Christmas is a time for giving, celebrating, family and magic. But did you know it’s also a time for equations? Department store Debenhams has decided to honour this recent Christmas tradition by tasking at least two members of Sheffield University’s undergraduate maths society to come up with formulae for ‘a perfectly decorated Christmas tree‘, picked up by The Sun, The Metro and others.

Previous festive howlers include ‘the formula for the perfect family Christmas‘ (sponsored by The Children’s Society to promote a book) and a prior stab at ‘the equation for the ideal Christmas tree‘ (sponsored by B&Q), which are just nonsensical strings of abbreviations. However, unlike those examples of naff-ematics, the Sheffield tree-decorating equations make enough sense for me to take a critical, overly-serious look at them on their own merits, and show how you might begin to come up with something more rational.

Firstly, I can’t see much at fault with the equation:

\[\text{Height of star or fairy (in cm)} = \frac{1}{10} \times \text{height of tree (in cm)}.\]

The idea that the two heights should be proportional seems reasonable, though you may wish to quibble over the apparently arbitrary constant. More importantly, the dimensions on either side of the equation are consistent, relating one length to another. Most ‘formula for’ stories fail even a rudimentary dimensional analysis. However, the unit of measurement provided (centimetres) isn’t needed—you could just as easily perform the calculation in metres or inches.

This dimensional consistency doesn’t hold for the next equation:

\[\text{Number of baubles} = \frac{\sqrt{17}}{10} \times \text{height of tree (in cm)}.\]

To take this equation seriously you would have to believe that $\frac{\sqrt{17}}{10}$ is some sort of universal bauble density constant giving the ideal number per centimetre of tree height, independent of the tree’s width.

The principle that “it is better to be vaguely right than exactly wrong” ((“Logic: Deductive and Inductive” by Carveth Read)) applies here. Since you can’t be precise about aesthetics, leaving a square root in the equation is merely a ploy to seem more mathematical. Debenhams claim the formulae are being rolled out for use by their personal shoppers nationwide. If you really were creating an equation for the shop floor, you would want something simple enough to be worked out by mental arithmetic or quickly tapped into a calculator. The constant here could have been approximated to $\frac{2}{5}$ or $0.4$.

The reason that these types of press-released equations fall outside the realm of mathematics is not the subject matter, but the lack of any reasoning or logical justification. Some mathematicians happily publish recreational and rigorous papers about, for instance, the optimal arrangement of a dartboard—and are able to question each other’s assumptions.

If I were pressed to come up with a similar formula for the number of baubles, I would start by modelling the Christmas tree as a solid cone with radius $r$ and height $h$. I might then assume the number of baubles is proportional to the surface area of the tree (you could perhaps argue that the volume is a more realistic quantity, as baubles can be placed ‘inside’ the cone representing my tree). This would give:

\[\text{Number of baubles} \propto \pi r \sqrt{r^2 + h^2}.\]

If you simplistically assume your Christmas tree’s radius (which I think can actually vary quite a bit) is proportional to its height, then you can gather up all the nasty constants to give:

\[\text{Number of baubles} \propto h^2.\]

But to decide on a constant for baubles per unit surface area, you still need some decorated Christmas trees that you find aesthetically pleasing, whose measurements you can substitute into the equation. It’s not for me to decide. In this case, Debenhams may want to see how a professional designer decorated trees of different sizes to check how well the assumptions worked out.

The final two equations are:

\[\begin{align} \text{Length of lights} &= \pi \times \text{height of tree}; \\ \\ \text{Length of tinsel} &= \frac{13\pi}{8}\times\text{height of tree}. \end{align}\]

The fact that these formulae involve $\pi$ appears reasonable, given the circular cross-section of Christmas trees. However, if you wanted to model the length of tinsel or lights as a conical spiral with $n$ turns, the precise formula would be:

\[ \text{Arc length} = \frac{h}{2} \sqrt{1+r^2(1+2\pi n)^2} + \frac{h(1+r^2)}{2 \pi n} \sinh^{-1} \left(\frac{2 \pi n r}{\sqrt{1+r^2}} \right). \]

One nice thing about mathematics is that so much of it has been done before. This means that it’s easy to come up with a sensible equation for even the most frivolous of situations. Here, I didn’t have to derive the formula myself from the parametric equations for the curve: I put a bit of trust in Wolfram MathWorld, and adapted the arc length they give for a conical spiral, substituting the angular frequency for $\frac{2 \pi n}{h}$. It’s not particularly pretty, but at least I’ve provided enough information for you to check the equation yourself.

Simplicity is also important if you need to convince a non-expert to adopt your mathematical idea, without having to take it on faith. Though the equation I’ve given is complicated enough to warrant a computer program, the ideas behind it are hopefully easy enough to understand. By knowing what the assumptions are, you could say “I like my lights to dangle instead of following the spiral exactly, so I’ll add a bit more length”.

Mathematics in the media doesn’t have to be like this, though I’m guessing these students didn’t expect their work to be plastered all over the press. If you want more useful festive formulae, you could do worse than the maths for the perfect Christmas present wrapping (sponsored by Amazon.co.uk), which looks dubious when written up in a newspaper, but which does contain mathematical content.

Kudos though for whoever came up with the ‘treegonometry’ pun in the Christmas tree press release.

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