﻿ java – 如何确定多项式逼近中的这些系数？ - 代码日志

#### java – 如何确定多项式逼近中的这些系数？

``````/* @(#)k_sin.c 1.3 95/01/18 */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunSoft, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/

/* __kernel_sin( x, y, iy)
* kernel sin function on [-pi/4, pi/4], pi/4 ~ 0.7854
* Input x is assumed to be bounded by ~pi/4 in magnitude.
* Input y is the tail of x.
* Input iy indicates whether y is 0. (if iy=0, y assume to be 0).
*
* Algorithm
*  1. Since sin(-x) = -sin(x), we need only to consider positive x.
*  2. if x < 2^-27 (hx<0x3e400000 0), return x with inexact if x!=0.
*  3. sin(x) is approximated by a polynomial of degree 13 on
*     [0,pi/4]
*                   3            13
*      sin(x) ~ x + S1*x + ... + S6*x
*     where
*
*  |sin(x)         2     4     6     8     10     12  |     -58
*  |----- - (1+S1*x +S2*x +S3*x +S4*x +S5*x  +S6*x   )| <= 2
*  |  x                               |
*
*  4. sin(x+y) = sin(x) + sin'(x')*y
*          ~ sin(x) + (1-x*x/2)*y
*     For better accuracy, let
*           3      2      2      2      2
*      r = x *(S2+x *(S3+x *(S4+x *(S5+x *S6))))
*     then                   3    2
*      sin(x) = x + (S1*x + (x *(r-y/2)+y))
*/

#include "fdlibm.h"

#ifdef __STDC__
static const double
#else
static double
#endif
half =  5.00000000000000000000e-01, /* 0x3FE00000, 0x00000000 */
S1  = -1.66666666666666324348e-01, /* 0xBFC55555, 0x55555549 */
S2  =  8.33333333332248946124e-03, /* 0x3F811111, 0x1110F8A6 */
S3  = -1.98412698298579493134e-04, /* 0xBF2A01A0, 0x19C161D5 */
S4  =  2.75573137070700676789e-06, /* 0x3EC71DE3, 0x57B1FE7D */
S5  = -2.50507602534068634195e-08, /* 0xBE5AE5E6, 0x8A2B9CEB */
S6  =  1.58969099521155010221e-10; /* 0x3DE5D93A, 0x5ACFD57C */

#ifdef __STDC__
double __kernel_sin(double x, double y, int iy)
#else
double __kernel_sin(x, y, iy)
double x,y; int iy;     /* iy=0 if y is zero */
#endif
{
double z,r,v;
int ix;
ix = __HI(x)&0x7fffffff;    /* high word of x */
if(ix<0x3e400000)           /* |x| < 2**-27 */
{if((int)x==0) return x;}        /* generate inexact */
z   =  x*x;
v   =  z*x;
r   =  S2+z*(S3+z*(S4+z*(S5+z*S6)));
if(iy==0) return x+v*(S1+z*r);
else      return x-((z*(half*y-v*r)-y)-v*S1);
}
``````

njuffa有一个有用的想法,使用身份来进一步限制您的域.例如,sin(x)= 3 * sin(x / 3)-4 * sin ^ 3(x / 3).使用此操作可以让您将域限制为0≤x≤π/ 12.您可以使用它两次将您的域限制为0≤x≤π/ 36.这样做会使您的泰勒扩张更快地获得您所需的准确度.而不是试图获得(π/ 4)n-1 / n！的任意精确的π值,我建议将π四舍五入到直到1 / n！满足您所需的准确度(或3-n / n！或9-n / n！