You are subscribed to this thread Proving conjecture for recursive function

  1. 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. jcsd
  3. HallsofIvy

    HallsofIvy 41,264
    Staff Emeritus
    Science Advisor

    Since you have two statements:
    f(2n+1)= 0 and f(2n)= 1/(2n+1), why not try two separate induction proofs?
  4. 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
  5. D H

    Staff: Mentor

    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.
  6. 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
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