Question

Let f be a periodic function of period 8 defined by f(x) = { 1 if - 4 < x ? 0 x if 0 < x ? 4 } Then the Fourier series of f is a_0 + ?_{n=1}^{?} a_n cos(n?x/4) + b_n sin(n?x/4) where a_0 = 4 a_n = (4((-1)^n-1))/(n^2pi^2) b_n = -2/(npi)(1+(-1)^n)

          Let f be a periodic function of period 8 defined by
f(x) = {
  1 if - 4 < x ? 0
  x if 0 < x ? 4
}
Then the Fourier series of f is
a_0 + ?_{n=1}^{?} a_n cos(n?x/4) + b_n sin(n?x/4)
where a_0 = 4
a_n = (4((-1)^n-1))/(n^2pi^2)
b_n = -2/(npi)(1+(-1)^n)

Let f be a periodic function of period 8 defined by
f(x) =
1 if - 4 < x ? 0
x if 0 < x ? 4

Then the Fourier series of f is
a0 + ?n=1^? an cos(n?x/4) + bn sin(n?x/4)
where a0 = 4
an = (4((-1)^n-1))/(n^2pi^2)
bn = -2/(npi)(1+(-1)^n)

Added by Cristian F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

05/25/2023

Step 1

First, we are given a periodic function with period 8, defined by: f(x) = \begin{cases} x & \text{if } -4 < x < 0 \\ -x & \text{if } 0 < x < 4 \end{cases} The Fourier series of a function is given by: f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n Show more…

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Let f(x) be a periodic function of period 8 defined by: f(x) = 4 if -4 < x < 0 f(x) = 0 if 0 < x < 4 Then the Fourier series of f is: f(x) = (a0/2) + Σ(an*cos(nπx/4) + bn*sin(nπx/4)) where: a0 = (1/4) * ∫[0,4] f(x) dx an = (1/2) * ∫[0,4] f(x)*cos(nπx/4) dx bn = (1/2) * ∫[0,4] f(x)*sin(nπx/4) dx The coefficients are given by: a0 = 2 an = (4/((nπ)^2)) * (1 - (-1)^n) bn = (2/(nπ)) * (1 - (-1)^n)