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Time for Pi

Time for Pi


Drinking strong coffee and thinking about binomial series.

Whilst writing various binomial series approximations, (which can be found here : Binomial Approximations) I started thinking about how $\pi$ can be approximated.


A circle, with radius 1 and centred on the origin, has equation $x^2 + y^2 = 1$. The quarter of the curve in the first quadrant has equation $y = \sqrt{1 - x^2}$. The area contained between the axes and this arc is

$$A = \int_0^1 \sqrt{1 - x^2} dx = \frac{\pi}{4}$$

We can now use the binomial series to give an approximation to the function $\sqrt{1-x^2}$.

$$\sqrt{1 - x^2} = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{16}x^6 + \dots$$

The enclosed area, can now be estimated by integrating this series

$$ \begin{align} A &= \int_0^1 \sqrt{1 - x^2} dx \\ &\approx \int_0^1 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{16}x^6 dx \\ &= \left[x - \frac{x^3}{6} - \frac{x^5}{40} - \frac{x^7}{112}\right]_0^1 \\ &= 0.799405\end{align} $$

This gives an estimated value of pi of $4\times0.799405 = 3.198$, which is a particularly poor estimate. Even if we extend our approximation to 25 or so terms, we still only get an answer accurate to 3 decimal places. This should be a little unsurprising as we are using an approximation up to the end of the radius of convergence and so our rate of convergence will be pitifully slow.

We can however, using the same series derive a system which converges much quicker, without too much additional work.

From the second diagram, we can see that the area of the sector contained between $(0,1)$, $(0,0)$ and $P\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ is $\frac{\pi}{12}$. We can also see that $$ \int_0^{\frac{1}{2}}\sqrt{1-x^2}dx = \frac{\pi}{12} + \frac{\sqrt{3}}{8} $$

A bit of rearrangement gives us $$\frac{\pi}{12} =  \int_0^{\frac{1}{2}}\sqrt{1-x^2}dx  - \frac{\sqrt{3}}{8}    $$

Using our earlier approximation gives 

$$ \frac{\pi}{12} \approx \left[x - \frac{x^3}{6} - \frac{x^5}{40} - \frac{x^7}{112}\right]_0^{\frac{1}{2}}  - \frac{\sqrt{3}}{8}  \approx 3.1417 $$

Which gives us an answer correct to three decimal places. This can now be rapidly enhanced by using additional terms in the expansion and as each alternate coefficient in the expansion is zero, we should get a good rate of convergence for each additional term. 

Taking a ten term expansion, for example, gives us an approximation to $\pi$, accurate to nine decimal places.