Appendix D Binomial Series
For \(\alpha \in \mathbb{R}\) and \(|x| \lt 1\)
\begin{equation*}
(1+x)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} x^n
\end{equation*}
where \(\binom{\alpha}{n} = \frac{\alpha(\alpha-1)\cdots (\alpha -
n+1)}{n!}.\)
To prove it, first note that \(|\binom{\alpha}{n+1}/\binom{\alpha}{n}|
= |(\alpha-n)/(n+1)| \to 1\text{.}\) So, the radius of convergence of the series is \(1\text{.}\) Let \(f(x)\) be the function represented by the power series on \((-1,1)\text{.}\) It is straight-forward to verify for \(n \ge 1\text{,}\)
\begin{align}
& n\binom{\alpha}{n} = \alpha\binom{\alpha-1}{n-1},
\text{and}\tag{D.1}\\
& \binom{\alpha-1}{n} + \binom{\alpha-1}{n-1} =
\binom{\alpha}{n}.\tag{D.2}
\end{align}
Multiply second equation by \(\alpha x^n\) and sum over \(n \ge
1\text{,}\) we get
\begin{align*}
\sum_{n=1}^{\infty}\alpha\binom{\alpha-1}{n} x^n +
\sum_{n=1}^{\infty} \alpha \binom{\alpha-1}{n-1}x^n & =
\alpha\sum_{n=1}^{\infty} \binom{\alpha}{n}x^n
\end{align*}
Using the first equation, we get
\begin{align*}
\sum_{n=1}^{\infty}\alpha\binom{\alpha-1}{n} x^n +
\sum_{n=1}^{\infty} n \binom{\alpha}{n}x^n &= \alpha(f(x)-1) \\
\alpha + \sum_{n=1}^{\infty}\alpha\binom{\alpha-1}{n} x^n +
x \sum_{n=1}^{\infty} n \binom{\alpha}{n}x^{n-1} & = \alpha f(x)\\
\sum_{n=0}^{\infty}\alpha\binom{\alpha-1}{n} x^n +
x \sum_{n=0}^{\infty} n \binom{\alpha}{n}x^{n-1} & = \alpha f(x)\\
\sum_{n=0}^{\infty}\alpha\binom{\alpha-1}{n} x^n + xf'(x) = \alpha f(x)
\end{align*}
Then using C.1 again, we have
\begin{gather*}
\sum_{n=0}^{\infty}(n+1)\binom{\alpha}{n+1} x^n + xf'(x) = \alpha f(x) \\
\sum_{n=1}^{\infty}n\binom{\alpha}{n} x^{n-1} + xf'(x) = \alpha f(x)
\end{gather*}
So,
\begin{equation*}
(1+x)f'(x) = \alpha f(x).
\end{equation*}
And
\begin{align*}
\int \frac{f'(x)}{f(x)} dx &= \frac{\alpha}{1+x} dx\\
\ln(f(x)) & = \alpha\ln(1+x) +C
\end{align*}
Since \(f(0) = 1\text{,}\) so \(C=0\) and we have \(f(x) = (1+x)^\alpha\text{.}\)
