Hello, I can't seem to understand how to use continued fractions to find the value of x in: x^2 + x - 72 = 0
Can someone please explain how to solve this problem? Basically, I need to prove that x = -9 or 8.
Try this.
$\displaystyle x^{2}+x-72$
$\displaystyle x^{2}+x=72$
$\displaystyle x(x+1)=72$
$\displaystyle x+1=\frac{72}{x}$
$\displaystyle x=-1+\frac{72}{x}$
Now, perform the iterations.
Hopefully, it convegres to the solutions. It may not do it too fast, though.
Continued fractions converges always and very quickly.
Let $\displaystyle [a_0;a_1,a_2,..]$ be a continued fraction then, if $\displaystyle \frac{p_k}{q_k}$ represents that $\displaystyle k$-th approximation $\displaystyle k=0,1,2,...$ in reduced terms we have:
$\displaystyle \left| x - \frac{p_k}{q_k} \right| \leq \frac{1}{q_kq_{k+1}} $
But, $\displaystyle q_k$ are and increasing sequence of integers so $\displaystyle q_k \geq k$ (this is a theorem).
This means,
$\displaystyle \left| x - \frac{p_k}{q_k} \right| \leq \frac{1}{k^2}$
EDIT: Mistake fixed.
(I made a pathetic mistake instead of $\displaystyle \frac{1}{4^k}$ it should be $\displaystyle \frac{1}{k^2}$ for $\displaystyle k\geq 1$).
Consider $\displaystyle x = 1+\frac{1}{x}$ the positive solution is $\displaystyle \phi \approx 1.618$.
Now it means $\displaystyle x = \frac{1}{1+\frac{1}{x}} = \frac{1}{1+\frac{1}{1+\frac{1}{x}}}$
You form the continued fraction or simply 1's so $\displaystyle [1;1,1,1,1,...$.
Now let $\displaystyle C_k$ represent the fraction (approximation) after the $\displaystyle k$ terms.
So $\displaystyle C_1 = [1;1] = 1+\frac{1}{1} = 2$ and the difference between the real value is $\displaystyle |1.618 - 2| = .618 \leq \frac{1}{1^2}$.
Now $\displaystyle C_2 = [1;1,1] = 1+\frac{1}{1+\frac{1}{1}} = \frac{3}{2}$ and the difference between the real value is $\displaystyle |1.618 - 1.5| = .118\leq \frac{1}{2^2} = .25$.
This tells us the difference between the approximation and the actual value gets very close.