# Squarish polynomials

For some reason I wanted to construct polynomials approximating this piecewise constant function ${f}$:

Of course approximation cannot be uniform, since the function is not continuous. But it can be achieved in the sense of convergence of graphs in the Hausdorff metric: their limit should be the “graph” shown above, with the vertical line included. In concrete terms, this means for every ${\epsilon>0}$ there is ${N}$ such that for ${n\ge N}$ the polynomial ${p_n}$ satisfies

$\displaystyle |p_n-f|\le \epsilon\quad \text{ on }\ [0,2]\setminus [1-\epsilon,1+\epsilon]$

and also

$\displaystyle -\epsilon\le p_n \le 1+\epsilon\quad \text{ on }\ [1-\epsilon,1+\epsilon]$

How to get such ${p_n}$ explicitly? I started with the functions ${f_m(x) = \exp(-x^m)}$ when ${m}$ is large. The idea is that as ${m\rightarrow\infty}$, the limit of ${\exp(-x^m)}$ is what is wanted: ${1}$ when ${x<1}$, ${0}$ when ${x>1}$. Also, for each ${m}$ there is a Taylor polynomial ${T_{m,n}}$ that approximates ${f_m}$ uniformly on ${[0,2]}$. Since the Taylor series is alternating, it is not hard to find suitable ${n}$. Let’s shoot for ${\epsilon=0.01}$ in the Taylor remainder and see where this leads:

• Degree ${7}$ polynomial for ${\exp(-x)}$
• Degree ${26}$ polynomial for ${\exp(-x^2)}$
• Degree ${69}$ polynomial for ${\exp(-x^3)}$
• Degree ${180}$ polynomial for ${\exp(-x^4)}$
• Degree ${440}$ polynomial for ${\exp(-x^5)}$

The results are unimpressive, though:

To get within ${0.01}$ of the desired square-ness, we need ${\exp(-1.01^m)<0.01}$. This means ${m\ge 463}$. Then, to have the Taylor remainder bounded by ${0.01}$ at ${x=2}$, we need ${2^{463n}/n! < 0.01}$. Instead of messing with Stirling’s formula, just observe that ${2^{463n}/n!}$ does not even begin to decrease until ${n}$ exceeds ${2^{463}}$, which is more than ${10^{139}}$. That’s a … high degree polynomial. I would not try to ask a computer algebra system to plot it.

Bernstein polynomials turn out to work better. On the interval ${[0,2]}$ they are given by

$\displaystyle p_n(x) = 2^{-n} \sum_{k=0}^n f(2k/n) \binom{n}{k} x^k (2-x)^{n-k}$

To avoid dealing with ${f(1)}$, it is better to use odd degrees. For comparison, I used the same or smaller degrees as above: ${7, 25, 69, 179, 439}$.

Looks good. But I don’t know of a way to estimate the degree of Bernstein polynomial required to obtain Hausdorff distance less than a given ${\epsilon}$ (say, ${0.01}$) from the square function.