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Sequence Problem, Rudin's PMA Ch.3 #17

  1. Jun 23, 2011 #1
    So for those of you who don't have the book the problem goes like this:

    Fix a > 1, take x_1 > sqrt(a), and define:

    [tex]x_{n+1}=\frac{a+x_n}{1+x_n}=x_n+\frac{a-x_n^{2}}{1+x_n}[/tex]

    a) Prove that x_1 > x_3 > x_5 > ...
    b) Prove that x_2 < x_4 < x_6 < ...

    Basically my strategy is to show that the sequence r_n=x_n+2 - x_n alternates sign, with it giving negative terms for n odd and positive terms for n even. My proof so far is as follows:

    [tex]x_{n+1} - x_n = \frac{a - x_n^{2}}{1+x_n}[/tex] and [tex]x_{n+2} - x_{n+1} = \frac{a - x_{n+1}^{2}}{1+x_{n+1}}[/tex]

    adding these two equations and setting r_n = x_n+2 - x_n produces:

    [tex]r_n = x_{n+2} - x_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - x_{n+1}^{2}}{1+x_{n+1}}[/tex]

    now substituting in [tex]\frac{a+x_n}{1+x_n}[/tex] for [tex]x_{n+1}[/tex] gives:

    [tex]r_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - (\frac{a+x_n}{1+x_n})^2}{1+(\frac{a+x_n}{1+x_n})}[/tex]

    from here I simplify the right side until obtaining:

    [tex]r_n = \frac{2x_n(a-x_n^{2})+2a}{(1+x_n)(2x_n+a+1)}[/tex]

    I have checked my work multiple times on this simplification and I cannot find any error. Yet if we set n=1 than clearly the denominator is positive, but if one chooses x_1 to be just a tiny bit greater than sqrt(a) than the numerator can be made positive as well, which contradicts the desired conclusion for part a).

    I'm fairly certain my strategy is viable, hopefully someone can do the simplification and see if they are getting the same result. Its certainly possible that I'm missing something.
     
    Last edited: Jun 24, 2011
  2. jcsd
  3. Jun 23, 2011 #2

    tiny-tim

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    Hi Poopsilon! :smile:

    I'm confused :redface: … first you put rn = xn+2 - xn,

    then you put rn = xn+1 - xn :confused:

    Use xn+2 - xn = (xn+2 - xn+1) plus (xn+1 - xn). :wink:
     
  4. Jun 23, 2011 #3
    Sorry, you answered really quick! I just finished reediting my post to clean up all the errors, it should be good now.
     
  5. Jun 25, 2011 #4
    I'm going to bump this once, hopefully someone can give me some feedback on it. If further clarification on the problem is required, don't hesitate to ask, thanks.
     
  6. Jul 1, 2011 #5
    Hey,
    I hope this isn't too late for your (my?) assignment. First of all, I would suggest using the form (a+xn)/(xn+1); in my opinion it's easier to work through and see the form. Second, the easiest way I know of to solve this question requires a little lemma. Write Xn+1 in terms of Xn, and if you begin with x1 you know the relationship between X1 and a so you can find the relationship between x2 and sqrt(a). [ie (a+x1)/(1+x1)=x2. What does x1= from this expression? you know when you get it to x1 you will have an expression >sqrt(a), and then solve again for x2.)

    If you can see the relationship between x2 and sqrt(a), x3 and sqrt(a), x4 and sqrt(a)... and so on, and prove the general case (for even and for odd) then you will be able to approximate the form of xn+1 (if you write it in terms of Xn-1).

    Hope that helps!
     
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