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From: jvl@ocsystems.com (Joel VanLaven)
Subject: Math error in Annex G ???
Date: 1996/07/24
Message-ID: <1996Jul24.214556.24621@ocsystems.com>#1/1
X-Deja-AN: 170000631
organization: OC Systems, Inc.
newsgroups: comp.lang.ada
Date: 1996-07-24T00:00:00+00:00
List-Id: <comp.lang.ada>


  I was implementing complex types from Annex G in the LRM as a way to learn
some Ada95 when I ran into something that I am sure is incorrect.  It has
to do with exponentiation of a complex number by a negative integer by going
through polar.  Essentially it says that exponentiation can be significantly
improved by converting to polar, using a polar exponentiation trick, and
then converting back.  That is fine, and I knew about that even without the
LRM mentioning it.  The problem is that the method that they give for
exponentiation with the polar of a complex number is wrong for negative
exponents.

  The LRM says (liberty taken with spacing):
"... exponentiating the modulus by the given exponent;
 multiplying the argument by the given exponent,
   when the exponent is positive,
 or dividing the argument by the absolute value of the given exponent,
   when the exponent is negative; ..."

Translating to mathese:
if n > 0 then
   (r,theta)**n=(r**n,n*theta)
if n < 0 then
   (r,theta)**n=(r**n,theta/abs(n))

but i**(-1) = 1/i = -i
and i**(-1) = (1,pi/2)**(-1) ?=? (1**(-1),(pi/2)/abs(-1)) = (1.pi/2) = i
obviously something is wrong there.  

I am almost positive that the desired math is:
for all n
   (r,theta)**n=(r**n,n*theta)

this comes about from how * works for polar complex numbers, namely:

(r1,th1)*(r2,th2)=(r1*r2,th1+th2)

this will prove it by induction for positive n as follows:

-- base case
(r,th)**1 = (r,th) = (r**1,1*th)

--inductive case
if (r,th)**n=(r**n,n*th) then
  (r,th)**(n+1)=((r,th)**n)*(r,th) = (r**n,n*th)*(r,th) =
  ((r**n)*r,(n*th)+th) = (r**(n+1),(n+1)*th)
Proven.

for negative n, simply use the polar definition of * and the part
for positive n as follows:
(r',th') = (r,th)**(-n) = 1/((r,th)**n) = 1/(r**n,n*th)
(r',th')*(r**n,n*th) = 1 = (1,0)
(r'*(r**n),th'+(n*th)) = (1,0)
r'*(r**n) = 1  and  th'+(n*th) = 0
r' = 1/(r**n) = r**(-n)  and  th' = -(n*th) = (-n)*th
(r,th)**(-n) = (r',th') = (r**(-n),(-n)*th)
Proven.

I have purposely not dealt with 0 exponents as they were not included in
this part of the LRM but mathematically it works the same way.

I submitted this to ada-comment a week ago, but I have a feeling that
comp.lang.ada might be a little more active...

If anyone can show me that the LRM is correct I'd love to see it.
I am practically positive that the LRM is just plain wrong math-wise,
but all proofs can have flaws, so if you want to give your complex
arithmetic skills a whirl, have at it.

-- Joel