[CLN-list] Power function

[Joshua Friedman]

I think you need to compute exp(k ln(x) ) for x^k

Sent from my iPad

On Jan 14, 2011, at 6:00 AM, cln-list-request at ginac.de wrote:

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> Today's Topics:
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>   1. exponants of floating points (or do I need it ?) (Franck Sabati?)
> 
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> ----------------------------------------------------------------------
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> Message: 1
> Date: Fri, 14 Jan 2011 09:40:23 +0100
> From: Franck Sabati? <fsabatie at gmail.com>
> To: cln-list at ginac.de
> Subject: [CLN-list] exponants of floating points (or do I need it ?)
> Message-ID:
>    <AANLkTinG7GBxWw0We=8tGv_DNW7CoHK3L-UcFNqaU7iF at mail.gmail.com>
> Content-Type: text/plain; charset="iso-8859-1"
> 
> Hello,
> 
> I have a hard time seeing how to best implement the following : I have a
> complicated equation of 4 variables that I need evaluated with a certain
> number of digits in precision (30 or so) :
> 
> dummy = (15/4)*(1/(pow(xi,5)*(1+i-k)*(2+i-k)*(3+i-k)*(4+i-k)*(5+i-k)))*(
> ( (3*pow(3+i-k,2)-(1+i-k)*(5+i-k))*pow(xi*xi-xb,2)
> -2*(2+i-k)*(4+i-k)*(1-xi*xi)*(xb*xb-xi*xi) ) * ( pow((xb+xi)/(1+xi),3+i-k)-
> pow((xb-xi)/(1-xi),3+i-k) ) +
> 6*(3+i-k)*xi*(1-xb)*(xi*xi-xb)*(pow((xb+xi)/(1+xi),3+i-k)+
> pow((xb-xi)/(1-xi),3+i-k)) );
> 
> 
> where the variables are i, k, xb and xi. Although most of the equation is
> relatively easy to implement in CLN, I have a hard time figuring out how to
> deal with the "pow". There appears to be no such exponant function for
> floats ? Also, it is still unclear to me how to enter this equation
> painlessly into my code. Do I have to put cl_float(1,30) when I use 1 for
> instance ? (same for other numbers). Should I evaluate this in floating
> precision or with "real numbers" ? And finally how do I evaluate expressions
> such as (something)^(something else) with the (something else) NOT being an
> integer.
> 
> Sorry about the very basic questions but most of the demo examples deal with
> integers and the few that do not, have very little information in them.
> 
> Thank you for your time,
> Franck
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