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I'm studying a proof of the Weierstrass approximation theorem that requires an uniform approximation using polynomials of the function |x| in the interval $[-1,1]$ i.e. we need a sequence of polynomials that converge to |x| in the supreme norm. One way to do this is to define for $t\in[0,1]$ the following sequence:

First set $P_0(t)=0$ for all t, and then using induction define $$P_{n+1}(t)=P_n(t)+\frac{1}{2}(t-P_n(t)^2).$$

The next step would be to prove using induction that for all $t\in[0,1]$, $$0\leq\sqrt{t}-P_n(t)\leq\frac{2\sqrt{t}}{2+n\sqrt{t}},$$ but i have not been able to do this. Can someone help me???

martin_galo
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  • This must have something to do with the recursive approximation of the square root. – Giuseppe Negro Sep 16 '18 at 09:40
  • You are seeking a sequence of polynomials that converge to the absolute value of $x$ on the interval (-1,1). Obviously this function is even. Therefore the polynomials must also be even. Yet your recursive scheme can yield odd terms. For example, $P1(t) = t/2$. Clearly this formula can not be correct ! – M. Wind Sep 17 '18 at 05:11
  • Not true. The polynomials $P_n$ converge uniformly to $\sqrt{t}$ for $t\in[0,1]$. If you use that $|t|=\sqrt{t^2}$ you get uniform convergence to the absolute value in $[-1,1]$. – martin_galo Sep 17 '18 at 07:39
  • From your comment I gather that you use the substitution $t = x^2$. This important detail is not explained at all in your original question. If you want to ask a question at this forum, you should include such important facts. We can not read your mind, all we have to work with is what you care to write down. – M. Wind Sep 17 '18 at 14:16
  • that's not relevant to question. all i'm asking is the upper bound for $\sqrt{t}-P_n(t)$ – martin_galo Sep 17 '18 at 18:30

1 Answers1

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Proof by induction. Consider: \begin{align} \sqrt{t}-P_{n+1}(t) &= \sqrt{t}-P_n(t)-\frac{1}{2}(t−P_n(t)^2) \\ &=(\sqrt{t}-P_n(t))(1-\frac{1}{2}(\sqrt{t}+P_n(t))) \end{align} Now by induction hypothesis: \begin{align} \sqrt{t}-P_{n+1}(t) & \le\frac{2\sqrt{t}}{2+n\sqrt{t}}(1-\frac{1}{2}(\sqrt{t}+P_n(t))) \end{align} Now the last term, still by induction, is majored by $$ 1-\frac{1}{2}(\sqrt{t}+P_n(t))=1-\sqrt{t}+\frac{1}{2}(\sqrt{t}-P_n(t))\le1-\sqrt{t}+\frac{\sqrt{t}}{2+n\sqrt{t}} $$ Putting thing together we have: \begin{align} \sqrt{t}-P_{n+1}(t) & \le\frac{2\sqrt{t}}{2+n\sqrt{t}}(1-\sqrt{t}+\frac{\sqrt{t}}{2+n\sqrt{t}}) \\ & =\frac{2\sqrt{t}}{2+(n+1)\sqrt{t}}\underbrace{\left[\frac{2+(n+1)\sqrt{t}}{2+n\sqrt{t}}(1-\sqrt{t}+\frac{\sqrt{t}}{2+n\sqrt{t}})\right]}_{g_n(t)} \end{align} Now we must prove that $g_n(t)\le 1$. You can compute $g_n(t)$ derivative, but please use a formal calculus tool like Maple or Mathematica... You get:

$$ g'_n(t)=-\frac{n^3 t+n^2 \left(t+6 \sqrt{t}\right)+n \left(6 \sqrt{t}+8\right)+4}{2 \left(n \sqrt{t}+2\right)^3} $$

which is always negative hence its maximal value is for $t=0$ and $g_n(t=0)=1$. We get our expected result: $$ \sqrt{t}-P_{n+1}(t) \le\frac{2\sqrt{t}}{2+(n+1)\sqrt{t}} $$ This proof is not very nice (someone will certainly gives another one which is nicer).

Now, there is a much more elegant way for this problem.

1/ It s obvious that $0\le P_n(t)$. Now take $P_n(t)\le \sqrt{t}$ as induction hypothesis. This is true for $n=0$, as $P_0=0$.

Now as before write: \begin{align} \sqrt{t}-P_{n+1}(t) &= \sqrt{t}-P_n(t)-\frac{1}{2}(t−P_n(t)^2) \\ &=(\sqrt{t}-P_n(t))(1-\frac{1}{2}(\sqrt{t}+P_n(t))) \end{align} By induction hypothesis $P_n(t)\le\sqrt{t}$, hence $\frac{1}{2}(\sqrt{t}+P_n(t))\le 1$ which also implies $0 \le 1-\frac{1}{2}\sqrt{t}+P_n(t))$.

By consequence $$ \sqrt{t}-P_{n+1}(t) = (\sqrt{t}-P_n(t))(1-\frac{1}{2}(\sqrt{t}+P_n(t))) \ge 0 $$ hence $P_{n+1}(t)\le\sqrt{t}$ which is what we wanted to prove.

2/ Now back to: $P_{n+1}(t)=P_n(t)+\frac{1}{2}(t-P_n(t)^2)$, observe that: $$ P_{n+1}(t)-P_n(t)=\frac{1}{2}(t-P_n(t)^2)\ge 0 $$ hence your sequence $P_n$ is increasing.

3/ Observe that $\forall t$, the sequence $P_n(t)$ is increasing and bounded, hence it converges point wise. Now, as $[0,1]$ is compact, by the Dini theorem you can conclude that the convergence is also uniform. Now the limit is obtained solving $p(t)=p(t)-(t-p(t)^2)$ which gives $p(t)=\sqrt{t}$. Q.E.D

Picaud Vincent
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  • Thank you for your help – martin_galo Sep 16 '18 at 21:02
  • you are welcome. It is a tedious proof with a lot of computations... I hope that someone will give you a nicer one. Please note that the second approahc, the one using the Dini theorem is much more elegant :) – Picaud Vincent Sep 16 '18 at 21:05
  • Using Dini's theorem in point 3/ requires the sequence $P_n$ to converge pointwise to some continuous function. The conclusion of your first approach shows that this is indeed the case. This was not shown in the second that I could see. So the two parts of your answer seems to complement each other rather than being alternative solutions! – rndnr562 Feb 15 '23 at 22:46
  • @mdnr562, you are right, thank you for pointing this out. – Picaud Vincent Feb 16 '23 at 06:14