# Newton’s forward interpolation

#### Newton’s forward interpolation

Suppose $$y=f( x )$$ is tabulated for equally spaced values $$x_i=x_0+ih$$ for i=0,1,2,...,n for the entries
$$x=x_0,x_0+h,x_0+2h,...,x_0+nh$$
with the values
$$y=y_0,y_1,y_2,...,y_n$$
Here,
$$x_i =x_0+ih$$ for $$i=0,1,2,\ldots ,n$$
Now, assuming $$y=f( x )$$ is a polynomial of degree n in x, we write
$$y=a_0+a_1( x-x_0 )+a_2( x-x_0 )( x-x_1 )+\ldots +a_n( x-x_0 )\ldots ( x-x_{n-1} )$$

Putting, $$x=x_0$$, we get
$$y=a_0+a_1( x-x_0 )+a_2( x-x_0 )( x-x_1 )+\ldots +a_n( x-x_0 )\ldots ( x-x_{n-1} )$$
or $$y_0=a_0+a_1 ($$ $$x_0$$ $$-x_0 )+a_2($$ $$x_0$$ $$-x_0 )($$ $$x_0$$ $$-x_1 )+\ldots +a_n($$ $$x_0$$ $$-x_0 )\ldots ($$ $$x_0$$ $$-x_{n-1} )$$
or $$y_0=a_0+0+0+\ldots +0$$
or $$y_0=a_0$$

Putting, $$x=x_1$$, we get
$$y=a_0+a_1( x-x_0 )+a_2( x-x_0 )( x-x_1 )+\ldots +a_n( x-x_0 )\ldots ( x-x_{n-1} )$$
or $$y_1=a_0+a_1 ($$ $$x_1$$ $$-x_0 )+a_2($$ $$x_1$$ $$-x_0 )($$ $$x_1$$ $$-x_1 )+\ldots +a_n($$ $$x_1$$ $$-x_0 ) ($$ $$x_1$$ $$-x_1 ) \ldots ($$ $$x_1$$ $$-x_{n-1} )$$
or $$y_1=a_0+a_1 h+0+\ldots +0$$
or $$y_1=y_0+a_1h$$
or $$a_1h=y_1-y_0$$
or $$a_1=\frac{\Delta y_0}{h}$$

Putting, $$x=x_2$$, we get
$$y=a_0+a_1( x-x_0 )+a_2( x-x_0 )( x-x_1 )+\ldots +a_n( x-x_0 )\ldots ( x-x_{n-1} )$$
or $$y_2=a_0+a_1 ($$ $$x_2$$ $$-x_0 )+a_2($$ $$x_2$$ $$-x_0 )($$ $$x_2$$ $$-x_1 )+a_3($$ $$x_2$$ $$-x_0 ) ($$ $$x_2$$ $$-x_1 ) ($$ $$x_2$$ $$-x_2 )+...$$
or $$y_2=y_0+a_1(2h)+a_2(2h )( h )+0+...$$
or $$y_2=y_0+2(a_1h)+a_2(2h )( h )+0+...$$
or $$y_2=y_0+2(y_1-y_0)+2a_2h^2$$
or $$2a_2h^2=y_2-y_0-2(y_1-y_0)$$
or $$2a_2h^2=y_2-y_0-2y_1+2y_0$$
or $$2a_2h^2=y_2-2y_1+y_0$$
or $$2a_2h^2=y_2-y_1y_1+y_0$$
or $$2a_2h^2=(y_2-y_1)-(y_1-y_0)$$
or $$2a_2h^2=(\Delta y_1)-(\Delta y_0)$$
or $$2a_2h^2=\Delta^2 y_0$$
or $$a_2=\frac{\Delta^2 y_0}{2!h^2}$$

Similarly, we get
$$a_i=\frac{\Delta ^iy_0}{i!h^i}$$ for $$i=0,1,2,\ldots ,n$$

Setting $$x=x_0+ph$$ and substituting, $$a_i$$ for $$i=0,1,2,\ldots ,n$$, we get
$$y=a_0+a_1( x-x_0 )+a_2( x-x_0 )( x-x_1 )+\ldots +a_n( x-x_0 )\ldots ( x-x_{n-1} )$$
or $$y=y_0+\frac{\Delta y_0}{h}[ph]+\frac{\Delta ^2y_0}{2!h^2}[p( p-1 )h^2]+\ldots +\frac{\Delta ^ny_0}{n!h^n}[p( p-1 )\ldots ( p-n+1 )h^n]$$
or $$y=y_0+p\Delta y_0+\frac{p( p-1 )}{2!}{{\Delta }^{2}}y_0+\ldots +\frac{p( p-1 )\ldots ( p-n+1 )}{n!}{{\Delta }^n}y_0$$

NOTE

1. This is called Newton’s forward difference interpolation formula
2. This formula is used for interpolating the values of y near the beginning of a set of tabulated values and extrapolating values of y a little backward (i.e., to the left) of $$y_0$$
3. The first two terms of this formula give the linear interpolation while the first three terms give a parabolic interpolation and so on.

#### Example 1

The population of a town was as given below. Estimate the population for the year 1895.

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

Solution
According to given set of data values, we form difference table as below

 X Y 1st diff 2nd diff 3rd diff 4th diff 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

Here,
$$h=10$$ and $$x_0=1891$$
Thus, we have
$$x=x_0+ph$$
or $$1895=1891+p\times 10$$
or $$p=0.4$$
Using formula, we get
$$y=y_0+p\Delta y_0+\frac{p( p-1 )}{2!}{{\Delta }^{2}}y_0+\ldots +\frac{p( p-1 )\ldots ( p-n+1 )}{n!}{{\Delta }^n}y_0$$
or $$y_{1895}=46+(0.4).(20)+\frac{(0.4)(-0.6)}{2!} (-5)+\frac{(0.4)(-0.6)(-1.6)}{3!} (2)+\frac{(0.4)(-0.6)(-1.6)(-2.6)}{3!} (-3)$$

or $$y_{1895}=46+8+0.6+0.128+0.1248$$

or $$P_{1895}= 54.85$$
This completes.

#### Example 2

The following table gives values of tan(x) for $$0.10\le x\le 0.30$$. Find tan(0.12)

 x 0.1 0.15 0.2 0.25 0.3 tan(x) 0.1003 0.1511 0.2027 0.2553 0.3039

Solution
According to given set of data values, we form difference table as below

X Y 1st diff 2nd diff 3rd diff 4th diff
0.10 0.1003
0.0508
0.15 0.1511 0.0008
0.1511 0.0002
0.20 0.2027 0.0010 0.0002
0.0526
0.0004
0.25 0.2553 0.0014
0.0540
0.30 0.3093

Here,
$$h=0.05$$ and $$x_0=0.10$$
Thus, we have
$$x=x_0+ph$$
or $$0.12=0.10+p\times 0.05$$
or $$p=0.4$$
Using formula, we get
$$y=y_0+p\Delta y_0+\frac{p( p-1 )}{2!}{{\Delta }^{2}}y_0+\ldots +\frac{p( p-1 )\ldots ( p-n+1 )}{n!}{{\Delta }^n}y_0$$
or $$\tan 0.12=0.1003+0.4\times 0.0508+\frac{0.4( -0.6 )}{2!}0.0008+\frac{( 0.4 )( -0.6 )( -1.6 )}{3!}0.0002+\frac{( 0.4 )( -0.6 )( -1.6 )( -2.6 )}{4!}0.0002$$
or $$\tan 0.12=0.1205$$
This completes.

#### Example 3

Find cubic polynomial which takes the following values.
$$y( 0 )=1,y( 1 )=0,y( 2 )=1$$ and $$y( 3 )=10$$

Solution
According to given set of data values, we form difference table as below

 X Y 1st diff 2nd diff 3rd diff 0 1 -1 1 0 2 1 6 2 1 8 9 3 10

Here,
$$h=1$$ and $$x_0=0$$
Thus, we have
$$x=x_0+ph$$
or $$x=p$$
Now, cubic polynomial for given data value is
$$y=y_0+p\Delta y_0+\frac{p( p-1 )}{2!}{{\Delta }^2}y_0+\ldots +\frac{p( p-1 )\ldots ( p-n+1 )}{n!}{{\Delta }^n}y_0$$
or $$y=1+x( -1 )+\frac{x( x-1 )}{2!}2+\frac{x( x-1 )( x-2 )}{3!}6$$
or $$y=1-x+x( x-1 )+x( x-1 )( x-2 )$$
or $$y=1-x+{x^2}-x+x( {x^2}-3x+2 )$$
or $$y=1-x+{x^2}-x+{x^{3}}-3{x^2}+2x$$
or $$y={x^{3}}-2{x^2}+1$$
This completes.

#### Exercise

Use Newton's Forward difference formula to compute followings.
1. Find the value when x=-1 from the following table [Answer: -2]
 x 0 1 2 3 y 1 0 1 10
2. Find solution of f(3) of an equation $$x^2-x+1$$ using a = 2 and b= 10, step value h = 1.
3. Estimate the number of students who obtained marks between 40 and 45
 Marks 30-40 40-50 50-60 60-70 70-80 Number 31 42 51 35 31
4. Find cubic polynomial which takes the following values: y( 0 )=1,y( 1 )=2, y( 2 )=1 and y( 3 )=10