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# 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.

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

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

## Mathematics Subject Classification

11F30*no label found*

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## Attached Articles

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