Continuous version of binomial expansion Is there a continuous version of the binomial expansion? That is, if $n\in\mathbb{N}$ and $a,b\in\mathbb{R}$, then
$$
(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^{n-k}b^k.
$$
However, if $x\in\mathbb{R}$, is there anything we can say about
$$
(a+b)^x
$$
regarding its expansion? I'm using the term 'expansion' here in a rather apprehensive way, but can we make sense of it in any way?
 A: The binomial power series is defined for any exponent $\alpha\in\mathbb{R}$ by
$$(1+x)^{\alpha}=\sum_{k=0}^{\infty}\dfrac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}x^k$$
The fraction generalizes binomial coefficients.  If $\alpha$ is a non-negative integer the series is actually finite since eventually $\alpha=k$ for some value of $k$ and gives the usual binomial expansion. Otherwise it is infinite.
As this is a series one needs to worry about convergence. The main results are that the series is always convergent when $|x|<1$, always divergent when $|x|>1$ except when $\alpha\in\mathbb{N}$ (since the series is finite then). When $|x|= 1$ the behaviour of the series depends more subtly on $\alpha$ (this case is mostly relevant when we allow $x$ to be complex).
A slight modification along the lines of
$$(a+b)^x=a^x\left(1+\dfrac{b}{a}\right)^x$$
lets you handle the more general version that you have in mind.
You can find a derivation on the wiki page
https://en.wikipedia.org/wiki/Binomial_series
or in any text about power series, as the binomial series is a very standard and useful power series.
NB: Both $\alpha$ and $x$ can actually be complex in this discussion without much changes.
A: Yes: the binomial series, under mild hypotheses. Suppose $a, b > 0$ and $a<b$.
By Taylor's formula, as $\frac ab<1$ we have
\begin{align}(a+b)^x&=b^x\Bigl(1+\frac ab\Bigr)^x=b^x\Bigl(1+x\frac ab+\frac{x(x-1)}2\frac{b^2}{a^2}+\dots+\binom{x}{k}\frac{b^k}{k!\,a^k}+\dotsm\Bigr)\\
&=b^x+x\,ab^{x-1}+\binom x2   a^2b^{x-2}+\dots+\binom xk x^kb^{x-k}+\dotsm
\end{align}
where we denomte $\dfrac{x}k$ the generalised binomial coefficient:
$$\binom xk=\frac{x(x-1)\dots(x-k+1)}{k!}.$$
