0

Guest

Courses New

Dear Sir, I would like to know how we can find if the given problem should be done with Newtons forward interpolation formula or Newtons backward interpolation formula. Please give me the method to find it. Thank you. Anitha ashok.

Dear Sir,

I would like to know how we can find if the given problem should be done with Newtons forward interpolation formula or Newtons backward interpolation formula. Please give me the method to find it.

Thank you.

Anitha ashok.

Grade:Upto college level

1 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

Newton's Interpolation Formulae

As stated earlier, interpolation is the process of approximating a given function, whose values are known at $ N+1$ tabular points, by a suitable polynomial, $ P_N(x),$ of degree $ N$ which takes the values $ y_i$ 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 $ y = f(x),$ when the tabular points $ x_i$ 's are equally spaced. Let

 

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

 

where the polynomial $ P_N(x)$ 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 $ a_0,a_1,...a_N,$ to be determined using the fact that $ P_N(x_i) = y_i$ for $ i=0,1,\ldots,N.$

So, for $ i=0,$ substitute $ x = x_0$ in (11.4.1) to get $ P_N(x_0)=y_0.$ This gives us $ a_0=y_0.$ 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 $ i=2,$ $ \;\; 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,

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


As this uses the forward differences, it is called NEWTON'S FORWARD DIFFERENCE FORMULA for interpolation, or simply, forward interpolation formula.

Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More