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176Chapter 5.Evaluation of FunctionsRational FunctionsYou evaluate a rational function likeR(x) =(5.3.4)in the obvious way, namely as two separate polynomials followed by a divide. As amatter of convention one usually chooses q0 = 1, obtained by dividing numeratorand denominator by any other q0 . It is often convenient to have both sets ofcoefficients stored in a single array, and to have a standard function available fordoing the evaluation:double ratval(double x, double cof[], int mm, int kk)Given mm, kk, and cof[0..mm+kk], evaluate and return the rational function (cof[0] +cof[1]x + · · · + cof[mm]xmm )/(1 + cof[mm+1]x + · · · + cof[mm+kk]xkk ).{int j;double sumd,sumn;Note precision! Change to float if desired.for (sumn=cof[mm],j=mm-1;j>=0;j--) sumn=sumn*x+cof[j];for (sumd=0.0,j=mm+kk;j>=mm+1;j--) sumd=(sumd+cof[j])*x;return sumn/(1.0+sumd);}CITED REFERENCES AND FURTHER READING:Acton, F.S.
1970, Numerical Methods That Work; 1990, corrected edition (Washington: Mathematical Association of America), pp. 183, 190. [1]Mathews, J., and Walker, R.L. 1970, Mathematical Methods of Physics, 2nd ed. (Reading, MA:W.A. Benjamin/Addison-Wesley), pp. 361–363. [2]Knuth, D.E. 1981, Seminumerical Algorithms, 2nd ed., vol. 2 of The Art of Computer Programming(Reading, MA: Addison-Wesley), §4.6. [3]Fike, C.T. 1968, Computer Evaluation of Mathematical Functions (Englewood Cliffs, NJ: PrenticeHall), Chapter 4.Winograd, S. 1970, Communications on Pure and Applied Mathematics, vol. 23, pp.
165–179. [4]Kronsjö, L. 1987, Algorithms: Their Complexity and Efficiency, 2nd ed. (New York: Wiley). [5]5.4 Complex ArithmeticAs we mentioned in §1.2, the lack of built-in complex arithmetic in C is anuisance for numerical work. Even in languages like FORTRAN that have complexdata types, it is disconcertingly common to encounter complex operations thatproduce overflows or underflows when both the complex operands and the complexresult are perfectly representable. This occurs, we think, because software companiesassign inexperienced programmers to what they believe to be the perfectly trivialtask of implementing complex arithmetic.Sample 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).p 0 + p1 x + · · · + pµ x µPµ (x)=Qν (x)q 0 + q 1 x + · · · + q ν xν1775.4 Complex ArithmeticActually, complex arithmetic is not quite trivial. Addition and subtractionare done in the obvious way, performing the operation separately on the real andimaginary parts of the operands.
Multiplication can also be done in the obvious way,with 4 multiplications, one addition, and one subtraction,(5.4.1)(the addition before the i doesn’t count; it just separates the real and imaginary partsnotationally). But it is sometimes faster to multiply via(a + ib)(c + id) = (ac − bd) + i[(a + b)(c + d) − ac − bd](5.4.2)which has only three multiplications (ac, bd, (a + b)(c + d)), plus two additions andthree subtractions. The total operations count is higher by two, but multiplicationis a slow operation on some machines.While it is true that intermediate results in equations (5.4.1) and (5.4.2) canoverflow even when the final result is representable, this happens only when the finalanswer is on the edge of representability.
Not so for the complex modulus, if youare misguided enough to compute it as|a + ib| =pa2 + b 2(bad!)(5.4.3)whose intermediate result will overflow if either a or b is as large as the squareroot of the largest representable number (e.g., 1019 as compared to 1038 ). The rightway to do the calculation is|a + ib| =p|a|p 1 + (b/a)2|b| 1 + (a/b)2|a| ≥ |b||a| < |b|(5.4.4)Complex division should use a similar trick to prevent avoidable overflows,underflow, or loss of precision,[a + b(d/c)] + i[b − a(d/c)]a + ib c + d(d/c)=[a(c/d)+b] + i[b(c/d) − a]c + id c(c/d) + d|c| ≥ |d|(5.4.5)|c| < |d|Of course you should calculate repeated subexpressions, like c/d or d/c, only once.Complex square root is even more complicated, since we must both guardintermediate results, and also enforce a chosen branch cut (here taken to be thenegative real axis).
To take the square root of c + id, first compute0 sp2p |c| 1 + 1 + (d/c)2w≡spp|c/d| + 1 + (c/d)2 |d|2c=d=0|c| ≥ |d||c| < |d|(5.4.6)Sample 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).(a + ib)(c + id) = (ac − bd) + i(bc + ad)178Chapter 5.Evaluation of FunctionsThen the answer is√0 w=0dw+iw=6 0, c ≥ 02wc + id =w 6= 0, c < 0, d ≥ 0(5.4.7)w 6= 0, c < 0, d < 0Routines implementing these algorithms are listed in Appendix C.CITED REFERENCES AND FURTHER READING:Midy, P., and Yakovlev, Y. 1991, Mathematics and Computers in Simulation, vol.
33, pp. 33–49.Knuth, D.E. 1981, Seminumerical Algorithms, 2nd ed., vol. 2 of The Art of Computer Programming(Reading, MA: Addison-Wesley) [see solutions to exercises 4.2.1.16 and 4.6.4.41].5.5 Recurrence Relations and Clenshaw’sRecurrence FormulaMany useful functions satisfy recurrence relations, e.g.,(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)Jn+1 (x) =2nJn (x) − Jn−1 (x)x(5.5.1)(5.5.2)nEn+1 (x) = e−x − xEn (x)(5.5.3)cos nθ = 2 cos θ cos(n − 1)θ − cos(n − 2)θ(5.5.4)sin nθ = 2 cos θ sin(n − 1)θ − sin(n − 2)θ(5.5.5)where the first three functions are Legendre polynomials, Bessel functions of the firstkind, and exponential integrals, respectively.
(For notation see [1].) These relationsare useful for extending computational methods from two successive values of n toother values, either larger or smaller.Equations (5.5.4) and (5.5.5) motivate us to say a few words about trigonometricfunctions. If your program’s running time is dominated by evaluating trigonometricfunctions, you are probably doing something wrong. Trig functions whose argumentsform a linear sequence θ = θ0 + nδ, n = 0, 1, 2, .
. ., are efficiently calculated bythe following recurrence,cos(θ + δ) = cos θ − [α cos θ + β sin θ]sin(θ + δ) = sin θ − [α sin θ − β cos θ](5.5.6)Sample 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).|d|+ iw 2w |d| − iw2w.
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