# You are subscribed to this thread Proving conjecture for recursive function

1. Apr 2, 2008

### superdog

Hi,

1. The problem statement, all variables and given/known data

Just having some troubles with a proof i have been asked to do, (sorry for not knowing the math code)

basically, f(1)=0, f(2)=1/3 and f(n)= ((n-1)/(n+1))*f(n-2)

and i've come up with the conjecture that f(n) = 0 when n is odd, and = 1/(n+1) when n is even.

and i have to prove my conjecture, this is where i'm stuck,

2. Relevant equations

anyone care to point me in the right direction?

3. The attempt at a solution

not really sure what method to use, induction?

2. Apr 2, 2008

### HallsofIvy

Staff Emeritus
Since you have two statements:
f(2n+1)= 0 and f(2n)= 1/(2n+1), why not try two separate induction proofs?

3. Apr 2, 2008

### superdog

Hey,

thanks for your help,

"f(2n)= 1/(2n+1)" you mean 1/(n+1) right, (not being picky just making sure)

yeah, i kinda what your saying, ie: 2n is a generator for all evens and 2n+1 is for odds, however, i'm not really confident on how to actually go through and do a induction proof on either, the recursive function is knda scaring me at the moment

4. Apr 2, 2008

### D H

Staff Emeritus
Halls meant exactly what he said: $f(2n)=1/(2n+1)$. You conjecture is that when n is even, $f(n)=1/(n+1)$. Another way of saying n is even is saying that $n=2m$, where $m$ is an integer. Apply this to your conjecture: $f(n)=1/(n+1)\;\rightarrow\; f(2m) = 1/(2m+1)$.

To prove some conjecture by induction, you need to show two things:
• That the conjecture is true for some base case and
• That if the conjecture is true for some m then it is also true for m+1.

The conjecture is obviously true for $m=1$ as $f(2\cdot1) = 1/(2\cdot1+1) = 1/3$. All that remains is proving the recursive relationship.

5. Apr 2, 2008

### superdog

Ahhhh, ok, thanks so much for the help! its all clicking into place now. especially after walking away from the bloody thing for a bit :P

thanks again
Sam

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