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exp(3)                   BSD Library Functions Manual                   exp(3)


     exp, exp2, expm1 -- exponential functions


     #include <math.h>

     expf(float x);

     exp(double x);

     long double
     expl(long double x);

     exp2f(float x);

     exp2(double x);

     long double
     exp2l(long double x);

     expm1f(float x);

     expm1(double x);

     long double
     expm1l(long double x);

     __exp10f(float x);

     __exp10(double x);


     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 stan-
     dard, 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 pre-
     cision.  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 manipu-
     lation may be necessary to take advantage of the expm1() function.  Con-
     sider the following example, abstracted from a developer's actual produc-
     tion 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 cata-
     strophic 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 iden-
     tify a factor with the form exp(x) - 1.0, and substitute expm1(x) in its


     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.


     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);


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


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

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

Mac OS X 10.9.1 - Generated Tue Jan 7 19:40:23 CST 2014