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 Hail to thee, Sofia! (Posted on 2010-12-13)

Let p= 1/2*3/4*5/6*7/8……..*99/100.

Is p bigger than, equal to or less than 0.1 ?

The famous Sofia Kovalevskaya (1850-1891) solved this problem allegedly in no time, when she was only twelve years old.

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 What does not work... | Comment 1 of 10
It is pretty easy to rewrite the expression as
p = 100!/(2^100*(50!)^2)
then taking natural logarithm of both sides
ln p = ln 100! - 100ln 2 - 2 ln50!
and then using Stirlings Approximation
ln p ≈ 100ln 100 - 100 - 100ln2 - 2(50ln 50 - 50)
ln p ≈ 100ln 100 - 100ln 2 - 100ln 50
ln p ≈ 100(ln 100 - ln 2 - ln 50)
ln p ≈ 100(ln 100/(2*50))
ln p ≈ 100*0
ln p ≈ 0
p ≈ 1

This is clearly incorrect since p < 1/2

Edited on December 13, 2010, 5:43 pm
 Posted by Jer on 2010-12-13 17:35:22

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