Question #2f21a

1 Answer
Mar 22, 2017

The binomial theorem states that for nn integer:

(a+x)^n = sum_(k=0)^n ((n),(k)) x^ka^(n-k)(a+x)n=nk=0(nk)xkank

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

(a+x)^nu = sum_(k=0)^n((nu),(k)) x^ka^(nu-k)+o(x^n)(a+x)ν=nk=0(νk)xkaνk+o(xn)

Explanation:

The binomial theorem states that for an integer exponent n in NN:

(a+x)^n = sum_(k=0)^n ((n),(k)) x^ka^(n-k)

where ((n),(k)) is the binomial coefficient:

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

When the exponent is a generic real number nu notin NN we can introduce the generalized binomial coefficient:

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

Now developing the function f(x) = (a+x)^nu in MacLaurin series we have:

f(0) = a^nu

f'(x) = nu (a+x)^(nu-1) so f'(0) = nu*a^(nu-1)

f''(x) = nu(nu -1) (a+x)^(nu-2) so f''(0) = nu(nu -1)a^(nu-2)

and we can easily see that in general:

f^((n))(0) = nu (nu - 1) (nu -2)* ... (nu -n +1) 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:

(a+x)^nu = sum_(k=0)^oo((nu),(k)) x^ka^(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->0:

(a+x)^nu = sum_(k=0)^n((nu),(k)) x^ka^(nu-k)+o(x^n)

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