On Sunday, April 7, 2019 at 4:55:15 PM UTC+3, bassam king karzeddin wrote:
> On Sunday, March 31, 2019 at 11:06:32 PM UTC+3, Vinicius Claudino Ferraz wrote:
> > With my simplist O.D.E. method
> > you can solve p(x) = 0
> > for all polynomial p(x) = a_0 + a_1 x + ... + a_n x^n
> > for all n = deg p(x)
> > for all Real coefficients
> > for all Complex roots
> >
> > https://www.dropbox.com/s/vv6qgj16hgk1sch/Solving%20Any%20Quintic.pdf?dl=0
> >
> > \documentclass[12pt,a4paper]{article}
> > \usepackage{amsmath}
> > \usepackage{amssymb} %mathbb
> > \usepackage{graphicx}
> > \usepackage{hyperref}
> > \usepackage[top=1.0cm,bottom=1.3cm,left=1.0cm,right=1.0cm]{geometry}
> >
> > \begin{document}
> >
> > \begin{align}
> > &a,b,c,t\in \mathbb{R}\,;\,y:\mathbb{R}\to\mathbb{R}\,;\,r\in\mathbb{C} \\
> > a \cdot y''(t) + b\cdot y'(t) + c\cdot y(t) &= 0 \\
> > y &= c_1 ( \exp r_1 t) + c_2 ( \exp r_2 t ) \\
> > y' &= c_1 r_1 ( \exp r_1 t ) + c_2 r_2 (\exp r_2 t) \\
> > y'' &= c_1 r_1^2 (\exp r_1 t) + c_2 r_2^2 (\exp r_2 t) \\
> > ar^2 + br + c &= 0 \\
> > ax^2 + bx + c &= 0 \\
> > y(t) &= x^2 \\
> > y' &= 2x \\
> > y'' &= 2 \\
> > \cfrac{c}{2} \cdot y'' + \cfrac{b}{2}\cdot y' + a\cdot y &= 0 \\
> > a'\cdot y'' + b'\cdot y' + c'\cdot y &= 0 \\
> > z_1 &= y \\
> > z_2 &= y' \\
> > z_2' + b/c \cdot z_2 + 2a/c\cdot z_1 &= 0 \\
> > Z' &= \left(\begin{matrix}0 & 1 \\ -c'/a' & -b'/a' \end{matrix}\right) \cdot Z \\
> > Z &= \left[ \exp \left(\begin{matrix}0 & 1 \\ -c'/a' & -b'/a' \end{matrix}\right)t \right] \cdot C \\
> > y &= C_1 \cdot f_1 + C_2 \cdot f_2 \\
> > f_1 &= \exp r_1 t \\
> > \ln f_1 &= r_1\cdot t \\
> > r_1 &= \cfrac{1}{t}\cdot \ln f_1 \in \mathbb{C} \\
> > r_2 &= \cfrac{1}{t}\cdot \ln f_2 \\
> > \text{Better than } r &= \cfrac{-b' \pm \sqrt{b'^2 - 4a'c'}}{2a'} \\
> > c r^2 + br + 2 a &= 0 \text{ is solved.}
> > \end{align}
> >
> > \begin{align}
> > \text{We're ready to solve }a'x^5 + b' x^4 + c' x^3 + d' x^2 + e' x + f' &= 0 \\
> > y(t) &= x^5 \\
> > y' &= 5x^4 \\
> > y'' &= 20x^3 \\
> > y''' &= 60x^2 \\
> > y^{(4)} &= 120x \\
> > y^{(5)} &= 120 \\
> > \text{Drop the line instead of line line}&. \\
> > f/120 \cdot y^{(5)} + e/120\cdot y^{(4)}+ d/60\cdot y''' + c/20\cdot y'' + b/5\cdot y' + ay &= 0 \\
> > Z' = \left(\begin{matrix}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ -f'/a' & -e'/a' & -d'/a' & -c'/a' & -b'/a' \end{matrix}\right) &\cdot Z \\
> > Z &= [\exp Mt] \cdot C \\
> > y = C_1 \cdot f_1 + C_2 \cdot f_2 + C_3 \cdot f_3 + C_4 \cdot f_4 &+ C_5 \cdot f_5 \\
> > f_i &= \exp r_i t \\
> > r_i &= 1/t \cdot \ln f_i \in \mathbb{C} \\
> > \text{Better than Cardano's formula for cubics and quartics}&. \\
> > \text{Better than Newton's method who does not exhibit the exact complex number}&. \\
> > \text{Better than no existing formula in radicals}&.
> > \end{align}
> >
> > \section{OUT OF CHARITY THERE IS NO SALVATION AT ALL.}
> >
> > But it's not only that. Roots have multiplicity.
> >
> > \begin{align}
> > (x - r_1)^5 &= 0 \\
> > (x - r_1)^4 (x - r_2) &= 0 \\
> > (x - r_1)^3 (x - r_2)^2 &= 0 \\
> > (x - r_1)^3 (x - r_2)(x - r_3) &= 0 \\
> > (x - r_1)^2 (x - r_2)^2 (x - r_3) &= 0 \\
> > (x - r_1)^2 (x - r_2) (x - r_3)(x - r_4) &= 0 \\
> > \text{Suppose }r(t) &= t^{-1} \cdot \ln f \\
> > r'(t) &= -t^{-2} \cdot \ln f + t^{-1}\cdot 1/f \cdot f' = 0 \\
> > - f \ln f + t \cdot f' &= 0 \\
> > r\text{ is a root if and only if Id } &= \cfrac{f \ln f}{f'}
> > \end{align}
> >
> > \begin{align}
> > Z &= \left[ \exp\, t M \right] \cdot C \\
> > Z &= \left[ \exp\, t \left(\begin{matrix}0 & 1 \\ -c'/a' & -b'/a' \end{matrix}\right) \right] \cdot C \\
> > y &= C_1 \cdot f_1 + C_2 \cdot f_2 \\
> > y' &= C_1 \cdot f_1' + C_2 \cdot f_2' \\
> > \left(\begin{matrix}f_1 & f_2 \\ f_1' & f_2' \end{matrix}\right) &= Id + tM + \cfrac{t^2}{2!}\cdot M^2 + \cfrac{t^3}{3!}\cdot M^3 + \cfrac{t^4}{4!}\cdot M^4 + \cdots \\
> > \text{Let } &M_{ij}^n \text{ be the }(i,j)\text{-cell of the result of } M^n \\
> > f_1 &= 1 + tM_{11} + \cfrac{t^2}{2!}\cdot M^2_{11} + \cfrac{t^3}{3!}\cdot M^3_{11} + \cfrac{t^4}{4!}\cdot M^4_{11} + \cdots \\
> > f_1' &= 0 + tM_{21} + \cfrac{t^2}{2!}\cdot M^2_{21} + \cfrac{t^3}{3!}\cdot M^3_{21} + \cfrac{t^4}{4!}\cdot M^4_{21} + \cdots \\
> > f_2 &= 0 + tM_{12} + \cfrac{t^2}{2!}\cdot M^2_{12} + \cfrac{t^3}{3!}\cdot M^3_{12} + \cfrac{t^4}{4!}\cdot M^4_{12} + \cdots \\
> > f_2' &= 1 + tM_{22} + \cfrac{t^2}{2!}\cdot M^2_{22} + \cfrac{t^3}{3!}\cdot M^3_{22} + \cfrac{t^4}{4!}\cdot M^4_{22} + \cdots \\
> > r_1 &= \cfrac{1}{t}\cdot \ln f_1 \in \mathbb{C} \text{ is a root iff } t\cdot f_1' = f_1 \ln f_1 \\
> > r_2 &= \cfrac{1}{t}\cdot \ln f_2 \in \mathbb{C} \text{ is a root iff } t\cdot f_2' = f_2 \ln f_2 \\
> > &\text{Our method states only that: }r = \cfrac{-b' \pm \sqrt{b'^2 - 4a'c'}}{2a'}
> > \end{align}
> >
> > Now do the same for the quintic.
> >
> > \begin{align}
> > Z &= \left[ \exp\, t M \right] \cdot C \\
> > Z &= \left[ \exp\, t \left(\begin{matrix}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ -f'/a' & -e'/a' & -d'/a' & -c'/a' & -b'/a' \end{matrix}\right) \right] \cdot C \\
> > y &= C_1 \cdot f_1 + C_2 \cdot f_2 + C_3 \cdot f_3 + C_4 \cdot f_4 + C_5 \cdot f_5 \\
> > y' &= C_1 \cdot f_1' + \cdots + C_5 \cdot f_5' \\
> > y'' &= C_1 \cdot f_1'' + \cdots + C_5 \cdot f_5'' \\
> > y''' &= C_1 \cdot f_1''' + \cdots + C_5 \cdot f_5''' \\
> > y'''' &= C_1 \cdot f_1'''' + \cdots + C_5 \cdot f_5'''' \\
> > \left(\begin{matrix}f_1 & f_2 & f_3 & f_4 & f_5 \\ f_1' & f_2' & f_3' & f_4' & f_5' \\ f_1'' & f_2'' & f_3'' & f_4'' & f_5'' \\ f_1''' & f_2''' & f_3''' & f_4''' & f_5''' \\ f_1'''' & f_2'''' & f_3'''' & f_4'''' & f_5'''' \end{matrix}\right) &= Id + tM + \cfrac{t^2}{2!}\cdot M^2 + \cfrac{t^3}{3!}\cdot M^3 + \cfrac{t^4}{4!}\cdot M^4 + \cdots \\
> > \text{Let } &M_{ij}^n \text{ be the }(i,j)\text{-cell of the result of } M^n \\
> > f_1 &= 1 + tM_{11} + \cfrac{t^2}{2!}\cdot M^2_{11} + \cfrac{t^3}{3!}\cdot M^3_{11} + \cfrac{t^4}{4!}\cdot M^4_{11} + \cdots \\
> > f_2 &= 0 + tM_{12} + \cfrac{t^2}{2!}\cdot M^2_{12} + \cfrac{t^3}{3!}\cdot M^3_{12} + \cfrac{t^4}{4!}\cdot M^4_{12} + \cdots \\
> > f_3 &= 0 + tM_{13} + \cfrac{t^2}{2!}\cdot M^2_{13} + \cfrac{t^3}{3!}\cdot M^3_{13} + \cfrac{t^4}{4!}\cdot M^4_{13} + \cdots \\
> > f_4 &= 0 + tM_{14} + \cfrac{t^2}{2!}\cdot M^2_{14} + \cfrac{t^3}{3!}\cdot M^3_{14} + \cfrac{t^4}{4!}\cdot M^4_{14} + \cdots \\
> > f_5 &= 0 + tM_{15} + \cfrac{t^2}{2!}\cdot M^2_{15} + \cfrac{t^3}{3!}\cdot M^3_{15} + \cfrac{t^4}{4!}\cdot M^4_{15} + \cdots \\
> > r_1 &= \cfrac{1}{t}\cdot \ln f_1 \in \mathbb{C} \text{ is a root iff } t\cdot f_1' = f_1 \ln f_1 \\
> > r_2 &= \cfrac{1}{t}\cdot \ln f_2 \in \mathbb{C} \text{ is a root iff } t\cdot f_2' = f_2 \ln f_2 \\
> > r_3 &= \cfrac{1}{t}\cdot \ln f_3 \in \mathbb{C} \text{ is a root iff } t\cdot f_3' = f_3 \ln f_3 \\
> > r_4 &= \cfrac{1}{t}\cdot \ln f_4 \in \mathbb{C} \text{ is a root iff } t\cdot f_4' = f_4 \ln f_4 \\
> > r_5 &= \cfrac{1}{t}\cdot \ln f_5 \in \mathbb{C} \text{ is a root iff } t\cdot f_5' = f_5 \ln f_5 \\
> > \end{align}
> >
> > Our method states only that: $r$ are all the solutions of the quintic $a'x^5 + b' x^4 + c' x^3 + d' x^2 + e' x + f' = 0$.
> >
> > \end{document}
> Yes, you are right about throwing away Galios, Raffani and Able theorem as well, since all radicals higher than second DEGREE are simply not more than a primitive human brain fart
>
> However, I strongly suspect those made up, forged or well-fabricated stories about those little amateur kids making greatest theorems in too early age (that are still too difficult to digest by highest obtained degrees nowadays) then dying too early and suddenly were discovered genius mathematicians many years after their early death, Wonder! (seems like Hindi-Drama Films)
>
> However, this is the turn of honest historians acquiring the minimum talents in mathematics (if at all existing) to investigate seriously those many so unbelievable stories in the history of alleged genius mathematicians, especially that hundreds of topics mainly in public forums were suddenly arising and pointing out many serious flaws and very big holes into that nonsense pretending like theorems in modern mathematics, which is a truly well-documented shame upon the shoulders of historian mathematicians and journalists as well, FOR SURE
>
> May I encourage non-mathematicians to DIG under and uncover the full truth about those many invented-like stories, since they were proven wrong beyond any little doubt FOR SURE
>
> BKK