rules of calculus for derivative of polynomial


In this entry, we will derive the properties of derivatives of polynomials in a rigorous fashion. We begin by showing that the derivative exists.

Theorem 1.

If A is a commutative ring and p is a polynomialPlanetmathPlanetmath in A[x], then there exist unique polynomials q and A such that p(x+y)=p(x)+yq(x)+y2r(x,y).

Proof.

We will first show existence, then uniqueness. Define f(y)=p(x+y)-p(x). Since f is a polynomial in y with coefficients in the ring A[x] and f(0)=0, we must have y be a factor of f(y), so f(y)=yg(x,y) for some g in A[x,y]. By definition of f, this means that p(x+y)-p(x)=yg(x,y). 11We are here making use of the identification of A[x][y] with A[x,y] to write the polynomial g either as a polynomial in y with coefficients in A[x] or as a polynomial in x and y with coefficients in A. Define q(x)=g(x,0) and h(x,y)=g(x,y)-g(x,0). Regarding h as a polynomial in y with coefficients in A[x], we may, similiarly to what we did earlier, note that, since h(0)=0 by construction, y must be a factor of h(y). Hence there exists a polynomial r with coefficients in A[x,y] such that h(y)=yr(x,y). Combining our definitions, we conclude that p(x+y)=p(x)+yq(x)+y2r(x,y).

We will now show uniqueness. Assume that there exists polymonomials q,r,Q,R such that p(x+y)=p(x)+yq(x)+y2r(x,y) and p(x+y)=p(x)+yQ(x)+y2R(x,y). Subtracting and rearranging terms, y(q(x)-Q(x))=y2(R(x,y)-r(x,y)). Cancelling y22Note that, in general, the cancellation law need not hold. However, even if A has divisorsMathworldPlanetmathPlanetmath of zero, it still will be the case that the polynomial y cannot divide zero, so we may cancel it., we have q(x)-Q(x)=y(R(x,y)-r(x,y)). Substituting 0 for y, we have q(x)-Q(x)=0. Replacing this in our equation, y(R(x,y)-r(x,y))=0. Cancelling another y, R(x,y)-r(x,y)=0. Hence, we conclude that Q=q and R=r, so our is unique. ∎

Hence, the following is well-defined:

Definition 1.

Let A be a commutative ring and let p be polynomial in A[x]. Then p is the unique element of A[x] such that p(x+y)=p(x)+yp(x)+y2r[x,y] for some rA[x,y]

We will now derive some of the rules for manipulating derivatives familiar form calculus for polynomials using purely algebraic operations with no limits involved.

Theorem 2.

If A is a commutative ring and p,qA[x], then (p+q)=p+q.

Proof.

Let us write p(x+y)=p(x)+yp(x)+y2r(x,y) and q(x+y)=q(x)+yq(y)+y2s(x,y). Adding, we have

p(x,y)+q(x,y)=p(x)+q(x)+y(p(x)+q(x))+y2(r(x,y)+s(x,y)).

By definition of derivative, this means that (p+q)=p+q. ∎

Theorem 3.

If A is a commutative ring and p,qA[x], then (pq)=pq+pq.

Proof.

Let us write p(x+y)=p(x)+yp(x)+y2r(x,y) and q(x+y)=q(x)+yq(y)+y2s(x,y). Multiplying, grouping terms, and pulling out some common factors, we have

p(x+y)q(x+y) =p(x)q(y)+y(p(x)q(x)+p(x)q(x))
+y2(p(x)s(x,y)+q(x)r(x,y)+p(x)q(y)
+yp(x)s(x,y)+yq(x)r(x,y)+y2r(x,y)s(x,y)).

By definition of derivative, this means that (pq)=pq+pq. ∎

Theorem 4.

If A is a commutative ring and p,qA[x], then (pq)=(pq)q.

Proof.

Let us write p(x+y)=p(x)+yp(x)+y2r(x,y) and q(x+y)=q(x)+yq(y)+y2s(x,y). Composing, grouping terms, and pulling out some common factors, we have

p(q(x+y)) =p(q(x)+yq(y)+y2s(x,y))
=p(q(x))+(yq(y)+y2s(x,y))p(q(x))
+(yq(y)+y2s(x,y))2r(q(x),yq(y)+y2s(x,y))
=p(q(x))+yp(q(x))q(y)
+y2(s(x,y)p(q(x))+(q(y)+ys(x,y))2r(q(x),yq(y)+y2s(x,y)))

By definition of derivative, this means that (pq)=(pq)q. ∎

Title rules of calculus for derivative of polynomial
Canonical name RulesOfCalculusForDerivativeOfPolynomial
Date of creation 2013-03-22 18:20:05
Last modified on 2013-03-22 18:20:05
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 10
Author rspuzio (6075)
Entry type DerivationPlanetmathPlanetmath
Classification msc 13P05
Classification msc 11C08
Classification msc 12E05
Related topic ProofOfPropertiesOfDerivativesByPureAlgebra