# question about interval of convergence of Taylor series

• Feb 16th 2011, 09:09 AM
bmnorvil
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?
• Feb 16th 2011, 06:02 PM
Prove It
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 \displaystyle x = 2.1\$, the MacLaurin series would require evaluating and summing multiples of \$\displaystyle \displaystyle (2.1)^n\$, while the Taylor series centred around \$\displaystyle \displaystyle c = 2\$ would require evaluating and summing multiples of \$\displaystyle \displaystyle (2.1 - 2)^n = (0.1)^n\$, a much easier task...
• Feb 28th 2013, 06:15 AM
bmnorvil
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.