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[Updated on Nov.10, Dec.05.2023 and Feb.18.2024] Applying a similar approach followed in these MO questions, for Catalan's and Apery's Constants, i.e. Wilf-Zeilberger proofs starting from Dougall's sum, working together with nonlinear $_2F_1$ and $_3F_2$ hypergeometric transformations, I have found four convergent series for the Lemniscate Constant $s = 2\varpi$ and two for $s=\Gamma(\frac{1}{3})$. As far as I know, they are currently the fastest series to compute such constants with arbitrary precision (they belong to a highly efficient computing class -P2B3 class- specially fitted to apply the binary splitting algorithm).

More details can be found in these links (1), (2) and (3)

I am not sure if some of them has been already published, so the question is very simple. Is any of these series known?

NOTE As of Feb.18.2024 question remains open for Eqs.(1), (2), (3), (7) and (11) below.

We use the following notation, where the constant $s$ is expressed as $$s=\sum_{n=1}^\infty\,\rho^n\cdot\frac{p(n)}{r(n)}\cdot\left[\begin{matrix} a & b & c & ... & z \\ A & B & C & ... & Z \\ \end{matrix}\right]_n=\sum_{n=1}^\infty\frac{p(n)}{r(n)}\cdot\prod_{k=1}^n\frac{r(k)}{q(k)}$$ here $p(n),q(n),r(n)$ are polynomials non vanishing for $n\in\mathbb{N}$, $q(n)$ and $r(n)$ have the same degree $d$ and the convergence ratio $|\rho|$ is the absolute value of the ratio of the leading terms of $r(n)$ and $q(n)$. The ratio of products of Pochhammer's symbols (rising factorials) is written as $$\left[\begin{matrix} a & b & c & ... & z \\ A & B & C & ... & Z \\ \end{matrix}\right]_n=\frac{(a)_n(b)_n(c)_n ... (z)_n}{(A)_n(B)_n(C)_n ... (Z)_n} $$ where the degree $d$ is the number of elements in a row (they are the same for both rows) and $$(w)_n = \frac{\Gamma(w+n)}{\Gamma(w)}=w(w+1)(w+2)...(w+n-1)$$

The computational speed is measured through the binary splitting cost $$ C_s = - \frac{4d}{\log|\rho|}.$$ This allows to (asymptotically) rank, classify and compare different hypergeometric-type algorithms by performance. Series are:


A. For the lemniscate constant

$$2\varpi =\frac{\;\Gamma(\frac{1}{4})^2}{\sqrt{2\pi}}=5.2441151085842396209296791797822388273655...$$a trascendental number, See Finch, Steven R., Mathematical constants, Encyclopedia of Mathematics and Its Applications 94. Cambridge: Cambridge University Press (ISBN 0-521-81805-2/hbk). xx, 602 p. (2003). ZBL1054.00001. (Section 6.1)

I) The first one has a cost $C_s=-4\cdot8/\log\frac{1}{2^{24}}=1.9326.$. It is $$\begin{equation*}2\varpi=\sum_{n=1}^\infty\left(\frac{1}{2^{24}}\right)^n\cdot\frac{P(n)}{R(n)}\cdot\left[\begin{matrix} \frac{1}{8}&\frac{3}{8}&\frac{5}{8}&\frac{7}{8}&\frac{3}{16}&\frac{7}{16}&\frac{11}{16}&\frac{15}{16}\\ \frac{1}{32}&\frac{5}{32}&\frac{9}{32}&\frac{13}{32}&\frac{17}{32}&\frac{21}{32}&\frac{25}{32}&\frac{29}{32}\\ \end{matrix}\right]_n\tag{1}\label{1} \end{equation*}$$ where $$ P(n) = 750599893155840\,n^7 - 2465512126676992\,n^6 + 3305810396971008\,n^5 - 2327426831319040\,n^4 + 919690491432960\,n^3 - 201024828994048\,n^2 + 22012514018112\,n - 897181286400$$$$R(n)=(8n-1)(8n-3)(8n-5)(8n-7)(16n-1)(16n-5)(16n-9)(16n-13)$$

II) The second series gives the reciprocal of this constant with about $3.01$ decimal digits per term. It has a cost $C_s=1.5661$: $$\begin{equation*}\frac{1}{2\varpi}=C_0\cdot\sum_{n=1}^\infty\left(-\frac{2^9}{3^9\cdot7^6}\right)^n\cdot\frac{n^2\,P(n)}{R(n)}\cdot\left[\begin{matrix} \frac{1}{36}&\frac{7}{36}&\frac{13}{36}&\frac{19}{36}&\frac{25}{36}&\frac{31}{36}\\ 1&1&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\ \end{matrix}\right]_n\tag{2}\label{2} \end{equation*}$$ where $C_0=2\cdot3^{9/2}\cdot7^{5/6}$ and $$P(n) = 99446494228488\,n^5 - 296948949253092\,n^4 + 339735211540956\,n^3 - 185806427026662\,n^2 + 48479683290426n - 4840729282291$$ $$R(n)=-(36n-5)(36n-11)(36n-17)(36n-23)(36n-29)(36n-35)$$

III) The third one for this constant, performs pretty fast. It has a cost $C_s=1.5643$: $$\begin{equation*}\frac{1}{2\varpi}=C_1\cdot\sum_{n=1}^\infty\left(\frac{2^9}{11^9}\right)^n\,\frac{n^2\,P(n)}{R(n)}\,\left[\begin{matrix} \frac{1}{36}&\frac{5}{36}&\frac{13}{36}&\frac{17}{36}&\frac{25}{36}&\frac{29}{36}\\ 1&1&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\ \end{matrix}\right]_n\tag{3}\label{3} \end{equation*}$$ where $C_1 = 2\cdot3^{15/4}\cdot11^{7/4}$ and $$P(n) = 17680561647336\,n^5 - 52825631815620\,n^4 + 60473303319276\,n^3 - 33092086224942\,n^2 + 8638260598818\,n - 862864755643$$ $$R(n)=(36n-7)(36n-11)(36n-19)(36n-23)(36n-31)(36n-35)$$

IIIa) [Updated Nov.10.2023] The following fast series to compute the lemniscate constant

$$\begin{equation*}2\varpi=5C_2 \cdot\sum_{n=1}^\infty\left(\frac{1}{7^4\cdot23^4}\right)^n\cdot\frac{n^2\,P(n)}{R(n)}\cdot\left[\begin{matrix} \frac{1}{16}&\frac{5}{16}&\frac{9}{16}&\frac{13}{16}\\ 1&1&\frac{1}{2}&\frac{1}{2}\ \end{matrix}\right]_n\tag{4}\label{4} \end{equation*}$$ where $C_2 = 2^{33/4}\cdot7^{7/4}\cdot23^{7/4}\cdot\pi,\,$ and $$P(n) = 6636032\,n^2 - 6636192\,n + 1659109$$ $$R(n)=(16n-3)(16n-7)(16n-11)(16n-15)$$

has a cost 0.7872 but it needs to compute $\pi$ which has a cost (Chudnovsky) 0.3670. It gives $C_s=1.1542$.

According to Jesús Guillera (pers. comm. Nov. 2023) some of these series can be bisected or trisected. Particularly Eq.(4) can be bisected into a $\,_2F_1$ function as $$\begin{equation*}2\varpi=C_3\cdot\pi\cdot\,_2F_1\left(\frac{1}{8},\frac{5}{8};1;\frac{1}{7^2\cdot23^2}\right)\tag{5}\label{5}\end{equation*}$$ for some algebraic value $C_3$. Eq.(5) is a fast monotone series that has a companion $\,_2F_1$ fast alternating series by applying Euler or Pfaff hypergeometric transformation. This gives $$\begin{equation*}2\varpi=C_4\cdot\pi\cdot\,_2F_1\left(\frac{1}{8},\frac{3}{8};1;-\frac{1}{2^6\cdot3^4\cdot5}\right)\tag{6}\label{6}\end{equation*}$$ being $C_4$ an algebraic value. Note that $7^2\cdot23^2=25921$ and $2^6\cdot3^4\cdot5=25920$ and Eqs. (5) and (6) are dual or twin linearly convergent fast series having practically the same performance. These linked $2\varpi\propto\pi\cdot\,_2F_1$ series are proven by purely hypergeometric methods. In fact, you can start from Eq.(2) here, a $\,_3F_2$ series, and use Clausen's formula to get the (squared) Eq.(6) above. From this, Eq.(5) is obtained applying Euler or Pfaff transformation.

Note also that, if the convergence rate is fixed, Clausen's formula, whenever it can be applied, produces from a fast $\,_3F_2$ series a faster $\,_2F_1$ having a lower computational cost by a factor $\frac{1}{3}$ just by reducing the polynomial degree from $d=3$ to $d=2$.

By Googling "25921 2F1" Eq.(5) is found in Table 4 here where a different proof was obtained by A. Ebisu using special values of Appell's bivariate F1 function and Goursat hypergeometric transformation.

Therefore the bisected series IIIa Eq.(4) is indeed already known which makes a partial answer to this post.

IIIb) [New Nov.10.2023. Updated Dec.05.2023]

Eqs.(5-6) allow to derive the fastest lemniscate constant $2\varpi$ Ramanujan-type series that avoid to compute $\pi$. There are currently two fastest series to compute the lemniscate constant by binary splitting. The proofs of both series are obtained following Guillera's article "A method for proving Ramanujan series for 1/π" using Table 2 for level ℓ = 2, row d = 13 and also "WZ proofs of Ramanujan-type series (via 2F1 evaluations)". The first one was obtained by J. Guillera working from Eq.(6) above. It is found and proven in the Appendix of this last publication -Eq.(48)-.

I have found the other one working from Eq.(5). This is

$$\begin{equation*}\frac{1}{2\varpi}=C_5\cdot\sum_{n=0}^\infty\left(\frac{1}{7^2\cdot23^2}\right)^n\cdot(1728\,n+55)\cdot\left[\begin{matrix} \frac{1}{8}&\frac{5}{8}\\ 1&1\\ \end{matrix}\right]_n\tag{7}\label{7} \end{equation*}$$ with $C_5=5^{3/4}\cdot2^{-3/4}\cdot7^{-5/4}\cdot23^{-5/4}$. This series has a cost $C_s=0.7872$.

The proof follows straightforward from Guillera's formula by means of the mentioned Table 2, level ℓ = 2, row d = 13, using Ebisu's formula -Eq.(5) above and Eq.(A'''.1) in Table 4- and expanding the rhs after applying $z\frac{d}{dz}$ to both sides of Euler's hypergeometric transformation $\,_2F_1(a,b;c;z)=(1-z)^{-a}\,_2F_1(a,c-b;c;\frac{z}{z-1})$ with parameters $[a,b,c] = [\frac{1}{8},\frac{3}{8},1]$, $z=-\frac{1}{25920}$ and $\frac{z}{z-1}=\frac{1}{7^2\cdot23^2}$

Lemniscate Eq.(7) above and Guillera's Eq.(48) have been implemented (Dec.02.2023) in y-cruncher's version 0.8.3 Build 9530 as the default algorithms to compute this constant.


B. For the cube of $\Gamma(\frac{1}{3})$ constant

IV) [Updated Feb.18.2024]

Series is

$$\begin{equation*}\Gamma\left(\frac{1}{3}\right)^3=C_6\cdot\sum_{n=1}^\infty\left(\frac{3^4}{11^4\cdot23^4}\right)^n\cdot\frac{n^2\,P(n)}{R(n)}\cdot\left[\begin{matrix} \frac{1}{24}&\frac{7}{24}&\frac{13}{24}&\frac{19}{24}\\ 1&1&\frac{1}{2}&\frac{1}{2}\ \end{matrix}\right]_n\tag{8}\label{8} \end{equation*}$$ where $C_6 = 2^{7}\cdot6^{1/2}\cdot11^{11/6}\cdot23^{11/6}\cdot\pi^2,\,$ and $$P(n) = 4097152\,n^2 - 4097536\,n + 1024535$$ $$R(n)=(24n-5)(24n-11)(24n-17)(24n-23)$$

This series has cost 0.9020 and $\pi$ (Chudnovsky) provides 0.3670 giving a total cost $C_s=1.2690$.

Following @CarP24's answer -also mentioned by Jesús Guillera (pers. comm. Nov. 2023)- Eq.(8) can be bisected into a $\,_2F_1$ function as $$\begin{equation*}\Gamma\left(\frac{1}{3}\right)^3=C_7\cdot\pi^2\cdot\,_2F_1\left(\frac{1}{12},\frac{7}{12};1;\frac{3^2}{11^2\cdot23^2}\right)\tag{9}\label{9}\end{equation*}$$ for some algebraic value $C_7$. Eq.(9) is a fast monotone series that has a companion $\,_2F_1$ fast alternating series by applying Euler or Pfaff hypergeometric transformation. This gives $$\begin{equation*}\Gamma\left(\frac{1}{3}\right)^3=C_8\cdot\pi^2\cdot\,_2F_1\left(\frac{1}{12},\frac{5}{12};1;-\frac{3^2}{2^9\cdot5^3}\right)\tag{10}\label{10}\end{equation*}$$ being $C_8$ an algebraic value. Note that $11^2\cdot23^2=64009$ and $2^9\cdot5^3=64000$ and Eqs. (8) and (9) are dual or twin linearly convergent fast series having practically the same performance. These $\Gamma\left(\frac{1}{3}\right)^3\propto\pi^2\cdot\,_2F_1$ series are proven by purely hypergeometric methods. For instance, from Eq.(42) here, a $\,_3F_2$ series, use Clausen's formula to get the (squared) Eq.(10) above. From this, Eq.(9) is obtained applying Euler or Pfaff transformation.

As @CarP24 answered, the bisected series IV Eq.(8) is already known here which makes a partial answer to this post.

IVa) [New Feb.18.2024]

Eqs.(9-10) allow to derive slightly faster $\Gamma\left(\frac{1}{3}\right)^3$ Ramanujan-type series that just need to compute $\pi$ instead of $\pi^2$. There are currently two fastest series to compute this constant by binary splitting. The proofs of both series are obtained following Guillera's article "A method for proving Ramanujan series for 1/π" using Table 4 for level ℓ = 1, row d = 7 and also "WZ proofs of Ramanujan-type series (via 2F1 evaluations)". The first one was obtained by J. Guillera working from known Eq.(10) above.

I have found the other one working from Eq.(9) now that @CarP24 has placed it in his answer below. This is

$$\begin{equation*}\frac{\pi}{\Gamma\left(\frac{1}{3}\right)^3}=C_9\cdot\sum_{n=0}^\infty\left(\frac{3^2}{11^2\cdot23^2}\right)^n\cdot(4800\,n+147)\cdot\left[\begin{matrix} \frac{1}{12}&\frac{7}{12}\\ 1&1\\ \end{matrix}\right]_n\tag{11}\label{11} \end{equation*}$$ with $C_9=2^{-1/2}\cdot11^{-7/6}\cdot23^{-7/6}$.

The proof follows from Guillera by means of the mentioned Table 4, level ℓ = 1, row d = 7, using Hakimoglu-Brown formula -Eq.(9) above- and expanding the rhs after applying $z\frac{d}{dz}$ to both sides of Euler's hypergeometric transformation $\,_2F_1(a,b;c;z)=(1-z)^{-a}\,_2F_1(a,c-b;c;\frac{z}{z-1})$ with parameters $[a,b,c] = [\frac{1}{12},\frac{5}{12},1]$, $z=-\frac{9}{64000}$ and $\frac{z}{z-1}=\frac{3^2}{11^2\cdot23^2}$

A non-native implementation of each series has been done for y-cruncher software (a high performance platform to compute a huge number of digits of some classical constants, having set several records in this regard). Custom input files were prepared for each formula and series were tested under this special platform against some known formulas and methods that has been used to break the $2\varpi$ and $\Gamma(\frac{1}{3})$ digits number records. The above formulas outperform providing the fastest binary splitting algorithms to compute these constants.

As it is reported here Eq.(4) on Dec.12.2023 and Eq.(8) on Dec.15.2023 were used to beat up to $10^{12}$ the current record of known decimal digits of $\Gamma\left(\frac{1}{4}\right)$ and $\Gamma\left(\frac{1}{3}\right)$ respectively.

Q: Excluding series IIIa Eq.(4) and IV Eq.(8), is any of these series known?

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    $\begingroup$ This is crazy, you have my upvote $\endgroup$
    – Yuriy S
    Oct 25, 2023 at 0:16
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    $\begingroup$ Incredible work ! $\endgroup$ Oct 25, 2023 at 4:13
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    $\begingroup$ Thanks @User. Lets check the binary splitting cost. [ numberworld.org/y-cruncher/internals/binary-splitting.html ]. For (1) we have Cs = 20/log(8640/3^3/2^2) = 4.564, For (2) we have Cs = 16/log(6635520/4^4) = 1.574. But formula IV) has cost Cs = 0.9020. This one is much faster. II) and III) have costs 1.5661 and 1.5663 respectively. They are very similar to (2) but they do not need to compute Pi which makes about 10% overload. Anyway I will work on them trying to obtain a faster formula. If I get it I will update this post. $\endgroup$ Oct 28, 2023 at 13:37
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    $\begingroup$ @TitoPiezasIII That series has $d=3$ and $\rho=6^6/(3^4\cdot160^3)=9/64000$. It gives $C_s=-4d/\log{\rho}=1.3530$. To compute $\Gamma(\frac{1}{3})$ you have to add the cost to get $\pi$ that is 0.3670 (Chudnovsky). This gives a total $C_s=1.620$. This value corresponds to the total series computing cost by binary splitting. Other floating point operations like Nth roots take no significant time. $\endgroup$ Oct 31, 2023 at 12:58
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    $\begingroup$ @TitoPiezasIII. I know them. The first is not linearly convergent but it is Dougall. It can be accelerated using WZ to produce other highly efficient series. If WZ is applied on this, the best it yields is the same Gosper-Pilehrood series that has $C_s=3.07$ It is named Pilehrood-short here mathoverflow.net/questions/424055/… The second one has $C_s=12/\log8=5.77$ too far from $3.07$. Since $\pi$ is computed ultrafast I guess that $G/\pi$ or $G\cdot\pi$ can produce a faster formula than Gosper-Pilehrood. $\endgroup$ Nov 1, 2023 at 13:08

1 Answer 1

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Regarding the last section B, the gamma 1/3 cube:

$$\,_2F_1\left(\frac{1}{12},\frac{7}{12};1;\frac{3^2}{253^2}\right) = \frac{253^{1/6}\,\Gamma^3\big(\tfrac{1}{3}\big)}{2\pi^2\sqrt{6}} $$

where $253 = 11\times23.\,$ Testing it in Wolfram. Source: https://iamned.com/math.

This is derived by applying a cubic and quadratic hypergeometric transformation to the 2f1 elliptic integral series at a singular value.

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  • $\begingroup$ No wonder that 2F1 looked familiar. That has got to be related to the one in my post, $$_2F_1\left(\frac1{12},\frac7{12};\frac23;\frac{40^3}{253^2}\right) =253^{1/6}\times\frac23$$ especially since this is also connected to singular values. $\endgroup$ Feb 18 at 1:49
  • $\begingroup$ Great!! @CarP24. This $\,_2F_1$ is Eq.(8) bisected (Also noted by J. Guillera in a pers. comm.). It is also Euler or Pfaff Transformation of $z=-\frac{9}{64000}$ series that Tito Piezas III asked in a comment above. In fact $\frac{z}{z-1}=\frac{9}{64009}$. I did not know that it was previously published in that Hakimoglu-Brown site. It gives me an idea to look for a new fast $\Gamma(\frac{1}{3})$ series. I will update this post if I can find it. Thank you $\endgroup$ Feb 18 at 2:54

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