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Series converging to |x|

  • Thread starter erogard
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  • #1
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Hi everyone,

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.
 

Answers and Replies

  • #2
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(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

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

Can you show that

[tex]|x|-P_n(x)\leq \frac{2}{n+1}[/tex]

Which of these three doesn't work?
 
  • #3
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Ok well I was able to show that [tex] P_n(x) \leq |x| [/tex] 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.
 
  • #4
22,097
3,279
Ok well I was able to show that [tex] P_n(x) \leq |x| [/tex] 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

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

Writing out [itex]P_{n+1}(x)[/itex], you get that you need to show

[tex]P_n(x)\leq P_n(x)+\frac{x^2-P_n^2(x)}{2}.[/tex]

So, what can you do next??
 
  • #5
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Oh well that follows directly from the fact that [tex] P_n(x) \leq |x| [/tex] Ok thanks a lot. I will try to prove the second inequality now.
 
  • #6
62
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Still can't get the second part. Tried something with the binomial expansion to show that
[tex] |x| \left( 1-\frac{|x|}{2} \right)^n < \frac{2}{n+1}[/tex]
but I can't relate my result to 2/(n+1).

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

EDIT: Just got the above inequality, I will try again the one with 2/(n+1).
 
  • #7
22,097
3,279
Still can't get the second part. Tried something with the binomial expansion to show that
[tex] |x| \left( 1-\frac{|x|}{2} \right)^n < \frac{2}{n+1}[/tex]
but I can't relate my result to 2/(n+1).

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

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

[tex] |x| \left( 1-\frac{|x|}{2} \right)^n [/tex]

by calculus?? Calculate the derivative and see what the critical points are (= where the derivative is 0 or does not exist)
 
  • #8
62
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Why don't you try to find the maximum of

[tex] |x| \left( 1-\frac{|x|}{2} \right)^n [/tex]

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.
 

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