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Tham khảo tài liệu 'lập trình c# all chap "numerical recipes in c" part 48', công nghệ thông tin phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 4.6 Multidimensional Integrals 161 Golub 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 Stoer J. and Bulirsch R. 1980 Introduction to NumericalAnalysis New York Springer-Verlag 3.6. Johnson L.W. and Riess R.D. 1982 Numerical Analysis 2nd ed. Reading MA Addison-Wesley 6.5. 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 Integrals Integrals of functions of several variables over regions with dimension greater than one are not easy. There are two reasons for this. First the number of function evaluations needed to sample an N-dimensional space increases as the Nth power of the number needed to do a one-dimensional integral. If you need 30 function evaluations to do a one-dimensional integral crudely then you will likely need on the order of 30000 evaluations to reach the same crude level for a three-dimensional integral. Second the region of integration in N-dimensional space is defined by an N 1 dimensional boundary which can itself be terribly complicated It need not be convex or simply connected for example. By contrast the boundary of a one-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 to one-dimensional integrals by the formula iX dtn Îtn dtn-1 Ît3 dt2 Ît2 f ti dti J0 J0 J0 J0 T i X x - t n-1f t dt n - 1