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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 series1π=12651060k=0(2kk)kj=0(kj)31777038k+89418(10602)kwith the binomial coefficient (nk).

Entry 34

Here are some optimum prime-generating polynomials P(n)=n2+n+41P(n)=2n2+29P(n)=2n2+2n+19P(n)=3n2+3n+23P(n)=5n2+5n+13P(n)=6n2+6n+31P(n)=7n2+7n+17The first one is the most famous, being Euler's. Expressed as the quadratic P(n)=an2+bn+c, its discriminant is d=b24ac. Using the values a,d of the above, one gets
eπ/1163=6403203+743.9999999999992eπ/2232=3964104.0000001eπ/2148=(842)4+103.99997eπ/3267=3003+41.99997eπ/55×47=(1847)2+15.991eπ/66×118=10602+9.99992eπ/77×61=(397)2+9.995These 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 algorithm1π=12(640320)3/2k=0(6k)!(3k)!k!331631114806k+13591409(6403203)k we can use the cube root 36403203=640320 as the median,1π=12(640320+12)3/2k=0(2kk)kj=0(3)k3j(k3j)(2jj)(3jj)1631114806k+(135914098181)(64032012)k1π=12(64032012)3/2k=0(2kk)kj=03k3j(k3j)(2jj)(3jj)1631114806k+(13591409+8181)(640320+12)k where (nk) 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.