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