how do they work, does anyone know how to explain it cause i need to do a portfolio and give it in tomorow, if not: death

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- Nov 14th 2006, 06:13 AMCiliaContienued fractions
how do they work, does anyone know how to explain it cause i need to do a portfolio and give it in tomorow, if not: death

- Nov 14th 2006, 06:19 AMThePerfectHacker
- Nov 14th 2006, 06:27 AMCilia
look, i kinda know it sounds weird but i've asked ppl aound me 2 but, their not of the friendly type or do not have tha mathematical genius to understand it....so PLEASE help me!

- Nov 14th 2006, 06:32 AMCilia
haha, i seriusly do know what to do. if u got the basic continued fraction (not the algebraic one, but the 1 one) and its something like:

t1=1+1

t2=1+(1/(1+1))

t3=1+(1/1+(1+1)

...

and ur suposed to give a formula for tn+1 in terms of tn - Nov 14th 2006, 06:37 AMThePerfectHacker
- Nov 14th 2006, 06:52 AMCilia
how that, for large values of n, tn satisfies an equation which can be expressed in form tn^2-tn-1.

obtain an equation in tn for a continous fraction where ther is a 2+ instead of a 1+.

...then solve the equation to fin the EXACT value - Nov 14th 2006, 06:57 AMThePerfectHacker
It is relying on a theorem which is not so easy to show that all continued fractions converge (the ones with integers).

So we know that $\displaystyle x=[1;1,1,...,]$

Converges to some number.

But,

$\displaystyle x=[1;x]$

For it repeats.

Thus,

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

Thus,

$\displaystyle x^2=x+1$

Thus,

$\displaystyle x^2-x-1=0$

Thus,

$\displaystyle x=\frac{1\pm \sqrt{5}}{2}$

This tells us that this is one of these two possible values.

It cannot be negative thus,

$\displaystyle x=\frac{1+\sqrt{5}}{2}=\psi$

The Divine Proportion (My body is shaped in Divine Proportion). - Nov 14th 2006, 07:09 AMSoroban
Hello, Cilia!

This is a vast and intricate topic which could take months to explain.

If you know the very basics, I can show you a tiny sliver of the whole sprectrum.

Your example:

$\displaystyle t_1 \:=\:1 + 1\qquad t_2 \:=\:1 + \frac{1}{1+1}\qquad t_3 \:=\:1 + \frac{1}{1 + \frac{1}{1+1}} $

By inspection ("eyeballing" it),

. . we see that each term is: one plue one over the preceding term.

That is: .$\displaystyle t_{n+1} \;= \; 1 + \frac{1}{t_n} $

Assuming that the fraction goes on forever: .$\displaystyle 1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}}$

. . can we determine its value?

$\displaystyle \text{Let }x\:=\:1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}}

\begin{array}{ccc} \\ \\ \bigg\}\text{ this is }x \end{array}

$

Hence, we have: .$\displaystyle x \:=\:1 + \frac{1}{x}\quad\Rightarrow\quad x^2 - x - 1 \:=\:0$

Quadratic Formula: .$\displaystyle x \:=\:\frac{\text{-}(\text{-}1) \pm \sqrt{(\text{-}1)^2 - 4(1)(\text{-}1)}}{2(1)} \;=\;\frac{1 \pm \sqrt{5}}{2}$

Since $\displaystyle x$ is positive, we have: .$\displaystyle x \:=\:\frac{1 + \sqrt{5}}{2} \:= \:1.618033989... $

. . which happens to be the Golden Mean, $\displaystyle \phi$.

$\displaystyle \text{Therefore: }\;1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}} \;=\;\phi$

- Nov 14th 2006, 08:38 AMTriKri
A useful link: Continued fraction - Wikipedia, the free encyclopedia