Power Series, Continuous Binomial Theorem

Continuous Binomial Theorem

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 c^t.

The first derivative is t×(c+x)^t-1, hence f′(0) = tc^t-1.

The second derivative evaluated at 0 is t(t-1)c^t-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)c^t-kx^k

When t is a positive integer, such as n, the n^th 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.

Square Root

f = ∑(k=0,∞) (½:k)c^½-kx^k

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 2^k and find 1×-1×-3×-5… out to -2k+3.

Setting the signs aside for the moment, the product becomes (2m)! over (m!×2^m), where m = k-1.

Divide by 2^k (since we multiplied by 2^k earlier), and obtain the following.

(2m)! over m!×2^m×2^k

2×(2m)! over m!×4^k

2k×(2m)! over k!×4^k

(2k)! over (2k-1)×k!×4^k

Bring the denominator of k! back in, and get this.

(½:k) = (2k)! over (2k-1)×k!×k!×4^k

(½:k) = (2k:k) over (2k-1)×4^k

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)×4^k) × c^½-kx^k