Local minimum of $\sqrt[x]{n\over v-x+1}$

Question

I'm trying to find the local minimum of $\sqrt[x]{n\over v-x+1}$ with respect to $x$. The restrictions on $x$ are that it must be $\le v$ and $\ge 1$. Also, $v$ and $n$ are fixed, and $v<n$.

My try: differentiate the expression, admittedly using WolframAlpha, and set equal to zero; get a long expression, much of which can be divided through by since it can't be zero, due to restrictions on $x$; wind up with the equation ${x \over v-x+1}=\ln(\frac n{v-x+1})$, which I have no idea how to solve.

Suggestions?

score 2 · Answer 1 · answered Aug 18 '14 at 10:37

You need to solve for $x$ $${x \over v-x+1}=\ln(\frac n{v-x+1})$$ Make a change of variable $$\frac n{v-x+1}=y$$ so $$x=-\frac{n}{y}+v+1$$ and the equation to solve becomes $$-n \log (y)-n+(v+1)y=0$$ Any equation of the form $$A+B z+C \log(D+Ez)=0$$ has a solution in terms of Lambert function. In this specific case, the solution is given by $$y=-\frac{n W\left(-\frac{v+1}{e n}\right)}{v+1}$$ from which the formula already given by Han de Bruijn.

That adds some understanding to the answers. Thanks! – Han de Bruijn Aug 18 '14 at 19:47 — Han de Bruijn, Aug 18 '14 at 19:47

Han de Bruijn · Accepted Answer · 2014-08-18T12:51:00.483

1

Your final equation can be solved using the Lambert W function . MAPLE (another computer algebra system) says the following: $$ x = \left( v+1 \right) \left( {\it LambertW} \left( -{\frac { \left( v+1 \right) {e^{-1}}}{n}} \right) +1 \right) \left( {\it LambertW} \left( -{\frac { \left( v+1 \right) {e^{-1}}}{n}} \right) \right) ^{ -1} $$ Personally, I'd prefer a numerical solution. The latter can be found with a standard numerical method for finding zeroes of a function. But then you have to know specific values of $n$ and $v$ .

edited Aug 18 '14 at 12:51

answered Aug 17 '14 at 19:25

Han de Bruijn

17,070

If you are interested in such numerical solutions, let me know. – Han de Bruijn Aug 17 '14 at 19:27
Depending on the value of the argument $x$, there are very good approximations of $W(x)$. These could be useful for starting a Newton procedure. – Claude Leibovici Aug 18 '14 at 10:22
@ClaudeLeibovici: Indeed, Newton-Raphson is exactly what I had in mind. – Han de Bruijn Aug 18 '14 at 12:53
What else could we think : Newton or Newton or Newton ? I suppose you are familiar with the approximations of $W(x)$; if it is not the case, it is a fascinating area. Cheers :-) – Claude Leibovici Aug 18 '14 at 16:35

Local minimum of $\sqrt[x]{n\over v-x+1}$

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