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Can anybody explain this to me? (Analysis)

  1. Nov 22, 2012 #1
    For question 32.2 in this link:

    http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw11sum06.pdf [Broken]

    I did not understand how [itex]b^2/2 \leq U(f)[/itex]. We know that we have strict inequality in [itex]t_{k+1} > \frac{t_k + t_{k+1}}{2} [/itex]...so don't we need to have [itex]b^2/2 < U(f)[/itex] instead of [itex]b^2/2 \leq U(f)[/itex]?

    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 22, 2012 #2

    Ray Vickson

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    If a strict inequality folds for any finite partitions, all you can conclude is that the non-strict inequality holds in the limit, or when taking the sup, etc. In other words, we cannot have U(f) < b^2/2, because in order to have that we would have to have U_P(f) < b^2/2 for some finite partition P.

    To put it another way: we have 1/n > 0 for all positive integers n, but inf{1/n: n is a positive integer} = 0.

    RGV
     
    Last edited by a moderator: May 6, 2017
  4. Nov 22, 2012 #3
    Thanks, but...

    "we cannot have U(f) < b^2/2, because in order to have that we would have to have U_P(f) < b^2/2 for some finite partition P."...but if you look at the last part of the solution, it actually shows that U(f,P) < b^2/2 for all partitions P...
     
  5. Nov 22, 2012 #4

    Ray Vickson

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    OK, so I got the inequalities reversed. The argument is still the same.

    RGV
     
  6. Nov 22, 2012 #5
    No I wasn't saying that you got the inequalities reversed...

    I was just saying that (at the very end of the solution), it is shown that U(f,P)> b^2/2 for all P.
     
  7. Nov 22, 2012 #6

    haruspex

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    No, you had it right the first time.
    Artusartos, Ray is saying the argument runs like this.
    We know that for any finite partition U_P(f) > b2/2. Suppose in the limit U(f) < b2/2. That would imply U_P(f) < b2/2 for some finite partition P, contradicting what we know. Therefore U(f) ≥ b2/2.
    The OP says, in part:
    don't we need to have b2/2<U(f) ?​

    Therein lies a misunderstanding. The relationship between b2/2 and U(f) is not a condition that's needed, it's a relationship we are trying to deduce. The point is that we cannot deduce U(f) > b2/2. This is because it is the limit of a sequence, and a sequence in which every term is > x can be equal to x in the limit. Instead, we can deduce the weaker relationship U(f) ≥ b2/2.
     
  8. Nov 23, 2012 #7
    Thanks a lot, I think I understand it more now...but...

    The other poster says "In other words, we cannot have U(f) < b^2/2, because in order to have that we would have to have U_P(f) < b^2/2 for some finite partition P." So (according to this statement) if we know that U_P(f) < b^2/2 for "some finite" partition P, then we can say that U(f) < b^2/2. In the text that I attached, it says that U_P(f) < b^2/2 for "all" finite partitions P...so why can't we say the same thing? Since, if we have U(f)<b^2/2 for "all" partitions, then we definitely have it for "some" partitions...
     
  9. Nov 23, 2012 #8

    haruspex

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    No, that doesn't follow. You're turning A [itex]\Rightarrow[/itex] B into B [itex]\Rightarrow[/itex] A:
    If {U(f) < b^2/2} then {[itex]\exists[/itex] P s.t. U_P(f) < b^2/2}​
    is not the same as
    If {[itex]\exists[/itex] P s.t. U_P(f) < b^2/2} then {U(f) < b^2/2}​
     
  10. Nov 23, 2012 #9
    Oh, ok...I get it. Thanks :)
     
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