Determine the Taylor Series expansion of sin(2x) about base a = 0 (Maclaurin series)

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a) Determine the Taylor Series expansion of sin(2x) about base a = 0 (Maclaurin series). Show the process how you derive the series below. Use backside of page if needed.
b) Write a MATLAB script to calculate the value of sin(2x) for x = pi/4 using 6 terms. Display the difference between the above approximation and the exact value by function sin.

The Correct Answer and Explanation is :

Part (a): Taylor Series Expansion of sin⁡(2x)\sin(2x)

The Taylor Series expansion of a function f(x)f(x) about a=0a = 0 is given by: f(x)=f(0)+f′(0)x+f′′(0)x22!+f′′′(0)x33!+…f(x) = f(0) + f'(0)x + \frac{f”(0)x^2}{2!} + \frac{f”'(0)x^3}{3!} + \dots

For sin⁡(2x)\sin(2x), f(x)=sin⁡(2x)f(x) = \sin(2x), so we calculate its derivatives:

  1. f(x)=sin⁡(2x)f(x) = \sin(2x)
  2. f′(x)=2cos⁡(2x)f'(x) = 2\cos(2x)
  3. f′′(x)=−4sin⁡(2x)f”(x) = -4\sin(2x)
  4. f′′′(x)=−8cos⁡(2x)f”'(x) = -8\cos(2x)
  5. f(4)(x)=16sin⁡(2x)f^{(4)}(x) = 16\sin(2x)
  6. f(5)(x)=32cos⁡(2x)f^{(5)}(x) = 32\cos(2x)

Evaluate each derivative at x=0x = 0:

  1. f(0)=sin⁡(0)=0f(0) = \sin(0) = 0
  2. f′(0)=2cos⁡(0)=2f'(0) = 2\cos(0) = 2
  3. f′′(0)=−4sin⁡(0)=0f”(0) = -4\sin(0) = 0
  4. f′′′(0)=−8cos⁡(0)=−8f”'(0) = -8\cos(0) = -8
  5. f(4)(0)=16sin⁡(0)=0f^{(4)}(0) = 16\sin(0) = 0
  6. f(5)(0)=32cos⁡(0)=32f^{(5)}(0) = 32\cos(0) = 32

Substitute into the Taylor Series formula up to 6 terms: sin⁡(2x)≈0+2x−8×33!+32×55!=2x−4×33+4×515\sin(2x) \approx 0 + 2x – \frac{8x^3}{3!} + \frac{32x^5}{5!} = 2x – \frac{4x^3}{3} + \frac{4x^5}{15}

Part (b): MATLAB Script

% MATLAB script to calculate sin(2x) using 6 terms of Taylor Series
x = pi / 4; % Value of x
n_terms = 6;

% Compute Taylor Series approximation
taylor_approx = 0;
for n = 0:n_terms-1
    coefficient = (-1)^n * (2^(2*n+1)) / factorial(2*n+1);
    taylor_approx = taylor_approx + coefficient * x^(2*n+1);
end

% Exact value
exact_value = sin(2 * x);

% Difference
difference = abs(exact_value - taylor_approx);

% Display results
fprintf('Taylor Series Approximation: %.8f\n', taylor_approx);
fprintf('Exact Value using sin function: %.8f\n', exact_value);
fprintf('Difference: %.8f\n', difference);

Explanation (300 Words)

The Taylor Series expansion provides an approximation of a function as a polynomial, making it computationally efficient to evaluate certain functions. In this problem, we derived the Maclaurin series for sin⁡(2x)\sin(2x), which is a special case of the Taylor series expanded about x=0x = 0. By calculating the derivatives of sin⁡(2x)\sin(2x) and evaluating them at x=0x = 0, we established the first six terms of the series: 2x−4×33+4x5152x – \frac{4x^3}{3} + \frac{4x^5}{15}.

The MATLAB script implements this series to approximate sin⁡(2x)\sin(2x) for x=π4x = \frac{\pi}{4}. It uses a loop to calculate the terms up to n=5n = 5, where the coefficients are determined based on the pattern derived from the Taylor expansion. The exact value of sin⁡(2x)\sin(2x) is calculated using MATLAB’s built-in sin function, providing a reference to compare the approximation.

The difference between the approximation and the exact value is calculated as the absolute difference. This highlights how accurately the Taylor series approximates the function within a specific range of xx. For x=π4x = \frac{\pi}{4}, the error is small, showcasing the effectiveness of the Taylor series. The approximation improves as more terms are included, but diminishing returns set in due to floating-point limitations in computations.

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