= Legendre least-squares projections of a cubic
{c}
{title2=$f(x)=1+x+x^2+x^3$}
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> gives
$$
\begin{array}{c|c|c|c|c}
n&h_n&\langle f,P_n\rangle&c_n&c_n^2h_n\\\hline
0&2&8/3&4/3&32/9\\
1&2/3&16/15&8/5&128/75\\
2&2/5&4/15&2/3&8/45\\
3&2/7&4/35&2/5&8/175
\end{array}
$$
For example, $\langle f,P_2\rangle=\tfrac12\int_{-1}^1(3x^4+2x^2-1)\,dx=4/15$: 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
$$
f=\frac43P_0+\frac85P_1+\frac23P_2+\frac25P_3.
$$
The successive <orthogonal projections> onto degrees at most $m$ are therefore
$$
\begin{array}{c|l|c}
m&q_m(x)&E_m=\|f-q_m\|_2^2\\\hline
0&4/3&608/315\\
1&4/3+8x/5&352/1575\\
2&1+8x/5+x^2&8/175\\
3&1+x+x^2+x^3&0
\end{array}
$$
Here $\|g\|_2^2=\int_{-1}^1g(x)^2\,dx$ 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 $\tfrac23P_2$ changes the constant from $4/3$ to $1$, and adding $\tfrac25P_3$ changes the linear <coefficient> from $8/5$ to $1$. Independence belongs to the chosen <orthogonal polynomial> coordinates.
\Image[/legendre-nested-projections-and-residual-budget.png]
{title=Nested Legendre least-squares projections of a cubic and the exact squared residual contributed by each omitted mode}
The full squared <L2 norm> is $\|f\|_2^2=192/35$, equal to $32/9+128/75+8/45+8/175$. The residual budgets are the omitted terms:
$$
E_0=\frac{128}{75}+\frac8{45}+\frac8{175}=\frac{608}{315},\qquad E_1=\frac8{45}+\frac8{175}=\frac{352}{1575},\qquad E_2=\frac8{175},\qquad E_3=0.
$$
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>.
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