noncommutative partial fractions and continued fractions
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Noncommutative Partial Fractions and Continued Fractions
Darlayne Addabbo
Professor Robert Wilson
Department of Mathematics
Rutgers University
July 16, 2010
Overview
1) Existence of partial fraction decompositions over division rings
2) Relation between Laurent series and quasideterminants
3) Generalization of Galois’s result that every periodic continued fraction satisfies a quadratic equation
Partial Fractions in C (complex numbers)
• If f(x) is a polynomial over C of degree n with distinct roots , then
for some in C.€
λ1,...,λ n
€
1
f (x)=
A1
x − λ1
+ ...+Anx − λ n
€
A1,...,An
€
Example
€
1
x 2 −1=
1
2x −1
−
1
2x +1
We may also write as
€
1
f (x)=
A1
x − λ1
+ ...+Anx − λ n
€
( f (x))−1 = (x − λ1)−1A1 + ...+ (x − λ n )
−1An
The two expressions are the same over C. However, over a division ring,
is ambiguous. (It could be equal to
or .)
To avoid ambiguity, we prefer the second notation.
€
Aix − λ i
€
(x − λ i)−1Ai
€
Ai(x − λ i)−1
Can we generalize the method of partial fractions to an arbitrary
division ring, D?• Recall that a division ring satisfies all of the axioms of
a field except that multiplication is not required to be commutative.
• Over a field, if f(x) is a monic polynomial of degree n, with n distinct roots ,
• But this doesn’t work in a division ring. It doesn’t even work in the quaternions.
€
λ1,...,λ n
€
f (x) = (x − λ1)...(x − λ n )
(Recall) The Algebra of Quaternions
• The algebra of quaternions is a four dimensional vector space over R (the real numbers), with basis 1, i, j, k and multiplication satisfying:
• ij=-ji=k• jk=-kj=i• ki=-ik=j • 1 is the multiplicative identity
An example
has roots i+1 and 1+i+jCheck:
Similarly,
€
f (x) = x 2 − (2 + j)x + (2 + j − k)
€
(i +1)2 − (2 + j)(i +1) + 2 + j − k
= −1+ 2i +1− 2i + k − 2 − j + 2 + j − k
= 0
€
(1+ i + j)2 − (2 + j)(1+ i + j) + 2 + j − k = 0.
But
Instead,
Note that each root corresponds to the rightmost factor of one of these expressions. (This is due to a result of Ore.)
€
f (x) ≠ (x − (1+ i + j))(x − (1+ i))
€
f (x) = (x − (1− i + j))(x − (1+ i)) = (x − (1− i))(x − (1+ i + j))
Using the above to solve partial fractions
where , are elements of the quaternions
€
(x 2 − (2 + j)x + (2 + j − k))−1 = (x − (1+ i))−1A1 + (x − (1+ i + j))−1A2
€
x 2 − (2 + j)x + (2 + j − k) = (x − y)(x − (1+ i))
= (x − z)(x − (1+ i + j))
€
y
€
z
So which gives
and (since we can write and in terms of 1+i and 1+i+j),
Thus
So we have generalized the method of partial fractions.
€
1 = (x − y)A1 + (x − z)A2
€
A1 + A2 = 0
€
(1+ i)A1 + (1+ i + j)A2 =1
€
y
€
z
€
1 1
1+ i 1+ i + j
⎡
⎣ ⎢
⎤
⎦ ⎥A1
A2
⎡
⎣ ⎢
⎤
⎦ ⎥=
0
1
⎡
⎣ ⎢
⎤
⎦ ⎥
Theorem
Let , with
, have distinct roots,
.
Then there exist elements, such that
€
f (x) = x n + B1xn−1 + ...+ Bn
€
B1,...,Bn ∈ D
€
λ1,...,λ n ∈ D
€
A1,...,An ∈ D
€
( f (x))−1 = (x − λ1)−1A1 + ...+ (x − λ n )
−1An
1) First prove inductively that
€
(x − yn )(x − yn−1)...(x − y1) =
€
x n + (−1)( y i)xn−1 + (−1)2( y j1 y j0 )
j1 > j0
∑i=1
n
∑ x n−2 + ...+ (−1)n ynyn−1...y1
2) By the Gelfand-Retakh Vieta Theorem, we can write
€
f (x) = x n + B1xn−1 + B2x
n−2 + ...+ Bn
€
=(x − a1,1)(x − a1,2)...(x − a1,n )
€
=(x − a2,1)(x − a2,2)...(x − a2,n )
€
=...
€
=(x − an,1)(x − an,2)...(x − an,n )
where each is an element of our division ring, D, and for all
€
ai, j ,1 ≤ i ≤ n,1 ≤ j ≤ n,
€
ai,n = λ i
€
i,1 ≤ i ≤ n
3) Obtaining our set of equations
Recall that we want to find such that
We multiply both sides of our equation by f(x). Notice that our terms cancel.
€
Ais
€
(x − λ1)−1A1 + (x − λ 2)−1A2 + ...+ (x − λ n )
−1An
€
=(x n + B1xn−1 + B2x
n−2 + ...+ Bn )−1
€
(x − λ i)−1
4) Obtaining our System of Equations
Comparing terms on both sides of our equation, we obtain
Using the Generalized Cramer’s Rule, we can solve this system of equations .
€
1 1 1 ... 1
λ1 λ 2 λ 3 ... λ nλ1
2 λ 22 λ 3
2 ... λ n2
M M M M M
λ1n−1 λ 2
n−1 λ 3n−1 ... λ n
n−1
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥
A1
A2
A3
M
An
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥
=
0
0
0
M
1
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥
Continued Fractions
In the commutative case, periodic continued fractions are often written as follows
, a field.€
α =a0 +1
a1 +1
a2 +1
...an +1
α
€
a0,a1,...,an ∈ F
Continued Fractions
In the noncommutative case, we avoid ambiguity by writing our continued fractions using nested parentheses.
where , a division ring
€
α =a0 + (a1 + (...+ (an +α −1)−1...)−1
€
a0,a1,...,an ∈ D
There is a theorem by Galois which says that every periodic continued fraction is a solution to a quadratic equation. I generalized this to show that a periodic continued fraction over a division ring, D, satisfies
where
We can write in terms of , the repeating terms of our continued fraction
€
xAx + Bx + xC + E = 0
€
A,B,C,E ∈ D
€
a1,a2,...,an
€
A,B,C,E
Current Work
Galois showed that there is a relationship between the complex conjugate of a periodic continued fraction and the periodic continued fraction obtained when we write the repeating terms in reverse order. We want to generalize this to the non-commutative case.
References
Gelʹfand, I. M.; Retakh, V. S. Theory of noncommutative determinants, and characteristic functions of graphs. (Russian) Funktsional. Anal. i Prilozhen. 26 (1992), no. 4, 1--20, 96; translation in Funct. Anal. Appl. 26 (1992), no. 4, 231--246 (1993)
Holtz, Olga, and Mikhail Tyaglov. "Structured Matrices, Continued Fractions, and Root Localization of Polynomials.” http://www.cs.berkeley.edu/~oholtz/RF.pdf
Lauritzen, Neils. "Continued Fractions and Factoring.”http://www.dm.unito.it/~cerruti/ac/cfracfact.pdf
Wilson, Robert L. "Three Lectures on Quasideterminants." Lecture. http://www.mat.ufg.br/bienal/2006/mini/wilson.pdf
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