# Math Help - Continued Fractions

1. ## 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.

2. 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.

3. Actually, I'm totally new to the continued fractions thing. Could you explain what these iterations are?

4. 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)

5. Keep subbing what into x?

6. Oh, so sub -1 + 72/x into x, right?

And what do you mean keep subbing to infinity?

7. 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.

8. can you explain that to me in lay man's terms?

9. 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.

10. I still don't understand.. could you use my problem as an example? x^2 +x - 72 = 0