This document discusses Neville's algorithm for polynomial interpolation. Neville's algorithm uses divided differences and recursion relations to find a unique polynomial that passes through a given set of n+1 points. The algorithm works by first finding polynomials that interpolate smaller subsets of the points, then combining these to find the full interpolating polynomial. An example is provided to demonstrate applying the algorithm to find the value of an interpolating polynomial for a given x-value.

1. Neville’s Interpolation
Dr. Varun Kumar
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-4 1 / 6

2. Outlines
1 Introduction to Neville’s algorithm
Example
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-4 2 / 6

3. Introduction to Neville’s algorithm
Important points
⇒ Neville’s algorithm is an algorithm used for polynomial interpolation.
⇒ This algorithm was derived by the mathematician Eric Harold Neville.
⇒ Given n + 1 points, there is a unique polynomial of degree ≤ n which
goes through the given points.
⇒ Neville’s algorithm is based on the Newton form of the
interpolating polynomial and the recursion relation for the divided
differences.
Neville’s algorithm
⇒ Let x0, x1, ...., xn are n + 1 input points and y0, y1, ...., yn are
corresponding n + 1 output point.
p(x) =
(x − x1)(x − x2)...(x − xn)
(x0 − x1)(x0 − x2)...(x0 − xn)
y0 + ... +
(x − x0)(x − x1)...(x − xn−1)
(xn − x0)(xn − x1)...(xn − xn−1)
yn
(1)
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-4 3 / 6

4. Continued–
⇒ Some mathematical representation
p01 =
x − x1
x0 − x1
y0 +
x − x0
x1 − x0
y1 (2)
⇒ If only two input points are corresponding output points are given.
⇒ Let p234 is another polynomial that can be expressed as
p234 =
(x − x3)(x − x4)
(x2 − x3)(x2 − x4)
y2 +
(x − x2)(x − x4)
(x3 − x2)(x3 − x4)
y3 +
(x − x2)(x − x2)
(x4 − x2)(x4 − x3)
y4 (3)
Theorem : Let f be a function defined at x0, x1, ..., xk and let xi and xj
are two distinct number in this set
p(x) =
(x − xj )p0,1,...j−1,j+1,..k − (x − xi )p0,1,...i−1,i+1,..k
xi − xj
(4)
Ex-
p01234 =
(x − x0)p1234 − (x − x1)p0234
x1 − x0
(5)
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-4 4 / 6

5. Continued–
⇒ Since
p(x) =
(x − xj )p0,1,...j−1,j+1,..k − (x − xi )p0,1,...i−1,i+1,..k
xi − xj
(6)
⇒ Hence
p(xi ) =
(xi − xj )p0,1,...j−1,j+1,..k − 0
xi − xj
= f (xi ) = p0,1,...j−1,j+1,..k (7)
⇒ Similarly
p(xj ) =
0 − (xi − xj )p0,1,...i−1,i+1,..k
xi − xj
= f (xj ) = p0,1,...i−1,i+1,..k (8)
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-4 5 / 6

6. Continued–
Q- Find f(1.5), where a input-output table has been given below
x f
x0 1 0.7651 p0
x1 1.3 0.6200 p1
x2 1.6 0.4554 p2
x3 1.9 0.2818 p3
x4 2.2 0.1103 p4
p01(x) =
(x − x0)p1 − (x − x1)p0
x1 − x0
→ p01(1.5) = 0.5233
⇒ Similarly p12(1.5) = 0.5102, p23(1.5) = 0.5132, p34(1.5) = 0.5104
⇒ Similarly p012(1.5) = 0.5121, p123(1.5) = 0.5112, p234(1.5) = 0.5137
⇒ Similarly p0123(1.5) = 0.5118, p1234(1.5) = 0.5118 and p01234(1.5) = 0.5118
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-4 6 / 6