# What is an analytic function?

1. May 25, 2005

### Peter VDD

What does it mean "an analytic function", "analyticity" of a function?

I know we have seen it but i cannot find it back.

2. May 25, 2005

### matt grime

eh? where are you looking for the definition? mathworld will tell you.

f is analytic if f satisfies the cauchy riemann equations

3. May 25, 2005

### quasar987

The definition I have learned is that a function is analytic on an interval if it equals its MacLaurin serie in that interval.

So f(x) is analytic in (-a,a) if

$$\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^n=f(x) \ \forall x\in(-a,a)$$

4. May 25, 2005

### mathwonk

your definition is a classical one, and holds fopr both real and copmplex variables. In the compelx case, where matt is working, there are (at least) three equivalent ways to characterize these functions:

1) locally given by convergent power series, i.e. taylor series (maclaurin series)

2) smooth and satifies cauchy riemann equations, given by matt (and by riemann in the first few pages of his ianugural dissertation as a necessary and sufficient condition for the derivative dw/dz of a smooth function w(z), to be "independent of dz", i.e. to have no dzbar part).

3) has everywhere a single complex derivative. (osgood's theorem?)

1) "easily" implies 2) and 3); that 3) implies 1) usually uses the Cauchy integral formula for functions satisfying 3), since then one can expand the "kernel" in the integrand in a power series.

5. May 25, 2005

### matt grime

I think I suffer from a malaise called the Cambridge syndrome whereby X we mean what everyone else calls Y. In particular I think I take analytic for what other people take to mean holomorphic. Sorry.

6. May 25, 2005

### mathwonk

actually you are in accord with numerous US books, matt. this is one concept whose name is not fixed uniformly, and all permutations of the possibilities are found. I think i am using the one from my favorite book, by Henri Cartan, obviously french, so perhaps with a different cultural preference as to the meaning.

We get in a lot of trouble here trying to write a PhD quals questions on this. Imagine the delight of many students when asked to prove that a holomorphic function has a taylor series, when their definition of holomorphic is that it is a convergent power series.

7. May 25, 2005

### Galileo

I learned f to be analytic in a point z, if there exists a neighbourhood of z on which f is differentiable.

8. May 25, 2005

### Hurkyl

Staff Emeritus
I also learned a function was analytic if it was locally equal to a power series.

9. May 25, 2005

### gravenewworld

Analytic functions=good

10. May 25, 2005

### HallsofIvy

Staff Emeritus
I think most mathematicians would accept as the fundamental definition of "analytic at a": "There exist some neighborhood about a such that the function is equal to its Taylor Series expansion about a in that neighborhood" (quasar987, I would not say "MacLaurin series" since a function analytic at, say 1 may not even be differentiable at 0 and so would not have a MacLaurin series.)

Of course it then follows that a function is analytic at a if and only if it
satisfies the Cauchy-Riemann equations at a if and only if it is differentiable at a.

Those last two only apply to complex functions of a complex variable whereas the definition can be applied to real functions of a real variable.

11. May 25, 2005

### mathwonk

I am reading Riemanns inaugural dissertation at the moment, as I have said, and he proposes there that properties which are possessed by functions formed from "analytic operations" should be taken as models of well behaved complex functions w(z). In particular they have the property that their derivatives dw/dz can be expressed using the usual rules for differentiation in terms of z. In particular "the ratio dw/dz does not depend on the value of dz."

He then writes out an expression for the ratio dw/dz, when w(z) is any smooth function of z, and observes that independence of dz is equivalent to the cauchy riemnan equation equations being satisifed.

i.e. in modern language he computes that dw =
(dw/dz) dz + (dw/dzbar) dzbar,

and hence for dw/dz to be independent of dz (equivalently dzbar), he needs dw/dzbar to be zero, i.e. the riemann cauchy equations to be satisfied.

so to paraphrase gravenewworld,
dw/dz independent of dz, good; otherwise: bad.

so people assuming these equations as the characterization of analytic functions are in very good company.

By the way, these works of riemann are incredible, if you have not read them, as i had not for the past 40 years.

In 15 pages he done, starting from scratch, cauchy integral theorem, argument principle, stokes theorem, topology of bordered surfaces, branched covers of regions in the plane, etc... etc.....my mind is blown just trying to keep up. (but no power series yet.)

And with all I have learned in my life, some of this stuff from 1851 is still new to me!

for example i did not know until today, that a connected oriented bordered surface which is a branched cover of a region in the plane, and becomes simply connected and connected when n-1 "transverse" cuts are made, has at most n boundary components, and if less, then the number of boundary components differs from n by an even number.

I also did not know that if one makes r transverse cuts and reduces the surface to a union of s simply connected components, then the number r-s is an invariant of the surface.

I am not there yet, but presumably this leads to riemanns notion of the topological genus. I.e. a compact connected oriented surface of genus g, with a small disc removed, has connectivity 2g+1, in the sense that it takes 2g transverse cuts to reduce it to a connected, simply connected surface.

this guy was "out there". :rofl:

12. May 26, 2005

### Galileo

Satisfaction of the Cauchy-Riemann relations at a point may not guarantee the existence of f' at that point. If you add the assumption that all first order partial derivatives ($u_x,u_y,v_x,v_y$) are continuous at a point, then that's sufficient to guarantee the analyticy of f at that point.

13. May 26, 2005

### mathwonk

riemann always worked in an open set, and was not as picky as people are today. He was working out the essential features of functions formed from analytic expressions, then setting those up as the important things to focus on. Years later people made precise definitions designed to give the right, or minimal, hypotheses for his results.

It is a little foreign to us today, as there was a lot of discovery in it. he was telling us what is true, and later we gave the precise hypotheses to prove his statements.

Whereas today we say, here are the definitions and hypotheses, now what can be deduced, he sort of said here is what must be true in a reasonable case, now you figure out the limitations on these statements.

For example when he decomposes surfaces, he tacitly assumes all the "lines" or paths he is considering meet each other in a finite set of points. he never even defined "lines". But the proof arguments are so clear, that afterwards, if you are so inclined, you can try to formulate the precise definitions that make his proofs work.

nowadays we have many precise versions of analyticity, even distribution theoretic versions, and can even speak of analyticity of functions in this sense that are not even continuous, but they turn out to be equal almost everywhere to a unique classically analytic function.

as you probably know better than me, a function satisifes the cauchy riemann equations in the distribution sense if the cauchy riemann operator yields a function which kills all other smooth functions when integrated against them.

This makes sense by integration by parts, since you can transfer the derivative from the measurable function to the smooth (compactly supported) one, under the integral sign.

i.e. a functiom which is not even differentiable, is killed by a differential operator, if it integrates to zero against the result of applying the dual operatoir every smooth compactly supported function.

(This is probably wrong too, in many details, but correct in spirit.)

i.e. suppose for good fucntions we have dual operators d and d* such that

<df,g> = <f, d*g>. then we can say that df = 0, even for f where d is not defined, provided <f,d*g> = 0 for all good g.

i.e. at least for good f, that would say df = 0, since the <df,g> = 0 for all good g.

then one proves a representation theoirem that for nice operators such as th cauchy riemann operator, or any elliptic operator maybe, that if df = 0, in the sense that <f,d*g> = 0 for all good g, then in fact f is equivalent a.e. to some good function f, such that df = 0.

so distribution theory is nothing but integration by parts.

the advantage is it breaks the problem of solving a d.e. into two parts, finding a measurable solution and then showing that solution is actually smooth.

Last edited: May 26, 2005
14. May 26, 2005

### HallsofIvy

Staff Emeritus

Exactly right. Thanks for correcting me.