The Road To Reality

[WORLD-HEN]

Hi, has any one read "The Road To Reality" by Roger Penrose? It looks quite complicated if you browse through it but I started reading the book and in the preface Penrose says that in the first 17 chapter he develops and explains all the mathematics necessary for the rest of the book. Is this really true? Is it easy or hard? I need some feedback. I was literally drooling over this book at the book store, a fairly mathematical treatment of loop quantum gravity in the last chapter! and the author claims that all needed mathematics will be explained! its like a dream!. Well any way, all feedback is welcome, tell me if its impossibly hard to understand or fairly easy or whatever.

[Gecko]

i have it. i had to ship it from the uk to hawaii = P. its a truly amazing book. definitly a must have. it is not meant for those that are weak at math though. he does explain what you need to know, but unless you have a pretty good math background, its really confusing. you could glance through it and give it a try. i love it.

[PhilosophyofPhysics]

i saw this at the book store. I was tempted to buy it, but I saw the math and decided that i'd better not. I still want it though. heh

[James R]

I have only browsed through it, but it looks like Penrose takes a fair amount of math for granted in the book. I think it would be heavy going for somebody with no background in math.

My advice: if this is your first physics/math book, it is probably not a good place to start. You'll most likely get lost very quickly.

[Telos]

The book gives this website for finding the solutions to the exercises:

http://www.roadsolutions.ox.ac.uk/

But, when you click on "solutions" there's just a note from Penrose saying he hasn't got around to posting them and they might be there in "November." What.. November 2012?

Does anyone know if there's another place that has the solutions?

[alex caps]

I picked this up a week ago and I must say.. wow, GREAT book. I am only on page 90 right now, but it does a great job at starting from the beginning and covering most everything. Rather than just seing the "popular" belief, Penrose tries to give the reader many different opinions and ideas and he states each very nicely. This book is not one to read straight though like a novel, but if you have time and like to learn a thing or two, definitely check it out.

[James R]

I couldn't help myself. I went out and bought the book, too, and am reading through it. So far, I'm only up to around Chapter 9.

Penrose takes quite a bit for granted, and skips over a lot of things in his explanations. For example, he lost me with his discussion of Riemann surfaces, and this is something I'm supposed to know something about.

The book is good for getting a taste of what's out there in terms of maths and physics, but in no way is it a textbook. You couldn't use it as the basis for teaching a beginner or intermediate course in maths or physics - at least not without bringing in a lot of supplementary material.

It's always interesting to read a book which sets out the ideas of a deep thinker who has been working on stuff for years. This is a good book, but probably doesn't quite achieve its stated aim of being very accessible to the general public.

For confirmation, just look at some of the "exercises", mentioned above. It would be very difficult for somebody who only had Penrose's book to complete many of the exercises (particularly the ones he classifies as intermediate or hard).

[PhilosophyofPhysics]

it's definitely not for someone who has not had much mathematics. I was ok through chapter 7. I did not have much a clue on 8 and 9. I'm currently learning some calculus. I'm on calc II next chapter. I guess I have something to push me. I would like to understand as much as I can of this book. Penrose loves complex numbers. I never knew they were so cool.

[Telos]

Well, you're not supposed to learn anything about physics until after Ch. 16, right? Aren't the first 16 chapters just mathematical context?

And if anyone ever finds a web page for the solutions, please post it here. The site I mentioned above makes it look like Penrose just forgot.

[rick1138]

I have it. It is a great book, but fairly advanced. Do not be intimidated if you don't understand everything in the book - it is basically an informal discussion of graduate level math. As far as complex numbers go, the book he mentions, Visual Complex Analysis, written by one of his students, is absolutely fantastic - THE book on complex analysis.

[bombadillo]

I have the book and I like it, but I wouldn't recommend it to a layperson. Perhaps the ideal readership would be math graduates who would like to understand some physics. Such graduates should have taken courses in com-plex analysis (and a bit of Riemann surface theory would not go amiss), differential geometry, general topology (a bit of algebraic topology would also be helpful), group theory, some representation theory, classical mechanics and classical electrodynamics. The Road to Reality is not a text, nor could it be within the scope of a mere 800 pages. It gives a heuristic overview that will only be comprehensible to those who already know the mathematics.

I see the book has been favorably reviewed by the press, and I can't help wondering whether the journalists assigned to write the reviews understood the book at all.

The book by Needham, titled, Visual Complex Analysis, is interesting, and on my shelves, but not one I'd recommend for a standard first course in the subject. Penrose wrote a favorable review for it perhaps because Needham refers to him obseqiously every few pages.

In summary, if all you've taken are a couple of calculus courses, you might like to have the book for the purpose of inspiration, but don't think you'll understand what a connection is, or what Cartan's exterior calculus is.

[rick1138]

Penrose wrote a favorable review for it perhaps because Needham refers to him obseqiously every few pages.

Needham quotes Penrose obseqiously in his book - he was one of his students.


I see the book has been favorably reviewed by the press, and I can't help wondering whether the journalists assigned to write the reviews understood the book at all.


They didn't. I read the review in the New York Times, it was hilarious. There was one quote, I don't remember it exactly, but it was something like, "I must admit that I stuggled with the concept of a '(1,3) valent tensor'", or i.e. the reviewer didn't understand a word of what he read.

[inquire4more]

I seem to agree with all the posts so far. I am currently reading my girlfriend's copy of this book and am thoroughly enjoying it, but I am not moving through it at other than a snail's pace (in my defence, I always take my sweet time getting throught the intro/review chapters, getting bored, but still plod through them in the event there is something I was not aware of). I have a great respect for Prof. Penrose. I just think he may be a bit (read, way the hell) too optimistic about either the knowledge level, intelligence, patience, or interest of the general public. His treatments of the mathematics are obviously and understandably not as in depth as a textbook, but imho, still a bit much for joe armchair-physicist out there. But, as I said, I am enjoying this book.

[PhilosophyofPhysics]

I'm on chapter 26 now. I love Quantum Mechanics. Can't wait to read bout the other theories too. I'll probably never learn the math, especially not on my own, but at least I'll reconize terms now when browing the physics boards or when reading other physics books. Sometimes I had to push myself to read on in some of the math sections because I knew greater things were coming up. I can see how some of the math might be interesting though. I was glad one I got to the physics.

[neurocomp2003]

So the book is worth than i take it...i keep glancing at it here and there when i visit chapters....apparently its been taken off shelf and a new one is coming out 2005-10 or 2005-11

[Peter.E]

People keep mentioning how hard the maths is in the book, but what background of maths do they have?

Could people tell me how good they are at maths so I can compare with me, see if I would be able to understand it. I obviously won't stand a chance if you are all undergraduates :)

[PhilosophyofPhysics]

My background is only calculus I in college, a bit of Calc II from highschool and review lately, and I just learned a bit about vectors. Vectors are cool!

[James R]

I am at PhD level in physics. There is maths in the book that is vaguely familiar to me, but not explained in enough detail for me to fully understand it without looking at other references.

[neurocomp2003]

like cliffoard algebra?

[ron damon]

I see there is a new edition of the book coming out on October 25, 2005. Does anyone know if it'll include anything new? Like exercises with answers?

I really want to buy it, but don't want to end up double dipping when/if an improved version is released.

Also, for the people who have read it, how will an ignorant fool like myself only acquainted with math up to a little of differential equations and vector calculus fare with this mammoth?

[neurocomp2003]

no its softcover vs hardcover

softcover is Oct2005-and half the price

[selfAdjoint]

I see there is a new edition of the book coming out on October 25, 2005. Does anyone know if it'll include anything new? Like exercises with answers?

I really want to buy it, but don't want to end up double dipping when/if an improved version is released.

Also, for the people who have read it, how will an ignorant fool like myself only acquainted with math up to a little of differential equations and vector calculus fare with this mammoth?

If you were fairly cool with your DE and vector courses, you should be all right at least half way through the book, and you will learn some interesting stuff. The last half of the book is mostly physics, and it's no harder than any other serious intro to quantum, relativity and so forth. You should try it if it interests you.

[SpaceTiger]

I am at PhD level in physics. There is maths in the book that is vaguely familiar to me, but not explained in enough detail for me to fully understand it without looking at other references.

This is my problem with the book. It's good for getting a new perspective on stuff I already know, but the information on the new stuff isn't sufficiently detailed for me to get a full grasp of it without resorting to other texts. It is certainly interesting, though, so perhaps I will treat it as one of those books that I make my way through gradually as I learn the new material from other sources.

[Telos]

Also, for the people who have read it, how will an ignorant fool like myself only acquainted with math up to a little of differential equations and vector calculus fare with this mammoth?

Chapter 7 is on complex-number calculus, which I was not at all familiar with, so I'll use that as an example. After reading the chapter, I know in order for there to be a concept of "slope" in complex calculus a function must satisfy what's known as the Cauchy-Riemann equations. This allows for a different kind of integration called countour integration, which looks like this \oint, and can be used to make a "beautiful" formula for nth derivatives. But I do not know why this is beautiful so I have to take his word for it. He expands on this concept by explaining the difference between homology and homotopy and thus, introducing topology. I have always known that contour lines are those lines showing equal elevation on geopgraphical topology maps, but Penrose never confirms or denies that this is the same idea. Somewhere during this he talks about convergent power series and finishes by mentioning the Dirichlet series, the Riemann Zeta function, and the Riemann Hypothesis, things which I have seen before but still cannot mentally grasp. I have read the chapter twice and although I know that all of these things are related, I do not know exactly how.

(Somewhere in all of this I am screaming to know how this relates to physics and my understanding of reality, yet there are still 8 more chapters until physics is mentioned...)

It relies on all the previous chapters, so perhaps I did not understand something very well that he had written before, or perhaps I was just not very focused.

It has been a few months, and so far this book has only served to motivate me into reading actual textbooks. In that way perhaps it is good - it sort of tests your will power to understand. If you are not interested in learning more, you say, "okay that's fine" and put it down. If you're interested in learning more, you say, "no, that's not fine... Penrose you glib" and find a more comprehensive source, but this time with some ideas and relationships you didn't know before, giving you an idea of what you might learn.

However, Penrose's insistence on calling various concepts like complex numbers "magic" is extremely irritating, like he's speaking to a kindergartener. "The magical fact thus arises, that any complex function that is complex-smooth is necessarily analytic!" :confused:

Analytic functions are so dreamy! :!!)

[mathwonk]

i recommend reading riemann. this stuff about the riemann cauchy equations is discussed there in the fiurst few pages of his thesis, and then he continues witha complete foundation for compelx calculus, and much more.

[Jimmy Snyder]

He expands on this concept by explaining the difference between homology and homotopy and thus, introducing topology. I have always known that contour lines are those lines showing equal elevation on geopgraphical topology maps, but Penrose never confirms or denies that this is the same idea.
It's not. I think the explaination in the book is much better than what I can do, but I will try anyway. Unfortunately, I don't have the book in front of me now. The issue has nothing to do with complex analysis, but rather with topology (not topography either).

If you start with two distinct points on a piece of paper and draw a circle in such a way that the two points form the endpoints of a diameter of the circle. Consider that there are two paths from one point to the other, each path being half of the circle. You can imagine continuously deforming one of those semicircular paths until it coincides with the other. However, if there were a hole in the paper at the center of the circle, you could not make the deformation continuous because you can't drag the path across the hole (think of a line perpendicular to the paper through the hole, then the path would snag on the line as you tried to deform it).

In this description, I used two semicircles, however the same reasoning would apply if you compared any two paths between any two points. If you can continously deform one path into the other, then the two paths are homotopic. In nonprecise terms, if there are no holes in the region between the two paths, then they are homotopic.

Homologic is like homotopic except that if your path doubles back on itself (as in the picture in the book), then you can cancel the doubling part even though doing so is not a continuous operation. Because you are allowed this extra operation, it turns out that if two lines are homotopic, they are certainly homologic, but not vice versa.

[rick1138]

If you get a copy of Visual Complex Analysis, you will find that it discusses many of the topological issues you are asking about. It was written by a student of Penrose. A good very basic introduction to algebraic topology is A Combinatorial Introduction to Topology.

[Telos]

Ah, thank you! For the suggestions and the help.

jimmysnyder, your explanation was very clear and more straightforward than Penrose's, IMO. He seems to dance around it with pretty language while feeling a mathematician's obligation to tangle one's self in rigor.

I just thought of this... perhaps it might be useful for readers to write their own glossaries while reading this book (Penrose does not provide you with one). I have been keeping notes like I would with a math book, thinking I would need to spend more time studying formulae and doing the practice problems. But the solutions to the problems have not been available and will not be for some time, so it might be better to spend more time on the concepts. A small stack of notecards could be an indispensable tool.

Oh, and readers could understandably be intimidated by the title into thinking that this book would be better if it was kept clean of notes in the margin or underlined sentences. If you purchased the book, you should not hesitate to mark it up like a coloring book.

[Jimmy Snyder]

jimmysnyder, your explanation was very clear and more straightforward than Penrose's, IMO.
Thank you for your kind words. I'm glad if my small effort helped you. I intended it to be little more than a repeat of what Professor Penrose had written. Sometimes it helps to read two different descriptions of the same thing. As to which one is more straightforward, I would say that it is very subjective. Perhaps if you had read mine first, you would be praising Professor Penrose for clearing up my explanation for you.

He seems to dance around it with pretty language while feeling a mathematician's obligation to tangle one's self in rigor.
As a mathematician wannabe myself, I have to disagree that there is anything like mathematical rigor in Professor Penrose's book. That is the charm of his book for me. To me it seems that he is explaining the underlying ideas without the fussiness over details and edge conditions that I am used to reading.

it might be better to spend more time on the concepts
A small stack of notecards could be an indispensable tool.
you should not hesitate to mark it up like a coloring book
Three excellent ideas.

I reread your previous message and I came to the realization that you may have misinterpretted the word 'contour' in the phrase contour integral. In this context, the paths are not related to the lines of constant elevation (contour lines), but rather relate to the outer edge (contour) of regions.

[mathwonk]

the group of paths up to homology is the abelianization of the group of apths modulo homotopy, so thats why the equivalence relation is broader.

i.e. any path of form A.A^-1.B.B^-1 is homotopic to zero, but also A.B.A^-1.B^-1 is homologous to zero.

since integration of a compelx differential is constant on homotopy classes, and defiens a homomorphism to an abelian group, namely the numbers, it is also constant on homology classes.

the whole purpose of these equivalence relations on paths was to state the cauchy integral theorem, and you could actually say that two paths are homologous if and only if the integrals of all complex holomorphic differential forms over them are the same.

[mathwonk]

riemann actually approaches the theory a little differently, he says two paths are equivalent if togetehr they form the boundary of a piece of surface in the given space. Then they have the same integral for all holo forms by stokes theorem.

his equivalence relation is essentially the same for surfaces i think, but in general may be different and is now called cobordism.

[Telos]

In this context, the paths are not related to the lines of constant elevation (contour lines), but rather relate to the outer edge (contour) of regions.

Oh, I see! That simple disambiguity changes a great deal. Thank you again.

[mathwonk]

the magical formula for the nth derivative of f(z) is the integral of f(t) multiplied by the n+1 st power of 1/(t-z), wrt t, roughly.

The magic of this is that it only involves powers, which make sense not just for n an integer but also for n an irrational, or even a complex number. Hence one can use it to take the <pi>th derivative! or the ith derivative.

This too is explained in Riemann's original works on compelx analysis.

By the way, professional mathematicians and physicists are themselves like kindergartners in their enthusiasm for math and its beauty, and are not talking down to people when they display it.

One thing puzzles me: is Penrose some kind of cult hero like Harry Potter? Why are so many people here dead set on reading a book they obviously do not have the background to read?

I.e. the goal seems to be to read PENROSE, rather than to learn the material from a sutiable source.

[Telos]

Mathwonk, I suppose physicists and mathematicians differ greatly on the meaning of "magic," then, from others in the populous, who see it as something that defies a law of nature or as something more New Age.

It strikes me particularly irritating here because magic relies on both illusion and stupefication to achieve its effects. I'm reading the book to be disillusioned... it's called Road to Reality, for chrissakes.

By the way, professional mathematicians and physicists are themselves like kindergartners in their enthusiasm for math and its beauty, and are not talking down to people when they display it.

Then they appear to use "magic" as synonomous with beauty. I can't disagree more. Magic, insofar as it is the intentional mystification of something, is the antithesis of beauty, and I am simply bothered by its use here.

But now that you mention it, I apologize for my kindergartener remark. It has been my intention to remove prejudice of age from my language. "Talking to a kindergartener" should not be connected with the meaning "talking down." Although there are no kindergarteners here to offend (I think), we secondarily harm them nonetheless.

[Edit: Oh, and that's not how I meant it originally... when we speak to children we often find it amusing to toy with them and confuse them, especially when they ask questions like, "how did you do that?" and you say, "magic!" But that's now how I meant it either... Instead of communicating his entusiasm it seems like Penrose is trying to market the ideas the make them more palatable. He could do it much better, I think, by proclaiming that they are not magic! And reveal their pervasive beauty for the reader.]

[mathwonk]

in this case, he means the conclusion one gets is far stronger than anyone has any right to expect, based on the tiny hypothesis, so it feels almost like magic. or perhaps one should say merely amazing, rather than beautiful.

I.e. the result is beautiful but the kicker is that you get so much bang for the buck in this result. why in complex calculus should assuming one derivative imply there are actually infinitely many? this never happens in real calculus. so "unexpected" and "extremely fortuitous" are other possible choices of words.

you should really meet the gentleman some time though as i was extremely fortunate to do, as he is the most charming unoffensive and modestly brilliant man. I am sure he would relieve your concern on this point infinitely better than I. He would probably even apologize he is so genuinely nice and considerate.

[selfAdjoint]

I.e. the result is beautiful but the kicker is that you get so much bang for the buck in this result. why in complex calculus should assuming one derivative imply there are actually infinitely many? this never happens in real calculus. so "unexpected" and "extremely fortuitous" are other possible choices of words.


Isn't this because the ways of approaching the limit in the plane are so much greater that they put stronger constraints on differentiability? On the real line you have two ways of approaching a point, from left and from right. In the plane there are as many as there are paths ending at the point. So the situation isn't that differentiation of complex variables is so strong, but that differentiation of real variables is so weak, it's easier to get a real derivative, but you don't get as much when you have it.

[Jimmy Snyder]

Isn't this because the ways of approaching the limit in the plane are so much greater that they put stronger constraints on differentiability? On the real line you have two ways of approaching a point, from left and from right. In the plane there are as many as there are paths ending at the point. So the situation isn't that differentiation of complex variables is so strong, but that differentiation of real variables is so weak, it's easier to get a real derivative, but you don't get as much when you have it.
If you actually had to do the work of approaching the limit along each path, then you would have infinite work to prove infinite differentiability. As you say, the conclusion is not so surprising given the amount of work you have to put into it. No 'magic' here.

However, you don't need to do that. Just prove the continuity of the partial derivatives (two directions for each of the real and imaginary parts of the function, as in the real case, for a total of four conditions) and the Cauchy-Riemann conditions (two conditions also involving those partials). For the work of proving 6 conditions, all of them involving real functions of a real variable, you get infinite differentiability. This is where the 'magic' lies.

I agree that 'magic' is a poor choice of words for a scientist to use when describing science (but here he is describing math). But just replace that word with 'power and beauty' and everything is good, so I am willing to cut the author some slack.

[selfAdjoint]

If you actually had to do the work of approaching the limit along each path, then you would have infinite work to prove infinite differentiability. As you say, the conclusion is not so surprising given the amount of work you have to put into it. No 'magic' here.

However, you don't need to do that. Just prove the continuity of the partial derivatives (two directions for each of the real and imaginary parts of the function, as in the real case, for a total of four conditions) and the Cauchy-Riemann conditions (two conditions also involving those partials). For the work of proving 6 conditions, all of them involving real functions of a real variable, you get infinite differentiability. This is where the 'magic' lies.



Yes indeed, the Cauchy-Riemann equations. I was going to mention them as showing the really strong constraints a locally analytical function obeys in complex analysis - real functions have nothing like it. I decided not to discuss them but I'm glad you brought them up.

[Telos]

I see it now, mathwonk. I started rereading the book with the new tools and perspectives I've gained from speaking with you and jimmysnyder, and it's turning out to be much more rewarding. What I mistook for Penrose's fussiness about details was probably my own fussiness. I'm just too eager to learn, I suppose. :smile:

[vanesch]

For the work of proving 6 conditions, all of them involving real functions of a real variable, you get infinite differentiability. This is where the 'magic' lies.

Well, it is in fact not true that 6 real conditions are sufficient. The 6 real conditions have to be valid IN EACH POINT IN A NEIGHBOURHOOD of the point where you want to show infinite differentiability. That's an infinite set of conditions, about similar to what you suggested: proving that the limit exists in EACH DIRECTION.

cheers,
Patrick.

[Jimmy Snyder]

Well, it is in fact not true that 6 real conditions are sufficient. The 6 real conditions have to be valid IN EACH POINT IN A NEIGHBOURHOOD of the point where you want to show infinite differentiability. That's an infinite set of conditions, about similar to what you suggested: proving that the limit exists in EACH DIRECTION.

cheers,
Patrick.

http://mathworld.wolfram.com/ComplexDifferentiable.html

I'm sorry I haven't had time today until now to explain this. You are correct that the 6 conditions have to be valid at each point in a neighborhood. However, you are wrong that the 6 conditions are not sufficient. Because if you can show that if the conditions hold at a single point, then you are assured that they will hold in a neighborhood. This is the magic that professor Penrose is talking about.

[mathwonk]

the difference between the plane and the line is not quite sufficient to explain the much stronger result in complex calculus, as we also have differentiable functioins of two variables in real calculus, but they do not have the strong properties compelx differentiable functions.

a function oif two variables is real differentiable iuf and only if it has a good real lienar approximation ( as a function of two real variables), and is compelx differentiable if that real linear function of two variables is actually complex linear.

now for some reason, being complex linear is so much stronger than just being real linear, that a function with complex linear approximation in a whole nbhd of a point, is infinitely differentiable.

[selfAdjoint]

now for some reason, being complex linear is so much stronger than just being real linear, that a function with complex linear approximation in a whole nbhd of a point, is infinitely differentiable.

The complex numbers form a field, while cartesian coordinates in the plane don't. Perhaps that's the difference.

[mathwonk]

maybe, but it seems to be also that the complex differentiable, functions are locally i.e. infinitesimally, angle preserving.

[selfAdjoint]

Isn't that a derived property ? I guess what we need to do is display the derivation of the Cauchy-Riemann equations and look at what properties of coplex numbers are used in the hypotheses. I'll be back.

Here again. The derivation given in Mathworld (http://mathworld.wolfram.com/Cauchy-RiemannEquations.html) as far as I can see uses:

1) algebraic properties of i
2) existence of complex conjugate
3) requirement that derivative be independent of orientation.

I think you could compress (1) and (2), and I believe (3) is basically a topological constraint. What do you think?

[mathwonk]

It seems to me that a complex differentiable function is essentially a function which is infinitesimally approximable by a complex linear function, which means multiplication by a complex number, which means a rotation and a scaling, so since both scaling and rotation preserve angles, it preserves angles.

that is the essential content of the CR equations, I think.

mathworld is nice but riemann is perhaps better.

his argument is that df/dz should depend only on z and not on dz, and hence the term df/dzbar should be zero.

riemann is really very clear.

[vanesch]

I'm sorry I haven't had time today until now to explain this. You are correct that the 6 conditions have to be valid at each point in a neighborhood. However, you are wrong that the 6 conditions are not sufficient. Because if you can show that if the conditions hold at a single point, then you are assured that they will hold in a neighborhood. This is the magic that professor Penrose is talking about.

Ok, let us give this a try.

f(x,y) = u(x,y) + i v(x,y)

Now, consider u(x,y) = x + sqrt(x^2 + y^2) x^3

Both partial derivatives are continuous in (x,y) = (0,0), and du/dx = 1 and du/dy = 0 in (0,0)

Consider v(x,y) = y + sqrt(x^2+y^2) x^3
Same conclusions hold, except that dv/dx = 0 and dv/dy = 1 this time, so the Cauchy Rieman equations hold for (x,y) = (0,0)
But note that they don't hold in a NEIGHBOURHOOD of (x,y) !

So our 6 conditions are valid AT (x,y) = (0,0) but not in a neighbourhood, no ?
(and I think that f is not an analytic function...)

Ok, I put this example together rather quickly, maybe I made a mistake somewhere...

[Jimmy Snyder]

Ok, I put this example together rather quickly, maybe I made a mistake somewhere...
You are up against professors D'Alembert, Cauchy, and Riemann, not me. I'm just the messenger. You must get your ducks in a row before messing with these guys.

[vanesch]

You are up against professors D'Alembert, Cauchy, and Riemann, not me. I'm just the messenger. You must get your ducks in a row before messing with these guys.

Well, my cards are on the table: look at the example !
However, I've been looking up exactly what is stated, and it is stated that the 6 conditions AT A POINT are sufficient to define the complex limit
(f(z) - f(z0)) /(z-z0), also called the complex derivative of f(z) at z0. So the first derivative has a meaning when the 6 conditions are met at a point, and that's maybe what you're hinting at.
HOWEVER, in order for the function to be holomorphic, meaning, ALL derivatives to exist, you need the function to have a first derivative IN A NEIGHBOURHOOD of a point (so that you can use Cauchy's integral there with a pole of n-th order in order to find f(n)(z) ; but you need to integrate along a small circle AROUND the point to do so). So my point is still that in order to have ALL derivatives simply when you have a first derivative (the "magic"), you need to have that first derivative in a neighbourhood.

I really don't think that if the 6 conditions are satisfied in a point, that they automatically are satisfied in a neighbourhood ; I think my example is a counterexample of that. Maybe I'm mistaken, but I don't see where my example goes wrong...

[Jimmy Snyder]

vanesch,

I was simply using the article from mathworld whose link I repeat merely for convenience:

http://mathworld.wolfram.com/ComplexDifferentiable.html

Therefore, when I said you were up against D'Alembert, Cauchy, and Riemann, I was wrong. You are only up against Wolfram. I now believe there is an error in his web page, and not in your statement. I quote his page and then provide what I now believe is a necessary edit.


Let z = x + iy and f(z) = u(x, y) + iv(x, y) on some region G containing the point z_0. If f(z) satisfies the Cauchy-Riemann equations and has continuous first partial derivatives at z_0, then f'(z_0) exists and is given by

equation elided

and the function is said to be complex differentiable (or, equivalently, analytic or holomorphic).

I think this should have said:


Let z = x + iy and f(z) = u(x, y) + iv(x, y) on some region G containing the point z_0. If f(z) satisfies the Cauchy-Riemann equations on G, and has continuous first partial derivatives at z_0, then f'(z_0) exists and is given by

equation elided

and the function is said to be complex differentiable (or, equivalently, analytic or holomorphic).


I know I'm not quoting you vanesch, but I put your name on it because it is essentially what you have been saying. If this altered version is indeed correct, then the implications that you drew are also correct. It is not enough to simply verify 6 conditions, but rather 2 plus 4 times the number of points in a region \aleph_1. In the real case there are only \aleph_0 derivatives to verify. It would seem that there is more work to prove analytic in the complex case than there is to prove infinite differentiability in the real case. However, I expect that there are techniques that tilt the balance in a way different than this calculation would imply.

I am still not 100% sure that the edit is necessary and correct and would like the opinion of a mathematician. However, I provide this simple observation that makes it seem necessary. If differentiability at a point could be verified by simply looking at the values of the function along the vertical and horizontal lines passing through the point, then take any analytic function, throw away the values off those two lines, and replace them with arbitrary nonsense. I don't see how to prove continuity, let alone differentiability.

If the contributors to this thread agree that the edit is necessary and correct, then I will contact mathworld and suggest it to them.

[vanesch]

Hi again,

I think what you quoted in Mathworld is actually correct: the 6 conditions (partial derivatives continuous, and the Cauchy-Riemann equations) need only to be satisfied IN A POINT for the complex derivative f'(z) to exist IN THAT POINT. That's what I said too, in my previous post. But that, by itself is not a "miracle".

The miracle seems to come in when we have a complex FIRST derivative that exists, and then ALL HIGHER DERIVATIVES POP UP OUT OF NOTHING. Well, THIS is what is not happening, if the complex first derivative only exists in a point. In order for this to happen, the complex first derivative has to exist in an open domain ; THEN you get the higher derivatives for free, and the proof goes by using the Cauchy-Riemann integral formula (which needs to circle around the point where you want to calculate the higher derivative, so you need a domain for this integral path to be drawn in the first place).

So, to resume:
- concerning the existance of the FIRST complex derivative, 6 conditions in one point are enough.
- concerning the existance of infinite differentiability, 6 conditions need to hold in a domain.

cheers,
Patrick.

EDIT: and I would like to add that in the two cases, there is less of a miracle than it seems ; in my first post I was only addressing the second point.

[Jimmy Snyder]

Here is a definition of analytic from an on-line book on Complex Analysis.

A function f is said to be analytic at a point z_0 \in C if it is differentiable at every z in some \epsilon-neighborhood of the point z_0

http://www.maths.mq.edu.au/~wchen/lnicafolder/ica03-cd.pdf

However in the Wolfram article it says that complex differentiable is equivalent to analytic.

Either professor Chen is wrong, or professor Wolfram is.

[vanesch]

H
Either professor Chen is wrong, or professor Wolfram is.

Or they use incompatible definitions :-) I'd go with Chen on this one, though. However, if we stick with "Road to reality" I think Penrose defines things as follows:

1) Complex differentiable: f'(z0) exists

2) Holomorphic: f(n)(z0) exists for all n

3) Analytic: f(z) equals a power series a + b (z-z0) + c (z-z0)^2 ....

Now, to my understanding, 1) on an open domain D implies 2) on D and implies 3) on each disk completely contained in D.

I'm not a mathematician but it is my understanding that authors seem to interchange sometimes "analytic", "differentiable", "holomorphic" and so on, and as one implies the others on an open domain, it usually doesn't make much of a difference in practice, except for nitpickers like us :-)

[mathwonk]

dear jimmy snyder: how are you quarelling with mathworld in post 50? is that a definition they give? if so how can you quarrel with a definition? are you saying it disagrees with other statements made on that same webpage? or it differs from other peoples definition?

if so, so what?

I think we must allow anyone to make any definition they choose, mustn't we?

unless they then use that definition incorrectly.

I'm not sure I understand the controversy.

you guys are right that all those definitions are equivalent if they all hold on an open set, and there are even stronger "distribution theoretic" ones that are also equivalent as well.


I have a deeper question: several people have asked why my avatar is a pikachu. what's a pikachu? is it like a schmoo in al capp? ( a sort of universal food item for meat eaters.)

[Jimmy Snyder]

if so, so what?
You are right, of course. Mathworld should feel free to define analytic any way they want. But they have defined it in such a way that vanesch's monstrosity is analytic and I don't believe that they meant to do so. I find it easier to believe that they made a simple error as I indicated.

[Jimmy Snyder]

I would like to strenghthen that. They have defined analytic in such a way that it does not even imply continuity. This surely was not intentional. If I am wrong on this point, then it's my bad. But if Mathworld is wrong, I should think they would appreciate being told.

[mathwonk]

why do you say that their definition does not imply continuity? they say the partials are continuous at the given point, which implies they exist in a neighborhhod of the point (or else continuity would make little sense).

then the function has bounded partials in a neighborhood of that point hence is continuous in a neighborhood of that point, no?

In fact with mathworld's definition the function is not only continuous at the point, but also differentiable there as a function of two variables (see Spivak, calculus on manifolds, page 31.) in the sense of having a linear approximation which is a linear function of two variables.

furthermore the linear approximation is actually complex linear by virtue of the cauchy riemann equations holding at the point, hence it is completely correct for them to call their function at least complex differentiable at the given point. It might be even more accurate to actually require less, and only say that the real derivative exists and is actually complex linear.


I do disagree with their sue of the word "analytic" for this however as that toi me implies existence of a powers series represenattion that holds in an entire nbhd.

however due to the fact that all definitions agree when they hold in an open set, the terminology has never stabilized.\

the most important case however is for functions which are complex differentiable in an open set, and in that case there is no controversy.

[Jimmy Snyder]

In the Mathworld definition, the partials are assumed to be continuous as real valued functions of a real variable (remember that in the definition of a partial derivative, the 'other' variables are kept constant). In fact, I believe that if you read their definition carefully, you will find that they define complex differentiable at a point completely in terms of the values of the function on the vertical and horizontal lines passing through that point and in total disregard of the values of that function off of those two lines. That means that a function could meet all of the requirements of the definition and not even be defined in a neighborhood of the point. Since they also say that complex differentiable is equivalent to analytic, I believe they did not say what they meant to say. If they make the very simple change that I recommend, then, in my opinion, it will bring their definition into line with what they probably meant.

[mathwonk]

you do not seem to get it: I'll try again - they say the partials are continuous at the point, which means tacitly that they are also defined near the point, hence the values of the function on vertical and horizontal lines centered at points near the given point are involved, i.e the values of the function at all nearby points.

now do you see what you have been missing?

I agree it is a little subtle, but it is explained in many calculus books, for example Courant.

I have also given you the spivak reference with a proof of the statement I made above. Just look it up. or look in any calculus book of several variables.

[mathwonk]

i looked at your suggested change and again I think you are missing the point. the only thing they have not said, but which is understood in what they have said, is that the partials are defined in G, and then it is sufficient to assume that they are continuous at the one point z0, and further that the Cauchy Riemann equations hold at that one point.

There is no reason at all for the Cauchy Riemann euqtions to hold in all of G, to deduce differentiability at the given point.

The way to settle all such disputes in mathematics is to give a proof of your own claims, and a counterexample to theirs. (I suggest however you read the proof I have referenced before spending too much time looking for a counterexample.)

[Jimmy Snyder]

Here is a function (using the notational convention of the Mathworld definition)

f(z) = 0 if x = 0 or y = 0
f(z) is not defined if x = y (where x is not zero)
f(z) = 1 otherwise.

du/dx = 0, du/dy = 0, dv/dx = 0, dv/dy = 0 (all evaluated at (0, 0))

all partials are continuous, and the Cauchy-Riemann conditions hold at (0,0). The function is analytic at (0, 0) (according to Mathworld). The function is not continuous in a neighborhood of (0, 0). The function is not even defined in a neighborhood of (0, 0).

[mathwonk]

actually, rereading your quote from mathworld, I would say that they already intended it to mean exactly what your suggested change says, but i still claim that stronger meaning is unnecessary, and if they did mean that, I would wonder why.

[Jimmy Snyder]

Did you read message 61? Because of the timestamps I fear you may not have seen it.

[mathwonk]

so? what is the interest of that example? obviously a fucntion defiend on two lines cannot ni any reasonable sense by approximated by a linear function.

the first requirement for a fucntion defined on M to be approximable by a linear fucntion is that M be approximable by a loinear space, i.e. have atangent space, so M must be a manifold.

[mathwonk]

post 61 does not satisfy mathworlds definition because the partials are not continuous at the given point because they are not defined nearby because the function itself is not defined nearby.


mathwonk:

science advisor, homework helper, and grumpy old twit.


(I like to control my own titles)

[Jimmy Snyder]

The partials (e.g. du/dy (x, y) = 0) are entire functions.

[vanesch]

The partials (e.g. du/dy (x, y) = 0) are entire functions.
That's not the partial derivative of the function you proposed. It is an entire function which coincides with the partial derivative on the x=0 line, no ?

[Jimmy Snyder]

Here is a copy of the message that I sent to Mathworld this morning.

I think there is a problem with the definition of 'Complex Differentiable' Compare this to the Theorem at the top of page 379 in the Dover publication of "Elementary Real and Complex Analysis" by Georgi Shilov. He requires that the Cauchy-Riemann equations hold in a neighborhood and that the partials are continuous in a neighborhood. But your definition only requires this at a point.

I'll let you know how it goes.

[Jimmy Snyder]

u(x, y) = x^2 + y^2
v(x, y) = 0

du/dx = 2x
du/dy = 2y
dv/dx = 0
dv/dy = 0

All partials are continuous at (0, 0), in fact continuous everywhere.

The Cauchy-Reimann conditions hold at (0, 0) and nowhere else. Therefore, the radius of convergence at (0, 0) is zero and the function is not analytic at the point (0, 0).

[Telos]

Our very own Chronon wrote review for RtR on his site, which I just discovered today.

http://www.chronon.org/reviews/Road_to_reality.html

It came to some of the same conclusions we had and more. I wish I had read it before I started reading the book!

[vanesch]

u(x, y) = x^2 + y^2
v(x, y) = 0

du/dx = 2x
du/dy = 2y
dv/dx = 0
dv/dy = 0

All partials are continuous at (0, 0), in fact continuous everywhere.

The Cauchy-Reimann conditions hold at (0, 0) and nowhere else. Therefore, the radius of convergence at (0, 0) is zero and the function is not analytic at the point (0, 0).

Yes, that's in the same flavor as my example :-)

In fact, you can even do something further: take a REAL function g(x) with continuous g'(x);
Now, compose the complex function:

f(x,y) = g(x) + i g(y)

(so u(x,y) = g(x) and v(x,y) = + g(y))

Now, du/dx = g'(x), du/dy = 0 ;
dv/dx = 0, dv/dy = g'(y)

Clearly the partials are continuous, and Cauchy-Rieman is satisfied in (x,y) = (0,0) (but not around it, except on the line x=y)

If ever this would imply analyticity, (all derivatives existing), then it would simply mean that the same property holds for the real function g(x): that the existence and continuity of the first derivative of g(x) would imply all derivatives to exist of g(x) (as a real function) ; and/or that the series devellopment of g(x) would exist.

cheers,
Patrick.

[Jimmy Snyder]

On Chronon's review

I have only just started reading Chapter 13 on Group theory so I can only say that I agree with Chronon on his description of the Math part of the book. Hopefully I will remember to reread the review once I get into the Physics part. I certainly agree with the bit about reading the book twice. I only understood about 75% of what I read. I was especially weak in the matter of Riemann surfaces. Chronon encourages me to go on to the Physics, which is the part I am really interested in, even if I don't fully understand the Math.

[Jimmy Snyder]

I was informed by someone at Mathworld that the article on complex differentiation will be edited for the next update. I don't know when that is. I am going on vacation for a week starting tomorrow afternoon so I may not be able to follow up when they do.

[mathwonk]

i don't suppose you would take my word for it, but there is nothing at all wrong on the page of mathworld to which you have linked above.

shilov apparently has a different definition for the word "analytic" than mathworld does, thats all.

i.e. mathworld is using the concept of "analytic" at a point as meaning complex differentiable, and pesumably shilov is using the word for a function defined and representable by a powers eries on an open disc.

it is just a difference of terminology that's all.

if they change that as a result of your message, it will be a case of the blind leading the blind.

[Jimmy Snyder]

Yes, that's in the same flavor as my example :-)
I knew you were going to say that. The ONLY advantage of my example over yours is that I was able to take the partials with confidence. My ability to calculate is rather poor.


f(x,y) = g(x) + i g(y)

This is an excellent example. It shows just how much the current Mathworld definition is at odds with the rest of the math world ;). Mathwonk, you are correct. It was their right to do it, but I doubt it was their intention, especially as they now intend to change the definition. By the way, they accepted my first example function (the non-continuous one) as reason enough to make the edit. But vanesch's examples are far more forceful in my opinion.

My example, by the way, points to a theorem that I have never seen but may well have been published somewhere:

The only real valued analytic functions of a complex variable are constants.
Proof: dv/dx = dv/dy = 0 everywhere. So, in order to satisfy the C-R conditions, du/dx and du/dy must also be zero wherever the function is analytic.

The same goes for pure imaginary valued analytic functions.

[mathwonk]

well althopugh i have opposed your point, I will now admit that I personally prefer the meaning of "analytic" that you have advocated. Still the only appropriate change in their page would be to omit that one word, not to change their correct definition of compelx differentiable.

[Jimmy Snyder]

I see that the Mathworld article has been edited. The new definition is grammatically incorrect in that it says 'the neighborhood' where it should say "a neighborhood', but I don't intend to bother them about it. Also they added a reference to Shilov's book on Real and Complex analysis which I had cited when I first contacted them.

Mathwonk, I think that Mathworld was forced to make the change they did for two reasons. One is that the phrase 'complex differentiable' is used elsewhere in Mathworld definitions in a form that indicates that the meaning is the same as analytic. If they followed your advice, they would have to change those definitions as well. The second reason is that if they stuck to the old definition then they would lose the theorem that got all this started: If a function is complex differentiable, then it is infinitely complex differentiable.