<s>[INST] Find a polynomial $P$ of the smallest degree such that $\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$ Find a polynomial P of the smallest degree such that $$\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$$
I am ususally solve this kind of "best approximation" problems by taking using $GS$ but I think this time it should be solved by Chebyshev polynomials since that was the topic of the lecture, I am not sure this is the idea here but I would like for some help here (Im not even sure if I can tag it under "chebyshev polynomials"
 [/INST] First, let us try $\mathrm{deg} P(x) \le 2$.
Let $P(x) = ax^2 + bx + c$.
We have $$\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 
\frac{2}{5}a^2 + \frac{4}{3}ac + \frac{2}{3}b^2 + 2c^2 - \frac{12}{11}a - \frac{12}{5} c + \frac{6}{7}.$$
It is easy to minimize a quadratic form. The best $P(x)$
is given by $P(x) = \frac{9}{11}x^2 + \frac{18}{55}$
with $\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 384/21175 \approx 0.01813 > \frac{1}{100}$.
Similarly, we try $\mathrm{deg} P(x) \le 3$. The best $P(x)$ is the same as that of $\mathrm{deg} P(x) \le 2$.
Then we try $\mathrm{deg} P(x) \le 4$.
The best $P(x)$ is given by
$P(x) = -\frac{189}{187}x^4 + \frac{315}{187}x^2 + \frac{45}{187}$
with $\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 1536/244783 \approx 0.006275 < \frac{1}{100}$.
Thus, the smallest degree is $4$.
Remark: How to find the best $P(x)$ if $\mathrm{deg} P(x) \le 2n$?
Let $P(x) = a_{2n}x^{2n} + \cdots + a_0$. We set
$$\frac{\partial }{\partial a_k} 
\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 0, \ k=0, 1, \cdots, 2n.$$
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