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    $\begingroup$ I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. $\endgroup$
    – Asaf Karagila Mod
    Jul 18, 2012 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$
    – cardinal
    Jul 18, 2012 at 19:40
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    $\begingroup$ At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! $\endgroup$
    – Grace Note StaffMod
    Oct 5, 2012 at 14:45
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    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$
    – leo
    Dec 17, 2012 at 18:03
  • 3
    $\begingroup$ PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. $\endgroup$ Dec 2, 2015 at 14:07

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In this answer, I want to multiply the denominator by a polynomial $P(x)$ such that the denominator can only be written in terms of $x^{3n}$ :

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


We can construct the polynomial $\thinspace P(x)\thinspace $ as follows :

Let $\thinspace x=z\thinspace $ be one of the roots of $\thinspace x^2-x-1\thinspace .$ This implies that, $\thinspace z^2=z+1\thinspace .$ This leads to :

$$ \begin{align}z^6&=z^3+1+3z(z+1)\\ &=z^3+1+3z^3\\ &=4z^3+1\end{align} $$

Then, by quickly applying synthetic division , you have :

$$ \begin{align}&P(z)(z^2-z-1)=z^6-4z^3-1\\ \implies &P(z)=\frac {z^6-4z^3-1}{z^2-z-1}\\ \implies &P(z)=z^4+z^3+2z^2-z+1\end{align} $$

Putting $\thinspace x=\sqrt [3]{5}\thinspace $ and $\thinspace x^3=5\thinspace $, then we reach the following conclusion :

$$ \begin{align}\frac {1}{1+x-x^2}&=-\frac {x\left(x^3\right)+x^3+2x^2-x+1}{x^6-4x^3-1}\\ &=-\frac {2x^2+4x+6}{4}\\ &=-\frac 12 \big(3+2\sqrt [3]{5}+\sqrt [3]{25}\big)\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

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As @CyclotomicField said in his/her comment, there is a lot of nice structures and patterns, which makes the problem so sweet ;-) You might see the problem when following the idea of your table a bit further. However, while you count the number of "odd steps + even steps" First, I include negative numbers as well, and for your "number of steps" I use "exponent of 2" ($=A_{1,r}$ where $r$ idexes the row-number) such that $$w_{1,r} = 3 \cdot a_{1,r} + 1$$, then $$a_{2,r} = w_{1,k}/2^{A_{1,r}} $$ where $A_{1,r}$ is such that the next step becomes odd. And so on. A short image from an Excel sheet, where I've colored the $A_{c,r}$ for $a_{1,r} = -21 \ldots +21 $ looks like this

image

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  1. $x(g):(0,1)\mapsto[0,1]$
  2. $x(1^-)=\sqrt{b^2+2b}-b>0$
  3. $x'(1^-)=b\frac{\sqrt{b^2+2b}-b}{\sqrt{b^2+2b}+a}>0$
  4. $x''(1^-)=-\frac{(x'(1^-)+2b)x'(1^-)}{\sqrt{b^2+2b}+2a}<0$
  5. $a(1-g)x'=0.5x^2+bg(x-1)$
  6. $2a(1-g)x'=(x-bg)^2-(bg+1)^2+1$
  7. $x=-bg+\sqrt{(bg)^2+2bg+2a(1-g)x'}$
  8. $-\frac{bg}{a(1-g)}\leq x'\leq\frac1{2a(1-g)}$
  9. $x((x')^{-1}(0))=\sqrt{(bg)^2+2bg}-bg\in(0,1)$
  10. $a(1-g)x''=(x+a+bg)x'+b(x-1)$
  11. $a(1-g)x'''=(x+2a+bg)x''+(x'+2b)x'$
  12. $x''((x')^{-1}(0))=b\frac{\sqrt{(bg)^2+2bg}-bg-1}{a(1-g)}<0$
  13. $x'((x'')^{-1}(0))=b\frac{1-x((x'')^{-1}(0))}{x((x'')^{-1}(0))+a+b(x'')^{-1}(0)}>0$
  14. $x'(x^{-1}(1))=\frac1{2a(1-x^{-1}(1))}>0$ \begin{eqnarray} x'(x^{-1}(1))>0&\quad x^{-1}(1)\;\text{DNE even if}\\ x'((x'')^{-1}(0))>0&\quad\text{we assume an}\\ 1\text{ is a maximum of }x&\quad\text{inflexion point DNE} \end{eqnarray}
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(in T.Tao's Blog )

You can use Ellison’s estimate, $$|2^x-3^y| \geq 2^x e^{-x/10}\qquad , $$ $\hspace{ 150 pt}$ which holds for $x \geq 12$ with $ x \neq 13, 14, 16, 19, 27$ and all $y$.

This is based on results by Pillai and Baker.

Reference: “WJ Ellison, On a theorem of S. Sivasankaranarayana Pillai, S´eminaire de th´eorie des nombres de Bordeaux, 1970 (1971), pp. 1–10.”

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If we have a sequence $a_k$ which is built by a linear recurrence rule with a couple of coefficients $c_k$, say $$ a_{k+3}=c_0 a_{k} + c_1 a_{k+1} + c_2 a_{k+2} $$ or, containing a constant $C$ $$ a_{k+2} = C+ c_0 a_k + c_1 a_{k+1} $$ then we need only few consecutive entries to find the coefficients $c_0,c_1,c_2$ or $C,c_0,c_1$ (in the examples).

I extend my comment to show a matrix-based method to find the correct coefficients if a sufficient number of consecutive entries $a_k$ are known/given.

(.... further text to be inserted ....)

\\ Pari/GP     
recursionfromseq(vk,co=0)=if(co, recursionfromseq_C(vk),recursionfromseq_a(vk))

{recursionfromseq_a(vk,co=0)=my(a,N=#vk,rs,cs,rnk,Data,T);
  \\ no constant
         cs= N \2;                             \\ print(N," ",cs); 
         Data= matrix(cs+1,cs,r,c,vk[r+c-1])  ; \\ printp(Data);
   rnk=matrank(Data[1..cs,1..cs]);                   \\ print(rnk);
   T=Data[1..rnk,1..rnk]^-1 * Data[2..rnk+1,1..rnk]; \\ printp(T);
   return(T);
   }

{recursionfromseq_C(vk,co=0)=my(a,N=#vk,rs,cs,rnk,Data,T);
 \\ with constant
         cs= N \2 + (N%2);                                  print(N," ",cs); 
         Data= matrix(cs+1,cs,r,c,if(c==1,1,vk[r+c-2]))  ;  printp(Data);
   rnk=matrank(Data[1..cs,1..cs]);                          print(rnk);
   T=Data[1..rnk,1..rnk]^-1 * Data[2..rnk+1,1..rnk];        printp(T); 
   return(T);
   }

(....)
ssd

This answer is free for anyone to use.

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$$\frac{3^{a} - 2^{b}}{2^{c}} = 1 - \frac{3^{m}}{2^{n}}k$$ $$3^{a} - 2^{b} = 2^{c} - \frac{3^{m}}{2^{n-c}}k \qquad \implies \qquad k=j \cdot 2^{n-c}$$ $$3^a - 2^b = 2^c - j \cdot 3^m $$ $$3^a(1+j3^{m-a})=2^b(1 + 2^{c-b})$$ $${1+j3^{m-a}\over2^b }={1 + 2^{c-b} \over 3^a}$$ $$ \implies c-b = (1- \{a:2\})*(1+\{a,3\})$$ $$ \implies c-b = 0 \qquad \text{when }a = 2a' \text{ is even} $$ $$ \implies c-b = 1+\{2a'+1,3\} \qquad \text{when }a = 2a'+1 \text{ is odd} $$

This answer is free for anyone to use.

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One more idea; echoing your observations (3), focusing more properties of your variable $b$ and following the equations downto (4).

I've sometimes tried to formalize this, but today I took the patience to go through it. We find -just another- viewpoint frome where it is much suggestive that the non-trivial 1-cycle cannot exist; but I could not yet relate the critical observation to some lower or upper bounds... .

Let's see. We put the focus on the first unnumbered eq. after (3). We find $$ 2^B = { 3^n b -1\over 2^n b-1 } \tag1 $$ Here we restrict $B$ as derived from $B=m-n $ and hope, there is some positive integer $b$ such that the equation is satisfied. The idea which I foolowed today is to look, which $b$ make, for a given $n$ the expression in the numerator $3^n b -1$ such that it contains the factor $2$ to the power $B$.
We know, that -for any given $e$- $2^e || 3^f-1$ periodically with increasing $f$. Here, if $f$ is odd, $e$ can only assume the value $e=1$, but if $f=2^{e-2}x$ (with any $x$ odd, let's assume here $x=1$) then $e$ can assume all values $e \ge 3$ and we'll have $$ 2^B || 3^{2^{B-2}} - 1 \tag {2.1} $$ Now, $n$ is not equal to $2^{B-2}$ so we could rewrite $$ 2^B = { 3^{n+(2^{B-2}-n)} -1\over 2^n b-1 }= { 3^n \cdot 3^{2^{B-2}-n} -1\over 2^n b-1 } \tag {2.2} $$ and have sure, that for $B \ge 3$ the rhs has the factor $2^B$, and we could set the variable $b$ to the provisorcal value $$b=3^{2^{B-2}-n}$$ But that (extremely high) value is not needed - it is enough that $$ b \equiv 3^{2^{B-2}-n} \pmod {2 \cdot 2^B}$$ Here is a short table to show this: $$ \small \begin{array} {rr|rrr} n & b & \hat b=\log_2(b) & B & d=\hat b - B \\ \hline 5 & 3 & 1.584962501 & 3 & 1.415037499 \\ 6 & 25 & 4.643856190 & 4 & -0.6438561898 \\ 7 & 3 & 1.584962501 & 5 & 3.415037499 \\ 8 & 1 & 0.0 & 5 & 5.000000000 \\ 9 & 11 & 3.459431619 & 6 & 2.540568381 \\ 10 & 89 & 6.475733431 & 6 & -0.4757334310 \\ 11 & 179 & 7.483815777 & 7 & -0.4838157773 \\ 12 & 17 & 4.087462841 & 8 & 3.912537159 \\ 13 & 347 & 8.438791853 & 8 & -0.4387918526 \\ 14 & 201 & 7.651051691 & 9 & 1.348948309 \\ 15 & 67 & 6.066089190 & 9 & 2.933910810 \\ 16 & 1217 & 10.24911345 & 10 & -0.2491134527 \\ 17 & 1771 & 10.79034850 & 10 & -0.7903484968 \\ 18 & 2297 & 11.16553514 & 11 & -0.1655351414 \\ 19 & 83 & 6.375039431 & 12 & 5.624960569 \\ 20 & 5489 & 12.42232762 & 12 & -0.4223276240 \\ \end{array} $$ For example, we use $n=9$, then $B=m-n=6$ and by the table $b=11$ (we can use all $b_k=11+ k \cdot 2 \cdot 2^B$ as well). Of course no one gives an integer solution for the rhs in eq (1), and increasing $k$ in $b_k$ means only that the whole expression in the rhs goes towards the fractional value $1.5^n$

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Let $p \in \mathbb P$ denote the chosen problem-parameter for some analysis.
Let $q \in \mathbb P$ denote the largest prime below $p$.
Let $\text{ql}$ denote the list of primes below $p$ : $\text{ql}=\{2,3,...,q\}$
Then let's write one transformation from $a_1 \to a_2$ as $$ a_2 = {a_1 \cdot p +1 \over 2^{A_1} 3^{B_1} \cdots q^{H_1}} \tag 1 $$ $\qquad \qquad $ where we assume, that the $a_1$ and $a_2$ are free from all factors from the primes-list $\text{ql}$.


  • So far we can already discuss a little interesting thing: does (for any $p$) exist a one-step cycle $a_0 \to a_0$ ?
    We rewrite (eq 1): $$ a_1 = {a_1 \cdot p +1 \over 2^{A_1} 3^{B_1} \cdots q^{H_1}} $$ and then $$ a_1 \cdot (2^{A_1} 3^{B_1} \cdots q^{H_1} - p) = 1 \\ a_1 = { 1 \over 2^{A_1} 3^{B_1} \cdots q^{H_1} - p } $$ Then we ask: can a combination of exponents $[A_1,B_1,...,H_1]$ lead to $2^{A_1} 3^{B_1} \cdots q^{H_1} = p+1$.
    The answer is positive, and for all $p$: since in $1 \cdot p+1$ we can only have primefactors smaller than $p$ and in $2^{A_1} 3^{B_1} \cdots q^{H_1}$ we have

$$ a_2 = T(a_1,[A_1,B_1,...,H_1]) $$


$$ \hat s = (p+{ 1\over a_\mu})^N \\ \hat s^{1/N} - p = {1\over a_\mu} \\ a_\mu = {1 \over \hat s^{1/N} - p} $$ and as a consequence $$ a_1 < a_\mu = {1 \over \hat s^{1/N} - p} $$

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This answer is free for anyone to use.

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This answer is free for anyone to use.

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$$\begin{align} -a^2+b^2+c^2 &= \color{blue}{p^2}\\ a^2-b^2+c^2 &= \color{blue}{(nq)^2}\\ a^2+b^2-c^2 &=q^2 \\ \hline -c^2+d^2+e^2 &= (np)^2\\ c^2-d^2+e^2 &= \color{blue}{p^2}\\ c^2+d^2-e^2 &= \color{blue}{(nq)^2} \end{align}$$ $$ \begin{align} a^2+b^2+c^2 &= p^2+2m^2q^2 \\ c^2+d^2+e^2 & = 2(mp)^2+(nq)^2 \end{align}$$ $$\begin{array}{} 2c^2 &=p^2+(nq)^2 &=...\\ 2b^2 &=p^2+q^2 &=\\ 2a^2 &=(nq)^2+q^2 &=q^2(n^2+1)\\ \hline 2c^2&=p^2+(nq)^2 & &\text{ repeated } \\ 2d^2&=(np)^2+(nq)^2 &=n^2(p^2+q^2)\\ 2e^2&=(np)^2 + p^2 &=p^2(n^2+1) \end{array}$$


$$\begin{array}{} 2a^2+2e^2 & = (p^2+q^2)(n^2+1)&=2m^2(p^2+q^2)&=2m^2\cdot 2b^2 \\ a^2+e^2 & = 2 (mb)^2 \\ 2d^2 &= 2(nb)^2 \end{array}$$

$$\begin{array}{} a^2+e^2 & = 2 (mb)^2 \\ 2d^2 &= 2(nb)^2 \end{array}$$

assume $(n,m)=(239,169)$ $$\begin{align} a &= 169 q\\ 2b^2 &=p^2+q^2\\ 2c^2 &=p^2+(239q)^2\\ d &= 239b\\ e &= 169p \end{align}$$

$$\begin{align} a &= mq\\ b &=\sqrt{\frac{p^2+q^2}2}\\ c &=\sqrt{\frac{p^2+n^2q^2}2}\\ d &= nb\\ e &= mp \end{align}$$ This space is available for use.

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This answer is free for anyone to use.

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Let $M$ be an $R$-module.

I define a unital, associative algebra based on $M$.

Consider formal expressions of the following form:

Let a term be:

$$ +1 * r * m_1 * m_2 * \cdots * m_n $$

$$ -1 * r * m_1 * m_2 * \cdots * m_n $$

Let an element of the domain $D$ be:

$$ t_1 + t_2 + t_3 \cdots $$

With $0$ representing an empty sum.

Define multiplication of terms in the following way:

$$ (+1 * r * m_1 \cdots) * (+1 * r' * m'_1 \cdots) = ((+1 * +1) * (r * r') * m \cdots * m'_1 \cdots) $$

Define multiplication of elements by invoking the distributive rule.

$$ (t_1 + t_2 \cdots) * (u_1 + u_2 \cdots) = t_1 * u_1 \cdots $$

Define addition of terms as formal addition.

Next we define equivalence of terms.


Let rings be commutative and unital.

Given an $R$-module $M$, it is possible to define an $R$-algebra $A$ in the following way.

Let $A^*$ consist of finite formal sums of elements of the form $\sum_i \prod_j m_{ij}$ where $m_{ij}$ is in $M$.

Addition is performed in the obvious manner.

Multiplication is also performed in the obvious manner.

Scalar multiplication is also performed in the obvious manner.


Can you split a module into its free and torsion components by examining the module presentation (defined below) alone?

The immediate context here is that I'm reading Atiyah-MacDonald chapter 2, which overs modules, and trying to map it back to stuff I've seen before like completely free algebras and group presentations. The way I think of it, a group presentation is a special case of taking a completely free algebra and modding it out by a set of equations, or, equivalently, building a category of models of an equational theory and choosing the initial object.

You can do something similar for modules over a fixed ring $R$.

In a certain sense, I'm trying to work out how much of module theory you can describe with module presentations (defined below), since those are more familiar to me than exact sequences and other tools with a categorical flavor.

A completely free algebra modded out by a huge number of equations) is used explicitly in Atiyah-Macdonald on pages 24 and 25 to define the tensor product (although without using those words). However, the authors use it begrudgingly:

We shall never again need to use the construction of the tensor product given in (2.12), and the reader may safely forget it if he prefers. (Atiyah-Macdonald 25)


A module is a group with operators where the operators form a ring.

It is also possible to give a module a group presentation.

For example, here is the quotient $\mathbb{R}[x]/(x)$ as an $\mathbb{R}[x]$-module:

  • $ \langle t \mid (x)t = 0 \rangle_{\mathbb{R}[x]} $

The idea is that we have expressions of the form $p(x)t$ where $p(x)$ is a real polynomial and $t$ is a propertyless constant. We get the axioms of a $\mathbb{R}[x]$-module for free, but impose the additional rule that $(x)t = 0$, which means that scalar multiplying any $p(x)t$ by

This link is relevant:

https://mathoverflow.net/a/363772


Suppose I have a module $M$ being acted on by a group $G$.

  • $0 + a = a$
  • $a + 0 = a$
  • $--a = a$
  • $a - a = 0$
  • $a + (b + c) = (a + b) + c$
  • $a + b = b + a$
  • $r(a+b) = ra + rb$
  • $(r+s)(a) = ra + sa$
  • $(-r)(a) = -ra$
  • $(rs)(a) = rsa$
  • $rsa = sra$
  • $g(ra + sb) = gra + gsb$
  • $(gh)(ra) = g(hra)$

Let $M$ be $\mathbb{H}$ as a real module. Let $G$ be $\mathbb{H} \setminus \{0\}$ as a group.

$G$-module.


modal formula: $\lozenge^{(n+1)}p \to \lozenge^n p$ enforces a maximum path length of $n$.

modal formula: $\left(\bigwedge_{1 \le i \le (n+1)}\lozenge p_i\right) \to \lozenge \left( \bigvee_{1 \le i \neq j \le (n+1)} p_i \land p_j \right)$

The above enforces a maximum degree of $n$


Let $M$ be a simple $R$-module. Let $\Lambda$ be the lattice of submodules of $M$.

We can define a function $f : \Lambda \to \Lambda$ as:

$$ f(N) = \{x_1 : x \in N\} \times \{x_2 : x \in N\} \times \cdots \times \{x_n : x \in N \} $$

I.e. $f$ takes an ideal and sends it to the product of its components.


What are the ideals of non-associative rings?


I'm looking for an example of a non-UFD GCD besides the algebraic integers and the ring of entire functions on $\mathbb{C}$ or the monoid ring $(k, \mathbb{Q}_{\ge 0})$.

I'm curious whether this example works.

Consider the ring $R = \mathbb{Z}[a,b,c]/(b^2-a, c^2-a)$. The idea being that $a$ is a propertyless element with two square roots, $b$ and $c$.

$R$ is not a UFD. $a$ factors as $b \cdot b$ and as $c \cdot c$ and $b$ and $c$ are not equal.


Acting on a complex number:

$k$ is a real number. $i$ is $\sqrt{-1}$. $T_z$ is a shift operator. $I_z$ is a shift by $iz$. $F$ is conjugation.

  • $k \triangleright (a+ib) = ka + ikb$
  • $i \triangleright (a+ib) = -b + ia$
  • $T_{k} \triangleright (a+ib) = (k+a)+ib$
  • $I_{k} \triangleright (a+ib) = a + i(b+k) $
  • $F \triangleright (a+ib) = a - ib$
  • $g - h \triangleright (a+ib) = (g \triangleright a+ib) - (h \triangleright a+ib)$
  • $gh \triangleright (a+ib) = g \triangleright h \triangleright (a+ib) $

Definition of higher-level operators, rotations and reflections.

  • $R_{p, \theta}$ is $T_{p} R_\theta T_{-p}$.
  • $X_{0, 0}$ is $F$.
  • $X_{p, 0}$ is $T_p F T_{-p}$
  • $X_{p, \theta}$ is $T_p R_{\theta} F R_{-\theta} T_{-p}$

Let a matrix $M$ be quasisymmetric if there exists permutation matries $U$ and $V$ such that $UMV = (UMV)^T$.

If I look at character tables of finite cyclic groups, it appears that the character table, considered as a matrix, if quasisymmetric when the group has prime order.

Some groups like $\mathbb{Z}_2 \times \mathbb{Z}_2$ are also quasisymmetric, but they are not cyclic and I'm only interested in cyclic groups at the moment.


Consider noncommutative, nonunital rings $R$ where the underlying group $(R, +)$ is cyclic.

If $n$ is finite, there is one isomorphism class of ring $R$ with a cycligc additive grouop for each divisor of $n$.


How do you characterize the rings with exactly one cool ideal.

Let's define a cool ideal $I \subset R$ as a prime ideal that is not principal and is generated by two elements.

So, in the ring $\mathbb{C}[x, y]$, $(x, y)$ is a cool ideal. However, it is not the only one, since $(x, y+1)$ is also a cool ideal.

The ring $\mathbb{C}[x, y]/S$, where $S$ is the set of all elements $j(x, y)$ such that $x \not \mid j(x, y)$ and $y \not \mid j(x, y)$, does have exactly one cool ideal. (I think this construction is $\mathbb{C}[x, y]$ localized at $(x, y)$?)

I'm wondering if there's a nice characterization of all rings with exactly one cool ideal.


What follows is what I've done so far to try to solve this question on my own. It consists of:

  • A proof that $\mathbb{C}[x, y]$ contains the cool ideal $(x, y)$ .
  • A proof that $\mathbb{C}[x, y]$ contains the additional cool ideal $(x, y+1)$ .
  • A proof that $\mathbb{C}[x, y]/S$ contains exactly one cool ideal.

Definition 31: polynomial with no constant term

A polynomial $p(\vec{x})$ has no constant term if and only if it is $0$ or has at least one term, but none of the terms are constant.

This definition says very little, I'm just emphasizing that $0$ is considered to have no terms at all instead of having a constant term which is zero.

Lemma 41: $p(x, y)$ is in $(x, y)$ if and only if $p$ has no constant term.

RTL: Suppose $p(x, y)$ has no constant term. We can collect all the terms with at least one $x$ into $f(x)x$ and all the remaining terms into $g(y)y$.

For example, $x^2 + xy + 3y^2 = (x+y)x + (3y)y$.

Thus, $p(x, y)$ is in $(x, y)$. Note that when $p(x, y)$ is $0$, we choose $f(x) = 0$ and $g(y) = 0$.

LTR: Suppose $p(x, y)$ is in $(x, y)$.

Thus, $p(x, y)$ has the form $f(x, y)x + g(x, y)y$. There are no constant terms in $f(x, y)x$ and there are no constant terms in $g(x, y)y$. Therefore $p(x, y)$ has no constant terms, as desired.

End of proof of Lemma 41.

Lemma 51: $(x, y)$ is a cool ideal of $\mathbb{C}[x, y]$.

$(x, y)$ is an ideal.

$(x, y)$ is generated by two elements.

Let's prove that $(x, y)$ is a prime ideal by noting that its complement is multiplicatively closed. Suppose we have not necessarily distinct elements of the complement of $(x, y)$; call them $f(x, y)$ and $g(x, y)$. By Lemma 41, $f(x, y)$ and $g(x, y)$ each have no constant term.

If $a(x, y)$ is a term of $f(x, y)$ and $b(x, y)$ is a term of $g(x, y)$, then $(ab)(x, y)$ is nonconstant because it is a product of two nonconstant terms.

Therefore $(fg)(x, y)$ has no nonconstant terms and, by Lemma 41, is an element of the complement of $(x, y)$.

Thus the complement of $(x, y)$ is multiplicatively closed and $(x, y)$ is a prime ideal.

Suppose for contradiction that $(x, y)$ is principal and generated by $h(x, y)$. If $h(x, y)$ contains a term that contains $x$, then $h(x, y) \not\mid y$. If $h(x, y)$ contains a term that contains $y$, then $h(x, y) \not\mid x$. If $h(x,y)$ has a constant term, then $(h) = (1) = \mathbb{C}[x, y]$. If $h$ has no terms at all, then $h=0$ which is a contradiction since $(0) \neq (x, y)$.

Lemma 71: $\varphi$ which sends $x$ to $p(x, y)$ and $y$ to $q(x, y)$ is a ring homomorphism.

Every element of $\mathbb{C}[x, y]$ is equivalent to a sum of the form $\sum_{m, n}a_{m,n}x^m y^n$, and, as usual, all but finitely many values of $a_{m,n}$ are zero.

Here is the definition of $\varphi$, based on how it acts on the monomials $x$ and $y$.

$$ \varphi(\sum_{m,n}a_{m,n}x^m y^n) = \sum_{m,n}a_{m,n} \varphi(x)^m \varphi(y)^n $$

$\varphi$ respects $-$.

$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n) - (\sum_{m,n}b_{m,n}x^my^n)) = \varphi(\sum_{m,n}(a-b)_{m,n}x^my^n) = \sum_{m,n} (a-b)_{m,n} \varphi(x)^m \varphi(y)^n = \varphi(\sum_{m,n} a_{m,n} x^m) $$

$\varphi$ respects multiplication.

$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n)(\sum_{k,l} b_{k,l}x^k y^l)) $$

Lemma 81: $\varphi$ preserves ideals when $\varphi$ is a surjective ring homomorphism.

like $I$ be an ideal. I want to show that $\varphi I$ is an ideal.

Suppose $a$ and $b$ are in $I$, then $\varphi(a) + \varphi(b) = \varphi(a + b)$, therefore $\varphi I$ is additively closed.

By hypothesis, $I$ is an ideal. Thus is satisfies $ra \in I$ for each $r$ in $R$ and each $a$ in $I$. Therefore, $\varphi(ra) \in \varphi I$ for each $r$ in $R$ and $i$ and $I$. $\varphi(r)$ hits every element of $S$ by surjectivity of $\varphi$, thus $\varphi I$ is an ideal of $S$.

End of proof of Lemma 81.

Lemma 91: $\varphi$ preserves cool ideals when $\varphi$ is a surjective homomorphism.

TBD.

End of proof of Lemma 91.

Lemma 61: $(x, y+1)$ is a cool ideal of $\mathbb{C}[x, y]$.

The map $\varphi : \mathbb{C}[x, y] \to \mathbb{C}[x, y]$ that sends $x$ to $x$ and $y$ to $y+1$ is surjective.

$$ \varphi((x, y)) = (\varphi(x), \varphi(y)) = (x, y+1) $$

Therefore $(x, y+1)$ is a cool ideal.

End of proof of Lemma 61.

Lemma 101: the elements of $\mathbb{C}[x, y] / S$ has the form $ux^m y^n$ where $u$ is a unit.

Let $p(x, y)$ be a polynomial in $\mathbb{C}[x, y]$ such that $x \not\mid p(x, y)$ and $y \not\mid p(x, y)$ . By construction of $S$, $p(x, y)$ is in $S$. Therefore $\frac{1}{p(x, y)}$ is in $\mathbb{C}[x, y] / S$.

Again by construction of $S$, $x$ and $y$ do not have inverses.

$0$ doesn't have an inverse and isn't expected to. Nonzero constants in $\mathbb{C}$ do have inverses.

Every element of $\mathbb{C}[x, y]/S$ thus has the form $\frac{p(x, y)x^n y^m}{q(x, y)}$ where $p(x, y)$ and $q(x, y)$ are in $S$.

$\frac{p(x, y)}{q(x, y)}$ is also in $S$, therefore an arbitrary element of $\mathbb{C}[x, y]$ has the form $ux^n y^m$.


How do you characterize the rings with exactly one cool ideal.

Let's define a cool ideal $I \subset R$ as a prime ideal that is not principal and is generated by two elements.

So, in the ring $\mathbb{C}[x, y]$, $(x, y)$ is a cool ideal. However, it is not the only one, since $(x, y+1)$ is also a cool ideal.

The ring $\mathbb{C}[x, y]/S$, where $S$ is the set of all elements $j(x, y)$ such that $x \not \mid j(x, y)$ and $y \not \mid j(x, y)$, does have exactly one cool ideal. (I think this construction is $\mathbb{C}[x, y]$ localized at $(x, y)$?)

I'm wondering if there's a nice characterization of all rings with exactly one cool ideal.


What follows is what I've done so far to try to solve this question on my own. It consists of:

  • A proof that $\mathbb{C}[x, y]$ contains the cool ideal $(x, y)$ .
  • A proof that $\mathbb{C}[x, y]$ contains the additional cool ideal $(x, y+1)$ .
  • A proof that $\mathbb{C}[x, y]/S$ contains exactly one cool ideal.

Definition 31: polynomial with no constant term

A polynomial $p(\vec{x})$ has no constant term if and only if it is $0$ or has at least one term, but none of the terms are constant.

This definition says very little, I'm just emphasizing that $0$ is considered to have no terms at all instead of having a constant term which is zero.

Lemma 41: $p(x, y)$ is in $(x, y)$ if and only if $p$ has no constant term.

RTL: Suppose $p(x, y)$ has no constant term. We can collect all the terms with at least one $x$ into $f(x)x$ and all the remaining terms into $g(y)y$.

For example, $x^2 + xy + 3y^2 = (x+y)x + (3y)y$.

Thus, $p(x, y)$ is in $(x, y)$. Note that when $p(x, y)$ is $0$, we choose $f(x) = 0$ and $g(y) = 0$.

LTR: Suppose $p(x, y)$ is in $(x, y)$.

Thus, $p(x, y)$ has the form $f(x, y)x + g(x, y)y$. There are no constant terms in $f(x, y)x$ and there are no constant terms in $g(x, y)y$. Therefore $p(x, y)$ has no constant terms, as desired.

End of proof of Lemma 41.

Lemma 51: $(x, y)$ is a cool ideal of $\mathbb{C}[x, y]$.

$(x, y)$ is an ideal.

$(x, y)$ is generated by two elements.

Let's prove that $(x, y)$ is a prime ideal by noting that its complement is multiplicatively closed. Suppose we have not necessarily distinct elements of the complement of $(x, y)$; call them $f(x, y)$ and $g(x, y)$. By Lemma 41, $f(x, y)$ and $g(x, y)$ each have no constant term.

If $a(x, y)$ is a term of $f(x, y)$ and $b(x, y)$ is a term of $g(x, y)$, then $(ab)(x, y)$ is nonconstant because it is a product of two nonconstant terms.

Therefore $(fg)(x, y)$ has no nonconstant terms and, by Lemma 41, is an element of the complement of $(x, y)$.

Thus the complement of $(x, y)$ is multiplicatively closed and $(x, y)$ is a prime ideal.

Suppose for contradiction that $(x, y)$ is principal and generated by $h(x, y)$. If $h(x, y)$ contains a term that contains $x$, then $h(x, y) \not\mid y$. If $h(x, y)$ contains a term that contains $y$, then $h(x, y) \not\mid x$. If $h(x,y)$ has a constant term, then $(h) = (1) = \mathbb{C}[x, y]$. If $h$ has no terms at all, then $h=0$ which is a contradiction since $(0) \neq (x, y)$.

Lemma 71: $\varphi$ which sends $x$ to $p(x, y)$ and $y$ to $q(x, y)$ is a ring homomorphism.

Every element of $\mathbb{C}[x, y]$ is equivalent to a sum of the form $\sum_{m, n}a_{m,n}x^m y^n$, and, as usual, all but finitely many values of $a_{m,n}$ are zero.

Here is the definition of $\varphi$, based on how it acts on the monomials $x$ and $y$.

$$ \varphi(\sum_{m,n}a_{m,n}x^m y^n) = \sum_{m,n}a_{m,n} \varphi(x)^m \varphi(y)^n $$

$\varphi$ respects $-$.

$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n) - (\sum_{m,n}b_{m,n}x^my^n)) = \varphi(\sum_{m,n}(a-b)_{m,n}x^my^n) = \sum_{m,n} (a-b)_{m,n} \varphi(x)^m \varphi(y)^n = \varphi(\sum_{m,n} a_{m,n} x^m) $$

$\varphi$ respects multiplication.

$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n)(\sum_{k,l} b_{k,l}x^k y^l)) $$

Lemma 81: $\varphi$ preserves ideals when $\varphi$ is a surjective ring homomorphism.

like $I$ be an ideal. I want to show that $\varphi I$ is an ideal.

Suppose $a$ and $b$ are in $I$, then $\varphi(a) + \varphi(b) = \varphi(a + b)$, therefore $\varphi I$ is additively closed.

By hypothesis, $I$ is an ideal. Thus is satisfies $ra \in I$ for each $r$ in $R$ and each $a$ in $I$. Therefore, $\varphi(ra) \in \varphi I$ for each $r$ in $R$ and $i$ and $I$. $\varphi(r)$ hits every element of $S$ by surjectivity of $\varphi$, thus $\varphi I$ is an ideal of $S$.

End of proof of Lemma 81.

Lemma 91: $\varphi$ preserves cool ideals when $\varphi$ is a surjective homomorphism.

TBD.

End of proof of Lemma 91.

Lemma 61: $(x, y+1)$ is a cool ideal of $\mathbb{C}[x, y]$.

The map $\varphi : \mathbb{C}[x, y] \to \mathbb{C}[x, y]$ that sends $x$ to $x$ and $y$ to $y+1$ is surjective.

$$ \varphi((x, y)) = (\varphi(x), \varphi(y)) = (x, y+1) $$

Therefore $(x, y+1)$ is a cool ideal.

End of proof of Lemma 61.

Lemma 101: the elements of $\mathbb{C}[x, y] / S$ has the form $\sum ux^m y^n$ where the elements $u$ are units.

Let $p(x, y)$ be a polynomial in $\mathbb{C}[x, y]$ such that $x \not\mid p(x, y)$ and $y \not\mid p(x, y)$ . By construction of $S$, $p(x, y)$ is in $S$. Therefore $\frac{1}{p(x, y)}$ is in $\mathbb{C}[x, y] / S$.

Again by construction of $S$, $x$ and $y$ do not have inverses.

$0$ doesn't have an inverse and isn't expected to. Nonzero constants in $\mathbb{C}$ do have inverses.

Every element of $\mathbb{C}[x, y]/S$ thus has the form $\frac{p(x, y)x^n y^m}{q(x, y)}$ where $p(x, y)$ and $q(x, y)$ are in $S$.

$\frac{p(x, y)}{q(x, y)}$ is also in $S$, therefore an arbitrary element of $\mathbb{C}[x, y]$ has the form $ux^n y^m$.

Lemma 111: $\mathbb{C}[x, y]/S$ has exactly one cool ideal $(x, y)$.

$(x, y)$ is a prime ideal.

$(x, y)$ is not principal.

Suppose $I$ is a cool ideal. It is generated by $(ux^my^n, vx^ky^l)$.

$u$ and $v$ are both units and don't affect the ideal.

It can't be the case that $m = n = 0$ or likewise for $k = l = 0$ because then the ideal generated would be $(1)$.


Let $F$ be an algebraic signature and $T$ be an equational $F$-theory. Let $A$ be an $F$-structure such that $A \models T$.

Let's call the relation $=$ on $A$ equality and define a proper congruence as one that is not equality.

I want to show that $A$ has a maximal proper congruence.

Let $(\Gamma)$ where $\Gamma$ is a set of pairs be the congruence generated by the pairs $\Gamma$ defined as:

$$ (\Gamma) = \bigcap \{R : R \in \Lambda \;\;\text{and}\;\; \Gamma \subset \Lambda \} $$

The full relation exists and is a congruence so $\{R: \cdots\}$ is never empty.

Let $\alpha$ be an ordinal such that $A_\alpha$ is the elements of $A$ indexed by $\alpha$.


How do you prove that $\sin(1)$ is not algebraic?


Name for extending a propositional logic to a first-order logic with constant and relation symbols.

We need the rules making $=$ an equivalence relation:

$$ \vdash a = a $$ $$ a = b \vdash b = a $$ $$ a = b, b = c \vdash a = c$$

And the following rule making $=$ a congruence.

$$ \vec{x} = \vec{y}, R(\vec{x}) \vdash R(\vec{y}) $$


Determinant algorithm.

$$ \begin{vmatrix} 1 & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \mapsto \begin{vmatrix} \begin{vmatrix} 1 & b \\ d & e \end{vmatrix} & \begin{vmatrix} 1 & c \\ d & f \end{vmatrix} \\ \begin{vmatrix} 1 & b \\ g & h \end{vmatrix} & \begin{vmatrix} 1 & c \\ g & i \end{vmatrix} \end{vmatrix} $$


Every representation besides the identity representation is reducible.

As proof, all permutations of dimension $n$ fix a subspace generated by $\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}$.

I'm wondering whether it is the case that any irreducible representation of a finite group can be extended to a representation with dimension one greater. Conversely, I'm also wondering whether I can hit all the irreducible subrepresentations of a finite group by taking subrepresentations of permutation representations, but the $\textbf{NonPermutation} \longrightarrow \textbf{Permutation}$ direction is the focus of this particular question to keep it narrow.

My goal here, eventually, is to come up with a heuristic that might help me compute a row or two of the character table of a group by hand by looking at permutation representations that fix a one-dimensional subspace. It is logically a follow-up to this question that I asked in the recent past.

$S_3$ is generated by the derangement $(123)$ and the transposition $(12)$.

We have the sign representation that sends $(12)$ to $\begin{bmatrix} -1 \end{bmatrix}$ and $(123)$ to $\begin{bmatrix} 1 \end{bmatrix}$. This is a 1-dimensional representation.

We can "extend it" to the reducible permutation representation $(12) \mapsto \begin{bmatrix} & 1 \\ 1 & \end{bmatrix}\;\;,\;\; (123) \mapsto \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} $, which has dimension 1 greater.

The faithful representation $(12) \mapsto \begin{bmatrix} & 1 \\ 1 & \end{bmatrix} \;\;,\;\; (123) \mapsto \begin{bmatrix} \omega & \\ & \omega^2 \end{bmatrix}$ with $\omega = \exp\left(\frac{2\pi}{3}\right)$ also has dimension 2.

And it can be extended to $(12) \mapsto \begin{bmatrix} & 1 & \\ 1 & & \\ & & 1 \end{bmatrix} \;\;,\;\; (123) \mapsto \begin{bmatrix} & 1 & \\ & & 1 \\ 1 & & \end{bmatrix}$, which has dimension 1 higher.


Let $R$ be a commutative ring with unity. Let $\Lambda$ be its lattice of ideals.

Let me define a function $f : \Lambda \to \Lambda$ using a transfinite definition.

Let $E_\lambda$ be the elements of $R$ placed in a well-order and indexed by the ordinal $\lambda$.

I will now compute $f(X)$, where $X$ is an ideal. Let $(S)$ be the ideal generated by a set $S$ and likewise for mixtures of sets and elements such as $(S, e)$.

Let $g(0) = X \;\,\text{if}\;\, (E_0, X) = R \;\,\text{else}\;\, (E_0, X)$.

For successor ordinals, let $g(\alpha+1) = g(\alpha) \;\,\text{if}\;\, (E_\alpha, g(\alpha+1)) = R \;\,\text{else}\;\, (E_\alpha, g(\alpha+1))$.

For limit ordinals, let $g(\alpha) = (g(\beta) : \beta < \alpha) \;\,\text{if}\;\, (E_\alpha, (g(\beta) : \beta < \alpha)) = R \;\,\text{else}\;\, (E_\alpha, (g(\beta) : \beta < \alpha))$.

And finally, define $f(X)$ as $g(\lambda)$.

If $X$ is a unit ideal, then $f(X) = X$ since we will never hit an else branch.

If $X$ is not a unit ideal, then $f(X)$ will be a maximal ideal containing $X$. Suppose for contradiction that $f(X)$


I was wondering the other day why interpretations from one structure $\mathfrak{A}$ to another $\mathfrak{B}$ preserve the truth of sentences in $\mathfrak{A}$. I think I have a simple proof of this fact, but it has led me to an expected conclusion.

First, let $E$ be the class of all structures whose domain is $\{c\}$, the denotation of every function and constant symbol is completely determined, but each relation symbol has two possible denotations.

It seems to me that we can always interpret any structure inside a structure within $E$ by sending every element of the domain to $c$.

This interpretation will preserve the truth of sentences in the original structure, but loses information about the relationship between, for example, $\forall x \mathop. \varphi(x) \land \psi(x)$ and $\varphi(x)$ and $\psi(x)$ since it identifies all satisfiable formulas with the same number of free variables.

Basically, my question is about the definition of interpretations and whether we need to or want to rule out interpretations going to a structure in $E$ when defining "just an interpretation" and not a "definable interpretation" or otherwise.


Interpretability is defined in Marker's Model Theory: An Introduction on page 24, but interpretations themselves are not until we get to the notion of definable interpretability. Hodges' a shorter model theory does define an interpretation on page 107 and that definition is similar to the one below; it includes domain formulas and maps formulas to formulas. I'm, however, using the definition from Wikipedia, I'm just changing it slightly to be a map between definable sets subject to certain conditions rather than a map between elements that preserves definability.

Here's the definition from Wikipedia:

An interpretation of a structure $M$ in a structure $M$ with parameters (or without parameters, respectively) is a pair $(n, f)$ where n is a natural number and $f$ is a surjective map from a subset of $N^n$ onto $M$ such that the $f$-preimage (more precisely the $f^k$ preimage of every $X \subset M^k$ definable in $M$ by a first-order formula without parameters is definable (in $N$) by a first-order formula with parameters (or without parameters, respectively).

The following definition below is intentionally agnostic with respect to whether parameters are allowed or not.

An interpretation in model theory $i : \mathfrak{A} \to \mathfrak{B}$ is a map from the definable sets of $\mathfrak{A}$ to the definable sets of $\mathfrak{B}$ subject to the following constraints below:

  • There must exist a natural number $n$ and a function $f : A \to 2^{B^{\,n}}$ such that, for any definable set $X$ of dimension $k$, $i(X)$ is equal to $\cup \{ f(\vec{v}) : \vec{v} \in X\}$, where $f(\vec{v})$ is a tuple of length $kn$ defined by elementwise application of $f$ and then combining the results together using the Cartesian product.
  • Destinations must be unique. For any distinct $a, b$ in $A$, $f(a)$ and $f(b)$ must be disjoint.
  • Destinations must exist. $f(a)$ is not empty for any $a$.

Theorem: The Wikipedia definition and my definition are equivalent.

LTR: Let $(n, g)$ be an interpretation. Let $f(x)$ be defined as $g^{\leftarrow}(x)$, the preimage of $x$. Note that $x$ is in $A$ and $f(x)$ is in $B^n$.

$g$ is surjective, so $f$ never sends anything to $\varnothing$. $g$ is a function, so $f$ always sends distinct elements of $A$ to disjoint subsets.

RTL: Let $f : A \to 2^{B^{\,n}}$ be an interpretation. Let $C$ be the image of $f$ and let $D$ be $\bigcup C$. Let $g$ be a function $g : C \to A$ defined as: $g(x) = y$ if and only if $x \in f(y)$. Since $f$ sends distinct items to disjoint sets and doesn't send anything to the empty set, $g$ is a function. $f$ is a function, therefore $g$ is surjective.

Lemma: a sentence $\varphi$ is true if and only if its definable set is a singleton.

Suppose $\varphi$ is true in a structure $\mathfrak{M}$. Then $\{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ will consist of exactly the empty variable map.

Suppose $\varphi$ is not true in a structure $\mathfrak{M}$. Then $\{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ will be empty.

Theorem: $i$ preserves the truth of sentences.

Let $\varphi$ be a sentence. Suppose $\varphi$ is true. $X = \{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ is a singleton. Then $f$, applied elementwise to the empty tuple, the sole element of $X$, will also produce an empty tuple and thus $i(X)$ will be a singleton.

Suppose $\varphi$ is not true. Then $X = \{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ will be empty and thus $i(X)$ will also be empty.

Therefore $i$ preserves the truth of sentences.

Theorem: the map $i : \mathfrak{A} \to E_0$ sending every element of the domain to $c$ is an interpretation, where $E_0$ is a structure in $E$.

Let $f$ be the function $x \mapsto \{c\}$, it is a constant map

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$\def\F{\operatorname F}$ Even though the OP is a high schooler, it would be fun to exactly solve their equation. They likely want the real roots. Let’s apply Lagrange reversion after taking the multivalued inverse tangent:

$$\tan(x)=x^3\iff x=\left(k+\frac12\right)\pi+\frac i2\ln\left(\frac{2i}{x^3-i}+1\right)\mathop\iff^{} x_k=\left(k+\frac12\right)\pi+\sum_{n=1}^\infty \left(\frac i2\right)^n \frac{d^{{n-1}}}{dx^{n-1}}\left.\frac{\ln^n\left(\frac{2i}{x^3-i}+1\right)}{n!}\right|_{\left(k+\frac12\right)\pi}$$

for $0,-1\ne k\in\Bbb Z$ defining $x_0=0,x_{-1}=x_1$. Use a Stirling S1 expansion:

$$\frac{d^{{n-1}}}{dx^{n-1}}\frac{\ln^n\left(\frac{2i}{x^3-i}+1\right)}{n!}=\sum_{m=n}^\infty \frac{S_m^{(n)}(2 i)^m}{m!}\frac{d^{n-1}}{dx^{n-1}}(x^3-i)^{-m}$$

Binomial series on $\frac{d^{n-1}}{dx^{n-1}} x^{-3m}(1-ix^{-3})^{-m}$ uses hypergeometric $_3\F_2$:

$$$$

fix binom link


For Expressions for the inverse function of $f(x) = \ln(x)e^x$

On some applications of the generalized hyper-Lambert functions mentions a complicated Lagrange inversion series expansion referencing

Saks and Sygmund [8, pp. 201–202])

However their Analytic functions paper does not evaluate the derivatives for the coefficients. The following methods are applicable to solving $za_1^{\cdots^{a_j^z}}=a$.

$$f(z)=z\underbrace{e^{e^{\dots^z}}}_{k “e”\text s}\implies f^{-1}(z)=\sum_{n=1}^\infty\frac{z^ n}{n!}\frac{d^{n-1}}{dw^{n-1}}e^{-n \overbrace{e^{e^{\dots^z}}}^{k-1 “e”\text s}}\bigg|_0$$

Repeated Maclaurin Expansion:

$$\frac{d^{n-1}}{dw^{n-1}}e^{-n \overbrace{e^{e^{\dots^z}}}^{-1 “e”\text s}}\bigg|_0 $$

Stirling Expansion:

$$\frac{d^{n-1}}{dw^{n-1}}e^{-n \overbrace{e^{e^{\dots^z}}}^{-1 “e”\text s}}\bigg|_0 $$

Coefficients for All Branches:

[Evaluate?]


$\def\srt{\operatorname{srt}}$

$$\sqrt[k]z_s=\srt_k(z)=z-\sum_{n=1}^\infty\frac1{n!}\frac{d^{n-1}}{dt^{n-1}}e^{(n+1)t-e^t-ne^{t-e^{\dots}}}\bigg|_{\ln(-\ln(z))}$$

$\vdots$

Perhaps we change order of summation to get a closed form finite series for the coefficients. The convergence is about $0<|z|<1.3$ excluding where the 2 real branches of $\srt_k(z)$ start. This general formula gives:

$$\srt_3(z)=$$

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Hmm. For any cyle we need $$ 2^m = (3+1/x_1)(3+1/x_2) \cdots (3+1/x_n) $$ Dividing by $3^n$ gives $$ 2^m/3^n = (1+1/3/x_1)(1+1/3/x_2) \cdots (1+1/3/x_n) $$ Because the rhs is larger than $1$, this says that $2^m > 3^n$ is required, and more precisely, the smallest $m$ is $m = \lceil n \cdot \log_2(3) \rceil $. The "ceil()" can be decomposed to explicitely $m = n \cdot \log_2(3) - \{ n \cdot \log_2(3) \} + 1 $

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This answer is free for anyone to use.

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This answer box is now in use. Please do not use.

Let $\ \mathcal{O}_1,\mathcal{O}_2,\dots,\mathcal{O}_x\ $

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    $\begingroup$ I have a question, if two people edit a post at the same time, one of them will undoubtedly lose his work. $\endgroup$
    – user1034536
    Jan 17 at 4:25
  • $\begingroup$ @user1034536 That is why it is a good idea, to initially edit the posting with one or two lines, saying "This answer box is now in use. Please do not use." Then, you can immediately save this editing. In effect you are placing a temporary do not disturb sign around the answer box. $\endgroup$ Mar 27 at 14:20

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