x Turn on thread page Beta
 You are Here: Home >< Maths

# Taylor and Maclaurin Series watch

1. Currently first year undergrad, but I think this was on further maths, but sorry if this isn't still sixth form.

Basically, I understand that the Maclaurin series is simply just a type of Taylor series, where x is centered at x=0. But what does this value which x is "centered" at actually mean?

For example, if I work out the Maclaurin series for e^x, centering x at 0, but then also work out the Taylor series when x is centered at -1, I'll gain a different expansion. I do not understand the significance of what it means by x being centered at a value, and why it changes the series. Thanks for any help.
2. (Original post by No-Idea.)
Currently first year undergrad, but I think this was on further maths, but sorry if this isn't still sixth form.

Basically, I understand that the Maclaurin series is simply just a type of Taylor series, where x is centered at x=0. But what does this value which x is "centered" at actually mean?

For example, if I work out the Maclaurin series for e^x, centering x at 0, but then also work out the Taylor series when x is centered at -1, I'll gain a different expansion. I do not understand the significance of what it means by x being centered at a value, and why it changes the series. Thanks for any help.
Taylor series gives you a succession of polynomials which approximate the original function near a point. Of course, unless your original function was polynomial, the same polynomial isn't going to approximate the function near both 0 and -1 (except by bizarre coincidence). Hence you get a different expansion if you expand around 0 or -1.
3. (Original post by Smaug123)
Taylor series gives you a succession of polynomials which approximate the original function near a point. Of course, unless your original function was polynomial, the same polynomial isn't going to approximate the function near both 0 and -1 (except by bizarre coincidence). Hence you get a different expansion if you expand around 0 or -1.
If you kept taking more terms in the taylor series would it keep becoming more accurate for terms that are further and further away from the centered point?

If so why bother taking the series centered at some value as opposed to just taking it at 0 and extending it until it was sufficiently accurate for whatever value you are interested in?
4. But what is the relevance of the centered point? What do you actually mean when saying "expand around 0 or -1"?
5. Let z be the point we expand around, and some other point. Let f be the function we're expanding.

(Original post by poorform)
If you kept taking more terms in the taylor series would it keep becoming more accurate for terms that are further and further away from the centered point?

If so why bother taking the series centered at some value as opposed to just taking it at 0 and extending it until it was sufficiently accurate for whatever value you are interested in?
Only if a lies within the radius of convergence of the series taken around z. In the case of log, for instance, the expansion of log(1+x) around 0 doesn't converge for x>1, so to get an approximation of log(3) you'd have to expand around somewhere else.

(Original post by No-Idea.)
But what is the relevance of the centered point? What do you actually mean when saying "expand around 0 or -1"?
The series you end up with is a perfect approximation to f at z, because it takes the same value as f does there. It's a reasonable approximation to f at points in the neighbourhood of z (that is, if z-a is "small" for certain values of "small"). The approximation in general gets worse and worse as z-a gets larger.

The Taylor approximation essentially consists of taking lots of information about the various derivatives of f at a point z, and combining them in a way that produces a polynomial which looks a lot like f. However, "derivative" is a local property: it only tells you much about the function close to the point where you took the derivative. (Example: f(x) = e^x: the 5th derivative at 1 is approximately 148. What does that tell you about f at a=10? Not very much, really, without a lot more information: all you really know is that it's bigger than e, and to get that we also needed the previous derivatives to be positive.)

"Expand around 0" means "take an approximation to the function which we only guarantee to be vaguely correct near 0".
6. Thanks, I think I'm starting to understand it now.

Why is the Taylor series actually useful for us then? What is the need to find the series around a given point z?
7. (Original post by No-Idea.)
Thanks, I think I'm starting to understand it now.

Why is the Taylor series actually useful for us then? What is the need to find the series around a given point z?
So that you can easily evaluate the function. is really quite hard to evaluate, being an infinite sum, but if you can give me a polynomial which is close to it somewhere, then I can evaluate that polynomial super-fast.

Also it has uses in Analysis more generally: in complex analysis, for instance, all holomorphic functions locally have power series, so you know that holomorphic functions "aren't horrible" locally, and it lets you prove results about arbitrary weird holomorphic functions using easy things about power series.
8. (Original post by Smaug123)
So that you can easily evaluate the function. is really quite hard to evaluate, being an infinite sum, but if you can give me a polynomial which is close to it somewhere, then I can evaluate that polynomial super-fast.

Also it has uses in Analysis more generally: in complex analysis, for instance, all holomorphic functions locally have power series, so you know that holomorphic functions "aren't horrible" locally, and it lets you prove results about arbitrary weird holomorphic functions using easy things about power series.
Thanks a lot, appreciate it.
9. How do I find the higher derivatives of functions? i don't need to use the expansion right to find higher derivatives.

Whats the difference between Maclaurin and Taylor?

A thorough explanation will be appreciated

Thanks
10. (Original post by john122334455)
How do I find the higher derivatives of functions? i don't need to use the expansion right to find higher derivatives.

Whats the difference between Maclaurin and Taylor?

A thorough explanation will be appreciated

Thanks
Please don't bump ancient threads. Start a new one if you have a question.

You can find higher derivatives by repeated differentiation.

A Maclaurin expansion is a special case of a Taylor expansion centred at zero.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 8, 2015
Today on TSR

### Loughborough better than Cambridge

Loughborough at number one

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams