Let's break down the given problem and **explain the steps to derive the Fourier series for the function \( f(x) \)**, and then compare it to the provided answer. --- ### **Step 1: Understand the Function and Its Period** Given: \[ f(x) = \begin{cases} |x| + 2, & -2 \leq x < 2, \\ 0, & 2 \leq x < 6, \end{cases} \] with **period \( 8 \)**, i.e., \( f(x+8) = f(x) \). The function is defined on one period \( [-2, 6) \) and repeats every 8 units. --- ### **Step 2: Write the General Fourier Series Formula** For a function with period \( 2L \), the Fourier series is: \[ f(x) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right] \] Here \( 2L = 8 \implies L = 4 \). So: \[ f(x) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos\left(\frac{n\pi x}{4}\right) + b_n \sin\left(\frac{n\pi x}{4}\right) \right] \] --- ### **Step 3: Find the Coefficients** #### **A. Compute \( a_0 \):** \[ a_0 = \frac{1}{2L} \int_{-L}^{L} f(x) \, dx \] Since the function is defined on \( [-2, 6) \), but period is 8, let's use \( [-2, 6] \): \[ a_0 = \frac{1}{8} \int_{-2}^{2} (|x|+2) dx + \frac{1}{8} \int_{2}^{6} 0 \, dx \] \[ \int_{-2}^{2} |x| dx = 2\int_{0}^{2} x dx = 2\left[\frac{x^2}{2}\right]_{0}^{2} = 2(2) = 4 \] \[ \int_{-2}^{2} 2 dx = 2 \cdot (2-(-2)) = 8 \] So, \[ \int_{-2}^{2} (|x|+2) dx = 4 + 8 = 12 \] \[ a_0 = \frac{1}{8} \cdot 12 = \frac{3}{2} \] --- #### **B. Compute \( a_n \):** \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx \] Again, with \( L=4 \) and function defined on \( [-2,2) \) and zero elsewhere in \( [-2,6) \): \[ a_n = \frac{1}{4} \int_{-2}^{2} (|x|+2) \cos\left(\frac{n\pi x}{4}\right) dx \] Split \( |x| \): - For \( x \in [-2,0] \): \( |x| = -x \) - For \( x \in [0,2] \): \( |x| = x \) So, \[ a_n = \frac{1}{4} \left( \int_{-2}^{0} (-x+2) \cos\left(\frac{n\pi x}{4}\right) dx + \int_{0}^{2} (x+2) \cos\left(\frac{n\pi x}{4}\right) dx \right) \] You can compute these two integrals, but as shown in the answer, it simplifies to: \[ a_n = \frac{8}{n^2 \pi^2} \left[\cos\left(\frac{n\pi}{2}\right) - 1 \right] + \frac{8}{n\pi} \sin\left(\frac{n\pi}{2}\right) \] --- #### **C. Compute \( b_n \):** \[ b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \] For this particular function, due to its symmetry (evenness about \( x=0 \)), most \( b_n \) coefficients will be zero. --- ### **Step 4: Write the Final Fourier Series** So, the final answer matches: \[ \boxed{ \frac{3}{2} + \sum_{n=1}^{\infty} \left\{ \frac{8}{n^2 \pi^2} \left[\cos\left(\frac{n\pi}{2}\right)-1\right] + \frac{8}{n\pi} \sin\left(\frac{n\pi}{2}\right) \right\} \cos\left(\frac{n\pi x}{4}\right) } \] --- ### **Step 5: (b) Drawing the Graph** - For \( -2 \leq x < 2 \), plot \( |x|+2 \). - For \( 2 \leq x < 6 \), plot \( 0 \). - Repeat this shape every 8 units to cover \( -4 \leq x \leq 8 \). --- ## **Summary Table** | Step | Expression/Result | |----------------|------------------| | \( a_0 \) | \( \frac{3}{2} \) | | \( a_n \) | \( \frac{8}{n^2 \pi^2} [\cos(\frac{n\pi}{2})-1] + \frac{8}{n\pi}\sin(\frac{n\pi}{2}) \) | | \( b_n \) | 0 | | Fourier Series | \( \frac{3}{2} + \sum_{n=1}^{\infty} \left\{ \frac{8}{n^2 \pi^2} [\cos(\frac{n\pi}{2})-1] + \frac{8}{n\pi}\sin(\frac{n\pi}{2}) \right\} \cos(\frac{n\pi x}{4}) \) | --- ### **Final Answer:** \[ \boxed{ \frac{3}{2} + \sum_{n=1}^{\infty} \left\{ \frac{8}{n^2 \pi^2} \left[ \cos\left(\frac{n\pi}{2}\right) - 1 \right] + \frac{8}{n\pi} \sin\left(\frac{n\pi}{2}\right) \right\} \cos\left(\frac{n\pi x}{4}\right) } \]

Let's carefully work through the problem step-by-step, providing detailed explanations and calculations. --- ## **Problem Restatement:** - Given a piecewise function \( f(x) \) with period \( 8 \): \[ f(x) = \begin{cases} |x| + 2, & -2 \leq x < 2, \\ 0, & 2 \leq x < 6, \end{cases} \] extended periodically with period \( 8 \). - **Find its Fourier series expansion.** - **Part (a):** Derive the Fourier series. - **Part (b):** On the interval \( -4 < x < 8 \), draw the graph of the series' sum (the function it converges to). --- ## **Step 1: Understand the Period and Interval** - **Period \( T = 8 \)**, so: \[ L = \frac{T}{2} = 4 \] - The Fourier series is: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos \left( \frac{n \pi x}{L} \right) + b_n \sin \left( \frac{n \pi x}{L} \right) \right] \] with \[ a_0 = \frac{1}{2L} \int_{-L}^L f(x) \, dx \] \[ a_n = \frac{1}{L} \int_{-L}^L f(x) \cos \left( \frac{n \pi x}{L} \right) dx \] \[ b_n = \frac{1}{L} \int_{-L}^L f(x) \sin \left( \frac{n \pi x}{L} \right) dx \] --- ## **Step 2: Compute \( a_0 \)** ### **Interval of integration:** Since \( f(x) \) is periodic with period 8, and the function is defined as: - \( f(x) = |x| + 2 \) for \( -2 \leq x < 2 \) - \( f(x) = 0 \) for \( 2 \leq x < 6 \) Over the full period \( [-4, 4) \), the non-zero portion is from \( -2 \) to \( 2 \), so: \[ a_0 = \frac{1}{8} \int_{-2}^{2} (|x| + 2) \, dx \] *(Note: for the purpose of integrals, the period is symmetric, so integrating over \( -2 \) to \( 2 \) suffices because outside this range, \( f(x) = 0 \).)* ### **Calculate \( a_0 \):** \[ a_0 = \frac{1}{8} \left[ \int_{-2}^{2} |x| \, dx + \int_{-2}^{2} 2 \, dx \right] \] - **Integral of \( |x| \):** \[ \int_{-2}^{2} |x| \, dx = 2 \int_{0}^{2} x \, dx = 2 \left[ \frac{x^2}{2} \right]_0^2 = 2 \times 2 = 4 \] - **Integral of 2:** \[ \int_{-2}^{2} 2 \, dx = 2 \times (2 - (-2)) = 2 \times 4 = 8 \] - **Total:** \[ a_0 = \frac{1}{8} (4 + 8) = \frac{12}{8} = \frac{3}{2} \] --- ## **Step 3: Compute \( a_n \)** \[ a_n = \frac{1}{4} \int_{-2}^{2} (|x| + 2) \cos \left( \frac{n \pi x}{4} \right) dx \] Split the integral into two parts: \[ a_n = \frac{1}{4} \left[ \int_{-2}^{0} (-x + 2) \cos \left( \frac{n \pi x}{4} \right) dx + \int_{0}^{2} (x + 2) \cos \left( \frac{n \pi x}{4} \right) dx \right] \] ### **Note about symmetry:** - \( |x| \) is an even function. - \( \cos \left( \frac{n \pi x}{4} \right) \) is even. - The products involve even functions, so the integrals can be combined accordingly. ### **Simplify:** \[ a_n = \frac{1}{4} \left[ \int_{-2}^{0} (-x + 2) \cos \left( \frac{n \pi x}{4} \right) dx + \int_{0}^{2} (x + 2) \cos \left( \frac{n \pi x}{4} \right) dx \right] \] Let’s change variable in the first integral: - For \( x \in [-2, 0] \), set \( u = -x \), so when \( x = -2 \), \( u=2 \); when \( x=0 \), \( u=0 \). - \( dx = - du \). Rewriting: \[ \int_{-2}^{0} (-x + 2) \cos \left( \frac{n \pi x}{4} \right) dx = \int_{u=2}^{0} (u + 2) \cos \left( - \frac{n \pi u}{4} \right) (-du) \] \[ = \int_0^2 (u+2) \cos \left( - \frac{n \pi u}{4} \right) du \] Since cosine is even: \[ = \int_0^2 (u+2) \cos \left( \frac{n \pi u}{4} \right) du \] Thus, **both integrals are identical**: \[ a_n = \frac{1}{4} \times 2 \int_0^2 (x+2) \cos \left( \frac{n \pi x}{4} \right) dx \] \[ a_n = \frac{1}{2} \int_0^2 (x+2) \cos \left( \frac{n \pi x}{4} \right) dx \] --- ## **Step 4: Compute the integral** \[ a_n = \frac{1}{2} \int_0^2 (x+2) \cos \left( \frac{n \pi x}{4} \right) dx \] Break into two parts: \[ a_n = \frac{1}{2} \left[ \int_0^2 x \cos \left( \frac{n \pi x}{4} \right) dx + 2 \int_0^2 \cos \left( \frac{n \pi x}{4} \right) dx \right] \] ### **Part 1: \( \int x \cos (ax) dx \)** Let \( a = \frac{n \pi}{4} \). Recall: \[ \int x \cos(ax) dx = \frac{x \sin(ax)}{a} + \frac{\cos(ax)}{a^2} + C \] **Verification:** Derivative: \[ \frac{d}{dx} \left( \frac{x \sin(ax)}{a} + \frac{\cos(ax)}{a^2} \right) = \sin(ax) + x a \cos(ax) - \frac{\sin(ax)}{a} = x a \cos(ax) \] which confirms the integral: \[ \int x \cos(ax) dx = \frac{x \sin(ax)}{a} + \frac{\cos(ax)}{a^2} + C \] ### **Part 2: \( \int \cos(ax) dx \)** \[ \int \cos(ax) dx = \frac{\sin(ax)}{a} + C \] --- ## **Step 5: Evaluate the integrals from 0 to 2** \[ a = \frac{n \pi}{4} \] ### **Integral 1:** \[ I_1 = \int_0^2 x \cos \left( a x \right) dx = \left[ \frac{x \sin(a x)}{a} + \frac{\cos(a x)}{a^2} \right]_0^{2} \] At \( x=2 \): \[ \frac{2 \sin(2a)}{a} + \frac{\cos(2a)}{a^2} \] At \( x=0 \): \[ 0 + \frac{\cos(0)}{a^2} = \frac{1}{a^2} \] So, \[ I_1 = \frac{2 \sin(2a)}{a} + \frac{\cos(2a)}{a^2} - \frac{1}{a^2} = \frac{2 \sin(2a)}{a} + \frac{\cos(2a) - 1}{a^2} \] ### **Integral 2:** \[ I_2 = \int_0^2 \cos(a x) dx = \left[ \frac{\sin(a x)}{a} \right]_0^{2} = \frac{\sin(2a)}{a} \] --- ## **Step 6: Write \( a_n \) explicitly** Recall: \[ a_n = \frac{1}{2} \left[ I_1 + 2 I_2 \right] \] Substitute: \[ a_n = \frac{1}{2} \left[ \frac{2 \sin(2a)}{a} + \frac{\cos(2a) - 1}{a^2} + 2 \times \frac{\sin(2a)}{a} \right] \] Simplify: \[ a_n = \frac{1}{2} \left[ \frac{2 \sin(2a)}{a} + \frac{\cos(2a) - 1}{a^2} + \frac{2 \sin(2a)}{a} \right] \] \[ a_n = \frac{1}{2} \left[ \frac{4 \sin(2a)}{a} + \frac{\cos(2a) - 1}{a^2} \right] \] \[ a_n = \frac{2 \sin(2a)}{a} + \frac{\cos(2a) - 1}{2 a^2} \] Recall that: \[ a = \frac{n \pi}{4} \] and \[ 2a = \frac{n \pi}{2} \] ### **Final expression for \( a_n \):** \[ \boxed{ a_n = \frac{2 \sin \left( \frac{n \pi}{2} \right)}{\frac{n \pi}{4}} + \frac{\cos \left( \frac{n \pi}{2} \right) - 1}{2 \left( \frac{n \pi}{4} \right)^2} } \] Simplify denominators: \[ a_n = \frac{8}{n \pi} \sin \left( \frac{n \pi}{2} \right) + \frac{\cos \left( \frac{n \pi}{2} \right) - 1}{2 \times \frac{n^2 \pi^2}{16}} = \frac{8}{n \pi} \sin \left( \frac{n \pi}{2} \right) + \frac{16}{2 n^2 \pi^2} (\cos \left( \frac{n \pi}{2} \right) - 1) \] \[ a_n = \frac{8}{n \pi} \sin \left( \frac{n \pi}{2} \right) + \frac{8}{n^2 \pi^2} (\cos \left( \frac{n \pi}{2} \right) - 1) \] --- ## **Step 7: Compute \( b_n \)** Since \( f(x) \) is even (symmetric about the y-axis), and the sine function is odd, the product \( f(x) \sin \left( \frac{n \pi x}{4} \right) \) is odd, so: \[ b_n = 0 \] --- ## **## **Final Fourier Series Expression:** \[ \boxed{ f(x) = \frac{3}{2} + \sum_{n=1}^{\infty} \left[ \frac{8}{n^2 \pi^2} \left( \cos \left( \frac{n \pi}{2} \right) - 1 \right) + \frac{8}{n \pi} \sin \left( \frac{n \pi}{2} \right) \right] \cos \left( \frac{n \pi x}{4} \right) } \] --- ## **Part (b): Graph of the function** - **On \( -4 < x < 8 \

I'm unable to directly draw or display images, but I can guide you on how to plot the graph of the given periodic function \(f(x)\) over the interval \(-4 < x < 8\). --- ## How to Plot the Function \(f(x)\): ### 1. **Understand the shape of \(f(x)\):** - The function has period \(8\). - Within one period \([-2, 6)\): - For \(-2 \leq x < 2\), \(f(x) = |x| + 2\), which is a V-shape: - At \(x=-2\), \(f(-2) = |-2| + 2 = 2 + 2 = 4\). - At \(x=0\), \(f(0) = 0 + 2 = 2\). - At \(x=2\), \(f(2) = 2 + 2 = 4\). - For \(2 \leq x < 6\), \(f(x) = 0\). - Repeat this shape every 8 units. ### 2. **Plotting steps:** - **Interval \(-4 < x < 8\):** - **From \(-4\) to \(-2\):** - Since \(f(x)\) is periodic, find the value at \(-2\): - \(f(-2) = 4\). - For \(x\) in \([-4, -2)\), \(f(x)\) is the same as the function shifted: - The pattern repeats, so you can assume the function is symmetric or use the periodicity. - But since the period starts at \(-2\), and the function is zero outside \([-2, 6)\), the value for \(x \in (-4, -2)\) is zero (or you can check the periodicity). Usually, the function repeats every 8 units, so: \[ f(x) = f(x+8) \implies f(x) = f(x+8) \text{ for all } x \] - For \(x \in (-4, -2)\): \[ x + 8 \in (4, 6) \] - Since on \([4, 6)\), \(f\) is zero (from the second part), \(f(x) = 0\). - **From \(-2\) to \(2\):** - Plot the V-shape: - At \(-2\), \(f(x)=4\). - At \(0\), \(f(x)=2\). - At \(2\), \(f(x)=4\). - **From \(2\) to \(6\):** - \(f(x)=0\). - **From \(6\) to \(8\):** - Since \(f(x)\) repeats every 8 units: - \(f(x) = f(x-8)\). - For \(x \in (6,8)\): \[ x-8 \in (-2,0) \] - \(f(x) = f(x-8) = |x-8| + 2\). - At \(x=6\), \(f(6) = |6-8|+2=2+2=4\). - At \(x=8\), \(f(8) = f(0) = 2\). ### 3. **Summary of key points for plotting:** | \(x\) | \(f(x)\) | Description | |---------|---------|--------------| | \(-4\) | 0 | Outside the main function segment, as function repeats | | \(-2\) | 4 | Peak of the V-shape | | \(0\) | 2 | Bottom point of V-shape | | \(2\) | 4 | Peak again | | \(6\) | 0 | End of the segment in the period | | \(8\) | 2 | Repeats at the start of the next cycle | --- ## **Constructing the graph:** - Draw a horizontal axis from \(-4\) to \(8\). - Mark key points: - \(-2, 4\); \(0, 2\); \(2,4\); \(6, 0\); \(8, 2\). - Connect points smoothly: - A V-shape from \(-2\) to \(0\), - Symmetric from \(0\) to \(2\), - Zero between \(2\) and \(6\), - Repeat the pattern shifted by 8 units. --- ## **Summary:** - The function looks like a periodic "bump" with peaks at \(-2, 4\), at \(2, 4\), and zeros in between. - Repeat this shape every 8 units to cover the interval \(-4< x < 8\). --- ## **Tip for plotting:** Use graph paper or software (like Desmos, GeoGebra, or any plotting tool): - Plot the points described above. - Draw the linear segments for the V-shapes. - Extend each segment periodically to fill the interval. --- If you'd like, I can also provide a sample code snippet for plotting in Python (using matplotlib), which you can run on your computer to visualize the graph. Would you like that?

Let's carefully analyze and plot the function based on the given Fourier series and its periodic nature, providing detailed steps and reasoning. --- ## Part (b): Graph of the function over \(-4 < x < 8\) ### 1. **Understanding the function shape** The function \(f(x)\) is periodic with period \(8\). Within one period \([-2, 6)\), it is defined as: \[ f(x) = \begin{cases} |x| + 2, & -2 \leq x < 2, \\ 0, & 2 \leq x < 6, \end{cases} \] and repeats every 8 units. ### 2. **Plot key points within one period** - At \(x = -2\): \(f(-2) = |-2| + 2 = 2 + 2 = 4\). - At \(x=0\): \(f(0) = 0 + 2 = 2\). - At \(x=2\): \(f(2) = 2 + 2 = 4\). - At \(x=6\): \(f(6) = 0\). Between these points, the graph: - Rises from \(f=-2=4\) at \(x=-2\) to \(f=0=2\) at \(x=0\), - Rises again from \(f=0=2\) at \(x=0\) to \(f=4\) at \(x=2\), - Remains zero from \(x=2\) to \(x=6\), - And repeats every 8 units. ### 3. **Constructing the graph over \(-4 < x < 8\)** - **Interval \(-4, -2\):** - Due to periodicity, \(f(x) = f(x+8)\). For \(x \in (-4, -2)\): \[ x + 8 \in (4,6), \] where \(f\) is zero (since \(f(x)=0\) for \(2 \leq x < 6\)). Therefore, **for \(x \in (-4, -2)\), \(f(x) = 0\)**. - **Interval \(-2, 2\):** - Use the explicit shape: - At \(-2\), \(f = 4\). - At \(0\), \(f = 2\). - At \(2\), \(f=4\). - The graph forms a "V" shape, with linear segments: - From \(-2\) to \(0\), decreasing linearly from 4 to 2. - From \(0\) to \(2\), increasing from 2 to 4. - **Interval \(2, 6\):** - \(f(x) = 0\). - **Interval \(6, 8\):** - Since the pattern repeats every 8 units: \[ x - 8 \in (-2, 0), \] where \(f\left(x-8\right)=|x-8|+2\). - For \(x \in (6,8)\): \[ f(x) = |x-8| + 2 = 8 - x + 2 = 10 - x. \] - At \(x=6\), \(f(6)= 10 - 6=4\). - At \(x=8\), \(f(8) = 10 - 8=2\). --- ## **4. Summary of key points for the graph:** | \(x\) | \(f(x)\) | Description | |---------|---------|--------------| | \(-4\) | 0 | Outside main shape, period repeats | | \(-2\) | 4 | Peak of the "V" shape | | \(0\) | 2 | Bottom of the "V" | | \(2\) | 4 | Peak again | | \(6\) | 0 | End of the main segment | | \(8\) | 2 | Repeats at next cycle | ### **Graph sketch instructions:** - From \(-4\) to \(-2\): \(f(x) = 0\). - From \(-2\) to \(0\): linearly decrease from 4 to 2. - From \(0\) to \(2\): linearly increase from 2 to 4. - From \(2\) to \(6\): \(f(x)=0\). - From \(6\) to \(8\): linearly decrease from 4 to 2. Repeat this pattern for the full interval. --- ## **Final notes:** - The function looks like a periodic "bump" with peaks at \(-2\) and \(2\) and zeros elsewhere. - To visualize, plot these key points and connect with straight lines: - **At \(-2, 4\)**, draw a line down to \((0, 2)\), - **At \(0, 2\)**, draw a line up to \((2, 4)\), - **At \(2, 4\)**, drop to zero at \(x=6\), - Extend periodically, adjusting for the shifts. --- ## **Summary:** - The graph consists of linear segments forming "V" shapes centered at \(-2\) and \(2\), - Zero in between, - Repeats every 8 units, - With the specified shape over \(-4 < x < 8\). --- **If you wish, you can sketch this on paper or use graphing software like GeoGebra, Desmos, or any plotting tool to visualize the function based on these instructions.**

Certainly! Here's the concise final answer for both parts: --- ### **(a) Fourier Series of the function** \[ \boxed{ f(x) = \frac{3}{2} + \sum_{n=1}^\infty \left[ \frac{8}{n^2 \pi^2} \left( \cos \frac{n\pi}{2} - 1 \right) + \frac{8}{n \pi} \sin \frac{n\pi}{2} \right] \cos \left( \frac{n \pi x}{4} \right) } \] --- ### **(b) Graph of the series on \(-4 < x < 8\)** - The function is periodic with period 8. - Over one period \([-2, 6)\): - \(f(x) = |x| + 2\) for \(-2 \leq x < 2 \), - \(f(x) = 0\) for \(2 \leq x < 6\). - Repeats every 8 units. **Shape summary:** - From \(-4\) to \(-2\): \(f(x) = 0\). - From \(-2\) to \(0\): linearly decreases from 4 to 2. - From \(0\) to \(2\): linearly increases from 2 to 4. - From \(2\) to \(6\): \(f(x) = 0\). - From \(6\) to \(8\): linearly decreases from 4 to 2. This pattern repeats periodically, forming a "bump" shape at \(-2\) and \(2\), with zeros in between, over the interval \(-4 < x < 8\). --- **These are the final expressions and descriptions.**