# What is the difference between the Taylor and Maclaurin series?

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?

• The main difference is that outside of calculus courses nobody uses the term "Maclaurin series". Or at least nobody who works in mathematics uses that term, in my experience. It's just called the power series or Taylor series at $0$. – KCd Jan 31 '17 at 5:13
• Wouldn't it be easier to google such questions before asking them here? Wikipedia article will give an answer in a couple of seconds – Yuriy S Jan 31 '17 at 9:59
• @Yuriy S Yeah, sorry. I spent a fair amount of time googling it, but my mathematics education is below university level so far, and many explanations left me confused, this is much clearer to me, thank you. – smaude Jan 31 '17 at 15:14

## 3 Answers

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.

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.

• Thanks! Do you need to make any modifications to a series not centered at zero, though? I mean, is it equally valid to center a series anywhere? – smaude Jan 31 '17 at 0:51
• @smaude That is called "Taylor series centered at some other point". :-) – Simply Beautiful Art Jan 31 '17 at 0:55
• Oh, haha, so is that true then? – smaude Jan 31 '17 at 0:57
• @smaude provided the function is analytic at the desired point, yes. But be aware of possibly different regions of convergence of expansions around different points. – Ruslan Jan 31 '17 at 7:36

The Taylor series is a series of functions of the form:

$$f(x)=\sum_{n=0}^{\infty}a_{n}(x-a)^n,$$ where $a_n=\frac{f^{(n)}(a)}{n!}.$ This series is called the Taylor series of $f (x)$ around the point $x = a.$ In the particular case of $a = 0$ the series is called the Maclaurin series.