Numerical interpolation using Newton's Forward Difference formula calculator

SolutionHelp

1. The population of a town in decimal census was as given below. Estimate population for the year 1895.

Year	1891	1901	1911	1921	1931
Population (in Thousand)	46	66	81	93	101

2. Use the following table

X	0.10	0.15	0.20	0.25	0.30
tan(X)	0.1003	0.1511	0.2027	0.2553	0.3073

Find (1) tan 0.12 (2) tan 0.26

Example

1. Find Solution using Newton's Forward Difference formula
x f(x)
1891 46
1901 66
1911 81
1921 93
1931 101

x = 1895
Finding option 1. Value f(2)

Solution:
The value of table for

$x$ and

$y$

x	1891	1901	1911	1921	1931
y	46	66	81	93	101

Newton's forward difference interpolation method to find solution

Newton's forward difference table is

x	y	$Deltay$	$Delta^2y$	$Delta^3y$	$Delta^4y$
1891	46
		$66-46=20$
1901	66		$15-20=-5$
		$81-66=15$		$-3--5=2$
1911	81		$12-15=-3$		$-1-2=-3$
		$93-81=12$		$-4--3=-1$
1921	93		$8-12=-4$
		$101-93=8$
1931	101

The value of

$x$ at you want to find the

$f(x) : x = 1895$

$h = x_1 - x_0 = 1901 - 1891 = 10$

$p = (x - x_0)/h = (1895 - 1891)/10 = 0.4$

Newton's forward difference interpolation formula is

$y(x) = y_0 + p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_0 + (p(p - 1)(p - 2))/(3!) * Delta^3y_0 + (p(p - 1)(p - 2)(p - 3))/(4!) * Delta^4y_0$

$y(1895) = 46 + 0.4 xx 20 + (0.4 (0.4 - 1))/(2) xx -5 + (0.4 (0.4 - 1)(0.4 - 2))/(6) xx 2 + (0.4 (0.4 - 1)(0.4 - 2)(0.4 - 3))/(24) xx -3$

$y(1895) = 46 +8 +0.6 +0.128 +0.1248$

$y(1895) = 54.8528$

Solution of newton's forward interpolation method

$y(1895) = 54.8528$