As others have noted, a power series is a series $\sum_{n=1}^{\infty} a_n x^n$ (or sometimes with $x$ translated by some $x_0$, to become $(x - x_0)$). Normally
when one says Taylor series, one means the Taylor series of some particular smooth function $f$. (So in mathematical speech, one wouldn't usually say "consider a Taylor series". You might say "consider a power series", or
"consider the Taylor series of the function $f$". At least, this is my experience.)
One complication in making too much of a distinction is that any power series (say with real coefficients) is the Taylor series of a smooth function (this is a theorem of Borel). So the distinction is more terminological than logical.
Added: Borel's theorem is discussed here.
This will be rather difficult.
Let's first of all suppose there are no problems, i.e. the functions you are interested in are infinitely differentiable and continuous in the range of values you are interested in.
Then, even with the simple example that you know the Maclaurin series of $f(x)$ and want the Taylor series of $f(x)$ about another point $a$, this is a cumbersome endeavour. Here is the procedure.
With the known Maclaurin series (you memorized the $m_k$)
$f(x) = \sum\limits_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} x^k = \sum\limits_{k = 0}^{\infty} m_k x^k$
you require the Taylor series
$f_a(x) = \sum\limits_{k = 0}^{\infty} c_k (x-a)^k$
Now the coefficients are given by
$c_k = \frac{f^{(k)} (a)}{k!}$
which in turn you can get by differentiating your MacLaurin series and evaluate at $a$:
$c_k = \frac{f^{(k)} (a)}{k!} = \sum\limits_{n = k}^{\infty} \frac{n!}{k! (n-k)!} m_n a^{n-k}$
E.g.
$c_0 = \sum\limits_{n = 0}^{\infty} m_n a^{n}$
So you need to sum an infinite series to arrive at your new coefficients - big labour.
Another method is to identify the coefficients of all powers of $(x-a)$. To do so, write the MacLaurin series
$f(x) = \sum\limits_{k = 0}^{\infty} m_k ((x-a)+a)^k
= \sum\limits_{k = 0}^{\infty} m_k \sum_{j=0}^k {k \choose j} (x-a)^j (a)^{k-j}$
This leads to the same result.
Things get worse if you consider $g(f(x))$.
I guess this already answers your question: it's not a smooth technique to calculate the Taylor expansion from the known MacLaurin series.
Best Answer
No. If we consider an infinitely differentiable function $f(x)$, then we can write down it's Taylor series about the point $x = a$ as $$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n.$$ Now, if we consider the translation $h(x)=f(x+a)$ and compute its Maclaurin series, we get $$\sum_{n=0}^{\infty}\frac{h^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n.$$ Note that $$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \ne \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n$$ because they clearly disagree at $x = a$. What we can say, however, is that the Taylor series for $f(x)$ is the Maclaurin series for $g(x) = f(x+a)$ shifted to the right by $a$.