Question

Find f(x)=cos x as a half range Fourier sine series in the range 0<=x<=pi and sketch the function within and outside of the given range.

Answer 1

Answer on Question #44917, Engineering, SolidWorks | CosmoWorks | Ansys

Problem.

Find $f(x) = \cos x$ as a half range. Fourier sine series in the range $0 \leq x \leq p_i$ and sketch the function within and outside of the given

Solution.

Let $g(x) = f(x) - 1$ (we need to do this transformation to construct odd function).

We first extend $g(x)$ as an odd periodic function $G(x)$ of period $2\pi$:

$$G(x) = \begin{cases} 1 - \cos x, & -\pi < x < 0 \\ \cos x - 1, & 0 \leq x \leq \pi \end{cases}$$

Since $G(x)$ is odd, then $a_n = 0$, for $n \geq 0$. We turn our attention to the coefficients $b_n$. For any $n \geq 1$, we have

$$\begin{aligned} b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} G(x) \sin(nx) \, dx &= \frac{1}{\pi} \int_{0}^{\pi} 2(\cos x - 1) \sin(nx) \, dx \\ &= \frac{1}{\pi} \int_{0}^{\pi} (\sin((n + 1)x) + \sin((n - 1)x) - 2 \sin nx) \, dx \end{aligned}$$

Hence

$$\begin{aligned} b_1 = \frac{1}{\pi} \int_{0}^{\pi} (\sin 2x - 2 \sin x) \, dx &= \left(- \frac{1}{2\pi} \cos 2x + \frac{2}{\pi} \cos x\right) \Big|_{0}^{\pi} = -\frac{4}{\pi}, \\ b_{2n+1} = \frac{1}{\pi} \int_{0}^{\pi} (\sin((2n + 2)x) + \sin(2nx) - 2 \sin((2n - 1)x)) \, dx \\ &= \left(- \frac{1}{2\pi(n + 1)} \cos(2(n + 1)x) - \frac{1}{2\pi n} \cos(2nx)\right) \Big|_{0}^{\pi} \\ &\quad + \frac{2}{\pi(2n + 1)} \cos((2n + 1)x) \Big|_{0}^{\pi} = -\frac{4}{\pi(2n + 1)}, \end{aligned}$$ $$\begin{array}{l} b_{2n} = \frac{1}{\pi} \int_{0}^{\pi} (\sin((2n + 1)x) + \sin((2n - 1)x) - 2 \sin(2nx)) \, dx \\ = \left(- \frac{1}{\pi(2n + 1)} \cos((2n + 1)x) - \frac{1}{\pi(2n - 1)} \cos((2n - 1)x)\right) \Big|_{0}^{\pi} \\ + \frac{2}{2\pi n} \cos(2nx) \Big|_{0}^{\pi} = \frac{2}{\pi(2n + 1)} + \frac{2}{\pi(2n - 1)} = \frac{8n}{\pi(2n + 1)(2n - 1)} \end{array}$$

for all positive integer $n$.

Therefore

$$G(x) \sim \frac{4}{\pi} \left(- \sin x + \frac{2 \sin 2x}{1 \cdot 3} - \frac{\sin 3x}{3} + \frac{2 \sin 4x}{3 \cdot 5} + \cdots\right).$$

Then

$$f(x) \sim 1 + \frac{4}{\pi} \left(- \sin x + \frac{2 \sin 2x}{1 \cdot 3} - \frac{\sin 3x}{3} + \frac{4 \sin 4x}{3 \cdot 5} + \cdots\right)$$

Answer:

$$f(x) \sim -1 + \frac{2}{\pi} \left(- \sin x + \frac{4 \sin 2x}{1 \cdot 3} - \frac{\sin 3x}{3} + \frac{8 \sin 4x}{3 \cdot 5} + \cdots\right).$$