View
12
Download
0
Category
Preview:
Citation preview
3.3 Divided Differences
1
Representing πth Lagrange Polynomial
β’ If ππ π₯ is the πth degree Lagrange interpolating polynomial that agrees with π π₯ at the points {π₯0, π₯1, β¦ , π₯π}, express ππ π₯ in the form: ππ π₯ =π0 + π1 π₯ β π₯0 + π2 π₯ β π₯0 π₯ β π₯1 +β―+ππ π₯ β π₯0 π₯ β π₯1 π₯ β π₯2 β¦ π₯ β π₯πβ1
β’ ? How to find constants π0,β¦, ππ?
2
Divided Differences
β’ Zeroth divided difference: π π₯π = π(π₯π)
β’ First divided difference:
π π₯π , π₯π+1 =π π₯π+1 β π π₯ππ₯π+1 β π₯π
β’ Second divided difference:
π π₯π , π₯π+1, π₯π+2 =π π₯π+1, π₯π+2 β π π₯π , π₯π+1
π₯π+2 β π₯π
β’ πth divided difference: π π₯π , π₯π+1, β¦ , π₯π+π
=π π₯π+1, π₯π+2, β¦ , π₯π+π β π π₯π , π₯π+1, β¦ , π₯π+πβ1
π₯π+π β π₯π
3
β’ Finding constants π0,β¦, ππ.
1. π₯ = π₯0: π0 = ππ π₯0 = π π₯0 = π π₯0
2. π₯ = π₯1: π π₯0 + π1 π₯1 β π₯0 = ππ π₯1 =π π₯1
βΉ π1 =π π₯1 β π π₯0π₯1 β π₯0
= π π₯0, π₯1
3. In general: ππ = π π₯0, π₯1, β¦ , π₯π for π = 0,β¦ , π
4
Newtonβs Interpolatory Divided Difference Formula
ππ π₯= π π₯0 + π π₯0, π₯1 π₯ β π₯0+ π π₯0, π₯1, π₯2 π₯ β π₯0 π₯ β π₯1 +β―+ π π₯0, β¦ , π₯π π₯ β π₯0 π₯ β π₯1 β¦(π₯ β π₯πβ1)
Or ππ π₯= π π₯0
+ [π[π₯0, β¦ , π₯π] π₯ β π₯0 β¦(π₯ β π₯πβ1)]
π
π=1
5
Table for Computing
6
β¦
Algorithm: Newtonβs Divided Differences
Input: π₯0, π π₯0 , π₯1, π π₯1 , β¦ , π₯π, π π₯π Output: Divided differences πΉ0,0, β¦ , πΉπ,π //comment: ππ π₯ = πΉ0,0 + [πΉπ,π π₯ β π₯0 β¦(π₯ β π₯πβ1)]
ππ=1
Step 1: For π = 0,β¦ , π set πΉπ,0 = π(π₯π) Step 2: For π = 1,β¦ , π For π = 1,β¦ , π
set πΉπ,π =πΉπ,πβ1βπΉπβ1,πβ1
π₯πβπ₯πβπ
End End Output(πΉ0,0,β¦πΉπ,π,β¦,πΉπ,π) STOP. 7
Theorem. Suppose that π β πΆπ[π, π] and π₯0, π₯1, β¦ , π₯π are distinct numbers in [π, π]. Then
βπ β π, π with π π₯0, β¦ , π₯π =π π (π)
π!.
Remark: When π = 1, itβs just the Mean Value Theorem.
8
Example. 1) Complete the following divided difference table. 2) Find the interpolating polynomial.
9
Recommended