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

Hi,

1. Homework Statement

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. Homework Equations

anyone care to point me in the right direction?

3. The Attempt at a Solution

not really sure what method to use, induction?

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

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

"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

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

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