Given Information:
The provided data are,
x |
0 |
1 |
2 |
5.5 |
11 |
13 |
16 |
18 |
y |
0.5 |
3.134 |
5.3 |
9.9 |
10.2 |
9.35 |
7.2 |
6.2 |
Formula used:
The zero-order Newton’s interpolation formula:
f0(x)=b0
The first-order Newton’s interpolation formula:
f1(x)=b0+b1(x−x0)
The second- order Newton’s interpolating polynomial is given by,
f2(x)=b0+b1(x−x0)+b2(x−x0)(x−x1)
The n th-order Newton’s interpolating polynomial is given by,
fn(x)=b0+b1(x−x0)+b2(x−x0)(x−x1)+⋯+bn(x−x0)(x−x1)⋯(x−xn−1)
Where,
b0=f(x0)b1=f[x1,x0]b2=f[x2,x1,x0]⋮b2=f[xn,⋯,x2,x1,x0]
The first finite divided difference is,
f[xi,xj]=f(xi)−f(xj)xi−xj
And, the n th finite divided difference is,
f[xn,xn−1,...,x1,x0]=f[xn,xn−1,...,x1]−f[xn−1,...,x1,x0]xn−x0
Calculation:
First, order the provided value as close to 8 as below,
x0=5.5,x1=11,x2=13,x3=2,x4=1,x5=16,x6=0 and x6=18.
Therefore,
f(x0)=9.9f(x1)=10.2f(x2)=9.35f(x3)=5.3
And,
f(x4)=3.134f(x5)=7.2f(x6)=0.5f(x7)=6.2
The first divided difference is,
f[x1,x0]=f(x1)−f(x0)x1−x0=10.2−9.911−5.5=0.35.5=0.054545
And,
f[x2,x1]=f(x2)−f(x1)x2−x1=9.35−10.213−11=−0.852=−0.425
And,
f[x3,x2]=f(x3)−f(x2)x3−x2=5.3−9.352−13=−4.05−11=0.368182
Similarly,
f[x4,x3]=2.166f[x5,x4]=0.271067f[x6,x5]=0.41875f[x7,x6]=0.31667
The second divided difference is,
f[x2,x1,x0]=f[x2,x1]−f[x1,x0]x2−x0=−0.425−0.05454513−5.5=−0.4795457.5=−0.06394
And,
f[x3,x2,x1]=f[x3,x2]−f[x2,x1]x3−x1=0.368182−(−0.425)2−11=0.793182−9=−0.08813
And,
f[x4,x3,x2]=f[x4,x3]−f[x3,x2]x4−x2=2.166−0.3681821−13=1.797818−12=−0.14982
Similarly,
f[x5,x4,x3]=−0.13535f[x6,x5,x4]=−0.147683f[x7,x6,x5]=−0.05104
The third divided difference is,
f[x3,x2,x1,x0]=f[x3,x2,x1]−f[x2,x1,x0]x3−x0=(−0.08813)−(−0.06394)2−5.5=−0.02419−3.5=0.00691143
And,
f[x4,x3,x2,x1]=f[x4,x3,x2]−f[x3,x2,x1]x4−x1=(−0.14982)−(−0.08813)1−11=−0.06169−10=0.006169
Similarly,
f[x5,x4,x3,x2]=0.004822f[x6,x5,x4,x3]=0.0061665f[x7,x6,x5,x4]=0.005513
The fourth divided difference is,
f[x4,x3,x2,x1,x0]=f[x4,x3,x2,x1]−f[x3,x2,x1,x0]x4−x0=(0.006169)−(0.00691143)1−5.5=−0.00074243−4.5=0.000165
And,
f[x5,x4,x3,x2,x1]=f[x5,x4,x3,x2]−f[x4,x3,x2,x1]x5−x1=(0.004822)−(0.006169)16−11=−0.0013475=−0.0002694
And,
f[x6,x5,x4,x3,x2]=f[x6,x5,x4,x3]−f[x5,x4,x3,x2]x6−x2=(0.0061665)−(0.004822)0−13=0.0013445−13=0.0001034
And,
f[x7,x6,x5,x4,x3]=f[x7,x6,x5,x4]−f[x6,x5,x4,x3]x7−x3=(0.005513)−(0.0061665)18−2=−0.000653516=−0.00004084
The fifth divided difference is,
f[x5,x4,x3,x2,x1,x0]=f[x5,x4,x3,x2,x1]−f[x4,x3,x2,x1,x0]x5−x0=(−0.0002694)−(0.000165)16−5.5=−0.000434410.5=−0.0000414
And,
f[x6,x5,x4,x3,x2,x1]=f[x6,x5,x4,x3,x2]−f[x5,x4,x3,x2,x1,x0]x6−x1=(0.0001034)−(0.0002694)0−11=−0.000166−11=0.0000151
And,
f[x7,x6,x5,x4,x3,x2]=f[x7,x6,x5,x4,x3]−f[x6,x5,x4,x3,x2]x7−x2=(−0.00004048)−(0.0001034)18−13=−0.000143885=0.000028776
The sixth divided difference is,
f[x6,x5,x4,x3,x2,x1,x0]=f[x6,x5,x4,x3,x2,x1]−f[x5,x4,x3,x2,x1,x0]x6−x0=(0.0000151)−(−0.0000414)0−5.5=0.0000565−5.5=−0.0000103
And,
f[x7,x6,x5,x4,x3,x2,x1]=f[x7,x6,x5,x4,x3,x2]−f[x6,x5,x4,x3,x2,x1]x7−x1=(0.0000288776)−(0.0000151)18−11=0.000013787=0.000002
The seventh divided difference is,
f[x7,x6,x5,x4,x3,x2,x1,x0]=f[x7,x6,x5,x4,x3,x2,x1]−f[x6,x5,x4,x3,x2,x1,x0]x7−x0=(0.000002)−(−0.0000103)18−5.5=0.000012312.5=0.000001
Therefore, the difference table can be summarized as,
i
|
xi
|
f(xi)
|
First |
Second |
Third |
Fourth |
Fifth |
Sixth |
7th |
0 1 2 3 4 5 6 7 |
5.5 11 13 2 1 16 0 18
|
9.9 10.2 9.35 5.3 3.134 7.2 0.5 6.2
|
0.054545-0.425 0.3682 2.166 0.2711 0.4188 0.3167 |
−0.06394
−0.08813
−0.14982
−0.13535
−0.14768
−0.05104
|
0.0069 0.0062 0.0048 0.0062 0.0055 |
0.0002
−0.0003 0.0001
−0.00004
|
−0.00004 0.00002 0.00003
|
−0.00001 0.000002 |
0.00 |
Since, the divided difference of fifth order is nearly equals to zero. So, the fourth-order polynomial is the optimal.
Therefore, the zero-order Newton’s interpolation polynomial is,
f0(x)=9.9
Thus, the value of y at x=8 is,
y=9.9
The first-order Newton’s interpolation polynomial is:
f1(x)=9.9+0.054545(x−5.5)
Thus, the value of y at x=8 is,
y=9.9+0.054545(8−5.5)=9.9+0.054545×2.5=9.9+0.1363625=10.036
The second- order Newton’s interpolating polynomial is,
f2(x)=9.9+0.054545(x−5.5)−0.06394(x−5.5)(x−11)
Thus, the value of y at x=8 is,
y=9.9+0.054545(8−5.5)−0.06394(8−5.5)(8−11)=10.036−0.06394×2.5×(−3)=10.036+0.47955=10.516
The third-order Newton’s interpolating polynomial is,
f3(x)={9.9+0.054545(x−5.5)−0.06394(x−5.5)(x−11)+0.0069(x−5.5)(x−11)(x−13)}
Thus, the value of y at x=8 is,
y={9.9+0.054545(8−5.5)−0.06394(8−5.5)(8−11)+0.0069(8−5.5)(8−11)(8−13)}=10.516+0.0069×2.5×(−3)×(−5)=10.775
The fourth-order Newton’s interpolating polynomial is,
f4(x)={9.9+0.054545(x−5.5)−0.06394(x−5.5)(x−11)+0.0069(x−5.5)(x−11)(x−13)+0.000165(x−5.5)(x−11)(x−13)(x−2)}
Thus, the value of y at x=8 is,
y={9.9+0.054545(8−5.5)−0.06394(8−5.5)(8−11)+0.0069(8−5.5)(8−11)(8−13)+0.000165(8−5.5)(8−11)(8−13)(8−2)}=10.775+0.000165×2.5×(−3)×(−5)×6=10.812
Hence, the value of y at x=8 by zero order Newton’s interpolating polynomial is 9.9, by first order Newton’s interpolating polynomial is 10.036, by second order Newton’s interpolating polynomial is 10.516, by third order Newton’s interpolating polynomial is 10.775 and by fourth order Newton’s interpolating polynomial is 10.812