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  • 8
    $\begingroup$ 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
  • 21
    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$
    – cardinal
    Jul 18, 2012 at 19:40
  • 14
    $\begingroup$ 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
  • 3
    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$
    – leo
    Dec 17, 2012 at 18:03
  • 3
    $\begingroup$ PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. $\endgroup$ Dec 2, 2015 at 14:07

17 Answers 17


Using this to:

  • learn math proper notation (feedback appreciated)
  • learn mathJax syntax
  • take snippets to articles.

Random Expressions:

$$ n * T_i | n\in \Bbb N + \{0\}$$

$a \equiv a \mod (b)$


Is the following sieve and conclusion novel? All prime numbers are contained in periodic sequences as described below.


All prime numbers $(p \in \Bbb P)$ are contained in the sequence:

$$ \Bbb P \subset P_{0..^\infty} ∪ G$$


  • $P_{0..^\infty} = \bigcup_{n=0}^\infty n * T * \pmb I + P |n \in \Bbb N + \{0\}$

  • $G = \{2, 3, 5, 7, 11, ..., g\} \iff \prod_{g \in G} g = T$ ; contains all initial primes and determines the period of the relatively primes ($c$) sequence $P$.

  • $\pmb I$ identity vector.

  • $P = \{1, c_{1}, …, c_{n} \} | c_i < T ; i = \{1, …,n\}$ contains all relatively primes of primes in $G$ that are less than $T$.

  • $c$ is a coprime in $P$ of generator primes in $G$: $\{c \equiv r \mod g|r>0, g \in G, c \in P \}$

Note: Harmonic term is used to denote $P$ is a pattern or sequence that has a period $T$ and a frequency $1/T$

I think this proves that prime numbers are contained within the periodic sequences of relatively primes of ($g \in G$). And while their exact number can't be determined before remotion of non-primes from $P$ their distribution (location) can be iteratively refined by bigger periodic sequences $P$ of increasing period $\prod_{g \in G} g= T$. Showing their distribution is not random.


Sieve definition


$n$ Is the n-th iteration $n \in \Bbb N$.

$G_n$ sequence of initial contiguous primes (or generator primes);

$$G_n = \{g_1, g_2, g_3, g_4, g_5, ..., g_n\} = \{2, 3, 5, 7, 11, ..., g_n\} \iff \prod_{g \in G_n} g = T_n$$

$T_n$ Period of sequence $P_n$. Determines the period with which $P_n$ repeats. It's the minimum common multiple of all generator primes ($g \in G_n$).

$P_n$ is the sequence of relatively primes ($c$) of initial primes ($g$) in $G_n$ that repeats with period $T_n$:

$$P_n = \{1, c_{n_1}, …, c_{n_m} \} $$ $$| c_{n_i} \equiv r \mod g , r>0 \land i=\{1, ..., m \} $$ $$ \land (g \in G_n , c_{n_i} < T_n )$$

$|P_n|$ cardinality of $P_n$ (i.e. 1 + number of relatively primes of generator primes: $c_{n_i} \in P_n | i = \{1, ..., m\} \land c_{n_i} < T_n)$ ,

$g_{n+1}$ For any given iteration $n^{th}$, the very next prime:

$$ g_{n+1} = \begin{cases} min(p \in \Bbb P) | p > max(g \in G_n), & \text{For $|P_n| \equiv 1$ } | (n \equiv 1 \implies 2); (n \equiv 2 \implies 3)\\ c_{n_1}, & \text{For $|P_n| > 1$} \end{cases} $$


$repeat(P, m, o) = \bigcup_{i=0}^m(i * o * \pmb I + P) |m \in \Bbb N , o \in \Bbb N$

E.g. $$repeat(\{1, 5\}, 3, 6) = \{0+1, 0+5, 6+1, 6+5, 12+1, 12+5\} = \{1, 5, 7, 11, 13, 17\}$$

Sieve definition

Base case, initial pattern:

$$P_{0} = \{1\} ; |P_{0}| = 1 ; G_{0} = \{\}; T_0 = \prod_{g \in G} g = 1 \implies g_{i+1} = g_{1} = 2 $$

Sieve expansion:

$$i \in \Bbb N + \{0\}$$

$$P_{i+1} = repeat(P_i, g_{i+1}, T_i) - g_{i+1} * P_i $$

$$G_{i+1} = G_i \cup \{g_{i+1}\} $$

$$T_{i+1} = \prod_{g \in G_{i+1}} g$$

$$|P_{i+1}| = |P_i| * (g_{i+1} - 1)$$

A few iterations

$$P_{0} = \{1\} ; |P_{0}| = 1 ; G_{0} = \{\}; T_0 = \prod_{g \in G} g = 1 \implies g_{i+1} = g_{1} = 2 $$

$n=1$ $$P_{1} = repeat(P_{0}, g_{1}, T_0) - g_{1} * P_0 = repeat(\{1\}, 2, 1) - 2 * \{1\} = \{1, 2\} - \{2\} =\{1\}$$

$$G_{1} = G_{0} \cup \{g_{1}\} = \{\} \cup \{2\}= \{2\}$$

$$T_1 = \prod_{g \in G_1} g = \prod \{2\} = 2$$

$$|P_{1}| = |P_{0}| * (g_{1} - 1) = |\{1\}| * (2 - 1) = 1 * 1 = 1$$

$$\implies P_{1} = \{1\} ; |P_{1}| = 1 ; G_{1} = \{2\}; T_1 = \prod_{g \in G_1} g = \prod \{2\} = 2 \implies g_{2} = 3 $$


$$P_{2} = repeat(P_1, g_{2}, T_1) - g_{2} * P_1 = repeat(\{1\}, 3, 2) - 3 * \{1\} = \{1, 3, 5\} - \{3\} =\{1, 5\}$$

$$G_{2} = G_{1} \cup \{g_{2}\} = \{2\} \cup \{3\}= \{2, 3\}$$

$$T_2 = \prod_{g \in G_2} g = \prod \{2, 3\} = 6$$

$$|P_{2}| = |P_{1}| * (g_{2} - 1) = |\{1\}| * (3 - 1) = 1 * 2 = 2$$

$$\implies P_{2} = \{1,5\} ; |P_{2}| = 2 ; G_{2} = \{2, 3\}; T_1 = \prod_{g \in G_2} g = \prod \{2, 3\} = 6 \implies g_{3} = 5 $$

Note from now on, $g_{i+1}$ is always $c_{1}$ i.e. $2^{nd}$ element in sequence $P$


$$P_{3} = repeat(P_{2}, g_{3}, T_2) - g_{3} * P_2 = \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29\} - \{5, 25\} = \{1, 7, 11, 13, 17, 19, 23, 29\}$$

$$G_{3} = G_{2} \cup \{g_{3}\} = \{2, 3\} \cup \{5\}= \{2, 3, 5\}$$

$$T_2 = \prod_{g \in G_3} g = \prod \{2, 3, 5\} = 30$$

$$|P_{3}| = |P_{2}| * (g_{3} - 1) = |\{1, 5\}| * (5 - 1) = 2 * 4 = 8$$

$$\implies P_{3} = \{1, 7, 11, 13, 17, 19, 23, 29\} ; $$ $$|P_{3}| = 8 ; G_{3} = \{2, 3, 5\}; T_3 = \prod_{g \in G_3} g = \prod \{2, 3, 5\} = 30 \implies g_{3} = 7 $$


$$\implies P_{4} = \{\color{red}{1},\color{green}{11},13,17,19,23,\color{red}{29,31},37,41,43,47,53,\color{red}{59,61},67,71,73,79,83,\color{red}{89},97,101,103,107,109,113,\color{red}{121},127,131,137,139,143,\color{red}{149,151},157,163,167,169,173,\color{red}{179,181},187,191,193,197,\color{green}{199,}\color{red}{209}\} ; $$ $$|P_{4}| = 48 ; G_{4} = \{2, 3, 5, 7\}; T_4 = \prod_{g \in G_4} g = \prod \{2, 3, 5, 7\} = 210 \implies g_{5} = 11 $$

Note twin prime candidates are always around $T_n ± 1 $ in $\color{red}{red}$ twin prime candidates for $T_3 = 30$

Note max gap between primes is always around $(T_n + 1; \color{green}{T_n + c_1})$ and $(\color{green}{c_{k-1}}; T_n - 1) | k=|P_n|$ in $\color{green}{green}$ for $T_4 = 210$

$… n → ∞ ; |G_n| → ∞ ; |P_n| → ∞; T_n → ∞$

Key implications

  • $\bigcup_{i=0}^\infty(i * T_n * \pmb I + P_n) |i \in \Bbb N$ discards every number that’s not a relatively prime of primes in $G_n$ thus, only numbers not discarded by the perio sequence can be prime.

  • Guaranteed primes in $P_n$:

$$\{c_{n_i} \in \Bbb P | c_{n_i} \in P_n , 1 < c_{n_i} < \max_{g \in G_n}{g}^2\}$$

  • Sequences $P_n$ repeat because at $\{ n * T_n | n \in \Bbb N \}$ all the frequencies of generator primes $(G)$ coincide and start over. $T_n = {\rm lcm}_{g \in G}(g) $ ($T_n$is the least common multiple of $g \in G$).

  • Any increasing sequence of contiguous primes $$K = \{2,3,5,7,11, ..., g_m, ..., p \}| g_m < p \land p < T $$

Where: $$T = \prod_{g \in K | g \leq g_m} g$$

$$ G = \{2, 3, 5, 7, 11, …, g_m\}| G \subset K \land \{g \in G \land g \in K | g \leq g_m \}$$

$$J = \{2,3,5,7,11,…, j\}| G \subset J \subset K \land \{j \in J \land j \in K | j \leq \sqrt{T} \}$$


$$K - G + \{1\} = P^i$$

Where $P^i$ is the initial portion of a (incomplete) sequence of relatively primes of $\{ j \in J \}$ with period $\prod_{j \in J} j = T_{P^i}$.

Such that:

$$ \bigcup_{n=0}^\infty{n * T_{P^i} * \pmb I ± P^i} | n \in \Bbb N$$

Contains all primes in the intervals $(n * T_{P^i} - T, n * T_{P^i} + T) | n \in \Bbb N + {0}$

  • In contrast with other sieves, while extending $P_n \to P_{n+1}$ every non- relatively prime of $g_{n+1}$ can be removed from the sequence only once. As sequences from previous iterations had already eliminated all relatively primes of $\{g \in G_{n} \}$ for all $\Bbb N$. Then the product $ g_{n+1} * P_n $ will always hit a relatively prime of $g_{n+1} $ in $ \{ P_{n+1} \cup g_{n+1} * P_n \} = repeat(P_n, g_{i+1}, T_n)$. This property allows for efficient programmatic implementations of the sieve.

A Java implementation of the sieve

A Javascript implementation ready to try.

** concepts originally published here

  • $\begingroup$ jsfiddle.net/juanmf/Lz7nycxm/latest made a little Markup + MathJax editor for quick edits. Markup support is minimal, but enough to get a good idea of final result. Adds a textarea with raw html at the bottom. $\endgroup$
    – juanmf
    May 27 at 2:46

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.

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.

image1We 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 image2 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. image3The slightly jittering effect in the Ellison-curve is because its function uses the somehow jittering values $S$ instead that of $N$ for its argument.


Testing link embedding procedure. Link description


This answer is free for anyone to use


This answer is free for anyone to use.


This answer is free for anyone to use.


$$ a_1 (2^S - 3^N) = 3^N ( 1/3 + 1/3 \cdot 2^A/3 + 1/3 \cdot 2^A/3 \cdot 2^B/3 + ... 2^A/3 2^B/3 \cdots 2^M/3) \\ a_1 \left(\frac{2^S}{3^N} - 1\right) = \frac13 + \frac{2^A}3 \left(1 + \frac{2^B}3 \left( 1 + \cdots \frac{2^M}3 \right)\right) \\ a_1 \left(2^{ 1 - \{Nß\}} - 1\right) = \frac13\left(1 + \frac{2^A}3 \left(1 + \frac{2^B}3 \left( 1 + \frac{2^C}3 \left(1 + \frac{2^D}3 \right)\right)\right)\right) \\ $$ where $ S=\lceil {Nß} \rceil \gt A+B+C+D \ge N-1$ and all $\{A,B,C,D\}\ge 1$This answer is free for anyone to use.


Question: Berry curvature and a discontinuous wavefunction

Body: Apparently, the wavefunction of quantum mechanics can have discontinuities at a point, because when conjugate squared, these contribute nothing to the overall probability.

However, the phase of a wavefunction has been found to matter in [recent years] in the form of Berry curvature. Insert briefest description of Berry curvature here.

What would happen if the wavefunction were discontinuous at a point, and we wanted to calculate the Berry curvature in the vicinity? Would the discontinuioty still contribute nothing to the physics, and only be a mathematical curiosity, or would it have an observable effect?

Tags: berry-pancharatnam-phase wavefunction quantum-mechanics

Suppose I have a triangular lattice, currently containing two atoms near absolute zero which are very close to one another.

poor quality paint image of the system

There are two spin states available (up and down). In the two atom system, the configurations with the equally lowest energy are those with atom up and the other one down. If a third atom is added to the triangular lattice, there are more configurations of the system of the system with equally lowest energy than there were previously, and the average energy per atom has increased.

A phase transition is classically defined as when the free energy of the system suddenly changes. It can also be defined as when spontaneous symmetry breaking of an order parameter occurs, or in more recent years when a change in the systems topology occurs.

I want to know if adding the third atoms constitutes a phase transition of sorts, and if so whether it is a topological one?

What happens when a particle leaves orbit. Does it just keep building up potential energy?

The further it gets away, the more potential it will gain relative to where it started. I ask not just for GR, where I believe there is some debate over the conservation of energy, but for classical electrodynamics as well. Where does the potential energy go for an electron freed from the orbit of an atom? Does it just keep building up energy as it moves further away from it, and perhaps lose potential energy relative to another atom somewhere else?


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

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


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 InvT 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 a closed form for an evaluation of inverse beta regularized mentioned in the self answer, but it has a different evaluation of $\text I^{-1}_x(a,b)$, 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.


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

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, Multidimensional Theta functions, 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:

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


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

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


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

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:


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

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

Formula 5:

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

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

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

which is correct

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

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

which works with this identity

with Weierstrass Sigma $\sigma(x;a,b)$:

$$\implies $$

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


The last set of Weierstrass elliptic function formulas use $\text I^{-1}_x\left(\frac13,\frac12\right)$ and the second omega constant $\omega_2$:


Are there any formulas for the Weierstrass Utility functions?

Formula 6:

These identities do not give a special case of inverse beta regularized, but they generate many more special cases of it when applied to other formulas in this answer:

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

also, using this identity, one easily finds the following formulas where the $(a,a)$ hand side extends the domain of the $\left(a,\frac12\right)$ case:

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

therefore, we have some identities with InvT, from the question

Formula 7:

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




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

Formula 8:

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 9:

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 10:

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 11:

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:

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


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


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

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

  • 1
    $\begingroup$ In formula 10 perhaps the first formula should be with minus and the second one with plus? $\endgroup$
    – Anixx
    Jun 19 at 19:41
  • 1
    $\begingroup$ @TymaGaidash I do not know what the correct formulas are, but it seenms that formula $\mp Q^{-1}\left(1,x^{\pm 1}\right)=\ln(x)$ is enough and $Q^{-1}(1,x)=-\ln(x)$ is excessive because it reflects one case of the first formula. So, if you want two formulas, it would make sense to write one with one combination of signs and one with the other combination, without plus-minus sign, or only one fotmula with plus-minus. $\endgroup$
    – Anixx
    2 days ago

This answer is free for anyone to use.


$$\begin{array} {rl} 2^Ab^2-2^T+1&=b \\ 2^Ab^2-b&=2^T-1 \\ 4^Ab^2-2^Ab&=2^A(2^T-1) \\ 4 \cdot 4^Ab^2-4\cdot 2^Ab+1&=4 \cdot 2^A(2^T-1) +1\\ (2 \cdot 2^Ab-1)^2&=4 \cdot 2^A(2^T-1) +1\\ x^2 &= 2^{T+2+A} - 2^{2+A} +1 \\ x^2 &= 2^n - 2^m +1 \end{array} $$ This is the form of an equation of whicht Laszlo Szagay has shown that there is only three "sporadic" solutions:$(n,m,x)\in\{(5,3,5),(7,3,11),(15,3,181) \} $ which is here $(T,A,x)\in\{(2,1,5),(4,1,11),(12,1,181) \} $.
Insert those in the first equation $$\begin{array} {rlrrr} (2 \cdot 2^Ab-1)^2&=4 \cdot 2^A(2^T-1) +1\\ (2 \cdot 2^1b-1)^2&=4 \cdot 2^1(2^2-1) +1&=8 \cdot 3+1&=25=5^2\\ (2 \cdot 2^1b-1)^2&=4 \cdot 2^1(2^4-1) +1&=8 \cdot 15+1&=121=11^2\\ (2 \cdot 2^1b-1)^2&=4 \cdot 2^1(2^{12}-1) +1&=8 \cdot 4095+1&=32761=181^2\\ \end{array}$$ We find, that only in the second equation the value $b$ is an odd integer, namely $b=3$, and since $S=A+T$ we have $$ 2^S = (m+1)(m+1/3) \\ S=5 \implies m=5 \\ 2^5 = (5+1)(5+1/3) = (5+1)/3\cdot (15+1) = 2 \cdot 16 = 32 $$ a valid solutions which is also the only one.

$$ 2 \cdot 2^Ab-1=\sqrt{4 \cdot 2^A(2^T-1) +1}\\ b={\sqrt{4 \cdot 2^A(2^T-1) +1}+1 \over 2 \cdot 2^A}\\ $$

$$ b \overset{?}= \frac 1{2^A} + {2^T -1 \over 1+{2^T -1 \over \frac 1{2^A} + {2^T -1 \over 1+{2^T -1 \over b} }} } \\ $$

On odd $b$: $$ f(b) = b^2 - \{b-1\}_2 \overset{?}=2^S \tag 1 \\ f(b) = 4c^2+4c+1 - \{c\}_2 \overset{?}=2^S \\ b^2-2^S \overset{?}= {b-1\over 2^A} \\ 2^A(b^2-2^S) +1 \overset{?}= b \\ 2^A b \overset{?}= 1+{2^T -1 \over b} \\ b \overset{?}= \frac 1{2^A}+{2^T -1 \over 2^A b} \\ b \overset{?}= \frac 1{2^A}+{2^T -1 \over 2^A \left( \frac 1{2^A}+{2^T -1 \over 2^A b} \right)} \\ b \overset{?}= \frac 1{2^A} + {2^T -1 \over 1+{2^T -1 \over b} } \\ $$

This answer is free for anyone to use


This answer is free for anyone to use.


This answer is free for anyone to use.


This answer is free for anyone to use.


From Marker's model theory an introduction on page 41.

Alternative Proof that the theory of torsion-free divisible Abelian groups are $\ge \aleph_1$ categorical that uses less knowledge of alegebra.

Proposition 2.2.4. The theory of torsion-free divisible Abelian groups is $\kappa$-categorical for all $\kappa > \aleph_0$.

A group $G$ is Abelian if and only if $a + b = b + a$ for all $a$ and $b$ in $G$.

I can impose this condition with the first order sentence:

$$ \forall a b \mathop. a + b = b + a $$

An Abelian group is torsion-free if and only if every element $a$ of finite order is the identity element.

An Abelian group $G$ is divisible if and only if, for all $m$ in $\mathbb{Z} \setminus \{0\}$, $mG = G$.

The proof proceeds by claiming that these groups are in 1:1 correspondence with vector spaces over $\mathbb{Q}$.


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 $\sqrt{14}$


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