Final answer:
To find the second Taylor polynomial for f(x) = ex cos x about x0 = 0, we find the first two derivatives of the function and evaluate them at x = 0. Using the Taylor series formula, the second Taylor polynomial P2(x) is 1 + x^2/2. To approximate f(0.5), we substitute x = 0.5 in P2(x), and the upper bound for the error is e/48.
Step-by-step explanation:
To find the second Taylor polynomial, we need to find the first two derivatives of the function and evaluate them at x = 0. For f(x) = ex cos x, the first derivative is f'(x) = ex (cos x - sin x) and the second derivative is f''(x) = ex (-2 sin x). Evaluating these at x = 0 gives f'(0) = 1 and f''(0) = 0.
Using the Taylor series formula, the second Taylor polynomial P2(x) is given by:
P2(x) = f(0) + f'(0)x + f''(0)x^2/2 = 1 + x^2/2.
a. To approximate f(0.5), we substitute x = 0.5 in P2(x):
P2(0.5) = 1 + (0.5)^2/2 = 1.125.
Using the error formula |f(0.5) - P2(0.5)| <= M/(3!)(0.5)^3, where M is an upper bound for the absolute value of the third derivative, we can find the upper bound for the error. In this case, the third derivative is f'''(x) = ex (-3 cos x), and since |ex (-3 cos x)| <= ex for all x, we can use e as our value for M. Evaluating the error formula gives an upper bound of e/(3!)(0.5)^3 = e/48.