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+2
x
4
a
2
b
2
(
X
2
+
Y
2
+
Z
2
)+
b
2
(
X
2
+
Y
2
)+
a
2
Z
2
− a
2
b
2
=0
,
x
4
+2
x
4
a
2
+
b
2
+
x
4
4
a
2
b
2
+
a
4
+
b
4
a
2
(
X
2
+
Y
2
)
b
2
Z
2
−
−
−
(J.54)
a
2
b
2
a
4
b
4
2
x
4
(
X
2
+
Y
2
+
Z
2
)
a
4
b
4
b
2
(
X
2
+
Y
2
)+
a
2
Z
2
a
2
b
2
−
−
−
=0
.
a
6
b
6
x
1
(
x
4
)
,x
2
(
x
4
)
,x
3
(
x
4
)
:
x
1
=(1+
b
2
x
4
)
−
1
X, x
2
=(1+
b
2
x
4
)
−
1
Y, x
3
=(1+
a
2
x
4
)
−
1
Z.
Backward step. Substitute
{
}
(J.55)
Test :
Λ
1
=
Λ
2
=1+
b
2
x
4
>
0
,Λ
3
=1+
a
2
x
4
>
0if
Λ
1
=
Λ
2
>
0and
Λ
3
>
0then
end
.
Here, we used MATHEMATICA 2.2 for DOS 387. The executable command is “GroebnerBasis
[Polynomials, Variables]” in a specified ordering. The fourteen elements of the computed Grobner
basis can be interpreted as following. The first equation is a univariate polynomial of order four in
the Lagrange multiplier identical to (
J.53
). As soon as we substitute the admissible value
x
4
into
the linear equations (
J.61
), (
J.65
), and (
J.69
), we obtain the unknowns
{
x
1
,x
2
,x
3
}
=
{
x, y, z
}
.
Box J.7 (Closed form solution).
2
,
2
{
X, Y, Z
}∈
T
{
x
1
,x
2
,x
3
}∈
E
a,a,b
to
{
L, B, H
}
.
Pythagoras in three dimensions:
H
:=
(
X − x
1
)
2
+(
Y − x
2
)
2
+(
Z − x
3
)
2
.
(J.56)
Convert
{x
1
,x
2
,x
3
}
and
{X, Y,Z}
to
{L, B}
:
tan
L
=
Y
−
x
2
x
1
=
Y
−
y
Z
−
x
3
(
X
x
,
tan
B
=
X
−
X
−
−
x
1
)
2
+(
Y
−
x
2
)
2
Z − x
3
(
X − x
)
2
+(
Y − y
)
2
.
=
(J.57)
Box J.8 (Buchberger algorithm, Grobner basis for solving the normal equations of the con-
straint minimum distance mapping).
Ideal
I
:=
:= [
x
1
+
b
2
x
1
x
4
− X, x
2
+
b
2
x
2
x
4
− Y, x
3
+
a
2
x
3
x
4
− Z, b
2
x
1
+
b
2
x
2
−
a
2
b
2
]
Groebner basis
G
:=
a
2
x
3
−
x
1
+
b
2
x
1
x
4
X, x
2
+
b
2
x
2
x
4
Y, x
3
+
a
2
x
3
x
4
Z, b
2
x
1
+
b
2
x
2
:= [
{
−
−
−
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