π, π, Archimedes!

This week, another derivation for a famous formula: Archimedes’
formula for π.


Some time in the 3rd century BC, Archimedes used the perimeter of a regular polygon, starting with an hexagon and repeatedly doubling the number of sides, to estimate the value for π. He arrived at the approximation

\displaystyle 3\frac{10}{71}=\frac{223}{71}<\pi<\frac{22}{7}=3\frac{10}{70}.

How he arrived to this result is a bit mysterious until we completely understand how he got that result. Let’s see together how he did it.

First, we must understand what happens to the perimeter when we go from n sides to 2n, assuming that the polygon remains regular (that is, all its sides have the same length). Let’s see what happens, but instead of an hexagon, let’s use a square. So the first approximation is an inscribed square, whose perimeter is 4 r \sqrt{2}, for a circle of radius r. If we show this in a quarter of the disk, we get something like this:


Here, s=r\sqrt{2}. Then, if we split the side in two and push it against the circle, we get two segments of length s'. How do we find this new length? First, let us place the information we need on the diagram:


From this diagram, we conclude we need to find b, because $latex \frac{1}{2}s is known. We have that:

\displaystyle s=r\sqrt{2},

\displaystyle a=\sqrt{r^2-\left(\frac{s}{2}\right)^2},

\displaystyle b=r-a=r-\sqrt{r^2-\left(\frac{s}{2}\right)^2},

and, at last,

\displaystyle s'=\sqrt{\left(\frac{s}{2}\right)^2+b^2},


\displaystyle s'=\sqrt{\left(\frac{s}{2}\right)^2+\left(r-\sqrt{r^2-\left(\frac{s}{2}\right)^2}\right)^2}.

* *

This suggests the recurrence

\displaystyle s_0 = r\sqrt{2},

\displaystyle s_t = \sqrt{\left(\frac{s_{t-1}}{2}\right)^2+\left(r-\sqrt{r^2-\left(\frac{s_{t-1}}{2}\right)^2}\right)^2}.

* *

How ever, that scary-looking BAMF1 simplifies quite a lot, and lets us see a lot of regularity:

\displaystyle s_0 = r\sqrt{2},

\displaystyle s_1 = r\sqrt{2-\sqrt{2}},

\displaystyle s_2 = r\sqrt{2-\sqrt{2+\sqrt{2}}},

\displaystyle s_3 = r\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}},

\displaystyle s_4 = r\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}},


* *

So we know have a formula for the length of the side in an 2^{t+2}-sided regular polygon, but that’s not the perimeter… to find the parameter, we must multiply by the number of sides:

p_t = 2^{t+2}s_t.

* *

If we set r=\frac{1}{2}, we get a perimeter ever closer to the actual circumference, and we get a better and better approximation for π. However, it’s doesn’t converge all that fast:


After 10 iterations, we have only 6 digits after the point, and at 20, 12.

* *

Let’s take a step back and consider how much of an achievement it was in Archimedes’ time. They did not have convenient notation, they did not have a good number system, and they had no computer. Archimedes gave up after four iterations (because he used an 96-sided polygon starting from an hexagon, he must have divided it four times, since indeed 96=6\times{2^4}!)

1 For bombastic and magniloquent formula, of course. What did you expect?

8 Responses to π, π, Archimedes!

  1. TheMachine says:

    I wonder if 4 iterations’ accuracy was good enough for “practical” uses of the constant back in his time, or if the error in his calculation would’ve ever caused any issues.

    • I suppose it depends on scale. Archimedes’ result 22/7 is 3.142857 (and decimals repeat), while the fourth power-of-two iteration gives 3.14033… both are about as precise, a relative error of ±0.0004 or so. However, my personal favorite is


      which gives 6 digits precision. It is due to Zǔ Chōngzhī (祖沖之), (429–500 AD) : https://en.wikipedia.org/wiki/Zu_Chongzhi

      However, we must be aware that we can’t reason about this approximation as we usually do: we know about π with great precision, we know about its properties. Archimedes didn’t. He was exploring, he was trying to figure out things. While it was clear for him that it was a constant, I’m not sure he understood it was an “incommensurable”.

      • TheMachine says:

        Yeah, that’s a fair point. I suppose I was thinking in reverse, pondering if the inaccuracy would have ever creeped up in *practical* use. However, giving it some more thought, I highly doubt that it would have, for example, caused some error in an architectural calculation. Say that an architect used his calculated constant to calculate circumference of a circle to lay a border with a marking (rope or something) and he had noticed his calculated length of rope was insufficient or in excess. In other words, would the slight inaccuracy have ever been noticed based on the “real” value. Given the relative precision, I guess it’s more fun to think about the times it may have gone unnoticed.

        Sorry for the scrambled thoughts though. I’m usually reading your blog at awful hours with no caffeine in my system either haha.

  2. […] course, we could go with a special, ad hoc procedure for each constant. Just as Archimedes used a 96-sided polygon to get […]

  3. assaf says:

    there is a typo in the s’ formula – the square was ‘dropped’ when b was replaced.

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