From: Raymond Manzoni Subject: Re: Riemann zeta Date: Sat, 29 May 1999 11:33:46 +0200 Newsgroups: sci.math To: fred@rodan.moremagic.com Keywords: Evaluating zeta and finding zeros F.Baube(tm) wrote: > I asked where I could find the first nontrivial > zeroes of the Riemann zeta function, and Raymond > Manzoni kindly directed > me to these two pages: > > http://www.lacim.uqam.ca/piDATA/zeta100.html > http://www.research.att.com/~amo/zeta_tables/index.html > > I also found a very nifty page at > > http://www.math.nyu.edu/ms_students/jjm247/julia/julia.html > > entitled > > "Images of the Filled Julia Set for the Riemann Zeta Function" > > This last page also has a Mathematica formula for approximating > zeta, but it is very complex, and it does assume that one can > calculate with complex numbers. > > What I would like is an approximation for "zeta(s)" in terms of > (say) "t" and "u" where "s" is complex and is equal to "t + ui". > > Then I could verify the first non-trivial zeroes of the function > for myself. > > Can anyone point me to a URl with the formula I seek ? > > thanks, > > fred Hi again, I played myself near the critical line with a (great)pocket computer Ti 92 and I used it to draw the image of zeta along the line x=1/2+ui (and parallels). You'll find later an answer I gave last year concerning another quest. zeta(x) ~ sum(1/k^x, k=1..N) + 1/((x-1)*N^(x-1)) -1/(2*N^x)+x/(12*N^(x+1))-x*(x+1)*(x+2)/(720*N^(x+3))+x*(x+1)*(x+2)*(x+3)*(x+4)/(30240*N^(x+5))+... Well complex operations are involved, of course, but they may be converted to real ones if you know the rules for the 4 basis operations and if you remember that : k^x = exp(x*ln(k)) =exp((t + ui)*ln(k)) = exp(t*ln(k))*(cos(u*ln(k))+i*sin(u*ln(k))) or k^x = k^t * (cos(u*ln(k))+i*sin(u*ln(k))) Concerning the critical section, let's suppose that : zeta(1/2+ui) = r(u)*exp(phase(u)*i) (with r(u) and phase(u) real functions) then the phase may be written in function of the gamma function : phase(u) = u*ln(PI)/2-argument(gamma(1/4+u/2*i)) (modulo PI) I noted that an approximation of the phase was (I didn't verify my computations): phase(u) ~ C+1/2*[(ln(2*PI/u)+1)*u+ArcTan(exp(PI*u))-1/(24u)-7/(2880u^3)-31/(40320u^5)+...] where C is a constant I didn't search and where the ArcTan=tan^(-1) term may be considered as PI/2 when u>=10) So that you may simply study the real function r(u) by multiplying zeta(1/2+ui) by exp(-phase(u)*i). Of course the function r(u) has no known "so simple" expression! Well perhaps there's to much information here! If you have any more question, I'll try to answer them, Raymond Manzoni Further references : The Classical reference : Titchmarsh, Heath-Brown "The theory of the Riemann Zeta-function" see : http://www.amazon.com/exec/obidos/ASIN/0198533691/qid=927969894/sr=1-10/002-8202812-0518237 Free software for computing gamma in complex plane (and MUCH more to any precision) Pari/GP for very quick high precision operations at : http://hasse.mathematik.tu-muenchen.de/ntsw/pari/Welcome.html Mupad for symbolic operations of all kinds at : http://www.mupad.de/ Another fine site concerning Riemann zeta function (with many references) is available (somes days only) at : http://www.astro.virginia.edu/~eww6n/math/RiemannZetaFunction.html torquemada@my-dejanews.com wrote: > Is there some source code on line for evaluating the Riemann zeta function in > a reasonable chunk of the complex plane - in particular along the critical > line. It doesn't have to use arbitrary precision arithmetic or work for > arbitrarily large complex numbers - I'm not hunting for zeroes or anything > like that. > > Dan > > -----------== Posted via Deja News, The Discussion Network ==---------- > http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own Hi, I used the following formula for that purpose: zeta(x) ~ sum(1/k^x, k=1..N) + 1/((x-1)*N^(x-1)) -1/(2*N^x)+x/(12*N^(x+1))-x*(x+1)*(x+2)/(720*N^(x+3))+x*(x+1)*(x+2)*(x+3)*(x+4)/(30240*N^(x+5))+... The greater the N the better the result (but even N=10 is good) The accuracy is rather fine when |Im(x)| < 2*PI*N and becomes very bad after that ! You may find more terms using the more general Euler Mac-Laurin formula : This allows to explore better the negative values of x (in fact you will get the exact values for x=-1,-2, until -2 times the number of Bernoulli terms!). With more terms the precision in the imaginary plane is better under 2*PI*N and worse over it! f(x) ~ sum(f(k), k=1..N) + int(f(u),u=N..oo) -f(N)/2-B1/2!*f'(N)+B2/4!*f'''(N)-... where Bn means the nth Bernoulli number This formula was very fruitful for me even on my little TI 92. (N=100 was enough) Hope it will help you too. Raymond Manzoni Free time mathematician