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Transcript
Generalizing Continued Fractions
Darlayne Addabbo
Professor Robert Wilson
Department of Mathematics
Rutgers University
June 11, 2010
Partial Fractions in C
• If f(x) is a polynomial over C of degree n
with distinct roots 1,..., n , then
1
a1
an

 ...
f (x) x  1
x  n
for some a1,...,an in C.


Example
1
a1
a2


2
x 1 x  x1 x  x 2
1
Here a1  a2 
and x1  x2 1
2

Can we generalize this process to
arbitrary division rings?
• 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 1,..., n , f (x)  (x  1)...(x  n )
• But this doesn’t work in a division ring. It doesn’t even
work in the quaternions.


(Recall) The Algebra of
Quaternions
• The algebra of quaternions is a four
dimensional vector space over R, 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 of the difficulties found in
working over the quaternions
f (x)  x  (2  j)x  (2  j  k)
2
has roots i+1 and 1+i+j
2
(i

1)
 (2  j)(i  1)  2  j  k
Check:
 1 2i  1 2i  k  2  j  2  j  k
0


But f (x)  (x  (1 i  j))(x  (1 i))
However,
f (x)  (x  (1 i  j))(x  (1 i))  (x  (1 i))(x  (1 i  j))
Note that each of the above factorizations
contains a multiplicative of x  i where i
is a root of f(x).
This is due to the Gelfand-Retakh Vieta
Theorem


Using the above to solve partial
fractions
(x 2  (2  j)x  (2  j  k))1  (x  (1 i))1 a1  (x  (1 i  j))1 a2
x 2  (2  j)x  (2  j  k)  (x  y 2 )(x  (1 i))
 (x  z2 )(x  (1 i  j))
where y 2,z2 are elements of the
quaternions

So 1 (x  y2 )a1  (x  z2 )a2 which gives
a1  a2  0

and (since we can write y2 and z 2 in terms of
1+i and 1+i+j),
(1 i)a1  (1 i  j)a2 1
Generalized

We can use the
Cramer’s Rule
to solve this set of equations. In this case

a1  j  a2
So we have generalized the method of partial
fractions.

Our Conjecture
Given a polynomial of degree n, we can
generalize the above and obtain a
system of equations:
 1

 1

 n1
1
1
2
n1
2
1 a1  0
   
n a2  0

   
   
... n1
an  1
n 
...
...
which can be solved using the

generalized
Cramer’s Rule