```robert.w.adams wrote:
...
> Since you mentioned the Rienmann zeta function, the region of
> convergence appears to be re(s) > 1. Is this due to the "pole" at s =
> 1?

That's exactly the reason.

The sum representation of the zeta function converges for all complex
s with re(s) > 1. However, just as in the case with the geometric
series, it turns out that the zeta function can be analytically
continued to the whole complex plane minus the point 1, where it has a
simple pole (pole of first order).

We used to have a professor  who said that if you sum the natural
numbers you get -1/12. That is because zeta(-1) is -1/12. If you plug
a -1 into the sum representation of the zeta function, you get exactly
the sum of the natural numbers ...

> If so, then I wonder about the function 1/zeta(s). You might say that,
> since zeta(s) does not converge for s < 1, then 1/zeta(s) also does
> not converge. But on the other hand, 1/zeta(s) has a zero at s=1
> instead of a pole; so shouldn't the region of convergence be
> different?

Sure. 1/zeta(s) is holomorphic on the complex plane minus all the
zeros of the zeta function. To simplify, you can just view zeta(s) as
a meromorphic function (which means it's inverse is guaranteed to
exist and also meromorphic).

I think the main thing here is to remember that an analytic function
(eg. zeta function) can be represented in many ways, locally around a
z_0 by a power series, and these representations need not necessarily
have the same domains of definitions.

BTW: To show that two analytic functions are equal in their common
domain of definition, all you have to do is find a sequence of numbers
with an accumulation point where the two functions conincide.

Regards,
Andor
```
```"Randy Yates" <yates@ieee.org> wrote in message
> Liz wrote:
>
> > Thanks for all the inputs.
> >
> > This business of analytic continuation is exactly what I am trying to
> > understand.

> That's not too hard. Get a book on complex variables (I suggest
> the old chestnut by Churchill and Brown). Essentially, analytic
> continuation is based on Taylor's theorem. If a function is analytic in a
> certain region, and you have a known "piece" of the function in a
> continuous interval or a neighborhood, then you can construct the
> series expansion for the function using Taylor's theorem. Then you
> can use that series expansion to generate the function at other points
> outside the area in which it is currently available.

That sounds like how I was remembering it, too.    I still have my Saff and
Snider, and looked it up to see how they explain it.   You can evaluate the
function, and all its derivatives anywhere within the region of convergence
of the original series.   Then write a new Taylor series expansion around
that point, preferrably not near the poles of the original series.

The example they have is someone working with the Log z series, expanded
around (1,0), with a pole at (0,0), but not recognizing the Log series,
which we know is analytic except at (0,0).   In the next section, they do a
series of analytic continuations, until they get to (-1,0) on a y>0 path,
and also on a y<0 path.   The somewhat obvious result that analytic
continuation may end up on a different branch of the function than one
started on, for functions with branch cuts.

-- glen

```
```Liz wrote:

> Thanks for all the inputs.
>
> This business of analytic continuation is exactly what I am trying to
> understand.

Hey Bob,

That's not too hard. Get a book on complex variables (I suggest
the old chestnut by Churchill and Brown). Essentially, analytic
continuation is based on Taylor's theorem. If a function is analytic in a
certain region, and you have a known "piece" of the function in a
continuous interval or a neighborhood, then you can construct the
series expansion for the function using Taylor's theorem. Then you
can use that series expansion to generate the function at other points
outside the area in which it is currently available.

This is used in extrapolation of signals that have been corrupted
by dropouts.

At least I hope this is right - I'm going from memory, the books
and papers are at work.
--
%% Fuquay-Varina, NC            %       'cause no one knows which side
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO

```
```Thanks for all the inputs.

This business of analytic continuation is exactly what I am trying to
understand. In signal-processing terms, I wonder if it is the same
thing as saying that sometimes an infinite sum (an infinitely-long FIR
filter, for example) can be converted via a power-series relationship
to a recursive (IIR) structure that has a larger region of
convergence. I guess that the thing I don't quite understand is, if
there really is such a power-series equivalence, then why don't they
give the same answer for EVERY point in the complex plane? Is it
really just an issue of it not being possible to truly sum a series to
infinity?

Since you mentioned the Rienmann zeta function, the region of
convergence appears to be re(s) > 1. Is this due to the "pole" at s =
1?

If so, then I wonder about the function 1/zeta(s). You might say that,
since zeta(s) does not converge for s < 1, then 1/zeta(s) also does
not converge. But on the other hand, 1/zeta(s) has a zero at s=1
instead of a pole; so shouldn't the region of convergence be
different?

Regards

```
```On 22 Oct 2003 17:40:27 -0700, robert.w.adams@verizon.net (Liz)
wrote:

>>Now, when you give this to a Math person, they will find the "region
>>of convergence" of this function in terms of the complex variable s.
>>If you ask them what happens outside of this region, they will simply
>>say "it doesn't converge there, so why are you asking?" But, in the
>>same breath, they will say that the function has zeros in this region
>>(the region that supposedly does not converge).
>>
>>Can someone explain this to me?

I think that zeroes are only meaningful inside the ROC.  Using one
of the examples from Oppenheim and Schafer, if

x(n) = (a^n)u(n)

then
inf                   1        z
X(z) = SUM [az^(-1)]^n = --------- = ---
0                1-az^(-1)   z-a

This has a zero at z=0, but a ROC of |z|>|a|.  Well if z=0 then
there is no value of 'a' for which |z|>|a|.

For convergence of X(z) it is necessary that

inf
SUM [(az^(-1)]^n < infinity
0

If z=0, the sum blows-up.  So basically when we say that X(z) has
a zero at a location, we really mean that it would have a zero at
that location IF it converged there.

Greg Berchin
```
```Liz wrote:

> As a signal-processing person, I am trying to wade through some
> heavy-duty math papers and having a problem.
>
> Suppose that you have a signal-processing network that is
> represented by a transfer function (either Laplace or Z, doesn't
> matter right now). Suppose that this transfer function is of the
> form;
>
> H(s) = a(s) + b(s) + c(s) ... to infinity
>
> Assume, for example, that a(s), b(s), etc, represent weighted
> ideal delay elements, so you have some infinite sum of weighted
> delays. Now, since the sum is infinite, it's possible to have
> poles that come from the infinite summation, rather than the more
> usual 1/(1-f(s)) type of recursive pole.
>
> Now, when you give this to a Math person, they will find the
> "region of convergence" of this function in terms of the complex
> variable s. If you ask them what happens outside of this region,
> they will simply say "it doesn't converge there, so why are you
> asking?" But, in the same breath, they will say that the function
> has zeros in this region (the region that supposedly does not
> converge).
>
> Can someone explain this to me?
>

Hi Liz/Bob,
in fact, I cannot imagine such a H(s) as you describe.
However, maybe I can shed a piece of light to your question about
convergence region...

if you think of the fraction 1/n as a basic element, you'll
certainly agree that the sum of 1/n for all n=1..inf is infinite.
You can say, sum(1/n) doesn't converge.
If you take 1/(n^2) instead, you'll find out that its sum converges
near 1,64.

Now we have two "functions", one converges, the other doesn't.
If you sum both sums together, you'll certainly get a mixture.
If it converges, depends on the weight of the influence of both.
Let's say you take w1*(sum(1/n))+w2*(sum(1/n^2)), where w1 and w2
are the weights of both. Playing around with some values will show
that the region of convergence now depends from w1 and w2.
We can define the region of convergence to be roc=f(w1,w2).

In this example we can easily find one pair of values where it's
convergent: w1=0, w2=1
Some more reflecting reveals w1=0;w2=arbitrary
And an example of inconvergence: w1=1; w2=0 (works with any other
value instead of 1 for w1, too)

I hope it's not too complicated yet.

The point now is this:
if you imagine w1 and w2 to be coordinates of a cartesian system,
you can "draw" the region of convergence.
Leaving out the (interesting) point w1=0, w2=0 for a moment, it's
obvious (hehe), that one axis marks a region of convergence, the
other a region of inconvergence.
Certainly there will be a sharp border between both (which we don't
know yet, but possibly we could calculate or estimate it).
In our example there are two regions of convergence which are
separated by the special spot w1=w2=0.

Imagine that the functions are not so trivial, then the regions will
be trivial neither. But in the end, this is what you mentioned:
math people can derive the ROC (region of convergence).

Now let's look at w1=w2=0 in the above example.
Is the function convergent here or not?

Sometimes the answer to such a question is easy, sometimes it isn't.
Sometimes even resorting of the elements of a sum decides if the sum
converges or not. (I remember that fact, but I don't have an
example at hand - sorry). I mean: in some cases, it's difficult to
find an exact answer to such critical points.

At other times it's easy: Imagine   that the weight of another
function is f(n)*(1/(w3-1)). At w3=1, the function value would be
infinite, resulting in inconvergence of the sum. But only at this
point, not in it's environment. This would be a thing like a pole.

You think it's trivial, but only combine this case with the point
where w1=w2=0 from above, and things are going to be so complex
that you don't find the answer.

The nice thing is, that usually you can easily define the ROC,
though sometimes not in a closed formula but only in a list of
regions, (and sometimes it's even a list with an infinite number of
regions).
The nasty thing is, that you usually know almost nothing about the
rest where convergence is not guaranteed.

Usually there are poles causing the inconvergence, but this need not
be the case as shown above.
However, if you think of folding regions like in Z-Transform /
Schwarz-Christoffel /..., it's the same thing anyhow.

Therefore, it might be the best to join the math people in not
bothering about what happens inside these critical areas.

HTH

Bernhard

```
```robert.w.adams wrote:
...
> H(s) = a(s) + b(s) + c(s) ... to infinity
>
> Assume, for example, that a(s), b(s), etc, represent weighted ideal
> delay elements, so you have some infinite sum of weighted delays.

Hi Bob,

if you let s be a complex variable, then H(s) is a sum of complex
functions a_k(s), k from 0 to infinity (notice I renamed your
functions).

>
> Now, when you give this to a Math person, they will find the "region
> of convergence" of this function in terms of the complex variable s.

Ok.

> If you ask them what happens outside of this region, they will simply
> say "it doesn't converge there, so why are you asking?" But, in the
> same breath, they will say that the function has zeros in this region
> (the region that supposedly does not converge).

I'm not quite sure about your specific H(s) and where those zeros come
from.

But look at the following:
Let a_k(s) := s^k. Then H(s) is the geometric series and converges for
all complex s with |s| < 1. However, H has an analytic continuation,
namely H_e(s) = 1/(1-s), which is defined for all complex s =/= 1, and
H_e(s) = H(s) for |s| < 1. (This example does not have any zeros
outside the region of convergence of the original sum. There are other
examples, like the Riemann-zeta function, which do.)

However, H(s) is quite clearly only convergent for |s| < 1, and does
not converge outside the unit disc. It is wrong to say that H(s) has
any value outside its region of convergence (even by a math person :).
It is right to say that H has a analytic continuation H_e which can
have a larger domain of definition (and thus can possibly have zeros
outside the region of convergence of H).

> Can someone explain this to me?

fwiw

Regards,
Andor
```
```> Liz wrote:

hey Bob!  what's this "Liz" thingie?

>> As a signal-processing person, I am trying to wade through some
>> heavy-duty math papers and having a problem.
>>
>> Suppose that you have a signal-processing network that is represented
>> by a transfer function (either Laplace or Z, doesn't matter right
>> now). Suppose that this transfer function is of the form;
>>
>> H(s) = a(s) + b(s) + c(s) ... to infinity
>>
>> Assume, for example, that a(s), b(s), etc, represent weighted ideal
>> delay elements, so you have some infinite sum of weighted delays. Now,
>> since the sum is infinite, it's possible to have poles that come from
>> the infinite summation, rather than the more usual 1/(1-f(s)) type of
>> recursive pole.
>>
>> Now, when you give this to a Math person, they will find the "region
>> of convergence" of this function in terms of the complex variable s.
>> If you ask them what happens outside of this region, they will simply
>> say "it doesn't converge there, so why are you asking?" But, in the
>> same breath, they will say that the function has zeros in this region
>> (the region that supposedly does not converge).
>>
>> Can someone explain this to me?

no, but isn't the ROC of a normal recursive thingie, the stuff outside the
unit circle (or right-half plane for the "s" guys) and don't you get
non-minimum phase filters that have zeros in that non-ROC area?

yet it is kinda hard to grasp how something that doesn't converge at all can
be zero at the same time.

i dunno.

we gotta get this Bob Adams to post here on comp.dsp more often and tell us
all of his DSP secrets.

Yates at yates@ieee.org wrote on 10/22/2003 21:05:

> Could it be that the region-of-nonconvergence includes isolated
> "points-of-convergence," so that the region generally doesn't converge
> but at these isolated points? I can't think of a function that would
> do this off the top of my head.

why not a non-min phase filter?

still dunno.

r b-j

```
```Liz wrote:

> As a signal-processing person, I am trying to wade through some
> heavy-duty math papers and having a problem.
>
> Suppose that you have a signal-processing network that is represented
> by a transfer function (either Laplace or Z, doesn't matter right
> now). Suppose that this transfer function is of the form;
>
> H(s) = a(s) + b(s) + c(s) ... to infinity
>
> Assume, for example, that a(s), b(s), etc, represent weighted ideal
> delay elements, so you have some infinite sum of weighted delays. Now,
> since the sum is infinite, it's possible to have poles that come from
> the infinite summation, rather than the more usual 1/(1-f(s)) type of
> recursive pole.
>
> Now, when you give this to a Math person, they will find the "region
> of convergence" of this function in terms of the complex variable s.
> If you ask them what happens outside of this region, they will simply
> say "it doesn't converge there, so why are you asking?" But, in the
> same breath, they will say that the function has zeros in this region
> (the region that supposedly does not converge).
>
> Can someone explain this to me?
>

Could it be that the region-of-nonconvergence includes isolated
"points-of-convergence," so that the region generally doesn't converge
but at these isolated points? I can't think of a function that would
do this off the top of my head.
--
%% Fuquay-Varina, NC            %       'cause no one knows which side
%%% 919-577-9882                %                   the coin will fall."
%%%% <yates@ieee.org>           %  'Big Wheels', *Out of the Blue*, ELO

```
```As a signal-processing person, I am trying to wade through some
heavy-duty math papers and having a problem.

Suppose that you have a signal-processing network that is represented
by a transfer function (either Laplace or Z, doesn't matter right
now). Suppose that this transfer function is of the form;

H(s) = a(s) + b(s) + c(s) ... to infinity

Assume, for example, that a(s), b(s), etc, represent weighted ideal
delay elements, so you have some infinite sum of weighted delays. Now,
since the sum is infinite, it's possible to have poles that come from
the infinite summation, rather than the more usual 1/(1-f(s)) type of
recursive pole.

Now, when you give this to a Math person, they will find the "region
of convergence" of this function in terms of the complex variable s.
If you ask them what happens outside of this region, they will simply
say "it doesn't converge there, so why are you asking?" But, in the
same breath, they will say that the function has zeros in this region
(the region that supposedly does not converge).

Can someone explain this to me?