Stoneham's series

[Gerson W. Barbosa]

(06-11-2024 05:12 PM)Albert Chan Wrote:  I have trouble doing correction for √2 ... it just seems too complicated.

\(\sqrt{2} \approx \displaystyle
\left(2\prod _{k=1}^{n} \left( 1 - \frac {1/4}{(2k-1)^2} \right) \right)
\left({1-\frac{2}
{1+32n+\frac{(4^2-1)^2/(2^2-1)}
{32n+\frac{(8^2-1)^2/(4^2-1)}
{32n+… +\frac{{((4n)^2-1)^2/((2n)^2-1)}}{32n}}}}}
\right)\)

Very nice! I presume finding the pattern of the numerators may have been particular troublesome, as they are not trivial. Your correction factor gives about two or perhaps 25/12 correct decimal digits per iteration:

. n |2*Product(((4k - 2)^2 - 1)/(4k - 2)^2,{k,1,n})*(1 + -2/(32n + 1 + ContinuedFractionK[((4k)^2-1)^2/((2k)^2-1),32n,{k,1,n}]))
.---+---------------------------------------------------------------------------------------------------------------------------
. 1 | 1.4151193633952254641910 | 1067/754
. 2 | 1.4142087816013794108361 | 465863/329416
. 3 | 1.4142135946217028898547 | 43179041/30532192
. 4 | 1.4142135621369493256490 | 99253807/70183040
. 5 | 1.4142135623748906144019 | 664950544227/470191038976
. 6 | 1.4142135623730811143616 | 153235360737661/108353762695168
. 7 | 1.4142135623730951582795 | 3602252403912943/2547177102352384
. 8 | 1.4142135623730950479347 | 620602650775941779/438832342786015232
. 9 | 1.4142135623730950488086 | 15082567717216397339/10664985910549020672
.10 | 1.4142135623730950488016 | 15606594740681145876391/11035528972365946421248


Of course, this is not adequate for computing the decimal digits of π or any other mathematical constant for that matter. Anyway, just as another mathematical curiosity, I thought of combining the Wallis Product and the Stoneham's product for √2/2:

π/2=4/3.16/15.36/35.64/63.100/99.144/143. …

√2/2=3/4.35/36.99/100. …

π/2.√2/2=4/3.3/4.16/15.36/35.35/36.64/63.100/99.99/100.144/143. …

π√2/4=16/15.64/33.144/143. …

π=2.√2.Π[k=1,inf,(4k)^2/((4k)^2-1)]

and

π~2.√2.Π[k=1,n,(4k)^2/((4k)^2-1)]*c,

where 'c' is a correction factor.

Thanks for saving me the trouble to find a correction factor for this one! it turns out that yours for the √2 product almost exactly matches it:

. n |2√2*Product((4k)^2/((4k)^2 - 1),{k,1,n})*(1 + 2/(32n + 15 + ContinuedFractionK[((4k)^2-1)^2/((2k)^2-1),32n + 16,{k,1,n}]))
.---+--------------------------------------------------------------------------------------------------------------------------
. 1 | 3.1412407295336494743353 | 25888/11655*√2
. 2 | 3.1415952167705760328966 | 1863108608/838692855*√2
. 3 | 3.1415926338285612147454 | 2607005335552/1173564727335*√2
. 4 | 3.1415926537455103768956 | 86592236552192/38980201708305*√2
. 5 | 3.1415926535885530258848 | 42678860747884199936/19212237343166192325*√2
. 6 | 3.1415926535898031757455 | 1474308092195358988304384/663671815222966253879175*√2
. 7 | 3.1415926535897931585427 | 25754557013712857799386464256/11593623947422059715601253825*√2
. 8 | 3.1415926535897932391070 | 16248219196766078634894964031488/7314268425672804441304956579225*√2
. 9 | 3.1415926535897932384574 | 3482380024930917685304486445187072/1567621777752447501676104398875275*√2
.10 | 3.1415926535897932384627 | 897305682114824147754072834090261807104/403929473094228867300191362519982221425*√2