cse590b lecture 6 cubic polynomials · high class spam dear dr. blinn, from your article titled...
TRANSCRIPT
CSE590B Lecture 6
Cubic Polynomials
Covariants and Resultants
James F. Blinn JimBlinn.Com
http://courses.cs.washington.edu/courses/cse590b/13au/
Covariants of Cubic
2 3
2 2
2 2
2 3
3 2
2
2
3 2
J
J
J
J
A A D ABC B
B AC ABD B C
C ACD B D BC
D AD BCD C
C C Cx,w
x,w
x,w
3 2 2 33 3J J J JA x B x w C xw D w ,x w J
Cx,w
x,w
x,w
3 2 2 33 3Ax Bx w Cxw Dw ,x w C
C C
2
2
AC BB AD BC xx w
AD BC DB CC w
H(x,w) x,w x,w
Hessian of J
C C C C C C
H H H
J J
H H H
H
2
H
H
2 H Hessian of J is same as
Hessian of C (up to a scale)
Locus of constant H for type 111
H
H< 0
H
H
a 22b 2 HB B
Always
positiveLinear combos
of C and J
C
J
bCa JB
Locus of constant H for type 1 1/1
H
H> 0
H
H
a 22b 2 HB B
H
H
a 22b 2 0
H
H
2a b C
J(C)
LLL
MMM
H(C)
bCa JB
0B B
If
Triple Root
Locus of constant H for type 1 1/1
H
H> 0
H
H
a 22b 2 HB B
H
H
a 22b 2 0
H
H
2a b
dLg MB L
L
M
M
dLg LL
L
L
LB B g L M
L
L
M
Mg
M d LM
M
L
Lg M d M
M
M
M
Mg
C
J(C)
LLL
MMM
H(C)
Linear combo
of L3 and M
3
All cubics with
Hessian H
Locus of constant H for type 1 1/1
H
H> 0
H
H
a 22b 2 HB B
H
H
a 22b 2 0
H
H
2a b
C
J(C)
LLL
MMM
H(C)
Linear combo
of L3 and M
3
All cubics with
Hessian H
dLg MB L
L
M
M
gdB B L ML
L
M
M M L
The two places the line hits the triple root curve
are the cubes of the two factors of H
Locus of constant H for type 21
MCL
L L
M
L L
L
M
H LL
ML
L
M
J C
J LL
ML
L
MM
L
L
If C is type 21 (double factor L),
J is type 3 (triple factor L)
Transform based on H HT
HH HH HH
HH H H
TT
Trace: HT 0
So it’s an involution Trace=0 involution
(true only for 2x2 matrices)
Show via arc-swap:
HH HH1
2TT
So
Transform based on H HT
H C
H
H
H C
H
H
1
2
H
H CH
H
C
H
H
H
H
1
4
, ,x w x wT
C J
Transform C by T
Transform J by T
Roots of C, H, J
x
w
Root of C
Root of J(C)
Root of H(C)
x
w
x
w
Type 111 Type 21 Type
x
w
H=0
J=0
Type 3
H rot90 H mirror45
11
1
H nilpotent
Rank
Mathematics
Physics
A
B
A B
B C
A B B C
B C C D
1 0 0
0 0 0
0 0 0
1 0 0
0 1 0
0 0 0
1 0 0
0 1 0
0 0 1
1 2 3
1 2 3
Mathematical Rank of 2-Tensor
21 0
0 0
xx w x
w
Rank 1
Rank 2
2 21 0
0 1
xx w x w
w
2
2
2
xa abx w ax bw
wab b
2 2A B x
x w ax bw cx dwB C w
Mathematical Rank of 3-Tensor
31 0 0 0
0 0 0 0x
Rank 1
Rank 2
3 2 2 2
3
2 2 2 3
a a b a b abax bw
a b ab ab b
3 3
ax bw cx dw
Rank 3
3 3 3
ax bw cx dw ex fw
Rank of Cubic
Rank 1
Type 3
Rank 2
1Type 1
1
Rank 3
Type 111 (Not unique)
Rank 3
Type 21 (Not unique)
IEEE CG&A Articles
Part 1 – The Shape of the Discriminant
May/Jun 2006, pages 84–93
Part 2 – The 11bar Case
Jul/Aug 2006, pages 90–100
Part 3 – General Depression and a New
Covariant
Nov/Dec 2006, pages 92–102
Part 4 – The 111 case
Jan/Feb 2007,
Part 5 – Back to Numerics
May/Jun 2007
How to Solve a Cubic Equation
Solving Numerically
Translate to get B=0
1 0
x w x wB A
t u
x w x ws v
IEEE CG&A article
“General Depression and a New Covariant”
3 2 2 3, 3 3x w Ax Bx w Cxw Dw C
“Depress” C General transform to get B=0
High Class Spam
Dear Dr. Blinn,
From your article titled “How to solve a cubic equation, part
3: general depression and a new covariant.” (IEEE Comput
Graph Appl. 2006 Nov-Dec;26(6):92-102.), we learned of
your research in the area of Depression and thought you
might be interest in knowing that GenWay offers an ELISA
kit for the in-vitro diagnostic quantitative determination of
Serotonin in various biological fluids.
You can click the following link to our website in order to
view the datasheet: Serotonin ELISA Kit
Invariants of Two Quadratics
R Q
RQR
Q
R
QQ R
RQ
Q Q
RR,,,
Q R
RQ
Q R
RQ
= +
Q R
RQ
Q Q
RR
Q Q
RR
Q Q
RR
Q Q
RR
Q
Q
R
R
Q R
RQ
Q
R
R
Q
R Q
RQR
Q
R
Q
2 2
Nice
Invariants of Two Quadratics
R Q
RQR
Q
R
QQ R
RQ
Q Q
RR,,,
Q R
RQ
Q Q
RR
Q Q
RR Q
Q
R
R
R Q
RQR
Q
R
Q
2 2
Nice
Resultant of Two Quadratics (v1)
F/E
B/E
Q
R
R
R
r Q,R Q
R
Q
R
Q
Q
,Q Rr +2
R Q
RQR
Q
R
Q
R Q
RQR
Q
R
QQ R
RQ
2 2
Resultant of Two Quadratics (v1)
F/E
B/E
Q
R
R
R
r Q,R Q
R
Q
R
Q
Q
,Q Rr +2
R Q
RQR
Q
R
Q
R Q
RQR
Q
R
QQ R
RQ
2 2
Resultant of Two Quadratics (v1)
R K L L K
R
Q
K
Q
L L
Q
K
= + = 2 LQK R
R Q
Q
L
L Q
Q
K
K
= L
K Q
Q
L
K
+
K
L Q
Q
K
L
K
K Q
Q
L
L
++
R
Q
= 4
LQKR
QLQK
F/E
B/E
Q
R
,Q Rr +2
R Q
RQR
Q
R
Q
Resultant of Two Quadratics (v1)
R K L L K
R
R Q
Q
LL Q
QK K
=
LK Q
Q LK
+ 22R
Q
= 4LQK
R
Q
LQK
,Q Rr
LL Q
QK K
4
K is a factor of Q
L is a factor of Q
F/E
B/E
Q
R
,Q Rr +2
R Q
RQR
Q
R
Q
Resultant of Two Quadratics (v2)
F/E
B/E
Q
R
S
Q b a R=S
R R
rQ,R
Q R
Q R
-det
Q Q
=S QRa,b
QQR1,0 =
RQR0,1 =
Resultant of Two Quadratics (v2)
F/E
B/E
Q
R
S
R R
rQ,R
Q R
Q R
-det
Q Q
QQR1,0 =
RQR0,1 =
QR QR
R R
Q R
Q R
Q Q
QR QR
rQ,R
QR QR
Two Cubics, both with a root at 0
Two Cubics have a common (real) root
iff
The line connecting them is tangent to the discrim surface
X
Y
Slice thru X,Y,Z space at ZD/A0
Tangent to surface from each point
From each point C
There will be one plane tangent to the surface
for each real root of C
X
Y
Slice thru X,Y,Z space at ZD/A0
Distribute 3 each C and D
C
C
C
C
CC
C
C
C
C
C
C
C
C
C
C
C
C
D D
C C
D
C C
D D
C C
D
Examples:
But only need CD connected diagrams since:
Five Possible CD diagrams C
DD
C
D
C
T3 =
D
C D
C D
C
T2a T0b
C C
C
D
D D
2 I C DD3=
C
C
D
D
3
2 3 2 33 2I D I T
2 0 33 2a bT T T
It can be shown that:
Resultant of Two Cubics C
DD
C
D
C
T3 = T0b
C C
C
D
D D
L
MN
N
LM
L
NMN
ML
M
NL
M
LN
D
2 I C D
2 6I C L
N
MC N
N
N
r C L
L
L
C M
M
M
We are looking for:
After some symbolic programming
3
2
3
0
216
72
24 72 24b
I
T
T
a
g
a g r
C L
N
M
a C L
N
M
C L
N
M
C M
N
L
C M
N
M
C L
N
L
C
N
L
M
C
L
L
M
C
N
N
M
C
L
M
NC M
L
M
C
N
N
L
g
33
2 0 36 9 bI T Tr
C N
N
N
r C L
L
L
C M
M
M
Resultant of Two Cubics C
DD
C
D
C
T3 = T0b
C C
C
D
D D
2 I C D
C
D
C
D
C
D
216 r
D D
D
C
C C
+ 9 + 9
C C
C
D
D D
33
2 0 36 9 bI T Tr