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

[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.