Nonorthogonal basis functions

A function f(x), x \in \mathbb{R}, f \in \mathbb{R}, not necessarily periodic, can be approximated over an interval I as a finite linear combination of N independent functions \sigma_i (x):

f(x) \approx \sum_{i=0}^{N-1} \alpha_i \sigma_i (x)

The condition for the optimality of this approximation is

min_{\alpha_k} \int_I (f(x) - \sum_{i=0}^{N-1} \alpha_i \sigma_i (x))^2 dx, k \in 0 .. N-1

equivalent to the set of equations

\dfrac{\partial}{\partial \alpha_k} \int_I (f(x) - \sum_{i=0}^{N-1} \alpha_i \sigma_i (x))^2 dx=0, k \in 0 .. N-1

that is

\int_I (f(x) - \sum_{i=0}^{N-1} \alpha_i \sigma_i (x)) \sigma_k (x))dx=0, k \in 0 .. N-1

or

\sum_{i=0}^{N-1} \alpha_i \int_I \sigma_i (x) \sigma_k (x) dx = \int_I f(x) \sigma_k (x) dx, k \in 0 .. N-1 \label{eq:sistema}

This is a system of N linear equations in the N unknowns \alpha_i , and it is nonsingular as long as the basis functions \sigma_i (x) are independent; if the basis functions are orthogonal the system is completely uncoupled and its solution is immediately available as

\alpha_i = \frac{\int_I f(x) \sigma_i (x) dx}{\int_I \sigma_i ^2 (x)dx}

If, moreover, the basis functions are orthonormal, this reduces to

\alpha_i = \int_I f(x) \sigma_i (x) dx

But, even if the \sigma_i (x) are not orthogonal, the full system still determines a set of optimal \alpha_i.

An important difference is that, while the \alpha_i computed for orthogonal basis functions, as a consequence of the system being uncoupled, are independent of the order N of the approximation, those computed for nonorthogonal basis functions do change with N, even if the difference is relatively small if N is large.

As an example, determine the optimal approximation of a given function as a truncated power series over a finite interval [a,b]:

\sum_{i=0}^{N-1} \alpha_i \int_a^b x^{i+k}dx=\int_a^b f(x) x^k dx

or

\sum_{i=0}^{N-1} \alpha_i \frac{b^{i+k+1}-a^{i+k+1}}{i+k+1}=\int_a^b f(x) x^k dx

To put numbers into the symbols, let a=-1, b=1 and f(x)=\sin(\pi x); then

\int_a^b \sin(\pi x) x^{2q}dx=0, and \int_a^b \sin(\pi x) x^{2q+1}dx can be computed and introduced into the system; at last, truncating to five terms I find

\sin(\pi x) \approx x (3.14158 + x^2 (-5.16724 +x^2 (2.54601 + x^2 (-0.586621 + x^2 0.066297))))

with an error around 10^{-5} inside the interval; of course the error gets rapidly worse outside.

Note that \alphas have nothing to do with the Taylor coefficients; Taylor series are valid in the neighborhood of a point, not over a finite interval.

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