# Constants for Rolle's Theorem applied to polynomials

Rolle's Theorem states that $f(1/2)=f(-1/2)+f'(x)$ has a root in the open real interval $(-1/2,1/2)$ if $f$ is continuous and differentiable. How large can the absolute value of such a root $\xi$ be if $f$ is a polynomial of degree $d$? (We choose $\xi$ to be the real root of minimal norm in the case of several solutions.)

A few comments:

$\xi$ can be arbitrarily close to the boundary if there is no condition on the degree.

$\xi$ is always $0$ in degree $2$.

I have examples (using random polynomials) with $\xi=.28867...$ for $d=3$ (the correct value is perhaps $1/(2\sqrt{3})$ given for example with $f(x)=x^3$), and $\xi=.324...$ for $d=5$.

Considering for $f$ a complex polynomial, the equation $f(1/2)=f(-1/2)+f'(x)$ seems always to have a complex solution of norm at most 1/2. Is this true? If yes, does the real constant working for real polynomials of degree $d$ also work for complex polynomials of degree $d$? (I have no counterexample.)

A positive answer for complex polynomials would imply the statement for entire functions (where the smallest root could perhaps be on the boundary of the complex disc of radius $1/2$).

• For complex polynomials there is no chance: take the entire function $e^{2\pi i z}$. – fedja Nov 2 '11 at 19:01
• For an explicit complex polynomial where it fails, take $f(z)=(1/2+z+\xi(1/2-z))^d$, where $\xi=e^{2\pi i/d}$. We have $f(1/2)=f(-1/2)=1$, but the only root of $f'(z)=d(1-\xi)(1/2+z+\xi(1/2-z))^{d-1}$ is $(\xi+1)/2(\xi-1)$, which has absolute value about $d/2\pi$ for large $d$. – Emil Jeřábek Nov 2 '11 at 19:19
• Thanks for your examples. Fedja's remark is of course enough to kill the complex questions by continuity. I like however also Jerabek's explicit example. I should have let the computer run a little bit longer, I guess, before asking prematurously on MO. – Roland Bacher Nov 2 '11 at 19:31
• This question was asked long time ago, but a (real) polynomial of degree 3 can vanish at $1/2$ and $−1/2$ and have a derivative which vanishes at any (nonzero) point of the real line, so I don't understand the problem, neither the solution. – user111 Aug 6 '16 at 14:08

## 2 Answers

For real polynomials $\xi$ can be $1/2 - O(1/d^2)$ but no closer to $1/2$ than that. The value $1/\sqrt{12} = 0.288675\ldots$ is correct for $d=3$, and also for $d=4$; for larger $d=2n-1$ or $d=2n$, the supremum occurs at half the largest root of the $n$-th Legendre polynomial $P_n$, and is not attained because at the extremal points $f$ also has points of inflection satisfying the Rolle condition, as at $x=0$ for the quintic polynomial $f(x) = x^3-4x^5$ which shows the supremum $\sqrt{3/20} = .387298\ldots$. (The reason $n=2$ is different is that $P_2(x) = (3x^2-1)/2$, and indeed $1/\sqrt{12}$ is half the largest root $1\sqrt{3}$, but there are no inflection points because $P_2$ has no zeros in the interior of $[-1/\sqrt{3},+1/\sqrt{3}]$. For $n=1$ the Legendre polynomial vanishes only at zero, which we know must be a Rolle point when $f$ is quadratic.)

Here's a Wolfram Alpha plot for $d=11$, for which $\xi = .4662\ldots$: (image source)

Given $n$, let $p(x)=P_n(2x)$, so that $p$ is a degree-$n$ orthogonal polynomial on $[-1/2,+1/2]$; denote its roots by $x_i$ in increasing order, $x_1 < x_2 < x_3 < \cdots < x_n$, with each $x_i = -x_{n+1-i}$ by symmetry. In particular $x_1 = -x_n$. For any polynomial $f$ of degree $d \leq 2n$, let $g(x) = f'(x) - cx$ where $c = f(1/2) - f(-1/2)$. I claim $g$ cannot be positive on $[x_1,x_n]$. Indeed $\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx = 0$, but by Gaussian quadrature $\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx$ is a positive linear combination of the $g(x_i)$ — contradiction. On the other hand, $g$ can be nonnegative but not identically zero on $[x_1,x_n]$, which happens iff $$g(x) = (ax+b) \phantom. p(x)^2 / (x_n^2-x^2)$$ for some $b>0$ and $a \in [-b/x_n,b/x_n]$. (Note that $a=0$ deals with $d=2n-1$.) Indeed the same Gaussian-quadrature formula shows that $g$ must vanish on each $x_i$, with double roots except possibly at $x_1$ and $x_n$; this gives $g(x) = (ax+b) \phantom. p(x)^2 / (x_n^2-x^2)$, and then the inequalities on $a,b$ follow from nonnegativity on $[x_1,x_n]$. Conversely, if $g$ is of that form then it is indeed nonnegative on $[x_1,x_n]$, and $$\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx = \int_{-1/2}^{1/2} \phantom. p(x) \frac{(ax+b)\phantom. p(x)}{x_n^2-x^2} \phantom. dx = 0$$ because the integral is the inner product on $[-1/2,+1/2]$ of $p$ with a polynomial of degree less than $n$.

We can then take $f_0(x)$ to be an indefinite integral of $g(x)$, which satisfies $f_0(-1/2)=f_0(+1/2)$ and is strictly increasing on $[x_1,x_n]$. For each $\epsilon>0$ we can then perturb $f_0$ to get a polynomial $f$ of the same degree, still taking the same values at $-1/2$ and $+1/2$, but with a strictly positive derivative on $[x_1+\epsilon,\phantom, x_n-\epsilon]$. For instance we may take $f(x) = f_0(\theta x) + \delta\cdot x$ with $\theta$ slightly larger than $1$ and $\delta>0$ chosen so that $f(-1/2)=f(+1/2)$. This completes the proof.

It so happens that I recently needed a bound similar to the second part of your question (concerning complex polynomials).

As I already mentioned in a comment above, you cannot have a universal bound on the norm of a solution to $f(1/2)=f(-1/2)+f'(z)$, since there are polynomials where the smallest such solution has norm linear in the degree: taking $f(z)=(1/2+z+\xi(1/2-z))^d$ where $\xi=e^{2\pi i/d}$ is the primitive $d$th root of unity, we have $f(1/2)=f(-1/2)=1$, but the only root of $f'(z)=d(1-\xi)(1/2+z+\xi(1/2-z))^{d-1}$ is $(\xi+1)/2(\xi-1)$, which has absolute value about $d/2\pi$ for large $d$.

On the other hand, there is also a linear upper bound:

Proposition 1: If $f$ is a complex polynomial of degree $d$, then there exists $z$ such that $f(1/2)=f(-1/2)+f'(z)$ and $|z|<\frac1{2(\sqrt[d-1]2-1)}\le\frac{d-1}{\log 4}$.

It follows by applying the following statement to $f(z)-(f(1/2)-f(-1/2))z$:

Proposition 2: Let $f$ be a complex polynomial of degree $d$ such that $f'(z)$ has no roots of norm less than $R$. Then $f$ is injective on the closed disk $\{z:|z|\le\mu R\}$, where $\mu=\sqrt[d-1]2-1\ge\frac{\log 2}{d-1}$.

This can be proved by writing

$$f(v)-f(u)=\int_u^vf'(z)\,dz=(v-u)\left(f'(0)+\int_0^1f'((1-t)u+tv)-f'(0)\,dt\right)$$

(where $u\ne v$, $|u|,|v|\le\mu R$) and using the bound $|f'((1-t)u+tv)-f'(0)|< |f'(0)|$, which follows from

Lemma: If $g$ is a degree $d$ polynomial with no roots of norm less than $R$, and $m>0$, then $|g(z)-g(0)|< ((1+m)^d-1)|g(0)|$ for all $z$ such that $|z|< mR$.

To prove the lemma, write $g(z)=c\prod_{i=1}^d(z-\alpha_i)$. This gives

$$\frac{g(z)}{g(0)}=\prod_i\left(1-\frac z{\alpha_i}\right)=\sum_{I\subseteq\{1,\dots,d\}}\prod_{i\in I}\frac{-z}{\alpha_i},$$

hence

$$\left|\frac{g(z)}{g(0)}-1\right|\le\sum_{I\ne\varnothing}\prod_{i\in I}\frac{|z|}R< (1+m)^d-1,$$

using the assumption $R\le|\alpha_i|$.

EDIT: As I suspected, this is a classical result, and the example with linearly shifted $z^d$ attains the optimal value:

Theorem (Grace, Heawood): If $f$ is a degree $d$ polynomial and $f(u)=f(v)=0$, $u\ne v$, then its derivative $f'$ has a root $z$ in the disk $\left|z-\frac{u+v}2\right|\le\frac{|u-v|}2\cot\frac\pi d$.

Corollary: If $f$ is a degree $d$ polynomial such that $f'$ has no root of norm less than $R$, then $f$ is injective in the disk $\{z:|z|< R\sin\frac\pi d\}$.

Corollary: If $f$ is a degree $d$ polynomial, there exists $z$ such that $f(1/2)=f(-1/2)+f'(z)$ and $|z|\le\frac12\cot\frac\pi d$.

A proof of the Grace–Heawood theorem can be found in M. Marden, Geometry of polynomials, AMS, 1966.