We have to distinguish two cases:
(1) If # p in NN# then the derivatives of order grater than #p# of #f(x)# are identically null. In such case:
#(df)/dx = px^(p-1)#
...
#(df^k)/dx^k = p(p-1)...(p-k+1)x^(p-k) = (p!)/(p-k!)x^(p-k)#
and for #x=1#
#[(df^k)/dx^k ]_(x=1) = (p!)/(p-k!)#
so the Taylor series has a finite number of terms:
#x^p = sum_(k=0)^p (p!)/(p-k!)(x-1)^k/(k!) = sum_(k=0)^p ((p),(k)) (x-1)^k#
where #((p),(k)) = (p!)/((k!)(p-k)!)# is a binomial coefficient.
(2) If # p notin NN# then we can use the definition of generalized binomial coefficient:
#((p),(k)) = (p(p-1)(p-2)...(p-k+1))/(k!)#
where #p in RR# and #k in NN#
In this case the derivatives of all orders are:
#(df^k)/dx^k = p(p-1)...(p-k+1)x^(p-k) = ((p),(k)) k! x^(p-k)#
and for #x=1#:
#[(df^k)/dx^k ]_(x=1) = ((p),(k)) (k!)#
so the Taylor series is:
#x^p = sum_(k=0)^oo ((p),(k)) (k!)(x-1)^k/(k!) = sum_(k=0)^oo ((p),(k)) (x-1)^k#
Note that if we use the definition of generalized binomial coefficient, we have:
#((p),(k)) = (p(p-1)(p-2)...color(blue)((p-p))..(p-k))/(k!) = 0#
for #p,k in NN# and #k > p#
then we can unify the notation as:
#x^p = sum_(k=0)^oo ((p),(k)) (x-1)^k#
for any #p#.
