1 /* e_jnf.c -- float version of e_jn.c. 2 * Conversion to float by Ian Lance Taylor, Cygnus Support, ian@cygnus.com. 3 */ 4 5 /* 6 * ==================================================== 7 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. 8 * 9 * Developed at SunPro, a Sun Microsystems, Inc. business. 10 * Permission to use, copy, modify, and distribute this 11 * software is freely granted, provided that this notice 12 * is preserved. 13 * ==================================================== 14 */ 15 16 #include "math.h" 17 #include "math_private.h" 18 19 static const float 20 two = 2.0000000000e+00, /* 0x40000000 */ 21 one = 1.0000000000e+00; /* 0x3F800000 */ 22 23 static const float zero = 0.0000000000e+00; 24 25 float 26 jnf(int n, float x) 27 { 28 int32_t i,hx,ix, sgn; 29 float a, b, temp, di; 30 float z, w; 31 32 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) 33 * Thus, J(-n,x) = J(n,-x) 34 */ 35 GET_FLOAT_WORD(hx,x); 36 ix = 0x7fffffff&hx; 37 /* if J(n,NaN) is NaN */ 38 if(ix>0x7f800000) return x+x; 39 if(n<0){ 40 n = -n; 41 x = -x; 42 hx ^= 0x80000000; 43 } 44 if(n==0) return(j0f(x)); 45 if(n==1) return(j1f(x)); 46 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */ 47 x = fabsf(x); 48 if(ix==0||ix>=0x7f800000) /* if x is 0 or inf */ 49 b = zero; 50 else if((float)n<=x) { 51 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */ 52 a = j0f(x); 53 b = j1f(x); 54 for(i=1;i<n;i++){ 55 temp = b; 56 b = b*((float)(i+i)/x) - a; /* avoid underflow */ 57 a = temp; 58 } 59 } else { 60 if(ix<0x30800000) { /* x < 2**-29 */ 61 /* x is tiny, return the first Taylor expansion of J(n,x) 62 * J(n,x) = 1/n!*(x/2)^n - ... 63 */ 64 if(n>33) /* underflow */ 65 b = zero; 66 else { 67 temp = x*(float)0.5; b = temp; 68 for (a=one,i=2;i<=n;i++) { 69 a *= (float)i; /* a = n! */ 70 b *= temp; /* b = (x/2)^n */ 71 } 72 b = b/a; 73 } 74 } else { 75 /* use backward recurrence */ 76 /* x x^2 x^2 77 * J(n,x)/J(n-1,x) = ---- ------ ------ ..... 78 * 2n - 2(n+1) - 2(n+2) 79 * 80 * 1 1 1 81 * (for large x) = ---- ------ ------ ..... 82 * 2n 2(n+1) 2(n+2) 83 * -- - ------ - ------ - 84 * x x x 85 * 86 * Let w = 2n/x and h=2/x, then the above quotient 87 * is equal to the continued fraction: 88 * 1 89 * = ----------------------- 90 * 1 91 * w - ----------------- 92 * 1 93 * w+h - --------- 94 * w+2h - ... 95 * 96 * To determine how many terms needed, let 97 * Q(0) = w, Q(1) = w(w+h) - 1, 98 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2), 99 * When Q(k) > 1e4 good for single 100 * When Q(k) > 1e9 good for double 101 * When Q(k) > 1e17 good for quadruple 102 */ 103 /* determine k */ 104 float t,v; 105 float q0,q1,h,tmp; int32_t k,m; 106 w = (n+n)/(float)x; h = (float)2.0/(float)x; 107 q0 = w; z = w+h; q1 = w*z - (float)1.0; k=1; 108 while(q1<(float)1.0e9) { 109 k += 1; z += h; 110 tmp = z*q1 - q0; 111 q0 = q1; 112 q1 = tmp; 113 } 114 m = n+n; 115 for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t); 116 a = t; 117 b = one; 118 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n) 119 * Hence, if n*(log(2n/x)) > ... 120 * single 8.8722839355e+01 121 * double 7.09782712893383973096e+02 122 * long double 1.1356523406294143949491931077970765006170e+04 123 * then recurrent value may overflow and the result is 124 * likely underflow to zero 125 */ 126 tmp = n; 127 v = two/x; 128 tmp = tmp*logf(fabsf(v*tmp)); 129 if(tmp<(float)8.8721679688e+01) { 130 for(i=n-1,di=(float)(i+i);i>0;i--){ 131 temp = b; 132 b *= di; 133 b = b/x - a; 134 a = temp; 135 di -= two; 136 } 137 } else { 138 for(i=n-1,di=(float)(i+i);i>0;i--){ 139 temp = b; 140 b *= di; 141 b = b/x - a; 142 a = temp; 143 di -= two; 144 /* scale b to avoid spurious overflow */ 145 if(b>(float)1e10) { 146 a /= b; 147 t /= b; 148 b = one; 149 } 150 } 151 } 152 b = (t*j0f(x)/b); 153 } 154 } 155 if(sgn==1) return -b; else return b; 156 } 157 158 float 159 ynf(int n, float x) 160 { 161 int32_t i,hx,ix,ib; 162 int32_t sign; 163 float a, b, temp; 164 165 GET_FLOAT_WORD(hx,x); 166 ix = 0x7fffffff&hx; 167 /* if Y(n,NaN) is NaN */ 168 if(ix>0x7f800000) return x+x; 169 if(ix==0) return -one/zero; 170 if(hx<0) return zero/zero; 171 sign = 1; 172 if(n<0){ 173 n = -n; 174 sign = 1 - ((n&1)<<1); 175 } 176 if(n==0) return(y0f(x)); 177 if(n==1) return(sign*y1f(x)); 178 if(ix==0x7f800000) return zero; 179 180 a = y0f(x); 181 b = y1f(x); 182 /* quit if b is -inf */ 183 GET_FLOAT_WORD(ib,b); 184 for(i=1;i<n&&(u_int32_t)ib!=0xff800000;i++){ 185 temp = b; 186 b = ((float)(i+i)/x)*b - a; 187 GET_FLOAT_WORD(ib,b); 188 a = temp; 189 } 190 if(sign>0) return b; else return -b; 191 } 192