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Analysis - sequence proof

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm looking for a kick in the right direction of how to approach it from 1b onwards ('cos thankfully I can figure out 1a for myself). Please see the attached screenshot (I'm not good at making the formulas appear right).

    2. Relevant equations

    I'm really not sure. I've noticed looking back through the readings that the bit I've circled in b is one of the properties of the bisection process, e.g "the length of the interval is halved at each step". I'm wondering if this question is getting me to demonstrate that effect. Other thank that, this chapter covers intro to Bolzano-Weierstrass theorem, Cauchy sequences, and using these with the triangle inequality and a little bit of the telescoping property.

    3. The attempt at a solution

    Umm? I tried:
    Let sk+2 - 1/2(sk+1 + sk) for all k[tex]\geq[/tex]1 and suppose |sk+2 - sk+1|
    then |sk+2 - sk+1| = |1/2 (sk+1 - sk) - 1/2 (sk - sk-1)|
    but given k[tex]\geq[/tex]1, isn't sk-1 invalid? If k=1 then sk-1=s0 which isn't part of the sequence.
    So I stopped.
    then I thought how about then |sk+2 - sk+1| = |1/2 (sk+1 - sk) - sk+1|
    = |1/2(-sk+1 - sk)|
    but then the sk+1 is negative, and I couldn't think of a way of making just that bit positive.
    So I wondered if I was going about this completely the wrong way, and maybe I should be using the triangle inequality, except the answer isn't an inequality.
    Hence I'm stumped as to how to start in on this thing, so if someone could give me a push, that would be really helpful.

    Attached Files:

  2. jcsd
  3. Jul 21, 2010 #2


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    In order for that to make sense you would have to have [itex]k\ge 2[/itex] but that's okay- you can prove [itex]|s_{n+2}- s_{n+1}|= (1/2)|s_{n+1}- s_n|[/itex] for n= 1, that is, that [itex]|s_3- s_2|= (1/2)|s_2- s_1|[/itex], directly.

  4. Jul 21, 2010 #3


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    Homework Helper

    Take the modulus and you have what they're asking you.
  5. Jul 21, 2010 #4


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    This is a nice question by the way, we never had such nice questions when I was studying analysis.
  6. Jul 21, 2010 #5
    Oh yeah. Ack! I've been doing my head in over a typo! :blushing: Thanks.
  7. Jul 21, 2010 #6
    Directly as in the same way as below (i.e. without my + becoming a -), or by calculating the s3 and s2 explicitly, or some other way?
    Last edited: Jul 21, 2010
  8. Jul 21, 2010 #7


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    As HallsOfIvy says, you can get the first few terms by calculating explicitly (I would recommend this). This is the first part of the induction process, you have to say what values of k for equation is valid for and then complete the proof.

    It's not that hard if you've done induction before.
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