# Math Help - Contienued fractions

1. ## 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

2. 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.

3. 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!

4. 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

5. 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
$t_n=[1:1,1,1,....,1]$
$t_{n+1}=[1;t_n]$
Thus,
$t_{n+1}=1+\frac{1}{t_n}$

6. 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

7. 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 $x=[1;1,1,...,]$
Converges to some number.
But,
$x=[1;x]$
For it repeats.
Thus,
$x=1+\frac{1}{x}$
Thus,
$x^2=x+1$
Thus,
$x^2-x-1=0$
Thus,
$x=\frac{1\pm \sqrt{5}}{2}$
This tells us that this is one of these two possible values.
It cannot be negative thus,
$x=\frac{1+\sqrt{5}}{2}=\psi$
The Divine Proportion (My body is shaped in Divine Proportion).

8. 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.

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

By inspection ("eyeballing" it),
. . we see that each term is: one plue one over the preceding term.

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

Assuming that the fraction goes on forever: . $1 + \frac{1}{1 + \frac{1}{1+\frac{1}{1 + ...}}}$
. . can we determine its value?

$\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: . $x \:=\:1 + \frac{1}{x}\quad\Rightarrow\quad x^2 - x - 1 \:=\:0$

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

Since $x$ is positive, we have: . $x \:=\:\frac{1 + \sqrt{5}}{2} \:= \:1.618033989...$
. . which happens to be the Golden Mean, $\phi$.

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