 Legendre least-squares projections of a cubic

Take the polynomial $f(x)=1+x+x^2+x^3$ on $[-1,1]$ with unit weight. Using the four Legendre polynomials above, direct integration givesFor example, $⟨f,P_2⟩=\frac{1}{2}{\int }_{-1}^1(3x^4+2x^2-1)dx=4/15$: the omitted odd terms have zero integral. Dividing by $h_2=2/5$ gives $c_2=2/3$. Substituting all four coefficients proves the exact identity

The successive orthogonal projections onto degrees at most $m$ are thereforeHere $\Vert g{\Vert }_2^2={\int }_{-1}^{1}g(x{)}^{2}dx$ is the squared L2 norm, without division by the interval length. Each row retains the previous Legendre polynomial coefficients and appends one new term. The ordinary power coefficients can change: adding $\frac{2}{3}{P}_2$ changes the constant from $4/3$ to $1$, and adding $\frac{2}{5}{P}_3$ changes the linear coefficient from $8/5$ to $1$. Independence belongs to the chosen orthogonal polynomial coordinates.

The full squared L2 norm is $\Vert f{\Vert }_2^2=192/35$, equal to $32/9+128/75+8/45+8/175$. The residual budgets are the omitted terms:For a direct check independent of summing the budgets, $f-q_2=x^3-3x/5$, and integrating its square gives $2/7-12/25+6/25=8/175$. This residual is orthogonal to $1,x,x^2$, exactly the conditions for the best quadratic least-squares solution. An orthogonal projection need not be a polynomial interpolant: for example, $q_2(1)=18/5$ while $f(1)=4$. The approximation allocates error across the interval according to the chosen weighted inner product.