cse590b lecture 6 cubic polynomials · high class spam dear dr. blinn, from your article titled...

CSE590B Lecture 6

Cubic Polynomials

Covariants and Resultants

James F. Blinn JimBlinn.Com

http://courses.cs.washington.edu/courses/cse590b/13au/

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

Syzygy of J

C

C

C

C

C

p p

p

pC

p p

C

p p

p

C

C

C

C

C

p p

p

C

CC

pp

C

p p

C

C

p

p

1

2= +

1

2

Hessian of J

C C CJ =

C CH

MCL

L L

M

L L

L

M

Symmetrize? Like we did with:

J J ?

J is already symmetrical!

C C Ca

b

c

C C Ca

b

c

C C Ca

b

c

C C Ca

b

c

a

b

c

C

C

C

C C Cc

a

b

0

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)

Discriminant of J

J J

H

H

2 H

J J

J J

H

H

4H

H

H

H

Sign of Discrim(J)

is same as

sign of Discrim(C)

Hessian of Linear Combo

bCa JB

Ca J

Ca J

C a C a

J bJb

b

b

B B

CJ

C

C

J

C C C

C C CC

=

=

cancel

Hessian of Linear Combo

bCa JB

H

H

a 22b 2 HB B

H

H

2 HJJ

Ca J

Ca J

C a C a

J bJb

b

b

B B

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


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


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 H H

Apply to H

H H H1

2

HH HH1

2


H C

H

H

H C

H

H

H C

H

H

= 0

Apply to C

H H HH1

2

H C

H

H

= JH C

H

H

1

2

= J


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

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Resultants

Resultant of Two Quadratics

F/E

B/E

Q

R

R

R

rQ,R Q

R

Q

R

Q

Q

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


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


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


R

R

rQ,R Q

R

Q

R

Q

QF/E

B/E

Q

R

S

R R

rQ,R

Q R

Q R

-det

Q Q


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 =


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

Resultant of Two Cubics

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

Geometry of Resultants

F/E

B/E

Q=KLLL

KK

Quadratic Cubic


C D

DL M

N= C

L

L

L

C

M

M

M

C

N

N

N

r(C,D) =

C D

r(C,D) =C

C

D

D

Possible Topologies

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


C b a D=CDa,b

CD

D2 =

CD

D6 =

CD

CD

CD

CD

CD

CD

3

2 6, 11 36D Dr C D

CD

D0 =

CD

CD

CD

CD

CD

3

2 0, 2 72D Dr C D

3

2 0 68 4 0D D D

Among the many possible diagrams:

It can be shown that:

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