Analytic continuation

Mike James

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

Murray Arnow

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.

Chris H. Fleming

On Jul 27, 2:51=A0pm, Mike James <mike.ja...@infomaxgroup.co.uk> 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).

Igor Khavkine

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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Mike James

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

J. J. Lodder

Mike James <mike.james@infomaxgroup.co.uk> wrote:

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

Mike James

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

Murray Arnow

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

J. J. Lodder

Mike James <mike.james@infomaxgroup.co.uk> 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

Mike James

J. J. Lodder wrote:
> Mike James <mike.james@infomaxgroup.co.uk> 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

Mike James

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

Igor Khavkine

On Aug 3, 6:10 am, Mike James <mike.ja...@infomaxgroup.co.uk> 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

Mike James

Igor Khavkine wrote:
> On Aug 3, 6:10 am, Mike James <mike.ja...@infomaxgroup.co.uk> 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

p.kinsler@ic.ac.uk

Mike James <mike.james@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?

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/

Mike James

p.kinsler@ic.ac.uk wrote:
> Mike James <mike.james@infomaxgroup.co.uk> 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

J. J. Lodder

Mike James <mike.james@infomaxgroup.co.uk> wrote:

> p.kinsler@ic.ac.uk wrote:
> > Mike James <mike.james@infomaxgroup.co.uk> 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