Newton's Forward/Backward Difference Interpolation

1. Equal Spacing: Newton’s Forward and Backward
Difference Interpolation
Dr. Varun Kumar
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 1 / 9

2. Outlines
1 Equal Spacing: Newton’s Forward Difference Formulation
Example
2 Equal Spacing: Newton’s Backward Difference Formulation
Example
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 2 / 9

3. Equal Spacing: Newton’s Forward Difference Formula
Important points
⇒ As per the Newton’s divide difference interpolation formula
f (x)= f0 + (x − x0)f [x0, x1] + (x − x0)(x − x1)f [x0, x1, x2] + ...+
(x − x0)(x − x1)....(x − xn−1)f [x0, ..., xn]
(1)
or
f [x0, ...., xk] =
f [x1, ....., xk] − f [x0, ...., xk−1]
xk − x0
⇒ Above expression is valid for arbitrarily spaced nodes.
⇒ In most of the practical experimentation, it can be adopted.
⇒ For some instances, the interval may be equi-spaced, i.e,
x0, x1 = x0 + h, x2 = x0 + 2h, ....., xn = x0 + nh (2)
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 3 / 9

4. Continued–
⇒ First forward difference of f at xj by
4fj = fj+1 − fj
⇒ Second forward difference of f at xj by
42
fj = 4fj+1 − 4fj
⇒ The kth forward difference of f at xj by
4k
fj = 4k−1
fj+1 − 4k−1
fj ∀ k = 1, 2, ...
⇒ In case of regular spacing in input (x), then
f [x0, x1, ....., xk] =
1
k!hk
4k
f0 (3)
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 4 / 9

5. Continued–
⇒ For k = 1, from (3)
f [x0, x1] =
f1 − f0
x1 − x0
=
1
h
(f1 − f0) =
1
1!h
4f0 (4)
⇒ Let xk+1 = x0 + (k + 1)h
f [x0, ......, xk+1] =
f [x1, ..., xk+1] − f [x0, ..., xk]
xk+1 − x0
=
1
(k + 1)h
h 1
k!hk
4k
f1 −
1
k!hk
4k
f0
i
=
1
(k + 1)!hk+1
4k+1
f0
(5)
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 5 / 9

6. Continued–
⇒ Applying Newton’s forward difference interpolation formula
f (x) ≈ pn(x) = f0 + r4f0 +
r(r − 1)
2!
42
f0 + ...+
r(r − 1)...(r − n + 1)
n!
4n
f0
(6)
where r = x−x0
h
Q Compute cosh 0.56. Utilize the four values in the following table and
estimate the error.
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 6 / 9

7. Equal Spacing: Newton’s Backward Difference Formulation
⇒ First backward difference of f at xj by
∇fj = fj − fj−1
⇒ Second backward difference of f at xj by
∇2
fj = ∇fj − ∇fj−1
⇒ kth backward difference of f at xj by
∇k
fj = ∇k−1
fj − ∇k−1
fj−1 (7)
⇒ As per Newton’s backward difference interpolation formula
f (x) ≈ pn(x) = f0 + r∇f0 +
r(r + 1)
2!
∇2
f0 + ... +
r(r + 1)...(r + n − 1)
n!
∇n
f0
(8)
where r = x−x0
h
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 7 / 9

8. Example
Q Compute a 7D value of the Bessel function J0(x) for x = 1.72 from
the four values in the following table, using
(a) Newton’s forward formula
(b) Newton’s backward formula
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 8 / 9

9. Exercise Problem
1 Calculate the Lagrange polynomial p2(x) for the values
Γ(1.00) = 1.0000, Γ(1.02) = 0.9888, Γ(1.04) = 0.9784 of the gamma
function and from it approximate Γ(1.01) and Γ(1.03).
2 Find e−0.25 and e−0.75 by linear interpolation of e−x with x0 = 0,
x1 = 0.5, and x0 = 0.5, x1 = 0, respectively. Then find p2(x) by
quadratic interpolation of e−x with x0 = 0, x1 = 0.5, and x2 = 1 and
from it e−0.25 and e−0.75 compare the errors.
3 Solve the above problem using Newton’s divide and difference
interpolation method.
Dr. Varun Kumar (IIIT Surat) Unit 2 / Lecture-3 9 / 9