# Math Help - What exactly does it mean for a power series to be about x=c, where c is any number?

1. ## What exactly does it mean for a power series to be about x=c, where c is any number?

I'm feeling quite confused with series in general: power series, taylor series, maclaurin series, etc.

If I have a series that correctly approximates a function f(x) in any point and is about x = 0, for example, does that just mean that the limit from both sides of x = 0 converges to the same point or does does being about x = 0 mean that the further away from x the value you input for x in the series, the more innacurate it will be compared to the real function itself?

Any input would be greatly appreciated!

2. ## Re: What exactly does it mean for a power series to be about x=c, where c is any numb

A power series is always of the form $\sum a_n(x- c)^n$. That sum is "about x= c". Once can also show that a power series always has a "radius of convergence", a number, r, such that the series converges for x in (c- r, c+ r) and diverges outside [c- r, c+ r] (may or may not converge at the endpoints, c- r and c+ r) (for real coefficients- if we allow complex numbers, the "interval of convergence" is the disk $\{z| |z- c|< r\}$ and so really is a "radius").

If you are actually talking about "series"- the infinite sums, then, no, being further from c does not affect "accuracy". For a function "analytic" inside it radius of convergence of the entire infinite sum gives the exact function. (There are functions, f(x), that have Taylor's series that converges for all x (the "radius of convergence is infinite) but NOT to f(x). The simplest example is " $f(x)= e^{-1/x^2}$ if x is not 0, f(0)= 0".)

Of course, if you are using finite sums, to nth power, to approximate a function, then the Lagrange error formula is $\frac{M}{(n+1)!}|x- c|^{n+1}$ so that, yes, if you are further from c, the error may be greater because $|x- c|^{n+1}$ is larger. But that is an upper bound on the error- not necessarily the error.