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Tuesday, June 10, 2025

Entry 157

Given the Jacobi theta functions ϑn(0,q) which traditionally uses the nome q=eπiτ. Define the modular lambda function α,

α=16(η(τ/2)η(2τ))8+16=(2η(τ/2)η2(2τ)η3(τ))8=(ϑ2(0,q)ϑ3(0,q))4

Then we propose for appropriate τ such as τ=d that the ratios below are radicals,

(ϑ2(0,q)2F1(12,12,1,α))4?=α(ϑ4(0,q)2F1(12,12,1,α))4?=1α(ϑ3(0,q)2F1(12,12,1,α))4?=1

Adding the first two implies the third. Hence, after removing the common denominator

(ϑ2(0,q))4+(ϑ4(0,q))4=(ϑ3(0,q))4

which is known to be true. As eta quotients in the same order above, 

(2η2(2τ)η(τ))4+(η2(τ2)η(τ))4=(η5(τ)η2(τ2)η2(2τ))4

Also, the equalities ϑ3(0,q)=m=qm2=η5(τ)η2(τ2)η2(2τ)which has a cubic version given in Entry 156.

Entry 156

Given the square of the nome, so q=e2πiτ and the Borwein cubic theta functions a(q),b(q),c(q). Define,

β=(3(η(τ/3)η(3τ))3+3)3=(c(q)a(q))3

Then we conjecture,

(c(q)2F1(13,23,1,β))3?=β(b(q)2F1(13,23,1,β))3?=1β(a(q)2F1(13,23,1,β))3?=1

Adding the first two implies the third,

(c(q))3+(b(q))3=(a(q))3

which is a known relationship of the Borwein cubic theta functions. As eta quotients,

(3η3(3τ)η(τ))3+(η3(τ)η(3τ))3=(η3(τ)+9η3(9τ)η(3τ))3

Like in the previous entry, the RHS is also a sum,

a(q)=m,n=qm2+mn+n2=η3(τ)+9η3(9τ)η(3τ)

Entry 155

It turns out we can use the previous entries to parameterize the equation xn+yn=zn for n=2,3,4. Given the square of the nome q=e2πiτ and assume the functions A(q),B(q),C(q). Define

γ=(8(η(τ/2)η(2τ))8+8)2=(C(q)A(q))2

Then we conjecture,

(C(q)2F1(14,34,1,γ)2)2?=γ(B(q)2F1(14,34,1,γ)2)2?=1γ(A(q)2F1(14,34,1,γ)2)2?=1

where C(q),B(q),A(q) are defined by the relation and eta quotients below,

(C(q))2+(B(q))2=(A(q))2

(8η8(2τ)η4(τ))2+(η8(τ)η4(2τ))2=(η8(τ)+32η8(4τ)η4(2τ))2 Note also that A(q)=1+24n=1nqn1+qn=(ϑ2(0,q))4+(ϑ2(0,q))4=η8(τ)+32η8(4τ)η4(2τ) where, in this particular instance, we let the Jacobi theta functions ϑn(0,q) use q=e2πiτ

Monday, June 9, 2025

Entry 154

As an overview of the previous four entries, Ramanujan's theory of elliptic functions to alternative bases uses the hypergeometric function 2F1(a,b;c;z) with a+b=c=1 for the cases a=16,14,13,12. This can be related to the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145 for n=1,2,3,4. Consider the equations, 

2F1(16,56,1,1α1)2F1(16,56,1,α1)=τ12F1(14,34,1,1α2)2F1(14,34,1,α2)=τ22F1(13,23,1,1α3)2F1(13,23,1,α3)=τ32F1(12,12,1,1α4)2F1(12,12,1,α4)=τ4 Then the αn can be solved and expressed in terms of the jn(τ) as discussed in the said entries.

Entry 153

Ramanujan's theory of elliptic functions to alternative bases can be related to the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145. Define,

α4(τ)=16(η(τ)η(4τ))8+16=(2η(τ)η2(4τ)η3(2τ))8

Let α4=α4(τ). Then we conjecture that,

2F1(12,12,1,1α4)2F1(12,12,1,α4)=τ4 as well as

j4(τ)=16α4(1α4)=((η(τ)η(4τ))4+42(η(4τ)η(τ))4)2=(η2(2τ)η(τ)η(4τ))24=(2F1(12,12,1,α4)η2(τ)×η(τ)η(4τ))24/5

Example. Let τ=3. Then α4=(32)4(21)4 solves 2F1(12,12,1,1α4)2F1(12,12,1,α4)=4×3 Alternatively and more familiar

2F1(12,12,1,1λ(r))2F1(12,12,1,λ(r))=r where λ(τ) is the modular lambda function.

Entry 152

We relate Ramanujan's theory of elliptic functions to alternative bases to the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145. Define,

α3(τ)=27(η(τ)η(3τ))12+27=(3(η(τ/3)η(3τ))3+3)3

Let α3=α3(τ). Then we conjecture that,

2F1(13,23,1,1α3)2F1(13,23,1,α3)=τ3 as well as

j3(τ)=27α3(1α3)=((η(τ)η(3τ))6+33(η(3τ)η(τ))6)2=(2F1(13,23,1,α3)η2(τ)×η(τ)η(3τ))24/4

Example. Let τ=3. Then α3=1250(18717121/3+1822/3) solves 2F1(13,23,1,1α3)2F1(13,23,1,α3)=3×3

Entry 151

Ramanujan's theory of elliptic functions to alternative bases can be related to the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145. Define,

α2(τ)=64(η(τ)η(2τ))24+64=(8(η(τ/2)η(2τ))8+8)2

Let α2=α2(τ). Then we conjecture that,

2F1(14,34,1,1α2)2F1(14,34,1,α2)=τ2 as well as

j2(τ)=64α2(1α2)=((η(τ)η(2τ))12+26(η(2τ)η(τ))12)2=(2F1(14,34,1,α2)η2(τ)×η(τ)η(2τ))24/3

Example. Let τ=3. Then α2=13(2+3)2(13+43) solves 2F1(14,34,1,1α2)2F1(14,34,1,α2)=2×3

Entry 150

Ramanujan's theory of elliptic functions to alternative bases considers the hypergeometric function 2F1(a,b;c;z) with a+b=c=1 for the cases a=16,14,13,12. We can relate this to the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145 for n=1,2,3,4. Define,

α1(τ)=12(111728j1(τ))

where j=j1(τ) is the j-function. Let α1=α1(τ). Then we conjecture that,

2F1(16,56,1,1α1)2F1(16,56,1,α1)=τ1 as well as

j1(τ)=432α1(1α1)=((η(τ)η(2τ))8+28(η(2τ)η(τ))16)3=(2F1(16,56,1,α1)η2(τ))24/2

Example. Let τ=3. Then α1=155(1+52)5 solves 2F1(16,56,1,1α1)2F1(16,56,1,α1)=3 since τ1=3i×i=3.

Entry 149

This is the octic overview. Using the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145, if τ are complex quadratics such that jn(τ) is a radical, then the following octics have a solvable Galois group, hence solvable in radicals j1=(x2+5x+1)3(x2+13x+49)xj22=j2(7x4196x31666x23860x+49)+(x2+14x+21)4xj3=(x+1)6(x2+x+7)xj42=7j4(x+1)4+(x+1)7(x7)x

It would be good if the one for j2 can be simplified. They have discriminants (with another factor of D2 suppressed),

D1=77(j11728)4j14D2=77(j2256)4j26D3=77(j3108)3j35D4=77(j464)4j412

Examples. Let τ=1+41/32 so j3(τ)=(43)6. Let τ=1+89/32 so j3(τ)=(103)6. Then (x+1)6(x2+x+7)x=(43)6(x+1)6(x2+x+7)x=(103)6 are octics both solvable in radicals. And also eπ41/3=(43)6+41.993eπ89/3=(103)6+41.99997 There are infinitely many τ but special ones such that jn(τ) are integers can be found in Entry 145.

Entry 148

This is the 7th-deg overview though only results by Klein for the j-function j=j1(τ) are known. In Klein's "On the Order-Seven Transformations of Elliptic Functions", he gave two elegant resolvents of degrees 7 and 8 in pages 306 and 313. Translated to more understandable notation, we have,

x(x2+7(172)x+7(1+72)3)3=j

y8+14y6+63y4+70y27=yj1728

If τ are complex quadratics such that j=j1(τ) is a radical, then the two resolvents have a solvable Galois group, hence solvable in radicals

Example. Let τ=1+1632, then j=6403203 and x(x2+7(172)x+7(1+72)3)3=6403203 is solvable in radicals. There are infinitely many such τ and some can be found in Entry 145. Note also that (y4+14y3+63y2+70y7)2y+1728=(y2+5y+1)3(y2+13y+49)y where the octic on the RHS will appear in the next entry.

Sunday, June 8, 2025

Entry 147

This is the sextic overview. Using the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145, if τ are complex quadratics such that jn(τ) is a radical, then the following sextics have a solvable Galois group, hence solvable in radicals j1=(x2+10x+5)3xj2=(x+1)4(x2+6x+25)xj32=j3(2x6+29x5+85x4+50x3)+54x6(2x1)j4=(x+1)5(x+5)x

with the ones in red by Joachim König. (It would be nice if the one for j3 can be simplified.) They have discriminants,

D1=55(j11728)2j14D2=55(j2256)3j23D3=55(j3108)3j311D4=55(j464)2j44

Examples. Let τ=1+1632 so j1(τ)=6403203. Let τ=582 so j2(τ)=3964. Then (x2+10x+5)3x=6403203(x+1)4(x2+6x+25)x=3964 are sextics both solvable in radicals. There are infinitely many τ but special ones such that jn(τ) are integers can be found in Entry 145.

Entry 146

This is the quintic overview of Entries 140-144. Using the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145, if τ are complex quadratics such that jn(τ) is a radical, then the following simple quintics have a solvable Galois group hence solvable in radicals x5+5x4+40x3=j1x(x5)4=j2x3(x5)2=j3x5+5x=64j4j4x5+5x310x2=j6 Example. Let τ=1+1632 so j1(τ)=6403203. Let τ=582 so j2(τ)=3964. Then x5+5x4+40x3=6403203x(x5)4=3964 are quintics both solvable in radicals. Of course, it is well-known that eπ163=6403203+743.99999999999925eπ58=3964104.00000017There are infinitely many τ but special ones such that jn(τ) are integers can be found in Entry 145.

Entry 145

Summarizing the McKay-Thompson series of the Monster discussed in Entries 140-144, 

j1=((η(τ)η(2τ))8+28(η(2τ)η(τ))16)3j2=((η(τ)η(2τ))12+26(η(2τ)η(τ))12)2j3=((η(τ)η(3τ))6+33(η(3τ)η(τ))6)2j4=((η(τ)η(4τ))4+42(η(4τ)η(τ))4)2=(η2(2τ)η(τ)η(4τ))24j6=((η(τ)η(3τ)η(2τ)η(6τ))3+23(η(2τ)η(6τ)η(τ)η(3τ))3)2

where j1(τ) is the j-function. Let τ be complex quadratics τ=12d or τ=12+d such that the jn(τ) are radicals. For the following special τ, then jn(τ) are integers 

j1(τ)whereτ=dford=1,2,3,4,7,andτ=1+d2ford=1,3,7,11,19,127,43,67,163.

j2(τ)whereτ=d2ford=4,6,10,18,22,58,andτ=1+d2ford=5,7,9,13,25,37.

j3(τ)whereτ=d/32ford=4,8,16,20,andτ=1+d/32ford=5,9,17,25,41,49,89.

j4(τ)whereτ=d2ford=3,7,andτ=1+d2ford=1,2,4.

j6(τ)whereτ=d/32ford=10,14,26,34,andτ=1+d/32ford=7,11,19,31,59.

Saturday, June 7, 2025

Entry 144

Define the McKay-Thompson series of Class 6A for the Monster j6=j6(τ)=((η(τ)η(3τ)η(2τ)η(6τ))3+23(η(2τ)η(6τ)η(τ)η(3τ))3)2 and the quintic x55αx3+10α2xα2=0 where α=1j632. Alternatively (y2+15)2(y5)=32(j632) z5+5z310z2=j6

Conjecture: "If τ is a complex quadratic such that j6=j6(τ) is an algebraic number with j632, then the quintics above have a solvable Galois group."

Example: Let j6(159/32)=10602, then (y2+15)2(y5)=32(1060232) z5+5z310z2=10602 are solvable in radicals.

Entry 143

Define the McKay-Thompson series of Class 4A for the Monster j4=j4(τ)=((η(τ)η(4τ))4+42(η(4τ)η(τ))4)2=(η2(2τ)η(τ)η(4τ))24 and the Bring-Jerrard quintic y5+5y=(64j4j4)1/2

Conjecture: "If τ is a complex quadratic such that j4=j4(τ) is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let j4(127)=212, then y5+5y=63 is solvable in radicals.

Entry 142

Define the McKay-Thompson series of Class 3A for the Monster j3=j3(τ)=((η(τ)η(3τ))6+33(η(3τ)η(τ))6)2 and the Euler-Jerrard quintic x5+5αx2α=0 Alternatively y3(y5)2=j3

Conjecture: "If τ is a complex quadratic such that j3=j3(τ) is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let j3(1+89/32)=3003, then y3(y5)2=3003 is solvable in radicals.

Entry 141

Define the McKay-Thompson series of Class 2A for the Monster j2=j2(τ)=((η(τ)η(2τ))12+26(η(2τ)η(τ))12)2 and the Bring-Jerrard quintic x55αxα=0 Alternatively y(y5)4=j2

Conjecture: "If τ is a complex quadratic such that j2=j2(τ) is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let j2(1210)=124, then y(y5)4=124z55z12=0 are solvable in radicals.

Entry 140

Given the Dedekind eta function η(τ) and define the McKay-Thompson series of Class 1A for the Monster, or better known as the j-function j j=j1(τ)=((η(τ)η(2τ))8+28(η(2τ)η(τ))16)3 and the Brioschi quintic x510αx3+45α2xα2=0 where α=1j1728. Alternatively (y2+20)2(y5)=j1728 z5+5z4+40z3=jConjecture: "If τ is a complex quadratic such that j is an algebraic number j1728, then the quintics above have a solvable Galois group." 

Example. Let j(1+1632)=6403203 and α=16403203+1728, then x510αx3+45α2xα2=0 (y2+20)2(y5)=64032031728 z5+5z4+40z3=6403203 are quintics solvable in radicals. (In fact, they factor into a quadratic and a cubic.)

Entry 139

The general quintic can be reduced to the following one-parameter forms

x510αx3+45α2xα2=0

x55αxα=0

x5+5αx2α=0

x5+5x+(1α64α)1/2=0

x55αx3+10α2xα2=0 with the last found by yours truly. They have neat discriminants

D1=55(11728α)2α8D2=55(1256α)α4D3=55(1108α)α2D4=55(1+64α)2α1D5=55(136α)(132α)α8 The integers (1728,256,108,64) appear in Ramanujan's theory of elliptic functions to alternative bases and we will connect these quintics to the McKay-Thompson series of class 1A, 2A, 3A, 4A, 6A for the Monster in subsequent entries. 

Entry 138

This summarizes the last several entries. Let fundamental discriminant d=4m with class number h(d)=2k and even m=2p for prime p. Previously, p1mod4 and p3mod4 were distinguished, but we can have a more unified approach. Given the modular lambda function λ(τ) and define the three simple functions α(n)=(n+n21)2β(n)=(n22+n21)2γ(n)=(n+n2+1)2 If p3mod4, then 1λ(2p)=α(n)β(n),wheren=2λ+1λ2λ1/41λ If p1mod4, then 1λ(2p)=α(n)γ(n),wheren=λ+12λ1/41λ with λ=λ(τ) for simplicity and α(n),β(n),γ(n) are units. Examples. Let m=2p with class number 4

For p=7,23 1λ(14)=α(n)β(n),n=2(1+2)1λ(46)=α(n)β(n),n=2(13+92) For p=17,41 1λ(34)=α(n)γ(n),n=3(4+17)1λ(82)=α(n)γ(n),n=3(51+841) One can observe that n is an algebraic number of degree half that of the class number. For m=2p with class number 8, examples for p3mod4 were given in Entry 136. For p1mod4 like p=89, then n48886n3+648n210314n+5751=0 and so on for other p.

Friday, June 6, 2025

Entry 137

This continues Entry 136. Recall the function β(n)=(n22+n21)2 and the examples of n which were quartic roots. It turns out these n have additional properties which yield fundamental units Uk though I don't know why.

For p=31, let n±=2(1+2)2(1+32±21+42) or the two real roots of the quartic. Then β(n+)β(n)=U31U62=(1520+27331)(42+31) For p=47, let n±=2(1+2)3(9±29+82) or again the two real roots. Then β(n+)β(n)=U47U94=(48+747)(7322+15147) and so on for p=31,47,79,191,239,431.

Entry 136

Given fundamental discriminants d=4m with class number h(d)=8. Then there are exactly ten m=2p for prime p, namely p1mod4=89,113,233,281 and p3mod4=31,47,79,191,239,431. The modular lambda function λ(2p) for both is a root of a deg-16 equation, but the latter is easier to factor into two deg-8 equations. Define the two simple functions α(n)=(n+n21)2β(n)=(n22+n21)2 It seems for the p3mod4, then 1λ(2p)=α(n)β(n) where α(n),β(n) are octic units and n is just a quartic root given by n=2λ+1λ2λ1/41λ with λ=λ(τ) for simplicity. Examples,

Let p=31 and n=2(1+2)2(1+32+21+42) then 1λ(62)=α(n)β(n)=(n+n21)2(n22+n21)2 Let p=47 and n=2(1+2)3(9+29+82) then 1λ(94)=α(n)β(n)=(n+n21)2(n22+n21)2 and so on for p=31,47,79,191,239,431. P.S. Note that solutions to the Pell-like equation x22y2=p appear in the nested radicals x+y2 above, namely12242=3192282=47

Entry 135

Mathworld has a list of the modular lambda function λ(τ) with the particular case τ=14 as the rather complicated λ(14)=11822(2+2)5+42 +11+82(2+22+25+42) which is approximately 0.011208. It can be calculated in Mathematica or WolframAlpha as ModularLambda[tau]. However, we can simplify and factor that into two quartic units as 1λ(14)=128(8+37)(7+8)(21/4+4+2)8=7960.423255 It is then just a matter of getting the reciprocal and square roots. One can do so similarly for discriminants d=4m with class number h(d)=4 and even m=14,62,142 as described in Entry 134.

Entry 134

Given fundamental discriminants d=4m with class number h(d)=4, there are exactly five m=2p for prime p, namely p=7,17,23,41,71. The cases p=17,41 were in Entry 115 while p=7,23,71 which are p3mod4 will be tackled here. Define the two simple functionsα(n)=(n+n21)2β(n)=(n22+n21)2and letn1=2+22n2=26+182n3=1450+10262Then the first quartic unit can be an eighth power 

\begin{align}\alpha(n_1) &=\frac1{2^8}\Big(\sqrt{0+\sqrt2}+\sqrt{4+\sqrt2}\Big)^8\\ \alpha(n_2)  &=\frac1{2^8}\Big(\sqrt{4+3\sqrt2}+\sqrt{8+3\sqrt2}\Big)^8\\ \alpha(n_3)  &=\frac1{2^8}\Big(\sqrt{36+27\sqrt2}+\sqrt{40+27\sqrt2}\Big)^8 \end{align} while the second is a product of quadratic units 

\begin{align}\beta(n_1) &= U_{7}\,\sqrt{U_{14}}=\big(8+3\sqrt{2}\big) \big(2\sqrt2+\sqrt{7}\big) \\ \beta(n_2)  &= U_{23}\,\sqrt{U_{46}}=\big(25+5\sqrt{23}\big) \big(78\sqrt2+23\sqrt{23}\big) \\ \beta(n_3)  &= U_{71}\,\sqrt{U_{142}}=\big(3480+413\sqrt{71}\big) \big(1710\sqrt2+287\sqrt{71}\big) \\ \end{align} which gives the radical expressions of  \begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n_1)\,\beta(n_1),\quad n_1=2+2\sqrt2\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n_2)\,\beta(n_2),\quad n_2=26+18\sqrt2 \\ \frac1{\lambda(\sqrt{-142})} &= \alpha(n_3)\,\beta(n_3),\quad n_3=1450+1026\sqrt2 \end{align} And since \beta(n) = \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2, then they are also nested radicals that use only \sqrt2

Entry 133

Given fundamental discriminants d=4m with class number h(-d)=2, there are exactly four even m = 6, 10, 22, 58. The most well-known is m=58 because of the near-integer \quad e^{\pi\sqrt{58}} = 396^4-104.00000017\dots and the appearance of 396^4 in the denominator of Ramanujan's famous 1/\pi formula. These m=2p for prime p have other interesting properties. Recall the modular lambda function \lambda(\tau) also discussed in Entry 112 \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8 We focus on m=2p for prime p = 3\,\text{mod}\,4 hence \begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)(\sqrt2+\sqrt3)\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})(7\sqrt2+3\sqrt{11}) \end{align} with fundamental units U_n. However, special d with class number h(-d)=2^k surprisingly can be expressed by nested radicals using only the square root of 2. So,\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= (1+\sqrt2)^2+\sqrt{1+ (1+\sqrt2)^4}\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &=  (1+\sqrt2)^6+\sqrt{1+ (1+\sqrt2)^{12}} \end{align}\quad Similar behavior can also be observed for 2p for p=7,23,71 which now have class number 4.

Tuesday, June 3, 2025

Entry 132

Given _2F_1(a,b;c;z) and Dedekind eta function \eta(\tau) where \tau = \frac{1+n\sqrt{-3}}2 for positive integer n. Then for type a+b=c=\color{blue}{\tfrac56} \begin{align}&\,_2F_1\big(\tfrac12,\tfrac13;\tfrac56;(1-2\delta_1)^2\big),\quad\;\delta_1 = \delta_2\\ &\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;(1-2\delta_2)^2\big),\quad\;\frac1{\delta_2}-1=\sqrt{\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}}\\ &\,_2F_1\big(\tfrac14,\tfrac7{12};\tfrac56;(1-2\delta_3)^2\big),\quad\color{red}{\delta_3 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\;\frac1{\delta_4}-1\,=\,\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}\end{align} For this type, there are infinitely many hypergeometrics such that both (z_1, z_2) in _2F_1(a,b;c;z_1) = z_2 are algebraic numbers when n is a positive integer. Note that  _2F_1\big(\tfrac12,\tfrac13;\tfrac56;z\big) =\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;z\big) so the first form is superfluous. Examples: Let \tau = \frac{1+5\sqrt{-3}}2, _2F_1\Big(\frac13,\frac12;\frac56;\frac45\Big)=\;\frac35\sqrt5 \quad _2F_1\Big(\frac16,\frac23;\frac56;\frac{80}{81}\Big)=\frac35\,(9\sqrt5)^{1/3}

Entry 131

Given _2F_1(a,b;c;z) and Dedekind eta function \eta(\tau) where \tau = \frac{1+n\sqrt{-1}}2 for positive integer n. Then for type a+b=c=\color{blue}{\tfrac34} \begin{align}&\,_2F_1\big(\tfrac14,\tfrac12;\tfrac34;(1-2\gamma_1)^2\big), \quad\frac1{\gamma_1}-1=\sqrt{-\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}}\\ &\,_2F_1\big(\tfrac16,\tfrac7{12};\tfrac34;(1-2\gamma_2)^2\big),\quad\color{red}{\gamma_2 =\,?} \\ &\,_2F_1\big(\tfrac18,\tfrac58;\tfrac34;(1-2\gamma_3)^2\big),\quad\frac1{\gamma_3}-1\,=\, -\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}\\ &\,_2F_1\big(\tfrac1{12},\tfrac23;\tfrac34;(1-2\gamma_4)^2\big),\quad\color{red}{\gamma_4 =\,?}\end{align} For this type, there are infinitely many hypergeometrics such that both (z_1, z_2) in _2F_1(a,b;c;z_1) = z_2 are algebraic numbers when n is a positive integer. Examples: Let \tau = \frac{1+5\sqrt{-1}}2

_2F_1\Big(\frac14,\frac12;\frac34;\frac{80}{81}\Big)=\frac95

\quad _2F_1\Big(\frac18,\frac58;\frac34;\frac{25920}{25921}\Big)=\frac35\,161^{1/4}

Sunday, June 1, 2025

Entry 130

Given _2F_1(a,b;c;z) and j-function j = j(\tau) where \tau = \frac{1+n\sqrt{-3}}2 for positive integer n. Then for type a+b=c=\color{blue}{\tfrac23} \begin{align}&\,_2F_1\big(\tfrac14,\tfrac5{12};\tfrac23;(1-2\beta_1)^2\big),\qquad \color{red}{\beta_1 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac12;\tfrac23;(1-2\beta_2)^2\big),\qquad \frac{1}{\beta_2}-1=\sqrt{\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}}\\ &\,_2F_1\big(\tfrac18,\tfrac{13}{24};\tfrac23;(1-2\beta_3)^2\big),\qquad \color{red}{\beta_3 =\,?} \\ &\,_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;(1-2\beta_4)^2\big),\qquad \frac{1}{\beta_4}-1=\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}\\\end{align} For this type, there are infinitely many hypergeometrics such that both (z_1, z_2) in _2F_1(a,b;c;z_1) = z_2 are algebraic numbers when n is a positive integer. Examples: Let \tau = \frac{1+3\sqrt{-3}}2, _2F_1\Big(\frac16,\frac12;\frac23;\frac{125}{128}\Big) =\frac43\times2^{1/6} _2F_1\Big(\frac1{12},\frac7{12};\frac23;\frac{64000}{64009}\Big) =\frac23\times253^{1/6} Let \tau = \frac{1+5\sqrt{-3}}2, _2F_1\left(\frac16,\frac12;\frac23;\Big(\frac45\Big)^2\Big(\frac{15-\sqrt5}{11}\Big)^3\right) =\frac35(5+4\sqrt5)^{1/6}

Entry 129

Given _2F_1(a,b;c;z) where a+b=c. The previous four (Entries 125-128) discuss closed-forms and can be neatly summarized as type a+b=c=\color{blue}{\tfrac12}  \begin{align}&\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\qquad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\qquad \tau_1=n\sqrt{-4}\\ &\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\qquad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\qquad \tau_2=n\sqrt{-3}\\ &\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\qquad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\qquad \tau_3=n\sqrt{-2}\\ &\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad\; \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\qquad\tau_4=n\sqrt{-1}\end{align} For this type, there are infinitely many hypergeometrics such that both (z_1, z_2) in _2F_1(a,b;c;z_1) = z_2 are algebraic numbers when n is a positive integer. However, for the parameters c=\frac12, \frac23,\frac34. \frac56, it seems only the first is easy. For type a+b=c=\color{blue}{\tfrac23}, it's more challenging and will be discussed in the next entry.

Entry 128

Assume \tau=n\sqrt{-\color{blue}{4}} for some positive integer n. Given _2F_1(a,b;c;z) where a+b=c=\frac12 for the case a=\tfrac1{4}. Let z_1 = (1-2w)^2 where w is w=\frac{16}{16+\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8} Then (z_1, z_2) are algebraic numbers in

_2F_1\left(\tfrac14,\tfrac14;\tfrac12;z_1\right) = z_2

Examples: 

If n=2 so \tau=2\sqrt{-4}, then,

_2F_1\left(\frac14,\frac14;\frac12;\,\frac{9\,(3+\sqrt2)^2}{(1+\sqrt2)^6}\right)=\frac38\big(2+\sqrt2\big) 

If n=3 so \tau=3\sqrt{-4}, then,

\quad _2F_1\left(\frac14,\frac14;\frac12;\,\frac{8\sqrt3}{(2+\sqrt3)^2}\right)=\frac23\big(3+2\sqrt3\big)^{1/2}

Saturday, May 31, 2025

Entry 127

Assume \tau=n\sqrt{-\color{blue}{3}} for some positive integer n. Given _2F_1(a,b;c;z) where a+b=c=\frac12 for the case a=\tfrac1{6}. Let z_1 = (1-2w)^2 where w is w=\frac{27}{27+\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12}} Then (z_1, z_2) are algebraic numbers in

_2F_1\left(\tfrac16,\tfrac13;\tfrac12;z_1\right) = z_2

Example: 

If n=2 so \tau=2\sqrt{-3}, then,

_2F_1\left(\frac16,\frac13;\frac12;\,\frac{25}{27}\right)=\frac{3\sqrt3}{4} 

If n=4 so \tau=4\sqrt{-3}, then,

\qquad _2F_1\left(\frac16,\frac13;\frac12;\,\frac{45^2\,(\sqrt2+\sqrt3)^2}{(1+\sqrt6)^8}\right)=\frac58\big(1+\sqrt6\big)

Entry 126

Assume \tau=n\sqrt{-\color{blue}{2}} for some positive integer n. Given _2F_1(a,b;c;z) where a+b=c=\frac12 for the case a=\tfrac1{8}. Let z_1 = (1-2w)^2 where w is w=\frac{64}{64+\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24}} Then (z_1, z_2) are algebraic numbers in

_2F_1\left(\tfrac18,\tfrac38;\tfrac12;z_1\right) = z_2

Examples: 

If n=2 so \tau=2\sqrt{-2}, then,

_2F_1\left(\frac18,\frac38;\frac12;\frac{(70\sqrt{2})^2\,(1+\sqrt2)^2}{(2+3\sqrt2)^6}\right)=\frac34\left(\frac{2+3\sqrt2}2\right)^{1/2}

If n=3 so \tau=3\sqrt{-2}, then,

_2F_1\left(\frac18,\frac38;\frac12;\frac{2400}{2401}\right)=\tfrac23\cdot7^{1/2}

Entry 125

Assume \tau=n\sqrt{-\color{blue}{1}} for some positive integer n. Given _2F_1(a,b;c;z) where a+b=c=\frac12 for the case a=\tfrac1{12}. Let j(\tau) be the j-function and z_1 = (1-2w)^2 where w is a root of \frac{12^3}{4w(1-w)} =j(\tau) Or more simply z_1=1-\frac{1728}{j(\tau)}, then (z_1, z_2) are algebraic numbers in

\,_2F_1\left(\tfrac1{12},\tfrac5{12};\tfrac12;z_1\right) = z_2

Examples: 

If n=2 so \tau=2\sqrt{-1} and j(\tau)=66^3, then,

_2F_1\left(\frac1{12},\frac5{12};\frac12;\frac{1323}{1331}\right)=\frac34\cdot11^{1/4} 

If n=3 so \tau=3\sqrt{-1}, then,

\qquad _2F_1\left(\frac1{12},\frac5{12};\frac12;\frac{(14\sqrt6)^2\,(72+43\sqrt3)}{(21+20\sqrt3)^3}\right)=\frac23(21+20\sqrt3)^{1/4}

Entry 124

In Entry 123, the 24th power of the golden ratio \phi and Gauss' constant G was discussed. We will use it again and connect it to the Dedekind eta function \eta(\tau) and Watson's triple integral \begin{align}I_1 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x\cos y\cos z}\\ &=\frac{\Gamma^4(\frac{1}{4})}{4\pi^3}\\ &= 2\,G^2 = 4\,\eta(i)^4 = 1.393203\dots\end{align} where I_1 =\frac{\,25\phi^6}{\sqrt{\phi^{24}-4}}\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \left(\frac{-\phi^{16}}{4(\phi^{24}-4)}\right)^{3n}=1.393203\dots Notice that the 24th power of the golden ratio is off by 4.

Entry 123

In the previous entry, we had the unusual hypergeometric function, call it \beta \beta =\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) = 2^{1/4}\left(\Big(\tfrac{1+\sqrt5}2\Big)^{12}-\sqrt{\Big(\tfrac{1+\sqrt5}2\Big)^{24}-1}\right)^{1/4}G with Gauss's constant G. Ramanujan found the unusual equality \sqrt[8]{1+\sqrt{1-\left(\tfrac{-1+\sqrt{5}}{2}\right)^{24}}} = \tfrac{-1+\sqrt{5}}{2}\,\left(\tfrac{1+\sqrt[4]{5}}{\sqrt{2}}\right) and one can notice the 24th powers in both cases. After some manipulation, we find the similar \sqrt[8]{\left(\tfrac{1+\sqrt{5}}2\right)^{12}-\sqrt{\left(\tfrac{1+\sqrt{5}}2\right)^{24}-1}} = \sqrt{\tfrac{1+\sqrt{5}}2}\left(\tfrac{-1+\sqrt[4]{5}}{\sqrt{2}}\right)\quad therefore \beta =\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) = 2^{1/4}\left(\tfrac{1+\sqrt5}2\right) \left(\tfrac{-1+\sqrt[4]{5}}{\sqrt{2}}\right)^2 G

Entry 122

Given Gauss' constant G, the elliptic integral singular value K(k_r), \begin{align}G &=\frac{\sqrt2}{\pi} K(k_1) = \frac2{\pi}\int_0^{\pi/2}\frac{d\,\theta}{\sqrt{1+\sin^2\theta}}\\ &= \tfrac1{\sqrt2}\,_2F_1\big(\tfrac12,\tfrac12;1;\tfrac12\big) = \,_2F_1\big(\tfrac12,\tfrac12;1;-1\big)\\ &=(2\pi)^{-3/2}\,\Gamma^2\big(\tfrac14\big) = 0.834626\dots\end{align} and modular lambda function \lambda(\tau) calculated by Mathematica as ModularLambda[tau]. The two hypergeometrics can be expressed as \tfrac1{\sqrt2}\,_2F_1\big(\tfrac12,\tfrac12;1;\lambda(i)\big) = \,_2F_1\big(\tfrac12,\tfrac12;1;\lambda(1+i)\big) = G while more complicated ones are \begin{align}\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/1\big)\Big) &= 2^{1/4}G\\ \,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/3\big)\Big) &= 6^{1/4}\left((1+\sqrt{3})^2-\sqrt{(2+\sqrt3)^4-1}\right)^{1/4}G\\ \quad\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) &= 2^{1/4}\left(\Big(\tfrac{1+\sqrt5}2\Big)^{12}-\sqrt{\Big(\tfrac{1+\sqrt5}2\Big)^{24}-1}\right)^{1/4}G\end{align} and so on, with fundamental units U_3 = 2+\sqrt3 and U_5= \frac{1+\sqrt5}2, and where the \lambda(\tau) are actually radicals. Curiously, \lambda(i)\ = \lambda\big(\sqrt{3+4i}\big) = \frac12.

Entry 121

This is the case a=\frac13 of {_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a} we have \frac{1}{48^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+\color{blue}{4}x^3}}=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\color{blue}{4}\big)= \frac3{5^{5/6}}

\frac{1}{48^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+\color{blue}{27}x^3}}=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\color{blue}{27}\big)=\frac{4}{7} (To be continued.) 

Friday, May 30, 2025

Entry 120

This is the case a=\frac14 of {_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a} we have \frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{3}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}3\big) = \frac{2}{3^{3/4}}

\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{80}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}{80}\big) = \frac35 (To be continued.) 

Entry 119

This is the case a=\frac16 of {_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a} we have \frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{\tfrac{125}3}x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{\tfrac{125}{3}})=\frac{2}{3^{5/6}}

\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{2^7\phi^9}\, x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{2^7\phi^9})=\frac{3}{5^{5/6}}\phi^{-1} with golden ratio phi. (To be continued.) 

Thursday, May 29, 2025

Entry 118

If one has a palindromic quartic of form z^4-abz^3+(a^2+b^2-2)z^2-abz+1=0 then its roots can be factored as roots (x,y) of quadratics x^2+ax+1=0\\ y^2+by+1= 0 z = xy = \left(\tfrac{-a+\sqrt{a^2-4}}2\right) \left(\tfrac{-b+\sqrt{b^2-4}}2\right) hence are products of quadratic units. (To be continued.)

Entry 117

For fundamental discriminants d=4m with class number h(-d)=16, there are exactly 60 m that are even. The largest is m = 3502 = 2\times17\times103 hence has 2^3 = 8 divisors. But this set has no m with 32 divisors so it seems one can't express their modular lambda function \lambda(\sqrt{-m}) with 16 quadratic units. (Unlike for h(-d)=8 where we can express a few \lambda(\sqrt{-m}) with 8 quadratic units.) However, using another function, we can have four quartic units. Given the nome q = e^{\pi i\tau}, \tau=\sqrt{-n}, and the Ramanujan G and g functions \begin{align}2^{1/4}G_n &= q^{-\frac{1}{24}}\prod_{k>0}(1+q^{2k-1}) = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)}\\ 2^{1/4}g_n &= q^{-\frac{1}{24}}\prod_{k>0}(1-q^{2k-1}) = \frac{\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\end{align} discussed in Entry 116. Then \color{red}u  = (g_{3502})^4 = \small\left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^4 = \big(a+\sqrt{a^2-1}\big)^2 \big(b+\sqrt{b^2-1}\big)^2 \big(c+\sqrt{c^2-1}\big) \big(d+\sqrt{d^2-1}\big) \approx 1.43\times10^{13} where \tau = \sqrt{-3502} and (a,b,c,d) are \begin{align}a &= \tfrac{1}{2}(23+4\sqrt{34})\\ b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\ c &= (429+304\sqrt{2})\\ d &= \tfrac{1}{2}(627+442\sqrt{2})\end{align}A version of this was first found by Daniel Shanks in 1980 (Quartic Approximations for Pi) but this one is slightly different with smaller integers since I simplified the first two expressions as squares. These radicals imply a very close approximation to pi,\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2\color{red}u)^6+24\big) which differs by just 10^{-161}.

Wednesday, May 28, 2025

Entry 116

For fundamental discriminants d=4m with class number h(-d)=8, there are exactly 29 m that are even. The three m = 210, 330, 462 were discussed in Entry 114 since they have special properties. However, the four largest are m = 598, 658, 742, 862. Given the nome q = e^{\pi i\tau}, we can use the Ramanujan G and g functions \begin{align}2^{1/4}G_n &= q^{-\frac{1}{24}}\prod_{k>0}(1+q^{2k-1}) = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)}\\ 2^{1/4}g_n &= q^{-\frac{1}{24}}\prod_{k>0}(1-q^{2k-1}) = \frac{\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\end{align} with \tau=\sqrt{-n} where Ramanujan uses G_n and g_n for odd n and even n, respectively. For these four m = n, then \begin{align}(g_{598})^2 &=  \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2 = \Big(6+\sqrt{26}+\sqrt{(6+\sqrt{26})^2-1}\Big)(1+\sqrt2)^2\Big(\tfrac{3+\sqrt{13}}2\Big)\\ (g_{658})^2 &=  \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2 = \, ?? \\ (g_{762})^2 &= \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2 = \Big(\tfrac{11+\sqrt{106}}2+\sqrt{\big(\tfrac{11+\sqrt{106}}2\big)^2-1}\Big)\,(1+\sqrt2)^2\,\Big(\tfrac{7+\sqrt{53}}2\Big)\\ (g_{862})^2 &=  \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2  = \, ??\end{align} I know the octics for the other two, though I don't know how to factor them into quartic units.

Entry 115

For fundamental discriminants d=4m with class number h(-d)=4, there are exactly twelve m that are even. In Entry 113, the seven with 8 divisors were discussed. The remaining five are m = 14, 34, 46, 82, 142 which are of form m=2p for prime p=7, 17, 23, 41, 71. From experience, p\equiv 1\, (\text{mod}\, 4) are more well-behaved, hence for m=34,82 \begin{align}\frac1{\sqrt{\lambda(\sqrt{-34})}} &= (1+\sqrt2)^2\sqrt{35+6\sqrt{34}} \left(\sqrt{\frac{5+\sqrt{17}}4}+\sqrt{\frac{1+\sqrt{17}}4}\right)^4 \\ \frac1{\sqrt{\lambda(\sqrt{-82})}} &\,=\, (1+\sqrt2)^4\, (9+\sqrt{82})\,\left(\sqrt{\frac{7+\sqrt{41}}2}+\sqrt{\frac{5+\sqrt{41}}2}\right)^4 \end{align}\quad For m = 14, 46,142, presumably they may be products of two quartic units  \frac1{\sqrt{\lambda(\sqrt{-m})}} = \big(a+\sqrt{a^2\pm1}\big)\big(b+\sqrt{b^2\pm1}\big) where (a,b) are roots of quadratics, but I haven't figured out the correct values yet.

Entry 114

This continues Entry 113. Recall the modular lambda function \lambda(\tau) \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8calculated in Mathematica as ModularLambda[tau]. We now chose fundamental discriminants d=4m with class number h(-d)=8 for even m with 16 divisors. There are only three, namely m = 210, 330, 462, hence \begin{align}210 &= 2\times3\times5\times7\\ 330 &= 2\times3\times5\times11\\ 462 &= 2\times3\times7\times11\end{align} Ramanujan found m=210, though I'm not sure he did for the other two, so I went ahead and found them \small\begin{align}\frac1{\sqrt{\lambda(\sqrt{-210})}} &= (1+\sqrt{2})^2 (2+\sqrt{3}) (8+3\sqrt{7}) (3+\sqrt{10})^2 (4+\sqrt{15})^2 (6+\sqrt{35}) (\sqrt{6}+\sqrt{7})^2 (\sqrt{14}+\sqrt{15}) \\ \frac1{\sqrt{\lambda(\sqrt{-330})}} &= (1+\sqrt{2})^2(2+\sqrt{3})^3(3+\sqrt{10})^2(10+3\sqrt{11})(4+\sqrt{15})(65+8\sqrt{66}) (\sqrt{44}+\sqrt{45})^2(\sqrt{54}+\sqrt{55}) \\ \frac1{\sqrt{\lambda(\sqrt{-462})}} &= (2+\sqrt{3})^2(5+2\sqrt{6})^2 (8+3\sqrt{7})^2(10+3\sqrt{11})(15+4\sqrt{14})(76+5\sqrt{231})(7\sqrt{2}+3\sqrt{11})^2(\sqrt{21}+\sqrt{22})\end{align} They are products of eight fundamental units U_n. To find products of sixteen fundamental units U_n, I checked class number h(-d)=16 for even m with 32 divisors from the class number list. Surprisingly, there are none. So it seems the pattern of \lambda(\sqrt{-m}) as a product of \color{blue}{2^k} quadratic units U_n where class number h(-4m)=\color{blue}{2^k} stops at 8. Note that \quad e^{\pi\sqrt{462}} = \frac{16}{\lambda\big(\sqrt{-462}\big)}-8.0000000000000000000000000000094\dots or for 29 zeros.

Entry 113

This continues Entry 112. Recall the modular lambda function \lambda(\tau) \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8 We chose fundamental discriminants d=4m with class number h(-d)=4 for even m with eight divisors, hence only m = 30, 42, 78, 102, and m =70, 130, 190 which are evenly divisible by 3 and 5, respectively.  \begin{align}\frac1{\sqrt{\lambda(\sqrt{-30})}} &= (2+\sqrt3)(5+2\sqrt6)(4+\sqrt{15})\sqrt{11+2\sqrt{30}}\\  \frac1{\sqrt{\lambda(\sqrt{-42})}} &= (2+\sqrt3)^2(1+\sqrt2)^2(8+3\sqrt{7})\sqrt{13+2\sqrt{42}}\\ \frac1{\sqrt{\lambda(\sqrt{-78})}} &= (2+\sqrt3)^3(5+2\sqrt6)(25+4\sqrt{39})\sqrt{53+6\sqrt{78}}\\ \frac1{\sqrt{\lambda(\sqrt{-102})}} &=(2+\sqrt3)^2(5+2\sqrt6)^2(50+7\sqrt{51})\sqrt{101+10\sqrt{102}}\end{align} as well as \begin{align}\frac1{\sqrt{\lambda(\sqrt{-70})}} &= (8+3\sqrt7)(15+4\sqrt{14})(6+\sqrt{35})\sqrt{251+30\sqrt{70}}\\  \frac1{\sqrt{\lambda(\sqrt{-130})}} &= (1+\sqrt2)^4\,(3+\sqrt{10})^2\,(5+\sqrt{26})\,(57+5\sqrt{130})\\ \frac1{\sqrt{\lambda(\sqrt{-190})}} &= (170+39\sqrt{19})(37+6\sqrt{38})(39+4\sqrt{95})\sqrt{52021+3774\sqrt{190}}\end{align}

For the last, note that U_{190}=52021+3774\sqrt{190} = (51\sqrt{10}+37\sqrt{19})^2 and similarly for other U_n with composite n but I chose to retain the a+b\sqrt{n} form. Most of these do not appear in Mathworld's list. Arranged by class number, one can see some patterns like the m have eight divisors and \lambda(\tau) is a product of four U_n. Next are m that have sixteen divisors and \lambda(\tau) is a product of eight U_n.

Entry 112

Regarding the previous two entries, the more famous function with evaluations that involve fundamental units U_n is the modular lambda function \lambda(\tau). Define the McKay-Thompson series of class 4C for the Monster (A007248) j_{4C}(\tau) = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8+16 = \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\right)^8 compare the RHS to \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8 which solves \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)} = -\tau\,i We chose fundamental discriminants d=4m with class number h(-d)=2 for even m = 6, 10, 22, 58, hence \begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)\sqrt{5+2\sqrt6}\\ \frac1{\sqrt{\lambda(\sqrt{-10})}} &= U_2^3\,U_{10} = (1+\sqrt2)^2(3+\sqrt{10})\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})\sqrt{197+42\sqrt{22}}\\ \frac1{\sqrt{\lambda(\sqrt{-58})}} &=\, U_2^6\,U_{58}\; =\; (1+\sqrt2)^6(99+13\sqrt{58})\end{align} Note that these m have four divisors and \lambda(\tau) is a product of two U_n. Also \begin{align}U_6 &= 5+2\sqrt6 = (\sqrt2+\sqrt3)^2\\ U_{22} &= 197+42\sqrt{22} = (7\sqrt2+3\sqrt{11})^2\end{align} can be expressed as squares, which simplify things. For the next entry, the m have eight divisors and \lambda(\tau) is a product of four U_n.

Entry 111

Given q = e^{2\pi i \tau} and define the McKay-Thompson series of Class 10E for Monster (A138516) j_{10E}(\tau) =  \frac{\eta(2\tau)\,\eta^5(5\tau)}{\eta(\tau)\,\eta^5(10\tau)} + 1= \left(\frac{\eta^2(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta^2(10\tau)}\right)^2 On a hunch, I decided to test this since it seems similar to Class 6E of the previous entry. It turns out j_{10E}(\tau) is also product of fundamental units U_n for appropriate \tau. (Here is a sample Wolfram calculation for U_5.) Let d=20m with class number h(-d)=4 for even m = 6,14,26,38 and odd m=17\begin{align}j_{10E}\big(\tfrac{\sqrt{-6/5}}2\big) &= U_2\, U_ {10}\\ j_{10E}\big(\tfrac{\sqrt{-14/5}}2\big) &= U_2^3\, U_{10}\\ j_{10E}\big(\tfrac{\sqrt{-26/5}}2\big) &= U_{13}^3\, U_{65}\\ j_{10E}\big(\tfrac{\sqrt{-38/5}}2\big) &= U_2^4\, U_5^9 \\ j_{10E}\big(\tfrac{1+\sqrt{-17/5}}2\big) &= -U_5^6\, U_{17}\end{align}

Entry 110

Given q = e^{2\pi i \tau}. Define the McKay-Thompson series of Class 6E for Monster (A128633) \begin{align}j_{6E}(\tau) &=\frac1{\big(\text{K}_6(q)\big)^3}+1 \\ &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4\end{align} where \text{K}_6(q) is the cubic continued fraction. Just like the modular lambda function \lambda(\tau), it seems j_{6E}(\tau) is also a product of fundamental units U_n for appropriate \tau. (Here is a sample Wolfram calculation for U_5.) We choose d=12m with class number h(-d)=4 for even m = 10, 14, 26, 34 (also found in the previous entry)

\begin{align}j_{6E}\big(\tfrac{\sqrt{-10/3}}2\big) &= U_2^2\,U_5^6 \\ j_{6E}\big(\tfrac{\sqrt{-14/3}}2\big) &= U_6\,U_{14} \\ j_{6E}\big(\tfrac{\sqrt{-26/3}}2\big) &= U_2^4\,U_{26}^2 \\ j_{6E}\big(\tfrac{\sqrt{-34/3}}2\big) &= U_2^6\,U_{17}^2\end{align} and odd m = 7, 11, 19, 31, 59

\begin{align}j_{6E}\big(\tfrac{1+\sqrt{-7/3}}2\big) &= U_3\sqrt{U_{21}^3} \\ j_{6E}\big(\tfrac{1+\sqrt{-11/3}}2\big) &= U_6\sqrt{U_{33}} \\ j_{6E}\big(\tfrac{1+\sqrt{-19/3}}2\big) &= U_3^3\sqrt{U_{57}} \\ j_{6E}\big(\tfrac{1+\sqrt{-31/3}}2\big) &= U_3^3\sqrt{U_{93}^3} \\ j_{6E}\big(\tfrac{1+\sqrt{-59/3}}2\big) &= U_2^6\sqrt{U_{177}} \end{align} Why this factors nicely I don't know. Also, some fundamental units U_n for composite n may actually be squares. For example, \begin{align}U_{21} &= \tfrac{5+\sqrt{21}}2=\Big(\tfrac{\sqrt{3}+\sqrt{7}}2\Big)^2\\ U_{33} &= 23+4\sqrt{33}=\big(2\sqrt{3}+\sqrt{11}\big)^2\end{align} and so on, hence these \sqrt{U_n} simplifies a bit.

Tuesday, May 27, 2025

Entry 109

Let q = e^{2\pi i \tau} and Dedekind eta function \eta(\tau). Define the following \quad\text{K}_{12}(\tau) = \text{K}_{12}(q) = q\,\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\quad It is connected to the mod 6 version, or the cubic continued fraction \; \text{K}_{6}(\tau) = \text{K}_{6}(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\quad

by the quadratic relation 

\frac1{\text{K}_{12}(\tau)}+\text{K}_{12}(\tau) = \frac1{\text{K}_6(\tau)\,\text{K}_6(2\tau)} = \frac{\eta^3(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta^3(12\tau)} = \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^2+1 To find exact values of \text{K}_{12}(\tau), we use discriminants d = 12m with class number h(-d) = 4 for even m=10,14,26,34 to get the orderly \begin{align}\frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-10/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-10/3}}4\Big) &= 1+\sqrt3\left(2+\sqrt{10}+\sqrt{-1+(2+\sqrt{10})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-14/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-14/3}}4\Big) &= 1+\sqrt3\left(4+\sqrt{21}+\sqrt{1+(4+\sqrt{21})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-26/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-26/3}}4\Big) &= 1+\sqrt3\left(15+4\sqrt{13}+\sqrt{1+(15+4\sqrt{13})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-34/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-34/3}}4\Big) &= 1+\sqrt3\left(28+5\sqrt{34}+\sqrt{-1+(28+5\sqrt{34})^2}\right) \end{align} Update: It turns out all the quartic roots can be factored into fundamentals units U_n, for example 2+\sqrt{10}+\sqrt{-1+(2+\sqrt{10})^2}=(1+\sqrt2)\big(\tfrac{1+\sqrt5}2\big)^3 = U_2\,U_5^3 More on Entry 110.

Entry 108

Let q = e^{2\pi i \tau} and the Dedekind eta function \eta(\tau). Define the following \text{K}_{10}(\tau) = \text{K}_{10}(q) = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})} While no continued fraction is yet known for this, it is connected to the mod 5 version, R(\tau) =  \text{K}_{5}(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}

or the Rogers-Ramanujan continued fraction R(\tau) by the simple relation

\text{K}_{10}(q) = R(q)\,R(q^2)\\ \text{K}_{10}(\tau) = R(\tau)\,R(2\tau)

It is known that\frac1{R(\tau)}-R(\tau) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1 Thus a quadratic root R(\tau) = \frac{- \frac{\eta(\tau/5)}{\eta(5\tau)}-1+\sqrt{\left( \frac{\eta(\tau/5)}{\eta(5\tau)}+1\right)^2+4}}2 which allows for easily computation of \text{K}_{10}(\tau). For example, given the golden ratio \phi = \frac{1+\sqrt5}2 and \sqrt{-1}=i, \begin{align}R\big(\tfrac12 i\big) &= 0.511428\dots = \tfrac12(\sqrt{5\phi\,}-\phi^2)(+\sqrt[4]5\sqrt{\phi}+\phi^2)\\ R\big(i\big) &= 0.284079\dots = (\sqrt[4]5\sqrt{\phi}-\phi)\\ R\big(2i\big) &=  0.081002\dots =\tfrac12(\sqrt{5\phi\,}-\phi^2)(-\sqrt[4]5\sqrt{\phi}+\phi^2) \end{align}which implies the exact values\begin{align}\text{K}_{10}\big(\tfrac12 i\big) &= R\big(\tfrac12 i\big)\,R\big(i\big) = 0.145286\dots\\ \text{K}_{10}\big(i\big) &= R\big(i\big)\,R\big(2i\big) \,= 0.203011\dots \end{align}

Entry 107

Given q = e^{2\pi i \tau} and the two q-continued fractions \begin{align}\quad\text{K}_4(q) &= \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\ & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\ddots }}}}\quad\end{align} \qquad \begin{align}\text{K}_8(q)\, &=\,  q^{1/2}\,\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ &=  \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\ddots}}}}\end{align}

then we can evaluate them using the modular lambda function \lambda(\tau) as \text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}\quad \quad\frac1{\text{K}_8(\tau)}-\text{K}_8(\tau)  = \frac{2}{\big(\lambda(4\tau)\big)^{1/4}}Mathematica can calculate \lambda(\tau) as ModularLambda[tau] and there is a list of exact values in Mathworld. We have previously given examples for \text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}. For \text{K}_8(\tau), then for d = 4m with class number h(-d)=2 and m=10,58, \begin{align}\frac1{\text{K}_8\big(\tfrac14\sqrt{-10}\big)}-\text{K}_8\big(\tfrac14\sqrt{-10}\big) &= 2(1+\sqrt2)\sqrt{3+\sqrt{10}} \;=\; 2\,U_2\sqrt{U_{10}}\\ \frac1{\text{K}_8\big(\tfrac14\sqrt{-58}\big)}-\text{K}_8\big(\tfrac14\sqrt{-58}\big) &= 2(1+\sqrt2)^3\sqrt{99+13\sqrt{58}} =2\,U_2^3\sqrt{U_{58}}\end{align} where U_n are fundamental units.

Entry 106

Let q = e^{2\pi i \tau}. In Entry 99, we gave the cubic continued fraction \begin{align}\text{K}_6(q)  &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\ & = \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}\end{align} Define the McKay-Thompson series of Class 6E for Monster (A128633)\begin{align}j_{6E}(\tau) &=\frac1{\big(\text{K}_6(q)\big)^3}+1 \\ &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4\end{align} While it is possible to evaluate K_6(q), doing it for j_{6E}(\tau) seems more "easy" since it is a 4th power. Let d=3m with class number h(-d)=2 for m=17,41,89. Then for \tau_1=\tfrac{1+\sqrt{-17/3}}2,\quad \tau_2=\tfrac{1+\sqrt{-41/3}}2,\quad \tau_3=\tfrac{1+\sqrt{-89/3}}2we get the evaluations\begin{align} -\sqrt{\frac{-3\;}{j_{6E}(\tau_1)}}+\sqrt{-\frac13\, j_{6E}(\tau_1)} &= 2^3+2(-2+2\sqrt{17})^{2/3}+2(2+2\sqrt{17})^{2/3}\\ -\sqrt{\frac{-3\;}{j_{6E}(\tau_2)}}+\sqrt{-\frac13\, j_{6E}(\tau_2)} &= 4^3+4(-2+10\sqrt{41})^{2/3}+4(2+10\sqrt{41})^{2/3}\\ -\sqrt{\frac{-3\;}{j_{6E}(\tau_3)}}+\sqrt{-\frac13\, j_{6E}(\tau_3)} &= 10^3+10(-2+106\sqrt{89})^{2/3}+10(2+106\sqrt{89})^{2/3}\end{align} where the RHS are roots of cubics and all quadratic irrationals, for example 72+8\sqrt{17} = (2+2\sqrt{17})^2, are squares.

Entry 105

Let q = e^{2\pi i\tau}. For convenience, define the Rogers-Ramanujan continued fraction in terms of \tau, hence R(q) = R(\tau). Given the Dedekind eta function \eta(\tau), it is known that \frac1{R(\tau)}-R(\tau) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1Thus a quadraticR(\tau) = \frac{- \frac{\eta(\tau/5)}{\eta(5\tau)}-1+\sqrt{\left( \frac{\eta(\tau/5)}{\eta(5\tau)}+1\right)^2+4}}2 For example, let \tau = \sqrt{-1}, then the formula gives us Ramanujan's evaluation, R(\sqrt{-1}) = \sqrt[4]5\sqrt{\phi}-\phi = 0.284079\dots with golden ratio \phi = \frac{1+\sqrt5}2. The formula also works when \tau = \frac{1+\sqrt{-n}}2 though R(\tau) is now negative. Since it contains q^{1/5}, one has to be careful with 5th roots. We give some nice new evaluations for discriminants d=5m with class number h(-d)=2 for m=(3,7,23,47) also using the golden ratio \phi

\begin{align}\frac1{R^5\Big(\tfrac{1+\sqrt{-3/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-3/5}}2\Big)-11 &= -(\sqrt5)^3\phi\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-7/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-7/5}}2\Big)-11 &= -(\sqrt5)^3\phi^3\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-23/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-23/5}}2\Big)-11 &= -(\sqrt5)^3\phi^9\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-47/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-47/5}}2\Big)-11 &= -(\sqrt5)^3\phi^{15}\\ \end{align} Note the 5th powers R^5(\tau) and how well-behaved their evaluations are.

Monday, May 26, 2025

Entry 104

Given q = e^{2\pi i \tau} and the q-continued fraction discussed in Entry 95\begin{align}\quad\text{K}_4(q) &= \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\ddots }}}}\end{align}

and compare it to the modular lambda function \lambda(\tau) = \left(\frac{\sqrt2\,\eta(\tfrac12\tau)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8\quad therefore the two are related by \text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8} Mathematica can calculate \lambda(\tau) as ModularLambda[tau] and there is list of exact values in Mathworld which also implies exact values for \text{K}_4(\tau). But we will give a nice consistent form when d = 4m has class number h(-d)=2 for m=5,13,37 as \begin{align}\lambda\big(\sqrt{-5}\big) &= \frac12-\sqrt{\left(\tfrac{{-1}+\sqrt{5}}2\right)^3}\\ \lambda\big(\sqrt{-13}\big) &= \frac12-3\sqrt{\left(\tfrac{{-3}+\sqrt{13}}2\right)^3}\\ \lambda\big(\sqrt{-37}\big) &= \frac12-21\sqrt{\left({-6}+\sqrt{37}\right)^3}\end{align} as well as the complex values \begin{align}\frac1{\lambda\left(\frac{1+\sqrt{-5}}2\right)} &= \frac12-\sqrt{\left(\tfrac{{-1}-\sqrt{5}}2\right)^3}\\ \frac1{\lambda\left(\frac{1+\sqrt{-13}}2\right)} &= \frac12-3\sqrt{\left(\tfrac{{-3}-\sqrt{13}}2\right)^3}\\ \frac1{\lambda\left(\frac{1+\sqrt{-37}}2\right)} &= \frac12-21\sqrt{\left({-6}-\sqrt{37}\right)^3}\end{align}

Entry 103

Level 12. Given q = e^{2\pi i \tau},  the q-Pochhammer symbol (a;q)_n, and Ramanujan's functions f(a,b) and f(-q) discussed in Entry 92. We have the sum-product identities \begin{align}A(q) &= \frac{f(-q,-q^{11})}{f(-q^4)} = \sum_{n=0}^\infty \frac {q^{4n^2+4n}\,(q;q^2)_{2n+1}} {(q^4;q^4)_{2n+1}}  = \prod_{n=1}^\infty(1-q^{12n-1})(1-q^{12n-11})\,\alpha_{12}\\ B(q) &=\frac{f(-q^5,-q^{7})}{f(-q^4)} \,=\, \sum_{n=0}^\infty \frac {q^{4n^2}\,(q;q^2)_{2n}} {(q^4;q^4)_{2n}} \quad = \; \prod_{n=1}^\infty(1-q^{12n-5})(1-q^{12n-7})\,\alpha_{12}\end{align}

where \alpha_{12} = \dfrac{(1-q^{12n})}{(1-q^{4n})}

Let a = q^{7/8}A(q) and b = q^{-1/8}B(q) then (a,b) are radicals for appropriate \tau. Define their ratio a/b \text{K}_{12}(q) = q\,\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\quad\qquad Surprisingly, this has a continued fraction (studied by Naika) based on a general form from Ramanujan for Level 4m, though its form is not as simple as the others. It is connected to the mod 6 version, \quad\text{K}_{6}(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})} = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)} or the previously discussed cubic continued fraction \text{K}_{6}(q) by the quadratic relation 

j_{12I}=\frac1{\text{K}_{12}(q)}+\text{K}_{12}(q) = \frac1{\text{K}_6(q)\,\text{K}_6(q^2)} = \frac{\eta^3(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta^3(12\tau)} = \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^2+1

 where j_{12I} is the McKay-Thompson series of class 12I for the Monster (A187144).

Entry 102

Level 10. Given q = e^{2\pi i \tau},  the q-Pochhammer symbol (a;q)_n, and Ramanujan's functions f(a,b) and \varphi(q) discussed in Entry 92. We have the sum-product identities

\begin{align}A(q) &= \frac{f(-q,-q^9)}{\varphi(-q)} \;=\; \sum_{n=0}^\infty \frac {q^{n(n+3)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}}  = \prod_{n=1}^\infty\frac{\alpha_{10}}{(1-q^{10n-3})(1-q^{10n-7})}\\ B(q) &= \frac{f(-q^3,-q^7)}{\varphi(-q)} = \sum_{n=0}^\infty \frac {q^{n(n+1)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}}  = \prod_{n=1}^\infty\frac{\alpha_{10}}{(1-q^{10n-1})(1-q^{10n-9})}\end{align}

where \alpha_{10} = \dfrac{(1-q^{10n})^2}{(1-q^{n})(1-q^{5n})}

Let a = q^{4/5}A(q) and b = q^{1/5}B(q) then (a,b) are radicals for appropriate \tau. Define their ratio a/b, \text{K}_{10}(q) = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})} While no continued fraction is yet known for this, it is connected to the mod 5 version \text{K}_{5}(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}

simply as, \text{K}_{10}(q) = \text{K}_5(q)\,\text{K}_5(q^2)

where \text{K}_5(q)=R(q) is just the Rogers-Ramanujan continued fraction.

Entry 101

Level 8. Recall the Level 4 continued fraction \begin{align}\text{K}_4(q) &= \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ &= \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}}\end{align} Compare its similarity to the Level 8 version using the ratio of (a,b) from Entry 100 \begin{align}\text{K}_8(q) &\,=\,  q^{1/2}\,\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ &= \cfrac{q^{1/2}}{1+\cfrac{q+q^2}{1+\cfrac{q^4}{1+\cfrac{q^3+q^6}{1+\cfrac{q^8}{1+\ddots}}}}} = \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\cfrac{q^8}{1+q^9+\ddots}}}}}\end{align}

In fact, they have the quadratic relation \quad\frac1{\text{K}_8(q)}-\text{K}_8(q) =\left(\frac{\sqrt2}{\text{K}_4(q^2)}\right)^2 = \left(\frac{\eta^3(4\tau)}{\eta(2\tau)\eta^2(8\tau)}\right)^2 while \text{K}_4(q) which is an eta quotient can also be expressed another way\frac1{\big(\text{K}_4(q)\big)^2}-\big(\text{K}_4(q)\big)^2 =\frac12\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^4\quad

Entry 100

Level 8. Given q = e^{2\pi i \tau},  the q-Pochhammer symbol, and Ramanujan's functions f(a,b) and \psi(q) discussed in Entry 92. We have the Gollnitz-Gordon sum-product identities,

\begin{align}A(q) &=\frac{f(-q,-q^7)}{\psi(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+2n}\,(-q;q^2)_n} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{8n-3})(1-q^{8n-4})(1-q^{8n-5})}\\ B(q) &= \frac{f(-q^3,-q^5)}{\psi(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}\,(-q;q^2)_n} {(q^2;q^2)_n} \;\; = \;\; \prod_{n=1}^\infty\frac{1}{(1-q^{8n-1})(1-q^{8n-4})(1-q^{8n-7})}\end{align}

Let a=q^{7/16}A(q) and b= q^{-1/16}\,B(q). Then (a, b), for appropriate \tau, are actually radicals.

Entry 99

Level 6. Using the eta quotients (a,b) from Entry 98 their ratio \dfrac{a}{b} is \begin{align}\text{K}_6(q) &= q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\ &=q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{2n-1})}{\; (1-q^{6n-3})^3} = q^{1/3}\prod_{n=1}^\infty \frac{(1+q^{3n})^3}{(1+q^{n})\;}\\  &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\ & = \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}\end{align}

this is also known as the cubic continued fraction. For appropriate \tau, then (a, b, \text{K}_6) are radicals. The formula for the j-function using \text{K}_6(q) employs polynomial invariants of the tetrahedron and the integer 12 of b= q^{-1/12}B(q) in Entry 98 reflects the order 12 of the tetrahedral group. Note that the cube of the reciprocal of \text{K}_6(q) is the McKay-Thompson series of class 6E of the Monster (A105559)\left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4

Entry 98

Level 6. Given q = e^{2\pi i \tau}Ramanujan's theta function f(a,b), its one-parameter version f(-q) = f(-q,-q^2), and the q-Pochhammer symbol. We have the sum-product identities,

\begin{align}A(q) &= \frac{f(-q,-q^5)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}(q,q^2)_n} {(-q,q)_{2n+1} (q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_6}{(1-q^{6n-3})(1-q^{6n-3})} = q^{-1/4}\frac{\eta(\tau)\eta^2(6\tau)}{\eta^2(2\tau)\eta(3\tau)}\\ B(q) &=  \frac{f(-q^3,-q^3)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2}(q,q^2)_n} {(-q,q)_{2n} (q^2;q^2)_n} \;=\; \prod_{n=1}^\infty\frac{\alpha_6}{(1-q^{6n-1})(1-q^{6n-5})} = q^{1/12}\frac{\eta^2(3\tau)}{\eta(2\tau)\eta(6\tau)}\end{align} where \alpha_6 = \dfrac{(1-q^{n})(1-q^{3n})}{(1-q^{2n})^2} 

Let a=q^{1/4}A(q) and b= q^{-1/12}\,B(q). Then (a, b) for appropriate \tau are eta quotients like in Level 4 and are actually radicals.

Sunday, May 25, 2025

Entry 97

Level 5 In Entry 96, we asserted that a=q^{11/60}H(q) and b= q^{-1/60}\,G(q) are radicals for appropriate \tau. Proof: Given the j-function j(q) and let R(q)=r=a/b

\begin{align}a &= q^{11/60}H(q) = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{11/20}}{j(q)^{11/60}\,(r^{10}+11r^5-1)}\\ b &= q^{-1/60}G(q) =  \frac{j(q)^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}\end{align} Getting their ratio a/b, then it simplifies to

r = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3/5}}{j(q)^{1/5}\,(r^{10}+11r^5-1)} Raising to the fifth power and moving j(q) to the LHS 

j(q)  = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3}}{r^5(r^{10}+11r^5-1)^5} and we get the icosahedral equation for the j-function in terms of the Rogers-Ramanujan continued fraction R(q)=r. Thus if j(q) and R(q) are radicals, then so are (a, b) while the integer 60 of b= q^{-1/60}G(q) reflects the order 60 of the icosahedral group. 

Entry 96

Level 5. Given q = e^{2\pi i \tau}Ramanujan's theta function f(a,b), its one-parameter version f(-q) = f(-q,-q^2), and the q-Pochhammer symbol. We have the Rogers-Ramanujan sum-product identities,

\begin{align}H(q) &= \frac{f(-q,-q^4)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\\ G(q) &=  \frac{f(-q^2,-q^3)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n}  = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\end{align} Their ratio is the Rogers-Ramanujan continued fraction

\begin{align}R(q) &= \frac{a}{b} = \frac{q^{11/60} H(q)}{q^{-1/60}\,G(q)} = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-2})}{(1-q^{5n-2})(1-q^{5n-3})}\\ &=\cfrac{q^{1/5}} {1+\cfrac{q} {1+\cfrac{q^2} {1+\cfrac{q^3} {1+\ddots}}}}\end{align} Let a=q^{11/60}H(q) and b= q^{-1/60}\,G(q). Then (a, b) for appropriate \tau are actually radicals. Proof in Entry 97.

Entry 95

Level 4. Using the eta quotients (a,b) from Entry 94 we found that -a^8+b^8=c^8 while their ratio is \begin{align}\text{K}_4(q) &= \frac{a}{b} = \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ &= \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}}\end{align}

For appropriate \tau, then (a, b, \text{K}_4) are radicals. The formula for the j-function using \text{K}_4(q) employs polynomial invariants of the octahedron and the integer 24 of b= q^{-1/24}B(q) in Entry 94 reflects the order 24 of the octahedral group. Note that the 8th power of the reciprocal of \text{K}_4(q) without the \sqrt2 is the McKay-Thompson series of class 4C of the Monster (A007248)\left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\right)^8-16=\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8

Entry 94

Level 4. Given q = e^{2\pi i \tau}, Ramanujan's theta function f(a,b) and its one-parameter version as f(-q) = f(-q,-q^2). We have the sum-product identities,

\begin{align}A(q) &= \frac{f(-q,-q^3)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-2})(1-q^{4n-2})} = q^{-1/12}\frac{\eta(4\tau)}{\eta(2\tau)}\\ B(q) &= \frac{f(-q^2,-q^2)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q^2;q^2)_n}  = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-1})(1-q^{4n-3})} = q^{1/24}\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\end{align}

where \alpha_4 = \dfrac{(1-q^{2n})}{(1-q^{4n})}

Let a = \sqrt2\,q^{1/12}A(q) and b= q^{-1/24}B(q), then (a, b) are just eta quotients and for appropriate quadratic \tau are actually radicals. Furthermore, the difference of their 8th powers is also an 8th power -a^8+b^8 =-\left(\frac{\sqrt2\,\eta(4\tau)\,}{\;\eta(2\tau)}\right)^8+\left(\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\right)^8=\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8

Entry 93

Let q = e^{2\pi i \tau}.  Ramanujan's theta function f(a,b) is given by f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2}\; b^{n(n-1)/2} Then the following have nice q-continued fractions \begin{align}\text{K}_4(q) &= q^{1/8}\frac{f(-q,-q^3)}{f(-q^2,-q^2)} = q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\ \text{K}_5(q) &= q^{1/5}\frac{f(-q,-q^4)}{f(-q^2,-q^3)} = q^{1/5} \prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\ \text{K}_6(q) &= q^{1/3}\frac{f(-q,-q^5)}{f(-q^3,-q^3)} = q^{1/3} \prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\ \text{K}_8(q) &= q^{1/2}\frac{f(-q,-q^7)}{f(-q^3,-q^5)} = q^{1/2} \prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ \text{K}_{12}(q) &= q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)} = q^{1/1} \prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\end{align} with the most famous being \text{K}_5(q)=R(q) since this is the Rogers-Ramanujan continued fraction. Each will be discussed in subsequent entries.

Entry 92

Given the q-Pochhammer symbol, \begin{align}(a;q)_n &= \prod_{k=0}^{n-1}(1-aq^k)\\ (a;q)_\infty &= \prod_{k=0}^{\infty}(1-aq^k)\end{align}

as well as the Ramanujan theta function,

f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}

In his Notebooks, Ramanujan also defined four one-parameter versions he commonly used as,

\begin{align}\varphi(q) &= f(q,\, q)\\ f(-q) &= f(-q, -q^2)\\ \psi(q) &= f(q,\, q^3)\\ \chi(q) &=\frac{f(-q^2,-q^2)}{f(-q,-q^2)}\\ \end{align}

which we will also use in later entries. There are several simple relations between these four auxiliary functions, one of which I found using all four is the elegant Fermat curve of degree 8,

\big[f(-q)\,\chi(-q)\big]^8+\big[\sqrt2\,q^{1/8}\psi(-q)\big]^8 = \big[\varphi(-q^2)\big]^8

We normally assume q = e^{2\pi i \tau} unless otherwise specified. For simplicity, it will also be assumed \tau is an complex quadratic number so that certain functions later will also evaluate as radicals.

Saturday, May 24, 2025

Entry 91

Here is a "bizarre" continued fraction from Ramanujan involving e and \pi \sqrt{\frac{\pi\,e}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}} However, since \Gamma\big(\tfrac12\big) = \sqrt{\pi}, we can express the LHS in terms of gamma functions and find a cubic counterpart using \Gamma\big(\tfrac13\big) 

\sum_{n=1}^\infty \frac{1}{2^n (\frac12)_n} = \Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big) \sqrt{\frac{e}{2}} = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}}

\sum_{n=1}^\infty \frac{1}{3^n (\frac13)_n} = \Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}} = 1+\cfrac{2/3}{2+\cfrac{3/3}{3+\cfrac{4/3}{4+\cfrac{5/3}{5+\ddots}}}}

Ramanujan's identity then becomes a sum of two continued fractions with a cubic version,

\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}} \; \color{blue}+ \; \cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}

\Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}}= 1+\cfrac{2/3}{2+\cfrac{3/3}{3+\cfrac{4/3}{4+\cfrac{5/3}{5+\ddots}}}} \; \color{blue}+ \; \cfrac1{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{\color{red}5}{1+\ddots}}}}

where the 4th continued fraction (also by Ramanujan) is missing numerators P(n)=3n+1 = 4,7,10,13,\dots The four cfracs then have closed-forms as,

\begin{align}\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} &= \Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big)\sqrt{\frac{e}{2}} \color{blue}+ \Big(\Gamma\big(\tfrac12,\tfrac12\big)\Big)\sqrt{\frac{e}{2}} \\ \Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}} &= \Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}} \color{blue}+ \Big(\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}}\end{align}

Entry 90

The modular lambda function given by Mathematica as ModularLambda[tau] can solve \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1} In the previous entry, we used the tribonacci constant for the case \tau=\sqrt{-11}. More generally, to solve the higher Heegner numbers d = 11,19,43,67,163 we can employ a slightly different method. For example, \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-x\big)}{_2F_1\big(\tfrac12,\tfrac12,1,x\big)}  = \sqrt{163} This also has an exact solution expressible in radicals x \approx 6.094\times10^{-17} as the smaller root of the quadratic 16x^2-16x+1=y and y is the real root of the cubic y^3+\frac{640320^3}{256}y-\frac{640320^3}{256}=0 The big number should be familiar and appears in Ramanujan's constant e^{\pi\sqrt{163}} = 640320^3 + 743.99999999999925\dots For the other Heegner numbers d, one just replaces the big number with the cube nearest to e^{\pi\sqrt{d}}-744, like 5280^3 for \sqrt{67} and so on.

Entry 89

The modular lambda function \lambda(\tau)\lambda(\tau) = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 given by Mathematica as ModularLambda[tau] can solve, \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1} For the special case \tau=\sqrt{-11} \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-u\big)}{_2F_1\big(\tfrac12,\tfrac12,1,u\big)}  = \sqrt{11} we can can use the tribonacci constant T, the real root of T^3-T^2-T-1=0. The solution in terms of T is u=\lambda(\tau)=\frac14\left(2-\sqrt{\frac{2T+15}{2T+1}}\right) = \frac14\left(2-\sqrt{\frac{17S+2}{3S+2}}\right)= 0.00047741\dots or in terms of S as the nice nested cube roots S = \frac1{T-1}=\sqrt[3]{\frac12+\sqrt[3]{\frac12+\sqrt[3]{\frac12+\sqrt[3]{\frac12+\dots}}}} the cubic version of the golden ratio's \phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}

Entry 88

The modular lambda function \lambda(\tau) \lambda(\tau) = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 discussed in the previous post solves, among other things, \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1} For example, let \tau=\sqrt{-2} so \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = \sqrt{2}

and the Mathematica command ModularLambda[tau] yields a real number equal to \lambda\big(\tau) = (1-\sqrt2)^2. Other \tau=\sqrt{n} can be found in Mathworld's list. But we can use more general complex \tau. For example, let \tau=\frac{1+\sqrt{-2}}2 which is no longer in the list. So \frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)} = {-\left(\tfrac{1+\sqrt{-2}}2\right)} \sqrt{-1} and we find the complex number, \lambda(\tau) = 4\big({-1}+\sqrt2\big)^3\left(4+\sqrt{2(1-5\sqrt2)}\right) \approx 1.1370849 + 0.9905592 i which is a root of quartic with two real roots and two complex roots. And so on for other complex quadratic irrationals \tau.

Entry 87

Define d_k = \eta(k\tau) with the Dedekind eta function \eta(\tau). We have the nice \left(\frac{\sqrt2\,d_1\,d_4^2}{d_2^3}\right)^8+\left(\frac{d_1^2\,d_4}{d_2^3}\right)^8 = 1 Equivalently \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8=\left(\frac{d_1}{d_4}\right)^8+8 Focusing on the first term, note that the three similar Monster functions

j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8,\quad  j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad  j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2
being the McKay-Thompson series of class 4C, 8E, 16B, respectively, are necessary to the 9 dependencies found by Conway, Norton, and Atkins such that the moonshine functions span a linear space of 172-9=163 dimensions (discussed in previous entries). In fact, scaled and flipped over, it is an important function,  \lambda(\tau)=\left(\frac{\sqrt2\,d_{1/2}\,d_2^2}{d_1^3}\right)^8 = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 known as the modular lambda function \lambda(\tau).

Entry 86

As discussed in the previous entry, the eta quotients \left(\frac{\eta(\tau)}{\eta(k\tau)}\right) are useful for j-function formulas. The easy levels k are when m=24/(k-1) is an integer. For square k=(4,9,25) yields m=(8,3,1) which can be found in the exponents below. j(\tau) =\frac{(x^2-48)^3}{x^2-64}, \quad\text{with}\quad x=\left(\frac {\sqrt4\,\eta(4\tau)}{\eta(\tau)}\right)^8+8

j(\tau) =\frac{x^3(x^3-24)^3}{x^3-27}, \quad\text{with}\quad x=\left(\frac {\sqrt9\,\eta(9\tau)}{\eta(\tau)}\right)^3+3

j(\tau) = \frac{-(x^{20}+12 x^{15}+14 x^{10}-12 x^5+1)^3}{x^{25} (x^{10}+11 x^5-1)} , \quad\text{with}\;\; x^{-1}-x=\left(\frac {\sqrt{25}\,\eta(25\tau)}{\eta(\tau)}\right)^1+1

Alternatively,

j(\tau) =\frac{(x^2+192)^3}{(x^2-64)^2}, \quad\text{with}\quad x=\left(\frac {\eta(\tau/2)}{\eta(2\tau)}\right)^8+8

j(\tau) =\frac{x^3(x^3+216)^3}{(x^3-27)^3}, \quad\text{with}\quad x=\left(\frac {\eta(\tau/3)}{\eta(3\tau)}\right)^3+3

j(\tau) = \frac{-(r^{20}-228 r^{15}+494 r^{10}+228 r^5+1)^3}{r^5 (r^{10}+11 r^5-1)^5} , \quad\text{with}\;\; r^{-1}-r=\left(\frac {\eta(\tau/5)}{\eta(5\tau)}\right)^1+1 After some manipulation, these eta quotients are related to q-continued fractions with octahedral, tetrahedral, and icosahedral symmetries, with r=R(q) being the well-known Rogers-Ramanujan continued fraction.

Entry 85

The eta quotients \left(\frac{\eta(\tau)}{\eta(k\tau)}\right) discussed previously are useful for j-function formulas. The easy levels k are when m=24/(k-1) is an integer. For prime k=(2,3,5,7,13) yields m=(24,12,6,4,2) which are the exponents of x below. j(\tau) =\frac{(x-16)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt2\,\eta(2\tau)}{\;\eta(\tau)}\right)^{24}

\; j(\tau) =\frac{(x+3)^3(x+27)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt3\,\eta(3\tau)}{\;\eta(\tau)}\right)^{12}

j(\tau) \,=\,\frac{(x^2+10x+5)^3}x, \;\quad\text{with}\;\; x=\left(\frac {\sqrt5\,\eta(5\tau)}{\;\eta(\tau)}\right)^6

j(\tau) =\frac{(x^2+5x+1)^3(x^2+13x+49)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4

j(\tau) =\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x, \;\text{with}\;\; x=\left(\frac {\sqrt{13}\,\eta(13\tau)}{\;\eta(\tau)}\right)^2

Entry 84

Summarizing, we find tentative patterns for \text{I. Levels}\; 4n+1 = (5,13)\\ \text{II. Levels}\; 4n+2 = (6,10)\\ \text{III. Levels}\; 4n+3 = (3,7)\,The definitions for these functions are in previous entries, but it may be good to have a few selected values together to get an overall picture. Note that U_n are fundamental units, important to Pell equations. 

I. Levels 4n+3 = (3,7)

\quad \begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -(\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -(2\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &= -(4\sqrt3)^6 \\j_{3B}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -3^3U_{5}^2 =-3^3\left(\tfrac{1+\sqrt{5}}2\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -3^3U_{17}^2 =-3^3\left(4+\sqrt{17}\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &=-3^3U_{41}^2 =-3^3\left(32+5\sqrt{41}\right)^2 \end{align} and one can observe its similarity to Level 7 \begin{align}j_{7A}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -(\sqrt7)^2\\ j_{7A}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -(3\sqrt7)^2 \\ j_{7A}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -(39\sqrt7)^2\\ j_{7B}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -7\,U_{5}^{2} = -7\left(\tfrac{1+\sqrt5}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -7\,U_{13}^{2} = -7\left(\tfrac{3+\sqrt{13}}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -7\,U_{61}^{2} = -7\left(\tfrac{39+5\sqrt{61}}2\right)^{2}\end{align}

II. Levels 4n+1 = (5,13)

\begin{align}j_{5A}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &= -(6\sqrt{23})^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &= -(18\sqrt{47})^2\\ j_{5B}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &=  -(\sqrt5)^3\left(\tfrac{1+\sqrt{5}}2\right)^9\\ j_{5B}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &=  -(\sqrt5)^3\left(\tfrac{1+\sqrt{5}}2\right)^{15}\end{align} where we see the golden ratio above and the bronze ratio below \begin{align}j_{13A}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = -(\sqrt7)^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -(2\sqrt{31})^2\\ j_{13B}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = {-\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)}\\ j_{13B}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)^3 \end{align}

III. Levels 4n+2 = (6,10)

\quad \begin{align}j_{6A}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -20^2\\  j_{6A}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -1060^2\\ j_{6B}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -U_{11}^2 = -\big(10+3\sqrt{11}\big)^2\\ j_{6B}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -U_{59}^2 = -\big(530+69\sqrt{59}\big)^2 \end{align} compared to \begin{align}j_{10A}\big(\tfrac12\sqrt{-6/5}\big) &= 6^2\\ j_{10A}\big(\tfrac12\sqrt{-38/5}\big) &= 76^2\\ j_{10D}\big(\tfrac12\sqrt{-6/5}\big) &=\, U_{10}^{2}\, = \left(3+\sqrt{10}\right)^{2}\\ j_{10D}\big(\tfrac12\sqrt{-38/5}\big) &=\, U_{5}^{18}\, = \left(\tfrac{1+\sqrt5}2\right)^{18} \end{align}

Entry 83

Level 13. Define d_k = \eta(k\tau) with Dedekind eta function \eta(k\tau) and the McKay-Thompson series of class 13B j_{13B}(\tau) = \left(\frac{d_1}{d_{13}}\right)^2

Examples. Let d=13m with class number h(-d)=2 for m=7, 31 and we find a phenomenon similar to Level 5 \begin{align}j_{13B}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = -\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)\\ j_{13B}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)^3 \end{align}

But where Class 5B involves the fundamental unit U_5 or the golden ratio, Class 13B involves U_{13} = \tfrac{3+\sqrt{13}}2 also known as the bronze ratio.

Entry 82

Level 13. Define d_k = \eta(k\tau) with Dedekind eta function \eta(k\tau) and the McKay-Thompson series of class 13A for the Monster. j_{13A}(\tau) = \left(\frac{d_1}{d_{13}}\right)^2+13\left(\frac{d_{13}}{d_1}\right)^2+6

Examples. We select d=13m for class number h(-d)=2 with m=7, 31 as well as class number h(-d)=6 with m=19, 151 and find a well-behaved pattern just like for level 5 \begin{align}j_{13A}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = -(\sqrt7)^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -(2\sqrt{31})^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-19/13}}{2}\Big) & = -(x\sqrt{19})^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-151/13}}{2}\Big) & = -(y\sqrt{151})^2\end{align}

where (x,y) are roots of cubics 19x^3 - 38x^2 + 21x - 9 = 0\\ 151y^3 - 2567y^2 - 512y - 44 = 0 and so on for class number h(-d)=10, etc.

Entry 81

Level 10. Define d_k = \eta(k\tau) with Dedekind eta function \eta(k\tau) and the McKay-Thompson series of class 10D for the Monster (A132130). j_{10D}(\tau) = \left(\frac{d_2\,d_5}{d_1\,d_{10}}\right)^6

Examples. We select d=20m with class number h(-d)=4 and find odd m=17 as well as m=6, 14, 26, 38 such that the following are special quadratic irrationals \begin{align}j_{10D}\Big(\tfrac{1+\sqrt{-17/5}}{2}\Big) &= -U_{5}^{12} = -\left(\tfrac{1+\sqrt5}2\right)^{12}\\ j_{10D}\big(\tfrac12\sqrt{-6/5}\big) &=\, U_{10}^{2}\, = \left(3+\sqrt{10}\right)^{2}\\ j_{10D}\big(\tfrac12\sqrt{-14/5}\big) &=\, U_{2}^{6}\, = \left(1+\sqrt2\right)^{6}\\ j_{10D}\big(\tfrac12\sqrt{-26/5}\big) &=\, U_{13}^{6}\, = \left(\tfrac{3+\sqrt{13}}2\right)^{6}\\ j_{10D}\big(\tfrac12\sqrt{-38/5}\big) &=\, U_{5}^{18}\, = \left(\tfrac{1+\sqrt5}2\right)^{18} \end{align}

as they are fundamental units U_n. Since U_5=\phi is the golden ratio, then the first and last implies the integers \left(\sqrt{-\phi^{12}}-1/\sqrt{-\phi^{12}}\right)^2 = -18^2\\ \left(\sqrt{\phi^{18}}-1/\sqrt{\phi^{18}}\right)^2 = 76^2 and similarly for the others as discussed in the previous entry.

Entry 80

Level 10. Define d_k=\eta(k\tau) with the Dedekind eta function \eta(\tau) and the McKay-Thompson series of class 10A for the Monster. 
j_{10A}(\tau) = \left(\left(\frac{d_2\,d_5}{d_1\,d_{10}}\right)^3-\left(\frac{d_1\,d_{10}}{d_2\,d_5}\right)^3\right)^2 Examples. We select d=20m with class number h(-d)=4 and find odd m=17 as well as m=6, 14, 26, 38 such that the following are well-behaved integers \begin{align}j_{10A}\Big(\tfrac{1+\sqrt{-17/5}}{2}\Big) &= -18^2\\ j_{10A}\big(\tfrac12\sqrt{-6/5}\big) &= 6^2\\ j_{10A}\big(\tfrac12\sqrt{-14/5}\big) &= 14^2\\ j_{10A}\big(\tfrac12\sqrt{-26/5}\big) &= 36^2\\ j_{10A}\big(\tfrac12\sqrt{-38/5}\big) &= 76^2 \end{align} Note that the last is responsible for the prime-generating polynomial F(n)=10n^2+19 which has discriminant d = -5\times 38 = -190 and is prime for 19 consecutive values n=0 - 18.