Question Five: a. Define the following: a. Taylor Series b. Maclaurin Series b. Find the Maclaurin Series for the function f(x) = e^x. c. Find the Taylor series for the function f(x) = ln x about x = 2.

$$f(x) = e^x \implies f(0) = e^0 = 1$$ $$f'(x) = e^x \implies f'(0) = e^0 = 1$$ $$f''(x) = e^x \implies f''(0) = e^0 = 1$$

In general, for any $n \ge 0$:

$$f^{(n)}(x) = e^x \implies f^{(n)}(0) = 1$$

The Maclaurin series is $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$.

$$e^x = \frac{f(0)}{0!}x^0 + \frac{f'(0)}{1!}x^1 + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$$ $$e^x = \frac{1}{0!}x^0 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots$$ $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$

The Maclaurin series for $f(x) = e^x$ is $\boxed{\sum_{n=0}^{\infty} \frac{x^n}{n!}}$.

$$f(x) = \ln x \implies f(2) = \ln 2$$ $$f'(x) = x^{-1} \implies f'(2) = 2^{-1} = \frac{1}{2}$$ $$f''(x) = -x^{-2} \implies f''(2) = -2^{-2} = -\frac{1}{4}$$ $$f'''(x) = 2x^{-3} \implies f'''(2) = 2 \cdot 2^{-3} = \frac{2}{8} = \frac{1}{4}$$ $$f^{(4)}(x) = -6x^{-4} \implies f^{(4)}(2) = -6 \cdot 2^{-4} = -\frac{6}{16} = -\frac{3}{8}$$

In general, for $n \ge 1$:

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

The Taylor series is $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$.

$$f(x) = f(2) + \frac{f'(2)}{1!}(x-2) + \frac{f''(2)}{2!}(x-2)^2 + \frac{f'''(2)}{3!}(x-2)^3 + \dots$$ $$f(x) = \ln 2 + \frac{1/2}{1!}(x-2) + \frac{-1/4}{2!}(x-2)^2 + \frac{1/4}{3!}(x-2)^3 + \dots$$ $$f(x) = \ln 2 + \frac{1}{2}(x-2) - \frac{1}{8}(x-2)^2 + \frac{1}{24}(x-2)^3 - \dots$$

For $n \ge 1$, the general term is $\frac{f^{(n)}(2)}{n!}(x-2)^n = \frac{(-1)^{n-1} (n-1)! 2^{-n}}{n!} (x-2)^n = \frac{(-1)^{n-1}}{n \cdot 2^n} (x-2)^n$.
The Taylor series for $f(x) = \ln x$ about $x=2$ is $\boxed{\ln 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 2^n} (x-2)^n}$.