## Thursday, May 26, 2016

### 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) = 1/11/2 + 1/21/2 + 1/31/2 + 1/41/2 + ......

Therefore as the size of each term successively increases, the series 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/(rs – 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/(rn – 1) - 1/(0 – 1 ) = 1/– 1 =  – 1.

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

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

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

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 below 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.

 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

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.