Complex Analysis Complex Analysis Math Forum

October 18th, 2017, 11:25 AM   #41
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 Originally Posted by zylo From Post # 10 Again, assume a closed island in w where F(z) undefined. Let F(z) be arbitrarily close to the border. F(z+$\displaystyle \Delta$z) = (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z)= (z+$\displaystyle \Delta$z}(f(z)+f'(z)$\displaystyle \Delta$z) F(z+$\displaystyle \Delta$z)=F(z)+(f(z)+f'(z)$\displaystyle \Delta$z).
I already pointed out two specific problems with this calculation in the other thread that you didn't address. I note that you simply copy/pasted your '}' typo and then failed to correctly apply the distributive law. And, your equality is NOT an equality, at best it's only approximate equality. In the other thread I provided a demonstration of this fact.

Also I don't know what "Let F(z) be arbitrarily close to the border" means. Can you give me an example? Name a number "arbitrarily close" to 0.

Last edited by Maschke; October 18th, 2017 at 12:08 PM. November 7th, 2017, 08:04 AM #42 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Reference post #40 If $\displaystyle F(z)$ is close to border: $\displaystyle F(z+\Delta z)=F(z)+F'(z)\Delta z$ is a valid approximation UP TO border. If $\displaystyle F(z) = zf(z)$ $\displaystyle F(z+\Delta z)=F(z)+(f(z)+zf'(z))\Delta z$ is a valid approximation ACROSS border. Contradiction. F(z) maps to entire complex plane. $\displaystyle \rightarrow$ FTA November 7th, 2017, 12:19 PM   #43
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Quote:
 Originally Posted by zylo Reference post #40 If $\displaystyle F(z)$ is close to border: $\displaystyle F(z+\Delta z)=F(z)+F'(z)\Delta z$ is a valid approximation UP TO border. If $\displaystyle F(z) = zf(z)$ $\displaystyle F(z+\Delta z)=F(z)+(f(z)+zf'(z))\Delta z$ is a valid approximation ACROSS border. Contradiction. F(z) maps to entire complex plane. $\displaystyle \rightarrow$ FTA
It's been a while and I don't think you ever dealt with the concerns I brought up last time. Can you present a complete argument, and if two things $X$ and $Y$ are approximately equal, please write $X \approx Y$ for clarity. Thanks. November 8th, 2017, 10:29 AM   #44
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 Originally Posted by Maschke It's been a while and I don't think you ever dealt with the concerns I brought up last time. Can you present a complete argument, and if two things $X$ and $Y$ are approximately equal, please write $X \approx Y$ for clarity. Thanks.
The argument is complete.

Suppose f(x)=x$\displaystyle ^{3}$ for x>a and 0 otherwise.
I can calculate a derivative for any point up to, but not including, x=a.
If f(x) is close to a$\displaystyle ^{3}$ (the "border")
f(x+$\displaystyle \Delta$x} = f(x)+f'(x)$\displaystyle \Delta$x is a valid approximation ($\displaystyle \approx$) UP TO a$\displaystyle ^{3}$.

On the other hand, if f(x) = x$\displaystyle ^{3}$ for all x,
If f(x) is close to a$\displaystyle ^{3}$ (the "border")
f(x+$\displaystyle \Delta$x} = f(x)+f'(x)$\displaystyle \Delta$x is a valid approximation ($\displaystyle \approx$) ACROSS a$\displaystyle ^{3}$.

I use =, in conjunction with "valid approximation," to emphasize that something does, or does not, exist across the border.

EDIT:
The analogy is imperfect. Imagine the w plane with a hole in it. If F(z) is near border, the existence of F'(z) doesn't guarantee that F(z) exists beyond the border. If F(z) =zf(z), where f(z) is a degree lower than F(z) and therefore by induction maps to entire complex plane, F(z) exists across the border.

Last edited by zylo; November 8th, 2017 at 10:45 AM. November 8th, 2017, 12:09 PM   #45
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Quote:
 Originally Posted by zylo I use =, in conjunction with "valid approximation," to emphasize that something does, or does not, exist across the border.
Now you are redefining the equal sign. That's never going to convince anyone of anything.

But first, what do you mean by "border?" Do you mean topological boundary? The boundary of a point in the plane is the point, and the boundary of its complement is also that point. Is that what you mean?

The other problem you have is the ambiguity of the phrase, "across the border." If we are in the real line and trying to prove the intermediate value theorem, we can make a continuity argument to say that if a function attains the value -1 and also 1, it must have a zero.

But in the plane, a function can go around a given point. I have no idea what "across the border means."

So three issues here:

1) Please use standard notation. $=$ is equality, $\approx$ is approximate equality.

2) What do you mean by border?

3) In the plane, what does across the border mean, given that you can always go around a given point without crossing it? November 9th, 2017, 11:14 AM   #46
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 Originally Posted by Maschke 2) What do you mean by border? Boundary of area of w-plane F(z) doesn't map to. 3) In the plane, what does across the border mean, given that you can always go around a given point without crossing it?
The argument can be simplified (corrected).

Let $\displaystyle F(z)=z^{n}+a_{n-1}z^{n-1}+..+a_{1}z$

Show F(z) maps to entire complex plane w. $\displaystyle \rightarrow$ FTA
F(z) and F'(z) exist for all z.

Assume map of F(z) has a hole in w.
Let F(z$\displaystyle _{0}$) be near border.
F(z) defined in any neighborhood of z$\displaystyle _{0}$.
F(z$\displaystyle _{0}$+z)$\displaystyle \approx$ F($\displaystyle z_{0}$)+F'(z$\displaystyle _{0}$)$\displaystyle \Delta$z
F(z) crosses the border.*
FTA

*F(z) crosses border when |F'(z$\displaystyle _{0}$)$\displaystyle \Delta$z| is greater than distance of F(z$\displaystyle _{0}$), in direction
$\displaystyle \Delta$z, from border. November 9th, 2017, 07:35 PM   #47
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Quote:
 Originally Posted by zylo *F(z) crosses border when |F'(z$\displaystyle _{0}$)$\displaystyle \Delta$z| is greater than distance of F(z$\displaystyle _{0}$), in direction $\displaystyle \Delta$z, from border.
I'm percolating on this. I'll spend some time trying to unpack it and I'll come back with questions. That could take a while, if ever.

Meanwhile the sense I get is that you're hoping that if $f$ hits two points on opposite sides of the hole, it must hit the hole. Am I understanding your intent?

This is true in the reals but not necessarily in the complex numbers. I wanted to just toss out an example. Take the function $f(z) = e^z$. We know it's defined and complex differentiable on the entire complex plane; and that it hits everything except $0$.

Now $f(0) = 1$ and $f(i \pi) = -1$. But $f$ gets from $1$ to $-1$ by going counterclockwise around the unit circle as $z$ goes up the imaginary axis. It never gets any closer to $0$ than it was at the start.

I don't know if this is where your thinking is, that some sort of straight line idea might work, but in the complex numbers it doesn't. In the plane the image of a function could circle around the bad point and never get near it. I don't think you can constrain the image the way you want to.

I'm wondering if I'm understanding your general idea correctly, even if I haven't unpacked your symbology yet.

FWIW I found this really cool Geogebra page. https://www.geogebra.org/m/Ac43DNNP

You can move a point around on the $z$-plane and see where it goes on the $w$ plane. It really snaps a lot of the theory into focus when you can see it. I wish they'd had stuff like this when I studied math.

Also FWIW you can mark up math on this site using a single dollar sign at each end instead of the math and \math tags. If you used dollar signs you'd save some typing and it would be easier for me to quote your equations when I reply. Just a thought, ignore it if you prefer your style of markup.

Last edited by Maschke; November 9th, 2017 at 07:54 PM. November 10th, 2017, 08:25 AM #48 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 For a point, there is an F(z$\displaystyle _{0}$) whose neighborhood contains the point. F'($\displaystyle z_{0}$)$\displaystyle \Delta$z contains all points in a neighborhood of F(z$\displaystyle _{0}$). Real comparisons are deceptive: y=cosx has a derivative for all x but does not map to all of y. No solution for y=2. w=cosz has a derivative for all z and maps to all of w. A solution for any w. November 10th, 2017, 10:38 AM   #49
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Quote:
 Originally Posted by zylo For a point, there is an F(z$\displaystyle _{0}$) whose neighborhood contains the point.
Can you give an example of $z_0$ for $F(z) = e^z$ for the point $w = 0$? November 13th, 2017, 09:15 AM #50 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 F'(z) exists for all z implies F(z) exists for all z but not that F(z) maps to all of w. See Range of Analytic Function Post #20. FTA not proven by this train of thought. Tags algebra, fundamental, proof, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post AspiringPhysicist Abstract Algebra 3 October 2nd, 2014 01:19 AM king.oslo Algebra 2 September 19th, 2013 11:48 PM ray Algebra 6 April 22nd, 2012 04:50 AM ThatPinkSock Abstract Algebra 1 November 11th, 2011 05:25 AM johnny Algebra 11 December 21st, 2007 10:51 PM

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