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It is an interpolation technique which is used if the interval difference is not same for all sequence of values. These differences are symmetric with respect to the arguments which is independent of the order of arguments.

DERIVATION

Let y = f(x) is known for (n + 1) distinct arguments ${\mathrm{x}}_{\mathrm{j}}$ (j = 0,1,..., n), it may not be equispaced, and ${\mathrm{y}}_{\mathrm{j}}$ = f(${\mathrm{x}}_{\mathrm{j}}$) (j = 0,1,..., n) be the corresponding entries. 

Now we have to construct a polynomial φ(x) of degree less than or equal to n, such that φ(x) replaces f(x) on the set of arguments ${\mathrm{x}}_{\mathrm{j}}$ i.e. 

Where ${\mathrm{R}}_{\mathrm{n}+1}\left(\mathrm{x}\right)$ is the remainder or error in interpolating f(x) for a given x, 

Let us choose φ(x) as a polynomial of degree ≤ n as

Similarly, we get 

And we get 

And 

GRAPHICAL REPRESENTATION

INPUT: 

The value of x and f(x)

OUTPUT: 

An interpolating point for a given value of x.

PROCESS:

Step 1: [taking inputs from the user]

	Read n [the number of values]
		for i = 0 to n - 1 repeat
			Read x1[i], y[i]
		[End of ‘for’ loop]
	Read x [the value for which f(x) is to be calculated]

Step 2: [Newton’s divide difference interpolation]
	Step 2.1: Set s ← y[0]
		Step 2.2: for i = 0 to n - 2 repeat
					Set d[i] ← ((y[i+1]-y[i])/(x1[i+j] - x1[i]))
					Set y[i] ← d[i]
				[End of ‘for’ loop]
				Set c1 ← 1
				for i = 0 to j - 1 repeat
					Set  c1 ← c1 × (x - x1[i])
			[End of ‘for’ loop]
				Set  c2 ← c2 + (y[0] × c1)
				Set  n ← n - 1
				Set  j ← j + 1
				While n != 1 repeat step 2.2
		Step 2.3: Set s ← s + c2
		Step 2.4: print s

Step 3: Stop