# Function Question

• Apr 7th 2008, 04:06 PM
shawn
Function Question
• Apr 7th 2008, 04:36 PM
Plato
How much of this have you done?
Have you done any of it?
If you have, what pattern do you see?
• Apr 7th 2008, 04:49 PM
shawn
Quote:

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.
• Apr 7th 2008, 05:24 PM
topsquark
Quote:

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,
$f(n) = \left \{ \begin{array}{rr}n + 3,\text{ if n is even} \\ n - 1,\text{ if n is odd} \end{array} \right .$

So to calculate $f(f(n))$,
If n is even this is $f(f(n)) = f(n + 3) = (n + 3) - 1 = n + 2$
since n + 3 is odd.

If n is odd then this is $f(f(n)) = f(n - 1) = (n - 1) + 3 = n + 2$
since n - 1 is even.

So $f^2(n) = n + 2$ for all n.

You do the rest.

-Dan
• Apr 7th 2008, 05:35 PM
shawn
Quote:

Originally Posted by topsquark
Well,
$f(n) = \left \{ \begin{array}{rr}n + 3,\text{ if n is even} \\ n - 1,\text{ if n is odd} \end{array} \right .$

So to calculate $f(f(n))$,
If n is even this is $f(f(n)) = f(n + 3) = (n + 3) - 1 = n + 2$
since n + 3 is odd.

If n is odd then this is $f(f(n)) = f(n - 1) = (n - 1) + 3 = n + 2$
since n - 1 is even.

So $f^2(n) = n + 2$ for all n.

You do the rest.

-Dan

Ah okay got it..thanks.

So, $f^2(n) = n + 2$ when n is odd
and $f^2(n) = n + 2$ when n is even

and $f^3(n) = n + 5$ when n is odd
and $f^3(n) = n + 1$ when n is even

and $f^4(n) = n + 6$ when n is odd
and $f^4(n) = n + 4$ when n is even

I don't see a pattern for a simple formula?
• Apr 8th 2008, 09:23 PM
shawn
Inductive Proof
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!