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The maximumdegree of exactness one would expect to achieve would therefore be 2n + m − 1.The question is whether this maximum degree of exactness can actually be achievedin practice, when the abscissas are required to all lie inside (a, b). The answer tothis question is not known in general.Kronrod showed that if you choose n = m + 1, an optimal extension canbe found for Gauss-Legendre quadrature. Patterson [12] showed how to computecontinued extensions of this kind. Sequences such as N = 10, 21, 43, 87, . . .
arepopular in automatic quadrature routines [13] that attempt to integrate a function untilsome specified accuracy has been achieved.4.6 Multidimensional Integrals161Golub, G.H. 1973, SIAM Review, vol. 15, pp. 318–334. [10]Kronrod, A.S. 1964, Doklady Akademii Nauk SSSR, vol. 154, pp. 283–286 (in Russian). [11]Patterson, T.N.L. 1968, Mathematics of Computation, vol.
22, pp. 847–856 and C1–C11; 1969,op. cit., vol. 23, p. 892. [12]Piessens, R., de Doncker, E., Uberhuber, C.W., and Kahaner, D.K. 1983, QUADPACK: A Subroutine Package for Automatic Integration (New York: Springer-Verlag). [13]Carnahan, B., Luther, H.A., and Wilkes, J.O. 1969, Applied Numerical Methods (New York:Wiley), §§2.9–2.10.Ralston, A., and Rabinowitz, P.
1978, A First Course in Numerical Analysis, 2nd ed. (New York:McGraw-Hill), §§4.4–4.8.4.6 Multidimensional IntegralsIntegrals of functions of several variables, over regions with dimension greaterthan one, are not easy. There are two reasons for this. First, the number of functionevaluations needed to sample an N -dimensional space increases as the N th powerof the number needed to do a one-dimensional integral.
If you need 30 functionevaluations to do a one-dimensional integral crudely, then you will likely need onthe order of 30000 evaluations to reach the same crude level for a three-dimensionalintegral. Second, the region of integration in N -dimensional space is defined byan N − 1 dimensional boundary which can itself be terribly complicated: It neednot be convex or simply connected, for example. By contrast, the boundary of aone-dimensional integral consists of two numbers, its upper and lower limits.The first question to be asked, when faced with a multidimensional integral,is, “can it be reduced analytically to a lower dimensionality?” For example,so-called iterated integrals of a function of one variable f(t) can be reduced toone-dimensional integrals by the formulaZxZ t3Z t2dtn−1 · · ·dt2f(t1 )dt1000Z x1=(x − t)n−1 f(t)dt(n − 1)! 0Ztndtn0(4.6.1)Alternatively, the function may have some special symmetry in the way it dependson its independent variables.
If the boundary also has this symmetry, then thedimension can be reduced. In three dimensions, for example, the integration of aspherically symmetric function over a spherical region reduces, in polar coordinates,to a one-dimensional integral.The next questions to be asked will guide your choice between two entirelydifferent approaches to doing the problem. The questions are: Is the shape of theboundary of the region of integration simple or complicated? Inside the region, isthe integrand smooth and simple, or complicated, or locally strongly peaked? DoesSample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press.Programs Copyright (C) 1988-1992 by Numerical Recipes Software.Permission is granted for internet users to make one paper copy for their own personal use.
Further reproduction, or any copying of machinereadable files (including this one) to any servercomputer, is strictly prohibited. To order Numerical Recipes books,diskettes, or CDROMsvisit website http://www.nr.com or call 1-800-872-7423 (North America only),or send email to trade@cup.cam.ac.uk (outside North America).Stoer, J., and Bulirsch, R. 1980, Introduction to Numerical Analysis (New York: Springer-Verlag),§3.6.Johnson, L.W., and Riess, R.D. 1982, Numerical Analysis, 2nd ed. (Reading, MA: AddisonWesley), §6.5..
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