2. Find Solution of an equation x^3-x+1 using Newton's Divided Difference Interpolation formula
x1 = 2 and x2 = 4
x = 3.8
Step value (h) = 0.5
Finding f(2)
Solution:
Equation is `f(x)=x^3-x+1`.
The value of table for `x` and `y`
| x | 2 | 2.5 | 3 | 3.5 | 4 |
|---|
| y | 7 | 14.125 | 25 | 40.375 | 61 |
|---|
Numerical divided differences method to find solution
Newton's divided difference table is
| x | y | `1^(st)` order | `2^(nd)` order | `3^(rd)` order | `4^(th)` order |
| 2 | 7 | | | | |
| | 14.25 | | | |
| 2.5 | 14.125 | | 7.5 | | |
| | 21.75 | | 1 | |
| 3 | 25 | | 9 | | 0 |
| | 30.75 | | 1 | |
| 3.5 | 40.375 | | 10.5 | | |
| | 41.25 | | | |
| 4 | 61 | | | | |
The value of `x` at you want to find the `f(x) : x = 3.8`
Newton's divided difference interpolation formula is
`f(x)=y_0 +(x-x_0) f[x_0, x_1]+(x-x_0)(x-x_1) f[x_0, x_1, x_2]+(x-x_0)(x-x_1)(x-x_2) f[x_0, x_1, x_2, x_3]+(x-x_0)(x-x_1)(x-x_2)(x-x_3) f[x_0, x_1, x_2, x_3, x_4]`
`y(3.8) = 7 + (3.8 -2) xx 14.25 + (3.8 -2)(3.8 -2.5) xx 7.5 + (3.8 -2)(3.8 -2.5)(3.8 -3) xx 1 + (3.8 -2)(3.8 -2.5)(3.8 -3)(3.8 -3.5) xx 0`
`y(3.8) = 7 + (1.8) xx 14.25 + (1.8)(1.3) xx 7.5 + (1.8)(1.3)(0.8) xx 1 + (1.8)(1.3)(0.8)(0.3) xx 0`
`y(3.8) = 7 +25.65 +17.55 +1.872 +0`
`y(3.8) = 52.072`
Solution of divided difference interpolation method `y(3.8) = 52.072`
This material is intended as a summary. Use your textbook for detail explanation.
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