I am going to take a break from the Lebesgue stuff and maybe write more on that tomorrow.
My numerical analysis class just turned in some homework and some really have some misunderstanding about Taylor Series and Power Series. I’ll provide some helpful hints to perplexed students.
For the experts who might be reading this: my assumption is that we are dealing with functions which are real analytic over some interval. To students: this means that can be differentiated as often as we’d like, that the series converges absolutely on some open interval and that the remainder term goes to zero as the number of terms approaches infinity.
This post will be about computing such a series.
First, I’ll give a helpful reminder that is crucial in calculating these series: a Taylor series is really just a power series representation of a function. And if one finds a power series which represents a function over a given interval and is expanded about a given point, THAT SERIES IS UNIQUE, no matter how you come up with it. I’ll explain with an example:
Say you want to represent over the interval . You could compute it this way: you probably learned about the geometric series and that for .
Well, you could compute it by Taylor’s theorem which says that such a series can be obtained by:
If you do such a calculation for one obtains , , and plugging into Taylor’s formula leads to the usual geometric series. That is, the series can be calculated by any valid method; one does NOT need to retreat to the Taylor definition for calculation purposes.
Example: in the homework problem, students were asked to calculate Taylor polynomials (of various orders and about ) for a function that looked like this:
. Some students tried to calculate the various derivatives and plug into Taylor’s formula with grim results. It is much easier than that if one remembers that power series are unique! Sure, one CAN use Taylor’s formula but that doesn’t mean that one should. Instead it is much easier if one remembers that Now to get one just substitutes for and obtains: . Then and one subtracts off to obtain the full power series:
Now calculating the bound for the remainder after terms is, in general, a pain. Sure, one can estimate with a graph, but that sort of defeats the point of approximating to begin with; one can use thumb rules which overstate the magnitude of the remainder term.