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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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 re(2): What does not work..FIGURES CAN'T LIE . Comment 10 of 10 |
(In reply to re: What does not work..FIGURES CAN'T LIE . by Ady TZIDON)

Of course it is not admissible to manipulate approximations as I did (there were no infinities involved) since there was no indication of the size of the error terms.  Clearly it didnt work but it was interesting that everything cancelled out so I shared it anyway.

I still want to do it the hard way so here is what I have:

Formula (26) on
http://mathworld.wolfram.com/StirlingsApproximation.html

Gives a way that does work.
Simplified it says a(n) < n! < b(n)
My rewriting of the expression was x = 100!/(2^100(50!)^2)
On the left put the smaller in the numerator and the larger in the denominator.  On the right do the opposite to get.

a(100)/(2^100*b(50)^2) < x < b(100)/(2^100*a(50)^2)

which simplifies (WORK OMITTED) to

e^(-901/360300)/sqrt(50pi) < x < e^(-1799/721200)/sqrt(50pi)

Focus in the right side.   The numerator is clearly less than 1 and the denominator is greater than 10 so
x < 1/10

 Posted by Jer on 2010-12-14 10:39:09

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