# How do I find a natural log without a calculator?

Nov 4, 2015

You can approximate $\ln x$ by approximating ${\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$ using Riemann sums with the trapezoidal rule or better with Simpson's rule.

#### Explanation:

For example, to approximate $\ln \left(7\right)$, split the interval $\left[1 , 7\right]$ into a number of strips of equal width, and sum the areas of the trapezoids with vertices:

$\left({x}_{n} , 0\right)$, $\left({x}_{n} , \frac{1}{x} _ n\right)$, $\left({x}_{n + 1} , 0\right)$, $\left({x}_{n + 1} , \frac{1}{{x}_{n + 1}}\right)$

If we use strips of width $1$, then we get six trapezoids with average heights: $\frac{3}{4}$, $\frac{5}{12}$, $\frac{7}{24}$, $\frac{9}{40}$, $\frac{11}{60}$, $\frac{13}{84}$.

If you add these, you find:

$\frac{3}{4} + \frac{5}{12} + \frac{7}{24} + \frac{9}{40} + \frac{11}{60} + \frac{13}{84} \approx 2.02$

If we use strips of width $\frac{1}{2}$, then we get twelve trapezoids with average heights: $\frac{5}{6}$, $\frac{7}{12}$, $\frac{9}{20}$,...,$\frac{25}{156}$,$\frac{27}{182}$

Then the total area is:

$\frac{1}{2} \times \left(\frac{5}{6} + \frac{7}{12} + \frac{9}{20} + \ldots + \frac{25}{156} + \frac{27}{182}\right) \approx 1.97$

Actually $\ln \left(7\right) \approx 1.94591$, so these are not particularly accurate approximations.

Simpson's rule approximates the area under a curve using a quadratic approximation. For a given $h > 0$, the area under the curve of $f \left(x\right)$ between ${x}_{0}$ and ${x}_{0} + 2 h$ is given by:

$\frac{h}{3} \left(f \left({x}_{0}\right) + 4 f \left({x}_{0} + h\right) + f \left({x}_{0} + 2 h\right)\right)$

Try improving the approximation using Simpson's rule, using $h = \frac{1}{2}$ then we get six approximate areas to sum:

$\frac{h}{3} \left(\frac{25}{6} + \frac{73}{30} + \frac{145}{84} + \frac{241}{180} + \frac{361}{330} + \frac{505}{546}\right) \approx 1.947$

That's somewhat better.