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This is Exercise 6.7.7 from Abbott's Understanding analysis. For context, the section is about the Weierstrass approximation theorem, but this is a step towards proving it, so we cannot use WAT to prove this result. The exercise reads:

a) Use the fact that $\lvert a \rvert = \sqrt{a^{2}}$ to prove that, given $\epsilon > 0$, there exists a polynomial $q(x)$ satisfying $\lvert \lvert x \rvert - q(x) \rvert < \epsilon$ for all $x \in [-1,1]$.

b) Generalize this conclusion to an arbitrary interval $[a,b]$.

a) Throughout the section, Abbott motivates the approximation using the Taylor series of $\sqrt{1-x}$. In fact, in a previous exercise, I managed to prove that the Taylor series of $\sqrt{1-x}$ converge in $[-1,1]$ (link to the MSE I used for support).

My idea to prove this result is based on the manipulation of series. In particular, we can start with the Taylor series of $\sqrt{1-x} = \sum_{n=0}^{\infty}a_{n}x^{n}$, and modify the series to reach $\sqrt{x^{2}}$ on the left hand side of the equation, as suggested by the exercise.

$ \begin{align} \sqrt{1-x} &= \sum_{n=0}^{\infty}a_{n}x^{n} \\ \sqrt{1+x^{2}} &= \sum_{n=0}^{\infty}a_{n}(-1)^{n}x^{2n} \quad \text{substituting} \ x \ \text{for} \ -x^{2}. \\ \sqrt{x^{2}} &= \sum_{n=0}^{\infty}a_{n}(-1)^{n}(x^{2n}+1) \quad \text{centering x on the left hand side to}\ 0. \end{align} $

I am particularly unsure about the last step because I believe this should change the convergence of the series.

b) I am unsure of how to tackle this question, as from the way I proved the convergence of the Taylor series of $\sqrt{1-x}$, it was necessary that $x \in [-1,1]$.

I would appreciate it if someone could tell me whether what I did in a) is correct, or else hint me/give me a valid proof of it. Also, hints or solutions for b) are appreciated.

Edit 30/Mar/2022

Using @RobPratt suggestion, I would like to supplement his answer with part b) of the exercise.

We start with

\begin{align} \sqrt{x^{2}} &= \sum_{n=0}^{\infty} a_{n}(1-x^{2})^{n} \quad \text{for}\ x \in [-\sqrt{2}, \sqrt{2}] \\ \sqrt{\sqrt{2}x^{2}} &= \sum_{n=0}^{\infty} a_{n} \left(1-\sqrt{2}x^{2}\right)^{n} \quad \text{for}\ x \in [-1, 1] \\ \sqrt{\frac{\sqrt{2}x^{2}}{c}} &= \sum_{n=0}^{\infty} a_{n} \left(1-\frac{\sqrt{2}x^{2}}{c}\right)^{n} \quad \text{for}\ x \in [-c, c],\ \text{where}\ c=\max\{a,b\}. \\ \end{align} Since $c=\max\{a,b\}$, we know that $[a,b] \subseteq [-c,c]$. Alternatively, if we want to approximate exactly in $[a,b]$, we have to make the following transformation $y:=\sqrt{2}x^{2}\rightarrow \frac{by}{2}+\frac{(a+b)((1-y)}{4}$. Starting from $[-1,1]$.

Sergio
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    You want $c=\max{|a|,|b|}$, and then $x\in[-c,c]$ implies that $2x^2/c^2\in[0,2]$, so you can apply the $\sqrt{x}$ series with $2x^2/c^2$ in place of $x$. – RobPratt Mar 30 '22 at 18:48
  • @RobPratt Thanks. I clearly don't understand properly how to change the limits of series when manipulating them. Can you please explain the procedure or point me to some reference? – Sergio Mar 30 '22 at 21:23
  • Is it because in $\sqrt{x}$ we have $0<x<2$, and replacing $x$ for $2x^2/c^2$ in that inequality and solving for x gives us the desired interval? – Sergio Mar 30 '22 at 21:30

1 Answers1

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First replace $1-x$ with $x$ and then replace $x$ with $x^2$: \begin{align} \sqrt{1-x} &= \sum_{n=0}^\infty a_n x^n &&\text{for $x\in[-1,1]$} \\ \sqrt{x} &= \sum_{n=0}^\infty a_n (1-x)^n &&\text{for $x\in[0,2]$} \\ \sqrt{x^2} &= \sum_{n=0}^\infty a_n (1-x^2)^n &&\text{for $x\in[-\sqrt{2},\sqrt{2}]$} \\ \end{align}

RobPratt
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  • Thanks for your answer! It is relieving that I was on a good path. I can see that although we arrived at the same left-hand side of the equation, the right-hand side is not the same. 1) Can you please explain to me the reason for the order of your substitutions? 2) To answer the second question in the exercise should I just scale $x$? For instance, if I take $y=x/\sqrt{2}$ would that give me the approximation for $x\in[1,1]$? – Sergio Mar 27 '22 at 19:02
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    I'm not sure how to explain my motivation for that order except that it works. For part b), apply a linear transformation that maps $-\sqrt{2}$ to $a$ and $\sqrt{2}$ to $b$. – RobPratt Mar 27 '22 at 19:11
  • Thank you for your help! – Sergio Mar 27 '22 at 19:16
  • I was just re-writing your solution and I realized that I don't understand why on the third line you have convergence whenever $x\in[-\sqrt{2},\sqrt{2}]$. Is this a typo? If not, can you briefly explain why is that as opposed to $[0,\sqrt{2}]$? – Sergio Mar 29 '22 at 18:15
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    If $x\in[-\sqrt2,\sqrt2]$, then $x^2\in[0,2]$, so you can substitute $x^2$ in place of $x$ on both sides of the second equation. – RobPratt Mar 29 '22 at 18:31
  • Of course! I forgot that detail. – Sergio Mar 29 '22 at 19:24
  • I just added the answer to b) in the main question. I wanted to practice the change of the limits of the interval of convergence. Could you please take a quick look and tell me whether what I wrote is correct? – Sergio Mar 30 '22 at 17:33
  • I am particularly not sure about the second assertion about $[a,b]$, instead of $[-c,c]$. – Sergio Mar 30 '22 at 17:46