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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$f(5)=10$

$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$

find f(2009)

$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$

find f(2009)

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$f(5)=10$

$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$

find f(2009)

$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$

find f(2009)

$f(5)=10$

$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$

find f(2009)

because of periodicity: f(5)=10, f(8)=9/11, f(11)=-1/10, f(14)=-11/9; f(17)=10, ...

f(2009)=10

f(2009)=10

.

- Feb 13, 2012

- 1,704

An equivalent and more simple statement of the problem is: given the difference equation...$f(5)=10$

$f(n+3)=\dfrac {f(n)-1}{f(n)+1},\,\, for \,\, all \,\,n\in N$

find f(2009)

$$ a_{n+1}= \frac{a_{n}-1}{a_{n}+1}\ ,\ a_{0}=10\ (1)$$

... find $a_{668}$...

Before trying a 'direct attack' to the non linear d.e. (1) it is better to search that (1) has some periodic solution, i.e. a solution where $a_{n+k}=a_{n}$. Let's set $a_{n+k}=y$ and $a_{n}=x$. We start with k=1 we obtain ...

$$y= \frac{x-1}{x+1}\ (2)$$

... and imposing y=x we arrive to the equation...

$$ x^{2}+1=0\ (3)$$

... which has no real solutions. Setting k=2 we arrive to the [surprisingly simple...] equation...

$$y = - \frac{1}{x}\ (4)$$

... that pratically solves the problem. The (4) indeed indicates that the solution has periodicity 4 and, given $a_{0}$, we have...

$$a_{0}\ ,\ \frac{a_{0}-1}{a_{0}+1}\ ,\ - \frac{1}{a_{0}}\ ,\ -\frac {a_{0}+1}{a_{0}-1}\ , \ a_{0}\ ,\ ...\ (5)$$

In our case is $a_{0}=10$ so that the other terms are $a_{1}= \frac{9}{11}$, $a_{2}= - \frac{1}{10}$, and $a_{3}= - \frac{11}{9}$ . Now 668 is divisible by 4 so that is $a_{668}=a_{0}=10$...

Kind regards

$\chi$ $\sigma$