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According to Abel-Ruffini Theorem and Galois Theory, you cannot solve the general quintic equation by radicals - i.e., it is impossible to express its solutions by a finite number of additions, subtractions, multiplications and taking roots of any terms including the general quintic equation's coefficients.

But in some special cases of the quintic equation (I mean special cases also UNSOLVABLE by radicals) IS IT POSSIBLE to demonstrate the equation's unsolvability by a finite number of basic arithmetic operations between the terms including its coefficients by sketching some small part of the appropriate INFINITE formula? LBNL, is it possible to sketch some small part of the root-taking tower of infinite height that would be part of some infinite formula for the solutions? In other words, is it possible to sketch some small part of such infinite formula with root-taking towers of infinite height in a similar way, as we can sketch for instance some small parts of many infinite convergent series in the calculus?

(For such purpose, I would fancy sketching parts of infinite formulas for the simplest quintic equations like e.g. x^5 - x + 1 = 0. But of course, I may be wrong, and some other quintic equations unsolvable by radicals can have more easily sketchable parts of their respective infinite, coefficients including, formulas.)

Thank you!

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For equation $t+x+x^5=0$ let $Q^5=-t\implies Q=\bigg\{-t^{1/5}, (-1)^{1/5} t^{1/5}, -(-1)^{2/5} t^{1/5}, (-1)^{3/5} t^{1/5}, -(-1)^{4/5} t^{1/5}\bigg\}$.

Then

$x=\dfrac{Q^5 (1 + S)}{1 + S + 5^n Q^{4 n} (-3 + 5 Q^4)}$

where

$\displaystyle S=\sum^{n\equiv 1\pmod 4}_{i=1}{k(i) 5^{i + 1}Q^{4 i}}$

and

$k(i)=\begin{cases}1,&\text{if $i\equiv 0,1\pmod 4$}\\2,&\text{if $i\equiv 2,3\pmod 4$}\end{cases}$

Verifing in W.Mathematica:

t = RandomInteger[{-1000000, 1000000}];
P = t + x + x^5;
Print["Equation: ", P, "=0"];
Print["Solution by CAS:"];
Print[NSolve[P == 0, x, 16]];
Print["Solution by formula:"];
Q = {-t^(1/5), (-1)^(1/5) t^(1/5), -(-1)^(2/5) t^(1/5), (-1)^(3/5) t^(1/5), -(-1)^(4/5) t^(1/5)};
k[i_] := If[Mod[i, 4] == 2 || Mod[i, 4] == 3, 2, 1];
n = 4*RandomInteger[{1, 1000}] + 1;
S = Sum[k[i] 5^(i + 1) Q^(4 i), {i, 1, n}];
Print["x = ", N[(Q^5 (1 + S))/(1 + S + 5^n Q^(4 n) (-3 + 5 Q^4)), 16] // Sort];
Print["n = ", n]

The less $abs(t)$, the worse the convergence.

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