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Factorialroot Fruition (Posted on 2022-11-24) Difficulty: 3 of 5
Let us consider:

X= 99√(99!), and:

Y= 100√(100!)
Determine, without direct evaluation and using analytic/semi-analytic (simple calculator + p&p) techniques, which of these is greater:

• (i) X or Y

• (ii) X/99 or Y/100

*** Computer programs/excel solver/advanced calculators/online solvers can be used only for purposes of verification.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution Solution Comment 2 of 2 |
Define f(n) = (n!)^(1/n).  Then X=f(99) and Y=f(100)

99 and 100 are 'large' so Stirling's Approximation can be substituted in for n! without materially impacting the values.  n! ~= (n/e)^n * sqrt(2*pi*n).

Then f(n) = (n!)^(1/n) ~= ((n/e)^n * sqrt(2*pi*n))^(1/n) = n/e * (2*pi*n)^(1/(2n))

Then X = f(99) ~= 99/e * (198*pi)^(1/198) = 37.6228
And Y = f(100) ~= 100/e * (200*pi)^(1/200) = 37.9924
So the answer to part (i) is Y is greater than X.

Also, X/99 = 0.380028 and Y/100 = 0.379924
So the answer to part (ii) is X/99 is greater than Y/100.

Note: No advanced solvers were used; only a common scientific calculator was used.

  Posted by Brian Smith on 2022-11-24 13:55:49
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