Analytical Geometry> interpolation...

how to find if we should do the sum with newton's interpolation forward or by using backward formula by seeing the sum.

anitha ashok , 13 Years ago

Grade Upto college level

1 Answers

SAGAR SINGH - IIT DELHI

Last Activity: 13 Years ago

Newton's Interpolation Formulae

As stated earlier, interpolation is the process of approximating a given function, whose values are known at tabular points, by a suitable polynomial, of degree which takes the values at $ x=x_i$ for $i=0,1,\ldots,N.$ Note that if the given data has errors, it will also be reflected in the polynomial so obtained.

In the following, we shall use forward and backward differences to obtain polynomial function approximating when the tabular points 's are equally spaced. Let

$\displaystyle f(x) \approx P_N(x),$

where the polynomial is given in the following form:

$\displaystyle P_N(x)$	$\displaystyle =$	$\displaystyle a_0+a_1(x-x_0)+a_2(x-x_0)(x-x_1)+ \cdots+ a_k(x-x_0)(x-x_1)\cdots(x-x_{k-1})$
		$\displaystyle +a_N (x-x_0)(x-x_1)\cdots(x-x_{N-1}).$	(11.4.1)

for some constants to be determined using the fact that for $i=0,1,\ldots,N.$

So, for substitute $ x = x_0$ in (11.4.1) to get This gives us Next,

$\displaystyle P_N(x_1)=y_1 \Rightarrow y_1=a_0+(x_1-x_0)a_1.$

So, $a_1=\frac{y_1-y_0}{h}=\displaystyle\frac{\Delta y_0}{h}.$ For $\;\; y_2=a_0+(x_2-x_0)a_1+(x_2-x_1)(x_2-x_0)a_2,$ or equivalently

$\displaystyle 2h^2a_2= y_2 - y_0 - 2h(\frac{\Delta y_0}{h})=y_2-2y_1+y_0 = \Delta^2 y_0.$

Thus, $a_2=\displaystyle\frac{\Delta ^2 y_0}{2h^2}.$ Now, using mathematical induction, we get

$\displaystyle a_k=\frac{\Delta ^k y_0}{k! \, h^k}\; {\mbox{ for }} \; k = 0, 1, 2, \ldots, N.$

Thus,

$\displaystyle P_N(x)$	$\displaystyle =$	$\displaystyle y_0+\frac{\Delta y_0}{h}(x-x_0)+\frac{\Delta ^2 y_0}{2! \,h^2}(x-x_0)(x-x_1) + \cdots + \frac{\Delta ^k y_0}{k! \, h^k} (x-x_0)\cdots(x-x_{k-1})$
		$\displaystyle +\frac{\Delta ^N y_0}{N! \, h^N}(x-x_0)...(x-x_{N-1}).$