expm1(), expm1f()

Compute the exponential of a number, then subtract 1

Synopsis:

#include <math.h>

double expm1 ( double x );

float expm1f ( float x );

Arguments:

x

The number for which you want to calculate the exponential minus one.

Library:

libm

Use the -l m option to qcc to link against this library.

Description:

The expm1() and expm1f() functions compute the exponential of x, minus 1 (e^x - 1).

A range error occurs if the magnitude of x is too large.

The value of expm1( x ) may be more accurate than exp( x ) - 1.0 for small values of x.

The expm1() and log1p() functions are useful for financial calculations of (((1+x)**n)-1)/x, namely:

expm1(n * log1p(x))/x

when x is very small (for example, when performing calculations with a small daily interest rate). These functions also simplify writing accurate inverse hyperbolic functions.

Returns:

The exponential value of x, minus 1.

---

If an error occurs, these functions return 0, but this is also a valid mathematical result. If you want to check for errors, set errno to 0, call the function, and then check errno again. These functions don't change errno if no errors occurred.

---

Examples:

#include <stdio.h>
#include <errno.h>
#include <inttypes.h>
#include <math.h>
#include <fpstatus.h>


int main(int argc, char** argv) 
{
    double a, b;

    a = 2;
    b = expm1(a);
    printf("(e ^ %f) -1  is %f \n", a, b);

    return(0);
}

produces the output:

(e ^ 2.000000) -1  is 6.389056

Classification:

ANSI, POSIX 1003.1

See also:

exp(), log1p()