My Math Forum Fundamental Theorem of Algebra Proof

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 September 16th, 2017, 03:29 AM #11 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 What you have shown is that if you can find a single root of a polynomial equation then you can find all of them because you have reduced the problem to a polynomial equation of lower degree. It is finding that first root that is the hard part! Thanks from JeffM1
September 18th, 2017, 11:46 AM   #12
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Just want to clean up my end of this thread.

Quote:
 Originally Posted by zylo 1) $\displaystyle (5-b)-(3-a)a=0$ 2) $\displaystyle (7-c)-(3-a)b=0$ 3) $\displaystyle 9- (3-a)c=0$ from 3), $\displaystyle c=\frac{9}{3-a}$. Sub into 2) to get $\displaystyle b(3-a)^{2}+7(3-a)+3=0$, which gives $\displaystyle (3-a)=\frac{k}{b}, b=\frac{k}{3-a}$. Sub into 1) to get a cubic in $\displaystyle a$ which is solvable.
$\displaystyle b(3-a)^{2}+7(3-a)+3=0$ gives $\displaystyle (3-a)=\frac{-7\pm\sqrt{49-12b}}{2b}$which takes you nowhere.

Dividing $\displaystyle P_{n}$ by a lower order parametric polynomial and setting coefficients in the remainder to 0 to find parameters didn't work.

For example:
$\displaystyle P_{n}(x)$ by $\displaystyle x+a$
$\displaystyle P_{n}(x)$ by $\displaystyle x^{2}+ax+b$
$\displaystyle P_{n}(x)$ by $\displaystyle x^{3}+ax^{2}+bx+c$

It didn't work because I couldn't reduce the polynomial factor problem to lower order polynomials solvable by induction.

 September 19th, 2017, 07:10 AM #13 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 You made two errors when substituting into the second equation.
 September 28th, 2017, 11:58 AM #14 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Fundamental Theorem of Algebra- Proof by Induction. Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form $\displaystyle f(z)=z(z^{2}+pz+q)=C$. By induction, $\displaystyle f(z)=z(z-a)(z-b)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$. $\displaystyle \theta _{a}$ and $\displaystyle \theta _{b}$ constant. $\displaystyle f(z)_{\rho=k}$ describes a closed curve around origin which increases in size incrementally (continuously) with $\displaystyle \rho =|z|$. $\displaystyle f(z)$ maps $\displaystyle z$ into complex plane and hence there is a $\displaystyle z=d$ for any $\displaystyle C$ and hence $\displaystyle z-d$ is a factor. Argument identical form for $\displaystyle P_{n}(z)$.
September 30th, 2017, 05:42 AM   #15
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Quote:
 Originally Posted by zylo Fundamental Theorem of Algebra- Proof by Induction. Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form $\displaystyle f(z)=z(z^{2}+pz+q)=C$. By induction, $\displaystyle f(z)=z(z-a)(z-b)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$. $\displaystyle \theta _{a}$ and $\displaystyle \theta _{b}$ constant. $\displaystyle f(z)_{\rho=k}$ describes a closed curve around origin which increases in size incrementally (continuously) with $\displaystyle \rho =|z|$. $\displaystyle f(z)$ maps $\displaystyle z$ into complex plane and hence there is a $\displaystyle z=d$ for any $\displaystyle C$ and hence $\displaystyle z-d$ is a factor. Argument identical form for $\displaystyle P_{n}(z)$.
Actually, as z describes a circle, $\displaystyle \theta _{a}$ and $\displaystyle \theta _{b}$ are not constant.

Argument is still the same.

 September 30th, 2017, 07:16 AM #16 Senior Member   Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics I don't know what $\theta_a, \theta_b$ are supposed to be. The RHS is constant because it is simply the complex polar form of $-c$ which is constant. However, this is not a proof by induction. First of all, there is no induction anywhere. It was pointed out earlier why induction fails to prove the FTA. Making the polynomials complex does not fix the problem. Secondly, this isn't a proof because you have a number of incorrect claims. To point out the obvious ones assume $f$ is your polynomial. You claim that $f(z)$ is a closed curve when $|z| = k$. This is obviously false e.g. $f(z) = z^2$ You claim that $|f(z)|$ is increasing with $z = \rho$. I'm not sure what this means. If you mean that $|f(z)|$ is increasing as $\theta$ increases where $z = \rho^{i\theta}$ then this is obviously false e.g. $f(z) = z^2 + z, \theta = 0, \pi$. Alternatively, you may mean $|f(z)|$ increases as $\rho$ increases which I think is a misunderstanding of the maximum principle which states that $|f(z)|$ must take its maximum value on the boundary of $B_{\rho}$ for any $\rho$. However, it doesn't require $|f(z)|$ to take this maximum value on the entire boundary. There are proofs of the FTA using complex analysis but they don't come from naive algebraic factoring or induction.
September 30th, 2017, 08:09 AM   #17
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Quote:
 Originally Posted by zylo Fundamental Theorem of Algebra- Proof by Induction. Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form 1) $\displaystyle f(z)=z(z^{2}+pz+q)=C$. 2) By induction, $\displaystyle f(z)=z(z-a)(z-b)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$. $\displaystyle f(z)_{\rho=k}$ describes a closed curve around origin which increases in size incrementally (continuously) with $\displaystyle \rho =|z|$. $\displaystyle f(z)$ maps $\displaystyle z$ into complex plane and hence there is a $\displaystyle z=d$ for any $\displaystyle C$ and hence $\displaystyle z-d$ is a factor. Argument identical form for $\displaystyle P_{n}(z)$.
Quote:
 Originally Posted by SDK I don't know what $\theta_a, \theta_b$ are supposed to be. The RHS is constant because it is simply the complex polar form of $-c$ which is constant. However, this is not a proof by induction. First of all, there is no induction anywhere. It was pointed out earlier why induction fails to prove the FTA. Making the polynomials complex does not fix the problem. Secondly, this isn't a proof because you have a number of incorrect claims. To point out the obvious ones assume $f$ is your polynomial. You claim that $f(z)$ is a closed curve when $|z| = k$. This is obviously false e.g. $f(z) = z^2$ You claim that $|f(z)|$ is increasing with $z = \rho$. I'm not sure what this means. If you mean that $|f(z)|$ is increasing as $\theta$ increases where $z = \rho^{i\theta}$ then this is obviously false e.g. $f(z) = z^2 + z, \theta = 0, \pi$. Alternatively, you may mean $|f(z)|$ increases as $\rho$ increases which I think is a misunderstanding of the maximum principle which states that $|f(z)|$ must take its maximum value on the boundary of $B_{\rho}$ for any $\rho$. However, it doesn't require $|f(z)|$ to take this maximum value on the entire boundary. There are proofs of the FTA using complex analysis but they don't come from naive algebraic factoring or induction.
Induction allows you to factor the lower order polynomial in 1) as in 2)

$\displaystyle z-a=\rho _{a}e^{i\theta _{a}}$

If $\displaystyle f(z)=z^{2}, f(\rho ,\theta+2\pi)=f(\rho ,\theta)$
You have to map all z to entire complex plane to prove there is a z for any given C.

The trace of f(z) as z goes around a circle is a closed curve. It changes continuously to $\displaystyle f(z) = z^{3}$ in the limit so f(z) maps every z to the entire complex plane f(z).

October 2nd, 2017, 05:25 AM   #18
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Quote:
 Originally Posted by zylo Edited 10/1/17
Fundamental Theorem of Algebra- Proof by Induction.

Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form
$\displaystyle f(z)=z(z^{2}+pz+q)=C$.
By induction, $\displaystyle f(z)=z(z-a)(z-b)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$.

$\displaystyle f(z)_{\rho=k}$ describes a closed curve, as z describes a circle, which distorts in shape incrementally (continuously) from infinitesimally small circle to infinitesimally large circle as $\displaystyle \rho$ increases. $\displaystyle f(z)$ maps $\displaystyle z$ into complex plane and hence there is a $\displaystyle z=d$ for any $\displaystyle C$ and hence $\displaystyle z-d$ is a factor.

Argument identical form for $\displaystyle P_{n}(z)$.

$\displaystyle z=\rho e^{i\theta}, \quad z-a=\rho _{a}e^{i\theta _{a}}, \quad z-b=\rho _{b}e^{i\theta _{b}}$
$\displaystyle a=|a|e^{i\alpha},\quad b=|b|e^{i\beta}$

Limiting curves:
If $\displaystyle \rho > > |a|,|b|, \quad \rho _{a}=\rho _{b}=\rho, \quad \theta_{a}=\theta_{a}=\theta$
$\displaystyle f(z) = \rho ^{3}e^{i3\theta}$, circle traversed 3 times.

If $\displaystyle \rho < < |a|,|b|, \quad z-a=|a|e^{i\alpha}, \quad z-b=|b|e^{i\beta},\quad \theta_{a}=\alpha, \quad \theta _{b}=\beta$
f(z)=$\displaystyle \rho |a||b|e^{i(\theta +\alpha +\beta)}$, circle

The notion that continuously varying a small curve into a larger curve surrounding it covers all the area between is admittedly not rigorous but reasonable, and gives a "proof" without analysis acceptable with just a knowledge of complex numbers.

Last edited by zylo; October 2nd, 2017 at 05:33 AM.

 October 3rd, 2017, 08:29 AM #19 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Previous post gave topological proof of FTA. An algebraic proof follows: Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form $\displaystyle f(z)=z(z^{2}+pz+q)=C$ Assuming induction for the second degree polynomial, $\displaystyle f(z)=z(z-a)(z-b)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$ $\displaystyle z=\rho e^{i\theta} \quad z-a=\rho _{a}e^{i\theta _{a}}, \quad z-b=\rho _{b}e^{i\theta _{b}}$ $\displaystyle a=|a|e^{i\alpha},\quad b=|b|e^{i\alpha}$ For any fixed angle $\displaystyle \theta$, $\displaystyle f(z)=\rho\rho_{a}\rho_{b}e^{i\Theta}, \Theta$ constant. $\displaystyle |f(z)|=\rho\rho_{a}\rho_{b}$ It follows that for any fixed value of theta, $\displaystyle |f(z)|$ varies from 0 to $\displaystyle \infty$ and hence $\displaystyle f(z)$ maps to entire complex plane and $\displaystyle f(z)=z(z^{2}+pz+q)$=C has a solution for any C. Argument exactly same for nth degree polynomial.
October 3rd, 2017, 08:59 AM   #20
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Quote:
 Originally Posted by zylo Previous post gave topological proof of FTA. An algebraic proof follows:
No that's not true. Your previous post claimed to not use analysis, but you are relying on the completeness of the complex plane.

Your "algebraic" proof is no proof at all, for the same reason.

This is the state of failed proofs of FTA from the 1700's. Nobody had the vocabulary to express completeness.

How do you know that as your circle expands it must necessarily pass through zero? That's the key point.

And by the way didn't I already outline this exact argument in post #4?

Last edited by Maschke; October 3rd, 2017 at 09:03 AM.

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