**III. Icosahedral group**

Given the *Rogers-Ramanujan identities* (see also here),

I observed that,

where, as in the previous post, is the j-function, , , and . Since it is known that,

this implies that,

Example. Let , hence . Then,

Furthermore, since Ramanujan established that,

if we define the two functions,

then the counterpart hypergeometric identity is also beautifully simple and given by,

In the next post, we will use one of the hypergeometric formulas to solve the general *quintic*.

Amazingly, Ramanujan found 3 continued fractions that can be associated with each group. See this article for more details. It turns out there are also corresponding hypergeometric formulas, and the numbers 12, 24, and 60 naturally appears.

First though, define the *j-function* as,

where,

This can be conveniently calculated in Mathematica as,

** NOTE**: In the formulas below, it will be assumed that,

**I. Tetrahedral group**

Given,

and,

then we have the simple relationship,

Example. Let , hence , then,

and *c(q)* can then be easily solved for as a cubic equation.

**II. Octahedral group**

Let,

then,

Example. Still using , then,

**III. Icosahedral group**

(To be discussed in the next post.)

]]>where is a *Pochhammer symbol* is, for rational {*a, b, c, z*}, generally not a rational value. In the link given above, equations (18) and (19) are,

There are in fact an infinite number of such equalities. One given by M. Glasser is, let , then,

Another, based on eq. (42) of Vidunas’ “*Transformations of algebraic Gauss hypergeometric functions*” is, let , then,

A third which yields not a rational but an algebraic number is remarkable for its connection to the *Rogers-Ramanujan continued fraction*. Let where,

then,

If the polynomials are familiar, it is because they are invariants of the *icosahedron*. They also appear in the *j-function* formula,

where,

and

]]>It can be shown that,

Consider the following evaluations,

In general, given a *root of unity*, , then,

for integer *p* > 1, any *non-zero real or complex a*, and where is the *digamma function*. Thus, since roots of unity are involved, the formula uses complex terms even though, as the two examples show, the sum may be real. But it turns out for real *a* and ** even** powers

for even *p* and any non-zero *a* except *a* = 1, which is given by the special case,

But one can split the *cotangent function* into its real and imaginary parts as,

hence cancel out the conjugate terms and leave only the real parts. For example, we have,

and so on. It is reminiscent of the situation with the zeta function,

which has a closed-form solution only for even *p*, and is expressed by the real and *Bernoulli numbers*. It makes me wonder if there is a closed-form formula for involving the roots of unity.

In *Identities Inspired from Ramanujan’s Notebooks*, Simon Plouffe recounts how, based on Ramanujan’s,

he found,

and similar ones for other *s* = 4m+3. On a hunch, and using Mathematica’s *LatticeReduce* function, I found that,

etc.

**II. Functions**

If we define,

then Plouffe discovered integer relations between,

for odd *s*, with *s* = 3 being,

Eliminating leads to the 3-term equalities in the *Introduction*. See Chamberland’s and Lopatto’s *Formulas for Odd Zeta Values*. On the other hand, by defining the function,

I observed integer relations between,

also for odd *s*, with *s* = 3 as,

and so on. Eliminating leads to the 4-term equalities in the *Introduction*.

**III. Conjecture**

The 4-term equalities have coefficients that are simple except for one term. Recall that,

*Conjecture*:

*“Using the positive case of* *for s = 4m+3, and the negative for s = 4m+5, then in the equation,*

*is a rational number.”*

The first few for *s* = {3, 7, 11,…} are while for *s* = {5, 9, 13,…} are These rationals may have a closed-form expression in terms of *Bernoulli numbers*, but I do not yet know the exact formulation.

I didn’t think this was an isolated result so set about to find a generalization. I found its counterpart,

Note that,

We can demystify the sum a bit by splitting the log function into parts. After some algebraic manipulation, we find that the first one becomes,

Thus it can be expressed in the form,

where {} are roots of the same equation, {} are roots of another, and *r* is a rational. The fact that,

was my clue that trigonometric functions may be involved. Define,

then for *p* = 5,

*p* = 7

*p* = 9

with the constants {} easily ascertained as {}, and so on. On the other hand, their counterparts are easier as the exponent has the same subscript as the base. Still defining,

then for *p* = 5,

*p* = 7

*p* = 9

etc.

]]>However, using the *Inverse Symbolic Calculator*, the first and the third, plus another one, can also be expressed as,

where . Interesting similar forms, isn’t it?

Unfortunately, it doesn’t seem to generalize to for *p* = 8. However, there is still *p* = 3 and, based on the even case, I assumed perhaps *roots of unity* are also involved. First, given the *Euler-Mascheroni constant* , and the *digamma function*,

where we suppress the subscript for ease of notation. Define,

and the *p*th root chosen such that , then I found that *p* = 3 generalizes as,

and so on, though a rigorous proof is needed that it holds true for all odd numbers *p*.

P.S. Going back to even *p*, note that *p* = 2, 4, 6 can also be expressed by the digamma function since,

for .

]]>where *k* is an *even integer* then,

for appropriate *z* such that the sum converges. For the special case when,

then,

Note that the terms are complex, but the sum is a real number so they must come in conjugate pairs. The arcsin of a complex root of unity can be given as,

where,

With this transformation, it is now possible to have an expression all in real terms. The case *k* = 2, 4 was given in the previous post. For *k* = 6, we have the counterpart to Sprugnoli’s equality as,

Note that the prime factors of *65* are *5* and *13*, and the square root of both appear above. However, for *k* = 8, while the expression contains the fraction as expected, the argument of the log and arcsin *do not* factor over the quadratic extension , but rather only over . Furthermore, the argument of the log for both *k* = 6, 8 are no longer simply expressible in terms of the *Dedekind eta function*, so observations for lower *k* do not generalize to higher ones.

For the next step, Renzo Sprugnoli gave the Ramanujan-like identity,

(The sign of the third term has been changed by this author.) However, to make it more symmetrical, we can express the *arctan* in terms of the *log function*. Since,

then,

In this manner, it reduces to the concise,

where, and are the appropriate roots of,

I found that, curiously, *the argument of the log can be expressed in terms of the Dedekind eta function*, . Let,

then,

*Is this coincidence?* Furthermore, using these as the argument of the *polylogarithm*,

one can find a *polylogarithm ladder* to express *Apery’s constant.* For example, getting the square root and reciprocal of so that *z* < 1,

then,

A simpler one exists for the other argument. The next step, of course, is,

Since the first three *Fermat primes* 3, 5, 17 have already appeared, it should be interesting to conjecture if 257 will be next.

When *x* = 0, they reduce into,

However, there is a *third* single-term equality,

*so there might be a third identity* that reduces to this as the special case *x* = 0.

To compare, there are three identities such that as , then those zeta values are the respective limit. For *x* *integer*, then,

The first two were found by Leshchiner and Koecher, respectively, while the third is *Theorem 2* in the same paper by Bailey, Borwein, and Bradley. The function is given in Mathematica as,

while is the *Euler-Mascheroni constant*. So are the Bailey-Borwein-Bradley pair of binomial sum identities in fact *a triplet*?