Conference rusure::math

From:	ROLL::USENET       "USENET Newsgroup Distributor" 12-FEB-1984 22:23
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!genrad!grkermit!masscomp!clyde!floyd!harpo!seismo!rlgvax!cvl!umcp-cs!james
Subject: Re: what is i raised to the ith power?
Posted: Thu Feb  9 15:21:43 1984


Easy, it is:

	 __
	-|| /
	   / 2
	e			(which is exp(-pi/2)).

The way to get this result is:

Any complex number C may be written as exp(a+bi) for some real a,b.
	      i
Then exp(a+bi)  = exp((a+bi) * i) = exp(ai - b).  (Using exponent rule.)
					   i
So consider C = i.  i = exp(0 + pi/2) ==> i  = exp(0i - pi/2) = exp(-pi/2).

The actual value is approximately .20787958

  --Jim

Newsgroups: net.math
Path: decwrl!decvax!harpo!ihnp4!houxm!houem!agd
Subject: i to the i
Posted: Fri Feb 10 06:08:07 1984

I tried to send this solution to you Mike but I don't think
it made it.  Jim's answer was only partially correct.  The
complex exponential is a multi-valued function, so i^i
is also multi-valued.  The values of i^i are:



            i        -(4n+1)*pi/2
           i   =    e                   n=...,-2,-1,0,1,2,3...


                     a (set) of real numbers!

Note that for n=0, we get Jim's result.  The error in Jim's
solution is in the step where he says

        i = exp(0 +pi/2) ==> ....

The argument (arg) for a complex number is not unique.  Pi/2
is the principle arg but you must add 2n*pi for all positive and
negative n.  If you insert this factor in Jim's proof, you will
end up with the proper answer.

                  Art Deacon
                  AT&T Bell Laboratories.

From:	ROLL::USENET       "USENET Newsgroup Distributor" 14-FEB-1984 22:05
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!mcnc!unc!ulysses!mhuxl!houxm!houem!agd
Subject: More on i**i
Posted: Mon Feb 13 07:32:50 1984

In response to Seaman and Rentsch:

The value of i^i is very well defined.  I gave
the values in a previous article and they are:

        i       -(4n+1)*pi/2
       i   =   e                  for all integer n.
  

There is no problem in defining log(i) either:
for any complex number z


     ln(z) = ln|z| + i(theta + 2n*pi) for all integer n

where -pi < theta <= pi and |z| is the modulus of z.
Theta is called the principle
argument of the ln.  As you can see there are an infinite
number of values for ln(z).  Of course the ln function
cannot be extended continuously to the entire complex
plane because of the pole.  However, if you consider
the Riemann surface, it can be.

In the exp(x) sum, we need x=i*ln(i) to compute i^i,
so it does require something "funny".  Being clever
has nothing to do with it.  That's the way it is.

For Seaman:
           i       -(2n+1)*pi
       (-1)   =   e              for all integer n

and
           i       -(4n-1)*pi/2
       (-i)   =   e              for all integer n.


Art Deacon
AT&T Bell Labs

Newsgroups: net.math
Path: decwrl!decvax!mcnc!unc!ulysses!mhuxl!ihnp4!inuxc!pur-ee!CS-Mordred!Pucc-H:Pucc-I:ags
Subject: References on i ** i, "principal logs"
Posted: Thu Feb 16 07:19:48 1984


A brief survey of the literature reveals that there is no universal agreement
as to what the "principal value of the logarithm" means.  I quote from Ahlfors, 
"Complex Analysis" page 47:

	"By convention the logarithm of a positive number shall always mean
the real logarithm, unless the contrary is stated.  The symbol a ** b, where
a and b are arbitrary complex numbers except for the condition a <> 0, is
always interpreted as an equivalent of exp(b log a).  If a is restricted to
positive numbers, log a shall be real, and a ** b has a single value.
Otherwise log a is the complex logarithm, and a ** b has in general infinitely
many values which differ by factors exp(2*PI*i*n*b).  There will be a single
value if and only if b is an integer n, and then a ** b can be interpreted as
a power of a or a ** (-1).  If b is a rational number with the reduced form
p/q, then a ** b has exactly q values and can be represented as (a**p) ** 1/q."

Note that this specifically denies assigning any unique value to i ** i.
Ahlfors also points out that w = exp(i*y) has a unique solution with y in
[0,2*PI), but declines to identify this value as a "principle value".  Note
that the choice of interval here differs from (-PI,PI] which some people
on the net have claimed as the "universally recognized principal branch".

Other textbooks have a variety of definitions.  For instance, Knopp's
"Theory of Functions" defines the principle value of the logarithm of z as
a path integral, from 1 to z, of du/u.  The path is constrained to lie
in the complex plane cut by the negative real axis, thus avoiding winding
problems.  The integral is path-independent within this domain.  Note that
"the principal logarithm of -1" has no meaning according to this definition.
The same definition can be found in James and James, "Mathematics Dictionary".

Weinberger's "A First Course in Partial Differential Equations" also uses
the path integral definition, but does not specify any particular location
for the cut.  He merely points out that "If we make a cut extending from z=0
to infinity to prevent any winding, [the logarithm] is analytic in the
remainder of the z-plane."

I think the conclusion is clear.  There is no universal agreement on the
meaning of the "principal value" of the logarithm.  By contrast, try finding
a text which offers anything other than the conventional meaning for the
real square root function, i.e. the positive branch.
-- 

Dave Seaman
..!pur-ee!pucc-i:ags

"Against people who give vent to their loquacity 
by extraneous bombastic circumlocution."

From:	ROLL::USENET       "USENET Newsgroup Distributor" 22-FEB-1984 23:02
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!harpo!eagle!mhuxl!ihnp4!inuxc!pur-ee!CS-Mordred!Pucc-H:Pucc-I:ags
Subject: What does FORTRAN have to do with mathematics?
Posted: Mon Feb 20 08:48:08 1984


No one mentioned this during the recent discussion of i**i, but I think
I have found the reason that a number of people insisted that the
principle value of the logarithm is the one with imaginary part in
(-PI,PI].

It's defined that way in X3.9-1978 (the FORTRAN standard).

That definition is just as good as any other, but it is not a universally
accepted one.  It seems that FORTRANers are only interested in getting an
answer, not in understanding the problem.  Consider complex exponentiation
for example.  There are infinitely many values to choose from, but somehow
FORTRAN manages to compute a unique value.  In order to do this, you have
to give up such things as continuity.

I recall the time an irate user demanded to know why his program kept
bombing every time it tried to compute A**B, where A and B were real 
and A happened to be negative.  When I asked him what sort of answer 
he hoped to get in that situation, he said he didn't know.  He just 
wanted an answer so his program would run.
-- 

Dave Seaman
..!pur-ee!pucc-i:ags

"Against people who give vent to their loquacity 
by extraneous bombastic circumlocution."

Newsgroups: net.math
Path: decwrl!decvax!mcnc!unc!ulysses!burl!clyde!watmath!csc
Subject: Re: References on i ** i, "principal logs"
Posted: Tue Feb 21 09:18:14 1984

    The complex log fuction can be defined in a number of ways.
Perhaps the most intuitive is as the inverse of the exponential
function (which can be defined as the sum of an infinite series).
Hence we define log(z) to be a complex number such that exp(log(z))=z.
Now any non zero complex number z can be written as exp(r + it),
(r and t real numbers).  r is uniquely determined but t isn't.
If t is valid so is t+2n(pi) with n an integer.  Therefore there
are an infinitely many complex numbers g such that exp(g)=z.
(if z is non zero).  Hence log is multi valued.  There is no
complex number g such that exp(g)=0, therefore log(0) is undefined.
Hence log is a multi valued function defined on the complex plane
minus zero.
      It does not make sense to define log(0)=0 as then
exp(log(0))=1.  The article which argued that log(0)=0 contained
a division by zero which implied i*2n(pi)=i*2n(pi)exp(i*(pi)/2)
or as exp(i*pi/2)=i this implies i*2n(pi)=-2n(pi).  A contradiction.
It is because dividing by zero leads to such contradictions that
such division is not defined.
    One can define a single valued log function by choosing one
value of t for each z, usually done by restricting t to some half
open interval of length 2(pi).  However one cannnot do this in such
a way as to have the resulting function continuous on the complex
plane (minus zero).  Also such equations as log(ab)=log(a) + log(b)
and log(exp(z))=z cannot hold for all a,b,z. The usual practice is
to use whichever "branch" (ie. choice of an interval for t) that
is most convenient for the task at hand.
                                           William Hughes

Newsgroups: net.math
Path: decwrl!decvax!harpo!eagle!mhuxl!ulysses!unc!rentsch
Subject: Re: More on i**i
Posted: Mon Feb 13 15:19:59 1984


"That's the way it is"?  No way.  That's the way we chose it to be.
In order to give i**i meaning, some *one* decided to use as a definition

   x**y is identical to exp( y * ln(x) )

But there is nothing magic about that defition!  There is no reason
that log has to be involved at all.  Compare this to the Gamma function.
The most common form of the gamma function is as an improper integral

  gamma(x) = integral from 0 to 1/0 of t**x exp(-t) dt

which is, for the positive real numbers, a perfectly reasonable definition.
But what about the complex numbers?  Well, if you use the inifinite product
definition instead, then there is no problem, it converges for all z in
the complex plane (except of course for the negative real integers).

So one more time:  I understand that *if* the definition

      x**y is identical to exp( y * ln(x) )

is used, then i**i is multiple valued.  The question is, if a *different*
definition is used, can i**i be singled valued instead?  (Alternatively
can someone suggest a compelling reason for adopting the definition
given above?  Arguments like "it's very natural" are not compelling.
Arguments like "not adopting it results in a contradiction" are compelling,
though you have to be careful about what is being contradicted,
since it may be the same definition in a disguised form.)

Tim

Newsgroups: net.math,net.philosophy
Path: decwrl!decvax!harpo!ulysses!mhuxl!ihnp4!ihlts!rjnoe
Subject: Re: i**i, foundations and philosophy
Posted: Wed Feb 15 08:49:54 1984


To begin with, one must define elementary functions for complex numbers
in such a way as to be consistent with their real counterparts.  Among
other things, this means than ln must be the inverse of exp.  Further, the
same rules of logarithms and exponents must apply.  For example,
ln(a^b) == b*ln a, e^(x+y) == e^x * e^y, etc.  Any other technique
leads to inconsistencies, forcing one to treat numbers with complex
components as entities entirely different from real numbers.  This
defeats the purpose of complex numbers.  Similar problems are encountered
by treating negative integers as fundamentally different, subject to
different laws of mathematics than positive integers.  A better idea
was to treat them all as simply integers.  The same tendency to make
mathematics general-purpose can be seen in the integration of irrational
numbers with the rationals to make the reals.  It took many people a
long time to accept the concept of zero, then rational numbers, then
negative numbers, then irrationals, and finally complex numbers.  (Let's
skip extended reals for the moment.)  But the wisest course is to for-
mulate general rules in math to apply to all these sets of numbers.

Until we get to complex numbers, we have a problem in mapping.  With any
of the other subsets, elementary functions can easily give us results
that don't map into the same set; subtraction on positive numbers,
division on integers, and both square root and logarithms on reals.
But with extended complex numbers, there is no such problem.  Any
combination of elementary functions one can imagine will map onto this
same set.  What still causes people some problems is that not all of
these map one-to-one.

Mappings that are not bijections are really rather common, even when one
is not used to dealing with anything but reals.  The square root function
is bivalued on the reals (except at zero, of course), and the square
(x^2) function maps two-to-one.  It is not surprising to find that
logarithms on complex numbers are multivalued.  To reject this is
equivalent to rejecting the Euler formula:
		e^(i*t) = cos t  +  i * sin t
Because e^[i*(t+2*PI*n)] has a single value for all integers n.  Thus the
exponential function is an infinite-to-one mapping.  To reject this is to
reject the basic concept of complex numbers.  Real numbers are on a line,
so complex numbers are on a plane.  Such a concept is very useful, very
natural for our Euclidean minds.  Certainly one can come up with a
self-consistent, but very different concept of complex numbers, but
human beings would have a difficult time using it.  I think mathematics
exists independently of any intelligence, but that doesn't mean that
all MODELS of it are the same.  The model must be suited to our minds.

Accepting the Euler formula means accepting traditional rules for a real
raised to a complex power.  But this also means accepting that
		ln i = i * (2n + .5) * PI
because
		e^[i*(2n+.5)*PI] = i
for all integers n.  It is fallacious to presume that just because certain
mappings are one-to-one in the real domain that the same must hold true
in the complex domain.  Keeping the utility of mathematics, we must say
that ln(i^i) = i * ln i.  Similar rules for multiplication of complex
numbers require that i * i = -1, a pure real.  The fact that i^i is
infinitely multivalued is inescapable from the accepted model (or
representation) of mathematics.  One can come up with a different repre-
sentation, but that does not make it useful or even consistent.

The best approach, for those who have trouble with i^i, is to use the
principal value, where n=0.  Then the modulus of all complex numbers must
be within the interval (-PI, PI].  Of course, any single half-open real
interval with a 2*PI width works just fine.
		Roger Noe		ihnp4!ihlts!rjnoe
		AT&T Bell Laboratories