Analytic continuation
By Yatir Halevi on Thursday, June 13,
2002 - 04:44 pm:
What is z(-1)? I have two calculations; one giving me the value -1/6
and the other giving infinity.
By Dan Goodman on Thursday, June 13,
2002 - 07:53 pm:
Yatir, the z(-1)=-1/6 result is correct. That is,
the z function is defined to be
whenever Re(s) > 1 (i.e. real part is greater than 1). This function can be
extended to the whole complex plane (via analytic continuation), but the
summation formula only holds when Re(s) > 1, so
. So there is no contradiction here. What you've probably found is an
expression for the z function which is valid when s=-1, it might not be
valid everywhere, you have to be quite careful about this sort of thing.
Actually, I'm told that in physics sometimes you can use the z function
to give a meaning to otherwise meaningless things like
,
it's called something like z-regularisation.
By Yatir Halevi on Friday, June 14, 2002 -
02:09 pm:
Dan, When euler first described his zeta function, he didn't
do any 'analytic continuation', so did he define it for only
x> 0?
And when you expand it to the complex plain, how is it now
defined if not by the sum of 1/ns ? And how can the
definition change when we do 'analytic continuation'?
Yatir
By Dan Goodman on Friday, June 14, 2002
- 02:41 pm:
Yatir, Euler and all of that lot were not very rigorous in
their work. Euler was happy to write things like 1+2+3+4+...=-1/6, so as
far as he was concerned the sum of 1/ns formula held everywhere. However,
without knowing it, he was doing analytic continuation when computing the
value of zeta for Re(s) £ 1.
The reason is, if two analytic (that is, complex differentiable) functions
agree on a small disc, they agree everywhere. Analytic continuation means:
suppose you have an analytic function f(z) defined on some region D in the
complex plane (above we're using D to be set of all s with Re(s) > 1), and
you have another function g(z) defined on another region E. If f(z)=g(z)
whenever z is in both D and E (the intersection of the two regions) and
there is a small disc in the intersection of D and E, then for any function
h(z) that agrees with f(z) on D, it must agree with g(z) on E. In
other words, all ways of extending an analytic function from one region to a
larger one give the same result.
Euler almost certainly didn't know this, but you can make sense of his results
using it. He, presumably - I don't actually know the history that well, would
have shown that some function f(z) (defined on some region including the
point -1 and some points with Re(z) > 1) was equal to
when Re(z) > 1 (probably he wouldn't have assumed Re(z) > 1 but his argument
almost certainly only worked when Re(z) > 1 for technical reasons to do with
'locally uniform convergence', he wouldn't have worried about this sort of
thing).
Thus he was analytically continuing
without knowing
that he was doing it. Probably he thought that he was finding two formulae that
were both valid everywhere, but he was wrong about this.
One of the remarkable things about these early mathematicians is that their
work is remarkably consistent despite the fact that it was so logically
unsound. If you're interested in reading more about this, the second volume of
Morris Kline's excellent history "Mathematical Thought from Ancient to Modern
Times" goes into quite a lot of detail as to why their logically deficient
methods gave such good answers. It is quite a hard book though.
By Dan Goodman on Friday, June 14, 2002 -
02:58 pm:
Here's a little picture to explain the
analytic continuation thing:
So, we start with a function f(z) defined on the red and purple
regions, we find any function g(z) defined on the blue and purple
regions which agrees with f(z) on the purple regions, and then we
use this function g(z) to extend f(z) to all of the coloured
regions.
[Note: if you're doing this procedure more than once, you have to
be a bit more careful, but if you're only doing it once it's
OK.]
By Yatir Halevi on Friday, June 14, 2002 -
05:29 pm:
What you are saying is that if we have a function f(x) we can
find another function that agrees with f(x) on some region, and
use it to extend f(x)?
How do we know that we can even do this? Our extension is only by
definition...we don't know if it is really true...do we? because
euler's zeta is defined for re(x)< 1...it is just infinity ...
Doesn't g(x) have to agree with f(x) in all its region - the blue
region...) in order for us to extend f(x), or is there a way to
combine both regions? Or do we just use for the blue region f(x)
for the red region g(x) and for the purple region f(x) or
g(x)?
But I think I found the example..
consider ex . by expanding it using taylor's
expansion, we get a polynomial, that agrees with ex
for all real x, but we use the polynomial to extend it to all
complex x... Am I correct, I hope so!
How do we prove that we can do such a thing?
Yatir
By Dan Goodman on Friday, June 14, 2002
- 07:13 pm:
Yatir, sometimes you can't extend a function. For example,
converges for |z| < 1 but f(z) behaves
so badly when |z|=1 that it can't be extended any further. The Taylor series
of ex converges everywhere, but some don't.
But supposing we can do it: you say ''our extension is only by definition''
which is correct. The point is, we get the same result however we choose to
extend it if we agree to certain rules: the first rule is that we are trying
to extend it to the same new region (otherwise we get problems), the second
rule is that we have to have the function being complex differentiable, the
third rule is that it has to agree with the original function on some small
disc. Any way of extending the function that obeys these rules will be the
same, so if we can find some way of doing it, we have found the only way of
doing it.
To get back to the zeta function, you say that z(s)=¥ whenever
Re(s) < 1,but why? This can't be true if we want z to be complex
differentiable, because if z is complex differentiable, it has to be
at least continuous, so if z(s)=¥ whenever Re(s) < 1 then it must
also be true whenever Re(s)=1. So, as we get closer to the line Re(s)=1
the value of z(s) should go to ¥. But this doesn't happen
except near s=1 (I'll have a think about this and see if I can find a proof
of this that doesn't use analytic continuation!)
By Yatir Halevi on Monday, June 17, 2002 -
04:33 pm:
Why does it have to be complex differentiable?
Yatir
By Dan Goodman on Monday, June 17, 2002
- 10:54 pm:
Well, complex differentiable functions (also called
analytic functions) have some very nice properties (which are very useful when
doing number theory, which is the main use of the z function), and since
the summation formula for z(s) does give a complex differentiable
function for Re(s) > 1, it makes sense to see if you can extend it so that
it's complex differentiable everywhere.
By Yatir Halevi on Wednesday, June 19,
2002 - 11:04 pm:
Complex differentiable means that it is a complex function and
that it is differentiable?
the gamma function can be seen as an analytic continuation of the
factorial...
Then how is the factorial complex differentiable?
Or maybe it isn't and it means that we can extend it by more than
one way? (right?)
Can we extend number theoretic functions (phi,sigma...)?
Yatir
By Dan Goodman on Thursday, June 20,
2002 - 01:35 am:
Complex differentiable means slightly more than just
differentiable and complex, unfortunately. You can think of a complex function
as a 2D vector valued function of two real variables: If z=x+i y and
f(z)=u(x,y)+i.v(x,y) then f(z) is differentiable if u and v are both
differentiable in x and y. However, for f(z) to be complex
differentiable, you have to have that
lime®0 (f(z+e)-f(z))/e exists and gives the same
result whichever way e®0.
For example, if f(x+i y)=x (i.e. f(z)=Re(z)) then f is certainly
differentiable (because u(x,y)=x and v(x,y)=0 are both differentiable in
x and y), but it isn't complex differentiable. If e = t where t
is a real number then (1/e)(f(0+e)-f(0))=1, so
|
lim
e®0
|
(f(0+e)-f(0))/e = 1
|
. If e = i t
then (1/e)(f(0+e)-f(0))=0, so
|
lim
e®0
|
(f(0+e)-f(0))/e = 0
|
. For it to be complex
differentiable, these need to be the same.
The gamma function isn't an analytic continuation of the factorial, although
it is an analytic extension of the factorial, but only one among many. The
reason is that the factorial function isn't defined on any discs (it's only
defined at positive integers), so you can't start off the analytic
continuation procedure.
You can find an analytic extension (in fact, lots of them) of any function
defined on the integers, so you could certainly do it for j, s,
etc. However, it might not be very useful to do so. The gamma function is, for
reasons that aren't entirely clear (anyone know why?), the "correct" way to
extend the factorial function (perhaps it's the only analytic function to
satisfy z.f(z)=f(z+1) for all z?). There's no reason to suppose that there is
a similar "correct" way to extend other integer functions which have useful
consequences.