Let P_N(x) be the N^th degree polynomial through the (N+1) points x₀, x₁, . . ., x_N and E_N(x) is the error in the approximation of f(x) then : 

E_N(x) = f(x) - P_N(x)

Since both f(x) and P_N(x) have same value at the x_i, i = 0, 1, . . . N, the error E(x) can be written as

E_N(x) = f(x) - P_N(x) = (x - x₀)(x - x₁). . .(x - x_N) g(x)

where g(x) represents the E_N(x) at non tabulated points x. Obviously f(x) - P_N(x) - E_N(x) = 0

Þ f(x) - P_n(x) - (x - x₀)(x - x₁) . . . (x - x_n) g(x) = 0

let us construct an auxilary function W(t) such that

W(t) = f(t) - P_N(t) - (t - x₀)(t - x₁) . . . (t - x_N) g(x)

i.e., W(t) is actually a function of t and x.

Now at all t = x_i, i = 0, 1, . . . n W(t) is zero and also at t = x. Hence there are about (n+2) zero's of W(t).

If we impose the law of mean value on W(t) then W(t) must be continuous and differentiable. Also there exists a root of w'(t) between each of the (N+2) zeros of W(t), that is there and a total of (N+1) zeros of W(t). If there exists W''(t) then there are N zeros and similarly if the (N+1) derivative exists then there exist atleast a zero in the interval of (x₀,x_n) let us call this as t = x then

Example :

For the following data find f(0.15) accuracy for the maximum possible degree. Also obtain a bound on the truncation error for this approximation
 
 

xi	fi
0.1	0.09983
		0.9884
0.2	0.19867		-0.099499814
		0.9685		-0.16000074
0.3	0.29552		-0.14750004		0.008332171
		0.939		-0.15666787
0.4	0.38942		-0.1945004
		0.9001
0.5	0.47943

f(0.15) = 0.09983 + 0.9884( 0.15 - 0.1 )  - 0.099499814( 0.15 - 0.1 )( 0.15 - 0.2 )
                             -  0.16000074( 0.15 - 0.1 )( 0.15 - 0.2 )( 0.15 - 0.3 ) 
                             + 0.008332171( 0.15 - 0.1 )( 0.15 - 0.2 )  ( 0.15 - 0.3 )( 0.15 - 0.4 ) 

f(0.15) = 0.14943796

Error in the approximation
 
 

E5(x) =	(.15 - .1) (.15 - .2) (.15 - .3) (.15 - .4) (.15 - .5)	f6(x)
	6!

where 0.1 < x < 0.5,    f⁶(x) = - sin(x)

Max | f⁶(x) |_{0.1 < x < 0.5} = 0.47943

ÞE₅(x) = 2.1849e^-08