NEWTON’S BACKWARD INTERPOLATION FORMULA
Submitted by:-
Prem chand yadav MCA II sem.


Let y=f(x) be a function of x which assumes the values f(a), f(a + h), f(a + 2h), … ................, f(a + nh) for (n + 1) equidistant values a, a + h, a + 2h,………….,a + nh of the independent variable x.
Let Φ(x) denotes polynomial of nth degree as follows:-

Φ(x) =a0 + a1 (x-xn) + a2 (x- xn) (x- xn-1) + a3 (x- xn) (x- xn-1) (x- xn-2) + ………
……… + an (x- xn) (x- xn-1) (x- xn-2) ……………… (x – x1) ….. (1)

Where a0, a1, a2, a3, a4 …………… an are constants.

Here,
x1 - x0 = h, x2 - x0 = 2h, x3 - x0 = 3h, …….etc. … (2)

By putting x= xn in equation no. (1), we have

Φ (xn) = a0 = yn (say) … (3)

Again put x = xn-1

Φ (xn-1) = a0 + a1 (xn-1 – xn )
yn-1 = yn + a1 (-h) [from equation no. (2) & (3)]

so,
yn - yn-1 = a1 (+h)

=> a1 = yn / h

Again
a2 = (yn - 2 yn-1 + yn-2 ) / 2h2

=> a2 = 2 yn / 2! h2

Similarly,

a3 = 3 yn / 3! h3
.
.
an = n yn / n! hn

Therefore,

y = yn + yn (x- xn ) / h + 2yn (x- xn ) (x- xn-1 ) /2! h2 + 3 yn (x- xn) (x- xn-1) (x-

xn-2) / 3! h3 + ……….+ n yn (x- xn ) (x- xn-1 ) (x- xn-2 )………(x- x1 ) / n! hn
This equation is known as Newton’s backward interpolation formula.

On putting p = (x- xn ) / h, we have


y = yn + p. 1 yn + p (p+1) / 2! . 2 yn + p (p+1) (p+2) / 3! . 3 yn + ………..

+ p (p+1) (p+2)………(p+n-1) / n! . n yn



This formula is particularly useful for interpolating the value of f(x) near the end of the set of given values.


Example 1:-
The population of a town was as given. Estimate the population for the year 1925.
Year (x) : 1891 1901 1911 1921 1931
Population (y) : 46 66 81 93 101
(in thousands)

Sol. Here, h

x1 – x0

> h

1901 – 1891

10,

Therefore p

(x – xn) / h

> p

(1925 – 1931) / 10

- 0.6

Difference table is:


x y y 2 y 3 y 4 y

1891 46 20
1901 66 15 -5
1911 81 12 -3 2 -3
1921 93 8 -4 -1
1931 101

Applying Newton’s backward difference formula, we get

y1925 = y1931 + p . y1931 + p (p+1) / 2! . 2 y1931 + p(p+1)(p+2) / 3! .

3 y1931 + p(p+1)(p+2)(p+3) / 4! . 4 y1931

= 101 + (-0.6) (8) + (-0.6)(0.4) / 2 . (-4) + (-0.6)(0.4)(1.4) / 3x2 . (-1)

+ (-0.6)(0.4)(1.4)(2.4) / 4x3x2 . (-3)
y1925 = 96.8368 thousands.

Hence the population for the year 1925 = 96836.8 = 96837.
Example 2:-
From the following table of half-yearly premium for policies maturing at different ages, estimate the premium for policy maturing at the age of 63.

Age : 45 50 55 60 65
Premium : 114.84 96.16 83.32 74.48 68.48
(in rupees)

Sol. The difference table is:

Age premium y 2 y 3 y 4 y
(x) (in rupees)
(y)

45 114.84 -18.68 5.84 -1.84 .68
50 96.16 -12.84 4 -1.16
55 83.32 -8.84 2.84
60 74.48 -6
65 68.48

Here h

x1 – x0

> h

50 – 45

5,

Therefore p

(x – xn) / h

> p

(63 – 65) / 5

-0.4

By Newton’s backward difference formula

y63 = y65 + p . y65 + p(p+1) / 2 . 2 y65 + p(p+1)(p+2) / 3x2 . 3 y65 +

p(p+1)(p+2)(p+3) / 4x3x2 . 4 y65

= 68.48 + (-0.4)(-6) + (-0.4)(0.6) / 2 . (2.84) + (-0.4)(0.6)(1.6) / 6 . (-1.16)
+ (-0.4)(0.6)(1.6)(2.6) / 24 . (0.68)

y63 = 70.585152 rupees Ans.










ALGORITHM FOR NEWTON’S BACKWARD FORMULA


1. Input the values if xi, yi and the value of x for which t is calculated.
2. Obtain h by using xi – xi-1.
3. construct the difference table by using

n yi = n-1 yi - n-1 yi-1

4. Find p = (x – yn ) / h.
5. Put all values in Newton’s backward interpolation formula.
6. Print the results.
7. Stop.




PROGRAM FOR NEWTON’S BACKWARD FORMULA

#include<stdio.h>
#include<conio.h>
void main()
{
int n,i,j=0,f;
float x[10],y[10],h,p,z[9],x1,y1;
clrscr();
printf("\nEnter no. of given values of x : ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("\nEnter value of x%d : ",i);
scanf("%f",&x[i]);
}
for(i=0;i<n;i++)
{
printf("\nEnter value of y%d : ",i);
scanf("%f",&y[i]);
}
printf("\nEnter value of x at which u want to find value of y : ");
scanf("%f",&x1);
h=x[1]-x[0];
p=(x1-x[n-1])/h;
f=n;
while(j<n)
{
f--;
for(i=0;i<f;i++)
y[i]=y[i+1]-y[i];
z[j]=y[i];
j++;
}
y1=z[0];
f

1;h

1;
for(i=1;i<n;i++)
{
f=f*i;
h=h*p;
y1=y1+h*z[i]/f;
p=p+1;
}
printf("\nValue of y at %f is : %f",x1,y1);
getch();
}