|
|
COMMENTS
|
In general, denominators, a(k,n) and numerators, b(k,n), of continued
fraction convergents to sqrt((k+1)/k) may be found as follows:
a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n)=2*a(k,2n-1)+a(k,2n-2)
and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1);
b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n)=2*b(k,2n-1)+b(k,2n-2)
and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1).
For example, the convergents to sqrt(4/3) start 1/1, 11/10, 23/21,
241/220, 505/461.
In general, if a(k,n) and b(k,n) are the denominators and numerators,
respectively, of continued fraction convergents to sqrt((k+1)/k)
as defined above, then
k*a(k,2n)^2-a(k,2n-1)*a(k,2n+1)=k=k*a(k,2n-2)*a(k,2n)-a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1)-k*b(k,2n)^2=k+1=b(k,2n-1)^2-k*b(k,2n-2)*b(k,2n);
for example, if k=5 and n=3, then a(5,n)=a(n) and
5*a(5,6)^2-a(5,5)*a(5,7)=5*10121^2-4830*106040=5;
5*a(5,4)*a(5,6)-a(5,5)^2=5*461*10121-4830^2=5;
b(5,5)*b(5,7)-5*b(5,6)^2=5291*116161-5*11087^2=6;
b(5,5)^2-5*b(5,4)*b(5,6)=5291^2-5*505*11087=6.
sqrt(6/5) = 1.09544511501... = 2/2 + 2/(1*21) +
2/(21*461) + 2/(461*10121) + 2/(10121*222201) +
For k>0 and n>2, let m=4*k+2, m(1)=1, m(2)=m-1 and m(n)=
m*d(n-1)-d(n-2); for n>0, let d(n)=m(n)*m(n+1).
Then, in general,
sqrt((k+1)/k)=2/2+2/d(1)+2/d(2)+2/d(3)+....
For example, if k=5, then m=22, sqrt(7/6)=1.080123450...
and 2/2+2/d(1)+2/d(2)+2/d(3)= 1.080123450...
|
