Let's solve the problem step by step: --- ### **Given Function:** \[ f(x) = \begin{cases} 1-x & 0 < x < 1 \\ 0 & 1 \leq x < 2 \end{cases} \] --- ## **(b) Represent \( f(x) \) as a Fourier Cosine Series** #### **Step 1: Define the Interval** - The interval is \( 0 < x < 2 \), so the period \( T = 2 \). - We're asked for a **cosine series**, i.e., an **even extension** over \( [0,2] \). #### **Step 2: Fourier Cosine Series Form** For \( f(x) \) on \( 0 < x < 2 \): \[ f(x) = \sum_{n=0}^{\infty} a_n \cos\left(\frac{n\pi x}{L}\right) \] where \( L = 2 \). The Fourier cosine coefficients are: \[ a_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) dx \] With \( L = 2 \): \[ a_n = \int_0^2 f(x) \cos\left(\frac{n\pi x}{2}\right) dx \] --- #### **Step 3: Compute \( a_0 \):** \[ a_0 = \frac{1}{2} \int_0^2 f(x) dx \] But \( f(x) = 1-x \) for \( 0 < x < 1 \), and \( f(x) = 0 \) for \( 1 \leq x < 2 \): \[ a_0 = \frac{1}{2} \left[ \int_0^1 (1-x) dx + \int_1^2 0 \, dx \right] = \frac{1}{2} \int_0^1 (1-x) dx \] \[ = \frac{1}{2} \left[ x - \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \left(1 - \frac{1}{2}\right) = \frac{1}{2} \left(\frac{1}{2}\right) = \frac{1}{4} \] --- #### **Step 4: Compute \( a_n \) for \( n \geq 1 \):** \[ a_n = \int_0^2 f(x) \cos\left(\frac{n\pi x}{2}\right) dx = \int_0^1 (1-x) \cos\left(\frac{n\pi x}{2}\right) dx \] (because \( f(x)=0 \) for \( x \) from 1 to 2) Let’s split: \[ a_n = \int_0^1 \cos\left(\frac{n\pi x}{2}\right) dx - \int_0^1 x \cos\left(\frac{n\pi x}{2}\right) dx \] Let’s solve each part. --- ##### **First Integral:** \[ I_1 = \int_0^1 \cos\left(\frac{n\pi x}{2}\right) dx \] Let \( u = \frac{n\pi x}{2} \implies du = \frac{n\pi}{2} dx \implies dx = \frac{2}{n\pi} du \) When \( x = 0, u = 0 \); when \( x = 1, u = \frac{n\pi}{2} \). So, \[ I_1 = \int_{u=0}^{u=\frac{n\pi}{2}} \cos(u) \frac{2}{n\pi} du = \frac{2}{n\pi} [\sin(u)]_0^{\frac{n\pi}{2}} = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right) \] --- ##### **Second Integral:** \[ I_2 = \int_0^1 x \cos\left(\frac{n\pi x}{2}\right) dx \] Use integration by parts: Let \( u = x \), \( dv = \cos\left(\frac{n\pi x}{2}\right) dx \) Then, \( du = dx \), \( v = \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \) So, \[ I_2 = u v\Big|_0^1 - \int_0^1 v du = \left[ x \cdot \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_0^1 - \int_0^1 \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) dx \] The second integral: \[ \int_0^1 \sin\left(\frac{n\pi x}{2}\right) dx \] Let \( w = \frac{n\pi x}{2} \implies dx = \frac{2}{n\pi} dw \) When \( x = 0, w = 0 \); when \( x = 1, w = \frac{n\pi}{2} \): \[ \int_0^1 \sin\left(\frac{n\pi x}{2}\right) dx = \frac{2}{n\pi} \int_0^{\frac{n\pi}{2}} \sin(w) dw = \frac{2}{n\pi} \left[ -\cos(w) \right]_0^{\frac{n\pi}{2}} = \frac{2}{n\pi} \left[ -\cos\left(\frac{n\pi}{2}\right) + 1 \right] \] Therefore, \[ I_2 = \left[ \frac{2 x}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_0^1 - \frac{2}{n\pi} \cdot \frac{2}{n\pi} [1 - \cos\left(\frac{n\pi}{2}\right)] \] \[ = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right) - \frac{4}{n^2\pi^2} [1 - \cos\left(\frac{n\pi}{2}\right)] \] --- #### **Step 5: Combine Results** \[ a_n = I_1 - I_2 = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right) - \left( \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right) - \frac{4}{n^2\pi^2} [1 - \cos\left(\frac{n\pi}{2}\right)] \right) \] \[ = \frac{4}{n^2\pi^2} [1 - \cos\left(\frac{n\pi}{2}\right)] \] --- ## **Final Fourier Cosine Series:** \[ f(x) = \frac{1}{4} + \sum_{n=1}^\infty \frac{4}{n^2\pi^2} \left[1 - \cos\left(\frac{n\pi}{2}\right)\right] \cos\left(\frac{n\pi x}{2}\right) \] --- ### **(c) Sketch for 3 Periods** - The function is periodic with period 2. - For sketching 3 periods, plot the function from \( x = -2 \) to \( x = 4 \). - The original \( f(x) \) is 1 at \( x = 0 \), decreases linearly to 0 at \( x = 1 \), and is 0 from \( x = 1 \) to \( x = 2 \), and repeats every 2 units. #### **Sketch Steps:** 1. Draw \( f(x) = 1-x \) for \( 0 < x < 1 \). 2. Draw \( f(x) = 0 \) for \( 1 \leq x < 2 \). 3. Repeat for intervals \( 2 < x < 4 \), \( -2 < x < 0 \), etc. --- ## **Summary Table** | Step | Expression | |-------------------------|----------------------------------------------------------------------------| | Period | \(T = 2\) | | \(a_0\) | \(\frac{1}{4}\) | | \(a_n\) | \(\frac{4}{n^2\pi^2} [1 - \cos(\frac{n\pi}{2})]\) | | Fourier series | \(f(x) = \frac{1}{4} + \sum_{n=1}^\infty a_n \cos(\frac{n\pi x}{2})\) | --- ## **Final Answers** ### **Fourier Cosine Series:** \[ \boxed{ f(x) = \frac{1}{4} + \sum_{n=1}^\infty \frac{4}{n^2\pi^2} \left[1 - \cos\left(\frac{n\pi}{2}\right)\right] \cos\left(\frac{n\pi x}{2}\right) } \] ### **Sketch:** - Draw the function as described above for three periods. --- If you want a plotted sketch, let me know!

Final Fourier Cosine Series: \[ f(x) = \frac{1}{4} + \sum_{n=1}^\infty \frac{4}{n^2 \pi^2} \left[1 - \cos\left(\frac{n\pi}{2}\right)\right] \cos\left(\frac{n\pi x}{2}\right) \] Note: The period is \( 2 \). The coefficients are: - \( a_0 = \frac{1}{4} \) - \( a_n = \frac{4}{n^2 \pi^2} \left[1 - \cos\left(\frac{n\pi}{2}\right)\right] \) for \( n \geq 1 \).