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What does it mean for a series to be centered around a number? I'm taking complex analysis and am suddenly very confused. I didn't have this explanation, or proof of taylor and power series in calculus, and I'm thinking here, it grew out of complex analysis and not real. But, I'm lookin' through the book, tryin' to get at the proofs, and well, I can't seem to understand why the series is centered around a number.

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You might be interested in this. –  J. M. May 9 '12 at 3:49
Ha! Great. So, we can expand about any point, and use any point that we may think will give us the most efficiency/ desired results? –  user5262 May 9 '12 at 5:59
Yes, depending on the application, we expand with respect to whatever point is convenient. It just so happens that expanding with respect to $0$ (i.e. Maclaurin) happens rather frequently... –  J. M. May 9 '12 at 11:52

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A power series is by definition something of the form $\sum_{n=0}^{\infty} c_n (z - a)^n$ for some sequence of numbers $(c_n)_{n=0}^{\infty}$ and some number $a$. It is said to be "centered at" $a$.

You can regard this as a purely formal thing (ie, just a word that goes with the data defining a power series).

You can also motivate the terminology as follows: if you want to think of the series just written as a function of the complex variable $z$, you need to know the domain of the function: the set of $z$ for which the series converges. And it turns out that, no matter what the sequence $(c_n)_{n=0}^{\infty}$ is, the set of complex numbers $z$ for which the series $\sum_{n=0}^{\infty} c_n (z - a)^n$ converges is a disc centered at $a$. So, power series "centered at $a$".

(For this to make absolute sense and not just be a suggestive phrasing, you need to allow the single point $\{a\}$ and all of $\mathbb{C}$ as a "discs", and allow for junk on the boundary--- ie, not just "closed disc" or "open disc" but "open disc, plus perhaps some junk on the boundary"--- in your definition of "disc". I hope this helped.)

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Note that when you get to Laurent series centred at $a$ (i.e. series in powers of $z-a$ allowing both positive and negative integer powers), the region of convergence might be everything outside a disk centred at $a$, or between two concentric disks (an annulus). –  Robert Israel May 9 '12 at 4:49
But, Leslie, what I am trying to figure out, is exactly where this 'a' comes from. Is it arbitrary? I think I am beginning to understand how the 'a' functions within the definition, for the series, but... why not just always do a Maclaurin Series? I'm gonna get back at my book and see if something elucidates itself with the help of the answers to my posted question. Thx! –  user5262 May 9 '12 at 6:12
Where $a$ "comes from" is up to whoever is defining the power series. In a general discussion of power series, it will probably be arbitrary. In specific examples it will not be. One reason for considering nonzero $a$ is that one often wants series expressions of functions like $f(z) = 1/z$ or $f(z) = \ln(z)$ that are not defined at $0$ and hence do not have MacLaurin series. Of course you can let $g(z) = f(z + a)$ and consider a MacLaurin series for $g$ (note that if $g$ is a sum of powers of $z$ then $f(z) = g(z-a)$ is a sum of powers of $(z-a)$) ... –  leslie townes May 9 '12 at 7:01
so there is no analytical question about general power series that cannot be answered by considering only power series centered at $0$. But it is no "easier" to only consider series at $0$, and in many respects it is unnatural to single out $0$ as the place to do power series. So the definition of "power series" is more general (if only to keep people from having to shift everything to $0$ anytime they want to consider them). The generality offered by being able to change $a$ is mainly in having this flexibility. It does not add anything genuinely "new" to the theory of power series. –  leslie townes May 9 '12 at 7:07
Thank you very much! That precisely answered what I had mind (or didn't, actually)! –  user5262 May 9 '12 at 14:17

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