Gosper Formula for inv $\pi$, properties.

Question

I need to understand very good how the properties of this formula

$\frac{4}{\pi} = \frac{5}{4} + \sum_{N \geq 1} \left[ 2^{-12N + 1} \times(42N + 5)\times {\binom {2N-1} {N}}^3 \right] $

Taken from the paper "Radian Reduction for Trigonometric Function" (Hanek Payne Algorithm)

Some remarkable properties are stated, specifically these four ones

1. The $k^{th}$ term of the formula is exactly representable in $6k$ bits;
2. The first $n$ terms of the sum can be represented exactly in $12n$ bits;
3. The most significant bit of the $k^{th}$ term has weight at most $2^{1-6k}$ and hence each successive term increases the number of valid bits in the sum by at least $6$;
4. If $12k < m + 1 \leq 12(k+1)$, then the $m^{th}$ bit of $\frac{4}{\pi}$ may be computed using only terms beyond the $k^{th}$.

My questions are: 1. How to prove the formula? 2. How to prove the properties stated above?

PS. I guess with terms the paper means the generic term $a_N$ of the sum...

Answer 1

The formula in question can be rewritten as $$\frac {1}{\pi}=\sum_{j=0}^{\infty} \binom{2j}{j}^3\frac{42j+5}{2^{12j+4}}\tag{1}$$ It was first given by Ramanujan in his famous paper Modular equations and approximations to $\pi$ in 1914.

Luckily Ramanujan gave a proof of the general theory of such series and left the calculations for reader (except for the simplest of the series).

I have presented the details of Ramanujan's theory alongwith proofs in my blog posts (see part 1, part 2, part 3). Here I will present a brief picture and the details of calculation related to the series in question.

Let $k, l$ (called elliptic moduli) be numbers in interval $(0,1)$ and let $K, L, K', L'$ denote the complete elliptic integrals of first kind $$K(k) =\int_{0}^{\pi}\frac{dx}{\sqrt {1-k^2\sin^2x}},\\ K=K(k), L=K(l),k'=\sqrt {1-k^2},l'=\sqrt{1-l^2},K'=K(k'),L'=K(l')\tag{2}$$ It can proved that the as $K'/K$ is a strictly decreasing function of $k$ and maps $(0,1)$ to $(0,\infty) $. Thus if $n$ is a positive integer then for a given $k\in(0,1)$ there is a unique $l\in(0,1)$ such that $L'/L=nK'/K$. Jacobi also proved that in this case $l$ is an algebraic function of $k$ ie there is a polynomial $P(x, y) $ with integer coefficients such that $P(k, l) =0$ (called modular equation of degree $n$) and further the ratio $K/L$ can also be expressed as an algebraic function of $l, k$.

Let $q=\exp(-\pi K'/K) $ denote the nome corresponding to elliptic modulus $k$ and then $q^n=\exp(-\pi L'/L) $ corresponds to $l$. Ramanujan considered the function $$P(q) =1-24\sum_{j=1}^{\infty} \frac{jq^{2j}}{1-q^{2j}}\tag{2}$$ which is related to Dedekind eta function $$\eta(q) =q^{1/24}\prod_{j=1}^{\infty} (1-q^j)\tag{3}$$ via $$P(q) = 12q\frac{d}{dq}\{\log\eta(q^2)\}\tag{4}$$ and proved that $$nP(q^n) - P(q) =\frac{4KL}{\pi^2}F_n(k,l)\tag{5}$$ where $F_n$ is an algebraic function depending on $n$. Ramanujan gave explicit formulas for $F_n$ for many values of $n$.

Another key identity related to $P(q) $ is $$nP(e^{-\pi\sqrt{n}}) +P(e^{-\pi/\sqrt{n}})=\frac{6\sqrt{n}}{\pi}\tag{6}$$ Using $(5),(6)$ Ramanujan established that $$P(e^{-\pi\sqrt{n}}) - \frac{3}{\pi\sqrt{n}}=\left(\frac{2K}{\pi}\right)^2A_n\tag{7}$$ where $A_n$ is an algebraic number and elliptic integral $K$ corresponds to nome $q=e^{-\pi\sqrt{n}}$.

Next Ramanujan considered the well known formula $$\left(\frac{2K}{\pi}\right) ^2=1+\sum_{j=1}^{\infty} \left(\frac{1\cdot 3\cdot 5\dots(2j-1)}{2\cdot 4\cdot 6\dots(2j)}\right) ^3(2kk')^{2j}=\sum_{j=0}^{\infty} \binom{2j}{j}^32^{-6j}(2kk')^{2j}\tag{8}$$ It is also well known that $$\eta(q^2)=2^{-1/3}\sqrt{\frac {2K}{\pi}}(kk')^{1/6}\tag{9}$$ Using $(8),(9)$ we get $$\eta^4(q^2)=\left(\frac{kk'}{4}\right)^{2/3}\sum_{j=0}^{\infty} \binom{2j}{j}^32^{-6j}(2kk')^{2j}$$ Logarithmic differentiation of the above equation with respect to $k$ gives $$P(q) =(1-2k^2)\sum_{j=0}^{\infty} (3j+1)\binom{2j}{j}^32^{-6j}(2kk')^{2j}\tag{10}$$ Using $(7),(8),(10)$ one can get a series for $1/\pi$.

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Let us now do the calculations for $n=7$ and obtain the series in question. We start with the modular equation of degree $7$ given by $$(kl) ^{1/4}+(k'l')^{1/4}=1\tag{11}$$ If $k, l$ correspond to nomes $e^{-\pi\sqrt{7}},e^{-\pi/\sqrt {7}}$ then we have $l=k', k'=l$ and then from $(11)$ we get $kk'=1/16$ or $$k^2(1-k^2)=\frac{1}{256}$$ or $$k^2=\frac{1-\sqrt {1-(1/64)} }{2}=\frac{8-3\sqrt{7}}{16}$$ (we need the smaller root here as the larger root is the value of $k'^2$). Thus from equation $(10)$ we get $$P(e^{-\pi\sqrt {7}})=\frac{3\sqrt{7}}{8}\sum_{j=0}^{\infty}(3j+1)\binom{2j}{j}^32^{-12j}\tag{12}$$ Next Ramanujan derived that $$7P(q^7)-P(q)=\frac{12KL}{\pi}(1+kl+k'l')$$ Putting $q=e^{-\pi/\sqrt{7}}$ we get $$7P(e^{-\pi\sqrt{7}})-P(e^{-\pi/\sqrt {7}})=\frac{27\sqrt {7}}{8}\left(\frac{2K}{\pi}\right)^2\tag{13}$$ where $K$ corresponds to nome $e^{-\pi\sqrt{7}}$. Adding the above to equation $(6)$ (with $n=7$) we get $$P(e^{-\pi\sqrt{7}})=\frac{3}{\pi\sqrt{7}}+\frac{27\sqrt {7}}{112}\left(\frac{2K}{\pi}\right)^2$$ Comparing this equation with $(12)$ and using $(8)$ (with $2kk'=1/8$) we can see that $$\frac{3}{\pi\sqrt{7}}=\frac{3\sqrt {7}}{8}\sum_{j=0}^{\infty} \left(3j+1-\frac{9}{14}\right)\binom{2j}{j}^32^{-12j}$$ or $$\frac{1}{\pi}=\sum_{j=0}^{\infty} \binom{2j}{j}^3\frac{42j+5}{2^{12j+4}}$$

Answer 2

We can prove the formula through the theory of elliptic integrals and modular forms.

That boils down to proving that:

$$ \sum_{n\geq 1}\binom{2n-1}{n}^3 x^n = -\frac{1}{8}+\frac{1}{2\pi^2}\left(K\left(\frac{1-\sqrt{1-64x}}{2}\right)\right)^2 $$ where $K$ is the complete elliptic integral of the first kind.

The remaining parts follow from the fact that $\binom{2n-1}{n}$ behaves like $\frac{4^n}{2\sqrt{\pi}} $ by Stirling's approximation.