Find the best approximation of the function
$$ f(x)= \begin{cases}
1 – x \quad\text{ for } 0 \le x \le 1 \\
1 + x \quad\text{ for } -1 \le x \le 0 \\
\end{cases}$$
in the interval $[-1,~ 1]$ by a polynomial $p(x)$ of degree $\le 4$, meaning that the integral
$$\int_{-1}^{1} (|f(x) – p(x)|)^2\,{\rm d}x $$ is minimal possible.
We know that the best approximation is unque and coinced with the orthogonal projection. The minimal distance is equal to to length of orthogonal projection of $x$ into ${\rm ort}_{U}x$.
Best Answer
Given a function $f\in L^2[-1,1]$, the polynomial of degree $d$ (in your case $d=4$) minimizing that quantity is
$\sum_{k=0}^d \alpha_k P_k(x)$
Where:
$P_k(x)=\frac{1}{2^kk!}\frac{d^k}{dx^k}[(x^2-1)^k]$, also known as the k-th Legendre polynomial. It's a suitable multiple of the k-th derivative of $(x^2-1)^k$, and it has degree exactly $k$. Legendre polynomials are a complete orthogonal system fro $L^2[-1,1]$; they are not orthonormal, since $\|P_k\|_2=\sqrt{\frac{2}{2k+1}}$.
$\alpha_k=\frac{2k+1}{2}\int_{-1}^1f(x)P_k(x)dx$
Wikipedia provides $P_k(x)$ for $k=0,\ldots,10$, you need the first 5 of them:
$$P_0(x)=1\\P_1(x)=x\\P_2(x)=\frac{3x^2-1}{2}\\P_3(x)=\frac{5x^3-3x}{2}\\P_4(x)=\frac{35x^4-30x^2+3}{8}$$
The rest is a calculation.