Solution to Polynomials
Jon
Capital letters are vectors
Lower case letters are scalars
* signifies dot product if between vectors
* signifies multiplication if between scalars
SUM means to sum over the range given
view full screen in Courier
DEVELOPMENT
given the equation of the polynomial,
SUM a_p*x^p = 0 p=0,1,2,...,n
and the equation of the plane,
SUM a_p*x_p = -a_0 p=1,2,3,...,n
N=(a_1,a_2,a_3,...,a_n)
X=(x, x^2, x^3, ..., x^n)
The normal to the plane intersects the plane at
(-a_0/(||N||^2))*N
And the plane intersects the coordinate axes at,
(-a_0/a_1|)*I_1, (-a_0/a_2)*I_2, ... (-a_0/a_n)*I_n
where I_1=(1,0,0,..), I_2=(0,1,0,..), I_n=(0,0,...,0,1)
Then construct the orthogonal basis where
W_1 = N/||N||
(-a_0/a_i)*I_i - (-a_0/(||N||^2))*N
W_2 = ----------------------------------------
||(-a_0/a_i)*I_i - (-a_0/(||N||^2))*N||
(-a_0/a_j)*I_j - (-a_0/(||N||^2))*N - (((-a_0/a_j)*I_j -
(-a_0/(||N||^2))*N)*w_2)w_2
W_3
= ----------------------------------------------------------------------------------------
||(-a_0/a_j)*I_j - (-a_0/(||N||^2))*N - (((-a_0/a_j)*I_j -
(-a_0/(||N||^2))*N)*w_2)w_2||
.
.
W_k = see web site http://jons-math.bravehost.com/nthdegree.html
X=SUM (X*W_p)*W_p p=1,2,3,..., n
EXAMPLES
E^3
X=SUM (X*W_p)*W_p p=1,2,3
X=x(1,x,x^2,)
(1,x,x^2)=SUM (1,x,x^2)*W_p)*W_p p=1,2,3 quadradic
E^4
.
X=SUM (X*W_p)*W_p p=1,2,3,4
X=x(1,x,x^2,x^3)
(1,x,x^2,x^3)=SUM (1,x,x^2,x^3)*W_p)*W_p p=1,2,3,4 cubic. solve with
quadradic
E^5
X=SUM (X*W_p)*W_p p=1,2,3,4,5
X=x(1,x,x^2,x^3,x^4)
(1,x,x^2,x^3,x^4)=SUM (1,x,x^2,x^3,x^4)*W_p)*W_p p=1,2,3,..., n quartic.
solve with cubic
.
.
E^n
X=SUM (X*W_p)*W_p p=1,2,3,...,n
X=x(1,x,x^2,x^3,...,x^(n-1)) solve with (n-1)th
DISCUSSION
The path from the origin to X may take different routes. W_3(i,j) may be
revised to W_3(j,i). The two different paths still arrive at X, so they may
be equated and solved.
Jon Giffen (c) 2009
Brian S
Ok, you have examples for powers 2 through 4. Polynomials of those
powers are always solvable. Show me an example for the fifth power
equation x^5-x-1=0. Please show all the work to finding the (only)
real root.
J.B. Wood
> Ok, you have examples for powers 2 through 4. Polynomials of those
> powers are always solvable. Show me an example for the fifth power
> equation x^5-x-1=0. Please show all the work to finding the (only)
> real root.
Hello, and AFAIK it can be shown that only polynomials of n <= 4 degree
have an algebraic solution for the roots. By "algebraic" it is meant
that the solution must be in terms of algebraic operators (addition,
subtraction, multiplication, division and radicals). Closed form
non-algebraic solutions for the roots of n > 4 polynomials are possible
(e.g. using elliptic integrals). Sincerely,
--
John Wood (Code 5550) e-mail: wo...@itd.nrl.navy.mil
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337
Arturo Magidin
> Brian S wrote:
> > Ok, you have examples for powers 2 through 4. Polynomials of those
> > powers are always solvable. Show me an example for the fifth power
> > equation x^5-x-1=0. Please show all the work to finding the (only)
> > real root.
>
> Hello, and AFAIK it can be shown that only polynomials of n <= 4 degree
> have an algebraic solution for the roots. By "algebraic" it is meant
> that the solution must be in terms of algebraic operators (addition,
> subtraction, multiplication, division and radicals). Closed form
> non-algebraic solutions for the roots of n > 4 polynomials are possible
> (e.g. using elliptic integrals).
To be more precise, every polynomial of degree 1, 2, 3, or 4 can be
solved "by radicals", expressions as you describe on the coefficients
of the polynomial. Moreover, there is a "formula" on the coefficients
that yields a solution in terms of the coefficients for every
polynomials of the given degree (like the qudratic formula for degree
2). For n>4, there always exist polynomials of degree n that cannot be
solved by radicals at all; but there always exist polynomials of
degree n that *can*. For example, x^n-1 can be solved by radicals for
every n.
--
Arturo Magidin
J.B. Wood
Hello, and I probably should have prefaced my previous post by
indicating the general polynomial equation a_n * x^n + a_n-1*x^(n-1) +
...+ a_0 = 0 in which none of the coefficients are presumed to be zero.
Other than that, I believe what I posted is accurate. Sincerely,
g.r...@iit.cnr.it
"However, solutions of the general quintic equation may be given in
terms of Jacobi theta functions or hypergeometric functions in one
variable. Hermite and Kronecker proved that higher order polynomials
are not soluble in the same manner. Klein showed that the work of
Hermite was implicit in the group properties of the icosahedron.
Klein's method of solving the quintic in terms of hypergeometric
functions in one variable can be extended to the sextic, but for
higher order polynomials, either hypergeometric functions in several
variables or "Siegel functions" must be used (Belardinelli 1960, King
1996, Chow 1999). In the 1880s, Poincaré created functions which give
the solution to the nth order polynomial equation in finite form.
These functions turned out to be "natural" generalizations of the
elliptic functions. "
See also http://mathworld.wolfram.com/QuinticEquation.html
where explicit formulas are given for the quintic equation.
Giovanni
--
http://anagram.it : anagrams, alphametics, arithmogriphs,...
Frederick Williams
(x + 1)^5 = 0 has roots expressible in radicals and has no coefficients
zero.
--
Which of the seven heavens / Was responsible her smile /
Wouldn't be sure but attested / That, whoever it was, a god /
Worth kneeling-to for a while / Had tabernacled and rested.
J.B. Wood
> "J.B. Wood" wrote:
>> Hello, and I probably should have prefaced my previous post by
>> indicating the general polynomial equation a_n * x^n + a_n-1*x^(n-1) +
>> ...+ a_0 = 0 in which none of the coefficients are presumed to be zero.
>> Other than that, I believe what I posted is accurate. Sincerely,
>
> (x + 1)^5 = 0 has roots expressible in radicals and has no coefficients
> zero.
>
algebraic solution for the roots of a general polynomial of higher
degree than the quartic. Sincerely,
Axel Vogt
The general formula, covering the case of *non-radicals*,
is given by Umemura in Lectures of Tate (but not that I
can follow that proof). IIRC.