Question

Obtain the Fourier series for the following periodic function which has a period of 2pie:
f (x) = {-1/2 (pie - x) for (-pie) < x <0
{1/2 (pie + x) for 0 < x < (pie)

Answer 1

Answer on Question #50745 - Math - Calculus

Obtain the Fourier series for the following periodic function which has a period of $2\pi$:

$$f(x) = \begin{cases} -\frac{1}{2}(\pi - x), & -\pi < x < 0 \\ \frac{1}{2}(\pi + x), & 0 < x < \pi \end{cases}$$

Solution

Let's compute coefficients of the Fourier series:

$$\begin{aligned} a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx &= \frac{1}{2\pi} \left( \int_0^{\pi} (\pi + x) \, dx - \int_{-\pi}^{0} (\pi - x) \, dx \right) = \frac{1}{2\pi} \left( (\pi x + \frac{1}{2} x^2) \Big|_0^{\pi} - (\pi x - \frac{1}{2} x^2) \Big|_{-\pi}^{0} \right) = \\ &= \frac{1}{2\pi} \left( \frac{3}{2} \pi^2 - \frac{3}{2} \pi^2 \right) = 0 \end{aligned}$$

for $n \geq 1$:

$$\begin{aligned} a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \, dx &= \frac{1}{2\pi} \left( \int_0^{\pi} (\pi + x) \cos nx \, dx - \int_{-\pi}^{0} (\pi - x) \cos nx \, dx \right) = \\ &= \frac{1}{2\pi} \left( \int_0^{\pi} \pi \cos nx \, dx - \int_{-\pi}^{0} \pi \cos nx \, dx + \int_0^{\pi} x \cos nx \, dx + \int_{-\pi}^{0} x \cos nx \, dx \right) = \\ &= \frac{1}{2\pi} \left( \frac{\pi}{n} \sin nx \Big|_0^{\pi} - \frac{\pi}{n} \sin nx \Big|_{-\pi}^{0} + \int_0^{\pi} x \cos nx \, dx + \int_{-\pi}^{0} x \cos nx \, dx \right) = \frac{1}{2\pi} \left( \int_0^{\pi} x \cos nx \, dx + \int_{-\pi}^{0} x \cos nx \, dx \right) \end{aligned}$$

Due to integration by parts formula, where $u(x) = x$ and $dv(x) = \cos nx \, dx$, we obtain $du(x) = dx$ and $v(x) = \frac{1}{n} \sin nx$. Thus, $\int x \cos nx \, dx = \frac{1}{n} x \sin nx - \frac{1}{n} \int \sin nx \, dx = \frac{1}{n} x \sin nx + \frac{1}{n^2} \cos nx + C$, where $C$ is an arbitrary real constant.

That is why

$$\begin{aligned} \frac{1}{2\pi} \left( \int_0^{\pi} x \cos nx \, dx + \int_{-\pi}^{0} x \cos nx \, dx \right) &= \frac{1}{2\pi} \left( \left( \frac{1}{n} x \sin nx + \frac{1}{n^2} \cos nx \right) \Big|_0^{\pi} + \left( \frac{1}{n} x \sin nx + \frac{1}{n^2} \cos nx \right) \Big|_{-\pi}^{0} \right) = \\ &= \frac{1}{n^2 2\pi} \left( (-1)^n - 1 + 1 - (-1)^n \right) = 0 \end{aligned}$$

Thus, $a_n = 0$, for $n \geq 1$

$$\begin{aligned} b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \, dx &= \frac{1}{2\pi} \left( \int_0^{\pi} (\pi + x) \sin nx \, dx - \int_{-\pi}^{0} (\pi - x) \sin nx \, dx \right) = \\ &= \frac{1}{2\pi} \left( \int_0^{\pi} \pi \sin nx \, dx - \int_{-\pi}^{0} \pi \sin nx \, dx + \int_0^{\pi} x \sin nx \, dx + \int_{-\pi}^{0} x \sin nx \, dx \right) = \frac{1}{2n} \left( \cos nx \Big|_0^{\pi} - \cos nx \Big|_{-\pi}^{0} \right) + \\ &+ \frac{1}{2\pi} \left( \int_0^{\pi} x \sin nx \, dx + \int_{-\pi}^{0} x \sin nx \, dx \right) = \frac{1}{n} \left( (-1)^n - 1 \right) + \frac{1}{2\pi} \left( \int_0^{\pi} x \sin nx \, dx + \int_{-\pi}^{0} x \sin nx \, dx \right) \end{aligned}$$

Due to integration by parts formula, where $u(x) = x$ and $dv(x) = \sin nx \, dx$, we obtain $du(x) = dx$ and $v(x) = -\frac{1}{n} \cos nx$. Thus, $\int x \sin nx \, dx = -\frac{1}{n} x \cos nx + \frac{1}{n} \int \cos nx \, dx = -\frac{1}{n} x \cos nx + \frac{1}{n^2} \sin nx + C$, where $C$ is an arbitrary real constant.

That is why

$$\begin{array}{l} \frac {1}{2 \pi} \left(\int_ {0} ^ {\pi} x \sin n x d x + \int_ {- \pi} ^ {0} x \sin n x d x\right) = \frac {1}{2 \pi} \left(\left(- \frac {1}{n} x \cos n x + \frac {1}{n ^ {2}} \sin n x\right) \Big | _ {0} ^ {\pi} + \left(- \frac {1}{n} x \cos n x + \frac {1}{n ^ {2}} \sin n x\right) \Big | _ {- \pi} ^ {0}\right) = \\ = \frac {1}{2 n} \left(1 - (- 1) ^ {n} - 1 + (- 1) ^ {n}\right) = 0. \end{array}$$

Thus, $b_{n} = \frac{1}{n}\left((-1)^{n} - 1\right)$ , for $n \geq 1$

Therefore we obtain the Fourier series of the function $f(x)$ :

$$f (x) \sim \sum_ {n = 1} ^ {\infty} \frac {1}{n} \left((- 1) ^ {n} - 1\right) \sin n x$$

Answer: $f(x) \sim \sum_{n=1}^{\infty} \frac{1}{n} \left((-1)^{n} - 1\right) \sin nx$

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