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Real- and Complex- Analytic Functions.

  1. Jul 23, 2010 #1
    Hi Again:

    I don't know if this is obvious or not: An analytic complex function
    f(z)=u(x,y)+iv(x,y) , can be made into an analytic function
    f: R^2 -->R^2, since each of u(x,y) and v(x,y) is itself a real-analytic
    function, i.e., we can use a standard argument by component function.

    How about in the opposite direction, i.e., we have a real-analytic
    function f(x), analytic in an interval (a,b). When can we extend
    f(x) into a complex-analytic function.?. I suspect , thinking of power series,
    that we can use the radius of convergence to construct an analytic function, i.e.,
    if f(x) is analytic in (a-r,a+r), then f(x) can be extended to a complex-analytic
    function in |z-a|<r .

    Is this correct.?
  2. jcsd
  3. Jul 24, 2010 #2
    This can be seen as follows: if [tex]f(z)=u(x,y)+\mathrm{i} v(x,y)[/tex] is complex analytic, then [tex]u[/tex] and [tex]v[/tex] are harmonic hence (real) analytic.

    Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.
  4. Jul 24, 2010 #3
    While Anthony's statement is true, the answer to Bacle's question is "no". Consider

    [tex]f(x) = \frac{1}{1 + x^2}[/tex]

    in the interval (-1, 3), Using Bacle's notation, a = 1, r = 2. Then f is real-analytic in the interval (-1, 3), but f is not complex-analytic in |z - 1| < 2 due to the poles at +-i. The function f would be complex-analytic in some disk that didn't include the poles, however.
  5. Jul 24, 2010 #4
    Apologies, I only read the first bit! That will teach me.
  6. Jul 29, 2010 #5
    PeteK, Anthony: Thnaks for your posts. Please check on this; I would appreciate

    your comments:

    Just a question: I guess we could look at 1/(1+x^2) on the real line as agreeing

    with 1-x^2+x^4 -... =Sum_(x=0...n) x^2n on (-1,1). We can also see it as

    the restriction of 1/(1+z^2) , to the real axis, so that, by the identity theorem

    (using the fact that (-1,1) has limit points) , 1/(1+z^2) is the analytic extension

    of ( 1/(1+x^2)) to the complex plane, and :

    1/(1+z^2) = 1-z^2+z^4 -...... on |z|=1

    But then it seems like 1/(1+x^2) on (-1,1) can be extended to the disk |z|=1.

    And it seems like the above could be generalized to analytic functions defined on

    the real line; by the comparison criterion, if f(x) were analytic in (-r,r) (we can

    translate the function if necessary so that it converges in that interval) , then

    we would have -1< |(a_nx^n)/ (a_(n-1)x^n-1)|<1 in (-r,r), which I believe

    would imply that the analytic extension of f(x) to the complex plane would converge

    in that same interval.?
    real line
  7. Jul 29, 2010 #6
    Sorry, I forgot to ask another followup:

    Anthony: I understand that if f(z)=u+iv , then both u,v are harmonic, but
    how does it follow from u,v being harmonic, that each is real-analytic.?.

    I understand that from harmonicity, the second derivatives would be killed,
    but it seems many other mixed partials would not be killed by u (equiv. v)
    being harmonic.

    Would you please expand.?

  8. Jul 29, 2010 #7
    What's a real analytic function?
  9. Jul 29, 2010 #8
    First, let's consider Anthony's statement that

    It would be more precise to say that "Any real analytic function on some open set on the real line can be extended to a complex analytic function on some open set of the complex plane." The function [itex]f(x) = 1/(1 + x^2)[/itex] shows that a real-valued function can be analytic on all of R, but not have a continuation that is analytic on all of C.

    Next, in your above statement you probably mean the disk {z: |z| < 1} since the complex power series [itex]1 - x^2 + x^4 - ...[/itex] doesn't converge, for example, if z = 1.

    Finally, in your last paragraph, do you mean something like "If f(x) is analytic in (r,r), then it has a continuation to the complex numbers that is analytic in the disk {z: |z| < r}"? If so, it not true as shown by the example of f(x) as given above and with any r > 1. If you meant something else, please restate.
  10. Jul 29, 2010 #9
    See the Wikipedia article on analytic functions.
  11. Jul 29, 2010 #10
    Thanks, PeteK.

    Would you please give a hint for why 1/(1+x^2) is real-analytic in
    the interval (-1,3).? I can't see it. I can see how it agrees with:

    Sum_i=0...oo x^(2n) =1-x^2+x^4-........ in (-1,1)

    But I don't see how it is real-analytic beyond (-1,1).

    Would you please explain.?

    P.S: you were right, the radius of convergence of f(z)=1-z^2+z^4-....

    is 1 , and, yes, there are poles at +i, -i' my bad. Tho I think

    f(z) can be continued beyond the disk.

  12. Jul 29, 2010 #11


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    If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2.
  13. Jul 29, 2010 #12
    "If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2. "

    Yes, but that is precisely where I am stuck; I can see how :


    is defined on (-1,1), and it agrees with 1/(1+x^2) , but I cannot

    think of any other series defined outside of (-1,1), which agrees

    with 1/(1+x^2). Can you spare a paradigm.?
  14. Jul 29, 2010 #13
    Do you know Taylor's Theorem and Taylor Series?
  15. Jul 29, 2010 #14
    Yes, I do know both, but I have not seen them for a few years now;
    I don't know if it is more accurate to say that I knew them.

    Since the functions we are working with are real C<sup>oo</sup> ,
    I guess the part of the theorem that you may be referring to is that,

    I do know, e.g., that, in the real case, if the radius of convergence
    of the Taylor series T(f(x)) of f(x) ) is r (let's standardize to (-r,r)
    again, by translation) , then T(f(x)) does not converge to f(x)
    either at r, or at -r , or both. Same goes for the complex case; there
    must be a point of non-convergence in the boundary of the circle of
    convergence, like in the case of 1/(1+z^2) , where you pointed out
    that there are poles at +i and at -i . (tho we may be able to continue
    f(z) analytically in C-{i,-i} . )

    This is the property I was using for 1/(1+x^2) , which has radius of
    convergence one, i.e., it converges in (-1,1), and diverges at x=1 and
    at x=-1 .
  16. Jul 29, 2010 #15


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    The point is that a disk centered on 0 is probably not the best choice if you're looking for a disk that contains 2.
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