Pincherle derivative

Short description: Type of derivative of a linear operator

In mathematics, the Pincherle derivative [math]\displaystyle{ T' }[/math] of a linear operator [math]\displaystyle{ T: \mathbb{K}[x] \to \mathbb{K}[x] }[/math] on the vector space of polynomials in the variable x over a field [math]\displaystyle{ \mathbb{K} }[/math] is the commutator of [math]\displaystyle{ T }[/math] with the multiplication by x in the algebra of endomorphisms [math]\displaystyle{ \operatorname{End}(\mathbb{K}[x]) }[/math]. That is, [math]\displaystyle{ T' }[/math] is another linear operator [math]\displaystyle{ T': \mathbb{K}[x] \to \mathbb{K}[x] }[/math]

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

(for the origin of the [math]\displaystyle{ \operatorname{ad} }[/math] notation, see the article on the adjoint representation) so that

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

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 [math]\displaystyle{ S }[/math] and [math]\displaystyle{ T }[/math] belonging to [math]\displaystyle{ \operatorname{End}\left( \mathbb{K}[x] \right), }[/math]

1. [math]\displaystyle{ (T + S)^\prime = T^\prime + S^\prime }[/math];
2. [math]\displaystyle{ (TS)^\prime = T^\prime\!S + TS^\prime }[/math] where [math]\displaystyle{ TS = T \circ S }[/math] is the composition of operators.

One also has [math]\displaystyle{ [T,S]^{\prime} = [T^{\prime}, S] + [T, S^{\prime}] }[/math] where [math]\displaystyle{ [T,S] = TS - ST }[/math] 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

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

This formula generalizes to

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

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

[math]\displaystyle{ \partial = \sum a_n {{d^n} \over {dx^n} } = \sum a_n D^n }[/math]

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

When [math]\displaystyle{ \mathbb{K} }[/math] has characteristic zero, the shift operator

[math]\displaystyle{ S_h(f)(x) = f(x+h) \, }[/math]

can be written as

[math]\displaystyle{ S_h = \sum_{n \ge 0} {{h^n} \over {n!} }D^n }[/math]

by the Taylor formula. Its Pincherle derivative is then

[math]\displaystyle{ S_h' = \sum_{n \ge 1} {{h^n} \over {(n-1)!} }D^{n-1} = h \cdot S_h. }[/math]

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

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

The "discrete-time delta operator"

[math]\displaystyle{ (\delta f)(x) = {{ f(x+h) - f(x) } \over h } }[/math]

is the operator

[math]\displaystyle{ \delta = {1 \over h} (S_h - 1), }[/math]

whose Pincherle derivative is the shift operator [math]\displaystyle{ \delta' = S_h }[/math].

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.