inductive reasoning

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August 30th 2009, 08:17 AM
nh149

inductive 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)
August 31st 2009, 04:53 AM
HallsofIvy

Quote:

Originally Posted by nh149

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)

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.
August 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?
August 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.