# Why is the polynomial best approximation to an even function itself even?

I have seen this stated and it seems intuitively obvious but I cannot prove it. I have a feeling it may be because a non-even best approximant would not satisfy the equioscillation property of the best approximation (that if |f-p|=d, where p is the best polynomial approximation of order n, there are at least n+2 successive points in the interval under consideration where alternately |f-p|=+-d ).

Note: By "best approximation" I mean the polynomial that minimizes |f-p| on [-1,1] where |.| is the sup norm. So e.g. the best approximation of order n would be p in Pn={polynomials degree<=n} which does this.

• Are you centering the approximation at zero? Is it an approximate over all real numbers? Dec 23, 2013 at 3:05
• Sorry for my ignorance but how do you mean "centering at zero"? f(x) is even about x=0 and the polynomial approximants are just general polynomials. The interval under consideration is [-1,1]. Dec 23, 2013 at 3:08
• If you're approximating with a polynomial, you get to choose where the polynomial will approximate the best. Dec 23, 2013 at 3:11
• I'm looking for the polynomial p that minimizes |f-p| over [-1,1] where |.| is the supremum norm. Dec 23, 2013 at 3:13

First, we must pin down what we mean by "better approximate." If you are interested in the interval $[-1,1]$ and your given function is $f(x)$, one reasonable definition of approximation error by a second function $g(x)$ is $$\int_{-1}^1 [f(x)-g(x)]^2\,dx.$$

Now suppose $f(x)$ is an even function, and $e(x)$ an even polynomial approximation to $f(x)$. Can we improve the approximation by adding some odd polynomial terms $o(x)$? Let's check:

\begin{align*} \int_{-1}^1 [f(x)-e(x)-o(x)]^2\,dx &= \int_{-1}^1 [f(x)-e(x)]^2 - 2[f(x)-e(x)]o(x) + o(x)^2\,dx\\ &= \int_{-1}^1 [f(x)-e(x)]^2\,dx + \int_{-1}^1 o(x)^2\,dx -2\int_{-1}^1[f(x)-e(x)]o(x)\,dx. \end{align*} Now let's use the fact that $f$ and $e$ are even, and $o$ is odd: $$\int_{-1}^1[f(x)-e(x)]o(x)\,dx = -\int_0^1[f(x)-e(x)]o(x)\,dx + \int_0^1 [f(x)-e(x)]o(x)\,dx = 0.$$

Therefore $$\int_{-1}^1 [f(x)-e(x)-o(x)]^2\,dx = \int_{-1}^1 [f(x)-e(x)]^2 + \int_{-1}^1 o(x)^2\,dx \geq \int_{-1}^1 [f(x)-e(x)]^2$$ and you were better off without the odd terms.

EDIT: For different norms, you carry out different flavors of the same argument. For instance, for the $\sup$ norm, \begin{align*} \sup_{x\in [-1,1]} |f(x)-e(x)-o(x)| &= \sup_{x\in [0,1]} \max\left(|f(x)-e(x)-o(x)|,|f(x)-e(x)+o(x)|\right)\\ &\geq \sup_{x\in [0,1]} |f(x)-e(x)|\\ &= \sup_{x\in [-1,1]} |f(x)-e(x)|. \end{align*}

• Sorry, I should have been clearer to start. The approximation error is defined as |f-g| where |.| is the sup norm on [-1,1]. Dec 23, 2013 at 3:29
• @Monty123 See edit. Dec 23, 2013 at 3:36

If you're approximating an even smooth (analytic) function on $[-1,1]$ with a Maclaurin series, we need only use this fact:

Prove that the derivative of an even differentiable function is odd, and the derivative of an odd is even.

Along with the fact that even continuous functions have the property that $f(0)=0$. Then only even terms remain in the series.