Question

expand Fourier series for f(x)=e^-x in the interval (-1,1)​

Answer 1

Fourier series expansion for f(x)=e^-x in the interval (-1,1)​ is f(x) = (1/2) * [sin(πx) / (1 + π^2) - sin(2πx) / (1 + 4π^2) + sin(3πx) / (1 + 9π^2) - ...]

Given:

f(x)=e^-x

interval = (-1,1)​

To find:

expand Fourier series for f(x)=e^-x in the interval (-1,1)​ = ?

Solution:

It is necessary to first determine the coefficients a0, an, and bn in order to expand the Fourier series for f(x) = e(-x) on the range (-1,1):

a0 = (1/L) * ∫[−L,L] f(x) dx, where L = 1 since the interval is (-1,1)

a0 = (1/2) * ∫[−1,1] e^(-x) dx

a0 = (1/2) * [e^(-1) - e^(1)]

a0 = (1/2) * [(1/e) - (1/e)]

a0 = 0

an = (1/L) * ∫[−L,L] f(x) * cos(nπx/L) dx

an = (1/2) * ∫[−1,1] e^(-x) * cos(nπx/1) dx

an = (1/2) * [(-1)^n + 1] * [1/(1 + n^2π^2)]

bn = (1/L) * ∫[−L,L] f(x) * sin(nπx/L) dx

bn = (1/2) * ∫[−1,1] e^(-x) * sin(nπx/1) dx

bn = (1/2) * (-1)^n * [1/(1 + n^2π^2)]

As a result, f(x) = e^(-x) on the interval (-1,1) is:

f(x) = Σ[ n = -∞ to ∞ ] [(1/2) * (-1)^n * [1/(1 + n^2π^2)] * sin(nπx/1)]

This expression can be simplified to:

f(x) = (1/2) * [sin(πx) / (1 + π^2) - sin(2πx) / (1 + 4π^2) + sin(3πx) / (1 + 9π^2) - ...]

#SPJ1