pow function

[Magix]

#include <math.h>
double pow( double base, double exp );
will pow (2, -30) works fine ? (exp is negative value). If not, what are the
workaround for exp to be negative?

[Jack Klein]

On Tue, 19 Oct 2004 10:39:49 +0800, "Magix" <ma***@asia.com> wrote in
comp.lang.c:

#include <math.h>
double pow( double base, double exp );
will pow (2, -30) works fine ? (exp is negative value). If not, what are the
workaround for exp to be negative?

Do you really think posting to usenet is faster and/or better than
looking at your compiler's library documentation, man pages, or online
help? Really?

Well, if you insist, here is a part of the C standard's definition of
the pow() function:

"2 The pow functions compute x raised to the power y. A domain error
occurs if x is finite and negative and y is finite and not an integer
value. A domain error may occur if x is zero and y is less than or
equal to zero. A range error may occur."

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
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alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Scott J. McCaughrin]

Magix <ma***@asia.com> wrote:
: #include <math.h>
: double pow( double base, double exp );
: will pow (2, -30) works fine ? (exp is negative value). If not, what are the
: workaround for exp to be negative?

2^(-30) = 1.0/(2^30) = (1/2)^30, so try pow(0.5, 30.0).

[Richard Bos]

"Scott J. McCaughrin" <sj******@bluestem.prairienet.org> wrote:

Magix <ma***@asia.com> wrote:
: #include <math.h>
: double pow( double base, double exp );

: will pow (2, -30) works fine ? (exp is negative value). If not, what are the
: workaround for exp to be negative?

2^(-30) = 1.0/(2^30) = (1/2)^30, so try pow(0.5, 30.0).

Ah, but now you need to work out 1/2 by hand. This is feasible for
constants, but what if base is a variable? Luckily,

base ** (-exp) == base ** (-1*exp) == (base ** -1) ** exp

so you could do pow(pow(2, -1), 30)...

Richard

[Martin Ambuhl]

Richard Bos wrote:

"Scott J. McCaughrin" <sj******@bluestem.prairienet.org> wrote:

Magix <ma***@asia.com> wrote:
: #include <math.h>
: double pow( double base, double exp );

: will pow (2, -30) works fine ? (exp is negative value). If not, what are the
: workaround for exp to be negative?

2^(-30) = 1.0/(2^30) = (1/2)^30, so try pow(0.5, 30.0).

Ah, but now you need to work out 1/2 by hand. This is feasible for
constants, but what if base is a variable? Luckily,

base ** (-exp) == base ** (-1*exp) == (base ** -1) ** exp

so you could do pow(pow(2, -1), 30)...

What the hell are you guys smoking?
#include <math.h>
#include <stdio.h>
#include <float.h>

int main(void)
{
double x1, x2;
x1 = pow(2, -30);
x2 = pow(0.5, 30);
printf("x1 = pow(2,-30) = %.*g\n"
"x2 = pow(0.5,30) = %.*g\n"
"x1-x2 = %.*g\n",
DBL_DIG, x1, DBL_DIG, x2, DBL_DIG, x1 - x2);
return 0;
}

[output]
x1 = pow(2,-30) = 9.31322574615479e-10
x2 = pow(0.5,30) = 9.31322574615479e-10
x1-x2 = 0

[Richard Bos]

Martin Ambuhl <ma*****@earthlink.net> wrote:

Richard Bos wrote:

"Scott J. McCaughrin" <sj******@bluestem.prairienet.org> wrote:

Magix <ma***@asia.com> wrote:
: #include <math.h>
: double pow( double base, double exp );

: will pow (2, -30) works fine ? (exp is negative value). If not, what are the
: workaround for exp to be negative?

2^(-30) = 1.0/(2^30) = (1/2)^30, so try pow(0.5, 30.0).

Ah, but now you need to work out 1/2 by hand. This is feasible for
constants, but what if base is a variable? Luckily,

base ** (-exp) == base ** (-1*exp) == (base ** -1) ** exp

so you could do pow(pow(2, -1), 30)...

What the hell are you guys smoking?

I can't speak for Mr. McCaughrin, but I was smoking some of those man
pages Jack Klein mentioned...

Richard