# Show that $|E(x)|\le 0.65 (x_1-x_0)^3M$ for the error of interpolation of a 2nd degree polynomial

Show that on the polynomial interpolation of a function $$f \in C^3$$ in 3 points $$x_0 < x_1 < x_2$$ with $$x_2-x_1 = x_1 - x_0$$, the error of interpolation verifies $$|E(x)| \le 0.65(x_1-x_0)^3M, \forall x$$ between $$x_0$$ and $$x_2$$, where $$M$$ is the absolute value of the maximum of the 3rd derivative of $$f$$ on the interval $$[x_0,x_2]$$.

I did

\begin{align} |E(x)| &\le 0.65(h)^3M \\ |E(p_2(x))| &\le \max_{x_2,x_0}|(x-x_0)(x-x_1)(x-x_2)|M/3! = \\ \end{align} \begin{align} |(x_0+h/2-x_2)(x_0+h/2-x_1)(x_0+h/2-x_2)|M/6&\le 0.65(h)^3M = \\ |(h/2)(-h/2)(3h/2)|M/6 &\le 0.65h^3M \Leftrightarrow \\ 3h^3/8* M/6 &\le 0.65h^3M \end{align}

...q.e.d?

How I got to x being equal to $$x_0+h/2$$ where h = $$x_1-x_0$$: I drew some 3rd degree polynomials and "kind of" noticed they were symmetrical and that $$x_0$$, $$x_1$$ and $$x_2$$ were always the zeroes. I also noticed that $$f(x_1)$$ was always equal to zero and that it stood halfway between the other two, and between each, there was a maximum or minimum. So I chose $$x = x_0+h/2$$.

Is this correct?

• Although it is true that the max of the $|(x-x_0)(x-x_1)(x-x_2)|$ occurs in some point between $x_0$ and $x_1$, it is not ok to assume that it occurs at the midpoint $x_0+\frac h2$. – PierreCarre Feb 23 at 15:23
• @PierreCarre why not? – Segmentation fault Feb 23 at 19:12
• Because you can compute exactly where will the maximum occur and it is not $x_0+\frac h2$. – PierreCarre Feb 23 at 19:42
• If you write $x=x_1+sh$, $s\in[-1,1]$, you get $(x-x_0)(x-x_1)(x-x_2)=(s^3-s)h^3$. The cubic polynomial has derivative $3s^2-1$, which quite easily gives the positions of the extrema. – Lutz Lehmann Feb 23 at 20:08

## 1 Answer

So, you wrote the error formula $$|E_2(x)| \leq |(x-x_0)(x-x_2)(x-x_2)| \frac{M}{3!}.$$

Now you can simply maximize the 3rd degree polynomial... The maximum will be attained at a stationary point, that you can compute explicitly. The stationary points are $$x_0 + (1 \pm \frac{\sqrt{3}}{3})h$$, and the corresponding maximum value will be $$\frac{2h^3}{3 \sqrt{3}}$$. In general, your choice of $$x = x_0 +\frac h2$$ is wrong and does not yield the maximum.

Using this information,

$$|E_2(x)|\leq \frac{M h^3}{9 \sqrt{3}}\approx 0.06415 (x_1-x_0)^3 M.$$

Finally, if this estimate holds, so does the (much worse) proposed estimate.

Regarding the maximization of $$|(x-x_0)(x-x_1)(x-x_2)|$$ in the interval $$[x_0, x_2]$$, as noted by Lutz Lehmann, it is equivalent to the maximization of $$g(s)=|(s^3-s)h^3|$$ for $$s \in [-1,1]$$. Since $$g$$ is zero on the boundary and at the branching point for the absolute value, the maximum will occur at a point where $$g'(s)=0$$, i.e. $$s= \pm \frac{\sqrt{3}}{3}$$. Substituting back in the function, we conclude that the maximum is $$g(\pm \frac{\sqrt{3}}{3}) = \frac{2h^3}{3\sqrt{3}}$$.

• How do you know the maximum is at $x_0 + (1 \pm \frac{\sqrt{3}}{3})h$? – Segmentation fault Feb 23 at 19:28
• @Segmentationfault the maximum will occur at a point where the derivative of the polynomial is zero, and you can compute the zeros of the derivative (these are the zeros of a polynomial with degree 2). – PierreCarre Feb 23 at 19:43
• I'm trying to calculate the maxima. I got that the derivative is $(3s^2-1)h^2$ where $s \in [-1,1]$. It's zero when $s = \pm 1/\sqrt(3)$. Replacing back the maxima is $(\pm 1/3 \mp 1/\sqrt(3))$. I don't know what to do next. – Segmentation fault Feb 23 at 23:29
• @Segmentationfault I'll add to my answer. – PierreCarre Feb 24 at 9:44