Relation between roots and derivatives.

Statement

The arithmetic mean of roots of a polynomial in a single variable of degree $n$ with real and/or complex coefficients whose leading coefficient is not zero will always be the same as the arithmetic mean of roots of its $k^{th}$ derivatives (where $k \in \mathbb{N} : 1 \le k \le (n-1)$ ).

Proof

Consider a general polynomial $p(x)$ of degree $n$, with the leading coefficient not equal to $0$. Let us say,

$p(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0$

Here, $a_n \neq 0$.

Using Vieta's formulas,

Its sum of roots is given by: $\dfrac{-a_{n-1}}{a_n}$

The arithmetic mean of its roots is thus: $\dfrac{-a_{n-1}}{n a_n} \ (\bigstar)$

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Let us differentiate this polynomial w.r.t $x$ once. Let us denote the new polynomial as $p_1 (x)$. This polynomial has exactly $n-1$ roots.

$p_1 (x) = p'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + (n-2) a_{n-2} x^{n-3} + \cdots + 2 a_2 x + a_1$

The sum of roots of $p_1 (x)$ is: $\dfrac{-(n-1) a_{n-1}}{n a_n}$

The arithmetic mean of roots of $p_1 (x)$ is thus: $\dfrac{-(n-1) a_{n-1}}{n(n-1) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)$

Which is exactly the same as the arithmetic mean of roots of $p(x)$.

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Let us differentiate it one more time w.r.t $x$. Call this polynomial $p_2 (x)$. This polynomial has exactly $n-2$ roots.

$p_2 (x) = p''(x) = n(n-1) a_n x^{n-2} + (n-1)(n-2) a_{n-1} x^{n-3} + (n-2)(n-3) a_{n-2} x^{n-4} + \cdots + 2 a_2$

The sum of roots of $p_2 (x)$ is: $\dfrac{-(n-1)(n-2) a_{n-1}}{n(n-1) a_n} = \dfrac{-(n-2) a_{n-1}}{n a_n}$

The arithmetic mean of roots of $p_2 (x)$ is thus: $\dfrac{-(n-2) a_{n-1}}{n(n-2) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)$

Which is exactly the same as the arithmetic mean of roots of $p(x)$.

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Let us prove this for a general $k^{th}$ derivative, where $(k \in \mathbb{N} : 1 \le k \le (n-1))$. Call this polynomial $p_k (x)$. This polynomial has exactly $n-k$ roots.

$p_k (x) = p^{k} (x) = n(n-1)(n-2) \cdots (n-k+1) a_n x^{n-k} + (n-1)(n-2)(n-3)\cdots (n-k) a_{n-1} x^{n-k-1} + (n-2)(n-3)(n-4)\cdots (n-k-1) a_{n-2} x^{n-k-2} + \cdots + k! a_k$

$\\ \text{OR} \\$

$p_k (x) = p^{k} (x) = \left(^n P_k\right) a_n x^{n-k} + \left(^{n-1} P_k\right) a_{n-1} x^{n-k-1} + \left(^{n-2} P_k\right) a_{n-2} x^{n-k-2} + \cdots + \left(^{n-(n-k)}P_{k}\right) a_k$

Here, $^n P_r$ denotes the permutation function which is equal to $\dfrac{n!}{(n-r)!}$.

The sum of roots of $p_k (x)$ is: $\dfrac{- \left( ^{n-1}P_k \right) a_{n-1}}{ \left( ^n P_k \right) a_n} = \dfrac{-(n-k) a_{n-1}}{n a_n}$

The arithmetic mean of roots of $p_k (x)$ is thus: $\dfrac{-(n-k) a_{n-1}}{n(n-k) a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)$

Which is exactly the same as the arithmetic mean of roots of $p(x)$.

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Continuing differentiating further w.r.t. $x$, for $n-1$ times yields us a linear polynomial. Let us denote this polynomial as $p_{n-1} (x)$. This polynomial has exactly $1$ root.

$p_{n-1} (x) = p^{n-1} (x) = n! a_n x + (n-1)!$

Here,

Sum of roots of $p_{n-1} (x)$ $=$ Arithmetic Mean of roots of $p_{n-1} (x)$ $=$ The only root of $p_{n-1} (x)$.

Hence, the value of all three is equal to: $\dfrac{-(n-1)! a_{n-1}}{n! a_n} = \dfrac{-a_{n-1}}{n a_n} \ (\bigstar)$

Which is exactly the same as the arithmetic mean of roots of $p(x)$.

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However, if we differentiate $p(x)$ one more time, it will yield us a constant which is $n! a_n$. The following $p^n (x)$ will not be a polynomial which has any roots and thus this process fails upon further differentiation.

Therefore, we come to the conclusion that the arithmetic mean of roots of a polynomial in a single variable of degree $n$ with real and/or complex coefficients and the leading coefficient not equal to zero will always be the same as the arithmetic mean of roots of its $k^{th}$ derivatives, where $k \in \mathbb{N}$ and $1 \le k \le (n-1)$.

Let us look at an example to be familiar with one of its useful applications.

Problem

Given that the polynomial $f(x) = x^4 + 10 x^3 +35 x^2 + 50 x +24$. Find the sum of the roots of its $2nd$ derivative polynomial.

Solution

Instead of differentiating $2$ times, we can directly use our mean-value approach.

So, we have,

$\text{Mean of roots of} \ f(x) = \text{Mean of roots of} \ f''(x)$

$\implies \dfrac{-10}{1} \times \dfrac{1}{4} = \dfrac{\sigma}{2} \qquad \qquad \small \color{#3D99F6}{\text{where} \ \sigma \ \text{denotes the sum of roots of} \ f''(x)}$

$\implies \sigma = -5 \ \Box$

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We can check our answer as,

$f(x) = x^4 + 10 x^3 +35 x^2 + 50 x +24 \\ f'(x) = 4 x^3 + 30 x^2 + 70x + 50 \\ f''(x) = 12 x^2 + 60x + 70$

Where it is evident that,

$\text{Sum of roots of} \ f''(x) \ (\sigma) = \dfrac{-60}{12} = -5 \ \Box$