Let $f$ be a $2\pi$ periodic function such that $f(x) = \pi -|x|$ in $[-\pi,\pi]$. Find its harmonic of order $3$
I'm learning about orthogonal families of polynmials and discrete polynomial approximations by fourier functions. I've came upon this exercise, but I don't know what an harmonic of order $3$ means.
The goal is to find the harmonic of order $3$ of that function. They are talking about Fourier series. Look here to know more. We know $f$ is periodic, with period $T=2\pi$, so $f$ can be written as an infinite sum:
$$f(t) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty}\left[\cos\left(\dfrac{2n\pi}{T} \,t \right) + \sin\left(\dfrac{2n\pi}{T} \,t \right) \right],$$
with
$$\begin{align} a_0 &= \dfrac{2}{T} \int\limits_{-T/2}^{T/2} f(t) \, dt, \\ a_n &= \dfrac{2}{T} \int\limits_{-T/2}^{T/2} f(t) \cos \left( \dfrac{2n\pi}{T}\,t \right) dt, \\ b_n &= \dfrac{2}{T} \int\limits_{-T/2}^{T/2} f(t) \sin \left( \dfrac{2n\pi}{T}\,t \right) dt. \end{align}$$
From here you can find $a_0$, $a_n$ and $b_n$, you could do that or you could find $a_3$ and $b_3$ (which are the terms of the harmonic of order 3). For example, to find the harmonic of order 5:
$$ \begin{align} a_5 &= \dfrac{2}{2\pi} \int\limits_{-\pi}^{\pi} (\pi - |t|) \cos \left( \dfrac{2\cdot5 \pi}{2\pi}\,t \right) dt, \\ &=\dfrac{1}{\pi} \int\limits_{-\pi}^{0} (\pi + t) \cos \left(5t \right)\, dt + \dfrac{1}{\pi} \int\limits_{0}^{\pi} (\pi - t) \cos \left(5t \right)\, dt, \end{align}$$
and
$$ \begin{align} b_5 &= \dfrac{2}{2\pi} \int\limits_{-\pi}^{\pi} (\pi - |t|) \sin \left( \dfrac{2\cdot5 \pi}{2\pi}\,t \right) dt, \\ &=\dfrac{1}{\pi} \int\limits_{-\pi}^{0} (\pi + t) \sin \left(5t \right)\, dt + \dfrac{1}{\pi} \int\limits_{0}^{\pi} (\pi - t) \sin \left(5t \right)\, dt, \end{align}$$
which is
$$a_5 \cos(5t) + b_5 \sin(5t).$$
Again, it's your choice: $a_n$ and $b_n$ and then substitute $n=3$ or find $a_3$ and $b_3$.