Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.4: Orthogonality and Normalization

  • Page ID
  • [ "article:topic", "authorname:nwalet", "license:ccbyncsa" ]

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Consider the series

    \[\frac{a_0}{2} + \sum_{n=1}\bigg[a_n\cos\bigg(\frac{n\pi x}{L}\bigg) + b_n\sin\bigg(\frac{n\pi x}{L}\bigg)\bigg], \hspace{3cm} -L \leq x \leq L.\]

    This is called a trigonometric series. If the series approximates a function f (as will be discussed) it is called a Fourier series and a and b are the Fourier coefficients of f.

    In order for all of this to make sense we first study the functions

    \[\{1,\cos\bigg(\frac{n\pi x}{L}\bigg), \sin\bigg( \frac{n\pi x}{L}\bigg)\}, \hspace{3 cm} n=1,2,\dots,\]

    and especially their properties under integration. We find that

    \[ \int_{-L}^L 1\cdot 1 dx = 2L,\]

    \[ \int_{-L}^L 1 \cdot \cos\bigg(\frac{n\pi x}{L}\bigg) dx = 0\]

    \[ \int_{-L}^L 1 \cdot \sin\bigg(\frac{n\pi x}{L}\bigg) dx = 0\]

    \[ \begin{align} \int_{-L}^L \cos\bigg(\frac{m\pi x}{L}\bigg) \cdot \cos\bigg(\frac{n\pi x}{L}\bigg) dx & =  \frac{1}{2}\int_{-L}^L \cos\bigg(\frac{(m+n)\pi x}{L}\bigg) + \cos\bigg(\frac{(m-n)\pi x}{L}\bigg) dx\\ & =   \bigg\{ \begin{array}{lr} 0  & \mbox{if } n \leq m \\ L & \mbox{if } n=m \end{array}\end{align}, \]

    \[ \begin{align} \int_{-L}^L \sin\bigg(\frac{m\pi x}{L}\bigg) \cdot \sin\bigg(\frac{n\pi x}{L}\bigg) dx & =  \frac{1}{2}\int_{-L}^L \cos\bigg(\frac{(m+n)\pi x}{L}\bigg) + \cos\bigg(\frac{(m-n)\pi x}{L}\bigg) dx\\ & =   \bigg\{ \begin{array}{lr} 0  & \mbox{if } n \leq m \\ L & \mbox{if } n=m \end{array}\end{align}, \]

    \[\begin{align} \int_{-L}^L \cos\bigg(\frac{m\pi x}{L}\bigg) \cdot \sin\bigg(\frac{n\pi x}{L}\bigg) dx  & =  \frac{1}{2}\int_{-L}^L \cos\bigg(\frac{(m+n)\pi x}{L}\bigg) + \cos\bigg(\frac{(m-n)\pi x}{L}\bigg) dx\\ & =   \bigg\{ \begin{array}{lr} 0  & \mbox{if } n \leq m \\ L & \mbox{if } n=m \end{array} \end{align},\]

    If we consider these integrals as some kind of inner product between functions (like the standard vector inner product) we see that we could call these functions orthogonal. This is indeed standard practice, where for functions the general definition of inner product takes the form

    \[(f,g) = \int_a^b w(x)f(x)g(x)dx.\]

    If this is zero we say that the functions f and g are orthogonal on the interval [ab] with weight function w. If this function is 1, as is the case for the trigonometric functions, we just say that the functions are orthogonal on [ab].

    The norm of a function is now defined as the square root of the inner-product of a function with itself (again, as in the case of vectors),

    \[ \norm{f} = \sqrt{\int_a^b w(x)f(x)^2dx}.\]


    If we define a normalised form of f (like a unit vector) as \( f/\norm{f}\), we have

    \[ \norm{\frac{f}{\norm{f}}} = \sqrt{\frac{\int_a^bw(x)f(x)^2dx}{\norm{f}^2}}=\frac{\sqrt{\int_a^b w(x)f(x)^2dx}}{\norm{f}}=\frac{\norm{f}}{\norm{f}}=1.\]

    Exercise \(\PageIndex{1}\)

    What is the normalised form of \(\big\{1, \cos\big(\frac{n\pi x}{L}\big), \sin\big(\frac{n\pi x}{L}\big)\big\}?\)


    \( \big\{\frac{1}{\sqrt{2L}}, \big(\frac{1}{\sqrt{L}}\big)\cos\big(\frac{n \pi x}{L}\big),\big(\frac{1}{\sqrt{L}}\big)\sin\big(\frac{n \pi x}{L}\big) \big\}\)

     A set of mutually orthogonal functions that are all normalised is called an orthonormal set.