Question

Find the fourier series expansion of periodic function f(x) in given interval, the period is 2? f(x) = {-1, -? < x < 0; 2, 0 ? x < ?} f(x) = ?_{n=1}^{?} (3((-1)^{n+1} + 1)sin(nx)) / (?n) f(x) = 1/2 + ?_{n=1}^{?} (3((-1)^{n+1} + 1)sin(nx)) / (?n) f(x) = 1 + ?_{n=1}^{?} (3((-1)^{n+1} + 1)sin(nx)) / (?n) f(x) = 1/2 + ?_{n=1}^{?} (3((-1)^{n+1} + 1)cos(nx)) / (?n)

          Find the fourier series expansion of periodic function f(x) in given interval, the period is 2?
f(x) = {-1, -? < x < 0; 2, 0 ? x < ?}
f(x) = ?_{n=1}^{?} (3((-1)^{n+1} + 1)sin(nx)) / (?n)
f(x) = 1/2 + ?_{n=1}^{?} (3((-1)^{n+1} + 1)sin(nx)) / (?n)
f(x) = 1 + ?_{n=1}^{?} (3((-1)^{n+1} + 1)sin(nx)) / (?n)
f(x) = 1/2 + ?_{n=1}^{?} (3((-1)^{n+1} + 1)cos(nx)) / (?n)

Find the fourier series expansion of periodic function f(x) in given interval, the period is 2?
f(x) = -1, -? < x < 0; 2, 0 ? x < ?
f(x) = ?n=1^? (3((-1)^n+1 + 1)sin(nx)) / (?n)
f(x) = 1/2 + ?n=1^? (3((-1)^n+1 + 1)sin(nx)) / (?n)
f(x) = 1 + ?n=1^? (3((-1)^n+1 + 1)sin(nx)) / (?n)
f(x) = 1/2 + ?n=1^? (3((-1)^n+1 + 1)cos(nx)) / (?n)

Added by Olga K.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

03/31/2023

Step 1

The function is defined as: f(x) = Z * 3((-1)^n+1 + 1)sin(nx) We want to find the Fourier series expansion of this function. The general formula for the Fourier series of a function with period T is: f(x) = a_0 + Σ [a_n * cos(2πnx/T) + b_n * sin(2πnx/T)] where Show more…

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Find the Fourier series expansion of the periodic function f(x) in the given interval, where the period is T. f(x) = 0, for 0 < x < T f(x) = Σ[3((-1)^n+1)/n]sin(nx), for 0 < x < T f(x) = Σ[3((-1)^n+1)/n] + lsin(nx), for 0 < x < T f(x) = Σ[2(3((-1)^n+1)/n)sin(nx)], for 0 < x < T f(x) = 1 + Σ[2(3((-1)^n+1)/n) + lcos(nx)], for 0 < x < T, n=1