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I was asked to find the best linear approximation to $f(x)=x^2$ in $x \in [0,1]$ using chebyshev polynomials, meaning, using the known property that $2^{1-n}T_n(x)$ is the best approximation to $0$ at $x \in [-1,1]$. (Where $T_n$ is chebyshev polynomial of degree $n$).

Before getting on to showing what I did, I would like to show a short solved example, so you have a better idea of what I'm talking about:

Example: We want to find the best linear approximation to $g(x)=x^3$ in $x \in [-1,1]$, we are given $T_3(x) = 4x^3-3x$

Using the known property of chebyshev polynomials, we know that $|2^{1-3}T_3(x)| = |t^3-\frac{3}{4}x|$ is the best approximation to zero, and so $\frac{3}{4}x$ is the best linear approximation to $x^3$.

We want to solve the original question in the same method, so first I transfered to $[-1,1]$:

let $t=2x-1$, $t \in [-1,1]$, so our original $f(x)=x^2$ now becomes $f(t)=(\frac{t+1}{2})^2=\frac{t^2+2t+1}{4}$

We want to find the best linear approximation to $f(t)=\frac{t^2+2t+1}{4}$ in $t \in [-1,1]$ using chebyshev polynomials.

But I don't see the connection. the second chebyshev polynomial is $T_2(t)=2t^2-1$. so we know $|t^2-\frac{1}{2}|$ is the best apprximation to zero, so $\frac{1}{2}$ is the best linear approximation to $t^2$. But that doesn't say anything about our original function, we are not interested in $t^2$, we are interested in $\frac{t^2+2t+1}{4}$.

I know the correct answer is $x-\frac{1}{8}$ but I don't know how to reach that answer.

Edit: Managed to solve it on my own. Sorry for bothering everyone. Will accept an answer to close this question.

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  • $\begingroup$ Nevermind, managed to solve it. had a "Eureka!" moment. $\endgroup$ – Oria Gruber Jan 21 '15 at 15:37
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Managed to solve it.

We showed that $\frac{1}{2}$ is the best linear approximation to $t^2$. It follows that $\frac{1}{2}+2t+1=\frac{3}{2}+2t$ is the best linear approximation to $t^2+2t+1$, and so

$\frac{3}{8}+\frac{1}{2}t$ is the best linear approximation to $\frac{t^2+2t+1}{4}$. Since we have $t=2x-1$, we find that $\frac{3}{8}+\frac{1}{2}t=x-\frac{1}{8}$ which is the desired result.

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    $\begingroup$ Well done! I thought the expression would have to be expanded using chebyshev-polynomials. $\endgroup$ – Peter Jan 21 '15 at 15:43
  • $\begingroup$ The only thing I am unsure of is, if $\frac{1}{2}$ is the best linear approximation to $t^2$, then is it true that $\frac{1}{2}+2t+1$ is the best linear approximation to $t^2+2t+1$? Seems very intuitive and agree with common sense, but I have no formal proof. $\endgroup$ – Oria Gruber Jan 21 '15 at 15:46
  • $\begingroup$ I think, this is obvious and does not require a formal proof. $\endgroup$ – Peter Jan 21 '15 at 15:48

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