3.3 Divided Differences
1
Representing ππth Lagrange Polynomial
β’ If ππππ π₯π₯ is the ππth degree Lagrange interpolating
polynomial that agrees with ππ π₯π₯ at the points
{π₯π₯0 , π₯π₯1 , β¦ , π₯π₯ππ }, ππππ π₯π₯ can be expressed in the form:
ππππ π₯π₯ = ππ0 + ππ1 π₯π₯ β π₯π₯0 +
ππ2 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 +
ππ3 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 π₯π₯ β π₯π₯2 +
β¦ + ππππ π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 π₯π₯ β π₯π₯2 β¦ π₯π₯ β π₯π₯ππβ1
β’ ? How to find constants ππ0 ,β¦, ππππ ?
2
Finding constants ππ0 ,β¦, ππππ
Given interpolating polynomial ππππ π₯π₯ = ππ0 + ππ1 π₯π₯ β π₯π₯0 +
ππ2 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 +
ππ3 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 π₯π₯ β π₯π₯2 +
β¦ + ππππ π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 π₯π₯ β π₯π₯2 β¦ π₯π₯ β π₯π₯ππβ1
ο At π₯π₯0 : ππ0 = ππππ π₯π₯0 = ππ π₯π₯0
ο At π₯π₯1 : ππ π₯π₯0 + ππ1 π₯π₯1 β π₯π₯0 = ππππ π₯π₯1 = ππ(π₯π₯1 )
ο At π₯π₯2 : ππ π₯π₯0 +
ππ π₯π₯1 βππ π₯π₯0
π₯π₯1 βπ₯π₯0
ππππ π₯π₯2 = ππ(π₯π₯2 )
οβ¦
ππ π₯π₯1 β ππ π₯π₯0
βΉ ππ1 =
π₯π₯1 β π₯π₯0
π₯π₯2 β π₯π₯0 + ππ2 π₯π₯2 β π₯π₯0 π₯π₯2 β π₯π₯1 =
ππ π₯π₯2 β ππ π₯π₯1
ππ π₯π₯1 β ππ π₯π₯0
β
π₯π₯2 β π₯π₯1
π₯π₯1 β π₯π₯0
βΉ ππ2 =
π₯π₯2 β π₯π₯0
3
3
Newtonβs Divided Difference
οΆ Zeroth divided difference:
ππ π₯π₯ππ = ππ π₯π₯ππ .
οΆ First divided difference:
ππ π₯π₯ππ , π₯π₯ππ+1
ππ[π₯π₯ππ+1 ] β ππ[π₯π₯ππ ]
=
.
π₯π₯ππ+1 β π₯π₯ππ
οΆ Second divided difference:
ππ π₯π₯ππ , π₯π₯ππ+1 , π₯π₯ππ+2
ππ[π₯π₯ππ+1 , π₯π₯ππ+2 ] β ππ[π₯π₯ππ , π₯π₯ππ+1 ]
=
.
π₯π₯ππ+2 β π₯π₯ππ
οΆ Third divided difference:
ππ π₯π₯ππ , π₯π₯ππ+1 , π₯π₯ππ+2 , π₯π₯ππ+3
ππ[π₯π₯ππ+1 , π₯π₯ππ+2 , π₯π₯ππ+3 ] β ππ[π₯π₯ππ , π₯π₯ππ+1 , π₯π₯ππ+2 ]
=
.
π₯π₯ππ+3 β π₯π₯ππ
οΆ Kth divided difference:
ππ π₯π₯ππ , π₯π₯ππ+1 , β¦ , π₯π₯ππ+ππ
ππ[π₯π₯ππ+1 , π₯π₯ππ+2 , β¦ , π₯π₯ππ+ππ ] β ππ[π₯π₯ππ , π₯π₯ππ+1 , β¦ , π₯π₯ππ+ππβ1 ]
=
π₯π₯ππ+ππ β π₯π₯ππ
4
Finding constants ππ0 ,β¦, ππππ -revisited
Given interpolating polynomial ππππ π₯π₯ = ππ0 + ππ1 π₯π₯ β π₯π₯0 +
ππ2 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 +
ππ3 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 π₯π₯ β π₯π₯2 +
β¦ + ππππ π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 π₯π₯ β π₯π₯2 β¦ π₯π₯ β π₯π₯ππβ1
ο ππ0 = ππ(π₯π₯0 ) = ππ[π₯π₯0 ]
ο ππ1 =
ππ π₯π₯1 βππ π₯π₯0
π₯π₯1 βπ₯π₯0
ο ππ2 =
ο ππ3 =
ππ π₯π₯2 βππ π₯π₯1
π₯π₯2 βπ₯π₯1
=
ππ[π₯π₯1 ]βππ[π₯π₯0 ]
π₯π₯1 βπ₯π₯0
ππ π₯π₯1 βππ π₯π₯0
π₯π₯1 βπ₯π₯0
β
π₯π₯2 βπ₯π₯0
ππ π₯π₯1 ,π₯π₯2 ,π₯π₯3 βππ π₯π₯0 ,π₯π₯1 ,π₯π₯2
π₯π₯3 βπ₯π₯0
ο ππππ = ππ π₯π₯0 , π₯π₯1 , β¦ , π₯π₯ππ .
=
= ππ[π₯π₯0 , π₯π₯1 ].
ππ[π₯π₯1 ,π₯π₯2 ]βππ[π₯π₯0 ,π₯π₯1 ]
π₯π₯2 βπ₯π₯0
= ππ π₯π₯0 , π₯π₯1 , π₯π₯2 , π₯π₯3 .
= ππ π₯π₯0 , π₯π₯1 , π₯π₯2 .
5
Interpolating Polynomial Using Newtonβs Divided
Difference Formula
ππππ π₯π₯
= ππ π₯π₯0 + ππ π₯π₯0 , π₯π₯1 π₯π₯ β π₯π₯0
+ ππ π₯π₯0 , π₯π₯1 , π₯π₯2 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 + β―
+ ππ π₯π₯0 , β¦ , π₯π₯ππ π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 β¦ (π₯π₯ β π₯π₯ππβ1 )
Or
ππππ π₯π₯
ππ
= ππ π₯π₯0 + οΏ½ [ππ[π₯π₯0 , β¦ , π₯π₯ππ ] π₯π₯ β π₯π₯0 β¦ (π₯π₯ β π₯π₯ππβ1 )]
ππ=1
Remark: ππππ = ππ π₯π₯0 , π₯π₯1 , β¦ , π₯π₯ππ for ππ = 0, β¦ , ππ
6
Example 3.3.1 Use the data in the table to construct
interpolating polynomial.
7
Table for Computing
β¦
8
Theorem 3.6 Suppose that ππ β πΆπΆ ππ [ππ, ππ] and
π₯π₯0 , π₯π₯1 , β¦ , π₯π₯ππ are distinct numbers in [ππ, ππ]. Then
βππ β ππ, ππ with ππ π₯π₯0 , β¦ , π₯π₯ππ =
ππ ππ (ππ)
.
ππ!
Remark: When ππ = 1, itβs just the Mean Value
Theorem.
9
Illustration. 1) Complete the following divided difference
table. 2) Find the interpolating polynomial.
10
Algorithm: Newtonβs Divided Differences
Input: π₯π₯0 , ππ π₯π₯0 , π₯π₯1 , ππ π₯π₯1 , β¦ , π₯π₯ππ , ππ π₯π₯ππ
Output: Divided differences πΉπΉ0,0 , β¦ , πΉπΉππ,ππ
//comment: ππππ π₯π₯ = πΉπΉ0,0 + βππππ=1[πΉπΉππ,ππ π₯π₯ β π₯π₯0 β¦ (π₯π₯ β π₯π₯ππβ1 )]
Step 1: For ππ = 0, β¦ , ππ
set πΉπΉππ,0 = ππ(π₯π₯ππ )
Step 2: For ππ = 1, β¦ , ππ
For ππ = 1, β¦ , ππ
set πΉπΉππ,ππ =
πΉπΉππ,ππβ1 βπΉπΉππβ1,ππβ1
π₯π₯ππ βπ₯π₯ππβππ
End
End
Output(πΉπΉ0,0 ,β¦πΉπΉππ,ππ ,β¦,πΉπΉππ,ππ )
STOP.
11
Forward difference formula for equally spaced nodes
β’ Let the points π₯π₯0 , π₯π₯1 , β¦ , π₯π₯ππ be equally spaced. β = π₯π₯ππ+1 β
π₯π₯ππ , for each ππ = 0, β¦ , ππ β 1;
and π₯π₯ = π₯π₯0 + π π π .
β’ Then
ππππ π₯π₯
= ππ π₯π₯0 + ππ π₯π₯0 , π₯π₯1 π₯π₯ β π₯π₯0 + ππ π₯π₯0 , π₯π₯1 , π₯π₯2 π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1
+ β― + ππ π₯π₯0 , β¦ , π₯π₯ππ π₯π₯ β π₯π₯0 π₯π₯ β π₯π₯1 β¦ π₯π₯ β π₯π₯ππβ1
= ππ π₯π₯0 + π π π π π π₯π₯0 , π₯π₯1 + π π π π β 1 β2 ππ π₯π₯0 , π₯π₯1 , π₯π₯2 + β―
+ π π π π β 1 β¦ (π π β ππ + 1)βππ ππ π₯π₯0 , β¦ , π₯π₯ππ
Or
ππππ π₯π₯ = ππππ π₯π₯0 + π π π = ππ π₯π₯0
π π
π π
Where
=
ππ
π π β1 β¦(π π βππ+1)
ππ!
ππ
π π
+οΏ½
ππ! βππ ππ π₯π₯0 , π₯π₯1 , β¦ , π₯π₯ππ
ππ
ππ=1
12
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