🚀 go-pugleaf

Jon <jon8338@peoplepc.com> wrote in message news:<MSHMc.12907$f4.5254@newsread3.news.atl.earthlink.net>...
> a correction is applied below:
> 
> Jon wrote:
> > If T*N+a[0] is the nth degree polynomial where T=(t,t^2,t^3,...,t^n), 
> > N=(a[1],a[2],a[3],...,a[n]), if T is described from two orthogonal 
> > systems, its linear requirement precipitates.  T remains invariant
> > despite the coordinate system used to describe it. Consequently, my 
> > revised web site showing the details,
> > 
> > APPLICATION OF
> > http://mypeoplepc.com/members/jon8338/polynomial/id7.html
> > 
> > EXAMPLE
> > 

Alright Jon, its time for some strong criticism.  Take a look at this:
t^3 - 2t^2 - 2t + 4 = 0
t = 2, sqrt(2), -sqrt(2)

> > (t-2)(t-3)(t-4)=0
> > (t^2-5t+6)(t-4)=0
> >       t-4
> > -------------
> > t^3-5t^2+ 6t
> >    -4t^2+20t-24
> > ----------------
> > t^3-9t^2+26t-24
> > 

N = (1, -2, -2)

> > N=+/-(26,-9,1)
> > N*U[2]=0 at U[2]=(4,13,13)
> >               i    j  k
> > U[3]=NxU[2]= 26   -9  1 =(-130,-334,374)
> >               4   13 13
> > 
> > N/|N|=
> > u[1]=( .944,-.327,.036)=(u[11],u[12],u[13])
> > u[2]=( .212, .691,.691)=(u[21],u[22],u[23])
> > u[3]=(-.249,-.640,.717)=(u[31],u[32],u[33])
> > 

I have decided to try this polynomial with two sets of vectors one is
marked with primes and the other is not.

u[1] = (1/3, -2/3, -2/3)
u[2] = (2/3, -1/3, 2/3)
u[3] = (2/3, 2/3, -1/3)

u'[1] = (1/3, -2/3, -2/3)
u'[2] = (2/sqrt(5), 1/sqrt(5), 0)
u'[3] = (2/sqrt(5), 0, 1/sqrt(5))

> > (-a[0]/|N|)u[1]=(.8717)u[1]
> > Q = +/-(-.823,.285,-.0314)
> > 

Q = -(4)/3 * (1/3, -2/3, -2/3) = (-4/9, 8/9, 8/9)
Q' = -(4)/3 * (1/3, -2/3, -2/3) = (-4/9, 8/9, 8/9)

> >     .212  -.249       1 0 0
> > A = .691  -.640    I= 0 1 0
> >     .691   .717       0 0 1
> > 
    [ 2/3  2/3]
A = [-1/3  2/3]
    [ 2/3 -1/3]

     [2/sqrt(5) 2/sqrt(5)]
A' = [1/sqrt(5) 0        ]
     [0         1/sqrt(5)]

> > T = (AA^T -I)^-1 Q   =(1.085,6.552,6.607)=(t,t^2,t^3)

For the unprimed set of vectors, the expression T = (AA^T -I)^-1 Q
does not exist since (AA^T -I) is not invertible in this case.  This
is an EXTREME problem.  You can't even guarantee the equations
generated to be solvable!

For the primed set of vectors, T' = (4/9, -8/9, -8/9).  Now Jon, since
you did some rounding in your example, how would you round these to
the correct values of 2, 1.41421356237309504880168872421, and
-1.41421356237309504880168872421?

> > 
> > t = 1.085  throw out
> > t = 2.560~3
> > t = 1.876~2
> > 
> > 
> > EXAMPLE
> > 
> > n=8
> > 
> > T=[((AA^T-I)^-1)^T((AA^T-I)^-1)]^-1((AA^T-I)^-1)^T Q
> correct above to,
> T=[(AA^T-I)^T(AA^T-I)]^-1(AA^T-I)^T Q
> 
> > 
> > (Least Squares)
> > 
> >           LEAST SQUARES SOLUTION TO MATRIX               
> > k   a[k]                   SUMMARY           
> > 0  -30     t^k         |t^k|         t        ERROR   
> > 1    9   2.384E-01   2.384E-01   2.384E-01  -2.731E+01   
> > 2    9   2.031E+01   2.031E+01   4.507E+00   1.590E+06   
> > 3    1  -1.902E+01   1.902E+01   2.669E+00   2.557E+04   
> > 4    3  -4.038E+00   4.038E+00   1.418E+00   2.165E+02   
> > 5    6   4.742E+00   4.742E+00   9.485E-01   1.830E+00   
> > 6    1  -3.753E+01   3.753E+01   1.830E+00   1.416E+03   
> > 7    1   3.320E+01   3.320E+01   1.649E+00   6.553E+02   
> > 8    9   3.010E+01   3.010E+01   1.530E+00   3.787E+02   
> > 
This has to be the worst, YOUR ERROR VALUES ARE UP TO HALF A MILLION
TIMES LARGER THAN THE T VALUES!!!!!!!!!!  When an engineer would see
those kinds of error values, he would scream!

> > 
> > Jon Giffen
> > jon8338@peoplepc.com
> >