Approximating ζ(s) for 0 < s < 1

In the range 0 < s < 1, the Riemann zeta function clearly diverges from the conventional perspective.

So for example when s = .5,  ζ(.5) = 1/1^1/2 + 1/2^1/2 + 1/3^1/2 + 1/4^1/2 + ...

And as each term - apart from the first whci is the same - will be greater than the corresponding harmonic series 1/1 + 1/2 + 1/3+ 1/4 + ..., it thereby, like the harmonic, diverges to infinity from this standpoint.

However by analytic continuation the function can be given a finite value for all values of s between 0 and 1 (including 0).

So it is very interesting in these circumstances that the extended Fibonacci sequence can be used very simply to provide approximations for all these values (between 0 and 1).

The ratio in this case does not even require the generation of successive terms in the sequence, for values of s (between 0 and 1) correspond with the 1-step case.

Therefore to compute the approximation - say - of ζ(.5), all the terms in the 1-step sequence will increase at a constant rate of .5 i.e. 1, .5, .25, .125,.... Therefore the ratio stays fixed at .5.

So using the formula 1/(r_s – 1), we obtain the approximation 1/(.5 – 1) = 1/(– .5) = – 2.

This compares with the true value for ζ(.5) =  – 1.460...

Therefore, though our approximation does indeed include a negative sign, it is still not very accurate in percentage terms.

However there is a simple way to improve such accuracy.

At the one extreme of the interval between 0 and 1, the extended Fibonacci approximation is 100% accurate i.e. where s = 1. Here the approximation, which is ∞, concurs exactly with the corresponding value of ζ(1).

Then at the other extreme i.e. where s = 0, the extended Fibonacci approximation is 200% of correct value, for when s = 0, 1/(r_n – 1) - 1/(0 – 1 ) = 1/– 1 =  – 1.

However the true value for ζ(0) = – .5.

So what is required is to put in a correcting factor which varies over the interval from 1 to 0 by 2^k 
where k varies (in reverse fashion) from 0 to 1. 


So we simply divide the estimate - obtained by the extended Fibonacci approximation - by 2^k .

Therefore the estimate for ζ(.5) = – 2/(2^.5) = – 1.414..

So this estimate does now indeed compare favourably with the true estimate of – 1.460.

And to illustrate further, ζ(.3) is now approximated by {1/((.3  – 1)}/2^.7 = .879388


And even simpler way of obtaining these estimates for ζ(s), where s ranges from 0 to 1, can now be suggested.

This is given by the formula (s + 1)/2(s  – 1)

So again when s = .5, we obtain (.5 + 1)/2(.5 – 1) = 1.5/(2( – .5) =  –  1.5.

This again compares very favourably with the true value for ζ(.5)  =  – 1.460.

In the table at the bottom, I give the actual values for  ζ(s) from .1 to .9. I then give the approximations using the two methods (i.e. extended Fibonacci and simple formula). Finally I give the % accuracy for both approximations. 

So we can see that both approximations provide estimates for ζ(s) with a consistently high degree of accuracy over the interval from 0 to 1. 

In fact since the first estimate consistently undershoots the correct value, while  the second consistently overshoots it, a much more accurate estimate can be obtained by obtaining the average of the two estimates obtained.

Therefore for example for ζ(.5), our new estimate = (– 1.414213 – 1.500000)/2 = – 1.457106, which is now 99.8% accurate!

And again, we are not confined just to these estimates for 0 to .9. All values within the range of 0 and 1 can be calculated using the same basic formulae.

 

s	ζ(s)	Ex. Fibonacci approx.	% Relative accuracy	Simple Formula approx.	% Relative accuracy
0	– .500000	– .500000	100.0	– .500000	100.0
.1	–  603037	– .595429	98.7	– .611111	97.4
. 2	– .733920	– .717936	97.8	– .750000	97.9
.3	– .904559	– .879388	97.2	– .928571	97.4
.4	– 1.134797	– 1.099589	96.9	– 1.166666	97.3
.5	– 1.460354	– 1.414213	96.8	– 1.500000	97.3
.6	– 1.952661	– 1.894645	97.0	– 2.000000	97.6
.7	– 2.778388	– 2.707507	97.4	– 2.833333	98.1
.8	– 4.437538	– 4.352752	98.1	– 4.500000	98.6
.9	– 9.430114	– 9.330329	98.9	– 9.500000	99.2