Maclaurin and Taylor series

[Vay]

When you find the Maclaurin series form of f(x)= e^(-3x), you will get the summation from n=0 to infinity of (-1)^n * ((3^n)/n!) * x^n. The question is, what is the reason for assuming that the X factor is always x^n, how do we know its not x^(2n) or x^(3n)?

Edited December 13, 2012 by Vay

[Bignose]

It is in the definition of the Taylor series

 

[math]\sum^\infty_{n=0}\frac{f^{(n)}(a)}{n!}\left(x-a\right)^n[/math]

 

This definition is nice because it can be shown to typically have some good convergence over a range, hopefully a range that is large enough for the need at hand.

Edited December 13, 2012 by Bignose

[Vay]

Thanks, I get it now. I just referenced back to the power series and found that it was just its formulation.

 

Another thing has troubled me, it's the 'a' of the Taylor series. Can it be anything? It seems the book is taking random values. Does it mean we can pick any value of 'a', or is there a set center 'a' for which a series would work? It doesn't make a lot of sense to me that there is always a center for the intervals of convergence, and that we can pick this center anywere, at least for the series that converges.

Edited December 13, 2012 by Vay