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Recursive Cauchy Sequence

  1. Feb 1, 2008 #1

    Vid

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    1. The problem statement, all variables and given/known data
    Prove that the sequence [tex]x_{n+1} = \frac{(x_{n})^{2} + 2}{2x_{n}} [/tex] where [tex]x_{1} = 1[/tex] is Cauchy.


    2. Relevant equations The sequence comes from using Newton's Method on x^2 + 2, but we aren't supposed to use sqrt(2) anywhere in the solution. In fact, the next part of the problem is to prove that is a = lim xn as n goes to infinity, then prove a is an upper bound for the set of all rational numbers who square is less than two, and then prove its the least upper bound for that set. We only know a few basic theorems about Cauchy sequences so I'm thinking I'm supposed to use the standard definition of a Cauchy seq. to prove this. ie {Xn} is Cauchy iff |xn - xm| < e for all n,m greater than or equal to N.


    3. The attempt at a solution
    I've spent several hours trying various manipulations trying to get some insight, but I haven't had any luck, yet. I just want a push in the right direction to start and I could hand it from there.
     
    Last edited: Feb 1, 2008
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  3. Feb 1, 2008 #2

    Dick

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    Here's one approach. Let f(x)=(x^2+2)/(2x). So x_n+1=f(x_n). Now show x_n is increasing. Do this by showing f(x)>x if x>=1 and x^2<2. Now show x_n is bounded by showing (f(x))^2<2 if x>=1 and x^2<2. Do you know that a bounded increasing sequence has a limit and is therefore Cauchy?
     
    Last edited: Feb 1, 2008
  4. Feb 1, 2008 #3

    Vid

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    Actually, it seems like we're going over that theorem today when we discuss lim inf and lim sup so I think that should work.

    The book we're using is a bounded copy make specially for this class that's an early version of this book: http://www.amazon.com/Advanced-Calculus-Introduction-Linear-Analysis/dp/0470232889

    The author is a Professor here at LSU which is why we use it. It discusses Archimedian Ordered fields first then goes directly into convergent and Cauchy sequences to develop the real numbers using the completeness axiom and then uses that to develop inf and sup. It then goes onto algebraic combinations of sequences, Bolzano-Weierstrass, nested interval theorem, basic topology and the Heine-Borel covering theorem, then finally countability of the rationals before moving onto continuous functions in chapter two which included a section on banach spaces and the sup norm. The next chapter then goes on to the reimann integral.

    Anyway that point I was trying to make is that sequences are covered really early and used throughout the rest of this class so I don't have too many theorems to work with in this proof.
     
  5. Feb 1, 2008 #4

    HallsofIvy

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    It seems to me you have a serious difficulty here! Yes, I can see that "The sequence comes from using Newton's Method on x2 + 2" and that is precisely the difficulty. If the sequence is Cauchy, then it will converge and, of course, if it converges it must converge to a real number that satisfies x2+ 2= 0. Which is impossible! Are you sure you were supposed to use Newton's method on x2- 2? That would give you a sequence defined by [tex]q_{n+1}= \frac{x_n^2- 2}{2x_n}[/tex]?

     
  6. Feb 1, 2008 #5

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    The sequence still converges to sqrt(2). I checked in mathematica. My professor said the obvious way to see this is x = (x^2+2)/(2x). 2x^2 = x^2+2. x^2-2 = 0. Though this was at the very end of class and I didn't write down his full reasoning (kind of regretting it now) . He definitely did say it came from newton's method though, but I'm not sure how.

    Oh well, I've already written and turned in the proof using the method Dick suggested (the class was at 2:30). Though it seems flawed since it turns out its not an increasing sequence. x1 = 1
    x2 = 1.5, x3 = 1.41667. Ah, well.

    Here's the image from mathematica.

    http://img136.imageshack.us/img136/3761/sequenceny6.jpg
     
    Last edited: Feb 1, 2008
  7. Feb 1, 2008 #6

    NateTG

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    This is a bit ham-handed, but...
    [tex]x_{n+1}^2=\frac{x_n^4+4x_n^2+4}{4x_n^2}=\frac{x_n^2}{4}+1+\frac{1}{x_n^2}[/tex]
    so
    [tex]y_n=x_n^2[/tex]
    gives
    [tex]y_{n+1}=\frac{y_n}{4}+\frac{1}{y_n}+1[/tex]
    So if
    [tex]y_n=2-\epsilon[/tex]
    [tex]y_{n+1}=2-\frac{\epsilon}{4} + \frac{\epsilon}{4- 2\epsilon}=2-\epsilon\left(\frac{1}{4}+\frac{1}{4-2\epsilon}\right)[/tex]
    But its clear that the expression in the parens is less than or equal to [itex]\frac{3}{4}[/itex] as long as [itex]\epsilon \leq 1[/itex] so there is clearly convergence to at least one square root of two. Since [itex]x_n>0[/itex] (easy to check) we have that the sequence is convergent.

    P.S. Any proof of convergence to the positive square root of 2 must use the initial value since the sequence clearly won't cross zero.
     
  8. Feb 1, 2008 #7

    Dick

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    Ooops. I've gotta apologize for that one. If you look at (x^2+2)/(2x)-x that's (sqrt(2)-x)(sqrt(2)+x)/2x. So f(x) is decreasing for x>sqrt(2). It does increase at x=1 but then f(1)=3/2 throws you into the decreasing region. THEN it decreases towards sqrt(2). I wasn't being very careful, was I? Sorry again.
     
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