Register to reply

Another maclaurin vs. taylor series question

by eprparadox
Tags: maclaurin, series, taylor
Share this thread:
eprparadox
#1
Jul24-14, 11:25 PM
P: 84
Hey guys,

Struggling with understanding this taylor vs. maclaurin series stuff.

So a few questions. Let's say that we have some function f(x).

1. By saying that we want to find the power series of f(x) and nothing else, are we implicitly stating that we are looking for a maclaurin series or a taylor series? OR do we have to specify around what point we're looking for this power series?

2. I went on wolfram alpha and I found the taylor series of sin(x) at x =2 and the maclaurin series of sin(x) (at x = 0). and then I evaluated the answers at the same x value (x = 4). And I got different answers. I thought they should come out to the same value since we're still expanding the same initial function, sin(x). Ultimately, I'm finding it difficult to understand how these two seemingly different power series are converging to one function.

3. What does it mean for a power series to be centered around a value, other than for it to be the center of the circle of convergence.

Hope these questions made sense. I just want to get a really strong intuition for maclaurin vs. taylor series.

Thanks!
Phys.Org News Partner Mathematics news on Phys.org
Math journal puts Rauzy fractcal image on the cover
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Simon Bridge
#2
Jul25-14, 01:19 AM
Homework
Sci Advisor
HW Helper
Thanks
Simon Bridge's Avatar
P: 13,052
http://en.wikipedia.org/wiki/Taylor_series
A Maclearan series is a Taylor series, but not all Taylor series are Maclearan series.
Both are examples of power series.

Power series: $$f(x)=\sum_{n=0}^\infty a_nx^n$$

Maclearan series: $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$

Taylor series:$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$$

So:
1. if the type of series is not indicated, then it doesn't matter to the argument being made.
Where there may be confusion, following statements should clear it up.

2. the series expansion for sin(x) has infinitely many terms - you were evaluating two different approximations so you got different answers.

3. see the wikipedia article (above) for details.
The power series is most accurate close to the point x=a in the formula - the series is said to be centered around this value. The Taylor series makes that point explicit while for the general power series it may not be so clear.
eprparadox
#3
Jul25-14, 02:13 AM
P: 84
Hey, thanks so much for the response. I'm reading Mary Boas' "Mathematical Guide in the physical sciences" and in it, she states a few theorems on power series. One of them is this:

"The power series of a function is unique, that is, there is just one power series of the form [tex] f(x)=\sum_{n=0}^\infty a_nx^n[/tex] which converges to the given function."

So in our case, our function is the sin(x) but there is more than one power series that converges to that function (the maclaurin series and the taylor series). Are they saying there is only one unique power series for any given POINT on the function?

Thanks again for the help

Simon Bridge
#4
Jul25-14, 02:34 AM
Homework
Sci Advisor
HW Helper
Thanks
Simon Bridge's Avatar
P: 13,052
Another maclaurin vs. taylor series question

The quote refers to the whole power series. There are many series which will sum to an approximation to the function.

The taylor series about a=4 and about a=0 are two ways of arriving at the same power series.
The steps are different.


Register to reply

Related Discussions
Help with Taylor Series/Maclaurin Series Question Calculus & Beyond Homework 15
Confused about Taylor and Maclaurin Series Calculus 4
Calculus II Help : Taylor/Maclaurin Series Calculus & Beyond Homework 1
MacLaurin and Taylor series General Math 1
Can someone help me understand Taylor and MacLaurin series? Calculus 4