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The complex Fourier series is: $$\sum_{n=-\infty}^{\infty}(-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}\sinh(l)e^{in\pi x/l}$$ How can I derive the real Fourier series (sines and cosines) from this? Do I just take the real part of it, or what? The book I'm using (Strauss) has no examples, and it doesn't provide the answer as reference.

Taking the real part makes some logical sense to me, because the complex part of a real number is zero anyway. But I don't know how to justify this, or even if it's correct.

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  • $\begingroup$ Yes, just write $e^{in\pi x / l}=\cos\left(n\pi x/l\right) + i\sin\left(n\pi x/l\right)$, multiply the product of complex numbers, and only keep the real terms (as you say, the imaginary terms must sum to zero anyway). $\endgroup$ – mjqxxxx Mar 3 '14 at 19:04
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Thanks to the formula $e^{ix} = \cos(x) + i \sin(x)$ you can transform $$\sum_{n = -\infty}^\infty c_n(l) e^{in\pi x \, /\, l}$$ into the following: $$\sum_{n = -\infty}^\infty c_n(l) \cos( n \pi x \, / \, l) + \sum_{n = -\infty}^\infty i c_n(l) \sin(n\pi x\,/\,l).$$ Then, by exploiting the identities $\cos(-x)=\cos(x)$ and $\sin(-x) = -\sin(x)$, you get $$c_0(l) + \sum_{n = 1}^\infty [c_n(l)+c_{-n}(l)] \cos( n \pi x \, / \, l) + \sum_{n = 1}^\infty i [c_n(l)-c_{-n}(l)] \sin(n\pi x\,/\,l) $$

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