'Guessing' local extrema of a polynomial given its roots

Question

To start with let's assume that $p$ is a degree $n>1$ polynomial in $x$ and has $n$ distinct roots $\alpha_1, \ldots, \alpha_n$. Without loss of generality we can also stipulate that $0 = \alpha_1 < \cdots < \alpha_n = 1$. This guarantees there are $n-1$ local extrema, occurring at locations $x_1 \in (\alpha_1, \alpha_2)$, $x_2 \in (\alpha_2, \alpha_3)$, and so on. One might like to guess at the values of $x_i$ based on knowledge of the roots, and I'm curious about any heuristics that could do better than guessing the midpoint of each interval.

Of course, if $n$ is small, then there exists an explicit formula; e.g. when $p(x)=(x-\alpha_1)(x-\alpha_2)$ then $x_1 = (\alpha_1 + \alpha_2)/2$, the midpoint between the roots. But as early as $n=3$ things get murky; if $p(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$, then $$x_{1,2}=\frac{\alpha_1+\alpha_2+\alpha_3}{3}\pm\frac{\sqrt{(\alpha_1+\alpha_2+\alpha_3)^2-3(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)}}{3}$$ From this we see that the two extrema are centered around the mean of the roots. For example when $\alpha_2=1/2$, they are at $1/2 \pm \sqrt{3}/6$, so they're "pushed out" toward $0$ and $1$ (as opposed to being evenly distributed at $1/4$ and $3/4$).

Based on this idea, I can imagine a heuristic that says the extremal values will be nearer some roots and further away from others, where "nearer" and "further" should be taken in a relative sense; in the simple example above they would be nearer $\alpha_1=0$ and $\alpha_3=1$, and further from $\alpha_2 = 1/2$. Once there are more roots, do the roles of the roots alternate? (a "near" root, then a "far root", then a "near" root again?).

A more concrete question to ask is: suppose I just guess that the extrema are at the midpoints between each pair of roots, call them $m_1 = (\alpha_1+\alpha_2)/2$, $m_2=(\alpha_2+\alpha_3)/2$, and so on. Call the error term $E=\frac{1}{n-1}\sum|m_i - x_i|^2$. How does $E$ depend on the roots of $p$? Is it monotonic with $n$ in some sense?

This is an idle curiosity; I'm just trying to dream up interesting Calc I problems and found something that is a little too interesting.

Answer 1

In my opinion, it is wrong to try to look for extreme points through examination of either the polynomial $p(x)$ or any of its roots. The polynomial is going to be continuous, and have both a first and second derivative everywhere.

Therefore, the extreme points are going to be the (n-1, not necessarily distinct) roots of $p'(x) = 0$ [i.e. $\{r_1, r_2, \cdots, r_{(n-1)}\}$]. For each root, $r_i$, the sign (positive or negative) of $p''(r_i)$ will indicate whether it is a local minimum or maximum.

Edit In your example, under the assumption that $p(x)$ has $n$ distinct real roots, $p'(x)$ has to have $(n-1)$ distinct real roots, because $p(x)$ must change direction $(n-1)$ times.

As for whether you would expect (for example) that the $p'(x)$ root between $\alpha_k$ and $\alpha_{(k+1)}$ will in general be closer to $\alpha_k$ or $\alpha_{(k+1)}$, I consider that to be an enormously complicated (and perhaps relatively unexplored) question. I think that you would have to somehow diagnose the interplay between the coefficients of either $p(x)$, $p'(x)$, or both.

Answer 2

Given the polynomial $$ p_{\,n} (x) = \prod\limits_{k = 1}^n {\left( {x - r_{\,k} } \right)} \quad \left| {\;r_{\,k} \le r_{\,k + 1} } \right. $$

Putting $$ p_{\,n} (x) = \prod\limits_{k = 1}^n {\left( {x - r_{\,k} } \right)} \quad \left| {\;r_{\,k} \le r_{\,k + 1} } \right. $$ then it is clear that if we keep only the absolute values, we are converting all the extremals into maxima, preserving the abscissas where they occur and their absolute values.

Then consider to take the logarithm $$ \eqalign{ & \left| {\,p_{\,n} (x)\,} \right| = \prod\limits_{k = 1}^n {\left| {\,x - r_{\,k} \,} \right|} \cr & L_{\,n} (x) = \ln \left| {\,p_{\,n} (x)\,} \right| = \sum\limits_{k = 1}^n {\ln \left| {\,x - r_{\,k} \,} \right|} \cr} $$ the extremals of $p_n(x)$ will be the maxima of $L_n(x)$, occurring in between its poles.

Let's take for example five roots, as in the following graph.

The contribution of the first two roots alone is shown by the black curve. The maximum would occur at the average of the two roots.
The contribution of the remaining three roots is the blue curve, which at a sufficient distance may be approximated by the pink one, as if the three roots were concentrated at their barycenter.

Clearly the action of the roots on the right is such as to "push" leftward the maximum between $r_1,r_2$, yet not beyond $r_1$.
Same in the reverse direction, the effect the effect of the first couple on the right triple.

To estimate the deviation, we can develop in series the two set of functions around $s = (r_1 + r_2)/2$.

Let's put $$ s = {{r_{\,2} + r_{\,1} } \over 2}\quad d = {{r_{\,2} - r_{\,1} } \over 2}\quad t = {{r_{\,3} + r_{\,4} + r_{\,5} } \over 3} $$ so that for $$ 0 \le \left| {\,x - s\,} \right| \le d \le t - s $$ we can write $$ \eqalign{ & L_{\,a} (x) = \ln \left| {\,x - r_{\,1} \,} \right| + \ln \left| {\,x - r_{\,2} \,} \right| \cr & = \ln \left| {\,x - s + d\,} \right| + \ln \left| {\,x - s - d\,} \right| = \cr & = \ln \left( {d^{\,2} - \left( {x - s} \right)^{\,2} } \right) = \cr & = 2\ln d + \ln \left( {1 - \left( {{{x - s} \over d}} \right)^{\,2} } \right) = \cr & = 2\ln d - \left( {{{x - s} \over d}} \right)^{\,2} + O\left( {\left( {{{x - s} \over d}} \right)^{\,4} } \right) \cr & \cr & L_{\,b} (x) = 3\ln \left| {\,x - t\,} \right| = 3\ln \left( {t - x} \right) = \cr & = 3\ln \left( {\left( {t - s} \right) - \left( {x - s} \right)} \right) = 3\ln \left( {t - s} \right) + 3\ln \left( {1 - {d \over {t - d}}\left( {{{x - s} \over d}} \right)} \right) = \cr & = 3\ln \left( {t - s} \right) - 3{d \over {t - d}}\left( {{{x - s} \over d}} \right) - {3 \over 2}\left( {{d \over {t - d}}} \right)^{\,2} \left( {{{x - s} \over d}} \right)^{\,2} + O\left( {\left( {{{x - s} \over d}} \right)^{\,3} } \right) \cr} $$

Therefore the maximum between $r_1$ and $r_2$ will move of the following amount $$ \left( {{{x - s} \over d}} \right) = 0\quad \Rightarrow \quad \left( {{{x - s} \over d}} \right) \approx - {{3{d \over {t - d}}} \over {3\left( {{d \over {t - d}}} \right)^{\,2} + 2}} $$

Generalization

The example above straitghtly suggests how to generalize it.

Always given the $n$ roots $r_1,\, r_2, \, \ldots , \, r_n$ arranged in a non-decreasing order from the logarithm defined above $$ L_{\,n} (x) = \ln \left| {\,p_{\,n} (x)\,} \right| = \sum\limits_{k = 1}^n {\ln \left| {\,x - r_{\,k} \,} \right|} $$ we know that the positions of the extremes is given by the solutions to $$ 0 = L_{\,n} '(x) = \sum\limits_{k = 1}^n {{1 \over {\,x - r_{\,k} \,}}} = {{p_{\,n} '(x)} \over {p_{\,n} (x)}} $$ which actually is a $n-1$ degree polynomial.
If the roots were only two (or if they are quite isolated from the others) there would be an extreme right in the middle.
We want to estimate how much the extreme between a couple of roots is deviated from the middle due to the influence of the other (external) roots.

So let's consider a consecutive couple of roots $r_j, \, r_{j+1}$ and let's define $$ s_{\,j} = {{r_{\,j + 1} + r_{\,j} } \over 2}\quad d_{\,j} = {{r_{\,j + 1} - r_{\,j} } \over 2} $$ and write $$ \eqalign{ & L_{\,j,\,n} '(x) = {1 \over {\,x - \left( {s_{\,j} - d_{\,j} } \right)\,}} + {1 \over {\,x - \left( {s_{\,j} + d_{\,j} } \right)\,}} + \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,x - r_{\,k} \,}}} = \cr & = {{2\left( {x - s_{\,j} } \right)} \over {\,\left( {x - s_{\,j} } \right)^2 - d_{\,j} ^2 }} + \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\left( {x - s_{\,j} } \right) - \left( {r_{\,k} - s_{\,j} } \right)\,}}} = \cr & = {1 \over {d_{\,j} }}{{2\left( {{{x - s_{\,j} } \over {d_{\,j} }}} \right)} \over {\,\left( {\left( {{{x - s_{\,j} } \over {d_{\,j} }}} \right)^2 - 1} \right)}} + {1 \over {d_{\,j} }} \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\left( {{{x - s_{\,j} } \over {d_{\,j} }}} \right) - \left( {{{r_{\,k} - s_{\,j} } \over {d_{\,j} }}} \right)\,}}} = \cr & = {1 \over {d_{\,j} }}\left( {{{2\xi _{\,j} } \over {\,\left( {\xi _{\,j} ^2 - 1} \right)}} + \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\xi _{\,j} - \rho _{\,k,\,j} \,}}} } \right) \cr & \left| {\; - 1 \le \xi _{\,j} = {{x - s_{\,j} } \over {d_{\,j} }} \le 1 < \rho _{\,k,\,j} = {{r_{\,k} - s_{\,j} } \over {d_{\,j} }}} \right. \cr} $$

Since $$ {1 \over {x - a}} = - {1 \over a}{1 \over {\left( {1 - x/a} \right)}} = - {1 \over a}\left( {1 + {x \over a} + \left( {{x \over a}} \right)^{\,2} + O\left( {\left( {{x \over a}} \right)^{\,3} } \right)} \right) \quad \left| {\;\left| {{x \over a}} \right| < 1} \right. $$ and $$ {x \over {\,\left( {x^2 - 1} \right)}} = - x\left( {1 + O\left( {x^{\,2} } \right)} \right) \quad \left| {\;\left| x \right| < 1} \right. $$ we can approximate at various levels the expression for $L_{\,j,\,n} '(x)$, the very first being $$ \bbox[lightyellow] { \eqalign{ & 0 \approx - 2\xi _{\,j} - \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\rho _{\,k,\,j} }}} - \xi _{\,j} \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\rho _{\,k,\,j} ^{\,2} }}} \quad \Rightarrow \cr & \Rightarrow \quad \xi _{\,j} \approx - \; {{\sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\rho _{\,k,\,j} }}} } \over {2 + \sum\limits_{\left\{ {\matrix{ {k \ne j,\,j + 1} \cr {1\, \le \,k\, \le \,n} \cr } } \right.} {{1 \over {\,\rho _{\,k,\,j} ^{\,2} }}} }} \cr} }$$