GrassmannCalculus`
CommonFactorTheorem  |
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Details and Options
Examples
(1)
Basic Examples
(1)
In[1]:=
<<GrassmannCalculus`
The following example is taken from Section 3.6, Example: Applying the Common Factor Theorem.
In[2]:=
 | SetBookVectorAssociation |
We will calculate the common factor from the following two 2-vectors in 3-space. The result should be a 1-vector.
In[3]:=
v1=3⋀+2⋀+3⋀;v2=(5+7)⋀;
e
1
e
2
e
1
e
3
e
2
e
3
e
2
e
3
e
1
In[4]:=
Print["Regressive product"]step1=v1⋁v2Print["Expansion by CommonFactorTheorem"]step2=
[step1]Print["Expand and Simplify exterior and regressive products."]step2//
step3=%//
Print["Substitute the unit-basis for the n-basis. We might have added a CongruenceFactor here."]step4=step3/.⋀⋀
Print"Use ",HoldForm@GrassmannRule[3,11],": ",x_⋁xstep4/.GrassmannRule[3,11]
 | CommonFactorTheorem |
| ExpandAndSimplifyExteriorProducts |
| ExpandAndSimplifyRegressiveProducts |
e
1
e
2
e
3
1
 | ★n |
1
★n
Regressive product
Out[4]=
(3⋀+2⋀+3⋀)⋁(5+7)⋀
e
1
e
2
e
1
e
3
e
2
e
3
e
2
e
3
e
1
Expansion by CommonFactorTheorem
Out[4]=
(3⋀+2⋀+3⋀)⋀⋁(5+7)+(3⋀+2⋀+3⋀)⋀(-5-7)⋁
e
1
e
2
e
1
e
3
e
2
e
3
e
1
e
2
e
3
e
1
e
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e
1
e
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e
2
e
3
e
2
e
3
e
1
Expand and Simplify exterior and regressive products.
Out[4]=
(-11⋀⋀)⋁+(3⋀⋀)⋁(5+7)
e
1
e
2
e
3
e
1
e
1
e
2
e
3
e
2
e
3
Out[4]=
-11⋁⋀⋀+15⋁⋀⋀+21⋁⋀⋀
e
1
e
1
e
2
e
3
e
2
e
1
e
2
e
3
e
3
e
1
e
2
e
3
Substitute the unit-basis for the n-basis. We might have added a CongruenceFactor here.
Out[4]=
-11⋁+15⋁+21⋁
e
1
1
★n
e
2
1
★n
e
3
1
★n
Use GrassmannRule[3,11]: x_⋁x
1
★n
Out[4]=
-11+15+21
e
1
e
2
e
3
On the other hand, evaluates immediately with a and evaluates without the .
ToCommonFactor
CongruenceFactor
CongruenceSimplify
CongruenceFactor
In[5]:=
step1//
step1//
| ToCommonFactor |
| CongruenceSimplify |
Out[5]=
★c(-11+15+21)
e
1
e
2
e
3
Out[5]=
-11+15+21
e
1
e
2
e
3
Here we choose a 3-space and consider the regressive product of two 2-elements, one of which is explicitly expressed as an exterior product, while the other is a general 2-element.
When the product is expressed as chooses the
(x⋀y)⋁
β
2
CommonFactorTheorem
A
form and decomposes the first factor in order to perform the expansion. (We turn on Precedence to better show the grouping of terms.)In[6]:=
★A;
;A=
(x⋀y)⋁
| ★P |
| CommonFactorTheorem |
β
2
Out[6]=
(x⋀)⋁y+(y⋀)⋁(-x)
β
2
β
2
When the product is expressed as chooses the
-⋁(x⋀y)
β
2
CommonFactorTheorem
B
form and decomposes the second factor in order to perform the expansion.In[7]:=
B=
-⋁(x⋀y)
| CommonFactorTheorem |
β
2
Out[7]=
-(⋀(-x))⋁y-(⋀y)⋁x
β
2
β
2
You can check that the results are the same by simpifying them.
In[8]:=
| ★ |
Out[8]=
-x⋁(y⋀ β 2 β 2 |
-x⋁(y⋀ β 2 β 2 |
If you want to explore the Common Factor Theorem applied to more general expressions you can compose the simple form of any graded elements first.
In[9]:=
| CommonFactorTheorem |
 | ComposeSimpleForm |
α
2
β
2
Out[9]=
(⋀)⋁+(⋀)⋁(-)
α
2
β
2
α
3
α
3
β
2
α
2
In[10]:=
| ComposeSimpleForm |
α
2
β
2
| CommonFactorTheorem |
Out[10]=
(⋀)⋁
α
1
α
2
β
2
Out[10]=
(⋀)⋁+(⋀)⋁(-)
α
1
β
2
α
2
α
2
β
2
α
1
If both factors are decomposable and of different grade, will choose to decompose the one of lower grade.
CommonFactorTheorem
In[11]:=
| ★ℬ |
4
| ComposeSimpleForm |
α
3
| ComposeSimpleForm |
β
2
Out[11]=
(⋀⋀)⋁(⋀)
α
1
α
2
α
3
β
1
β
2
In[12]:=
| CommonFactorTheorem |
Out[12]=
(⋀⋀⋀(-))⋁+(⋀⋀⋀)⋁
α
1
α
2
α
3
β
1
β
2
α
1
α
2
α
3
β
2
β
1
In[13]:=
| ★★P |
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