In general, a power series of the form #\sum_{n=0}^{\infty}c_{n}(x-a)^{n}=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+c_{3}(x-a)^{3}+\cdots# is said to be "centered at #x=a#". The reason for this terminology is that a power series like this will converge on some interval centered at #x=a# (though it might just converge at #x=a# and it can also converge over the entire real line #RR#, which is not, technically, centered anywhere). In other words, it will converge on some interval with #a# at the midpoint of the interval (again, excluding the cases where it only converges at #x=a# or where it converges for all #x\in RR#).
For example, the real power series #\sum_{n=0}^{\infty} 1/3^{n}(x-7)^{n}=1+1/3 (x-7)+ 1/9 (x-7)^2+ 1/27 (x-7)^3+\cdots# is a geometric series and can be shown to converge for all #x# in the open interval #(4,10)#. Note that this interval is centered at #a=7#.
Furthermore, in this last example, you can actually take #x# to be a complex number and the series converges for all #x\in CC# with the property that #|x-7|<3# (the distance between #x# and 7 is less than 3), which is an open "disk" (the inside of a circle) in the complex plane centered at 7.
Taylor series are special kinds of power series. Given a function #f# that is infinitely differentiable over some open interval #(c,d)# containing the number #x=a#, you can compute the corresponding Taylor series for #f# centered at #x=a# as the following expression:
#\sum_{n=0}^{\infty} f^{(n)}(a)/(n!)(x-a)^{n}=f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!) (x-a)^3+\cdots#
As before, this will converge on some interval centered at #x=a#.
For the geometric series example above, it happens to be the Taylor series for #f(x)=3/(10-x)# centered at #a=7#.
