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Tuesday, October 4, 2016

Entry 37

There is an elegant continued fraction associated with the Ramanujan G and g functionsGn=21/4η2(τ)η(τ2)η(2τ) gn=21/4η(τ2)η(τ)where τ=n. Recall that (Gngn)8(G8ng8n)=14. So perhaps it is no surprise that they can be expressed by the octic continued fraction. Given the nome q=eπiτ, thenβ(τ)=(1+G12n±1G12n2)1/4=2q1/81+q1+q+q21+q2+q31+q3+where the ± sign changes beyond a bound n. Alternatively, β(τ)=(g12n+g24n+1)1/4. For example, since G1/4=(1+2)1/423/16 this implies the identity(14+21/4(1+2)31421/4(1+2)3)1/4=11+2 and is the value of the continued fraction when q=eπi1/4.

Monday, October 3, 2016

Entry 36

Given the nome p=eπiτ and τ=n, then the Ramanujan functions Gn and gn are
Gn=21/4p1/24k=1,3,5,(1+pk)gn=21/4p1/24k=1,3,5,(1pk) These are ubiquitous in Ramanujan's Notebooks. Well-known values are G5=ϕ1/4 and G25=ϕ, where ϕ is the golden ratio. However, given the nome's square q=e2πiτ, note also the three Weber modular functions,
f(τ)=q1/48n=1(1+qn1/2)f1(τ)=q1/48n=1(1qn1/2)f2(τ)=2q1/24n=1(1+qn) Notice the similarity of the definitions. Mathematica doesn't have built-in functions for these, but fortunately can be expressed by the more familiar Dedekind eta function η(τ) with τ=n, Gn=21/4f(τ)=21/4η2(τ)η(τ2)η(2τ) gn=21/4f1(τ)=21/4η(τ2)η(τ)The two functions Gn and gn obey(Gngn)8(G8ng8n)=14which is consequence of Weber'sf1(τ)8+f2(τ)8=f(τ)8Ramanujan calculated many explicit values for Gn and gn, one of which is the remarkable G125=3.63352G125φ5/4=1+15(1+22/5φ1/5(545+(φ5)3/2+545(φ5)3/2))2 with the reciprocal golden ratio φ=1+520.61803.

Sunday, October 2, 2016

Entry 35

This continues Entry 34. Similar to its more famous cousin, Euler's prime-generating polynomial, the formulaP(n)=6n26n+31 yields consecutive prime values from n=029. Solving for P(n)=0, one gets τ=6+70812=3+1776. Plugging this into β(τ)=((η(2τ)η(3τ)η(τ)η(6τ))6(η(τ)η(6τ)η(2τ)η(3τ))6)2 with Dedekind eta function η(τ), we find that it exactly yields an integerβ(3+1776)=10602such thate2π/6177=10602+9.999992and the level-6 Ramanujan-Sato series\frac{1}{\pi}=\frac{1}{265\cdot1060}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}^3\,\frac{177\cdot7038k+89418}{(-1060^2)^k}with the binomial coefficient \binom{n}{k}.

Entry 34

Here are some optimum prime-generating polynomials \begin{aligned} P(n) &=n^2+n+41\\ P(n) &=2n^2+29\\ P(n) &=2n^2+2n+19\\ P(n) &=3n^2+3n+23\\ P(n) &=5n^2+5n+13\\ P(n) &=6n^2+6n+31\\ P(n) &=7n^2+7n+17 \end{aligned}The first one is the most famous, being Euler's. Expressed as the quadratic P(n) = an^2+bn+c, its discriminant is d=b^2-4ac. Using the values a,d of the above, one gets
\begin{aligned} &e^{\pi/1\,\sqrt{163}} = 640320^3 +743.9999999999992\dots\\ &e^{\pi/2\,\sqrt{232}} = 396^4 -104.0000001\dots\\ &e^{\pi/2\,\sqrt{148}} = (84\sqrt{2})^4 +103.99997\dots\\ &e^{\pi/3\,\sqrt{267}} = 300^3 + 41.99997\dots\\ &e^{\pi/5\,\sqrt{5\times47}} = (18\sqrt{47})^2 + 15.991\dots\\ &e^{\pi/6\,\sqrt{6\times118}} = 1060^2 + 9.99992\dots\\ &e^{\pi/7\,\sqrt{7\times61}} = (39\sqrt{7})^2 + 9.995\dots \end{aligned}These approximations (actually, the exact values of certain eta quotients) can be used as denominators in pi formulas known as Ramanujan-Sato series to be discussed in Entry 35.
 

Entry 33

In a similar manner to Entry 32, given the Chudnovsky algorithm\frac{1}{\pi} = \frac{12}{(640320)^{3/2}} \sum^\infty_{k=0} \frac{(6k)!}{(3k)!\,k!^3} \frac{3\cdot163\cdot1114806k + 13591409}{(-640320^3)^k} we can use the cube root \sqrt[3]{-640320^3}=-640320 as the median,\begin{aligned}\frac{1}{\pi}&=\frac{12}{(640320+12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\frac{163\cdot1114806k+(13591409-8\cdot181)}{(-640320-12)^k}\\ \frac{1}{\pi}&=\frac{12}{(640320-12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\,3^{k-3j}\,\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\,\frac{163\cdot1114806k+(13591409+8\cdot181) }{(-640320+12)^k}\end{aligned} where \binom{n}{k} is the binomial coefficient. These are now level-9 Ramanujan-Sato series

Saturday, October 1, 2016

Entry 32

It is well-known that\frac{1}{\pi} =\frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot70\cdot13\,k+1103}{(396^4)^k}But it turns out we can also use the square root \sqrt{396^4}=\pm396^2 as a median point for two other pi formulas. First express the above as\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{2\cdot58\cdot15015k+72798}{(396^4)^k}then\begin{aligned}\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2-16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798+37)}{(-396^2+16)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\end{aligned} where \binom{n}{k} is the binomial coefficient. Note that they have a beautifully symmetric form and how the same integers (which figure in Pell equations as discussed in Entry 1) appear in all three formulas. These two are level-8 Ramanujan-Sato series.