Newton's Discovery of the Binomial Series
Following in Wallis's footsteps, Newton recognized that the key to
calculating 
was finding a way of evaluating 
If that exponent were an integer instead of 1/2,
life would be easy. Like Wallis, Newton begins by comparing what he
has to what he can evaluate. He looks at the expansions of 
for integer values of m.
He considered a table of the cofficients and tried to guess what coefficients
would correspond to an exponent of m
= 1/2.
m |
|
|
|
|
|
|
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1/2 |
|
|
|
|
|
|
1 |
1 |
1 |
0 |
0 |
0 |
0 |
3/2 |
|
|
|
|
|
|
2 |
1 |
2 |
1 |
0 |
0 |
0 |
5/2 |
|
|
|
|
|
|
3 |
1 |
3 |
3 |
1 |
0 |
0 |
7/2 |
|
|
|
|
|
|
4 |
1 |
4 |
6 |
4 |
1 |
0 |
9/2 |
|
|
|
|
|
|
5 |
1 |
5 |
10 |
10 |
5 |
1 |
It is easy to guess that the values in the first column are all 1,
and the values in the second column must equal m.
What about the third column?
Newton would have been very familiar with the sequence 1,
3, 6, 10, …, the triangular numbers. The
jth
triangular number is the sum of the integers from 1
to j,
and it equals j(j+1)/2.
The exponent m
corresponds to the (m
– 1)st triangular number, so the formula to use in the
third column is m(m
– 1)/2.
If the values in the first column are constant, the values in the second
column increase linearly, and the values in the third column are given
by a quadratic formula, then it makes sense to look for a cubic polynomial
for the fourth column, a quartic polynomial for the fifth, and so on.
Armed with this assumption, it is not difficult to determine what these
polynomials must be.
We know that the cubic polynomial in m
that fits the coefficients of 
must have roots at m
= 0, 1,
and 2. This
cubic polynomial must be c
m(m-1)(m-2) for some
still to be determined constant c.
We can find c
by using the fact that that this polynomial is 1
when m
= 3:
This polynomial is m(m
– 1)(m – 2)/6.
A similar argument shows us that the polynomial in the next column
should be m(m
– 1)(m – 2)(m – 3)/4!
and the polynomial in the fifth column should be m(m
– 1)(m – 2)(m – 3)(m
– 4)/5!.
In general, the column that corresponds to 
will have zeros at m
= 0, 1, 2, …, k – 1, and a 1
at m
= k. The corresponding polynomial is
Since this is defined for any value of m,
we can fill in the table.
m |
|
|
|
|
|
|
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1/2 |
1 |
1/2 |
–1/8 |
1/16 |
–5/128 |
7/256 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
3/2 |
1 |
3/2 |
3/8 |
–1/16 |
3/128 |
–3/256 |
2 |
1 |
2 |
1 |
0 |
0 |
0 |
5/2 |
1 |
5/2 |
15/8 |
5/16 |
–5/128 |
3/256 |
3 |
1 |
3 |
3 |
1 |
0 |
0 |
7/2 |
1 |
7/2 |
35/8 |
35/16 |
35/128 |
–7/256 |
4 |
1 |
4 |
6 |
4 |
1 |
0 |
9/2 |
1 |
9/2 |
63/8 |
105/16 |
315/128 |
63/256 |
5 |
1 |
5 |
10 |
10 |
5 |
1 |
All of this has been supposition. Newton did not know how to prove it,
but he did realize that he needed some independent confirmation. What
he had discovered suggested that
|