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

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


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?

  • $\begingroup$ 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 $x_0+\frac h2$. $\endgroup$ Commented Feb 23, 2021 at 15:23
  • $\begingroup$ @PierreCarre why not? $\endgroup$ Commented Feb 23, 2021 at 19:12
  • 1
    $\begingroup$ Because you can compute exactly where will the maximum occur and it is not $x_0+\frac h2$. $\endgroup$ Commented Feb 23, 2021 at 19:42
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Feb 23, 2021 at 20:08

1 Answer 1


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 $g(\pm \frac{\sqrt{3}}{3}) = \frac{2h^3}{3\sqrt{3}}$.

  • $\begingroup$ How do you know the maximum is at $x_0 + (1 \pm \frac{\sqrt{3}}{3})h$? $\endgroup$ Commented Feb 23, 2021 at 19:28
  • $\begingroup$ @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). $\endgroup$ Commented Feb 23, 2021 at 19:43
  • $\begingroup$ 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. $\endgroup$ Commented Feb 23, 2021 at 23:29
  • $\begingroup$ @Segmentationfault I'll add to my answer. $\endgroup$ Commented Feb 24, 2021 at 9:44

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