TITLE: What is the difference between the Taylor and Maclaurin series?
QUESTION [26 upvotes]: What is the difference between the Taylor and the Maclaurin series? Is the series representing sine the same both ways? Can someone describe an example for both?

REPLY [37 votes]: A Taylor series centered at $x=x_0$ is given as follows:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
while a Maclaurin series is the special case of being centered at $x=0$:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n$$
You may find this very similar to a power series, which is of the form
$$f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$$
Particularly where $a_n=\frac{f^{(n)}(x_0)}{n!}$.  If a function is equal to it's Taylor series locally, it is said to be an analytic function, and it has a lot of interesting properties.  However, not all functions are equal to their Taylor series, if a Taylor series exists.
One may note that most of the most famous Taylor series are a Maclaurin series, probably since they look nicer.  For example,
$$\sin(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
or,
$$\sin(x)=\sum_{n=0}^\infty\frac{(-1)^n(x-2\pi)^{2n+1}}{(2n+1)!}$$
Which is trivially due to the fact that $\sin$ is a periodic function.  So, if you had to choose, you'd probably choose the first representation.  Just a convention.
The geometric series is a rather beautifully known Maclaurin series, which one may derive algebraically without taking derivatives:
$$\frac1{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+x^3+\dots$$
However, it gets a little bit more involved when you try to take the Taylor series at a different point.