how would you do this?
Find the Taylor series for ln(x) about 2.
Just for the record, let $\displaystyle f(x) = ln(x)$
$\displaystyle f(x) = ln(x)$ ==> $\displaystyle f(2) = ln(2)$
$\displaystyle f^{\prime}(x) = \frac{1}{x}$ ==> $\displaystyle f^{\prime}(2) = \frac{1}{2}$
$\displaystyle f^{\prime \prime}(x) = -\frac{1}{x^2}$ ==> $\displaystyle f^{\prime \prime}(2) = -\frac{1}{4}$
etc.
So
$\displaystyle f(x) \approx ln(2) + \frac{1}{1!}\frac{1}{2}(x - 2) - \frac{1}{2!}\frac{1}{4}(x - 2)^2 + ~ ...$
$\displaystyle f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...$
-Dan
Crap! I goofed!
The general form for the Taylor series of f(x) about x = a is:
$\displaystyle f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + ~ ... ~ + \frac{1}{n!}f^{(n)}(a)(x - a)^n + ~ ...$
So for a = 2 the first coefficient is
$\displaystyle \frac{1}{1!}f^{\prime}(2) = 1 \cdot \frac{1}{2} = \frac{1}{2}$
The second coefficient is
$\displaystyle \frac{1}{2!}f^{\prime \prime}(2) = \frac{1}{2} \cdot -\frac{1}{4} = -\frac{1}{8}$
So the series should be:
$\displaystyle f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...$
You can work the general formula for the nth term. It's easy enough.
-Dan