Lecture3 Interpolation
Lecture3 Interpolation
Lecture3 Interpolation
Here f xs f s E f 0
s
1
1 s
f 0 1
s
f0
Example: Given
x : 3 4 5 6 7 8
f(x) : 27 64 125 216 343 512. Estimate (7.5).
Note: As we have seen that Newton’s forward difference formula gives better result when xs
lies in the beginning of the data table. Similarly, Newton’s backward difference formula is best
suited when xs lies in the end of data table. Then, it can be seen that for those values of xs ,
which lie almost in the middle of the data table neither N.Fw or N.Bw retains sufficient number
of terms in the formula and thus the interpolated value will not be sufficiently accurate. To
overcome this problem we derive two more formulae named as Stirling's formula and Bessel’s
formula. More commonly Stirling's formula is termed as central difference formula.