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Denominators cannot be zero. in .NET Printer PDF 417 in .NET Denominators cannot be zero.




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Denominators cannot be zero. use vs .net pdf417 integrated toget pdf417 2d barcode for .net Microsoft Word 6.9 Fresnel Integrals, Cosine and Sine Integrals cc=Cadd(b, Cdiv(Complex(a,0.0),cc)); del=Cmul(cc,d); h=Cmul(h,del); if (fabs(del.r-1.

0)+fabs(del.i) < EPS) break; } if (k > MAXIT) nrerror("cf failed in frenel"); h=Cmul(Complex(ax,-ax),h); cs=Cmul(Complex(0.5,0.

5), Csub(ONE,Cmul(Complex(cos(0.5*pix2),sin(0.5*pix2)),h))); *c=cs.

r; *s=cs.i; } if (x < 0.0) { *c = -(*c); *s = -(*s); } } Use antisymmetry.

. Cosine and Sine Integrals The cosine and sine integrals are de ned by Ci(x) = + ln x + cos t 1 dt t Si(x) =. sin t dt t (6.9.6).

Here 0 .5772 . .

. is Euler s constant. We only need a way to calculate the functions for x > 0, because Si( x) = Si(x), Ci( x) = Ci(x) i (6.

9.7). Once again VS .NET barcode pdf417 we can evaluate these functions by a judicious combination of power series and complex continued fraction. The series are x3 x5 + 3 3! 5 5! x4 x2 + Ci(x) = + ln x + 2 2! 4 4! Si(x) = x The continued fraction for the exponential integral E1 (ix) is E1 (ix) = Ci(x) + i[Si(x) /2] = e ix = e ix 1 1 2 2 1 ix + 1 + ix + 1 + ix + 12 22 1 1 + ix 3 + ix 5 + ix (6.

9.9). (6.9.8).

The even .NET PDF-417 2d barcode form of the continued fraction is given in the last line and converges twice as fast for about the same amount of computation. A good crossover point from the alternating series to the continued fraction is x = 2 in this case.

As for the Fresnel integrals, for large x the precision may be limited by the precision of the sine and cosine routines.. 6. . Special Functions #include & lt;math.h> #include "complex.h" #define EPS 6.

0e-8 #define EULER 0.57721566 #define MAXIT 100 #define PIBY2 1.5707963 #define FPMIN 1.

0e-30 #define TMIN 2.0 #define TRUE 1 #define ONE Complex(1.0,0.

0). Relative e .net framework PDF 417 rror, or absolute error near a zero of Ci(x). Euler s constant .

Maximum number of iterations allowed. /2. Close to smallest representable oating-point number.

Dividing line between using the series and continued fraction.. void cisi( float x, float *ci, float *si) Computes the cosine and sine integrals Ci(x) and Si(x). Ci(0) is returned as a large negative number and no error message is generated. For x < 0 the routine returns Ci( x) and you must supply the i yourself.

{ void nrerror(char error_text[]); int i,k,odd; float a,err,fact,sign,sum,sumc,sums,t,term; fcomplex h,b,c,d,del; t=fabs(x); if (t == 0.0) { Special case. *si=0.

0; *ci = -1.0/FPMIN; return; } if (t > TMIN) { Evaluate continued fraction by modi ed b=Complex(1.0,t); Lentz s method ( 5.

2). c=Complex(1.0/FPMIN,0.

0); d=h=Cdiv(ONE,b); for (i=2;i<=MAXIT;i++) { a = -(i-1)*(i-1); b=Cadd(b,Complex(2.0,0.0)); d=Cdiv(ONE,Cadd(RCmul(a,d),b)); Denominators cannot be zero.

c=Cadd(b,Cdiv(Complex(a,0.0),c)); del=Cmul(c,d); h=Cmul(h,del); if (fabs(del.r-1.

0)+fabs(del.i) < EPS) break; } if (i > MAXIT) nrerror("cf failed in cisi"); h=Cmul(Complex(cos(t),-sin(t)),h); *ci = -h.r; *si=PIBY2+h.

i; } else { Evaluate both series simultaneously. if (t < sqrt(FPMIN)) { Special case: avoid failure of convergence sumc=0.0; test because of under ow.

sums=t; } else { sum=sums=sumc=0.0; sign=fact=1.0; odd=TRUE; for (k=1;k<=MAXIT;k++) { fact *= t/k; term=fact/k; sum += sign*term; err=term/fabs(sum); if (odd) { sign = -sign; sums=sum; sum=sumc; } else { sumc=sum; sum=sums;.

6.10 Dawson s Integral } if (err < EPS) break; odd=!odd; } if (k > MAXIT) nrerror("maxits exceeded in cisi"); } *si=sums; *ci=sumc+log(t)+EULER; } if (x < 0.0) *si = -(*si); }. CITED REFE pdf417 for .NET RENCES AND FURTHER READING: Stegun, I.A.

, and Zucker, R. 1976, Journal of Research of the National Bureau of Standards, vol. 80B, pp.

291 311; 1981, op. cit., vol.

86, pp. 661 686. Abramowitz, M.

, and Stegun, I.A. 1964, Handbook of Mathematical Functions, Applied Mathematics Series, Volume 55 (Washington: National Bureau of Standards; reprinted 1968 by Dover Publications, New York), s 5 and 7.

. 6.10 Dawson s Integral Dawson s Integral F (x) is de ned by F (x) = e x et dt (6.10.1).

The functi on can also be related to the complex error function by i z2 e [1 erfc( iz)] . F (z) = 2 A remarkable approximation for F (x), due to Rybicki [1], is 1 F (z) = lim h 0 e (z nh) n. (6.10.2).

(6.10.3).

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