Question #2f21a

1 Answer
Mar 22, 2017

The binomial theorem states that for #n# integer:

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

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)#

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.