roots to *any* polynomial
Jon
This solution for the roots of *any* polynomial requires some manipulation
of complex numbers. An example for the solutions to the Pentic is provided.
g.r...@iit.cnr.it
Your formulas are obviously wrong.
Take for example the quintic x^5+x+1=0.
According you your formulas, when a=b=c=1,
the term 1-(a+b)c/(a^2+b^2) vanishes and we are left with
[1/2 + (1/2)^(1/10)]^(1/4) which clearly (being a positive real...)
cannot be a root of x^5+x+1=0.
I suggest you to check your formulas in a similar way
(before posting them): it will take you just few seconds.
Moreover the last formula cannot be right because it expresses
the general solution of a polynomial equation by means of radicals
and the 4 operations and it is a well known proved fact that this
in general is not possible.
giovanni
--
http://anagram.it : anagrams, alphametics, arithmogriphs, ...
Jon Slaughter
> On Nov 26, 7:32 pm, "Jon" <jon8...@peoplepc.com> wrote:
>> http://jons-math.bravehost.com/rootformula.html
>>
>> This solution for the roots of *any* polynomial requires some
>> manipulation of complex numbers. An example for the solutions to the
>> Pentic is provided.
>
> Your formulas are obviously wrong.
> Take for example the quintic x^5+x+1=0.
> According you your formulas, when a=b=c=1,
> the term 1-(a+b)c/(a^2+b^2) vanishes and we are left with
> [1/2 + (1/2)^(1/10)]^(1/4) which clearly (being a positive real...)
> cannot be a root of x^5+x+1=0.
>
> I suggest you to check your formulas in a similar way
> (before posting them): it will take you just few seconds.
>
Why would he want to do that? Do you really think he cares about truth?
g.r...@iit.cnr.it
> g.re...@iit.cnr.it wrote:
> > I suggest you to check your formulas in a similar way
> > (before posting them): it will take you just few seconds.
>
> Why would he want to do that? Do you really think he cares about truth?
Sorry, being new to this newsgroup I did not realize the nature of the
guy.
Brian S
Go ahead and keep replying to Jon. Someone has to tell him the
obvious wrongness of his idea.
Dik T. Winter
Take for your quintic a = b = 1, c = -2. Your formula gives:
x = {{3^(1/5) - 2}^(1/4)
which not even close to the actual root of 1.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Ostap S. B. M. Bender Jr.
> http://jons-math.bravehost.com/rootformula.html
>
> This solution for the roots of *any* polynomial requires some manipulation
> of complex numbers.
>
Really? How about this one:
ax + b
or
x^2 - a^2
?
>
> An example for the solutions to the Pentic is provided.
>
Proof by example?
Heissenburger
http://www.wolframalpha.com/
it (tries to) solves everything!
just type your question!
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