Re: roots of polynomials
From: BASSAM KARZEDDIN (bassam_at_ahu.edu.jo)
Date: 07/24/04
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Date: Sat, 24 Jul 2004 02:21:50 +0000 (UTC)
On 23 Jun 2004 17:17:16 GMT, Robert Israel wrote:
>In article <wqyc33lllytp@legacy>,
>bassam mizied karzeddin <bassam@ahu.edu.jo.or.karzeddin@yahoo.co.uk>
wrote:
>|>On 23 Jun 04 02:17:38 -0400 (EDT), osama ebraheem ababnah wrote:
>|>>find out one real root for the following uneven degree equation:
>
>|>>f(x)=x^n+ax^m+b=0
>
>|>>n is uneven positive integer
>|>>m is positive integer less than n
>|>>f(x) is a rational integeral function of x
>|>>(a&b) are real rational cooefficients
>
>|>I found one real root for the case (a=b=1)& m is uneven positive
>|>integer,here is the solution:
>
>|>x=-{1-1/n+(2m-n+1)/(2!n^2)-(3m-2n+1)(3m-n+1)/(3!n^3)
>|> +(4m-3n+1)(4m-2n+1)(4m-n+1)/(4!n^4)
>|> -(5m-4n+1)(5m-3n+1)(5m-2n+1)(5m-n+1)/(5!n^5)+...}
>
>A very interesting formula. It comes, I think, from a series in
powers
>of t for a root of x^n + t x^m - 1. According to Maple:
>[view in fixed-width font]
>
>> series(RootOf(x^n + t*x^m - 1, x), t);
>
>
> 1 n - 1 - 2 m 2 (2 n - 1 - 3 m) (n - 1 - 3 m) 3
>1 - - t - ----------- t - ----------------------------- t
> n 2 3
> 2 n 6 n
>
> (3 n - 1 - 4 m) (n - 1 - 4 m) (2 n - 1 - 4 m) 4
> - --------------------------------------------- t
> 4
> 24 n
>
> (n - 1 - 5 m) (3 n - 1 - 5 m) (2 n - 1 - 5 m) (4 n - 1 - 5 m) 5
> - ------------------------------------------------------------- t
> 5
> 120 n
>
>University of British Columbia
subject:obtain one real root for the following odd degree
f(x)=x^n+ax^m+b=0
solution:case-1)
let, m be odd positive integer less than n
let, r=[(b^k)/(a^n)]^(1/m) ,where, r is the airthmetical mth root of
let s=[(a^n)/(b^k)]^(1/n) ,where, s is the airthmetical nth root of
(a^n)/(b^k) ,(b=/0)
if, (b^k)/{abs(a)}^n < (m^m)*(k^k)/(n^n)
x=-{(b/a)^(1/m)}*[1-r/m+(2n-m+1)r^2/(2!m^2)
if,(b^k)/{abs(a)}^n > (m^m)*(k^k)/(n^n)
let, m be even positive integer less than n
Neither,in the third world nor in the first could I convey these ideas
ANY COMMENTS OR ANY COUNTER EXAMPLE IS WELCOMED AT THIS SITE
Thanking to all who care about the the unborn & undying facts.
Bassam King Karzeddin
> / 6\
> + O\t /
>
>I don't know about the convergence of this series.
>
>Note that if n and m are odd, x^n + t x^m - 1 = 0 iff
>(-x)^n + t (-x)^m + 1 = 0. And for a root of z^n + a z^m + b = 0
>you can take z = -b^(1/n) x where x is a root of
>x^n + a b^(m/n-1) x^m - 1 = 0.
>
>Robert Israel israel@math.ubc.ca
>Department of Mathematics <a
href="http://www.math.ubc.ca/~israel">http://www.math.ubc.ca/~israel>
>Vancouver, BC, Canada V6T 1Z2
equation,that is continious rational function of x
let, (k=n-m)
(b^k)/(a^n) ,(a=/0)
-(3n-2m+1)(3n-m+1)r^3/(3!m^3)
+(4n-3m+1)(4n-2m+1)(4n-m+1)*r^4/(4!*m^4)
-(5n-4m+1)(5n-3m+1)(5n-2m+1)(5n-m+1)r^5/(5!m^5)+...]
x=-{b^(1/n)}*[1-s/n+(2m-n+1)s^2/(2!n^2)-(3m-2n+1)(3m-n+1)s^3/(3!n^3)
+(4m-3n+1)(4m-2n+1)(4m-n+1)*s^4/(4!*n^4)
-(5m-4n+1)(5m-3n+1)(5m-2n+1)(5m-n+1)s^5/(5!n^5)+...]
case-2
let x=1/y ,(y=/0) in the above equation,then you get equation of the
previous solved form for one real root.
In fact,this is nothing but a very little introduction to obtain all
roots to any general polynomials and hopefuly by radicals, soon.
for so many years,because Iam a (civil Engineer) not a profisional
mathematician(I know nothing about Maple or Mathematica ,nor that good
in computer science and English languge) that is why,I find this site,
the best to learn and convey ideas.
P.O.Box 20
Engineering & Project Department
AL-HUSSEIN BIN TALAL UNIVERSITY
MA,AN-JORDAN
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