diff options
author | Andreas Jaeger <aj@suse.de> | 2001-05-15 07:57:46 +0000 |
---|---|---|
committer | Andreas Jaeger <aj@suse.de> | 2001-05-15 07:57:46 +0000 |
commit | ef25b29e9a1fd54625c98af36d767f353cd45488 (patch) | |
tree | ea1956cad07d19a3da879d1e26a9ad9558183df9 | |
parent | 083973f34e1c70e05e117db37239716995127cfd (diff) | |
download | glibc-ef25b29e9a1fd54625c98af36d767f353cd45488.tar.gz glibc-ef25b29e9a1fd54625c98af36d767f353cd45488.tar.xz glibc-ef25b29e9a1fd54625c98af36d767f353cd45488.zip |
expm1 implementation for 128-bit long double.
-rw-r--r-- | sysdeps/ieee754/ldbl-128/s_expm1l.c | 145 |
1 files changed, 145 insertions, 0 deletions
diff --git a/sysdeps/ieee754/ldbl-128/s_expm1l.c b/sysdeps/ieee754/ldbl-128/s_expm1l.c new file mode 100644 index 0000000000..f662ee9093 --- /dev/null +++ b/sysdeps/ieee754/ldbl-128/s_expm1l.c @@ -0,0 +1,145 @@ +/* expm1l.c + * + * Exponential function, minus 1 + * 128-bit long double precision + * + * + * + * SYNOPSIS: + * + * long double x, y, expm1l(); + * + * y = expm1l( x ); + * + * + * + * DESCRIPTION: + * + * Returns e (2.71828...) raised to the x power, minus one. + * + * Range reduction is accomplished by separating the argument + * into an integer k and fraction f such that + * + * x k f + * e = 2 e. + * + * An expansion x + .5 x^2 + x^3 R(x) approximates exp(f) - 1 + * in the basic range [-0.5 ln 2, 0.5 ln 2]. + * + * + * ACCURACY: + * + * Relative error: + * arithmetic domain # trials peak rms + * IEEE -79,+MAXLOG 100,000 1.7e-34 4.5e-35 + * + */ + +/* Copyright 2001 by Stephen L. Moshier */ + + +#include "math.h" +#include "math_private.h" + +/* exp(x) - 1 = x + 0.5 x^2 + x^3 P(x)/Q(x) + -.5 ln 2 < x < .5 ln 2 + Theoretical peak relative error = 8.1e-36 */ + +static long double + P0 = 2.943520915569954073888921213330863757240E8L, + P1 = -5.722847283900608941516165725053359168840E7L, + P2 = 8.944630806357575461578107295909719817253E6L, + P3 = -7.212432713558031519943281748462837065308E5L, + P4 = 4.578962475841642634225390068461943438441E4L, + P5 = -1.716772506388927649032068540558788106762E3L, + P6 = 4.401308817383362136048032038528753151144E1L, + P7 = -4.888737542888633647784737721812546636240E-1L, + Q0 = 1.766112549341972444333352727998584753865E9L, + Q1 = -7.848989743695296475743081255027098295771E8L, + Q2 = 1.615869009634292424463780387327037251069E8L, + Q3 = -2.019684072836541751428967854947019415698E7L, + Q4 = 1.682912729190313538934190635536631941751E6L, + Q5 = -9.615511549171441430850103489315371768998E4L, + Q6 = 3.697714952261803935521187272204485251835E3L, + Q7 = -8.802340681794263968892934703309274564037E1L, + /* Q8 = 1.000000000000000000000000000000000000000E0 */ +/* C1 + C2 = ln 2 */ + + C1 = 6.93145751953125E-1L, + C2 = 1.428606820309417232121458176568075500134E-6L, +/* ln (2^16384 * (1 - 2^-113)) */ + maxlog = 1.1356523406294143949491931077970764891253E4L, +/* ln 2^-114 */ + minarg = -7.9018778583833765273564461846232128760607E1L, big = 2e4932L; + + +long double +__expm1l (long double x) +{ + long double px, qx, xx; + int32_t ix, sign; + ieee854_long_double_shape_type u; + int k; + + /* Overflow. */ + if (x > maxlog) + return (big * big); + + /* Minimum value. */ + if (x < minarg) + return (4.0 / big - 1.0L); + + /* Detect infinity and NaN. */ + u.value = x; + ix = u.parts32.w0; + sign = ix & 0x80000000; + ix &= 0x7fffffff; + if (ix >= 0x7fff0000) + { + /* Infinity. */ + if (((ix & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3) == 0) + { + if (sign) + return -1.0L; + else + return x; + } + /* NaN. */ + return (x + x); + } + + /* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */ + xx = C1 + C2; /* ln 2. */ + px = __floorl (0.5 + x / xx); + k = px; + /* remainder times ln 2 */ + x -= px * C1; + x -= px * C2; + + /* Approximate exp(remainder ln 2). */ + px = (((((((P7 * x + + P6) * x + + P5) * x + P4) * x + P3) * x + P2) * x + P1) * x + P0) * x; + + qx = (((((((x + + Q7) * x + + Q6) * x + Q5) * x + Q4) * x + Q3) * x + Q2) * x + Q1) * x + Q0; + + xx = x * x; + qx = x + (0.5 * xx + xx * px / qx); + + /* exp(x) = exp(k ln 2) exp(remainder ln 2) = 2^k exp(remainder ln 2). + + We have qx = exp(remainder ln 2) - 1, so + exp(x) - 1 = 2^k (qx + 1) - 1 + = 2^k qx + 2^k - 1. */ + + px = ldexpl (1.0L, k); + x = px * qx + (px - 1.0); + return x; +} + +weak_alias (__expm1l, expm1l) +#ifdef NO_LONG_DOUBLE +strong_alias (__expm1, __expm1l) weak_alias (__expm1, expm1l) +#endif |