Question

F (0) = 0

f'(0) = 1
f(ⁿ⁺¹)(0)=(−n)fⁿ(0) for all n≥1
A function f has derivatives of all orders for −1 (b) Determine whether the Maclaurin series described in Part A converges absolutely, converges conditionally, or diverges at x=1. Explain your reasoning

Answer 1

The Maclaurin series converges absolutely at x=1.

Step-by-step explanation:

The Maclaurin series described in Part A converges absolutely at x=1. In order to determine this, we need to use the Taylor series expansion of the function f(x) around x=0. Since f(x) has derivatives of all orders for x=-1 (as mentioned in the question), we can write the Maclaurin series as:

f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...

For the given function, we have f(0)=0 and f'(0)=1. Therefore, the Maclaurin series is:

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

This series converges absolutely at x=1 because the absolute value of each term in the series approaches 0 as the term number increases.