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 a_{0} , a_{1}, .... 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 = x_{0} in (11.4.1) to get P_{N} (x_{0}) = y_{o }This gives us a_{0} = y_{0 }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
b_{0 }= y_{N}
b_{1} =
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
Obtain the Newton's forward interpolating polynomial, P_{5 }(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 x_{0} = 0, h = 0.001 and we get
P_{5}(x) =
Thus,
P_{5}(0.0045) = P_{5} (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).
0.70  72  0.74  0.76  0.78 
 
0.84229  0.87707  0.91309  0.95045  0.98926 

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:

 
0.70  0.84229  0.03478  0.00124  0.0001  0.00001 


0.72  0.87707  0.03602  0.00134  0.00011 



0.74  0.91309  0.03736  0.00145 




0.76  0.95045  0.03881 





0.78  0.98926 






In the above table, we note that Δ^{3}y is almost constant, so we shall attempt 3^{rd} degree polynomial interpolation.
Note that x_{0 }= 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) x_{0} = 9.0, (ii) x_{0 }= 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 x_{0} = 9.0, h = 1.0 and x = 10.3, we have This gives,
f (10.3) ≈
= 5.559
2. for x_{0 }= 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 x_{0} = 10.0 to be a better approximation.
for x_{0} = 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
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
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.
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.
Following data gives the temperatures (in ^{0}C ) 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.