answer:The five-point formula for the second derivative is given by: f''(x) ≈ (f(x-2h) - 2f(x-h) + 2f(x+h) - f(x+2h)) / (2h^2) We are given the function f(x) = cos(x), the point x = π/4, and the step size h = 0.1. Let's calculate the second derivative using the five-point formula: f''(π/4) ≈ (cos(π/4 - 2*0.1) - 2cos(π/4 - 0.1) + 2cos(π/4 + 0.1) - cos(π/4 + 2*0.1)) / (2 * 0.1^2) Now, we need to calculate the cosine values for each of the points: cos(π/4 - 2*0.1) ≈ cos(0.7854 - 0.2) ≈ cos(0.5854) ≈ 0.8378 cos(π/4 - 0.1) ≈ cos(0.7854 - 0.1) ≈ cos(0.6854) ≈ 0.7648 cos(π/4 + 0.1) ≈ cos(0.7854 + 0.1) ≈ cos(0.8854) ≈ 0.6361 cos(π/4 + 2*0.1) ≈ cos(0.7854 + 0.2) ≈ cos(0.9854) ≈ 0.5567 Now, plug these values into the five-point formula: f''(π/4) ≈ (0.8378 - 2*0.7648 + 2*0.6361 - 0.5567) / (2 * 0.1^2) f''(π/4) ≈ (-0.1804) / 0.02 f''(π/4) ≈ -9.02 So, the estimated second derivative of f(x) = cos(x) at x = π/4 using the five-point formula with a step size of h = 0.1 is approximately -9.02.

answer:The central finite difference method is a numerical method to approximate the first derivative of a function. The formula for the central finite difference method is: f'(x) ≈ (f(x + h) - f(x - h)) / (2h) Given the function f(x) = cos(x), x = pi/4, and h = 0.1, we can plug these values into the formula: f'(pi/4) ≈ (cos(pi/4 + 0.1) - cos(pi/4 - 0.1)) / (2 * 0.1) Now, we can calculate the values: cos(pi/4 + 0.1) ≈ cos(0.886) ≈ 0.636396 cos(pi/4 - 0.1) ≈ cos(0.686) ≈ 0.769286 Plugging these values back into the formula: f'(pi/4) ≈ (0.636396 - 0.769286) / (2 * 0.1) ≈ -0.13289 / 0.2 ≈ -0.66445 So, the first derivative of the function f(x) = cos(x) at x = pi/4 using the central finite difference method is approximately -0.66445. Now, let's approximate the error using the formula for the truncation error. The truncation error for the central finite difference method is given by: Error ≈ (h^2 / 6) * f'''(c) where c is a value between x - h and x + h. Since the third derivative of cos(x) is sin(x), we can write: Error ≈ (h^2 / 6) * sin(c) We know that h = 0.1 and sin(c) is bounded by -1 and 1, so the maximum error can be: Error ≈ (0.1^2 / 6) * 1 ≈ 0.001667 Thus, the approximate error in our answer using the central finite difference method is 0.001667.

answer:To find the numerical approximation of the derivative of f(x) = 5x^3 - 2x^2 + 3x - 1 at x = 2 using the forward, backward, and central difference methods with a step size of h = 0.1, we'll first define the function and then apply each method. Function: f(x) = 5x^3 - 2x^2 + 3x - 1 Forward difference method: f'(x) ≈ (f(x + h) - f(x)) / h f'(2) ≈ (f(2.1) - f(2)) / 0.1 f(2.1) = 5(2.1)^3 - 2(2.1)^2 + 3(2.1) - 1 = 47.259 f(2) = 5(2)^3 - 2(2)^2 + 3(2) - 1 = 33 f'(2) ≈ (47.259 - 33) / 0.1 = 142.59 Backward difference method: f'(x) ≈ (f(x) - f(x - h)) / h f'(2) ≈ (f(2) - f(1.9)) / 0.1 f(1.9) = 5(1.9)^3 - 2(1.9)^2 + 3(1.9) - 1 = 20.759 f'(2) ≈ (33 - 20.759) / 0.1 = 122.41 Central difference method: f'(x) ≈ (f(x + h) - f(x - h)) / (2h) f'(2) ≈ (f(2.1) - f(1.9)) / 0.2 f'(2) ≈ (47.259 - 20.759) / 0.2 = 132.5 Now, let's find the exact derivative and the error for each method: Exact derivative: f'(x) = 15x^2 - 4x + 3 f'(2) = 15(2)^2 - 4(2) + 3 = 135 Error estimation: Forward difference error: |142.59 - 135| = 7.59 Backward difference error: |122.41 - 135| = 12.59 Central difference error: |132.5 - 135| = 2.5 So, the numerical approximations of the derivative at x = 2 using the forward, backward, and central difference methods are 142.59, 122.41, and 132.5, respectively, with errors of 7.59, 12.59, and 2.5.

answer:First, let's solve the derivative using the central difference method. The formula for the central difference method is: f'(x) ≈ (f(x + h) - f(x - h)) / (2 * h) Given f(x) = sin(x), x = 0.5, and h = 0.1, we have: f'(0.5) ≈ (sin(0.5 + 0.1) - sin(0.5 - 0.1)) / (2 * 0.1) f'(0.5) ≈ (sin(0.6) - sin(0.4)) / 0.2 f'(0.5) ≈ (0.564642 - 0.389418) / 0.2 f'(0.5) ≈ 0.87512 Now, let's estimate the integral using the trapezoidal rule. The formula for the trapezoidal rule is: ∫(a to b) f(x) dx ≈ (h / 2) * (f(a) + 2 * Σf(x_i) + f(b)) Given f(x) = sin(x), a = 0, b = 1, and h = 0.1, we have: ∫(0 to 1) sin(x) dx ≈ (0.1 / 2) * (sin(0) + 2 * (sin(0.1) + sin(0.2) + sin(0.3) + sin(0.4) + sin(0.5) + sin(0.6) + sin(0.7) + sin(0.8) + sin(0.9)) + sin(1)) ∫(0 to 1) sin(x) dx ≈ (0.1 / 2) * (0 + 2 * (0.09983 + 0.19867 + 0.29552 + 0.38942 + 0.47943 + 0.56464 + 0.64422 + 0.71736 + 0.78333) + 0.84147) ∫(0 to 1) sin(x) dx ≈ (0.1 / 2) * (0 + 2 * 4.37242 + 0.84147) ∫(0 to 1) sin(x) dx ≈ (0.1 / 2) * 10.08629 ∫(0 to 1) sin(x) dx ≈ 0.50431 So, the approximate derivative of f(x) = sin(x) at x = 0.5 using the central difference method is 0.87512, and the approximate integral of f(x) over the interval [0, 1] using the trapezoidal rule is 0.50431.