The
following is a proof of the Binomial Theorem for all values, claiming
to be algebraic. Because we use limits, it could be claimed to be
another calculus proof in disguise.
We assume: [1.1] That is, we assume the Principle of Indeterminate Coefficients, that (1+x)n is a polynomial of the stated kind. The coefficients of x are assumed to be a function of n, and not a function of x.
We use a number y instead of x: [1.2] And the same result should follow.
Subtracting 1.2 from 1.1 [1.3]
Divide the left-hand side by (1+x)-(1+y), and the right-hand side by (x-y), its equal: [1.4]
When r is a non-negative integer, the terms will become zero when k=r+1.
When
r is not a non-negative integer, then, after k exceeds r, the
subsequent terms will alternate in sign, each one being the negative of
the previous one, and the series will continue infinitely.
The validity of our argument depends on the validity of the Principle of Indeterminate Coefficients, which depends on x not being infinite. As the x-value of the kth term is xk,
then for the terms to remain finite as k approaches infinity, |x|<1,
when r is a not a non-negative integer. The value |x|<1 ensures the
terms approach zero as k tends to infinity (which is what we need for
the Principle of Indeterminate Coefficients) and it means the series
could be convergent (which it actually is), but we have not proved
this, yet.
[1.1]
[1.2]
[1.3]
[1.4]
[1.5]
[1.6]
[1.7]
[1.8]
[1.9]
[1.10]
[1.11]
[1.12]
[1.13]
[1.13.1]
[1.14]
[1.15]
[1.15]
[2.1]