# Continued Fractions

• Oct 21st 2007, 03:22 PM
vperera
Continued Fractions
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.
• Oct 21st 2007, 04:08 PM
galactus
Try this.

$x^{2}+x-72$

$x^{2}+x=72$

$x(x+1)=72$

$x+1=\frac{72}{x}$

$x=-1+\frac{72}{x}$

Now, perform the iterations.

Hopefully, it convegres to the solutions. It may not do it too fast, though.
• Oct 21st 2007, 04:19 PM
vperera
Actually, I'm totally new to the continued fractions thing. Could you explain what these iterations are?
• Oct 21st 2007, 04:22 PM
galactus
See what x equals?. See the x under the 72?. Keep subbing it in, ad infinitum.

Do it enough and it should converge to the solution(s)
• Oct 21st 2007, 04:33 PM
vperera
Keep subbing what into x?
• Oct 21st 2007, 04:44 PM
vperera
Oh, so sub -1 + 72/x into x, right?

And what do you mean keep subbing to infinity?
• Oct 21st 2007, 05:37 PM
ThePerfectHacker
Quote:

Originally Posted by galactus
Hopefully, it convegres to the solutions. It may not do it too fast, though.

Continued fractions converges always and very quickly.

Let $[a_0;a_1,a_2,..]$ be a continued fraction then, if $\frac{p_k}{q_k}$ represents that $k$-th approximation $k=0,1,2,...$ in reduced terms we have:
$\left| x - \frac{p_k}{q_k} \right| \leq \frac{1}{q_kq_{k+1}}$
But, $q_k$ are and increasing sequence of integers so $q_k \geq k$ (this is a theorem).
This means,
$\left| x - \frac{p_k}{q_k} \right| \leq \frac{1}{k^2}$

EDIT: Mistake fixed.
• Oct 21st 2007, 06:08 PM
vperera
can you explain that to me in lay man's terms? :)
• Oct 21st 2007, 06:17 PM
ThePerfectHacker
Quote:

Originally Posted by vperera
can you explain that to me in lay man's terms? :)

(I made a pathetic mistake instead of $\frac{1}{4^k}$ it should be $\frac{1}{k^2}$ for $k\geq 1$).

Consider $x = 1+\frac{1}{x}$ the positive solution is $\phi \approx 1.618$.

Now it means $x = \frac{1}{1+\frac{1}{x}} = \frac{1}{1+\frac{1}{1+\frac{1}{x}}}$

You form the continued fraction or simply 1's so $[1;1,1,1,1,...$.

Now let $C_k$ represent the fraction (approximation) after the $k$ terms.

So $C_1 = [1;1] = 1+\frac{1}{1} = 2$ and the difference between the real value is $|1.618 - 2| = .618 \leq \frac{1}{1^2}$.

Now $C_2 = [1;1,1] = 1+\frac{1}{1+\frac{1}{1}} = \frac{3}{2}$ and the difference between the real value is $|1.618 - 1.5| = .118\leq \frac{1}{2^2} = .25$.

This tells us the difference between the approximation and the actual value gets very close.
• Oct 21st 2007, 07:13 PM
vperera
I still don't understand.. could you use my problem as an example? x^2 +x - 72 = 0