calculus with polynomials

In mathematics, polynomials are perhaps the simplest functions with which to do calculus. Their derivatives and integrals are given by the following rules:

\frac{d}{dx} \sum^n_{k=0} a_k x^k = \sum^n_{k=0} ka_kx^{k-1}

\int \sum^n_{k=0} a_k x^k\,dx= \sum^n_{k=0} \frac{a_k x^{k+1}}{k+1}  + c.

Hence the derivative of x¹⁰⁰ is 100x⁹⁹ and the integral of x¹⁰⁰ is x¹⁰¹/101 + c.

Proof

Because differentiation is linear, we have:

\frac{d\left( \sum_{r=0}^n a_r x^r \right)}{dx} =

\sum_{r=0}^n \frac{d\left(a_r x^r\right)}{dx} = \sum_{r=0}^n a_r \frac{d\left(x^r\right)}{dx}. So it remains to find

\frac{d\left(x^r\right)}{dx}

for any natural number r. There is a proof by induction using the product rule; this depends only on the case r = 1.

Generalisations

\frac{d}{dx} \left(ax^k\right) = akx^{k-1}

is generally true for all values of k where x^k is meaningful. In particular it holds for all rational k for values of x where x^k is defined. Similarly for integration, see table of integrals. If one has polynomials with coefficients that are not real or complex numbers (perhaps they are integers, or numbers modulo a prime number) then one can formally define the derivative according to the rules given above. This is useful, for example, in determining whether a polynomial will have multiple roots: compute the greatest common divisor of the polynomial and its formal derivative. If this polynomial is zero, then the original polynomial cannot have any multiple roots.

References

Larson, Ron; Hostetler, Robert P.; and Edwards, Bruce H. (2003). Calculus of a Single Variable: Early Transcendental Functions (3rd edition). Houghton Mifflin Company. ISBN 061822307X.