expm1, expm1f, expm1l - cppreference.com

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# expm1, expm1f, expm1l

From cppreference.com
< c‎ | numeric‎ | math

Defined in header `<math.h>` |  |
float expm1f( float arg ); | (1) | (since C99)
double expm1( double arg ); | (2) | (since C99)
long double expm1l( long double arg ); | (3) | (since C99)
Defined in header `<tgmath.h>` |  |
#define expm1( arg ) | (4) | (since C99)

1-3) Computes the e (Euler's number, `2.7182818`) raised to the given power `arg`, minus 1.0. This function is more accurate than the expression std::exp(arg)-1.0 if `arg` is close to zero.
4) Type-generic macro: If `arg` has type long double, `expm1l` is called. Otherwise, if `arg` has integer type or the type double, `expm1` is called. Otherwise, `expm1f` is called.

### [edit] Parameters

arg | - | floating point value

### [edit] Return value

If no errors occur earg
-1 is returned.

If a range error due to overflow occurs, `+HUGE_VAL`, `+HUGE_VALF`, or `+HUGE_VALL` is returned.

If a range error occurs due to underflow, the correct result (after rounding) is returned.

### [edit] Error handling

Errors are reported as specified in math_errhandling

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

- If the argument is ±0, it is returned, unmodified
- If the argument is -∞, -1 is returned
- If the argument is +∞, +∞ is returned
- If the argument is NaN, NaN is returned

### [edit] Notes

The functions `expm1` and log1p are useful for financial calculations, for example, when calculating small daily interest rates: (1+x)n
-1 can be expressed as expm1(n * log1p(x)). These functions also simplify writing accurate inverse hyperbolic functions.

For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg

### [edit] Example

Run this code

    #include <stdio.h>
    #include <math.h>
    #include <float.h>
    #include <errno.h>
    #include <fenv.h>
    #pragma STDC FENV_ACCESS ON
    int main(void)
    {
        printf("expm1(1) = %f\n", expm1(1));
        printf("Interest earned in 2 days on on $100, compounded daily at 1%\n"
               " on a 30/360 calendar = %f\n",
               100*expm1(2*log1p(0.01/360)));
        printf("exp(1e-16)-1 = %g, but expm1(1e-16) = %g\n",
               exp(1e-16)-1, expm1(1e-16));
        // special values
        printf("expm1(-0) = %f\n", expm1(-0.0));
        printf("expm1(-Inf) = %f\n", expm1(-INFINITY));
        //error handling
        errno = 0; feclearexcept(FE_ALL_EXCEPT);
        printf("expm1(710) = %f\n", expm1(710));
        if(errno == ERANGE) perror("    errno == ERANGE");
        if(fetestexcept(FE_OVERFLOW)) puts("    FE_OVERFLOW raised");
    }

Possible output:

    expm1(1) = 1.718282
    Interest earned in 2 days on on $100, compounded daily at 1%
     on a 30/360 calendar = 0.005556
    exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16
    expm1(-0) = -0.000000
    expm1(-Inf) = -1.000000
    expm1(710) = inf
        errno == ERANGE: Result too large
        FE_OVERFLOW raised

### [edit] See also

expexpfexpl (C99)(C99) | computes e raised to the given power (ex) (function) [edit]
exp2exp2fexp2l (C99)(C99)(C99) | computes 2 raised to the given power (2x) (function) [edit]
log1plog1pflog1pl (C99)(C99)(C99) | computes natural (base-e) logarithm of 1 plus the given number (ln(1+x)) (function) [edit]
C++ documentation for expm1

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