Created: 2020-09-05T23:08:00+00:00
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Newton interpolation maps x to f(x) using a table of known values.
| | x1 | x2 | x3 | x4 | |------+----+----+----+----| | x | -5 | -1 | 0 | 2 | | f(x) | -2 | 6 | 1 | 3 | |------+----+----+----+----| | | f1 | f2 | f3 | f4 |
| x | f(x) | f(x)[1] 1st | f(x)[2] 2nd | f(x)[3] 3rd | |----+------+-----------------+-----------------------+-----------------------| | -5 | -2 | (f2-f1)/(x2-x1) | (f2[1]-f1[1])/(x3-x1) | (f2[2]-f1[2])/(x4-x1) | | -1 | 6 | (f3-f2)/(x3-x2) | (f3[1]-f2[1])/(x4-x2) | | | 0 | 1 | (f4-f3)/(x4-x3) | | | | 2 | 3 | | | |
$ f3(x)=b_1 + b_2(x-x1) + b_3(x-x1)(x-x2) + b_4(x-x1)(x-x2)(x-x3)
The topmost values from the differences table constitute the "b values."
| f(x) | f(x)[1] | f(x)[2] | f(x)[3] | | b1 | b2 | b3 | b4 |