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Taken from Lang's R & F Analysis (p.60). For some reason I can't see why, for $t \in [0,1]$, the following is true for all natural numbers $n$ (by an inductive argument): $$0 \leq \sqrt{t} - P_n(t) \leq \frac{2\sqrt{t}}{2 + n\sqrt{t}},$$ where $$P_{n+1}(t) = P_n(t) +\frac{1}{2}\left(t-P_n(t)^2\right),$$ and $P_0(t) = 0$.

Soyuz
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  • There is no explanation in the book? – Crostul Apr 01 '16 at 08:48
  • @Crostul Nope, it's actually an exercise (problem 7 in $\text{III, §4}$). – Soyuz Apr 01 '16 at 17:29
  • Are you interested in a more intuitive solution to finding such a function, or are you set on using this method to define a sequence of polynomials? – Disintegrating By Parts Apr 02 '16 at 20:49
  • @TrialAndError I am just interested in this method. In particular, I feel that there should be a relatively straightforward inductive proof for this problem, but I can't seem to do it. – Soyuz Apr 02 '16 at 20:51
  • Substitute $\sqrt{x} - P_n(x) = Q_n(x)$ which satisfies the same recursion as here http://math.stackexchange.com/questions/822495/show-that-p-nx-rightrightarrows-0-qquad-x-in0-1 – r9m Jun 29 '16 at 12:18

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I can't comment, so I answer. I have a very similar problem in proving stone weierstrass. Those sequence of polynomials in $[0,1]$ converge to square root function from below. Show inductevely that they are less than the square root, and show they for a fixed $x$ the sequence $p_n(x)$ is increasing, and thus has a limit, and the limit has to satisfy a equation analogus to the definition of the recursion of $ P_n$. That suffices to prove that they converge square root function from below, that possibly helps. I know it doesn't solve the exercise.

Santropedro
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