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Answer 1

Since the post is quite long I immediately precise that I substantially ask to prove the inclusion $\eqref{eq:simple3} $ using the inclusion $ \eqref{eq:simple2} $ and if you want you may not read the marked parts at a first look. So let's we well illustrate the issue.

Definition

A topological group is a group $(X,*)$ equipped with a topology $\cal T$ with respect to which the functions $$ p:X\times X\ni (x_1,x_2)\longrightarrow x_1*x_2\in X\quad\text{and}\quad s:X\ni x\longrightarrow x^{-1}\in X $$ are continuous.

So with respect the last definition I am trying to prove that any topological group $(X,*,\cal T)$ is completely regular using the following Munkres procedure.

So first of all let's we assume that for any closed set $C$ disjoint from the identity $e$ there exists a continuous function $f$ from $X$ to $[0,1]$ such that $$ f(e)=0\quad\text{and}\quad f[C]\subseteq\{1\} $$ and thus let's we prove that the statement woulds holds.

So if $x_0\in X$ is different from $e$ then we observe that the function $$ f_0: X\ni x\longrightarrow x*x_0^{-1}\in X $$ is a homeomorphism such that $$ f_0(x_0)=e $$ so that if $x_0\notin C$ then by assumption there exists a continuous function $f$ from $x$ to $[0,1]$ such that $$ f(e)=0\quad\text{and}\quad f\big[f_0[C]\big]\subseteq\{1\} $$ and thus finally $f\circ f_0$ is a continuous function from $X$ to $[0,1]$ such that $$ (f\circ f_0)(x_0)=0\quad\text{and}\quad(f\circ f_0)[C]\subseteq\{1\} $$ which proves that the complete regularity of $X$ follows proving the existence of a function $f$ from $X$ to $[0,1]$ which separated $e$ from any closed set not containing it.

Now it is a well know result that for any neighborhood $U$ of $e$ there exists a neighborhood $V$ of $e$ such that $$ V^2\subseteq U $$ so that let's we make a sequence $(V_n)_{n\in\omega}$ of neighborhood of $e$ such that $$ V_0\subseteq X\setminus C $$ and such that $$ V_{n+1}*V_{n+1}\subseteq V_n $$ for each $n\in\omega$. Now let be $$ \Delta:=\Big\{\frac m{2^n}:m,n\in\Bbb Z\Big\} $$ the set of dyadic numbers which is countable and dense: so let's we put $$ \begin{equation}\tag{1}\label{eq:simple1}U_{\frac m n}:=\begin{cases}\emptyset,\,\text{if }\frac m n\le 0\\V_{n},\,\text{if } \frac m n=\frac 1{2^n}\le 1\\V_{n+1}*U_{\frac h{2^n}}\,\text{if } \frac m n=\frac{2h+1}{2^{n+1}}<1\\ X,\,\text{if }\frac m n>1\end{cases}\end{equation} $$ for each $m,n\in\Bbb Z$. Therefore let's we verify that $U_\delta$ is defined for each $\delta\in\Delta$ but clearly $U_\delta$ is surely defined when $\delta\notin(0,1)$ so that we have only to prove that $U_\delta$ is defined for $\delta\in(0,1)$.

So we observe that $2^n$ is positive for each $n\in\Bbb Z$ so that the inequality $$ 0<\frac m{2^n}<1 $$ for $m,n\in\Bbb Z$ holds if and only if also the inequality $$ 0<m<2^n $$ holds but if $n\le 0$ then $2^n$ is less than $1$ so that the last inequality can holds only for $n>0$ and thus let's we prove inductively over $n$ that $U_{\frac m{2^n}}$ is well defined for any $m<2^n$: so if $n=1 $ then $m$ can only be equal to $1$ and $U_{\frac 1 2}$ is equal to $V_1$ so that let's we assume that the statement holds for any $n$; thus we observe that if $m<2^{n+1}$ is even then there exists $h\in\Bbb N$ such that $$ U_{\frac m{2^{n+1}}}=U_{\frac{2h}{2^{n+1}}}=U_{\frac h {2^n}} $$ so that by the inductively step $U_{\frac m{2^{n+1}}}$ is defined if $h<2^n$ - but also if $2^n\le h$; otherwise if $m<2^{n+1}$ is odd then there exists $h\in\Bbb N$ such that $$ U_{\frac m{2^{n+1}}}=U_{\frac {2h+1}{2^{n+1}}}=V_{n+1}*U_{\frac h{2^n}} $$ but by the inductively set $U_{\frac h{2^n}}$ is defined if $h<2^n$ - and also if $2^n\le h$ - so that we conclude that $U_{\frac m{2^{n+1}}}$ is defined.

Now if $n$ is negative then $\frac m{2^n}$ is greater than $1$ or less than $0$ so that the inclusion $$ \begin{equation}\tag{2}\label{eq:simple2}V_n*U_{\frac k{2^n}}\subseteq U_{\frac{k+1}{2^n}}\end{equation} $$ trivially holds ans thus let's we prove inductively that it holds also for $n$ positive assuming that the inequality $$ 0<k<2^n $$ holds because otherwise $\eqref{eq:simple2}$ trivially holds.

So first of all we observe that $\eqref{eq:simple2}$ holds trivially for $n=0$. Therefore let's we assume that the inclusion holds for any $n\in\Bbb N$. So we observe that if $k$ is even then there exists $h\in\Bbb N$ such that $$ V_{n+1}*U_{\frac k{2^{n+1}}}=V_{n+1}*U_{\frac{2h}{2^{n+1}}}=V_{n+1}*U_{\frac h{2^n}}=U_{\frac{2h+1}{2^{n+1}}}=U_{\frac{k+1}{2^{n+1}}} $$ where we used $\eqref{eq:simple1}$; otherwise if $k$ is odd then there exist $h\in\Bbb N$ such that $$ V_{n+1}*U_{\frac k{2^{n+1}}}=V_{n+1}*U_{\frac{2h+1}{2^{n+1}}}=\\V_{n+1}*V_{n+1}*U_{\frac h{2^n}}\subseteq V_n*U_{\frac h{2^n}}\subseteq U_{\frac{h+1}{2^n}}=\\U_{\frac{2(h+1)}{2^{n+1}}}=U_{\frac{(2h+1)+1}{2^{n+1}}}=U_{\frac{k+1}{2^{n+1}}} $$ where we used $\eqref{eq:simple1}$ and the inductive hypotesis: thus we conclude that the above inclusion holds for $n+1$ and so finally by induction for any $n\in\Bbb N$.

Now let's we assume $\eqref{eq:simple2}$ implies that for any $\delta_1,\delta_2\in\Delta$ such that $\delta_1\le\delta_2$ the inclusion $$ \begin{equation}\tag{3}\label{eq:simple3}\operatorname{cl}U_{\delta_1}\subseteq U_{\delta_2}\end{equation} $$ holds so that let's we prove that the position $$ \begin{equation}\tag{4}\label{eq:simple4}f(x):=\inf\{\delta\in\Delta:x\in U_\delta\}\end{equation} $$ defines a continuous function $f$ from $X$ to $[0,1]$ such that $$ f(e)=0\quad\text{and}\quad f[C]\subseteq 1 $$

So by Archimedean property we observe that for any $\delta\in\Delta$ there exists $n\in\Bbb N$ such that $$ \frac 1 n<\delta $$ so that by $\eqref{eq:simple3}$ we conclude that $$ e\in V_{n}=U_{\frac 1 n}\subseteq U_\delta $$ which proves that $$ f(e)=0 $$ Moreover by $\eqref{eq:simple1}$ and $\eqref{eq:simple3}$ we observe that for any $\delta\in\Delta$ with $\delta\le 1$ the inclusion $$ U_\delta\subseteq U_1=V_0\subseteq X\setminus C $$ holds so that also the identity $$ f[C]\subseteq\{1\} $$ holds.

So we have only to prove continuity showing that $f^{-1}\big[[0,a)\big]$ and $f^{-1}\big[(a,1]\big]$ are open for any $a\in(0,1)$.

First of all for any $x\in X$ we let put $$ \Delta_x:=\{\delta\in\Delta:x\in U_\delta\} $$

• Now let be $$ U_-:=\bigcup_{\delta<a}U_\delta $$ and so we observe that if $x\in U_-$ then $x\in U_\delta$ with $\delta<a$ which implies that $$ \inf \Delta_x\le \delta<a $$ so that $x\in f^{-1}\big[[0,a)\big]$; conversely if $x\in f^{-1}\big[[0,a)\big]$ then $\inf\Delta_x<a$ so that there exists $\delta_a\in\Delta_x$ such that $x\in U_\delta$ with $\delta_a<a$ and so $x\in U_-$: thus we conclude that $$ f^{-1}\big[[0,a)\big]=U_- $$ which proves that $f^{-1}\big[[0,a)\big]$ is open.

• Now let be $$ U_+:=\bigcup_{\delta>a}X\setminus\operatorname{cl}U_\delta $$ and so we observe that if $x\in f^{-1}\big[(a,1]\big]$ then $x\in U_+$ because if $x\notin U_1$ then $\eqref{eq:simple3}$ implies that $x\notin\operatorname{cl}U_\delta$ for any $\delta\in\Delta$ with $\delta<1$ and because if $x\in U_1$ then the inequality $$ \inf\Delta_x=f(x)>a $$ implies that $x\not\in\operatorname{cl}U_\delta$ for any $\delta\in\Delta$ with $a<\delta<f(x)$ since $\Delta$ is dense so that there exists $d\in\Delta$ such that $$ a<\delta<d<f(x) $$ and so by $\eqref{eq:simple3}$ if $x\in\operatorname{cl}U_\delta$ then $x\in U_d$ and this would disagree with the inequality $$ d<\inf\Delta_x $$ on the contrary if $x\in U_+$ then $x\in X\setminus\operatorname{cl}U_\delta$ for $\delta>a$ but if $d\in\Delta$ is less than $\delta$ then by $\eqref{eq:simple3}$ the inclusion $$ \operatorname{cl}U_d\subseteq U_\delta\subseteq\operatorname{cl}U_\delta $$ holds and so it must also be $$ X\setminus\operatorname{cl}U_\delta\subseteq X\setminus\operatorname{cl}U_d $$ so that necessarily $x\notin U_d$ for any $d\in\Delta$ with $d\le\delta$ and so if $x\in U_d$ then necessarily $\delta<d$ so that $$ a<\delta\le\inf\Delta_x=f(x) $$ that is $x\in f^{-1}\big[(a,1]\big]$: so we finally conclude that $$ f^{-1}\big[(a,1]\big]=U_+ $$ which proves that $f^{-1}\big[(a,1]\big]$ is open.

So as you can see I was not able to prove the inclusion $\eqref{eq:simple3}$ so that I ask to do it and moreover I ask if I well proved the marked parts with a grey bar since Munkres did not prove them. So could someone help me, please?

Answer 2

(still editing)

The lower bounds of distances of perfect powers $\Delta=\mid a^A-b^B \mid$ is one of those attractive, simple-to-state, but difficult-to-solve problem. For me it occured in the question of cycles in the Collatz-problem, where we ask for the lower bound of distances of perfect powers of $2$ and $3$: $$\Delta(N,S)=|2^S - 3^N|$$ or $$\Lambda(N,S)=|S \cdot \ln2-N \cdot \ln 3| \qquad .$$
What I've found so far is

• the estimate by J.W.Ellison formulated in the 1970'ies (adressing $\Delta(N,S)$), and
• the estimate by G.Rhin formulated in 1987 (adressing $\Lambda(N,S)$), as reported in J.Simons & deWeger[2007] on the existence of "m-cycles" in the Collatz-problem.

The lower bounds from those estimates are -in the view of an amateur- astonishingly far away from the actual lowest distances -as checked for $N$ of about $1$ million digits. (Of course, the only relevant $N$ are selected according to the convergents of the continued fraction of $\log_2(3)$.)

Often discussions of this topic state the general form of estimation-formulae, but don't give explicite values for the constants involved - for instance in the well-visited blog of T. Tao on the "separation of powers of 2 and 3". Naive functional estimates for such lower bounds based on heuristics would suggest small values for such constants, for instance while G.Rhin gives something like $1/457/N^{13.3}$ a naiive fitting suggests something like $.../N^1$ or $.../N^{1.01}$ instead, valid for $N$ in my mentioned range of search.
A recent guess, using $.../N/\ln(N)$ instead, leaving aside any $1+\varepsilon$-exponent at all seems to be a really good estimate for the lower bounds. However, the famous estimate for the number of primes below some number $n$ as value of the integrallogarithm $Li()$ as proposed by the young K.F.Gauss has later been proven to be not really a lower bound, but where the first counterexample could only be upper-limited by something like $10^{10^{10^{34}}}$(WP:Skewes-Number) (modern computation reduced that upper bound to $1.39 \cdot 10^{316}$, see there).

Actual computations with large exponents diverge extremely far from that given estimates so I once tried to find an own algebraical pathway to such an lower-bound estimate. My current attempt provides a lower bound which is much nearer to the empirical values than the two other estimates - unfortunately this is not based on an analytical argument but only on heuristic finding.
Here I intend to create a "biglist" of immediately usable such lower bounds and invite other readers to add information and / or alternatives to the two established estimates and to try to improve my proposed estimate with analytical arguments.

Preliminaries:
While Ellison immediately looks at $\Delta(N,S) = |2^S - 3^N|$ and gives $$\Delta(N,S) \gt \exp(S \cdot (\ln 2 - 0.1)) \qquad S\gt 27 \tag {EL}$$

G. Rhin looks at the difference of logarithms $\Lambda(N,S)=| S \ln2 - N \ln 3 |$ and gives $$\Lambda(N,S) \gt {1\over 457 N ^{13.3} }| \tag{RH}$$ I've looked as well at $\Lambda(N)$ and propose $$\Lambda(N,S) \gt {1\over 10 N \ln N } \tag{Lb}$$ To compare that bounds, we must norm for the data of $\Delta()$ or for that of $\Lambda()$. The best adaption is surely dependent on the question that one has - either using a formula involving the exponential expression or one involving the logarithmic expression. So the discussion of this should be done in the single answers of the big-list where we reference some question in MSE.
To trigger the interest, here some pictures, how the three estimates behave in contrast to the empirical values. Here I adapted the Ellison-estimate to a $\Lambda()$-version.

We see, that the empirical $\Lambda()$ jitter between $0$ and $1/2 \cdot \ln 2$ (blue dots), the values at $N=\{2,5,12,53,...\}$ show very small distances, and even decreasing towards zero. The idea is to have a continuous function $f(N)$ which can supply a lower bound, such that we can say $\Lambda(N,S) \gt f(N)$ .
The red curve for the Ellison-estimate is below of the empirical values only for $N \gt 17$ while the green curve for the Rhin-estimate is so near the zero-line, that we barely see it. My estimate shown by the brown curve is always below the $\Lambda()$ .
To see a bit more detail, and to get aware of the basic characteristics of the estimates, a picture in log()-scalings is more appropriate. Picture 2: x,y logarithmic scaling Here we see the characteristics better. All three methods of estimates work for $N$ towards $=200$. Ellison took a much different shape compared with Rhin's, and while Rhin's lower bound is unfortunate small over the whole range, the Ellison's is initally too large, then better than Rhin's but from about $N \approx 700$ it decreases much faster than Rhin's. So, for the larger $N$, Rhin's lower bound should be preferred, but for the moderate values in $N$ Ellison's estimate gives a simpler formulation and a better lower bound.
But the "LBLam"-function has the best characteristic; it is nearly parallel to a lower hullcurve, and also is valid from the smallest $N=2$ on. That this property holds forever, that means the formula is analytically useful for all $N$, has not been proved, but empirical heuristic show that it holds for all $N$ up to $1$ million decimal digits and is always tight to the most extreme small values of $\Lambda()$...
A last picture, for the aesthetical impression is the following. I rescaled the values of $\Lambda(N,S)$ to the open interval $-1[...]1$ by $w(N,S)=4 \cdot \Lambda(N,S) / \ln 2 -1$ and then show the values $Y(N,S)=\tanh^{-1}(w(N,S))$ and the accordingly rescaled values for the three lower-bound-estimates. The slightly jittering effect in the Ellison-curve is because its function uses the somehow jittering values $S$ instead that of $N$ for its argument.

Answer 3

Testing link embedding procedure. Link description

Answer 4

The coefficients in your first equation (multinomial theorem) are the multinomial coefficients (of the first kind). They count integer combinations.

For $a_i=i$, your considered formula is the sum of the multinomial coefficients of the second kind (Stirling numbers of the first kind ($s$) [Abramowitz/Stegun 1970]). They count integer permutations.

$$\sum_{1l_1+...+kl_k=s}\frac{s!}{1^{l_1}l_1!...k^{l_k}l_k!}=\sum_{r=1}^s\left[^s_r\right]=s!$$

The generating function for the coefficients is:

$$\sum_{s=m}^\infty\left(\sum_{1l_1+...+kl_k=s}\frac{s!}{1^{l_1}l_1!...k^{l_k}l_k!}x_1^{l_1}...x_k^{l_k}\right)\frac{t^s}{s!}=\frac{1}{m!}\left(\sum_{i=1}^\infty\frac{1}{i}x_it^i\right)^m.$$

For $a_i=i!$, your considered formula is the sum of the multinomial coefficients of the third kind (Stirling numbers of the second kind [Abramowitz/Stegun 1970]), the Bell number $B_s$. They count set partitions.

$$\sum_{1l_1+...+kl_k=s}\frac{s!}{1!^{l_1}l_1!...k!^{l_k}l_k!}=\sum_{r=1}^s\left\{^s_r\right\}=B_s$$

The generating function for the coefficients is:

$$\sum_{s=m}^\infty\left(\sum_{1l_1+...+kl_k=s}\frac{s!}{1!^{l_1}l_1!...k!^{l_k}l_k!}x_1^{l_1}...x_k^{l_k}\right)\frac{t^s}{s!}=\frac{1}{m!}\left(\sum_{i=1}^\infty\frac{1}{i!}x_it^i\right)^m.$$
$\ $

[Abramowitz/Stegun 1970] Abramowitz, M.; Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standard 1970. p. 823

Answer 5

This box is free for everyone to use

Answer 6

While generally we find ourselves checking for whether there is something common between the $2$ events to check for mutual exclusiveness

Here's another definition of mutually exclusive events:

If there is a set of events such that if any one of them occurs, none of the others can occur, the events are said to be mutually exclusive.

A counterexample to both of the above: (1,2] and [2,3) are mutually exclusive yet have a common outcome 2.

If event $a$ and $b$ are totally not related, can we still add them

Nope, not until you know that the events are mutually exclusive.

???

If the events $a$ and $b$ are not related then we say that they are independent

This characterisation is vague and potentially misleading (see below).

A and B are different students and they give different exams and aren't related in any way.

events $a$ and $b$ are independent by common knowledge (not mathematical but common sense) so the $P(a\cap b)=P(a)\cdot P(b)=0.42$.

A and B aren't events so it doesn't make sense to discuss their independence. Secondly, the event of A passing test 1 and the event of B passing test 2 are not necessarily independent even if they reside in different cities and are taking tests that aren't ostensibly related! As pointed out above, the word "related" is misleadingly causing confusion. Please refer to my answer.

In your example too it seems quite intuitive enough that events are independent so we'll consider/assume them that way.

he p(b) is probability that we go to jail if we rob a bank, p(a) is the probability of jack eat an apple today,

the jail probability p(b) is dependent on the rob probability, is it right that we add p(b) to p(a) to say they are not mutually exclusive?

So, they are not dependent and your intuition is correct that they are indeed independent (and not mutually exclusive).

Re-read the OP's question; it's not actually making sense: are they trying to compare (1) jail & jobbery or (2) jailrobbery & apple. For both cases, independence does not necessarily follow; for Case 2, theoretically, mutual exclusiveness is possible.

Answer 7

Let $S$ be a set, $\wp(S)$ its powerset, and $D_{0}, D_{1}$ decision problems;

Follows that $D_{0} \in NP \iff \exists \ x \in \wp(S) \ \land \exists \ D_{1} \in P \ |$

Answer 8

This box is free for everyone to use.

Answer 9

The Wolfram Language’s Inverse Beta Regularized $\text I^{-1}_z(a,b)$ is a quantile function. This applicable yet obscure function appears in Excel as BETA.INV and a special case of it as the Inverse T distribution function InvT(x=area,d=degrees of freedom) in AP Statistics. It’s special cases have not been well investigated yet, but they include elliptic, polynomial root, transcendental solution, and other functions.

The main definition of this function is:

$$\frac{\int_0^z t^{a-1} (1-t)^{b-1}dt}{\int_0^1 t^{a-1} (1-t)^{b-1}dt}=x \implies z=\text I^{-1}_x(a,b);0\le x\le 1 \text{ and }a,b>0$$

Note that there are many other “special cases” of inverse beta regularized which do not have closed forms, so they have no connections to other functions. Therefore, finding more special cases of the Incomplete Beta function will show it’s relationship to other (special) functions. Using properties of the Gauss Hypergeometric function and inverting also helps find special cases.

What are some other closed forms or named constants of special cases and limits of Inverse Beta Regularized not in the self answer? Are there any identities or transformations for Inverse Beta Regularized?

If you found an alternate form or simplification of an identity in the self answer then please answer too.

The self answer also has guiding questions which are part of the block-quoted question

---

Formula $1$:

$$x^p+cx+a=0\implies x=\frac{ap}{c(1-p)}\text I^{-1}_\frac{ a^{p-1}(p-1)}{c^p\left(\frac1p-1\right)^p}(-p,2)$$

with a tester of the formula.

and

$$x^r+ax+b=0\implies x=\frac{b r}{a(1-r)\text I^{-1}_\frac{b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(r-1,2)}$$

If $0< r< 1$, simply substitute $x\to x^\frac 1r$. Test the formula here

In short, $$\text I^{-1}_z (n\in\Bbb N, b)$$ solves for some power function’s root which should have a closed form in terms of hypergeometric functions, the Bring Radical, Radicals, and Elliptic functions, but the relationships between these functions and Inverse Beta Regularized is complicated, so which ones are there? Also, there may be a way to use the result to find more than one solution too, but how?

Formula 2: A quarter period of a trigonometric function:

$$\sqrt{\text I^{-1}_\frac{2x}\pi\left(\frac12,\frac12\right)}=\sin(x)\implies\text I^{-1}_x\left(\frac12,\frac12\right)=\sin^2\left(\frac\pi 2x\right)$$

Test the identity here

Formula 3:

A constant named the Dottie Number:

$$\text{Dottie Number}=\text D=\sin^{-1}\left(1-2\text I^{-1}_\frac12\left(\frac12,\frac32\right)\right)=\sqrt{1-\left(1-2 \text I^{-1}_\frac12\left(\frac12,\frac32\right)\right)^2}\implies\text I^{-1}_\frac12\left(\frac12,\frac32\right)=\frac{1-\sqrt{1-\text{D}^2}}{2}=\frac{1-\sin(\text D)}2$$

See the results here.

Formula 4: Notice the Jacobi Amplitude $\text{am}(x,k)$ with parameter $k$ which can derive other Jacobi Elliptic function identities where $\text L_1$ is the First Lemniscate Constant:

$$\sin^{-1}\sqrt[4]{\text I^{-1}_\frac{x}{\text L_1}\left(\frac14,\frac12\right)}=\text{am}(x,-1)$$

also note Jacobi SN, Jacobi NC, and the lemniscate case half period constant $\omega$

$$\text I^{-1}_x\left(\frac14,\frac12\right)=\text{sn}^4\left(\text L_1 x,-1\right)=\text{nc}^2\left(2\omega i(1-x),\frac12\right)$$

where there is a true identity. Next, with the Jacobi CN function:

$$\text I^{-1}_x\left(\frac12,\frac14\right)=4\text{sn}^2(\text L_1 x,2) \text{cn}^2(\text L_1 x,2)$$

and

$$\sec^{-1}\sqrt[4]{\text I^{-1}_{1-\frac x{2\omega }}\left(\frac14,\frac12\right)}=\text{am}\left(ix,\frac12\right)$$

which is also true and finally,

$$\frac12\sin^{-1}\sqrt{\text I^{-1}_\frac x{\text L_1}\left(\frac12,\frac14\right)}=\text{am}(x,2)$$

which is correct

For the derivative of the Jacobi amplitude, we have Jacobi DN:

$$\sqrt{\sqrt{\text I^{-1}_\frac x{\text L_1}\left(\frac14,\frac12\right)}+1}=\text{dn}(x,-1)\implies\text I^{-1}_x\left(\frac14,\frac12\right)=\left(\text{dn}^2(\text L_1x,-1)-1\right)^2$$

which is true.

$$\sqrt[4]{1-\text I^{-1}_\frac{x}{\text L_1}\left(\frac12,\frac14\right)}=\text{dn}(x,2)\implies \text I^{-1}_x\left(\frac12,\frac14\right)=1-\text{dn}^4\left(\text L_1x,2\right)$$

which works similarly

*simplify?:

$$\frac{\sqrt{\frac1{\sqrt{\text I^{-1}_{1-\frac x{2\omega}}\left(\frac14,\frac12\right)}}+1}}{\sqrt2}=\text{dn}\left(ix,\frac12\right)\implies\text I^{-1}_x\left(\frac12,\frac14\right)=1-\frac1{\left(2\text{dn}^2\left(2\omega ix,\frac12\right)-1\right)^2}$$

which works. Finally, Jacobi Epsilon $\varepsilon(x,k)$ with the Incomplete Beta function $\text B_z(a,b)$

$$\frac14 \text B_{\text I^{-1}_\frac x{\text L_1}\left(\frac12,\frac14\right)}\left(\frac12,\frac34\right)=\varepsilon(x,2)\implies \text B_{\text I^{-1}_x\left(\frac12,\frac14\right)}\left(\frac12,\frac34\right) =4\varepsilon(\text L_1x,2)$$ which is true. Similarly,

$$x+\frac14 \text B_{\text I^{-1}_\frac x{\text L_1}\left(\frac14,\frac12\right)}\left(\frac34,\frac12\right)=\varepsilon(x,-1)\implies\text B_{\text I^{-1}_x\left(\frac14,\frac12\right)}\left(\frac34,\frac12\right)=4\varepsilon(\text L_1x,-1)-4\text L_1x $$

which is also true

finally, @Bertrand87’s answer and the ubiquitous constant=U:

*simplify:

$$\frac{\text L_1}{\sqrt2}+\frac{\text U}{2}-\frac{\text B_{\frac1{\text I^{-1}_{1-\frac{\sqrt2x}{\text L}}\left(\frac14,\frac12\right)}}\left(\frac12,\frac14\right)+ \text B_{\frac1{\text I^{-1}_{1-\frac{\sqrt2x}{\text L}}\left(\frac14,\frac12\right)}}\left(\frac12,\frac34\right)}{4\sqrt 2}=\varepsilon\left(i x,\frac12\right)\implies$$

Compare the value of this value with EllipticE(am(i,1/2),1/2) for numerical proof

Formula 5: Please note that sections for Weierstrass Zeta and Sigma, defined later, can also use these transformations. There are multiple formulas for completion and understanding compositions of functions on Inverse Beta Regularized

The Weierstrass $\wp(x;a,b)$ function has this first basic formula $$\frac{\sqrt a}{2\sqrt{\text I^{-1}_{-\frac{\sqrt[4]a x}{2\omega}}\left(\frac14,\frac12\right)}}=\wp(x;{a,0})\implies\text I^{-1}_x\left(\frac14,\frac12\right)=\frac a{4\wp^2\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)}$$

which works. There is also the Weierstrass $\wp’(x;a,b)$:

$$\frac{a^\frac34\left(1-2\text I^{-1}_{-\frac{\sqrt[4]a x}{2\omega}}\left(\frac14,\frac14\right)\right)}{\sqrt 2\text I^{-1}_{-\frac{\sqrt[4]a x}{2\omega}}\left(\frac14,\frac12\right)^\frac34}=\wp’(x,{a,0})\implies\frac{\left(\text I^{-1}_x\left(\frac14,\frac12\right)-1\right)^2}{\text I^{-1}_x\left(\frac14,\frac12\right)^3}=\frac4{a^3}\wp’^4\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)$$

which is correct

using Weierstrass Zeta $\zeta(x;a,b)$, the Incomplete Elliptic Integral of the Second Kind $\text E(x,k)$, the Second Lemniscate Constant$=\text L_2$, and the Ubiquitous Constant=U:

\begin{align}-\frac{\sqrt[4]a}{4\sqrt 2}\text B_{\text I^{-1}_{\frac{\sqrt[4]ax}{2\omega}}\left(\frac14,\frac12\right)}\left(-\frac14,\frac12\right)=-\frac{\sqrt[4]a}{\sqrt2}\left(\text E\left(\frac12\cos^{-1}\sqrt {\text I^{-1}_{\frac{\sqrt[4]ax}{2\omega}}\left(\frac14,\frac12\right)},2\right) +\frac{2\text I^{-1}_{\frac{\sqrt[4]ax}{4\omega}}\left(\frac14,\frac14\right)-1}{\sqrt[4]{\text I^{-1}_{\frac{\sqrt[4]ax}{2\omega}}\left(\frac14,\frac12\right)}}\right)-\frac{\text U}2=\zeta(x;a,0)\implies \text B_{\text I^{-1}_x\left(\frac14,\frac12\right)}\left(-\frac14,\frac12\right)=-\frac{4\sqrt2}{\sqrt[4]a}\zeta\left(\frac{2\omega x}{\sqrt[4]a};a,0\right),\text E\left(\frac12\cos^{-1}\sqrt {\text I^{-1}_x\left(\frac14,\frac12\right)},2\right) +\frac{\sqrt{1-\text I^{-1}_x\left(\frac14,\frac12\right)}}{\sqrt[4]{\text I^{-1}_x\left(\frac14,\frac12\right)}}=\frac1{\sqrt[4]a}\left(\sqrt2\zeta\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)- \text L_2 \right)\end{align}

which works with this identity. Also with Weierstrass Sigma $\sigma(x;a,b)$, MeijerG, The $\,_3\text F_2$ Hypergeometric function, Inverse Jacobi NS, Beta Regularized $\text I_x(a,b)$, and the complex conjugate:

\begin{align}= \frac{\sqrt2}{\sqrt[4]a}\exp\left(\frac{\text G_{3,3}^{3,2}\left(^{0,\frac34,1}_{0,0,\frac14}\bigg|\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)\right)}{4\Gamma^2\left(\frac14\right)}+\frac{\sqrt[4]a\text U}2x\right) = \frac{\sqrt2}{\sqrt[4]a}\exp\left({\frac{\,_3\text F_2\left(1,1,\frac54;\frac74,2;\frac1{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}\right)}{12 \text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}-\frac\pi4\text I_\frac1{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}\left(\frac14,\frac12\right)+\frac{\sqrt[4]a\text U}2x}\right) =\frac{\sqrt2}{\sqrt[4]a}\exp\left(\lim_{b\to0}{\frac{\,_2\text F_1\left(b,\frac14;\frac34;\frac1{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}\right)-1}{4a}-\text L_2\overline{\text{ns}^{-1}\left(\sqrt[4]{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)},-1\right)}+\frac{\sqrt[4]a\text U}2x}\right)=\sigma(x;a,0)\implies \text G_{3,3}^{3,2}\left(^{0,\frac34,1}_{0,0,\frac14}\bigg|\text I^{-1}_x\left(\frac14,\frac12\right)\right)= 4\Gamma^2\left(\frac14\right)\left[\ln\left(\frac{\sqrt[4]a}{\sqrt2}\sigma\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)\right)-\frac{\sqrt[4]a\text U}2\cdot \frac{2\omega x}{\sqrt[4]a}\right],,\end{align}

which is correct with this identity. Does $\lim_{b\to0}\frac{\,_2\text F_1\left(b,\frac14;\frac34;x\right)-1}{4a} $ have a form in terms of simpler functions?

There is also a formula including $\text I^{-1}_x\left(\frac16,\frac12\right)$ and the second omega constant $\omega_2$:

$$\frac{\sqrt[3]b}{2^\frac23 \sqrt[3]{\text I^{-1}_\frac{\sqrt[6]ax}{\omega_2}\left(\frac16,\frac12\right)}}=\wp(x;0,b)\implies\text I^{-1}_x\left(\frac16,\frac12\right)=\frac b{4\wp^3\left(\frac{\omega_2x}{\sqrt[6]b};0,b\right)}$$ which is true

and

$$-\sqrt{\frac b{{\text I^{-1}_\frac{\sqrt[6]bx}{\omega_2}\left(\frac16,\frac12\right)}}-1}=\wp’(x;0,b)\implies \text I^{-1}_x\left(\frac16,\frac12\right)=\frac b{\wp’^2\left(\frac{\omega_2x}{\sqrt[6]b};0,b\right)+b}$$

which works

these formulas should have a form with Inverse Weierstrass P $\wp^{-1}(x;a,b)$:

$$-\frac{\sqrt[6]b\text B_{\text I^{-1}_\frac{\sqrt[6]bx}{\omega_2}\left(\frac16,\frac12\right)}\left(-\frac16,\frac12\right)}{6\sqrt[3]2}=\zeta(z;0,b)\implies \text B_{\text I^{-1}_x\left(\frac16,\frac12\right)}\left(-\frac16,\frac12\right)=-\frac{6\sqrt[3]2}{\sqrt[6]b}\zeta\left(\frac{\omega_2 x}{\sqrt[6]b};0,b\right)$$

which is true

finally with the first omega constant$=\omega_1$ * $$\frac1{\sqrt[6]b}\exp\left(\frac{\pi \sqrt[6]bx}{4\text{Im}(\omega_1)}-\frac{\text G_{3,3}^{3,2}\left(^{0,\frac23,1}_{0,0,\frac16}\bigg|\text I^{-1}_\frac{\sqrt[6]bx}{\omega_2}\left(\frac16,\frac12\right)\right)}{24 \sqrt\pi2^\frac23 \text{Im}(\omega_1) }-\frac\pi{6\sqrt3}\right) =\sigma(x;0,b)\implies ,,$$

which works

The last set of Weierstrass elliptic function formulas use $\text I^{-1}_x\left(\frac13,\frac12\right)$. To avoid complex numbers, $-B$ is used in a weierstrass function’s parameter:

$$-\frac{\sqrt[3]{B\text I^{-1}_{\frac{3\sqrt[6]Bx}{2\text{Im}(\omega_1)}-2}\left(\frac13,\frac12\right)}}{2^\frac23}=\wp(x;0,-B)\implies\text I^{-1}_x\left(\frac13,\frac12\right)=-\frac4B\wp^3\left(\frac{2\text{Im}(\omega_1)(x+2)}{3\sqrt[6]B};0,-B\right)$$

which is true

also,

$$\sqrt B\left(2\text I^{-1}_{\frac{3\sqrt[6]Bx}{4\text{Im}(\omega_1)}-1}\left(\frac13,\frac13\right)-1\right)=\wp’(x;0,-B)\implies=\text I^{-1}_x\left(\frac13,\frac13\right)=\frac{\wp’\left(\frac{4\text{Im}(\omega_1)(x+1)}{3\sqrt[6]B};0,-B\right)}{2\sqrt B}+\frac12$$

which works

with Baxter’s Four-Coloring Constant=B,

$$\sqrt[6]B\left(\frac{\text B_{\text I^{-1}_{\frac{3\sqrt[6]Bx}{2\text{Im}(\omega_1)}-2}\left(\frac13,\frac12\right)}\left(\frac23,\frac12\right)}{6\sqrt[3]2}+\frac1{\text B}\right)=\zeta(x;0,-B)\implies \text B_{\text I^{-1}_x\left(\frac13,\frac12\right)}\left(\frac23,\frac12\right)=6\sqrt[3]2\left(\frac{\zeta\left(\frac{2\text{Im}(\omega_1)(x+2)}{3\sqrt[6]B};0,-B\right)}{\sqrt[6]B}-\frac1{\text B}\right)$$

which is correct

\begin{align}\implies\end{align}

Are there any formulas for the Weierstrass utility functions or modular forms in terms of Inverse Beta Regularized? Formula 6:

Although possibly not a special case of $\text I^{-1}_x (a,b)$, its definition shows that it gives a symmetric part of a period of a hyperelliptic function. What is inverse beta regularized in terms of Riemann Theta function? For more information, please see this question.

Formula 7:

This identity does not give a special case of inverse beta regularized, but it generates many more special cases when applied to other formulas in this answer:

$$\text I^{-1}_x(a,b)=1-\text I^{-1}_{1-x}(b,a)$$

InvT, from the question, represents most special cases of inverse beta regularized in this answer with the sign function

$$\sqrt d\text{sgn}\left(x-\frac12\right)\sqrt{\frac1{1-\left(2\text I^{-1}_x\left(\frac d2,\frac d2\right)-1\right)^2}-1}=\text{InvT}(x,d)$$

which works . Therefore, with this identity, we have these conversion formulas and identities applicable to other formulas in this answer

$$\text I^{-1}_x(a,a)=\frac12-\frac12\sqrt{1-\text I^{-1}_{2x}\left(a,\frac12\right)}=\frac12\left(\frac{\text{sgn}\left(x-\frac12\right)}{\sqrt{\frac{2a}{\text{InvT}(x,2a)}+1}}+1\right)\implies\text I^{-1}_x\left(a,\frac12\right)=1-\left(2\text I^{-1}_\frac x2(a,a)-1\right)^2=1-\frac1{\frac{2a}{\text{InvT}^2\left(\frac x2,2a\right)}+1}$$

which is true

Formula 8:

Remember to appropriately transform Inverse Beta Regularized if you want another period of a periodic function like in formulas $2,4,5$. Another way to extend the function is using these different formulas using

$$$$

similarly,

$$$$

which can be used to find this result among many others.

$$$$

and

$$$$ Formula 9:

The first limit of $\text I^{-1}_x(a,b)$ with a closed form is

$$1-\lim_{a\to0}\text I^{-1}_{-ax}(1,a),\lim_{a\to0}\frac1{\text I^{-1}_{1-ax}(a,1)}=e^x\implies\lim_{a\to0}\text I^{-1}_{ax}(1,a)=1-e^{-x}$$

which is true

similarly with the Lambert W function

$$-\lim_{a\to0}\text I^{-1}_{a\ln(-x)+a+1}(a,2)=\text W(x)\implies\lim_{a\to0}\text I^{-1}_{ax}(a,2)=\text W\left(-e^{-x-1}\right)$$

which work. Next,

$$2\lim_{a\to0}\text I^{-1}_{ax+\frac12}(a,a)-1=\tanh(x)\implies\lim_{a\to0}\text I^{-1}_{ax}\left(\frac12,a\right)=\tanh^2\left(\frac x2\right)$$

which are correct

Formula 10:

Inverse beta regularized generalizes Inverse Gamma Regularized $Q^{-1}(a,x)$ from this limit:

$$\lim_{b\to\infty}b\,\text I^{-1}_x(a,b)=Q^{-1}(a,1-x)\implies\lim_{b\to\infty}b\,\text I^{-1}_x(b,a)=1-Q^{-1}(a,x)$$

which is correct

Formula 11:

One special case of $Q^{-1}(a,x)$, and therefore a special case of $\text I^{-1}_x(a,b)$, is

$$\mp Q^{-1}\left(1,x^{\pm 1}\right)=\ln(x)$$

where one of each sign is taken which works. Next with the $-1$st branch of the Lambert W function,

$$-Q^{-1}(2,-ex)-1=\text W_{-1}(x)\implies Q^{-1}(2,x)=-\text W_{-1}\left(-\frac xe\right)-1$$

which is true. Also with the Inverse Error function,

$$\lim_{a\to\infty}\sqrt a\left(2\text I^{-1}_\frac{x+1}2(a,a)-1\right)=\text{erf}^{-1}(x)$$

which works and with inverse erfc,

$$Q^{-1}\left(\frac12,x\right)=\text{erfc}^{-1}(x)^2,0<x\le 1$$

which works and one of each sign is taken.

Formula 12:

While possibly not closed forms, maybe these unique limits of Inverse Gamma Regularized are named inverting the Exponential Integral function $\text{Ei}(x)$ and Logarithmic Integral function $\text{li}(x)$ for the specified branch:

$$y=-\lim_{a\to0}Q^{-1}(a,-ax)\implies \text{Ei}(y)=x,y<0$$

and

$$y=\lim_{a\to0}e^{-Q^{-1}(a,-ax)}\implies \text{li}(y)=x,0\le y<1 $$

therefore,

$$y=\lim_{a\to0}Q^{-1}(a,ax)\implies -\text{Ei}(-y)=x,y>0$$

which work. Are there any other closed form particular cases or identities for Inverse Gamma Regularized?

Answer 10

(Feel free to use this for your drafts.)

Question: Polytopes in vector spaces

Tags: number-systems

Body: Aside from the complex numbers, the triplex numbers, and their isomorphisms, I'd like to know which other finite-dimensional normed vector spaces over reals have multiplication rules that allow for the existence of elements whose positive integer powers form vertices of certain polytopes. To narrow things down, here are some rules:

1. The polytopes must be regular and convex.
2. The vertices of each polytope must all be unit vectors that are roots of unity (i.e. $v^n=1$ for some $n>0$ based on the multiplication rules).
3. The positive integer powers of a vertex must also be vertices of the same polytope instance. With the second rule, this means the multiplicative identity must always be a vertex.

Since only $n$-simplexes, $n$-orthoplexes, and $n$-orthotopes have regular convex variants in all dimensions $n$, we can restrict ourselves to those polytope families, but let us add a fourth rule:

4. If the vector space is $n$-dimensional, all regular convex simplexes, orthoplexes, and orthotopes of $n$ dimensions or less must have instances in the space following the above rules. We can ignore $n<2$ as a given.

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For $n=2$, as implied, $\Bbb{C}$ is one such vector space. The imaginary unit $i$ and its powers form a square (which is both the 2-orthoplex and the 2-orthotope), and the third roots of unity in the complex plane form an equilateral triangle (2-simplex).

I don't think dual numbers and split-complex numbers have roots of unity that form an equilateral triangle, but please correct me if I'm wrong.

For $n=3$, I believe the triplex numbers (which, I've been told, is isomorphic to $\mathbb R\times \mathbb C$) is another such vector space, as I'll try to demonstrate below.

To lessen confusion, let's use the notation $j$ and $k$ for the non-real parts of the triplex basis. We find the following polytopes in triplex space:

$3$-orthoplex

$1$, $j$, $k$ and their negatives form an octahedron. $j$ and $k$ are primitive third roots of unity and are the squares of each other. $-j$ and $-k$ are primitive sixth roots of unity and their powers coincide with all six vertices of the octahedron.

$2$-orthoplex/$2$-orthotope

The non-unit circle that touches $1$, $j$, and $k$ seems to share one important property with the unit circle on the complex plane: For any regular convex polygon whose vertices lie on the circle and includes $1$, all their vertices are $n$th roots of unity. And in fact, $s=\frac{1}{3}+\frac{1-\sqrt{3}}{3}j+\frac{1+\sqrt{3}}{3}k$ and its powers are on this circle, form a square, and are fourth roots of unity. $s^2$ is a primitive second root of unity.

$3$-orthotope

The square we just described above is a face of a cube centered on the origin in triplex space. The other vertices of the cube are the negatives of $s$ and its powers. In particular, $-s$ can in fact be considered as the $c$ I mentioned in this question about cubic numbers, so you could check some more details there.

$3$-simplex

The powers of $-s$ (a.k.a $c$) form one of the demicubes of the cube described above, and are fourth roots of unity. The demicube is of course a regular tetrahedron.

$2$-simplex

This one's the easiest: The third roots of unity $1$, $j$, and $k$ form a suitable equilateral triangle.

You could check out the triplex polytopes in this geogebra link. (The point labeled "A" is meant to be the multiplicative identity.)

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Questions:

Are there any other vector spaces that fit the above criteria, particularly for $n=4$? I know $\Bbb{H}$ won't fit because quaternion powers are always co-planar and wouldn't be able to form regular polyhedra and polychora.

Also, what property do complex and triplex numbers have that allow them to form regular polytopes as described above? Is there an algorithm to construct suitable vector spaces in higher dimensions?

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Motivation:

I'm mainly interested in the creation of $n$-simplexes in higher dimensions because it relates to other ideas I've had in the past, but I reckon adding the other polytope families will make this exercise potentially more useful in general.

Answer 11

This space is available for use.

Answer 12

update2
To show a method how to at least arrive at some estimates for the coefficients and get meaningful approximations for the values in the $\varphi^{\circ -1}()$ function.
I follow here the ansatz of A. Robbins.

First we observe, that in (eq 6d) $ \varphi(\varphi^{\circ -1}(x)+1)=\exp(x)$ we can $\varphi^{\circ -1}()$ apply on each side. (Of course we must assume that the powerseries/dotproducts have some finite radius of convergence). We get then the equation $$ \varphi^{\circ -1}(x)+1=\varphi^{\circ -1}(\exp(x)) \tag {7a}$$ In matrix-notation (let's write for notational comfort $Q=\phi^{-1}$):

$$ V(x) \cdot Q \cdot P = V(x) \cdot B \cdot Q $$ and then the identity of coefficients in the matrix-compositions alone $$ Q \cdot P = B_b \cdot Q \tag {7b} $$ We assume the inverse $\varphi()$-function to be $$ Q(x)= q_1 x + q_2 x^2 + q_3 x^3 + ...\tag {7c} $$ and that we have that coefficients in the second column in the (Carleman-) matrix $Q$: $$ Q_{[,1]} = [0,q_1,q_2,q_3,...] \tag {7d} $$ Then, if we look again at (eq 7b) we can reduce the equality to the essential part $$ (Q \cdot P)_{[,1]} = (B_b \cdot Q)_{[,1]} \tag {7e} $$ which is $$ V(0) + Q_{[,1]} = B_b \cdot Q_{[,1]} \tag {7f}$$ Now we can rearrange the left $Q_{[,1]}$ to the right $$ V(0) = B_b\cdot Q_{[,1]} -Q_{[,1]} \\ V(0) = (B_b - I)\cdot Q_{[,1]} \tag {7g}$$

Now, using a finite size, say $n=8$ we could feed this in a solver, like matsolve(Bb-I,V(0)) but unfortunately the first column in the expression $B_b - I$ is completely zero (because $B_b$ has the form of a Carlemanmatrix) and is thus not invertible.

The idea of Robbins was here,

• to cheat a bit: to remove the first column and the last row to have again a square matrix, but which can now be inverted. If this works at all it might later be justified by the goal that we want to work with matrices of infinite size where such a transition might be possible and still meaningful.
• to try this for increasing size of the matrices and to look, whether that increasing would make the leading coefficients converge to something.

I've reproduced that idea, and because the improving of approximation is slow, document only the doubled sizes: $n=2^1,2^2,2^3,...,2^7$ (for the shortened $B_b-I$!) and got the following output:

Table $7h$: approximations to coefficients $q_k$ of powerseries $Q(x)$ $$ \small \begin{array} {r|rrrrrrrrrrrrr} n & q_0 & q_1 & q_2 & q_3 & q_4 & q_5 & q_6 & q_7 & ...\\ \hline 2&0 & 1.0000000 & 0 & . & . & . & . & . \\ 4&0 & 0.92307692 & 0.24615385 & -0.18461538 & 0.015384615 & . & . & . \\ 8&0 & 0.91644296 & 0.24850496 & -0.11395859 & -0.093197376 & 0.020154115 & 0.040614135 & -0.021681561 \\ 16&0 & 0.91595862 & 0.24921856 & -0.11061148 & -0.093582187 & 0.010588156 & 0.035629206 & 0.0054841446 \\ 32& 0 & 0.91594399 & 0.24933929 & -0.11046111 & -0.093883524 & 0.010030915 & 0.035820110 & 0.0064801129 \\ 64& 0 & 0.91594562 & 0.24935378 & -0.11046270 & -0.093932220 & 0.010001216 & 0.035889177 & 0.0065707651 \\ 128& 0 & 0.91594601 & 0.24935461 & -0.11046450 & -0.093936126 & 0.010002714 & 0.035897333 & 0.0065738120 \end{array} $$

This values are also documented in the Robbins-article, however he wrote them as $n$'th derivatives, and his values must be divided by factorials. (After that I got differences to Robbins' data in about $7$'th decimal digit):
Table $7i$: comparision of coefficients of computed $Q(x)$ and Robbins' documentation: $$ \small \begin{array} {r|r} \text{reproduction} & \text{Robbins}\\ n=128 & n=160 \\ \hline 0 & . \\ 0.91594601 & 0.91594603 \\ 0.24935461 & 0.24935462 \\ -0.11046450 & -0.11046461 \\ -0.093936126 & -0.093936220 \\ 0.010002714 & 0.010002913 \\ 0.035897333 & 0.035897641 \\ 0.0065738120 & 0.0065737081 \\ -0.012305809 & -0.012306288 \end{array} $$ Robbins claims then additionally that a meaningful assumption would be $q_0=-1$ , because then we have normalized $Q_R(0)=-1$ and $Q_R(1)=0$ and $Q_R(e)=1$ (let's write that modified function $Q()$ indexed as $Q_R()$ from now). $Q_R(x)$ is then the function $\text{slog}()$ in Robbins' article.

Calculations by this style give then meaningful numbers with accuracy of some $6$ to $10$ digits precision, such that $Q_R(e^x)-Q_R(x)\approx 1$ and $Q_R(1)=0$ and for (small!) values $Q_R(\,^he) = h$. For instance using that value that Robbins gives for $ x =\,^{0.5}e \approx 1.6463546427466649$ and using this value in the $Q_R()$ function gives $Q_R( x ) = 0.50000026$ which I find quite a nice approximation (using $n=64$ terms/matrices). Note that this $x$ seems to exceed the range of convergence, but since $Q()$ has coefficients with (roughly) alternating signs the sum can be done by Euler-summation with small order. (With that Eulersummation we find by binary search the more precise value $x_{0.5}=1.64635422897119922045=\,^{0.5}e $ and $Q_R(x_{0.5})=0.5$ to $20$ digits)

Robbins does not give the coefficients $c_k$ (by the inversion of the powerseries $Q_R(x)$).
But assuming the $Q(x)$ function with $q_0=0$ we can do inversion and get coefficients: $$ \hat \varphi(x)= 1.0917674 x - 0.32449483 x^2 + 0.34983599 x^3 - 0.23085419 x^4 + 0.20133057 x^5 + O(x^6) \tag {7j} $$ _{(where I write $\hat \varphi()$ to denote that this is not taken from the original definition of $\varphi()$ above)}

Looking at this with $64$ coefficients it is much likely that the radius of convergence is small, but it seems, that we can subtract the coefficients of the $\log(1+x)$-Mercator series from this, and get a better series even with larger range of convergence, say $$ \hat \varphi_2(x) = \hat \varphi(x) - \log(1+x) \tag {7k} $$ The first few coefficients are then $$ \hat \varphi_2(x)= 0.091767404 x + 0.17550517 x^2 + 0.016502661 x^3 + 0.019145810 x^4 + 0.0013305742 x^5 + 0.0023142725 x^6 - 2.1040292 E-5 x^7 + O(x^8) \tag {7L} $$

With this I get the following results:
Table $7m$ - tetrations of $e$ to height $x$ $$ \begin{array} {} x & y=\hat \varphi_2(x)+\log(1+x) & \exp(y) &\text{comment}\\ \hline 0& 0& 1\\ 0.5 & 0.49856329 & 1.6463542 & 13 \text{ dig correct to } x_{0.5}\\ \pi/4 & 0.77623646 & 2.1732776 & \\ 1 & 1.0000000 & 2.7182818 & 22 \text{ dig correct}\\ 1.5 & 1.6463542 & 5.1880309 & 16 \text{ dig good} \\ 2.0 & 2.7182818 & 15.154262 & \text{$9$ and $8$ dig good} \end{array}$$

This looks very good, and leaves only, to add the exponentiation to the operation to have a tetration-function for the basis $e$. We can thus fractionally tetrate to base $e$ and the final form should then be: $$ \hat {\text{tet}}(x) = \exp(\hat \varphi_2(x) +\log(1+x)) \\ = \,^x e \tag {7n}$$

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Well, this seems to have now escaped from your construction of the $c_k$ and your $\varphi(x)$-function. At the moment, I don't see how to establish a clearer relation, perhaps I can come back to this later...

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Answer 13

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Answer 14

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Answer 15

This Answer to How to avoid downvotes for beginner's questions? is relevant, as upvotes usually accompany Good Questions. Here I build off that Answer, giving instructions with examples.

Advice	Examples/comments
1 Please use Mathjax! Tutorial link | Reference Link $$ \rule{10em}{0em} %this is to space the table out in portrait mode on mobile. sorry for the hack$$	Good: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Bad: x=-b+-sqrt(b^2-4ac)/2a,
8 Use a descriptive title that is easy to read (and don't use only Mathjax).	If your title doesn't grab the attention of the right people, you won't get an answer. Also, "Does $b^2-4ac<0$ imply that $P$ has no roots?" is better than "$b^2-4ac<0\implies \{x:P(x)=0\}=\emptyset?$".
3 After writing a title, look through the auto-generated list of 'similar questions' to avoid posting a duplicate.	If you found a related post but you can't understand it, don't just say "I saw this Q on this site before"; give the link (see point 5). More on searching
9 Get to the point quickly; details can come after the "punchline". Also pay attention to paragraphing and punctuation.	Questions that are easier (and shorter) to read attract more Answers. This is somewhat of an essay-writing skill, but no more so than writing any request for assistance.
6 Give the source of, or motivation for, the question.	"in Generatingfunctionology by H. Wilf, 2nd ed, page 234..." is better than "I saw it in my combinatorics textbook". Also, if it smells like a contest question and has no source, I closevote and downvote.
7 Describe your level of math education, and other things that you suspect are relevant background material.	e.g., "I took one year of undergraduate mathematics in the UK, but had to take a break", "We just proved Heine-Borel so perhaps we need to use it."
2 Try to solve the problem yourself first, and then summarise your efforts and what went wrong.	If your question is about a calculation, consider typing out the whole calculation.
4 Ask about specific issues that are possible to answer authoritatively (in principle).	Good: "Why does the proof fail in three dimensions?" Bad: "How can I solve it?" with no further context.
5 Don't require users to click on any external link.	Key parts of the Question ought to be typed out, and any necessary illustration embedded (if you have 10 reputation). Supplementary links are fine.
10 Proofread to remove typos and ambiguities.	Taking a 5-minute break then re-reading your post out loud may be helpful.

Other opinions:

11. Some questions of a more exploratory type aren't so suited here, even if they follow the advice above or elsewhere. For instance, "Here's something I created (e.g. I generalised a definition) Is it useful?". If you don't know if it is useful, asking "Has this been done before?" is also bad. Perhaps consider writing a blog.

12. There are certain topics that generate a bad kneejerk reaction here on Math.SE, due to e.g. a certain math video going viral (or even Getting Things Wrong). In my experience, these are usually questions on Logic, open problems, and divergent series (beyond merely proving divergence). In these types of questions, you should distance yourself from the popsci and address the actual mathematical problem (in particular, be very rigourous). If you must ask directly about the popsci, consider if somewhere else like the more informal Mathematics chatroom, or a math-related part of reddit is more appropriate.

Answer 16

Assume $x^2-14y^2 = 2 $ with $\gcd(x,y)=1$. Show, that then $x/y$ is a convergent of the continued fraction of $\sqrt{14}$.

Reformulating the expression towards that in Khinchin's theorem: $$\begin{array} {} {x^2 \over y^2} - 14 &= {2 \over y^2} \\ ({x \over y} - \sqrt{14})({x \over y} + \sqrt{14}) &= {2 \over y^2} \\ ({x \over y} - \sqrt{14})(\delta + 2\sqrt{14}) &= {2 \over y^2} \\ \left| {x \over y} - \sqrt{14} \right| &= {2 \over y^2 (\pm \delta + 2\sqrt{14})} \\ \left| {x \over y} - \sqrt{14} \right| &= {1 \over y^2 (\pm \delta/2 + \sqrt{14})} &\lt {1 \over 2y^2 } \\ \end{array} $$ The last inequality satisfies Khinchin's theorem for $x/y$ being a convergent of $\sqrt{14}$

Answer 17

Adding to Brian's comment-Answer. Several months ago, even though StackExchange doesn't seem amenable to subsite-specific changes, I was triggered to idealistically shoot off the following feedback via the Contact link. Their reply was so stock that I don't remember what it said.

This feedback will need to be fleshed out (for example, I've not explained what's unique about our situation, and focused exclusively on PSQs) if it is to be properly submitted as a feature request. Perhaps it could be collated and summarised. I have redrafted it (no mention of the EoQS as it is punitive, non-universal and non-timeless), and am leaving it here for now.

Please also refer to Martin's comments here.

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Feedback: Mathematics StackExchange acutely needs a site-specific notice

To a casual visitor who has just stumbled on Mathematics StackExchange, it is simply a forum for asking mathematics questions (yet it is really not); understandably, students in search of homework help continually deluge this site with “problem-statement questions” (PSQs) that very quickly get downvoted, closed, and deleted by the community who additionally then have to manually issue, in the Comments section, friendly reminders to the culprit.

Now, few people actually read Terms & Conditions or instruction manuals; likewise, it is unrealistic to expect that visitors will have read the Site Tour or anysuch before posting their PSQ.

And while first-time posters are currently flashed a generic “Asking a good question” modal window after clicking Submit, the message may be ineffectual, appearing too late—a user, ready to fire off their PSQ, is inclined to wave it off as another T&C—and failing to communicate the key point about site standards, that is, that PSQs, being deleterious, are unwelcome.

To be clear: PSQs severely degrade the quality of Mathematics StackExchange, as they add noise and obstruct good questions from surfacing on Google.

This issue continues to consume a great amount of the collective resources of the community: the abovementioned routine needlessly repeats many times per hour.

To address the above, it is imperative to install on this subsite's homepage a single-line notice—visible before users start composing any PSQ—pointing to such a table of Guidelines for Attracting Answers, which practically sums up Mathematics StackExchange's ethos and communicates that its goal of being a useful repository of mathematical Q&A is incompatible with being a homework service. (Such a notice would neither be glaring nor break the visual consistency across the StackExchange sites.)