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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
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    $\begingroup$ 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 Maybe someday we can get rid of this outdated crutch.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

17 Answers 17

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Background

In the following MO question, I asked along what curve an ellipse rolls down fastest. It was pointed out that one first needs to find any curve the ellipse rolls down from, without losing contact from the curve and jumping up and down.

Now, it turns out that are various curves along which an ellipse can roll horizontally. These are described in this video by Morphocular. It turns out that, if one wants to describe the equation of the curve along which an ellipse with width $2a$ and height $2b$ rolls down horizontally with the center as its axle point, the x-coordinate of this equation comes down to an elliptic integral (see 5:04 of the video):

$$x = \int \frac{b}{\sqrt{1-\epsilon^{2} \cos^{2}(\theta)}} d \theta \label{1}\tag{1} $$

This is not an easy expression to deal with. However, there is a way out of one chooses the foci as the axle points, instead of the center of the ellipse. Assume that the ellipse has widht $1$ and height $\sqrt{2}$. In this case, we can describe the curve with a simple sine wave, with period $\pi$ and amplitude $1/2$. Here is a picture (a snapshot from the video at 7:39):

                                               enter image description here

The equation of this curve can thus be described succinctly as $$y = \frac{1}{2} \sin(2x) \label{2}\tag{2}$$

enter image description here

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Background

The classical Riemann integral of a function $f : [a,b] \to \mathbb{R}$ can be defined by setting $$\int_{a}^{b} f(x) \ dx := \lim_{\Delta x \to 0} \sum f(x_{i}) \ \Delta x. $$ Here, the limit is taken over all partitions of the interval $[a,b]$ whose norms approach zero.

We can do something roughly similar with product integrals. They take the limit over a product instead of a sum, and can be interpreted as continuous analogues of discrete products.

There are multiple types of product integrals. Type I is often refered to as Volterra's integral. It is defined as follows:

\begin{align*} \prod_{a}^{b} \left(1+f(x) \ dx \right) &:= \lim_{\Delta x \to 0} \left(1 + f(x_{i}) \ \Delta x \right) \newline &= \exp \left( \int_{a}^{b} f(x) \ dx \right). \tag{1} \label{1} \end{align*}

However, this is not a multiplicative operator. As an alternative, there is also Type II, the geometric integral. It is defined as

\begin{align*} \prod_{a}^{b} f(x)^{dx} &:= \lim_{\Delta x \to 0} \prod f(x_{i})^{\Delta x} \newline &= \exp \left( \int_{a}^{b} \ln f(x) \ dx \right). \tag{2} \label{2} \end{align*}

Question

I wonder whether something similar can be done with other kinds of expressions (infinite or finite). In particular, I am curious whether we can obtain a continuous analogue of the discrete simple continued fraction. In other words, I am looking for a way to complete the following table, by finding a definition of the bottom right cell:

\begin{array}{|c|c|c|} \hline & \text{additive} & \text{multiplicative} & \text{continued fraction} \\ \hline \text{discrete} & \sum_{i=a}^{b} f(i) & \prod_{i=a}^{b} f(i) & \underset{i=a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(i)} \\ \hline \text{continuous} & \int_{a}^{b} f(x) \ dx & \prod_{a}^{b} f(x)^{dx} & \underset{a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(x)} \overline{dx} \\ \hline \end{array}

I think we can define it by setting

$$ \underset{a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(x)} \overline{dx} := \lim_{\Delta x \to 0} \large{\mathrm K} \frac{1}{f(x_{i})} \Delta x .$$

I am not sure, however, how this would translate into a formula that is similar to \eqref{1} or \eqref{2}.

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  • $\begingroup$ In this case, I've used the space for a question, as the usual previewer doesn't render the mathematics properly in some cases. $\endgroup$
    – Max Muller
    Mar 10 at 14:25
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(Originally from MSE, but crossposted here upon suggestion from the comments)

In this MSE post, user Moishe Kohan provides an example of a non-continuous open and closed ("clopen") function $\mathbb R^n \to \mathbb R^n$ for $n\geq 3$ by citing and using the following propositions of David Wilson:

Wilson Setup: for $(X,d)$ a metric space, some subsets $A,B \subseteq X$, and $\scr G$ a collection of subsets of $X$, we use the notation

  • $\mathscr G^*:= \bigcup_{G\in \mathscr G} G$;
  • $N_\epsilon(A):= \{x\in X: \exists a\in A \text{ s.t. } d(x,a)<\epsilon\},$ the $\epsilon$-neighborhood of $A$;
  • $\newcommand{\diam}{\operatorname{diam}} \diam[A]:= \sup_{x,y\in A} d(x,y)$ is the diameter of $A$;
  • $d[A,B]:= \inf\{\epsilon>0: A \subseteq N_\epsilon(B) \text{ and } B \subseteq N_\epsilon(A)\}$ is the Hausdorff distance;
  • $\mu(\mathscr G) := \max\{\diam[G]:G\in \scr G\}$.

Wilson Prop. 1: let $Z_1,Z_2$ be compact metric spaces with $X\subseteq Z_1$ and $Y\subseteq Z_2$ compact subsets (of course themselves metric spaces, with metrics induced by $Z_1,Z_2$ resp.). Suppose there exist 2 sequences $(\mathscr F_n)_{n=1}^\infty, (\mathscr G_n)_{n=1}^\infty$ where each $\mathscr F_n, \mathscr G_n$ is a finite collections of compact subsets of $Z_1,Z_2$ resp. (i.e. $\mathscr F_n = \{F_{n,1},\ldots, F_{n, m_n}\}$ for compact $F_{n,i} \subseteq Z_1$; analogous for $\mathscr G_n$), satisfying the following 5 properties:

  1. [The $\mathscr F_n$ are a sequence of covers decreasing down to $X$]: $$Z_1 \supseteq \mathscr F_n^* \supseteq \mathscr F_{n+1}^* \supseteq X \text{ for all $n$,} \quad \text{ and } \bigcap_{n=1}^\infty \mathscr F_n^* = X.$$

  2. [The $\mathscr G_n$ are a sequence of covers decreasing down to $Y$]:

    same as (1.) but with $Z_2$, $\mathscr G_n$, and $Y$ replacing $Z_1, \mathscr F_n, X$ resp.

  3. [The covers $\mathscr G_n$ are made up of ever smaller compact sets]: $$\forall \epsilon>0, \exists N>0 \quad \text{ s.t. } \quad n>N \implies \mu(\mathscr G_n)<\epsilon.$$

  4. [There exist functions $\phi_n : \mathscr F_n \to \mathscr G_n$, preserving inter-level nestings and nontrivial intra-level intersections]:

    1. for compact sets $F_{n-1} \in \mathscr F_{n-1}, F_n \in \mathscr F_n$, $$F_{n-1}\supseteq F_n \implies \phi_{n-1}(F_{n-1}) \supseteq \phi_n(F_n)$$
    2. If $F_n, F_n' \in \mathscr F_n$, then $$F_n \cap F_n' \neq \varnothing \implies \phi_n(F_n) \cap \phi_n(F_n') \neq \varnothing$$
  5. [Nested sequence of compact $F_n$ around each point]: $$\forall x\in X, \quad \exists (F_n)_{n=1}^\infty \in (\mathscr F_n)_{n=1}^\infty \quad \text{ s.t. } \forall n, \quad x \in F_n \subseteq F_{n-1}.$$

----THEN:---- there exists a continuous function $g:X\to Y$ defined by using (5.) to choose a nested sequence $(F_n)_n\in (\mathscr F_n)_n$ around $x$, and setting $$g\left(\bigcap_{n=1}^\infty F_n\right) := \bigcap_{n=1}^\infty \phi_n(F_n),$$ i.e. setting $g$ on the intersection $\bigcap_{n=1}^\infty F_n$ to be the point in the singleton set $\bigcap_{n=1}^\infty \phi_n(F_n)$ (by (3.), the $\phi_n(F_n)$ are necessarily shrinking in diameter).

Propositions 1 and 3 of

D. Wilson, Open mappings of the universal curve onto continuous curves. Trans. Amer. Math. Soc. 168 (1972), 497–515.

[Moishe Kohan's] Proposition [cooked up from Wilson's propositions]. Let $I^n$ be the closed $n$-dimensional cube, $n\ge 3$, and $J^n\subset int(I^n)$ is a closed subcube. Let $Q$ denote the interior of $J^n$. Then, there exists an open continuous map $g: I^n\setminus Q\to I^n$ which equals the identity on the boundary of $I^n$ and sends $\partial Q$ to the interior of $I^n$.

I am interested in the $2$-dimensional case, for which I propose the following conjecture:

Conjecture: there are no continuous open (w.r.t. subspace topologies) maps $f$ from the closed annulus $\mathbb A:= \{x\in \mathbb R^2: 1\leq |x|\leq 2\}$ to the closed disk $\mathbb D := \overline{B(0,2)}$, which restricts to the identity map on the outer boundary circle $\partial \mathbb D = C(0,2)\subseteq \mathbb A$, and sends the inner boundary circle $C(0,1)$ to the interior of $\mathbb D$.

I was not able to prove or disprove this conjecture for even the Simpler Case: where $f$ maps the inner boundary circle $C(0,1)$ to the single point $0\in \mathbb D$.




Some attempts I made on the Simpler Case.

Attempt 1: Because we want $f$ to be open, we in particular want any open neighborhood in $\mathbb A$ of any boundary point $b\in C(0,1)$ to map to an open set containing $0\in \mathbb D$, in particular containing some $B(0,\epsilon)$.

My idea was map circles $C(0,1+\eta)$ to circles $C(0,\eta)$, and to to "swirl"/"smear" the circles $C(0,1+\eta)$ more and more extremely as $\eta \searrow 0$, so that even a very small neighborhood $B(b,\delta)\cap \mathbb A$ of $b\in C(0,1)$ containing just a $\approx \frac{2\delta}{2\pi}$ fraction of the circles $C(0,1+\eta)$ for small enough $\eta>0$ would get "swirled"/"smeared" to contain the entire circle $C(0,\eta)$.

So something like $f$ maps $z:=re^{i\theta}\in \mathbb A$ (thinking of $\mathbb R^2$ as the complex plane) to $(r-1)e^{i\theta\cdot \frac{1}{r-1}}$, mapping an arc of the circle $C(0,1+\eta)$ of arclength $\ell$ to an arc of the circle $C(0,\eta)$ with arclength $\ell \cdot \frac{1}{r-1}$. Unfortunately, the corresponding formula for $f$ would be $$f(z)=\frac{|z|-1}{|z|^\frac{1}{|z|-1}} \cdot z^{\frac{1}{|z|-1}},$$ which maybe looks fine, until one remembers that raising a complex number to a non-integer power needs (non-continuous) branch cuts to define. :(


Attempt 2: One can make a continuous "swirling only" (no "spreading", i.e. no multiplicative factor in the argument variable) by mapping $r e^{i\theta} \mapsto (r-1) e^{i\theta + i\cdot \frac{1}{r-1}}$, i.e. $$f(z) = \frac{|z|-1}{|z|} \cdot z \cdot e^{\frac{1}{|z|-1}},$$ which I think is continuous, and does "swirl" a very small neighborhood $N_b:= B(b,\delta)\cap \mathbb A$ to something that does sort of "wrap around" $0 \in \mathbb D$, but in doing so has a lot of holes: this $f$ preserves that proportion of the arclength of $N_b \cap C(1+\eta)$, meaning $\text{arclength}(N_b \cap C(1+\eta)) = \text{arclength}(f(N_b) \cap C(0,\eta))$, so $f$ can't possibly map $N_b$ to something containing $B(0,\epsilon)$.


Attempt 3: Finally, because of this "monodromy problem" of defining this "swirling/smearing" map on the entire circle $C(0,1+\eta)$, I had an idea of cutting up $C(0,1+\eta)$ into $\frac 1\eta$ many pieces (restricting to $\eta \in \{\frac 1n: n=2, 3, 4, \ldots\}$), making $f$ "swirl/smear" each of those pieces into the entirety of $C(0,\eta)$, thus guaranteeing that $f(N_b)$ contains complete circles $C(0,\eta)$ arbitrarily close to $0\in \mathbb D$.

More precisely, I would partition $C(0,1+\frac 1n)$ into $2n$ equally sized, equally spaced pieces, and define $\{K_{n, i}\}_{i=1}^n$ to be the closures of the say odd-indexed pieces. I can map $K_{n,i}$ continuously to cover the entirety of $C(0,\frac 1n)$. Then for any $N_b:= B(b,\delta)\cap \mathbb A$, it does contain some $K_{n,i}$ for all $n$ sufficiently large, and hence $f(N_b)$ contains $C(0,\frac 1n)$ for all $n$ sufficiently large. I can extend this $f$ defined on $C(0,1) \cup \bigcup_{n=2}^\infty\bigcup_{i=1}^n K_{n,i}$ via the Tietze extension theorem to a continuous function $\mathbb A \to \mathbb D$, but again I really doubt it is a an open map.




So as it stands regarding the Simpler Case, I can't even find a continuous function $f:\mathbb A\to \mathbb D$ that maps the neighborhood $B(b,\delta)\cap \mathbb A$ (for even a single fixed $\delta>0$) of any point $b\in C(0,1)$ to an open set containing $0\in \mathbb D$; nor can I disprove its existence.

---(EDIT: user1789 posted an answer that was later found to be incorrect and hence deleted, but if I recall correctly, that answer was able to map the inner boundary in "an open manner", but pushed the issue into the interior of the annulus. Those with sufficiently high reputation should be able to see this attempt.)---

Of course, one possible route of proving the Conjecture is if one accomplishes the original MSE post's goal of proving that every clopen function $\mathbb R^2 \to \mathbb R^2$ must be continuous (as for example is true in the $\mathbb R^1 \to \mathbb R^1$, as I showed here), but I suspect that problem is even harder than this one.

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$$ k 2^B = rk + 2^B-1 \\ k (2^B-r) = 2^B-1 \\ $$

The first means $n = 2^Ak - 1$. But I don't know how to use the second.

It looks like a Diophantine equation $2ᴺ⁻ᴬn - 3ᴬm = -1$, which could be so

$$(2^N - 3^A)n = 3^A - 2^A \\ (2^N - 3^A)(n+1) = 2^N - 2^A = 2^A ( 2^B -1) \\ \implies n=2^Ak-1 \\ (2^N - 3^A)k = 2^B -1 \\ \implies k \lt 2^B $$ Now let's decode $3^A$ modular in terms of $2^B$: $3^A = m2^B+r$ where $0<r<2^B$ $$ (2^N - m2^B-r)k = 2^B -1 \\ (2^B2^A - m2^B-r)k = 2^B -1 \\ 2^B(2^A - m)k-2^B = rk -1 \\ (2^A - m)k = {rk -1 \over 2^B}+1 = {r \over2^B }k +1 - 1/2^B \\ \text{and in the rhs } \qquad {r \over2^B }k +1 - 1/2^B <k+1 $$ so also the lhs must be smaller than $k+1$ and thus must be $k$ and thus $$ m = 2^A - 1$$ But this means, a perfect power $3^A$ is of the form $(2^A-1)2^B+r$ and in binary representation this means $$ 3^A = \text{'111...1 00...0'}_2 + r $$

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Maximum dimension of a subspace of $M_n(\mathbb C)$ containing no invertible matrix

I would want someone to cross check my proof.

Let $\mathbb F$ be a field.
Let $V$ be a subspace of $\mathbb F^{n\times n}$ in which there is no invertible matrix.

We want to find the maximum possible dimension of $V$.

None of the matrices in $V$ is full rank. Using this, I can explicitly construct a $V$ which has dimension $n(n-1)$. $$\left\{\begin{pmatrix}R_1\\ R_2\\ \vdots \\ R_{n-1}\\ \mathbf 0_{1\times n}\end{pmatrix}:R_i\in\mathbb F^{1\times n}\right\}$$ Notice that all of them have a common eigenvalue $0$ associated with a common eigenvector $(0,0)$

Now suppose if there is a vector space where evr$$

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This post is free to use for anyone.

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First of all let's we remember the following result.

Theorem

Let be $\lambda$ and ordinal: a predicate $\mathbf P$ is true for any $\alpha$ in $\lambda$ when the truth of $\mathbf P$ for any $\beta$ in $\alpha$ implies that for $\alpha$.

Let be now $f$ a function from an order $(X,\mathcal U)$ into an order $(Y,\mathcal V)$: we will say that $f$ is monotone if for any $a$ and $b$ in $X$ the inequality $$ a\preccurlyeq_\mathcal U b $$ implies the inequality $$ f(a)\preccurlyeq_\mathcal V f(b) $$ So let's we prove the following result.

Conjecture

A function $f$ from an ordinal $\lambda$ into an order $(X,\mathcal O)$ is monotone if and only if the following holds:

  • for any $\delta$ in $\lambda$ the inequality \begin{equation} f(\delta)\preccurlyeq_\mathcal O f(\delta+1) \tag{1} \label{successive imagine} \end{equation} holds;
  • if $\alpha$ in $\lambda$ is limit then the inequality \begin{equation} f(\beta)\preccurlyeq_\mathcal O f(\alpha) \tag{2} \label{limit imagine} \end{equation} holds for any $\beta$ in $\alpha$.

Proof Let's we assume that ineq. \eqref{successive imagine}-\eqref{limit imagine} hold and thus let's we prove even the proposition \begin{equation}(\forall\alpha)\Biggl((\alpha\in\lambda)\to\biggl((\forall\beta)\Big((\beta\in\alpha)\to\big(f(\beta)\preccurlyeq_\mathcal O f(\alpha)\big)\Big)\biggl)\Biggl) \tag{3} \label{monotonia} \end{equation} holds.

   So let be $\alpha$ an element of $\lambda$ and thus let's we assume the proposition
    \begin{equation}
        (\forall\gamma)\Big((\gamma\in\beta)\to\big(f(\gamma)\preccurlyeq_\mathcal O f(\beta)\big)\Big)
        \tag{4}
        \label{ipotesi induttiva}
    \end{equation}
    for all $\beta$ in $\alpha$. Well if there exist an ordinal $\delta$ such that
    $$
    \alpha=\delta+1
    $$
    then any for any $\beta$ in $\alpha$ the inequality
    $$
    \beta\le\delta
    $$
    holds so that by inductive hypothesis and ineq. \eqref{successive imagine} the inequality
    $$
    f(\beta)\preccurlyeq_\mathcal O f(\delta)\preccurlyeq_\mathcal O f(\delta+1)=f(\alpha)
    $$
    holds; after all if $\alpha$ is limit then by ineq. \eqref{limit imagine} surely \label{ipotesi induttiva} holds for $\alpha$ too: so we conclude \ref{ipotesi induttiva} holds for $\alpha$ always and thus by transfinite induction it generally holds for all $\alpha$ in $\lambda$ so that \ref{monotonia} holds.  

Conversely if $f$ is monotone then ineq. \eqref{successive imagine}-\eqref{limit imagine} trivially holds -by monotonicity definition.

So I ask if the conjecture is actually true and thus if I well proved: could someone help me, please?

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$$(1/u)^{u-1/u}=2y$$ $$e^{ln(1/u)(u-1/u)}=2y$$ $u\to e^x$: $$e^{-x(e^x-e^{-x}))}=2y$$

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(Editing just now please do not interfere -------------)

part 4
We'll start with the assumption in (3.4c), ($S = \lceil N \cdot \log_2(3) \rceil $) and for convenience we reformulate this as $$ \alpha \cdot 2^S = \alpha \cdot 3^N + Q(E_{N:S}) \tag {4.1} $$


part 1
First a bit about notation.

The Collatz-transformation is sometimes equivalently expressed in the Syracuse-form : $$ \text{over positive odd integers $a_k$ and }\\ \text{positive integers $A_k$} \tag {Syracuse notation}$$ $$ \mathcal R: \qquad a_{k+1}={ 3a_k+1 \over 2^{A_k} } \tag {1} $$ I use capital letters in the exponents at $2$: $ A_k=\nu_2(3a_k+1)$ and as well $N$ for the number of all odd steps (= length of the orbit) and $S$ for the sum-of-exponents/number-of-divisions-by-$2$.

part 2
An iterated transformation can be memorized by the vector of exponents alone $$E_{N,S}:=[A_k]_{k=1...N} \tag {2.1}$$ and has the well known expression: $$ a_N = a_0 \cdot {3^N \over 2^S} + {Q([A_1,A_2,...,A_N]) \over 2^S} \tag {2.2}$$ where the short-form-notation $Q()$ expanded is the similarly known sum $$ Q([A_1,A_2,...,A_N]) = 3^{N-1} + 3^{N-2}2^{A_1} + 3^{N-3}2^{A_1+A_2} + ... + 3^0 2^{A_1+A_2 + ... + A_{N-1}} \tag {2.3} $$


part 3 Cycles and solutions over rational numbers

From $(2.2)$ we can write the equation for some transformation $a_0 \to a_N$: $$ 2^S a_N = 3^N a_0 + Q(E_{N:S}) \tag {3.1} $$ To form a cycle we demand for this $a_0 = a_N$ . We can derive a determination for $a_0$: $$ a_0 = { Q(E_{N:S}) \over 2^S - 3^N } \tag {3.2} $$ We know only one cycle in the Collatz-problem over positive odd integers ($a_0=1$) and only 3 more in the negative odd integers ($a_0 \in \{-1,-5,-17\}$).
But obviously there are solutions over the rational numbers. To memorize that generalization to rationals we use now the letter $\alpha$ for such a rational or integer solution: $$ \alpha = { Q(E_{N:S}) \over 2^S - 3^N } \tag {3.3} $$ This little equation shows that to have $\alpha$ in the positive numbers we need, that the denominator must be positive, and thus we need $$ 2^S > 3^N \tag {3.4a}$$ and by this, $$S \ge \lceil N \cdot \log_2(3) \rceil \tag {3.4b} $$ or as expressed in the OP $$S \ge \lfloor 1+ N \cdot \log_2(3) \rfloor $$


The cycle- and stoppingtime-discussions usually assume, that in (3.4b) we can always apply the assumption of equality: $$S = \lceil N \cdot \log_2(3) \rceil \tag {3.4c} $$ and the OP puts this assumption in question, targetting the possibility that perhaps $$S = 1+\lceil N \cdot \log_2(3) \rceil \tag {3.4d} $$ to calculate the correct stopping time.

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for real $u,y$:

$$\left(\frac{1}{u}\right)^{u-\frac{1}{u}}=2y$$ $$e^{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)}=2y$$

We can press many expression into a form for applying hyper Lambert W:

$$ue^{f(u)e^u}=e^{g(u)}$$ $$f(u)=\frac{g(u)-\ln(u)}{e^u}$$ $\ $

$$g(u)=\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)$$ $$f(u)=\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u}$$ $$ue^\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u}=2y$$ $\ $

$$G\left(\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u};u\right)=2y$$ $$u=HW\left(\frac{\ln\left(\frac{1}{u}\right)\left(u-\frac{1}{u}\right)-\ln(u)}{e^u};2y\right)$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

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This post is free for anyone to use.

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$\def\dn{\operatorname{dn}}$

This goal is to understand how to expand inverses of non-elementary functions as a series. For example the Jacobi dn Fourier cosine series from Paramanand’s blogspot:

$$\dn(u,k)=\frac{a_0}2+\sum_{n=1}^\infty a_n\cos(2nz),a_n=\frac1\pi\int_{-\pi}^\pi\dn(u,k)e^{-2inz}dz,z=\frac{\pi u}{2K(k)}$$

using residue calculus to find $a_n=\frac{2\pi}{K(k)(q^{-n}(k)+q^n(k)}$ with the nome $q(k)$ and complete elliptic integral of the first kind $K(k)$. However, for someone not knowing the residue theorem, it would be hard to derive this result.


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?]

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  • $\begingroup$ The editing sandbox is very slow to type in. $\endgroup$ Apr 2 at 20:36
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Free for use. ${}{}{}{}{}{}{}{}$

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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, 2023 at 4:25
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    $\begingroup$ @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, 2023 at 14:20
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With a polynomial with real coefficients $p(x)$, what's the largest and smallest imaginary part you can get evaluating $p$ at a quaternion $\vec{q}$ giving $p(\vec{q})$?


What is $\mathbb{Q} \otimes \mathbb{R}$ as Abelian groups?

$q, q'$ are in $\mathbb{Q}$. The rest $a,b,c,d,r,r'$ are in $\mathbb{R}$.

$$ \begin{bmatrix} q & r \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} q' \\ r' \end{bmatrix} = \begin{bmatrix} qa+rc & qb+rd \end{bmatrix} \begin{bmatrix} q' \\ r' \end{bmatrix} = qaq' + rcq' + qbr' +rdr' $$

This map $\mathbb{Q} \times \mathbb{R} \to \mathbb{R}$ is multilinear.


Integral domains with exactly two units.

We have $\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$, but no other finite ones.

Also $\mathbb{Z}[x]$ and $(\mathbb{Z}/3\mathbb{Z})[x]$.


Prove completeness of Routley-Meyer semantics using

equational logic (kinda) with $\Delta$ which sends truth values to $2$, e.g. modus ponens is the following:

$$ \Delta(a) \land \Delta(a \to b) \to \Delta(b) $$

The idea is to make a unifying logic that can express both the inference rules of the relevance logic $B$ and worlds and propositional variables for the semantics. It might not work.


Derivative operator defined in terms of forward difference where you pass in the nominal degree of the function in question.

Also, $\mathbb{Z}/p\mathbb{Z}$ as an almost $\mathbb{Q}$-vector space. Make $1/p$ behave like $0$. Then can we define stuff like $\sin$ using its power series?


Matrix algebras generated by:

$$ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \;\; \text{and} \;\; \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$


Four-place distance relation $Sabcd$

with the intuitive meaning $(a, b) \simeq (c, d)$, in prose $a$ is the same distance from $b$ as $c$ is from $d$.

With this you can defined lines and circles ... and triangles and other stuff like that.

How do you axiomatize the Euclidean version?


  • $\varphi : R \to \text{End}(S)$
  • $\psi : S \to \text{End}(T)$

We can make $\alpha : R \times S \to \text{End}(T)$ as follows:

  • $\alpha(r, s, t) = \psi(\varphi(r, s), t)$

Let rings be commutative but not necessarily unital.

Let multiplicatively closed sets do not necessarily contain $1$.

Are there any infinite rings such that all multiplicatively closed sets are finitely generated?

In the infinite ring $\mathbb{Z}$, the ring itself is not finitely generated as a multiplicatively closed set, since there are infinitely many prime numbers by Euclid's theorem.

If we assume temporarily that we have a ring multiplicatively generated by $x$, so the elements are $\{x, x^2, x^3, \cdots\}$ with addition somehow defined, then all of its sets are finitely multiplicatively generated.

Let $F \subset \mathbb{N}$ and consider the set $X$ multiplicatively generated by $\{x^k : k \in F \}$.


Ultranaive compactification.

Add a new point, say that its only open neighborhood is the entire space.

Now, any open cover of the space MUST include $\varnothing^c$.

Is this a valid compactification?


A logic where $\varphi \to \psi$ maps to $\bigwedge_{v \in \text{Vars}(v)} \lozenge(v) \land \square(\varphi \to \psi) $ in the classical modal logic S5.

The idea is that this logic is potentially badly behaved and interacts poorly with substitution.


We work in NFU with choice, infinity, and pairs.

$$ [\forall xyzw]((x, y) = (z, w) \leftrightarrow (x=z \land z=w)) $$

Let $P(z)$ hold of $z$ if and only if it is a pair:

$$ P(z) \;\; \text{is defined as} \;\; [\exists r][\exists w](z = (r, w)) $$

Let $R(x)$ hold of $x$ if it is a relation:

$$ R(x) \;\;\text{is defined as}\;\; [\forall z \in x](P(x)) $$

Let $B(x)$ hold of $x$ if and only if it is a bijection of the whole space:

$$ B(x) \;\;\text{is defined as}\;\; R(x) \land [\forall z](\cdots) $$


What are the maximal ideals of this ring?

Let $X$ be an infinite set with a distinguished point $x_0$.

Let $F$ be a family of sets $\{ a : a \in 2^X \;\text{and $a$ is finite} \}$ .

Let $a + b$ be $\{ w : w \in a \not\leftrightarrow w \in b \}$.

Let $a\cdot b$ be $(a \cap b) \setminus \{x_0\}$.

$0$ is the annihilator for $\cdot$.

Additionally, $(a+b)\cdot(c + d) = ac + bc + ad + cd$ and the other distributive properties hold.


Best effort substructure

One somewhat obvious fact is that given an arbitrary subset of the domain of a structure, one cannot always take a substructure with that subset as its domain. Relation symbols do not present an obstacle here, but function symbols do.

(Also, what to do about partial congruences?) Partial congruences are interesting, they also give us partial ideals.

Let $L$ be a language and let $A$ be an $L$-structure. Let $B$ be a subset of $A$.


green's relations on semigroups.


Fake Burnside ring.

This is based on intentionally misunderstanding how the product (defined on $G$-sets) in the Burnside ring is defined.

Suppose we have $\boxed{1}$ and $\boxed{3}$ and

$\boxed{1}\boxed{1}$ is $\boxed{1}$

$\boxed{1}\boxed{3}$ is $\boxed{3}$

$\boxed{3}\boxed{1}$ is $\boxed{3}$

$\boxed{3}\boxed{3}$ is $9\,\boxed{1}$

multiplication is associative.

Equivalent to ring $\mathbb{Z}[b] / (b^2 - 9)$.

It gets weirder if we get rid of $\boxed{1}$.

Which rules correspond to actual groups?

The split complexes have a fake Burnside representation if we allow $\boxed{-1}$.

The complex numbers however do not have a fake Burnside representation.


https://hsm.stackexchange.com/questions/6256/did-gauss-know-jacobis-four-squares-theorem/12593

Where does this function $P(x, y) = \sum_{k} x^{(k^2)}(y^k + y^{-k})$ come from?


Annihilator as a maximal ideal

Let all rings be commutative and unital.

Lemma 51: Let $G$ be a finite group. If $I$ is an ideal of $R$, then $I[G]$ is an ideal of $R[G]$.

Proof. Suppose $I$ is an ideal of $R$.

Consider an element of $I[G]$. It always has the form $\sum_i a_ig_i$ where $a_i$ is an element of $I$ and $g_i$ is an element of $G$. $I[G]$ is obviously closed under subtraction.

Consider an element of $R[G]$ it halways has the form $\sum_j r_jg_j$ where $r_j$ is an element of $R$.

Their product is $\sum_{ij}r_ja_ig_ig_j$.

I note that $r_ja_i$ is in $I$.

Therefore $I[G]$ is closed under multiplication by $R[G]$, as desired.


Lemma 61: Let $G$ be a finite group. If $I$ and $J$ are ideals of $R$ and $I \subsetneq J$, then $I[G] \subsetneq J[G]$ and furthermore $J \setminus I[G]$ is nonempty.

$G$ contains an identity element $e$. Let $j$ be an element of $J$ that isn't in $I$. $je$ is in $J[G]$ but not $I[G]$, as desired.

$je = j$, additionally, is in $J$.


Note that $S$ is an $R$-module in addition to being an $R[G]$-module.

$I$ is the annihilator in $R$ of $S$, i.e.

$$ I = \{ r : rS = 0 \} $$

Suppose that $I$ is not a maximal ideal of $R$, i.e. there exists a $J$ such that $I \subsetneq J \subsetneq R$.

$(I[G])S$ is $0$.

$(R[G])S$ is the whole module $S$ because $1$ is in there.

Now let's look at $JS$.

$JS$ cannot be $0$, because $J$ is not a subset of the annihilator of $S$.


ways of distinguishing standard topology and half-open topology

  1. half-open is a disjoint union of open sets.

Ideals of cylindric algebras.

If you tweak the signature of a cylindric algebra you can get the following:

  1. The constants $0$ and $1$.
  2. xor $+$ and and $*$
  3. The universal quantifiers $u_k$ corresponding to $[\forall x_k](\cdots)$ in the traditional syntax of first-order logic.
  4. The constants $d_{kl}$ corresponding to $x_k = x_l$ in the traditional syntax.

Neglecting 4, this makes a cylindric algebra into a boolean ring.

Let $I$ be an ideal of a cylindric algebra.

Let an ideal $I$ be universally-closed if and only if $u_k(I) \subset I$ for all $k$.


Constructing a point set for a topology given by a lattice.

Let $\Lambda$ be a lattice with arbitrary meets and finite joins. Is there a way to convert it to a topology whose lattice of closed sets is isomorphic to it?


multi-valued algebra.

Let $f, g, h, \cdots$ be multi-valued functions ($n+1$-place relations).

Let $f(g(x)) \approx h(x)$ insist that $h(x)$ and $g(x)$ are equal at all values and that, for all values of $x$ $h(x)$ takes on zero or one values and likewise for $f(g(x))$.

This is one possible idea for a multi-valued alegbra.


this is wrong. we want the rank of $\kappa$

How do you axiomatize ZFC + Worldly Cardinals?

How do I tell when I've defined ZFC + Worldly Cardinals instead of something weaker like ZFC + Con(ZFC)?


Given the definition of a worldly cardinal, I can see how to extend ZFC with enough axioms to make a constant symbol $\kappa$ a worldly cardinal.

Let $M$ be a $(\kappa, \in)$-structure that we're considering as a possible model of ZFCW.

Let $\kappa$ be a constant symbol, intended to name our worldly cardinal.

Here's an axiomatization of ZFCW.

Let $P$ be a list of parameters.

Let $\varphi(P)$ be a sentence with parameters that's an axiom of ZFC (possibly instantiated from an axiom schema).

Let $\varphi^*(P)$ be the following $P \in \kappa \to \varphi'(P)$ where $\varphi'$ is $\varphi$ but with $\forall x (\cdots)$ replaced with $\forall x (x \in \kappa \to \cdots)$ and likewise for $\exists$. The notation $P \in k$ abbreviates $(P_1 \in \kappa \land P_2 \in \kappa \land \cdots \land P_n \in \kappa)$.

At this point, we only know that $\kappa$ is a standard model of ZFC.

If I keep going and insist that $\kappa$ is transitive and totally ordered by the elementhood relation, then I think I've only insisted that $\kappa$ is a standard and transitive model.

At this point, I'm out of ideas for what additional axioms to toss in to make $\kappa$ worldly enough.


I changed this question completely since I first wrote it. This answer completely solved my original question of how parameters work in the axiomatization of ZFC. Basically, when assessing the truth of $M \models \text{ZFC}$ for some $(\in)$-structure $M$, the parameters used in expansions of axiom schemas of $\text{ZFC}$ are taken directly from the carrier of $M$.


transfinite induction:

Let $A$ be a set of ordinals.

$$ f(0) = \cdots \\ f(\alpha^+) = \cdots \\ f(\lim A) = \cdots $$

Definition of $+$. Let $\kappa + B$ abbreviate $\{x : x \in B \land x+\kappa \}$

$$ 0+\kappa = \kappa \\ \kappa+0 = \kappa \\ \kappa + \beta^+ = (\kappa+\beta)^+ \\ \kappa + \lim(B) = \lim(B + \kappa)$$


Is the "ordinalification" of ZFC equiconsistent with ZFC?

Let's define $T$, a two-sorted theory ($E, F$) with the following relations:

  • $ \in : E \times E \to 2 $
  • $ \in : E \times F \to 2 $

and the following function:

  • $ (,) : E \times E \to E $

Let $\Delta$ consist of all well-formed formulas $\varphi(\vec{x})$ with parameters in the language of ZFC such that:

  1. Every parameter in $\vec{x}$ is an ordinal.
  2. All bound variables introduced by quantifiers are restricted to be ordinals, i.e. $\forall x ( \cdots )$ does not occur outside of $\forall x (\mathrm{ordinal}(x) \to \cdots)$ and likewise for $\exists$.
  3. $\varphi(\vec{x})$ is a theorem of ZFC.

Let $T$ consist of $\Delta$ together with:

  1. Comprehension: For all well-formed formulas with parameters $\varphi(x, \cdots)$, it holds that $\exists f \mathop ( x \in f \leftrightarrow \varphi(x, \cdots))$.
  2. Extensionality for the sort $F$.
  3. The defining property of ordered pairs (and no other properties about ordered pairs). $\forall x \forall y \forall z \forall w ((x, y) = (z, w) \leftrightarrow (x = z) \land (y = w))$

These axioms are inspired by the first-order theory called second-order arithmetic.

Consistency of ZFC implies consistency of $T$.

Assume ZFC is consistent.

ZFC has a model $A$.

Let's construct a new model of $T$, $B$.

Take $A$ and attach an arbitrarily chosen bijection between $\text{Ord} \times \text{Ord}$ and $\text{Ord}$ as the interpretation of $(,)$.

Take the powerset of $\text{dom}(A)$ as the interpretation of the sort $F$.

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