[352] | 1 | /* origin: FreeBSD /usr/src/lib/msun/src/e_jn.c */
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| 2 | /*
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| 3 | * ====================================================
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| 4 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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| 5 | *
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| 6 | * Developed at SunSoft, a Sun Microsystems, Inc. business.
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| 7 | * Permission to use, copy, modify, and distribute this
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| 8 | * software is freely granted, provided that this notice
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| 9 | * is preserved.
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| 10 | * ====================================================
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| 11 | */
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| 12 | /*
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| 13 | * jn(n, x), yn(n, x)
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| 14 | * floating point Bessel's function of the 1st and 2nd kind
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| 15 | * of order n
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| 16 | *
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| 17 | * Special cases:
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| 18 | * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
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| 19 | * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
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| 20 | * Note 2. About jn(n,x), yn(n,x)
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| 21 | * For n=0, j0(x) is called,
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| 22 | * for n=1, j1(x) is called,
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| 23 | * for n<=x, forward recursion is used starting
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| 24 | * from values of j0(x) and j1(x).
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| 25 | * for n>x, a continued fraction approximation to
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| 26 | * j(n,x)/j(n-1,x) is evaluated and then backward
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| 27 | * recursion is used starting from a supposed value
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| 28 | * for j(n,x). The resulting value of j(0,x) is
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| 29 | * compared with the actual value to correct the
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| 30 | * supposed value of j(n,x).
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| 31 | *
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| 32 | * yn(n,x) is similar in all respects, except
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| 33 | * that forward recursion is used for all
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| 34 | * values of n>1.
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| 35 | */
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| 36 |
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| 37 | #include "libm.h"
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| 38 |
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| 39 | static const double invsqrtpi = 5.64189583547756279280e-01; /* 0x3FE20DD7, 0x50429B6D */
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| 40 |
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| 41 | double jn(int n, double x)
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| 42 | {
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| 43 | uint32_t ix, lx;
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| 44 | int nm1, i, sign;
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| 45 | double a, b, temp;
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| 46 |
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| 47 | EXTRACT_WORDS(ix, lx, x);
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| 48 | sign = ix>>31;
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| 49 | ix &= 0x7fffffff;
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| 50 |
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| 51 | if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */
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| 52 | return x;
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| 53 |
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| 54 | /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
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| 55 | * Thus, J(-n,x) = J(n,-x)
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| 56 | */
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| 57 | /* nm1 = |n|-1 is used instead of |n| to handle n==INT_MIN */
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| 58 | if (n == 0)
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| 59 | return j0(x);
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| 60 | if (n < 0) {
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| 61 | nm1 = -(n+1);
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| 62 | x = -x;
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| 63 | sign ^= 1;
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| 64 | } else
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| 65 | nm1 = n-1;
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| 66 | if (nm1 == 0)
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| 67 | return j1(x);
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| 68 |
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| 69 | sign &= n; /* even n: 0, odd n: signbit(x) */
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| 70 | x = fabs(x);
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| 71 | if ((ix|lx) == 0 || ix == 0x7ff00000) /* if x is 0 or inf */
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| 72 | b = 0.0;
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| 73 | else if (nm1 < x) {
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| 74 | /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
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| 75 | if (ix >= 0x52d00000) { /* x > 2**302 */
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| 76 | /* (x >> n**2)
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| 77 | * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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| 78 | * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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| 79 | * Let s=sin(x), c=cos(x),
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| 80 | * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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| 81 | *
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| 82 | * n sin(xn)*sqt2 cos(xn)*sqt2
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| 83 | * ----------------------------------
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| 84 | * 0 s-c c+s
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| 85 | * 1 -s-c -c+s
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| 86 | * 2 -s+c -c-s
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| 87 | * 3 s+c c-s
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| 88 | */
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| 89 | switch(nm1&3) {
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| 90 | case 0: temp = -cos(x)+sin(x); break;
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| 91 | case 1: temp = -cos(x)-sin(x); break;
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| 92 | case 2: temp = cos(x)-sin(x); break;
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| 93 | default:
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| 94 | case 3: temp = cos(x)+sin(x); break;
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| 95 | }
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| 96 | b = invsqrtpi*temp/sqrt(x);
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| 97 | } else {
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| 98 | a = j0(x);
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| 99 | b = j1(x);
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| 100 | for (i=0; i<nm1; ) {
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| 101 | i++;
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| 102 | temp = b;
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| 103 | b = b*(2.0*i/x) - a; /* avoid underflow */
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| 104 | a = temp;
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| 105 | }
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| 106 | }
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| 107 | } else {
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| 108 | if (ix < 0x3e100000) { /* x < 2**-29 */
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| 109 | /* x is tiny, return the first Taylor expansion of J(n,x)
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| 110 | * J(n,x) = 1/n!*(x/2)^n - ...
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| 111 | */
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| 112 | if (nm1 > 32) /* underflow */
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| 113 | b = 0.0;
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| 114 | else {
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| 115 | temp = x*0.5;
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| 116 | b = temp;
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| 117 | a = 1.0;
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| 118 | for (i=2; i<=nm1+1; i++) {
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| 119 | a *= (double)i; /* a = n! */
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| 120 | b *= temp; /* b = (x/2)^n */
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| 121 | }
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| 122 | b = b/a;
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| 123 | }
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| 124 | } else {
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| 125 | /* use backward recurrence */
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| 126 | /* x x^2 x^2
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| 127 | * J(n,x)/J(n-1,x) = ---- ------ ------ .....
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| 128 | * 2n - 2(n+1) - 2(n+2)
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| 129 | *
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| 130 | * 1 1 1
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| 131 | * (for large x) = ---- ------ ------ .....
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| 132 | * 2n 2(n+1) 2(n+2)
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| 133 | * -- - ------ - ------ -
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| 134 | * x x x
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| 135 | *
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| 136 | * Let w = 2n/x and h=2/x, then the above quotient
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| 137 | * is equal to the continued fraction:
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| 138 | * 1
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| 139 | * = -----------------------
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| 140 | * 1
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| 141 | * w - -----------------
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| 142 | * 1
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| 143 | * w+h - ---------
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| 144 | * w+2h - ...
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| 145 | *
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| 146 | * To determine how many terms needed, let
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| 147 | * Q(0) = w, Q(1) = w(w+h) - 1,
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| 148 | * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
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| 149 | * When Q(k) > 1e4 good for single
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| 150 | * When Q(k) > 1e9 good for double
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| 151 | * When Q(k) > 1e17 good for quadruple
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| 152 | */
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| 153 | /* determine k */
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| 154 | double t,q0,q1,w,h,z,tmp,nf;
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| 155 | int k;
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| 156 |
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| 157 | nf = nm1 + 1.0;
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| 158 | w = 2*nf/x;
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| 159 | h = 2/x;
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| 160 | z = w+h;
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| 161 | q0 = w;
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| 162 | q1 = w*z - 1.0;
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| 163 | k = 1;
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| 164 | while (q1 < 1.0e9) {
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| 165 | k += 1;
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| 166 | z += h;
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| 167 | tmp = z*q1 - q0;
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| 168 | q0 = q1;
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| 169 | q1 = tmp;
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| 170 | }
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| 171 | for (t=0.0, i=k; i>=0; i--)
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| 172 | t = 1/(2*(i+nf)/x - t);
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| 173 | a = t;
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| 174 | b = 1.0;
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| 175 | /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
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| 176 | * Hence, if n*(log(2n/x)) > ...
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| 177 | * single 8.8722839355e+01
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| 178 | * double 7.09782712893383973096e+02
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| 179 | * long double 1.1356523406294143949491931077970765006170e+04
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| 180 | * then recurrent value may overflow and the result is
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| 181 | * likely underflow to zero
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| 182 | */
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| 183 | tmp = nf*log(fabs(w));
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| 184 | if (tmp < 7.09782712893383973096e+02) {
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| 185 | for (i=nm1; i>0; i--) {
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| 186 | temp = b;
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| 187 | b = b*(2.0*i)/x - a;
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| 188 | a = temp;
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| 189 | }
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| 190 | } else {
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| 191 | for (i=nm1; i>0; i--) {
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| 192 | temp = b;
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| 193 | b = b*(2.0*i)/x - a;
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| 194 | a = temp;
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| 195 | /* scale b to avoid spurious overflow */
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| 196 | if (b > 0x1p500) {
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| 197 | a /= b;
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| 198 | t /= b;
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| 199 | b = 1.0;
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| 200 | }
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| 201 | }
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| 202 | }
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| 203 | z = j0(x);
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| 204 | w = j1(x);
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| 205 | if (fabs(z) >= fabs(w))
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| 206 | b = t*z/b;
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| 207 | else
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| 208 | b = t*w/a;
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| 209 | }
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| 210 | }
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| 211 | return sign ? -b : b;
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| 212 | }
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| 213 |
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| 214 |
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| 215 | double yn(int n, double x)
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| 216 | {
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| 217 | uint32_t ix, lx, ib;
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| 218 | int nm1, sign, i;
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| 219 | double a, b, temp;
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| 220 |
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| 221 | EXTRACT_WORDS(ix, lx, x);
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| 222 | sign = ix>>31;
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| 223 | ix &= 0x7fffffff;
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| 224 |
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| 225 | if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */
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| 226 | return x;
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| 227 | if (sign && (ix|lx)!=0) /* x < 0 */
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| 228 | return 0/0.0;
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| 229 | if (ix == 0x7ff00000)
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| 230 | return 0.0;
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| 231 |
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| 232 | if (n == 0)
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| 233 | return y0(x);
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| 234 | if (n < 0) {
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| 235 | nm1 = -(n+1);
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| 236 | sign = n&1;
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| 237 | } else {
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| 238 | nm1 = n-1;
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| 239 | sign = 0;
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| 240 | }
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| 241 | if (nm1 == 0)
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| 242 | return sign ? -y1(x) : y1(x);
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| 243 |
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| 244 | if (ix >= 0x52d00000) { /* x > 2**302 */
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| 245 | /* (x >> n**2)
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| 246 | * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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| 247 | * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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| 248 | * Let s=sin(x), c=cos(x),
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| 249 | * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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| 250 | *
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| 251 | * n sin(xn)*sqt2 cos(xn)*sqt2
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| 252 | * ----------------------------------
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| 253 | * 0 s-c c+s
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| 254 | * 1 -s-c -c+s
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| 255 | * 2 -s+c -c-s
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| 256 | * 3 s+c c-s
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| 257 | */
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| 258 | switch(nm1&3) {
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| 259 | case 0: temp = -sin(x)-cos(x); break;
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| 260 | case 1: temp = -sin(x)+cos(x); break;
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| 261 | case 2: temp = sin(x)+cos(x); break;
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| 262 | default:
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| 263 | case 3: temp = sin(x)-cos(x); break;
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| 264 | }
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| 265 | b = invsqrtpi*temp/sqrt(x);
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| 266 | } else {
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| 267 | a = y0(x);
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| 268 | b = y1(x);
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| 269 | /* quit if b is -inf */
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| 270 | GET_HIGH_WORD(ib, b);
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| 271 | for (i=0; i<nm1 && ib!=0xfff00000; ){
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| 272 | i++;
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| 273 | temp = b;
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| 274 | b = (2.0*i/x)*b - a;
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| 275 | GET_HIGH_WORD(ib, b);
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| 276 | a = temp;
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| 277 | }
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| 278 | }
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| 279 | return sign ? -b : b;
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| 280 | }
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