使用一些与阶乘相关的杂散运算，我发现阶乘的多项式展开与正弦的多项式展开之间的联系非常相似。
在创建了一个小程序以生成阶乘的多项式展开系数之后（在这里实际上是没有意义的，因为它需要计算阶乘才能生成阶乘的：P），我能够将Pi与乘法相乘并创建一个sin（x）的O（n ^ 2）算法。我想将其与其他用于计算罪孽的算法进行比较。
我在网上看了一下，但是关于罪恶的计算却找不到很多。我找到了CORDIC，但找不到关于高效实现的任何信息，所以最终我放弃了，来到这里问，因为这对任何对计算机中三角函数的计算感兴趣的人都是一个好问题。
那么在软件中找到sin（x）的典型和/或最佳方法是什么，我的算法如何站立？
答案 0 :(得分：3)
Google搜索的最佳方法是弄清楚某种流行语言是如何完成的，例如搜索std::sin implementation，这将为您提供C ++实现。
以下是GCC标准数学库（从here提取）使用的一种实现。
/*******************************************************************/
/* An ultimate sin routine. Given an IEEE double machine number x */
/* it computes the correctly rounded (to nearest) value of sin(x) */
/*******************************************************************/
#ifndef IN_SINCOS
double
SECTION
__sin (double x)
{
double t, a, da;
mynumber u;
int4 k, m, n;
double retval = 0;
SET_RESTORE_ROUND_53BIT (FE_TONEAREST);
u.x = x;
m = u.i[HIGH_HALF];
k = 0x7fffffff & m; /* no sign */
if (k < 0x3e500000) /* if x->0 =>sin(x)=x */
{
math_check_force_underflow (x);
retval = x;
}
/*--------------------------- 2^-26<|x|< 0.855469---------------------- */
else if (k < 0x3feb6000)
{
/* Max ULP is 0.548. */
retval = do_sin (x, 0);
} /* else if (k < 0x3feb6000) */
/*----------------------- 0.855469 <|x|<2.426265 ----------------------*/
else if (k < 0x400368fd)
{
t = hp0 - fabs (x);
/* Max ULP is 0.51. */
retval = copysign (do_cos (t, hp1), x);
} /* else if (k < 0x400368fd) */
/*-------------------------- 2.426265<|x|< 105414350 ----------------------*/
else if (k < 0x419921FB)
{
n = reduce_sincos (x, &a, &da);
retval = do_sincos (a, da, n);
} /* else if (k < 0x419921FB ) */
/* --------------------105414350 <|x| <2^1024------------------------------*/
else if (k < 0x7ff00000)
{
n = __branred (x, &a, &da);
retval = do_sincos (a, da, n);
}
/*--------------------- |x| > 2^1024 ----------------------------------*/
else
{
if (k == 0x7ff00000 && u.i[LOW_HALF] == 0)
__set_errno (EDOM);
retval = x / x;
}
return retval;
}
此呼叫
/* Given a number partitioned into X and DX, this function computes the sine of
the number by combining the sin and cos of X (as computed by a variation of
the Taylor series) with the values looked up from the sin/cos table to get
the result. */
static inline double
__always_inline
do_sin (double x, double dx)
{
double xold = x;
/* Max ULP is 0.501 if |x| < 0.126, otherwise ULP is 0.518. */
if (fabs (x) < 0.126)
return TAYLOR_SIN (x * x, x, dx);
mynumber u;
if (x <= 0)
dx = -dx;
u.x = big + fabs (x);
x = fabs (x) - (u.x - big);
double xx, s, sn, ssn, c, cs, ccs, cor;
xx = x * x;
s = x + (dx + x * xx * (sn3 + xx * sn5));
c = x * dx + xx * (cs2 + xx * (cs4 + xx * cs6));
SINCOS_TABLE_LOOKUP (u, sn, ssn, cs, ccs);
cor = (ssn + s * ccs - sn * c) + cs * s;
return copysign (sn + cor, xold);
}
和
/* Given a number partitioned into X and DX, this function computes the cosine
of the number by combining the sin and cos of X (as computed by a variation
of the Taylor series) with the values looked up from the sin/cos table to
get the result. */
static inline double
__always_inline
do_cos (double x, double dx)
{
mynumber u;
if (x < 0)
dx = -dx;
u.x = big + fabs (x);
x = fabs (x) - (u.x - big) + dx;
double xx, s, sn, ssn, c, cs, ccs, cor;
xx = x * x;
s = x + x * xx * (sn3 + xx * sn5);
c = xx * (cs2 + xx * (cs4 + xx * cs6));
SINCOS_TABLE_LOOKUP (u, sn, ssn, cs, ccs);
cor = (ccs - s * ssn - cs * c) - sn * s;
return cs + cor;
}
为
#define SINCOS_TABLE_LOOKUP(u, sn, ssn, cs, ccs) \
({ \
int4 k = u.i[LOW_HALF] << 2; \
sn = __sincostab.x[k]; \
ssn = __sincostab.x[k + 1]; \
cs = __sincostab.x[k + 2]; \
ccs = __sincostab.x[k + 3]; \
})
因此，您的问题的答案是，一个聪明的实现将泰勒级数展开与表查找结合起来。
这是Sun使用的另一种方法：
/* @(#)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);
}
其他方法包括使用汇编指令，以便在硬件中完成计算（尽管有时会导致problems）。