The Road To Reality - Is It Easy or Hard to Understand?

[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 existence of the FIRST complex derivative, 6 conditions in one point are enough.
- concerning the existence 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 function defiend on two lines cannot ni any reasonable sense by approximated by a linear function.

the first requirement for a function defined on M to be approximable by a linear function 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 development 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, that's 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.