Alt L2 Function

Just as the L1 function can be given two expressions i.e. sum over the positive integers and product over the primes respectively, equally the L2 function can be given two expressions.

In previous blog entries I have referred to this latter expression (which has its roots in the L1 function) as the Alt L2 function.

Now with respect to the Riemann zeta function, a special case occurs where s = 1, yielding the harmonic series,

1/1 + 1/2  + 1/3 + 1/4 + …

This can then be expressed in terms of combinations as

1/1C₀ + 1/2C₁ + 1/3C₂ + 1/4C₃  + …

Now, with respect to the denominator, we can define a new series where the nth term represents the sum of the n terms of the previous series.

So we thereby obtain

1/1 + 1/(1 + 2) + 1/(1 + 2 + 3) + 1/(1 + 2 + 3 + 4) + …

= 1 + 1/3 + 1/6 + 1/10 + …

Now this can equally be expressed in terms of combinations,

i.e. 1/2C₀ + 1/3C₁ + 1/4C₂ + 1/5C₃  + …

So the get this new series, we only need to increase the starting number of each combination (in relation to the previous harmonic series) by 1.

Now if we look at this later combination expression (which starts with 1/2C₀), we find that its value is equal to the corresponding L2 function i.e. 1 + x¹ + x²  + x³  +…, where x = 1/2.

So 1 + 1/2 + 1/2²  + 1/2³ + …   = 2, and  

1 + 1/3 + 1/6 + 1/10 + …  = 2.

And this is universally the case for any integer k (k > 1, where x = 1/k) .

So when x = 1/k, then the L2 function i.e.

1/(1 – 1/k) = 1 + 1/k + 1/k²  + 1/k³ + …  has a corresponding Alt L2 expression as,

1/kC₀ + 1/{(k + 1)C₁}  + 1/{(k + 2)C₂} + 1/{(k + 3)C₃} + …

Therefore when for example k = 4,

1/(1 – 1/4) = 1 + 1/4 + 1/4²  + 1/4³ + …  = 4/3.

So this represents the appropriate L2 function in this case where x = 1/2² = 1/4.

And this can be given an Alt L2 expression as,

1/4C₀ + 1/5C₁ + 1/6C₂ + 1/7C₃ + …

= 1 + 1/5 + 1/15 + 1/35 + …  = 4/3

In fact a fascinating product type expression can also be given for this Alt L2 function.

So when as in this case the starting value of the combination is 4, we let the numerator and denominator of the 1st term = 4 * 3 * 2 * 1.

Thus the 1^st term = (4 * 3 * 2 * 1)/(4 * 3 * 2 * 1) = 1.

Then to get the next term we simply add 1 to each value in the denominator while holding the numerator fixed.

So the 2^nd term = (4 * 3 * 2 * 1)/(5 * 4 * 3 * 2) = 1/5.

And for each subsequent term, we keep repeating by adding 1 to each previous value in the denominator (while holding the numerator fixed).

So the 3^rd term = (4 * 3 * 2 * 1)/(6 * 5 * 4 * 3)  = 1/15,

and the 4^th term = (4 * 3 * 2 * 1)/(7 * 6 * 5 * 4)  = 1/35 and so on.

And the existence here of 4 product terms (in relation to numerator and denominator) coincides with the fact that the value of k (with respect to the L2 function) = 4.