# Contienued fractions

• Nov 14th 2006, 06:13 AM
Cilia
Contienued 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 AM
ThePerfectHacker
Quote:

Originally Posted by Cilia
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

It just an alternative way of expressing a number (rational or irrational). This have a fabolous property that they from the best possible convergent of a number with such a denominator.
• Nov 14th 2006, 06:27 AM
Cilia
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 AM
Cilia
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 AM
ThePerfectHacker
Quote:

Originally Posted by Cilia
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

$\displaystyle t_n=[1:1,1,1,....,1]$
$\displaystyle t_{n+1}=[1;t_n]$
Thus,
$\displaystyle t_{n+1}=1+\frac{1}{t_n}$
• Nov 14th 2006, 06:52 AM
Cilia
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 AM
ThePerfectHacker
Quote:

Originally Posted by Cilia
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

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 AM
Soroban
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.

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

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 AM
TriKri