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Waldronjw 2022-07-07 Answered
Growth rate of 1 / ( log ( x ) log ( x 1 ) )
Let x > 1 be a real number. Let y = 1 log ( x ) log ( x 1 )
My question: Approximately how fast does y grow (asymptotically) in terms of x? (e.g. linear, polynomial, exponential)?
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Answers (2)

vrtuljakc6
Answered 2022-07-08 Author has 16 answers
log ( b ) log ( a )
= a b d log ( x ) d x d x
= a b 1 x d x
Since a < x < b, we have 1 b < 1 x < 1 a , so
a b 1 b d x < a b 1 x d x < a b 1 a d x
Therefore,
b a b < log ( b ) log ( a ) < b a a
Putting a = x 1 and b = x gives
1 x < log ( x ) log ( x 1 ) < 1 x 1
So the answer is that y is about x
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dikcijom2k
Answered 2022-07-09 Author has 6 answers
There is another way to look at this problem. If "x" is large, the function can writey = - 1 / Log[1 - 1 / x]. Computing a Taylor series around an infinite value of "x" leads to y = x - 1/2 - 1/ (12 x) - 1 / (24 x^2) + .... So, "y" varies linearly just as "x". You could easily check that this is a very good approximation for x=2. If you compute the second derivayive of "y" with respect to "x", you will also notice that its value is very small, then very low curvature.
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