The Formal Derivative Of A Polynomial

The Formal Derivative of a Polynomial

Recall from the Simple and Multiple Roots of Polynomials page that if $K$ is a field, $f(x) \in K[x]$, and $F$ is a splitting field of $f(x)$ over $K$ then we can write $f(x)$ as a splitting of linear factors:

(1)
\begin{align} \quad f(x) = (x - r_1)^{m_1}(x - r_2)^{m_2}...(x - r_n)^{m_n} \end{align}

Where $r_1, r_2, ..., r_n \in F$ are the roots of $f(x)$ and $m_1, m_2, ..., m_n \in \mathbb{N}$ are the multiplicities of the roots. We said that a root of $f(x)$ is a simple root if it has multiplicity $1$, and a multiple root if it has multiplicity greater than or equal to $2$.

We now define the formal derivative of a polynomial.

 Definition: Let $K$ be a field and let $f(x) \in K[x]$ with $f(x) = a_0 + a_1x + ... + a_nx^n$. The Formal Derivative of $f(x)$ is the polynomial $f'(x) = a_1 + 2a_2x + ... + na_nx^{n-1}$ where for each $k \in \{ 1, 2, ..., n \}$ we define $ka_k := (\underbrace{a_k + a_k + ... + a_k}_{k\:\mathrm{times}})$.

The following theorem tells us that the sum, difference, and product rule hold for the formal derivatives of polynomials.

 Theorem 1: Let $K$ be a field and let $f(x), g(x) \in K[x]$. Then: a) $[f(x) + g(x)]' = f'(x) + g'(x)$. b) $[f(x) - g(x)]' = f'(x) - g'(x)$. c) $[f(x)g(x)]' = f(x)g'(x) + f'(x)g(x)$.

We will only prove (a). The remaining proofs are very similar.

• Proof of a) Let $f(x) = a_0 + a_1x + ... + a_nx^n$ and let $g(x) = b_0 + b_1x + ... + b_mx^m$. Without loss of generality, assume that $n \geq m$. Then:
(2)
\begin{align} \quad f(x) + g(x) = (a_0 + b_0) + (a_1 + b_1)x + ... + (a_m + b_m)x^m + a_{m+1}x^{m+1} + ... + a_nx^n \end{align}
• Taking the formal derivative yields:
(3)
\begin{align} \quad [f(x) + g(x)]' &= (a_1 + b_1)x + ... + m(a_m + b_m)x^{m-1} + (m+1)a_{m+1}x^m + ... + na_n^{n-1} \\ &= [a_1x + ... na_n^{n-1}] + [b_1x + ... + mb_mx^{m-1}] \\ &= f'(x) + g'(x) \quad \blacksquare \end{align}

The following theorem generalizes an important result regarding multiple roots of polynomials.

 Theorem 2: Let $K$ be a field and let $f(x) \in K[x]$. Then $f(x)$ has no multiple roots if and only if $\mathrm{gcd}(f(x), f'(x)) = 1$.