stylistic:tone
Calculus

Calculating derivatives and integrals of polynomials is particularly simple, compared to other kinds of functions.The derivative of the polynomial P = a n x n + a n − 1 x n − 1 + ⋯ + a 2 x 2 + a 1 x + a 0 = ∑ i = 0 n a i x i {\displaystyle P=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots +a_{2}x^{2}+a_{1}x+a_{0}=\sum _{i=0}^{n}a_{i}x^{i}} with respect to x is the polynomial n a n x n − 1 + (n − 1) a n − 1 x n − 2 + ⋯ + 2 a 2 x + a 1 = ∑ i = 1 n i a i x i − 1. {\displaystyle na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+\dots +2a_{2}x+a_{1}=\sum _{i=1}^{n}ia_{i}x^{i-1}.} Similarly, the general antiderivative (or indefinite integral) of P {\displaystyle P} is a n x n + 1 n + 1 + a n − 1 x n n + ⋯ + a 2 x 3 3 + a 1 x 2 2 + a 0 x + c = c + ∑ i = 0 n a i x i + 1 i + 1 {\displaystyle {\frac {a_{n}x^{n+1}}{n+1}}+{\frac {a_{n-1}x^{n}}{n}}+\dots +{\frac {a_{2}x^{3}}{3}}+{\frac {a_{1}x^{2}}{2}}+a_{0}x+c=c+\sum _{i=0}^{n}{\frac {a_{i}x^{i+1}}{i+1}}} where c is an arbitrary constant. For example, antiderivatives of x + 1 have the form 1 / 3 x + x + c.

For polynomials whose coefficients come from more abstract settings (for example, if the coefficients are integers modulo some prime number p, or elements of an arbitrary ring), the formula for the derivative can still be interpreted formally, with the coefficient ka understood to mean the sum of k copies of a. For example, over the integers modulo p, the derivative of the polynomial x + x is the polynomial 1.