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Comparison theorem.

  1. Mar 20, 2012 #1
    Just as talked about in stewart in strategy for integration.

    I found notes online that also say:

    g(x) >= f(x) >= 0, then you want to prove convergence on g. If f(x) >= g(x) >= 0, then you want to prove divergence on g. Im pretty sure I follow the logic here, but how exactly does one pick g?? I've been working on hours picking g that results in the opposite of required case (ex. divergence for g(x) >= f(x) >= 0) which does not then prove anything about f(x)!! :(
     
  2. jcsd
  3. Mar 20, 2012 #2

    Office_Shredder

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    The typical strategies for simple problems are to note that rational functions behave like their highest degree terms, log grows slower than power functions which grow slower than exponentials, and sine and cosine are bounded by one. If you have some specific examples in mind we can work through finding the comparison
     
  4. Mar 20, 2012 #3
    I think that will help well enough. Thank you.
     
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