# Another maclaurin vs. taylor series question

 Homework Sci Advisor HW Helper Thanks 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.
 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 $$f(x)=\sum_{n=0}^\infty a_nx^n$$ 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