Newton's Interpolation - Interpolation and Extrapolation, Business Mathematics and Statistics | Business Mathematics and Statistics - B Com PDF Download

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,,...,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

where the polynomial P_N (x) is given in the following form: 

P_N (x)

= 

 (11.4.1)

for some constants a₀ , a₁, .... a_N to be determined using the fact that P_N (x_i) = y_i for i = 0,1,...,N

So, for i = o substitute x = x₀ in (11.4.1) to get  P_N (x₀) = y_o This gives us a₀ = y₀ Next,

So,  For   or equivalently

Thus,  Now, using mathematical induction, we get

Thus, 

P_N (x) = 

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

EXERCISE 11.4.1   Show that

   and

and in general,

For the sake of numerical calculations, we give below a convenient form of the forward interpolation formula.

Let  then

With this transformation the above forward interpolation formula is simplified to the following form: 

P_N (u) = 

= 

 (11.4.2)

If N =1, we have a linear interpolation given by

 (11.4.3)

For N = 2, we get a quadratic interpolating polynomial:

(11.4.4)

and so on.

It may be pointed out here that if f(x) is a polynomial function of degree N then P_N (x) coincides with f(x) on the given interval. Otherwise, this gives only an approximation to the true values of f(x)

If we are given additional point x_N+1 also, then the error, denoted by  is estimated by

Similarly, if we assume, P_N(x) is of the form

then using the fact that P_N (x_i = y_i) we have

b₀ = y_N

b₁ = 

b_2 ⁼ 

.

.

.

b_k ⁼ 

Thus, using backward differences and the transformation  x= x_N + h_u we obtain the Newton's backward interpolation formula as follows:

(11.4.5)

EXERCISE 11.4.2   Derive the Newton's backward interpolation formula (11.4.5) for N = 3

Remark 11.4.3   If the interpolating point lies closer to the beginning of the interval then one uses the Newton's forward formula and if it lies towards the end of the interval then Newton's backward formula is used.

Remark 11.4.4   For a given set of n tabular points, in general, all the n points need not be used for interpolating polynomial. In fact N is so chosen that N^th forward/backward difference almost remains constant. Thus N is less than or equal to n.

EXAMPLE 11.4.5  

1. Obtain the Newton's forward interpolating polynomial, P₅ (x) for the following tabular data and interpolate the value of the function at x = 0.0045.x

   x	0	0.001	0.002	0.003	0.004	0.005
   y	1.121	1.123	1.1255	1.127	1.128	1.1285

   
   Solution: For this data, we have the Forward difference difference table

    

   0	1.121	0.002	0.0005	-0.0015	0.002	-.0025
   .001	1.123	0.0025	-0.0010	0.0005	-0.0005
   .002	1.1255	0.0015	-0.0005	0.0
   .003	1.127	0.001	-0.0005
   .004	1.128	0.0005
   .005	1.1285

   Thus, for  x = x_o + h_u, where  x₀ = 0, h = 0.001 and  we get 
   
   P₅(x) = 
   

   Thus, 

P₅(0.0045) = P₅ (0.001x4.5)

= 

= 1.12840045.

2. Using the following table for  tan x, approximate its value at 0.71 Also, find an error estimate (Note  tan (0.71) = 0.85953).

Solution: As the point x = 0.71 lies towards the initial tabular values, we shall use Newton's Forward formula. The forward difference table is:

In the above table, we note that  Δ³y is almost constant, so we shall attempt 3^rd degree polynomial interpolation.

Note that  x₀ = 0.70, h = 0.02 gives  Thus, using forward interpolating polynomial of degree 3, we get

Thus, tan(0.71) ≈ 

= 0.859535

An error estimate for the approximate value is

Note that exact value of  tan (0.71) (upto  decimal place) is 0.85953 and the approximate value, obtained using the Newton's interpolating polynomial is very close to this value. This is also reflected by the error estimate given above.

3. Apply 3^rd degree interpolation polynomial for the set of values given in Example 11.2.15, to estimate the value of  f (10.3) by taking

(i) x₀ = 9.0,    (ii) x₀ = 10.0

Also, find approximate value of  f (13.5)

Solution: Note that x = 10.3x is closer to the values lying in the beginning of tabular values, while x =13.5 is towards the end of tabular values. Therefore, we shall use forward difference formula for x = 10.3 and the backward difference formula for x = 13.5 Recall that the interpolating polynomial of degree 3 is given by

Therefore,

1. for x₀ = 9.0, h = 1.0 and x = 10.3, we have  This gives, 

f (10.3) ≈ 

= 5.559

2. for x₀ = 10.0, h=1.0 and x = 10.3 we have  This gives, 

f(10.3)  ≈  

= 5.54115.

Note: as x = 10.3x is closer to  x = 10.0, we may expect estimate calculated using x₀ = 10.0 to be a better approximation.

1. for x₀ = 13.5, we use the backward interpolating polynomial, which gives,

   

    

   Therefore, taking x_N = 14, h = 1.0 and  x = 13.5, we have  This gives, 
   f(13.5) ≈ 
   
   = 7.8125 
   
    

EXERCISE 11.4.6  

1. Following data is available for a function y = f(x)

    

   x	0	0.2	0.4	0.6	0.8	1.0
   y	1.0	0.808	0.664	0.616	0.712	1.0

   Compute the value of the function at x = 0.3 and x = 1.1

2. The speed of a train, running between two station is measured at different distances from the starting station. If x is the distance in km from the starting station, then v (x) the speed (in km/hr ) of the train at the distance x is given by the following table:

   x	0	50	100	150	200	250
   v(x)	0	60	80	110	90	0

   Find the approximate speed of the train at the mid point between the two stations.

3. Following table gives the values of the function  at the different values of the tabular points x,

   x	0	0.04	0.08	0.12	0.16	0.20
   S(x)	0	0.00003	0.00026	0.00090	0.00214	0.00419

   Obtain a fifth degree interpolating polynomial for S(x) Compute S (0.02) and also find an error estimate for it.

4. Following data gives the temperatures (in ⁰C ) between 8.00 am to 8.00 pm. on May 10, 2005 in Kanpur:

   Time	8 am	12 noon	4 pm	8pm
   Temperature	30	37	43	38

   Obtain Newton's backward interpolating polynomial of degree  to compute the temperature in Kanpur on that day at 5.00 pm.