# 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 for 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

where the polynomial is given in the following form:
 (11.4.1)

for some constants to be determined using the fact that for

So, for substitute in (11.4.1) to get This gives us Next,

So, For or equivalently

Thus, Now, using mathematical induction, we get

Thus,

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:
 (11.4.2)

If =1, we have a linear interpolation given by

 (11.4.3)

For we get a quadratic interpolating polynomial:

 (11.4.4)

and so on.

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

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

Similarly, if we assume, is of the form

then using the fact that we have

Thus, using backward differences and the transformation 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

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 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, for the following tabular data and interpolate the value of the function at

 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 where and we get

Thus,

2. Using the following table for approximate its value at Also, find an error estimate (Note ).
 0.7 72 0.74 0.76 0.78 0.84229 0.87707 0.91309 0.95045 0.98926
Solution: As the point 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 is almost constant, so we shall attempt degree polynomial interpolation.

Note that gives Thus, using forward interpolating polynomial of degree we get

An error estimate for the approximate value is

Note that exact value of (upto decimal place) is 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 degree interpolation polynomial for the set of values given in Example 11.2.15, to estimate the value of by taking

Also, find approximate value of
Solution: Note that is closer to the values lying in the beginning of tabular values, while is towards the end of tabular values. Therefore, we shall use forward difference formula for and the backward difference formula for Recall that the interpolating polynomial of degree is given by

Therefore,
1. for and we have This gives,

2. for and we have This gives,

Note: as is closer to we may expect estimate calculated using to be a better approximation.
3. for we use the backward interpolating polynomial, which gives,

Therefore, taking and we have This gives,

EXERCISE 11.4.6
1. Following data is available for a function

 x 0 0.2 0.4 0.6 0.8 1 y 1 0.808 0.664 0.616 0.712 1

Compute the value of the function at and

2. The speed of a train, running between two station is measured at different distances from the starting station. If is the distance in from the starting station, then the speed (in ) of the train at the distance 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 0 0.04 0.08 0.12 0.16 0.2 S(x) 0 3e-05 0.00026 0.0009 0.00214 0.00419

Obtain a fifth degree interpolating polynomial for Compute and also find an error estimate for it.

4. Following data gives the temperatures (in ) 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.

A K Lal 2007-09-12