Let's solve the problem step by step. ## 1. **Understanding the Problem** Given: \( f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2 - x, & 1 < x \leq 2 \end{cases} \) You are asked for its **sine series expansion** and a sketch. --- ## 2. **Setting Up the Sine Series** Since the function is defined on \(0 < x < 2\), we will use the **Fourier sine series** on the interval \([0,2]\): \[ f(x) = \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi x}{2}\right) \] where \[ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \] Here, \(L = 2\). --- ## 3. **Calculating \(b_n\)** \[ b_n = \int_{0}^{2} f(x) \sin\left(\frac{n\pi x}{2}\right) dx \] Split the integral at \(x=1\): \[ b_n = \int_{0}^{1} x \sin\left(\frac{n\pi x}{2}\right) dx + \int_{1}^{2} (2-x) \sin\left(\frac{n\pi x}{2}\right) dx \] **Let’s calculate each part:** --- ### **A. First Integral:** \[ I_1 = \int_{0}^{1} x \sin\left(\frac{n\pi x}{2}\right) dx \] Use integration by parts: Let \(u = x\), \(dv = \sin\left(\frac{n\pi x}{2}\right) dx\) - \(du = dx\) - \(v = -\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\) So, \[ I_1 = \left[-\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\right]_{0}^{1} + \int_{0}^{1} \frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right) dx \] The second integral: \[ \int \cos(ax) dx = \frac{1}{a} \sin(ax) \] So, \[ \int_{0}^{1} \cos\left(\frac{n\pi x}{2}\right) dx = \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{0}^{1} \] Thus, \[ I_1 = \left[-\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\right]_{0}^{1} + \frac{2}{n\pi} \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{0}^{1} \] \[ = -\frac{2}{n\pi} \left[ 1 \cdot \cos\left(\frac{n\pi}{2}\right) - 0 \cdot 1 \right] + \frac{4}{(n\pi)^2} \left[ \sin\left(\frac{n\pi}{2}\right) - 0 \right] \] \[ = -\frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] --- ### **B. Second Integral:** \[ I_2 = \int_{1}^{2} (2-x) \sin\left(\frac{n\pi x}{2}\right) dx \] Let’s break it up: \[ I_2 = 2 \int_{1}^{2} \sin\left(\frac{n\pi x}{2}\right) dx - \int_{1}^{2} x \sin\left(\frac{n\pi x}{2}\right) dx \] First term: \[ 2 \int_{1}^{2} \sin\left(\frac{n\pi x}{2}\right) dx = 2 \left[ -\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right) \right]_{1}^{2} = -\frac{4}{n\pi} \left[ \cos\left(n\pi\right) - \cos\left(\frac{n\pi}{2}\right) \right] \] Second term (integration by parts, similar to \(I_1\)), let’s use the same steps as above: Let \(u = x\), \(dv = \sin\left(\frac{n\pi x}{2}\right) dx\) - \(du = dx\) - \(v = -\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\) So, \[ \int x \sin\left(\frac{n\pi x}{2}\right) dx = -\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi x}{2}\right) \] So, \[ \int_{1}^{2} x \sin\left(\frac{n\pi x}{2}\right) dx = \left[ -\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi x}{2}\right) \right]_{1}^{2} \] Calculate at \(x=2\) and \(x=1\): - \(x=2\): \(\cos(n\pi) = (-1)^n\), \(\sin(n\pi)=0\) - \(x=1\): \(\cos(\frac{n\pi}{2})\), \(\sin(\frac{n\pi}{2})\) So, \[ = \left( -\frac{4}{n\pi} \cos(n\pi) + 0 \right) - \left( -\frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \right) \] \[ = -\frac{4}{n\pi} (-1)^n + \frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) - \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] Now, plug back to \(I_2\): \[ I_2 = -\frac{4}{n\pi} \left[ \cos(n\pi) - \cos\left(\frac{n\pi}{2}\right) \right] - \left[ -\frac{4}{n\pi} (-1)^n + \frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) - \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \right] \] \[ = -\frac{4}{n\pi} \left[ (-1)^n - \cos\left(\frac{n\pi}{2}\right) \right] + \frac{4}{n\pi} (-1)^n - \frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] \[ = -\frac{4}{n\pi} (-1)^n + \frac{4}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{n\pi} (-1)^n - \frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] \[ = \left( -\frac{4}{n\pi} (-1)^n + \frac{4}{n\pi} (-1)^n \right) + \left( \frac{4}{n\pi} \cos\left(\frac{n\pi}{2}\right) - \frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) \right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] \[ = 0 + \frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] --- ### **C. Add \(I_1\) and \(I_2\):** \[ b_n = \left(-\frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right)\right) + \left(\frac{2}{n\pi} \cos\left(\frac{n\pi}{2}\right) + \frac{4}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right)\right) \] \[ b_n = 0 + \frac{8}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \] --- ## **Final Sine Series:** \[ \boxed{ f(x) = \sum_{n=1}^\infty \frac{8}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi x}{2}\right) } \] --- ## **Graph Sketch** - For \(0 \leq x \leq 1\), \(f(x) = x\) (a straight line from (0,0) to (1,1)) - For \(1 < x \leq 2\), \(f(x) = 2-x\) (a straight line from (1,1) to (2,0)) This is a symmetric "tent" function over \([0,2]\), peaking at \(x=1\).  --- ### **Summary Table** | Step | Expression | |------|-------------| | Sine Series | \( f(x) = \sum_{n=1}^\infty \frac{8}{(n\pi)^2} \sin\left(\frac{n\pi}{2}\right) \sin\left(\frac{n\pi x}{2}\right) \) | | Key Values | \( f(0) = 0, f(1) = 1, f(2) = 0 \) | | Shape | Triangle, peak at \(x=1, f(1)=1\) | --- **Let me know if you want to see the first few nonzero terms explicitly!**

Let's carefully derive the sine series expansion of the given function step-by-step, providing detailed calculations and explanations. --- ### **Problem Restatement** Given the piecewise function on the interval \(0 < x < 2\): \[ f(x) = \begin{cases} x, & 0 < x \leq 1 \\ 2 - x, & 1 < x < 2 \end{cases} \] We are asked to find the **sine series expansion** of \(f(x)\), assuming the Fourier sine series on \([0, 2]\). --- ### **Step 1: Fourier Sine Series on \([0, 2]\)** The Fourier sine series for \(f(x)\) over \([0, 2]\) is: \[ f(x) = \sum_{n=1}^\infty b_n \sin \left( \frac{n \pi x}{2} \right) \] where the coefficients \(b_n\) are: \[ b_n = \frac{2}{L} \int_0^L f(x) \sin \left( \frac{n \pi x}{L} \right) dx \] with \(L=2\). So, \[ b_n = \int_0^2 f(x) \sin \left( \frac{n \pi x}{2} \right) dx \] --- ### **Step 2: Split the integral** Since \(f(x)\) is piecewise: \[ b_n = \int_0^1 x \sin \left( \frac{n \pi x}{2} \right) dx + \int_1^2 (2 - x) \sin \left( \frac{n \pi x}{2} \right) dx \] Let’s call these: \[ b_n = I_1 + I_2 \] --- ### **Step 3: Compute \(I_1 = \int_0^1 x \sin \left( \frac{n \pi x}{2} \right) dx \)** **Method:** Integration by parts. - Set: \[ u = x \quad \Rightarrow \quad du = dx \] \[ dv = \sin \left( \frac{n \pi x}{2} \right) dx \] - Find \(v\): \[ v = -\frac{2}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \] **Applying integration by parts:** \[ I_1 = uv \big|_0^1 - \int_0^1 v du \] \[ I_1 = -\frac{2x}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \big|_0^1 + \frac{2}{n \pi} \int_0^1 \cos \left( \frac{n \pi x}{2} \right) dx \] Calculate boundary term at \(x=1\): \[ -\frac{2 \times 1}{n \pi} \cos \left( \frac{n \pi}{2} \right) = - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) \] At \(x=0\): \[ -\frac{2 \times 0}{n \pi} \cos 0 = 0 \] Now, compute the integral: \[ \int_0^1 \cos \left( \frac{n \pi x}{2} \right) dx \] - The antiderivative: \[ \frac{2}{n \pi} \sin \left( \frac{n \pi x}{2} \right) \] - Evaluate from 0 to 1: \[ \frac{2}{n \pi} \left[ \sin \left( \frac{n \pi}{2} \right) - 0 \right] = \frac{2}{n \pi} \sin \left( \frac{n \pi}{2} \right) \] **Putting it all together:** \[ I_1 = - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{2}{n \pi} \times \frac{2}{n \pi} \sin \left( \frac{n \pi}{2} \right) \] \[ I_1 = - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 4: Compute \(I_2 = \int_1^2 (2 - x) \sin \left( \frac{n \pi x}{2} \right) dx \)** Break into two parts: \[ I_2 = 2 \int_1^2 \sin \left( \frac{n \pi x}{2} \right) dx - \int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx \] --- ### **Part 4.1: \( 2 \int_1^2 \sin \left( \frac{n \pi x}{2} \right) dx \)** - Antiderivative: \[ - \frac{2}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \] - Evaluate from 1 to 2: \[ - \frac{2}{n \pi} \left[ \cos(n \pi) - \cos \left( \frac{n \pi}{2} \right) \right] \] --- ### **Part 4.2: \( \int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx \)** Similar to \(I_1\), use integration by parts: - \(u = x\), \(du = dx\) - \(dv = \sin \left( \frac{n \pi x}{2} \right) dx\), \(v = - \frac{2}{n \pi} \cos \left( \frac{n \pi x}{2} \right)\) - So, \[ \int x \sin \left( \frac{n \pi x}{2} \right) dx = - \frac{2x}{n \pi} \cos \left( \frac{n \pi x}{2} \right) + \frac{2}{n \pi} \int \cos \left( \frac{n \pi x}{2} \right) dx \] - The integral: \[ \int \cos \left( \frac{n \pi x}{2} \right) dx = \frac{2}{n \pi} \sin \left( \frac{n \pi x}{2} \right) \] - Evaluated between 1 and 2: \[ - \frac{2x}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \bigg|_1^2 + \frac{2}{n \pi} \times \frac{2}{n \pi} \left[ \sin \left( \frac{n \pi x}{2} \right) \right]_1^2 \] At \(x=2\): \[ - \frac{4}{n \pi} \cos(n \pi) \quad \text{(since } \cos(n \pi) = (-1)^n) \] At \(x=1\): \[ - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) \] Similarly, the sine term: \[ \frac{4}{(n \pi)^2} \left[ \sin(n \pi) - \sin \left( \frac{n \pi}{2} \right) \right] \] But \(\sin(n \pi) = 0\), so this simplifies to: \[ - \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] **Putting it all together:** \[ \int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx = - \frac{4}{n \pi} \cos(n \pi) + \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) - \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 5: Sum \(I_2\)** Recall: \[ I_2 = 2 \left[ - \frac{2}{n \pi} \left( \cos(n \pi) - \cos \left( \frac{n \pi}{2} \right) \right) \right] - \left[ - \frac{4}{n \pi} \cos(n \pi) + \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) - \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \right] \] Simplify each: \[ I_2 = - \frac{4}{n \pi} \left( \cos(n \pi) - \cos \left( \frac{n \pi}{2} \right) \right) + \frac{4}{n \pi} \cos(n \pi) - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] Distribute: \[ I_2 = - \frac{4}{n \pi} \cos(n \pi) + \frac{4}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{n \pi} \cos(n \pi) - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] Note that \(- \frac{4}{n \pi} \cos(n \pi) + \frac{4}{n \pi} \cos(n \pi) = 0\). These cancel each other. Remaining terms: \[ I_2 = \frac{4}{n \pi} \cos \left( \frac{n \pi}{2} \right) - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] \[ I_2 = \left( \frac{4}{n \pi} - \frac{2}{n \pi} \right) \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] \[ I_2 = \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 6: Total coefficient \(b_n\)** Recall: \[ b_n = I_1 + I_2 \] Insert the expressions: \[ b_n = \left( - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \right) + \left( \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \right) \] The \(\cos\) terms cancel: \[ - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) = 0 \] Sum the \(\sin\) terms: \[ \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) = \frac{8}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ## **

Let's carefully derive the Fourier sine series expansion of the given piecewise function, providing a complete step-by-step solution with detailed calculations. --- ### **Problem Restatement** Given the function: \[ f(x) = \begin{cases} x, & 0 < x \leq 1 \\ 2 - x, & 1 < x < 2 \end{cases} \] defined on the interval \([0,2]\), find its **Fourier sine series expansion** over \([0,2]\). The series has the form: \[ f(x) = \sum_{n=1}^\infty b_n \sin \left( \frac{n \pi x}{2} \right) \] where the coefficients \(b_n\) are given by: \[ b_n = \int_0^2 f(x) \sin \left( \frac{n \pi x}{2} \right) dx \] --- ### **Step 1: Computing the Fourier sine coefficients \(b_n\)** Since \(f(x)\) is piecewise, split the integral: \[ b_n = \int_0^1 x \sin \left( \frac{n \pi x}{2} \right) dx + \int_1^2 (2 - x) \sin \left( \frac{n \pi x}{2} \right) dx \] Denote these as: \[ b_n = I_1 + I_2 \] --- ### **Step 2: Compute \(I_1 = \int_0^1 x \sin \left( \frac{n \pi x}{2} \right) dx\)** Use **integration by parts**: - Let \(u = x\), \(du = dx\) - Let \(dv = \sin \left( \frac{n \pi x}{2} \right) dx\) Calculate \(v\): \[ v = - \frac{2}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \] Applying integration by parts: \[ I_1 = uv \big|_0^1 - \int_0^1 v du = - \frac{2x}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \big|_0^1 + \frac{2}{n \pi} \int_0^1 \cos \left( \frac{n \pi x}{2} \right) dx \] Evaluate the boundary term at \(x=1\): \[ - \frac{2}{n \pi} \times 1 \times \cos \left( \frac{n \pi}{2} \right) = - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) \] At \(x=0\): \[ 0 \] Now, compute the integral: \[ \int_0^1 \cos \left( \frac{n \pi x}{2} \right) dx = \frac{2}{n \pi} \sin \left( \frac{n \pi x}{2} \right) \big|_0^1 = \frac{2}{n \pi} \left( \sin \left( \frac{n \pi}{2} \right) - 0 \right) = \frac{2}{n \pi} \sin \left( \frac{n \pi}{2} \right) \] Putting it all together: \[ I_1 = - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{2}{n \pi} \times \frac{2}{n \pi} \sin \left( \frac{n \pi}{2} \right) = - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 3: Compute \(I_2 = \int_1^2 (2 - x) \sin \left( \frac{n \pi x}{2} \right) dx\)** Break into two parts: \[ I_2 = 2 \int_1^2 \sin \left( \frac{n \pi x}{2} \right) dx - \int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx \] --- ### **Part 3.1: Compute \(2 \int_1^2 \sin \left( \frac{n \pi x}{2} \right) dx\)** Antiderivative: \[ - \frac{2}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \] Evaluate from \(x=1\) to \(x=2\): \[ - \frac{2}{n \pi} \left[ \cos(n \pi) - \cos \left( \frac{n \pi}{2} \right) \right] \] --- ### **Part 3.2: Compute \(\int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx\)** Again, use **integration by parts**: - \(u = x\), \(du = dx\) - \(dv = \sin \left( \frac{n \pi x}{2} \right) dx\), \(v = - \frac{2}{n \pi} \cos \left( \frac{n \pi x}{2} \right)\) Applying parts: \[ \int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx = - \frac{2x}{n \pi} \cos \left( \frac{n \pi x}{2} \right) \big|_1^2 + \frac{2}{n \pi} \int_1^2 \cos \left( \frac{n \pi x}{2} \right) dx \] Calculate boundary term: At \(x=2\): \[ - \frac{4}{n \pi} \cos(n \pi) \] At \(x=1\): \[ - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) \] Integral: \[ \int_1^2 \cos \left( \frac{n \pi x}{2} \right) dx = \frac{2}{n \pi} \left[ \sin \left( \frac{n \pi x}{2} \right) \right]_1^2 = \frac{2}{n \pi} \left( \sin(n \pi) - \sin \left( \frac{n \pi}{2} \right) \right) \] But \(\sin(n \pi) = 0\), so: \[ \int_1^2 \cos \left( \frac{n \pi x}{2} \right) dx = - \frac{2}{n \pi} \sin \left( \frac{n \pi}{2} \right) \] Putting all together: \[ \int_1^2 x \sin \left( \frac{n \pi x}{2} \right) dx = - \frac{4}{n \pi} \cos(n \pi) + \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) - \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 4: Sum \(I_2\)** Recall: \[ I_2 = - \frac{2}{n \pi} \left[ \cos(n \pi) - \cos \left( \frac{n \pi}{2} \right) \right] - \left[ - \frac{4}{n \pi} \cos(n \pi) + \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) - \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \right] \] Simplify: \[ I_2 = - \frac{2}{n \pi} \cos(n \pi) + \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{n \pi} \cos(n \pi) - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] Notice that: \[ - \frac{2}{n \pi} \cos(n \pi) + \frac{4}{n \pi} \cos(n \pi) = \frac{2}{n \pi} \cos(n \pi) \] and \[ \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) = 0 \] Thus, \[ I_2 = \frac{2}{n \pi} \cos(n \pi) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 5: Total coefficient \(b_n\)** Now, summing \(I_1\) and \(I_2\): \[ b_n = I_1 + I_2 = \left( - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \right) + \left( \frac{2}{n \pi} \cos(n \pi) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \right) \] Combine like terms: - The \(\cos\) terms: \[ - \frac{2}{n \pi} \cos \left( \frac{n \pi}{2} \right) + \frac{2}{n \pi} \cos(n \pi) \] - The \(\sin\) terms: \[ \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) + \frac{4}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) = \frac{8}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] So, \[ b_n = \frac{2}{n \pi} \left( \cos(n \pi) - \cos \left( \frac{n \pi}{2} \right) \right) + \frac{8}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) \] --- ### **Step 6: Simplify the coefficients** Recall the values: \[ \cos(n \pi) = (-1)^n \] and \[ \cos \left( \frac{n \pi}{2} \right) \] has known values depending on \(n \bmod 4\): | \(n \bmod 4\) | \(\cos \left( \frac{n \pi}{2} \right) \) | |--------------|-------------------------------------| | 1 | 0 | | 2 | -1 | | 3 | 0 | | 0 | 1 | Similarly, \(\sin \left( \frac{n \pi}{2} \right)\): | \(n \bmod 4\) | \(\sin \left( \frac{n \pi}{2} \right) \) | |--------------|-------------------------------------| | 1 | 1 | | 2 | 0 | | 3 | -1 | | 0 | 0 | But for a general expression, it's best to leave the coefficients as: \[ \boxed{ b_n = \frac{2}{n \pi} \left( (-1)^n - \cos \left( \frac{n \pi}{2} \right) \right) + \frac{8}{(n \pi)^2} \sin \left( \frac{n \pi}{2} \right) } \] --- ### **### **Final Fourier sine series expansion:** \[ \boxed{ f(x) = \sum_{n=1}^\infty \left[ \frac{2}{n \pi} \left( (-1)^n - \cos \left( \frac{n \pi}{2} \right) \right) + \frac{8}{(n \pi)^2}