Find Solution of an equation 2x^3-4x+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)=2x^3-4x+1`.
The value of table for `x` and `y`
| x | 2 | 2.5 | 3 | 3.5 | 4 |
|---|
| y | 9 | 22.25 | 43 | 72.75 | 113 |
|---|
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 | 9 | | | | |
| | 26.5 | | | |
| 2.5 | 22.25 | | 15 | | |
| | 41.5 | | 2 | |
| 3 | 43 | | 18 | | 0 |
| | 59.5 | | 2 | |
| 3.5 | 72.75 | | 21 | | |
| | 80.5 | | | |
| 4 | 113 | | | | |
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) = 9 + (3.8 -2) xx 26.5 + (3.8 -2)(3.8 -2.5) xx 15 + (3.8 -2)(3.8 -2.5)(3.8 -3) xx 2 + (3.8 -2)(3.8 -2.5)(3.8 -3)(3.8 -3.5) xx 0`
`y(3.8) = 9 + (1.8) xx 26.5 + (1.8)(1.3) xx 15 + (1.8)(1.3)(0.8) xx 2 + (1.8)(1.3)(0.8)(0.3) xx 0`
`y(3.8) = 9 +47.7 +35.1 +3.744 +0`
`y(3.8) = 95.544`
Solution of divided difference interpolation method `y(3.8) = 95.544`
This material is intended as a summary. Use your textbook for detail explanation.
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