NAME

exp, exp2, expm1 — exponential functions

SYNOPSIS

#include <math.h>

float
expf(float x);

double
exp(double x);

long double
expl(long double x);

float
exp2f(float x);

double
exp2(double x);

long double
exp2l(long double x);

float
expm1f(float x);

double
expm1(double x);

long double
expm1l(long double x);

float
__exp10f(float x);

double
__exp10(double x);

DESCRIPTION

The exp() function computes e**x, the base-e exponential of x.

The exp2() function computes 2**x, the base-2 exponential of x.

The __exp10() function computes 10**x; it is not defined in the C standard, and therefore may not be portable to other platforms. It is provided as a convenience to programmers because it may be computed more efficiently than pow(10,x).

If x is nearly zero, then the common expression exp(x) - 1.0 will suffer from catastrophic cancellation and the result will have little or no precision. The expm1() function provides an alternative means to do this calculation without the risk of significant loss of precision. If you find yourself using this function, you are likely to also be interested in the log1p() function.

Note that computations numerically equivalent to exp(x) - 1.0 are often hidden in more complicated expressions; some amount of algebraic manipulation may be necessary to take advantage of the expm1() function. Consider the following example, abstracted from a developer's actual production code in a bug report:

double z = exp(-x/y)*(x*x/y/y + 2*x/y + 2) - 2

When x is small relative to y, this expression is approximately equal to:

double z = 2*(exp(-x/y) - 1)

and all precision of the result is lost in the computation due to catastrophic cancellation. The developer was aware that they were losing precision, but didn't know what to do about it. To remedy the situation, we do a little algebra and re-write the expression to take advantage of the expm1() function:

  exp(-x/y)*(x*x/y/y + 2*x/y + 2) - 2
= (2*exp(-x/y) - 2) + exp(-x/y)*((x*x)/(y*y) + 2*x/y)

This transformation allows the result to be computed to a high degree of accuracy as follows:

const double r = x/y;
const double emrm1 = expm1(-r);
double z = 2.0*emrm1 + (1.0 + emrm1)*(2.0 + r)*r;

It is not always easy to spot such opportunities for improvement; if an expression involving exp() seems to be suffering from an undue loss of accuracy, try a few simple algebraic operations to see if you can identify a factor with the form exp(x) - 1.0, and substitute expm1(x) in its place.

SPECIAL VALUES

exp(±0) and exp2(±0) return 1.

exp(-infinity) and exp2(-infinity) return +0.

exp(+infinity) and exp2(+infinity) return +infinity.

expm1(±0) returns ±0.

expm1(-infinity) returns -1.

expm1(+infinity) returns +infinity.

For all these functions, a range error occurs if the magnitude of x is too large.

VECTOR OPERATIONS

If you need to apply the exp() functions to SIMD vectors or arrays, using the following functions provided by the Accelerate.framework may give significantly better performance:

#include <Accelerate/Accelerate.h>

vFloat vexpf(vFloat x);
vFloat vexpm1f(vFloat x);
void vvexpf(float *y, const float *x, const int *n);
void vvexp(double *y, const double *x, const int *n);
void vvexpm1f(float *y, const float *x, const int *n);
void vvexpm1(double *y, const double *x, const int *n);
void vvexp2f(float *y, const float *x, const int *n);
void vvexp2(double *y, const double *x, const int *n);

STANDARDS