for the function f1(x)= 2x-1

______

x+1

fn+1(x)= f1(fn(x)), n>=1

it can be shown that f35 = f5.

Find f28.

how can i get this?(Worried)

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- Aug 30th 2009, 08:17 AMnh149inductive reasoning
for the function f1(x)= 2x-1

______

x+1

fn+1(x)= f1(fn(x)), n>=1

it can be shown that f35 = f5.

Find f28.

how can i get this?(Worried)

- Aug 31st 2009, 04:53 AMHallsofIvy
Since 35= 5(7) and 28= 4(7), I think you need to calculate $\displaystyle f_7$ directly.

Do you know how to do that? Since $\displaystyle f_1(x)= \frac{2x-1}{x+1}$, $\displaystyle f_2(x)= f_1(f_1(x))= \frac{2\frac{2x-1}{x+1}- 1}{\frac{2x-1}{x+1}+1}$ and multiplying both numerator and denominator by x+ 1 gives $\displaystyle f_2(x)= \frac{4x-2-x-1}{2x-1+x+1}= \frac{3x-3}{3x}$$\displaystyle = \frac{x-1}{x}$.

Once you get to $\displaystyle f_7$, I think you will find the rest is easy. - Aug 31st 2009, 08:13 AMnh149i've been thinking of..
f28=f4 since f35=f5

because 35=5(7) and 28=4(7)..

is it right? - Aug 31st 2009, 09:30 AMOpalg
If $\displaystyle y = \frac{2x-1}{x+1}$ then $\displaystyle x = \frac{1+y}{2-y}$. So the inverse function to $\displaystyle f_1$ is given by $\displaystyle f_{-1}(x) = \frac{1+x}{2-x}$.

But if $\displaystyle f_{35} = f_5$ then the maps $\displaystyle f_n$ repeat with period 30. Therefore $\displaystyle f_{28} = f_{-2}$, where $\displaystyle f_{-2}(x) = f_{-1}(f_{-1}(x))$. I think that's the least painful way to calculate $\displaystyle f_{28}$.

You can check this by using HallsofIvy's calculation for $\displaystyle f_2(x) = \frac{x-1}x$ to see that $\displaystyle f_4(x) = f_2(f_2(x)) = {\frac{x-1}x - 1 \above1pt \frac{x-1}x}$. This turns out to be the same as $\displaystyle f_{-2}$. So in fact the $\displaystyle f_n$s repeat with period 6.