Question 2f21a

Mar 22, 2017

The binomial theorem states that for $n$ integer:

${\left(a + x\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {x}^{k} {a}^{n - k}$

Based on the MacLaurin theorem we can generalize for non integer exponents:

${\left(a + x\right)}^{\nu} = {\sum}_{k = 0}^{n} \left(\begin{matrix}\nu \\ k\end{matrix}\right) {x}^{k} {a}^{\nu - k} + o \left({x}^{n}\right)$

Explanation:

The binomial theorem states that for an integer exponent $n \in \mathbb{N}$:

${\left(a + x\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {x}^{k} {a}^{n - k}$

where $\left(\begin{matrix}n \\ k\end{matrix}\right)$ is the binomial coefficient:

((n),(k)) = (n!)/(k!(n-k)!)

When the exponent is a generic real number $\nu \notin \mathbb{N}$ we can introduce the generalized binomial coefficient:

((nu ),(k)) = (nu * ( nu -1) ( nu - 2 ) * ... * (nu - k+1))/(k!)

Now developing the function $f \left(x\right) = {\left(a + x\right)}^{\nu}$ in MacLaurin series we have:

$f \left(0\right) = {a}^{\nu}$

$f ' \left(x\right) = \nu {\left(a + x\right)}^{\nu - 1}$ so $f ' \left(0\right) = \nu \cdot {a}^{\nu - 1}$

$f ' ' \left(x\right) = \nu \left(\nu - 1\right) {\left(a + x\right)}^{\nu - 2}$ so $f ' ' \left(0\right) = \nu \left(\nu - 1\right) {a}^{\nu - 2}$

and we can easily see that in general:

${f}^{\left(n\right)} \left(0\right) = \nu \left(\nu - 1\right) \left(\nu - 2\right) \cdot \ldots \left(\nu - n + 1\right) {a}^{\nu - n}$

so that the MacLaurin series is:

(a+x)^nu = sum_(k=0)^oo (nu (nu - 1) (nu -2)* ... (nu -k +1))/(k!) a^(nu-k) x^k#

that is:

${\left(a + x\right)}^{\nu} = {\sum}_{k = 0}^{\infty} \left(\begin{matrix}\nu \\ k\end{matrix}\right) {x}^{k} {a}^{\nu - k}$

Using the MacLaurin theorem we can state that if we truncate the series at the $n$-th term, the rest is an infinitesimal of order higher then ${x}^{n}$ for $x \to 0$:

${\left(a + x\right)}^{\nu} = {\sum}_{k = 0}^{n} \left(\begin{matrix}\nu \\ k\end{matrix}\right) {x}^{k} {a}^{\nu - k} + o \left({x}^{n}\right)$

which can be seen as a generalization of the binomial formula.