0,0 → 1,282 |
/* Copyright (C) 1994 DJ Delorie, see COPYING.DJ for details */ |
/* @(#)e_jn.c 5.1 93/09/24 */ |
/* |
* ==================================================== |
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. |
* |
* Developed at SunPro, a Sun Microsystems, Inc. business. |
* Permission to use, copy, modify, and distribute this |
* software is freely granted, provided that this notice |
* is preserved. |
* ==================================================== |
*/ |
|
#if defined(LIBM_SCCS) && !defined(lint) |
static char rcsid[] = "$Id: e_jn.c,v 1.6 1994/08/18 23:05:37 jtc Exp $"; |
#endif |
|
/* |
* __ieee754_jn(n, x), __ieee754_yn(n, x) |
* floating point Bessel's function of the 1st and 2nd kind |
* of order n |
* |
* Special cases: |
* y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal; |
* y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal. |
* Note 2. About jn(n,x), yn(n,x) |
* For n=0, j0(x) is called, |
* for n=1, j1(x) is called, |
* for n<x, forward recursion us used starting |
* from values of j0(x) and j1(x). |
* for n>x, a continued fraction approximation to |
* j(n,x)/j(n-1,x) is evaluated and then backward |
* recursion is used starting from a supposed value |
* for j(n,x). The resulting value of j(0,x) is |
* compared with the actual value to correct the |
* supposed value of j(n,x). |
* |
* yn(n,x) is similar in all respects, except |
* that forward recursion is used for all |
* values of n>1. |
* |
*/ |
|
#include "math.h" |
#include "math_private.h" |
|
#ifdef __STDC__ |
static const double |
#else |
static double |
#endif |
invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */ |
two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */ |
one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */ |
|
#ifdef __STDC__ |
static const double zero = 0.00000000000000000000e+00; |
#else |
static double zero = 0.00000000000000000000e+00; |
#endif |
|
#ifdef __STDC__ |
double __ieee754_jn(int n, double x) |
#else |
double __ieee754_jn(n,x) |
int n; double x; |
#endif |
{ |
int32_t i,hx,ix,lx, sgn; |
double a, b, temp, di; |
double z, w; |
|
/* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) |
* Thus, J(-n,x) = J(n,-x) |
*/ |
EXTRACT_WORDS(hx,lx,x); |
ix = 0x7fffffff&hx; |
/* if J(n,NaN) is NaN */ |
if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x; |
if(n<0){ |
n = -n; |
x = -x; |
hx ^= 0x80000000; |
} |
if(n==0) return(__ieee754_j0(x)); |
if(n==1) return(__ieee754_j1(x)); |
sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */ |
x = fabs(x); |
if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */ |
b = zero; |
else if((double)n<=x) { |
/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */ |
if(ix>=0x52D00000) { /* x > 2**302 */ |
/* (x >> n**2) |
* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) |
* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) |
* Let s=sin(x), c=cos(x), |
* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then |
* |
* n sin(xn)*sqt2 cos(xn)*sqt2 |
* ---------------------------------- |
* 0 s-c c+s |
* 1 -s-c -c+s |
* 2 -s+c -c-s |
* 3 s+c c-s |
*/ |
switch(n&3) { |
case 0: temp = cos(x)+sin(x); break; |
case 1: temp = -cos(x)+sin(x); break; |
case 2: temp = -cos(x)-sin(x); break; |
case 3: temp = cos(x)-sin(x); break; |
} |
b = invsqrtpi*temp/sqrt(x); |
} else { |
a = __ieee754_j0(x); |
b = __ieee754_j1(x); |
for(i=1;i<n;i++){ |
temp = b; |
b = b*((double)(i+i)/x) - a; /* avoid underflow */ |
a = temp; |
} |
} |
} else { |
if(ix<0x3e100000) { /* x < 2**-29 */ |
/* x is tiny, return the first Taylor expansion of J(n,x) |
* J(n,x) = 1/n!*(x/2)^n - ... |
*/ |
if(n>33) /* underflow */ |
b = zero; |
else { |
temp = x*0.5; b = temp; |
for (a=one,i=2;i<=n;i++) { |
a *= (double)i; /* a = n! */ |
b *= temp; /* b = (x/2)^n */ |
} |
b = b/a; |
} |
} else { |
/* use backward recurrence */ |
/* x x^2 x^2 |
* J(n,x)/J(n-1,x) = ---- ------ ------ ..... |
* 2n - 2(n+1) - 2(n+2) |
* |
* 1 1 1 |
* (for large x) = ---- ------ ------ ..... |
* 2n 2(n+1) 2(n+2) |
* -- - ------ - ------ - |
* x x x |
* |
* Let w = 2n/x and h=2/x, then the above quotient |
* is equal to the continued fraction: |
* 1 |
* = ----------------------- |
* 1 |
* w - ----------------- |
* 1 |
* w+h - --------- |
* w+2h - ... |
* |
* To determine how many terms needed, let |
* Q(0) = w, Q(1) = w(w+h) - 1, |
* Q(k) = (w+k*h)*Q(k-1) - Q(k-2), |
* When Q(k) > 1e4 good for single |
* When Q(k) > 1e9 good for double |
* When Q(k) > 1e17 good for quadruple |
*/ |
/* determine k */ |
double t,v; |
double q0,q1,h,tmp; int32_t k,m; |
w = (n+n)/(double)x; h = 2.0/(double)x; |
q0 = w; z = w+h; q1 = w*z - 1.0; k=1; |
while(q1<1.0e9) { |
k += 1; z += h; |
tmp = z*q1 - q0; |
q0 = q1; |
q1 = tmp; |
} |
m = n+n; |
for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t); |
a = t; |
b = one; |
/* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n) |
* Hence, if n*(log(2n/x)) > ... |
* single 8.8722839355e+01 |
* double 7.09782712893383973096e+02 |
* long double 1.1356523406294143949491931077970765006170e+04 |
* then recurrent value may overflow and the result is |
* likely underflow to zero |
*/ |
tmp = n; |
v = two/x; |
tmp = tmp*__ieee754_log(fabs(v*tmp)); |
if(tmp<7.09782712893383973096e+02) { |
for(i=n-1,di=(double)(i+i);i>0;i--){ |
temp = b; |
b *= di; |
b = b/x - a; |
a = temp; |
di -= two; |
} |
} else { |
for(i=n-1,di=(double)(i+i);i>0;i--){ |
temp = b; |
b *= di; |
b = b/x - a; |
a = temp; |
di -= two; |
/* scale b to avoid spurious overflow */ |
if(b>1e100) { |
a /= b; |
t /= b; |
b = one; |
} |
} |
} |
b = (t*__ieee754_j0(x)/b); |
} |
} |
if(sgn==1) return -b; else return b; |
} |
|
#ifdef __STDC__ |
double __ieee754_yn(int n, double x) |
#else |
double __ieee754_yn(n,x) |
int n; double x; |
#endif |
{ |
int32_t i,hx,ix,lx; |
int32_t sign; |
double a, b, temp; |
|
EXTRACT_WORDS(hx,lx,x); |
ix = 0x7fffffff&hx; |
/* if Y(n,NaN) is NaN */ |
if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x; |
if((ix|lx)==0) return -one/zero; |
if(hx<0) return zero/zero; |
sign = 1; |
if(n<0){ |
n = -n; |
sign = 1 - ((n&1)<<2); |
} |
if(n==0) return(__ieee754_y0(x)); |
if(n==1) return(sign*__ieee754_y1(x)); |
if(ix==0x7ff00000) return zero; |
if(ix>=0x52D00000) { /* x > 2**302 */ |
/* (x >> n**2) |
* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) |
* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) |
* Let s=sin(x), c=cos(x), |
* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then |
* |
* n sin(xn)*sqt2 cos(xn)*sqt2 |
* ---------------------------------- |
* 0 s-c c+s |
* 1 -s-c -c+s |
* 2 -s+c -c-s |
* 3 s+c c-s |
*/ |
switch(n&3) { |
case 0: temp = sin(x)-cos(x); break; |
case 1: temp = -sin(x)-cos(x); break; |
case 2: temp = -sin(x)+cos(x); break; |
case 3: temp = sin(x)+cos(x); break; |
} |
b = invsqrtpi*temp/sqrt(x); |
} else { |
u_int32_t high; |
a = __ieee754_y0(x); |
b = __ieee754_y1(x); |
/* quit if b is -inf */ |
GET_HIGH_WORD(high,b); |
for(i=1;i<n&&high!=0xfff00000;i++){ |
temp = b; |
b = ((double)(i+i)/x)*b - a; |
GET_HIGH_WORD(high,b); |
a = temp; |
} |
} |
if(sign>0) return b; else return -b; |
} |