Consider the following finite-difference scheme for approximating the derivative of a function f(x)f'(x)≈a*f(x)+b*f(x+h)+c*f(x+2h)+d*f(x+3h)/h(2.1) Apply Taylor’s (Lagrange remainder) theorem about the point x to write down expressions for f(x + h), f(x + 2h) and f(x + 3h) in terms of h, f(x), f'(x), f''(x), etc, with remainder terms thatare O(h^4).(2.2) Substitute these expressions into (2.1) and hence determine a system of linear equations that must be satisfied by the coefficients a, b, c and d for the error of (2.1) to be O(h^3) as h → 0.(2.3) Solve the system from part (2.2) for the coefficients a, b, c and d. You may use technology to solve the system, but the answers must be exact.(2.4) Use the finite-difference formula (2.1) together with the coefficients you found in part (2.3) to estimate the derivative of f(x) = e^x at x = 0 for h = 0.5. What is the magnitude of the error in this case? How much smaller would h need to be to decrease the error by a factor of 1000?

(2.1) The given finite-difference scheme for approximating the derivative of a function f(x) is:

f'(x) ≈ af(x) + bf(x+h) + cf(x+2h) + df(x+3h) / h

(2.2) Applying Taylor’s (Lagrange remainder) theorem about the point x, we can write down expressions for f(x + h), f(x + 2h) and f(x + 3h) in terms of h, f(x), f'(x), f''(x), etc, with remainder terms that are O(h^4).

f(x + h) = f(x) + hf'(x) + h^2/2f''(x) + h^3/6*f'''(x) + O(h^4)

f(x + 2h) = f(x) + 2hf'(x) + 2^2h^2/2f''(x) + 2^3h^3/6*f'''(x) + O(h^4)

f(x + 3h) = f(x) + 3hf'(x) + 3^2h^2/2f''(x) + 3^3h^3/6*f'''(x) + O(h^4)

(2.3) Substituting these expressions into (2.1), we get:

f'(x) ≈ [af(x) + b(f(x) + hf'(x) + h^2/2f''(x) + h^3/6f'''(x)) + c(f(x) + 2hf'(x) + 2^2h^2/2f''(x) + 2^3h^3/6f'''(x)) + d(f(x) + 3hf'(x) + 3^2h^2/2f''(x) + 3^3h^3/6*f'''(x))] / h

This simplifies to:

f'(x) ≈ [a + b + c + d]f(x) + h[b + 2c + 3d]f'(x) + h^2[b/2 + 2c + 9d/2]f''(x) + h^3[b/6 + 4c/3 + 9d/2]*f'''(x)

For the error of (2.1) to be O(h^3) as h → 0, the coefficients a, b, c and d must satisfy the following system of linear equations:

a + b + c + d = 0 b + 2c + 3d = 1 b/2 + 2c + 9d/2 = 0 b/6 + 4c/3 + 9d/2 = 0

(2.4) Solving this system of equations, we get:

a = -3/2, b = 2, c = -3/2, d = 1

Using the finite-difference formula (2.1) together with these coefficients to estimate the derivative of f(x) = e^x at x = 0 for h = 0.5, we get:

f'(0) ≈ (-3/2)e^0 + 2e^0.5 - 3/2*e^1 + e^1.5 / 0.5 = 1.297

The exact derivative of f(x) = e^x at x = 0 is 1, so the error is |1 - 1.297| = 0.297.

To decrease the error by a factor of 1000, h would need to be approximately 0.0005.