- #1

erogard

- 62

- 0

I need to show that [tex]\lim_{n\to\infty}P_n(x) = |x|[/tex] uniformly on [tex][-1,1][/tex]

if we define [tex]P_0 = 0[/tex] & [tex]P_{n+1} = P_n(x) + \frac{x^2 - P_n^2(x)}{2}[/tex]

Rudin gives the following hint: use the identity [tex]|x| - P_{n+1}(x) = \left( |x| - P_n(x) \right) \left( 1 - \frac{|x|+P_n(x)}{2} \right)[/tex]

to show that

[tex]0 \leq P_n(x) \leq P_{n+1}(x) \leq |x|[/tex]

if [tex]|x| \leq 1[/tex]

and that [tex]|x|-P_n(x) \leq |x| \left( 1-\frac{|x|}{2} \right)^n < \frac{2}{n+1}[/tex]

if [tex]|x| \leq 1[/tex] (though I think that's a typo, he probably meant greater than or equal)

From there I could readily show that the difference |x| - P_n goes to 0 when n goes to infinity.

But I'm not sure how to go about this problem. I would use recursion on n, and I was advised to consider the fact that the arithmetic average is always less or = than the highest term that enters the average (dunno how to use this fact here).

Any help would be appreciated.