Analytic continuation

In my previous post, I promised to say something about analytic continuation. To do so, let me first recall Taylor’s theorem: if we have some function f(z) that is analytic in some domain D, then we can write this function as an infinite power series:

f(z) = ∑ an(z-z0)n

with n = 0, 1, 2,… going all the way to infinity (n = 0 → ∞) and the successive coefficients an equal to  an = [f(n)(z0)]/n! (with f(n)(z0) denoting the derivative of the nth order, and n! the factorial function n! = 1 x 2 x 3 x … x n).

I should immediately add that this domain D will always be an open disk, as illustrated below. The term ‘open’ means that the boundary points (i.e. the circle itself) are not part of the domain. This open disk is the so-called circle of convergence for the complex function f(z) = 1/(1-z), which is equivalent to the (infinite) power series f(z) = 1 + z + z+ z3 + z4 +… [A clever reader will probably try to check this using Taylor’s theorem above, but I should note the exercise involves some gymnastics. Indeed, the development involves the use of the identity 1 + z + z+ z3 + … + zn = (1 – zn+1)/(1 – z).]

singularity

This power series converges only when the absolute value of z is (strictly) smaller than 1, so only when ¦z¦ < 1. Indeed, the illustration above shows the singularity at the point 1 (or the point (1, 0) if you want) on the real axis: the denominator of the function 1/(1-z) effectively becomes zero there. But so that’s one point only and, hence, we may ask ourselves why this domain should be bounded a circle going through this one point. Why not some square or rectangle or some other weird shape avoiding this point? That question takes a few theorems to answer, and so I’ll just say that this is just one of the many remarkable things about analytic functions: if a power series such as the one above converges to f(z) within some circle whose radius is the distance from z(in this case, zis the origin and so we’re actually dealing with a so-called Maclaurin expansion here, i.e. an oft-used special case of the Taylor expansion) to the nearest point zwhere f fails to be analytic (that point zis equal to 1 in this case), then this circle will actually be the largest circle centered at zsuch that the series converges to f(z) for all z interior to it.

PuffThat’s quite a mouthful so let me rephrase it. What is being said here is that there’s usually a condition of validity for the power series expansion of a function: if that condition of validity is not fulfilled, then the function cannot be represented by the power series. In this particular case, the expansion of f(z) = 1/(1-z) = 1 + z + z+ z3 + z4 +… is only valid when ¦z¦ < 1, and so there is no larger circle about  z0 (i.e. the origin in this particular case) such that at each point interior to it, the Taylor series (or the Maclaurin series in this case) converges to f(z).

That being said, we can usually work our way around such singularities, especially when they are isolated, such as in this example (there is only this one point 1 that is causing trouble), and that is where the concept of analytic continuation comes in. However, before I explain this, I should first introduce Laurent’s theorem, which is like Taylor’s theorem but it applies to functions which are not as ‘nice’ as the functions for which Taylor’s theorem holds (i.e. functions that are not analytic everywhere), such as this 1/(1-z) function indeed. To be more specific, Laurent’s theorem says that, if we have a function f which is analytic in an annular domain (i.e. the red area in the illustration below) centered at some point z(in the illustration below, that’s point c) then f(z) will have a series representation involving both positive and negative powers of the term (z – z0).

256px-Laurent_series

More in particular, f(z) will be equal to f(z) = ∑ [an(z – z0)n] + ∑ [bn/(z-z0)n], with n = 0, 1,…, ∞ and with an and  bn coefficients involving complex integrals which I will not write them down here because WordPress lacks a good formula editor and so it would look pretty messy. An alternative representation of the Laurent series is to write f(z) as f(z) = ∑ [cn(z – z0)n], with c= (1/2πi) ∫C [f(z)/(z – z0)n+1]dz (n = 0, ±1, ±2,…, ±∞). Well – so here I actually did write down the integral. I hope it’s not too messy 🙂 .

It’s relatively easy to verify that this Laurent series becomes the Taylor series if there would be no singularities, i.e. if the domain would cover the whole disk (so if there would be red everywhere, even at the origin point). In that case, the cn coefficient becomes (1/2πi) ∫C [f(z)/(z – z0)-1+1]dz for n = -1 and we can use the fact that, if f(z) is analytic everywhere, the integral ∫C [f(z)dz will be zero along any contour in the domain of f(z). For n = -2, the integrand becomes f(z)/(z – z0)-2+1 = f(z)(z – z0) and that’s an analytic function as well because the function z – zis analytic everywhere and, hence, the product of this (analytic) function with f(z) will also be analytic everywhere (sums, products and compositions of analytic functions are also analytic). So the integral will be zero once again. Similarly, for n = -3, the integrand f(z)/(z – z0)-3+1 = f(z)(z – z0)2 is analytic and, hence, the integral is again zero. In short, all bn coefficients (i.e. all ‘negative’ powers) in the Laurent series will be zero, except for n = 0, in which case bn = b0 = a0. As for the acoefficients, one can see they are equal to the Taylor coefficients by using what Penrose refers to as the ‘higher-order’ version of the Cauchy integral formula: f(n)(z0)/n! = (1/2πi) ∫C [f(z)/(z – z0)n+1]dz.

It is also easy to verify that this expression also holds for the special case of a so-called punctured disk, i.e. an annular domain for which the ‘hole’ at the center is limited to the center point only, so this ‘annular domain’ then consists of all points z around z0 for which 0 < ¦z – z0¦ < R. We can then write the Laurent series as f(z) = ∑ an(z-z0)+ b1/(z – z0) + b2/(z – z0)+…+ bn/(z – z0)+… with n = 0, 1, 2,…, ∞.

OK. So what? Well… The point to note is that we can usually deal with singularities. That’s what the so-called theory of residues and poles is for. The term pole is a more illustrious term for what is, in essence, an isolated singular point: it has to do with the shape of the (modulus) surface of f(z) near this point, which is, well… shaped like a tent on a (vertical) pole indeed. As for the term residue, that’s a term used to denote this coefficient b1 in this power series above. The value of the residue at one or more isolated singular points can be used to evaluate integrals (so residues are used for solving integrals), but we won’t go into any more detail here, especially because, despite my initial promise, I still haven’t explained what analytic continuation actually is. Let me do that now.

For once, I must admit that Penrose’s explanation here is easier to follow than other texts (such as the Wikipedia article on analytic continuation, which I looked at but which, for once, seems to be less easy to follow than Penrose’s notes on it), so let me closely follow his line of reasoning here.

If, instead of the origin, we would use a non-zero point z0 for our expansion of this function f(z) = 1/(1-z) = 1 + z + z+ z3 + z4 +… (i.e. a proper Taylor expansion, instead of the Maclaurin expansion around the origin), then we would, once again, find a circle of convergence for this function which would, once again, be bounded by the singularity at point (1, 0), as illustrated below. In fact, we can move even further out and expand this function around the (non-zero) point zand so on and so on. See the illustration: it is essential that the successive circles of convergence around the origin, z0, zetcetera overlap when ‘moving out’ like this.

analytic continuation

So that’s this concept of ‘analytic continuation’. Paraphrasing Penrose, what’s happening here is that the domain D of the analytic function f(z) is being extended to a larger region D’ in which the function f(z) will also be analytic (or holomorphic – as this is the term which Penrose seems to prefer over ‘analytic’ when it comes to complex-valued functions).

Now, we should note something that, at first sight, seems to be incongruent: as we wander around a singularity like that (or, to use the more mathematically correct term, a pole of the function) to then return to our point of departure, we may get (in fact, we are likely to get) different function values ‘back at base’. Indeed, the illustration below shows what happens when we are ‘wandering’ around the origin for the log z function. You’ll remember (if not, see the previous posts) that, if we write z using polar coordinates (so we write z as z = reiθ), then log z is equal to log z = lnr + i(θ + 2nπ). So we have a multiple-valued function here and we dealt with that by using branches, i.e. we limited the values which the argument of z (arg z = θ) could take to some range α < θ < α + 2π. However, when we are wandering around the origin, we don’t limit the range of θ. In fact, as we are wandering around the origin, we are effectively constructing this Riemann surface (which we introduced in one of our previous posts also), thereby effectively ‘gluing’ successive branches of the log z function together, and adding 2πto the value of our log z function as we go around. [Note that the vertical axis in the illustration below keeps track of the imaginary part of log z only, i.e. the part with θ in it only. If my imaginary reader would like to see the real part of log z, I should refer him to the post with the entry on Riemann surfaces.]

Imaginary_log_analytic_continuation

But what about the power series? Well… The log z function is just like any other analytic function and so we can and do expand it as we go. For example, if we expand the log z function about the point (1, 0), we get log z = (z – 1) – (1/2)(z – 1)+ (1/3)(z – 1)– (1/4)(z – 1)+… etcetera. But as we wonder around, we’ll move into a different branch of the log z function and, hence, we’ll get a different value when we get back to that point. However, I will leave the details of figuring that one out to you 🙂 and end this post, because the intention here is just to illustrate the principle, and not to copy some chapter out of a math course (or, at least, not to copy all of it let’s say :-)).

If you can’t work out it out, you can always try to read the Wikipedia article on analytic continuation. While less ‘intuitive’ than Penrose’s notes on it, it’s definitely more complete, even if does not quite exhaust the topic. Wikipedia defines analytic continuation as “a technique to extend the domain of a given analytic function by defining further values of that function, for example in a new region where an infinite series representation in terms of which it is initially defined becomes divergent.” The Wikipedia article also notes that, “in practice, this continuation is often done by first establishing some functional equation on the small domain and then using this equation to extend the domain: examples are the Riemann zeta function and the gamma function.” But so that’s sturdier stuff which Penrose does not touch upon – for now at least, but I expect him to develop such things in later Road to Reality chapters).

Post scriptum: Perhaps this is an appropriate place to note that, at first sight, singularities may look like no big deal: so we have a infinitesimally small hole in the domain of function log z or 1/z s or whatever, so what? Well… It’s probably useful to note that, if we wouldn’t have that ‘hole’ (i.e. the singulariy), any integral of this function (I mean the integral of this function along any closed contour around that point, i.e. ∫C f(z)dz, would be equal to zero, but when we do have that little hole, like for f(z) = 1/z, we don’t have that result. In this particular case (i.e. f(z) = 1/z), you should note that the integral ∫C (1/z)dz, for any closed contour around the origin, equals 2πi, or that, just to give one more example here, that the value of the integral ∫C [f(z)/(z – z0)]dz is equal to 2πf(z0). Hence, even if f(z) would be analytic over the whole open disk, including the origin, the ‘quotient function’ f(z)/z will not be analytic at the origin and, hence, the value of the integral of this ‘quotient function’ f(z)/z around the origin will not be zero but equal to 2πi times the value of the original f(z) function at the origin, i.e. 2πi times f(0). Vice versa, if we find that the value of the integral of some function around a closed contour – and I mean any closed contour really – is not equal to zero, we know we’ve got a problem somewhere and so we should look out for one or more infinitesimally small little ‘holes’ somewhere in the domain. Hence, singularities, and this complex theory of poles and residues which shows us how we can work with them, is extremely relevant indeed: it’s surely not a matter of just trying to get some better approximation for this or that value or formula or so. 🙂

In light of the above, it is now also clear that the term ‘residue’ is well chosen: this coefficient bis equal to ∫C f(z)dz divided by 2π(I take the case of a Maclaurin expansion here) and, hence, there would be no singularity, this integral (and, hence, the coefficient b1) would be equal to zero. Now, because of the singularity, we have a coefficient b≠ 0 and, hence, using the term ‘residue’ for this ‘remnant’ is quite appropriate.

No royal road to reality

I got stuck in Penrose’s Road to Reality in chapter 5 already. That is not very encouraging – because the book has 34 chapters, and every chapter builds on the previous one, and usually also looks more difficult than the previous one.

In Chapter 5, Penrose introduces complex algebra. As I tried to get through it, I realized I had to do some more self-study. Indeed, while Penrose claims no other books or courses are needed to get through his, I do not find this to be the case. So I bought a fairly standard course in complex analysis (James Ward Brown and Ruel V. Churchill, Complex variables and applications) and I’ve done chapter 1 and 2 now. Although these first two chapter do not nothing else than introduce the subject-matter, I find the matter rather difficult and the terminology confusing. Examples:

1. The term ‘scalar’ is used to denote real numbers. So why use the term ‘scalar’ if the word ‘real’ is available as well? And why not use the term ‘real field’ instead of ‘scalar field’? Well… The term ‘real field’ actually means something else. A scalar field associates a (real) number to every point in space. So that’s simple: think of temperature or pressure. The term ‘scalar’ is said to be derived from ‘scaling’: a scalar is that what scales vectors. Indeed, scalar multiplication of a vector and a real number multiplies the magnitude of the vector without changing its direction. So what is a real field then? Well… A (formally) real field is a field that can be extended with a (not necessarily unique) ordering which makes it an ordered field. Does that help? Somewhat I guess. But why the qualifier ‘formally real’? I checked and there is no such thing as an ‘informally real’ field. I guess it’s just to make sure we know what we are talking about, as ‘real’ is a word with many meanings.

2. So what’s a field in mathematics? It is an algebraic structure: a set of ‘things’ (like numbers) with operations defined on it, including the notions of addition, subtraction, multiplication, and division. As mentioned above, we have scalar fields and vector fields. In addition, we also have fields of complex numbers. We also have fields with some less likely candidates for addition and multiplication, such as functions (one can add and multiply functions with each other). In short, anything which satisfies the formal definition of a field – and here I should note that the above definition of a field is not formal – is a field. For example, the set of rational numbers satisfies the definition of a field too. So what is the formal definition? First of all, a field is a ring. Huh? Here we are in this abstract classification of algebraic structures: commutative groups, rings, fields, etcetera (there are also modules – a type of algebraic structure which I had never ever heard of before). To put it simply – because we have to move on of course – a ring (no one seems to know where that word actually comes from) has addition and multiplication only, while a field has division too. In other words, a ring does not need to have multiplicative inverses. Huh?  It’s simply really: the integers form a ring, but the equation 2x = 1 does not have a solution in integers (x = ½) and, hence, the integers do not form a field. The same example shows why rational numbers do.

3. But what about a vector field? Can we do division with vectors? Yes, but not by zero – but that is not a problem as that is understood in the definition of a field (or in the general definition of division for that matter). In two-dimensional space, we can represent vectors by complex numbers: z = (x,y), and we have a formula for the so-called multiplicative inverse of a complex number: z-1 = (x/x2+y2, -y/x2+y2). OK. That’s easy. Let’s move on to more advanced stuff.

4. In logic, we have the concept of well-formed formulas (wffs). In math, we have the concept of ‘well-behaved’: we have well-behaved sets, well-behaved spaces and lots of other well-behaved things, including well-behaved functions, which are, of course, those of interest to engineers and scientists (and, hence, in light of the objective of understanding Penrose’s Road to Reality, to me as well). I must admit that I was somewhat surprised to learn that ‘well-behaved’ is one of the very few terms in math that have no formal definition. Wikipedia notes that its definition, in the science of mathematics that is, depends on ‘mathematical interest, fashion, and taste’. Let me quote in full here: “To ensure that an object is ‘well-behaved’ mathematicians introduce further axioms to narrow down the domain of study. This has the benefit of making analysis easier, but cuts down on the generality of any conclusions reached. […] In both pure and applied mathematics (optimization, numerical integration, or mathematical physics, for example), well-behaved means not violating any assumptions needed to successfully apply whatever analysis is being discussed. The opposite case is usually labeled pathological.” Wikipedia also notes that “concepts like non-Euclidean geometry were once considered ill-behaved, but are now common objects of study.”

5. So what is a well-behaved function? There is actually a whole hierarchy, with varying degrees of ‘good’ behavior, so one function can be more ‘well-behaved’ than another. First, we have smooth functions: a smooth function has derivatives of all orders (as for its name, it’s actually well chosen: the graph of a smooth function is actually, well, smooth). Then we have analytic functions: analytic functions are smooth but, in addition to being smooth, an analytic function is a function that can be locally given by a convergent power series. Huh? Let me try an alternative definition: a function is analytic if and only if its Taylor series about x0 converges to the function in some neighborhood for every x0 in its domain. That’s not helping much either, is it? Well… Let’s just leave that one for now.

In fact, it may help to note that the authors of the course I am reading (J.W. Brown and R.V. Churchill, Complex Variables and Applications) use the terms analytic, regular and holomorphic as interchangeable, and they define an analytic function simply as a function which has a derivative everywhere. While that’s helpful, it’s obviously a bit loose (what’s the thing about the Taylor series?) and so I checked on Wikipedia, which clears the confusion and also defines the terms ‘holomorphic’ and ‘regular’:

“A holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of every point in its domain. The existence of a complex derivative in a neighborhood is a very strong condition, for it implies that any holomorphic function is actually infinitely differentiable and equal to its own Taylor series. The term analytic function is often used interchangeably with ‘holomorphic function’ although the word ‘analytic’ is also used in a broader sense to describe any function (real, complex, or of more general type) that can be written as a convergent power series in a neighborhood of each point in its domain. The fact that the class of complex analytic functions coincides with the class of holomorphic functions is a major theorem in complex analysis.”

Wikipedia also adds following: “Holomorphic functions are also sometimes referred to as regular functions or as conformal maps. A holomorphic function whose domain is the whole complex plane is called an entire function. The phrase ‘holomorphic at a point z0’ means not just differentiable at z0, but differentiable everywhere within some neighborhood of z0 in the complex plane.”

6. What to make of all this? Differentiability is obviously the key and, although there are many similarities between real differentiability and complex differentiability (both are linear and obey the product rule, quotient rule, and chain rule), real-valued functions and complex-valued functions are different animals. What are the conditions for differentiability? For real-valued functions, it is a matter of checking whether or not the limit defining the derivative exists and, of course, a necessary (but not sufficient) condition is continuity of the function.

For complex-valued functions, it is a bit more sophisticated because we’ve got so-called Cauchy-Riemann conditions applying here. How does that work? Well… We write f(z) as the sum of two functions: f(z) = u(x,y) + iv(x,y). So the real-valued function u(x,y) yields the real part of f(z), while v(x,y) yields the imaginary part of f(z). The Cauchy-Riemann equations (to be interpreted as conditions really) are the following: ux = vy and uy = -v(note the minus sign in the second equation).

That looks simple enough, doesn’t it? However, as Wikipedia notes (see the quote above), differentiability at a point z0 is not enough (to ensure the existence of the derivative of f(z) at that point). We need to look at some neighborhood of the point z0 and see if these first-order derivatives (ux, uy, vx and vy) exist everywhere in that neighborhood and satisfy these Cauchy-Riemann equations. So we need to look beyond the point z0 itself hen doing our analysis: we need to  ‘approach’ it from various directions before making any judgment. I know this sounds like Chinese but it became clear to me when doing the exercises.

7. OK. Phew!  I got this far – but so that’s only chapter 1 and 2 of Brown & Churchill’s course !  In fact, chapter 2 also includes a few sections on so-called harmonic functions and harmonic conjugates. Let’s first talk about harmonic functions. Harmonic functions are even better behaved than holomorphic or analytic functions. Well… That’s not the right way to put it really. A harmonic function is a real-valued analytic function (its value could represent temperature, or pressure – just as an example) but, for a function to qualify as ‘harmonic’, an additional condition is imposed. That condition is known as Laplace’s equation: if we denote the harmonic function as H(x,y), then it has to have second-order derivatives which satisfies Hxx(x,y) + Hyy(x,y) = 0.

Huh? Laplace’s equation, or harmonic functions in general, plays an important role in physics, as the condition that is being imposed (the Laplace equation) often reflects a real-life physical constraint and, hence, the function H would describe real-life phenomena, such as the temperature of a thin plate (with the points on the plate defined by the (x,y) coordinates), or electrostatic potential. More about that later. Let’s conclude this first entry with the definition of harmonic conjugates.

8. As stated above, a harmonic function is a real-valued function. However, we also noted that a complex function f(z) can actually be written as a sum of a real and imaginary part using two real-valued functions u(x,y) and v(x,y). More in particular, we can write f(z) = u(x,y) + iv(x,y), with i the imaginary number (0,1). Now, if u and v would happen to be harmonic functions (but so that’s an if of course – see the Laplace condition imposed on their second-order derivatives in order to qualify for the ‘harmonic’ label) and, in addition to that, if their first-order derivatives would happen to satisfy the Cauchy-Riemann equations (in other  words, f(z) should be a well-behaved analytic function), then (and only then) we can label v as the harmonic conjugate of u.

What does that mean? First, one should note that when v is a harmonic conjugate of u in some domain, it is not generally true that u is a harmonic conjugate of v. So one cannot just switch the functions. Indeed, the minus sign in the Cauchy–Riemann equations makes the relationship asymmetric. But so what’s the relevance of this definition of a harmonic conjugate? Well… There is a theorem that turns the definition around: ‘A function f(z) = u(x,y) + iv(x,y) is analytic (or holomorphic to use standard terminology) in a domain D if and only if v is a harmonic conjugate of u. In other words, introducing the definition of a harmonic conjugate (and the conditions which their first- and second-order derivatives have to satisfy), allows us to check whether or not we have a well-behaved complex-valued function (and with ‘well-behaved’ I mean analytic or holomorphic).    

9. But, again, why do we need holomorphic functions? What’s so special about them? I am not sure for the moment, but I guess there’s something deeper in that one phrase which I quoted from Wikipedia above: “holomorphic functions are also sometimes referred to as regular functions or as conformal maps.” A conformal mapping preserves angles, as you can see on the illustration below, which shows a rectangular grid and its image under a conformal map: f maps pairs of lines intersecting at 90° to pairs of curves still intersecting at 90°. I guess that’s very relevant, although I do not know why exactly as for now. More about that in later posts.

342px-Conformal_map.svg