As usual, the first n in the table is zero, which isn't a natural
number.
Because Δ_{3 }is a constant, the sum is a cubic of
the form
an^{3}+bn^{2}+cn+d, [1.0]
and
we can find the coefficients using simultaneous equations, which we can
make as we wish, as we know how to add squares to the table and to sum
them, even if we don't know the formula.
In the table below,
we create three equations, noting that d=0 from the first one
(revealing the reason for the non-natural number zero)
Using 1.1 with 1.2 and 1.3 we can make two new equations:
6a+2b=3 [1.4]
24a+6b=11 [1.5]
By subtracting 3x Equation 1.4 from 1.5, we get:
6a=2
So a=1/3 [Also noting, by the way, that Δ_{3}/3!=1/3]
Substituting a=1/3 in 1.4 gives
2b=1
So b=1/2
Finally, substituting these values into 1.1 we get
1/3+1/2+c=1
So c=1/6
[Actually, with the sum of the powers, the sum of the coefficients in
the formula is always 1]
So,
we can substitute our values into 1.0, to get the sum of the squares of
the first n natural numbers (or first n positive integers):
n^{3}/3+n^{2}/2+n/6
Or, in various forms:
Sum of the squares
of the first n natural numbers using summation
We tried this with the sum
of the natural numbers using summation,
and fell flat on our faces, so this time we will go straight into
setting up for the sum of the cubes, in the hope we will find our
formula for the squares. (There is no reason stated here why
this
method should continue to work, however. Sometimes such approaches work
only in specific cases. Fortunately, this approach does work for any
sum of the powers of the natural numbers).
As before, we set up as follows:
Saying the sum to n is one term less than the sum to (n+1)
We expand the sums:
As expected, the cubic terms cancel, and we rearrange the formula to
have the sum of the squares on the left:
Expanding the cube and summing the sums:
Adding like terms:
Dividing throughout by 3 gives us the formula for the sum of the
squares:
Or:
Sum of Natural
Numbers Squared Using Errors
When we did this with the natural
numbers,
we found there was little work to do. With the squares, we have to go a
little further. While the error on each term for the sum of the numbers
is constant, the error on the squares depends on the term.
The graph below of y=x^{2},
has the squares of the natural numbers represented by rectangles, and
the area under the graph is approximately the sum of the squares (areas
of the rectangles).
By integrating the x^{2}, we find the area under
the graph to be x^{3}/3,
so the area of the rectangles (sum of squares) is the area under the
graph plus the error. If the number of squares is n, we can write n^{3}/3
as the approximation:
[3.1]
Where
is the sum of the squares and E_{n} is the error on
approximating to n^{3}/3.
The sum of the n-1 squares and the error E_{n-1},
gives us:
[3.2]
We note that the difference between 3.1 and 3.2 is;
[3.3]
being the nth square
The difference between 3.1 and 3.2 is also:
[3.4]
Expanding (n-1)^{3} in 3.4:
[3.5]
Rounding up similar terms:
[3.6]
From 3.3, we get:
[3.7]
Expressing the errors in terms of n:
[3.8]
This is a recursion formula. If we know one of the errors we can find
the other. We do know what E_{0} is, because it is
the error on the 0th term, which is zero. So, letting n=1, we have:
E_{1}-E_{0}=1-1/3
Letting n=2, and changing the relationship a bit:
E_{2}=E_{1}+2-1/3
E_{2}=1-1/3+2-1/3
It
is now evident that the sum of the squares on both sides will cancel
out. We could get a sum for the natural numbers out of this, but we can
do that directly
and much more easily.
After
some thought, we can conclude that the last term must be negative for
us to find an expression for the sum of the squares. It was negative
for the sum of the natural numbers, and it will be negative again for
the sum of the cubes, but will not work for the sum of the powers of
four. That is, this method works for the squares of the odd numbers.
The expression corresponding to 5.4 for the sum of the cubes is:
[5.5]
Here
the sum of the cubes on the right-hand-side is negative. I also note
that in 5.4 and 5.5 the coefficients correspond to the binomial
coefficients of the square and the cubic