Analytic continuation

By Yatir Halevi on Thursday, June 13, 2002 - 04:44 pm:

What is

ζ (- 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

ζ (- 1) = - 1 / 6

result is correct. That is, the

ζ

function is defined to be

ζ (s) = \sum_{n = 1}^{\infty} 1 / n^{s}

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

ζ (- 1) \neq \sum_{n = 1}^{\infty} n

. So there is no contradiction here. What you've probably found is an expression for the

ζ

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 $ζ$ function to give a meaning to otherwise meaningless things like $\sum_{n = 1}^{\infty} n$ , it's called something like $ζ$ -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/n^s ? 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 / n^{s}

formula held everywhere. However, without knowing it, he was doing analytic continuation when computing the value of zeta for

Re (s) \leq 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 $\sum_{n = 1}^{\infty} 1 / n^{z}$ 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 $\sum_{n = 1}^{\infty} 1 / n^{z}$ 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:

Analytic Continuation diagram

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 e^x . by expanding it using taylor's expansion, we get a polynomial, that agrees with e^x 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,

f (z) = \sum_{n = 0}^{\infty} z^{2^{n}}

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

e^{x}

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 $ζ (s) = \infty$ whenever $Re (s) < 1$ ,but why? This can't be true if we want $ζ$ to be complex differentiable, because if $ζ$ is complex differentiable, it has to be at least continuous, so if $ζ (s) = \infty$ 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 $ζ (s)$ should go to $\infty$ . 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

ζ

function), and since the summation formula for

ζ (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

\lim_{ε \to 0} (f (z + ε) - f (z)) / ε

exists and gives the same result whichever way

ε \to 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 $ε = t$ where $t$ is a real number then $(1 / ε) (f (0 + ε) - f (0)) = 1$ , so $lim_{ε \to 0} (f (0 + ε) - f (0)) / ε = 1$ . If $ε = i t$ then $(1 / ε) (f (0 + ε) - f (0)) = 0$ , so $lim_{ε \to 0} (f (0 + ε) - f (0)) / ε = 0$ . 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 $φ$ , $σ$ , 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.