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Next: Newton-Gregory Backward Difference
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Newton Interpolation polynomial with equidistant points:
Gregory-Newton Forward Difference Approach:
Very often it so happens in practice that the given data set
correspond to a sequence
of
equally spaced points. Here we can assume that
![](img199a.gif) |
(1) |
where
![$ x_{0}$](img200.png)
is the starting point (sometimes, for convenience,
the middle data point is taken as
![$ x_{0}$](img200.png)
and in such a case the
integer
![$ i$](img201.png)
is allowed to take both negative and positive values.)
and
![](img202a.gif)
is the step size. Further it is enough to calculate simple
differences rather than the divided differences as in the
non-uniformly placed data set case. These simple differences can
be forward differences
![$ (\Delta f_{i})$](img203.png)
or backward differences
![$ (\nabla f_{i})$](img204.png)
. We will first look at forward differences and
the interpolation polynomial based on
forward differences.
The first order forward difference
![$ \Delta f_{i}$](img205.png)
is defined as
![$\displaystyle \Delta f_{i}=f_{i+1}-f_{i}$](img206.png) |
(7.1) |
The second order forward difference
is defined
as
![$\displaystyle \Delta^{2}f_{i}=\Delta f_{i+1}-\Delta f_{i}$](img208.png) |
(7.2) |
The
order forward difference
is
defined as
![$\displaystyle \Delta^{k}f_{i}=\Delta^{k-1}f_{i+1}-\Delta^{k-1}f_{i}$](img211.png) |
(7.3) |
Since we already know Newton interpolation polynomial in terms of
divided differences, to derive or generate Newton interpolation
polynomial in terms of forward differences it is enough to express
forward differences in terms of divided differences.
Recall the definition of first divided difference
![$ f[x_{0},x_{1}]$](img212.png)
,
![% latex2html id marker 2670
$\displaystyle \therefore \Delta f_{0}=hf[x_{0},x_{1}]$](img214.png) |
(8.1) |
Similarly we can get
![$\displaystyle \Delta f_{1}= hf[x_{1},x_{2}]$](img215.png) |
(8.2) |
By the definition of second order forward difference
, we get
In a similar way, in general, we can show that
![$\displaystyle \Delta^{k}f_{i}=k!h^{k}f[x_{i},x_{i+1},x_{i+2}...x_{i+k}]$](img223.png) |
(8.4) |
![% latex2html id marker 2701
$\displaystyle \therefore f[x_{i},x_{i+1},...x_{i+k}]=\frac{\Delta^{k}f_{i}}{k!h^{k}}$](img224.png) |
(8.5) |
For
,
![$\displaystyle f[x_{0},x_{1}...x_{k}]=\frac{\Delta^{k}f_{0}}{k!h^{k}}$](img226.png) |
(8.6) |
Now using (6.1) & (8.6) the Newton forward difference interpolation
polynomial may be written as follows:
![](img227b.gif) |
(9) |
To rewrite (9) in a simpler way let us set
i.e |
![](img234a.gif) |
(10) |
where
This is known as Newton-Gregory forward difference interpolation
polynomial. For convenience while constructing (10) one can first
generate a forward difference table and use the
![$ \Delta^{k}f_{i}$](img210.png)
from the table. Suppose we have data set
![$ (x_{i}f_{i})$](img236.png)
,
![$ i=0,1,2,3,4$](img237.png)
then forward difference table looks as follows:
Example 1:
Given the following data, estimate
using Newton-Gregory forward difference interpolation polynomial:
i |
0 |
1 |
2 |
3 |
4 |
![$ x_{i}$](img42.png) |
1.0 |
3.0 |
5.0 |
7.0 |
9.0 |
![$ f_{i}$](img240.png) |
0 |
1.0986 |
1.6094 |
1.9459 |
2.1972 |
Solution:
Here we have five data points i.e
Let us first generate the forward difference table.
Newton Gregory forward difference interpolation
polynomial is given by:
Example 2:
Given the following data estimate f(4.12)
using Newton-Gregory forward difference interpolation
polynomial:
i |
0 |
1 |
2 |
3 |
4 |
5 |
![$ x_{i}$](img42.png) |
0 |
1 |
2 |
3 |
4 |
5 |
![$ f_{i}$](img240.png) |
1 |
2 |
4 |
8 |
16 |
32 |
Solution:
Let us first generate the Newton-Gregory forward
difference table:
![](rat9.gif)
Here
We know that the forward difference interpolation polynomial is
given by:
Exercise: Calculate
using Newton-Gregory forward difference formula for the following data
1)
x |
10 |
20 |
30 |
40 |
50 |
f(x) |
0.1736 |
0.3420 |
0.5000 |
0.6428 |
0.7660 |
and ![](rat6.gif)
2)
x |
1.0 |
2.0 |
3.0 |
4.0 |
f(x) |
0.0 |
0.6931 |
1.0986 |
1.3863 |
and ![](ra7.gif)
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