Growth rate of 1 <mrow class="MJX-TeXAtom-ORD"> / </mrow> ( log &#x2061;<!--

Waldronjw

Waldronjw

Answered question

2022-07-07

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)?

Answer & Explanation

vrtuljakc6

vrtuljakc6

Beginner2022-07-08Added 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
dikcijom2k

dikcijom2k

Beginner2022-07-09Added 6 answers

There is an alternative perspective on this issue. The function can write a huge "x" = - 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". It would be simple to verify that this approximation for x=2 is highly accurate. 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.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?