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Which method would you recommend for error estimation of the following approximation? $$\frac{1}{K} \sum_{j=0}^{K-1}\frac{cos(2\pi\frac{j}{K}u)}{P_{n}(\cos[\pi\frac{j}{K}])}\approx\int_{0}^{1}\frac{cos(2\pi xu)}{P_{n}(\cos[\pi x])}dx$$ Here $P_{n}$ some polynomial $u=1,2...K/2$

$\frac{1}{12k^2}f''(\psi)$ is a very bad estimator

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Perhaps explain why the second derivative one is a bad estimator? – Willie Wong Jul 8 '10 at 12:39
I think the function is not smooth enough. Specially for u close to K/2. But even for u=1 the real error is a lot smaller then f''(x) – vilvarin Jul 8 '10 at 13:17
maybe I should look at it like at fourier series? – vilvarin Jul 8 '10 at 14:27
Do you have an explicit formula for the Polynomial? Or at least do you know where and to what order the roots are? – Willie Wong Jul 8 '10 at 17:21
Function $P_n(cos(\pi x))$ depends on some parameter . It can be whether convex, bounded with 0.5 and 1 (this case is more interesting for me) or concave, bounded with 1 and some A>1(depends on the parameter) – vilvarin Jul 9 '10 at 23:03
up vote 3 down vote accepted

My first guess would be to use the Euler-Maclaurin summation formula (Wikipedia article). This proves, amongst other things, that the error goes down exponentially if the integrand is a periodic function on [0,1].

Added: After thinking about it a bit more, I'm wondering about some things. Firstly, the formula given in the question is not the trapezoidal rule (as promised in the title and suggested by the result for the error), but it is the rectangle rule which is only first order. Secondly, if the integrand has poles in [0,1] (that is, if $P_n(\cos(\pi x))=0$ for some $x\in[0,1]$), then the error estimate becomes meaningless; in this case you probably need different techniques like complex analysis to prove anything. A final remark: perhaps you can use the elementary techniques explained in: Weideman, "Numerical integration of periodic functions: a few examples", Amer. Math. Monthly 109 (2002), no. 1, 21-36 (MathSciNet).

I think I need some more background in order to have further help. In particular, do you know anything about the polynomials $P_n$, and what kind of result do you hope to get?

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Isn't the second derivative esimate essentially the Euler-Maclaurin one? – Willie Wong Jul 8 '10 at 15:20
@Willie: The Euler-Maclaurin formula gives an asymptotic series and if you truncate after the first term you get the second derivative estimate. Truncating later gives other estimates involving higher derivatives. @vilvarin: I'm not sure what your problem is. I would have thought that P_n is the main problem, while u is easy to handle. Perhaps you can give more details on what you have done up to now? – Jitse Niesen Jul 9 '10 at 9:50

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