Function Composition. (1 Viewer)

RealiseNothing · Oct 3, 2012

For $f_n(2012)$ it cycles between:

$2012$

$\frac{1}{-2011}$

$\frac{2011}{2012}$

Sy123 · Oct 3, 2012

$f_n(x)=f_0(f_{n-1}(x))=f_0(f_0...(f_0(x)))$

Where there are an n number of f_0

$f_0(f_0(x))=\frac{-1}{x}+1$
Through algebra

$f_0(f_0(f_0(x)))=x$

Therefore for every multiple of 3 of f_0, it will equal x,
So 6 number of f_0 is x
2010 number of f_0 is x

Therefore:

$f_0(f_0...f_0(x))=\frac{1}{1-x} $ ... for 2011 number of $ f_0(x)$

So then finding f_2012 (x)

$f_0(f_0...(f_0(2012)))=\frac{1}{-2012}+1 $ for 2012 number of $ f_(0) \\ \\ =\frac{2011}{2012}$

Function Composition. (1 Viewer)

Fus Ro Dah

Member

RealiseNothing

what is that?It is Cowpea

Sy123

This too shall pass

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