Pi Formulas, Ramanujan, and the Baby Monster Group
Keywords: Pi formulas, class polynomials, j-function, modular functions, Ramanujan, finite groups.
I. Introduction
II. Pi Formulas
III. Baby Monster Group
IV. Conclusion
In 1914, Ramanujan wrote a fascinating article in the Quarterly Journal of Pure and Applied Mathematics. The title was “Modular equations and approximations to p” and he discusses certain methods to derive exact and approximate evaluations. In fact, in his Notebooks, he gave several formulas, seventeen in all, for p (or to be more precise, 1/p). For a complete list, see http://mathworld.wolfram.com/PiFormulas.html. He gave little explanation on how he came up with them, other than saying that there were “corresponding theories”. One of the more famous formulas is given by,
1/(pÖ8) = 1/992 S r (26390n+1103)/3964n
where r = (4n)!/(n!4) and the sum S (from this point of the article onwards) is to go from n = 0 to ¥.
How did Ramanujan find such beautiful formulas? We can only speculate at the heuristics he used to find them, since they were only rigorously proven to be true in 1987 by the Borwein brothers. However, one thing we do know: Ramanujan wrote down in his Notebooks, among the others, the approximation1,
epÖ58 » 24591257751.9999998222…
which can also be stated as,
epÖ58 » 3964 – 104
The appearance of 3964 in both the exact formula and the approximation is no coincidence. The connection can be highlighted even further when you realize that 26390 = 5*7*13*58. What are these numbers 58 and 3964? The former is a squarefree discriminant d of a quadratic form ax2+bxy+cy2 and 396 can be derived from the class invariant,
g582 = (5+Ö29)/2
which in turn is related to the fundamental solution (x1 = 5, y1 = 1) of the Pell equation x2-29y2 = -4.
How can a formula for p have something to do with Pell equations? That, I believe, is what makes a mathematical journey so rewarding: the unexpected and fascinating connections we stumble on along the way. We will not be able to give an answer to that question in this article but we can do so for a more tractable one: How do we derive 396 or, in general, other algebraic numbers involved in pi formulas, from class invariants? It turns out the answer has to do with transforming roots of Weber class polynomials into roots of Ramanujan class polynomials, polynomials which we have mentioned in “Ramanujan’s Constant (epÖ163) and its Cousins” [1]. In that article, we also discussed Hilbert class polynomials and pi formulas derived from them and that is where we will start.
Similar to epÖ58 but more famous, is epÖ163 which can be shown to be approximately,
epÖ163 » 6403203 + 743.99999999999925…
The reason for such close approximation is because of the j-function, j(t), which has the q-series expansion,
j(t) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + …
where q = e2pit and t is the half-period ratio defined as,
Case 1: For d = 4m,
t = Ö(-m)
Case 2: For d = 4m+3,
t = (1+Ö(-d))/2
where m is a positive integer, and d is the unsigned discriminant.
In [1], we were quite liberal with our use of what we called the j-function but we will be more strict here. First, while in some older texts the convention was to label the j-function as j(q), it seems the modern one is to label it as j(t), or “j-tau”. Second, an obvious observation is that since d is positive, then t is imaginary. We are evaluating j(t) at imaginary arguments yet it will have a real value at those points. However, it can be positive or negative. For case 1, for all d > 0, then j(t) is always positive. For case 2, for d > 0, there are three cases,
d < 3, j(t) is positive
d = 3, j(t) is zero
d > 3, j(t) is negative
From this point, what we will call as j(q) will refer to the absolute value of the j-function, while j(t) will refer to the signed value. This value is an algebraic integer of degree n where n is the class number of d. We have also observed that j(t) was a good approximation for numbers of the form epÖd.
Case 1: If d = 4m, t = Ö(-m), then epÖ(4m) » j(t) – 744, for d > 12.
Example 1. Let m = 5, d = 20 (class number 2),
j(Ö-5) = 23(25+13Ö5)3
so,
epÖ20 » 23(25+13Ö5)3 -744
Example 2. Let m = 10, d = 40 (class number 2),
j(Ö-10) = 63(65+27Ö5)3
so,
epÖ40 » 63(65+27Ö5)3 - 744
Case 2: If d = 4m+3, t = (1+Ö-d)/2, then epÖd » | j(t)| + 744, for d > 12.
Example 1. Let d = 163 (class number 1),
j((1+Ö-163)/2) = -6403203
so,
epÖ163 » 6403203 + 744
Technically, case 2 is also defined for other d. For example, for d = 4m+2, say, d = 2, we have,
j((1+Ö-2)/2) = 103(19-13Ö2)3
However, for the purposes of this article, we will limit case 2 only to form 4m+3. One should also take note that j(Ö-d) and j((1+Ö-d)/2) for d = 4m+3 are different, as illustrated by,
j(Ö-7) = 2553
and,
j((1+Ö-7)/2) = -153
which can easily be confirmed by an adequate computer algebra system. Anyway, now that we have the preliminaries out of the way, we can go to the pi formulas. To recall, we pointed out that there are exactly nine discriminants d with class number 1, the absolute values given as 3, 4, 7, 8, 11, 19, 43, 67, 163, with the larger values giving the better approximations:
eπ√7 ≈ 153 + 697
eπ√11 ≈ 323 + 738
eπ√19 ≈ 963 + 744
eπ√43 ≈ 9603 + 744
eπ√67 ≈ 52803 + 744
eπ√163 ≈ 6403203 + 744
We also have the formulas due to the Chudnovsky brothers (which had an their inspiration Ramanujan’s formula cited in the opening paragraph):
Let, c = (6n)!/((n!)3(3n)!),
1/(3p) = S c (-1)n (63n+8)/(153)n+1/2
1/(4p) = S c (-1)n(154n+15)/(323)n+1/2
1/(12p) = S c (-1)n (342n+25)/(963)n+1/2
1/(12p) = S c (-1)n (16254n+789)/(9603)n+1/2
1/(12p) = S c (-1)n (261702n+10177)/(52803)n+1/2
1/(12p) = S c (-1)n (545140134n+13591409)/(6403203)n+1/2
The formulas obviously use the j(q)’s of the given approximations. The factorization of a term in the numerator also indicate what d is involved,
63 = 32 *7
154 = 2*7*11
342 = 2*32 *19
16254 = 2*33 *7*43
261702 = 2*32 *7*31*67
545140134 = 2*32 *7*11*19*127*163
Now why is this? It turns out the answer is in the general form of the formula for d = 4m+3, d > 3, (implying j(t) as negative),
1/p = S c (-1)n (An+B)/(C)n+1/2
where,
A = Ö(d*(1728-j(t))); C = -j(t)
Since, 1728 = 123 (remember the taxicab anecdote?2) then,
7(123 + 153) = 72 *272
11(123 + 323) = 42 *112 *142
19(123 + 963) = 122 *192 *182
43(123 + 9603) = 122 *432 *3782
67(123 + 52803) = 122 *672 *39062
163(123 + 6403203) = 122 *1632 *33444182
For these d, the expression d*(1728-j(t)) is a perfect square! So there are beautiful Diophantine relationships behind these pi formulas as well, which explains why the factorization of A contains d. However, we are still missing the constant B and that is the fly in the ointment. Technically, there is a closed-form expression for B in terms of Eisenstein series but it gets too complicated for this article. The interested reader is referred to “Ramanujan, Modular Equations, and Approximations to Pi” by Bailey, Borwein, and Borwein which contains a side article on the Chudnovskys’ approach as well as their own which we shall discuss later. The other way to find B (since A and C are already known) is simply to solve the equation,
1/p = S c (-1)n (An+B)/(C)n+1/2
using enough n terms which should give B to a reasonable accuracy. For these six d, B conveniently is an integer, given by 3*8, 4*15, 12*25, 12*789, 12*10177, and 12*13591409, respectively. Since A and B have a common factor, then one can see why 1/p is divided by either 3, 4, or 12. We have also pointed out that there are non-maximal or non-fundamental discriminants with class number 1 as shown by the approximations,
eπ√16 ≈ 663 - 744
eπ√28 ≈ 2553 - 744
We can also use these two j(q)’s to find more pi formulas. The general form (with some small changes) for d = 4m is given by,
1/p = S c (An+B)/(C)n+1/2
where,
A = Ö(-d*(1728-j(t))); C = j(t)
Note that there is no (-1)n for d = 4m. The presence of (-1)n in the formula for d = 4m+3 is simply the negative sign of j(t). For d = 4m, since j(t) is positive, then there is no need for it. The only difference really is in A since d is negated (though the product ends up positive). Solving for B, we then have two additional formulas,
1/p = 24Ö2 S c (63n+5)/(663)n+1/2
1/p = 162 S c (133n+8)/(2553)n+1/2
We can, of course, use d with a higher class number n, though B would now be an algebraic number, usually of degree n. For the interested reader who wishes to find pi formulas using the method outlined above, one can find the defining polynomial of the real number B using the Integer Relations applet at http://www.cecm.sfu.ca/projects/IntegerRelations/.
In [1], I remarked that I was not aware of pi formulas that used the roots of Weber class polynomials. In fact there are, and they were there all along in Ramanujan’s 17 formulas! However, the roots were barely discernable and it took some time to recognize them. To recall, the Weber modular function, or w(q) had the q-series expansion,
w(q) = 1/q + 24 + 276q + 2048q2 + 11202q3 + …
and w(q), just like j(q), are algebraic integers for suitable arguments involving imaginary quadratic irrationals. These algebraic integers are defined by the Weber class polynomials. For an extensive list of these polynomials, the reader is referred to Annegret Weng’s website “Class Polynomials of CM-Fields”, http://www.exp-math.uni-essen.de/zahlentheorie/classpol/class.html.
The w(q) also provide excellent approximations to the transcendental numbers epÖd for d > 0. However, we will limit our discussion to odd d. Since case 1 was for d = 4m, this restricts us to case 2 for d = 4m+3. Case 2 can be divided into d of form 8n+3 and 8n+7, a distinction made in [1]. However, we can further refine this as to whether d is divisible by 3, hence giving us four cases. Let x be the appropriate real root of the Weber class polynomial:
A. For 8n+3: Degree of poly is 3n, where n is the class number of d. (With the exception of d = 3 which has a Weber class polynomial of degree 1.)
Case 1. If 3 doesn’t divide d, then epÖd » x24 – 24.
Example: Let d =163, n = 1,
epÖ163 » x24 – 24; (x3-6x2+4x-2=0)
Case 2. If 3 divides d, then epÖd » 28x8 – 24.
Example: Let d = 51, n = 2,
epÖ51 » 28x8 – 24; (x6-8x5-3x4+6x3+9x2+2x+1=0)
B. For 8n+7: Degree of poly is n, where n is the class number of d.
Case 3. If 3 doesn’t divide d, then epÖd » 212x24 – 24.
Example: Let d = 23, n = 3,
epÖ23 » 212x24 – 24; (x3-x-1=0)
Case 4. If 3 divides d, then epÖd » 212x8 – 24.
Example: Let d =39, n = 4,
epÖ39 » 212x8 – 24; (x4-3x3-4x2-2x-1=0)
Again, having the preliminaries out of the way, we can go to the pi formulas. In the list given by Mathworld, the first three of Ramanujan’s formulas are,
4/p = S (6n+1)(1/2)n3 /(4n(n!)3)
16/p = S (42n+5)(1/2)n3 /(64n(n!)3)
32/p = S (42Ö5n+5Ö5+30n-1)(1/2)n3 ((Ö5-1)/2)8n /(64n(n!)3)
where (a)n is the rising factorial, aka Pochhammer symbol, such that (a)n = (a)(a+1)(a+2)…(a+n-1). In this guise, it is hard to see the Weber class polynomial root, though the presence of (Ö5-1)/2 is a clue. One way then is to reformulate them in terms of ordinary factorials.
Let, h = ((2n)!/(n!2))3. Then,
1/p = 1/4 S h (6n+1) /28n
1/p = 1/16 S h (42n+5) /212n
1/p = 1/32 S h (42Ö5n+5Ö5+30n-1) /(212((1+Ö5)/2)8)n
Compare to,
epÖ3 » 28(1)8 – 25.23 (Case 2)
epÖ7 » 212(1)24 – 24 (Case 3)
epÖ15 » 212((1+Ö5)/2)8 – 24 (Case 4)
since w(q)3 = 1, w(q)7 = 1, w(q)15 = ((1+Ö5)/2)8. The denominators of these formulas use the w(q)’s, just like the j(q)’s found in the Chudnovskys’ formulas! (Though the w(q)’s are multiplied by appropriate powers of 2.) The general form of the formula is then,
1/p = S h (An+B) /(2yw(q))n
In Weng’s website one find the polynomials that define w(q) for d up to 422500, so one can find analogous formulas, though Ramanujan found the smaller and easier cases. (Since w(q) = xk for some power k, one has to find k first.) However, unlike the formulas derived from the Hilbert class polynomials, I haven’t come across any explicit references how to derive A and B from w(q), other than in a passing mention in a paper (see paper by Chan, Gee, and Tan) specifically for odd d. It should be interesting to find other formulas like these, especially one that uses d = 163.
The last class polynomials we discussed in [1] were the Ramanujan class polynomials. To recall, the r-function r(q) has the q-series expansion,
r(q) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + …
and the appropriate root of the class polynomial gives the r(q)’s. These will useful in understanding Ramanujan’s last ten pi formulas in Mathworld’s list. This time, the formulas involve even discriminants d = 4m, where d has class number 2 (both fundamental and non-fundamental). Analogous to the Hilbert class polynomials, the approximation is,
eπ√m ≈ r(q) –/+ 104
for even and odd m, respectively. For even m, we have,
eπ√6 ≈ (4Ö3)4 - 106
eπ√10 ≈ 124 -104
eπ√18 ≈ 284 -104
eπ√22 ≈ (12Ö11)4 -104
eπ√58 ≈ 3964 -104
where d = 4*18 is non-fundamental. In the list, the pi formulas were in terms of rising factorials. To make the connection more clear, one can just express it in terms of ordinary factorials. Let, r = (4n)!/(n!4),
1/(8p) = Ö3 S r (8n+1)/(4Ö3)4n+2
1/(16p) = Ö8 S r (10n+1)/124n+2
1/(16p) = 3Ö3 S r (40n+3)/284n+2
1/(8p) = Ö11 S r (280n+19)/(12Ö11)4n+2
1/(16p) = Ö8 S r (26390n+1103)/3964n+2
One can easily see the r(q)’s in the denominators. For odd m, we have,
eπ√5 ≈ (4Ö2)4 + 100
eπ√13 ≈ (12Ö2)4 + 104
eπ√37 ≈ (84Ö2)4 + 104
and non-fundamental discriminants d = 4*9, 4*25,
eπ√9 ≈ 3*84 + 104
eπ√25 ≈ 20*244 + 104
giving us,
1/(4p) = S r (-1)n(20n+3)/(4Ö2)4n+2
1/(4p) = S r (-1)n(260n+23)/(12Ö2)4n+2
1/(4p) = S r (-1)n(21460n+1123)/(84Ö2)4n+2
1/(4p) = 3 S r (-1)n(28n+3)/(3*84)n+1/2
1/(4p) = 5 S r (-1)n(644n+41)/(20*244)n+1/2
and the r(q)’s are apparent in the denominators as well. How do we find formulas like these? It turns out the answer lies in transforming Weber polynomials into Ramanujan class polynomials, or w(q) ® r(q). Thus, the transformation not only gives us pi formulas but allows us to verify the polynomials we derived for r(q). The transformations are given by,
For even m,
r(q)4m = (4ÖC)4, where C = (gm12 + gm-12)/2
For odd m,
r(q)4m = (4ÖC)4, where C = (Gm12 - Gm-12)/2
where gm and Gm are the Ramanujan g- and G-functions. See http://mathworld.wolfram.com/Ramanujang-andG-Functions.html. These are basically identical to the w(q) as given by the relationships,
w(q)4m = gm24, w(q)4m = Gm24
Thus, let d = 4m and the transformation is, for even m,
r(q)d = (4ÖC)4, where C = (w(q)d1/2 + w(q)d-1/2)/2
and for odd m,
r(q)d = (4ÖC)4, where C = (w(q)d1/2 – w(q)d-1/2)/2
Before we go into the pi formulas, we can give an example. Since we have,
eπ√58 ≈ 26((5+√29)/2)12 + 24
Then,
w(q)232 = ((5+√29)/2)12
Or,
C = (((5+√29)/2)6 + ((5+√29)/2)-6) /2 = 992
Since,
r(q)232 = (4Ö(992))4 = 3964
Thus,
eπ√58 ≈ 3964 -104
as expected. In [1], we pointed out that Ramanujan gave r(q) for d with class number 2 and 4. However, we managed to find some for d = 4m with class number 6, for m = 26, 106, 202, 298 using the Integer Relations applet, the first one being,
eπ√26 ≈ (4y)4 – 104; (y3-13y2-9y-11=0)
With this transformation, we can verify if indeed these are correct. From Weng’s website, we find that the polynomial for d = 4*26 = 104 is given by,
x6-2x5-2x4+2x2-2x-1 = 0 (eq.1)
such that,
eπ√26 ≈ 26x12 + 24
with x the appropriate real root. Then w(q)104 = x12. Or,
C = (x6 + x -6) /2
It should be pointed out that it will not always be the case that w(q)4m = x12, where x is the appropriate root of the listed polynomial. In general, it is w(q)4m = xk, where,
eπ√m ≈ 26xk +/- 24
so,
C = (xk/2 +/- x -k/2) /2
for even and odd m, respectively, so as was stated earlier one has to determine the power k first. In [1] we gave examples for k and rules when m = 2q and m = p for p, q prime, but it gets trickier if they are composite. It is so much easier with Hilbert class polynomials with root j(t) since it is automatically,
eπ√(4m) ≈ |j(t)| +/- 744
though the disadvantage now is these polynomials have terribly large coefficients. Continuing with our example, we will assume C to be a perfect square, C = y2,
y2 = (x6 + x -6) /2 (eq.2)
A second point to remember, it will not always be the case that C will be a perfect square. Our example involves m = 2q where q is prime. Anyway, by eliminating the variable x between eq.1 and eq.2, we get,
(y3+13y2-9y+11)2(y3-13y2-9y-11)2 = 0
since,
r(q)d = (4ÖC)4, and C = y2
then r(q)104 = (4y)4. Or,
eπ√26 ≈ (4y)4 – 104; (y3-13y2-9y-11=0)
for the real root of y, which is precisely the same relation as the one above! So satisfying to know that the applet was correct. The same way can be used to prove the three relations for the other d. And not only proving, but finding other r(q) as well. In [1], we asked for the explicit equation for the r(q) for d with class number 8, with d = 4m, and m = 178, 226, 466, 562 (which again are form 2q for prime q). From the same very useful website, we find that the Weber polynomial for d = 4*178 = 712 has the coefficient list,
{1, -21, -51, -30, 2, -30, -51, -21, 1}
which is a palindromic octic equation (in the variable x). We find that,
eπ√178 ≈ 26x12 + 24
with x the appropriate real root, so w(q)712 = x12. Using the same approach as the one above, we then have,
eπ√178 ≈ (12y)4 – 104; (y4-2962y3+72y2-382y-71=0)
with y the appropriate real root. The same method can be used for the other d as well as for those with even higher class numbers. For the third time, now that we have the preliminaries out of the way, we can discuss the pi formulas. The general form is given by, let r = (4n)!/(n!4),
1) Even m.
1/(16p) = S r (An + B)/(4ÖC)4n+2
where,
A = ((gm12 – gm-12)/2)Öm; C = (gm12 + gm-12)/2
or, equivalently, let w(q)d = xk,
A = ((xk/2 – x -k/2)/2)Öm; C = (xdk/2 + x -k/2)/2
2) Odd m.
1/(16p) = S r (-1)n(An + B)/(4ÖC)4n+2
where,
A = ((Gm12 + Gm-12)/2)Öm; C = (Gm12 – Gm-12)/2
hence there are two differences: first, for the latter, it is an alternating sum and second, there is an exchange of arithmetic operations in the expressions for A and C. There is an explicit expression for B but again it is too complicated for this article. The curious reader is referred to the paper by Bailey and the Borweins for the details. The alternative method that we will use is again to solve for B using enough terms in the summation and to use the Integer Relations applet to determine its defining polynomial. For example, since we know that w(q)232 = xk = ((5+√29)/2)12, then,
A = ((xk/2 – x -k/2)/2)Ö58 = 52780Ö2; C = (xk/2 + x -k/2)/2 = 992
and solving for B up to a sufficient accuracy, the applet gives us the quadratic polynomial which has a root B = 2206Ö2. So,
1/(16p) = S r (52780Ö2n + 2206Ö2)/3964n+2
or, moving the common factor 2Ö2,
1/(16pÖ8) = S r (26390n+1103)/3964n+2
and we have Ramanujan’s formula. In [1] we wondered if we could use higher class numbers and we certainly can. Using d = 4m with class number 4 for m = 34, 82, we have the formulas,
1/(8pÖ2) = (4+Ö17) S r (v1n+v2)/(4v3)4n+2
1/(16pÖ2) = (32+5Ö41) S r (w1n+w2)/(4w3)4n+2
where,
v1 = 8(165+41Ö17)1/2Ö17; v2 = (411+163Ö17)1/2; v3 = 3(4+Ö17);
and,
w1 = 20(1347+211Ö41)1/2Ö82; w2 = (51099+9097Ö41)1/2; w3 = 3(51+8Ö41)
Note that (4+Ö17) and (32+5Ö41) are fundamental units, involved in the solutions to the Pell equations x2 – 17y2 = -1 and x2 – 41y2 = -1. They were factored ffrom the original expressions of the vi and wi for i = 1, 2 which were unwieldy. For class number 6 for m = 26, we have,
1/(16pÖ8) = S r (x1n+x2)/(4x3)4n+2
where,
x1 = (x3-338x2-364x-1352)1; x2 = (8x3-168x2-27x-27)1; x3 = (x3-13x2-9x-11)1
or x1, x2, x3 are the real roots of the cubics above, respectively, which anyway are just one-real root cubics. These are just examples and there are certainly other d we can use with class number 4, 6, 8, ad infinitum.
What is the Baby Monster? Richard Borcherds, who proved the Monstrous Moonshine Conjecture, had a cute answer to this question. In a Notice of the AMS, he wrote a short article, “What is the Monster?” As to what the Baby Monster is, he jokingly wrote that when he was a grad student, his supervisor John Conway3 would bring his one-year old son to the department, who was soon known as the baby monster.4 For the full article, see http://math.berkeley.edu/~reb/papers/whatismonster/whatismonster.pdf.
Joking aside, the Baby Monster, also known as Fischer’s baby monster group, is the second largest of the sporadic finite groups. Many of its properties were described by B. Fischer, who also contributed to the discovery of the largest sporadic finite group, namely the Monster Group, also known as Fischer-Griess Monster.
What is the connection between these groups and the modular functions j(q), w(q), and r(q) that we have discussed? In [1], we have pointed out that the coefficients of their expansions are in fact McKay-Thompson series for the Monster. Another way to illustrate the connection is to express the coefficients of the normalized j-function j(t) in terms of the irreducible representations of the Monster. To recall,
j(t) - 744 = 1/q + 196884q + 21493760q2 + 864299970q3 + …
Since the degrees of the irreducible representations of Monster Group “M” are given by 1, 196883, 21296876, 842609326, etc…(A001379, Sloane’s Online Encyclopedia of Integer Sequences), we observe that,
1 = 1
196884 = 1 + 196883
21493760 = 1 + 196883 + 21296876
864299970 = 1(2) + 196883(2) + 21296876 + 842609326
or the coefficients of j(t) are simple linear sums of the representations of the Monster. (The second identity, which led to the Monstrous Moonshine Conjecture, was observed by John McKay in the late 1970’s. J.G. Thompson carried it further to the other coefficients.)
What I noticed was that there seemed to be an analogous relationship between Ramanujan’s function r(q), (which is also defined by the transformation rule w(q) ® r(q) that we have given) and the Baby Monster. The normalized r(q) would be,
r(q) - 104 = 1/q + 4372q + 96256q2 + 1240002q3 + …
(See A007267.) Given the degrees of irreducible representations of Baby Monster Group “B”: 1, 4371, 96255, 1139374, etc… (A001378), then,
1 = 1
4372 = 1 + 4371
96256 = 1 + 4371(0) + 96255
1240002 = 1(2) + 4371 + 96255 + 1139374
Is this coincidence? Will the relationship exist only for the first few terms, or does it go on for the whole sequence? The relationship between j(t) and the Monster has already been proven so does this automatically imply this relationship? After all, the coefficient list of r(q) is also the McKay-Thompson series of class 2A for Monster, and since the double cover of the Baby Monster is a subgroup of the Monster, then r(q) and the Baby Monster should be connected via the Monster group, though that is just a guess.
This article is meant to be some sort of a sequel to Ramanujan’s Constant and Its Cousins [1], as there were some loose ends left hanging there. I hope we have clarified some points though I believe there are still others that can be explored further. The conclusions we have stated here are to supersede those of the earlier article, especially with regards to pi formulas, since this is the expanded version of that topic after all.
Because of space constraints, there were a lot of intriguing avenues that I didn’t go into in this article. For example, while in [1] we discussed the modular functions j(q), w(q), and r(q) and here we discussed the transformation w(q) ® r(q), there is in fact also a second transformation j(q) ® r(q). However, for the purposes of this article, since Weng’s website has w(q) polynomials galore and there is no corresponding website for j(q) (and understandably, as these have LARGE coefficients), the first transformation was more useful. However, there is still a third transformation w(q) ® j(q) and that might be interesting to explore, though perhaps we can do so for another time.
To sum up the main theme of this article, which was on pi formulas, we have discussed three formulas that use the modular function w(q) and ten formulas that use r(q). That makes thirteen. But Ramanujan had seventeen formulas. What about the other four? It turns out that they use still another modular function.
But that is another story.
Footnotes:
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© 2005
Titus Piezas III
Mar 31, 2005
www.oocities.org/titus_piezas/ramanujan.html ¬ Click here for an index of articles.
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