expm1(), expm1f(), expm1l()

Compute the exponential of a number, then subtract 1

Synopsis:

#include <math.h>

double expm1 ( double x );

float expm1f ( float x );

long double expm1l ( long double 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(), expm1f(), and expm1l() functions compute the exponential of x, minus 1 (ex - 1).

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.

To check for error situations, use feclearexcept() and fetestexcept(). For example:

Returns:

The exponential value of x, minus 1.

If x is: These functions return: Errors:
±0.0 0.0, with the same sign as x
A value that would cause overflow Inf FE_OVERFLOW
-Inf -1
Inf Inf
NaN NaN

These functions raise FE_INEXACT if the FPU reports that the result can't be exactly represented as a floating-point number.

Examples:

#include <stdio.h>
#include <math.h>
#include <fenv.h>
#include <stdlib.h>

int main( void )
{
    int except_flags;
    double a, b;

    feclearexcept(FE_ALL_EXCEPT);

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

    except_flags = fetestexcept(FE_ALL_EXCEPT);
    if(except_flags) {
        /* An error occurred; handle it appropriately. */
    }

    return EXIT_SUCCESS;
}

produces the output:

(e ^ 2.000000) -1  is 6.389056

Classification:

ANSI, POSIX 1003.1

Safety:  
Cancellation point No
Interrupt handler No
Signal handler No
Thread Yes