You are subscribed to this thread Proving conjecture for recursive function

[superdog]

Hi,

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,

Homework Equations

anyone care to point me in the right direction?

The Attempt at a Solution

not really sure what method to use, induction?

 

[HallsofIvy]

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

 

[superdog]

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]

"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.

 

[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