Question

For the following function, find the Taylor series centered at x = π and then give the first 5 nonzero terms of the Taylor series and the open interval of convergence.

f(x) = cos(x).

Answer 1

To find the Taylor series centered at x = π for the function f(x) = cos(x), use the formula for the Taylor series and plug in the values for f(x) and its derivatives. The first 5 nonzero terms of the series are -1, 0(x - π)^2/2!, 0(x - π)^4/4!, 0(x - π)^6/6!, 0(x - π)^8/8!, and the open interval of convergence is (-∞, ∞).

Step-by-step explanation:

To find the Taylor series centered at x = π for the function f(x) = cos(x), we need to use the formula for the Taylor series:

Taylor series = f(c) + f'(c)(x - c)/1! + f''(c)(x - c)^2/2! + f'''(c)(x - c)^3/3! + ...

For f(x) = cos(x), we have f(π) = cos(π) = -1, f'(x) = -sin(x), f''(x) = -cos(x), f'''(x) = sin(x), and so on. Plugging these values into the Taylor series formula gives:

cos(x) = -1 + 0(x - π)/1! + (-1)(x - π)^2/2! + 0(x - π)^3/3! + ...

The first 5 terms of this series are: -1 + 0(x - π)^2/2! - 0(x - π)^4/4! + 0(x - π)^6/6! - 0(x - π)^8/8!.

The open interval of convergence for this series is (-∞, ∞) since the Taylor series for cos(x) converges for all real numbers x.