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Why does the integral for determining the coefficients of a complex fourier series contain a negative sign in the exponent:

$$c_n = \int_{-a}^{a} f(x)e^{-in\pi x/a} \,dx$$

scouring the internet for an answer, I stumbled upon a proof, and it looks good. However, to my intuition it would make sense to define the $c_n$ as the inner product of $f(x)$ and choice of basis $e^{in\pi x/a}$. Like this:

$$c_n = \frac{\langle f(x),e^{in\pi x/a} \rangle}{\langle e^{in\pi x/a},e^{in\pi x/a} \rangle}$$ But this of course does not have the negative sign in the exponent. Why is this a case, is there something wrong in my reasoning?

MPW
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juggernugger
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    That inner product in the numerator does have the minus sign in the exponent, because the inner product (in US conventions, oppositely in French) is complex-conjugate-linear in the second argument. – paul garrett Jul 25 '22 at 17:43
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    Paul, I think the choice is less US vs French but rather physics (engineering, applied...) vs math. I studied in France and depending on the type of book you'll have the complex conjugate to the left (in physics, it's the <bra| which is in the dual and the |ket> which is in the original vector space) or to the right (the canonical bilinear form on E x E*). – Max Jul 25 '22 at 17:48
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    You're missing a $2/a$ in front of that integral. – eyeballfrog Jul 25 '22 at 17:51

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