Exercise 1.4.9 (Finite differences)

Let f ( x ) be a function of a real variable, and let Δ f be the function Δ f ( x ) = f ( x + 1 ) − f ( x ) . For k > 1 , put Δ k f = Δ ( Δ k − 1 f ) . The function Δ k f ( x ) is called the k th forward difference of f . Show that

Δ k f ( x ) = ∑ j = 0 k ( − 1 ) j ( k j ) f ( x + k − j ) .

Answers

Proof.

Let f = R → ℝ be a function of a real variable.

The fonction Δ f is defined by

Δ f ( x ) = f ( x + 1 ) − f ( x ) ( x ∈ ℝ ) .

The function Δ k f is defined inductively by

Δ 0 f = f , Δ k f = Δ ( Δ k − 1 f ) ( k ≥ 1 ) .

Consider the proposition defined for k ∈ ℕ by

𝒫 ( k ) ⟺ ∀ ⁡ x ∈ ℝ , Δ k f ( x ) = ∑ j = 0 k ( − 1 ) j ( k j ) f ( x + k − j ) .

∙ If k = 0 , Δ 0 f = f = ( − 1 ) 0 ( 0 0 ) f ( x ) = ∑ ⁡ j = 0 0 ( − 1 ) j ( k j ) f ( x + k − j ) , thus 𝒫 ( 0 ) is true.

∙ Assume that 𝒫 ( k ) is true for some integer k ∈ ℕ . Then, for all x ∈ ℝ ,

Δ k + 1 f ( x ) = Δ k f ) ( x + 1 ) − Δ k f ( x ) = ∑ j = 0 k ( − 1 ) j ( k j ) f ( x + 1 + k − j ) − ∑ j = 0 k ( − 1 ) j ( k j ) f ( x + k − j ) = f ( x + k + 1 ) + ∑ j = 1 k ( − 1 ) j ( k j ) f ( x + 1 + k − j ) − ∑ j = 0 k − 1 ( − 1 ) j ( k j ) f ( x + k − j ) − ( − 1 ) k f ( x ) = f ( x + k + 1 ) + ∑ j = 1 k ( − 1 ) j ( k j ) f ( x + 1 + k − j ) − ∑ i = 1 k ( − 1 ) i − 1 ( k i − 1 ) f ( x + k − i + 1 ) − ( − 1 ) k f ( x ) = f ( x + k + 1 ) + ∑ j = 1 k ( − 1 ) j ( k j ) f ( x + 1 + k − j ) + ∑ j = 1 k ( − 1 ) j ( k j − 1 ) f ( x + 1 + k − j ) + ( − 1 ) k + 1 f ( x ) = f ( x + k + 1 ) + ∑ j = 1 k ( − 1 ) j [ ( k j ) + ( k j − 1 ) ] f ( x + 1 + k − j ) + ( − 1 ) k + 1 f ( x ) = f ( x + k + 1 ) + ∑ j = 1 k ( − 1 ) j ( k + 1 j ) f ( x + 1 + k − j ) + ( − 1 ) k + 1 f ( x ) = ∑ j = 0 k + 1 ( − 1 ) j ( k + 1 j ) f ( x + k + 1 − j )

This gives 𝒫 ( k + 1 ) . The induction is done: so

for all k ∈ ℕ , and for all x ∈ ℝ ,

Δ k f ( x ) = ∑ j = 0 k ( − 1 ) j ( k j ) f ( x + k − j ) .

□

User profile picture
2024-07-26 10:34
Comments