Question

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that Rn(x) -> 0.] Find the associated radius of convergence R. f(x) = 2(1 - x)-2

Answer 1

To find the Maclaurin series for f(x) = 2(1 - x)^-2, we can use the definition of a Maclaurin series. The Maclaurin series is f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ..., where f'(x), f''(x), f'''(x), etc. are the derivatives of f(x). Substituting the derivatives of f(x) into the Maclaurin series, we find that the Maclaurin series for f(x) is 2 + 4x + 12x^2 + 48x^3 + .... The associated radius of convergence R is infinite, meaning the Maclaurin series converges for all values of x.

Step-by-step explanation:

To find the Maclaurin series for f(x) = 2(1 - x)^{-2}, we can use the definition of a Maclaurin series. The Maclaurin series is given by:

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

First, let's find the derivatives of f(x):

f'(x) = 4(1 - x)^{-3}

f''(x) = 12(1 - x)^{-4}

f'''(x) = 48(1 - x)^{-5}

Substituting these derivatives into the Maclaurin series, we get:

f(x) = 2 + 4x + 12x^2 + 48x^3 + ...

The associated radius of convergence R is infinite, which means the Maclaurin series converges for all values of x.

Answer 2

Hello from MrBillDoesMath!

Answer:

When I submitted the answer in the attachment I got "It appears that your answer contains either a link or inappropriate words. Please correct and submit again!"

Baloney! Please review the attachment which contains a screenshot of what I worte.

MrB.