Solve x in Periodic Continued Fraction Expansion

[*FaerieLight*]

Homework Statement

Find the real positive number x with periodic continued fraction expansion [0;a,b,a,b,...] where a and b are positive integers.

Homework Equations

The Attempt at a Solution

I have determined that x definitely involves a root.

 

[HallsofIvy]

It's been a while since I have done these kinds of problems but I think the notation "[0;a,b,a,b,...]" means a continued fraction of the form
x= \frac{a}{b+ \frac{a}{b+ \frac{a}{b+\cdot\cdot\cdot}}}

If that is correct, then that is the same as
x= \frac{a}{b+ x}
since the "continued" fraction in the denominator is still just "x".

Of course, that is the same as the quadratic equation x(b+ x)= x^2+ bx= a or x^2+ bx- a= 0. Use the quadratic formula to solve that.

 

[fzero]

If \alpha=[0;a,b,a,b,...], what do you get by computing

\frac{1}{\alpha} -a?

Basically you want to continue performing operations on \alpha until you get \alpha back. The resulting equation will always be quadratic.

Edit: HallsofIvy, not quite, the continued fractions in the [a_0;a_1,\ldots] notation are the so-called simple continued fractions, which always have unit numerators. So

[0;a,b,a,b,...] = 0 + \frac{1}{a+\frac{1}{b+\frac{1}{a+\cdots}}}

Your method still works for this of course, the result is just a bit more complicated.

 

[*FaerieLight*]

Thanks HallsofIvy!

I was taught that [0;a,b,a,b,a,...] denotes
\frac{1}{a+\frac{1}{b+\frac{1}{a+...}}}
The quadratic then becomes ax²+abx-b=0

Thank you very much! I actually looked at the continued fractions expansions for \sqrt{n} for n=1,2,3,...,20 and tried to find patterns. I did find some interesting ones, but I just couldn't manage to use them to work out this problem. I suspected the solution was a simple one, and indeed thank you very much for showing me this beautifully simple solution.

 

[*FaerieLight*]

Thanks to you too, fzero!