Find the Fourier series of $x-x^2 , f(x+2π)=f(x)$

Question

Find a Fourier series to represent $x−x^2$ from $x=−π$ to $x=π$ using complex Fourier series.

I got $C0$ = $-π^2/3$ I am getting $Cn$=$π^2/n^2 + π/in^2 + π^2/2in - π/2n^2 $

After simplification I am getting i in numerator. Kindly help.

Answer 1

Let us compute these desired coefficients. We set $f:\mathbb{R} \to \mathbb{R}$ to be the $2\pi$-periodic function such that for $x \in ]-\pi,\pi], \, f(x) = x-x^2$.

Let $n \in \mathbb{Z}^*$, $$\begin{align*} 2\pi \cdot c_n(f) &= \int_{-\pi}^{\pi} e^{-inx}(x-x^2) dx = \frac{i}{n}\left[ e^{-inx}(x-x^2)\right]_{-\pi}^{\pi} - \frac{i}{n} \int_{-\pi}^{\pi} e^{-inx}(1-2x) dx \\ &= \frac{i}{n}(-1)^n (\pi-\pi^2 -(-\pi-\pi^2)) \\ &-\left( \frac{i^2}{n^2} (-1)^n \left[ e^{-inx}(1-2x)\right]_{-\pi}^{\pi} \frac{i^2}{n^2} - \int_{-\pi}^{\pi} e^{-inx}\cdot (-2) dx\right)\\ &= (-1)^n\frac{2i\pi}{n} + (-1)^n\frac{1}{n^2} (1-2\pi -(1+2\pi)) + 0 \\ &= 2\pi (-1)^n\left( -\frac{2}{n^2} + \frac{i}{n} \right) \end{align*}$$

So $$c_n(f) = -(-1)^n\frac{2}{n^2} + (-1)^n\frac{i}{n}$$

Now $c_0(f) = \frac{1}{2\pi} \int_{-\pi}^{\pi} (x-x^2) dx = \frac{1}{2\pi}\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{-\pi}^{\pi} = \frac{1}{2\pi}\left(\frac{\pi^2}{2} - \frac{\pi^3}{3} - (\frac{\pi^2}{2} + \frac{\pi^3}{3})\right) = -\frac{\pi^3}{3} $

Since $c_{-n} = \bar{c_n}$, this gives us $$\begin{align*} f(x) &= \sum_{n \in \mathbb{Z}} c_n e^{inx} \\ &= c_0 + \sum_{n=1}^\infty c_n e^{inx} + \sum_{n=1}^\infty c_{-n} e^{-inx}\\ &= c_0 + \sum_{n=1}^\infty c_n e^{inx} + \sum_{n=1}^\infty \overline{c_{n} e^{inx}} \\ &= c_0 + \sum_{n=1}^\infty c_n e^{inx} + \overline{c_{n} e^{inx}} \\ &= c_0 + \sum_{n=1}^\infty 2 \Re(c_n e^{inx}) \\ &= c_0 + \sum_{n=1}^\infty a_n \cos(nx) + \sum_{n=1}^\infty b_n \sin(nx) \end{align*}$$ where $a_n = 2 \Re(c_n) = -(-1)^n\frac{4}{n^2}$ and $b_n = -2\Im(c_n) = -(-1)^n\frac{2}{n}$.