NEWTON'S DIVIDED DIFFERENCE FORMULA

 Let us assume that the function f(x) is linear then we have f(xi) - f(xj) (xi - xj)
where xi and xj are any two tabular points, is independent of xi and xj. This ratio is called the first divided difference of f(x) relative to xi and xj and is denoted by f [xi, xj]. That is
 f [xi, xj] = f(xi) - f(xj) =  f [xj, xi] (xi - xj)
Since the ratio is independent of  xi and xj we can write   f [x0, x] = f [x0, x1]
 f(x) - f(x0) = f [x0, x1] (x - x0)

f(x) = f(x0) + (x - x0) f [x0, x1]

 = 1 | f(x0) x0 - x | f1 - f0 f0x1 - f1x0 = x + x - x0 f(x1) x1 - x x1 - x0 x1 - x0
So if f(x) is approximated with a linear polynomial then the function value at any point x can be calculated by using f(x) @ P1(x) = f(x0) + (x - x1) f [x0, x1]
where f [x0, x1] is the first divided difference of  relative to x0 and x1.
Similarly if f(x) is a second degree polynomial then the secant slope defined above is not constant but a linear function of x. Hence we have
 f [x1, x2] - f [x0, x1] x2 - x0
is independent of x0, x1 and x2. This ratio is defined as second divided difference of f relative to x0, x1 and x2. The secind divided difference are denoted as
 f [x1, x2] - f [x0, x1] f [x0, x1, x2] = x2 - x0
Now again since f [x0, x1,x2] is independent of x0, x1 and x2 we have
f [x1, x0, x] = f [x0, x1, x2]
 f [x0, x] - f [x1, x0] = f [x0, x1, x2] x - x1

f [x0, x] = f [x0, x1] + (x - x1) f [x0, x1, x2]

 f [x] - f [x0] = f [x0, x1] + (x - x1) f [x0, x1, x2] x - x0

f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2]

This is equivalent to the second degree polynomial approximation passing through three data points
 x0 x1 x2 f0 f1 f2
So whenever f(x) is approximated with a second degree polynomial, the value of f(x) at any point x can be computed using the above polynomial.
In the same way if we define recursively kth divided difference by the relation
 f [x1, x2, . . ., xk] - f [x0, x1, . . ., xk-1] f [x0, x1, . . ., xk] = xk - x0
The kth degree polynomial approximation to f(x) can be written as
f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2
+ . . . + (x - x0) (x - x1) . . . (x - xk-1) f [x0, x1, . . ., xk].
This formula is called Newton's Divided Difference Formula. Once we have the divided differences of the function f relative to the tabular points then we can use the above formula to compute f(x) at any non tabular point.
Computing divided differences using divided difference table:  Let us consider the points (x1, f1), (x2, f2), (x3, f3) and (x4, f4) where x1, x2, x3 and x4 are not necessarily equi-distant points then the divided difference table can be written as
xi
fi
f [xi, xj
f [xi, xj, xk
f [xi, xj, xk, xl]

x1
f1

 f [x1, x2] = f2 - f1 x2 - x1
x2
f2
 f [x2, x3] - f [x1, x2] f [x1, x2,x3] = x3 - x1

 f [x2, x3] = f3 - f2 x3 - x2
 f [x2, x3,x4] - f [x1, x2,x3] f [x1, x2,x3,x4] = x4 - x1
x3
f3
 f [x3, x4] - f [x2, x3] f [x2, x3,x4] = x4 - x2

 f [x3, x4] = f4 - f3 x4 - x3
x4
f4
Example : Compute f(0.3) for the data
 x 0 1 3 4 7 f 1 3 49 129 813
using Newton's divided difference formula.
Solution : Divided difference table
 xi fi 0 1 2 1 3 7 23 3 3 49 19 80 3 4 129 37 228 7 813
Now Newton's divided difference formula is

f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2] +  (x - x0) (x - x1) (x - x2)f [x0, x1,  x2, x3]

f(0.3) = 1 + (0.3 - 0) 2 + (0.3)(0.3 - 1) 7 + (0.3) (0.3 - 1) (0.3 - 3) 3

= 1.831

Since the given data is for the polynomial f(x) = 3x3 - 5x2 + 4x +1 the analytical value is f(0.3) = 1.831
The analytical value is matched with the computed value because the given data is for a third degree polynomial and there are five data points available using which one can approximate any data exactly upto fourth degree polynomial.
Properties :
1. If f(x) is a polynomial of degree N, then the Nth divided difference of f(x) is a constant.
Proof : Consider the divided difference of xn
 (x1+ h)n - xn n h xn-1+ . . . Dxn = = x + h - x h

=  a polynomial of degree (n - 1)

Also since divided difference operator is a linear operator, D of any Nth degree polynomial is an (N-1)th degree polynomial and second D is an (N-2) degree polynomial, so on  the Nth divided difference of an Nth degree polynomial is a constant.
2. If   x0, x1, x2 . . . xn are the (n+1) discrete points then the Nth divided difference is equal to
 f0 fn f[x0, x1, x2 . . . xn] = + . . . + (x0 - x1) . . . (x0 - xn) (xn - x0) . . . (xn - xn-1)
Proof : If n = 0 Þ f(x0) = f(x0) hence the result is true let us assume that the result is valid upto n = k
 f0 fk Þ f[x0, x1, . . . xk] = + . . . + (x0 - x1) . . . (x0 - xk) (xk - x0) . . . (xk - xk-1)
Consider the case n = k + 1
 f[x1, x2, . . .  xk+1] - f[x0, x1, . . . xk] Þ f[x0, x1, . . . xk, xk+1] = (xk+1 - x0)

 1 [ f1 fk+1 ] = + . . . + (xk+1 - x0) (x1 - x2) . . . (x1 - xk+1) (xk+1 - x1) . . . (xk+1 - xk)

 1 [ f0 fk ] - + . . . + (xk+1 - x0) (x0 - x1) . . . (x0 - xk) (xk - x0) . . . (xk - xk-1)

 f0 + f1 ( 1 1 ) fk+1 = - +...+ (x0-x1)...(x0-xk+1) (x1-x2)...(x1 - xk)(xk+1 - x0) x1-xk+1 x1-x0 (xk+1-x0)...(xk+1-xk)

 f0 + f1 fk+1 = + . . . + (x0 - x1) . . . (x0 - xk+1) (x1 - x0) (x1 - x2) . . . (x1 - xk+1) (xk+1 - x0) . . . (xk+1 - xk)
3. Sheppard Zigzag rule :
Consider the divided difference table for the data points (x0, f0), (x1, f1), (x2, f2) and (x3, f3)
In the difference table the dotted line and the solid line give two differenct paths starting from the function values to the higher divided difference's posssible to the function values. The Sheppard's zigzag rule says the function value at any non-tabulated from the dotted line or from the solid line are same provided the order of xi are taken in the direction of the zigzag line. That is any f(x) through the dotted line can be approximated as

f(x) = f0 + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1,x2]  + (x - x0) (x - x1) (x - x2)f [x0, x1, x2, x3].

Similarly f(x) over the solid line is euivalent to

f(x) = f2 + (x - x2) f [x1, x2] + (x - x2) (x - x1) f [x1, x2,x3]  + (x - x2) (x - x1) (x - x3)f [x0, x1, x2, x3].

Example : Find f(1.5) from the data points

 x 0 0.5 1 2 f(x) 1 1.8987 3.7183 11.3891

f(1.5) along the dotted line is
f(1.5) = 1 + 1.5 x 1.7974 + 1.5 (1) x 1.8418 + (1.5) (1) (0.5) x 0.4229
= 6.770

Similarly f(1.5) along the solid line is
f(1.5) = 3.7183+(1.5 - 1)x3.6392+(1.5 - 1)(1.5 - 0.5)x2.6877+(1.5 -1)(1.5 - 0.5)(1.5 - 2)x0.4229
= 6.770

The data is given for f(x) = x2 + ex and the analytical value for f(1.5) = 6.7317

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