rules of calculus for derivative of formal power series

In this entry, we will show that the rules of calculushold for derivatives of formal power series. Whilethis could be verified directly in a manner analogousto what was done for polynomials in the parent entry,we will take a different tack, deriving the resultsfor power series from the corresponding results forpolynomials. The basis for our approach is theobservation that the ring of formal power series canbe expressed as a limit of quotients of the ring ofpolynomials:

$$A[[x]]=\underset{k\to \mathrm{\infty }}{lim}A[x]/⟨x^k⟩$$

Thus, we will proceed in two steps, first extending thederivative operation to the quotient rings and showing thatits properties still hold there, then extending it to thelimit and showing that its properties hold there as well.

We begin by noting that the derivative is well-definedas a map from $A[x]/⟨x^{k+1}⟩$ to$A[x]/⟨x^k⟩$ for all integers $k\ge 0$.

Theorem 1.

Suppose that $A$ is a commutative ring, $k$ is a non-negativeinteger, and that $p$ and $q$ are elements of $A\mathit{}\mathrm{[}x\mathrm{]}$ suchthat $p\mathrm{\equiv }q$ modulo $x^{k\mathrm{+}\mathrm{1}}$. Then $p^{\mathrm{\prime }}\mathrm{\equiv }{q}^{\mathrm{\prime }}$modulo $x^k$.

Proof.

By definition of congruence, $p(x)=q(x)+x^{k+1}r(x)$ forsome polynomial $r\in A[x]$. Taking derivaitves, $p^{\prime }(x)=q^{\prime }(x)+x^k(kr(x)+x{r}^{\prime }(x))$, so $p^{\prime }$ and $q^{\prime }$ areequivalent modulo $x^k$.∎

It is easy to verify that the sum and product rules holdin this new context:

Theorem 2.

If $A$ is a commutative ring, $k$ is a non-negative integer,and $f\mathrm{,}g$ are elements of $A\mathit{}\mathrm{[}x\mathrm{]}\mathrm{/}\mathrm{⟨}{x}^{k\mathrm{+}\mathrm{1}}\mathrm{⟩}$,then ${\mathrm{(}f\mathrm{+}g\mathrm{)}}^{\mathrm{\prime }}\mathrm{=}{f}^{\mathrm{\prime }}\mathrm{+}{g}^{\mathrm{\prime }}$.

Proof.

Let $p,q$ be representatives of the equivalence classes$f,g$. Then we have ${(p+q)}^{\prime }=p^{\prime }+q^{\prime }$ by the correspondingtheorem for polynomials. Hence, by definition of quotient,we have ${(f+g)}^{\prime }=f^{\prime }+g^{\prime }$.∎

Theorem 3.

If $A$ is a commutative ring, $k$ is a non-negative integer,and $f\mathrm{,}g$ are elements of $A\mathit{}\mathrm{[}x\mathrm{]}\mathrm{/}\mathrm{⟨}{x}^{k\mathrm{+}\mathrm{1}}\mathrm{⟩}$,then ${\mathrm{(}f\mathrm{\cdot }g\mathrm{)}}^{\mathrm{\prime }}\mathrm{=}{f}^{\mathrm{\prime }}\mathrm{\cdot }g\mathrm{+}f\mathrm{\cdot }{g}^{\mathrm{\prime }}$.

Proof.

Let $p,q$ be representatives of the equivalence classes$f,g$. Then we have ${(p\cdot q)}^{\prime }=p^{\prime }\cdot q+p\cdot q^{\prime }$by the corresponding theorem for polynomials. Hence, bydefinition of quotient, we have${(f\cdot g)}^{\prime }=f^{\prime }\cdot g+f\cdot g^{\prime }$.∎

When considering the chain rule, we need to note that compositiondoes not always pass to the quotient, so we need to restrict theoperands to obtain a well-defined operation. In particular, wewill consider the following two cases:

Theorem 4.

If $A$ is a commutative ring, $p\mathrm{,}q\mathrm{,}r$ is a n element of $A\mathit{}\mathrm{[}x\mathrm{]}$,and $q\mathrm{\equiv }r$ modulo $x^k$ for some integer $k\mathrm{>}\mathrm{0}$, then$p\mathrm{\circ }q\mathrm{\equiv }p\mathrm{\circ }r$ modulo $x^k$.

Theorem 5.

If $A$ is a commutative ring, $k$ is a non-negative integer, and$p\mathrm{,}q\mathrm{,}P\mathrm{,}Q$ are elements of $A\mathit{}\mathrm{[}x\mathrm{]}$ such that $p\mathrm{\equiv }P$ modulo$x^k$, $q\mathrm{\equiv }Q$ modulo $x^k$ and $p\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{0}$, then $p\mathrm{\circ }q\mathrm{\equiv }P\mathrm{\circ }Q$ modulo $x^k$.

[More to come]