## Analytic continuation

I have a very simple question but one I've not been able to answer by
looking in the usual sources.

If I have a function f defined by a power series limited by a pole at a
say, and I analytically continue the function with another suitably
located series then....

is it possible for the region of validity of second series to include
the pole of the first series and hence eliminate the pole?

mikej

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 Mike James wrote: >I have a very simple question but one I've not been able to answer by >looking in the usual sources. > >If I have a function f defined by a power series limited by a pole at a >say, and I analytically continue the function with another suitably >located series then.... > >is it possible for the region of validity of second series to include >the pole of the first series and hence eliminate the pole? > >mikej > Stated simply: the radius of convergence of a power series is the the distance to the nearest singularity. You can redefine your origins to construct a power series that valid for an arbitrary region, but its radius of convergence is still constrained by the origin's distance to the nearest singularity. The answer to your question is no.
 On Jul 27, 2:51=A0pm, Mike James wrote: > I have a very simple question but one I've not been able to answer by > looking in the usual sources. > > If I have a function f defined by a power series limited by a pole at a > say, and I analytically continue the function with another suitably > located series then.... > > is it possible for the region of validity of second series to include > the pole of the first series and hence eliminate the pole? > > mikej You need to use a rational approximant (Pade) instead of a series approximant (Taylor).

## Analytic continuation

On Jul 27, 2:51 pm, Mike James <mike.ja...@infomaxgroup.co.uk> wrote:

> If I have a function f defined by a power series limited by a pole at a
> say, and I analytically continue the function with another suitably
> located series then....
>
> is it possible for the region of validity of second series to include
> the pole of the first series and hence eliminate the pole?

No, that's impossible. That would imply that analytic continuation is
not unique. The only wiggle room is in the multi-sheetedness of the
function's Riemann surface. That is, the two expansions will approach
the same point of the complex plane along different sheets of the
Riemann surface. But I don't think that's what you had in mind.

Hope this helps.

Igor
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 Igor Khavkine wrote: > On Jul 27, 2:51 pm, Mike James wrote: >> is it possible for the region of validity of second series to include >> the pole of the first series and hence eliminate the pole? > > No, that's impossible. That would imply that analytic continuation is > not unique. The only wiggle room is in the multi-sheetedness of the > function's Riemann surface. That is, the two expansions will approach > the same point of the complex plane along different sheets of the > Riemann surface. But I don't think that's what you had in mind. > > Hope this helps. > > Igor Yes, at least it has helped me work out what it is that is troubling me about the idea. If you think of an analytic function as and object that exists independently of its representation by a formula then the pole of a given series expansion might be regarded as a flaw in the representation - as in the case of a co-ordinate chart that doesn't cover a particular point say. If this is the case then the pole isn't part of the analytic function, its an anomaly of the formula representing it. If on the other hand the pole is part of the function and this causes the function representing it to fail then clearly another representation must also have a pole in the same place. Continuity arguments suggest to me that the pole is part of the function and must exist in any representation. However this is slightly worrying to me in that the information within a disk of convergence contains enough information to construct the (analytic) function anywhere in the complex plane - including all of its poles(?) mikej
 Mike James wrote: > Igor Khavkine wrote: > > On Jul 27, 2:51 pm, Mike James wrote: > > >> is it possible for the region of validity of second series to include > >> the pole of the first series and hence eliminate the pole? > > > > No, that's impossible. That would imply that analytic continuation is > > not unique. The only wiggle room is in the multi-sheetedness of the > > function's Riemann surface. That is, the two expansions will approach > > the same point of the complex plane along different sheets of the > > Riemann surface. But I don't think that's what you had in mind. > > > > Hope this helps. > > > > Igor > > Yes, at least it has helped me work out what it is that is troubling me > about the idea. > > If you think of an analytic function as and object that exists > independently of its representation by a formula then the pole of a > given series expansion might be regarded as a flaw in the representation > - as in the case of a co-ordinate chart that doesn't cover a particular > point say. > > If this is the case then the pole isn't part of the analytic function, > its an anomaly of the formula representing it. > > If on the other hand the pole is part of the function and this causes > the function representing it to fail then clearly another representation > must also have a pole in the same place. > > Continuity arguments suggest to me that the pole is part of the function > and must exist in any representation. > > However this is slightly worrying to me in that the information within a > disk of convergence contains enough information to construct the > (analytic) function anywhere in the complex plane - including all of its > poles(?) mikej Perhaps it helps you to understand that analytic functions satisfy a system of differential equations. (Cauchy-Riemann) In the case of differential equations it's not surprising at all that you may find the function 'everywhere' from its values in a small region. Jan
 J. J. Lodder wrote: > Perhaps it helps you to understand that analytic functions > satisfy a system of differential equations. > (Cauchy-Riemann) > In the case of differential equations it's not surprising at all > that you may find the function 'everywhere' > from its values in a small region. > > Jan > Yes indeed - but including points where the function doesn't have a derivative?! I can see that there is no problem for a smooth analytic function with derivatives of all order defined at a point even being extended to the entire real line (or in the complex case plane) but the idea that the derivatives also encode the poles of a function is more worrying. mikej
 Mike James wrote: >J. J. Lodder wrote: > >> Perhaps it helps you to understand that analytic functions >> satisfy a system of differential equations. >> (Cauchy-Riemann) >> In the case of differential equations it's not surprising at all >> that you may find the function 'everywhere' >> from its values in a small region. >> >> Jan >> >Yes indeed - but including points where the function doesn't have a >derivative?! >I can see that there is no problem for a smooth analytic function with >derivatives of all order defined at a point even being extended to the >entire real line (or in the complex case plane) but >the idea that the derivatives also encode the poles of a function is >more worrying. >mikej > As they should be. Any analytic continuation must dance around these singular points. That's why your power series can't be simply defined everywhere.[1] But using the distance to the nearest singularity to define the power series' domain of convergence is still very useful, if not of immediate interest regarding your problem. That's how the radius of convergence of the power-series representations of elliptic functions is determined. I've used it to find the radius of convergence of some pretty nasty power-series myself.[2] Finally, you may want to look at some texts on elliptic functions. This type of problem has been dealt with quite successfully there. I, personally, have found some very useful hints in solving problems inside these books. [1] As I recall the original post asked about finding a valid power series representation. [2] "Particle trajectories in 1/r fields", M. Arnow, J. Math Physics, V. 21, 1980
 Mike James wrote: > J. J. Lodder wrote: > > > Perhaps it helps you to understand that analytic functions > > satisfy a system of differential equations. > > (Cauchy-Riemann) > > In the case of differential equations it's not surprising at all > > that you may find the function 'everywhere' > > from its values in a small region. > > > > Jan > > > Yes indeed - but including points where the function doesn't have a > derivative?! You can find the function wherever it is well-defined. It is worse than that however, for analytic functions may have natural boundaries, beyond which they cannot be continued. > I can see that there is no problem for a smooth analytic function with > derivatives of all order defined at a point even being extended to the > entire real line (or in the complex case plane) One derivative will do. If differentiable, it will be infinitely differentiable. > but > the idea that the derivatives also encode the poles of a function is > more worrying. Why? The poles are just as much part of the function as the regular parts. You can't have an analytic function that behaves as 1/z away from z=0 without having the pole at z=0 To clarify: my remark about satisfying the differential equations was meant to relativize the importance of the process of continuation by successive power serie expansions. These may be seen as just a convenient way of solving the Cauchy-Riemann equations. Best, Jan
 J. J. Lodder wrote: > Mike James wrote: > >> Yes indeed - but including points where the function doesn't have a >> derivative?! > > You can find the function wherever it is well-defined. > It is worse than that however, for analytic functions may have natural > boundaries, beyond which they cannot be continued. Although not really connected to my initial question - can you give me an example of a "natural boundary"? >> the idea that the derivatives also encode the poles of a function is >> more worrying. > > Why? The poles are just as much part of the function > as the regular parts. > You can't have an analytic function that behaves as 1/z > away from z=0 without having the pole at z=0 Put like that - yes I can't see what I'm worried about - 1/z behaviour away from the pole is perfectly reasonable and, as you say implies the pole's existence at z=0. (I guess my problem is due to not really taking on board the difference between a Taylor and a Laurent series.) I think I now understand - analytic continuation can't "remove" poles as information about them is contained in any region that you are using to continue the function. They really are part of the function and not any particular formula that you are using to encode the function. Many thanks mikej
 Murray Arnow wrote: > As they should be. Any analytic continuation must dance around these > singular points. That's why your power series can't be simply defined > everywhere.[1] Yes I agree but my question was really about the poles that were far enough away not to effect the power series in use - i.e. well outside the radius of convergence. What I really wanted to know is whether or not these are "real" and part of the function - and the answer is obviously (now it has been explained to me) yes. Thanks mikej
 On Aug 3, 6:10 am, Mike James wrote: > Although not really connected to my initial question - can you give me > an example of a "natural boundary"? A class of examples are so-called lacunary series. Take f(z) = sum_{n=0 to oo} z^(2^n) = z + z^2 + z^4 + z^8 + ... The radius of convergence of this series is equal to 1. However, unlike a series with a pole at distance 1 from the origin, this series diverges as every point on the boundary of the unit disc. In fact, every point with |z|=1 is an essential singularity. The proof is sketched in the Wikipedia page discussion the class of lacunary series: http://en.wikipedia.org/wiki/Lacunary_function Hope this helps. Igor
 Igor Khavkine wrote: > On Aug 3, 6:10 am, Mike James wrote: > >> Although not really connected to my initial question - can you give me >> an example of a "natural boundary"? > > A class of examples are so-called lacunary series. Take > > f(z) = sum_{n=0 to oo} z^(2^n) = z + z^2 + z^4 + z^8 + ... > > The radius of convergence of this series is equal to 1. However, > unlike a series with a pole at distance 1 from the origin, this series > diverges as every point on the boundary of the unit disc. In fact, > every point with |z|=1 is an essential singularity. > > The proof is sketched in the Wikipedia page discussion the class of > lacunary series: > > http://en.wikipedia.org/wiki/Lacunary_function > > Hope this helps. > > Igor > Oh yes it does... it helps to know that such terrible things exist :-) mikej
 Mike James wrote: > If I have a function f defined by a power series limited by a pole at a > say, and I analytically continue the function with another suitably > located series then.... > is it possible for the region of validity of second series to include > the pole of the first series and hence eliminate the pole? You could multiply your existing series by a factor which eliminates the pole you don't like; and try to do (express) everything with the new pole-free series instead. I can't guarantee this will help whatever you in want to do, however. -- ---------------------------------+--------------------------------- Dr. Paul Kinsler Blackett Laboratory (PHOT) (ph) +44-20-759-47734 (fax) 47714 Imperial College London, Dr.Paul.Kinsler@physics.org SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
 p.kinsler@ic.ac.uk wrote: > Mike James wrote: >> is it possible for the region of validity of second series to include >> the pole of the first series and hence eliminate the pole? > > You could multiply your existing series by a factor which eliminates > the pole you don't like; and try to do (express) everything with the > new pole-free series instead. I can't guarantee this will help whatever > you in want to do, however. > > Yes this is what prompted the question - I'm looking at various methods of regularisation and suddenly realised that, despite thinking that I did understand analytic continuation, I really didn't.... I think I do now so thanks everyone. And I might have some questions on regularisation when I had time to think about it a bit more. mikej
 Mike James wrote: > p.kinsler@ic.ac.uk wrote: > > Mike James wrote: > > >> is it possible for the region of validity of second series to include > >> the pole of the first series and hence eliminate the pole? > > > > You could multiply your existing series by a factor which eliminates > > the pole you don't like; and try to do (express) everything with the > > new pole-free series instead. I can't guarantee this will help whatever > > you in want to do, however. > > > > > > > Yes this is what prompted the question - I'm looking at various methods > of regularisation and suddenly realised that, despite thinking that I > did understand analytic continuation, I really didn't.... > I think I do now so thanks everyone. You do analytic regularization by setting up everything in such a way that all your expressions are analytic functions in some variable. (warning: this step is in no way unique, except by definition) Then you expand everything in a Laurent series around the pole, and discard all pole parts. The result is arbitrary by a 'finite regularization', that is, an additive constant. (as are all other regularization methods, unless you fix that constant by definition) The simplest way to see that this must be the case is by noting that instead of Regu (f(z)) you may also take 1/g(z) Regu (g(z)f(z)), with g(z) some arbitrary entire function. The linear part of g(z) and the pole part of f(z) multiply to the arbitrary extra constant. Best, Jan