Question

find the maclaurin series for \\(f(x)\\) using the definition of a maclaurin series. assume that \\(f\\) has a power series expansion, but you do not need to show \\(r_n(x) \to 0\\).

\\f(x) = 7(1 - x)^{-2}\\

\\f(x) = \sum_{n = 0}^{\infty} \left( \dots \
ight)\\

find the associated radius of convergence \\(r\\).

\
= 1\\

Explanation:

⚡ Using what you learned: Taylor and Maclaurin Series

Step 1: Find derivatives of \( f(x) \)

$$ f(x) = 7(1 - x)^{-2} $$

$$ f'(x) = 7 \cdot (-2)(1 - x)^{-3} \cdot (-1) = 7 \cdot 2(1 - x)^{-3} $$

$$ f''(x) = 7 \cdot 2 \cdot 3(1 - x)^{-4} $$

$$ f^{(n)}(x) = 7 \cdot (n+1)!(1 - x)^{-(n+2)} $$

Step 2: Evaluate derivatives at \( x = 0 \)

$$ f(0) = 7 $$

$$ f'(0) = 7 \cdot 2 $$

$$ f''(0) = 7 \cdot 6 $$

$$ f^{(n)}(0) = 7(n+1)! $$

Step 3: Write the Maclaurin series

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $$

$$ f(x) = \sum_{n=0}^{\infty} \frac{7(n+1)!}{n!} x^n $$

$$ f(x) = \sum_{n=0}^{\infty} 7(n+1)x^n $$

Answer:

$$ 7(n+1)x^n $$