Results 1 to 10 of 10

Math Help - Complex analysis, infinite series expansion of a function

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Complex analysis, infinite series expansion of a function

    I'm studying from my class notes and I don't understand a step. The function \log (x+1) is analytic over \mathbb{C}-\mathbb{R}^{\leq 1}.
    Furthermore \log (x+1)=\sum _{n=1}^{+\infty} \frac{ f^{n} (0) z^n}{n!}=\sum _{n=1}^{\infty} \frac{(-1)^{n-1}z^n}{n}. From this I get that the radius of convergence of the function is 1, ok but my problem is how could they write the function log(x+1) as an infinite series whose "center" is 0 while the function isn't analytic there. I know I'm not expressing myself well. I mean why do they used f^{n}(0) instead of say any number greater than 2 (since the radius of convergence is 1 and the function isn't analytic for any real number lesser or equal than 1).
    So it would mean that any function, regardless of analytic or not, can be written in terms of infinite series? It seems to contradict a theorem that I can't find at the moment.
    Any help is greatly appreciated, as usual.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I think you're confusing +1 with -1. The argument of the logarithm function needs to be in the region \mathbb{C}-\mathbb{R}^{\le 0}, to use your notation. I think you'll see that this implies x+1 must be in the region \mathbb{C}-\mathbb{R}^{\le -1}. As a check: f(0)=\ln(0+1)=\ln(1)=0, which is perfectly well-defined, and in an analytic region of the logarithm function.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Ok thank you, you're right indeed.
    I have another question(s) then. Is it possible to write the function log(x+1) in terms of a Laurent series centered at -10 (on the real axis), at -1? I hear/read about poles, singularities, etc. But in the case of a "straight line" of undefinition of this function, are each points of the line considered as essential singularities? Is it still possible to get a Laurent series centered at any point in the complex plane of this function in particular and of any function?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I think you could get a Laurent series expansion about -10 if you took a different branch of the logarithm function - one that doesn't have a branch cut discontinuity on the negative real axis. I'm a bit hazy about different branches, I'm afraid. Others on the forum might be able to help out a bit more on that.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by arbolis View Post
    Ok thank you, you're right indeed.
    I have another question(s) then. Is it possible to write the function log(x+1) in terms of a Laurent series centered at -10 (on the real axis), at -1? I hear/read about poles, singularities, etc. But in the case of a "straight line" of undefinition of this function, are each points of the line considered as essential singularities? Is it still possible to get a Laurent series centered at any point in the complex plane of this function in particular and of any function?
    As Ackbeet said, you would have to take a different branch of the function. The branch cut can be placed pretty much anywhere; in fact, any non-intersecting curve joining the two branch points (in this case -1 and \infty) will do the job.

    A branch of a multivalued function is simply not defined on the branch cut; it is impossible to assign any value to a point there in a way that will result in the function being analytic at that point. You have to pick a different branch! The points of the branch cut, should you assign some values to them, would be neither poles, nor removable singularities, nor essential singularities, but they'd still be singular points.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Bruno J. View Post
    As Ackbeet said, you would have to take a different branch of the function. The branch cut can be placed pretty much anywhere; in fact, any non-intersecting curve joining the two branch points (in this case -1 and \infty) will do the job.

    A branch of a multivalued function is simply not defined on the branch cut; it is impossible to assign any value to a point there in a way that will result in the function being analytic at that point. You have to pick a different branch! The points of the branch cut, should you assign some values to them, would be neither poles, nor removable singularities, nor essential singularities, but they'd still be singular points.
    Ok thanks a lot. I'm not sure I'm getting a right intuition. I understand that the main branch of the function cut at -1 on the real axis and that the function isn't defined there. I remember from my class notes that the definition of an essential singularity is "a singularity that isn't a pole nor a removable singularity". So now I think the definition is just wrong if you said that the point -1 is a singularity that is neither a pole, removable singularity nor an essential singularity.
    I still don't understand why it's meaningless to write a Laurent series or a function for points where there are singularities that are neither poles, essential/removable singularities.
    I know I can't assign any value out of the domain of the function log (x+1) that could make it analytic; this is why the points on the real axis lesser or equal to -1 are not removable singularities.
    All in all, a Laurent series can only be used to write a function over a domain where the singularities (poles, removables and essentials ones) are isolated?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Many different questions there!

    I remember from my class notes that the definition of an essential singularity is "a singularity that isn't a pole nor a removable singularity". So now I think the definition is just wrong if you said that the point -1 is a singularity that is neither a pole, removable singularity nor an essential singularity.
    We have to be a bit picky here. For a point to qualify as a removable singularity of a function (in the usual sense), the function must first be holomorphic on a punctured open neighbourhood of this point. This is not the case for a point on a branch cut; the function is not even continuous along the cut! But we can extend the notion of a removable singularity to points which lie on a branch cut (other than the branch points); because we can, indeed, define the function properly there (by moving the branch cut, or, what is equivalent, by analytic continuation).

    So I'd say that points on a branch cut (other than the branch points) qualify as removable singular points in some extended sense. They definitely do not qualify as essential singularities or poles. (An essential singularity is essentially a pole of infinite order.)

    Essentially, what I'm saying is that if you look at the function \log(1+x) as a single-valued function defined on its Riemann surface, then there are only two troublesome points, -1 and \infty, which are essential singularities. Branch cuts are only a construct which allow us to "fit" inside the plane a function which is "too big" for the plane.



    I still don't understand why it's meaningless to write a Laurent series or a function for points where there are singularities that are neither poles, essential/removable singularities.
    It isn't! You can expand your function in a Laurent series (in fact, in a Taylor series) around every point other than -1 and \infty. You just have to remember that you will not obtain a single-valued function in this way.

    All in all, a Laurent series can only be used to write a function over a domain where the singularities (poles, removables and essentials ones) are isolated?
    Yes; a function has no Laurent expansion around a nonisolated singular point (which is necessarily an essential singularity).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Bruno J. View Post
    Many different questions there!



    We have to be a bit picky here. For a point to qualify as a removable singularity of a function (in the usual sense), the function must first be holomorphic on a punctured open neighbourhood of this point. This is not the case for a point on a branch cut; the function is not even continuous along the cut! But we can extend the notion of a removable singularity to points which lie on a branch cut (other than the branch points); because we can, indeed, define the function properly there (by moving the branch cut, or, what is equivalent, by analytic continuation).

    So I'd say that points on a branch cut (other than the branch points) qualify as removable singular points in some extended sense. They definitely do not qualify as essential singularities or poles. (An essential singularity is essentially a pole of infinite order.)

    Essentially, what I'm saying is that if you look at the function \log(1+x) as a single-valued function defined on its Riemann surface, then there are only two troublesome points, -1 and \infty, which are essential singularities. Branch cuts are only a construct which allow us to "fit" inside the plane a function which is "too big" for the plane.





    It isn't! You can expand your function in a Laurent series (in fact, in a Taylor series) around every point other than -1 and \infty. You just have to remember that you will not obtain a single-valued function in this way.



    Yes; a function has no Laurent expansion around a nonisolated singular point (which is necessarily an essential singularity).
    Thanks a lot for your time, I appreciate it very much. Several things are a bit over my head for now.
    I have a doubt. You wrote
    Yes; a function has no Laurent expansion around a nonisolated singular point (which is necessarily an essential singularity)
    . Does this mean that the point -10 (on the real axis) for the function log (x+1) with the main branch as usual is an essential singularity?
    And when you said
    The points of the branch cut, should you assign some values to them, would be neither poles, nor removable singularities, nor essential singularities, but they'd still be singular points.
    , do you mean that if I assign any value to f(-1) it would be a singular point that is not removable/essential nor a pole? (I don't understand enough English to understand the perfect meaning of your sentence, hence my question for a clarification).
    Thanks for all.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by arbolis View Post
    Thanks a lot for your time, I appreciate it very much.
    You're welcome!

    Does this mean that the point -10 (on the real axis) for the function log (x+1) with the main branch as usual is an essential singularity?
    No; it's a removable singularity in the extended sense which I described above. There are two possible values which you could reasonably assign to this point, depending on whether you move the branch cut just a little above or just a little below the point.

    do you mean that if I assign any value to f(-1) it would be a singular point that is not removable/essential nor a pole?
    No! The points -1 and \infty, if anything, are essential singularities. Other points of the cut are less troublesome, because we can move the cut away from them and define the function analytically there.

    All of this business might seem quite artificial and contrived; but this is because we are working in the plane. If you learn a bit about Riemann surfaces, you will get a better idea of what is going on.

    You should not take this discussion as a good explanation of what singularities are. In a sense, it doesn't even make sense to ask whether points on a branch cut are singular, because to qualify as such, the function would first have to be defined in a punctured neighbourhood of these points. What I've been saying is that, if we really had to call them singular points, then the branch points (in this example) would be essential singularities, and the other points of the cut, removable ones.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Oct 2012
    From
    United States
    Posts
    11

    Re: Complex analysis, infinite series expansion of a function

    Try the old standard trick. We know that log(x+1) can be derived from the geometric series 1/1-(-x)) = sum n=0 to inf (1/(1-(-x)^n)_ (HELP, where are symbols like sums, integrals etc. hiding?) The sum converges absolutely and thus can be integrated term-by-term yielding a new sum which represents -1log(1-x), after a little tiny bit ofalgebra, you're done.

    Is there a drop down panel somewhere with necessary symbols to use in posts? They must be available as I see them in others' posts, beautifully expressed! Can someone show me how to generate such expressions as are seen in the original above. It sure would make for easier responses to
    folks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Infinite series (Analysis)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 17th 2012, 03:09 AM
  2. Mittag-Leffler Cotangent expansion proof Complex analysis
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 26th 2010, 09:44 AM
  3. Nonlinear ODE by an infinite series expansion
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 8th 2009, 02:59 PM
  4. expansion of infinite series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 26th 2009, 10:57 AM
  5. Analysis, infinite series
    Posted in the Calculus Forum
    Replies: 10
    Last Post: October 29th 2008, 12:23 PM

Search Tags


/mathhelpforum @mathhelpforum