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Sorry for the undeveloped question---it seems it should be elementary, but I just can't find anything useful! A nudge in the direction of relevant theorems would be greatly appreciated.

Really the title says it all. I have $p(x) \neq q(x)$, both of degree $n$ and convex on $\mathbb{R}$ with $p(0) = q(0) = 0$. Is there any better lower bound on the number of intersections than just $n$?

Does this question even make sense?

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Could you please edit your message? What's $p(x)\neq q(x)$? Do you mean $p\neq q$? The "lower bound for intersections" is clearly $1$. Namely at $0$. Example: $p(x)=x$, $q(x)=x^2$. Do you mean "upper bound"? –  Alexandre Eremenko Nov 13 '12 at 1:01
    
Is the question about the number of intersections of graphs of $p$ and $q$? –  Dima Pasechnik Nov 13 '12 at 3:49

1 Answer 1

Let $f(x)$ and $g(x)$ be two polynomials of degree $n$, for $n$ even, that intersect at $n$ points, including $0$. Clearly these exist. Let

$p(x)=f(x)+ N (x^2+x^{n})-f(0)$

$q(x)=g(x) + N (x^2+x^n) -g(0)$

for $N$ sufficiently large.

Then $p(x)$ and $q(x)$ are convex, of degree $n$, still have $n$ intersections, and $p(0)=q(0)=0$.

So $n$ is best possible.

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