Prime Number Theorem and Its Expansion: A Puzzling Equation

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SUMMARY

The discussion centers on the Prime Number Theorem and its mathematical expansions, specifically the equations involving logarithmic identities. The key equations presented are \(\frac{1}{\log(2x)} = \frac{1}{\log x} - \frac{\log 2}{\log^{2} x} + O\left(\frac{1}{\log^{3} x}\right)\) and \(\frac{1}{\log^{2}(2x)} = \frac{1}{\log^{2} x} + O\left(\frac{1}{\log^{3} x}\right)\). The discussion highlights the use of Taylor series expansion for the term \(\frac{1}{1 + u}\) where \(u = \frac{\log 2}{\log x}\), applicable for \(x > 2.

PREREQUISITES
  • Understanding of logarithmic functions and properties
  • Familiarity with Taylor series and asymptotic notation
  • Knowledge of the Prime Number Theorem
  • Basic calculus concepts related to limits and expansions
NEXT STEPS
  • Study the derivation of the Prime Number Theorem
  • Learn about Taylor series expansions in mathematical analysis
  • Explore asymptotic notation and its applications in number theory
  • Investigate advanced logarithmic identities and their proofs
USEFUL FOR

Mathematicians, students of number theory, and anyone interested in advanced mathematical concepts related to the Prime Number Theorem and logarithmic expansions.

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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 let's them draw this conclusion.
Any help is appreciated!
 
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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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