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I have a function f(x,n) can be expressed as a cubic function of x with coefficients that are functions of n. For example x^3 + (n-2)x^2 + (3n-6)x + n.

I want to prove that for every positive value of n, there exists a real, positive value of x such that f(x,n)=0.

I know this is true for the function I have in mind but I am not sure how to go about proving it.

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closed as too localized by Steven Landsburg, Anton Petrunin, Goldstern, Misha, Suvrit May 21 '13 at 1:28

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What you're trying to prove is not true for the example you gave, since all the terms in it are positive for positive $x$ as soon as $n\gt2$. – Barry Cipra May 20 '13 at 16:16
Yes Barry, this example will not hold. Do you know if there is a general method of proving this (if it holds) for any cubic equation of this form? – Austen May 20 '13 at 16:35
The canonical answer is Sturm sequences (see for example the Wikipedia entry ), which let one count the roots of a polynomial of any degree in ${\bf R}$ or in an interval. I might have written more about it, but the question was already closed so this will have to do. – Noam D. Elkies May 21 '13 at 18:53
up vote 2 down vote accepted

For convenience, write the cubic function as

$$f(x) = x^3 + 3ax^2 + 3bx + c,$$

where $a$, $b$, and $c$ are (polynomial?) functions of $n$. As has been noted in comments, if $c<0$, you're guaranteed a positive real zero $x$, so the only question is what to do for values of $n$ for which $c\ge0$.

The only way you can have a positive real root when $c\ge0$ is if $f$ has a local minimum at a positive $x$ and takes a non-positive value there. To check for this, look at the derivative

$$f'(x) = 3(x^2+2ax+b),$$

note that you need $a^2\ge b$ to have a local minimum at all, and then you need $x = \sqrt{a^2-b}-a \gt 0$ to have the local minimum at a positive $x$. (For example, if $a\gt0$, then you need $b\lt0$.) You now need only check whether $f(\sqrt{a^2-b}-a)\le0$.

What's unclear is how easy or hard it is to check the various inequalities for the coefficient functions the OP has in mind.

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I am assuming you have some different example in mind since as Barry Cipra pointed out your objective is not true in your example (and any example that eventually has all positive coefficients). First of all every cubic has a real root by the intermediate value theorem. Hence if you want to prove the positivity I recommend you start with the general solution to the cubic. It is ugly, but you could perhaps derive positivity (if it is true) in your class of examples.

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Thanks Stephen. Yes I have a different function in mind. I had hoped there was some elegant method to do this. – Austen May 20 '13 at 16:33
It seems like you should just post your specific polynomial – David Benson-Putnins May 20 '13 at 16:36
Yes, I see what you mean. I'll start from the general solution, fix in my f(n) coefficients and work my way down until I can show that x will be positive for all n>0. – Austen May 20 '13 at 16:38
David, this is one of them. x^3 + (2-n)x^2 + (1-2n)x -n -8 – Austen May 20 '13 at 16:40
Assume without loss of generality that the leading coefficient is 1. Then a sufficient condition (which applies in your example) is certainly that $f(0,n)<0$. This only involves the constant coefficient. Another sufficient condition is that $f(2013)<0$. Another sufficient condition is that $f(r)<0$, where $r$ the geometric mean of the absolute values of the second and third coefficient. Etc, etc. None of these conditions is necessary, however. (This is why this is a comment, not an answer.) – Goldstern May 20 '13 at 17:19

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