EXP(3)							   BSD Library Functions Manual 						    EXP(3)

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 pro-
     vided 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 advan-
     tage 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);

SEE ALSO

     log(3), pow(3), math(3)

STANDARDS

     The exp(), exp2(), and expm1() functions conform to ISO/IEC 9899:2011.

4th Berkeley Distribution					  August 16, 2012					 4th Berkeley Distribution