October 18th, 2017, 11:25 AM  #41  
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  Quote:
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
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  
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  Quote:
 
November 8th, 2017, 10:29 AM  #44  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91  Quote:
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  
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  Quote:
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  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91  Quote:
Let $\displaystyle F(z)=z^{n}+a_{n1}z^{n1}+..+a_{1}z$ 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.* Contradiction 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  
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  Quote:
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
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 
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  
November 13th, 2017, 09:15 AM  #50 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
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. 

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