Approximating the natural logarithm

If we define $f:x\rightarrow(-\infty,1)\rightarrow (-\infty,\infty)$ by $f(x) \ln(1-x)$ and look at the definition of derivative for f'(0) we see that

$$f'(0) = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} = \lim_{h\rightarrow 0}\frac{\ln(1-h)}{h}$$

On the other hand, we can compute f'(0) from the chain rule and get f'(0) = -1 so we can conclude that for |x|<<1 that

$$\frac{\ln(1-x)}{x} \approx -1$$

or

$$\ln(1-x) \approx -x$$

or

$$\ln(1-x) + x\approx 0$$

when |x| <<1. If we want to improve our approximation this time we should try comparing with x² for $\ln(1-x) + x$ is neither even nor odd. Hence we must consider the behaviour of

$$\frac{\ln(1-x) + x}{x^2}$$

as x approaches . L'Hopital's rule can be tried here. The ratio of the derivatives is

$$\frac{\frac{-1}{1-x} + 1}{2x} = -\frac{1}{2}\frac{1}{x-1}$$

which has a limit of -1/2 as x approaches . Therefore

$$\lim_{x\rightarrow 0}\frac{\ln(1-x) + x}{x^2} = -\frac{1}{2}$$

and we may conclude that for |x|<<1 that

$$\ln(1-x) \approx -x - \frac{1}{2}x^2$$

Here is a picture of $\ln(1-x) - (-x - \frac{1}{2}x^2)$:

You should not be too impressed. We can repeat the procedure twice more and get $\ln(1-x) \approx -x - x^2/2 - x^3/3 - x^4/4$. Then the differences look like

and you see we are doing much better for negative values of x than for the positive values. Why?

We shall discuss at a later time how to tell just how good these approximations are.