mathemagicio.us » Christmas http://mathemagicio.us Ramblings and musings about math, science and truth Wed, 14 Oct 2009 01:02:47 +0000 http://wordpress.org/?v=2.8.4 en hourly 1 Happy eighth day of Christmas! http://mathemagicio.us/2009/01/01/happy-eighth-day-of-christmas/ http://mathemagicio.us/2009/01/01/happy-eighth-day-of-christmas/#comments Thu, 01 Jan 2009 17:55:50 +0000 Mark Przepiora http://mathemagicio.us/?p=36 As we all celebrate another revolution around the sun and come halfway closer to Epiphany, let’s explore a holiday-themed math problem.

The number of gifts one receives at the end of the Twelve Days Of Christmas is ironically 364—a gift for each day of the year except Christmas itself.

But what about the inhabitants of Gaia-n, who celebrate not 12 but n days of Christmas?

On the first day, such an inhabitant receives 1 present; on the second day, 1 + 2; on the third, 1 + 2 + 3; etc. Then, thanks to a theorem by some clever little kid, we can write the number of gifts he receives on day k as T_k = k(k+1)/2, the kth triangular number, and thus the total number of presents he receives in all ndays as the sum of triangular numbers C_n = \sum_{k=1}^n T_k (where C is for Christmas.)

T3 + T4 is a square

A geometric argument easily shows that T_{k-1} + T_{k} = k^2, so (assuming n is even)

C_n = \sum_{k=1}^{n/2} (2k)^2 = 4\sum_{k=1}^{n/2} k^2

which is four times a pyramid number! Substituting the closed expression for the sum and simplifying gives,

C_n = n^3/6 + n^2/2 + n/3 = \frac{1}{6} n(n+1)(n+2).

We can easily verify that this expression also gives the correct answer when n is odd, and moreover C_{12} = 12\cdot 13\cdot 14/2 = 364 as we hoped!

On other planets, inhabitants may celebrate a fraction of a number of days of Christmas, and they may wish to extend the expression C_n to any real (or complex!) number, i.e.

C(z) = \frac{1}{6} z(z+1)(z+2)

which we naturally call the Christmas polynomial.

We may also wonder how many gifts someone will receive who has relatives on each of the planets Gaia-1, Gaia-2, …, up to Gaia-n. That is, what is the value of the sum Z_n = \sum_{k=1}^n C_n?

Because we may write T_n = \binom{n+1}{2} and C_n = \binom{n+2}{3}, it is a reasonable guess that Z_n = \binom{n+3}{4} . This indeed turns out to be the case, and so we may naturally define a generalized Christmas function

\chi_{m,n} = \binom{n+m}{m+1}

(where χ is for Xmas.) This results in the sequences

\chi_{0,n} = n
\chi_{1,n} = T_n
\chi_{2,n} = C_n

and so on, allowing us to calculate a great number of Christmas gift-giving scenarios.

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