Higher Order Derivatives — Question 6

Question 6 - Solution

We are given: $$f(x) = \frac{1}{1 - x} = (1 - x)^{-1}$$

(a) Compute derivatives:

Using the chain rule, $$\frac{d}{dx}(1 - x)^{-1} = -1(1 - x)^{-2} \cdot (-1) = (1 - x)^{-2}$$ $$f'(x) = \frac{1}{(1 - x)^2}$$

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

$$f^{(3)}(x) = \frac{d}{dx}\left( \frac{2}{(1 - x)^3} \right) = 2 \cdot 3 (1 - x)^{-4} = \frac{6}{(1 - x)^4}$$

$$f^{(4)}(x) = \frac{d}{dx}\left( \frac{6}{(1 - x)^4} \right) = 6 \cdot 4 (1 - x)^{-5} = \frac{24}{(1 - x)^5}$$

(b) General formula:

Each derivative introduces a factorial factor and increases the power of the denominator by 1: $$\boxed{f^{(n)}(x) = \frac{n!}{(1 - x)^{n + 1}}}$$

(c) Evaluate at $x = 0$:

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

Thus the Taylor series centered at $x = 0$ is: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} x^n$$

$$\boxed{f(x) = \sum_{n=0}^{\infty} x^n \quad \text{for } |x| < 1}$$