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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1863.0. "Should have kept awake in those lin. alg. lectures" by IOSG::CARLIN (Dick Carlin IOSG, Reading, England) Fri Mar 25 1994 17:26

    How would you approach the following?
    
    Given an integral 2x2 matrix with equal diagonal terms and det = 1, eg
    
    		A  =	( 5  12)
    			( 2   5)
    
    Find a similar, but rational, matrix X such that
    
    	X^n = A  for some n.
    
    By using some results from another field I have been looking at, I can
    predict possible results (eg n=6 and rational divisor =23) giving
    
    	X = (25/23 24/23)   and   X^6 = A
    	    ( 4/23 25/23)
    
    But I would rather it were the other way around. Is there a standard
    way of attacking this problem so I can use it to *derive* results in my
    other field of study?
    
    Sorry if that's not as clear as I intended.      Dick
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1863.1correctionIOSG::CARLINDick Carlin IOSG, Reading, EnglandTue Mar 29 1994 08:4312
    Sorry, there was an error in .0
    
    A doesn't have a sixth root, but it does have the cube root
    
    	Y = (49/35 84/35)
    	    (14/35 49/35)
    
    X is merely close to being a sixth root.
    
    The original question still stands.
    
    Dick
1863.2RTL::GILBERTFri Apr 01 1994 00:1524
    I find
    
    	(a b)^6   ( 5 12 )   ( A B )
    	(c a)   = ( 2  5 ) = ( C A )
    
    where
    	a = 0.391435775293197
        c = 0.438410379701368
        b = c * B/C
                                (A B)             (a b)
    In general, the 6th root of (C A) is given by (c a), where
    a and c are solutions to:
    
    	(C*a^2 + B*c^2) * (B^2*c^4 + 14*B*C*a^2*c^2 + C^2*a^4) = A*C^3
    	2*a*c * (C*a^2 + 3*B*c^2) * (3*C*a^2 + B*c^2) = C^3
    and
    	b = c * B/C
    
    This is worth noting:
    
    	( sqrt(2)  sqrt(12) )^2  ( 5 12 )
        ( sqrt(0.5) sqrt(2) )  = ( 2  5 )
    
    Finally, the cube root in .1 is incorrect.
1863.3There's a law against thatVMSDEV::HALLYBFish have no concept of fireFri Apr 01 1994 13:216
>    This is worth noting:
>    
>    	( sqrt(2)  sqrt(12) )^2  ( 5 12 )
>       ( sqrt(0.5) sqrt(2) )  = ( 2  5 )
    
    Really? For what value of sqrt(6)?
1863.4RTL::GILBERTMon Apr 04 1994 16:119
    Oops.  There was a typo.  It should read:
    
    	( sqrt(2)  sqrt(18) )^2  ( 5 12 )
    	( sqrt(0.5) sqrt(2) )  = ( 2  5 )
    
    Also,
    	( sqrt(3)  sqrt(12) )^2  ( 5 12 )
    	( sqrt(1/3) sqrt(3) )  = ( 2  5 )
    
1863.5Maybe it generalizes?VMSDEV::HALLYBFish have no concept of fireMon Apr 04 1994 18:5718
1863.6This reply might be bug-free!IOSG::CARLINDick Carlin IOSG, Reading, EnglandSun Apr 24 1994 21:0848
    >                   <<< Note 1863.2 by RTL::GILBERT >>>
    >    Finally, the cube root in .1 is incorrect.

    Yes Peter you are right, I was guilty of two careless mistakes in a
    row. But let me expand on what led up to my false conclusions
    
    I was looking at the equation x^2 - 6y^2 = 1150 which arose in 1837.* .
    Using the reasoning in 1837.4 we only need to find the "generator"
    solutions with y < sqrt(1150*2*2) and the transformation
    
    	A = (5 12)
            (2  5)
    
    will provide the rest.
    
    The generator solns are (34,1) (38,7) (50,15) (70,25) (106,41) (158,63)
    (s1 to s6 say)
    and the first generated solution is A*(34,1) = (182,73).
    (s7 say)
    
    I was looking for relationships between the generator solutions (to try
    and cut down the number of generators) and the
    following appeared:
    
    s2 = X*s1   s4 = X*s2   s6 = X*s4
    s3 = X'*s2   s5 = X'*s3   !s7 = X'*s5!
    
    Where  X = (25/23 24/23)   and   X' = (25.4/23 26.4/23)
               ( 4/23 25/23)              ( 4.4/23 25.4/23)
    
    implying that Y = X'*X is a cube root of A.
    
    My first mistake was to confuse X' with X (must have been a really bad
    day) and the second was to believe the statement in !...!
    
    Other variations on the original equation are more obliging. For
    example
    
    x^2 - 5y^2 = 4 has the transform matrix ( 9 20)
                                            ( 4  9)
    
    which has the cube root (3/2 5/2)
                            (1/2 3/2)
    
    which lets all solutions be derived from the single generator (2,0).
    
    Anyway, sorry about the red herrings. I've obviously got some more
    analysis to do.