Update 24 Nov 2015: It is solved. Please refer to this arXiv paper.

For polynomials with real coefficients, I am trying to prove the following version of fundamental theorem of algebra, which avoids using complex numbers in the proof. Existence of complex roots will be a corollary of this theorem, if proven successfully.

Sorry for the long post. This is my own method and I could not make it shorter right now.

For every polynomial with real coefficients of order greater than 2, there exists a quadratic polynomial with real coefficients which factorizes it.

Why? Every real odd polynomial can be factored using only one variable, which produces a real root. Odd and even numbers interlace, but real even polynomials cannot always be factorized using one variable. Maybe two variables can factorize them always.

My question:
In the quotient formalism, if we divide a monic polynomial with positive coefficients with divisor $q(x)=x^{2}-ax-b$, the remainder is of the form $P(a,b)x+Q(a,b)$. I can show that $P(a,b)=0$ has one solution for large positive $b=b2$ and $n-1$ roots for large negative $b=-b1$ using Sturm chain. Equivalently, using Sturm's chain I can show that $Q(a,b)=0$ has two asymptotic solutions for $b=0$ on the lower half $a/b$ plane, as well as $n-2$ solutions for $Q(a,b)=0$ for the large negative $b=-b1$. I am certain that for large negative $b=-b1$, the solutions of $P(a,b)=0$ and $Q(a,b)=0$ interlace. Since the zero contours form connected curves, applying Jordan's curve theorem in the rectangle ABCD, A=[-a1 -b1] and C=[a2 b2] for will force at least once intersection of $P(a,b)=0$ and $Q(a,b)=0$ inside the rectangle, thus proving the theorem. **I have not being able to show the interlacing yet. Can someone please help me with relevant ideas or existing previous work regarding the interlacing?

enter image description here Please open the image in new tab for easier viewing

Procedure of my proof:

Part 1. Given any monic polynomial of order even order $n$, shifting it by $s$ gives the polynomial

$$f(x+s)=\sum_{k=0}^{n} \frac{d^{k} f(s) }{k! \ d x^{k}}x^{n-k}=\sum_{k=0}^{n} C^{n}_{k} \ (1+\epsilon(k,s)) \ s^{n-k} \ x^{n-k} \ ....(1)$$ $\lim_{s \to \infty} \epsilon(k,s)=0$
$C^{n}_{k}$ is the permutation formula.

Now, if any polynomial has a quadratic factor, any shift of origin still retains the factorization. So it suffices to prove the proposed theorem for transformed polynoials of the form $(1)$, which have positive and increasing coeffcients with decreasing power of $x$.

Part 2. We can derive the quotient by repeatedly replacing $x^{2}=ax+b$. For any power $m$, if we write the quotient formula as $x^{m}\equiv p_{m}x+q_{m}$ , we can use the recursion formula $p_{m}=ap_{m-1}+q_{m}$ and $q_{m}=b q_{m-1}$.

Using this method to write the quotient formula for $f(x+s)$, we get $$P(a,b)=\sum_{k=1}^{n-1} l_{k} \ g(k,b) \ a^{n-k}$$ $$Q(a,b)=\sum_{k=2}^{n} p_{k} \ h(k,b) \ a^{n-k}$$

$g(k,b)$ is monic polynomial with positive coefficients of $b$ with highest power $ \lfloor {\frac{k}{2}} \rfloor $
$h(k,b)$ is monic polynomial with positive coefficients of $b$ with highest power $ 1+\lfloor {\frac{k}{2}} \rfloor $

$l_{k}$ and $p_{k}$ are positive increasing sequences of $k$ (Which can be ensured by using large $s$ for $f(x+s)$). For sufficiently large positive $b$, applying Sturm chain method directly shows that there is only one root for $P(a,b)=0$.

$Q(a,b=0)=Constant \ne 0$. So $Q(a,b=0)=0$ does not have any solution for $a$. However, by using the transformation $b=ma$ and using the recusrion formula, we can show that $$\lim_{m \to 0} [P(a,m)-n \ (1+\epsilon(n-1,s)) s^{n-1}] = \lim_{m \to 0} [Q(a,m)- \frac{ \ (1+\epsilon(n,s)) \ s^{n}}{m}]$$ With some work, this directly shows that $Q(a,b)=0$ has two asymptotic solutions along $+a$ and $-a$ in the lower half $a/b$ plane.

Using Sturm chain, it can be proved that large negative $b$, $P(a,b)=0$ has $n-1$ roots and $Q(a,b)=0$ has $n-2$ roots. This happens because the coefficients of the polynomial of $a$ increase in magnitude with decreasing power and changes sign after every second power I have a feeling this is the basic structure behind the Fundamental Theorem of Algebra, since sign change over four consecutive powers is what complex numbers are cpable of producing

The vertical sides of the rectangle ABCD can be directly estimated by taking a lower and upper limit of the roots of $P(a,-b1<b<b2)=0$, in presence of interlacing.

Part 3. What is left We need to show that for large negative $b$, the $n-1$ roots of $P(a,b)=0$ and $n-2$ roots of $Q(a,b)=0$ interlace. I am certain it can be used by comparing $l_{k}$ and $p_{k}$ and using the recursion formula. If there is an easy way to prove this from the recursion formula, it will be great.

Update: Here is the outline of the proof.
$$f(x)=\sum_{k=0}^{n} c_{k} \ x^{n-k} $$ $$g_{0}=P(f(x))=P(\sum_{k=0}^{n} c_{k} \ x^{n-k} )= \sum_{k=0}^{n} c_{k} \ P(x^{n-k}) ....(1)$$ $$ \frac{g_{1}}{b} =\frac{Q(f(x))}{b}=P(\sum_{k=0}^{n-1} c_{k} \ x^{n-k-1} )= \sum_{k=0}^{n-1} c_{k} \ P(x^{n-k-1})+ \frac{c_{0}}{b}....(1)$$

Using $g_{0}$ and $g_{1}$ as the first two entries of of Sturm chain, we get $$ \frac{-1^{i-1} g_{i}}{b^{i}} \approx \sum_{k=0}^{n-i} c_{k} \ P(x^{n-k-i})....(1)$$ for large $b$.

Using the recursion formula between $g_{i-1},g_{i}$ and $g_{i+1}$, the intermediate value theorem and the growth properties of real polynomials, if the roots of $g_{i}$ and $g_{i+1}$ interlace, it can be shown that $g_{i-1}$ and $g_{i}$ have interlacing roots for large negative $b$. The quadratic and linear terms of $g$ have interlacing roots for large negative $b$. This implies that $P(a,b)$ and $Q(a,b)$ have interlacing roots for large negative $b$.

The quadratic term of $g$ does not have a root for large positive $b$.Using the same technique as above, it can be shown that $P(a,b)$ has one root and $Q(a,b)$ has no root for large positive $b$.

I will follow up with a full write up over the weekend.

  • 1
    $\begingroup$ Hi! Er.. well... I think your solution should be an answer.. And not part of the question. $\endgroup$ – Physicist137 Aug 9 '16 at 2:09
  • $\begingroup$ Hi! Thanks for the suggestion. This gives me a chance to provide updates, which I completely forgot to do. The procedure I left in my question was not complete, so I left it there for possible further discussions. The actual answer follows from the procedure but is much more complicated. The complete proof is under review for publication now and will upload a link when the publication comes out. $\endgroup$ – sobasu Aug 9 '16 at 9:01
  • $\begingroup$ Here is a link to the complete proof uploaded on arxiv. I am struggling to improve the writing and hoping to publish it in a journal soon. $\endgroup$ – sobasu Jun 27 '17 at 9:50
  • 1
    $\begingroup$ See also the proof at math.stackexchange.com/a/2244195/124095. $\endgroup$ – mweiss Feb 26 '18 at 5:17
  • $\begingroup$ Thanks @mweiss for sharing. I cannot evaluate the shared proof, as it flew way over my head. I am only a novice in math :) $\endgroup$ – sobasu Feb 27 '18 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.