Binomial Theorem/General Binomial Theorem/Proof 3

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Theorem

Let $\alpha \in \R$ be a real number.

Let $x \in \R$ be a real number such that $\size x < 1$.


Then:

\(\ds \paren {1 + x}^\alpha\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\alpha^{\underline n} } {n!} x^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dbinom \alpha n x^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\prod_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } x^n\)
\(\ds \) \(=\) \(\ds 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } {2!} x^2 + \dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } {3!} x^3 + \cdots\)

where:

$\alpha^{\underline n}$ denotes the falling factorial
$\dbinom \alpha n$ denotes a binomial coefficient.


Proof

The series is the Maclaurin series expansion of the function $\map f x = \paren {1 + x}^\alpha$.

The derivatives of $f$ are:


\(\ds \map {f^{\paren 0} } x\) \(=\) \(\ds \paren {1 + x}^\alpha\)
\(\ds \map {f^{\paren 1} } x\) \(=\) \(\ds \alpha \paren {1 + x}^{\alpha - 1}\)
\(\ds \map {f^{\paren 2} } x\) \(=\) \(\ds \alpha \paren {\alpha - 1} \paren {1 + x}^{\alpha - 2}\)
\(\ds \map {f^{\paren n} } x\) \(=\) \(\ds \alpha \paren {\alpha - 1} \cdots \paren {\alpha - n + 1} \paren {1 + x}^{\alpha - n}\)
\(\ds \) \(=\) \(\ds \alpha^{\underline n} \paren {1 + x}^{\alpha - n}\)


Evaluated at $x = 0$, we have:

\(\ds \map {f^{\paren 0} } x\) \(=\) \(\ds \alpha^{\underline n} \paren {1 + 0}^{\alpha - n}\)
\(\ds \) \(=\) \(\ds \alpha^{\underline n}\)


The Maclaurin series of $f$ is:

\(\ds \map f x)\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \map {f^{\paren n} } 0\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \alpha^{\underline n}\) substituting derivative at $0$
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\alpha^{\underline n} } {n!} x^n\) rearranging

$\blacksquare$