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.

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- Oct 21st 2007, 03:22 PMvpereraContinued 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 PMgalactus
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. - Oct 21st 2007, 04:19 PMvperera
Actually, I'm totally new to the continued fractions thing. Could you explain what these iterations are?

- Oct 21st 2007, 04:22 PMgalactus
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 PMvperera
Keep subbing what into x?

- Oct 21st 2007, 04:44 PMvperera
Oh, so sub -1 + 72/x into x, right?

And what do you mean keep subbing to infinity? - Oct 21st 2007, 05:37 PMThePerfectHacker
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. - Oct 21st 2007, 06:08 PMvperera
can you explain that to me in lay man's terms? :)

- Oct 21st 2007, 06:17 PMThePerfectHacker
(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. - Oct 21st 2007, 07:13 PMvperera
I still don't understand.. could you use my problem as an example? x^2 +x - 72 = 0