Compute the exponential of a number, then subtract 1
#include <math.h> double expm1 ( double x ); float expm1f ( float x ); long double expm1l ( long double x );
The expm1(), expm1f(), and expm1l() functions compute the exponential of x, minus 1 (e^{x} - 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:
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.
#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
Safety: | |
---|---|
Cancellation point | No |
Interrupt handler | No |
Signal handler | No |
Thread | Yes |