# What exactly is an “analytic function”?

This is for real analysis so I'm not worried about complex analytic functions.

The definition in my book just says: "A function f(x) which is represented by a power series with a positive radius of convergence is said to be 'real analytic at the origin', or simply 'analytic'."

To me it seems like this definition ends prematurely. Like when I was reading it I expected there to be an "if" and then a list of qualifications.

I guess I just don't really understand the rationale behind creating this definition in the first place. What would even be the point of representing a function by a power series that didn't have a positive radius of convergence? If the power series doesn't converge for any values of x, it's not really representative of anything, right?

I feel like this is actually obvious and I'm just overthinking it, but is this definition literally just giving a name to functions that can be represented as power series that actually converge for x values in some radius? There was an exercise earlier in the chapter before analytic functions were defined that asked us to find an infinite power series which represented a function; was that function analytic as well?

Thanks in advance and sorry if this is a silly question.

• Yes, you have it exactly right. The point isn't that you would represent a non-analytic function with a power series; as you observed, that would be silly. The point is that some functions can't be so represented, and it's useful to have a special term to distinguish functions that can be so represented from functions that can't. – MJD Apr 28 '16 at 5:44
• This might be an English sentence parsing issue; in the definition, try replacing 'which is represented' by 'which can be represented'. – Ted Apr 28 '16 at 6:22
• Actually, power series with zero radius of convergence can be useful: see asymptotic series. Just not as nice as those that have positive radius of convergence. – Robert Israel Apr 28 '16 at 6:25
• The definition point out that the radius of convergence is $\rho>0$ and we have to discard $\rho=0$. And if you like the "if" formulation, you can also rephrase as: "A function f(x) is said to be real analytic if it is represented by a power series with a positive radius of convergence". For analytic functions, we can di§erentiate and integrate them termwise, add them by adding their power series, and multiply them by multiplying their power series. Derivatives, antiderivatives, sums, and products of analytic functions are all analytic. And the power series is the Taylor series. – alexjo Apr 28 '16 at 6:40
• Yes, any function can be approximated (interpolated) increasingly exact by polynomial (finite power series) of increasing degree. But here "power series around $x$" want the series to have same derivative as the function. To find it, you don't depend on $f$ in whole domain, but on how f behave near x. So if $f(x)=a_0x^0+a_1x^1+...$ at 0 then $a_0=f(0),a_1=f'(0),a_2=\frac{f''(0)}2,$ and so on. It does not depend on how $f$ behave at other points other than 0. And if the series equal to $f$, then we say that $f$ is analytic with the radius of convergence equal to how far from 0 the series exact. – user202729 May 26 '16 at 15:45

Some functions can't be represented by a power series while others can. We have this definition to put functions into this analytic or not analytic category -- we don't have to say "$f$ can be expanded into a power series", we just have to say "$f$ is analytic." Another thing, a lot of the time we can have equivalent definitions of the same thing. This way we can avoid confusion by calling it by one name.