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?
2026-04-20 02:46:24.1776653184
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What is the difference between the Taylor and Maclaurin series?
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A MacLaurin series is a special occurrence of the Taylor Series where the series is constructed around x=0.
MacLaurin series are generally used if able to.
For example, you can estimate $f(x)=\sin{x}$ with a Maclaurin series.
However, you can't estimate $f(x) = \frac{1}{x}$ with a Maclaurin series because $f(x)$ is undefined when $x=0$, so most people choose to center it around $x=1$. Usage is all about preference.
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.