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

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.

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Do you mean the $\cos/\sin$ Fourier series, or the exponential Fourier series. For the latter, this is really straghtforward as $a_ne^{in(x-a)}=a_ne^{-ina}e^{inx}$. For the former, there are about two extra steps, which are: trigonometric formula + linearity. –  1015 Apr 12 '13 at 4:25

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 $f$.

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How do you actually solve integral (f (t-a) dt) from -pi to +pi? –  Mir Rafsan Jami Apr 25 '13 at 0:46
@MirRafsanJami : if I understand the question correctly, you are not told what $f$ actually is, so the most you can do is express the Fourier coefficients of $g$ in terms of $a$ and the Fourier coefficients of $f$. And of course in my answer I assume you were using the sine/cosine Fourier series, not the complex Fourier series. –  Stefan Smith Apr 26 '13 at 23:01

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 $g$...

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Oh no. The fourier coefficients of f are not given. I think we have to solve it for the general case. –  Mir Rafsan Jami Apr 11 '13 at 22:40