Results 1 to 6 of 6

Math Help - Mean value theorem & Cauchy sequence

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Mean value theorem & Cauchy sequence





    [note: also under discussion in s.o.s. math board]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post




    [note: also under discussion in s.o.s. math board]
    Show that the proper hypotheses are made. Then we have that on the interval [a_{n},a_{n+1}] that there exists some c such that \frac{\cos(a_n)-\cos(a_{n+1})}{a_n-a_{n+1}}=\frac{a_{n+1}-a_{n+2}}{a_n-a_{n+1}}=-\sin(c). Thus, we have that \left|a_{n+1}-a_{n+2}\right|=|\sin(c)||a_n-a_{n+1}|\leqslant|a_n-a_{n+1}|.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by Drexel28 View Post
    Show that the proper hypotheses are made. Then we have that on the interval [a_{n},a_{n+1}] that there exists some c such that \frac{\cos(a_n)-\cos(a_{n+1})}{a_n-a_{n+1}}=\frac{a_{n+1}-a_{n+2}}{a_n-a_{n+1}}=-\sin(c). Thus, we have that \left|a_{n+1}-a_{n+2}\right|=|\sin(c)||a_n-a_{n+1}|\leqslant|a_n-a_{n+1}|.
    But when you write the interval [a(n),a(n+1)] and apply the mean value theorem, it is implictly assumed that a(n)<a(n+1), which is not true for our sequence, e.g. a0=0, a1=1, a2=0.54, a3=0.86,...

    How can we remove the assumption a(n)<a(n+1) while reaching the same conclusion that |a(n+2)-a(n+1)| ≤ 0.85|a(n+1)-a(n)|?

    thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    But when you write the interval [a(n),a(n+1)] and apply the mean value theorem, it is implictly assumed that a(n)<a(n+1), which is not true for our sequence, e.g. a0=0, a1=1, a2=0.54, a3=0.86,...

    How can we remove the assumption a(n)<a(n+1) while reaching the same conclusion that |a(n+2)-a(n+1)| ≤ 0.85|a(n+1)-a(n)|?

    thanks.
    I do leave holes on purpose you know? Do you not think I didn't realize this. You are supposed to do some of this. This is to help you learn.

    But, I'll give you a hint. Does it matter which is bigger than the other? Clearly it does if we want to define an interval in the conventional way, but does it matter if we have the itneval I gave you or the reverse one? Is the result any different?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    OK, I got it now.
    Actually I got confused before about the "for ALL n" part in "a(n+1)>a(n) for ALL n" . I thought we need montone sequences, but acutally we don't.

    Given any two consecutive terms a(n) and a(n+1), either a(n)<=a(n_1) or a(n)>a(n+1), and in either case the MVT gives the result. Done.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Subsequence of a Cauchy Sequence is Cauchy
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: September 30th 2010, 02:29 AM
  2. cauchy sequence, contractive sequence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 25th 2010, 07:25 AM
  3. Cauchy Sequence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 10th 2010, 09:18 PM
  4. cauchy sequence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 3rd 2009, 07:06 PM
  5. Cauchy Sequence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 14th 2008, 01:33 AM

Search Tags


/mathhelpforum @mathhelpforum