2
$\begingroup$

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

Set $g(x)=f(x+\pi)$ for all $x \in \mathbb{R}$ and let \begin{equation} \frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n}\cos{nx} + B_{n}\sin{nx} \right] \tag{**} \end{equation} 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}

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – John Hughes Nov 27 '13 at 16:29
  • $\begingroup$ From that it would seem as though $A_{n}=-a_{n}$ $\endgroup$ – user111731 Nov 27 '13 at 16:40
0
$\begingroup$

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 $B_n=(-1)^nb_n$.

$\endgroup$
0
$\begingroup$

Hint: You need the identity

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.