Talk:Rouché's theorem

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[edit] Recent changes

Hi Dysprosia. I saw your changes to the page. I was just writing to you here, when I got your message :)

If I understand it correctly, your motivation was to explain my dirty hack. I hate dirty hacks myself, so I see your point.

I would just like to explain why I think your change did not make things better.

First, you did not make my hack really go away. You replaced it with the hack

{f'(z) \over f'(z)}+{A \over 1+g(z)/f(z)}

and you said "let's pretend this is how things will come up, and let's calculate A". So, ultimately it is not that the solution got more natural. The thing

{A \over 1+g(z)/f(z)}

still looks out of the blue.

Second, this came with the heavy price of lots of new formulas, the method of underterminate coefficients, etc. Not a pleasant sight for average Joe.

So I would just live with a dirty short hack rather than a equally dirty but more laborious hack. But ultimately it does not matter so much. I would leave it up to you to decide how the solution will look like. Oleg Alexandrov 00:53, 15 Jan 2005 (UTC)

Well, it's a bit more than just pretending, since we want the answer in a certain form anyway. But of course it's still not the best way of setting out the proof. Perhaps we could just explain the hack by stating it and just proving it after, which doesn't improve palatability, but it's probably a little more readable and lucid that way.
Or maybe we should find an entirely new proof altogether. PlanetMath has one, but I don't know how suitable it is. Dysprosia 03:40, 15 Jan 2005 (UTC)
I thought of this for a while. Could we just go back to your log proof, with my cleanup? But then you need to add one more line there, going back from the log to the fraction form so that the argument principle is applicable. Oleg Alexandrov 04:07, 15 Jan 2005 (UTC)

[edit] Your message

Thanks for your efforts to clean up Rouche's theorem! It's always better when another's looking and editing a page. Dysprosia 10:10, 14 Jan 2005 (UTC)

I do prefer the proof method that simply relied on using the rules of differentiation and a little basic algebra. Regardless, I have tried to derive

{h'(z) \over h(z)}=\frac{f'(z)}{f(z)}+\frac{\left(1+g(z)/f(z)\right)'}{1+g(z)/f(z)},

in the article, but I don't believe it is the best, nor an absolutely adequate explanation. It would be best if you personally filled in your derivation. Could you do so? Thanks Dysprosia 00:39, 15 Jan 2005 (UTC)

I assume you mean you liked your solution better. My problem with it was that you used logarithms, and I don't think it is easy to prove that the log (f+g) is well defined. Recall that the log is not defined on a ray coming from the origin.
There is no rationale behind my way. I just took a shortcut from your previous solution with the log. You see, that was a dirty hack too. Ultimately, the question of why on earth one has to deal with 1+g(z)/f(z) is not answered. I don't think it can be answered. This is just not one of those proofs where one cannot understand why things happen the way they happen. Oleg Alexandrov 01:06, 15 Jan 2005 (UTC)

[edit] some problems

Here are a few problems that you guys should be able to fix easily.

  • In the theorem statement, it says "the same number of zeros in C". I guess that should be "inside C".
  • f and g are assumed to be analytic inside and on C, but they might have zeros on C. That gives trouble with integrating the function f'/f along C. I don't have my books handy but I guess that f is supposed to be nonzero on C as well. (Or else, the proof has to be modified.)

--Zero 01:35, 15 Jan 2005 (UTC)

I made the mistake with "in C". My lame excuse is that I don't like this "inside C" terminolgy to start with, I would like more "inside the region delimited by C". I will fix it.

Now, that f is non-zero on C comes for free, because of |g(z)|<|f(z)|. Oleg Alexandrov 01:43, 15 Jan 2005 (UTC)

Yes, good reasoning. My next concern is the meaning of "inside" when C is complicated (consider a figure-of-eight). Why not restrict it to a simple closed contour and save the hassle? We don't need to have the most general possible statement in WP. This goes for Argument principle too, which seems to assume a simple closed contour without saying so. In the case of a simple closed contour, the Jordan Curve Theorem says that "inside" is well-defined. --Zero 11:22, 15 Jan 2005 (UTC)
Well Zero, you are sharp. Not only it has to be a simple closed contour, it also has to go counter-clockwise, otherwise the signs turn out wrong. I will fix that soon. By the way, even before you said the above, I had added a picture on the Argument principle article, which shows the situation. (As they say, a picture is worth... :)

Oleg Alexandrov 17:55, 15 Jan 2005 (UTC)

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