Let's tackle **problem 3(a)** in two steps, as requested. --- ### **Step 1: Compute the full Fourier series for \( f(x) = e^x \) on \( -1 < x < 1 \)** The **Fourier series** for a function \( f(x) \) on \( -L < x < L \) is: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right] \] with: \[ a_0 = \frac{1}{L}\int_{-L}^{L} f(x)dx \] \[ a_n = \frac{1}{L}\int_{-L}^L f(x) \cos\left(\frac{n\pi x}{L}\right)dx \] \[ b_n = \frac{1}{L}\int_{-L}^L f(x) \sin\left(\frac{n\pi x}{L}\right)dx \] Here, \( L = 1 \), so the formulas become: \[ a_0 = \int_{-1}^{1} e^x dx = [e^x]_{-1}^{1} = e^1 - e^{-1} \] \[ a_n = \int_{-1}^{1} e^x \cos(n\pi x) dx \] \[ b_n = \int_{-1}^{1} e^x \sin(n\pi x) dx \] #### Calculation of \( a_n \) and \( b_n \): - Use integration by parts or the known result: \[ \int e^{ax}\cos(bx)dx = \frac{e^{ax}}{a^2 + b^2}(a\cos(bx) + b\sin(bx)) \] \[ \int e^{ax}\sin(bx)dx = \frac{e^{ax}}{a^2 + b^2}(a\sin(bx) - b\cos(bx)) \] For our case, \( a=1 \), \( b = n\pi \): So, \[ a_n = \left. \frac{e^x}{1 + (n\pi)^2} \left( \cos(n\pi x) + n\pi \sin(n\pi x) \right) \right|_{-1}^{1} \] \[ b_n = \left. \frac{e^x}{1 + (n\pi)^2} \left( \sin(n\pi x) - n\pi \cos(n\pi x) \right) \right|_{-1}^{1} \] **Explanation:** Here, we've set up all the formulas for the Fourier series and calculated the explicit integrals for the coefficients \( a_0, a_n, b_n \). The next step is to write out the truncated series and discuss the graph. --- ### **Step 2: Write the truncated series (n = 0,1,2,3) and explain the graph** - **\( n=0 \) term:** \[ \frac{a_0}{2} = \frac{e - e^{-1}}{2} \] - **\( n=1 \) term:** \[ a_1 = \left. \frac{e^x}{1 + \pi^2} \left( \cos(\pi x) + \pi \sin(\pi x) \right) \right|_{-1}^{1} \] \[ = \frac{1}{1+\pi^2} \left[ e^1(\cos \pi + \pi \sin \pi) - e^{-1}(\cos(-\pi) + \pi \sin(-\pi)) \right] \] But \(\cos(\pi) = -1\), \(\sin(\pi) = 0\), \(\cos(-\pi) = -1\), \(\sin(-\pi) = 0\): \[ a_1 = \frac{1}{1+\pi^2} [e^1(-1) - e^{-1}(-1)] = \frac{-e + e^{-1}}{1+\pi^2} \] - **Similarly, calculate \( b_1 \), and so on for \( n=2,3 \) using the same approach.** **Truncated series (up to \( n=3 \))**: \[ f(x) \approx \frac{a_0}{2} + a_1 \cos(\pi x) + b_1 \sin(\pi x) + a_2 \cos(2\pi x) + b_2 \sin(2\pi x) + a_3 \cos(3\pi x) + b_3 \sin(3\pi x) \] **Graph explanation:** - The truncated Fourier series will closely match \( f(x) = e^x \) on \( -1 < x < 1 \). - Outside this interval (\( |x| > 1 \)), the Fourier series will repeat its periodic extension, so you will see discontinuities at \( x = \pm1, \pm3, \ldots \). - You may observe the Gibbs phenomenon at these discontinuities. **Final Answer:** The full Fourier series for \( f(x) = e^x \) on \( -1 < x < 1 \) is: \[ f(x) = \frac{e-e^{-1}}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos(n\pi x) + b_n \sin(n\pi x) \right] \] where \[ a_n = \left. \frac{e^x}{1 + (n\pi)^2} \left( \cos(n\pi x) + n\pi \sin(n\pi x) \right) \right|_{-1}^{1} \] \[ b_n = \left. \frac{e^x}{1 + (n\pi)^2} \left( \sin(n\pi x) - n\pi \cos(n\pi x) \right) \right|_{-1}^{1} \] The truncated series (up to \( n=3 \)) approximates \( f(x) \) well on \( -1 < x < 1 \), and extends periodically outside this interval, so the graph will show the exponential on \( -1 < x < 1 \) and its periodic copies elsewhere. The y-axis scale should be large enough to accommodate \( e^1 \approx 2.718 \) and \( e^{-1} \approx 0.368 \). --- Let me know if you want the explicit values for \( n=2,3 \), or want to continue to (b)!

Let's proceed step-by-step for each part, providing explanations and reasoning. --- ### **Part (a): Compute the full Fourier series for \(f(z) = e^z\) on \(-1 < z < 1\)** **Step 1: Find Fourier coefficients** For \(f(z) = e^z\) on \(-1<z<1\), the Fourier series is: \[ f(z) = \frac{a_0}{2} + \sum_{n=1}^\infty [a_n \cos(n\pi z) + b_n \sin(n\pi z)] \] with: \[ a_0 = \int_{-1}^1 e^z dz = e^1 - e^{-1} \] \[ a_n = \int_{-1}^1 e^z \cos(n\pi z) dz \] \[ b_n = \int_{-1}^1 e^z \sin(n\pi z) dz \] **Explanation:** We need to evaluate these integrals to get the coefficients. **Step 2: Compute \(a_n\) and \(b_n\)** Using standard integral results: \[ \int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) \] Similarly, \[ \int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) \] Set \(a=1\), \(b = n\pi\), evaluate from \(-1\) to 1: \[ a_n = \left[\frac{e^z}{1 + (n\pi)^2} \left( \cos(n\pi z) + n \pi \sin(n\pi z) \right) \right]_{z=-1}^{z=1} \] \[ b_n = \left[\frac{e^z}{1 + (n\pi)^2} \left( \sin(n\pi z) - n \pi \cos(n\pi z) \right) \right]_{z=-1}^{z=1} \] **At \(z=1\):** \[ e^1 = e \] \[ \cos(n\pi) = (-1)^n, \quad \sin(n\pi) = 0 \] **At \(z=-1\):** \[ e^{-1} = e^{-1} \] \[ \cos(-n\pi) = \cos(n\pi) = (-1)^n, \quad \sin(-n\pi) = 0 \] Calculate \(a_n\): \[ a_n = \frac{e}{1+(n\pi)^2} [(-1)^n + n\pi \times 0] - \frac{e^{-1}}{1+(n\pi)^2} [(-1)^n + n\pi \times 0] = \frac{(-1)^n (e - e^{-1})}{1+(n\pi)^2} \] Similarly, \(b_n\): \[ b_n = \frac{e \times 0 - n \pi e (-1)^n}{1+(n\pi)^2} - \frac{e^{-1} \times 0 - n \pi e^{-1} (-1)^n}{1+(n\pi)^2} \] \[ = - \frac{n \pi (-1)^n (e - e^{-1})}{1+(n\pi)^2} \] **Summary:** \[ a_0 = e - e^{-1} \] \[ a_n = \frac{(-1)^n (e - e^{-1})}{1 + (n\pi)^2} \] \[ b_n = - \frac{n \pi (-1)^n (e - e^{-1})}{1 + (n\pi)^2} \] --- ### **Final Fourier series (full):** \[ \boxed{ f(z) = \frac{e - e^{-1}}{2} + \sum_{n=1}^\infty \left[ \frac{(-1)^n (e - e^{-1})}{1 + (n\pi)^2} \cos(n\pi z) - \frac{n \pi (-1)^n (e - e^{-1})}{1 + (n\pi)^2} \sin(n\pi z) \right] } \] --- ### **Part (b): Fourier cosine series for \(g(z) = \cosh z\) on \(0 < z < 1\)** **Why no calculation needed?** Because \(\cosh z\) is *even* on \([0,1]\), and the Fourier cosine series naturally represents even functions on \([0,L]\). **The Fourier cosine series:** \[ g(z) = a_0/2 + \sum_{n=1}^\infty a_n \cos(n\pi z) \] with \[ a_0 = \frac{2}{1} \int_0^1 \cosh z \, dz = 2 \sinh 1 \] \[ a_n = 2 \int_0^1 \cosh z \cos(n \pi z) dz \] **Note:** - Since \(\cosh z\) is even, the cosine series suffices. - No need to compute these integrals explicitly for the purpose of understanding the approximation. **Graph and explanation:** - The partial sum with \(n=0,1,2,3\) terms will approximate \(\cosh z\) on \([0,1]\). - Extending to the domain \(-3 \leq z \leq 3\), the Fourier cosine series extension will produce a smooth, periodic approximation. - The graph will show \(\cosh z\) on \([0,1]\) and its periodic extension with oscillations closer to the original function as more terms are added. --- ### **Part (c): Fourier sine series for \(h(z) = \sinh z\) on \(0 < z < 1\)** **Why no computation needed?** Because \(\sinh z\) is *odd* on \([0,1]\), and the sine series naturally represents odd functions on \([0,L]\). **The Fourier sine series:** \[ h(z) = \sum_{n=1}^\infty b_n \sin(n\pi z) \] where \[ b_n = 2 \int_0^1 \sinh z \sin(n \pi z) dz \] - The partial sum includes \(n=1,2,3\). - Extending to \(-3 \leq z \leq 3\), the sine series creates an odd periodic extension of \(\sinh z\). **Graph and explanation:** - The partial sine series approximates \(\sinh z\) on \([0,1]\). - Outside \([0,1]\), the periodic odd extension will oscillate and approximate \(\sinh z\) with Gibbs phenomena near discontinuities at the period boundaries. - The approximation improves as more sine terms are included. --- ### **Part (d): Does it make sense to find Fourier sine/cosine series for \(g(z)\) or \(f(z)\)?** - **Fourier sine series** are suited for functions defined on \([0,L]\) that are *odd* extensions; they are ideal if the original function is odd or naturally extended as odd. - **Fourier cosine series** are suited for functions that are *even* or symmetric about the y-axis. **For \(g(z) = \cosh z\):** - Since \(\cosh z\) is even, a cosine series makes sense. - A sine series would not be appropriate because \(\cosh z\) is not odd, and the sine series would not represent the function accurately over the interval. **For \(f(z) = e^z\):** - No particular symmetry; both sine and cosine series could be used depending on the boundary conditions. - But the full Fourier series (combining sine and cosine) is most general. **Conclusion:** - It **does** make sense to find a Fourier cosine series for an even function like \(\cosh z\). - It **does not** make sense to find a Fourier sine series for \(\cosh z\), nor to find a sine series for \(f(z) = e^z\) unless symmetry conditions are specified. --- **Summary:** | Function | Appropriate Fourier Series | Reasoning | |---|---|---| | \(f(z) = e^z\) | Full Fourier series | No symmetry; general case | | \(g(z) = \cosh z\) | Fourier cosine series | Even function | | \(h(z) = \sinh z\) | Fourier sine series | Odd function | --- **Final notes:** - Visualizing these series on the extended domain helps understand how Fourier series approximate functions periodically. - The choice of sine or cosine series depends on the symmetry of the original function. Let me know if you'd like detailed calculations of specific coefficients or graphs!

**Final comprehensive answer for all parts (a), (b), (c), and (d):** --- ### **(a) Fourier Series for \(f(z) = e^z\) on \(-1 < z < 1\)** **Coefficients:** \[ a_0 = \int_{-1}^1 e^z dz = e - e^{-1} \] \[ a_n = \frac{(-1)^n (e - e^{-1})}{1 + (n \pi)^2} \] \[ b_n = - \frac{n \pi (-1)^n (e - e^{-1})}{1 + (n \pi)^2} \] **Fourier series:** \[ \boxed{ f(z) = \frac{e - e^{-1}}{2} + \sum_{n=1}^\infty \left[ \frac{(-1)^n (e - e^{-1})}{1 + (n\pi)^2} \cos(n\pi z) - \frac{n \pi (-1)^n (e - e^{-1})}{1 + (n\pi)^2} \sin(n\pi z) \right] } \] **Graph explanation:** The partial sum with \(n=0,1,2,3\) terms approximates \(e^z\) well within \(-1<z<1\). Extending periodically, the graph shows repeated exponential “bumps” with discontinuities at the interval edges, illustrating the Gibbs phenomenon. --- ### **(b) Fourier Cosine Series for \(g(z) = \cosh z\) on \(0<z<1\)** - Since \(\cosh z\) is even, the Fourier cosine series is appropriate and straightforward: \[ a_0 = 2 \int_0^1 \cosh z \, dz = 2 \sinh 1 \] \[ a_n = 2 \int_0^1 \cosh z \cos(n \pi z) dz \] - No explicit calculation of these integrals is needed now; the series is: \[ g(z) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(n\pi z) \] - The partial sum with \(n=0,1,2,3\) terms provides a smooth approximation on \([0,1]\), extended periodically to \([-3,3]\). The graph will show the smooth \(\cosh z\) shape in \([0,1]\) and a periodic extension with oscillations outside. **Note:** Because of the symmetry, the cosine series captures \(\cosh z\) accurately, and no sine terms are necessary. --- ### **(c) Fourier Sine Series for \(h(z) = \sinh z\) on \(0<z<1\)** - \(\sinh z\) is odd, so the sine series is suitable: \[ b_n = 2 \int_0^1 \sinh z \sin(n \pi z) dz \] - The series: \[ h(z) = \sum_{n=1}^\infty b_n \sin(n \pi z) \] - Partial sum with \(n=1,2,3\) terms approximates \(\sinh z\). Extending periodically, the graph will show the odd extension of \(\sinh z\), with oscillations near the boundaries and better approximation as more terms are added. **Note:** We don’t need to explicitly compute \(b_n\) here; the structure and symmetry justify the series form. --- ### **(d) Appropriateness of Fourier series types for \(g(z)\) and \(f(z)\)** - **For \(g(z) = \cosh z\):** It makes sense to use a **Fourier cosine series** because \(\cosh z\) is *even*. The cosine series naturally captures even symmetry. - **For \(f(z) = e^z\):** Since \(e^z\) is neither even nor odd, a **full Fourier series** (sine + cosine) is appropriate; or you can choose based on boundary conditions. It **does not** make sense to use a sine series alone because \(e^z\) is not odd, nor a cosine series alone because it’s not even. - **Summary:** - Fourier cosine series is suitable for even functions like \(\cosh z\). - Fourier sine series is suitable for odd functions like \(\sinh z\). - For functions without symmetry, the full Fourier series is appropriate. --- **In conclusion:** **(a)** Full Fourier series for \(e^z\): \[ \boxed{ f(z) = \frac{e - e^{-1}}{2} + \sum_{n=1}^\infty \left[ \frac{(-1)^n (e - e^{-1})}{1 + (n \pi)^2} \cos(n\pi z) - \frac{n \pi (-1)^n (e - e^{-1})}{1 + (n \pi)^2} \sin(n\pi z) \right] } \] **(b)** Fourier cosine series for \(\cosh z\): \[ a_0= 2 \sinh 1, \quad a_n = 2 \int_0^1 \cosh z \cos(n\pi z) dz \] **(c)** Fourier sine series for \(\sinh z\): \[ b_n= 2 \int_0^1 \sinh z \sin(n\pi z) dz \] **(d)** - Fourier cosine series makes sense for \(\cosh z\) (even). - Fourier sine series makes sense for \(\sinh z\) (odd). - For \(e^z\), the full Fourier series is appropriate.