Newton's forward and backward difference interpolation

NEWTON'S FORWARD AND BACKWARD DIFFERENCE INTERPOLATION

(a) Newton's forward interpolation formula for equal intervals.

Theorem : Let, the function y = f (x) take the values y₀, y₁, … y_n at the points x₀,x₁, x₂, ... x_n, where x_i = x₀ + ih. Then, Newton's forward interpolation polynomial is given by

Equation (5) is known as Gregory-Newton forward interpolation formula.

(b) Newton's Backward interpolation formula [A.U M/J 2012]

Theorem: Let the function y = f (x) take the values y₀, y₁, ..., y_n at the points x₀ , x₁ x_n, where x_i = x₀ + ih. Then Newton's Backward interpolation polynomial is given by

Equation (6) is known as Gregory-Newton backward difference interpolation formula.

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1. Using Newton's forward interpolation formula, find the polynomial f(x) satisfying the following data. Hence, evaluate y at x = 5. [AU, March, 1996, M/J 2006, A.U Trichy A/M. 2010, A.U. Tvli N/D 2010, A.U. N/D 2012] [A.U M/J 2012]

Solution: We form the difference table

2. Using Newton's forward inte olation formula, find the polynomial f(x) Hence, find f(2). satisfying the following data.

Solution: We form the difference table

3. A third degree polynomial passes through the points (0, -1), (1, 1), (2, 1) and (3, -2), using Newton's forward interpolation formula find the polynomial. Hence, find the value at 1.5[A.U. May, 2000]

Solution: We form the difference table

4. Using Newton's forward interpolation formula find the cubic polynomial which takes places the following values :

Evaluate, f(4) using Newton's backward formula. Is it the same as obtained from the cubic polynomial found above?

[AU, May 2000] [A.U M/J 2013] [A.U N/D 2014 NM]

Solution: We form the difference table

There are only 4 data given. Hence, the polynomial of degree 3.

Newton's forward formula is

The cubic polynomial same for both the cases.

5. Use Newton's backward difference formula to construct an interpolating polynomial of degree 3 for the data:

f (−0.75) = −0.07181250, f (-0.5) = -0.02475

f(-0.25) = 0.33493750, f (0) = 1.10100. Hence find f (- 1/3)

 [A.U. April/May 2003] [A.U N/D 2019 R-13]

Solution:

We form the difference table

Newton's backward difference formula is

6. From the following data, find at x = 43 and x = 84  [A.U CBT N/D 2010] [A.U N/D 2016

Solution:

Since six data are given, P (x) is x = 43, use forward interpolation and to find interpolation formula.

7. From the data given below, find the number of students whose weight is between 60 to 70. [A.U. N/D 2003, A/M 2010, M/J 2012]

Solution:

Difference table

Let us calculate the number of students whose weight is less than 70.

We will use forward difference formula

Number of students whose weight is between 60 and 70

= y (70) - y (60) 424 – 370 = 54

8. The following data are taken from the steam table :

Find the pressure at temperature t = 142° and t = 175°.

[AU M/J 2009, N/D 2009] [A.U N/D 2015 R-13]

Solution:

We form the difference table:

9. Using Newton's forward formula, find sin (0.1604) from the following table

Solution:

The difference table is

10. From the following table, find the value of tan 45° 15' by Newton's re forward interpolation formula. [A.U M/J 2007] [A.U N/D 2010] [A.U N/D 2021 R-17]

Solution:

We use forward interportation formula: also h = 1

The difference table is

11. The population of a town is as follows :

Estimate the population increase during the period 1946 to 1976. [A.U N/D 2011]

Solution:

The difference table is

12. Find y(2.25) using Newton's backward difference formula from the following data: [A.U N/D 2010]

Solution:

The difference table is

13. From the given table, the values of y are consecutive terms of a series of which 23.6 is the 6th term. Find the first and tenth terms of the series. [A.U N/D 2007]

Solution:

The difference table is