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Properties of Summations and Integrals question

  1. Aug 16, 2011 #1
    Let's say we have the statement [itex]\sum^{\infty}_{0}f(x)=\frac{\sum^{\infty}_{0}g(x)}{\sum^{\infty}_{0}h(x)}[/itex] does this imply that
    [itex]\int^{\infty}_{0}f(x)=\frac{\int^{\infty}_{0}g(x)}{\int^{\infty}_{0}h(x)}[/itex]?

    Also if [itex]\sum^{\infty}_{0}f(x)=\sum^{\infty}_{0}g(x)[/itex] does this imply that [itex]f(x)=g(x)[/itex], or just that f(x)~g(x) (asymptotically equivalent)?

    Thanks.
     
    Last edited: Aug 16, 2011
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  3. Aug 16, 2011 #2

    hunt_mat

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    Your first statement makes no sense, what are you summing over?
     
  4. Aug 16, 2011 #3
    Sorry, thought it was pretty obvious, but I guess not. I didn't think it was necessary to add dx to the integrals or add that summation was over x as it was implied, and I didn't want to typeset anymore than I had to because it is new to me.
     
  5. Aug 16, 2011 #4

    micromass

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    Are you summing over all the integers x, or are you summing over all the reals x? If you assume the latter, it doesn't make much sense. If you're meaning the former, then consider a function like

    [tex]g(x)=\left\{\begin{array}{c}1/x^2~\text{if x is an integer}\\ 0~\text{otherwise}\end{array}\right.[/tex]

    that should be the basis of a counterexample for both statements.
     
  6. Aug 17, 2011 #5
    Summing over integers. Also, are these statements true for continuous functions that aren't piecewise?
     
  7. Aug 17, 2011 #6

    micromass

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    I don't think so. Similar counterexamples hold. Try to find a continuous version of my counterexample!
     
  8. Aug 17, 2011 #7
    The first statement is false. But I am correct in assuming that the second statement implies that the infinite sums of f(x) and g(x) are asymptotically equivalent (that is, [itex]lim_{n->\infty}\frac{\sum^{n}_{x=0}f(x)}{\sum^{n}_{x=0}g(x)}=1[/itex]).
     
    Last edited: Aug 17, 2011
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