Document the method used to compute expf. Taken from exp, with

changes to reflect differences in computation between the two.
This commit is contained in:
Warner Losh 2012-10-19 22:47:44 +00:00
parent aa616aa22d
commit 527a7c56f7

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@ -21,6 +21,68 @@ __FBSDID("$FreeBSD$");
#include "math.h"
#include "math_private.h"
/* __ieee754_expf
* Returns the exponential of x.
*
* Method
* 1. Argument reduction:
* Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658.
* Given x, find r and integer k such that
*
* x = k*ln2 + r, |r| <= 0.5*ln2.
*
* Here r will be represented as r = hi-lo for better
* accuracy.
*
* 2. Approximation of exp(r) by a special rational function on
* the interval [0,0.34658]:
* Write
* R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ...
* We use a special Remes algorithm on [0,0.34658] to generate
* a polynomial of degree 2 to approximate R. The maximum error
* of this polynomial approximation is bounded by 2**-27. In
* other words,
* R(z) ~ 2.0 + P1*z + P2*z*z
* (where z=r*r, and the values of P1 and P2 are listed below)
* and
* | 2 | -27
* | 2.0+P1*z+P2*z - R(z) | <= 2
* | |
* The computation of expf(r) thus becomes
* 2*r
* expf(r) = 1 + -------
* R - r
* r*R1(r)
* = 1 + r + ----------- (for better accuracy)
* 2 - R1(r)
* where
* 2 4
* R1(r) = r - (P1*r + P2*r)
*
* 3. Scale back to obtain expf(x):
* From step 1, we have
* expf(x) = 2^k * expf(r)
*
* Special cases:
* expf(INF) is INF, exp(NaN) is NaN;
* expf(-INF) is 0, and
* for finite argument, only exp(0)=1 is exact.
*
* Accuracy:
* according to an error analysis, the error is always less than
* 0.5013 ulp (unit in the last place).
*
* Misc. info.
* For IEEE float
* if x > 8.8721679688e+01 then exp(x) overflow
* if x < -1.0397208405e+02 then exp(x) underflow
*
* Constants:
* The hexadecimal values are the intended ones for the following
* constants. The decimal values may be used, provided that the
* compiler will convert from decimal to binary accurately enough
* to produce the hexadecimal values shown.
*/
static const float
one = 1.0,
halF[2] = {0.5,-0.5,},