newton interpolation problem

with w(x)=(x-x(0))(x-x(1))...(x-x(n)) prove that

f[x(0),x(1),x(2),..,x(n))]=summation(0,n) (f(x(i))/(derivative(w(x(i))))

and hence calculate the limit for formula f[x(0),x(1),x(2),..,x(n))] when x(2)->x(1) while all other points remain fixed

i have done the first part but stuck in the second part of calculating the limit.please help

note:f[x(0),x(1),x(2),..,x(n))] is the newton forward difference function

Re: newton interpolation problem

Quote:

Originally Posted by

**prasum** with w(x)=(x-x(0))(x-x(1))...(x-x(n)) prove that

f[x(0),x(1),x(2),..,x(n))]=summation(0,n) (f(x(i))/(derivative(w(x(i))))

and hence calculate the limit for formula f[x(0),x(1),x(2),..,x(n))] when x(2)->x(1) while all other points remain fixed

i have done the first part but stuck in the second part of calculating the limit.please help

note:f[x(0),x(1),x(2),..,x(n))] is the newton forward difference function

it sure would be nice if you learned to use LaTex. It's not that hard.

Is this what you mean?

$w(x)=\displaystyle \prod_{i=1}^n \left(x-x_i\right)$ prove that

$f\left(x_0, x_1,\dots, x_n\right)=\displaystyle{\sum_{i=0}^n} \dfrac{f(x_i)}{\dfrac{dw}{dx_i}}$

and calculate

$\displaystyle \lim_{x_2 \to x_1}f\left(x_0, x_1,\dots, x_n\right)$

Re: newton interpolation problem

yes can u please help me with the second part of calculating the limit