Re: Exp() function reloaded

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Date	Sun, 15 Apr 2012 09:30:05 -0400
From	pete <pfiland@mindspring.com>
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Subject	Re: Exp() function reloaded
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Cross-posted to 3 groups.

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hopcode wrote:
> 
> Il 15.04.2012 07:26, pete ha scritto:
> >
> > 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.
> >
> > ...
> >      for (square = 0; x>  1; x /= 2) {
> >      while (-1>  x) {
> >          x /= 2;
> >      do {
> >          b /= n++;
> >          b /= n++;
> > ...
> 
> that is a waster divisional toy, pete.
> replace it with something better. ;-)

The point that I was trying make,
is that if you can do Taylor's for (-1.0 <  x),
then it may not be necessary to consider
e^-x being essentially 1 / e^x.

-- 
pete

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Exp() function reloaded hopcode <hopcode@invalid.de> - 2012-04-15 04:31 +0200
  Re: Exp() function reloaded pete <pfiland@mindspring.com> - 2012-04-15 01:26 -0400
    Re: Exp() function reloaded hopcode <hopcode@invalid.de> - 2012-04-15 12:18 +0200
      Re: Exp() function reloaded pete <pfiland@mindspring.com> - 2012-04-15 09:30 -0400
    Re: Exp() function reloaded "io_x" <a@b.c.invalid> - 2012-04-16 09:01 +0200
      Re: Exp() function reloaded "io_x" <a@b.c.invalid> - 2012-04-16 11:50 +0200
        Re: Exp() function reloaded "io_x" <a@b.c.invalid> - 2012-04-16 15:46 +0200
          Re: Exp() function reloaded "io_x" <a@b.c.invalid> - 2012-04-16 16:07 +0200
            Re: Exp() function reloaded "io_x" <a@b.c.invalid> - 2012-04-17 09:36 +0200
              Re: Exp() function reloaded phreda <pabloreda@gmail.com> - 2012-04-19 19:43 -0700
                Re: Exp() function reloaded hopcode <hopcode@invalid.de> - 2012-04-24 14:48 +0200
                Re: Exp() function reloaded phreda <pabloreda@gmail.com> - 2012-04-25 12:36 -0700
        Re: Exp() function reloaded pete <pfiland@mindspring.com> - 2012-04-16 18:12 -0400

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