I'm taking my first real analysis course and I'm trying to get a better feel about analytic functions. My understanding is that an analytic function is one which can be written as a power series. My understanding is that a power series is one of the form $\sum_n a_nx^n$. I was thinking back to Fourier series and I'm pretty sure they don't fit this form.

I'm a little curious about these trigonometric series. Are they analytic? They don't fit the form $\sum_n a_nx^n$. If they are, or can be, what are the circumstances making that so?

My big question is, what type of functions are not analytic? I know of examples such as the absolute value function and such, but why EXACTLY can they not be represented as power series? Does it have to do with smoothness? And given one that's not analytic on all of R - like, I'm guessing a Fourier series with sharp points - how can one represent a smooth piece of it in the form $\sum_n a_nx^n$?

Davide Giraudo
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  • Certainly any analytic function is smooth, but not all smooth functions are analytic. The typical example of a smooth non-analytic function is the *bump function*, defined by $f(x)=e^{-1/(1-x^2)}$ for $|x|<1$ and $f(x)=0$ elsewhere. – Alex Becker Jan 11 '12 at 05:03
  • Are you only considering functions of real variables, or are you also considering complex variables? – Jonas Meyer Jan 11 '12 at 05:04
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    To pick up on one point: $\sin(x)$ is a trigonometric function with a simple Fourier series: it is also a (real-)analytic function at every point, using the power-series expansions for $\sin$ and $\cos$ –  Jan 11 '12 at 05:09
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    Just real variables for right now. (But my ears are open to complex values!) I'm just interested in finding out more about the subject. And Alex, thanks for the reply! Could you explain why the bump function is non-analytic? Can it not be represented as a power series? If so, why not, exactly? Thanks so much for the replies, everyone. I'm glad I joined. – N.G. Jan 11 '12 at 05:10
  • @J.Thompson: The function in Alex Becker's comment has all derivatives equal to $0$ at $x=1$, so its Taylor series about $1$ is $0$. However, $f$ is positive for values arbitrarily close to $1$, while an analytic function would equal its Taylor series in an interval centered at $1$. – Jonas Meyer Jan 11 '12 at 05:16
  • @J.Thompson If the bump function $f(x)$ were analytic we would have some power series which is equivalent to $f(x)$ in some neighborhood around $1$. But to the right of one it must match the power series for $0$ (which is $0$) and to the left it must match the power series for $e^{-1/(1-x^2)}$ (which isn't), a contradiction. More generally, any real analytic function which is zero on some open set (in this case $(-\infty,-1)\cup (1,\infty)$) must be $0$ everywhere. – Alex Becker Jan 11 '12 at 05:17
  • I know that sin(x) is an analytic function. But what about sum(sin(nx)/n)? I guess I should have been more explicit. I thought Fourier series were infinite sums of sins and cosines in the form Sum(a_n*cos(nx) + b_n*sin(nx)) Then we're talking about "trigonometric power series," so the variable "x" doesn't appear as x^n, rather it's "inside" sin or cos. And by the way...how does THAT effect stuff? – N.G. Jan 11 '12 at 05:17
  • @J.Thompson A finite sum is an infinite sum where all but finitely many summands are $0$, so $\sin(x)$ is certainly a Fourier series. – Alex Becker Jan 11 '12 at 05:19
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    Two related threads: http://math.stackexchange.com/questions/47430/is-fourier-series-an-inverse-of-taylor-series http://math.stackexchange.com/questions/7301/connection-between-fourier-transform-and-taylor-series – t.b. Jan 11 '12 at 05:35
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    A little detail that bothers me about what you said: An analytic function is not simply one that can be written as a power series. Rather, is one such that, for each point $a$ of its domain, there is an open interval centered at $a$, where the function can be written as a power series about $a$. It is almost the same, but it explains why $1/(1-x)$ is analytic in its domain ${\mathbb R}\setminus\{1\}$. [Note that the series $\sum x^n$ only converges in $(-1,1)$.] – Andrés E. Caicedo Jan 11 '12 at 06:48

1 Answers1


It is certainly not the case that all Fourier series are analytic; they can represent much more general functions, even discontinuous functions. Wikipedia's Fourier series page has plenty of examples along these lines.

Mark McClure
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    At the same time, $x$ has no Fourier series representation but is clearly analytic, so the two concepts are different but neither more general than the other. – Alex Becker Jan 11 '12 at 05:43
  • If we think of the domain of the Fourier series as an interval with the ends identified, then it makes perfect sense to think of $x$ as having a Fourier series; sometimes, this is called "Fourier series on a circle". But your point is well taken. – Mark McClure Jan 11 '12 at 05:58
  • I see you're point, but I do not think the OP thinks of the domain of Fourier series this way. – Alex Becker Jan 11 '12 at 06:00