Let's solve **part (a)** step by step: ### Given function and interval: \( f(x) = e^{-x}, \quad 0 < x < 1 \) --- ## **Step 1: Fourier Sine Series Expansion** The Fourier sine series for \( f(x) \) on \( (0, 1) \) is: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(n\pi x) \] where \[ b_n = 2 \int_0^1 f(x) \sin(n\pi x) dx = 2 \int_0^1 e^{-x} \sin(n\pi x) dx \] **Let's compute \( b_n \):** Let’s use integration by parts or a standard integral: \[ \int_0^1 e^{-x} \sin(n\pi x) dx \] Let \( I = \int_0^1 e^{-x} \sin(n\pi x) dx \). The standard result is: \[ \int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) \] Here, \( a = -1 \), \( b = n\pi \): So, \[ I = \left[ \frac{e^{-x}}{1 + (n\pi)^2} \left( -\sin(n\pi x) - n\pi \cos(n\pi x) \right) \right]_0^1 \] At \( x = 1 \): - \( e^{-1} \) - \( \sin(n\pi) = 0 \) - \( \cos(n\pi) = (-1)^n \) At \( x = 0 \): - \( e^{0} = 1 \) - \( \sin(0) = 0 \) - \( \cos(0) = 1 \) Evaluating: \[ I = \frac{1}{1 + (n\pi)^2} \left[ e^{-1} (0 - n\pi(-1)^n) - (0 - n\pi(1)) \right] \] \[ = \frac{1}{1 + (n\pi)^2} \left[ -e^{-1} n\pi(-1)^n + n\pi \right] \] \[ = \frac{n\pi}{1 + (n\pi)^2} \left[ 1 - (-1)^n e^{-1} \right] \] So: \[ b_n = 2I = \frac{2 n \pi}{1 + (n\pi)^2} \left[ 1 - (-1)^n e^{-1} \right] \] **Explanation:** We used the formula for the sine series coefficients and calculated the integral using standard results for integrating products of exponentials and sines. --- ## **Step 2: Fourier Cosine Series Expansion** The Fourier cosine series for \( f(x) \) on \( (0, 1) \) is: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(n\pi x) \] where \[ a_n = 2 \int_0^1 e^{-x} \cos(n\pi x) dx \] Again, use the standard integral: \[ \int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) \] With \( a = -1 \), \( b = n\pi \): \[ I_c = \int_0^1 e^{-x} \cos(n\pi x) dx = \left[ \frac{e^{-x}}{1 + (n\pi)^2} (-\cos(n\pi x) + n\pi \sin(n\pi x)) \right]_0^1 \] At \( x=1 \): - \( e^{-1} \) - \( \cos(n\pi) = (-1)^n \) - \( \sin(n\pi) = 0 \) At \( x=0 \): - \( e^0 = 1 \) - \( \cos(0) = 1 \) - \( \sin(0) = 0 \) Evaluating: \[ I_c = \frac{1}{1 + (n\pi)^2} \left[ e^{-1} ( -(-1)^n + 0 ) - ( -1 + 0 ) \right] \] \[ = \frac{1}{1 + (n\pi)^2} \left[ -e^{-1}(-1)^n + 1 \right] \] \[ = \frac{1}{1 + (n\pi)^2} \left[ 1 - (-1)^n e^{-1} \right] \] Thus, \[ a_n = 2 I_c = \frac{2}{1 + (n\pi)^2} \left[ 1 - (-1)^n e^{-1} \right] \] **Explanation:** We used the formula for the cosine series coefficients and calculated the integral using the standard result for exponentials times cosines. --- ## **Final Answers** ### **(a)** #### **Fourier Sine Series:** \[ f(x) = e^{-x} = \sum_{n=1}^{\infty} \frac{2 n \pi}{1 + (n\pi)^2} \left[ 1 - (-1)^n e^{-1} \right] \sin(n\pi x) \] #### **Fourier Cosine Series:** \[ f(x) = e^{-x} = \sum_{n=1}^{\infty} \frac{2}{1 + (n\pi)^2} \left[ 1 - (-1)^n e^{-1} \right] \cos(n\pi x) \] --- If you would like the solution for part (b), let me know!