Binomial expansion for $(x+a)^n$ for non-integer n I finally figured out that you could differentiate $x^n$ and get $nx^{n-1}$ using the derivative quotient, but that required doing binomial expansion for non-integer values.
The most I can find with binomial expansion is the first, second, last, and second to last terms.
So how do I find something like $(x+a)^{\pi}$?  When differentiating in calculus, I didn't need to find terms after the second because I knew they would all cancel out, but how do you find these terms?
Do they work for negative exponents as well?
And does this work for complex exponents?
Which came first, Euler's method for complex exponents or binomial expansion for complex exponents?
 A: The Binomial theorem for any index $n\in\mathbb{R}$ with $|x|<1,$ is
$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\ldots$
For $(x+a)^\pi$ one could take $x$ or $a$ common according as if $|a|<|x|$ or $|a|<|x|$ and use Binomial theorem for any index. i.e., $x^\pi(1+a/x)^\pi$ in case $|a|<|x|.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
& \color{#44f}{\left.\rule{0pt}{5mm}\pars{a + x}^{n}
\,\right\vert_{\ \verts{x/a}\ <\ 1}}\,\,\, =\
a^{n}\pars{1 + {x \over a}}^{n} = a^{n}\sum_{k = 0}^{\color{red}{\large\infty}}
{\Gamma\pars{n + 1} \over \Gamma\pars{k + 1}\Gamma\pars{n - k + 1}}
\pars{x \over a}^{k}
\\[5mm] = & \
\bbx{\color{#44f}{\sum_{k = 0}^{\infty}
{\Gamma\pars{n + 1} \over \Gamma\pars{k + 1}\Gamma\pars{n - k + 1}}
\, a^{n - k}\,x^{k}}} \\ &
\end{align}
By the way, $\ds{{\Gamma\pars{n + 1} \over \Gamma\pars{k + 1}\Gamma\pars{n - k + 1}} = {n \choose k}}$: It's the generalized binomial expression. When $\ds{n \in \mathbb{N}_{\geq\ 0}}\,$, the factor $\ds{\Gamma\pars{n - k + 1}}$ truncates the sum at $\ds{n}$.
