[2006-06-23] Mathematical Digressions

In my previous post, I said that the sum of the geometric series:

1 + x + x2 + ... + xn-1

is (xn - 1)/(x - 1). One of my beliefs (as also shown by that post) is that you should never take a formula at its face value. Here I quickly attempt to show why this particular formula should be correct using a very commonly used proof. Let us assume that S denotes the desired sum. That is:

S = 1 + x + x2 + ... + xn-1

If we multiply both sides of this equation by x, we get:

x×S = x + x2 + x3 + ... + xn

If we subtract each side of the first equation from the corresponding side of the second equation we get:

(x - 1)×S = x + x2 + x3 + ... + xn - 1 - x - x2 - ... - xn-1

This simplifies to:

(x - 1)×S = xn - 1

which gives:

S = (xn - 1)/(x - 1)

as we had claimed.

By the way, while attending a conference call last night where it was particularly hard to concentrate on what was being discussed, I began to doodle and stumbled upon a simple result which might have been interesting about a millennium ago and is probably just an idle curiosity now. I am also sure it is contained somewhere in our high school text books for mathematics, but I cannot recall seeing it before. (Despite what might be indicated by these blog posts, I do not have much of an interest in mathematics itself and have never pursued it for its own sake.)

Anyways, looking at the factorisation of (x2 - y2) into (x - y)×(x + y) and (x3 - y3) into (x - y)×(x2 + x×y + y2), I wondered if (x - y) is always a factor of (xn - yn). By playing around a little bit with the terms, I could write:

xn - yn = (x - y)×(xn-1 + xn-2×y + ... + x×yn-2 + yn-1)

confirming my hypothesis. If you set y=1, this reduces to:

xn - 1 = (x - 1)×(xn-1 + xn-2 + ... + x + 1)

which can be rewritten as:

(xn - 1)/(x - 1) = xn-1 + xn-2 + ... + x + 1

which is the same formula that we derived just a while back for calculating the sum of a geometric series!

(Originally posted on Blogspot.)

Other Posts from 2006