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Replies: 11   Last Post: Apr 17, 2012 3:36 AM

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 pete Posts: 190 Registered: 12/11/04
Posted: Apr 15, 2012 1:26 AM

hopcode wrote:
>
> Hi,
> here my method for the exponentianal function e^x .
> it is a mix of 2 fundamentals:
> - Taylor series running at a 10 steps for decimals -1.0 < x < +1.0
> - a couple of basic rules of the logarithms.

> it is ~40 lines of code (my Taylors's exp() code + main routine)
> it accepts only +numbers for now, and makes no exaustive check
> on the floats. for those and negative values i leave it to the
> reader's creativity ( being e^-x essentially 1 / e^x ).

I've code Taylors's exp(x) for both positive and negative x in C,
using -1.0 < x < +1.0,
but without making use of e^-x being essentially 1 / e^x.

/* BEGIN new.c output */

fs_expl(20.3) is 654904512.153239
fs_expl(20.3) - 654904512.153230 is 8.583069e-006

fs_expl(-20.3) is 1.526940e-009
fs_expl(-20.3) - 1.526940e-009 is 1.591266e-016

/* END new.c output */

/* BEGIN new.c */

#include <stdio.h>

long double fs_expl(long double x)
{
long unsigned n, square;
long double b, e, old;

for (square = 0; x > 1; x /= 2) {
++square;
}
while (-1 > x) {
++square;
x /= 2;
}
e = b = n = 1;
do {
b /= n++;
b *= x;
e += b;
b /= n++;
b *= x;
old = e;
e += b;
} while (e > old);
while (square-- != 0) {
e *= e;
}
return e;
}

int
main(void)
{
puts("/* BEGIN new.c output */\n");
printf("fs_expl(20.3) is %lf\n",
fs_expl(20.3));
printf("fs_expl(20.3) - 654904512.153230 is %le\n\n",
fs_expl(20.3) - 654904512.153230);

printf("fs_expl(-20.3) is %le\n",
fs_expl(-20.3));
printf("fs_expl(-20.3) - 1.526940e-009 is %le\n\n",
fs_expl(-20.3) - 1.526940e-009);

puts("/* END new.c output */");
return 0;
}

/* END new.c */

--
pete

Date Subject Author
4/14/12 hopcode
4/15/12 Dave Dodson
4/15/12 hopcode
4/15/12 pete
4/15/12 hopcode
4/15/12 pete