# Pincherle derivative

{{ safesubst:#invoke:Unsubst||$N=Unreferenced |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} In mathematics, the Pincherle derivative T’ of a linear operator T:K[x] → K[x] on the vector space of polynomials in the variable x over a field K is the commutator of T with the multiplication by x in the algebra of endomorphisms End(K[x]). That is, T’ is another linear operator T’:K[x] → K[x]

${\displaystyle T':=[T,x]=Tx-xT=-\operatorname {ad} (x)T,\,}$

so that

${\displaystyle T'\{p(x)\}=T\{xp(x)\}-xT\{p(x)\}\qquad \forall p(x)\in \mathbb {K} [x].}$

This concept is named after the Italian mathematician Salvatore Pincherle (1853–1936).

## Properties

The Pincherle derivative, like any commutator, is a derivation, meaning it satisfies the sum and products rules: given two linear operators ${\displaystyle \scriptstyle S}$ and ${\displaystyle \scriptstyle T}$ belonging to ${\displaystyle \scriptstyle \operatorname {End} \left(\mathbb {K} [x]\right)}$

The usual derivative, D = d/dx, is an operator on polynomials. By straightforward computation, its Pincherle derivative is

${\displaystyle D'=\left({d \over {dx}}\right)'=\operatorname {Id} _{\mathbb {K} [x]}=1.}$

This formula generalizes to

${\displaystyle (D^{n})'=\left({{d^{n}} \over {dx^{n}}}\right)'=nD^{n-1},}$

by induction. It proves that the Pincherle derivative of a differential operator

${\displaystyle \partial =\sum a_{n}{{d^{n}} \over {dx^{n}}}=\sum a_{n}D^{n}}$

is also a differential operator, so that the Pincherle derivative is a derivation of ${\displaystyle \scriptstyle \operatorname {Diff} (\mathbb {K} [x])}$.

The shift operator

${\displaystyle S_{h}(f)(x)=f(x+h)\,}$

can be written as

${\displaystyle S_{h}=\sum _{n=0}{{h^{n}} \over {n!}}D^{n}}$

by the Taylor formula. Its Pincherle derivative is then

${\displaystyle S_{h}'=\sum _{n=1}{{h^{n}} \over {(n-1)!}}D^{n-1}=h\cdot S_{h}.}$

In other words, the shift operators are eigenvectors of the Pincherle derivative, whose spectrum is the whole space of scalars ${\displaystyle \scriptstyle {\mathbb {K} }}$.

If T is shift-equivariant, that is, if T commutes with Sh or ${\displaystyle \scriptstyle {[T,S_{h}]=0}}$, then we also have ${\displaystyle \scriptstyle {[T',S_{h}]=0}}$, so that ${\displaystyle \scriptstyle T'}$ is also shift-equivariant and for the same shift ${\displaystyle \scriptstyle h}$.

The "discrete-time delta operator"

${\displaystyle (\delta f)(x)={{f(x+h)-f(x)} \over h}}$

is the operator

${\displaystyle \delta ={1 \over h}(S_{h}-1),}$

whose Pincherle derivative is the shift operator ${\displaystyle \scriptstyle {\delta '=S_{h}}}$.