Tight estimate for a log-linear inequality Given q>0 and p, how do we get a tight estimate for the smallest x such that xlog(x)+px>=q? (such an x always exists).

tun1ju2k1ki

tun1ju2k1ki

Answered question

2022-09-30

Tight estimate for a log-linear inequality
Given q > 0 and p, how do we get a tight estimate for the smallest x such that x log ( x ) + p x q? (such an x always exists).

Answer & Explanation

Krista Arroyo

Krista Arroyo

Beginner2022-10-01Added 4 answers

Consider the function
f ( x ) = x log ( x ) + p x q
The first derivative
f ( x ) = log ( x ) + 1 + p
cancels at a single place x = e ( p + 1 ) and at this point the value of the function is ( q + e ( p + 1 ) ). The second derivative being always positive, then this point corresponds to a minimum.
So, the function (which is defined for x 0) starts from q, goes to a minimum and increases again. This means that the inequality
x log ( x ) + p x q
will hold for any x larger than the root of f ( x ) = 0
The solution of this equation is given by
x = q W ( q e p )
where appears Lambert function.
Since q > 0, the solution is larger than x 0 = e p and the first iterate of Newton method would be x 1 = q + e p . But f ( x 0 ) f ( x 0 ) < 0 makes x 1 an overestimate of the solution (from Darboux theorem).
Edit
Since, in a comment, you tell that your problem is to find n such as
2 n n ! < ϵ
If ound in old notes of mine a similar problem that I adapted to your. Considering ϵ = 10 k , I was able to find this simple correlation
n 3.45433 + 2.52846 k 0.746311
which I established for 5 k 50
So, if you want to refine more, use this as an estimate from which you start Newton method using Stirling approximation for n !

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