continued fraction
Mathworld





a

0


+









b

1












a

1


+









b

2












a

2


+









b

3












a

3


+
⋱

















{\displaystyle a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+\ddots }}}}}}}


The task is to write a program which generates such a number and prints a real representation of it. The code should be tested by calculating and printing the square root of 2, Napier's Constant, and Pi, using the following coefficients:

For the square root of 2, use 




a

0


=
1


{\displaystyle a_{0}=1}

 then 




a

N


=
2


{\displaystyle a_{N}=2}

. 




b

N




{\displaystyle b_{N}}

 is always 



1


{\displaystyle 1}

.







2


=
1
+








1









2
+








1









2
+








1









2
+
⋱

















{\displaystyle {\sqrt {2}}=1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+\ddots }}}}}}}


For Napier's Constant, use 




a

0


=
2


{\displaystyle a_{0}=2}

, then 




a

N


=
N


{\displaystyle a_{N}=N}

. 




b

1


=
1


{\displaystyle b_{1}=1}

 then 




b

N


=
N
−
1


{\displaystyle b_{N}=N-1}

.





e
=
2
+








1









1
+








1









2
+








2









3
+








3









4
+
⋱






















{\displaystyle e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {2}{3+{\cfrac {3}{4+\ddots }}}}}}}}}


For Pi, use 




a

0


=
3


{\displaystyle a_{0}=3}

 then 




a

N


=
6


{\displaystyle a_{N}=6}

. 




b

N


=
(
2
N
−
1

)

2




{\displaystyle b_{N}=(2N-1)^{2}}

.





π
=
3
+








1









6
+








9









6
+








25









6
+
⋱

















{\displaystyle \pi =3+{\cfrac {1}{6+{\cfrac {9}{6+{\cfrac {25}{6+\ddots }}}}}}}




See also

   Continued fraction/Arithmetic for tasks that do arithmetic over continued fractions.


require 'bigdecimal'
 
# square root of 2
sqrt2 = Object.new
def sqrt2.a(n); n == 1 ? 1 : 2; end
def sqrt2.b(n); 1; end
 
# Napier's constant
napier = Object.new
def napier.a(n); n == 1 ? 2 : n - 1; end
def napier.b(n); n == 1 ? 1 : n - 1; end
 
pi = Object.new
def pi.a(n); n == 1 ? 3 : 6; end
def pi.b(n); (2*n - 1)**2; end
 
# Estimates the value of a continued fraction _cfrac_, to _prec_
# decimal digits of precision. Returns a BigDecimal. _cfrac_ must
# respond to _cfrac.a(n)_ and _cfrac.b(n)_ for integer _n_ >= 1.
def estimate(cfrac, prec)
  last_result = nil
  terms = prec
 
  loop do
    # Estimate continued fraction for _n_ from 1 to _terms_.
    result = cfrac.a(terms)
    (terms - 1).downto(1) do |n|
      a = BigDecimal cfrac.a(n)
      b = BigDecimal cfrac.b(n)
      digits = [b.div(result, 1).exponent + prec, 1].max
      result = a + b.div(result, digits)
    end
    result = result.round(prec)
 
    if result == last_result
      return result
    else
      # Double _terms_ and try again.
      last_result = result
      terms *= 2
    end
  end
end
 
puts estimate(sqrt2, 50).to_s('F')
puts estimate(napier, 50).to_s('F')
puts estimate(pi, 10).to_s('F')