# Fourier coefficients

Let $\mathbb{T}^{n}=\mathbb{R}^{n}/(2\pi\mathbb{Z})^{n}$ be the $n$-dimensional torus, let $\{\phi_{k}(x)\}_{k\in\mathbb{Z}^{n}}$ be an orthonormal basis for $L^{2}(\mathbb{T}^{n})$, and suppose that $f(x)\in L^{2}(\mathbb{T}^{n})$.

We can expand $f$ as a Fourier series

 $\displaystyle\sum_{k\in\mathbb{Z}^{n}}\hat{f}(k)\phi_{k},$

and we call the numbers $\hat{f}(k)$ the Fourier coefficients of $f$ with respect to the given basis. In particular, the Fourier series for $f$ converges to $f$ in the $L^{2}$ norm.

The most basic incarnation of this is finding the Fourier coefficients of a Riemann integrable function with respect to the orthonormal basis given by the trigonometric functions:

Let $f$ be a Riemann integrable function from $[-\pi,\pi]$ to $\mathbb{R}$. Then the numbers

 $\displaystyle a_{0}$ $\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx,$ $\displaystyle a_{n}$ $\displaystyle=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx,$ $\displaystyle b_{n}$ $\displaystyle=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx.$

are called the Fourier coefficients of the function $f.$

The above can be repeated for a Lebesgue-integrable function $f$ if we use the Lebesgue integral in place of the Riemann integral. This is the usual setting for modern Fourier analysis.

The trigonometric series

 $a_{0}+\sum_{n=1}^{\infty}(a_{n}\cos(nx)+b_{n}\sin(nx))$

is called the trigonometric series of the function $f$, or Fourier series of the function $f.$

Title Fourier coefficients FourierCoefficients 2013-03-22 13:57:07 2013-03-22 13:57:07 mathcam (2727) mathcam (2727) 19 mathcam (2727) Definition msc 11F30 GeneralizedRiemannLebesgueLemma FourierSeriesOfFunctionOfBoundedVariation DirichletConditions Fourier series trigonometric series