# minimality property of Fourier coefficients

Let $f$ be a Riemann integrable periodic real function with period $2\pi$ and $n$ a positive integer.  Among all “trigonometric polynomials”

 $\varphi(x)\,:=\,\frac{\alpha_{0}}{2}\!+\!\sum_{j=1}^{n}(\alpha_{j}\cos{jx}+% \beta_{j}\sin{jx}),$

the polynomial with the coefficients $\alpha_{j}$ and $\beta_{j}$ being the Fourier coefficients

 $\alpha_{j}=a_{j}:=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos{jx}\,dx$

and

 $\beta_{j}=b_{j}:=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{jx}\,dx$

for the Fourier series of $f$ produces the minimal value of the mean square deviation

 $\frac{1}{2\pi}\int_{-\pi}^{\pi}[f(x)\!-\!\varphi(x)]^{2}\,dx.\\$

Proof.  For any fixed number $n$, it’s a question of giving the least value to the definite integral

 $\displaystyle m\;:=\;\frac{1}{2\pi}\int_{-\pi}^{\pi}\left[f(x)-\frac{\alpha_{0% }}{2}\!-\!\sum_{j=1}^{n}(\alpha_{j}\cos{jx}+\beta_{j}\sin{jx})\right]^{2}dx% \quad(\geqq 0).$ (1)

Expanding $m$ and integrating termwise yields

 $\displaystyle m\;=$ $\displaystyle\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x))^{2}\,dx-\frac{\alpha_{0}}{2% \pi}\int_{-\pi}^{\pi}f(x)\,dx$ $\displaystyle-\frac{1}{\pi}\sum_{j=1}^{n}\alpha_{j}\int_{-\pi}^{\pi}f(x)\cos{% jx}\,dx-\frac{1}{\pi}\sum_{j=1}^{n}\beta_{j}\int_{-\pi}^{\pi}f(x)\sin{jx}\,dx+% \frac{1}{2\pi}\frac{\alpha_{0}^{2}}{4}\int_{-\pi}^{\pi}dx$ $\displaystyle+\frac{1}{2\pi}\sum_{j=1}^{n}\alpha_{j}^{2}\int_{-\pi}^{\pi}\cos^% {2}{jx}\,dx+\frac{1}{2\pi}\sum_{j=1}^{n}\beta_{j}^{2}\int_{-\pi}^{\pi}\sin^{2}% {jx}\,dx$ $\displaystyle+\frac{\alpha_{0}}{2\pi}\sum_{j=1}^{n}\alpha_{j}\int_{-\pi}^{\pi}% \cos{jx}\,dx+\frac{\alpha_{0}}{2\pi}\sum_{j=1}^{n}\beta_{j}\,\int_{-\pi}^{\pi}% \sin{jx}\,dx+\frac{1}{\pi}\sum_{j=1}^{n}\sum_{k=1}^{n}\alpha_{j}\beta_{k}\!% \int_{-\pi}^{\pi}\cos{jx}\,\sin{kx}\,dx$ $\displaystyle+\frac{1}{\pi}\sum_{j=1}^{n}\sum_{k\neq j}\alpha_{j}\alpha_{k}% \int_{-\pi}^{\pi}\cos{jx}\,\cos{kx}\,dx+\frac{1}{\pi}\sum_{j=1}^{n}\sum_{k\neq j% }\beta_{j}\beta_{k}\int_{-\pi}^{\pi}\sin{jx}\,\sin{kx}\,dx.$

Here, we have the Fourier coefficients

 $\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\,dx\;=\;a_{0},\quad\frac{1}{\pi}\int_{-\pi}% ^{\pi}f(x)\cos{jx}\,dx\;=\;a_{j},\quad\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{% jx}\,dx\;=\;b_{j}.$

Furthermore,

 $\int_{-\pi}^{\pi}\cos^{2}{jx}\,dx=\int_{-\pi}^{\pi}\sin^{2}{jx}\,dx\;=\;\pi,% \quad\int_{-\pi}^{\pi}\cos{jx}\,\sin{kx}\,dx\;=\;0$

and

 $\int_{-\pi}^{\pi}\cos{jx}\,\cos{kx}\,dx=\int_{-\pi}^{\pi}\sin{jx}\,\sin{kx}\,% dx\;=\;0\quad\mbox{for\;\;}k\neq j.$

Using all these we can write

 $m\;=\;\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x))^{2}\,dx-\frac{\alpha_{0}a_{0}}{2}-% \sum_{i=1}^{n}(\alpha_{i}a_{i}+\beta_{i}b_{i})+\frac{a_{0}^{2}}{4}+\frac{1}{2}% \sum_{i=1}^{n}(\alpha_{i}^{2}+\beta_{i}^{2}).$

Adding and subtracting still the sum  $\frac{a_{0}^{2}}{4}+\frac{1}{2}\sum_{i=1}^{n}(a_{i}^{2}+b_{i}^{2})$  yields finally the form

 $m\;=\;\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x))^{2}\,dx-\frac{a_{0}^{2}}{4}-\frac{% 1}{2}\sum_{i=1}^{n}(a_{i}^{2}+b_{i}^{2})+\frac{1}{4}(\alpha_{0}-a_{0})^{2}+% \frac{1}{2}\sum_{i=1}^{n}[(\alpha_{i}-a_{i})^{2}+(\beta_{i}-b_{i})^{2}].$

The three first addends of this sum do not depend on the choice of the quantities $\alpha_{i}$ and $\beta_{i}$.  The other addends are non-negative, and their sum is minimal, equal 0, when

 $\alpha_{i}=a_{i},\quad\beta_{i}=b_{i}\quad\forall i.$

Accordingly, the mean square deviation $m$, i.e. (1), is minimal when one uses the Fourier coefficients. Q.E.D.

## References

• 1 N. Piskunov: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele.  Kirjastus Valgus, Tallinn (1966).
Title minimality property of Fourier coefficients MinimalityPropertyOfFourierCoefficients 2013-03-22 18:22:00 2013-03-22 18:22:00 pahio (2872) pahio (2872) 13 pahio (2872) Theorem msc 42A16 msc 42A10 CommonFourierSeries UniquenessOfFourierExpansion