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:

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.

and

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:

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:

- 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 table0 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,

- Using the following table for
approximate its value
at
Also, find an error estimate (Note
).
0.70 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 - Apply
degree interpolation polynomial for the set of
values given in Example 11.2.15, to estimate the
value of
by taking
**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- for
and
we have
This gives,

- for
and
we have
This gives,

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

- for
and
we have
This gives,

- Following data is available for a function
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 and

- 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.

- 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 0.20 S(x) 0 0.00003 0.00026 0.00090 0.00214 0.00419 Obtain a fifth degree interpolating polynomial for Compute and also find an error estimate for it.

- 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.