Post on 15-Oct-2014
A polynomial p(x) of degree less than or equal to n is an interpolating polynomial for f(x) on the interval [x0,xn] if it satisfies the condition
NodesInterpolation points or Knots or Data Points
p(x) interpolates f(x)
iii yxpxf )()( ni ,...,2,1,0
nxxxx ,...,,, 210
),(),...,,(),,(),,( 221100 nn yxyxyxyx
Consider the following data points (0,-1), (1,2),(2,7) for a
function f(x). Let . Show that p(x) interpolates f(x).
Let f(x)=sinx. Show that the polynomial
interpolates f at nodes .
12)( 2 xxxp
xxxp44
)( 22
,2,0
where
Note:
Remark
n
iii xLyxp
0
)()(
))...()()...()((
))...()()...()((
)(
)()(
1110
1110
,0
niiiiii
nii
n
jij ji
ji
xxxxxxxxxx
xxxxxxxxxx
xx
xxxL
1)( ii xL jixL ji ,0)(
)( ii xfy
Using the nodes x0=2, x1=2.5, x2=4, find the second degree interpolating polynomial for the function f(x)=1/x. Use the interpolating polynomial to approximate f(3).
Since
Solve the unknowns by Gauss Jordan Elimination Method.
nnxaxaxaaxp ...)( 2
210
nixpxf ii ,...,2,1,0),()(
naaaa ,...,,, 210
Use the method of undetermined coefficients to find an interpolating polynomial for the function that passes through the points (-1,8),(0,5), (1,2) and (2,5).
33
2210)( xaxaxaaxp
Augmented Matrix
3
3210
45)(
1,0,4,5
xxxp
aaaa
Consider an interpolating polynomial of the form
NEWTON’S FORMDivide successively p(x) by the linear factors
Then work backwards to express it in
nnxaxaxaaxp ...)( 2
210
))...((...
))(()()(
10
102010
nn xxxxc
xxxxcxxccxp
))...((),(),( 1210 nxxxxxxxx
Find the Newton’s Form of the interpolating polynomial
obtained from the interpolation points (0,-1),(1,2) and (2,7).
Solution: We divide successively by (x-0) and (x-1). To get
12)( 2 xxxp
Working backwards, we write
The coefficients ck are called divided differences.
Zero-th Divided Differences are the corresponding function values of the nodes.
First Divided Differences wrt to and
Kth divided difference relative to nodes
ii
iiii xx
xfxfxxf
1
11
][][,
)(][ ii xfxf
ix 1ix
kiii xxx ,...,, 1
iki
kikiiikikiiikikiii xx
xxxxfxxxxfxxxxf
],,...,,[],,...,,[,,...,, 11111111
nk xxxxfc ,...,,, 210
When we have nodes
If nodes are re-ordered as , then we can write
the interpolating polynomial as
))...((,...,,,...
))((,,)(,)(
0210
102100100
nn xxxxxxxxf
xxxxxxxfxxxxfxfxp
nk xxxxfc ,...,,, 210
))...((,...,,,...
))((,,)(,)(
0210
1121
xxxxxxxxf
xxxxxxxfxxxxfxfxp
nn
nnnnnnnnn
nxxxx ...,,, 210
021 ...,,, xxxx nnn
Example: Take as the function to be interpolated, using the nodes
xxf 2log)(
.4,2,1 210 xxx
kkx kxf
0
1
2 4
2
1
1
2 2
1
0
1
6
1
)8)(1(6
1)2)(1(
6
1)1)(1(0)(2 xxxxxxp
1, kk xxf 21,, kkk xxxf
Table of Divided Difference Method
Example: Take as the function to be interpolated, using the nodes
xxf 2log)(
.4,2,1 210 xxx
kkx
0
1
2 4
2
1
1
01
22
16
1
)8)(1(6
1)2)(4(
6
1)4(
2
12)(2
xxxxxxp
kxf 1, kk xxf 21,, kkk xxxf
EXAMPLEGiven the divided difference table below, find
the interpolating using Newton’s interpolatory divided difference formula and use this to approximate f(1.5)
f(1.5)~p4(1.5)=.511820