*Compute the exponential of a number, then subtract 1*

#include <math.h> double expm1 ( doublex); float expm1f ( floatx);

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

`libm`

Use the ` -l m` option to

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.

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. |

#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

Safety: | |
---|---|

Cancellation point | No |

Interrupt handler | No |

Signal handler | No |

Thread | Yes |