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I'm done being confused by Galois theory and am back to being confused by elementary calculus.

I have a polynomial of degree $m-1$ that is bounded by the curve $y = x^m$ and intersects it at a nonzero number of points. Does each of these intersection points represent a change in concavity (or two, given that it has to go down and up again)? Also, am I right in saying there can be at most $m-3$ changes in concavity? Thanks, I kind of forgot this stuff.

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I believe the curve $2x^3 - 3x^2$ is bounded by $x^4$ and intersects it at the origin, but doesn't change concavity there. – MartianInvader Jul 12 '11 at 20:00
ok, thanks! i guess that settles it then – badatmath Jul 12 '11 at 20:21
up vote 1 down vote accepted

Say you have a polynomial $y = p_{m-1}(x)$ and there exist some real solutions of $p_{m-1}(x) = x^m$.

A function $f(x)$ is concave at point $x$ if $f''(x)<0$ and convex if $f''(x)>0$. The change in concavity occurs when $f''(x)=0$.

If you are asking whether that implies that $p''_{m-1}(x)$ is zero at those points, than I think it does not. Consider example of $m=4$ and $p_3(x) = x^3-x+1$. The equation $p_3(x) = x^4$ admits two real solutions $x=\pm1$. The second derivative of the polynomial is $6x$ and is not zero at those points.

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thanks...I fail at calculus :P – badatmath Jul 12 '11 at 18:45
There was the condition "bounded by $\dots$" – André Nicolas Jul 12 '11 at 18:46
oh wait...yeah I guess this is still an open question then, good call :) – badatmath Jul 12 '11 at 18:56

About the "also" part of the question:

It is clear that if $m < 3$ there are no changes in concavity.

Suppose that $m \ge 3$. Your polynomial has degree $m-1$. So its second derivative has degree $m-3$.

Everywhere that there is a change in concavity, the second derivative is $0$. But a polynomial of degree $k$ has at most $k$ zeros. So the function can change concavity at most $m-3$ times.

Side Comment: The $0$ polynomial does not have degree $0$, or at least not many mathematicians say that it does. One convention is that it does not have a degree. Another convention is that it has degree $-\infty$. I have also seen $-1$ offered as the degree of the $0$ polynomial.

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Thank you for your answer! – badatmath Jul 13 '11 at 5:04

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