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authorAndreas Jaeger <aj@suse.de>2001-05-15 07:57:46 +0000
committerAndreas Jaeger <aj@suse.de>2001-05-15 07:57:46 +0000
commitef25b29e9a1fd54625c98af36d767f353cd45488 (patch)
treeea1956cad07d19a3da879d1e26a9ad9558183df9
parent083973f34e1c70e05e117db37239716995127cfd (diff)
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expm1 implementation for 128-bit long double.
-rw-r--r--sysdeps/ieee754/ldbl-128/s_expm1l.c145
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