In the case of differentiation, we first write the interpolating formula on the interval and the differentiate the polynomial term by term to get an approximated polynomial to the derivative of the function. When the tabular points are equidistant, one uses either the Newton's Forward/ Backward Formula or Sterling's Formula; otherwise Lagrange's formula is used. Newton's Forward/ Backward formula is used depending upon the location of the point at which the derivative is to be computed. In case the given point is near the mid point of the interval, Sterling's formula can be used. We illustrate the process by taking (i) Newton's Forward formula, and (ii) Sterling's formula.
Recall, that the Newton's forward interpolating polynomial is given by
Differentiating (13.2.1), we get the approximate
value of the first derivative at
as
Thus, an approximation to the value of first derivative at i.e. is obtained as :
(13.2.4) |
1.73 | 1.74 | 1.75 | 1.76 | 1.77 | |
1.772844100 | 1.155204006 | 1.737739435 | 1.720448638 | 1.703329888 |
Hence compute from following table the value of the derivative of at :
1.05 | 1.15 | 1.25 | |
2.8577 | 3.1582 | 3.4903 |
Note the error between the computed value and the true value is
Hence compute the derivative of at from the following table:
1.15 | 1.20 | 1.25 | |
3.1582 | 3.3201 | 3.4903 |
Hence compute from following table the value of the derivative of at :
1.05 | 1.10 | 1.15 | 1.20 | 1.25 | |
2.8577 | 3.0042 | 3.1582 | 3.3201 | 3.4903 |
0.00 | 0.05 | 0.10 | 0.15 | 0.20 | 0.25 | |
0.00000 | 0.10017 | 0.20134 | 0.30452 | 0.41075 | 0.52110 |
In particular, the value of the derivative at is given by
0.4 | 0.6 | 0.7 | |
3.3836494 | 4.2442376 | 4.7275054 |
Note: The reader is advised to derive the above expression.
A K Lal 2007-09-12