Numerical Analysis/Newton form example

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We'll find the interpolating polynomial passing through the points $(1,-6) = (x_{0},y_{0})$ , $(2,2) = (x_{1},y_{1})$ , $(4,12) = (x_{2},y_{2})$ , using the Newton form of the interpolation polynomial.

The Newton form is given by the formula $p(x) = \sum_{j=0}^{k}a_{j}n_{j}(x)$ , where $a_{j} = [y_{0},\ldots,y_{j}]$ and $n_{j}(x) = \prod_{i = 0}^{j-1}(x-x_{i})$ , with $n_{0}(x) = 1$ . We start by finding each $n_{j}(x)$ .

$n_{0}(x) = 1$

$n_{1}(x) = x-1$

$n_{2}(x) = (x-1)(x-2) = x^{2} -3x +2$

Next, we find the necessary divided differences. First, $[y_{0}] = -6$ , $[y_{1}] = 2$ , and $[y_{2}] = 12$ . For the next level, we have:

$[y_{0},y_{1}] = \frac{2+6}{2-1} = 8$

$[y_{1},y_{2}] = \frac{12-2}{4-2} = 5$

Finally, we can find:

$[y_{0},y_{1},y_{2}] = \frac{5-8}{4-1} = -1$ .

Now, we can find the coefficients $a_j$ .

$a_{0} = [y_{0}] = -6$

$a_{1} = [y_{0},y_{1}] = 8$

$a_{2} = [y_{0},y_{1},y_{2}] = -1$

Substituting and simplifying, we get our interpolating polynomial:

$p(x) = -6 +8(x-1) - (x^{2} -3x +2) = -x^{2} + 11x -16$ .

Adding a point

Now let's add the point $(3,-10) = (x_{3},y_{3})$ to our data set and find the new polynomial using the same method. Due to the formula for the Newton form, we only have to add the term $a_{3}n_{3}(x)$ to our previous interpolating polynomial.

First, we have

$n_{3}(x) = (x-1)(x-2)(x-4) = x^{3} - 7x^{2} +14x -8$ .

Now to find $a_{3}$ we calculate some more divided differences.

$[y_{3}] = -10$

$[y_{2},y_{3}] = \frac{-10-12}{3-4} = 22$

$[y_{1},y_{2},y_{3}] = \frac{22-5}{3-2} = 17$

$a_{3} = [y_{0},\ldots,y_{3}] = \frac{17+1}{3-1} = 9$

So, our new interpolating polynomial is:

$p(x) = -x^{2} + 11x -16 + 9(x^{3}-7x^{2}+14x-8) = 9x^{3}-64x^{2}+137x-88$ .

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