Re: Pascal's triangle, origin? (was: Re: Why should -1 multiplied by -1 be plus 1 and not -1)

From: Ken Pledger (Ken.Pledger_at_mcs.vuw.ac.nz)
Date: 08/06/04 

Date: Fri, 06 Aug 2004 14:36:31 +1200

In article <2ne2n0FvpgbvU1@uni-berlin.de>,
 Herman Jurjus <h.jurjus@hetnet.nl> wrote:

> Kenneth Doyle wrote:
> [snip]
>
> > Consider the example of Pascal's triagnle. It was discovered that you can
> > arrange numbers in a pattern that was sort of interesting in itself, but
> > had no apparant utility. Only much later was it discovered that that
> > pattern was apparant in a numerical abstraction of some particular aspect
> > of the real world. I remember that it was that example that convinced me
> > to persevere with math even when I couldn't work out how it related to
> > anything concrete.
>
> Although i don't have a concrete reference to contradict it,
> i'd be really surprised if Pascal's triangle was 'invented' in the way
> you say, as something without utility.
> In the least, i'd expect that it was discovered from calculating (a+b)^n
> or some combinatorics problem.
>
> Does anyone know more about this?

      Moslem and Chinese mathenmaticians introduced the triangle (long
before Pascal's time) as an aid in finding powers of binomials. You'll
know that the expansion of (a + b)^n has coefficients from a row of
the table.

      Here's how they constructed each row from the row before it.
Suppose you already knew that the expansion of (a + b)^4 has
coefficients 1, 4, 6, 4, 1, and you obtained (a + b)^5 by
multiplying that expansion by another factor (a + b). The new term in
(a^3)(b^2), for example, then comes from two places: the old 4(a^3)b
multiplied by b, and the old 6(a^2)(b^2) multiplied by a. Adding
these gives (4 + 6)(a^3)(b^2) = 10(a^3)(b^2). The same applies
elsewhere in the triangle, giving the well-known recurrence relation
which adds two successive numbers in a row to give a number in the row
below them, as in my example 4 + 6 = 10.

      Many present-day students' understanding of the triangle stalls at
that level, alas! They find the coefficients for, say, (a + b)^6 by
starting at the top and using the recurrence relation to work
laboriously down the triangle row by row, scattering arithmetical errors
in their wake.

      However, Pascal in 1654 had much deeper insights, including a
direct calculation of each binomial coefficient which we now write as
the formula n(n - 1)(n - 2)...(n - r + 1)/(r!). With this formula you
can jump straight into any position in the triangle without climbing
down rung by rung from the top.

      Then Newton in 1665 started using the infinite binomial series
which comes from using values of n other than natural numbers; but
that takes us beyond Pascal's triangle.

            Ken Pledger.

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