Suppose that f(x) = 1 for 0 < x < 5 and f(x) = 0 for 5 < x < 10. Write f(x) as follows: f(x) = ∑ C_n * e^(inπx/5) from n = -∞ to ∞ Then C0 = ? C1 = ? Equivalently, we can write... Suppose that f(x) = 1 for 0 < x < 5 and f(x) = 0 for 5 < x < 10. Write f(x) as follows: f(x) = ∑ C_n * e^(inπx/5) from n = -∞ to ∞ Then C0 = ? C1 = ? Equivalently, we can write f(x) in the form f(x) = ∑ (A_n * sin(nπx/5) + B_n * cos(nπx/5)) + B0 from n = 1 to ∞ where An = ? Bn (with n ≠ 0) = ? B0 = ? Read more

Steps to Solve

1. Calculate $C_0$

$C_0$ is the average value of the function $f(x)$ over one period. The period is $10$.

$C_0 = \frac{1}{10} \int_{0}^{10} f(x) dx = \frac{1}{10} \int_{0}^{5} 1 dx + \frac{1}{10} \int_{5}^{10} 0 dx = \frac{1}{10} [x]_0^5 = \frac{1}{10} (5-0) = \frac{5}{10} = \frac{1}{2}$

2. Calculate $C_1$

$C_1 = \frac{1}{10} \int_{0}^{10} f(x) e^{-i x 2\pi/10} dx $

$C_1 = \frac{1}{10} \int_{0}^{5} 1 \cdot e^{-i x \pi/5} dx + \frac{1}{10} \int_{5}^{10} 0 \cdot e^{-i x \pi/5} dx $

$C_1 = \frac{1}{10} \int_{0}^{5} e^{-i x \pi/5} dx = \frac{1}{10} [\frac{e^{-i x \pi/5}}{-i \pi/5}]_0^5 = \frac{1}{10} \frac{5}{-i\pi} [e^{-i x \pi/5}]_0^5 = \frac{1}{-2i\pi} [e^{-i\pi} - e^{0}] = \frac{i}{2\pi} [e^{-i\pi} - 1]=\frac{i}{2\pi} [-1-1] = \frac{i}{2\pi} [-2]=\frac{-i}{\pi}$

3. Calculate $A_n$

$A_n = \frac{2}{10} \int_{0}^{10} f(x) \sin(n x 2\pi / 10) dx = \frac{1}{5} \int_{0}^{5} \sin(nx\pi/5) dx + \frac{1}{5} \int_{5}^{10} 0 \cdot \sin(nx\pi/5) dx$

$A_n = \frac{1}{5} \int_{0}^{5} \sin(nx\pi/5) dx = \frac{1}{5} [-\frac{5}{n\pi} \cos(nx\pi/5)]_0^5 = -\frac{1}{n\pi} [\cos(n\pi) - \cos(0)] = -\frac{1}{n\pi} [\cos(n\pi) - 1]$

If $n$ is even, $\cos(n\pi) = 1$, so $A_n = -\frac{1}{n\pi} [1-1] = 0$ If $n$ is odd, $\cos(n\pi) = -1$, so $A_n = -\frac{1}{n\pi} [-1-1] = \frac{2}{n\pi}$

So, $A_n = \begin{cases} 0, & \text{if } n \text{ is even} \ \frac{2}{n\pi}, & \text{if } n \text{ is odd} \end{cases}$

4. Calculate $B_n$ (with $n \ne 0$)

$B_n = \frac{2}{10} \int_{0}^{10} f(x) \cos(n x 2\pi / 10) dx = \frac{1}{5} \int_0^5 \cos(nx\pi/5) dx + \frac{1}{5} \int_5^{10} 0 \cdot \cos(nx\pi/5) dx$

$B_n = \frac{1}{5} \int_0^5 \cos(nx\pi/5) dx = \frac{1}{5} [\frac{5}{n\pi} \sin(nx\pi/5)]_0^5 = \frac{1}{n\pi} [\sin(n\pi) - \sin(0)] = \frac{1}{n\pi} [0 - 0] = 0$

5. Calculate $B_0$

$B_0$ represents the DC component or the average value of the function $f(x)$.

$B_0 = \frac{1}{10} \int_{0}^{10} f(x) dx = \frac{1}{10} \int_{0}^{5} 1 dx + \frac{1}{10} \int_{5}^{10} 0 dx = \frac{1}{10} [x]_0^5 = \frac{1}{10} (5-0) = \frac{5}{10} = \frac{1}{2}$