Convergence of Series to |x|: A Calculus Approach

[erogard]

Hi everyone,

I need to show that $$\lim_{n\to\infty}P_n(x) = |x|$$ uniformly on $$[-1,1]$$
if we define $$P_0 = 0$$ & $$P_{n+1} = P_n(x) + \frac{x^2 - P_n^2(x)}{2}$$

Rudin gives the following hint: use the identity $$|x| - P_{n+1}(x) = \left( |x| - P_n(x) \right) \left( 1 - \frac{|x|+P_n(x)}{2} \right)$$
to show that
$$0 \leq P_n(x) \leq P_{n+1}(x) \leq |x|$$
if $$|x| \leq 1$$
and that $$|x|-P_n(x) \leq |x| \left( 1-\frac{|x|}{2} \right)^n < \frac{2}{n+1}$$
if $$|x| \leq 1$$ (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.

 

[micromass]

(though I think that's a typo, he probably meant greater than or equal)

It's not a typo. You are always working with x in [-1,1].

So, where exactly are you stuck?? Can you show that "idenitity"? Can you show that

$$0\leq P_n(x)\leq P_{n+1}(x)\leq |x|$$

Can you show that

$$|x|-P_n(x)\leq \frac{2}{n+1}$$

Which of these three doesn't work?

 

[erogard]

Ok well I was able to show that $$P_n(x) \leq |x|$$ by recursion on n for all natural numbers n by using the given identity and the fact about the average that I mentioned. But not that P_n is less than P_n+1 yet, couldn't get the inductive step to work out the way I want (my inductive hypothesis being that there exist a natural number k such that P_k < P_k+1 and from there I'm trying to show that P_k+1 < P_k+2.

Also, my bad regarding the inequality with |x|, it's obv not a typo considering the interval we're working on.

 

[micromass]

Ok well I was able to show that $$P_n(x) \leq |x|$$ by recursion on n for all natural numbers n by using the given identity and the fact about the average that I mentioned. But not that P_n is less than P_n+1 yet, couldn't get the inductive step to work out.

OK, so you need to show that

$$P_n(x)\leq P_{n+1}(x)$$

Writing out $P_{n+1}(x)$, you get that you need to show

$$P_n(x)\leq P_n(x)+\frac{x^2-P_n^2(x)}{2}.$$

So, what can you do next??

 

[erogard]

Oh well that follows directly from the fact that $$P_n(x) \leq |x|$$ Ok thanks a lot. I will try to prove the second inequality now.

 

[erogard]

Still can't get the second part. Tried something with the binomial expansion to show that
$$|x| \left( 1-\frac{|x|}{2} \right)^n < \frac{2}{n+1}$$
but I can't relate my result to 2/(n+1).

Also would you use recursion to show that $$|x|-P_n(x) \leq |x| \left( 1-\frac{|x|}{2} \right)^n$$ ?

EDIT: Just got the above inequality, I will try again the one with 2/(n+1).

 

[micromass]

Still can't get the second part. Tried something with the binomial expansion to show that
$$|x| \left( 1-\frac{|x|}{2} \right)^n < \frac{2}{n+1}$$
but I can't relate my result to 2/(n+1).

Also would you use recursion to show that $$|x|-P_n(x) \leq |x| \left( 1-\frac{|x|}{2} \right)^n$$ ?

EDIT: Just got the above inequality, I will try again the one with 2/(n+1).

Why don't you try to find the maximum of

$$|x| \left( 1-\frac{|x|}{2} \right)^n$$

by calculus?? Calculate the derivative and see what the critical points are (= where the derivative is 0 or does not exist)

 

[erogard]

Why don't you try to find the maximum of

$$|x| \left( 1-\frac{|x|}{2} \right)^n$$

by calculus?? Calculate the derivative and see what the critical points are (= where the derivative is 0 or does not exist)

Got it, thanks a lot for your help.