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**chisigma** If a function $\displaystyle f(t)$ is periodic of period $\displaystyle 2 \tau$, then it can be expanded in Fourier series as...

$\displaystyle f(t)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n}\ \cos \frac{\pi\ n\ t}{\tau} + b_{n}\ \sin \frac{\pi\ n\ t}{\tau})$ (1)

... where...

$\displaystyle a_{n}= \frac{1}{\tau}\ \int_{0}^{2 \tau} f(t)\ \cos \frac{\pi\ n\ t}{\tau}\ dt$

$\displaystyle b_{n}= \frac{1}{\tau}\ \int_{0}^{2 \tau} f(t)\ \sin \frac{\pi\ n\ t}{\tau}\ dt$ (2)

If we consider the integral...

$\displaystyle F(t)= \int_{0}^{t} f(x)\ dx $ (3)

... it is easy to see that if $\displaystyle a_{0}=0$ then $\displaystyle F(t)$ is also periodic of period $\displaystyle 2 \tau$. Now $\displaystyle \varphi_{i} (t) = (-1)^{\lfloor 2^{i}\ t \rfloor}$ is periodic of period $\displaystyle 2^{1-i}$ and for it is $\displaystyle a_{0}=0$, so that also...

$\displaystyle f_{i} (t)= \int_{0}^{t} \varphi_{i} (x)\ dx$ (4)

... is periodic of period $\displaystyle 2^{1-i}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$