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In an exam I was given this question:

let $f(x)=x^3$. We want to find the best linear approximation (best in the sense that the maximal error is minimized) of $f$ in the interval $[-1,1]$ using chebyshev polynomials, we are also given as a hint that $T_3(x)=4x^3-3x$

Here is what the teacher did:

Chebyshev has proved that for any monic polynomial on the interval $[-1,1]$:

$max: |p(x)| \geq 2^{1-n}$ and we have an equality only if $p(x)=2^{1-n}T_n(x)$

so if $deg(p(x))=3$, then $max: |p(x)|$ is minimized when

$p(x)=2^{1-3}T_3(x)=\frac{1}{4}T_3(x)=x^3-\frac{3}{4}x$

So among all polynomials of degree 3, $max: |x^3-\frac{3}{4}x|$ is the best. So our best linear approximation for $x^3$ is $\frac{3}{4}x$.

Here is what I don't understand:

I completely understand what the teacher did, but it is purely coincidental that it worked. It only worked because he asked for a linear approximation, but what if he would ask for an approximation of degree $2$? Then that chebyshev polynomial we have provides no information for us.

How can we apply the same method, to solve the question: give the best quadratic approximation of $f(x)=x^3$ given that $T_3(x)=4x^3-3x$. We would have a similar problem if we were given $f(x)=x^2$. Basically the third chebyshev polynomial doesn't give us any information in those cases, is there a way to generalize this method so it will work for all cases?

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