Limit of Newton's Method on polynomial $Ax^4 + Bx^3+ Cx^2 + Dx + E$?

Question

So if you took the function $f(x) = Ax^4 + Bx^3+ Cx^2 + Dx + E$ and did Newton's Method repeatedly, you would get a sequence that converges to at most $4$ roots. I was wondering what would happen if you took the limit of that sequence, would you get the formula for the roots of a quartic equation? And what if you do it for a polynomial of nth degree. I know it's been proven that there can't be an explicit formula for polynomials of degree higher than $4$ so I'm not sure what would happen if you tried to find it through Newton's Method. Would it not diverge because there are multiple roots and the initial term (the guess) determines which root you find? But shouldn't that just get you a formula for the root in terms of the initial guess? I tried looking at the first few terms for Newton's Method but they became too much to work with so I was wondering if there was a program that could do Newton's Method for functions with unknown coefficients. I think I could do it with Python but it would be a lot of work to do it from scratch so if there's a python library that can support operations with variables, I would then just have to write a program for Newton's Method.

Answer 1

If the initial point $x_0$ is close enough to a root, the iterations will indeed converge to that root. However, the formulas for successive iterations (as expressions in the coefficients $A,B,C,D,E$ and $x_0$) just get more and more complicated, and there is no clear way to get from them to a formula for the root. For a very simple example, take $f(x) = x^4 - A$; the Newton iteration is the map $$N(x) = \frac{A+3x^4}{4 x^3}$$ The next couple of iterations give you $$ \eqalign{N(N(x)) &= \frac{243 x^{16}+580 A x^{12}+162 A^2 x^{8}+36 A^3 x^{4}+3 A^4}{16 x^{3} \left(3 x^{4}+A\right)^{3}}\cr N(N(N(x))) &= \frac{10460353203 x^{64}+134696910096 A x^{60}+524761849512 A^{2} x^{56}+1030291232688 A^{3} x^{52}+1172817593940 A^{4} x^{48}+847334588688 A^{5} x^{44}+439534043928 A^{6} x^{40}+171389329200 A^{7} x^{36}+51770245554 A^{8} x^{32}+12287794608 A^{9} x^{28}+2311181208 A^{10} x^{24}+346123152 A^{11} x^{20}+41168340 A^{12} x^{16}+3822640 A^{13} x^{12}+262440 A^{14} x^{8}+11664 A^{15} x^{4}+243 A^{16}}{64 x^{3} \left(243 x^{16}+580 A x^{12}+162 A^{2} x^{8}+36 A^{3} x^{4}+3 A^{4}\right)^{3} \left(3 x^{4}+A \right)^{3}} }$$ and you really don't want to see $N(N(N(N(x))))$.

Answer 2

This has been studied:

Doyle, Peter; McMullen, Curt, Solving the quintic by iteration, Acta Math. 163, No. 3-4, 151-180 (1989). ZBL0705.65036.