Final answer:
The Taylor series expansion of the function f(x) = e^(-x) about the point a = 1 is e^(-1) + (-e^(-1))(x-1) + e^(-1)(x-1)^2/2! + e^(-1)(x-1)^3/3! + ... The radius of convergence is 0 and the interval of convergence is x = 1.
Step-by-step explanation:
The Taylor series expansion of the function f(x) = e^(-x) about the point a = 1 can be found by taking derivatives of the function at a and plugging them into the Taylor series formula. The formula is given by:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
For the function f(x) = e^(-x) about the point a = 1, the derivatives can be found as:
f(a) = e^(-1)
f'(a) = -e^(-1)
f''(a) = e^(-1)
Substituting these values into the formula gives:
f(x) = e^(-1) + (-e^(-1))(x-1) + e^(-1)(x-1)^2/2! + e^(-1)(x-1)^3/3! + ...
The radius of convergence of the series is the distance from the center of the series (a) to the nearest point where the series converges, which is the point x = a. So, in this case, the radius of convergence is 0. The interval of convergence is the range of values of x for which the series converges. Since the radius of convergence is 0, the interval of convergence is just the point x = a, which in this case is x = 1.