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Given is a sequence of polynomials $P_n$, defined as follows: $P_0(x)=0, P_{n+1}(x) = P_n(x) + \frac{x-P_n^2(x)}{2}. $, n= 0,1,2,..., and x is real.

Proving that for all non-negative integers n and x at [0;1] this holds:

$0\leq \sqrt(x)-P_n(x) \leq \frac{2}{n+1} . $.

I checked for small cases n=0,1,2 the hypothesis, and it turns out true fot all x at [0;1]! But how do we proceed? Setting some recurrence? I tried induction, but can't find ties.

Suzu Hirose
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Jackie Poehler
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    make the substitution $Q_n(x) = \sqrt{x} - P_n(x)$ here and you'll get $Q_0(x)=0, Q_{n+1}(x) = Q_n(x) + \frac{x-Q_n^2(x)}{2}$, the answers already established stronger inequalities on upperbound of $Q_n(x)$ :-) – r9m Dec 13 '14 at 10:13

1 Answers1

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first I prove right hand:

let $$a_{n}=\sqrt{x}-P_{n}(x)$$ then we have $$a_{n+1}(x)=a_{n}(x)\left[1-\dfrac{\sqrt{x}+P_{n}(x)}{2}\right] \le a_{n}(x)\left(1-\dfrac{\sqrt{x}}{2}\right),x\in[0,1]$$ so $$a_{n}(x)\le a_{0}\left(1-\dfrac{\sqrt{x}}{2}\right)^n =\sqrt{x}\left(1-\dfrac{\sqrt{x}}{2}\right)^n$$ note $$g(t)=t\left(1-\dfrac{t}{2}\right)^n\le\dfrac{2}{n+1}$$ because we let $$f(x)=\ln{x}+n\ln{(1-\dfrac{x}{2})}\Longrightarrow f'(x)=\dfrac{1}{x}+\dfrac{n}{x-2}=0$$ $$x=\dfrac{2}{n+1}$$ so $$g(t)=t(1-\dfrac{t}{2})^n\le t=\dfrac{2}{n+1},0<t$$ so we have $$a_{n}(x)\le\dfrac{2}{n+1}$$

and the left is clear

math110
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    After "then we have", is it $1-\dfrac{x+P_{n}(x)}{2}$ or $1-\dfrac{\sqrt{x}+P_{n}(x)}{2}$ ? – Kelenner Dec 13 '14 at 07:30
  • Beauty And mathematical gourgeousness – Jackie Poehler Dec 13 '14 at 07:41
  • How do you prove that line with the g function? – Jackie Poehler Dec 13 '14 at 08:52
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    @Jackie one can use $\displaystyle g(t) = t\left(1-\frac{t}{2}\right)^n \le \frac{2}{n}\left(\frac{\frac{nt}{2}+n\left(1-\frac{t}{2}\right)}{n+1}\right)^n = \frac{2}{n\left(1+\frac{1}{n}\right)^{n+1}}$ from Am-Gm inequality, and since $\displaystyle \left(1+\frac{1}{n}\right)^{n+1} \ge e$, we get $\displaystyle g(t) \le \frac{2}{ne}$. – r9m Dec 13 '14 at 12:57