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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
    Commented Jul 18, 2012 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$
    – cardinal
    Commented 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
    Commented 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
    Commented 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 Maybe someday we can get rid of this outdated crutch. $\endgroup$ Commented Dec 2, 2015 at 14:07

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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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    – Max Muller
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könnt Ihr mir bitte helfen?

Mein Ziel ist es, möglichst umfassende für die Praxis des Gleichungslösens in geschlossener Form nützliche Sätze zu formulieren und Beweise dafür zu finden.

Sind meine Vermutung und ihr Beweis unten korrekt? Sind sie gut formuliert? Wie können sie verbessert werden?

Eine Funktion $f$ heißt genau dann Funktion in den komplexen Zahlen, wenn $f\colon D\subseteq\mathbb{C}\to\mathbb{C}$.

Vermutung:
Seien
$\mathbb{K}\in\{\mathbb{Q},\mathbb{C}\}$,
$n\in\mathbb{N}_{>1}$,
$f_1,...,f_n$ $\mathbb{K}$-algebraisch unabhängige Funktionen in den komplexen Zahlen,
$z$ komplexe Lösungsvariable.
Wenn $P(z_1,...,z_n)\in\mathbb{K}(z_1,...,z_n)\setminus\mathbb{K}(z_1)\setminus ...\setminus\mathbb{K}(z_n)$ $\mathbb{K}$-irreduzibel ist, dann lässt sich die Gleichung $$P(f_1(z),...,f_n(z))=0\tag{1}$$ nicht durch alleiniges Anwenden einer endlichen Anzahl $\mathbb{K}$-algebraischer Operationen und anschließende einmalige Anwendung der Umkehrrelation einer der Funktionen $f_1,...,f_n$ auf die Gleichung lösen.

Beweis:
Wir führen einen Widerspruchsbeweis.
Damit die Gleichung durch einmalige Anwendung der Umkehrrelation einer der Funktionen $f_1,...,f_n$ auf die Gleichung gelöst werden kann, muss die Gleichung nach einem $f_i(z)$, mit $i\in\{1,...,n\}$, aufgelöst werden.
Da die Gleichung $\mathbb{K}$-irreduzibel ist und $f_1,...,f_n$ $\mathbb{K}$-algebraisch unabhängig sind, lässt sich durch alleiniges Anwenden einer endlichen Anzahl $\mathbb{K}$-algebraischer Operationen keins der $f_1(z),...,f_n(z)$ aus der Gleichung eliminieren.

$$P(f_1(z),...,f_n(z))=0$$ $$f_i(z)=B(f_1(z),...,f_{i-1}(z),f_{i+1} (z)...,f_n(z))$$ $$z=f^{-1}(B(f_1(z),...,f_{i-1}(z),f_{i+1}(z)...,f_n(z)))$$

q.e.d.

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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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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
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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. $\endgroup$ Commented Mar 27, 2023 at 14:20
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This equation has many variables, so methods like Lagrange inversion or transforms would be hard to apply. However, series reversion works well. First, use @Claude Leibovici’s answer to notice that $x\sim a=n+\sqrt n+2$, set $x=v+a$ and subtract $v_0$ so it contains $(0,0)$:

$$x^x=(x-n)^{x+n}\iff f(v)=(v+a)\ln(v+a)-(v+a+n)\ln(v+a-n)-v_0=-v_0,v_0= a\ln(a)-(a+n)\ln(a-n)$$

with $f^{-1}(v)=y(v)$. We use a power series method. Firstly, find a differential equation for $y$:

$\frac1{y’}=\ln(y+a)-\ln(y+a-n)-\frac{2n}{y+a-n}\iff -a(a-n)^2y’’= (a-n)(3a-n)yy’’ +n(a+n)y’^3+ (3a-2n)y^2y’’ +nyy’^3+y^3y’’,y(0)=0,y’(0)=\frac{a-n}{(a-n)(\ln(a)-\ln(a-n))-2n}$

Substituting $y=\sum\limits_{p=1}^\infty a_px^p$:

$$-a(a-n)^2\sum_{p=1}^\infty a_pp(p-1)x^{p-2}=(a-n)(3a-n)\sum_{m=1}^\infty\sum_{k=1}^\infty a_ma_k k(k-1) x^{m+k-2}+n(a+n) \sum_{m=1}^\infty\sum_{k=1}^\infty\sum_{j=1}^\infty a_ma_ka_jmkj x^{m+k+j-3}+(3a-2n) \sum_{m=1}^\infty\sum_{k=1}^\infty\sum_{j=1}^\infty a_ma_ka_jj(j-1)x^{m+k+j-2}+n \sum_{m=1}^\infty\sum_{k=1}^\infty\sum_{j=1}^\infty\sum_{r=1}^\infty a_ma_ka_ja_rmkjx^{m+k+j+r-3}+ \sum_{m=1}^\infty\sum_{k=1}^\infty\sum_{j=1}^\infty\sum_{r=1}^\infty a_ma_ka_ja_r r (r-1) x^{m+k+j+r-2}$$

Gathering powers of $x$, index shifting, and combining sums gives a recurrence relation for the coefficients. We can use t to invert $f(v)=(v+a)\ln(v+a)-(v+a+n)\ln(v+a-n)-v_0$ or $x^x(x-n)^{-x-n}=b$ equivantly. However, the solution to the question’s equation is $x=a+y(-v_0)$. Therefore:

$$\begin{aligned}x^x=(x-n)^{x+n}\iff x=a+\sum_{p=1}^\infty a_p((a+n)\ln(a+n)-a\ln(a))^p\\\\-a p(p-1) (a-n)^2 a_p=\sum_{m=1}^{p-1}ma_m a_{p-m} (a_1 n (a+n)(p-m)+(m-1)(a-n)(3a-n))+\sum_{m=1}^{p-1}\sum_{k=1}^{p-m-1}a_ma_{p-m-k}(j(k-1)(3a-2n)a_k+mn(a+n)(k+1)(p-m-k)a_{k+1})+\sum_{m=1}^{p-1}\sum_{k=1}^{p-m-1}\sum_{j=1}^{p-m-k}ja_ma_ka_j(nmk a_{p-m-k-j+1}+(j-1)a_{p-m-k-j}),a_1=\\a_p=\left\{,,,\dots\right\}\end{aligned}$$

Redefining a better initial $a$ will make the series converge more quickly.


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

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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