Numerical Analysis/Newton form example

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:

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