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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

1048.0. "What is wrong with this?" by HPSTEK::XIA () Wed Apr 05 1989 17:55

    
    
    I saw this in the MAA news letter:
    
    
     ix       ix (2pi/2pi)       2*i*pi (x/2pi)       (x/2pi)
    e    =  (e  )            = (e      )         = (1)        = 1
    
    Eugene
T.RTitleUserPersonal
Name
DateLines
1048.1AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoWed Apr 05 1989 18:5717
	The step from

		  ix (2pi/2pi)
		(e  )

	to

		  2*i*pi (x/2pi)
		(e      )

	isn't correct.

	They are taking "laws of exponents" worked out for real numbers
	(and even there with some restrictions) and trying to apply them
	to complex numbers.

	Dan
1048.2Bait & Switch with the roots of unity.DWOVAX::YOUNGSharing is what Digital does best.Fri Apr 07 1989 10:0059
Re .1:

>	They are taking "laws of exponents" worked out for real numbers
>	(and even there with some restrictions) and trying to apply them
>	to complex numbers.

Actually, I would even go a little bit farther that they are taking the Laws 
of Exponents as worked out for Real-valued exponents of POSITIVE Real-valued
bases.  In particular these laws allow at most only one possible solution 
to an exponential expression by ignoring the problems of the multiple roots
of Unity.  

This expression starts out with a non-positive root and then manipulates
it with rules that only allow positive roots.  This is all hidden by the 
use of the old  e^(i*pi)  expression which hardly anybody has an intuitive
feel for.  But the same trick can be applied to a much simpler non-complex 
problem that makes it easier to see whats going wrong.

Instead of

     ix       ix (2pi/2pi)       2*i*pi (x/2pi)       (x/2pi)
    e    =  (e  )            = (e      )         = (1)        = 1
    
lets use:

      x        x (2/2)             2 (x/2)             (x/2)        
    -1   =  (-1 )            =  (-1 )            =  (1)       = 1

     ^        ^                   ^                  ^          ^
     |        |                   |                  |          |
    (1)      (2)                 (3)                (4)        (5)

Now what is happening here is that expression (3) is introducing roots that 
did not exist before and are therefore not valid.  Then step (5) elimnates
all the non-positive roots leaving only "1", which is the valid root ONLY 
when "x" is an even positive(?) integer.


To elucidate this, lets take the simplest example, where  x=1  :

      1        1 (2/2)             2 (1/2)             (1/2)        
    -1   =  (-1 )            =  (-1 )            =  (1)       = 1

     ^        ^                   ^                  ^          ^
     |        |                   |                  |          |
    (1)      (2)                 (3)                (4)        (5)


Now (1) is clearly equal to -1, so this is the only valid root.  But the
transformation from (2) to (3) is now calling for the Square Root of 1,
which has 2 answers, -1 AND 1.  Since we know that only -1 is valid, then
clearly we have introduced a bogus root.  Finally in the transformation
from (4) to (5) we are asserting that the only valid solution to the
Square Root of One is 1.  This looks right when manipulating exponents, 
but we know that in fact 1 has TWO square roots.  This transformation is
therefore incorrectly throwing away one of the 2 roots, and in fact the
root that is thrown away was the one that was originally valid.

--  Barry