Discrete Splines and Vision - Bezier and Splines in Image Processing and Machine Vision - page 144

m

ζ k = Ψ j,k ( y )[( x − t j ) Q j,k− 1 ( x )+( t j + k − x ) Q j +1 ,k− 1 ( x )] .

j =1

Since x [ t k ,t m +1 ), we have Q 1 ,k− 1 ( x )= Q m +1 ,k− 1 ( x ) = 0. Hence, ζ k can

be written as

∈

m

ζ k = γ j,k ( x, y ) Q j,k− 1 ( x ) , (7.21)

j =2

where

γ j,k ( x, y )= Ψ j,k ( y )( x − t j )+ Ψ j− 1 ,k ( y )( t j + k− 1 − x ) .

But it can be shown in a straightforward way that

γ j,k ( x, y )=( y − x )( t j + k− 1 − t j ) Ψ j,k− 1 ( y ) .

Therefore, ζ k can be written as

m

ζ k =( y − x ) Ψ j,k− 1 ( y )( t j + k− 1 − t j ) Q j,k− 1 ( x ) .

j =2

Since, ( t j + k− 1 −

t j ) Q j,k− 1 ( x )= N j,k− 1 ( x )and N 1 ,k− 1 ( x ) = 0, we get ζ k =

( y − x ) ζ k− 1 (from equation (7.18)). But then ζ k =( y − x ) k− 1 ζ 1 . since ζ 1 =1,

equation (7.18), follows.
Lemma 2 :
Let φ j and a j be as in Theorem 1. For any y ∈ t and any x ∈ [ t k ,t m +1 ),

m

x ) k− 1
+

( y − = φ j,k ( y ) N j,k ( x ) . (7.22)

j =1

Proof :
Let us fix x and μ be such that t μ ≤ x<t μ +1 . Since N j,k ( x )=0for x/ ∈

[ t j ,t j + k ), we have to show that

μ

de =

x ) k− 1
+

( y − = σ k φ j,k ( y ) N j,k ( x ) . (7.23)

j = μ − k +1

Assume y = t μ . Since, φ j,k ( t μ ) contains a factor t μ − t μ for j = μ − k +1 , ··· ,μ −

1, we have σ k = φ μ,k ( t μ ) N μ,k ( x ). But φ μ,k ( t μ ) = 0 since a μ ∈ [ t μ ,t μ + k ).

x ) k− + and equation (7.23) follows in this case. Sim-
ilarly, if y = t μ− 1 , then σ k = φ μ− 1 ,k ( t μ− 1 ) N μ− 1 ,k ( x )+ φ μ,k ( t μ− 1 ) N μ,k ( x )=
0=( t μ− 1 −

Hence, σ k =0=( t μ −

x ) k− + . Continuing in this way, we see that equation (7.23)
holds for y = t s and s μ . Next let us assume y = t μ +1 . Then σ k =

φ μ−k +1 ,k ( t μ +1 ) N μ−k +1 ,k ( x ). But φ μ−k +1 ,k ( t μ +1 )= Ψ μ−k +1 ,k ( t μ +1 ) and equa-
tion (7.23) follows from equation (7.18). Similarly, equation (7.23) follows from
equation (7.18) for y = t s and s

≤

t μ +1 .
We shall now turn to the proof of Theorem 1.

≥