3b1b Fourier Calculation

Question

At the end of this video, 3b1b gives an exercise asking the viewer to show how the notion of a complex Fourier series presented in the video is equivalent to this alternative real-valued formulation:

So what he showed in the video was that the Fourier series of a function (with real domain $[0,1]$) could be computed as: $$f(t)=c_{0}+\sum_{n=1}^{\infty}(c_{n}e^{n2\pi{i}t}+c_{-n}e^{-n2\pi{i}t})$$ where the $n$th coefficient $c_{n}$ is given by: $$c_{n}=\int_{0}^{1}f(t)e^{-n2\pi{i}t}dt$$ Using the hints he gives, here is what I did: $$f(t)=\frac{a_{0}}{2}+\frac{b_{0}}{2}i+\sum_{n=1}^{\infty}\biggr((\frac{a_{n}}{2}+\frac{b_{n}}{2}i)(\cos(n2\pi{t})+i\sin(n2\pi{t}))+(\frac{a_{-n}}{2}+\frac{b_{-n}}{2}i)(\cos(-n2\pi{t})+i\sin(-n2\pi{t}))\biggr)$$ We may observe that $b_{0}=0$, $a_{n}=a_{-n}$, and $-b_{n}=b_{-n}$, so the equation can be simplified to: $$f(t)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos(n2\pi{t})-b_{n}\sin(n2\pi{t}))$$ which is almost what the video shows, except for that minus sign where there should be a plus sign...where did I go wrong?

Answer 1

If you jump back to the explanation at around 12 minutes in, for a purely real-valued function, $c_{-n} = c_{n}^*$, where $^*$ represents the complex conjugate (if $z = a + bi$, then $z^* = a - bi$). I don't know if it was stated in these terms explicitly, I can only find him talking about the little circles going in the opposite direction.

Using this, the definitions of the coefficients given in the video, and the definition of the complex exponential, we can find that it isn't actually technically $c_n = \frac{a_n}{2} + \frac{b_n}{2} i$ but it's really $c_{-n} = \frac{a_n}{2} + \frac{b_n}{2} i$, so actually $c_n = \frac{a_n}{2} - \frac{b_n}{2} i$

\begin{align*} c_n &= \int_0^1 f(t) \mathrm{e}^{-2\pi i n t} \, \mathrm{d}t \\ &= \int_0^1 f(t) (\cos(-2\pi n t) + i \sin(-2\pi n t)) \, \mathrm{d}t \\ &= \int_0^1 f(t) (\cos(2\pi n t) - i \sin(2\pi n t)) \, \mathrm{d}t \\ &= \int_0^1 f(t) \cos(2\pi n t) \, \mathrm{d}t - i \int_0^1 f(t) \sin(2\pi n t) \, \mathrm{d}t \\ &= \frac{a_n}{2} - \frac{b_n}{2} i \end{align*}

It's a small mistake in the video. Excellent question, and good work actually attempting the challenges, and catching the errata.