Originally Posted by
topsquark A Maclauren series for a function is just the Taylor series at an argument of 0.
So for $\displaystyle f(x) = ln(1 + x)$:
$\displaystyle f(x) = ln(1 + x) \implies f(0) = 0$
$\displaystyle f^{\prime}(x) = \frac{1}{1 + x} \implies f^{\prime}(0) = 1$
$\displaystyle f^{\prime \prime}(x) = -\frac{1}{(1 + x)^2} \implies f^{\prime \prime}(0) = -1$
etc.
So
$\displaystyle f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + \frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ... $
$\displaystyle f(x) \approx 0 + \frac{1}{1} \cdot 1 \cdot x + \frac{1}{2} \cdot (-1) \cdot x^2 + ~ ...$
$\displaystyle f(x) \approx x - \frac{1}{2}x^2 + ~ ...$
-Dan