Find the Fourier series of $g (x) = f (x-a)$, where $f$ is $2\pi$-periodic and $a$ is a real number.

Question

Find the Fourier series of $g (x) = f (x-a)$, where $f$ is $2\pi$-periodic and $a$ is a real number.

This is for real analysis so I cannot use Euler's formula to compute the Fourier coefficients.

Answer 1

If you have the Fourier coefficients for $f$, you can substitute $x \mapsto x - a$ in its expansion, and use the sum formulas for $\sin$ and $\cos$ to get those for $g$...

Answer 2

Hint: for $a_n$, for example,

$$a_n = \frac{1}{\pi} \int_{-\pi}^\pi g(x)\cos(nx)\,dx=\frac{1}{\pi}\int_{-\pi}^\pi f(x-a)\cos(nx)\,dx =\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(n(x+a))\,dx =\ldots $$

$$\ldots \frac{1}{\pi}\int_{-\pi}^\pi f(x)(\cos(nx)\cos(na)-\sin(nx)\sin(na))\,dx=\ldots$$

$$\ldots = \cos(na)*\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx)\,dx -\sin(na)*\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(nx)\,dx =\...$$

You say the Fourier coefficients of $f$ are not given, but you will find the Fourier coefficients of $g$ in terms of those of $f$.