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```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 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
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);

```

```     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
```

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