1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with Recurrence Formula

  1. May 12, 2012 #1
    Hi guys. I'm having a bit of trouble with what I thought was a simple math question.

    1. The problem statement, all variables and given/known data
    [itex]x_{n}[/itex] = [itex]\int_0^1 \frac{t^n}{t+7}dt[/itex]

    Show that [itex]x_0[/itex]=ln(8/7) and [itex] x_n = n^{-1} - 7x_{n-1} [/itex]

    2. The attempt at a solution

    Showing x0 = ln(8/7) is a vanilla textbook log question. I'm having trouble with the second part. I am using integration by parts on the form:
    [itex]\int^1_0 t \frac{t^{n-1}}{t+7}dt [/itex]
    and letting u=t and dv=[itex] \frac{t^{n-1}}{t+7}dt [/itex]
    This give:
    [itex]tx_{n-1}|^1_0 - \int_0^1 x_{n-1} dt\\
    = x_{n-1} - \int_0^1 x_{n-1} dt[/itex]

    at which point I'm stuck. I'm not sure if I've used the right IBP substitution or if I'm just almost there and it's just a case of simplifying what I have into a more general case (but can't see it).

    Thanks
     
  2. jcsd
  3. May 12, 2012 #2
    I'm still trying to figure it out using IBP, but try using long division on the integrand instead to arrive at a recursive relationship for the quotient. Then integrate.
     
    Last edited: May 13, 2012
  4. May 12, 2012 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Very astute! I don't think integration by parts leads anywhere. You might change your post to just the hint, "try long division" instead of giving the whole solution. It's more subtle.
     
  5. May 12, 2012 #4
    Thanks Dick. And yes, I suppose you're right. I just edited it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with Recurrence Formula
  1. Recurrence formula (Replies: 1)

  2. Recurrence Help (Replies: 0)

Loading...