Orthogonal functions: Difference between revisions

Revision as of 10:42, 8 October 2023

Latest revision as of 05:51, 24 December 2024

(One intermediate revision by one other user not shown)

Line 1:

{{Short description|Type of function}}

In [[mathematics]], '''orthogonal functions''' belong to a [[function space]] that is a [[vector space]] equipped with a [[bilinear form]]. When the function space has an [[interval (mathematics)|interval]] as the [[___domain of a function|___domain]], the bilinear form may be the [[integral]] of the product of functions over the interval:

:<math> \langle f,g\rangle = \int \overline{f(x)}g(x)\,dx .</math>

The functions <math>f</math> and <math>g</math> are [[bilinear form#Reflexivity and orthogonalityOrthogonality_(mathematics)|orthogonal]] when this integral is zero, i.e. <math>\langle f, \, g \rangle = 0</math> whenever <math>f \neq g</math>. As with a [[basis (linear algebra)|basis]] of vectors in a finite-dimensional space, orthogonal functions can form an infinite basis for a function space. Conceptually, the above integral is the equivalent of a vector [[dot product]]; two vectors are mutually independent (orthogonal) if their dot-product is zero.

Suppose <math> \{ f_0, f_1, \ldots\}</math> is a sequence of orthogonal functions of nonzero [[L2-norm|''L''²-norm]]s <math display="inline"> \left\| f_n \right\| _2 = \sqrt{\langle f_n, f_n \rangle} = \left(\int f_n ^2 \ dx \right) ^\frac{1}{2} </math>. It follows that the sequence <math>\left\{ f_n / \left\| f_n \right\| _2 \right\}</math> is of functions of ''L''²-norm one, forming an [[orthonormal sequence]]. To have a defined ''L''²-norm, the integral must be bounded, which restricts the functions to being [[square-integrable function|square-integrable]].