How can I find a polynomial approaching |x| such that the maximum error is 0.1?
Well the Stone-Weirestrass theorem states that this is possible. However, in most proofs I have seen you never actually see how to construct such polynomials.
Though not all hope is lost. Look at Bernstein Polynomials. Here is a link to the Wikipedia page on them. Also a while back I wrote a blog entry on just this topic, here is a link to the post. I don't think that I mentioned this in the post say using the method described in the post you end up with a polynomial of degree $\displaystyle n$, there might be other polynomials of lower degree that also give you the desired error. This is because of some very conservative estimations that are made.
Obviously, you only need even powers. If you tried, say $\displaystyle ax^4+ bx^2$, then the error, for given x is $\displaystyle |ax^4+ bx^2- |x||$. Because of symmetry, it is sufficient to look at x> 0 so that is $\displaystyle |ax^4+ bx^2- x|$. Is it possible to choose a and b so that is never larger than .1 for all x between -1 and 1?
The function to be round up in the interval $\displaystyle [-1,1]$ is $\displaystyle y(x) = |x|$ and it is an even function, so that we search an even polinomial of degree $\displaystyle 2n$ written as $\displaystyle p(x)= \sum_{k=0}^{n} p_{k}\cdot x^{2k}$ that approximates $\displaystyle y(x)$ in 'min-max' sense. If we indicate the 'error function' with $\displaystyle e(x)= y(x)-p(x)$, the 'min-max condition' force that to be...
$\displaystyle e(\frac{k}{n+1}) = (-1)^{k}\cdot e$ , $\displaystyle k=0,1,...,n+1$ (1)
The (1) is a system of linear equations in the unknown variables $\displaystyle p_{0},p_{1}, ... , p_{n}, e$ that can be solved in standard way. In the case $\displaystyle n=1$ it becomes...
$\displaystyle p_{0} + e = 0$
$\displaystyle p_{0} + \frac{p_{1}}{4} - e = \frac{1}{2}$
$\displaystyle p_{0} + p_{1} + e = 1$ (2)
... the solution of which is $\displaystyle p_{0}= \frac{1}{8}, p_{1}= 1, e=-\frac{1}{8}$ , so that the 'min-max' polynomial is $\displaystyle p(x)= \frac{1}{8} + x^{2}$ and the error is alternatively $\displaystyle +.125$ and $\displaystyle -.125$. You require an error not grater than $\displaystyle .1$, so that the min-max polynomial is of order $\displaystyle 4$ , i.e. $\displaystyle n=2$ and (1) becomes...
$\displaystyle p_{0} + e =0$
$\displaystyle p_{0} + \frac{p_{1}}{9} + \frac {p_{2}}{81} - e = \frac{1}{3}$
$\displaystyle p_{0} + \frac{4}{9}\cdot p_{1} + \frac{16}{81}\cdot p_{2} + e = \frac{2}{3} $
$\displaystyle p_{0} + p_{1} + p_{2} - e = 1$ (3)
... the solution of which is left as exercise ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The 'solution' of the 'exercise' I have proposed in the last post is $\displaystyle p_{0}=\frac{1}{16}, p_{1}= 2, p_{2}= -\frac{9}{8}, e= -\frac{1}{16}$ , so that the min-max polynomial of degree 4 gives an approximation with maximum error $\displaystyle e= .0625$...
If greater precision is required it is necessary to increase $\displaystyle n$ and the maximum error I suppose is $\displaystyle e= \frac{1}{2^{n+2}}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$