Posts Tagged ‘pi’

A new continued fraction for Zeta(3)?

Continuing the discussion from the previous post, Ramanujan also gave a continued fraction for \zeta(3) as,

\zeta(3) = 1+\cfrac{1}{u_1+\cfrac{1^3}{1+\cfrac{1^3}{u_2+\cfrac{2^3}{1+\cfrac{2^3}{u_3 + \ddots}}}}}

where the u_n, starting with n = 1, are given by the linear function,

u_n = 4(2n-1) = 4, 12, 20, 28, \dots

(Notice the difference from the other version since this one has the cubes twice used as numerators.)  Using a similar approach to Apery’s of finding a faster converging version, I found via Mathematica that,

\zeta(3) = \cfrac{6}{v_1 + \cfrac{1^3}{1 + \cfrac{1^3}{v_2 + \cfrac{2^3}{1 + \cfrac{2^3}{v_3 +\ddots}}}}}

where the v_n are now given by the cubic function,

v_n = 4(2n-1)^3 = 4, 108, 500, 1372, \dots

Of course, a more rigorous mathematical proof is needed that indeed the equality holds.


Update:  J.M. from proved that the continued fraction DOES converge to \zeta(3) by connecting it to Apery’s version!  One consequence of his analysis is that Apery’s generating polynomial can be seen as,

(2n-1)(17n^2-17n+5) = n^3 + (n-1)^3 + 4(2n-1)^3

Continued fractions for Zeta(2) and Zeta(3)

It seems there is a nice “pattern” between the continued fractions for the Riemann zeta function,

\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\dots

at s = 2, and s = 3.  First though, a little introduction.

The origins of this function go back to 1644 when, at the tender age of 18, the Italian mathematician Pietro Mengoli (1626-1686) first proposed what would be later known as the Basel Problem, namely to determine the exact value of the sum of the reciprocals of the squares.  Euler would later find that, for n a positive integer, then \zeta(2n) is a rational multiple of \pi^{2n}, with the first case s = 2 as,

\zeta(2) = \frac{\pi^2}{6}

which obviously is irrational.  However, the status of odd s was not as easy to resolve.  It was only in 1979 that Apery proved that \zeta(3) is irrational, which henceforth was called Apery’s constant.

I. Zeta(2)

One of its infinite number of continued fractions can be given as,

m\,\zeta(2) = \cfrac{1}{u_1+\cfrac{1^4}{u_2+\cfrac{2^4}{u_3+\cfrac{3^4}{u_4+\ddots}}}}

where m = \frac{1}{2} and the u_n, starting with n = 1, are generated by,

u_n = 2n-1 = 1, 3, 5, 7,\dots

or simply the odd numbers.  The convergence is slow, but Apery found it can accelerated by using a quadratic function,

u_n = 11n^2-11n+3 = 3, 25, 69, 135,\dots

but now m = \frac{1}{5}.

II. Zeta(3)

Likewise, this also has an infinite number of continued fractions (see Ramanujan’s versions here), but one important form is,

m\,\zeta(3) = \cfrac{1}{v_1-\cfrac{1^6}{v_2-\cfrac{2^6}{v_3-\cfrac{3^6}{v_4-\ddots}}}}

where m = 1 and the v_n, again starting with n = 1, are,

v_n = (n-1)^3+n^3 = (2n-1)(n^2-n+1) = 1, 9, 35, 91, \dots

Apery again found an accelerated version,

v_n = 34n^3-51n^2+27n-5 = (2n-1)(17n^2-17n+5) = 5, 117, 535, 1463, \dots

where now m = \frac{1}{6}, and established that its rate of convergence was such that \zeta(3) could not be a ratio of two integers.

III. Connection between Zeta(2) and Zeta(3)

Define the two sequences,

B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251,\dots

A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, \dots

then it was established that these Apery numbers have the recurrence relations,

n^2 B_n = (11n^2-11n+3)B_{n-1}+(n-1)^2B_{n-2}

n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2}

Interesting that the same polynomials pop up, isnt’ it?  Furthermore, their limiting ratios have the common form,


thus for b = 1, 2,

\lim_{n \to \infty} \frac{B_{n+1}}{B_n} = \left(\frac{1+\sqrt{5}}{2}\right)^5 = 11.0901\dots

\lim_{n \to \infty} \frac{A_{n+1}}{A_n} = \left(1+\sqrt{2}\right)^4 = 33.9705\dots

hence the golden ratio and the silver ratio surprisingly turn up in the continued fractions for \zeta(2) and \zeta(3), respectively.  It is easy to check other sequences,

C_n = \sum_{k=0}^n {\binom n k}^p {\binom {n+k}k}^q

for some small {p, q}, but there are no other recurrence relations similar in form to the two above.

IV. Zeta(5)

There is an orderly non-simple continued fraction for all \zeta(s), with the next odd s as,

m \zeta(5) = \cfrac{1}{w_1-\cfrac{1^{10}}{w_2-\cfrac{2^{10}}{w_3-\cfrac{3^{10}}{w_4-\ddots}}}}

where m = 1 and the w_n are,

w_n = (n-1)^5+n^5 = (2n-1)(n^4-2n^3+4n^2-3n+1) = 1, 33, 275, 1267, \dots

Unfortunately, no one has yet found an accelerated version where the w_n are generated by a 5th degree (or higher) polynomial, and m is rational.

Can you find one?

For further reading, refer to Alfred van der Poorten’s excellent, A Proof That Euler Missed.

Some of Ramanujan’s continued fractions for pi

The digits of pi go on forever apparently with no discernible pattern. However, there are beautifully simple patterns in its (ironically) non-simple continued fraction expansions. Examples are,

\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \ddots} }}}

known by Lord Brouncker (1620-1684), and,

\pi = 3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \ddots} }}

One can see the affinity between the two.  They in fact belong to the same family.  Given complex numbers {n, x} with Re[x] > 0, then,

F(n,x) = \Large{\frac{\Gamma\left(\frac{x+n+1}{4} \right)\Gamma\left(\frac{x-n+1}{4} \right) }{\Gamma\left(\frac{x+n+3}{4} \right)\Gamma\left(\frac{x-n+3}{4} \right)}} = \cfrac{4}{x + \cfrac{1^2-n^2}{2x + \cfrac{3^2-n^2}{2x + \cfrac{5^2-n^2}{2x + \ddots} }}}

where \Gamma(\tau) is the gamma function.  This is Entry 25 in Ramanujan’s Second Notebook (Chapter 12) though this result was also known by Euler. For the case = 0, the continued fraction assumes the form of the examples and the function simplifies as,

F(0,x) = \frac{\Gamma^2\left(\frac{x+1}{4}\right)}{\Gamma^2\left(\frac{x+3}{4}\right)}

For x an odd integer, then (0, x) is a rational multiple of \pi or 1/\pi.  Specifically, for x = {1, 3, 5, 7, 9}, we have,

F(0,x) = \pi,\; \frac{4}{\pi}, \frac{\pi}{4}, \frac{16}{9\pi}, \frac{9\pi}{64}

More generally,

F(0,4m+1) = \big(\frac{(2m)!}{m!^2}\big)^2\, \frac{\pi}{2^{4m}}

F(0,4m+3) = \big(\frac{m!^2}{(2m+1)!}\big)^2\, \frac{2^{4m+2}}{\pi}

See p. 178 of Annie Cuyt’s HandBook of Continued Fractions for Special Functions.  For more examples, see also this article Ramanujan’s Continued Fractions, Apery’s Constant, and more.