# Fourier Series and periodicity

Let $f$ be a $2 \pi$-periodic piecewise continuous function and let $$f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n}\cos{nx}+b_{n}\sin{nx} \right] \tag{*}$$ denote its Fourier series.

Set $g(x)=f(x+\pi)$ for all $x \in \mathbb{R}$ and let $$\frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n}\cos{nx} + B_{n}\sin{nx} \right] \tag{**}$$ denote the Fourier series of $g$. Express $A_{n},B_{n}$ in terms of $a_{n},b_{n}$.

All I can think of is to use the definition of Fourier series:

\begin{align} a_{n} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos{nx}dx, \quad n=0,1,2,\dots \\ b_{n} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{nx}dx, \quad n=1,2, \dots\end{align}

which \begin{align} A_{n}&=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\cos{nx}dx \\ &=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x+\pi)\cos{nx}dx \\ &=\frac{1}{\pi}\int_{-2\pi}^{0}f(u)\cos{n(u-\pi)du} \\ &=??\end{align}

• You might want to start with the function $f(x) = 1 + \cos(2 \pi x) + 3 \sin (2 \pi x)$ and see what $g$ looks like, using your trig skills. You should then be able to generalize. – John Hughes Nov 27 '13 at 16:29
• From that it would seem as though $A_{n}=-a_{n}$ – user111731 Nov 27 '13 at 16:40

I did it with a $2\ell$-periodic function, but just take $\ell=\pi$ throughout:

From what you wrote, $$A_n={1\over \ell}\int_{-\ell}^\ell g(x)\cos(n\pi x/\ell)\,dx={1\over \ell}\int_{-\ell}^\ell f(x+\ell)\cos(n\pi x/\ell)\,dx.$$ The change of variables $u=x+\ell$ followed by $\cos(a-b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ leads to

\begin{align} {1\over \ell}\int_{-\ell}^\ell f(x+\ell)\cos(n\pi x/\ell)\,dx&={1\over \ell}\int_{0}^{2\ell} f(u)\cos({n\pi\over \ell}(u-\ell))\,dx\\ &={1\over \ell}\int_{0}^{2\ell} f(u)\cos({n\pi u\over \ell})(-1)^n\,dx\\ &=(-1)^n a_n. \end{align}

A similar argument shows $B_n=(-1)^nb_n$.

Hint: You need the identity

$$\sin(a+b)=\sin(a)\cos(b)+\sin (b)\cos(a)$$

and the one for $\cos(x)$ too.