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Real Analysis

  1. Sep 20, 2010 #1
    View attachment 28380 Have really no idea where to start with this one. Anyone wanna help me get started?

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  2. jcsd
  3. Sep 20, 2010 #2


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    well that is about as complicated a problem i have seen, to bash out a simple idea.

    basically it says to try 1/2. if that doesn't work then you try next either 1/4 or 3/4, and more precisely you try 1/4 if 1/2 was too big, and you try 3/4 if 1/2 was too small.

    say 1/2 was too small and you next try 3/4. then eiother it works or it doesn't.

    if 3/4 is too small you try next 7/8, and if 3/4 is too big try next 5/8.

    get it? eventually (after the end of the world that is) you have an infinite number of tries x all getting nearer each other and also ll values x^n getting nearer to y.
    so you work to prove that the x's converge to something whose mth power must converge to y.

    for this you need an axiom that tells yiou when a sequence of =reals converges.

    or as this problem puts it, you need an axiom that says a shrinking sequence of nested, bounded, closed intervals contains at least one common point, and exactly one point if they shrink to zero in length.

    or maybe it is rigged to use the fact that a bounded monotone sequence of reals converges.

    to do the problem as asked just put your nose down and try to follow the steps one at a time blindly.

    i myself dislike such problems. i mean what do you learn from them? maybe a little technical skill.
  4. Sep 20, 2010 #3
    THANK YOU. You have made that much clearer. Tomorrow I will work on it and post my work.
  5. Sep 21, 2010 #4
    OK, I am confused, the first 3 parts seem trivial and i can't imagine doing anything but going through the process a few times and showing that they are all true by induction...
    Part (d) however, i do not see how this is true. How am I to know how far away an and bn are from each other? I would have thought they could be anything. I specifically dont see how/why |a1-b1|=1/2
  6. Sep 21, 2010 #5


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    well without looking back i recall the next numbers ere the average of the previous ones so they seem to get closer by one half each time.
  7. Sep 23, 2010 #6

    Would this theorem be the Bolzano-Weierstrass Theorem??(Every bounded infinite subset of R^k has a limit point in R^k). Specifically for part (e).
  8. Sep 23, 2010 #7


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    You start with b0=1 and a0=0. Then we check to see if [tex]\left( \frac{1}{2}^n\right)=y[/tex]. If so, we're done. If not, there are two possibilities: we make a1=1/2 or we make b1=1/2, and leave the other number alone depending on whether [tex] \left(\frac{1}{2}\right)^n[/tex] is too big or too small.

    Then |a1-b1| is either |0-1/2| or |1/2-1|
  9. Sep 23, 2010 #8
    I would like to post my work on this and see if anybody has a problem. I am very unsure on how to do this...
    (a)initially, an[tex]\leq[/tex]bn and an[tex]\leq[/tex]c[tex]\leq[/tex]bn. If cm>y we set an=an-1 and bn=c so that we still have an-1[tex]\leq[/tex]an[tex]\leq[/tex]bn[tex]\leq[/tex]bn-1. If cm<y then we we set an=c and bn=bn-1 so that we still have an-1[tex]\leq[/tex]an[tex]\leq[/tex]bn[tex]\leq[/tex]bn-1 and the sequence satisfies part (a).

    (b)Since all bn and an are reals, they are ordered. Since a0[tex]\leq[/tex]a1[tex]\leq[/tex]...[tex]\leq[/tex]an[tex]\leq[/tex]bn[tex]\leq[/tex]...[tex]\leq[/tex]b1[tex]\leq[/tex]b0, then [an,bn][tex]\subseteq[/tex][an-k,bn-k] for any positive integer k[tex]\leq[/tex]n.

    (c) Note that an<c and bn>c. Since in our process, if cm[tex]\neq[/tex]y, we shift our nested intervals(infinitely many times if necessary) so that y1/m is contained in all of them, then the bounds of these sequences, namely an and bn for all n satisfy an<y1/m<bn[tex]\Rightarrow[/tex] amn<y<bmn for all n.

    (d)Our process of setting either bn=c or an=c depending on whether ym>c or ym<c makes our interval smaller by 1/2 each time, where initially, the interval is of length 1 ( |0-1| ). Hence |bn-an|=2-n

    (e) I would like to just use the Bolzano Weierstrass Theorem here.... This is as far as I have done..
  10. Sep 23, 2010 #9


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    There are two parts to point e: First is that the intersection is non-empty, and the second is that there is only one point. You can use a convergence argument to get that it's not empty: the sequence an is bounded above and increasing, so converges to something we'll call a; the sequence bn is bounded below and decreasing, so has a limit also called b. By preservation of weak inequalities, [tex] a\leq b[/tex] (make sure this is true! otherwise you're stuck in the mud)

    Prove that [a,b] is in the infinite intersection, and use that to prove that a=b also
  11. Sep 23, 2010 #10
    I think the buzzword for this kind is the bisection method.
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