$\displaystyle \sqrt{5} $
the floor of sqrt[5] is 2
$\displaystyle \sqrt{5} = 2 + \sqrt{5} -2 = 2 + \frac{(\sqrt{5}-2)(\sqrt{5}+2)}{\sqrt{5}+2}= 2 + \frac{5 -4}{\sqrt{5}+2} = 2 + \frac{1}{\sqrt{5}+2} $
[2 ;
$\displaystyle \sqrt{5}+2 = 4 + \sqrt{5} - 2 $
[2;4
$\displaystyle \sqrt{5} - 2 = \frac{(\sqrt{5}-2)(\sqrt{5}+2)}{\sqrt{5}+2} = \frac{1}{\sqrt{5} +2 } $
$\displaystyle \sqrt{5} +2 = 4 + \sqrt{5} - 2 $
[2;4,4
and so on
$\displaystyle \sqrt{7} $ the has the floor 2
$\displaystyle \sqrt{7} = 2 + \frac{\sqrt{7} -2}{1} = 2 + \frac{3}{\sqrt{7}+2} $
turned over $\displaystyle \frac{3}{\sqrt{7}+2}$
$\displaystyle \frac{\sqrt{7}+2}{3} $ we say before $\displaystyle \sqrt{7} = 2 +\sqrt{7}-2 $
so
$\displaystyle \frac{2+\sqrt{7} -2 +2 }{3} = \frac{4 + \sqrt{7}-2}{3} = 1 + \frac{\sqrt{7} -1}{3}= 1 + \frac{7-1}{3(\sqrt{7}+1)}= \frac{6}{3(\sqrt{7}+1)} $
turned over $\displaystyle \frac{6}{3(\sqrt{7}+1)} $
$\displaystyle \frac{\sqrt{7}+1}{2} $ multiply with the conjugate and so on
i find [2;1
can u continue ?
Okay, first off, what's the point of turning it over to get
$\displaystyle \frac{45}{\sqrt{7}+2}$ and $\displaystyle \frac{6}{3(\sqrt{7}+1)} $ (respectively) ?
Then, how did you get from
$\displaystyle \frac{4 + \sqrt{7}-2}{3} $ to $\displaystyle 1 + \frac{\sqrt{7} -1}{3} $ ?
Lastly, when did you know to place 2 and 1 in the list like you did ( [2;1 ) ?
in fact I'm just doing what in here
the way is for example we want to find continued fraction of sqrt[s]
let f=floor of sqrt[s] ok
then it is trivial that
sqrt[s] = f + sqrt[s] - f, is it ok for now if it then we find the first number
[f;
now we say sqrt[s] = f + (sqrt[s] - f ) the expression in the brackets multiply it with it is conjugate
$\displaystyle \frac{(\sqrt{s}-f)(\sqrt{s}+f)}{\sqrt{s}+f} = \frac{s - f^2}{\sqrt{s}+f} $
turn over it
$\displaystyle \frac{\sqrt{s}+f}{s - f^2} $ we say that sqrt[s] = f + sqrt[s] - f
so
$\displaystyle \frac{\sqrt{s}+f}{s - f^2} = \frac{2f + \sqrt{s} - f }{s - f^2 } $
suppose that 2f/(s-f^2) = r + t/(s-f^2) , r is integer so
$\displaystyle \frac{2f + \sqrt{s} - f }{s - f^2 } = r + \frac{ \sqrt{s} - f +t}{s-f^2} $
[f;r,
$\displaystyle \frac{ \sqrt{s} - f +t}{s-f^2} $ multiply with the conjugate then turn over and so on
Here's my code:
Which gives output for $\displaystyle \displaystyle \sqrt{5}$:Code:import java.util.ArrayList; public class ContFracSqrt { static boolean v=true; // verbosity public static void main(String[] args) { print(cfe(5)); } static void print(ArrayList<Integer> a) { int i; String s=a.toString(); if(a.size()>1) for(i=2;;i++) if(s.charAt(i)==',') { s=s.substring(0,i)+"; ("+s.substring(i+2); s=s.substring(0,s.length()-1)+")]"; break; } System.out.println(s); } static ArrayList<Integer> cfe(int n) { ArrayList<Integer> x=new ArrayList<Integer>(); int a=(int)Math.sqrt(n),b=a,c=1,d,e,f,g; x.add(a); if(a*a==n) return x; if(v) System.out.println("\\sqrt{"+n+"}="+a+"+\\dfrac{\\sqrt{"+n+"}-"+a+"}{1}"); if(v) System.out.print("\\dfrac{1}{\\sqrt{"+n+"}-"+a+"}="); while(true) { d=c; c=n-b*b; g=gcd(c,d); c/=g; d/=g; b=-b; f=a-c; for(e=0;b<=f;e++) b+=c; x.add(e); if(v) System.out.println(e+"+\\dfrac{\\sqrt{"+n+"}-"+b+"}{"+c+"}"); if(b==a&&c==1) return x; if(v) System.out.print("\\dfrac{"+c+"}{\\sqrt{"+n+"}-"+b+"}="); } } static int gcd(int a,int b) { return (b==0)?a:gcd(b,a%b); } }
$\displaystyle \sqrt{5}=2+\dfrac{\sqrt{5}-2}{1}$
$\displaystyle \dfrac{1}{\sqrt{5}-2}=4+\dfrac{\sqrt{5}-2}{1}$
$\displaystyle [2; (4)]$
and for $\displaystyle \displaystyle \sqrt{7}$:
$\displaystyle \sqrt{7}=2+\dfrac{\sqrt{7}-2}{1}$
$\displaystyle \dfrac{1}{\sqrt{7}-2}=1+\dfrac{\sqrt{7}-1}{3}$
$\displaystyle \dfrac{3}{\sqrt{7}-1}=1+\dfrac{\sqrt{7}-1}{2}$
$\displaystyle \dfrac{2}{\sqrt{7}-1}=1+\dfrac{\sqrt{7}-2}{3}$
$\displaystyle \dfrac{3}{\sqrt{7}-2}=4+\dfrac{\sqrt{7}-2}{1}$
$\displaystyle [2; (1, 1, 1, 4)]$
Let me know how this works for you. I decided against ASCII mode since LaTeX is so much nicer.