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

[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.

 

[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 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 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!