Pincherle derivative

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]

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

so that

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

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 Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle S

and Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle  T 
belonging to Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.):  \scriptstyle \operatorname{End} \left( \mathbb K[x] \right) 

1. Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ (T + S)^\prime = T^\prime + S^\prime }

;

1. Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ (TS)^\prime = T^\prime\!S + TS^\prime }

where Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ TS = T \circ S}
is the composition of operators ;

One also has Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ [T,S]^\prime = [T^\prime , S] + [T, S^\prime ] }

where Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ [T,S] = TS - ST}
is the usual Lie bracket, which follows from the Jacobi identity.

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

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

This formula generalizes to

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

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

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

is also a differential operator, so that the Pincherle derivative is a derivation of Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle \operatorname{Diff}(\mathbb K [x]) .

The shift operator

$S_h(f)(x) = f(x+h) \,$

can be written as

$S_h = \sum_{n=0} {{h^n} \over {n!} }D^n$

by the Taylor formula. Its Pincherle derivative is then

$S_h' = \sum_{n=1} {{h^n} \over {(n-1)!} }D^{n-1} = h \cdot S_h.$

In other words, the shift operators are eigenvectors of the Pincherle derivative, whose spectrum is the whole space of scalars Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ \mathbb K } .

If T is shift-equivariant, that is, if T commutes with S_h or Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ [T,S_h] = 0} , then we also have Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ [T',S_h] = 0} , so that $\scriptstyle T'$ is also shift-equivariant and for the same shift $\scriptstyle h$.

The "discrete-time delta operator"

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

is the operator

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

whose Pincherle derivative is the shift operator Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \scriptstyle{ \delta ' = S_h } .

See also

• Commutator
• Delta operator
• Umbral calculus

External links

• Weisstein, Eric W. "Pincherle Derivative". From MathWorld—A Wolfram Web Resource.
• Biography of Salvatore Pincherle at the MacTutor History of Mathematics archive.