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Basic Intro. Analysis Sequence Proof

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that x0[tex]\in[/tex] (-1,0) and xn=[tex]\sqrt{xn-1+1}[/tex] -1 for n[tex]\in[/tex]N. Prove that xn[tex]\uparrow[/tex]0 as n[tex]\rightarrow[/tex][tex]\infty[/tex].

    the formatting is odd so read outloud, the fucntion for x sub n is "x sub n equals sqrt(x sub (n-1) + 1) -1. If it is still unclear, ask and I can post an image of it typed in maple or something.

    2. Relevant equations

    There are several (many) theorems which could be applied but all of them are well known and prety trivial/obvious.

    3. The attempt at a solution

    Basically I'm having trouble with the "second" part of this. I proved by induction that the sequence is monotonically increasing but I am unable to prove that the limit is 0. I've tried directly proving it using the definition of a limit of a sequence (given [tex]\epsilon[/tex]>0, there is an n.....) but am unable to do so because of the fact that this is a recursive function and not a function of n. I've also tried proving this logically by trying to show that 0=sup{xn} but again the fact that this is a recursive function inhibits this. One more note: this is problem number 1 in my text and usually (always) the problems get harder as you go along which makes me think I am missing something very trivial here. Any ideas?
     
    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 9, 2009 #2

    LCKurtz

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    Given that you can show all xn > 0, they are going to have a hard time converging to 0 if the sequence is increasing, eh?
     
    Last edited: Oct 9, 2009
  4. Oct 9, 2009 #3
    Ahh looks like I mistyped it. x0 should be within (-1,0).
     
  5. Oct 9, 2009 #4

    LCKurtz

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    Well, so far you have the sequence increasing. Is it bounded above?
    If so, what does that imply?

    If you knew xn had a limit, what would happen to your recursion formula if you let n --> oo?
     
  6. Oct 9, 2009 #5
    My claim is that yes it is bounded above namely by 0 and this can imply alot of things; that there exists a supremum of the set {xn}, that by the Monotone Convergence Property a finite limit exists etc. but I don't know how this helps to show not that the limit exists but that it equals zero.

    I'm not quite sure what you mean by your last question...I do know the limit exists and as n goes to infinity, the elements of the sequence approach this limit.

    Thank you for your quick replies.
     
  7. Oct 9, 2009 #6

    LCKurtz

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    If you know xn has a limit of x as n --> oo, what happens if you take the limit of both sides of the equality?

    [tex]x_n = \sqrt{x_{n-1}+1}-1[/tex]
     
  8. Oct 11, 2009 #7
    hmm...once again I'm not exactly sure what you're asking since in the problem statement we are given that xn does have a limit, namely 0. So if you take the limit of both sides of the equality

    [tex]x_n = \sqrt{x_{n-1}+1}-1[/tex]

    You're left with 0 on the left and some other thing on the right. But again, I'm not sure how this helps to prove that n approaches 0 in the first place. Clearly I'm not understanding what you're trying to say. Thank you for the quick replies though, I have been unable to get back to the thread sooner due to massive amounts of homework in other classes.
     
  9. Oct 11, 2009 #8

    LCKurtz

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    "You're left with 0 on the left and some other thing on the right." Really? What other thing?

    Anyway, no, you weren't given that {xn} has a limit. You were asked to prove that xn --> 0. Your previous argument that {xn} is increasing and bounded above by 0 is what establishes that a limit exists: xn --> x for some limit x, but you haven't established what x is, even though the statement of the problem claims it is 0. So, once again:

    If you know xn has an unknown limit of x as n --> oo, what happens if you take the limit of both sides of the equality?.
    [tex]
    x_n = \sqrt{x_{n-1}+1}-1
    [/tex]
     
  10. Oct 12, 2009 #9
    I think that before reading this I was doing what you were talking about all along. Basically I took the limit of both sides of
    [tex]
    x_n = \sqrt{x_{n-1}+1}-1
    [/tex]
    assuming that a limit existed and I called it "a." In doing so, on the left we get "a=..." and on the right we get some other stuff. Now here's where I'm fairly certain I am now doing it correctly but we'll see...

    So i can easily construct a subsequence of xn which converges (by a theorem in the book) to the same number a. This subsequence is xn-1. So by a little algebra (is this valid) I get that a=sqrt(a+1)-1 and then a little algebra later we get that a=0 or a=-1. It's easy to show a cannot equal -1 thus a=0 which is the limit and we're done. I guess my only question is that algebraic/limit steps I took; when taking the limit of the right side, did you do it the way I described above or was I incorrect? Or should I add one to the left side and then square both sides and take the limit of the equality then? I'll be working on this throughout the night and refreshing this page every ten minutes or so but thanks again for all the help so far.
     
  11. Oct 12, 2009 #10

    LCKurtz

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    I think you're on the right track now.
     
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