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

[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.