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Defined in header <math.h> | ||
float expm1f(float arg); | (1) | (since C99) |
double expm1(double arg); | (2) | (since C99) |
longdouble expm1l(longdouble arg); | (3) | (since C99) |
Defined in header <tgmath.h> | ||
#define expm1( arg ) | (4) | (since C99) |
2.7182818) raised to the given powerarg, minus1.0. This function is more accurate than the expressionexp(arg)-1.0 ifarg is close to zero.arg has typelongdouble,expm1l is called. Otherwise, ifarg has integer type or the typedouble,expm1 is called. Otherwise,expm1f is called.Contents |
| arg | - | floating-point value |
If no errors occurearg
-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.
Errors are reported as specified inmath_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
The functionsexpm1 andlog1p are useful for financial calculations, for example, when calculating small daily interest rates:(1+x)n
-1 can be expressed asexpm1(n*log1p(x)). These functions also simplify writing accurate inverse hyperbolic functions.
For IEEE-compatible typedouble, overflow is guaranteed if709.8 < arg.
#include <errno.h>#include <fenv.h>#include <float.h>#include <math.h>#include <stdio.h>// #pragma STDC FENV_ACCESS ONint main(void){printf("expm1(1) = %f\n", expm1(1));printf("Interest earned in 2 days 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 valuesprintf("expm1(-0) = %f\n", expm1(-0.0));printf("expm1(-Inf) = %f\n", expm1(-INFINITY));//error handlingerrno=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) = 1.718282Interest earned in 2 days on $100, compounded daily at 1% on a 30/360 calendar = 0.005556exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16expm1(-0) = -0.000000expm1(-Inf) = -1.000000expm1(710) = inf errno == ERANGE: Result too large FE_OVERFLOW raised
(C99)(C99) | computese raised to the given power (\({\small e^x}\)ex) (function)[edit] |
(C99)(C99)(C99) | computes2 raised to the given power (\({\small 2^x}\)2x) (function)[edit] |
(C99)(C99)(C99) | computes natural (base-e) logarithm of 1 plus the given number (\({\small \ln{(1+x)} }\)ln(1+x)) (function)[edit] |
C++ documentation forexpm1 | |
