Final answer:
The Taylor series for f(x) = cos(x) + 1 centered at x = π can be found using the Maclaurin series for cos(x) and the general form of the Taylor series. The series is f(x) = -cos(π)/2!)(x - π)^2 + (sin(π)/3!)(x - π)^3 + ...
Step-by-step explanation:
The Taylor series for a function f(x) centered at x = a can be found using the Maclaurin series and the general form of the Taylor series. The Maclaurin series for cos(x) is given by:
cos(x) = 1 - (x^2/2!) + (x^4/4!) - (x^6/6!) + ...
Now, we can use the fact that -cos(x) = cos(x - π) to find the Taylor series for f(x) = cos(x) + 1 centered at x = π. The general form of a Taylor series is:
f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3 + ...
For x = π, a = π, and f(a) = f(π) = cos(π) + 1 = -1 + 1 = 0. Differentiating f(x) = cos(x) + 1 with respect to x, we get f'(x) = -sin(x). Evaluating f'(π), we get f'(π) = -sin(π) = 0. Similarly, f''(x) = -cos(x), f'''(x) = sin(x), and so on. Substituting these values into the general form of the Taylor series, we get:
f(x) = 0 + 0(x - π) + (-cos(π)/2!)(x - π)^2 + (sin(π)/3!)(x - π)^3 + ...
Simplifying this expression, we have:
f(x) = -cos(π)/2!)(x - π)^2 + (sin(π)/3!)(x - π)^3 + ...