# Some approximations for $\arccos(1/(1+x))$

I was trying to calculate the maximum ground distance you can see on mountains, with your elvation given.

After some simple geometry, I was able to come up with the following formula:

Let $h$ be your elevation, $d(h)$ be the maximal distance you can see, then

$$d(h)=2\pi R\arccos\frac{R}{R+h}$$

where R is the radius of earth. We take $R=6378100m$ as its value.

But when I plot it in excel, here's what I got: The unit for the vertical axis is km while the unit for the horizontal axis is m.

Amazingly, for $d\in(0,20000m)$ (essentially the maximum elevation you can achieve without paying millions to board a spaceship) , $d(h)$ can be approximated by

$$d(h)\approx22345\sqrt{h}$$

here

with a $r^{2}$ value of 1!.

Only when $h>5\cdot10^{5}m$ dose the the approximation begin to deviate away significantly.

Does any one have a explanation of this from a numerical prespective?

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The Taylor series for the cosine is $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dotsb$$

Truncating this series after the $x^2$ term gives the rather good approximation $$\cos x \approx 1 - \frac12 x^2,$$ Substituting $\sqrt{2x}$ for $x$ we get $$\arccos (1-x) \approx \sqrt{2x}.$$

Since $\dfrac{1}{1+x} = 1 - \dfrac{x}{1+x} \approx 1 - x$ for small values of x, this is also a good approximation for $$\arccos \dfrac{1}{1+x}$$.

To verify this approximation, consider that original equation

$$d(h)=2\pi R\arccos\frac{R}{R+h}$$

$$=2\pi R\arccos\frac{1}{1+\frac{h}{R}}$$

Since $h/R$ is small in this case,

$$d(h)=\approx2\pi R\sqrt{2h/R}$$

$$=\left(\pi\sqrt{8R}\right)\sqrt{h}$$

$$=22441\sqrt{h}$$

Which only differs from Excel's calculation of the coefficient by $0.42\%$

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And for the expansion to make sense, $x = h/R$ must be small. (+1) –  user26872 Jul 4 '12 at 18:40