Another Neat Piece Of Algebra - Series Summation
MartinPacker 11000094DH Visits (3708)
Here's another neat piece of algebra: A technique for summing series.
You know what b - a + c - b + d - c is. Right?
Suppose I were to write the same sum as:
(b - a) +
(c - b) +
(d - c)
The answer is still d - a. Right?
Now, suppose I re-label with s0 = a, s1 = b, s2 = c and s3 = d. We end up with:
(s1 - s0) +
(s2 - s1) +
(s3 - s2)
This is actually pretty scalable terminology as you can write sr for any arbitrary value of r. And that's one of the strengths of algebra: generalisation.
So let's do that up to n:
(s1 - s0) +
(s2 - s1) +
(sn - sn-1)
which is, of course, sn - s0.
But what has that got to do with summing series?
If we can replace each (sr - sr-1) by a single term ur you may see the relevance...
u1 + u2 + ... + un = sn - s0.
The series summation boils down to a "simple" subtraction. The trick is to find these s terms, given the u terms. Let's try it with an example.
Summing The Integers
This is the series 1, 2, 3, ... , n.
The r'th u term is just r. ur = r. So we now have to find the sr term. Remember sr - sr-1 has to equal r.
Try sr = r (r + 1).
Then sr-1 = (r - 1) r or r (r - 1).
So sr - sr-1 = [(r + 1) - (r - 1)] r or 2r. Not quite what we wanted. But we know - dividing by the factor of 2 - we should've guessed sr = ½ r (r + 1).
So sn - s0 = ½ n (n + 1) - ½ 0 (0 + 1) = ½ n (n + 1) - 0 = ½ n (n + 1).
The sum of the first n integers being ½ n (n + 1) is a well-known result. Admittedly it could've been done another way. But it's simple enough to show the method.
Another Example - Summing The Squares Of The Integers
This is the series 1, 4, 9, ... , n² .
In this case we need to do something that will appear slightly perverse:
Rewrite r² as r ( r + 1) - r.
If you can split each of the terms in a series into two terms you can sum these sub terms. I just did the split. We already know how to sum the "r" portion. It's ½ n (n + 1). So we need to sum the r (r+1) portion and subtract ½ n (n+1) from the result.
Try sr = r (r + 1) (r + 2).
Again we need to find sr-1.
(r - 1) r (r + 1)
r (r+1) (r-1)
Sosr - sr-1 = [(r + 2) - (r - 1)] r (r + 1) or 3r (r + 1).
This is 3 times what we want so we should've guessed sr = 1/3 r (r + 1) (r + 2).
So this portion of the sum is 1/3 n (n + 1) (n + 2) - 1/3 0 (0 + 1)(0 + 2) or 1/3 n (n + 1) (n + 2).
But we need to subtract ½ n (n + 1) from this:
1/3 n (n + 1) (n +2) - ½ n (n + 1) = 2/6 n (n + 1) (n +2) - 3/6 n (n + 1)
1/6 n (n + 1) [ 2 (n + 2) - 3] = 1/6 n (n + 1)( 2 n + 1) .
If you try it for a few values you'll see it's right. This isn't such a well known result as for the sum of the integers.
I'm conscious there's been some fiddliness here - which is where I normally fall down.
But I think the "sum a series by converting it to a single subtraction" trick is a neat one - which is why I share it with you.