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October 5th, 2017, 04:38 PM   #31
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 SDK, I apply FTA by induction to the lower order polynomial f(z), regardless of degree. I map on to circles and lines, not to circles and lines.
I don't see where you have proved this. If you mean what I think you mean then it isn't true.

Quote:
 1) Given 3rd degree complex polynomial $\displaystyle z(z^{2}+pz+q)=zf(z)=c$ By induction applied to the second degree polynomial, $\displaystyle f(z)$ maps z to entire complex plane.
This is not true because your argument in (2) is flawed.

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 2) $\displaystyle zf(z)$ maps continuously to entire complex plane (from 0 to infinity).
Prove this. If you think its obvious consider the following claim:
$f(x) = x$ maps continuously to the entire real line. Therefore $xf(x) = x^2$ must also map to the entire real line.

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 3) $\displaystyle zf(z)=c$ has a solution (root) z=d for any c.
You have stated this about 14 ways and proved it in 0 ways.

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 4) Same argument for any degree.
Same problem for any degree.

Quote:
 I have googled FTA. The complexity and abstraction are not my cup of tea. I prefer a simple, transparent, comprehensible, easy to remember, proof, where possible. If it bothers you, be consoled by fact that it is virtually unnoticed and writ in water, and the others will be around a long time. Anyhow, thanks for your interest.
The complexity is related to the fact that the FTA is far less obvious than you seem to think. You think its obvious because you are familiar with this fact since grade school. Much like 6+6 = 12 is obvious until one learns how to do modular addition and then it isn't. I think you should try to more carefully write down your ideas and you will see it isn't so clear that FTA must be true.

 October 6th, 2017, 07:02 AM #32 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Post #18 "Topological" proof Post #26 "Mapping" proof ---------------------------------------------------------------------------------------- Post # 28: 1) Given 3rd degree complex polynomial $\displaystyle z(z^{2}+pz+q)=zf(z)=c$ 2) By induction applied to the second degree polynomial, $\displaystyle f(z)$ maps z to entire complex plane. 3) $\displaystyle zf(z)$ maps continuously to entire complex plane (from 0 to infinity). 4) $\displaystyle zf(z)=c$ has a solution (root) z=d for any c. 4) Same argument for any degree. --------------------------------------------------------------------------------- =============================================== Post # 29 Given any n degree complex polynomial written as: 1) $\displaystyle F_{n}(z)=-a_{0}$ 2) $\displaystyle F_{n}(z)$ is continuous for all $\displaystyle z$ and varies continuously from complex $\displaystyle 0$ to complex $\displaystyle \infty$ (eventually $\displaystyle z^{n}$ predominates). 3) It follows $\displaystyle F_{n}(z)=-a_{0}$ has a root for any $\displaystyle a_{0}$.[/QUOTE] =============================================== If $\displaystyle F_{n}(z)=u+iv$, The ray $\displaystyle \theta=constant$ maps to continuous curve in u,v plane which goes from 0 to infinity. Induction: If $\displaystyle P_{1}$ is true and $\displaystyle P_{n-1} \rightarrow P_{n}, P_{n}$ is true for all n. CORE ASSERTIONS: 28-3 and 29-2 Last edited by zylo; October 6th, 2017 at 07:41 AM. Reason: remove comment about 28-3
 October 6th, 2017, 08:28 AM #33 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics I agree that the statement in 28-3 is the crux of your proof. If that is your point then I misunderstood you as I thought you were claiming to have proved this assertion. Just to be clear, you haven't proved that claim anywhere and it appeared to me that you were claiming it to be obvious. My point was that this fact is highly nontrivial and far from obvious. The counterexample above shows it can fail on $\mathbb{R}$. However, if you want an example which is complex, consider $f(z) = \bar{z}$ which is clearly surjective. However, $zf(z) = \left|z \right|^2 \in \mathbb{R}$ is clearly not. If you want to prove that $zf(z)$ is a surjective map onto $\mathbb{C}$, then it becomes very important that $f$ is a polynomial should be used somewhere. However, this is nothing more than a restatement of the FTA and proving this claim is equivalent to proving FTA.
 October 6th, 2017, 02:06 PM #34 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Given the n-degree polynomial F(z)=F$\displaystyle _{n}$(z)=c 1) F(z) is continuous and F(0)=0 and $\displaystyle \lim_{z\rightarrow \infty}$F(z)=$\displaystyle \infty$ 2) Assume there is no z st F(z)=c where c is a point in w plane, and F(z) exists in a neighborhood of c. Then z exists st F(z)=c by continuity. 3) If there is a hole in w plane, let c be a border point. Then F(z) exists in a neighborhood of c by continuity. Contradiction. 4) FTA
October 6th, 2017, 02:41 PM   #35
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 Originally Posted by zylo Given the n-degree polynomial F(z)=F$\displaystyle _{n}$(z)=c
Are you saying the polynomial is constant here? I'm confused.

Quote:
 1) F(z) is continuous and F(0)=0 and $\displaystyle \lim_{z\rightarrow \infty}$F(z)=$\displaystyle \infty$
Is this an assumption? I'm fine with it either way, since it's true for $zf(z)$ whenever $f$ is a polynomial.

Quote:
 2) Assume there is no z st F(z)=c where c is a point in w plane, and F(z) exists in a neighborhood of c. Then z exists st F(z)=c by continuity. 3) If there is a hole in w plane, let c be a border point. Then F(z) exists in a neighborhood of c by continuity. Contradiction.
There is no justification for your assumption that $F(z)$ maps onto a neighborhood of $c$. The assumption that it does is completely equivalent to FTA. Once again, this isn't a proof or even an outline of a proof. It is literally a restatement of FTA in different language. Also, what is a w plane?

Last edited by skipjack; October 7th, 2017 at 02:11 AM.

 October 7th, 2017, 03:39 AM #36 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 A complex polynomial always has a solution because F(z)=$\displaystyle F_{n}(z)=z^{n}+a_{n-1}z^{n-1}+.....+a_{1}z$ is continuous everywhere and maps z from 0 to $\displaystyle \infty$. Proof: If there were any "holes" in the map, F(z) would be discontinuous there. For example, The ray $\displaystyle \theta$ = constant maps to a curve in the w plane (F(z)) which goes from 0 to $\displaystyle \infty$. If there were a break in the curve, F(z) would be discontinuous there.
October 9th, 2017, 09:13 PM   #37
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 Originally Posted by zylo A complex polynomial always has a solution because F(z)=$\displaystyle F_{n}(z)=z^{n}+a_{n-1}z^{n-1}+.....+a_{1}z$ is continuous everywhere and maps z from 0 to $\displaystyle \infty$. Proof: If there were any "holes" in the map, F(z) would be discontinuous there. For example, The ray $\displaystyle \theta$ = constant maps to a curve in the w plane (F(z)) which goes from 0 to $\displaystyle \infty$. If there were a break in the curve, F(z) would be discontinuous there.
I refuse to believe that you know this many concepts and still think what you have written is even remotely true or even feasible. I suspect this is just trolling. If by some freak chance this isn't the case, I suggest you spend more time reading. Anyways enjoy your "proof".

October 10th, 2017, 08:01 AM   #38
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Quote:
 Originally Posted by zylo A complex polynomial always has a solution because F(z)=$\displaystyle F_{n}(z)=z^{n}+a_{n-1}z^{n-1}+.....+a_{1}z$ is continuous everywhere and maps z from 0 to $\displaystyle \infty$. Proof: If there were any "holes" in the map, F(z) would be discontinuous there. For example, The ray $\displaystyle \theta$ = constant maps to a curve in the w plane (F(z)) which goes from 0 to $\displaystyle \infty$. If there were a break in the curve, F(z) would be discontinuous there.
$\displaystyle z^{n}+a_{n-1}z^{n-1}+...+a_{1}z$ takes on all values in the complex plane.
$\displaystyle \therefore z^{n}+a_{n-1}z^{n-1}+...+a_{1}z=-a_{0}$ always has a solution.
Just in case anyone else missed it.

All the difficulties and complexities of the other solutions are due to the fact that they start with $\displaystyle z^{n}+a_{n-1}z^{n-1}+...+a_{1}z+a_{0}=0$ and try to show it has a solution.
My unique approach makes it virtually trivial. All you have to see is that z^{n}+a_{n-1}z^{n-1}+...+a_{1}z maps continuously to the entire complex plane (from 0 to $\displaystyle \infty$).

October 10th, 2017, 12:50 PM   #39
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 Originally Posted by zylo $\displaystyle z^{n}+a_{n-1}z^{n-1}+...+a_{1}z$ takes on all values in the complex plane.[/MATH]).
All your most recent post does is show that this statement implies FTA. But that doesn't really help much - it is essentially a restatement of FTA! You still need to prove this statement; I've looked through your posts and all of your attempts are badly flawed.

Quote:
 Originally Posted by zylo All the difficulties and complexities of the other solutions are due to the fact that they start with $\displaystyle z^{n}+a_{n-1}z^{n-1}+...+a_{1}z+a_{0}=0$
Are you sure? The problem is to show that there is a $z$ such that $z^{n}+a_{n-1}z^{n-1}+...+a_{1}z+a_{0}=0$ - how could a proof start by assuming such a $z$ existed?

In fact, the two proofs I can currently remember (one uses Liouville's theorem, the other uses the fundamental group of the circle) start with the opposite assumption, that no such $z$ exists, to get a contradiction. So the difficulties/complexities of those proofs certainly couldn't be down to the reason you've given.

Last edited by cjem; October 10th, 2017 at 12:52 PM.

 October 18th, 2017, 09:52 AM #40 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Fundamental Theorem of Algebra: Every complex polynomial has a root. $\displaystyle P_{n}(z)=z^{n}+a_{n-1}z^{n-1}+..+a_{1}z+a_{0}=z(z^{n-1}+a_{n-1}z^{n-2}+..+a_{2}z+a_{1})+a_{0}$ $\displaystyle P_{n}(z)=zP_{n-1}(z)+a_{0}$ Example: $\displaystyle P_{3}(z)=z^{3}+a_{2}z^{2}+a_{1}z+a_{0}=z(z^{2}+a_{ 2}z+a_{1})+a_{0}=zP_{2}(z)+a_{0}.$ Proof by induction: $\displaystyle P_{2}=0$ has a solution. $\displaystyle P_{n-1}(z)=0$ has a solution $\displaystyle \rightarrow P_{n}(z)=0$ has a solution. $\displaystyle P_{n}(z)=0$ has a solution if $\displaystyle zP_{n-1}(z)$ maps all z to entire complex plane. $\displaystyle P_{n-1}(z)$ maps to entire complex plane (given). It is shown* that $\displaystyle zP_{n-1}(z)$ maps to entire complex plane if $\displaystyle P_{n-1}(z)$ does. Note complex polynomials have a derivative for all z. -------------------------------------------------------------------------------- * Range of Analytic Function From Post #8 Assume F(z) has a derivative for all z and maps to w plane with a hole in it (area where F(z) doesn't map to). The hole is closed because F'(z) is undefined on it's border. In this case, F'(z) is defined and F'(z)$\displaystyle \Delta$z exists in F(z) for $\displaystyle \Delta$z sufficiently small no matter how close you get to the border. For example, F(z) could map to the inside of a circle for all z and F'(z) exist everywhere inside the circle. The only way out would be to show that at a point near the border F'(z)$\displaystyle \Delta$z is greater than the distance to the border. If F(z) = zf(z) where f(z) exists, has a derivative everywhere, and maps to entire complex plane, then so does F(z). 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) where f(z) and f'(z) exist and $\displaystyle \Delta$z is associated with f'(z) so not restricted by the border. It follows F(z) can cross the border, a contradiction, so the island doesn't exist and F(z) maps to entire complex plane. Example: Proof of FTA by induction where f(z) and zf(z) are polynomials. Comment: In other words, F(z) can't cross border (no contradiction) but zf(z) can (contradiction).

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