Answer to Question #206786 in Calculus for Leela

Question #206786

Find Fourier series expansion for the function f(x)={x+4 for 0<x<π,-x-π for -π<x<0}




1
Expert's answer
2021-06-15T03:41:00-0400

 f(x)={x+4 for 0<x<π,

-x-π for -π<x<0}




The Fourier series expansion of any function f(x) is given by


f(x) = "\\dfrac{ a_0}{2}" + "a_1 cosx" + "a_2 cos2x" + ........ + "b_1 sin x + b_2 sin 2x" + ............ (1)



where,


a0 = "\\dfrac{1}{\\pi}" "\\int_{-\\pi}^\\pi" f(x) dx


a0 = "\\dfrac{1}{\\pi}" ("\\int^0_{-\\pi}" ( - x - "\\pi" ) dx + "\\int^\\pi_{0}" (x + 4))dx

a0 = "\\dfrac{1}{\\pi}"[ "(\\dfrac{-x^2}{2} - 4x)_{-\\pi}^0" +"(\\dfrac{x^2}{2} + 4x)^\\pi_0" ]


a0 = "\\pi" ...........................................(2)



an = "\\dfrac{1}{\\pi}" "\\int_{-\\pi}^\\pi" f(x) cos nx dx


an = "\\dfrac{1}{\\pi}" ("\\int^0_{-\\pi}" ( - x - "\\pi" ) dx + "\\int^\\pi_{0}" (x + 4)) cos nx dx


an = "\\dfrac{1}{\\pi}" "(\\dfrac{(-x-\\pi) sin nx}{n} - \\dfrac{cos nx}{n^2})^0_{-\\pi}" + "(\\dfrac{(x + 4)sin nx}{n^2} + \\dfrac{cos nx}{n^2})^\\pi_0"


an = "\\dfrac{(-1)^n - 1}{\\pi*n^2}" ...............................................(2)



From equation(2) we have



a1 = "\\dfrac{-2}{\\pi}" , a2 = 0 , a3 = "\\dfrac{-2}{9\\pi}"




bn = "\\dfrac{1}{\\pi}" "\\int_{-\\pi}^\\pi" f(x) sin nx dx


bn = "\\dfrac{1}{\\pi}" ("\\int^0_{-\\pi}" ( - x - "\\pi" ) dx + "\\int^\\pi_{0}" (x + 4)) sin nx dx


bn = "\\dfrac{1}{\\pi}" "(\\dfrac{(-x-\\pi) sin nx}{n} - \\dfrac{cos nx}{n^2})^0_{-\\pi}" + "(\\dfrac{(x + 4)sin nx}{n^2} + \\dfrac{cos nx}{n^2})^\\pi_0"


bn = "\\dfrac{1}{n\\pi}" "[\\pi(1 + (-1)^n - 4(1 - (-1)^n)]" ............................(3)


From equation (3) we have


b1 = "\\dfrac{-8}{\\pi}" , b2 = 1 , b3 = "\\dfrac{-8}{3\\pi}"



f(x) = "\\dfrac{\\pi}{2}" + "\\dfrac{-2cos x}{\\pi} + \\dfrac{-2cos3x}{9\\pi}" + ......... + "\\dfrac{-8 sin x}{\\pi} + sin 2x + \\dfrac{-8 sin 3x}{3\\pi}" + ........









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