Generalization of Difference of Powers Identity

Question

For integers $n\geq2,$ we have the formula $$a^n-b^n=(a-b)\sum_{j=0}^{n-1}a^{n-j-1}b^j.$$ Is there a similar formula for reals $n>0?$ Perhaps where the sum becomes a series, as is in the case of the binomial theorem?

Answer 1

As @Greg Martin suggested, it is actually possible to use binomial series to obtain a certain formula, although it may not the form that you want. For simplicity, we will consider $(x^s - 1) / (x - 1)$ for real $s > 0$, then the result will be obtained by putting $x = a / b$. Now put $x = 1 + t$ and we can expand the equation as

$$ \frac{x^s - 1}{x - 1} = \frac{(1 + t)^s -1}{t} = \frac{1}{t} \sum_{i=1}^{\infty} \binom{s}{i}t^i = \sum_{i=1}^{\infty} \binom{s}{i} t^{i-1} = \sum_{i=1}^{\infty} \binom{s}{i}(x - 1)^{i-1}= \sum_{i=0}^{\infty} \binom{s}{i+1}(x-1)^i \\ = s + \frac{s(s-1)}{2!}(x-1) + \frac{s(s-1)(s-2)}{3!} (x-1)^2 + \cdots $$ where $\binom{s}{i}$ is the generalized binomial coefficient. So we get $$ \frac{a^s - b^s}{a - b} = \frac{(a/b)^s - 1}{(a/b)-1} \cdot b^{s-1} = b^{s-1} \sum_{i=0}^{\infty} \binom{s}{i+1}\left(\frac{a}{b} - 1\right)^{i}. $$

I believe that what you may want is a series centered at $x = 0$, not $x = 1$. The behavior of $x^s$ at $x = 0$ is not that good, especially they are not analytic at $x = 0$ for $s \not \in \mathbb{Z}$ (they aren't even smooth). But when $s = n / m$ is rational, then we can expand it as Puiseux series

$$ \frac{x^{n/m} - 1}{x - 1} = \frac{1 - x^{n/m}}{1 - x} = (1 - x^{n/m}) \sum_{j=0}^{\infty} x^j = \sum_{j=0}^{\infty} (x^{j} - x^{j + n/m}). $$

For example,

$$ \frac{x^{3/2} -1 }{x - 1} = 1 + x - x^{3/2} + x^2 - x^{-5/2} + x^3 - \cdots. $$