From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: integrating periodic analytic functions
Date: 24 Jan 2001 17:44:27 -0500
Newsgroups: sci.math.num-analysis

In article <94nhsb$sru$1@nnrp1.deja.com>,  <f_nastov@my-deja.com> wrote:
:Hi,
:
:Is it true that the trapezoid method is exact for numerically
:integrating periodic analytic functions? Is there a minimum
:number of points that need to be used? If so, are the number of
:points related to the number of terms in the Fourier series.
:
:If I'm totally off on the above, is there an accurate method
:for integrating periodic functions that requires few points.

It's not true in general. What is true:

The error E(N) for number of points N converges to zero 
faster than any power 1/N^m. A little more precisely:

If the periodic function extends analytically to a closed 
(not necessarily maximal) strip abs(Im(z)) <= b then the 
errors are bounded by a multiple of exp(-c*N) where c>0 
depends in an increasing way on b. (Someone will find a 
specific form, I hope.)

Hope it helps, ZVK(Slavek)
==============================================================================

From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: integrating periodic analytic functions
Date: 25 Jan 2001 10:39:33 GMT
Newsgroups: sci.math.num-analysis

In article <94nhsb$sru$1@nnrp1.deja.com>,
 f_nastov@my-deja.com writes:
|> Hi,
|> 
|> Is it true that the trapezoid method is exact for numerically
|> integrating periodic analytic functions? Is there a minimum number
|> of points that need to be used? If so, are the number of points
|> related to the number of terms in the Fourier series.

in general: _no_
look at the asymptotic development of the error in the fixed stepsize
trapezoid rule 
T(h) def= (h/2)*(f(a)+2*sum_{i=1,n-1}f(a+i*h)+f(b)) where h=(b-a)/n

then if f is of arbitrarily high differentiability on [a,b]
(that means that all derivatives exist in ]a,b[ and have finite limits 
on the boundaries), then

               for arbitrary m in |N

T(h) = I + \sum_{i=1,m} h^{2*i} B_{2i}/((2i)!)(f^{2i-1}(b)-f^{2i-1}(a)) 
         + (b-a)*h^{2*m+2} B_{2m+2}/((2m+2)!) f^{2m+2}(xsi_m)

with some xsi_m in ]a,b[. Here B_{2i} are the Bernoulli numbers. 
I is the true integral.

Since

    2/((2*pi)^{2i}) <= | B_{2i}/((2i)!) | <= 4/((2*pi)^{2i})

it follows that in the case  

    (**)   |f^{2k}(x)|<=c^{2k} for all x in [a,b] and all k
           with some fixed c>0 
     and 
          h*c/(2*pi) < 1  

     and f periodic with period b-a

    it follows that 

     T(h)=I.

(your desired result).
The condition (**) is an extremely strong one. As far as I know it is valid only
for 
entier functions (from Pringsheims theorem, first time correctly proved by
Halmos).

hope that helps
peter