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Log expansion

  1. Apr 30, 2012 #1
    Hi there,

    working on Prime Number Theorem and the book gives an equality that I probably should know...

    [itex]\frac{1}{log(2x)}= \frac{1}{logx}- \frac{log2}{log^{2}x} + O(\frac{1}{log^{3}x})[/itex]

    and

    [itex]\frac{1}{log^{2}2x} = \frac{1}{log^{2}x} + O(\frac{1}{log^{3}x}) [/itex]

    Not sure what kind of expansion or what lets them draw this conclusion.
    Any help is appreciated!
     
    Last edited: Apr 30, 2012
  2. jcsd
  3. May 1, 2012 #2

    mathman

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    Gold Member

    1/log2x = 1/(log2 + logx) = (1/logx)(1/(1 + {log2/logx})

    1/(1 + u) = 1 - u + u2 - u3 + ...
    Let u=log2/logx in the first line. x > 2 is condition.
     
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