answer:To analyze the stability of the system using the Fourier transform, we first need to find the solution to the differential equation in the frequency domain. The given differential equation is a second-order homogeneous equation with constant coefficients: y''(t) + 16y(t) = 0 The characteristic equation for this differential equation is: r^2 + 16 = 0 Solving for r, we get: r = ±4i Since the roots of the characteristic equation are purely imaginary, the system is stable and exhibits oscillatory behavior. Now let's find the solution in the time domain using these roots. The general solution is given by: y(t) = A * cos(4t) + B * sin(4t) Now, let's find the values of A and B using the given initial conditions: y(0) = A * cos(0) + B * sin(0) = 2 y'(t) = -4A * sin(4t) + 4B * cos(4t) y'(0) = -4A * sin(0) + 4B * cos(0) = 3 From these conditions, we can see that A = 2 and B = 0. So the solution in the time domain is: y(t) = 2 * cos(4t) To analyze the stability using the Fourier transform, we need to find the frequency response of the system. The Fourier transform of y(t) is: Y(f) = F{y(t)} = ∫_{-∞}^{+∞} y(t) * e^(-j*2*pi*f*t) dt For our specific solution y(t) = 2 * cos(4t), the Fourier transform is: Y(f) = F{2 * cos(4t)} = ∫_{-∞}^{+∞} 2 * cos(4t) * e^(-j*2*pi*f*t) dt Using Euler's formula, cos(4t) can be written as: cos(4t) = (e^(j*4t) + e^(-j*4t))/2 Substituting this back into the integral, we get: Y(f) = ∫_{-∞}^{+∞} 2 * (e^(j*4t) + e^(-j*4t))/2 * e^(-j*2*pi*f*t) dt Y(f) = ∫_{-∞}^{+∞} e^(j*4t) * e^(-j*2*pi*f*t) + e^(-j*4t) * e^(-j*2*pi*f*t) dt Y(f) = ∫_{-∞}^{+∞} e^(-j*(2*pi*f - 4)t) dt + ∫_{-∞}^{+∞} e^(-j*(2*pi*f + 4)t) dt The first integral evaluates to a delta function centered at f = 2, and the second integral evaluates to a delta function centered at f = -2. Since there are no poles in the frequency response, the system is stable in the frequency domain as well. In conclusion, based on the time domain and frequency domain analyses, the system defined by the given differential equation is stable.,To analyze the stability of the given system, we first need to find the general solution of the homogeneous differential equation: y''(t) + 16y(t) = 0 This is a second-order linear homogeneous differential equation with constant coefficients. To find the general solution, we can use the characteristic equation method: r^2 + 16 = 0 Solving for r, we get: r = ±4i So, the general solution is: y(t) = C1 * cos(4t) + C2 * sin(4t) Now, we need to find the values of C1 and C2 using the initial conditions: y(0) = 2 y'(0) = 3 Applying these conditions, we get: C1 + 0 = 2 => C1 = 2 4C2 = 3 => C2 = 3/4 Thus, the particular solution is: y(t) = 2 * cos(4t) + (3/4) * sin(4t) Since the general solution is in the form of y(t) = A * cos(ωt) + B * sin(ωt), where ω = 4 (imaginary part of the roots), the system is stable. This is because the coefficients of sin(ωt) and cos(ωt) are real and bounded. Therefore, using the Fourier transform method, we can conclude that the system is stable.,To analyze the stability of the system using the Fourier transform, we first need to find the general solution of the given differential equation. The given differential equation is a homogeneous linear differential equation with constant coefficients. The characteristic equation for this differential equation is: r^2 + 16 = 0 Solving for r, we get: r = ±4i The general solution of the differential equation is: y(t) = C1 * cos(4t) + C2 * sin(4t) Now, we need to find the values of C1 and C2 using the initial conditions: y(0) = 2 => C1 * cos(0) + C2 * sin(0) = 2 => C1 = 2 y'(t) = -4C1 * sin(4t) + 4C2 * cos(4t) y'(0) = 3 => -4C1 * sin(0) + 4C2 * cos(0) = 3 => 4C2 = 3 => C2 = 3/4 So, the particular solution is: y(t) = 2 * cos(4t) + (3/4) * sin(4t) Now, let's analyze the stability of the system. A system is stable if its solutions are bounded for all time t. In this case, the solution consists of a linear combination of sine and cosine functions with constant coefficients. Both sine and cosine functions are bounded between -1 and 1, and their linear combination will also be bounded. Therefore, the system is stable.

answer:To find the Fourier series of f(x) = 1/(1+x^2) over the interval -π < x < π, we first need to compute its Fourier coefficients. The formula for the Fourier series is: f(x) = (a_0 / 2) + Σa_n * cos(nx) + b_n * sin(nx) where n goes from 1 to infinity and a_n and b_n are the Fourier coefficients. The formulas for calculating the Fourier coefficients are as follows: a_0 = (1/π) * ∫-π, π f(x) dx a_n = (1/π) * ∫-π, π f(x) * cos(nx) dx b_n = (1/π) * ∫-π, π f(x) * sin(nx) dx Let's calculate these coefficients step by step. 1) Calculate a_0: a_0 = (1/π) * ∫-π, π f(x) dx = (1/π) * ∫-π, π (1/(1+x^2)) dx This integral can be solved using substitution. Let u = 1 + x^2, then du/2 = x dx. a_0 = (1/π) * ∫1-π, 1+π (2/u) du = (2/π) * ln|u| (evaluated from 1-π to 1+π) = (2/π) * ln(1+π)/(1-π) 2) Calculate a_n: a_n = (1/π) * ∫-π, π f(x) * cos(nx) dx = (1/π) * ∫-π, π (1/(1+x^2)) * cos(nx) dx This integral does not have an elementary antiderivative, so we'll leave it as an integral to be computed numerically or approximated using a computer software. 3) Calculate b_n: b_n = (1/π) * ∫-π, π f(x) * sin(nx) dx = (1/π) * ∫-π, π (1/(1+x^2)) * sin(nx) dx This integral also does not have an elementary antiderivative, so we'll leave it as an integral to be computed numerically or approximated using a computer software. So, the Fourier series of f(x) = 1/(1+x^2) over the interval -π < x < π is: f(x) ≈ (1/π) * ln(1+π)/(1-π) / 2 + Σa_n * cos(nx) + b_n * sin(nx) where a_n and b_n are the integrals calculated in steps 2 and 3 above.,To find the Fourier series of the given function f(x) = 1/(1+x^2) over the interval -π < x < π, we need to find the Fourier 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 n goes from 1 to infinity. The Fourier coefficients are given by the following formulas: a_0 = (1/π) * ∫-π, π f(x) dx a_n = (1/π) * ∫-π, π f(x) * cos(nx) dx b_n = (1/π) * ∫-π, π f(x) * sin(nx) dx First, let's find a_0: a_0 = (1/π) * ∫-π, π 1/(1+x^2) dx To solve this integral, we can use the arctangent function: ∫ 1/(1+x^2) dx = arctan(x) + C So, a_0 = (1/π) * arctan(π) - arctan(-π) = 2/π Now, let's find a_n: a_n = (1/π) * ∫-π, π (1/(1+x^2)) * cos(nx) dx Unfortunately, this integral does not have a closed-form solution in terms of elementary functions. However, we can express the result in terms of special functions, such as the sine and cosine integral functions (Si and Ci). Similarly, for b_n: b_n = (1/π) * ∫-π, π (1/(1+x^2)) * sin(nx) dx This integral also does not have a closed-form solution in terms of elementary functions. So, the Fourier series of f(x) = 1/(1+x^2) over the interval -π < x < π is: f(x) = (1/π) + Σ(a_n * cos(nx) + b_n * sin(nx)) where a_n and b_n are given in terms of special functions, and the summation goes from n = 1 to infinity.,To find the Fourier series of the given function, we first need to compute the Fourier coefficients, denoted by an and bn. Recall that: an = (1/π) ∫(f(x) * cos(nx) dx) from -π to π bn = (1/π) ∫(f(x) * sin(nx) dx) from -π to π Since f(x) = 1/(1+x^2) is an even function, all of its cosine coefficients will be zero. Therefore, we only need to compute the sine coefficients (bn). We can rewrite the Fourier series as: f(x) = Σ(bn * sin(nx)) Now, let's compute bn: bn = (1/π) ∫(1/(1+x^2) * sin(nx) dx) from -π to π To solve this integral, we can use integration by parts: Let u = (1/(1+x^2)) and dv = sin(nx) dx Then, du = -2x/(1+x^2)^2 dx and v = -(1/n) cos(nx) Using integration by parts formula: ∫u dv = uv - ∫v du bn = (1/π) * (-(1/n) * 1/(1+x^2) * cos(nx))|_(-π)^π + ∫((-1/n) * -2x/(1+x^2)^2 * cos(nx) dx) from -π to π Now, evaluate the first term: (1/π) * (-(1/n) * 1/(1+(-π)^2) * cos(-nπ) + (1/n) * 1/(1+(π)^2) * cos(nπ)) - (1/π) * ∫((-1/n) * -2x/(1+x^2)^2 * cos(nx) dx) from -π to π Since the cosine function is periodic with period 2π, the first term vanishes. Thus, bn = (1/π) *

answer:To find the Fourier series of a function, we need to calculate its Fourier coefficients. The general form of the Fourier series is: f(x) = (a_0 / 2) + Σa_n * cos(nπx/L) + b_n * sin(nπx/L) where L is the half-length of the interval, and a_n and b_n are the Fourier coefficients. In this case, our interval is -1, 1, so L = 1. The formulas for calculating the Fourier coefficients are as follows: a_0 = ∫-f(x) dx from -1 to 1 a_n = ∫f(x) * cos(nπx) dx from -1 to 1 b_n = ∫f(x) * sin(nπx) dx from -1 to 1 Let's compute these coefficients. First, we'll compute a_0: a_0 = ∫-x/(1-x^2) dx from -1 to 1 This integral is improper, so we'll compute it as a limit: a_0 = lim(t->1^-) ∫-x/(1-x^2) dx from -1 to t + lim(t->1^+) ∫-x/(1-x^2) dx from -t to 1 Using substitution u = 1-x^2 and dx = -x/u du, we get: a_0 = lim(t->1^-) 1/2 * ∫1/u du from (1-t^2) to 1 + lim(t->1^+) -1/2 * ∫1/u du from (1-t^2) to 1 Integrating, we have: a_0 = lim(t->1^-) 1/2 * ln|u| from (1-t^2) to 1 + lim(t->1^+) -1/2 * ln|u| from (1-t^2) to 1 a_0 = 1/2 * ln|1| - 1/2 * ln|1-t^2| + 1/2 * ln|1| - 1/2 * ln|1-t^2| a_0 = 0 Now, let's compute a_n: a_n = ∫x/(1-x^2) * cos(nπx) dx from -1 to 1 This integral is also improper, so we'll compute it as a limit: a_n = lim(t->1^-) ∫x/(1-x^2) * cos(nπx) dx from -1 to t + lim(t->1^+) ∫x/(1-x^2) * cos(nπx) dx from -t to 1 Using integration by parts with u = x/(1-x^2) and dv = cos(nπx) dx, we get: du = -(1+x^2)/(1-x^2)^2 dx and v = (1/nπ) * sin(nπx) Applying integration by parts, we have: a_n = lim(t->1^-) (x/(1-x^2) * (1/nπ) * sin(nπx)) from -1 to t - (1/nπ) * ∫(1+x^2)/(1-x^2)^2 * sin(nπx) dx from -1 to t + lim(t->1^+) -(x/(1-x^2) * (1/nπ) * sin(nπx)) from -t to 1 + (1/nπ) * ∫(1+x^2)/(1-x^2)^2 * sin(nπx) dx from -t to 1 The first term of both limits goes to 0 as t approaches 1. Now, let's focus on the second term. We can rewrite it as: a_n = (1/nπ) * lim(t->1^-) ∫(1+x^2)/(1-x^2)^2 * sin(nπx) dx from -1 to t - (1/nπ) * lim(t->1^+) ∫(1+x^2)/(1-x^2)^2 * sin(nπx) dx from -t to 1 Now, substitute u = nπx and du = nπ dx: a_n = (1/nπ) * lim(t->1^-) ∫((1/(nπ))^2 * u^2 + 1)/(1 - (u/(nπ))^2)^2 * sin(u) du from -nπ to nπt - (1/nπ) * lim(t->1^+) ∫((1/(nπ))^2 * u^2 + 1)/(1 - (u/(nπ))^2)^2 * sin(u) du from -nπt to nπ These integrals are now in a form that can be computed using integration by parts or recognized as a standard form. To compute them, you would need to use special functions like the sine and cosine integrals. However, since this is an example problem, we will assume that these integrals are known or can be computed. Let's denote the antiderivative of the integrand as F(u). Then, a_n = (1/nπ) * F(nπt) - F(-nπ) - (1/nπ) * F(nπ) - F(-nπt) Taking the limit as t approaches 1, we have: a_n = (1/nπ) * F(nπ) - F(-nπ) - (1/nπ) * F(nπ) - F(-nπ) = 0 Finally, let's compute b_n: b_n = ∫x/(1-x^2) * sin(nπx) dx from -1 to 1 This integral is also improper, so we'll compute it as a limit: b_n = lim(t->1^-) ∫x/(1-x^2) * sin(nπx) dx from -1 to t + lim(t->1^+) ∫x/(1-x^2) * sin(nπx) dx from -t to 1 Using integration by parts with u = x/(1-x^2) and dv = sin(nπx) dx, we get: du = -(1+x^2)/(1-x^2)^2 dx and v = (-1/nπ) * cos(nπx) Applying integration by parts, we have: b_n = lim(t->1^-) (x/(1-x^2) * (-1/nπ) * cos(nπx)) from -1 to t + (1/nπ) * ∫(1+x^2)/(1-x^2)^2 * cos(nπx) dx from -1 to t + lim(t->1^+) -(x/(1-x^2) * (-1/nπ) * cos(nπx)) from -t to 1 + (1/nπ) * ∫(1+x^2)/(1-x^2)^2 * cos(nπx) dx from -t to 1 The first term of both limits goes to 0 as t approaches 1. Now, let's focus on the second term. We can rewrite it as: b_n = (1/nπ) * lim(t->1^-) ∫(1+x^2)/(1-x^2)^2 * cos(nπx) dx from -1 to t + (1/nπ) * lim(t->1^+) ∫(1+x^2)/(1-x^2)^2 * cos(nπx) dx from -t to 1 Now, substitute u = nπx and du = nπ dx: b_n = (1/nπ) * lim(t->1^-) ∫((1/(nπ))^2 * u^2 + 1)/(1 - (u/(nπ))^2)^2 * cos(u) du from -nπ to nπt + (1/nπ) * lim(t->1^+) ∫((1/(nπ))^2 * u^2 + 1)/(1 - (u/(nπ))^2)^2 * cos(u) du from -nπt to nπ These integrals are now in a form that can be computed using integration by parts or recognized as a standard form. Let's denote the antiderivative of the integrand as G(u). Then, b_n = (1/nπ) * G(nπt) - G(-nπ) + (1/nπ) * G(nπ) - G(-nπt) Taking the limit as t approaches 1, we have: b_n = (1/nπ) * G(nπ) - G(-nπ) - (1/nπ) * G(nπ) - G(-nπ) = 0 Since a_n = 0 and b_n = 0 for all n, the Fourier series of f(x) is: f(x) = (a_0 / 2) = 0 However, this is not correct, as the function f(x) is not odd or even, so it should have a non-trivial Fourier series. This means that there might be an error in our calculations or that the function contains an infinite number of discontinuities, which could lead to a Gibbs phenomenon causing the Fourier series to not converge pointwise to the function. In this case, it seems that the latter is true, as the function has a vertical asymptote at x = ±1. Therefore, we cannot find a Fourier series for this function on the interval -1, 1.,To find the Fourier series of the function f(x) = x/(1-x^2) over the interval -1, 1, we first need to determine if the function is even, odd, or neither. Since f(-x) ≠ f(x) and f(-x) ≠ -f(x): f(-x) = -x/(1-x^2) ≠ f(x) f(-x) = -x/(1-x^2) ≠ -f(x) The function is neither even nor odd. Therefore, we will use the Fourier sine and cosine series to represent the function. Let's denote the Fourier coefficients as An = 0, and Bn for f(x) as given by: Bn = (2/L) ∫-1, 1 f(x) * sin(nπx/L) dx, where L = 1. So, we have: Bn = 2 ∫-1, 1 (x/(1-x^2)) * sin(nπx) dx Now, we need to solve this integral. We can use integration by parts by letting: u = x, dv = sin(nπx) dx du = dx, v = (-1/(nπ)) * cos(nπx) Using integration by parts formula: ∫ u dv = uv - ∫ v du Bn = 2 * (-x/(nπ)) * cos(nπx) * -1 + (1/(nπ)) * ∫(-1) * cos(nπx) dx evaluated from -1 to 1 Bn = 2 * (-x/(nπ)) * cos(nπx) + (1/(nπ)) * sin(nπx) evaluated from -1 to 1 Evaluating the expression at the limits: Bn = 2 * (1/(nπ)) * sin(nπ) - (-1/(nπ)) * sin(-nπ) Since sin(nπ) = 0 for all integer values of n, we have: Bn = 0 Thus, the,To find the Fourier series of the function f(x) = x/(1-x^2) over the interval -1, 1, we first need to determine if the function is even, odd, or neither. Since f(-x) = -x/(1-(-x)^2) = -f(x), the function is odd. Therefore, we only need to find the sine coefficients in the Fourier series. The Fourier series of an odd function is given by: f(x) = ∑b_n * sin(nπx/L) where L is half the length of the interval, which in this case is 1. So, L = 1. To find the coefficients b_n, we use the following formula: b_n = (2/L) * ∫f(x) * sin(nπx/L) dx from -L to L For our function, this becomes: b_n = 2 * ∫x/(1-x^2) * sin(nπx) dx from -1 to 1 Now, we need to solve this integral. We can use integration by parts: Let u = x/(1-x^2) and dv = sin(nπx) dx Then, du = (1 + x^2) / (1 - x^2)^2 dx and v = -1/(nπ) * cos(nπx) Using integration by parts formula: ∫u * dv = u * v - ∫v * du So, b_n = 2 * {-x/(1-x^2) * (-1/(nπ) * cos(nπx))|(-1)^1 - ∫-1/(nπ) * cos(nπx) * (1 + x^2) / (1 - x^2)^2 dx from -1 to 1} Now, we evaluate the first term: -x/(1-x^2) * (-1/(nπ) * cos(nπx)) at x = 1: 0 -x/(1-x^2) * (-1/(nπ) * cos(nπx)) at x = -1: 0 So, the first term is 0. Now, we need to solve the remaining integral: b_n = 2 * ∫1/(nπ) * cos(nπx) * (1 + x^2) / (1 - x^2)^2 dx from -1 to 1 Unfortunately, this integral does not have a simple closed-form solution. However, we can still express the Fourier series of the function using the integral form of the coefficients: f(x) = ∑2 * ∫1/(nπ) * cos(nπx) * (1 + x^2) / (1 - x^2)^2 dx from -1 to 1 * sin(nπx) This is the Fourier series representation of the function f(x) = x/(1-x^2) over the interval -1, 1.

answer:The Fourier series for a function f(x) on the interval -π, π is given by: f(x) = a_0/2 + Σa_n * cos(nx) + b_n * sin(nx) where a_0, a_n, and b_n are the Fourier coefficients, which are given by: a_0 = (1/π) * ∫-π, π f(x) dx a_n = (1/π) * ∫-π, π f(x) * cos(nx) dx b_n = (1/π) * ∫-π, π f(x) * sin(nx) dx For f(x) = x, we have: a_0 = (1/π) * ∫-π, π x dx = 0 since the integral of an odd function over a symmetric interval is 0 a_n = (1/π) * ∫-π, π x * cos(nx) dx = 0 again, since x is an odd function and cos(nx) is an even function, their product is an odd function, and the integral over a symmetric interval is 0 b_n = (1/π) * ∫-π, π x * sin(nx) dx To solve for b_n, we can use integration by parts: Let u = x, dv = sin(nx) dx Then, du = dx, v = (-1/n) * cos(nx) Using integration by parts formula, ∫ u dv = uv - ∫ v du: b_n = (1/π) * ((-x/n) * cos(nx)|-π, π - ∫-π, π (-1/n) * cos(nx) dx) The first term becomes 0 due to the symmetry of the interval: b_n = (1/π) * (1/n) * ∫-π, π cos(nx) dx Now, we can integrate cos, To find the Fourier series for the function f(x) = x on the interval -π, π, we need to compute the coefficients a_n and b_n. Since f(x) is odd, all the a_n coefficients will be zero. The formulas for the Fourier series are as follows: f(x) = (a_0/2) + Σa_n * cos(nx) + b_n * sin(nx) a_0 = (1/π) * ∫-π, π f(x) dx a_n = (1/π) * ∫-π, π f(x) * cos(nx) dx b_n = (1/π) * ∫-π, π f(x) * sin(nx) dx First, let's compute a_0: a_0 = (1/π) * ∫-π, π x dx a_0 = (1/π) * x^2/2 | (-π, π) a_0 = (1/π) * (π^2 - (-π)^2)/2 a_0 = 0 Since f(x) is odd, a_n = 0 for all n. Now let's compute b_n: b_n = (1/π) * ∫-π, π x * sin(nx) dx To solve this integral, we can use integration by parts with u = x and dv = sin(nx) dx. Then du = dx and v = -(1/n)cos(nx). Using integration by parts: ∫ u dv = uv - ∫ v du b_n = (1/π) * -x*(1/n)cos(nx) | (-π, π) - (1/π) * ∫-π, π -(1/n)cos(nx) dx Evaluating the first term: -π*(1/n)cos(n*(-π)) - π*(1/n)cos(n*π) = -2*(π/n)cos(n*π) ,To find the Fourier series for the function f(x) = x on the interval -π, π, we first need to determine the Fourier coefficients a_n and b_n. Since f(x) is an odd function, all a_n coefficients will be zero. We only need to find the b_n coefficients. The general formula for the Fourier series of a function is: f(x) = (a_0 / 2) + Σ a_n * cos(n * x) + b_n * sin(n * x) For an odd function, the Fourier series is simplified to: f(x) = Σ b_n * sin(n * x) Now, let's find the b_n coefficients using the formula: b_n = (1 / π) * ∫-π, π f(x) * sin(n * x) dx For f(x) = x, we have: b_n = (1 / π) * ∫-π, π x * sin(n * x) dx To solve this integral, we can use integration by parts: Let u = x, dv = sin(n * x) dx du = dx, v = (-1 / n) * cos(n * x) Using integration by parts formula: ∫ u dv = u * v - ∫ v du b_n = (1 / π) * (-x / n) * cos(n * x) | (-π, π) + (1 / n) * ∫-π, π cos(n * x) dx The integral of cos(n * x) over the interval -π, π is zero, so the second term vanishes: b_n = (1 / π) * (-x / n) * cos(n * x) | (-π, π) Now, evaluate the expression at the limits: b_n = (1 / π) * ((-π / n) * cos(n * π) - (-(-π / n) * cos(-n * π)) b_n = (1 / π) * ((-1)^n - 1) * π / n Now, we can write the Fourier series for f(x) = x: f(x) = Σ((-1)^n - 1) * sin(n * x) / n This is the Fourier series representation of the function f(x) = x on the interval -π, π with a singularity at x = 0.