqrt{n}\left( \frac{S_n}{n} - \mu\right) \xrightarrow{n \to+\infty} N(0,\sigma^2)$$

Apply the delta method with $g(x) = \frac{1}{x}$, you get

$$\sqrt{n}\left( \frac{n}{S_n} -  \frac{1}{\mu}\right) \xrightarrow{n \to+\infty} N(0,\sigma^2 \frac{1}{\mu^4})$$


答:
Exactly and only those values not coprime to $26$: even values and $13$.

Perhaps more important is why there must be a multiplicative inverse for all the co-prime values. This hinges on knowing that there is only one multiplicative identity (that is, $1$) and so that $a\cdot x = x \implies a = 1$. Then we can build a running product of any coprime number $n$ (taking successive powers $n^2, n^3, \ldots$), which also can only be coprime values, and eventually we will reach $n^k=1$ - so the step before that, $n^{k-1}$, gave us the multiplicative inverse of $n$.


问:
Pointwise convergence of absolute value function
Suppose that $f_1, f_2,..$ be a sequence of function from $\mathbb{R} \to \mathbb{R}$.

Suppose also that the sequence $f_n$ converges pointwise to $f$.

Is it true that the sequence $|f_n|$ also converges pointwise to $|f|$ ? Could you please provide a proof ?

P/s:

I understand that to prove the statement (if it is true), we need to prove that for every $x \in \mathbb{R}$, for every $\epsilon > 0$, there exists a number $N > 0$ such that $||f_n(x)| - |f(x)|| < \epsilon $ for all $n > N$.

However, I don't know how to make appear the absolute value $|f_n(x)|$ as well as $|f(x)|$ from the hypothesis of pointwise convergence of $f_n$.

答:
Hint:

The reverse triangle inequality states that, if $x,y\in \mathbb{R}$ then:
$$
\left||x|-|y|\right| \leq |x-y|.
$$

Can you apply this to your question?


答:
Remember that the units of the variance are $m^2.$
If you want an intuitive understanding [in addition to @BenjaminWang's Answer (+1)], it may help
to compare standard deviations.

Simulation: First, multiplying single observations by $5.$
With a million iterations one can expect about two significant digits
of accuracy, maybe three.

```<code>set.seed(225)
x = rnorm(10^6, 150, 4)
mean(x);  var(x);  sd(x)
[1] 149.994    # aprx E(X) = 150
[1] 16.04013   # aprx Var(X) = 16
[1] 4.005013   # aprx SD(X) = 4
</code>```

Multiplying by $5:$

```<code>mean(5*x);  var(5*x);  sd(5*x)
[1] 749.9667  # aprx E(5X) = 5*150 = 750
[1] 400.3069  # aprx Var(5X) = 25Var(X) = 25(16) = 400
[1] 20.00767  # aprs SD(5X) = 5SD(X) = 5(4) = 20

hdr = "Five Times Individual Obs: NORM(750, 20)"
hist(5*x, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dnorm(x, 750, 20), add=T, col="orange", lwd=2)
</code>```

Second looking at $S = X_1 + X_2 + \cdots X_5:$

```<code>s = replicate(10^6, sum(rnorm(5, 150, 4)))
mean(s);  var(s);  sd(s)
[1] 750.0015  # aprx E(S) = 150 + ... + 150 = 750
[1] 80.09643  # aprx Var(S) = 16 + ... + 16 = 80
[1] 8.949661  # aprx 80^.5 = 8.944
sqrt(80)
[1] 8.944272  # exact

hdr = "Sum of 5 Indep Obs: NORM(750, 8.94)"
hist(s, prob=T, br=60, xlim=c(650,850), col="skyblue2", main=hdr)
 curve(dnorm(x, 750, sqrt(80)), add=T, col="orange", lwd=2)
</code>```

Title: The Variance Rule of $V(cX) = c^2V(X)$

Body: As a statistics major, I learned a while back that the $V(cX) = c^2 V(X)$, where $V(X)$ is the variance of $X$. However, I never quite understood why that was the case. Consider this: We have an event $X$ such that each event is independent of any previous events. Take for instance, a roll of copper wire. We know that the mean length is $150 \text{m}$ and the standard deviation is $4\text{m}$. If we want to find the variance of $5$ rolls of copper wire, we could write this as $V(5X) = 25V(X) = 400\text{m}^2$. However, an alternative way to write the problem is doing $V(X + X + X + X + X) = V(X) + \ ... \ + V(X)$ - as they are all independent. Thus, we can simplify to $5V(X)$ and get $80\text{m}^2$. This second way seems like it should be the same as the first way, but they produce different answers. Can anyone explain this? I was helping a friend out with a similar problem and the second way actually produces the correct answer, so can anyone explain why this is the case?

Body:
Remember that the units of the variance are $m^2.$
If you want an intuitive understanding [in addition to @BenjaminWang's Answer (+1)], it may help
to compare standard deviations.

Simulation: First, multiplying single observations by $5.$
With a million iterations one can expect about two significant digits
of accuracy, maybe three.

set.seed(225)
x = rnorm(10^6, 150, 4)
mean(x);  var(x);  sd(x)
[1] 149.994    # aprx E(X) = 150
[1] 16.04013   # aprx Var(X) = 16
[1] 4.005013   # aprx SD(X) = 4

Multiplying by $5:$

mean(5*x);  var(5*x);  sd(5*x)
[1] 749.9667  # aprx E(5X) = 5*150 = 750
[1] 400.3069  # aprx Var(5X) = 25Var(X) = 25(16) = 400
[1] 20.00767  # aprs SD(5X) = 5SD(X) = 5(4) = 20

hdr = "Five Times Individual Obs: NORM(750, 20)"
hist(5*x, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dnorm(x, 750, 20), add=T, col="orange", lwd=2)

Second looking at $S = X_1 + X_2 + \cdots X_5:$

s = replicate(10^6, sum(rnorm(5, 150, 4)))
mean(s);  var(s);  sd(s)
[1] 750.0015  # aprx E(S) = 150 + ... + 150 = 750
[1] 80.09643  # aprx Var(S) = 16 + ... + 16 = 80
[1] 8.949661  # aprx 80^.5 = 8.944
sqrt(80)
[1] 8.944272  # exact

hdr = "Sum of 5 Indep Obs: NORM(750, 8.94)"
hist(s, prob=T, br=60, xlim=c(650,850), col="skyblue2", main=hdr)
 curve(dnorm(x, 750, sqrt(80)), add=T, col="orange", lwd=2)


问:
quadratic over Z7 no solutions?
I was given the question
Solve this formula for X1 and X2 over Z7 (mod7)

$$
3x^2+x+1 = 0
$$

So I went on and converted it to the quadratic formula over Z7 like and solved it like so:
$$
\frac{6\pm \sqrt{1+3*3*1}}{6} =
\frac{-1\pm \sqrt{1-4*3*1}}{6} =
\frac{-1\pm \sqrt{3}}{6} =
$$

And... now what?
Can this even be solved? is it correct to say that there is no solution as $$\sqrt{3}$$ does not exist in Z7?
Am I missing something?

答:
MOD 7

$$ 5(3x^2 + x + 1) \equiv x^2 +5x+5 \equiv (x-1)^2 - 3 \pmod 7  $$
and $(3|7) = -1;$ $3$ is not a square mod 7. The squares are $0,1,2,4 \pmod 7$

MOD 5

$$ 2(3x^2 + x + 1) \equiv x^2 +2x+2 \equiv (x+1)^2 + 1 \equiv (x+1)^2 - 4 \pmod 5  $$
and $4$ has square roots $2,-2 \equiv 2,3 \pmod 5,$ these are $x+1$ so $x \equiv 1,2 \pmod 5$

Title: quadratic over Z7 no solutions?


问:
Intuition for Strange Exponential Limit
I am an undergrad math student, and I am just looking for intuition or even just some reason why

$$\lim_{n\to\infty}\Big(1+\frac{x}{n}\Big)^n=e^x$$

I know you can justify the answer with L'Hospital's Rule or power series, but those don't illuminate why this would be true in a purely logical, mathematical sense.

Edit: I understand this is sort of difficult to answer and is very open ended. I think that I can clear it up a bit. Simply plugging in $n=\infty$ into the limit would, hypothetically, give $1^\infty$, which of course is an indeterminate form. So then why is this indeterminate, and what does that have to do with $e$? Or is this really just the definition of $e$?

答:
You can get this easily from the elementary definition of $e$, if that's what you want. Note:

$$\lim_{n\to\infty}\bigg(1 + \frac{x}{n}\bigg)^{n}\ \stackrel{t= \frac{n}{x}}{=}\ \lim_{t\to \infty}\bigg(\bigg(1 + \frac{1}{t}\bigg)^{t}\bigg)^{x}=\bigg(\lim_{t\to \infty}\bigg(1 + \frac{1}{t}\bigg)^{t}\bigg)^{x} = \boxed{e^{x}}$$

Title: Intuition for Strange Exponential Limit


问:
A quick question about showing that a sequence is increasing
For a sequence $a_n$ can we show that: $$a_n \leq a_{n+1} \iff \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n} \geq 1$$

Or a proof like this wouldn't be sufficient?

答:
I'll assume your statement, written more precisely, is:

$$\exists N \in \mathbb{N} \text{ such that }\ a_n \leq a_{n+1}\ \forall n \geq N \iff \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n} \geq 1.$$

$\ \Rightarrow\ $ is false because the limit may not exist. e.g. $1,1,2,2,4,4,8,8,16,16,32,32,64,64,...$. Here, $ \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}$ does not exist (it alternates between $1$ and $2$).

$\Leftarrow\ $ is also false, e.g. consider $(a_n)_{n \in \mathbb{N}} = (\frac{1}{n})_{n \in \mathbb{N}}.$ Then $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n} \geq 1,\ $ but $\nexists N \in \mathbb{N} \text{ such that }\ a_n \leq a_{n+1}\ \forall n \geq N.$


问:
Nabla and its rules
I am currently discovering vector analysis and was asked to prove that $(\vec{u} . \nabla) f \neq ( \nabla . \vec{u})f$ with $\vec{u}$ a vector field and $f$ a scalar field. I really struggle to understand how this nabla thing works. Can someone give me examples/rules ?

I found the correct proof but I don't understand why it works :
$(\vec{u} . \nabla) f = (u_x \partial_x + u_y \partial_y + u_z \partial_z)f $ but why is $u_x \partial_x \neq \partial_x u_x$,  and what is $u_x \partial_x$, why is this "$\partial_x$" not containing anything inside ? Shouldn't it be differentiating a function ? Sorry for all these questions, this is very new to me.

答:
"Nabla" is a symbolic "vector differential operator".  It can be written, symbolically, $\nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial Y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$.

Just as there are three kinds of vector products, "scalar product", which multiplies a vector and a scalar to get a vector, "dot product", which multiplies two vectors to get a scalar, and "cross product", which multiplies two vectors to get a vector, so there are three kinds of "products" with nabla.

"grad" (gradient) which applies to a scalar valued function, f(x, y, z), to get a vector function:
$\nabla f= (\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k})f(x,y,z)= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$"

"div" (divergence) which applies to a vector valued function, $\vec{F(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ to get a scalar valued function: $\nabla\cdot \vec{F}= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$.

"curl" which applies to a vector valued function, $\vec{F(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, to get a vector valued function:
$\nabla\times F= $$\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y,z) & g(x,y,z) & h(x,y,z)\end{array}\right|$

$= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$.

Title: Nabla and its rules


问:
Does the series $\sum\limits_{k=1}^{\infty} \int\limits_0^1 \frac{\cos(2 k \pi x) \cos(2 n \pi x)}{k}\,\mathrm dx$ equal $0$ for $n \in \Bbb N$?
Could someone confirm if my answer is correct?
$$\sum_{k=1}^{\infty} \int_0^1 \frac{\cos(2 k \pi x) \cos(2 n \pi x)}{k}\,\mathrm dx=0, \quad n\in \mathbb{N}$$</p>

答:
$$I:=\int\limits_0^1\frac{\cos^22n\pi x}{n}dx$$

$$u:=2n\pi x\implies dx=\frac{du}{2n\pi}\implies$$

$$I=\frac1{2n^2\pi}\int\limits_0^{2n\pi}\cos^2u\,du=\left.\frac1{2n^2\pi}\left(\frac{u+\sin u\cos u}2\right)\right|_0^{2n\pi}=\frac1{2n^2\pi}\left(\frac{2n\pi}2\right)=\frac1{2n}$$


答:
To begin with, we better have $a_n>0$ for all $n$, or else we may have (additional) problems with some inequalities turning over.

Even then your suggested equivalence is far from true:

If $a_n\le a_{n+1}$ for all $n$, then either $\lim_{n\to\infty}\frac{a_{n+1}}{a-n}\ge 1$ or the limit doesn't exist.

If $\lim_{n\to\infty}\frac{a_{n+1}}{a-n}> 1$, then $a_n<a_{n+1}$ for all but finitely many $n$

If $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}= 1$, then it may be the case that $a_n\le a_{n+1}$ doesn't hold for any $n$ at all.

Concrete conunter examples:

With $a_n=2^{\lfloor n/2\rfloor}$ we have $a_n\le a_{n+1}$ but $\frac{a_{n+1}}{a_n}$ alternates between $1$ and $2$, hence the limit does not exist.

With $a_n=n!$ we have $a_n\le a_{n+1}$ but $\frac{a_{n+1}}{a_n}=n\to\infty$ and again there i sno convergence. However, you way indeed want to allow $\lim=+\infty$ so this is perhaps not  a counterexample.

For some (big) $N\in\Bbb N$, let $a_n=\max\{2^N-n,2^n\}$. Then  $\lim\frac{a_{n+1}}{a_n}=2>1$, but $a_{n+1}<a_n$ for $1\le n<N-1$.

Let $a_n=1+\frac1n$. Then $a_n>a_{n+1}$ for all $n$, but  $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}= 1$.

Title: A quick question about showing that a sequence is increasing

Body:
For a sequence $a_n$ can we show that: $$a_n \leq a_{n+1} \iff \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n} \geq 1$$

Or a proof like this wouldn't be sufficient?


问:
If $S\subset \mathbb{R}^n$ is closed, then $\bigcap_{k=1}^{\infty}(1+\frac{1}{k})S=S$
Based on the comment I edited the question.

Let $S \subset \mathbb{R}^n$ be a bounded symmetric convex set.

Question: How to show that if $S$ is closed, then $\bigcap_{k=1}^{\infty}(1+\frac{1}{k})S$ is equal to $S$.

I do not know how to use the closed property to prove this lemma.

Thanks for any help.

答:
If $S$ is nonempty, symmetric about the origin and convex, then it's easy to see that $0\in S$.
Now, if $U$ is convex and $0\in U$, then $U \subset \alpha U$ for all $\alpha\geqslant 1$, so clearly $S$ is contained in the intersection (regardless of $S$ being closed).

On the other hand, if $x\in\bigcap_{k=1}^{\infty}(1+\frac{1}{k})S$, then $x\in (1+\frac{1}{k})S$ for each $k\geqslant 1$.
This means that for each $k\geqslant 1$ there is some $x_k \in S$ with $x = (1+\frac1k)x_k$.

In other words, the sequence $(x_k)_{k\geqslant 1}$ given by

$$x_k = \left(\frac 1{1+\frac1k}\right)x = \left(\frac{k}{k+1}\right)x$$

lies in $S$.

Now, does this sequence converge? And if $S$ is closed, what else can you say...?


答:
If you ask a mathematician what the number $e$ is and what makes it so special, he will either say it's properties in analysis, such as
$$\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x,\quad \int e^x\, \mathrm{d}x = e^x+c $$
or it's beatiful combination with $i,\pi,1$ and $0$
$$e^{i\pi}+1=0,$$
the famous Euler's identity, which is usually proven by the also beatiful power series for $e$ that Euler discovered:
$$\sum_{n=0}^{\infty}\frac{1}{n!}.$$

None of these formulas is the one you stated. But still, this formula has it's own history. It is the definition of $e$, how it all started, how Euler discovered further properties. Below are two interesting YouTube videos that tell you everything about the definition of $e$:

More maths (starts at 1:18)

More history (Numberfile)

Addendum: I remember quite well how I asked myself the exact same question ;).

Title: Intuition for Strange Exponential Limit

Body:
I am an undergrad math student, and I am just looking for intuition or even just some reason why

$$\lim_{n\to\infty}\Big(1+\frac{x}{n}\Big)^n=e^x$$

I know you can justify the answer with L'Hospital's Rule or power series, but those don't illuminate why this would be true in a purely logical, mathematical sense.

Edit: I understand this is sort of difficult to answer and is very open ended. I think that I can clear it up a bit. Simply plugging in $n=\infty$ into the limit would, hypothetically, give $1^\infty$, which of course is an indeterminate form. So then why is this indeterminate, and what does that have to do with $e$? Or is this really just the definition of $e$?


问:
Weak stationary white noise sequence that are identically distributed but not independent
Let $\{Z_t\}$ be a sequence of random variables that are weakly stationary, i.e., $\mathbb{E}[Z_t]=0$ and $Var[Z_t]=\sigma^2$ and $Z_t$'s are uncorrelated, i.e., $cov(Z_s,Z_t)=0$ for all $t\neq s$. Can we create such a sequence of $Z_t$ that are not independent? Also, can we have the other way around, that is, $Z_t$'s are independent but not identically distributed?

My try:
For the first question, I am thinking of $Z_t = (-1)^tX_t$ where $X_t \sim N(0,1)$ but do not know how to show they are dependent.

答:
If one chooses to define discrete-time white noise $\{Z_n\}$ as a sequence of zero-mean random variables that are uncorrelated but not necessarily independent, then there is nothing left to prove; the definition says it all. The $Z_n$'s are uncorrelated but need not be independent.

A far superior deinition of discrete-time white noise is as a sequence of zero-mean random variables that are independent (preferably identically distributed as well). The key point here is that with uncorrelated but dependent random variables, one cannot make a linear prediction (other than $0$) of the value of $Z_{n+i}$ given $Z_n$ (say) but it might be possible to make a nonlinear prediction.  In fact, the minimum mean-square error predictor of $Z_{n+i}$ when we know $Z_n$ is $E[Z_{n+i}\mid Z_n]$ which is, in general, a nonlinear function of $Z_n$. It sticks in my craw that one should be able to predict the value of white noise in any way (not just linear prediction), and the requirement of independence for white noise sequences eliminates this possibility:  if $Z_{n+i}$ and $Z_n$ are independent, then $E[Z_{n+i}\mid Z_n] = E[Z_{n+i}] = 0$.

Now, if one wants a sequence of zero-mean random variables that are uncorrelated but not necessarily independent, consider the following.

Let $\{X_{2n}\colon n\in \mathbb Z\}$ be a sequence
of independent identically distributed zero-mean standard normal random variables, and let $\{B_n \colon n\in \mathbb Z\}$ be a sequence of independent discrete random variables taking on values $+1$ and $-1$ with equal probability $\frac 12$ and independent of the $X_{2n}$'s. Now set $X_{2n+1} = X_{2n}B_n$ for each integer $n\in \mathbb Z$.
Verify that $X_{2n+1}$ is also a standard normal random variable and that $X_{2n}$ and $X_{2n+1}$ are uncorrelated normal random variables (hint: $E[X_{2n}X_{2n+1}] = E[X_{2n}^2]E[B_n] =0$). But note that $X_{2n}$ and $X_{2n+1}$ are not independent random variables; indeed, conditioned $X_{2n}=\alpha$ we know that the conditional distribution of $X_{2n+1}$ is that of a discrete random variable taking on values $\pm\alpha$ with equals probability instead of serenely continuing to be standard normal as independence demands.
Thus, $\{X_{m}\colon m\in \mathbb Z\}$ is a sequence of uncorrelated standard normal random variables (discrete-time white noise process according to the OP's definition) in which pairs $(X_{2n},X_{2n+1})$ are not independent (though they are uncorrelated). Call such pairs doublets. More generally, $(X_m, X_k)$ are either part of the same doublet (and hence uncorrelated but not independent) or belong to different doublets and thus are uncorrelated as well as independent.

Turning to the OP's other questions to whether the $Z_n$'s can be independent but not identically distributed, it is trivial to create such a "white noise process" according to the OP's definition. All we need is to have a sequence of independent zero-mean random variables with varying distribution functions, but all having the same finite variance $\sigma^2$; for example. $Z_n \sim N(0,\sigma^2)$ if $n$ is odd and
$Z_n \sim U\left(-\sqrt{3}\sigma. +\sqrt{3}\sigma\right)$ if $n$ is even.It is not clear to me that this type of process has any particularly important role to play in any situation, but ymmv just as much as the OP's seems to be doing.


问:
Why is the genus (in terms of Euler characteristic) the maximum number of disj. closed curves that can be embedded in a surface without separating it?
Let $S$ be a compact surface and $g$ its genus defined in terms of Euler characteristic.
Why is $g$ the maximum number of disjoint simple closed curves that can be embedded in the surface without disconnecting it?
I would be grateful if anyone could tell me how to prove it or give me a reference. Thanks.

答:
Starting with the orientable case, consider a closed, connected, oriented surface $S$ of genus $g$. The genus is related to its Euler characterstic $\chi(S)$ by the formula $\chi(S)=2-2g$, and this number $g = 1 - \frac{\chi(S)}{2}$ is equal to that maximal number of pairwise disjoint circles in $S$ that do not separate $S$. Here's the proof.

When you cut along a circle, the Euler characteristic does not change. So if you cut $S$ along $k$ circles, and assuming those circles do not disconnect $S$, you will get a compact, connected, orientable surface-with-boundary that I'll denote $A_k$, that has $2k$ boundary circles, and that has Euler characterstic
$$\chi(A_k) = \chi(S)
$$
You can then glue a single disc to each of those $2k$ circles, to get a closed connected surface that I'll denote $B_k$. The same logic (gluing along circles does not change the Euler characteristic), together with the fact that each individual disc has Euler characterstic $1$, gives the result
$$\chi(B_k) = \chi(A_k) + 2k = \chi(S) + 2k
$$
It is known that the maximum Euler characteristic of a connected, closed, orientable surface is the Euler characteristic of the sphere which is equal to $2$, and therefore
$$\chi(S) + 2k \le 2
$$
Substituting $\chi(S)=2-2g$ one obtains $g \le k$.

To show that the maximum $k=g$ is obtained, start with a sphere whose Euler characteristic equals $2$, remove $2g$ discs which reduces the Euler characterstic to $2-2g$, and now glue those circles in pairs to give a closed oriented surface of the same Euler characterstic $2-2g$ as your surface $S$, together with a system of $g$ circles that do not separate the surface. Since the Euler characteristic is a complete topological invariant of closed, connected, oriented surfaces, and knowing that $\chi(S)=2-2g$, this surface is homeomorphic to the original surface $S$.

Now for the non-orientable case. Let $S$ be a closed, connected, nonorientable surface. The terminology here is less standard, so I'll be careful and refer to the nonorientable genus $g$ of $S$, which is related to Euler characterstic by the formula $\chi(S)=2-g$, and is again equal to the maximal number of pairwise disjoint circles in $S$ that do not separate $S$.

The proof is a bit trickier in the nonorientable case, due to the fact that there are two kinds of circles in $S$: two-sided and one-sided (in an orientable surface, every circle is two-sided). So, suppose that we have a pairwise disjoint collection of $k$ circles in $S$ that do not separate $S$. We may partition this collection into $k_1$ one-sided circles and $k_2$ two-sided circles, with $k = k_1 + k_2$. Again let $A$ be the connected surface-with-boundary obtained by cutting along these circles. Again $\chi(A)=\chi(S)$. But in this case the number of boundary circles of $A$ is equal to $k_1 + 2 k_2$. So, gluing in discs produces a closed surface $B$ with $\chi(B)=\chi(A)+k_1 + 2k_2 = \chi(S) + k_1 + 2 k_2$. We therefore have
$$2 - g + k = \chi(S) + k_1 + k_2 \le \chi(S) + k_1 + 2 k_2 = \chi(B) \le 2
$$
and so
$$k \le g
$$
To show that the maximum is achieved, one can get a hint by examining the above chain of inequalities. First, if $k_2 \ge 1$ then we get strict inequality $k < g$. Also, if $B$ is not the sphere then we again get strict inequality $k < g$. So, the only way that the maximum can be achieved is by a collection of $g$ one-sided circles which, when cut along, give a sphere with $g$ discs removed.

So, start from a sphere with $g$ discs removed, having $g$ boundary circles. Identify each boundary circle to itself by the antipodal map. The result is a non-orientable surface of Euler characteristic $2-g$. Since closed, connected, non-orientable surfaces are classified by their Euler characterstic, the resulting surface is homeomorphic to your original surface.

Here's one last comment relating the two concepts of genus. Given a closed, connected, orientable surface $S$, its genus is also equal to the number of toruses ($T^2 = S^1 \times S^1$) needed so that their connected sum is homeomorphic to $S$. And given a closed, connected, nonorientable surface $S$, its nonorientable genus equals the number of projective planes needed so that their connected sum is homeomorphic to $S$. Often these are the descriptions used to define genus and nonorientable genus.


问:
Find two LI continuous differentiable functions $f,g: I \in \mathbb{R} \rightarrow \mathbb{R}$ such that $(f\cdot g')(t) - (f'\cdot g)(t)) = 0$.
Find two linear independent continuous differentiable functions $f,g: I \in \mathbb{R} \rightarrow  \mathbb{R}$ with  $(f\cdot g')(t) - (f'\cdot g)(t)) = 0$.

Any idea on how can I find such functions? I imagine that, if $(f\cdot g')(t) - (f'\cdot g)(t)) = 0$, then:

$(f\cdot g')(t) = (f'\cdot g)(t)) \Longrightarrow \frac{f'}{f}(t) = \frac{g'}{g}(t) \Longrightarrow \ln(f(t)) = \ln(g(t)) + C \Longrightarrow f(t) = C \cdot g(t)$.

This implies $f,g$ are L.D., a contradiction. Maybe the trick is with choosing the right interval $I$ to define the functions. I don't know.

Any help? Thanks!

Body: Hint: Let $I=[-1,1]$. $f(t)=t^2$, $g(t)=t\left|t\right|$.


问:
Quotient of Cayley graph of the free group on two generators by a subgroup.
If $F=F(\{a,b\})$ is the free group on two generators $a$ and $b$ and $G$ is the subgroup $$G=\:\langle b^n a b^{-n}|\: n\in \mathbb{N}\rangle \leq F$$
I am trying to work out what the quotient graph $\Delta / G$ looks like, where $\Delta = \Delta(F;\{a,b\})$, the Cayley graph of F with respect to it's generators.

I think that $\Delta$ is the 4-regular tree, and I think I understand that if you quotient out the Cayley graph by the group itself then all of the vertices become one and so you get a wedge of $|S|$ circles where $S$ is the generating set, but clearly when we quotient out by a smaller group, we see that not all vertices become one.

In fact, $G$ is not finitely generated (as far as I can tell), and so do we get a sort of infinite wedge of circles? I assume we would if we looked at the quotient $\Delta(G;S)/G$ for $S$ the set of generators of $G$, but we are taking a bigger initial group so I would think that it would still be an infinite wedge of circles, but with a 'bigger' infinite number... In that case its only vertex would have infinite valence.

Or do we just think of it as the Cayley graph of the finitely generated group with generators $a,b$ but an infinite set of relations (those given by the elements of $G$?) So is it just an infinite 4-regular graph?

答:
If you think of $F$ as acting by left multiplication on the vertices of $\Delta$, then the natural interpretation of the quotient graph is as a graph whose vertices are the distinct cosets $Gg$ of $G$ in $F$, with an edge (labelled $x$) from $Gg \to Ggx$ for each $x \in \{a,a^{-1},b,b^{-1}\}$.

So, if you ignore the labels, it is still a an infinite 4-regular graph, but it is no longer a tree, and it is not a simple graph, because there are edges labelled $a$ and $a^{-1}$ from the vertex $Gb^n$ to itself for each $n \in {\mathbb N}$. (Does your ${\mathbb N}$ include $0$?)

Note that if you take $n \in {\mathbb Z}$ rather than $n \in {\mathbb N}$, then you get a normal subgroup, and the only vertices are $Hb^n$.


问:
Arc length equal to distance travelled by a particle in a cycloid
How can I prove that the arc length is equal to the distance travelled by a particle in a cycloid.

I know the definition of arc length and that I can get it by multiplying the angle (radians) and the radius.

And with that I can get $x = r(t-\sin(t))$ and $y = r(1-\cos(t))$.

But I want to know, using math, why the distance travelled in the $x$-axis is equal to the length of the arc.

Thanks.

Cycloid

答:
I'm not sure where you got that the distance traveled in the $x$-direction is equal to the arc length of a cycloid.

The distance traveled in the $x$-dimension is just the circumference of the generating circle, or $2\pi r$. We can confirm this using the parametrization in the question. We're given $x(t) = r(t-\sin(t))$, so through one rotation, we go from $t = 0$ to $t = 2\pi$, and we have:

$$ x(2\pi) - x(0) = 2r(2\pi-\sin(2\pi))-2r(0-\sin(0)) = 2\pi r .$$

To get to the arc length $S$ of one arch, we can integrate the differential $dS = \sqrt{{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}} $ from $t = 0$ to $t = 2\pi$:

$$ S = \int_0^{2\pi} \sqrt{{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}} \ dt $$
$$ = \int_0^{2\pi} {\sqrt{{\left(r-r\cos(t)\right)^2 + \left(r\sin(t)\right)^2}}} \ dt $$
$$ = \int_0^{2\pi} {\sqrt{r^2-2r^2\cos(t) + r^2\cos^2(t) + r^2\sin^2(t)}} \ dt $$
$$ = \int_0^{2\pi} {\sqrt{2r^2-2r^2\cos(t)}} \ dt $$
$$ = r\int_0^{2\pi} {\sqrt{2-2\cos(t)}} \ dt $$
$$ = 2r\int_0^{2\pi} {\sin\left(\frac{t}{2}\right)} \ dt $$
$$ = 4r \left[-\cos\left(\frac{t}{2}\right) \right]_0^{2\pi} $$
$$ = 8r . $$

Then, $2\pi r \neq 8r$, so the distance traveled in the $x$-dimension is not equal to the arc length.


答:
Alternative intuition

$e$ is the real number such that $\frac{d}{dx} e^x = e^x.$

$\frac{d}{dx} (1 + [x/n])^n = n \times (1 + [x/n])^{(n-1)} \times [1/n]
= (1 + [x/n])^{(n-1)}.$

By binomial expansion: $\lim_{n\to\infty} (1 + [x/n])^n ~=~ \lim_{n\to\infty} (1 + [x/(n+1)])^n~=~ \lim_{n\to\infty} (1 + [x/n])^{(n-1)}.$

Title: Intuition for Strange Exponential Limit


答:
You just have the show the limit of the natural log of the expression   is $x$.

Indeed, from a high school limit,
$$\ln\biggl(1+\frac xn\biggr)^{\!n}=n\ln\Bigl(1+\frac xn\Bigr)=x\,\underbrace{\frac{\ln\Bigl(1+\frac xn\Bigr)}{\frac xn}}_{\substack{\downarrow \rlap{\:n\to\infty}\\1}} $$
tends to $x$ as $n$ tends to $\infty$.

Title: Intuition for Strange Exponential Limit


问:
Do there exist uniform triangular prisms with all vertices in $\mathbb Z^3$?
It's quite easy to find a regular square prism (cube) or a regular triangular antiprism (octahedron) with vertices in $\mathbb Z^3$. Take for instance, take the convex hulls $$ \begin{align*}
&\operatorname{hull}(\{(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}), \text{ and}
\\
&\operatorname{hull}(\{(0,0,1),(0,1,0),(1,0,0),(0,0,-1),(0,-1,0),(-1,0,0)\})
\end{align*}
$$
respectively.

For a much more interesting example of embedding the cube, see Table 1 from Ionascu and Obando: $$
\operatorname{hull}(\{(0, 3, 2), (1, 1, 4), (2, 2, 0), (2, 5, 3), (3, 0, 2), (3, 3, 5), (4,4, 1), (5, 2, 3)\}).
$$

Question

Do there exist uniform triangular prisms with all vertices in $\mathbb Z^3$ (equivalently $\mathbb Q^3$)?
If not, can it be done in $\mathbb Z^4$, $\mathbb Z^5$, etc?

(I've checked all examples in $\mathbb Z^3$ where the triangular faces have side lengths up to $\sqrt{700}$. This did not produce any examples, but there are plenty of near-misses.)

答:
Suppose that there exists a triangular prism in $\mathbb{Z}^3$ of side length $s$. From a given vertex $V$, let $a$ and $b$ be the vectors on the triangular face containing $V$ and $c$ the vector from $V$ to its pair on the other triangular face. Note that $a,b$, and $c$ are all of length $s$.

Now, consider the integer vector $a\times b$; since $a$ and $b$ are $60^\circ$ apart, it has length $s^2\cdot\frac{\sqrt{3}}2$. Up to a change of sign, this vector is parallel to $c$, so $c$ is a rational multiple of $a\times b$ (since both lie in $\mathbb{Z}^3$). Thus, $s$ and $\frac{s^2\sqrt{3}}2$ differ by a rational factor, so we conclude that $s$ is a rational multiple of $\sqrt{3}$.

This means that we can scale the prism so that its side length is an integer multiple of $\sqrt{3}$. Then scale it by a further factor of $2$, so that the midpoints of each edge also have integer coordinates. Let $k\sqrt{3}$ be the integer distance from a vertex to a midpoint of an edge.

But now consider the integer vector from the midpoint of a non-triangle edge to a non-adjacent vertex: it has length $(k\sqrt{3})\cdot \sqrt{5}$.

So, the squared distance $15k^2$ between these two integer points is the sum of three squares. Note that $k^2$ is of the form $4^n(8a+1)$, so $15k^2$ is of the form $4^n(8b+7)$. But, by Legendre's three-square theorem, such integers are not expressible as the sum of three squares! So no such prism exists.

There do exist triangular prisms in $\mathbb{Z}^5$: Take the first three coordinates to be any permutation of $(0,0,1)$, and the last two to be either $(0,0)$ or $(1,1)$. This gives a triangular prism of side length $\sqrt{2}$.

I'm not sure yet about $\mathbb{Z}^4$ - my guess is no, but checking some small examples seems worthwhile.


问:
Which triangle follows the condition of $2\tan \frac{A}{2}=\frac{\sin A}{\sin B\sin C}$?
The problem is as follows:

Let a triangle $A$, $B$ and $C$, where it follows this condition.

$2\tan \frac{A}{2}=\frac{\sin A}{\sin B\sin C}$

On which of the following triangles such condition is met?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{Isosceles triangle}\\
2.&\textrm{Right triangle}\\
3.&\textrm{Acute triangle}\\
4.&\textrm{Right isosceles triangle}\\
\end{array}$

This problem is somewhat complicated to my understanding. As I don't know exactly where to begin with.

$2\csc A - 2\cot A= \frac{\sin A}{\sin B \sin C}$

Then:

$2\frac{1}{\sin A}-2\frac{\cos A}{\sin A} =\frac{\sin A}{\sin B \sin C}$

But that's where I could only reach because from further expanding the earlier equation it doesn't seem to get useful information.

Can someone help me here, on how to arrive to the conclusion of which of those triangles mentioned would match that peculiar rule?.

It would help a lot an answer which could explain the most details as possible so to avoid getting confused with the steps.

答: You have: $2\tan(\frac{A}{2}) = \dfrac{\sin A}{\sin B\sin C}\implies 2\dfrac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}= \dfrac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin B\sin C}\implies \sin B\sin C = \cos^2(\frac{A}{2})= \dfrac{1+\cos A}{2}\implies 1+\cos A = 2\sin B\sin C= \cos(B-C)-\cos(B+C)= \cos(B-C)+\cos A\implies \cos(B-C) = 1\implies B-C=0\implies B = C$. Thus triangle $ABC$ is isosceles.


问:
choosing maximum number of sets
So I am still into this infinite theory setting and I am stuck at a question again. I am currently reading about the topic regarding the power set of a countably infinite set like ($\mathbb{Q},\mathbb{N}$). It is obvious that the power set is continuum. So the initial question was this(this one I solved):

Given a countably infinite set $H$, at most how many subsets can you choose from the power set $P(H)$ such that every countably infinite many have infinite elements in common, however every continuum many have finite elements in common.

Now the answer to the previous question is that you can only choose countably infinite many subsets this way. However there is a problem related to this which I cannot seem to find a solution to. The problem is as follows:

Given a countably infinite set $H$. At most how many subsets can you choose from $P(H)$ such that every $n$ set have infinite elements in common, however every $n+1$ set have finite many in common.($n$ is a positive integer at least 2)

This one seems easier at first read, but I personally find it much harder, and I can't solve it. I am really interested in the answer, and what technique works to prove it.(or if the answer is continuum many then what motivated the construction)

答:
For each $n$ we can certainly get countably infinitely many sets.

Since $H$ is countably infinite, we can index it by the $(n+1)$-element subsets of $\Bbb N$ as

$$H=\{h_F:F\subseteq\Bbb N\text{ and }|F|=n+1\}\,.$$

For each $k\in\Bbb N$ let $H_k=\{h_F\in H:k\in F\}$; clearly $H_k$ is an infinite subset of $H$. If $G\subseteq\Bbb N$, and $|G|=n$, then

$$\begin{align*}
\bigcap_{k\in G}^nH_k&=\{h_F\in H:G\subseteq F\}\\
&=\{h_{G\cup\{k\}}:k\in\Bbb N\setminus G\}
\end{align*}$$

is infinite. If $|G|=n+1$, however, then

$$\bigcap_{k\in G}^nH_k=\{h_F\in H:G\subseteq F\}=\{h_G\}$$

is finite, and if $|G|>n+1$, then

$$\bigcap_{k\in G}^nH_k=\{h_F\in H:G\subseteq F\}=\varnothing\,.$$

Added 5 March 2022: I can now improve this to get a family $\mathscr{H}$ of $\mathfrak{c}=2^\omega$ subsets of $H$ with the desired property.

Let $T$ be the complete binary tree of height $\omega$, and for $k\in\omega$ let $L_k$ be the $k$-th level of $T$. Let $\mathscr{F}$ be the family of all $n$-element sets $F\subseteq T$ such that $F\subseteq L_k$ for some $k\in\omega$; $\mathscr{F}$ is countably infinite, so we can index $H$ as $H=\{h_F:F\in\mathscr{F}\}$.

Let $\mathscr{B}$ be the set of branches of $T$; $|\mathscr{B}|=2^\omega$, and for each finite $\mathscr{C}\subseteq\mathscr{B}$ there is a minimum $m_\mathscr{C}\in\omega$ such the the branches in $\mathscr{C}$ are disjoint at level $m_\mathscr{C}$. For each $B\in\mathscr{B}$ let

$$H_B=\{h_F\in H:F\cap B\ne\varnothing\}\,,$$

and let $\mathscr{H}=\{H_B:B\in\mathscr{B}\}$; clearly $|\mathscr{H}|=2^\omega$. Suppose that $\mathscr{H}_0=\{H_B:B\in\mathscr{B}_0\}$, where $\mathscr{B}_0\subseteq\mathscr{B}$, and $|\mathscr{B}_0|=n$. For $k\ge m_{\mathscr{B}_0}$ let $F_k=L_k\cap\bigcup\mathscr{B}_0$; then $F_k\in\mathscr{F}$, and $F_k\cap B\ne\varnothing$ for each $B\in\mathscr{B}_0$, so $h_{F_k}\in\bigcap\mathscr{H}_0$, and it follows that $\bigcap\mathscr{H}_0$ is infinite.

Now suppose instead that $|\mathscr{B}_0|=n+1$. If $h_F\in\bigcap\mathscr{H}_0$ for some $F\in\mathscr{F}$, then there is a $k\in\omega$ such that $F\subseteq L_k$. Clearly $k<m_{\mathscr{B}_0}$, as otherwise the $n+1$ branches in $\mathscr{B}_0$ are disjoint at level $k$. Thus, $F$ is a subset of the finite set $\bigcup_{k<m_{\mathscr{B}_0}}L_k$, and $\bigcap\mathscr{H}_0$ must be finite.

Title: choosing maximum number of sets


问:
For valid probability densities $p(x)$ and $q(x), \int p(x)\ln(q(x))~dx = 0$. Can you catch my mistake?
I was trying to solve the following integral by parts, $$ \int p(x)\ln(q(x))dx$$ for valid probability density functions $p(x)$ and $q(x)$,

$$ \int p(x)\ln(q(x))dx = \ln(q(x)) - \int\frac{q'(x)}{q(x)}dx,$$ since $p(x)$ is assumed second function in the product and $\int p(x)dx = 1$.
$$ \int p(x)\ln(q(x))dx = \ln(q(x)) - \int\frac{q'(x)}{q(x)}dx = \ln(q(x)) - \ln(q(x)) = 0.$$
Is this wrong even if the limits are assumed to be entire $\mathbb{R}$. I'm sure this is very elementary and I'm missing something very fundamental here.

答:
It would help to specify some bounds here, say $a$ and $b$. Note that assuming the existence of $\ln(q(x))$ requires $q(x)$ to be positive on the interval (while there may be probability densities on the interval that don't have this property, we don't care about them right now).

Let $P$ be an antiderivative of $p$ on $[a,b]$, i.e. a function on $[a,b]$ with $P'(x) = p(x)$ for all $x \in [a,b]$.

By integration by parts with "$u$" taken to be $\ln(q(x))$ and "$dv$" taken to be $p(x)$ we have $du = q'(x)/q(x)$ and $v = P(x)$, so
$$
\begin{align*}
\int_a^b p(x) \ln(q(x)) \, dx & = \ln(q(x)) P(x) \Big \vert_a^b - \int_a^b P(x) \frac{q'(x)}{q(x)} \, dx \\
& = \ln(q(b)) P(b) - \ln(q(a)) P(a) - \int_a^b P(x) \frac{q'(x)}{q(x)} \, dx.
\end{align*}
$$
which looks fine to me, but no cancellation takes place.

The issue is that the integration by parts formula requires an antiderivative of $p$, which is a function, and which is definitely not the number $\int_a^b p(x) \, dx$.  (If this number were an antiderivative of $p$, then $p$ would be constant and not a valid probability distribution.)  The distinction is blurred if you omit bounds everywhere and write $\int p(x) dx$ for both objects.


问:
What is the best way to prove general differentiation product rule w/o going back to limit interpretation?
I am trying  to put some flow in how concepts and rules are established in basic differentiation: which comes first and which comes second.  Particularly, my question is what the best way is  to prove product rule in general w/o going to the very definition of differentiation. Here is my approach.

The 4 basic foundations needed are below.

Definition of differentiation: $f' (x)=\lim\limits_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Chain rule: $f'(x) = \frac {d f} {dx}
= \frac {df_1} {df_2} \frac {df_2} {df_3} \frac {df_3} {dx}$

Derivative of $ e^x:   \frac{d}{dx} e^x = e^x$

Derivative of $\ln x: \frac{d}{dx}\ln x = \frac{1}{x}$

Then we prove  product rule : $(uv)'=u'v+uv'$ in the following steps.

Let $y=u\cdot v$

$ln(y) = \ln(u)+\ln(v) $            - property of logarithm

$\frac{1}{y}y'=\frac{1} {u}u'+\frac{1} {v}u'$ - differentiation on both sides  with  chain rule, $\frac{d}{dy} ln (y) =\frac{1}{y} $

$y'=y\frac{1} {u}u'+y\frac{1} {v}u'$ - multiply both sides by $y$

$(uv)'=u'v+uv'$ - replace $y$ with $uv$

The above process can be easily applied to  $(uvw)'=u'vw+uv'w+uvw^\prime$ and so on...

Any one has suggestions for  different and better approaches?

答:
One good way of doing differential rules is to replace each variable with "variable + differential".  Then, at the end, any differential multiplied by a differential can be eliminated (you can think of it as being an infinitely small decimal multiplied by another infinitely small decimal - the value is then infinitely-infinitely small).

So, below, we start off with $z$ and then solve for $dz$.  Step 4 eliminates $z$ itself by subtracting the first equation.

$$z = x\cdot y\\
z + dz = (x + dx)(y + dy) \\
z + dz = xy + x\,dy + y\,dx + dx\,dy \\
dz = x\,dy + y\,dx + dx\,dy \\
dz \simeq x\,dy + y\,dx
$$


问:
Looking for Advanced Single Variable Calculus textbooks (NOT REAL ANALYSIS)
Does anyone know any other good advanced calculus textbooks that have much more advanced or rigorous problems?

Textbooks like: Spivak, JEE Main, Advanced Calculus Explored, Problems in Mathematical Analysis (Demidovich), AoPS Calculus, Examples & Problems in Advanced Calculus, Advanced Calculus of a Single Variable, I.A. Maron, etc.

Textbooks that aren't what I'm looking for: Rudin, Apostol, Courant, Folland, G.H. Hardy, etc.

Again, I'm not looking for Real Analysis, I'm looking for like a very "honors" type calculus book without proofs. Please, no proofs, just a calculus book with challenging or advanced problems.

I tried looking through other common Mathexchange questions and they're just proofs and answer the same exact thing that I'm not looking for.

答: I think the MIT calculus book may fit your needs.</p>


问:
Showing that $y$ is irreducible in $\mathbb R[x,y]/(x^2+y^2-1)$.
Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ which is called the coordinate ring of the unit circle.

I am trying to prove that $y$ is irreducible in $B.$

I have the following information about $B$:

1- $B$ is an integral domain.

2- $B = \mathbb R[x] \oplus \mathbb R[x]y$ and multiplication by $y$ is a map of $B$ as an $\mathbb R[x]$-module. And $y$ is integral over $\mathbb R[x]$ and so there is a ring inclusion $B \rightarrow M_2(\mathbb R[x]),$ called a representation of $B$ described below:

$$\phi:B \rightarrow M_{2}(\mathbb R[x]), \phi (a_{1} + a_{2}y) =  \begin{pmatrix}
a_{1} & a_{2}(1-x^2) \\
a_{2} & a_{1}
\end{pmatrix} $$

3- Also, I know that $B^* = \mathbb R^*$

I am trying to use  Eisenstein criterion for showing that $y$ is irreducible.

Here is my try:

We need to find a prime $p$ such that $p$ divides $0$ because we have only 1 coefficient which is the leading coefficient and every other coefficients are $0$ and any prime can divide zero trivially. Also any prime does not divide 1. But now I do not know why $p^2$ does not divide $a_0$ which is zero in our case ...I need $p^2$ to divide $a_0$ to fullfill the conditions for Eisenstein Criterion. Could anyone explain this to me please?

Also, is there any use for the fact that $y$ is integral in this proof?

答:
Define $N:B\to\mathbb R[x]$ by $N(a_1+b_1y)=\det\phi(a_1+b_1y)$. Notice that $N$ behaves like a norm!

Let $z_i=a_i+b_iy$, $i=1,2$, such that $y=z_1z_2$. Then $N(y)=N(z_1)N(z_2)\Leftrightarrow x^2-1=N(z_1)N(z_2)$. Now we have the following cases:

(i) $\deg N(z_1)=0\Leftrightarrow z_1\in\mathbb R^*$,

(ii) $\deg N(z_1)=2$ $\Leftrightarrow$ $\deg N(z_2)=0$ $\Leftrightarrow$ $z_2\in\mathbb R^*$,

(iii) $\deg N(z_1)=\deg N(z_2)=1$. If $N(z_1)=x-1$, then $a_1^2(x)+b_1^2(x)(x^2-1)=x-1\Rightarrow x-1\mid a_1(x)\Rightarrow\exists a'_1\in\mathbb R[x]$ such that $a_1=(x-1)a'_1$ and plugging this in the foregoing equation we get $a'_1(x)^2(x-1)+b_1^2(x)(x+1)=1$. Looking now at the dominant coefficients of $a'_1$ and $b_1$ we find that one of these is zero (false!) or the sum of their square is zero (false!).


问:
An example of an algebraic loop which has different L and R inverses?
Can anyone point me toward a simple example of a non-associative algebraic loop (i.e. a quasigroup with an identity) for which at least one element has a left inverse which is not equal to its right inverse?

That's the end of the question, but here is a little extra detail in case it helps:

I can find plenty of non-associative loops for which the opposite condition is true, i.e. for which the left and right inverses for each element are identical to each other.  The simplest such example I have come across had this Cayley table:

```<code>0   1   2   3   4   5   6   7
1   7   5   0   6   2   4   3
2   6   7   5   0   3   1   4
3   0   6   7   5   4   2   1
4   5   0   6   7   1   3   2
5   2   3   4   1   7   0   6
6   4   1   2   3   0   7   5
7   3   4   1   2   6   5   0
</code>```

which I found here: http://ericmoorhouse.org/pub/bol/htmlfiles8/8_1_4_0.html  This example is clearly a loop as it has (1) the Latin-square property required by quasigroups, and (2) an identity, which in this case is zero. It is also non-associative (which I require) since 3.(4.5)=3.1=0 while (3.4).5=5.5=7. However: all the zeros are spaced symmetrically about the leading diagonal so left and right inverses are identical.  That's not what I want!

A good answer to my question would be a similar Cayley table which has both an identity element and the Latin square property and which is not associative, but which has the identity elements (i.e. the zeros) not distributed symmetrically about the leading diagonal.

Note that an identity (for the purposes of our question, and Loops in general) must be both a left and right identity. Therefore, the following Cayley table is not a valid answer (even though it is a Latin square with non-symmetrically placed zeros) because it demonstrates only a left identity not a right identity (and because it is probably also associative .. though I didn't check this):

```<code>0 2 1
1 0 2
2 1 0
</code>```

The (not necessarily authoritative!) wikipedia page https://en.wikipedia.org/wiki/Quasigroup#Loops suggests that Loops of the form I am looking for do exist.  But if wikipedia is wrong, and there are no non-associative Loops with differing left and right inverses, then a proof of that would also be acceptable as an answer!

P.S.:
My question is almost the opposite of a very old one which wanted to know about loops which had two-sided inverses: Examples of loops which have two-sided inverses.
That question strikes me as strange, in retrospect, because EVERY example of a non-associative loop I have found in the last hour would have answered that question -- almost.

Comment added after this question was successfully answered:

The accepted answer referred to the following Cayley table. I copy it here so that future readers will not have to rely on the stability of a hyperlink in order to be able to read the answer.

```<code>∘   0   1   2   3   4
0   0   1   2   3   4
1   1   4   0   2   3
2   2   3   4   1   0
3   3   0   1   4   2
4   4   2   3   0   1
</code>```

答:
Pointer:  https://math.stackexchange.com/a/1444602/123905

The referenced MSE answer exhibits a five element example of a non-associative loop having at least one element with distinct left- and right-inverses.


问:
Find asymptotics of function "n+n/2+n/4..+1 = 2n-1"
I need to find asymptotics of the function
$f(n) = 2^k + 2^{k-1} + 2^{k-2} + ... + 2 + 1 = 2n - 1$, When $n$ is $2^k$

I have to find constants $a$ and $b$, so that $f(n) \sim a * n^b$ when
$n \to \infty$

Or if $f(n)$ grows exponentially, I need to find $a$ and $b$, so that
$f(n) \sim a * b^n$

I really can't figure out how to do this, could someone help me out?

答:
Welcome to MSE!

It seems to me you're already done.

You've said that $f(n) = 2n - 1$... But this grows like $2n^1$, since

$$\lim_{n \to \infty} \frac{f}{2n} = 1 - \frac{1}{2n} \to 1$$

So $f$ grows polynomially, with $a = 2$ and $b = 1$.

I hope this helps ^_^