Transforming an equation with logarithms

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    Logarithms
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SUMMARY

The discussion focuses on transforming the logarithm of the Binomial coefficient, specifically log(n! / ((n - m)! m!), using the Stirling approximation. The initial expression derived is n log n - m log m - (n - m) log(n - m). The desired transformation is to achieve the form (n - m) log(n/(n - m)) + m log(n/m). The user seeks assistance in manipulating the original equation by adding and subtracting m log n to facilitate this transformation.

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  • Understanding of logarithmic properties and transformations
  • Familiarity with the Binomial coefficient and its logarithmic representation
  • Knowledge of Stirling's approximation for factorials
  • Basic algebraic manipulation skills
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  • Explore Stirling's approximation and its applications in combinatorics
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Pnin
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I want to approximate the logarithm of the Binomial coefficient log (n!/ ((n - m)! m!) with the the Stirling approximation log x! ≈ x log x - x

I got

n log n - m log m - (n - m) log(n - m)

but I want

(n - m) log (n/(n - m)) + m log (n/m)

Can someone help how to transform the first equation into the latter?
 
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Try to add and subtract ##m \log{n}## in your first expresion
 
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thanks!
 

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