You are subscribed to this thread Proving conjecture for recursive function

superdog · Apr 2, 2008

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?

HallsofIvy · Apr 2, 2008

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

superdog · Apr 2, 2008

HallsofIvy said: ↑

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

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

D H · Apr 2, 2008

superdog said: ↑

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

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

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 [itex]m=1[/itex] as [itex]f(2\cdot1) = 1/(2\cdot1+1) = 1/3[/itex]. All that remains is proving the recursive relationship.

superdog · Apr 2, 2008

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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superdog

HallsofIvy

superdog

D H

superdog

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