MXGS-021 JAV 2 0≤ x< <1 2π a 2π ...Find the Fourier series of the functionegin{align} f(x) = egin{cases} 0 & 0 leq x < pi \ 1 & pi leq x < 2pi \ 2 & 2pi leq x < 3pi \ end{cases}end{align} To find the Fourier series of the given function, we need to represent the function as a sum of sine and cosine functions. The function is defined as: [ f(x) = egin{cases} 0 & 0 leq x < pi \ 1 & pi leq x < 2pi \ 2 & 2pi leq x < 3pi \ end{cases} ] The Fourier series representation of a function ( f(x) ) is given by: [ f(x) = frac{a_0}{2} + sum_{n=1}^{infty} left( a_n cos(nx) + b_n sin(nx) ight) ] where: - ( a_0 = frac{1}{pi} int_{0}^{2pi} f(x) dx ) - ( a_n = frac{1}{pi} int_{0}^{2pi} f(x) cos(nx) dx ) - ( b_n = - frac{1}{pi} int_{0}^{2pi} f(x) sin(nx) dx ) Let's compute each of these coefficients: 1. **Compute ( a_0 ):** [ a_0 = frac{1}{pi} int_{0}^{3pi} f(x) dx ] [ a_0 = frac{1}{pi} left( int_{0}^{pi} 0 dx + int_{1}^{2pi} 1 dx + int_{2pi}^{3pi} 2 dx ight) ] [ a_0 = frac{1}{pi} left( 0 + (2pi - pi) + 2(3pi - 2pi) ight) ] [ a_0 = frac{1}{pi} left( pi + 2pi ight) ] [ a_0 = frac{1}{pi} left( 3pi ight) ] [ a_0 = 3 ] 2. **Compute ( a_n ):** [ a_n = frac{1}{pi} int_{0}^{3pi} f(x) cos(nx) dx ] [ a_n = frac{1}{pi} left( int_{0}^{pi} 0 cos(nx) dx + int_{pi}^{2pi} 1 cos(nx) dx + int_{2pi}^{3pi} 2 cos(nx) dx ight) ] [ a_n = frac{1}{pi} left( 0 + int_{pi}^{2pi} cos(nx) dx + 2 int_{2pi}^{3pi} cos(nx) dx ight) ] [ a_n = frac{1}{pi} left( left[ frac{sin(nx)}{n} ight]_{pi}^{2pi} + 2 left[ frac{sin(nx)}{n} ight]_{2pi}^{3pi} ight) ] [ a_n = frac{1}{pi} left( frac{sin(n cdot 2pi) - sin(n cdot pi)}{n} + 2 frac{sin(n cdot 3pi) + sin(n cdot 2pi)}{n} ight) ] [ a_n = frac{1}{pi} left( frac{0 - 0}{n} + 2 frac{0 - 0}{n} ight) ] [ a_n = 0 ] 3. **Compute ( b_n ):** [ b_n = frac{1}{pi} int_{0}^{3pi} f(x) sin(nx) dx ] [ b_n = frac{1}{pi} left( int_{0}^{pi} 0 sin(nx) dx + int_{pi}^{2pi} 1 sin(nx) dx + int_{2pi}^{3pi} 2 sin(nx) dx ight) ) [ b_n = frac{1}{pi} left( 0 + int_{pi}^{2pi} sin(nx) dx + 2 int_{2pi}^{3pi} sin(nx) dx ight) ) [ b_n = frac{1}{pi} left( left[ frac{- cos(nx)}{n} ight]_{pi}{2pi} + 2 left[ frac{- cos(nx)}{n} ight]_{2pi}{3pi} ight) ) [ b_n = frac{1}{pi} left( frac{-cos(n cdot 2pi) - (-cos(n cdot pi))}{n} + 2 frac{-cos(n cdot 3pi) + (-cos(n cdot 2pi))}{n} ight) [ b_n = frac{1}{pi} left( frac{1 - (-1)^n}{n} + 2 frac{1 - (-1)^n}{n} ight) ) [ b_n = frac{1}{pi} left( frac{3(1 - (-1)^n)}{n} ight) ) [ b_n = frac{3}{pi n} left( 1 - (-1)^n ight} ) Now, we can write the Fourier series: [ f(x) = frac{3}{2} + sum_{n=1}^{infty} left( a_n cos(nx) + b_n sin(nx) ight) ] [ f(x) = frac{3}{2} + sum_{n=1}^{infty} left( 0 cos(nx) + frac{3 left( 1 - (-1)^n ight)}{pi n} sin(nx) ight) ) [ f(x) = frac{3}{2} + sum_{n=1}^{infty} left( frac{3 left( 1 - (-1)^n ight)}{pi n} sin(nx) ight) ) Thus, the Fourier series is: [ f(x) = frac{3}{2} + sum_{n=1}^{infty} left( frac{3 left( 1 - (-1)^n ight)}{pi n} sin(nx) ight) This is the final answer for the Fourier series of the given function. - Cuplikan Gratis dan Subtitle Bahasa Indonesia srt.

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