# inductive reasoning

• Aug 30th 2009, 08:17 AM
nh149
inductive reasoning
for the function f1(x)= 2x-1
______
x+1

f
n+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 AM
HallsofIvy
Quote:

Originally Posted by nh149
for the function f1(x)= 2x-1
______
x+1

f
n+1(x)= f1(fn(x)), n>=1
it can be shown that f35 = f5.
Find f28.
how can i get this?(Worried)

Since 35= 5(7) and 28= 4(7), I think you need to calculate $f_7$ directly.

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

Once you get to $f_7$, I think you will find the rest is easy.
• Aug 31st 2009, 08:13 AM
nh149
i'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 AM
Opalg
Quote:

Originally Posted by nh149
for the function $f_1(x)= \frac{2x-1}{x+1}$
$f_{n+1}(x)= f_1(f_n(x))$, n>=1
it can be shown that $f_{35} = f_5$.
Find $f_{28}$.

If $y = \frac{2x-1}{x+1}$ then $x = \frac{1+y}{2-y}$. So the inverse function to $f_1$ is given by $f_{-1}(x) = \frac{1+x}{2-x}$.

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

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