Decoupled coefficients of orthogonal polynomial projections ($c_n=⟨f,p_n{⟩}_w/h_n$)

Let $f$ be a real function with $\Vert f{\Vert }_w^2={\int }_a^{b}f(x{)}^{2}w(x)dx<\infty$, and let $V_m$ be the vector space of real polynomials of degree at most $m$. Its orthogonal projection is the least-squares polynomial in an orthogonal-polynomial basisEach coefficient depends on just one weighted inner product with $f$ and one squared norm. In particular, the coefficients $c_0,\dots ,c_m$ stay the same when the allowed polynomial degree increases. This statement assumes a fixed weight, interval and normalization of the orthogonal polynomials.

To see why the least-squares normal equations decouple, first write a trial polynomial as $q={\sum }_{j=0}^m{a}_j{\varphi }_j$ in any basis of $V_m$. The residual must be orthogonal to each permitted direction, givingThe Gram matrix usually couples the unknown coefficients. With ${\varphi }_j=p_j$, it becomes $G_{ij}=h_i{\delta }_{ij}$, leaving the separate equations $h_i{a}_i=b_i$. For the monomial basis on $[-1,1]$ with $w=1$, by contrast, $G_{ij}$ is zero when $i+j$ is odd and $2/(i+j+1)$ when $i+j$ is even. Distinct even powers, or distinct odd powers, remain coupled.

There is also a proof of the minimum that uses no differentiation. The defining coefficients give $⟨f-q_m,p_n{⟩}_w=0$ for $0\le n\le m$. Thus $f-q_m$ is orthogonal to all of $V_m$. For every $v\in V_m$, the Pythagorean theorem in an inner-product space givesThe second term is nonnegative and vanishes exactly when $v=q_m$, proving that the least-squares solution is unique. This is an integrated squared-error optimum; it does not assert a smallest error at every point or a smallest uniform norm.

For $N>m$, the nested orthogonal projections satisfy the exact residual budgetIndeed, $q_N-q_m$ is a linear combination of mutually orthogonal polynomials, and $f-q_N$ is orthogonal to it. Every newly retained component removes exactly $c_n^2{h}_n$ from the squared error. No infinite expansion or completeness assumption is needed for this finite identity. The Legendre least-squares projections of a cubic make every term explicit.

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.