Building newton interpolaton polynomial

Question

Let $x_k, y_k \in \Bbb R, \ x_k = kh, \ k = 0, 1, 2$ and $h \in (0,1)$. I want to build an interpolation polynomial $p \in P_2$ (so a polynomial of degree $2$) with the help of the Newton base. Following the formula, I receive

$$p(x) = y_0 + {{y_1 - y_0} \over h}(x-x_0) + {{2y_1 - y_0 - y_2} \over 2h^2}(x-x_0)(x-x_1).$$

Although we have $p(x_0) = y_0$, we don't have $p(x_1) = y_1$ and $p(x_2) = y_2$, and I just don't see where I am mistaken here. It should be a simple calculation actually. Can anybody help me out?

Have you seen a presentation of the divided differences tableau? — hardmath, Jun 28 '17 at 15:32
since $x_0 = 0, x_1 = h, x_2 = 2h$ we get $p(x_1) = y_0 + \frac{y_1 - y_0}{h}(x_1) = y_0 + \frac{y_1 - y_0}{h}(h) = y_0 + y_1 - y_0 = y_1$ — Chiray, Jun 28 '17 at 15:35

Prajwal Kansakar · Accepted Answer · 2017-06-28T15:42:46.510

3

You got the coefficient of $(x-x_0)(x-x_1)$ wrong. It should be $$\frac{\frac{y_2-y_1}{x_2-x_1}-\frac{y_1-y_0}{x_1-x_0}}{x_2-x_0}=\frac{y_0+y_2-2y_1}{2h^2}.$$

edited Jun 28 '17 at 15:42

answered Jun 28 '17 at 15:35

Prajwal Kansakar

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Building newton interpolaton polynomial

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