freebsd-skq/lib/msun/man/exp.3

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.\"     from: @(#)exp.3	6.12 (Berkeley) 7/31/91
.\" $FreeBSD$
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.Dd April 5, 2005
.Dt EXP 3
.Os
.Sh NAME
.Nm exp ,
.Nm expf ,
.\" The sorting error is intentional.  exp and expf should be adjacent.
.Nm exp2 ,
.Nm exp2f ,
.Nm expm1 ,
.Nm expm1f ,
.Nm log ,
.Nm logf ,
.Nm log10 ,
.Nm log10f ,
.Nm log1p ,
.Nm log1pf ,
.Nm pow ,
.Nm powf
.Nd exponential, logarithm, power functions
.Sh LIBRARY
.Lb libm
.Sh SYNOPSIS
.In math.h
.Ft double
.Fn exp "double x"
.Ft float
.Fn expf "float x"
.Ft double
.Fn exp2 "double x"
.Ft float
.Fn exp2f "float x"
.Ft double
.Fn expm1 "double x"
.Ft float
.Fn expm1f "float x"
.Ft double
.Fn log "double x"
.Ft float
.Fn logf "float x"
.Ft double
.Fn log10 "double x"
.Ft float
.Fn log10f "float x"
.Ft double
.Fn log1p "double x"
.Ft float
.Fn log1pf "float x"
.Ft double
.Fn pow "double x" "double y"
.Ft float
.Fn powf "float x" "float y"
.Sh DESCRIPTION
The
.Fn exp
and the
.Fn expf
functions compute the base
.Ms e
exponential value of the given argument
.Fa x .
.Pp
The
.Fn exp2
and the
.Fn exp2f
functions compute the base 2 exponential of the given argument
.Fa x .
.Pp
The
.Fn expm1
and the
.Fn expm1f
functions compute the value exp(x)\-1 accurately even for tiny argument
.Fa x .
.Pp
The
.Fn log
and the
.Fn logf
functions compute the value of the natural logarithm of argument
.Fa x .
.Pp
The
.Fn log10
and the
.Fn log10f
functions compute the value of the logarithm of argument
.Fa x
to base 10.
.Pp
The
.Fn log1p
and the
.Fn log1pf
functions compute
the value of log(1+x) accurately even for tiny argument
.Fa x .
.Pp
The
.Fn pow
and the
.Fn powf
functions compute the value
of
.Ar x
to the exponent
.Ar y .
.Sh ERROR (due to Roundoff etc.)
The values of
.Fn exp 0 ,
.Fn expm1 0 ,
.Fn exp2 integer ,
and
.Fn pow integer integer
are exact provided that they are representable.
.\" XXX Is this really true for pow()?
Otherwise the error in these functions is generally below one
.Em ulp .
.Sh RETURN VALUES
These functions will return the appropriate computation unless an error
occurs or an argument is out of range.
The functions
.Fn pow x y
and
.Fn powf x y
raise an invalid exception and return an \*(Na if
.Fa x
< 0 and
.Fa y
is not an integer.
An attempt to take the logarithm of \*(Pm0 will result in
a divide-by-zero exception, and an infinity will be returned.
An attempt to take the logarithm of a negative number will
result in an invalid exception, and an \*(Na will be generated.
.Sh NOTES
The functions exp(x)\-1 and log(1+x) are called
expm1 and logp1 in
.Tn BASIC
on the Hewlett\-Packard
.Tn HP Ns \-71B
and
.Tn APPLE
Macintosh,
.Tn EXP1
and
.Tn LN1
in Pascal, exp1 and log1 in C
on
.Tn APPLE
Macintoshes, where they have been provided to make
sure financial calculations of ((1+x)**n\-1)/x, namely
expm1(n\(**log1p(x))/x, will be accurate when x is tiny.
They also provide accurate inverse hyperbolic functions.
.Pp
The function
.Fn pow x 0
returns x**0 = 1 for all x including x = 0, \*(If, and \*(Na .
Previous implementations of pow may
have defined x**0 to be undefined in some or all of these
cases.
Here are reasons for returning x**0 = 1 always:
.Bl -enum -width indent
.It
Any program that already tests whether x is zero (or
infinite or \*(Na) before computing x**0 cannot care
whether 0**0 = 1 or not.
Any program that depends
upon 0**0 to be invalid is dubious anyway since that
expression's meaning and, if invalid, its consequences
vary from one computer system to another.
.It
Some Algebra texts (e.g.\& Sigler's) define x**0 = 1 for
all x, including x = 0.
This is compatible with the convention that accepts a[0]
as the value of polynomial
.Bd -literal -offset indent
p(x) = a[0]\(**x**0 + a[1]\(**x**1 + a[2]\(**x**2 +...+ a[n]\(**x**n
.Ed
.Pp
at x = 0 rather than reject a[0]\(**0**0 as invalid.
.It
Analysts will accept 0**0 = 1 despite that x**y can
approach anything or nothing as x and y approach 0
independently.
The reason for setting 0**0 = 1 anyway is this:
.Bd -ragged -offset indent
If x(z) and y(z) are
.Em any
functions analytic (expandable
in power series) in z around z = 0, and if there
x(0) = y(0) = 0, then x(z)**y(z) \(-> 1 as z \(-> 0.
.Ed
.It
If 0**0 = 1, then
\*(If**0 = 1/0**0 = 1 too; and
then \*(Na**0 = 1 too because x**0 = 1 for all finite
and infinite x, i.e., independently of x.
.El
.Sh SEE ALSO
.Xr fenv 3 ,
.Xr math 3