I am looking for a proof that does not use derivatives of the elementary theorem given in the title:

Theorem: A polynomial $p:\mathbb{R}\to\mathbb{R}$ of degree $n$ cannot have more than $n-1$ local extrema.

Of course, the proof is really easy using derivatives, but for teaching reasons (and curiosity) I am interested in finding a proof that avoids them and uses only elementary facts about polynomials. I found one that uses derivatives only in disguise (see below), but I'd be glad if someone has a better one.

Here goes the proof. First, it is possible to define a `formal derivative' $p'(x)$ of a given polynomial $p(x)$ only by its action on the monomials, that is: $a\cdot x^k$ becomes $ka\cdot x^{k-1}$. (It seems that it is exactly what Rolle did when proving the first version of Rolle's Theorem in 1691, which predates calculus and was done only for polynomials.) Then, it is easy to show that the formal derivative obeys the usual product rule and that $p'(x)=q'(x)$ if $b\in\mathbb{R}$ and $q(x)=p(x)+b$. Now, observe that $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff $(p(x) - p(a)) = (x-a)^2\cdot q(x)$ for some $q(x)$. The reason is that $p(x)-p(a)$ touches but does not cross the horizontal axis at $a$, so it must be divisible by $(x-a)^2$. By the product rule, $p'(a) = 0$, hence there is no more than $n-1$ local extrema since $p'(x)$ is of degree $n-1$.

While we don't need to define conceptually the derivative for this proof to work, it feels like an ad-hoc cheat to use them anyway, and we must take the pain of proving the product rule. But maybe there is no way out.

  • $\begingroup$ One writes "this extremum is" or "these extrema are", but "extremas" is not a correct form. (But confused students use lots of different forms for this.) ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 18 '15 at 20:11
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    $\begingroup$ Thanks, I'll correct this. Or these. (But certainly not theses.) $\endgroup$ – Mathieu Baillif Aug 18 '15 at 20:13
  • $\begingroup$ This isn't fully flushed out — just a thought that could lead somewhere but needs a real proof. Suppose a polynomial $f$ has degree $n$ and has $n$ extrema. Qualitatively, it is clear that you can draw a curve, $g$, which passes through $f$ exactly $n+1$ times (on both sides of each extremum) and $g\neq f$. Then $f-g$ is a polynomial of degree (at most) $n$ but has $n+1$ distinct roots, a contradiction $\endgroup$ – Elliot G Aug 18 '15 at 20:30
  • $\begingroup$ Nice question! You will have to somehow make sense of continuity and $p$ being a polynomial. To me this sounds like the only hope is the intermediate value theorem in combination with the real version of the Fundamental Theorem of Algebra (every real polynomial can be written as the product of linear factors and quadratic factors with real coefficients). Maybe induction on the number of quadratic factors would work? $\endgroup$ – Damian Reding Aug 18 '15 at 21:02
  • $\begingroup$ @Elliot G : It is not clear to me a priori how you can justify that $g$ can be chosen to be $\not= f$. $\endgroup$ – Mathieu Baillif Aug 18 '15 at 22:12

Let's try a concrete example:

$f(x)$ is a $3$rd-degree polynomial in $x$ and has a maximum at $x=5$ and a minimum at $x=8$. Could there be another extremum?

First, let us see that the two extrema described above imply $f(x)\to+\infty$ as $x\to+\infty$. To see this, suppose the other alternative holds: $f(x)\to-\infty$ as $x\to+\infty$. Then the curve that starts going upward when $x$ increases beyond $x=8$ would turn around and go downward. Draw the picture and you'll see that that means some line intersects the curve four times, whereas a $3$rd-degree polynomial can intersect a straight line only three times.

Now let's ask about other extrema. How about, for example, a minimum at $x=3$? Then for some small number $\delta$, you have $f(3-\delta)>f(3)<f(3+\delta)$. Draw the straight line that passes through the point $(3-\delta, f(3))$ and point at which $x=3$ and $y$ is slightly smaller than $f(5)$. If you've drawn the graph consistently with the information above, showing $f(x)\to+\infty$ as $x\to+\infty$.

Six possibilities arise:

  • A local minimum somewhere in $(-\infty,5)$. This is the case discussed in the paragraph above.
  • A local maximum in that interval.
  • A local minimum somewhere in $(5,8)$.
  • A local maximum in that interval.
  • A local minimum somewhere in $(8,\infty)$.
  • A local maximum in that interval.

Each of those leads to a graph that is intersected by a straight line more than three times.

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  • $\begingroup$ That's a good example, and a good idea. But I do not see how to easily generalize it to higer degrees, or to obtain a convincing conceptual general proof. $\endgroup$ – Mathieu Baillif Aug 18 '15 at 22:09

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