From: Sam Kong <sam.s.kong@...>
Date: 2006-02-03T08:23:18+09:00
Subject: Re: Work around for "Bignum out of Float range"?


Nuralanur@aol.com wrote:
> Dear Sam,
>
> your problem is somewhat insolvable, since the square root of 5 has
> an infinite number of digits, so it cannot be represented correctly
> on any computer (in decimal or binary notation).
> For arbitrary precision calculations with Fixnum exponents, there's
> bigdecimal.
>
> require "bigdecimal"
>
> def calc(n,prec)
>    # prec is the precision of the sqrt  calculations
> res=(BigDecimal.new("2") **  n)*BigDecimal("5").sqrt(prec)
> end
>
>
>
>
> puts calc(10000) => 0.44610[several lines of digits]*10^3011 (for  prec=10)
>
>
> However, don't believe in too many of these digits...
> If you still want to multiply by square roots accurately, it might be a  good
> idea to look at continued fractions - every square root has a  continued
> fraction representation that eventually ends in a periodic pattern.
> There is an introduction to arithmetic with them at
> _http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html_
> (http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html)
>
> Hope that  helps,
>
> Axel

Thank you, Axel.
Actually I came across this problem while reading an article about
fibonacci.
See
http://epsilondelta.wordpress.com/2006/01/29/programming-like-a-mathematician-i-closures/
..
There's a formula for fibonacci and the writer showed 1000000th
fibonacci.
I wanted to calculate it using the formula in Ruby.

I'll read the docs you mentioned.

Thanks.
Sam