Let c be a constant and let f be the differentiable function (c+x)t, where t is a positive real number. What is the power series for f?
Evaluate f at 0 and get ct.
The first derivative is t×(c+x)t-1, hence f′(0) = tct-1.
The second derivative evaluated at 0 is t(t-1)ct-2, and so on.
Divide these derivatives by k! to get the taylor coefficients. This gives a factor of the form t×(t-1)×(t-2)×…(t-k+1) over k!. Use the binomial coefficient (t:k) to represent this expression. When t is a positive integer n, the result is the traditional binomial coefficient (n:k). We are simply generalizing the notation to any real number t.
Now we are ready to write the power series for f.
f = ∑{k=0,∞} (t:k)ct-kxk
When t is a positive integer, such as n, the nth derivative of (c+x)n is a constant for all x, and all higher derivatives are 0. The power series truncates to a finite polynomial, as one would expect when f itself is a polynomial. In fact, the power series reproduces the traditional binomial theorem.
Return to (c+x)t, and write it as exp(t×log(c+x)). this is analytic at 0, hence f agrees with its power series on a circle of convergence. Since log(c+x) converges out to |x| = c, and exp() converges everywhere, f converges at least as far as c. If x is even slightly larger than c, say c+ε, move far out in the power series and consider the ratio of adjacent terms, (t-k)/(k+1) times (c+ε)/c. By the ratio test, this series does not converge, hence the radius of convergence is c.
Note, When t = n, f is a polynomial that converges across the entire plane.
Since f is analytic it extends uniquely to complex numbers, i.e. (c+z)t. And the exponent t could be complex as well. This is the continuous binomial theorem in its full generality.
f = ∑(k=0,∞) (½:k)c½-kxk
Is there a convenient formula for (½:k)?
Set the denominator of k! aside for the moment, and consider the product of 1/2 times -1/2 times -3/2 times -5/2 etc. Multiply by 2k and find 1×-1×-3×-5… out to -2k+3.
Setting the signs aside for the moment, the product becomes (2m)! over (m!×2m), where m = k-1.
Divide by 2k (since we multiplied by 2k earlier), and obtain the following.
(2m)! over m!×2m×2k
2×(2m)! over m!×4k
2k×(2m)! over k!×4k
(2k)! over (2k-1)×k!×4k
Bring the denominator of k! back in, and get this.
(½:k) = (2k)! over (2k-1)×k!×k!×4k
(½:k) = (2k:k) over (2k-1)×4k
This isn't quite right when k = 0, as it produces -1 instead of 1. But remember, we set the signs aside. Let's bring them back in now. When k is 1 the sign is positive, and it alternates from there, so multiply by (-1)k+1, and it all works out.
sqrt(c+x) = ∑(k=1,∞) (-1)k+1 × (2k:k) / ((2k-1)×4k) × c½-kxk
If you like, pull sqrt(c) out of the sum, and replace c-kxk with (x/c)k.