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 polynomial in $A[x]$, then there exist unique polynomials $q$ and $A$ such that $p(x+y)=p(x)+y\,q(x)+y^{2}\,r(x,y)$.

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 $f(y)=y\,g(x,y)$ for some $g$ in $A[x,y]$. By definition of $f$, this means that $p(x+y)-p(x)=y\,g(x,y)$. 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 $h(y)=y\,r(x,y)$. Combining our definitions, we conclude that $p(x+y)=p(x)+y\,q(x)+y^{2}\,r(x,y)$.

We will now show uniqueness. Assume that there exists polymonomials $q,\,r,\,Q,\,R$ such that $p(x+y)=p(x)+y\,q(x)+y^{2}\,r(x,y)$ and $p(x+y)=p(x)+y\,Q(x)+y^{2}\,R(x,y)$. Subtracting and rearranging terms, $y(q(x)-Q(x))=y^{2}(R(x,y)-r(x,y))$. Cancelling $y$22Note that, in general, the cancellation law need not hold. However, even if $A$ has divisors 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 $p^{\prime}$ is the unique element of $A[x]$ such that $p(x+y)=p(x)+y\,p^{\prime}(x)+y^{2}\,r[x,y]$ for some $r\in A[x,y]$

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 $p,q\in A[x]$, then $(p+q)^{\prime}=p^{\prime}+q^{\prime}$.

Proof.

Let us write $p(x+y)=p(x)+y\,p^{\prime}(x)+y^{2}\,r(x,y)$ and $q(x+y)=q(x)+y\,q^{\prime}(y)+y^{2}\,s(x,y)$. Adding, we have

 $p(x,y)+q(x,y)=p(x)+q(x)+y(p^{\prime}(x)+q^{\prime}(x))+y^{2}(r(x,y)+s(x,y)).$

By definition of derivative, this means that $(p+q)^{\prime}=p^{\prime}+q^{\prime}$. ∎

Theorem 3.

If $A$ is a commutative ring and $p,q\in A[x]$, then $(p\cdot q)^{\prime}=p^{\prime}\cdot q+p\cdot q^{\prime}$.

Proof.

Let us write $p(x+y)=p(x)+y\,p^{\prime}(x)+y^{2}\,r(x,y)$ and $q(x+y)=q(x)+y\,q^{\prime}(y)+y^{2}\,s(x,y)$. Multiplying, grouping terms, and pulling out some common factors, we have

 $\displaystyle p(x+y)q(x+y)$ $\displaystyle=p(x)q(y)+y(p^{\prime}(x)q(x)+p(x)q^{\prime}(x))$ $\displaystyle+y^{2}(p(x)s(x,y)+q(x)r(x,y)+p^{\prime}(x)q^{\prime}(y)$ $\displaystyle\quad+y\,p^{\prime}(x)s(x,y)+y\,q^{\prime}(x)r(x,y)+y^{2}\,r(x,y)% s(x,y)).$

By definition of derivative, this means that $(p\cdot q)^{\prime}=p^{\prime}\cdot q+p\cdot q^{\prime}$. ∎

Theorem 4.

If $A$ is a commutative ring and $p,q\in A[x]$, then $(p\circ q)^{\prime}=(p^{\prime}\circ q)\cdot q^{\prime}$.

Proof.

Let us write $p(x+y)=p(x)+y\,p^{\prime}(x)+y^{2}\,r(x,y)$ and $q(x+y)=q(x)+y\,q^{\prime}(y)+y^{2}\,s(x,y)$. Composing, grouping terms, and pulling out some common factors, we have

 $\displaystyle p(q(x+y))$ $\displaystyle=p\left(q(x)+y\,q^{\prime}(y)+y^{2}\,s(x,y)\right)$ $\displaystyle=p(q(x))+\left(y\,q^{\prime}(y)+y^{2}\,s(x,y)\right)p^{\prime}(q(% x))$ $\displaystyle\quad+\left(y\,q^{\prime}(y)+y^{2}\,s(x,y)\right)^{2}r\left(q(x),% y\,q^{\prime}(y)+y^{2}\,s(x,y)\right)$ $\displaystyle=p(q(x))+y\,p^{\prime}(q(x))q^{\prime}(y)$ $\displaystyle\quad+y^{2}\left(s(x,y)p^{\prime}(q(x))+\left(q^{\prime}(y)+y\,s(% x,y)\right)^{2}r\left(q(x),y\,q^{\prime}(y)+y^{2}\,s(x,y)\right)\right)$

By definition of derivative, this means that $(p\circ q)^{\prime}=(p^{\prime}\circ q)\cdot q^{\prime}$. ∎

Title rules of calculus for derivative of polynomial RulesOfCalculusForDerivativeOfPolynomial 2013-03-22 18:20:05 2013-03-22 18:20:05 rspuzio (6075) rspuzio (6075) 10 rspuzio (6075) Derivation msc 13P05 msc 11C08 msc 12E05 ProofOfPropertiesOfDerivativesByPureAlgebra