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!

Homework Help: Cauchy sequence question

  1. Oct 16, 2008 #1
    Let a_1 and a_2 be arbitrary real number that are not equal. For [tex]n \geq 3[/tex], define a_n inductively by,

    [tex]a_n = \frac{1}{2} (a_{n-1}+a_{n+2} )[/tex]

    I cannot get the result that the book gets. I proceed,

    [tex]a_{n+1} - a_{n} = \frac{1}{2}(a_n + a_{n-1} ) - \frac{1}{2} (a_{n-1} + a_{n-2} ) = \frac{1}{2} ( a_n - a_{n-2}
    )= \frac{1}{2}(a_n + a_{n-1} ) [/tex]

    The book got the answer,

    [tex] a_{n+1} - a_n = \frac{-1}{2} (a_n - a_{n-1} ) [/tex]

    Any help for me?
  2. jcsd
  3. Oct 16, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    You've got some typos in there that are making this really confusing. The original recursion is a_n=(1/2)*(a_n-1+a_n-2), right? That says 2*a_n-a_n-1=a_n-2. I agree with the book.
  4. Oct 16, 2008 #3
    I can't see it. I've done some algebra and I still can't get anywhere from your last step.
  5. Oct 16, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    You've got a_n+1-a_n=(1/2)(a_n-a_n-2), right? Put the expression for a_n-2 into that. If you still aren't getting it show your work.
  6. Oct 16, 2008 #5
    I got it. Thanks Dick
  7. Oct 17, 2008 #6


    User Avatar
    Science Advisor

    Here's a "cheat":
    Since [itex]\sum x^n/n![/itex] is the Taylor's series for ex, 1+ 1/2!+ 1/3!+ ... converges to e. Since it converges, it is a Cauchy sequence!

    Or you could just note that the numeric series converges by the ratio test. Again, since it converges, it is a Cauchy sequence.

    But I doubt that either of those answers is what was wanted!
  8. Oct 17, 2008 #7
    Those are nice things to put the problem into perspective though. Sometimes I (and hopefully others) get lost in the algebra of the problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook