Whenever you are trying to come up with a formula or whenever starting out a mathematical induction proof, it is worthwhile to try the first few values of n to see what is happening. So starting with the function definition:
Let's look at the first few values of fn (x):
So this function cycles. The proof is almost done already, so I'll leave the rest to you.
Stephen La Rocque.>
RJ wrote back
I understand everything you showed me in the answer, but how does that show me the proof by mathematical induction. I don’t understand, all it seems like is that you are plugging in one equation for the other. I'm still a little confused. So if you could give me a little more insight I would appreciate it.
What Sue did was to show that
fo(x) = 2/(2-x)
f1(x) = (2-x)/(1-x)
f2(x) = (2x - 2)/x and
f3(x) = x
From here it follows that
f4 (x) = fo o f3 (x) = fo(x) = 2/(2-x) and it follows that
f5(x) = f1(x)
f6(x) = f2(x)
f7(x) = f3(x)
and the pattern repeats. The essence of the induction argument is to be more formal with the clause the pattern repeats.
I am going to change the indexing so that it starts at 1 rather than 0.
Let gn = fn-1for n = 1, 2, ... For each n there are integers k and i with i = 1, 2, 3 or 4 so that n = 4k + i. The pattern I see and want to prove is that for each k
g4k+1(x) = 2/(2-x)
g4k+2(x) = (2-x)/(1-x)
g4k+3(x) = (2x - 2)/x and
g4k+4(x) = x
I am going to prove this using induction on k. Sue proved this for k = 0. This is the first step in the induction argument. I want to show that the pattern repeats. To do this I assume that I have proved this for some particular value of k and I want to verify it is true for the next value of k. So I am assuming the 4 equations above are true for some particular value of k but not necessarily for any larger k.
Using the definition in the problem
g4(k+1)+1(x) = g(4k+4)+1 (x) = f0(g4k+4(x)) = f0(x) = 2/(2-x)
In a similar fashion you can show that
g4(k+1)+2(x) = (2-x)/(1-x)
g4(k+1)+3(x) = (2x - 2)/x and
g4(k+1)+4(x) = x
Hence the pattern does hold for k+1.
Thus, by induction the formula I found for fn is valid for all n = 0, 1, 2, ...