## Newton’s forward/backward difference

### Newton Forward Difference:

If the x value of a data set are uniformly spaced then let us define for the data set.

$$S= \frac {x-x_0}{\Delta x}$$

This variable is called interpolating variable. i.e. s is linear in ‘x’.

Now the unique nth degree polynomial can also be represented as,

$$P_n(x)=f_0+s\Delta f_0+ \frac {s(s-1)}{2!}\Delta^2 f_0 + \frac {s(s-1)(s-2)}{3!}\Delta^3 f_0+ \cdots + \frac {s(s-1)(s-2)\cdots (s-(n-1))}{n!}\Delta^n f_0$$

This is Newton’s forward difference polynomial.

You can also represent this forward difference polynomial using binomial coefficient representation

$$\text{i.e.} \binom{s}{i}= \frac {s(s-1)(s-2)\cdots (s-(i-1))}{i!}$$
$$\therefore P_n(x)= f_0+ \binom{s}{1}\Delta f_0+ \binom{s}{2}\Delta^2 f_0+\cdots$$

### Example:

Find Solution using Newton’s Forward Difference formula(x = 1895)

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

### Solution:

Newton’s forward difference table is

x y Δy Δ2y Δ3y Δ4y
1891 46
20
1901 66 -5
15 2
1911 81 -3 -3
12 -1
1921 93 -4
8
1931 101
$$Δx=x_1-x_0=1901-1891=10$$
$$S= \frac {x-x_0}{\Delta x}$$
$$S= \frac {1895-1891}{10}$$
$$S= 0.4$$

Newton’s forward difference interpolation formula is

$$P_n(x)=f_0+s\Delta f_0+ \frac {s(s-1)}{2!}\Delta^2 f_0 + \frac {s(s-1)(s-2)}{3!}\Delta^3 f_0+ \cdots + \frac {s(s-1)(s-2)\cdots (s-(n-1))}{n!}\Delta^n f_0$$
$$=46+0.4\times 20+\frac {0.4(0.4-1)}{2}\times(-5)+\frac {0.4(0.4-1)(0.4-2)}{6}\times(2)\frac {0.4(0.4-1)(0.4-2)(0.4-3)}{24}\times(-3)$$
$$=46+8+0.6+0.128+0.1248$$
$$=54.8528$$

### Newton Backward Difference Polynomial:

As explained for Newton’s forward difference polynomials, we can also use backward difference formulas to develop polynomials.

We can use backward differences

For a uniformly spaced data, if xn is the base in an (n+1) data (x0,f0),(x1,f1),..,(xn,fn)

$$\Delta f_n-f_n-f_{n-1}$$
$$S= \frac {x-x_n}{\Delta x}$$

We have the unique nth degree polynomial in the power series as:

$$P_n(x)=f_n+s\Delta f_n+ \frac {s(s+1)}{2!}\Delta^2 f_n + \frac {s(s+1)(s+2)}{3!}\Delta^3 f_n+ \cdots + \frac {s(s+1)(s+2)\cdots (s+(n-1))}{n!}\Delta^n f_n$$

### Example:

Find Solution using Newton’s Backward Difference formula(x = 1925)

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

### Solution:

Newton’s backward difference table is

x y Δy Δ2y Δ3y Δ4y
1891 46
20
1901 66 -5
15 2
1911 81 -3 -3
12 -1
1921 93 -4
8
1931 101
$$Δx=x_1-x_0=1901-1891=10$$
$$S= \frac {x-x_n}{\Delta x}$$
$$S= \frac {1925-1931}{10}$$
$$S= -0.6$$

Newton’s backward difference interpolation formula is

$$P_n(x)=f_n+s\Delta f_n+ \frac {s(s+1)}{2!}\Delta^2 f_n + \frac {s(s+1)(s+2)}{3!}\Delta^3 f_n+ \cdots + \frac {s(s+1)(s+2)\cdots (s+(n-1))}{n!}\Delta^n f_n$$
$$=101+(-0.6)\times 8+\frac {-0.6(-0.6+1)}{2}\times(-4)+\frac {-0.6(-0.6+1)(-0.6+2)}{6}\times(-1)\frac {-0.6(-0.6+1)(-0.6+2)(-0.6+3)}{24}\times(-3)$$
$$=101-4.8+0.48-0.056+0.1008$$
$$=96.8368$$