Tuesday, December 31, 2019

Ramanujan's Tau Function and Euler's Pentagonal Number Theorem

Ramanujan found that when the infinite series,

x(1 – x)24(1 – x2) 24(1 – x3)24(1 – x4)24  is multiplied out we obtain

1x 24x2 + 252x3 – 1472x4 + 4830x5 – 6048x6 – 16744x7 + 84480x8 – … 

This is now known as the Ramanujan tau function.

Now if we divide by x then

(1 – x)24(1 – x2)24(1 – x3)24(1 – x4)24  

= {(1 – x)(1 – x2)(1 – x3)(1 – x4) …}24  

 = 1 24x + 252x2 – 1472x3 + 4830x4 – 6048x5 – 16744x6 + 84480x7 – …  

So the tau values for each term thereby remain unchanged.

Interestingly, Euler earlier discovered that

(1 – x)(1 – x2)(1 – x3)(1 – x4) …

= x0 – x– x2  + x+ x– x12 – x15 + x22 + x26 – …

So the powers of the x terms in this infinite series comprise the generalised pentagonal numbers

i.e. 0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, …

This thereby means that we can show a fascinating connection as between these generalised pentagonal numbers and the corresponding Ramanujan tau values.

Thus {x0 – x– x2  + x+ x– x12 – x15 + x22 + x26 – …}24

= 1 24x + 252x2 – 1472x3 + 4830x4 – 6048x5 – 16744x6 + 84480x7 – …  

Then a related Euler formula shows that

(1 – x)(1 – x2)(1 – x3)(1 – x4) …      =  1 – {x/(1 – x)} + x3/{(1 – x)(1 – x2)}
– x6/{(1 – x)(1 – x2)( 1 – x3)} + …

So this latter expression involves (in the sole x term of each numerator) the triangular numbers 1, 3, 6, 10, 15, …

Therefore we can equally show a fascinating connection as between the Ramanujan tau values and the triangular numbers.