# Log expansion

1. Apr 30, 2012

### autobot.d

Hi there,

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

$\frac{1}{log(2x)}= \frac{1}{logx}- \frac{log2}{log^{2}x} + O(\frac{1}{log^{3}x})$

and

$\frac{1}{log^{2}2x} = \frac{1}{log^{2}x} + O(\frac{1}{log^{3}x})$

Not sure what kind of expansion or what lets them draw this conclusion.
Any help is appreciated!

Last edited: Apr 30, 2012
2. May 1, 2012

### mathman

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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