# Local Maximum of a 4th degree polynomial (Interpolation)

I am trying to determine the following: $$max_{x_{1} \leq x \leq x_{4}} \mid (x - x_{1})(x-x_{2})(x-x_{3})(x-x_{4}) \mid$$ but in terms of $h = x_{4}-x_{3} = x_{2}- x_{2}$ etc. The differences between the four points along the x-axis are all $h$ due to this being equally spaced interpolation.

So through the use of $\textbf{Sympy}$ I managed to determine values for this local maximum in terms of $h$. Just easy substitution and basic derivatives.

What I got though is the following: $$max = -h^{4} \mbox{ or } \frac{9}{16}h^{4}$$ So, all that bugs me is the negative sign in the first expression. Is it that one due to the fact that one is looking for the maximum value which is of an absolute nature. Which one is the correct maximum value?

Just to build on what is going on, for a 3rd degree polynomial the following is found: $$max = \frac{2}{3\sqrt{3}} h^{3}$$ And for a 2nd degree polynomial: $$max = \frac{1}{4}h^{2}$$ The reason for all of this is to determine the maximum error in interpolation of functions. Thank you so much for your time.

• $-h^4$ is the value of $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ (without the $|\cdot |$) at the position where $|(x-x_1)(x-x_2)(x-x_3)(x-x_4)|$ attains its maximum. – Reinhard Meier Mar 7 '17 at 11:43

Symmetry suggests the substitution $y = x-\cfrac{x_1+x_4}{2}= x - x_1-\cfrac{3h}{2} \in \left[-\cfrac{3h}{2}, \,\cfrac{3h}{2}\right]\,$, then the expression can be written in terms of $y$ as:
$$\left(y+\cfrac{3h}{2}\right)\left(y+\cfrac{h}{2}\right)\left(y-\cfrac{h}{2}\right)\left(y-\cfrac{3h}{2}\right)= \left(y^2-\cfrac{9h^2}{4}\right)\left(y^2-\cfrac{h^2}{4}\right)$$
The latter is a quadratic in $y^2\in \left[0, \,\cfrac{9h^2}{4}\right]$ which attains its extrema on the interval at $y=0$ for a maximum value of $\cfrac{9}{16}h^4\,$ and at $y=\cfrac{\pm\sqrt{5}}{2}h$ for a minimum value of $-h^4\,$. Since $\left|-h^4\right| \gt \left|\cfrac{9}{16}h^4\right|\,$, the maximum absolute value of the original expression on the given interval is $h^4\,$.