tun1ju2k1ki

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 $x\mathrm{log}(x)+px\ge q$? (such an $x$ always exists).

Given $q>0$ and $p$, how do we get a tight estimate for the smallest $x$ such that $x\mathrm{log}(x)+px\ge q$? (such an $x$ always exists).

Krista Arroyo

Beginner2022-10-01Added 4 answers

Consider the function

$$f(x)=x\mathrm{log}(x)+px-q$$

The first derivative

$${f}^{\prime}(x)=\mathrm{log}(x)+1+p$$

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 $x\ge 0$) starts from $-q$, goes to a minimum and increases again. This means that the inequality

$$x\mathrm{log}(x)+px\ge q$$

will hold for any $x$ larger than the root of $f(x)=0$

The solution of this equation is given by

$$x=\frac{q}{W\left(q\phantom{\rule{thinmathspace}{0ex}}{e}^{p}\right)}$$

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 $f({x}_{0}){f}^{\u2033}({x}_{0})<0$ 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

$$\frac{{2}^{n}}{n!}<\u03f5$$

If ound in old notes of mine a similar problem that I adapted to your. Considering $\u03f5={10}^{-k}$, I was able to find this simple correlation

$$n\approx 3.45433+2.52846\phantom{\rule{thinmathspace}{0ex}}{k}^{0.746311}$$

which I established for $5\le k\le 50$

So, if you want to refine more, use this as an estimate from which you start Newton method using Stirling approximation for $n!$

$$f(x)=x\mathrm{log}(x)+px-q$$

The first derivative

$${f}^{\prime}(x)=\mathrm{log}(x)+1+p$$

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 $x\ge 0$) starts from $-q$, goes to a minimum and increases again. This means that the inequality

$$x\mathrm{log}(x)+px\ge q$$

will hold for any $x$ larger than the root of $f(x)=0$

The solution of this equation is given by

$$x=\frac{q}{W\left(q\phantom{\rule{thinmathspace}{0ex}}{e}^{p}\right)}$$

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 $f({x}_{0}){f}^{\u2033}({x}_{0})<0$ 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

$$\frac{{2}^{n}}{n!}<\u03f5$$

If ound in old notes of mine a similar problem that I adapted to your. Considering $\u03f5={10}^{-k}$, I was able to find this simple correlation

$$n\approx 3.45433+2.52846\phantom{\rule{thinmathspace}{0ex}}{k}^{0.746311}$$

which I established for $5\le k\le 50$

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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