# Re: Work around for "Bignum out of Float range"?

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

[email protected]n.invalid wrote:

def calc(n,prec)
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.
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

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