# Weierstrass approximation theorem - absolute value

I am really struggling on this exercise, and don't even know where to start.

(a) Use the fact that $|a|=\sqrt{a^2}$ to prove that, given $\epsilon>0$, there exists a polynomial $q(x)$ satisfying $||x|-q(x)|<\epsilon$ for all $x$ in $[-1,1]$.

(b) Generalize this conclusion to an arbitrary interval $[a,b]$.

I recognize that this is the conclusion of the Weierstrass approximation theorem. But I don't know how to use it.

Thank you so much in advance for your help!

• Notice that the absolute value function is continuous. So it follows immediately from the theorem. – mucciolo Feb 26 '17 at 3:21
• For an explicit construction, find a polynomial $p$ such that $|p(x)-\sqrt x |<\epsilon$ for $x\in [0,1].$ Then for $x\in [-1,1]$ you have $|p(x^2)-|x|\;|=|p(x^2)-\sqrt {x^2}|<\epsilon.$ – DanielWainfleet Feb 26 '17 at 4:09
• But if I simply say "the absolute value function is continuous, so the the WAT says ||x|−q(x)|<ϵ", is it enough? I mean, they are asking on [−1,1] first an then on any [a,b]... It can not be that simple right? – mathz Feb 26 '17 at 13:50
• And if I use you reasoning with leading to |p(x2)−x2‾‾√|<ϵ... Am I suppose to find an explicit expression of p(x)? I think I am really confuse about what the problem is asking... – mathz Feb 26 '17 at 13:51
• The exercise as written is just asking for the assurance of the existence of such polynomial. We do not need to know the explicit formula of such polynomial. That is, this is just a straightfoward implication of WAT. What statement of WAT you have in mind? – mucciolo Feb 27 '17 at 0:27