In effort to solve ax^5+bx+c=0

[Jon]

This is in effort to solve,

ax^5+bx+c=0

however a flaw in my answer occurs when a = -b

other than that, the evolution of the solution is at my web page,
http://jons-math.bravehost.com/pentic.html

I haven't tested it out.

[Sam Wormley]

http://www.wolframalpha.com/input/?i=ax%5E5%2Bbx%2Bc%3D0+%2C+a%3D-b

[Jon]

N=(a[1],a[2],a[3],...,a[n])
X=(x,x^2,x^3,...,x^n)

N*X+a[0]=0 polynomial

a[1]x[1]+a[2]x[2]+a[3]x[3]+...+a[n]x[n]+a[0]=0
plane

construct an orthogonal basis from N,
j[1]=N/|N|
j[2]=(-a[0]/a[n])i[n]-(-a[0]/|N|^2)N
j[3]=...
j[4]=...
.
.
j[n]=

j[2],j[3],j[4],...,j[n] all lie on the plane and are
mutually orthogonal.

Permutate the couples of j[i] with j[1], project X onto
the resulting planes and solve as with,

http://jons-math.bravehost.com/pentic.html

.

"Jon" <jon...@peoplepc.com> wrote in message
news:h_2dnSvNJv3RtLnW...@earthlink.com...

[Frederick Williams]

There's no need to.

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prey to intercept suspicious pigeons in mid-air.
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[Peter Hill]

On Sun, 13 Dec 2009 14:00:33 +0000, Frederick Williams
<frederick...@tesco.net> wrote:

>Jon wrote:
>>
>> This is in effort to solve,
>>
>> ax^5+bx+c=0
>>
>> however a flaw in my answer occurs when a = -b
>>
>> other than that, the evolution of the solution is at my web page,
>> http://jons-math.bravehost.com/pentic.html
>>
>> I haven't tested it out.
>
>There's no need to.

http://en.wikipedia.org/wiki/Bairstow%27s_method

http://home.att.net/~srschmitt/zenosamples/zs_linbairstow.html

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