Tennis maths, anyone?

A few weeks ago we wrote about seeding at the French Open tournament, very concerned that top seed Novak Djokovic was being treated unfairly. In the end it didn't matter: predictably, Djokovic and second seed Rafael Nadal won through to the final and, very predictably, Nadal was the victor. That was despite the umpire unsettling Rafa, by demanding that the finalists play on through hell and absurdly high water.

Now it's on to Wimbledon and more wonderful weather. However the English actually have the sense to come in out of the rain. This means that Djocky and Rafa will probably have lots of time to while away in the locker room. So, how might they occupy themselves? Well, maths is always a good option.

Here's an easy question to start off our tennis champs: in total, how many matches will be played in the Men's Singles at Wimbledon?

There are 128 players in the draw, which means there are 64 matches in the first round. The 64 winners then meet in the 32 matches of the second round, and so on until, inevitably, Djokovic and Nadal are left to play the final match. So, we just have to sum 64 + 32 + 16 + 8 + 4 + 2 + 1.

Simple stuff. However, as we noted in our solutions to last year's Summer Quiz, there's an even easier method to obtain the answer. There are 128 players, and all but one of them (probably Djokovic) will be eliminated along the way, one match at a time. So, there must be 127 matches in total. Very easy!

The original sum is what is known as a geometric series: each number we add on is a fixed multiple of the previous number. In our case the fixed multiple is 1/2. (Or, if you want to write the sum in reverse, starting at 1 and finishing with 64, you can think of the fixed multiple as 2.) And, Wimbledon has provided us an with easy method to find the sum of our particular geometric series:

Can we do others? Wimbledon begins with 128 players, which is 2⁷, and definitely other powers of 2 can be handled similarly. If we begin with 2^N players, who compete in pairs in the usual manner, then counting the number of matches will show that

What about other powers? In a short and lovely article, mathematician Vincent Schielack used the tournament scenario to derive formulas for other geometric sums.

To apply Vincent's very clever argument, let's imagine we have a new television show, Voicing with the Stars. The show begins with 2187 contestants, who are separated into 729 groups of three. Each group has a sing-off to see who advances. In the next round (yawn!), 729 contestants are left, who are again grouped into threes for an elimination sing-off. Eventually, just one singer - our new Kylie Minogue - remains. So, how many sing-offs would we have had along the way?

As with Wimbledon, there are two different ways to count the number of "matches". Summing directly, we have 729 sing-offs in the first round, then 243 sing-offs in the second round, and so on, dividing by 3 each time. However, overall we know that all but one contestant will be eliminated, with two contestants eliminated in each sing-off. Equating the results of the two calculations, we can see that

An analogous formula holds if we begin with 3^N singers. And it's just as easy to begin with M^N singers, with sing-offs between groups of M singers, and where one member from each group progresses to the next round. Arguing as we have above, you should be able to convince yourself that

So, for any positive whole number M, we have a formula for the geometric sum with constant multiple 1/M (or, reading the sum from right to left, constant multiple M). But what about other values of M? Is the formula always true?

If M = 1 then the formula has a troublesome 0 in the denominator, but in this case we hardly need a formula: the sum reduces to adding 1 to itself a bunch of times. For any other value of M the formula is indeed true. How can we see that?

If M is a fraction then we can still argue with tournaments or TV shows to prove the formula. Alternatively, there is a well-know algebraic proof that will justify the formula for all M (except 1).

However, as Victor points out, we don't really need to argue further. Given our geometric sum formula is true for all positive whole numbers (except 1), it will automatically be true for all numbers (except 1). That's not overly difficult to see, but we don't want to give the whole game away: let's leave something for Djocky and Rafa to ponder.

Puzzle to Ponder: Can you help Djokovic and Nadal complete our argument: given we know our geometric formula is true for M = 2, 3, 4 ..., can you explain why the formula must then be true for all M (other than 1)?

Burkard Polster teaches mathematics at Monash and is the university's resident mathemagician, mathematical juggler, origami expert, bubble-master, shoelace charmer, and Count von Count impersonator.

Marty Ross is a mathematical nomad. His hobby is smashing calculators with a hammer.

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