October 5th, 2017, 04:38 PM  #31  
Senior Member Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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$f(x) = x$ maps continuously to the entire real line. Therefore $xf(x) = x^2$ must also map to the entire real line. Quote:
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October 6th, 2017, 07:02 AM  #32 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
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 $\displaystyle P_{n1} \rightarrow P_{n}, P_{n}$ is true for all n. CORE ASSERTIONS: 283 and 292 Last edited by zylo; October 6th, 2017 at 07:41 AM. Reason: remove comment about 283 
October 6th, 2017, 08:28 AM  #33 
Senior Member Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics 
I agree that the statement in 283 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, $zf(z) = \leftz \right^2 \in \mathbb{R}$ 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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
Given the ndegree 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  
Senior Member Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics  Are you saying the polynomial is constant here? I'm confused. Quote:
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Last edited by skipjack; October 7th, 2017 at 02:11 AM.  
October 7th, 2017, 03:39 AM  #36 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
A complex polynomial always has a solution because F(z)=$\displaystyle F_{n}(z)=z^{n}+a_{n1}z^{n1}+.....+a_{1}z$ 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  
Senior Member Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
October 10th, 2017, 08:01 AM  #38  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
$\displaystyle \therefore z^{n}+a_{n1}z^{n1}+...+a_{1}z=a_{0}$ 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 $\displaystyle z^{n}+a_{n1}z^{n1}+...+a_{1}z+a_{0}=0$ 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_{n1}z^{n1}+...+a_{1}z maps continuously to the entire complex plane (from 0 to $\displaystyle \infty$).  
October 10th, 2017, 12:50 PM  #39  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 198 Thanks: 59 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
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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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
Fundamental Theorem of Algebra: Every complex polynomial has a root. $\displaystyle P_{n}(z)=z^{n}+a_{n1}z^{n1}+..+a_{1}z+a_{0}=z(z^{n1}+a_{n1}z^{n2}+..+a_{2}z+a_{1})+a_{0}$ $\displaystyle P_{n}(z)=zP_{n1}(z)+a_{0}$ 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. $\displaystyle P_{n1}(z)=0$ has a solution $\displaystyle \rightarrow P_{n}(z)=0$ has a solution. $\displaystyle P_{n}(z)=0$ has a solution if $\displaystyle zP_{n1}(z)$ maps all z to entire complex plane. $\displaystyle P_{n1}(z)$ maps to entire complex plane (given). It is shown* that $\displaystyle zP_{n1}(z)$ maps to entire complex plane if $\displaystyle P_{n1}(z)$ 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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