answer:To find the Fourier series of the given function, we first need to find the coefficients a_n and b_n. The general form of a Fourier series is: f(x) = a_0/2 + Σ(a_n * cos(nx) + b_n * sin(nx)) where the summation is from n=1 to infinity. We can find a_0, a_n, and b_n using the following formulas: a_0 = (1/π) * ∫[f(x)]dx over the interval (-π, π) a_n = (1/π) * ∫[f(x) * cos(nx)]dx over the interval (-π, π) b_n = (1/π) * ∫[f(x) * sin(nx)]dx over the interval (-π, π) First, let's find a_0: a_0 = (1/π) * [∫(2)dx from -π to 0 + ∫(-1)dx from 0 to π] a_0 = (1/π) * [2x|(-π, 0) - x|(0, π)] a_0 = (1/π) * [2(0) - 2(-π) - (π - 0)] a_0 = (1/π) * (2π - π) a_0 = 1 Now, let's find a_n: a_n = (1/π) * [∫(2 * cos(nx))dx from -π to 0 + ∫(-1 * cos(nx))dx from 0 to π] a_n = (1/π) * [(2/n) * sin(nx)|(-π, 0) - (1/n) * sin(nx)|(0, π)] a_n = (1/π) * [(2/n) * (sin(0) - sin(-nπ)) - (1/n) * (sin(nπ) - sin(0))] a_n = (1/π) * [(2/n) * 0 - (1/n) * 0] a_n = 0 Finally, let's find b_n: b_n = (1/π) * [∫(2 * sin(nx))dx from -π to 0 + ∫(-1 * sin(nx))dx from 0 to π] b_n = (1/π) * [(-2/n) * cos(nx)|(-π, 0) + (1/n) * cos(nx)|(0, π)] b_n = (1/π) * [(-2/n) * (cos(0) - cos(-nπ)) + (1/n) * (cos(nπ) - cos(0))] b_n = (1/π) * [(-2/n) * (1 - (-1)^n) + (1/n) * ((-1)^n - 1)] For n = odd, b_n = (1/π) * [(-2/n) * (1 - (-1)) + (1/n) * ((-1) - 1)] = (1/π) * [(-2/n) * 2 + (1/n) * (-2)] = -4/nπ For n = even, b_n = 0 Now we can write the Fourier series for the given function: f(x) = 1/2 + Σ((-1)^((n-1)/2) * (4/nπ) * sin(nx)) for n odd, and 0 for n even This is the Fourier series representation of the given function with a periodic discontinuity.

answer:To find the Fourier series of the given function f(x) = 2x - 1 on the interval [0, 2π] with a periodic discontinuity at x = π, we need to compute the Fourier coefficients a_n and b_n for the sine and cosine terms. The general form of a Fourier series is: f(x) = a_0 + Σ[a_n * cos(n * x) + b_n * sin(n * x)] where the summation is from n = 1 to infinity, and a_0, a_n, and b_n are the Fourier coefficients. First, let's find the constant term a_0: a_0 = (1/π) * ∫[f(x)]dx from 0 to 2π Since the function has a periodic discontinuity at x = π, we need to split the integral into two parts: a_0 = (1/π) * [∫[f(x)]dx from 0 to π + ∫[f(x)]dx from π to 2π] For the interval [0, π], f(x) = 2x - 1: a_0 = (1/π) * [∫[(2x - 1)]dx from 0 to π + ∫[f(x)]dx from π to 2π] Now, for the interval [π, 2π], since the function is periodic with a discontinuity at x = π, we can write f(x) as f(x) = 2(x - π) - 1: a_0 = (1/π) * [∫[(2x - 1)]dx from 0 to π + ∫[(2(x - π) - 1)]dx from π to 2π] Now, we can evaluate the integrals: a_0 = (1/π) * [(x^2 - x)| from 0 to π + ((x^2 - 2πx + π^2 - x)| from π to 2π)] a_0 = (1/π) * [(π^2 - π) + (4π^2 - 4π^2 + π^2 - π)] a_0 = (1/π) * (π^2 - π) Now, let's find the coefficients a_n and b_n: a_n = (1/π) * ∫[f(x) * cos(n * x)]dx from 0 to 2π b_n = (1/π) * ∫[f(x) * sin(n * x)]dx from 0 to 2π Again, we need to split the integrals into two parts due to the periodic discontinuity at x = π: a_n = (1/π) * [∫[(2x - 1) * cos(n * x)]dx from 0 to π + ∫[(2(x - π) - 1) * cos(n * x)]dx from π to 2π] b_n = (1/π) * [∫[(2x - 1) * sin(n * x)]dx from 0 to π + ∫[(2(x - π) - 1) * sin(n * x)]dx from π to 2π] Now, we can integrate by parts: For a_n: Let u = 2x - 1, dv = cos(n * x)dx du = 2dx, v = (1/n) * sin(n * x) For b_n: Let u = 2x - 1, dv = sin(n * x)dx du = 2dx, v = -(1/n) * cos(n * x) Now, we can compute the integrals: a_n = (1/π) * [(u * v)| from 0 to π + (u * v)| from π to 2π - ∫[v * du] from 0 to 2π] a_n = (1/π) * [((2x - 1) * (1/n) * sin(n * x))| from 0 to π + ((2(x - π) - 1) * (1/n) * sin(n * x))| from π to 2π - (1/n) * ∫[sin(n * x) * 2]dx from 0 to 2π] b_n = (1/π) * [(u * v)| from 0 to π + (u * v)| from π to 2π - ∫[v * du] from 0 to 2π] b_n = (1/π) * [((2x - 1) * (-1/n) * cos(n * x))| from 0 to π + ((2(x - π) - 1) * (-1/n) * cos(n * x))| from π to 2π + (1/n) * ∫[cos(n * x) * 2]dx from 0 to 2π] Now, we can evaluate the integrals and simplify the expressions for a_n and b_n. Finally, we can write the Fourier series for the given function: f(x) = (π^2 - π)/π + Σ[a_n * cos(n * x) + b_n * sin(n * x)]

answer:First, let's find the period of the function. Since the function is defined on the interval [-4, 4], the period is 2L = 8, so L = 4. Now, let's find the coefficients a_0, a_n, and b_n. a_0 = frac{1}{4} int_{-4}^4 f(x) dx = frac{1}{4} left[ int_{-4}^0 1 dx + int_0^4 -1 dx right] = frac{1}{4} left[ (x Big|_{-4}^0) - (x Big|_0^4) right] = frac{1}{4} left[ (0 - (-4)) - (4 - 0) right] = frac{1}{4} left[ 4 - 4 right] = 0 a_n = frac{1}{4} int_{-4}^4 f(x) cosleft(frac{npi x}{4}right) dx = frac{1}{4} left[ int_{-4}^0 1 cdot cosleft(frac{npi x}{4}right) dx + int_0^4 -1 cdot cosleft(frac{npi x}{4}right) dx right] = frac{1}{4} left[ int_{-4}^0 cosleft(frac{npi x}{4}right) dx - int_0^4 cosleft(frac{npi x}{4}right) dx right] Now, integrate the cosine terms: = frac{1}{4} left[ frac{4}{npi} sinleft(frac{npi x}{4}right) Big|_{-4}^0 - frac{4}{npi} sinleft(frac{npi x}{4}right) Big|_0^4 right] = frac{1}{npi} left[ sinleft(frac{npi (0)}{4}right) - sinleft(frac{npi (-4)}{4}right) - sinleft(frac{npi (4)}{4}right) + sinleft(frac{npi (0)}{4}right) right] = frac{1}{npi} left[ - sinleft(npiright) + sinleft(-npiright) right] Since sin(-x) = -sin(x) and sin(nπ) = 0 for all integer values of n, the expression simplifies to: a_n = 0 Now, let's find b_n: b_n = frac{1}{4} int_{-4}^4 f(x) sinleft(frac{npi x}{4}right) dx = frac{1}{4} left[ int_{-4}^0 1 cdot sinleft(frac{npi x}{4}right) dx + int_0^4 -1 cdot sinleft(frac{npi x}{4}right) dx right] = frac{1}{4} left[ int_{-4}^0 sinleft(frac{npi x}{4}right) dx - int_0^4 sinleft(frac{npi x}{4}right) dx right] Now, integrate the sine terms: = frac{1}{4} left[ -frac{4}{npi} cosleft(frac{npi x}{4}right) Big|_{-4}^0 + frac{4}{npi} cosleft(frac{npi x}{4}right) Big|_0^4 right] = frac{1}{npi} left[ cosleft(frac{npi (0)}{4}right) - cosleft(frac{npi (-4)}{4}right) + cosleft(frac{npi (4)}{4}right) - cosleft(frac{npi (0)}{4}right) right] = frac{1}{npi} left[ 1 - cosleft(npiright) + cosleft(npiright) - 1 right] Since cos(nπ) = (-1)^n, the expression simplifies to: b_n = frac{1}{npi} left[ 1 - (-1)^n + (-1)^n - 1 right] = frac{1}{npi} left[ 2 - 2(-1)^n right] Now, we can write the Fourier series for the function f(x): f(x) = frac{a_0}{2} + sum_{n=1}^infty left[ a_n cosleft(frac{npi x}{4}right) + b_n sinleft(frac{npi x}{4}right) right] = 0 + sum_{n=1}^infty left[ 0 cdot cosleft(frac{npi x}{4}right) + frac{1}{npi} left( 2 - 2(-1)^n right) sinleft(frac{npi x}{4}right) right] = sum_{n=1}^infty frac{1}{npi} left( 2 - 2(-1)^n right) sinleft(frac{npi x}{4}right) This is the Fourier series for the given function f(x).

answer:To find the Fourier series of f(x), we need to determine the coefficients a_n and b_n in the following series representation: f(x) = a_0 + Σ [a_n * cos(nπx) + b_n * sin(nπx)] where the summation is from n = 1 to infinity. First, let's find a_0: a_0 = (1/L) * ∫[f(x)]dx from -L to L, where L is half the period of the function (in this case, L = 1). a_0 = (1/1) * [∫[1]dx from -1 to 0 + ∫[-1]dx from 0 to 1] a_0 = [x] from -1 to 0 + [-x] from 0 to 1 a_0 = (0 - (-1)) + (1 - 0) a_0 = 1 + 1 a_0 = 2 Now, let's find a_n: a_n = (1/L) * ∫[f(x) * cos(nπx)]dx from -L to L a_n = (1/1) * [∫[1 * cos(nπx)]dx from -1 to 0 + ∫[-1 * cos(nπx)]dx from 0 to 1] a_n = [∫[cos(nπx)]dx from -1 to 0 + ∫[-cos(nπx)]dx from 0 to 1] a_n = [sin(nπx)/nπ] from -1 to 0 + [-sin(nπx)/nπ] from 0 to 1 a_n = (0 - sin(-nπ)/nπ) + (-sin(nπ)/nπ - 0) a_n = (sin(nπ)/nπ) - (sin(nπ)/nπ) a_n = 0 Now, let's find b_n: b_n = (1/L) * ∫[f(x) * sin(nπx)]dx from -L to L b_n = (1/1) * [∫[1 * sin(nπx)]dx from -1 to 0 + ∫[-1 * sin(nπx)]dx from 0 to 1] b_n = [∫[sin(nπx)]dx from -1 to 0 + ∫[-sin(nπx)]dx from 0 to 1] b_n = [-cos(nπx)/nπ] from -1 to 0 + [cos(nπx)/nπ] from 0 to 1 b_n = (-cos(0)/nπ - (-cos(-nπ)/nπ)) + (cos(nπ)/nπ - cos(0)/nπ) b_n = (1/nπ - (-1)^n/nπ) + ((-1)^n/nπ - 1/nπ) b_n = 2/nπ (1 - (-1)^n) So, the Fourier series of f(x) is: f(x) = 2 + Σ [2/nπ (1 - (-1)^n) * sin(nπx)] for n = 1 to infinity This is the Fourier series representation of the given function f(x) over the interval [-1, 1].