Tuesday, October 4, 2016
Entry 37
There is an elegant continued fraction associated with the Ramanujan G and g functionsGn=2−1/4η2(τ)η(τ2)η(2τ) gn=2−1/4η(τ2)η(τ)where τ=√−n. Recall that (Gngn)8(G8n−g8n)=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+G−12n±√1−G−12n2)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)3−√14−21/4(1+√2)3)1/4=√11+√2 and is the value of the continued fraction when q=eπi√−1/4.
Monday, October 3, 2016
Entry 36
Given the nome p=eπiτ and τ=√−n, then the Ramanujan functions Gn and gn are
Gn=2−1/4p−1/24∞∏k=1,3,5,…(1+pk)gn=2−1/4p−1/24∞∏k=1,3,5,…(1−pk) 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)=q−1/48∞∏n=1(1+qn−1/2)f1(τ)=q−1/48∞∏n=1(1−qn−1/2)f2(τ)=√2q1/24∞∏n=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=2−1/4f(τ)=2−1/4η2(τ)η(τ2)η(2τ) gn=2−1/4f1(τ)=2−1/4η(τ2)η(τ)The two functions Gn and gn obey(Gngn)8(G8n−g8n)=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=3.6335…2G125φ−5/4=−1+1√5(1+22/5φ1/5(5√4−√5+(φ√5)3/2+5√4−√5−(φ√5)3/2))2 with the reciprocal golden ratio φ=−1+√52≈0.61803.
Gn=2−1/4p−1/24∞∏k=1,3,5,…(1+pk)gn=2−1/4p−1/24∞∏k=1,3,5,…(1−pk) 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)=q−1/48∞∏n=1(1+qn−1/2)f1(τ)=q−1/48∞∏n=1(1−qn−1/2)f2(τ)=√2q1/24∞∏n=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=2−1/4f(τ)=2−1/4η2(τ)η(τ2)η(2τ) gn=2−1/4f1(τ)=2−1/4η(τ2)η(τ)The two functions Gn and gn obey(Gngn)8(G8n−g8n)=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=3.6335…2G125φ−5/4=−1+1√5(1+22/5φ1/5(5√4−√5+(φ√5)3/2+5√4−√5−(φ√5)3/2))2 with the reciprocal golden ratio φ=−1+√52≈0.61803.
Sunday, October 2, 2016
Entry 35
This continues Entry 34. Similar to its more famous cousin, Euler's prime-generating polynomial, the formulaP(n)=6n2−6n+31 yields consecutive prime values from n=0→29. 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/6√177=10602+9.999992…and the level-6 Ramanujan-Sato series1π=1265⋅1060∞∑k=0(2kk)k∑j=0(kj)3177⋅7038k+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=b2−4ac.
Using the values a,d of the above, one gets
eπ/1√163=6403203+743.9999999999992…eπ/2√232=3964−104.0000001…eπ/2√148=(84√2)4+103.99997…eπ/3√267=3003+41.99997…eπ/5√5×47=(18√47)2+15.991…eπ/6√6×118=10602+9.99992…eπ/7√7×61=(39√7)2+9.995…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.
eπ/1√163=6403203+743.9999999999992…eπ/2√232=3964−104.0000001…eπ/2√148=(84√2)4+103.99997…eπ/3√267=3003+41.99997…eπ/5√5×47=(18√47)2+15.991…eπ/6√6×118=10602+9.99992…eπ/7√7×61=(39√7)2+9.995…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 algorithm1π=12(640320)3/2∞∑k=0(6k)!(3k)!k!33⋅163⋅1114806k+13591409(−6403203)k we can use the cube root 3√−6403203=−640320 as the median,1π=12(640320+12)3/2∞∑k=0(2kk)k∑j=0(−3)k−3j(k3j)(2jj)(3jj)163⋅1114806k+(13591409−8⋅181)(−640320−12)k1π=12(640320−12)3/2∞∑k=0(2kk)k∑j=03k−3j(k3j)(2jj)(3jj)163⋅1114806k+(13591409+8⋅181)(−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.
Subscribe to:
Posts (Atom)