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📄 log1p.c

📁 操作系统SunOS 4.1.3版本的源码
💻 C
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#ifndef lintstatic	char sccsid[] = "@(#)log1p.c 1.1 92/07/30 SMI";#endif/* * Copyright (c) 1987 by Sun Microsystems, Inc. *//* LOG1P(x)  * RETURN THE LOGARITHM OF 1+x * IEEE DOUBLE PRECISION  * CODE BASED ON 4.3BSD, MODIFIED BY K.C. NG, 6/29/87.  *  * Required system supported functions: *	scalbn(x,n)  *	copysign(x,y) *	ilogb(x)	 *	finite(x) * * Required kernel function: *	log__L(z) * * Method : *	1. Argument Reduction: find k and f such that  *			1+x  = 2^k * (1+f),  *	   where  sqrt(2)/2 < 1+f < sqrt(2) . * *	2. Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) *		 = 2s + 2/3 s**3 + 2/5 s**5 + ....., *	   log(1+f) is computed by * *	     		log(1+f) = 2s + s*log__L(s*s) *	   where *		log__L(z) = z*(L1 + z*(L2 + z*(... (L6 + z*L7)...))) * *	   See log__L() for the values of the coefficients. * *	3. Finally,  log(1+x) = k*ln2 + log(1+f).   * *	Remarks 1. In step 3 n*ln2 will be stored in two floating point numbers *		   n*ln2hi + n*ln2lo, where ln2hi is chosen such that the last  *		   20 bits (for VAX D format), or the last 21 bits ( for IEEE  *		   double) is 0. This ensures n*ln2hi is exactly representable. *		2. In step 1, f may not be representable. A correction term c *	 	   for f is computed. It follows that the correction term for *		   f - t (the leading term of log(1+f) in step 2) is c-c*x. We *		   add this correction term to n*ln2lo to attenuate the error. * * * Special cases: *	log1p(x) is NaN with signal if x < -1; log1p(NaN) is NaN with no signal; *	log1p(INF) is +INF; log1p(-1) is -INF with signal; *	only log1p(0)=0 is exact for finite argument. * * Accuracy: *	log1p(x) returns the exact log(1+x) nearly rounded. In a test run  *	with 1,536,000 random arguments on a VAX, the maximum observed *	error was .846 ulps (units in the last place). * * Constants: * The hexadecimal values are the intended ones for the following constants. * The decimal values may be used, provided that the compiler will convert * from decimal to binary accurately enough to produce the hexadecimal values * shown. */#include <math.h>#include "libm.h"double log1p(x)double x;{	static double zero=0.0, negone= -1.0, one=1.0, 		      half=1.0/2.0, small=1.0E-20;   /* 1+small == 1 */	double log__L(),z,s,t,c;	int ilogb();	int k,finite();	if(!finite(x)) return Inf+x;	/* x is +-INF or NaN */	else { if( x > negone ) {	   /* argument reduction */	      if(fabs(x)<small) {		   dummy(fmax-fabs(x));	/* raise inexact if x is not zero */		   return(x);	      }	      if (x>1e300) k=ilogb(x); else k=ilogb(one+x); 	      z=scalbn(x,-k); t=scalbn(one,-k);	      if(z+t >= sqrt2 ) 		  { k += 1 ; z *= half; t *= half; }	      t += negone; x = z + t;	      c = (t-x)+z ;		/* correction term for x */ 	   /* compute log(1+x)  */              s = x/(2+x); t = x*x*half;	      c += (k*ln2lo-c*x);	      z = c+s*(t+log__L(s*s));	      x += (z - t) ;	      return(k*ln2hi+x);	   }	/* end of if (x > negone) */	    else {		if ( x == negone ) return( negone/zero );	        else return ( zero / zero );	   }	}}#ifdef VAX	/* VAX D format (56 bits) *//* static double *//* L1     =  6.6666666666666703212E-1    , Hex  2^  0   *  .AAAAAAAAAAAAC5 *//* L2     =  3.9999999999970461961E-1    , Hex  2^ -1   *  .CCCCCCCCCC2684 *//* L3     =  2.8571428579395698188E-1    , Hex  2^ -1   *  .92492492F85782 *//* L4     =  2.2222221233634724402E-1    , Hex  2^ -2   *  .E38E3839B7AF2C *//* L5     =  1.8181879517064680057E-1    , Hex  2^ -2   *  .BA2EB4CC39655E *//* L6     =  1.5382888777946145467E-1    , Hex  2^ -2   *  .9D8551E8C5781D *//* L7     =  1.3338356561139403517E-1    , Hex  2^ -2   *  .8895B3907FCD92 *//* L8     =  1.2500000000000000000E-1    , Hex  2^ -2   *  .80000000000000 */static long        L1x[] = { 0xaaaa402a, 0xaac5aaaa};static long        L2x[] = { 0xcccc3fcc, 0x2684cccc};static long        L3x[] = { 0x49243f92, 0x578292f8};static long        L4x[] = { 0x8e383f63, 0xaf2c39b7};static long        L5x[] = { 0x2eb43f3a, 0x655ecc39};static long        L6x[] = { 0x85513f1d, 0x781de8c5};static long        L7x[] = { 0x95b33f08, 0xcd92907f};static long        L8x[] = { 0x00003f00, 0x00000000};#define       L1    (*(double*)L1x)#define       L2    (*(double*)L2x)#define       L3    (*(double*)L3x)#define       L4    (*(double*)L4x)#define       L5    (*(double*)L5x)#define       L6    (*(double*)L6x)#define       L7    (*(double*)L7x)#define       L8    (*(double*)L8x)#else	/* IEEE double */static doubleL1     =  6.6666666666667340202E-1    , /*Hex  2^ -1   *  1.5555555555592 */L2     =  3.9999999999416702146E-1    , /*Hex  2^ -2   *  1.999999997FF24 */L3     =  2.8571428742008753154E-1    , /*Hex  2^ -2   *  1.24924941E07B4 */L4     =  2.2222198607186277597E-1    , /*Hex  2^ -3   *  1.C71C52150BEA6 */L5     =  1.8183562745289935658E-1    , /*Hex  2^ -3   *  1.74663CC94342F */L6     =  1.5314087275331442206E-1    , /*Hex  2^ -3   *  1.39A1EC014045B */L7     =  1.4795612545334174692E-1    ; /*Hex  2^ -3   *  1.2F039F0085122 */#endifstatic double log__L(z)double z;{#ifdef VAX    return(z*(L1+z*(L2+z*(L3+z*(L4+z*(L5+z*(L6+z*(L7+z*L8))))))));#else	/* IEEE double */    return(z*(L1+z*(L2+z*(L3+z*(L4+z*(L5+z*(L6+z*L7)))))));#endif}static dummy(x)double x;{	return 1;}

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