question about interval of convergence of Taylor series

If a Taylor approximation has an interval of convergence of all real numbers, why would we need to rebuild it at any other x value? If we create e^x at c = 0 to approximate e^0.1, why would we need to redo it at c = 2 to approximate e^2.1? Or am I misunderstanding what interval of convergence tells me?

Quote:

Originally Posted by bmnorvil

If a Taylor approximation has an interval of convergence of all real numbers, why would we need to rebuild it at any other x value? If we create e^x at c = 0 to approximate e^0.1, why would we need to redo it at c = 2 to approximate e^2.1? Or am I misunderstanding what interval of convergence tells me?

The point I think is trying to be made is that certain series may be easier to evaluate at different points than others.

E.g. at the point $\displaystyle x = 2.1$ , the MacLaurin series would require evaluating and summing multiples of $\displaystyle (2.1)^n$ , while the Taylor series centred around $\displaystyle c = 2$ would require evaluating and summing multiples of $\displaystyle (2.1 - 2)^n = (0.1)^n$ , a much easier task...

Re: question about interval of convergence of Taylor series

Thanks for this. It's years later and I understand MUCH more than I did then.