Thread: Function Question

Thanks in advance!

How much of this have you done?
Have you done any of it?
If you have, what pattern do you see?

Originally Posted by Plato

How much of this have you done?
Have you done any of it?
If you have, what pattern do you see?

For part a, for f2(x) i get f(f(x)), so what would f(f(n)) be? Without knowing n, how would I calculate this? maybe i'm just confused about what to provide for an answer.

Originally Posted by shawn

For part a, for f2(x) i get f(f(x)), so what would f(f(n)) be? Without knowing n, how would I calculate this? maybe i'm just confused about what to provide for an answer.

Well,


So to calculate ,
If n is even this is
since n + 3 is odd.

If n is odd then this is
since n - 1 is even.

So for all n.

You do the rest.

-Dan

Originally Posted by topsquark

Well,


So to calculate ,
If n is even this is
since n + 3 is odd.

If n is odd then this is
since n - 1 is even.

So for all n.

You do the rest.

-Dan

Ah okay got it..thanks.

So, when n is odd
and when n is even

and when n is odd
and when n is even

and when n is odd
and when n is even

I don't see a pattern for a simple formula?

I'm doing part c now,
Basis is easy when k = 3.

Inductive step

Let x >=3 be an integer.
Our IH is (from the formula's I found for part b)
f^x (n) = n + x when x is even, and
f^x (n) = n + x + 2(-1)^n when x is odd

We want to prove that
f^x+1(n)=n+x+1 when x+1 is even, and
f^x+1(n)=n+x+1+2(-1)^n when x+1 is odd

Is the above correct?

So we have two cases from here:

Case 1: when x+1 is even, x is odd
Now, f^x+1(n) = f(f^x(n)) = f(n +x + 2(-1)^n)

Now, I think we have two more sub cases but I'm not sure?

Case 2: when x+1 is odd, x is even
Now, f^x+1(n) = f(f^x(n)) = f(n + x)
Now, I think we have two more sub cases but I'm not sure?


Can someone help me finish the sub cases?

Thanks!