How to realize Riemann Siegel Theta Function with MATLAB？

How to realize Riemann Siegel Theta Function with MATLAB？

Answer 1

[John D'Errico]

You start by writing the formula for it. That can be found on virtually the first line in this link.

https://en.wikipedia.org/wiki/Riemann–Siegel_theta_function

However in MATLAB, the gamma function seems to be defined only for a real variable, unless you use the symbolic toolbox version of gamma. So this should work:

t = (1:5)';
[t,angle(double(gamma(sym(1/4 + i*t/2)))) - log(pi)/2*t]
ans =
                         1         -1.76754795281229
                         2         -2.52591091881613
                         3         -2.99456469601083
                         4         -3.29063503121648
                         5         -3.45962037536346

That is valid for real arguments t. I'd need to do some reading to know if this works for complex arguments, though I assume it does not, since the wiki link I gave indicates it is valid for real arguments. Hopefully you care only about real arguments, and are uninterested in an extension to the complex plane.

[yifei wang]

Thank you for your reply. Because we have restrictions on using Google, I have not found it before.

There is a new question now. When I use webMathematica to calculate Riemann_Siegel theta funcion, the results of ± t are the same, but this formula is different. Which one do you think is more accurate?

[John D'Errico]

My apologies. I never think of others having a restriction on how they can search. A strange world we live in.

I generally trust Wolfram pretty highly for special function computations. Here, at Wolfram Alpha:

https://www.wolframalpha.com/input/?i=riemannsiegeltheta%282%29

it too yields a slightly different value from what I showed. Then I looked more carefully at what I had written, and I sleepily put in a paren in the wrong place.

This fixes it, and agrees with the other computations, at least for the values of t I used.

[t,angle(double(gamma(sym(1/4 + i*t/2)))) - log(pi)/2*t]
ans =
                         1         -1.76754795281229
                         2         -2.52591091881613
                         3         -2.99456469601083
                         4         -3.29063503121648
                         5         -3.45962037536346

As you can see, it now fully agrees with the Wolfram result.

I'll change my answer to reflect the correct set of parens in the formula so others will not be confused.

[yifei wang]

I got it. You are a great help. Thank you.

[兰花 西]

Excuse me, when I set the parameter to a relatively large imaginary number（-1.93097838020833 + 38142819340462.3 i） in the above formula, I cannot get a valid result. Do you know how to solve it? Thank you !