<s>[INST] Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$. 
My (wrong) solution I thought that I would get the solution by calculating the orthogonal projection $w=a+bx+cx^2$ of $v=\sin x$ on subspace $W=\langle u_1,u_2,u_3 \rangle=\langle 1,x,x^2\rangle$ using Gramm matrix. Then I have
$$\begin{pmatrix}\langle u_1, u_1\rangle&\langle u_1, u_2\rangle&\langle u_1, u_3\rangle\\
\langle u_2, u_1\rangle&\langle u_2, u_2\rangle&\langle u_2, u_3\rangle\\
\langle u_3, u_1\rangle&\langle u_3, u_2\rangle& \langle u_3, u_3\rangle\end{pmatrix}
\begin{pmatrix} a\\b\\c\end{pmatrix}=
\begin{pmatrix} \langle u_1, v\rangle\\
\langle u_2,v\rangle\\
\langle u_3, v\rangle \end{pmatrix}$$
So then
$$\begin{pmatrix} \int_0^\pi 1&\int_0^\pi x&\int_0^\pi x^2\\
\int_0^\pi x& \int_0^\pi x^2& \int_0^\pi x^3 \\
\int_0^\pi x^2&\int_0^\pi x^3&\int_0^\pi x^4\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} \int_0^\pi \sin x\\\int_0^\pi x\sin x\\\int_0^\pi x^2 \sin x \end{pmatrix}$$
But solving these equations didn't give me any good answer. So my question is - is my way of solving it completely wrong (if so, can you give me hints how to do it otherwise)?
Thank you
Edit
Then$$ 
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
\frac{\pi^2}{2}&\frac{\pi^3}{3}&\frac{\pi^4}{4}\\
\frac{\pi^3}{3}&\frac{\pi^4}{4}&\frac{\pi^5}{5}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\\pi\\\pi^2-4 \end{pmatrix}$$
$$ 
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&\frac{\pi^4}{12}&\frac{4\pi^5}{45}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
$$ 
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&0&\frac{\pi^5}{180}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
Edit II My way (I have found the mistake I did) will get the result
$$f(x)=\dfrac{60(\pi^2-12)}{\pi^5}x^2-\dfrac{60(\pi^2-12)}{\pi^4}x+\dfrac{12(\pi^2-10)}{\pi^3}$$
which I hope is the right answer with error approx. $0,000936$
 [/INST] The answer has already been given.
In addition, the numerical solving and the graph are shown below, in order to compare the results.


Note that the result corresponds to the mean least square absolute deviation. A small modification of the method and formulas would give the result corresponding to the mean least square relative deviation, which could be more convenient in the low range of $\sin(x)$.
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