Oracle Scratchpad

March 30, 2012


Filed under: Philosophy — Jonathan Lewis @ 5:56 pm GMT Mar 30,2012

Here’s a wonderful lesson from Cary Millsap – be very careful if you ever want to sell him anything – that reminded me of a Powerpoint slide I had produced for a presentation a few years ago. It took me a little time to track it down but I finally found the slide, reproduced below, in a presentation called: “The Burden of Proof” that I had given for the Ann Arbor Oracle User Group in 2002. (The picture of the Earth is the Apollo 17 image from NASA):

Dominic Delmolino wrote a follow-up to Cary’s post, ending with a question about generalising the formula for a circle – and listing a few results relating to regular polygons (or N-gons), observing that the results he got seemed to be approaching Cary’s circle result as N – the number of sides of the polygon – increased. So is there an opportunity for a proof lurking there somewhere ? Yes and writing a note about it seemed to be much more entertaining than doing real work, so here is it.

We start with a regular N-gon, and clarify what operation we are doing to “raise the string above the surface”. (The sketch is a little crude, I was using Powerpoint at the time.)

The diagram shows a smaller hexagon inside a larger hexagon. To construct the larger hexagon from the smaller we draw a line from the centre of the original to the middle of one of the sides, and extend it outwards. We then slide the side in the direction of this line a distance h, leaving the side parallel to its original position. Repeat for all six sides of the hexagon.

We now have an “exploded” hexagon. To complete the larger hexagon, we need to take each of its sides and extend them at both ends by a distance d. The change in the perimeter from the smaller hexagon to the larger hexagon is then “2d * number of sides” (12d in the case of the hexagon), and we have to ask ourselves if we can work out a way of expressing d in terms of other values that we already know.

Looking at the diagram (and allowing for the poor quality of drawing) you should be able to note the similar triangles that allow us to say that d/h = tan(x) — sorry, you have to know some geometry and trigonometry at this point.

Since we have a regular N-gon, we know that x = pi/N radians — sorry, you have to work in radians, not degrees.

Substituting we have:

change in perimeter =

2d * N =
2 * h * tan(x) * N =
2h * N tan(pi/N)

Note that this formula is independent of the size of the original N-gon.

And now we get to the interesting bit – can we relate this formula to Cary’s formula where, for a circle, the change in perimeter was 2 * change in radius * pi. The answer is yes – but only if you know your Maclaurin series. The Maclaurin series for tan(x) – where x is measured in radians and has a modulus less than pi/2 – is:

 x + x3 + 2x5/15 + …

In our case we have x = pi/N, which is going to make our formula a little messy:

pi/N + pi3/3N3 + 2 pi5/15N5 + …

If we slot this into our  formula for the change in perimeter above we get:

2h * N * (pi/N + pi3/3N3 + 2 pi5/15N5 + …)

and now we can multiply out N through the series to get:

2h * (pi + pi3/3N2 + 2 pi5/15N4 + …)

Consider then that a circle is the limit of a regular N-gon as N tends to infinity. Let N tend to infinity in the final formula and every term except the first one vanishes – so at the limit the formula becomes 2h * pi, which is exactly the formula that Cary derived in his blog.


I don’t think I got to Maclaurin series until I was about 15 – but I wouldn’t be particularly surprised if Cary’s children get there a little bit sooner.


  1. […] there’s curiosityJonathan Lewis on There’s convincing and then there’s curiosityProof « Oracle Scratchpad on There’s convincing and then there’s […]

    Pingback by Oracle Musings » The proof is out there — April 7, 2012 @ 8:48 pm GMT Apr 7,2012 | Reply

  2. Funny, I drew a complete blank on Maclaurin, but I remember Taylor series, I guess ’cause there were times I would use the concept to approximate interest charges in my head. Couldn’t do it now, ‘puters make me lazy. My older son solved Cary’s problem faster than I ever could have. It is such an excellent demonstration against intuition.

    Comment by jgarry — April 9, 2012 @ 6:32 pm GMT Apr 9,2012 | Reply

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